diff --git "a/stack-exchange/math_stack_exchange/shard_11.txt" "b/stack-exchange/math_stack_exchange/shard_11.txt" deleted file mode 100644--- "a/stack-exchange/math_stack_exchange/shard_11.txt" +++ /dev/null @@ -1,19579 +0,0 @@ -TITLE: Prove that $\int_0^1t^{p-1}(1-t)^{q-1}\,dt=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}$ for positive $p$ and $q$ -QUESTION [13 upvotes]: I'm trying to prove that for $p,q>0$, we have $$\int_0^1t^{p-1}(1-t)^{q-1}\,dt=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}.$$ - -The hint given suggests that we express $\Gamma(p)\Gamma(q)$ as a double integral, then do a change of variables, but I've been unable thus far to express it as a double integral. -Can anyone get me started or suggest an alternate approach? -Note: This wasn't actually given to me as the $\Gamma$ function, just as a function $f$ satisfying $$f(p)=\int_0^\infty e^{-t}t^{p-1}\,dt$$ for all $p>0$, but I recognized that. This is in the context of an advanced calculus practice exam. - -REPLY [9 votes]: Indeed, to show this identity, one can start from the double integral -$$ -\Gamma(p)\cdot\Gamma(q)=\int_0^\infty \mathrm e^{-x}x^{p-1}\,\mathrm dx\cdot\int_0^\infty \mathrm e^{-y}y^{q-1}\,\mathrm dy=\iint_{[0,+\infty)^2} \mathrm e^{-x-y}x^{p-1}y^{q-1}\,\mathrm dx\,\mathrm dy. -$$ -The change of variable $x=ts$, $y=(1-t)s$, with $0\leqslant t\leqslant1$ and $s\geqslant0$, whose Jacobian yields $\mathrm dx\,\mathrm dy=s\,\mathrm ds\,\mathrm dt$, shows that -$$ -\Gamma(p)\cdot\Gamma(q)=\int_0^{+\infty}\int_0^1\mathrm e^{-s}t^{p-1}s^{p-1}(1-t)^{q-1}s^{q-1}s\,\mathrm ds\,\mathrm dt. -$$ -Finally, -$$\Gamma(p)\cdot\Gamma(q)=\int_0^{+\infty}\mathrm e^{-s}s^{p+q-1}\,\mathrm ds\cdot\int_0^1t^{p-1}(1-t)^{q-1}\,\mathrm dt=\Gamma(p+q)\cdot\int_0^1t^{p-1}(1-t)^{q-1}\,\mathrm dt, -$$ -and this is it. The last integral above is called the beta number $\mathrm B(p,q)$.<|endoftext|> -TITLE: Does $\lim_{h\rightarrow 0}\ [f(x+h)-f(x-h)]=0$ imply that $f$ is continuous? -QUESTION [19 upvotes]: Suppose $f$ is a real function defined on $\mathbb{R}$ which satisfies - $$\lim_{h\rightarrow 0}\ [f(x+h)-f(x-h)]=0.$$ - Does this imply that $f$ is continuous? - -Source: W. Rudin, Principles of Mathematical Analysis, Chapter 4, exercise 1. - -REPLY [17 votes]: For those interested in more about this property, functions satisfying it are called symmetrically continuous functions. It is known that if a function is symmetrically continuous at every point in an interval, then the set of points at which the function is not continuous is small in both the measure sense (has Lebesgue measure zero) and the Baire category sense (is a first category set). However, the non-continuity set of such a function can have cardinality continuum. Indeed, it can have cardinality continuum in every open subinterval of the domain interval. -See Marcin Szyszkowski’s 2000 Ph.D. Dissertation under Chris Ciesielski (at West Virginia University), Symmetrically Continuous Functions, which is freely available on the internet, for much more about these functions. -Incidentally, one way to answer Rudin's question (definitely not what he had in mind) is to observe that there are $c$ many continuous functions and $2^c$ many symmetrically continuous functions (see Miroslav Chlebík's 1991 Proc. AMS paper for this last result). -(added next day) This morning I had a chance to look at Brian S. Thomson's Symmetric Properties of Real Functions (1994). Given a function $f:{\mathbb R} \rightarrow {\mathbb R}$ that is symmetrically continuous at each point, let $D(f)$ be the set of points at which $f$ is not continuous in the usual sense. As of 1994 (and even now, I believe), no exact characterization is known for those subsets of $\mathbb R$ that are equal to $D(f)$ for some symmetrically continuous function $f.$ However, it is known that if $Z$ is a countable union of $N$-sets, then there exists a symmetrically continuous function $f$ such that $Z \subseteq D(f)$ (i.e. $f$ is not continuous at each point in $Z,$ and perhaps also at some points not in $Z$). -A corollary of this is that there exists a symmetrically continuous function $f$ such that $D(f)$ has Hausdorff dimension $1$ in every open interval (i.e. $D(f)$ is "everywhere of Hausdorff dimension $1").$ Note that this is much stronger than $D(f)$ having cardinality $c$ in every open interval, the result I mentioned previously. The "everywhere of Hausdorff dimension $1$" result follows from the fact that there exist, in every open interval, $N$-sets with Hausdorff dimension arbitrarily close to $1,$ and hence each open interval contains a $\sigma$-$N$-set with Hausdorff dimension equal to $1,$ and therefore we can put a $\sigma$-$N$-set of Hausdorff dimension $1$ in each interval of the form $(r,s)$ where $r -TITLE: non-equivalent bundles -QUESTION [5 upvotes]: Is it possible to find a specific example of two fiber bundles with the same base, group, fiber and homeomorphic total spaces but these bundles are not equivalent/isomorphic, if so -should I find a bundle map F between two bundles, inducing identity on the common base but F does not preserve fibers? (I don't know what it means, got mixed up) or -should I define the action of group on fibers differently? - -REPLY [5 votes]: In the paper "K-theory doesn't exist," (J. Pure Appl. Alg., 1978 vol 12) Akin explains that if $p: P\rightarrow B$ is a non-trivial principal $G$-bundle, the map $P\times G\rightarrow B$ (project to the first factor and then apply $p$) can be made into a principal $G\times G$-bundle in two different ways, by specifying two different actions of $G\times G$ on $P\times G$. -In general, these are not isomorphic as principal bundles (Akin shows that if they were always isomorphic, the $K$-theory of $B$ would be trivial, which is the origin of the paper's title).<|endoftext|> -TITLE: Schwarz Lemma - like exercise -QUESTION [5 upvotes]: There's this exercise: let $\,f\,$ be analytic on $$D:=\{z\;\;;\;\;|z|<1\}\,\,,\,|f(z)|\leq 1\,\,,\,\,\forall\,z\in D$$ and $\,z=0\,$ a zero of order $\,m\,$ of $\,f\,$. -Prove that $$\forall z\in D\,\,,\,\,|f(z)|\leq |z|^m$$ -My solution: Induction on $\,m\,$: for $\,m=1\,$ this is exactly the lemma of Schwarz, thus we can assume truth for $\,k1\,$ . Since $\,f(z)=z^mh(z)\,\,,\,h(0)\neq 0\,$ analytic in $\,D\,$ , put -$$g(z):=\frac{f(z)}{z}=z^{m-1}h(z)$$ -Applying the inductive hypothesis and using Schwarz lemma $\,\,(***)\,\,$ we get that -$$|g(z)|=\left|\frac{f(z)}{z}\right|=|z|^{m-1}|h(z)|\stackrel{ind. hyp.}\leq |z|^{m-1}\Longrightarrow |f(z)|\leq |z^m|$$ -and we're done...almost: we still have to prove $\,|g(z)|\leq 1\,$ for all $\,z\in D$ in order to be able to use the inductive hypothesis and this is precisely the part where I have some doubts: this can be proved as follows (all the time we work with $\,z\in D\,$): -$(1)\,\,$ For $\,f(z)=z^mh(z)\,$ we apply directly Schwarz lemma and get -$$|f(z)|=|z|^m|h(z)|\leq |z|\Longrightarrow |z|^{m-1}h(z)|\leq 1$$ -And since now the function $\,f_1(z)=z^{m-1}h(z)\,$ fulfills the conditions of S.L. we get -$(2)\,\,$ Applying again the lemma, -$$|f_1(z)|=|z|^{m-1}|h(z)|\leq |z|\Longrightarrow |z^{m-2}h(z)|\leq 1$$and now the function $\,f_2(z):=z^{m-2}h(z)\,$ fulfills the conditions of them lemma so...etc. -In the step$\,m-1\,$ we get -$$|z||h(z)|\leq |z|\Longrightarrow {\color{red}{\mathbf{|h(z)|\leq 1}}}\,$$ -and this is what allows us to use the inductive hypothesis in $\,\,(***)\,\,$ above. -My question: Is there any way I can't see right now to deduce directly, or in a shorter way, that $\,|h(z)\leq 1\,$ ? - -REPLY [7 votes]: For $0< r<1$, let $D_r=\{z\in\mathbb{C}:|z|\le r\}$. -The function $g(z)=\dfrac{f(z)}{z^m}$ is analytic on $D$ (see Removable Singularity) and $|g(z)|\le\frac{1}{r^m}$ on $\partial D_r$. The maximum modulus principle says that -$$ -|g(z)|\le\frac{1}{r^m}\text{ for }z\in D_r\tag{1} -$$ -Since $(1)$ holds for all $r<1$, we have that $|g(z)|\le1$ for $z\in D$, and therefore, -$$ -|f(z)|\le|z^m|\tag{2} -$$<|endoftext|> -TITLE: CFT via Brauer groups vs via ideles -QUESTION [16 upvotes]: I am interested in the relationship between the following two versions of CFT: -Version 1: (Brauer Group Version) -Let $K$ be a number field. One constructs, for every finite place $v$ of $K$, a map $inv_v:Br(K_v)\rightarrow \mathbb{Q}/\mathbb{Z}$ in a (fairly straightforward) cohomological manner. Then the short sequence: -$$1\rightarrow Br(K)\rightarrow \oplus Br(K_v)\rightarrow \mathbb{Q}/\mathbb{Z}\rightarrow 1$$ -where $\oplus Br(K_v)\rightarrow \mathbb{Q}/\mathbb{Z}$ is given by $\sum inv_v$, is exact. -Version 2: (Idele Version) -We can construct a map $$K^{\times}\backslash \mathbb{I}_K/\prod O_v^{\times} \to Gal(K^{ab}/K),$$ -where $\mathbb{I}_K$ are the ideles associated with $K$, by $$(1,...,1,\pi_v,1,...,1)\mapsto Frob_v.$$ Furthermore, this map induces an isomorphism between the profinite completion of $K^{\times}\backslash \mathbb{I}_K/\prod O_v^{\times}$ and $Gal(K^{ab}/K)$. -My question is: How do these two formulations of Class Field Theory relate to one another. Does one imply the other and vice versa? How does one get from one statement to the other? I have never quite been able to square this circle in my mind, even though I've been exposed to CFT for years. Any help would be greatly appreciated. - -REPLY [2 votes]: I think that the most powerful and comprehensive link lies in the cohomological machinery of "class formations" introduced by Artin-Tate in their book "Class field theory", chapter 14. This is almost the last chapter, so nobody expects it to be easily understandable. -But let me try to give an idea by considering first the easier case of local CFT: so how do we relate your 2 versions for a local field K ? Version $1$ gives the "invariant" isomorphism $inv_{K}$ between Br(K) and Q/Z. Version $2$ gives the "reciprocity isomorphism" between the profinite completion of K* and Gal($K^{ab}$/K). Accept for one moment that the profinite topology and the normic toplogy on K* are the same (this is a thechnical point in the proof of the so called "existence theorem"). Then version $2$ is equivalent to saying that , given a finite abelian extension L/K with Galois group G, there is a canonical isomorphism, the "local reciprocity"map, between K/NL* and G (where N denotes the norm from L to K). Here comes cohomology : Br(K) can be viewed as $H^{2}$($K_s$/K, $K_s$*), and in our case this can be shown to be $H^{2}$($K_{nr}$/K, $K_{nr}$*), where $K_{nr}$ is the maximal unramified extension of K, whose Galois group over K is known to be the profinite completion of (Z, +). This gives at once that $H^{2}$($L$/K, $L$*) of order n = [L:K] and there is an monomorphism $inv_{L/K}$ from to $H^{2}$($L$/K, $L$*), as well as a generator $u_{L/K}$ which is sent to $1/n$ . But general cohomological machinery says that the cup product by $u_{L/K}$ gives an isomorphism from $H^{-2}$(L/K, Z) to $H^{0}$(L/K, L$^{*}$) (Tate cohomology) , so we are done, because $H^{0}$(L/K, L$^{*}$) is by definition equal to $K*$/N$L*$. -For a number field K, the proof follows exactly the same steps, replacing the multiplicative group $L*$ by the idèle class C$_{L}$ of L. The technical difficulties lie in the computation of the cohomology of C$_{L}$ (whereas the cohomology of the idèles is easy). Finally, notice that L* and C_${L}$ are the two prototypical examples of class formations. Ref.: for local (resp. global) CFT, see e.g. chapter 6 (resp. 7) of Cassels-Fröhlich's book "Algebraic number theory". -PS: I'm sorry, but something went wrong with my typing.<|endoftext|> -TITLE: Show $f$ is constant if $|f(x)-f(y)|\leq (x-y)^2$. -QUESTION [23 upvotes]: Problem: Let $f$ be defined for all real $x$, and suppose that -$$|f(x)-f(y)|\le (x-y)^2$$ -for all real $x$ and $y$. Prove $f$ is constant. -Source: W. Rudin, Principles of Mathematical Analysis, Chapter 5, exercise 1. - -REPLY [30 votes]: For any $x\in\mathbb{R}$, -$$ -\begin{align} -|f'(x)| -&=\lim_{h\to0}\frac{|f(x+h)-f(x)|}{|h|}\\ -&\le\lim_{h\to0}\frac{h^2}{|h|}\\ -&=0 -\end{align} -$$ -Therefore, $f$ is constant.<|endoftext|> -TITLE: Explanation for why $1\neq 0$ is explicitly mentioned in Chapter 1 of Spivak's Calculus for properties of numbers. -QUESTION [11 upvotes]: During the first few pages of Spivak's Calculus (Third edition) in chapter 1 it mentions six properties about numbers. - -(P1) If $a,b,c$ are any numbers, then $a+(b+c)=(a+b)+c$ -(P2) If $a$ is any number then $a+0=0+a=a$ -(P3) For every number $a$, there is a number $-a$ such that $a+(-a)=(-a)+a=0$ -(P4) If $a$ and $b$ are any numbers, then $a+b=b+a$ -(P5) If $a,b$ and $c$ are any numbers, then $a\cdot(b\cdot c)=(a\cdot b)\cdot c$ -(P6) If $a$ is any number, then $a\cdot 1=1\cdot a=a$ - -Then it further states that $1\neq 0$. In the book it says that it was an important fact to list because there is no way that it could be proven on the basis of the $6$ properties listed above - these properties would all hold if there were only one number, namely $0$. -Questions: -1) How does one rigorously prove that $1\neq0$ cannot be proven from the $6$ properties listed? -2) It says that "these properties would all hold if there were only one number, namely $0$." Is a reason as to why this is explicitly mentioned is to avoid this trivial case where we only have the number $0$? Is there another deeper reason as to why this sentence was mentioned in relation to $1\neq 0$? -NB: Can someone please check if the tags are appropriate and edit if necessary? Thanks. - -REPLY [5 votes]: (1) How does one rigorously prove that 1≠0 cannot be proven from the 6 properties listed? -(2) It says that "these properties would all hold if there were only one number, namely 0 ." Is a reason as to why this is explicitly mentioned is to avoid this trivial case where we only have the number 0 ? Is there another deeper reason as to why this sentence was mentioned in relation to 1≠0 ? - -Any equational algebraic theory whose axioms are all universal, i.e. that assert equalities of terms composed of operations, variables, and constants, for all values of the variables, necessarily has a one element model. Indeed, defining all of the constants to be the one element (say $0)$ and defining all the operations to have value $0$ makes all axioms true, since they evaluate to $\,0 = 0.$ -Hence $\,1\ne 0\,$ is not deducible from your axioms since it is not true in a one element model. -The reason that $\rm\:1\ne 0\:$ is adjoined as an axiom for fields (and domains) is simply a matter of convenience. For example, it proves a very convenient target for proofs by contradiction, which often conclude by deducing $\rm\:1 = 0.\:$ Also, it avoids the inconvenience of needing to explicitly exclude in proofs motley degenerate cases that occur in one element rings, e.g. that $\rm\:0\:$ is invertible, since $\rm\:0\cdot 0 = 1\, (= 0).\:$ Much more so than proofs by contradiction, this confuses many students (and even some experienced mathematicians) as witnessed here in the past, e.g. see the long comment threads here and here (see esp. my comments in Hendrik's answer).<|endoftext|> -TITLE: congruent to mod p $1^{p-2}+2^{p-2}+\cdots+\left(\frac{p-1}{2}\right)^{p-2}\equiv\frac{2-2^p}{p}\pmod p.$ -QUESTION [5 upvotes]: Let $p$ be an odd prime.How to prove that -$$1^{p-2}+2^{p-2}+\cdots+\left(\frac{p-1}{2}\right)^{p-2}\equiv\frac{2-2^p}{p}\pmod p.$$ - -REPLY [2 votes]: $$2^p=(1+1)^p=\sum_{k=0}^p\binom {p}{k}=\left[\binom{p}{0}+\binom{p}{p}\right]+\ldots+\left[\binom{p}{\frac{p-1}{2}}+\binom{p}{\frac{p+1}{2}}\right]=$$ -$$=2\left[\binom{p}{0}+\ldots+\binom{p}{\frac{p-1}{2}}\right]$$ -and etc.<|endoftext|> -TITLE: If $n\ne 4$ is composite, then $n$ divides $(n-1)!$. -QUESTION [33 upvotes]: I have a proof and need some feedback. It seems really obvious that the statement is true but it is always the obvious ones that are a little trickier to prove. So I would appreciate any feedback. Thank you! -Here is what I am asked to prove: - -If $n$ is composite then $(n-1)! \equiv 0 \pmod n$. - -Proof: -$n$ is composite $\implies n=ab$ where $a,b \in \mathbb{Z}$ and $0 2$, then $2p < p^2 = n$, and thus we may choose $a = p$ and $b = 2p$ to show that $2n \mid (n-1)!$, and therefore also that $n \mid (n-1)!$. -Finally, the fact that the result does not hold for any prime $n$ follows easily from the fundamental theorem of arithmetic, as the prime factorization of $(n-1)!$ will not contain $n$ if it is prime. (I'm sure there are weaker lemmas that could be used to prove this, but why bother? The FToA does it cleanly and easily.) Thus, for integers $n > 1$, $n \nmid (n-1)!$ if and only if $n$ is either prime or $4$. -(In fact, the result holds trivially for $n = 1$ too, at least under the usual definition of $0! = 1$, but that does not follow from the argument above — $1$ is not composite in the sense needed for the argument.)<|endoftext|> -TITLE: Cech cohomology of $\mathbb A^2_k\setminus\{0\}$ -QUESTION [6 upvotes]: I'm trying to prove, via the Cech cohomology, that $S=\mathbb A^2_k\setminus\{0\}$ with the induced Zariski topology is not an affine variety. Consider the structure sheaf $\mathcal O_{\mathbb A^2_k}\big|_S:=\mathcal O_S$ (which is quasi coherent), i must show that $\exists n$ such that $\check H^n(S,\mathcal O_S)\neq 0$. It is enough to prove that $\check H^n(\mathcal U,\mathcal O_S)\neq0$ for a certain affine cover of $S$ (and a certain $n$); so let's choose $\mathcal U=\{D(X), D(Y)\}$ where $D(X)=\{(x,y)\in S\,:\, x\neq 0\}$ and $D(Y)=\{(x,y)\in S\,:\, y\neq 0\}$. Clearly for $n\ge 2$ we have that $\check H^n(S,\mathcal O_S)=0$, so i must show that $\check H^1(\mathcal U,\mathcal O_S)\neq0$. The Cech complex is: -$$\mathcal O_S(D(X))\times\mathcal O_S(D(Y))=\Gamma(S)_X\times\Gamma(S)_Y\longrightarrow \mathcal O_S(D(X)\cap D(Y))=\Gamma(S)_{XY}\longrightarrow 0\cdots$$ -with the homomorphism: $d^0: (f,g)\mapsto g|_{{D(X)\cap D(Y)}}-f|_{{D(X)\cap D(Y)}}$. To complete the proof i should conclude that $d^0$ is not surjective, but why is this true? -thanks - -REPLY [11 votes]: First notice that the restriction morphism $\Gamma(\mathbb A^2_k,\mathcal O_{ \mathbb A^2_k})\to \Gamma(S, \mathcal O_S)$ is bijective because the affine plane $A^2_k$ is normal ("Hartogs phenomenon"). -Hence we may identify $\Gamma(S, \mathcal O_S)$ with the polynomial ring $k[X,Y]$ -a) The open set $D(X)$ is isomorphic to $\mathbb G_m\times \mathbb A^1_k$ where $\mathbb G_m=\operatorname {Spec} k[T,T^{-1}]$, the affine line with origin deleted. -Hence $\Gamma(D(X),\mathcal O_{ A^2_k})=k[X,X^{-1},Y]$. -b) Similarly $D(Y)$ is isomorphic to $\mathbb A^1_k \times \mathbb G_m$. -Hence $\Gamma(D(Y),\mathcal O_{ A^2_k})=k[X,Y, Y^{-1}]$. -c) Finally the open set $D(X)\cap D(Y)$ is isomorphic to the product $\mathbb G_m\times_k \mathbb G_m$ . -Hence $\Gamma(D(X)\cap D(Y),\mathcal O_{ A^2_k})=k[X,X^{-1}]\otimes _k k[Y,Y^{-1}]= k[X,X^{-1},Y,Y^{-1}]$. -d) With these identifications established, the first cohomology group $\check H^1(\mathcal U,\mathcal O_S)$ of the structural sheaf is the cohomology of the complex -$$ k[X,X^{-1},Y]\times k[X,Y,Y^{-1}] \to k[X,X^{-1},Y,Y^{-1}] \to 0 $$ where the non trivial map is $$(f(X,X^{-1},Y),g(X,Y,Y^{-1}))\mapsto g(X,Y,Y^{-1})-f(X,X^{-1},Y)$$ -e) Hence we see that the required cohomology is the following infinite dimensional $k$-vector space , spectacularly violating vanishing of cohomology for affine schemes, which $S$ is thus not. -Final result -$$ \check H^1(\mathcal U,\mathcal O_S)=\check H^1(S,\mathcal O_S)=\oplus _{i,j\gt 0} \; k\cdot X^{-i} Y^{-j} $$<|endoftext|> -TITLE: $|2^x-3^y|=1$ has only three natural pairs as solutions -QUESTION [10 upvotes]: Consider the equation $$|2^x-3^y|=1$$ in the unknowns $x \in \mathbb{N}$ and $y \in \mathbb{N}$. Is it possible to prove that the only solutions are $(1,1)$, $(2,1)$ and $(3,2)$? - -REPLY [5 votes]: Case 1: If $2^x=3^y+1$ then $x \ge 1$. If $y=0$ then $x=1$. -If $y \ge 1$ then $3^y+1 \equiv 1 \pmod{3}$. Therefore $2^x \equiv 1 \pmod{3}$. Hence $x=2x_1$ with $x_1 \in \mathbb{N}^*$. The equation is equivalent to $$3^y= (2^{x_1}-1)(2^{x_1}+1)$$ -Since $2^{x_1}+1>2^{x_1}-1$ and $\gcd (2^{x_1}-1,2^{x_1}+1) \ne 3$ then $2= 3^{m} (3^{n-m}-1)$ with $2^{x_1}-1=3^m,2^{x_1}+1=3^n$. Thus, $m=0,n=1$. We obtain $x_1=1$ or $x=2$. Thus, $y=1$. -Case 2. If $3^y=2^x+1$. If $x=0$ then there is no natural number $y$. If $x=1$ then $y=1$. If $x \ge 2$ then $3^y \equiv 1 \pmod{4}$. Thus, $y$ is even, let $y=2y_1$ with $y_1 \in \mathbb{N}^*$. The equation is equivalent to $$2^y=(3^{y_1}-1)(3^{y_1}+1)$$ -Thus, $2=2^n(2^{m-n}-1)$ with $2^m=3^{y_1}+1,2^n=3^{y_1}-1$. It follows that $n=1,m=2$. Thus, $y_1=1$ or $y=2$. We get $x=3$. -The answer is $\boxed{ (x,y)=(1,0),(1,1),(2,1),(3,2)}$.<|endoftext|> -TITLE: Is locally free sheaf of finite rank coherent? -QUESTION [10 upvotes]: Let $\mathcal{F}$ be a locally free sheaf of finite rank of scheme $X$, is $\mathcal{F}$ coherent? -By the definition of locally free sheaf, there exists an open cover {$U_i$} of $X$ such that $\mathcal{F}|_{U_i}$ is isomorphic to the sheaf $\widetilde{\mathcal{O}(U_i)^n}$. But we don't know each $U_i$ affine or not! -So, it that true or not? -How about $X$ being locally noetherian? It $X$ is, we can find $V_{ij} \subset U_i$ s.t. $V_{ji} = Spec(A_{ji})$. And $\mathcal{F}|_{V_{ij}} = \mathcal{F}|_{U_i}|_{V_{ij}}$...? -For example, $X(\Delta)$ is a toric varity with $\Delta$ consists of strongly convex polyhedral cones. -Thank you very much!! - -REPLY [18 votes]: The exact condition for locally free sheaves on a ringed space $(X,\mathcal O_X)$ to be coherent is exactly that $\mathcal O_X$ be coherent. -a) The condition is clearly necessary since $\mathcal O_X$ is locally free. -b) It is sufficient because if the structure shaf is coherent, then coherence is a local property and because a direct sum of coherent sheaves is coherent: apply to $\mathcal F \mid U_i \cong (\mathcal O\mid U_i)^{\oplus r} $ -And when is $\mathcal O_X$ coherent? -There is, to my knowledge, no very good non-tautological criterion. -However, for locally noetherian schemes, it is the case that $\mathcal O_X$ is coherent, so for these schemes, yes, locally free sheaves are coherent.<|endoftext|> -TITLE: Example of non-decomposable ideal -QUESTION [5 upvotes]: An ideal $I$ of a commutative unital ring $R$ is called decomposable if it has a primary decomposition. - -Can you give an example of an ideal that is not decomposable? - -All the examples I can think of are decomposable. Thanks. - -REPLY [8 votes]: Zero ideal in $C[0,1]$ is not decomposable. -More generally, - -If $X$ is an infinite compact Hausdorff space then the zero ideal of $C(X)$, the ring of real valued continuous functions on $X$, is not decomposable. - -Proof : First let us note that every maximal ideal of $C(X)$ is of the form $M_x=\lbrace f \in C(X) : f(x)=0\rbrace$, for some $x \in X$. Note that, if the zero ideal of $C(X)$ were decomposable, then there -would be only finitely many minimal prime ideals of $C(X)$. This certainly -sounds strange since every maximal ideal $M_x$ of $C(X)$ contains a minimal -prime ideal as every maximal ideal is also prime ideal. Hence to show that the zero ideal of $C(X)$ is not decomposable it is enough to show that if $x\not= y\in X$ then any two minimal prime ideals -$P_1\subseteq M_x, P_2\subseteq M_y$ are different: Since $X$ is Hausdorff and normal, there is an open set $U$ -such that $x\in U$ and $y\not\in\overline{U}$. By Urysohn's lemma there are$ f,g\in C(X)$ -such that $f(U) = 0, f(y) = 1, g(x) = 1,$ and $g(X \setminus U) = 0.$ So $fg = 0$. Therefore $fg \in P_1$ but $g \not\in P_1$, because $g(x)\not=0$, hence $f \in P_1$. But $f \not\in P_2$, because $f(y)\not=0$. Hence $P_1\not=P_2$.<|endoftext|> -TITLE: Why do the endpoints of the Maclaurin series for arcsin converge? -QUESTION [7 upvotes]: The series $$\sum_{n=0}^\infty {{-\frac {1} 2} \choose n} \frac{(-1)^n}{2n+1}$$ is an endpoint for the Maclaurin series for arcsin(x). (The other endpoint is just the negative of this one.) I played around with this a bit and turned it into three (potentially useful) forms : -$\sum_{n=0}^\infty \frac12\frac34\frac56\cdots\frac{2n-1}{2n} \frac{1}{2n+1}$ -$\sum_{n=0}^\infty \left( 1-\frac 1 2 \right) \left( 1-\frac 1 4 \right)\cdots \left( 1-\frac 1 {2n} \right)\frac{1}{2n-1}$ -$\sum_{n=0}^\infty \dfrac{(2n)!}{2^{2n}(n!)^2}\dfrac{1}{2n+1}$, -but I'm not sure where one could go from here. - -REPLY [2 votes]: Another approach would be Stirling's closedly related Wallis' formula. It gives -$$\lim_{n \to \infty} \frac{(2n)!}{(2n-1)!}\frac{1}{\sqrt n}=\sqrt{\pi}$$ -This means that your $n$th term is asymptotically equal to -$$ {\sqrt{\frac 1 {\pi n}}}\frac{1}{2n+1}$$ -which explains why the series converges. In particular, -$$\lim_{n \to \infty} \frac{n^{3/2}}{\sqrt{n}(2n+1)}=\frac 1 2 $$ so the convergence of $$\sum_{n >0} n^{-3/2} $$ implies that of your series.<|endoftext|> -TITLE: How to derive the equation for a bézier curve -QUESTION [7 upvotes]: So, I remember a while back there was a maths competition and we were given a curve that we needed to write an equation for. I just skipped the question since I didn't even know where to begin. I remember it was one among the last few questions of the paper and it was worth a lot of points. -I don't really remember what the curve looked like; it was something spirally, but I can't recall it to save my life right now. -So, I drew this curve in Inkscape (it's a Bézier curve. Or a few of them linked together, according to Wikipedia. If it's required I will post the whole path). And I would like to write the equation for it (with someone's help, obviously). - - -I was always a bit bad with curves, graphs and lines, but I want to understand them better. So, I was hoping someone could explain the process of deriving the equation for a curve. -P.S: I'd like it if you could use another curve (it can be something simpler, but try avoiding something overly complicated) so I can crack this one on my own, but if you feel like using this curve as an example I won't mind. - -EDIT -So have been browsing the internet, read a few Wikipedia entries about Bazier curves, and I understand how they're drawn (mostly the GIFs helped, haha), but I am still stumped when it comes to mathematically representing a Bézier curve. Also, I will add this image, which is the path and its control points (at the end of the blue lines; I didn't paint them in): - -And also, the contents of the .tex file for the shape. -%LaTeX with PSTricks extensions -%%Creator: 0.48.2 -%%Please note this file requires PSTricks extensions -\psset{xunit=.5pt,yunit=.5pt,runit=.5pt} -\begin{pspicture}(451.46875,34.25392151) - { - \newrgbcolor{curcolor}{1 0 0} - \pscustom[linewidth=3,linecolor=curcolor] - { - \newpath - \moveto(450.48448,1.10834551) - \curveto(404.89404,41.45133951)(333.34998,42.21654151)(281.90128,9.03018551) - \curveto(258.09407,-6.32636849)(228.42388,9.91159551)(202.75741,15.38398551) - \curveto(145.68728,27.55199551)(85.852286,40.32786151)(28.08402514,26.23698551) - \curveto(18.5710181,23.91656551)(9.403556,20.24334551)(0.681686,15.78116551) - } - } - \end{pspicture} - -Thanks! - -REPLY [6 votes]: Linear Bézier curve is simply a line given by parametric equation $R(t) = A+t(AB)$ , A being initial point and B being final point. -For Quadratic Bézier curve, take a look at the following picture. - -Let the point between $P_1$ and $P_0$ be $Q_1$ and $P_1$ and $P_2$ be $Q_2$. Let our path be traced by $Q_0$. Then from above figure. -$$ \frac{P_0Q_1}{P_0P_1} = \frac{P_1Q_2}{P_1P_2} = \frac{Q_1Q_0}{Q_1Q2} = t \text{ (say)} $$ -$$Q_1 = P_0 + t(P_0P_1), Q_2 = P_1 + t(P_1P_2)$$ -So we have -$$Q_0 = Q_1 + t(Q_1Q_2) = P_0 + t(P_0P_1) + t(P_1 + t(P_1P_2) - (P_0 + t(P_0P_1)))$$ -Have a look at more elaborate article on Wikipedia.<|endoftext|> -TITLE: Is this kind of simplicial complex necessarily homotopy equivalent to a wedge of spheres? -QUESTION [5 upvotes]: Suppose $d \ge 2$ and $S$ is a finite simplicial complex of dimension $2d$, such that -(1) $S$ is simply connected, i.e. $\pi_1 ( S) = 0$, and -(2) all the homology of $S$ is in middle degree, i.e. $\widetilde{H}_i ( S, \mathbb{Z}) = 0,$ unless $i = d$, and -(3) homology $\widetilde{H}_d(S, \mathbb{Z})$ is torsion-free. -Does it necessarily follow that $S$ is homotopy equivalent to a wedge of spheres of dimension $d$? -If $S$ can be shown to be homotopy equivalent to a $d$-dimensional cell complex, for example, this would follow by standard results. - -REPLY [4 votes]: I will also assume that $S$ is connected, as is usually implied by simply connected, though your definition does not explicitly state this. Since $S$ is a finite simplicial complex, $H_d(S, \mathbb Z) \cong \bigoplus_{i=1}^k \mathbb Z$. By the Hurewicz theorem, the Hurewicz map $\pi_d(S) \to H_d(S)$ is an isomorphism, so we can pick maps $p_i: S^d \to S$ such that $P = \bigvee_i p_i : H_d(\bigvee_i S^d) \to H_d(S)$ is an isomorphism, of course, we also see that $P : H_k(\bigvee_i S^d) \to H_k(S)$ is an isomorphism for $k\neq d$. Since both spaces are simply connected, this implies by the Hurewicz theorem again that $P$ is a weak equivalence, and by the Whitehead Theorem, $P$ is a homotopy equivalence.<|endoftext|> -TITLE: Open math problems which high school students can understand -QUESTION [49 upvotes]: I request people to list some moderately and/or very famous open problems which high school students,perhaps with enough contest math background, can understand, classified by categories as on arxiv.org. Please include statement of the theorems,if possible, and if there are specific terms, please state what they mean. -Thank you.I am quite inquisitive to know about them and I asked this question after seeing how Andrew J.Wiles was fascinated by Fermat's last theorem back in high school. - -REPLY [4 votes]: An open problem I find surprising, the PAC (Perimeter to Area Conjecture) due to Keleti (1998): - -Conjecture: The perimeter to area ratio of the union of finitely many unit squares in the plane does not exceed 4. - -See for example Bounded - Yes, but 4? and references therein.<|endoftext|> -TITLE: Characterization of real functions which have limit at each point -QUESTION [10 upvotes]: The following problem is Exercise 7.K from the book van Rooij-Schikhof: A Second Course on Real Functions and it is very close to a question which was recently discussed in chat. -So I thought that sharing this interesting problem with other MSE users could be useful. Here's the problem: - -Let $L$ be the set of all functions $f\colon [0,1]\to\mathbb R$ that have the property that $\lim\limits_{x\to a} f(x)$ exists for all $a \in [0, 1]$. - Show that: - (i) $L$ is a vector space. Each $f \in L$ is bounded. - (ii) For each $f \in L$, define $f^c(x): = \lim\limits_{y\to x} f(y)$ ($x \in [0, 1]$). $f^c$ is continuous. - (iii) '$f^c =0$' is equivalent to 'there exist $x_1, x_2, \dots$ in $[0,1]$ and $a_1, a_2,\dots$ in $U$ with - $\lim\limits_{n\to\infty} a_n = 0$, such that $f(x_n) = a_n$ for every $n$, and $f=0$ elsewhere'. - (iv) Describe the general form of an element of $L$. Show that every $f\in L$ is Riemann integrable. - -The original question in the chat was about functions $\mathbb R\to\mathbb R$, but it does not change much in the parts (iii) and (iv). - -REPLY [2 votes]: I'll post my solution, too. - -(i) It is obvious that $L$ is a vector space. -If we fix some $\varepsilon>0$ then we have for each $a\in[0,1]$ an open neighborhood $U_a$ such that $\operatorname{diam} U_a<\varepsilon$. The sets $U_a$ form an open cover of the compact set $[0,1]$, so we have a finite subcover. Therefore the range of $f$ is covered by finitely many bounded sets and it is bounded itself. - -(ii) Let us try to show the continuity of the function $f^c$ defined by $f^c(x)=\lim\limits_{y\to x} f(y)$. -Suppose that $f^c$ is not continuous at some point $a$. Let us denote $b:=f^c(a)$. Then there exists a sequence $x_n$ such that $x_n\to a$, $x_n\ne a$ and $f^c(x_n)$ does not converge to $b$. -By definition of $f^c$ we can find for each $x_n$ a point $y_n$ such that $|y_n-x_n|<1/2^n$ and $|f(y_n)-g(x_n)|<1/2^n$. We can assume that, in addition to this, $y_n\ne a$. -For such sequence we have $y_n\to a$ and $f(y_n)\not\to b=f^c(a)$, which contradicts the definition of $f^c$. - -(iii) We want to show $f^c=0$ $\Leftrightarrow$ $f$ has the form described in the part (iii). -The implication $\boxed{\Leftarrow}$ is easy: We have $f^c(x)=\lim\limits_{n\to\infty} f(x_n)$ for any sequence $x_n\to x$, $x_n\ne x$. If we choose any $\varepsilon$, there are only finitely many $n$'s such that $|f(x_n)|>\varepsilon$. This implies that the limit must be zero. -$\boxed{\Rightarrow}$ If suffices to show that the set $M_\varepsilon:=\{x\in\mathbb R; |f(x)|>\varepsilon\}$ is finite for each $\varepsilon>0$. (Once we have shown this, we can get all non-zero points as the union $\bigcup\limits_{n=1}^\infty M_{1/n}$, so there is only countably many of them. And they can also be ordered in the required way.) -Suppose that for some $\varepsilon>0$ the set $M_\varepsilon:=\{x\in\mathbb R; |f(x)|>\varepsilon\}$ is infinite. Then this set has an accumulation point $a\in[0,1]$. We have $|f_c(a)|\ge\varepsilon$, contradicting the assumption that $f^c$ is zero. - -(iv) Now let $f$ be an arbitrary function from $L$. -It is relatively easy to see that if we denote $g:=f-f^c$, then $g^c$ is identically zero. (I.e., $\lim\limits_{x\to a} g(x)=0$ for each $a\in[0,1]$.) Note that $f=f^c+g$. -So we get that each function from $L$ is a sum of a continuous function and a function which has the form described in the part (iii) of this exercise. -Now to show that $f$ is Riemann integrable we could use Lebesgue's criterion for Riemann integrability and the fact that every countable set has Lebesgue measure zero. -But this can be also shown directly from the definition: For any given $\varepsilon>0$ we have only finitely many points such that $|f(x)|\ge\varepsilon$. If we cover them with small enough intervals, we will get that the Riemann sum over these intervals is in absolute value at most $\varepsilon$. For arbitrary partition containing the small intervals, the absolute value of the Riemann sum over the remaining intervals is at most $\varepsilon$. So we have $|R(f,\Delta)|\le 2\varepsilon$ whenever the norm of the partition is chosen small enough to ensure that the intervals around the "exceptional" points (the points with $|f(x)|\ge\varepsilon$) are small enough.<|endoftext|> -TITLE: Is there any famous number theory conjecture proven impossible to be find out the truth or false? -QUESTION [7 upvotes]: Is there any famous number theory conjecture proven undecidable? -Is there any history about it? -i would like to know any number theory conjecture by the types of undecidable. - -REPLY [6 votes]: [references added, at bottom] -You know the Collatz problem, right? Start with a positive integer, if it's even, divide by 2, if it's odd, multiply by 3 and add 1, iterate and see what happens? The possibilities being that you eventually go into a cycle, or that you never go into a cycle? -Well, you can fit that problem into a family of problems where you fix a modulus $m$, and $m$ different linear functions $f_0,f_1,f_{m-1}$, and then given a positive integer you look at what the remainder would be if you were to divide it by $m$, and if that remainder would be $r$ you apply the function $f_r$ to the positive integer; you iterate this procedure, and again the question is whether you ever get into a cycle. The functions $f_i$ have to be chosen so that you always get integers out of the computations. -It has been proved that there is a modulus $m$ and a collection of functions $f_0,f_1,f_{m-1}$ such that the cycle question is undecidable. -I don't know whether this qualifies. For the original Collatz, the standard conjecture is that you always wind up in the cycle $4,2,1,4,2,1,\dots$. But original Collatz has not been proved undecidable. I don't know whether anyone has written down an explicit example of a problem in this family for which undecidablity can be proved, and I don't know what conjectures are out there about members of the family. -This is probably covered in Lagarias' book about $3x+1$. -EDIT: You can get the flavor of the argument here, although that looks like a preprint that needs some work. The published version of that paper is Kurtz, S. A. and Simon, J. "The Undecidability of the Generalized Collatz Problem," in Theory and Applications of Models of Computation: Proceedings of the 4th International Conference (TAMC 2007) held in Shanghai, May 22-25, 2007 (Ed. J.-Y. Cai, S. B. Cooper, and H. Zhu), Springer, Berlin, pp. 542-553, 2007. The published version is available at doi:10.1007/978-3-540-72504-6_49 but I think that's behind a pay wall. -Conway's paper on this is Conway, J. H., Unpredictable iterations, Proc. 1972 Number Th. Conf., University of Colorado, Boulder, Colorado, pp. 49-52, 1972. It's in the Lagarias book.<|endoftext|> -TITLE: Taking stalk of a product of sheaves -QUESTION [13 upvotes]: Let $(\mathscr{F}_\alpha)_\alpha$ be a family of sheaves on $X$, and $\prod_\alpha\mathscr{F}_\alpha$ the product sheaf. If $x\in X$, is it true that -$$\left(\prod_\alpha\mathscr{F}_\alpha\right)_x\simeq\prod_\alpha(\mathscr{F}_\alpha)_x \ ?$$ -I think $(\oplus_\alpha\mathscr{F}_\alpha)_x\simeq\oplus_\alpha(\mathscr{F}_\alpha)_x$ may be true, but not the product sheaf. - -REPLY [21 votes]: You are right. It is true for direct sums, even arbitrary colimits (since colimits commute with colimits, and remember the description of the stalk as a colimit; or represent it as a pullback functor to the point), and also for finite products, even more generally for finite limits (since these commute with filtered colimits in, say, algebraic categories, in which the sheaves should live). But it is not true for infinite products. -There is always a canonical map $(\prod_{\alpha} F_{\alpha})_x \to \prod_{\alpha} (F_{\alpha})_x$. But it doesn't have to be injective, even for very nice spaces $X$ and sheaves $F_{\alpha}$. Take $X=\mathbb{R}$ and $F_{\alpha}$ the sheaf of continuous function for $\alpha \in \mathbb{N}$, and $x=0$. Let $f_{\alpha} : \mathbb{R} \to \mathbb{R}$ be a continuous function which vanishes on $]-1/(\alpha+1),+1/(\alpha+1)[$, but does not vanish at $1/{\alpha}$. Then $(f_{\alpha})_{\alpha}$ represents an element in the kernel of the canonical map, which is not trivial.<|endoftext|> -TITLE: Elements of finite order in $SL_2(\mathbb{Z})$ -QUESTION [8 upvotes]: Assume $A \in SL_2(\mathbb{Z})$ has finite order, and let $N_0 \in \mathbb{N}$ be the smallest natural number such that $A^{N_0} = I$. I want to show that the only possible values of $N_0$ are $1,2,3,4$ or $6$. I have a proof, but I worry it is incomplete, so my (two part) question is: -$$ -\text{Is the following proof correct? If not, where does the proof break down?} -$$$$ -\text{Does there exist a proof using only first principles? } -$$ -Basically, my proof follows from Corollary 2.4 here, which says "Any homomorphism $SL_2(\mathbb{Z}) \to \mathbb{C}^\times$ has image in the 12th roots of unity" and Lagrange's theorem. Namely, let $N_0$ be as above, and let $\phi: SL_2(\mathbb{Z}) \to \mathbb{C}^\times$ be a homomorphism. Then, since $\phi(I) = 1$, -$$ -\phi(I)=\phi(A^{N_0}) = \phi(A)^{N_0} = 1. -$$ -By Lagrange's theorem, and Corollary 2.4, $N_0 | 12$, so $N_0=1,2,3,4$ or $6$. -\qed -My trouble is that although $N_0$ is the order of $A$ in $SL_2(\mathbb{Z})$, it does \emph{not} need to be the order of $\phi(A)$ in the 12th roots of unity, in general, but I think I used this implicitly when I used Lagrange's theorem. If $\phi(A)^{N_0} = 1$ then all we can conclude is that the order of $\phi(A)$ divides $N_0$, not that it equals $N_0$. If we were still in $SL_2(\mathbb{Z})$, then this implies it does equal $N_0$ by assumption of minimality, but like I said, I don't see why this needs to hold in $\mathbb{C}^\times$. For exmaple, what's stopping $N_0 = 8$ while $\phi(A)$ has order 4 in the 12th roots of unity? This doesn't seem to yield an immediate contradiction. -Whether or not this proof is correct, I'd like to see how (if) my troubles can be resolved, and if a proof in this way can work. I would also like to know if there is a proof using only the definition of $SL_2(\mathbb{Z})$, and perhaps some ingenuity, since I wouldn't consider Corollary 2.4 a standard fact (to, say, a beginning graduate student). - -REPLY [13 votes]: An element of finite order in $SL_2(\mathbb C)$ has roots of unity as eigenvalues. Thus an element in $SL_2(\mathbb Z)$ of finite order has as eigenvalues a pair of roots of unity as eigenvalues, say $\zeta$ and $\zeta'$. Since the determinant of the element $= 1$, we have $\zeta \zeta' = 1,$ and so $\zeta' = \overline{\zeta}$. Since our matrix is in $SL_2(\mathbb Z)$, it has integral trace, and so $2 \Re(\zeta)$ is an integer. -It is not hard to check that the only roots of unity satisfying $\zeta + \overline{\zeta}$ being an integer are $\zeta = \pm 1, \pm i, (\pm 1 \pm \sqrt{-3})/2$. (We are looking for an angle $\theta = 2\pi/n$ such that $\cos \theta$ is an integer or a half-integer, and the only possibilities are thus $\cos \theta = 0, \pm (1/2), \pm 1$, giving the listed roots of unity.) The order of such an element (which corresponds to the order of its eigenvalues, in the group of all roots of unity) is thus $1,$ $2,$ $3$, $4$, or $6$. - -REPLY [3 votes]: You are absolutely correct that the order of $A$ need not agree with the order of $\phi(A)$. Try thinking about what the characteristic polynomial of $A$ can be.<|endoftext|> -TITLE: Description of flipping tableau for inversions in reduced decompositions of permutations -QUESTION [10 upvotes]: Short version: Is there a graphical description of the possible orders in which inversions can appear in a reduced decomposition of a permutation? -Something akin to the definition of standard Young tableaux. -Inversions of longest permutation version -A permutation of degree $n$ is a bijection on $\{1,2,\cdots,n\}$. An inversion of a permutation $\pi$ of degree $n$ is a pair $(i,j)$ such that $1 \leq i < j \leq n$ but $\pi(i) > \pi(j)$. A reduced decomposition of a permutation $\pi$ of degree $n$ is an expression $\pi = g_1 \cdots g_m$ so that (1) each $g_k = (i,i+1)$ is an adjacent transposition for some $i$, (2) $m$ is the number of inversions of $\pi$. The longest element $\pi_0$ of degree $n$ is the permutation in which every pair is an inversion; $\pi_0 = (1,n)(2,n-1)\cdots$ is the "reversal". -In a reduced decomposition, the set of inversions of $g_1 \cdots g_{k+1}$ has precisely one more element $u_k = \{i,j\}$ than the set of inversions of $g_1 \cdots g_k$. The flipping tabelau of the decomposition is the tableau whose $j,i$ entry is $k$ where $u_k=\{i,j\}$ with $i -TITLE: Tensors: Acting on Vectors vs Multilinear Maps -QUESTION [21 upvotes]: I have the feeling like there are two very different definitions for what a tensor product is. I was reading Spivak and some other calculus-like texts, where the tensor product is defined as -$(S \otimes T)(v_1,...v_n,v_{n+1},...,v_{n+m})= S(v_1,...v_n) * T(v_{n+1},...,v_{n+m}) $ -The other definition I read in a book on quantum computation, its defined for vectors and matrices and has several names, "tensor product, Kronecker Product, and Outer product": http://en.wikipedia.org/wiki/Outer_product#Definition_.28matrix_multiplication.29 -I find this really annoying and confusing. In the first definition, we are taking tensor products of multilinear operators (the operators act on vectors) and in the second definition the operation IS ON vectors and matrices. I realize that matrices are operators but matrices aren't multilinear. Is there a connection between these definitions? - -REPLY [42 votes]: Let's first set some terminology. -Let $V$ be an $n$-dimensional real vector space, and let $V^*$ denote its dual space. We let $V^k = V \times \cdots \times V$ ($k$ times). - -A tensor of type $(r,s)$ on $V$ is a multilinear map $T\colon V^r \times (V^*)^s \to \mathbb{R}$. -A covariant $k$-tensor on $V$ is a multilinear map $T\colon V^k \to \mathbb{R}$. - -In other words, a covariant $k$-tensor is a tensor of type $(k,0)$. This is what Spivak refers to as simply a "$k$-tensor." - -A contravariant $k$-tensor on $V$ is a multilinear map $T\colon (V^*)^k\to \mathbb{R}$. - -In other words, a contravariant $k$-tensor is a tensor of type $(0,k)$. - -We let $T^r_s(V)$ denote the vector space of tensors of type $(r,s)$. So, in particular, - -$$\begin{align*} -T^k(V) := T^k_0(V) & = \{\text{covariant $k$-tensors}\} \\ -T_k(V) := T^0_k(V) & = \{\text{contravariant $k$-tensors}\}. -\end{align*}$$ -Two important special cases are: -$$\begin{align*} -T^1(V) & = \{\text{covariant $1$-tensors}\} = V^* \\ -T_1(V) & = \{\text{contravariant $1$-tensors}\} = V^{**} \cong V. -\end{align*}$$ -This last line means that we can regard vectors $v \in V$ as contravariant 1-tensors. That is, every vector $v \in V$ can be regarded as a linear functional $V^* \to \mathbb{R}$ via -$$v(\omega) := \omega(v),$$ -where $\omega \in V^*$. - -The rank of an $(r,s)$-tensor is defined to be $r+s$. - -In particular, vectors (contravariant 1-tensors) and dual vectors (covariant 1-tensors) have rank 1. - -If $S \in T^{r_1}_{s_1}(V)$ is an $(r_1,s_1)$-tensor, and $T \in T^{r_2}_{s_2}(V)$ is an $(r_2,s_2)$-tensor, we can define their tensor product $S \otimes T \in T^{r_1 + r_2}_{s_1 + s_2}(V)$ by -$$(S\otimes T)(v_1, \ldots, v_{r_1 + r_2}, \omega_1, \ldots, \omega_{s_1 + s_2}) = \\ -S(v_1, \ldots, v_{r_1}, \omega_1, \ldots,\omega_{s_1})\cdot T(v_{r_1 + 1}, \ldots, v_{r_1 + r_2}, \omega_{s_1 + 1}, \ldots, \omega_{s_1 + s_2}).$$ -Taking $s_1 = s_2 = 0$, we recover Spivak's definition as a special case. -Example: Let $u, v \in V$. Again, since $V \cong T_1(V)$, we can regard $u, v \in T_1(V)$ as $(0,1)$-tensors. Their tensor product $u \otimes v \in T_2(V)$ is a $(0,2)$-tensor defined by -$$(u \otimes v)(\omega, \eta) = u(\omega)\cdot v(\eta)$$ - -As I suggested in the comments, every bilinear map -- i.e. every rank-2 tensor, be it of type $(0,2)$, $(1,1)$, or $(2,0)$ -- can be regarded as a matrix, and vice versa. -Admittedly, sometimes the notation can be constraining. That is, we're used to considering vectors as column vectors, and dual vectors as row vectors. So, when we write something like $$u^\top A v,$$ -our notation suggests that $u^\top \in T^1(V)$ is a dual vector and that $v \in T_1(V)$ is a vector. This means that the bilinear map $V \times V^* \to \mathbb{R}$ given by -$$(v, u^\top) \mapsto u^\top A v$$ -is a type $(1,1)$-tensor. -Example: Let $V = \mathbb{R}^3$. Write $u = (1,2,3) \in V$ in the standard basis, and $\eta = (4,5,6)^\top \in V^*$ in the dual basis. For the inputs, let's also write $\omega = (x,y,z)^\top \in V^*$ and $v = (p,q,r) \in V$. Then -$$\begin{align*} -(u \otimes \eta)(\omega, v) & = u(\omega) \cdot \eta(v) \\ -& = \begin{pmatrix} - 1 \\ - 2 \\ - 3 -\end{pmatrix} (x,y,z) -\cdot -(4,5,6) \begin{pmatrix} - p \\ - q \\ - r -\end{pmatrix} \\ -& = (x + 2y + 3z)(4p + 5q + 6r) \\ -& = 4px + 5 qx + 6rx \\ -& \ \ \ \ \ 8py + 10qy + 12py \\ -& \ \ \ \ \ 12pz + 15qz + 18rz \\ -& = (x,y,z)\begin{pmatrix} - 4 & 5 & 6 \\ - 8 & 10 & 12 \\ - 12 & 15 & 18 -\end{pmatrix}\begin{pmatrix} - p \\ - q \\ - r -\end{pmatrix} \\ -& = \omega \begin{pmatrix} - 4 & 5 & 6 \\ - 8 & 10 & 12 \\ - 12 & 15 & 18 -\end{pmatrix} v. -\end{align*}$$ -Conclusion: The tensor $u \otimes \eta \in T^1_1(V)$ is the bilinear map $(\omega, v)\mapsto \omega A v$, where $A$ is the $3 \times 3$ matrix above. -The Wikipedia article you linked to would then regard the matrix $A$ as being equal to the tensor product $u \otimes \eta$. - -Finally, I should point out two things that you might encounter in the literature. -First, some authors take the definition of an $(r,s)$-tensor to mean a multilinear map $V^s \times (V^*)^r \to \mathbb{R}$ (note that the $r$ and $s$ are reversed). This also means that some indices will be raised instead of lowered, and vice versa. You'll just have to check each author's conventions every time you read something. -Second, note that there is also a notion of tensor products of vector spaces. Many textbooks, particularly ones focused on abstract algebra, regard this as the central concept. I won't go into this here, but note that there is an isomorphism -$$T^r_s(V) \cong \underbrace{V^* \otimes \cdots \otimes V^*}_{r\text{ copies}} \otimes \underbrace{V \otimes \cdots \otimes V}_{s \text{ copies}}.$$ -Confusingly, some books on differential geometry define the tensor product of vector spaces in this way, but I think this is becoming rarer.<|endoftext|> -TITLE: Number of well-ordering relations on a well-orderable infinite set $A$? -QUESTION [8 upvotes]: Given a well-orderable infinite set $A$, can we always say that the set $$\left\{R\subset A\times A:\langle A,R\rangle\, \text{is a well-ordering}\right\}$$ has cardinality $2^{|A|}$? How much Choice is required for the proof of this? -I believe that where $A$ is countably infinite, we can proceed without any use of Choice. Is that correct? - -REPLY [8 votes]: This is a delicate matter. Do you mean all the well orders or just up to isomorphism? -For the former observe, for example, that if a set $A$ has one well-ordering then any permutation induces a different well-ordering, although of the same order type. For the natural numbers there are $2^{\aleph_0}$ many permutations so there are at least continuum many well-orders, and that is just of one isomorphism type! On the other hand, there can only be continuum many relations, so we have exhausted the cardinality. -We can continue by induction on the $\aleph$-cardinals, in fact, this is the only way we can resume. Why? Well, if a set has any well-ordering then it has to be in bijection with an ordinal, and if it is infinite this ordinal can be an $\aleph$-number. The argument for $\omega_1$, $\omega_2$ and so on is the same as above and this would require no choice at all. -If a set cannot be well-ordered, well... it has no well-ordering! However if we assume the axiom of choice then every set can be well-ordered and going through cardinals is enough. So arguing this claim for all sets is to require the full axiom of choice. - -On the other hand, if you are interested in order types rather than mere orders than the claim that there are $2^A$ many is in fact to assert the Generalized Continuum Hypothesis, since for a set of cardinality $\kappa$ there are only $\kappa^+$ many ordinals of cardinality $\kappa$, and therefore only $\kappa^+$ many order types. -Whether or not $2^\kappa=\kappa^+$ is undecided in ZFC.<|endoftext|> -TITLE: Proof $\lim\limits_{n \rightarrow \infty} {\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}=2$ using Banach's Fixed Point -QUESTION [7 upvotes]: I'd like to prove $\lim\limits_{n \rightarrow \infty} \underbrace{\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}_{n\textrm{ square roots}}=2$ using Banach's Fixed Point theorem. -I think I should use the function $f(x)=\sqrt{2+x}$. This way, if I start the iterations for example with $x_0=0$, I will have $x_1=\sqrt2$. When I calculate $x_2$ I will get $\sqrt{2+\sqrt{2}}$. And $x_3 = \sqrt{2+\sqrt{2+\sqrt{2}}}$ and so on. I can see that these iterations are monotone increasing, but how can I show that this converges to 2? -Pseudo-related formula I found: http://en.wikipedia.org/wiki/Vieta_formula -Many thanks in advance! - -Following clark's advice, here's my proof this is a contraction. I'm using the interval $D=[0, 2]$. -$f'(x)=\frac{1}{2\sqrt{x+2}}$, which is monotone decreasing. This means its highest value in $D$ is $0$. $f'(0)=\frac{1}{2\sqrt{2}} < 1$. The rate $M$ of the contraction is then $\frac{1}{2\sqrt{2}}$. - -REPLY [2 votes]: This doesn't answer your question, but it might be of interest: -Define que sequence $\{x_n\}$ by -$$\begin{cases} x_0=0 \cr x_n = \sqrt{k+x_{n-1}}\end{cases}$$ -with $k>0$ -I claim that $$\lim_{n \to \infty}x_n=r$$ -where $r$ is the positive root of the equation -$$\tag A x^2-x-k=0 $$ -PROOF -$(1)$ The sequenece is increasing. By induction: -It is true for $x_0=0,x_1=\sqrt k$. Assume true for $k=1,2,\dots,n$, then -$$x_n > x_{n-1} \Rightarrow x_n+k > x_{n-1}+k \Rightarrow$$ -$$\Rightarrow \sqrt{x_n+k} > \sqrt{x_{n-1}+k} \Rightarrow x_{n+1} > x_n$$ -$(2)$ The sequence is bounded above by $r$. By induction: -It is true for $n=0,1$. Assume true for $k=1,2,\dots,n$, then -$$x_{n} < r$$ -$$x_{n}+k < r+k$$ -$$\sqrt{x_{n}+k} < \sqrt{r+k}=r$$ -since $r$ satisfies $(A)$. -Then by the Monotone Convergence Theorem, the sequence has a limit. In particular, this means that $\ell = \lim x_n = \lim x_{n-1}$, so that -$$\lim_{n \to \infty} x_n = \lim_{n \to \infty}\sqrt{x_{n-1}+k} $$ -$$\lim_{n \to \infty} x_n = \sqrt{ \lim_{n \to \infty} x_{n-1}+k} $$ -$$\ell = \sqrt{\ell+k} $$ -$$\ell^2-\ell -k = 0 $$ -Then either -$$\ell_1 = \frac{1+\sqrt{1+4k}}{2}$$ -or -$$\ell_2 = \frac{1-\sqrt{1+4k}}{2}$$ -But the latter is impossible since $\ell_2 <0$. It follows that -$$\ell_1 = r$$ the positive root of the equation $x^2-x-k=0$. $\blacktriangle$ -Your problem is the special case $k=2$, which yields -$$\ell = \frac{1+\sqrt{1+4\cdot 8 }}{2}=2$$<|endoftext|> -TITLE: Evaluation of $\sum_{n=1}^\infty \frac{1}{\Gamma (n+s)}$ -QUESTION [9 upvotes]: I want to try and evaluate this interesting sum: -$$\sum_{n=1}^\infty \frac{1}{\Gamma (n+s)}$$ -where $0 \le s < 1$ -WolframAlpha evaluates this sum to be -$$\sum_{n=1}^\infty \frac{1}{\Gamma (n+s)} = e\left(1-\frac{\Gamma(s, 1)}{\Gamma(s)}\right)$$ -Some notable cases of this sum would be when $s=0$ (producing the Taylor's series for $e$) and when $s=\frac{1}{2}$: -$$\sum_{n=1}^\infty \frac{1}{\Gamma (n+\frac{1}{2})} = e \operatorname {erf}(1)$$ -I would be very interested to know the steps of how one would evaluate this interesting sum. - -REPLY [12 votes]: We find an integral expression for the sum (the $u$ integral below) without appealing to the properties of special functions. -We have -$$\begin{eqnarray*} -\sum_{n=1}^\infty \frac{1}{\Gamma(n+s)} -&=& \frac{1}{\Gamma(s+1)} -\underbrace{\left(1+\frac{1}{s+1}+\frac{1}{(s+1)(s+2)} + \ldots\right)}_{f(s)}. -\end{eqnarray*}$$ -The series $f(s)$ is a simple example of an inverse factorial series. -Such series were studied even in the 18th century by Nicole and Stirling and are dealt with, for example, in Whittaker and Watson's A Course of Modern Analysis. -One way to develop such a series is by successively integrating by parts the right hand side of -$$f(s) = \int_0^1 d\xi\, s(1-\xi)^{s-1} F(\xi),$$ -where $F(\xi)$ is some analytic function of $\xi$ and -$\int_0^1$ is shorthand for $\lim_{\epsilon\to 0^+}\int_0^{1-\epsilon}$. -One finds -$$\begin{eqnarray*} -f(s) &=& F(0) + \frac{F'(0)}{s+1} + \frac{F''(0)}{(s+1)(s+2)} +\ldots. -\end{eqnarray*}$$ -For details on the restrictions on $F(\xi)$, see Whittaker and Watson's 4th edition, $\S 7.82$. -For this problem we have $F^{(n)}(0) = 1$, so $F(\xi) = e^\xi$. -Then -$$\begin{eqnarray*} -\sum_{n=1}^\infty \frac{1}{\Gamma(n+s)} &=& \frac{f(s)}{\Gamma(s+1)} \\ -&=& \frac{1}{\Gamma(s+1)} \int_0^1 d\xi\, s(1-\xi)^{s-1} e^\xi \\ -&=& \frac{e}{\Gamma(s)} \int_0^1 du\, u^{s-1} e^{-u} - \hspace{10ex}(\textrm{let }u=1-\xi) \\ -&=& \frac{e}{\Gamma(s)} \gamma(s,1), -\end{eqnarray*}$$ -where $\gamma(s,x)$ is the lower incomplete gamma function. -Note that $\gamma(s,x) = \Gamma(s) - \Gamma(s,x)$, where $\Gamma(s,x)$ is the upper incomplete gamma function. -Therefore, -$$\begin{eqnarray*} -\sum_{n=1}^\infty \frac{1}{\Gamma(n+s)} -&=& e\left(1-\frac{\Gamma(s,1)}{\Gamma(s)}\right), -\end{eqnarray*}$$ -as claimed. -Thanks for the interesting question!<|endoftext|> -TITLE: Find the sum of this series :$ \frac{1}{{1!2009!}} + \frac{1}{{3!2007!}} + \cdots + \frac{1}{{1!2009!}}$ -QUESTION [8 upvotes]: Find the sum of this series : -$$\sum\limits_{\scriptstyle 1 \leqslant x \leqslant 2009 \atop - {\scriptstyle x+y=2010 \atop - \scriptstyle {\text{ }}x,y{\text{ odd}} }} {\frac{1}{{x!y!}}} = \frac{1}{{1!2009!}} + \frac{1}{{3!2007!}} + \cdots + \frac{1}{{1!2009!}}$$ -I tried converting it into binomial coefficients and I'm getting sort of $\dfrac{2^{2009}}{2009!}$ -Please help me. - -REPLY [8 votes]: By cancelling the even terms and doubling up the odd terms and dividing by $2$, the sum is -$$ -\begin{align} -&\frac{1}{2010!}\frac12\left(\sum_{k=0}^{2010}\binom{2010}{k}-\sum_{k=0}^{2010}(-1)^k\binom{2010}{k}\right)\\[6pt] -&=\frac{1}{2010!}\frac12\left((1+1)^{2010}-(1-1)^{2010}\right)\\[6pt] -&=\frac{2^{2009}}{2010!} -\end{align} -$$<|endoftext|> -TITLE: A weak converse of $AB=BA\implies e^Ae^B=e^Be^A$ from "Topics in Matrix Analysis" for matrices of algebraic numbers. -QUESTION [18 upvotes]: It is a well known fact that if $A,B\in M_{n\times n}(\mathbb C)$ and $AB=BA$, then $e^Ae^B=e^Be^A.$ -The converse does not hold. Horn and Johnson give the following example in their Topics in Matrix Analysis (page 435). Let $$A=\begin{pmatrix}0&0\\0&2\pi i\end{pmatrix},\qquad B=\begin{pmatrix}0&1\\0&2\pi i\end{pmatrix}.$$ Then $$AB=\begin{pmatrix}0&0\\0&-4\pi^2\end{pmatrix}\neq\begin{pmatrix}0&2\pi i\\0&-4\pi^2\end{pmatrix}=BA.$$ We have $$e^A=\sum_{k=0}^{\infty}\frac 1{k!}\begin{pmatrix}0&0\\0&2\pi i\end{pmatrix}^k=\sum_{k=0}^{\infty}\frac 1{k!}\begin{pmatrix}0^k&0\\0&(2\pi i)^k\end{pmatrix}=\begin{pmatrix}e^0&0\\0&e^{2\pi i}\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}.$$ -For $S=\begin{pmatrix}1&-\frac i{2\pi}\\0&1 \end{pmatrix},$ we have $$e^B=e^{SAS^{-1}}=Se^AS^{-1}=S\begin{pmatrix}1&0\\0&1\end{pmatrix}S^{-1}=\begin{pmatrix}1&0\\0&1\end{pmatrix}.$$ -Therefore, $A,B$ are such non-commuting matrices that $e^Ae^B=\begin{pmatrix}1&0\\0&1\end{pmatrix}=e^Be^A.$ -It is clear that $\pi$ is important in this particular example. In fact, the authors say what follows. - -It is known that if all entries of $A,B\in M_n$ are algebraic numbers and $n\geq 2,$ then $e^A\cdot e^B=e^B\cdot e^A$ if and only if $AB=BA.$ - -No proof is given. How does one go about proving that? - -REPLY [2 votes]: A broad-brush approach is as follows. We want to prove that if $e^A$ and $A$ have the same number of eigenvalues, then $e^A$ is a polynomial in $A$. This is true for diagonal matrices and hence for diagonalizable matrices. Since the diagonalizable matrices are dense, it is true for all matrices. -We can avoid the density argument though. Any square matrix $A$ can be written in -exactly one way as $A=S+N$ where $S$ and $N$ are polynomials in $A$ and $S$ is diagonalizable and $N$ is nilpotent. Using the series expansion for $e^A$ we find that -$$ -e^A = e^D(I+N+\cdots+N^{k-1}) = e^D + M -$$ -with $M=e^D(N+\cdots+N^{k-1})$. Note that $M^k=0$. As $e^D$ and $M$ are polynomials in $A$ and $N$, they are polynomials in $A$; they are also the diagonalizable and nilpotent parts in the decomposition of $e^A$. -Now suppose $e^A$ and $e^B$ commute. Using the above we have -$$ -e^A = e^{D_A}+M_A,\quad e^B = e^{D_B}+M_B. -$$ -Then any two of the six matrices here commute (because the first three matrices are polynomials in $e^A$ and the second three polynomials in $e^B$) and therefore $e^{D_A}$ and $e^{D_B}$ commute. Since $e^A=e^{D_A}(I-N_A)^{-1}$ etc we see that that -$(I-N_A)^{-1}$ and $(I-N_B)^{-1}$ commute, which implies that $N_A$ and $N_B$ commute. So if $D_A$ and $D_B$ are polynomials in $e^{D_A}$ and $e^{D_B}$ respectively, then $AB=BA$. As $D_A$ and $D_B$ are diagonalizable, we are OK -provided the number of distinct eigenvalue of $e^{D_A}$ is equal to the number of distinct eigenvalues of $D_A$ (and ditto for $e^{D_B}$ and $D_B$).<|endoftext|> -TITLE: Does the specification of a general sequence require the Axiom of Choice? -QUESTION [8 upvotes]: Many results in elementary analysis require some form of the Axiom of Choice (often weaker forms, such as countable or dependent). My question is a bit more specific, regarding sequences. -For example, consider a standard proof of the boundedness theorem which states that a function continuous on a closed interval $I$ is bounded on that interval. In the first step of the proof, one specifies a sequence as follows: -Suppose for contradiction that $f$ is unbounded. Then for every $n\in\mathbb{N}$ there exists $x_n$ such that $f(x_n) > n$. This specifies a sequence $(x_n)$. -I'm not sure if the above example requires choice. To me, it certainly feels like it does. More specifically, I think that we are specifying a sequence of sets $$A(n) = \left\{x\in I\mid f(x) > n\right\}$$ and claiming the existence of a choice function $g$ such that $g(n) \in A(n)$ so that this example specifically requires the axiom of countable choice. Please clarify whether my reasoning is correct. More specifically, does the construction of any general sequence (such as one defined as above, or perhaps recursively) then require some form of choice? Thanks for any help. - -REPLY [5 votes]: Actually, if $f$ is continuous from a closed interval (say $[0,1]$) into $\mathbb R$ then you do not need the axiom of choice to prove it is bounded. - -First observe that closed and bounded intervals are still compact even without the axiom of choice. The proof of this is quite nice and simple, let $\mathcal B$ be an arbitrary open cover of $[0,1]$, simply consider $x=\sup\{y\in[0,1]\mid [0,y]\text{ has a finite subcover in }\mathcal B\}$, deduce that $[0,x]$ is finitely covered as well, and then argue that we have to have $x=1$ (by the same reason). -We can deduce from the above that a subset of $\mathbb R$ is compact if and only if it is closed and bounded. If it is compact it cannot be unbounded, and it has to be closed since $\mathbb R$ is a Hausdorff space; on the other hand, if it is closed and bounded it is a subset of a closed interval, therefore closed in a compact space and thus compact. -It is still true that if $f$ is a continuous function from a compact set into a metric space then its image is compact. To see this simply note that every open cover of the image can be translated into an open cover of the compact domain, therefore we can take a finite subcover, and this translate to a finite subcover of the image. - -Therefore the image of a continuous image of a closed interval is compact and the image attains minimal and maximal values (since the image is a closed set). -Also note that $A(n)$ as you specified it is simply intersection of $I$ with an open set which is the preimage of $(n,\infty)$. Choosing from open sets is doable without the axiom of choice [3]. - -In general, when we simply produce one sequence we can sometimes avoid choice if we have a method of calculating the next element in a uniform way (induction is not a uniform way!). If we simply "take another element" then we end up using choice, but we can sometimes avoid these things (for example, instead of arbitrary $\delta$ take $\frac1k$ for the least $k$ fitting). It may even be possible, when needed just one sequence, to use the rational numbers. Those are countable and in particular well-orderable and we can choose from those as much as we want. -However sometimes we want to argue that non-trivial sequences exist, and for this we indeed have to have some choice. For example the proof that $\lim_{x\to a}f(x)=f(a)$ implies $f\colon A\to\mathbb R$ is continuous at $a$ may break, because we need to argue for all sequences and not produce just one. -It may be the case that $A$ itself is Dedekind-finite, and every sequence has only finitely many terms (at least one of those repeating, of course) so in the above case $x_n\to a$ implies that almost always $x_n=a$, but we can make sure that $f$ is not continuous at $a$. Indeed in such $A$ there are only finitely many rational numbers, and pulling the trick of choosing rationals no longer works. - -Further reading: - -Continuity and the Axiom of Choice -Axiom of choice and calculus -Open Sets of $\mathbb{R}^1$ and axiom of choice<|endoftext|> -TITLE: Non-$C^{*}$ Banach algebras? -QUESTION [7 upvotes]: It suddenly occurred to me almost every Banach algebra I know is actually a $C^{*}$ algebra. Several kinds of function algebras are definitely $C^{*}$ algebras. So is the matrix algebra. Although one gets a non-$C^*$ algebra by focusing on the upper triangular matrices, the norm still satisfies the $C^*$ identity. -The algebra of operators on a general banach space is not $C^*$, but at least for me this is a too abstract class that do not provide much intuition. -Thus I wonder whether someone has some good examples of banach algebras that fail the $C^*$ identity but are on the other hand elementary enough to provide intuition and direct computation, like the function algebras. -Thanks! - -REPLY [5 votes]: I must say I find it odd (and a bit worrisome) that one can find textbooks whereby you learn the definition of a Banach algebra (and even a Banach *-algebra) without seeing examples that are not $C^*$-algebras. There is more to life than $B(H)$... -Anyway, some commutative examples which are naturally algebras of functions. In every case the involution is just conjugation of functions. -1) For $G$ a locally compact abelian group (think ${\mathbb Z}^k$ or ${\mathbb T}^k$ or ${\mathbb R}^k$) with dual group $\Gamma$, take -$$A(G) = \{ f\in C_0(G) \mid \widehat{f} \in \ell^1(\Gamma) \} $$ -the so-called Fourier algebra of $G$. (One can define $A(G)$ for arbitrary locally compact groups but the definition is more technical.) -2) Algebras of Lipschitz/H\"older functions. Take your favourite compact metric space $(X,d)$, take some $0<\alpha<1$, and define -$$ L_\alpha(f) = \sup_{x,y\in K; x\neq y} \frac{ \vert f(x)-f(y) \vert }{d(x,y)^\alpha} $$ -then take -$$ {\rm Lip}_\alpha(X,d) = \{ f: X\to {\mathbb C} \mid L_\alpha(f)<\infty \} $$ -equipped with the norm $\Vert f \Vert_\alpha := \Vert f\Vert_{\infty} + L_\alpha(f)$. -3) The algebra $C^k[0,1]$ of $k$-times continuously differentiable functions on $[0,1]$ (for $k\geq 1$), equipped with the natural norm built out of the sup-norms of the derivatives. -If you are willing to consider Banach algebras without involution then there are ${\rm many}^{\rm many}$ more examples.<|endoftext|> -TITLE: Odd-dimensional complex skew-symmetric matrix has eigenvalue $0$ -QUESTION [9 upvotes]: There is the standard proof using $$\det(A)=\det( A^{T} ) = \det(-A)=(-1)^n \det(A)$$ -I would like a proof that avoids this. Specifically, there is the proof that for $A$ a $\bf{real} $ matrix, the transpose is the same as the adjoint, which gives (using the complex inner product) $\lambda \|x\|^2 =\langle Ax, x \rangle= \langle x, -Ax \rangle=-\overline{\lambda } \|x\|^{2}$, so any eigenvalue is purely imaginary. Then we conclude that, since any odd-dimensional real matrix has a real eigenvalue, that eigenvalue must be zero. This argument doesn't work for a general complex skew-symmetric matrix. Is there something I'm missing, is there a way to modify this argument to get that zero is an eigenvalue for the complex case? Also, can somebody please give a geometric reason why odd-dimensional skew-symmetric matrices have zero determinant (equiv., a zero eigenvalue)? -Thanks! - -REPLY [4 votes]: The proof you wrote down works in any field of characteristic not equal to $2$. Here is a similar such proof which avoids determinants due to Ian Frenkel. -Pass to the algebraic closure. Find $P$ such that $A = PDP^{-1}$ with $D$ upper triangular (for example using Jordan normal form, but you don't need to work this hard; it follows from the existence of eigenvectors). Then the diagonal entries of $D$ are the eigenvalues of $A$. Moreover, -$$A^T = (P^T)^{-1} D^T P^T$$ -where $D^T$ is lower triangular, so the diagonal entries of $D$ are also the eigenvalues of $A^T$. Since $A^T = -A$ it follows that the multiset of eigenvalues of $A$ is closed under negation. Since there are an odd number of them there must be an eigenvalue which is its own negative, so it must be zero.<|endoftext|> -TITLE: Question of an isomorphism of $\epsilon_ 0$ and a subset of the rationals. -QUESTION [5 upvotes]: I don't know if this question is appropriated for this site. Anyway, I'm searching for an isomorphism of order $f:K \longrightarrow \epsilon_o $, such that $(K, \leq)$ is a subset(proper or not) of $(\mathbb{Q}, \leq)$ and $\epsilon_o = \sup\{\omega, \omega^{\omega}, \omega^{\omega^{\omega}}, ... \}$. Actually, I'm not finding neither an isomorphism between $K$ and $\omega^{\omega}$. - Thanks in advanced. - -REPLY [4 votes]: Here is a method for constructing a subset of the rationals order isomorphic to $\varepsilon_0$. -First, you need a fundamental sequence for each limit ordinal $\alpha$ less than $\varepsilon_0$. (A fundamental sequence for $\alpha$ is an increasing sequence of ordinals whose supremum is $\alpha$.) Here is the standard definition in case you don't know it. This is an inductive definition over $\alpha$. Represent $\alpha$ in Cantor normal form, -$\alpha = \omega^{\alpha_1} + \omega^{\alpha_2} + \ldots + \omega^{\alpha_n}$. -If $\alpha_n = \beta + 1$, define $\alpha [i] = \omega^{\alpha_1} + \omega^{\alpha_2} + \ldots + \omega^{\beta} * i$. -If $\alpha_n$ is a limit ordinal, define $\alpha [i] = \omega^{\alpha_1} + \omega^{\alpha_2} + \ldots + \omega^{\alpha_n [i]}$. -For $\varepsilon_0$, we can use the fundamental sequence $\varepsilon_0[0] = 1, \varepsilon_0[1] = \omega, \varepsilon_0[2] = \omega^\omega, \varepsilon_0[3] = \omega^{\omega^\omega}, \ldots$ . -Now that we have fundamental sequences, we can start the construction. We will associate ordinals to various points in the interval [0, 1]. First, as a setup, place 0 at 0, and $\varepsilon_0$ at 1 (Note: 1 will not be included in $K$; I'm just placing $\varepsilon_0$ temporarily.) Next, we iterate the following rule $\omega$ times: -RULE: For each limit ordinal $\alpha$ we have already placed, we place the fundamental sequence for that $\alpha$ in the interval between $\alpha$ and the previously placed ordinal. -For the first iteration, there is just one limit ordinal, $\varepsilon_0$. We have placed $\varepsilon_0$ at 1, and the previous ordinal is at 0, so we are to place the fundamental sequence for $\varepsilon_0$ in the interval [0, 1]. We can choose any infinite sequence of points in [0, 1]; we will use the sequence 1/2, 3/4, 7/8, ... . So place $\varepsilon_0[0]$ at 1/2, $\varepsilon_0[1]$ at 3/4, $\varepsilon_0[2]$ at 7/8, and so on. -For the second iteration, we now have infinitely many limit ordinals placed; for each limit ordinal, we apply the previous procedure. For example, take $\varepsilon_0[3] = \omega^{\omega^\omega}$, which was placed at 15/16. The previous ordinal was placed at 7/8, so we want to insert the fundamental sequence for $\omega^{\omega^\omega}$ into the interval [7/8, 15/16]. Now, there is a slight problem; we have $\omega^{\omega^\omega}[0] = \omega$ and $\omega^{\omega^\omega}[1] = \omega^\omega$, neither of which are greater than the previously placed ordinal, $\omega^\omega$. So we simply drop any elements of the fundamental sequence that are not greater than the previously placed ordinal. So, we place $\omega^{\omega^\omega}[2] = \omega^{\omega^2}$ at 15/16 - 1/32, $\omega^{\omega^\omega}[3] = \omega^{\omega^3}$ at 15/16 - 1/64, and $\omega^{\omega^\omega}[n] = \omega^{\omega^n}$ at $\frac{15}{16} - \frac{1}{2^{n+3}}$. We repeat this procedure for all limit ordinals previously placed. -After iterating the rule $\omega$ times, we will have placed every ordinal $\le \varepsilon_0$. So we set $K$ equal to the set of all placed points excluding 1, and we have a set of order type $\varepsilon_0$. -This may seem more complicated than what you wanted, but I believe it is as simple as can be expected for an ordinal as complicated as $\varepsilon_0$. -The same procedure can be used for any ordinal of the form $\omega^\alpha$ for which we can define a collection of fundamental sequences; and we can define a collection of fundamental sequences for any ordinal which we can define an ordinal notation. So, for example, we can explicitly define a subset of the rationals of order type $\omega^{CK}_1$, by using Kleene's $O$ to define fundamental sequences. (This subset will not be recursive, of course.)<|endoftext|> -TITLE: a) Prove that $f$ has a removable singularity if $f'$ does; b) Evaluate $\int_0^\infty\frac{\log x}{(1+x)^3}\,dx$ -QUESTION [7 upvotes]: a) Let $\,f\,$ be an analytic function in the punctured disk $\,\{z\;\;;\;\;0<|z-a|1\}\,$ minus one quarter of the circle $\,\{z\;\;;\;\;|z|=\epsilon\,\,,0<\epsilon< -TITLE: Functions between topological spaces being continuous at a point? -QUESTION [11 upvotes]: Given metric spaces $B$ and $P$, a function $q: B \to P$ is continuous at $c \in B$ if for every $\epsilon > 0$, there exists $\delta > 0$ such that -$$d_B(x, c) < \delta \implies d_P(q(x), q(c)) < \epsilon$$ -But if $B$ and $P$ happen to be topological spaces, $q$ is continuous if the preimage of every open subset of $P$ is open in $B$. So in this case, what would it mean for $q$ to be continuous at $c \in B$? - -REPLY [12 votes]: $q$ is continuous at $c\in B$ if and only if for every neighborhood $W$ of $q(c)$ there exists a neighborhood $U$ of $c$ such that $q(U)\subseteq W$. You may replace "neighborhood" with "open set that contains". -If you translate what this means in the case of the topology induced by a metric, you will find that it is exactly the usual $\epsilon$-$\delta$ definition.<|endoftext|> -TITLE: Set of convergence is measurable. -QUESTION [8 upvotes]: Possible Duplicate: -pointwise convergence in $\sigma$-algebra - -Problem: Prove that the set of points at which a sequence of measurable real functions converges is a measurable set. (I believe the problem means functions from the reals to the reals.) -Source: W. Rudin, Real and Complex Analysis, Chapter 1, exercise 5. -I have posted a proposed solution in the answers. - -REPLY [6 votes]: Let the sequence of functions be $\{f_n(x)\}$. The $\lim \inf$ and $\lim\sup$ of this sequence of functions are measurable (extended-valued) functions. Denote them $h(x)$ and $g(x)$. The set $A$ where $g$ and $h$ are both positive infinity or both negative infinity is measurable, as they are each measurable functions. -Consider the function $p(x)=h\chi_{\mathbb{R}-A}-g\chi_{\mathbb{R}-A}$. It is zero precisely where the original sequence of functions has a limit. Then $E=p^{-1}(\{0\})$ is measurable, so and $E\cup A$ is measurable, and it is the set of points where the sequence has a limit, so we are done.<|endoftext|> -TITLE: Hensel lifting square roots $\!\bmod p\,$ to $\!\bmod p^2$ -QUESTION [6 upvotes]: I've been working on this problem for a while, but hit a dead end. -Here's the problem: -Suppose $p$ is an odd prime. Also let $b^2 \equiv a \pmod p$ and $p$ does not divide $a$. Prove there exists some $k \in \mathbb{Z}$ such that $(b+kp)^2 \equiv a \pmod {p^2}$. -Here's what I've tried so far: -$$(b+kp)^2 \equiv b^2 + 2bkp + k^2p^2 \equiv a \pmod {p^2}$$ -Here, I need to find such $k$ that satisfy this congruence. Equivalently, I need to find such $k$ so that $p^2$ divides $(b^2-a) + 2bkp + p^2k^2$ or equivalently show that $p^2$ divides $(b^2-a) + 2bkp$ for some $k \in \mathbb{Z}$. -So far, since $p$ divides $(b^2-a)$, then by definition, there exists some $x \in \mathbb{Z}$ such that $b^2-a = px$. I got stuck here, and I've tried some examples, but I haven't seen any pattern that pertains to this problem. -Any insight would be helpful. - -REPLY [4 votes]: Generally, as here, suppose $\rm\:f(x) \in \mathbb Z[x]\:$ and $\rm\:f(x_1)\equiv 0\pmod p.\,$ -By Taylor, $\rm\ 0\equiv f(x_1\!+kp) \equiv f(x_1) + f'(x_1)\, kp \pmod{p^2}$ -$\rm\qquad \iff p^2\:|\:p\,(f(x_1)/p + f'(x_1)k)\iff p\:|\:f(x_1)/p + f'(x_1)\, k$ -If $\rm\: p\nmid f'(x_1)\:$ this has a unique solution $\rm\ k \equiv -f(x_1)/(p\, f'(x_1))\pmod p\:$ -Hence $\rm\:f(x_2)\equiv 0\pmod{p^2}\:$ for $\rm\:x_2 = x_1\! + kp \equiv x_1 - \dfrac{f(x_1)}{f'(x_1)}\pmod p$ -If you know calculus you may recognize this as Newton's method (here called Hensel's Lemma). -In your case $\rm\:f(x) = x^2\! - a,\:$ so $\rm\:f'(x) = 2x,\:$ so $\rm\:\bbox[5px,border:1px solid #c00]{x_2 = x_1 - (x_1^2-a)/(2x_1)}$<|endoftext|> -TITLE: Burnside's Lemma -QUESTION [10 upvotes]: I've been trying to understand what Burnside's Lemma is, and how to apply it, but the wiki page is confusing me. The problem I am trying to solve is: - -You have 4 red, 4 white, and 4 blue identical dinner plates. In how - many different ways can you set a square table with one plate on each - side if two settings are different only if you cannot rotate the table - to make the settings match? - -Could someone explain how to use it for this problem, and if its not too complicated, try to explain to me what exactly it is doing in general? - -REPLY [8 votes]: There are four possible rotations of (clockwise) 0, 90, 180 and 270 degrees respectively. Let us denot the 90 degree rotation by $A$, so the other rotations are then its powers $A^i,i=0,1,2,3$. The exponent is only relevant modulo 4, IOW we have a cyclic group of 4 elements. These rotations act on the set of plate arrangements. So if $RWBR$ denotes the arrangement, where there is a Red plate on the North side, White on the East side, Blue on the South, and another Red plate on the Western seat, then -$$ -A(RWBR)=RRWB, -$$ -because rotating the table 90 degrees clockwise moves the North seat to East et cetera. -The idea in using Burnside's lemma is to calculate how many arrangements are fixed under the various rotations (or whichever motions you won't count as resulting in a distinct arrangement). Let's roll. -All $3^4=81$ arrangements are fixed under not doing anything to the table. So $A^0$ has $81$ fixed points. -If an arrangement stays the same upon having a 90 degree rotation act on it, then the plate colors at North and East, East and South, South and West, West and North must all match. IOW we use a single color only. Therefore the rotation $A^1$ has only $3$ fixed points: RRRR, WWWW and BBBB. -The same applies to the 270 degree rotation $A^3$. Only 3 fixed points. -The 180 degree rotation $A^2$ is more interesting. This rotation swaps the North/South and East/West pairs of plates. For an arrangement to be fixed under this rotation, it is necessary and sufficient that those pairs of plates had matching colors, but we can use any of the three colors for both N/S and E/W, so altogether there are 9 arrangements stable under the 180 degree rotation. -The Burnside formula then tells that the number of distinguishable table arrangements is -$$ -\frac14(81+3+9+3)=\frac{96}4=24. -$$<|endoftext|> -TITLE: Why is the absence of zero divisors not sufficient for a field of fractions to exist? -QUESTION [10 upvotes]: I've recently begun to read Skew Fields: The General Theory of Division Rings by Paul Cohn. -On page 9 he writes, - -Let us now pass to the non-commutative case. The absence of zero-divisors is still necessary for a field of fractions to exist, but not sufficient. The first counter-example was found by Malcev [37], who writes down a semigroup whose semigroup ring over $\mathbb{Z}$ is an integral domain but cannot be embedded in a field. Malcev expressed his example as a cancellation semigroup not embeddable in a group, and it promped him to ask for a ring $R$ whose set $R^\times$ of nonzero elements can be embedded in a group, but which cannot itself be embedded in a field. - -The cited paper [37] is On the immersion of an algebraic ring in a skew field, Math. Ann 113 (1937), 686-91. (EDIT by M.S: doi: 10.1007/BF01571659, -GDZ.) -I've had no luck finding this freely available online, nor at the library. Does anyone have reference to this paper, or at least the part where Malcev demonstrates these two parts of his counter-example? I would greatly appreciate seeing it. Thanks. - -REPLY [7 votes]: You can find in most good noncommutative algebra books a discussion of how the normal "field of fractions" construction fails in some noncommutative domains. Your keywords to look for are "Ore condition" and "right Ore domain". If you haven't had luck downloading Cohn's example, there will be many under these keywords. -In short, the right Ore condition is necessary and sufficient for the normal field of fractions definition to work. Without it, it may be impossible to add or multiply fractions, because there will problems finding common denominators. -Here's what I mean: If you carry out the normal equivalence relation in an effort to create a right division ring of fractions, you use $(x,y)\sim(w,z)$ if there exists nonzero $s$ and $t$ such that $ys=zt\neq0$ and $xs=wt$. This allows you to bring things to common denominators, but notice you are only allowed to introduce things on the right. -Suppose you want to add $(a,b)+(c,d)$ where $b,d$ are nonzero. You would like to define this as $(stuff,bd)$. You can form $(ad,bd)\sim (a,b)$ but you are unable to form $(bc,bd)$ because you cannot introduce $b$ on the left. - -If you have access through googlebooks or otherwise, Lam's Lectures on Modules and Rings recounts Mal'cev's example of a domain which is not embeddable into a division ring on page 290.<|endoftext|> -TITLE: Prove that $\frac{1}{\sin^2 z } = \sum\limits_{n= -\infty} ^ {+\infty} \frac{1}{(z-\pi n)^2} $ -QUESTION [7 upvotes]: I have the following problem: - -Find the constants $c_n$ so that -$$ \frac{1}{\sin^2 z } = \sum_{n= -\infty} ^ {+\infty} \frac{c_n}{(z-\pi n)^2} $$ -and the series converges uniformly on every bounded set after dropping finitely many terms. Justify all your claims. -Hint: Use Liouville's theorem to prove the equality. - -Let $c_n = 1 $ for all $n$. -Let $\displaystyle f(z) := \frac{1}{\sin^2 z}$ and $\displaystyle g(z) := \sum_{n= -\infty} ^ {+\infty} \frac{1}{(z-\pi n)^2} $. -Begin by showing that $h$ is an analytic function which converges uniformly on compact subsets of $\mathbb{C} \backslash \mathbb{Z}$. -Suppose $K$ is a compact set which contains no integers. Define -$$ \delta_n = \inf_{z \in K} |\pi-z/n| =\frac{1}{n} \inf_{z \in K} |\pi n-z|, $$ -where the infimum exists because $K$ is compact. Also, compactness implies boundedness. Thus as $n \to \pm \infty$, we have $\delta_n \to \pi$. -Therefore, for sufficiently large $n$, we have $\delta_n > 2$, and -$$ \frac{1}{|z \pm n | ^2} \leq \frac{1}{\delta_n ^2 n^2} < \frac{1}{4n^2}. $$ -By the Weierstrass M-test, $g$ converges absolutely uniformly on $K$. Since each term is analytic on $K$, we conclude that the series converges to an analytic function on $\mathbb{C} \backslash \mathbb{Z}$. -Clearly the only poles of $g(z)$ are at $\pi n$ for each integer $n$, with corresponding principal part $\frac{1}{(z-\pi n)^2}$. -For each integer $n$, we have $\sin^2 (\pi n) = \frac{d}{dz} \sin^2(\pi n) = 0$ and $\frac{d^2}{dz^z} \sin^2 (\pi n) \neq 0$, so $f(z) = 1/ \sin^2 (z)$ also has a pole of order two at each integer multiple of $\pi$, and no other poles. -Furthermore, the principal part of $f(z)$ is, using the Laurent formulas and contour integration, equal to $\displaystyle \frac{1}{(z-n\pi)^2}$. -Thus $h(z) := f(z)-g(z)$ has removable singularities at the points $n\pi$. -Note that both $f$ and $g$ are periodic with period $\pi$. That is, -$$ f(z) = f(z +\pi) \quad \text{ and } \quad g(z) = g(z+\pi) \quad \text{ for all } z \in \mathbb{C}\backslash\mathbb{Z}. $$ -Thus, since $h$ is bounded on the square $\{ z: |\operatorname{Re} z| < \pi, |\operatorname{Im} z |< \pi \}$ and periodic, we may conclude that $h$ is bounded on the set $\{ z: |\operatorname{Im} z |< \pi \}$. To show boundedness on the entire plane, we show that it holds on the vertical strip (with center removed) -$$ S = \{ z: 0 \leq \operatorname{Re} z \leq \pi, | \operatorname{Im} z | \geq \pi \}. $$ -For $z$ in $S$, -\begin{align*} -\sum_{n= -\infty} ^ {+ \infty} |z-\pi n | ^{-2} -&= \sum_{n= -\infty} ^ {+ \infty} \frac{1}{(x-\pi n)^2 + y^2} -&\leq \sum_{n= -\infty} ^ {+ \infty} \frac{1}{(\pi (n-1))^2 +y^2} -\end{align*} -\begin{align*} -& = \sum_{n= -\infty} ^ {+ \infty} \frac{1}{(\pi n)^2 +y^2} -&< 2\sum_{n=0} ^ {+ \infty} \frac{1}{(\pi n)^2 +y^2} -&\leq 2\sum_{n=0} ^ {+ \infty} \frac{1}{(\pi n)^2 +\pi ^2}. -\end{align*} -Thus $g$ is bounded on the set. It is easy to show that $f$ is also bounded on $S$. Thus the difference $f-g$ is also bounded. By the periodicity of $h$, the function is bounded on the entire plane, and is constant by Liouville's theorem. -I'm wondering if my reasoning so far is valid; and if so, how to show that the relevant constant is in fact zero. -Thanks. - -REPLY [2 votes]: Equation $(8)$ from this answer says -$$ -\begin{align} -\pi\cot(\pi z) -&=\frac1z+\sum_{k=1}^\infty\frac{2z}{z^2-k^2}\\ -&=\frac1z+\sum_{k=1}^\infty\left(\frac{1}{z-k}+\frac{1}{z+k}\right)\tag{1} -\end{align} -$$ -The convergence is uniform on compact subsets of $\mathbb{C}$, so we can differentiate to get -$$ --\pi^2\csc^2(\pi z)=-\frac{1}{z^2}-\sum_{k=1}^\infty\left(\frac{1}{(z-k)^2}+\frac{1}{(z+k)^2}\right)\tag{2} -$$ -A wee bit o' manipulation, and a change of variables, gives us -$$ -\csc^2(z)=\sum_{k=-\infty}^\infty\frac{1}{(z-k\pi)^2}\tag{3} -$$<|endoftext|> -TITLE: An infinite product for $\left(\frac{\eta(13\tau)}{\eta(\tau)}\right)^2$? -QUESTION [13 upvotes]: Given the Dedekind eta function, -$$\eta(\tau) = q^{1/24} \prod_{n=1}^\infty (1-q^n)$$ -where $q = \exp(2\pi i\tau)$. Consider the following "family", -$\begin{align} -\left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{24} &= \frac{u^8}{(-1+16u^8)^2},\;\;\; u = q^{1/8} \prod_{n=1}^\infty \frac{(1-q^{4n-1})(1-q^{4n-3})}{(1-q^{4n-2})^2}\\[2.5mm] -\left(\frac{\eta(3\tau)}{\eta(\tau)}\right)^{12} &= \frac{c^3}{(1+c^3)(-1+8c^3)^2},\;\;c = q^{1/3} \prod_{n=1}^\infty \frac{(1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-3})^2}\\[2.5mm] -\left(\frac{\eta(5\tau)}{\eta(\tau)}\right)^{6}\; &= \frac{r^5}{(r^5+u_5^5)(r^5-u_5^{-5})},\quad r\; =\; q^{1/5} \prod_{n=1}^\infty \frac{(1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})}\\[2.5mm] -\left(\frac{\eta(7\tau)}{\eta(\tau)}\right)^{4}\;&= \frac{h(h-1)}{1+5h-8h^2+h^3},\quad h = 1/q\, \prod_{n=1}^\infty \frac{(1-q^{7n-2})^2(1-q^{7n-5})^2(1-q^{7n-3})(1-q^{7n-4})}{(1-q^{7n-1})^3(1-q^{7n-6})^3}\\[2.5mm] -\left(\frac{\eta(13\tau)}{\eta(\tau)}\right)^{2} &=\frac{s}{(s-u_{13})(s+u_{13}^{-1})},\quad\; s =\; ???\\ -\end{align}$ -with fundamental units $u_n$ as $u_5 = \frac{1+\sqrt{5}}{2}$ and $u_{13} = \frac{3+\sqrt{13}}{2}$. The second-to-the-last appears in Chap 10 (10.2) of Duke's Continued Fractions and Modular Functions. -Question: What is the analogous infinite product, if any, for $\left(\frac{\eta\,(13\tau)}{\eta\,(\tau)}\right)^2$ similar to the ones above? -Postscript: This question has been modified before, as it was a bit unclear. - -REPLY [6 votes]: Michael Somos just today found the identity, -$$ -\left(\frac{\eta(13\tau)}{\eta(\tau)}\right)^{2} = \frac{s}{s^2-3s-1}$$ -where, -$$s=\frac{1}{q}\; \prod_{n=1}^\infty \frac{ (1-q^{13n-2})(1-q^{13n-5})(1-q^{13n-6})(1-q^{13n-7})(1-q^{13n-8})(1-q^{13n-11}) }{(1-q^{13n-1})(1-q^{13n-3})(1-q^{13n-4})(1-q^{13n-9})(1-q^{13n-10})(1-q^{13n-12})} $$ -thus completing the family for $N = 2,3,5,7,13$. -Does this address Matt E and Loeffler's comments? Is "s" a modular function?<|endoftext|> -TITLE: Problem book on differential forms wanted -QUESTION [7 upvotes]: I want to get used to differential forms. Thus I would like to solve a bunch of problems, especially on integration of differential forms. -So I need a collection of problems with answers/solutions, starting from really elementary ones. No theorem proving, just straightforward calculations. - -REPLY [7 votes]: You may like Chapter 10 "Differential Forms, Integral Formulae, De-Rham Cohomology" of Mishchenko, Solovyev, Fomenko "Problems in Differential Geometry and Topology" (English translation, Mir Publishers, 1985). I have not seen a newer version of it that may be even better: A.T. Fomenko, A.S. Mischenko, Y.P. Solovyev: Selected problems in differential geometry and topology. Cambridge Scientific Publishers, Cambridge, 2006 -Update 1. There is also a very interesting collection of problems by Prof. W.-H. Steeb. See Ch.4 Differential Forms and Applications in "Problems and Solutions in Differential Geometry" -Update 2. A comprehensive set of problems on differential geometry can be found in Analysis and Algebra on Differentiable Manifolds: A Workbook for Students, by P. M. Gadea, J. Munoz Masqué, see Ch.2 "Tensor Fields and Differential Forms".<|endoftext|> -TITLE: Have there been efforts to introduce non Greek or Latin alphabets into mathematics? -QUESTION [68 upvotes]: As a physics student, often I find when doing blackboard problems, the lecturer will struggle to find a good variable name for a variable e.g. "Oh, I cannot use B for this matrix, that's the magnetic field". -Even ignoring the many letters used for common physical concepts, it seems most of the usual Greek and Latin letters already have connotations that would make their usage for other purposes potentially confusing, for instance one would associate $p$ and $q$ with integer arguments, $i,j,k$ with indices or quaternians, $\delta$ and $\varepsilon$ with small values, $w$ with complex numbers and $A$ and $B$ with matrices, and so forth. -It then seems strange to me that there's been no effort to introduce additional alphabets into mathematics, two obvious ones, for their visual clarity, would be Norse runes or Japanese katakana. -The only example I can think of offhand of a non Greek or Latin character that has mainstream acceptance in mathematics would be the Hebrew character aleph ($\aleph$), though perhaps there are more. -My question then, is have there been any strong mainstream efforts, perhaps through using them in books, or from directly advocating them in lectures or articles, to introduce characters from other alphabets into mathematics? If there have been, why have they failed, and if there haven't been, why is it generally seen as unnecessary? -Thank you, and sorry if this isn't an appropriate question for math.stackexchange.com, reading the FAQ made it appear as if questions of this type were right on the borderline of acceptability. - -REPLY [7 votes]: have there been any strong mainstream efforts, perhaps through using them in books, or from directly advocating them in lectures or articles, to introduce characters from other alphabets into mathematics? - -I had fun introducing some in my PhD thesis: - -This has nothing to do with notation, but forced me to redner several non-Latin, non-Greek alphabets. I strived to obtain all the quotes in their original language (even when rendering them would have been quite tricky); at the moment I count: English, Italian, French, Spanish, Japanese, ancient Hebrew, German, Russian. Some are still missing, but at some point I will add Egyptian (plus, a Japanese anonymous quote, as well as one from Ueshiba O Sensei's writings, is still reported in English). I'm working on that. -In Prop. 4.2.7 the "tower of $f$" symbol is the \rook command, the symbol that denotes the rook in the game of chess. -page 107, I "adoptt an extremely compact notation, referring to a (donné de) recollement with the [...] letter "rae" of the Georgian -alphabet. -footnote 4, page 113: The symbol [...] (pron. "glue") recalls the alchemical token describing the process of amalgamation between two or more elements (one of which is often mercury): although amalgamation is not recognized as a proper stage of the "Magnum Opus", several sources testify that it belongs to the alchemical tradition. -footnote 4, page 150: The symbol [...] (pron. retort) recalls the alchemical token for an alembic; here the term hints at the double meaning of the word retort. -Notation 7.1.1. The set of slicings is denoted 切(C), as the Japanese verb 切る (“kiru”, to cut) is formed from the radical 切 (the same of katana). -The masonic "pigpen cipher" is used, but only as a graphical substitute for the "pullback" or "pushout" or "pullback and pushout" symbols. - -Also, but the idea is not mine, the Yoneda embedding appears here as the letter "yo" in hiragana alphabet. -I have a list of desiderata of alphabets or notation I would like to include: if I manage to do Mathematics for a sufficiently long time, I'll try to include all of them in a meaningful way. - -Arabic -Devanagari -Quenya -Cuneiform alphabet -Phoenician -Tamil -Nsibidi script (still looking for a "grammar" for that); I also suspect there's still no way to employ the ideograms as a ready-to-use font. This means I have to write it. Not scared of that, but I need some time. -Inuktitut -Obviously, Ithkuil (but I foresee some difficulties). -Obviously, the imaginary alphabet invented by L. Serafini for the Codex Seraphinianus -I'd like to find a way to meaningfully employ the "palatal click" as an unary operator on symbols, so natural that it becomes usual and famous. -How could I even forget mentioning musical alphabet? "Clefs" for example.<|endoftext|> -TITLE: Moment Generating Function for Sum of Independent Random Variables -QUESTION [8 upvotes]: I'm taking a graduate course in probability and statistics using Larsen and Marx, 4th edition and looking specifically at estimation methods this week. I ran into a homework problem that is related to moment generating functions and I can't quite connect the dots on how they arrived at the solution. -If you have three independent random variables $$Y_{1}, Y_{2}, Y_{3}$$ and you would like to determine the moment-generating function of $$W = Y_{1} + Y_{2} + Y_{3}$$ knowing that each of the three independent random variables have the same pdf $$f_{y} = \lambda y e^{-\lambda y}, y \geq 0$$ -The easy part of the this problem is applying the theorem that says for $$W = W_{1} + W_{2} + W_{3}$$ the moment generating function of the sum is: $$M_{W}(t) = M_{W_{1}}(t)* M_{W_{2}}(t)* M_{W_{3}}(t)$$ -Where I run into trouble is getting the individual moment generating functions for the Y's. The problem directs you to apply yet another theorem where you would let, for example, another random variable V equal to $$aY_{1}+b$$ and it follows that $$M_{V}(t) = e^{bt}M_{W}(at)$$ -The solution states that if you allow $$V = (1/\lambda)*W$$ then the pdf of V then becomes $$f_{V}(y) = ye^{-y}, y \geq 0$$ and subsequently, you can get the moment generating function using a simple integration by parts but I can't quite follow the application of the theorem used to get to the pdf of V. -Any insight? Likely a fundamental property I missed along the way... - -REPLY [2 votes]: You might want to check your pdf of $Y$ for typographical errors. -Is it a valid pdf? -First, if $Y = aX+b$, then $$f_Y(y) = \frac{1}{|a|}f_X\left(\frac{y-b}{a}\right)$$ -and so if $f_X(x) = xe^{-x}$ for $x \geq 0$ $b = 0$, $a = \lambda^{-1}$, then -$f_Y(y) = \lambda (\lambda y) e^{-\lambda y}$ for $y \geq 0$. (This is the -correct pdf, not what you have, but for an alternative, see the answer -by Marvis). Thus, if you can figure -out $M_X(t)$, the MGF of $X$, -then you can deduce the MGF of $Y$ as $M_Y(t) = M_X(\lambda^{-1}t)$, from which it follows that $M_W(t) = [M_X(\lambda^{-1}t)]^3$. -So what is $M_X(t)$? As Marvis points out to you, -$$M_X(t) = E[e^{tX}] = \int_0^\infty e^{tx} x e^{-x}\,\mathrm dx -= \int_0^\infty x e^{(t-1)x}\,\mathrm dx$$ -which integral you should be able to compute via integration by parts.<|endoftext|> -TITLE: Is there any number in $\mathbb Z[\sqrt{5}]$ with norm equals 2? -QUESTION [5 upvotes]: Is there any number $a+b\sqrt{5}$ with $a,b \in \mathbb{Z}$ with norm (defined by $|a^2−5b^2|$) equal 2? - -REPLY [4 votes]: Hint $\rm\,\ 2\:|\:a^2\!-\!5b^2\! = (a\!-\!b)(a\!+\!b)\!-\!4b^2\Rightarrow\, 2\:|\:a\pm b\:\Rightarrow\:2\:|\:a\mp b\:\Rightarrow\:4\:|\:a^2\!-\!5b^2$<|endoftext|> -TITLE: History of Dual Spaces and Linear Functionals -QUESTION [17 upvotes]: Does anyone know or can anyone give a reference explaining how the concepts of a linear functional and particularly that of a dual space developed? I know Riesz published his famous representation theorem in the first decade of the $1900$s, but did he use these concepts then? -My main reason for asking this question is motivation: These two concepts seem to arise out of thin air in all linear algebra books that I have looked at. It would be nice to find a motivating (non-contrived) example that forces you to look at the dual space of a vector space. - -REPLY [9 votes]: Dieudonne's History of functional analysis gives a very thorough account of the development of these notions. -The theory of linear functionals and related ideas comes out of the theory of linear equations. One paticular direction in which this was generalized, which was of particular importance for later developments, was that of the theory of integral equations, in part by thinking of an integral equation as a limit of a system of linear equations in an increasing number of unknowns. Over time, this set of ideas developed into modern functional analysis. See Dieudonne's book for (many) more details. - -REPLY [5 votes]: I found two very interesting articles that explain the motivation behind many things including dual spaces and linear functionals. Here they are: -On the Hahn-Banach Theorem -The Hahn-Banach Theorem: The Life and Times -It's annoying that the history of these things is not discussed in most textbooks. A lot of things are making a lot more sense to me now.<|endoftext|> -TITLE: Lexicographical order - posets vs preorders -QUESTION [6 upvotes]: I found the following definition for lexicographical ordering on Wikipedia (and similar definitions in other places): -Given two partially ordered sets $A$ and $B$, the lexicographical order on the Cartesian product $A \times B$ is defined as $(a,b) \le (a',b')$ if and only if $a < a'$ or ($a = a'$ and $b \le b'$). The result is a partial order. If $A$ and $B$ are totally ordered, then the result is a total order as well. -Does this definition work equally well if $A$ and $B$ are preorders, rather than posets? In other words, does the anti-symmetric property of the ordering relations on A and B make any difference here? - -REPLY [8 votes]: Yes, if $\langle A,\le_A\rangle$ and $\langle B,\le_B\rangle$ are preorders, the same construction yields a preorder $\langle A\times B,\preceq\rangle$. Specifically, for $\langle a,b\rangle,\langle c,d\rangle\in A\times B$ define $\langle a,b\rangle\preceq \langle c,d\rangle$ iff $a<_A c$, or $a\le_A c$ and $b\le_B d$. Clearly $\preceq$ is reflexive, so you have only to check that it’s transitive. - -Added: I should have said explicitly what I mean by $a<_A c$ for the preorder $\le_A$. I don’t mean that $a\le_A c$ and $a\ne c$. Rather, I mean that $a\le_A c$ and $c\not\le_A a$. Equivalently, I mean that $a\le_A c$ and $a\not\sim_A c$, where $a\sim_A c$ iff $a\le_A c$ and $c\le_A a$. Here $\sim_A$ is the equivalence relation on $A$ whose equivalence classes are sets of $\le_A$-indistinguishable members of $A$, and the relation induced on $A/\sim_A$ by $\le_A$ is a partial order. The point is that if $a\sim_A a'$, I want to treat $a$ and $a'$ exactly alike. -If we were dealing with partial orders $\le_A$ and $\le_B$, I could define $\langle a,b\rangle\preceq \langle c,d\rangle$ iff $a<_A c$, or $a=c$ and $b\le_B d$, using the usual definition of lexicographic order. Note, though, that for partial orders that definition is equivalent to saying that $\langle a,b\rangle\preceq \langle c,d\rangle$ iff $a<_A c$, or $a\le_A c$ and $b\le_B d$: if $\langle a,b\rangle\preceq \langle c,d\rangle$ by the latter definition, then either $a<_A c$, in which case $\langle a,b\rangle\preceq \langle c,d\rangle$ by the usual definition as well, or $a\le_A c$, $a\not<_A c$, and $b\le_B d$, in which case $a=c$ and $b\le_B d$, and again $\langle a,b\rangle\preceq \langle c,d\rangle$ by the usual definition. For preorders the two are not equivalent, because a preorder need not be antisymmetric. For preorders the equivalent formulation is that $\langle a,b\rangle\preceq \langle c,d\rangle$ iff $a<_A c$, or $a\sim_A c$ and $b\le_B d$. - -To that end suppose that $\langle a,b\rangle\preceq\langle c,d\rangle\preceq\langle e,f\rangle$; clearly $a\le_A c\le_A e$. If either $a<_A c$ or $c<_A e$, then $a<_A e$, and $\langle a,b\rangle\preceq\langle e,f\rangle$. Otherwise we have $b\le_B d$ and $d\le_B f$, so $b\le_B f$, and again $\langle a,b\rangle\preceq\langle e,f\rangle$.<|endoftext|> -TITLE: Let $Y$ be an ordered set in the order topology with $f,g:X\rightarrow Y$ continuous, show that $\{x:f(x)\leq g(x)\}$ is closed in $X$ -QUESTION [12 upvotes]: Let $Y$ be an ordered set in the order topology with $f,g:X\rightarrow Y$ continuous. Show that the set $A = \{x:f(x)\leq g(x)\}$ is closed in $X$. -I am completely stumped on this problem. As far as I can tell I've either got the task of proving $A$ gathers its limit points, or showing that there is a closed set $V \subset Y$ such that $f^{-1}(V)$ or $g^{-1}(V)$ is equal to A, which would prove $A$ closed since both $f$ and $g$ are continuous. -The latter strategy seems like the more likely to succeed. Unfortunately I can't find any way of constructing this set $V$. The fact that I don't know if $f$ or $g$ are injective means I keep running into the problem of having $f^{-1}(V)$ or $g^{-1}(V)$ give me extra points not in $A$. And even if I were able to construct $V$ this wouldn't guarantee it was closed. -I suspect I may need to use the continuity of both $f$ and $g$ together in some way but I can't see how. Can anyone give me some guidance on this problem? Thanks. - -REPLY [3 votes]: I feel like this question could be quite difficult for a lot of beginners, so, I will post a perhaps more detailed answer than the ones I see here. -If you only wanted hints for this problem, look no further -Otherwise, here we go: -Proof: If we let $M = \{x \in X: f(x) \leq g(x)\}$, then it suffices to show that $X - M$ is open. If $X-M$ is empty, then we are done, so suppose not. Let $x \in X-M$. Then, it must be the case that $f(x) > g(x)$. We will now show the existence of a neighborhoods $U_x$ and $V_x$ about $f(x)$ and $g(x)$, respectively, that are not only disjoint but also satisfy the property that every element in $U_x$ is larger than every element of $V_x$. -Case 1: -$Y$ contains no smallest or largest element -In this case, we note that there must be $a < g(x)$ and $b> f(x)$. If there is no $y \in Y$ satisfying $g(x) < y < f(x)$, then $V_x = (a, f(x))$ and $U_x = (g(x), b)$ are open sets in $Y$ satisfying the properties outlined above. If there is such $y \in Y$, then we can simply let $U_x = (y, b)$ and $V_x = (a, y)$. -Case 2: $Y$ contains a smallest element but not a largest element -We will note that this case will also cover, via a similar argument, the case where $Y$ contains a largest but not a smallest element. If $g(x)$ is not the smallest element, then there is a smallest element $a_0$ in $Y$ with $a_0 < g(x)$. Since $Y$ has no largest element, then there is $b > f(x)$. If there is no $y \in Y$ satisfying $g(x) < y < f(x)$, then we simply let $V_x = [a_0, f(x))$ and $U_x = (g(x), b)$. If there is such $y$, then let $U_x = (y,b)$ and $V_x = [a, y)$. Now, if $g(x)$ is the smallest element of $Y$, then if there is no $y$ with $g(x) < y < f(x)$, let $V_x = [g(x), f(x))$ and let $U_x = (g(x), b)$. If there is such $y$, then we simply let $V_x = [g(x), y)$ and $U_x = (y, b)$. -Case 3: $Y$ contains a largest and smallest element -If neither $g(x)$ nor $f(x)$ are the largest or smallest element, then one can proceed, as in case 1, to find $U_x$ and $V_x$. If either $g(x)$ or $f(x)$, but not both is a smallest/largest element, then one can proceed as in case 2. Thus, we consider the case where $g(x)$ is the smallest element and $f(x)$ is the largest element. If there is $y$ satisfying $g(x) < y < f(x)$, then we simply let $U_x = (y, f(x)]$ and we can let $V_x = [g(x), y)$. If no such $y$ exists, then let $U_x = (f(x), g(x)]$ and let $V_x = [f(x), g(x))$. -Since we have shown that neighborhoods $U_x$ and $V_x$ of $f(x)$ and $g(x)$, respectively, exists such that every element of $U_x$ is larger than every element of $V_x$, we see by the continuity of $f$ and $g$ that $f^{-1}(U_x)$ and $g^{-1}(V_x)$ are open in $X$. Thus, $f^{-1}(U_x) \cap g^{-1}(V_x)$ is open in $X$. If $m \in f^{-1}(U_x) \cap g^{-1}(V_x)$, then $f(m) \in U_x$ and $g(m) \in V_x$, so $f(m) > g(m)$, so $m \in X-M \implies f^{-1}(U_x) \cap g^{-1}(V_x) \subseteq X-M$. Since $x \in X-M$ was arbitrary, this shows us that for every element $x$ of $X-M$, there is a neighborhood $B_x$ of $x$ contained in $X-M$. Thus, $$X-M = \bigcup_{x \in X-M} B_x$$ - So, $X-M$ is open, as desired. -As a side note, perhaps individuals with more experience will see that in the middle of the proof, we also proved that every ordered set $Y$ equipped with the order topology is Hausdorff.<|endoftext|> -TITLE: How to prove if a function is bijective? -QUESTION [28 upvotes]: I am having problems being able to formally demonstrate when a function is bijective (and therefore, surjective and injective). Here's an example: -How do I prove that $g(x)$ is bijective? -\begin{align} -f &: \mathbb R \to\mathbb R \\ -g &: \mathbb R \to\mathbb R \\ -g(x) &= 2f(x) + 3 -\end{align} -However, I fear I don't really know how to do such. I realize that the above example implies a composition (which makes things slighty harder?). In any case, I don't understand how to prove such (be it a composition or not). -For injective, I believe I need to prove that different elements of the codomain have different preimages in the domain. Alright, but, well, how? -As for surjective, I think I have to prove that all the elements of the codomain have one, and only one preimage in the domain, right? I don't know how to prove that either! -EDIT -f is a bijection. Sorry I forgot to say that. - -REPLY [4 votes]: First show that $g$ is injective ($1$-$1$) by showing that if $g(x)=g(y)$, then $x=y$. This isn’t hard: if $g(x)=g(y)$, then $2f(x)+3=2f(y)+3$, so by elementary algebra $f(x)=f(y)$. By hypothesis $f$ is a bijection and therefore injective, so $x=y$. -Now show that $g$ is surjective. To do this, you must show that for each $y\in\Bbb R$ there is some $x\in\Bbb R$ such that $g(x)=y$. That requires finding an $x\in\Bbb R$ such that $2f(x)+3=y$ or, equivalently, such that $f(x)=\frac{y-3}2$. But $f$ is known to be a bijection and hence a surjection, so you know that there is such an $x\in\Bbb R$. -In general this is one of the two natural ways to show that a function is bijective: show directly that it’s both injective and surjective. The other is to construct its inverse explicitly, thereby showing that it has an inverse and hence that it must be a bijection. You could take that approach to this problem as well: -$$g^{-1}(y)=f^{-1}\left(\frac{y-3}2\right)\;,$$ -since -$$\begin{align*} -g\left(f^{-1}\left(\frac{y-3}2\right)\right)&=2f\left(f^{-1}\left(\frac{y-3}2\right)\right)+3\\ -&=2\left(\frac{y-3}2\right)+3\\ -&=y\;, -\end{align*}$$ -and since $f$ is a bijection, $f^{-1}\left(\frac{y-3}2\right)$ exists for every $y\in\Bbb R$. -Added: As Marc reminds me, this is only half the job: if you take this approach, you must either show directly that $g$ is injective, as I did above, or verify that the function that I called $g^{-1}$ above is a two-sided inverse, i.e., that $g^{-1}\big(g(x)\big)=x$ for $x\in\Bbb R$. This is not particularly difficult in this case: -$$\begin{align*} -g^{-1}\big(g(x)\big)&=g^{-1}\big(2f(x)+3\big)\\ -&=f^{-1}\left(\frac{\big(2f(x)+3\big)-3}2\right)\\ -&=f^{-1}\big(f(x)\big)\\ -&=x\;, -\end{align*}$$ -since $f$ is a bijection.<|endoftext|> -TITLE: Poincaré Lemma Contractible Hypothesis -QUESTION [8 upvotes]: Poincaré's Lemma is often stated as saying that a closed differential form on a star-shaped domain is exact. More generally, it is true that a closed differential form on a contractible domain is exact. -What I am wondering is if there is an easy example of a closed differential form on a simply connected domain which is not exact. - -REPLY [2 votes]: Because specific examples have already been given, I will make some general statements. -I will assume (without proof) that an $n$ dimensional manifold $M$ is orientable if and only if there is a nonvanishing $n$ form on $M$. -Let $M$ be a compact, orientable, $n$ dimensional manifold without boundary. Then, it is a simple application of Stokes' theorem to show that the nonvanishing form is closed but not exact. To answer your question, we just need to find a simply connected manifold with these properties, for example, $S^n$ for $n \geq 2$. -For an even more general theorem (which is much harder to prove), if $M$ is a connected $n$ dimensional manifold, $H^n_{de}(M) = 0$ if $M$ is non-orientable or non-compact and $H^n_{de}(M) = \mathbb{R}$ otherwise (if it is both orientable and compact.)<|endoftext|> -TITLE: Actually calculating an Intersection of Variety and Divisor -QUESTION [6 upvotes]: This has been bugging me for a while now. -Say I have a projective variety given by some polynomial $P$ and the canonical divisor of the projective space. How can I concretly calculate the Intersection of the two? -And by concretly I mean, actually get a result that is not purely abstract? (Like actual Intersection points, degree, etc...) - -REPLY [5 votes]: Unfortunately, you will run into a basic difficulty in your desire to make this example concrete, which is that the canonical bundle on projective space is anti-ample, and so the canonical bundle is not effective. (Rather, it is its dual that is an effective divisor.) -More precisely, if $H$ is the linear equivalence class of a hyperplane in $\mathbb P^n$, then the canonical divisor $K$ is equal to $-(n+1)H$. -So if you want to intersect $K$ with $V(P)$ (the variety cut out by the polynomial $P$) you will have to use at least a little bit of theory, even if only to interpret what you are doing. -You might be better of starting with $H$ itself. Then $H \cap V(P)$ is the linear equivalence class of a hyperplane section of $V(P)$. Assuming that $V(P)$ is smooth, then Bertini's theorem says that a generic member of the linear equivalence class $H\cap V(P)$ will be smooth, and even irreducible if the dimension of $V(P)$ is at least two (i.e. if $n \geq 3$). -Then one way to write $K \cap V(P)$ is simply as $-(n+1) \bigl( H \cap V(P) \bigr)$. -Alternatively, one could consider the linear equivalence class $(n+1)\bigl( H \cap V(P)\bigr).$ This is the class of intersections of $V(P)$ with a degree $n+1$ hypersurface, and again a generic member is smooth (and irreducible if $V(P)$ is of dimension at least $2$). Then you think of $K \cap V(P)$ as being -the negative of this class.<|endoftext|> -TITLE: Is the zero map (between two arbitrary rings) a ring homomorphism? -QUESTION [9 upvotes]: I was looking at the definition under Wikipedia, which states that for arbitrary rings $R,S$, a ring homomorphism $f:R\to S$ must satisfy $f(1)=1.$ Here, I assume they mean $1$ as the multiplicative identity. Certainly then, this implies the zero map is not a ring homomorphism? -This seems somehow intuitionally false; that is, we would want the zero map to be a ring homomorphism, as it is a group homomorphism between groups, a continuous function between reals, a smooth function between manifolds, etc. Could someone help explain why the zero map in the category of rings seems to be an exception to this pattern? - -REPLY [2 votes]: A variety in the sense of a universal algebra is just a bunch of algebraic structures equipped with functions of certain arities and certain relations. For example, the variety of monoids has two functions, one of arity zero, a constant $1$, and one of arity two, a binary operation $\cdot$, satisfying the relations $1 \cdot m = m \cdot 1 = m$ and $m \cdot (n \cdot k) = (m \cdot n) \cdot k$. A homomorphism between two objects of a variety is a map preserving all the functions. For example, a homomorphism between two monoids $M,N$ is a map $f : M \to N$ of the underlying sets such that $f(1)=1$ and $f(m \cdot n) = f(m) \cdot f(n)$ for all $m,n \in M$. This is not a matter of convention as many might think; it follows the general paradigm that homomorphisms preserve the whole structure. -In fact, with this notion, every variety constitutes a category. You always have to remember in which category you are! When you consider a monoid $M$ as a semigroup $U(M)$, i.e. you forget the unit $1$, you have got a different object. You may regard $U(-)$ as the forgetful functor from the category of monoids to the category of semigroups. It is important to keep in mind that this is not the identity; although many authors etc. treat it as such. This causes many confusions. But when you keep in mind that $U$ takes you into another category, it is clear as crystal: A homomorphism $U(M) \to U(N)$ is by definition a map of the underlying sets which preserves $\cdot$, whereas a homomorphism $M \to N$ is by definition a map of the underlying sets which preserves $\cdot$ and $1$. -Now the same story of rings and rngs (a rng is an abelian group with a distributive and associative multiplication; so roughly a ring without the requirement of a unit). By abuse of notation, let us denote the forgetful functor from rings to rngs again by $U$. Let $R,S$ be rings. Then, by definition, a homomorphism $f : R \to S$ is a map preserving all the structure, i.e. $f(0)=0$, $f(x+y)=f(x)+f(y)$, $f(-x)=-f(x)$, $f(x \cdot y) = f(x) \cdot f(y)$, $f(1)=1$ for all $x,y \in R$. Now the condition $f(x+y)=f(x)+f(y)$ for all $x,y \in R$ already implies $f(0)=0$ and $f(-x)=-f(x)$ for all $x \in R$. Many authors are motivated by this to drop these two conditions from the definition, which in my opinion is a bad idea. It is just a lemma that you only have to check that $+$ is is preserved; the correct definition is just a special case of the more general one in universal algebra, which you should even keep in mind when you don't study universal algebra because algebraic structures never come alone. Of course, this discussion already applies to the variety of groups, see Hurkyl's answer. In contrast to that definition of homomorphism of rings, a homomorphism of rngs is a map preserving $0,+,-,\cdot$ (there is no way to talk about $1$). In particular, a homomorphism $R \to S$ yields a homomorphism $U(R) \to U(S)$, but not vice versa. -Now it is easy to answer the question whether the zero map is a homomorphism. Namely, it depends on the category in which you are working. If a homomorphism of rings $f : R \to S$ happens to be the zero map, we get $1 = f(1) = 0$, thus $S = 0$. So yes, there is a zero homomorphism, but only when the target ring is trivial. In fact, something stronger is true: When $R = 0$, then also $S = 0$. One says that the zero ring is a strict initial object. In contrast to that, for all rngs $P,Q$, the zero map $P \to Q$ is a homomorphism. -By the way, it is quite interesting for rings $R,S$ to study homomorphisms $U(R) \to U(S)$; these correspond to a decomposition $S \cong S_1 \times S_2$ and a homomorphism $R \to S_1$ (Hint: The image of $1$ is idempotent). See also the answers at MO/34332, especially the one by James Borger is very enlightening.<|endoftext|> -TITLE: "The deepest and most important of the fundamental properties of the real numbers" according to Edmund Landau -QUESTION [5 upvotes]: In his book Differential and Integral Calculus, Edmund Landau gives an introductory chapter, and before starting it, he assumes as true some theorems. Among them, it is one he calls: "The deepest and most important of the fundan1ental properties of the real numbers" - -Let there be given any division of all the real numbers into two classes, having the following properties: -$(a)$ Neither class is empty. -$(b)$ Every number of the first class is smaller than every number of the - second class. (In other words, if $a \xi$ belongs to the second class. - -What is this theorem called? Should it be a consequence of the least upper bound property of the real numbers? -If we change "real" by "rational" in the first sentence, this is the approach to Dedekind cuts and thus one of the possible constructions of the real numbers. The last sentence can be stated as "each of the division determines a one (two) unique set(s) of rational numbers" or something of the sort. - -REPLY [4 votes]: What is this theorem called? - -The name of this property is "Dedekind completeness". And yes, it is a consequence of the least upper bound (LUB) property. In fact, in the presence of the ordered field axioms, it is equivalent to the LUB property. -It's actually a good exercise to work out the proofs both ways: using the LUB property to derive Dedekind completeness, and then vice versa.<|endoftext|> -TITLE: proof of Zassenhaus formula for exponentials of linear operators -QUESTION [9 upvotes]: The Zassenhaus formula is -$$e^{t(X+Y)}= e^{tX}~ e^{tY} ~e^{-\frac{t^2}{2} [X,Y]} ~ -e^{\frac{t^3}{6}(2[Y,[X,Y]]+ [X,[X,Y]] )} ~ -e^{\frac{-t^4}{24}([[[X,Y],X],X] + 3[[[X,Y],X],Y] + 3[[[X,Y],Y],Y]) } \cdots$$ -from this Wikipedia page. -$X$ and $Y$ are linear operators, and $[X,Y]$ is their commutator. -I mostly want to prove it for the case where the commutator of $X$ and $Y$ is a constant, or simply the general proof. -What I'm looking for is either a reference to a place where it's proven or at -least some shove in the right direction. -So far I've tried to just expand the exponential to see if I see anything -but have no ideas so far. - -REPLY [4 votes]: If the case $[X,Y]=c$ ($c$ a constant) is sufficient for you: -notice that -$$\frac{d}{dt}e^{tX}e^{tY}e^{-t^2c/2}=Xe^{tX}e^{tY}e^{-t^2c/2}+e^{tX}Ye^{tY}e^{-t^2c/2} --tc\,e^{tX}e^{tY}e^{-t^2c/2} =(*).$$ -As $e^{tX}Ye^{-tX}=Y+ct$ (to see it: the derivative of both sides is $c$ and the value for $t=0$ is $Y$), we have -$$(*)=(X+Y)e^{tX}e^{tY}e^{-t^2c/2}.$$ -If $F(t)=e^{tX}e^{tY}e^{-t^2c/2}$, we have $F(0)=1$ and $dF/dt=(X+Y)F(t)$, which are the defining relations for $e^{tX+tY}$.<|endoftext|> -TITLE: Isotopy to the identity on disk -QUESTION [5 upvotes]: Let $D^2 \subset \mathbb{R}^2$ the unit disk and $f: D^2 \rightarrow D^2$ a -homeomorphism with the property that $f$ restricted to the boundary $\partial D^2$ -is the identity. Then $f$ is ambient isotopic to the identity. -I know the Annulus Theorem and I can use it to ambient isotope $f$ to the -identity on any circle inside $D^2$, but I have no clue how to extend it such that it turns f to the identity on an open set around this circle or even construct the -isotopy that works for the whole $D^2$. - -REPLY [3 votes]: Subdivide the disk $D^2$ into an annulus and a smaller disk $D_r^2$ of radius $r$. Now let $f_r$ be the identity on the annulus. On the smaller disk take $f_r$ to be a rescaled $f$. -Now $f_0$ is the identity and $f_1$ is f and $f_r$ is an isotopy.<|endoftext|> -TITLE: Generalized Nakayama over a local ring with an almost nilpotent ideal -QUESTION [5 upvotes]: Let $(A,\mathfrak{m})$ be a local ring. Let us call $\mathfrak{m}$ almost nilpotent if for every sequence $a_1,a_2,\dotsc$ in $\mathfrak{m}$ there is some $n \geq 1$ such that $a_1 \cdot \dotsc \cdot a_n = 0$. Let $M$ be an $A$-module with $M = \mathfrak{m} M$. I would like to prove $M = 0$. -This is an exercise in Lenstra's Galois Theory for Schemes I've been struggling with quite some time. If $M$ is finitely generated, it is trivial (Nakayama). Now let's say $M$ is countably generated, by $m_1,m_2,m_3,\dotsc$. By elimination, we may then assume that $m_i \in \langle m_{i+1},m_{i+2},\dotsc \rangle$. If we had $m_i \in \langle m_{i+1} \rangle$, it would be easy to conclude: Choose $a_i \in \mathfrak{m}$ with $m_i = a_i m_{i+1}$. Now apply the assumption to the sequence $(a_i)$, this shows $m_1 = 0$. Since we could choose $m_1$ arbitrary, $M=0$. However, in general case, the equations become quite horrible and sums, not just products, are involved. If you draw a tree representing the linear combinations, I know that every path must end in a zero eventually, but not that the whole tree must end eventually. My gut feeling is that there may be universal counterexamples ... - -REPLY [3 votes]: Lemma 28.3 from Anderson & Fuller, Rings and Categories of Modules, solves your question completely.<|endoftext|> -TITLE: Proof that convex open sets in $\mathbb{R}^n$ are homeomorphic? -QUESTION [20 upvotes]: This is an exercise from Kelley's book. Could someone help to show me a proof? -It seems very natural, and it is easy to prove by utilizing the arctan function in $\mathbb{R}^1$. -Thanks a lot. - -REPLY [3 votes]: Here is a fairly complete proof for the existence of a -diffeomorphism in the star-shaped case. This was gleaned -from old, incomplete answers (not on math stackexchange) -to this question. Getting a diffeomorphism is only -slightly harder than getting a homeomorphism. -The idea is to push points outwards radially relative -to the star point so that the boundary reaches infinity -and all problems with non-smoothness of the original -boundary disappear. -First, if the region is unbounded, map it diffeomorphically -to a bounded region by pushing points in radially relative -to the star point using a nice dilation function. Almost -any $C^\infty$ diffeomorhism from $[0,\infty)$ to $[0,1)$ -will do. Make it equal to the identity map near $0$ to -avoid complications with differentiability at the star point. -After this, we may assume that the region is bounded. -One idea that doesn't quite work in any obvious way is to -use $r(x)$ = radius in direction $x$ (where $x$ is in the -unit circle) to push radially outwards using a suitable -modification of $r$, since $r$ may be discontinuous. -To avoid discontinuities, start instead with $d(x)$ = distance -from $x$ to the complement of the region, where $x$ is now in -the region. It is an easy exercise that $d$ is continuous. -$d$ doesn't have the mapping properties that we need. Adjust it -by averaging the inverse of it along rays from the star point: - $$e(x) = \int_0^1 1/d(s + t*(x-s))dt,$$ -where s is the star point. The final mapping (if we only want -a homeomorphism) is: - $$m(x) = e(x)*(x-s).$$ -The main points to show for $m$ being a homeomorphism are: -$m$ is one to one because $\|m\|$ is strictly increasing on each ray -(because $1/d > 0$); the image of $m$ is the whole of $\mathbb R^n$ -because $e(x)$ approaches $\infty$ as $x$ approaches the boundary -along each ray ($1/d$ certainly does, and a simple estimate shows -that integration preserves this); the inverse is continuous because -the map is proper (the dilation factor is bounded from below). -Minor modifications turn $m$ into a $C^\infty$ diffeomorphism. The -above gives a continuous $d$ which is strictly positive and approaches -$0$ sufficiently rapidly at the boundary. An easy partition of unity -argument gives a strictly positive $C^\infty$ $d$ that is smaller than -the continuous $d$ so it has the same boundary property. $e$ and $m$ -are then obviously also $C^\infty$. The inverse of $m$ is $C^\infty$ -for much the same reason that $m$ is one to one: the partial -derivative of $\|m\|$ along each ray is invertible because it is -$1/d > 0$. The full derivative at $s$ is $Dm(s) = e(s)1_n$, where $1_n$ -is the identity map on (the tangent space of) $\mathbb R^n$. The full -derivative at points other than $s$ is best calculated in polar -coordinates $(r,\theta$) where $r = \|x-s\|$ and $\theta$ is in the -unit sphere: -$$m(r,\theta) = (r*e(r,\theta), \theta);$$ -$$Dm(r,\theta) = \left( \begin {array} {ccc} - 1/d(r,\theta) & 0 \\ * & 1_{n-1} \end {array} \right).$$ -These are obviously invertible everywhere, so the inverse of $m$ is -$C^\infty$. -By the Riemann mapping theorem, the diffeomorphism can be chosen to be -analytic in dimension $2$. I don't know if this is true in any dimension. -There is no obvious way to adapt the above to give an analytic $m$ even -in dimension $2$. -The integration trick is from an answer by Robin Chapman in mathforum -in 2005. I couldn't find any better answer on the internet. But this -answer says to use any smooth strictly positive function that vanishes -on the complement for $d$. This has the following flaw: $1/d$ might not -approach $\infty$ at the boundary rapidly enough for its radial integrals -to also approach $\infty$. For example, let the region be the unit disk -and $d(x) = (1-\|x\|^2)^\frac 1 2$. -This question is an exercise in many books. I first saw it in Spivak's -Differential Geometry Volume 1 (exercise 25 p.322). But later, Spivak -gives a proof for a much easier case (Lemma 11.17 p.592 is for when $r$ is -$C^\infty$; Lemma 11.18 p.593 shows that $r$ is always lower-semicontinuous -(Spivak says upper-smicontinuous, but this is backwards); -the preamble to Lemma 11.19 p.593 says that proving the general case is -"quite a feat", so Lemma 11.19 only proves an isomorphism of (De Rham) -cohomology). But the integration trick seems to make the feat small. -It is exercise 8 p.20 in Hirsch's Differential Topology. This exercise -is starred, but has no hints. -Other answers give further references to textbooks. Most books don't -prove it. Some say that it is hard and others give it as an exercise. -It is exercise 7 p.86 in Brocker and Janich's Introduction to -Differential Topology. This exercise gives a hint about using the -flow of a radial vector field. It becomes clear that the integration -trick is a specialized version of this general method.<|endoftext|> -TITLE: Constructing a Galois extension field with Galois group $S_n$ -QUESTION [18 upvotes]: Given a field $F$, can you necessarily construct a field extension $E \supset F$ such that $\operatorname{Gal}(E/F) = S_n\,$? - -REPLY [34 votes]: We will prove that there exists a finite Galois extension $K/\mathbb{Q}$ such that $S_n$ = $Gal(K/\mathbb{Q})$ for every integer $n \geq 1$. -We will follow mostly van der Waerden's book on algebra. -You can also see his proof on Milne's course note on Galois theory. -However, Milne refers to his book for a crucial theorem(Proposition 1 below) whose proof uses multivariate polynomials. -Instead, we will use elementary commutative algebra to prove this theorem. -Notations -We denote by |S| the number of elements of a finite set S. -Let $K$ be a field. -We denote by $K^*$ the multiplicative group of $K$. -Let $\tau$ = $(i_1, ..., i_m)$ be a cycle in $S_n$. -The set {$i_1, ..., i_m$} is called the support of $\tau$. -Let $\sigma \in S_n$. -Let $\sigma$ = $\tau_1\dots\tau_r$, where each $\tau_i$ is a cycle of length $m_i$ and they have mutually disjoint supports. -Then we say $\sigma$ is of type [$m_1, ..., m_r$]. -Definition 1 -Let $F$ be a field. -Let $f(X)$ be a non-constant polynomial of degree n in $F[X]$. -Let $K/F$ be a splitting field of $f(X)$. -Suppose $f(X)$ has distinct $n$ roots in $K$. -Then $f(X)$ is called separable. -Since the splitting fields of $f(X)$ over $F$ are isomorphic to each other, -this definition does not depend on a choice of a splitting field of $f(X)$. -Definition 2 -Let $F$ be a finite field. -Let $|F| = q$. -Let $K/F$ be a finite extension of $F$. -Let $\sigma$ be a map: $K \rightarrow K$ defined by $\sigma(x) = x^q$ for each $x \in K$. -$\sigma$ is an automorphism of $K/F$. -This is called the Frobenius automorphism of $K/F$. -Definition 3 -Let $G$ be a permutation group on a set $X$. -Let $G'$ be a permutation group on a set $X'$. -Let $f:X \rightarrow X'$ be a bijective map. -Let $\lambda:G \rightarrow G'$ be an isomophism. -Suppose $f(gx) = \lambda(g).f(x)$ for any $g \in G$ and any $x \in X$. -Then G and G' are said to be isomorphic as permutation groups. -Lemma 1 -Let $F$ be a field. -Let $f(X)$ be a separable polynomial of degree $n$ in $F[X]$. -Let $K/F$ be a splitting field of $f(X)$. -Let $G = Gal(K/F)$. -Let $S$ be the set of roots of $f(X)$ in K. -Then $G$ acts transitively on $S$ if and only if $f(X)$ is irreducible in $F[X]$. -Proof: -If $f(X)$ is irreducible, clearly $G$ acts transitively on $S$. -Conversely, suppose $f(X)$ is not irreducible. -Let $f(X) = g(X)h(X)$, where $g$ and $h$ are non-constant polynomials in $F[X]$. -Let $T$ be the set of roots of $g(X)$ in $K$. -Since $G$ acts on $T$ and $S \neq T$, $S$ is not transitive. -QED -Lemma 2 -Let $F$ be a field. -Let $f(X)$ be a separable polynomial in F[X]. -Let $f(X) = f_1(X)...f_r(X)$, where $f_1(X), ..., f_r(X)$ are distinct irreducible polynomials in $F[X]$. -Let $K/F$ be a splitting field of $f(X)$. -Let $G = Gal(K/F)$. -Let $S$ be the set of roots of $f(X)$ in $K$. -Let $S_i$ be the set of roots of $f_i(X)$ in $K$ for each $i$. -Then $S = \cup S_i$ is a disjoint union and each $S_i$ is a $G$-orbit. -Proof: -This follows immediately from Lemma 1. -Lemma 3 -Let $F$ be a finite field. -Let $K/F$ be a finite extension of $F$. -Then $K/F$ is a Galois extension and $Gal(K/F)$ is a cyclic group generated by the Frobenius automorphism $\sigma$. -Proof: -Let $|F| = q$. -Let $n = (K : F)$. -Since $|K^*| = q^n - 1$, $x^{q^n - 1} = 1$ for each $x \in K^*$. -Hence $x^{q^n} = x$ for each $x \in K$. -Hence $\sigma^n = 1$. -Let m be an integer such that $1 \leq m < n$. -Since the polynomial $X^{q^m} - X$ has at most $q^m$ roots in $K$, $\sigma^m \neq 1$. -Hence $\sigma$ generates a subgroup $G$ of order n of $Aut(K/F)$. -Since $n = (K : F)$, $G = Aut(K/F)$. -Since $|Aut(K/F)| = n$, $K/F$ is a Galois extension. -QED -Lemma 4 -Let $F$ be a finite field. -Let $f(X)$ be an irreducible polynomial of degree $n$ in $F[X]$. -Let $K/F$ be a splitting field of $f(X)$. -Let $\sigma$ be the Frobenius automorphism of $K/F$. -Then $Gal(K/F)$ is a cyclic group of order $n$ generated by $\sigma$. -Proof: -Let $\alpha$ be a root of $f(X)$ in $K$. -By Lemma 3, $F(\alpha)/F$ is a Galois extension. Hence $K = F(\alpha)/F$ -By Lemma 3, $Gal(F(\alpha)/F)$ is a cyclic group of order $n$ generated by $\sigma$. -QED -Lemma 5 -Let $F$ be a finite field. -Let $f(X)$ be an irreducible polynomial of degree $n$ in $F[X]$. -Let $K/F$ be a splitting field of $f(X)$. -Let $G = Gal(K/F)$. -Let $\sigma$ be the Frobenius automorphism of $K/F$. -Let $S$ be the set of roots of $f(X)$. -We regard $G$ as a permutation group on $S$. -Then $\sigma$ is an $n$-cycle. -Proof: -This follows immediately from Lemma 4. -Lemma 6 -Let $F$ be a finite field. -Let $f(X)$ be a separable polynomial in F[X]. -Let $f(X) = f_1(X)...f_r(X)$, where $f_1(X), ..., f_r(X)$ are distinct irreducible polynomials in $F[X]$. -Let $m_i$ = deg $f_i(X)$ for each $i$. -Let $K/F$ be a splitting field of $f(X)$. -Let $G = Gal(K/F)$. -Let $\sigma$ be the Frobenius automorphism of $K/F$. -Let $S$ be the set of roots of $f(X)$. -We regard $G$ as a permutation group on $S$. -Then $\sigma$ is a permutation of type $[m_1, ..., m_r]$. -Proof: -This follows immediately from Lemma 2, Lemma 3 and Lemma 5. -Lemma 7 -$S_n$ is generated by $(k, n)、k = 1、．．．、n - 1$. -Proof: -Let $(a, b)$ be a transpose on {$1, ..., n$}. -If $a \neq n$ and $b \neq n$, then $(a, b) = (a, n)(b, n)(a, n)$. -Since $S_n$ is generated by transposes, we are done. -QED -Lemma 8 -Let $G$ be a transitive permutation group on a finite set $X$. -Let $n = |X|$. -Suppose $G$ contains a transpose and a $(n-1)$-cycle. -Then $G$ is a symmetric group on X. -Proof: -Without loss of generality, we can assume that $X$ = {$1, ..., n$} and $G$ contains -a cycle $\tau$ = $(1, ..., n-1)$ and transpose $(i, j)$. -Since $G$ acts transitively on $X$, there exists $\sigma \in G$ such that $\sigma(j)$ = $n$. -Let $k$ = $\sigma(i)$. -Then $\sigma(i, j)\sigma^{-1}$ = $(k, n) \in G$. -Taking conjugates of $(k, n)$ by powers of $\tau$, we get $(m, n), m = 1, ..., n - 1$. Hence, by Lemma 7, $G = S_n$. -QED -Lemma 9 -Let $F$ be a finite field. -Let $n \geq 1$ be an integer. -Then there exists an irreducible polynomial of degree $n$ in $F[X]$. -Proof: -Let $|F| = q$. -Let $K/F$ be a splitting field of the polynomial $X^{q^n} - X$ in $F[X]$. -Let $S$ be the set of roots of $X^{q^n} - X$ in $K$. -It's easy to see that $S$ is a subfield of $K$ containing $F$. -Hence $S = K$. -Since $X^{q^n} - X$ is separable, $|S| = q^n$. -Hence $(K : F) = n$. -Since $K^*$ is a cyclic group, $K^*$ has a generator $\alpha$. -Let $f(X)$ be the minimal polynomial of $\alpha$ over $F$. -Since $K = F(\alpha)$, the degree of $f(X)$ is $n$. -QED -Lemma 10 -Let $f(X) \in \mathbb{Z}[X]$ be a monic polynomial. -Let $p$ be a prime number. -Suppose $f(X)$ mod $p$ is separable in $\mathbb{Z}/p\mathbb{Z}[X]$. -Then $f(X)$ is separable in $\mathbb{Q}$. -Proof: -Suppose $f(X)$ is not separable in $\mathbb{Q}$. -Since $\mathbb{Q}$ is perfect, there exists a monic irreducible $g(X) \in \mathbb{Z}[X]$ such that $f(X)$ is divisible by $g(X)^2$. Then $f(X)$ (mod $p$) is divisible by $g(X)^2$ (mod $p$). This is a contradiction. -QED -Proposition 1 -Let $A$ be an integrally closed domain and let $P$ be a prime ideal of $A$. -Let $K$ be the field of fractions of A. -Let $\tilde{K}$ be the field of fractions of $A/P$. -Let $f(X) ∈ A[X]$ be a monic polynomial without multiple roots. -Let $\tilde{f}(X) \in (A/P)[X]$ be the reduction of $f(X)$ mod $P$. -Suppose $\tilde{f}(X)$ is also wihout multiple roots. -Let $L$ be the splitting field of $f(X)$ over $K$. -Let $G$ be the Galois group of $L/K$. -Let S be the set of roots of $f(X)$ in $L$. -We regard $G$ as a permutation group on $S$. -Let $\tilde{L}$ be the splitting field of $\tilde{f}(X)$ over $\tilde{K}$. -Let $\tilde{G}$ be the Galois group of $\tilde{L}/\tilde{K}$. -Let $\tilde{S}$ be the set of roots of $\tilde{f}(X)$ in $\tilde{L}$. -We regard $\tilde{G}$ as a permutation group on $\tilde{S}$. -Then there exists a subgroup $H$ of $G$ such that $H$ and $\tilde{G}$ are isomorphic as permutation groups. -Proof: -See my answer here. -Corollary -Let $f(X) \in \mathbb{Z}[X]$ be a monic polynomial of degree $m$. -Let p be a prime number. -Suppose $f(X)$ mod $p$ is separable in $\mathbb{Z}/p\mathbb{Z}[X]$. -Suppose $f \equiv f_1...f_r$ (mod $p$), where each $f_i$ is monic and irreducible of degree $m_i$ in $\mathbb{Z}/p\mathbb{Z}[X]$. -Let $K/\mathbb{Q}$ be a splitting field of $f(X)$. -Let $M$ be the set of roots of $f(X)$. -$G = Gal(K/\mathbb{Q})$ can be regarded as a permutation group on $M$. -Then $G$ contains an element of type [$m_1, ..., m_r$]. -Proof: -By Lemma 10, $f(X)$ is separable in $\mathbb{Q}[X]$. -Let $F_p$ = $\mathbb{Z}/p\mathbb{Z}[X]$. -Let $\tilde{f}(X) \in F_p[X]$ be the reduction of $f(X)$ mod $p$. -Let $\tilde{K}/F_p$ be a splitting field of $\tilde{f}(X)$. -Let $\tilde{G}$ be the Galois group of $\tilde{K}/F_p$. -Let $\tau$ be the Frobenius automorphism of $\tilde{K}/F_p$. -Let $\tilde{M}$ be the set of roots of $\tilde{f}(X)$. -We regard $\tilde{G}$ as a permutation group on $\tilde{M}$. -By Lemma 6, $\tau$ is a permutation of type $[m_1, ..., m_r]$. -Hence the assertion follows by Proposition 1. -QED -Theorem -There exists a finite Galois extension $K/\mathbb{Q}$ such that $S_n$ = $Gal(K/\mathbb{Q})$ for every integer $n \geq 1$. -Proof(van der Waerden): -By Lemma 9, we can find the following irreducible polynomials. -Let $f_1$ be a monic irreducible polynomial of degree $n$ in $\mathbb{Z}/2\mathbb{Z}[X]$. -Let $g_0$ be a monic polynomial of degree 1 in $\mathbb{Z}/3\mathbb{Z}[X]$. -Let $g_1$ be a monic irreducible polynomial of degree $n - 1$ in $\mathbb{Z}/3\mathbb{Z}[X]$. -Let $f_2 = g_0g_1$. -If $n - 1 = 1$, we choose $g_1$ such that $g_0 \ne g_1$. -Hence $f_2$ is separable. -Let $h_0$ be a monic irreducible polynomial of degree 2 in $\mathbb{Z}/5\mathbb{Z}[X]$. -If $n - 2$ is odd, Let $h_1$ be a monic irreducible polynomial of degree $n - 2$ in $\mathbb{Z}/5\mathbb{Z}[X]$. -Let $f_3 = h_0h_1$. -Since $h_0 \neq h_1$, $f_3$ is separable. -If $n - 2$ is even, $n - 2 = 1 + a$ for some odd integer $a$. -Let $h_1$ and $h_2$ be monic irreducible polynomials of degree $1$ and $a$ respectively in $\mathbb{Z}/5\mathbb{Z}[X]$. -Let $f_3 = h_0h_1h_2$. -If a = 1, we choose $h_2$ such that $h_1 \ne h_2$. -Hence $f_3$ is separable. -Let $f = -15f_1 + 10f_2 + 6f_3$. -Since each of $f_1, f_2, f_3$ is a monic of degree $n$, f is a monic of degree $n$. -Then, -$f \equiv f_1$ (mod 2) -$f \equiv f_2$ (mod 3) -$f \equiv f_3$ (mod 5) -Since $f \equiv f_1$ (mod 2), $f$ is irreducible. -Let $K/\mathbb{Q}$ be the splitting field of $f$. -Let $G = Gal(K/\mathbb{Q})$. -Let $M$ be the set of roots in $K$. -We regard $G$ as a permutation group on $M$. -Since $f$ is irreducible, $G$ acts transitively on $M$. -Since $f \equiv f_2$ (mod 3), $G$ contains a $(n-1)$-cycle by Corollary of Proposition 1. -Similarly, since $f \equiv f_3$ (mod 5), $G$ contains a permutation $\tau$ of type [$2, a$] or [$2, 1, a$], -where $a$ is odd. -Then $\tau^a$ is a transpose. -Hence $G$ contains a transpose. -Hence, $G$ is a symmetric group on $M$ by Lemma 8. -QED<|endoftext|> -TITLE: While proving that every vector space has a basis, why are only finite linear combinations used in the proof? -QUESTION [13 upvotes]: Statement: Every vector space has a basis - -Standard Proof:It is observed that a maximal linearly independent set is a basis. - Let $\mathscr{Y}$ be a chain of linearly independent subsets of a vector space $\mathscr{V}$. The union of such a set can serve as an upper bound for it.To apply Zorn's lemma,we have to check whether the union is linearly independent? Well, if $t_1,\dots,t_n$ belong to the union, then each $t_i$ belongs to some linearly independent set $L_i\in \mathscr{Y}$. Because $\mathscr{Y}$ is a chain, one of these sets $L_i$ contains all the others. If that is $L_j$, then the linear independence of $L_j$ implies that no non-trivial linear combination of $t_1,\dots,t_n$ can be zero, which proves that the union of the sets in $\mathscr{Y}$ is linearly independent. Therefore, by Zorn’s lemma, there is a maximal linearly independent set and hence a basis. - -My question: Why are we using only finite linear combinations to show that the union is linearly independent. Surely, if the union is infinite,then there do exist many infinite linear combinations of elements of the union, which cannot be proven to be linearly independent by the same reasoning. I suspect, that perhaps we are not concerned with infinite linear combinations due to issues of convergence, but I'm not sure. -Clear answers will be appreciated. - -REPLY [22 votes]: The definition of linear independence is that any finite linear relation is trivial. -Vector spaces in general do not have any concept of an infinite sum at all. For those vector space where the usual concept of an infinite sum of reals can be generalized, one may speak of a different kind of span/basis where one allows infinite linear combinations in addition to finite ones. That gives rise to a separate concept, different from the usual kind of linear-combinations basis. -When one needs to distinguish between the different notions of basis, an ordinary basis that works by finite linear combinations is called a "Hamel basis" or "algebraic basis", and one that needs infinite linear combinations to span everything is called a "Schauder basis" (though strictly speaking the latter name implies some additional conditions). - -REPLY [8 votes]: A vector space has no notion of convergence or order. Since these are required for assigning meaning to an infinite sum, we are restricted to finite sums. -In the study of normed linear spaces, the notion of Schauder basis is an object of study.<|endoftext|> -TITLE: On the Diophantine equation $a^2+b^2 = c^2+k$ -QUESTION [13 upvotes]: Given the Diophantine equation, -$$a^2+b^2=c^2+k$$ -where k is a constant integer. Let $0 < a \le b$, and $\Delta_k(N)$ be the number of primitive solutions with $0 < c < N$ for some bound N. For example, for k = 0 and $N = 10^5 $, there are exactly 15919 primitive Pythagorean triples, hence $\Delta_0(10^5) = 15919$. -"Conjecture: -$$\lim_{N\to\infty}\frac{\Delta_k(N)}{N} = \text{constant}$$ -where the constant, for negative k, involves $\frac{1}{\sqrt{-k}}$." -In the table below, the first column gives selected k, subsequent columns give (rounded to 5 decimal places) the ratio $\frac{\Delta_k(N)}{N}$ up to $N = 10^7$, its conjectured limit L as $N \to \infty$, and error difference of L and ratio at $N = 10^7$. -$$\begin{array}{ccccc} -k&10^5&10^6&10^7&\infty&\text{Error diff}\\ -0&0.15919&0.15914&0.15916& \to 0.15915 \;=\; \frac{1}{2\pi}&0.00001\\ --1&0.12517&0.12497&0.12499& \to 0.12500 \;=\; \frac{1}{8}&0.00001\\ --2&0.17697&0.17679&0.17680& \to 0.17677 = \frac{1}{4\sqrt{2}}&0.00003\\ --3&0.28871&0.28868&0.28864& \to 0.28867 = \frac{1}{2\sqrt{3}}&0.00003\\ --5&0.22354&0.22367&0.22357& \to 0.22360 = \frac{1}{2\sqrt{5}}&0.00003\\ --7&0.37772&0.37780&0.37799& \to 0.37796 = \frac{1}{\sqrt{7}}&0.00003\\ -\end{array}$$ -Lehmer (1900) proved the case k = 0. Question: Can you prove the conjectured limits are valid/invalid? -Similar simple limits can be found for other negative k, but not for positive k. Code by Daniel Lichtblau of Wolfram Research to find the number of solutions $\le N$ can be found here, though $N = 10^7$ already takes more than an hour, and beyond that takes MUCH more. Do you know of a faster code? - -REPLY [8 votes]: (this is not a proof) -Apparently, for squarefree $k$, the limit is $l(k) = \frac 1 2 \prod_{\text {prime }p} f_k(p)$, where : -$$ f_k(p) = \left\{ \begin{array}{ll} \lim_{m \to \infty} E[ \#\{c \in \mathbb Z/p^m\mathbb Z, a^2+b^2 = c^2+k \}]_{(a,b)\in \mathbb (Z/p^m\mathbb Z)^2} & \text {for } p=2 \\ -\lim_{m \to \infty} E[1+v_p(c^2+k)]_{c \in \mathbb Z/p^m\mathbb Z} & \text {for } p\equiv 1 \mod 4\\ -\lim_{m \to \infty} E[\frac 1 2 (1 + (-1)^{v_p(c^2+k)})]_{c \in \mathbb Z/p^m\mathbb Z} & \text {for } p\equiv 3 \mod 4 \end{array} \right.$$ -The argument for this formula is the following : the number of ways to write an integer $n$ as a sum of two squares is the number of elements of norm $n$ in $\mathbb Z[i]$. Modulo the units of $\mathbb Z[i]$, it is the product of the $(1+v_p(n))$ for $n \equiv 1 \mod 4$ and of the $\frac 1 2 (1 + (-1)^{v_p(n)})$ for $n \equiv 3 \mod 4$ (it is a direct consequence of the ideal group structure of $\mathbb Z[i]$). So the factors with $p>2$ express the expected value of the number of elements whose norm is a "randomly chosen" integer. The factor for $p=2$ is there to correct the conspiracy of $a^2+b^2$ not hitting things uniformly in $\mathbb Z_{(2)}$ (for example it never hits things of the form $4q+3$). Finally, the $\frac 1 2$ in front of the product is there because we are counting elements of norm $c^2+k$ not only up to units, but also up to conjugation. -Let $\chi(p) = 1$ if $p \equiv 1 \mod 4$ and $\chi(p) = -1$ if $p \equiv 3 \mod 4$, and I will suppose that $k$ is squarefree. -If $p$ is an odd prime dividing $k$, then $v_p(c^2+k) = \min (v_p(c^2,1)) = 1$ if $p$ divides $c$, $= 0$ otherwise. In this case, $f_k(p) = \frac{p-1}p + \frac 2 p = \frac {p+1} p$ when $\chi(p)=1$ ; and $f(p) = \frac{p-1}p$ when $\chi(p)=-1$ : $f_k(p) = \frac{p+\chi(p)}p$ -If $p$ is an odd prime not dividing $k$ and if $-k$ is a square mod $p$, then $-k$ has two distinct square roots in $\mathbb Z_{(p)}$ and we have $v_p(c^2+k) = v_p(c+\sqrt{-k}) + v_p(c- \sqrt{-k})$, so that the "probability" of $v_p(c^2+k) \ge l$ is $2/p^l$. Thus, we get $f_k(p) = 1 + 2\chi(p)/p + 2/p^2 + 2\chi(p)/p^3 + \ldots = 2/(1- \chi(p)\frac 1p)-1 = \frac{p+\chi(p)}{p-\chi(p)}$ -If $p$ is an odd prime not dividing $k$ and if $-k$ is not a square mod $p$, then $c^2+k$ can never be a multiple of $p$, so $v_p(c^2+k) = 0$, and $f_k(p) = 1$. -Let $\chi_d(p) = 1$ if $d$ is a square mod $p$, $-1$ if it is not, and $0$ if $p$ divides $d$. We see that we always have $$f_k(p) = \frac{p+\chi(p)}{p-\chi_{-k}(p)\chi(p)} = \frac{p+\chi(p)}{p-\chi_k(p)} = \frac{p^2-1}{(p-\chi_k(p))(p-\chi(p))} $$ -Then we recognize an Euler product, which converges (except when $k = 1$, where $\chi_k$ is the trivial character) and which we can compute thanks to the class number formula (be careful because so far we implicitly want to have $2$ in the conductor of $\chi_k$) $$\prod f_k(p) = \frac{f_k(2) L(\chi_k,1)L(\chi,1)}{\frac 34 \zeta(2)} = \frac{f_k(2) h(k) R(k) \phi_2(k) }{(w_k/2) \sqrt{|k|} \phi_\infty(k)}$$ -where $h(k)$ is the class number of $\mathbb Q(\sqrt{k})$, $R(k)$ is its regulator,$w_k$ its number of roots of unity, $\phi_2(k) = 3$ if $k \equiv 5 \mod 8$, $1$ otherwise, $\phi_\infty(k) = \frac \pi 2$ if $k>0$, $1$ otherwise. -Now, $f_k(2)$ is not difficult to compute, because the expectation of the number of solutions stabilizes very quickly. The formula agrees with the results presented so far -and also works for positive $k$ : -$\begin{array}{ccccccccccccc} k & -13 & -11 & -10 & -7 & -6 & -5 & -3 & -2 & -1 & +2 & +3 & +5 \\ -f_k(2) & 1/2 & 1 & 1/2 & 2 & 1/2 & 1/2 & 1 & 1/2 & 1/2 & 1/2 & 1/2 & 1 \\ -l(k) & \frac 1 {2 \sqrt{13}} & \frac 3 {2 \sqrt {11}} -& \frac 1 {2 \sqrt{10}} & \frac 1 {\sqrt 7} & \frac 1 {2 \sqrt{6}} - & \frac 1 {2 \sqrt{5}} & \frac 1 {2 \sqrt{3}} & \frac 1 {4 \sqrt{2}} & \frac 18 - & \frac {\log(1+\sqrt 2)}{2 \pi \sqrt 2} & \frac {\log(2+\sqrt 3)}{2 \pi \sqrt 3} - & \frac {3 \log(\frac {1+\sqrt 5}2)}{\pi \sqrt 5}\end{array} $<|endoftext|> -TITLE: For which categories we can solve $\text{Aut}(X) \cong G$ for every group $G$? -QUESTION [28 upvotes]: It is usually said that groups can (or should) be thought of as "symmetries of things". The reason is that the "things" which we study in mathematics usually form a category and for every object $X$ of a (locally small) category $\mathcal{C}$, the set of automorphisms (symmetries) of $X$, denoted by $\text{Aut}_{\mathcal{C}}(X)$, forms a group. -My question is: Which categories that occur naturally in mathematics admit all kinds of symmetries? More precisely, for which categories we can solve the equation (of course up to isomorphism) $$\text{Aut}_{\mathcal{C}}(X) = G$$ for every group $G$? -I will write what I could find myself about this, which also hopefully illustrates what kind of answers that would interest me: - Negative for $\mathsf{Set}$: Infinite sets have infinite symmetry groups and for finite sets we get $S_n$'s. So if we let $G$ to be any finite group which is not isomorphic to some $S_n$, the equation has no solution. -Negative for $\mathsf{Grp}$: -No group can have its automorphism group a cyclic group of odd order. -Positive for $\mathsf{Grph}$ (category of graphs): -Frucht's theorem settles this for finite groups. Also according to the wikipedia page, the general situation was solved independently by de Groot and Sabidussi. -An obvious necessary condition is that $\mathcal{C}$ should be a large category. - This paper shows that the equation can be solved if $\mathcal{C}$ is the category of Riemann surfaces with holomorphic mappings and $G$ is countable. -If we take $\mathcal{C}$ to be the category of fields with zero characteristic, I guess the equation relates to the inverse Galois problem. Edit: This may be much easier than the inverse Galois problem, as Martin Brandenburg commented. - -REPLY [8 votes]: Every finite group arises as the automorphism group of a finite poset. This is the subject of Automorphism groups of finite posets by J. A. Barmak and E. G. Minian, available online. -Since the category of finite posets is isomorphic(!) to the category of finite $T_0$ spaces, in particular every finite group is the automorphism group of a topological space. More generally, it is proven in Spaces with given homeomorphism groups by Thornton, also available online, that every finitely generated group is the automorphism group of a topological space. I wonder if we can realize every group? -The paper Representing a Profinite Group as the Homeomorphism Group of a Continuum -by K. H. Hofmann and S. A. Morris, available online, deals with the question whether every profinite group is isomorphic to the automorphism group of a compact connected space (endowed with the compact open topology).<|endoftext|> -TITLE: Calculating the norm of an element in a field extension. -QUESTION [9 upvotes]: Given a number field $\mathbb{Q}[\beta]$, where the minimal polynomial of $\beta$ in $\mathbb[Z][x]$ has degree $n$, I would like to calculate the norm of the general element $$a_0+a_1\beta+\cdots+a_{n-1}\beta^{n-1}.$$ -In particular, here is my attempt when $\alpha=2^\frac{1}{3}$: - -Let $K=Q[2^\frac{1}{3}]$. Using the definition of the norm, it is the determinant of the linear transformation. Consider $\alpha =a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c$ acting by multiplication on the element $d+2^{\frac{1}{3}}e+2^{\frac{2}{3}}f$. Since $$\left(a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c\right)\left(d+2^{\frac{1}{3}}e+2^{\frac{2}{3}}f\right)=ad+2bf+2ce+2^{\frac{1}{3}}\left(ae+bd+2cf\right)+2^{\frac{2}{3}}\left(af+dc+be\right) $$ in the basis $[1,2^{\frac{1}{3}},2^{\frac{2}{3}}$ we may view multiplication by $\alpha$ as a linear transform $$\alpha\left[\begin{array}{c} -d\\ -e\\ -f -\end{array}\right]=\left[\begin{array}{c} -ad+2bf+2ce\\ -ae+bd+2cf\\ -af+dc+be -\end{array}\right].$$ Using the above, we see that $$\alpha=\left[\begin{array}{ccc} -a & 2c & 2b\\ -b & a & 2c\\ -c & b & a -\end{array}\right] $$ in this basis. Taking the determinant we find $$\det \left[\begin{array}{ccc} -a & 2c & 2b\\ -b & a & 2c\\ -c & b & a -\end{array}\right] =a\left(a^{2}-2bc\right)-2c\left(ba-2c^{2}\right)+2b(b^{2}-ac) $$ $$=a^{3}+2b^{3}+4c^{3}-6abc. $$ - This means we have shown that $N_K(\alpha)=a^{3}+2b^{3}+4c^{3}-6abc$ for $\alpha=a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c$ - -Questions: -(1) Was the above calculation correct? Can we conclude that the norm of a general element in that space is $a^{3}+2b^{3}+4c^{3}-6abc$? -(2) Is there a better way to do this computation? What about if the extension is Galois? - -REPLY [4 votes]: 1) Yes, the calculation was correct. A decent way to check you haven't made any arithmetic errors is to try some small integers for $a,b,c,d,e,f$ and check the norm is multiplicative. -2) This is probably the best way to do the computation. If we have a Galois extension, then the norm of an element is the product of all of its Galois conjugates. There are two ways I can think of that we can use this fact: - -Produce the elements of $G(K/\mathbb{Q}).$ Sometimes one has to do this anyway, so in those cases that could be a short cut. -Assume for now your general element has the relevant coefficient non-zero so that its degree is the degree of the extension. Then since the list of distinct Galois conjugates being $\alpha_1, \alpha_2, \cdots \alpha_n$ implies the minimal polynomial is $m(x)=(x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_n),$ we can read the norm of $\alpha$ off the constant term of the minimal polynomial. Then the norm for lower degree elements follows by a limit argument.<|endoftext|> -TITLE: How to prove $I + t X$ is invertible for small enough $ | t | ?$ -QUESTION [7 upvotes]: Let $X \in \text{GL}_n(\mathbb{R})$ be an arbitrary real $n\times n$ matrix. How can we prove rigorously: -$$ \underset{b>0} {\exists} : \underset{|t|\le b} {\forall} : \det (I + t X) \neq 0 $$ -If necessary, we could also assume that $t \ge 0.$ - -REPLY [3 votes]: $\det(A)$ is a polynomial function in the entries of $A$. The solution set to $\det(A) = 0$ is a closed subset, and so the set of invertible matrices is an open set. In particular, there is an open neighborhood $U$ of $I$ such that every matrix in $U$ is invertible. By choosing $t$ sufficiently small, we guarantee $I + tX \in U$. - -$\det(I + tX)$ is a polynomial function of $t$, and so $\det(I + tX) = 0$ has finitely many roots, and $t=0$ is not a root. Therefore, we can find an open interval containing 0 such that $I + tX$ is invertible on that interval. - -We can compute the Taylor series for $(I + tX)^{-1}$ about 0: -$$ (I + tX)^{-1} = I - t X + t^2 X^2 - t^3 X^3 + \cdots $$ -It's not difficult to see that the right hand side is convergent in every component of the matrix (e.g. ratio test along with an upper bound on the entries for $X^n$) on an interval of positive radius. By multiplying through by $(I + tX)$, we can check that the sum really is the inverse.<|endoftext|> -TITLE: Existence of a semi-martingale that matches given densities -QUESTION [6 upvotes]: Let $\{p_t\}_{t \geq 0}$ be a family of densities. Is there any result concerning the existence of a semi-martingale $\{X_t\}_{t \geq 0}$ such that for all $t\geq 0$, the density of $X_t$ is $p_t$ ? - -REPLY [2 votes]: After some investigations I found this theorem due to Kellerer. - -Theorem (Kellerer, 1972) $-$ Let $(\mu_t)_{t\in[0,T]}$ be a family of probability measures of $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ with first moments, such that for $s -TITLE: Sum of powers and prime numbers -QUESTION [13 upvotes]: I'm not able to find solutions of the following equation: -$$2^k+3^k=p$$ -where $p$ is a prime number and $k \in N$. -It's easy to show that we have a solution when $k=1,2,4$. -Is it possible to find any other solutions or the only possible values of $k$ are the previous ones? -Thank you in advance. - -REPLY [5 votes]: We begin with two simplifying observations: -Observation 1: If $m=2^k+3^k$ where $k=ab$ where $a$ is odd, then $m$ is divisible by $2^b+3^b$. -Proof: Let $s=2^b$ and $t=3^b$. We have \[2^{ab}+3^{ab}=s^a+t^a=(s+t)(s^{a-1}-s^{a-2} \cdot t+\cdots-s \cdot t^{a-2}+t^{a-1}).\] (See e.g. MathWorld.) (QED) -[NB. A comment on Sloane's A082101 asserts that it must be divisible by $5$, which is incorrect (e.g. $2^{10}+3^{10}=60073$).] -Hence, for $m$ to be prime, we must have that $k=0$ or $k$ is a power of $2$. So we will switch to studying $n=2^{2^k}+3^{2^k}$. -Observation 2: If $n=2^{2^k}+3^{2^k}$, and $p$ is an odd prime divisor of $n$, then $p=2^{k+1}q+1$ for some $q \in \mathbb{Z}^+$. -Proof: Let $x$ be a primitive root modulo $p$. Define $t,s,r$ to be the minimum positive integers such that $2 \equiv x^t \pmod p$, $r \equiv x^s \pmod p$ and $-1 \equiv x^r \pmod p$. -We know $x^{2r} \equiv 1 \pmod p$. Hence $|\mathbb{Z}_p^*|=\varphi(p)=p-1$ divides $2r$ (since $x$ is a primitive root). Hence $r=c \cdot (p-1)/2$ for some $c \in \mathbb{Z}^+$. We know $r$ divides $|\mathbb{Z}_p^*|=\varphi(p)=p-1$ by Lagrange's Theorem. Hence $c \in \{1,2\}$. We conclude $c=1$ (as otherwise, $r=p-1$ and $x^r \equiv -1 \pmod p$ contradicts Fermat's Little Theorem). Hence $r=(p-1)/2$. -Since $p$ divides $n$, we know $x^{2^k t} \equiv x^{r+2^k s} \pmod p$, or equivalently $2^k t \equiv r+2^k s \pmod {p-1}$. We rearrange this to give $2^k (t-s) \equiv r \pmod {p-1}$. Importantly, $2^k (t-s) \not\equiv 0 \pmod {p-1}$. Thus, if $2^w$ is the largest power of $2$ dividing $p-1$, then $2^k (t-s) \not\equiv 0 \pmod {2^w}$, since $2^w$ does not divide $r=(p-1)/2$. Hence $w \geq k+1$. (QED) -(Note similar observations can be shown for Fermat numbers.) -I used these observations to perform trial division, which came up with the following: -2^(2^1)+3^(2^1) is prime: 13 -2^(2^2)+3^(2^2) is prime: 97 -2^(2^3)+3^(2^3) has a proper divisor: 17 -2^(2^4)+3^(2^4) has a proper divisor: 3041 -2^(2^5)+3^(2^5) has a proper divisor: 1153 -2^(2^6)+3^(2^6) has a proper divisor: 769 -2^(2^7)+3^(2^7) has a proper divisor: 257 -2^(2^8)+3^(2^8) has a proper divisor: 72222721 -2^(2^9)+3^(2^9) has a proper divisor: 4043777 -2^(2^10)+3^(2^10) has a proper divisor: 2330249132033 -2^(2^11)+3^(2^11) has a proper divisor: 625483777 -2^(2^12)+3^(2^12) has a proper divisor: 286721 -2^(2^13)+3^(2^13) has a proper divisor: 14496395542529 -2^(2^14)+3^(2^14) has a proper divisor: 2752513 -2^(2^15)+3^(2^15) has a proper divisor: 65537 -2^(2^16)+3^(2^16) has a proper divisor: 319291393 -2^(2^17)+3^(2^17) has a proper divisor: 54498164737 -2^(2^18)+3^(2^18)=n does not satisfy 2^(n-1) = 1 (mod n) [Fermat's primality test.] -2^(2^19)+3^(2^19) has a proper divisor: 7340033 -2^(2^20)+3^(2^20) has a proper divisor: 23068673 -2^(2^21)+3^(2^21) has a proper divisor: 2878894768129 -2^(2^22)+3^(2^22) has a proper divisor: 453236490241 -2^(2^23)+3^(2^23) has a proper divisor: 106216554497 -2^(2^24)+3^(2^24) has a proper divisor: 342456532993 - -2^(2^27)+3^(2^27) has a proper divisor: 488015659009 - -For $k=18$, my computer didn't easily find a divisor, so I ran a base $2$ Fermat primality test, which showed that $2^{(2^{18})}+3^{(2^{18})}$ is definitely not a prime. So now the smallest open case is $2^{(2^{25})}+3^{(2^{25})}$, which has 16009533 digits. If it were prime (and it almost certainly won't be), it would be the 4-th largest known (see "The Largest Known Primes--A Summary"). -Heuristic: Assume, for the sake of argument, that the probability of a natural number $n$ being prime is independent and approximately $1/\ln n$ (which is vaguely justified by the prime number theorem). Then the expected number of primes of the form $2^{2^k}+3^{2^k}$ is $O(1)$. -Non-proof: The expected number of primes of the form $2^{2^k}+3^{2^k}$ is $\sum_{k \geq 0} \mathrm{Pr}[n=2^{2^k}+3^{2^k} \text{ is prime}] = \sum_{k \geq 0} 1/\ln n$. We observe that $n \geq e^{2^k}$, so $\sum_{k \geq 0} 1/\ln n \leq 1/2^k=2$. (QED) -If we start the sum at $k=25$, we can estimate the probability of another prime of the desired form existing at $5.96 \times 10^{-8}$. -Of course, the independence assumption is invalid (e.g. if $n$ is prime and $n \geq 3$ then $n+1$ is not prime). But this gives a feeling that it would be possible not to have any more primes of this form. (Similar heuristics are used to argue that there's probably no more Fermat primes.) -Answer: I think the only possible answer to your question will be "don't know, but probably not" for a long time (similar to Fermat Numbers). -(I also wish to acknowledge the use of Primeform (OpenPFGW), along with my own code, for the computations in this answer.)<|endoftext|> -TITLE: Lower semicontinuous function as the limit of an increasing sequence of continuous functions -QUESTION [11 upvotes]: Let $ f:\mathbb{R}^m \rightarrow (-\infty,\infty] $ be lower semicontinuous and bounded from below. Set $f_k(x) = \inf\{f(y)+k d( x,y ): y\in \mathbb{R}^m\} $ , where $d(x,y)$ is a metric. It is easy to see that each $f_k$ is continuous and $f_1 \leq f_2\leq ...\leq f \\$. However, I don't know how to prove that $ \lim_{k \rightarrow \infty}f_k(x) = f(x) $ for every $x\in\mathbb{R}^m $. - -REPLY [5 votes]: Sea $-M$ una cota inferior de $f$ (con $M > 0$).\ -Tomemos $x\in X$, para todo $\epsilon > 0$ existe $\delta > 0$ tal que si $y\in B(x,\delta)$ entonces $f(y) > f(x) - \epsilon$.\ - Por otro lado, para todo $k\geq 1$ existe $y_k \in X$ tal que - $$f(y_k) + kd(x,y_k) < f_k(x) + \epsilon \leq f(x) + \epsilon$$ - Como $f(y_k) \geq f_k(x) \geq -M$ la desigualdad $kd(x,y_k) < f(x) + \epsilon + M$ es satisfecha. Luego, dividiendo por $k$ se tiene que $0\leq d(x,y_k) < \dfrac{(f(x) + \epsilon + M)}{k}$, como esto sucede para todo $k\geq 1$ deducimos que $d(x,y_k)$ converge a $0$. Luego existe $k_0 \geq 1$ para el cual $d(x,y_k) < \delta$ cuando $k\geq k_0$. De esto, se cumple que $f(y_k) > f(x) - \epsilon$ para todo $k\geq k_0$.\ - Luego $0\leq f(x) - f_k(x) < f(x) - (f(y_k) + kd(x,y_k) - \epsilon) = (f(x) - f(y_k)) - kd(x,y_k) + \epsilon < 2\epsilon$ para todo $k\geq k_0$. Por lo tanto $f_k(x)$ converge a $f(x)$.<|endoftext|> -TITLE: A sequence of polynomials that converges to $0$ pointwise except at $z=0$ -QUESTION [6 upvotes]: This is an exercise from John Conway's book on complex analysis: - -Investigate if there exists a sequence of polynomials $(P_n)$ that fulfills the conditions $P_{n}(0)=1$ for all natural numbers $n$ and $\lim_{n\rightarrow\infty}P_n(z)=0$ for all $z\neq0$ - -Polynomials obey the maximum principle, but I don't see how to apply it if all we know is point-wise convergence. (Uniform convergence would imply $|P_n|<\epsilon$ on the unit circle for large $n$, contradicting $P_n(0)=1$.) - -REPLY [3 votes]: I think this can be done using Mergelyan's theorem: Let $K_n$ be a compact set of the form $\{0\}\cup \{z: 1/n\le |z|\le n, |\arg z-\theta_n|\ge \epsilon_n \}$. On this set you can approximate the function $\chi_{\{0\}}$ by a polynomial $P_n$ within $1/n$. Make sure to choose $\theta_n$ and $\epsilon_n$ so that no point of the plane belongs to $K_n^c$ for infinitely many $n$.<|endoftext|> -TITLE: Determinant of a real skew-symmetric matrix is square of an integer -QUESTION [15 upvotes]: Let $A$ be a real skew-symmetric matrix with integer entries. Show that $\operatorname{det}{A}$ is square of an integer. - -Here is my idea: If $A$ is skew-symmetric matrix of odd order, then $\operatorname{det}{A}$ is zero. So, take $A$ to be of even order and non-singular. Since all the eigenvalues of $A$ are of the form $ia$ and its conjugate (where $a$ is real number), we see that $\operatorname{det}{A}$ is square of a real number. But I am not getting how to show it is square of an integer. - -REPLY [2 votes]: For a skew symmetric $A$, $\det(A)={\rm pfaffian}(A)^2$ where pfaffian is an integral polynomial function of the entries of the matrix $A$. For the case of an integer matrix the pfaffian is therefore an integer. Hence the result you want.<|endoftext|> -TITLE: Is it possible to combine two integers in such a way that you can always take them apart later? -QUESTION [12 upvotes]: Given two integers $n$ and $m$ (assuming for both $0 < n < 1000000$) is there some function $f$ so that if I know $f(n, m) = x$ I can determine $n$ and $m$, given $x$? Order is important, so $f(n, m) \not= f(m,n)$ (unless $n=m$). - -REPLY [2 votes]: f(m,n) = m +(1/n) -If there are numbers to the right of the decimal, -m =number to the left of decimal. -n = the reciprocal of the numbers to the right of the decimal -example -f(m,n) =10.5 then m =10, n=1/(0.5) =2 -If there are no numbers to right of the decimal [f(m,n) is itself an integer] - then m = f(m,n) -1, - n =1 -example -f(m,n) =20 then m=19, n =1<|endoftext|> -TITLE: Under what conditions does $(\frac{3}{p})(\frac{-1}{p})=1?$ Two ways, different results. -QUESTION [7 upvotes]: I have quite a problem, two methods, different results. something's wrong. -I'm trying to find under what conditions the Legendre symbol for $(\frac{3}{p})(\frac{-1}{p})=1$. -First Way: $(\frac{3}{p})(\frac{-1}{p})=(\frac{3}{p})\cdot(-1)^{(p-1)/2}$. -For $p\equiv1\pmod4$, I get that $(\frac{3}{p})=(\frac{p}{3})$ and it is $1$ iff $p\equiv1\pmod3$ and $p\equiv1\pmod4$. Otherwise I get that $(\frac{3}{p})=-(\frac{p}{3})$ so it has to be that $p\equiv1\pmod3$, in order to get $-1\cdot -1 \cdot 1=1$. The Chinese remainder theorem tells me that the $p\equiv1,7 \pmod{12}$. -Second way (I believe that it is the problematic one but yet I can't see what's wrong): -$$\left(\frac{3}{p}\right)\left(\frac{-1}{p}\right)=\left(\frac{3}{p}\right)\cdot(-1)^{(p-1)/2}=\left(\frac{p}{3}\right)(-1)^{((p-1)/2)\cdot((3-1)/2)}\cdot(-1)^{(p-1)/2}=\left(\frac{p}{3}\right),$$ this equals $1$ only for $p \equiv 1 \pmod 3$. -What's wrong with the second way? I believe it's all legal. -Thanks a lot! - -REPLY [7 votes]: Nothing's wrong. The answer is that for a prime $p > 3$, $\left( \frac{3}{p} \right) \left( \frac{-1}{p} \right) = 1 \iff p \equiv 1 \pmod 3$. Note that the possible residue classes modulo $12$ for a prime $p > 3$ are the ones which are coprime to $12$, i.e., $1,5,7,11$. Of these the ones congruent to $1$ modulo $3$ are $1$ and $7$. -Let me give you yet a third way to solve this problem. For an odd prime $p$, set $p^* = (-1)^{\frac{p-1}{2}} p$. Then an equivalent statement of quadratic reciprocity is that for distinct odd primes $p,q$, -$\left( \frac{p}{q} \right) = \left( \frac{q^*}{p} \right)$. -(Admittedly, verifying this is similar in nature to what you're doing! The point though is that this is a formula worth proving and then remembering.) -Applying this with $q = 3$, we get $\left( \frac{3}{p} \right) \left( \frac{-1}{p} \right) = \left( \frac{-3}{p} \right) = \left(\frac{p}{3} \right)$, which is $1$ iff $p \equiv 1 \pmod{3}$. - -REPLY [5 votes]: They are the same thing: $p\equiv 1, 7\pmod {12}$ is the same as $p\equiv 1\pmod 3$ when $p$ is prime. That's because $p\equiv 4,10\pmod {12}$ is not possible when $p$ is prime.<|endoftext|> -TITLE: If $\int_0^1 f(x)x^n \ dx=0$ for every $n$, then $f=0$. -QUESTION [11 upvotes]: Possible Duplicates: -Nonzero $f \in C([0, 1])$ for which $\int_0^1 f(x)x^n dx = 0$ for all $n$ -Slight generalization of an exercise in (blue) Rudin -What can we say about $f$ if $\int_0^1 f(x)p(x)dx=0$ for all polynomials $p$? - -I found a nice problem I would like to share. -Problem: If $f$ is continuous on $[0,1]$, and if -$$\int_0^1 f(x)x^n \ dx =0$$ -for every positive integer $n$, prove that $f(x)=0$ on $[0,1]$. -Source: W. Rudin, Principles of Mathematical Analysis, Chapter 7, Exercise 20. -I have posted a proposed solution in the answers. - -REPLY [20 votes]: We will show that the integral of $f^2$ over $[0,1]$ is 0. This will show that $f$ is zero, because if $f$ were not identically equal to zero, $f^2$ would be positive on some interval (by continuity) and have a nonzero integral. -Using the Stone-Weierstrass theorem, we can approximate $f$ uniformly by polynomials $P_n$ so that $||P_n-f||<1/n$ in the $\sup$ norm. The given condition obviously implies that the integral of $f(x)P(x)$ is zero for any polynomial $P$, by linearity. Note that -$$\left|\int_0^1 f(x)f(x) \ dx \right|=\left|\int_0^1 f(x)f(x) \ dx - \int_0^1 f(x)P_n(x)\right|\le \int_0^1 |f(x)| |f(x)-P_n(x)|\ dx \\\le \frac{1}{n}\int_0^1 |f(x)| \ dx.$$ -The last integral is constant, so taking $n$ arbitrarily large completes the proof.<|endoftext|> -TITLE: Help in self-studying mathematics. -QUESTION [5 upvotes]: Is this a reasonable list for who seek to learn mathematics by "self learning" program? -and is it a well sorted list to follow? -http://www.math.niu.edu/~rusin/known-math/index/index.html - -REPLY [6 votes]: Our very own Pete Clark, now of the University of Georgia faculty, was once upon a time a very talented undergraduate student at the University of Chicago. Back then, he and several of his fellow U of C honor students compiled a list of their recommended textbooks at various levels and posted it online. It does need to be updated, but I think you'll find it quite helpful. In fact,I've been asking Pete to revise it for a few months now. I hope he'll eventually find the time to do so. -Here it is: -http://www.ocf.berkeley.edu/~abhishek/chicmath.htm<|endoftext|> -TITLE: Difference between congruence and similarity transformations -QUESTION [5 upvotes]: I am trying to understand the difference between a "congruence" or "similarity" transformation for two $n \times n$ matrices (which for the sake of simplicity, we'll assume are real). From what I can glean, a similarity transformation represents a change of basis from one orthogonal basis in $\mathbf{R}^n$ to another. My understanding is that a congruence transformation is an isometry, and so, it seems it would represent some geometrical operation like a (rigid) rotation, reflection, etc which preserves angles ad distances (but not necessarily orientation). -If someone can tell me if this is correct or correct any mistakes in my interpretation, I'd be most appreciative. -Thanks in advance. - -REPLY [6 votes]: From what I can glean, a similarity transformation represents a change of basis from one orthogonal basis in $\mathbb R^n$ to another. - -This is incorrect. A similarity transformation $A\mapsto P^{-1}AP$ indeed amounts to a change of basis (in the sense that both matrices represent the same linear map in different bases), but neither the original nor the new basis need be orthogonal. They can be any two bases. - -a congruence transformation is an isometry - -This is incorrect, or at least imprecise. Matrix congruence transformation $A\mapsto P^T AP$ also amounts to a change of basis, but in the sense that both matrices represent the same quadratic form in different bases. I think you confused the matrix congruence with the notion of congruence of sets in geometry. These are two different things with the same name (and there are more).<|endoftext|> -TITLE: If $f$ is positive and increasing, then $f'(x)x^2$ is increasing? -QUESTION [8 upvotes]: If $f: (0, \infty)\to (0, \infty)$ is increasing, is it true that the function $x\longmapsto f'(x) \cdot x^2 $ is increasing? We can assume that $f$ is twice differentiable. -Can someone provide a counter-example, a function $f$ which is increasing and positive, but $f'(a)\cdot a^2 < f'(b)\cdot b^2$, for some $ a>b $ ? - -REPLY [9 votes]: $$ -f(x) = 1-e^{-x}. -$$ -We have $x^2 f'(x)>0$ for all $x>0$ and $x^2 f'(x)\to0$ as $x\to\infty$, so $x^2 f'(x)$ cannot continue increasing. (It increases for small enough values of $x$.)<|endoftext|> -TITLE: Nilpotent Groups as Perturbations of the Identity -QUESTION [5 upvotes]: In Terry Tao's blog, specifically this post, he says that nilpotent groups can be thought of in algebraic geometry and analysis as modeling infinitesimal perturbations of the identity. Can someone provide a reference where this point of view is used? - -REPLY [3 votes]: The idea of nilpotents representing infinitesimal change dates back at least to the invention/discovery of calculus by Newton and Leibniz! In deformation theory of schemes at least, the whole story begins with the study of flat families over the spectrum of $k[t]/t^2,$ which are viewed as infinitesimal perturbations of the "central" fibre parametrized by the nilpotent parameter $t.$ This is mentioned in Hartshorne II.8 in the exercises I believe, but there is an entire field of work in deformation theory dedicated to this beautiful concept. For example, Harthorne has a recent introductory book called "Deformation Theory".<|endoftext|> -TITLE: Markov chains- recurrence and transience -QUESTION [10 upvotes]: This is an exercise in Durrett's probability book. - -$p$ is the transition probability for a markov chain on a countable space. $f$ is said to be superharmonic if $f(x)\geq\sum_y p(x,y)f(y)$, or equivalently $f(X_n)$ is a supermartingale. Suppose $p$ is irreducible. If every nonnegative superharmonic function is constant, show that $p$ is recurrent. - -It's not that easy to use the statement "every nonnegative superharmonic function is constant", so I tried 2 ways to reformulate the statement. -1.For all $f\geq 0$ nonconstant, there exists an $x$ s.t. $f(x)<\sum_y p(x,y)f(y)$. Show that $p$ is recurrent. -2.If $p$ is transient, show that there exists a nonnegative superharmonic function which is nonconstant. -But I haven't got a clue about how to prove this. - -REPLY [10 votes]: I think the easiest way is through strategy 2). Let $X$ be a transient irreducible Markov Chain. Consider an arbitrary $x_{0}$ and define: -$$\tau = \inf\{n \geq 0: X_{n} = x_{0}\}$$ -Define -$$f(x) = P(\tau < \infty|X_{0}=x)$$ -By construction, $f(x) \in [0,1]$ and $f(x_{0}) = 1$. Since $X$ is transient, there exists $y$ such that $f(y) < 1$. Finally, by Markovianity, for any $x \neq x_{0}$, -$$f(x) = P(\tau < \infty|X_{0}=x) = \sum_{y}{p(x,y)P(\tau < \infty|X_{1}=y)} = \sum_{y}{p(x,y)f(y)}$$ -and -$$f(x_{0}) = 1 \geq \sum_{y}{p(x,y)f(y)}$$ -Hence, $f$ is super-harmonic and non-constant. -ps: the condition is actually necessary and sufficient. Can you prove the reverse?<|endoftext|> -TITLE: Corollary of Cauchy's theorem about finite groups. -QUESTION [5 upvotes]: Cauchy's Theorem: Let $G$ be a finite group and let $p$ divide the order of $G$. Then $G$ has an element of order $p$ and consequently a subgroup of order $p$ (of course the cyclic subgroup generated by the aforementioned element of order $p$). -Corollary: Let $G$ be a finite group, and $p$ be a prime. Then $G$ is a $p$-group if and only if the order of $G$ is a power of $p$. -Relevant Definition: $G$ is a $p$-group (for a prime $p$) if every element of $G$ has order $p^m$ for some $m\geq 1$. -The proof of the theorem was a beautiful application of using group actions to count. Now I'm trying to use the theorem to prove the corollary which is stated in my text leaving the proof as an exercise, but I am stuck because there doesn't seem to be a natural way to apply the theorem. Please accept my 'request for a hint'. -Thanks very much! - -REPLY [4 votes]: If the order $n$ of the group is not a power of $p$, there is a prime $q\ne p$ that divides $n$. But then $\dots$.<|endoftext|> -TITLE: Finding a splitting field of $x^3 + x +1$ over $\mathbb{Z}_2$ -QUESTION [9 upvotes]: Finding a splitting field of $x^3 + x +1$ over $\mathbb{Z}_2$. -Ok so originally I messed around with $x^3 + x +1$ for a bit looking for an easy way to factor it and eventually decided that the factors are probably made up of really messy nested roots. So then I tried looking at the quotient field $\mathbb{Z}_2[x]/x^3 + x + 1$ to see if I would get lucky and it would contain all three roots but it doesn't. Is there a clever way to easily find this splitting field besides using the cubic formula to find the roots and then just directly adjoining them to $\mathbb{Z}_2$? -Edit: Ok it turns out I miscalculated in my quotient field, $\mathbb{Z}_2[x]/x^3 + x + 1$ does contain all three roots. - -REPLY [5 votes]: Two good answers already here, but I wanted to emphasize the usefulness of the Frobenius endomorphism. Let $f(X) = X^3 + X + 1$. If I let $\alpha$ denote the image of $X$ in $k = \mathbb F_2[X]/(f(X))$ then applying Frobenius to $0 = f(\alpha)$ gives -\[ -0 = (\alpha^3 + \alpha + 1)^2 = (\alpha^2)^3 + \alpha^2 + 1. -\] -Hence $\alpha^2$ is also a root of $f$, and $\alpha^2 \neq \alpha$ because $\alpha \neq 0, 1$. Since $k$ contains two roots of the cubic $f$, it contains three. -To bring in slightly more technology: extensions of finite fields are Galois and in particular normal, and this implies that if you adjoin one root of an irreducible polynomial over a finite field then you've actually adjoined all of them. This is, of course, not true in general!<|endoftext|> -TITLE: How does one graduate from Hecke Operators to Hecke Correspondences? -QUESTION [19 upvotes]: I've read (skimmed heartily) basic books on the topic of modular forms. (The last being Silverman's Advanced Topics in the Arithmetic of Elliptic Curves.) -I strive for an understanding which is as mathematically mature as possible. (Read: strive for a Langlands-program-ish understanding.) Alas, I am still far from succeeding. -I have heard on many an occasion people referencing "Hecke correspondences". I am aware of the definition of a correspondence, but I'm at a loss of how to think about Hecke correspondences! What made them arise? How are they helpful? What suggested to anyone that they should define them? How do they relate to Hecke operators? Is this thing helpful towards Langlands? -Ach... Hopefully this is within the realm of mathstackexchange (or is this more appropriate to mathoverflow?). This has been gnawing at me for months. -References are also welcome. - -REPLY [18 votes]: It might help to go back to the definition of Hecke operators in level $1$ in Serre's Course in arithmetic. For a prime $p$ and a lattice $\Lambda$, the $p$the Hecke corresondence (I forget if Serre uses exactly this terminology) takes -$\Lambda$ to $\sum \Lambda'$, where $\Lambda'$ runs over all index $p$ sublattices of $\Lambda$. -This is a multi-valued function from lattices to lattices (it is $1$-to-$p+1$-valued). -Now lattices (mod scaling) are just elliptic curves: $\Lambda \mapsto \mathbb C/\lambda$. And so we can also think of this as a multi-valued map from the moduli space of ellitic curves (i.e. the $j$-line, or $Y_0(1)$ if you like) to itself. -How to describe a multi-valued map more geometrically? Think about its graph -inside $Y_0(1) \times Y_0(1)$. The graph of a function has the property -that its projection onto the first factor is an isomorphism. The graph of a $p+1$-valued function has the property that its projection onto the first -factor is of degree $p+1$. -This graph has an explicit description: it is just $Y_0(p)$ (the modular curve -of level $\Gamma_0(p)$). Remember that $Y_0(p)$ parameterizes pairs $(E,E')$ of $p$-isogenous curves. We embed it into $Y_0(1) \times Y_0(1)$ in the obvious way, by mapping the pair $(E,E')$ (thought of as an element of $Y_0(p)$) to $(E,E')$ (thought of as an element of the product). -In terms of the upper half-plane variable $\tau$, one can think of this map as -being $\tau \bmod \Gamma_0(p)$ maps to $\bigl(\tau \bmod SL_2(\mathbb Z), p\tau -\bmod SL_2(\mathbb Z) \bigr).$ -So we have recast Serre's description of the $p$th Hecke operator in terms of a correspondence on lattices in the geometric language of correspondences on curves: i.e. the $p$th Hecke operator is given by a mutli-valued morphism -from $Y_0(1)$ to itself, rigorously encoded by its graph thought of as a curve -in the product surface $Y_0(1) \times Y_0(1)$, which is in fact isomorphic to -$Y_0(p)$. -We can easily compactify the situation, to get $X_0(p)$ embedding as the graph -of a correspondence on $X_0(1) \times X_0(1)$. -[Caveat: Actually the map $Y_0(p) \to Y_0(1) \times Y_0(1)$ need not be an embedding; it is a birational map onto its image, but the image can be singular -(and the same applied with $X$'s instead of $Y$'s). This is because the point on $Y_0(p)$ is not just the pair $(E,E')$, but the additional data of the $p$-isogeny $E\to E'$, which is not uniquely determined up to isomorphism in some -exceptional cases. But this is a technical point which is not worth fussing about at the beginning.] - -The advantage of having a geometric correspondence in sight is that whenever -we apply any kind of linearization functor to our curve, the correspondence will turn into a genuine single valued operator. -The point is that if we have a multi-valued function from one abelian group to another, we can just add up the values to get a single-valued function. -So the correspondence $T_p$ induces genuine maps from the Jacobian of $X_0(1)$ to itself, or from the cohomology of $X_0(1)$ to itself, or from the space of holomorphic differentials on $X_0(1)$ to itself. -Now actually in the case of $X_0(1)$, which has genus zero, the Jacobian and the space of holomorphic differentials are trivial. But we can do everything with $X_0(N)$ or $X_1(N)$ in place of $X_0(1)$ for any $N$, and all the same remarks apply. -Remembering that the holomorphic differentials on $X_0(N)$ are the weight two cuspforms of level $N$, one can compute that the $p$th Hecke correspondence gives rise to the usual $p$th Hecke operator on cuspforms in this way. - -What's the point of considering the correspondence? There are many; here's one: -if we reduce everything mod $p$, we get a mod $p$ correspondence on the mod $p$ reduction of $X_0(N)$, whose graph is the mod $p$ reduction of $X_0(Np)$. But this latter reduction is well-known to be singular, and in fact reducible; it is the union of two copies of $X_0(N)$. Thus the $p$th Hecke correspondence mod $p$ decomposes as the sum of two simpler correspondences, which one checks to be the Frobenius morphism from $X_0(N)$ Mod $p$ to iself, and its dual. -This is the Eichler--Shimura congruence relation (in some form it actually goes back to Kronecker), and it underlies the relationship between $T_p$-eigenvalues and the trace of Frobenius in the $2$-dimensional Galois reps. attached to Hecke eigenforms. - -Some MO posts which are vaguely relevant: -The map on differentials induced by a correspondence -The Eichler --Shimura relation<|endoftext|> -TITLE: Average number of times it takes for something to happen given a chance -QUESTION [12 upvotes]: Given a chance between 0% and 100% of getting something to happen, how would you determine the average amount of tries it will take for that something to happen? -I was thinking that $\int_0^\infty \! (1-p)^x \, \mathrm{d} x$ where $p$ is the chance would give the answer, but doesn't that also put in non-integer tries? - -REPLY [2 votes]: Here's an easy way to see this, on the assumption that the average actually exists (it might otherwise be a divergent sum, for instance). Let $m$ be the average number of trials before the event occurs. -There is a $p$ chance that it occurs on the first try. On the other hand, there is a $1-p$ chance that it doesn't happen, in which case we have just spent one try, and on average it will take $m$ further tries. -Therefore $m = (p)(1) + (1-p)(1+m) = 1 + (1-p)m$, which is easily rearranged to obtain $m = 1/p$.<|endoftext|> -TITLE: Does Brownian motion visit every point uncountably many times? -QUESTION [25 upvotes]: Let $B_t$ be a one-dimensional standard Brownian motion. - -Is it true that, almost surely, for every $x \in \mathbb{R}$ the set $\{t : B_t = x\}$ is uncountable? - -Let $A_x$ be the event that $\{t : B_t = x\}$ is uncountable. It is well known that $\mathbb{P}(A_0) = 1$. By recurrence and the strong Markov property, we also have $\mathbb{P}(A_x) = 1$ for every $x$. I am asking about the probability of $A = \bigcap_{x \in \mathbb{R}} A_x$. Because of the uncountable intersection, it is not even obvious that this set is measurable. -George Lowther in this comment says the answer is yes, but I don't see how to prove it. - -REPLY [9 votes]: Actually there is an almost sure lower bound on the Hausdorff dimension of the level sets of Brownian motion that follows from the Ray-Knight theorem. -$$ a.s \quad \forall a\in \mathbb{R}, \quad \text{dim}\{t \geq 0 | B(t) =a\} \geq \frac{1}{2}. $$ -This is a theorem (6.48, p. 170) in the book, "Brownian Motion" by Peres and Morters.<|endoftext|> -TITLE: Physical interpretation: weighted eigenvalues of the Laplacian with a potential -QUESTION [7 upvotes]: I've already posted this question on Physics.SE, but I thougth it could be useful to ask also here. -No problem if moderators will ask me to cancel this thread... But, please, have mercy! :-D - -Let $\Omega \subseteq \mathbb{R}^N$ be a domain and let $V,m:\Omega \to \mathbb{R}$ be two measurable and sufficiently summable functions. -When one considers the eigenvalue problem for the operator $\mathcal{L}:=-\Delta +V$ w.r.t. the weight $m$, i.e.: -$$\tag{P} \begin{cases} --\Delta u(x) + V(x)\ u(x) = \lambda\ m(x)\ u(x) &\text{, in } \Omega\\ -u(x)=0 &\text{, on } \partial \Omega , -\end{cases}$$ -the function $V$ is usually called potential and the function $m$ is called weight. -Then, a weighted eigenvalue of $\mathcal{L}$ w.r.t. $m$ is any number $\lambda \in \mathbb{R}$ s.t. (P) has at least one nontrivial weak solution $u\in H_0^1(\Omega)$, i.e.: -$$\forall \phi \in C_c^\infty(\Omega),\quad \int_\Omega \nabla u\cdot \nabla \phi\ \text{d} x + \int_\Omega V\ u\ \phi\ \text{d} x = \lambda\ \int_\Omega m\ u\ \phi\ \text{d} x\; .$$ -My questions are: - - -Is there any reasonable physical interpretation of those eigenvalues? And what is it? -Why have the functions $V$ and $m$ those names? - - -Moreover, I heard that the $p$-laplacian (i.e., $\Delta_p u := \operatorname{div} (|\nabla u|^{p-2}\ \nabla u)$, which reduces to the usual laplacian when $p=2$) can be used to model nonlinear elasticity or something like that; therefore I have also the following question: - -What about any possible physical meaning of the nonlinear weighted eigenvalues coming from the problem: - $$\tag{Q} \begin{cases} --\Delta_p u(x) + V(x)\ |u(x)|^{p-2}\ u(x) = \lambda\ m(x)\ |u(x)|^{p-2}\ u(x) &\text{, in } \Omega\\ -u(x)=0 &\text{, on } \partial \Omega , -\end{cases}$$ - where $1 < p < \infty$? - -Many thanks in advance, guys! - -REPLY [6 votes]: The left hand side $-F=-\Delta u + Vu$ models force in a material where points try to pull their neighbors towards their local value in a spring-like manner, but also get pulled down by an external force that increases linearly with displacement (for example, other springs or long range gravity). -Now suppose $m$ is understood as a mass (density), and consider Newton's law $F=ma=mu_{tt}$. We see that solving $-\Delta u + Vu=\lambda m u$ is finding modes such that -$$-u_{tt}=\lambda u.$$ -In other words, modes that will stay the same shape, but simply grow (complex-)exponentially in time. -Here is a 1-dimensional diagram: - -Edit: To clarify, the extension to the p-laplacean, $\nabla \cdot |\nabla u|^{p-2} \nabla u=\nabla \cdot k(u,x) \nabla u$ models a material where the force of molecules pulling on their neighbors is p-nonlinear in the displacement gradient. In other words, the "springs" in the above diagram are not ideal.<|endoftext|> -TITLE: proving that $2^{m-1}$ has reminder $1$ when divided by $m$ -QUESTION [7 upvotes]: Let $$m = \frac{4^p - 1}{3}$$ -Where $p$ is a prime number exceeding $3$. how to prove that $2^{m-1}$ has reminder $1$ when divided by $m$ - -REPLY [8 votes]: $2^{2p}=4^p=3m+1\equiv 1 \pmod m$ so the result follows if $2p\mid m-1$. -Since $m$ is odd $2\mid m-1$, and by Fermat's Little Theorem $p\mid 4^p-4=3(m-1)$. Since $p>3$ is prime we must have $p\mid m-1$.<|endoftext|> -TITLE: All integer solutions for $x^4-y^4=15$ -QUESTION [7 upvotes]: I'm trying to find all the integer solutions for $x^4-y^4=15$. -I know that the options are $x^2-y^2=5, x^2+y^2=3$, or $x^2-y^2=1, x^2+y^2=15$, or $x^2-y^2=15, x^2+y^2=1$, and the last one $x^2-y^2=3, x^2+y^2=5$. -Only the last one is valid. $x^2+y^2=15$ is not solvable since the primes which have residue $3$ modulo $4$ is not of an equal power. -One particular solution for $x^2-y^2=3, x^2+y^2=5$, is $x_0=2, y_o=1$. How do I get to an expression of a general solution for this system? -Thanks! - -REPLY [7 votes]: First, assume $x$ and $y$ are strictly positive. -$x^4 - y^4 = (x-y)(x+y)(x^2+y^2)$. Of these three factors, only the first can equal 1. So they must be 1, 3, and 5. From 1 and 3 you already get $x=2, y=1$, and then you just have to check that this is a valid solution. -Now we allow the negative solutions, to get $x = \pm 2, y = \pm 1$.<|endoftext|> -TITLE: Last few digits of $n^{n^{n^{\cdot^{\cdot^{\cdot^n}}}}}$ -QUESTION [7 upvotes]: I want to compute last few digts (as much as possible ) of the following number -$$ N:=n^{n^{n^{\cdot^{\cdot^{\cdot^n}}}}}\!\!\!\hspace{5 mm}\mbox{ if there are $k$ many $n$'s in the expression and $n\in\mathbb{N}$ }$$ -I have seen many particular cases of this problem. I think for odd $n$ the units digit is $n^3\mbox{ mod } 10 $ and for even $n$ the units digit is 6, for all $k\geq 3$ . How much can we say about the other digits ? - -REPLY [2 votes]: Here's a little bit of computational knowledge... -If we want the first $d$ digits, we can calculate the result by modular arithmetic. In other words, modulo $10^d=2^d5^d$. -The more time-consuming portion is calculating the result modulo $5^m$. We can note that $$k^m \mod n \equiv k^{m \mod \phi(n)} \mod n$$ where $\phi(n)$ is Euler's Totient function. -We can apply this function recursively, i.e. -$$m^m \mod n \equiv m^{m \mod \phi(n)} \mod n$$ -$$m^{m^m} \mod n \equiv m^{\left(m^m \mod \phi(\phi(n))\right) \mod \phi(n)} \mod n$$ -$$\dots$$ -where $n=5^d$. Therefore, the most extensive operation is exponentiation modulo $n$. This can be done in $O(\log(n))$ operations via exponentiation by squaring or binary exponentiation. -This operation is done at most $n$ times, so we get a conservative bound of $O(n \log(n))$ or, really, $O(5^d \log(5^d))$ operations.<|endoftext|> -TITLE: How may I prove this integral inequality? -QUESTION [9 upvotes]: Prove the following inequality: -$$\frac{\sqrt{\pi}}{2}\le\int_{0}^{1} \left({\log(\csc(x))}\right)^{1/3} dx$$ -What should i start with? (it's not a homework but a hobby related activity) - -REPLY [7 votes]: Note that $\sqrt{\pi}/2=\Gamma(3/2)$. -For $01/x$ so -$$ -\int_0^1 (\log(\csc(x)))^{1/3}~dx > \int_0^1(-\log(x))^{1/3}~dx -$$ -Substitute $y=-\log(x)$ to get -$$ -\int_0^1(-\log(x))^{1/3}~dx = \int_0^\infty y^{1/3}e^{-y}~dy=\Gamma(4/3) -$$ -So your inequality follows from $\Gamma(4/3)>\Gamma(3/2)$. Unfortunately $\Gamma(4/3)$ does not have a simple expression, so I don't yet see a way to show this last step except by evaluating it numerically.<|endoftext|> -TITLE: Fraction of two binomial coefficients -QUESTION [6 upvotes]: In an exercise I was asked to simplify a term containing the following fraction: $${\binom{m}{k}\over\binom{n}{k}}$$ -The solution does assume the following is true in the first step, without explaining why. I unfortunately cannot reconstruct the step: -$${\binom{m}{k}\over\binom{n}{k}} = {\binom{n-k}{m-k}\over\binom{n}{m}}$$ -Has anyone a good explanation why this transformation is possible? - -REPLY [2 votes]: The equivalence of the 2 expressions above has been explained by previous posters. -By simplification, I wonder if you mean produce an equivalent expression that will be more likely to compute without causing register overflow for factorials of large numbers ? Or do you just want to simplify it in a purely mathematical sense, i.e. write it in its most simple form ? -If the former, it's best to write it as a product of a series of quotients. -In the case of the expression $${\binom{m}{k}},$$ this works out as the product of k individual quotients. -For the full deal, i.e. $${\binom{m}{k}\over\binom{n}{k}},$$ you also have a similar number of quotients under the line. -So the overall ratio also works out as the product of k quotients, each of which will not be too big a number. Computing this, the m, n and k are transformed to floating point before evaluating each quotient and then forming the product of all k of them. The result is then rounded to give the true integer result. -If the latter, I do not know of a simpler form, except perhaps mPk / nPk (mPk = number of k sized permutations possible for m objects). -But this does not look much better. -You could also develop the right hand side of the equivalence above and get : -$$\frac{\displaystyle{\frac{m!}{n!}}}{\displaystyle{\hspace 5pt\frac{(m-k)!}{(n-k)!}}\hspace 5pt}$$ -This then reduces to $$\frac{m(m-1)(m-2)\cdots(m-n+1)}{(m-k)(m-k-1)(m-k-2)\cdots(m-n+1)}$$ -which is a product of (m-n) individual quotients. -$$\prod_{i=0, j=0}^{i=n-1, j=n-k-1}\frac{m-i}{m-k-j}$$ -So you just loop through this process for (m-n) times, each time calculating each quotient and multiplying it to the existing product. -If m and n are both much bigger than k, then there should be less computations doing it this way as well as less risk of register overflow. -Anyone knowing Latex better than me please edit this mess. -While it's a bit simplified computationally, it's been complicated expressionally . . .<|endoftext|> -TITLE: infinite number of irreducible polynomials in $\mathbb{Z}/2{\mathbb Z}[X]$ -QUESTION [5 upvotes]: For $A= \mathbb{Z}/2{\mathbb Z}[X]$ ring of polynomials with coefficient in the field $\mathbb{Z}/2{\mathbb Z},$ I need to show that there are infinite number of irreducible polynomials in $A.$ -How do I show that? I didn't come to any conclusion. I though of series of polynomials but since it it modulo $2$ they were not suitable. -Any direction? -(And: does any one have a link to a web where I can choose Latex symbols and see how they are written? I had one, but I lost it, and can't find it in Google) - -REPLY [4 votes]: Besides Euclid's classical method, here's another approach. Recall that the sequence of polynomials $\rm\:f_n = (x^n\!-\!1)/(x\!-\!1)\:$ is a strong divisibility sequence, i.e. $\rm\:(f_m,f_n) = f_{(m,n)}$ in $\rm\mathbb Z[x].\:$ Hence the subsequence with prime indices yields an infinite sequence of pairwise coprime polynomials. Further the linked proof shows the gcd has linear (Bezout) form $\rm\:(f_m,f_n) = f_{(m,n)}\! = g\, f_m + h\, f_n,\,$ $\rm\, g,h\in\mathbb Z[x],\:$ so said coprimality persists mod $2;\,$ indeed $\rm\:(p,q)=1\:$ for primes $\rm\:p\ne q,$ so -$$\rm\,mod\ 2\!:\ \ d\:|\:f_p,f_q\ \Rightarrow\ d\:|\:g\,f_p\!+\!h\,f_q = f_{(p,q)}\! = f_1 = 1.\,$$ -Thus, for each prime $\rm\:p,\:$ choosing a prime factor of $\rm\:f_p\:$ yields infinitely many prime polynomials mod $2,\,$ none associate (being pairwise coprime). Note that this argument works quite generally.<|endoftext|> -TITLE: A question concerning on the axiom of choice and Cauchy functional equation -QUESTION [6 upvotes]: The Cauchy functional equation: -$$f(x+y)=f(x)+f(y)$$ -has solutions called 'additive functions'. If no conditions are imposed to $f$, there are infinitely many functions that satisfy the equation, called 'Hamel' functions. This is considered valid if and only if the Zermelo's axiom of choice is accepted as valid. -My question is: suppose we don't consider valid the axiom of choice, this means that we have a finite number of solutions? Or maybe the 'Hamel' functions are still valid? -Thanks for any hints ore answer. - -REPLY [14 votes]: What you wrote is not true at all. The argument is not valid "if and only if the axiom of choice holds". - -Note that there are always continuous functions of this form, all look like $f(x)=ax$, for some real number $a$. There are infinitely many of those. -The axiom of choice implies that there are discontinuous functions like this, furthermore a very very weak form of the axiom of choice implies this. In fact there is very little "choice" which can be inferred from the existence of discontinuous functions like this, namely the existence of non-measurable sets. -Even if the axiom of choice is false, it can still hold for the real numbers (i.e. the real numbers can be well-ordered even if the axiom of choice fails badly in the general universe). However even if the axiom of choice fails at the real numbers it need not imply that there are no such functions in the universe. -We know that there are models in which all functions which have this property must be continuous, for example models in which all sets of real numbers have the Baire property. There are models of ZF in which all sets of reals have the Baire property, but there are non-measurable sets. So we cannot even infer the existence of discontinuous solutions from the existence of non-measurable sets. -Observe that if there is one non-discontinuous then there are many different, since if $f,g$ are two additive functions then $f\circ g$ and $g\circ f$ are also additive functions. The correct question is to ask whether or not the algebra of additive functions is finitely generated over $\mathbb R$, but to this I do not know the answer (and I'm not sure if it is known at all). - -More: - -Is there a non-trivial example of a $\mathbb Q-$endomorphism of $\mathbb R$?<|endoftext|> -TITLE: Calculating the following definite integral -QUESTION [6 upvotes]: How may I approach and compute the following integral? Could polylogarithm functions and complex numbers be avoided? -$$ \int_0^{\frac{\pi}{2}} \frac{x}{\tan x} dx$$ - -REPLY [3 votes]: Let's take $t=\tan x$ Then -$$I= \int_0^{\frac{\pi}{2}} \frac{x}{\tan x} dx=\int_{0}^{\infty}\frac{\arctan t}{t(1+t^2)}dt $$ Consider -$$I(a)= \int_{0}^{\infty}\frac{\arctan (at)}{t(1+t^2)}dt $$ Then -$$\frac{dI}{da}=\int_{0}^{\infty}\frac{1}{(1+a^2t^2)(1+t^2)}dt =\frac{\pi}{2(1+a)}$$ and -$$I(a)=\frac{\pi}{2}\ln (1+a)+const$$ $const=0$ because $I(0)=0$ -Thus -$$I(a)=\frac{\pi}{2}\ln (1+a)$$ and $I=I(1)=\frac{\pi}{2}\ln (2)$<|endoftext|> -TITLE: Measurable homomorphism of $\mathbb{T}$ into $\mathbb{C}^\times$ -QUESTION [6 upvotes]: Working through Katznelson's An Introduction to Harmonic Analysis and have been stumped by the following problem for the past few days: Show that a measurable homomorphism of $\mathbb{T}=\mathbb{R}/2\pi\mathbb{Z}$ into $\mathbb{C}^\times$ is actually a map into the unit circle. By previous exercises, it suffices to show that the image is compact or that the homomorphism is continuous. I found a similar exercise in Rudin's Real and Complex Analysis which asks to show that every Lebesgue measurable character on $\mathbb{R}$ is continuous. After spending a good deal of time playing around with the solution to that exercise, I realized that Rudin defines a character as a complex homomorphism having modulus 1, presupposing the result in a fundamental way. -I succeeded in showing (by copying the proof of Rudin's Theorem 9.23 and the exercise), given that a measurable homomorphism must have modulus 1, it must also be continuous and therefore given by an exponential. This is a later problem in Katznelson's book. A hint to the original problem would be appreciated! - -REPLY [7 votes]: Here's a proof that doesn't use any tools. Let $\chi:\mathbb{T}\to\mathbb{C}^{\times}$ be a measurable homomorphism. Consider the sets $C_n=\{z\in\mathbb{C}\mid 1/n<|z| -TITLE: Determinant of an $n\times n$ complex matrix as an $2n\times 2n$ real determinant -QUESTION [18 upvotes]: If $A$ is an $n\times n$ complex matrix. Is it possible to write $\vert \det A\vert^2$ as a $2n\times 2n$ matrix with blocks containing the real and imaginary parts of $A$? -I remember seeing such a formula, but can not remember where. Any details, (and possibly references) for such a result would be greatly appreciated. - -REPLY [7 votes]: Let $A=B+iC$, where $B$ and $C$ are $n\times n$ real matrices, and let -$$ -\widetilde{A}=\left( -\begin{array}{rr} -B & -C \\ -C & B -\end{array} -\right). -$$ -Then $\det\widetilde{A}=|\det A|^2$. -Proof. -$$ -\begin{align*} -\det\widetilde{A}&=\det\left( -\begin{array}{cc} -B+iC & -C+iB \\ - C & B -\end{array} -\right)= -\det\left( -\begin{array}{cc} -B+iC & 0 \\ - C & B-iC -\end{array} -\right)= -\det -\left( -\begin{array}{ll} -A & 0 \\ -C & \overline{A} -\end{array} -\right)\\ -&= -(\det A)(\det\overline{A})=(\det A)(\overline{\det A})=|\det A|^2. -\end{align*} -$$<|endoftext|> -TITLE: Cauchy's Residue Theorem on a Singularity Outside a Contour -QUESTION [6 upvotes]: I recently ran into the following exercise: - -Evaluate - $$\oint_\Gamma\frac{\cos z}{(z-\pi)^2}dz,$$where $\Gamma$ is a complete circuit of the circle $|z|=1$. - -Clearly, the singularity lies outside the contour: -                                     -However, recall Cauchy's Residue Theorem: - -If $\Gamma$ is a simple closed positively oriented contour and $f$ is analytic inside and on $\Gamma$ except at the points $z_1$, $z_2$, ..., $z_n$, then - $$\int_\Gamma f(z)dz=2\pi i\sum_{j=1}^{n}\text{Res}(z_j),$$ - where - $$\text{Res}(f;z_0)=\lim_{z\to z_0}\frac{1}{(m-1)!}\frac{d^{m-1}}{dz^{m-1}}[(z-z_0)^mf(z)].$$ - -If I use this to evaluate the integral, I will have that -$$\text{Res}(f;\pi)=\lim_{z\to\pi}\frac{d}{dz}[\cos z]=0,$$and hence the value of the integral will evaluate to $0$, which is the correct answer. -Is this purely coincidental? Because, as I understand it, for Cauchy's theorem to hold, the singularities must lie within the contour (or perhaps not?). Thanks in advance! - -REPLY [5 votes]: $$\oint_\Gamma\frac{\cos z}{(z-\pi)^2}dz=0$$ -by Cauchy-Goursat Theorem because the function being integrated is holomorphic/analytic on $\Gamma$. -That's it - no residues need to be computed!<|endoftext|> -TITLE: Why is full- & faithful- functor defined in terms of Set properties? -QUESTION [9 upvotes]: Wikipedia entry or Roman's "Lattices and Ordered Sets" p.286, or Bergman's General Algebra and Universal Constructions, p.177 and in fact every definition of full and/or faithful functor is defined in terms of the Set-theoretical properties: surjective and injective on (compatible) arrows. -Why aren't full-/faithful- defined in terms of epic and monic, in other words, in terms of algebraic invertibility or cancellation properties, eg if it is required to consider not a set of arrows but a topology or order (or any other category) of them. -Is this an historical accident awaiting suitable generalization, or is there some fundamental reason why Set seems to always lurk in the background? - -REPLY [8 votes]: Full and Faithful can be easily defined in general with no reference to Set. -Just state: -Faithful functor F -$\forall (f,g: A \to B)$: $Ff = Fg$ implies $f = g$ -Full functor F -$\forall (h: FA\to FB)$ $\exists (f: A \to B):Ff = h$ -Just FOL, no set theory or category of sets. -You can find this in CWM chapter 1 section Functors<|endoftext|> -TITLE: viscosity solution vs. weak solution -QUESTION [23 upvotes]: viscosity solution vs. weak solution -I am confused between the two. Is one a subset of the other or they are the same/completely different notions? Suppose I have an equation, $u_t=\mathcal{L}u$ for an elliptic operator $\mathcal{L}$ with bad coefficients so that it doesn't have a strong solution, but I can find a weak solution. What is the relation to viscosity in such a case? - -REPLY [14 votes]: The viscosity solutions was first introduced in the context of the Hamilton-Jacobi equation by the vanishing viscosity method. It would be difficult to apply the notion of distributional weak solution in this context, because the derivatives occur inside a nonlinear function. Further complications would arise because a distributional weak solution is not necessarily unique (for degenerate elliptic equations or nonlinear first order equations). Because of the nonlinearity, a Hamilton-Jacobi equations often doesn't have a classical solution, even if the Hamiltonian is an analytic function. -The viscosity solution was later generalized to degenerate elliptic equations, where the vanishing viscosity method itself no longer works. In case the equation is linear in the derivatives, both distributional weak solutions and viscosity solutions are defined, and we can investigate their relationship. My guess would be that the viscosity solutions turns out to also be a weak solutions in this context, but that there will in general be also weak solutions which are not viscosity solutions.<|endoftext|> -TITLE: Lines in the plane and recurrence relation -QUESTION [10 upvotes]: I am trying to solve the following problem from Cohen's Basic Techniques of Combinatorial Theory: - -A collection of $n$ lines in the plane are are said to be in general position if no two are parallel and no three are concurrent. Let $a_n$ be the number of regions into which $n$ lines in general position divide the plane. How big is $a_n$? - -The recurrence relationship for $a_n$ is immediate, namely -$$a_{n+1}=a_n+n.$$ -I have been trying to apply generating functions to solve this recurrence, but I am making a mistake somewhere. My attempt is as follows: -Let $G(X)=\sum_{k\geq 1} a_k X^k$. Multiplying both sides of the recurrence by $X^k$ and summing over all valid $k$ gives -$$\sum_{k\geq 1}a_{k+1}X^k=\sum_{k\geq 1}a_kX^k+\sum_{k\geq 1}kX^k.$$ -Using the fact that $\frac{1}{(1-x)^2}=\sum_{k\geq 1} kX^k$, I get -$$\sum_{k\geq 1}a_{k+1}X^k=G(X)+\frac{1}{(1-x)^2}.$$ -The left hand side is equal to -$$\frac{G(X)-a_1}{X}=\frac{G(X)-2}{X}$$ -since $a_1=2$. Solving for $G(X)$, I get -$$G(X)=\frac{2-4X+3x^2}{(1-x)^3}.$$ -Expanding with partial fractions gives -$$G(X)=\frac{3}{1-x}-\frac{2}{(1-x)^2}+\frac{1}{(1-x)^3}.$$ -Transforming this into power series gives -\begin{align} -G(X)&=3\sum_{k\geq 0}X^k-2\sum_{k\geq 0} (k+1)X^k+\sum_{k\geq 0}(k+1)(k+2)X^k\\ -&=\sum_{k\geq 0} (3-2(k+1)+(k+1)(k+2))X^k, -\end{align} -which is not correct. I have tried starting over, but I seems to be making a pretty serious mistake. Any thoughts? -Solution: -The real recurrence relation is $a_{n+1}=a_n+(n+1)$. Let $G(X)=\sum_{k\geq 1} a_k X^k$. Multiplying both sides of the recurrence by $X^k$ and summing over all valid $k$ gives -$$\sum_{k\geq 1}a_{k+1}X^k=\sum_{k\geq 1}a_kX^k+\sum_{k\geq 1}(k+1)X^k.$$ -Solving in terms of $G(X)$ gives -$$G(X)=\frac{x(2-2x+x^2)}{(1-x)^3}.$$ -By expanding with partial fractions we see -$$G(X)=\frac{x(2-2x+x^2)}{(1-x)^3}=\frac{x}{1-x}+\frac{x}{(1-x)^3}.$$ -Using the power series identities $\frac{1}{2}\sum_{k\geq 1}k(k-1)X^{k-1}=\frac{x}{(1-x)^3}$ and $\sum_{k\geq 1}X^k=\frac{x}{1-x}$. Substituting these power series in gives -\begin{align} -G(X)&=\frac{1}{2}\sum_{k\geq 1}k(k-1)X^{k-1}+\sum_{k\geq 1}X^k\\ -&=\frac{1}{2}\sum_{k\geq 1}k(k+1)X^{k}+\sum_{k\geq 1}X^k \quad \dagger\\ -&=\sum_{n\geq 1} (\binom{k+1}{2}+1)X^k. -\end{align} -Therefore, by equating coefficients, we have $$a_n=\binom{n+1}{2}+1.$$ -The step $\dagger$ follows from the fact that the first coefficient in the power series is $0$. -Note: This is NOT homework, I am merely working through exercises during summer vacation. I'm sure there are other ways to approach the problem, but I am trying to practice using generating functions. - -REPLY [8 votes]: As an alternative to generating functions (which is usually a very nice way to solve problems), let me mention the following. Assume no three lines are concurrent. -Each time a line is added, it adds a region for each region it passes through, i.e. it divides that region in two. The number of regions through which a line passes is equal to the number of lines crossed plus one. Each time the line crosses another line, it creates a point of intersection, so we could count the number of lines crossed plus one to be the number of points of intersection added plus the number of lines added. Thus, the number of regions is one (the plane) plus the number of points of intersection plus the number of lines. -If we have $n$ lines, there are $\binom{n}{2}$ points of intersection. Thus, the number of regions is -$$ -\binom{n}{2}+\binom{n}{1}+\binom{n}{0} -$$<|endoftext|> -TITLE: Does this intuition for "calculus-ish" continuity generalize to topological continuity? -QUESTION [11 upvotes]: In the past, I've always motivated continuity of a function from (some subset of) $\mathbb R$ to $\mathbb R$ based on the (incomplete) definition $\lim_{x \to c} f(x) = f(c)$; continuity at isolated points was never really too hard to explain. The main tool behind my explanation is a "using your common sense, what should the value here be?" prompt. -My usual example is this: You have a video camera, you're taking a video of the projectile motion of a ball or something. At precisely 3 minutes, your battery dies. Your camera is so awesome that it records stuff up to 2.999...[however many (finitely many) 9's] minutes. Common sense, i.e. "how the real world works", should tell you that this is enough for you to find out where the ball is at 3 minutes, even without any footage at that specific instant. If it were just a little bit higher/lower than "the right answer", then the ball would have to magically teleport from one point to another in [snaps fingers], which doesn't make any sense. -Then I get to the punchline - that continuity is required for us to "predict" things within "reasonable expectations". Things don't magically jump/teleport from one point to another, neither does temperature, brightness, etc. At isolated points, there's not enough "surrounding data" for us to make predictions, so any value can potentially be "reasonable", because if you claim that my prediction is "too high/low", I can always argue that it's conceivable that there might be a spike/dip in the surrounding data, so who's to say my prediction is definitely wrong? -(If this is a bad explanation, please let me know how I can improve it.) - -Fast-forward to a rather different scenario, now I'm trying to explain that the product topology is the coarsest topology that makes all projections continuous. I got countered with "but why should all projections be continuous?" I was stumped. I'd never really thought about the intuition behind continuity outside of $\mathbb R$. So my question is, can I generalize this "prediction" thing to motivate why projections of a product space onto its factor spaces should always be continuous? If so, how? If not, why not? -I have read the answers here, but I think/hope my question is a little different. -Also, feel free to yell at me if I've gotten continuity wrong for the last few years (and poisoned the minds of many others). But I would appreciate it if you could follow up with suggestions. Thanks. - -REPLY [5 votes]: As countinghaus pointed out in the comments, your intuition for continuity is correct as long as the topological space is second countable and Hausdorff. The intuition to which I am referring is the condition $\lim_{x\to c} f(x)=f(c)$ for all $c\in X$. -However, for general topological spaces, we encounter problems with this intuition. If we remove the condition that the space be Hausdorff, then the problem is that the question: "using your common sense, what should the value here be?" may have multiple (even infinitely many) correct answers (where correct means "makes the function continuous"). For example, if we equip a set $Y$ with the trivial topology, then every function $f:X\to Y$ is continuous (here $X$ denotes an arbitrary topological space). Suppose we look at a function $f:X\to Y$ where $f(x)=c$ for all $x\not=x'$ with $c\in Y$, $x'\in X$. Then we can ask the question "using your common sense, what should the value at zero be?", we might expect $f(x')=c$ to be the correct answer (it makes $f$ a constant function), but this is not a unique answer because any value for $x'$ makes $f$ continuous! -The necessity of the second countability condition is more subtle. Let $X$ denote the long line. If you are unfamiliar with this space, I recommend reading about it because it is great for finding counterexamples like this one! The idea behind the long line is to "paste" together uncountably many copies of the half-open interval $[0,1)$. Note that the real line $\mathbb{R}$ is homeomorphic to a countable such pasting. The term "long line" comes from the idea that we are simply pasting more of these same intervals, so the resulting space should be "longer" than the standard real line. -Now, consider the function $f:X\to\mathbb{R}$ where $f$ is $\sin(2\pi x)$ on each $[0,1)$ interval. I agree that I did not rigorously define this function but that is because I chose not to rigorously define the long line. Suffice it to say that it is not hard to rigorously define the function $f$, but the intuition is clear. Now, We might expect that $f$ is continuous (for if we did this with a countable pasting, we would obtain the function $\sin(2\pi x)$ on $\mathbb{R}$, which is continuous). However, it can be shown that every continuous function whose domain is the long line is eventually constant. That is, for intervals far enough down the line, the function maintains a constant value. Since our function $f$ is not eventually constant, it cannot be continuous. -This is why we need our topological space to be second countable and Hausdorff in order for your intuition to be the correct one. As for your question on the product topology specifically, I believe Brian Scott has handled this superbly in his answer.<|endoftext|> -TITLE: Roots of an irreducible polynomial in a finite field -QUESTION [12 upvotes]: Given a irreducible polynomial $f \in K[x]$ where $|K|=q$ is a finite field and $\deg(f)=n$. If $\alpha$ is a root of $f$ why are $\alpha, \alpha^q, \dots, \alpha^{q^{n-1}}$ the only possible candidates for the roots of $f$? - -REPLY [12 votes]: Amended in response to Arturo Magidin's comments. -$\beta^q = \beta$ for all $\beta \in \mathbb F_q$. Also, in any field -that includes $\mathbb F_q$ as a subfield, $(a-b)^q = a^q - b^q$. -Now suppose that $\alpha$ is an element in a field -that contains $\mathbb F_q$ as a subfield, -and that $f(\alpha)= 0$. Then, -$$ -0 = [f(\alpha)]^q = \left[\sum_{i=0}^n f_i \alpha^i\right]^q -= \sum_{i=0}^n (f_i)^q (\alpha^i)^q -= \sum_{i=0}^n f_i (\alpha^q)^i -= f(\alpha^q) -$$ -and so $f(\alpha) = 0 \implies f(\alpha^q) = 0$. It follows -that the elements -$\alpha, \alpha^q, \alpha^{q^2}, \alpha^{q^3}\cdots $ are roots -of $f(x)$. How many of these are distinct? Suppose that -$\alpha^{q^0}, \alpha^{q^1}, \cdots, \alpha^{q^{m-1}}$ are distinct elements -but $\alpha^{q^m}$ is a repeat, that is, $\alpha^{q^m} = \alpha^{q^i}$ for some $i \in \{0, 1, \ldots, m-1\}$. If $i$ were greater than $0$, -then we would have that -$$\left(\alpha^{q^{i-1}}\right)^q = \alpha^{q^i} -= \alpha^{q^m} = \left(\alpha^{q^{m-1}}\right)^q -\implies \left(\alpha^{q^{i-1}}\right)^q - \left(\alpha^{q^{m-1}}\right)^q -= 0 \implies \left(\alpha^{q^{i-1}} - \alpha^{q^{m-1}}\right)^q = 0$$ -and so $\alpha^{q^{i-1}} = \alpha^{q^{m-1}}$ in contradiction to the -hypothesis that $\alpha^{q^0}, \alpha^{q^1}, \cdots, \alpha^{q^{m-1}}$ -are distinct elements. We conclude that $\alpha^{q^m} = \alpha$. -Now consider the polynomial $g(x)$ defined as -$$g(x) = \prod_{i=0}^{m-1}\left(x - \alpha^{q^i}\right) = \sum_{j=0}^m g_jx^j$$ -whose roots are the $m$ distinct elements -$\alpha^{q^0}, \alpha^{q^1}, \cdots, \alpha^{q^{m-1}}$. Then -$$\begin{align*} -[g(x)]^q &= \left[\prod_{i=0}^{m-1}\left(x - \alpha^{q^i}\right)\right]^q -= \prod_{i=0}^{m-1}\left(x - \alpha^{q^i}\right)^q -= \prod_{i=0}^{m-1}\left(x^q - \alpha^{q^{i+1}}\right)\\ -&= \prod_{k=1}^{m}\left(x^q - \alpha^{q^{k}}\right) -= \prod_{i=0}^{m-1}\left(x^q - \alpha^{q^i}\right) = g(x^q). -\end{align*}$$ -Thus, -$\displaystyle [g(x)]^q = \left[ \sum_{j=0}^m g_jx^j\right]^q -= \sum_{j=0}^m (g_j)^q(x^j)^q -= \sum_{j=0}^m (g_j)^q(x^q)^j$ -equals $\displaystyle g(x^q) = \sum_{j=0}^m g_j(x^q)^j$, that is, -proving that $(g_j)^q = g_j$ for $0 \leq j \leq m$. Therefore, -we see that -$g(x) \in \mathbb F_q[x]$. But, $g(x)$ is a divisor of $f(x)$ which -is given to be irreducible over $\mathbb F_q$. So it must be -that $m = n$ and $f(x)$ is a scalar multiple of $g(x)$. -Thus, $\alpha^{q^0}, \alpha^{q^1}, \cdots, \alpha^{q^{n-1}}$ -are precisely the $n$ distinct roots of the degree-$n$ irreducible -polynomial $f(x) \in \mathbb F_q[x]$.<|endoftext|> -TITLE: A proof of Cartan-Dieudonné's theorem -QUESTION [5 upvotes]: As an assignment, we were asked to prove a theorem by Cartan and Dieudonné in the following form (the Euclidean space $\mathbb{E}^n$ is meant to be the usual $\mathbb{R}^n$ endowed with the usual inner product): -Every isometry of the Euclidean space $\mathbb{E}^n$ is the product of (at most) $n$ reflections about hyperplanes. -I proved it in the following way and I'd like to know whether it's a valid proof or not. I'd also like to know about other proofs (I'm sure there are...) : -We use induction on $n$. Let's assume the proposition for $n-1$. Let $\phi: \mathbb{E}^n\rightarrow \mathbb{E}^n$ be an isometry $\neq id$, and $\{e_1,\ldots, e_n\}$ be the canonical basis of $\mathbb{E}^n$. Consider the reflection $\sigma$ about the hyperplane orthogonal to the vector $e_1-\phi(e_1)$, which we assume non-zero, without loss of generality since $\phi\neq id$. -Clearly $\sigma(e_1)=\phi(e_1)$, so $e_1$ is ok. Since $\{\phi(e_2),\ldots, \phi(e_n)\}$ and $\{\sigma(e_2),\ldots, \sigma(e_n)\}$ are orthormal bases of the subspace $\langle \sigma(e_1)\rangle ^{\perp}$, there is an isometry $\psi: \mathbb{E}^n\rightarrow \mathbb{E}^n$ such that $\psi(\sigma(e_1))=\sigma(e_1)$ and $\psi(\sigma(e_j))=\phi(e_j)$, ($j\neq 1$). Since $\psi$ induces an isometry on $\langle \sigma(e_1)\rangle ^{\perp}$, by the inductive hypothesis, it can be written as a product of reflections about subspaces of $\langle \sigma(e_1)\rangle ^{\perp}$ and dimension $n-2$. Each of these subspaces can be extended to an hyperplane of $\mathbb{E}^n$ so that $\sigma(e_1)$ belongs to that and hence remains unchanged under those reflections. $\square$ - -REPLY [3 votes]: While this is a very old question, I find both the proofs, one proposed in the question itself and one by Stillwell cited in the answer by Drew Armstrong, seriously lacking from a formal point of view, even though the basic argument (the same in both proofs) can be made to work well. -Typos aside, here are a few defects. Proofs by induction need (to mention) a base case. There is a fundamental difference between a Euclidean vector space of dimension$~n$ and $\Bbb R^n$ equipped with the standard inner product (although the latter is a (standard) instance of the former): a hyperplane in an $n$-dimensional Euclidean vector space is an $n-1$ dimensional Euclidean vector space, but a hyperplane in $\Bbb R^n$ by no stretch of imagination is $\Bbb R^{n-1}$, unless one wants to entertain the idea that a same set can be subset of another set in many different ways (injective maps were invented to overcome that difficulty). So if one is going to do induction via hyperplanes, it pays to state the theorem for general Euclidean vector spaces rather than just for the spaces $\Bbb R^n$. An essential point is that a reflection of a hyperplane is not a reflection of the whole space, although it can be uniquely extended to one; one needs to prove that the desired properties carry over during this extension (OP does this correctly, Stillwell completely ignores this). Finally, in such a proof by induction one should not exclude the identity from the statement to be proved, since the induction works towards the identity, and would therefore fatally hit upon a case where the induction hypothesis cannot be applied. This is a problem in the OP answer; the Stillwell proof does not exclude the identity form the statement, but simply excludes it from its proof. In both cases the implicit convention is that the identity gives a product of no reflections at all, but this is never said aloud. -But the main obstruction to a nice proof by induction is too weak a statement of the theorem; we need some induction loading. N'en déplaise à Cartan et Dieudonné, the theorem should give the precise number of reflections needed, which is the codimension of the subspace $\ker(\phi-I)$ of vectors fixed by $n$ (clearly this is the minimal number possible, as a product of $k$ reflections fixes at least the intersection of their hyperplanes, which has codimension at most$~k$). -Theorem. Let $\phi$ be an automorphism (orthogonal endomorphism) of an Euclidean vector space $E$ of dimension$~n$; put $d=\dim(\ker(\phi-I_E))$. Then $\phi$ can be written as a product of $n-d$ orthogonal (hyperplane) reflections. -We can prove this by induction on $n-d$. The base case is $n-d=0$, so $\phi=I_E$ which is a product of $0$ reflections. For the inductive case assume $n-d>0$, and the theorem proved for smaller values of $n-d$. Since $\ker(\phi-I_E)\neq E$, one can choose a vector $v\notin\ker(\phi-I_E)$; after doing so, put $H=(\phi(v)-v)^\perp$, a hyperplane, $\sigma_H$ the orthogonal reflection with respect to$~H$, and $\phi'=\sigma_H\circ\phi$. We have $\phi'(v)=v$. Also $\ker(\phi-I_E)\subseteq H$ (since $(w\mid\phi(v)-v)=0$ when $\phi(w)=w)$), so $\ker(\phi'-I_E)\supseteq\ker(\phi-I_E)+\langle v\rangle\supset\ker(\phi-I_E)$. Since on the other hand $\ker(\phi-I_E)\supseteq\ker(\phi'-I_E)\cap H$, it follows that $\dim(\ker(\phi'-I_E))=\dim(\ker(\phi-I_E))+1=d+1$. This shows that the induction hypothesis applies to $\phi'$, and expresses it as a product of $n-d-1$ reflections. Then $\phi=\sigma_H\circ\phi'$ is written as a product of $n-d$ reflections $\square$.<|endoftext|> -TITLE: Normal subgroups of infinite symmetric group -QUESTION [23 upvotes]: I recently took a course on group theory, which mentioned that the following proposition is equivalent to the continuum hypothesis: "The infinite symmetric group (i.e. the group of permutations on the set $\mathbb{N}$) has exactly 4 normal subgroups." Does anyone have any references or explanation for this? - -REPLY [6 votes]: The widely known reference for this result is Schreier-Ulam (1933). But this was previously proved by Luigi Onofri (1929), in the third of a series of 3 articles, of 1927, 1928, and 1929, not mentioned in Schreier-Ulam's article, in which he also introduces (§4 of the first opus) its natural topology on the group of permutations of a countable set. -L. Onofri. Teoria delle sostituzioni che operano su una infinità numerabile di elementi. Memoria 3a. Annali di Matematica Pura ed Applicata -December 1929, Volume 7, Issue 1, pp 103–130. (restricted access: - Springerlink). -This is in §141, p124: - -I gruppi $G_1$ e $G_2$ sono gli unici sottogruppi invarianti del totale $G$. (The groups $G_1$ and $G_2$ are the unique invariant subgroups of the whole group $G$.) - -Here $G$ is the group of permutations of an infinite countable set $I$, $G_1$ its subgroup of finitely supported permutations ("substitutions operating on finitely many elements" in his language) and $G_2$ the subgroup of even elements in $G_1$, and "invariant" means invariant under conjugation. -Onofri's proof has 3 steps (I try to use italics when I translate in English without translating in modern math language): -(a) he defines (§100, p104) a subset $H$ of $G$ to be infinitely transitive of infinite grade if it is transitive by pre-composition of injective self-maps whose image with infinite complement, or, to be closer to his language, if for any two infinite systems of (distinct) elements $[x_1,x_2,\dots]$, $[y_1,y_2,\dots]$ in $I$ such that the complements (called residual systems) of both $\{x_i:i\ge 1\}$ and $\{y_i:i\ge 1\}$ are infinite, there exists $s\in H$ ("a substitution $s$ of $H$") such that $s(x_i)=y_i$, $(i=1,2,\dots)$. Here "infinite grade" refers to infiniteness of the complement. -He proves (§110, p108) that the only subgroup of the group of permutations of $I$ that is infinitely transitive of infinite grade, is the whole group of permutations. -(b) He proves (§136, p121) that the only normal subgroup not contained in the subgroup of finitely supported permutations, is the whole group, and his strategy consists in proving that such a subgroup is infinitely transitive of infinite grade. -His statement is actually Un complesso C contenente una sostituzione h su infiniti elementi e le sue trasformate mediante G, coincide con il totale. -A literal translation is: A complex containing a substitution h on infinitely many elements and its conjugate transforms G coincides with the total. -An adapted translation, from my understanding, is: "A subsemigroup (of $G$, the group of permutations of $I$) containing an infinitely supported permutation as well as its conjugates coincides with the whole group of permutations." -Here I should explain why I'm translating complesso (literally "complex") into "subsemigroup": "complesso" is evoked, not defined in the first memoir (1927, §25, p89): indeed he defines a somewhat inefficient terminology for subsemigroups of the group of permutations "(sub)groups" for subgroups, "pseudogroups" those subsemigroups that are not subsemigroups, and even the splits pseudogroups as "simple" and "composite", where "simple" means that no element is invertible. Then he seems to use "complex" means something which either a group, a "simple" pseudogroup, or a "composite pseudogroup". I'm still a bit confused since he occasionally uses "complesso" for cosets, but anyway the proof is convincing if one interprets as "subsemigroup". -(c) The hardest part is done, since the result now reduces to classify the normal subgroups contained in the subgroup of finitely supported permutations; this short concluding step is §141, p124. Actually, he then (§142) observes that the alternating subgroup is the "only invariant subgroup" of the group of finitely supported permutations while the alternating subgroup itself has "no normal subgroup except the identity". It's a bit sloppy on whether he considers the trivial/whole subgroups as normal ("invariant") subgroups, but the proof is totally correct as far as I could check. -He, by the way, observes (§139) after step (b) the corollary that the group of permutations of $I$ has no proper subgroup of finite index, and in particular (§140) that there is no way to coherently extend parity of finite permutations to arbitrary permutations (since this would produce a subgroup of index 2).<|endoftext|> -TITLE: Level sets of convex functions -QUESTION [5 upvotes]: Let $f:\mathbb{R}^2\rightarrow\mathbb{R}$ be a convex function. For $t\in\mathbb{R}$, consider the corresponding level set -$$f^{-1}\{t\}=\{(x,y)\in\mathbb{R}^2: f(x,y)=t\}.$$ -For the application I have in mind, it will suffice to assume that all the level sets of $f$ are compact curves. In that case, denoting arc length measure on the plane by $\ell(\cdot)$, when does an estimate of the form -$$\Big|\ell(f^{-1}\{t\})-\ell(f^{-1}\{t'\})\Big|\leq C |t-t'|^{\delta}$$ -hold for some $C<\infty$ and $\delta>0$? And how does the optimal exponent $\delta$ depend on $f$? -The intuition is that the curves $f^{-1}\{t\}$ and $f^{-1}\{t'\}$ should be "close to each other" whenever $t$ is close to $t'$. In which other ways can this be made rigorous? -Thank you. - -REPLY [2 votes]: Here is an approach in terms of "intuitive calculus": -Let -$$\gamma_v:\quad s\mapsto z(s)=\bigl(x(s),y(s)\bigr)\qquad(0\leq s\leq L_v)$$ -be the arc length parametrization of the level set $f^{-1}(v)$. For each $s$ the tangent vector $\dot\gamma_v(s)$ is orthogonal to the gradient $\nabla f$ at $z(s)$. If we orient $\gamma_v$ such that $\nabla f$ points to the outside we can therefore write $\nabla f\bigl(z(s)\bigr)$ in the form $$\nabla f\bigl(z(s)\bigr)=\rho\bigl(z(s)\bigr)\bigl(\dot y(s),-\dot x(s)\bigr)\ ,\qquad\rho(z):=\bigl|\nabla f(z)\bigr|\ .$$ -For an infinitesimal $\epsilon>0$ the level set $\gamma_{v+\epsilon}:=f^{-1}(v+\epsilon)$ has the parametrization -$$\gamma_{v+\epsilon}:\quad s\mapsto\Bigl(x(s)+\epsilon{ \dot y(s)\over\rho(z(s))},\ y(s)-\epsilon{ \dot x(s)\over\rho(z(s))}\Bigr)\qquad(0\leq s\leq L_v)\ ,$$ -where we have neglected higher order terms in $\epsilon$. Thus we get -$$\dot\gamma_{v+\epsilon}(s)\doteq \Bigl(\dot x(s)+\epsilon{ \ddot y\rho-\dot y\dot\rho\over\rho^2},\ \dot y(s)-\epsilon{ \ddot x\rho-\dot x\dot\rho\over\rho^2}\Bigr)$$ -and therefore -$$\bigl|\dot\gamma_{v+\epsilon}\bigr|\doteq1+2\epsilon{\dot x\ddot y-\ddot x\dot y\over\rho}$$ -or -$$\bigl|\dot\gamma_{v+\epsilon}(s)\bigr|\doteq 1+\epsilon{\kappa(s)\over\rho(z(s))}\ ,$$ -where we have denoted the curvature of $\gamma_v$ by $\kappa$. From this we deduce the following formula for the derivative of $L_v$ with respect to $v$: -$${dL_v\over dv}=\int_0^{L_v}{\kappa(s)\over\bigl|\nabla f(z(s))\bigr|}\ ds\ .\qquad(*)$$ -Now that we have this formula it seems intuitively pretty obvious: When we let $v$ increase then the length $L_v$ increases most in parts of $\gamma_v$ where the curvature is large, and does not change along straight parts of $\gamma_v$. -The integrand in formula $(*)$ can be expressed in terms of the partial derivatives of $f$. One obtains (see, e.g., Bronstein-Semendjajew) -$${dL_v\over dv}=\int_0^{L_v}{f_{xx}f_y^2-2f_{xy}f_xf_y+f_{yy}f_x^2\over (f_x^2+f_y^2)^4}\Biggr|_{z(s)}\ ds\ .$$<|endoftext|> -TITLE: Not every metric is induced from a norm -QUESTION [79 upvotes]: I have studied that every normed space $(V, \lVert\cdot \lVert)$ is a metric space with respect to distance function -$d(u,v) = \lVert u - v \rVert$, $u,v \in V$. -My question is whether every metric on a linear space can be induced by norm? I know answer is no but I need proper justification. -Edit: Is there any method to check whether a given metric space is induced by norm ? -Thanks for help - -REPLY [6 votes]: One possible way of showing that a metric does not arise from a norm is to show that it is bounded, as it then cannot be homogeneous. - -Proof: -Take $d$ a bounded metric on a space $X$: i.e. $\exists D \in \mathbb{R}_+$ such that $\forall (x,y) \in X^2, d(x,y) \leq D$. Suppose now for contradiction that $d$ arises from a norm, i.e. the exists a norm $ \|\cdot\|$ such that $d(x,y) = \|x - y\|$. Recall that the distance must then must be homogeneous, for we have $d(\lambda x, \lambda y) = \|\lambda x - \lambda y\| = |\lambda| \cdot \|x - y\| = |\lambda| d(x,y)$. -Take now arbitrary $(x_0, y_0) \in X^2$, then we must have that $d\left( \frac{D+1}{d(x_0, y_0)} \cdot x_0, \frac{D+1}{d(x_0, y_0)} \cdot y_0 \right) = \frac{D+1}{d(x_0, y_0)} \cdot d(x_0, y_0) = D + 1 > D $ which contradicts the upper bound of the metric.<|endoftext|> -TITLE: Curry-Howard correspondence -QUESTION [6 upvotes]: I read that the Curry-Howard correspondence introduces an isomorphism between typed functions and logical statements. For example, supposedly the function -$$\begin{array}{l} -I : \forall a. a \to a\\ -I = \lambda x. x -\end{array}$$ -can be interpreted as: -$$p \implies p \text{ for any proposition } p$$ -Is that correct? How might I interpret a function $K : \forall ab. a \to (b \to a)$? What about $\text{succ} : \mathbb{N} \to \mathbb{N}$, where $\text{succ} = \lambda x. x + 1$? -If the scope of this question is too large, can someone recommend an introductory book that would cover this? A proof of the isomorphism would probably be enlightening, too: can someone point me to one? - -REPLY [2 votes]: The short version of the Curry-Howard correspondence is: to interpret a type as a proposition, interpret the function type $a\to b$ as "$a$ implies $b$". Interpret the product type $a\times b$ as "Both $a$ and $b$". Interpret the disjoint union type $a \sqcup b$ as "Either $a$ or $b$ (or both)". -Then for example the type of the $K$ combinator, $a\to(b\to a)$, is interpreted as the proposition that if we know $a$, then if we know $b$ we can conclude $a$. Or the type of the program which gets a pair $(a,b)$ and returns $b$ is $a\times b\to b$, and this is the theorem that says that if you can prove $a$ and $b$, then you have a proof of $b$, usually written $a\land b\to b$. -(In the Haskell language, $a\times b$ is notated (a,b) and $a\sqcup b$ is notated Either a b.) -The successor function is a bit silly, because its type is a trivial theorem. The type $\Bbb N$ is known to be inhabited, because it contains constants like 0. Such types correspond to propositions that are known to be true. For example, there is a program, written 0, with type $\Bbb N$, and this is indeed a theorem. So the successor function's type, ${\Bbb N}\to{\Bbb N}$, just says that $\Bbb N$ implies $\Bbb N$, which is not very interesting. Perhaps slightly more interesting is the theorem $a\to \Bbb N$, which corresponds to the program $\lambda a. 13$. -There is more to the correspondence than that types are theorems of logic. More important is that a program with a certain type corresponds to a proof of the corresponding theorem. Evaluation of the program corresponds to elimination of the "cut" rule from the proof. Purely functional programs correspond to theorems of intuitionistic logic, but imperative programs correspond to theorems of classical logic, which is stronger. Double-negation corresponds to exception handling. Glivenko's theorem, which says that any theorem of classical logic has an intuitionistic analogue, corresponds to the implementation of an imperative program in a functional language in continuation-passing style. -I recommend that you read Morten Heine Sørensen and Paweł Urzyczyn's Lectures on the Curry-Howard Isomorphism for complete details. -Also potentially interesting: this answer constructs a program with type $\lnot\lnot(P\lor\lnot P)$.<|endoftext|> -TITLE: How to solve infinite repeating exponents -QUESTION [8 upvotes]: How do you approach a problem like (solve for $x$): -$$x^{x^{x^{x^{...}}}}=2$$ -Also, I have no idea what to tag this as. -Thanks for any help. - -REPLY [11 votes]: I'm just going to give you a HUGE hint. and you'll get it right way. Let $f(x)$ be the left hand expression. Clearly, we have that the left hand side is equal to $x^{f(x)}$. Now, see what you can do with it.<|endoftext|> -TITLE: What is the simplest way to prove that the logarithm of any prime is irrational? -QUESTION [25 upvotes]: What is the simplest way to prove that the logarithm of any prime is irrational? -I can get very close with a simple argument: if $p \ne q$ and $\frac{\log{p}}{\log{q}} = \frac{a}{b}$, then because $q^\frac{\log{p}}{\log{q}} = p$, $q^a = p^b$, but this is impossible by the fundamental theorem of arithmetic. So the ratio of the logarithms of any two primes is irrational. Now, if $\log{p}$ is rational, then since $\frac{\log{p}}{\log{q}}$ is irrational, $\log{q}$ is also irrational. So, I can conclude that at most one prime has a rational logarithm. -I realize that the rest follows from the transcendence of $e$, but that proof is relatively complex, and all that's left to show is that no integer power of $e$ is a prime power (because if $\log p$ is rational, then $e^a = p^b$ has a solution). It is easy to prove that $e$ is irrational ($e = \frac{a}{b!} = \sum{\frac{1}{n!}}$, multiply by $b!$ and separate the sum into integer and fractional parts) but I can't figure out how to generalize this simple proof to show that $e^x$ is irrational for all integer $x$; it introduces a $x^n$ term to the sum and the integer and fractional parts can no longer be separated. How to complete this argument, or what is a different elementary way to show that $\log{p}$ is always irrational? - -REPLY [3 votes]: I'm not sure if this qualifies as a very elementary proof (and I have not worked out all the details myself), but there appears to be quite a nice proof that any rational power of $e$ is irrational in Chrystal's Algebra, which might well be available online, as it is certainly old enough to be out of copyright. Interestingly, Chrystal refers to his "Algebra" as "An Elementary Text-Book". -In my copy (Vol II. published in 1889) the result appears as Corollary 3, on page 495, in chapter XXXIV on "General Continued Fractions". He first proves a result about general continued fractions, to the effect that: -If $a_2, a_3, \dots$ and $b_2, b_3, \dots$ are all positive integers, then the continued fraction $$\cfrac{b_2}{a_2 + \cfrac{b_3}{a_3 + \cfrac{b_4}{a_4 + \dots}}}$$ -converges to an irrational limit provided that after some value of $n$ the condition $a_n\nless b_n$ be always satisfied. -He then deduces the above Corollary 3, from the expansion -$$\tanh x = \cfrac{x}{1 + \cfrac{x^2}{3 + \cfrac{x^2}{5 + \dots}}},$$ -although he does not explain the deduction in details, but simply states that it can be deduced in a similar way to a previous result he gives concerning $\pi$ and $\pi^2$.<|endoftext|> -TITLE: The ring $\mathbb Z[\sqrt{-2}]= \{a+b\sqrt{-2} ; a\in \mathbb Z,b\in \mathbb Z \}$ has a Euclidean algorithm -QUESTION [8 upvotes]: I need to prove that the ring $\mathbb Z[\sqrt{-2}]= \{a+b\sqrt{-2} ; a\in \mathbb Z,b\in \mathbb Z \}$ has a Euclidean algorithm, and to decide whether there are infinitely many primes in this ring. - -How do I approach this kind of question? -Thank you. - -REPLY [10 votes]: It is the same argument as the one for the Gaussian integers. -Let $w$ and $z$ be in our ring. We want to show that there exist numbers $q$ and $r$ in our ring such that $w=zq+r$, and $N(r) \lt N(z)$, where $N$ is the usual norm. -Consider the complex number $\frac{w}{z}$. There are real numbers $s$ and $t$ such that $\frac{w}{z}=r+s\sqrt{2}i$. Let $a$ be the nearest integer to $r$, and let $b$ be the nearest integer to $s$. Then $a+b\sqrt{2} i$ is in our ring. -Since $|r-a|\le \frac{1}{2}$, and $|s-b|\le \frac{1}{2}$, the norm of $(r+s\sqrt{2}i)-(a+b\sqrt{2}i)$ is $\le \left(\frac{1}{2}\right)^2 + 2\left(\frac{1}{2}\right)^2$, which is $\le \frac{3}{4}$. -Finally, let $q=a+b\sqrt{2}i$, and $r=w-qz$. Then $r=\frac{r}{z}z=\left(\frac{w}{z}-q\right)z$, and therefore -$$N(r)=N\left(\frac{w}{z}-q\right)N(z)\le \frac{3}{4}N(z).$$ -For infinitely many irreducibles, the usual "Euclid" argument works with minor changes, and the irreducibles are prime. Or else, more simply, show that for any ordinary prime $p_i$, there is a non-unit irreducible $w_i$ that divides $p_i$. This produces infinitely many primes in our ring. It is useful to observe that by Bezout's Theorem, any two ordinary integers that are relatively prime in the ordinary sense do not have a common non-unit divisor in our ring.<|endoftext|> -TITLE: Expansion lemma for paracompact hausdorff space. -QUESTION [5 upvotes]: [Munkres, Section41, Ex5] Let $X$ be paracompact. We proved a "shrinking lemma" for arbitrary indexed open coverings of $X$. Here is - an "expansion lemma" for arbitrary locally finite indexed families in - $X$. -Lemma. Let $\{ B_ \alpha \} _ { \alpha \in J}$ be a locally finite - indexed family of subsets of the paracompact Hausdorff space $X$. Then - there is a locally finite indexed family $\{ U_ \alpha \} _ { \alpha - \in J}$ of open sets in $X$ such that $ B_ \alpha \subset U_ \alpha$ - for each $\alpha$. - -In the proof of the shrinking lemma(Munkres lemma41.6), the author uses the collection of all open sets such that the closure is contained in the given set. So I used the collection of all open sets containing $B_ \alpha$ for some $ \alpha$. But this seems to not work. How can I prove this? - -REPLY [5 votes]: For each $x\in X$ let $V(x)$ be an open nbhd of $x$ that meets only finitely many members of $\mathscr{B}$. Let $\mathscr{W}$ be a locally finite open star refinement of $\{V(x):x\in X\}$. -For each $\alpha\in J$ let $U_\alpha=\bigcup\{W\in\mathscr{W}:W\cap B_\alpha\ne\varnothing\}$; clearly $B_\alpha\subseteq U_\alpha$, and $U_\alpha$ is open. Let $\mathscr{U}=\{U_\alpha:\alpha\in J\}$. -Let $W_0\in\mathscr{W}$ be arbitrary; there is a $y\in X$ such that $$\operatorname{st}(W_0,\mathscr{W})=\bigcup\{W\in\mathscr{W}:W\cap W_0\ne\varnothing\}\subseteq V(y)\;.$$ Thus, if $F_0=\{\alpha\in J:B_\alpha\cap\operatorname{st}(W_0,\mathscr{W})\ne\varnothing\}$, then $F_0$ is finite. Suppose that $W_0\cap U_\alpha\ne\varnothing$ for some $\alpha\in J$; then there is a $W\in\mathscr{W}$ such that $W\cap B_\alpha\ne\varnothing\ne W\cap W_0$, so $B_\alpha\cap\operatorname{st}(W_0,\mathscr{W})\ne\varnothing$, and $\alpha\in F_0$. That is, each $W\in\mathscr{W}$ meets only finitely many $U\in\mathscr{U}$. -Now let $x\in X$ be arbitrary; $\mathscr{W}$ is locally finite, so $x$ has an open nbhd $G$ that meets only finitely many members of $\mathscr{W}$. We’ve just seen that each $W\in\mathscr{W}$ meets only finitely many members of $\mathscr{U}$, so $G$ meets only finitely many members of $\mathscr{U}$. Since $x$ was arbitrary, $\mathscr{U}$ is locally finite, as desired.<|endoftext|> -TITLE: Duality of a finitely generated projective modules -QUESTION [9 upvotes]: Let $M$ and $N$ be a finitely generated projective module over a ring $R$. Suppose we have a non degenerate bilinear pairing $\langle \ \cdot \ ,\ \cdot\ \rangle: M \times N \to R$. -I want to show $M$ is isomorphic to the dual $N^*$ of $N$. -The injectivity of $M$ into $N^*$ follows from the non degeneracy of the pairing by definiting a map $x \mapsto \langle x, \cdot\rangle$. What I cannot prove is surjectivity. -If I impose a condition that $R$ is an injective module then I think surjectivity also follows. But I want to prove it without any condition on $R$ -Any help is appriciated. - -REPLY [8 votes]: If by non-degenerate you mean that $\langle x , - \rangle = 0 \Leftrightarrow x = 0$ (or $\langle - , y \rangle \Leftrightarrow y = 0$), your question is equivalent to: Is every injective homomorphism between finitely generated projective $R$-modules an isomorphism? The answer is no, even for maps $R \to R$. -If by non-degenerate you mean that $\langle x , - \rangle = 0 \Leftrightarrow x = 0$ and $\langle - , y \rangle \Leftrightarrow y = 0$, then your question is equivalent to: If $M \to N$ is an injective homomorphism of f.g. proj. modules such that $N^* \to M^*$ is also injective, then $M \to N$ is an isomorphism. But again this is false for maps $R \to R$. -The "correct" definition of a perfect pairing $M \times N \to R$ is the following: Both maps $M \to N^*$ and $N \to M^*$ are isomorphisms.<|endoftext|> -TITLE: Proving that there exists $a_i \in \{a_1,\dots,a_k\}$ so that for any positive integer $n$ $F(n)$ is divisible by $a_i$ -QUESTION [12 upvotes]: Given $F(x) \in \mathbb{Z}{[x]}$. For all positive integer $n$ , $F(n)$ is divisible by one of $a_1 , a_2 , \dots , a_k\in \mathbb{Z}$ .how to Prove that there exists $a_i \in \{a_1,\dots,a_k\}$ so that for any positive integer $n$ $F(n)$ is divisible by $a_i$ - -REPLY [7 votes]: Suppose that no $a_i$ divides $F(n)$ for every $n$. -For each $a_i$, there is a power of prime factor of $a_i$ that does not divide $F(n)$ for every $n$. This forms a set of prime powers from which we can extract the minimum elements with respect to divisibility: $p_1^{e_1},\dots,p_m^{e_m}$, with $p_1,\dots,p_m$ distinct primes. Call $q_j$ any number such that $p_j^{e_j} \nmid F(q_j)$. -By the Chinese remainder theorem, there is some $n_0$ such that for every $j$: -$$n_0\equiv q_j \pmod{p_j^{e_j}}$$ -Therefore: -$$F(n_0)\equiv F(q_j)\pmod {p_j^{e_j}}$$ -so by definition of $q_j$, $p_j^{e_j} \nmid F(n_0)$. As a consequence no $a_i$ can divide $F(n_0)$: this contradicts the hypothesis. -So there must be some $a_i$ that divides $F(n)$ for every $n$.<|endoftext|> -TITLE: Calculate $\int f(x) f''(x)dx$ -QUESTION [5 upvotes]: I have a curiosity. If -$\int f(x) f'(x)dx=\int f(x) df(x)=\frac{\left(f(x)\right)^{2}}{2}+C$ -what is the result of: -$\int f(x) f''(x)dx$ - -REPLY [2 votes]: Using by parts $\int f(x)f''(x)dx= f(x)f'(x)-\int f'^2(x)dx$. - -REPLY [2 votes]: $$ -f(x)f'(x)-\int f'(x)^2 \, dx \ ? -$$<|endoftext|> -TITLE: Verifying Ito isometry for simple stochastic processes -QUESTION [6 upvotes]: It is known that stochastic integral must satisfy the isometry property which is -$$ -\mathbb{E}\left[ \left( \int_0^T X_t~dB_t\right)^2 \right] = \mathbb{E} \left[ \int_0^T X^2_t~dt \right] . -$$ -I am trying to prove this property for a simple stochastic process. What I said so far that is -$$ -\mathbb{E}\left[\sum_{i=0}^{n-1} X_i \left(B(t_{i+1})-B(t_i)\right)\right]^2, -$$ -then I am stuck. I know that we should to write the square sum as double sum to continue the proof but I couldn't do it. Any help please! - -REPLY [5 votes]: I take it that a simple process is a process of the form -$$ -Y_t=\sum_{i=0}^{n-1} X_i 1_{]t_i,t_{i+1}]}(t),\quad t\geq 0, -$$ -where $0\leq t_1 -TITLE: Deducing results in linear algebra from results in commutative algebra -QUESTION [14 upvotes]: Here are two examples of results which can be deduced from commutative algebra: - -Any $n\times n$ complex matrix is conjugate to a Jordan canonical matrix (can be proven using the structure theorem for modules over a PID, in this case $\mathbf{C}[T]$ - see for example these course notes). -Commuting matrices have a common eigenvector (this can be seen as a consequence of Hilbert's Nullstellensatz, according to Wikipedia). - -My question is, does anyone know of other examples of results in linear algebra which can be deduced (non-trivially*) from results in commutative algebra? -More precisely: Which results about modules over fields can be optained via modules over more general commutative rings? (Thanks to Martin's comment below for suggesting this precision of the question). -.* by "non-trivially," I mean you have to go deeper than simply applying module theory to modules over a field. - -REPLY [4 votes]: This answer doesn't contain anything new and it was already alluded via the JNF in the question, but let me elaborate this point of view: -Given a vector space $V$ over a field $K$, then an endomorphism $f : V \to V$ is the same as giving $V$ the structure of a left $K[T]$-module (which restricts to the given $K$-module structure). The multiplication with $T$ on the left is just $f$. This is just the most natural gadget when you think about polynomials $f^n + a_{n-1} f^{n-1} + \dotsc + a_0$ and their action on $V$. Many notions of linear algebra can be understood in more concise terms with the help of this module (and actually this can be used as a motivation for an introduction to modules in a linear algebra class): - -A $f$-invariant subspace of $V$ is just a $K[T]$-submodule of $V$ -That $V$ is $f$-cyclic just means that $V$ is cyclic as a $K[T]$-module, i.e. generated by one element, i.e. isomorphic to $K[T]/(p)$ for some $p$. We have that $V$ is finite-dimensional iff $p \neq 0$. -It immediately follows that $f$-invariant subspaces of $f$-cyclic spaces are also $f$-cyclic. Try to prove this without the language of modules. -That $V$ is $f$-indecomposable just means that it is indecomposable as a $K[T]$-module -More general and concisely: The arrow category of $\mathsf{Vect}_K$ is isomorphic to the category of $K[T]$-modules. -The minimal polynomial $p$ of $f$ is just the unique normed generator of the kernel of the ring homomorphism $K[T] \to \mathsf{End}_K(V)$ given by the module. -When you think about the endomorphism of $K[T]/(p)$ which multiplies with $T$ in terms of the canonical basis, you will arrive automatically at the companion matrix of $p$ and the relation with $f$-cyclic vector spaces. You don't have to "learn" this, it is just there. -The structure theorem for finitely generated modules over principal ideal domains implies the general normal form (which holds over every field $K$): Every endomorphism of a finite-dimensional vector space is similar to a direct sum of cyclic endomorphisms. In coordinate language: Every matrix is similar to a block matrix of companion matrices. -If $K$ is algebraically closed, you may reduce to polynomials of the form $(T-\lambda)^n$, but $K[T]/(T-\lambda)^n \cong K[T]/T^n$ has a very simple matrix, namely the Jordan block. Hence, the general normal form implies the Jordan normal form: Every matrix is similar to a block matrix of Jordan blocks.<|endoftext|> -TITLE: Does this example fail the CLT? -QUESTION [7 upvotes]: Let $X_i, i \in \mathbb{N}$ be independent random variables with $E[X_i] = \mu_i$ and $\mathrm{var}(X_i) = \sigma_i^2 < \infty$. -Define $S_n := \sum_{i=1}^{n} X_{i}$, and $s_n := \sqrt{\sum_{i=1}^{n} \sigma_{i}^2}$ . -Under the above assumptions, following are two different groups of sufficient conditions for $$ \frac{ S_n - \sum_{i=1}^n\mu_i}{s_n} \to N(0,1) \text{ in distribution}.$$ - -Lindeberg CLT from Wikipedia - -$$\text{Lindeberg condition: }\forall \epsilon >0, \quad \lim_{n \to \infty} \frac{\sum_{i=1}^{n} \mathrm{E} [(X_{i} - \mu_{i})^2 I_{\{|X_{i} - \mu_{i}| > \epsilon s_n\}} ] }{s_n^2} = 0.$$ - -Theorem D.19 of William Greene's Econometric Analysis (p112 of his appendix D file or Theorem 11 on p14 of this note ) - -$$\text{Feller-Levy condition: } \lim_{n\to\infty} \frac{\max_{i=1,\dots,n}\sigma_i^2}{s_n^2} = 0$$ - $$\text{name-unknown condition: }\lim_{n\to\infty} \frac{s_n^2}{n} < \infty$$ - -Note: - -I have changed the notation a bit. -In the book, instead of $\frac{\max_{i=1,\dots,n}\sigma_i^2}{s_n^2}$, it writes $ \frac{\max_{i=1,\dots,n}\sigma_i}{\sqrt{n} s_n}$. This is said to be a typo, and it should be $ \frac{\max_{i=1,\dots,n}\sigma_i}{ s_n}$. And $ \frac{\max_{i=1,\dots,n}\sigma_i}{ s_n}$ converges to finite, if and only if $ \frac{\max_{i=1,\dots,n}\sigma_i^2}{s_n^2}$ converges to finite. (Note: I have changed the notation a bit.) -The author said Feller's (1968) Introduction to Probability Theory and Its Applications was the original source for that result. I know it has two volumes, and after some search, I didn't find this version of CLT in either volume. I wonder if I am missing something? (As a side note, in Feller's 1968 Introduction to Probability Theory and Its Applications Volume 1, although the "name-unknown condition" doesn't appear for CLT, it appears for Law of Large Number (see page 254 formula (5.5)). Right below it is Lindeberg condition for Lindeberg CLT, which makes me wonder if the "name-unknown condition" in Greene's Theorem D.19 is a typo?) - - -Here is an example from Robert Isreal's earlier reply, which seems to indicate something might be wrong in the second one, i.e. the Greene's. -Let $X_i$ be independent with $$P(X_i = 2^i) = P(X_i = -2^i) = 2^{-2i-1}, P(X_i = 0) = 1 - 2^{-2i}.$$ Thus $\mu_i = 0$ and $\sigma_i^2 = 1$. - -Since $$P(S_n = 0) > P(X_i = 0 \text{ for all }i) > 1 - -\sum_{i=1}^\infty 2^{-2i} = 2/3,$$ it shows directly that $S_n/s_n$ -does not go to $N(0,1)$ in distribution. A normal distribution has -$P(Z=0)=0$. -The Lindeberg condition in Wikipedia's is not satisfied by the example (take $ϵ=1$ for -example). So it is not conclusive that $S_n/s_n$ should or should not converge to -$N(0,1)$ in distribution. -The two conditions in Greene's are satisfied, since $s_n^2 = n$. So -$S_n/s_n$ should converges to $N(0,1)$ in distribution. - -So the result from Greene's is not consistent with the results from other ways. I wonder what has gone wrong? Thanks! - -REPLY [2 votes]: A bit too long for a comment. This feels nothing short of a conspiracy. First off, I picked up a copy of Greene's Econometrics and the result there is $\lim_{n\rightarrow\infty}\max_i\frac{\sigma_i}{n\bar{\sigma}_n}=0$, where he defines $\bar{\sigma}_n^2=\frac{1}{n}(\sigma_1^2+\ldots+\sigma_n^2)$ in addition to $\bar{\sigma}^2=\lim_{n\rightarrow\infty}\bar{\sigma}_n^2$, which does not match your lecture notes, where I see the condition to be $\max_{i}\frac{\sigma_i^2}{n\bar{\sigma}^2_n}\rightarrow 0$. However, I still believe this is completely false, regardless of whose condition is correct, Greene or your lecture notes. -Let me call Lindberg Feller Condition '(L)', call $\lim_{n\rightarrow\infty} \sigma_k^2/s_n^2=0$ '(F)' and the statement "the central limit theorem holds" as '(CLT)'. We have that $(L)\Leftrightarrow (CLT)\ and\ (F) $. (F) here is the so called "Feller Condition." Normally, one has $(L)\Rightarrow (CLT)\ and \ (F)$ but the point is that (CLT) does NOT imply (L), unless we also have (F), or vice versa that (F) does not alone imply (L). -I do know that there's something called the Hajek Sirlak Theorem which says that if $X_i$ are i.i.d. with mean zero and variance $\sigma^2$ then $S_n=a_1X_1+\ldots+a_nX_n$ converges upon normalization to a normal distribution if the condition $\max_{i\leq n} a_i/\sqrt{\sum_i^n a_i^2}\rightarrow 0$ holds. The issue is I think you need the i.i.d. condition to hold.<|endoftext|> -TITLE: Which do you recommend for learning how to write proofs — How to Prove it by Velleman, or How to Solve it by Polya? -QUESTION [17 upvotes]: Which of these two books is suited for a student looking to learn how to write proofs? -I have a working knowledge of calculus and linear algebra, but I'm not good at writing proofs. My intention is to learn proofs in general, not necessarily for the calculus and linear algebra. - -How to Prove it by Daniel Velleman - -How to Solve It by George Polya - - -I ask because the latter is suggested on a highly voted question here, but the former has a more apt name. The reviews aren't helping. You can suggest other books. - -REPLY [3 votes]: Another book that you should check out is Gary Chartrand's Mathematical Proofs: A Transition to Advanced Mathematics. It has an instructor solutions manual.<|endoftext|> -TITLE: Criterion for convergence of the sequence of powers of a linear operator to $0$ -QUESTION [9 upvotes]: Let $T$ be a linear operator in a Banach space $\mathbf{B}$. Suppose that for every $x \in \mathbf{B}$ there exists some real numbers $c_x>0$ and $a_x<1$ such that $||T^nx|| \leq ca^n$, for all $n \in \mathbb{N}$. Prove that $||T^n|| \to 0$ as $n \to \infty$. -I could only deduce that $T^nx \to 0$ for all $x \in \mathbf{B}$, but I don't think it's enough. It's easy to show, using the uniform boundedness principle, that $||T^n||$ is bounded, but again that's not enough. - -REPLY [7 votes]: Apply uniform boundedness to the family of bounded operators $\{ nT^n \}$. Since $\| T^n x \|$ decays so rapidly, we still have $\| nT^n x\| \to 0$ for any fixed $x$, so $\|nT^n\|$ is bounded by some constant $c$, and so $\|T^n\| \le c/n$.<|endoftext|> -TITLE: Why does the "zig-zag comb" weakly deformation retract onto a point? -QUESTION [9 upvotes]: I am starting to read Hatcher's book on Algebraic Topology, and I am a little stuck with exercise 6 in Chapter 0. - -Let $Z$ be the zigzag subspace of $Y$ homeomorphic to $\mathbb{R}$ indicated by the heavier line in the picture: - -(see here for picture and definitions) -Show there is a deformation retraction in the weak sense of $Y$ onto $Z$, but no true deformation retraction. - -It's easy to show no true deformation retract is possible, but how does one show that a weak deformation retract is possible? Clearly we must deformation retract onto a disconnected subspace of of $Z$; however, it would appear that all open neighborhoods of every point are disconnected. - -REPLY [10 votes]: This is an elaboration of mixedmath's response. -For each point $a$ in Y, there is a natural path leading from $Y$ off to the right. For example, if $a$ is already on the zigzag $Z$, then $a$ just travels rightward along the zigzag. If $a$ is on one of the bristles, then first $a$ travels towards the zigzag, and then subsequently off to the right. -Let $f_a: [0, \infty) \to Y$ be this path, where the point travels at constant speed 1. -Consider the map $H: Y \times I \to Y$ defined by $H(a, t) = f_a(t)$, for $0 \leq t \leq 1$. I claim this is our desired homotopy. The only part that isn't clear is continuity; there are several cases to check here but none of them are hard.<|endoftext|> -TITLE: Difference between supremum and upper bounds and between infimum and lower bounds -QUESTION [7 upvotes]: I'm having some difficulties catching the difference between upper bound and supremum and, similarly, between lower bound and infimum. -Let's take a look at this set: -$A=\{x\in \mathbb Q | 0 -TITLE: Solving $(t^2+1)(y''-2y+1)=e^t$ with the initial conditions: $y(0)=y'(0)=1$ -QUESTION [10 upvotes]: Since it is important to me I would like to award a user who would kindly explain me what are my mistakes and what is the correct way to solve the whole problem with 500 points. -I'd really like your help with understanding how to solve this Cauchy problem: $(t^2+1)(y''-2y+1)=e^t$ with the initial conditions: $y(0)=y'(0)=1$. -I see a lot of methods and I am completely confused about what are the steps for solving this equation. First I wrote $$(y''-2y+1)=\frac{e^t}{(t^2+1)}.$$ -I read that I need to solve first the homogeneous equation $(y''-2y+1)=0$. Do I use Abel to reduce the order of the equation? I know that I need particular solution, so $y=0.5$ would do. -Now as far as I understand I need to use Wronskian determinant $$\begin{vmatrix} -0.5 &y \\ -0 &y' -\end{vmatrix}=c\cdot e^{\int^t_0-(2)/1 ds}=c\cdot e^{-2t}=0.5y'$$ -so $y'=2ce^{-2t}$, here I can use the data given me in the beginning so $c=0.5$ and $y'=e^{-2t}$ and $y=-0.5e^{-2t}+d$ and from the initial data again $d=1.5$ and $y=-0.5e^{-2t}+1.5$. -so now $y_h=-0.5e^{-2t}+1.5$ now I need to find $y_p=-0.5e^{-2t}u_1(t)+1.5u_2(t)$. -Then, I wrote the floowing :$$\begin{bmatrix} --0.5e^{-2t} &1.5 \\ - e^{-2t}& 0 -\end{bmatrix}\begin{pmatrix} -u_1'\\ -u_2' -\end{pmatrix}=\begin{pmatrix} -0\\ -\frac{e^t}{(t^2+1)} -\end{pmatrix}.$$ By using this system, I found $u_1$ and $u_2$ and the final solution is $y_=y_h+y_p$. -Please tell me- am I right? Is this basically the way to do that? Was I allowed to divide the original solution with $(t^2+1)$ or should I had to solve $(t^2+1)(y''-2y+1)$ as a homogeneous equation? -Thank you!! - -REPLY [9 votes]: Setting $x=(y,y')^T$, the problem becomes -$$ -\dot{x}=Ax+f(t)e_2, \ x(0)=(1,1)^T, -$$ -with $e_2=(0,1)^T$, $f(t)=e^t/(t^2+1)-1$ and -$A=\left[\begin{array}{cc}0&1\cr2&0\end{array}\right]$. -Thus -$$ -x(t)=e^{tA}\left[x(0)+\int_0^tf(s)e^{-sA}e_2ds\right]=e^{tA}x(0)+\int_0^tf(s)e^{(t-s)A}e_2, -$$ -and $y$ corresponds to the first component of $x$. -Since $A^2=2I$, it follow that -\begin{eqnarray} -e^{sA}&=&\sum_{k=0}^{\infty}\frac{s^{2k}}{(2k)!}A^{2k}+\sum_{k=0}^{\infty}\frac{s^{2k+1}}{(2k+1)!}A^{2k+1} -=\sum_{k=0}^{\infty}\frac{2^ks^{2k}}{(2k)!}I+\sum_{k=0}^{\infty}\frac{2^ks^{2k+1}}{(2k+1)!}A\cr -&=&\cosh(\sqrt{2}s)I+\frac{1}{\sqrt{2}}\sinh(\sqrt{2}s)A -=\left[\begin{array}{cc}\cosh(\sqrt{2}s)&\frac{\sinh(\sqrt{2}s)}{\sqrt{2}}\cr\sqrt{2}\sinh(\sqrt{2}s)&\cosh(\sqrt{2}s)\end{array}\right]. -\end{eqnarray} -So we have -\begin{eqnarray} -y(t)&=&\cosh(\sqrt{2}t)+\frac{1}{\sqrt{2}}\sinh(\sqrt{2}t)+\frac{1}{\sqrt{2}}\int_0^tf(s)\sinh(\sqrt{2}(t-s))ds\cr -&=&\cosh(\sqrt{2}t)+\frac{1}{\sqrt{2}}\sinh(\sqrt{2}t)+\frac{1}{\sqrt{2}}\int_0^t\left(\frac{e^s}{s^2+1}-1\right)\sinh(\sqrt{2}(t-s))ds\cr -&=&\cosh(\sqrt{2}t)+\frac{1}{\sqrt{2}}\sinh(\sqrt{2}t)+\frac{1}{2}(1-\cosh(\sqrt{2}t))+g(t)\cr -&=&\frac{1}{2}+\frac{1}{2}\cosh(\sqrt{2}t)+\frac{1}{\sqrt{2}}\sinh(\sqrt{2}t)+g(t), -\end{eqnarray} -where -$$ -g(t):=\frac{1}{\sqrt{2}}\int_0^t\frac{e^s}{s^2+1}\sinh(\sqrt{2}(t-s))ds -$$ -cannot be expressed with elementary functions.<|endoftext|> -TITLE: The image of an ideal under a homomorphism may not be an ideal -QUESTION [5 upvotes]: This is an elementary question about ideals. Consider a ring homomorphism -$$ -f: \mathbb{Z} \rightarrow \mathbb{Z}[x], -$$ -and consider the ideal $\left< 2\right>$ in $\mathbb{Z}$. When why is it that $f(\left< 2\right>)$ is not an ideal? -Some websites say that $f(\left< 2\right>)$ is not an ideal because it -does not contain nonconstant polynomials. That still doesn't make sense on why it is not an ideal. -Thank you all. - -REPLY [4 votes]: I just want to reflect this in words, as I think the ideas are quite simple. -Your title refers to the extension of an ideal. In fact, you have extended the ring, but have not extended the ideal. As others point out there is an ideal generated by the image of your original ideal under the inclusion, and this is the proper notion of the extension of your original ideal. -So the main reason that $\left< 2\right>$ is not an ideal in the extended ring $\mathbb Z[x]$ is that it is not big enough. If you think how much you can extend a ring, your ideal can't really hope to keep up unless it is extended too. -It is not always the case that the extension will lead to a proper ideal - think about the inclusion $f: \mathbb Z \to \mathbb Q $<|endoftext|> -TITLE: If any triangle has area at most 1 , points can be covered by a rectangle of area 2. -QUESTION [15 upvotes]: I am working on this problem for some time, and I am not able to finish the argument: -There is a finite number of points in the plane, such that every triangle has area at most 1. Prove that the points can be covered by a rectangle of area 2. -My progress so far: -Consider the two points furthest apart, A and B. Draw lines through A and B, perpendicular to AB. Then every point has to be between the two lines, since AB has maximal length among our pair of points. Now take the points C and D, furthest apart from the line AB, on each of the two halfplanes determined by AB. Every other point in the plane has to be between the two lies through C and D, parallel to AB . -The rectangle determined by the four lines gives us an area of at most 4, but I feel like this argument can be improved somehow to give us the required area of 2. -Thank you! (Edit: thank you for the advice on posting problems here ) - -REPLY [5 votes]: I believe that the theorem you stated is false, but there's a simple variation of your statement that you can actually prove! - revised theorem: Given a finite number, n, of points in a plane such that any three points form a triangle of area 1 or less, then there is a triangle of area 4 in the plane that contains all $n$ given points. -A quick outline of why this is true: --We have finitely many points in the plane, and hence finitely many ways to form a triangle. So there is a least upper bound $\alpha$ on the areas of the triangles, and at least one triangle has area $\alpha$. Start by drawing this triangle, which vertices we will call A,B and C, and which side we will call (AB), (BC), and (CA). --From triangle ABC, we can draw a bigger triangle and in so doing complete the proof. -Draw the line (AB)' which is parallel to (AB) and passes through vertex C. -Similarly draw (BC)' which is parallel to (BC) and passes through vertex A, -and (CA)' which is parallel to (CA) and passes through vertex B. -The intersection of (AB)',(BC)', and (CA)', is a triangle of area $4*\alpha -\leq 4 $, which we'll call ABC'. --The important point is that, by construction, ABC' must contain all n of our original points. Indeed, pick a point D distinct from A, B, C. We could draw the triangles ABD,ACD, or BCD, each of which has area less than or equal to $\alpha$. Keeping in mind that $Area=\frac{bh}{2}$, we see that area(ABD)$\leq$area(ABC) implies that the shortest distance between D and (AB) must be less than or equal to the shortest distance between C and (AB). In other words, D must lie between (AB) and (AB)' or between (AB) and the reflection of (AB)' over (AB). Likewise, the shortest distances between D and (CA) and between D and (BC) must be respectively less than or equal to the shortest distances between B and (CA) and A and (BC). -So D must lie also between (CA) and (CA)' or (CA) and the reflection of (CA)' over (CA), and D must lie between (BC) and (BC)' or (BC) the reflection of (BC)' over (BC). -This restriction on where the point D can lie can be shown quite easily to be equivalent to the restriction that D must lie in the triangle ABC'. So there is a triangle of area $4*\alpha$, which is less than or equal to 4 by assumption, which contains every point of our plane. --Informally, this proves the revised version of the theorem you gave. Sorry it's a couple years late, hope it's still interesting or helpful for somebody.<|endoftext|> -TITLE: set in $\mathbb{R}$ which is not a Borel-set -QUESTION [7 upvotes]: Possible Duplicate: -Lebesgue measurable but not Borel measurable -Constructing a subset not in $\mathcal{B}(\mathbb{R})$ explicitly - -if i start from the topology of $\mathbb{R}$, i.e. all open sets, and then build the closure under countable union and complement i get the so called Borel-$\sigma$-Algebra, the smallest set which contains all open sets (i.e the topology) of $\mathbb{R}$ and is a $\sigma$-algebra? do you know any sets on $\mathbb{R}$ which are not cointained in this Borel-$\sigma$-algebra? - -REPLY [19 votes]: There are $2^{2^{\aleph_0}}$ subset of $\mathbb{R}$. There are $2^{\aleph_0}$ Borel subsets. Hence there certainly exists subsets of $\mathbb{R}$ that are not Borel. -However, there are even some very nicely defined sets that are not Borel. For example, there is a Lebesgue Measureable set that is not Borel. The cantor set has measure zero and is uncountable. Hence every subset of the Cantor set is Lebesgue Measureable and by a cardinality argument, there exists one which is not Borel. Analytic sets can be defined to be continuous images of the real line. There even exists analytic sets which are not Borel. - -REPLY [11 votes]: Endow $[0,1]$ with mod 1 real arithmetic. Let $K$ denote the set of all rationals in $[0,1]$; this is a subgroup of $[0,1]$. Define $x\sim y$ if $x - y\in K$. (Note: mod 1 arithmetic). This is an equivalence relation. -Denote by $E$ a system of direct representatives under this equivalence relation; to wit, $E$ represents a choice of one element from each $\sim$-equivalence class. -We have the countable disjoint union -$$[0,1] = \bigcup_{k\in K} (E + k).$$ -Were this to be Lebesgue measurable we would have -$$1 = |[0,1]| = \sum_{k\in K} |E + k| $$ -by $\sigma$-additivity of Lebesgue measure. -However, by translation-invariance, we have $|E + k| = |E|$ for $k \in K$. -This is not possible. -What has gone wrong? We operated under the premise that $E$ is Lebesgue measurable. By contradiction, it is not. Since Lebesgue measure is the completion of Borel measure, it is not a Borel set either.<|endoftext|> -TITLE: Possibilities of an action of $S^1$ on a disk. -QUESTION [6 upvotes]: I'm dealing with actions of the circle over differentiable manifolds. In the book I'm reading, they use the fact that an action of $S^1$ over a disk has to be equivalent (there has to exist an equivariant diffeomorphism) to a rotation. Can someone give me a hint to prove this? Or maybe a reference where I can consult this result? It seems to be a "well known fact" but I haven't been able to find a place where they prove it. - -REPLY [5 votes]: Your question is a little imprecise but I'll take it to be the assertion that a smooth action of $S^1$ on $D^n$, i.e. a smooth homomorphism $S^1 \to Diff(D^n)$ is conjugate via a diffeomorphism of $D^n$ to a homomorphism $S^1 \to SO_n$. -First off, the generator of the motion is a vector field on $D^n$ which is tangent to $\partial D^n = S^{n-1}$. Since its tangent on the boundary, you can perturb the vector field near the boundary to be outward-pointing. Poincare-Hopf kicks in and tells you this vector field needs to have a zero on the interior, so your original vector field has a zero on the interior. -So in the orbit decomposition of $D^n$ you have a non-empty fixed point set. So all we need to do is show the fixed point set is isotopic to a linear subspace -- once you have that you know that the disc $D^n$ is just an $S^1$-equivariant tubular neighbourhood of that fixed point set, and so its characterized by its behaviour near the fixed point set, which is linear. -Hmm, this part seems likely to not be true. Apparently there are actions of circles on discs with precisely two fixed points in the interior. Ref I don't have access to the paper from home so I haven't looked at anything other than the first page, but this appears to be a non-linear action. Perhaps in the reference you're referring to they're content with a local result rather than a global result? You might want to check the wording carefully. Locally it's certainly true by the above argument.<|endoftext|> -TITLE: Quantization of angular momentum: is Dirac's proof wrong? -QUESTION [11 upvotes]: I'm trying to understand the physicist's proof of the theorem on the spectral structure of angular momentum operators (I'm being told that this proof is due to Dirac). I will refer to Ballentine's book, chap. 3, section 1. -Angular momentum operators $J_x, J_y, J_z$, whatever definition we adopt, are bound to satisfy the commutation relations -$$\tag{AMCR} [J_x, J_y]=iJ_z,\ [J_y, J_z]=iJ_x,\ [J_z, J_x]=iJ_y,$$ -which imply -$$[J^2, J_z]=0.$$ -Following Ballentine's book, we claim that from those relations alone we can infer: - -$J^2$ and $J_z$ have pure point spectrum; -the possibile eigenvalues for $J^2$ are of the form $j(j+1)$ for a half-integer $j$; -for every eigenvalue $j(j+1)$ of $J^2$, $J_z$ has the corresponding eigenvalues $m=-j, -j+1 \ldots j$. Those eigenvalues are simultaneous, meaning that for each one of them there exists a ket $\lvert j, m\rangle$ such that $J^2\lvert j, m\rangle=j(j+1)\lvert j, m\rangle$ and $J_z\lvert j, m\rangle=m\lvert j, m\rangle$. $J_z$ hasn't got any more simultaneous eigenvalues with $J^2$: in particular the spectrum of both operators is discrete. - - -I'll give an outline of the proof, stressing the troublesome parts in italic. If necessary, you can follow the complete proof at the link above. -Let $\lvert \beta, m\rangle$ be a simultaneous eigenket of $J^2, J_z$ respectively. -Step 1) We prove that $\beta \ge m^2$. -Step 2) Let $j$ be the maximum allowable value of $m$ for fixed $\beta$. Then, applying a ladder operator $J_+$ to $\lvert \beta, j\rangle$, we see that $J_+\lvert \beta, j\rangle=0$. (*) -Step 3) Turns out that $\beta=j(j+1)$ and that $-j=\min(m)$. -Step 4) Because of the existence of ladder operators, eigenvalues for $J_z$ are integer spaced (!). In particular, the difference between $j$ and $-j$ must be integer and so $j$ must be half integer. $\square$ -Step 2 and Step 4 are troublesome. Step 4 seems to assume implicitly that if $\lvert j, m'\rangle$ is an eigenket then it can be reached by successive applications of ladder operators. For example, if $m'>m$ then it is assumed implicitly that $\lvert j, m'\rangle=J_+J_+\ldots J_+\lvert j,m\rangle$. I don't understand why this should be the case. -I would gladly accept any hint or reference on the question. Thank you. - -(*) EDIT There used to be a question here, which is now solved. See Peter Taylor's comment and AlbertH answer below. Question was: "To do so the author uses implicitly the fact that $j$ is an eigenvalue of $J^2$ and so that an eigenket $\lvert \beta, j\rangle$ exists. This doesn't look obvious to me". end of edit - -REPLY [5 votes]: I don't know the mentioned book. -However, about Step 2, $J_x, J_y, J_z$ are generators of an unitary representation of $SO(3)$ that is a compact Lie group. By Peter-Weyl theorem all irreducible unitary representation of a compact Lie group are finite-dimentional. So $J_z, J^2$ are made up of self-adjoint finite-dimensional operators and as such they admits at least an eigenvector. -Edit -Let $J_x, J_y, J_z$ be an irreducible representation of the above commutation rules. -By Schur's lemma, $[J^2, J_i] = 0$ for each $i\in\{x, y, z\}$, implies -$$ -J^2 = \beta \mathbb I -$$ -$J_z$ is a finite-dimensional self adjoint operator, hence diagonalizable. -Let's take an eigenvector, $\lvert m \rangle$, of $J_z$, with repeated applications of $J_-$ and $J_+$ we can construct a sequence of eigenvectors -$$ -s := \{\lvert m - h \rangle, \dotsc, \lvert m - 1\rangle, \lvert m \rangle, \lvert m + 1 \rangle, \dotsc,\lvert m + k \rangle \} -$$ -Above sequence is finite because its elements are independent vectors of the finite-dimensional representation space $V$. -Let's call $S$ the subspace spanned by vectors in $s$. The equalities -$$ -J_x = \frac 1 2 (J_+ + J_-)\\ -J_y = \frac 1 {2i} (J_+ - J_-) -$$ -show that $S$ is an invariant space for $J_x, J_y, J_z$. Since the representation is irreducible $S$ must coincide with $V$ and no other (indipendent) eigenvectors of $J_z$ can exist.<|endoftext|> -TITLE: Unitisation of $C^{*}$-algebras via double centralizers -QUESTION [9 upvotes]: In most of the books I read about $C^{*}$-algebras, the author usually embeds the algebra, say, $A$, as an ideal of $B(A)$, the algebra of bounded linear operators on $A$, by identifying $a$ and $M_a$, the left multiplication of $a$. -However, in Murphy's $C^{*}$-algebras and operator theory, $A$ is embedded as an ideal of the space of 'double centralizers'. See p39 of his book. -I do not quite understand why we need this complicated construction since the effect is almost the same as the usual embedding. The author remarked that this construction is useful in certain approaches to K-theory, which further confuses me. -Can somebody give a hint? -Thanks! - -REPLY [8 votes]: There is a minimal way to imbed a nonunital $C^*$-algebra $A$ into a unital $C^*$-algebra. As a $*$-algebra this is $A\oplus \mathbb C$ with componentwise addition, with multiplication $(a,s)(b,t)=(ab+ta+sb,st)$, and with involution $(a,s)^*=(a^*,\overline s)$. But to give this algebra a $C^*$ norm, one method is to identify it with $\{L_a:a\in A\}+\mathbb C\mathrm{id}_A\subset B(A)$, where $L_a:A\to A$ is defined by $L_ab=ab$. One can then check that the operator norm of this algebra as a subspace of $B(A)$ is a $C^*$ norm. -There is also a maximal way to imbed a nonunital $C^*$-algebra $A$ into a unital $C^*$-algebra as an ideal in an "essential" way. The essentialness is captured by stipulating that every nonzero ideal in the unitization intersects $A$ nontrivially. This is equivalent to the condition that $bA=\{0\}$ implies $b=0$. As mland mentioned, this maximal unitization is the multiplier algebra of $A$, $M(A)$. The double centralizer approach is one particular concrete description, but $M(A)$ has other decriptions and is characterized by a universal property: For every imbedding of $A$ as an essential ideal in a $C^*$-algebra $B$, there is a unique $*$-homomorphism from $B$ to $M(A)$ that is the identity on $A$. -t.b. has already mentioned that in the commutative case this runs parallel to one-point versus Stone–Čech compactification. -Here is another example. The algebra $K(H)$ of compact operators on an infinite dimensional Hilbert space $H$ has minimal unitization (isomorphic to) $K(H)+\mathbb CI_H$, and multiplier algebra (isomorphic to) $B(H)$. -One reason we may want to go all the way to $M(A)$ is to better understand automorphisms of $A$. Conjugation by a unitary element of $M(A)$ is an automorphism of $A$. In the case of $K(H)\subset B(H)\cong M(K(H))$, every automorphism is of this form, and you couldn't get most of these automorphisms by only conjugating by unitaries in the minimal unitization $K(H)+\mathbb C I_H$. -The approach mentioned by mland of identifying $M(A)$ with the algebra of adjointable operators on $A$ can be found in Lance's Hilbert C*-modules or in Raeburn and Williams's Morita equivalence and continuous trace C*-algebras with a lot more useful introductory information in each. I agree with mland that for the basics of K-theory you do not need to get into multiplier algebras, but you can learn more about their importance in K-theory from Blackadar's K-theory for operator algebras. Chapter VI is described as a collection of "all the results needed for Ext-theory and Kasparov theory," and it starts with a review of multiplier algebras and examples.<|endoftext|> -TITLE: Differentiation under the integral sign. -QUESTION [6 upvotes]: I'm working through an integral suggested for practice at the end of the Wikipedia article on differentiation under the integral sign, and I'm stuck. -I am attempting to evaluate this integral: -$$\int_0^{\pi/2} \frac{x}{\tan x} \ dx.$$ -The article suggests the following parameterization: -$$F(a)=\int_0^{\pi/2} \frac{\tan^{-1}(a\tan (x))}{\tan x} \ dx.$$ -Differentiating with respect to $a$, we get -$$F'(a)=\int_0^{\pi/2} \frac{1}{1+a^2\tan^2 x} \ dx.$$ -I can't find a way to evaluate this, and neither can Wolfram Alpha. The special values $a=0,1$ are easy, but I fail to see how they help. -How can I finish evaluating this integral? -Edit: I think it's just substitution, maybe. I'll update the post accordingly soon. -Edit 2: -Indeed, the substitution $u=\tan x$ and identity $\sec^2 = 1 + \tan^2$ transform the above integral into -$$F'(a) = \int_0^{\pi/2} \frac{1}{(1+u^2)(1+a^2u^2)}.$$ -This can be solved with partial fractions. - -REPLY [3 votes]: You have that -$$\frac{1}{{\left( {1 + {u^2}} \right)\left( {1 + {a^2}{u^2}} \right)}} = \left( {\frac{A}{{1 + {u^2}}} + \frac{B}{{1 + {a^2}{u^2}}}} \right)$$ -Thus you want (after cross mult.) -$$1 = A + A{a^2}{u^2} + B + B{u^2}$$ -This is -$$\eqalign{ - & A + B = 1 \cr - & A{a^2} + B = 0 \cr} $$ -Which gives -$$A = \frac{1}{{1 - {a^2}}}$$ -and in turn -$$B = 1 - A = \frac{{{a^2}}}{{{a^2} - 1}}$$ -which means -$$\frac{1}{{\left( {1 + {u^2}} \right)\left( {1 + {a^2}{u^2}} \right)}} = \frac{1}{{{a^2} - 1}}\left( {\frac{{{a^2}}}{{1 + {a^2}{u^2}}} - \frac{1}{{1 + {u^2}}}} \right)$$ -Can you move on?<|endoftext|> -TITLE: Find a single-valued analytic branch of $\sqrt{z^2-1}$ in $\mathbb{C} \backslash [-1,1]$. -QUESTION [5 upvotes]: I have the following question: - -Show there is a single-valued analytic branch $f(z)$ for $\sqrt{z^2-1}$ in $\mathbb{C}\backslash [-1,1]$ such that $f(x) < 0$ for $x>1$. Here $[-1,1]$ denotes a closed interval in $\mathbb{R}$. - -The following solution mimics a derivation from Moore and Hadlock's text. -Note that $z^2-1 = (z-1)(z+1)$. Consider the following four functions: -\begin{align*} -w_1(z) &= \sqrt{|z-1|} e^{(i\operatorname{arg_1}(z-1))/2} &\text{ where } \operatorname{arg_1}(z-1) \in [0,2\pi) \\ -w_2(z) &= \sqrt{|z+1|} e^{(i\operatorname{arg_1}(z+1))/2} &\text{ where } \operatorname{arg_1}(z+1) \in [0, 2\pi) \\ -w_3(z) &= \sqrt{|z-1|} e^{(i\operatorname{arg_2}(z-1))/2} &\text{ where } \operatorname{arg_2}(z-1) \in [-\pi,\pi) \\ -w_4(z) &= \sqrt{|z+1|} e^{(i\operatorname{arg_2}(z+1))/2} &\text{ where } \operatorname{arg_2}(z+1) \in [-\pi, \pi) -\end{align*} -Note that -$w_1$ is analytic on the set $\{ z \in \mathbb{C}: \operatorname{arg}(z-1) \neq 0\} = \mathbb{C} \backslash [1, \infty)$; -$w_2$ is analytic on the set $\{ z \in \mathbb{C}: \operatorname{arg}(z+1) \neq 0\} = \mathbb{C} \backslash [-1, \infty)$; -$w_3$ is analytic on the set $\{ z \in \mathbb{C}: \operatorname{arg}(z-1) \neq \pi\} = \mathbb{C} \backslash (-\infty, 1]$; and -$w_4$ is analytic on the set $\{ z \in \mathbb{C}: \operatorname{arg}(z+1) \neq \pi \} = \mathbb{C} \backslash (-\infty, -1]$. -Therefore $w_1w_2$ is analytic on the set $\mathbb{C} \backslash [-1, \infty)$ and $w_3w_4$ is analytic on the set $\mathbb{C} \backslash (-\infty, 1]$. -Furthermore, on the set $\mathbb{C} \backslash (-\infty, 1]$, we have $w_1w_2 = w_3w_4$, as we see below: -If $x > 1$, then $\operatorname{arg_1}(x \pm 1) = \operatorname{arg_2}(x \pm 1) = 0$, and thus -$$w_1w_2(x)= w_3w_4(x) = \sqrt{x^2-1} \cdot e^{i \cdot 0} = \sqrt{x^2-1}. $$ -If $\operatorname{Im}(z) > 0$, then $w_1w_2(z) = w_3w_4(z)$ since $\operatorname{arg}(z-1), \operatorname{arg}(z+1) \in (0, \pi)$. -If $\operatorname{Im}(z) < 0$, then $\operatorname{arg_1}(z\pm 1) = 2\pi + \operatorname{arg_2}(z\pm 1)$. Then -$$w_1w_2(z) = \sqrt{|z^2-1|} e^{i(\operatorname{arg_1}(z-1) +\operatorname{arg_1}(z-1))/2} = \sqrt{|z^2-1|} e^{i(\operatorname{arg_2}(z-1) +\operatorname{arg_2}(z-1))/2 + 2\pi i } $$ -$$ = \sqrt{|z^2-1|} e^{i(\operatorname{arg_2}(z-1) +\operatorname{arg_2}(z-1))/2} = w_3w_4(z).$$ -This implies that $w_1w_2$ is analytic not only on $\mathbb{C} \backslash [-1, \infty)$, but on the larger set $\mathbb{C} \backslash [-1, 1]$. This is a single-valued analytic branch of $\sqrt{z^2-1}$. However, for $x >1$, we have $w_1w_2(x) >0$. -To address this, multiply $w_1w_2$ by $e^{i\pi}$. The resulting function is a single-valued analytic branch with the desired property. -I understand the computations here, but I'm trying to grasp the motivation behind the definitions, and see how it might generalize. Is there a way to attack general problems of this type, or a simple explanation of why these particular $w$ functions were chosen? - -REPLY [3 votes]: EDIT: -A more straightforward solution is to note that $z^2 - 1$ is in $(-\infty,0]$ exactly when $z$ is in the interval $[-1,1]$ or the imaginary axis. If we use the principal branch of $\sqrt{}$, $\sqrt{z^2-1}$ will be analytic everywhere else. This would be positive on $[1,\infty)$, so instead we take $-\sqrt{z^2-1}$. To avoid having a branch cut on the imaginary axis, we switch to $+\sqrt{z^2-1}$ in the left half plane. Thus -$$f(z) = \cases{-\sqrt{z^2-1} & for $\text{Re}(z) \ge 0$\cr - +\sqrt{z^2-1} & for $\text{Re}(z) < 0$\cr}$$<|endoftext|> -TITLE: Detailed diagram with mathematical fields of study -QUESTION [17 upvotes]: Some time ago, I was searching for a detailed diagram with mathematical fields of study the nearest one I could find is in this file, second page. -I want something that shows information like: "Geometry leads to I topic, Geometry and Algebra leads do J topic and so on. -Can you help me? - -REPLY [5 votes]: http://www.math-atlas.org/ love this page! It appears that is has not been updated in a long time. To date i have not found its equal<|endoftext|> -TITLE: Intersection between a cylinder and an axis-aligned bounding box -QUESTION [6 upvotes]: Given a 3D axis-aligned bounding box (represented as its minimum point and maximum point) and a 3D cylinder of infinite length, what's the best way to test for intersection? - -REPLY [5 votes]: Suppose you are given a box defined by -$$x_{min}\leq x\leq x_{max},y_{min}\leq y\leq y_{max},z_{min}\leq z\leq z_{max}$$ -and a cylinder of radius $r$ along the vector $\vec v=(v_1,v_2,v_3)^T$ with its center passing through $\vec p=(p_1,p_2,p_3)^T$. First construct the projection matrix $P$ which maps vectors onto their projection in the plane perpendicular to $\vec v$. I assume $v_1\neq 0$, otherwise the matrix will have to be constructed slightly differently. We have -$$P=\begin{pmatrix} -0 & 1 & 0\\ -0 & 0 & 1\\ -\end{pmatrix}\begin{pmatrix} -v_1 & 0 & 0\\ -v_2 & 1 & 0\\ -v_3 & 0 & 1\\ -\end{pmatrix}^{-1}=\begin{pmatrix} -\frac{-v_2}{v_1} & 1 & 0\\ -\frac{-v_3}{v_1} & 0 & 1\\ -\end{pmatrix}$$ -and thus a vector $\vec x$ lies in the cylinder iff $\|P(\vec x-\vec p)\|\leq r$. Note that the projection of the box is a convex set and is equal to the convex hull of the images of its vertices, and that either the image of center of the cylinder is contained in the image of the box or the boundary of the image of the box intersects the image of the cylinder. -In the first case, we have that the line $\vec p + t\vec v$ intersects one of the three planes $x=x_{min},y=y_{min},z=z_{min}$ within the boundary of the box. The points of intersection are given by $$\begin{pmatrix} -x_{min}\\ -p_2+\frac{x_{min}-p_1}{v_1}v_2\\ -p_3+\frac{x_{min}-p_1}{v_1}v_3\\ -\end{pmatrix}, \begin{pmatrix} -p_1+\frac{y_{min}-p_2}{v_2}v_1\\ -y_{min}\\ -p_3+\frac{y_{min}-p_2}{v_2}v_3\\ -\end{pmatrix}, \begin{pmatrix} -p_1+\frac{z_{min}-p_3}{v_3}v_1\\ -p_2+\frac{z_{min}-p_3}{v_3}v_2\\ -z_{min}\\ -\end{pmatrix}$$ -and are nonexistant if $v_1,v_2$ or $v_3$ respectively are $0$. It is easy to check whether these are contained in the box. If any of them are, the intersection is nonempty. -In the second case, let $\vec x_1+t\vec y_1,\ldots,\vec x_{12}+t\vec y_{12}$ be the lines which form the edges of the box (and are exactly the edges when $0\leq t\leq 1$), which are easy to determine from the vertices. We want to know whether there is a solution to $\|P(\vec x_i+t\vec y_i-\vec p)\|\leq r$ for any $i$ with $0\leq t\leq 1$. Observe that $\|P(\vec x_i+t\vec y_i-\vec p)\|^2-r^2$ is a quadratic equation in $t$, so we can differentiate it to get a linear equation and solve for $0$ to get a value $t_i$. We then check for $i=1,\ldots,12$ whether $\|P(\vec x_i+t_i\vec y_i-\vec p)\|\leq r$ and $0\leq t_i\leq 1$. If this holds for some $i$, then the intersection is nonempty. Otherwise we test the endpoints $t=0$ and $t=1$ for each $i$ to see whether any of them satisfy $\|P(\vec x_i+t\vec y_i-\vec p)\|\leq r$. If so, the intersection is nonempty. -Finally, if all the above tests have failed then the intersection is empty. -In general, problems like this are studied in the field of Real Semialgebraic Geometry. More complicated questions of this form are approached by a method known as Cylindrical Algebraic Decomposition or CAD, which is similar to the method I employed but would have reduced the number of needed calculations by determining the six lines which bound the image rather than using all twelve lines that bound the box, but is more difficult to grasp.<|endoftext|> -TITLE: Is '$10$' a magical number or I am missing something? -QUESTION [281 upvotes]: It's a hilarious witty joke that points out how every base is '$10$' in its base. Like, -\begin{align} - 2 &= 10\ \text{(base 2)} \\ - 8 &= 10\ \text{(base 8)} -\end{align} -My question is if whoever invented the decimal system had chosen $9$ numbers or $11$, or whatever, would this still be applicable? I am confused - Is $10$ a special number which we had chosen several centuries ago or am I missing a point? - -REPLY [5 votes]: I've always assumed it was the number of fingers on the human hand that originated the decimal system. I sometimes make people feel better about their age by saying something like, "Hey, if humans has 6 fingers on each hand you'd still be in your thirties."<|endoftext|> -TITLE: Extending isometries between compact subspaces of Cantor space -QUESTION [6 upvotes]: Let $\omega$ be the set of natural numbers. $2^\omega$ is the Cantor space. -Suppose $K$, $L \subset 2^\omega$ are compact, and there is an isometry $f: K \to L$. Then how could one extend $f$ to an isometry from $2^\omega$ to $2^\omega$? Here we are considering $2^\omega$ with the minimum difference metric, which gives the standard product topology; i.e. -$ d(x,y) = 2^{-\min \{ n : x(n) \neq y(n) \}}. $ - -REPLY [2 votes]: When $u$ is a finite binary word, $u\cdot 2^\omega$ means the interval of $2^\omega$ formed by infinite words with prefix $u$ ($\cdot$ denotes word concatenation). -Let $A_{fin}$ be the set of finite prefixes of $A\subseteq 2^\omega$. -We say that $F: A_{fin}\to B_{fin}$ is an isometry if: - -$|F(x)|=|x|$; -$x\preceq y$ implies $F(x)\preceq F(y)$ where $\preceq$ is the prefix relation; -$F$ is injective, or equivalently $F(x0)\ne F(x1)$ when both are in the domain of $F$. - -Side-note: When $A_{fin}=B_{fin}$ (or more generally, when the multiset of word lengths of $A_{fin}$ and $B_{fin}$ coincide), the last condition can also be replaced by "$F$ is bijective". The first condition then becomes redundant. - -Lemma: The relation - $$\forall u\in A_{fin}, \quad f(u\cdot 2^\omega)\subseteq F(u)\cdot 2^\omega$$ - is a bijection between isometries $f: A\to B$ and isometries $F: A_{fin}\to B_{fin}$. - -Proof: - -When $uv\in A$, $F(u)$ must be the first $|u|=n$ symbols of $f(uv)$. This constructs $F$ in a well-defined way because $d(f(uv),f(uw))=d(uv,uw)\le 2^{-n}$. When $x\ne y$, if $|x|\ne|y|$ then of course $F(x)\ne F(y)$; otherwise $d(f(x),f(y))=d(x,y)>2^{-|x|}$ therefore $F(x)\ne F(y)$. -Conversely, $f(x)=\lim_{u\preceq x} F(u)$ is well-defined by monotony of $F$, and if $d(x,y)=2^{-n}$ we can let $u0,u1$ be the prefixes of length $n+1$ of $x,y$: $2^{-n}=d(u0,u1)=d(F(u0),F(u1))=d(f(x),f(y))$. - -Side-note: this also gives the following statement, which was not a priori obvious: - -Corollary: An isometry $f: A\to A$ on $2^\omega$ is a bijection. - - - -Corollary: Isometries $A_{fin}\to B_{fin}$ as defined above are precisely the isometries $A_{fin}\to B_{fin}$ under the minimum difference metric and under word length, where the metric is extended to finite words by $d(x,u)=\max d(x,u0^\omega),d(x,u1^\omega)$. This justifies the use of the word "isometry". - - - -Theorem: Let $A,B$ be arbitrary subsets of $2^\omega$. An isometry $f: A\to B$ can be extended to an isometry $g: 2^\omega\to 2^\omega$. As could also be shown by topological arguments, the extension is unique if and only if $A_{fin}=(2^\omega)_{fin}$, that is iff $A$ is dense in $2^\omega$. - -Proof: We just have to extend an isometry $F:A_{fin}\to B_{fin}$. When $u$ is non-empty, let $\sigma_F(u)$ be the last symbol of $F(u)$. We have that $\sigma:\{0,1\}^+\to\{0,1\}$ is the last symbol of an isometry $(2^\omega)_{fin}\to (2^\omega)_{fin}$ if and only if $\sigma(x1)=\neg \sigma(x0)$. So we can define $$\sigma(x0)=\begin{cases} -\sigma_F(x0) & x0\in A_{fin}\\ -\neg\sigma_F(x1) & x1\in A_{fin}\\ -0 & \text{else} -\end{cases}$$ -so that the relations $\sigma(x1)=\neg \sigma(x0)$ and $G(ua)=G(u)\sigma(ua)$ uniquely define an isometry $G: (2^\omega)_{fin}\to (2^\omega)_{fin}$. $G$ extends $F$, so the isometry $g: 2^\omega\to 2^\omega$ extends $f$.<|endoftext|> -TITLE: What is Kosambi-Cartan-Chern (KCC) theory? -QUESTION [7 upvotes]: I'm reading a paper that's basically about stability analysis of the Lane-Emden differential equation. The authors make use of "Kosambi-Cartan-Chern (KCC) theory". I've been trying to find out what this "theory" is about and haven't really gotten anywhere except for a another paper (PDF) that refers to this KCC theory as an established thing. -Given that the papers discuss the theory as an established thing, I expected there to be a Wikipedia or MathWorld article or something. But no luck. References are given to a series of papers by Kosambi, Cartan and Chern from the 1930's but I haven't hit any textbooks that describe it. For interest, the relevant references appear to be - -Cartan, E. Observations sur le mémoir précédent, Math. Zeitschrift 37 (1933), -619-622. -Chern, S.S., Sur la géometrie d'un système d'equations differentialles du second -ordre, Bull. Sci. Math. 63 (1939), 206-212 -Kosambi, D.D. Parallelism and path-space, Math. Zeitschrift 37 (1933), 608-618 - -REPLY [7 votes]: The name 'KCC-theory' was introduced in a book of Antonelli, Ingarden and Matsumoto entitled The Theory of Sprays and Finsler Spaces with Applications in Physics -and Biology published in 1993. It is mostly used in the physics and biology fields. -The KCC theory concerns the following. Consider a second order differential equation $(d^2x_i/dt^2) + g_i(x,x',t) = 0$, for $i = 1, ... , n$, where $x = (x_1, ..., x_n)$, t is the time parameter, $x'$ denotes $((dx_1/dt), ..., (dx_n/dt))$, and $g_i$'s are smooth functions of $(x, x', t)$ defined on a domain in the $(2n + 1)$-dimensional Euclidean space. The aim is to understand what geometric properties of the system of integral curves - the paths associated with the system of differential equations - remain invariant under nonsingular transformation of the coordinates involved. The theory describes certain invariants, which are specific tensors depending on the $g_i$'s, which characterize the geometry of the system, in the sense that two such systems can be locally transformed into each other if and only if the corresponding invariants are equivalent tensors. In particular, a given system as above can be transformed into one for which the $g_i$'s are identically 0, so that the integral curves are all straight lines, if and only if the associated tensor invariants are all zero. -The problem may be viewed also as that of realising the integral curves of a second order differential equation as geodesics for an associated linear connection on the tangent bundle. Kosambi introduced a method using calculus of variations, which involves realizing the paths as extremals of a variational principle; this is related to finding a 'metric' for the path space. -By associating a non-linear connection and a Berwald type connection to the dynamical system, five geometrical invariants are obtained, with the second invariant giving the Jacobi stability of the system. -These review articles may be of help: - -http://emis.icm.edu.pl/journals/AMAPN/vol24_1/amapn24_16.pdf -http://www.desy.de/~mnrw2/publications/2009_pre79_046606.pdf<|endoftext|> -TITLE: Formula of this map (Escher's work) -QUESTION [5 upvotes]: As you know, the picture below is one of M.C. Escher's works. - -I have thought a lot about what would be the rule of the function drawn on the surface of sphere? Is it a $\mathbb R\rightarrow\mathbb R^3$ or $\mathbb R^3\rightarrow\mathbb R^3$ function? Honestly, the idea that he drew it is new now and I don’t think his works be outdated ever. Thanks for sharing your knowledge about this great picture with me. :) - -REPLY [5 votes]: It seems to me that the bands are supposed to twist indefinitely when they approach the poles (as observed by Erick Wong). That effect is achieved by using loxodromes as boundary lines. The loxodromes are roughly the spherical equivalents of the logarithmic spirals in that a loxodrome makes a constant angle with all the latitudes that it crosses - very much like the logarithmic spirals that have constant angles between their tangents and the radii. -The attaced image is generated by Mathematica. I used $m=1/2$ (see the link for the meaning of this parameter), so e.g. one of the bands was generated by - -ParametricPlot3D[{Cos[t+u]/Cosh[t/2],Sin[t+u]/Cosh[t/2],Tanh[t/2]},{t,-8,8},{ - u,0,Pi/4},PlotPoints->{81,5}] - -and the other parts came out with the same formula, but the range of the variable $u$ shifted by an integer multiple of $\pi/2$. The range of $t$ is should be symmetric, but the endpoints (here $\pm8$) are, again, a largely arbitrary choice of mine. I did a linear change of parameters to that on the Wikipage in order to make sure that all the loxodromes cover the same interval of latitudes, and also to make one half of the boundaries of the tiles to have constant latitude. -Here's the image. To get a better match with Escher's painting it might be necessary to fine-tune the value of $m$ further. - -To see the "endless twisting" here is a close-up of the polar region. The image is necessarily quite flat now. The formula is the same, but the range of the parameter $t$ is now $4\le t\le 16$. - -Edit: Here's an animated version<|endoftext|> -TITLE: Formula to this pattern? $1$, $11$, $21$, $1211$, $111221$, $\ldots$ -QUESTION [6 upvotes]: I have this pattern: -1 -11 -21 -1211 -111221 - -I'm guessing it's a fibo pattern, been at it for hours now. Anyone know? - -REPLY [16 votes]: It's known as the look-and-say sequence. - -REPLY [2 votes]: In the first row, there is one 1 (11). In the second, two 1s (21). In the third, there is one 2 then one 1 (1211). In the fourth, there is one 1 then one 2 then two 1s (111221). It continues in this way, so the next row would be 312211.<|endoftext|> -TITLE: Meaning of commutative diagram -QUESTION [12 upvotes]: What is the meaning of a commutative diagram in mathematics? - -For example, if a map translate an object, then rotate it around the origin and then translate it again, is this a commutative diagram? - -REPLY [9 votes]: Here is an example of a commutative diagram: - -Here $A$, $B$, and $C$ are mathematical "objects": perhaps sets, groups, or spaces, and $f$, $g$, and $h$ are "arrows", which are some sort of mapping between the objects that preserves their structure. -The prototypical example is that the objects are sets and the arrows are functions, but the idea itself is extremely general and encompasses objects and arrows that are nothing at all like sets and functions. -The diagram above means that $f$ is an arrow from $B$ to $A$, $g$ is an arrow from $A$ to $C$, and $h$ is an arrow from $B$ to $C$. But the most important part of its meaning is that the arrow you get by going from $B$ to $C$ along the top path is the same arrow as the one you get by going along the bottom path. That is, $g$ and $f$ can be composed, and $$g\circ f = h.$$ -Diagrams of other shapes are similar: any time there is more than one path between two objects, the diagram asserts that the arrows along the paths can be composed and yield the same result. -Here's another example: - -This asserts that $$h\circ f = k\circ g.$$ -There are a lot of additional twists to the notation: a dotted shaft on an arrow often means that the indicated arrow is unique; a little cross in the corner of a square asserts that the square is a "pullback square", and so on. But the main point is simply to assert an equation between arrows.<|endoftext|> -TITLE: Why is $d\theta/dx$ necessarily $\cos \theta$ in this physics problem? Or am I wrong? -QUESTION [5 upvotes]: I'm asking this on the math stack exchange because it seems that the key part of this physics problem I'm asking for help on is more related to the geometry of it than the physics of it. -I'm independently going through Physics for Scientists and Engineers, 3rd Edition, by Raymond A. Serway, Chapter 3, problem 80, which, to paraphrase the problem, goes like this: -A boy scout ties a rope to his food, throws the rope over a branch, and starts walking at a constant velocity $v_0$ and the food starts going straight up. The food starts at the level of his hands and at a height $h$ above that is the branch. The distance he is away from the vertical rope at any given time is $x$. (Here's my drawing of the situation below, the book has a fancier picture of a real human and a tree, but I've stripped it down to the essentials). -The problem goes on to say: -(a) Prove that the food's velocity $v$ when the boy's position is $x$ is equal to: -$$v = v_0 \cdot x \cdot (h^2 + x^2)^{-1/2}.$$ -(b) Prove that the food's acceleration a when the boy's position is $x$ is equal to: -$$a = v_0^2 \cdot h^2 \cdot (h^2 + x^2)^{-3/2}.$$ -I can solve part (a) by drawing a triangle down by the boy's hands as shown and understanding that the food's velocity wraps around the branch in the direction along $l$, and $v_0$ is then the hypotenuse, so $v$ is $v_0 \sin \theta$, which $\sin \theta$ is then equivalent to $x / (h^2 + x^2)^{1/2}$, so $v$ is $v_0 \cdot x \cdot (h^2 + x^2)^{-1/2}$, which is the answer. -I can also see that a is the time derivative of $v$. I can imagine using the chain rule to say that $a = dv/dt = dv/d \theta \cdot d \theta /dx \cdot dx/dt$. I can see that $dv/d \theta = v_0 \cos \theta $ and I can see that $dx/dt$ is the constant $v_0$ that the boy is moving at. I can look at the answer I am supposed to arrive at for part (b) and what I have so far and see that $d \theta /dx$ is supposed to be $\cos \theta$, but that's what I can't see as natuarally following. -Can someone help me visualize or analytically prove why in this setup, $d \theta /dx$ is necessarily $\cos \theta$? - -REPLY [8 votes]: Note (from the big triangle) that $\tan\theta=\frac{x}{h}$. Differentiating both sides with respect to $x$, and using the Chain Rule, we get -$$\sec^2\theta\frac{d\theta}{dx}=\frac{1}{h},$$ -and therefore -$$\frac{d\theta}{dx}=\frac{1}{h}\cos^2\theta.$$ -Not quite the $\cos\theta$ that you expected, but kind of close. -Remark: The angle is a very useful thing from the point of view of the physical intuition. However, let $y$ be the distance of the food from the branch, and let $k$ be the total length of the rope. Then the height of the food is $h-y$, so if we know everything about $y$, we know everything about the height of the food. -Note that in the big triangle, the hypotenuse is $k-y$. It follows by the Pythagorean Theorem that -$$(k-y)^2=h^2+x^2.$$ -Differentiating with respect to $t$, we get -$$-2(k-y)\frac{dy}{dt}=2x\frac{dx}{dt}.\tag{$1$}$$ -But $k-y=(h^2+x^2)^{1/2}$ and $\frac{dx}{dt}=v_0$, so we get -$$-\frac{dy}{dt}=v_0x(h^2+x^2)^{-1/2}.$$ -To find the rate at which the food is rising, just change the sign. -For the acceleration, differentiating both sides of $(1)$ with respect to $t$ works very nicely. Since $x$ changes at a constant rate, the derivative of $x\frac{dx}{dt}$ with respect to $t$ is just $\left(\frac{dx}{dt}\right)^2$. - -REPLY [3 votes]: Now $d\theta$ is so small that we can approximate the arc length to $dx\cos \theta$ -$$r.d\theta=dx \cos \theta$$ -So $$\frac{d\theta}{dx}=\frac{\cos \theta}{r}$$ -where $r=\sqrt{x^2+h^2}$<|endoftext|> -TITLE: How to explain what it means to say a function is "defined" on an interval? -QUESTION [5 upvotes]: I am having difficulty in explaining the terminology "defined" to the students I am assisting. Here is the sentence: If a real-valued function $f$ is defined and continuous on the closed interval $[a,b]$ in the real line, then $f$ is bounded on $[a,b]$. Can I have some thoughts on how to explain the word "defined" used in the sentence? Thank you. - -REPLY [2 votes]: I support the point made by countinghaus that confusing a function with a formula representing a function is a really common error. Later on when things are complicated, you need to be able to think very clearly about these things. However, I also guess from other comments made that there is a bit of a fuzzy notion present in precalculus or basic calculus courses along the lines of 'the set of real numbers at which this expression can be evaluated to give another real number'....? -Anyhow, if we are to be proper and mathematical about this, it seems to me that the issue with understanding what it means for a function to be defined on a certain set is with whatever definition of `function' you are using. The way I was taught, functions are things that have domains. I agree with pritam; It's just something that's included. For example, a measure space is actually three things all interacting in a certain way: a set, a sigma algebra on that set and a measure on that sigma algebra. Given the sigma algebra, you could recover the "ground set" by taking the union of all the sets in the sigma-algebra. -A function is a domain $A$ and a codomain $B$ and a subset $f \subset A\times B$ with the property that if $(x,y)$ and $(x,y')$ are both in $f$, then $y=y'$ and that for every $x \in A$ there is some $y \in B$ such that $(x,y) \in f$. We write $f : A \to B$. If $(x,y) \in f$, we write $f(x) = y$. We may say, for any set $S \subset A$ that $f$ is defined on $S$.<|endoftext|> -TITLE: the sum: $\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n}=\ln(2)$ using Riemann Integral and other methods -QUESTION [16 upvotes]: I need to prove the following: -$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+(-1)^{n+1}\frac{1}{n}+\cdots=\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n}=\ln(2)$$ -Method 1:) -The series $\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n}$ is an alternating series, thus it is convergent, say to $l$. Therefore, both $s_{2n}$ and $s_n$ are convergent to the same limit $l$. -$$ -\begin{align} -s_{2n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots-\frac{1}{2n} & =\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{2n}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2n}\right) \\[10pt] -& =\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} -\end{align} -$$ -It is an easy exercise to prove that: $$\lim_{n \to \infty }s_{2n}=\lim_{n \to \infty }s_n =\lim_{n \to \infty }\left [ \frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} \right ]=\ln(2)$$ which implies that the given alternating series converges to $l=\ln 2$ -However, I am interested to see a proof of this problem using the definition of the Riemann Integral as a sum of infinitely many rectangles of widths tending to zero. I tried to come up with a proof for this, but I couldn't. Can anyone share please? -Also, I am interested to see other methods of solving this problem (other than my method and the Riemann method). If anyone of you is aware of any other methods, please share :) - -REPLY [3 votes]: You can use the following geometric series -$$\sum_{n=0}^\infty x^n=\frac{1}{1-x},|x|<1.$$ -Then integrate both sides to get -$$\sum_{n=1}^\infty \frac{x^n}{n}=\int_0^x\frac{1}{1-t}dt=-\ln (1-x),|x|<1.$$ -Since the LHS converges at $x=-1$, letting $x\to-1$ will give us -$$\sum_{n=1}(-1)^{n+1}\frac{1}{n}=\ln2.$$<|endoftext|> -TITLE: diagonalizing a matrix over the $\ell$-adics -QUESTION [10 upvotes]: Let $M$ be a $2 \times 2$ matrix with coefficients in $\mathbb{Z}_{\ell}$ whose characteristical polynomial is -$$ -P(T) = T^2- (a+d) T + (ad-bc). -$$ -I've encountered the following assertion: If $P(T)$ factors over $\mathbb{Z}/\ell \mathbb{Z}$ as -$$ -P(T) = (T-\lambda_1)(T-\lambda_2) -$$ -with $\lambda_1 \not\equiv \lambda_2 \pmod{\ell}$, then in fact $M$ is diagonalizable over $\mathbb{Z}_{\ell}$. -I was able to prove this using a straightforward argument by diagonalizing over $\mathbb{Z}/\ell \mathbb{Z}$ and then explicitly lifting. However, I was wondering if there was a more conceptual way to explain this in terms of Hensel's Lemma or something (i.e. a more conceptual way to package this lifting argument). - -REPLY [4 votes]: Let $R$ be a local henselian ring (e.g. complete local rings) with residue field $k$. Let $\phi$ be an endomorphism of $V=R^n$ such that its characteristic polynomial $P(T)\in R[T]$ is completely split and separable in $k[T]$. Then $\phi$ is diagonalizable over $R$. - -First by Hensel property, $P(T)$ is completely split and separable in $R[T]$ (this is the only place we use henselian hypothesis). Denote by $a_1, \dots, a_n\in R$ the eigenvalues of $\phi$ and by $V_{a_i}$ the eigensubspace $\mathrm{ker}(\phi - a_i I_n)\subseteq V$. The point is to prove that $V=\oplus_{1\le i\le n} V_{a_i}$. As the submodules $V_{a_i}$ are in direct sum (note that $a_i-a_j$ is an unit in $R$ if $i\ne j$), it is enough to show that -$V=\sum_{1\le i\le n} V_{a_i}$. -Consider the sub-$R$-algebra $A=R[\phi]$ of $\mathrm{End}(V)$ generated by $\phi$. For any $i\le n$, let -$$ e_i=\prod_{j\ne i} (\phi-a_j I_n)\in A.$$ -We have $\phi e_i=(\phi-a_i I_n)e_i + a_i e_i=P(\phi)+a_ie_i=a_ie_i$. So -$e_iV\subseteq V_{a_i}$ and it is enough to show that $I_n\in \sum_i e_iA$ or, equivalently, that $e_1, \dots, e_n$ generate the unit ideal of $A$. But all these properties are true over the residue field $k$ of $R$ by the diagonalizability condition over $k$. So -$$A\otimes_R k=\sum_{1\le i\le n} \bar{e}_i A\otimes_R k.$$ -(Of course this can also be checked directly over $k$.) Thus -$$ A/(\sum_i e_iA) \otimes_R k=0$$ -and $A=\sum_i e_iA$ by Nakayama's lemma.<|endoftext|> -TITLE: How many Fourier coefficients are enough ( in discrete fourier transform) -QUESTION [7 upvotes]: Probably there's a similar question, but I could't find it through all questions about FT -As the title says, how many Fourier coefficients are enough, to be able to "resume" the original function, using inverse discrete Fourier transform? -For example, in the definition from Wikipedia, it looks like we need N coefficients, where N is the number of given points from the original discrete function. I also noticed, that for FFT (fast Fourier transform), the number of calculated coefficients is the same as the number of given points. -Is this always like this? Or we may have fewer coefficients? Or more? And is there a way to estimate this number of necessary coefficients? -I ran some test with randomly generated "control points" of a discrete function and applied DFT and IDFT (in this order) and all control points were recreated. - -REPLY [2 votes]: The number of coefficients needed to reconstitute a given signal depends on signal itself, and on how you're sampling it. -Here's an intuitive answer. The discrete Fourier transform tries to represent your signal as a sum of frequencies, from $0$ frequency (or DC - a flat response), up to some higher frequency that depends on the granularity of your sampling. When in your simulations you increase the sampling rate of your signal (make your sample points closer together), you necessarily increase the frequency range that your signal can represent, and hence you will have more Fourier coefficients for higher frequencies. -However, you might make your signal very slow (have only low frequencies). In this case, you won't need all the Fourier coefficients to reproduce your signal (try making your test signal a slow sign wave, for instance). On the other hand, if you make your signal very fast (have lots of high frequencies), you won't be able to drop coefficients and still reproduce your signal. There is a limit on how "fast" your signal can be and still be reproduced at a given sampling rate. It's called the Nyquist Rate. For certain very special signals, you might be able to do better than the Nyquist rate. This is related to the field of compressed sensing.<|endoftext|> -TITLE: entire function is constant -QUESTION [5 upvotes]: Let $ f $ a complex entire function such that: -$$ |f(z)| \leq \sqrt{2|z|} + \frac{1}{\sqrt{2|z|}} \quad \forall z \neq 0 $$ -Prove that $ f$ is constant. -Thank's in advance! - -REPLY [7 votes]: Let $g(z)=\frac{f(z)-f(0)}{z}$. Then $g(z)$ is entire and for all $z$ with $|z| \geq 1$ you have -$$ \left| g(z) \right| \leq \left|\frac{f(z)}{z} \right| + \left|\frac{f(0)}{z} \right| -\leq \left|\frac{\sqrt{2|z|} + \frac{1}{\sqrt{2|z|}}}{z} \right| + \left|f(0) \right| $$ -Now from here it is easily to prove that $g$ is bounded (note that by continuity $g$ is bounded on $|z| \leq 1$), thus, since entire is constant. -This proves that $f(z)=az+b $ for some $a,b$. Plug this in your equation and see that $a =0$. - -REPLY [6 votes]: Since $f(z)$ is entire, from Cauchy integral formula, we have $$f^{(n)}(0) = \dfrac{n!}{2 \pi i}\oint_{C_r} \dfrac{f(z)}{z^{n+1}} dz$$ -Now argue out why $f^{(n)}(0) = 0$ for $n > 0$ by letting $r \to \infty$ and looking at what happens to the integral.<|endoftext|> -TITLE: Calculate integrals involving gamma function -QUESTION [15 upvotes]: What are the usual ways to follow in order to solve the integrals given below? -$$\begin{align*} -I&=\int_0^1 \ln\Gamma(x)\,dx\\ -J&=\int_0^1 x\ln\Gamma(x)\,dx -\end{align*}$$ - -REPLY [6 votes]: The integral $I$ was mentioned on chat recently, and my solution is different than those given before. -Since $x\Gamma(x)=\Gamma(x+1)$, we have -$$ -\int_0^n\log(\Gamma(x))\,\mathrm{d}x+\int_0^n\log(x)\,\mathrm{d}x -=\int_1^{n+1}\log(\Gamma(x))\,\mathrm{d}x\tag{1} -$$ -Subtracting $\int_1^n\log(\Gamma(x))\,\mathrm{d}x$ from $(1)$ gives -$$ -\int_0^1\log(\Gamma(x))\,\mathrm{d}x+\int_0^n\log(x)\,\mathrm{d}x -=\int_n^{n+1}\log(\Gamma(x))\,\mathrm{d}x\tag{2} -$$ -Stirling's approximation says -$$ -\log(\Gamma(x))=x\log(x)-x-\frac12\log(x)+\frac12\log(2\pi)+o(1)\tag{3} -$$ -Integrating $(3)$ between $n$ and $n+1$ yields -$$ -\begin{align} -&\int_n^{n+1}\log(\Gamma(x))\,\mathrm{d}x\\ -&=\left[\frac12x^2\log(x)-\frac14x^2-\frac12x^2-\frac12x\log(x)+\frac12x\right]_n^{n+1}+\frac12\log(2\pi)+o(1)\\ -&=n\log(n)-n+\frac12\log(2\pi)+o(1)\tag{4} -\end{align} -$$ -Furthermore, -$$ -\int_0^n\log(x)\,\mathrm{d}x=n\log(n)-n\tag{5} -$$ -In light of $(2)$, subtracting $(5)$ from $(4)$ gives -$$ -\begin{align} -\int_0^1\log(\Gamma(x))\,\mathrm{d}x -&=\frac12\log(2\pi)+o(1)\\ -&=\frac12\log(2\pi)\tag{6} -\end{align} -$$<|endoftext|> -TITLE: Why does $L^2$ convergence not imply almost sure convergence -QUESTION [9 upvotes]: What's wrong with this argument? - -Let $f_n$ be a sequence of functions such that $f_n \to f$ in $L^2(\Omega)$. This means $$\lVert f_n - f \rVert_{L^2(\Omega)} \to 0,$$ i.e., -$$\int_\Omega(f_n - f)^2 \to 0.$$ -Since the integrand is positive, this must mean that $f_n \to f$ a.e. - -Why is this not true? Apparently this only true for a subsequence $f_n$ (and in all $L^p$ spaces). - -REPLY [16 votes]: Consider the following sequence of characteristic functions $f_n \colon [0,1] \to R$ defined as follows: -$f_1 = \chi[0, 1/2]$ -$f_2 = \chi[1/2, 1]$ -$f_3 = \chi[0, 1/3]$ -$f_4 = \chi[1/3, 2/3]$ -$f_5 = \chi[2/3, 1]$ -$f_6 = \chi[0, 1/4]$ -$f_7 = \chi[1/4, 2/4]$ -and so on. -Then $f_n \to 0$ in $L^2$, but $f_n$ does not converge pointwise.<|endoftext|> -TITLE: Showing that an entire function is a polynomial -QUESTION [7 upvotes]: Let $f(z)$ be an entire function, $R_n$ a sequence of positive real numbers tending to $\infty$ such that $f(z) \neq 0$ on $|z|=R_n$ and there exists $M>0$ such that -$$\int_{|z|=R_n} \left|\frac{f'(z)}{f(z)}\right| ~dz -TITLE: Inequality between chromatic number and number of edges of a graph -QUESTION [8 upvotes]: I have not been able to find a proof to the statement that if a graph $G$ has $\chi(G)=k$, then it must have at least $\binom{k}{2}$ edges. Would you be able to show me a simple proof? - -REPLY [19 votes]: Suppose we have a $k$-coloring of our graph, with $\chi(G)=k$. Then for any two colors, call them red and blue, there must be some edge that connects them; if there weren't, we could paint every red vertex blue, and we would have a $(k-1)$-coloring of our graph. There are $\binom{k}{2}$ pairs of colors, and any given edge cannot connect more than $2$ colors, so there must be at least $\binom{k}{2}$ edges in our graph.<|endoftext|> -TITLE: Value of $n$ for which an improper integral is convergent. -QUESTION [5 upvotes]: A question from the Calculus book that I'm self-studying is asking me to determine the value of $n$ for which the improper integral below is convergent: -$$\int_1^{+\infty}\left( \frac{n}{x+1} - \frac{3x}{2x^2 + n} \right ) dx$$ -My attempt is below: -Using the definition of improper integral: -$$\lim_{b\to+\infty} \int_1^{b}\left( \frac{n}{x+1} - \frac{3x}{2x^2 + n} \right ) dx$$ -I will first solve the indefinite integral: -$$\int\left( \frac{n}{x+1} - \frac{3x}{2x^2+n} \right )dx = \int \frac{n}{x+1}dx-\int \frac{3x}{2x^2+n}dx$$ -For the first integral, $\int \frac{n}{x+1}dx=n\ln (x+1) + \text{constant}$. For the second integral, substituting $u = 2x^2+n$ (and therefore $dx=\frac{du}{4x}$), $\int \frac{3x}{2x^2+n}dx=\frac{3}{4}\int \frac{1}{u}du=\frac{3}{4}\ln(2x^2+n) + \text{constant}$. So, the result for the whole integral is: -$$\int\left( \frac{n}{x+1} - \frac{3x}{2x^2+n} \right )dx = n\ln(x+1) - \frac{3}{4}\ln(2x^2+n) + \text{constant} = \ln\left(\frac{(x+1)^n}{(2x^2+n)^{3/4}}\right) + \text{constant}$$ -Thus, the value of the definite integral from $x = 1$ to $x = b$ is: -$$\ln\left(\frac{(b+1)^n}{(2b^2+n)^{3/4}}\right)-\ln\left(\frac{(2)^n}{(2+n)^{3/4}}\right)$$ -Thus, the value of the improper integral is: -$$\lim_{b\to+\infty} \left[\ln\left(\frac{(b+1)^n}{(2b^2+n)^{3/4}}\right)-\ln\left(\frac{2^n}{(2+n)^{3/4}}\right)\right]$$ -For this limit to exist, it only depends on the limit of the term $\ln\left(\dfrac{(b+1)^n}{(2b^2+n)^{3/4}}\right)$, because the term $\ln\left(\frac{2^n}{(2+n)^{3/4}}\right)$ is a constant. To calculate the limit of $\ln\left(\dfrac{(b+1)^n}{(2b^2+n)^{3/4}}\right)$ as $b\to +\infty$, I suppose we can neglect the terms $1$ in $(b+1)^n$ and $n$ in $(2b^2+n)^{3/4}$, because they become negligible as $b$ gets larger. Thus we get, for the limit of this term: -$$\lim_{b\to+\infty} \ln\left(\frac{(b)^n}{(2b^2)^{3/4}}\right)$$ -For the above limit to exist, it seems that $n = 3/2$ is the only possible value, because $\lim_{b\to+\infty} \ln\left(\dfrac{(b)^{3/2}}{(2b^2)^{3/4}}\right) = \lim_{b\to+\infty} \ln\left(\dfrac{b^{3/2}}{2^{3/4}b^{3/2}}\right) = \lim_{b\to+\infty} \ln\left(\dfrac{1}{2^{3/4}}\right)$. If any different value for $n$ were chosen, the logarithm would approach $-\infty$ or $+\infty$, and therefore the limit would not exist. -For $n = 3/2$, the value of the limit is: -$$\lim_{b\to+\infty} \left[\ln\left(\frac{(b)^{3/2}}{(2b^2)^{3/4}}\right)-\ln\left(\frac{2^{3/2}}{(2+3/2)^{3/4}}\right)\right]=\lim_{b\to+\infty} \left[\ln\left(\frac{1}{2^{3/4}}\right)-\ln\left(\frac{2^{3/2}}{(7/2)^{3/4}}\right)\right]=\frac{3}{4}\ln\frac{7}{16}$$ -This is equal to the answer given by the book. So, this informal argument of neglecting the terms $1$ and $n$ worked. -But I would like to ask if the reasoning above is correct, and if this informal way of finding the value of $n$ is good. Is there a more formal way? - -Edit: A more formal way of showing that $n$ should be $3/2$ is suggested by David Mitra in the comments to this question below: -$$\lim_{b\to+\infty} \ln \left( {(b+1)^n\over (2b^2+n)^{3/4} } \right) = \lim_{b\to+\infty} \ln \left( {b^n\over b^{3/2}} \cdot {\bigl(1+{1\over b}\bigr)^n\over \bigl(2 +{n\over b^2}\bigr)^{3/4}} \right)$$ -It is clear from above that $n$ should equal $3/2$. If it were a different value, the limit would not exist, because the expression inside the logarithm would approach either $0$ (causing the logarithm to approach $-\infty$) or $+\infty$ (causing the logarithm to tend to $+\infty$). - -REPLY [4 votes]: Your method is correct and it is easy to formalize your neglecting of minor terms. -We can see that $$\ln\left(\dfrac{(b+1)^n}{(2b^2+n)^{3/4}}\right) - \ln\left( \frac{b^n}{(2b^2)^{3/4}}\right)= \ln \left( (1+1/b)^n (1+n/(2b^2))^{-3/4} \right) \to \ln 1=0$$ -as $b\to \infty.$ So the limit of the first term equals the limit of the second term if either limit exists. -However, this is probably not the best way to do this problem. The point of these types of exercises is to get you used to estimating the value of an integral, by doing something to get a feel for the growth of the integrand. Often we can't explicitly integrate the term. -Here, if you put the terms on a common denominator you get $$ \frac{n^2 + 2nx^2 - 3x(x+1)}{(x+1)(2x^2+n)}.$$ -The critical fact used often is: $\displaystyle \int^{\infty}_1 \frac{1}{x^a} dx $ converges only for $a>1.$ -So the $\frac{n^2}{(x+1)(2x^2+n)}$ part clearly converges. So we only need to worry about $$\frac{2nx^2 - 3x(x+1) }{(x+1)(2x^2+n)}.$$ If $n=3/2$ then $2nx^2=3x^2$ is able to cancel with the $-3x^2$ term in the next term, so we have a linear term in the top, cubic denominator, so behaviour like $1/x^2$ which is convergent. But if not, then we always get some quadratic term in the numerator, so the behaviour is like some constant term times $1/x$, which diverges. It is instructive to formalize these ideas with some inequalities, which I will leave you to try.<|endoftext|> -TITLE: Determining the minimum number of pixels on the boundary of a circle drawn in discrete space -QUESTION [6 upvotes]: I am trying to draw a circle in discrete space (actual image pixel space). I have the center (x,y) and radius r of a circle that I am supposed to draw. The manner in which I draw this circle is the following: -Starting from the center position (x,y), I have a for loop over angles $\theta \in \{0,2\pi\}$. Lets say the angle is incremented by $\Delta\theta$ in every iteration. In each iteration, I calculate an x-deviation and a y-deviation based on, -$$\Delta x = r cos(\theta)\\ \Delta y = r sin(\theta).$$ -The point on the circumference of the circle is then calculated as -$$x' = \text{round}(x + \Delta x)\\ y'= \text{round}(y + \Delta y).$$ -This gives a location $(x', y')$ in discrete space at which I can color a pixel. How do I determine for a given radius, what is the minimum number of discrete "pixels" I will have along the circumference. -In other words, lets say if I have a radius of 10, then how many unique discrete points would I have along the boundary of the circle? Is this problem well defined? I know there is a pitfall here of what consists of a discrete circumference. I consider every connected pixel to be a circumference point. - -REPLY [7 votes]: You should probably be using Bresenham's Circle Algorithm? -The "distance" between adjacent pixels is either 1 or $\sqrt{2}$, i.e., two adjacent pixels either share a row or column or they are diagonal. This limits how much of the circumference consecutive pixels can consume. So the number of pixels needed, $x$, to complete the circumference of a circles with a radius of $r$ pixels is bounded $\sqrt{2} \pi r \leq x \leq 2 \pi r$, or roughly $4.443 \times r \leq x \leq 6.283\times r$. This gives a first approximation, by which to check the later estimate. -To get more accuracy we see that, by symmetry, only the pixels drawn in the first octant (i.e., $\pi/4$) need to be computed; the remaining can be found by using the same offsets but in different directions. With each consecutive pixel in the first octant progressing upward, there are at least $\lfloor r \sin(\pi/4)\rfloor$ pixels. Multiply this by 8, and you get roughly $5.657\times r$, which is within the earlier bounds.<|endoftext|> -TITLE: How can I intuit the role of the central limit theorem in breaking the curse of dimensionality for Monte Carlo integration -QUESTION [13 upvotes]: I would like to more intuitively understand where the power of Monte Carlo integration comes from for large-dimensional domains of integration. -Other questions on this site have referenced the proof that the scaling of the error in a MC integration goes as $N^{-1/2}$, where $N$ is the number of function evaluations, and does not depend on the dimension $d$ of the domain of integration. On the other hand, the scaling of the error in a uniform sampling integration goes as some power of $N^{-1/d}$. Consequently, for a sufficiently large $d$, one may achieve a desired level of accuracy using a fewer number of function evaluations with a Monte Carlo method than with a uniform numerical quadrature method. The essence of the proof is an invocation of the central limit theorem. I understand the proof on a formal level. -However, I have no intuition for why this proof works. It still seems like "cheating" to me that by either randomly or quasi-randomly selecting the locations to evaluate the function being integrated, you can achieve a more accurate integration than you would by choosing to evaluate the points in a uniform set of spacings (provided that the dimensionality of the integral is large enough). -I have tried to construct an extremely non-rigorous analogy to conducting a statistical survey. One could either poll people at regular distance spacings in a neighborhood, or instead poll the same number of people in completely randomized locations. If there were no correlation between a person's response and the location where that person lives, then I would be able to conclude that my sampling error would be the same using either method. If there were such a correlation after all, then I might bias my results. Is this sort of reasoning at all on the right track to building intuition for the proof? -To phrase my question another way, it seems that when performing a high-dimensional integral you gain accuracy by including a variety of length scales in the points where you evaluate the functions. Why is that? And, is the role of the randomness essentially to ensure that you use a wide range of length scales? - -REPLY [6 votes]: As you correctly point out, the error in a uniform sampling integration goes as some power of $N^{−1/d}$. The problem of uniform sampling is indeed similar to a non-representative statistical survey. Many of the points in a uniform tensor grid don't add much new information. For example, assume that the function you want to integrate looks like $F(x,y)\approx f(x)+g(y)$ in a local square. If you sample this square with a $k\times k$ tensor grid, you will have used $k^2$ evaluations of $F$, but only received information about $f$ and $g$ for $k$ different arguments. A Taylor expansion of $F$ in a local square as $F(x,y)=ax+by+c+O(x^2+y^2)$ indicates that it is indeed not unusual for $F$ to locally look like that. (Note that no term related to $xy$ occurs in this expression.) -To analyze the connection between uniform tensor grids and the Taylor expansion further, let's look at multivariate polynomials of a fixed degree $k$ in $d$ variables. The function values on a $d$-dimensional uniform tensor grid with $k+1$ points in each direction uniquely determine a polynomial where the max degree of each individual variable (in each term) is smaller or equal than $k$. However, the degree (of an individual term) of a polynomial is defined as the sum of the degrees of the individual variables, so that the determined polynomial has degree $dk$ and $(k+1)^d$ individual terms (=degrees of freedom). On the other hand, a general multivariate polynomial of degree $k$ in $d$ variables only has $\frac{(d+k)!}{d!k!}$ individual terms (=degrees of freedom). For fixed $k$, we have $\frac{(d+k)!}{d!k!}=\frac{d^k}{k!} + O(d^{k-1})$, which is polynomial in $d$ (i.e. not exponential). -What has all this to do with randomness and Monte Carlo integration? It might help explain why the naive tensor product approach doesn't break down for Monte Carlo integration, while it does break down for uniform sampling. Unfortunately, there is no straightforward way to turn an effective (naive) randomized algorithm into an effective (naive) deterministic algorithm. However, we know from experience that problems which can be solved efficiently by randomized algorithms normally can also be solved efficiently by (more complicated) deterministic algorithms. The deterministic high-dimensional integration methods I know work by exploiting recursion (divide and conquer) in a clever way. However, this is not the place to explain hierarchical bases, hierarchical subspace splitting or the Smolyak method. A more black-box approach would be to use space-filling curves as a building block instead, where the recursion is already built into the construction of the curve. I don't know how well this would work for integration, but it has been asked here before. At least I know that the research groups investigating applications of sparse grids often also investigate applications of space filling curves.<|endoftext|> -TITLE: Difference between permutation and combination? -QUESTION [8 upvotes]: Permutation: -$$P(n,r) = \frac{n!}{(n-r)!}$$ -Combination: -$$C(n,r) = \frac{n!}{(n-r)!r!}$$ -Apparently, you use combination when the order doesn't matter. Great. I see how a combination will give you all the possible well, combinations. However, I don't see what exactly does a permutation do then. - -REPLY [2 votes]: There is some common confusion between these two terms. -The word "permutation" in general refers to one of three things depending on context. It can mean the order (arrangement) of a set as in combinatorics. Or it can refer to an arrangement of a subset of a given size as also in combinatorics. Or it can refer to an OPERATION of REarrangement in a space of -such operations as in group theory, which is a very different thing. -"Combination" refers only to the makeup (constituency) of a subset with no concept of order. -The term "combinations" refers to the number of subsets of a given size containing different constituents. -The combinatorial meanings of both of these terms are still valid when applied to sets with duplicate elements. A permutation (arrangement or rearrangement) can apply to a set or subset that contains duplicates. But "combination" usually assumes distinct elements in the subset, though the original set can contain duplicates. Of course, if so indicated, a combination could also contain duplicates. The presence of duplicates affects the combinatorial formulas for all of these.<|endoftext|> -TITLE: Krull dimension of $\mathbb{C}[x_1, x_2, x_3, x_4]/\left< x_1x_3-x_2^2,x_2 x_4-x_3^2,x_1x_4-x_2 x_3\right>$ -QUESTION [14 upvotes]: Krull dimension of a ring $R$ is the supremum of the number of strict inclusions in a chain of prime ideals. - -Question 1. Considering $R = \mathbb{C}[x_1, x_2, x_3, x_4]/\left< x_1x_3-x_2^2,x_2 x_4-x_3^2,x_1x_4-x_2 x_3\right>$, how does one calculate the Krull dimension of $R$? This variety is well-known as the twisted cubic in $\mathbb{P}^3$. -Question 2. In general for any ring $R$, how are the Krull dimension of $R$ and the dimension of Spec$(R)$ related? - -Thank you. - -REPLY [21 votes]: I'll prove the following result: - -$$K[x_1, x_2, x_3, x_4]/\left< x_1x_3-x_2^2,x_2 x_4-x_3^2,x_1x_4-x_2 x_3\right>\simeq K[s^3,s^2t,st^2,t^3],$$ where $K$ is a field. - -Let $\varphi: K[x_1,x_2,x_3,x_4]\to K[s, t]$ be the ring homomorphism that maps $x_1\mapsto s^3$, $x_2\mapsto s^2t$, $x_3\mapsto st^2$ and $x_4\mapsto t^3$. Obviously $\operatorname{Im}\varphi=K[s^3,s^2t,st^2,t^3]$; this is a subring of $K[s,t]$ and the extension $K[s^3,s^2t,st^2,t^3]\subset K[s,t]$ is integral, hence $\dim K[s^3,s^2t,st^2,t^3]= \dim K[s,t]=2.$ -It remains to prove that $\ker\varphi=\left< x_1x_3-x_2^2, x_2 x_4-x_3^2, x_1x_4-x_2 x_3\right>$. By definition $\varphi(f(x_1,x_2,x_3,x_4))=f(s^3,s^2t,st^2,t^3)$. In particular, this shows that the polynomials $g_1=x_1x_3-x_2^2$, $g_2=x_2 x_4-x_3^2$ and $g_3=x_1x_4-x_2 x_3$ belong to $\ker\varphi$. -Conversely, let $f\in\ker\varphi$, i.e. $f(s^3,s^2t,st^2,t^3)=0$. We want to show that $$f\in\left< x_1x_3-x_2^2, x_2 x_4-x_3^2, x_1x_4-x_2 x_3\right>.$$ The initial monomials of $g_1$, $g_2$, resp. $g_3$ with respect to the lexicographical order are $x_1x_3$, $x_2x_4$, resp. $x_1x_4$. The remainder on division of $f$ to $G=\{g_1,g_2,g_3\}$, denoted by $r$, is a $K$-linear combination of monomials none of which is divisible by $x_1x_3$, $x_2x_4$, resp. $x_1x_4$. This shows that the monomials of $r$ can have one the following forms: $x_1^ix_2^j$ with $i\ge 1$ and $j\ge 0$, $x_2^kx_3^l$ with $k\ge 1$ and $l\ge 0$, respectively $x_3^ux_4^v$ with $u,v\ge 0$. In order to prove that $f\in\left< x_1x_3-x_2^2, x_2 x_4-x_3^2, x_1x_4-x_2 x_3\right>$ it is enough to show that $r=0$. But we know that $f(s^3,s^2t,st^2,t^3)=0$ and therefore $r(s^3,s^2t,st^2,t^3)=0$. The monomials (in $s$ and $t$) of $r(s^3,s^2t,st^2,t^3)$ are of the following types: $s^{3i+2j}t^j$, $s^{2k+l}t^{k+2l}$, respectively $s^ut^{2u+3v}$. Now check that there is no possible cancelation between these monomials (because they can't be equal), so $r=0$. -Now it follows that $\dim R=2$.<|endoftext|> -TITLE: $f, g$ entire functions with $f^2 + g^2 \equiv 1 \implies \exists h $ entire with $f(z) = \cos(h(z))$ and $g(z) = \sin(h(z))$ -QUESTION [7 upvotes]: I am studying for a qualifier exam in complex analysis and right now I'm solving questions from old exams. I am trying to prove the following: - -Prove that if $f$ and $g$ are entire functions such that $f(z)^2 + g(z)^2 = 1$ for all $z \in \mathbb{C}$, then there exists an entire function $h$ such that $f(z) = \cos(h(z))$ and $g(z) = \sin(h(z))$. - - -My Attempt -The approach that occurred to me is the following. Since $f(z)^2 + g(z)^2 = 1$ then we have $(f(z) + ig(z))(f(z) - ig(z)) = 1$. Then each factor is nonvanishing everywhere in $\mathbb{C}$ and thus by the "holomorphic logarithm theorem" we know that since $\mathbb{C}$ is simply connected, there exists a holomorphic function $H:\mathbb{C} \to \mathbb{C}$ such that -$$e^{H(z)} = f(z) + ig(z)$$ -and then we can write $\exp(H(z)) = \exp\left(i\dfrac{H(z)}{i} \right) = \exp(ih(z))$, -where $h(z) := \dfrac{H(z)}{i}$. -Thus so far we have an entire function $h(z)$ that satisfies -$$e^{ih(z)} = f(z) + ig(z)$$ -On the other hand, we also know that $e^{iz} = \cos{z} + i \sin{z}$ for any $z \in \mathbb{C}$, thus we see that -$$e^{ih(z)} = \cos{(h(z))} + i \sin{(h(z))} = f(z) + ig(z)$$ -Thus at this point I would like to conclude somehow that we must have $f(z) = \cos(h(z))$ and $g(z) = \sin(h(z))$, but I can't see how and if this is possible. - -My questions - -Is the approach I have outlined a correct way to proceed, and if so how can I finish my argument? -If my argument does not work, how can this be proved? - -Thanks for any help. - -REPLY [5 votes]: You approach appears to be correct, and it can be finished with the following thought: not only do complex exponentials split into combinations of trigonometric functions, but trig functions also split into combinations of complex exponentials. Indeed: -$$\cos\alpha=\frac{e^{i\alpha}+e^{-i\alpha}}{2},\quad \sin\alpha=\frac{e^{i\alpha}-e^{-i\alpha}}{2i}.$$ -This is applicable for not just real $\alpha$, but complex as well. You've deduced $e^{ih(z)}=f(z)+ig(z)$ for some entire function $h$, and taking inverses gives $e^{-ih(z)}=f(z)-ig(z)$, so averaging these two will give you $\cos h(z)=f(z)$ (and similarly, $\sin h(z)=g(z)$).<|endoftext|> -TITLE: Decomposing polynomials with integer coefficients -QUESTION [50 upvotes]: Can every quadratic with integer coefficients be written as a sum of two polynomials with integer roots? (Any constant $k \in \mathbb{Z}$, including $0$, is also allowed as a term for simplicity's sake.) - -(In other words, is any given $P(x) = A + Bx + Cx^2$ expressible as -$$P(x) = \color{red}{k(x-r_1)(x-r_2)\cdots(x-r_n)} + \color{blue}{\ell(x-s_1)(x-s_2)\cdots(x-s_m)}$$ -where all variables other than $x$ are integers?) As an example of such a decomposition, if $C = 1$ then $P(x) = (A - Ax) + (Ax + Bx + x^2) = \color{red}{-A(x-1)} + \color{blue}{(x)(x+A+B)}$. The "two polynomials" restriction is essential; expressions like $P(x) = \color{red}{(A)} + \color{green}{(Bx)} + \color{blue}{(Cx^2)}$ don't count. -I've been contemplating this statement for a while and could use some help. I'm having trouble whether trying to prove it or find a (verifiable) counterexample. (Note that the components can have arbitrarily high degrees $n,m$ but cancel out to give $P(x)$.) Variations on completing the square didn't help. -If the answer is affirmative, I would also be interested in the following generalizations: - - -In addition to quadratics, can higher-order polynomials be decomposed into two polynomials? -(Refinement of the above if it is true) If two polynomials do not suffice for $P(x)$ of arbitrary degree, is there a finite number $N$ that does? - - -Thanks in advance for any ideas or help. -Note: I have used the colors I can most easily distinguish in the question, but if they cause other people difficulty please feel free to change them or remove them. - -REPLY [12 votes]: (EDIT: Gerry Myerson pointed out a serious oversight in my previous answer. In the following answer, I consider polynomials with integer roots to have degree at least one, meaning nonzero constants are considered as a separate case. I believe this will address the gap found by Gerry.) -$1 + 9x + 27x^2$ cannot be expressed as either the sum of a constant and a polynomial with integer roots, or the sum of two polynomials with integer roots. -First, we'll show that $1 + 9x + 27x^2$ cannot be decomposed as a constant plus a polynomial with integer roots. Suppose to the contrary that $1 + 9x + 27x^2 = a + c(x+r_1)(x+r_2)$ for some integers $a, c, r_1, r_2$. Then $1 + 9x + 27x^2 = (a + cr_1r_2) + c(r_1 + r_2)x + cx^2$, and so $c = 27$. However, no integer values of $r_1, r_2$ will make $c(r_1 + r_2) = 27(r_1 + r_2) = 9$, meaning $1 + 9x + 27x^2$ has no such decomposition. -To prove the second assertion, we'll establish that there are families of quadratics which, if written as the sum of two polynomials with integer roots, can only be decomposed as the difference of two cubics. We can then demonstrate the impossibility of decomposing $1 + 9x + 27x^2$ in this way using modular arithmetic and brute force. First it will be useful to prove some intermediate results. -Lemma 1.1. A polynomial $f$ where the leading coefficient does not divide $f(n)$ for any $n$ cannot be expressed as the sum of two polynomials with integer roots of differing degrees. -Proof. Note that $f$ cannot have integer roots, otherwise the leading coefficient would divide $f(n) = 0$ for some $n$. Now suppose $f$ has a decomposition as: -$$f(x) = k(x + r_1)(x + r_2)\cdots(x + r_i) + l(x + s_1)(x + s_2)\cdots(x + s_j)$$ -with $i \ne j$. WLOG, let $i > j$. Since $f$ has no integer roots, both $k$ and $l$ are not zero. Consequently, $i$ must equal the degree of $f$ and $k$ must equal the leading coefficient of $f$. However, by evaluating the polynomial at $x = -s_1$, we see that $k$ divides $f(-s_1)$, a contradiction. So there is no such decomposition. $\square$ -Lemma 1.2. A polynomial $f$ which is odd for all $f(n)$, when decomposed into the sum of two polynomials with integer roots, must be the sum of one polynomial with all even roots and the other with odd roots. -Proof. Note that $f$ has no integer roots, and so $k$ and $l$ are not zero. Now we examine the decomposition of $f$ as -$$f(x) = k(x + r_1)(x + r_2)\cdots(x + r_i) + l(x + s_1)(x + s_2)\cdots(x + s_j)$$ -Given that $f(0)$ is odd, exactly one of the terms on the RHS is odd for the substitution $x = 0$. WLOG, suppose the first term is odd; all factors are also odd, and so $k, r_1, r_2, \ldots, r_i$ are odd. Given that $f(1)$ is odd, the first term is even for the substitution $x = 1$, and by repeating our previous analysis we discover that $l, s_1 + 1, s_2 + 1, \ldots, s_j + 1$ are odd. This implies that $s_1, s_2, \ldots, s_j$ are even. So one polynomial in the decomposition of $f$ must have all even roots, and the other odd. $\square$ -Lemma 1.3. If a quadratic polynomial $f(x) = a + bx + cx^2$ with odd coefficients such that $\gcd(b, c) \not\mid a$ can be decomposed into two polynomials with integer roots, each polynomial in the decomposition must be cubic. -Proof. For quadratic polynomials, $\gcd(b, c) \not\mid a$ is equivalent to the condition that $c$ never divides $f(n)$ for any $n$, so we can apply lemma (1.1). Since $f(n)$ is also odd for all $n$ we may also use lemma (1.2). Then consider the decomposition of $f$ as -$$f(x) = a + bx + cx^2 = k(x + r_1)(x + r_2)\cdots(x + r_i) + l(x + s_1)(x + s_2)\cdots(x + s_j)$$ -From lemma (1.1), we know that $i = j$. From lemma (1.2), we can suppose WLOG that the first polynomial on the RHS has all odd roots. Reinterpreting the decomposition modulo 2, we have -$$ -\begin{align} -f(x) \equiv 1 + x + x^2 & \equiv (x+1)^i + x^j \pmod 2 \\ -& \equiv (x+1)^j + x^j \pmod 2 \\ -& \equiv (x+1)^j - x^j \pmod 2 -\end{align} -$$ -The coefficients of $(x+1)^j$ are the binomial coefficients; so the coefficient of the linear term is ${j \choose j-1} = {j \choose 1} = j$, which for even $j$ does not match the parity of the linear term of $f$. For odd $j$, the coefficient of the $x^{j-1}$ term is also ${j \choose 1}$, but since all terms with degree greater than two have coefficients equal to zero, $j$ must equal three. So if quadratics with odd coefficients which satisfy lemma (1.1) can decomposed into two polynomials with integer roots, they can only be expressed as the sum (difference) of two cubics with integer roots. $\square$ -Remark. With some trial and error, we can demonstrate polynomials which are not decomposable as the sum of two polynomials with integer roots on the strength of lemma (1.2) alone. For example, no values of $i$, $j$ will make $(x+1)^i + x^j \equiv 1 + x^3 + x^5 \pmod 2$. - -Now, specifically, we aim to show that $1 + 9x + 27x^2$ cannot be decomposed into two cubics, and therefore into two polynomials with integer roots. -The easiest approach is bruteforce calculation: since $f(x) = 1 + 9x + 27x^2$ satisfies all criteria of lemma (1.3), we consider a decomposition for $f$ as the difference of two cubics -$$ -\begin{align} -f(x) & = & 1 + 9x + 27x^2 & = (x + r_1)(x + r_2)(x + r_3) - (x + s_1)(x + s_2)(x + s_3) \\ -& \equiv & 1 & \equiv (x + r_1)(x + r_2)(x + r_3) - (x + s_1)(x + s_2)(x + s_3) \pmod 9 -\end{align} -$$ -and attempt to find suitable values of $r_1, r_2, r_3, s_1, s_2, s_3$ such that the RHS remains equivalent to one (modulo 9) for all values of $x$. -The following Javascript function will generate an array of all the unique triples $r_1, r_2, r_3$ for a given modulus, which can be taken to be the roots of a cubic under the same modulus, and the order of elements doesn't matter. Then it tries all pairs of triples hoping to find a pair whose difference remains equivalent to the target (with respect to the modulus) for all substitutions of $x$. - -function pairsOfCubicsDifferenceEquivalentToXmoduloY(target, modulus) { - target = mod(target, modulus); - - /* Generate all unique triples. */ - for (var i = 0, triplets = []; i < modulus; ++i) { - for (var j = i; j < modulus; ++j) { - for (var k = j; k < modulus; ++k) { - triplets.push([i,j,k]); - } - } - } - - /* Brute force. Try all pairs of triples. */ - for (var i = 0, result = []; i < triplets.length; ++i) { - for (var j = 0; j < triplets.length; ++j) { - for (var x = 0; x <= modulus; ++x) { - if ( - mod( (x+triplets[i][0])*(x+triplets[i][1])*(x+triplets[i][2]) - - (x+triplets[j][0])*(x+triplets[j][1])*(x+triplets[j][2]) - , modulus) != target) { - break; - } - /* Full circle! A solution. */ - if (x == modulus) { - result.push( [triplets[i], triplets[j]] ); - } - } - } - } - - return result; - - /* Auxiliary: make remainder a nonnegative value. */ - function mod(x, modulus) { - return ((x % modulus) + modulus) % modulus; - } -} - -Anticlimactically, evaluating pairsOfCubicsDifferenceEquivalentToXmoduloY(1, 9) will return no results, and so $1 + 9x + 27x^2$ cannot be expressed as the difference of two cubics, or therefore as the sum of two polynomials with integer roots. $\square$ - -(One!) further thought: the following observations let us give another exposition of a result by RicardoCruz, where he shows that quadratics $f(x) = a + bx + cx^2$ where $\gcd(b, c) \mid a$ can be decomposed systematically. -Observation 2.1 (Shifting.) If a quadratic $f(x)$ has a decomposition -$$f(x) = a + bx + cx^2 = k(x + r_1)(x + r_2)\cdots(x + r_i) + l(x + s_1)(x + s_2)\cdots(x + s_j)$$ -then it has a decomposition $f(x+n)$ for any integer $n$. -$$f(x+n) = f(n) + (2cn+b)x + cx^2 = k(x + n+r_1)(x + n+r_2)\cdots(x + n+r_i) + l(x + n+s_1)(x + n+s_2)\cdots(x + n+s_j)$$ -Observation 2.2 (Linear terms are absorbent.) For a quadratic $f(x) = a + bx + cx^2$, if the leading coefficient divides the constant term, we can decompose $f$ into polynomials with integer roots. Let $a = cpq$ and rewrite $f$ as -$$ -\begin{align} -f(x) & = a + bx + cx^2 \\ -& = cpq + bx + cx^2 \\ -& = c(x^2 + pq) + bx \\ -& = c(x + p)(x + q) - c(p+q)x + bx \\ -f(x) & = c(x + p)(x + q) + (b - c[p+q])x -\end{align} -$$ -Returning now to our initial motivation. If $f(x) = a + bx + cx^2$ and $\gcd(b, c) \mid a$, we can factor out $\gcd(b, c)$ from each term, so after pulling out a constant we may assume that $\gcd(b, c) = 1$. Now we can find an $n$ such that $c \mid f(n)$, by considering the quadratic modulo $c$. -$$ -\begin{align} -f(n) & \equiv 0 \pmod c \\ -a + bn + cn^2 & \equiv 0 \pmod c \\ -bn & \equiv -a \pmod c \\ -\end{align} -$$ -where $b$ is invertible since $\gcd(b, c) = 1$. If we "shift" $f(x)$ by $n$ using observation (2.1), then the constant term $f(n)$ will be divisible by $c$, and we can apply observation (2.2) to find a decomposition for $f(x+n)$, then shift back by $-n$ to get a decomposition for $f(x)$. -Let's take an example from RicardoCruz's answer, where $f(x) = 2 + 57x + 31x^2$. -$$ -\begin{align} -f(n) & \equiv 0 \pmod {31} \\ -2 + 57n + 31n^2 & \equiv 0 \pmod {31} \\ --5n & \equiv -2 \pmod {31} \\ -n & \equiv -12 \pmod {31} -\end{align} -$$ -and then -$$ -\begin{align} -f(x) & = 2 + 57x + 31x^2 \\ -f(x-12) & = 2 + 57(x-12) + 31(x-12)^2 \\ -& = 3782 - 687x + 31x^2 \\ -& = 31(x^2 + 122) - 687x \\ -\end{align} -$$ -We can pick any divisors of $122$, so why not $-2$ and $-61$? -$$ -\begin{align} -f(x-12) & = 31(x^2 + [-2][-61]) - 687x \\ -& = 31(x - 2)(x - 61) + (31 \cdot 63)x - 687x \\ -& = 31(x - 2)(x - 61) + 1266x \\ -f([x+12] - 12) & = 31([x+12] - 2)([x+12] - 61) + 1266[x+12] \\ -f(x) & = 31(x + 10)(x - 49) + 1266(x+12) -\end{align} -$$ -which agrees with RicardoCruz's analysis.<|endoftext|> -TITLE: The minus Laplacian operator is positive definite -QUESTION [5 upvotes]: In a textbook of functional analysis I found this equation derived from Green's first identity -$$\int _{ \Omega }^{ }{ u{ \nabla }^{ 2 }ud\tau } =\int _{ \partial \Omega }^{ }{ u\frac { \partial u }{ \partial n } ds } -\int _{ \Omega }^{ }{ \left| \nabla u \right| ^{2}d\tau }$$ -Then it goes on saying that if the boundary conditions on u are such that the integral over the boundary vanishes then the operator $ -\nabla^{2}$ is positive definite. -Why ? -What I can see is that $$\int _{ \Omega }^{ }{ u{ \nabla }^{ 2 }u+{ \left| \nabla u \right| }^{ 2 }d\tau } =0$$ -and what I'd need to declare that the operator is positive definite is : $$\left< -{ \nabla }^{ 2 }u,u \right> >0\Leftrightarrow \int _{ \Omega }^{ }{ -{ \nabla }^{ 2 }u\bar { u } d\tau } >0$$ -So far I don't see how to prove that the operator is positive definite... -Thanks for any kind of help. - -REPLY [5 votes]: Green's identity reads: -$$\int_U \left( \psi \nabla^{2} \varphi + \nabla \varphi \cdot \nabla \psi\right)\, dV = \oint_{\partial U} \psi \left( \nabla \varphi \cdot \mathbf{n} \right)\, dS$$ -Select $\psi=\bar{u}$ and $\varphi=u$ and negate: -$$-\int_U\bar{u}\nabla^2u+\nabla \bar{u}\cdot\nabla u\; dV=-\oint_{\partial U}\bar{u}(\nabla u\cdot\mathbf{n})dS.$$ -Of course $\nabla u\cdot\mathbf{n}=0$ by hypothesis on the boundary conditions, so we may rearrange this to -$$\left\langle -\nabla^2 u,u\right\rangle=\int_U \overline{\nabla u}\cdot \nabla u\;dV.$$ -The integrand on the right is $\sum_i|\partial u/\partial x_i|^2$, so of course it is nonnegative, and is in fact only zero when $\nabla u=0$. In fact the integral on the right displays a means to defining an inner product for complex-valued vector functions, hence $\langle\nabla u,\nabla u\rangle $ in Andrew's answer, and knowing this a priori would provide a very direct means to seeing positive definiteness. The inner product is -$$\langle \mathbf v,\mathbf w\rangle =\int_U \mathbf v\cdot \overline{\mathbf w}\; dV.$$ -It is somewhat unclear to me if the statement about $-\Delta^2$'s positive definiteness is in the context of real-valued or complex-valued functions $u$. In the former situation $u=\bar{u}$ so you already had all you needed, and in the latter situation the identity it had was slightly off (didn't involve a complex conjugate) for the purpose at hand, albeit a slight modification was all that was necessary.<|endoftext|> -TITLE: Evaluating $\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\, \mathrm{d}x$ -QUESTION [20 upvotes]: I have to evaluate: -$$\int_{0}^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\, \mathrm{d}x. $$ -I can't get the right answer! So please help me out! - -REPLY [37 votes]: Let $I$ denote the integral and consider the substitution $u= \frac{\pi }{2} - x.$ Then $I = \displaystyle\int_0^{\frac{\pi }{2}} \frac{\sqrt{\cos u}}{\sqrt{\cos u } + \sqrt{\sin u }} du$ and $2I = \displaystyle\int_0^{\frac{\pi }{2}} \frac{\sqrt{\cos u} + \sqrt{\sin u }}{\sqrt{\cos u } + \sqrt{\sin u }} du = \frac{\pi }{2}.$ Hence $I = \frac{\pi }{4}.$ -In general, $ \displaystyle\int_0^a f(x) dx = \displaystyle\int_0^a f(a-x) $ $dx$ whenever $f$ is integrable, and $\displaystyle\int_0^{\frac{\pi }{2}} \frac{\cos^a x}{\cos^a x + \sin^a x } dx = \displaystyle\int_0^{\frac{\pi }{2}} \frac{\sin^a x}{\cos^a x + \sin^a x } dx = \frac{\pi }{4}$ for $a>0$ (same trick.) - -REPLY [23 votes]: Note that $\sin(\pi/2-x)=\cos x$ and $\cos(\pi/2-x)=\sin x$. The answer will exploit the symmetry. -Break up the original integral into two parts, (i) from $0$ to $\pi/4$ and (ii) from $\pi/4$ to $\pi/2$. -So our first integral is -$$\int_{x=0}^{\pi/4} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx.\tag{$1$} $$ -For the second integral, make the change of variable $u=\pi/2-x$. Using the fact that $\sin x=\sin(\pi/2-u)=\cos u$ and $\cos x=\cos(\pi/2-u)=\sin u$, and the fact that $dx=-du$, we get after not much work -$$\int_{u=\pi/4}^{0} -\frac{\sqrt{\cos u}}{\sqrt{\cos u}+\sqrt{\sin u}}\,du$$ -Change the dummy variable of integration variable to the name $x$. Also, do the integration in the "right" order, $0$ to $\pi/4$. That changes the sign, so our second integral is equal to -$$\int_{x=0}^{\pi/4} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx.\tag{$2$}$$ -Our original integral is the sum of the integrals $(1)$ and $(2)$. Add, and note the beautiful cancellation $\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}+ \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}=1$. Thus our original integral is equal to -$$\int_0^{\pi/4}1\,dx.$$ -This is trivial to compute: the answer is $\pi/4$. -Remark: Let $f(x)$ and $g(x)$ be any reasonably nice functions such that $g(x)=f(a-x)$. Exactly the same argument shows that -$$\int_0^a\frac{f(x)}{f(x)+g(x)}\,dx=\frac{a}{2}.$$<|endoftext|> -TITLE: Boyd & Vandenberghe, question 2.31(d) — stuck on simple problem regarding interior of a dual cone -QUESTION [6 upvotes]: In Boyd & Vandenberghe's Convex Optimization, question 2.31(d) asks to prove that the interior of the dual cone $K^*$ is equal to -$$\text{int } K^* = \{ z \mid z^\top x > 0, \,\forall x \in K \} \tag{1}$$ -Recall that the dual cone of a cone $K$ is the set: -$$K^* = \{ y \mid y^\top x \ge 0 , \forall x \in K \}.$$ -I've spent a solid chunk of time trying to prove this simple and seemingly evident statement about the interior of the dual cone but keep getting stuck on the problem of bounding the inner product of $x$ with a perturbed version of $z$ (details provided below). Frustratingly, the proof in the book's answer key (available online) takes as given this very fact that that I am stuck on proving. -My work in proving statement (1) is given below. I hope someone can show me the piece of mathematical technology I'm missing. Thanks! - -Question 2.31(d): -Let $K$ be a convex cone and $K^* = \{ y \mid y^\top x \ge 0 $ for all $ x \in K \}$ be its dual cone. -Let $S = \{ z \mid z^\top x > 0 $ for all $ x \in K \}.$ Show that $S = \text{int } K^*.$ -Certainly $S \subseteq K^*.$ Now consider some arbitrary point $z_0 \in S$. For all $x \in K$ we have $z_0^\top x > 0$. It's clear that we need to find an $\epsilon$ such that for all $z' \in D(z_0, \epsilon)$, -$$z'^\top x > 0 ,\,\forall x \in K$$ -Unfortunately, I don't know how to show $z'^\top x > 0$ for $z' \in D(z_0, \epsilon)$ when $\epsilon$ is chosen sufficiently small. I do know we can write $z'$ as $z_0 + \gamma u$ where $\|u\| = 1$ and $\gamma \in (0,\epsilon)$. And I do know that -$$z_0^\top x - \gamma \|x\| ~\le~ z_0^\top x + \gamma u^\top x ~\le~ z_0^\top x + \gamma \|x\|$$ -where $z_0^\top x > 0$ and $\gamma \|x\| \ge 0$. However, I don't know how to show the critical piece, that -$$z_0^\top x - \gamma \|x\| > 0$$ -when $\epsilon$ is chosen sufficiently small since $x$ can range over $K$ and therefore $\|x\|$ can be arbitrarily large. Frustratingly, the solution in B&V just takes it for granted that for sufficiently small $\epsilon$, -$$z_0^\top x + \gamma u^\top x > 0$$ -I've looked online for some matrix perturbation theory results to apply but nothing I've found has been useful. -Any help is greatly appreciated. - -REPLY [3 votes]: The question has an additional bug: $K$ should be assumed to be closed. For instance, take $K$ to be the positive orthant together with $0$, i.e., $K=\{x\in\mathbb{R}^{n}:x_{i}>0,i=1,\ldots n\} \cup \{0\}$. Then $K^{\ast}=\mathbb{R}_{+}^{n}=\{x\in \mathbb{R}^{n}:x_{i}\geq0,i=1,\ldots n\}$. Every element $z\in K^{\ast}\backslash \{0\}$ satisfies $z^{T}x>0$ for all $x\in K\backslash \{0\}$, even those at the boundary of $K^{\ast}$.<|endoftext|> -TITLE: Construction of $\Bbb R$ from $\Bbb Q$ -QUESTION [6 upvotes]: As it is true that we can construct all rational numbers $\Bbb Q$ from the set of integers $\Bbb Z$, is it possible to construct the set of real numbers $\Bbb R$ from $\Bbb Q$? If yes, how? Is there any procedure? And if no, is there any proof that we can't ? Thanks! - -REPLY [3 votes]: To complete the list from Spivak, we can also define the reals 'naively' as infinite decimal sequences, but this is essentially picking a representative element from an equivalence class of Cauchy sequences. -(eg $\pi$ should be the limit of the sequence 3, 3.1, 3.14, 3.141, ...)<|endoftext|> -TITLE: Is the Mellin transform useful to solve differential equations? -QUESTION [5 upvotes]: The Mellin transform is defined as: -$$F(\mu)=\int_0^\infty f(x)x^{\mu-1}dx$$ -The derivative of the Mellin transform is: -$$F'(\mu)=-(\mu-1)F(\mu-1)$$ -Applying this property, for example to the Bessel equation: -$$x^2 y''+xy'+(x^2-\nu^2)y$$ we can transform it in the complex difference equation: -$$Y(\mu + 2) = ±(\mu − \nu)(\mu + \nu)Y(\mu)$$ -where $Y(\mu)$ is the Mellin transform of $y(x)$. -Is this method useful in general to solve ODE? -Thanks - -REPLY [8 votes]: Here we give a simple example of the use of the Mellin transform to solve a differential equation. -Consider the differential equation -$$\begin{equation*} -f'(t) + f(t) = 0.\tag{1} -\end{equation*}$$ -On taking the Mellin transform we find that the -corresponding complex difference equation is -$-(s-1)\phi(s-1) + \phi(s) = 0$, or -$$\phi(s+1) = s \phi(s).$$ -This is the recurrence relation for the gamma function, so -$$\phi(s) = \Gamma(s)$$ -up to an overall numerical factor. -The solution to (1) must be -$$\begin{equation*} -f(t) = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} ds\, t^{-s} \Gamma(s),\tag{2} -\end{equation*}$$ -where $c$ picks out some suitable contour avoiding the poles of $\Gamma$. -Recall that $\Gamma$ is meromorphic with simple poles at $0, -1, -2, \ldots$. -Existence of the inverse transform (2) can be shown by appealing to the asymptotic behavior of $\Gamma$, -$$|\Gamma(x+i y)| \sim \sqrt{2\pi} |y|^{x-1/2} e^{-|y|\pi/2},$$ -in the limit $|y|\to \infty$. -We choose $c>0$. -With this choice of contour the integral (2) is the Cahen-Mellin integral. -Pushing the contour to the left, we pick up all the poles of the gamma function. -To calculate the residue, recall that near $s=-n$ -$$\Gamma(s) = \frac{(-1)^n}{n!}\frac{1}{s+n} + O(1).$$ -This is a straightforward consequence of the recurrence relation for $\Gamma$. -Thus, -$$\begin{eqnarray*} -f(t) &=& \sum_{n=0}^\infty \mathrm{Res}_{s=-n} t^{-s} \Gamma(s) \\ -&=& \sum_{n=0}^\infty t^{n}\frac{(-1)^n}{n!} \\ -&=& e^{-t}. -\end{eqnarray*}$$<|endoftext|> -TITLE: trivial Picard group -QUESTION [8 upvotes]: let $S=\operatorname{Spec}(A)$ be an affine scheme. For which ring $A$, not field is it known that $H^1(S,\mathcal{O}_S^{*})$ is trivial? -If $X\to S$ is a finite map and $H^1(S,\mathcal{O}_S^{*})$ is trivial, is it true that also $H^1(X,\mathcal{O}_X^{*})$ is trivial? -Thanks - -REPLY [11 votes]: Sample answers to your first question: If $S$ is Spec of a local ring, or of a UFD, then $H^1(S, \mathcal O_S^{\times})$ is trivial. -The answer to your second question is no: $X = $Spec $\mathbb C[x,y]/(y^2 - x^3 +x) \to S =$ Spec $\mathbb C[x]$ gives a counterexample of a geometric nature, and $X =$Spec $\mathbb Z[\sqrt{-5}] \to S =$ Spec $\mathbb Z$ gives a counterexample of an arithmetic nature.<|endoftext|> -TITLE: Infinite-dimensional translation-invariant measure -QUESTION [5 upvotes]: Why is there no translation-invariant measure on an infinite-dimensional Euclidean space? Is there a reasonably short, insightful proof? -I am interested in an infinite-dimensional space with a definite inner product but not necessarily complete in the corresponding topology. Thus it need not be a Hilbert space. - -REPLY [14 votes]: The statement in the question "Why is there no translation-invariant measure on an infinite-dimensional Euclidean space?" is not correct. -(i) A counting measure defined in infinite-dimensional Euclidean space is an example of such measure which is translation-invariant. -(ii)There does not exist a translation-invariant Borel measure in an infinite-dimensional Euclidean space $\ell_2$ which gets the value 1 on the unit ball. Indeed, assume the contrary and let $\mu$ be such a measure. let $(e_k)_{k \in N}$ be a standard basis with $||e_k||=1$ for $k \in N$. Let $B_k$ be an open ball with center at $\frac{e_k}{2}$ and with radius $r$ less than $\frac{\sqrt{2}}{4}$. Then $(B_k)_{k \in N}$ is a family of pairwise disjoint open balls with radius $r$. On the one hand, $\mu$ measure of $B_k$ must be zero because in other case the $\mu$ measure of the unit ball will be $+\infty$. On the other hand, since $\ell_2$ is separable, $\ell_2$ can be covered by countably many translations of $B_1$ which together with an invariance of $\mu$ implies that $\mu$ measure of $\ell_2$ is zero. This is a contradiction and assertion (ii) is proved. -(iii) There exists a translation-invariant measure on an infinite-dimensional Euclidean space $\ell_2$ which gets the value 1 on the parallelepiped $P$ defined by $P=\{x : x \in \ell_2 ~\&~ ||\le \frac{1}{2^k}\}$. -Let $\lambda$ be infinite-dimensional Lebesgue measure in $R^{\infty}$ (see, Baker R., ``Lebesgue measure" on~$\mathbb{R}^{\infty}$,Proc. Amer. Math. Soc., vol. 113, no. 4, 1991, -pp.1023--1029). We set -$$ -(\forall X)(X \in {\cal{B}}(\ell_2) \rightarrow \mu(X)=\lambda(T(X))) -$$ -where ${\cal{B}}(\ell_2)$ denotes the $\sigma$-algebra of Borel subsets of $\ell_2$ -and the mapping $T : \ell_2 \to R^{\infty} $ is defined by: $T(\sum_{k \in N}a_ke_k)=(2^{k-1}a_k)_{k \in N}$. -Then $\mu$ satisfies all conditions participated in (iii). -P.S. There exist many interesting translation-invariant non-sigma finite Borel measures in infinite-dimensional separable Banach spaces(see, for example, G.Pantsulaia , On generators of shy sets on Polish topological vector spaces, New York J. Math.,14 ( 2008) , 235 – 261)<|endoftext|> -TITLE: Inclusion of $L^p$ spaces -QUESTION [10 upvotes]: Let $X \subset L^1(\mathbb{R})$ a closed linear subspace satisfying \begin{align} -X\subset \bigcup_{p>1} L^p(\mathbb{R})\end{align} - Show that $X\subset L^{p_0}(\mathbb{R})$ for some $p_0>1.$ - -I guess the problem is that in infinite measure spaces the inclusion $L^p\subset L^q$ only holds for $p=q$. Is it maybe possbile to apply Baire's Theorem in some way? - -REPLY [8 votes]: The comments posted below are related to a previous answer, which wasn't good. Now it's a new version, which is, I think, correct. -I think I have an approach which uses Baire's categories theorem. We define for an integer $k$ -$$F_k:=\{f\in X: \lVert f\rVert_{L^{1+1/k}}\leq k\}.$$ - -$F_k$ is closed (for the $L^1$ norm). Indeed, let $\{f_j\}\subset F_k$ which converges in $L^1$ to $f$. A subsequence $\{f_{j'}\}$ converges to $f$ almost everywhere, hence -$$\int_{\Bbb R}|f|^{1+1/k}dx=\int_{\Bbb R}\liminf_{j'}|f_{j'}|^{1+1/k}dx\leq - \liminf_{j'}\int_{\Bbb R}|f_{j'}|^{1+1/k}dx\leq k.$$ -We have $X=\bigcup_{k\geq 1}F_k$. Indeed, take $f\in X$; then $f\in L^p$ for some $p>1$. For $k$ large enough, $1+1/k\leq p$ and breaking the integral on the sets $\{|f|<1\}$, $\{|f|\geq 1\}$ -$$\lVert f\rVert_{L^{1+1/k}}^{1+1/k}\leq \lVert f\rVert_{L^1}+\lVert f\rVert_{L^p}^p,$$ -so -$$\lVert f\rVert_{L^{1+1/k}}\leq \left(\lVert f\rVert_{L^1}+\lVert f\rVert_{L^p}^p\right)^{1-\frac 1{k+1}}.$$ -The RHS converges to $\lVert f\rVert_{L^1}+\lVert f\rVert_{L^p}^p$, so it's smaller than two times this quantity for $k$ large enough. Now, just consider $k$ such that -$$2\left(\lVert f\rVert_{L^1}+\lVert f\rVert_{L^p}^p\right)\leq k.$$ - -By Baire's categories theorem, we get that a $F_{k_0}$ has a non-empty interior. That is, we can find $f_0\in F_{k_0}$ and $r_0>0$ such that if $\lVert f-f_0\rVert_{L^1}\leq r_0$ then $f\in F_{k_0}$. Consider $f\neq 0$ an element of $X$. Then $f_0+\frac{r_0f}{2\lVert f\rVert_{L^1}}\in F_{k_0}$. We have that -$$\left\lVert \frac{r_0f}{2\lVert f\rVert_{L^1}}\right\rVert_{L^{1+1/k_0}}\leq -\left\lVert f_0+ \frac{r_0f}{2\lVert f\rVert_{L^1}}\right\rVert_{L^{1+1/k_0}}+\lVert f_0\rVert_{L^{1+1/k_0}}\leq 2k_0,$$ -hence -$$\lVert f\rVert_{1+1/k_0}\leq \frac{4k_0}{r_0}\lVert f\rVert_{L^1},$$ -which proves the embedding. -For an example where the space is infinite dimensional, look at the answers here. - -A remark: we didn't use the fact that we worked on $\Bbb R$, and it seems it works for each measured space with a non-negative measure. That is, if $(S,\mathcal A,\mu)$ is a measured space with $\mu$ non-negative, and if $X$ is a closed subspace of $L^1(S,\mu)$ contained in $\bigcup_{p>1}L^p(X,\mu)$, then we can find $p_0$ such that $X\subset L^{p_0}(X,\mu)$.<|endoftext|> -TITLE: How to compute a generator of this cyclic quadratic residue group? -QUESTION [5 upvotes]: This question refers to the article: -G. Ateniese et al., "Remote data checking using provable data possession" -Let $p = 2p' + 1$ and $q = 2q' + 1$ be large primes. -$N = pq$. -Let $g$ be a generator of $QR_N$ (the set of quadratic residues modulo $N$). $QR_N$ is the unique cyclic group of $Z_N^*$ of order $p'q'$. -In the article, authors says: - -"We can obtain $g$ as $g = a^2 \mod N$, where $a \overset{R}{\leftarrow} Z_N^*$ such that $gcd(a \pm 1, N) = 1$" - -where $\overset{R}{\leftarrow}$ denotes the operation of "picking an element at random". -Actually I have two questions: -1) First, I do not exactly understand how to interpret the author's sentence. -Is it: -"[...] such that $gcd(a+1,N) = 1$ and $gcd(a-1,N) = 1$" ? -Or rather it is: -"[...] such that $gcd(a+1,N) = 1$ or $gcd(a-1,N) = 1$" ? -I think the second one (or) is the correct interpretation, but I'm not sure, and in any case I don't see why it is so. -Can any one provide me a proof? -2) Second, I'm wondering how to efficiently choose $a$ at random. -For a prime $p$ it is trivial to choose an element in $Z_p^*$ since $Z_p^* = \lbrace 1, \dots, p - 1\rbrace$. -But here $N$ is not prime. The only algorithm that comes to my mind is: - -repeat - pick a random number a in [1, N-1] -until gcd(a, N) = 1 - -Is there a more efficient way to compute $a$? -Thank you in advance! - -REPLY [4 votes]: The author means and. -You want $g$ to have multiplicative order $p'q'$ mod $N$, so $g$ is the square of an element $a$ of order $2p'q'$. Clearly we need to have $a$ relatively prime to $N$. -Necessity of condition -Calculating a gcd with $N$ is fancy talk for "Does $p$ or $q$ divide it?" If $p$ divides $a^2-1$, then consider $a^{q-1} -1$. It is divisible by $q$ by Fermat's little theorem, but it is also divisible by $p$ since $q-1 = 2q'$ and $a^2$ is congruent to 1 mod $p$. Hence if $p$ divides $a^2-1$, then $a$ only has order dividing $2q'$. Since $p$ is prime, $p$ divides $a^2-1 = (a-1)(a+1)$ if and only if it divides $a-1$ or $a+1$. Hence you need to make sure $p$ does not divide either of them. -So it is necessary that $\gcd(N,a\pm1)=1$ (where we mean and). -Sufficiency of condition -The multiplicative group mod $N$ has structure $C_2 \times C_{2p'q'}$ so the only orders other than $2p'q'$ and $p'q'$ are $2, p', q', 2p', 2q'$. We need to rule out these last 5 possible orders. -If the order of $a$ is 2, then $\gcd(a^2-1,N)=N$. -If $a^{p'} \equiv 1 \mod N$, then $a^{p'} \equiv 1 \mod q$, but $p'$ and $q-1=2q'$ are relatively prime (assuming $p\neq q$; never use $N=p^2$), so $a \equiv 1 \mod q$ and so $\gcd(a^2-1,N)=q$. -The other cases follow similarly. -Finding one quickly -We take advantage of the fact that almost every element works: -isGoodNumber = function( a, p, q ) { - local aModP, aModQ; - aModP = a mod p; - if ( ( aModP == 0 ) || ( aModP == 1 ) || ( aModP == (p-1) ) ) - return false; - aModQ = a mod q; - if ( ( aModQ == 0 ) || ( aModQ == 1 ) || ( aModQ == (q-1) ) ) - return false; - return true; -} - -How many bad numbers are there? Well $q=N/p$ are divisible by $p$, $p=N/q$ are divisible by $q$, so that is at most $3q + 3p$ bad numbers (the factor of 3 being "0, 1, or -1"). The actual number is a little lower. How many numbers are there total? $pq$. What percentage are good numbers? $\frac{pq - 3q - 3p}{pq} = 1 - \tfrac3p - \frac3q \approx 100\%$ as long as $p$ and $q$ are large. -In other words: -repeat a := Random( [1..N] ); -until (a mod p notin [0,1,p-1]) and (a mod q notin [0,1,q-1]); - -Is probably not actually going to repeat. It will probably find you a good $a$ on the first try. Don't forget to square it to get $g$. (If it was already a $g$, then squaring it won't hurt it.)<|endoftext|> -TITLE: Irreducible factors of $X^p-1$ in $(\mathbb{Z}/q \mathbb{Z})[X]$ -QUESTION [6 upvotes]: Is it possible to determine how many irreducible factors has $X^p-1$ in the polynomial ring $(\mathbb{Z}/q \mathbb{Z})[X]$ has and maybe even the degrees of the irreducible factors? (Here $p,q$ are primes with $\gcd(p,q)=1$.) - -REPLY [9 votes]: It has one factor of degree $1$, namely $x-1$. All the remaining factors have the same degree, namely the order of $q$ in the multiplicative group $(\mathbb{Z}/p \mathbb{Z})^*$. To see it: this is the length of every orbit of the action of the Frobenius $a\mapsto a^q$ on the set of the roots of $(x^p-1)/(x-1)$ in the algebraic closure.<|endoftext|> -TITLE: Need some help on a non-example of equicontinuity -QUESTION [20 upvotes]: In an attempt to better understand the definition of an equicontinuous family of continuous functions, I want to find a simple non-example. -My intuition says that the family $\{f_n\colon[0,1]\to\mathbb R\}_{n\in\mathbb N}$ given by $f_n(x)=x^n$ is not equicontinuous, but I do not know how to show this. -Any help is appreciated. - -REPLY [4 votes]: The problem is at $1$. Indeed, assume that $\{f_n\}$ is equi-continuous at $1$. Then we can find a $\delta>0$ such that if $|1-x|\leq \delta$ and $0\leq x\leq 1$ then for each integer $n$, $|f_n(x)-f_n(1))|\leq 1/3$, hence $|x^n-1|\leq 1/3$ and $x^n\geq 2/3$. If $1/{n^2}\leq \delta$ then we should have $\left(1-\frac 1{n^2}\right)^n\geq 2/3$. It's impossible, since the RHS converges to $0$. -In fact, we know by Arzela-Ascoli-'s theorem that a uniformly bounded sequence of equi-continuous functions on a compact interval admit a converging subsequence for the uniform norm (and this limit is continuous). We can check that the pointwise limit of $\{f_n\}$ is not continuous.<|endoftext|> -TITLE: Are the eigenvalues of the Hecke operators always real? -QUESTION [5 upvotes]: Are the eigenvalues of the Hecke operators $T_n$ for $M_k(\text{SL}_2(\mathbb{Z}))$ always real? I think I have an answer but I am not confident with my arguments. -If $f$ is a normalized eigenform, then $f$ have real Fourier coefficients. And if $f$ is a normalized eigenform, then its coefficients are precisely the eigenvalues of $T_n$. Thus $T_n$ must have real eigenvalues. Is this correct? - -REPLY [6 votes]: Yes, in this context (and more generally for newforms on $\Gamma_0(N)$) the eigenvalues of the $T_n$ must be real. What your argument shows is that -this is*equivalent* to the statement that a normalized eigenform has real Fourier coefficients. If you accept that the Fourier expansion is necessarily real, you -are done. -On the other hand, I would normally prove that the Fourier expansion is real by first proving that the Hecke eigenvalues are real. For this, I would use the fact that $T_n$ is self-adjoint for the Petersson inner product, and that the eigenvalues of a self-adjoint operator are necessarily real.<|endoftext|> -TITLE: Law of large numbers for Plancherel random Young diagrams -QUESTION [6 upvotes]: Do you know a reference book on the law of large numbers for random Plancherel Young diagrams ? I know the book of Kerov, but actually, it is only a compilation of his articles, and i need something more detailed. -Kerov-Vershik and Logan-Shepp independently proved in 1977 a law of large numbers for random Young diagrams ditributed according to the Plancherel measure. I am interested in the asymptotics and working on the article of Vershik-Kerov : "Asymptotic of the largest and the typical dimensions of irreducible representations of a symmetric group" 1985. -In particular, I need for my master's thesis to control the $\| . \|_{\infty}$ norm of the difference between the limit shape and the normalized shape of a Young diagram with n boxes. There is an interesting result proved in the paper that I mentioned (Theorem 3), but I'm not convinced by the proof, because the authors assume that the shape has a support in $[-a, a]$ where $a$ is a constant. But in my opinion, the length of the support can be of order $O(\sqrt{n})$. - -REPLY [2 votes]: A book on this subject: -"The Surprising Mathematics of Longest Increasing Subsequences", -A book in progress by Dan Romik. -You are interested in Section 1.17 (version 1.1 of the draft).<|endoftext|> -TITLE: If the diagram is commutative, $f$ is one-one and $g$ is onto. -QUESTION [7 upvotes]: Let $f:A \to B$ , $g:B \to A$, and $\operatorname{id}:A \to A$ be the identity. If the diagram -$\hspace{4cm} $ -commutes, prove that ${ f}$ is one-one and ${ g}$ is onto. -THEOREM If the diagram above commutes, then $ f$ is one-one and $g$ is onto. -PROOF The commutativity of the diagram implies that -$${g f}(x)=x \;; \forall x\in A$$ -thus $gf $ is one-one and onto. -$(1)$ $ f$ is one-one. Assume that ${ f}(x)={ f}(y)$.Then -$${ f}(x)={ f}(y)$$ -$${ gf}(x)={ gf}(y)$$ -$$x=y$$ -so that $ { f}$ is one-one. -$(2)$ $g$ is onto. If it wasn't the case $gf $ wouldn't be onto, which is impossible. - -I don't know wether the last "proof" is right. I can see why it must be the case but I'm not able to exlain it and give a clear proof like that of $f$ being one-one which seems very clear. - -REPLY [9 votes]: Theorem. Let $A$ and $B$ be sets. Consider the following three statements: - -$f\colon A\to B$ is one-to-one. -There exists $g\colon B\to A$ such that $gf = \mathrm{id}_A$ ($f$ has a left inverse). -For every set $C$ and every functions $h,k\colon C\to A$, if $fh = fk$ then $h=k$ ($f$ is left-cancellable). - -Then 2$\implies$3$\iff$1. Moreover, if $A$ is nonempty, then 1$\implies$2 so all three are equivalent. -Proof. -2$\implies$3. Let $h,k\colon C\to A$ be such that $fh = fk$. Let $g$ be the function guaranteed by 2. Then -$$h = \mathrm{id}_Ah = (gf)h = g(fh) = g(fk) = (gf)k = \mathrm{id}_Ak = k.$$ -Thus, $f$ is left-cancellable. -3$\implies$1. Let $a,a'\in A$ be such that $f(a)=f(a')$. We need to prove that $a=a'$. Let $C=\{0\}$, $h\colon C\to A$ be given by $h(0)=a$, and $k\colon C\to A$ be given by $k(0)=a'$. Then $fh(0) = f(a) = f(a') = fk(0)$. so $fh=fk$. Since $f$ is left-cancellable, we conclude that $h=k$. Hence $a = h(0) = k(0) = a'$, proving that $f$ is one-to-one. -1$\implies$3. Let $C$ be a set and $h,k\colon C\to A$ be such that $fh = fk$. We need to show that $h=k$. Let $c\in C$. Then $f(h(c)) = f(k(c))$; since $f$ is one-to-one, we conclude that $h(c)=k(c)$. Since $h(c)=k(c)$ for all $c\in C$, it follows that $h=k$. -1$\implies$2 if $A$ is nonempty and $B$ are nonempty. Since $A$ is nonempty, there exists $a_0\in A$. Define $g\colon B\to A$ as follows: -$$g(b) = \left\{\begin{array}{ll} -a &\text{if }b\in f(A)\text{ and }f(a)=b;\\ -a_0&\text{if }b\notin f(A). -\end{array}\right.$$ -This is well-defined, since $f(a)=f(a')=b$ implies $a=a'$. And if $b\in f(A)$, then there exists $a\in A$ such that $f(a)=b$. Now, let $a\in A$. Then $g(f(a)) = a$, so $gf = \mathrm{id}_A$, as desired. $\Box$ -In particular, given a function $f\colon A\to B$, if a $g$ exists that makes the diagram commute, then $f$ is one-to-one. Conversely, if $f$ is one-to-one and $A$ is nonempty (or $A$ and $B$ are both empty), then we can find a $g$ that makes the diagram commute. -Here's the dual: -Theorem. Let $A$ and $B$ be sets. Consider the following three statements: - -$f\colon A\to B$ is onto. -There exists $g\colon B\to A$ such that $fg=\mathrm{id}_B$ ($f$ has a right inverse). -For every set $C$ and every functions $h,k\colon B\to C$, if $hf = kf$, then $h=k$ ($f$ is right-cancellable). - -Then 2$\implies$3$\iff$1. Moreover, the implication 1$\implies$2 is equivalent to the Axiom of Choice. -Proof. 2$\implies$3 Let $C$ be a set and let $h,k\colon B\to C$ be such that $hf=kf$. Let $g$ be the function guaranteed by 2; then: -$$h = h\mathrm{id}_B = h(fg) = (hf)g =(kf)g = k(fg) = k\mathrm{id}_B = k.$$ -3$\implies$1. Let $C=\{0,1\}$ and define $h\colon B\to C$ by $h(b)=1$ for all $b$, and -$$k(b) = \left\{\begin{array}{ll} -1 & \text{if }b\in f(A);\\ -0 &\text{if }b\notin f(A). -\end{array}\right.$$ -Then $hf = kf$, hence by 3 we have $h=k$. Therefore, $k(b)=1$ for all $b\in B$, hence $f(A)=B$; that is, $f$ is onto. -1$\implies$3. Suppose that $C$ is a set and $h,k\colon B\to C$ are functions such that $hf = kf$. Let $b\in B$; we need to show $h(b)=k(b)$. Since $f$ is onto, there exists $a\in A$ such that $f(a)=b$. Therefore, -$$h(b) = h(f(a)) = hf(a) = kf(a) = k(b).$$ -Thus, $h=k$. -If the Axiom of Choice holds, then 1$\implies$2: If $f$ is onto, then for every $b\in B$, the set $f^{-1}(b) = \{a\in A\mid f(a)=b\}$ is nonempty. By the Axiom of choice, there exists a function $g\colon B\to \cup_{b\in B} f^{-1}(b)$ such that $g(b)\in f^{-1}(b)$ for each $b\in B$. I claim that $fg=\mathrm{id}_B$. Indeed, for every $b\in B$, $g(b)\in f^{-1}(b)$, so $f(g(b))=b$. -If 1$\implies$2 holds, then the Axiom of Choice holds. Let $\mathcal{X}=\{A_i\}_{i\in I}$ be a nonempty family ($I\neq\varnothing$) of nonempty sets ($A_i\neq\varnothing$ for each $I\in I$). We need to show that there exists a function $g\colon I\to\cup_{i\in I}A_i$ such that $g(i)\in A_i$ for each $i\in I$. -Let $B_i = A_i\times\{i\}$. Note that the family $\mathcal{Y}=\{B_i\}_{i\in I}$ consists of pairwise distinct sets. Let $Y=\cup_{i\in I}B_i$, an define $f\colon Y\to I$ by $f(b_i,i) = i$ (projection onto the second component). The map is onto, since each $A_i$ is nonempty, so $B_i\neq\varnothing$. By our assumption that 1$\implies$2, there exists $h\colon I\to Y$ such that $h(i)\in B_i$ for each $i\in I$. Let $\pi_i\colon B_i\to A_i$ be the projection onto the first coordinate, $\pi_i(b_i,i) = b_i$. Define $g\colon I\to \cup_{i\in I}A_i$ by -$$g(i) = \pi_i\circ h (i).$$ -Since $h(i)\in B_i=A_i\times\{i\}$, then $\pi_i\circ h(i) \in A_i$. This holds for each $i$, so $g$ is the desired choice function. $\Box$ -In particular, if the diagram commutes, then $g$ is onto. Conversely, assuming the axiom of choice, if $g$ is onto, then we can find an $f$ that fits into the diagram and makes it commute.<|endoftext|> -TITLE: Explain Carmichael's Function To A Novice -QUESTION [8 upvotes]: I understand that the Carmichael Function (I'm going to call λ) is essentially the smallest positive integer m, where $a^m$ is congruent $1 \pmod n$ for all $a$ co-prime to $n$ and less than $n$. -6 makes sense to me. The only co-prime is 5 and $5^2 = 25\equiv 1 \pmod 6$. So $λ(6) = 2$. -I also know that the answer to $λ(49)$ is $42$. I'm not clear on what the "shortcut" is though. I realized I could brute force the answer, but that's not feasible for any significant values of n. I understand there is some sort of mathematical "shortcut" to derive this answer... something to do with LCM of prime powers. I would appreciate an explanation on this shortcut in terms an avg 14 year old could understand. No Fermat or Euler. - -REPLY [15 votes]: The first thing we want is this: - -If $\gcd(a,b) = 1$ and $a|bc$, then $a|c$. - -This is fairly easy to establish. For example, you can use the Bezout identity: $a$ and $b$ are coprime if and only if there exist $x$ and $y$ such that $ax+by=1$. Multiplying through by $c$ we get $c = acx + bcy$. Since $a$ divides both $acx$ and $bcy$ (the latter by virtue of dividing $bc$), then $a$ divides $c$. -From this we have: - -Suppose that $a$ and $n$ are coprime integers, and $x$ and $y$ are integers such that $ax\equiv ay\pmod{n}$. Then $x\equiv y\pmod{n}$. - -Indeed, if $ax\equiv ay\pmod{n}$, the $n$ divides $ax-ay = a(x-y)$. Since $\gcd(n,a)=1$ by assumption, it follows that $n|x-y$, so $x\equiv y\pmod{n}$. -Hence, for each $a$ that is coprime to $n$, there is at least one positive integer $k$ such that $a^k\equiv 1\pmod{n}$: - -If $\gcd(a,n)=1$, then there exists a positive integer $k$ such that $a^k\equiv 1\pmod{n}$. - -There are only finitely many positive integers smaller than $n$ that are coprime to $n$. Let's say there are $t$ of them. Then the numbers $a\bmod n$, $a^2\bmod n$, $a^3\bmod n,\ldots, a^{t+1}\bmod n$ cannot all be distinct, because each of them is coprime to $n$, and there are $t+1$ of them, all smaller than $n$. So there must exist exponents $i$ and $j$, $i\lt j$, such that $a^i\equiv a^j\pmod{n}$. Write $j=i+\ell$, and we have -$$a^i(1) \equiv a^ia^{\ell}\pmod{n}.$$ -Since $\gcd(a^i,n)=1$, then we can cancel and we get $1\equiv a^{\ell}\pmod{n}$, which shows there must be some exponent that works. - -If $a^k\equiv 1\pmod{n}$, and $r$ is a multiple of $k$, then $a^r\equiv 1\pmod{n}$. - -This is fairly straightforward: we can write $r=kt$ for some $t$. Then using the laws of exponents, we have: -$$a^r = a^{kt} = (a^k)^t \equiv 1^t \equiv 1\pmod{n}.$$ -So, given a positive integer $n$, why does there exist a smallest $m$ such that $a^m\equiv 1\pmod{n}$ for every positive $a$ that is smaller than $n$ and coprime to $m$? First, note that there has to exist some $m$ that works: there are only finitely many positive integers coprime to $m$; call them $a_1,\ldots,a_f$. For each of them, we know there exists some positive integer $k_1$, $k_2,\ldots,k_f$, with the property that $a_i^{k_i}\equiv 1\pmod{n}$. Now, if $k=k_1k_2\cdots k_f$, then $k$ is a multiple of each $k_i$, so $a_i^k\equiv 1\pmod{n}$. -That is, there certainly are positive integers that "work" for all $a$ coprime to $n$, positive, and smaller than $n$. Since there are such integers, there must be a smallest one. That smallest one is the value of the Carmichael function. -Using the division algorithm, it is easy to check that in fact any number that "works" is a multiple of the smallest one. -Now, as to the "short-cut": it relies on two facts: - -We know what the value is for prime powers. And -The Carmichael function satisfies the following: if $m$ and $n$ are relatively prime, then $C(mn) = \mathrm{lcm}(C(m),C(n))$. - -The latter assertion follows from the Chinese Remainder Theorem: the positive integers smaller than $mn$ that are coprime to $mn$ are in one-to-one correspondence to pairs of integers $(a,b)$ with $0\lt a\lt m$, $0\lt b\lt n$, and $\gcd(a,m)=\gcd(b,n)=1$ (namely, given $x$, $0\lt x\lt mn$ coprime to $mn$, set $a=x\bmod m$ and $b=x\bmod n$ (the Chinese Remainder Theorem guarantees that this is a bijection). And $x^r\equiv 1\pmod{mn}$ if and only if $a^r\equiv 1\pmod{m}$ and $b^r\equiv 1\pmod{n}$. Thus, any number that is a multiple of both $C(m)$ and $C(n)$ will "work" for $mn$, and any number that "works" for $mn$ must be a multiple of $C(m)$ and of $C(n)$. Hence, the set of integers that "work" are the common multiples of $C(m)$ and $C(n)$, and so the smallest integer that works is the least common multiple. -As to the value in the prime factors, for odd primes, this is just Euler's Theorem; it takes a bit of work, but one can show that if $p$ is an odd prime, then $C(p^r) = (p-1)p^{r-1}$. -For $p=2$, we have that $C(2)=1$, $C(4) = 2$, and $C(2^{r}) = 2^{r-2}$ if $r\geq 3$. -These formulas give you a way to compute $C$ for any integer that you can factor into primes.<|endoftext|> -TITLE: Functions that take rationals to rationals -QUESTION [47 upvotes]: What is known about $\mathcal C^\infty$ functions $\mathbb R\to\mathbb R$ that always take rationals to rationals? Are they all quotients of polynomials? If not, are there any that are bounded yet don't tend to a limit for $x\to +\infty$? If there are, then can we also require them to be analytic? -(This is basically just a random musing after I found myself using trigonometric functions twice in unrelated throwaway counterexamples. It struck me that it was kind of conceptual overkill to whip out a transcendental function for the purpose I needed: basically that it had to keep wiggling forever. I couldn't think of a nice non-trancendental function to do this job, however, and now wonder whether that is because they can't exist). - -REPLY [16 votes]: The references given by Dave Renfro and Cocopuffs did not completely answer my question, but gave enough inspration that I think I've got it now. After several false starts: -Theorem. There exists an analytic function $\xi:\mathbb R\to[0,1]$ such that - -$\xi(\mathbb Q)\subseteq \mathbb Q$. -$\xi(x)$ does not tend to a limit for $x\to\infty$. - -Proof. Let $q_1, q_2, \ldots$ be a fixed enumeration of the rationals. Define by simultaneous induction a sequence of functions $f_0, f_1, f_2\ldots$ and integers $m_0 -TITLE: Is the set of all rationals with denominators less than $10^6$ closed in $\mathbb{R}$? -QUESTION [5 upvotes]: I'm wondering if the set of all rationals with denominators less than $10^6$ is closed in the real number system. I think it is not, so that's what I've been trying to prove. I've tried looking at its complement and showing that it is open, but I didn't progress much. My last resort is to show that there is a sequence of rationals in my set converging to an irrational number (or to a rational whose denominator is greater than $10^6$). The problem is, how do I ensure that all simplified fractions in my sequence will have denominator less than $10^6$? - -REPLY [10 votes]: Of course it's closed. Just blow it up by a factor of $(10^6)!$, and all its elements are integers.<|endoftext|> -TITLE: $f$ has an essential singularity in $z_0$. What about $1/f$? -QUESTION [14 upvotes]: Let $\Omega$ be a non-empty, open subset of $\mathbb C$. Consider an holomorphic function $f:\Omega \setminus \{z_0\} \to \mathbb C$ and suppose we know $z_0$ is an essential singularity of $f$. -I am wondering what we can say about the function $\tilde{f}:=\frac{1}{f}$ and its singularity in $z_0$. Do you know any theorem that answers to this question? -Actually, I can't prove anything, since I do not know the answer: I've studied some examples. For instance, if you take $f(z)=e^{\frac{1}{z}}$ then $\tilde{f}$ will still have an essential singularity, hasn't it? -On the other side, if we take $f(z)=\sin(\frac{1}{z})$ then I think that $z_0=0$ becomes a limit point of poles for $\tilde{f}$ (so we can't classify it, because it isn't an isolated singularity). -Wha do you think? Do you know any useful theorem concerning this? Thank you in advance. - -REPLY [13 votes]: Your two examples essentially span all the possibilities. By the Big Picard Theorem, we know that $f$ assumes all but one value in $\Bbb{C}$ infinitely often, in any neighborhood of $z_0$. -If the missed value is $0$, then $\frac{1}{f}$ has an isolated singularity at $z_0$, and it must clearly be essential (since otherwise $f$ itself would have a pole or removable singularity). -If the missed value is not $0$, or if there is no missed value, then $\frac{1}{f}$ will have a sequence of poles that converges to $z_0$, and hence the singularity at $z_0$ will not be isolated. So, thinking of $\frac{1}{f}$ as a holomorphic function on its largest possible domain (which will omit some sequence of points converging to $z_0$), the singularity at $z_0$ is technically unclassifiable. -On the other hand, even in this case you can regard $\frac{1}{f}$ as a meromorphic function from $\Omega \backslash \{z_0\}$ to the extended complex plane $\hat{\Bbb{C}}$, and when considered as such it will have an essential singularity at $z_0$. This is a natural enough thing to do that I suspect most people would be happy just saying "$\frac{1}{f}$ has an essential singularity at $z_0$."<|endoftext|> -TITLE: Irreducible polynomial not attaining squares over finite field -QUESTION [6 upvotes]: Is it possible to construct an irreducible polynomial $f$ over $\mathbb{F}_{q}$ such that $f(x)$ is a non-square for any $x \in \mathbb{F}_{q}$? -I can prove the existence of irreducible polynomials (Euclid's argument), and I can construct polynomials with no square values (for example by Lagrange interpolation through non-squares), but satisfying these 2 conditions feels difficult. -Motivation: I am trying to construct an hyper-elliptic curve $y^2 = f(x)$ with no rational points. -EDIT: Can you construct such an $f$ that will not be constant on the ground field $\mathbb{F}_{q}$? - -REPLY [3 votes]: Let $h$ be any function $\mathbb{F}_q \to \mathbb{F}_q \setminus \{ 0 \}$. Then there is an irreducible polynomial $f$ such that $f(x) = h(x)$ for all $x \in \mathbb{F}_q$. In particular, if we choose the values of $h$ to be non-squares, then there is an irreducible polynomial achieving these values. For $p \geq 5$, this answers the question of whether we can make $f$ nonconstant. -Proof By Lagrange interpolation, there is some polynomial $b(x)$ which coincides with the function $h$ on $\mathbb{F}_q$. Consider polynomials of the form $f(x) = a(x) (x^q-x) + b(x)$. Any such $f$ is equal to $h$ on $\mathbb{F}_q$. We have $GCD(x^q-x, b(x))=1$, since $h$ is nonzero. Dirichlet's theorem on primes in arithmetic progressions holds for $\mathbb{F}_q[x]$ (see Chapter 4 in Rosen's Number theory in function fields), so there is an irreducible polynomial of the form $(x^q-x) a(x) + b(x)$.<|endoftext|> -TITLE: For a trigonometric polynomial $P$, can $\lim \limits_{n \to \infty} P(n^2) = 0$ without $P(n^2) = 0$? -QUESTION [10 upvotes]: Disclaimer: The original version of this question focused on $2^n$ in lieu of $n^2$. It is in the hope that the question is easier with $n^2$ that I changed it. -I have an always-nonnegative (on the nonnegative integers) trigonometric polynomial¹ $P$: -$$P(n) = \mathcal{R}\left(\sum_{j=1}^k a_j e^{i \theta_j n}\right),$$ -with $\lim \limits_{n \to \infty} P(n^2) = 0$. -I want to show that $P(n^2) = 0$. -Here are a few basic facts on trigonometric polynomials that may help. A -trigonometric polynomial is an almost-periodic function. This implies in -particular that: - -For $\epsilon > 0$, there is a number $L=L(\epsilon)$ such that any -interval of length $L$ on the real line has an $\epsilon$-translation -integer, that is, an integer $t$ such that $|P(n) - P(n +t)| < \epsilon$ -for any $n$. -From any sequence $\{P(n + m_k)\}$, one can extract a subsequence which -converges uniformly with respect to $n$. Moreover the function to which it -converges is almost-periodic itself. ($\{m_k\}$ is an arbitrary sequence -of numbers). -In particular, for any sequence $\{m_k\}$, for any $\epsilon$, there exist -$i \neq j$ such that $m_i - m_j$ is an $\epsilon$-translation number. - -An even simpler case is to consider that $\lim \limits_{n \to \infty} P(n) = -0$. A proof that it implies that $P(n) = 0$ for any $n$ is the following. -Suppose there is an $m$ such that $P(m) > 0$, and set $\epsilon = P(m)/2$. -Now for any $N$, there exists an $\epsilon$-translation number $t$ of $P$ with $t -> N$ an integer. Then $|P(m) - P(m+t)| < \epsilon$, thus $P(m+t) > -\epsilon$. Thus $\lim \limits_{n \to \infty} P(n)$ is not null. -¹: Trigonometric polynomial is taken in the sense of, say, Corduneanu (in -Almost Periodic Functions). Wikipedia seems to have a different definition. - -REPLY [2 votes]: [Edit: This answer concerns a previous version of the question.] -The following is a counter-example (assuming "positive" is meant to be read "non-negative"): -$$P(x)=\left(1+\cos\pi(x-1)+\cos(\sqrt2-1)\pi(x-1)+\cos\sqrt2\pi(x-1)\right)^2$$ -It is trivially non-negative for all $x\in\Bbb R$, since it is a square. For all $k\in\Bbb N_0$ we have: -$$\begin{align}P(2k)&=0\tag{1}\\P(2k+1)&=16 \cos^4k \sqrt2\pi\tag{2}\end{align}$$ -which we can show using the usual trigonometric formulas: -$$\begin{align}P(2k)&=\left(1+\cos\pi(2k-1)+\cos(\sqrt2-1)\pi(2k-1)+\cos\sqrt2\pi(2k-1)\right)^2\\&=\left(1+(-1)+\cos(\sqrt2\pi(2k-1)-\pi(2k-1))+\cos\sqrt2\pi(2k-1)\right)^2\\&=\left(\cos(\sqrt2\pi(2k-1)+\pi)+\cos\sqrt2\pi(2k-1)\right)^2\\&=\left(-\cos(\sqrt2\pi(2k-1))+\cos\sqrt2\pi(2k-1)\right)^2\\&=0\\ \\P(2k+1)&=\left(1+\cos\pi2k+\cos(\sqrt2-1)\pi2k+\cos\sqrt2\pi2k\right)^2\\&=\left(1+1+\cos(2\sqrt2k\pi-2k\pi)+\cos2\sqrt2k\pi\right)^2\\&=\left(2+2\cos2\sqrt2k\pi\right)^2\\&=\left(4\cos^2\sqrt2k\pi\right)^2\\&=16 \cos^4 \sqrt2k\pi\end{align}$$ -Also, $P$ is not a periodic function, which I think we can most likely show by choosing an arbitrary $T\neq 0$ and showing that $P(x+T)-P(x)$ is non-zero for some $x\in\Bbb R$. This seems a bit tedious, so I shall supply the calculation later if I have some more time. [Edit: As user10676 notes, this calculation is not necessary. $P$ is not periodic because $P(x)=16$ only for $x=1$.] -It is quite clear though that the period cannot be an integer, which can be seen directly by examining $(1)$ and $(2)$ and using the irrationality of $\sqrt2$. Since this is what the question is asking about, we have indeed found a relevant counter-example.<|endoftext|> -TITLE: Gerrymandering on a high-genus surface/can I use my powers for evil? -QUESTION [77 upvotes]: Somewhat in contrast to this question. -Let's say the Supreme Court has just issued a ruling that the upper and lower roads of an overpass need not be in the same congressional district. This makes states with lots of overpasses into high-genus surfaces for the purpose of gerrymandering. I, an unscrupulous mathematician, get a call from my state legislature, asking me to consult for them about how best to place new roads in order to achieve their desired district layout, in exchange for a generous fee. -Specifically, my state is a rectangle with no existing roads or overpasses. It has $n$ congressional districts. The state legislature has in mind a collection $\{P_{i,j}\}$ of convex polygons with nonempty interior which tile the state, where $i$ ranges from $1$ to $n$ and $j$ from $1$ to at most $m$ for each $i$, and wishes for each $P_{i,j},P_{i',j}$ to be in the same congressional district, thus must be connected by roads (smooth curves). Any time a road passes through a polygon of a different congressional district, an overpass must be built across it to avoid disrupting the other district. Any time two roads meet, an overpass must also be built. Note that for any fixed $i$, some but not all of the $P_{i,j}$ may be empty. The legislature wants to know the minimum number $\mathcal O(n,m)$ of overpasses that need to be built in order to facilitate their gerrymandering, in the worst-case scenario. -Equivalently, take a tiling of the unit square $S$ by convex polygons $P_{i,j}$, and $n$-color them such that at most $m$ polygons have the same color. For each color $i$, choose smooth curves $\lambda_{i1},\ldots,\lambda_{ir_i}:[0,1]\to S$ such that -$$\bigcup\limits_{j=1}^m P_{i,j}\cup \bigcup\limits_{k=1}^{r_i}\mathrm{im}(\gamma_{ik})$$ is connected. For each color $i$, choose points $p_{i1},\ldots,p_{is_i}$ such that -$$\left(\bigcup\limits_{j=1}^m P_{i,j}\setminus \left(\bigcup\limits_{i'\neq i}\bigcup\limits_{k=1}^{r_{i'}}\mathrm{im}(\gamma_{i'k})\right)\right)\cup \{p_{i1},\ldots,p_{is_i}\}$$ -is connected. If we can choose the curves and points freely, then $\mathcal O(n,m)$ is the worst-case (among all choices of tiling and coloring) minimum possible value of $\sum\limits_{i=1}^n s_i$. -Clearly $\mathcal O(n,1)=0$. A very rough upper bound is $\mathcal O(n,m)\leq n(m-1)(nm-1)$ since we can connect the necessary districts by drawing $n(m-1)$ curves, none of which touch the boundary and each of which intersects at most all $\leq nm-1$ other districts, and intersect each polygon in straight lines. A rough lower bound comes from dividing the square into $nm$ vertical strips and coloring them cyclically, i.e. red, yellow, blue. red, yellow, blue etc. Then any path which connects all districts of the same color must pass through at least all but one district of each other color, so $\mathcal O(n,m)\geq (n-1)(m-1)$. -Can these bounds be improved upon, or even an exact formula be found? -Note: For those of you not in the United States, this question won’t make much sense unless you’re familiar with gerrymandering. -Edit: This question was inspired by my state legislature (Kansas), which tried to make a congressional district which consisted of the conservative western third of the state, a road going across the state, and two liberal cities along the eastern border. -Edit 2: For the purposes of this problem, I assume that at most two roads meet at any overpass. - -REPLY [2 votes]: Can we not ignore multiplicities of overpasses, at least in the second formulation of the problem? We know that each polygon with a path through it requires at least one overpass. But if we have a polygon with any number of paths through it, such as: - -Can we not replace these path segments with a piecewise linear path the goes from one boundary point, which remains the same, to some central overpass, to another boundary point. This would give the following: - -Which doesn't affect anything outside of the polygon and reduces all paths inside to one overpass. This gives an upper bound of $mn$.<|endoftext|> -TITLE: Finding Asymptotes of Hyperbolas -QUESTION [8 upvotes]: To find a asymptote its either b2/a2 or a2/b2 depending on the way the equation is written. -With the problem -$$\frac{(x+1)^2}{16} - \frac{(y-2)^2}{9} = 1$$ -The solutions the sheet I have is giving me is $3/4x - 3/4$ and $3/4 x + 5/4$ -I thought it was just supposed to be $\pm 3/4x$. - -REPLY [5 votes]: For the hyperbola $$\dfrac{(x - h)^2}{a^2} - \dfrac{(y - k)^2}{b^2} = 1$$ -The asymptotes are $y - k = \pm \dfrac{b}{a}(x - h)$. -You could leave your answer as $y - 2 = \pm \dfrac{3}{4}(x + 1)$, or write two separate equations. - -Edit... If you do write separate equations, you'll have -$y - 2 = \dfrac{3}{4}(x + 1)$ and $y - 2 = - \dfrac{3}{4}(x + 1)$, which are, in "slope-intercept form": -$y = \dfrac{3}{4} x + \dfrac{11}{4}$ and $y = - \dfrac{3}{4}x + \dfrac{5}{4}$<|endoftext|> -TITLE: How does a harmonic oscillator with nonlinear damping behave? -QUESTION [13 upvotes]: It is well known that for a harmonic oscillator with linear damping, -$$\ddot x+c\dot x+x=0$$ -with positive $c$, the amplitude of the oscillations decays exponentially when $c<2$. If it is higher than $2$, the system fails to oscillate at all and is said to be overdamped. -Suppose the damping is nonlinear instead, following a power law -$$\ddot x+c\lvert \dot x\rvert^{p-1}\dot x+x=0.$$ -For example, $p=1$ recovers linear damping, while $p=2$ gives quadratic damping which can model aerodynamic drag. I assume that in general a closed-form solution is not possible due to the presence of the absolute value signs. What can be said about the asymptotic behaviour of the system? -Edit: While @doraemonpaul's comment and @mjqxxx's answer are very nice, I am more interested in stronger results than merely the existence or absence of overdamping. For comparison, consider a first-order nonlinear decay equation, -$$\dot x+\lvert x\rvert^{p-1}x=0.$$ -The solution to this has the form $x = \pm(p-1)(t-t_0)^{1/(1-p)}$ with certain conditions on $t_0$. When $p<1$, the solution drops to zero in finite time; when $p>1$, it decays roughly as $t^{-1/(p-1)}$ which is much slower than exponential. What are the corresponding characterizations of how the amplitude of the nonlinearly damped harmonic oscillator behaves? What is the exponent of the decay when $p > 1$? Can the system come to rest in finite time if $p < 1$? - -REPLY [4 votes]: So, following the usual treatment of the harmonic oscillator, note that the equations of motion for $x$ and $v$ are -$$ -\begin{eqnarray} -\dot{x} &=& +v \\ -\dot{v} &=& -x - c|v|^{p}v, -\end{eqnarray} -$$ -where the damping term is clearly small for small $v$ when $p>1$. In the two-dimensional phase space, the equations of motion become particularly simple in radial coordinates: letting $x=r\cos\theta$ and $v=r\sin\theta$, we have -$$ -\begin{eqnarray} -\dot{r}\cos\theta - r\dot\theta\sin\theta &=& +r\sin\theta \\ -\dot{r}\sin\theta + r\dot\theta\cos\theta &=& -r\cos\theta - cr^p\lvert\sin\theta\rvert^{p-1}\sin\theta, -\end{eqnarray} -$$ -or, solving the system, -$$ -\begin{eqnarray} -\dot{r} &=& -c r^{p}\lvert\sin\theta\rvert^{p-1}\sin^{2}\theta\\ -\dot{\theta} &=&-1-cr^{p-1}\lvert\sin\theta\rvert^{p-1}\sin\theta\cos\theta. -\end{eqnarray} -$$ -For $p>1$, the motion is always underdamped. Since $r$ is always decreasing, there comes a time where $r$ is small enough that we can guarantee that $|\dot{r}| -TITLE: Nilpotent Lie group homeomorphic to $\mathbb R^n$? -QUESTION [7 upvotes]: Is it true that a connected, simply connected, nilpotent $n$-dimensional Lie group $G$ is homeomorphic to $\mathbb R^n$? -EDIT: -Maybe a possible argument is the following: Since $G$ is simply connected, $G$ cannot contain any non-trivial maximal compact subgroups. By a theorem associated to Iwasawa and Malcev, all maximal compact subgroups are conjugate and thus have the same dimension. By a theorem by Hochschild(?), $G/K$ is diffeomorphic to $\mathbb R^n$, where $K$ is a(ny) maximal compact subgroup. But for $G$ simply connected, connected, nilpotent, $K$ must be trivial, whence $G$ itself is diffeomorphic to $\mathbb R^n$. - -REPLY [2 votes]: Let $G$ be a simply connected nilpotent Lie group with Lie algebra $\mathfrak g$. The nilpotency of $G$ implies that the Baker-Campbell-Hausdorff formula has only finitely many terms, so it converges everywhere on $\mathfrak g$. So it can be used to define a group law $\times$ on the Lie algebra $\mathfrak g$ such that $(\mathfrak g,\times)$ is a Lie group whose Lie algebra is isomorphic to $\mathfrak g$. Since two simply connected Lie groups with the same Lie algebra are isomorphic, $G$ is isomorphic to $(\mathfrak g,\times)$. In particular, $G$ is homeomorphic to $\mathfrak g$.<|endoftext|> -TITLE: Graduate school self-doubt (currently an undergraduate) -QUESTION [25 upvotes]: I don't know if this is the right place for such a question, so please redirect me to a more appropriate place if necessary. -My main question is, if you go to a small school, are you doomed to fail right from the get-go? My secondary question is, what are my options then, if I am indeed doomed to "fail"? The following is more of a personal description, so obviously it doesn't apply to everyone, which would make the question "not general enough", but I suspect that there are many others caught in a similar situation. -I'm going to be a senior (fourth-year) in the fall (August) and am currently doing my "homework" on graduate school admissions. I go to a small liberal arts college you've probably never heard of. From what I've read and heard, I'm extremely discouraged and I'd say I'm borderline depressed about my future, but the truth is, I don't really want to do anything else other than learn more math (and get a PhD in the process if I'm capable of it). -I like to think that I've worked very hard (but everyone says the same thing). Classes I've taken include topology, complex analysis, algebra, real analysis, statistics, graph theory, number theory, and Galois theory. I've been told that none of my coursework means much to graduate schools because my school is relatively unknown. Furthermore, much as I can get some good recommendation letters, I've heard that since none of my letter writers is "relatively well-known" in their area, and everyone who applies to graduate schools gets glowing reviews anyway, I can't compete with other applicants at all. -I've gone through all of Herstein, a little more than half (i.e., as much as I could) of Rudin, most of Hoffman/Kunze, and the first half of Munkres. None of these texts are actually used at my school. In other words, I don't think I've given any less than I'm physically capable of in the last three years, and if that isn't enough, I really shudder to think what is. -I haven't taken the GRE yet but am currently studying for it. But it's no secret that the GRE is a relatively silly measure of grad school success. A bad score would condemn me to eternal hell, but a good score isn't a direct ticket to success. -I've heard the "anything can happen ... who's to say that you can't succeed ... people from small schools have gotten into big schools" argument, but in all realisticness, there's absolutely no reason to pick me over a Princeton graduate who has the exact same credentials, and there are tons of them. So what are my best options given this reality? -Any advice/encouragement/experiences would be very much appreciated. Thank you. - -REPLY [2 votes]: Haha, you're worried waaay too much! If you're studying out of those books and have a good solid foundation on what you've learned, then you should be fine. However, the probability of you being accepted, and as a consequence passing, any PhD program in Mathematics is up to you. Allow me to explain what I mean (though this was posted 3 years ago, it may be beneficial to anyone who has the same question now). -You're an undergraduate now. What do you have to show for the fact that you are actually a mathematician and that you can not only pass graduate course classes and advance the field? Here are just some example questions you may want to consider: - -How are your grades in your courses? Is your GPA high? -How much did you retain from your courses? For example, if I asked you what the definition of a vector space is, could you list out the criterion for me? -What research have you done in the field? This is often a really big determining factor on acceptance in graduate school, especially if you've done research and made rediscoveries of your own. -Was your research useful? If you're working in a particular field, did your rediscovery contribute to the field itself? -Could you actually handle the course-load? This isn't just a factor of timing or difficulty. Recruiters will base this off of your GPA, and your GRE, since those are the most official ways to test your knowledge and ingenuity. However, you will be asked what you're going to shoot for in the future. Make sure you do your homework on the school, what kind of mathematics the school mostly centers around, and the results they've recently ascertained through research. Speaking of which, -Are you up-to-date on the most recent mathematical discoveries and unsolved mathematical problems? This isn't necessarily a big issue, but it may help out with decision making, along with some of the factors above. - -Most students do not have all of this criteria, and make it into graduate school in mathematics, but the biggest things are your thesis and your GPA. The books you're using is fine. If it helps, talk to your research advisor and get his/her advice on it as well. Hope this helps!<|endoftext|> -TITLE: Why is a strict $p$-ring whose residue ring is a field necessarily local? -QUESTION [7 upvotes]: Let $A$ be a strict $p$-ring. Recall that this means $A$ is $p$-adically separated and complete, $p:A\rightarrow A$ is injective, and $A/pA$ is a perfect $\mathbf{F}_p$-algebra. If $A/pA$ is a field, then $A$ is known to be a discrete valuation ring. The only way I have seen this proved is to use the actual construction of the $p$-Witt ring $W(A/pA)$, and I'm wondering if there is a way to prove this using only the definition of a strict $p$-ring. -In general, a ring $B$ is a discrete valuation ring if and only if it is local, max-adically separated, and its maximal ideal is principal and non-nilpotent (this is basically Proposition 2 in Serre's Local Fields, although in place of the separated hypothesis, he has a Noetherian hypothesis). If $A$ is a strict $p$-ring with $A/pA$ a field, then $pA$ is maximal, and non-nilpotent by definition. Also, $A$ is $p$-adically separated by definition. But I can't think of a simple way to prove that $A$ is local, i.e., that $pA$ is the unique maximal ideal of $A$. I suppose one could work out the universal formulas which define the ring operations on such a ring and use them to characterize the non-units as precisely the elements of $pA$, but I was hoping there might be a way to avoid this. -The reason I expect such a thing might be possible is that Serre claims that "a complete discrete valuation ring, absolutely unramified, with perfect residue field $k$, is nothing other than a strict $p$-ring with residue ring $k$." He doesn't give any argument for why a strict $p$-ring whose residue ring is a field is local. This leads me to believe I'm missing a potentially obvious argument. -I apologize if this question doesn't seemed well-defined. The ring operations on a strict $p$-ring can be derived from the definitions (starting by proving the existence of a multiplicative system of representatives, then getting series expansions, etc.), so, proving locality from this, strictly speaking, is just using the definitions, but I would say there's considerable work involved (or at least considerable messiness). Basically I'm trying to avoid writing down the polynomials that give the ring structure. - -REPLY [6 votes]: If $x\not\in pA$, then there is some $y$ such that $xy\equiv1$(mod $p$). Then $x$ is invertible mod $p^n$ for any $n$ since $(xy)^{p^n}\equiv 1$(mod $p^n$). Thus $x$ is invertible.<|endoftext|> -TITLE: As a $\mathbb{Z}$-module, is $\mathbb{R}\otimes_\mathbb{Z} \mathbb{R}$ isomorphic to $\mathbb{R}$? -QUESTION [8 upvotes]: Is it true that $\mathbb{R}\otimes_\mathbb{Z} \mathbb{R}$ (the tensor product of $\mathbb{R}$ and $\mathbb{R}$ over $\mathbb{Z}$) is not isomorphic to $\mathbb{R}$ as a $\mathbb{Z}$-module? Please give proof. -It is easy to prove that the tensor product of $\mathbb{Q}$ and $\mathbb{Q}$ over $\mathbb{Z}$ is isomorphic to $\mathbb{Q}$ as $\mathbb{Z}$-modules. - -REPLY [12 votes]: As an abelian group, there is an isomorphism $\mathbb R\cong\mathbb Q^{(I)}$ with $I$ a set of cardinality equal to that of $\mathbb R$; here $\mathbb Q^{(I)}$ denotes a direct sum of $|I|$ copies of $\mathbb Q$, as usual. This is a consequence of the fact that $\mathbb R$ is a $\mathbb Q$-vector space, and it has a basis as such. -Since the tensor product distributes over arbitrary direct sums, $$\mathbb R\otimes_{\mathbb Z}\mathbb R\cong \mathbb Q^{(I)}\otimes_{\mathbb Z}\mathbb Q^{(I)}\cong(\mathbb Q\otimes \mathbb Q)^{(I\times I)}.$$ Now, since $I\times I$ has the cardinality of $I$ and you know that $\mathbb Q\otimes \mathbb Q\cong\mathbb Q$, we have $$(\mathbb Q\otimes \mathbb Q)^{(I\times I)}\cong\mathbb Q^{(I)}\cong\mathbb R$$.<|endoftext|> -TITLE: Chern Class = Degree of Divisor? -QUESTION [14 upvotes]: Is the first chern class the same as the degree of the Divisor? -Say, $C$ is some divisor on $M$, is $c_1(\mathcal O (C)) = \text{deg }C$? -And say I have some Divisor $D$ with first chern class $c_1(\mathcal{O}(D)) = k[S]$ where $[S]$ is some class in $H^2(M)$. Is it true that the integral first chern class is just $k\cdot \text{deg }S$? - -REPLY [23 votes]: If $M$ is a complex manifold of dimension $n$, the exact sequence of sheaves -$$0\to\mathbb Z\to \mathcal O_M\to \mathcal O^*_M\to 0$$ -yields a morphism of groups in cohomology (part of a long exact sequence) $$c_1:H^1(M,\mathcal O^*_M)\to H^2(M,\mathbb Z)$$ -Since $H^1(M,\mathcal O^*_M)$ can be identified to $Pic(M)$, the group of line bundles on $M$, we get the morphism $$c_1:Pic(M)\to H^2(M,\mathbb Z)$$ -This morphism coincides with the first Chern class defined (for $C^\infty$line bundles) in differential geometry in terms of curvature of a connection. -If now $M$ is a compact Riemann surface (= of dimension $1$), we may evaluate a cohomology class $c\in H^2(M,\mathbb Z) $ on the fundamental class$[M]$ of the Riemann surface $M$, thereby obtaining an isomorphism $I: H^2(M,\mathbb Z) \xrightarrow {\cong} \mathbb Z: c\mapsto \langle c,[M] \rangle$, which composed with the first Chern class yields the degree for line bundles: $$deg=I\circ c_1:Pic(M)\to \mathbb Z: L \mapsto deg(L)=\langle c_1(L),[M] \rangle$$ -Finally, if $L$ is associated to the divisor$D$ i.e. $ L=\mathcal O(D)$, we have the pleasantly down to earth (but highly non tautological!) formula $$deg(\mathcal O(D))=deg (D)$$ -which says that the degree on the left, obtained by sophisticated differential geometry, can be computed in a childishly simply way by computing the degree of the divisor $D\in \mathbb Z^{(X)}$ seen purely formally as an element of the free abelian group on $X$. -All this is explained in Griffiths-Harris, Principles of algebraic geometry, Chapter 1.<|endoftext|> -TITLE: How to learn from proofs? -QUESTION [102 upvotes]: Recently I finished my 4-year undergraduate studies in mathematics. During the four years, I met all kinds of proofs. Some of them are friendly: they either show you a basic skill in one field or give you a better understanding of concepts and theorems. -However, there are many proofs which seem not so friendly: the only feeling I have after reading them is "how can one come up with that", "how can such a long proof be constructed" or "why does it look so confusing". What's worse, most of those hard proofs are of those important or famous theorems. All that I can do with these hard proofs is work hard on reciting them, leading me to forget them after exams and learn nothing from them. This makes me very frustrated. -After failing to find the methodology behind those proofs, I thought, "OK, I may still apply the same skill to other problems." But again, I failed. Those skills look so complicated and sometimes they look problem-specific. And there are so many of them. I just don't know when to apply which one. Also, I simply can't remember all of them. -So my questions are: How to learn from those hard proofs? What can we learn from them? What if the skill is problem-specific? (How do I find the methodology behind them?) -I need your advice. Thank you! -P.S. Threre are a lot of examples. I list only four below. -Proof of Sylow Theorem in Algebra -Proof of Theroem 3.4 in Stein's Real Analysis. - -Theorem 3.4 If $F$ is of bounded variation on $[a,b]$, then $F$ is differentiable almost everywhere. - -Proof of Schauder fixed point theorem in functional analysis. -Proof of open mapping theorem in functional analysis. - -REPLY [15 votes]: I find it very helpful to revisit proofs after some time. I find there's two levels of understanding a proof: ---> line-by-line: you find the proof written in a particular textbook. for each sentence in the proof, you understand how it follows from the prior one. ---> conceptually: you can reproduce the proof because at each step it's clear to you what to do. (maybe there's a calculation or two that involves a random trick that you have to look up.) -Obviously the goal is to reach the second arrow. -Reading a proof over and over until you have reached the first arrow is sort of like doing an exercise, and, like all exercises, it helps acclimate your mind to the ideas at hand, and, in this way, it helps you get to the second arrow. But there's also an essential ingredient of time. Come back in three months, see how you feel. -You might think the second arrow entails the first, but actually it can still be hard to read an author's particular way of writing a proof, even when you understand the proof of a theorem. For instance, I think I could write down the proof of the Baire category theorem (don't quiz me), but if I were to go read a proof in a book, I'd probably get confused about the particulars of the notation without spending a little time, or, worse, convince myself that I was following. "You just do the obvious thing right? Yeah, ok, so wait, why is he saying $x_n \in B_n$, wait, what's $B_n$, oh, whatever, I know this."<|endoftext|> -TITLE: Idea behind factoring an operator? -QUESTION [5 upvotes]: Suppose if we have an operator $\partial_t^2-\partial_x^2 $ what does it mean to factorise this operator to write it as $(\partial_t-\partial_x) (\partial_t+\partial x)$ When does it actually make sense and why ? - -REPLY [6 votes]: In the abstract sense, the decomposition $x^2-y^2=(x+y)(x-y)$ is true in any ring where $x$ and $y$ commute (in fact, if and only if they commute). -For sufficiently nice (smooth) functions, differentiation is commutative, that is, the result of derivation depends on the degrees of derivation and not the order in which we apply them, so the differential operators on a set of smooth ($C^\infty$) functions (or abstract power series or other such objects) form a commutative ring with composition, so the operation makes perfect sense in that case in a quite general way. -However, we only need $\partial_x\partial_t=\partial_t\partial_x$, and for that we only need $C^2$. Of course, differential operators on $C^2$ do not form a ring (since higher order derivations may not make sense), but the substitution still is correct for the same reasons. You can look at the differentiations as polynomials of degree 2 with variables $\partial_\textrm{something}$. -For some less smooth functions it might not make sense.<|endoftext|> -TITLE: Non-Galois cubic extensions -QUESTION [7 upvotes]: Is there a necessary and sufficient condition for when a cubic extension of $\mathbb{Q}$ is not a Galois extension? - -REPLY [15 votes]: Theorem. Let $\alpha$ be a primitive element of a cubic extension $\mathbb{Q}(\alpha)$ over $\mathbb{Q}$. Then $\mathbb{Q}(\alpha)$ is Galois over $\mathbb{Q}$ if and only if the discriminant of the irreducible polynomial of $\alpha$ is a rational square. -Proof. Let $f(x)$ be the monic irreducible of $\alpha$, which must be a - cubic. If $\alpha=\alpha_1$, $\alpha_2$, and $\alpha_3$ are the roots of $f(x)$, then the discriminant of $f(x)$ is equal to -$$\Delta = (r_1-r_2)^2(r_1-r_3)^2(r_2-r_3)^2;$$ -the splitting field contains $\sqrt{\Delta}=(r_1-r_2)(r_1-r_3)(r_2-r_3)$. If the extension if Galois, then all three roots lie in $\mathbb{Q}(\alpha)$, hence so does $\sqrt{\Delta}$. Since $[\mathbb{Q}[\sqrt{\Delta}]\colon\mathbb{Q}]$ is either $1$ or $2$, and would have to divide $[\mathbb{Q}(\alpha)\colon\mathbb{Q}]=3$, it follows that $\sqrt{\Delta}\in\mathbb{Q}$; that is $\Delta$ is a rational square as claimed. -Conversely, suppose that $\mathbb{Q}(\alpha)$ is not Galois. Then the splitting field of $\alpha$ over $\mathbb{Q}$ has degree $6$, and the cubic extension is not normal. Hence, the Galois group of $f(x)$ over $\mathbb{Q}$ is isomorphic to $S_3$ (it cannot be cyclic of order $6$, because then the unique subextension $\mathbb{Q}(\alpha)$ of degree $3$ would be Galois). But that means that there is an automorphism of the splitting field $K$ that transposes $r_1$ and $r_2$ and fixes $r_3$; this automorphism does not fix $\sqrt{\Delta}=(r_1-r_2)(r_1-r_3)(r_2-r_3)$; therefore, $\sqrt{\Delta}\notin\mathbb{Q}$, so $\Delta$ is not a rational square, as claimed. $\Box$<|endoftext|> -TITLE: Expression for $\sum\limits_{k=0}^n \frac1{k+1}\left\{ n \atop k\right\}$? -QUESTION [6 upvotes]: It is known that $\sum\limits_{k=0}^n \left\{ n \atop k\right\} k = \varpi(n+1) - \varpi(n)$. Any ideas for computing $\sum\limits_{k=0}^n \frac1{k+1}\left\{ n \atop k\right\}$ ? ($\left\{ n \atop k\right\}$ denotes the Stirling numbers of the second kind and $\varpi(n)$ the $n$-th Bell number) - -REPLY [2 votes]: The error in leonbloy's approximation $\sum_{k=0}^n\frac1{k+1}\left\{ n \atop k\right\} = B_{n-1} + E_n$ is exactly $$E_n = - \sum_{k=0}^{n-1}\left\{ n-1 \atop k\right\} \frac1{(k+1)(k+2)}.$$ -Moreover, the asymptotic can be improved to $$\sum_{k=0}^n\frac1{k+1}\left\{ n \atop k\right\} \sim B_{n-1} - B_{n-3},$$ -(and probably further, for anyone who wants to continue the process below). - -Theorem 4 of my paper "On Solutions to a General Combinatorial Recurrence" (Journal of Integer Sequences, 14 (9): Article 11.9.7, 2011), with $\left| {n \atop k} \right| = \left\{ n \atop k\right\}$ and $f(k,m) = \frac{1}{k+1}$ says that -$$\begin{align}\sum_{k=0}^n\frac1{k+1}\left\{ n \atop k\right\} &= \sum_{k=0}^{n-1}\left\{ n-1 \atop k\right\} (k+1) \frac1{k+1} - \sum_{k=0}^{n-1}\left\{ n-1 \atop k\right\} \frac1{(k+1)(k+2)} \\ -&= B_{n-1} - \sum_{k=0}^{n-1}\left\{ n-1 \atop k\right\} \frac1{(k+1)(k+2)}. \end{align} $$ -Applying Theorem 4 again, this time with $f(k,m) = \frac{1}{(k+1)(k+2)}$, yields (after some simplification) -$$\begin{align} &\sum_{k=0}^{n-1}\left\{ n-1 \atop k\right\} \frac1{(k+1)(k+2)} \\ -&= \sum_{k=0}^{n-2}\left\{ n-2 \atop k\right\} \frac1{k+1} - \sum_{k=0}^{n-2}\left\{ n-2 \atop k\right\} \frac1{(k+1)(k+2)} - 2 \sum_{k=0}^{n-2}\left\{ n-2 \atop k\right\} \frac1{(k+1)(k+2)(k+3)}. \end{align}$$ -The first term on the right-hand side dominates, and with leonbloy's approximation $$\sum_{k=0}^{n-2}\left\{ n-2 \atop k\right\} \frac1{k+1} \sim B_{n-3},$$ -we get $$\sum_{k=0}^n\frac1{k+1}\left\{ n \atop k\right\} \sim B_{n-1} - B_{n-3}.$$ -By comparison with leonbloy's results (again, after taking logarithms) -n exact B_{n-1} - B_{n-3} B_{n-1} -4 1.5114 1.3863 1.6094 -8 6.7348 6.7154 6.7765 -16 21.0308 21.0273 21.0475 -32 57.5872 57.5866 57.5935<|endoftext|> -TITLE: An extension of the birthday problem -QUESTION [9 upvotes]: Th birthday problem (or paradox) has been done in many way, with around a dozen thread only on math.stackexchange. The way it is expressed is usually the following: -"Let us take $n$ people "independently" (no twins, etc.). What is the probability that no two people share the same birthday?" -It is abstracted in the following way: -"Let $X_1$, $\cdots$, $X_n$ be $n$ i.i.d. random variables taken uniformly in $[[1, 365]]$. What is the probability that all the $X_i$'s are distinct?" -There are many generalizations, for instance: -"Let $n$, $d$ be two positive integers, $n \leq d$. Let $X_1$, $\cdots$, $X_n$ be $n$ i.i.d. random variables taken uniformly in $[[1, d]]$. What is the probability $p(n,d)$ that all the $X_i$'s are distinct?" -One can show that in the regimen $1 \ll n \ll d$, the probability $p(n,d)$ is logarithmically equivalent to something like $e^{-\frac{n^2}{2d}}$ (Wikipedia) or $e^{-\frac{n^2}{d}}$ (my computations)*. This problem can be reduced to simple combinatorics, and Stirling's formula (for instance) gives the solution. -However, in the real world, the birthdays are not distributed that way. One might also want to estimate the probability that two peoples are born the same half-day, the same hour, etc. The following generalization seems natural: -"Let $\mu$ be a probability measure on $[0,1]$ absolutely continuous with respect to the Lebesgue measure. Let $n$, $d$ be two positive integers. For $k \in [[0,d-1]]$, let $a_k := [k/d, (k+1)/d]$. Let $X_1$, $\cdots$, $X_n$ be $n$ i.i.d. random variables in $[0,1]$ with distribution $\mu$. What is the probability $p(n,d)$ that all the $X_i$'s lie in different elements of the partition?" -I would expect the solution to be something like $e^{-C(\mu) \frac{n^2}{d}}$, with perhaps some explicit expression of $C(\mu)$. But the combinatorial solutions do not work as well in this setting, and all I can get are very crude bounds when the density of $\mu$ is bounded. In addition, I would expect $C(\mu)$ to be minimal when $\mu$ is the Lebesgue measure, but I don't know how to prove it. One might wonder what happens when $\mu$ is no longer absolutely continuous, but this might be a bit too broad of a generalization. -I am sure this problem has been done to death, but I don't have any access to the literature right now, and quick search didn't yield anything (the generalizations of the birthday problem I found are quite different). Any result/proof/reference related to the problems above would be nice. -.* By the way, any rigorous proof of either of the two facts (or of any similar-sounding result) is appreciated. I don't know which I can trust more, between my computations and Wikipedia. - -REPLY [3 votes]: Let's rephrase the problem: -There are $m$ bins, and $n$ items placed independently into a random bin with probability $p_k$ of going into bin $k\in\{1,\ldots,m\}$. What is the likelihood that no to items go into the same bin? -Let's start with an approximate solution which is good for giving some intuition about the problem and works well as long as $n\ll m$. -For any to items, the probability of them going into the same bin is $q=\sum_{k=1}^m p_k^2$. Hence, the probability that they do not go into the same bin is $1-q$. Since there are $n(n-1)/2$ different pairs of item, if we make the approximation that any two pairs of items are independent (true if pairs have no items in common and almost true if the two pairs have one item in common), we find the likelihood that no pair go into the same bin to be $\approx(1-q)^{n(n-1)/2}\approx e^{qn(n-1)/2}$. -Now, let's do this a bit more formally. If we define the polynomial -$$F(x)=\prod_{k=1}^m (1+p_kx)=\sum_{j=0}^m \frac{f_j x^j}{j!},$$ -the coefficient $f_n$ is the likelihood that that $n$ items go into $n$ distinct bins: sums over all different ways to pick $n$ bins and the likelihood that the $n$ items be placed into these bins in any order (the factor $n!$). -We can now make an approximation: -$$ -\begin{split} -F(x) =&\prod_{k=1}^m (1+p_kx) = \exp\left\{\sum_{j=1}^m\ln(1+p_jx)\right\}\\ -=&\exp\left\{\sum_{j=1}^m p_jx-\frac{p_j^2x^2}{2}+\frac{p_j^3x^3}{3}-\cdots\right\}\\ -=&\exp\left\{x-\frac{q_2x^2}{2}+\frac{q_3x^3}{3}-\cdots\right\}\\ -=&e^x\cdot e^{-q_2x^2/2}\cdot e^{q_3x^3/3}\cdot e^{-q_4x^4/4}\cdots\\ -\end{split} -$$ -where $q_r=\sum_{k=1}^m p_k^r$ (so the above $q=q_2$ while $q_1=1$). We can show that $q_k\le q_2^{k-1}$ (e.g. from $q_r^2\ge q_{r-1}q_{r+1}$) which tells us $q_2$ is the dominant adjustment. -If we make the expansion -$$ -\begin{split} -\exp\left\{-\frac{q_2x^2}{2}+\frac{q_3x^3}{3}-\cdots\right\} -=&\sum_{k=0}^\infty\frac{1}{k!} - \left(-\frac{q_2x^2}{2}+\frac{q_3x^3}{3}-\cdots\right)^k\\ -=&1-a_2x^2+a_3x^3-a_4x^3+\cdots\\ -\end{split} -$$ -we get $a_2=q_2/2$, $a_3=q_3/3$, $a_4=q_4/4-q_2^2/8$, etc. Entering these into the power expansion for $F(x)$, we get -$$ -\begin{split} -F(x)=&e^x\cdot\left(1-a_2x^2+a_3x^3-a_4x^3+\cdots\right)\\ -=&\sum_{n=0}^\infty\left(\frac{1}{n!} - -\frac{a_2}{(n-2)!}-\frac{a_3}{(n-3)!}-\cdots\right)\cdot x^n\\ -=&\sum_{n=1}^\infty\big(1-a_2n(n-1)+a_3n(n-1)(n-2)-\cdots\big)\cdot\frac{x^n}{n!}\\ -\end{split} -$$ -so the effect of $a_rx^r$ is an adjustment $a_rn(n-1)\cdots(n-r+1)$ which for $r\ll n$ has the same magnitude as $a_rn^2$. -If we ignore $q_r$ for $r>2$ and take the effect of $a_{2r}x^{2r}$ to be $\approx a_{2r}[n(n-1)]^r$ (which is true for $r=1$ but only approximate for $r>1$, we get the original approximation: $e^{-q_2n(n-1)/2}$.<|endoftext|> -TITLE: Show that $\left(1+\dfrac{1}{n}\right)^n$ is monotonically increasing -QUESTION [38 upvotes]: Show that $U_n:=\left(1+\dfrac{1}{n}\right)^n$, $n\in\Bbb N$, defines a monotonically increasing sequence. - -I must show that $U_{n+1}-U_n\geq0$, i.e. $$\left(1+\dfrac{1}{n+1}\right)^{n+1}-\left(1+\dfrac{1}{n}\right)^n\geq0.$$ -I am trying to go ahead of this step. - -REPLY [4 votes]: My two cents (as later similar question closed): -$$\frac{a_{n+1}}{a_n}=\frac{\left( 1+ \frac{1}{n+1} \right)^{n+1}}{\left( 1+ \frac{1}{n} \right)^{n}}=\frac{n+1}{n}\left( 1- \frac{1}{(n+1)^2} \right)^{n+1}>\\ ->\frac{n+1}{n}\left( 1- \frac{1}{(n+1)} \right)=1$$<|endoftext|> -TITLE: Evaluate $\lim_{x\to 0} (\ln(1-x)-\sin x)/(1-\cos^2 x)$ -QUESTION [8 upvotes]: I've got this limit: $$\displaystyle\lim_{x\to 0} \frac{\ln(1-x)-\sin x}{1-\cos^2 x}$$ and the problem is that it doesn't exist. But I am not very perceptive and I didn't avoid catching in a trap and I started to trying solve this with L'Hôpital's rule. And my question is: are there any ways to notice that given limit doesn't exist in time? If I had been given such a limit on a test, what is the ideal way to solve it? - -REPLY [2 votes]: \begin{align} -f(x) & = \dfrac{\log(1-x) - \sin(x)}{\sin^2(x)} = \dfrac{\left(-x - \dfrac{x^2}2 - \dfrac{x^3}3 - \cdots \right) - \left( x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots \right)}{\left( x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots \right)^2}\\ -& = \dfrac{-2x + \mathcal{O}(x^2)}{x^2 + \mathcal{O}(x^3)} = -\dfrac{2+\mathcal{O}(x)}{x+\mathcal{O}(x^2)} -\end{align} -Hence, $$\lim_{x \to 0^+} f(x) = - \infty$$ $$\lim_{x \to 0^-} f(x) = \infty$$<|endoftext|> -TITLE: $(a+b)^\beta \leq a^\beta +b^\beta$ for $a,b\geq0$ and $0\leq\beta\leq1$ -QUESTION [6 upvotes]: It seems that $(a+b)^\beta \leq a^\beta +b^\beta$ for $a,b\geq0$ and $0\leq\beta\leq1$. However, I could not prove this nor the same result for a general concave and increasing function (for which it might not hold). If the inequality is true, does it follow from some general inequality or is there some other simple proof? - -REPLY [8 votes]: When $a=b=0$, the result is obvious. Assume that $(a,b)\ne(0,0)$. Then, $t=a/(a+b)$ is in $[0,1]$ hence $t^{\beta}\geqslant t$ because $\beta\leqslant1$. Likewise, $1-t=b/(a+b)$ is in $[0,1]$ hence $(1-t)^{\beta}\geqslant 1-t$. Summing these two inequalities yields $t^{\beta}+(1-t)^{\beta}\geqslant 1$, which is your result.<|endoftext|> -TITLE: Maximum-likelihood estimation for continuous random variable with unknown parameter -QUESTION [5 upvotes]: Let $X$ be a random variable with the unknown parameter $\lambda$ and the following pdf - $$f(t)=2\lambda t\cdot\mathrm e^{-\lambda t^2}\cdot\textbf{1}_{[0,\infty)}(t)$$ - where $\textbf{1}_A(x)$ is an indicator function with - $$\textbf{1}_A(x)=\begin{cases}1,&\text{if }x\in A,\\0,&\text{else.}\end{cases}$$ - Let $\vec x=(x_1,\ldots,x_n)$ be a sample of $X$. Determine the maximum-likelihood estimator $\widehat{\lambda}$ such that the following is true for the likelihood-function $\mathcal L(\vec x;\lambda)$: - $$\forall \lambda\;:\;\mathcal L(\vec x;\lambda)\leq \mathcal L(\vec x;\widehat\lambda)$$ - -For the sake of simplicity my first thoughts were to get the log-likelihood this way: -$$\mathcal L(\vec x;\lambda)=\prod\limits_{i=1}^nf(x_i)\implies \ln(\mathcal L(\vec x;\lambda))=\sum\limits_{i=1}^n\ln(f(x_i))$$ -This is the point where I'm stuck: i don't know how to compute the derivative to maximize the function $$\frac{\mathrm d \ln(\mathcal L(\vec x;\lambda))}{\mathrm d\lambda}\overset{!}{=}0.$$ -Any hints on how to derive the sum would be appreciated. - -REPLY [4 votes]: We have -$$L(\vec x,\lambda)=\prod_{j=1}^n(2\lambda x_j)\cdot e^{-\lambda x_j^2}=2^n\left(\prod_{k=1}^nx_k\right)\lambda ^n\exp\left(-\lambda \lVert x\rVert^2\right)\chi_{x_j\geq 0\forall j},$$ -hence assuming the $x_j> 0$. -$$\log L(\vec x,\lambda)=n\log 2+\sum_{j=1}^n\log x_j+n\log\lambda-\lambda\lVert x\rVert^2.$$ -Now taking the derivative with respect to $\lambda$, we get -$$\partial_{\lambda}\log L(\vec x,\lambda)=\frac n{\lambda}-\lVert x\rVert^2.$$ -I let you finish the computation.<|endoftext|> -TITLE: Show that the limit of functions is continuous -QUESTION [8 upvotes]: Let $f_n$ be a sequence of not necessarily continuous functions $\mathbb{R} \rightarrow \mathbb{R}$ such that $f_n(x_n) \rightarrow f(x)$ whenever $x_n \rightarrow x$. Show that f is continuous. -What I am trying to do is to show that whenever we have $x \in \mathbb{R}$ and $x_n \rightarrow x$, then $f(x_n) \rightarrow f(x)$, when $n \rightarrow \infty$. -These types of things are usually showed by using the triangle inequality. I know I can make $|f(x) - f_n(x_n)|$ as small as possible by choosing a big enough n. I can also make $|f(x) - f_n(x)|$ as small as possible. But I am not able to combine these to prove that $|f(x) - f(x_n)|$ can be made as small as possible. - -REPLY [5 votes]: First note that the hypothesis implies that the $f_n$ converge pointwise to $f$. To see this, consider the constant sequence $\langle x_n:n\in\Bbb N\rangle$ where $x_n=x$ for each $n\in\Bbb N$: $$\langle f_n(x):n\in\Bbb N\rangle=\langle f_n(x_n):n\in\Bbb N\rangle\to f(x)\;.$$ -Now suppose that $f$ is not continuous at $x$, and let $\langle x_n:n\in\Bbb N\rangle\to x$ be such that $\langle f(x_n):n\in\Bbb N\rangle$ does not converge to $f(x)$. Then there is an $\epsilon>0$ such that $|f(x_n)-f(x)|\ge\epsilon$ for infinitely many $n\in\Bbb N$, so you can find a subsequence $\langle x_{n_k}:k\in\Bbb N\rangle$ such that $|f(x_{n_k})-f(x)|\ge\epsilon$ for every $k\in\Bbb N$. Since $\langle x_{n_k}:k\in\Bbb N\rangle\to x$, you might as well assume from the start that you have a sequence $\langle x_n:n\in\Bbb N\rangle$ and an $\epsilon>0$ such that $\langle x_n:n\in\Bbb N\rangle\to x$ and $|f(x_n)-f(x)|\ge\epsilon$ for all $n\in\Bbb N$. -By hypothesis $\langle f_n(x_n):n\in\Bbb N\rangle\to f(x)$. Choose $n_0\in\Bbb N$ so that $|f_n(x_0)-f(x_0)|<\epsilon/2$ for all $n\ge n_0$; we can do this, since the $f_n$’s converge pointwise to $f$. Now choose $n_1>n_0$ so that $|f_n(x_1)-f(x_1)|<\epsilon/2$ for all $n\ge n_1$. Continue in this way to construct an increasing sequence $\langle n_k:k\in\Bbb N\rangle$ such that $|f_n(x_k)-f(x_k)|<\epsilon/2$ for all $n\ge n_k$. -Now form a new sequence $\langle y_n:n\in\Bbb N\rangle$ as follows: -$$y_n=\begin{cases} -x_0,&\text{if }n\le n_0\\ -x_k,&\text{if }n_{k-1} -TITLE: Construction of special $\omega_1$-Aronszajn tree -QUESTION [5 upvotes]: Problem from Kunen II.40: -The definition is the following: An $\omega_1$-Aronszajn tree $T$ called special iff $T$ is the union of $\omega$ antichains. -Need to prove that $T$ is special iff there is a map $f: T \rightarrow \mathbb{Q}$ such that for $x,y \in T, x < y \rightarrow f(x) < f(y)$, and show that a special Aronszajn tree exist. -The hint is to construct $T$ and $f$ simultaneously by induction. -It seems like I should somehow "pack" the antichains (equivalence classes?), to achieve a map from a "large" tree to a relatively small set. Couldn't advance any furher than that, though. -Any help? -Thanks in advance. - -REPLY [2 votes]: Suppose $T$ is a countable union of antichains. We are going to construct a map $g: T\to 2^\omega$ so that range of $g$ is countable and it is strictly increasing (with respect to the lexicographical ordering on $2^\omega$). Pick a function $f: T\to \omega$ so that $f^{-1}(n)$ is an antichain for all $n$. For $t\in T$, define $g(t)=x$ by: $x(n)=1$ if and only if $n\leq f(t)$ and $\{ s\in T: s\leq t\}\cap f^{-1}(n)\ne\varnothing$. It is easy to verify that $g$ is as required. -To construct a special A-tree. Consider the subtree of the A-tree constructed in Kunen's book which consists of nodes of successor height.<|endoftext|> -TITLE: Representation of Cyclic Group over Finite Field -QUESTION [8 upvotes]: The post Irreducible representations of a cyclic group over a field of prime order discusses the irreducible representations of a cyclic group of order $N$ over a finite field $\mathbb{F}_p$ where $N$ does not divide $p$. -Where can I find information about the irreducible representations in the case where $p$ does divide $N$? -(I'm interested in this because I'm wondering if, given a finite dimensional vector space $\mathbb{F}_{p}^{M}$ if there are examples of particularly simple invertible linear operators $T$ such that $T$ does not preserve a subspace. Since the vector space contains finitely many elements it seems that this is the same thing as an irreducible representation of a cyclic group. -However, based on the post I read above, where p does not divide N, it seems like T is just multiplication by a primitive element that generates the extension $\mathbb{F}_{q^{M}}$.) - -REPLY [2 votes]: In general, if a finite group $G$ has a normal subgroup $U$ whose order is power of a prime $p,$ and $G$ acts irreducibly on finite dimensional vector space $V$ over a field $F$ of characteristic $p,$ then $U$ has a non-zero fixed vector on $V,$ while on the other hand, the fixed-point subspace $V^{U}$ of $U$ on $V$ is easily checked to be $G$-invariant, so we must have $V^{U} = V$ by irreducibility as $V^{U}$ is non-zero. This is a well-known fact, and a more general version of the remark at the end of Qiaochu's answer. The argument when $V$ has finite cardinality is easier considering orbit lengths on non-zero vectors. However, in general, to prove that $V^{U} \neq \{0 \}$ (without further assumption) we can proceed by induction on $|U|.$ The case that $U$ has order $p$ is just linear algebra, as a generator of $U$ has the eigenvalue $1$. If $|U| >p,$ let $Z$ be a subgroup of $Z(U)$ of order $p,$ which is possible as $Z(U) \neq 1$ (since $U$ is $p$-group). If $Z \neq U,$ then we still have $Z \lhd U$ as $Z$ is characteristic in $U.$ By induction $V^{Z} = V,$ and then we can replace $G$ by $G/U$ and $U$ by $U/Z$ to obtain the desired inclusion by induction.<|endoftext|> -TITLE: integration of a continuous function $f(x) $ and $xf(x)$ is zero -QUESTION [6 upvotes]: Possible Duplicate: -Prove that $\exists a0$ then there will be a neighborhood of $a$ say $(a-\epsilon,a+\epsilon)$ where $f(x)>0$ and hence the integral will not be equal to $0$, but I don't know where I am using the other integral condition. am I wrong anywhere in my proof? please help. - -REPLY [16 votes]: Define $F(x):= \int_0^xf(t)\,dt$ for $x \in [0,1]$. Then the second integral tells us -$$ -\begin{aligned} -0 = \int_0^1xf(x)\,dx \, & = \, xF(x)\Bigr|_0^1-\int_0^1F(x)\,dx -\\ -& = 1\,F(1) - 0\,F(0) - \int_0^1F(x)\,dx -\\ -& = \int_0^1f(x)dx - \int_0^1F(x)\,dx -\\ -& = -\int_0^1F(x)\,dx. -\end{aligned} -$$ -By the mean value theorem and continuity of $F$, which follows from continuity of $f$, this tells us that there is $c \in (0,1)$ such that $F(c)=0$. That is, -$$ -\int_0^cf(x)\,dx = 0. -$$ -It follows that -$$ -\int_c^1f(x)\,dx = \int_0^1f(x)\,dx-\int_0^cf(x)\,dx = 0 - 0 = 0 -$$ -as well. -We then apply the mean value theorem two more times, using continuity of $f$, for the intervals $[0,c]$ and $[c,1]$ to find $a \in (0,c)$ and $b \in (c,1)$ respectively such that $f(a)=f(b)=0$.<|endoftext|> -TITLE: Partition of Unity question -QUESTION [5 upvotes]: I am starting to read the book "Differential Forms in Algebraic Topology" by Bott and Tu. -In the proof of the exactness of the Mayer - Vietoris sequence (Proposition 2.3, page 22 - 23) a partition of unity $\{\rho_U,\rho_V\}$ subordinate to an open cover of two open sets $U,V$ is applied to a function $f$ which is defined on the intersection $U \cap V$. -Then the authors emphasize that $\rho_V \,f$ is a function on $U$. I struggle to see why this is true, I thought $\rho_V$ has support contained in $V$, and so from the picture in the book I have the impression that the function $\rho_U\,f$ is a function with support in $U$. I am aware this must be false, the authors even stress the fact that one has to multiply by the partition function of the other open set! Any help to understand this would be great, many thanks !! - -REPLY [5 votes]: The support (Wikipedia article) of $\rho_V$ is contained in $V$, so that $\rho_V(x)=0$ for any $x\notin V$. Thus, even though $f$ is only defined on $U\cap V$, we can extend $\rho_Vf$ to $U\cap V^c$ by declaring $\rho_Vf(x)=0$ for all $x\in U\cap V^c$ (this extending of the domain of the function is being done implicitly by the authors). This defines $\rho_Vf$ on all of $U=(U\cap V)\cup (U\cap V^c)$.<|endoftext|> -TITLE: Product norm on infinite product space -QUESTION [7 upvotes]: Today I proved that if $V$ is a normed space with norm $\|\cdot\|$ then I can define a norm on $V \times V$ that induces the same topology as the product topology as follows: $\| (v,w) \|_{V \times V} = \|v\| + \|w\|$. -I think I can do the same for an infinite product $V^{\mathbb N}$ by defining $\|(v_n)\|_{\mathbb N} = \sum_{n=0}^\infty \frac{1}{2^n} \|v_n\|$ and I proved it using the proof of the case $V \times V$ and changing some minor things. -Can you confirm that this result is correct? Thanks. - -REPLY [15 votes]: Proposition. If $X_i, i \in \mathbb{N}$ is any sequence of nonzero topological vector spaces, then the product topology on $X = \prod_i X_i$ is not normable. -Proof. Suppose to the contrary there is a norm $\|\cdot\|$ on $X$ which induces the product topology. Let $B \subset X$ be the open unit ball of $\|\cdot\|$. $B$ is open in the product topology and contains 0, so we can find a basic open neighborhood of 0 which is contained in $B$. That is, there is a set of the form $U = U_1 \times U_2 \times \dots \times U_n \times X_{n+1} \times \cdots$, where $U_i \subset X_i$ is open and nonempty, such that $0 \in U \subset B$. Choose any nonzero $x_{n+1} \in X_{n+1}$ and set $x = (0,\dots, 0, x_{n+1}, 0, \dots)$. Then for any $t \in \mathbb{R}$ we have $tx \in U \subset B$. If we take $t = 2/\|x\|$, we have $\|tx\| = 2$ which is absurd since $B$ is the unit ball. -In short, the problem is that every open set of the product topology contains a line, and balls of a norm do not.<|endoftext|> -TITLE: Can someone help me to evaluate or at least to approximate the following sum. -QUESTION [6 upvotes]: Can someone help me to evaluate the following sum. It appeared while calculating the number of elements of intersections of some sets: -$$\sum_{d|n} \frac{d}{\phi(d)^2}.$$ - -REPLY [8 votes]: Since $\dfrac{n}{\phi(n)^2}$ is a multiplicative function, we know that -$$ -\varphi(n)=\sum_{d|n}\frac{d}{\phi(d)^2}\tag{1} -$$ -is also multiplicative. On the power of a prime, $p^j$ and $j>0$, $\phi(p^j)=p^j\frac{p-1}{p}$, so -$$ -\begin{align} -\varphi(p^k) -&=1+\sum_{j=1}^k\frac{p^2}{p^j(p-1)^2}\\ -&=1+\frac{p}{(p-1)^2}\sum_{j=0}^{k-1}\frac{1}{p^j}\\ -&=1+\frac{p^k-1}{p^{k-2}(p-1)^3}\tag{2} -\end{align} -$$ -Thus, for $n$ whose prime factorization is -$$ -n=\prod_jp_j^{k_j}\tag{3} -$$ -we have -$$ -\varphi(n)=\prod_j\left(1+\frac{p_j^{k_j}-1}{p_j^{k_j-2}(p_j-1)^3}\right)\tag{4} -$$ -Estimation: -Note that for $k>0$, -$$ -1+\frac1p\left(\frac{p}{p-1}\right)^2\le1+\frac{p^k-1}{p^{k-2}(p-1)^3}\lt1+\frac1p\left(\frac{p}{p-1}\right)^3\tag{5} -$$ -For larger primes, $(5)$ might help to approximate the term in $(4)$.<|endoftext|> -TITLE: support of a differential form on manifold -QUESTION [8 upvotes]: In the book "Differential forms in Algebraic Topology" by Bott and Tu, the support of a differential form $\omega$ on a manifold $M$ is defined to be "the smallest closed set $Z$ so that $\omega$ restricted to $Z$ is not $0$." (page 24). -I am a little confused, suppose we let $M = \mathbb{R}$ with the trivial atlas $\{ \mathbb{R}, \text{Id} \}$ and consider the $0-$form $\omega = x$. Then any non - zero point would constitute a set on which the restriction of $\omega$ is non - zero. But I expect the authors want the support to be the smallest closed set containing all the points at which $\omega$ is non - zero. Where is my misunderstanding ? Lots of thanks for help! - -REPLY [11 votes]: Your expectation is the correct definition: the support of a differential form $\omega \in \Omega^k(M)$ is the set -$$\mathrm{supp}(\omega) = \overline{\{p \in M : \omega_p \not\equiv 0\}}.$$ -I looked in my copy of Bott and Tu and indeed they defined it incorrectly. Perhaps a better way to phrase their sentence would be "the support of $\omega$ is the smallest closed set $Z$ so that $\omega$ restricted to any point in the interior of $Z$ is not identically $0$," or, as Micah suggests in the comments, "the support of $\omega$ is the smallest closed set $Z$ such that $\omega$ restricted to $M \setminus Z$ is identically $0$."<|endoftext|> -TITLE: successful absurd formalities -QUESTION [16 upvotes]: Has anyone published in print or on a web site or elsewhere a compilation of successful illogical formal arguments? By those I mean arguments that follow a form in disregard of the legality of its applicability in the circumstances to which it is applied, and get a right answer. -One notable example occurred when a formula was discovered for finding solutions of third-degree algebraic equations with real coefficients, when it was known that a real root exists. It was necessary to take square roots of negative numbers. These "imaginary" numbers canceled out and then (the punch line): when the resulting values were substituted into the equation, it checked. -Many things that Euler did could be considered examples. It has been noted here in stackexchange that Euler's derivation of the product formula for the sine function considered the infinitely large leading coefficient of the polyonomial that was the product (which is an infinite product and so has no leading coefficient). -Paul Dirac's delta function was seen to lead to demonstrably correct results at a time when it was not known to to have any logically rigorous justification. -Here's an example from Paul Halmos' article Does Mathematics Have Elements: - - -In the general ring theory question there are no numbers, no absolute values, no inequalities, and no limits - those concepts are totally inappropriate and cannot be brought to bear. Nevertheless an impressive-sounding classical phrase, "the principle of permanence of functional form", comes to the rescue and yields an analytically inspired proof in pure algebra. The idea is to pretend that $1/(1-ba)$ can be expanded in a geometric series (which is utter nonsense), so that $$(1-ba)^{-1} = 1 + ba + baba + bababa + \cdots.$$ - It follows (it doesn't really, but it's fun to keep pretending) that - $$(1-ba)^{-1} = 1 + b(1 + ab + abab + ababab + \cdots )a,$$ - and, after one more application of the geometric series pretense, this - yields - $$(1-ba)^{-1} = 1 + b(1-ab)^{-1}a.$$ - Now stop the pretense and verify that, despite its unlawful derivation, the - formula works. If, that is, $c = (1-ab)^{-1}$, so that $(1-ab)c = c(1-ab) = 1$, then $1 + bca$ is the inverse of $1 - ba$. Once the statement is put this way, its proof becomes a matter of (perfectly legal) mechanical computation. - Why does it all this work? What goes on here? Why does it seem that - the formula for the sum of an infinite geometric series is true even for an - abstract ring in which convergence is meaningless? What general truth does - the formula embody? I don't know the answer, but I note that the formula - is applicable in other situations where it ought not to be,[ . . . ] - - -Question: Has anyone published in print or on a web site or elsewhere a compilation of examples of this phenomenon? (Maybe annotated with explanations of the resolution of the seeming paradox in cases where it is known.) - -REPLY [2 votes]: I think the MathOverflow thread on "jokes in the sense of Littlewood" is in the same spirit.<|endoftext|> -TITLE: Question in do Carmo's book Riemannian geometry -QUESTION [6 upvotes]: This is a question on Do Carmo's book "Riemannian Geometry" (question 7 from chapter 7): -Let $f:M\to \bar{M}$ be a diffeomorphism beetwen two riemannian manifolds. Suppose $\bar{M}$ complete and that there is $c>0$ such that: $$|v|\geq c|df_pv|$$ for every $p\in M$ and $v\in T_pM$. Prove that $M$ is complete. -I think one approach could be using the Hopf-Rinow Theorem, because this question is in a chapter about this theorem.Thanks. - -REPLY [8 votes]: Let $\{x_n\}$ be a Cauchy sequence in $M$. Lets see that $\{x_n\}$ is convergent on $M$. This shows $M$ is a complete metric space. By Hopf-Rinow's theorem, $M$ is complete, in geodesic sense. -Claim: $d_{\bar{M}}(y_m,y_n)\leqslant \frac{1}{c}d_M(x_m,x_m)$ -Proof: given a curve $\gamma$ in $M$, we have -$$\ell(\gamma)=\int|\gamma'|dt\geqslant c\int|df_{\gamma}\gamma'|=c\cdot \ell(f(\gamma))$$ -Once $f$ is a diffeomorfism, the differentiable curves of $M$ and $\bar{M}$ are in bijection. So, considering the curves joinning $x_m$ to $x_n$ (and its images joinning $y_m$ to $y_n$), we can take infimum and it leads us to the result. -Claim: $y_n=f(x_n)$ is Cauchy in $\bar{M}$. -Proof: it follows directly from the claim above. -Hence, once $\bar{M}$ is complete, we have $y_n\rightarrow y$ in $\bar{M}$. But $f$ is a diffeomorphism. Then $x_n=f^{-1}(y_n)\rightarrow f^{-1}(y)$ in $M$.<|endoftext|> -TITLE: Motivation for adjoint operators in finite dimensional inner-product-spaces -QUESTION [15 upvotes]: Given a finite dimensional inner-product-space $(V,\langle\;,\rangle)$ and an endomorphism $A\in\mathrm{End}(V)$ we can define its adjoint $A^*$ as the only endomorphism such that $\langle Ax, y\rangle=\langle x, A^*y\rangle $ for all $x,y\in V$. -While all of this lets us prove some stuff about unitary, normal and hermitian matrices, I'd like to know if there's some other motivation behind its introduction (Is there a geometric interpretation? Any other algebraic remark?) - -REPLY [8 votes]: Here is an algebraic approach to adjoint operators. Let us strip away the existence of an inner product and instead take two vector spaces $V$ and $W$. Furthermore, let $V^*$ and $W^*$ be the linear duals of $V$ and $W$, that is, the collection of linear maps $V\to k$ and $W\to k$, where $k$ is the base field. If you're working over $\mathbb R$ or $\mathbb C$, or some other topological field, you might want to work with continuous linear maps between topological vector spaces. -Given a linear operator $A: V\to W$, we can define a dual map $A^*: W^* \to V^*$ by $(A^*(\phi))(v)=\phi(A(v))$. It is straight forward to verify that this gives a well defined linear map between the vector spaces. This dual map is the adjoint of $A$. For most sensible choices of dual topologies, this map should also be continuous. -The question is, how does this relate to what you are doing with inner products? Giving an inner product on $V$ is the same as giving an isomorphism between $V$ and $V^*$ as follows: -Given an inner product, $\langle x, y \rangle$, we can define an isomorphism $V\to V^*$ via $x\mapsto \langle x, - \rangle$. This will be an isomorphism by nondegeneracy. Similarly, given an isomorphism $\phi:V\to V^*$, we can define an inner product by $\langle x,y\rangle =\phi(x)(y)$. The "inner products" coming from isomorphisms will not in general be symmetric, and so they are better called bilinear forms, but we don't need to concern ourselves with this difference. -So let $\langle x,y \rangle$ be an inner product on $V$, and let $\varphi$ be the corresponding isomorphism $\varphi:V\to V^*$ defined above. Then given $A:V\to V$, we have a dual map $A^*:V^* \to V^*$. However, we can use our isomorphism to define a different dual map (also denoted $A^*$, but which we will denote by $A^{\dagger}$ to prevent confusion) by $A^{\dagger}(v)=\varphi^{-1}(A^*\phi(v))$. This is the adjoint that you are using. -Let us see why. In what follows, $x\in V, f\in V^*$. Note that $\langle x, \varphi^{-1} f \rangle = f(x)$ and so we have -$$ \langle Ax, \varphi^{-1}f \rangle = f(Ax)=(A^*f)(x)=\langle x, \varphi^{-1}(A^* f) \rangle $$ -Now, let $y=\varphi^{-1}f$ so that $\varphi(y)=f$ Then we can rewrite the first and last terms of the above equality as -$$\langle Ax, y \rangle = \langle x, \varphi^{-1}(A^* \phi(y)) \rangle = \langle x, A^{\dagger}y \rangle $$<|endoftext|> -TITLE: Solve for integer matrix such that $R^gu=v$ given $u$ and $v$ -QUESTION [8 upvotes]: Given non-negative integer $n$-vectors $u$ and $v$, how does one find all $n \times n$ non-negative integer matrices $R$ and powers $g$ such that $R^gu=v$? - -REPLY [2 votes]: All solutions can be found using an exhaustive algorithmic approach. Note: -$$ Ru=v=\left[ \begin {array}{ccc} R_{{1,1}}&R_{{1,2}}&...\\ -...&...&...\\ ...&R_{{n,n-1}}&R_{{n,n}} -\end {array} \right]\left[ \begin {array}{c} u_{{1}}\\ ... -\\ u_{{n}}\end {array} \right]=\left[ \begin {array}{c} v_{{1}}\\ ... -\\ v_{{n}}\end {array} \right], - $$ -implies: -$$\sum_{j=1}^{n}R_{i,j}u_j=v_i\,:\,i=1..n,\,\,\,\,\,\,\,\,\,\,(1)$$ -so the problem is equivalent to: -Find all (non-negative) solutions to $n$ independent linear Diophatine equations in $n$ variables, $R_{i,1}..R_{i,n}$ (Handbook of Discrete and Combinatorial Mathematics, Rosen & Michaels). - -Solution: a particular solution is found using the Euclidean Algorithm (or some educated guessing) and from there all solutions are found in an algorithmic fashion. -With that said: - -no solutions exist unless $\text{gcd}(u_1,u_2,..,u_n)|v_i,\forall i$. Proof: $\text{gcd}(u_1,u_2,..,u_n)$ divides the left hand side of Eqn (1) so it must divide the right. -only non-negative solutions (values for $R_{i,j}$) are aloud so, when $u_j>0 \,\,\forall j$, the number of solutions to each Diophantine equation is finite and thus the number of distinct matrix solutions is finite. Proof: non-negativity bounds the numbers from below and the finite size of $v_i$ bounds them from above. -if $u_j=0$ for some $j$, then $R_{i,j}$ can be made arbitrary $\forall j$ and thus, if any solutions exists the number of distinct matrix solutions is infinite. - - -Example: n=2 -Consider the simple linear Diophantine equation for the $n=2$ case: -$$ax+by=c\,:\, \text{gcd}(a,b)|c$$ -if $x_0,y_0$ is a particular solution then all solutions are of the form: -$$x=x_0+\frac{b}{\text{gcd}(a,b)}k,y=y_0-\frac{a}{\text{gcd}(a,b)}k\,:\,k\in\mathbb{Z}$$ -Proof: Subbing in the solutions shows they are solutions. To prove they are the only solutions change variables to $x=X+x_0,y=Y+y_0$ to obtain: $\frac{a}{\text{gcd}(a,b)}X+\frac{b}{\text{gcd}(a,b)}Y=0$ then because $\frac{b}{\text{gcd}(a,b)}\nmid\frac{a}{\text{gcd}(a,b)}$ we have $\frac{b}{\text{gcd}(a,b)}|X$ so $X=k\frac{b}{\text{gcd}(a,b)},k\in \mathbb{Z}$ and similarly $Y=-k\frac{a}{\text{gcd}(a,b)},k\in \mathbb{Z}$. - -Apply this to the following example: -$$ \left[ \begin {array}{cc} R_{{1,1}}&R_{{1,2}}\\ R_{ -{2,1}}&R_{{2,2}}\end {array} \right] -\left[ \begin {array}{c} 6\\ 9 -\end {array} \right] =\left[ \begin {array}{c} 45\\ 51 -\end {array} \right] $$ -divide through by $\text{gcd}(6,9)=3$ to simplify things and obtain: -$$2\,x+3\,y=15:(x,y)=(R_{1,1},R_{1,2})$$ -$$2\,s+3\,t=17:(s,t)=(R_{2,1},R_{2,2})$$ -educated guess leads to particular solutions: -$$(x_0=-15,y_0=15),(s_0=-17,t_0=17)$$ -general solutions for the first equation are: -$$x=-15+3k,y=15-2k$$ -only positive solutions occur for $k=5,6,7$ which give: -$$(R_{1,1},R_{1,2}):(0,5),(3,3),(6,1)$$ -a similar reasoning leads to: -$$(R_{2,1},R_{2,2}):(1,5),(4,3),(7,1)$$ -and thus the only matrices that solve the problem are formed by all combinations of the row vectors found above, $9$ in total: -$$ \left[ \begin {array}{cc} 0&5\\ 1&5\end {array} - \right],\left[ \begin {array}{cc} 0&5\\ 4&3\end {array} - \right],\left[ \begin {array}{cc} 0&5\\ 7&1\end {array} - \right],\left[ \begin {array}{cc} 3&3\\ 1&5\end {array} - \right],\left[ \begin {array}{cc} 3&3\\ 4&3\end {array} - \right],\left[ \begin {array}{cc} 3&3\\ 7&1\end {array} - \right],\left[ \begin {array}{cc} 6&1\\ 1&5\end {array} - \right],\left[ \begin {array}{cc} 6&1\\ 4&3\end {array} - \right],\left[ \begin {array}{cc} 6&1\\ 7&1\end {array} - \right]. $$ -If $n>2$ an iterative procedure is needed to solve the Diophantine equations (see reference). Because of the equivalence of this problem with such a well studied research area this is probably the best, if not the only approach. -To see if $R$ and $R^g, g>1$, both solve the equation find the row vectors in the $g=1$ case following the method outlined above and check for any closure when raising to a power, i.e. check if any of the allowed matrices raised to the power are also one of the found solutions as the set of solutions for the $g=1$ case contains all allowed solutions (as noted in the comments by @ˈjuː.zɚ79365 ). -But one important restriction should be noted. If: $$Ru=R^gu=v,\,\,\,\,\,\,\,\,\,(2)$$ -then multiplying by $R^{g-1}$ gives: -$$R^gu=R^{2g-1}u,$$ which by $(2)$ implies: -$$R^{2g-1}u=Ru=v,$$ -and multiplying by $R^{g-1}$ again will show that $R^{3g-2}u=v$ e.t.c, continuing this process shows $(2)$ implies:$$\lim_{m \to \infty}R^{m(g-1)+1}u=v.$$ -Now instead of non-negative $(\ge0)$, suppose all elements of $R$ are positive integers $\ge1$. Then by the definition of matrix multiplication: -$$(AB)_{i,j}=\sum_{k=1}A_{i,k}B_{k,j},$$ -raising $R$ to higher powers will be adding and multiplying positive integers $\ge1$ which will mean matrix elements of $R^{m(g-1)+1}$ will be increasing to infinity as $m\rightarrow \infty$. Eventually for $m$ large enough: $$R^{m(g-1)+1}_{i,j}>\text{max}(v)\,\forall \,i,j,$$ and so when $R^{m(g-1)+1}$ is multiplied by $u$, a vector with non-negative elements, there is no way it can equal $v$. This leads to a contradiction and thus a theorem is obtained. - -Theorem: If $R$ is a positive integer matrix, $u$ and $v$ are non-negative integer vectors, $v$ is not the zero vector and: -$$Ru=v,$$ -then: $$R^gu\ne v\,\,:\,\,\forall\,g>1\in \mathbb{Z}.$$ -This is very restrictive. While it does not extend to all non-negative matrices it does rule out any solution in $R$ for $Ru=R^gu=v,g>1$, if when raised to a high enough power $R$ becomes a positive matrix (such a matrix is known as "primitve") as we could then just apply the theorem.<|endoftext|> -TITLE: Sharkovskii-type results in other topological spaces? -QUESTION [6 upvotes]: I recently came across Sharkovskii's Theorem which asserts that if $f:\mathbb{R} \to \mathbb{R}$ is continuous and has a cycle of length $m$, then $f$ has a cycle of length $n$ for any $n$ which comes after $m$ in the following ordering: -$$3 < 5 < 7 < 9 < ...$$ -$$ < 2\cdot3 < 2\cdot5 < 2\cdot7 < 2 \cdot 9 < ...$$ -$$...$$ -$$ < 2^n\cdot3 < 2^n\cdot5 < 2^n\cdot7 < 2^n \cdot 9 < ...$$ -$$...$$ -$$... < 2^n < 2^{n-1} < ... <2 < 1$$ -I was wondering how strongly this theorem depends on the structure of the underlying topological space? For example, the theorem is clearly not true in any discrete space, but it is true for a closed interval. Does anyone know of a fairly general set of a criteria a topological space $X$ must meet for a Sharkovskii-type result to hold for continuous functions $f:X\to X$? -Thanks in advance. - -REPLY [2 votes]: I suggest you look at a recent paper in Discrete and Continuous Dynamical Systems-A by Conner, Grant and Meilstrup (http://www.aimsciences.org/journals/displayArticlesnew.jsp?paperID=7409) for more examples of spaces which are Sharkovsky spaces. For example, covering spaces of the Warsaw circle, among others.<|endoftext|> -TITLE: Proving a metric space to be compact -QUESTION [5 upvotes]: I have the following metric space: The set $X$ of all sequences with members from the set $\{1,2,\ldots, n\}$, together with the metric $$d(x,y)=\frac{1}{\min\{j\in\mathbb{N}:x_j\ne y_j\}}.$$ I wish to prove two things about this space: - -1) $X$ is compact. -2) If $T:X\to X$ is defined by $Tx_n=x_{n+1}$ then $T$ is continuous. - -Its many years since I studied topology so if someone can help me even a little bit I will be very grateful. Thanks. - -REPLY [3 votes]: To prove that $X$ is compact, it suffice to show that it is totally bounded and complete. -To show that it is totally bounded, let $\epsilon\gt 0$. Let $m\gt 0$ be such that $\frac{1}{m}\lt\epsilon$. Now, there are only $n^m$ possible finite sequences of length $m$ with values in $\{1,2,\ldots,n\}$. For each such sequence $s$, let $\mathcal{O}_s$ consist of all elements of $X$ that agree with $s$ in the first $m$ points. The set is open, since for every $(x_n)\in \mathcal{O}_s$, any sequence within $\frac{1}{m}$ of $(x_n)$ must agree with $x_n$ on the first $m$ terms, and hence have $s$ as its initial term. Moreover, the distance between any two elements of the set is less than $\epsilon$, so the radius of $\mathcal{O}_s$ is less than $\epsilon$. Since $X$ is the union of the $\mathcal{O}_s$, we see that $X$ is totally bounded. -Now suppose that $(x_n)_m$ is a Cauchy sequence. Then for every $k\in\mathbb{N}$ there exists $M(k)\gt 0$ such that if $j,\ell\geq M(k)$, then $d((x_n)_i,(x_n)_j)\lt \frac{1}{k}$. That means that $(x_n)_i$ and $(x_n)_j$ agree in the first $k$ terms. Define the sequence $(y_n)$ by letting $y_k$ be the common $k$th term to all sequences with index greater than $M(k)$. Show that $(x_n)_m\to (y_n)$. -Thus, $X$ is compact.<|endoftext|> -TITLE: Problem on Euler's Phi function -QUESTION [8 upvotes]: Let $S(n)$ be $S(n)=\left\{k\;\left|\;\left\{\frac{n}{k}\right\}\right.\geq \frac{1}{2}\right\}$,where $\{x\}$ is the fractional part of $x$ -Prove that : -\begin{align} -\sum_{k\in S(n)} \varphi(k)=n^2 -\end{align} -maybe the fact -\begin{align} -\sum_{k \leq 2n}\left\lfloor{\frac{n}{k}+\frac{1}{2}}\right\rfloor \varphi(k)=\frac{3n^2+n}{2}\end{align} -and -\begin{align} -\sum_{k \leq 2n}\left\lfloor{\frac{n}{k}}\right\rfloor \varphi(k)=\frac{n^2+n}{2}\end{align} -can help, but I can't prove these two equations - -Edited:Sorry that my question is to prove the two equation before, because I've already known how to relate the problem with these two equations, and I've also known how to prove the second equation from Identity involving Euler's totient function: $\sum \limits_{k=1}^n \left\lfloor \frac{n}{k} \right\rfloor \varphi(k) = \frac{n(n+1)}{2}$ -but I can't solve the first one with the similar method. - -REPLY [7 votes]: I'll recall (as is more or less indicated in the link you provided) that for the second identity -$$ -\sum_{0 -TITLE: Find functions family satisfying $ \lim_{n\to\infty} n \int_0^1 x^n f(x) = f(1)$ -QUESTION [18 upvotes]: I wonder what kind of functions satisfy -$$ \lim_{n\to\infty} n \int_0^1 x^n f(x) = f(1)$$ -I suppose all functions must be continuous. - -REPLY [5 votes]: The equation is true for any integrable function $f$ on $[0,1]$ so that $1$ is a Lebesgue point for $f$, in the sense that $\lim_{y\to1}\bar f(y)=f(1)$, where $$\bar f(y)=\frac1{1-y}\int_y^1 f(y)\,dy.$$ -Indeed, using $x^n=(n+1)\int_0^x y^n\,dx$ in the integral and interchanging the order of integration, we find after a bit of calculation $$n\int_0^1x^nf(x)\,dx=n(n+1)\int_0^1(1-y)y^{n+1}\bar f(y)\,dy,$$ -in which we notice that $n(n+1)(1-y)y^{n+1}$ is a delta sequence, in this case converging to the delta function at $y=1$. -Noting that $\bar f$ is continuous in $[0,1)$, and can be extended continuously to $[0,1]$ when $1$ is a Lebesgue point for $f$, completes the proof. - -REPLY [3 votes]: Certainly all the continuous functions do satisfy this property. But I think we will have to see whether there are classes of functions which are not continuous but satisfy this property. -To show that continuous functions satisfy this, we shall use the well-known Stone Weierstrass Theorem (SWT) which states that for any continuous function $f:[a,b] \to \mathbb{R} $ , there exist a sequence of polynomials which uniformly converge to $f$. -Define, $$n \int_0^1 x^n f(x)dx=L_n(f)$$ -Then,$L_n(x^k) = \ n \int_0^1 x^n x^k dx= \frac{n}{n+k+1} $ and $\lim_{n\to\infty} L_n(x^k)=1$ for all non-negative integers $k$. -Now if $P(x)$ is an arbitrary polynomial then from above it follows that $\lim_{n\to\infty} L_n(P)=P(1)$ -Let $\epsilon>0.$ Given a continuous function $f$ and a polynomial $P$ we have that $$|L_n(f)-L_n(P)| \leq L_n(|f-P|) \leq L_n( \epsilon/3)=\epsilon/3$$ provided that $d(f,P) \leq \epsilon/3$ and indeed by the SWT we can choose such a polynomial $P$ -And also there exist $N \in \mathbb{N}$ such that for all $n > N, |L_n(P)-P(1)|<\epsilon/3$ -Now for such an $n$, -$$|L_n(f)-f(1)| \leq|L_n(f)-L_n(P)|+|L_n(P)-P(1)|+|P(1)-f(1)|<\epsilon$$ -This proves that -$\lim_{n\to\infty}L_n(f)=f(1)$ for all continuous functions.<|endoftext|> -TITLE: Is the adjoint representation of SO(4) self-dual? -QUESTION [6 upvotes]: The adjoint representation (over the complex numbers) of SO(4) is 6-dimensional. Is this representation self-dual? -Other than the adjoint representation and its dual, are there other irreducible 6-dimensional representations of SO(4) over the complex numbers? - -REPLY [8 votes]: The adjoint representation of any semisimple Lie group $G$ is self-dual. This follows from the fact that the Killing form gives a nondegenerate $G$-invariant bilinear map $\mathfrak{g} \times \mathfrak{g} \to \mathbb{R}$ for such $G$ (and we can then extend scalars to $\mathbb{C}$ to get a corresponding such map for $\mathfrak{g} \otimes_{\mathbb{R}} \mathbb{C}$). -The adjoint representation of $\text{SO}(4)$ is also not irreducible (and this is the only $n \ge 2$ for which this is true). This follows from the fact that $\mathfrak{so}(4) \cong \mathfrak{su}(2) \oplus \mathfrak{su}(2)$ as a Lie algebra and so the adjoint representation of $\mathfrak{so}(4)$ on itself is a direct sum of two $3$-dimensional representations. -Edit: In fact $\text{SO}(4)$ admits no irreducible $6$-dimensional representations. Its universal cover is $\text{Spin}(4) \cong \text{SU}(2) \times \text{SU}(2)$ which admits two irreducible $6$-dimensional representations given by $V \otimes S^2(V)$ and $S^2(V) \otimes V$ where $V$ is the defining representation of $\text{SU}(2)$. The kernel of the covering map $\text{Spin}(4) \to \text{SO}(4)$ is generated by $(-1, -1)$ and this acts nontrivially in both of the above representations, so neither descends to a representation of $\text{SO}(4)$.<|endoftext|> -TITLE: How many irreducible 3-dimensional representations does SO(4) have? -QUESTION [6 upvotes]: EDITED: I learned something from Qiaochu Yuan's answer to my other question on SO(4). I clarify my original question with more information below. My original question is: Does $SO(4)$ have exactly two irreducible 3-dimensional representations that are dual to each other? -The universal cover of SO(4) is ${\mathrm{Spin}}(4)\cong {\mathrm{SU}}(2)\times {\mathrm{SU}}(2)$. An irreducible representation of ${\mathrm{Spin}}(4)$ thus takes the form $V\otimes W$ where $V$ and $W$ are irreducible representations of SU(2). Here I consider representations over the complex numbers. There is exactly one irreducible representation of SU(2) in each dimension. Thus there are exactly two 3-dimensional irreducible representations of ${\mathrm{Spin}}(4)$, namely $V\otimes {\mathbf{1}}$ and ${\mathbf{1}}\otimes V$, where $V$ is the standard representation on SU(2) and ${\mathbf{1}}$ is the trivial representation. -Does the representation $V\otimes {\mathbf{1}}$ descend to $SO(4)$? From the above, we can conclude that either SO(4) has no irreducible 3-dimensional representations or has two of them. Furthermore, if SO(4) has two of them, then these two representations are dual to each other. - -REPLY [2 votes]: Yes, we have that $SO(4)=$Spin$(4)/\mathbb Z_2$ and so we need only worry about whether $-1=-1\otimes1\in $Spin$(4)$ acts trivially, which indeed it does on both the $\mathbf 1$ and $\mathbf 3$.<|endoftext|> -TITLE: Is there a way to define the "size" of an infinite set that takes into account "intuitive" differences between sets? -QUESTION [6 upvotes]: The usual way to define the "size" of an infinite set is through cardinality, so that e.g. the sets $\{1, 2, 3, 4, \ldots\}$ and $\{0, 1, 2, 3, 4, \ldots\}$ have the same cardinality. However, is this the only way to define a useful "size" of an infinite set? Could one conceivably define a "size" where the two example sets have different sizes? - -REPLY [12 votes]: The question is what do you want to capture in the notion of cardinality, and what is your settings. -For example, if you only care about subsets of the natural numbers you can say that $A$ has cardinality larger than $B$ if either $A$ is strictly larger than $B$ or the complement of $A$ is strictly smaller. -In this sense, $|\{0,1,2,\ldots\}|$ is larger than $|\{1,2,3,4\ldots\}|$, as the former is everything is the latter has a non-empty complement (namely, $\{0\}$). This notion is not a linear order, which may seem a bit bad but then again - without the axiom of choice cardinalities are not linearly ordered anyway. - -Another example is if you fix for every set $A$ a linear ordering $\leq_A$. Now we say that $|A|\leq|B|$ if and only if $(A,\leq_A)$ is embedded into $(B,\leq_B)$. This notion of size is also nice and has the benefit of $\mathbb N$ being strictly smaller than $\mathbb Z$ and both smaller than $\mathbb Q$ (as order types, of course). -On the other hand, this notion of cardinality lacks the Cantor-Bernstein property, namely anti-symmetry: the order type of $\mathbb Q\cap[0,1]$ and that of $\mathbb Q\cap(0,1)$ are not equal, but either one embeds into the other. -This is also a bit bad, but there is a natural ordering on cardinals which also lacks this property without the axiom of choice: $|A|\leq^\ast|B|$ if and only if $A=\varnothing$ or there exists a surjection $f\colon B\to A$. Without the axiom of choice it is consistent that there are two sets which can be mapped onto one another, but not bijectively. - -The above can be generalized to any form of structure. Simply fix for every set in the universe some sort of structure and consider the embedding as a natural order. However in many cases you do lose something in the sense that this is no longer acting as we would expect from cardinality. Or perhaps we are expecting the wrong things... -Further reading: - -Cardinality != Density? -Comparing the sizes of countable infinite sets<|endoftext|> -TITLE: Proof that every finite dimensional normed vector space is complete -QUESTION [43 upvotes]: Can you read my proof and tell me if it's correct? Thanks. -Let $V$ be a vector space over a complete topological field say $\mathbb R$ (or $\mathbb C$) with $\dim(V) = n$, base $e_i$ and norm $\|\cdot\|$. Let $v_k$ be a Cauchy sequence w.r.t. $\|\cdot\|$. Since any two norms on a finite dimensional space are equivalent, $\|\cdot\|$ is equivalent to the $l^1$-norm $\|\cdot\|_1$ which means that for some constant $C$, $\varepsilon > 0$, $k,j$ large enough, -$$ \varepsilon > \|v_j - v_k\| \geq C \|v_j - v_k\|_1 e_i= C \sum_{i=1}^n |v_{ji} - v_{ki}| \geq |v_{ji} - v_{ki}|$$ -for each $1 \leq i \leq n$. Hence $v_{ki}$ is a Cauchy sequence in $\mathbb R$ (or $\mathbb C$) for each $i$. $\mathbb R$ (or $\mathbb C$) is complete hence $v_i = \lim_{k \to \infty} v_{ki} $ is in $\mathbb R$ (or $\Bbb C$) for each $i$. Let $v = (v_1, \dots , v_n) = \sum_i v_i e_i$. Then $v$ is in $V$ and $\|v_k - v\| \to 0$: -Let $\varepsilon > 0$. Then -$$ \|v_k - v\| \leq C \|v_k - v\|_1 = C \sum_{i=1}^n |v_{ki} - v_i| \leq C^{'}n \varepsilon$$ -for $k$ large enough. - -REPLY [14 votes]: I think there is a simple way to demonstrate this. -Let $ \left(E, \| \| \right)$ be a $\mathbb{K}$ finite dimensionial vector space. -Consider the following application: -\begin{aligned} -\mathbb { K }^ { n } & \rightarrow E \\ -\left( \lambda _ { 1 } , \ldots , \lambda _ { n } \right) & \mapsto \sum _ { i = 1 } ^ { n } \lambda _ { i } e _ { i } -\end{aligned} -which is a linear isometric bijection (thus, a homeomorphism) between $ \left(\mathbb{ K }^ { n }, \| \| _\infty\right)$ and $ \left(E, \| \|_{\infty} \right)$. -The pre-image of a complete space by a uniformly continuous and bijective function being complete and $ \left(\mathbb{ K }^ { n }, \| \| _\infty\right)$ being a complete space, then $ \left( E, \| \| _\infty\right)$ is complete. Since all norms on a finite dimensional space are equivalent, it folows that $ \left(E, \| \| \right)$ is complete.<|endoftext|> -TITLE: Pullbacks of categories -QUESTION [22 upvotes]: Let $\mathfrak{Cat}$ be the 2-category of small categories, functors, and natural transformations. Consider the following diagram in $\mathfrak{Cat}$: -$$\mathbb{D} \stackrel{F}{\longrightarrow} \mathbb{C} \stackrel{G}{\longleftarrow} \mathbb{E}$$ -There are several notions of pullback one could investigate in $\mathfrak{Cat}$: - -The ordinary pullback in the underlying 1-category $\textbf{Cat}$: these exist and are unique, by ordinary abstract nonsense. Explicitly, $\mathbb{D} \mathbin{\stackrel{1}{\times}_\mathbb{C}} \mathbb{E}$ has objects pairs $(d, e)$ such that $F d = G e$ (evil!) and arrows are pairs $(k, l)$ such that $F k = G l$. This evidently an evil notion: it is not stable under equivalence. For example, take $\mathbb{C} = \mathbb{1}$: then we get an ordinary product; but if $\mathbb{C}$ is the interval category $\mathbb{I}$, we have $\mathbb{1} \simeq \mathbb{I}$, yet if I choose $F$ and $G$ so that their images are disjoint, we have $\mathbb{D} \mathbin{\stackrel{1}{\times}_\mathbb{C}} \mathbb{E} = \emptyset$, and $\emptyset \not\simeq \mathbb{D} \times \mathbb{E}$ in general. -The strict 2-pullback is a category $\mathbb{D} \mathbin{\stackrel{s}{\times}_\mathbb{C}} \mathbb{E}$ and two functors $P : \mathbb{D} \mathbin{\stackrel{s}{\times}_\mathbb{C}} \mathbb{E} \to \mathbb{D}$, $Q : \mathbb{D} \mathbin{\stackrel{s}{\times}_\mathbb{C}} \mathbb{E} \to \mathbb{E}$ such that $F P = G Q$, with the following universal property (if I'm not mistaken): for all $K : \mathbb{T} \to \mathbb{D}$ and $L : \mathbb{T} \to \mathbb{E}$ such that $F K = G L$, there is a functor $H : \mathbb{T} \to \mathbb{D} \mathbin{\stackrel{s}{\times}_\mathbb{C}} \mathbb{E}$ such that $P H = K$ and $Q H = L$, and $H$ is unique up to equality; if $K' : \mathbb{T} \to \mathbb{D}$ and $L' : \mathbb{T} \to \mathbb{E}$ are two further functors such that $F K' = G L'$ and $H' : \mathbb{T} \to \mathbb{D} \mathbin{\stackrel{s}{\times}_\mathbb{C}} \mathbb{E}$ satisfies $P H' = K'$ and $Q H' = L'$ and there are natural transformations $\beta : K \Rightarrow K'$ and $\gamma : L \Rightarrow L'$, then there is a unique natural transformation $\alpha : H \Rightarrow H'$ such that $P \alpha = \beta$ and $Q \alpha = \gamma$. So $\mathbb{D} \mathbin{\stackrel{s}{\times}_\mathbb{C}} \mathbb{E} = \mathbb{D} \mathbin{\stackrel{1}{\times}_\mathbb{C}} \mathbb{E}$ works, and in particular, strict 2-pullbacks are evil. -The pseudo 2-pullback is a category $\mathbb{D} \times_\mathbb{C} \mathbb{E}$, three functors $P : \mathbb{D} \times_\mathbb{C} \mathbb{E} \to \mathbb{D}$, $Q : \mathbb{D} \times_\mathbb{C} \mathbb{E} \to \mathbb{E}$, $R : \mathbb{D} \times_\mathbb{C} \mathbb{E} \to \mathbb{C}$, and two natural isomorphisms $\phi : F P \Rightarrow R$, $\psi : G Q \Rightarrow R$, satisfying the following universal property: for all functors $K : \mathbb{T} \to \mathbb{D}$, $L : \mathbb{T} \to \mathbb{E}$, $M : \mathbb{T} \to \mathbb{C}$, and natural isomorphisms $\theta : F K \Rightarrow M$, $\chi : G L \Rightarrow M$, there is a unique functor $H : \mathbb{T} \to \mathbb{D} \times_\mathbb{C} \mathbb{E}$ and natural isomorphisms $\tau : K \Rightarrow P H$, $\sigma : L \Rightarrow Q H$, $\rho : M \Rightarrow R H$ such that $\phi H \bullet F \tau = \rho \bullet \theta$ and $\psi H \bullet G \sigma = \rho \bullet \chi$ (plus some coherence axioms I haven't understood); and some further universal property for natural transformations. -By considering the cases $\mathbb{T} = \mathbb{1}$ and $\mathbb{T} = \mathbb{2}$, it seems that $\mathbb{D} \times_\mathbb{C} \mathbb{E}$ can be taken to be the following category: its objects are quintuples $(c, d, e, f, g)$ where $f : F d \to c$ and $g : G e \to c$ are isomorphisms, and its morphisms are triples $(k, l, m)$ where $k : d \to d'$, $l : e \to e'$, $m : c \to c'$ make the evident diagram in $\mathbb{C}$ commute. The functors $P, Q, R$ are the obvious projections, and the natural transformations $\phi$ and $\psi$ are also given by projections. -Question. This seems to satisfy the required universal properties. Is my construction correct? -Question. What are the properties of this construction? Is it stable under equivalences, in the sense that $\mathbb{D}' \times_{\mathbb{C}'} \mathbb{E}' \simeq \mathbb{D} \times_\mathbb{C} \mathbb{E}$ when there is an equivalence between $\mathbb{D}' \stackrel{F'}{\longrightarrow} \mathbb{C}' \stackrel{G'}{\longleftarrow} \mathbb{E}'$ and $\mathbb{D} \stackrel{F}{\longrightarrow} \mathbb{C} \stackrel{G}{\longleftarrow} \mathbb{E}$? -Finally, there is the non-strict 2-pullback, which as I understand it has the same universal property as the pseudo 2-pullback but with "unique functor" replaced by "functor unique up to isomorphism". -Question. Is this correct? - -General question. Where can I find a good explanation of strict 2-limits / pseudo 2-limits / bilimits and their relationships, with explicit constructions for concrete 2-categories such as $\mathfrak{Cat}$? So far I have only found definitions without examples. (Is there a textbook yet...?) - -REPLY [9 votes]: Zhen, I am not sure I understand your question. - -Every stict 2-limit is obviously also a 1-limit in the underlying 1-category, so these are not really different concepts (a 2-limit is a strengthen version of a limit; BTW, since Cat is 2-complete then every 1-limit in Cat is automatically a 2-limit). -Your construnction of a 2-pseudo pullback is fine. However, it is easy to verify that it is not stable under equivalence of categories ([Added] in the sense that: $C$ is a limit of $F$ and $D$ is equivalent to $C$ does not imply that $D$ is a limit of $F$). All of the mentioned limits are defined in terms of "strict" adjunctions (or, more acurately, in terms of strict universal properties), i.e. there is a natural isomorphism: -$$\Delta(C) \rightarrow F \approx C \rightarrow \mathit{lim}(F)$$ -To obtain a concept that is stable under equivalences, you have to replace this natural isomorphism by a natural equivalence of categories (plus perhaps some additional coherence equations). Of course, every strict 2-limit is also a "weak" 2-limit in the above sense (because every isomorphism is an equivalence), so again in a complete 2-category you will not get anything new. - -[Added] -3. Let $\mathbb{W}$ be a 2-category, and $X$ a 1-category. There are three types of 2-cateogrical cones in $\mathbb{W}$ of the shape of $X$: - -$\mathit{Cone}$ --- objects are strict functors $X \rightarrow \mathbb{W}$, 1-morphisms are strict natural transformations between functors, and 2-morphisms are modifications between natural transformations -$\mathit{PseudoCone}$ --- objects are pseudo functors $X \rightarrow \mathbb{W}$, 1-morphisms are pseudo natural transformations between functors, and 2-morphisms are modifications between natural transformations -$\mathit{LaxCone}$ --- objects are lax functors $X \rightarrow \mathbb{W}$, 1-morphisms are lax natural transformations between functors, and 2-morphisms are modifications between natural transformations - -A limit of a strict functor $F \colon X \rightarrow \mathbb{W}$ is a 2-representation of: -$$\mathit{Cone}(\Delta(-), F)$$ -where $\Delta$ is the usual diagonal functor. A pseudolimit of $F$ is a representation of: -$$\mathit{PseudoCone}(\Delta(-), F)$$ -And a lax limit is a representation of: -$$\mathit{LaxCone}(\Delta(-), F)$$ -In each case if you take equivalent functors, then you get equivalent representations. However, in each case the notion of equivalent functors is different. Perhaps your problem is that you are using the equivalence from $\mathit{PseudoCone}$ in the context of $\mathit{Cone}$ -[Added^2] -I have missed one of your questions: - -Finally, there is the non-strict 2-pullback, which as I understand it has the same universal property as the pseudo 2-pullback but with "unique functor" replaced by "functor unique up to isomorphism". - -If by a non-strict pullback you mean a weak (pseudo)pullback in the above sense, then the universal property is much more subtle --- it does not suffice to say that there is a functor $f \colon X \rightarrow \mathit{Lim}(F)$ that is unique up to a 2-isomorphism (just like in the definition of a limit you do not say that there is an object which is unique up to 1-isomorphism), you have to say that for every cone $\alpha \colon \Delta(X) \rightarrow F$ there exists $f \colon X \rightarrow \mathit{Lim}(F)$ such that for any cone $\beta$ on $X$ with its $g \colon X \rightarrow \mathit{Lim}(F)$ and every family of 2-morphism $\tau \colon \alpha \rightarrow \beta$ that is compatible with $F$ there exists a unique 2-morphism $f \rightarrow g$ such that everything commutes. -However, if by a non-strict pullback you mean a lax pullback, then the construction is similar to your construction of a pseudopullback --- without requirement that your $f$ and $g$ are isomorphisms. -You have aslo asked: - -Where can I find a good explanation of strict 2-limits / pseudo 2-limits / bilimits and their relationships, with explicit constructions for concrete 2-categories such as $\mathfrak{Cat}$? So far I have only found definitions without examples. (Is there a textbook yet...?) - -I do not know of any good textbook, but can provide you with two examples. -There is a simple general procedure to construct strict/pseudo/lax limits and colimits in $\mathbf{Cat}$. You shall notice that to give a monad is to give a lax functor $T \colon 1 = 1^{op} \rightarrow \mathbf{Cat}$. Then the lax colimit of $T$ is the Kleisli category for the monad $T$, and the lax limit of $T$ is the Eilenberg-Moore category for the monad $T$. This idea may be pushed a bit further: you may think of a lax functor $\Phi \colon \mathbb{C}^{op} \rightarrow \mathbf{Cat}$ as of a kind of a "multimonad". Then its multi-Kleisli resolution is given by the Grothendieck construction $\int \Phi$ (this construction gives a fibration precisely when $\Phi$ is a pseudofunctor). And similarly its multi-Eilenberg-Moore category is given by a suitable collection of (ordinary) algebras. In these construction if you impose a reguirement on cartesian morphisms / algebras to be isomorphisms, then you get a pseudocolimit / pseudolimit of $\Phi$, and if you impose identities instead of isomorphisms you get a colimit / limit. -What is more, the Grothendieck construction works also for the bicategory of distributors; and because there is a duality on the bicategory of distributors you may construct (lax/pseudo/strict) limits in this bicategory via the Grothendieck construction as well.<|endoftext|> -TITLE: Truth Table for If P then Q -QUESTION [12 upvotes]: Possible Duplicate: -In classical logic, why is (p -> q) True if both p and q are False? - -The Logic table for If P then Q is as follows: -P Q If P then Q -T T T -T F F -F T T -F F T - -What I don't understand is, How can there be a truth table for this? -As far as I understand, If p then Q means "if P is true, Q has to be true. Any other case, I don't know" -So, from what I understand, the first 2 rows of the truth table state that "If P is true and Q is true, the outcome is correct and If P is true and Q is false, the outcome is incorrect (F)" -What about the last 2 rows? - -REPLY [2 votes]: You might find this hand-out for beginning logic students helpful too, http://www.logicmatters.net/resources/pdfs/Conditionals.pdf<|endoftext|> -TITLE: Physical interpretation of the Lie Bracket -QUESTION [6 upvotes]: I've come accross this physical interpretation for $ [X,Y] $ which I don't understand : - -Follow $X$ for some time $\epsilon$; -Follow $Y$ for $\epsilon$; -Follow -X for $\epsilon$; -Follow -Y for $\epsilon$; - -In the limit as $\epsilon$ approaches 0, the result of the above motion approaches the Lie Bracket $[X,Y]$. -Maybe someone can elucidate this for me? - -REPLY [2 votes]: The phrase 'flow $g$ along $Y$ a small distance $\epsilon$' is very nice, but all it means ultimately is just 'take the directional derivative of the function (e.g. surface) $g$ along $Y$'. Thus $X(Y)g$ means 'take the directional derivative of $Yg$ along $X$'. But $Yg$ is itself a directional derivative. Thus, $X(Y)g$ means 'take the directional derivative along $X$ of the directional derivative of $g$ along $Y$'. -Intuitively this is a generalisation of $\frac{\partial^2 g}{\partial x \partial y}$, since in the Lie bracket the two vector fields $X$ and $Y$ do not have to be orthogonal. -The second half of the Lie bracket then subtracts the same derivations in reverse order. If the two derivations commute, the Lie bracket is zero. -The vector flow terminology has definite aesthetic appeal (my original background is in fluid mechanics), but it remains difficult for me to visualise intuitively how the four arcs shrinking to a point as $\epsilon\rightarrow 0$ end up with this double derivation. The slope of a surface $g(x,y)$ in a particular direction at a given point $(x,y)$, on the other hand, is immediately obvious; and how that slope then may change in the direction of the other vector field is also intuitively clear. Once this has been established, the fact that these two double derivations may differ holds no mysteries. -Of course, the picture of the four arcs and whether they close or not remains the best visualisation of the Lie bracket once what it means has been understood.<|endoftext|> -TITLE: If $p$ is a prime and $x,y \in \mathbb{Z}$, then $(x+y)^p \equiv x^p+y^p \pmod{p}$ -QUESTION [5 upvotes]: I want to prove that if $p$ is a prime and $x,y \in \mathbb{Z}$, then $$(x+y)^p \equiv x^p+y^p \pmod{p}$$ -So far I know that -$$(x+y)^p = \sum_{k=0}^{p} \dbinom{p}k x^{p-k} y^k$$ -A part of the above equation is supposed to cancel, I think, but I can't figure out a way how to make it cancel. - -REPLY [3 votes]: HINT -If $p$ is a prime, then $p$ divides $\dbinom{p}{k}$ for all $k \in \{1,2,\ldots,p-1\}$. -The identity you have i.e. $(x+y)^p = x^p + y^p$, is referred to as Freshman's dream. -Move your mouse over the gray area for the complete answer. - - Note that $$\dbinom{p}{k} = \dfrac{p \times (p-1) \times (p-2) \times \cdots \times (p-k+2) \times (p-k+1)}{k \times (k-1) \times (k-2) \times \cdots \cdots \times 2 \times 1}$$ is an integer. Since $k \in \{1,2,\ldots,p-1\}$, and $p$ is a prime, none of $k, k-1, k-2, \ldots, 2$ divide $p$. Hence, we can factor $p$ from the numerator to get that $$\dbinom{p}{k} = \dfrac{p \times (p-1) \times (p-2) \times \cdots \times (p-k+2) \times (p-k+1)}{k \times (k-1) \times (k-2) \times \cdots \cdots \times 2 \times 1} = p \times M$$ where $M \in \mathbb{Z}$.<|endoftext|> -TITLE: Generalized PNT in limit as numbers get large -QUESTION [9 upvotes]: If $\pi_k(n)$ is the cardinality of numbers with k prime factors (repetitions included) less than or equal n, the generalized Prime Number Theorem (GPNT) is: -$$\pi_k(n)\sim \frac{n}{\ln n} \frac{(\ln \ln n)^{k-1}}{(k-1)!}.$$ -The qualitative appearance of the actual distribution of $\pi_k(n)$ for k = 1,2,3,..., agrees very well with the GPNT, for numbers $n$ within reach of my laptop. But I noticed that as $n$ and $k$ get large, "most" of the numbers less than $n$ seem to have relatively few factors. -Writing $n = 2^m$ and replacing $k$ by $x$ we can graph -$$f(x) =\frac{2^m (\ln\ln 2^m)^{x-1}}{\ln 2^m (x-1)!}$$ -from $x = 1$ to $m$ (since no number will have more than m factors) and see that for relatively small fixed $m$, most of the area under the curve f is is contained in a steep bell-shaped curve on the far left of the image. -I take this to suggest that as we consider very large sets, $S_m = \{ 1,2,3,...,2^m\},$ almost all elements of these sets have a "very small" number of factors (including repetitions). -Can this idea be (or has it been) quantified? The phrase "very small" is frustrating, and I think we might be able to say something more concrete about, for example, concentration of the proportion of area as a function of x and m...? Thanks for any suggestions. -Edit: the answer Eric Naslund gave below is splendid and I won't neglect to accept it. In response to the answer, I wonder if there is any reason not to be able to get something like that answer from the expression $f(x)$? -After all, $f(x)$ appears to be a Poisson-like curve with a mean near the average number of prime factors. If I let m = 100 and then 500 (i.e., we're using $2^{100},2^{500}$), $f'(x) = 0$ at $x \approx 4.73, 6.34$, respectively, while $\ln\ln 2^m$ is respectively 4.23, 5.84. If f is a valid expression for the asymptotic behavior of $\pi_k(n)$, wouldn't we expect it to give us this additional information? Can we not prove it? - -REPLY [5 votes]: I am not quite sure that I understand exactly what you're asking, but I think that some of the following points may shed some light on the matter. -Suppose we fix $n$ and let $k$ vary. Studying $\pi_k(n)/n$ is then equivalent to looking at the distribution of the number of prime factors of integers up to $n$ (by the way, this turns out not to be very sensitive to the distinction between "with repetition" and "without repetition"). -We have to be a bit careful as the asymptotic for $\pi_k(n)$ is intended for $k$ fixed and $n$ growing (but we can still use it as long as $k$ grows sufficiently slowly with $n$). With that caveat in mind, the approximate distribution given by GPNT can be rewritten as: -$$\frac{\pi_{k+1}(n)}{n} \approx \frac{\lambda^{k} e^{-\lambda}}{k!},\quad k=0,1,2,\ldots$$ -where $\lambda = \log \log x$, and I have shifted the value of $k$ by 1. By rewriting it this way, we can recognize the right hand side exactly as a Poisson distribution with parameter $\lambda = \log \log x$. -Loosely interpreted, every number (above 1) is guaranteed to have at least one prime factor, but the additional primes beyond that behave Poissonly with a rate of $\log \log x$. The peak of any Poisson distribution always occurs at $\lambda$. You don't need to take derivatives to see this, just look at the ratio of the $k-1$ and $k$ terms: -$$\frac{\lambda^k e^{-\lambda}}{k!} \cdot \frac{(k-1)!}{\lambda^{k-1} e^{-\lambda}} = \frac{\lambda}{k}.$$ -So clearly the Poisson distribution is increasing in $k$ for $k < \lambda$ and decreasing for $k > \lambda$, and the peak occurs at $\lfloor\lambda\rfloor$. -Moreover, it is well-known (see the linked Wikipedia article) that the normal distribution with equal mean and variance $\lambda$ approximates the Poisson distribution when $\lambda$ is large. Perhaps this explains the bell curve you observed. -Finally, note that equal mean and variance $\log \log x$ is precisely the situation that arises from the Erdős-Kac theorem in Eric Naslund's answer, although let me reiterate that I am deliberately ignoring the relative sizes of $n$ and $k$ for which these separate facts are known to be true.<|endoftext|> -TITLE: Does existence of anti-derivative imply integrability? -QUESTION [17 upvotes]: If $f$ has an anti-derivative in $[a,b]$ does it imply that $f$ is Riemann integrable in $[a,b]$? - -REPLY [17 votes]: Take $f(x)=\begin{cases} x^2\sin (1/x^2), &x\ne 0, \\ 0, &x=0. \end{cases}\quad$ Then $g=f'$ exists everywhere but is unbounded over $[-1,1]$. $g$ thus has a primitive but is not Riemann integrable.<|endoftext|> -TITLE: Finding a function that fits the "lowest points" of another one -QUESTION [5 upvotes]: I came up with this problem, which I cannot solve myself. -Consider the function: -$\displaystyle f(x) = x^{\ln(|\pi \cos x ^ 2| + |\pi \tan x ^ 2|)}$, which has singularities at $\sqrt{\pi}\sqrt{n + \dfrac{1}{2}}$, with $n \in \mathbb{Z}$. -Looking at its graph: - -we can see it is globally increasing: - -I was wondering if there exists a function $g(x)$, such that $f(x) - g(x) \ge 0, \forall x \in \mathbb{R^{+}}$ and that best fits the "lowest points" of $f(x)$. -Sorry for the inaccurate terminology but I really don't know how to express this concept mathematically. Here is, for example, $g(x) = x ^ {1.14}$ (in red): - -Actually $g(x)$ is not correct because for small values of $x$ it is greater than $f(x)$. - -Is it possible to find such a $g(x)$, given that the "nearest" is $g(x)$ to $f(x)$'s "lowest points" the better it is? Again, sorry for my terminology, I hope you could point me in the right direction. -Thanks, - -REPLY [2 votes]: As $a^{\ln b}=\exp(\ln a\cdot\ln b)=b^{\ln a}$ the function $f$ can be written in the following way: -$$f(x)=\bigl(\pi|\cos(x^2)|+\pi|\tan(x^2)|\bigr)^{\ln x}\ .$$ -Now the auxiliary function -$$\phi:\quad{\mathbb R}\to[0,\infty],\qquad t\mapsto \pi(|\cos(t)|+|\tan(t)|)$$ -is periodic with period $\pi$ and assumes its minimum $\pi$ at the points $t_n=n\pi$. The function $$\psi(x):=\phi(x^2)=\pi|\cos(x^2)|+\pi|\tan(x^2)|\bigr)$$ assumes the same values as $\phi$; in particular it is $\geq\pi$ for all $x\geq0$ and $=\pi$ at the points $x_n:=\sqrt{n\pi}$ $\ (n\geq0)$. Therefore -$$f(x)=\bigl(\psi(x)\bigr)^{\ln x}\geq \pi^{\ln x}=x^{\ln\pi}\qquad(x\geq1)$$ -and $=x^{\ln\pi}$ at the $x_n>1$. For $0 -TITLE: Prime spiral distribution into quadrants -QUESTION [14 upvotes]: Is it known that the primes on the Ulam prime spiral distribute themselves equally -in sectors around the origin? To be specific, say the quadrants? -(Each quadrant is closed on one axis and open on the other.) -For example, in the $50 \times 50$ spiral below, I count the number of primes in the four quadrants to be $(103,96,88,86)$ ($\sum=373$), leading to ratios $(0.276,0.257,0.236,0.231)$: -      - -For the $500 \times 500$ spiral, I count $(5520,5553,5535,5469)$ ($\sum=22077$) leading to -$(0.250,0.252,0.251,0.248)$. Empirically there is a convergence to $\frac{1}{4}$, but I wonder if this has been proven? -Thanks! - -REPLY [2 votes]: Community wiki answer so that this question can be marked as resolved: -An affirmative answer has been given on MathOverflow.<|endoftext|> -TITLE: How find the value of this integral -QUESTION [6 upvotes]: How can I compute $$\int_{-\pi}^\pi\frac{\sin(13x)}{\sin x}\cdot\frac1{1+2^x}\mathrm dx?$$ - -REPLY [8 votes]: Hint 1: -$$ -\int\limits_{-a}^a f(x)dx=\int\limits_{-a}^a \frac{f(x)+f(-x)}{2}dx -$$ -Hint 2: -$$ -\frac{\sin (n x)}{\sin x}= -\frac{(e^{ix})^n-(e^{-ix})^{n}}{e^{ix}-e^{-ix}}= -\sum\limits_{k=0}^{n-1} (e^{ix})^{n-k-1}(e^{-ix})^k -$$ -Hint 3: -$$ -\int\limits_{-\pi}^{\pi} e^{ikx}dx= -\begin{cases} -2\pi&\text{ if }\quad k=0\\ -0 &\text{ if }\quad k\in\mathbb{Z}\setminus\{0\} -\end{cases} -$$<|endoftext|> -TITLE: Secretary problem for unknown n? -QUESTION [5 upvotes]: So one of my good friends is starting to date again (after being out of the country for two years), and I think that it might be helpful, or at least fun, to keep track of her dates in a ranked fashion so that we can always be on the look-out for the optimum stopping point (i.e. who she should marry) in a semi-rigourous fashion (yes, we're nerding out about this). So I understand what the procedure is for the secretary problem with a known n, but since we're going to be doing this on the fly, how do we know when to accept the new best ranked guy as the one? Thanks! - -REPLY [3 votes]: As asked, you should estimate how many candidates there will be, then divide by e. It is clearly not 1,000,000 and probably not 10, either. I think if you study it, the optimum is rather flat, so being off somewhat is not that big a deal. There are many "real life" things that modify the problem. The two largest that I think of are first, that as you meet candidates, you get an idea of the distribution, so can make a more informed decision and second, there is an opportunity cost of waiting, which should bias you early.<|endoftext|> -TITLE: Are Polish space and LCCB space related? -QUESTION [8 upvotes]: It seems these spaces are the most useful ones for doing probabilities. Are LCCB (locally compact with countable basis) somewhat more general spaces that when endowed with a metric become Polish? I think I once knew the answer to this question. Thanks - -REPLY [5 votes]: Theorem. Every locally-compact second-countable Hausdorff space is a Polish space. -I thought I'd quote a sketch of proof of above fact from somewhere else. -The following sketch is from https://golem.ph.utexas.edu/category/2008/08/polish_spaces.html - -if X is second countable locally compact Hausdorff, then the one-point - compactification X+ is metrizable and compact, hence complete (under - any metric), and of course second countable, hence X+ is Polish. An - open subspace of a Polish space is Polish, hence X is Polish. -Bourbaki (which I found as a Google Book result) provided invaluable - assistance. - -As for why an open subspace of a Polish space is Polish, again from the same link: - -I had no idea that an open set of a Polish space was Polish — it - seemed pretty tough removing that extra point and finding a complete - metric on what’s left. -I now see how easy this is: we start with a metric on the one-point - compactification X+, remove the point at infinity, and ‘stretch’ the - metric near the removed point to get a complete metric on X. - -As for why the one-point compactification is metrizable, the selected answer from the following thread explains how to build a local countable basis at infinity: -If $X$ is locally compact, second countable and Hausdorff, then $X^*$ is metrizable and hence $X$ is metrizable<|endoftext|> -TITLE: Help find hard integrals that evaluate to $59$? -QUESTION [122 upvotes]: My father and I, on birthday cards, give mathematical equations for each others new age. This year, my father will be turning $59$. -I want to try and make a definite integral that equals $59$. So far I can only think of ones that are easy to evaluate. I was wondering if anyone had a definite integral (preferably with no elementary antiderivative) that is difficult to evaluate and equals $59$? Make it as hard as possible, feel free to add whatever you want to it! - -REPLY [135 votes]: compact : $$\int_0^\infty \frac{(x^4-2)x^2}{\cosh(x\frac{\pi}2)}\,dx$$<|endoftext|> -TITLE: Improper integral about exp appeared in Titchmarsh's book on the zeta function -QUESTION [7 upvotes]: May I ask how to do the following integration? -$$\int_0^\infty \frac{e^{-(\pi n^{2}/x) -(\pi t^2 x)}}{\sqrt{x}} dx $$ -where $t>0$, $n$ a positive integer. -This came up on page 32 (image) of Titchmarsh's book, The Theory of the Riemann Zeta-Function. Specifically for the sum involving $b_n$, I am wondering how to - -multiply by $e^{-\pi t^{2} x}$ and integrate over $(0,\infty)$ - -REPLY [6 votes]: $$\begin{eqnarray*} -\int_0^\infty \frac{dx}{\sqrt{x}}\, \exp\left(-\frac{\pi n^2}{x} - \pi t^2 x\right) -&=& 2\sqrt{\frac{n}{t}} \int_0^\infty dz\, \exp\left(-\pi n t(z^2 + z^{-2})\right) - \hspace{5ex} (\textrm{let } x=z^2 n/t) \\ -&=& \sqrt{\frac{n}{t}} \int_{-\infty}^\infty ds\, e^{s/2} - \exp\left(-2 \pi n t \cosh s\right) - \hspace{5ex} (\textrm{let } z=e^{s/2}) \\ -&=& 2\sqrt{\frac{n}{t}} \int_0^\infty ds\, \cosh\left(\frac{s}{2}\right) - \exp\left(-2 \pi n t \cosh s\right) \\ -&=& 2\sqrt{\frac{n}{t}} K_{\frac{1}{2}} (2\pi n t) - \hspace{5ex} (\textrm{modified Bessel function, 2nd kind}) \\ -&=& \frac{e^{-2\pi n t}}{t} -\end{eqnarray*}$$ -Addendum: An approach not involving special functions. -$$\begin{eqnarray*} -\int_0^\infty \frac{dx}{\sqrt{x}}\, - \exp\left(-\frac{\pi n^2}{x} - \pi t^2 x\right) -&=& 2\sqrt{\frac{n}{t}} \int_0^\infty dz\, - \exp\left(-\pi n t(z^2 + z^{-2})\right) - \hspace{5ex} (\textrm{as before}) \\ -&=& 2\sqrt{\frac{n}{t}} \int_0^\infty dz\, - \exp\left(-\pi n t(z-z^{-1})^2 - 2\pi n t\right) \\ -&=& \sqrt{\frac{n}{t}} e^{-2\pi n t} \int_{-\infty}^\infty du\, - \left(1+\frac{u}{\sqrt{u^2+4}}\right) e^{-\pi n t u^2} - \hspace{4ex} (z-z^{-1}=u) \\ -&=& \frac{e^{-2\pi n t}}{t} - \hspace{5ex} (\textrm{odd integral vanishes; Gaussian left over}) -\end{eqnarray*}$$<|endoftext|> -TITLE: Is the free commutative monoid on $n$ generators essentially the same as a polynomial ring in $n$ variables? -QUESTION [5 upvotes]: I let $M$ denote the free commutative monoid generated by some elements $x_1,\dots, x_n$. Suppose too that $R$ is a commutative ring. Then I can construct the free algebra of $M$ over $R$, denote it $R[M]$. -Recall that $R[M]$ is then the set of mappings $f\colon M\to R$ such that $f(m)\neq 0$ for only finitely many $m\in M$. Then $+$ and $\cdot$ are given by $(f+g)(m)=f(m)+g(m)$, $(fg)(m)=\sum_{pq=m}f(p)q(p)$, $0(m)=0$, and $1(1)=1$ and $1(m)=0$ if $m\neq 1$. Moreover, $R[M]$ has a subring isomorphic to $R$ by associating $r\in R$ with $r'\in R[M]$ such that $r'(1)=r$ and $r'(m)=0$ otherwise. It also has a submonoid of the multiplicative monoid isomorphic to $M$ by associating $m\in M$ with $m'\in R[M]$ given by $m'(m)=1$ and $m'(n)=0$ otherwise. -Viewing $M$ and $R$ as being in $R[M]$, I know that any element of $R[M]$ can be written as $\sum r_im_i$. So I'm curious, is $R[M]$ isomorphic to $R[x_1,\dots,x_n]$? To me it appears that the elements of the sets look more or less the same, as does the $+$ operation, but I'm not sure if they actually are isomorphic or not. Thanks for any explanation. - -REPLY [2 votes]: As Arturo Magidin said in comments, the monoid ring $R[\mathbb N_0^n]$ and the ring of polynomials in $n$ variables $R[x_1,\dots,x_n]$ are indeed naturally isomorphic. (I write $\mathbb N_0$ to indicate that $0$ is included, as it's not always considered an element of $\mathbb N$.) Wikipedia article Monoid Ring gives this as a primary example.<|endoftext|> -TITLE: Orthogonal Trajectories -QUESTION [6 upvotes]: I am asked to show that the given families of curves are orthogonal trajectories of each other. -$$x^2+y^2=ax$$ -$$x^2+y^2=by$$ -I know that two functions are called orthogonal if at every point their tangents lines are perpendicular to each other. If I differentiate both of these functions, and the resulting expressions are reciprocals of one another, I have shown that they are orthogonal trajectories of each other. -1. -$$x^2+y^2=ax$$ -$$2x+2yy'=a'x+x'a$$ -$$y'=\frac{a-x}{2y}$$ - - -$$x^2+y^2=by$$ -$$2x+2yy'=b'y+y'b$$ -$$y'=\frac{b-2x}{y}$$ - -The results I get from differentiating these two functions don't seem to be reciprocals of each other. I am wondering if I have differentiated these two functions incorrectly, or if there is a point substitution that will show these two are reciprocals. - -REPLY [6 votes]: Suppose your two curves are defined by $x^2+y^2=ax$ and $x^2+y^2=by$ , with $a,b$ real constants. To compare the tangent line slopes at given points for each curve, we differentiate the first equation to find that -$$ -2x + 2yy' = a~~\text{and so}~~y' = \frac{a-2x}{2y}~~, -$$ -and the second to find that -$$ -2x + 2yy' = by' ~~\text{and so}~~ y' = \frac{2x}{b-2y}~~. -$$ -These answers are different from the ones you got -- note that what I did was group terms containing $y'$ on one side of the equation, and divide through by whatever factor accompanied it. -We are in business so long as the product of these two quantities, for a given pair $(x,y)$ where the curves intersect, is $-1$. So, multiply one by the other and we get -$$ -\frac{2x(a-2x)}{2y(b-2y)} = \frac{x(a-2x)}{y(b-2y)} ~~. -$$ -I believe the problem you ran into is that it is not clear (in an algebraic sense anyways) that these factors should cancel in any way. But remember -- since we are looking at a point where the two curves we were given intersect, we may apply both of those equations. Where does $a-2x$ appear? -Well we have $x^2 + y^2 = ax$ , so that $ax - x^2 = y^2$ , $ax - 2x^2 = y^2 - x^2$ , and $x(a-2x) = y^2-x^2$. Now bear with me, while this may not look simpler, observe that by the same token, -$$ -x^2 + y^2 = by ~~\text{means that}~~ by-2y^2 = x^2-y^2 ~~\text{and so}~~ y(b-2y) = x^2-y^2 ~~. -$$ -The factors on top and bottom are indeed the same aside for a sign switch, so the two slopes are negative reciprocal. -BTW: A helpful, and even pretty exercise is to actually plot some of these orthogonal curves. In this case, you'll notice that the first family is circles with an $x$-offset of the center from the origin, while the second family is circles with a $y$-offset. Is it clear why these are orthogonal? - -REPLY [5 votes]: Start from the first family of curves like this: $$x^2+y^2=ax \longrightarrow 2x+2yy'=a$$ $$\longrightarrow y'=\frac{a-2x}{2y}$$ we see that $a=\frac{x^2+y^2}{x}$ so put it to the last result above. we get:$$y'=\frac{y^2-x^2}{2xy}$$ Now if you want to find orthogonal trajectories of the first family you should solve: $$y'=-\frac{2xy}{y^2-x^2}$$ which is homogenous equation. Solve it and you will find the second family of curves with choosing suitable constant $b$ in the last solution. - -REPLY [3 votes]: generally two equation of curves are perpendicular to each other,if product of their slopes is $-1$. -from first we have -$2*x+2*y*y'=a$ so $2*y*y'=a-2*x$ finally we have $y'=(a-2*x)/(2*y)$ -from second curve -$2*x+2*y*y'=b*y'$ or $2*x=b*y'-2*y*y'$. -factor out y',we will have $2*x=y'(b-2*y)$ and $y'=(2*x)/(b-2*y)$ -if they are orthogonal,it depend what points $(x,y)$ we have,simple you need points to put and see if product of slopes is $-1$<|endoftext|> -TITLE: RSA public key cryptosystem -QUESTION [6 upvotes]: I got stuck on a homework question. If anyone could help me with this certain problem, I would be grateful. I'll state what the problem say and some relevant theorem (i believe) that I used to partly prove the problem. -Problem: A decryption exponent for an RSA public key $(N,e)$ is an integer $d$ with the property that $a^{de} \equiv a \mod N$ for all integers $a$ such that $\gcd(a,N) = 1$. Note that $N = pq$ but $p,q$ is unknown where $p,q$ are distinct primes and $e$ is the encryption exponent. -Suppose I can find a decryption exponent for a given $(N,e)$ where $N$ is a fixed modulus and for a large number $e$ that is not $N$. How can I factor $N$? -Attempt: According to the RSA public key cryptosystem, primes $p$, $q$ are picked such that they are distinct and an encryption exponent $e$ is chosen such that $\gcd(e, (p-1)(q-1)) = 1$. -So here's what I did: Suppose that I am given two public keys with both public keys having the same modulus, so I have $(N, e_1)$ and $(N,e_2)$. By hypothesis, I can find $d_1$ and $d_2$ such that $e_1d_1 \equiv 1 \mod (p-1)(q-1)$ and $e_2d_2 \equiv 1 \mod (p-1)(q-1)$. By definition of congruences, there exists $k_1, k_2 \in \mathbb{Z}$ such that $e_1d_1 - 1 = k_1[(p-1)(q-1)]$ and $e_2d_2 - 1 = k_2[(p-1)(q-1)]$. Then I took the $\gcd(e_1d_1-1,e_2d_2 -1) = \gcd(k_1,k_2)[(p-1)(q-1)]$. -Consider the case when $\gcd(k_1,k_2) = 1$. Then we found the value of $(p-1)(q-1)$, so $(p-1)(q-1) = pq - (p+q) + 1 = N - (p+q) + 1$. This implies that $(p+q) = N + 1 - (p-1)(q-1)$. So we use the quadratic formula to solve $X^2 - (p+q)X + N$ since this equation equals $(X-q)(X-p)$. Hence, I found the factors of $N$. -But if the case that $\gcd(k_1,k_2) > 1$ seems to got me stuck. If anyone can help me with this by providing a way to think about this problem, that'll be great. - -REPLY [2 votes]: I would try finding non-trivial square roots of $1$ modulo $N$. This is a bit -probabilistic, but works in practice reasonably quickly I think. -You know that $de-1=\ell\,\mathrm{lcm}\{(p-1),(q-1)\}$. Note that both $p-1$ and -$q-1$ are even, so they are not coprime. Write -$$ -de-1=2^km, -$$ -with $m$ odd (this is easy as you can just keep dividing by two). -Let $a$ be a random integer. Most likely it is coprime to $N$ (check with Euclid- if not , then you found a factor of $N$ and can quit). Compute the power $z=a^m$ modulo $N$. -Keep squaring $z$ (modulo $N$). -We know that $z^{2^k}\equiv 1\pmod N$, because -$$ -z^{2^k}\equiv a^{2^km}\equiv 1\pmod N, -$$ -as $2^km$ is divisible by both $p-1$ and $q-1$. -Let $k_1$ be the smallest integer, $0\le k_1\le k$, such that -$z^{2^{k_1}}\equiv 1\pmod N$. If $k_1=0$, then we are out of luck, -and must try another $a$. Otherwise let us examine -$$ -x\equiv z^{2^{k_1-1}} \pmod N. -$$ -Then $x^2\equiv 1\pmod N$. If $x\equiv -1$, then, again we are out of luck, and -should try another $a$. If not, then we are done, because $(x-1)$ (and $(x+1)$) -will have a non-trivial common factor with $N$ that we can again find with Euclid's -algorithm. -Can we be denied success by a string of bad luck? Not really! Let $U_{q,2}$ be -the group of residue classes modulo $q$ of order that is a power of two (=2-Sylow -subgroup of $\mathbb{Z}/q\mathbb{Z}^*$), and similarly $U_{p,2}$. Then $z$ is -equally likely to land on any element of $U_{q,2}$ (resp. $U_{p,2}$), and by -CRT these two choices are independent from one another. The process fails, if and only -if $z$ has the same order $2^{k_1}$ in both groups, because that is when -$z^{2^{k_1-1}}\equiv-1$ modulo both $p$ and $q$ resulting in $x\equiv -1\pmod{N}$. The chance of this happening is -at most one half. If $|U_{q,2}|=|U_{p,2}|$, then we succeed at least when $a$ is -a quadratic residue modulo exactly one of the factors $p$ and $q$, because $z$ is in the maximal proper subgroup of $U_{q,2}$ (resp. $U_{p,2}$), iff it is a quadratic residue -modulo $q$ (resp. modulo $p$). If $|U_{q,2}|\neq|U_{p,2}|$, the situation is even better, -because success is guaranteed, when $z$ is of maximal order in the bigger group. -So with at least an even money chance of success in each round, we will succeed sooner rather than later.<|endoftext|> -TITLE: stopped filtration = filtration generated by stopped process? -QUESTION [9 upvotes]: I am interested in a proof of the following statement which seems intuitive, but is somehow really tricky: -Let $X$ be a stochastic process and let $(\mathcal{F}(t) : t \geq 0)$ be the filtration that it generates (unaugmented). Let $T$ be a bounded stopping time. Then we have -$\mathcal{F}(T) = \sigma(X(T \wedge t) : t \geq 0)$ -I have a proof at hand (Bain and Crisan, Fundamentals of Stochastic Filtering, page 309), but in my opinion there is a major gap. I will try to explain the idea of proof. -Let $V$ be the space of functions $[0,\infty) \rightarrow \mathbb{R}$ equipped with the sigma algebra generated by the cylinder sets. Consider the canonical map $X^T:\Omega \rightarrow V$ which maps $\omega$ to the trajectory $t \mapsto X(t \wedge T(\omega),\omega)$. Then we have $\sigma(X(T \wedge t) : t \geq 0) = \sigma(X^T)$. -The difficult part is $\subseteq$. Let $A \in \mathcal{F}(T)$. We want to find a measurable map $g:V \rightarrow \mathbb{R}$ such that $1_A = g \circ X^T$, then we're done. It is now straightforward to show that $1_A$ is constant on sets where the sample paths of $X^T$ are constant. (To be more precise: for $\rho \in \Omega$ consider the set $\mathcal{M}(\rho) = \lbrace \omega : X(\omega,t) = X(\rho,t), 0 \leq t \leq T(\rho) \rbrace$. Then $T$ and $1_A$ are constant on every set of this form). -The problem is: this is not sufficient! It suffices to construct a map $g$ such that $1_A = g \circ X^T$, but how we can we know that $g$ is measurable? This is where the proof of Bain and Crisan comes up short IMO. -I can show this result only under the assumption that the map $X:\Omega \rightarrow V$ be surjective: Since $A \in \mathcal{F}(\infty)$, we have a measurable map $g$ such that $1_A = g \circ X$. Let $x \in V$. Then $T$ and $1_A$ are constant on the preimage of $x$ under $X$. Therefore, $g(x)$ does not depend on the values of $x$ after time $T$ (which is constant on the preimage of $x$). Since $X$ is surjective, we have $g(x) = g(K^Tx)$, where $K$ is the killing functional $K^tx(s) = x(t \wedge s)$. Hence, $g \circ X = g \circ X^T$, and we are done. -I think that this result could be a little bit deeper. I have seen two proofs of this for the special case that $X$ is the coordinate process on $C[0,\infty)$, one is given in the book of Karatzas & Shreve, Lemma 5.4.18. The fact that Karatzas proves this late in the book only in this special case somehow makes me think that the general case is not so easy. -I would really appreciate any comment or other reference for this result. - -REPLY [4 votes]: I've been worried about the same thing. Here's what I came up with: -Assume that $X$ is progressively measurable (e.g. cadlag), then the inclusion $\sigma (X_s^T : s \le t)\subset\mathcal{X}_T^0$ trivial. Without this condition, this inclusion won't hold in general, I believe. For the other direction: -$A \in \mathcal{X}_T^0 \Leftrightarrow \mathbb{1}_A$ is $\mathcal{X}_T^0$ measurable $\Leftrightarrow \mathbb{1}_A = Y_T$ for some $\mathcal{X}_t^0$-optional process $Y_t$. -Thus it suffices to show that for all optional processes $Y_t$, $Y_T$ is $\sigma (X_s^T: s \ge 0)$ measurable. -Now, the optional processes are generated by stochastic intervals of the form $[\sigma,\infty)$, so using a functional class argument it is enough to show that $\{\sigma \le T\} \in \sigma (X_s^T : s \ge 0)$ for all $\mathcal{X}_t^0$-stopping times $\sigma$, $T$. Do this via discretisation of $\sigma$, $T$ and taking limits. (Similarly we can prove that a $\mathcal{X}_t^0$-stopping time $T$ is a $\sigma (X_s^T : s \ge 0)$ stopping time, I believe.) -EDIT: I spoke too soon, this discretisation argument doesn't work unless the filtration is right continuous.... I have no idea how to proceed. -EDIT2: It appears that the proof is IV.100 of Probabilities and Potential (Dellacherie and Meyer)　though I do not currently have this to hand.<|endoftext|> -TITLE: Prove that this recurrence always cycles -QUESTION [5 upvotes]: If $n$ is a nonnegative integer, let $S_n=\{0, 1, 2, \dots, 2n+1\}$. For $t\in S_n$ repeatedly perform - if t is even - t = t/2 - else - t = (n + 1 + ⌊t/2⌋) - -It seems that ultimately $t$ becomes equal to the initially chosen number. -For instance, if $n = 15$ (so $S_{15} = \{0, 1, 2, \dots, 31\}$): -For $t=1$; - $ 1 - 16- 8- 4- 2 - 1$ -For $t=3$; - $3 - 17 - 24 - 12 - 6 - 3$ - -Prove that this property is always true for any $n$ and $t$ where - $n\geq 0$ and $0\leq t\leq (2n+1)$. Also, For a given value of $T$, - how can I find a number in $S_n$ which will be never be - encountered while following this property for $T$? - -REPLY [3 votes]: You already have two good answers to your first question, that your operation -$$ -t\rightarrow -\begin{cases} -t/2 & \text{if $t$ is even}\\ -n+1+\lfloor t/2\rfloor & \text{if $t$ is odd} -\end{cases} -$$ -will generate a sequence that will eventually return to whatever starting value you use. For your second question, Jyrki has noted that this is closely related to the prefect shuffle operation. For example, with $n=7$, starting with $t=3$ your transformation would yield the sequence -$$ -3\rightarrow 9\rightarrow 12\rightarrow 6\rightarrow 3 -$$ -Consider the set $S_7=\{0, 1, \dots, 15\}$ as a deck of 16 cards, initially ordered. A perfect out-shuffle, as it's known, takes that deck, splits it into two equal-sized piles, $\langle\; 0, 1, 2, 3, 4, 5, 6, 7\;\rangle$ and $\langle\; 8, 9, 10, 11, 12, 13, 14, 15\;\rangle$, and then reassembles them in a new order by alternately interleaving the two piles, giving you -$$ -\begin{array}{cccccccccccccccc} -0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15\\ -0 & 8 & 1 & 9 & 2 & 10 & 3 & 11 & 4 & 12 & 5 & 13 & 6 & 14 & 7 & 15 -\end{array} -$$ -where the second row is the card values and the top row is their positions in the new deck order. We can view this permutation as indicating that the card in position 3 will be placed in position 9, the card in position 9 will go in position 12, the card in position 12 will be placed in position 6 and the card in position 6 will be placed in position 3, where we started. Doing this for all cards gives a disjoint cycle decomposition, -$$ -(0)(1\ 8\ 4\ 2)(3\ 9\ 12\ 6)(5\ 10)(7\ 11\ 13\ 14)(15) -$$ -It's no coincidence (and it's fairly easy to show) that the $(3\ 9\ 12\ 6)$ cycle is exactly the same as your transformation produced and that this holds in general. -This, then, is a long-winded way of answering your second question: given any $T$, repeated iteration of your transformation will miss all the values not in the cycle containing $T$. In particular, for any $n$, the cycle starting with $0$ will miss value $2n+1$, the cycle starting with $2n+1$ will miss $0$ and any other starting value will miss both $0$ and $2n+1$ (which, I'll admit, doesn't actually need any of the results I've shown, but it's too pretty not to mention).<|endoftext|> -TITLE: finding all integer $n$ such that $ n\mid2^{n!}-1$ -QUESTION [5 upvotes]: how to find all integer $n$ such that $ n\mid2^{n!}-1$ -I find: -Of course $2 \nmid n$. We prove that, if $2 \nmid n$ then $n \mid 2^{n!}-1$. $2 \nmid n \iff n = 2k+1 , k \ge 0$, we'll prove: -$2^{(2k+1)!} \equiv 1\pmod{2k+1}$ -Let $n = p_1^{a_1}\cdot p_2^{a_2} \cdot ... \cdot p_s^{a_s}$, we'll prove that: -$2^{(2k+1)!} \equiv 1 \pmod{p_1^{a_1}}$ -Let $t = ord_{p_1^{a_1}}2 \iff 2^t \equiv 1\pmod{p_1^{a_1}}$, so: -$t \mid (2k+1)! \iff (2k+1)! = l\cdot t \iff l = \frac{(2k+1)!}{t} \in \mathbb{Z}_{+}$ -And $2^t \equiv 1\pmod{p_1^{a_1}}/^l \Rightarrow 2^{(2k+1)!} \equiv 1\pmod{p_1^{a_1}}$ -Analogously we show divisibility $2^{(2k+1)!}-1$ by $p_2^{a_2} , ... , p_s^{a_s}$ - -REPLY [2 votes]: $n$ needs to be odd as $2^{n!}-1$ is odd.Let $n=2k+1$ for some $k\in \Bbb Z.$ Now, $2^{\phi(2k+1)}\equiv1 \pmod {2k+1}$ as $\gcd(2,2k+1)=1$. Therefore, $2^{(2k+1)!}\equiv 1\pmod{2k+1}$ to hold $\phi(2k+1)$ must divide $(2k+1)!$. As we know $\phi(2k+1)\lt 2k+1$, therefore, it exists somewhere in the product $1\times 2\times ... \times (2k+1)$. Hence, $\phi(2k+1)$ divides $(2k+1)!\implies $$\frac{2^{n!}-1}{n}$ is an integer for all odd $n\in\Bbb Z$.<|endoftext|> -TITLE: Is a countable subspace of a metric space closed? -QUESTION [6 upvotes]: Let $(X,d)$ be a metric space and $a_1,a_2,\ldots\in X$. Define $A=\{a_n:n\in\mathbf N\}$. Is $A$ closed in $(X,d)$? -And is $A$ closed when X is a topological space? - -REPLY [2 votes]: Let $X$ be an infinite topological space. Then: -1) If $X$ is discrete, then (all subsets are closed, so) every countably infinite subset is closed. -2) If $X$ is non-discrete, separated and first-countable -- in particular, if $X$ is non-discrete and metrizable -- then there is a non-isolated point $x$ and a sequence $\{x_n\}_{n=1}^{\infty}$ of distinct elements of $X \setminus \{x\}$ converging to $x$. Thus $A_1 = \{ x_n: n \in \mathbb{Z}^+ \}$ is a countably infinite subset which is not closed, and $A_2 = A_1 \cup \{x\}$ is a countably infinite subset which is closed. -3) Above the hypothesis of first countability can be somewhat weakened: it is enough to assume instead that $X$ is sequential. But it cannot be dropped entirely: let $X$ be an uncountable set endowed with the cofinite topology, in which a proper subset is closed if and only if it is finite. Then $X$ is a separated space which has no countably infinite closed subsets. -At the moment I am not seeing a clean necessary and sufficient condition on a general topological space for it to admit a countably infinite closed subset. Maybe someone else can do better?<|endoftext|> -TITLE: Relation between linearity and injectivity of an entire function -QUESTION [7 upvotes]: Given $f$ entire function on $\mathbb C$ and $f$ one-one. Is it true that $f$ is linear? -At least among polynomials the only such functions are linear! - -REPLY [20 votes]: The link given by @Patience leads to a proof, but one can avoid the heavier things like Picard, Casorati-Weierstrass and the very notion of essential singularity. Liouville's theorem is enough. -Pick a point $a$ such that $f\,'(a)\ne 0$. (I don't even want to argue that $f\,'$ never vanishes). Normalize so that $a=0$, $f(0)=0$, and $f\,'(0)=1$. Since $f$ is an open map, there exists $r>0$ such that $\{w:|w| -TITLE: to show $f(t)=g(t)$ for some $t\in [0,1]$ -QUESTION [7 upvotes]: Let $f,g:[0,1] \rightarrow \mathbb{R}$ be non-negative, continuous functions so that -$$\sup_{x \in [0,1]} f(x)= \sup_{x \in [0,1]} g(x).$$ -We need To show $f(t)=g(t)$ for some $t\in [0,1].$ -Thank you for help. - -REPLY [7 votes]: Let -$$ -f(x_1)=\max_{0 \leq x \leq 1} f(x), \quad g(x_2) = \max_{0 \leq x \leq 1} g(x). -$$ -If $x_1=x_2$, we are done. Otherwise, suppose that $x_1g(x_1)$. By the intermediate value theorem, there must exist a point $\bar{x} \in (x_1,x_2)$ such that $f(\bar{x})=g(\bar{x})$.<|endoftext|> -TITLE: Weak*-convergence of regular measures -QUESTION [6 upvotes]: Let $K$ be a compact Hausdorff space. Denote by $ca_r(K)$ the set of all countably additive, signed Borel measures which are regular and of bounded variation. Let $(\mu_n)_{n\in\mathbb{N}}\subset ca_r(K)$ be a bounded sequence satisfying $\mu_n\geq 0$ for all $n\in\mathbb{N}$. Can we conclude that $(\mu_n)$ (or a subsequence) converges in the weak*-topology to some $\mu\in ca_r(K)$ with $\mu\geq 0$? - -REPLY [5 votes]: We cannot. -Let $K = \beta \mathbb{N}$ be the Stone-Cech compactification of $\mathbb{N}$, and let $\mu_n$ be a point mass at $n \in \mathbb{N} \subset K$. Suppose to the contrary $(\mu_n)$ has a weak-* convergent subsequence $\mu_{n_k}$. Define $f : \mathbb{N} \to \mathbb{R}$ by $f(n_k) = (-1)^k$, $f(n) = 0$ otherwise. Then $f$ has a continuous extension $\tilde{f} : K \to \mathbb{R}$. By weak-* convergence, the sequence $\left(\int \tilde{f} d\mu_{n_k}\right)$ should converge. But in fact $\int \tilde{f} d\mu_{n_k} = \tilde{f}(n_k) = (-1)^k$ which does not converge. -If $C(K)$ is separable, then the weak-* topology on the closed unit ball $B$ of $C(K)^* = ca_r(K)$ is metrizable. In particular it is sequentially compact and so in that case every bounded sequence of regular measures has a weak-* convergent subsequence. As Andy Teich points out, it is sufficient for $K$ to be a compact metrizable space. Also, since there is a natural embedding of $K$ into $B$, if $B$ is metrizable then so is $K$. -One might ask whether it is is possible for $B$ to be sequentially compact without being metrizable. I don't know the answer but I suspect it is not possible, i.e. that metrizability of $B$ (and hence $K$) is necessary for sequential compactness. -We do know (by Alaoglu's theorem) that closed balls in $C(K)^*$ are weak-* compact, so what we can conclude in general is that $\{\mu_n\}$ has at least one weak-* limit point. However, as the above example shows, this limit point need not be a subsequential limit.<|endoftext|> -TITLE: Evaluating $\int_{0}^{\infty}(\ln \tan^2 bx)/(a^2+x^2)\ dx$ -QUESTION [11 upvotes]: Some time ago I came across one of the integrals, which still goes over my mind: -$$\int_{0}^{\infty}\frac{\ln \tan^2(bx)}{a^2+x^2}dx$$ a and b are parameters. -I would be interested in possible solutions with complex analysis and without it as well. - -REPLY [8 votes]: If one can prove that the given integral converges, it's not hard to compute its value. Let's assume from now that the integral does converge. Since $\tan^2(-bx)=\tan^2(bx)$ and $(-a)^2=a^2$, there is no loss of generality in assuming that $a,b>0$. Then -$$ -I(a,b)=\int_0^\infty\frac{\ln\tan^2(bx)}{a^2+x^2}dx=2b\int_0^\infty\frac{\ln|\tan x|}{a^2b^2+x^2}dx=b\int_\mathbb{R}\frac{\ln|\tan x|}{a^2b^2+x^2}dx. -$$ -Consider the function -$$ -f: \mathbb{C} \to \mathbb{C},\ f(z)=b\frac{\ln|\tan z|}{a^2b^2+z^2}. -$$ -Given $n \in \mathbb{N}$, with $0<1/n -TITLE: Exhaustion of open sets by closed sets -QUESTION [15 upvotes]: Let $X$ be a topological space with topology $\tau$. Let $U\in \tau$. Say that $U$ can be countably exhausted by closed sets if there exists a family of sets $F_n \subset U$ indexed by $n\in\mathbb{N}$ such that - -$F_n \subset V \subset F_{n+1}$ for some $V\in \tau$; -$X\setminus F_n \in \tau$; and -$\bigcup\limits_{n\in\mathbb{N}}F_n = U$. - -Is there a charaterisation of topologies for which every open set can be countably exhausted by closed sets? - -Trivially if every open set is also closed (the discrete topology, say) then every open set can be exhausted. Just take $U = V = F_n$. -For an example in which exhaustion is impossible, consider the real line with the co-countable topology. If $U\subsetneq \mathbb{R}$ is open, and if $F_n$ is a non-empty closed set, we have that necessarily $F_{n+1}$ contains a non-empty open set, is closed, and hence must be $\mathbb{R}$. Hence exhaustion is impossible in this topology. -The first condition seems to suggest that the topology being normal may be part of a sufficient condition. (Indeed, normal + separable + Hausdorff seems to be enough; but how much weaker can we go?) - -REPLY [11 votes]: A $T_1$ space has the property that every open set can be countably exhausted by closed sets iff it is perfectly normal. -Assume that $X$ is $T_1$ and that every $U\in\tau$ can be countably exhausted by closed sets. Let $x\in U\in\tau$. There are sequences $\langle F_n:n\in\Bbb N\rangle$ of closed sets and $\langle V_n:n\in\Bbb N\rangle$ of open sets such that -$$F_n\subseteq V_n\subseteq\operatorname{cl}V_n\subseteq F_{n+1}\tag{1}$$ -for each $n\in\Bbb N$ and $U=\bigcup_{n\in\Bbb N}F_n$. Lemma 1.5.14 of Engelking’s General Topology ensures that $X$ is $T_4$ (normal and $T_1$); the proof is similar to the proof that regular Lindelöf spaces are normal. -It’s also clear from $(1)$ that every open set in $X$ is a countable union of regular closed sets. In particular, every open set is an $F_\sigma$, so $X$ is even perfectly normal. It’s well-known that in a $T_1$ space perfect normality is equivalent to each of the following properties: - -Open sets are cozero-sets. -Closed sets are zero-sets. -If $H$ and $K$ are disjoint closed sets, there is a continuous $f:X\to[0,1]$ such that $H=f^{-1}[\{0\}]$ and $K=f^{-1}[\{1\}]$. - -Conversely, suppose that $X$ is a perfectly normal $T_1$ space, and let $U$ be a non-empty open subset of $X$. There is a continuous $f:X\to[0,1]$ such that $U=f^{-1}\big[(0,1]\big]$. For $n\in\Bbb N$ let -$$F_n=f^{-1}\left[\left[\frac1{2^n},1\right]\right]\text{ and }V_n=f^{-1}\left[\left(\frac1{2^{n+1}},1\right]\right]\;;$$ -clearly the sequences $\langle F_n:n\in\Bbb N\rangle$ and $\langle V_n:n\in\Bbb N\rangle$ satisfy $(1)$, and $U=\bigcup_{n\in\Bbb N}F_n$.<|endoftext|> -TITLE: Fractional Binomial Coefficients -QUESTION [9 upvotes]: I recently examined the binomial coefficient $\binom{\frac{1}{2}}{k}$ and found that the denominator was always a power of two. The same is true of $\binom{\frac{1}{3}}{k}$, where the denominator is always a power of three. While the first proof was simple, the case of $\frac{1}{3}$ was messy and involved counting the powers of 3 in the numerator and denominator. The case of $\frac{1}{p}$ for prime $p$ can be done in the same bashy way. Is it possible there is a more elegant proof? - -REPLY [5 votes]: Here’s a purely conceptual proof, more advanced, that has the advantage of involving no computation and no induction: - -Let $N$ be a positive number, and let $P_N(t)$ be the polynomial function that tells you the coefficient of $x^N$ in $(1+x)^t$. You know that $P_N$ has rational coefficients, and takes integer values when $t$ is evaluated to a positive integer. -Now consider any prime number $q$. The polynomial $P_N$ has its coefficients in the field ${\mathbb{Q}}_q$ of $q$-adic numbers, is continuous, and takes values in the $q$-adic integers ${\mathbb{Z}}_q$ whenever $t$ is evaluated to a positive integer. -But any rational number without $q$ in its denominator is a $q$-adic integer, so $q$-adically approximable by ordinary integers, and in fact by positive integers. Thus if $\alpha$ is a rational with no $q$ in its denomiator, we get $P_N(\alpha)\in{\mathbb{Z}}_q$, by continuity of $P_N$. -In particular, the coefficients of $(1+x)^{1/p}$ are rational numbers that are in ${\mathbb{Z}}_q$ for each $q\ne p$: the only denominators are powers of $p$. - -A more advanced argument allows you to say just how divisible by $p$ the denominators of the coefficients of $(1+x)^{1/p}$ will be, but that’s a story for another day.<|endoftext|> -TITLE: How to calculate $\int_{-a}^{a} \sqrt{a^2-x^2}\ln(\sqrt{a^2-x^2})\mathrm{dx}$ -QUESTION [11 upvotes]: Well,this is a homework problem. -I need to calculate the differential entropy of random variable -$X\sim f(x)=\sqrt{a^2-x^2},\quad -a -TITLE: Proving :$\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1$ -QUESTION [8 upvotes]: Let $a,b,c>0$ be real numbers such that $a+b+c=3$,how to prove that? : -$$\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1$$ - -REPLY [2 votes]: By AM-GM one has $\,abc\leq\left(\frac{a+b+c}{3}\right)^3=1$. And with the -Cauchy–Bunyakovsky–Schwarz inequality we obtain -$$\sum_{cyc}\frac{1}{2a^2b+1}=\sum_{cyc}\frac{c^2}{2a^2c^2b+c^2}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(2a^2b^2c+a^2)}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(2ab+a^2)}=1.$$<|endoftext|> -TITLE: About positive semidefiniteness of one matrix -QUESTION [10 upvotes]: It is not clear how to prove that the matrix $(\min(i,j))_{i,j=1,\dots,n}$ is (or is not) positive semidefinite. There are some facts from Horn and Johnson's book Matrix Analysis: if $A \in M_n$ is positive semidefinite, then $a_{ii}a_{jj} \ge a_{ij}^2, i,j=1,2,\dots, n$ (source). It seems that this condition is not enough for the matrix to be positive semidefinite. Are there any techniques to check that all eigenvalues of the matrix are positive, or that the minor determinants are non-negative? - -REPLY [5 votes]: Another alternative solution: it can be shown that the inverse of the min-matrix is the perturbed Toeplitz tridiagonal matrix -$$\begin{pmatrix}2&-1&&&\\-1&2&-1&&\\&-1&\ddots&\ddots&\\&&\ddots&2&-1\\&&&-1&1\end{pmatrix}$$ -whose characteristic polynomial is -$$(-1)^n\left((x-1)U_{n-1}\left(\frac{x-2}{2}\right)-U_{n-2}\left(\frac{x-2}{2}\right)\right)$$ -(where $U_n(x)$ is the Chebyshev polynomial of the second kind), and has the eigenvalues -$$\mu_k=2+2\cos\left(\frac{2k\pi}{2n+1}\right),\qquad k=1\dots n$$ -Thus, the eigenvalues of the min-matrix are the reciprocals of the $\mu_k$, -$$\lambda_k=\frac14\sec^2\left(\frac{k\pi}{2n+1}\right)$$ -and these are easily seen to be positive, thus showing that the min-matrix is positive definite.<|endoftext|> -TITLE: Weak-* sequential compactness and separability -QUESTION [12 upvotes]: Let $X$ be a Banach space, and let $B$ be the closed unit ball of $X^*$, equipped with the weak-* topology. Alaoglu's theorem says that $B$ is compact. If $X$ is separable, then $B$ is metrizable, and in particular it is sequentially compact. - -What about the converse? If $B$ is sequentially compact, must $X$ be separable? - -This question was inspired by this one. - -REPLY [4 votes]: For an explicit counterexample: let $H$ be a non-separable Hilbert space, and suppose we have a sequence of vectors $x_n$ in the unit ball of $H$. Let $H_0$ be the closed linear span of the $x_n$, so that $H_0$ is a separable Hilbert space. Then the $x_n$ have a subsequence converging weakly in $H_0$. But because of the orthogonal decomposition $H = H_0 \oplus H_0^\perp$, it is easy to see that this subsequence also converges weakly in $H$.<|endoftext|> -TITLE: Too many topics taught in class? Frustrated student in need of advice and encouragement. -QUESTION [8 upvotes]: Location: New York CUNY (as education systems might be different in other places) -I started my life studying philosophy and psychology and then at 22 transitions to computer science. It took me a long time to understand the importance of mathematics. -I was placed in calculus1 and while it was hard to remember math I managed to pull a B+. -However there is a recurrent problem I have with math classes I am taking, namely too many topics to really understand everything. -In calculus2 and probability and statistics we covered so many topics that I felt that any attempt to understand the material would only hurt me on the test. Constant new topics without any chance to play around and really grasp the material. Moreover textbooks rarely focus on the WHY of things, nor do professors have the time to explain due to computational emphasized department finals and curriculum requirements. I feel like every class has artificially inflated amount of material and it becomes especially obvious when teachers end up rushing up to two topics a class at the end of the semester. -For example we went over possion distribution and later waiting-time possion distribution and besides showing us how to do the problems 0 emphasis was put on why it works and where it came from. I am VERY frustrated with this frenzy of meaningless formulas. -I lately feel that perhaps I am an idiot or something but I just don't see any of the textbooks explain things properly. When they do attempt at explanation it is just a soup of symbols without any intuition. It is like they just copy pasted proofs to make it seem rigorous. The emperor is naked. -Is it my laziness or stupidity or a known problem in education curriculum in early undergraduate math classes? It is killing my recently gained joy for math and lowering my self esteem. -Perhaps people can advice on some math books that would fill the gaps and actually explain things. - -REPLY [7 votes]: It is a well-known phenomenon that certain beginning undergrad math classes can be crowded with topics, to the point of making it almost impossible to explain them all with the degree of detail and clarity that one would like (and I am speaking as someone who has been on the other side of it --- i.e. teaching these classes). There are various reasons for this; one is that certain topics must be covered (due to demands of various later courses, both within the math dept. and in other depts. for which these courses acts as service courses), and there aren't enough separate course slots available to separate them out into different courses. Well-designed curricula try to minimize this phenomenon, but it's not easy; courses and curricula have momenta of their own, and are not as easy to change or redirect as you might think. -In any case, given the situation you describe, it is probably not realistic to learn everything in your courses to the degree of precision and understanding you would like; like many other things in college and in life, there will have to be a compromise between the ideal and the realistic in your learning. What is possible, I would think, would be to learn some part of your curriculum more carefully and in greater depth. -To this end, I would suggest that you choose one part of your course that you found the most intriguing, and that you would most like to learn, and ask a specific question about that part of your curriculum. (E.g. based on your complaint about the discussion of the Poisson distribution, maybe you would like to understand better the different probablility distributions, where they come from and why we study them, and you could ask as question about "Resources for a beginner to learn and understand different probablility distributions".) -Try this with one topic at a time, and try to balance your study between "keeping up with current topic in class" and "learning topics of interest for -personal development/understanding". As (or if) you move onto more advanced math -classes, these two threads of your study will start to become more closely -entwined, because the pace of introduction of new topics will slow, and you will get the chance to study each topic in more depth.<|endoftext|> -TITLE: Compute $\lim\limits_{n\to\infty} \prod\limits_2^n \left(1-\frac1{k^3}\right)$ -QUESTION [25 upvotes]: I've just worked out the limit $\lim\limits_{n\to\infty} \prod\limits_{2}^{n} \left(1-\frac{1}{k^2}\right)$ that is simply solved, and the result is $\frac{1}{2}$. After that, I thought of calculating $\lim\limits_{n\to\infty} \prod\limits_{2}^{n} \left(1-\frac{1}{k^3}\right)$, but I don't know how to do it. According to W|A, the result is pretty nice, but I don't see how W|A gets that. (See here.) Is there any easy way to get the answer? - -REPLY [9 votes]: The last step of Andrew getting -\begin{align}\lim_{n\to \infty}\prod _{k=2}^n \left(1-\frac{1}{k^3}\right)= \frac{\cosh \frac{\sqrt{3} \pi }{2} \Gamma \left(n-\frac{i \sqrt{3}}{2}+\frac{3}{2}\right) \Gamma \left(n+\frac{i \sqrt{3}}{2}+\frac{3}{2}\right)}{3 \pi n^3 \Gamma^2 (n)}\end{align} -was a bit ambigous. -using another method, note that -\begin{align*}\Gamma(z)=\frac{1}{z e^{\gamma z}}\prod_{k=1}^{\infty}\frac{k e^{\frac{z}{k}}}{z+k} -\end{align*} -holds for all complex number $z$ except negative integer, we obtain -\begin{align}g(z)=\prod_{k=1}^{\infty} (1+\frac{z}{k})e^{\frac{-z}{k}}=\frac{1}{z\Gamma(z)e^{\gamma z}}\end{align} -Thus -\begin{align} -g(\omega)g(\omega^2)=\prod_{k=1}^{\infty}\frac{k^2+k+1}{k^2}e^{-\frac{1}{k}}=\frac{1}{\Gamma(\omega)\Gamma(\omega^2) e^{\gamma}}=\frac{3}{e}\prod_{k=2}^{\infty}\frac{k^2+k+1}{k^2}e^{-\frac{1}{k}} -\end{align} -where $-\omega$ is the root of $x^3=1$ -From -\begin{align} -\prod_{k=2}^{\infty}(1-\frac{1}{k})e^{\frac{1}{k}}=\lim_{n\to \infty}\frac{1}{n} e^{\frac{1}{2}+\cdots+\frac{1}{n}}=e^{\gamma -1} -\end{align} -Thus -\begin{align} -\prod_{k=2}^{\infty}\left(1-\frac{1}{k^3}\right)=\prod_{k=2}^{\infty}\frac{k^2+k+1}{k^2}e^{-\frac{1}{k}}\prod_{k=2}^{\infty}(1-\frac{1}{k})e^{\frac{1}{k}}=\frac{1}{3\Gamma(\omega)\Gamma(\omega^2)} -\end{align} -and hence the result -By the similar way we may get $\prod_{k=2}^{\infty}(1-\frac{1}{k^n})$<|endoftext|> -TITLE: Is it possible to capture a light ray in a solar panel? -QUESTION [9 upvotes]: A while ago I was wondering how we could use mathematics to increase the efficiency of solar panels. The kind of mathematics I was thinking about in particular was Dynamical Billiards. Though I think it is improbable that it is currently technologically feasible to create the solar panel I am thinking about, I guess the following question could be interesting from a mathematical point of view. -We first need some definitions: -An $(k+l)$-setting is a collection of two sets of points in $\mathbb{R^d}$, $\{n_1,...,n_k \}, \{m_1,...,m_l \}$ such that each of the points $m_i \in \{m_1,...,m_{l-1} \}$ is connected with the point $m_{i+1}$, by some continuous function (the continuous function may differ for each pair of points $(m_i,m_{i+1})$). -For example, this is an $(k+l)$-setting in $\mathbb{R^2}$: - -I hope it's somewhat readable. In this case, we have $k=2$ and $l=5$. -Furthermore, we say that a $(k+l)$-setting is good, if it is possible create a circle, such that the points $m_1,...,m_l$ are in the circle, but the points $n_1,...,n_k$ are not in the circle. If a $(k+l)$-setting isn't good, it's bad. For example: - -Please notice that the first example of a $(k+l)$-setting is bad. No matter how you draw the circle, the point $n_2$ is always contained in it. -Now, the point of these definitions is that I would like to launch a light rays from the the points $n_1,...,n_k$, that bounces against the continuous functions. These continuous functions act as a mirror, causing the light ray to reflect according to the laws of reflection. We assume that the light ray loses no energy with each reflection, thereby maintaining its intensity on it whole course. The continuous functions that act as a boundary of the "solar panel" aren't affected by a reflection either. - -Question 1: Does there exist a good $(1+l)$-setting in $\mathbb{R^2}$, such that we can send a light ray from point $n_1$ into the circle that encloses the points $m_1,...,m_l$, in such a way that the light ray never leaves the circle? - -If you can answer the question in the affirmative, I have a number of follow-up questions: - -Question 2.1: If there exists such a setting, how is it visualised? -Question 2.2: What about the case $k>1$ ? -Question 2.3: What about the case $d>2$ ? - -If you answer the question in the negative, I also have some follow-up questions: - -Question 3.1: Why does such a setting not exist? Can you prove it cannot exist? -Question 3.2: What if $d>2$ ? - -Please notice that, when $d=3$, the continuous functions between the points become continuous surfaces, and when $d=4$, they become volumes, etc. -By the way, it would be great if someone told me how I could center the images, or places them in the middle themselves. - -REPLY [6 votes]: This has been studied, under various assumptions. -For the case of a parallel beam of light rays, -there is a nice description in Serge Tabachnikov's 2005 book -Geometry and Billiards, pp. 52-53: - - - - - - - -Going deeper is a paper by Daswon et al., "Light Traps" (PDF link). -They show it is possible to trap a pencil of rays (rays through a point), -but it is impossible to trap diffuse light. -Their main negative theorem - -states that if the initial light has nonzero measure - in the phase space, it cannot be trapped.<|endoftext|> -TITLE: How to think of the field $F(\alpha)$ -QUESTION [6 upvotes]: The way I learned it was given a field extension $F \subset E$, and an element $\alpha \in E$ -$$F(\alpha) := \{p(\alpha)/q(\alpha) : p(x), q(x) \in F[x] ,q(\alpha) \not = 0\} $$ -Is there an easier way to think about the field $F(\alpha)$ - -REPLY [2 votes]: For an extension $E/F$ and $\alpha\in E$, if $\alpha$ is algebraic over $F$, then the extension $F(\alpha)\subseteq E$ can also be described in a way which is intrinsic to $F$ (i.e. which involves no mention of $E$). -To say that $\alpha$ is algebraic means there exists a non-zero (hence non-constant) polynomial $f\in F[X]$ with $f(\alpha)=0$. There is a unique $F$-algebra map $\varphi:F[X]\rightarrow E$ satisfying $X\mapsto(\alpha)$. The kernel is non-zero (because $f$ is in it) and proper (since this is a map of non-zero rings), so is of the form $(g)$ for a unique monic $g$, say of degree $d\geq 1$. One can characterize $g$ as the monic polynomial of minimal degree having $\alpha$ as a root. The homomorphism $\varphi$ gives rise to an isomorphism $F[X]/(g)\cong\mathrm{im}\varphi$. The image of $\varphi$ is $F[\alpha]\subseteq E$, the $F$-subalgebra of $E$ generated by $\alpha$, which can concretely be described as the set of polynomials in $\alpha$ with coefficients in $F$. This is a domain, being a subring of a field. One can also show that the elements $1+(g), X+(g),\ldots,X^{d-1}+(g)$ form a basis for $F[X]/(g)\cong F[\alpha]$ as an $F$-vector space. In particular, $F[\alpha]$ is of dimension $d$ as an $F$-vector space. From this it follows that $F[\alpha]$ is a field. Indeed, consider any non-zero element $\beta$ of $F[\alpha]$ and the $F$-endomorphism $x\mapsto\beta x:F[\alpha]\rightarrow F[\alpha]$. This is injective, hence surjective, since source and target have the same finite dimension, so there is $x$ with $x\beta=1$, and $\beta$ is a unit. Since, by definition, $F(\alpha)$ is the smallest subfield of $E$ containing $F$ and $\alpha$, and $F[\alpha]$ is such a field, we must have $F[\alpha]=F(\alpha)$. -In summary, we have constructed an isomorphism (of $F$-algebras) $F[X]/(g)\cong F(\alpha)$ (along the way we showed that $F(\alpha)$ consists of all polynomials in $\alpha$ with coefficients in $F$). Since $F[X]/(g)$ is a field, $(g)$ must be a maximal ideal, and $g$ is irreducible. It follows that $g$ is the unique monic irreducible polynomial in $F[X]$ having $\alpha$ as a root. In this way you get a description of $F(\alpha)$ as $F[X]/(g)$ which doesn't directly involve $E$. -Conversely, if $g\in F[X]$ is a monic irreducible polynomial, then $(g)$ is maximal, so $E:=F[X]/(g)$ is a field containing $F$, and if $\alpha:=X+(g)$, then $\alpha$ is a root of $g$ in $E$, and $E=F(\alpha)$. -Note that this is only valid for $\alpha$ algebraic over $F$. As pointed out in one of the other answers, if $\alpha$ is transcendental over $F$, then the map $\varphi:F[X]\rightarrow E$ given by $X\mapsto\alpha$ is injective (this is exactly what it means for $\alpha$ to be transcendental over $F$), so one has $F[X]\cong F[\alpha]$ as $F$-algebras. This isomorphism of domains then extends uniquely to an $F$-algebra isomorphism of fraction fields $F(X)\cong F(\alpha)$.<|endoftext|> -TITLE: How to approach integrals as a function? -QUESTION [5 upvotes]: I'm trying to solve the following question involving integrals, and can't quite get what am I supposed to do: -$$f(x) = \int_{2x}^{x^2}\root 3\of{\cos z}~dz$$ -$$f'(x) =\ ?$$ -How should I approach such integral functions? Am I just over-complicating a simple thing? - -REPLY [7 votes]: For this problem, you will ultimately use a version of the Fundamental Theorem of Calculus: If $f$ is continuous, then the function $F$ defined by $F(x)=\int_a^x f(z)\,dz$ is differentiable and $F'(x)=f(x)$. -So for instance, for $F(x)=\int_0^x\root3\of{\cos z}\,dz$, we have $F'(x)=\root3\of{\cos x}$. -One can combine this with the chain rule, when it applies, to differentiate a function whose rule is of the form $F(x)=\int_a^{g(x)} f(z)\,dz$. Here, we recognize that $F$ is a composition of the form $F=G\circ g$ with $G(x)=\int_a^x f(z)\,dz$. The derivative is $F'(x)=\bigl[ G(g(x))\bigr]'=G'(g(x))\cdot g'(x)=f(g(x))\cdot g'(x)$. -For example, for $F(x)=\int_0^{x^2}\root3\of{\cos z}\,dz$, we have $F'(x)=\root3\of{\cos x^2}\cdot(x^2)'=2x\root3\of{\cos x^2} $. -Now to tackle your problem proper and take advantage of these rules, we just "split the integral": -$$\tag{1} -\int_{2x}^{x^2}\root3\of{\cos z}\,dz= -\int_{2x}^{0}\root3\of{\cos z}\,dz+ -\int_{0}^{x^2}\root3\of{\cos z}\,dz. -$$ -But wait! We can only use the aforementioned differentiation rules for functions defined by an integral when it's the upper limit of integration that is the variable. The first integral in the right hand side of $(1)$ does not satisfy this. Things are easily remedied, though; write the right hand side of $(1)$ as: -$$ --\int_{0}^{2x}\root3\of{\cos z}\,dz+ -\int_{0}^{x^2}\root3\of{\cos z}\,dz; -$$ -and now things are set up to use our rule (of course, you'll also use the rule $[cf+g]'=cf'+g'$).<|endoftext|> -TITLE: Quaternion Rings -QUESTION [12 upvotes]: Let $R$ be a commutative ring. Define the Hamilton quaternions $H(R)$ over $R$ to be the free $R$-module with basis $\{1, i, j, k\}$, that is, -$$H(R)=\{a_0+a_1i+a_2j+a_3k\;\;:\;\;a_l\in R\}.$$ -and multiplication is defined by: $i^2=j^2=k^2=ijk=-1$. -Is well-known that over a field $F$ (with char $F\neq 2$) the ring $H(F)$ is a division ring or isomorphic to $M_2(F)$. What can we say about the Hamilton quaternions over an arbitrary commutative ring $R$? Is still true that $H(R)$ is a division ring or isomorphic to $M_2(R)$? We must impose some conditions to the ring to make it happen? - -REPLY [9 votes]: (Your definition of the multiplication looks slightly strange to me. I would prefer to write $k = ij = -ji$. This implies your relations; as Arturo shows in his comment below, your relations also imply mine, but to my mind it's sort of mixing two different ways to give an $R$-algebra: you could either give it via generators and relations, or if it's a free $R$-module you can give an explicit basis and the structure constants.) -If $R$ is a field of characteristic different from $2$, then $H(R)$ is by definition the quaternion algebra $\left( \frac{-1,-1}{R} \right)$. In general, for $a,b \in R^{\times}$, the quaternion algebra -$\left( \frac{a,b}{R} \right) = R\langle i,j \rangle/(i^2 = a, j^2 = b, ji = -ij)$ -is indeed always a division algebra or isomorphic to $M_2(R)$: see e.g. $\S 5.1$ of these notes. -(If $R$ is a field of characteristic $2$, then $H(R)$ is a commutative ring, so is not what you want. There are analogues of quaternion algebras in characteristic $2$ defined slightly differently.) -If $R$ is a domain, then the center of $H(R)$ is $R$. It follows that if $R$ is not a field, $H(R)$ is not a division ring. Note also that if $R \subset S$ is an extension of domains, then $H(S) \cong H(R) \otimes_R S$. -In particular $H(\mathbb{Z})$ is not a division ring. Neither is it isomorphic to $M_2(\mathbb{Z})$: if so, then $H(\mathbb{\mathbb{R}})$ would be isomorphic to $M_2(\mathbb{Z}) \otimes_{\mathbb{Z}} \mathbb{R} \cong M_2(\mathbb{R})$, but $H(\mathbb{R}) = \left( \frac{-1,-1}{\mathbb{R}}\right)$ is the classical Hamiltonian quaternions, which is well-known to be a division ring. -In general, if $R$ is a domain of characteristic different from $2$ with fraction field $K$, then $H(R)$ is an order in the quaternion algebra $H(K)$. This is more a definition than anything else, but it gives you a name. There is a vast literature on quaternion orders: for just a little bit, see e.g. these notes.<|endoftext|> -TITLE: Finding a primitive element for the field extension $\mathbb{Q}(\sqrt{p_{1}},\sqrt{p_{2}},\ldots,\sqrt{p_{n}})/\mathbb{Q}$ -QUESTION [5 upvotes]: Let $p_1,\ldots,p_n\in\mathbb{N}$ be different prime numbers, it can be shown - that $[\mathbb{Q}(\sqrt{p_1},\ldots,\sqrt{p_n}):\mathbb{Q}]=2^n$ and in any case it is clearly finite since $[\mathbb{Q}(\sqrt{p_1},\ldots,\sqrt{p_n}):\mathbb{Q}]\leq2^n$. -Since $char(\mathbb{Q})=0$ then $\mathbb{Q}$ is perfect hence every field extension is separable, in particular $\mathbb{Q}(\sqrt{p_{1}},\sqrt{p_{2}},\ldots,\sqrt{p_{n}})/\mathbb{Q}$ is separable. -Since $\mathbb{Q}(\sqrt{p_{1}},\sqrt{p_{2}},\ldots,\sqrt{p_{n}})/\mathbb{Q}$ is a finite and separable field extension, by the primitive element theorem, it holds that there exist $\alpha\in\mathbb{Q}(\sqrt{p_{1}},\sqrt{p_{2}},\ldots,\sqrt{p_{n}})$ s.t $\mathbb{Q}(\alpha)=\mathbb{Q}(\sqrt{p_{1}},\sqrt{p_{2}},\ldots,\sqrt{p_{n}})$. -I wish to find such element $\alpha$ (i.e. a primitive element, that we know exist). -I know how to do this in the case $n=2$, I tried to generalize and prove this claim by induction, in the induction step I need to prove: - -$\sqrt{p_{1}}+\cdots+\sqrt{p_{n-1}}\in\mathbb{Q}(\sqrt{p_{1}}+\sqrt{p_{2}}+\sqrt{p_{n}})$ -$\sqrt{p_{n}}\in\mathbb{Q}(\sqrt{p_{1}}+\sqrt{p_{2}}+\sqrt{p_{n}})$ - -What I tried to do is to look at : -$$ -\begin{align} -& (\sqrt{p_{1}}+\sqrt{p_{2}}+\sqrt{p_{n}})(\sqrt{p_{1}}+\sqrt{p_{2}}-\sqrt{p_{n}}) \\ -& =((\sqrt{p_{1}}+\sqrt{p_{2}}+\sqrt{p_{n-1}})+\sqrt{p_{n}})((\sqrt{p_{1}}+\sqrt{p_{2}}+\sqrt{p_{n-1}})-\sqrt{p_{n}}) \\ -& =(\sqrt{p_{1}}+\cdots+\sqrt{p_{n-1}})^{2}-p_{n} -\end{align} -$$ -If $n=2$ then this product is in $\mathbb{Q}$ hence in $\mathbb{Q}(\sqrt{p_{1}}+\sqrt{p_{2}})$ hence -$\sqrt{p_{1}}-\sqrt{p_{2}}\in\mathbb{Q}(\sqrt{p_{1}}+\sqrt{p_{2}})$ so adding we get $\sqrt{p_1}\in\mathbb{Q}(\sqrt{p_1}+\sqrt{p_2})$ hence $\sqrt{p_2}\in\mathbb{Q}(\sqrt{p_1}+\sqrt{p_2})$ and we have proven $(2)$ -So the reason I fail here is that I can't manage to show $$\sqrt{p_{1}}+\sqrt{p_{2}}-\sqrt{p_{n}}\in\mathbb{Q}(\sqrt{p_{1}}+\sqrt{p_{2}}+\sqrt{p_{n}}).$$ -Can someone please help me find a primitive element, or help complete the proof I am trying to do here ? help is very much appriciated! - -REPLY [6 votes]: I believe this is the solution Dylan was hinting at. To show that $\alpha=\sum \sqrt{p_i}$ generates $E=\mathbb Q(\{\sqrt{p_i}\})$ it suffices to show that $\alpha$ is not fixed by any automorphism of $E$. Notice that any automorphism of $E$ maps $\sqrt{p_i}$ to $\pm \sqrt{p_i}$. So it suffices to demonstrate that -$$\sum \sqrt{p_i} \neq \sum s_i\sqrt{p_i}$$ -for any choice of $s_i$ such that at least one $s_i$ is $-1$. But this is immediate because by cancelling the positive $s_i$ we would have -$$\sum \sqrt{p_j}=-\sum \sqrt{p_j}.$$ -It follows that $\alpha$ is not fixed by any automorphism of $E$ and so $E=\mathbb Q(\alpha)$. -This is pretty much the same argument as Geoffs, but maybe it's a little bit clearer.<|endoftext|> -TITLE: A compact operator is completely continuous. -QUESTION [7 upvotes]: I have a question. -If $X$ and $Y$ are Banach spaces, we have to prove that a compact linear operator is completely continuous. - A mapping $T \colon X \to Y$ is called completely continuous, if it maps a weakly convergent sequence in $X$ to a strongly convergent sequence in -$Y$ , i.e., $x_n\underset{n\to +\infty}\rightharpoonup x$ implies $\lVert Tx_n- -Tx\rVert_Y\to 0$. - -REPLY [6 votes]: Since it's a homework question I will just give some steps. - -By linearity, we can assume that $x=0$. -We have to show that for each subsequence of $\{Tx_n\}$, we can extract a further subsequence which converges to $0$ in norm in $Y$. -A weakly converging sequence is bounded. -$T$ maps bounded sets to sets with a compact closure. - -Once the second steps is shown, we can conclude. Indeed, assume that $Tx_n$ doesn't converge to $0$. Then we are able to find $\delta>0$ and $A$ an infinite subset of the natural numbers such that $\lVert Tx_k\rVert_Y\geq\delta$ for each element of $A$. We can consider it as a subsequence, and we can't extract a further subsequence which converges to $0$, a contradiction.<|endoftext|> -TITLE: How many elements in a ring can be invertible? -QUESTION [9 upvotes]: If $R$ is a finite ring (with identity) but not a field, let $U(R)$ be its group of units. Is $\frac{|U(R)|}{|R|}$ bounded away from $1$ over all such rings? - -It's been a while since I cracked an algebra book (well, other than trying to solve this recently), so if someone can answer this, I'd prefer not to stray too far from first principles within reason. - -REPLY [2 votes]: The bound given by navigetor23 is tight for example in the case $R=\mathcal{O}/\langle p^2\rangle$, where $\mathcal{O}$ is the ring of integers of a finite unramified extension of the $p$-adic numbers of degree $n$: $|R|=p^{2n}$ and there are $p^n$ non-units consisting of the cosets in $p\mathcal{O}$.<|endoftext|> -TITLE: opposite of disjoint -QUESTION [13 upvotes]: Sets whose intersection is the empty set are called disjoint. What is the opposite of a disjoint set? For example the sets $\{1,2\}$ and $\{2,3\}$ satisfy this condition. I know that you can just say not-disjoint. But I was wondering if there was a specific term for it. I ask this because someone told me that this term existed but that he could`nt remeber. I have searched extensively but I haven't found it. - -REPLY [8 votes]: You seem to be looking for an adjective meaning the opposite of disjoint. Given two sets that aren't disjoint, I would just call them intersecting sets, where intersecting is an adjectival participle derived from the verb to intersect.<|endoftext|> -TITLE: How popular and used were logarithm tables? -QUESTION [6 upvotes]: I've heard that, for a time, logarithm tables "sold more than the Bible". Can someone produce some reliable documentation about how prevalent they were ? Would a common shopkeep have one ? Would a common merchant ship have one ? -(they would be used to make multiplications and divisions faster) -I am interested in information regarding any time period (though, as my wording suggests, the question was brought to my attention in the context of the 15th century) - -REPLY [5 votes]: I used them in secondary school (high school) in Ireland in the 70's. Not everyone could afford calculators, but log books were provided at public exams. Slide rules were allowed. These books also had tables for trigonometric functions as well. -In engineering school, we used steam tables. -I still have log tables I 'borrowed' from my last exam. -Now I use tax tables (what's wrong with a formula?).<|endoftext|> -TITLE: What is so wrong with polynomial hierarchy collapsing -QUESTION [7 upvotes]: Many computational complexity researchers believe that finite-level collapse of polynomial hierarchy is unlikely. Why do they believe like this? - -REPLY [6 votes]: It's possible to generalize to this conclusion essentially directly from the opinion that P $\neq$ NP. Let's think of NP-completeness via SAT, the problem of satisfying a Boolean formula in $n$ variables. In the polynomial hierarchy PH, we construct complete problems similarly by allowing quantifier alternations e.g. $\forall X_1 \exists X_2 ... \forall X_n \varphi(X_1,...,X_n)$. Here the $X_i$ might be sets of variables, not just individuals, in the Boolean formula $\varphi$. The problem of satisfying the kind of formula given is $\Pi_n^P$ complete, while if I'd started with $\exists$ it would have been $\Sigma_n^P$ complete. -OK, now suppose PH collapses, so that say for all $n > n_0: \Pi^P_n, \Sigma^P_n \subset \Sigma^P_{n_0}$. Then TQBF$_{n_0},$ the complete problem described above for $\Sigma^P_{n_0}$, is actually complete for PH as a whole. Conversely, if PH has any complete problem, it must lie at some finite level, in which all higher levels must then be contained, so we see PH collapses if and only if it has a complete problem. This bothers people in the same way as the idea that NP $\subset$ P. In the latter case, we'd have to be able in general to find a needle in a haystack without having to look at each piece of hay, while decades of work seem to indicate that NP problems can't generally be approached any more efficiently than via brute force. For higher polynomial complexity classes, the parallel supposition is that we can't find out, for example, whether a $\Sigma^P_2$ formula $\exists X_1 \forall X_2 \varphi(X_1,X_2)$ is satisfiable any more efficiently than by iterating through all possible assignments to the variables $X_1$ and, on each step of that loop, checking whether we can satisfy $\varphi$ with each possible assignment to the $X_2$. -So, if you're skeptical about the assumption that P $\neq$ NP, that's stronger than being skeptical that PH collapses to any particular finite level, but in essence this holds conversely, too: if I can drop off all but $k$ alternations of quantifiers on PH problems, I am going to have to take much more seriously the idea that I can drop off all the quantifiers-namely the idea that P = NP. -The last issue worth mentioning regards PSPACE. The quantified Boolean formula problem TQBF with arbitrarily many quantifier alternations is PSPACE complete, which shows that at least PSPACE $\supset$ PH. If PH=PSPACE, then PH collapses, because then TQBF sits at some finite level of the hierarchy and is PH-complete. On the other hand, if PH collapses, it might suggest PH=PSPACE, because otherwise we'd be in the very strange situation where each fixed TQBF$_n$ was reducible to TQBF$_{n_0}$ for some $n_0$ while the unbounded TQBF was not.<|endoftext|> -TITLE: Normal subgroups of free groups: finitely generated $\implies$ finite index. -QUESTION [27 upvotes]: I am looking at what should be a simple exercise in geometric group theory. I have reduced the problem to just completing an exercise from Hatcher, Section 1.B page 87: - -7. If $F$ is a finitely generated free group and $N$ is a nontrivial normal subgroup of - infinite index, show, using covering spaces, that $N$ is not finitely generated. - -A finitely generated free group can be realised as the fundamental group of a wedge of circles, so it seems I should be looking at the covering space of this bouquet induced by the infinite-index normal subgroup $N$. Since it is a normal subgroup, I know the group of deck transformations of my covering space is naturally isomorphic to the subgroup itself. Supposing that $N$ is finitely generated, I would like to lift its generating loops to the covering space, I will get, because of the infinite-index, loops starting at all the fibers of my base point. I would like from this to get that the group of deck transformations is finitely generated, but I can't see it. - -REPLY [5 votes]: Let $F$ be the finitely generated free group, say with $n$ generators, so that if $X$ is the wedge of $n$ circles with wedge point $x_0$, $\Pi_1(X,x_0)\cong F$. We may then construct the covering space $\tilde{X}$ corresponding to $N$, which we know to be a connected graph, where each vertex projects to the wedge point and each edge projects to one of the circles in $X$. As usual, we may take $T$ a maximal tree inside of $\tilde{X}$ and use the fact that the quotient $\tilde{X}/T$ is homotopoy equivalent to $\tilde{X}$, and so has the same fundamental group, namely $N$. -Everything we have done so far may be done in complete generality fo any $N\leq \Pi_1(X)$. We now prove that if $N$ is as given, it is not finitely generated. -The first step is to recall that the fundamental group of $\tilde{X}/T$ is always free with cardinality equal to the cardinality of edges in $\tilde{X}\setminus T$ (since the quotient collapses all of the verticies and edges inside $T$ down to a single point, so all that is left is a wedge of circles, with each circle corresponding to an edge not in $T$). So if $N$ were finitely generated, there would be only finitely many edges in $X\setminus T$. We will now show that if $N$ is also normal and non-trivial and of infinite index, then $T$ contains a loop, contradicting the fact that $T$ is a tree, so we win. -Since $N$ is normal, for any vertex $v \in\tilde{X}$, $p_*\Pi_1(\tilde{X},v)=N$, and since $N$ is non-trivial, there is some non-trivial loop $\gamma$ in $X$ based at $x_0$ corresponding to an element of $N$. This may be expressed in terms of a reduced word, say of length $k\geq 1$. (We count e.g. $a^2$ as having length $2$.) We will find a vertex $v$ such that the lift $\tilde{\gamma}$ of $\gamma$ at $v$ lies inside of $T$. But then since $[\gamma] \in p_*\Pi_1(\tilde{X},v) = N$ is non-trivial, $\tilde{\gamma}$ is a non-trivial loop in $T$. (It is a loop since $\gamma$ lies in $p_*\Pi_1(\tilde{X},v)$). -To find $v$, let $V_0$ be the set of verticies such that one of the edges of that vertex lies outside of $T$. Let $V_i$ be the set of verticies of distance at most $i$ from some element of $V_0$ (i.e. the set of verticies that can be reached from following a path starting at $V_0$ that travels no more than $i$ edges). Since $N$ is of infinite index, $\tilde{X}$ has infinitely many verticies, but since each vertex has finitely many edges (each vertex must have one arrow going in and one going out for each generator of $F$, so has at most $2n$ edges) $V_i$ is always a finite set. Thus we may take any $v \in \tilde{X} \setminus V_{k}$. Since $\gamma$ is a word of length $k$, the lift of $\gamma$ at $v$ is a path of length $k$ and so never hits a vertex in $V_0$. But this means that all of the edges travelled by $\tilde{\gamma}$ lie inside of $T$, as claimed.<|endoftext|> -TITLE: convergence of infimum -QUESTION [5 upvotes]: I have a question that I encountered during my internship: -Consider a convergent sequence of continuous, convex functions $\{f_n(x)\}_n$ defined in $\mathbb{R}^M$. These functions are uniformly Lipschitz continuous, that is, $\exists C\in\mathbb{R}$ such that: -$$\forall x,y \in \mathbb{R^M},\forall n\ge1\quad |f_n(x)-f_n(y)|\le C|x-y|.$$ -Furthermore, each function $f_n(x)$ has a minimizer. The properties of simple convergence and uniform Lipschitz continuity allow us to prove that the convergence is uniform in any compact of $\Bbb R^M$. -My question is: -Can we demonstrate that $\inf_{\Bbb R^M}f_n(x)$ converges to $\inf_{\Bbb R^M}f_{\infty}(x)$ as $n\rightarrow\infty$, -where $f_{\infty}(x)$ is the limit of $f_n(x)$ and it is supposed that $\inf_{\mathbb{R^M}}f(x)$ is finite? -Thanks a lot! - -REPLY [5 votes]: No. Take the sequence -$$ -f_n : x\in \mathbb R \to \frac 1 n \lvert x - n \rvert -1 \in \mathbb R -$$ -We have -$$ -\lim_n f_n(x) = 0 \quad\forall x\in \mathbb R\\ -\inf_{\mathbb R} f_n(x) = f_n(n) = -1\\ -\lvert f_n(x) - f_n(y) \rvert = \frac 1 n \big\lvert \lvert x - n \rvert - \lvert y - n\rvert \big\rvert \leq \frac 1 n \lvert x - y\rvert \leq \lvert x -y \rvert -$$ -the functions $f_n$ converge punctually to $f_\infty = 0$, they are convex and uniformly Lipschitz continuous but -$$ -\lim_n \left(\inf_{\mathbb R} f_n(x) \right) = -1 \neq \inf_{\mathbb R} f_\infty(x) = 0 -$$<|endoftext|> -TITLE: Order of a product of subgroups. Prove that $o(HK) = \frac{o(H)o(K)}{o(H \cap K)}$. -QUESTION [43 upvotes]: Let $H$, $K$ be subgroups of $G$. Prove that $o(HK) = \frac{o(H)o(K)}{o(H \cap K)}$. -I need this theorem to prove something. - -REPLY [2 votes]: This is similar to Martin Brandenburg's answer. It is from Steve Roman's Fundamentals Of Group Theory. -Define a function $f: H \times K \to HK$ which maps $(h, k) \mapsto hk$. This is a surjective function and the fibers $f^{-1}(y)$ partition $H \times K$ into $|HK|$ disjoint sets. Since $|H \times K| = |H| \cdot |K|$, all that remains is to show that each fiber $f^{-1}(y)$ has size $|H \cap K|$. -Fix $y = hk \in HK$. Every element in $H \times K$ can be written as $(hu, vk)$ for some $u \in H$ and $v \in K$. But -$$f(hu, vk) = hk \text{ iff } huvk = hk \text{ iff } uv = 1 \text{ iff } v = u^{-1} $$ -which means -$$f^{-1}(y) = \{(hu, u^{-1}k) \: : \: u \in H \cap K\}$$ -Thus, $f^{-1}(y)$ indeed has size $|H \cap K|$.<|endoftext|> -TITLE: How to prove: $S=\frac{4}{3}\sqrt{ m(m-m_a)(m-m_b)(m-m_c)}$ -QUESTION [5 upvotes]: If $$m_a, m_b, m_c$$ are the medians of a triangle and let $$m=\frac{m_a+ m_b+ m_c}{2}$$ -then Area $S$ of triangle is given by -$$S=\frac{4}{3}\sqrt{ m(m-m_a)(m-m_b)(m-m_c)}$$ This looks very similar to Heron's formula. How to prove this formula? - -REPLY [2 votes]: A simple way to prove this is given here without much computation: http://jwilson.coe.uga.edu/emt725/Medians.Triangle/Area.Medians.Tri.html<|endoftext|> -TITLE: proper subgroups of finite p-groups are properly contained in the normalizer -QUESTION [9 upvotes]: I am trying to prove the following, - -Let $G$ be a finite $p$-group and let $H$ be a proper subgroup. Then there exists a subgroup $H'$ such that - $$ -H\lneq H'\leq G -$$ - and $H\triangleleft H'$. - -Obviously, the natural choice for $H'$ would be the normalizer $N_G(H)$ of $H$ in $G$. However, one needs to prove then that $H\lneq N_G(H)$ in finite $p$-groups. I am aware of a proof of this fact by induction on the order of $G$. However, I was wondering if there was another proof which only used group actions? - -REPLY [2 votes]: In fact, for a finite group $\,G\,$, to be nilpotent is equivalent to the normalizer condition, namely: every proper subgroup $\,H\leq G\,$ is properly contained in its normalizer, and here enters what Geoff mentioned at the end of his answer. -Since a (non-trivial) finite $\,p-\,$group is trivially nilpotent (as its center is always non-trivial), we have -$$1=:Z_0\leq Z_1\leq\ldots\leq Z_n=G\,\,,\,Z_i:=Z\left(G/Z_{i-1}\right)$$ the upper central series. -It follows, among other things, that $\,[G,Z_i]\leq Z_{i-1}\,\,,\,\forall i=1,2,...,n$ -So, if $\,H\lneq G\,$ then there exists $\,0\leq i\leq n\,\,s.t.\,\,Z_{i-1}\leq H\lneq Z_i\,$ , so that -$$\exists\,z\in Z_i-H\Longrightarrow [G:z]\in [G:Z_i]\leq [G:Z_{i-1}]\leq Z_{i-1}\leq H\Longrightarrow$$ -$$\Longrightarrow\,\forall\,g\in G\,\,,\,[g,z]:=g^{-1}z^{-1}gz\in H\Longrightarrow h^{-1}z^{-1}hz\in H\,\,,\,\,\forall\,h\in H\Longrightarrow z\in N_G(H)$$ -and from this it follows at once that $\,H\lneq N_G(H)\,$<|endoftext|> -TITLE: Trivial Restriction of Line Bundles -QUESTION [8 upvotes]: Say I have some projective space $\mathbb{P}^n$ and some line bundle $L=\mathcal{O}(-k)$. -Now, I want to have a subvariety $Y$ in $\mathbb{P}^n$ such that $L\vert_Y$ is trivial. -When is this the case? I can only think of trivial solutions, like when $Y$ is just a point and I can't seem to find a standard treatment of this in literature - -REPLY [10 votes]: Let's exclude the other trivial solution: $k=0$ and any $Y$. -Suppose now that $k\ne 0$. As $O(-k)|_Y$ trivial is equivalent to $O(k)|_Y$ (isomorphic to the dual of $O(-k)|_Y$) trivial, we can restrict to the case $k<0$. Then $L$, hence $L|_Y$, are ample. If $L|_Y$ is moreover trivial, then $O_Y$ is ample, which implies that $Y$ is affine. So necessarily $Y$ is affine. If $Y$ is a closed subvariety, this forces $Y$ to be a finite set. - -Conclusion: if $Y$ is a closed subvariety such that $L|_Y$ is trivial with $k\ne 0$, then $Y$ is a finite subset.<|endoftext|> -TITLE: When a scheme theoretical fiber is reduced? -QUESTION [9 upvotes]: I'd like to ask some basic things about algebraic geometry. -Suppose I have a map $\phi:V\to W$, between affine varieties over $k=\mathbb{C}$. -For any point $y \in W$, the scheme theoretical fiber is defined to be $Spec(k[V]\otimes_{k[W]} k )$. The question is when is this reduced? -I'm seeking an answer like: For the generic point in the image of $\phi$, this is true. -I also like to ask a proper reference for this (hope not in EGA, but sections in some textbook. such as Hartshorne). -Thank you very much! - -REPLY [12 votes]: If you are over a field of char. $0$, then yes, it is true that the fibre over a generic point of the image is reduced. To see this, first replace $W$ with the closure of the image of $\phi$, so as to assume that $\phi$ is dominant. Then the morphism $\phi$ corresponds to an injection of finite-type domains over $k$, say $A \hookrightarrow B$. (The injectivity follows from the dominance of $\phi$.) Now let's look at the generic fibre of $\phi$: this is Spec of the tensor product $K(A) \otimes_A B$ (here $K(A)$ denotes the fraction field of $A$), which is a localization of $B$, and so a domain (and hence reduced). Now if we are in char. zero, the reduced $K(A)$-algebra $K(A)\otimes_A B$ is necessarily geometrically reduced (i.e. stays reduced after extending to an algebraic closure of $K(A)$), and the property of fibres being geometrically reduced is open on the base, i.e. on Spec $A$; thus the fibres over an open subset of the image of $\phi$ will be reduced. (See e.g. Thm. 2.2 of these notes. You can find this in many places (though mabye not Hartshorne); these notes are just what turned up near the top of a quick google search.) -In char. $p$, this need not be true. E.g. consider the Frobenius map $\mathbb A^1 \to \mathbb A^1$, given by $x \mapsto x^p$. Then the fibre over every (closed) point is non-reduced. (The fibre over the generic point is reduced, but not geometrically reduced.) Nevertheless, if the generic fibre (in the scheme-theoretic sense) is geometrically reduced, then the fibre over an open set of closed points will be reduced too.<|endoftext|> -TITLE: Motivation for studying quadratic algebras, Koszul algebras, Koszul duality -QUESTION [8 upvotes]: I'm trying to gain a practical understanding of Koszul duality in different areas of mathematics. Searching the internet, there's lots of homological characterisations and explanations one finds, but I would like concrete examples of where Koszul duality is either a nice way of looking at a result, or where it provides us with new results. If someone were to be able to explain for example: - -Why is the Ext-algebra (Yoneda-algebra) (and the way it characterizes Koszulity) useful? -Is there a more elementary way of explaining some of the results and applications in the Beilinson-Ginzburg-Soergel paper Koszul duality patterns in representation theory? -I find it hard to grasp Manin's motivations for viewing quadratic algebras as noncommutative spaces, cfr. "Quantum groups and noncommutative geometry". Can anyone comment on this? -What are the applications of Koszul algebras? Example: I know that by a result of Fröberg every projective variety has a homogeneous coordinate ring that is Koszul, but what does this do for us really? - -Those are some of the examples that spring to mind, but feel free to add other phenomena. Also, let me know if you think there's a way to improve this question. - -REPLY [2 votes]: The Koszul complex and quadratic data arise in deformation quantization of quadratic Poisson structures: you can prove an important result on quantized realization via generators and relations of quadratic data, derived Morita theory of $A_{\infty}$-modules categories or introduce Koszul duality of exotic derived categories of dg structures. -If you are interested in these topics you can read -http://arxiv.org/abs/0911.4377 (generators and relations) -http://arxiv.org/abs/1206.2846 (derived Morita equivalence in the $A_{\infty}$ setting) -http://arxiv.org/abs/0905.2621 (exotic derived categories and Koszul duality)<|endoftext|> -TITLE: Prove that there does not exist any positive integers pair $(m, n)$ satisfying: $n(n + 1)(n + 2)(n + 3) = m{(m + 1)^2}{(m + 2)^3}{(m + 3)^4}$ -QUESTION [12 upvotes]: How to prove that, there does not exist any positive integers pair $(m,n)$ satisfying: -$n(n + 1)(n + 2)(n + 3) = m{(m + 1)^2}{(m + 2)^3}{(m + 3)^4}$. - -REPLY [17 votes]: This is an edited version of a partial answer that I posted sometime ago and subsequently deleted (not sure if resurrecting an answer is the correct thing to do after it has been up-voted and then deleted, perhaps someone will advise). If anyone can suggest where any of this can be improved, or point out any mistakes, I would be grateful. -Consider the equation -$$n(n+1)(n+2)(n+3) = m(m+1)^2(m+2)^3(m+3)^4$$ -To avoid some trivialities later on, it is easy to check that there are no solutions with $m=1$ or $m=2$. -Using the fact that $n(n+1)(n+2)(n+3)$ is almost a square, we have -$$(n^2+3n+1)^2-1 = m(m+2)\times[(m+1)(m+2)(m+3)^2]^2$$ -Putting $N = n^2+3n+1$ and $M = (m+1)(m+2)(m+3)^2$, this becomes -$$N^2-1 = m(m+2)M^2$$ -so that -$$N^2-1 = [(m+1)^2-1]M^2.$$ -Our approach now is to write this as -$$N^2 - [(m+1)^2-1]M^2 = 1,$$ -which is Pell's equation, with $d = (m+1)^2-1$. In this case there is a particularly nice solution for the Pell equation, as the continued -fraction is very simple in this instance. For convenience we change notation slightly and -use $k = m+1$, so that we are looking at solutions of -$$x^2 - (k^2-1)y^2 = 1,$$ -and bear in mind that for any solution $(x,y)$ we also -require $$y = (m+1)(m+2)(m+3)^2 = k(k+1)(k+2)^2.$$ -So we investigate the properties of solutions of the Pell equation above by looking at the -standard continued fraction method. -We have $$\sqrt{k^2-1} = (k-1)+\cfrac{1}{1+\cfrac{1}{(2k-2)+\cfrac{1}{1+\cfrac{1}{(2k-2) + \ddots}}}}$$ -which gives the first few solutions $(x_n,y_n)$ as $(1,0), (k,1), (2k^2-1,2k), \dots$. -Looking at the solutions for $y$, we see that they are generated by the recurrence relation -$$y_{n+2} = 2ky_{n+1} - y_n, \mbox{ with } y_0 = 0, y_1 = 1.$$ -Recalling that we also need $y = k(k+1)(k+2)^2$, it is enough to prove that this last expression cannot be one -of the $y_n$ from the recurrence relation as follows. -Clearly, the values $y_n$ are strictly increasing, and we claim that -$$y_4 < k(k+1)(k+2)^2 < y_5$$ -A bit of algebra gives $$y_4 = 8k^3-4k$$ so that -\begin{equation*}k(k+1)(k+2)^2-y_4 = k^4-3k^3+8k^2+8k = k^3(k-3)+8k^2+8k\end{equation*} -which is clearly positive for $k\geq 3$ and is easily checked to be positive for $k=1,2$. -For the right-hand inequality above, we have -$$y_5 = 16k^4-12k^2+1$$ -and then -$$y_5 - k(k+1)(k+2)^2 = 16k^4-12k^2+1 - (k^4+5k^3+8k^2+4k)$$ -$$ = 15k^4-5k^3-20k^2-4k+1$$ -and by examining the graph (because I can't see an elegant way to do this bit) we see that this is -positive for $k\geq 2$, and we know that $k=1$ (corresponding to $m=2$) is not a solution of -the original equation. -This shows that $k(k+1)(k+2)^2$ cannot be one of the $y_n$, so that no solution of the original equation is possible. -I am sure that there ought to be a simpler solution, but I have been unable to find one.<|endoftext|> -TITLE: Fermat's theorem on sums of two squares -QUESTION [5 upvotes]: There's Fermat's theorem on sums of two squares. -As the prime numbers that are $1\bmod4$ can be divided into the sum of two squares, will the squared numbers be unique? -For example, $41=4^2+5^2$ and the squared numbers will be 4 and 5. - -REPLY [4 votes]: Primes of the form $p=4k+1\;$ have a unique decomposition as sum of squares $p=a^2+b^2$ with $0 -TITLE: Uniqueness of morphism in definition of category theory product (etc) -QUESTION [9 upvotes]: I'm trying to understand the categorical definition of a product, which describes them in terms of existence of a unique morphism that makes such-and-such a diagram commute. I don't really feel I've totally understood the motivation for this definition: in particular, why must that morphism be unique? What's the consequence of omitting the requirement for uniqueness in, say, Set? - -REPLY [8 votes]: This is a question which you will be able to answer yourself after some experience ... anyway: -The cartesian product $X \times Y$ of two sets $X,Y$ has the property: Every element of $X \times Y$ has a representation $(x,y)$ with unique elements $x \in X$ and $y \in Y$. This is the important and characteristic property of ordered pairs. In other words, if $*$ denotes the one-point set: For every two morphisms $x : * \to X$ and $y : * \to Y$ there is a unique morphism $(x,y) : * \to X \times Y$ such that $p_X \circ (x,y) = x$ and $p_Y \circ (x,y) = y$. But since we can do everything pointwise, the same holds for arbitrary sets instead of $*$: For every two morphisms $x : T \to X$ and $y : T \to Y$ (which you may think of families of elements in $X$ resp. $Y$, also called $T$-valued points in the setting of functorial algebraic geometry), there is a unique morphism $(x,y) : T \to X \times Y$ such that $p_X \circ (x,y) = x$ and $p_Y \circ (x,y) = y$. Once you have understood this in detail, this motivates the general definition of a product diagram in a category. After all, these appear everywhere in mathematics (products of groups, vector spaces, $\sigma$-algebras, topological spaces, etc.). Of course, the uniqueness statement is essential. Otherwise, the product won't be unique and in the case of sets you will get many more objects instead of the usual cartesian product. In general, an object $T$ satisfies the universal property of $X \times Y$ without the uniqueness requirement iff there is a retraction $T \to X \times Y$. So for example in the category of sets every set larger than the cartesian product will qualify.<|endoftext|> -TITLE: which of the spaces are Locally Compact -QUESTION [7 upvotes]: [NBHM_2006_PhD Screening Test_Topology] - -which of the spaces are Locally Compact - -$A=\{(x,y): x,y \text{ odd integers}\}$ -$B=\{(x,y): x,y\text{ irrationals}\}$ -$C=\{(x,y): 0\le x<1, 05\}$ - - -A topological space $X$ is locally compact if every point has a neighborhood which is contained in a compact set. -well, I can prove that $\mathbb{Q}$ is not locally compact, so 1,2, are not Locally Compact, 3 is clearly locally compact. I am not ssure about 4. thank you. - -REPLY [5 votes]: For 4): -All open or closed subsets of a locally compact Hausdorff space are locally compact in the subspace topology. $R^2$ is locally compact and Hausdorff and $D = p^{-1}((5, \infty))$ is the inverse image of an open set under a continuous function $p(x,y) = x^2+103xy+7y^2$.<|endoftext|> -TITLE: The completion of a noetherian local ring is a complete local ring -QUESTION [5 upvotes]: We have defined the completion of a noetherian local ring $A$ to be $$\hat{A}=\left\{(a_1,a_2,\ldots)\in\prod_{i=1}^\infty A/\mathfrak{m}^i:a_j\equiv a_i\bmod{\mathfrak{m}^i} \,\,\forall j>i\right\}.$$ -I have a slight problem trying to understand the proof that then $\hat{A}$ is a complete local ring with maximal ideal $\hat{\mathfrak{m}}=\{(a_1,a_2,\ldots)\in\hat{A}:a_1=0\}$. -Proof. If $(a_1,a_2,\ldots)\in\hat{\mathfrak{m}}$, then $a_i\equiv 0\bmod{\mathfrak{m}}$ for all $i$, i.e., $a_i\in\mathfrak{m}$. Hence $$\hat{\mathfrak{m}}^i=\left\{(a_1,a_2,\ldots)\in\hat{A}:a_j=0\,\,\forall j\leq i\right\}.$$ -So the canonical map $\hat{A}\to A/\mathfrak{m}^i$, $(a_1,a_2,\ldots)\mapsto a_i$, is surjective with kernel $\hat{\mathfrak{m}}^i$. Thus $\hat{A}/\hat{\mathfrak{m}}^i\cong A/\mathfrak{m}^i$. In particular, $\hat{\mathfrak{m}}$ is a maximal ideal. But why is it the only one in $\hat{A}$? -If $(a_1,a_2,\ldots)\not\in\hat{\mathfrak{m}}$, we have $a_1\neq 0$, hence $a_1\not\in\mathfrak{m}$, hence it is a unit. By the defining property of the completion, $a_j$ is a unit for all $j$. So I could choose a candidate for the inverse of $(a_1,a_2,\ldots)$ by choosing inverse elements of the $a_j$. Why would this candidate be in $\hat{A}$ then, i.e. how can I choose it properly such that the congruences on the right hand side of the definition of the completion would be fulfilled? -Regards! - -REPLY [2 votes]: Inverses are unique. [If $R$ is a ring in which $y, z$ are inverses for $x$ then $y = y(xz) = (yx)z = z$.] So if $b_2 \in A/\mathfrak m^2$ is the inverse for $a_2$ then its homomorphic image in $A/\mathfrak m$ must be the inverse of $a_1$, and so on. -You could think of this in a different way: any element not in $\hat{\mathfrak m}$ can be written as $a + x$ for some $a \in A \setminus \mathfrak m$ and $x \in \hat{\mathfrak m}$, and hence as $a(1 - y)$ for some $y \in \hat{\mathfrak m}$. Then $a^{-1}(1 + y + y^2 + \cdots )$ is an inverse.<|endoftext|> -TITLE: Polynomials representing primes -QUESTION [10 upvotes]: Suppose over $\mathbb{Z}$ we are given an irreducible polynomial $p(x)$. -Can we say that at the very least that $p(x)$ represents a prime as $x$ runs through integers? -Thanks in advance. - -REPLY [8 votes]: Let $P(x)$ be a polynomial of degree $\ge 1$, with integer coefficients, such that no $d \gt 1$ divides all the coefficients. -If $P(x)$ has degree $1$, then $P$ represents at least one prime. This is a consequence of Dirichlet's Theorem on primes in arithmetic progression (and easily implies that Theorem). -As has been pointed out, for degree $\ge 2$, irreducibility is not enough to ensure that a polynomial represents a prime. For some irreducible polynomials $P(x)$, there exists a $d \gt 1$ such that $d$ divides $P(n)$ for every integer $n$. -However, that can only happen for relatively simple congruential reasons. So let us focus attention on polynomials $P(x)$ for which there is no such universal $d$. Unfortunately, it is an open problem whether such a polynomial must necessarily represent at least one prime. -Example: There is a good deal of evidence that there are infinitely many primes of the form $x^2+1$. However, whether or not there are infinitely many is a long-standing open problem, often called the Hardy-Littlewood Conjecture. If we could show that for all $a\ne 0$, (or even infinitely many $a$) there exists $x$ such that $(2ax)^2+1$ is prime, that would settle the Hardy-Littlewood Conjecture. (Conversely, the Hardy-Littlewood Conjecture implies that there are infinitely many such $a$.) -So the question you raised seems to be extremely difficult even for polynomials of degree $2$!<|endoftext|> -TITLE: Does there exist a universal pushdown automaton? -QUESTION [10 upvotes]: Let $\Sigma$ be a fixed alphabet and let $PDA(\Sigma)$ be the set of all Push-Down-Automata (PDA's) having input alphabet $\Sigma$. Is there an alphabet $S$ and a function $f:PDA(\Sigma) \to S^∗$ such that the set $\{(f(A),w) : w \in L(A)\}$ is accepted by a non-deterministic push-down automaton? If not, is the answer positive if we replace PDA's with Finite Automata (keeping the universal recognizer a PDA)? - -REPLY [6 votes]: This is not possible, assuming $(f(A),w)$ means writing the code of $A$ and then writing input $w$. Context-free (and regular) languages have a pumping property, that is certain substrings can be repeated. I assume you have seen the concept. The length-bounds on the repeated parts are determined by the language you are interested in, and can be arbitrary large. The "universal PDA" on the other hand has its own fixed pumping constant that cannot match arbitrary large values in the coded languages. This argument however only works if we can enforce having to avoid pumping in the $f(A)$ part of the description. For regular languages that is not a big problem, for context-free we need a rather strong pumping result. -Sorry, no details here. Not enough space ....<|endoftext|> -TITLE: If $V \times W$ with the product norm is complete, must $V$ and $W$ be complete? -QUESTION [11 upvotes]: Let $V,W$ be two normed vector spaces (over a field $K$). Then their product $V \times W$ with the norm $\|(x,y)\| = \|x\|_V + \|y\|_W$ is a normed space. -Using this norm it's easy to show that if $V,W$ are complete then so is $V \times W$. To see this, let the limit of the sequence $(x_n , y_n)$ be $(x,y) = (\lim x_n, \lim y_n)$. Then for $n$ large enough, both $\|x - x_n\|_V$ and $\|y - y_n\|_W$ are less than $\varepsilon / 2$ and hence $\|(x,y) - (x_n, y_n)\|< \varepsilon$. -The other direction does (probably) not hold. Can someone show me an example of a space $V \times W$ that is complete but either $V$ or $W$ (or both) are not? - -REPLY [13 votes]: If $V$ is not complete and $V\times W$ complete, take $\{v_n\}$ a Cauchy sequence which doesn't converge in $V$. Then $(v_n,0)$ is a Cauchy sequence in $V\times W$ and converges to $(v,w)\in V\times W$. We have -$$\lVert (v_n,0)-(v,w)\rVert_{V\times W}=\lVert v_n-v\rVert+\lVert w\rVert\to 0$$ -hence $v_n\to v$ in $V$, a contradiction. -Hence $V\times W$ is complete if and only if so are $V$ and $W$. - -REPLY [11 votes]: I believe that the spaces $V$ and $W$ must be complete whenever $V\times W$ is complete. -Closed subspace of a complete normed space is complete. -The space $V$ is isometrically isomorphic to the closed subspace $V\times\{0\}$ of $V\times W$.<|endoftext|> -TITLE: If $F$ is a formally real field then is $F(\alpha)$ formally real? -QUESTION [6 upvotes]: Let us call a field $F$ $\textit{ formally real }$ if $-1$ is not expressible as a sum of squares in $F$. Now suppose $F$ is a formally real field and $f(x)\in F[x]$ be an irreducible polynomial of odd degree and $\alpha $ is a root of $f(x)$. Is it true that $F(\alpha)$ is also formally real ? - -REPLY [4 votes]: Yes, and here is a proof. ${} {} {}$<|endoftext|> -TITLE: Deciding a problem: is it in $NP$, $NPC$ or $P$? -QUESTION [7 upvotes]: I'd like your help with understanding whether the following problem is in $P$, $NP$, $NPC$. -The problem $B$: -Input: a $3CNF$ formula which contains more than one clause. -output: Can we divide the formula to two $3CNF$ satisfiable clauses? -I'd really like to learn how to analyze this kind of questions. -First I know that in order that the problem would be in $NP$ I need to find a polynomial verifier to the problem, so I can check all the options for partitions and see if it's satisfies the problem. so the problem is probably in $NP$, in order to know whether it is complete, I need to find a reduction from a NP-hard problem, first one came to my head is $3SAT$, but how a suitable reduction would look like? or maybe it is simply in $P$? what is the way for deciding this? -I'd really appreciate some explanations. -Thanks a lot! - -REPLY [6 votes]: The following program solves problem $B$ in constant time: -Input B. -Output "yes". - -To see that "yes" is always the right answer, consider that we can divide the input into one set of clause where each clause contains at least one positive literal, and another set where each clause contains at least one negative literal. The first set can be satisfied by making everything true; the second can be satisfied by making everything false. -(In the special case where one of the sets just described is empty, the original 3CNF formula must be satisfiable, and so any partition of it will work). - -The above solution assumes that you're allowed to mix and match between the clauses in the input. If you must preserve the order of the clauses and just divide the initial 3CNF into a "front end" and a "back end", then the problem is NP-complete, by reduction from 3SAT itself: Given any 3CNF formula, choose a fresh variable $x$ and append the two clauses $x\lor x\lor x$ and $\bar x\lor \bar x\lor \bar x$. The only way to split the extended formula into satisfiable parts is to split between the new clauses, which means that the extended formula is in $B$ exactly if the original one was satisfiable.<|endoftext|> -TITLE: Possible cup product structures on a manifold -QUESTION [9 upvotes]: I am studying for a qualifying exam, and I came across this problem: -Let $M$ be a closed orientable connected 4-manifold with $H^1(M) = H^3(M) = 0$ and $H^2(M) \cong H^4(M) \cong \mathbb Z$. What are the possible cup product structures on $H^*(M)$? -My thoughts: -Just using properties of graded rings, we see that the product will be determined by $\alpha^2 \in H^4$, where $\alpha$ is a generator of $H^2$. If we fix $\beta$ a generator of $H^4$, then $\alpha^2 = n\beta$, so it seems like there is a distinct structure for each $n \in \mathbb Z$. -My question is: is there anything special about the cohomology ring $M$ that restricts the possibilities further? Somehow my answer seems too easy. - -REPLY [11 votes]: I will denote by $LH^*$ the free part of the cohomology ring. (That means that $LH^r(M)$ is $H^r(M)/\mathrm{torsion part}$.) For a connected 4-manifold, the symmetric form $I : LH^2(M) \times LH^2(M) \to \mathbb Z$ defined by $\alpha \smile \beta = I(\alpha, \beta) [M]$ is nondegenerate (that's one of the consequences of Poincaré duality). By definition, this means that the map $LH^2(M,\mathbb Z) \to \mathrm{Hom}(LH^2(M),\mathbb Z)$ is a bijection: in more pedestrian terms, that means that the matrix of $I$ has determinant $\pm 1$). So, in your case, $n$ is either $1$ or $-1$. -Both cases happen: you get $n=1$ with $\mathbb P^2(\mathbb C)$ and -$n=-1$ with the same manifold with the orientation reversed.<|endoftext|> -TITLE: A maximal ideal is always a prime ideal? -QUESTION [28 upvotes]: A maximal ideal is always a prime ideal, and the quotient ring is always a field. In general, not all prime ideals are maximal. 1 - -In $2\mathbb{Z}$, $4 \mathbb{Z} $ is a maximal ideal. Nevertheless it is not prime because $2 \cdot 2 \in 4\mathbb{Z}$ but $2 \notin 4\mathbb{Z}$. What is that is misunderstand? - -REPLY [51 votes]: Let $R$ be a ring, not necessarily with identity, not necessarily commutative. -An ideal $\mathfrak{P}$ of $R$ is said to be a prime ideal if and only if $\mathfrak{P}\neq R$, and whenever $\mathfrak{A}$ and $\mathfrak{B}$ are ideals of $R$, then $\mathfrak{AB}\subseteq \mathfrak{P}$ implies $\mathfrak{A}\subseteq \mathfrak{P}$ or $\mathfrak{B}\subseteq \mathfrak{P}$. -(The condition given by elements, $ab\in P$ implies $a\in P$ or $b\in P$, is stronger in the case of noncommutative rings, as evidence by the zero ideal in the ring $M_2(F)$, with $F$ a field, but is equivalent to the ideal-wise definition in the case of commutative rings; this condition is called "strongly prime" or "totally prime". Generally, with noncommutative rings, "ideal-wise" versions of multiplicative ideal properties are weaker than "element-wise" versions, and the two versions are equivalent in commutative rings). -When the ring does not have an identity, you may not even have maximal ideals. But here is what you can rescue; recall that if $R$ is a ring, then $R^2$ is the ideal of $R$ given by all finite sums of elements of the form $ab$ with $a,b\in R$ (that is, it is the usual ideal-theoretic product of $R$ with itself, viewed as ideals). When $R$ has an identity, $R^2=R$; but even when $R$ does not have an identity, it is possible for $R^2$ to equal $R$. -Theorem. Let $R$ be a ring, not necessarily with identity, not necessarily commutative. If $R^2=R$, then every maximal ideal of $R$ is also a prime ideal. If $R^2\neq R$, then any ideal that contains $R^2$ is not a prime ideal. In particular, if $R^2\neq R$ and there is a maximal ideal containing $R^2$, this ideal is maximal but not prime. -Proof. Suppose that $R^2=R$. Let $\mathfrak{M}$ be a maximal ideal of $R$; by assumption, we know that $\mathfrak{M}\neq R$. Now assume that $\mathfrak{A},\mathfrak{B}$ are two ideals such that $\mathfrak{A}\not\subseteq \mathfrak{M}$ and $\mathfrak{B}\not\subseteq\mathfrak{M}$. We will prove that $\mathfrak{AB}$ is not contained in $\mathfrak{M}$ (we are proving $\mathfrak{M}$ is prime by contrapositive). Then by the maximality of $\mathfrak{M}$, it follows that $\mathfrak{M}+\mathfrak{A}=\mathfrak{M}+\mathfrak{B}=R$. -Then we have: -$$\begin{align*} -R &= R^2\\ - &= (\mathfrak{M}+\mathfrak{A})(\mathfrak{M}+\mathfrak{B})\\ -&= \mathfrak{M}^2 + \mathfrak{AM}+\mathfrak{MB}+\mathfrak{AB}\\ -&\subseteq \mathfrak{M}+\mathfrak{M}+\mathfrak{M}+\mathfrak{AB}\\ -&=\mathfrak{M}+\mathfrak{AB}\\ -&\subseteq R, -\end{align*}$$ -hence $\mathfrak{M}\subsetneq\mathfrak{M}+\mathfrak{AB}=R$. Therefore, $\mathfrak{AB}\not\subseteq\mathfrak{M}$. Thus, $\mathfrak{M}$ is a prime ideal, as claimed. -Now suppose that $R^2\neq R$ and $\mathfrak{I}$ is an ideal of $R$ that contains $R^2$. If $\mathfrak{I}=R$, then $\mathfrak{I}$ is not prime. If $\mathfrak{I}\neq R$, then $RR\subseteq \mathfrak{I}$, but $R\not\subseteq \mathfrak{I}$, so $\mathfrak{I}$ is not prime. In particular, if $\mathfrak{M}$ is a maximal ideal containing $R^2$, then $\mathfrak{M}$ is not prime. $\Box$ -In your example, we have $R=2\mathbb{Z}$, $R^2=4\mathbb{Z}\neq R$, so any ideal that contains $R^2$ (in particular, the ideal $R^2$ itself) is not prime. And since $4\mathbb{Z}$ is a maximal ideal containing $R^2$, exhibiting a maximal ideal that is not prime. (In fact, $2\mathbb{Z}$ has maximal ideals containing any given ideals; this can be proven directly, or invoking the fact that it is noetherian)<|endoftext|> -TITLE: Eigen value of a block matrix -QUESTION [5 upvotes]: I have following block matrices: -$$M_1 = \left(\begin{array}{cc}A & B\\B' & D\end{array}\right)$$ -and -$$M_2 = \left(\begin{array}{cc}A & -B\\-B' & D\end{array}\right)$$ -I want to show that $\mathrm{eig}(M_1) = \mathrm{eig}(M_2)$. -How can I prove that? - -REPLY [2 votes]: Let me explicate Jyrki's answer, somewhat (feel free to upvote my answer, but I think you should give his answer precedence as far as accepting an answer). First of all, recall that the eigenvalues of a matrix $M$ are the zeroes of the characteristic polynomial $\mathrm{char}(P):=\det(M-tI)$, where $I$ is the identity matrix of the same size as $M$. Secondly, recall that $\det(AB)=\det(A)\det(B)$, whenever $A,B$ are same-sized square matrices. Thirdly, if $P$ is an invertible matrix (that is, $\det(P)\neq 0$), then $\det(P^{-1})=\frac{1}{\det(P)}$ (this actually follows from the second fact, since an identity matrix has determinant of $1$. -Now, as a consequence of all that, we see that if $P$ is any invertible matrix of the same size as a matrix $M$, then we have -$\begin{eqnarray*} -\mathrm{char}(PMP^{-1}) & = & \det(PMP^{-1}-tI)\\ -& = & \det(PMP^{-1}-tPP^{-1})\\ -& = & \det\bigl(P(M-tI)P^{-1}\bigr)\\ -& = & \det(P)\det(M-tI)\det(P^{-1})\\ -& = & \det(M-tI)\\ -& = & \mathrm{char}(M), -\end{eqnarray*}$ -so "conjugating" a matrix will not change its characteristic polynomial, and so will not change its eigenvalues. -In particular, let's suppose that $A$ is a $k\times k$ matrix and $D$ is a $m\times m$ matrix, and let $I_k$ and $I_m$ denote the $k\times k$ and $m\times m$ identity matrices, respectively. Jyrki is suggesting, then, that you set $$P=\left(\begin{array}{cc} I_k & 0_{k\times m}\\0_{m\times k} & -I_m\end{array}\right),$$ then observe that $M_2=PM_1P^{-1}$. From this, the desired conclusion follows by the work above. - -In a comment below, p.s. pointed out yet another way to show that conjugation doesn't change eigenvalues, and it's so nice that I'm going to go ahead and append it to my answer. Suppose $(\lambda,x)$ is an eigenpair of $M$, and that $P$ is an invertible matrix of the same size as $M$. Putting $y=Px$, we see that $$(PMP^{-1})y=PM(P^{-1}P)x=PMx=P(\lambda x)=\lambda(Px)=\lambda y,$$ so every eigenvalue of $M$ is an eigenvalue of any given conjugate of $M$. Since $M=P^{-1}(PMP^{-1})(P^{-1})^{-1}$, then by similar reasoning, every eigenvalue of a conjugate of $M$ is also an eigenvalue of $M$. -The only drawback to this approach (that I can see) is that it doesn't make immediately obvious the fact that the respective algebraic multiplicities of the eigenvalues are conjugation-invariant, as well as the eigenvalues, themselves. On the other hand, this approach shows much more readily that the geometric multiplicities of the eigenvalues is conjugation-invariant. Both of these are, of course, more information than you were actually seeking, but I wanted to point out that these stronger results hold.<|endoftext|> -TITLE: For which topological spaces $X$ can one write $X \approx Y \times Y$? Is $Y$ unique? -QUESTION [15 upvotes]: Under what conditions on topology of a given space $X$ it can be written as product space $Y\times Y$ for some other topological space $Y$. Lets call such space $Y$ a square root of X. Then given that square root of $X$ exists is it unique (within topological isomorphism) ? - -REPLY [4 votes]: Here is a necessary condition for a space to have a square root. By the Künneth theorem, if $X=Y\times Y$, then with respect to a field $F$, we must have $H_k(X;F)=\bigoplus_{i+j=k}H_i(Y;F)\otimes H_j(Y;F)$. Therefore, associating to the space $X$ the sequence of betti numbers $b_i(X;F)=\dim_F H_i(X;F)$, we must have that the sequence $(b_i)$, if consisting of finite numbers, must be the convolution square of another sequence of non-negative integers. -If there are only finitely many non-zero betti numbers for $X$, then the problem of existence of a convolution square root is the same as determining whether a polynomial has a polynomial square root with non-negative integer coefficients. If there are infinitely many non-zero betti numbers, the problem becomes one related to square roots of power series, and I'm not sure if there is an easy way to tackle that. -In any event, this gives the nice result that an odd dimensional compact manifold cannot have a square root.<|endoftext|> -TITLE: Why don't analysts do category theory? -QUESTION [103 upvotes]: I'm a mathematics student in abstract algebra and algebraic geometry. Most of my books cover a great deal of category theory and it is an essential tool in understanding these two subjects. -Recently, I started taking some functional analysis courses and I discovered that there is almost no category theory done in these courses. But since most of the spaces studied in functional analysis are objects in categories (e.g. the normed spaces form a category), I find it rather strange that the books leave the category theory out. -Is there a reason for this? - -REPLY [67 votes]: There are two questions here, in reality, I think. -First, in brief, I am told by many people that I "do functional analysis in the theory of automorphic forms", and I certainly do find a categorical viewpoint very useful. Second, in brief, it is my impression that the personality-types of many people who'd style themselves "(functional) analysts" might be hostile to or disinterested in the worldview of any part of (even "naive") category theory. -In more detail: as a hugely important example, I think the topology on test functions on $\mathbb R^n$ is incomprehensible without realizing that it is a (directed) colimit (direct limit). The archetype of incomprehensible/unmotivated "definition" (rather than categorical characterization) is in Rudin's (quite admirable in many ways, don't misunderstand me!) "Functional Analysis"' definition of that topology. -That is, respectfully disagreeing with some other answers, I do not think the specific-ness of concrete function spaces reduces the utility of a (naive-) categorical viewpoint. -From a sociological or psychological viewpoint, which I suspect is often dominant, it is not hard to understand that many people have a distaste for the structuralism of (even "naive", in the sense of "naive set theory") category theory. And, indeed, enthusiasm does often lead to excess. :) -I might claim that we are in a historical epoch in which the scandals of late 19th and early 20th century set theory prey on our minds (not to mention the mid-19th century scandals in analysis), while some still react to the arguable excesses of (the otherwise good impulses of) Bourbaki, react to certain exuberances of category theory advocates... and haven't yet reached the reasonable equilibrium that prosaically, calmly, recognizes the utilities of all these things. -Edit: since this question has resurfaced... in practical terms, as in L. Schwartz' Kernel Theorem in the simplest case of functions on products of circles, the strong topology on duals of Levi-Sobolev spaces is the colimit of Hilbert space topologies on duals (negatively-indexed Levi-Sobolev spaces) of (positively-indexed) Levi-Sobolev spaces. As I have remarked in quite a few other places, it was and is greatly reassuring to me that a "naive-categorical" viewpoint immediately shows that there is a unique (up to unique isomorphism) reasonable (!) topology there... -Similarly, for pseudo-differential operators, and other "modern" ideas, it is very useful to recast their description in "naive-categorical" terms, thereby seeing that the perhaps-seeming-whimsy in various "definitions" is not at all whimsical, but is inevitable. -A different example is characterization of "weak/Gelfand-Pettis" integrals: only "in my later years" have I appreciated the unicity of characterization, as opposed to "construction" (as in a Riemann/Bochner integral).<|endoftext|> -TITLE: Closure of the span in a Banach space -QUESTION [10 upvotes]: Let $X$ be a Banach space, and $S$ a subset. Is it true that $\overline {\operatorname{span}(S)}$ is equal to the set of the elements of $X$ that are obtained as norm convergent infinite sums of the scalar multiples of the elements of $S$? I can see that the infinite sums are in the closure of the span, and also that it would suffice to see that the collection of infinite sums of elements of $S$ forms a norm-closed set. I just can't show that. - -REPLY [10 votes]: I'll add an indirect argument from analysis. Let $X$ be the space of continuous functions on $[0,1]$ with the supremum norm, and take $S$ to be the set of monomials. Weierstrass tells us that the closed span of $S$ is the entire space. Yet, a function cannot be written as the sum of a uniformly convergent power series unless it extends to a complex analytic function in the open unit disk (which nonsmooth functions obviously don't), -[Added to address the concern about rearrangement] -Power series can be rearranged with impunity. Indeed, suppose $f(x)=\sum_{j=1}^\infty c_j x^{n_j}$ in $C[0,1]$. Since the sum converges at $x=1$, the coefficients are bounded: $|c_j|\le M$. Therefore, on every disk $\{ x\in \mathbb C: |x|\le r\}$, $r<1$, the j-th term is bounded by $Mr^{n_j}$. Hence the series converges uniformly and absolutely by the M-test of the same Weierstrass guy. We can rearrange it now, but don't really have to, because the limit of any uniformly convergent series of polynomials in a complex domain is a holomorphic function. -The same argument works if we replace $C[0,1]$ with the Hilbert space $L^2[0,1]$. Indeed, the $L^2$ norm of $c_j x^{n_j}$ is $|c_j|/\sqrt{2n_j+1}$, and this must be bounded if the series converges. On any disk of radius less than 1 the supremum of $c_j x^{n_j}$ is bounded by $Mr^{n_j}\sqrt{2n_j+1}$; these form a convergent numerical series and the M-test applies again.<|endoftext|> -TITLE: $\mathcal{B}^{-1}_{s\to x}\{e^{as^2+bs}\}$ and $\mathcal{L}^{-1}_{s\to x}\{e^{as^2+bs}\}$ , where $a\neq0$ -QUESTION [9 upvotes]: http://en.wikipedia.org/wiki/Integral_transform#Table_of_transforms claims than the integral form of inverse bilateral Laplace transform and inverse Laplace transform are both the same. But are they true in reality? -For $\mathcal{B}^{-1}_{s\to x}\{e^{as^2+bs}\}$ , since http://en.wikipedia.org/wiki/Lists_of_integrals#Definite_integrals_lacking_closed-form_antiderivatives has a formula that $\int_{-\infty}^\infty e^{-(ax^2+bx+c)}~dx=\sqrt{\dfrac{\pi}{a}}e^{\frac{b^2-4ac}{4a}}$ and http://eqworld.ipmnet.ru/en/auxiliary/inttrans/FourCos3.pdf has a formula that $\int_{-\infty}^\infty e^{-ax^2}\cos ux~dx=\dfrac{1}{2}\sqrt{\dfrac{\pi}{a}}e^{-\frac{u^2}{4a}}$ , it is clear that $\mathcal{B}^{-1}(e^{as^2+bs})(x)$ exist a close-form whenever $a<0$ or $a>0$ . -However, for $\mathcal{L}^{-1}_{s\to x}\{e^{as^2+bs}\}$ , it is troublesome. Since http://eqworld.ipmnet.ru/en/auxiliary/inttrans/laplace3.pdf has a formula that $\int_0^\infty e^{-(ax^2+px)}~dx=\dfrac{1}{2}\sqrt{\dfrac{\pi}{a}}e^{\frac{p^2}{4a}}\mathrm{erfc}\left(\dfrac{p}{2\sqrt{a}}\right)$ , it seems that $\mathcal{L}^{-1}_{s\to x}\{e^{as^2+bs}\}$ difficult to have close-form. The difficulty of $\mathcal{L}^{-1}_{s\to x}\{e^{as^2+bs}\}$ seems to be far away from $\mathcal{B}^{-1}_{s\to x}\{e^{as^2+bs}\}$ , and make me doubt the accuracy in http://en.wikipedia.org/wiki/Integral_transform#Table_of_transforms. -So is it possible to express $\mathcal{L}^{-1}_{s\to x}\{e^{as^2+bs}\}$ in terms of integrals with real lower limit and upper limit? -HINTS: -By $\int_{-\infty}^\infty e^{-(ax^2+bx+c)}~dx=\sqrt{\dfrac{\pi}{a}}e^{\frac{b^2-4ac}{4a}}$ , -$e^{as^2}=\dfrac{1}{\sqrt{4a\pi}}\int_{-\infty}^\infty e^{-\frac{x^2}{4a}-sx}~dx$ -$=\dfrac{1}{2\sqrt{a\pi}}\int_{-\infty}^\infty e^{-\frac{x^2+4asx}{4a}}~dx$ -$=\dfrac{1}{2\sqrt{a\pi}}\int_{-\infty}^\infty e^{-\frac{x^2+4asx+4a^2s^2-4a^2s^2}{4a}}~dx$ -$=\dfrac{1}{2\sqrt{a\pi}}\int_{-\infty}^\infty e^{-\frac{(x+2as)^2-4a^2s^2}{4a}}~dx$ -$=\dfrac{1}{2\sqrt{a\pi}}\int_{-\infty}^\infty e^{-\frac{x^2-4a^2s^2}{4a}}~dx$ -$=\dfrac{1}{\sqrt{a\pi}}\int_0^\infty e^{-\frac{x^2-4a^2s^2}{4a}}~dx$ -$=\dfrac{1}{\sqrt{a\pi}}\int_{-2as}^\infty e^{-\frac{(x+2as)^2-4a^2s^2}{4a}}~dx$ -$=\dfrac{1}{\sqrt{a\pi}}\int_{-2as}^\infty e^{-\frac{x^2}{4a}-sx}~dx$ - -REPLY [5 votes]: The difference comes about because of the different analytic structure of the function in the frequency space (and I guess there is some mistake on the Wiki-page). -The inverse of the one-sided Laplace transform is indeed given by the Bromwich integral which coincides with what is written on the page. Here, the real constant $c$ is large enough such that the contour is to the right of all singularities. The reason is that if the one-sided Laplace transform exists, it converges for all $s$ with $c\leq \mathop{\rm Re}s$. The constant $c$ depends on the function in time-domain. If $f(t) \in O(e^{c t})$ then $f(s)$ is analytic for real part larger than $c$. -On the other hand, the inverse of the two-sided Laplace transform the constant $c$ should be chosen differently from what is claimed in the sentence after the table. In general, the Laplace transformed function will be only converge and thus be analytic in a strip with $c_-\leq\mathop{\rm Re}s \leq c_+$. The inverse transform has to be integrated within this strip. -What is important to notice: given a function in the frequency domain, the inversion process might not be unique. One always need the function and the strip of analyticity (i.e., the region where the original Laplace transform did converge). For the one-sided Laplace transform this remark is unimportant as the strip is always to the right of the last singularity. But to invert the two-sided Laplace transform, one need to integrate in between two singularities and thus there might be multiple choices. So a Laplace transformed function always need to be given with the region of analyticity: -e.g. consider the functions -$$f_1(t) = \begin{cases} 0, & t<0\\ -e^{2t}-e^{-3t},& t> 0\end{cases}$$ -and -$$f_1(t) = \begin{cases} -e^{2t}, & t<0\\ --e^{-3t},& t> 0\end{cases}.$$ -The two-sided Laplace transform of the two functions coincides! -$$\mathcal{B} (f_1) =\mathcal{B} (f_2) = \frac{5}{s^2+s - 6}.$$ -However, the region of convergence of the first function is $\mathop{\rm Re}s >2$. -Whereas the second function converges on the strip with $-3 < \mathop{\rm Re}s < 2$ in the complex plane. -The inverse of the one-sided Laplace transform will of course be $f_1$ (as by definition of the transform does not care about $t<0$).<|endoftext|> -TITLE: Why are restrictions important? -QUESTION [5 upvotes]: When simplifying expressions, why do we add on restrictions for the simplified form if the original form was undefined at a certain point? The simplified form is defined at those points, so why should it be restricted? -An example of what I mean: -$$\frac{x^2−1}{x−1}=x+1, x≠1$$ - -REPLY [6 votes]: In short, it's because a function is all about the rule and domain of definition. -For example, let's consider the real-valued functions $f(x)=\frac{x^2}{x}$ and $g(x)=x$, with their maximal real domains of definition. In particular, then, $\mathrm{dom}(f)=\Bbb R\smallsetminus\{0\}$ and $\mathrm{dom}(g)=\Bbb R$. That means that $f$ and $g$ are not the same function, since their domains are not the same. Now, where both are defined, the rule is the same. However, this isn't enough for them to be the same function. Now, $f$ is a restriction of $g$ (specifically, to $\Bbb R\smallsetminus\{0\}$), so they are certainly related functions. -Edit: In the context of your previous question (regarding limits and restrictions), it is worth noting that $f$ can be continuously extended to $g$, since for any $\varepsilon>0$ there exists $\delta>0$ (in particular, in this case, $\delta=\varepsilon$) such that for any $0<|x|<\delta$ we have $\left|f(x)-0\right|<\varepsilon$.<|endoftext|> -TITLE: When is $\mathbb{F}_p[x]/(x^2-2)\simeq\mathbb{F}_p[x]/(x^2-3)$ for small primes? -QUESTION [11 upvotes]: I've been considering the rings $R_1=\mathbb{F}_p[x]/(x^2-2)$ and $R_2=\mathbb{F}_p[x]/(x^2-3)$, where $\mathbb{F}_p=\mathbb{Z}/(p)$. -I'm trying to figure out if they're isomorphic (as rings I suppose) or not for primes $p=2,5,11$. -I don't think they are for $p=11$, since $x^2-2$ is irreducible over $\mathbb{F}_{11}$, so $R_1$ is a field. But $x^2-3$ has $x=5,6$ as solution in $\mathbb{F}_{11}$, so $R_2$ is not even a domain. -For $p=5$, both polynomials are irreducible, so both rings are fields with $25$ elements. I know from my previous studies that any two finite fields of the same order are isomorphic, but I'm curious if there is a simpler way to show the isomorphism in this case, without resorting to that theorem. -For $p=2$, neither ring is even a domain as $x=0$ is a solution of $x^2-2$ and $x=1$ is a solution for $x^2-3$, but I'm not sure how to proceed after that. Thank you for any help. - -REPLY [2 votes]: As you say, it really depends how much theory you are prepared to use. If $2$ and $3$ are both quadratic non-residues (mod $p$), then the polynomials $x^{2}-2$ and $x^{2}-3$ are both irreducible in $\mathbb{F}_{p}[x],$ and both quotient rings are fields with $p^{2}$ elements. When $p$ is a prime and $n$ is a positive integer, the unique field up to isomorphism with $p^{n}$ elements is the splitting field for $x^{p^{n}}-x$ over $\mathbb{F}_{p}$, which is not so difficult to verify. As a matter of interest, the odd primes $p$ for which $2$ is a quadratic residue are those congruent to $\pm 1$ (mod 8), and the odd primes for which $3$ is a quadratic residue are primes congruent to $\pm 1$ (mod 12) (and 3 itself). -As you have observed, if one of $2,3$ is a quadratic residue (mod $p$) and the other is not, then one factor ring is a field and the other is not an integral domain, so they are certainly not isomorphic rings. -If both $2$ and $3$ are quadratic residues (mod $p$) and $p \not \in \{2,3\},$ then the rings are again isomorphic. Both are isomorphic to a ring direct sum of two copies of $\mathbb{F}_{p}$. For note that if $c^{2} = 2,$ then the image of $\frac{c-x}{2c}$ is an idempotent element of the quotient ring, and a similar argument works with 3 when 3 is a non-zero square. I leave the case $p \in \{2, 3\}$ for you to consider.<|endoftext|> -TITLE: "Probability" of a large integer being prime -QUESTION [5 upvotes]: Someone once told me (rather testily) that we cannot speak of the "probability that a number is prime" because the sequence is deterministic. I think I understood his point but would like to make sure. There is a theorem in Stopple's Primer of Analytic Number Theory (p. 97): - -The probability that a large integer $N$ is prime is about $\dfrac{1}{\log N}$. - -Of course, a large integer N is either prime or it is not. Its status is completely determined by its predecessors. -As long as we are careful to define the sample space, is there anything here that is controversial? Are there other probability-related objections to Stopple's theorem? -Thanks for any insight. -Edit: This was a pedagogical device, not a theorem, as the answers below (and Stopple) make clear. - -REPLY [7 votes]: Speaking of the "probability that a number is prime" is less of a specific fact but more of a guiding principle. A lot of things can be quickly derived by applying this principle, and when a pleasing (alleged) fact is discovered, one can look for a more rigorous proof if one so desires. -As an example, consider computing the sum -$$ \sum_{p \leq n} \frac{1}{p}. $$ -Applying the principle that a number $k$ is prime with probability $1 / \log k$, then the 'expected' value of the sum is -$$ \sum_{p \leq n} \frac{1}{p} \approx \sum_{k = 2}^{n} \frac{1}{k \log k} -\approx C_1 + \int_2^n \frac{dx}{x \log x} = C_2 + \log \log n $$ -where $C_1$ and $C_2$ are some constants. This turns out to be the correct answer, in the sense that the difference of the two sides converges to $0$ if $C_2$ is chosen to be the Meissel-Mertens constant.<|endoftext|> -TITLE: Some examples of virtually cyclic groups -QUESTION [8 upvotes]: The only virtually cyclic groups (ie. groups containing $\mathbb{Z}$ as subgroup of finite index) I really know are : the groups $F \times \mathbb{Z}$, where $F$ is a finite group, and the infinite dihedral group $D_{\infty}$ (isomorphic to $\mathbb{Z}_2 \ast \mathbb{Z}_2$). -But all these groups are finitely presented, just-infinite (ie. their proper quotients are finite) and residually finite (ie. for all element $g$, there exists a morphism $\varphi$ onto a finite group such that $\varphi(g) \neq 1$). -So I am looking for examples of virtually cyclic groups without one of these properties. I only know that there exists a virtually abelian group not just-infinite but without having an explicit example. -As other virtually abelian groups, there is also the generalized dihedral groups $\text{Dih}(G)$ where $G$ is an infinite finitely generated abelian group, but I don't know them really. Are they virtually cyclic ? -NB: The groups I consider are finitely generated. - -REPLY [2 votes]: The existing answer is restricted to virtually-cyclic groups, but more general things can be said: Finite presentability and residual finiteness are both preserved when moving from finite index subgroups to the big group. That is, suppose $H$ is a finite index subgroup of $G$. Then, - -If $H$ is finitely presentable then so is $G$. This can be proven using covering spaces. -If $H$ is residually finite then so is $G$. The way to prove this is to remember that a group $T$ is residually finite if for each $x\in T$ there exists a subgroup $K_x$ of finite index in $T$ such that $x\not\in K_x$. So, suppose $x\in G$ and we shall find such a finite index subgroup of $G$. if $x\not\in H$ then we are done, by taking $K_x=H$, while if $x\in H$ then there exists some $K_x\in H$ such that $x\not\in K_x$ and $K_x$ has finite index in $H$. As $K_x$ has finite index in $H$ it also has finite index in $G$, as required. - -Therefore, every virtually-cyclic subgroup is both finitely presentable and residually finite. So the groups you are looking for do not exist!<|endoftext|> -TITLE: Sword, pizza and watermelon -QUESTION [7 upvotes]: Suppose that we have a sword and cut a pizza and watermelon. What is the maximum number of pieces of pizza or watermelon obtained after 10 cuts. Is there a general formula - -REPLY [3 votes]: You can get 1024 of each by rearranging pieces between cuts to all be cut in half with the next cut. In fact, this is exactly how I make my famous samurai-ninja-watermelon-pizza salad, though I recommend skinning the rind from the watermelon first. -Note: I've also found that you can get even more pieces of pizza if you fold it like a kid's paper snowflake between cuts, but I've never personally made it over 8,192 pieces with this method...<|endoftext|> -TITLE: Does every Noetherian ring contain at least one maximal ideal? -QUESTION [6 upvotes]: I want to prove that a noetherian ring $R \neq \{0\}$ contains at least one maximal ideal. - -My idea is to consider $\langle 0 \rangle$ and $\langle 1 \rangle$: -If there is no ideal $I$ with $\langle 0 \rangle \subsetneq I \subsetneq\langle 1 \rangle$ then $\langle 0 \rangle$ is maximal. -Otherwise for each infinite chain $\langle 0 \rangle \subset I_1 \subset \cdots$ there exists $i \in \mathbb{N}$ such that for all $j>i$, $I_i = I_j$. Then $I_i$ is maximal. -Is that correct? - -REPLY [10 votes]: You have the right idea but your proof isn't correct as stated. It isn't true that each stabilizing chain ends in a maximal ideal. For example, you could just find a Noetherian ring where $0$ isn't maximal and just repeat $0 \subset 0 \subset 0 \subset \dots$ -What you should do is define this process: -Start with $0$. If $0$ is maximal then define $I_i = 0$ for $i \geq 1$. Otherwise, find some proper ideal $I_1$ that contains $0$ and put it in the list. -$0 \subset I_1$ -If $I_1$ is maximal let $I_i = I_1$ for $i \geq 2$. Otherwise find some proper ideal $I_2$ that contains $I_1$ and put it in the list. -Inductively we have $0 \subset I_1 \subset I_2 \subset \dots \subset I_n \subset \dots$ -By the Noetherian property, this chain stabilizes. The ideal it stabilizes to is maximal, or else by construction we would have chosen an ideal properly containing it to succeed it in the chain. -Also note that by a Zorn's lemma argument, every non zero ring has a maximal ideal. They key here is that for Noetherian rings you don't need Zorn's lemma. -EDIT: -I was informed that this argument uses the axiom of dependent choice so I will rewrite it here to make this clear. -The axiom of dependent choice states that for any nonempty set $X$ and any entire binary relation $T$ on $X$, there exists a sequence $(x_n)$ such that for all $n \geq 0$, $x_n T x_{n+1}$. A binary relation $T$ on $X$ is entire if for all $x \in X$, there exists $y \in X$ such that $xTy$. -Let $X$ be the set of all proper ideals of a nonzero noetherian ring $R$. Since $R \neq 0$, $X$ is nonempty. Consider the binary relation "<" of strict inclusion. If $R$ has no maximal ideals, then "<" is entire. So by the axiom of dependent choice, if $R$ has no maximal ideals, then we may choose a sequence $(x_n)$ such that -$x_1 < x_2 < \dots < x_n < \dots$ -This contradicts the Noetherian property. Hence $R$ has a maximal ideal.<|endoftext|> -TITLE: Inequality $\frac{a^2+b^2+c^2}{a^5+b^5+c^5}+\cdots+\frac{d^2+a^2+b^2}{d ^5+a^5+b^5}\le\frac{a+b+c+d}{abcd}$ -QUESTION [5 upvotes]: Let:$a,b,c,d>0$ be real numbers ,how to prove that : -$$\frac{a^2+b^2+c^2}{a^5+b^5+c^5}+\frac{b^2+c^2+d^2}{b^5+c^5+d^5}+\frac{c^2+d^2+a^2}{c^5+d^5+a^5}+\frac{d^2+a^2+b^2}{d ^5+a^5+b^5}\le\frac{a+b+c+d}{abcd}$$. -Edit : I think I proved it. From Cauchy inequality we have -$$ -\left(\sum\limits_{cyc}x^5\right)\left(\sum\limits_{cyc}x\right)\geq \left(\sum\limits_{cyc}x^3\right)^2 -$$ -From Chebyshev inequality it follows -$$ -\left(\sum\limits_{cyc}x\right)\left(\sum\limits_{cyc}x^2\right)\leq 3\left(\sum\limits_{cyc}x^3\right)^2 -$$ -hence -$$ -\frac{\left(\sum\limits_{cyc}x^5\right)}{\left(\sum\limits_{cyc}x^2\right)}= -\frac{\left(\sum\limits_{cyc}x^5\right)\left(\sum\limits_{cyc}x\right)}{\left(\sum\limits_{cyc}x\right)\left(\sum\limits_{cyc}x^2\right)}\leq -\frac{3}{\left(\sum\limits_{cyc}x^3\right)}\leq -\frac{1}{xyz} -$$ -In the last step I used AM-GM inequality. The rest is clear. -Is there a different way to prove it ? - -REPLY [4 votes]: This question can be solved only using the AM-GM inequality. Building on an idea that Pan Yang suggests in his comment, it suffices to show the following. -$$\frac{a^2+b^2+c^2}{a^5+b^5+c^5}\le\frac{d}{abcd}=\frac{1}{abc}$$ -This is equivalent to showing that -$$a^3 bc+b^3ca+c^3ab\le a^5+b^5+c^5 $$ -However, this is true by taking a weighted AM-GM as such (I'm writing it in full): -$$\frac{1}{5}a^5+\frac{1}{5}a^5+\frac{1}{5}a^5+\frac{1}{5}b^5+\frac{1}{5}c^5\ge(a^{15}b^{5}c^{5})^{\frac{1}{5}}=a^3bc$$ -and similarly for the terms $b^3ca, c^3ab$. This completes the proof.<|endoftext|> -TITLE: Can two topological spaces surject onto each other but not be homeomorphic? -QUESTION [30 upvotes]: Let $X$ and $Y$ be topological spaces and $f:X\rightarrow Y$ and $g:Y\rightarrow X$ be surjective continuous maps. Is it necessarily true that $X$ and $Y$ are homeomorphic? I feel like the answer to this question is no, but I haven't been able to come up with any counter example, so I decided to ask here. - -REPLY [4 votes]: Just to be different, here’s an example that isn’t related to the Hahn-Mazurkievicz theorem. The example is originally due (with a different purpose) to K. Sundaresan, Banach spaces with Banach-Stone property, Studies in Topology (N.M. Stavrakas & K.R. Allen, eds.), Academic Press, New York, 1975, pp. 573-580; the argumentation is mine, On an example of Sundaresan, Top. Procs. 5 (1980), pp. 185-6. The surjections are $1$-$1$ save at a single point each, where they are $2$-$1$. -Let $X=\omega^*\cup(\omega\times 2)$, where $\omega^*=\beta\omega\setminus\omega$, and $2$ is the discrete two-point space, let $\pi:X\to\beta\omega$ be the obvious projection, and endow $X$ with the coarsest topology making $\pi$ continuous and each point of $\omega\times 2$ isolated. Let $N=\omega\times 2$, for $n\in\omega$ let $P_n=\{n\}\times 2$, and let $\mathscr{P}=\{P_n:n\in\omega\}$. A function $f:X\to X$ preserves pairs if $f[P_n]\in\mathscr{P}$ for all but finitely many $n\in\omega$. - -Lemma. Let $f:X\to X$ be an embedding; then $f$ preserves pairs. -Proof. Suppose that $f$ does not preserve pairs. Since $f$ is injective, an easy recursion suffices to produce an infinite $M\subseteq\omega$ such that $(\pi\circ f)\upharpoonright\bigcup\{P_n:n\in M\}$ is injective. Let $M_i=M\times\{i\}$ for $i\in 2$. Then $$\left(\operatorname{cl}_XM_i\right)\setminus N=\left(\operatorname{cl}_{\beta\omega}M\right)\setminus\omega\ne\varnothing$$ for $i\in 2$, so $$\left(\operatorname{cl}_Xf[M_0]\right)\setminus N=\left(\operatorname{cl}_Xf[M_1]\right)\setminus N\ne\varnothing\;.$$ But $$\left(\operatorname{cl}_Xf[M_i]\right)\setminus N=\left(\operatorname{cl}_{\beta\omega}f[M_i]\right)\setminus\omega$$ for $i\in 2$, $\pi\big[f[M_0]\big]\cap\pi\big[f[M_1]\big]=\varnothing$, and disjoint subsets of $\omega$ have disjoint closures in $\beta\omega$, so $\operatorname{cl}_Xf[M_0]\cap\operatorname{cl}_Xf[M_1]=\varnothing$; this is the desired contradiction. $\dashv$ - -Now let $p$ be any point not in $X$, and let $Y=X\cup\{p\}$, adding $p$ to $X$ as an isolated point. - -Proposition. $Y$ is not homeormorphic to $X$. -Proof. Suppose that $h:Y\to X$ is a homeomorphism; it follows from the lemma that $h\upharpoonright X$ preserves pairs. Let $$A=\bigcup\Big\{P_n\in\mathscr{P}:h[P_n]\in\mathscr{P}\Big\}\cup\omega^*\;.$$ Then $\big|X\setminus h[A]\big|$ is finite and even, $|Y\setminus A|$ is finite and odd, and $h\upharpoonright(Y\setminus A)$ is a bijection between these two sets, which is absurd. $\dashv$ - -Finally, the maps -$$f:Y\to X:y\mapsto\begin{cases} -y,&\text{if }y\in X\\ -\langle 0,0\rangle,&\text{if }y=p -\end{cases}$$ -and -$$g:X\to Y:x\mapsto\begin{cases} -x,&\text{if }x\in\omega^*\\ -p,&\text{if }\pi(x)=0\\ -\langle n-1,i\rangle,&\text{if }x=\langle n,i\rangle\text{ and }n>0 -\end{cases}$$ -are continuous surjections. -By the way, each of $X$ and $Y$ embeds in the other, so these spaces witness the lack of a Schröder-Bernstein-like theorem for compact Hausdorff spaces and embeddings.<|endoftext|> -TITLE: Is there a categorical definition of submetry? -QUESTION [172 upvotes]: (Updated to include effective epimorphism.) - -This question is prompted by the recent discussion of why analysts don't use category theory. It demonstrates what happens when an analyst tries to use category theory. - -Consider the category CpltMet in which the objects are complete metric spaces and morphisms are 1-Lipschitz maps; the maps $f:X\to Y$ such that $d_Y(f(a),f(b))\le d_X(a,b)$ for all $a,b\in X$. Note that in this category monomorphisms are injective 1-Lipschitz maps, and epimorphisms are 1-Lipschitz maps with dense range, but not necessarily surjective. -An isometric embedding is a map $f:X\to Y$ such that $d_Y(f(a),f(b)) = d_X(a,b)$ for all $a,b\in X$. I can describe such maps in the arrow-speak as follows. $f:X\to Y$ is an isometric embedding iff the following holds: whenever $f$ factors through an epimorphism $g:X\to Z$ (meaning $f=h\circ g$ for some $h:Z\to Y$), $g$ is an isomorphism. (Proof is given in Note 1.) If there is a better categorical description of isometric embeddings, I'd like to see it. -A submetry is a map $f:X\to Y$ such that for every $a\in X$ and every $r\ge 0$ we have $f(B_X(a,r))=B_Y(f(a),r)$ where $B$ denotes a closed ball. (See Note 2 about the definition). To appreciate this definition, consider the following. - -isometric embeddings are characterized by the condition $f^{-1}(B_Y(f(a),r))=B_X(a,r)$, mirroring the definition of submetry. -among 1-Lipschitz maps, submetries are characterized by the 2-point lifting property: -for every $y_0,y_1\in Y$ and every $x_0\in f^{-1}(y_0)$ there exists $x_1\in f^{-1}(y_1)$ such that $d_X(x_0,x_1)=d_Y(y_0,y_1)$. -for linear operators between Banach spaces, the adjoint of an isometric embedding is a submetry (the proof is an exercise with Hahn-Banach). - - -My question is: can the submetries be defined categorically, preferably in a way that makes them a dual class to isometric embeddings? - -The problem is that reversing the arrows in the above definition of an isometric embedding gives a wider class of maps than submetries. Indeed, the reversed definition is: $f:Y\to X$ does not factor through any monomorphism $g:Z\to X$ unless $g$ is an isomorphism. But this holds, for example, for the function $f:\mathbb R\to \mathbb R$ defined by -$$ -(*)\qquad \qquad f(x)=\begin{cases} x+1,\quad &x\le -1 \\ 0,\quad &|x|\le 1 \\ x-1,\quad &x\ge 1 -\end{cases}$$ -which is not a submetry in the standard metric of the real line. (See Note 3 for the proof.) Maybe I'm reversing arrows in a "wrong" description of isometric embeddings. - -Note 1. If $f$ preserves distances, then so does $g$; having dense range, $g$ must be onto because $X$ is complete; hence, $g$ is an isomorphism. Conversely, if $f$ decreases distances somewhere, let $Z$ be the same set as $X$ with the metric $(d_X(a,b)+d_Y(f(a),f(b)))/2$. The identity map $g:X\to Z$ is an epimorphism, $f$ factors through it, but $g$ is not an isomorphism. -Note 2. I am following the original definition of submetry given by Berestovskii ("Submetries of space-forms of nonnegative curvature", 1987). If one uses open balls instead of closed, the class is enlarged to weak submetries. In Riemannian Geometry by Petersen the term submetry is used for more general maps, which I would call weak local submetries. -Note 3. Proof: Suppose $f=g\circ h$ where $g: Z\to \mathbb R$ is a monomorphism. Then $h$ maps $[-1,1]$ into a single point $z\in Z$. When $a\le -1$ and $b\ge 1$, the triangle inequality yields $d_Z(h(a),h(b))\le |a-b|-2=|f(a)-f(b)|$. Hence, $g$ must be an isomorphism. -Note 4. Following the immersion:submersion terminology of differential geometry, I'd like to call an isometric embedding an immetry, but I'm not sure that the neologism would catch on. - -Following the suggestion by @t.b., I considered the concept of an effective epimorphism. Unfortunately, the undesirable map defined by (*) appears to be effective. Indeed, let $R=\{(x,y)\in\mathbb R^2: f(x)=f(y)\}$. The orthogonal projections $\pi_x,\pi_y : R\to \mathbb R$ are 1-Lipschitz and satisfy $f\circ \pi_x=f\circ \pi_y$ by construction. As far as I can tell, $\pi_x$ and $\pi_y$ qualify as a kernel pair for which $f$ is a coequalizer. - -REPLY [6 votes]: A common problem, whenever one tries to express analysis concepts in terms of category theory, is that analysis is done with inequalities rather than equalities. So, in general, ordinary category theory is not very well-suited to talk about analysis. On the other hand, a slight generalization of categories, namely order-enriched categories (or even quantale-enriched) gives a much better language to talk about analysis. It can be thought of as a "category theory with inequalities". However, the translation often requires some work. - -Now to the question. In the context of enriched category theory, a metric space can be seen as a particular enriched category (see for example here). Complete metric spaces can be described categorically in quite an elegant way (here), and the correspondent enriched notion of functor gives exactly 1-Lipschitz maps. -(In enriched category theory, distances are usually not required to be symmetric or finite, but of course one can make that requirement, and basically all the results restrict to ordinary metric spaces.) -Treating then your category as a category of enriched categories, submetries are precisely the proper maps in the sense of Definition 3.1.1 of this book: - -Gavin J. Seal, Dirk Hofmann and Walter Tholen, Monoidal Topology, Cambridge University Press, 2014. - -(Not to be confused with the different, but related, notion of proper map in topology.) -I won't copy the definition since it requires quite some buildup - but let's see what it means for our case. -I paraphrase from the book above, Chapter V, Example 3.13(3): - -For metric spaces, a 1-Lipschitz map $f:X\to Y$ is proper if and only if - $$ -d_Y(f(x),y) \;=\; \inf \big\{d_X(x,x')\;|\;x'\in f^{-1}(y)\big\} . -$$ - for all $x\in X$ and $y\in Y$. - -The two notions agree for $f:X\to Y$ 1-Lipschitz, let's show this explicitly. - -Suppose that $f$ is a submetry. Let $r>0$, $x\in X$ and $y\in Y$. If $d_Y(f(x),y)\le r$, then -$$ -y \;\in\; B_Y(f(x),r) \;=\; f(B_X(x,r)) , -$$ -therefore for all $x'$ with $f(x')=y$, we have $d_X(x,x')\le r$. In other words, -$$ -\inf \big\{d_X(x,x')\;|\;x'\in f^{-1}(y)\big\} \;\le\; r. -$$ -This is true for all $r$, so -$$ -\inf \big\{d_X(x,x')\;|\;x'\in f^{-1}(y)\big\} \;\le\; d_Y(f(x),y). -$$ -The reverse inequality holds by the fact that $f$ is 1-Lipschitz. Therefore $f$ is proper in the sense above. -Conversely, suppose that $f$ is proper in the sense given above. Then for $x\in X$ and $r>0$, -$$ -f(B_X(x,r)) \;\subseteq B(f(x),r). -$$ -To prove equality suppose that $y\in B(f(x),r)$, i.e. that $d_Y(f(x),y)\le r$. Then -$$ -\inf \big\{d_X(x,x')\;|\;x'\in f^{-1}(y)\big\} \;\le\; r, -$$ -so for all $x'$ such that $f(x')=y$, $d_X(x,x')\le r$. So $y\in f(B_X(x,r))$. This means that $f$ is a submetry. - -(You can also see what happens if we drop the 1-Lipschitz requirement.) - -By the way, we've noticed this equivalence of notions during our work (arXiv:1808.09898, see Appendix A therein), which indeed is about a categorical study of some structures in analysis and functional analysis (such as the stochastic order of probability measures and ordered Banach spaces). -(Thanks to Tobias Fritz, Slava Matveev and Walter Tholen for the pointers.)<|endoftext|> -TITLE: Further reading on the $p$-adic metric and related theory. -QUESTION [6 upvotes]: In his book Introduction to Topology, Bert Mendelson asks to prove that -$$(\Bbb Z,d_p)$$ -is a metric space, where $p$ is a fixed prime and -$$d_p(m,n)=\begin{cases} 0 \;,\text{ if }m=n \cr {p^{-t}}\;,\text{ if } m\neq n\end{cases}$$ -where $t$ is the multiplicty with which $p$ divides $m-n$. Now, it is almost trivial to check the first three properties, namely, that -$$d(m,n) \geq 0$$ -$$d(m,n) =0 \iff m=n$$ -$$d(m,n)=d(n,m)$$ -and the only laborious was to check the last property (the triangle inequality). I proceeded as follows: -Let $a,b,c$ be integers, and let -$$a-b=p^s \cdot k$$ -$$b-c=p^r \cdot l$$ -where $l,k$ aren't divisible by $p$. -Then $$a-c=(a-b)+(b-c)=p^s \cdot k+p^r \cdot l$$ -Now we have three cases, $s>r$, $r>s$ and $r=s$. We have respectively: -$$a-c=(a-b)+(b-c)=p^r \cdot(p^{s-r} \cdot k+ l)=p^r \cdot Q$$ -$$a-c=(a-b)+(b-c)=p^{s} \cdot( k+p^{r-s} \cdot l)=p^s \cdot R$$ -$$a-c=(a-b)+(b-c)=p^s \cdot (k+l)=p^s \cdot T$$ -In any case, -$$d\left( {a,c} \right) \leqslant d\left( {a,b} \right) + d\left( {b,c} \right)$$ -since -$$\eqalign{ - & \frac{1}{{{p^r}}} \leqslant \frac{1}{{{p^s}}} + \frac{1}{{{p^r}}} \cr - & \frac{1}{{{p^s}}} \leqslant \frac{1}{{{p^s}}} + \frac{1}{{{p^r}}} \cr - & \frac{1}{{{p^s}}} \leqslant \frac{1}{{{p^s}}} + \frac{1}{{{p^s}}} \cr} $$ -It might also be the case $k+l=p^u$ for some $u$ so that the last inequality is -$$\frac{1}{{{p^{s + u}}}} \leqslant \frac{1}{{{p^s}}} + \frac{1}{{{p^s}}}$$ -$(1)$ Am I missing something in the above? The author asks to prove that in fact, if $t=t_p(m,n)$ is the exponent of $p$, that -$$t\left( {a,c} \right) \geqslant \min \left\{ {t\left( {a,b} \right),t\left( {b,c} \right)} \right\}$$ -That seems to follow from the above arguement, since if $s \neq r$ then -$$t\left( {a,c} \right) = t\left( {a,b} \right){\text{ or }}t\left( {a,c} \right) = t\left( {b,c} \right)$$ -and if $s=r$ then -$$t\left( {a,c} \right) \geqslant t\left( {a,b} \right){\text{ or }}t\left( {a,c} \right) \geqslant t\left( {b,c} \right)$$ -$(2)$ Is there any further reading you can suggest on $p$-adicity? - -REPLY [4 votes]: Your argument looks fine to me. - -A very good online introduction to $p$-adics are these notes; it covers modular arithmetic leading up to Hensel's, basic analysis with the numbers, the very strange metric topology of $\Bbb Q_p$ (every point inside of a ball is a center, there are locally but not globally constant functions, etc.), and a bit of field theory and algebraic number theory. -I don't really have any recommendation for number-theoretically in-depth stuff. Also, I tend to stick to online documents because the web is where I spend all my time anyway. -In actuality, my favorite discussion is p-adic integration and the theory of groups, which involves category theory and abstract algebra (the $p$-adics are constructed as an inverse limit of topological rings, for example), as well as measure theory and group theory. This source perhaps takes more background to digest its contents satisfactorily, but it hits my buttons well. -Just for fun, I also suggest Pictures of Ultrametric Spaces. Related entertainment: (working with adeles) the character group of $\Bbb Q$ and (working with profinite integers) profinite Fibonacci numbers.<|endoftext|> -TITLE: Parabolic linear PDE with continuous coefficients; how to solve and explanation of text needed -QUESTION [7 upvotes]: I have the following PDE -$$u_t = a(x,t)u_{xx} + b(x,t)u_x + c(x,t)u - f(x,t)$$ -$$u|_{t=0} = u_0$$ -over the domain $S^1 \times [0,T)$. The coefficients and $f$ are in $C^{k,\alpha}$ for some $k$ (and are $2\pi$ periodic, ignore if this doesn't make sense). Also $a$ is such that the equation is uniformly parabolic. -My questions: -1) To get an a-priori estimate for this equation -$$\lVert u \rVert_{C^{k+2, \alpha}} \leq C(\lVert f \rVert_{C^{k, \alpha}} + \lVert u_0 \rVert_{C^{k+2, \alpha}})$$ -what do I do? The only thing I know of is multiplying by a test function and integrating and using Gronwall but this gives me norms in Sobolev spaces, I believe. -2) Apparently, the following is true, but I need some explanation: - -There is a unique solution $u \in C^{k+2, \alpha}$. Proof: first solve the Cauchy problem in the smooth category by means of separation of variables. Then use an approximation argument coupling with the global a-priori estimate above to get the general result. - -What's the Cauchy problem (Wikipedia doesn't help. It just says the domain is a manifold)? What's smooth category? What's the approximation argument thing? Sorry if these questions are stupid but I have never heard of this stuff. -Any references or help would be appreciated. - -REPLY [3 votes]: As Andrew commented, obtaining Hölder estimates can sometimes be labour intensive. So I'll ignore part (1) for now. -For part (2): - -The term "Cauchy problem" is just another name for "initial(-boundary) value problem". In your case there is no boundary since the spatial domain is a closed manifold. -Smooth category just means that given smooth initial data, try to solve the equation with smooth solution. Usually this is obtained by separation of variables or explicitly integrating against some Green's function. -Approximation argument is the idea that for most reasonable function (Banach) spaces $X$ defined on a manifold $M$, the space $C^\infty_0(M)$ of smooth functions with compact support is dense in $X$ in the Banach norm. Hence we can approximate a, say, $C^{k,\alpha}$ initial data with a sequence of $C^\infty$ initial data. Now if there is a good a priori estimate available (such as the Hölder estimate you quoted in your question), then we can transfer the convergence of the initial data to a convergence of the sequence of smooth solutions in $C^{k+2,\alpha}$ norm, obtaining then a $C^{k+2,\alpha}$ solution as the limit of a sequence fo smooth solutions, which we derived from the sequence of smooth initial data approximating the $C^{k,\alpha}$ initial data.<|endoftext|> -TITLE: Universal property of the completion of rings / modules -QUESTION [10 upvotes]: If $A$ is a noetherian local ring and $M$ an $A$-module, then we define the completion $\hat{M}$ of $M$ with respect to the stable $\mathfrak{m}$-filtration $\{M_n\}$ by $$\hat{M}=\left\{(a_1,a_2,...)\in\prod_{i=1}^\infty M/M_i:a_j\equiv a_i\bmod{M_i}\,\,\forall j>i\right\}.$$ -See also my previous question. -Now in the book I use (A SINGULAR Introduction to Commutative Algebra by Greuel/Pfister), there is no universal property of this completion mentioned, but once it uses something that looks like one: We have a map from $K[x_1,...,x_n]_{\langle x_1,...,x_n\rangle}$ to some complete ring, hence we got a map from $K[[x_1,...,x_n]]$ to it. Is that the 'universal property of the completion of a ring / module', and if yes, is it somehow obvious from my definition of the completion, so that we could use it directly? -How to prove this property with the above definition? (If it really works for modules; I don't know, at least for rings I guess it should be something like: If $A\to B$ is a ring homomorphism and $B$ is complete, then there is a unique map $\hat{A}\to B$ extending it). Well, I think I maybe got a clue right now, and you could perhaps tell me if this is the correct way (I'd still like to know if this is 'the' universal property of the completion): -If $A\to B$ is a ring homomorphism ($A$ and $B$ noetherian local rings; does this homomorphism have to be local, too?), and $B$ is complete, then I get and induced map $\hat{A}\to\hat{B}=B$ as wanted. - -REPLY [9 votes]: As you are considering modules endowed with stable $m$-filtrations, these are simply $A$-modules with $m$-adic topology (defined by the filtration $(m^nM)_n$). -The completion $\hat{M}$ satisfies the following universal property in the category of $m$-adic $A$-modules: - -There exists an $A$-linear map $i: M\to \hat{M}$, -For any complete $A$-module $N$ for any $A$-linear map $f : M\to N$, there exists a unique factorization of $f$ as -$i : M\to \hat{M}$ and $\hat{f} : \hat{M} \to N$. - -Proof: (1) the map $i$ is given by $i(x)=(x \mod mM, x \mod m^2M, ...)$. (2) The map $f$ induces $M/m^nM \to N/m^nN$. Passing to the limit we get a map $\hat{f} : \hat{M}\to \hat{N}=N$. As $i(M)$ is dense in $\hat{M}$, for any factorization as in (2), $\hat{f}$ is uniquely determined by $f$. -For homomorphisms of local rings $\rho : A\to B$ you have to require the maps to be local to insure continuity, then $\rho$ induces -$$ A/m_A^n \to B/m_B^n $$ -for all $n\ge 1$. Passing again to the limit we get a ring homomorphism $\hat{A}\to \hat{B}$ which is uniquely determined by $\rho$ because $A$ is dense in $\hat{A}$.<|endoftext|> -TITLE: Infinite product -QUESTION [6 upvotes]: How do I solve the infinite product of $$\prod_{n=2}^\infty\frac{n^3-1}{n^3+1}?$$ -I know that I have to factorise to $$\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)},$$ -but how do I do the partial product? -Thanks a lot in advance. -If I'm not mistaken the Answer is 2/3 - -REPLY [9 votes]: Hint: Let $f(x)=x^2+x+1$. Then $f(n)=n^2+n+1$, and $f(n-1)=n^2-n+1$. This will enable you to "telescope" the terms $\frac{n^2+n+1}{n^2-n+1}$. A whole lot of cancelling going on. -The $\frac{n-1}{n+1}$ terms also telescope. -I would suggest that you write down the terms you are multiplying, for $n=2$, $3$, $4$, even $5$. Express each term in the factored form mentioned in the post. For example, for $n=2$ we will have $\frac{1\cdot 7}{3\cdot 3}$. The collapse will be visually clear. - -REPLY [7 votes]: After factorization, the product looks like $(\frac{1}{3}\frac{2}{4}\frac{3}{5}\frac{4}{6}\frac{5}{7}\cdots)(\frac{7}{3}\frac{13}{7}\frac{21}{13}\frac{31}{21}\frac{43}{31}\cdots)=(2)(1/3)=2/3$.$$$$ Here terms in first () are from expression $\frac{n-1}{n+1}$ and terms in second () from expression $\frac{n^2+n+1}{n^2-n+1}$ .<|endoftext|> -TITLE: Is convolution operator compact? -QUESTION [6 upvotes]: I know convolution is not a Hilbert–Schmidt integral operator, but it needs more to tell if convolution is compact or not. - -REPLY [12 votes]: Let us use $T_f$ to denote the operator which convolutes with $f$ and let $M_\phi$ to denote the operator which multiplies by $\phi$. Since Young's inequality provides sharp bounds, we necessarily have $f \in L^1$, if we want $T_f:L^2\to L^2$ to even be bounded. Consider $FT_f$, the composition of the Fourier transform with $T_f$. Since $F$ is unitary, $T_f$ is compact if and only if $FT_f$ is. By the convolution theorem we have -$$FT_fg = \hat{f}\cdot\hat{g} = M_{\hat{f}}Fg$$ -so that $FT_f = M_{\hat{f}}F$ is a multiplication operator composed with the Fourier transform. Again, the operator on the right hand side is compact if and only if the multiplication operator $M_{\hat{f}}$ is. Now, $\hat{f} \in C_0$ and hence $\hat{f} \in L^\infty$. But the only compact multiplication operator on $L^2$ induced by a bounded measurable function is the operator that is identically zero. Hence $\hat{f}$ and therefore $f$ must be identically zero. It follow that the only compact convolution operator on $L^2$ is the operator which is identically zero. -To see that there are no non-trivial compact multiplication operators, suppose that $f \in L^\infty$ and $f \neq 0$. Then there exists a set $E$ of positive measure such that $|f| \geq \epsilon > 0$ on $E$. Thus $M_f$ would be compact with a bounded left inverse on a subspace isomorphic to $L^2(E)$ which is impossible.<|endoftext|> -TITLE: Pythagorean Theorem for imaginary numbers -QUESTION [10 upvotes]: If we let one leg be real-valued and the other leg equal $bi$ then the Pythagorean Theorem changes to $a^2-b^2=c^2$ which results in some kooky numbers. -For what reason does this not make sense? Does the Theorem only work on real numbers? Why not imaginary? - -REPLY [4 votes]: In the vein of @AndréNicolas' comment that "ideas like this can be useful" ... -Heron's formula for the area, $W$, of a (non-obtuse) triangle in terms of the lengths of its sides is equivalent to the Pythagorean Theorem for Right-Corner Tetrahedra: -$$W^2 = X^2 + Y^2 + Z^2$$ -where $W$ is the "hypotenuse-face" opposite three mutually-perpendicular edges that form right triangles with areas $X$, $Y$, $Z$. -Specifically, if we position the right corner at the origin in 3-space, choose the other vertices at $(x,0,0)$, $(0,y,0)$, $(0,0,z)$, and write -$$\begin{align} -a^2 = y^2 + z^2 \qquad b^2 &= z^2 + x^2 \qquad c^2 = x^2 + y^2 \\ -X = \frac{1}{2} y z \qquad Y &= \frac{1}{2} z x \qquad Z = \frac{1}{2} x y -\end{align}$$ -then -$$x^2 = \frac{1}{2}\left(-a^2 + b^2 + c^2\right) \qquad y^2 = \frac{1}{2}\left(a^2-b^2+c^2\right) \qquad z^2 = \frac{1}{2}\left(a^2+b^2-c^2\right)$$ -whence -$$\begin{align} -W^2 = X^2 + Y^2 + Z^2 &= \frac{1}{4}\left(y^2 z^2+z^2x^2+x^2y^2\right)\\ -&=\frac{1}{16}\left(-a^4-b^4-c^4+2a^2b^2+2b^2c^2+2c^2a^2\right) \\ -&=\frac{1}{16}\left(a+b+c\right)\left(-a+b+c\right)\left(a-b+c\right)\left(a+b-c\right) -\end{align}$$ -which is Heron's Formula for $W$ in terms of side-lengths $a$, $b$, $c$. -Note that I wrote the equivalence is for "non-obtuse" triangles. This is because "hypotenuse-face" $W$ cannot have an obtuse angle; to see this, write $\theta$ for the angle opposite side $a$, so that -$$\cos \theta =\frac{-a^2+b^2+c^2}{2b c} = \frac{x^2}{bc}$$ -This value is non-negative (and, hence, $\theta$ is non-obtuse) for any edge-length $x$ ... well, any real edge-length $x$. However, if we allow $x$ to be an imaginary length, then $\theta$ can be obtuse, and then Heron's Formula can be seen to apply to all triangles.<|endoftext|> -TITLE: Numerical methods book -QUESTION [8 upvotes]: I'm looking for an introductory book on numerical methods. -I'm beginning to learn to program (in Haskell, a functional language, if that would affect the recommendations). The reason I want such a book is to practice my programming skills by implementing easy math-related algorithms. For example, calculations of transcendental or special functions (perhaps the Gamma function and Bessel functions) and interesting constants, solving differential equations (by Runge-Kutta and series solutions, for example), implementing things like Newton-Raphson approximation, primality testing (Miller-Rabin, for example) and so on. -I would prefer something with a wide sampling of algorithms. I'm looking for breadth, not depth. My intention is to use this as sort of a programming exercise book. I care about having a lot of different, fun, and hopefully somewhat simple algorithms to implement. -I would say I have a fairly good mathematical background, so don't shy away from recommending something that is proof based or theoretical at times. If the book includes a lot of mathematics, great! I'm always up for learning more math. I don't have formal computer science training, but I would be willing to do a small amount of reading on the basics of algorithms and data structures if I had to. If the book included this, that would be great. -I should mention that I haven't taken more than an introductory course in ODEs and have no experience with PDEs other than solving the basic heat and wave equations, so I would not be able to understand or appreciate advanced material in these areas (I get the impression that solving PDEs numerically is a big, important field). However, I do know some harmonic analysis, so seeing fast Fourier transforms might be fun. - -REPLY [7 votes]: I really enjoyed reading Forman Acton's Numerical Methods that Usually Work. He writes with a lot of personality and presents interesting problems. I don't know that this is the best book to learn the field from because (1) it was written in 1970 and last revised in 1990 and (2) I am not an expert, so I simply don't know how to evaluate it. But it sounds like these might not be major obstacles from your perspective.<|endoftext|> -TITLE: Which of the following metric spaces are complete? -QUESTION [7 upvotes]: [NBHM_2006_PhD Screening test_Topology] - -Which of the following metric spaces are complete? - -$X_1=(0,1), d(x,y)=|\tan x-\tan y|$ -$X_2=[0,1], d(x,y)=\frac{|x-y|}{1+|x-y|}$ -$X_3=\mathbb{Q}, d(x,y)=1\forall x\neq y$ -$X_4=\mathbb{R}, d(x,y)=|e^x-e^y|$ - - -$2$ is complete as closed subset of a complete metric space is complete and the metric is also equivalent to our usual metric. -$3$ is also complete as every Cauchy sequence is constant ultimately hence convergent. -$4$ is not complete I am sure but not able to find out a counter example, not sure about 1.thank you for help. - -REPLY [10 votes]: For (1), consider the sequence $\left\langle\frac1{2^n}:n\in\Bbb N\right\rangle$. Is it $d$-Cauchy? Does it converge to anything in $X_1$? -For (4), what about $\langle -n:n\in\Bbb N\rangle$?<|endoftext|> -TITLE: Is the empty set partially ordered ? Also, is it totally ordered? -QUESTION [10 upvotes]: I am not sure on how to go about this. Please provide clear explanations. - -REPLY [4 votes]: The above answers confirm that $\emptyset$ with the empty ordering is in fact a linear order. I would just like to add that it is in fact also well-ordered under this ordering since it is linearly ordered and every non-empty subset of $\emptyset$ has a least element. This is true vacuously since there are no nonempty subsets at all.<|endoftext|> -TITLE: Write down the sum of sum of sum of digits of $4444^{4444}$ -QUESTION [19 upvotes]: Let $A = 4444^{4444}$; -Then sum of digits of $A = B$; -Then sum of digits of $B = C$; -Then sum of digits of $C = D$; -Find $D$. -What should be the approach here? - -REPLY [5 votes]: It's well-known that, because $10 \equiv 1 \bmod 9$, and therefore $10^k = 1 \bmod 9$ for all $k>0$, we have that the sum of digits of $n$, $S(n) \equiv n \bmod 9$. -So what is $4444^{4444} \bmod 9$? The above equivalence gives us that $4444 \equiv 7 \bmod 9$. -Now we need the order of $7$ modulo $9$, the smallest $s$ such that $7^s \equiv 1 \bmod 9$. This is easy to find by examination: $7^2 \equiv 4 \bmod 9$ and so $7^3 \equiv 28 \equiv 1 \bmod 9$. -So $4444^{4444} \equiv 7^{4444} \equiv 7 \cdot (7^3)^{1481} \equiv 7 \ \bmod 9$, and the same equivalence is true for $B,C$ and $D$. -Now roughly how big is $A$? Since $\log_{10}4444 \approx 3.64777$, we know that $\log_{10}A \approx 16210.7$, that is $A$ has $16211$ digits. This gives us that $B \le 16211\times 9 = 145899$. So if $B>100000$, the first two digits sum to no more than $5$, which means that $C \le 45$ (that maximum being when $B=99999$). -Finally we can use our knowledge that $C \equiv 7 \bmod 9$ and observe that $C$ must be one of the values $\{7,16,25,34,43\}$ and thus that $S(C) = \fbox{D = 7}$. - -Additional thoughts: -We get exactly the same answer, $D=7$, for $A=55555^{55555}$ . Impressively, a slight variation on the same process also works to get $D=9$ for the case $A=999999999^{999999999}$ (nine nines).<|endoftext|> -TITLE: Continuous extensions of continuous functions on dense subspaces -QUESTION [8 upvotes]: I thought that if I have a function $f: \mathbb Q \to \mathbb R$ that is continuous then I can (uniquely) extend it to a continuous function $F: \mathbb R \to \mathbb R$ as follows: for $r \in \mathbb R \setminus \mathbb Q$ pick a sequence $q_n$ converging to $r$ and then define $F(r) = \lim_{n \to \infty} f(q_n)$. -So I thought there must be a theorem saying that given a continuous function $f: D \to Y$ where $D$ is a dense subset of a metric space $X$ one can uniquely extend it to $F: X \to Y$. -Instead I found a theorem stating this but with the additional requirement that $f$ has to be uniformly continuous. Now I'm confused: is my example above wrong? Where does uniform continuity come in here? -Thanks. - -REPLY [15 votes]: Uniform continuity ensures that the Cauchy sequence $(q_n)$ in $\mathbb Q$ is mapped to a Cauchy (and hence convergent) sequence $\bigl(f(q_n)\bigr)$ in $\mathbb R$. If $f$ is just continuous, $\bigl(f(q_n)\bigr)$ needn't converge (remember: $f$ is just continuous on $\mathbb Q$, so you can't argue, that convergent sequences $q_n \to r \not\in\mathbb Q$ are mapped onto convergent sequences by $f$). For example $x\mapsto \frac 1{x - \sqrt 2}$ is continuous (not uniformly!) on $\mathbb Q$ and can't be extendend. -In fact, you only need the property, that Cauchy sequences are mapped to Cauchy sequences, which is a little weaker than uniform continuity, and a little stronger than continuity (for some properties of such functions, see here for start).<|endoftext|> -TITLE: Why is $(3,x^3-x^2+2x-1)$ not principal in $\mathbb{Z}[x]$? -QUESTION [5 upvotes]: I have a small hitch in showing $(3,x^3-x^2+2x-1)$ is not principal in $\mathbb{Z}[x]$. Towards the contrary, I suppose $(3,x^3-x^2+2x-1):=(3,f)=(g)$ is principal. Then $3\in (g)$, so $3=gh$ for some $g,h\in\mathbb{Z}[x]$. Thus $g,h$ must be constant, and $g\mid 3$, so $g=1,3$. But $g$ cannot be $3$, since $f\neq 3p$ for any $p\in\mathbb{Z}[x]$, since the coefficients are not all divisible by $3$. -If $g=1$, then $(3,f)=\mathbb{Z}[x]$. I don't think this is true, but I don't know how to make it rigorous. I tried supposing $1=pf+3r$ where $p,r\in\mathbb{Z}[x]$ to reach a contradiction and show that $1\notin(3,f)$, but I don't how to more formally prove it. What can I do? Thanks. - -REPLY [2 votes]: Hint $\rm\,\ fg = 1 + 3h\ \Rightarrow\ mod\ 3\!:\ fg\equiv 1\:\Rightarrow\:deg(fg) = 0\:\Rightarrow\: deg(f) = 0\:$ contra $\rm\: f \equiv x^3 +\:\cdots$ -Remark $\ $ The same proof works for $\rm\:3\to m > 1\:$ and any $\rm\:f\:$ both monic and nonconstant mod $\rm\:m.$ Generally over any ring R, a polynomial $\rm\:f\in R[x]\:$ is a unit iff $\rm\,f_0 = f(0)\,$ is a unit in R and all higher coefficients are nilpotent in R, i.e. for all $\rm\:i>0,\,\ f_i^n = 0\:$ for some $\rm\:n\in \Bbb N.$ In particular, if $\rm\:f\in \mathbb Z[x]\:$ has degree $> 1$ and leading coefficient $\rm\:c\:$ coprime to $\rm\:m>1\:$ then in $\rm\:R = \Bbb Z/m\:$ the leading coefficient of $\rm\:f\:$ becomes a unit, so $\rm\:f\:$ remains a nonunit over R (as the hint shows more simply).<|endoftext|> -TITLE: Any affine algebraic group is linear. -QUESTION [12 upvotes]: It is a well-known result that any affine algebraic group is a closed subgroup of some $\mathrm{Gl}_n(\Bbbk)$. However, I would like to see a proof for that, so I looked it up in various books, more precisely in - -Procesi, Lie Groups - An Approach through Invariants and Representations, the theorem on page 172 -Borel, Linear Algebraic Groups, 2nd Edition, Proposition 1.10 -Alexander H.W. Schmitt, Geometric Invariant Theory and Decorated Principal Bundles, Theorem 1.1.3.3 - -The proof is always more or less the same and always lacks an important argument (in my opinion). Let me give you a quick overview: -If $\mu:G\times G\to G$ denotes the multiplication morphism and $\Bbbk[G]=\Bbbk[f_1,\ldots,f_n]$ then we may assume that the $f_i$ span a $G$-invariant subspace $V\subseteq\Bbbk[G]$. One can show that the comorphism $\mu^\sharp:\Bbbk[G]\to\Bbbk[G]\otimes\Bbbk[G]$ maps $V$ into $\Bbbk[G]\otimes V$ and thus, -$$ \mu^\sharp(f_i) = \sum_{j=1}^n \psi_{ij} \otimes f_j $$ -for certain $\psi_{ij}\in\Bbbk[G]$. One then maps $g$ to the matrix $\Psi_g$ with entries $\psi_{ij}(g)$. -Now my question is: Why is this a homomorphism of groups? -I can tell you what I have done so far. One may consider the triple multiplication morphism $\nu:G\times G\times G\to G$ which is just $\nu:=\mu\circ(\mathrm{id}\times\mu)$, so we get -$$\begin{align*} -\nu^\sharp(f_i) &= (\mathrm{id}\times\mu)^\sharp\left(\sum\nolimits_{j=1}^n \psi_{ij} \otimes f_j\right) = - \sum_{j=1}^n \psi_{ij} \otimes \mu^\sharp(f_j) -\\ &= -\sum_{j=1}^n \psi_{ij} \otimes \sum_{k=1}^n \psi_{jk} \otimes f_k -= \sum_{k=1}^n \left(\sum\nolimits_{j=1}^n \psi_{ij} \otimes \psi_{jk}\right) \otimes f_k -\end{align*}$$ -so -$f_i(abc) = \sum_{k=1}^n (\Psi_a\Psi_b)_{ik} \cdot f_k(c)$. Conversely, we may also understand $abc$ as the product of $ab$ with $c$, so -$$\begin{align*} -\sum_{k=1}^n (\Psi_a\Psi_b)_{ik} \cdot f_k(c) = f_j(abc) &= \sum_{k=1}^n (\Psi_{ab})_{ik} \cdot f_k(c). -\end{align*}$$ -Now, what we need are elements $c_j\in Z(f_k\mid k\ne j)\setminus Z(f_j)$ to show that $\Psi_a\Psi_b=\Psi_{ab}$. I have a vague feeling that $f_j$ can not be contained in $(f_k\mid k\ne j)$ because $G$ acts linearly on $V$, but for some reason I am stuck. So my question is: - -Can you finish my proof, i.e. show that $Z(f_k\mid k\ne j)\setminus Z(f_j)\ne \emptyset$ or equivalently, $f_j\notin(f_k\mid k\ne j)$? -If not, can you give an alternative proof? - -REPLY [10 votes]: Well, I think the idea is this. What does it mean to find an injective homomorphism $G\to \operatorname{GL}(W)$? It means to find a faithful representation $W$ of $G$. Somehow, the intuition tells me that the $V$ you are talking about is the right $W$, because it is spanned by elements that describe well $G$. -In fact, this is right: if an element $g$ fixes $V$, it fixes all of $k[G]$, hence it must be the identity. So we have an injective homomorphism (of groups, not talking of algebraic structure yet) -$$\rho:G\to \operatorname{GL}(V)$$ -Now, what coefficients do the matrices of $\rho$ have? Well, we may suppose that the $f_i$'s are a basis of $V$ (they generate, so I can extract a basis). Now, -$$(\rho(g)f_i)(h)=f_i(hg)=\sum_j\psi_{i,j}\otimes f_j(g,h)=\sum_j\psi_{i,j}(g)f_j(h)$$ -hence -$$\rho(g)f_i=\sum_j\psi_{i,j}(g)f_j$$ -So $\rho(g)=(\psi_{i,j}(g))_{1\leq i,j\leq n}$, and $\rho$ is in fact an injective homomorphism of algebraic groups.<|endoftext|> -TITLE: What is the sum of sum of digits of $4444^{4444^{4444}}$? -QUESTION [38 upvotes]: A recent question asked about the sum of sum of sum of digits of $4444^{4444}$. The solution there works mainly because the number chosen is small enough for the sum of sum of sum to be equal to the repeated sum: i.e. if we sum digits further, the result does not change. Since finding repeated sums of digits is just a matter of elementary number theory, this solves the problem. -It seems the following question might be much harder: what is the sum of sum of digits of $$4444^{4444^{4444}}?$$ -In other words, let $f:\Bbb N_0\to\Bbb N_0$ be the function defined by $f(n)=\textrm{sum of decimal digits of }n$. - -What is the value of $f\left(f\left(4444^{4444^{4444}}\right)\right)$? - -In this question, we have not yet reached a single-digit number, which at least seems to make it much harder. -Some estimates: the number of decimal digits of $4444^{4444^{4444}}$ is equal to $$\left\lfloor\log_{10}4444^{4444^{4444}}\right\rfloor+1,$$ which implies $$f\left(4444^{4444^{4444}}\right)\le9\left(\log_{10}4444^{4444^{4444}}+1\right).$$ -Next, the number of digits of this last number is at most $$\left\lfloor\log_{10}\left(9\left(\log_{10}4444^{4444^{4444}}+1\right)\right)\right\rfloor+1,$$ which is $16213$, according to Wolfram|Alpha. Therefore, $$f\left(f\left(4444^{4444^{4444}}\right)\right)\leq9\cdot16213=145917.$$ -So the number we are looking for has at most $6$ digits. This makes it very feasible to express in decimal notation, but possibly hard to find. -We might be further interested in numbers like $$f\left(f\left(f\left(4444^{4444^{4444^{4444}}}\right)\right)\right),$$ so a related question would be: - -Is there any hope for a general method of evaluating such functions or is the behaviour of the $k$-fold composition $f^k$ completely chaotic? - -REPLY [6 votes]: You can find an upper bound for it even without using computer or any calculator: -$$ -f(N) < 9 (4444^{4444} \times \log_{10} 4444 + 1) < 9 \times 4 \times 4444^{4444} + 9 -$$ -$$ -f(f(N)) < 9 ( \log_{10}9 + \log_{10}4 + 4444 \log_{10}4444 + 1) < 9 (3 + 4444 \times 4) = 9 \times 17779 = 160011 -$$ -so -$$ f(f(N))<160011 $$ -this is a large range but it can be smaller with calculator. (Note that you should have computed base 10 logarithm instead of natural logarithm) -the range consists of 160011 numbers, and by knowing the reminder of 9, only 17,779 numbers are left, and answer is one of them. -Of course this is not an exact answer, but it is straightforward! -Edit: I just misused a formula which is about something different, sorry for that!<|endoftext|> -TITLE: Prove by mathematical induction that $2n ≤ 2^n$, for all integer $n≥1$? -QUESTION [5 upvotes]: I need to prove $2n \leq 2^n$, for all integer $n≥1$ by mathematical induction? -This is how I prove this: -Prove:$2n ≤ 2^n$, for all integer $n≥1$ -Proof: $2+4+6+...+2n=2^n$ -$i.)$ Let $P(n)=1 - P(1): 2(1)=2^1\implies 2=2$. - Hence, $P(1)$ is true. -$ii.)$ -Assume that $P(n)$ is true for $n=k$, i.e, $2+4+6+...+2k=2^k$, and prove that $P(n)$ is also true for $n=k+1$, i.e, $2+4+6...+2(k+1)=2^{(k+1)}$ -from the assumption add $2(k+1)$ on both sides so we have - $2+4+6...2k+2(k+1)=2^k+2(k+1)$ -I'm confused with $2^k+2(k+1)$, I don't know how to make $2^k$ be equivalent to $2^{k+1}$. -I feel i'm doing something wrong. -Any help would be appreciated! - -REPLY [2 votes]: Recall that, by induction, -$$ -2^n = \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \ldots + \binom{n}{n-1} + \binom{n}{n}. -$$ -All the terms are positive; observe that -$$ -\binom{n}{1} = n, \quad \binom{n}{n-1} = n. -$$ -Therefore, -$$ -2^n \geq n+n=2n. -$$ -Remark: I suggest this proof since the plain inductive proof of your statement has been given in many answers. I also believe that an inductive proof of the binomial expansion is an instructive exercise, probably more instructive than the one you are trying to solve.<|endoftext|> -TITLE: Hilbert's Nullstellensatz without Axiom of Choice -QUESTION [5 upvotes]: Motivation -This question came from my efforts to solve this problem presented by Andre Weil in 1951. -Can we prove the following theorem without Axiom of Choice? -Theorem -Let $A$ be a commutative algebra of finite type over a field $k$. -Let $I$ be an ideal of $A$. -Let $\Omega(A)$ be the set of maximal ideals of $A$. -Let $V(I)$ = {$\mathfrak{m} \in \Omega(A)$; $I \subset \mathfrak{m}$}. -Let $f$ be any element of $\cap_{\mathfrak{m} \in V(I)} \mathfrak{m}$. -Then there exists an integer $n \geq 1$ such that $f^n \in I$. -EDIT -So what's the reason for the downvotes? -EDIT -What's wrong with trying to prove it without using AC? -When you are looking for a computer algorithm for solving a mathematical problem, such a proof may provide a hint. At least, you can be sure that there is a constructive proof. -EDIT -To Martin Brandenburg, I think this thread also answers your question. - -REPLY [2 votes]: Lemma 1 -Let $A$ be a commutative algebra of finite type over a field $k$. -Then there exists a maximal ideal $P$ of $A$ such that $A/P$ is a finite $k$-module. -Proof: -If every element of $A$ is algebraic over $k$, then $A$ is a finite $k$-module. Therefore the lemma is trivial. -Hence we can assume otherwise. -By Noether normalization lemma (this can be proved without AC), there exist algebraically independent elements $x_1,\dots, x_n$ in $A$ such that $A$ is a finitely generated module over the polynomial ring $A' = k[x_1,\dots, x_n]$. -Let $\mathfrak{m} = (x_1,\dots, x_n)$ be the ideal of $A'$ generated by $x_1,\dots, x_n$. -Clearly $\mathfrak{m}$ is a maximal ideal of $A'$. -By the answer by QiL to this question, there exists a prime ideal $P$ of $A$ lying over $\mathfrak{m}$. -Since $A/P$ is a finitely generated module over $k = A'/m$, $P$ is a maximal ideal. -QED -Lemma 2 -Let $A$ be a commutative algebra of finite type over a field k. -Let $f$ be a non-nilpotent element of $A$. -Then there exists a maximal ideal $P$ of $A$ such that $f \in A - P$. -Proof: -Let $S$ = {$f^n; n = 1, 2, \dots$}. -Let $A_f$ be the localization with respect to $S$. -By Lemma 1, there exists a maximal ideal $\mathfrak{m}$ of $A_f$ such that $A_f/\mathfrak{m}$ is a finite $k$-module. -Let $P$ be the inverse image of $\mathfrak{m}$ by the canonical morphism $A \rightarrow A_f$. -$A/P$ can be identified with a subalgebra of $A_f/\mathfrak{m}$. -Since $A_f/\mathfrak{m}$ is a finite $k$-module, $A/P$ is also a finite $k$-module. -Hence $P$ is a maximal ideal. -Clearly $f \in A - P$. -QED -The title theorem follows immediately from the following lemma by replacing $A$ with $A/I$. -Lemma 3 -Let $A$ be a commutative algebra of finite type over a field k. -Let $\Omega(A)$ be the set of maximal ideals of $A$. -Let $f$ be any element of $\cap_{\mathfrak{m} \in \Omega(A)} \mathfrak{m}$. -Then $f$ is nilpotent. -Proof: -This follows immediately from Lemma 2.<|endoftext|> -TITLE: how to understand the tensor product canonical line bundle $\otimes$ dual bundle -QUESTION [8 upvotes]: Suppose we have a Riemann surface $M$ together with a holomorphic vector bundle $E \to M$ of rank n. let $K$ denote the canonical line bundle and let $E^*$ denote the dual bundle -I am trying to understand the tensor product $K \otimes E^*$. -I have lots of trouble, because I need to do this on my own and my background in differential geometry and multilinear algebra is not strong. -I shall try to explain how I would describe $K \otimes E^*$ locally, if somebody could comment on what goes wrong that would be immensely helpful! - -The canonical bundle $K$ over a Riemann surface $M$ is the cotangent bundle, or the bundle of holomorphic $1$-forms. In local coordinates $(z,\overline{z})$ an element of $K$ can be written as - $$ -\omega = \frac{i}{2} f\,dVol_h -$$ - where $f$ is a holomorphic function on $M$ (or at least defined locally) and $dVol_h = \frac{i}{2}\, h dz\wedge d\overline{z}$, the Volume element induced by the metric $h\,dz\,d\overline{z}$ on $M$ (here again $h$ is a holomorphic function on $M$). - -Question 1: Is this a correct way to describe the canonical bundle locally ? - -The dual bundle $E^*$ can be described locally given a choice of basis $\{e_1,\dots, e_n\}$ for the bundle $E$. We take the dual basis $\{\phi^1,\dots\phi^n\}$ (so $\phi_k (e_l) = \delta^k_l$) and write - $$ - \sigma = \sum^n_{k = 1} g_k\,\phi_k -$$ - where the coeficients $g_k$ are holomorphic functions locally defined on $M$. - -Question 2: is this description sufficient ? i fear the locality of what I want to show is not emphasized enough but i am not sure how the local coordinates $(z,\overline{z})$ on a patch $V \subset M$ (say) should be mentioned here. do i need to invoke local coordinates on $E$, or is this done by specifying the basis? - -Therefore the bundle $K \otimes E^*$ consists of tensor fields which have local description - $$ -\omega \otimes \sigma = \sum^n_{k = 1} (\frac{i}{2}\,h\,f\,g_k)\, dz \wedge d\overline{z} \otimes \phi_k -$$ - -Question 3: does this formula makes sens ? I am "very unsure" here - I have a wedge product and a tensor symbol and don't know how to write this in a correct way. Below I attempt to understand such an object, maybe my last lines give away more of my misunderstandings: - -an element in $K \otimes E^*$ can be interpret as a map $TM \otimes E \to \mathbb{C}$. alternatively we can also think of it as something that can be integrated, that which would amount to a contraction of the tensor. - -Question 4: do these interpretations make sense ? how would I write them out as rigorous definitions? -Many thanks for comments and help!!! - -REPLY [15 votes]: Your first and biggest misconception is that you seem to be mixing up real and complex dimensions. A Riemann surface is a $2$-dimensional real manifold, but as a complex manifold it is $1$-dimensional. Hence you have just one local complex coordinate on $M$, namely $z$. The local coordinates are not $(z, \bar{z})$. -There is more structure on the tangent bundle of $M$ and the bundle of $k$-forms on $M$. Here's the general picture. Let $X$ be a $2n$-dimensional complex manifold, and consider local coordinates $(x_1, \dots, x_n, y_1, \dots, y_n)$. Using the complex structure on the tangent bundle (which we denote $i$) we can locally define vector fields -$$\frac{\partial}{\partial z_j} = \frac{1}{2} \left(\frac{\partial}{\partial x_j} - i \frac{\partial}{\partial y_j} \right),$$ -$$\frac{\partial}{\partial \bar{z}_j} = \frac{1}{2} \left( \frac{\partial}{\partial x_j} + i \frac{\partial}{\partial y_j} \right).$$ -We also locally have the dual basis of $1$-forms -$$dz_j = dx_j + i ~dy_j,$$ -$$d\bar{z}_j = dx_j - i ~dy_j.$$ -Then we get a splitting of the cotangent bundle -$$T^\ast M = \Lambda^{1,0} T^\ast M \oplus \Lambda^{0,1} T^\ast M,$$ -where pointwise $\Lambda^{1,0} T^\ast M$ is spanned by the $dz_j$ and $\Lambda^{0,1} T^\ast M$ is spanned by the $d\bar{z}_j$. More generally, we have the splitting -$$\Lambda^k T^\ast M = \Lambda^{k,0} T^\ast M \oplus \Lambda^{k-1, 1} T^\ast M \oplus \cdots \oplus \Lambda^{1, k-1} T^\ast M \oplus \Lambda^{0,k} T^\ast M,$$ -where for $p + q = k$, $\Lambda^{p,q} T^\ast M$ is spanned pointwise by the $dz_{j_1} \wedge \cdots \wedge dz_{j_p} \wedge d\bar{z}_{j_{p+1}} \wedge \cdots \wedge d\bar{z}_{j_k}$. Sections of $\Lambda^{p,q} T^\ast M$ are called $(p,q)$-forms, sections of $\Lambda^{k,0} T^\ast M$ are called holomorphic $k$-forms, and sections of $\Lambda^{0,k} T^\ast M$ are called antiholomorphic $k$-forms. -On a Riemann surface, the canonical bundle is the bundle of holomorphic $1$-forms, i.e. -$$K = \Lambda^{1,0} T^\ast M.$$ -Therefore in terms of a local coordinate $z$, a section $\alpha$ of $K$ is described by -$$\alpha(z) = f(z) ~dz$$ -for some holomorphic function $f$. Your expression of the form -$$\omega = F ~dz \wedge d\bar{z}$$ -would locally describe a section of $\Lambda^{1,1} T^\ast M$, not a holomorphic $1$-form. -Your local description is closer to being ok. Here I would let $\{e_1, \dots, e_n\}$ be a local frame for $E$, i.e. a set of $n$ local sections that form a basis for each fiber $E_z$ of $E$ for $z$ in our coordinate neighborhood. Then $\{\phi_1, \dots, \phi_n\}$ would be the dual frame defined by -$$\phi_i(z)(e_j(z)) = \delta_{ij}$$ -for all $z$ in our coordinate neighborhood. Then locally a section $\sigma$ of $E^\ast$ would look like -$$\sigma(z) = \sum_{k = 1}^n g_k(z) ~\phi_k(z) \tag{$\ast$}$$ -for each $z$ in our coordinate neighborhood. -Putting the above together, a section of $K \otimes E^\ast$ is a linear combination of sections of the form $\alpha \otimes \sigma$, where $\alpha$ is a section of $K$ and $\sigma$ is a section of $E^\ast$, and locally such a $\alpha \otimes \sigma$ looks like -$$(\alpha \otimes \sigma)(z) = \sum_{k = 1}^n f(z)g_k(z) ~dz \otimes \phi_k(z)$$ -for each $z$ in our coordinate neighborhood. -If $TM$ is the full holomorphic tangent bundle of $M$, then a section $\alpha \otimes \sigma$ of $K \otimes E^\ast$ can be considered as a map from sections of $TM \otimes E$ to the space of holomorphic functions on $M$ as follows. Let $\alpha \otimes \sigma$ have the form $(\ast)$ determined above. A section of $TM \otimes E$ is a linear combination of sections of the form $v \otimes s$, where $v$ is a section of $TM$ and $s$ is a section of $E$. Locally we have -$$(v \otimes s)(z) = \left( a(z) \frac{\partial}{\partial z} + b(z) \frac{\partial}{\partial \bar{z}} \right) \otimes \left( \sum_{j = 1}^n h_j(z) ~e_j(z) \right).$$ -Then we have -\begin{align*} -(\alpha \otimes \sigma)(v \otimes s)(z) & = \alpha(v)(z) \sigma(s)(z) \\ - & = f(z) ~dz\left( a(z) \frac{\partial}{\partial z} + b(z) \frac{\partial}{\partial \bar{z}} \right) \sum_{k = 1}^n g_k(z) ~\phi_k(z) \left( \sum_{j = 1}^n h_j(z) ~e_j(z) \right) \\ - & = f(z) \left( a(z) \cdot 1 + b(z) \cdot 0 \right) \sum_{k = 1}^n \sum_{j = 1}^n \delta_{kj} g_k(z) h_j(z) \\ - & = f(z)a(z) \sum_{k = 1}^n g_k(z)h_k(z). -\end{align*} -When the above is considered over a single point, we see how to map an element of $TM \otimes E$ to $\Bbb C$ using an element of $K \otimes E^\ast$.<|endoftext|> -TITLE: Show that the product of the $\phi(p-1)$ primitive roots of $p$ is congruent modulo $p$ to $(-1)^{\phi(p-1)}$ -QUESTION [6 upvotes]: If $p$ is a prime, show that the product of the $\phi(p-1)$ primitive roots of $p$ is congruent modulo $p$ to $(-1)^{\phi(p-1)}$. - -I know that if $a^k$ is a primitive root of $p$ if gcd$(k,p-1)=1$.And sum of all those $k's$ is $\frac{1}{2}p\phi(p-1)$,but then I don't know how use these $2$ facts to show the desired result. -Please help. - -REPLY [4 votes]: If $a^k$ is a primitive root modulo $p$, then so is $a^{-k}$. Thus, if $p-1$ is even, then the sum of coprime integers between 1 and $p-1$ with $p-1$ must be =$\phi(p-1)(p-1)/2$, and hence the product of $\phi(p-1)$ primitive roots modulo p must be $a^{\phi(p-1)(p-1/2)}\equiv(-1)^{\phi(p-1)} \pmod p$. If $p-1$ is odd, then the result is trivial, as $p=2$.<|endoftext|> -TITLE: Proving that a linear isometry on $\mathbb{R}^{n}$ is an orthogonal matrix -QUESTION [8 upvotes]: I wish to prove that if $T:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is defined by $T(v)=Av$ (where -$A\in M_{n}(\mathbb{R})$) is an isometry then $A$ is an orthogonal -matrix. -I am familiar with many equivalent definition for $A\in M_{n}(\mathbb{R})$ -to be orthogonal, and it doesn't matter to me which one to show. -What I tried to do is the following: -$||x-y||=||Ax-Ay||\implies\langle x-y,x-y\rangle=\langle Ax-Ay,Ax-Ay\rangle\implies\langle x-y,x-y\rangle=\langle x-y,A^{t}A(x-y)\rangle$, -from here I thought that I will be able to deduce $A^{t}A=I$ and -complete the proof, but I was unable to do so. -How can I complete the proof, or prove this in another fashion ? Help -is appreciated! - -REPLY [8 votes]: We know that $\,\langle\, x,y\,\rangle =0\,\,\,\forall\,y\Longleftrightarrow x=0\,$ , so -$$\forall x,y\,\,:\,\langle\,x,y\,\rangle=\langle\,Ax,Ay\,\rangle=\langle\,x,A^tAy\,\rangle\Longrightarrow \langle\,x,(A^tA-I)y\,\rangle=0$$ -$$\Longrightarrow (A^tA-I)y=0\,\,\,\,\forall y\,\,\Longrightarrow A^tA-I=0\Longrightarrow A^tA=I$$<|endoftext|> -TITLE: Find the minimum value of this integral. -QUESTION [5 upvotes]: Find the minimum value of this integral: For what value of $k > 1$ is -$$ - \int_k^{k^2} \frac 1x \log\frac{x-1}{32}\, \mathrm dx -$$ -minimal? -After applying Newton-Leibniz, I got $k = 3$ and then did 2nd derivative test, it gave me positive result, thus 3 is the answer but I want to know if there's a smarter/slicker way to do? - -REPLY [3 votes]: $J=\int_k^{k^2}dx \frac{1}{x}\ln\frac{x-1}{32}=-\operatorname{dilog}(k)-\ln(\frac{k}{32}-\frac{1}{32})\ln(k)+\operatorname{dilog}(k^2)+\ln(\frac{k^2}{32}-\frac{1}{32})\ln(k^2)$ -where: -$$\operatorname{dilog}(x)=\int_1^x \, dt \ln(\frac{t}{1-t})$$ -The derivative respect to $k$ is: -$$\frac{\partial J}{\partial k}=-\frac{1}{k}[5 \ln(2)+\ln(k-1)-2\ln((k-1)(k+1))$$ -putting: -$$\frac{\partial J}{\partial k}=0$$ -the result is $k=3$<|endoftext|> -TITLE: If $H\unlhd G$ with $(|H|,[G:H])=1$ then $H$ is the unique such subgroup in $G$. -QUESTION [7 upvotes]: Here is a problem from "An introduction to the Theory of Groups" by J.J.Rotman: - -Let $G$ be a finite group, and let $H$ be a normal subgroup with $(|H|,[G:H])=1$. Prove that $H$ is the unique such subgroup in $G$. - -I assumed there was another normal subgroup like $H$, say $K$, such that $(|K|,[G:K])=1$. My aim was to show that $[K: K\cap H]=1 $ that was not held if I didn’t suppose $|H|=|K|$ . My question is if my last assumption about two subgroups is right? If it isn’t, please guide me. Thanks. - -REPLY [2 votes]: To make things explicit: Assume $\lvert H \rvert = \lvert K \rvert$. -As a consequence of the 2nd isomorphism theorem, $HK \le G$ and -$$\lvert HK \rvert -= \frac{\lvert H \rvert \lvert K \rvert}{\lvert H \cap K \rvert} -= \frac{\lvert H \rvert^2}{\lvert H \cap K \rvert}$$ -By Lagrange's Theorem: -$$\lvert G \rvert = [G:H] \lvert H \rvert -= [G:HK] \lvert HK \rvert $$ -We derive -$$[G:H] = [G:HK] \frac{\lvert H \rvert}{\lvert H \cap K \rvert}$$ -If $\lvert H \cap K \rvert < \lvert H \rvert$ then there is some factor $n \ne 1$ of $\lvert H \rvert$ on the RHS (note: $\lvert H \vert /\lvert H \cap K \rvert - = n \in \mathbb N$). Thus $n \mid [G:H] $, a contradiction that $\operatorname{gcd}(\lvert H \rvert, [G:H]) = 1$. -We conclude $\lvert H \cap K \rvert = \lvert H \rvert$ and since $\lvert H \rvert = \lvert K \rvert$ we can conclude $H = K$.<|endoftext|> -TITLE: Changing the order of $\lim$ and $\sup$ -QUESTION [7 upvotes]: Suppose that $f_n:X\to [0,1]$ where $X$ is some arbitrary set. Suppose that -$$ - f_n(x)\geq f_{n+1}(x) -$$ -for all $x\in X$ and all $n = 0,1,2,\dots$ so there exists $\lim_n f_n(x)$ point-wise, let's call it $f(x)$. -Define $f^*_n:=\sup\limits_{x\in X}f_n(x)$, $f^*:=\sup\limits_{x\in X}f(x)$ and $\hat f:= \lim\limits_n f^*_n$. I wonder when $f^* = \hat f$, i.e. -$$ - \lim\limits_n \sup\limits_{x\in X}f_n(x) = \sup\limits_{x\in X}\lim\limits_n f_n(x). -$$ -I was googling the topic, but strangely have not found any information, strangely because I expected it to be available as for changing the order of limits or of integration. -Some simple facts: $\hat f\geq f^*$ and the reverse is true at least when $f_n$ converges uniformly to $f$. This does not hold in general, e.g. when $f_n = 1_{[n,\infty)}$. -I would appreciate any other ideas that you can advise me. Also related to this. A similar question was asked here. - -REPLY [2 votes]: Your problem is equivalent to asking if -$$\begin{equation}\inf_n \sup_x f_n(x) = \sup_x \inf_n f_n(x) \tag{*}\end{equation}.$$ -This question is the subject of minimax theorems, e.g. Sion's minimax theorem. Minimax theorems are applied frequently in game theory. -Generally speaking, (*) is true if $n \mapsto f_n(x)$ is (quasi-)concave and $x \mapsto f_n(x)$ is (quasi-)convex. You will also need some sort of (semi-)continuity and a "nice" structure for the sets in the infimum and supremum (for instance $x \in \mathbb{R}$ and $f_n(x) = \frac1n |g(x)|$).<|endoftext|> -TITLE: $GL_n(k)$ (General linear group over a algebraically closed field) as a affine variety? -QUESTION [5 upvotes]: In the context of linar algebraic groups, I read in my notes from the lecture that's already some while ago that $GL_n(k)$ is an algebraic variety because $GL_n=D(\det)$, $ \det \in k [ (X_{ij})_{i,j} ]$. -Now, $k$ is an algebraically closed field, $\det$ is the determinant and $k [ (X_{ij})_{i,j} ]$ are the polynomials in unkwnowns $X_{ij}$. But I cannot find what $D$ meant, maybe it's also a typo or uncommon notation -.- -How can I interpret $GL_n (k)$ as a variety? -(Or what does this $D$ stand for?) - -REPLY [9 votes]: I think $D$ becomes a more common notation when working with schemes, but here $D(f)$ for $f \in k[X_1, \ldots, X_n]$ should mean $\{a \in k^n : f(a) \neq 0\}$. So it's the open set where $f$ does not vanish. The important thing is that $D(f)$ is isomorphic to the algebraic set defined by $fX_{n + 1} - 1$ in $k^{n + 1}$ via the map $(a_1, \ldots, a_n) \mapsto (a_1, \ldots, a_n, 1/f(a_1, \ldots, a_n))$.<|endoftext|> -TITLE: Two notions of uniformizer -QUESTION [5 upvotes]: Let $X$ be a projective algebraic curve and consider a `uniformizing' map $h:X \rightarrow \mathbb{P}^1$. Is there any connection between this notion of uniformizer and a uniformizer of the maximal ideal of the local ring at a point $P \in X$? - -REPLY [5 votes]: There is no connection because, as far as I know, there is no such thing as a "uniformizing map $X\to \mathbb P^1$". -Assume $X$ is a smooth irreducible projective curve over an algebrically closed field $k$ of arbitrary characteristic. -Then a morphism $h:X\to \mathbb P^1$ (which is not the constant map with value $\infty$) is exactly the same as a rational function $f\in Rat(X)$. -Each point $P\in X$ has an associated local ring $k\subsetneq \mathcal O_P\subsetneq k(T)$, which is is a discrete valuation ring by smoothness of $X$. -The function $h$ is said to be a uniformizer of $X$ at $P$ if $h$ is a generator of the maximal ideal $\mathfrak m_P$ of $\mathcal O_P$. -This is equivalent to saying that $h$ is non-ramified (or in alternative terminology étale) at $P$. -The interpretation of "uniformizing map" in your question might be that $h$ be a uniformizer at each $P\in X$. -However such a situation only occurs if $h:X \xrightarrow {\simeq} \mathbb P^1$ is an isomorphism: this result follows from Riemann-Hurwitz and is the algebraic geometer's way of saying that $\mathbb P^1_k$ is simply connected.<|endoftext|> -TITLE: Showing $\mathbb{Q}(\sqrt[4]{2},i)=\mathbb{Q}(\sqrt[4]{2}+i)$ using the Galois orbit of $\sqrt[4]{2} + i$ -QUESTION [6 upvotes]: The following is a problem from I. Martin Isaac's Algebra. Let $E=\mathbb{Q}(\sqrt[4]{2}+i)$. I am trying to show $\mathbb{Q}(\sqrt[4]{2},i)=E$ with the following hint: - -Find at least five different elements in the orbit of $i+\sqrt[4]{2}$ under $\text{Gal}(E/\mathbb{Q})$. - -I have solved the problem in the usual manner i.e. by showing $\mathbb{Q}(\sqrt[4]{2}+i)\subseteq \mathbb{Q}(\sqrt[4]{2},i)$ and $\mathbb{Q}(\sqrt[4]{2},i)\subseteq \mathbb{Q}(\sqrt[4]{2}+i)$ with a few calculations. -My question is the following: - -What is the theoretical framework behind Isaacs' hint? - -REPLY [5 votes]: It's easy to see that $\mathbb{Q}(\sqrt[4]{2} + i) \subseteq \mathbb{Q}(\sqrt[4]{2}, i)$. Additionally, $\mathbb{Q}(\sqrt[4]{2}, i)$ has degree $8$ over $\mathbb{Q}$. So the degree of $\mathbb{Q}(\sqrt[4]{2} + i)$ over $\mathbb{Q}$ must be a divisor of $8$. If you can show that it has degree at least $5$, you know that it has to be $8$ and so the fields are equal. - -REPLY [5 votes]: Let $K/k$ be a Galois extension with group $G$. Let $\alpha$ be an element of $K$ with minimal polynomial $f$ over $k$. Then $G$ acts transitively on the set of roots of $f$ in $K$, which all have multiplicity $1$ because our extension is separable. Hence -$$ -[k(\alpha) : k] = \deg f = \#(G\alpha) \quad \text{divides} \quad \#(G) = [K : k]. -$$ -Is clear to you how one might find five conjugates in this case, by the way?<|endoftext|> -TITLE: relations between distance-preserving, norm-preserving, and inner product-preserving maps -QUESTION [13 upvotes]: An inner product can induce a norm by defining $\|x\| = \sqrt{\langle x,x \rangle}$, the a norm can induce a metric by setting $d(x,y) = \|x - y\|$. But not every norm (metric) is induced from inner product (norm), unless the parallelogram law (homogeneity and translation invariance conditions) is (are) satisfied. -Suppose an inner product induces a norm, which then induces a metric, using these defined inner product, norm and metric, can we tell what are the relations between distance-preserving, norm-preserving, and inner product-preserving maps? I just know one: if an isometry (distance-preserving), which is injective, is also surjective, then it's unitary (bijective), which means the isometry is also inner product-preserving. For example, distance-preserving maps on a compact metric space are also inner product-preserving. Are there any other relations? - -REPLY [15 votes]: I am not completely sure what your question is. Here are some thoughts which seem to be relevant: -Let E be a normed space with induced metric, and $f:E\to E$ a set-map. - -If f preserves distance and f(0)=0, then it preserves norm: $\|fv\|=d(fv,f0)=d(v,0)=\|v\|$. -If f preserves norm and addition, then it preserves distance: $d(fv,fw)=\|f(v-w)\|=\|v-w\|=d(v,w)$. -If f preserves distance and is surjective, then it is affine, by Mazur–Ulam (thus linear if moreover $f(0)=0$). - -Let F be an inner-product space, with induced norm and metric, and $f:F\to F$ a set-map. - -If f is linear, then to preserve distance, norm, or inner-product are three equivalent conditions: for distance and norm this follows from the above, and for norm and inner product this follows by polarization.<|endoftext|> -TITLE: Laplace transform of integrated geometric Brownian motion -QUESTION [9 upvotes]: Is there any closed form of the Laplace transform of an integrated geometric Brownian motion ? -A geometric Brownian motion $X=(X_t)_{t \geq 0}$ satisifies $dX_t = \sigma X_t \, dW_t$ where $W=(W_t)_{t \geq 0}$ denotes a Brownian motion and the associated integrated Brownian motion is $\int_0^t X_s \, ds$. The Laplace transform of an integrated gometric Brownian motion is thus -$$ \mathcal{L}(\lambda) = \mathbb{E}\left[e^{-\lambda \int_0^t X_s ds } \right]$$ - -REPLY [3 votes]: What is known is explained in C. Albanese, S. Lawi, Laplace transform of an integrated gometric Brownian motion, MPRF 11 (2005), 677-724, in particular in the paragraph of the Introduction beginning by A separate class of models...<|endoftext|> -TITLE: proving :$\frac{(ab+b)(2b+1)}{(ab+a)(5b+1)}+\frac{(bc+c)(2c+1)}{(bc+b)(5c+1)}+\frac{(ca+a)(2a+1)}{(ca+c)(5a+1)}\ge\frac{3}{2}$ -QUESTION [7 upvotes]: Let $a,b,c, >0$ be real numbers such that $$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3$$ -How to prove that : -$$\frac{(ab+b)(2b+1)}{(ab+a)(5b+1)}+\frac{(bc+c)(2c+1)}{(bc+b)(5c+1)}+\frac{(ca+a)(2a+1)}{(ca+c)(5a+1)}\ge\frac{3}{2}$$ - -REPLY [2 votes]: Let $a=\frac{1}{x} ,b=\frac{1}{y} ,c=\frac{1}{z} $ -$ \frac{(x+1)(y+2)}{(y+1)(y+5)} + \frac{(y+1)(z+2)}{(z+1)(z+5)}+\frac{(z+1)(x+2)}{(x+1)(x+5)} \geq \frac{3}{2} $ -note that $\frac{y+2}{(y+1)(y+5)}\ge \frac{3}{4(y+2)} $ -then is enough to prove that $\frac{x+1}{y+2} + \frac{y+1}{z+2} + \frac{z+1}{x+2} \geq 2$ -by Cauchy-Schwarz -$\frac{x+1}{y+2} + \frac{y+1}{z+2} + \frac{z+1}{x+2} =$ -$\frac{(x+1)^2}{ (x+1)(y+2) }+\frac{(y+1)^2}{(y+1)(z+2)}+\frac{(z+1)^2}{(z+1)(x+2)}\geq \frac{((x+1)+(y+1)+(z+1))^2}{(x+1)(y+2) +(y+1)(z+2) +(z+1)(x+2) } \geq 2$<|endoftext|> -TITLE: Prove that the dihedral group $D_4$ can not be written as a direct product of two groups -QUESTION [9 upvotes]: I like to know why the dihedral group $D_4$ can't be written as a direct product of two groups. It is a school assignment that I've been trying to solve all day and now I'm more confused then ever, even thinking that the teacher might have missed writing out that he means normal subgroups. -On another thread it was stated (as the answer to this question) that the direct product of two abelian groups is again abelian. If we consider the direct product of abelian subgroups $H$,$K\in G$ where $HK=G$ (for all $g \in G$, $g=hk$ $h \in H$, $k \in K$.) I can't understand why this would imply $g=kh$? It is not stated anywhere that $H$,$K$ has to be normal! But if this implication is correct I do understand why $D_4$ (that is non-abelian) can't be written as a direct product of two groups. But if it's not, as I suspect, I need some help! -We know that all groups of order 4 and 2 are abelian, (since $4=p^2$), but only 4 of the subgroups of $D_4$ are normal: -Therefor I can easily show that $D_4$ can't be a direct product of normal subgroups: -The only normal subgroups of $D_4$ is the three subgroups of order 4, (index 2 theorem): $\{e,a^2,b,a^2b\}$, $\langle a\rangle$, $\{e,a^2,ab,a^3b\}$ and the center of $D_4=\{e,a^2\}$ We can see that these are not disjoint. So $D_4$ can't be a direct product of normal subgroups. The reason for this being that the center is non-trivial. But why can't $D_4$ be a direct product of any two groups? -If we write the elements of $D_4$ as generated by $a$ and $b$, $a^4=e$, $b^2=e$, $ba=a^3b$ why isn't $D_4=\langle b\rangle\langle a\rangle$ ? I calculated the products of the elements of these two groups according to the rule given above and ended up with $D_4$, and also $\langle a\rangle$, $\langle b\rangle$ is disjoint...? Why is this wrong? Very thankful for an answer! - -REPLY [8 votes]: For $G$ to be the direct product of $H$ and $K$, by definition we must meet the following conditions: - -$H$ and $K$ are normal subgroups of $G$; -$G=HK$; and -$H\cap K=\{e\}$. - -Note that you do not take all three conditions in your first paragraph explicitly. The normality of $H$ and $K$ is implicit when you say that it is a direct product. This is what you are missing: the definition of "direct product" requires normality of the two subgroups. -In general, if $H$ and $K$ are normal and $H\cap K=\{e\}$, then $hk=kh$ for all $h\in H$ and $k\in K$: indeed, consider the element $hkh^{-1}k^{-1}$. Writing it as $(hkh^{-1})k$, it is a product of two elements of $K$ (by normality of $K$), so it lies in $K$. But writing it as $h(kh^{-1}k)$, then, by normality of $H$, it is a product of two elements of $H$, so it lies in $H$. Hence, $hkh^{-1}k^{-1}$ lies in $H\cap K=\{e\}$. So $hkh^{-1}k^{-1}=e$. Multiplying by $kh$ on the right, we get $hk=kh$. -In particular, if $H$ and $K$ are abelian, then $G$ is abelian: given $hk$ and $h'k'$ in $G$ (using the fact that $G=HK$), then -$$(hk)(h'k') = h(kh')k' = h(h'k)k' = (hh')(kk') = (h'h)(k'k) = h'(hk')k = h'(k'h)k = (h'k')(hk)$$ -so $G$ is abelian. -You are correct, however, that if $G$ is a product (not a direct product, but just a product) of two abelian subgroups $H$ and $K$ (which only requires that $HK=G$), then one cannot conclude that $G$ itself is abelian. For example, consider the nonabelian group of order $27$ and exponent $3$, presented by -$$\Bigl\langle a,b,c \;\Bigm|\; a^3=b^3=c^3=e,\ ba=abc,\ ac=ca,\ ab=ba\Bigr\rangle.$$ -Let $H= \langle a,c\rangle$ and $K=\langle b\rangle$. Then $H$ and $K$ are each abelian, and $G=HK$, but $G$ is not abelian. - -So, if you are asking whether $D_4$ can be written as a direct product of two proper subgroups, you agree that it cannot, because "direct product" necessarily requires $H$ and $K$ to be normal in $G$, and that would necessarily make $G$ abelian, which it is not. -Now, a separate question is: can we write $G$ as a product, not necessarily direct, of two subgroups? This only requires $G=HK$, and it does not even require $H\cap K=\{e\}$. -In this case, the answer is "yes": you can. You can even do it with $H\cap K=\{e\}$. For example, writing -$$D_4 = \Bigl\langle r,s\;\Bigm|\; r^4 = s^2 = e,\ sr=r^3s\Bigr\rangle$$ -then we can take $H=\langle e, r, r^2, r^3\rangle$, and $K=\{e,s\}$. Then $HK$ has: -$$|HK| = \frac{|H|\,|K|}{|H\cap K|} = \frac{4\times 2}{1} = 8$$ -elements, hence $HK=D_4$. -In fact, $D_4$ is a semidirect product of $C_4$ by $C_2$, which is what I exhibit above; in order for $G$ to be an (internal) semidirect product of $H$ and $K$, we require $H$ and $K$ to be subgroups such that: - -$H\triangleleft G$; -$G=HK$; and -$H\cap K=\{e\}$. - -In particular, every expression of $G$ as a direct product of two subgroups is also an expression as a semidirect product, but not conversely; and every semidirect product is also an expression as a product, but not conversely.<|endoftext|> -TITLE: Realization of graded algebras with Poincaré duality -QUESTION [5 upvotes]: Question: Given a finite dimensional positively graded algebra $A$ - over some ring $R$ that satisfies Poincaré duality in some dimension - $n$, is there necessarily a topological space $X$ such that $H^*(X;R) \cong A$? - -I recognise this is some sort of realization question but I don't know much algebraic topology. - -The case I am most interested in is when $R$ is a field. As vague motivation, I'm interested in whether, given such an $A$ over $\mathbb{Q}$, there is an elliptic Sullivan algebra $(\Lambda V, d)$ such that $H(\Lambda V, d) \cong A$. The converse appears in the textbook Rational Homotopy Theory by Felix et. al.: - -Theorem: If $(\Lambda V,d)$ is an elliptic Sullivan algebra (i.e. $V$ - and $H(\Lambda V, d)$ are finite dimensional vector spaces) over a - field of characteristic 0, then $H(\Lambda V, d)$ satisfies Poincaré - duality. - -There is at least some Sullivan algebras $(\Lambda V, d)$ quasi-isomorphic to $A$ (since $A^0 \cong R$) but whether any of them are elliptic is the question. I may make this another post later. - -REPLY [3 votes]: The answer is, in general, no. -For example, the following is corollary 4L.10 of Hatcher's Algebraic Topology book (freely available): - -If $H^\ast(X;\mathbb{Z})$ is a polynomial algebra $ \mathbb{Z}[\alpha]$, possibly truncated by $\alpha^m = 0$ for $m > 3$, then $|\alpha| = 2$ or $4$. - -Here, $|\alpha|$ denotes the degree of $\alpha$, meaning $\alpha \in H^{|\alpha|}(X;\mathbb{Z})$.<|endoftext|> -TITLE: Can any Polynomial be factored into the product of Linear expressions? -QUESTION [10 upvotes]: Specifically I am wondering if... -Given a Polynomial of n degree in one variable with coefficients from the Reals. -Will every Polynomial of this form be able to be factored into a product of n linear (first degree) Polynomials, with the coefficients of these factors not being constrained to $\mathbb{R}$ but to $\mathbb{C}$ instead. -By a product of linear polynomials I mean something of the form: -$$(Ax+a)(Bx+b)(Cx+c)...$$ -I am also interested in the general behaviour of polynomials if we also play around with the parameters of my question. Such as factoring polynomials with complex coefficients, or coefficients from any set for that matter. As well Polynomials in any (Positive Integer?) number of variables. - -REPLY [21 votes]: The Fundamental Theorem of Algebra states precisely that: - -Fundamental Theorem of Algebra. Every nonzero polynomial $p(x)$ with coefficients in $\mathbb{C}$ can be factored, in essentially a unique way, as a product of a constant and linear terms, in the form - $$p(x) = a(x-r_1)\cdots(x-r_n)$$ - where $a$ is the leading coefficient of $p(x)$ and $n$ is the degree of $p(x)$. - -This result does not hold in general. In some cases, you cannot factor them (you cannot in $\mathbb{R}$, for instance, where $x^2+1$ cannot be written as a product of linear terms; over $\mathbb{Q}$, there are polynomials of arbitrarily high degree that cannot be factored at all, let alone into a product of linear terms). In some cases, the factorization is not unique: in the quaternions, the polnomial $x^2+1$ can be factored in infinitely many essentially distinct ways as a product of linear terms, e.g., $x^2+1 = (x+i)(x-i) = (x+j)(x-j) = (x+k)(x-j) = \cdots$. -A field $F$ is said to be algebraically closed if and only if every nonconstant polynomial $p(x)$ with coefficients in $F$ has at least one root. It is then easy to verify that $F$ is algebraically closed if and only if every nonzero polynomial can be factored into a product of linear terms, as above. Subject to some technical assumptions (The Axiom of Choice), every field is contained in an algebraically closed field, and for every field there is a "smallest" algebraically closed field that contains it, so there is a "smallest" larger field $K$ where you can guarantee that every polynomial factors. This does not hold for arbitrary sets of coefficients (e.g., the quaternions, because they are non-commutative; or rings with zero divisors; and others). -Once you go beyond one variable, it is no longer true that a polynomial can always be factored into terms of degree one, even in $\mathbb{C}$. For example, the polynomial $xy-1$ in $\mathbb{C}[x,y]$ cannot be written as a product of polynomials of degree $1$, $(ax+by+c)(rx+sy+d)$. If you could, then $ar=bs=0$, and we cannot have $a=b=0$ or $r=s=0$, so without loss of generality we would have $b=0$ and $r=0$, so we would have $(ax+c)(sy+d) = xy-1$. Then $ad=0$, so $d=0$ (since $a\neq 0)$, but then we cannot have $cd=-1$. So no such factorization is possible.<|endoftext|> -TITLE: Is $M(x)=O(x^σ)$ possible with $σ≤1$ even if the Riemann hypothesis is false? -QUESTION [7 upvotes]: The wiki page on Mertens conjecture and the Connection to the Riemann hypothesis says - -Using the Mellin inversion theorem we now can express $M$ in terms of 1/ζ as - $$ - M(x) = \frac{1}{2 \pi i} \int_{\sigma-i\infty}^{\sigma+i\infty} \frac{x^s}{s \zeta(s)}\, ds -$$ - which is valid for $\color{blue}{1} < σ < 2$, and valid for $\color{red}{1/2} < σ < 2$ on the Riemann hypothesis. ... From this it follows that - $$ - M(x) = O(x^{\color{red}{1/2}+\epsilon}) -$$ - for all positive ε is equivalent to the Riemann hypothesis, ... - -The $\color{red}{\text{red}}$ color indicates the question My question changed, due to anon's comment's, to: - - -If Riemann was false, would this imply a bound of $M(x)=O(x^{\color{blue}{1}+\epsilon})\! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \!\! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \!\! \! \! \! \! \! \! \! \! \!\! \! \! \! \! \! \! \! \! \! ---------\;\;$? $ \phantom{------------------------------------------------}$ - Is $M(x)=O(x^σ)$ possible with $σ≤1$ even if RH is false? - - -A look at Mertens function, makes me think that it should be easy to prove this. - -But I still don't have a clue... - -REPLY [15 votes]: There is a trivial bound of $|M(x)| \le x$ for all $x\ge 0$, because the Möbius function is bounded by $1$. So we already have $M(x) = O(x^1)$ regardless of whether RH is true or false. -This turns out to be rather close to the best known unconditional bound on $M(x)$, which looks like -$$M(x) = O(x \exp( -c\, \log^{0.6} x \log\log^{-0.2} x))$$ -(for instance see this paper of Nathan Ng). In other words, we do not even have a proven upper bound of the form $O(x^{0.999})$. -Because the Dirichlet series for $\mu(x)$ is just $1/\zeta(s)$, bounds on $M(x)$ can be obtained by Perron's formula using knowledge of the poles of $1/\zeta(s)$, in other words the zeros of $\zeta(s)$. (It is worth noting that Granville and Soundararajan have recently developed a new approach to many such problems without intimate knowledge of $\zeta(s)$ in the critical strip.) -The fact that we are still rather ignorant about the zeros of $\zeta(s)$ is the reason we don't know a significantly better bound for $M(x)$ than the trivial one. At the same time, RH being false is not a very strong statement: it just means there is some zero of $\zeta(s)$ with $\Re(s) > 1/2$. While this does preclude a bound of $M(x) = O_\epsilon(x^{1/2+\epsilon})$, it does not rule out $M(x) = O(x^{3/4+\epsilon})$, if all the zeros of $\zeta(s)$ happen to lie to the left of $\Re(s) = 3/4$. -One last comment: it is somewhat naive to use the observed differences as evidence of how easy it is to prove a bound. For a more striking example, try graphing the prime gaps function $d(n) = p_{n+1} - p_n$. You will find that $d(n)$ appears to be $O(\log^2 n)$ (with a pretty small constant), but even assuming RH we don't know how to prove $d(n) = O(\sqrt{n}).$<|endoftext|> -TITLE: Formula for partial sum of Riemann zeta function -QUESTION [6 upvotes]: Possible Duplicate: -Finite Sum of Power? - -Suppose $f(s,k) = \sum_{n=1}^k n^{-s}$ is the Riemann zeta function truncated at the k-th term. I read on mathoverflow that there is a formula for $f(s,k)$ in terms of Bernoulli numbers, but I can't find it on the web. Would someone happen to know it or could point to a link? I am primarily interested in the case when $s$ is a negative real number. -Thanks! - -REPLY [4 votes]: Perhaps that you are thinking at the formula proposed by Woon 'A New Representation of the Riemann Zeta Function zeta(s)'. - -REPLY [3 votes]: Note that -$$\sum_{n=1}^k n^{-s}=H^{(s)}_k$$ -where $H^{(s)}_k$ is the generalized harmonic number. -Some identities are mentioned here, e.g. -$$H^{(s)}_k=\frac{(-1)^{-s}B_{-s+1}+B_{-s+1}(k+1)}{-s+1}$$ - -REPLY [2 votes]: Presumably mathoverflow was talking about Faulhaber's formula -$$ -\sum_{k=1}^n k^p = \frac{B_{p+1}(n+1)-B_{p+1}(0)}{p+1} -$$ -in terms of Bernoulli polynomials. If $p$ is a positive integer, then the coefficients of the Bernoulli polynomials are essentially Bernoulli numbers.<|endoftext|> -TITLE: How to see that the shift $x \mapsto (x-c)$ is an automorphism of $R[x]$? -QUESTION [8 upvotes]: In the process of studying irreducibility of polynomials, I encountered the criterion that $p(x)$ is irreducible if and only if $p(x-c)$ is irreducible. When trying to determine what properties of the ring were preserved under this map $x \mapsto (x-c)$, which appears sometimes to be called the shift isomorphism, I read that it was an isomorphism of the polynomial ring $R[x]$, but my attempts to prove that fact only led me through some difficult calculations, at which I generally fail. -So how does one prove that the map is an isomorphism of $R[x]$? Is it an isomorphism for all rings $R$ and for any number of variables? -Is this just a specific case of a more general phenomenon? - -REPLY [9 votes]: Hmmm...what about directly? -$$f(x)=\sum_{n=0}^\infty a_nx^n\,,\,g(x)=\sum_{n=0}^\infty b_nx^n\,\in R[x]$$ -with $\,\,a_n=b_n=0 \,\,\text{ for all but a finite number of indices}$ . Since -$$f(x)+g(x)=\sum_{n=0}^\infty (a_n+b_n)x^n\,\,,\,f(x)g(x)=\sum_{n=0}^\infty\left(\sum_{k=0}^na_kb_{n-k}\right)x^n$$ -we get that, denoting the map by$\,\phi(x):=x-c\,$, we have -$$\phi(f+g)=\sum_{n=0}^\infty (a_n+b_n)(x-c)^n=\sum_{n=0}^\infty a_n(x-c)^n+\sum_{n=0}^\infty b_n(x-c)^n=\phi(f)+\phi(g)$$ -$$\phi(fg)=\sum_{n=0}^\infty\left( \sum_{k=0}^n a_kb_{n-k}\right)(x-c)^n=\sum_{n=0}^\infty a_n(x-c)^n\sum_{n=0}^\infty b_n(x-c)^n=\phi(f)\phi(g)$$ -Remember: all the above are in fact finite sums, so we can re-arrange them as we wish. Also, $\,\phi(r)=r\,\,\,\forall\,r\in R\,$, so in particular $\,\phi(1)=1\,$ .<|endoftext|> -TITLE: Kernel of $T$ is closed iff $T$ is continuous -QUESTION [10 upvotes]: I know that for a Banach space $X$ and a linear functional $T:X\rightarrow\mathbb{R}$ in its dual $X'$ the following holds: -\begin{align}T \text{ is continuous } \iff \text{Ker }T \text{ is closed}\end{align} -which probably holds for general operators $T:X\rightarrow Y$ with finite-dimensional Banach space $Y$. I think the argument doesn't work for infinite-dimensional Banach spaces $Y$. Is the statement still correct? I.e. continuity of course still implies the closedness of the kernel for general Banach spaces $X,Y$ but is the converse still true? - -REPLY [4 votes]: No, this is not true. Here is a counterexample for every infinite-dimensional Banach space $Y$. -Take a discontinuous functional $\phi\in Y^*$ (this is always possible if $Y$ is infinite-dimensional, by the axiom of choice). Let $A:=\ker \phi$, and $B$ any algebraic supplement. Note that $A$ is non-closed, and $B$ is closed in view of $\dim B=\text{codim} A=1$. -Consider the projection $p:Y\to Y$ onto $A$, w.r.t. the algebraic direct sum decomposition $Y=A\oplus B$. Now $\ker p=B$ is closed, but $\ker(1-p)=A$ is not, so $p$ is discontinuous.<|endoftext|> -TITLE: Solve $ \left( \log_3 x \right)^2 + \log_3 (x^2) + 1 = 0$ -QUESTION [5 upvotes]: I'm new to logarithms and I am having trouble solving this equation -$$ \left( \log_3 x \right)^2 + \log_3 (x^2) + 1 = 0.$$ -How would I solve this? A step-by-step response would be appreciated. -Also, I know how to solve it with assigning $\log_3 (x)$ as $x$ and solving $x^2 + x + 1 = 0$, and getting the answer from there. I am looking to see how I would do it with just logarithms and no quadratics. -Thanks - -REPLY [4 votes]: You should get the answer $x = \frac{1}{3}$. -Here is why. Solving the quadratic equation $w^2 + 2w + 1=0$ gives $w=-1$ as a repeated root. Substituting $w =\log_3(x)$ in $w = -1$ gives $\log_3(x) = -1$ -Which implies $3^{\log_3(x)} = 3^{-1} \implies x = \frac{1}{3}$. (we used the inverse function $3^x$ of our function $\log_3 (x)$ ).<|endoftext|> -TITLE: Proof if $a \vec v = 0$ then $a = 0$ or $\vec v = 0$ -QUESTION [10 upvotes]: I'm kicking myself over this one, but I just can't seem to make the argument rigorous. From Axler's Linear Algebra Done Right: -for a vector space $V$ with an underlying field $F$: -Take an element $a$ from $F$ and $\vec{v}$ from $V$. $a\vec{v}=\vec{0}\implies a=0 $ or $ \vec{v}=\vec{0}$ -After only being able to come up with half of a direct proof, I tried doing this by proving the contrapositive $a\neq 0 \wedge \vec{v} \neq \vec{0} \implies a\vec{v}\neq \vec{0}$ -Say $a\vec{v}=\vec{u}$. -Since $a$ is non-zero, we can divide both sides by $a$. -$$\vec{v}=\frac 1 a \vec{u}$$ -If $\vec{u}$ were $\vec{0}$ then by -$$\frac 1 a \vec{0}=\frac 1 a (\vec{0}+\vec{0})\implies\frac 1 a \vec{0}=\vec{0}$$ -$v$ would be $0$ as well. Since it isn't by assumption, $\frac 1 a \vec{u}$ cannot be zero and so $\vec{u}$ cannot be as well. - -Is this fully rigorous? It seems like a very simple question, but I'm not sure about it. Namely, the last step of $\frac 1 a \vec{u}\neq 0 \implies \vec{u}\neq 0$ doesn't seem obvious. I think I need to use the $1\vec{v}=\vec{v}$ axiom, but I'm not sure how. -Is there a more direct proof? This whole contrapositive business seems a bit clunky for something so simple. - -REPLY [2 votes]: We can also show the contrapositive: If $a \neq 0$ and $v \neq 0$ implies $av \neq 0.$ -Assume $a\neq 0$ and $v \neq 0.$ Let $V$ has dimension $n,$ and express $v$ in terms of the standard basis vectors: $$v = v_1 e_1 + \ldots + v_n e_n.$$ Then by our assumption that $v \neq 0$ and by the definition of linear independence, at least one of the $v_i$'s is non-zero. Now, multiply $v$ by $a$ to get $$av = (av_1) e_1 + \ldots + (av_n) e_n.$$ But since $a \neq 0,$ at least one of the $av_i$'s will be non-zero, and hence the non-zero linear combination $$(av_1) e_1 + \ldots + (av_n) e_n$$ cannot equal to the $0$ vector. So, $av \neq 0.$<|endoftext|> -TITLE: knowledge needed to understand Fermat's last theorem proof -QUESTION [9 upvotes]: It's a very soft question. -So, Wiles proved Fermat's last theorem, and what knowledge of which area would be required to understand the theorem? - -REPLY [25 votes]: To understand the proof, you need to have a good background in arithmetic and algebraic geometry (including but certainly not limited to the theory of elliptic curves), commutative algebra, algebraic number theory, and modular forms. -If you have an understanding of these things at a graduate level (say a second -or third year graduate student who is beginning research on these sorts of topics), then you will be able to understand the strategy of the proof and some -aspects in more detail. If you are at the level of having successfully written a thesis in these sorts of areas, you will be able to have a substantial understanding of the proof. -Having a complete understanding is perhaps even more difficult (as Kevin Buzzard points out in a remark in one of the MO threads linked to above) because the proof uses base change results in the theory of automorphic forms (due primarily to Langlands) which are quite -technical to prove, and rely on techniques quite different to most of the techniques that Wiles himself uses. -The standard introduction to the argument is the graduate text "Modular forms and Fermat's Last Theorem" (edited by Cornell, Silverman, and Stevens). Another is provided by the long article titled "Fermat's Last Theorem" of Darmon, Diamond, and Taylor. Of course there are many introductions at a more basic level (see the MO threads linked above), but the two references I've given actually contain many details of the proof. -The upshot is that it's probably not realistic to expect to understand much more than the very big picture strategy of the proof unless you are a graduate student focussing on this part of number theory.<|endoftext|> -TITLE: Can two sets have same AM, GM, HM? -QUESTION [5 upvotes]: Can two set of numbers(same size) have same arithmetic , geometric, and harmonic mean ? -When I say different set they must differ by at-least $1$ element and also what if set is not be of distinct elements? - -REPLY [6 votes]: Yes. In fact, you can find solutions in the integers. -$\{-4, -3, 1, 3, 4\}$ -$\{-6, -2, 1, 2, 6\}$<|endoftext|> -TITLE: A continuous, injective function $f: \mathbb{R} \to \mathbb{R}$ is either strictly increasing or strictly decreasing. -QUESTION [27 upvotes]: I would like to prove the statement in the title. -Proof: We prove that if $f$ is not strictly decreasing, then it must be strictly increasing. -So suppose $x < y$. -And that's pretty much how far I got. Help will be appreciated. - -REPLY [19 votes]: Consider $g\colon \{(x,y)\mid x -TITLE: Solving a Nonlinear Recursion -QUESTION [15 upvotes]: In the course of some research computations I have been doing, I run up against a recursion -$$ a_{n+3} = a_{n+2}a_{n+1} - a_n $$ -I've tried to find out if it's possible to solve recursions of this form, but can't find much since it's nonlinear. Does anyone know of methods that might be applicable; or failing that, if there are any assumptions on the initial conditions which might make it solvable? -Hope this is clear; I don't have any background in number theory or discrete math. Thanks. - -REPLY [14 votes]: Let $F_1,F_2,\dots$ be a Fibonacci-like sequence, such that $F_{n}=F_{n-1} + F_{n-2}$, and let -$$a_{n} = e^{F_{n}} + e^{-F_{n}} = 2\cosh F_n.$$ -Then -$$ -\begin{eqnarray} -a_{n-1}a_{n-2} &=& \left(e^{F_{n-1}} + e^{-F_{n-1}}\right) \left(e^{F_{n-2}} + e^{-F_{n-2}}\right) \\ -&=& e^{F_{n-1} + F_{n-2}} + e^{F_{n-1} - F_{n-2}} + e^{-F_{n-1} + F_{n-2}} + e^{-F_{n-1} - F_{n-2}} \\ -&=& e^{F_{n}} + e^{F_{n-3}} + e^{-F_{n-3}} + e^{-F_{n}} \\ -&=& a_{n} + a_{n-3}, -\end{eqnarray} -$$ -which is exactly your recursion. This family of solutions, parametrized by $(F_1, F_2)$, only covers part of the full range of initial conditions $(a_1, a_2, a_3)$. In particular, it will cover those cases where -$$ -a_3 = \frac{1}{2} a_1 a_2 \pm \frac{1}{2}\sqrt{\left(a_1^2-4\right)\left(a_2^2-4\right)}. -$$<|endoftext|> -TITLE: What is the volume of this 3d shape? -QUESTION [5 upvotes]: I'm wondering if there is an equation that represents the volume of an arbitrary 3d primitive -matching this description: -1.) Point at center of sphere -2.) Each edge is the length of the radius -3.) 3 flat sides, 1 arc side -Image: - -So it's kind of a sector of a sphere, but instead of a conical shape it's more of a tetrahedral shape, but with a curved end. - -REPLY [2 votes]: The volume is $\frac13\Omega r^3$, where $\Omega$ is the solid angle subtended at the centre. Wikipedia has several formulas for the solid angle, including the one in Rahul's answer.<|endoftext|> -TITLE: Proving Inequalities using Induction -QUESTION [21 upvotes]: I'm pretty new to writing proofs. I've recently been trying to tackle proofs by induction. I'm having a hard time applying my knowledge of how induction works to other types of problems (divisibility, inequalities, etc). I've been checking out the other induction questions on this website, but they either move too fast or don't explain their reasoning behind their steps enough and I end up not being able to follow the logic. -I do understand how to tackle a problem which involves a summation. -This is the one I just did (the classic "little gauss" proof): -Prove $1+2+3+\dots+n = n(n+1)/2$ -I. Basis -$1=(1+1)/2$ -$1=1$ -II. Induction -Assume the expression holds for an arbitrary $n=k$ such that -$1+2+3+\dots+k = k(k+1)/2$ -Show that the expression holds for $n=k+1$ -$1+2+3+...+n+k+(k+1) = k+1[(k+1)+1]/2$ -And this is done mainly by observing that we already have a formula for 1 through k on the LHS, so the equation can be rewritten as -$k(k+1)/2 + (k+1) = k+1[(k+1)+1]/2$ -NOTE: I believe this is using the inductive hypothesis. Please correct me if I'm wrong. -Anyway, finding common denominators on the left hand side and distributing on the right, you eventually show that it's true. This (so far) has worked for every proof I've attempted that involves a summation on the left hand side. - -Now, I start losing it when the format changes. For example, this inequality proof I'm trying to write. I'll post what I have here: -$n^2 \ge 2n$ for all $n>1$ -I. Basis -$2^{2} \ge 2(2)$ -$4 \ge 4$ -II. Induction -Assume the inequality holds for an arbitrary $n=k$, such that -$k^2 \ge 2(k)$ -Show that the expression holds for $n=k+1$ such that -$(k+1)^2 \ge 2(k+1)$ -This is where I get lost andI know I'm supposed to invoke the IH somewhere in the expression. But unlike the summation problem earlier, I'm not sure where to begin. Could anyone point me in the right direction? - -REPLY [10 votes]: We have to prove $2^n \geq 2n$ for $n>1$ -Basis: -$n=2$ which satisfies the above relation. -Induction hypothesis: -Here we assume that the relation is true for some $k$ i.e. $P(k)\colon 2^k \geq2k$. -Now we have to prove that the relation also holds for $k+1$ by using the induction hypothesis. -This means that we have to prove $$P(k+1)\colon 2^{k+1} \geq 2(k+1)$$ -So the general strategy is to reduce the expressions in $P(k+1)$ to terms of $P(k)$. So, -$$2^{k+1}=2^k\cdot 2=2^k+2^k \geq2^k+2 \quad \quad \{\text{Since }n>1\}$$ -Now we will use induction hypothesis that $2^k\ge 2k$ which gives us -$$ 2^{k+1}\geq 2^k+2 \geq 2k+2 =2(k+1)$$ -which was required. Hence we have proved $P(k+1).$<|endoftext|> -TITLE: Uncountable disjoint union of $\mathbb{R}$ -QUESTION [12 upvotes]: I'm doing 1.2 in Lee's Introduction to smooth manifolds: Prove that the disjoint union of uncountably many copies of $\mathbb{R}$ is not second countable. -So first, let $I$ be the set over which we are unioning. Then I believe the disjoint union is just $\mathbb{R}\times I$. Then I believe that sets of the form $\cup_{x\in A}(x,i)$ is open if and only if $A\subset\mathbb{R}$ is open (I know the open sets are defined with the canonical injection though). At first I thought if I let $I=\mathbb{R}$, then $\mathbb{R}\times I=\mathbb{R^2}$, but now I am thinking maybe they just have the same elements, but the topology is different, and this is why $\mathbb{R}\times I$ is not second countable? - -REPLY [3 votes]: An attempt to simplify Arthur Fischer's argument: -Each copy of the base space is an open set, and these are pairwise disjoint. Since there uncountably many of them, the countable chain condition - is not satisfied, therefore the union is not separable or second countable.<|endoftext|> -TITLE: Nth derivative of $\tan^m x$ -QUESTION [10 upvotes]: $m$ is positive integer, -$n$ is non-negative integer. -$$f_n(x)=\frac {d^n}{dx^n} (\tan ^m(x))$$ -$P_n(x)=f_n(\arctan(x))$ -I would like to find the polynomials that are defined as above -$P_0(x)=x^m$ -$P_1(x)=mx^{m+1}+mx^{m-1}$ -$P_2(x)=m(m+1)x^{m+2}+2m^2x^{m}+m(m-1)x^{m-2}$ -$P_3(x)=(m^3+3m^2+2m)x^{m+3}+(3m^3+3m^2+2m)x^{m+1}+(3m^3-3m^2+2m)x^{m-1}+(m^3-3m^2+2m)x^{m-3}$ -I wonder how to find general formula of $P_n(x)$? -I also wish to know if any orthogonal relation can be found for that polynomials or not? -Thanks for answers -EDIT: -I proved Robert Isreal's generating function. I would like to share it. -$$ g(x,z) = \sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^n}{dx^n} \tan^m(x) -= \tan^m(x+z) $$ -$$ \frac {d}{dz} (\tan^m(x+z))=m \tan^{m-1}(x+z)+m \tan^{m+1}(x+z)=m \sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^n}{dx^n} \tan^{m-1}(x)+m \sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^n}{dx^n} \tan^{m+1}(x)= \sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^n}{dx^n} (m\tan^{m-1}(x)+m\tan^{m+1}(x))=\sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^n}{dx^n} (\dfrac{d}{dx}(\tan^{m}(x)))=\sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^{n+1}}{dx^{n+1}} (\tan^{m}(x))=\sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^{n+1}}{dx^{n+1}} (\tan^{m}(x))$$ - -$$ \frac {d}{dz} ( \sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^n}{dx^n} \tan^m(x) -)= \sum_{n=1}^\infty \dfrac{z^{n-1}}{(n-1)!} \dfrac{d^n}{dx^n} \tan^m(x) = \sum_{n=1}^\infty \dfrac{z^{n-1}}{(n-1)!} \dfrac{d^n}{dx^n} \tan^m(x)=\sum_{k=0}^\infty \dfrac{z^{k}}{k!} \dfrac{d^{k+1}}{dx^{k+1}} \tan^m(x)$$ -I also understood that it can be written for any function as shown below .(Thanks a lot to Robert Isreal) -$$ \sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^n}{dx^n} h^m(x) -= h^m(x+z) $$ -I also wrote $P_n(x)$ as the closed form shown below by using Robert Israel's answer. -$$P_n(x)=\frac{n!}{2 \pi i}\int_0^{2 \pi i} e^{nz}\left(\dfrac{x+\tan(e^{-z})}{1-x \tan(e^{-z})}\right)^m dz$$ -I do not know next step how to find if any orthogonal relation exist between the polynomials or not. Maybe second order differential equation can be found by using the relations above. Thanks for advice. - -REPLY [2 votes]: I have been working on the problem of finding the nth derivative and the nth anti derivative of elementary and special functions for years. You are asking a question regarding a class of functions I have called "the class of meromorphic functions with infinite number of poles. I refer you to the chapter in my Ph.D. thesis (UWO, 2004) where you can find some answers.<|endoftext|> -TITLE: Some method to solve $\int \frac{1}{\left(1+x^2\right)^{2}} dx$ and some doubts. -QUESTION [9 upvotes]: First approach. -$\int \frac{1}{1+x^2} dx=\frac{x}{1+x^2}+2\int \frac{x^2}{\left(1+x^2\right)^2} dx=\frac{x}{1+x^2}+2\int \frac{1}{1+x^2}dx-2\int \frac{1}{\left(1+x^2\right)^2}dx$ -From this relationship, I get: -$2\int \frac{1}{\left(1+x^2\right)^2}dx=\frac{x}{1+x^2}+\int \frac{1}{1+x^2}dx$ Then: -$\int \frac{1}{\left(1+x^2\right)^2}dx=\frac{1}{2}\left[\frac{x}{1+x^2}+\arctan x\right]+C$ This is a recursive solution. -Second approach. -$x=\tan t$ in $t\in (- \pi/2, \pi/2)$, i.e. $t=\arctan x$, then $dx=(1+x^2) dt$. -$\int \frac{1}{\left(1+x^2\right)^2}dx=\int \frac{1}{1+x^2}dt=\int \frac{\cos^2t}{\sin^2t+\cos^2t}dt=\int \cos^2t dt=\frac{1}{2}\int \left(1+\cos 2t \right) dt=\frac{t}{2}+\frac{1}{4}\sin 2t$ -This result can be rewritten (using trigonometric formulas): -$\frac{t}{2}+\frac{1}{4}\sin 2t=\frac{t}{2}+\frac{1}{2}\sin t \cos t$ -From $\cos^2 t=\frac{1}{1+x^2}$, I have: -$|\cos t|=\sqrt{\frac{1}{1+x^2}}$ but in $t\in (- \pi/2, \pi/2)$, $|\cos t|=\cos t$. So: -$\cos t=\sqrt{\frac{1}{1+x^2}}$. Now I have a problem: $|\sin t|=\sqrt{\frac{1}{1+x^2}}$, but $|\sin t|\neq \sin t$ for $t\in (- \pi/2, \pi/2)$. Any suggestions, please? This integral can be solved in other ways? -Thanks. - -REPLY [4 votes]: This integral can be evaluated in that way too as shown below : -$$\int \frac{1}{a^2+x^2} dx=\frac{1}{a} \arctan \frac{x}{a} +c(a) $$ -$$\frac {d}{da} (\int \frac{1}{a^2+x^2} dx)= \frac {d}{da}(\frac{1}{a} \arctan \frac{x}{a} +c(a) ) $$ -$$\int \frac{-2a}{(a^2+x^2)^2} dx= \frac{-1}{a^2} \arctan \frac{x}{a} + \frac{1}{a} \frac{-x/a^2}{(1+x^2/a^2)}+c_1(a) ) $$ -for $a=1$ -$$-2\int \frac{1}{(1+x^2)^2} dx= - \arctan x + \frac{-x}{(1+x^2)}+c_1(1) $$ -$$\int \frac{1}{(1+x^2)^2} dx= \frac{1}{2} \arctan x + \frac{x}{2(1+x^2)}+k $$<|endoftext|> -TITLE: Solving ODEs: The Frobenius Method, worked examples -QUESTION [16 upvotes]: I find the Frobenius Method quite beautiful, and I would like to be able to apply it. In particular there are three questions in my text book that I have attempted. In each question my limited understanding has stopped me. Only one of these questions (the last) is assigned homework. The rest are examples I found interesting*. -1) $ L[y] = xy'' + 2xy' +6e^xy = 0 $ (1) -The wikipedia article begins by saying that the Frobenius method is a way to find solutions for ODEs of the form -$ x^2y'' + xP(x) + Q(x)y = 0 $ -To put (1) into that form I might multiply across by x, giving me -$ x^2y'' + x[2x]y' + [6xe^x]y = 0 $ (2) -But is that OK? The first step in the method seems to be dividing by $ x^2 $, so can't I just leave the equation in its original form? I'll assume I can. -Now we let $ y_1 = \sum _{n=0}^{\infty} a_n x^{r+n} $ -then, $ y_1' = \sum _{n=0}^{\infty} (r+n)a_nx^{r+n-1} $ -and, $ y_1'' = \sum _{n=0}^{\infty} (r+n)(r+n-1)a_nx^{r+n-2} $ -substituting into (2) we get, -$ x\sum _{n=0}^{\infty}(r+n)(r+n-1)a_nx^{r+n-2} + 2x\sum _{n=0}^{\infty}(r+n)a_nx^{r+n-1} + 6e^x\sum _{n=0}^{\infty}a_nx^{r+n} = 0 $ -But now what? I am aware that $ 6e^x = 6\sum _{n=0}^{\infty}x^n/n! $, but my powers stop there. Can I multiply the two series together? I would have to multiply each term in one series by every term in the other, and I don't know how to deal with that. The text provides no worked examples in which P(x) or Q(x) are not polynomials... so for now my work stops here. -2) $ L[y] = x(x-1)y'' + 6x^2y' + 3y = 0 $ -Again, I will leave the question in its original form, rather than try to get that x^2 in front (I realise I am not checking that the singular point is a regular singular point, but checking the answer in the back of the book, x = 1 and x = 0 are indeed regular points). With two regular singular points, I expect I will get 2 sets of answers: one near x = 1 and the other near x = 0. Is it enough to just proceed with one case and then the next? I will assume so, and begin with the case close to x = 0. -Again, letting $ y_1 = \sum _{n=0}^{\infty} a_n x^{r+n} $, and taking the appropriate derivatives, we find by substitution, -$ x(x-1)\sum _{n=0}^{\infty}(r+n)(r+n-1)a_nx^{r+n-2} + 6x^2\sum _{n=0}^{\infty}(r+n)a_nx^{r+n-1} + 3\sum _{n=0}^{\infty}a_nx^{r+n} = 0 $ -$ x^2\sum _{n=0}^{\infty}(r+n)(r+n-1)a_nx^{r+n-2} - x\sum _{n=0}^{\infty}(r+n)(r+n-1)a_nx^{r+n-2} + 6x^2\sum _{n=0}^{\infty}(r+n)a_nx^{r+n-1} + 3\sum _{n=0}^{\infty}a_nx^{r+n} = 0 $ -$ \sum _{n=0}^{\infty}(r+n)(r+n-1)a_nx^{r+n} - \sum _{n=0}^{\infty}(r+n)(r+n-1)a_nx^{r+n-1} + \sum _{n=0}^{\infty}6(r+n)a_nx^{r+n+1} + \sum _{n=0}^{\infty}3a_nx^{r+n} = 0 $ -we shift the indexes of the above sums, so that everything will be in terms of the same power of x. -$ \sum _{n=1}^{\infty}(r+n-1)(r+n-2)a_{n-1}x^{r+n-1} - \sum _{n=0}^{\infty}(r+n)(r+n-1)a_nx^{r+n-1} + \sum _{n=2}^{\infty}6(r+n-2)a_{n-2}x^{r+n-1} + \sum _{n=1}^{\infty}3a_{n-1}x^{r+n-1} = 0 $ -we synchronise the indexes in order to group like terms, by extracting early terms from each series, -$ r(r-1)a_0x^r + \sum _{n=2}^{\infty}(r+n-1)(r+n-2)a_{n-1}x^{r+n-1} - r(r-1)a_0x^{r-1} - r(r+1)a_1x^r - \sum _{n=2}^{\infty}(r+n)(r+n-1)a_nx^{r+n-1} + \sum _{n=2}^{\infty}6(r+n-2)a_{n-2}x^{r+n-1} + 3a_0x^{r-1} + \sum _{n=2}^{\infty}3a_{n-1}x^{r+n-1} = 0 $ -rearranging, we get -$ r(r-1)a_0x^r - r(r-1)a_0x^{r-1} - r(r+1)a_1x^r + 3a_0x^{r-1} + \sum _{n=2}^{\infty}(r+n-1)(r+n-2)a_{n-1}x^{r+n-1} - \sum _{n=2}^{\infty}(r+n)(r+n-1)a_nx^{r+n-1} + \sum _{n=2}^{\infty}6(r+n-2)a_{n-2}x^{r+n-1} + \sum _{n=2}^{\infty}3a_{n-1}x^{r+n-1} = 0 $ -At this point I expect the indicial equation to emerge, and I expect it to be similar to an Euler Equation. That is, I expect a polynomial that I can solve to get two 'exponents at the singularity'. Unfortunately, I cannot see an indicial equation and am at a loss to know precisely why. -3) $ L[y] = xy'' + y = 0 $ -Finally we come to the assigned question, which I have been able to manipulate into an almost final form. -Again, letting $ y_1 = \sum _{n=0}^{\infty} a_n x^{r+n} $, taking derivatives, and substituting into L, we get -$ x\sum _{n=0}^{\infty} (r+n)(r+n-1)a_nx^{r+n-2} + \sum _{n=0}^{\infty} a_n x^{r+n} = 0 $ -$ \sum _{n=0}^{\infty} (r+n)(r+n-1)a_nx^{r+n-1} + \sum _{n=0}^{\infty} a_n x^{r+n} = 0 $ -Now shifting indexes, -$ \sum _{n=0}^{\infty} (r+n)(r+n-1)a_nx^{r+n-1} + \sum _{n=1}^{\infty} a_{n-1} x^{r+n-1} = 0 $ -and extracting the $ 0^{th} $ term of the first sum, -$ r(r-1)a_0x^{r-1} + \sum _{n=1}^{\infty} (r+n)(r+n-1)a_nx^{r+n-1} + \sum _{n=1}^{\infty} a_{n-1} x^{r+n-1} = 0 $ -$ r(r-1)a_0x^{r-1} + \sum _{n=1}^{\infty}[(r+n)(r+n-1)a_n + a_{n-1}]x^{r+n-1} = 0 $ -And voila! We have an indicial term with solutions $r_1 = 1$ and $r_2 = 0$, and a recurrence relation. From my text I expect that -$y_1 = |x|^{r_1}[1+\sum _{n=0}^{\infty}a_nx^n]$ -and since $ r_1 - r_2 \in \mathbb{Z} $, -$y_2 = ay_1ln|x| + |x|^{r_2}[1 + \sum _{n=0}^{\infty}b_nx^n]$ -to find $a_n$ we observe the recurrence relation with $ r = r_1 = 1 $, -$ (r+n)(r+n-1)a_n + a_{n-1} $ -$ a_n = -a_{n-1}/n(n+1) $ -so, $ a_1 = -a_0/2*1 $ -$ a_2 = -a_1/2*3 = a_0/3*2*1*2*1 = a_0/3!2! $ -$ a_3 = -a_2/3*4 = -a_0/4!3! $ -and in general, $ a_n = (-1)^na_0/n!(n+1)! $ -so we have $ y_1 = |x| + \sum _{n=0}^{\infty} (-1)^na_0x^{n+1}/n!(n+1)! $ -Not so easily done with r = r_2 = 0, I'm afraid... -since the relation becomes $ a_n = -a_{n-1}/n(n-1) $, which means we can't have a_1 for fear of division by zero. Never the less, starting at n = 2 we get, -$ a_2 = -a_1/2*1 $ -$ a_3 = -a_2/2*3 = a_1/3*2*1*2*1 = a_1/3!2! $ -$ a_4 = -a_3/3*4 = -a_1/4!3! $ -and in general, $ a_n = (-1)^{n-1}a_1/n!(n-1)! $ -so we have $ y_2 = ay_1ln|x| + 1 + \sum _{n=0}^{\infty} (-1)^{n-1}a_1x^{n+1}/n!(n-1)! $ -Which I feel may not be correct... and even if it is, how should one man solve for a in a single lifetime? -Thanks everyone for looking at this. I want to stress that I am not just a student looking for help in his homework: I would really like to understand this method because it appeals to me. I particularly like the way we extract the indicial expression from the sums, in order to synchronise them. That is so cool. And how you get 1 recurrence relation that you can use for both solutions: neat. -PS sorry if my Latex is not perfect? I'm just getting started with it. - -questions taken from "Elementary Differential Equations and Boundary Value Problems" by William E. Boyce and Richard C. DiPrima (9th ed), from sectin 5.6 pp 290 - -REPLY [10 votes]: but my powers stop there. Can I multiply the two series together? I would have to multiply each term in one series by every term in the other, and I don't know how to deal with that. - -I cracked my head about that a little while and I figured out how you can multiply two infinite series without getting lost. Here's an animation explaining my technique:<|endoftext|> -TITLE: what exactly is mathematical rigor? -QUESTION [17 upvotes]: I am a programmer moving up to computer science and even in my past life as a mechanical engineer I used sufficient maths to get by but I have always wondered what exactly is this thing called rigor. -Please make me understand it. - -REPLY [16 votes]: In mathematical logic, you can define a system in which statements are strings of symbols, and there's a list of axioms (assumed to be true) and deductive rules (which derive a statement from other statements). In this system, a proof of a statement is a sequence of steps which starts with the axioms, uses the deductive rules, and ends with that statement. -A fully rigorous result is one with a proof in this form. -In practice, we don't have time to carry out proofs at this level of detail. Instead we rely on our intuition, and do only as much checking as we need to convince ourselves and our colleagues that we could make the proof fully rigorous if we wanted to. This is just a judgement call that you will learn to make. Study proofs that have been published, and you'll get a good idea of the level of rigor that's expected. -Occasionally it turns out that we were wrong and a published proof has a hole in it. Also, it turns out that there are subtleties about choosing a plausible logical system which can trip us up. But generally this approach works well. -The reason we need to worry about rigor is simply that otherwise, it's too easy to delude ourselves about what we know. We are often surprised by what turns out to be true, or by what we cannot manage to prove.<|endoftext|> -TITLE: Example of a function continuous at only one point. -QUESTION [12 upvotes]: Possible Duplicate: -Find a function $f: \mathbb{R} \to \mathbb{R}$ that is continuous at precisely one point? - -I want to know some example of a continuous function which is continuous at exactly one point. -We know that $f(x)=\frac{1}{x}$ is continuous everywhere except at $x=0$. But i think this in reverse manner but i dont get any example. So please help me out! - -REPLY [31 votes]: One standard example is the function -$$f(x)=\begin{cases} -x,&\text{if }x\in\Bbb Q\\ -0,&\text{if }x\in\Bbb R\setminus\Bbb Q\;. -\end{cases}$$ -That is, $f(x)=x$ if $x$ is rational, and $f(x)=0$ if $x$ is irrational. This function is continuous only at $x=0$. -Added: The same basic idea can be used to build a function that is continuous at any single specified point. With a little more ingenuity, you can use it to get, for instance, a function that is continuous just at the integers: -$$f(x)=\begin{cases} -\sin\pi x,&\text{if }x\in\Bbb Q\\ -0,&\text{if }x\in\Bbb R\setminus\Bbb Q\;. -\end{cases}$$ -This works because $\sin\pi x=0$ if and only if $x\in\Bbb Z$. - -REPLY [12 votes]: Just take something like the Dirichlet function: -$$f : \mathbb R \ni x \mapsto \begin{cases} x&\text{if}\; x\in \mathbb Q\\0&\text{otherwise}\end{cases}$$ -Then $f$ is continuous only at $x=0$.<|endoftext|> -TITLE: When is matrix multiplication commutative? -QUESTION [130 upvotes]: I know that matrix multiplication in general is not commutative. So, in general: -$A, B \in \mathbb{R}^{n \times n}: A \cdot B \neq B \cdot A$ -But for some matrices, this equations holds, e.g. A = Identity or A = Null-matrix $\forall B \in \mathbb{R}^{n \times n}$. -I think I remember that a group of special matrices (was it $O(n)$, the group of orthogonal matrices?) exist, for which matrix multiplication is commutative. -For which matrices $A, B \in \mathbb{R}^{n \times n}$ is $A\cdot B = B \cdot A$? - -REPLY [3 votes]: Another commuting example: -ANY two square matrices that, are inverses of each other, commute. - A B = I -inv(A)A B = inv(A) # Premultiplying both sides by inv(A) -inv(A)A B A = inv(A)A # Postmultiplying both sides by A - B A = I # Canceling inverses - -QED -There are lots of "special cases" that commute. -The multiplication of two diagonal matrices, for example. -Aside: for any two square invertible matrices, A, B, there is something -that can be said about AB vs. BA -If AB = C -then BA = inv(A) C A = B C inv(B) - -(Proof: substitute AB for C in the result, and cancel inverses)<|endoftext|> -TITLE: Center of a ring isomorphic to endomorphism ring of identity functor -QUESTION [5 upvotes]: I am having trouble verifying the following (this is self-study): - -There is an isomorphism between the center of a ring $A$ and the ring of endomorphisms of the identity functor of the category of (right) $A$-modules. - -The map $\Psi\ \colon Z(A) \to \mathrm{End}(1_{Mod\ A})$ sends -$a \in Z(A)$ to multiplication by $a$ (let's write $\Psi(a) = \theta^{a}$). -This poses no trouble. -It seems reasonable to define the inverse map -$\Xi\ \colon \mathrm{End}(1_{Mod\ A}) \to Z(A)$ as sending -$ \eta \in \mathrm{End}(1_{Mod\ A})$ to $\eta_A(1)$, since -from this we readily get -$\Xi \circ \Psi (a) = \Xi(\theta^a)=\theta^{a}_{A}(1)=1a=a$. -However I can't see why $\Psi \circ \Xi = 1_{\mathrm{End}(1_{Mod\ A})}$. -Unfolding, we have to prove that -$\Psi \circ \Xi (\eta) = \theta^{\eta_A(1)} = \eta$, that is, for any right $A$-module $L$, we need to have $\theta^{\eta_A(1)}_L = \eta_L$. -I'm at a loss on this. I tried to find the right components on which to use -naturality (that's how I succeeded to prove that $\eta_A(1)$ does lie in $Z(A)$, -which wasn't obvious to me from the outset), but to no avail. -Some googling led me to this blog post by Q. Yuan (see the "sub-example") where the direction of the map $\Xi$ is deduced from the fact that $A$ is a generator of $\mathrm{Mod}\ A$. However I can't quite see explicitly why (in the perspective adopted there, I believe this is just rephrasing the issue I'm having) the lifting of a central element to an endomorphism of $1_{\mathrm{Mod}\ A}$ is inverse to the composition of the two isomorphisms with the natural injection $\mathrm{End}(1_{\mathrm{Mod}\ A}) \to Z(\mathrm{End}_A(A))$. - -REPLY [5 votes]: As suggested in the comments, break it into pieces. -Step 1. The identity functor is the same as the functor $\text{Hom}_A(A, \text{_})$ so you just need to know the endomorphisms of this latter functor. By Yoneda, the endomorphisms of this functor are the same as the endomorphisms of $A$; this is cheating a little since Yoneda is about the functor to sets and not the functor to $A$-modules, so you have to chase through the proof of Yoneda to see that $A$-linear endomorphisms of this functor correspond to $A$-linear endomorphisms of $A$. -Step 2. The $A$-linear endomorphisms of $A$ are $Z(A)$. This step is easy; clearly $Z(A)$ gives rise to $A$-linear endomorphisms of $A$ and any endomorphism is determined by $f(1)$, which must be central.<|endoftext|> -TITLE: Annihilators in matrix rings -QUESTION [5 upvotes]: Let $R$ be a finite commutative ring. For $n>1$ consider the full matrix ring $M_n(R)$. For a matrix $A\in M_n(R)$ is true that the cardinality of the left annihilator (in $M_n(R)$) of $A$ equals the cardinality of the right annhilator? - -REPLY [3 votes]: No, this isn't true in general, but it's true in principal ideal rings. Consider $R=\mathbb{F}_2[x,y]/(x,y)^2$ (a finite ring that isn't a principal ideal ring, which I found in this answer by Zev). The matrix -$$A=\pmatrix{x&x\\y&y}$$ -annihilates the $16$ vectors that have no constant term, both on the left and on the right, but on the right it also annihilates the $16$ vectors that have a constant term in both components, for a total of $32$. Thus the left annihilator in $M_n(R)$ has $256$ elements, whereas the right annihilator has $1024$. -However, if $R$ is an elementary divisor ring, we can bring $A$ into Smith normal form using invertible matrices. Since this is diagonal, its left and right annihilators are the same up to transposition, and the invertible matrices don't change their cardinality. (See this MO question.) -According to Theorem 15.9 in Matrices over commutative rings, all principal ideal rings are elementary divisor rings. (Wikipedia only claims this for principal ideal domains.) According to this paper, a necessary condition for a ring to be an elementary divisor ring is that all finitely generated ideals are principal. Since in a finite ring all ideals are finitely generated, a finite elementary divisor ring must be a principal ideal ring. Thus a finite ring is an elementary divisor ring if and only if it is a principal ideal ring.<|endoftext|> -TITLE: Why is $\mathbb{C}^{g}$ the universal cover of any connected compact complex Lie group of dimension g? -QUESTION [6 upvotes]: This question came up while I was studying the book "Complex Abelian Varieties" by Lange/Birkenhake. -More precisely, the authors prove in Lemma 1.1 that every connected compact complex Lie group of dimension $g$, $g$ a positive integer, is a complex torus. At one point they are using the fact that the universal cover of any such Lie group is a complex vector space of dimension $g$. They even give a reference (Theorem 18.4.1, "The Structure of Lie Groups" by Gerhard Hochschild). However, Theorem 18.4.1 (which is instead listed as Proposition 18.4.1 in the cited book) reads -"If a semisimple analytic group has a faithful finite-dimensional continuous representation then its center is finite" -I really don't have a clue how this should apply to my problem, because a complex torus is abelian, hence in particular not semisimple. On the other hand my knowledge of Lie groups is very very rudimentary so I might overlook something. -Can anyone show me how Proposition 18.4.1 in Hochschild's book helps me, or instead give me a reference to a proof of the statement in the title of my question? Of course any outline of a proof would also be appreciated, however I'd rather favor a reference. A reference to a direct proof of the above cited Lemma 1.1 would be very helpful too. -Thank you in advance - -REPLY [3 votes]: The following steps lead to a solution. Let me first define standard notation. If $G$ is a Lie group and if $g\in G$, then the map $\phi_{g}:G\to G$ defined by the rule $\phi_g(x)=gxg^{-1}$ for $x\in G$ is referred to as conjugation by $g$. If $f:M\to N$ is a smooth map, then we denote by $T_{p}(f):T_{p}(M)\to T_{f(p)}(N)$ the differential of $f$ at $p\in M$. -Exercise 1: Let $G$ be a complex Lie group. The adjoint representation of $G$ is the map $\rho:G\to \text{GL}(\mathfrak{g})$ defined by the rule $\rho(g)=T_{e}(\phi_{g})\in \text{GL}(\mathfrak{g})$ where $e\in G$ is the identity element. Prove that $\rho:G\to \text{GL}(\mathfrak{g})$ is a complex (holomorphic) representation of $G$. -Exercise 2: In the context of Exercise 1, assume that $G$ is in addition compact and connected. Prove that the adjoint representation $\rho:G\to \text{GL}(\mathfrak{g})$ is trivial. (Hint: use the maximum modulus principle.) Conclude that $G$ is an abelian group. -We have now set the stage to prove that the exponential map $\text{exp}:\mathfrak{g}\to G$ is a covering map. -Exercise 3: Let $G$ be a Lie group. If $X,Y\in \mathfrak{g}$ commute, i.e., $[X,Y]=0$, then prove that $e^{X}e^{Y}=e^{X+Y}=e^{Y}e^{X}$. In the context of Exercise 2, conclude that $\text{exp}:\mathfrak{g}\to G$ is a homomorphism. -Exercise 4: Use the Hopf-Rinow theorem (or look it up if necessary) to prove that $\text{exp}:\mathfrak{g}\to G$ is surjective in the context of Exercise 2. -Exercise 5: Prove that $\text{exp}:\mathfrak{g}\to G$ is a covering map in the context of Exercise 2. (Hint: use Exercise 3 and Exercise 4. Recall that $\text{exp}:\mathfrak{g}\to G$ is a local homeomorphism at the origin of $\mathfrak{g}$ because its differential at the origin of $\mathfrak{g}$ is the identity map.) -I hope this helps!<|endoftext|> -TITLE: How many triangles with integral side lengths are possible, provided their perimeter is $36$ units? -QUESTION [13 upvotes]: How many triangles with integral side lengths are possible, provided their perimeter is $36$ units? -My approach: -Let the side lengths be $a, b, c$; now, -$$a + b + c = 36$$ -Now, $1 \leq a, b, c \leq 18$. -Applying multinomial theorem, I'm getting $187$ which is wrong. -Please help. - -REPLY [19 votes]: The number of triangles with perimeter $n$ and integer side lengths is given by Alcuin's sequence $T(n)$. The generating function for $T(n)$ is $\dfrac{x^3}{(1-x^2)(1-x^3)(1-x^4)}$. Alcuin's sequence can be expressed as -$$T(n)=\begin{cases}\left[\frac{n^2}{48}\right]&n\text{ even}\\\left[\frac{(n+3)^2}{48}\right]&n\text{ odd}\end{cases}$$ -where $[x]$ is the nearest integer function, and thus $T(36)=27$. See this article by Krier and Manvel for more details. See also Andrews, Jordan/Walch/Wisner, these two by Hirschhorn, and Bindner/Erickson.<|endoftext|> -TITLE: Sylow 2-subgroups of the group $\mathrm{PSL}(2,q)$ -QUESTION [7 upvotes]: What is the number of Sylow 2-subgroups of the group $\mathrm{PSL}(2,q)$? - -REPLY [8 votes]: When $q$ is a power of $2,$ we have ${\rm PSL}(2,q) = {\rm SL}(2,q)$ and a Sylow $2$-normalizer is a Borel subgroup of order $q(q-1).$ Hence there are $q+1$ Sylow $2$-subgroups as ${\rm SL}(2,q)$ has order $(q-1)q(q+1)$. When $q$ is odd, the order of ${\rm PSL}(2,q)$ is $\frac{q(q-1)(q+1)}{2}.$ A Sylow $2$-subgroup of ${\rm SL}(2,q)$ is (quaternion or) generalized quaternion and a Sylow $2$-subgroup of ${\rm PSL}(2,q)$ is either a Klein $4$-group or a dihedral $2$-group with $8$ or more elements. In all these cases, a Sylow $2$-subgroup of ${\rm SL}(2,q)$ contains its centralizer, and some elementary group theory allows us to conclude that the same is true in ${\rm PSL}(2,q).$ -The outer automorphism group of a dihedral $2$-group with $8$ or more elements is a $2$-group. Hence a Sylow $2$-subgroup of ${\rm PSL}(2,q)$ is self-normalizing when $q \equiv \pm 1$ (mod 8), and in that case the number of Sylow $2$-subgroups of ${\rm PSL}(2,q)$ is $q(q^{2}-1)_{2^{\prime}}$ where $n_{2^{\prime}}$ denotes the largest positive odd divisor of the positive integer $n.$ When $q \equiv \pm 3$ (mod 8), then a Sylow $2$-normalizer of ${\rm PSL}(2,q)$ must have order $12$ ( a Sylow $2$-subgroup is a self-centralizing Klein $4$-group, but there must be an element of order $3$ in its normalizer by Burnside's transfer theorem). In this case, the number of Sylow $2$-subgroups of ${\rm PSL}(2,q)$ is $q(\frac{q^{2}-1}{24})$<|endoftext|> -TITLE: Connectedness of the boundary -QUESTION [28 upvotes]: My question is about the following claim: - -For $n \geq 2$, let $A\subset \mathbb R^n$ be a non-empty, open, bounded set. Assume $A$ and its complement are connected and $\text{int}(\text{cl}(A)) = A$. Then $\partial A$ is connected. - -Without the assumption $\text{int}(\text{cl}(A)) = A$, the statement is false (just take any open set and remove an interior point). This condition is not necessary but I think it is sufficient to get the claim. -This seems a trivial matter, but I cant's find a proof using only basic topological tools. Does anyone know something about this ? Thank you very much. - -REPLY [9 votes]: This is true in general: a connected open set $A$ in $\mathbb{R}^n$ has disconnected complement if and only if it has disconnected boundary. -The "only if" follows from $\mathbb{R}^n$ being normal. -As for the "if", consider that $\partial A$ has two components, $X$ and $Y$. Neither of them can lie in the interior of closure of $A$ (it is then a clopen set in the complement of $A$, and we assumed it to be connected). So assume to the contrary, and this actually means that both $X$ and $Y$ have points from the closure of interior of complement of $A$ (don't worry, that was the longest:), say $x$ and $y$. Now you can connect $x$ to $y$ by a curve through $A$ (we can do such things in $\mathbb{R}^n$), and then connect $y$ to $x$ with a curve through the complement of $A$ (we need that nonempty interior of complement for that). This gives a loop that cannot be homotoped to constant, yet $\mathbb{R}^n$ is simply connected, a contradiction. -This cannot be done non-homotopically. The equivalence as I stated it characterises (is equivalent) to the ambient space having trivial first cohomology. The map $f$ from Deian Govc's answer cannot exist precisely because it would give a nontrivial cohomology class.<|endoftext|> -TITLE: proving convergence of a sequence and then finding its limit -QUESTION [7 upvotes]: For every $n$ in $\mathbb{N}$, let: $$a_{n}=n\sum_{k=n}^{\infty }\frac{1}{k^{2}}$$ -Show that the sequence $\left \{ a_{n} \right \}$ is convergent and then calculate its limit. -To prove it is convergent, I was thinking of using theorems like the monotone convergence theorem. Obviously, all the terms $a_{n}$ are positive. So, if I prove that the sequence is decreasing, then by the monotone convergence theorem it follows that the sequence itself is convergent. $a_{n+1}-a_{n}=-\frac{1}{n}+\sum_{k=n+1}^{\infty }\frac{1}{k^{2}}$. But, I can't tell from this that the difference $a_{n+1}-a_{n}$ is negative. -If anybody knows how to solve this problem, please share. - -REPLY [5 votes]: $$ -n\sum_{k=n}^\infty\frac1{k^2}=\sum_{k=n}^\infty\frac{n^2}{k^2}\frac1n\tag{1} -$$ -is a Riemann sum ($x=\frac{k}{n}$ and $\mathrm{d}x=\frac1n$) for -$$ -\int_1^\infty\frac1{x^2}\mathrm{d}x=1\tag{2} -$$ -Although $(2)$ is an improper Riemann integral, because $\frac1{x^2}$ is decreasing, we can still use the rectangular upper and lower approximations to $(2)$: -$$ -\sum_{k=n+1}^\infty\frac{n^2}{k^2}\frac1n<\int_1^\infty\frac1{x^2}\mathrm{d}x< \sum_{k=n}^\infty\frac{n^2}{k^2}\frac1n\tag{3} -$$ -where the sums in $(3)$ differ by $\frac1n$. Therefore, combining $(1)$-$(3)$ yields -$$ -1 -TITLE: Pointwise convergence implies $L^p$ convergence? -QUESTION [5 upvotes]: Let $f: X \to [0, \infty) \subset \mathbb R$ measurable where $X$ is a measure space. Let $f_n : X \to [0, \infty) $ be simple functions (i.e. linear combinations of characteristic functions of measurable sets) such that for each $x \in X$, $f_n(x) \leq f_{n+1}(x)$ and $f_n(x)$ converges to $f(x)$. -How can I prove that $$ \|f_n - f \|_p = \left ( \int_X |f - f_n|^p d \mu\right )^{1/p} \xrightarrow{n \to \infty} 0$$ -I don't think this is right but if for $n > N_x$, $|f_n(x) - f(x)| \leq \varepsilon$, we can let $N = \sup_{x \in X} N_x$ to get $\|f_n - f\|_\infty \leq \varepsilon$ and then - $$ \|f_n - f \|_p = \left ( \int_X \|f - f_n\|^p d \mu\right )^{1/p} \leq \left ( \int_X \varepsilon^p d \mu\right )^{1/p} = \mu(X)^{1/p} \varepsilon $$ -But $\mu(X)$ could be infinite so I'm not sure what to do. - Thanks. -Edit What assumptions do I need to make this true? - -REPLY [4 votes]: The answer to the question in the title is: No, even on finite measured spaces. -For an example, consider $X=(0,1)$ endowed with the Lebesgue measure, and $f_n=2^n\cdot\mathbf 1_{(0,1/n)}$.<|endoftext|> -TITLE: How do we know Taylor's Series works with complex numbers? -QUESTION [11 upvotes]: Euler famously used the Taylor's Series of $\exp$: -$$\exp (x)=\sum_{n=0}^\infty \frac{x^n}{n!}$$ -and made the substitution $x=i\theta$ to find -$$\exp(i\theta) = \cos (\theta)+i\sin (\theta)$$ -How do we know that Taylor's series even hold for complex numbers? How can we justify the substitution of a complex number into the series? - -REPLY [7 votes]: The series $f(z)=\sum_{n=0}^{\infty} \frac{z^n}{n!}$ is convergent on $C$, and thus it defines an Analytic function. -Now there are few different ways to convince yourself that this has to be $e^z$. -For once, it is the only Analytic continuation of $e^x$ to the complex plane... -Or, alternately, you can prove that $f(z_1+z_2)=f(z_1)f(z_2)$ and $f(1)=e$. Also, you can show that it is the only differentiable function satisfying these two relations. -If you prefer differential equations, $f'(z)=f(z)$ and $f(1)=e$ uniquely determine a solution, and bot $e^z$ and $f(z)$ are solutions....<|endoftext|> -TITLE: Admissible ordinals... -QUESTION [6 upvotes]: a little question about admissible sets: -Is every $\mathfrak{M}$-admissible ordinals an admissible ordinal ? where $\mathfrak{M}$ is a -$L$-structure over $L=\{R_1,\dots,R_k \}$. -Thanks. - -REPLY [2 votes]: Removing this question from the unanswered list: - -Yes, admissibility relativizes downward. For a transitive structure to be admissible, it must be amenable, and satisfy $\Sigma_0$-collection. Both are conditions that keep holding if you "remove parameters". This is obvious for amenability. For collection, a little argument is needed. If you have access to Devlin's "Constructibility" book, this is at the beginning of II.7. (Andres Caicedo)<|endoftext|> -TITLE: Convergence of $a_{n}=\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{1}{\sqrt{k}}$? -QUESTION [7 upvotes]: For $n$ in $\mathbb{N}$, consider the sequence $\left \{ a_{n} \right \}$ defined by: -$$a_{n}=\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{\sqrt{k}}$$ -I would like to prove whether this sequence is convergent, and if so what its limit is. -I can prove by induction that $\sum\limits_{k=1}^{n}\frac{1}{\sqrt{k}}\geqslant \sqrt{n}$ For any $n$ in $\mathbb{N}$. Hence, $a_{n}\geqslant 1$. I wanted to prove that the sequence is decreasing and then use the monotone convergence theorem to prove it is convergent. However, I couldn't come up with a proof for this part. Anyone know how to prove convergence and find the limit? -I had another proof based on using Riemann sums, but I am looking for another proof using onne of the tricks used to prove convergence of sequences. Here is my proof: -$$ -a_{n}=\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{\sqrt{k}}=\sum_{k=1}^{n}\frac{1}{n}\frac{1}{\sqrt{\frac{k}{n}}}. -$$ -Hence, $$\lim_{n \to \infty }a_{n}=\int_{0}^{1}\frac{dx}{\sqrt{x}}=2$$ - -REPLY [9 votes]: Using only elementary inequalities and no (improper) integral: -For every $k$, $\sqrt{k+1}-\sqrt{k}=\dfrac1{\sqrt{k+1}+\sqrt{k}}$ and $2\sqrt{k}\lt\sqrt{k+1}+\sqrt{k}\lt2\sqrt{k+1}$, hence -$$ -\sqrt{k+1}-\sqrt{k}\lt\dfrac1{2\sqrt{k}}\lt\sqrt{k}-\sqrt{k-1}. -$$ -Summing the rightmost inequality yields -$$ -a_n\lt\frac2{\sqrt{n}}\sum_{k=1}^n\sqrt{k}-\sqrt{k-1}=2. -$$ -Likewise, summing the leftmost inequality yields -$$ -a_n\gt\frac2{\sqrt{n}}\sum_{k=1}^n\sqrt{k+1}-\sqrt{k}=2\frac{\sqrt{n+1}-1}{\sqrt{n}}\gt2-\frac2{\sqrt{n}}. -$$ -Since $\dfrac2{\sqrt{n}}\to0$, this proves that $\lim\limits_{n\to\infty}a_n=2$.<|endoftext|> -TITLE: Does anything precede incrementation in the operator "hierarchy?" -QUESTION [9 upvotes]: I here define the hierarchy of basic mathematical operators and their respective "inverse" operation (see hyperoperation). -$$ -\begin{array}{c|c|c|} - & \text{Operator} & \text{"Inverse"} \\ \hline -\text{Incrementation} & a+1 & a-1 \\ \hline -\text{Addition} & a+b & a-b \\ \hline -\text{Multiplication} & ab & \frac{a}{b} \\ \hline -\text{Exponentiation} & a^b & \sqrt[b]{a} \\ \hline -\text{Tetration} & ^ba & \sqrt[b]{a}_s \\ \hline -\text{} \vdots & \vdots & \vdots \\ \hline -\end{array} -$$ -Now, it is clear that each "level" is simply the previous one except the process is done several times. Adding two integers is like incrementing one integer many times. Multiplication is addition of the same number many times. -My question is: does anything precede incrementation? - -REPLY [2 votes]: To expand on MphLee's answer: -Successor can also be viewed as a binary function $H_0(a,b) = S(b)$. Then we have that $H_0(a, Sb) = H_{-1}(a,H_0(a,b)) = H_{-1}(a,Sb)$ so $H_{-1}$ agrees with $H_0$ when the second argument is greater than $0$. However, if we stop there, $H_{-1}(a,0)$ can be specified freely. -If we continue further, we can show that $H_{-k}(a, b) = S(b)$ for $b \ge k$: -If $b \ge k$, then $SSb = H_{-k}(a, Sb) = H_{-k-1}(a, H_{-k}(a,b)) = H_{-k-1}(a, Sb)$. -However, then $2 = H_{-1}(a, 1) = H_{-2}(a, H_{-1}(a,0))$, so if $f(a) := H_{-1}(a,0) \ge 2$, we get that $2 = f(a) + 1$, so $f(a) = 1$, a contradiction. Therefore, $f(a) \le 1$. -If we continue further it gets more complicated. I think if you extend it infinitely, they all have to be successor, even though they don't in a finite extension.<|endoftext|> -TITLE: Weak convergence and weak star convergence. -QUESTION [9 upvotes]: If region $\Omega$ is bounded and $u_n$ has weak star convergence in $L^\infty ( \Omega)$ to some $u\in L^\infty(\Omega)$ , does it imply that $u_n$ converges weakly in any $L^p(\Omega) $ ? -I think i got it : If $sup$ of a function is finite then integral over a bounded region is finite with any $p$ norm . is it right ? - -REPLY [13 votes]: $\{u_n\}\subset L^\infty(\Omega)$ converges in the weak star topology to $u\in L^\infty(\Omega)$ if -$$ -\lim_{n\to\infty}\int_\Omega u_n\phi\,dx=\int_\Omega u\,\phi\,dx\quad\forall\phi\in L^1(\Omega). -$$ -Since $\Omega$ is bounded, $L^\infty(\Omega)\subset L^p(\Omega)\subset L^1(\Omega)$ for all $p\ge1$. It follows that $u_n$ converges weakly to $u$ in $L^p(\Omega)$ for all $p\in[1,\infty)$.<|endoftext|> -TITLE: Holoedric isomorphism? -QUESTION [6 upvotes]: While trying to read the following article - -Schottenfells, Ida May. Two Non-Isomorphic Simple Groups Of The Same Order 20,160. - -I found the term "holoedrically isomorphic". In an abstract for the article, I also came across the claim that "Holoedric isomorphism is the only isomorphism that can exist between two simple groups." -I haven't been able to find a definition for this term anywhere. Presumably it is a stronger notion than that of a normal isomorphism of groups? What does it mean? - -REPLY [7 votes]: Isomorphism did not used to mean 1-1, just onto. Holoedric isomorphisms are both 1-1 and onto. See page 381 of Burnside's Theory of Groups (1ed). The standard english term was "simply isomorphic", the French term was isomorphisme holoédrique. -The corresponding term for non-injective epimorphism (so onto, not 1-1) was "multiply isomorphic" in English, and isomorphisme meriédrique in French. -"édrique" appears to be about the same as "edral" as in "dihedral". - -Schottenfels, Ida May. "Two non-isomorphic simple groups of the same order 20,160." -Ann. of Math. (2) 1 (1899/00), no. 1-4, 147–152. -MR1502265 -DOI:10.2307/1967281<|endoftext|> -TITLE: Simple application of Stone-Weierstrass -QUESTION [6 upvotes]: I was looking for a simple application of the Stone-Weierstrass theorem. -First I thought that if $X$ is any compact measure space then Stone-Weierstrass implies that $C_c(X)$ is dense in $L^p$. -But I have to assume that $X$ is compact otherwise I don't have $1$ in $C_c(X)$. That of course makes it a boring example since then $C_c(X) = C(X)$. Can someone show me a slightly more interesting but still simple example? Thank you. - -REPLY [4 votes]: This is an example I learned only recently from one of my professors (I don't know much about probability, so I'm not sure whether this is an old hat): -We can use Stone-Weierstrass to prove the Kolmogorov extension theorem (or a version thereof). Namely, given a collection $X = \{\mu_\alpha\}_{\alpha}$ of probability measures on $[0,1]$, there exists a probability measure $\mu$ on $\prod_{\alpha} [0,1]$ such that $$\mu(\pi_{\alpha_1}^{-1}(A_1) \cap \dots\cap \pi_{\alpha_n}^{-1}(A_n) ) = \prod_{i=1}^n \mu_{\alpha_i}(A_i)$$ -for all $\alpha_1, \dots, \alpha_n$, where $\pi_\alpha: X \to [0,1]$ denotes projection. -Sketch of proof: The set $Q \subset C(X)$ of continuous maps, which depend only on "finitely many components", i.e. those $f\in C(X)$ which can be written in the form $$f(x) = g(\pi_{\alpha_1}(x), \dots, \pi_{\alpha_n}(x)) \qquad \text{ for some }\alpha_1, \dots, \alpha_n,\, n\in \mathbb N \text{ and } g\in C([0,1]^n)$$ -is a unital algebra which separates points. By Tychonoff's theorem we know that $X$ is compact, so $Q$ is dense by Stone-Weierstrass. -We can now define a continuous linear functional $I: C(X) \to \mathbb R$ as follows: Given $f(x) = g(\pi_{\alpha_1}(x), \dots, \pi_{\alpha_n}(x)) \in Q$ as above, we define -$$I(f) = \int_{[0,1]^n} g(x_1, \dots, x_n) \, dx_1 \dots dx_n$$ -Then $I$ is well-defined and is a bounded linear functional on the dense subspace $Q\subset C(X)$. Therefore it can be extended uniquely to a bounded linear functional on all of $C(X)$. -Finally, Riesz' representation theorem for compact Hausdorff spaces shows that to every such functional $I$ there corresponds a unique Radon measure $\mu$ such that $I(f) = \int_X f\, d\mu$. -It is now not hard to show that this $\mu$ satisfies $\mu(\pi_{\alpha_1}^{-1}(A_1) \cap \dots\cap \pi_{\alpha_n}^{-1}(A_n) ) = \prod_{i=1}^n \mu_{\alpha_i}(A_i)$ by finding a suitable sequence of continuous functions approximating the indicator function of $\pi_{\alpha_1}^{-1}(A_1) \cap \dots\cap \pi_{\alpha_n}^{-1}(A_n) $. $\square$ -I had seen other proofs of this extension theorem (which I didn't like too much, because I couldn't easily see what's going on), so I was really amazed when I first heard of the above argument. I especially like how various big theorems suddenly pop up and fit together so well! =)<|endoftext|> -TITLE: Why convolution of integrable function $f$ with some sequence tends to $f$ a.e. -QUESTION [9 upvotes]: Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be integrable with$\int_{\mathbb{R}}g(x)dx=1$ and $|g(x)| \leq \frac{C}{(1+|x|)^{1+h}}$ for $x \in \mathbb{R} $, where $C, h>0$ are constants. -Let -$g_t(x)=\frac{1}{t} g(\frac{x}{t})$ for $x \in \mathbb{R}$, $t>0$. -I want to show that: -If $f\in L^p$, where $1\leq p\leq \infty$, then $f*g_t(x) \rightarrow f(x)$ a.e. -I have tried in this way: -Let $x\in \mathbb{R}$ be the Lebesgue point of $f$, that is $lim_{r\rightarrow 0} \frac{1}{r} \int_{B(x,r)} |f(y)-f(x)|dx=0$, then -$$ -|f*g_t(x)-f(x)|\leq \int_{\mathbb{R}} g_t(x-y)|f(y)-f(x)|dy =I_1+I_2, -$$ -where -$I_1=\int_{B(x,t)} g_t(x-y)|f(y)-f(x)|dy \leq\frac{1}{t} \int_{B(x,t)} -\frac{C}{(1+\|\frac{x-y}{t}\|)^{1+h}} |f(y)-f(x)|dy $ -$ \leq C\frac{1}{t}\int_{B(x,t)} |f(y)-f(x)|dy \rightarrow 0 \ as \ t \rightarrow 0;$ -$I_2=\int_{\mathbb{R}\setminus B(x,t)} g_t(x-y)|f(y)-f(x)|dy .$ -I don't know how to estimate the integral $I_2$. - -REPLY [7 votes]: Since it is a local result, we may assume without loss of generality that $p=1$. -The Hardy-Littlewood centered maximal function of $f$ is defined as -$$ -f^*(x)=\sup_{r>0}\frac1{2\,r}\int_{x-r}^{x+r}|f(y)|\,dy=\sup_{r>0}\frac1{2\,r}\int_{|y|\le r}|f(x-y)|\,dy. -$$ -It satisfies the weak one-one inequality -$$ -|\{x:f^*(x)>\delta\}|\le\frac{M}{\delta}\|f\|_1\quad\forall\delta>0, -$$ -where $M>0$ is a constant independent of $f$ and $\delta$. We first bound $g_t\ast f(x)$ in terms of $f^*(x)$. -$$\begin{align*} -|g_t\ast f(x)|&\le \frac{C}{t}\int_{|y|\le t}\frac{|f(x-y)|}{(1+y/t)^{1+h}}\,dy+\frac{C}{t}\sum_{k=0}^\infty\int_{2^kt<|y|\le 2^{k+1}t}\frac{|f(x-y)|}{(1+y/t)^{1+h}}\,dy\\ -&\le\frac{C}{t}\int_{|y|\le t}|f(x-y)|\,dy+\frac{C}{t}\sum_{k=0}^\infty\int_{|y|\le 2^{k+1}t}\frac{|f(x-y)|}{(1+2^k)^{1+h}}\,dy\\ -&\le 2\,C\,f^*(x)+C\Bigl(\sum_{k=0}^\infty\frac{2^{k+2}}{(1+2^k)^{1+h}}\Bigr)f^*(x)\\ -&\le K\,f^*(x), -\end{align*}$$ -where -$$ -K=2\,C+C\sum_{k=0}^\infty\frac{2^{k+2}}{(1+2^k)^{1+h}}<\infty. -$$ -Now we mimic the proof of the Lebesgue differentiation theorem. Given $\epsilon>0$ choose a continuous function $\phi$ with compact support such that $\|f-\phi\|_1<\epsilon$. Then -$$ -g_t\ast f(x)-f(x)=g_t\ast(f-\phi)(x)+\bigl(g_t\ast\phi(x)-\phi(x)\bigr)+(\phi(x)-f(x)), -$$ -from where -$$ -|g_t\ast f(x)-f(x)|\le K(f-\phi)^*(x)+|g_t\ast\phi(x)-\phi(x)|+|\phi(x)-f(x)| -$$ -Since $\phi$ is continuous, the middle term converges to $0$ as $t\to0$ for all $x$. Let $\delta>0$. Then -$$ -\{x:\limsup_{t\to0}|g_t\ast f(x)-f(x)|>2\,\delta\}\subset\{x:(f-\phi)^*(x)>\frac\delta{K}\}\cup\{x:|\phi(x)-f(x)|>\delta\}. -$$ -By Chebychev's inequality -$$ -|\{x:|\phi(x)-f(x)|>\delta\}|\le\frac1\delta\|\phi-f\|_1\le\frac\epsilon\delta. -$$ -By the weak $1$-$1$ inequality for the maximal function -$$ -|\{x:(f-\phi)^*(x)>\frac\delta{K}\}|\le\frac{K\,M}\delta\|\phi-f\|_1\le\frac{K\,M\,\epsilon}\delta. -$$ -Finally -$$ -|\{x:\limsup_{t\to0}|g_t\ast f(x)-f(x)|>2\,\delta\}|\le\frac{K\,M+1}{\delta}\,\epsilon. -$$ -Since $\epsilon$ was arbitrary, it follows that -$$ -|\{x:\limsup_{t\to0}|g_t\ast f(x)-f(x)|>2\,\delta\}|=0\quad\forall\delta>0. -$$<|endoftext|> -TITLE: In Taxicab Geometry, what is the solution to d(P, A) = 2 d(P, B) for two points, A and B? -QUESTION [5 upvotes]: Taxicab and Euclidean geometry differ a great deal, due to the modified metric function: -$$d_T(A,B)=|x_a-x_b|+|y_a-y_b|$$ -(Note that this means when measuring distance, it is not the length of the hypotenuse, but the sum of the legs of the same right triangle.) -My Main Problem -In Euclidean geometry, the answer to the question "Find the locus of points $X$ such that: $d(X, A) = 2 * d(X, B)$" yields a regular, Euclidean circle. A little bit of algebra makes this very trivial. -But what is the answer to the same problem, but for $d_T$? -What I Know So Far -This kind of geometry actually has a very interesting property, namely that as things rotate, their measures change. Consider the cases where points share either one of their coordinates. Many times, those situations yield the same answers as do their Euclidean counterparts. -Some things are noticeably different, though. For instance, a circle, as defined as the set of points a fixed distance from one point, actually comes out as a square, rotated 45 degrees. It is also trivial to illustrate that. -It did occur to me that the answer to this problem could be analogous to Euclidean geometry, and the solution may simply be a Taxicab circle (a square). But this didn't seem to work out. Plus, I worked out the solution for the points sharing an x or y coordinate, I end up with two mirror-image line segments. But the general case, where the two points are corners of any rectangle still eludes me. My second educated guess was that the solution could be a Euclidean circle, but this didn't work out either. -Lastly, some constructions seemed to differ depending on whether the points I chose formed the opposite diagonal corners of a general rectangle, or a square. E.g. (0,0) and (3, 3) seem to be a yet different type of exception. -Any thoughts on this problem would be greatly appreciated! - -REPLY [7 votes]: EDIT: alright, got it. There is a special case, when $AB$ are both on a horizontal or a vertical line. Then the shape asked for is a convex kite shape, two orthogonal edges of equal length with slopes $\pm 1,$ the other two edges with slopes out of $3,-3, \frac{1}{3},\frac{-1}{3}, $ the whole figure symmetric across the $AB$ line, as one would expect. Otherwise, draw the rectangle with $AB$ as the endpoints of one diagonal. If this rectangle is a square, or the longer edge is no more than twice the length of the other, the shape is an isosceles trapezoid, as described below. If the longer edge is more than twice as long as the other, the shape is a nonconvex hexagon, as in Rahul's image. Furthermore, if we fix the longer edge and call it length $1,$ as the shorter edge goes to length $0$ the resulting hexagon shape comes to resemble the kite shape, which is its limit. Anyway, there are just the three possibilities. Note that, in the cases when $AB$ are not in a line parallel to the $x$ or $y$ axis, we can color the nine regions of the tic-tac-toe board as a chess or checkerboard; then regions the same color as the rectangle have segments (if any) of slope $\pm 1,$ while any segments in the regions that are the other color have slope among $3,-3, \frac{1}{3},\frac{-1}{3}. $ It is all pretty rigid. -ORIGINAL: A little to add to what Ross and Rahul pointed out. One segment is automatically there, the segment inside the rectangle passing through the 2/3 point along the $AB$ diagonal, but with slope $\pm 1,$ and closer to $B.$ This segment is part of a "circle" around $A$ as well as a "circle" around $B.$ There is another such with the same slope, passing through a point we might as well call $A',$ which is the reflection of $A$ along the line $AB,$ the same distance from $B$ as $A$ is but on the other side. This can be the longest edge involved, as it continues through one of Ross's tic-tac-toe regions. -There may also be common "circle" segments rotated $90^\circ$ from those, as in Rahul's diagram. So one ought to check for those first, in the nine regions, any segments with slope $\pm 1$ that are the overlap of a circle around $A$ and circle around $B.$ Rahul has shown that you can get three such segments, and there are automatically at least two, but I do not see how one could get four such. So I think the diagram you are asking about is likely to be either a quadrilateral (an isosceles trapezoid) or a non-convex hexagon, being an isosceles trapezoid with one vertex replaced by an extra triangle, one edge of which is orthogonal to the two parallel edges of the trapezoid part. For that matter, there are really only two genuinely distinct slopes allowed, $\pm 1$ and $3,-3, \frac{1}{3},\frac{-1}{3}. $<|endoftext|> -TITLE: Motivation to understand double dual space -QUESTION [52 upvotes]: I am helping my brother with Linear Algebra. I am not able to motivate him to understand what double dual space is. Is there a nice way of explaining the concept? Thanks for your advices, examples and theories. - -REPLY [11 votes]: This is an old post and the existing answers are good, but I feel like I can add another perspective. It looks fairly abstract at first, but if you're dealing with double duals, you just got to accept some abstraction. (By the way, I vividly remember solving the exercise in my undergrad linear algebra class showing the natural isomorphism $V \simeq V^{**}$. It felt like I had just warped my mind to the highest level of abstraction possible for human beings. The dual of a dual, woah ...) -Let $V$ be our vector space (over a field $K$, say). Let's assume we already understand the dual $V^*$. Then we will, after a short consideration, accept that the map -$$V^* \times V \rightarrow K$$ -$$(l,v) \mapsto l(v)$$ -is bilinear. We call it a bilinear form because its values are just scalars. -Now imagine we have any other vector space $W$ (over the same field) with a bilinear form -$$f:W \times V \rightarrow K.$$ -We want to know how far this $W$ and $f$ are from the "ideal" form $V^* \times V \rightarrow K$ above. Note that for any such $W$ and $f$ we get a map, let's call it "c" for "comparison", -$$c: W \rightarrow V^*$$ -$$w \mapsto f(w, \cdot)$$ -(i.e. the image of $w$ is the linear form in $V^*$ which sends a given $v$ to $f(w,v)$). So how do $W$ and $V^*$ "compare"? - -Easy exercise: $c$ is injective $\Leftrightarrow$ for all $0\neq w \in W$, there is $v\in V$ such that $f(w,v) \neq 0$ -Trickier exercise: $c$ is surjective $\Leftrightarrow$ for all $0\neq v \in V$, there is $w\in W$ such that $f(w,v) \neq 0$. - -Note that the "trickier" part needs that $\dim(V) < \infty$. -So the vector space $W$ and the bilinear form $f$ are "ideal" (more technical term: "perfect pairing") if and only if the bilinear form is non-degenerate (which just summarises the criteria on the RHS of the above exercises). -But now look at that non-degeneracy criterion: It is symmetric in the first and second component. In vague terms, $W$ is the dual of $V$ iff that form is non-degenerate; but if that form is non-degenerate, then so is the form that you just get from flipping the components; which means that then $V$ is also the dual of $W$. I.e. $V = V^{**}$. -This whole thinking of duality in terms of non-degenerate bilinear forms, as abstract as it seems on first encounter, becomes incredibly helpful in some applications. E.g. in representation theory it's very neat to translate all the time between self-duality of certain representations and existence of certain $G$-invariant forms on them. Or, a generalisation of this kind of thinking basically leads to adjunction of tensor products and Hom, and all that wonderful general algebra stuff, I recently used that e.g. here to motivate the definition of certain Lie algebra actions.<|endoftext|> -TITLE: Does an arbitrary function preserve arbitrary intersections? -QUESTION [5 upvotes]: Let $S,T$ be sets and let $\Lambda$ be an index-set. Let $f:S\to T$ and $\{A_\lambda\}_{\lambda\in\Lambda}$ be a collection of subsets in $S$ (i.e. $A\lambda\subset S$ for all $\lambda\in\Lambda$). Does $f$ satisfy $$f\left[\bigcap_{\lambda\in\Lambda} A_\lambda\right]\subset\bigcap_{\lambda\in\Lambda} f[A_\lambda]\text{ and }f\left[\bigcup_{\lambda\in\Lambda} A_\lambda\right]=\bigcup_{\lambda\in\Lambda} f[A_\lambda]?$$ - -REPLY [4 votes]: Sure; the arguments are exactly the same as for the finite case. -Suppose that $x\in\bigcap_{\lambda\in\Lambda}A_\lambda$; then $x\in A_\lambda$ for each $\lambda\in\Lambda$, so $f(x)\in f[A_\lambda]$ for each $\lambda\in\Lambda$, and therefore $f(x)\in\bigcap_{\lambda\in\Lambda}f[A_\lambda]$. It follows immediately that -$$f\left[\bigcap_{\lambda\in\Lambda}A_\lambda\right]\subseteq\bigcap_{\lambda\in\Lambda}f[A_\lambda]\;.$$ -Now suppose that $x\in\bigcup_{\lambda\in\Lambda}A_\lambda$; then there is some $\lambda_0\in\Lambda$ such that $x\in A_{\lambda_0}$. Clearly $f(x)\in f[A_{\lambda_0}]\subseteq\bigcup_{\lambda\in\Lambda}A_\lambda$, so -$$f\left[\bigcup_{\lambda\in\Lambda}A_\lambda\right]\subseteq\bigcup_{\lambda\in\Lambda}f[A_\lambda]\;.$$ -Conversely, if $y\in\bigcup_{\lambda\in\Lambda}f[A_\lambda]$, then $y\in f[A_{\lambda_0}]$ for some $\lambda_0\in\Lambda$, and therefore there is an $x\in A_{\lambda_0}$ such that $f(x)=y$. Clearly $x\in\bigcup_{\lambda\in\Lambda}A_\lambda$, so $f(x)\in f\left[\bigcup_{\lambda\in\Lambda}A_\lambda\right]$, and it follows that -$$f\left[\bigcup_{\lambda\in\Lambda}A_\lambda\right]\supseteq\bigcup_{\lambda\in\Lambda}f[A_\lambda]$$ and hence that -$$f\left[\bigcup_{\lambda\in\Lambda}A_\lambda\right]=\bigcup_{\lambda\in\Lambda}f[A_\lambda]\;.$$<|endoftext|> -TITLE: Is $10^{8}!$ greater than $10^{10^9}$? -QUESTION [9 upvotes]: My question is: -$10^8! > 10^{10^9}$ ? -I know that factorial is greater than exponential, but I am not sure about this specific case. -Thanks, - -REPLY [26 votes]: If we expand $10^8!$, we have the product of $10^8$ positive numbers, all of which are less than or equal to $10^8$. So $10^8!<(10^8)^{10^8}=10^{8\cdot 10^8}<10^{10^9}$. - -REPLY [5 votes]: By Stirling’s approximation, -$$\begin{align*} -10^8!&\approx\sqrt{2\cdot10^8\pi}\left(\frac{10^8}e\right)^{10^8}\\ -&<10^5\cdot\left(10^8\right)^{10^8}\\ -&<\left(10^8\right)^{10^8+1}\\ -&=10^{8\cdot10^8+8}\\ -&<10^{10^9}\;. -\end{align*}$$ -Alternatively, -$$\begin{align*} -\ln 10^8!&=\sum_{k=1}^{10^8}\ln k\\ -&<\int_1^{10^8}\ln x\,dx\\ -&=\left[x\ln x-x\right]_1^{10^8}\\ -&=8\cdot10^8\ln 10-10^8+1\\ -&<10^9\ln 10\\ -&=\ln 10^{10^9}\;, -\end{align*}$$ -and $f(x)=\ln x$ is strictly increasing, so $10^8!<10^{10^9}$. - -REPLY [5 votes]: $10^8!=\prod \limits_{n=1}^{10^8}n<\prod \limits_{n=1}^{10^8}10^8=(10^8)^{10^8}=10^{8\times10^8}<10^{10\times10^8}=(10)^{10^9}$ - -REPLY [4 votes]: If we use Stirling's approximation, $10^8! \approx (10^8)^{(10^8)}e^{-10^8}=10^{8\cdot10^8}e^{-10^8}\lt 10^{10^9}$ by a factor of about $(100e)^{10^8}$ For numbers of this size, $\sqrt {2\pi10^8}$ is negligible.<|endoftext|> -TITLE: Convergence of $\sum_{n=1}^{\infty }\frac{a_{n}}{1+na_{n}}$? -QUESTION [9 upvotes]: I need to prove the convergence/divergence of the series $\sum_{n=1}^{\infty }\frac{a_{n}}{1+na_{n}}$ based on the convergence/divergence of the series $\sum_{n=1}^{\infty }a_{n}$. It is given that $a_{n}> 0$, $\forall n\in \mathbb{N}$ -If the series $\sum_{n=1}^{\infty }a_{n}$ is convergent, then from $\frac{a_{n}}{1+na_{n}}< a_{n}$ and the comparison test, we conclude that $\sum_{n=1}^{\infty }\frac{a_{n}}{1+na_{n}}$ is convergent. However, if the series $\sum_{n=1}^{\infty }a_{n}$ is divergent, I have no idea how to prove the convergence/divergence of $\sum_{n=1}^{\infty }\frac{a_{n}}{1+na_{n}}$. It is definitely divergent (just take $a_{n}=\frac{1}{n}$), but I have no clue how to prove it. - -REPLY [5 votes]: Another easy example for the convergence of $ \sum_{n=1}^{\infty}\dfrac{a_{n}}{1+na_{n}} $ even though $ \sum_{n=1}^{\infty}a_{n} $ is divergent: -Define $ a_n $ as follows. -For each $ n\in \mathbb{N} $, -\begin{equation*} - a_{n} = \begin{cases} - 1 & \text{; if } n=k^2 \text{ for some }k\in \mathbb{N}\text{ (i.e. n is a square number) }\\\\ - 0 & \text{; otherwise }.\\ -\end{cases} -\end{equation*} -Then obviously $$ \sum_{n=1}^{\infty}a_{n} \text{ is divergent.}$$ -Notice that $$ \sum_{n=1}^{\infty}\dfrac{a_{n}}{1+na_{n}}= \sum_{k=1}^{\infty}\dfrac{1}{1+k^2}\leq \sum_{k=1}^{\infty}\dfrac{1}{k^2}. $$Therefore $$ \sum_{n=1}^{\infty}\dfrac{a_{n}}{1+na_{n}}\text{ is convergent.}$$<|endoftext|> -TITLE: The probabilty of a new arrangement of 52 cards deck? -QUESTION [7 upvotes]: I just read this article, it claims that if you just shuffle a 52 card deck, you will mostly be creating an arrangement that no human have ever seen before. -But this doesn't seem right since every time we create a new arrangement, this one is now a candidate to happen again, so every time we fail, the odds increase. -How can we mathematically review this claim? - -REPLY [17 votes]: The number of arrangements is $52!$. Using a tool like Wolfram Alpha, or by using the Stirling approximation,, or even just a scientific calculator, we find that $52!$ is larger than $8\times 10^{67}$. -Assume, outrageously, that there have been $10$ billion people on Earth for $10000$ years, each shuffling and dealing a deck of cards every second. In $10000$ years, there are fewer than $3.2\times 10^{11}$ seconds. Multiplying by $10$ billion gets us $3.2\times 10^{21}$ shuffled decks, far short of $8\times 10^{67}$, which is a seriously large number. So a very tiny fraction of the possible hands has been dealt. -It is now probably intuitively reasonable that if all orderings are equally likely, then with probability close to $1$, all orderings have been different. -But the intuition can be unreliable (witness the Birthday Problem). So we make a more detailed estimate. It turns out that the relevant fact is that the square of $3.2\times 10^{21}$, though huge, is a very tiny fraction of the number of possible deals. -Mathematical details: Suppose that there are $N$ different arrangements of cards, and that we shuffle and deal the cards out independently $n$ times, where $n$ is smaller than $N$. Then the probability that all the results are different is -$$\frac{N(N-1)(N-2)\cdots(N-n+1)}{N^n}.$$ -The top is bigger than $N-n$, so the probability is bigger than -$$\left(1-\frac{n}{N}\right)^n.$$ -Using the fact that $\left(1-\frac{x}{k}\right)^k$ is approximately $e^{-x}$, we find that the probability the results are all different is $\gt e^{-n^2/N}$. -Let $N=52!$, and let $n$ be our absurdly high number of $3.2\times 10^{21}$ shuffles. If $x$ is close to $0$, then $e^{-x}$ is approximately $1-x$. Thus the probability the shuffles are all different is greater than $1-\frac{n^2}{N}$. (A better estimate gives that it is approximately $1-\frac{n^2}{2N}$.) This is a number which is very close to $1$. The probability there has been one or more repetition, even with $n$ grossly overestimated, is less than $10^{-25}$. -The probability of at least one match grows rapidly as $n$ grows. Already at $n=10^{33}$, it is roughly $6\times 10^{-3}$, not large but certainly not negligible.<|endoftext|> -TITLE: How to find all naturals $n$ such that $\sqrt{1 {\underbrace{4\cdots4}_{n\text{ times}}}}$ is an integer? -QUESTION [10 upvotes]: How to find all naturals $n$ such that $\sqrt{1\smash{\underbrace{4\cdots4}_{n\text{ times}}}}$ is an integer? - -REPLY [7 votes]: For $n \geq 4$ your number is equal to $4444$ modulo $10000$, and in particular modulo $16$. If it were a square, then $4444$ would be a square modulo $16$, implying $1111$ is a square modulo $4$. But $1111=3$ mod $4$, contradiction.<|endoftext|> -TITLE: Time complexity to calculate a digit in a decimal -QUESTION [8 upvotes]: As we know, it is quiet fast to calculate any digit in a rational number. For example, if I'm given 1/7 (0.142857 142857 ...) and any integer K, I could easily return the Kth digit of 1/7, by doing a integral modular. -But this kind of easy things won't happen on some other numbers, like PI. (this wiki page http://en.wikipedia.org/wiki/Calculating_pi#Digit_extraction_methods says that it costs O(n^2) to calculate the Nth digit in PI) -After that observation I make a rudimentary assertions (if we assume that modular operation is constant time complexity): - -For any rational number (and its decimal form), it costs constant time to calculate its any digit; -Conversely, given a number, if it costs constant time to calculate its any digit, it is a rational number. - -I have tried and I think it's close to prove the first one, but I don't have any idea about the second. Have anyone got any idea about that? Or have I missed any articles describing this kind of problems? -Moreover, if we call the time complexity to calculate a digit in a decimal as its digit time complexity (seems we need a better name), we could partition all real numbers by its digit time complexity. Then how to calculate the cardinal for each subset? - -REPLY [3 votes]: This is basically the Hartmanis-Stearns Conjecture, and it is a major open question. The basic difference between your question and the Conjecture is basically a formalization of "constant time". See -http://rjlipton.wordpress.com/2012/06/04/transcendental-aspects-of-turing-machines/ and -http://rjlipton.wordpress.com/2012/06/15/why-the-hartmanis-stearns-conjecture-is-still-open/ for lots of discussion on the topic. -In a nutshell, the Hartmanis-Stearns Conjecture considers the complexity of computing the next digit in the decimal expansion. Rather than taking $n$ as an input and outputting the $n$-th digit (as you propose), the machine prints the decimal expansion forever. The catch is that it must print the digits at a constant rate. This is known as being computable in "real time". -The HS Conjecture says that any such number is either rational or transcendental. - -Let's take a look at your two claims. - -For any rational number (and its decimal form), it costs constant time to calculate its any digit; - -The intuition behind this is that any rational number's decimal expansion is eventually periodic. In other words, there is a fixed prefix, followed by a repeating pattern forever. So to output the $n$th digit, we have two cases. If $n$ is within the initial prefix, we output that digit (from a pre-computed table). Otherwise, we compute $n \bmod p$ and return that element of the repeating pattern. -Unfortunately, this is not technically "constant time". Both checking if $n$ is within the prefix and computing $n\bmod p$ require time $O(\log n)$ since you have to read every bit. -In the HS version, the machine can remember its location between digits. It therefore only has to increment a pointer -- either move to the next spot in the pre-computed table or move to the next location in the repeating pattern. These are both actually constant time operations. - -Conversely, given a number, if it costs constant time to calculate its any digit, it is a rational number. - -Again, we run into technicalities of what exactly constant-time means. Under many reasonable encoding schemes for $n$ (ie, little-endian binary) this does indeed imply rationality. -Other reasonable encodings (ie, big-endian binary) yield transcendental numbers. Consider the number whose $n$-th binary digit is 0 iff the second bit in $n$ (written in big-endian binary) is 0. This number is clearly not periodic, since the runs keep increasing in length, so it's not rational. In fact, it's actually transcendental. - -Note that the Hartmanis-Stearns conjecture does not claim that every transcendental number can be printed in real time. I don't have a counter-example handy (please leave a comment if you do!) -Another issue is that the HS conjecture allows for extremely-complicated calculations for some digits, provided that they are preceded by a sufficiently-long trivial stretch. Converting the big-endian example above into the real-time formalism uses this trick, for example. However, if you view the theorem as a negative statement about computing irrationals, then the loophole actually strengthens the conjecture (since more potential algorithms are legal). -Another way of formalizing constant time is the word-model. In this model, basic integer operations ($+$, $-$, $<$, $=$, etc) are considered constant time. I don't know what the status of the conjecture is in that model. (Pointers greatly appreciated!)<|endoftext|> -TITLE: Why is the product of all units of a finite field equal to $-1$? -QUESTION [15 upvotes]: Suppose $F=\{0,a_1,\dots,a_{q-1}\}$ is a finite field with $q=p^n$ elements. I'm curious, why is the product of all elements of $F^\ast$ equal to $-1$? I know that $F^\ast$ is cyclic, say generated by $a$. Then the product in question can be written as -$$ -a_1\cdots a_{q-1}=1\cdot a\cdot a^2\cdots a^{q-2}=a^{(q-2)(q-1)/2}. -$$ -However, I'm having trouble jumping from that product to $-1$. What is the key observation to make here? Thanks! - -REPLY [2 votes]: The $q-1$ elements of $F^*$ are roots of the polynomial $x^{q-1}-1$. Hence, -$$-1 = \prod_{\alpha \in F^{*}} (-\alpha) -= (-1)^{q-1}\prod_{\alpha \in F^{*}} \alpha -\Rightarrow \prod_{\alpha \in F^{*}} \alpha = (-1)^q -= \begin{cases}-1,&q ~ \text{odd},\\ -+1,&q ~ \text{even},\end{cases}$$ -but, as André reminded you, $-1 = +1$ if the characteristic of the field is $2$, -and so we need not make special cases but just write that -$\displaystyle \prod_{\alpha \in F^{*}} \alpha = -1$.<|endoftext|> -TITLE: Prove that , any primitive root $r$ of $p^n$ is also a primitive root of $p$ -QUESTION [15 upvotes]: For an odd prime $p$, prove that any primitive root $r$ of $p^n$ is also a primitive root of $p$ - -So I have assumed $r$ have order $k$ modulo $p$ , So $k|p-1$.Then if I am able to show that $p-1|k$ then I am done .But I haven't been able to show that.Can anybody help me this method?Any other type of prove is also welcomed. - -REPLY [2 votes]: Exercise: If $X$ is a generating set for a group $G$ and $\phi:G\to H$ a homomorphism, then $\phi(X)$ is a generating set for the image of $G$ under $\phi$; in particular this means that if $\phi$ is surjective/onto then $\phi(X)$ is a generating set for $H$. -Now consider $G=(\Bbb Z/p^n\Bbb Z)^\times$ and $H=(\Bbb Z/p\Bbb Z)^\times$ with $\phi$ the mod $p$ map (see that it's well-defined and onto!) with a singleton generating set $X=\{a\}$ for $G$.<|endoftext|> -TITLE: What is the rotation axis and rotation angle of the composition of two rotation matrix in $\mathbb{R}^{3}$ -QUESTION [6 upvotes]: I was told in class that a rotation matrix is defined by a rotation -angle and rotation axis, if we call the rotation axis $v$ and take -a basis of $\mathbb{R}^{3}=\{v\}\bigoplus\{v\}^{\perp}$ then the -matrix is similar by an orthogonal matrix to a matrix of the form -$$\begin{pmatrix}\cos\theta & -\sin\theta\\ -\sin\theta & \cos\theta\\ - & & 1 -\end{pmatrix}$$ -I asked my self the following question: If I rotate in the $xy$ plain -(i.e. rotation axis is $z$) in angle $\theta$, and then rotate in -the $yz$ plain (i.e. rotation axis is $x$) in angle $\varphi$ -, what rotation matrix I get ? -I tried multiplying the corresponding matrices but that did not produce -anything useful, I can't also thing of a vector $v\in\mathbb{R}^{3}$that -is invariant under the composition... -What is the rotation axis, and the rotation angle of these two compositions -? Help is appreciated! - -REPLY [2 votes]: I assume you want the quick-and-ready answer Gibbs was teaching his maritime-inclined spherical-law-of-cosines Yale students in the 1880s. With the advent of Pauli matrices for the fundamental rep of SU(2), this reduced to the standard group composition law of SU(2) , albeit the final answer is always a mess in terms of the original variables. -A rotation by angle θ around an axis $\hat{e}$ is represented by a Gibbs vector $\vec{f}=\hat{e} \tan (\theta/2)$. So, for your first rotation, $ \vec{f}=\hat{z} \tan (\theta/2)$, and for your second one, $\vec{g}=\hat{x} \tan (\phi/2)$. -The composition of the two rotations then amounts to -$$ -\frac{\vec{f}+\vec{g}-\vec{f}\times\vec{g}}{1-\vec{f}\cdot \vec{g}} ~. -$$ -In your specific case, the dot product in the denominator vanishes, while the cross product in the numerator is in the pure y direction, which might illustrate why the eigenvector with eigenvalue 1 of your 3×3 matrix product appeared messy. - -In any case, (Olinde Rodrigues, 1840), the axis boils down to the (un-normalized!) half-angle vector -$$ -\hat{x} \sin\phi/2 \cos \theta/2+\hat{z} \cos\phi/2 \sin \theta/2 -\hat{y}\sin \phi/2 \sin\theta/2 . -$$ - -Nevertheless, the combined rotation angle $\gamma$ is much simpler, as you see from the Pauli-matrix WP expression, namely a degenerate spherical law of cosines (the spherical analog of the Pythagorean theorem), -$$ -\cos \gamma/2 = \cos\theta/2 \cos \phi/2 , -$$ -basically admiralty stuff. Cf. this answer, and this chapter. - -To sum up, your new rotation axis is, indeed, left unchanged by the succesion of your two rotations, -$$\begin{pmatrix} 1 & \\ -&\cos\phi & -\sin\phi\\ - &\sin \phi & \cos \phi -\end{pmatrix} \begin{pmatrix}\cos\theta & -\sin\theta\\ -\sin\theta & \cos\theta\\ - & & 1 -\end{pmatrix} \begin{bmatrix}\cos\theta/2 \sin\phi/2\\ --\sin\theta/2 \sin\phi/2\\ - \sin\theta/2 \cos\phi/2 -\end{bmatrix} =\begin{bmatrix}\cos\theta/2 \sin\phi/2\\ --\sin\theta/2 \sin\phi/2\\ - \sin\theta/2 \cos\phi/2 -\end{bmatrix} .$$ - -Reality check: Take $\theta=\phi=\pi/2$ and watch $(1,-1,1)\mapsto (1,1,1)\mapsto(1,-1,1)$ in two steps.<|endoftext|> -TITLE: Are countable intersections of convex sets convex? -QUESTION [6 upvotes]: Let $X$ be a Banach space $\{C_n\colon n\in\mathbb N\}$ a collection ofconvex sets in $X$. Is the set $$C=\bigcap_{n\in\mathbb N}C_n$$ convex? - -REPLY [2 votes]: In fact, any intersection of convex sets in this setting is convex. Suppose that $\mathcal{E}$ is a Banach space and that $\mathcal{K}$ is any family of convex sets. If $\bigcap\mathcal{K}$ is void, it is vacuously convex. Otherwise, take $x,y\in\bigcap\mathcal{K}$, $\lambda\in[0,1]$ and $K\in \mathcal{K}$. Since $K$ is convex, we have -$$(1-\lambda)x + \lambda y\in K.$$ -But this holds for any $K\in \mathcal{K}$, so -$$(1-\lambda)x + \lambda y\in \bigcap \mathcal{K}.$$ -There is no induction or sequential property here. In fact, the argument generalizes to any real vector space.<|endoftext|> -TITLE: Is there a simple method to prove that the square of the Sorgenfrey line is not normal? -QUESTION [6 upvotes]: Is there a simple method to prove that the square of the Sorgenfrey line is not normal? -The method in the book is a little complex. -Could someone help me? - -REPLY [11 votes]: I always use Jones' lemma. It's a handy tool to show non-normality of other spaces as well. You need some basic facts: - -Suppose $X$ is normal. For every pair $A$, $B$ of closed disjoint non-empty subsets in $X$, there is a continuous function $f: X \rightarrow [0,1]$ such that $f(x) = 0$ for $x \in A$ and $f(x) = 1$ for $x \in B$. This is often called Urysohn's lemma. - -If $f,g: X \rightarrow Y$ are continuous, and $Y$ is Hausdorff, and for some dense subset $D$ of $X$ we have $f(x) = g(x)$ for all $x \in D$, then $f(x) = g(x)$ for all $x \in X$. (Proof sketch: if not for some $x$, pull back disjoint open neighbourhoods of $f(x)$ and $g(x)$, both of these intersect $D$ and $f$ and $g$ cannot agree on those points.) -This implies: - - -2') The function $R$ that maps a continuous function $f$ from $X$ to $Y$ to a -continuous function $R(f)$ from $D$ to $Y$ by restricting $f$ to $D$, is 1-1. -Now, -Jones' Lemma: If $X$ is normal and $D$ is dense and infinite in $X$ and $C$ is closed and discrete (in the subspace topology) in $X$ then (as cardinal numbers) $2^{|C|} \le 2^{|D|}$. -Proof: for every non-trivial subset $A$ of $C$, $A$ and $C \setminus A$ are disjoint, closed in $X$ (both are closed in $C$, as $C$ is discrete, and closed subsets of a closed set are closed in the large set.), so by 1) there is a continuous function $f_A$ on $X$ that maps $A$ to $0$ and $C \setminus A$ to $1$. -Note that this defines a family of distinct continuous functions (if $A \neq B$ then we -can find a point in $A\setminus B$ or $B \setminus A$ that shows that $f_A \neq f_B$) from -$X$ to $[0,1]$. But from 2' we know that there is a 1-1 mapping from the set of all continuous functions from $X$ to $[0,1]$ to the set of all continuous functions from $D$ to $[0,1]$ and the latter set is bounded in size by $[0,1]^D = (2^{|N|})^{D} = 2^{|N||D|} = 2^{|D|}$, and the last step holds as $D$ is infinite. -As we have a family of size $2^{|C|}$ (all non-trivial, i.e. non-empty, non-$C$, subsets of $C$) we conclude that $2^{|C|} \le 2^{|D|}$, and this concludes the proof. -Applications: -a) The Sorgenfrey plane: using the antidiagonal $C = \{(x, -x): x \in \mathbb{R} \}$ and $D = \mathbb{Q} \times \mathbb{Q}$ as dense subset. As $2^{|C|} = 2^\mathfrak{c} > \mathfrak{c} = 2^{|D|}$, Jones' lemma says that $X$ cannot be normal. -b) The Niemytzki plane (or Moore plane) (see e.g. here) is not normal, with a similar computation, using $C$ the $x$-axis and $D$ the rational points in the upper halfplane.<|endoftext|> -TITLE: Do we have always $f(A \cap B) = f(A) \cap f(B)$? -QUESTION [10 upvotes]: Suppose $A$ and $B$ are subsets of a topological space and $f$ is any function from $X$ to another topological space $Y$. Do we have always $f(A \cap B) = f(A) \cap f(B)$? -Thanks in advance - -REPLY [6 votes]: I don't know about topological spaces. But given $f:X\rightarrow Y$ and $A,B \subseteq X$ here's an attempt for proving that $$ f \: \text{injective} \rightarrow f(A \cap B) = f(A) \cap f(B),$$ where -$$ (f(A \cap B) = f(A) \cap f(B)) -\leftrightarrow -( -\underbrace{f(A \cap B) \subseteq f(A) \cap f(B)}_{(i)} -) -\wedge -( -\underbrace{f(A \cap B) \supseteq f(A) \cap f(B)}_{(ii)} -). -$$ -(i) (contrapositive) $y \notin f(A) \cap f(B)$ is equivalent to saying that $y \notin f(A)$ and $y \notin f(B)$. Given $y \notin f(A)$ it is also not in $f(A \cap B)$, since $f(A) \supseteq f(A \cap B).$ Accordingly, $y \notin f(B)$ requires that $y \notin f(A \cap B).$ -(ii) $y \in f(A)$ and $y \in f(B)$ implies, since $(\forall x,z \in X) f(x) = f(z) \rightarrow x = z$, that $y \in \{ f(x) : x \in A \wedge x \in B \}$ , or $y \in f(A \cap B)$. In short -$$ (y \in f(A) \wedge y \in f(B) \rightarrow y \in f(A \cap B)) \leftrightarrow f(A \cap B) \supseteq f(A) \cap f(B). $$ $\Box$ - - -See also this page.<|endoftext|> -TITLE: About the inverse of the Jacobian matrix -QUESTION [6 upvotes]: I have a doubt on Jacobian matrices. Consider the non linear transformation -$$ -\left[ -\begin{array}{c} -x\\ -y\\ -z -\end{array}\right] - = \mathbf{G}\left( - \left[ -\begin{array}{c} -\hat{x}\\ -\hat{y}\\ -\hat{z} -\end{array}\right] -\right) = - \left[ -\begin{array}{c} -\hat{x}g(\hat{z})\\ -\hat{y}g(\hat{z})\\ -\hat{z} -\end{array}\right] -$$ -whose Jacobian reads -$$ -\text{J} = -\left[ -\begin{array}{ccc} -g & 0 & \hat{x}g'\\ -0 & g & \hat{y}g'\\ -0 & 0 & 1 -\end{array} -\right] -$$ -If I invert this matrix I get -$$ -\text{J}^{-1} = -\left[ -\begin{array}{ccc} -1/g & 0 & -\hat{x}g'/g\\ -0 & 1/g & -\hat{y}g'/g\\ -0 & 0 & 1 -\end{array} -\right] -$$ -which I thought should be the same as the Jacobian of the inverse transformation. However, solving for $\hat{x}, \hat{y}, \hat{z}$ in the definition of the transformation, I get -$$ -\left[ -\begin{array}{c} -\hat{x}\\ -\hat{y}\\ -\hat{z} -\end{array}\right] - = \mathbf{G}^{-1}\left( - \left[ -\begin{array}{c} -x\\ -y\\ -z -\end{array}\right] -\right) = - \left[ -\begin{array}{c} -x/g(z)\\ -y/g(z)\\ -z -\end{array}\right] -$$ -whose Jacobian now reads -$$ -\text{J}^{-1} = -\left[ -\begin{array}{ccc} -1/g & 0 & -\hat{x}g'/g^2\\ -0 & 1/g & -\hat{y}g'/g^2\\ -0 & 0 & 1 -\end{array} -\right] -$$ -which is slightly different. My question is: which one is the correct Jacobian for the inverse? Weren't they supposed to be the same? If so, where's my mistake? -Thank you in advance! - -REPLY [6 votes]: $G$ maps a point $p$ to $G(p)$. The Jacobian maps a tangent vector at $p$ to one at $G(p)$. The inverse is the Jacobian for $G^{-1}$ at $G(p)$. So, in the second formula you should substitute $x g(z)$ for $x$, $yg(z)$ for $y$, and $z$ for $z$ to recover the first. All consistent (modulo the typo I mentioned in the comment), well done!<|endoftext|> -TITLE: Inclusion of $\mathbb{L}^p$ spaces, reloaded -QUESTION [6 upvotes]: I have a follow-up from this question. It was proved that, if $X$ is a linear subspace of $\mathbb{L}^1 (\mathbb{R})$ such that: - -$X$ is closed in $\mathbb{L}^1 (\mathbb{R})$; -$X \subset \bigcup_{p > 1} \mathbb{L}^p (\mathbb{R})$, - -then $X \subset \mathbb{L}^p (\mathbb{R})$ for some $p>1$. -I was wondering whether one could find a subspace $X$ satisfying these hypotheses and which is infinite-dimensional. It turns out this is possible. If one chooses a bump function, and considers the closure for the $\mathbb{L}^1 (\mathbb{R})$ norm of the space generated by the translates by integers of this bump function, one can emulate the $\ell^1$ space. The resulting $X$ will be closed, and included in $\mathbb{L}^p (\mathbb{R})$ for all $p>0$. To avoid this phenomenon, I'll restrict myself to smaller spaces. - -Is there a linear, closed, infinite-dimensional subspace $X$ of $\mathbb{L}^1 ([0,1])$ which is included in $\mathbb{L}^p ([0,1])$ for some $p>1$? - -The problem is that any obvious choice of countable basis will very easily generate all of $\mathbb{L}^1 ([0,1])$ (polynomials, trigonometric polynomials...), or $\mathbb{L}^1 (A)$ for some $A \subset [0,1]$, or at least one function which is in $\mathbb{L}^1$ but not in $\mathbb{L}^p$ for $p>1$... - -REPLY [3 votes]: There exists such space. Let $\{R_i\}$ be the set of Rademacher functions on $[0,1]$. Specifically, let -$$R_k(x) = sign(\sin(\pi k x)).$$ -For every $\alpha = (\alpha_1, \dots, \alpha_k, \dots) \in \ell_2$, let $f_{\alpha}(x) = \sum \alpha_k R_k(x)$. Note that by the Khintchine inequality, for every $p\in [1,\infty)$, -$$A_p \|\alpha\|_2 \leq \|f_{\alpha}\|_p \leq B_p \|\alpha\|_2$$ -(where $A_p$ and $B_p$ depend only on $p$). In particular, $\|f_{\alpha}\|_1 < \infty$ for all $\alpha\in \ell_2$. Therefore, $f_{\alpha} \in L_1[0,1]$ for every $\alpha \in \ell_2$. Consider $V = \{f_{\alpha}: \alpha\in \ell_2\}$. -Clearly, $V$ is a linear subspace of $L_1[0,1]$. Moreover, $V$ is a closed subspace. Indeed if a sequence $f_{\alpha^{(1)}},f_{\alpha^{(2)}}, f_{\alpha^{(3)}}, \dots$ converges to $g\in L_1[0,1]$, then $f_{\alpha^{(k)}}$ is a Cauchy sequence in $L_1[0,1]$. Therefore, by the Khinchine Inequality, $\alpha^{(k)}$ is a Cauchy sequence in $\ell_2$. Hence it converges to some $\alpha^* \in \ell_2$. Then $f_{\alpha^{(k)}} \to f_{\alpha^*}$, and $g = f_{\alpha^*}$ (a.e.). -The Khintchine inequality implies that $V$ is a subspace of every $L_p[0,1]$ (where $1\leq p < \infty$).<|endoftext|> -TITLE: Convention on non-negative singular values? -QUESTION [6 upvotes]: In the literature I have on disposal it is stated that singular values are non-negative values, and that, for a symmetric matrix $A$, the SVD and EVD coincide. This would mean that singular values of $A$ are the eigenvalues of $A$, but the eigenvalues of $A$ can be negative, regardless of $A$ being symmetric. -So, I wonder if the choice of singular values being exclusively positive is some kind of convention? If so, how degenerate that is given the above observation the equivalence of SVD and EVD for symmetric matrices? - -REPLY [3 votes]: You can factor a (not necessarily square) matrix as orthogonal times diagonal times orthogonal, and the diagonal entries need not all be non-negative. But multiplying a row or a column of an orthogonal matrix by $-1$ still gives an orthogonal matrix, and you can do that and change a minus to a plus in the diagonal matrix. In that way, the two orthogonal matrices can be chosen so that the diagonal entries in that matrix are all non-negative. Those are what are taken to be the singular values. -It's a convention to define it that way. But I suspect there are theorems that say that's the only way to define it that makes it have specified nice properties, and those theorems would not be mere conventions.<|endoftext|> -TITLE: "Sum equals integral" identities similar to $\int_0^1 \frac{dx}{x^x} = \sum_{n = 1}^{\infty} \frac{1}{n^n}$ -QUESTION [28 upvotes]: It is quite a well known fact that: -$$\int_0^{+\infty} \frac{\sin x}{x} \, dx = \frac{\pi}{2}$$ -also the value of related series is very similiar: -$$\sum_{n = 1}^{+\infty} \frac{\sin n}{n} = \frac{\pi - 1}{2}$$ -Combining these two identities and using ${\rm sinc}$ function we get: -$$\int_{-\infty}^{+\infty} {\rm sinc}\, x \, dx = \sum_{n = -\infty}^{+\infty} {\rm sinc}\, n = \pi$$ -What is more interesting is the fact that the equality: -$$\int_{-\infty}^{+\infty} {\rm sinc}^k\, x \, dx = \sum_{n = -\infty}^{+\infty} {\rm sinc}^k\, n$$ -holds for $k = 1,2,\ldots, 6$. There are some other nice identities with ${\rm sinc}$ where sum equals integral but moving on to other functions we have e.g.: -$$\sum_{n = -\infty}^{+\infty} \binom{\alpha}{n} e^{int} = \int_{-\infty}^{+\infty} \binom{\alpha}{n} e^{itx} \, dx = (1+e^{it})^\alpha, \; \alpha >-1$$which is due to Pollard & Shisha. -And finally the identity which is related to the famous Sophomore's Dream: -$$\int_0^1 \frac{dx}{x^x} = \sum_{n = 1}^{+\infty} \frac{1}{n^n}$$ -Unfortunately in this case the summation range is not even close to the interval of integration. -Do you know any other interesting identities which show that "sum = integral"? - -REPLY [15 votes]: Several papers are dedicated to the subject of integrals of functions that equal the sum of the same function, primarily for estimation purposes. -Boas and Pollard (1973) has some interesting sum-integral equalities: -$$\pi/\alpha=\sum_{n=-\infty}^\infty \frac{\sin^2 (c+n)\alpha}{(c+n)^2}=\int_{-\infty}^\infty \frac{\sin^2 (c+n)\alpha}{(c+n)^2}\, \text{d}n$$ -$$\pi\operatorname{sgn} a=\sum_{n=-\infty}^\infty \frac{\sin (n+c)\alpha}{n+c}=\int_{-\infty}^\infty \frac{\sin (n+c)\alpha}{n+c}\, \text{d}n$$ -It also gives several general formulae for functions that suffice: -$$\sum_{n=-\infty}^\infty f(n)=\int_{-\infty}^\infty f(n) \, \text{d}n$$ -mainly with Fourier analysis. - -This paper gives an equality with the Bessel J function: -$$\int_{-\infty}^\infty \frac{J_y (at) J_y(bt)}{t}\, \text{d}t=\sum_{t=-\infty}^\infty \frac{J_y (at) J_y(bt)}{t}$$ -and some more references: - -There have been a number of studies of this kind of sum-integral -equality by various groups, for example, Krishnan & Bhatia in the -1940s (Bhatia & Krishnan 1948; Krishnan 1948a,b; Simon 2002) and Boas, -Pollard & Shisha in the 1970s (Boas & Stutz 1971; Pollard & Shisha -1972; Boas & Pollard 1973). - - -See also Surprising Sinc Sums and Integrals which has some other equalities. This paper also states that (paraphrasing) -If $G$ is of bounded variation on $[−\delta, \delta]$, vanishes outside $(−α, α)$, is Lebesgue integrable over $(−α, α)$ with $0 < α < 2\pi$ and has a Fourier transform of $g$, then -$$\sum_{n=-\infty}^\infty g(n)=\int_{-\infty}^\infty g(x)\, \text{d}x+\sqrt{\frac{\pi}{2}}(G(0-)+G(0+))$$ - -Ramanujan's second lost notebook contains some sums of functions that equal the integral of their functions (Chapter 14, entries 5(i), 5(ii), 16(i), 16(ii)). - -If you want, even more references with examples are in the papers I have mentioned.<|endoftext|> -TITLE: Infinite Sum Axioms in Tohoku -QUESTION [6 upvotes]: In his Tohoku paper, section 1.5, Grothendieck states the following axioms that an abelian category might satisfy: -AB4)Infinite sums exist, and the direct sum of monomorphisms is a monomorphism. -AB5)Infinite sums exist, and the and if $A_i$ (indices in some possibly infinite set $I$) is a filtrated family of subsets of some object A in the category, and B another subset of A, then $(\sum A_i)\cap B = \sum (A_i\cap B)$ -(A subset is what I am translating sous-truc as meaning... I am not sure if this is the correct English notation for this notion.) -Grothendieck states that AB5 is stronger than AB4, without proof. I cannot prove it myself; can someone enlighten me as to why this is true? - -REPLY [5 votes]: So let us suppose we have a family of monomorphisms $A_i \to B_i$, where $i$ varies over an indexing set $I$. Let $A = \bigoplus_i A_i$ and $B = \bigoplus_i B_i$; we want to show that $A \to B$ is also a monomorphism. Let $K$ be the kernel, so that we have an exact sequence -$$0 \longrightarrow K \longrightarrow A \longrightarrow B$$ -Let $\mathcal{J}$ be the system of all finite subsets of $I$: this is a filtered poset, and if we define $A_j = \bigoplus_{i \in j} A_i $ and $B_j = \bigoplus_{i \in j} B_i$ for each finite subset $j$ of $I$, we get a filtered system. In any abelian category, given a diagram -$$\begin{alignedat}{3} -0 \longrightarrow \mathord{} & K_j & \mathord{} \longrightarrow \mathord{} & A_j & \mathord{} \longrightarrow \mathord{} & B_j \\ -& \downarrow && \downarrow && \downarrow \\ -0 \longrightarrow \mathord{} & K & \mathord{} \longrightarrow \mathord{} & A & \mathord{} \longrightarrow \mathord{} & B \\ -\end{alignedat}$$ -if the rightmost vertical arrow is a monomorphism and both two rows are exact, then the left square is a pullback square; since $A_j \to A$ is also a monomorphism, this amounts to saying that $K_j = A_j \cap K$. Since $j$ is a finite set, $A_j \to B_j$ is automatically a monomorphism, so $K_j = 0$. Taking the filtered colimit over $\mathcal{J}$, we obtain -$$\sum_j K_j = \left( \sum_j A_j \right) \cap K = A \cap K = K$$ -so $K = 0$. Thus, $A \to B$ is a monomorphism. - -In fact, AB5 is strictly stronger than AB4. Take $\mathcal{A} = \textbf{Ab}^\textrm{op}$. The opposite of an abelian category is an abelian category, and it is not hard to show that $\mathcal{A}$ satisfies AB3 and AB4; but $\mathcal{A}$ does not satisfy AB5. This is the same example used by Grothendieck in his paper.<|endoftext|> -TITLE: Counting zero-digits between 1 and 1 million -QUESTION [7 upvotes]: I just remembered a problem I read years ago but never found an answer: - -Find how many 0-digits exist in natural numbers between 1 and 1 million. - -I am a programmer, so a quick brute-force would easily give me the answer, but I am more interested in a pen-and-paper solution. - -REPLY [10 votes]: Just to show there is more than one way to do it: -How many zero digits are there in all six-digit numbers? The first digit is never zero, but if we pool all of the non-first digits together, no value occurs more often than the others, so exactly one-tenth of them will be zeroes. There are $9\cdot 5\cdot 10^5$ such digits all of all, and a tenth of them is $9\cdot 5 \cdot 10^4$. -Repeating that reasoning for each possible length, the number of zero digits we find between $1$ and $999999$ inclusive is -$$\sum_{n=2}^6 9(n-1)10^{n-2} = 9\cdot 54321 = 488889 $$ -To that we may (depending on how we interpret "between" in the problem statement) need to add the 6 zeroes from 1,000,000 itself, giving a total of 488,895. - -REPLY [7 votes]: This solution might appeal to a programmer's mind. Let's say we zero-pad (to $6$ digits) the million numbers from $0$ to $999999$. Then each of the $6$ positions uses each digit equally, for a total of $6000000/10 = 600000$ zeros. -Now note that compared to the actual question, we included $000000$, and excluded $1000000$, which balances out nicely. We just need to undo the zero-padding: there are $9$ positive one-digit numbers which were padded by 5, $90$ two-digit numbers padded by $4$, etc. What remains is -$$600000 - 5\cdot9 - 4\cdot90 - 3\cdot900 - 2\cdot9000 - 1\cdot90000 = 488895.$$<|endoftext|> -TITLE: Geometric or intuitive proof of the symmetry of second partial derivatives -QUESTION [9 upvotes]: What was given in my calc book is a "consider the function" proof. That is, the author gives a function out of the blue and would deduce all the nice properties from it. I'd prefer a proof which is motivated (perhaps, intuitive) - you see how the proof is crafted in the mind of the person. So my question is a geometric or intuitive proof of -$$\frac{\partial ^2 f}{\partial x \, \partial y} = \frac{\partial^2 f}{\partial y \, \partial x}$$ - -REPLY [8 votes]: One intuitive explanation for this is the following simple geometrical fact: if you first shift the $xy$-plane to the right and then shift it up, you end up with the same result as if you had first shifted everything up and then to the right. -Using some fancy mathematical language: the equation $\frac{\partial}{\partial x}\frac{\partial}{\partial y}=\frac{\partial}{\partial y}\frac{\partial}{\partial x}$ says that the standard vector field in $x$ direction commutes with the standard vector field in $y$ direction (their Lie bracket is zero). One can prove that this is the same as saying that their flows commute which is the statement I gave above.<|endoftext|> -TITLE: Does $\sum_{n\ge1} \sin (\pi \sqrt{n^2+1}) $ converge/diverge? -QUESTION [14 upvotes]: How would you prove convergence/divergence of the following series? -$$\sum_{n\ge1} \sin (\pi \sqrt{n^2+1}) $$ -I'm interested in more ways of proving convergence/divergence for this series. Thanks. -EDIT -I'm going to post the solution I've found here: -$$a_{n}= \sin (\pi \sqrt{n^2+1})=\sin (\pi (\sqrt{n^2+1}-n)+n\pi)=(-1)^n \sin (\pi (\sqrt{n^2+1}-n))=$$ -$$ (-1)^n \sin \frac{\pi}{\sqrt{n^2+1}+n}$$ -The sequence $b_{n} = \sin \frac{\pi}{\sqrt{n^2+1}+n}$ monotonically decreases to $0$. Since our series is an alternating series then it converges. - -REPLY [14 votes]: $$ -\sum_{n\ge1} \sin\left(\pi\sqrt{n^2+1}\right) = \sum_{n\ge1} \pm\sin\left(\pi\left(\sqrt{n^2+1}-n\right)\right) -$$ -(Trigonometric identity. Later we'll worry about "$\pm$".) -Now -$$ -\sqrt{n^2+1}-n = \frac{1}{\sqrt{n^2+1}+n} -$$ -by rationalizing the numerator. -So we have the sum of terms whose absolute values are -$$ -\left|\sin\left(\frac{\pi}{\sqrt{n^2+1}+n}\right) \right| \le \frac{\pi}{\sqrt{n^2+1}+n} \to0\text{ as }n\to\infty.\tag{1} -$$ -But the signs alternate and the terms decrease in size, so this converges. (They decrease in size because sine is an increasing function near $0$ and the sequence inside the sine decreases.) -It does not converge absolutely, since $\sin x\ge x/2$ for $x$ small and positive, and the sum of the terms asserted to approach $0$ in $(1)$ above diverges to $\infty$. - -REPLY [12 votes]: It converges as an alternating series. For each $n$, we have $\sin((n+\delta)\cdot\pi)$ for some small $\delta$ which approaches $0$ at the limit and decreases monotonically (as the $1$ in $\sqrt{n^2+1}$ becomes less significant compared to the $n^2$. -For even $n$, this expression will take on smaller and smaller positive values, as $\delta$ shrinks and $(n+\delta)\cdot\pi$ gets closer and closer to a zero at $2m\pi$ for some natural $m$ from the right. -Similarly, it will take on shrinking negative values for odd $n$. -The absolute value of each term tends to $0$ and decreases monotonically, so we have convergence by the alternating series test.<|endoftext|> -TITLE: Can the order of learning be changed? -QUESTION [12 upvotes]: I have been advised by many people to learn from scratch. So I decided to learn it. But I have the following questions in my mind. - -Can we skip "the Trinity" and learn something else directly? ("the Trinity" refers to analysis, topology and algebra) -Why are these things so important and why are they taught for 3 years? -Are there any books that give a integrated introduction about this subject and take us to the advanced level? -What are the most important concepts that one needs to master in them? I saw many mathematical problems that are applied, like finding integrals etc. Can they be skipped? - -Thanks a lot. - -REPLY [2 votes]: I have come to realize there is a marked difference between a math/scientific education and a liberal arts education which might be applicable to your question. In a liberal arts education, you can hop on the train at any station. There is nothing stopping you from looking at the great problems of Shakespeare long before you confront Chaucer. In many ways this dynamic devalues the liberal education. -Whereas in math, science or technical fields, there is a cumulative progression.This imputes a sense of rigor into the dynamic, as many topics assumes an explicit familiarity with prerequisites - and they mean it. Granted there is something very satisfying in studying, e.g., a number theory text rather than a text with a litany of theorems on commutative algebra. That is, you get to use something you know more so than accumulating theorems. -But without a sound foundation, you're not in the game. And will at some point wish you had cultivated more capacity when you had the energy and opportunity to do so.<|endoftext|> -TITLE: Separated schemes and unicity of extension -QUESTION [10 upvotes]: In point set topology, we have the following result, which is easily proved. -Theorem. Let $Y$ be Hausdorff space and $f,g:X \to Y$ be continuous functions. If there exists a set $A\subset X$ such that $\bar{A} = X$ and $f|_A = g|_A$, then $f=g$. -I was trying to understand what would be the natural generalization of this fact in the category of schemes. We know that the correct analogous of a Hausdorff space is a separated scheme. So I was thinking in a statement like this: -"Let $Y$ be a separated scheme and $f,g:X \to Y$ be morphisms of schemes. If there exists a set $A\subset X$ such that $\bar{A} = X$ and $f|_A = g|_A$, then $f=g$." -The first problem is that we must be careful with this restriction "$f|_A$", since $f$ is a morphism and I want to consider the morphism of sheaves also. Then I saw that Liu's book on Algebraic Geometry has the following statement: -"Let $Y$ be a separated scheme, $X$ a reduced scheme, and $f,g:X \to Y$ morphisms of schemes. If there exists a dense open subset $U$ such that $f|_U=g|_U$, then $f=g$." -Now this makes sense, since we are dealing with open subsets now. But I still find this result too restrictive. So I came up with this: -"Let $Y$ be a separated scheme, $X$ a reduced scheme, and $f,g:X \to Y$ morphisms of schemes. If there exists a morphism $\varphi:S \to X$ such that $\varphi(S)$ is dense in $X$ and $f\circ \varphi = g\circ \varphi$, then $f=g$." -It's easy to see that Liu's proof of the result concerning only the open set also applies to this context. Finally, let's go to the questions: - -Is this really the best generalization? Is there any other results in this direction that are at least slightly different? -I can see where the "reduced" hypothesis enters in the proof, but I found it a little strange. Is it just a technical point or can be "understanded" in some sense? Maybe counterexamples of this fact when this hypothesis isn't valid would help to clarify , but I didn't think of any. - -P.S. Sorry for the bad english. - -REPLY [3 votes]: To answer your question (1): Let $X=\mathrm{Spec}(B)$ be such that $B\to O_X(U)$ is not injective for some dense open subset $U$ (this can't happen if $B$ is reduced), let $Y=\mathrm{Spec}\mathbb Z[t]$. Fix an element $b\in B$ non-zero such that $b|_U=0$. -Let $\varphi: \mathbb Z[t] \to B$ be any ring homomorphism and let $\psi : \mathbb Z[t] \to B$ be defined by $\psi(t)=\varphi(t)+b$. Then the corresponding morphisms $f, g : X\to Y$ coincide on $U$ but are not equal. -Standard example of such $B$: $B=k[x,y]/(x^2, xy)$ and $U=X\setminus \{ (0,0)\}=D(y)$. -For your question (1), you can replace the hypothesis $X$ reduced by $S\to X$ schematically dominant. When $X$ is noetherian, this means that the image of $S\to X$ contains the associated points of $X$ (= maximal points of $X$ if the latter is reduced).<|endoftext|> -TITLE: Is every algebraic curve birational to a planar curve -QUESTION [13 upvotes]: Let $X$ be an algebraic curve over an algebraically closed field $k$. -Does there exist a polynomial $f\in k[x,y]$ such that $X$ is birational to the curve $\{f(x,y)=0\}$? -I think I can prove this using Noether Normalization Lemma. -Is this correct? If yes, is it too much? That is, is there an easier argument? - -REPLY [8 votes]: The result holds over an arbitrary ground field $k$ if we assume that the curve is geometrically regular: i.e., if $X \otimes_k \overline{k}$ is regular, or equivalently if the extension $k(X)/k$ is separable, as in QiL's comment above. I believe that QiL's proof works here verbatim. Also Qiaochu Yuan's argument works, as there is a separable Noether normalization theorem: the fraction field of a geometrically regular variety can be written as a finite separable extension of a rational function field. (See Corollary 16.18 in Eisenbud's text on commutative algebra.) -Note that it is often possible to prove more: in $\S 5$ of these notes, I give a (detailed) sketch of a proof that any geometrically regular curve $C$ over an infinite field is birational to a plane curve with only ordinary double points as singularities. Essentially I follow Hartshorne's proof and explain why the hypothesis therein that "$k$ is algebraically closed" can be replaced by "$k$ is infinite". -This stronger conclusion however need not hold over a finite field: if a curve $C$ can be "immersed" in $\mathbb{P}^2$ with only double point singularities, then $\# C(\mathbb{F}_q) \leq 2 \# \mathbb{P}^2(\mathbb{F}_q)$. But a curve over a finite field can have arbitrarily many $\mathbb{F}_q$-rational points.<|endoftext|> -TITLE: convergence of alternating series — weakening a hypothesis -QUESTION [6 upvotes]: A comment below this answer inspires this question. -Suppose $a_n\in\mathbb{R}$ for $n=1,2,3,\ldots$ and $|a_n|\to0$ as $n\to\infty$. -Further suppose the terms alternate in sign. -If moreover the sequence $\{|a_n|\}_{n=1}^\infty$ is decreasing, then $\displaystyle\sum_{n=1}^\infty a_n$ converges. -How much can the hypothesis that it is decreasing be weakened while still being strong enough that the sum must converge? And are there any interesting or useful weaker hypotheses? - -REPLY [3 votes]: Here is one of my favorite counter-examples for when you don't assume that the sequence $(|a_n|)$ is decreasing. Let us put, for all $n \geq 2$, -$$a_n := \ln \left( 1 + \frac{(-1)^n}{\sqrt{n}} \right).$$ -This sequence converges to $0$, and is alternating. However, since \ln (1+x) = $x-x^2/2 + O (x^3)$, we get: -$$a_n := \frac{(-1)^n}{\sqrt{n}} - \frac{1}{2} \left( \frac{(-1)^n}{\sqrt{n}} \right)^2 + O (n^{-\frac{3}{2}}) = \frac{(-1)^n}{\sqrt{n}} - \frac{1}{2n} + O (n^{-\frac{3}{2}}).$$ -The series whose general term is $\frac{(-1)^n}{\sqrt{n}}$ is convergent, since it is alternating. The $O (n^{-\frac{3}{2}})$ term is summable, by comparison with Riemann sums. What is left is $\frac{1}{2n}$, whose corresponding series is divergent. Hence, $\sum_{k=0}^{n-1} a_k$ diverges to $- \infty$. -More generally, "alternating but not summable" + "non-negative sequence, which decays faster but is still not summable" gives a sequence which is equivalent to the initial alternating sequence, but whose sum does not converges. Something like $\frac{(-1)^n}{\sqrt{n}} + \frac{1}{n}$ is a typical example (but I prefer the sequence $(a_n)$, where the trap is concealed - it shows that you have to be careful).<|endoftext|> -TITLE: Why is the inradius of any triangle at most half its circumradius? -QUESTION [11 upvotes]: Is there any geometrically simple reason why the inradius of a triangle should be at most half its circumradius? I end up wanting the fact for this answer. -I know of two proofs of this fact. -Proof 1: -The radius of the nine-point circle is half the circumradius. Feuerbach's theorem states that the incircle is internally tangent to the nine-point circle, and hence has a smaller radius. -Proof 2: -The Steiner inellipse is the inconic with the largest area. The Steiner circumellipse is the circumconic with the smallest area, and has 4 times the area of the Steiner inellipse. Hence the circumcircle has at least 4 times the area of the incircle. -These both feel kind of sledgehammerish to me; I'd be happier if there were some nice Euclidean-geometry proof (or a way to convince myself that no such thing is likely to exist, so the sledgehammer is necessary). -EDIT for ease of future searching: The internet tells me this is often known as "Euler's triangle inequality." - -REPLY [3 votes]: It's enough to show that $OI^2=R(R-2r) \iff R^2-OI^2=2Rr$, where $O$,$I$,$R$, and $r$ is the circumcenter, incenter, circumradius, and inradius respectively. By power of a point, $R^2-OI^2=AI\times IL$, so it's enough to show that $AI\times IL =2Rr \iff \frac{AI}{r}=\frac{2R}{IL}$. This is trivial since $\triangle AFI \sim \triangle KBL$ and $LI=BL$ (By Incenter-Excenter lemma)<|endoftext|> -TITLE: A sequence with infinitely many radicals: $a_{n}=\sqrt{1+\sqrt{a+\sqrt{a^2+\cdots+\sqrt{a^n}}}}$ -QUESTION [12 upvotes]: Consider the sequence $\{a_{n}\}$, with $n\ge1$ and $a>0$, defined as: -$$a_{n}=\sqrt{1+\sqrt{a+\sqrt{a^2+\cdots+\sqrt{a^n}}}}$$ -I'm trying to prove here 2 things: a). the sequence is convergent; b). the sequence's limit when n goes to $\infty$. I may suppose that there must be a proof for this general case. I saw this problem with the case $a=2$ (where it was required to prove only the convergence), but this is just a particular case. The generalization seems to be much more interesting. - -REPLY [7 votes]: Here is a full answer to part (a) and a partial answer to part (b). Call $(a_n(a))_{n\geqslant1}$ the sequence when the value of the parameter is $a$. -One has $a_0(1)=1$ and $a_{n+1}(1)=u(a_n(1))$ for every $n\geqslant0$ with $u(x)=\sqrt{1+x}$. Hence the usual technique shows that the sequence $(a_n(1))_{n\geqslant0}$ is increasing to $a_\infty(1)=\alpha$ where $\alpha$ solves the equation $\alpha=u(\alpha)$, that is, $\color{red}{\alpha=\frac12(1+\sqrt5)}$. -When $a\lt1$, $a_n(a)\leqslant a_n(1)$ and $(a_n(a))_{n\geqslant0}$ is increasing hence $(a_n(a))_{n\geqslant0}$ converges to a finite limit $a_\infty(a)$ with $\color{red}{\sqrt{1+\sqrt{a}}\lt a_\infty(a)\leqslant \alpha}$. -When $a\gt1$, $\sqrt{1+\sqrt{aa_{n-1}(1)}}\leqslant a_n(a)\leqslant\sqrt{1+\sqrt{a}a_{n-1}(1)}$ and $(a_n(a))_{n\geqslant0}$ is increasing hence it converges to a finite limit $a_\infty(a)$ with $\color{red}{\sqrt{1+\sqrt{\alpha a}}\lt a_\infty(a)\leqslant\sqrt{1+\alpha\sqrt{a}}}$. -To show the upper bound on $a_n(a)$, one carries over every power of $a$ to the left until it reaches the position of $\sqrt{a}$. Crossing a square root sign halves the exponent and $a\gt1$ hence the power of $a$ which just crossed a square root sign is smaller than the preceding one. For example, the first step of the proof uses $a^{n/2}\leqslant a^{n-1}$ to deduce -$$ -\sqrt{a^{n-1}+\sqrt{a^n}}=\sqrt{a^{n-1}+a^{n/2}\sqrt1}\leqslant\sqrt{a^{n-1}(1+\sqrt1)}=a^{(n-1)/2}\sqrt{1+\sqrt1}, -$$ -the second step uses $a^{(n-1)/2}\leqslant a^{n-2}$, and so on, until $a^{3/2}\leqslant a^2$ and $a^{2/2}\leqslant a$. A similar reasoning yields the lower bound. -Finally, the map $a\mapsto a_\infty(a)$ is nondecreasing from $\color{red}{a_\infty(0)=1}$ to $\color{red}{a_\infty(+\infty)=+\infty}$.<|endoftext|> -TITLE: Sobolev Spaces and Weak Derivatives -QUESTION [5 upvotes]: As you can probably guess, I'm currently studying about differential operators and functional analysis. -We've studied the following theorem: - -A function $f \in L^2 (\Omega) $ lies in $ W^{1,2} ( \Omega) $ if and only if there exists a function $g \in L^2 ( \Omega ) $ such that: - $$\int_\Omega f \left\{ b_0 (x) \phi(x) - \sum_{i=1}^n \frac{ \partial(b_i (x) \phi(x)}{\partial x_i} \right\} d^n x = \int_\Omega g(x) \phi(x) d^n x $$ for every choice of functions $b_i (x) \in C^\infty (\bar{\Omega} ) $, and $\phi \in C_c^\infty (\Omega ) $ . - -Can someone help me use this theorem in order to prove that the function $f(x)= \frac{x_1}{|x|} $ is in $W^{1,2}( \{ x \in \mathbb{R} ^n : |x| <1 \} ) $? What should be my $g$ and how can I prove it? -I really need your help ! -Thanks ! - -REPLY [3 votes]: I looked this up in the Davies' book that you are reading. The functions $b_i$ are coefficients of a differential operator $D_b$, and $g$ depends on these coefficients (precisely, $g=D_b f$). Anyway, it would be strange to use this property of $W^{1,2}$ functions here. We just want to show that $f(x)=x_1/|x|$ is in $W^{1,2}(\Omega)$ where $\Omega=\{x:|x|<1\}$. -Here is one approach, not necessarily the quickest but instructive. Let $\phi:[0,\infty)\to[0,1]$ be a $C^\infty$ function such that $\phi(t)=0$ when $t\le 1$, $\phi(t)=1$ when $t\ge 2$, and $-2\le \phi'\le 0$ always. Introduce the smooth approximations $f_k=f\psi_k$ where $\phi_k(x)=\phi(k|x|)$, $n=1,2,\dots$. It is clear that $f_k\to f$ in $L^2(\Omega)$. I claim that $(\nabla f_k)$ is Cauchy in $L^2(\Omega)$. Indeed, for $k -TITLE: A little integration paradox -QUESTION [11 upvotes]: The following integral can be obtained using the online Wolfram integrator: -$$\int \frac{dx}{1+\cos^2 x} = \frac{\tan^{-1}(\frac{\tan x}{\sqrt{2}})}{\sqrt{2}}$$ -Now assume we are performing this integration between $0$ and $2\pi$. Hence the result of the integration is zero. -On the other hand when looking at the integrand, $\displaystyle \frac{1}{1+\cos^2 x}$, we see that it is a periodic function that is never negative. The fact that it is never negative guarantees that the result of the integration will never be zero (i.e., intuitively there is a positive area under the curve). -What is going on here? - -REPLY [3 votes]: If $\tan(y)=a\tan(x)$, then $\tan(y-x)=\frac{(a-1)\tan(x)}{1+a\tan^2(x)}$, and therefore, -$$ -y=x+\tan^{-1}\left(\frac{(a-1)\tan(x)}{1+a\tan^2(x)}\right)\tag{1} -$$ -Since $\left|\frac{(a-1)\tan(x)}{1+a\tan^2(x)}\right|\le\frac{\left|a-1\right|}{2\sqrt{a}}$, $(1)$ is a nice continuous function of $x$. -In this question, $a=\frac1{\sqrt2}$, so we get -$$ -\int\frac{\mathrm{d}x}{1+\cos^2(x)}=\frac{x-\tan^{-1}\left(\frac{(\sqrt2-1)\tan(x)}{\sqrt2+\tan^2(x)}\right)}{\sqrt2}\tag{2} -$$<|endoftext|> -TITLE: The analogous generalization for the commutativity of unions. -QUESTION [6 upvotes]: Let $\{I_j\}$ be a family of sets indexed by $J$ and let $$K=\bigcup_{j\in J}I_j$$ -Then let $\{A_k\}$ be a family of sets indexed by $K$. The generalization of the associative law for unions is that -$$\bigcup_{k\in K}A_k=\bigcup_{j\in J}\left(\bigcup_{i\in I_j}A_i \right)$$ -What I interpret this as is: "To take the union over $K$, pick an $I_j \in K$, perform the union of all $A_i$ such that $i\in I_j$, and for each $j\in J$ unite all this possible unions to get $\bigcup_{k\in K}A_k$. What this is saying is that the order in which the $j$ and thus the $I_j$ are picked is of no importance in the ultimate union. The above is a generalization of $$(A\cup B)\cup C=A\cup (B\cup C)$$ -How can I find the analogous generalization for $$A \cup B=B \cup A?$$ - -REPLY [3 votes]: I'm trying to answer the question from the point of view of someone who has read just the Halmos' book up to the page 35 and doesn't have any other sources of knowledge about the Set Theory. -At first, let's make a more concise understanding of what we are trying to accomplish. It's important because I couldn't figure out the answer on the aforementioned question about the Associative Law by myself if there were no explanations in the related thread there. -So, the unions were introduced by the Axiom of Specification in the Section 4 of Halmos' book: - -$$U = \{x: x \in X \text{ for some $X$ in $\mathfrak{C}$}\}$$ - this set is called the union of the collection $\mathfrak{C}$ of sets - -thus, being sets in the first place, "plain" unions of collections of sets don't have a notion of order in which we pick the unified sets and thus no property of commutativity. -The notion of commutativity appears when Halmos introduced the notation for the special case of a union of a pair of sets: - -$$\bigcup\{X:X \in \{A, B\}\} = A \cup B$$ - -and when we use that notation, it follows from the definition that: -$$ A \cup B = B \cup A $$ -Though we're asked by Halmos to generalize that idea using the notion of families of sets, it seems appealing to stop there and at first try to simply express the very same property of commutativity of pairs of sets using the notion of families of sets. -It doesn't seem to be feasible though as the difference between $A \cup B$ and $B \cup A$ is purely notational. Those two notations correspond to the very same object of a union of pair of sets. -So the commutativity that Halmos is asking as to prove should be something else than the order in which we write the elements of the pair that we are applying the union operation to. -When we read the Section 4 further, we encounter the way to generalize pairs to define (unordered) triples, and, probably, n-tuples like - -$$\{a, b, c\} = \{a\}\cup \{b\}\cup \{c\}$$ - -and it is here where we can find an "order" to apply the notion of "commutativity" to. It seems that we can list the singletons in the r.h.s. in any order and still obtain the same n-tuple in the l.h.s. provided the set of the singletons is the same. The "order" is defined by the left associativity of our $\cup$-notation for unions of pairs. -To concoct the generalized version that we're aiming for I used the example instance of it similar to the on provided by Brian M. Scott in the related thread: -$$(A_0 \cup A_1) \cup A_2 = (A_0 \cup A_2) \cup A_1 $$ -the "generalization" means that this and similar equalities should hold as long as the "set" of the unified terms is $\{A_0, A_1, A_2\}$. -Up to the moment where the question appears there is only one place where "the order" is mentioned in the book which are sets of the form (it relates to the left hand side order of pair-unions in our example, so it is given a name): -$$OrderSet_2 = \{\{2\}, \{2, 1\}, \{2, 1, 0\}\}$$ -which, for instance defines "order" of $2-1-0$. The element which is the member of a singleton-element $\{2\}$ is the first element of the defined "order": $2$, by applying the similar procedure to the set of relative complements of other elements of the $OrderSet_2$ with the singleton: -$$OrderSet_1 = \{\{1\}, \{1, 0\}\}$$ -we obtain the next element of the order: $1$. -$OrderSet_2$ defines the order of commutations in a "reversed" way in a sense that the elements of it which come first correspond to the "outer" unified terms, i.e. the order of -$$2-1-0$$ -corresponds to the order of commutations: -$$(A_0 \cup A_1) \cup A_2$$ -We see that union-of-pair is an operation on two arguments so each union in our generalization will consist of just two terms. We are using the notation of families, so that each union will be represented as a union of a range of a family, hence terms united should correspond to the "terms" (according to the Halmos' terminology in the beginning of the chapter), and if we want to have two terms in each union, the index set of each family should contain two elements. -To start with, the resulting union (in the l.h.s.) $(A_0 \cup A_1) \cup A_2$ which we may denote as $U$ or also as $U_2$: -$$U = U_2 = (A_0 \cup A_1) \cup A_2$$ -is the union of the union denoted as $U_1$: -$$U_1 = \bigcup_{X \in S_1}X \:\text{where}\: S_1 = \{A_0, A_1\}$$ and the set $A_2$ -so -$$U_2 = U_1 \cup A_2$$ -We can express it in a form of a family with the domain $I_2$ that we can construct as consisting of two indices: -$$I_2 = \{i_{A_2}, i_{U_1}\}$$ and the function (family in Halmos' terminology) -$$X_2(i_{A_2}) = A_2 \quad X_2(i_{U_1}) = U_1$$ -as: -$$U_2 = \bigcup_{i \in I_2}{\{{X_2}_i\}}$$ -we can express $U_1$ in a similar way (by an expense of introducing of more notation). -$U_1 = \bigcup_{X \in S_1}X$ where $S_1 = \{A_0, A_1\}$, thus it can be written as the union of the set denoted as $U_0$ -$$U_0 = \bigcup_{X \in S_0}X \:\text{where}\: S_0 = \{A_0, \varnothing \}$$ and the set $A_1$ -so -$$U_1 = U_0 \cup A_1$$ -We can express it in a form of a family with the domain $I_1$ that we can construct as consisting of two indices: -$$I_1 = \{i_{A_1}, i_{U_0}\}$$ and the function -$$X_1(i_{A_1}) = A_1 \quad X_1(i_{U_0}) = U_0$$ -as: -$$U_1 = \bigcup_{i \in I_1}{\{{X_1}_i\}}$$ -Finally, in a similar way, the set $U_0 = \bigcup_{X \in S_0}X \:\text{where}\: S_0 = \{A_0, \varnothing \}$ can be constructed as a family with the domain $$I_0 = \{i_{A_0}, i_{\varnothing}\}$$ and the function -$$X_0(i_{A_0}) = A_0 \quad X_0(i_{\varnothing}) = \varnothing$$ -as: -$$U_0 = \bigcup_{i \in I_0}{\{{X_0}_i\}}$$ -Thus the construction of a family of sets $U = U_2$ seems to be finished. That family corresponded to a particular order of the union of pairs operation defined by the set $OrderSet_2$ in a way it was constructed which can be generalized as follows: -if $j$ is some "order set" which defines an order among the unified sets, -$$U_{j} = \bigcup_{i \in I_j}\{{X_j}_i\}$$ -where -$$I_j = \{{i_U}_j, {i_A}_j\}$$ -where the term $X_j({i_U}_j)$ is either "$U_{j-1}$" which is defined similarly based on the order set obtained by removing the singleton and the corresponding elements from the other sets of $j$ in a way $OrderSet_1$ was obtained from $OrderSet_2$ above, or $\varnothing$ if $j$ is the "last" order set to consider, i.e. is a singleton of a singleton. -the term $X_j({i_A}_j)$ is an element of a singleton of $j$ (the "first" element of a given order). -Using that generalization the statement of generalized commutativity of unions can be formulated like "unions obtained by the aforementioned procedure from any two order sets which define an order of elements of the same set are equal".<|endoftext|> -TITLE: Find a big-O estimate for $f(n)=2f(\sqrt{n})+\log n$ -QUESTION [7 upvotes]: While self-studying Discrete Mathematics, I found the following question in the book "Discrete Mathematics and Its Applications" from Rosen: - -Suppose the function $f$ satisfies the recurrence relation $f(n)=2f(\sqrt{n})+\log n$ whenever $n$ is a perfect square greater than 1 and $f(2)=1$. -Find a big-O estimate for $f(n)$ [Hint: Make the substitution $m=\log n$.] - -(Note: here, $\log n$ stands for base 2 logarithm of $n$.) -In this solution, I am supposed to use the following variant of the Master Theorem: - -Master Theorem. Let $f$ be an increasing function that satisfies the recurrence relation -$$f(n)=af(n/b)+cn^d$$ -whenever $n=b^k$, where $k$ is a positive integer, $a\geq 1$, $b$ is an integer greater than 1, and $c$ and $d$ are real numbers with $c$ postive and $d$ nonnegative. Then -$$\begin{align}f(n)&\text{is }\begin{cases} O(n^d)&\text{ if $a < b^d$,} \\ O(n^d\log n)&\text{ if $a = b^d$,} \\ O(n^{\log_b a})&\text{ if $a > b^d$.} \end{cases}\end{align}$$ - -I solved it as follows, but I'm not sure if this is correct: -Following the hint given, I made the substitution $m = \log n$. Then, $n=2^m$. Rewriting the recurrence relation with this value for $n$: -$$f(2^m)=2f(\sqrt{2^m})+\log (2^m)\text{ with }f(2)=1$$ -$$f(2^m)=2f(2^{m/2})+m\text{ with }f(2)=1$$ -(because $\sqrt{2^m}=2^{m/2}$ and $\log (2^m)=m$.) -To simplify the analysis, I will rewrite the recurrence relation above for $T(m)=f(2^m)$: -$$T(m)=2T(\dfrac{m}{2})+m\text{ with }T(1)=1$$ -Now I will apply the Master Theorem for $T(m)$. In this case, $d=1$, $b=2$ and $a=2$, this recurrence relation meets the condition $a=b^d$ in the Master Theorem; therefore: -$$T(m)\text{ is }O(m^d\log m)=O(m\log m)$$ -Now I will rewrite the estimate above in terms of $f(n)$, substituting $T(m)=f(2^m)=f(2^{\log n})=f(n)$ and $m=\log n$: -$$f(n)\text{ is } O(\log n\log \log n)$$ -Is this solution correct? If yes, is there any way to simplify it further? - -REPLY [9 votes]: Let's use recursion trees (otherwise known as "the proof of the Master theorem")! - -The root of the recursion tree for $f(n)$ has value $\log n$. -Each child of the root has value $\log \sqrt{n} = (\log n)/2$, so the total value of all nodes at depth $1$ is also $\log n$. -(Sanity check:) Each grandchild of the root has value $\log \sqrt{\sqrt{n}} = (\log n)/4$. So the total value of all $4$ nodes at level $2$ is also $\log n$. -An easy inductive argument implies that for any $\ell$, the total value of the $2^\ell$ nodes at level $\ell$ is $\log n$. -Thus, all level sums are identical, which implies that $f(n) = \Theta(L\log n)$, where $L$ is the number of levels. -Another easy inductive argument implies that each subtree rooted at level $L$ represents the function $f(n^{2^{-L}})$. So the recursion bottoms out at the smallest level $L$ such that $n^{2^{-L}} \le 2$. (There's nothing special about the number 2 here; any constant larger than 1 will do.) We have -$$ -n^{2^{-L}} \le 2 \iff 2^{-L}\log n \le 1 \iff \log n \le 2^L \iff L \le \log\log n. -$$ -We conclude that $f(n) = \Theta(\log n \log \log n)$.<|endoftext|> -TITLE: Egoroff's theorem in Royden Fitzpatrick (comparison with lemma 10) -QUESTION [6 upvotes]: Hi math stackexchangers, -I have a question about the difference between two math statements (for reference they can be found in Royden Fitzpatrick pages 64-65). - -Egoroff's Theorem: Assume $E$ has finite measure. Let $\{f_n\}$ be a sequence of measurable functions on $E$ that converges pointwise on $E$ to the real-valued function $f$. Then for each $\varepsilon > 0$, there is a closed set $F$ contained in $E$ for which - $$\{f_n\} \to f\text{ uniformly on }F\text{ and }m(E \setminus F) < \varepsilon.$$ -Lemma 10: Under the assumptions of Egoroff's Theorem, for each $\eta > 0$ and $\delta > 0$, there is a measurable subset $A$ of $E$ and an index $N$ for which - $$|f_n - f| < \eta\text{ on }A\text{ for all }n \geq N -\text{ and }m(E \setminus A) < \delta.$$ - -My question is this. What is the difference between Egoroff's Theorem and Lemma 10? Am I misunderstanding uniform convergence? To me it looks like Lemma 10 provides uniform convergence. Is the difference that Egoroff's Theorem ensures that $F$ is closed but Lemma 10 doesn't ensure that $A$ is closed? -Thanks - -REPLY [7 votes]: A significant difference was not mentioned in the comments: In the statement of Lemma 10, $A$ depends on $\eta$. Let us forget about closedness for a minute (I'll come back to that later) and compare the two statements: -Egoroff: $$(\forall \varepsilon>0) (\exists A\subseteq E), m(E\setminus A)<\varepsilon \boldsymbol{\large( \forall \eta>0)}$$ $$(\exists N\in\mathbb N)(\forall x\in A)(\forall n\geq N) |f_n(x)-f(x)|<\eta.$$ -Lemma 10: $$(\forall \varepsilon>0) \boldsymbol{\large( \forall \eta>0)} (\exists A\subseteq E), m(E\setminus A)<\varepsilon $$ $$(\exists N\in\mathbb N)(\forall x\in A)(\forall n\geq N) |f_n(x)-f(x)|<\eta.$$ -I know this isn't a pretty way to write it, but I hope it makes clear the important difference between the two. Notice that in Egoroff's theorem, $A$ had to be chosen once and for all to work for all (subsequently chosen) $\eta>0$, whereas in Lemma 10 $A$ only needs to work for a previously fixed $\eta$. -I remember when I was first learning measure theory from Royden doing an exercise showing how Egoroff's theorem can be proved using Lemma 10 (or whatever it was numbered in my edition). It was pretty straightforward because the exercise told me what sequences of $\delta$s and $\eta$s to use. - -As for closedness, note that Egoroff's theorem without the assumption of closedness applies to all finite measure spaces (where there need not even be a topology in general) and the closedness can be added using inner regularity. For example, a subset $A$ of $\mathbb R$ is Lebesgue measurable if and only if for all $\varepsilon>0$ there exists a closed set $F\subseteq A$ such that $m^*(A\setminus F)<\varepsilon$ (where $m^*$ denotes Lebesgue outer measure). Uniform convergence on $A$ implies uniform convergence on F, and $E\setminus F = (E\setminus A)\cup (A\setminus F)$.<|endoftext|> -TITLE: Looking for a 'second course' in logic and set theory (forcing, large cardinals...) -QUESTION [8 upvotes]: I'm a recent graduate and will likely be out of the maths business for now - but there are a few things that I'd still really like to learn about - forcing and large cardinals being two of them. -My background is what one would probably call a 'first graduate course' in logic and set theory (some intro to ZFC, ordinals, cardinals, and computability theory). Can you recommend any books or online lecture notes which are accessible to someone with my previous knowledge? -Thanks a lot! - -REPLY [5 votes]: I would recommend the following as excellent graduate level introductions to set theory, including forcing and large cardinals. - -Thomas Jech, Set Theory. -Aki Kanamori, The higher infinite. See the review I wrote of it for Studia Logica. - -I typically recommend to my graduate students, who often focus on both forcing and large cardinals, that they should read both Jech and Kunen (mentioned in Francis Adams answer) and play these two books off against one another. For numerous topics, Jech will have a high-level explanation that is informative when trying to understand the underlying idea, and Kunen will have a greater level of notational detail that helps one understand the particulars. Meanwhile, Kanamori's book is a great exploration of the large cardinal hierarchy. -I would also recommend posting (and answering) questions on forcing and large cardinals here and also on mathoverflow. Probably most forcing questions belong on mathoverflow.<|endoftext|> -TITLE: Why use the derivative and not the symmetric derivative? -QUESTION [21 upvotes]: The symmetric derivative is always equal to the regular derivative when it exists, and still isn't defined for jump discontinuities. From what I can tell the only differences are that a symmetric derivative will give the 'expected slope' for removable discontinuities, and the average slope at cusps. These seem like extremely reasonable quantities to work with (especially the former), so I'm wondering why the 'typical' derivative isn't taken to be this one. What advantage is there to taking $\lim\limits_{h\to0}\frac{f(x+h)-f(x)} h$ as the main quantity of interest instead? Why would we want to use the one that's defined less often? - -REPLY [9 votes]: Following my comment on the Mean Value Theorem. Since MVT fails, anything we prove from MVT is likely to fail as well. For example: -Find the minimum of the (symetrically) differentiable function -$f(x) = x+2|x|$ on the interval $[-1,1]$. -Usual solution: find where the derivative is zero. Answer: nowhere! -Since $f'(x) = -1$ on $[-1,0)$, $f'(0)=1$, -and $f'(x)=3$ on $(0,1]$.<|endoftext|> -TITLE: Are commutative C*-algebras really dual to locally compact Hausdorff spaces? -QUESTION [55 upvotes]: Several online sources (e.g. Wikipedia, the nLab) assert that the Gelfand representation defines a contravariant equivalence from the category of (non-unital) commutative $C^{\ast}$-algebras to the category of locally compact Hausdorff (LCH) spaces. This seems wrong to me. -The naive choice is to take all continuous maps between LCH spaces. This doesn't work. For example, the constant map $\mathbb{R} \to \bullet$ does not come from a morphism $\mathbb{C} \to C_0(\mathbb{R})$, the problem being that composing with the map $\bullet \to \mathbb{C}$ sending $\bullet$ to $1$ gives a function on $\mathbb{R}$ which doesn't vanish at infinity. It is necessary for us to restrict our attention to proper maps. -But this still doesn't work. If $A, B$ are any commutative $C^{\ast}$-algebras we can consider the morphism -$$A \ni a \mapsto (a, 0) \in A \times B.$$ -This morphism does not define a map on Gelfand spectra; if $\lambda : A \times B \to \mathbb{C}$ is a character factoring through the projection $A \times B \to B$, then composing with the above morphism gives the zero map $A \to \mathbb{C}$. This contradicts the nLab's claim that taking Gelfand spectra gives a functor into locally compact Hausdorff spaces (if one requires that the morphisms are defined everywhere on the latter category). -The correct statement appears to be that commutative $C^{\ast}$-algebras are contravariantly equivalent to the category $\text{CHaus}_{\bullet}$ of pointed compact Hausdorff spaces; the functor takes an algebra to the Gelfand spectrum of its unitization (we adjoin a unit whether or not the algebra already had one). There is an inclusion of the category of LCH spaces and proper maps into this category but it is not an equivalence because maps $(C, \bullet) \to (D, \bullet)$ in $\text{CHaus}_{\bullet}$ may send points other than the distinguished point of $C$ to the distinguished point of $D$. -So do sources mean something else when they claim the equivalence with locally compact Hausdorff spaces? - -REPLY [11 votes]: The following categories are contravariantly equivalent: - -locally compact Hausdorff spaces with proper continuous maps -commutative C$^*$-algebras with non-degenerate $*$-homomorphisms - -Here, a $*$-homomorphism $f : A \to B$ is non-degenerate if the following equivalent conditions are satisfied: - -The ideal generated by the set-theoretic image of $f$ is dense in $B$. -For every approximative unit $(u_i)$ in $A$ its image $f(u_i)$ is an approximative unit in $B$. -For some approximative unit $(u_i)$ in $A$ its image $f(u_i)$ is an approximative unit in $B$. - -I don't think that multiplier algebras are necessary ...<|endoftext|> -TITLE: Clarifying a comment of Serre -QUESTION [12 upvotes]: Let $\rho_{\ell}$ be the "mod $\ell$" Galois representation associated to an elliptic curve $E/K$ (i.e., corresponding to the action of Galois on the $\ell$-torsion points). Serre proved that in the case where the image of Galois is the normalizer of a nonsplit Cartan subgroup, this defines a quadratic extension of $K$ which is actually unramified. -In the course of the proof, he makes the following remark, which I cannot decipher. If $E$ has multiplicative reduction at a prime $v$ not dividing $\ell$, then the theory of Tate curves gives the exact sequence -$$ -0 \rightarrow \mu_{\ell} \rightarrow E_{\ell} \rightarrow \mathbb{Z}/\ell \mathbb{Z} \rightarrow 0, -$$ -which is compatible with the action of the inertia group at $v$, denoted $I_v$. Therefore, the image of $I_v$ under this Galois representations is either trivial or cyclic of order $\ell$. -Now, I see why this must be the case: since $v \nmid \ell$, the inertia acts trivially on $\mu_{\ell}$ (the $\ell$th roots of unity). According to the theory of tate curves, the $\ell$-torsion points are generated by $\mu_{\ell}$ and $q^{1/\ell}$; this is either a degree $\ell$ extension or trivial. -However, this doesn't seem to have anything to do with Serre's exact sequence, and I figure that learning how Serre sees this little fact could be useful. Can someone tell me what he means? - -REPLY [7 votes]: As you write, when $E$ is a Tate curve, the $\ell$-torsion $E[\ell]$ contains -$\mu_{\ell}$ as a Galois-sub-module, with the quotient $E[\ell]/\mu_{\ell}$ -being generated by the image of $q^{1/\ell}$. Note that any Galois conjugate of $q^{1/\ell}$ is equal to $\zeta q^{1/\ell}$ for some $\zeta \in \mu_{\ell}$, and so the Galois action on $q^{1/\ell}$ modulo $\mu_{\ell}$ is trivial. This gives the exact sequence $0 \to \mu_{\ell} \to E[\ell] \to \mathbb Z/\ell \to 0.$<|endoftext|> -TITLE: A isometric map in metric space is surjective? -QUESTION [19 upvotes]: Possible Duplicate: -Isometries of $\mathbb{R}^n$ - -Let $X$ be a compact metric space and $f$ be an isometric map from $X$ to $X$. Prove $f$ is a surjective map. - -REPLY [65 votes]: Here is an alternative to the proof linked to in the comments: -Suppose there existed $x \in X\setminus f(X)$. Then $x$ has positive distance $d$ from the compact set $f(X)$. Now consider the recursively defined sequence $$x_0 := x, \qquad x_n := f(x_{n-1}) \quad \forall \, n>0$$ -We have $d(x_0, x_n)\ge d$ for all $n>0$, by assumption on $x$. This implies that we also have $d(x_k, x_{k+n}) = d(x_0, x_n) \ge d$ for all $k,n>0$ (here we use that $f$ is an isometry). Therefore $d(x_n, x_m) \ge d$ for all $m\ne n$, which is in contradiction to sequential compactness of $X$.<|endoftext|> -TITLE: The Frobenius endomorphism -QUESTION [6 upvotes]: Let $\mathbf F$ be a field of prime characteristic $p$. It is known that the Frobenius map $c\phi=c^p~~\forall c\in\mathbf F$ is an endomorphism of $\mathbf F$. Moreover, since the only ideals of $\mathbf F$ are $\{0\}$ and $\mathbf F$, we know that $\ker(\phi)=\{0\}$. This implies that $\phi$ is injective. It is known that $\phi$ is not surjective in general. However, if we consider the following argument below, it seems to me that $\phi$ must be an automorphism (i.e., a bijective endomorphism). -"We claim that $\mathbf F/\{0\}\cong\mathbf F$. Consider the homomorphism $\theta:\mathbf F/\{0\}\rightarrow\mathbf F$ given by $(\{0\}+r)\theta=r$, where -$r\in\mathbf F$. Then it is easy to show that it is both injective and surjective; and so our claim holds. However, we also know by the First Ring Isomorphism Theorem that $\mathbf F/\ker(\phi)\cong \text{im}(\phi)$. So we conclude that $\text{im}(\phi)\cong\mathbf F$. But $\text{im}(\phi)\subseteq\mathbf F$ so necessarily $\text{im}(\phi)=\mathbf F$. Therefore $\phi$ is an automorphism of $\mathbf F$." -Could someone tell me where the mistake lies? - -REPLY [9 votes]: Your mistake comes in assuming that because $\textrm{im} \phi \subseteq \Bbb{F}$ and $\textrm{im} \phi \cong \Bbb{F}$ then $\textrm{im} \phi = \Bbb{F}$. Here's a good example to consider. I will produce for you an example of two fields $E,H$ one of which is contained in the other with $E \cong H$ (as rings) but $E \neq H$. -Consider $F = \Bbb{Z}/2\Bbb{Z}$ and $t$ an indeterminate. Let $E = F(t)$, $H = F(t^2)$. Then you can see that $H \subseteq E$ and $H \cong E$ (as rings) by the map $f$ that is constant on $F$ and sends $t \mapsto t^{2}$. However clearly $H \neq E$. To see this begin by noticing that $F(t^2)$ is the fraction field of $F[t^2]$. If $t \in F(t^2)$, we have that $\frac{p(t^2)}{q(t^2)} = t$ for some polynomials $p$ and $q$. Then you have $tq(t^2) = p(t^2)$. But then the guy on the left has all odd powers while the guy on the left only even powers, a contradiction.<|endoftext|> -TITLE: Why does $\{1\cdot a\! \! \pmod p, 2\cdot a\! \! \pmod p,\ldots, (p-1)\cdot a\! \! \pmod p\}$ $= \{1, 2,\ldots, p-1\}$ when $a$ and $p$ are coprime? -QUESTION [5 upvotes]: Why is it that $\{1\cdot a \pmod p, 2\cdot a \pmod p,\ldots, (p-1)\cdot a \pmod p\} = \{1, 2,\ldots, p-1\}$ (albeit in a different order) when a and p are coprimes? -I can't figure this out and I've been beating my head for the whole weekend. -Googling around I've found mention of Fermat's Little Theorem (e.g. here), but I can't see how it helps me. -I've verified it by hand, it seems perfectly believable to me (mostly because I find myself thinking of the way the circle of fifths works), but I can't come up with a good proof. -Any help, pretty please? -Thanks a lot. -P.S.: Pardon my English. I'm from the land of pizza and mandolins. - -REPLY [3 votes]: This can be interpreted as a group-theoretic result: the set of numbers coprime to $n$ form a group under multiplication modulo $n$. The hardest part of this is showing that inverses exist. This follows from Bézout's theorem, which states that for any $a$ and $n$ there exist $x$ and $y$ coprime such that $ax + yn = \gcd(a,n)$. From this we can see that if $a$ and $n$ are coprime, i.e. $\gcd(a,n)=1$, then $ax$ is $1$ mod $n$, and $x$ must be coprime to $n$ too (because $\gcd(r,s)$ divides any combination of $r$ and $s$, and here we have a combination of $x$ and $n$ that gives 1), so $a$ has a multiplicative inverse modulo $n$. -Anyway, once you've got that the set of numbers coprime to $n$ form a group under multiplication modulo $n$, that's essentially equivalent to saying the map "multiply by $a$" is a bijection modulo $n$. That's essentially the same as your original statement. -(Technicality: we showed that multiply-by-$a$ is invertible in the group of numbers coprime to $n$ modulo $n$, when really what we wanted was for it to be invertible in the set of numbers modulo $n$, but it's not hard to see that the argument above extends to that case. I suppose what I'm really doing is showing that $a$ is a unit in the ring of numbers modulo $n$, and the group I mentioned is the group of units in that ring). -This answer uses some more heavyweight machinery than necessary, but I think it's a neat way of looking at the result.<|endoftext|> -TITLE: Lie algebra representation induced from homomorphism between spin group and SO(n,n) -QUESTION [5 upvotes]: Consider the spin group, we know it is a double cover with the map: -$\rho: Spin(n,n)\longrightarrow SO(n,n)$ s.t $\rho(x)(v)= xvx^{-1}$ where $v$ is an element of 2n dimensional vector space V and $x$ is an element of spin group (multiplications are Clifford multiplication). I read that this map induces a lie algebra representation given by: -$d\rho:so(n,n) \longrightarrow so(n,n)$ s.t $d \rho_{x}(v)=xv-vx$ here $x$ is an element of $so(n,n)$ and $v$ is again an element of V. -I cannot understand the derivation of this lie algebra representation. Can anyone help me? :) - -REPLY [4 votes]: This basically comes from the product rule of differentiation. Recall that for a general Lie algebra homomorphism $\rho : G\to H$ you can compute its derivative $d\rho: Lie(G) \to Lie(H)$ by the formula -$$ -d\rho(X) = \frac{d}{dt}\vert_{t=0} \rho(\exp(tX)). -$$ -In the case you have, this gives -$$ -d\rho(x) (v) = \frac{d}{dt}\vert_{t=0} \exp(tx) v \exp(-tx), -$$ -where $x \in so(n,n)$, which is identified with the second filtration of the Clifford algebra. Now you can think of the right hand side is taking place in the Clifford algebra and the product rule of differentiation holds so you get -$$ -d\rho(x)(v) = \frac{d}{dt} \vert_{t=0} \exp(tx) v \exp(0x) + \exp(0x) v \frac{d}{dt}\vert_{t=0} \exp(-tx) = xv +v(-x). -$$ -Notice that this formula does not make sense if you think of $so(n,n)$ in terms of $2n \times 2n$ matrices since right multiplying a column vector by a matrix doesn't make sense. The $xv - vx$ is taking place in the Clifford algebra.<|endoftext|> -TITLE: How to evaluate $\int_{-\infty}^\infty \frac{\cos x}{\cosh x}\,\mathrm dx$ by hand -QUESTION [17 upvotes]: How can I evaluate$$\int_{-\infty}^\infty \frac{\cos x}{\cosh x}\,\mathrm dx\text{ and }\int_0^\infty\frac{\sin x}{e^x-1}\,\mathrm dx.$$ -Thanks in advance. - -REPLY [2 votes]: Let us start from the Weierstrass product for the sine function: -$$ \sin(x) = x \prod_{n\geq 1}\left(1-\frac{x^2}{\pi^2 n^2}\right)\tag{1} $$ -This identity over the real line can be proved by exploiting Chebyshev polynomials, but it holds over $\mathbb{C}$ and it is equivalent to -$$ \sinh(x) = x \prod_{n\geq 1}\left(1+\frac{x^2}{n^2\pi^2}\right).\tag{1'} $$ -If we apply $\frac{d}{dx}\log(\cdot)$ (the logarithmic derivative) to both sides of $(1)$ we end up with -$$ \cot(x)-\frac{1}{x} = \sum_{n\geq 1}\frac{2x}{x^2-n^2 \pi^2} \tag{2}$$ -$$ \coth(x)-\frac{1}{x} = \sum_{n\geq 1}\frac{2x}{x^2 + n^2 \pi^2} \tag{2'}$$ -and by expanding the main term of the RHS of $(2)$ as a geometric series we have -$$ \frac{1-x\cot x}{2} = \sum_{n\geq 1}\frac{\frac{x^2}{n^2 \pi^2}}{1-\frac{x^2}{n^2\pi^2}} =\sum_{n\geq 1}\sum_{m\geq 1}\left(\frac{x^2}{n^2 \pi^2}\right)^m = \sum_{m\geq 1}x^{2m}\frac{\zeta(2m)}{\pi^{2m}} \tag{3}$$ -$$ \frac{1-x\coth x}{2} = \sum_{m\geq 1}(-1)^m x^{2m}\frac{\zeta(2m)}{\pi^{2m}}\tag{3'} $$ -$(3')$ alone is enough to prove that $\zeta(2m)$ is always a rational multiple of $\pi^{2m}$, related to a coefficient of the Maclaurin series of $\frac{z}{e^z-1}=-\frac{z}{2}+\frac{z}{2}\coth\frac{z}{2}$, i.e. to a Bernoulli number, but that is not the point here. -Let us deal with the second integral: -$$ \int_{0}^{+\infty}\frac{\sin z}{e^z-1}\,dz = \sum_{m\geq 1}\int_{0}^{+\infty}\sin(z) e^{-mz}\,dz = \sum_{m\geq 1}\frac{1}{m^2+1} $$ -can be evaluated through $(2')$ by picking $x=\pi$. In a similar fashion -$$ \int_{0}^{+\infty}\frac{\cos z}{e^z+e^{-z}}\,dz = \sum_{m\geq 0}(-1)^m\int_{0}^{+\infty}\cos(z)e^{-(2m+1)z}\,dz = \sum_{m\geq 0}(-1)^m \frac{(2m+1)}{(2m+1)^2+1} $$ -can be computed from the Weierstrass product of the cosine function, -$$ \cos(x) = \prod_{m\geq 0}\left(1-\frac{4x^2}{(2m+1)^2\pi^2}\right)\tag{4} $$ -$$ \cosh(x) = \prod_{m\geq 0}\left(1+\frac{4x^2}{(2m+1)^2\pi^2}\right)\tag{4'} $$ -leading by logarithmic differentiation to -$$ \sum_{m\geq 0}\frac{x}{x^2+(2m+1)^2} = \frac{\pi}{4}\tanh\left(\frac{\pi x}{2}\right)\tag{5} $$ -and also to -$$ \sum_{m\geq 0}\frac{(-1)^m(2m+1)}{x^2+(2m+1)^2} = \frac{\pi}{4}\operatorname{sech}\left(\frac{\pi x}{2}\right).\tag{6}$$<|endoftext|> -TITLE: Prove that: $ \int_{0}^{1} \ln \sqrt{\frac{1+\cos x}{1-\sin x}}\le \ln 2$ -QUESTION [11 upvotes]: I plan to prove the following integral inequality: -$$ \int_{0}^{1} \ln \sqrt{\frac{1+\cos x}{1-\sin x}}\le \ln 2$$ -Since we have to deal with a convex function on this interval i thought of considering the area of the trapeze that can be formed if we unify the points $(0, f(0))$ and $(1, f(1))$, where the function $f(x) =\ln \sqrt{\frac{1+\cos x}{1-\sin x}}$, but things are ugly even if the method itself isn't -complicated. So, -I'm looking for something better if possible. - -REPLY [8 votes]: Here's a (hopefully) corrected proof that uses convexity along with the trapezoid rule: You can rewrite what you're trying to prove as -$$ \int_{0}^{1} \ln {\frac{1+\cos x}{1-\sin x}}\,dx\le 2\ln 2$$ -Let $f(x) = \ln {\frac{1+\cos x}{1-\sin x}} = \ln(1 + \cos x) - \ln (1 - \sin x)$. -Then -$$f'(x) = -\frac{\sin x}{1 + \cos x} + \frac{\cos x}{1 - \sin x}$$ -Using the tangent half-angle formula, this is the same as -$$-\tan(x/2) + \tan(x/2 + \pi/4)$$ -Therefore -$$f''(x) = -(1/2)\sec^2(x/2) + 1/2\sec^2(x/2 + \pi/4)$$ -Since $\sec$ is increasing on $(0,1/2 + \pi/4)$, we see that $f''(x) > 0$. So the integrand is convex. When applied to a convex function, the trapezoid rule always gives a result larger than the integral. But already with $2$ pieces, the trapezoid rule here gives -$$1/4(\ln(1 + \cos(0)) - \ln(1 - \sin(0)) + 2(\ln(1 + \cos(1/2)) - \ln(1 - \sin(1/2)))$$ -$$ +\ln(1 + \cos(1)) - \ln(1 - \sin(1)) )$$ -$$= 1.3831395912690787...$$ -This is slightly less than $2\ln2 = 1.3862943611198906...$, so the original integral is less than $\ln 2$ as needed.<|endoftext|> -TITLE: Is trying to prove a theorem without Axiom of Choice useless? -QUESTION [7 upvotes]: Suppose there is a well-known theorem whose usual proof uses Axiom of Choice. -Is trying to prove it without Axiom of Choice useless? -What merits can such a proof have? - -REPLY [37 votes]: It is often not useless at all. First let me give out a few reasons why it is useful: - -It gives us a better understanding of how well-behaved some objects are. For example, vector spaces are not well-behaved in general, but in particular cases they are well-behaved (e.g. finitely generated ones). -On the other hand, compact metric spaces are quite well-behaved and many of their properties remain valid even without the axiom of choice. -It can be the case that a proof without the axiom of choice is a much more constructive one, and allows us to examine the features of an object which was proved only to exist if the axiom of choice were used. -One example coming to mind is the compactness of closed and bounded subsets of $\mathbb R$, another is the uniform continuity of a continuous function from a compact metric space into a metric space. -If a proof fails it allows us a better understanding of how much choice is needed for a certain assertion. Dependent Choice for Baire's Category theorem; Countable Choice for the equivalence between different topological properties; etc. - -On the other hand, it is somewhat useless to try and prove a theorem without the axiom of choice if other parts of the theory already assume it. If you already assume that every vector space has a basis, showing that a certain proposition relying on this property holds without the axiom of choice is moot. -One example for this is the ultrafilter theorem which, together with the Krein-Milman theorem, imply the axiom of choice [4], so if you end up using both these propositions there is no use in avoiding choice anymore. It's there. -Regardless of the above, one should remember that not all things in modern mathematics are true in the absence of choice (that is, aside from the axiom of choice itself) and often reformulation and distinction between equivalent definitions is required. In those cases one can, and should, ask themselves how much choice is needed for a particular result. -For example, the assertion "$\mathbb R$ is not a countable union of countable sets" requires the axiom of choice, but it is provable from a vastly weaker statement, "countable unions of countable sets are countable". - -Further reading: - -Why worry about the axiom of choice? (MathOverflow) -Axiom of choice - to use or not to use -Advantage of accepting the axiom of choice -Is Banach-Alaoglu equivalent to AC? -Motivating implications of the axiom of choice?<|endoftext|> -TITLE: Why algebraic closures? -QUESTION [28 upvotes]: Let me begin by summarizing the question: - -Why do we care about fields closed under rational exponentiation, and less about fields closed under other operations? - -Historically the solution for polynomials was important, and people were trying to find a good way to test when a certain polynomial has a root. This led to talking about algebraically closed fields, where polynomials have roots. I will particularly focus on the rationals from now on. -How do we construct the rational numbers? We begin with $\{0,1\}$ and we say that if we add $1+1+\ldots+1$ we never have $0$. We begin by closing this set under addition, then under subtraction and then under division. -However we can consider the alternative, we iterate from $\{0,1\}$ and at every step we add solutions all four operations on the elements we have thus far, so the construction would go like: - -$\{0,1\}$, we begin. Next we add additive inverse, a term for $1+1$ and multiplicative inverse for $1$: -$\{-1,0,1,2\}$. Now we add the additive inverse for $2$, the addition of $1+2$, and multiplicative inverse for $2$: -$\{-2,-1,0,\frac12,1,2,3\}$. Now we add the sums possible with the elements we have so far, the missing inverses, and so on: -$\{-3,-2,-1\frac12,-1,-\frac12,0,\frac14,\frac13,\frac12,1,1\frac12,2\frac12,3,3\frac12,4,5\}$. We continue ad infinitum. - -It is not very hard to see that any rational number is in this set, and that this set forms a field (indeed the rationals). -Consider now the family of operations $\exp_q(x)=x^q$ defined for $x\geq 0$ and for a rational number $q$. If we reiterate the above algorithm from $\mathbb Q$ and close it under $\exp_q(x)$ for positive $q$ we end up with a subfield of the real-closure of $\mathbb Q$. The result, while not the entire real algebraic numbers, is radically closed. Any number in the field has a root of any rational order. If we only take closure under a limited collection $\exp_q$ functions we will get a subfield of this field (e.g. close only under $\exp_{0.5}$). -Consider now the Sine-closure of $\mathbb Q$: - -$Q_0=\mathbb Q$, we begin with the rationals. -$Q_1=Q_0\cup\{\sin(x)\mid x\in Q_0\}$, the rationals were closed under field operations, so we only need to add $\sin$'s. -$Q_2$ is the collection of all sums and multiplication of pairs from $Q_1$, adding additive and multiplicative inverses, and adding $\sin(x)$ for $x\in Q_1$. -$Q_3$ constructed the same. - -We finish by taking $\mathbb Q_{\sin}=\bigcup_{n=0}^\infty Q_n$. This is a field which extends $\mathbb Q$ and is closed under the function $\sin$. This field contains transcendental elements and is countable, so it is a non-algebraic subfield of $\mathbb R$. - -So why are we mostly interested in algebraically closed fields, and not in fields like the Sine-closure of $\mathbb Q$ (or perhaps other construction similar to this one)? -Is the reason historical, is it because algebra and analysis are somewhat disjoint in their purposes and analysis takes $\mathbb{R,C}$ to begin with? - -REPLY [11 votes]: As far as I can tell, the first field you have constructed is not the real algebraic numbers; you only get real algebraic numbers whose minimal polynomial has solvable Galois group (and you don't even get all those, e.g. because of casus irreducibilis). -Anyway, dealing with transcendental numbers is hard! As far as I know, there is no algorithm for deciding equality of elements of your field (I don't know if this is open or known to be undecidable). Deciding equality of algebraic numbers, on the other hand, is straightforward: compute minimal polynomials, then compute the two numbers you're interested in to sufficiently high precision until their digits start to disagree. -There is also a lot one can say about algebraic numbers, and this is already good enough for many applications (e.g. algebraic closures suffice to give us eigenvalues of linear operators on finite-dimensional vector spaces). One can also talk about the algebraic closure very generally of any field, whereas operations like $\sin x$ are specific to characteristic $0$ and require the existence of a topology relative to which the defining series converges. -The natural objects appearing in mathematics closed under lots of operations like $\sin x$ are usually not fields (although model theorists are interested in exponential fields), but rather things like the smooth algebra $C^{\infty}(M)$ of smooth functions on a smooth manifold $M$ or Banach algebras. The keyword here is "functional calculus," as in holomorphic functional calculus.<|endoftext|> -TITLE: Proof that operator is compact -QUESTION [8 upvotes]: Prove that the operator $T:\ell^1\rightarrow\ell^1$ which maps $x=(x_1,x_2,\dots)$ to $\left(x_1,\frac{x_2}{2},\frac{x_3}{3},\dots\right)$ is compact. - -For an arbitrary sequence $x^{(N)}\in\ell^1$ one would have extract a convergent subsequence of $T x^{(N)}$. Maybe via the diagonal argument? - -REPLY [7 votes]: Define -$$T_j(x):=\left(x_1,\frac{x_2}2,\dots,\frac{x_j}j,0,\dots,0\right).$$ -It's a compact operator (because it's finite ranked) and -$$T(x)-T_j(x)=\left(0,\dots,0,\frac{x_{j+1}}{j+1},\dots\right),$$ -hence -$$\lVert T(x)-T_j(x)\rVert_{\ell^1}=\sum_{k=j+1}^{+\infty}\frac{|x_k|}{k}\leq \frac 1{j+1}\sum_{k=j+1}^{+\infty}|x_k|\leq \frac 1{j+1}\lVert x\rVert_{\ell^1},$$ -which proves that $\lVert T-T_j\rVert\leq \frac 1{j+1}$. -To conclude, notice that a norm limit of compact operators is compact. - -REPLY [2 votes]: Check that it's the limit of the following finite rank operators and invoke the theorem that a uniform limit of finite rank operators on a Banach Space is a compact operator. -$$T_n(x_1,x_2....)=(x_1,\frac{x_2}2...\frac{x_n}n,0,0,0,0)$$ -You can verify this by checking that $T_n$ is a Cauchy sequence in $L(\mathcal l_1)$<|endoftext|> -TITLE: A permutation in $S_n$ is 'regular' if and only if it is a power of an n-cycle? -QUESTION [5 upvotes]: Define a permutation $\alpha\in S_n $ to be regular if either $\alpha$ has no fixed points and it is the product of disjoint cycles of the same length, or $\alpha=(1)$. -Prove that $\alpha$ is regular if and only if $\alpha$ is a power of an n-cycle. - -It is a homework question in J.J.Rotman's book: A first course in abstract algebra with applications. -I have just begun reading the book, and I find this question confusing----I've tried using induction, but couldn't figure out a good way. I would really appreciate your help. - -REPLY [7 votes]: First, note that a power of an $n$-cycle is necessarily regular: if $\sigma=(a_0,a_2,\ldots,a_{n-1})$, then $\sigma^k$ maps $a_i$ to $a_{i+k\bmod n}$. If $\gcd(k,n)=1$, then $\sigma^k$ is an $n$-cycle again, hence regular. If $\gcd(k,n)=d$, then you should verify that $\sigma^k$ is a product of $d$ cycles, each of length $\frac{n}{d}$. This can be done by thinking about the additive cyclic group of order $n$ and the cosets of the subgroup generated by $d$. -For the converse, suppose that $\alpha = (a_0,\ldots,a_{d-1})(a_{d},\ldots,a_{2d-1})\cdots(a_{(k-1)d},\ldots,a_{kd-1})$ where $kd=n$. with the experience of the "if" part, you should be able to cook up an $n$-cycle $\sigma$ such that $\sigma^d = \alpha$. -To give you an example. Let's take $n=6$. Let $\sigma=(a_0,a_1,a_2,a_3,a_4,a_5)$. The powers of $\sigma$ are: -$$\begin{align*} -\sigma &= (a_0,a_1,a_2,a_3,a_4,a_5) &(\gcd(1,6)=1)\\ -\sigma^2&= (a_0,a_2,a_4)(a_1,a_3,a_5) &(\gcd(2,6)=2)\\ -\sigma^3&= (a_0,a_3)(a_1,a_4)(a_2,a_5) &(\gcd(3,6)=3)\\ -\sigma^4 &= (a_0,a_4,a_2)(a_1,a_5,a_3) &(\gcd(4,6)=2)\\ -\sigma^5&= (a_0,a_5,a_4,a_3,a_2,a_1) &(\gcd(5,6)=1)\\ -\sigma^6 &= \mathrm{id}. -\end{align*}$$ -and you can see that each power is regular. -Now, conversely, suppose we are given, say, $\alpha=(1,4)(2,3)(5,6)$. If it is a power of a $6$ cycle, it must be a cube of a $6$-cycle. Which $6$-cycle? There are a couple of possibilities; one is $(1,2,5,4,3,6)$.<|endoftext|> -TITLE: lim sup of sequence of continuous function from $[0,1]\rightarrow [0,1]$ -QUESTION [5 upvotes]: $f_n:[0,1]\to [0,1]$ be a continuous function and let $f:[0,1]\to [0,1]$ be defined by $$f(x)=\operatorname{lim\;sup}\limits_{n\rightarrow\infty}\; f_n(x)$$ Then $f$ is - -continuous and measurable - -continuous, but need not be measurable - -measurable, but need not be continuous - -need not be measurable or continuous. - - - -I guess $3$ is correct, but I'm not able to prove it. - -REPLY [6 votes]: Note that $$\lim\sup f_n(x)=\inf_{n\geq 1} \sup_{k\geq n} f_k(x)$$ -Let $g_n(x)=\sup_{k\geq n}f_k(x)$, then -$$g_n^{-1}(-\infty,a]=\lbrace x :g_n(x)\leq a\rbrace=\lbrace x :\sup_{k\geq n}f_k (x)\leq a\rbrace=\lbrace x: f_k(x)\leq a\mbox{ for all } k\geq n\rbrace$$ -Hence $$g_n^{-1}(-\infty,a]=\bigcap_{k\geq n}f_k^{-1}(-\infty,a]$$ -It follows that $g_n^{-1}(-\infty,a]$ is a measurable set (being intersection of measurable sets) and so $g_n$ is measurable.. -In a similar fashion we can also prove that $\inf_{n\geq 1}g_n$ is also measurable (try), so we have $\lim\sup f_n(x)$ is measurable. -For the second part $f_n(x)=x^n$ converges pointwise to $g$ where $g(x)=0$ when $0\leq x<1$ and $g(1)=1$ and surely $g$ is not continous.<|endoftext|> -TITLE: How many decimal places are needed for incremental average calculation? -QUESTION [6 upvotes]: If using the following formula to incrementally calculate average value: -$$m_n = m_{n-1} + \frac{a_{n}-m_{n-1}}{n}$$ -And the number of decimal places to where you can save $m_n$ is limited to $d$ decimal digits, how many $n$ values can you use to incrementally calculate average before the rounding error becomes larger than $10^{-p}$, if the value of $a_n$ is always exactly $p$ decimal digit precise? - -REPLY [3 votes]: You have used $base-10$ notation and hence, I am doing the analysis in the same, but to do the actual analysis for a computer, I believe, it would be best for you to do it using the $base-2$. -Since we can have only $d$ decimal places, in every division which is not done by $10$ and the odd case where the mantissa when represented without the point and hence as an integer is a multiple of the divisor, you introduce an error of $10^{-d}$ at every stage (for binary computers, you would lose $2^{-d}$). This is only for this particular formula, since here divisor is integer and hence, no error comes from the divisor. -Also, suppose at some stage $m_{k} = c_{k} \times 10^{q_{k}}$ and $a_{k} = d_{k} \times 10^{r_{k}}$, then adding or subtracting them introduces an error of $10^{q_{k}-r_{k}-d}$ assuming $q > r$. So, now we can add up these operations. So, now your total error accumulated would be -$\varepsilon = \sum_{k=0}^{k=n} 2 \times10^{\left(q_{k}-r_{k}-d\right)} + 10^{-d}$ -So, then going by the rule-of-thumb in the best case where $q_{k} = r_{k}$, I guess, you will be done in at around 4 steps after which your error is of the order of $10^{-d+1}$ -So, this is a rough hand calculation without assuming anything about the system and hence the maximum error I will say but you can get some detailed analysis at this document by Oracle on error calculation.<|endoftext|> -TITLE: $\tbinom{2p}{p}-2$ is divisible by $p^3$ -QUESTION [6 upvotes]: The problem is as follows: -Let $p>3$ be a prime. Show that $\tbinom{2p}{p}-2$ is divisible by $p^3$. The only thing I can think of is that $(2p)!-2(p!)^2$ is divisible by $p^2$ which doesn't help me much. Can someone point me in the right direction? Is there a combinatorial approach to this problem? -Thanks - -REPLY [2 votes]: $\tbinom{2p}{p} = \frac{(p+1)(p+2)...(p+p-1)(p+p)}{1.2...(p-1)p} = \frac{(p+1)(p+2)...(p+p-1)2}{1.2...(p-1)}$ -$\tbinom{2p}{p} -2 $ will be divisible by $p^3$ -iff $ \frac{(p+1)(p+2)...(p+p-1)2}{1.2...(p-1)} -2 $ is divisible by $p^3$ -iff $ ((p+1)(p+2)...(p+p-1)) -(p-1)! $ is divisible by $p^3$ as (2,p)=1 and (p,(p-1)!)=1 -Let f(x)= $\prod_{1≤r≤p-1}(x+r) = \sum _{0≤r≤p-1}a_rx^r$ . -Then (x+1)f(x+1)=(x+p)f(x). -Putting the values of f(x) and f(x+1) and comparing the coefficients of the different powers of x, -for $x^p$, 1=1 -for $x^{p-1}, p+a_{p-1}=pC_1+a_{p-1}$ -and so on. -Clearly,p |$a_r$ and by Wolstenholme's Theorem $p^2|a_1$ -So in $\prod_{1≤r≤p-1}(p+r)$, -$a_0$ is (p-1)!, -the co-efficient of p($a_1$) is the sum of product of r taken 2 at a time, -the co-efficient of $p^2$($a_2$) is the sum of product of r taken 3 at a time and the rest terms are divisible by $p^3$. -As $p|a_2$ and $p^2|a_1$, the result follows.<|endoftext|> -TITLE: Throwing balls into $b$ buckets: when does some bucket overflow size $s$? -QUESTION [8 upvotes]: Suppose you throw balls one-by-one into $b$ buckets, uniformly at random. At what time does the size of some (any) bucket exceed size $s$? - -That is, consider the following random process. At each of times $t=1, 2, 3, \dots$, - -Pick up a ball (from some infinite supply of balls that you have). -Assign it to one of $b$ buckets, uniformly at random, and independent of choices made for previous balls. - -For this random process, let $T = T(s,b)$ be the time such that - -At time $T-1$ (after the $T-1$th ball was assigned), for each bucket, the number of balls assigned to it was $\le s$. -At time $T$ (after the $T$th ball was assigned), there is some bucket for which the number of balls assigned to it is $s + 1$. - - -What can we say about $T$? If we can get the distribution of $T(s,b)$ that would be great, else even knowing its expected value and variance, or even just expected value, would be good. -Beyond the obvious fact that $T \le bs+1$ (and therefore $E[T]$ exists), I don't see anything very helpful. The motivation comes from a real-life computer application involving hashing (the numbers of interest are something like $b = 10000$ and $s = 64$). - -REPLY [2 votes]: I just wrote some code to find the rough answer (for my particular numbers) by simulation. -$ gcc -lm balls-bins.c -o balls-bins && ./balls-bins 10000 64 -... -Mean: 384815.56 Standard deviation: 16893.75 (after 25000 trials) - -This (384xxx) is within 2% of the number ~377xxx, specifically -$$ T \approx b \left( (s + \log b) \pm \sqrt{(s + \log b)^2 - s^2} \right) $$ -that comes from the asymptotic results (see comments on the question), and I must say I am pleasantly surprised. -I plan to edit this answer later to summarise the result from the paper, unless someone gets to it first. (Feel free!)<|endoftext|> -TITLE: Countably compact paracompact space is compact -QUESTION [6 upvotes]: The proof that I have seen for the result "countably compact paracompact implies compact" involves metacompactness, which follows from paracompactness. -I wonder if it can be proved without going through metacompactness. - -REPLY [11 votes]: Perhaps the easiest argument uses the following lemmata. - -Lemma 1. Let $\mathscr{A}$ be a locally finite family of sets in a space $X$; then $\mathscr{A}$ is closure-preserving, i.e., $\bigcup\{\operatorname{cl}A:A\in\mathscr{A}\}$ is closed. -Proof. Let $F=\bigcup\{\operatorname{cl}A:A\in\mathscr{A}\}$, and suppose that $x\in X\setminus F$. $\mathscr{A}$ is locally finite, so $x$ has an open nbhd $G$ such that $\mathscr{A}(x)=\{A\in\mathscr{A}:G\cap A\ne\varnothing\}$ is finite. Let $$V=G\setminus\bigcup\{\operatorname{cl}A:A\in\mathscr{A}(x)\}\;;$$ then $V$ is an open nbhd of $x$ disjoint from $F$, and $F$ is closed. $\dashv$ -Lemma 2. If $X$ is countably compact, every locally finite family of non-empty subsets of $X$ is finite. -Proof. Suppose that there is an infinite locally finite family $\mathscr{A}$ of non-empty subsets of $X$. Any subcollection of $\mathscr{A}$ is still locally finite, so we may assume that $\mathscr{A}=\{A_n:n\in\omega\}$ is countably infinite. For $n\in\omega$ let $$F_n=\bigcup_{k\ge n}\operatorname{cl}A_k\;;$$ by Lemma 1 each $F_n$ is closed. Clearly each $F_n$ is non-empty, $F_0\supseteq F_1\supseteq F_2\supseteq\ldots\,$, and $\bigcap_{n\in\omega}F_n=\varnothing$. But then $\{X\setminus F_n:n\in\omega\}$ is an open cover of $X$ with no finite subcover, and $X$ is not countably compact. $\dashv$ - -Now the desired result is trivial: if $X$ is paracompact, every open cover of $X$ has a locally finite open refinement, and if $X$ is also countably compact, that locally finite open refinement must in fact be finite. - -I should point out that the usual proof doesn’t really use metacompactness: it just uses the fact that a point-finite open cover has an irreducible subcover and the obvious fact that a locally finite open cover is point-finite. One could easily combine all of this into a direct proof. - -Let $X$ be paracompact, $T_1$, and countably compact, and let $\mathscr{U}$ be an open cover of $X$. Let $\mathscr{V}$ be a locally finite open refinement of $\mathscr{U}$, and index $\mathscr{V}=\{V_\xi:\xi<\kappa\}$ for some cardinal $\kappa$. Let $\mathscr{W}_0=\varnothing$. Suppose that we’ve already chosen $W_\xi\in\mathscr{V}$ for $\xi<\eta$. Let $\mathscr{W}_\eta=\{W_\xi:\xi<\eta\}$. If $\mathscr{W}_\eta$ covers $X$, stop. Otherwise, $A_\eta=\{\xi<\kappa:V_\xi\nsubseteq\bigcup\mathscr{W}_\eta\}\ne\varnothing$, and we set $W_\eta=V_{\min A_\eta}$ and continue. Clearly this construction must stop at some $\mathscr{W}_\eta$. -By construction, for each $\xi<\eta$ there is a point $x_\xi\in W_\xi\setminus\bigcup\mathscr{W}_\xi$; let $D=\{x_\xi:\xi<\eta\}$. $\mathscr{W}_\eta$ is locally finite, so each $x\in X$ has an open nbhd $G_x$ such that $G_x\cap D$ is finite. $X$ is $T_1$, so if $x\notin D$, then $G_x\setminus D$ is an open nbhd of $x$ disjoint from $D$, and if $x\in D$, then $\{x\}\cup(G_x\setminus D)$ is an open nbhd of $x$ disjoint from $D\setminus\{x\}$. Thus, $D$ is closed and discrete in $X$, and since $X$ is countably compact, $D$ must be finite. But then $\mathscr{W}_\eta$ is a finite open refinement of $\mathscr{U}$, which therefore has a finite subcover, and $X$ is compact. $\dashv$ - -There’s no mention here of metacompactness, and no explicit mention of irreducible covers, but it’s really the same proof: I’ve just rolled all of the pieces into a single argument. -Yet, another argument uses the (non-trivial) result that if $X$ is a paracompact Hausdorff space, then every open cover of $X$ has a $\sigma$-discrete open refinement: - -Let $\mathscr{U}$ be an open cover of $X$, and let $\mathscr{V}=\bigcup_{n\in\omega}\mathscr{V}_n$ be an open refinement of $\mathscr{U}$ such that each $\mathscr{V}_n$ is discrete. Since $X$ is countably compact, each $\mathscr{V}_n$ is finite, and $\mathscr{V}$ is therefore a countable open cover of $X$. But then $\mathscr{V}$ has a finite subcover, which is a finite open refinement of $\mathscr{U}$, so $\mathscr{U}$ has a finite subcover, and $X$ is compact. $\dashv$ - -That looks nice and short, but the characterization of paracompactness that it uses is harder than the first the argument that I gave.<|endoftext|> -TITLE: $\int\frac{\sin\left(x\right)}{\cos\left(x\right)}\,\mathrm{d}x$ by substitution -QUESTION [8 upvotes]: I'm trying to solve the following integral: -$$\int\frac{\sin\left(x\right)}{\cos\left(x\right)}\,\mathrm{d}x$$ -Using the substitution method with the substitution $u = \sin\left(x\right)$. -The exercise has two parts: the first one is using the substitution $u = \cos\left(x\right)$. No problem. -I'm having difficulties with the second part, which is using the substitution $u = \sin\left(x\right)$. -I spent a couple of hours with the exercise before asking here, and after some trials I got this: -$$\int f\left(g\left(x\right)\right)g'\left(x\right)\,\mathrm{d}x = \int f\left(u\right)\,\mathrm{d}u$$ -$$g\left(x\right) = \sin\left(x\right)$$ -$$g'\left(x\right) = \cos\left(x\right)$$ -$$f\left(x\right) = \frac{x}{\cos^2\left(\arcsin(x)\right)} = \frac{x}{1 - x^2}$$ -$$\int f\left(u\right)\,\mathrm{d}u = -\frac{1}{2}\log|1 - u^2| + C = -\frac{1}{2}\log|1 - \sin^2\left(x\right)| + C$$ -$$1 - \sin^2\left(x\right) = \cos^2\left(x\right)$$ -$$\int\frac{\sin\left(x\right)}{\cos\left(x\right)}\,\mathrm{d}x = -\frac{1}{2}\log|\cos^2\left(x\right)| + C = -\log|\cos\left(x\right)| + C$$ -But it feels too complicated, $f\left(x\right)$ was really hard for me to discover. What am I missing? - -REPLY [3 votes]: What you did is fine. One can view it as carrying out the standard strategy when we have a product of integer powers of sines and cosines, with at least one of the powers odd. In this case, both powers are odd. -Note that -$$\frac{\sin x}{\cos x}=\frac{\sin x\cos x}{\cos^2 x}=\frac{\sin x\cos x}{1-\sin^2 x}.$$ -Then the substitution $u=\sin x$ transforms our integral to -$$\int \frac{u}{1-u^2}\,du.$$ -Remark: If we had instead something like $\int\frac{\sin^4 x}{\cos x}\,dx$, then the same strategy of substituting for $\sin x$ would work. The substitution $u=\cos x$ would be less attractive.<|endoftext|> -TITLE: Stuck with handling of conditional probability in Bishop's "Pattern Recognition and Machine Learning" (1.66) -QUESTION [6 upvotes]: I've just started working through the book, and I'm stuck with how the author handles conditional probability in (1.66). -The context is as follows. In this chapter we are working with a curve fitting task: we try to fit a polynomial $\sum w_ix^i$ to a training set $\{\mathbf {x}, \mathbf {t}\}$ with an assumption of Gaussian noise, i.e. for a single observed value $t$ for $x$, $p(t|x,\mathbf{w},\beta)=N(t|\sum w_ix^i,\beta^{-1})$, and assuming independence of data points, the likelihood is $p(\mathbf{t}|\mathbf{x},\mathbf{w},\beta)=\prod N(t_n|\sum w_ix^i,\beta^{-1}).$ Prior distribution of $\mathbf{w}$ is a multivariate Gaussian: $p(\mathbf{w}|\alpha)=N(\mathbf{0},\alpha^{-1}\mathbf{I})$. -Now, the author states: "Using Bayes’ theorem, the posterior distribution for w is proportional to the product of the prior distribution and the likelihood function: -$$p(\mathbf{w}|\mathbf{x},\mathbf{t},\alpha,\beta)\propto p(\mathbf{t}|\mathbf{x},\mathbf{w},\beta)p(\mathbf{w}|\alpha) \tag{1.66}$$ -Later, this proportionality is used to maximize the probability on the left to obtain MAP value of $\mathbf{w}$, so significant factors cannot be simply omitted on the right. -This is where I'm stuck. The problem is I don't understand how he applies Bayes here. I tried to derive it, and that's what I've got: -$$p(\mathbf{w}|\mathbf{x},\mathbf{t},\alpha,\beta)\propto p(\mathbf{x},\mathbf{t},\mathbf{w},\alpha,\beta)=p(\mathbf{t}|\mathbf{x,w},\alpha,\beta)p(\mathbf{x,w},\alpha,\beta)\tag{A}$$ -where the latter equals to $p(\mathbf{w}|\alpha,\beta,\mathbf{x})p(\alpha,\beta,\mathbf{x})$. I see that we can get rid of the second factor here because it is not really interesting if we want to maximize the expression — these are just model parameters or the data which is given. Also, I see that we can rewrite the first factor as $p(\mathbf{w}|\alpha)$. That's why: say we have $p(A|BC)\text{, then } p(A|BC)=\frac {p(ABC)}{p(BC)}=\frac {p(AB)p(C)}{p(B)p(C)}=p(A|B)$, which holds if both ($AB$ and $C$) and ($B$ and $C$) are independent, and indeed both $\mathbf{w}$ and $\alpha$ are independent with both $\beta\text{ and }\mathbf{x}$. -But I don't see why we can "remove" $\alpha$ from the first factor of (A). For one, $\mathbf{w}$ and $\alpha$ are not independent. Maybe I don't understand the problem well enough? Maybe this probability can be factorized so that we can say "this factor is irrelevant,let's hide it under $\propto$"? -So, the questions: -1) Please help with understanding of the proportionality (1.66). -2) It's hard for me to see the benefit of using conditional distributions on things like $\alpha$ and $\beta$. Is, for example, $p(\mathbf{w})$ of any interest in this task? I don't see how any meaning can be attributed to it. Could someone explain that? -3) Is there some common knowledge about conditioning on several random variables, like the one I used ($p(A|BC)=p(A|B)$ if both pairs ($AB$ and $C$) and ($B$ and $C$) are independent), that make it obvious? - -REPLY [5 votes]: Let me answer to these questions in a different order. -3) $p(A|BC)=p(A|B)$ also when $A$ is conditionally independent of $C$ given $B$. Actually, this is taken to be as a definition of conditional independence in that same book, chapter 8.2 "Conditional Independence." -1) The proportionality holds because of 3): in this case it is implicitly assumed that $\mathbf{t}$ is conditionally independent of $\alpha$ given $(\mathbf{x},\mathbf{w},\beta)$. -Moreover, if we look at the the Bayesian network in Figure 8.7, we can see that $\alpha$ and $\mathbf{t}$ are d-separated by a set of nodes $\{\mathbf{x},\mathbf{w},\beta\}$. Indeed, there are only two paths between $\alpha$ and $\mathbf{t}$: - -$\alpha \rightarrow \mathbf{w} \rightarrow \mathbf{t}$ blocked by $\mathbf{w}$ (head-to-tail); -$\alpha \rightarrow \mathbf{w} \rightarrow t \leftarrow \beta \rightarrow \mathbf{t}$ blocked by $\mathbf{w}$ and $\beta$ (tail-to-tail). - -That's it! I think that the author should have at least mentioned why this implicit leaving out of $\alpha$ is justified, even if it is explained further in the text. -2) Seems like it is the root of the Bayesian approach — use probabilities everywhere, adding conditional probability in case of functional dependency, and hope that the joint probability of all involved variables will factorize nicely thanks to conditional independence of some variables. -At this place in the book, I feel that it has a downside that it becomes hard to think of probability spaces of the involved variables in an intuitive way. See that same question in the original, well, question: "what is $p(\mathbf{w})$?"<|endoftext|> -TITLE: Why is $\log(b,n) = \lfloor \log_b(n) \rfloor$ primitive recursive? -QUESTION [5 upvotes]: I read in an introduction to primitive recursive function and Wikipedia that -$$\log(b,n) = \lfloor \log_b(n) \rfloor$$ is primitive recursive. But how can that be? Is there any easy proof (and therefore a definition of the function using only constants, projection, composition and primitive recursion)? -Thanks in advance! - -REPLY [4 votes]: I'll try and sketch a construction. Firstly, note that a function defined by primitive recursive cases from primitive recursive functions is still primitive recursive (rather easy to prove). -So, we attempt to define this by recursion on $n$. Since $\log_b(0)$ is not something we want to consider, we'll assume $b,n > 0$. Let -$$\log(b,1) = 0$$ -Which is certainly primitive recursive. Then define -$$\log(b,n+1) = F(\log(b,n),b,n)$$ -Where $F$ is the following function, defined by cases. $F$ takes $\log(b,n)+1$, and checks if -$$b^{(\log(b,n) + 1)} > n+1$$ -In other words, it sees if $\log(b,n) + 1$ is still shooting too high. If it is, then we stick with what we've got: $\log(b,n)$. Otherwise, the time has finally come to move on up to $\log(b,n) + 1$, and so $F$ outputs that. -It's not hard to see that $F$ is defined by primitive recursive cases from primitive recursive functions, and so have shown that $\log(b,n)$ is primitive recursive.<|endoftext|> -TITLE: Recovering the topology of an affine scheme from the specialization preorder -QUESTION [9 upvotes]: Let $A$ be a commutative ring. The specialization preorder on $\mathrm{Spec}(R)$ is given by $\mathfrak{p} \prec \mathfrak{q} \Leftrightarrow \mathfrak{p} \in \overline{\{\mathfrak{q}\}} \Leftrightarrow \mathfrak{q} \subseteq \mathfrak{p}$. Is it possible to recover the topology on $\mathrm{Spec}(A)$ from this preorder? -If $A$ is noetherian (or just $A_{red}$ noetherian, which has the same spectral space), then the closed subsets are the finite unions of irreducible closed subsets, and the irreducible closed subsets are precicely those of the form $\{\mathfrak{q} : \mathfrak{q} \prec \mathfrak{p}\}$ for some $\mathfrak{p}$. Thus, in this case, we may recover the topology. -Is it possible to do so in general? More formally, assume that $X$ is a set, endowed with two spectral topologies. Assume that their specialization preorders are the same. Does this imply that the topologies coincide? Probably not. But what are interesting additional assumptions on these topologies (not on the rings) which make it true? - -REPLY [6 votes]: Here is an example of two spectral topologies with the same specialization order. -Every boolean space is homeomorphic to the prime spectrum of some commutative ring (boolean spaces are spectral, and Hochster showed that spectral spaces are precisely the spaces arising as prime spectra of commutative rings). -The specialization order for boolean spaces is trivial: every point is closed. -So two boolean spaces will have the same specialization order if they have the same cardinality. -So take two countable boolean spaces that are not homeomorphic, for example -$$X = \{0\} \cup \{ 1/n : n = 1,2,3,\dots \} \subseteq \mathbb{R}$$ -and $Y = X \sqcup X$. -The associated commutative rings in this example will be non-noetherian of Krull dimension zero.<|endoftext|> -TITLE: Equivalent Definitions of the Operator Norm -QUESTION [76 upvotes]: How do you prove that these four definitions of the operator norm are equivalent? -$$\begin{align*} -\lVert A\rVert_{\mathrm{op}} &= \inf\{ c\;\colon\; \lVert Av\rVert\leq c\lVert v\rVert \text{ for all }v\in V\}\\ -&=\sup\{ \lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert\leq 1\}\\ -&=\sup\{\lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert = 1 \}\\ -&=\sup\left\{ \frac{\lVert Av\rVert}{\lVert v\rVert}\;\colon\; v\in V\text{ with }v\neq 0\right\}. -\end{align*}$$ - -REPLY [4 votes]: Another equivalent definition, which I can rarely see: -$$\lVert A\rVert=\sup_{\lVert x \rVert < 1} \lVert Ax \rVert$$ -It's easy to see that -$$\sup_{\lVert x \rVert < 1} \lVert Ax \rVert \leqslant \sup_{\lVert x \rVert \leqslant 1} \lVert Ax \rVert$$ -For the other direction, let $(r_n)$ be a sequence of real numbers, with $0 \leqslant r_n < 1$ and $r_n \to 1$, and $x \in V$ with $\lVert x \rVert \leqslant 1$. Then, we have that $\lVert r_n x\rVert<1$, which implies that $\lVert A(r_n x) \rVert \leqslant \sup_{\lVert x \rVert < 1} \lVert Ax \rVert$, but $\lVert A(r_n x) \rVert \to \lVert A x\rVert$, which implies that $\lVert Ax \rVert \leqslant \sup_{\lVert x \rVert < 1} \lVert Ax \rVert$.<|endoftext|> -TITLE: $-\iint_{A}(y+x)\,dA$ Evaluating -QUESTION [5 upvotes]: I am bit unsure about the following problem: -Evaluate the double integral: -$$-\iint_{A}(y+x)\,dA$$ -over the triangle with vertices $(0,0), (1,1), (2,0)$ -OK, so I figured here that I would do this by first evaluating the integral over the region bounded by the vertices $(0,0), (1,1), (1,0)$ and then evaluate the integral over the region bounded by the vertices $(1,0), (1,1), (2,0)$ before adding the two answers together, and then reversing the sign of this answer (since there is a minus sign in front of the original double integral). Thus, I begin by finding: -$$\int_{0}^{1}dx \int_{0}^{x}(y+x)\,dy$$ -When solved this gives me the answer $\frac{1}{2}$. -Next I solve: -$$\int_{1}^{2}dx \int_{1}^{2-x}(y+x)\,dy$$ -When solved this gives me the answer $-\frac{7}{6}$. -I have verified both the integrals in Wolframalpha, and they give me the same answer. I would therefore believe that the final answer should be: -$$-(\frac{1}{2} - \frac{7}{6}) = \frac{2}{3}$$ -However, the final answer should, according to the book, be $-\frac{4}{3}$. -Thus, obviously I do something wrong here. If anyone can help me out, I would greatly appreciate it. Is it perhaps that it is not allowed to "split up" this into two separate integrals? I couldn't find a way to solve this without doing this. - -REPLY [2 votes]: Your second integral should be $$\int_{1}^{2}dx \int_{0}^{2-x}(y+x)dy.$$ -Your lower $y$ limit was 1 instead of 0. -Draw the triangle to see the area you are integrating over.<|endoftext|> -TITLE: Motivation for definition of logarithm in Feynman's Lectures on Physics -QUESTION [11 upvotes]: I'm not sure if the title is descriptive enough; feel free to change it if you come up with something better. -I've been reading through Feynman's Lectures on Physics. In the first volume, he dedicates a chapter to just math. He starts with the natural numbers and addition by 1, and builds his way to the complex numbers, with the purpose of proving Euler's formula $e^{ix} = \cos x + i \sin x$. It's a very nice read, but there's a part in the middle I'm having trouble understanding. -After having introduced irrational numbers, he begins to explain how to calculate (or define, rather) irrational powers as succesive approximations of rational powers and how to calculate logarithms, which is a related problem. In particular, he gives as an example how to find solutions to the equations $x = 10^\sqrt{2}$ and $x = \log_{10} 2$. To do that, he makes a table of successive square roots of ten, by calculating $10^s$ for $s = 1, \frac1{2}, \frac1{4}, \frac1{8}, \cdots , \frac1{1024}$. He remarks that this is enough to calculate some logarithms, because if we already calculated $10^s$ and we want its logarithm, it is simply $s$. -He also notices that as we take square roots of number that get closer and closer to $1$, there is a pattern: $\sqrt{1+\delta}$ is approximately $1+\delta/2$, so, for numbers that are already close to $1$, (such as $10^{1/1024}$, which is the last square root he calculated) instead of keeping on doing square roots we can just "guess" at the result with a pretty good accuracy. -Now, here's the part I don't understand: Having calculated $10^{1/1024}$ to be approximately $1.0022511$, he says the following: - -[...] it is clear that, to an excellent approximation, if we take - another root, we shall get 1.00112 something, and rather than actually - take all the square roots, we guess at the ultimate limit. When we take a small fraction $\Delta$ of $1024$ as $\Delta$ approaches zero, - what will the answer be? Of course it will be some number close to - $0.0022511 \Delta$. Not exactly $0.0022511 \Delta$, however -- we can - get a better value by the following trick: we substract the $1$, and - then divide by the power $s$. This ought to correct all the excesses - to the same value. - -He then adds another column: for each $s$, in addition to $10^s$ there's $\frac{10^s-1}{s}$, and it looks like this converges to something as $s$ gets smaller. I recognize this as one of the usual formulas for the logarithm, but I don't follow why he introduced it. Later he uses this to make an approximation formula: $10^\Delta \approx 1+\Delta \ln 10$. I understand this, but I don't get where he got that from. Could someone clarify this? -I wasn't sure about asking this question because I thought it might be hard to understand if you've never read this chapter. If this is the case, let me know and I'll try to edit it a bit. - -REPLY [13 votes]: Thanks for bringing this to my attention. I checked the lecture tape, and found that what Feynman originally said was, "When I took a small fraction of 1024, as the fraction went to zero, what would the answer be here?" This was edited in 1961 by Robert Leighton into the form it now takes in The Feynman Lectures on Physics. I agree with you that the current version is a bit confusing, and will recommend that it be changed from, "When we take a small fraction Δ of 1024 as Δ approaches zero..." to, "When we take a small fraction Δ/1024 as Δ approaches zero..." -Mike Gottlieb [Editor, The Feynman Lectures on Physics New Millennium Edition]<|endoftext|> -TITLE: Learning differential/Riemannian geometry for PDEs -QUESTION [11 upvotes]: I know there have been threads on which books to learn DG/RG from but hopefully this is sufficiently different to avoid closure. -Can anyone recommend a book to learn DG/RG (whichever is appropriate) so that I can do PDEs on manifolds? At the moment I am reading through John M Lee's Introduction to Smooth Manifolds and I am wondering whether I really need to learn all the topics in it since it doesn't touch RG which I believe is more used in the theory of PDEs. Maybe there is a better text. -Also, any topics to particularly study or avoid would be useful. -Thanks - -REPLY [3 votes]: If you're assuming the Riemannian manifold has a fixed metric, then the most introductory source I've found is Folland, Introduction to Partial Differential Equations, which discusses aspects of PDEs on hypersurfaces and the Laplace-Beltrami operator, for example. -If you're looking for something more advanced, but which avoids getting into curvatures, -then Grigor'yan's Heat kernel and analysis on Manifolds (AMS,2009) is excellent IMHO.<|endoftext|> -TITLE: What is a separator object? -QUESTION [5 upvotes]: Let $S$ be an object of category $C$. We say $S$ is a separator object of $C$ if whenever -$$Y \stackrel{f_1} \longleftarrow X \stackrel{f_2}{\longrightarrow} Y $$ -$(\forall x[S\stackrel{x}\longrightarrow X \Rightarrow f_1x=f_2x]) \Rightarrow f_1 =f_2$ -This reminds of the definition of an epimorphism. I am wondering if we could define separator object as an object which all of its morphisms are epic. - -REPLY [5 votes]: The definition of a separator object appears to enshrine "collective" right-cancellation, but not the individual right-cancellation that is mandated by epicness. IOW, an object is a separator object when $f_1x=f_2x$ for specific $f_1,f_2$ can be right-cancelled if it holds for all $x$, whereas a specific $x$ is epic if $f_1x=f_2x$ can be right-cancelled for any $f_1,f_2$. These do not mean the same thing (which arrows are being universally quantified vs. individually specified is distinct) and aren't necessarily compatible, so we can't define a separator object as one with all outbound morphisms epic. -Consider the following situation: given $S$, there exist arrows from $S$ to $X$, $S\xrightarrow{x}X$ and $S\xrightarrow{y}X$ , and arrows from $X$ to $Y$, $X\xrightarrow{f}Y$ and $X\xrightarrow{g}Y$, such that $fx=gx$ but $fy\ne gy$ and $f\ne g$. This is not in contradiction with the definition of a separator object (so $S$ could still be a separator), while it does preclude $x$ from being an epimorphism. We can even devise a category with precisely these objects and arrows (plus $1_S,1_X,1_Y$), and $S$ will be a separator because the definition is vacuously fulfilled. -On the other hand, if an object $S$ has the property that all outbound morphisms are epic, then it will also be a separator object, so they are related.<|endoftext|> -TITLE: $F, G \in k[X_1, \dots , X_n]$ homogeneous of degrees $r$ and $r+1$ $\implies$ $F+G$ is irreducible -QUESTION [5 upvotes]: I have a question about Exercise 2-34 from William Fulton's Algebraic Curves book. The exercise is as follows. - - -Suppose that $F, G \in k[X_1, \dots , X_n]$ are forms (i.e. homogeneous polynomials) of degree $r$ and $r+1$ respectively, without common factors (where $k$ is a field). Prove that $F + G$ is irreducible. - - -I'm having some trouble trying to prove this. First, since $F, G$ are arbitrary I can't think of a particular irreducibility criterion to apply here, so my only idea was to try by contradiction. - -Notation -I use Fulton's notation for the homogenization and dehomogenization of a polynomial with respect to the variable $X_{n+1}$. Say if $f \in k[X_1, \dots , X_n]$ has degree $d$, then its homogenization with respect to $X_{n+1}$ is denoted by $f^* = X_{n+1}^d f\left ( \frac{X_1}{X_{n+1}} , \dots , \frac{X_n}{X_{n+1}} \right)$. -Similarly, if $F \in k[X_1, \dots , X_{n+1}]$ is homogeneous, then its dehomogenization with respect to $X_{n+1}$ is denoted by $F_* = F(X_1, \dots , X_n, 1)$. - -My attempt -Thus I assume by contradiction that $F + G$ is reducible, then there are polynomials $H, R \in k[X_1, \dots , X_n]$ such that $F + G = HG$ and also $ 0 < \deg{H},\deg{G} < r + 1$. Then the homogenization is -$$(F + G)^* = X_{n+1} F + G = (HR)^* = H^* R^*$$ -Now I'm somewhat stuck. I've tried to play around with this but without luck. I suppose that maybe I should try to get a contradiction to the assumption that $F$ and $G$ have no common factors, but I don't know how. - -My questions - -How can this be proved? -If my approach by contradiction is correct, how can the argument be finished? - -As always, thank you for any help. - -REPLY [7 votes]: Say that a non-zero polynomial $F$ has width $w\in\mathbb N_0$ if the degree of the monomial of highest degree appearing in $F$ with non-zero coefficient minus the degree of the monomial of lowest degree appearing in $F$ with non-zero coefficient is $w$. For example, if $F=F_{n+1}+F_n$ with $F_{n+1}$ and $F_n$ non-zero and homogeneous of degrees $n+1$ and $n$, the width of $F$ is $1$; notice that the polyonomials of width $0$ are exactly the non-zero homogeneous polynomials. - -Show that if $F$ and $G$ are polynomials with widths $w_F$ and $w_G$, then the product $FG$ has width $w_F+w_G$. -Use that to prove that if a polynomial $F$ of width $1$, as in your case, is reducible it necessarily has an non-constant homogeneous factor. -Finally, use that to prove what you want.<|endoftext|> -TITLE: Szemerédi's Regularity Lemma -QUESTION [10 upvotes]: I am trying to understand the statement of the Szemerédi's Regularity Lemma as explained here, but am not able to wrap my head around to what it really means. In particular the concept of $\epsilon$-regularity and how it relates to drawing edges at random is not clear to me. If someone can explain the statement and the meaning of the theorem (not the proof) I would be most obliged. -Thanks. - -REPLY [8 votes]: In a sense $\epsilon-$regularity is one way of capturing what it means for a graph to be "random-like". For comparison, let's suppose that $G$ is a random bipartite graph on two large subsets $A$ and $B$ having edge probability $p$ (meaning that every edge appears with probability $p$ independently; here $p$ is fixed). Here's several ways in which you might try and say $G$ really does "look random". --The edges of $G$ are spread out evenly -- there aren't any huge clusters or huge empty spaces in the graph. --Each small bipartite graph $H$ appears in $G$ about the number of times you expect it to (roughly $p^{E(H)}$ times the number of appearances in the complete bipartite graph). --Almost all vertices in $A$ have degree roughly $p |B|$, and almost all pairs of vertices in $A$ have roughly $p^2 |B|$ common neighbors (being a neighbor of one vertex doesn't make you significantly more or less likely to be a neighbor of another vertex). The same holds if you swap $A$ and $B$. -As it turns out, the above three characterizations are actually equivalent, if quantified in the right way (this is a phenomenon known as "quasirandomness"). The idea of $\epsilon-$regularity is to quantify the first of these properties. -For each $S \subseteq A$ and $T \subseteq B$, our random graph is expected to have $p|S| |T|$ edges connecting $S$ with $T$. Rewording this slightly, the density $d(S,T):=\frac{|E(S,T)|}{|S||T|}$ is on average equal to $p$. So we might hope to make our "random-like" definition something like -"For every $S$ and $T$, we have -$|d(S,T)-p|<\epsilon$" -But there's a problem here...if $S$ and $T$ are too small, there's no hope for this to hold (imagine the extreme case where $|S|=|T|=1$). So we'll add one additional condition: We only want the above bound to hold for subsets of size at least $\epsilon n$. The idea is that if a small subset is off, it won't have too much impact on things like subset counts anyway, so we might as well worry only about the large subsets. This modified definition is what is used for $\epsilon$-regularity in Szemerédi's Lemma. -What the Lemma itself says is that any dense graph can be approximated by a bounded number of these $\epsilon-$regular graphs (depending only on $\epsilon$, not on the size of the original graph). "Approximated by" here means that you may need to delete a small number ($\epsilon n^2$) edges, but once you delete them you'll be left with a graph on $k$ subsets $A_1, A_2, \dots, A_k$ of vertices, so that for every $i$ and $j$ the graph between $A_i$ and $A_j$ is random-like in all of the ways above [The value of $p$ may be different for each pair $(i,j)$].<|endoftext|> -TITLE: Finding two numbers given their sum and their product -QUESTION [11 upvotes]: Which two numbers when added together yield $16$, and when multiplied together yield $55$. - -I know the $x$ and $y$ are $5$ and $11$ but I wanted to see if I could algebraically solve it, and found I couldn't. -In $x+y=16$, I know $x=16/y$ but when I plug it back in I get something like $16/y + y = 16$, then I multiply the left side by $16$ to get $2y=256$ and then ultimately $y=128$. Am I doing something wrong? - -REPLY [3 votes]: If $x+y=16$ and $xy=55$, then, since $\dfrac{x+y}{2}=8$, we can let $x = 8-z$ and $y=8+z$ and it won't hurt to suppose that $z$ is positive.. -Then -\begin{align} - xy &= 55 \\ - (8-z)(8+z) &= 55 \\ - 64 - z^2 &= 55 \\ - z^2 &= 9 \\ - z &= 3 \\ - x &= 8-3 = 5 \\ - y &= 8+3 = 11 -\end{align}<|endoftext|> -TITLE: If two polynomials are equal as functions, are they necessarily equal as polynomials? -QUESTION [6 upvotes]: Say you have a finite field $F$ of order $p^k$. Suppose that $f,g\in F[X_1,\dots,X_m]$, such that the degree of each $X_i$ is strictly less than $p^k$ in both $f$ and $g$. I'm putting this condition to avoid things like $f=X_1X_2X_3^{p^k}$ and $g=X_1X_2X_3$ which technically define the same polynomial function over $F$ since $X_i^{p^k}-X_i$ is in the kernel of the evaluation homomorphism, but are not equal in the polynomial ring. -Under this condition, if $f$ and $g$ define the same polynomial function over $F$, are they equal as polynomials? By equality of polynomial functions, I mean they are equal as sets of ordered pairs. I feel like restricting the degree of each indeterminate should force this to be so, but how can it actually be proven? - -REPLY [5 votes]: Formalizing Jacob's answer (+1) and generalizing it to $m$ variables as in did's comment. -Let $F=GF(q)$ be the field of $q$ elements, and let $P=(a_1,a_2,\ldots,a_m)\in F^m$ be an arbitrary point. Let us fix an index $j$, $1\le j\le m$. The polynomial -$$ -f_j(x)=\prod_{a\in F, a\neq a_j}(x-a)=\frac{x^q-x}{x-a_j} -$$ -has the properties that $f_j(a_j)\neq0$, and $f_j(a)=0$ for all $a\in F, a\neq a_j.$ Therefore the product -$$ -f_P(x_1,x_2,\ldots,x_m):=f_1(x_1)f_2(x_2)\cdots f_m(x_m)=\prod_{j=1}^m f_j(x_j) -$$ -vanishes at all other points of $F^m$ except at $P$. Furthermore, the polynomial $f_P$ -is of degree $\le q-1$ with respect to all the variables $x_j,j=1,2,\ldots,m.$ -Let $V_m$ be the space of polynomials in $m$ variables over $F$ such that no variable -appears with degree $\ge q$. Let $V_m'$ be the space of $F$-valued functions on $F^m$. -There is a natural evaluation mapping $ev:V_m\rightarrow V_m'$. Our earlier calculations show that the functions $ev(f_P)$, with $P$ ranging over the points of $F^m$, form a basis -of $V_m'$. Therefore $ev$ is surjective. Because $V_m$ and $V_m'$ both have dimension $q^m$ as vector spaces over $F$, the mapping $ev$ must also be injective (rank-nullity). -This answers your question in the affirmative.<|endoftext|> -TITLE: If $m\geq2$ is an integer, then $\sum\limits_{n=1}^{\infty}m^{-n^2}$ is irrational -QUESTION [9 upvotes]: Let $m \geq2$ be an integer. I want to ask how to prove that the sum of the following series is irrational: -$$\sum _{n=1}^{\infty} \frac{1}{m^{n^2}}$$ - -REPLY [16 votes]: To get a grasp on what is going on here, we start with a simple, seemingly unrelated, problem, namely, we show that the decimal representation of $x=1/7$ is periodic. -We know how to do that, right? One should perform the long division of $1$ by $7$ and very soon the digits produced by the division will begin to cycle. The result is $$x=0.14285714285714285714\ldots,$$ usually abbreviated as $x=0.\overline{142857}$. Thus, the decimal representation is indeed periodic... but now we might want to ask two questions: - -What causes this periodic expansion? -What does it mean? - -Well, the meaning is clear: the expansion says that $$x=n\cdot10^{-6}+n\cdot10^{-12}+n\cdot10^{-18}+\cdots$$ with $n=142857$, that is, that $$x=n\cdot10^{-6}\cdot\sum\limits_{i=0}^{+\infty}10^{-6i}=\frac{n\cdot10^{-6}}{1-10^{-6}}=\frac{n}{10^6-1}.$$ -In other words, we started from a representation of $x$ as a fraction, namely $x=1/7$, and we reached another representation of $x$ as a fraction, namely $x=142857/(10^6-1)$. If the goal is to compute the decimal representation of $x$, the second fraction is actually quite nice because, due to the simple fact that $142857\lt10^7$, the periodicity of the representation of $x$ becomes obvious. But we still have two mysteries to solve: - -Why did one obtain $10^6-1$ as the denominator? -Why did one obtain $142857$ as the numerator? - -At this point, one could note that, since the expression of every rational number as a reduced fraction is unique, one better have $10^6-1=7\cdot n$. In particular, $10^6-1$ should be a multiple of $7$, that is $10^6=1\pmod{7}$. Now, this rings a bell! One knows that, as soon as $k$ and $b$ are relatively prime, $k^{\phi(b)}=1\pmod{b}$. Since $10$ and $7$ are relatively prime and $\phi(7)=6$, indeed $10^6-1$ is a multiple of $7$, and the remark also explains the appearance of $6$ as the exponent of $10$. More importantly, it suggests a reason why the whole shebang holds and, at the same time, a way to vastly generalize our observations to any rational number. -So, we now consider any rational number $x=a/b$. We assume without loss of generality that $a\geqslant1$, $b\geqslant2$ (otherwise $x$ is an integer), $a\leqslant b-1$ (otherwise, shift $x$ by an integer), and that $b$ has no factor $2$ or $5$ (otherwise, multiply $b$ by powers of $5$ or $2$ to get a power of $10$, then multiplying $x$ by this power of $10$ simply shifts the expansion of $x$). Thus, $10$ and $b$ are relatively prime and $10^c=1\pmod{b}$, for some positive integer $c$. This means that $10^c=bd+1$ for some integer $d$, which implies that $x=n/(10^c-1)$ with $n=ad\lt10^c-1$. One gets $x=0.\overline{n_{c-1}n_{c-2}\cdots n_1}$, where $n=\sum\limits_{i=0}^{c-1}n_i\cdot10^i$ with $n_i$ in $\{0,1,\ldots,9\}$ is the decimal representation of $n$. This proves that the decimal representation of $x$ is indeed periodic and, at the same time, yields a way to compute this representation. -(When $a=1$ and $b=7$, indeed one gets $c=6$, $d=999999/7=142857$ and $n=142857$.) - -Coming back finally to the question asked, one sees that there is nothing specific to the base $10$ here. Thus, consider any integer $m\geqslant2$ and some real number $x$ in $[0,1]$. Thus, $x=\sum\limits_{i=1}^{+\infty}x_im^{-i}$ for some sequence $(x_i)_{i\geqslant1}$ with values in $\{0,1,\ldots, m-1\}$. Then, $x$ is a rational number if and only if the sequence $(x_i)_{i\geqslant1}$ is ultimately periodic. -In the case at hand, $x_i=1$ when $i$ is a square and $x_i=0$ otherwise, hence the gaps between the digits $1$ are unbounded. This forbids ultimate periodicity, hence $x$ is irrational.<|endoftext|> -TITLE: Inner product is jointly continuous -QUESTION [5 upvotes]: I'm attempting another exercise from my notes: -Show that an inner product on an inner product space is jointly continuous with respect to the induced norm:if $v_n \to v$ and $w_n \to w$ as $n \to \infty$, then $\langle v_n, w_n\rangle \to \langle v,w \rangle$ as $n \to \infty$. -I'd not heard jointly continuous before so I googled and found the following definition in Kelley: -Let $P: F \times X \to Y$ defined by $(f,x) \mapsto f(x)$. Each topology on $F$ gives rise to a product topology on $F \times X$. A topology for $F$ is said to be jointly continuous iff $P$ is continuous. -I think in the case of the inner product, $F$ is the one point space $\{ \langle \cdot, \cdot \rangle \}$ and $X = V \times V$. Then there is only one topology on $F$ and jointly continuous just means that the inner product is continuous. Question 1: Is that correct so far? -So to show that $\langle \cdot, \cdot \rangle$ is continuous we need to show that if $(x_n, y_n) \to (x,y) $ then $\langle x_n, y_n \rangle \to \langle x, y \rangle$ in $\mathbb R$. -On $V \times V$ we can define the norm $\|(a,b)\| = \max(\|a\|, \|b\|)$. Question 2: How do I show that this norm induces the same topology as the product topology? -Now I want to show that for $\varepsilon > 0$ there is an $N$ such that for $n > N$, $| \langle x_n, y_n \rangle - \langle x, y \rangle | < \varepsilon$. In the max norm, I have $\|x_n - x\|^2 = \langle x_n, x_n \rangle - \langle x_n, x \rangle - \langle x, x_n \rangle + \langle x, x \rangle < \varepsilon^2$, for both $x_n,x$ and $y_n,y$. Question 3: How can I use this to show $| \langle x_n, y_n \rangle - \langle x, y \rangle | < \varepsilon$? - -REPLY [3 votes]: We will use the infinity or max norm on $V\times V$, that is, for $(u,v)\in V\times V$, $\lVert (u,v)\rVert_{V\times V}=\max\{\lVert u\rVert_V,\lVert v\rVert_V\}$, where the norm on $V$ is the norm induced by the inner product $\langle\cdot,\cdot\rangle$. I can't speak with authority on why this is equivalent to the product topology, but I don't think it should be too difficult to show for a finite product of normed spaces - those are some well-behaved constructs. For a function to be jointly continuous, it means simply that it is continuous in the product topology, rather than only being continuous in each variable separately, with the others fixed. That is, for $f:X\times Y\to Z$, $f$ is jointly continuous if, for all $(x_0,y_0)\in X\times Y$, given any $\epsilon>0$, there exists a $\delta>0$ such that, for all $(x,y)\in X\times Y$, $d_{X\times Y}((x,y),(x_0,y_0))<\delta$ implies $d_Z(f(x,y),f(x_0,y_0))<\epsilon$. So let's do it, using the induced norm as our metric. -Fix an $x_0,y_0\in V$, and let $\epsilon>0$ be given. Choose $\delta=\min\{1,\frac{\epsilon}{2(\lVert y_0\rVert + 1)},\frac{\epsilon}{2(\lVert x_0\rVert+1)}\}$, and note $\lvert\lVert y\rVert_V-\lVert y_0\rVert_V\rvert\leq \lVert y-y_0\rVert_V<1$, by the reverse triangle inequality and our choice of $\delta$. This implies $\lVert y\rVert_V<1+\lVert y_0\rVert_V$. -Let $x,y\in V$, and suppose $\lVert (x,y)-(x_0,y_0)\rVert_{V\times V}=\lVert(x-x_0,y-y_0)\rVert_{V\times V}<\delta$, implying $\lVert x-x_0\rVert_V<\delta$ and $\lVert y-y_0\rVert_V<\delta$. Then -$\lvert\langle x,y\rangle-\langle x_0,y_0\rangle\rvert = \lvert\langle x,y\rangle-\langle x_0,y\rangle+\langle x_0,y\rangle-\langle x_0,y_0\rangle\rvert$ -$\leq \lvert\langle x,y\rangle-\langle x_0,y\rangle\rvert+\lvert\langle x_0,y\rangle-\langle x_0,y_0\rangle\rvert$ -$= \lvert\langle x-x_0,y\rangle\rvert+\lvert\langle x_0,y-y_0\rangle\rvert$ -$\leq \lVert x-x_0\rVert_V\lVert y\rVert_V + \lVert x_0\rVert_V \lVert y-y_0\rVert_V$ (by Cauchy-Bunyakovsky-Schwarz inequality) -$< (\frac{\epsilon}{2(\lVert y_0\rVert+1)})(1+\lVert y_0\rVert_V) + (\lVert x_0\rVert_V+1)(\frac{\epsilon}{2(\lVert x_0\rVert+1)})$, (since $\lVert x_0\rVert_V<\lVert x_0\rVert_V+1$, and by our choice of $\delta$) -$= \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon$.<|endoftext|> -TITLE: A proof of Wolstenholme's theorem -QUESTION [25 upvotes]: This was inspired by this question. I tried to use the identity -$${2n \choose n}=\sum_{k=0}^n {n \choose k}^2$$ -(see this question) to prove that $$\binom{2p}p\equiv2\pmod{p^3}$$ if $p\gt3$ is prime. (Wikipedia gives a combinatorial proof that this special case of Wolstenholme's theorem implies the general case.) Since $\binom p0=\binom pp=1$, we want -$$\sum_{k=1}^{p-1}\binom pk^2\equiv0\pmod{p^3}\;.$$ -The binomial coefficients are all divisible by $p$, so we can write this as -$$\sum_{k=1}^{p-1}\left(\frac1p\binom pk\right)^2\equiv0\pmod{p}\;.$$ -But -$$\frac1p\binom pk=\frac{(p-1)\cdots(p-(k-1))}{k(k-1)\cdots1}\equiv\frac1k\frac{(-1)\cdots(-(k-1))}{(k-1)\cdots1}\equiv\pm\frac1k\pmod p\;,$$ -and thus -$$\sum_{k=1}^{p-1}\left(\frac1p\binom pk\right)^2\equiv\sum_{k=1}^{p-1}\left(\pm\frac1k\right)^2\equiv\sum_{k=1}^{p-1}\left(\frac1k\right)^2\pmod{p}\;.$$ -As $k$ traverses the non-zero residues $\bmod p$, so does $1/k$, so this is just twice the sum of the quadratic residues $\bmod p$, which is zero for $p\gt3$ as required. -I couldn't find this proof of Wolstenholme's theorem anywhere, so I'm wondering: - -Is there a mistake in it? -If not, is it known? -If not, is it relevant? - -[Update:] -In the meantime I came across this paper, which does make extensive use of the sum-of-squares identity in the context of Wolstenholme's theorem, though not for proving the theorem itself. -Regarding the use of the sum over reciprocals in proving the harmonic-number version of the theorem, as in the paper linked to in Don's answer: I was aware of that use, but the main reason I thought that this alternative proof might nevertheless be of interest is that to prove the congruence for the binomial coefficient from the two congruences for the (generalized) harmonic numbers requires two sums, whereas I'm only evaluating a single sum. - -REPLY [11 votes]: I think your proof is fine, but you can take a peek at this paper where it is proved that the numerator of -$$H(p-1):=1+\frac{1}{2}+...+\frac{1}{p-1}=\sum_{k=1}^{p-1}\frac{1}{k}$$ -is divisible by $\,p^2\,\,,\,p>3$ a prime. -Claims 1-2 in this paper are pretty nice (claim 3 is Geoff's remark), and the proof flows smoothly, imo.<|endoftext|> -TITLE: Limit inferior/superior of sequence of sets -QUESTION [8 upvotes]: Let $(\Omega, \mathcal{A}, \mu)$ be a measure space, where $\mu(\Omega)< \infty$. Further $(A_n)_{n \in \mathbb{N}}$ is a a sequence of $\mathcal{A}$-measurable sets. I want to prove, that -$$ \mu ( \liminf_{n \rightarrow \infty} A_n) \leq \liminf_{n \rightarrow \infty} \mu (A_n) \leq \limsup_{n \rightarrow \infty} \mu (A_n) \leq \mu (\limsup_{n \rightarrow \infty} A_n)$$ -holds for any sequence $(A_n)_{n \in \mathbb{N}}$. -I have no experience working with the limit superior/inferior. Clearly -$$\mu ( \liminf_{n \rightarrow \infty} A_n) \leq \mu (\limsup_{n \rightarrow \infty} A_n)$$ -holds, since it is easy to prove that the one is a superset of the other. Also -$$\liminf_{n \rightarrow \infty} \mu (A_n) \leq \limsup_{n \rightarrow \infty} \mu (A_n)$$ -holds, for any sqeuence. But I am stuck how to show the connection. I could use the Definitions, then I get -$$ \mu (\bigcup_n^\infty \bigcap_{k=n}^\infty A_n) \leq \lim_{n \rightarrow \infty} \inf_{k \geq n} \mu(A_n) \leq \inf_{n \geq 0} \sup_{k \geq n} \mu(A_n) \leq \mu (\bigcap_n^\infty \bigcup_{k=n}^\infty A_n) $$ -But I don't know if this helps. Anyone got a hint how to go on? - -REPLY [5 votes]: Consider sets -$$ -B_n=\bigcap\limits_{k=n}^\infty A_k -$$ -Obviously, $B_n\subset A_k$ for all $k\geq n$, so $\mu(B_n)\leq \mu(A_k)$ for all $k\geq n$ -After taking infimum we get $\mu(B_n)\leq\inf_{k\geq n}\mu(A_k)$. Since $B_n\subset B_{n+1}$ for all $n\in\mathbb{N}$ then the sequence $\{\mu(B_n):n\in\mathbb{N}\}$ is non-decreasing, so there exist $\lim_{n\to\infty}\mu(B_n)$. Similarly the sequence $\{\inf_{k\geq n}\mu(A_k):n\in\mathbb{N}\}$ is non-decreasing hence there exist $\lim_{n\to\infty}\inf_{k\geq n}\mu(A_k)$. Since existence of limits is justified we can say -$$ -\lim\limits_{n\to\infty}\mu(B_n)\leq -\lim\limits_{n\to\infty}\inf\limits_{k\geq n}\mu(A_k)=\liminf\limits_{n\to\infty}\mu(A_n)\tag{1} -$$ -Again recall that $B_n\subset B_{n+1}$ for all $n\in\mathbb{N}$, so -$$ -\mu\left(\bigcup\limits_{n=1}^\infty B_n\right)=\lim\limits_{n\to\infty}\mu(B_n)\tag{2} -$$ -It is remains to note that -$$ -\liminf\limits_{n\to\infty}A_n= -\bigcup\limits_{n=1}^\infty \bigcap\limits_{k=n}^\infty A_k= -\bigcup\limits_{n=1}^\infty B_n\tag{3} -$$ -From $(1)$, $(2)$ and $(3)$ we have -$$ -\mu\left(\liminf\limits_{n\to\infty}A_n\right)\leq \liminf\limits_{n\to\infty}\mu(A_n) -$$ -Now try to prove in the similar way the second inequality.<|endoftext|> -TITLE: Concrete example of calculation of $\ell$-adic cohomology -QUESTION [11 upvotes]: Let $p$ and $\ell$ be distinct prime numbers. -Consider in the affine plane $\mathbb{A}^2_{\mathbb{F}_p}$ with coordinates $(x,y)$ the union $L$ of the axes $x = 0$ and $y = 0$. -How does one compute the $\ell$-adic cohomology groups with compact support $H^i_c(L,\mathbb{Q}_\ell)$? I thought I had some idea of what $\ell$-adic cohomology is, but I don't even manage to do this example... - -REPLY [6 votes]: (1) All we really need to know to make the computation is that cohomology with compact support $H^*_c$ (with values in $K$-vector spaces) satisfies the following properties: - -(Localization sequence) If $i: Z \hookrightarrow X$ is a closed immersion and $j:U\hookrightarrow X$ the complementary open immersion, then for any sheaf $F$ we have a long exact sequence -$$ - \to H^{i}_c(U) \to H^{i}_c(X) \to - H^{i}_c(Z) \to H^{i+1}_c(U) \to -$$ -(Cohomology of the affine space) $ - H^{i}_c(\mathbb{A}^n) = \begin{cases} K & if~ i=2n \cr 0 & else \end{cases} -$ - -(2) Now we can play with the long exact sequences -Writing the localization sequence for $\mathbb{A}^1 = \{0\} \coprod \mathbb{G}_m$ one finds -$ - H^{i}_c(\mathbb{G}_m) = \begin{cases} K & if~ i=1,2 \cr 0 & else \end{cases} -$ This is dual to $H^{2-i}(\mathbb{G}_m)$ as expected. -Then the localization sequence for $L = \mathbb{A}^1 \coprod \mathbb{G}_m$ reduces to -$$ - 0\to H^{0}_c(L) \to 0 \to - K \to H^{1}_c(L) \to 0 \to - K \to H^{2}_c(L) \to K \to - 0 -$$ -so -$ - H^{i}_c(L) = \begin{cases} K & if~ i=1 \cr K^2 & if~ i=2 \cr 0 & else \end{cases} -$ -(3) I should add a few word about the localizing sequence. For any immersion $j: U\hookrightarrow X$, we have 3 functors on sheaves - -the restriction functor $j^*$ -its right adjoint: the classical direct image $j_*$ -the extension by zero $j_!$. - -Facts (see J.S. Milne's lecture notes for example): - -$j_!$ is always exact -If $j$ is a closed immersion then $j_! = j_*$. -If $i: Z \hookrightarrow X$ is a closed immersion and $j:U\hookrightarrow X$ the complementary open immersion, then for any sheaf $F$ we have an exact sequence -$$ - 0 \to j_!j^*F \to F \to i_*i^*F \to 0 -$$ - -Now if the variety $X$ is not proper, you can always find a dense open immersion $u:X\hookrightarrow \overline{X}$ into a proper variety. Since $u_!$ is exact, that $u_!i_* = u_!i_! = (ui)_!$ and applying $H^*(\overline{X},-)$ we obtain a long exact sequence -$$ - \to H^{i}_c(U,j^*F) \to H^{i}_c(X,F) \to - H^{i}_c(Z,i^*F) \to H^{i+1}_c(U,j^*F) \to -$$ -For the computation $H^*_c(\mathbb{A}^n)$ it reduces to that of $H^*_c(\mathbb{P}^n)$ by the localizing sequence. But since $\mathbb{P}^n$ is proper, we have $H^{*}_c(\mathbb{P}^n) = H^*(\mathbb{P}^n) = K[h]/(h^{n+1})$ where $h$ is the class of any hyperplane and has degree 2. Moreover, the morphism $\mathbb{P}^n \hookrightarrow \mathbb{P}^{n+1}$ induces the natural projection.<|endoftext|> -TITLE: Where are good resources to study combinatorics? -QUESTION [24 upvotes]: I am an undergraduate wiht basic knowledge of combinatorics, but I want to obtain sound knowledge of this topic. Where can I find good resources/questions to practice on this topic? -I need more than basic things like the direct question 'choosing r balls among n' etc.; I need questions that make you think and challenge you a bit. - -REPLY [6 votes]: Lots of good suggestions here. Another freely available source is Combinatorics Through Guided Discovery. It starts out very elementary, but also contains some interesting problems. And the book is laid out as almost entirely problem-based, so it useful for self study.<|endoftext|> -TITLE: Finding the radical of an integer -QUESTION [11 upvotes]: Given a number $x = p_1^{e_1}\cdots p_n^{e_n}$ with different primes $p_i$ and exponents $e_i \ge 1$, is there an efficient way to find $p_1\cdots p_n$? - -I ask this because for polynomials it's easy: with $K$ a field of characteristic $0$ and $$f = g_1^{e_1} \cdots g_n^{e_n} \in K[X]$$ irreducible we have $$g_1 \cdots g_n = \frac{f}{\mathrm{gcd}(f,f\,')}$$ where $f\,'$ is the formal derivative. -This proof can't be used for integers, unless there's a trick that I don't see. - -REPLY [15 votes]: Currently, no feasible (polynomial time) algorithm is known for -recognizing squarefree integers or for computing the squarefree -part of an integer. In fact it may be the case that this problem -is no easier than the general problem of integer factorization. -Computing the radical $\rm\:rad(n)\:$ is equivalent to computing the squarefree part $\rm\:sf(n)\:$ because -$$\rm rad(n)\, =\, sf(n)\, sf(n/rad(n)) $$ -This problem is important because one of the main tasks -of computational algebraic number theory reduces to it (in -deterministic polynomial time). Namely the problem of computing -the ring of integers of an algebraic number field depends upon -the square-free decomposition of a polynomial discriminant -when computing an integral basis, e.g. [2] S.7.3 p.429 or [1] -This is due to Chistov [0]. See also Problems 7,8, p.9 in [3], -which lists 36 open problems in number theoretic complexity. -The primary reason that such problems are simpler in function fields versus number fields -is due to the availability of derivatives. This opens up a powerful -toolbox that is not available in the number field case. For example -once derivatives are available so are Wronskians - which provide powerful -measures of dependence in transcendence theory and diophantine approximation. -A simple yet stunning example is the elementary proof of the polynomial case of Mason's ABC theorem, which yields as a very special case a high-school-level proof of FLT for polynomials, cf. -my MO post and my old sci.math post [4]. -Such observations have motivated searches for "arithmetic analogues of derivations". For example, see Buium's paper by that name in Jnl. Algebra, 198, 1997, 290-99, and see his book Arithmetic differential equations. -[0] A. L. Chistov. The complexity of constructing the ring of integers -of a global field. Dokl. Akad. Nauk. SSSR, 306:1063--1067, 1989. -English Translation: Soviet Math. Dokl., 39:597--600, 1989. 90g:11170 -http://citeseerx.ist.psu.edu/showciting?cid=854849 -[1] Lenstra, H. W., Jr. Algorithms in algebraic number theory. -Bull. Amer. Math. Soc. (N.S.) 26 (1992), no. 2, 211--244. 93g:11131 -http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.105.8382 -[2] Pohst, M.; Zassenhaus, H. Algorithmic algebraic number theory. -Cambridge University Press, Cambridge, 1997. -[3] Adleman, Leonard M.; McCurley, Kevin S. -Open problems in number-theoretic complexity. II. -Algorithmic number theory (Ithaca, NY, 1994), 291--322, -Lecture Notes in Comput. Sci., 877, Springer, Berlin, 1994. 95m:11142 -http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.48.4877 -[4] Dubuque, Bill. $ $ sci.math.research post, 1996/07/17 -poly FLT, abc theorem, Wronskian formalism [was: Entire solutions of f^2+g^2=1] -http://groups.google.com/group/sci.math/msg/4a53c1e94f1705ed -http://groups.google.com/groups?selm=WGD.96Jul17041312@berne.ai.mit.edu<|endoftext|> -TITLE: An irreducible polynomial $f \in \mathbb R[x,y]$, whose zero set in $\mathbb A_{\mathbb R}^2$ is not irreducible -QUESTION [14 upvotes]: This is an exercise on Page 8 of Hartshone's Algebraic Geometry: - -Give an example of an irreducible polynomial $f \in \mathbb R[x,y]$, whose zero set $Z(f)$ in $\mathbb A_{\mathbb R}^2$ is not irreducible. - -I think such an example must come from the fact that $\mathbb R$ is not algebraically closed. But I have no idea as to finding a concrete one. -Thanks very much. - -REPLY [4 votes]: Added: This answer interprets "irreducible" to mean with respect to the classical (analytic) topology on $C(\mathbb{R})$, not the (relative) Zariski topology. -Consider the elliptic curve $y^2 = x^3 + Ax + B$, for $A,B \in \mathbb{R}$ with -$\Delta = 4A^3 + 27B^2 \neq 0$. -Then the polynomial $f(x,y) = x^3 + Ax + B - y^2$ is irreducible. However the real zero set of $f$ has two components iff $x^3 + Ax + B$ has three real roots (see e.g. this picture for a good idea of what's going on here). Thus taking for instance $A = -1, B = 0$ gives an answer to the question. -If you want to think about things this way, a natural followup result is Harnack's Theorem: if $C/\mathbb{R}$ is a smooth, geometrically integral projective curve of genus $g$, then the real locus $C(\mathbb{R})$ has at most $g+1$ connected (hence irreducible, by smoothness) components, and all numbers of components between $0$ and $g+1$ are possible. Thus the above example is in some sense the simplest.<|endoftext|> -TITLE: Coin tosses until I'm out of money -QUESTION [9 upvotes]: The question I think is a simple one, but I've been unable to answer or find an answer for it yet: -There's a simple game: if you flip heads you win a dollar (from the house), but if you flip tails you lose a dollar (to the house). -If I start with n dollars (and the house has infinite money), how many flips can I expect to do before I've lost all my money? This is different than the common question of how many flips can I do before I have a run of length 'n'. In this case you can lose your money by never having a run of length more than 2, for example, simply by repeating win 1, lose 2, win 1, lose 2, etc... -I can write out a decision tree on this, but I haven't been able to generalize it into a formula yet. - -REPLY [4 votes]: Your question can be described via a simple random walk on $\mathbb Z$: Choose i.i.d. random variables $X_1, X_2, \ldots$ such that -$$P[X_i = +1] = P[X_i = -1] = \frac 12$$ -$X_i$ describes the outcome of the $i$-th toss of a coin: $+1$ means we win one dollar, $-1$ we lose one dollar. Furthermore, let $S_k = \sum_{i=1}^k X_i$ be the amount of money lost or won until time $k$. -Now, for a positive integer $n$ (which describes the initial amount of money we own), define $$T_{-n} = \inf\{k\in \mathbb N\mid S_k = -n\}$$ to be the number of coin tosses until we go bankrupt. Then we have - -Lemma: $P[T_{-n} > k] = \Theta(1/\sqrt k) \qquad \text{ as }k\to \infty$ - -which is to say: The probability of not being bankrupt after the $k$-th toss of a coin decreases like $1/\sqrt{k}$ (in particular, this goes to zero; so we will go bankrupt, eventually). On the other hand the calculation $$E[T_{-n}]= \sum_{k=1}^\infty P[T_{-n}>k] \approx \sum_{k=1}^\infty \frac{1}{\sqrt{k}} = \infty$$ shows - -Corollary: $E[T_{-n}] = \infty$ - -which says, we may well have to wait a very very long time before we go bankrupt. - - -In an attempt to be self-contained. Here is -Proof of the Lemma: The proof involves two steps: - -Claim 1: We have $$P[S_n = 2k-n] = \binom{n}{k}2^{-n}$$ for $k\le n$, and $P[S_n = x] = 0$ for all other $x$. - -and - -Claim 2: For $k>0$ we have $$P[T_{-n} \le k] = P[S_k \notin (-n, n]\,] $$ - -For Claim 1 just note that in order for $S_n = 2k-n$, we need to win $k$ times and lose $n-k$ times and there are $\binom nk$ possibilites to choose $k$ winners out of $n$. -For Claim 2 write $P[T_{-n} \le k] = \sum_{b = -\infty}^\infty P[T_{-n} \le k, S_k = b]$ and notice that for $b > -n$ we have $$P[T_{-n} \le k, S_k = b] = P[S_k = -2n - b]$$ The last assertion is obtained by reflecting each path visiting $-n$ and ending at $b$ at the first time it hits $-n$. So that for the second part of its path (after first hitting $-n$), all the values of the $X_i = \pm 1$ get swapped to $X_i = \mp 1$. This gives a one-to-one correspondence between paths to $b>-n$, which visit $-n$ and paths to $-2n-b$. (I'm sure this explanation for Claim 2 is hardly understandable, but I can't come up with a better explanation... This is sometimes called the reflection principle). So -\begin{align} -P[T_{-n} \le k] &= \sum_{b = -\infty}^\infty P[T_{-n} \le k, S_k = b] \\ -&= \sum_{b\le -n} P[S_k = b] + \sum_{b> -n} P[S_k = -2n -b] \\ -&= P[S_k = -n] + 2P[S_k < -n] \\[6pt] -&= P[S_k \notin (-n,n]\, ] -\end{align} -Therefore (even values of $k$ suffice by monotonicity) -\begin{align} -P[T_{-n} > 2k] &\ge P[S_{2k} = 0\, ] = \binom{2k}{k} 2^{-2k} \sim \frac{1}{\sqrt{\pi k}} \\ -P[T_{-n} > 2k] &\le 2n \cdot P[S_{2k} = 0\, ] = 2n\cdot \binom{2k}{k} 2^{-2k} \sim \frac{2n}{\sqrt{\pi k}} -\end{align} -q.e.d.<|endoftext|> -TITLE: Hypergeometric formulas for the Rogers-Ramanujan identities? -QUESTION [9 upvotes]: Let $q = e^{2\pi i \tau}$. Given the j-function, -$$j = j(q) = 1/q + 744 + 196884q + 21493760q^2 + \dots$$ -and define, -$$k = j-1728$$ -Let $\tau =\sqrt{-N}$, where $N > 1$. Anybody knows how to prove the RHS of these conjectured relations?: -$$\begin{align}q^{-1/60} G(q) = q^{-1/60} \prod_{n=1}^\infty \frac{1}{(1-q^{5n-1})(1-q^{5n-4})} &= j\,^{1/60}\,_2F_1\left(\tfrac{19}{60},\tfrac{-1}{60},\tfrac{4}{5},\tfrac{1728}{j}\right)\\ -&= k\,^{1/60}\,_2F_1\left(\tfrac{29}{60},\tfrac{-1}{60},\tfrac{4}{5},\tfrac{-1728}{k}\right)\\[2.5mm] -q^{11/60} H(q) = q^{11/60} \prod_{n=1}^\infty \frac{1}{(1-q^{5n-2})(1-q^{5n-3})} &= j\,^{-11/60}\,_2F_1\left(\tfrac{31}{60},\tfrac{11}{60},\tfrac{6}{5},\tfrac{1728}{j}\right)\\ -&= k\,^{-11/60}\,_2F_1\left(\tfrac{41}{60},\tfrac{11}{60},\tfrac{6}{5},\tfrac{-1728}{k}\right)\end{align}$$ - -REPLY [5 votes]: Let -$$\begin{align} - g(\tau) &= q^{-1/60} G(q) -\\ h(\tau) &= q^{11/60} H(q) -\end{align}$$ -First of all, the equalities between the hypergeometric $j$ and $k$ expressions -follow from the eulerian transformation -$${}_2F_1(a,b;c;z) = (1-z)^{-b}\,{}_2F_1\left(c-a,b;c;\frac{z}{z-1}\right)$$ -Therefore it suffices to prove the identities between $g$ resp. $h$ -and the corresponding hypergeometric $j$ expressions. -I will translate those to more familiar identities. -We will use the Rogers-Ramanujan continued fraction -(RRCF), -$$\rho(\tau) = \frac{h(\tau)}{g(\tau)} = q^{1/5}\frac{H(q)}{G(q)}$$ -Formula $(22)$ from the above MathWorld entry on RRCF essentially states that -$$\frac{1}{\rho^{5}} - 11 - \rho^5 = \frac{1}{g^6\,h^6}$$ -(Use the product representation of $g$ and $h$ to identify the right-hand side). -From this, we can easily deduce -$$\begin{align} - g &= \frac{1} - {\left(\rho - 11\,\rho^6 - \rho^{11}\right)^{1/12}} -\\ h &= \frac{\rho} - {\left(\rho - 11\,\rho^6 - \rho^{11}\right)^{1/12}} -\end{align}$$ -assuming that the arguments to the radicals are small positive reals, -which should follow from the restrictions you have placed on $\tau$. -Furthermore, formula $(46)$ from the above Mathworld entry on RRCF -gives the relation of $\rho$ with Klein's $j$: -$$j = \frac -{\left(1+228\,\rho^5+494\,\rho^{10}-228\,\rho^{15}+\rho^{20}\right)^3} -{\left(\rho - 11\,\rho^6 - \rho^{11}\right)^5}$$ -which allows us to write -$$\begin{align} - g\,j^{-1/60} &= -\left(1+228\,\rho^5+494\,\rho^{10}-228\,\rho^{15}+\rho^{20}\right)^{-1/20} -\\ h\,j^{11/60} &= \frac -{\left(1+228\,\rho^5+494\,\rho^{10}-228\,\rho^{15}+\rho^{20}\right)^{11/20}} -{1 - 11\,\rho^5 - \rho^{10}} -\end{align}$$ -again assuming that the arguments to the radicals are small positive reals. -I need a deus ex machina now, and Raimundas Vidūnas arXiv:0807.4808v1 comes to the rescue. -His formulae $(59)$ and $(61)$ -in section 6.3 ("icosahedral hypergeometric equations", p. 20) -state precisely that -$$\begin{align} -{}_2F_1\left(\tfrac{19}{60},-\tfrac{1}{60};\tfrac{4}{5};\varphi_1(x)\right) - &= \left(1-228\,x+494\,x^2+228\,x^3+x^4\right)^{-1/20} -\\{}_2F_1\left(\tfrac{31}{60},\tfrac{11}{60};\tfrac{6}{5};\varphi_1(x)\right) - &= \frac{\left(1-228\,x+494\,x^2+228\,x^3+x^4\right)^{11/20}} - {1 + 11\,x - x^2} -\\\text{where}\qquad - \varphi_1(x) &= \frac{-1728\,x\,(1+11\,x-x^2)^5} - {\left(1-228\,x+494\,x^2+228\,x^3+x^4\right)^3} -\end{align}$$ -And your conjecture follows from setting $x=-\rho^5$ which implies -$\varphi_1(x) = \frac{1728}{j}$. -Summarizing, there is an algebraic relation between $j$ and $g$ resp. $h$, -and the hypergeometric ${}_2F_1$ expressions are designed to solve those -algebraic relations for $g\,j^{-1/60}$ resp. $h\,j^{11/60}$, given $j$.<|endoftext|> -TITLE: Solving Recurrence $T_n = T_{n-1}*T_{n-2} + T_{n-3}$ -QUESTION [5 upvotes]: I have a series of numbers called the Foo numbers, where $F_0 = 1, F_1=1, F_2 = 1 $ - then the general equation looks like the following: -$$ -F_n = F_{n-1}(F_{n-2}) + F_{n-3} -$$ -So far I have got the equation to look like this: -$$T_n = T_{n-1}*T_{n-2} + T_{n-3}$$ -I just don't know how to solve the recurrence. I tried unfolding but I don't know if i got the right answer: -$$ -T_n = T_{n-i}*T_{n-(i+1)} + T_{n-(i+2)} -$$ -Please help, I need to describe the algorithm which I have done but the analysing of the running time is frustrating me. - -REPLY [7 votes]: Numerical data looks very good for -$$F_n \approx e^{\alpha \tau^n}$$ -where $\tau = (1+\sqrt{5})/2 \approx 1.618$ and $\alpha \approx 0.175$. Notice that this makes sense: When $n$ is large, the $F_{n-1} F_{n-2}$ term is much larger than $F_{n-3}$, so -$$\log F_n \approx \log F_{n-1} + \log F_{n-2}.$$ -Recursions of the form $a_n = a_{n-1} + a_{n-2}$ always have closed forms $\alpha \tau^n + \beta \tau^{-n}$. - -Here's a plot of $\log \log F_n$ against $n$, note that it looks very linear. A best fit line (deleting the first five values to clean up end effects) gives that the slope of this line is $0.481267$; the value of $\log \tau$ is $0.481212$. -Your sequence is in Sloane but it doesn't say much of interest; if you have anything to say, you should add it to Sloane.<|endoftext|> -TITLE: What is the importance of functions of class $C^k$? -QUESTION [9 upvotes]: In all calculus textbooks, after the part about successive derivatives, the $C^k$ class of functions is defined. -The definition says : - -A function is of class $C^k$ if it is differentiable $k$ times and the $k$-th derivative is continuous. - -Wouldn't be more natural to define them to be the class of functions that are differentiable $k$ times? -Why is the continuity of the $k$th derivative is so important so as to justify a specific definition? - -REPLY [4 votes]: You can certainly consider $k$-times differentiable functions on, say, $[a,b]\subset \mathbb R$ and give them a notation, like $D^k[a,b]$. -The point however is that many well-known and interesting theorems, true for $C^k[a,b]$, will fail for $D^k[a,b]$or won't even make sense. Here are three examples from elementary calculus: -a) Integration by parts for $u,v\in C^1[a,b]$: $$\int_{a}^{b}u(x)v^{\prime }(x)dx=\left( -u({b})v(b)-u(a)v(a)\right) -\int_{a}^{b}u^{\prime -}(x)v(x)dx$$ The integrals don't even make sense a priori if $u', v'$ are not continuous. -b) Taylor's formula -$$f(x)=\sum_{i=0}^k\frac{f^{(i)}(a)}{i!}(x-a)^i+\int_a^x\frac{f^{(k+1)}(t)}{k!}(x-t)^kdt\quad (x\in [a,b])$$ is valid for $f\in C^{k+1}([a,b])$ and again doesn't make sense for $f\in D^{k+1}([a,b])$ -c) The change of variables formula $$ \int_a^b f(\phi (t))\phi'(t)dt=\int_{\phi(a)}^{\phi(b)} f(x)dx $$ again necessitates that $\phi$ be a $C^1$-function and not merely a differentiable one.<|endoftext|> -TITLE: An entire function is identically zero? -QUESTION [14 upvotes]: I'm preparing for a PhD prelim in Complex Analysis, and I encountered this question from an old PhD prelim: -Suppose $f(z)$ is an entire function such that $|f(z)| \leq \log(1+|z|) \forall z$. Show that $f \equiv 0$. -Well, for $z=0$, $|f(0)| \leq 0$. On the other hand, for $z \neq 0$, $\log(1+|z|) > 0$, a positive constant. I'm guessing this would mean that $f$ turns out to be a bounded entire function, so then by Liouville's theorem, $f$ is constant, but this doesn't necessarily mean that $f \equiv 0$, does it? Am I wrong somewhere? Some guidance would be much appreciated! - -REPLY [4 votes]: We already know that $f(0) = 0$. Now from $|f(z)| \leq \log(1 + |z|)$ and we get that on every circle of radius $R$ about the origin, -$$|f^{(n)}(0)| \leq \frac{\log(1 + |R|) \times n!}{R^n} \hspace{5mm} \forall n \geq 1$$ -by the Cauchy estimate. Since $R^n$ grows faster than $\log (1+ |R|)$ when $n \geq 1$, we see that $f^{(n)}(0) = 0$ for all $n \geq 1$. But we also know $f(0) = 0$ so that $f^{(n)}(0) = 0$ for all $n \geq 0$. Since $f$ is entire the identity principle implies $f \equiv 0$.<|endoftext|> -TITLE: Dolbeault cohomology of the complex projective space. -QUESTION [8 upvotes]: Let $X=\mathbb{CP}^n$. We proved using the hodge decomposition that $H^0(X,\Omega^p)=0$ if $p\neq 0$. But I do not understand why I cannot have global holomorphic differential p-forms not even constants. - -I want to understand why $H^0(X,\Omega^p)=0$ without using the Hodge decomposition. -And without using GAGA I would like to see a proof of $H^0(Proj(\mathbb{C}[x_0,..,x_n]),\Omega_{X/\mathbb{C}}^p)=0$ for $p\neq 0$ - -REPLY [7 votes]: I'll flesh out my hint. Suppose that $\alpha$ is a global holomorphic $p$-form on $\mathbb{P}^{n-1}$. Pullback $\alpha$ along the map $\mathbb{C}^{n} \setminus \{ 0 \} \to \mathbb{P}^{n-1}$; let the pulled back form be $\beta = \sum f_I(x_1, \ldots, x_n) d x_{i_1} \wedge \cdots d x_{i_p}$. The functions $f_I$ are holmorphic functions on $\mathbb{C}^{n} \setminus \{ 0 \}$; by Hartog's Lemma, they extend to holomorphic functions on $\mathbb{C}^n$. -The fact that $\beta$ is pulled back from $\mathbb{P}^{n-1}$ means that $\beta$ must be invariant under dilation. I.e. $\delta_t^{\ast} \beta = \beta$ where $\delta_t$ is the map $(x_1, \ldots, x_n) \mapsto (t x_1, \ldots, t x_n)$. Translating into specific equations, this shows that $f_I(t x_1, \ldots, t x_n) = t^{-p} f_I(x_1, \ldots, x_n)$. -If $f_I(x_1, \ldots, x_n)$ is nonzero, this implies that $\lim_{t \to 0} f_I(t x_1, \ldots, t x_n) = \infty$. This contradicts that $f_I$ extends to a homolorphic function at $0$.<|endoftext|> -TITLE: Difficulty of functional analysis exam -QUESTION [5 upvotes]: I've just written the final exam in my introductory course to functional analysis (2nd year bachelor degree). I felt quite well prepared but nevertheless found the exam pretty challenging in the timeframe of two hours. I'd appreciate any comments about what you think about it! The exam can be found here: http://www.math.lmu.de/~michel/SS12_FA_Final_Test_01.pdf -In particular I'd appreciate any hints how to solve 6 ii), which states: - -Use the fourier-series of $$f(x)=\begin{cases}x^2 &\mbox{ for } x\in[0,\pi] \\ (2\pi-x)^2 &\mbox{ for } x\in (\pi,2\pi ]\end{cases} $$ to calculate $\sum_{n\in\mathbb{N}}\frac{(-1)^{n-1}}{n^2}$ - -REPLY [2 votes]: I would find the exam 'challenging' to do in 2 hours. The idea part is fine for me, the computation too demanding for me in that time. -The Fourier series for $f$ are $\hat{f}_0 = \frac{\pi^2}{3}$, $\hat{f}_n = 2 \frac{(-1)^n}{n^2}$, for $n \neq 0$. Then since $f$ is Lipschitz at $0$, we have -$f(0) = \sum_n \hat{f}_n$. Since $f(0) = 0$, we get $\hat{f}_0 = - \sum_{n \neq 0} \hat{f}_n$. Since $\hat{f}_n = \hat{f}_{-n}$, we have $\hat{f}_0 = - 2\sum_{n > 0} \hat{f}_n$, from which it follows that -$$ \sum_{n > 0} \frac{(-1)^{n-1}}{n^2} = \frac{\pi^2}{12}.$$ -(I used wxmaxima to do the fairly elementary integration, I tried by hand but made too many mistakes!)<|endoftext|> -TITLE: Is the continued fraction of the square root of a base $\phi$ (golden ratio) number periodic when the continued fraction is expressed in base $\phi$? -QUESTION [5 upvotes]: I have been looking at concise ways to represent irrational numbers using only integers. -I was thinking about base $\phi$ (golden ratio base) and how it can represent the quadratic extension of the rationals with $\sqrt 5$ (i.e. $\mathbb Q[\sqrt 5]$) in a finite (potentially infinite but periodic) representation. For example, $5 + \sqrt 5 = 10000.01$ exactly. -I also noticed that $\sqrt a$ always results in a periodic continued fraction (CF) when $a$ is a square-free integer. The golden ratio $\phi$ is the least accurate irrational number to approximate by truncating its continued fraction, as it is [1, 1, 1, ...] (see Hurwitz's theorem and the Wikipedia Continued Fractions page). -Since truncating $\phi$'s CF is not as accurate a rational approximation as for other irrationals, and given that base $\phi$ allows us to represent $\phi$ concisely ($\phi = 10$) and allows a representation of integers (e.g. $2 = 10.01$), which is what CF's already use, - -Are there any potential problems with using base $\phi$ as the base for a CF representation considering it gets around the accuracy issue for $\phi$ and allows continued fractions involving $\sqrt 5$ to be finite (I think)? -Is the continued fraction of the square root of a base $\phi$ number periodic when the CF is expressed in base $\phi$, in much the same manner as a square free integer? -Are there any other advantages/disadvantages to representing continued fractions in this base instead of another more typical base? - -This is my first post, so please let me know if I need to correct anything. - -REPLY [2 votes]: Steven has given an excellent answer, but I fear many people will not read all the comments and decide that the answer is basically no instead of yes due to the loss of a canonical form of continued fractions. -I will now try to demonstrate how to go about finding a good representation for base-$\phi$ square roots and why this might be useful. -It turns out that the sine or cosine of any integer multiple of $\pi/60$ can be represented as a linear combination of the following 6 factors (see Exact Trigonometric Constants): -$1$ -$\sqrt 2$ -$\sqrt 3$ -$\sqrt 5$ -$\sqrt{5+\sqrt 5}$ -$\sqrt{5-\sqrt 5}$ -Except for 1, these are basically quadratic extensions of $\mathbb Q$. -I want to represent a subset of the constructable numbers that use only sines and cosines of angles -$\dfrac{n\pi}{60}$ where $n \in \mathbb Z$. -Generalized continued fractions (GCF) allow me to represent square roots using the following formula (see Methods of Computing Square Roots): -$\sqrt z = \sqrt{x^2 + y} = x + \dfrac{y}{2x + \dfrac{y}{2x + \dfrac {y}{2x + \dfrac{y}{\ddots}}}}$ -where $x$ and $y$ are positive real numbers, thus guaranteeing convergence as long as the product of successive $2x$'s grows faster than the product of successive $y$'s (see the Convergence Problem). Generally we want $x$ to be as large as possible and $y$ to be as small as possible for faster convergence. -For $\sqrt 2$, $x = 1$ and $y = 1$. -For $\sqrt 3$, x = 1 and $y = 2$ does not converge quickly. Let's take a look in base-$\phi$. -$3 = 100.01 = \phi^2 + \phi^{-2}$ so $x = \phi$ and $y = \phi^{-2}$ will give much faster convergence. You might wonder if $\sqrt 2$ has a similar faster convergence, but as $2 = 10.01 = \phi + \phi^{-2}$ we don't have an easy way to represent $x$ as $\sqrt \phi$. If we re-write it as $2 = 1.11 = 1 + \phi^{-1} + \phi^{-2}$ we can take the $1$ for $x$ leaving $y = 0.11 = 1.00 = 1$ for $y$, which simply recovers $x = 1$ and $y = 1$. -$\sqrt 5$ is already directly representable in base-$\phi$, but if we wanted to use integers, $x = 2$ and $y = 1$ works just fine, and base-$\phi$ for $5$ doesn't give us a better convergence. -Now for the fun part. -$5 + \sqrt 5 = 10000.01 = \phi^4 + \phi^{-2}$, so $x = \phi^2$ and $y = \phi^{-2}$ will give good convergence, and -$5 - \sqrt 5 = 100.0001 = \phi^2 + \phi^{-4}$, so $x = \phi$ and $y = \phi^{-4}$ will give good convergence, both in a finitely representable periodic form. -Combinations of these factors like $\sqrt 6 = \sqrt 2\sqrt 3$ (for $15^\circ$) and $\sqrt 2\sqrt{5 + \sqrt 5}$ (for $12^\circ$) are also periodic GCF's. -This gives us a nice finite way to represent vectors made from sines and cosines of integer multiples of $3^\circ$. -So to answer my original questions: - -Yes. One problem is that uniqueness is lost for CF representations. There are infinitely many ways to represent the same CF. This is related to the problem base-$\phi$ has in that there are infinitely many finite ways to represent the same number (using the 100 = 011 conversion). This is not a problem as long as we can find a consistent representation for numbers we are interested in. -Yes, at the cost of uniqueness. -One advantage is potentially faster convergence of a GCF than with only integers. Another is the ability to represent quadratic extensions (square roots) of $\mathbb Z[\phi]$ in a periodic form.<|endoftext|> -TITLE: Closed forms for various expectations involving the standard normal CDF -QUESTION [7 upvotes]: Suppose that $X\sim\mathcal{N}\left(0,1\right)$ (i.e., $X$ is a standard normal random variable) and $a,b,$ and $c$ are some real constants. Does any of the following expectations have a closed-form? - -$\mathbb{E}\left[\log\Phi\left(aX\right)\right]$ -$\mathbb{E}\left[\Phi\left(aX\right)\log\Phi\left(aX\right)\right]$ -$\mathbb{E}\left[\Phi\left(aX\right)\Phi\left(bX+c\right)\right]$ - -I've tried the usual derivative trick followed by Stein's lemma, but the expressions didn't get much simpler. For the third one if $a=b$ and $c=0$ the closed-form solution is $$\mathbb{E}\left[\left(\Phi\left(aX\right)\right)^2\right]=\frac{\tan^{-1}\left(\sqrt{1+2a^2}\right)}{\pi}$$ but I couldn't reach at any generalization. - -REPLY [5 votes]: Cases 1. and 2. when $a=\pm1$ follow from the following remark: for every suitable function $u$, -$$ -\mathrm E(u'(\Phi(X)))=\int_{-\infty}^{+\infty} u'(\Phi(x))\varphi(x)\,\mathrm dx=\left[u(\Phi(x))\right]_{x=-\infty}^{x=+\infty}=u(1)-u(0). -$$ -For example, $u(t)=t\log(t)-t$ yields $u'(t)=\log t$, $u(1)=-1$ and $u(0)=0$ hence -$$ -\mathrm E(\log\Phi(aX))=-1,\quad a=\pm1. -$$ -Likewise, $u(t)=\frac12t^2\log t-\frac14t^2$ yields $u'(t)=t\log t$, $u(1)=-\frac14$ and $u(0)=0$ hence -$$ -\mathrm E(\Phi(aX)\log\Phi(aX))=-\tfrac14,\quad a=\pm1. -$$ -About case 3., $u(t)=\frac13t^3$ yields $u'(t)=t^2$, $u(1)=\frac13$ and $u(0)=0$ hence -$$ -\mathrm E(\Phi(aX)^2)=\tfrac13,\quad a=\pm1. -$$ - -Another approach is to differentiate with respect to the parameter $a$ and to use Stein's lemma. Consider $v(a)=\mathrm E(u(\Phi(aX)))$ for some suitable function $u$, then $v(0)=u(\frac12)$ and -$$ -v'(a)=\mathrm E(X\varphi(aX)u'(\Phi(aX)))=\mathrm E(Xg(X))$$ where $$ g(x)=\varphi(ax)u'(\Phi(ax)). -$$ -Stein's lemma and the identities $\varphi'(s)=-s\varphi(s)$ and $\Phi'(s)=\varphi(s)$ yield -$$ -v'(a)=\mathrm E(g'(X))=\mathrm E(-a^2X\varphi(aX)u'(\Phi(aX))+a\varphi(aX)^2u''(\Phi(aX))), -$$ -hence -$$ -v'(a)=-a^2v'(a)+a\mathrm E(\varphi(aX)^2u''(\Phi(aX))), -$$ -and -$$ -v'(a)=\frac1{1+a^2}\mathrm E(\varphi(aX)^2u''(\Phi(aX))). -$$ -If $u(t)=t^2$, $u''(t)=2$ hence one gets $v(0)=u(\frac12)=\frac14$ and -$$ -\mathrm E(\varphi(aX)^2)=\int_{-\infty}^{+\infty}\varphi(ax)^2\varphi(x)\,\mathrm dx=\frac1{2\pi}\int_{-\infty}^{+\infty}\varphi(\sqrt{1+2a^2}x)\,\mathrm dx=\frac1{2\pi\sqrt{1+2a^2}} -$$ -hence -$$ -v'(a)=\frac{a}{\pi (1+a^2)\sqrt{1+2a^2}}. -$$ -Integrating this, one gets -$$ -v(a)=\frac14+\frac1{\pi}\int_0^a\frac{x\mathrm dx}{(1+x^2)\sqrt{1+2x^2}}=\frac14+\frac1{\pi}\left[\arctan\sqrt{1+2x^2}\right]_{x=0}^{x=a}, -$$ -and finally, -$$ -\mathrm E(\Phi(aX)^2)=\frac1\pi\arctan\sqrt{1+2a^2}. -$$ -Likewise, $u(a)=\mathrm E(\Phi(aX)\Phi(bX))$ yields -$$ -u'(a)=\mathrm E(X\varphi(aX)\Phi(bX))=\mathrm E(Xg(X))$$ where $$ g(x)=\varphi(ax)\Phi(bx). -$$ -Stein's lemma and the formula $g'(x)=-a^2x\varphi(ax)\Phi(bx)+b\varphi(ax)\varphi(bx)$ yield -$$ -(1+a^2)v'(a)=b\mathrm E(\varphi(aX)\varphi(bX)), -$$ -hence -$$ -v'(a)=\frac{b}{2\pi (1+a^2)\sqrt{1+a^2+b^2}}. -$$ -Finally, -$$ -\mathrm E(\Phi(aX)\Phi(bX))=\frac1\pi\arctan\sqrt{1+2b^2}+\frac{b}{2\pi}\int_b^a\frac{\mathrm dx}{(1+x^2)\sqrt{1+x^2+b^2}}. -$$ -Amongst several equivalent formulations, this means that - -$$ -\mathrm E(\Phi(aX)\Phi(bX))=\frac14+\frac1{2\pi}\arctan\left(\frac{ab}{\sqrt{1+a^2+b^2}}\right). -$$ - -The case $\mathrm E(\Phi(aX)\Phi(bX+c))$ might also be solvable with this method.<|endoftext|> -TITLE: Prove $\frac{\sin A \cos A}{\cos^2 A - \sin^2 A} = \frac{\tan A}{1-\tan^2 A}$ -QUESTION [5 upvotes]: How would I simplify this difficult trigonometric identity: -$$\frac{\sin A \cos A}{\cos^2 A - \sin^2 A} = \frac{\tan A}{1-\tan^2 A}.$$ -I am not exactly sure what to do. -I simplified the right side to -$$\frac{\frac{\sin A}{\cos A}}{1-\frac{\cos^2 A}{\sin^2 A}}$$ -But how would I proceed. - -REPLY [3 votes]: $$RHS = \frac{\tan A}{1-\tan^2 A} = \frac{\frac{\sin A}{\cos A}}{1-\frac{\sin^2 A}{\cos^2 A}}$$ -$$=\frac{\frac{\sin A}{\cos A}}{\frac{\cos^2 A - \sin^2 A}{\cos^2 A}}\cdot \frac{\cos^2 A}{\cos^2 A}$$ -$$=\frac{\sin A \cos A}{\cos^2 A - \sin^2 A} = LHS$$<|endoftext|> -TITLE: Closedness of sets under linear transformation -QUESTION [10 upvotes]: Let $Y$ be a closed subset of $\mathbb{R}^m$ (in fact $Y$ is convex and compact, but I think the extra assumptions are irrelevant). Let $A \in \mathbb{R}^{n \times n}$ be a non-singular matrix (so $A^{-1}$ exists). Let $C \in \mathbb{R}^{m \times n}$ be any matrix. Is the set -$$ Y' = \{ C A x \in \mathbb{R}^m \, : \, x \in \mathbb{R}^n, C x \in Y\} $$ -also closed? -Note: just to be clear, the definition of $Y'$ means $Y' = C A X = \{ C A x \in \mathbb{R}^m \, : \, x \in X \}$ where $X = \{ x \in \mathbb{R}^n \, : \, C x \in Y \}$. - -REPLY [9 votes]: I think we do need $Y$ to be compact. Otherwise take $Y=\{(x,y): x>0, y\ge 1/x\}$, let $C:\mathbb R^3\to\mathbb R^2$ be the projection $(x,y,z)\mapsto (x,y)$, and $A:\mathbb R^3\to\mathbb R^3$ be the reflection $(x,y,z)\mapsto (x,z,y)$. Now $X=\{(x,y,z): x>0, y\ge 1/x\}$, hence $AX=\{(x,y,z): x>0, z\ge 1/x\}$, and $CAX=\{(x,y): x>0\}$. -Assuming $Y$ is compact, we can argue as follows. Note that $X=X_0+\ker C$ where $X_0=\{x\in (\ker C)^\perp: Cx\in Y\}$ is compact. Hence for any linear map $T$ we have $TX=TX_0+T(\ker C)$, where $TX_0$ is compact and $T(\ker C)$ is a linear subspace. Any set of the form (compact set)+(linear subspace) is closed, being the preimage of a closed set under projection.<|endoftext|> -TITLE: Expressing the wave equation solution by separation of variables as a superposition of forward and backward waves. -QUESTION [5 upvotes]: (From an exercise in Pinchover's Introduction to Partial Differential Equations). -$$u(x,t)=\frac{A_0 + B_0 t}{2}+\sum_{n=1}^{\infty} \left(A_n\cos{\frac{c\pi nt}{L}}+ B_n\sin{\frac{c\pi nt}{L}}\right)\cos{\frac{n\pi x}{L}}$$ -is a general (and formal, at least) solution to the vibrating string with fixed ends. How to write this as a superposition of a forward and a backward wave? That is, as $f(x+ct)+f(x-ct)$ for some $f$. (No need to worry about rigour here, an heuristic will do.) -I know, by elementary trigonometry, that $$\left(A_n\cos{\frac{c\pi nt}{L}}+ B_n\sin{\frac{c\pi nt}{L}}\right)\cos{\frac{n\pi x}{L}} =\\= (1/2)(A_n\cos +B_n\sin)\left(\frac{c\pi nt}{L} + \frac{n\pi x}{L}\right)+(1/2)(A_n\cos +B_n\sin)\left(\frac{c\pi nt}{L} - \frac{n\pi x}{L}\right), $$ but this doesn't seem to work because the variable $x$ is the one that changes sign, so apparently this cannot be interpreted as a sum of forward and backward waves. -Is there a workaround to this? -EDIT. The second wave is from another function $g$. The answer is then straightforward after oen's comment. - -REPLY [3 votes]: Notice that $\cos(t-x) = \cos(x-t)$ and $\sin(t-x) = -\sin(x-t)$.<|endoftext|> -TITLE: A strangely connected subset of $\Bbb R^2$ -QUESTION [21 upvotes]: Let $S\subset{\Bbb R}^2$ (or any metric space, but we'll stick with $\Bbb R^2$) and let $x\in S$. Suppose that all sufficiently small circles centered at $x$ intersect $S$ at exactly $n$ points; if this is the case then say that the valence of $x$ is $n$. For example, if $S=[0,1]\times\{0\}$, every point of $S$ has valence 2, except $\langle0,0\rangle$ and $\langle1,0\rangle$, which have valence 1. -This is a typical pattern, where there is an uncountable number of 2-valent points and a finite, possibly empty set of points with other valences. In another typical pattern, for example ${\Bbb Z}^2$, every point is 0-valent; in another, for example a disc, none of the points has a well-defined valence. -Is there a nonempty subset of $\Bbb R^2$ in which every point is 3-valent? I think yes, one could be constructed using a typical transfinite induction argument, although I have not worked out the details. But what I really want is an example of such a set that can be exhibited concretely. -What is it about $\Bbb R^2$ that everywhere 2-valent sets are well-behaved, but -everywhere 3-valent sets are crazy? Is there some space we could use instead of $\Bbb R^2$ in which the opposite would be true? - -REPLY [8 votes]: I claim there is a set $S \subseteq {\mathbb R}^2$ that contains exactly three points in every circle. -Well-order all circles by the first ordinal of cardinality $\mathfrak c$ as $C_\alpha, \alpha < \mathfrak c$. By transfinite induction I'll construct sets $S_\alpha$ with -$S_\alpha \subseteq S_\beta$ for $\alpha < \beta$, and take -$S = \bigcup_{\alpha < {\mathfrak c}} S_\alpha$. These will have the following properties: - -$S_\alpha$ contains exactly three points on every circle $C_\beta$ for $\beta \le \alpha$. -$S_\alpha$ does not contain more than three points on any circle. -$\text{card}(S_\alpha) \le 3\, \text{card}(\alpha)$ - -We begin with $S_1$ consisting of any three points on $C_1$. -Now given $S_{<\alpha} = \bigcup_{\beta < \alpha} S_\beta$, consider the circle $C_\alpha$. -Let $k$ be the cardinality of $C_\alpha \cap S_{<\alpha}$. By property (2), $k \le 3$. If $k = 3$, take $S_\alpha = S_{<\alpha}$. -Otherwise we need to add in $3-k$ points. Note that there are fewer than ${\mathfrak c}$ circles determined by triples of points in $S_{<\alpha}$, all of which are different from $C_\alpha$, and so there are fewer than $\mathfrak c$ points of $C_\alpha$ that are -on such circles. Since $C_\alpha$ has $\mathfrak c$ points, we can add in a point $a$ of $C_\alpha$ that is not on any of those circles. If $k \le 1$, we need a second point $b$ not to be on the circles determined by triples in $S_{<\alpha} \cup \{a\}$, and if $k=0$ a third point $c$ not on the circles determined by triples in $S_{<\alpha} \cup \{a,b\}$. Again this can be done, and it is easy to see that properties (1,2,3) are satisfied. -Finally, any circle $C_\alpha$ contains exactly three points of $S_\alpha$, and no -more than three points of $S$ (if it contained more than three points of $S$, it would have more than three in some $S_\beta$, contradicting property (2)).<|endoftext|> -TITLE: infinite sums of trigonometric functions -QUESTION [5 upvotes]: Find the sum of the series: -$$\sum_{n = 1}^\infty \left( \sin \left(\frac{1}{n}\right) - \sin\left(\frac{1}{n+1} \right) \right).$$ -By the convergence test the limit of this function is $0$ but I'm not sure how to prove whether or not this function converges or diverges. - -REPLY [6 votes]: Write the sum as $\sin(\frac{1}{1}) - \sin(\frac{1}{2}) + \sin(\frac{1}{2}) - \sin(\frac{1}{3}) + \sin(\frac{1}{3}) - \sin(\frac{1}{4}) + \cdots$. All terms but the first cancel and we are left with $\sin(1)$. You have already established that the limit of the terms is $0$, so the limit of the sum is $\sin(1)$.<|endoftext|> -TITLE: How to find convergence region of $\sum_{n\geqslant 0, m \geqslant 0} x^n y^m \binom{n+m}{n}^2$ -QUESTION [7 upvotes]: The following two series are special cases of Appell $F_3$ and $F_4$, namely: -$$ - \mathcal{S}_1 = \sum_{n \geqslant 0, m \geqslant 0} \frac{x^n y^m}{\binom{n+m}{n}} -$$ -and -$$ - \mathcal{S}_2 = \sum_{n \geqslant 0, m \geqslant 0} \binom{n+m}{n}^2 x^n y^m -$$ -How would one establish that $\mathcal{S}_1$ converges for $\{ (x,y)\colon -1 -TITLE: Showing that a CW space is contractible if it is endowed with a certain binary operation -QUESTION [9 upvotes]: I am having trouble with the following homework problem, and was hoping someone could provide me with a hint: I am given a connected CW space $X$ which has a continuous associative operation $(x,\ y)\mapsto x\circ y$. It is also given that $x\circ x=x$ for any $x\in X$ and that there exists some $e\in X$ such that $x\circ e=e\circ x$ for all $x\in X$. I need to show that $X$ is contractible. -I have noted that by Whitehead's theorem, it is enough to show that all homotopy groups of $X$ vanish. So I was thinking of using the opertation $\circ$ to show that any map $f:(S^n,\sigma_0)\to (X,\ e)$ is homotopic to the constant map to $e$ (somehow this distinguished point $e$ seems suspicious). -I considered, for instance, $\phi :S^n\times I\to X$ by $(x,\ t)\mapsto f(x)\circ \gamma_x (t)$, where $\gamma_x $ is a path from $e$ to $f(x)$. At $t=0$ we get the map $x\mapsto f(x)\circ e$, and at t=1 we get the map $x \mapsto f(x)\circ f(x)=f(x)$. If somehow I knew how to choose $\phi$ to be continuous (i.e., if I knew how to choose the $\gamma_x$), and if I knew that $e\circ x=e$ for all $e$, then $\phi$ would be the desired homotopy. -But I have no idea how to make $\phi $ continuous, and see no reason why $e\circ x=e$ should hold. Furthermore, I have not yet used the associativity of $\circ$ or the fact that $e$ commutes will all points of $x$. In short, I highly doubt that I am on the right track. -Any hint would be highly appreciated! -Thanks, -Roy - -REPLY [2 votes]: Isn't the operation $\circ$ inherited by the $n$th homotopy groups based at $e$? Such a homotopy group $G$ will have an associative operation $\circ$ satisfying the interchange law with the group operration, and so by the Eckmann-Hilton argument, $\circ$ is the group operation. But then for $x$ in the group, $x=x\circ x=xx$, and so $x=1$. -oops! As Jason pointed out, one needs more argument! -I like to display the interchange law as a matrix -$$\matrix{x& y\\z&w}$$ -where, say, the horizontal composition is $\circ$ and the vertical is the group multiplication. So the above reads $(x\circ y)(z \circ w)=(xz) \circ (yw)$. Since you wanted a hint, try the matrices -$$\matrix{e&1\\1&e} \qquad \qquad\matrix{1&e\\e&1}$$ -and make deductions. -By the way, in the more usual case where both compositions have identities, associativity comes for free, as follows from the matrix -$$\matrix{y &1 \\ z&w}$$ -Note added later: this is not best organised, so let me know if you have problems with it. But note that the two matrices involving $e$ are seemingly the only ones giving real information.<|endoftext|> -TITLE: Question about proof of Going-down theorem -QUESTION [8 upvotes]: I have written a proof of the Going-down theorem that doesn't use some of the assumptions so it's false but I can't find the mistake. Can you tell where it's wrong? -*Going-down*$^\prime$: Let $R,S$ be rings such that $S \subset R$ and $R$ is integral over $S$. Let $q_1$ be a prime ideal in $R$ such that $p_1 = q_1 \cap S$ is a prime ideal in $S$. Let $p_2 \subset p_1$ be another prime ideal in $S$. Then there exists a prime ideal $q_2$ in $R$ s.t. $q_2 \cap S = p_2$. (Note: the missing assumptions are that $R,S$ are integral domains and $S$ is integrally closed) -Theorems we use for the false proof: -1.(5.10) Let $A \subset B$ be rings, $B$ integral over $A$, and let $p$ be a prime ideal of $A$. Then there exists a prime ideal $q$ of $B$ such that $q \cap A = p$. -2.(5.6.)If $S$ is a multiplicatively closed subset of $A$ and $B$ is integral over $A$ then $S^{-1}B$ is integral over $S^{-1}A$. -3.(3.11 iv)) The prime ideals of $S^{-1}A$ are in one-to-one correspondence ($p \leftrightarrow S^{-1}p$) with prime ideals of $A$ which don't meet $S$. - -False proof: -By (5.6), $R_{q_1}$ is integral over $S_{p_1}$. By (3.11), $\overline{p_2} = {p_2}_{p_1}$ is prime in $S_{p_1}$ since $p_2 \subset p_1$ by assumption. By (5.10) there exists a prime ideal $\overline{q_2}$ in $R_{q_1}$ such that $\overline{q_2} \cap S_{p_1} = \overline{p_2}$. We know that the following diagram commutes: -$$ -\begin{matrix} -S & \xrightarrow{i} & R \\ -\left\downarrow{\psi}\vphantom{\int}\right. & & \left\downarrow{\varphi}\vphantom{\int}\right.\\ -S_{p_1}& \xrightarrow{i_{p_1}} & R_{q_1} -\end{matrix} -$$ -We claim that $q_2 = \varphi^{-1}\overline{q_2}$ is an ideal such that $q_2 \subset q_1$ and $q_2 \cap S = p_2$. The claim $q_2 \subset q_1$ follows by construction (or from (3.11)). We also have -$$ \varphi^{-1} (\overline{q_2}) \cap S = i^{-1}\varphi^{-1}(\overline{q_2}) = \psi^{-1}i^{-1}_{p_1}(\overline{q_2}) = \psi^{-1}(S_{p_1} \cap \overline{q_2} ) = \psi^{-1}(\overline{q_2}) = p_2$$ - -REPLY [5 votes]: You begin your proof with "By (5.6), $R_{q_1}$ is integral over $S_{p_1}$" -But this recourse to (5.6) is illegitimate: the multiplicative set here is $\Sigma=S\setminus \mathfrak p_1 $ . -However $R_{\mathfrak q_1}=T^{-1}R $ where $T=R\setminus \mathfrak {q}_1$ and we only have $\Sigma \subset T$, not $\Sigma=T$. -(Don't think, draw a Venn diagram!) -So $S_{\mathfrak p_1}\to \Sigma^{-1} R$ is indeed integral, but $S_{p_1}\to R_{q_1}$ has no reason to be integral. -Edit -In Exercise 4 of the same chapter 5 the authors give an example where indeed $R_{q_1}$ is not integral over $S_{p_1}$, so that it is not just that one cannot apply (5.6), but the result (unsurprisingly) is impossible to deduce from the too weak hypothesis.<|endoftext|> -TITLE: Multiple choice question - number of real roots of $x^6 − 5x^4 + 16x^2 − 72x + 9$ -QUESTION [6 upvotes]: The equation $x^6 − 5x^4 + 16x^2 − 72x + 9 = 0$ has -(A) exactly two distinct real roots -(B) exactly three distinct real roots -(C) exactly four distinct real roots -(D) six distinct real roots - -REPLY [8 votes]: You have: -$f(x)=x^6-5x^4+16x^2-72x+9$ -$f'(x)=6x^5-20x^3+32x-72$ -$f''(x)=30x^4-60x^2+32$ -If you notice that -$$f''(x)=30(x^4-2x^2+1)+2=30(x^2-1)^2+2 \ge 2 >0,$$ -you can see that $f'(x)$ is strictly increasing. Together with $\lim\limits_{x\to-\infty} f'(x)=-\infty$ and $\lim\limits_{x\to\infty} f'(x)=\infty$ this implies that there is exactly one root $x_0$ of $f'(x)$. -Thus $f(x)$ is decreasing on $(-\infty,x_0)$ and increasing on $(x_0,\infty)$. -Since $\lim\limits_{x\to\infty} f(x)=\infty$ and $f(1)=1-5+16-72+9=-51$, we see that $f(x)$ has both positive and negative values. -Thus $f(x)$ must have two real roots, one of them in the interval $(-\infty,x_0)$ and another one in the interval $(x_0,\infty)$. - -You can check the behavior of $f(x)$ and the behavior of $f'(x)$ at WolframAlpha.<|endoftext|> -TITLE: Is there a way to reverse factorials? -QUESTION [23 upvotes]: Is there any way I can 'undo' the factorial operation? JUst like you can do squares and square roots, can you do factorials and factorial roots (for lack of a better term)? -Here is an example: 5! = 120. Is there a way I can work out the number that must be factorialed (??) to give the answer 120? - -REPLY [26 votes]: You can just divide the "answer" by consecutive positive integers, and when the result is 1, the last number you divided by is the number that the "answer" is factorial of. For example: 120 / 2 = 60, 60 / 3 = 20, 20 / 4 = 5, 5 / 5 = 1, so the number that 120 is the factorial of is 5.<|endoftext|> -TITLE: Category Theory usage in Algebraic Topology -QUESTION [24 upvotes]: First my question: - -How much category theory should someone studying algebraic topology generally know? - -Motivation: I am taking my first graduate course in algebraic topology next semester, and, up to this point, I have never taken the time to learn any category theory. I've read that category theory helps one to understand the underlying structure of the subject and that it was developed by those studying algebraic topology. Since I do not know the exact content which will be covered in this course, I am trying to find out what amount of category theory someone studying algebraic topology should generally know. -My university has a very general outline for what the course could include, so, to narrow the question a bit, I will give the list of possible topics for the course. -Possible Topics: - -unstable homotopy theory -spectra -bordism theory -cohomology of groups -localization -rational homotopy theory -differential topology -spectral sequences -K-theory -model categories - -All in all, I am well overdue to learn the language of categories, so this question is really about how much category theory one needs in day to day life in the field. -Update -I emailed the professor teaching the course and he said he hopes to cover the following (though maybe it is too much): - -homotopy, homotopy equivalences, mapping cones, mapping cylinders -fibrations and cofibrations, and homotopy groups, and long exact homotopy sequences. -classifing spaces of groups. -Freudenthal theorem, the Hurewicz and the Whitehead theorem. -Eilenberg-MacLane spaces and Postnikov towers. -homology and cohomology theories defined by spectra. - -REPLY [10 votes]: I agree with Paul Siegel's very nice answer, and would just like to add one thing that's a little too long for a comment. -Depending on what direction you take, algebraic topology can become practically synonymous with higher category theory. This can come in multiple ways. First, the category of topological spaces has spaces as its objects, continuous maps as its morphisms, homotopies as its 2-morphisms, homotopies between homotopies as its 3-morphisms, etc. Stated perhaps a little too cavalierly, the point of model categories is essentially to set up a general framework for studying higher categories that may (or may not) look like that of spaces. But then, higher categories themselves are also a lot like spaces. In this analogy, functors are like continuous maps, natural transformations are like homotopies, etc. (The fact that there are these two totally distinct ways that spaces and categories interact really blindsided me the first time I saw it.) -Anyways, the point is that if you pursue algebraic topology seriously, you may eventually have to get very friendly with category theory and be okay with using ridiculous and scary phrases like "homotopy left Kan extension" and such. It seems that usage of higher category theory in algebraic topology is very much on the rise, so it's possible that in twenty years, algebraic topologists will have no choice but to become well-versed in all this stuff. (I'm certainly not. Not yet, at least.) Just a heads-up.<|endoftext|> -TITLE: Limit of a sequence involving root of a factorial: $\lim_{n \to \infty} \frac{n}{ \sqrt [n]{n!}}$ -QUESTION [6 upvotes]: I need to check if -$$\lim_{n \to \infty} \frac{n}{ \sqrt [n]{n!}}$$ converges or not. Additionally, I wanted to show that the sequence is monotonically increasing in n and so limit exists. Any help is appreciated. I had tried taking log and manipulating the sequence but I could not prove monotonicity this way. - -REPLY [16 votes]: What you have is actually an indefinite integral in disguise. Let's -first consider the reciprocal of what you have: -\begin{eqnarray*} -\lim_{n\to\infty}\frac{(n!)^{1/n}}{n} & = & e^{{\displaystyle \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\ln\left(\frac{k}{n}\right)}}\\ - & = & e^{{\displaystyle \int_{0}^{1}\ln xdx}}\\ - & = & e^{-1}. -\end{eqnarray*} -Thus we get that -$$ -\lim_{n\to\infty}\frac{n}{(n!)^{1/n}}=e. -$$<|endoftext|> -TITLE: limit of an integral with a Lorentzian function -QUESTION [6 upvotes]: We want to calculate the $\lim_{\epsilon \to 0} \int_{-\infty}^{\infty} \frac{f(x)}{x^2 + \epsilon^2} dx $ for a function $f(x)$ such that $f(0)=0$. We are physicist, so the function $f(x)$ is smooth enough!. -After severals trials, we have not been able to calculate it except numerically. -It looks like the normal Lorentzian which tends to the dirac function, but a -$\epsilon$ is missing. -We wonder if this integral can be written in a simple form as function of $f(0)$ or its derivatives $f^{(n)}(0)$ in 0. -Thank you very much. - -REPLY [2 votes]: Let $f$ be smooth with compact support. Consider the double layer potential (up to a constant) -$$ -u(x_1,x_2)=-2\pi\int_{-\infty}^{\infty}\frac{\partial}{\partial x_2}\Gamma(x_1-y_1,x_2)f(y_1)\,dy_1= -$$ -$$ -\int_{-\infty}^{\infty} \frac{x_2f(y_1)}{x^2 + x_2^2} dx, -$$ -where $\Gamma(x)=-\frac1{2\pi}\log|x|$ is a fundamental solution for the Laplace equation. -As is known $u$ is smooth up to the boundary for smooth $f$. We have -$$ -\frac{\partial u(0,0)}{\partial x_2}= -\lim_{\epsilon \to 0+} \frac{u(0,\epsilon)-u(0,0)}\epsilon= -\lim_{\epsilon \to 0+} \frac{u(0,\epsilon)-f(0)}\epsilon= -\lim_{\epsilon \to 0+}\int_{-\infty}^{\infty} \frac{ f(y_1)}{x^2 + \epsilon^2} dx, -$$ -which is the required value. -To calculate it note that -$$ -\frac{\partial u(x_1,x_2)}{\partial x_2} = --2\pi\int_{-\infty}^{\infty}\frac{\partial^2}{\partial x_2^2}\Gamma(x_1-y_1,x_2)f(y_1)\,dy_1= -$$ -$$ -2\pi\int_{-\infty}^{\infty}\frac{\partial^2}{\partial y_1^2}\Gamma(x_1-y_1,x_2)f(y_1)\,dy_1= -2\pi\int_{-\infty}^{\infty}\Gamma(x_1-y_1,x_2)f''(y_1)\,dy_1 -$$ -because $\Gamma$ satisfies the Laplace equation. -The last integral converges uniformly for $|x|\le 1$ so taking the limit $x\to0$ gives -$$ -\frac{\partial u(0,0)}{\partial x_2}=2\pi\int_{-\infty}^{\infty}\Gamma(0-y,0)f''(y)\,dy_1=-\int_{-\infty}^{\infty}\log|y|f''(y)\,dy. -$$<|endoftext|> -TITLE: General form of Integration by Parts -QUESTION [13 upvotes]: This is a question just out of interest to know the power of integration by parts. There are various level of integration by parts. What are some of the most general form of integration by parts? -I have encountered it very often in PDE's. I look forward to gaining more insights on it. -Thank you for your ideas, help and discussions. - -REPLY [21 votes]: All versions of integration by parts that I have seen boil down to two things. - -Stokes' Theorem: if $\omega$ is an $n-1$ form on $M$, and $n$-manifold with boundary $\partial M$, then -$$ \int_M \mathrm{d}\omega = \int_{\partial M} \omega $$ -Leibniz rule for differential forms: -$$ \mathrm{d}(\eta \wedge \omega) = \mathrm{d}\eta \wedge \omega + (-1)^{\text{degree}(\eta)}\eta\wedge \mathrm{d}\omega $$ - -The only other ingredient that is sometimes needed is some basic housecleaning coming from Riemannian and/or differential geometry: things like how covariant or coordinate partial derivatives relate to the exterior derivative, and how to write the divergence of a vector field as equivalently the exterior derivative of its dual $n-1$ form.<|endoftext|> -TITLE: Some Good Algebraic Topology Exercises -QUESTION [10 upvotes]: I am teaching a topology prep course for first year graduate students taking their qualifying exams. I have been able to think of about ten days' worth of exercises, but am running out of ideas. Does anyone have any good questions or a place to find them? I am looking for exercises involving singular homology that are not just "Compute homology of" type questions. In particular I need some good Euler characteristic, degree of mapping, and Jordan-Alexander Complement problems. Though, any questions at all are surely welcome. The class they took is equivalent to Chapter 2 of Hatcher's 'Algebraic Topology' book. - -REPLY [2 votes]: I like questions of the sort: -1) Suppose $p: \mathbb{R}P^2 \to X$ is a covering map. Prove that $p$ is a homeomorphism. -2) Suppose $p: \mathbb{C}P^2 \to X$ is a covering and suppose that $X$ is a manifold. Again show that $p$ is a homeomorphism. -(the first one works only by looking at the euler characteristic, for the second one I guess one needs Poincaré duality).. -3) Show that there is no map $f: S^n\to S^1$ which is $\mathbb{Z}/2\mathbb{Z}$ equivariant (that is $f(-x) = -f(x)$).<|endoftext|> -TITLE: How many points does Stone-Čech compactification add? -QUESTION [12 upvotes]: I would like to know how Stone-Čech compactification works with simple examples, like $(0,1)$, $\mathbb{R}$, and $B_r(0)$ (the open ball of $R^2)$. I've studied the one-point compactification and this is way more difficult to understand. All the texts I've found till now start immediately with functions and closure of functions. If somebody could give me some "visible" ideas, I would bear those in mind and understanding the theory would be a little easier. - -REPLY [14 votes]: There are spaces like $[0,\omega_1)$ with the order topology where the Stone-Cech compactification is the same as the one-point compactification. -On the other hand, the Stone-Cech compactification of the natural numbers $\mathbb N$ has cardinality $2^{\mathfrak c}$. So the answers depends a lot on the space.<|endoftext|> -TITLE: Whats the probability a subset of an $\mathbb F_2$ vector space is a spanning set? -QUESTION [13 upvotes]: Let $V$ be an $n$-dimensional $\mathbb F_2$ vector space. Note that $V$ has $2^n$ elements and $\mathcal P(V)$ has $2^{2^n}$. -I'm interested in the probability (under a uniform distribution) that an element of $\mathcal P(V)$ is a spanning set for $V$. Equivalently a closed form formula (or at least one who's asymptotics as $n\rightarrow \infty$ are easy to work out) for the number of spanning sets or non-spanning sets. -It's not hard to show that the probability is greater than or equal to $1/2$. Since any subset of size greater than $2^{n-1}$ must span the space. I calculated the proportion of spanning sets of size $n$ for $n$ up to $200$ which seem to be going to a number starting with $.2887$. This leads me to believe that the probability exceeds $1/2$. I couldn't nail down a formula for arbitrary sized subsets though to continue experimental calculations. -I feel like this is something that's been done before, but googling I've mostly found things concerning counting points on varieties over finite fields or counting subspaces of finite fields. Any references would be appreciated. - -REPLY [5 votes]: It's worth mentioning that there is an explicit formula for the probability that $N$ random vectors drawn from $\mathbb{F}_q^n$ span. (Note: I am doing sampling with replacement. If you want to require the $N$ vectors to be distinct, you can do this using inclusion-exclusion, but the result will be much messier.) Of course, the answer is $0$ if $N -TITLE: Why is an alternating $2$-form decomposable if and only if its self-wedge vanishes? -QUESTION [6 upvotes]: Given a vector space $V$, and a $2$-tensor $w$ in the second exterior power $\Lambda^2 V$. Assume that $w \wedge w=0$. Why is $w$ decomposable? -Thanks for your help! - -REPLY [3 votes]: A proof not requiring a basis and canonical form: Evaluate the interior product -with any vector $V$ for which $i_Vw\ne0$: $0=i_V(w\wedge w)=(2i_Vw)\wedge w$. So -$w=(i_Vw)\wedge u$, for some 1-form $u$.<|endoftext|> -TITLE: Combinatorial Interpretation of Fractional Binomial Coefficients -QUESTION [10 upvotes]: My question is a bit imprecise - but I hope you like it. I even strongly think it has a proper answer. -The binomial coefficient $\binom{\frac{1}{2}}{n}$ is strongly related to Catalan numbers - the expression $(1-4x)^{\frac{1}{2}}$ appears when calculating the generating function of the Catalan numbers and solving a quadratic equation. -I am trying to find some combinatorial interpretation of $\binom{\frac{1}{k}}{n}$ for non-zero integer $k$. I feel it must exist - I don't know if it is because of intuition or because I've seen something similar and forgot. I want an elementary interpretation, maybe related to trees (since Catalan numbers count binary trees). -So, can anyone find a combinatorial interpretation of those coefficients (possibly multiplied by some power such as $k^n$)? -My motivation: I can show p-adically that $\binom{\frac{1}{k}}{n}$ is $p$-integral for any prime $p$ not dividing $k$. I am looking for a combinatorial proof of this property, and a combinatorial interpretation of $k^m \binom{\frac{1}{k}}{n}$ (for some integer $m$) will suffice for this. - -REPLY [4 votes]: Propp's Exponentiation and Euler Measure contains such an interpretation.<|endoftext|> -TITLE: $\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \mathrm {d}x$ Evaluate Integral -QUESTION [22 upvotes]: Here is a fun integral I am trying to evaluate: -$$\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \ dx=\frac{\pi \binom{2n}{n}}{2^{2n+1}}.$$ -I thought about integrating by parts $2n$ times and then using the binomial theorem for $\sin(x)$, that is, using $\dfrac{e^{ix}-e^{-ix}}{2i}$ form in the binomial series. -But, I am having a rough time getting it set up correctly. Then, again, there is probably a better approach. -$$\frac{1}{(2n)!}\int_{0}^{\infty}\frac{1}{(2i)^{2n}}\sum_{k=0}^{n}(-1)^{2n+1-k}\binom{2n}{k}\frac{d^{2n}}{dx^{2n}}(e^{i(2k-2n-1)x})\frac{dx}{x^{1-2n}}$$ -or something like that. I doubt if that is anywhere close, but is my initial idea of using the binomial series for sin valid or is there a better way?. -Thanks everyone. - -REPLY [2 votes]: I am just adding the proof of the identity for those who have interest: -$$ \sin^{2n+1} x = \frac{1}{4^n}\sum_{k=0}^{n}(-1)^{n-k}\binom{2n+1}{k}\sin\left(\left(2(n-k)+1\right)x\right). $$ -Using the complex representation and the Binomial Theorem, we have -$$\begin{aligned} -\sin^{2n+1}x&=\left(\frac{\mathrm{e}^{ix}-\mathrm{e}^{-ix}}{2i}\right)^{2n+1}\\ -&=\frac{(-1)^n}{2^{2n+1}i}\sum_{k=0}^{2n+1}\binom{2n+1}{k}\mathrm{e}^{i(2n+1-k)x}(-1)^k\mathrm{e}^{i(-kx)}\\ -&=\frac{(-1)^n}{2^{2n+1}i}\sum_{k=0}^{2n+1}(-1)^k\binom{2n+1}{k}\mathrm{e}^{i(2(n-k)+1)x}\\ -&=\frac{(-1)^n}{2^{2n+1}i}\sum_{k=0}^{2n+1}(-1)^k\binom{2n+1}{k}\left[\cos\left(\left(2(n-k)+1\right)x\right) + i\sin\left(\left(2(n-k)+1\right)x\right)\right]\\ -&=\frac{(-1)^n}{2^{2n+1}}\sum_{k=0}^{2n+1}(-1)^k\binom{2n+1}{k}\left[\sin\left(\left(2(n-k)+1\right)x\right) - i\cos\left(\left(2(n-k)+1\right)x\right)\right] -\end{aligned} -$$ -Now, observe that -$$\begin{aligned} -\sum_{k=0}^{2n+1} a_{k} &= \sum_{k=0}^{n}a_{k}+\sum_{k=n+1}^{n+n+1}a_{k}\\ -&=\sum_{k=0}^{n}a_{k}+\sum_{k=0}^{n}a_{n+1+k}\\ -&=\sum_{k=0}^{n}a_{k}+\sum_{k=0}^{n}a_{n+1+n-k}\\ -&=\sum_{k=0}^{n}\left(a_{k}+a_{2n+1-k}\right) -\end{aligned} -$$ -Apply with $a_{k}=(-1)^{k}\binom{2n+1}{k}\left[\sin\left(\left(2(n-k)+1\right)x\right) - i\cos\left(\left(2(n-k)+1\right)x\right)\right]$, so -$$\begin{aligned} -a_{2n+1-k}&=-(-1)^{k}\binom{2n+1}{2n+1-k}\left[-\sin\left(\left(2(n-k)+1\right)x\right) - i\cos\left(\left(2(n-k)+1\right)x\right)\right]\\ -&=(-1)^{k}\binom{2n+1}{k}\left[\sin\left(\left(2(n-k)+1\right)x\right) + i\cos\left(\left(2(n-k)+1\right)x\right)\right]. -\end{aligned} -$$ -Then, -$$ a_{k}+a_{2n+1-k}=2(-1)^{k}\binom{2n+1}{k}\sin\left(\left(2(n-k)+1\right)x\right). $$ -Therefore, -$$\begin{aligned} \sin^{2n+1} x&=\frac{1}{4^{n}}\sum_{k=0}^{n}(-1)^{n-k}\binom{2n+1}{k}\sin\left(\left(2(n-k)+1\right)x\right)\\ -&=\frac{1}{4^n}\sum_{k=0}^{n}(-1)^k\binom{2n+1}{n-k}\sin\left((2k+1)x\right)\\ -&=\frac{1}{4^n}\sum_{k=0}^{n}(-1)^k\binom{2n+1}{n+k+1}\sin\left((2k+1)x\right), -\end{aligned}$$ -as desired.<|endoftext|> -TITLE: Use Stokes's Theorem to show $\oint_{C} y ~dx + z ~dy + x ~dz = \sqrt{3} \pi a^2$ -QUESTION [5 upvotes]: I am a little stuck on the following problem: -Use Stokes's Theorem to show that -$$\oint_{C} y ~dx + z ~dy + x ~dz = \sqrt{3} \pi a^2,$$ -where $C$ is the suitably oriented intersection of the surfaces $x^2 + y^2 + z^2 = a^2$ and $x + y + z = 0$. -OK, so Stokes's Theorem tells me that: -$$\oint_{C}\vec{F} \cdot d\vec{r} = \iint_{S}\operatorname{curl} \vec{F} \cdot \vec{N} ~dS$$ -I have calculated: -$$\operatorname{curl} \vec{F} = -\vec{i} - \vec{j} - \vec{k}.$$ -I then figured that on the surface $S$ we must have: -$$\vec{N}dS = \vec{i} + \vec{j} + \vec{k}dxdy$$ -since this follows from the equation of the given plane. -However this will then give me: -$$\operatorname{curl} \vec{F} \cdot \vec{N} = -1 -1 -1 = -3$$ -And thus I would get, if project this onto the $xy$-plane: -$$\iint_{S} \operatorname{curl} \vec{F} \cdot \vec{N} ~dS = -3 \iint_{A} dA = -3 \pi a^2$$ -which is obviously not correct. -I would greatly appreciate it if someone could help me with this. I actually had multivariable calculus a few years ago, and I know that I knew this stuff then. However, now that I need it again I notice that I've become quite rusty. -Thanks in advance :) - -REPLY [4 votes]: This is an effort to get this question off from the unanswered queue. -Also here I use a rather general approach than the classical Kelvin-Stokes theorem. -Stokes theorem reads: -$$ -\int_{\partial M} \omega = \int_{M} d\omega. -$$ -Hence -$$ -\int_C y\,dx + z\,dy +x\,dz = \pm \int_S d(y\,dx + z\,dy +x\,dz) \\ -= \pm\int_S dy\wedge dx + dz\wedge dy + dx\wedge dz. -$$ -For simplicity we choose $S$ to be planar surface bounded by the sphere, not the hemisphere bounded by the plane. On the plane $z = -x-y$, hence above integral is -$$ -\pm 3\int_S dx\wedge dy.\tag{$\star$} -$$ -Using the natural parametrization of the plane, this integral can be interpreted as the area of projected area $S$ on to the $xy$-plane (or notice $\star dz = dx\wedge dy$). The intersection of $x+y+z = 0$ with $x^2 + y^2 + z^2 =a^2$, projected onto the $xy$-plane, is an ellipse. -The end points for this projected ellipse are: minor axis end points are projection of $(-1/\sqrt{6},-1/\sqrt{6},2/\sqrt{6} )a$, and $(1/\sqrt{6},1/\sqrt{6},-2/\sqrt{6})a$, achieved when we set $x=y$. We can compute the length of the minor axis is $b = \sqrt{3}a/3$. The major axis end points are achieved by setting $x+y=0$, $(\sqrt{2}/2, -\sqrt{2}/2, 0)a$ and $(\sqrt{2}/2, -\sqrt{2}/2, 0)a$, the length of the major axis is just $a$. Hence the project area is $\sqrt{3}\pi a^2/3$ and plugging back to $(\star)$ yields -$$ -\int_C y\,dx + z\,dy +x\,dz = \pm \sqrt{3} \pi a^2. -$$ -The sign depends on $C$ is chosen to be rotated counter-clockwisely or clockwisely with respect to the normal vector to the plane $x+y+z=0$. - -To address your own question: - -Why I got $\displaystyle \iint_{S} \operatorname{curl} \vec{F} \cdot \vec{n} ~dS = -3 \iint_{A} dA = -3 \pi a^2$ if I use $dS = \sqrt{3}dA$ so that $n\,dS = (1/\sqrt{3},1/\sqrt{3},1/\sqrt{3})\sqrt{dS} = (1,1,1)$? - -This is not correct, for that if you use the area element $dS = \sqrt{3} \,dx dy = \sqrt{3} \,dA$, the new $A$ is the projected ellipse on the $xy$-plane, having area $\sqrt{3}\pi a^2/3$ (please see the argument above), not having the same area $\pi a^2$ with $S$. - -Standard way using Kelvin-Stokes as Arturo Magidin pointed out in the comments: choosing $C$ rotates counter-clockwisely with respect to the unit vector $n = -(1,1,1)/\sqrt{3}$ normal to the plane $x+y+z=0$: -$$ -\int_C y\,dx + z\,dy +x\,dz = \int_{S} \nabla \times (y,z,x)\cdot n\,dS -\\ -= \int_{S} (-1,-1,-1)\cdot (-1,-1,-1)/\sqrt{3} \,dS = \sqrt{3}|S| = \sqrt{3}\pi a^2. -$$<|endoftext|> -TITLE: Help me evaluate $\int_0^1 \frac{\log(x+1)}{1+x^2} dx$ -QUESTION [20 upvotes]: I need to evaluate this integral: $\int_0^1 \frac{\log(x+1)}{1+x^2} dx$. -I've tried $t=\log(x+1)$, $t=x+1$, but to no avail. I've noticed that: -$\int_0^1 \frac{\log(x+1)}{1+x^2} dx = \int_0^1\log(x+1) \arctan'(x)dx =\left. \log(x+1)\arctan(x) \right|_{x=0}^{x=1} - \int_0^1\frac{\arctan(x)}{x+1}dx$ -But can't get further than this. -Any help is appreciated, thank you. - -REPLY [5 votes]: $\displaystyle A=\int_0^1\dfrac{\Big(\log(1+x)\Big)^2}{1+x^2}dx$ -Perform the change of variable $y=\dfrac{1-x}{1+x}$ -$\displaystyle A=\int_0^1\dfrac{\left(\log\left(\dfrac{2}{1+x}\right)\right)^2}{1+x^2}dx=\int_0^1\dfrac{\Big(\log 2-\log(1+x)\Big)^2}{1+x^2}dx$ -$\displaystyle A=\int_0^1\dfrac{\Big(\log 2\Big)^2}{1+x^2}dx-2\int_0^1\dfrac{\log 2\log(1+x)}{1+x^2}dx+A$ -Therefore, -$\displaystyle \int_0^1\dfrac{2\log 2\log(1+x)}{1+x^2}dx=\int_0^1\dfrac{\Big(\log 2\Big)^2}{1+x^2}dx$ -Finally, -$\displaystyle \int_0^1\dfrac{\log(1+x)}{1+x^2}dx=\int_0^1\dfrac{\log 2}{2(1+x^2)}dx=\dfrac{\log 2}{2}\Big[\arctan x\Big]_0^1=\dfrac{\pi\log 2}{8}$<|endoftext|> -TITLE: Proof of Chebyshev Inequality -QUESTION [6 upvotes]: I was going through the proof of the Chebyshev Inequality here . -And I seem to be facing some trouble in the approximation stage. I can't seem to follow how $\epsilon$ has been approximated to $(t-\mu)$. - -REPLY [5 votes]: $\epsilon$ is not being approximated by $t - \mu$. What is happening is that for $t$ outside the interval $(\mu - \epsilon, \mu + \epsilon)$, $\epsilon^2 \leq (t - \mu)^2$ and hence the two integrals can be underestimated. -The reasoning "since $t \leq \mu - \epsilon \Rightarrow \epsilon \leq | t - \mu | \Rightarrow \epsilon^2 \leq (t - \mu)^2$" on the page you refer to, applies to the rewriting of the left integral, c.q., $t$ on the left of the interval $(\mu - \epsilon, \mu + \epsilon)$. A similar reasoning applies to the right integral, c.q., the right of that interval.<|endoftext|> -TITLE: Evaluating :$\int \frac{1}{x^{10} + x}dx$ -QUESTION [22 upvotes]: $$\int \frac{1}{x^{10} + x}dx$$ -My solution : -$$\begin{align*} -\int\frac{1}{x^{10}+x}\,dx&=\int\left(\frac{x^9+1}{x^{10}+x}-\frac{x^9}{x^{10}+x}\right)\,dx\\ -&=\int\left(\frac{1}{x}-\frac{x^8}{x^9+1}\right)\,dx\\ -&=\ln|x|-\frac{1}{9}\ln|x^9+1|+C -\end{align*}$$ -Is there completely different way to solve it ? - -REPLY [6 votes]: Let we generalise the problem with a slightly different way. Consider -$$\int\frac{\mathrm dx}{x^n+x}$$ -Making substitution $x=\frac{1}{y}$ and $z=y^{n-1}$ then reverse it back we get -\begin{align} -\int\frac{\mathrm dx}{x^n+x}&=-\int\frac{y^{n-2}}{1+y^{n-1}}\mathrm dy\\[9pt] -&=-\frac{1}{n-1}\int\frac{\mathrm dz}{1+z}\\[9pt] -&=-\frac{\ln |1+z|}{n-1}+C\\[9pt] -&=\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\large\ln |x|-\frac{\ln \left|1+x^{n-1}\right|}{n-1}+C}} -\end{align} -In your case -$$\int\frac{\mathrm dx}{x^{10}+x}=\ln |x|-\frac{\ln \left|1+x^{9}\right|}{9}+C$$<|endoftext|> -TITLE: General solution of $C_{n+2}(x)=xC_n(x)+nC_{n-1}(x)$ -QUESTION [5 upvotes]: Airy differential equation. -$y''(x)=xy(x)$ -$y'''(x)=y(x)+x y'(x)$ -$y'^v(x)=x^2y(x)+2 y'(x)$ -$y^v(x)=4xy(x)+x^2 y'(x)$ -$y^{(6)}(x)=(x^3+4)y(x)+6x y'(x)$ -. -. -$y^{(n)}(x)=A_n(x)y(x)+B_n(x) y'(x)$ -Where $A_n(x)$ and $B_n(x)$ are polynomials -($y^{(n)}(x)$ means $n$-th order derivative of $y(x)$ ) -I have found that -$C_{n+2}(x)=xC_n(x)+nC_{n-1}(x)$ -If $C_2(x)=x$ ;$C_3(x)=1$ ; $C_4(x)=x^2$ are initial condition -$C_n(x)=A_n(x)$ -If $C_2(x)=0$;$C_3(x)=x$;$C_4(x)=2$ are initial condition -$C_n(x)=B_n(x)$ -How can we find the general solution of $C_{n+2}(x)=xC_n(x)+nC_{n-1}(x)$ ? -Can we express $C_n(x)$ as Airy function? -Thanks for answers. -$EDIT:$ -I tried generating function method as Sam recommended. -$$g(z,x)=\sum_{n=3}^\infty z^n C_n(x)$$ -$$g(z,x)=z^3C_3(x)+z^4C_4(x)+\sum_{n=5}^\infty z^n C_n(x)$$ -$$g(z,x)=z^3C_3(x)+z^4C_4(x)+\sum_{n=3}^\infty z^{n+2} C_{n+2}(x)$$ -$$g(z,x)=z^3C_3(x)+z^4C_4(x)+\sum_{n=3}^\infty z^{n+2}(xC_n(x)+nC_{n-1}(x))$$ -$$g(z,x)=z^3C_3(x)+z^4C_4(x)+xz^2\sum_{n=3}^\infty z^{n}C_n(x)+z^4\sum_{n=3}^\infty nz^{n-2}C_{n-1}(x)$$ -$$g(z,x)=z^3C_3(x)+z^4C_4(x)+xz^2\sum_{n=3}^\infty z^{n}C_n(x)+3z^5C_2(x)+z^4\sum_{n=4}^\infty nz^{n-2}C_{n-1}(x)$$ -$$g(z,x)=z^3C_3(x)+z^4C_4(x)+3z^5C_2(x)+xz^2\sum_{n=3}^\infty z^{n}C_n(x)+z^4\sum_{n=3}^\infty (n+1)z^{n-1}C_{n}(x)$$ -$$g(z,x)=z^3C_3(x)+z^4C_4(x)+3z^5C_2(x)+xz^2\sum_{n=3}^\infty z^{n}C_n(x)+z^4\sum_{n=3}^\infty nz^{n-1}C_{n}(x)+z^3\sum_{n=3}^\infty z^{n}C_{n}(x)$$ -We can get that -$$\frac{\partial g(z,x)}{\partial z}=\sum_{n=3}^\infty n z^{n-1} C_n(x)$$ -And finally, -$$g(z,x)=z^3C_3(x)+z^4C_4(x)+3z^5C_2(x)+xz^2g(z,x)+z^4\frac{\partial g(z,x)}{\partial z}+z^3g(z,x)$$ - -$$-z^4\frac{\partial g(z,x)}{\partial z}- (z^3+xz^2-1)g(z,x)=z^3C_3(x)+z^4C_4(x)+3z^5C_2(x)$$ -$$\frac{\partial g(z,x)}{\partial z}+\frac{ z^3+xz^2-1}{z^4}g(z,x)=-\frac{1}{z}C_3(x)-C_4(x)-3zC_2(x)$$ -I will let you know after solving the first order linear differential equation. - -REPLY [5 votes]: Let $G(z)$ be the exponential generating function defining $G(z) = \sum_{n=0}^\infty C_n \frac{z^n}{n!}$. The recurrence equation $C_{n+2} = x C_n + n C_{n-1}$ translates into a differential equation: -$$ - G^{\prime\prime}(z)- (x+z) G(z)= C_2 - x C_0 -$$ -This equation admits a closed form solution: -$$ - G(z) = \kappa_1 Ai(x+z) + \kappa_2 Bi(x+z) + (x+z)^2(C_2 - x C_0) h(x+z) -$$ -where -$$ - h(u) = \frac{1}{2} \cdot {}_{0}F_{1}\left(\frac{2}{3}, \frac{u^3}{9}\right) \cdot {}_1 F_{2}\left(\frac{2}{3}; \frac{4}{3}, \frac{5}{3}; \frac{u^3}{9}\right) - - {}_{0}F_{1}\left(\frac{4}{3}, \frac{u^3}{9}\right) \cdot {}_1 F_{2}\left(\frac{1}{3}; \frac{2}{3}, \frac{4}{3}; \frac{u^3}{9}\right) -$$ -The initial condition $G(0) = C_0$ and $G^\prime(0) = C_1$ determines $\kappa_1$ and $\kappa_2$. - -Added: After substitution of $C_0 = 1$, $C_1=0$, $C_2=x$ for $A_n(x)$ we get: -$$ - \sum_{n=0}^\infty \frac{z^n}{n!} A_n(x) = \pi \left( Bi^\prime(x) Ai(x+z) - Ai^\prime(x) Bi(x+z) \right) -$$ -Similarly, for $B_n(x)$ with $C_0=0$, $C_1=1$, $C_2=0$ we obtain: -$$ - \sum_{n=0}^\infty \frac{z^n}{n!} B_n(x) = \pi \left( Ai(x) Bi(x+z) - Bi(x) Ai(x+z) \right) -$$ -The simplification is made possible by simple expression of ${}_0F_1$ in terms of Airy functions: -$$\begin{eqnarray} - {}_{0}F_1\left(\frac{2}{3}; \frac{u^3}{9}\right) &=& -\frac{\Gamma\left(-\frac{1}{3}\right)}{ 3^{5/6}} \frac{1}{2} \left( \sqrt{3} Ai(u) + Bi(u) \right) \\ - {}_{0}F_1\left(\frac{4}{3}; \frac{u^3}{9}\right) &=& -\frac{\Gamma\left(-\frac{2}{3}\right)}{ 3^{5/6}} \frac{1}{u} \left( \sqrt{3} Ai(u) - Bi(u) \right) -\end{eqnarray} $$ -Of course, appearance of $y(x+z)$ should be anticipated, knowing the origin of the problem: -$$ - \left(\sum_{n=0}^\infty A_n(x) \frac{z^n}{n!} \right) y(x) + \left(\sum_{n=0}^\infty B_n(x) \frac{z^n}{n!} \right) y^\prime(x) = \sum_{n=0}^\infty \frac{z^n}{n!} y^{(n)}(x) = y(x+z) -$$<|endoftext|> -TITLE: Interpretation of limsup-liminf of sets -QUESTION [44 upvotes]: What is an intuitive interpretation of the 'events' -$$\limsup A_n:=\bigcap_{n=0}^{\infty}\bigcup_{k=n}^{\infty}A_k$$ -and -$$\liminf A_n:=\bigcup_{n=0}^{\infty}\bigcap_{k=n}^{\infty}A_k$$ -when $A_n$ are subsets of a measured space $(\Omega, F,\mu)$. -Of the first it should be that 'an infinite number of those events is verified', but -I don't see how to explain (or interpret this). Thanks for any help! - -REPLY [11 votes]: For the begining note that -$$ -s\in\bigcap\limits_{\lambda\in\Lambda} S_\lambda\Longleftrightarrow -\forall\lambda\in\Lambda\quad s\in S_\lambda -$$ -$$ -s\in\bigcup\limits_{\lambda\in\Lambda} S_\lambda\Longleftrightarrow -\exists \lambda\in\Lambda\quad s\in S_\lambda -$$ -Now we can show more formal explanation of what $\limsup$ is. -$$ -x\in\limsup\limits_{n\to\infty}A_n -\Longleftrightarrow -x\in\bigcap\limits_{n=0}^\infty\bigcup\limits_{k=n}^\infty A_k -\Longleftrightarrow -\forall n\in\mathbb{N}\quad x\in\bigcup\limits_{k=n}^\infty A_k -\Longleftrightarrow -$$ -$$ -\forall n\in\mathbb{N}\quad\exists k\geq n\quad x\in A_k -$$ -This means exactly that $x$ occurs in the sequence of sets $\{A_n:n\in\mathbb{N}\}$ infinitely many times. -Here is a formal explanation for $\liminf$ -$$ -x\in\liminf\limits_{n\to\infty}A_n -\Longleftrightarrow -x\in\bigcup\limits_{n=0}^\infty\bigcap\limits_{k=n}^\infty A_k -\Longleftrightarrow -\exists n\in\mathbb{N}\quad x\in\bigcap\limits_{k=n}^\infty A_k -\Longleftrightarrow -$$ -$$ -\exists n\in\mathbb{N}\quad\forall k\geq n\quad x\in A_k -$$ -This means exactly that $x$ occurs in all the sets of sequence $\{A_n:n\in\mathbb{N}\}$ except maybe in the first $n$ sets. - -REPLY [10 votes]: Read 'forall' with intersection, 'exists' with union. I prefer to think of events as follows: -$x \in\limsup A_n$ iff $\forall n$ $\exists k \geq n$ such that $x \in A_k$. -$x \in\liminf A_n$ iff $\exists n$ $\forall k \geq n$ we have $x \in A_k$.<|endoftext|> -TITLE: Map Surjective on a Disk -QUESTION [9 upvotes]: I've got another question from a student that has stumped me: Let $D^{n+1}$ be the $n+1$-disk, with boundary sphere $S^n$. Suppose $f:D^{n+1}\longrightarrow \mathbb{R}^{n+1}$ is a map such that $f(S^n)\subseteq S^n$. Furthermore, suppose that $f|_{S^n}$ has nonzero degree. Show that $f(D^{n+1})$ contains $D^{n+1}$. -I have to admit, I'm at a loss to even start this problem. - -REPLY [4 votes]: Alright, under the general heading of the Hopf Degree Theorem, we have the Extension Theorem. I'm looking at Guillemin and Pollack, pages 145-146, in the smooth category actually. -BUT see WOOKIE for the continuous case: -A map $f: \mathbb S^n \rightarrow \mathbb S^n$ is extendable to a map $F: \mathbb D^{n+1} \rightarrow \mathbb S^n$ if and only if $\deg(f)=0.$ -The Extension Theorem is the same with the preimage replaced by any compact connected oriented $W$ with boundary $\partial W$ of dimension $n+1$ and $n.$ -Anyway, the $f|_{S^n}$ you are given has nonzero degree, so there is no extension $F$ that maps all of the closed ball to the sphere. Meanwhile, assume that there is a point $U$ in the open ball that is not in the image of $f.$ Compose $f$ with central projection from $U$ onto $\mathbb S^n.$ This new map takes $\mathbb D^{n+1}$ to $\mathbb S^n,$ so it is an extension. This is a contradiction. -Note that it was not really necessary to have the missing point $U$ be at the origin, as the sphere is star-shaped around any point of the open ball. Furthermore, as the original $f\left(\mathbb D^{n+1}\right)$ is compact, the distance from it to $U$ is bounded from below. This seems necessary for concluding that the composed map is continuous.<|endoftext|> -TITLE: Can this function be rewritten to improve numerical stability? -QUESTION [14 upvotes]: I'm writing a program that needs to evaluate the function $$f(x) = \frac{1 - e^{-ux}}{u}$$ often with small values of $u$ (i.e. $u \ll x$). In the limit $u \to 0$ we have $f(x) = x$ using L'Hôpital's rule, so the function is well-behaved and non-singular in this limit, but evaluating this in the straightforward way using floating-point arithmetic causes problems when $u$ is small: then the exponential is quite close to 1, and when we subtract it from 1, the difference doesn't have great precision. Is there a way to rewrite this function so it can be computed more accurately both in the small-$u$ limit and for larger values of $u$ (i.e. $u \approx x$)? -Of course, I could add a test to my program that switches to computing just $f(x) = x$, or perhaps $f(x) = x - ux^2/2$ (second-order approximation) when $u$ is smaller than some threshold. But I'm curious to see if there's a way to do this without introducing a test. - -REPLY [3 votes]: As J.M. pointed out in a comment to another answer, some platforms and programming languages have a primitive that computes $2^x-1$ or $e^x-1$ in a single step with without precision loss for small $x$. For example, in C99 there is the expm1() function, which ought to map to the F2XM1 instruction on the x86 platform (via a scaling of $x$ by $\frac{1}{\log2}$).<|endoftext|> -TITLE: Normalizers of automorphism groups -QUESTION [30 upvotes]: In abstract groups $\Gamma$ the normalizer $N_\Gamma(S)$ of a subset $S\subseteq\Gamma$ is the subgroup of all $x \in \Gamma$ that commute with $S$, i.e. $xS = Sx$, i.e. $x\ y\ x^{-1} \in S $ for all $y \in S$. -Among the permutations $S_n$ of the vertices of a graph $G$ of order $n$ (or any other kind of structure) there is one distinguished subgroup: the automorphisms $\text{Aut}(G)$, that reflect the symmetries of $G$: -$$\alpha \in \text{Aut}(G)\quad \Leftrightarrow \quad \alpha G = G$$ -To give $\alpha G = G$ a proper meaning, identify $G$ with an adjacency matrix, for example. -The normalizer of $\text{Aut}(G)$ is another distinguished subgroup: it consists of those permutations $\pi$ of the vertices, such that $\text{Aut}(\pi G) = \text{Aut}(G)$, i.e. that respect the symmetries of $G$, as can be shown like this: -$\quad \pi \in N_{S_n}(\text{Aut}(G))\\ -\Leftrightarrow \pi^{-1}\alpha\ \pi \in \text{Aut}(G)\ \text{for all}\ \alpha \in \text{Aut}(G)\\ -\Leftrightarrow \pi^{-1}\alpha\ \pi\ G = G\ \text{for all}\ \alpha \in \text{Aut}(G)\\ -\Leftrightarrow \alpha\ \pi\ G = \pi\ G\ \text{for all}\ \alpha \in \text{Aut}(G)\\ \Leftrightarrow\alpha \in \text{Aut}(\pi G)\ \text{for all}\ \alpha \in \text{Aut}(G) $ -Note that $\text{Aut}(\pi G)$ and $ \text{Aut}(G)$ are of course isomorphic for every $\pi \in S_n$: -$$\text{Aut}(\pi G) \simeq \text{Aut}(G)$$ -but this is not the matter of concern. The matter of concern is -$$\text{Aut}(\pi G) = \text{Aut}(G)$$ -My first question now is: - -Does the normalizer of the automorphisms of a structure has an established name on its own? - -Something like symmetry preserving rearrangements (compared to adjaceny preserving rearrangements [what automorphisms are] or structure preserving rearrangements [what general permuations - of labels - are])? Note, that and how the following permutation is (i) symmetry preserving and (ii) an element of the normalizer of $\text{Aut}(G)$ and that (iii) most other permuations are not: - - -Where is the normalizer of the automorphisms of a structure investigated in its own right - or plays an explicit role, e.g. in a theorem? - -More specific: - -How can the normalizer of the automorphisms of a structure be - defined/characterized without reference to (and prior definition of) the - latter? - -REPLY [3 votes]: Some background first. -Given a (left) action of a group $S$ on a set $\Gamma$, we call $\Gamma$ a $S$-set. A morphism of $S$-sets $\phi:\Gamma\to\Lambda$ is a function such that $\phi(s\gamma)=s\phi(\gamma)$ for all $s\in S$ and $\gamma\in\Gamma$. In this way, the class of all $S$-sets becomes a category. -If $A\le S$ is a subgroup, then the left coset space $S/A$ is canonically an $S$-set, with action given by left multiplication, i.e. $\sigma(sA)=(\sigma s)A$ (meaning $\sigma$ sends $sA$ to $(\sigma s)A$). -If $\Gamma$ is a transitive $S$-set, meaning for all $\gamma_1,\gamma_2\in\Gamma$ there exists a $s\in S$ such that $s\gamma_1=\gamma_2$, and we have an element $\gamma\in\Gamma$ with point-stabilizer $A=\mathrm{Stab}_S(\gamma)$, then the map $S/A\to\Gamma$ given by $sA\mapsto s\gamma$ is well-defined, is a bijection, and is a morphism of $S$-sets. In other words, $\Gamma\cong S/A$ are isomorphic in the category of $S$-sets. (This is essentially a categorified version of the orbit-stabilizer theorem, which is usually just a numerical equality of cardinalities.) -The automorphism group $\mathrm{Aut}(\Gamma)$ of $\Gamma$ in the category of $S$-sets (i.e. the group of $S$-equivariant bijections $\Gamma\to\Gamma$) will be a direct product of wreath products of automorphism groups of orbits, so without loss of generality we would want to compute $\mathrm{Aut}(\Gamma)$ when $\Gamma=S/A$ is an orbit. -Suppose $\phi:S/A\to S/A$ is $S$-equivariant. Pick $\tau$ with $\phi(A)=\tau A$, then since $aA=A$ or all $a\in A$ we need $a\phi(A)=\phi(A)$ or in other words $a\tau A=\tau A$ for all $a\in A$, which is equivalent to $\tau\in N_S(A)$. Conversely, if $\tau\in N_S(A)$ then $\tau A=A\tau$ so really $\phi(sA)=sA\tau$ is just right-multiplication-by-$\tau$, which is evidently commutes with the left $S$-action. -However, there is some redundancy. Every $\tau\in N_S(A)$ defines an automorphism of $\Gamma$, but different elements $\tau\ne \tau'$ may define the same automorphism. The map $N_S(A)\to\mathrm{Aut}_S(S/A)$ can, however, be conceived of as a group homomorphism, where $\tau\mapsto R_{\tau^{-1}}$ with $R_{\tau^{-1}}(sA)=sA\tau^{-1}$ (we need to use $\tau^{-1}$ in order for $R_{\tau \tau'}=R_\tau\circ R_{\tau'}$). The kernel of $N_S(A)\to\mathrm{Aut}_S(S/A)$ is easily seen to be $A$ itself, so we conclude $\mathrm{Aut}_S(S/A)\cong N_S(A)/A$. -If we want to realize all of $N_S(A)$ as a symmetry group, instead of $N_S(A)/A$, we need to equip $\Gamma=S/A$ with more than just an $S$-action. We need to make it a groupoid. -A groupoid is, in one definition, a category with only isomorphisms. More concretely, a groupoid is a set of states and a set of transitions between states that can be composed. For example, the fifteen puzzle has many different configurations of pieces on the board, and many ways of transitioning between different states (a transition is a sequence of shifts of pieces) which can be composed together. The same goes for the Rubik's cube toy. -There is something special about the Rubik's cube's associated groupoid: it is an action groupoid. To understand this, first forget the colors on the cube (so it's entirely gray, say), and take its symmetry group $S$ of layer rotations and have it act on the set of states of the Rubik's cube. -Formally, if $S$ acts on $\Gamma$, the action groupoid $\Gamma/\!/S$ has $\Gamma$ as its set of states, and one transition $\gamma_1\to\gamma_2$ for every $s\in S$ with $s\gamma_1=\gamma_2$, which we can write as ordered triples $(s,\gamma,s\gamma)$. We can depict this with a Cayley graph: think of $\Gamma$ as a bunch of vertices, and draw an arrow $\gamma\to s\gamma$ labelled by $s$ for all $\gamma\in\Gamma,s\in S$. -Composing transitions $\gamma_1\xrightarrow{s}\gamma_2$ and $\gamma_2\xrightarrow{s'}\gamma_3$ gives $\gamma_1\xrightarrow{s's}\gamma_3$. -Note that $S$ canonically acts by groupoid automorphisms of $\Gamma/\!/S$. Applying $s'$ sends state $\gamma$ to state $s\gamma$ and sends transition $\gamma_1\xrightarrow{s'}\gamma_2$ to transition $s\gamma_2\xrightarrow{ss's^{-1}}s\gamma_2$. This identifies $S$ with a subgroup of $\mathrm{Aut}_{\mathrm{Gpd}}(\Gamma/\!/S)$ -Exercise. The group of groupoid automorphisms of $\Gamma/\!/S$ that intertwine with the $S$-action is isomorphic to the normalizer $N_S(A)$, where $\Gamma=S/A$. -(Note that it will not be the same subset as $N_S(A)\subseteq S\subset\mathrm{Aut}_{\mathrm{Gpd}}(\Gamma/\!/S)$.) -Now let's relate this to graphs. Actually, we can do this with any kind of labelled combinatorial structure on a finite set. Formally, a combinatorial species is an endofunctor of the category of finite sets with bijections. For example, on a finite set $V$, we can build labelled graphs on $V$, labelled trees, partial orderings, lattice orderings, linear orderings, permutations, permutations with a certain cycle structure, set partitions, set partitions with specified block sizes, $k$-subsets for some whole number $k$, power sets, power sets of power sets, and so forth. -The fact that a species $F$ is a functor means any "label exchange" $\pi:V\to W$ (we think of $V$ and $W$ as a set of labels we can slap onto some combinatorial structure, and grocers frequently exchange one set of labels with another if, for instance, a sale takes place) induces a function $F(\pi):FV\to FW$ between combinatorial structures. -For example, if $F$ is the species of graphs, then this has the following interpretation in semi-colloquial language. Graph with labels from $V$ can be turned into graphs with labels from $W$, because $\pi$ exchanges labels from $V$ for labels from $W$, and this induces a map from the collection $FV$ of graphs with labels from $V$ to the collection $FW$ of graphs with labels from $FW$. -In particular, if $V=W$, then $\pi\in S_V$ induces a $F(\pi)\in S_{FV}$. That is, we have a group homomorphism $S_V\to S_{FV}$, which can be interpreted as an action of $S_V$ on the collection of all $F$-structures on $V$. The orbits are the isomorphic $S_V$-structures, and the relabellings $\pi\in S_V$ provide the isomorphisms between isomorphic $F$-structures. -Suppose we work again with the species $F$ of graphs. A graph on $V$ can be formalized as a collection of edges $E$, or in other words a collection of $2$-subsets (unordered pairs). (This means that $FV=\mathcal{P}(\binom{V}{2})$ is a composition of the power set species with the $2$-subsets species.) The collection $\Gamma$ of all graphs on $V$ isomorphic to a given one with edge set $E$ is an orbit under the action of $S_V$, and the automorphism group $A$ of the particular graph is stabilizer of $E\in\Gamma$. -Now $S=S_V$ acts by groupoid automorphisms of the action groupoid $\Gamma/\!/S$, and the groupoid symmetries that commute with the $V$-relabelling action is $N_S(A)$. -Exercise. Draw all cycle graphs on four given points in the plane labelled $1,2,3,4$. (You should get three: a quadrilateral and two hourglass figures.) Draw arrows between these cycle graphs and label then with the permutations in $S_4$ to get the action groupoid. (You should have $24$ arrows total, with four from each figure to itself.) The fact that the automorphism group $V_4$ is normal in $S_4$ corresponds to the fact all permutations give relabelling-intertwining groupoid symmetries. -We can do this with directed graphs too. -Exercise. Do the same as the previous exercise, but with three points and directed cycle graphs. -Exercise. Do the same with rooted ternary trees on four points.<|endoftext|> -TITLE: A Haar measure via the Lebesgue measure on $\Bbb R^d$ -QUESTION [11 upvotes]: $\newcommand{\d}{\mathrm{d}}$ -This the Exercise 3, Chapter 11 of the Gerald B. Folland book Real Analysis: - -Let $G$ be a locally compact group that is homeomorphic to an open subset $U$ of $\Bbb R^d$ in such a way that, if we identify $G$ with $U$, left translation is an affine map -- that is, $xy=A_x(y)+b_x$ where $A_x$ is a linear transformation of $\Bbb R^d$ and $b_x\in\Bbb R^d$. Then $|\det A_x|^{-1}\d x$ is a left Haar measure on $G$, where $\d x$ denotes Lebesgue measure on $\Bbb R^d$. - -Let me know if I understand this. -What the problem gives us is $G$ a locally compact group, an open set $U\subseteq\Bbb R^d$, and a bijection $\varphi:G\to U$ such that $\varphi$ and $\varphi^{-1}$ are both continuous and satisfy that given $u\in G$ there is a linear operator $A_{\varphi(u)}:\Bbb R^d\to \Bbb R^d$ and $b_{\varphi(u)}\in\Bbb R^d$ so that $$\varphi(uv)=A_{\varphi(u)}(\varphi(v))+b_{\varphi(u)}.$$ - -Is this right - -If it is, define $f:\Bbb R^d\to\Bbb [0,\infty[$ given by $$f(x)=|\det A_x|^{-1}\quad\text{i.e.}\quad f(x)=|\det A_{\varphi(\varphi^{-1}(x))}|^{-1}.$$ - -Is the problem asking if the measure $\mu$ in $G$ given by $$\mu(E)=\int_{\varphi(E)}f(x)\d x$$ is a left Haar measure? - -This reminds me the formula $$\int_E f(y)\d y=|\det T|\int_{T^{-1} E} f(Tx)\d x,$$ -but I don't know what can I do. -Any advice in how to interpret the problem or on how to proceed is very appreciated. - -REPLY [7 votes]: The notation $xy=A_xy+b_x$ is quite confusing indeed. It appears on the left hand side, that $x,y\in G$; however, on the right hand side, it would appear that $y\in\mathbb{R}^d$. Your introduction of the homeomorphism, $\varphi:G\mapsto U\subset\mathbb{R}^d$, is the right thing to do. - -Is this right - -Yes, that is a restatement of part of the hypothesis with the homeomorphism given explicitly by $\varphi$. Since $\varphi$ is a homeomorphism, we could simply write -$$ -\varphi(uv)=A_u\varphi(v)+b_u -$$ - -Is the problem asking if the measure $\mu$ in $G$ given by $$\mu(E)=\int_{\varphi(E)}f(x)\mathrm{d} x$$ is a left Haar measure? - -Since $f(x)$ is meaningless for $x\not\in U$, I would define $f:U\mapsto[0,\infty)$ by -$$ -f(\varphi(u))=|\det A_u|^{-1} -$$ -and then define $\mu$ as you do above: -$$ -\begin{align} -\mu(E) -&=\int_{\varphi(E)}f(x)\,\mathrm{d}x -\end{align} -$$ -Now note that -$$ -\begin{align} -\varphi(uvw) -&=\color{#C00000}{A_{uv}}\varphi(w)+\color{#00A000}{b_{uv}}\\ -&=A_u\varphi(vw)+b_u\\ -&=A_u(A_v\varphi(w)+b_v)+b_u\\ -&=\color{#C00000}{A_uA_v}\varphi(w)+\color{#00A000}{A_ub_v+b_u} -\end{align} -$$ -Therefore, we get $A_{uv}=A_uA_v$. Thus, we get -$$ -\begin{align} -f(\varphi(uv)) -&=|\det(A_{uv})|^{-1}\\ -&=|\det(A_uA_v)|^{-1}\\ -&=|\det(A_u)\det(A_v)|^{-1}\\ -&=f(\varphi(u))f(\varphi(v)) -\end{align} -$$ -At this point, to show the translation invariance, -$$ -\begin{align} -\mu(uE) -&=\int_{\varphi(uE)}f(x)\,\mathrm{d}x\\ -&=\int_{\varphi(E)}\color{#C00000}{f(\varphi(u\varphi^{-1}(x)))}\,\color{#00A000}{\mathrm{d}(A_{u}x+b_{u^{-1}})}\\ -&=\int_{\varphi(E)}\color{#C00000}{f(\varphi(u))f(x)}\color{#00A000}{|\det A_{u}|\,\mathrm{d}x}\\ -&=\int_{\varphi(E)}|\det A_{u}|^{-1}f(x)|\det A_{u}|\,\mathrm{d}x\\ -&=\int_{\varphi(E)}f(x)\,\mathrm{d}x\\ -&=\mu(E) -\end{align} -$$<|endoftext|> -TITLE: Notation for element-wise division of vectors -QUESTION [11 upvotes]: I am wondering if there is any standard notation for the element-wise division of vectors. -I am going to use $\oslash$ for this purpose, similar to $\odot$ that is used in some texts for element-wise multiplication. For example, assuming $\vec{u}$ and $\vec{v}$ are vectors of length $k$, then $\vec{x} = \vec{u} \oslash \vec{v}$ if $\vec{x}(i) = \vec{u}(i) / \vec{v}(i)$. Would that be strange to use this in a scientific paper? - -REPLY [8 votes]: That is known as The Hadamard division and is documented here: -https://en.wikipedia.org/wiki/Hadamard_product_(matrices)#Analogous_operations -as: -$\begin{align} - C &= A {\oslash} B \\ - C_{ij} &= A_{ij} / B_{ij} -\end{align}$ -There are some other useful Hadamard operations and notations listed there. -It is noteworthy that the Hadamard multiplication uses the symbol "$\circ$" not "$\otimes$" or "$\odot$".<|endoftext|> -TITLE: Given a matrix A, how to find B such that AB=BA -QUESTION [10 upvotes]: Let $A = \begin{pmatrix} -1 & 1& 1\\ -1 & 2 &3 \\ -1 &4 & 5 -\end{pmatrix}$ and $D = \begin{pmatrix} -2 & 0& 0\\ -0 & 3 &0 \\ -0 &0 & 5 -\end{pmatrix}$. -It is found that right-multiplication by D multiplies each column of A by the corresponding diagonal entry of D, whereas left-multiplication by D multiplies each row of A by the corresponding diagonal entry of D. -Construct a 3 x 3 matrix B, not the identity matrix or zero matrix, such that $AB=BA$. - -REPLY [3 votes]: Notice that $A$ is diagonalizable, so any matirx with the same eigenvectors will commute with $A$, for example -$$B = \begin{pmatrix} -0.2481 & 0.31385 & -0.034765 \\ -0.48816 & -0.65821 & 0.76725 \\ --0.20907 & 1.0811 & 0.28335 -\end{pmatrix}$$<|endoftext|> -TITLE: Are there any elegant methods to classify of the Gaussian primes? -QUESTION [11 upvotes]: Out of curiosity, are there any relatively quick classifications of all the Gaussian primes, the primes in $\mathbb{Z}[i]$? -I found a classification here, but the process comes off as rather tedious. No doubt the end classification is nice, but is there a quicker and cleaner process to classify them all? -If it matters at all, I have a decent grasp on elementary number theory/quadratic reciprocity and ring/field theory but not much in the way of algebraic number theory yet. - -REPLY [12 votes]: Lattice theorem for rings: For any ring homomorphism $\psi:A\to B$ and any ideal $I$ of $B$, the set $\psi^{-1}(I)$ is an ideal of $A$. This map of ideals is inclusion-preserving, i.e. if $I\subseteq J$ are ideals of $B$, then $\psi^{-1}(I)\subseteq \psi^{-1}(J)$. Furthermore, for any prime ideal $P\subset B$, we have that $\psi^{-1}(P)$ is a prime ideal of $A$. - -We have an inclusion homomorphism $j:\mathbb{Z}\to\mathbb{Z}[i]$. Our strategy is to classify the (non-zero) prime ideals $P$ of $\mathbb{Z}[i]$ according to the "value" of $j^{-1}(P)=P\cap \mathbb{Z}$. -First, we will re-express the ring $\mathbb{Z}[i]$ in a more convenient form: - -Consider the ring homomorphism $\operatorname{ev}:\mathbb{Z}[x]\to\mathbb{Z}[i]$ defined by $\phi(f)=f(i)$. This function is surjective, and the kernel of $\operatorname{ev}$ is the ideal $(x^2+1)$ of $\mathbb{Z}[x]$. Therefore, by the first isomorphism theorem, $\operatorname{ev}$ descends to an isomorphism $\phi:R\to\mathbb{Z}[i]$, where $R=\mathbb{Z}[x]/(x^2+1)$. - -For the remainder of this post, we will now be interested in classifying the prime ideals of $R$. The lattice theorem guarantees that $\phi$ sets up a bijective, order-preserving correspondence between the prime ideals of the two rings. I can explain more about why this is okay if you'd like. -Consider the inclusion homomorphism $k:\mathbb{Z}\to R$ (of course $k=\phi^{-1}\circ j$); we will pretend for the sake of expediency that $\mathbb{Z}$ is actually contained as a subset of $R$. -For each (non-zero) prime ideal $(q)$ of $\mathbb{Z}$, we want to classify the prime ideals $Q$ of $R$ such that $k^{-1}(Q)=(q)$, i.e. the prime ideals $Q$ of $R$ that contain $(q)$. Note that a prime ideal $Q$ of $R$ will contain the $\mathbb{Z}$-ideal $(q)$ if and only if $Q$ contains the element $q$, which is the case if and only if $Q$ contains the $R$-ideal $qR$. By the lattice theorem, the prime ideals of $R$ containing $qR$ are in order-preserving bijection with the prime ideals of $$R/qR=(\mathbb{Z}[x]/(x^2+1))/q(\mathbb{Z}[x]/(x^2+1))\cong \mathbb{Z}[x]/(q,x^2+1)\cong \mathbb{F}_q[x]/(x^2+1).$$ -If $q=2$, then in $\mathbb{F}_2$ we have $x^2+1=(x+1)^2$, so that -$$R/2R\cong \mathbb{F}_2[x]/(x+1)^2$$ -has one prime ideal, namely $(x+1)\mathbb{F}_2[x]/(x+1)^2$. Unwinding our various isomorphisms, this corresponds to $i+1$ in $\mathbb{Z}[i]$. -If $q\equiv 1\bmod 4$, then $x^2+1$ is reducible (by quadratic reciprocity there is some $a\in \mathbb{F}_q$ such that $a^2=-1$ in $\mathbb{F}_q$), so that -$$R/qR\cong \mathbb{F}_q[x]/(x-a)(x+a)$$ -which has two prime ideals, $(x-a)\mathbb{F}_q[x]/(x-a)(x+a)$ and $(x+a)\mathbb{F}_q[x]/(x-a)(x+a)$. Unwinding our various isomorphisms, this corresponds to a conjugate pair of Gaussian primes $\pi$, $\overline{\pi}$ with norm $q\equiv 1\bmod 4$ in $\mathbb{Z}[i]$. -Lastly, if $q\equiv 3\bmod 4$, then $x^2+1$ is irreducible (by quadratic reciprocity there is no $a\in \mathbb{F}_q$ such that $a^2=-1$ in $\mathbb{F}_q$), so that -$$R/qR\cong \mathbb{F}_q[x]/(x^2+1)\cong\mathbb{F}_{q^2}$$ -is a field and therefore has one prime ideal, namely the zero ideal. Unwinding our various isomorphisms, this corresponds to a Gaussian prime $q\equiv 3\bmod 4$ that lives in $\mathbb{Z}$. - -Needless to say, this method is well-known and not original to me; for example I'm pretty sure that Neukirch does precisely this at the outset of his book. I'll see if I can find some good references later. Also, I'd just like to comment that algebro-geometrically, what's happening is we're looking at the fibers of $\operatorname{Spec}\mathbb{Z}[i]$ over $\operatorname{Spec}\mathbb{Z}$; look at the curve in $\operatorname{Spec}\mathbb{Z}[x]$ corresponding to the ideal $(x^2+1)$ in Mumford's famous sketch: - -I have to get to sleep, so I'm signing off for now, but I'll respond to any comments or questions when I can.<|endoftext|> -TITLE: Possibly false proof in AM -QUESTION [8 upvotes]: Here is the excerpt of the book where I suspect a mistake (page 66): - -Where they say "The restriction to $A$ of the natural homomorphism $A^\prime \to k^\prime$" I think we don't want a restriction. We start with the quotient map $\pi: A[x^{-1}] \to A[x^{-1}] /m$ where $m$ is a maximal ideal containing $x^{-1}$. We take an algebraic closure $\Omega$ of the field $A[x^{-1}] /m$ and consider the map $i \circ \pi: A[x^{-1}] \to \Omega$. Then by the previous theorem, (5.21), we can extend $i \circ \pi$ to some valuation ring $B$ of $K$ containing $A[x^{-1}]$: $g: B \to \Omega$ such that $g|_{A[x^{-1}]} = i \circ \pi$. Then $g(x^{-1}) = 0$. Hence $x^{-1} \in ker(g)$ and since the kernel is a proper ideal of $B$, $x^{-1}$ is not a unit in $B$ and hence $x$ is not in $B$. -Do you agree with my version and that what is written in Atiyah-Macdonald is not correct? Thank you. - -REPLY [4 votes]: Congratulations for spotting the difficulty and correcting it, Clark: you are absolutely right and I completely agree with your version! -As a slightly different formulation for the proof that $x\notin B$, I would just remark that if we had $x\in B$, we would deduce the absurd conclusion $$1=g(1)=g(x\cdot x^{-1})=g(x)\cdot g(x^{-1})=g(x)\cdot 0=0$$<|endoftext|> -TITLE: Prove that $ 1.462 \le \int_0^1 e^{{x}^{2}}\le 1.463$ -QUESTION [12 upvotes]: Prove the following integral inequality: -$$ 1.462 \le \int_0^1 e^{{x}^{2}}\le 1.463$$ -This is a high school problem. So far i did manage to prove that the integral is bigger than $1.462$ by using Taylor expansion, namely: -$$1.462\le 1.4625=\int_0^1 1+x^2+\frac{x^4}{2}+\frac{x^6}{6}+\frac{x^8}{24}+\frac{x^{10}}{120}\le \int_0^1 e^{{x}^{2}}$$ -For the right bound i'm still looking for a way. However, i wonder if there is an elegant way to solve both sides. - -REPLY [5 votes]: Integrating the power series for $e^{x^2}$, term by term, gives -$$ -\int_0^1e^{x^2}\,\mathrm{d}x=\sum_{k=0}^\infty\frac1{(2k+1)k!} -$$ -Since -$$ -\begin{align} -\sum_{k=n+1}^\infty\frac1{(2k+1)k!} -&\le\frac1{(2n+3)(n+1)!}\left[1+\frac1{n+2}+\frac1{(n+2)^2}+\dots\right]\\ -&=\frac1{(2n+3)(n+1)!}\frac{n+2}{n+1}\\ -&\le\frac1{(2n+1)(n+1)!} -\end{align} -$$ -If we use $n=5$ in -$$ -\sum_{k=0}^n\frac1{(2k+1)k!}\le\int_0^1e^{x^2}\,\mathrm{d}x\le\frac1{(2n+1)(n+1)!}+\sum_{k=0}^n\frac1{(2k+1)k!} -$$ -we get -$$ -1.4625300625\le\int_0^1e^{x^2}\,\mathrm{d}x\le1.4626563252 -$$<|endoftext|> -TITLE: Why is every irreducible matrix with period 1 primitive? -QUESTION [6 upvotes]: In a certain text on Perron-Frobenius theory, it is postulated that every irreducible nonnegative matrix with period $1$ is primitive and this proposition is said to be obvious. However, when I tried to prove the theorem by myself, I found that I was unable to come up with a formal proof. Intuitively, of course, I am sure that the theorem holds (and the converse statement is in fact easy to prove), but I am unable to prove it. -According to the text, the theorem should be obvious from the graph-theoretic interpretation of these notions. For a given nonnegative matrix, it is possible to construct a digraph as follows: There is an edge from the $i$-th vertex to the $j$-th vertex if and only if the entry $(i,j)$ of the matrix is positive. Thus, the matrix is irreducible if and only if its digraph is strongly connected. The period is defined to be the greatest common divisor of lengths of cycles (more precisely, the closed paths) of the graph. And finally, the matrix is said to be primitive if there is a positive integer $n$ such that for each pair of vertices of the graph there is a path of length $n$ interconnecting these vertices. -The theorem to be proved can thus be restated as follows: For every strongly connected digraph with the greatest common divisor of the lengths of closed paths equal to $1$, there is a positive integer $n$ such that for every pair of vertices of the digraph there is a path of length $n$ interconnecting these vertices. -It seems to me that the theorem might be proved by means of number theory, but I have not been able to find a proof up to now. To be more specific, I am looking for a proof without the use of the Perron-Frobenius theorem (the proposition is used in the text to prove the Perron-Frobenius theorem). Any ideas? -Thank you in advance. - -REPLY [9 votes]: Here is the argument given in section 1.3 of Gregory F. Lawler's Introduction to Stochastic Processes. It treats stochastic matrices $P$, -but I think the argument applies to general non-negative matrices. -For each state $i$ define $J_i=\{n: P^n(i,i)>0\}$. This is a semigroup and since -we have assumed that $P$ is aperiodic, we have $\gcd(J_i)=1$ and it follows that -$J_i$ contains all sufficiently large integers. -That is, there is some integer $M(i)$ so that for all $n\geq M(i)$ we have -$P^n(i,i)>0$. -Since $P$ is irreducible, there exists some $m(i,j)$ such that $P^{m(i,j)}(i,j)>0$. - Hence for $n\geq M(i)$, -$$P^{n+m(i,j)}(i,j)\geq P^n(i,i)\,P^{m(i,j)}(i,j)>0.$$ -Let $M$ be the maximum value of $M(i)+m(i,j)$ over all pairs $(i,j)$. Then -for $n\geq M$, $P^n(i,j)>0$ for all states $i,j$. -Essentially the same argument is found in section 1.3 of - Markov Chains and Mixing Times by Levin, Peres, and Wilmer. So it looks like probabilists have not found a better proof.<|endoftext|> -TITLE: Chain rule for Hessian matrix -QUESTION [12 upvotes]: Given $f\colon \mathbb{R}^n\rightarrow \mathbb{R}$ smooth and $\phi \in GL(n)$. What is the Hessian matrix $H_{f\circ \phi} = \left(\frac{\partial ^2 (f\circ \phi)}{\partial x_i\partial x_j}\right)_{ij}$? - -REPLY [11 votes]: Denote $H_g(x)$ the Hessian matrix of a function $g$. Denote $g=f\circ \phi$. By the chain rule, we have -$$D(f\circ\phi)(x)\cdot h=D(f(\phi x))\cdot D\phi\cdot h=D(f(\phi x))\cdot \phi\cdot h$$ -hence $D(g)(x)=D(f(\phi x))\cdot \phi$. In particular, -$$\partial_j g(x)=\sum_{k=1}^n\partial_kf(\phi x)a_{kj},$$ -where $a_{kj}$ is the $(k,j)$-th entry of $\phi$.We can do the same, for a fixed $k$, for the map $x\mapsto \partial_kf(\phi x)$. We get -\begin{align} -\partial_{ij}f(\phi x)&=\sum_{k,l=1}^n(H_f(\phi x))_{lk}a_{li}a_{kj}\\ -&=\sum_{k=1}^n(\phi^tH_f(\phi x))_{ik}a_{kj}\\ -&=(\phi^tH_f(\phi x)\phi)_{ij}. -\end{align}<|endoftext|> -TITLE: Value of $P(12)+P(-8)$ if $P(x)=x^{4}+ax^{3}+bx^{2}+cx+d$, $P(1)=10$, $P(2)=20$, $P(3)=30$ -QUESTION [5 upvotes]: What will be the value of $P(12)+P(-8)$ if $P(x)=x^{4}+ax^{3}+bx^{2}+cx+d$ - provided that $P(1)=10$, $P(2)=20$, $P(3)=30$? - -I put these values and got three simultaneous equations in $a, b, c, d$. What is the smarter way to approach these problems? - -REPLY [2 votes]: Other way doing this: -We try to find reals $e,f,g$ such that $P(12)+P(-8)=eP(1)+fP(2)+gP(3)$. So, if we try to equal "$x^k$ evaluated", we gain a system of equations: -$$ -\left\{\begin{array}{ccc} -1^ke+2^kf+3^kg&=&12^k+(-8)^k -\end{array}\right.,\quad k=0,\cdots,4 -$$ -In particular, -$$ -\left\{\begin{array}{ccc} -e+f+g&=&1+1\\ -e+2f+3g&=&12+(-8)\\ -e+2^2f+3^2g&=&12^2+(-8)^2 -\end{array}\right. -$$ -and we obtain $e=100,f=-198,g=100$. Verifying the others values for $k$: -$$ -\left\{\begin{array}{ccc} -100+2^3(-198)+3^3\cdot100&=&12^3+(-8)^3\\ -100+2^4\cdot(-198)+3^4\cdot100&=&12^4+(-8)^4 + 19800 -\end{array}\right. -$$ -So, $P(12)+P(-8)=100P(1)-198P(2)+100P(3)+1980=19840$<|endoftext|> -TITLE: The "need" for cohomology theories -QUESTION [21 upvotes]: In many surveys or introductions, one can see sentences such as "there was a need for this type of cohomology" or "X succeeded in inventing the cohomology of...". -My question is: why is there a need to develop cohomology theories ? What does it bring to the studies involved ? -(I have a little background in homological algebra. Apart from simplicial homology and the fact that it allows to "detect holes", assume that I know nothing about more complicated homology or cohomology theories) - -REPLY [10 votes]: The point is that different cohomology theories are applicable in different situations and are computed from different data. For example, simplicial/singular cohomology is computed from a triangulation (or the map of a simplex) into your space, while, for example, Cech cohomology is computed from just the different open covers of your space. -Also, for certain algebraic-geometry applications, no matter how we define the topology on a variety, we never seem to get enough open sets to do any kind of homological computations with, and so the development of etale cohomology solved this obstacle and let us use easy to compute 'topological' tools to investigate geometric objects.<|endoftext|> -TITLE: Projection matrices -QUESTION [6 upvotes]: I have found these two apparently contradicting remarks about projection matrices: - -A matrix $P$ is idempotent if $PP = P$. An idempotent matrix that is also Hermitian is called a projection matrix. - -$P$ is a projector if $PP = P$. Projectors are always positive which implies that they are always Hermitian. - - -Which of both is correct? Is a matrix $P$ that verifies $PP=P$ always Hermitian? - -REPLY [6 votes]: The fact that a projection matrix is Hermitian or not depends on your definition of projection matrices. Usually, if $P$ satisfies $PP = P$, then $P$ is idempotent, and is called a projection matrix, no matter it's Hermitian or not. If $P$ is also Hermitian, then it's called orthogonal projection, otherwise it's oblique projection. But some authors only define Hermitian idempotent matrix (orthogonal projection) as projection. See here to find more.<|endoftext|> -TITLE: Evaluating $\int_{0}^{\infty}\frac{\arctan \sin^2x}{x}dx$ -QUESTION [6 upvotes]: Seems to be a hard nut: -$$I=\int_{0}^{\infty}\frac{\arctan \sin^2x}{x}dx$$ Any hint? - -REPLY [12 votes]: It is not hard to show that -\begin{equation}\int_{0}^{\infty} \frac{\sin^2 x}{x} \; dx = \infty, \quad \cdots \quad (1)\end{equation} -and it is also easy to show that -$$\arctan x \geq Cx \quad \text{for} \quad 0 \leq x \leq 1 \quad \cdots \quad (2)$$ -for some positive constant $C > 0$. Now it is clear that these together imply -\begin{equation}\int_{0}^{\infty} \frac{\arctan \sin^2 x}{x} \; dx \geq C \int_{0}^{\infty} \frac{\sin^2 x}{x} \; dx = \infty\end{equation} - -Indeed, we first show that $(1)$ diverges. It suffices to show that -$$ \int_{2012}^{\infty} \frac{\sin^2 x}{x} \; dx = \infty. $$ -By integration by parts, we have -$$ \begin{align*} -\int_{2012}^{R} \frac{\sin^2 x}{x} \; dx -&= \left[ \frac{1}{2} - \frac{\sin 2x}{4x}\right]_{2012}^{R} + \int_{2012}^{R} \left( \frac{1}{2x} - \frac{\sin 2x}{4x^2}\right) \; dx\\ -&= \frac{1}{2}\log R + O(1), -\end{align*}$$ -which proves $(1)$ by letting $R\to\infty$. -Now we prove $(2)$. by examining second derivative of arc-tangent function, we find that it is concave on $[0, 1]$. Thus on this interval we have -$$\arctan x \geq (\arctan 1) x,$$ -which proves $(2)$ with $C=\arctan 1 =\frac{\pi}{4}$.<|endoftext|> -TITLE: What are rational integer coefficients? -QUESTION [13 upvotes]: I have a question about the following excerpt from Atiyah-Macdonald (page 30): - - -“A ring $A$ is said to be finitely generated if it is finitely generated as a $\mathbb Z$-algebra. This means that there exist finitely many elements $x_1,\dotsc,x_n$ in $A$ such that every element of $A$ can be written as a polynomial in the $x_i$ with rational integer coefficients.” - -I suspect one should delete "rational" and then it says that $A$ is called finitely generated if $A = \mathbb Z [a_1, \dots a_n]$ for some $a_i \in A$, that is, every element in $A$ can be written as a polynomial in $a_i$ with integer coefficients. -If this is a typo it is not mentioned on MO but perhaps it is not and I misunderstand the definitions. If I do: What are rational integer coefficients? - -REPLY [13 votes]: In algebra there are lots of different situations where "integers" arise, as -in the context of algebraic integers. The term rational integer just distinguishes the normal integers from any other sort of algebraic integer. - -REPLY [10 votes]: It is common to use the word "integer" to refer not only to elements of $\mathbb{Z}$, but also to a wider class of complex numbers called "algebraic integers": those numbers that are roots of a monic polynomial with integer coefficients. -In this context, "rational integer" is used for elements of $\mathbb{Z}$: it means an integer which is also a rational number. -The word "rational" also gets generalized sometimes. If we have a particular base field of interest, we call elements of that field "rational" and elements of its algebraic extensions "irrational". This reduces to the usual meaning when we take $\mathbb{Q}$ as the base field. -I see this generalization most commonly used when talking about the base field in the abstract, or when working over a particular finite field. e.g. if our base field is $\mathbb{F}_3$, then when discussing the elements of, say, $\mathbb{F}_{27}$, we would call $0,1,2$ rational, and the remaining 24 elements irrational.<|endoftext|> -TITLE: Proof by Induction $n^2+n$ is even -QUESTION [6 upvotes]: I'm not entirely sure if I'm going about proving $n^2+n$ is even for all the natural numbers correctly. -$P(n): = n^2+n$ - -$P(1) = 1^2+1 = 2 = 0$ (mod $2$), true for $P(1)$ - -Inductive step for $P(n+1)$: - -$\begin{align}P(n+1) &=& (n+1)^2+(n+1)\\ -&=&n^2+2n+1+n+1\\ -&=&n^2+n+2(n+1)\end{align}$ - -Does this prove $n^2+n$ is even as it's divisible by $2$? Thanks! - -REPLY [13 votes]: I see other answers provide different (possibly simpler) proofs. To finish off your proof: -by the induction hypothesis $n^2+n$ is even. Hence $n^2 + n = 2k$ for some integer $k.$ We have $$n^2+n+2(n+1) = 2k + 2(n+1) = 2(k+n+1) = 2\times\text{an integer} = \text{even}.$$ - -Does this prove $n^2+n$ is even as it's divisible by $2$? - -The key here is to remember stating & using the induction hypothesis.<|endoftext|> -TITLE: What are possibilities to disprove the Collatz Conjecture? -QUESTION [13 upvotes]: I was thinking about the Collatz Conjecture yesterday, and as opposed to trying to prove it, I was considering what would make the conjecture false. There were only two cases I could think of: - -We find a number that begins a sequence that trends out to infinity ($3n+1$ dominates the $\frac{n}{2}$) -We encounter a sequence of numbers that always results in itself/a completely isolated sequence. - -I would assume the primary contradiction to look for would be number one, but what sort of research could be done into the second idea? Is the second idea even possible? -Are there any other failures of the Collatz Conjecture that I haven't thought of? - -REPLY [9 votes]: In the "syracuse"-formulation of the collatz-transformation $T:= a_{n+1}={3a_n+1 \over 2^{A_n}}$ on odd a where $A_n$ is the greatest value that keeps $a_{n+1}$ integer we can eleganly write a formula to approach the cycle-problem. -Let's write in the following example $a,b,c,...$ for $a_1,a_2,a_3,...$ and $A,B,C,...$ for $A_1,A_2,A_3,...$ then we have for the example-assumtion of a 3-step-cycle -$$ b= {3a+1\over2^A} \qquad c= {3b+1\over2^B} \qquad a= {3c+1\over2^C} \qquad $$ and the equation must hold: -$$ bca = \left({3a+1\over2^A}\right)\left({3b+1\over2^B}\right)\left({3c+1\over2^C}\right)$$ - Reordering gives the "critical equation" for the general cycle problem (it is obvious how it can be extended to as many assumed steps as wished?) with $S=A+B+C+...$ and $N=(number-of-steps)$ -$$ 2^S = \left(3+{1\over a}\right)\left(3+{1\over b}\right)\left(3+{1\over c}\right) \\ =3^N \left(1+{1\over 3a}\right)\left(1+{1\over 3b}\right)\left(1+{1\over 3c}\right) $$ - One can immediately see, that the S must be so that a perfect power of 2 is slightly greater then the N 'th perfect power of 3. But we find empirically, that perfect powers of 2 and 3 are not very near - so the parentheses must be big and thus the elements $a_k$ of the cycle must be small ! You might try to find yourself possible $a,b,c$ for an assumed 3-step cycle which gives the critical equation -$$ 32 = 27 \left(1+{1\over 3a}\right)\left(1+{1\over 3b}\right)\left(1+{1\over 3c}\right) $$ -and only assuming the most general facts, that $a,b,c$ are all odd, distinct and not divisible by 3. Can you disprove the 3-step-cycle? -Well, since we know, that all $a_k \lt 2^{50} $ are not member of a cycle, the fractions in the parentheses are very small and we need a lot of parentheses and thus a high N, to increase $3^N$ to the next perfect power $2^S$ - by this formula we can derive a general lower bound for the number of steps required given a lower bound for the elements of the cycle $a_k$ . -This "critical equation" does not suffice for the disproof of cycles alone, because for any lower bound of a we can find a lower bound of N which lets the rhs reach and "overtake" the next perfect power of 2. The next step of possible research seems then to re-introduce the remaining modular requirements on subsequent $a_k$ . There has been real progress on the general "critical equation" assuming certain structures in the $a_k$, namely that of an "1-cycle" and "m-cycle" structure, specific simple and restricted forms of assumed cycles, which could then be disproved by the known distance-characteristic of powers of 2 to powers of 3, using the continued fraction of $\log(3)/ \log(2)$ and theory of rational approximation of linear forms in logarithms. -(See the wikipedia article where I also linked to online available articles of John Simons and Benne de Weger. The basic article of Ray Steiner is unfortunately not freely available online - this is really a grief!)<|endoftext|> -TITLE: How can I determine which series comparison test to use? -QUESTION [8 upvotes]: In my textbook, there is a section of questions that's instructions reads "Test for convergence or divergence, using each one of the following tests once," and the test choices it gives me are - -nth-Term Test -p-Series Test -Integral Test -Limit Comparison Test -Geometric Series Test -Telescoping Series Test -Direct Comparison Test - -Now, I was able to get most of the problems in this section right on the first try, but when I compared my answers to those in the solutions manual, I found that often the way I chose was not the same as theirs. Sometimes, my solution was simpler, but most of the time, the method they chose was quicker. For one of the problems, I did the limit comparison test and I ended up with a $b_{n}$ that was one, so I was really just doing some convoluted mix between the Limit Comparison Test and the nth-Term Test. -How can I develop skills to help me to decide which method to choose? Is there some "trick" to deciding which method will be the simplest? - -REPLY [4 votes]: The p-series and geometric series tests are for specific types of sequences, and it is clear when you can apply them. -Use the integral test for positive, decreasing functions or negative, increasing functions only (do not forget this condition). -Telescoping series always look like $\sum f(x+1)-f(x)$, so like the other series, they are for a particular type of series but watch out for the series $\sum \frac{1}{n(n+1)}$ and similar series that can be made into a telescoping series using partial fractions. -The best thing to do is to simply practice exercises until you have mastered their usage.<|endoftext|> -TITLE: Swatting flies with a sledgehammer -QUESTION [24 upvotes]: Prompted by a recent exchange with Gerry Myerson, I was wondering if anyone has a favorite example of a relatively simple problem with a rather elementary (though perhaps complicated) answer for which there's another answer that relies on an elegant use of a powerful result that's almost certainly beyond the background of the poser of the question. - -REPLY [11 votes]: OK, here's my proof that $49 < 50$. -There's one point in the first quadrant where the circle $x^2+y^2=1$ intersects the line $x=y$, and the tangent line to the circle at that point has slope $-1$ and $x$-intercept $\sqrt{2}$. Every line with the same slope and a slightly smaller $x$-intercept intersects the circle twice, and those with a larger intercept do not intersect the circle. -Now suppose we use $7/5$ as an approximation to $\sqrt{2}$, and draw the line with slope $-1$ and that $x$-intercept. Lo and behold, it passes through the points $(4/5,3/5)$ and $(3/5,4/5)$, which are on the circle. So it's a secant line, not a tangent line. Therefore we conclude that -$$ -\frac75<\sqrt{2}. -$$ -Therefore -$$ -49 = 7^2 < 5^2\cdot2=50. -$$<|endoftext|> -TITLE: Determine if it is possible to fit 2 circles in a rectangle -QUESTION [9 upvotes]: I have the following problem: - -Given a Rectangle with $L$ length and $W$ width and $2$ circles with $r_1$ and $r_2$ radius, determine if it's possible to fit these two circles inside the rectangle. - -I realized that: - -If $2r_1 > L$ or $2r_1 > W$ or $2r_2 > L$ or $2r_2 > W,$ then it is not possible to fit the circles in the rectangle. Thus if $2 r_1 + 2r_2 \leq L$ and $2r_1 + 2r_2 \leq W,$ they can fit vertically or horizontally. - -My doubt is to check if they can fit diagonally? How do I get it ? - -REPLY [6 votes]: Here is my thought process here: -It's true that the best way to proceed will always be to stick one circle 'in one corner' of the rectangle and to try to fit the other circle 'in the opposite corner.' I take this for granted. It is very easy to understand where the center of a circle will go when the circle is 'tucked in a corner,' as both sides of the rectangle will be tangent to the circles. For example, if the rectangle is at coordinates $(0,0), (0,h), (l,0),$ and $(l,h)$, and a circle of radius $r_1$ is in the bottom-left corner, then the coordinates of its center will be $(r_1, r_1)$. If another circle, this one of radius $r_2$, is tucked in the top-right corner, the coordinates of it's center will be at $(l-r_2, h-r_2)$. -So one has to check two things: one has to make sure that both circles fit in the rectangle itself (without worrying about overlap), and then one has to make sure that the distance between the centers is greater than the sum of the radii. -So $2r_1, 2r_2 \leq l,h$ and $d[(r_1,r_1), (l-r_2,h-r_2)] \geq r_1 + r_2$ should be necessary and sufficient, where I let $d(\cdot, \cdot)$ denote the distance function.<|endoftext|> -TITLE: Coherence Conditions and Strict Monoidal Equivalences -QUESTION [6 upvotes]: Consider the following: two monoidal categories $({\cal C},\otimes)$, and $({\cal D},\odot)$, and a functor $F:{\cal C} \to {\cal D}$, that gives an equivalence (of ordinary categories) between ${\cal C}$ and ${\cal D}$. Moreover, let us assume that we also have -$$ -F(X \otimes Y) = F(X) \odot F(Y), ~~~~~ \text{ for } X,Y \in {\cal C}. -$$ -Does it automatically follow from this that $F$ is a (strict) equivalence of mondoidal categories? That is to say, do the necessary coherence conditions hold automatically? -To me this seems obviously true. However, there have times when what were "obvious truths" to me in general category theory, turned out to be anything but. So, I'd like a voice of confirmation. Thanks! - -REPLY [2 votes]: I haven't been able to come up with an explicit counterexample, but I believe one exists. The problem amounts to finding a category $\mathcal{C}$ equipped with a tensor product $\otimes$ and a unit $I$ such that there are two distinct coherence structures $(\lambda, \rho, \alpha)$. We make $\mathcal{C}$ into a monoidal category using the first coherence structure, and we take $\mathcal{D}$ to be same as $\mathcal{C}$ but with the second coherence structure $(\lambda', \rho', \alpha')$ instead. Suppose $\alpha \ne \alpha'$. Then, the "identity" functor $F : \mathcal{C} \to \mathcal{D}$ is an isomorphism of categories but fails to be a strong monoidal functor under the obvious identity transformations $I \to F I$, $F A \otimes F B \to F (A \otimes B)$ because the diagram below fails to commute: -            -(A similar argument can be made for the case $\lambda \ne \lambda'$ or $\rho \ne \rho'$.) - -Basically, what I would like to say is that because the theory of monoidal categories with two coherence structures is a consistent essentially algebraic theory, such a category $\mathcal{C}$ exists: simply take the free such category generated by one object. Unfortunately, one still has to prove that this category has two distinct coherence structures!<|endoftext|> -TITLE: Residue of Rankin-Selberg L-function for non-trivial nebentypus -QUESTION [9 upvotes]: Let $f\in S_k(\Gamma_0(N),\chi)$ be a normalized holomorphic newform (i.e. weight $k$, level $N$, nebentypus $\chi$) and write its Fourier expansion as -$$ f(z)=\sum_{n\ge 1} \lambda_f(n)n^{(k-1)/2}e^{2\pi i n z}$$ -for $\Im z >0$ where $\lambda_f(1)=1$ and, by Deligne's bound, $|\lambda_f(n)|\le d(n)$. Here $d(n)$ denotes the number of positive divisors of $n$. Define the Rankin-Selberg convolution $L$-function as -$$ -L(s,f\times \bar{f}) = \sum_{n=1}^\infty \frac{|\lambda_f(n)|^2}{n^s} -$$ -for $\Re s >1$. This $L$-function has a simple pole at $s=1$. -Questions: -(1) What is this residue? -(2) How does one compute this residue? - -REPLY [3 votes]: As you have said, let $f$ we a weight $k$ form of nebentypus $\chi$. Then in particular, $f(\gamma z) = (cz + d)^k \chi(d) f(z)$, where $\gamma = \left(\begin{smallmatrix} a&b\\c&d \end{smallmatrix}\right)$ is a matrix in our congruence subgroup du jour. -Notice that $\lvert f(z) \rvert^2 y^k$ is invariant under the slash operator, as -$$ \begin{align} -\lvert f(\gamma z) \rvert^2 (\gamma y)^k &= f(\gamma z) \overline{f(\gamma z)} \frac{y^k}{\lvert cz + d \rvert^{2k}} \\ -&= (cz+d)^k \overline{(cz + d)^k} \chi(d) \overline{\chi(d)} f(z) \overline{f(z)} \frac{y^k}{\lvert cz + d\rvert^{2k}} \\ -&= \lvert f(z) \rvert^2 y^k. -\end{align}$$ -This means that it is meaningful to take the inner product against the normal Eisenstein series -$$ E(z,s) = \sum_{\gamma \in \Gamma_\infty \backslash \Gamma_0(N)} \text{Im}(\gamma z)^s.$$ -Performing an unfolding of the integral along the critical strip, we can see that -$$ \langle \lvert f \rvert^2 y^k, E(z,s) \rangle = \frac{\Gamma(s + k - 1)}{(4 \pi)^{s + k - 1}} \sum_n \frac{a(n)^2}{n^{s + k - 1}},$$ -where $a(n) = \lambda(n)n^{(k-1)/2}$ are the Fourier coefficients of $f$. Notice the sum on the right is the Rankin-Selberg $L$-function, and I deliberately avoid indicating any normalization anywhere. -As we understand the analytic behaviour of the Eisenstein series and the $\Gamma$ function, we can understand the analytic behavior of the $L$-function. Most importantly, since the Eisenstein series has a single pole at $s = 1$ of known residue, we understand the pole of the $L$-function, and it has residue -$$ \langle fy^{k/2}, fy^{k/2} \rangle R \frac{(4\pi)^k}{\Gamma(k)},$$ -where $R$ is the residue of the Eisenstein series attached to the congruence subgroup (and often looks something like $\frac{3}{\pi}$).<|endoftext|> -TITLE: There are infinitely many $m$ such that $m^4 + 1$ has large prime factors -QUESTION [6 upvotes]: How do I prove that there is infinite set of numbers $m$ such that the biggest prime divisor of $m^4+1$ is bigger than $2m$? - -REPLY [6 votes]: Let $p$ be a prime congruent to 1 modulo 8 --- there are infinitely many such. For each such prime, there are 8 numbers $a$, $0\lt a\lt p$, such that $p$ divides $a^8-1=(a^4-1)(a^4+1)$. So there are four numbers $m$ such that $p$ divides $m^4+1$. They come in pairs that add up to $p$, so two of them are less than $p/2$. Either one is thus an $m$ with a prime divisor exceeding $2m$. -Each $m$ has only finitely many $p$ dividing $m^4+1$, so there are infinitely many such $m$.<|endoftext|> -TITLE: What does order topology over Ordinal numbers look like, and how does it work? -QUESTION [11 upvotes]: Space of ordinal numbers are one of the favorite examples of my professor in general topology. I quite understand the idea at the base of ordinal numbers (few things or nothing about the concept of ordinal itself in set theory). What we were told was just about constructing bigger and bigger ordinals like putting boxes into boxes into boxes...and that starting at a certain ordinal number, there are finite steps backwards and infinite forward. -So I still don't have a clear what a finite and what a successor ordinal are, I know the definitions and I have a vague idea, but I can't link it in a successful way with the proposition I reported at the end of the previous paragraph. -If somebody could explain these concepts in a better way, maybe then I could understand how to verify or disprove base-numerability, compactness and other important topological properties on the space. I'm not now interested in the ZF theory but in understanding how to manage and work on this space, in particular on $[0,\Omega]$ or $[0,\Omega)$, where $\Omega$ id the first uncountable ordinal. - -REPLY [24 votes]: I’m going to use the more standard notation $\omega_1$ for the first uncountable ordinal. -The most important thing to understand is that the linear order $\le$ on the set $[0,\omega_1]$ is a well-order: if $A$ is any non-empty subset of $[0,\omega_1]$, $A$ has a smallest element with respect to $\le$. That is, there is some $\alpha\in A$ such that $\alpha\le\beta$ for every $\beta\in A$. I’ll use this fact in a bit, but first let’s look a little closer at the elements of $[0,\omega_1]$. -The first countably infinitely many elements are the finite ordinals; you can think of these as being simply the non-negative integers, $0,1,2,3,\ldots$; this is, so to speak, the low end of the order $\le$. Now let $A=\{\alpha\in[0,\omega_1]:\alpha\text{ is not a finite ordinal}\}$. The set of finite ordinals is countable, and $[0,\omega_1]$ is uncountable, so $A\ne\varnothing$, and therefore $A$ has a least (or smallest) element; we call this element $\omega$. The set $\{0,1,2,\dots,\}\cup\{\omega\}$ is still countable, so the set $$[0,\omega_1]\setminus\Big(\{0,1,2,\dots,\}\cup\{\omega\}\Big)$$ is non-empty and therefore has a least element; we call this element $\omega+1$. This $\omega+1$ is the smallest ordinal after $\omega$: it comes right after $\omega$ in the order, so it’s the successor of $\omega$, just as $2$ is the successor of $1$. At this point we have a low end of $[0,\omega_1]$ that looks like this: -$$0,1,2,3,\dots,\omega,\omega+1$$ -It should be intuitively clear that we can repeat this argument countably infinitely many times to produce $\omega+2,\omega+3,\dots\,$, and indeed $\omega+n$ for every finite ordinal $n$. Now we have an initial segment of $[0,\omega_1]$ that looks like this: -$$0,1,2,3,\dots,\omega,\omega+1,\omega+2,\omega+3,\dots$$ -The only ordinals in this set that are not successors are $0$, since there’s nothing before it at all, and $\omega$, since there is nothing immediately before it: no matter what finite ordinal $n$ you consider, $n+1\ne=\omega$. -But this set is still countable, so there is a smallest ordinal in -$$[0,\omega_1]\setminus\{0,1,2,\dots,\omega,\omega+1,\omega+2\dots\}\;;$$ -this ordinal is denoted by $\omega\cdot 2$, and like $\omega$, it’s not a successor: it is not $\alpha+1$ for any $\alpha$. In other words, it’s a limit ordinal, as is $\omega$. ($0$ is a bit anomalous: it’s not a successor ordinal, but it’s also not a limit ordinal.) -After this the ordinals and their standard notations start getting a bit complicated, and we don’t really have to go into them to understand the topological properties of $[0,\omega_1)$ and $[0,\omega_1]$. It’s also true that the usual set-theoretic definition of the ordinals makes each ordinal equal to the set of its predecessors: -$$1=\{0\},2=\{0,1\},3=\{0,1,2\},\dots,\omega=\{0,1,2,\dots\},\omega+1=\{0,1,2,\dots,\omega\}\,,$$ -and so on; this is the boxes within boxes that you mentioned. However, you don’t need to worry about this, either. -Now let’s see why every strictly decreasing sequence in $[0,\omega_1]$ is finite. Suppose that we had an infinite sequence $\langle\alpha_n:n\in\Bbb N\rangle$ such that $\alpha_0>\alpha_1>\alpha_2>\ldots\,$; then the set $A=\{\alpha_n:n\in\Bbb N\}$ would be a non-empty subset of $[0,\omega_1]$ with no least element, contradicting the fact that $[0,\omega_1]$ is well-ordered. Infinite increasing sequences are no problem at all, however: for each $\alpha\in[0,\omega_1)$, the set $[0,\alpha]$ is countable, so $[0,\omega_1)\setminus[0,\alpha]\ne\varnothing$, so there are elements of $[0,\omega_1)$ bigger than $\alpha$. The smallest of these is $\alpha+1$, the successor of $\alpha$. Thus, starting at any $\alpha\in[0,\omega_1$ I can form an infinite increasing sequence $\langle\alpha,\alpha+1,\alpha+2,\dots\rangle$ whose members are all still in $[0,\omega_1)$. -Next, let’s see why $[0,\omega_1)$ is first countable. Let $\alpha\in[0,\omega_1)$. Suppose first that $\alpha$ is a successor ordinal, say $\alpha=\beta+1$; then $(\beta,\alpha+1)=[\beta+1,\alpha+1)=[\alpha,\alpha+1)=\{\alpha\}$ is an open nbhd of $\alpha$ in the order topology, so $\alpha$ is an isolated point, and $\big\{\{\alpha\}\big\}$ is certainly a countable local base at $\alpha$! Note that $0$ behaves like a successor ordinal: $[0,1)=\{0\}$ is an open nbhd of $0$, so $0$ is also an isolated point. -Now suppose that $\alpha$ is a limit ordinal. For each $\beta<\alpha$ the set $(\beta,\alpha+1)=(\beta,\alpha]$ is an open nbhd of $\alpha$. Every open nbhd of $\alpha$ contains an open interval around $\alpha$, which in turn contains one of these intervals $(\beta,\alpha]$, so $$\mathscr{B}_\alpha=\Big\{(\beta,\alpha]:\beta<\alpha\Big\}$$ is a local base at $\alpha$. Finally, $\alpha<\omega_1$, and $\omega_1$ is the first ordinal with uncountably many predecessors, so there are only countably many $\beta<\alpha$, and $\mathscr{B}_\alpha$ is therefore countable. Thus, every point of $[0,\omega_1)$ has a countable local base, and $[0,\omega_1)$ is therefore first countable. -Note that $[0,\omega_1]$ is not first countable, because there is no countable local base at $\omega_1$: if $\big\{(\alpha_n,\omega_1]:n\in\Bbb N\big\}$ is any countable family of open intervals containing $\omega_1$, let $A=\bigcup_{n\in\Bbb N}[0,\alpha_n]$. Then $A$, being the union of countably many countable sets, is a countable subset of $[0,\omega_1)$, so $[0,\omega_1)\setminus A\ne\varnothing$. Pick any $\beta\in[0,\omega_1)\setminus A$; then $(\beta,\omega_1]$ is an open nbhd of $\omega_1$ that does not contain any of the sets $(\alpha_n,\omega_1]$, and therefore the family $\big\{(\alpha_n,\omega_1]:n\in\Bbb N\big\}$ is not a local base at $\omega_1$. That is, no countable family is a local base at $\omega_1$, so $[0,\omega_1]$ is not first countable at $\omega_1$. -Finally, let’s look at compactness. Suppose that $\mathscr{U}$ is an open cover of $[0,\omega_1]$. Then there is some $U_0\in\mathscr{U}$ such that $\omega_1\in U_0$. This $U_0$ must contain a basic open nbhd of $\omega_1$, so there must be an $\alpha_1<\omega_1$ such that $(\alpha_1,\omega_1]\subseteq U_0$. $\mathscr{U}$ covers $[0,\omega_1]$, so there is some $U_1\in\mathscr{U}$ such that $\alpha_1\in U_1$. This $U_1$ must contain a basic open nbhd of $\alpha_1$, so there is some $\alpha_2<\alpha_1$ such that $(\alpha_2,\alpha_1]\subseteq U_2$. Continuing in this fashion, we can construct a decreasing sequence $\alpha_1>\alpha_2>\alpha_3>\ldots\,$, which, as we saw before, must be finite. Thus, there must be some $n\in\Bbb Z^+$ such that $\alpha_n=0$, and at that point $\{U_0,\dots,U_n\}$ is a finite subcover of $\mathscr{U}$. -The space $[0,\omega_1)$, on the other hand, is not compact. It is countably compact, however. The easiest way to prove this is to show that $[0,\omega_1)$ has no infinite, closed, discrete subset. Suppose that $A$ is a countably infinite subset of $[0,\omega_1)$. Let $\beta$ be the smallest element of $[0,\omega_1)$ that is bigger than infinitely many elements of $A$. (You’ll have to explain why $\beta$ exists, using the fact that $A$ is countable and $[0,\omega_1)$ is well-ordered.) Finally, show that $\beta$ is a limit point of $A$. Then either $\beta\notin A$, in which case $A$ isn’t closed, or $\beta\in A$, in which case $A$ isn’t discrete.<|endoftext|> -TITLE: History of Quadratic Formula -QUESTION [5 upvotes]: My wife is planning a lesson on the quadratic formula for high school students, who have previously learned how to complete the square. It would be nice to open the lesson with some historical background. -Does anybody know of any nice articles, anything from a few paragraphs to page or two long, that discusses the history of the formula and what problems its original discoverers were trying to solve. We would prefer an article that does not spill the beans about the formula's derivation. - -REPLY [2 votes]: Here is a collection of answers from comments, so that this question can be put to bed. -J.M. says: - -You could start with looking at this and this. If memory serves, apart from the Babylonians, the Chinese and Hindus also knew about (a special form of) the quadratic formula, but I don't have my references right now. - -Robert Israel says: - -Also look at http://www-history.mcs.st-and.ac.uk/HistTopics/Quadratic_etc_equations.html - -Andre Nicolas says: - -For the "completing the square part" see al-Khwarizmi, who was completing an honest to goodness square. For example, to explain solution of $x^2+2bx=d$ ($b$, $d$ positive) imagine a square house of side $x$, with porches of width $b$, length $x$ on north and east. Complete this to a square by adding $b\times b$ square. Big square has side $x+b$, area $d+b^2$, so $x=\sqrt{d+b^2}−b$. -The teacher may be interested in the following simpler derivation of the quadratic formula. The main pedagogical advantage is that one ends up avoiding division until the very end. The trick is very old, going back to Sridhara, but is in my opinion insufficiently used. - -Dave L. Renfro says: - -I'll send you some .pdf files by email shortly, but the following (freely available on the internet) will probably be more useful: Solving Quadratic Equations By analytic and graphic methods; Including several methods you may never have seen by Pat Ballew (2007).<|endoftext|> -TITLE: People on a Cube -QUESTION [6 upvotes]: On a cuboidal planet with side length 1, either people live on each of the eight vertices. One day, each of them decides to move to a vertex that is a distance of $\sqrt{2}$ away from him. What is the probability that no two of them occupy the same vertex after this move? - -REPLY [11 votes]: Hint: you can color the vertices with two colors. People stay on their own color and can move to any other vertex of that color. So you can consider the problem on a tetrahedron, and square the result. Then you are looking for a derangement on the four items (why?)<|endoftext|> -TITLE: Books like Grundlagen der Analysis in French -QUESTION [7 upvotes]: I am looking for some recommendations for a mathematics (text)book written in French. I am hoping to learn to read and write mathematics in French since I expect to take some mathematics courses that will be taught in French next year. -Basically, I would like a book whose linguistic and mathematical level approaches that of Landau's book Grundlagen der Analysis- the language should be clear and simple (read: not unnecessarily fancy or ornamental) and the mathematics contained in the book should be comprehensible to someone who has taken a course or two in analysis and algebra. It would be a huge plus if you can recommend a book that also contains a French-English vocabulary at the back like in Landau's book but the recommendation should primarily be based on the clarity of the text and the level of mathematical depth. -Please note that the book you recommend need not be a textbook- it could be lecture notes that are particularly well-written or monographs exploring some topic or even a popular account of mathematics that is relatively easy to understand. Also, the level of the book need not be at the university level. Depending on the book, it might even be preferable if it was at the high school level. -So, any suggestions? - -REPLY [4 votes]: Here a collection of French mathematics books choose the ones you want they are among the most famous textbooks and problems solving books. and i hope that will help you. -Cours (Text-books) -Cours de mathematiques Tome 1 - Algèbre, J.M.Arnaudiès, P.Delezoide, H.Fraysse -Cours de Mathematiques Tome 2 - Analyse, J.M.Arnaudiès, P.Delezoide, H.Fraysse -Cours de Mathematiques Tome 3 - Complements d'analyse, J.M.Arnaudiès, P.Delezoide, H.Fraysse -Cours de mathematiques Tome 4 - Algebre bilineaire et geometrie, J.M.Arnaudiès, P.Delezoide -Cours de mathématiques spéciales Tome 1 - Algèbre, E.Ramis, C.Deschamps, J.Odoux -Cours de mathématiques spéciales Tome 2 - Algèbre et applications à la géométrie, E.Ramis, C.Deschamps, J.Odoux -Cours élémentaire de mathématiques supérieures Tome 1 - Algèbre, J.Quinet -Cours élémentaire de mathématiques supérieures Tome 2 - Fonctions usuelles, J.Quinet -Cours élémentaire de mathématiques supérieures Tome 3 - Calcul intégral et séries, J.Quinet -Cours élémentaire de mathématiques supérieures Tome 4 - Equations différentielles, J.Quinet -Cours élémentaire de mathématiques supérieures Tome 5 - Géométrie, J.Quinet -Eléments d'analyse Tome 1 - Fondements de l'analyse moderne, Jean Dieudonné -Eléments d'analyse Tome 2 - Analyse fonctionnelle linéaire, Jean Dieudonné -Eléments d'analyse Tome 3 - Analyse sur les variétés, Jean Dieudonné -Eléments d'analyse Tome 4 ,Jean Dieudonné -Eléments d'analyse Tome 5 - Groupes de Lie compacts et semi-simples, Jean Dieudonné -Eléments d'analyse Tome 6 - Analyse harmonique, Jean Dieudonné -Eléments d'analyse Tome 7 - Equations fonctionnelles linéaires, 1ère partie - Opérateues pseudo-différentiels, Jean Dieudonné -Eléments d'analyse Tome 8 - Equations fonctionnelles linéaires, 2ème partie - Problèmes aux limites, Jean Dieudonné -Eléments d'analyse Tome 9 - Topologie algébrique, topologie différentielle élémentaire, Jean Dieudonné -Mathématiques 3 - Analyse cours et Exercices, E.Azoulay, J.Avignant -Mathématiques 4 - Algèbre, E.Azoulay, J.Avignant -Mathématiques L3 - Algèbre cours complet avec 400 tests et exercices corrigés, A.Szpirglas -Mathématiques L3 - Analyse Cours complet avec 600 tests et exercices corrigés, Jean-Pierre Marco -Roger Godement - Analyse Mathématique 1, Convergence, fonctions élémentaires -Roger Godement - Analyse Mathématique 2, Calcul différentiel et intégral,séries de Fourier,fonctions holomorphes -Roger Godement - Analyse Mathématique 3, Fonctions analytiques, différentielles et variétés, surfaces de Riemann -Roger Godement - Analyse Mathématique 4, Intégration et théorie spectrale, analyse harmonique, le jardin des délices modulaires -Mathematique 2eme annee Cours et Exercices Corriges, C.Deschamps, A.Warusfel -Mathématiques Algebre et geometrie 50% et 50% exercices, G.Auliac, J.Delcourt, R.Goblot -Mathématiques générales 1 ère année cours et exercices corrigés, J.Vélu -Mathématiques méthodes et exercices MP, J.M Monier -Mathématiques TOUT-EN-UN 1ère année cours et exercices corigés MPSI, PCSI, Série E.Ramis, C.Deschamps, A.Warusfel -TD Analyse, 70% Applications 30% Cours, Jean-Pierre Lecoutre, Philippe Pilibossian -Topologie et analyse fonctionnelle cours de licence et 240 exercices et 30 problèmes -corrigés, Y.Sonntag -Calcul diffrentiel et calcul intégral 3eme année Cours et exercices avec solutions, Marc -Chaperon -Complement d'analyse, K.Arbenz, A.Wohlhauser -Cours d'Algébre avec énoncés 40 Exercices et 300 Problèmes, J.Querré -Cours de Mathematiques MP-MP', Jean Voedts -Problems solving (Exercices resolus) -Algèbre Exercices avec solutions, E. Ramis, C. Deschamps, J. Odoux -Analyse Tome 1 - Exercices avec solutions, E. Ramis, C. Deschamps, J. Odoux -Analyse Tome 2 - Exercices avec solutions, E. Ramis, C. Deschamps, J. Odoux -Exercices resolus Tome 1 - Algebre du cours de mathematiques Tome 1, J.M.Arnaudiès -Exercices resolus Tome 2 - Analysee du cours de mathematiques, J.M.Arnaudiès, P.Delezoide -Exercices resolus Tome 3 - Complements d'analyse du cours de mathematiques, J.M.Arnaudiès -Exercices resolus Tome 4 - Algebre bilineaire et geometrie du cours de mathematiques, -J.M.Arnaudiès, P.Delezoide, H.Fraysse -Exercice d'algebre 1, B.Calvo, J.Doyen, A.Calvo, F.Boschet -Exercices d'analyse 1, B.Calvo, J.Doyen, A.Calvo, F.Boschet -Problèmes d'Analyse Tome 1 - Nombres réels suites et séries, Exercices corrigés,Wieslawa -J.Kaczor, Maria T.Nowak -Problèmes d'Analyse Tome 2 - Continuité et dérivabilité, Exercices corrigés,Wieslawa J.Kaczor, -Maria T.Nowak -Problèmes d'Analyse Tome 3 - Intégration, Exercices corrigés,Wieslawa J.Kaczor, Maria T.Nowak -275 Exercices et Problèmes Résolus de Mathématiques Supérieures, A.Abouhazim, A.Abkari, -S.R.Nsiri, M.El Mountassir<|endoftext|> -TITLE: Why does Maclaurin get his own polynomial? -QUESTION [8 upvotes]: Why is a Taylor polynomial centered around $0$ called a Maclaurin polynomial? It's only a special case of the Taylor polynomial, and it is calculated the exact same way as a Taylor polynomial centered at any number. It doesn't seem to carry the same weight as other named concepts such as Euler's number, which has special properties when you differentiate, integrate, etc. - -REPLY [15 votes]: Stigler's Law: No scientific discovery is named after its original discoverer (this was discovered by Merton). - -REPLY [8 votes]: It's called a Maclaurin series because Colin Maclaurin made extensive use of them to make advancements in the field of geometry. He also covered this case of Taylor series extensively in his treatise of fluxions. -If you want a really unfair example, you should see l'Hôpital's rule. This rule was discovered by Johann Bernoulli but it is named l'Hôpital's rule because a guy called Guillaume de l'Hôpital published it in his book on differential calculus.<|endoftext|> -TITLE: Can someone resolve my confusion about uniqueness of diagonalization? -QUESTION [5 upvotes]: I am a bit confused about diagonalization. I have $A$ which I know is diagonalizable. I want to find $P$ such that $A = P \Sigma P^{-1}$ where $\Sigma$ is diagonal. Under what circumstances is $P$ unique, if ever? If it is not unique, is it at least unique up to some operation? - -REPLY [4 votes]: For any diagonalizible matrix $n\times n$, for $n\geq 2$, $P$ is not unique. If the eigenvalues of $A$ are not all equal, then $\Sigma$ is not unique as well. If you change the columns of $P$ that will correspond to changing the appropriate columns in $\Sigma$. Even if, suppose, you fix $\Sigma$, even then you can change the columns in $P$ that correspond to eigenvectors of the same eigenvalue, or you can multiply them by scalar or make linear combinations. In short, in this case (when you fix $\Sigma$), $P$ will be unique up to elementary matrix column operations, but only between columns that correspond to the same eigenvalue. - -REPLY [4 votes]: In general, $P$ won't be unique. You can always: - -Change the order of different eigenvalues in $\Sigma$; that is, the values along the main diagonal. This will produce changes in the order of the corresponding eigenvectors; that is, the columns of $P$. -Even keeping $\Sigma$ untouched, if you have eigenvalues with multiplicity greater than 1, this will produce subspaces of eigenvectors with dimension greater than one. Any basis of which can be used for the columns of $P$. - -For instance, take $A$ to be the diagonal matrix -$$ -A = -\begin{pmatrix} -1 & 0 & 0 \\ -0 & 1 & 0 \\ -0 & 0 & 2 -\end{pmatrix} -$$ -Then you can obviously take $\Sigma = A$ and $P$ to be the unit matrix, but any matrix of the form -$$ -P = -\begin{pmatrix} -a & b & 0 \\ -c & d & 0 \\ -0 & 0 & e -\end{pmatrix} -$$ -will do (as long as $ad - bc \neq 0$ and $e \neq 0$). -Or, for the same $A$, you could also take -$$ -\Sigma = -\begin{pmatrix} -2 & 0 & 0 \\ -0 & 1 & 0 \\ -0 & 0 & 1 -\end{pmatrix} -$$ -and any -$$ -P = -\begin{pmatrix} -e & 0 & 0 \\ -0 & a & b \\ -0 & c & d -\end{pmatrix} -$$ -with the same restrictions as before is ok.<|endoftext|> -TITLE: relation between $W^{1,\infty}$ and $C^{0,1}$ -QUESTION [6 upvotes]: I know that $f \in C^{0,1}_{loc}(U)\Leftrightarrow f \in W^{1,\infty}_{loc}(U)$ and I have a reference for this. I would like a reference or a explanation for $C^{0,1} = W^{1,\infty}$ on domain convex. - -REPLY [7 votes]: Suppose $f\in C^{0,1}(U)$. Then $f$ is Lipschitz on every segment parallel to coordinates axis (and on other segments, too). Hence, it is absolutely continuous on every segment, with bounded derivative. This qualifies it as a member of $W^{1,\infty}(U)$. -Conversely, suppose $W^{1,\infty}(U)$. This means that $f$ is absolutely continuous on almost every coordinate-aligned segment, with bounded derivative. That is to say, $f$ is Lipschitz on such segments, with a uniform bound on Lipschitz constant. Let $E$ be the union of these "good" segments. Because $U$ is convex, any two points $(x_1,\dots,x_n)$ and $(y_1,\dots,y_n)$ in $E$ can be connected by a polygonal line contained within $E$ with total length at most -$$2\sum |x_i-y_i|\le 2\sqrt{n} \|x-y\|$$ -Therefore, $f$ is Lipschitz on $E$. Since $U\setminus E$ is a null set, we can redefine $f$ there to make it Lipschitz on $U$ (extending $f$ to $U$ by continuity). -Reference: Theorem 4.1 in Lectures on Lipschitz Analysis by Heinonen. -See also Sobolev embedding for $W^{1,\infty}$？<|endoftext|> -TITLE: What makes elementary row operations "special"? -QUESTION [16 upvotes]: This is probably a stupid question, but what makes the three magical elementary row operations, as taught in elementary linear algebra courses, special? In other words, in what way are they "natural" (as opposed to "arbitrary")? -It seems that they're always presented in a somewhat haphazard manner ("these are the three legendary elementary row operations, don't ask why, they just are"). From what I understand, they satisfy some nice properties, such as the inverse of each being an operation of the same type, etc. But is there something that characterizes them, i.e. is there some definition of what constitutes "elementary" that's only satisfied by the three types of elementary matrices, and no other matrix? - -REPLY [6 votes]: By the way, it may be amusing to note that you don't really need the row-interchange operation. Thus, instead of having a special operation to interchange rows $i$ and $j$, you could: - -add row $j$ to row $i$ -subtract row $i$ from row $j$ -multiply row $j$ by $-1$ -subtract row $j$ from row $i$ - -and it would have the same effect. In terms of the elementary matrices, -$$ \pmatrix{1 & -1\cr 0 & 1\cr} \pmatrix{1 & 0\cr 0 & -1\cr} \pmatrix{1 & 0\cr -1 & 1\cr} -\pmatrix{1 & 1\cr 0 & 1\cr} = \pmatrix{0 & 1\cr 1 & 0\cr}$$ -Similarly, you don't need to be able to add an arbitrary nonzero multiple of row $j$ to row $i$, as long as you can add row $j$ to row $i$: to add $t$ times row $j$ to row $i$, you first multiply row $j$ by $t$, then add row $j$ to row $i$, then multiply row $j$ by $1/t$. -So you could get by with just - -add a row to another row -multiply a row by a nonzero scalar - -Not that I'd want to do this in practice...<|endoftext|> -TITLE: How to show differentiability implies continuity for functions between Euclidean spaces -QUESTION [5 upvotes]: A function $f: \mathbb{R^n} \to \mathbb{R^m}$ is differentiable at $a$ if there exists a linear map $ \lambda: \mathbb{R^n} \to \mathbb{R^m}$ such that -$$\lim_{h \to 0} \frac{\|f(a+h) - f(a) - \lambda(h)\|}{\|h\|} = 0$$ -So clearly, if $f$ is differentiable at $a$ then $\lim_{h \to 0} f(a+h) - f(a) - \lambda(h) = 0$, but where do you proceed from here? - -REPLY [11 votes]: More explicitly: -The limit shows that for any $\epsilon>0$, there exists a $\delta>0$ so that if $\|h\| < \delta$, then $\|f(a+h) - f(a) - \lambda(h)\| \leq \epsilon \|h\|$. $\lambda$ is continuous, hence bounded, so we have $\|\lambda(h)\| \leq K \|h\|$, for some $K$. -Then we have $\|f(a+h) - f(a)\| \leq \|f(a+h) - f(a) - \lambda(h)\| + \|\lambda(h)\|\leq (\epsilon + K ) \|h\|$. -Now let $\eta >0$, then if $\|h\| < \min(\delta, \frac{\eta}{\epsilon+K})$, we have $\|f(a+h) - f(a)\| < \eta$, which shows that $f$ is continuous at $a$.<|endoftext|> -TITLE: When antidifferentiating, are we impliclty restricting to an interval? -QUESTION [5 upvotes]: I was asked this question by a student I am tutoring and I was left a little puzzled because his textbook only defines antiderivatives on intervals (which leads me to believe its author would answer the question in the title in the affirmative). -To my understanding, finding an antiderivative of $f$ means finding a function $F$ with $F' = f$. It does not matter if the domain of $f%$ is not connected. For example, $\int \frac{dx}{x} $ denotes an antiderivative on all of $\mathbb{R} - 0$, and not just on some arbitrary interval $I \subseteq \mathbb{R} - 0 $. Am I mistaking? - -REPLY [7 votes]: You can use whatever conventions you want; the author is free to choose hers and you are free to choose yours. -One issue with defining antiderivatives on (nice) subsets of $\mathbb{R}$ which are not connected is that they are no longer just only unique up to constants; they are only unique up to locally constant functions. For example, on $\mathbb{R} \setminus \{ 0 \}$ any function which has one constant value when $x$ is negative and another when $x$ is positive has zero derivative. This is the kind of subtlety that I suspect it would be a good idea to avoid in a calculus course. -Another issue is that you'd like to write antiderivatives down using definite integrals, but for example if a function $f$ is not defined in the interval $(-1, 1)$ then it is unclear what an integral such as $\int_{-2}^x f(x) \, dx$ would mean for $x > 1$...<|endoftext|> -TITLE: Why is associativity required for groups? -QUESTION [18 upvotes]: Why is associativity required for groups? -I'm doing a linear algebra paper and we're focusing on groups at the moment, specifically proving whether something is or is not a group. There are four axioms: - -The set is closed under the operation. -The operation is associative. -The exists and identity in the group. -Each element in the group has an inverse which is also in the group. - -Why does the operation need to be associative? -Thanks - -REPLY [50 votes]: It is not that associativity is required for groups... That is quite backwards: the truth is actually that groups are associative. -Your question seems to come from the idea that people decided how to define groups and then began to study them and find them interesting. In reality, it happened the other way around: people had studied groups way before actually someone gave a definition. When a definition was agreed upon, people looked at the groups they had at hand and saw that they happened to be associative (and that that was a useful piece of information about them when working with them) so that got included in the definition. -If I may say so, it is this which is important to understand. The way we teach abstract algebra nowdays somewhat obscures this fact, but this is how essentially everything comes to be.<|endoftext|> -TITLE: Groups such that every finitely generated subgroup is isomorphic to the integers -QUESTION [6 upvotes]: What are examples of groups such that every finitely generated subgroup is isomorphic to $\mathbb{Z}$? - -REPLY [12 votes]: Suppose any nontrivial finitely generated subgroup of $G$ is isomorphic to $\Bbb{Z}$, and that $G$ itself is nontrivial. Choose some nonzero $e \in G$. For any $g \in G$, the subgroup $\langle e, g\rangle$ is isomorphic to $\Bbb{Z}$, so in particular $ne=mg$ for some integers $m,n$. Define a map $f:G \to \Bbb{Q}$ by $g \mapsto n/m$. Then $f$ is a well-defined group homomorphism; since $G$ is torsion-free and $e$ nonzero, it is straightforward to check that $f$ is injective. So $G$ is isomorphic to a nontrivial subgroup of $\Bbb{Q}$, which are classified, e.g., in this paper.<|endoftext|> -TITLE: Braid groups and the fundamental group of the configuration space of $n$ points -QUESTION [12 upvotes]: I am giving a lecture on Braid Groups this month at a seminar and I am confused about how to understand the fundamental group of the configuration space of $n$ points, so I will define some terminology which I will be referring to and then ask a few questions. -The configuration space of $n$ points on the complex plane $\mathbb{C}$ is: -$$C_{o,\hat{n}}(\mathbb{C})=\{(z_{1},...,z_{n}) \in \mathbb{C} \times \cdot \cdot \cdot \times \mathbb{C} \text{ | } z_{i} \neq z_{j} \text{ if } i \neq j\}$$ -A point on $C_{o,\hat{n}}$ is denoted by a vector $\vec{z} = (z_{1},...,z_{n})$. The group action $S_{n} \times C_{o,\hat{n}}(\mathbb{C}) \rightarrow C_{o,\hat{n}}(\mathbb{C})$ has the property that $\forall \vec{z} \in C_{o,\hat{n}}(\mathbb{C})$, $\sigma\vec{z}=\vec{z}$ implies $\sigma = e$, where $\sigma \in S_{n}$ and $e$ is the identity permutation. Thus, the symmetric group acts freely on $C_{o,\hat{n}}$, permuting the coordinates in each $\vec{z} \in C_{o,\hat{n}}$. According to the notes I am reading, the orbit space $S_{n}(\vec{z})=\{\sigma\vec{z} \text{ | } \sigma \in S_{n}\}$ with $\vec{z} \in C_{o,\hat{n}}(\mathbb{C})$ is equivalent to the configuration space modulo the permutation group. That is, -$$C_{o,n} = S_{n}(\vec{z}) = C_{o,\hat{n}}(\mathbb{C})\Big/\Sigma_{n}$$ -is the orbit space of the group action and we let $\tau: C_{o,\hat{n}} \rightarrow C_{o,n}$ be the orbit space projection. Then, letting $\vec{p}=(p_{1},...,p_{n})$ be a fixed base point, we define the pure braid group $\textbf{P}_{n}$ on $n$ strands and the braid group $\textbf{B}_{n}$ on $n$ strands to be the following fundamental groups, respectively. -$$\textbf{P}_{n} = \pi_{1}(C_{o,\hat{n}},\vec{p})$$ -$$\textbf{B}_{n} = \pi_{1}(C_{o,n},\tau(\vec{p}))$$ -EDIT: I have been able to answer some of my own questions recently by further thinking and research but I have one remaining question. - -We place a base point $\vec{p} = (p_{1},...,p_{n})$ in our configuration space $C_{o,\hat{n}}$ and then we want to consider loops in configuration space. I can visualize how if we place our base point as a configuration of collinear points that we obtain a braid on $n$ strands, but I am not sure why there is a non-trivial loop. My understanding of homotopy is that if there is no "hole" in the space, then we can continuously retract our loop in $C_{o,\hat{n}}$ back to our base point $\vec{p}$. Why can we not do this in this case? - -REPLY [12 votes]: Sorry, but my first answer was completely and utterly wrong. Since this question is one of the top search results for the query "braid group fundamental group configuration space" I think it's high time I've updated with a correct explanation! :-) - - -I am not sure why there is a non-trivial loop. My understanding of homotopy is that if there is no "hole" in the space, then we can continuously retract our loop back to our base point. Why can we not do this in this case? - -Short answer. You're thinking of $B_n(\Bbb R)$ when you should be thinking of $B_n(\Bbb C)$. -Long answer. Let $X$ be a "nice" topological space (say, a manifold). Define $F_n(X)$ to be the subspace of $X^n$ comprised of tuples with distinct coordinates. The symmetric group $S_n$ acts on it freely, and we can form the $n$-configuration space as the quotient $SF_n(X):=F_n(X)/S_n$. Then we define the braid group as $B_n(X)=\pi_1(SF_n(X))$. (Of course, $SF_n(X)$ should be path-connected...) -If you take $X=\Bbb R$ then the connected components of $F_n(X)$ are blocks for the action of $S_n$. Given any two tuples $(x_1,\cdots,x_n)$ and $(y_1,\cdots,y_n)$ with $x_12$? In configuration space (which has $2n$ real dimensions, so is hard to visualize) a single point, a "configuration," represents $n$ distinct points in a plane (which is easy to visualize). And a path in configuration space represents each of the $n$ points in the plane having a path in and out of it. -Thus, imagine a continuum (indexed by $[0,1]$) of copies of $\Bbb C$ (resting flat) piled on top of each other. If one lets the altitude represent time, then the paths traced out between the points represent strings, and if one looks at this picture from the side one sees braid diagrams! Example: -$\hskip 1.3in$ -Since we can choose our basepoint for $\pi_1$ to be anything, without loss of generality we may assume it is $\{1,2,\cdots,n\}\subset\Bbb C$ for the purpose of visualization. Tuples in $\Bbb C^n$ with nondistinct coordinates represent two strings intersecting at the same point, which is why wemust delete this subspace from $\Bbb C^n$: to prevent collisions. -A path in $\Bbb C^n$ ending where it started means each colored string above would have to go back to its original point, and this defines a pure braid. If we quotient by the action of $S_n$, we essentially allow the path in configuration space to go to any of the permuted configurations, which means the strings in the braid diagram can connect different dots. - -There is another way to visualize braids that is also very interesting: mapping classes of the closed unit disk with $n$ points inside deleted. I recently asked a question about generalizing this idea to generalized braid groups. Mappings can warp the unit disk like a sheet of rubber, but the rubber is attached to the boundary (the unit circle) which must remain fixed pointwise. When you delete $n$ points, that essentially means your mappings of the disk must restrict to a permutation of those $n$ points. -To visualize what such mappings look like, for $B_2$ imagine putting two fingers on the two points in the disk, then using your two fingers to warp the rubber disk by turning it one way or the other with your fingers. Remember the border of the rubber sheet is stuck in place, so you'll be twisting the inside of the rubber relative to the outside rather than lamely rotating it. In general for $n$ points, you can do the same thing by twisting the rubber around two points with two fingers. -There two ways to twist two marked points around each other with your fingers (clockwise or counterclockwise), corresponding to which string goes over/under which in the braid diagram. The paths that the marked points take throughout the twisting process essentially trace out a braid diagram. Intuitively, we should be able to "lift" any braid diagram into a composition of such twistings of the unit disk. More detail is given in the link.<|endoftext|> -TITLE: Reference for upper and lower bounds on $e^x$ -QUESTION [8 upvotes]: I'm looking for a reference for deriving the following commonly used upper and lower bounds for $e^x$: -$$1 - x \le e^{-x}$$ -and, assuming $x \le 1/2$, -$$1 - x \ge e^{-2x}. -$$ - -REPLY [8 votes]: First Inequality: Let $f(x)=e^{-x}-(1-x)$. Then $f'(x)=-e^{-x}+1$. This is $0$ at $x=0$, negative if $x \lt 0$, and positive for $x \gt 0$. So $f(x)$ reaches an absolute minimum at $x=0$. The minimum value is $0$, so $f(x)\ge 0$ for all $x$. -Second Inequality: Let $g(x)=(1-x)-e^{-2x}$. Then $g'(x)=-1+2e^{-2x}$. This is $0$ for $x=(\ln 2)/2$. -We find that $g'(x)$ is positive for $x\lt (\ln 2)/2$, and negative for $x \gt (\ln 2)/2$. The function $g(x)$ is increasing up to $x=(\ln 2)/2$, then decreasing. -We have $g(0)=0$. Since $g(x)$ is increasing until $(\ln 2)/2$, it follows that $g(x)$ is negative when $x$ is negative, so our inequality fails if $x \lt 0$. -For positive $x$, $g(x)$ increases up to $x=(\ln 2)/2$, and then decreases. After a while it will become negative. Let us check whether $g(x)$ is still positive at $x=\frac{1}{2}$. Calculate. We get $g(1/2)\approx 0.132$. So $g(x) \ge 0$ at least in the interval $[0,1/2]$. In fact $g(x) \ge 0$ from $0$ to a bit beyond $0.79$. -Remark: There will be few standard North American calculus books that don't do the first inequality. Usually it is done in the equivalent version $1+x \le e^x$. It is an early application of the connection between increasing functions and positive derivatives. The inequality (among many others) is mentioned here. -The second inequality uses the same method, but the details involve more work.<|endoftext|> -TITLE: Show that the curve $y^2 = x^3 + 2x^2$ has a double point, and find all rational points -QUESTION [5 upvotes]: Show that the curve $y^2 = x^3 + 2x^2$ has a double point. Find all rational points on this curve. -By implicit differentiation of $x$, $-3x^2 - 4x$ vanishes iff $x = -4/3$ and $0$. -By implicit differentiation of $y$, $2y$ vanishes iff $y = 0$. -Taking the second derivative, I got $-6x-4$ and then using the point on the curve $(0,0)$ I got $-4$. Is this my double point? -Thanks for any help! - -REPLY [2 votes]: You may find the following fact useful: suppose that a plane curve is described by a polynomial equation $F(x,y)=0$. Write -$$ -F(x,y)=\sum_{d\geq0}F_d(x,y)\qquad\qquad(*) -$$ -where in $F_d(x,y)$ you collect all the monomials in $F$ of degree exactly $d$. Then: - -$(0,0)$ belongs to the curve if and only if $F_0\equiv0$ (by which, I mean that is $0$ identically). -$(0,0)$ is a singular point of the curve if and only if $F_0\equiv0$ and $F_1\equiv0$. - -Moreover if $d_0\geq2$ is the smallest degree appearing in $(*)$, then $F_d=0$ is the tangent cone at $(0,0)$. Note that this answers immediately and trivially to your first question. -Finally, the method applies to just any point $(x_0,y_0)$ after a simple change of variables $X=x-x_0$, $Y=y-y_0$.<|endoftext|> -TITLE: Group of groups -QUESTION [8 upvotes]: The product $\times$ of two groups is associative and commutative and there's a neutral element $\{1\}$. Let's say I create "virtual groups" which are inverses with respect to $\times$ (like getting $\mathbb{Z}$ from $\mathbb{N}$). Then I have a group $G$ whose elements are all groups. -This isn't allowed, though, is it? Since a set can't be a member of itself. - -REPLY [2 votes]: In addition to the problems mentioned by Brian Scott, there is another problem. -You seem to be trying to mimic the construction of the enveloping group of a semigroup (this is the image of the adjoint functor to the forgetful functor from $\mathsf{Group}$ to $\mathsf{Semigroup}$ or to $\mathsf{Monoid}$); in the commutative case, this is often called the Grothendieck group. Generally, you do not have an embedding of the original semigroup into the enveloping; you need the semigroup to be cancellative for that to hold. -But in this setting, there are groups $A$, $B$, and $C$, pairwise nonisomorphic, such that $A\times B\cong A\times C$. That would mean that in the enveloping group, we would need to identify $A$ with the trivial group, or $B$ with $C$. You are going to have a lot of collapse in this resulting object. For example, if your class of groups includes the infinite direct product of copies of $\mathbb{Z}$, and this group is not identified with the trivial group, then every finite rank free abelian group will be identified with each other (a less than desirable outcome). -You could get away with doing this for, say, finite groups. You can even do it within ZFC, by restricting the groups to having underlying set contained in a particular infinite set. Then, under the equivalence relation of "isomorphism" you have a cancellative commutative monoid under cartesian product, so you can construct the Grothendieck group by adjoining "formal inverses". I don't think you get anything particularly interesting, though: the monoid is free abelian on the indecomposable groups, and so the Grothendieck group is just free abelian on the indecomposable groups.<|endoftext|> -TITLE: Is there a monotonic function discontinuous over some dense set? -QUESTION [32 upvotes]: Can we construct a monotonic function $f : \mathbb{R} \to \mathbb{R}$ such that there is a dense set in some interval $(a,b)$ for which $f$ is discontinuous at all points in the dense set? What about a strictly monotonic function? - -My intuition tells me that such a function is impossible. -Here is a rough sketch of an attempt at proving that such a function does not exist: we could suppose a function satisfies these conditions. Take an $\epsilon > 0$ and two points $x,y$ in this dense set such that $x m$ then $$B(x_m) = \{n: \psi(n) < x_m < x_k < x\} \subset \{n: \psi(n) < x_k < x\} = B(x_k).$$ By the -finiteness of $\nu$ we have that by upward measure continuity -$$\lim_{k \to \infty} H(x_k) = \lim_{k \to \infty} \nu(B(x_k)) = \nu\left(\bigcup_{k=1}^\infty B(x_k) \right) = \nu(\left \{n: \psi(n) < x\}\right ) = H(x).$$ -Note that if $ n \in \bigcup B(x_k)$ there is an $K$ so that $\psi(n) < x_k < x $ so any $n$ with $\psi(n) < x$ is in $\bigcup B(x_k).$ -Next we claim that $H$ is only right continuous only when $x$ is irrational. Take a sequence of $x_k \to x$ from the right and rearrange the sequence to be strict montonic. It follows that if $k > m$ then $$B(x_m) = \{n: \psi(n) < x_m\} \supset \{n: \psi(n) < x_k < x_m\} = B(x_k).$$ By finiteness of $\nu$ and downard measure continuity -$$\lim_{k \to \infty} H(x_k) = \lim_{k \to \infty} \nu(B(x_k)) = \nu\left(\bigcap_{k=1}^\infty B(x_k) \right) = \nu(\left \{n: \psi(n) < x_k\ \forall k\}\right ).$$ -If $x$ is irrational then $m \in \{n: \psi(n) < x_k\ \forall k\}$ implies that $\psi(m) < x$ and if $\psi(m) < x$ then $\psi(m) < x_k$ for all $k$ so $\{n: \psi(n) < x_k\ \forall k\} = B(x)$ and $H(x_k) \to H(x)$ from the right. If $x$ is rational then $x= \psi(q)$ for some -$q \in \mathbb{N}.$ Thus $x < x_k \forall k$ implies that $\{n: \psi(n) < x_k\ \forall k\} = B(x) + \{q\} = D$. It follows that $\nu(D) = \nu(B(x)) + 2^{-q} > H(x)$. So $H(x_k) \to H(x) + 2^{-q} \neq H(x)$ from the right, and so $H$ is not right continuous at the rationals. -We have thus shown that for any $x \in [0,1] \setminus \mathbb{Q}$, any sequence $x_k \to x$ has the property $\lim H(x_k) = x$ from the left and the right, and if $x \in [0,1] \cap \mathbb{Q}$ then if $x_k \to x$, $\lim H(x_k)$ does not exist. Therefore $H$ is continuous at every irrational and discontinuous at every rational. -You can then repeat this construction along the whole real line by adding the nearest integer each time.<|endoftext|> -TITLE: Calculating equidistant points around an ellipse arc -QUESTION [17 upvotes]: As an extension to this question on equiangular fisheye distortion, how can I calculate equidistant points around an ellipse (or 1/4 segment of) given it's aspect ratio? -When it's circular, I can use a simple angle increment around the central point, but as the aspect ratio gets smaller, the length of the arc gets shorter and the angles are no longer equidistant. - -REPLY [17 votes]: I assume that by “equidistant” you mean that the arc length along the ellipse between any two adjacent points is the same, not the Euclidean distance between the points in the plane. -Complete circumference -Computing the circumference of an ellipse can be formulated using the complete elliptic integral of the second kind. More precisely, the circumference is -$$c=4a\,E(m)\qquad m=1-b^2/a^2$$ -at least if you formulate the elliptic integral in terms of the parameter $m=1-b^2/a^2$. If you instead define it in terms of the elliptic modulus $k=\sqrt m$, then you plug that into the formula. So now you know the total circumference, and can divide that by some integer $n$ depending on how many points you want along the perimeter. Then integral multiples of that length will yield equidistant points. -Point at given arc length -Next you need to turn arc length back into position on the ellipse. This means you'll have to compute the inverse of the incomplete elliptic integral of the second kind. So for the $k$-th point (for $0\le k -TITLE: A question about a proof of a weak form of Hilbert's Nullstellensatz -QUESTION [8 upvotes]: I'm trying to prove the following (corollary 5.24 page 67 in Atiyah-Macdonald): -Let $k$ be a field and let $B$ be a field that is a finitely generated $k$-algebra, i.e. there is a ring homomorphism $f: k \to B$ and $B = k[b_1, \dots , b_n]$ for $b_i \in B$. Then $B$ is an algebraic (and hence, in this case, finite) extension of $k$. -There is a proof in Atiyah-Macdonald but it's more like a hint and I'm not sure I understand the details. Can you tell me if this detailed version of the proof is correct? Here goes (thanks!): -We need to show that $b_i$ are algebraic over $k$. Since $k$ is a field we know that $f$ is injective so we may view $k$ as a subfield of $B$ ($f$ is our embedding). Then $k \subset B$ are integral domains and $B$ is finitely generated so we are in the position to apply proposition 5.23 which tells us the following: -If $b$ is a non-zero element in $B$ then we can find a non-zero element $c$ in $k$ such that if $f: k \to \Omega$ is a homomorphism into an algebraically closed field $\Omega$ such that $f(c) \neq 0$ then there exists an extension $g: B \to \Omega$ of $f$ such that $g(b) \neq 0$. -We observe that $1$ is a non-zero element of $B$. The inclusion $i: k \hookrightarrow$ of $k$ into its algebraic closure $\overline{k}$ is a ring homomorphism such that $i(1) \neq 0$. By the previously stated proposition we hence can find a ring homomorphism $g: B \to \overline{k}$ such that $g(1) \neq 0$. Although the fact that $g(1) \neq 0$ doesn't interest us. But since $g$ is a ring homomorphism defined on a field we know that it's injective hence we can view $B$ as a subfield of $\overline{k}$. And now we are done since we have $k \subset B \subset \overline{k}$, hence $B$ is contained in the algebraic closure of $k$ and hence every element of $B$ is algebraic over $k$. In particular, $b_i$. -I wonder why it's called "Nullstellensatz". It doesn't seem to have anything to do with roots of polynomials. -Here is an image of proposition 5.23: - -REPLY [6 votes]: First, instead of "a weak form of Hilbert's Nullstellensatz" I prefer the term Zariski's Lemma for the result you're asking about, for instance because (i) it's more informative (it was indeed introduced by Zariski in a 1947 paper) and (ii) it's not so "weak": it is a statement valid over an arbitrary field from which Hilbert's Nullstellensatz -- which holds only over algebraically closed fields -- can be rather easily deduced via the Rabinowitsch Trick. -If I may, I recommend my own commutative algebra notes for a treatment of Zariski's Lemma and the Nullstellensatz. I first introduce Zariski's Lemma in $\S 11$ and prove it using the Artin-Tate Lemma, which I have come to regard as the simplest, most transparent proof. But then throughout the notes I come back to give several other proofs of ZL: -$\bullet$ In $\S 12.2$ using Hilbert Rings. -$\bullet$ In $\S 14.5.3$ using Noether's Normalization Lemma (left as an exercise). -$\bullet$ In $\S 17.4$ using valuation rings. -The last of these is the one you're reading about now from Atiyah-Macdonald. In my opinion it is also the most complicated: you might have better luck with one of the other three...<|endoftext|> -TITLE: Transition Kernel of a Markov Chain -QUESTION [8 upvotes]: Supposing $X_t$ is a Markov Process, can the transition kernel be defined by -$$K_t(x,A):= P(X_{t+1} \in A | X_t = x)?$$ Assume that $X_t : \Omega \to \mathbb{R}^n$. -The issue is that under the normal definition of conditional probability, r.h.s is defined as $$P(X_{t+1} \in A | X_t = x) =\frac{P( (X_{t+1} \in A) \cap (X_t = x))}{P(X_t = x)}$$ and the denominator is zero for most random variables. Even -if this is assumed to be $E[I_A(X_{t+1}) | \sigma(X_t = x)]$, the conditional -expectation can take arbitrary values on the set $\{X_t =x\}$ if $P(X_t = x) =0$. -Another definition I could gather from the web is that $K_t(x,A)$ is called a -transition kernel if $$K_t(X_t(\omega),A) := E[I_A(X_{t+1})|X_t](\omega)~\forall -\omega \in \Omega.$$ -Also, $K_t(x,\cdot)$ should be a probability measure so that the conditional -expectations should be regular (if I am not wrong). -The book (page 18) I am reading uses the first definition given above. -Thanks for the help. - -REPLY [6 votes]: The transition kernel $K_t$ is defined by some measurability conditions and by the fact that, for every measurable Borel set $A$ and every (bounded) measurable function $u$, -$$ -\mathrm E(u(X_t):X_{t+1}\in A)=\mathrm E(u(X_t)K_t(X_t,A)). -$$ -Hence, each $K_t(\cdot,A)$ is defined only up to sets of measure zero for the distribution of $X_t$, in the following sense: if $K_t$ is a transition kernel for $X_t$ and if, for every measurable Borel set $A$, $X_t$ is almost surely in $C_A$, where -$$ -C_A=\{x\in\mathbb R^n\,\mid\,K_t(x,A)=\tilde K_t(x,A)\}, -$$ -then $\tilde K_t$ is also a transition kernel for $X_t$.<|endoftext|> -TITLE: Integral-Summation inequality. -QUESTION [11 upvotes]: The following question was in an entrance exam: - -Show that, if $n\gt0$, then: $$\int_{{\rm e}^{1/n}}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}=\frac{2}{n^2{\rm e}}$$ You are allowed to assume $\lim_{x\to\infty}{\frac{\ln{x}}{x}}=0$. Hence explain why, if $1\lt a\lt b$, then: $$\int_{b}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}\lt\int_{a}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}$$ -Deduce that: $$\sum_{n=1}^{N}{\frac{1}{n^{2}}}<\frac{\rm e}{2}\int_{{\rm e}^{1/N}}^{\infty}{\left(\frac{1-x^{-N}}{x^{2}-x}\right)\ln{x}\:dx},$$ Where $N\in\mathbb{N}:N\gt1$. - -The first part I believe I integrate by parts, such that: -$$\int{\frac{\ln{x}}{x^{n+1}}\:dx}=\frac{1}{n}\int{x^{-n-1}\:dx}-\frac{x^{-n}\ln{x}}{n}$$ -Clearly $\int{x^{-n-1}\:dx}=-\frac{x^{-n}}{n}$, so we have: -$$\int_{{\rm e}^{1/n}}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}=\left.\frac{x^{-n}(n\ln{x}+1)}{n^{2}}\right|_{{\rm e}^{1/n}}^{\infty}=\frac{{\rm e}^{-\frac{n}{n}}(\frac{n}{n}+1)}{n^{2}}-0=\frac{2}{n^{2}{\rm e}}$$ -As required. To prove the next inequality, all that is required is to demonstrate that $\left.\frac{\ln{x}}{x^{n+1}}\right|_{x=1}\geq0$, $\lim_{x\to\infty}{\frac{\ln{x}}{x^{n+1}}}\geq0$, and $\left(\frac{\ln{x}}{x^{n+1}}\right)'\neq0$, $\forall x\in[1,\infty)$ and $\forall n\gt0$. -The first inequality is verified simply by observing that $\ln{1}=0$, therefore $\frac{\ln{1}}{1^{n+1}}=0$, $\forall n$.The second inequality can be written as: -$$\lim_{x\to\infty}{\left(\frac{1}{x^{n}}\frac{\ln{x}}{x}\right)},$$ -And as we know $\lim_{x\to\infty}{\frac{\ln{x}}{x}}=0$, and $\lim_{x\to\infty}{\frac{1}{x^{n}}}=0$, the second inequality must also be true. To verify the second inequality we simply differentiate using the quotient rule and look for critical points: -$$\frac{d}{dx}{\left(\frac{\ln{x}}{x^{n+1}}\right)}=\frac{x^{n}-(n+1)x^{n}\ln{x}}{x^{2n+2}}=\frac{1-(n+1)\ln{x}}{x^{n+2}}$$ -As $\ln{x}$ is a monotonically increasing function, and at $x=1$, $\ln{x}=0$, we can show that $\forall n\gt 0$, and $x\gt1$, $\left(\frac{\ln{x}}{x^{n+1}}\right)\leq 0$, therefore, the function is positive for all values of $x\gt 0$, which means by the fundamental theorem of calculus that if $1\lt a\lt b$, then $\forall n\gt0$: -$$\int_{b}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}<\int_{a}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}$$ -As required. -However, I am stuck for the final part of the question. My first thought was that as $\frac{1}{n^{2}}$ is monotonically decreasing, $\forall n\gt0$; any integral performed over the region $(0,\infty)$ will have a positive error term, therefore, we can replace the summation with $\int_{1}^{N}{\frac{1}{n^2}\:dn}=\left.-\frac{1}{n}\right|_{1}^{N}=-\frac{1}{N}+1$. I am not sure how to perform the second integral, however. -Thanks in advance! - -REPLY [10 votes]: The equation $$\int_{{\rm e}^{1/n}}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}=\frac{2}{n^2{\rm e}}$$ can be rewritten as $$\frac{1}{n^2}=\frac{\rm e}2\int_{{\rm e}^{1/n}}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}$$ -Now, simply sum this for $n=1,2,\ldots,N$ and get: $$\begin{align}\sum_{n=1}^N\frac{1}{n^2}=&\sum_{n=1}^N\frac{\rm e}2\int_{{\rm e}^{1/n}}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}<\sum_{n=1}^N\frac{\rm e}2\int_{{\rm e}^{1/N}}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}=\\=&\frac{\rm e}2\int_{{\rm e}^{1/N}}^{\infty}\sum_{n=1}^N{\frac{\ln{x}}{x^{n+1}}\:dx}=\frac{\rm e}2\int_{{\rm e}^{1/N}}^{\infty}\ln{x}\sum_{n=1}^N{\frac{1}{x^{n+1}}\:dx}\end{align}$$ -(To get the inequality, we used that the integrand is positive on $(1,\infty)$, which shows that $\int_{{\rm e}^{1/n}}^{\infty}\leq\int_{{\rm e}^{1/N}}^{\infty}$ for $N\geq n$ and equality holds if and only if $N=n$.) The sum here is just the sum of a finite geometric progression. Thus, we have $$\sum_{n=1}^N{\frac{1}{x^{n+1}}}=\sum_{n=1}^N{(x^{-1})^{n+1}}=x^{-2}\sum_{n=1}^{N}{(x^{-1})^{n-1}}=x^{-2}(\frac{(x^{-1})^N-1}{x^{-1}-1})=\frac{x^{-N}-1}{x-x^2}$$ -This gives us $$\sum_{n=1}^{N}{\frac{1}{n^{2}}}<\frac{\rm e}{2}\int_{{\rm e}^{1/N}}^{\infty}{\left(\frac{1-x^{-N}}{x^{2}-x}\right)\ln{x}\:dx},$$ which is exactly what we wanted.<|endoftext|> -TITLE: How to prove floor identities? -QUESTION [6 upvotes]: I'm trying to prove rigorously the following: -$\lfloor x/a/b \rfloor$ = $\lfloor \lfloor x/a \rfloor /b \rfloor$ for $a,b>1$ -So far I haven't gotten far. It's enough to prove this instead -$\lfloor z/c \rfloor$ = $\lfloor \lfloor z \rfloor /c \rfloor$ for $c>1$ -since we can just put $z=\lfloor x/a \rfloor$ and $c=b$. - -REPLY [4 votes]: This is best done universally, i.e. using the universal definition of floor -$\begin{align} k\,&\le\, \lfloor x\rfloor \color{#c00}{\iff} k\,\le\, x,\quad {\rm for\ all}\,\ k\in\Bbb Z\\[1em] -{\bf Lemma}\qquad\quad\color{#0a0}{\lfloor r/n\rfloor}\,& =\, \lfloor{\lfloor r\rfloor}/n\rfloor\ \ {\rm -for}\ \ 00\\[.4em] -\color{#c00}\iff\ \ nk\ &\le\,\ \ \ r\qquad\ \ {\rm by}\,\ n\in\Bbb Z\\[.4em] -\iff\ \ \ \ k\ &\le\:\ \ \ r/n\quad\ {\rm by}\,\ n>0\\[.4em] -\color{#c00}\iff\ \ \ \ k\ &\le\ \ \color{#0a0}{\lfloor r/n\rfloor}\quad {\small\bf QED} -\end{align}$ -Yours is special case $\ r = x/a,\,\ n = b.$ -See the links in comments below for more general (category-theoretic) viewpoints.<|endoftext|> -TITLE: A counterexample in topology -QUESTION [19 upvotes]: Semi-local simple connectedness is a property that arises in Algebraic Topology in the study of covering spaces, namely, it is a necessary condition for the existence of the universal cover of a topological space X. It means that every point $x \in X$ has a neighborhood $N$ such that every loop in $N$ is nullhomotopic in $X$ (not necessarily through a homotopy of loops in $N$). The way I see it, the prefix "semi-" is refers more to "simply connected" than to "locally", since if such $N$ exists, all other neighborhoods of $x$ inside $N$ also have the property, so each point has a fundamental system of (open) neighborhoods for which the property holds (EDIT see Qiaochu's comment). Instead it isn't true that a semi-local simply connected space is locally simply connected (i.e each point has a fundamental system of open, simply connected neighborhoods): take the space $$ X = \frac{H \times I}{ \sim } $$ where $H$ is the "Hawaiian earring", or infinite earring (which is an example of non semi-locally simply connected space) and $\sim$ is the equivalence that identifies $H \times \{0\}$ to one point. -However I was interested in finding another type of counterexample. Consider the topological property (call it $*$) consisting in the existence, for all $x \in X$, of a simply connected, not necessarily open, neighborhood of $x$. We have $$ \text{semi-local simple connectedness} \Leftarrow * \Leftarrow \text{local simple connectedness} \wedge \text{simple connectedness} $$ I am wondering if $$ \text{semi-local simple connectedness} \Rightarrow * $$ holds. Intuitively it shouldn't, but I'm having trouble finding a counterexample. For example, the space $X$ described above won't work because it is simply connected (even contractible). It seems to me that, if a counterexample does exist, it must have local pathologies (to ensure that a certain point $x$ doesn't have a simply connected neighborhood), and globally the space should allow loops close to $x$ to be nullhomotopic, but in such a way that every neighborhood $N$ of $x$ contains a small enough loop that will not contract in $N$. EDIT Also, I am looking for a counterexample which is a locally path connected space (a previous answer showed a counterexample without this property.. but @answerer it was interesting anyway, you shouldn't have deleted it!) -But perhaps I'm wrong, and the two assertions are equivalent, or maybe I am missing something very simple. If anybody has any ideas please share, thank you! - -REPLY [7 votes]: Call the "interesting" point in the Hawaiian earring $q$ (the point where all the circles intersect). In your space $X$, glue the quotiented point $\{H \times \{0\}\}$ to the point $(q,1) \in H \times I$, essentially bending your cone around in a loop. Call the resulting quotient space $Y$, and the glued point $p$. Note that $p$ could be described either as $(q,1)$ or as $\{H \times \{0\}\}$. ($Y$ can also be described as the Mapping Torus of the map that takes the entire Hawaiian earring to the point $q$). -$Y$ is semi-locally simply connected but doesn't satisfy property $∗$. -To see $Y$ is semi-locally simply connected, note that any point has a neighborhood that is disjoint from a set of the form $H \times \{x\}$ for some $x \in I$. The inclusion of such a neighborhood into $Y$ is nullhomotopic (simply retract the portion contained in $H \times [0,x)$ to $p$, then follow the strong deformation retraction of $X$ to $\{H \times \{0\}\}$), which certainly implies that any loop contained in it is nullhomotopic in $Y$. -But $p$ itself has no simply-connected neighborhood. Any such neighborhood $N$ contains a loop $L \times \{1\}$ in $H \times \{1\}$, which could only be contracted in $N$ by including all of $L \times I$ in $N$. In particular, you'd have to include all of $\{q\} \times I$ in $N$, But $\{q\} \times I$ is itself a loop which isn't nullhomotopic in $Y$, and so also isn't nullhomotopic in $N$. Thus, $N$ cannot be simply connected.<|endoftext|> -TITLE: Explain why $E(X) = \int_0^\infty (1-F_X (t)) \, dt$ for every nonnegative random variable $X$ -QUESTION [49 upvotes]: Let $X$ be a non-negative random variable and $F_{X}$ the corresponding CDF. Show, - $$E(X) = \int_0^\infty (1-F_X (t)) \, dt$$ - when $X$ has : a) a discrete distribution, b) a continuous distribution. - -I assumed that for the case of a continuous distribution, since $F_X (t) = \mathbb{P}(X\leq t)$, then $1-F_X (t) = 1- \mathbb{P}(X\leq t) = \mathbb{P}(X> t)$. Although how useful integrating that is, I really have no idea. - -REPLY [4 votes]: The function $x1[x>0]$ has derivative $1[x>0]$ everywhere except for $x=0$, so by a measurable version of the Fundamental Theorem of Calculus -$$ -x1[x>0]=\int_0^{x}1[t>0]\ dt=\int_0^{\infty}1[x>t]\ dt,\qquad \forall x\in\mathbb R. -$$ -Applying this identity to a non-negative random variable $X$ yields -$$ -X=\int_0^{\infty}1[X>t]\ dt,\quad a.s. -$$ -Taking expectations of both sides and using Fubini to interchange integrals, -$$ -\mathbb EX=\int_0^{\infty}\mathbb P(X>t)\ dt. -$$<|endoftext|> -TITLE: Subgroups between $S_n$ and $S_{n+1}$ -QUESTION [14 upvotes]: Lets look at $S_n$ as subgroup of $S_{n+1}$. -How many subgroups $H$, $S_{n} \subseteq H \subseteq S_{n+1}$ there are ? - -REPLY [16 votes]: None. -Let $S_n -TITLE: Proof of Hilbert's Nullstellensatz -QUESTION [5 upvotes]: I'm working through my notes and I'm stuck in the middle of the proof of Hilbert's Nullstellensatz. -(Hilbert's Nullstellensatz) Let $k$ be an algebraically closed field. Let $J$ be an ideal in $k[x_1, \dots , x_n]$. Let $V(J)$ denote the set of $x$ in $k$ such that all $f$ in $J$ vanish on them. Let $U \subset k^n$ and let $I(U)$ denote the set of $f$ in $k[x_1, \dots , x_n]$ that vanish on $U$. Then -$$ r(J) = I(V(J))$$ -where $r$ denotes the radical of $J$. - -Let me go through the proof as far as I understand it: -$\subset$: Easy. Let $p \in r(J)$. Then $p^k \in J$ which means $p(x)^k = 0$ for all $x$ in $V(J)$. Hence $p(x) = 0$ for $x$ in $V(J)$ hence $p \in I(V(J))$. -$\supset$: Assume $f \notin r(J)$. Then for all $k>0$, $f^k \notin J$. We know that there exists a prime ideal $p$ such that $J \subset p$ and $f^k \notin p$ for all $k>0$. To see this we use the same argument used in the proof of proposition 1.8. on page 5 in Atiyah-MacDonald: Let $\Sigma$ be the set of all ideals that do not contain any power of $f$. We order $\Sigma$ by inclusion and use Zorn's lemma to get a maximal element $p$. We claim $p$ is prime. Assume neither $x \notin p$ nor $y \notin p$ (then we want to show $xy \notin p$). Then $p + (x), p + (y)$ are ideals properly containing $p$ hence neither of them is in $\Sigma$ hence $f^n \in p + (x)$ and $f^m \in p + (y)$. Now $f^{n+m} \in (p + (x)) (p + (y)) = p^2 + (x)\cdot p + (y)\cdot p + (xy) \subset p + (xy)$ so $p + (xy) \notin \Sigma$. Hence $p$ is properly contained in $p + (xy)$ hence $xy$ cannot lie in $p$. -So we have $p$ is a prime ideal containing $J$. Now consider the map -$$ k[x_1, \dots, x_n] \xrightarrow{\pi_1} k[x_1, \dots, x_n]/p \xrightarrow{i} (k[x_1, \dots, x_n]/p) [\overline{f}^{-1}] =: B([\overline{f}^{-1}]) \xrightarrow{\pi_2} B[\overline{f}^{-1}] /m$$ -where $\overline{f}$ denotes $\pi_1 (f)$ and $m$ is some maximal ideal in $B[\overline{f}^{-1}]$. We may assume $f \neq 0$ so that $\overline{f} \neq \overline{0}$. $\overline{f}^{-1}$ is an element of the field of fractions of $B$ so we may adjoin it to $B$ to get a new ring. -Since we only adjoined one element and otherwise only took quotients, the thing coming out on the RHS is a finitely generated $k$-algebra (because $ k[x_1, \dots, x_n]$ is). -Now by theorem 5.24 in Atiyah-MacDonald we know that $k \cong B[\overline{f}^{-1}] /m$. -The proof now finishes as follows: -"Let $t_1, \dots, t_n$ denote the images of $x_1 , \dots, x_n$ under this composite ring homomorphism. -(*)By construction, $g \in J \implies g \in p \implies g(t_1, \dots, t_n) = 0 \implies (t_1 , \dots, t_n ) \in V(J)$. -(**)On the other hand, $f(t_1 , \dots, t_n )$ is precisely the image of $f$ in $B[\overline{f}^{-1}] /m$, which is a unit. $\implies f(t_1 , \dots, t_n ) \neq 0 \implies f \notin I(V(J))$." -Question 1: What is the line (*) showing? I think we want to show $f \notin I(V(J))$, where does this come in here? -Question 2: Why is $f(t_1 , \dots, t_n )$ a unit? -Thank you for your help. - -REPLY [5 votes]: This is a comment made into an answer. Suppose $f$ is any polynomial not in $r(J)$. -Question 2 $f(t_1,\dots,t_n)$ is a unit because $f$ is (by construction) invertible in $B[f^{-1}]$ and remains so in the homomorphic image $B[f^{-1}]/m$. But the image of $f$ is precisely $f(t_1,\dots,t_n)$. -Question 1 The point of $(*)$ is that there is one distinguished point $(t_1,\dots,t_n)$ in $B[f^{-1}]/m×⋯×B[f^{-1}]/m≃k^n$ that kills all the polynomials in $J$. Put another way, -$$\mathrm{the~point~ of~}(*)\mathrm{~ is~ that~}(t_1,\dots,t_n)\in V(J)$$ -and thus all the polynomials in $I(V(J))$ must be killed by it. By $(**)$, $f$ does not take this point to $0$ and so -$$\mathrm{the~point~ of~}(**)\mathrm{~ is~ that~}f\notin I(V(J)).$$ -By now the authors have shown that -$f\notin r(J)\Rightarrow f\notin I(V(J))$ i.e. $k[x_1,\dots,x_n]\setminus r(J)\subset k[x_1,\dots,x_n]\setminus I(V(J))$ i.e. -$$I(V(J))\subset r(J).$$<|endoftext|> -TITLE: How to solve the $C^\alpha$ Poisson equation on closed Riemannian manifolds? -QUESTION [5 upvotes]: To be specific, suppose $M$ is a closed oriented manifold, $g$ is a Riemannian metric of $M$. -Let $\Delta_g$ be the Laplace-Beltrami operator w.r.t. $g$. -Prove: Suppose $f\in C^\alpha(M)$ satisfies $\int_M f\, dVol_g=0$, then there exists a function $u\in C^{2,\alpha}(M)$ such that $\Delta_g u=f$ in $M$, and $u$ is unique up to plus a constant, here $0<\alpha<1$. -My attempt is that, firstly use $D(u):=\int_M(\frac{1}{2} |\nabla u|^2+fu)dVol_g$ is a convex functional with a lower bound on $W_0^{1,2}(M)$ to show that there exists a weak solution $u\in W^{1,2}(M)$, next use the $L^2$-regularity theory to show that $u\in W^{2,2}(M)$, but I don't know how to improve the regularity of $u$ further. (Actually, I can use the method to prove that if $f$ is $C^\infty$, then $u$ is also $C^\infty$, but I cannot extend this result to $C^\alpha$ case.) -Another attempt is Schauder estimate. However, in Gilbarg and Trudinger's book they assume that $u\in C^{2,\alpha}(M)$ already to get some interior derivative norm bound of $u$, while I don't know how to establish $u\in C^{2,\alpha}(M)$. They give a continuity method to ensure that, but it seems their discussion works for domains in Euclidean space, not for manifolds. Therefore, I want to split the question into coordinate charts, but I failed, because I don't know how to use the condition $\int_M f\, dVol_g=0$ and how to give boundary conditions in every coordinate charts. -Since I'm a novice in PDE, my presentation of the problem might have some errors. Please correct them by comments or answers. Also, any comments or answers are welcome. -Thanks for your help. - -REPLY [2 votes]: I think I've got an answer to this question. -First, use the functional $D(u)=\int_M(\frac{1}{2}|\nabla u|^2+fu)dVol_g$, we can find a weak solution $u\in W^{1,2}_0(M)$, since $f\in C^\alpha(M)$ hence in $L^2(M)$. Next we shall show that $u\in C^{2,\alpha}(M)$. -We cite theorem 6.14 from Gilbarg-Trudinger's book: - -Let $L$ be strictly elliptic in a bounded domain $\Omega$, with $c\leq0$, and let $f$ and the coefficients of $L$ belong to $C^\alpha(\overline{\Omega})$. Suppose that $\Omega$ is a $C^{2,\alpha}$ domain and that $\phi\in C^{2,\alpha}(\overline\Omega)$. Then the Dirichlet problem, - $$ -Lu=f \textrm{ in }\Omega,\quad u=\phi\textrm{ on }\partial\Omega, -$$ - has a unique solution lying in $C^{2,\alpha}(\overline\Omega)$. - -The proof relies on the solvability of classical Dirichlet problem of laplacian, continuity method and Schauder estimates. -Suppose $U$ is a coordinate chart s.t. $\overline{U}$ is diffeomorphic to a closed ball, then by the theorem cited above, $\Delta_g v=f$ in $U$, $v = 0$ on $\partial U$ is solvable, and $v\in C^{2,\alpha}(\overline{U})$. Therefore, $u-v$ is a weak solution of $\Delta_g (u-v)=0$. By $L^2$-regularity, we have $u-v\in C^\infty(U)$, thus $u\in C^{2,\alpha}(U)$. Since we can choose finitely many $U$ to cover $M$, finally we have $u\in C^{2,\alpha}(M)$.<|endoftext|> -TITLE: Question about integral closures and localizations -QUESTION [5 upvotes]: Suppose $A$ is an integral domain with integral closure $\overline{A}$ (inside its fraction field), $\mathfrak{q}$ is a prime ideal of $A$, and $\mathfrak{P}_1,\ldots,\mathfrak{P}_k$ are the prime ideals of $\overline{A}$ lying over $\mathfrak{q}$. Show that $\overline{A_\mathfrak{q}} = \bigcap\overline{A}_\mathfrak{P_i}$ (note that the LHS is the integral closure of a localization, whereas the RHS is the intersection of localizations of integral closures of $A$). -If it would help, I suppose we could assume that $A$ has dimension 1, so that $\overline{A}$ is Dedekind, though I don't think that assumption is required. -(Geometrically, we're comparing the integral closure of a local ring at a singular point Q of some variety with the intersection of local rings at points in the normalization mapping to Q). -Thanks. - -REPLY [4 votes]: First notice that without extra hypothesis, the integral closure needs not be finite over $A$, and there might be infinitely many prime ideals lying over a given one in $A$. Neverthless, the equality holds. -It is easy to see that the LHS is contained in the RHS because the latter is integrally closed and contained in the field of fractions of $A_q$. Let $f$ be an element of the RHS (possibly infinite intersection). Denote by $B=\overline{A}$. Consider the denominator ideal -$$ I=\{ b\in B \mid bf\in B\}$$ -of $f$. I claim that $I\cap A$ is not contained in $q$. Admitting this, let $s\in I\cap A\setminus q$, then $sf\in B$ and $f$ is integral over $A_q$. -Now we prove the claim. Suppose that $I\cap A\subseteq q$. Then, geometrically, $q\in V(I\cap A)\subseteq \mathrm{Spec}(A)$. As $A/(I\cap A)\to B/I$ is injective and integral, $\mathrm{Spec}(B/I)\to \mathrm{Spec}(A/(I\cap A))$ is surjective. Identify $V(I)$ with $\mathrm{Spec}(B/I)$, this implies that there exists $p\in V(I)$ such that $p\cap A=q$. So $p$ is a prime ideal of $B$ lying over $q$ and containing $I$. This contradicts the hypothesis $f\in B_p$.<|endoftext|> -TITLE: Set of maximal chains in $\langle\mathcal{P}(\mathbb{N}),\subseteq\rangle$ -QUESTION [7 upvotes]: Let $S$ be the set of all maximal chains in the poset $\langle$$\mathcal{P}$$($$\mathbb{N}$$),$$\subseteq$$\rangle$ partitioned into equivalence classes by their order types. -How many different order types are there? Can you give me some insight into the structure of $S$ when quasi-ordered by embedding of order types? What is the supremum of ordinals that can be embedded into this quasi-order? - -REPLY [4 votes]: Somehow nobody dares to answer this question. (Edit: Thomas' answer didn't show up on my computer in time. We have the same result. I am talking about jumps, he talks about successors.) -Let $L$ be a linear order. Two points $x,y\in L$ form a jump -if $x -TITLE: Calculating the minimum of $\cos x \sin y$ -QUESTION [5 upvotes]: I am about to start university in October, to study computer science, and have been asked by my university to complete a number of problem sheets. I have become stuck on the following question, and therefore would appreciate any help possible. -The numbers $x$ and $y$ are subject to the constraints $x+y=\pi$. Find the values of $x$ and $y$ for which $\cos x\sin y$ takes its minimum value. -Using this question as a starting point towards a solution, I have the following steps attempted so far. -\begin{align*} -\Lambda(x, y, \lambda)&=\cos x\sin y+\lambda(x+y-\pi)\\ -\frac{\partial\Lambda}{\partial x}&=-\sin x\sin y+\lambda=0\\ -\frac{\partial\Lambda}{\partial y}&=\cos x\cos y+\lambda=0\\ -\frac{\partial\Lambda}{\partial\lambda}&=x+y-\pi=0\\ -x&=\cos^{-1}\left(\frac{\lambda}{\sin y}\right)\\ -y&=\cos^{-1}\left(\frac{-\lambda}{\cos x}\right)\\ -\cos^{-1}\left(\frac{\lambda}{\sin y}\right)+\cos^{-1}\left(\frac{-\lambda}{\cos x}\right)&=\pi\\ -\end{align*} -Unfortunately, such maths is way beyond my abilities, as I have only studied A-Level Maths and Further Maths, and I am working from the first answer in the referenced question and the Wikipedia articles on Lagrange Multipliers and Partial Derivatives. -I'm unsure of the correct tags to apply, so any help there would also be wonderful. -Edit: After receiving a number of hints, this is part of my solution, however, I'm not sure on how to properly phrase the last bit of the question with respect to properly solving the inequality or expressing values for $y$. -\begin{align*} -\sin x\cos y&=\sin x\cos(\pi-x)\\ -&=-\sin x\cos x\\ -&=\sin x\cos x\\ -&=\frac12\sin2x\\ -\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac12\sin2x\right)&=\cos2x\\ -\cos2x&=0\Rightarrow x=\frac12\left(n\pi-\frac\pi2\right),n\in\mathbb{Z}\\ -\frac{\mathrm{d}}{\mathrm{d}x}\cos2x&=-2\sin2x\\ --2\sin2x>0&\Rightarrow\sin2x<0\\ -\end{align*} - -REPLY [9 votes]: You've made this unnecessarily complicated. Note that the constraint $x+y=\pi$ means that you can substitute $y=\pi-x$ into your original expression. Thus, you need only minimize the $1$-variable function $\cos(x)\sin(\pi-x)$, which can be achieved through methods of basic calculus. -Edit: One can make this even simpler by using the identities $\sin(\pi-x)=\sin(x)$ and $2\cos(x)\sin(x)=\sin(2x)$.<|endoftext|> -TITLE: Can continuity of inverse be omitted from the definition of topological group? -QUESTION [15 upvotes]: According to Wikipedia, a topological group $G$ is a group and a topological space such that -$$ (x,y) \mapsto xy$$ and -$$ x \mapsto x^{-1}$$ -are continuous. The second requirement follows from the first one, no? (by taking $y=0, x=-x$ in the first requirement) -So we can drop it in the definition, right? - -REPLY [4 votes]: There is even a term for a group endowed with a topology such that multiplication is continuous (but inversion need not be): a paratopological group.<|endoftext|> -TITLE: Is R with $j_d$ topology totally disconnected? -QUESTION [5 upvotes]: Let consider the topological space $R_j=(\mathbb{R},j_r)$ where $j_r$ is generated by right side open intervals, i.e $[a,b)$ for $a,b \in \mathbb{R}$; note that this topology includes the euclidean topology. -$R_j$ is not a connected space, because given $x \in \mathbb{R}$ we have that $(-\infty,x)$ and $[x,+\infty)$ provide a disconnection or $R_j$. -Given the arbitrariness in choosing $x$, is it possible that $R_j$ is totally disconnected? Let consider a certain $y:x0.$ if we suppose there is a certain limit in this reduction, then we would have to accept that $[x-\varepsilon /2,x+\varepsilon /2)$ is a connected component, which is a contradiction with the beginning of this discussion. -What can we conclude, from this? - -REPLY [8 votes]: This space is usually called the Sorgenfrey line. You can also prove directly that it’s totally disconnected. If $A\subseteq\Bbb R$ has at least two points, choose $x,y\in A$ with $x -TITLE: What are the differences between class, set, family, and collection? -QUESTION [97 upvotes]: In school, I have always seen sets. I was watching a video the other day about functors, and they started talking about a set being a collection, but not vice-versa. I also heard people talking about classes. What is their relation? Some background would be nice. -It has to do with something called Russell's paradox, but I don't know what that is. -I think that the difference between a family and a set is that the former is a function and the latter is a set. Is this right? - -REPLY [116 votes]: The idea behind a "collection" is simply a notion of a bunch of mathematical objects which are collected into one big pile. Think of it as a big bin full of trash, diamonds and empty bottles of beer, it doesn't have to make sense what is in this collection, it's just a collection. -One of the problem to explain these things to people who are not mathematicians (or trying to "outsmart a set theorist", as I ran into several of those) is that the notion of a collection is not fully formal unless you already know what sets and class are, and even then it's not exactly what we mean. -Let me start over now. Doing mathematics we often have an idea of an object that we wish to represent formally, this is a notion. We then write axioms to describe this notion and try to see if these axioms are self-contradictory. If they are not (or if we couldn't prove that they are) we begin working with them and they become a definition. Mathematicians are guided by the notion but they work with the definition. Rarely the notion and the definition coincide, and you have a mathematical object which is exactly what our [the mathematicians] intuition tells us it should be. -In this case, a collection is a notion of something that we can talk about, like a mystery bag. We might know that all the things inside this mystery bag are apples, but we don't know which kind; we might know they are all Granny Smith, but we cannot guarantee that none of them is rotten. A collection is just like that. We can either know something about its elements or we don't, but we know that it has some. -Mathematician began by describing these collections and calling them sets, they did that in a relatively naive way, and they described the axioms in a rather naive way. To the non-mathematician (and to most of the non-set theorists) everything is still a set, and we can always assume that there is a set theorist that assured that for what we need this is true. Indeed, if we just wanted to discuss the real numbers, there is no worry at all we can assume everything we work with is a set. -This naive belief can be expressed as every collection is a set. It turned out that some collections cannot be sets, this was expressed via several paradoxes, Cantor's paradox; Russell's paradox; and other paradoxes. The exact meaning is that if we use that particular axiomatic description of "what is a set" then we can derive from it contradiction, which is to say that these axioms are inconsistent. -After this happened several people began working on ways to eliminate this problem. One method in common was to limit the way we can generate collections which are sets. This means that you can no longer derive such contradiction within the theory, namely you cannot prove that such collection even exists - or rather you can prove it doesn't. -The common set theory nowadays called ZFC (named after Zermelo and Fraenkel, the C denotes the axiom of choice) is relatively close to the naive way from which set theory emerges, and it still allows us to define collections which are not sets, though, for example "the collection of all sets". These collections are called classes, or rather proper classes. -What is definable? This is a whole story altogether, but essentially it means that we can describe it with a single formula (perhaps with parameters) of one free variable. "$x$ is taller than 1.68m" is an example to such formula, and it defines the class of all people taller than said height. -So in ZFC we may define a collection which is not a set, like the collection of all the singletons, or the collection of all sets. These are not sets because they are too big, in some sense, to be sets, but they are classes, proper classes. We can talk about collections which are not definable but that requires a lot more background in logic and set theory to get into. - -To sum up - -Classes are collections which can be defined, sets are particular classes which are relatively small and there are classes which are not sets. Collections is a notion which is expressed via both these mathematical objects, but need not be well-defined otherwise. -Of course when we say defined we mean in the context of a theory, e.g. ZFC. In this sense, sets are things which "really exist" whereas classes are collections we can talk about despite their possible nonexistence. - - -One last thing remains, families. Well, as you remarked families are functions. But functions are sets, so families are sets. We can make a slight adjustment to this, and we can, in fact, talk about class functions, and an index which is not a set but a proper class. We, therefore, can talk about families which are classes. -Generally, speaking, if so, a family is a correspondence from one collection into another which uses one collection as indices for elements from another collection. - -To read more - -What is the difference between a class and a set? - -Why is "the set of all sets" a paradox, in layman's terms?<|endoftext|> -TITLE: Galois Field GF(4) -QUESTION [7 upvotes]: Question: -Why is the table of $GF(4)$ look like the one below? I know it has to do with the fact that 4 is composite -Let $GF(4) = \{0,1,B,D\}$ -Addition: -$$ -\begin{array}{c|cccc} -+ & 0& 1& B & D \\ -\hline - 0& 0 & 1 & B & D \\ -1 & 1 & 0 & D & B \\ -B & B & D & 0 & 1 \\ -D & D & B & 1 & 0 \end{array} -$$ -Multiplication: -$$\begin{array}{c|cccc} -\cdot & 0 & 1 & B & D \\ -\hline -0 & 0 & 0 & 0 & 0 \\ -1 & 0 & 1 & B & D \\ -B & 0 & B & D & 1 \\ -D & 0 & D & 1 & B - \end{array}$$ - -REPLY [19 votes]: This is the only way to define operations on a four-element set making it a field (up to a permutation of elements). I will use some general properties of fields in the following. - -First, multiplication and addition are commutative, which saves us some guesswork (we only need to determine half the tables). -Furthermore, there's got to be 0 and 1. Multiplication by 1 and 0 works the same in any field, so that takes cares of two rows in the multiplication table. -In any field, elements distinct from zero with multiplication form an abelian group. In case of $\mathbf F_4$, it is a three-element group, and there is only one such group, so it must be the cyclic group of order three, hence $B^2=D$, $BD=1$, $D^2=B$, so we're done with multiplication. -The characteristic of a field is always a prime number, so it must be $2$ in case of $\mathbf F_4$, so there must be zeroes on diagonal of additive table. -Additive identity and inverse are unique, so $B+1\neq B,1,0$, so it must be $D$, similarly $D+1=B$ -Knowing the above, we can easily see that $B+D=B+B+1=0+1=1$, so we're done.<|endoftext|> -TITLE: Is the coordinate ring of SL2 a UFD? -QUESTION [18 upvotes]: Is the ring $K[a,b,c,d]/(ad-bc-1)$ a unique factorization domain? - -I think this is a regular ring, so all of its localizations are UFDs by the Auslander–Buchsbaum theorem. However, I know there are Dedekind domains (which are regular; every local ring is a PID, so definitely UFD) that are not UFDs, so being a regular ring need not imply the ring is a UFD. -With the non-UFD Dedekind domains (at least the number rings), I can usually spot a non-unique factorization, but I don't see any here in this higher dimensional example. - -REPLY [8 votes]: Let $R=K[X,Y,Z,T]/(XY+ZT-1)$. It's easily seen that $R$ is an integral domain. -In the following we denote by $x,y,z,t$ the residue classes of $X,Y,Z,T$ modulo the ideal $(XY+ZT-1)$. -First note that $x$ is prime: $R/xR\simeq K[Z,Z^{-1}][Y]$. Then observe that $R[x^{-1}]=K[x,z,t][x^{-1}]$ and $x$, $z$, $t$ are algebraically independent over $K$. This shows that $R[x^{-1}]$ is a UFD and from Nagata's criterion we get that $R$ is a UFD.<|endoftext|> -TITLE: Universal Cover of projective plane glued to Möbius band -QUESTION [10 upvotes]: This is the second part of exercise 1.3.21 in page 81 of Hatcher's book Algebraic topology, and the first part is answered here. -Consider the usual cell structure on $\mathbb R P^2$, with one 1-cell and one 2-cell attached via a map of degree 2. Consider the space $X$ obtained by gluing a Möbius band along the 1-cell via a homeomorphism with its boundary circle. - -Compute $\pi_1(X)$, describe the universal cover of $X$ and the action of $\pi_1(X)$ on the universal cover. - -Using van Kampen's theorem, $\pi_1(X)=\langle x,y \mid x^2=1, x=y^2\rangle=\langle y\mid y^4=1\rangle=\mathbb Z_4$. -I have tried gluing spheres and Möbius strips in various configurations, but have so far not been successful. Any suggestions? - -REPLY [8 votes]: Let $M$ be the Möbius band and let $D$ be the $2$-cell of $RP^2$. Then $X$ is the result of gluing $M$ to $D$ along a map $\partial D\rightarrow \partial M$ of degree $2$. Hence $\pi_1(X)$ has a single generator $\gamma$, that comes from the inclusion $M\rightarrow X$, and the attached disc $D$ kills $\gamma^4$, hence $\pi_1(X)\cong {\mathbb Z}/4$. Alternatively, take the homotopy $M\times I\rightarrow M$ that shrinks the Möbius band to its central circle $S\subset M$; this gives a homotopy from $X = M\cup D$ to the space ${\mathbb S}^1\cup_f D$, where $f\colon \partial D\rightarrow {\mathbb S}^1$ is a map of degree $4$. -The universal cover of the space ${\mathbb S}^1\cup_f D$ is described in Hatcher's Example 1.47, and it is the homeomorphic to the quotient of $D\times\{1,2,3,4\}$ by the relation $(x,i)\sim (y,j)$ iff $x=y$ and $x\in \partial D\times \{i\}$. -Now let $D$ denote de closed unit disc in ${\mathbb R}^2$. The universal cover of the space $X$ is homeomorphic to the quotient of $D\times \{a,b,c,d\}\cup S^1\times [-1,1]$ by the relations - -$(x,a)\sim (x,b)\sim (x,1)$ for all $x\in S^1 = \partial D$ -$(x,c)\sim(x,d)\sim (x,-1)$ for all $x\in S^1=\partial D$ - -and $\pi_1(X)\cong {\mathbb Z}/4$ acts as follows: - -$(re^{2\pi i\theta},a)\to (re^{2\pi i(\theta + 1/4)}, c)\to (re^{2\pi i(\theta+1/2)},b)\to (re^{2\pi i (\theta + 3/4)},d)\rightarrow (re^{2\pi i\theta},a)$ for the points in the discs $D\times \{a,b,c,d\}$ -$(e^{2\pi i \theta},t)\to (e^{2\pi i (\theta+1/4)},-t)\to (e^{2\pi i (\theta + 1/2)},t)\to (e^{2\pi i(\theta + 3/4)},-t)\rightarrow (e^{2\pi i \theta},t)$ for the points in $S^1\times [-1,1]$. - -and the map $\tilde{X}\rightarrow X$ sends the four discs to the disc and the cylinder to the Möbius band.<|endoftext|> -TITLE: Matrices whose Linear Combinations are All Singular -QUESTION [10 upvotes]: I'd like to know if the following problem of elementary linear algebra is already solved / solvable. -For two (singular) $n\times n$ matrices $P$ and $Q$, if $\det(\lambda P+\mu Q)=0$ for any $\lambda,\mu\in\mathbb{R}$, what are conditions on $P$ and $Q$? - -REPLY [2 votes]: This is not a complete answer, just some sufficient and some necessary conditions. -Let's look at the opposite condition. Fix $u,v\in\mathrm{End}_k(V)$ two singular (non proportional) endomorphisms of the finite dimensional $k$-vector space $V$: what conditions are necessary or sufficient for there to be a scalar $\lambda$ such that $u+\lambda v\in\mathrm{GL}(V)$? -The following conditions are necessary: - -$\mathrm{Ker}(u)\oplus\mathrm{Ker}(v)$; -$\mathrm{Im}(u)+\mathrm{Im}(u)=V$; -$v\big(\mathrm{Ker}(u)\big)\oplus u\big(\mathrm{Ker}(v)\big)$. - -The first is @chaochuang's answer, and the second is @Marc van Leeuwen' comment. To see the necessity of the third condition, write down matrices in a basis adapted to $V=\mathrm{Ker}(u)\oplus\mathrm{Ker}(v)\oplus S$. -The following conditions are sufficient (at least when $k$ is big enough, for instance if $\mathbb Q\subset k$): - -The induced morphism $\tilde{v}:\mathrm{Ker}(u)\rightarrow V\rightarrow V/\mathrm{Im}(u)$ is an isomorphism i.e. $v(\mathrm{Ker}(u))\oplus\mathrm{Im}(u)$; -The induced morphism $\tilde{u}:\mathrm{Ker}(v)\rightarrow V\rightarrow V/\mathrm{Im}(v)$ is an isomorphism i.e. $u(\mathrm{Ker}(v))\oplus\mathrm{Im}(v)$. - -To see the sufficiency of the first point, write down the matrix for $u+\lambda v$ where the initial base of $V$ is adapted to $S\oplus \mathrm{Ker}(u)=V$, and the final base is the images under $u$ of the basis vectors chosen in $S$ and the images under $v$ of the basis vectors in $\mathrm{Ker}(u)$, and let $\lambda\rightarrow 0$: the determinant of this matrix is a polynomial in $\lambda$ of the form $\lambda^d(1+\cdots)$ where $d=\dim~\mathrm{Ker}(u)$. -Going back to the initial problem, we get $3$ sufficient conditions for the span of $u$ and $v$ to be completely singular, and $2$ necessary conditions.<|endoftext|> -TITLE: Any finitely generated subgroup of $(\mathbb{Q},+)$ is cyclic. -QUESTION [9 upvotes]: $\fbox{1}$ Prove that any finitely generated subgroup of $(\mathbb{Q},+)$ is cyclic. -$\fbox{2}$ Prove that $\mathbb{Q}$ is not isomorphic to $\mathbb{Q}\times \mathbb{Q}$. - -Any hints would be appreciated. - -REPLY [12 votes]: Let $\frac{m_1}{n_1},...,\frac{m_k}{n_k}$ be the generators of $G\leq\Bbb Q$ and $g\in G$. Then we have $g=a_1\frac{m_1}{n_1}+...+a_k\frac{m_k}{n_k}=\frac{a_1m_1n_2...n_{k}+...+a_km_kn_1...n_{k-1}}{n_1...n_k}$ for some integers $a_1,...,a_k$. Thus $G$ is a subgroup of the cyclic group $<\frac{1}{n_1...n_k}>$. Hence $G$ must be cyclic.<|endoftext|> -TITLE: Homeomorphisms of X form a topological group -QUESTION [12 upvotes]: So I'm just learning about the compact-open topology and am trying to show that for a compact, Hausdorff space ,$X$, the group of homeomorphisms of $X$, $H(X)$, is a topological group with the compact open topology. This topology has a subbasis of sets $\{f\in H(X):f(C)\subseteq V\}=S(C,V)$ for compact $C$, open $V$. This is my first attempt at getting to know this topology, so I'd appreciate some help with the part of a proof I have so far, possibly a better way to go about proving this, and any other help understanding this topology. -First, let $c:H(X)\times H(X)\rightarrow H(X)$ be composition. For $f,g\in H(X)$, let $S(C,V)$ be a subbasis set with $f\circ g\in S(C,V)$. So $f(g(C)\subseteq V$ which means $f\in S(g(C),V)$ and $g\in S(C,f^{-1}(V)$. The product of these open sets works since if $h_1(C)\subseteq f^{-1}(V)$ and $h_2(g(C))\subseteq V$, then $h_2\circ h_1(C)\subseteq V$. -Then let $i:H(X)\rightarrow H(X)$ be inversion. Take a subbasis set $O=\{g:g(C)\subseteq U\}$ and consider $i^{-1}(O)=\{g^{-1}:g(C)\subseteq U\}$. If $h^{-1}\in i^{-1}(O)$, then one thing we have is that $h(C)\subset U$, but I'm not totally sure what to do with this. This part seems like it should be easier, but I am just not seeing it. -Thanks. - -REPLY [8 votes]: Here you don't need much topology, it boils down to pure manipulation of sets and bijective functions, and the two following facts: $(*)$ closed subsets of compact spaces are compact, and $(**)$ compact subsets of Hausdorff spaces are closed. Just take complements and use the fact that $h$ is by definition a bijection, so -$$\begin{array}{RCL} -h\in S(C,U) & \Longleftrightarrow & h(C)\subset U\\ -& \Longleftrightarrow & X\setminus U\subset \underbrace{X\setminus h(C)}_{=h(X\setminus C)}\\ -& \Longleftrightarrow & h^{-1}(X\setminus U)\subset X\setminus C \\ -& \Longleftrightarrow & h^{-1}\in S(X\setminus U,X\setminus C) -\end{array}$$ -(which is a subbasic open neighborhood) i.e. -$$\mathrm{inv}^{-1}( S(C,U))=\mathrm{inv}(S(C,U))=S(X\setminus U,X\setminus C)$$ -which proves the inversion map $\mathrm{inv}$ to be continuous.<|endoftext|> -TITLE: How to select an field of study? -QUESTION [8 upvotes]: As my question implies, I am a young, developing mathematician, with concerns about my future math career. In particular, I'd like to know how to select a future field of study. Through my courses and readings, I've started to entertain the possibility of studying Lie theory. I think this field has some nice, deep structure and is particularly well-suited to my minor interest in physics. Of course, other fields are also appealing, so I am having a hard time narrowing my choices. -My question is, then, how does one select a field of study? I'd love to hear some personal experiences that relate to this question. -Some of the considerations for choosing a field of study, which I have been asking myself, are - -What appeals to you? What do you enjoy doing? -What do you want to know? What questions do you want to answer for yourself? -What does the world want to know? What fields have important, pending questions that should be answered? -To what fields is your chosen field connected? Do these connections play a large role in your chosen field? Are you interested in these possible connections? -Which fields are are on the rise? Which fields are currently exciting and moving in a good direction? -What are you good at? What areas seem to be intuitive to you? - -If someone could talk about some of these considerations in regard to their own experience, I would be particularly grateful. Furthermore, if there are any considerations that I am missing, please feel free to add to the list. - -REPLY [13 votes]: Don't worry about what fields are "hot" or not (point 5). For one thing, that can change very quickly when you least expect it. New results or sudden applications may catapult a particular area to the fore, and answers to questions can doom a field to obscurity. It's going to take you several years from now until you are done, and there is no guarantee that what is "hot" or "on the rise" right now will still be hot or on the rise when you are done. -I would also advice to give preponderance to 1 and 2. If you become a mathematician, you will spend your time thinking about these things, and they better be things that you want to spend time thinking about. If doing your research becomes a chore, you are doomed, no matter how good you are at it, or how hot the field is. You'll be looking for excuses not to do it. -Point 4 is not really something that can go into deciding what field to study, because you sort of need to know a bit more about the field to really be able to answer it. -Mainly, in my opinion, you'll want to do something that you personally find exciting, interesting, and fun, for whatever reason. Sure, it's nice for the people who are working in hot areas and can easily get grants, since it is always better to be rich and healthy than sick and poor, but remember: if things go well, you'll be doing this for 30-50 years. You want to enjoy it.<|endoftext|> -TITLE: Orthogonal complement of a vector bundle -QUESTION [6 upvotes]: Let $E \rightarrow X$ be a vector bundle with an inner product. If $F$ is a sub-bundle, we can define an orthogonal complement bundle $F^\perp$ (see http://www.math.cornell.edu/~hatcher/VBKT/VB.pdf for the construction, and the source of the problem). I am trying to show that $F^\perp$ is independent of the choice of inner product, up to isomorphism. I have tried to do a local construction, defining the isomorphism in each locally trivial neighborhood, but I do not see a way to patch these together into a global isomorphism. Any suggestions? - -REPLY [6 votes]: Hint: First show that for an inner product space $(V, \langle \cdot, \cdot \rangle)$ and a subspace $W \subseteq V$, -$$W^\perp \cong V/W.$$ -Then, using the quotient bundle construction of the previous exercise (exercise 2), show that for a Euclidean vector bundle $E \longrightarrow X$ with sub-bundle $F$, -$$F^\perp \cong E/F.$$ -Since $E/F$ is independent of the bundle metric, this will show that $F^\perp$ does not depend on the bundle metric (up to isomorphism). -Let me know if you want me to provide the details of the argument.<|endoftext|> -TITLE: Solve $\, \mathrm dy/\, \mathrm dx = e^{x^2}$ -QUESTION [7 upvotes]: I want to solve $$\frac{\, \mathrm dy}{\, \mathrm dx}=e^{x^{2}}.$$ i using variable separable method to solve this but after some stage i stuck with the integration of $\int e^{x^{2}}\, \mathrm dx$. i dont know what is the integration of $\int e^{x^{2}}\, \mathrm dx$. Please help me out! - -REPLY [4 votes]: $$ -\frac{dy}{dx}=e^{x^{2}} -$$ -has no elementary solution. The error function (also called the Gauss error function) is a special function (non-elementary) of sigmoid shape which occurs in probability, statistics and partial differential equations. It is defined as: -$$ - \operatorname{erf}(x) = \frac{2}{\sqrt{\pi}}\int_{0}^x e^{-t^2} dt. -$$ -See the link for reference and more information and thus, J.M. ...?<|endoftext|> -TITLE: Prove that for $n \in \mathbb{N}, \sum\limits_{k=1}^{n} (2k+1) = n^{2} + 2n $ -QUESTION [5 upvotes]: I'm learning the basics of proof by induction and wanted to see if I took the right steps with the following proof: -Theorem: for $n \in \mathbb{N}, \sum\limits_{k=1}^{n} (2k+1) = n^{2} + 2n $ -Base Case: -let $$ n = 1 $$ Therefore $$2*1+1 = 1^{2}+2*1 $$ Which proves base case is true. -Inductive Hypothesis: -Assume $$\sum_{k=1}^{n} (2k+1) = n^{2} + 2n $$ -Then $$\sum_{k=1}^{n+1} (2k+1) = (n+1)^{2} + 2(n+1) $$ -$$\iff (2(n+1) +1)+ \sum_{k=1}^{n} (2k+1) = (n+1)^{2} + 2(n+1) $$ -Using inductive hypothesis on summation term: -$$\iff(2(n+1) +1)+ n^{2} + 2n = (n+1)^{2} + 2(n+1) $$ -$$\iff 2(n+1) = 2(n+1) $$ -Hence for $n \in \mathbb{N}, \sum\limits_{k=1}^{n} (2k+1) = n^{2} + 2n $ Q.E.D. -Does this prove the theorem? Or was my use of the inductive hypothesis circular logic? - -REPLY [4 votes]: Your proof is fine, but for me, it didn't look like the clearest presentation. For instance, I think it'd be better to drop the biconditionals, and for your inductive step just write -$$\begin{eqnarray} -\sum_{k=1}^{n+1} (2k+1) &=& n^2 + 2n + 2(n+1) + 1 \\ -&=& n^2 + 2n + 1 + 2(n + 1)\\ -&=& (n+1)^2 + 2(n+1)\\ -\end{eqnarray}$$<|endoftext|> -TITLE: When does Poincaré inequality hold? -QUESTION [6 upvotes]: Poincaré inequality is given by $$\int_\Omega u^2\le C\int_\Omega|\nabla u|^2dx ,$$ where $\Omega$ is bounded open region in $\mathbb R^n$. -However this inequality is not satisfied by all the function. Take for example a constant function $u=10$ in some region. -Happy to have have some discussions about it. -Thanks for your help. - -REPLY [3 votes]: This inequality is, as you have shown by a simple example, not valid as stated. Things can be fixed up in various ways: -In the Wikipedia article on the Poincaré inequality it is assumed that the mean value $u_\Omega$ of $u$ on $\Omega$ is zero, resp., the left side of the inequality is replaced by $\int_\Omega|u-u_\Omega|^2\ {\rm d}x$. -In the quoted source the author refers to a certain Sobolev space which is the completion of $C_0^1(\Omega)$. Here the ${}_0$ means that compact support is assumed; in particular these functions vanish on the boundary.<|endoftext|> -TITLE: Product of Riemannian manifolds? -QUESTION [5 upvotes]: Given two Riemannian manifolds $(M,g^M)$ and $(N,g^N)$ is there a natural way to combine them to be a Riemannian manifold? Some kind of $(M \times N, g^{M \times N})$. - -REPLY [7 votes]: Yes. Using the natural isomorphism $T(M \times N) \cong TM \times TN$, define the metric on $T(M \times N)$ as follows for each $(p,q) \in M \times N$. -$$g^{M\times N}_{(p,q)} \colon T_{(p,q)}(M \times N) \times T_{(p,q)}(M \times N) \to \mathbb{R},$$ -$$((x_1,y_1),(x_2,y_2)) \mapsto g^M_p(x_1,x_2) + g^N_q(y_1,y_2).$$ -Alternately, if you think of the metrics as vector bundle isomorphisms $g^M \colon TM \to T^* M$ and $g^N \colon TN \to T^* N$, then the metric on $M \times N$ is just the induced vector bundle isomorphism -$$g^M \oplus g^N \colon TM \oplus TN \to T^*M \oplus T^* N,$$ -$$(x,y) \mapsto (g^M(x), g^N(y))$$ -Here, $TM \oplus TN$ is just $TM \times TN$ as a set, and the base manifold of $TM \oplus TN$ is $M \times N$.<|endoftext|> -TITLE: Showing that a homogenous ideal is prime. -QUESTION [17 upvotes]: I'm trying to read a proof of the following proposition: -Let $S$ be a graded ring, $T \subseteq S$ a multiplicatively closed set. Then a homogeneous ideal maximal among the homogeneous ideals not meeting $T$ is prime. -In this proof, it says - -"it suffices to show that if $a,b \in S$ are homogeneous and $ab \in \mathfrak{p}$, then either $a \in \mathfrak{p}$ or $b \in \mathfrak{p}$" - -where $\mathfrak{p}$ is our maximal homogeneous ideal. I don't know how to prove this is indeed sufficient. If I try writing general $a$ and $b$ in terms of "coordinates": -$$a=a_0+\cdots+a_n$$ where $a_0 \in S_0,$ -then I can see it working for small $n$, but it seems to get so complicated I wouldn't know how to write down a proof. Is there a better way to attack this problem? - -REPLY [25 votes]: We wish to prove: - -If $S$ is a $\mathbb{Z}$-graded ring and $\mathfrak{p}$ is a homogeneous ideal of $S$ satisfying $ab \in \mathfrak{p}$ implies $a$ or $b$ in $\mathfrak{p}$ for homogeneous $a$ and $b$, then $\mathfrak{p}$ is prime. - -So take 2 general elements $a,b \in \mathfrak{p}$ and assume $ab \in \mathfrak{p}$ but neither $a$ nor $b$ is in $\mathfrak{p}$. Let $a = \sum a_d$ and $b = \sum b_d$ be their homogeneous decompositions. Since $a \not \in \mathfrak{p}$, then some $a_d \not \in \mathfrak{p}$, and since all but finitely many $a_d$ are $0$, there exists a largest integer $d$ such that $a_d \not \in \mathfrak{p}$. Similarly, there exists a largest integer $e$ such that $b_e \not \in \mathfrak{p}$. -Since $ab \in \mathfrak{p}$ and $\mathfrak{p}$ is a homogeneous ideal, then all the components of $ab$ are in $\mathfrak{p}$. The $d+e$ component of $ab$ is $\sum a_i b_j$ where we sum over all pairs $(i,j)$ with $i+j = d+e$. But each such pair $(i,j)$, other than $(d,e)$, must have either $i>d$ or $j>e$, and hence (by the maximality of $d$ and $e$) we have $a_i b_j \in \mathfrak{p}$. Thus $a_d b_e \in \mathfrak{p}$ also, yet neither $a_d$ nor $b_e$ is in $\mathfrak{p}$, which contradicts the original assumption about $\mathfrak{p}$ for the homogeneous elements $a_d$ and $b_e$. -In short: If $a,b$ is a general counterexample for the primality of $\mathfrak{p}$, then $a_d, b_e$ is a homogeneous counterexample.<|endoftext|> -TITLE: Strict inequalities in LP -QUESTION [10 upvotes]: How should we deal with strict inequalities in a linear programming problem? For example: -inequalities such as $ax< b$; - -REPLY [10 votes]: In general strict inequalities are not treated in linear programming problems, since the solution is not guaranteed to exist on corner points. -Consider the $1$-variable LPP: $Max$ $x$ subject to $x<3$. Now there does not exist any value of $x$ for which maximum is achieved and which lies in the feasible region.<|endoftext|> -TITLE: On certain decomposition of unitary symmetric matrices -QUESTION [9 upvotes]: It is well known that a symmetric matrix over field $\Bbb F$ is congruent to a diagonal matrix, i.e., there exists some A s.t. $A^TUA=D$ with $U$ symmetric and $D$ diagonal. If $\Bbb F=\Bbb C$ then we can make $D=I$. -Recently I learned that if $U$ is unitary that we can do one step further by requiring $A$ to be unitary too. A similar result holds for unitary skew matrices. But I fail to figure out a proof myself. -Can anyone provide a proof of this or at least help me to locate some references? Many thanks! - -REPLY [5 votes]: Let $U$ be complex symmetric unitary, i.e., $U^T = U$ and $U^* U = I$. From Takagi factorization we know that $U = V D V^T$, i.e., -$$D = V^{-1} U V^{-T},$$ -for some unitary $V$ and real diagonal $D$. Here, we do not have $D = I$, but we do have the unitarity of $V$. -Using the unitarity of $U$ and $V$, we see that -\begin{align*} -D^* D &= (V^{-T})^* U^* V^{-*} V^{-1} U V^{-T} = \overline{V^{-1}} U^* (V V^*)^{-1} U V^{-T} = \overline{V^{-1}} U^* U V^{-T} \\ -&= \overline{V^{-1}} V^{-T} = \overline{(V^* V)^{-1}} = {\rm I}. -\end{align*} -In other words, $D$ is real diagonal unitary, which means that it has diagonal elements $-1$ and $1$. Define a complex diagonal matrix $X = \operatorname{diag}(x_1, x_2, \dots, x_n)$: -$$x_k = \begin{cases} -1, & D_{kk} = 1, \\ -{\rm i}, & D_{kk} = -1. -\end{cases}$$ -Obviously, $D = X X^T$. Defining $A = VX$ we get -$$U = V D V^T = (VX) (VX)^T = A {\rm I} A^T,$$ -which is the required form. Notice that $X$ is unitary, so $A$ is unitary as well.<|endoftext|> -TITLE: Example of Artinian module that is not Noetherian -QUESTION [10 upvotes]: I've just learned the definitions of Artinian and Noetherian module and I'm now trying to think of examples. Can you tell me if the following example is correct: -An example of a $\mathbb Z$-module $M$ that is not Noetherian: Let $G_{1/2}$ be the additive subgroup of $\mathbb Q$ generated by $\frac12$. Then $G_{1/2} \subset G_{1/4} \subset G_{1/8} \subset \dotsb$ is a chain with no upper bound hence $M = G_{1/2}$ as a $\mathbb Z$-module is not Noetherian. -But $M$ is Artinian: $G_{1/2^n}$ are the only subgroups of $G_{1/2}$. So every decreasing chain of submodules $G_i$ is bounded from below by $G_{1/2^{\min i}}$. -Edit In Atiyah-MacDonald they give the following example: - -Let $G$ be the subgroups of $\mathbb{Q}/\mathbb{Z}$ consisting of all elements whose order is a power of $p$, where $p$ is a fixed prime. - Then $G$ has exactly one subgroup $G_n$ of order $p^n$ for each $n \geq 0$, and $G_0 \subset G_1 \subset \dotsb \subset G_n \subset \dotsb$ (strict inclusions) so that $G$ does not satisfy the a.c.c. - On the other hand the only proper subgroups of $G$ are the $G_n$, so that $G$ does satisfy d.c.c. -(Original images here and here.) - -Does one have to take the quotient $\mathbb{Q}/\mathbb{Z}$? - -REPLY [13 votes]: Fix a prime $p$ and let $M_p={\Bbb Z}(\frac1p)/{\Bbb Z}$. -It is not difficult to see that the only submodules of $M_p$ are those generated by $\frac1{p^k}+{\Bbb Z}$ for $k\geq0$. From this it follows that $M_p$ is Artinian but not Noetherian.<|endoftext|> -TITLE: Evaluating the product $\prod\limits_{k=1}^{n}\cos\left(\frac{k\pi}{n}\right)$ -QUESTION [22 upvotes]: Recently, I ran across a product that seems interesting. -Does anyone know how to get to the closed form: -$$\prod_{k=1}^{n}\cos\left(\frac{k\pi}{n}\right)=-\frac{\sin(\frac{n\pi}{2})}{2^{n-1}}$$ -I tried using the identity $\cos(x)=\frac{\sin(2x)}{2\sin(x)}$ in order to make it "telescope" in some fashion, but to no avail. But, then again, I may very well have overlooked something. -This gives the correct solution if $n$ is odd, but of course evaluates to $0$ if $n$ is even. -So, I tried taking that into account, but must have approached it wrong. -How can this be shown? Thanks everyone. - -REPLY [15 votes]: If $n$ is even, then the term with $k=n/2$ makes the product on the left $0$ and $\sin\left(\frac{n}{2}\pi\right)=0$. So assume that $n$ is odd. -$$ -\begin{align} -\prod_{k=1}^n\cos\left(\frac{k\pi}{n}\right) -&=-\prod_{k=1}^{n-1}\cos\left(\frac{k\pi}{n}\right)\tag{1}\\ -&=-\prod_{k=1}^{n-1}\frac{\sin\left(\frac{2k\pi}{n}\right)}{2\sin\left(\frac{k\pi}{n}\right)}\tag{2}\\ -&=\frac{-1}{2^{n-1}}\frac{\prod\limits_{k=\frac{n+1}{2}}^{n-1}\sin\left(\frac{2k\pi}{n}\right)}{\prod\limits_{k=1}^{\frac{n-1}{2}}\sin\left(\frac{(2k-1)\pi}{n}\right)}\tag{3}\\ -&=\frac{(-1)^{\frac{n+1}{2}}}{2^{n-1}}\frac{\prod\limits_{k=1}^{\frac{n-1}{2}}\sin\left(\frac{(2k-1)\pi}{n}\right)}{\prod\limits_{k=1}^{\frac{n-1}{2}}\sin\left(\frac{(2k-1)\pi}{n}\right)}\tag{4}\\ -&=-\frac{\sin\left(n\frac\pi2\right)}{2^{n-1}}\tag{5} -\end{align} -$$ -$(1)$: $\cos(\pi)=-1$ -$(2)$: $\sin(2x)=2\sin(x)\cos(x)$ -$(3)$: cancel $\sin\left(\frac{j\pi}{n}\right)$ in the numerator and denominator for even $j$ from $2$ to $n-1$ -$(4)$: in the numerator, change variable $k\mapsto k+\frac{n-1}{2}$ and use $\sin(x+\pi)=-\sin(x)$ -$(5)$: for odd $n$, $\sin\left(n\frac\pi2\right)=(-1)^{\frac{n-1}{2}}$<|endoftext|> -TITLE: Inverse problem from pdes -QUESTION [5 upvotes]: A linear inverse problem is given by: -$\ \mathbf{d}=\mathbf{A}\mathbf{m}+\mathbf{e}$ -where d: observed data, A: theory operator, m: unknown model and e: error. -To minimize the effect of the noise; a Least Square Error (LSE) model estimate is commonly used: -$\ \mathbf{\tilde{m}}=(\mathbf{A^\top A})^{-1}\mathbf{A^\top d}$ -As an example problem I am considering the model: -$\ T(x,z) = 0.5*sin(2\pi*x) - z$ -My observed data is the noisy partial derivatives of T: -$\ P(x,z)=T_x+e$ -$\ Q(x,z)=T_z+e$ - -By introducing matrix finite difference operators I can formulate these two pdes as linear equations: -$\ \mathbf{P=D_x*T} $ -$\ \mathbf{Q=D_z*T} $ -Below is the LSE solution of these equations: - -There is one big problem with this solution: -in the model the line where T=0 goes between z=-0.5 and z=0.5, -whereas in the estimate it goes between z=-1 and z=1. -Some "scaling" of the z coordinate seems to be taking place. -Why is this happening? -Other than this the solution is quite good except for the blocky pattern. -This pattern seems to be related to the sampling; the finer the sampling the finer the blocky pattern. -What is the explanation for this pattern? -In order to get a smoother solution I tested with a smoothing regularization constraint: -$\ a*\mathbf{L*T}=0$ -where a is a scaling parameter and L is the finite difference Laplacian matrix. -I noticed a trade-off between smoothness and correctness depending on the scaling parameter. For large scaling parameters the sine wave pattern of the model turns into a square wave pattern: - -How can I determine a suitable a? -Inverse problems are often troubled by instabilities which make it necessary to introduce a dampening regularization term. -This does not seem to be the case for this problem. -However since A is not square I could not find its condition number using cond() in Matlab. -Is there some way in Matlab to find if an LSE problem with a nonsquare matrix is unstable? -BTW Here is the Matlab code I used: -% number of samples in x and z direction: -n = 101; - -% amount of noise added to partial derivatives: -e = 0.5; - -x = linspace(-1,1,n); -z = linspace(-1,1,n); -dx = x(2) - x(1); -dz = z(2) - z(1); -[X,Z] = meshgrid(x,z); - -% Hidden model: -subplot(1,3,1); -T = 0.5*sin(2*pi*X) - Z; -imagesc(x,z,T); -xlabel('x'); -ylabel('z'); -title('T = 0.5*sin(2*pi*x) - z'); -colorbar; - -% Observed noisy data: -P = pi*cos(2*pi*X) + e*randn(n,n); % Tx -Q = -ones(n,n) + e*randn(n,n); % Tz - -subplot(1,3,2); -imagesc(x,z,P); -xlabel('x'); -ylabel('z'); -title('P = Tx + e'); -colorbar; -subplot(1,3,3); -imagesc(x,z,Q); -xlabel('x'); -ylabel('z'); -title('Q = Tz + e'); -colorbar; - -% Central difference matrices: -E = sparse(2:n, 1:n-1, 1, n, n); -D = (E' - E) / 2; -I = speye(n); -Dx = kron(D,I)/dx; -Dz = kron(I,D)/dz; - -A = []; b = []; -A = [A; Dx]; b = [b; P(:)]; -A = [A; Dz]; b = [b; Q(:)]; - -% Least Squares Solution: -figure; -subplot(1,2,1); -imagesc(x,z,T); -xlabel('x'); -ylabel('z'); -title('T'); -colorbar; - -tlse = lsqr(A, b, 1e-3, 1000*100); -Tlse = reshape(tlse, [n n]); -subplot(1,2,2); -imagesc(x,z,Tlse); -xlabel('x'); -ylabel('z'); -title('Tlse'); -colorbar; - -figure; -% Least Squares Solution regularized by smoothing Laplace operator: -D2 = E + E' + sparse(1:n, 1:n, -2, n, n); -Dxx = kron(D2,I)/(dx^2); -Dzz = kron(I,D2)/(dz^2); -L = Dxx + Dzz; -ns = 3; -si = 1; -for si = 1 : ns; - subplot(1,ns,si); - % regularization: - a = (si - 1)*5e-4; - Ar = [A; a*L]; br = [b; zeros(n^2,1)]; - tlse = lsqr(Ar, br, 1e-3, 1000*100); - Tlse = reshape(tlse, [n n]); - imagesc(x,z,Tlse); - str = sprintf('Tlse a=%g',a); - title(str); - si = si + 1; -end - -EDIT: -My central difference matrix D had an edge problem. As an example for size n=5: -was: -D = - - 0 0.5000 0 0 0 - -0.5000 0 0.5000 0 0 - 0 -0.5000 0 0.5000 0 - 0 0 -0.5000 0 0.5000 - 0 0 0 -0.5000 0 - -Rewrote this to -D = - - -1.0000 1.0000 0 0 0 - -0.5000 0 0.5000 0 0 - 0 -0.5000 0 0.5000 0 - 0 0 -0.5000 0 0.5000 - 0 0 0 -1.0000 1.0000 - -With more correct difference matrices I get much nicer results: - -And this is without any regularization. Sorry for the sloppiness. I learned a lot tough :-). Thanks for your answers! - -REPLY [4 votes]: I would also consider whether your approximations to the first derivatives aren't introducing any odd-even decoupling through the central differencing you use - switch the differencing out to a one-sided approximation to see if that improves your unregularised solution.<|endoftext|> -TITLE: Given a number $11 \leqslant n\leqslant 99$, how to write a couple of numbers which total to $n$ -QUESTION [5 upvotes]: Yesterday, a friend of mine asked me for a number between 11 and 99 (not 100% sure about the boundaries). I had no idea what he was up to and called 38, about half a minute later he had written down the following: -$$\begin{pmatrix} 18 & 1 & 12 & 7\\ 11&8&17&2\\5&10&3&20\\4&19&6&9 \end{pmatrix}$$ -At first, I thought it was already pretty impresive that all rows, columns and the two diagonals sum up to 38. Then he showed me that also the four 2x2 squares in the corners sum up to 38, and so do -$$\begin{pmatrix} 1 & 12\\ 8 & 17 \end{pmatrix}, \begin{pmatrix} 8 & 17 \\ 10 & 3 \end{pmatrix} \text{ and } \begin{pmatrix} 10 & 3\\ 19 & 6 \end{pmatrix}.$$ -Does this "thing" have a name? How does one produce such a matrix in such a short amount of time? - -REPLY [3 votes]: (A compilation of some comments...) -These are called "Magic Squares". -Wikipedia -Here's how to create a magic square for any number<|endoftext|> -TITLE: A question about a proof of Noetherian modules and exact sequences -QUESTION [13 upvotes]: I proved part (i) of the following: -Proposition 6.3. Let $0 \to M' \xrightarrow{\alpha} M \xrightarrow{\beta} M'' \to 0$ be an exact sequence of $A$-modules. Then -i) $M$ is Noetherian $\iff$ $M'$ and $M'"$ are Noetherian; -ii) $M$ is Artinian $\iff$ $M'$ and $M''$ are Artinian -Proof. We shall prove i); the proof of ii) is similar. -$\implies$: An ascending chain of submodules of $M'$ (or $M''$) gives rise to a chain in $M$, hence is stationary. -Can you tell me if my proof of $(i)\Longleftarrow$ is correct? Here goes: -Let $M^\prime$ and $M^{\prime \prime}$ be Noetherian. Let $L_n$ be an ascending chain of submodules in $M$. Then $\alpha^{-1}(L_n)$ is an ascending chain of submodules in $M^\prime$. Hence $\alpha^{-1}(L_n)$ is stationary, that is, $\alpha^{-1}(L_n) = \alpha^{-1}(L_{n+1})$ for $n$ large enough. Hence $L_n = L_{n+1}$ for $n$ large enough since $\alpha$ is injective hence $L_n$ is stationary. -The proof given in Atiyah-Macdonald is the following but I don't understand why they need both maps, $\alpha$ and $\beta$: -$\Longleftarrow$: Let $(L_n)_{n \ge 1}$ be an ascending chain of submodules of $M$; then $(\alpha^{-1}(L_n))$ is a chain in $M'$, and $(\beta(L_n))$ is a chain in $M''$. For large enough $n$ both these chains are stationary, and it follows that the chain $(L_n)$ is stationary. $\Box$ - -REPLY [19 votes]: If this proof worked then we would be able to conclude that a module is Noetherian if it has a Noetherian submodule. This is certainly not the case — as a silly example, the trivial submodule is always Noetherian. If we forget about $\beta$ then it is true that $(\alpha^{-1}(L_n))_n$ will stabilize, but it's not as if $\alpha(\alpha^{-1}(L_n)) = L_n$, since $\alpha$ need not be surjective. -To use Atiyah and MacDonald's proof, you want to show that if $N_1 \subset N_2$ are submodules of $M$ such that $\alpha^{-1}(N_1) = \alpha^{-1}(N_2)$ and $\beta(N_1) = \beta(N_2)$, then $N_1 = N_2$. For this, take $x \in N_2$. By assumption there is a $y \in N_1$ such that $\beta(y) = \beta(x)$. By exactness, then, there is a $z \in M'$ such that $\alpha(z) = x - y \in N_2$. Thus $z \in \alpha^{-1}(N_2) = \alpha^{-1}(N_1)$. Do you see how to conclude that $x \in N_1$? -[We do need that $N_1 \subset N_2$, by the way. Think about $M = \mathbb R^2$ with $M' = \text{the $x$-axis}$ and the family of lines $y = mx$ for $m \neq 0$.]<|endoftext|> -TITLE: What surfaces in $\mathbb R^3$ are such that every planar section (with more than 1 point) has nontrivial symmetry? -QUESTION [6 upvotes]: In $\mathbb R^3$ , the intersection of a plane and a sphere (e.g. $x^2 + y^2 + z^2 = 1$) is either empty, a single point, or a circle. All isometries of those circles are realized by isometries of the full sphere. In contrast, every plane intersects a cone (e.g. $x^2 + y^2 - z^2 = 0$) in a conic section which has reflection symmetries along one, two, or more axes. The one reflection is realized by an isometry of the cone, but the second, in general, is not. -What surfaces in $\mathbb R^3$ , or general subsets of $\mathbb R^3$ , are such that all non-empty planar intersections are either 1 point or have nontrivial symmetry? When are those symmetries not extensible to a symmetry of the full surface? - -REPLY [3 votes]: I suspect the answer is that the surface in $\mathbb R^3$ must be a homogeneous quadratic form in $(x,y,z)$ set equal to a constant, as in -$$ a x^2 + b y^2 + c z^2 + r y z + s z x + t x y = k. $$ -For you, all of $a,b,c,r,s,t,k$ are real numbers. There is little benefit to allowing lower order terms as in $\alpha x + \beta y + \gamma z,$ a translation takes the result to center on the origin again. For that matter, if you know enough linear algebra, a rotation takes this surface into a diagonalized one, -$$ a_1 x_1^2 + b_1 y_1^2 + c_1 z_1^2 = k_1, $$ where some of $a_1,b_1,c_1,k_1$ may be positive, some $0,$ some negative. You might try graphing these with all $a_1,b_1,c_1,k_1 \in \{-1,0,1 \},$ there are 16 possibilities but there is repetition, so maybe take $a_1 \leq b_1 \leq c_1$ but any $k_1.$ -Note that the intersection of any plane with this surface is, in coordinates appropriate for that plane, and orthogonal coordinates if we demand it, a quadratic in two variables $u,v$ call it $A u^2 + B u v + C v^2 + D u + Ev = F.$ This has symmetries, or may be a single point, or a line or pair of lines, and so on. -Well, these examples work. I do not expect there will be any others. For most planes through one of these surfaces, the symmetry of the figure within the plane will not extend to the whole surface. Finally, a proof that only these surfaces work would be pretty elaborate. -EDIT: user mjqxxxx gave some I had not considered, the one that is connected and new is: take any curve in the $xy$ plane that has a symmetry, a reflection or $180^\circ$ rotation or something. Then construct the cylinder over it, meaning take the same figure with arbitrary $z.$ Any plane slice preserves the symmetry. Now, my current opinion is that if we take something in the plane that has only a $120^\circ$ rotation symmetry, some kind of pinwheel, after making the cylinder most slanted planes through the cylinder will not preserve that symmetry. Need to thinnmmnnkk. That would be a nice result, though, connected examples are either full-dimensional quadratic things or cylinders over lower-dimensional examples. EDDITTT: more in comments below. This is getting out of hand. The second example that is essentially lower-dimensional is take any curve in the $xy$ plane with $y \geq 0$ and rotate it around the $x$-axis.<|endoftext|> -TITLE: does the uniform continuity of $f$ implies uniform continuity of $f^2$ on $\mathbb{R}$? -QUESTION [10 upvotes]: my question is if $f:\mathbb{R}\rightarrow\mathbb{R}$ is uniformly continuous, does it implies that $f^2$ is so?and in general even or odd power of that function? - -REPLY [9 votes]: By $f^2(x)$, do you mean $(f(x))^2$, or $f(f(x))$? (The notation $f^2(x)$ can mean both of these; which is standard depends on the area of mathematics you work in.) -If you mean $(f(x))^2$, then as other answers have pointed out, this may not be uniformly continuous. -However, if you mean $f(f(x))$, then yes, this is uniformly continuous whenever f is. -Proof. Given $\epsilon > 0$, we need some $\delta > 0$ such that whenever $|x - y| < \delta$, $|f(f(x)) - f(f(y))| < \epsilon$. -We know, by uniform continuity of $f$ applied to $\epsilon$, that there’s some $\epsilon_1$ such that whenever $|x - y| < \epsilon_1$, $|f(x) - f(y)| < \epsilon$. By the uniform continuity of $f$ again, applied to $\epsilon_1$, there’s $\delta$ such that whenever $|x - y| < \delta$, $|f(x) - f(y)| < \epsilon_1$. -So putting these together, if $|x - y| < \delta$, then $|f(x) - f(y)| < \epsilon_1$, and so $|f(f(x)) - f(f(y))| < \epsilon$; so $\delta$ is as required.  $\square$<|endoftext|> -TITLE: What exactly is a Haar measure -QUESTION [7 upvotes]: I've come across at least 3 definitions, for example: - -Taken from here where $\Gamma$ is a topological group. Apparently, this definition doesn't require the Haar measure to be finite on compact sets. -Or from Wikipedia: -"... In this article, the $\sigma$-algebra generated by all compact subsets of $G$ is called the Borel algebra...." -Then $\mu$ defined on this sigma algebra is a Haar measure if it's outer and inner regular, finite on compact sets and translation invariant. -So I gather the important property of a Haar measure is that it's translation invariant. -Question 1: What I don't gather is, what do I get if I define it on the Borel sigma algebra as opposed to defining it on the sigma algebra generated by compact sets (as they do on Wikipedia)? -Question 2: Can I put additional assumptions on $G$ so that I can drop the requirement that $\mu$ has to be finite on compact sets? -Question 3: As you can guess from my questions I'm poking around in the dark trying to find out how to define a Haar measure suitably. Here suitably means, I want to use it to define an inner product so I can have Fourier series. Are there several ways to do this which lead to different spaces? By this I mean, if I define it on the Borel sigma algebra, can I do Fourier series for a different set of functions than when I have a measure on the sigma algebra generated by compact sets? Or what about dropping regularity? Or dropping finiteness on compact sets? -Thanks. - -REPLY [5 votes]: Summarizing some comments, and continuing: the main point is that Haar measure is translation-invariant (and for non-abelian groups, in general, left-invariant and right-invariant are not identical, but the discrepancy is intelligible). -Unless you have intentions to do something exotic (say, on not-locally-compact, or not-Hausdorff "groups", which I can't recommend), you'll be happier later to have a regular measure, so, yes, the measure of a set is the inf of the measures of the opens containing it, and is the sup of the measures of the compacts contained in it, and, yes, the measure of a compact is finite. Probably you will also want completeness, especially when taking products, so subsets of measure-zero sets have measure zero. -Probably you'll want your groups to be countably-based, too, to avoid some measure-theoretic pathologies. -Then, for abelian topological groups (meaning locally compact, Hausdorff, probably countably-based), the basics of "Fourier series/transforms" work pretty well, as in Pontryagin and Weil. The non-abelian but compact case also turns out very well.<|endoftext|> -TITLE: Even numbers have more factors than odd numbers... -QUESTION [8 upvotes]: This was an exercise to show that, in a sense, the even numbers have more prime factors than the odds, but--if it's right-- I still have a question. -As an heuristic calculation, we could take a large interval (1, 2N) on which the average number of prime factors with repetitions is $\mu$ (for sufficiently large N, $\mu $ is about $ \ln \ln 2N$; see answer to this problem). WLOG N is even, then the set $S_2 = \{N+2, N+4, N+6, ..., 2N \} $ corresponds to a sequence $S_1 = \{\frac{N}{2}+1, \frac{N}{2}+ 2, ..., N \}$. The average number of primes in $S_1$ is only slightly less than $\mu$ for large N$^{(1)}$, and so the average number of primes $\mu_2$ in $ S_2$ is $\mu+1$ primes. Let $\mu_o$ be the average number of primes of odd numbers in $[N,2N]$. -Since on $[N,2N]$ the average number of primes is also about $\mu$, we have that $$\mu =\frac{( \mu_2 + \mu_o)}{2} = \frac{(\mu + 1 + \mu_o )}{2},$$ and so the average number of primes for the odd numbers $^{(2)} $ is $$\mu_o = \mu - 1 = \mu_2 - 2 $$ so -$$\mu_2 - \mu_o \approx 2.$$ -This argument has I think a somewhat complicated generalization. Using the same reasoning for multiples of $3, 5,...,p_k $ and so on, the net result would be an average number of primes $\mu_{p_k} $ for multiples of the set $P_k = \{ 2,3,5, ..., p_k \}$ with an increasing number of numbers that are multiples of more than one such prime, so that $\mu_{p_k} > \mu + 1$ and $\mu_{p_k}$ is an increasing function of N (or k ). So if we call $\mu_n$ the average number of primes for non-multiples of $P_k$, I expect that for large N the difference -$$\mu_{p_k} - \mu_n > 2 . $$ My question is whether we reach a point beyond which $$ \mu_{p_k} = \mu + \beta$$ with $\beta(N)>1$ and a,b constants of proportionality with $a > b$ , because multiples of $P_k$ take up more than half the interval, so that $$\mu = a\mu_{p_k} + b\mu_n = a(\mu + \beta) + b\mu_n$$ and $$\mu(1-a) = \mu b = a\beta + b\mu_n$$ and finally $$\mu_n = \mu -\frac{a\beta}{b} < 2.$$ -I am guessing so, but that the asymptotic relationships involved make it a somewhat weak assertion? -Hopefully the question is clear. Thanks. -(1) Because $\ln \ln 2N = \ln (\ln 2 + \ln N) \sim \ln \ln N.$ -(2) On [1, N] for N = 2,500,000, $\mu_2 - \mu_o$ is about 1.9. - -REPLY [5 votes]: If I understand the question correctly, it's mainly a matter of the order of limits. -There are two numbers going to infinity here, $N$ and $k$. If you keep $k$ fixed and let $N$ go to infinity, your argument goes through just as it did when you considered only $2$ as a factor, and as you say, asymptotically (for $N\to\infty$) the discrepancy between $\mu_{p_k}$ and $\mu_n$ will be greater than $2$; it increases with $k$ (if you let $k$ go to $\infty$ after $N$), and it does so without bound, as does the average number of prime factors itself. -However, that doesn't mean you can write $\beta(N)$, increase $k$ at finite $N$ and conclude from the growing discrepancy that the average number of prime factors for the remaining numbers will eventually fall below $2$; that would only happen if you increase $k$ to a point where $\log\log N$ begins to change significantly from the factors you're pulling out. -To calculate the discrepancy (for fixed $k$ and $N\to\infty)$, note that the reason we get a discrepancy of $2$ considering only the prime factor $2$, and not the $1$ that one might have expected, is that after pulling out a factor of $2$ from the even numbers they still have a chance of containing further factors of $2$, whereas the odd numbers don't. We can express this as -$$\mu=\mu_0+\frac12+\frac14+\frac18+\dotso=\mu_0+\frac1{2-1}=\mu_0+1\;,$$ -since all numbers on average have $\mu_0$ prime factors other than $2$, half of them half one factor of two, a quarter have two, and so on. We can do the same thing with the other primes up to $p_k$, and the "probabilities" for different primes will asymptotically be independent, so we get -$$\mu=\mu_n+\sum_{m=1}^k\frac1{p_m-1}\;.$$ -The sum diverges for $k\to\infty$, so you can make the discrepancy arbitrarily large; but you then have to go to very high $N$ to actually see it, since you're pulling out more and more factors.<|endoftext|> -TITLE: Compute the trigonometric integrals -QUESTION [7 upvotes]: How would you compute the following integrals? -$$ I_{n} = \int_0^\pi \frac{1-\cos nx}{1- \cos x} dx $$ -$$ J_{n,m} = \int_0^\pi \frac{x^m(1-\cos nx)}{1- \cos x} dx $$ -For instance, i noticed that the first integral is convergent for any value of $n$ since $\lim_{x\to0} \frac{1- \cos nx}{1 - \cos x}= n^2$. This fact may also be extended to the second integral, as well. - -REPLY [3 votes]: Another generating function approach is to set $z=e^{ix}$ in -$$ -\begin{align} -&\sum_{k=0}^\infty\frac{(it)^m}{m!}\int_0^{\pi}x^m\frac{1-\cos(nx)}{1-\cos(x)}\,\mathrm{d}x\\ -&=\int_0^{\pi}z^t\left(\frac{z^n-1}{z-1}\right)^2\frac{\mathrm{d}z}{iz^n}\\ -&=\frac1i\int_0^{\pi}\left(\sum_{k=1-n}^{n-1}(n-|k|)z^{k+t-1}\right)\mathrm{d}z\\ -&=\frac1i\sum_{k=1-n}^{n-1}(n-|k|)\frac{(-1)^ke^{\pi it}-1}{k+t}\\ -&=in\frac{1-e^{\pi it}}{t}-2it\sum_{k=1}^{n-1}\frac{n-k}{k^2-t^2}\left(1-(-1)^ke^{\pi it}\right)\tag{1} -\end{align} -$$ -As $t\to0$, it is easy to see that we get $\pi n$.<|endoftext|> -TITLE: Connected nice topological spaces that are transitive but not 2-transitive under homeomorphisms -QUESTION [6 upvotes]: I was thinking about a question on here earlier, and came up with this question. -[Added Hausdorff note, below.] -It is easy to see that the group of self-homeomorphisms of the real line acts $2$-transitively on the space, but not $3$-transitively. -Likewise, it is clear that, for example, $\mathbb R\setminus \{0\}$ is a space on which the group of self-homemorphisms is $1$-transitive but not $2$-transitive. -If you take the real line with the open sets of the form $(-\infty,a)$ for some $a$, then I guess this gives you an example, but that space is not Hausdorff. -I can't seem to think of an example of a connected Hausdorff space where the self-homeomorphisms are $1$-transitive but not $2$-transitive. - -REPLY [5 votes]: Let $X=\big(\omega_1\times[0,1)\big)\setminus\{\langle 0,0\rangle\}$ ordered lexicographically. Clearly $X$ is connected and Hausdorff; indeed, $X$ is even hereditarily collectionwise normal. For each $x\in X$ the set $(\leftarrow,x)$ is homeomorphic to $\Bbb R$ with the Euclidean topology, so $X$ is transitive under autohomeomorphisms, but it’s clear that no autohomeomorphism of $X$ can reverse the order of a pair of points: that would require embedding $\omega_1^*$ in an initial segment of $X$.<|endoftext|> -TITLE: Necessary and sufficient conditions for a matrix to be a valid correlation matrix. -QUESTION [10 upvotes]: It's not too hard to see that any correlation matrix must have certain properties, such as all entries in the range -1 to 1, symmetric, positive semi-definite (excluding pathological cases like singular matrices for the moment). -But I'm wondering about the other direction. If I write down some matrix that is positive semi-definite, is symmetric, has 1's along the main diagonal, and all entries are between -1 and 1, does this guarantee that there exists a set of random variables giving rise to that correlation matrix? -If the answer is easy to see, a helpful hint about how to define the set of random variables that would give rise to a given matrix would be appreciated. - -REPLY [11 votes]: Yes, although the restriction that all entries are between $-1$ and $1$ is implied by the other properties (and so is not needed). -Let $\Sigma$ be a $n \times n$, symmetric, positive semidefinite matrix with $1$'s along the main diagonal. -First, $\Sigma$ is a covariance matrix. Since $\Sigma$ is symmetric and positive semidefinite, $\Sigma$ has a nonnegative symmetric square root $\Sigma^{1/2}$. Let $X$ be a $n$-vector of independent random variables, each with variance $1$. (For example, $X$ could be an $n$-vector of independent $N(0,1)$ random variables.) Construct the $n$-vector $\Sigma^{1/2} X$. Then, by properties of covariance matrices, $$\text{cov} (\Sigma^{1/2} X) = \Sigma^{1/2} \text{cov}(X) \Sigma^{1/2} = \Sigma^{1/2} I \Sigma^{1/2} = \Sigma.$$ Thus $\Sigma$ is the covariance matrix for the random vector $\Sigma^{1/2} X$. (This derivation is on the Wikipedia article for covariance matrices.) -Second, since $\Sigma$ has $1$'s on its diagonal, the standard deviation of each random variable in $\Sigma^{1/2} X$ is $1$. Thus the correlation matrix $R$ for the random vector $\Sigma^{1/2} X$ is $$R = I^{-1} \Sigma I^{-1} = \Sigma.$$ Thus $\Sigma$ is a correlation matrix as well.<|endoftext|> -TITLE: Loopspace of Eilenberg Mac Lane space -QUESTION [5 upvotes]: Is the loop space of the Eilenberg-MacLane space $K(G,1)$ dependent only on the cardinality of $G$? For instance, is the loop space of $K(\mathbb{Z}_4, 1)$ homotopy equivalent to that of $K(\mathbb{Z}_2 \times \mathbb{Z}_2, 1)$? - -REPLY [3 votes]: I think it's important to note that loop spaces have more structure than just their topology. Concantenation makes them into H-spaces ($A_\infty$ spaces), and if you use the Moore loop space, we can even make them into strictly associative monoids. J.P. May's book "The Geometry of Iterated Loop Spaces" basically says that every associative monoid ($A_\infty$ space) is equivalent as an associative monoid ($A_\infty$ space) to a loop space. Thus, the homotopy theory of loop spaces is essentially the same as the homotopy theory of associative monoids ($A_\infty$ spaces). Thus, when considering loop spaces we ought to consider the monoid structure as well. The monoids $G$ and $\Omega B G$ are weakly equivalent (here $BG = K(G, 1)$). So, $\Omega B G$ is really $G$ as a monoid, and $\Omega B\mathbb Z /4$ and $\Omega B(\mathbb Z /2 \oplus \mathbb Z/2)$ are not equivalent as monoids. -Edit: The key thing to note about maps of monoids, is that they induce maps of monoids on $\pi_0$, which any old continuous map will not do. If $\Omega B\mathbb Z /4$ and $\Omega B(\mathbb Z /2 \oplus \mathbb Z/2)$ were equivalent as monoids, then on $\pi_0$ we would get an isomorphism of monoids $\mathbb Z /4 \cong \mathbb Z/2 \oplus \mathbb Z/2$.<|endoftext|> -TITLE: Intersection of Simply-Connected Sets -QUESTION [5 upvotes]: Let $U,V$ be two simply connected subsets of a topological space. -Prove or disprove: -$U \cap V$ is simply connected. - -REPLY [8 votes]: Let $S^1$ be the circle in $\mathbb R^2$, $U=\{(x,y)\in S^1: x\geq 0\}$ and $V=\{(x,y)\in S^1: x\leq 0\}$. Then $U$ is the right half of a circle and $V$ is the left half, both of which are simply connected. What is $U\cap V$?<|endoftext|> -TITLE: Proving $\sum\limits_{l=1}^n \sum\limits _{k=1}^{n-1}\tan \frac {lk\pi }{2n+1}\tan \frac {l(k+1)\pi }{2n+1}=0$ -QUESTION [50 upvotes]: Prove that $$\sum _{l=1}^{n}\sum _{k=1}^{n-1}\tan \frac {lk\pi } {2n+1}\tan \frac {l( k+1) \pi } {2n+1}=0$$ - - -It is very easy to prove this identity for each fixed $n$ . For example let $n = 6$; writing out all terms in a $5 \times 6$ matrix, we obtain: -$\begin{matrix} -\tan \dfrac {\pi } {13}\tan \dfrac {2\pi } {13} - & -\tan \dfrac {2\pi } {13}\tan \dfrac {3\pi } {13} -& -\tan \dfrac {3\pi } {13}\tan \dfrac {4\pi } {13} -& -\tan \dfrac {4\pi } {13}\tan \dfrac {5\pi } {13} -& -\tan \dfrac {5\pi } {13}\tan \dfrac {6\pi } {13} - \\ -\tan \dfrac {2\pi } {13}\tan \dfrac {4\pi } {13} - & - \tan \dfrac {4\pi } {13}\tan \dfrac {6\pi } {13} -& -\tan \dfrac {6\pi } {13}\tan \dfrac {8\pi } {13} -& -\tan \dfrac {8\pi } {13}\tan \dfrac {10\pi } {13} -& -\tan \dfrac {10\pi } {13}\tan \dfrac {12\pi } {13} -\\ -\tan \dfrac {3\pi } {13}\tan \dfrac {6\pi } {13} -& -\tan \dfrac {6\pi } {13}\tan \dfrac {9\pi } {13} -& -\tan \dfrac {9\pi } {13}\tan \dfrac {12\pi } {13} -& -\tan \dfrac {12\pi } {13}\tan \dfrac {15\pi } {13} -& -\tan \dfrac {15\pi } {13}\tan \dfrac {18\pi } {13} -\\ -\tan \dfrac {4\pi } {13}\tan \dfrac {8\pi } {13} -& -\tan \dfrac {8\pi } {13}\tan \dfrac {12\pi } {13} -& -\tan \dfrac {12\pi } {13}\tan \dfrac {16\pi } {13} -& -\tan \dfrac {16\pi } {13}\tan \dfrac {20\pi } {13} -& -\tan \dfrac {20\pi } {13}\tan \dfrac {24\pi } {13} -\\ -\tan \dfrac {5\pi } {13}\tan \dfrac {10\pi } {13} -& -\tan \dfrac {10\pi } {13}\tan \dfrac {15\pi } {13} -& -\tan \dfrac {15\pi } {13}\tan \dfrac {20\pi } {13} -& -\tan \dfrac {20\pi } {13}\tan \dfrac {25\pi } {13} -& -\tan \dfrac {25\pi } {13}\tan \dfrac {30\pi } {13} -\\ -\tan \dfrac {6\pi } {13}\tan \dfrac {12\pi } {13} -& -\tan \dfrac {12\pi } {13}\tan \dfrac {18\pi } {13} -& -\tan \dfrac {18\pi } {13}\tan \dfrac {24\pi } {13} -& -\tan \dfrac {24\pi } {13}\tan \dfrac {30\pi } {13} -& -\tan \dfrac {30\pi } {13}\tan \dfrac {36\pi } {13} - \end{matrix}$ -one can notice then, that the first column vanish the fourth one : -$\tan \dfrac {\pi } {13}\tan \dfrac {2\pi } {13}=-\tan \dfrac {12\pi } {13}\tan \dfrac {15\pi } {13}$ -$\tan \dfrac {2\pi } {13}\tan \dfrac {4\pi } {13}=-\tan \dfrac {24\pi } {13}\tan \dfrac {30\pi } {13}$ -$\tan \dfrac {3\pi } {13}\tan \dfrac {6\pi } {13}=-\tan \dfrac {16\pi } {13}\tan \dfrac {20\pi } {13}$ -$\tan \dfrac {4\pi } {13}\tan \dfrac {8\pi } {13}=-\tan \dfrac {4\pi } {13}\tan \dfrac {5\pi } {13}$ -$\tan \dfrac {5\pi } {13}\tan \dfrac {10\pi } {13}=-\tan \dfrac {8\pi } {13}\tan \dfrac {10\pi } {13}$ -$\tan \dfrac {6\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {20\pi } {13}\tan \dfrac {25\pi } {13}$ -and the third column vanish the fifth one : -$\tan \dfrac {3\pi } {13}\tan \dfrac {4\pi } {13}=-\tan \dfrac {30\pi } {13}\tan \dfrac {36\pi } {13}$ -$\tan \dfrac {6\pi } {13}\tan \dfrac {8\pi } {13}=-\tan \dfrac {5\pi } {13}\tan \dfrac {6\pi } {13}$ -$\tan \dfrac {9\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {25\pi } {13}\tan \dfrac {30\pi } {13}$ -$\tan \dfrac {12\pi } {13}\tan \dfrac {16\pi } {13}=-\tan \dfrac {10\pi } {13}\tan \dfrac {12\pi } {13}$ -$\tan \dfrac {15\pi } {13}\tan \dfrac {20\pi } {13}=-\tan \dfrac {20\pi } {13}\tan \dfrac {24\pi } {13}$ -$\tan \dfrac {18\pi } {13}\tan \dfrac {24\pi } {13}=-\tan \dfrac {15\pi } {13}\tan \dfrac {18\pi } {13}$ -while the second column is self-vanishing: -$\tan \dfrac {2\pi } {13}\tan \dfrac {3\pi } {13}=-\tan \dfrac {10\pi } {13}\tan \dfrac {15\pi } {13}$ -$\tan \dfrac {4\pi } {13}\tan \dfrac {6\pi } {13}=-\tan \dfrac {6\pi } {13}\tan \dfrac {9\pi } {13}$ -$\tan \dfrac {8\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {12\pi } {13}\tan \dfrac {18\pi } {13}$ . -So the equality occurs. -But how to generalize the proof? - -REPLY [24 votes]: Starting with -$$ -\tan(x-y) = \frac{\tan(x)-\tan(y)}{1+\tan(x)\tan(y)}\tag{1} -$$ -we get -$$ -\tan(x)\tan(y)=\frac{\tan(x)-\tan(y)}{\tan(x-y)}-1\tag{2} -$$ -Thus, -$$ -\tan\left(\frac{l(k+1)\pi}{2n+1}\right)\tan\left(\frac{lk\pi}{2n+1}\right) -=\frac{\tan\left(\frac{l(k+1)\pi}{2n+1}\right)-\tan\left(\frac{lk\pi}{2n+1}\right)}{\tan\left(\frac{l\pi}{2n+1}\right)}-1\tag{3} -$$ -Therefore, because of the telescoping sum, -$$ -\begin{align} -\sum_{k=1}^{n-1}\tan\left(\frac{l(k+1)\pi}{2n+1}\right)\tan\left(\frac{lk\pi}{2n+1}\right) -&=\frac{\tan\left(\frac{ln\pi}{2n+1}\right)-\tan\left(\frac{l\pi}{2n+1}\right)}{\tan\left(\frac{l\pi}{2n+1}\right)}-(n-1)\\ -&=\frac{\tan\left(\frac{ln\pi}{2n+1}\right)}{\tan\left(\frac{l\pi}{2n+1}\right)}-n\tag{4} -\end{align} -$$ -Note that -$$ -\frac{\tan\left(\frac{(2n+1-l)n\pi}{2n+1}\right)}{\tan\left(\frac{(2n+1-l)\pi}{2n+1}\right)} -=\frac{\tan\left(\frac{ln\pi}{2n+1}\right)}{\tan\left(\frac{l\pi}{2n+1}\right)}\tag{5} -$$ -so that by replacing the odd $l$s with even $2n+1-l$s and using $\tan(2x)=\frac{2\tan(x)}{1-\tan^2(x)}$, we get -$$ -\begin{align} -\sum_{l=1}^n\frac{\tan\left(\frac{ln\pi}{2n+1}\right)}{\tan\left(\frac{l\pi}{2n+1}\right)} -&=\sum_{l=1}^n\frac{\tan\left(\frac{2ln\pi}{2n+1}\right)}{\tan\left(\frac{2l\pi}{2n+1}\right)}\\ -&=-\sum_{l=1}^n\frac{\tan\left(\frac{l\pi}{2n+1}\right)}{\tan\left(\frac{2l\pi}{2n+1}\right)}\\ -&=\frac12\sum_{l=1}^n\left(\tan^2\left(\frac{l\pi}{2n+1}\right)-1\right)\tag{6} -\end{align} -$$ - -Using contour integration, we will compute -$$ -\sum_{l=1}^n\tan^2\left(\frac{\pi l}{2n+1}\right)=n(2n+1)\tag{7} -$$ -Note that -$$ -\frac{(2n+1)/z}{z^{2n+1}-1}\tag{8} -$$ -has simple poles. It has residue $1$ at $z=e^{\large\frac{2\pi li}{2n+1}}$ for each $l$ and residue $-(2n+1)$ at $z=0$. -Furthermore, at $z=e^{i\theta}$, -$$ --\left(\dfrac{z-1}{z+1}\right)^2=\tan^2(\theta/2)\tag{9} -$$ -Because the total residue of -$$ -f(z)=\left(\frac{z-1}{z+1}\right)^2\frac{(2n+1)/z}{z^{2n+1}-1}\tag{10} -$$ -is $0$, we get that the sum of its residues at $z=0$ and $z=-1$ equals -$$ -\sum_{l=1}^{2n}\tan^2\left(\frac{\pi l}{2n+1}\right)=2\sum_{l=1}^n\tan^2\left(\frac{\pi l}{2n+1}\right)\tag{11} -$$ -First, -$$ -\mathrm{Res}_{z=0}f(z)=-(2n+1)\tag{12} -$$ -Next, -$$ -\begin{align} -\mathrm{Res}_{z=-1}f(z) -&=\mathrm{Res}_{z=0}f(z-1)\\ -&=\mathrm{Res}_{z=0}(2n+1)\left(\frac{z-2}{z}\right)^2\frac1{1-z}\frac1{1+(1-z)^{2n+1}}\\ -&=\mathrm{Res}_{z=0}(2n+1)\left(1-\frac4z+\frac4{z^2}\right)(1+z+\dots)\frac12\left(1+\frac{2n+1}{2}z+\dots\right)\\ -&=\mathrm{Res}_{z=0}\frac{2n+1}{2}\left(1-\frac4z+\frac4{z^2}\right)\left(1+\frac{2n+3}{2}z+\dots\right)\\ -&=\mathrm{Res}_{z=0}\frac{4n+2}{z^2}+\frac{(2n+1)^2}{z}+\dots\\ -&=(2n+1)^2\tag{13} -\end{align} -$$ -Combining $(11)$, $(12)$, and $(13)$, yields $(7)$. - -Combining $(4)$, $(6)$, and $(7)$ yields -$$ -\sum_{k=1}^{n-1}\tan\left(\frac{l(k+1)\pi}{2n+1}\right)\tan\left(\frac{lk\pi}{2n+1}\right)=0\tag{14} -$$<|endoftext|> -TITLE: Automorphism group of a formal power series ring -QUESTION [9 upvotes]: Let $A$ be a commutative ring. -Let $A[[x]]$ be the ring of formal power series in one variable. -Can we determine the structure of the automorphism group of $A[[x]]$ over $A$? -This is a related question. - -REPLY [8 votes]: A continuous automorphism $\psi$ of $A[[x]]$ is determined by what it does to $x$ and what it does to the elements of $A$. If $\psi$ leaves elements of $A$ fixed, as I believe you are specifying, then the only question is which elements of $A$ may occur as $x^\psi$ (I write the image of $x$ under $\psi$ this way to avoid later confusion). Since $x$ itself is “analytically nilpotent”, by which I mean that its powers converge to $0$, continuity implies that $x^\psi$ must be a.n. as well. As you know from the comments to your recent post, any such series $u(x)=\sum\alpha_ix^i\in A[[x]]$ will definitely determine a homomorphism from $A[[x]]$ to itself. -Then two questions remain: precisely which are the analytically nilpotent elements of $A[[x]]$, and precisely what characterizes invertible series (to make $\psi$ bijective). If $A$ has discrete topology, then the set of a.n. elements of $A[[x]]$ is just $xA[[x]]$, but for a general topological ring, I’m not at all sure what the answer might be. But suppose $A$ is complete under a topology given by the powers of some ideal $I$ for which $\bigcap_n I^n=(0)$. Then when we call $\sqrt I$ the radical of $I$, that is the set of all elements of $A$ for which some positive power lands in $I$ (so that $\sqrt I$ is an ideal of $A$), then $\sqrt{I}+xA[[x]]$ is the set of all a.n. elements of $A[[x]]$. -What about invertibility of a series $u(x)=\sum_0^\infty\alpha_ix^i$? You’ve already seen that if $A$ has discrete topology and $\alpha_0=0$, $u$ is invertible if and only if $\alpha_1$ has a reciprocal in $A$. Again, in the general case of a topological ring, I don’t know the answer, but in the interesting case that $A$ is complete under the $I$-adic topology mentioned before, all that’s necessary again is that $\alpha_1$ should have reciprocal in $A$. The only proof of this that I know involves Weierstrass Preparation, but I don’t doubt that others have more direct arguments. -As an example of a nontrivial, maybe even nonobvious automorphism of the most general type, consider the $\psi$ for which $x^\psi=u(x)=p -x + x^p\in{\mathbb{Z}}_p[[x]]$. The constant ring is complete under the $(p)$-adic topology, so everything I said before applies. The issue of what kind of noncontinuous automorphisms of $A[[x]]$ there might be, I leave to you.<|endoftext|> -TITLE: Simple group of order $660$ is isomorphic to a subgroup of $A_{12}$ -QUESTION [13 upvotes]: Prove that the simple group of order $660$ is isomorphic to a subgroup of the alternating group of degree $12$. - -I have managed to show that it must be isomorphic to a subgroup of $S_{12}$ (through a group action on the set of Sylow $11$-subgroups). Any suggestions are appreciated! - -REPLY [7 votes]: Let $G$ be our simple group of order $660$, contained in $S_{12}$. Since $A_{12}$ is normal in $S_{12}$, the subgroup $A_{12} \cap G$ is normal in $G$. Since $G$ is simple and any subgroup of $S_{12}$ with more than two elements contains an even permutation, we get $A_{12} \cap G = G$. -To generalize, suppose $G$ is a simple group with $|G| > 2$. With the same idea as before, you can show that if $G$ is isomorphic to a subgroup $H$ of $S_n$, then $H$ is a subgroup of $A_n$.<|endoftext|> -TITLE: show that if $a | c$ and $b | c$, then $ab | c$ when $a$ is coprime to $b$. -QUESTION [5 upvotes]: Given two numbers $a$ and $b$, where $a$ is co-prime to $b$, - -Show that for any number $c$, if $a|c$ and $b | c$ then $ab| c$. -Is the reverse also true? In other words, if $ab |c$ and $a$ is co-prime to $b$, then do we have $a | c$ as well as $b|c$? - -REPLY [11 votes]: The second property holds, but the assumption that $\gcd(a,b)=1$ is superfluous. In general, if $x|y$ and $y|z$, then $x|z$. Since $a|ab$ and $b|ab$, then $ab|c$ implies $a|c$ and $b|c$, without any conditions on $\gcd(a,b)$. -On the other hand, $\gcd(a,b)=1$ is required for the first. -There are a number of ways of doing the proof. One is to use the Bezout identity: for any integers $a$ and $b$, there exist integers $x$ and $y$ such that $\gcd(a,b)=ax+by$. If $\gcd(a,b)=1$, then we can write $1 = ax+by$. Multiplying through by $c$, we get $c=axc + byc$. Since $a|c$, we can write $c=ak$; and since $b|c$, we can write $c=b\ell$. So we have -$$c = axc+byc = ax(b\ell) + by(ak) = ab(x\ell + yk),$$ -so $ab|c$. -Another is to use the following property: -Lemma. If $a|xy$ and $\gcd(a,x)=1$, then $a|y$. -This can be done using the Bezout identity, but here is a proof that avoids it and only uses the properties of the gcd (so it is valid in a larger class of rings than Bezout rings): -$$\begin{align*} -r|\gcd(a,xy) & \iff -r|a,xy\\ - &\iff r|ay, xy,a\\ -&\iff r|\gcd(ay,xy),a\\ -&\iff r|y\gcd(a,x),a\\ -&\iff r|y,a\\ -&\iff r|\gcd(a,y) -\end{align*}$$ -In particular, $a=\gcd(a,xy)$ divides $\gcd(a,y)$, which divides $y$, hence $a|y$. -With that Lemma on hand, we obtain that the result you want as follows: If $a|c$, then $c=ak$ for some $k$. Then $b|ak$, $\gcd(a,b)=1$. so $b|k$. Hence $k=b\ell$; thus, $c=ak=ab\ell$, so $ab|c$. - -REPLY [7 votes]: $\rm\, a\:|\:c\:\Rightarrow\:ab\:|\:bc\:\Rightarrow\: \color{#0A0}{\bf a}\:|\:\color{blue}{\bf b}\,(c/b)\color{#C00}\Rightarrow\, a\:|\:c/b\:\Rightarrow\:ab\:|\:c\ \, $ by $\rm\,\ (\color{#0A0}{\bf a},\color{blue}{\bf b})=1\,$ and $\rm\color{#C00}{Euclid's\ Lemma}$ -The converse is easy: $\rm\:a\:|\:ab\:|\:c\:\Rightarrow\: a\:|\:c\:$ by transitivity of "divides".<|endoftext|> -TITLE: Showing that the universal enveloping algebra of some $\mathfrak{g}$ is isomorphic to $\mathbb{C}[x_i,\partial/\partial x_i]$ -QUESTION [5 upvotes]: The universal enveloping algebra $U(\mathfrak{g})$ of a Lie algebra $\mathfrak{g}$ over $\mathbb{C}$ is defined to be -$$ -\dfrac{\mathbb{C}\oplus\mathfrak{g}\oplus ( \mathfrak{g}\otimes \mathfrak{g})\oplus (\mathfrak{g}\otimes \mathfrak{g}\otimes \mathfrak{g})\oplus\ldots}{\langle a\otimes b-b\otimes a -[a,b]\rangle}, -$$which means it is an associative tensor algebra generated by the vector space $\mathfrak{g}$ mod out all elements of the form $ a\otimes b-b\otimes a -[a,b]$, where $a,b\in \mathfrak{g}$. -Suppose $\mathfrak{g}=\mathbb{C}[e,p,q]/\langle [p,q]=e,[p,e]=0,[q,e]=0 -\rangle$. Then we have a (ring) isomorphism -$$ -U(\mathfrak{g})/\langle e-1\rangle \cong \mathbb{C}[x, \partial/\partial x] -$$ where we can define the map to be $U(\mathfrak{g})\stackrel{\phi}{\rightarrow}\mathbb{C}[x,\partial/\partial x]$ -by sending -$$ -p \mapsto \partial/\partial x -\mbox{ and } -q \mapsto x. -$$ -One can check that since -$$ -\phi([p,q])=[\partial/\partial x,x]= 1-x\dfrac{\partial}{\partial x}, -$$ we take $e=x\dfrac{\partial}{\partial x}$, which gives us the isomorphism $U(\mathfrak{g})/\langle 1-e\rangle \cong \mathbb{C}[x,\partial/\partial x]$. - -Now what should $\mathfrak{g}$ be so that $U(\mathfrak{g})/I\cong \mathbb{C}[x_1,x_2,\ldots, x_n,\partial /\partial x_1,\partial /\partial x_2, \ldots, \partial /\partial x_n]$? - -I could be wrong but I'm guessing that if we take $\mathfrak{g}$ to be the algebra $\mathbb{C}[p_i,q_i,e_i]/I$ where $I$ is generated by brackets of the form -$$ -[p_i,q_j]= \left\{ \begin{aligned} -e_i &\mbox{ if } i=j\\ --q_j p_i &\mbox{ if } i\not= j \\ -\end{aligned} -\right. -$$ -$$ -[p_i,e_j]= \left\{ \begin{aligned} -0 &\mbox{ if } i=j\\ -- p_i &\mbox{ if } i\not= j \\ -\end{aligned} -\right. -$$ -$$ -[q_i,e_j]= \left\{ \begin{aligned} -0 &\mbox{ if } i=j\\ -q_i &\mbox{ if } i\not= j, \\ -\end{aligned} -\right. -$$ -then $U(\mathfrak{g})/\langle e_i-1\rangle\cong \mathbb{C}[x_1,x_2,\ldots, x_n,\partial /\partial x_1,\partial /\partial x_2, \ldots, \partial /\partial x_n]$ where we take the map to be -$$ -p+\langle e_i-1\rangle \mapsto \partial/\partial x_i -$$ -and -$$ -q+\langle e_i-1\rangle \mapsto x_i. -$$ - -Also, since the order of multiplication matters, i.e., - $$ -(\overbrace{\partial/\partial x_1 + \partial/\partial x_2}^{\deg -1?})(\overbrace{x_1}^{\deg 1}) = 1+0=\overbrace{1}^{\deg 0} -$$ while - $$ -x_1(\partial/\partial x_1 + \partial/\partial x_2) = x_1 \partial /\partial x_1 + x_1\partial/\partial x_2, -$$ how should one think about multiplication in the skew polynomial algebra? - -Finally, I just noticed that $x_i\dfrac{\partial}{\partial x_i}$ form mutually orthogonal idempotent elements in $\mathbb{C}[x_i,\partial/\partial x_i]$. That is, -$$ -\left(\sum_i x_i \dfrac{\partial}{\partial x_i}\right)^2 = \left(\sum_i x_i \dfrac{\partial}{\partial x_i}\right).$$ I am guessing that these are the only non-scalar idempotent elements in the algebra. - -Do these idempotent elements have a geometric interpretation? - -REPLY [4 votes]: $\mathfrak{g}$ should be a Heisenberg Lie algebra. A nice coordinate-invariant way to describe these is as $V \oplus \mathbb{C}e$ where $V$ is a symplectic vector space and the symplectic form gives the Lie bracket, which takes values in $\mathbb{C}e$ (which is central). This is precisely the Lie algebra generated by the operators $x_1, x_2, ... x_n, \frac{\partial}{\partial x_1}, ... \frac{\partial}{\partial x_n}$ acting on $\mathbb{C}[x_1, ... x_n]$. -I do not believe that the elements you wrote down are idempotent.<|endoftext|> -TITLE: The minimum value of $(\frac{1}{x}-1)(\frac{1}{y}-1)(\frac{1}{z}-1)$ if $x+y+z=1$ -QUESTION [5 upvotes]: $x, y, z$ are three distinct positive reals such that $x+y+z=1$, then the minimum possible value of $(\frac{1}{x}-1) (\frac{1}{y}-1) (\frac{1}{z}-1)$ is ? -The options are: $1,4,8$ or $16$ -Approach: $$\begin{align*} -\left(\frac{1}{x} -1\right)\left(\frac{1}{y}-1\right)\left(\frac{1}{z}-1\right)&=\frac{(1-x)(1-y)(1-z)}{xyz}\\ -&=\frac{1-(x+y+z)+(xy+yz+zx)-xyz}{xyz}\\ -&=\frac{1-1+(xy+yz+zx)-xyz}{xyz}\\ -&=\frac{xy+yz+zx}{xyz} - 1 -\end{align*}$$ -Now by applying $AM≥HM$, I got the least value of $(xy+yz+zx)/xyz$ as $9$, so I got final answer as $8$. Is it correct? - -REPLY [6 votes]: If we put no constraint on $x$, $y$, and $z$ apart from $x$, $y$, $z$ positive and $x+y+z=1$, then indeed your calculation, and the one by Patrick Da Silva, show that the minimum value is $8$, attained at $x=y=z=\frac{1}{3}$. -However, the problem specifies that $x$, $y$ and $z$ are distinct real numbers. If we take that constraint into account, there is no minimum. We can get arbitrarily close to $8$ (but above $8$) by choosing $x$, $y$, and $z$ distinct and close to $\frac{1}{3}$, but we cannot attain $8$ with distinct $x$, $y$ and $z$. - -REPLY [2 votes]: If you look at the statement of the AM-HM inequality correctly, you'd see that -$$ -\frac 1x + \frac 1y + \frac 1z \ge 9 -$$ -and equality only happens when $x=y=z$. Therefore we can assume equality happens to find the minimum value, but $x+y+z = 1$ implies $x=y=z=1/3$. Therefore the minimum is $8$ and is attained uniquely at the point $(1/3, 1/3, 1/3)$. Does that clarify the doubts you had? -Hope that helps,<|endoftext|> -TITLE: Examples of stable curves $g\geq 2$? -QUESTION [6 upvotes]: I'd like to get my hands on some simple examples of families of stable curves. Ideally these would come in the form of a projective curve $C$ over a 1 dimensional base $B$, say $B = \mathbb{A}^1$. The generic fiber would be smooth and the special fiber would be nodal. -For genus 0 there is the nice example of the closure in $\mathbb{P}^2_{\mathbb{A}^1}$ of Spec $k[x,y,t]/(xy - t)$ where $t$ is the coordinate on $\mathbb{A}^1$. -For genus 1 there is the closure in $\mathbb{P}^2_{\mathbb{A}^1}$ of Spec $k[x,y,t]/(y^2 = x(x-t)(x+1))$. -I know for genus $2$ you can't expect to have an example in $\mathbb{P}^2_{\mathbb{A}^1}$ but what about something in $\mathbb{P}^3_{\mathbb{A}^1}$? -Whenever $g = (d-1)(d-2)/2$ I would think you can get an example in $\mathbb{P}^2_{\mathbb{A}^1}$, is there anyway to control the number of nodes you get in the special fiber? - -REPLY [3 votes]: For curves of genus 2 (or more generally hyperelliptic curves), it is easier to describe them as normalization of $\mathbb P^1$ in a quadratic extension than writting explictely equations in $\mathbb{P}^3$. Let $g\ge 2$. If $X$ is the normalization of $\mathbb P^{1}_{k[t]}$ (with $\mathrm{char}(k)\ne 2$) in the extension -$$L:=k(t)(x)[y]/(y^2-(x^2-t)\prod_{1\le i\le 2g}(x-\lambda_i)) $$ -where the $\lambda_i\in k^*$ are pairwise distinct, then $X$ is stable, with only one singular fiber (above $t=0$). It can be obtained by glueing two affine open subsets -$$ \mathrm{Spec} (k[t][x,y]), \quad \mathrm{Spec} (k[t][1/x, y/x^{g+1}]). $$ -In the first chart, $x,y$ are related by the above equation. In the second one, $1/x, y/x^{g+1}$ satisfy the relation -$$ (y/x^{g+1})^2=(1-t(1/x)^2)\prod_{1\le i\le 2g}(1-\lambda_i (1/x)).$$ -They are respectively the normalization of $\mathrm{Spec}(k[t][x])$ and $\mathrm{Spec}(k[t][1/x])$ in $L$. -For plane curves of degree $d\ge 3$ (in $\mathrm{char}(k)=0$ or $>d$), take the equation -$$ (y^2-x^2)(x^{d-2}+y^{d-2}+z^{d-2})+t(x^{d}+y^{d}+z^{d}) $$ -which defines a relative curve in $\mathbb P^2_{k[t]}$ over $k[t]$. The generic fiber is smooth by the Jacobian criterion. The fiber above $t=0$ is stable with three irreducible components: two straight lines and a Fermat curve of degree $d-2$. -There might be other singular non-stable fibers, I did'nt check. -Edit I forgot the second part of the question (not appearing in the title). -The answer in the general case is : - -If $X$ is a stable curve of genus $g\ge 2$, then the number of singular points is bounded by $3g-3$. This bound is reached for all $g$. - -I don't think being plane curve improves the bound. -To prove this bound, there is either a direct elementary way or a theoretical way. The elementary one is by induction on the number of singular points. If $X$ is irreducible, then the number of singular points in bounded by $g$. Now suppose $X$ is reducible and fix an intersection point $s\in X$. Let $Y$ be the normalization of $X$ at $s$. It consists in separate the two irreducible components $\Gamma_1, \Gamma_2$ meeting at $s$. Then $Y$ has one or two connected components. Each of these connected components might fail to be stable. This happens if $\Gamma_1$ or $\Gamma_2$ are projective lines and meets the other irreducible components at exactly two points (plus $s$). Then we contract such a component, compute the new genus/genera, and the new number of singular points, and we check that $3g-3$ is an acceptable bound. -The bound $3g-3$ is actually reached for any $g\ge 2$. Construct a stable curve with $2g-2$ rational smooth irreducible components $\Gamma_1,\dots, \Gamma_{g-1}$, $\Gamma'_1,\dots, \Gamma'_{g-1}$, and make $\Gamma_i$ intersect $\Gamma'_i$, $\Gamma_{i-1}$ and $\Gamma_{i+1}$ with the convent $\Gamma_{i+g-1}=\Gamma_{i}$. Similarly $\Gamma'_i$ intersects (only) $\Gamma_i$, $\Gamma'_{i\pm 1}$. This construction can cause problem for $g=2$. Then take two projective lines meeting at three points. -Now some ideas for the theoretical proof. Note that $3g-3$ is the dimension of the moduli space of stable curves of genus $g$. Then use the computation of the dimension of deformation spaces as in Deligne-Mumford (The irreducibility of the space of curves of given genus), §1.<|endoftext|> -TITLE: Question about whether axiom of choice is needed in this proof -QUESTION [10 upvotes]: Do I need axiom of choice in this proof here? -I think not: at each step we choose one element from a set $N - \langle g_1, \dots, g_k \rangle $. So while there is indeed a countable number of sets involved from which we choose elements, I could also think of the process as follows: -Assume $N$ is generated uncountably. Let $C = \{g_1 , g_2, \dots \}$ be a countable subset of the generators of $N$. (So far we have not used choice, right?) We may write $C$ as a countable union of singleton sets $\bigcup_n \{g_n\}$. But now we can write down an explicit choice function: Let $c(\{x\}) = x$. Since the union is countable, we may consider the choice function $\tilde{c}: \mathbb N \to C, n \mapsto g_n$. -From here, we can finish the argument as follows: Then the following is an increasing chain of submodules: $\langle \tilde{c}(1) \rangle \subset \langle \tilde{c}(1), \tilde{c}(2) \rangle \subset \dots$, avoiding the axiom of choice. -Would someone tell me where my argument is flawed? Thanks. - -REPLY [13 votes]: There is a delicate point about the definition of Noetherian which requires the axiom of choice. -Let $X$ be an amorphous set, namely an infinite set that cannot be partitioned into two disjoint infinite subsets. Such $X$ has the interesting property that if ${\cal A\subseteq P}(X)$ is a chain, then $\cal A$ is finite. -Now let $M=\bigoplus_X\mathbb Z$ as a module over $\mathbb Z$. Every chain of modules is finite, simply since it defines a chain of subsets of $X$ and every such chain is finite. -On the other hand, it is clear that $M$ is not finitely generated and therefore not Noetherian in the definition that "every submodule is finitely generated", simply take $M$ itself to be that submodule. We also have that the family of finitely generated submodules does not have a maximal element. - -It should be remarked that the equivalence between "every module is finitely generated" and "every non-empty family of submodules has a maximal element" requires the axiom of choice (specifically it requires Dependent Choice, which amongst other things implies that every infinite set has a countably infinite subset). -More: - -Can one avoid AC in the proof that in Noetherian rings there is a maximal element for each set? -Where is the Axiom of choice used?<|endoftext|> -TITLE: How to prove the function $y = \sin x$ is not a closed function? -QUESTION [6 upvotes]: I came across a question: - -Suppose that $f(x) = \sin x$ is a function from $\mathbb R$ to $[-1,1]$. How do I prove the function $f(x) = \sin x$ is not a closed function? - -By "closed function", I mean a function such that the image of any closed set is closed. - -REPLY [10 votes]: Closed function is a function such that image of every closed set is closed. -It is relatively easy to see that, for any $\varepsilon>0$, every $\varepsilon$-discrete subset of real line is closed. (A subset $A$ of a metric space $(X,d)$ is called $\varepsilon$-discrete if for any two distinct points $x,y\in A$ we have $d(x,y)\ge\varepsilon$. For subsets of real line, this condition means $|x-y|\ge\varepsilon$.) -Can you find a sequence $(x_n)$ with the following properties? - -$x_n\in(2n\pi,(2n+1)\pi)$ (which implies that $\{x_n; n\in\mathbb N\}$ is an $\varepsilon$-discrete subset for any $\varepsilon<\pi$) -$\lim\limits_{n\to\infty} \sin x_n =y$ but $y\notin\{\sin x_n; n\in\mathbb N\}$ - -If $(x_n)$ fulfills the above properties, then $A=\{x_n; n\in\mathbb N\}$ is a closed set, but the image of this set is not closed.<|endoftext|> -TITLE: Uniform boundedness principle statement -QUESTION [6 upvotes]: Consider the uniform boundedness principle: -UBP. Let $E$ and $F$ be two Banach spaces and let $(T_i)_{i \in I}$ be a family (not necessarily countable) of continuous linear operators from $E$ into $F$. Assume that $\sup_{i \in I} \|T_ix \| < \infty$ for all $x \in E$. Then $\sup_{i \in I} \|T_i\|_{\mathcal{L}(E,F)} < \infty$. -I don't understand the statement of the UBP. The assumption tells us that, fixed an element $u$, we surely find a $\|T_ku\|< \infty$ (in particular, for that fixed $u$ each other $T$ is limited in $u$ too). The conclusion tells us that the sup over the $i$'s of the set -$$ -\biggl\{ \sup_{\|x\|\leq 1} \|Tx\| \biggr\} -$$ -is limited. But isn't that clear from the assumption? I mean, if each $T$ is bounded, a fortiori the conclusion must hold... please explain me where I am wrong. - -REPLY [9 votes]: To illustrate the point which Davide explained in his answer, let us have a look at a concrete example. -Let us take $E=$ the set of all real sequences with finite support (i.e., only finitely many terms are non-zero). Let us use the norm $\|x\|=\sup_n |x_n|$. The space $E$ is a linear normed space, but it is not a Banach space. (So the assumptions of Banach-Steinhaus theorem are not fulfilled.) -Let us take $F=\mathbb R$ and $T_n(x)=\sum_{k=1}^n x_k$. -For every $x\in E$ we have $|T_n(x)| \le \sum_{k=1}^n |x_k|$, which is a finite number. So $\sup_n |T_n(x)|<+\infty$ for any fixed $x\in E$. -But if we take $x_n=(\underset{\text{$n$-times}}{\underbrace{1,\dots,1,}}0,0,\dots)$, then $T_n(x_n)=n$ and $\|x_n\|\le 1$. So we see that $T_n(x)$ is not bounded on the unit ball, i.e. -$$\sup_{\|x\|\le 1, n\in\mathbb{N}} T_n(x)=+\infty.$$ - -REPLY [5 votes]: The problem is that $\sup_{i\in I}\lVert T_ix\rVert=:c(x)$ depend on $x$. The fact that it's bounded of each $x$ of the unit ball doesn't imply that it can be bounded uniformly on the unit ball (otherwise we would deduce that each linear map is continuous). -So the aim of the UBP, is, its name suggests, to show $\sup_{\lVert x\rVert=1}c(x)$ is finite.<|endoftext|> -TITLE: Endomorphisms preserve Haar measure -QUESTION [7 upvotes]: I am having trouble following the argument in page 21 of P. Walters, Intro. to ergodic theory, of the following statement: - -Any continuous endomorphism on a compact group preserves Haar measure. - -Obviously, this is not true as stated, as the trivial homomorphism does not preserve Haar measure. Even restricting ourselves to nontrivial homomorphisms, I am still unable to follow the argument, which is as follows: -Let $A : G \rightarrow G$ be the endomorphism and let $m$ denote the Haar measure. Define a probability measure on Borel sets $\mu(E) = m(A^{-1}(E))$ . Now, -$$ \mu(Ax.E) = m(A^{-1}(Ax.E)) = m(x.A^{-1}E) = \mu(E) .$$ -I agree with the first equality by definition, and the last equality because of $m$ being a Haar measure. But I am unable to fathom why the middle equality is true. -Moreover, even if the said equality is true, the rest of the proof goes as follows: We see that $\mu$ is rotation invariant, and we use the uniqueness of Haar measure to prove that $\mu = m$. In this, I am unable to see how the above equality assures that $\mu$ is rotation-invariant. -Help would be appreciated. - -REPLY [5 votes]: The moral is that you have to assume surjectivity of $A$. -For the middle equality: we have $A^{-1}(Ax.E)=x.A^{-1}E$ as sets: -$y\in A^{-1}(Ax.E)$ means $Ay\in Ax.E$ means $A(x^{-1}y)\in E$, while -$y\in x.A^{-1}E$ means $x^{-1}y\in A^{-1}E$. -For the rotation-invariance: we now see $\mu(Ax.E)=\mu(E)$ for all $x\in G$. In other words, $\mu(y.E)=\mu(E)$ for all $y\in \text{im} A\subset G$. By surjectivity we have $\text{im} A=G$.<|endoftext|> -TITLE: Question about direct sum of Noetherian modules is Noetherian -QUESTION [6 upvotes]: Here is a corollary from Atiyah-Macdonald: - -Question 1: The corollary states that finite direct sums of Noetherian modules are Noetherian. But they prove that countably infinite sums are Noetherian, right? (so they prove something stronger) -Question 2: I have come up with the following proof of the statement in the corollary, can you tell me if it's correct? Thank you: -Assume $M_i$ are Noetherian and let $(\bigoplus_i L_i)_n$ be an increasing sequence of submodules in $\bigoplus_i M_i$. Then in particular, $L_{in}$ is an increasing sequence in $M_i$ and hence stabilises, that is, for $n$ greater some $N_i$, $L_{in} = L_{in+1} = \dots $. Now set $N = \max_i N_i$. Then $(\bigoplus_i L_i)_n$ stabilises for $n> N$ and is equal to $\bigoplus_i L_i$, where $L_i = L_{iN_i}$. -This proves that finite direct sums of Noetherian modules are Noetherian so it's a bit weaker. But if it's correct it proves the corollary. - -REPLY [8 votes]: As pointed out by others, the submodules of $M=M_1\oplus M_2$ are not necessarily direct sums of submodules of $M_1$ and $M_2$. Nevertheless, you always have the exact sequence -$$0\rightarrow M_1\rightarrow M\rightarrow M_2\rightarrow 0$$ -Then $M$ is Noetherian if (and only if) $M_1$ and $M_2$ are Noetherian. One direction is trivial (if M is Noetherian then $M_1$ and $M_2$ are Noetherian). I prove the other direction here: -Assume $M_1$ and $M_2$ are Noetherian. Given nested submodules $N_1\subset N_2\subset \cdots$ in $M$, we can see their images stabilize in $M_1$ and $M_2$. More precisely, the chain -$$(N_1\cap M_1)\subset (N_2\cap M_1)\subset\cdots$$ -terminates in $M_1$, say at length $j_1$ and so does -$$(N_1+M_1)/M_1\subset (N_2+M_1)/M_1\subset\cdots$$ -in $M_2$, say at length $j_2$. Set $j=\max\{j_1,j_2\}$ to get -$$(N_{j}\cap M_1)=(N_{j+1}\cap M_1)=\cdots$$ -in $M_1$ and -$$(N_{j}+M_1)/M_1=(N_{j+1}+M_1)/M_1=\cdots$$ -in $M_2$. But $N_j$'s are nested modules. Hence the above equalities can occur if and only if $N_j=N_{j+1}=\cdots$. To check this claim, pick $n\in N_{j+1}$. Then $m:=n'-n\in M_1$ for some $n'\in N_{j}$. But $m\in N_{j+1}$ as well. Hence $m\in N_{j+1}\cap M_1$, that is $m\in N_{j}\cap M_1$ by the first equality above. So $m\in N_j$, that is $n\in N_j$, giving us $N_{j+1}\subset N_j$. -So the chain $N_1\subset N_2\subset ...$ terminates in $M$.<|endoftext|> -TITLE: Nullstellensatz for polynomials generating the same hypersurface -QUESTION [5 upvotes]: Let $X$ be an affine variety and $f,g\in \mathcal{O}(X)$ such that $\lbrace f=0\rbrace = \lbrace g= 0\rbrace$. How does it follow by the Nullstellensatz that $f^n = g^m$ for some $m,n\in \mathbb{N}$? This is claimed in a proof, but I don't see the reason. -If it is not true in general, are there additional conditions under which the statement is true? -Many thanks in advance! - -REPLY [6 votes]: It doesn't follow because it's not true. For example, let $f = x^2 y, g = x y^2$. (This shows that even the weaker claim $(f^n) = (g^m)$ for some $n,m$ is false. The strong claim is false for silly reasons like the coefficients not matching up; for example, let $f = x, g = 2x$ and let the underlying field have characteristic zero.) -The correct statement is that the Nullstellensatz implies that $(f), (g)$ have the same radical. In particular, $f^n \in (g)$ and $g^m \in (f)$ for some $n, m$.<|endoftext|> -TITLE: Number of Points on the Jacobian of a Hyperelliptic Curve -QUESTION [5 upvotes]: Consider a genus 2 hyperelliptic curve $X$ over a finite field $\mathbb{F}_{p^{k}}$ for $k \leq 4$. Let $J$ be the Jacobian of $X$. Is there a relation between the zeta function of $X/\mathbb{F}_{p^{k}}$ and $\#J(\mathbb{F}_{p^{k}})$? - -REPLY [5 votes]: In fact, you do not have to assume anything about the genus, nor is it relevant that the curve is hyperelliptic, nor does the cardinality of the finite field matter: -Theorem. Let $C$ be a smooth projective curve of genus $g$ over a finite field $\mathbb F_q$ and let -$$ Z(C;t)=\frac{L(t)}{(1-t)(1-qt)} $$ -denote the zeta function of $C$, where $L(t)\in\mathbb Z[t]$. Then the number of $\mathbb F_q$-rational points on the Jacobian $J$ of $C$ equals $L(1)$. -Proof. Let $\ell$ denote a prime distinct from $\operatorname{char}\mathbb F_q$. The $q$-power Frobenius endomorphism of $C$ induces a purely inseparable isogeny $\varphi$ of degree $q$ of $J$, and thereby an endomorphism $T_\ell(\varphi)$ of the $\ell$-adic Tate module $T_\ell(J)$. The number of $\mathbb F_q$-rational points of $J$ is the number of fixed points of $\varphi$, and since $1-\varphi$ is a separable isogeny, we have -$$ \#J(\mathbb F_q) = \#\ker(1-\varphi) = \deg(1-\varphi) = \det(1-T_\ell(\varphi)) \text. $$ -By definition, this equals $\chi_\varphi(1)$, where $\chi_\varphi(t)$ is the characteristic polynomial of $T_\ell(\varphi)$. -Now note that the Tate module is a special case of $\ell$-adic cohomology: There is a natural isomorphism -$$ T_\ell(J)\otimes_{\mathbb Z_l}\mathbb Q_\ell \cong H^1(C,\mathbb Q_\ell) \text, $$ -hence we may apply the Lefschetz trace formula for $\ell$-adic cohomology (and some linear algebra) to deduce -$$ L(t) = \det(1-t\varphi^\ast\mid H^1(C,\mathbb Q_\ell)) \text. $$ -Since $H^1(C,\mathbb Q_\ell)$ is $2g$-dimensional, this implies -$$ L(t) = t^{2g}\det(1/t-\varphi^\ast\mid H^1(C,\mathbb Q_\ell) = t^{2g}\chi_\varphi(1/t) \text. $$ -Note this is the "reverse polynomial" of $\chi_\varphi(t)$, i.e., it has the same coefficients in reversed order. -Together with the above, evaluating this at $1$ shows the claim -$$ \#J(\mathbb F_q) = L(1) \text. \tag*{$\square$}$$ -A reference for this is Section 5.2.2 and 8.1.1 of Cohen and Frey's Handbook of Elliptic and Hyperelliptic Curve Cryptography, 1st ed.<|endoftext|> -TITLE: Proving $(1+a_{2})^{2}(1+a_{3})^{3}\cdots(1+a_{n})^{n}\ge n^n$ for positive reals $a_2,\ldots, a_n$ whose product is $1$ -QUESTION [5 upvotes]: Let $n \ge3$ be an integer, and let $a_{2},a_{3}, ... ,a_{n}$ be positive real numbers such that $a_{2} a_{3}\cdots a_{n}=1.$ Prove that: - $$(1+a_{2})^{2}(1+a_{3})^{3}\cdots(1+a_{n})^{n}\ge n^n$$ - -This is the 2nd problem of the 53rd IMO and seems pretty interesting. How would we solve that? - -REPLY [9 votes]: Another approach that I thought of : -Set $a_2 =\frac{x_2}{x_3}, a_3=\frac{x_3}{x_4},\ldots, a_n=\frac{x_n}{x_2}$. This is a very useful substitution that we use in cases when we have a product equal to one like in this one $ a_2 a_3 \cdots a_n=1 $. -Now we need to prove that $$ (x_2+x_3)^2 (x_3+x_4)^3 \cdots (x_n+x_2)^n > n^n x_3^2 x_4^3 \cdots x_{n}^{n-1}x_2^n$$ -which become obvious since for each $k$ by applying the Arithmetic-Geometric Mean we have that: -$$ (x_k+x_{k+1})^{k}=\left(x_k+(k-1)\frac{x_{k+1}}{k-1}\right)^k\geqslant k^k x_k\frac{x_{k+1}^{k-1}}{(k-1)^{k-1}} $$ -Just multiply for $k$ from $2$ to $n$.<|endoftext|> -TITLE: Poset of idempotents -QUESTION [6 upvotes]: Let $A$ be an associative algebra and consider the poset $I$ of idempotents in $A$, where as usual if $e,f \in A$ we say $e \leq f$ provided that $e = fef$. Clearly $0 \in I$ is minimal, but in general I can't say anything else about this poset. -This leads me to ask: are there any other general properties of $I$ without more assumptions on $A$? In particular, if I am given some poset with a minimal element, is it possible to construct an algebra with the prescribed poset of idempotents? -Edit: I should have said I'm working over a field (or at least a ring with no nontrivial idempotents), since otherwise idempotents in the base ring complicate things. Also: bonus points for producing examples where $A = \cup_{e \in I} eAe$. - -REPLY [5 votes]: The poset of idempotents is equipped with a relative complement operation: $e \le f$ if and only if $ef = fe = e$, and then we compute that -$$(f - e)^2 = f - e$$ -hence $f - e$ is also an idempotent. This operation, which I"ll write as $e \Rightarrow f$, satisfies the following axioms: - -$e \Rightarrow f$ is order-preserving in the second variable and order-reversing in the first, -$(0 \Rightarrow f) = f$, -$((e \Rightarrow f) \Rightarrow f) = e$, -$\inf(e, e \Rightarrow f) = 0$. - -In particular, for fixed $f$ it defines an order-reversing isomorphism from the interval $[0, f]$ to itself. Consequently, any poset with a minimal element $0$ and an element $e$ such that the interval $[0, e]$ is not isomorphic to its opposite cannot admit a relative complement and so cannot be the poset of idempotents of a ring. A poset with this property of minimal size is the poset $\{ 0 \lt a \lt b, c \lt 1 \}$. -Another minimal counterexample is the total order $3 = \{ 0 \lt 1 \lt 2 \}$, which admits a unique map $e \Rightarrow f$ satisfying the first, second, and third axioms above, and this map does not satisfy the fourth axiom. - -Another thing you can say about the idempotents in a rng is that they are equipped with an interesting relation, namely commutativity. If $e, f$ are commuting idempotents, then $ef$ gives their inf and $e + f - ef$ gives their sup. More generally, any commuting subposet of the idempotents is a lattice.<|endoftext|> -TITLE: Stalk of a pushforward sheaf in algebraic geometry -QUESTION [13 upvotes]: Excuse me if this is a naive question. Let $f : X \to Y$ be a morphism of varieties over a field $k$ and $\mathcal{F}$ a quasi-coherent sheaf on $X$. I know that for general sheaves on spaces not much can be said about the stalk $(f_*\mathcal{F})_y$ at $y \in Y$ (let's just talk about closed points), but does anything nice happen in this situation? If $f$ is proper then the completion of the stalk is described by the formal function theorem, but I'm interested in the honest stalk. -Suppose, for instance, that $X = \text{Spec } A$ is affine (so $\mathcal{F} = \widetilde{M}$ for some $A$-module $M$) and that the scheme-theoretic fiber of $f$ over the closed point $y \in Y$ is reduced. Write this fiber as a union of irreducible components $Z_1 \cup \cdots \cup Z_r$ corresponding to prime ideals $\mathfrak{p}_1,\cdots,\mathfrak{p}_r \subset A$, so there is an associated semi-local ring $S^{-1}A$ with $S = A \setminus (\mathfrak{p}_1 \cup \cdots \cup \mathfrak{p}_r)$. Hopefully here $(f_*\mathcal{F})_y$ is just $S^{-1}M$ regarded as an $\mathcal{O}_{Y,y}$-module via the natural map $\mathcal{O}_{Y,y} \to S^{-1}A$. -Does this make sense? Is something like this true when $X$ is not affine and/or the fiber is nonreduced? -Edit: As Georges's answer shows, this cannot possibly be true in general. I wonder if there is still hope when $f$ is proper? My example in the comments below (the squaring map $\mathbb{A}^1 \to \mathbb{A}^1$ with $\mathcal{F} = \mathcal{O}$ and $y = 1$) is actually consistent with my guess above: $(f_*\mathcal{F})_y$ is the localization of $k[t]$ at $k[t^2] \setminus (t^2-1)k[t^2]$. It is not hard to see that this coincides with the localization at $k[t] \setminus ((t-1) \cup (t+1))$, i.e. the semilocal ring at $\{ \pm 1 \}$. - -REPLY [10 votes]: No, I'm afraid what you say does not work, even in the simplest situation: -a) Let $k$ be a field , $X=\mathbb A^1_k=Spec (k[t])$ , $Y=Spec (k)=\lbrace y\rbrace$ and let $f:X\to Y$ be the obvious morphism. -Then for $\mathcal F=\mathcal O_X$ we have $(f_*\mathcal F)_y=k[t]$. -b) On the other hand, still in your notation, $X=Z_1$, $\mathfrak p_1=(0)$, $M=k[t]$, $S=k[t]\setminus \lbrace 0\rbrace $ and thus $S^{-1}M=k(t)$ -c) Conclusion: $$(f_*\mathcal F)_y=k[t]\neq S^{-1}M=k(t)$$ - -REPLY [4 votes]: I think the best we can do is the following. The question is local on the base, so we may as well assume $Y = \text{Spec } A$ and $X = \text{Spec } B$ are affine and $\mathcal{F} = \widetilde{M}$ where $M$ is a $B$-module. Then if $y = \mathfrak{p} \in \text{Spec } A$ then $(f_*\mathcal{F})_y$ is the $A_{\mathfrak{p}}$-module $M_{\mathfrak{p}}$, where we think of $M$ now as an $A$-module.<|endoftext|> -TITLE: $e^x(\ln x-c) =\sum \limits_{k=0}^\infty \frac{ x^{k} \Gamma'(k+1)}{ (k!)^2}$ Is it correct result? -QUESTION [7 upvotes]: $e^x=\sum \limits_{k=0}^\infty \frac{x^k}{k!}$ -We can write $e^x=\sum \limits_{k=0}^\infty \frac{x^k}{ \Gamma(k+1)}$ -Where $\Gamma(x)$ is Gamma function -$\Gamma(k+1)=k\Gamma(k)$ -$\frac{\Gamma(k+1)}{\Gamma(k)}=k$ -$\frac{\Gamma(1)}{\Gamma(0)}=0$ -$\frac{1}{\Gamma(0)}=0$ -$\frac{\Gamma(-1+1)}{\Gamma(-1)}=\frac{\Gamma(0)}{\Gamma(-1)}=-1$ -$\frac{1}{\Gamma(-1)}=\frac{-1}{\Gamma(0)}=-1.\frac{1}{\Gamma(0)}=0$ -If we continue in that way, we get result -for $m $ is non-positive integer, $\frac{1}{\Gamma(m)}=0$ -Thus we can write $e^x=\sum \limits_{k=-\infty}^\infty \frac{x^k}{ \Gamma(k+1)}$ -Then we extended $e^x$ for n is integer -(Equation 1): $$e^x=\sum \limits_{k=-\infty}^\infty \frac{x^{k+n}}{ \Gamma(k+n+1)}$$ -$n \in Z $ {...,-2,-1,0,1,2,...} -It is possible to extend the defination to $z \in C$. -$f(x)=\sum \limits_{k=-\infty}^\infty \frac{x^{k+z}}{ \Gamma(k+1+z)}$ -$$\frac{d(f(x))}{dx}=\frac{d}{dx}(\sum \limits_{k=-\infty}^\infty \frac{x^{k+z}}{ \Gamma(k+1+z)})= \sum \limits_{k=-\infty}^\infty (k+z)\frac{x^{k+z-1}}{ (k+z)\Gamma(k+z)}= \sum \limits_{k=-\infty}^\infty \frac{x^{k+z-1}}{ \Gamma(k+z)}=\sum \limits_{k=-\infty}^\infty \frac{x^{k+z}}{ \Gamma(k+1+z)}=f(x)$$ -$$\frac{d(f(x))}{dx}=f(x)$$ -$$f(x)=c(z)e^x$$ -$c(z)e^x=\sum \limits_{k=-\infty}^\infty \frac{x^{k+z}}{ \Gamma(k+1+z)}$ -According to Equation 1, $c(z) = 1$ for $z \in Z$ but I noticed I need to find what is $c(z)$ for $z \in C$. (Thanks to Norbert for his contribution) -After that we can find the result: -$$\frac{\partial(c(z)e^x)}{\partial z}=\sum \limits_{k=-\infty}^\infty \frac{\partial}{\partial z}(\frac{x^{k+z}}{ \Gamma(k+1+z)})$$ -$$c'(z)e^x=\sum \limits_{k=-\infty}^\infty \frac{\partial}{\partial z}(\frac{x^{k+z}}{ \Gamma(k+1+z)})$$ -$$c'(z)e^x=\sum \limits_{k=-\infty}^\infty (\frac{\ln x . x^{k+z}}{ \Gamma(k+1+z)})-\sum \limits_{k=-\infty}^\infty (\Gamma'(k+1+z)\frac{ x^{k+z}}{ \Gamma^2(k+1+z)})$$ -$$c'(z)e^x=\ln x \sum \limits_{k=-\infty}^\infty (\frac{ x^{k+z}}{ \Gamma(k+1+z)})-\sum \limits_{k=-\infty}^\infty (\Gamma'(k+1+z)\frac{ x^{k+z}}{ \Gamma^2(k+1+z)})$$ -$$e^x(c(z)\ln x-c'(z)) =\sum \limits_{k=-\infty}^\infty (\frac{ x^{k+z} \Gamma'(k+1+z)}{ \Gamma^2(k+1+z)})$$ -If we take $z=0$, we get an interesting result. -$$e^x(c(0)\ln x -c'(0)) =\sum \limits_{k=0}^\infty \frac{ x^{k} \Gamma'(k+1)}{ (k!)^2}$$ -$c(0)=1$ according to Equation 1 -Thus -$$e^x(\ln x -c'(0)) =\sum \limits_{k=0}^\infty \frac{ x^{k} \Gamma'(k+1)}{ (k!)^2}$$ -I do not know what $c'(0) is$. -Acoording to Norbert's answer. $c'(0) \approx -0.596347$ -I have not seen that result in other place. Is it known result?Please let me know if my results are correct or not. -Can we extend all such functions that include $\Gamma(x)$ in denominator? -Thanks for advice - -REPLY [4 votes]: Let -$$\begin{eqnarray*} -g(x) &=& \sum \limits_{k=0}^\infty \frac{ x^{k} \Gamma'(k+1)}{ (k!)^2} \\ -&=& \sum_{k=0}^\infty \frac{x^k}{k!}\psi(k+1), -\end{eqnarray*}$$ -where $\psi$ is the digamma function. -Using the recurrence formula for the digamma, $\psi(x+1) = \psi(x) + 1/x$, one can show that $g(x)$ satisfies the differential equation -$$\begin{equation*} -g'(x) - g(x) = \frac{1}{x}(e^x-1). -\tag{1} -\end{equation*}$$ -The boundary condition is $g(0) = \psi(1) = -\gamma$, where $\gamma$ is the Euler-Mascheroni constant. -(We interpret $g(0)$ as $\lim_{x\to 0^+} g(x)$.) -The differential equation (1) can be solved using standard techniques. -See below for a derivation. -The solution is $g(x) = e^x[\log x + E_1(x)]$, -where $E_1(x) = \int_x^\infty dt\, e^{-t}/t$ is the exponential integral. -(We assume for the moment that $|\mathrm{Arg}(x)| < \pi$.) -Thus, -$$\begin{equation*} -\sum \limits_{k=0}^\infty \frac{ x^{k} \Gamma'(k+1)}{ (k!)^2} -= e^x[\log x + E_1(x)]. \tag{2} -\end{equation*}$$ -This is in agreement with @Pink Elephants result. -Using the series representation for $E_1(x)$ for $|\mathrm{Arg}(x)| < \pi$ -we find -$$\begin{equation*} -\sum \limits_{k=0}^\infty \frac{ x^{k} \Gamma'(k+1)}{ (k!)^2} -= -\gamma e^x + e^x \sum_{k=1}^\infty \frac{(-1)^{k+1}x^k}{k k!}. \tag{3} -\end{equation*}$$ -Notice the logarithmic singularities cancel one another, so the apparent branch cut in (2) is illusory. -The sum on the right of (3) converges for all complex $x$ and satisfies the boundary condition by inspection. -Derivation of (2) -Using the integrating factor technique, we find the particular solution -$$\begin{eqnarray*} -g_p(x) &=& e^{x-a} \int_{a}^x dt\, e^{-(t-a)} \frac{1}{t}(e^t-1) \\ -&=& e^x[\log x + E_1(x)] - e^x[\log a + E_1(a)], -\end{eqnarray*}$$ -where $|\mathrm{Arg}(a)|<\pi$. -To satisfy the boundary condition we must add the homogeneous solution -$$g_h(x) = e^x[\log a + E_1(a)],$$ -so -$$\begin{equation*} -g(x) = e^x[\log x + E_1(x)]. -\end{equation*}$$ -Here we use the fact that $\lim_{x\to 0^+} E_1(x) = -\log x - \gamma$, so -$\lim_{x\to 0^+} g(x) = -\gamma$, as required.<|endoftext|> -TITLE: Projective and injective modules; direct sums and products -QUESTION [11 upvotes]: I need two counterexamples. -First, a direct sum of $R$-modules is projective iff each one is projective. - But I need an example to show that, “an arbitrary direct product of projective modules need not be a projective module.” -If I let $R= \mathbb Z$ then $\mathbb Z$ is a projective $R$-module, but the direct product $\mathbb Z \times \mathbb Z \times \cdots$ is not free, hence it is not a projective module. We have a theorem which says that every free module over a ring $R$ is projective. Am I correct? -Second, a direct product of $R$-modules is injective iff each one is injective -but I need an example to show that the direct sum of injective modules need not be injective. - -REPLY [9 votes]: As for the first question: yes, $P = \prod_{i=1}^{\infty} \mathbb{Z}$ is a direct product of free $\mathbb{Z}$-modules which is not free. Since $\mathbb{Z}$ is a PID, $P$ is also not projective. The proof that $P$ is not free is nontrivial, but I believe it has already been given either here or on MathOverflow. -As for the second question: the Bass-Papp Theorem asserts that a commutative ring $R$ is Noetherian iff every direct sum of injective $R$-modules is injective. Thus every non-Noetherian ring carries a counterexample. The proof of the result -- given for instance in $\S 8.9$ of these notes -- is reasonably constructive: if -$I_1 \subsetneq I_2 \subsetneq \ldots \subsetneq I_n \subsetneq \ldots$ -is an infinite properly ascending chain of ideals of $R$, then for all $n$ let $E_n = -E(R/I_n)$ be the injective envelope (see $\S 3.6.5$ of loc. cit.) of $R/I_n$, and let $E = \bigoplus_{n=1}^{\infty} E_n$. Then $E$ is a direct sum of injective modules and (an argument given in the notes shows) that $E$ is not itself injective.<|endoftext|> -TITLE: an integer sum of products of tangents -QUESTION [22 upvotes]: This question arose from my initial attempts at answering this question. I later found a way to transform the desired sum into a sum of squares of tangents, but before I did, I found numerically that apparently -$$ -\sum_{l=1}^n\tan\frac{jl\pi}{2n+1}\tan\frac{kl\pi}{2n+1}=m_{jkn}(2n+1) -$$ -with integer factors $m_{jkn}$, for which I haven't been able to find an explanation. If $j$ or $k$ is coprime to $2n+1$, we can sum over $jl$ or $kl$ instead, so most cases (in particular all for $2n+1$ prime) can be reduced to the case $j=1$. Here are the numerically determined factors $m_{1kn}$ for $n\le18$ (with $n$ increasing downward and $k$ increasing to the right): -$$ -\begin{array}{r|rr} -&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18\\\hline1&1\\ -2&2&0\\ -3&3&-1&1\\ -4&4&0&1&0\\ -5&5&-1&1&1&1\\ -6&6&0&2&-2&0&0\\ -7&7&-1&2&-1&1&0&1\\ -8&8&0&2&0&2&2&2&0\\ -9&9&-1&3&1&1&-3&1&-1&1\\ -10&10&0&3&-2&2&-1&1&0&1&0\\ -11&11&-1&3&-1&1&-1&1&3&-1&1&1\\ -12&12&0&4&0&2&0&0&-4&0&0&0&0\\ -13&13&-1&4&1&3&0&1&-1&1&1&3&0&1\\ -14&14&0&4&-2&2&2&2&0&2&4&0&0&2&0\\ -15&15&-1&5&-1&3&-3&3&-1&3&-5&1&1&1&-1&1\\ -16&16&0&5&0&2&-1&2&0&1&-2&1&1&-2&-2&1&0\\ -17&17&-1&5&1&3&-1&2&-1&1&-1&1&5&1&0&1&1&1\\ -18&18&0&6&-2&4&0&2&0&2&-2&2&-6&0&0&4&2&0&0\\ -\end{array} -$$ -(See also the table in this answer to the other question, which shows the case $j=k+1$; in that case the rows of the table sum to $0$ because of the identity that's the subject of the other question.) -The values $m_{11n}=n$ reflect the sum of squares of tangents that I determined in my answer to the other question. I have no explanation for the remaining values. I've tried using the product formula for the tangent; multiplying by a third tangent to use the triple tangent product formula; and finding a polynomial whose roots are the products being summed; but none of that worked out. -This vaguely reminds me of character theory; the values $\tan\frac{kl\pi}{2n+1}$ for fixed $k$ are like characters, and their dot products are integer multiples of the "group order" $2n+1$; though if they were characters the dot products couldn't be negative. -I'd appreciate any insight into this phenomenon, and of course ideally a way to calculate the $m_{jkn}$. -[Update:] -I've verified the periodicities that Brian observed in comments up to $n=250$: -$$m_{1,k,n+k} = m_{1kn}+[k \text{ odd}]\;,$$ -$$m_{1,k+4d+2,k+4d+2+d}=m_{1,k,k+d}\;,$$ -where the bracket is the Iverson bracket. - -REPLY [8 votes]: [Just as I was finishing this I saw user8268's answer. I suspect the explanations are related.] -I wasn't intending to answer my own question in this case, but I've now found an explanation. The character theory analogy turned out to be more useful than I expected. Thinking of the values $\tan\frac{kl\pi}{2n+1}$ for fixed $k$ as vectors composed of integer multiples of mutually orthogonal vectors made me wonder what these mutually orthogonal vectors might be. A natural choice was a Fourier-style set of sines or cosines, and indeed it turns out that -$$ -\sum_{l=1}^{2n}\sin\frac{2jl\pi}{2n+1}\tan\frac{kl\pi}{2n+1}= -\begin{cases} -\pm(2n+1)&\gcd(k,2n+1)\mid j\;,\\ -0&\text{otherwise.} -\end{cases} -$$ -I found this surprising at first, but it's actually not too difficult to explain. We have -$$ -\def\ex#1{\mathrm e^{#1}}\def\exi#1{\ex{\mathrm i#1}}\def\exm#1{\ex{-\mathrm i#1}} -\ex{2n\mathrm i\phi}-\ex{-2n\mathrm i\phi}=\left(\exi{\phi}+\exm{\phi}\right)\left(\ex{(2n-1)\mathrm i\phi}-\ex{(2n-3)\mathrm i\phi}+\dotso+\ex{-(2n-3)\mathrm i\phi}-\ex{-(2n-1)\mathrm i\phi}\right)\;, -$$ -so -$$ -\sin(2j\phi)=2\cos\phi\left(\sin((2j-1)\phi)-\sin((2j-3)\phi)+\dotso+(-1)^{j+1}\sin\phi\right)\;. -$$ -Thus, for $k=1$ the cosine in the denominator of the tangent is cancelled, and the remaining sine picks out the last term in the sum of alternating sines with odd frequencies, which yields -$$ -\sum_{l=1}^{2n}\sin\frac{2jl\pi}{2n+1}\tan\frac{l\pi}{2n+1}=(-1)^{j+1}(2n+1)\;. -$$ -But since the integers $jl$ and $kl$ in the arguments of both factors only matter $\bmod2n+1$, if $k$ is coprime to $2n+1$, we can sum over $kl$ instead of $l$ and will get the result for $k^{-1}j\bmod(2n+1)$, so for $k$ coprime to $2n+1$ -$$ -\sum_{l=1}^{2n}\sin\frac{2jl\pi}{2n+1}\tan\frac{kl\pi}{2n+1}=(-1)^{\sigma_k(j)+1}(2n+1)\;, -$$ -where $\sigma_k$ is the permutation effected by multiplication with $k^{-1}\bmod(2n+1)$. If $1\lt\gcd(k,2n+1)\mid j$, the sum reduces to $\gcd(k,2n+1)$ identical copies, whereas if $\gcd(k,2n+1)\nmid j$, cancellation lets the sum vanish. -Thus, the $m_{jkn}$ are integers because the vectors $\left\{\tan\frac{kl\pi}{2n+1}\right\}_l$ are integer linear combinations of vectors $\left\{\sin\frac{2jl\pi}{2n+1}\right\}_l$ whose dot products are all either $0$ or $\pm1$, and these values can be obtained using only elementary number theory, namely permutations induced by multiplicative inverses.<|endoftext|> -TITLE: Elements of $\mathbb{F}_p$ having cube roots in $\mathbb{F}_p$ -QUESTION [6 upvotes]: Let $p$ be a prime number, and let $\mathbb{F}_p$ be the field with $p$ elements. How many elements of $\mathbb{F}_p$ have cube roots in $\mathbb{F}_p$? - -I had this question on an exam and after reviewing I am still not sure. Any help would be appreciated. - -REPLY [3 votes]: Consider the $3$-power map $f$ from the multiplicative subgroup $(\mathbb{Z}/p\mathbb{Z})^\times \longrightarrow (\mathbb{Z}/p\mathbb{Z})^\times$, given by $f(x) = x^3 \bmod p$. This is a homomorphism, so to find the size of the image it suffices to find the size of the kernel. -This group is cyclic, so there is some generator $g$ such that $(\mathbb{Z}/p\mathbb{Z})^\times = \langle g \rangle$. The kernel of the map $f$ in terms of $g$ consists of those elements $g^n$ such that $g^{3n} \equiv 1 \bmod p$. This occurs exactly when $3n$ is a multiple of $\varphi(p) = (p-1)$ (Fermat's Little Theorem and Carmichael's Theorem). Thus the kernel of the map $f$ consists of elements $g^n$ such that $3n \equiv 0 \bmod (p-1)$, or $3n = \alpha (p-1)$ for some integer $\alpha$. We may assume $1 \leq n \leq p-1$, as each element of $(\mathbb{Z}/p\mathbb{Z})^\times$ is given by one such $g^n$. -If $\gcd(3, p-1) = 1$, then necessarily $(p-1) \mid n$, which implies that $n = (p-1)$. In this case the only $g^n$ such that $g^{3n} \equiv 1 \bmod p$ is $g^{p-1} \equiv 1 \bmod p$. Thus the size of the kernel is $1$, and so the image has size $p-1$. All numbers mod $p$ are $3$th powers. -More generally, if $\gcd(3, p-1) = d$, then $f(g^{\frac{p-1}{d}}) = 1$, and this is also true of the $d$ powers of $g^{\frac{p-1}{d}}$ up to $g^{p-1}$. The size of the kernel is $d$, and thus the size of the image is $\frac{p-1}{d}$. -Here, this means if $\gcd(3, p-1) = 3$, then $f(g^{\frac{p-1}{3}}) = 1$, and it's precisely the three powers of $g^{\frac{p-1}{3}}$ in the kernel. -(This clearly generalizes to $k$-powers as well).<|endoftext|> -TITLE: How to avoid perceived circularity when defining a formal language? -QUESTION [51 upvotes]: Suppose we want to define a first-order language to do set theory (so we can formalize mathematics). -One such construction can be found here. -What makes me uneasy about this definition is that words such as "set", "countable", "function", and "number" are used in somewhat non-trivial manners. -For instance, behind the word "countable" rests an immense amount of mathematical knowledge: one needs the notion of a bijection, which requires functions and sets. -One also needs the set of natural numbers (or something with equal cardinality), in order to say that countable sets have a bijection with the set of natural numbers. -Also, in set theory one uses the relation of belonging "$\in$". -But relation seems to require the notion an ordered pair, which requires sets, whose properties are described using belonging... -I found the following in Kevin Klement's, lecture notes on mathematical logic (pages 2-3). -"You have to use logic to study logic. There’s no getting away from it. -However, I’m not going to bother stating all the logical rules that are valid in the metalanguage, since I’d need to do that in the metametalanguage, and that would just get me started on an infinite regress. -The rule of thumb is: if it’s OK in the object language, it’s OK in the metalanguage too." -So it seems that, if one proves a fact about the object language, then one can also use it in the metalanguage. -In the case of set theory, one may not start out knowing what sets really are, but after one proves some fact about them (e.g., that there are uncountable sets) then one implicitly "adds" this fact also to the metalanguage. -This seems like cheating: one is using the object language to conduct proofs regarding the metalanguage, when it should strictly be the other way round. -To give an example of avoiding circularity, consider the definition of the integers. -We can define a binary relation $R\subseteq(\mathbf{N}\times\mathbf{N})\times(\mathbf{N}\times\mathbf{N})$, where for any $a,b,c,d\in\mathbf{N}$, $((a,b),(c,d))\in R$ iff $a+d=b+c$, and then defining $\mathbf{Z}:= \{[(a,b)]:a,b\in\mathbf{N}\}$, where $[a,b]=\{x\in \mathbf{N}\times\mathbf{N}: xR(a,b)\}$, as in this question or here on Wikipedia. In this definition if set theory and natural numbers are assumed, then there is no circularity because one did not depend on the notion of "subtraction" in defining the integers. -So my question is: - -Question Is the definition of first-order logic circular? - If not, please explain why. - If the definitions are circular, is there an alternative definition which avoids the circularity? - -Some thoughts: - -Perhaps there is the distinction between what sets are (anything that obeys the axioms) and how sets are expressed (using a formal language). -In other words, the notion of a set may not be circular, but to talk of sets using a formal language requires the notion of a set in a metalanguage. -In foundational mathematics there also seems to be the idea of first defining something, and then coming back with better machinery to analyse that thing. -For instance, one can define the natural numbers using the Peano axioms, then later come back to say that all structures satisfying the axioms are isomorphic. (I don't know any algebra, but that seems right.) -Maybe sets, functions, etc., are too basic? Is it possible to avoid these terms when defining a formal language? - -REPLY [38 votes]: I think an important answer is still not present so I am going to type it. This is somewhat standard knowledge in the field of foundations but is not always adequately described in lower level texts. -When we formalize the syntax of formal systems, we often talk about the set of formulas. But this is just a way of speaking; there is no ontological commitment to "sets" as in ZFC. What is really going on is an "inductive definition". To understand this you have to temporarily forget about ZFC and just think about strings that are written on paper. -The inductive definition of a "propositional formula" might say that the set of formulas is the smallest class of strings such that: - -Every variable letter is a formula (presumably we have already defined a set of variable letters). -If $A$ is a formula, so is $\lnot (A)$. Note: this is a string with 3 more symbols than $A$. -If $A$ and $B$ are formulas, so is $(A \land B)$. Note this adds 3 more symbols to the ones in $A$ and $B$. - -This definition can certainly be read as a definition in ZFC. But it can also be read in a different way. The definition can be used to generate a completely effective procedure that a human can carry out to tell whether an arbitrary string is a formula (a proof along these lines, which constructs a parsing procedure and proves its validity, is in Enderton's logic textbook). -In this way, we can understand inductive definitions in a completely effective way without any recourse to set theory. When someone says "Let $A$ be a formula" they mean to consider the situation in which I have in front of me a string written on a piece of paper, which my parsing algorithm says is a correct formula. I can perform that algorithm without any knowledge of "sets" or ZFC. -Another important example is "formal proofs". Again, I can treat these simply as strings to be manipulated, and I have a parsing algorithm that can tell whether a given string is a formal proof. The various syntactic metatheorems of first-order logic are also effective. For example the deduction theorem gives a direct algorithm to convert one sort of proof into another sort of proof. The algorithmic nature of these metatheorems is not always emphasized in lower-level texts - but for example it is very important in contexts like automated theorem proving. -So if you examine a logic textbook, you will see that all the syntactic aspects of basic first order logic are given by inductive definitions, and the algorithms given to manipulate them are completely effective. Authors usually do not dwell on this, both because it is completely standard and because they do not want to overwhelm the reader at first. So the convention is to write definitions "as if" they are definitions in set theory, and allow the readers who know what's going on to read the definitions as formal inductive definitions instead. When read as inductive definitions, these definitions would make sense even to the fringe of mathematicians who don't think that any infinite sets exist but who are willing to study algorithms that manipulate individual finite strings. -Here are two more examples of the syntactic algorithms implicit in certain theorems: - -Gödel's incompleteness theorem actually gives an effective algorithm that can convert any PA-proof of Con(PA) into a PA-proof of $0=1$. So, under the assumption there is no proof of the latter kind, there is no proof of the former kind. -The method of forcing in ZFC actually gives an effective algorithm that can turn any proof of $0=1$ from the assumptions of ZFC and the continuum hypothesis into a proof of $0=1$ from ZFC alone. Again, this gives a relative consistency result. - -Results like the previous two bullets are often called "finitary relative consistency proofs". Here "finitary" should be read to mean "providing an effective algorithm to manipulate strings of symbols". -This viewpoint helps explain where weak theories of arithmetic such as PRA enter into the study of foundations. Suppose we want to ask "what axioms are required to prove that the algorithms we have constructed will do what they are supposed to do?". It turns out that very weak theories of arithmetic are able to prove that these symbolic manipulations work correctly. PRA is a particular theory of arithmetic that is on one hand very weak (from the point of view of stronger theories like PA or ZFC) but at the same time is able to prove that (formalized versions of) the syntactic algorithms work correctly, and which is often used for this purpose.<|endoftext|> -TITLE: morphism from a local ring of a scheme to the scheme -QUESTION [15 upvotes]: Let $X$ be a scheme, and $x \in X.$ Let $U=\text{Spec}(A)$ be an open affine subset containing $x,$ then we have the natural morphism $\mathcal{O}_X(U) \to \mathcal{O}_{X,x}$ inducing a morphism $ \text{Spec} \;\mathcal{O}_{X,x} \to U$ and by composing it with the open immersion $U \hookrightarrow X$ we get a morphism $f: \text{Spec} \;\mathcal{O}_{X,x} \to X.$ - -Why this definition does not depend on the choice of $U?$ and -What is the image of $f?$ - -REPLY [3 votes]: Here's a point of view that combines aspects of both of the current (good) answers but makes things more explicit. -Take $U$ an affine open piece of $X$ containing $x$, $U \cong \operatorname{Spec}(A)$, with $x$ corresponding to $P < A$. Then the localisation map $\phi:A\rightarrow A_P$ induces a morphism $\operatorname{Spec}(\mathcal{O}_{X,x})\rightarrow \operatorname{Spec}(A)\rightarrow X$. The image of this morphism is the set of primes in $A$ that are themselves contained in $P$. (Using properties of localisation of rings of morphisms between affine schemes.) This is equal to the intersection of every open subset of $\operatorname{Spec}(A)$ containing $P$, as is easily verified by checking for distinguished affine open subsets of $U$, $D(f)$ for $f \in A\setminus P$. -Now suppose $f \in A/ P$. Then the image of $\phi$ lies inside the distinguished open affine piece $D(f)$. Equivalently, $\phi$ factors through $A_f$, i.e. there is $\phi_f:A_f\rightarrow A_P$, such that if $i_f:A\rightarrow A_f$ is the localisation map, $\phi = \phi_f\circ i_f$. This means that the morphism $\operatorname{Spec}(\mathcal{O}_{X,x})\rightarrow\operatorname{Spec}(A_f)\rightarrow X$ induced by taking affine open piece $U' \cong \operatorname{Spec}(A_f)$ of $X$ is equal to the morphism induced by $U$. (Here we are implicitly using that for a manipulatively closed subset $S$ of a ring $R$, with $P$ a prime not meeting $S$, $(S^{-1}R)_{S^{-1}P}$ is naturally isomorphic to $R_P$.) -Thus if $V \cong \operatorname{Spec}(B)$ is another open affine piece of $X$ containing $x$, then taking open $W \subset U\cap V$ a distinguished open affine piece of both $U$ and $V$ (i.e. there is $a \in A, b \in B$ such that $W \cong D(a)$ and $W \cong D(b)$), the morphism $\operatorname{Spec}(\mathcal{O}_{X,x})\rightarrow X$ induced by $A$ is equal to the morphism induced by $A_a$ by the above paragraph, which is equal to the morphism induced by $B_b$ from the construction of $W$, which is then equal to the morphism induced by $B$ using the above paragraph again. -So we have shown that the morphism is independent of choice of open affine piece. Given this, the image is immediately seen to be the intersection of every open set about $x$, and (interestingly) this view is completely independent of the choice of affine piece.<|endoftext|> -TITLE: What real numbers are in the Mandelbrot set? -QUESTION [10 upvotes]: The Mandelbrot set is defined over the complex numbers and is quite complicated. It's defined by the complex numbers $c$ that remain bounded under the recursion: -$$ z_{n+1} = z_n^2 + c,$$ -where $z_1 = 0$. -If $c$ is real, then above recursion will remain real. So for what values of $c$ does the recursion remain bounded? - -REPLY [2 votes]: If z is a complex number whose distance to origin is bigger than $|c|$ and 2 then z is a point than scape for the iteration of the function $z^2+c$. It's easy to demostrate this, then the recursion remain bounded inside the closed ball of radio 2, but we can find the mandelbrot set inside $[-2,0.7]\times[-1.2,1.2]$.<|endoftext|> -TITLE: Positivity of the norm of an element of a cyclotomic field -QUESTION [6 upvotes]: Let $l$ be an odd prime number and $\zeta$ be an $l$-th primitive root of unity in $\mathbb{C}$. -Let $\mathbb{Q}(\zeta)$ be the cyclotomic field and $\alpha$ be a non-zero element of $\mathbb{Q}(\zeta)$. -There exists a polynomial $f(X) \in \mathbb{Q}[X]$ such that $\alpha = f(\zeta)$. Let $N(\alpha) = f(\zeta)f(\zeta^2)...f(\zeta^{l-1})$. -From $\bar\zeta = \zeta^{-1}$ it follows that $\bar f(\zeta) = f(\zeta^{-1})$. -Likewise, $\bar f(\zeta^i) = f(\zeta^{-i})$ for $i = 1,2,\cdots,l - 1$. -Since $f(\zeta^i)\bar f(\zeta^i) = |f(\zeta^i)|^2 > 0$, it follows that -$N(\alpha) > 0$. -We used the fact that the field of complex numbers $\mathbb{C}$ has an $l$-th primitive root of unity. It seems to me that this fact can only be proved by using some (elementary) analysis. My question is: - -Can we prove $N(\alpha) > 0$ purely algebraically? - -In other words, can we prove $N(\alpha) > 0$ without using the field of real numbers? Please note that $\mathbb{Q}(\zeta) \cong \mathbb{Q}[X]/(1 + X + ... + X^{l-1})$ can be constructed purely algebraically from $\mathbb{Q}$. - -REPLY [3 votes]: Yes, this can be proved using "only elementary properties of $\mathbb{Q}$", since the original argument can be rewritten in this manner. Let $K=\mathbb{Q}[X]/(1+X+X^2+\dots+X^{\ell-1})$, and let $\zeta$ be the image of $X$ under the natural map $\mathbb{Q}[X]\to K$. Let $L=\mathbb{Q}(\zeta+1/\zeta)$. All that needs to be done is to define a total ordering on $L$ which makes $L$ into an ordered field, and for which $-2\le\zeta+1/\zeta\le 2$. For, if this is accomplished then the classical proof works: any element of $K$ is $a+b\zeta$ with $a,b\in L$, and -$$ -N_{K/L}(a+b\zeta)=(a+b\zeta)(a+b/\zeta)=a^2+b^2+ab(\zeta+1/\zeta)\ge a^2+b^2-2ab = (a-b)^2 \ge 0, -$$ -so for any $h(X)\in\mathbb{Q}[X]$ it follows that -$$ -N_{K/\mathbb{Q}}(h(\zeta)) = \prod_{i=1}^{(l-1)/2} h(\zeta^i) h(1/\zeta^i) -= \prod_{i=1}^{(l-1)/2} N_{K/L}(h(\zeta)) \ge 0. -$$ -It remains only to define a total ordering on $L$ which is compatible with the field operations, and for which $-2\le\zeta+1/\zeta\le 2$. If $\ell=3$ then $\mathbb{Q}(\zeta+1/\zeta)=\mathbb{Q}$, and there is nothing to prove. So assume $\ell>3$. Let $f(X)$ be the minimal polynomial of $\zeta+1/\zeta$ over $\mathbb{Q}$. For any rational number $\beta$, define $\zeta+1/\zeta$ to be less than $\beta$ if $\beta$ is greater than all rational numbers which are smaller than all positive rational numbers $\gamma$ for which $f(\gamma)\cdot f(0)<0$. If $\zeta+1/\zeta$ is not less than $\beta$, then define $\zeta+1/\zeta$ to be greater than $\beta$. Exercise for the OP: verify that this ordering has a unique extension to a total ordering on $L$ which is compatible with the field operations. QED<|endoftext|> -TITLE: Deciding whether $2^{\sqrt2}$ is irrational/transcendental -QUESTION [7 upvotes]: Is $2^\sqrt{2}$ irrational? Is it transcendental? - -REPLY [10 votes]: According to Gel'fond's theorem, if $\alpha$ and $\beta$ are algebraic numbers (which $2$ and $\sqrt 2$ are) and $\beta$ is irrational, then $\alpha^\beta$ is transcendental, except in the trivial cases when $\alpha$ is 0 or 1. -Wikipedia's article about the constant $2^{\sqrt 2}$ says that it was first proved to be transcendental in 1930, by Kuzmin.<|endoftext|> -TITLE: Evaluate :$\int \frac{\sin^{-1} \sqrt{x} -\cos^{-1} \sqrt{x}}{\sin^{-1} \sqrt{x} +\cos^{-1} \sqrt{x}} dx$ -QUESTION [9 upvotes]: How to evaluate -$$ -\int \frac{\sin^{-1} \sqrt{x} -\cos^{-1} \sqrt{x}}{\sin^{-1} \sqrt{x} +\cos^{-1} \sqrt{x}} dx -$$ -I know that $\sin^{-1} \sqrt{x} +\cos^{-1} \sqrt{x}=\frac{\pi}{2}$ but after that I have no idea, so please help me. Thanks in advance. -I tried this way, -$$ -\int \frac{\sin^{-1} \sqrt{x} -\cos^{-1} \sqrt{x}}{\frac{\pi}{2}}dx -$$ -then I put the value $\sin^{-1} \sqrt{x} +\cos^{-1} \sqrt{x}=\frac{\pi}{2}$ so -$$ -\frac{2}{\pi}\int\left(\sin^{-1} \sqrt{x} -\left(\frac{\pi}{2}-\sin^{-1} \sqrt{x}\right)\right)dx -$$ -Is this right? -after that I integrate by part and get, -$$ \int \frac{\sqrt{x}}{\sqrt{1-x}}$$ now,what can i do? - -REPLY [9 votes]: Let $$ I_0=\int \frac{\sin^{-1} \sqrt{x} -\cos^{-1} \sqrt{x}}{\sin^{-1} \sqrt{x} +\cos^{-1} \sqrt{x}} dx $$$$\Rightarrow I_0=\int\frac{\frac{\pi}{2}-2\cos^{-1}\sqrt{x}}{\frac{\pi}{2}}dx$$$$\Rightarrow I_0=\int (1-\frac{4}{\pi}\cos^{-1}\sqrt{x})dx$$ $$\Rightarrow I_0=x-\frac{4}{\pi}\int\cos^{-1}\sqrt{x})dx$$ Now Consider $$I_1= \int\cos^{-1}\sqrt{x}dx$$ $$\Rightarrow I_1=\int 2z\cos^{-1} zdz$$ Where $$ x=z^2$$ Hence Integrating by parts we get $$ I_1 = 2zcos^{-1}z+ \int \frac{z^2}{\sqrt{1-z^2}}dz$$ $$I_1 = 2zcos^{-1}z+ \int \frac{1}{\sqrt{1-z^2}}dz-\int\sqrt{1-z^2}dz$$ $$\int \frac{1}{\sqrt{1-z^2}}dz=-\cos^{-1}z$$ $$\int\sqrt{1-z^2}dz=\frac{z\sqrt{1-z^2}}{2}+\frac{1}{2}\sin^{-1}z$$<|endoftext|> -TITLE: Is this metric space complete? -QUESTION [5 upvotes]: No, it is not complete metric space: by Stone-Weierstrass thm we know that $|x|$ can be uniformly approximated by sequence of polynomials which are clearly $\mathcal{C}^1[0,1]$, but $|x|$ is not $\mathcal{C}^1$. -Is my argument correct? - -REPLY [3 votes]: [Added: By $C[0,1]$ I mean the set of continuous functions $f: [0,1] \rightarrow \mathbb{R}$ endowed with the metric $d(f,g) = \max_{x \in [0,1]} |f(x)-g(x)|$. This is a complete metric space by the Cauchy Criterion for Uniform Convergence.] -Yes. To recap it: you have a complete metric space, $\mathcal{C}[0,1]$, and a subspace, $\mathcal{C}^1[0,1]$, which is not closed (rather, it is proper and dense). Therefore $\mathcal{C}^1[0,1]$ cannot be complete. -Added: As t.b. points out, the absolute value function is $C^1$ on the interval $[0,1]$, so you should pick something else (e.g. what t.b. says). I also agree that Weierstrass Approximation is much more than you need here. For instance, in Example 7 of these notes I show -- in an intentionally clunky, hands-on fashion -- that the absolute value function is a uniform limit of $C^1$-functions on $[-1,1]$.<|endoftext|> -TITLE: Degree of Toric Divisors -QUESTION [6 upvotes]: Is it possible to calculate the degree of a toric divisor directly from the fan of the toric -variety? -If so, how is this done? Or is there some alternative way to calculate the degree of these divisors? - -REPLY [3 votes]: Let the toric variety be $X$; let $d = \mathrm{dim} \ X$. For any Cartier $T$-divisor $D$, we can describe $\dim H^0(X, \mathcal{O}(D))$ as the number of lattice points in a certain polytope $P_D$. See Section 5.3 of Fulton's Introduction to Toric Varieties. The degree of $D$ is $d! \mathrm{Vol}(P_D)$. -Proof: Let $h(n)$ be the Hilbert polynomial $h(n) = \dim H^0(X, \mathcal{O}(nD))$. We know that the leading term of $h(n)$ is $\mathrm{deg}(\mathcal{O}(D)) n^d/d!$. But also, the number of lattice points in $n \cdot P_D$ is $\mathrm{Vol}(P_D) n^d + O(n^{d-1})$. Equating these two, $\mathrm{deg}(\mathcal{O}(D)) = d! \mathrm{Vol}(P_D)$. $\square$<|endoftext|> -TITLE: how exactly did calculus change our understanding of the world? -QUESTION [11 upvotes]: I am taking calculus course and I keep wondering if this is really necessary. I know it is the cornerstone of modern science but what I don't understand is why and how. -Was it impossible to pursue physics and engineering had calculus not been invented, or without it being applied? What exactly has calculus opened up for us that pre calculus math hadn't? - -REPLY [5 votes]: Newton's laws, discovered more than 300 years ago, describe the motion of all objects that are not too fast, small, or heavy. -The discovery of Newton's laws signalled the beginning of physics as a truly predictive science. -Newton's second law is a second order differential equation${}^\dagger$, -\begin{equation*} -\frac{d^2{\bf x}}{dt^2} = \frac{{\bf F}}{m}.\tag{1} -\end{equation*} -In words, the second derivative of the position of an object with respect to time, the acceleration, is directly proportional to the net force impressed upon the -object, and inversely proportional to the mass of the object. -Without calculus and the machinery of differential equations, (1) is one of nature's incomprehensible secrets. -Given the object's initial position and velocity, Newton's laws can be used to predict its subsequent motion. -If the initial conditions and the forces are known with great certainty this prediction will describe everything about the motion of the object far into the future. -Every day countless physicists and engineers rely on Newton's laws to find out where the artillery round will land, how will the car crumple on impact, whether the bridge will fall down in a strong wind, and so on. - - -${}^\dagger$Differential equations relate derivatives of a function to the function itself.<|endoftext|> -TITLE: Distance between a point and a m-dimensional space in n-dimensional space ($m -TITLE: Why use a Kalman filter instead of keeping a running average? -QUESTION [24 upvotes]: I've been trying to understand Kalman filters. Here are some examples that have helped me so far: - -http://bilgin.esme.org/BitsBytes/KalmanFilterforDummies.aspx -https://scipy-cookbook.readthedocs.io/items/KalmanFiltering.html - -These use the algorithm to estimate some constant voltage. How could using a Kalman filter for this be better than just keeping a running average? Are these examples just oversimplified use cases of the filter? -(If so, what's an example where a running average doesn't suffice?) -EDIT: -For example, consider the following Java program and output. The Kalman output doesn't match the average, but they're very close. Why pick one over the other? -int N = 10; // Number of measurements - -// measurements with mean = .5, sigma = .1; -double z[] = -{ - 0, // place holder to start indexes at 1 - 0.3708435, 0.4985331, 0.4652121, 0.6829262, 0.5011293, - 0.3867151, 0.6391352, 0.5533676, 0.4013915, 0.5864200 -}; - -double Q = .000001, // Process variance - R = .1*.1;// Estimation variance - -double[] xhat = new double[N+1],// estimated true value (posteri) -xhat_prime = new double[N+1], // estimated true value (priori) -p = new double[N+1], // estimated error (posteri) -p_prime = new double[N+1],// estimated error (priori) -k = new double[N+1]; // kalman gain - -double cur_ave = 0; - -// Initial guesses -xhat[0] = 0; -p[0] = 1; - -for(int i = 1; i <= N; i++) { - // time update - xhat_prime[i] = xhat[i-1]; - p_prime[i] = p[i-1] + Q; - - // measurement update - k[i] = p_prime[i]/(p_prime[i] + R); - xhat[i] = xhat_prime[i] + k[i]*(z[i] - xhat_prime[i]); - p[i] = (1-k[i])*p_prime[i]; - - // calculate running average - cur_ave = (cur_ave*(i-1) + z[i])/((double)i); - - System.out.printf("%d\t%04f\t%04f\t%04f\n", i, z[i], xhat[i], cur_ave); -} - -output: - Iter Input Kalman Average - 1 0.370844 0.367172 0.370844 - 2 0.498533 0.432529 0.434688 - 3 0.465212 0.443389 0.444863 - 4 0.682926 0.503145 0.504379 - 5 0.501129 0.502742 0.503729 - 6 0.386715 0.483419 0.484227 - 7 0.639135 0.505661 0.506356 - 8 0.553368 0.511628 0.512233 - 9 0.401392 0.499365 0.499917 - 10 0.586420 0.508087 0.508567 - -REPLY [14 votes]: In fact, they are the same thing in certain sense, I will show your something behind Kalman filter and you will be surprised. -Consider the following simplest problem of estimation. We are given a series of measurement -$z_1, z_2, \cdots, z_k$, of an unknown constant $x$. We assume the additive model -\begin{eqnarray} -z_i= x + v_i, \; i=1,2, \cdots, k ~~~~~~~~~~~ (1) -\end{eqnarray} -where $v_i$ are measurement noises. If nothing else is known, then everyone will agree that a reasonable estimate of $x$ given the $k$ measurements can be given by -\begin{eqnarray} -\hat{x}_k= \frac{1}{k} \sum_{i=1}^{k} z_i ~~~~~~~~~~~ ~~~~~~~~~~~ (2) -\end{eqnarray} -this is average. -Now we can re-write above eq.(2) by simple algebraic manipulation to get -\begin{eqnarray} -\hat{x}_k= \hat{x}_{k-1} + \frac{1}{k} (z_k-\hat{x}_{k-1}) ~~~~~~~~~~~ (3) -\end{eqnarray} -Eq.(3) which is simply Eq.(2) expressed in recursive form has an interesting interpretation. It says that the best estimate of $x$ after $k$ measurement is the best estimate of $x$ after $k-1$ measurements plus a correction term. The correction term is the difference between what you expect to measure based on $k-1$ measurement, i.e., and what you actually measure $z_k$. -If we label the correction $\frac{1}{k}$ as $P_k$, then again simply algebraic manipulation can write the recursive form of $P_k$ as -\begin{eqnarray} -P_k=P_{k-1}-P_{k-1}(P_{k-1}+1)^{-1}P_{k-1} ~~~~~~~~~~~ (4) -\end{eqnarray} -Believe it or not, Eqs.(3-4) can be recognized as the Kalman filtering equations for this simple case. -Any discussion is welcomed. -Reference: -Explaining Filtering (Estimation) in One Hour, Ten Minutes, One Minute, and One Sentence by Yu-Chi Ho<|endoftext|> -TITLE: Updates on Lehmer's Totient Problem -QUESTION [12 upvotes]: As I read here and in many books on the Theory of Numbers, we are yet to prove or disprove the existence of any composite $n$ such that $\phi(n)\mid n-1$. -Is there progress in this line? - -REPLY [10 votes]: Depends what you call progress. -Grau Ribas and Luca, Cullen numbers with the Lehmer property, Proc. Amer. Math. Soc. 140 (2012), no. 1, 129–134, MR2833524 (2012e:11002), prove there are no counterexamples of the form $k2^k+1$. -Burcsi, Czirbusz, and Farkas, Computational investigation of Lehmer's totient problem, Ann. Univ. Sci. Budapest. Sect. Comput. 35 (2011), 43–49, MR2894552, prove that if $n$ is composite and $k\phi(n)=n-1$ and $n$ is a multiple of 3 then $n$ has at least 40000000 prime divisors, and $n\ge10^{360000000}$. -There's more. If you have access to MathSciNet, just type in Lehmer and totient, and see what comes up.<|endoftext|> -TITLE: Boundary of product manifolds such as $S^2 \times \mathbb R$ -QUESTION [7 upvotes]: Simple question but I am confused. -What is the boundary of $S^2\times\mathbb{R}$? Is it just $S^2$? -What would be the general way to evaluate the boundary of a product manifold? -Thanks for the replies! - -REPLY [6 votes]: If you are interested in differentiable manifolds, a variation of what joriki said is still true. If $X$ is a manifold without boundary and $Y$, a manifold with boundary $\partial Y$, then $X\times Y$ is a manifold with boundary $X\times \partial Y$. For more details, you can look at many places, e.g. Guillemin and Pollack Differential Topology, Chapter 2.<|endoftext|> -TITLE: Schwarz inequality for unital completely positive maps -QUESTION [6 upvotes]: I came across the following form of Schwarz inequality for completely positive maps in Arveson's paper: - -Let $\delta:\mathcal{A}\to\mathcal{B}$ be a unital completely positive linear map between two $C^*$-algebras, then \begin{equation}\delta(A)^*\delta(A)\le \delta(A^*A),\end{equation} - -which is a crucial part in proving proposition 3. -I do not know this inequality and thus searched on wiki, something related is - -(Kadison-Schwarz) If $\phi$ is a unital positive map, then \begin{equation}\phi(a^*a)\ge\phi(a^*)\phi(a)\end{equation} for all normal elements $a$. - -However, I cannot find a proof of Kadison-Schwarz. Also, since Kadison-Schwarz works only for normal elements, there seems to be a gap between Kadison-Schwarz and Schwarz. -I wonder where I can find a proof to the first inequality. -Thanks! - -REPLY [7 votes]: Note that Kadison-Schwarz is stated for positive (not necessarily cp) maps; that's why the restriction to normals is required (the C$^*$-algebra generated by a normal is abelian, and then any positive map is completely positive). -Also, for Schwarz inequality all that is required is 2-positivity (which of course is implied by complete positivity). -This is the proof of the inequality as in Paulsen's book: -We have -$$ -\begin{bmatrix}1&a \\ a^*& a^*a\end{bmatrix}=\begin{bmatrix}1&a\\ 0&0\end{bmatrix}^*\begin{bmatrix}1&a\\ 0&0\end{bmatrix}\geq0. -$$ -Then -$$ -0\leq\delta^{(2)}\left(\begin{bmatrix}1&a \\ a^*& a^*a\end{bmatrix}\right) -=\begin{bmatrix}1&\delta(a) \\ \delta(a)^*& \delta(a^*a)\end{bmatrix} -$$ -Applying this in particular to the vector $\begin{bmatrix}-\delta(a)\eta\\ \eta\end{bmatrix}$, we get -$$ -0\leq\left\langle\begin{bmatrix}1&\delta(a) \\ \delta(a)^*& \delta(a^*a)\end{bmatrix}\begin{bmatrix}-\delta(a)\eta\\ \eta\end{bmatrix},\begin{bmatrix}-\delta(a)\eta\\ \eta\end{bmatrix}\right\rangle=\langle(\delta(a^*a)-\delta(a)^*\delta(a))\eta,\eta\rangle. -$$ -As the vector $\eta$ can be chosen arbitrarily, we conclude that $\delta(a^*a)-\delta(a)^*\delta(a)\geq0$.<|endoftext|> -TITLE: Continuous Actions and Homomorphisms -QUESTION [5 upvotes]: I am learning about the compact-open topology and have a small proposition I am struggling to prove. Let $G$ be a topological group, $X$ a compact, Hausdorff space, and $H(X)$, the homeomorphisms of $X$, have the compact open topology. I want to show that an action of $G$ on $X$, call it $\gamma_1:G\times X\rightarrow X$ is continuous iff its associated homomorphism $\gamma_2:G\rightarrow H(X)$, where $g\mapsto \phi_g$ (left-translation by $g$), is continuous. -I can do the $\Leftarrow$ direction: assuming $\gamma_2$ is continuous, the map $(g,x)\mapsto (\phi_g,x)$ is continuous. Since the evaluation map $(\phi_g,x)\mapsto \phi_g(x)$ is continuous, and $\gamma_1$ is the composition of these two, $\gamma_2$ is continuous. -The other direction is where I'm not sure how to proceed. We can take $\phi_g\in H(X)$ and a subbasis set $S(C,U)=\{f:f(C)\subseteq U\}$ for $C$ compact, $U$ open, such that $\phi_g\in S(C,U)$. So $\gamma_2^{-1}(S(C,U))=\{h\in G:h\cdot C\subseteq U\}$. This is all true, but I am not sure it is helpful. I am not sure what the right approach is; in particular I don't see how/when to leverage the fact that $\gamma_1$ is continuous. -Thanks for any hints, direction, insight, etc. - -REPLY [4 votes]: I think, it is quite important to carefully write down what one is supposed to do in a situation like this one, as it is really easy to get confused. -You are given a subbasis element $S(C,U)$ with $C\subset X$ compact, and $U\subset X$ open. To prove continuity of $\gamma_2$ it suffices to show that for any $g\in G$ satisfying $g(C)\subset U$ there exists a neighbourhood $V\subset G$ of $g$ such that -$$ h\in V \implies h(C) \subset U$$ -Luckily we are only given one information about the action of $G$ on $X$, so we immediately now where to start: Fix $g\in G$ as above. We know that for any $x\in C$ we have $g(x) \in U$. Also fix $x$ for the moment. -By assumption on the continuity of $\gamma_1$ there are neighbourhoods $W_x \subset X$ of $x$ and $V_x\subset G$ of $g$ such that -$$(h,y)\in V_x\times W_x \implies h(y) = \gamma_1(h,y)\in U$$ -Now finitely many of these $W_x$ cover $C$, say $C\subset W_{x_1}\cup\dots\cup W_{x_n}$. -Let $V = V_{x_1}\cap \dots \cap V_{x_n}$. Then $V$ is an open neighbourhood of $g$ in $G$ and I'll leave it to you to verify that $\gamma_2(V)\subset S(C,U)$.<|endoftext|> -TITLE: Is $\pmb{\eta}\cdot\pmb{\omega_1} = (\pmb{\eta} + \pmb{1})\cdot\pmb{\omega_1}$? -QUESTION [11 upvotes]: $\pmb{\eta}$ - order type of $\mathbb{Q}$. -$\pmb{1}$ - order type of a singleton set. -$\pmb{\omega_0}$ - order type of $\mathbb{N}$. -$\pmb{\omega_1}$ - order type of the first uncountable ordinal. - -It is easy to see that $\pmb{\eta}\cdot\pmb{\omega_0} = (\pmb{\eta} + \pmb{1})\cdot\pmb{\omega_0}$, in fact, both sides are $\pmb{\eta}$. -Question: -Is $\pmb{\eta}\cdot\pmb{\omega_1} = (\pmb{\eta} + \pmb{1})\cdot\pmb{\omega_1}$? - -REPLY [6 votes]: Here is a proposed solution, I have not verified all the details, but I believe this should work. -Let $\mathbb Q^\ast$ be the rational numbers plus an endpoint, let $A,B$ be a partition of this set to intervals such that $A$ has order type $\eta$. In $\mathbb Q$ fix some partition into two parts $X,Y$ such that both are intervals and $X$ is of order type $\eta+1$. -For $\alpha<\omega_1$ we write $A_\alpha,B_\alpha,X_\alpha,Y_\alpha$ to be the corresponding parts in the $\alpha$-th copies of $\mathbb Q,\mathbb Q^\ast$. -Now we define by induction: - -For $\alpha=0$ simply send $A_0+B_0$ into $X_0$, and $A_1$ into $Y_0$. -If $\alpha$ is a limit ordinal, do the same. Namely $A_\alpha+B_\alpha$ into $X_\alpha$ and $A_{\alpha+1}$ into $Y_\alpha$. -If $\alpha=\beta+1$, send $B_\alpha$ into $X_\alpha$ and $A_{\alpha+1}$ into $Y_\alpha$. - -It is clear that arriving at any limit ordinal $\alpha$ we have an isomorphism of $(\eta+1)\cdot\alpha$ into $\eta\cdot\alpha$, so the step taken at the limit ordinal itself is well-defined (we do not need to worry about embedding $A_\alpha$ in a prior step). -It is clearly an order isomorphism, and it is a bijection for obvious reasons too.<|endoftext|> -TITLE: Proving $\int_{1}^{\infty}\frac{\sin x}{\left(\log x\right)^{\frac{1}{2}}}dx$ converges -QUESTION [6 upvotes]: Prove that the following improper integral converges $$\int_{1}^{\infty}\frac{\sin x}{\left(\log x\right)^{\frac{1}{2}}}dx.$$ - -I see that you can show this using Dirichlet's Convergence Test, but how would you show it not using this test? - -REPLY [6 votes]: You have two issues to address. First, is $\int_1^{1.1}\frac{\sin x}{\sqrt{\log x}}\,dx$ convergent? Second, is $\int_{1.1}^{\infty}\frac{\sin x}{\sqrt{\log x}}\,dx$ convergent? -The first integral blows up at its left end point. But we can show that for the some constant $C$, $\frac{\sin x}{\sqrt{\log x}}<\frac{C}{\sqrt{x-1}}$ for $x$ in $(1,1.1)$. Then by direct comparison, this piece is convergent. The numerator $\sin(x)$ is clearly positive and less than $1$. So we'd need to show that $\log x>k(x-1)$ for some constant $k$, for all $x\in(1,1.1)$. You can do this in the standard way by noting both are equal at $x=0$ and comparing derivatives. -The second integral breaks up into a sequence of alternating integrals $$\int_{1.1}^{\pi}\frac{\sin x}{\sqrt{\log x}}\,dx+\sum_{n=1}^{\infty}\int_{n\pi}^{n\pi+\pi}\frac{\sin x}{\sqrt{\log x}}\,dx$$ By the alternating test for integrals, we need to check that $\lim_{n\to\infty}\left|\int_{n\pi}^{n\pi+\pi}\frac{\sin x}{\sqrt{\log x}}\,dx\right|=0$. This is true, since -$$\begin{align} -\left|\int_{n\pi}^{n\pi+\pi}\frac{\sin x}{\sqrt{\log x}}\,dx\right|&<\frac{1}{\sqrt{\log(n\pi)}}\left|\int_{n\pi}^{n\pi+\pi}\sin(x)\,dx\right|\\ -&=\frac{2}{\sqrt{\log(n\pi)}}\end{align}$$ -Joriki reminded me that the alternating series test also requires that the absolute values of the integrals be decreasing. (Otherwise you could have say, the positive integrals going down like $1/n$ with the negative ones going down like $1/2^n$, and the whole thing diverges.) So we should check that -$$\begin{align} -\left|\int_{n\pi}^{n\pi+\pi}\frac{\sin x}{\sqrt{\log x}}\,dx\right|&>\left|\int_{n\pi+\pi}^{n\pi+2\pi}\frac{\sin x}{\sqrt{\log x}}\,dx\right|\end{align}$$ which is true since the numerator function just repeats another period with the same values, while the denominator function becomes larger.<|endoftext|> -TITLE: Research done by high-school students -QUESTION [59 upvotes]: I'm giving a talk soon to a group of high-school students about open problems in mathematics that high-school students could understand. To inspire them, I would like to give them examples of high-school students who have made original contributions in mathematics. One example I have is the 11th-grader from Hawai'i named Kang Ying Liu who in 2010 "discover[ed] nine new geometric formulas for describing triangle inequalities." -Do you have any other examples of high-school students who have made original contributions in mathematics? - -REPLY [5 votes]: Sylvain Cappell's paper, "The Theory of Semi-cyclical Groups with Special Reference to Non-Aristotelian Logic," won him the 1963 Westinghouse Talent Search when he was 16 years old, as recollected in this Boy's Life article. This was the first of Sylvain's many important mathematical discoveries. Tangentially interesting is that the runner-up in this contest, Sylvain's rival, was a certain William Leonard Pickard.<|endoftext|> -TITLE: If the covariance matrix is $\Sigma$, the covariance after projecting in $u$ is $u^T \Sigma u$. Why? -QUESTION [6 upvotes]: I read in this answer that: - -If covariance matrix is $\Sigma$, the covariance after projecting in - $u$ is $u^T \Sigma u$. - -I fail to see this, how do I get the covariance of a set of points after projecting those points along the direction $u$ as a function of $u$ and $\Sigma$ ? - -REPLY [10 votes]: The covariance matrix for a vector quantity $x$ is $\langle xx^\top\rangle-\langle x\rangle\langle x^\top\rangle$. The covariance for the projection $u^\top x$ is -$$\langle u^\top xx^\top u\rangle-\langle u^\top x\rangle\langle x^\top u\rangle=u^\top\langle xx^\top\rangle u-u^\top\langle x\rangle\langle x^\top\rangle u=u^\top\left(\langle xx^\top\rangle-\langle x\rangle\langle x^\top\rangle\right)u\;.$$ -The point is basically that you can pull $u$ out of all the expectation values because it's a constant.<|endoftext|> -TITLE: Is $(\pmb{1} + \pmb{\eta})\cdot\pmb{\omega_1} = \pmb{1} + \pmb{\eta}\cdot\pmb{\omega_1}$? -QUESTION [17 upvotes]: $\pmb{\eta}$ - order type of $\mathbb{Q}$. -$\pmb{1}$ - order type of a singleton set. -$\pmb{\omega_0}$ - order type of $\mathbb{N}$. -$\pmb{\omega_1}$ - order type of the first uncountable ordinal. - -It is easy to see that $\pmb{\eta}\cdot\pmb{\omega_0} = (\pmb{\eta} + \pmb{1})\cdot\pmb{\omega_0} = \pmb{\eta}$ and $(\pmb{1} + \pmb{\eta})\cdot\pmb{\omega_0} = \pmb{1} + \pmb{\eta}\cdot\pmb{\omega_0} = \pmb{1} + \pmb{\eta}$. -As it was shown in the answer to my last question, we also have $(\pmb{\eta} + \pmb{1})\cdot\pmb{\omega_1} = (\pmb{\eta}+\pmb{1})\cdot\pmb{\omega_0}\cdot\pmb{\omega_1}= \pmb{\eta}\cdot\pmb{\omega_1} $. -Question: -Is $(\pmb{1} + \pmb{\eta})\cdot\pmb{\omega_1} = \pmb{1} + \pmb{\eta}\cdot\pmb{\omega_1}$? - -REPLY [16 votes]: Here is a somewhat relaxed way to see that the orders $1+\eta\cdot\omega_1$ and $(1+\eta)\cdot\omega_1$ are not isomorphic, by finding a topological feature of one order not exhibited in the other. -First, observe that $(1+\eta)\cdot\omega_1$ has a closed subset of -order type $\omega_1$, namely, the points corresponding to the -initial point of each of the intervals $1+\eta$ used to create it. This suborder contains limit points for all its increasing $\omega$-sequences. -But meanwhile, $1+\eta\cdot\omega_1$ has no such suborder. This is because this order contains no limits of sequences of points from distinct intervals (the $\eta$ intervals used to build it), but any subset -of order type $\omega_1$ will have increasing sequences of points from distinct intervals. -So they are not isomorphic.<|endoftext|> -TITLE: A binomial identity -QUESTION [15 upvotes]: I was wandering if someone knows an elementary proof of the following identity: -$$ -\frac{(a)_n (b)_n}{(n!)^2} = \sum_{k=0}^n (-1)^k {1-a-b \choose k} -\frac{(1-a)_{n-k}(1-b)_{n-k}}{((n-k)!)^2}\ , -$$ -where $a,b$ are arbitrary real numbers, -$$(x)_0=1,\quad -(x)_n:=x(x+1)\cdots(x+n-1) \quad -\mbox{for $n\geq 1$} -$$ -is the Pochhammer's symbol, and -$$ -{x\choose 0}=1,\quad -{x\choose k} = \frac{x(x-1)\cdots (x-k+1)}{k!} \quad -\mbox{for $k\geq 1$}$$ -is a binomial coefficient. -The proof that I know uses the Hypergeometric differential equation. One has to continue analytically the solutions along a path connecting two singular points. This could be done by some well known integral representation of the Hypergeometric function. -I think that there should be a combinatorial proof. Since this is an identity between polynomials in $a$ and $b$, it is enough to prove it for $a$ and $b$ negative integers, i.e., we may assume that $a=-p$ and $b=-q$ where $p,q\in \mathbb{Z}_{\geq 0}$. The identity then turns into -$$ -{p\choose n}{q\choose n} = -\sum_{k=0}^n (-1)^k {1+p+q\choose k}{p+n-k\choose n-k}{q+n-k\choose n-k}. -$$ -I did not try very hard to proof the above identity and I did not search the literature that much, but since it comes from an interesting subject I think that it is worth finding an alternative proof. - -REPLY [2 votes]: Suppose we seek to evaluate -$$\sum_{k=0}^n (-1)^k -{1+p+q\choose k} {p+n-k\choose n-k} {q+n-k\choose n-k}$$ -which is claimed to be -$${p\choose n}{q\choose n}.$$ -Introduce -$${p+n-k\choose n-k} -= \frac{1}{2\pi i} -\int_{|z_1|=\epsilon} \frac{(1+z_1)^{p+n-k}}{z_1^{n-k+1}} -\; dz_1$$ -and -$${q+n-k\choose n-k} -= \frac{1}{2\pi i} -\int_{|z_2|=\epsilon} \frac{(1+z_2)^{q+n-k}}{z_2^{n-k+1}} -\; dz_2.$$ -Observe that these integrals vanish when $k\gt n$ and we may extend -$k$ to infinity. -We thus obtain for the sum -$$\frac{1}{2\pi i} -\int_{|z_1|=\epsilon} -\frac{(1+z_1)^{p+n}}{z_1^{n+1}} -\frac{1}{2\pi i} -\int_{|z_2|=\epsilon} -\frac{(1+z_2)^{q+n}}{z_2^{n+1}} -\\ \times \sum_{k\ge 0} {1+p+q\choose k} (-1)^k -\frac{z_1^k z_2^k}{(1+z_1)^k (1+z_2)^k} -\; dz_2\; dz_1.$$ -This is -$$\frac{1}{2\pi i} -\int_{|z_1|=\epsilon} -\frac{(1+z_1)^{p+n}}{z_1^{n+1}} -\frac{1}{2\pi i} -\int_{|z_2|=\epsilon} -\frac{(1+z_2)^{q+n}}{z_2^{n+1}} -\\ \times -\left(1-\frac{z_1 z_2}{(1+z_1)(1+z_2)}\right)^{p+q+1} -\; dz_2\; dz_1$$ -or -$$\frac{1}{2\pi i} -\int_{|z_1|=\epsilon} -\frac{(1+z_1)^{n-q-1}}{z_1^{n+1}} -\frac{1}{2\pi i} -\int_{|z_2|=\epsilon} -\frac{(1+z_2)^{n-p-1}}{z_2^{n+1}} -(1+ z_1 + z_2)^{p+q+1} -\; dz_2\; dz_1$$ -Supposing that $p\ge n$ and $q\ge n$ this may be re-written as -$$\frac{1}{2\pi i} -\int_{|z_1|=\epsilon} -\frac{1}{z_1^{n+1} (1+z_1)^{q+1-n}} -\frac{1}{2\pi i} -\int_{|z_2|=\epsilon} -\frac{1}{z_2^{n+1} (1+z_2)^{p+1-n}} -\\ \times (1+ z_1 + z_2)^{p+q+1} -\; dz_2\; dz_1$$ -Put $z_2 = (1+z_1) z_3$ so that $dz_2 = (1+z_1) \; dz_3$ to get -$$\frac{1}{2\pi i} -\int_{|z_1|=\epsilon} -\frac{1}{z_1^{n+1} (1+z_1)^{q+1-n}} -\frac{1}{2\pi i} -\int_{|z_2|=\epsilon} -\frac{1}{(1+z_1)^{n+1} z_3^{n+1} (1+(1+z_1)z_3)^{p+1-n}} -\\ \times (1+ z_1)^{p+q+1} (1+z_3)^{p+q+1} -\; (1+z_1) \; dz_3\; dz_1$$ -which is -$$\frac{1}{2\pi i} -\int_{|z_1|=\epsilon} -\frac{(1+z_1)^p}{z_1^{n+1}} -\frac{1}{2\pi i} -\int_{|z_2|=\epsilon} -\frac{1}{z_3^{n+1} (1+ z_3 + z_1 z_3)^{p+1-n}} -\\ \times (1+z_3)^{p+q+1} -\; dz_3\; dz_1 -\\ = \frac{1}{2\pi i} -\int_{|z_1|=\epsilon} -\frac{(1+z_1)^p}{z_1^{n+1}} -\frac{1}{2\pi i} -\int_{|z_2|=\epsilon} -\frac{(1+z_3)^{n+q}}{z_3^{n+1} (1 + z_1 z_3 /(1+z_3))^{p+1-n}} -\; dz_3\; dz_1$$ -Extracting the residue for $z_1$ first we obtain -$$\sum_{k=0}^n {p\choose n-k} -\frac{(1+z_3)^{n+q}}{z_3^{n+1}} -{k+p-n\choose k} -(-1)^k \frac{z_3^k}{(1+z_3)^k}.$$ -The residue for $z_3$ then yields -$$\sum_{k=0}^n (-1)^k {p\choose n-k} -{k+p-n\choose k} {n-k+q\choose n-k}.$$ -The sum term here is -$$\frac{p!\times (p+k-n)!\times (q+n-k)!} -{(n-k)! (p+k-n)! \times k! (p-n)! \times (n-k)! q!}$$ -which simplifies to -$$\frac{p!\times n! \times (q+n-k)!} -{(n-k)! \times n!\times k! (p-n)! \times (n-k)! q!}$$ -which is -$${n\choose k} {p\choose n}{q+n-k\choose q}$$ -so we have for the sum -$${p\choose n} -\sum_{k=0}^n {n\choose k} (-1)^k {q+n-k\choose q}.$$ -To evaluae the remaining sum we introduce -$${q+n-k\choose q} -= \frac{1}{2\pi i} -\int_{|v|=\epsilon} \frac{(1+v)^{q+n-k}}{v^{q+1}} \; dv$$ -getting for the sum -$${p\choose n} -\frac{1}{2\pi i} -\int_{|v|=\epsilon} -\frac{(1+v)^{q+n}}{v^{q+1}} -\sum_{k=0}^n {n\choose k} (-1)^k \frac{1}{(1+v)^k} \; dv -\\ = {p\choose n} -\frac{1}{2\pi i} -\int_{|v|=\epsilon} -\frac{(1+v)^{q+n}}{v^{q+1}} -\left(1-\frac{1}{1+v}\right)^n \; dv -\\ = {p\choose n} -\frac{1}{2\pi i} -\int_{|v|=\epsilon} -\frac{(1+v)^{q}}{v^{q-n+1}} \; dv -= {p\choose n} {q\choose q-n}$$ -which is $${p\choose n} {q\choose n}.$$ -This concludes the argument.<|endoftext|> -TITLE: Linearly ordered sets "somewhat similar" to $\mathbb{Q}$ -QUESTION [11 upvotes]: $\pmb{\eta}$ - order type of $\mathbb{Q}$. -$\pmb{\eta} + \pmb{1}$ - order type of $\mathbb{Q}\cap(0, 1]$. -$\pmb{\omega_1}$ - order type of the first uncountable ordinal. - -Let's say that a linear order type is q-like if every proper initial segment of an instance of this type is order-isomorphic to $\pmb{\eta}$ or $\pmb{\eta}+\pmb{1}$. For example, $\pmb{\eta}$, $\pmb{\eta}+\pmb{1}$, $\pmb{\eta}\cdot\pmb{\omega_1}$ and $\pmb{\eta}+(\pmb{1}+\pmb{\eta})\cdot\pmb{\omega_1}$are q-like. -Question: How many distinct q-like order types exist? - -REPLY [14 votes]: This is a great question! -There are a large uncountable number of these orders. -Consider as a background order the long rational line, the order -$\mathcal{Q}=([0,1)\cap\mathbb{Q})\cdot\omega_1$. For each -$A\subset\omega_1$, let $\mathcal{Q}_A$ be the suborder obtained -by keeping the first point from the $\alpha^{\rm th}$ interval -whenever $\alpha\in A$, and omitting it if $\alpha\notin A$. That -is, -$$\mathcal{Q}_A=((0,1)\cap\mathbb{Q})\cdot\omega_1\ \cup\ \{(0,\alpha)\mid\alpha\in -A\}.$$ -This corresponds to adding $\omega_1$ many copies of $\eta$ or -$1+\eta$, according to the pattern specified by $A$ as a subset of $\omega_1$. -I claim that the isomorphism types of these orders, except for the question of a least element, correspond precisely to -agreement-on-a-club for $A\subset\omega_1$. -Theorem. $\mathcal{Q}_A$ is isomorphic to $\mathcal{Q}_B$ if -and only if $A$ and $B$ agree on having $0$ and also agree -modulo the club filter, meaning that there is a closed unbounded set $C\subset\omega_1$ -such that $A\cap C=B\cap C$. In other words, this is if and only -if $A$ and $B$ agree on $0$ and are equivalent to $B$ in $P(\omega_1)/\text{NS}$, as subsets -modulo the nonstationary ideal. -Proof. If $A$ and $B$ agree on $0$ and agree on a club $C$, then we may build an -isomorphism between $\mathcal{Q}_A$ and $\mathcal{Q}_B$ by -transfinite induction. Namely, for each $\alpha\in C$, we will -ensure that $f$ restricted to the cut below $(0,\alpha)$ is an -isomorphism of the segment in $\mathcal{Q}_A$ to that in -$\mathcal{Q}_B$. The point is that $(0,\alpha)$ is actually a -point in $\mathcal{Q}_A$ if and only if it is point in -$\mathcal{Q}_B$, and so these points provide a common frame on -which to carry out the transfinite recursion. If we have an -isomorphism up to such a point, we can continue it to the next -point, since this is just adding a copy of $\eta$ or of $1+\eta$ -on top of each (the same for each), and at limits we take the -union of what we have built so far, which still fulfills the -property because $C$ is closed. So $\mathcal{Q}_A\cong\mathcal{Q}_B$. -Conversely, if $f:\mathcal{Q}_A\cong\mathcal{Q}_B$, then $A$ and $B$ must agree on $0$. Let $C$ be -the set of closure ordinals of $f$, that is, the set of $\alpha$ -such that $f$ respects the cut determined by the point -$(0,\alpha)$. This set $C$ is closed and unbounded in $\omega_1$. -Furthermore, it is now easy to see that $(0,\alpha)\in -\mathcal{Q}_A$ if and only if $(0,\alpha)\in \mathcal{Q}_B$ for -$\alpha\in C$, since this point is the supremum of that cut, and -it would have to be mapped to itself. Thus, $A\cap C=B\cap C$ and -so $A$ and $B$ agree modulo the club filter. QED -Corollary. There are $2^{\aleph_1}$ many distinct q-like linear orders up to isomorphism. -Proof. The theorem shows that there are as many different q-like linear -orders as there are equivalence classes of subsets of $\omega_1$ -modulo the non-stationary ideal. So the number of such orders is -$|P(\omega_1)/\text{NS}|$. This cardinality is $2^{\aleph_1}$ -because we may split $\omega_1$ into $\omega_1$ many disjoint -stationary sets, by a theorem of Solovay and Ulam, and the union -of any two distinct subfamilies of these differ on a stationary -set and hence do not agree on a club. -So there are at least $2^{\aleph_1}$ many distinct q-like orders -up to isomorphism, and there cannot be more than this, since every -such order has cardinality at most $\omega_1$. QED -Finally, let me point out, as Joriki mentions in the comments, that every uncountable q-like linear order is isomorphic to $\mathcal{Q}_A$ for some $A$. If $L$ is any such order, then select an unbounded $\omega_1$ sequence in $L$, containing none of its limits, chop $L$ into the corresponding intervals these elements and define $A$ according to whether these corresponding intervals have a least element or not. Thus, we have a complete characterization of the q-like linear orders: the four countable orders, and then the orders $\mathcal{Q}_A$ with two $A$ from each class modulo the nonstationary ideal, one with $0$ and one without.<|endoftext|> -TITLE: Is there a name for the "most square" factorization of an integer? -QUESTION [7 upvotes]: For the definition that follows, I'm curious to know if there's a known name (to enable a literature search relating to algorithms). -Definition. Given an integer $n$, the maximally square factorization consists of the integers $\{a,b\}$ such that $n=ab$ and the difference $|a-b|$ is minimized. Formally: -$$\{a,b\} = \arg \min_{\{x,y : x|n, y|n, xy=n\}} |x-y|.$$ -Examples: - -For $n=16$, we get $\{a,b\}=\{4,4\}$. -For $n=1300$, we get $\{a,b\}=\{26,50\}$. - -Questions: -1) Does this concept have a name? Any closely related concepts are also of interest. Number theory is not my strength. -2) Is there some clever way for finding $\{a,b\}$ that doesn't require fully factorizing $n$ and taking the two integers closest to $\sqrt{n}$? - -REPLY [7 votes]: Erdos conjectured that almost all integers have a pair $d,d'$ of divisors satisfying $d\lt d'\le2d$. ["Almost all" means that if you take the number of integers less than $n$ having this property, and divide by $n$, and then let $n$ go to infinity, the quotient will approach 1.] -This was proved, in a strengthened form, in Maier and Tenenbaum, On the set of divisors of an integer, Invent. Math. 76 (1984), no. 1, 121–128, MR0739628 (86b:11057). -Many other papers have taken the Maier-Tenenbaum paper as a starting point for further investigations. There's a whole book, Hall and Tenenbaum, Divisors, Cambridge Tracts in Mathematics, 90, Cambridge University Press, Cambridge, 1988, MR0964687 (90a:11107).<|endoftext|> -TITLE: Difference between "space" and "algebraic structure" -QUESTION [14 upvotes]: What is the difference between a "space" and an "algebraic structure"? For example, metric spaces and vector spaces are both spaces and algebraic structures. Is a group a space? Is a manifold a space or an algebraic structure, both or neither? - -REPLY [3 votes]: There's no precise definition; anyway, the way I look at it: - -There exist many different types of spaces (e.g. sets, posets, graphs, digraphs, metric spaces, uniform spaces, topological spaces, manifolds, etc.) There's no precise definition, but anyway, spaces usually form a distributive category. Furthermore, if $U : \mathbf{C} \rightarrow \mathbf{Set}$ denotes the relevant forgetful functor, then this usually preserves coproducts, and it usually has a left-adjoint $F$ such that the inclusion $UFS \leftarrow S$ is an isomorphism for each set $S$. -(For this reason, I wouldn't consider vector spaces to really be "spaces".) -We may consider spaces that are equipped with further algebraic structure, like a set equipped with the structure of a group, or a topological space equipped with the structure of a group, or a manifold equipped with the structure of a group. Etc. These can often be defined as finite-product preserving functors out of an appropriately chosen Lawvere theory. Categories of algebraic structures usually don't form a distributive category, despite that all relevant products and coproducts often exist, and the relevant forgetful functor to the relevant category of spaces usually doesn't preserve coproducts. But I think the biggest giveaway that we're not dealing with spaces is that if $U$ and $F$ are the relevant free and forgetful functors, there tend to exist spaces $S$ such that the inclusion $UFS \leftarrow S$ is not an isomorphism.<|endoftext|> -TITLE: Is every set a subset? -QUESTION [15 upvotes]: Is every set a subset of a larger set? In other words, for an arbitrary set S, can one always construct a set S' such that S is a proper subset of S'? -Is this question even meaningful? - -REPLY [12 votes]: You can even prove it without the Axiom of Regularity. -Lemma. Given a set $A$, there exists a set $B$ such that $B\notin A$. -Proof. Let $B=\{a\in A\mid a\notin a\}$. This is a set, by the Axiom of Separation. -I claim that $B\notin A$. Indeed, if $B\in A$ and $B\in B$, then $B\notin B$ by the definition of $B$. But if $B\in A$ and $B\notin B$, then $B\in B$ by the definition of $B$. This contradiction arises from the assumption that $B\in A$, hence $B\notin A$. $\Box$ -Now, given the set $A$, let $B$ be a set such that $B\notin A$. By the Axiom of Power Set, there is a set $P$ such that $B\in P$; by the Axiom of Separation, we obtain the set $\{B\}$. -Now we have $A$ and $\{B\}$ are sets. By the Axiom of Pairs, we have a set that consists exactly of $A$ and $\{B\}$, $X=\{A,\{B\}\}$. -Finally, by the Axiom of Union, there is a set $Y$ such that $Y=\cup\{A,\{B\}\} = \{x\mid x\in A\text{ or }x\in\{B\}\}$ (normally, we would write $Y = A\cup\{B\}$, but this is the notation in the Axiom of Union). -Now, $A\subseteq Y$ is immediate. And since $B\in Y$ but $B\notin A$, we have $A\neq Y$. Thus, $A\subsetneq Y$, as desired. -Note. If we assume the Axiom of Regularity, then $a\notin a$ holds for all sets $a$, hence $B=A$, and the construction just yields the set given by Brian Scott.<|endoftext|> -TITLE: How to get principal argument of complex number from complex plane? -QUESTION [17 upvotes]: I am just starting to learn calculus and the concepts of radians. Something that is confusing me is how my textbook is getting the principal argument ($\arg z$) from the complex plane. i.e. for the complex number $-2 + 2i$, how does it get $\frac{3\pi}{4}$? (I get $\frac{\pi}{4}$). -The formula is $\tan^{-1}(\frac{b}{a})$, and I am getting $\frac{\pi}{4}$ when I calculate $\tan^{-1}(\frac{2}{-2})$. When I draw it I see that the point is in quadrant 2. -So how do you compute the correct value of the principal argument? - -REPLY [3 votes]: I've come up with this recipe for principal argument. It saves messing around adding or subtracting $\pi$. -$$\text{Arg} (z) = n\ \text{cos}^{-1} \left(\frac{x}{z}\right)$$ -in which n = 1 if y ≥ 0 but n = -1 if y < 0. -I've tried to 'automate' the value of n, but the best I can do is -$$\text{Arg} (z) = \frac{y}{|y|}\ \text{cos}^{-1} \left(\frac{x}{z}\right).$$ -Unfortunately this fails for y = 0 (real z), so the y = 0 case would still have to be catered for separately. -Edit: A very ugly self-contained recipe would be -$$\text{Arg} (z) = \text{sgn}\left(\text{sgn}(y) + \frac{1}{2}\right)\ \text{cos}^{-1} \left(\frac{x}{z}\right).$$<|endoftext|> -TITLE: Elementary proof of the Hurwitz formula -QUESTION [9 upvotes]: I am aware of two forms of the Hurwitz formula. The first is more common, and deals only with the degrees. So if $f:X \rightarrow Y$ is a non-constant map of degree $n$ between two projective non-singular curves, with genera $g_X$ and $g_Y$, then -$$ -2(g_X-1) = 2n(g_Y-1) + \deg(R), -$$ -where $R$ is the ramification divisor of $f$. -The proof of this was given to me as an exercise when I started my PhD, and I am very happy with it. -However, in some other work that I was doing it appeared that one could strengthen this to say rather that if $K_X={\rm div}(f^*(dx))$ and $K_Y={\rm div}(dx)$ are canonical divisors of $X$ and $Y$, then -$$ -K_X = n\cdot K_Y + R. -$$ -I have found this alluded to in a number of places, and even stated in Algebraic Curves Over Finite Fields by Carlos Moreno. However, this was without proof, and every idea of a proof that I have seen is in sheaf theoretic language. I am slowly getting through sheaves and schemes, but I am currently trying to prove this in an elementary manner (fiddling around with orders of $dx$ etc.), in the wildly ramified case (the tamely ramified case is fine). -I would like to know if - -It is possible to prove this without using sheaves etc in the wildly ramified case -If so, are there any references that would help with this. - -edit: Also, is there a different name for the "more specific" Hurwitz formula? - -REPLY [8 votes]: You might like Griffiths's proof of Riemann-Hurwitz (Introduction to Algebraic Curves page 91, theorem (8.5) ) for a morphism $f:X\to Y$ . -It is very natural: he starts from a meromorphic differential form $\omega \in \Omega(Y)$ (whose existence he provisionally admits ) , lifts it to $f^*\omega \in \Omega(X)$, then computes and compares the divisors $div(f^*\omega), f^*(div(\omega))$ and $R$ locally in coordinates. -Riemann-Hurwitz is not difficult but somewhat explains the slightly confusing result that taking divisors and lifting do not commute for differential forms. -Edit -Stichtenoth gives a proof for fields of any characteristic in Corollary 3.4.13 in the language of function fields, but only under a separability assumption. -The most general formula I'm aware of (but I'm not in the least a specialist) and which does not require separability assumptions is Tate's Genus Formula. -You can find it as Corollary 9.5.20 in Villa's Topics in the Theory of Algebraic Function Fields.<|endoftext|> -TITLE: Proof of the divisibility rule of 17. -QUESTION [13 upvotes]: Rule: Subtract 5 times the last digit from the rest of the number, if the - result is divisible by 17 then the number is also divisible by 17. - -How does this rule work? Please give the proof. - -REPLY [2 votes]: Write the original number as 10x+y to separate out the last digit (y) from the number without the last digit (x). Recognize that divisibility by a 17 means you can write the number as 17n for a positive integer n. So our statement is if x-5y=17n then 10x + y = 17m, where x,y,m,n are positive integers. -Assume x - 5y = 17n. Now we try to get the left-hand side to look like our original number. First multiply both sides by 10. (10x - 50y = 170n), and then add 51y to both sides. (10x+ y = 170n + 51y). Factor 17 from the right hand side. (10x+y=17(10n+3y)). Recognize that 10n+3y is a positive integer as n and y are integers, let's define it as m, proving that 10x+y=17m meaning that the original number is divisible by 17 if x-5y was divisible by 17.<|endoftext|> -TITLE: Evaluating complicated sum -QUESTION [5 upvotes]: Evaluate for a fixed $m\neq 1$ ( $m\in \mathbb{N}$ ) -$$\sum _{k=1}^{n}\left[\left( \sum _{i=1}^{k}i^{2}\right) \left(\sum _{k_{1}+k_{2}+...+k_{m}=k}\dfrac {\left( k_{1}+k_{2}+\ldots +k_{m}\right) !} {k_{1}!k_{2}!...k_{m}!}\right)\right]$$ - -REPLY [3 votes]: As Shaktal showed earlier it is remains to find closed form for -$$ -\sum\limits_{k=1}^n\frac{m^k k^3}{3} -$$ -Consider the following equality -$$ -\sum\limits_{k=1}^n x^k=\frac{x(x^n-1)}{x-1} -$$ -After triple differentiation we get -$$ -\sum\limits_{k=1}^n kx^{k-1}=\frac{d}{dx}\frac{x(x^n-1)}{x-1}\\ -\sum\limits_{k=1}^n k(k-1)x^{k-2}=\frac{d^2}{dx^2}\frac{x(x^n-1)}{x-1}\\ -\sum\limits_{k=1}^n k(k-1)(k-2)x^{k-3}=\frac{d^3}{dx^3}\frac{x(x^n-1)}{x-1} -$$ -Now let's make a small trick -$$ -\sum\limits_{k=1}^n\frac{k^3 x^k}{3}= -\frac{1}{3}\sum\limits_{k=1}^n (k (k-1)(k-2)+3k(k-1)+k)x^k= -\frac{1}{3}\left(x^3\sum\limits_{k=1}^n k (k-1)(k-2)x^{k-3} +3x^2\sum\limits_{k=1}^n k(k-1)x^{k-2} +x\sum\limits_{k=1}^n k x^{k-1} \right)= -\frac{1}{3}\left(x^3\frac{d^3}{dx^3}\frac{x(x^n-1)}{x-1} +3x^2\frac{d^2}{dx^2}\frac{x(x^n-1)}{x-1} +x\frac{d}{dx}\frac{x(x^n-1)}{x-1} \right) -$$ -It is remains to simplify this expression (i.e. differentiante) and then substitute $x=m$<|endoftext|> -TITLE: Does the identity $\det(I+g^{-1})\det(I+g)=|\det(g-I)|^2$ hold for $g \in U(n)$? -QUESTION [5 upvotes]: In a paper (corollary 1, p.14) the following identity is used: - -Let g be a unitary matrix. Then: -$$\det(I+g^{-1})\det(I+g)=|\det(g-I)|^2 \text{ for }g \in U(n)$$ - -Now my question is why this holds -I calculated: -$$\det(I+g^{-1})\det(I+g)=\overline{\det(I+g^t)}\det(I+g)=\overline{\det(I+g)}\det(I+g)=|\det(I+g)|^2$$ -Where the second equality holds as $I$ has only entries in the diagonal ($I$ is of course the unit matrix). But this is not the same as on the right side. -(I also thought that maybe there was a typo on the left side where should be minus-signs. However in the paper itself it is needed that there are plus-signs.) -Thanks for any hints. -Edit: -This equality was in the scope of an integral: $$\int_{U(n)}\prod_{l=1}^{k}det(I+g^{-1})\prod_{l=1}^{k}\det(I+g)dg=\int_{U(n)}|\det(g-I)|^{2k}dg$$ -With a change of variable it was solved with my calculation done above. See Giuseppe's answer. - -REPLY [6 votes]: Dear AndreasS I have given a look at the paper. -I think that the calculation $$\det(I+g^{-1})\det(I+g)=|\det(I+g)|^2 \text{ for }g \in U(n)$$ is correct. -But in order to obtain the result stated in Corollary 1, you just need the change of variable $g\mapsto -g$ in the integral over $U(n)$, so that $$\int_{U(n)}|\det(g-I)|^{2k}dg=\int_{U(n)}|\det(I+g)|^{2k}dg.$$ -Then in the paper you find how factorize $|\det(I+g)|^{2k}.$ -I hope that it helps.<|endoftext|> -TITLE: 2x2 Matrices and Differences of Fractions -QUESTION [5 upvotes]: Consider the difference of two arbitrary fractions, $\frac{a}{b}$ and $\frac{c}{d}$. -$$\frac{a}{b}-\frac{c}{d}=\frac{ad-bc}{bd}$$ -The numerator is the determinant of the 2x2 matrix $$ \left( \begin{array}{ccc} -a & c \\ -b & d \\ \end{array} \right)$$ -Is there any reason for this? Are the two related in any way? - -REPLY [6 votes]: Think of the determinant as an expression for an area or volume spanned by the vectors $(a,b)^T$ and $(c,d)^T$. If the ratios, which represent the direction of the vector are equal, i.e. $a/b=c/d$, then the area/volume is $0$.<|endoftext|> -TITLE: The prime spectrum of a Dedekind Domain -QUESTION [24 upvotes]: Let $A$ be a Dedekind Domain, let $X = \operatorname{Spec}(A)$. Are all open sets in $X$ basic open sets? Thinking about the Zariski topology (in the classical sense) of a non-singular affine curve, if I had to guess, I would say "yes" but I can't think of a proof or counterexample. So far I have only been able to prove the result is true for PID's. -Any help is much appreciated. -Cheers - -REPLY [19 votes]: No, it is not true that for a Dedekind ring $A$ all open subsets of $X = \operatorname{Spec}(A)$ are principal. Here is a counter-example: -a) Let $\bar X$ be an elliptic curve over $\mathbb C $ and $P\in \bar X$ a non-torsion point (this means that for all $n\geq 1 $ we have $n\cdot P\neq O$ in the abelian group $\bar X$, where $O$ is the zero element of $\bar X$ ). -Let $X=\bar X\setminus \lbrace O \rbrace$. - The ring $A=\Gamma(X,\mathcal O) $ is Dedekind-because $X$ is affine and smooth of dimension one-and we have $X=\operatorname{Spec}(A)$. -b) I claim however that $X_P= X\setminus \lbrace P \rbrace$ is a non-principal open subset of $X$. -Indeed if we had $X_P=D(f)$ for some $f\in A$, the divisor of $f$ seen as a rational function on $\bar X$ would be of the form $div(f)=nP-nO\in \operatorname {Div}(\bar X)$. -But then, by Abel-Jacobi's theorem, we would have in the group $\bar X$ the relation $n\cdot P-n\cdot O=n\cdot P=O$, contradicting the choice of $P$. -Edit -While I was typing my answer, Bruno gave a perfect criterion for every open subset of $X = \operatorname{Spec}(A)$ to be principal: that $Cl(A)=Pic(X)$ be torsion. -So, for which Dedekind rings $A$ is that true? -My counter-example of course is one whereit is not true. -Actually for any elliptic curve $\bar X$, only denumerably many points of the group $\bar X ( \mathbb C)$ are torsion and this implies that $Pic(X) \quad (X=\bar X\setminus \lbrace O \rbrace)$ has continuum many non-torsion elements. -But this is small beer. -Now comes the real surprise: Claborn has proved that given any abelian group $G$ whatsoever , there exists a Dedekind ring $A$ with $Cl(A)=G$. -So, take any abelian group $G$ with a non-torsion element and Claborn's theorem provides you with a Dedekind ring $A$ whose spectrum $\operatorname{Spec}(A)$ has a non-principal open subset!<|endoftext|> -TITLE: Further examples of 1-transitive but not 2-transitive groups of self-homeomorphisms -QUESTION [6 upvotes]: In a question from last week, I was searching for "nice" spaces on which the group of self-homeomorphisms acted $1$-transitively but not $2$-transitively on the space. In that case, I had the condition that the space be connected and Hausdorff. -The example given in answer to that question was related to the "Long Line." -Now I'm asking more concretely: -Can we find a connected infinite sub-space of Euclidean space, $X\subset\mathbb R^n$ for some $n$, on which the entire group of self-homeomorphisms of $X$ acts $1$-transitively on $X$ but not $2$-transitively? Ideally, examples without the axiom of choice, although examples using the axiom of choice would be interesting, too. -[In the last question, someone asked for the definition of $n$-transitivity. Wolfram's MathWorld has a slightly flawed definition, but it works for infinite sets.] - -REPLY [4 votes]: I believe I have found an example as the following argument should show. In case someone finds a mistake, or further clarifications are required, comments are of course welcome. -Let $S^1\subseteq\Bbb C$ be the unit circle and let $f:\Bbb R\to S^1\times S^1$ be defined by $f(t)=(e^{it},e^{i\sqrt{2}t})$. Define $X=f(\Bbb R\setminus\{0\})$. (With the usual identification of $\Bbb C=\Bbb R^2$, we have that $X$ is a subspace of $\Bbb R^4$ and since the torus embeds into $\Bbb R^3$ we can also regard $X$ as embedded into $\Bbb R^3$). -Now we shall prove some lemmas to show that $X$ has the properties we seek. -Lemma 1. $X$ is connected. -Proof. Clearly, $X$ has at most two components, since the sets $A=f((-\infty,0))$ and $B=f((0,\infty))$ are connected. To see $X$ is connected, it is enough to show that $A$ and $B$ are not separated. To see this, examine points of the form $f(2k\pi)=(1,e^{2\sqrt{2}k\pi i})$, where $k\in\Bbb N$. These are dense in $\{1\}\times S^1$, because $\sqrt{2}\pi$ is an irrational multiple of $\pi$. The same holds for points of the form $f(-2k\pi)$, where $k\in\Bbb N$. So we have found a subset of $A$ and a subset of $B$ that are both dense in $\{1\}\times S^1$. This implies that $A$ and $B$ are not separated. $\square$ -Lemma 2. The action of $\operatorname{Homeo}(X)$ on $X$ is $1$-transitive. -Proof. Write $x\sim y$ whenever there is a homeomorphism $h:X\to X$ such that $h(x)=y$. As we know, this is an equivalence relation. We will show that every two points $x,y\in X$ are equivalent, i.e. in the same orbit of the action. This follows from several simple cases: -Case 1. If there is a $k_0\in\Bbb Z$ such that $a\in f((2k_0\pi,2(k_0+1)\pi))$, then $a\sim f(2k_0\pi+\pi)$. -First, for every $r\in(0,2\pi)$ define a map $g_r:\Bbb R\to\Bbb R$ as follows: $$g_r(t)=\begin{cases}2k\pi+\frac{t-2k\pi}{r}\pi;& \text{if }\exists k\in\Bbb Z:t\in[2k\pi,2k\pi+r]\\2k\pi+\pi+\frac{(t-2k\pi-r)}{2\pi-r}\pi;& \text{if }\exists k\in\Bbb Z:t\in[2k\pi+r,2(k+1)\pi]\end{cases}$$ -The relevance of $g_r$ is that it is a homeomorphism of $\Bbb R$ that for $k\in\Bbb Z$ has fixed points $2k\pi$ and $g_r(2k\pi+r)=2k\pi+\pi$ and, most importantly, $g_{r}(t+2k\pi)=g_{r}(t)+2k\pi$ for all $t\in\Bbb R$ and $k\in\Bbb Z$. Now let $a=f(t_0)$, let $r_0=t_0-2k_0\pi$ and define $h:X\to X$ by $$h(f(t))=f(g_{r_0}(t))$$ -This is a homeomorphism of $X$. To see this, we define $H:S^1\times S^1\to S^1\times S^1$ by $$H(e^{it},e^{is})=(e^{ig_{r_0}(t)},e^{i(\sqrt{2}g_{r_0}(t)+s-\sqrt{2}t)})$$ This is a well-defined map: suppose $T=t+2k\pi$ and $S=s+2l\pi$ for some $k,l\in\Bbb Z$. We will use the crucial property of $g_{r_0}$ that $g_{r_0}(t+2k\pi)=g_{r_0}(t)+2k\pi$ for all $k\in\Bbb Z$. Let's calculate: $$e^{ig_{r_0}(T)}=e^{ig_{r_0}(t+2k\pi)}=e^{ig_{r_0}(t)+i2k\pi}=e^{ig_{r_0}(t)}$$ and $$\begin{align}e^{i(\sqrt{2}g_{r_0}(T)+S-\sqrt{2}T)}&=e^{i(\sqrt{2}g_{r_0}(t+2k\pi)+s+2l\pi-\sqrt{2}t-2\sqrt{2}k\pi)}\\&=e^{i(\sqrt{2}g_{r_0}(t)+2\sqrt{2}k\pi+s+2l\pi-\sqrt{2}t-2\sqrt{2}k\pi)}\\&=e^{i(\sqrt{2}g_{r_0}(t)+s-\sqrt{2}t)+i2l\pi}\\&=e^{i(\sqrt{2}g_{r_0}(t)+s-\sqrt{2}t)}\end{align}$$ This shows that $H$ is well defined. Its inverse $H^{-1}:S^1\times S^1\to S^1\times S^1$ is defined by $$H^{-1}(e^{iu},e^{iv})=(e^{ig^{-1}_{r_0}(u)},e^{i(\sqrt{2}g^{-1}_{r_0}(u)+v-\sqrt{2}u)})$$ This is again well defined. The proof is basically the same as for $H$, except that this time we use that $g^{-1}_{r_0}(t+2k\pi)=g^{-1}_{r_0}(t)+2k\pi$. This proves that $H$ is a bijection. As both $H$ and $H^{-1}$ are continuous, $H$ is a homeomorphism. -Now, we notice that $h$ is just the restriction of $H$ to $X$. Since the restriction of a homeomorphism is a homeomorphism (onto its image, which in this case is precisely $X$), $h:X\to X$ is itself a homeomorphism. -Case 2. If there is a $k_0\in\Bbb Z$ such that $a\in f((2k_0\frac{\pi}{\sqrt2},2(k_0+1)\frac{\pi}{\sqrt2}))$, then $a\sim f(2k_0\frac{\pi}{\sqrt2}+\frac{\pi}{\sqrt2})$. -The proof of this case is basically the same as in the first case, except that instead of $g_r$ we use a homeomorphism that has fixed points at $2k\frac{\pi}{\sqrt2}$, etc. For example, $t\mapsto \frac1{\sqrt{2}} g_r(\sqrt{2}t)$ should do the job. (The proof is analogous because the $S^1$'s in $S^1\times S^1$ appear in a completely symmetric manner.) -Case 3. If $a=f(t_0)$ for some $t_0>0$ then $a\sim f(-t_0)$. -Here we simply use the map $h:X\to X$ defined by $h(f(t))=f(-t)$. -Now, cases 1 and 2 show that every two points in the $f$-image of an interval of the form $(2k\pi,2(k+1)\pi)$ or $(2k\frac\pi{\sqrt2},2(k+1)\frac\pi{\sqrt2})$ are equivalent. Because of the way how such intervals overlap, we see that every two points in $A$ are equivalent and every two points in $B$ are equivalent. Case 3 then shows that some two points from $A$ and $B$ are equivalent. Conclusion: all points of $X$ are equivalent. $\square$ -Lemma 3. The action of $\operatorname{Homeo}(X)$ on $X$ is not $2$-transitive. -Proof. To see this, let $x,y\in A$ and $z\in B$. Then there cannot be any homeomorphism that takes $x$ to $x$ and $y$ to $z$. To see this, it suffices to show that $A$ and $B$ are precisely the path components of $X$. Since they are clearly path-connected, it is enough to show that $X$ is not path connected. -To do this, we shall first define $U_-=X\setminus(S^1\times\{-1\})$ and $U_+=X\setminus(S^1\times\{1\})$. Then the following holds: - -connected components of $U_+$ are precisely the sets $f((2k\frac\pi{\sqrt2},2(k+1)\frac\pi{\sqrt2}))$ for $k\in\Bbb Z$ -connected components of $U_-$ are the sets $f((2k\frac\pi{\sqrt2}-\frac\pi{\sqrt2},2k\frac\pi{\sqrt2}+\frac\pi{\sqrt2}))$ for $k\in\Bbb Z\setminus\{0\}$ and because $f(0)\notin X$ two further components: $f((-\frac\pi{\sqrt2},0))$ and $f((0,\frac\pi{\sqrt2}))$ - -We will now show how to prove the first of these claims. The second one has a similar proof. -Clearly, the sets of the form $f((2k\frac\pi{\sqrt2},2(k+1)\frac\pi{\sqrt2}))$ are connected. They are indeed subsets of $U_+$ because $e^{i\sqrt2t}=1$ if and only if $t=2l\frac\pi{\sqrt2}$ for some $l\in\Bbb Z$ and there is no such $t$ in the interval $(2k\frac\pi{\sqrt2},2(k+1)\frac\pi{\sqrt2})$. Now we just have to show that these are not contained in any larger connected set. -We will show that these are indeed the quasi-components of $U_+$. To show this, define for each $x,y\in\Bbb R$ an open set $U_{x,y}$ as follows: $$U_{x,y}=U_+\cap\{(e^{iw}e^{it},e^{i\sqrt2t})|\;t\in(0,2\frac{\pi}{\sqrt2}),w\in(x,y)\}$$ This is an open set by definition. In fact, it is clopen for most choices of $x$ and $y$, since $(e^{iw}e^{it},e^{i\sqrt2t})\in X$ is only possible if $w=2k\pi+2l\frac{\pi}{\sqrt2}$ for some $k,l\in\Bbb Z$. Indeed, the set of points $(x,y)$ for which $U_{x,y}$ is clopen, is dense in $\Bbb R^2$. -Now, let $C=f((2k\frac\pi{\sqrt2},2(k+1)\frac\pi{\sqrt2}))$ for some $k\in \Bbb Z$. There is a unique $L\in[0,2\pi)$ such that $C=\{(e^{iL}e^{it},e^{i\sqrt2t})|\;t\in(0,2\frac{\pi}{\sqrt2})\}$. By the above, we may thus choose a sequence $(x_n,y_n)_n$ such that for each $n\in\Bbb N$ we have $x_n -TITLE: Sequences, subsequences, and continuity of functions -QUESTION [7 upvotes]: It's been a few years since I studied point-set topology, and I'm a bit rusty on the basics. Would appreciate help with the following question. -Suppose $f:X\rightarrow Y$ is a map between two topological spaces, and I know that for any sequence $x_n\rightarrow x$ in $X$, there is a subsequence $x_{n_k}$ such that $f(x_{n_k})\rightarrow f(x)$. Does it follow that $x_n\rightarrow x\Rightarrow f(x_n)\rightarrow f(x)$? In the case when $X$ is first countable, the first condition is enough to imply that $f$ is continuous (I believe), and so the second condition must hold. I suspect this isn't true for general $X$, but I can't come up with a counterexample. -Thanks. - -REPLY [10 votes]: If $x_n\to x$ and $f(x_n)$ doesn't converge to $f(x)$, take $V$ a neighborhood of $f(x)$ such that for infinitely many $k$, $f(x_k)\notin V$. We can write this infinitely many as a subsequence $x_{n_k}$. We have $x_{n_k}\to x$, but we can't extract a subsequence of $\{f(x_{n_k})\}$ which is convergent to $f(x)$.<|endoftext|> -TITLE: orientability of riemann surface -QUESTION [7 upvotes]: Could any one tell me about the orientability of riemann surfaces? -well, Holomorphic maps between two open sets of complex plane preserves orientation of the plane,I mean conformal property of holomorphic map implies that the local angles are preserved and in particular right angles are preserved, therefore the local notions of "clockwise" and "anticlockwise" for small circles are preserved. can we transform this idea to abstract riemann surfaces? -thank you for your comments - -REPLY [7 votes]: Holomorphic maps not only preserve nonoriented angles but oriented angles as well. Note that the map $z\mapsto\bar z$ preserves nonoriented angles, in particular nonoriented right angles, but it does not preserve the clockwise or anticlockwise orientation of small circles. -Now a Riemann surface $S$ is covered by local coordinate patches $(U_\alpha, z_\alpha)_{\alpha\in I}\ $, and the local coordinate functions $z_\alpha$ are related in the intersections $U_\alpha\cap U_\beta$ by means of holomorphic transition functions $\phi_{\alpha\beta}$: One has $z_\alpha=\phi_{\alpha\beta}\circ z_\beta$ where the Jacobian $|\phi'_{\alpha\beta}(z)|^2$ is everywhere $>0$. It follows that the positive ("anticlockwise") orientation of infinitesimal circles on $S$ is well defined througout $S$. In all, a Riemann surface $S$ is oriented to begin with. -For better understandng consider a Moebius band $M:=\{z\in {\mathbb C}\ | \ -1<{\rm Im}\, z <1\}/\sim\ ,$ where points $x+iy$ and $x+1-iy$ $\ (x\in{\mathbb R})$ are identified. In this case it is not possible to choose coordinate patches $(U_\alpha, z_\alpha)$ covering all of $M$ such that the transition functions are properly holomorphic. As a consequence this $M$ is not a Riemann surface, even though nonoriented angles make sense on $M$.<|endoftext|> -TITLE: Inverse image of a union equals the union of the inverse images -QUESTION [19 upvotes]: I just wonder if my following solution is true. -Let $X,Y$ be sets, let $f:X\to Y$ be a function, let $\{Y_i\}_{i\in I}$ be a family of subsets of $Y$. (Note: I use equalities instead of mutual containment) -$$\begin{align}f^{-1}\left[\bigcup_{i\in I} Y_i\right] - &= \{x \in X: \mbox{there exists an}\quad i \in I\mbox{ such that } y \in Y_i,f(x)=y\} - \\&=\bigcup_{i \in I} f^{-1}\left[Y_i\right] -\end{align}$$ -I initially do not know how to get from the left to right, but when I put both sets in set notation, they turn out to be the same, hence the one line proof. Something go ultimately wrong? - -REPLY [33 votes]: The statement is true, and your argument is essentially right, but I would say that you are skipping steps to achieve that identification. (Also, you should not have those $\infty$'s on top of the union symbol). I would instead add: -$$\begin{align*} -f^{-1}\left[\bigcup_{i\in I}Y_i\right] &= \left\{ x\in X\;\left|\; f(x)\in\bigcup_{i\in I}Y_i\right\}\right.\\ -&=\Biggl\{x\in X\;\Biggm|\; \exists i\in I\text{ such that }f(x)\in Y_i\Biggr\}\\ -&= \bigcup_{i\in I}\{x\in X\mid f(x)\in Y_i\}\\ -&= \bigcup_{i\in I}f^{-1}[Y_i]. -\end{align*}$$ -The first equality is by definition of inverse image; the second by definition of the union; the third is by definition of union; and the fourth by definition of inverse image.<|endoftext|> -TITLE: an infinite series involving odd zeta -QUESTION [13 upvotes]: I ran across a cool series I have been trying to chip away at. -$$\sum_{k=1}^{\infty}\frac{\zeta(2k+1)-1}{k+2}=\frac{-\gamma}{2}-6\ln(A)+\ln(2)+\frac{7}{6}\approx 0.0786\ldots$$ -where A = the Glaisher-Kinkelin constant. Numerically, it is approx. $1.282427\ldots$ -I began by writing zeta as a sum and switching the summation order -$$\sum_{n=2}^\infty \sum_{k=1}^\infty \frac{1}{(k+2)n^{2k+1}}$$ -The first sum is the series for $-n^3\ln(1-\frac{1}{n^2})-n-\frac{1}{2n}$ -So, we have $-\sum_{n=2}^\infty \left[\ln(1-\frac{1}{n^2})+n+\frac{1}{2n}\right]$ -This series numerically checks out, so I am onto something. At first glance the series looks like it should diverge, but it does converge. -Another idea I had was to write out the series of the series: -$$1/3(1/2)^{3}+1/4(1/2)^{5}+1/5(1/2)^{7}+\cdots -+1/3(1/3)^{3}+1/4(1/3)^{5}+1/5(1/3)^{7}+\cdots +1/3(1/4)^{3}+1/4(1/4)^{5}+1/5(1/4)^{7}+\cdots$$ -and so on. -This can be written as -$$1/3x^{3}+1/4x^{5}+1/5x^{7}+\cdots +1/3x^{3}+1/4x^{5}+1/5x^{7}+\cdots + 1/3x^{3}+1/4x^{5}+1/5x^{7}+\cdots $$ -where $x=1/2,1/3,1/4,\ldots$ -This leads to the series representation for: -$$\frac{-\ln(1-x^2)}{x^3}-\frac{1}{x}-\frac{x}{2}$$ -Since $x$ is of the form $1/n$, we end up with the same series as before. -Now, my quandary. How to finish?. Where in the world does that Glaisher-Kinkelin constant come in, and how can that nice closed from be obtained?. Whether from the series I have above or some other means. As usual, it is probably something I should be seeing but don't at the moment. -The GK constant has a closed form of $$e^{\frac{1}{12}-\zeta^{'}(-1)}$$. -Which means an equivalent closed form would be $\frac{-\gamma}{2}+\ln(2)+6\zeta^{'}(-1)+\frac{2}{3}$ -Thanks all. - -REPLY [9 votes]: We have -$$ -\begin{eqnarray*} -&&\sum_{n=2}^\infty\left( -n^3\log(1-\frac{1}{n^2})-n-\frac{1}{2n}\right)\\ -&=&\lim_{N\to\infty}\sum_{n=2}^N\left( -n^3\log(1-\frac{1}{n^2})-n-\frac{1}{2n}\right)\\ -&=&\lim_{N\to\infty}\left(\sum_{n=2}^N -n^3\log(1-\frac{1}{n^2})-\left(\frac{N^2+N-2}{2}\right)-\left(\frac{\log(N)}{2}-\frac{1}{2}+\frac{\gamma}{2}+O\left(\frac{1}{N}\right)\right)\right)\\ -&=&\lim_{N\to\infty}[\sum_{n=2}^N (2n^3\log(n)-n^3\log(n+1)-n^3\log(n-1))-\left(\frac{N^2+N-2}{2}\right)-\left(\frac{\log(N)}{2}-\frac{1}{2}+\frac{\gamma}{2}\right)] -\end{eqnarray*} -$$ -In the sum on the last line, we may gather together the coefficients of each logarithm (terms at the boundary of the sum are a little funny), giving -$$ -\begin{eqnarray*} -&&\lim_{N\to\infty}[\sum_{n=2}^N(-6n\log(n))+\log(2)+(N^3+3N^2+3N+1)\log(N)-N^3\log(N+1)-\left(\frac{N^2+N-2}{2}\right)-\left(\frac{\log(N)}{2}-\frac{1}{2}+\frac{\gamma}{2}\right)]\\ -&=&\lim_{N\to\infty}[\sum_{n=2}^N(-6n\log(n))+\log(2)-N^3\log\left(1+\frac{1}{N}\right)+(3N^2+3N+1)\log(N)-\left(\frac{N^2+N-2}{2}\right)-\left(\frac{\log(N)}{2}-\frac{1}{2}+\frac{\gamma}{2}\right)]\\ -&=&\lim_{N\to\infty}[\sum_{n=2}^N(-6n\log(n))+\log(2)-N^3\left(\frac{1}{N}-\frac{1}{2N^2}+\frac{1}{3N^3}+O\left(\frac{1}{N^4}\right)\right)+(3N^2+3N+1)\log(N)-\left(\frac{N^2+N-2}{2}\right)-\left(\frac{\log(N)}{2}-\frac{1}{2}+\frac{\gamma}{2}\right)]\\ -&=&\lim_{N\to\infty}[\sum_{n=2}^N(-6n\log(n))+\left(3N^2+3N+\frac{1}{2}\right)\log(N)+\left(\frac{-3}{2}N^2+\frac{7}{6}-\frac{\gamma}{2}+\log(2)\right)]\\ -&=&-6\log\left(\lim_{N\to\infty}\left(\frac{\prod_{n=1}^N n^n}{N^{N^2/2+N/2+1/12}e^{-N^2/4}}\right)\right)+\frac{7}{6}-\frac{\gamma}{2}+\log(2)\\ -&=&-6\log(A)+\frac{7}{6}-\frac{\gamma}{2}+\log(2) -\end{eqnarray*} -$$ -Here, I am taking -$$ -A=\lim_{N\to\infty}\frac{\prod_{n=1}^N n^n}{N^{N^2/2+N/2+1/12}e^{-N^2/4}} -$$ -as the definition of the Glaisher-Kinkelin constant.<|endoftext|> -TITLE: Diagonalizing Matrices over UFDs -QUESTION [6 upvotes]: Suppose $R$ is a UFD, $K$ its field of fractions. If a matrix $M$ with entries in $R$ has distinct eigenvalues, then it is certainly diagonalizable over $K$. If those eigenvalues are actually in $R$, must it be diagonalizable over $R$? -If it helps: the specific case I'm interested in is when $K$ is a local field of characteristic $0$ and $R$ its integers. - -REPLY [8 votes]: No. Consider the matrix -$$A=\left(\begin{array}{rr} -3&-1\\-1 & 3\end{array}\right)$$ -over $\mathbb{Z}$. It is certainly diagonalizable over $\mathbb{Q}$, and the eigenvalues are $2$ and $4$, in $\mathbb{Z}$; the eigenspaces are $(a,-a)$ and $(b,b)$. Any diagonalizing matrix over $\mathbb{Q}$ would then be of the form -$$P = \left(\begin{array}{rr}a&b\\-a&b\end{array}\right)\qquad\text{or}\qquad \left(\begin{array}{rr}a & b\\a&-b -\end{array}\right),$$ -so that $P^{-1}AP$ is diagonal. But the determinant of $P$ is either $2ab$ or $-2ab$. If $a$ and $b$ are nonzero integers, then the inverse of $P$ cannot be a matrix with integer coefficients, since the determinant is not $\pm 1$. -Added. By the way: the assumption that the eigenvalues lie in $R$ is unnecessary: if $M$ has coefficients in $R$, then the characteristic polynomial is monic and has coefficients in $R$. In particular, if it splits over $K$, then it splits over $R$ (by the Rational Root Theorem, or by Gauss's Lemma). So all you need to assume is that the minimal polynomial splits over $K$ and is squarefree; the polynomial will then necessarily split over $R$ as well.<|endoftext|> -TITLE: An order type $\tau$ equal to its power $\tau^n, n>2$ -QUESTION [14 upvotes]: In this question we are concerned only with linear (aka total) order types. By a cardinality of an order type we understand a cardinality of an instance of this type, which obviously does not depend on selection of such particular instance. Some of the order types that occur most often and have particularly nice properties are ordinal numbers (which are order types of well-orders) and their well-known and prominent subset, finite ordinals (aka natural numbers $\mathbb{N}$). -Recall that: - -$0, 1, 2, \dots$ — unique linear order type for each finite cardinality. -$\omega$ — the order type of $\mathbb{N}$ ordered by magnitude, the smallest infinite (denumerable) ordinal. -$\omega_1$ — the order type of the set of all countable ordinals, ordered by "is initial segment of" relation (or, equivalently, by $\in$ relation). It is the smallest uncountable ordinal, the initial ordinal of the 2nd infinite cardinal $\aleph_1$. -$\eta$ — the dense countable order type of rational numbers $\mathbb{Q}$ ordered my their magnitude, which is also an order type of any dense denumerable linear order without first and last elements (e.g. the set of positive algebraic numbers). - -The sum and product of order types are natural generalizations of (and are consistent with) the sum and product of ordinals, which we consider well-known. These operations are associative, but in general, not commutative. For example: $\eta+1 \ne 1+\eta\ne\eta$, but $\eta+1+\eta=\eta+\eta=\eta=\eta\cdot\eta=\eta\cdot\omega=(\eta+1)\cdot\omega\ne\omega\cdot\eta$ (the last order being not dense). Of course $\eta\cdot\omega_1\ne\eta$ because of different cardinality, although every proper initial segment of $\eta\cdot\omega_1$ is $\eta$ or $\eta+1$. Note that $\eta\cdot\omega_1=(\eta + 1)\cdot\omega_1\ne(1+\eta)\cdot\omega_1$, and even $(1+\eta)\cdot\omega_1\ne1+\eta\cdot\omega_1$, although $(1+\eta)\cdot\omega=1+\eta$. We agree that a positive integer power of an order type is just a syntactic shortcut for repeated multiplication. -There are some order types satisfying $\tau^2=\tau$, for example: $0, 1, \eta, \omega\cdot\eta$ and $\omega^2\cdot\eta$. - -Question: Is there a linear order type $\tau$ such that $\tau^2\ne\tau$, but $\tau^n=\tau$ for some integer $n>2$? - -REPLY [5 votes]: Such questions have been considered in the past. From W. Sierpiński's Cardinal and Ordinal Numbers, second edition revised, Warszawa 1965, p. 235: "We do not know so far any example of two types $\varphi$ and $\psi$, such that $\varphi^2=\psi^2$ but $\varphi^3\ne\psi^3$, or types $\gamma$ and $\delta$ such that $\gamma^2\ne\delta^2$ but $\gamma^3=\delta^3$. Neither do we know any type $\alpha$ such that $\alpha^2\ne\alpha^3=\alpha$." Also, from p. 254: "We do not know whether there exist two different denumerable order types which are left-hand divisors of each other. Neither do we know whether there exist two different order types which are both left-hand and right-hand divisors of each other." Of course, if $\tau^n=\tau$ for some integer $n\gt2$, then $\tau^2$ and $\tau=\tau^2\tau^{n-2}=\tau^{n-2}\tau^2$ are both left-hand and right-hand divisors of each other. -For what it's worth, here is a partial answer to your question, for a very special class of order types. By "order type" I mean linear order type. An order type $\xi$ is said to have a "first element" if it's the type of an ordered set with a first element, i.e., if $\xi=1+\psi$ for some $\psi$; the same goes for "last element". -Proposition. If $\alpha$ is a countable order type, and if $\alpha\xi=\alpha$ for some order type $\xi$ with no first or last element, then $\alpha\beta=\alpha$ for every countable order type $\beta\ne0$. -Corollary. If $\tau$ is a countable order type with no first or last element, and if $\tau^n=\tau$ for some integer $n\gt1$, then $\tau^2=\tau$. -The corollary is obtained by setting $\alpha=\beta=\tau$ and $\xi=\tau^{n-1}$ in the proposition. -The proposition is proved by a modified form of Cantor's back-and-forth argument. Namely, let $A$ be an ordered set of type $\alpha=\alpha\xi$, and let $B$ be an ordered set of type $\alpha\beta$. An isomorphism between $A$ and $B$ will be constructed as the union of a chain of partial isomorphisms $f_k$ of the following form. The domain of $f_k$ is $I_1\cup I_2\cup\dots\cup I_k$, where $I_1,\dots,I_k$ are intervals in $A$ of order type $\alpha$; $I_1\lt\dots\lt I_k$; the interval in $A$ between $I_j$ and $I_{j+1}$ ($1\le j\lt k$), as well as the interval to the left of $I_1$ and the interval to the right of $I_k$, have order types which are nonzero right multiples of $\alpha$. The range of $f_k$ is $J_1\cup\dots\cup J_k$ where $J_1,\dots,J_k$ are intervals in $B$ of type $\alpha$, etc. etc. etc., and $f(I_1)=J_1,\dots,f(I_k)=J_k$.<|endoftext|> -TITLE: Is $\mathbb{Z}[\sqrt{2},\sqrt{3}]$ flat over $\mathbb{Z}[\sqrt{2}]$? -QUESTION [5 upvotes]: Is $\mathbb{Z}[\sqrt{2},\sqrt{3}]$ flat over $\mathbb{Z}[\sqrt{2}]$? The definitions doesn't seem to help. An idea of how to look at such problems would be helpful. - -REPLY [6 votes]: $\mathbb Z[\sqrt{2},\sqrt{3}]$ is freely generated as a $\mathbb Z[\sqrt{2}]$-module (exercise). Free modules are flat. QED<|endoftext|> -TITLE: Injective linear transformation from a vector space to the dual space of its dual space -QUESTION [7 upvotes]: I am currently trying to understand some concepts from some Linear Algebra. I seem to be having quite some difficulty understanding dual spaces and their dual spaces. I found this problem and was wondering how to get started on it. -Let $V$ be a vector space over the field $F$. Let $V^{*}$ be the dual space of $V$ and let $V^{**}$ be the dual space of $V^{*}$. Show that there is an injective linear transformation -$\phi : V \rightarrow V^{**}$. - -REPLY [3 votes]: The canonical answer is the one given by anon above. The point is that when you write $f(v)$, for $v\in V$, $f\in V^*$, you can see it as "$f$ acting on $v$", or you can also see it as "$v$ acting on $f$"; this second point of view defines the injection you are looking for. The physicists write $f(v)$ as $\langle f,v\rangle$ to emphasize this duality, and that's where the word "dual" comes from.<|endoftext|> -TITLE: Aftermath of the incompletness theorem proof -QUESTION [5 upvotes]: This is somewhat of a minor point about the incompletness theorem, but I'm always a little unsure: -So one proves that there is a formula which is unprovable in the theory of consideration. Okay, at this point one is done. -Then, as that unproven sentence contains the claim that this (the unprovable-ness) would happen, one is in some sense justified to say "So the statement really is true, but still, it's unprovable within the theory". Now here is my problem: I'm very unsure in which sense this notion of truth which somewhat comes from outside the theory is sensible. There is a note on that point on the wikipedia page but I don't really comprehend it. - -The word "true" is used disquotationally here: the Gödel sentence is true in this sense because it "asserts its own unprovability and it is indeed unprovable" (Smoryński 1977 p. 825; also see Franzén 2005 pp. 28–33). It is also possible to read "GT is true" in the formal sense that primitive recursive arithmetic proves the implication Con(T)→GT, where Con(T) is a canonical sentence asserting the consistency of T (Smoryński 1977 p. 840, Kikuchi and Tanaka 1994 p. 403) - -So I can see that if one is technically aware that one is now talking in a meta language, that one has introduced a new "true". But then again (a) if one reflects on the fact that one draws such technicals conclusions outside of the initial freamework, then it seems to me one should really introduce another meta-meta-language. And (b) isn't it really just a little ambiguous to say "Gödels incompleteness theorems says that there are true statements, which can't be proven within a certain strong theory"? -I'd be thankful if someone could elaborate on that. - -REPLY [8 votes]: This is an extension of a comment below Arthur Fischer's answer. For concreteness we will work with the theory PA but in principle any sufficiently strong theory would act the same. -Working in PA, or even in the weaker theory PRA, we can formally prove the implication -$$ -\text{Con}(PA) \to G_{PA} -$$ -where $G_{PA}$ is the Gödel sentence of PA. This leads to two conclusions: - -Because we know from the first incompleteness theorem that $G_{PA}$ is not provable in PA, and because $\text{Con}(PA) \to G_{PA}$ is provable in PA, $\text{Con}(PA)$ must not be provable in PA. This is the standard way to prove the second incompleteness theorem. -If we were working in a setting where we had already assumed $\text{Con}(PA)$, and we have access to the normal resources of PRA, then we can prove $G_{PA}$. In particular, when we are proving the incompleteness theorems we are working in normal mathematics, which includes much more than just PRA, and we assume $\text{Con}(PA)$ when we are proving that theorem. Under that assumption we can prove $G_{PA}$ in normal mathematics. Thus the Gödel sentence is "true" in exactly the same sense that $\text{Con}(PA)$ is true when we assume it as a hypothesis to prove the incompleteness theorem. - -The key point in the second bullet is that we don't prove $G_{PA}$ starting with nothing. We prove $G_{PA}$ starting with the knowledge or assumption that $\text{Con}(PA)$ holds. If we did not assume the truth of $\text{Con}(PA)$, or have a separate proof of $\text{Con}(PA)$, the argument in that bullet would be useless. But once we do assume $\text{Con}(PA)$ is true, it only takes a very weak theory (PRA) to deduce that $G_{PA}$ is also true under that assumption.<|endoftext|> -TITLE: Quotient of a free $\mathbb{Z}$-module -QUESTION [5 upvotes]: I'm trying to find the quotient of a free $\mathbb{Z}$-module, but somehow I don't really find the right procedure on how to get the right quotient (nor have I found any sources). I've read What does it mean here to describe the structure of this quotient module? already, but I didn't manage to reproduce the second matrix from that specific post. -In my specific example I have $\mathbb{Z}^3$ as a left-$\mathbb{Z}$ module. Lets call the generators $e_1, e_2, e_3$. My submodule is given by -$$K:=\operatorname{span}_{\mathbb{Z}}\left$$ -My idea to get the quotient module $\mathbb{Z}^3/K$ thus far was to use the equalities -$$i_1 = -2e_2+e_3\\ -i_2 = -2e_1+e_3 -$$ -Which yields (for $z_1,z_2,z_3,l_1,l_2\in\mathbb{Z}$): -$$ z_1e_1+z_2e_2+z_3e_3 \equiv z_1e_1+z_2e_2+z_3e_3+l_1i_1+l_2i_2\\ - = z_1e_1+z_2e_2+z_3e_3+l_1(-2e_2+e_3)+l_2(-2e_1+e_3)\\ - = (z_1-2l_1)e_1 + (z_2-2l_2)e_2 + (z_3+l_1+l_2)e_3$$ -So for $z_1$ even, the first summand should be identical to zero. For $z_1$ uneven it should be identical to one. Same goes for $z_2$ and the second summand, whereas the third summand should be identical to zero in any case (though i'm not really sure about how to deal with that one, since I've already fixed $l_1$ and $l_2$). Hence I'd say my quotient should be $\mathbb{Z}_2^2$. However, letting sage do the calculations, it claims that the result should be $\mathbb{Z}\oplus\mathbb{Z}_2$ -So my question is: Is my result wrong? If so: Where's my mistake and is there a better way to do it? - -REPLY [3 votes]: The computational way to calculate this kind of quotient uses the Smith normal form. However, for small modules we can easily figure out a change of basis that immediately reveals the quotient structure. Let $a,b,c$ denote the standard basis of $\mathbb Z^3$, we know that we can add/substract basis elements to/from other basis elements and obtain another basis. So $a, b-a, c-2b$ is another basis of $\mathbb Z^3$. Since your submodule is -$$ -\langle c-2b, c-2a \rangle = \langle c-2b, 2(b-a)\rangle, -$$ -the quotient is -$$ -\frac{\langle a,b,c\rangle}{\langle c-2b, c-2a\rangle} = \frac{\langle a, b-a, c-2b\rangle}{\langle c-2b, 2(b-a)\rangle} \cong \mathbb Z \oplus \mathbb Z_2. -$$<|endoftext|> -TITLE: Counterexample of Compactness -QUESTION [5 upvotes]: Let $X$ be a metric space and $E\subset X$. -Let {$G_i$} be an open cover of $E$ - -For every open cover {$G_i$}, there exists a finite subcover {$G_{i_n}$} of $E$ such that $G_{i_n} \in${$G_i$}. -For every open cover {$G_i$}, there exists {$M_n$}, a finite family of open sets, such that $E\subset$$\bigcup M_n \subset \bigcup G_i$. - -As you know, if 1 is true, then $E$ is compact. -I think 1 and 2 both have the same meaning, but can't prove the equivalence. (1→2 is trivial, but 2→1 is not) -If 1&2 are not equivalent, please give me some counterexamples.. - -REPLY [5 votes]: $2$ does not imply $1$. Take for example $E=M=\mathbb{R}$, with the open cover $\{ G_n=(-\infty,n)|n \in \mathbb{N}\}$. This cover has no finite subcover, but we can just take $M=\mathbb{R}$ and $\{ M \}$ is a set satisfying the conclusion of $2$.<|endoftext|> -TITLE: Classification of prime ideals of $\mathbb{Z}[X]$ -QUESTION [136 upvotes]: Let $\mathbb{Z}[X]$ be the ring of polynomials in one variable over $\Bbb Z$. - -My question: Is every prime ideal of $\mathbb{Z}[X]$ one of following types? -If yes, how would you prove this? - -$(0)$. - -$(f(X))$, where $f(X)$ is an irreducible polynomial. - -$(p)$, where $p$ is a prime number. - -$(p, f(X))$, where $p$ is a prime number and $f(X)$ is an irreducible polynomial modulo $p$. - -REPLY [118 votes]: Let $\mathfrak{P}$ be a prime ideal of $\mathbb{Z}[x]$. Then $\mathfrak{P}\cap\mathbb{Z}$ is a prime ideal of $\mathbb{Z}$: this holds whenever $R\subseteq S$ are commutative rings. Indeed, if $a,b\in R$, $ab\in R\cap P$, then $a\in P$ or $b\in P$ (since $P$ is prime). (More generally, the contraction of a prime ideal is always a prime ideal, and $\mathfrak{P}\cap\mathbb{Z}$ is the contraction of $\mathfrak{P}$ along the embedding $\mathbb{Z}\hookrightarrow\mathbb{Z}[x]$). -Thus, we have two possibilities: $\mathfrak{P}\cap\mathbb{Z}=(0)$, or $\mathfrak{P}\cap\mathbb{Z}=(p)$ for some prime integer $p$. -Case 1. $\mathfrak{P}\cap\mathbb{Z}=(0)$. If $\mathfrak{P}=(0)$, we are done; otherwise, let $S=\mathbb{Z}-\{0\}$. Then $S\cap \mathfrak{P}=\varnothing$, $S$ is a multiplicative set, so we can localize $\mathbb{Z}[x]$ at $S$ to obtain $\mathbb{Q}[x]$; the ideal $S^{-1}\mathfrak{P}$ is prime in $\mathbb{Q}[x]$, and so is of the form $(q(x))$ for some irreducible polynomial $q(x)$. Clearing denominators and factoring out content we may assume that $q(x)$ has integer coefficients and the gcd of the coefficients is $1$. -I claim that $\mathfrak{P}=(q(x))$. Indeed, from the theory of localizations, we know that $\mathfrak{P}$ consists precisely of the elements of $\mathbb{Z}[x]$ which, when considered to be elements of $\mathbb{Q}[x]$, lie in $S^{-1}\mathfrak{P}$. That is, $\mathfrak{P}$ consists precisely of the rational multiples of $q(x)$ that have integer coefficients. In particular, every integer multiple of $q(x)$ lies in $\mathfrak{P}$, so $(q(x))\subseteq \mathfrak{P}$. But, moreover, if $f(x)=\frac{r}{s}q(x)\in\mathbb{Z}[x]$, then $s$ divides all coefficients of $q(x)$; since $q(x)$ is primitive, it follows that $s\in\{1,-1\}$, so $f(x)$ is actually an integer multiple of $q(x)$. Thus, $\mathfrak{P}\subseteq (q(x))$, proving equality. -Thus, if $\mathfrak{P}\cap\mathbb{Z}=(0)$, then either $\mathfrak{P}=(0)$, or $\mathfrak{P}=(q(x))$ where $q(x)\in\mathbb{Z}[x]$ is irreducible. -Case 2. $\mathfrak{P}\cap\mathbb{Z}=(p)$. -We can then consider the image of $\mathfrak{P}$ in $\mathbb{Z}[x]/(p)\cong\mathbb{F}_p[x]$. The image is prime, since the map is onto; the prime ideals of $\mathbb{F}_p[x]$ are $(0)$ and ideals of the form $(q(x))$ with $q(x)$ monic irreducible over $\mathbb{F}_p[x]$. If the image is $(0)$, then $\mathfrak{P}=(p)$, and we are done. -Otherwise, let $p(x)$ be a polynomial in $\mathbb{Z}[x]$ that reduces to $q(x)$ modulo $p$ and that is monic. Note that $p(x)$ must be irreducible in $\mathbb{Z}[x]$, since any nontrivial factorization in $\mathbb{Z}[x]$ would induce a nontrivial factorization in $\mathbb{F}_p[x]$ (since $p(x)$ and $q(x)$ are both monic). -I claim that $\mathfrak{P}=(p,p(x))$. Indeed, the isomorphism theorems guarantee that $(p,p(x))\subseteq \mathfrak{P}$. Conversely, let $r(x)\in\mathfrak{P}(x)$. Then there exists a polynomial $s(x)\in\mathbb{F}_p[x]$ such that $s(x)q(x) = \overline{r}(x)$. If $t(x)$ is any polynomial that reduces to $s(x)$ modulo $p$, then $t(x)p(x)-r(x)\in (p)$, hence there exists a polynomial $u(x)\in\mathbb{Z}[x]$ such that $r(x) = t(x)p(x)+pu(x)$. Therefore, $r(x)\in (p,p(x))$, hence $\mathfrak{P}\subseteq (p,p(x))$, giving equality. -Thus, if $\mathfrak{P}\cap\mathbb{Z}[x]=(p)$ with $p$ a prime, then either $\mathfrak{P}=(p)$ or $\mathfrak{P}=(p,p(x))$ with $p(x)\in\mathbb{Z}[x]$ irreducible. -This proves the desired classification.<|endoftext|> -TITLE: Good problem book on Abstract Algebra -QUESTION [21 upvotes]: I am currently self-studying abstract algebra from Artin. In that background, I am looking for a problem book in a spirit somewhat similar to Problems in Mathematical Analysis by AMS so that I have a lot of problems to solve. - -REPLY [6 votes]: For a first pass through the material, I really enjoyed Pinter's A Book of Abstract Algebra, as it presented the material quite well (although it did not go very deep). Also, I started out hating Algebra: Pure and Applied by Papantonopoulou, but I actually quite enjoyed it by the end of the course I used it for. Tons of problems in that one. - -REPLY [5 votes]: While I think that doing all the problems in Artin is more than enough, you might want to look through the problems on algebra in the Berkeley Problems in Mathematics and see whether you can solve them reasonably quickly. The Berkeley Problems is not a book to look for fascinating algebra problems, though. Unfortunately, I don't know of any problem books in algebra. I think Artin actually has very good, not-so-standard problems for undergraduate algebra.<|endoftext|> -TITLE: Classification of prime ideals of $\mathbb{Z}[X]/(f(X))$ -QUESTION [8 upvotes]: Let $\mathbb{Z}[X]$ be the ring of polynomials in one variable. -Let $f(X) \in \mathbb{Z}[X]$ be a monic irreducible polynomial. -Let $A = \mathbb{Z}[X]/(f(X))$. -Let $\theta$ = $X$ (mod $f(X)$). -My question: Is the following proposition correct? If yes, how would you prove this? -Proposition -Let $P$ be a non-zero prime ideal of $A$. -Then the following assertions hold. -(1) $P$ contains a prime number $p$. -(2) One of the following two cases occurs. -a. If $f(X)$ is irreducible mod $p$, then $P = (p)$. -b. If $f(X)$ is not irreducible mod $p$, then $P = (p, g(\theta))$, where $g(X)$ is an irreducible factor of $f(X)$ mod $p$. -(3) $P$ is a maximal ideal and $A/P$ is a finite field of characteristic $p$. -This is a related question. - -REPLY [8 votes]: Let us denote by $P$ the corresponding ideal of $\mathbb{Z}[X]$ as well. It will contain polynomials that are not multiples of $f(X)$. Let $g(X)$ be one such. Because $f(X)$ is irreducible, the greatest common divisor (in the Euclidean domain $\mathbb{Q}[X]$) must be equal to 1. By Bezout's identity there exist polynomials $u(X),v(X)\in \mathbb{Q}[X]$ such that -$$ -u(X)f(X)+v(X)g(X)=1. -$$ -Multiplying this by the least common multiple $m$ of the denominators of the coefficients of $u(X)$ and $v(X)$ we see that there exists polynomials $U(X)=mu(X),V(X)=mv(X)\in \mathbb{Z}[X]$ such that -$$ -m=U(x)f(X)+V(X)g(X)\in P. -$$ -So the ideal $P$ contains non-zero integer constants. Because $P$ is a prime ideal, the intersection $P\cap\mathbb{Z}$ is also a (non-zero) prime ideal, and hence it contains a prime number $p$. Because $P$ is non-trivial, $p$ is the only prime number in -$P$. Part (1) is settled. -From (1) we deduce that the quotient ring $A/P$ is finite. As $A$ is an integral domain, and $P$ is a prime ideal, the ring $A/P$ is also an integral domain. A finite integral domain is always a field, so (3) is proven. -Clearly $A/P$ is generated (as a ring) by $\theta$. So $A/P=\mathbb{F}_p[\theta]$, where I again abuse notation and denote $X+P$ also with $\theta$. Let $g(X)\in\mathbb{F}_p[X]$ be the minimal polynomial of $\theta$ over the prime field. Obviously $g(X)\mid \overline{f}(X)$, where $\overline{f}(X)$ stands for the reduction of $f(X)$ modulo $p$. The claim (2) follows from this relatively easily.<|endoftext|> -TITLE: Generating random numbers with the distribution of the primes -QUESTION [7 upvotes]: I would like to generate random numbers whose distribution mimics that of the primes. -So the number of generated random numbers less than $n$ should grow like $n / \log n$, -most intervals $[n,n+n^\epsilon]$ should contain approximately $n^\epsilon / \log n$ -generated numbers (Selberg's short intervals), etc. Does anyone know of a computationally feasible method for generating such "look-a-like primes"? -Addendum. I implemented Henry and Xoff's suggestions. -Here are several instances of the first ten "pseudoprimes": -$$ 4, 5, 9, 10, 17, 23, 27, 28, 31, 44 $$ -$$ 7, 8, 9, 10, 12, 15, 18, 19, 27, 34 $$ -$$ 6, 11, 15, 16, 23, 26, 27, 29, 45, 49 $$ -And here is the cumulative distribution pseudoprimes up to $10^6$ (red), together with -a plot of $n / \log n$ (purple), for one random run: - -REPLY [4 votes]: Not sure if this fits the bill, but how about the following process. -Since the (normal) primes can be generated by a sieve (sieve out all multiples of 2, then 3, then 5, etc.), how about using a similar sieving process, but choose a random offset for each run through the sieving process? -For example, first sieve out every second number starting randomly with 1 or 2. -Then sieve out all multiples of 3 starting randomly at 1, 2 or 3. -Repeat until you get to the square root of the limit you want to go up to. -Since the standard Eratosthenes sieve is pretty fast when it gets going, this might be a good process if you want a reasonable-sized block of "primes".<|endoftext|> -TITLE: Why is $U(n,n)= GL(2n, \mathbb{C}) \cap O(2n,2n)$? -QUESTION [6 upvotes]: I have read the statement above. I tried to prove it but I could not. For me, $O(2n,2n)$ already a subgroup of $GL(2n,\mathbb{C})$. - -REPLY [2 votes]: Usually, $O(2n,2n)$ is a group of $4n$-by-$4n$ matrices, so both $GL(2n,\mathbb C)$ and $U(n,n)$ would need to be mapped to $GL(4n,\mathbb R)$. This is not a problem, though can cause some cognitive dissonance. Any choice of isomorphism $\mathbb C^{2n}\rightarrow\mathbb R^{4n}$ will do, giving imbeddings to $GL(4n,\mathbb R)$. -A common trick with classical groups is the following: let $(,)$ be the hermitian form of signature $n,n$ left invariant by $U(n,n)$. Let $\langle u,v\rangle=\Re(u,v)$ be the real part. Using our chosen isomorphism $\mathbb C^{2n}\rightarrow \mathbb R^{4n}$ gives a symmetric $\mathbb R$-valued from $\langle,\rangle$ on $\mathbb R^{4n}$. By design, the image of $U(n,n)$ preserves $\langle,\rangle$. Also, the signature of $\langle,\rangle$ is simply "double" that of $(,)$, so is $2n,2n$. -Thus, for simple reasons $U(n,n)$ injects to $O(2n,2n)$. Then a dimension-count (looking at the Lie algebra) shows that the dimensions are equal, so at least the connected components of the identity are the same.<|endoftext|> -TITLE: Methods to solve differential equations -QUESTION [10 upvotes]: We are given the equation -$$\frac{1}{f(x)} \cdot \frac{d\left(f(x)\right)}{dx} = x^3.$$ -To solve it, "multiply by $dx$" and integrate: -$\frac{x^4}{4} + C = \ln \left( f(x) \right)$ -But $dx$ is not a number, what does it mean when I multiply by $dx$, what am I doing, why does it work, and how can I solve it without multiplying by $dx$? -Second question: -Suppose we have the equation $$\frac{d^2f(x)}{dx^2}=(x^2-1)f(x)$$ -Then for large $x$, we have $\frac{d^2f(x)}{dx^2}\approx x^2f(x)$, with the approximate solution $ke \cdot \exp \left (\frac{x^2}{2} \right)$ -Why is it then reasonable to suspect, or assume, that the solution to the original equation, will be of the form $f(x)=e^{x^2/2} \cdot g(x)$, where $g(x)$ has a simpler form then $f(x)$? When does it not work? -Third question: -The method of replacing all occurences of $f(x)$, and its derivatives by a power series, $\sum a_nx^n$, for which equations does this work or lead to a simpler equation? -Do we lose any solutions this way? - -REPLY [2 votes]: More Deapth for Question 1: -Andr'e noted that the idea of multiplying by dx is "mathematically dubious", and in reality there are many situations where this technique will get you into trouble if you apply it haphazardly without understanding what's going on in the background. To the end of providing that background I'm going to go throug the solution process of your first example with as much detail about each step as possible. -$$ -{{1}\over{f(x)}}{{df}\over{dx}} = {{x^3}} -$$ -casting this to an integration problem: -$$ -\int {{f'(x)} \over {f(x)}}dx = \int {{x^3}} dx -$$ -now here's where things change from simply thinking about "multiplying" by a differential. in stead of using that (as mentioned before) problematic through process we'll use integration by substitution, but for a very special case. -What this process really means geometrically is using the infinitesimal breakdown of distances along the function itself as the measure for our calculated area (in stead of the distance along the real line). -Literally stating: -$$ -du = \lim_{h \to 0} {{f(x)-f(x+h) \over {h}}dx} -$$ -which when plugged into the Riemann integral becomes: -$$ -{\int {{f'(x)} \over {f(x)}}dx} = \lim_{max(x_i - x_0) \to 0}\sum_{i=1}^{n}{{f(x_i)-f(x_0)}\over {f(x_i)(x_i-x_0)}}{(x_i - x_0))} -$$ -$$ -= \lim_{max(x_i - x_0) \to 0}\sum_{i=1}^{n}{{1}\over {f(x_i)}}{(f(x_i)-f(x_0))} -$$ -by taking the observation that our limit can be changed to: -$$ -= \lim_{max(f(x_i) - f(x_0)) \to 0}\sum_{i=1}^{n}{{1}\over {f(x_i)}}{(f(x_i)-f(x_0))} -$$ -wihtout altering the meaning (this result is actually non-trivial but I won't get into that here) we have demonstrated that the integrals are equivalent. -This then results in the integral: -$$ -\int {{1} \over {f(x)}}df(x) = \int {{x^3}} dx -$$ -As a digression; often in calculus classes you're taught this using a variable as a stand in for your function, such as $u = f(x)$, and then do something like $du = f'(x)dx$ to make things easier on your sensibilites for the time (as using a function as a variable isn't usually as easy to grasp, and comes along with other issues). In this case I'll follow both concepts through so you can see how they connect. -The subsitution method with $u = f(x)$ and $du = f'(x)dx$ would then look like: -$$ -\int {{f'(x)dx} \over {u}} = \int {{du} \over {u}} = \int {{x^3}} dx -$$ -and from here we either get (if I combine the constants for simplicity): -$$ -ln(|f(x)|+c = {{x^4} \over {4}} -$$ -or with the substitution: -$$ -ln(|u|+c = {{x^4} \over {4}} -$$ -where we need to put the f(x) back in for u (since they're equal) and get: -$$ -ln(|f(x)|+c = {{x^4} \over {4}} -$$ -either way. -From here it's just cancellations and algebra until you get your final result. -There are some important things of note here for a general problem of the type: -$$ -g(f(x))f'(x) = h(x) -$$ -Since the integral is actually some $ \int g(f(x)) df(x)$ if the integral of the function g fails to exist on the region of concern the solution cannot be found using this method. Somtimes by simply "cancelling" differentials, or "multiplying through" by them this gets obfuscated.<|endoftext|> -TITLE: Degree of a Divisor = Self Intersection? -QUESTION [9 upvotes]: This has been bugging me for some time now. I know that in certain cases that the self-intersection of divisors and it's degree are the same. Like hyerplanes in projective space. I sometimes read that certain degrees are actually defined as the self-intersection. -Is this always the case, are self-intersection and degree basically the same thing? And how are they related? -Thanks in Advance! - -REPLY [12 votes]: I think I know what Mike meant to ask. I think the question was: Let $D \subset X$ be a divisor, and embed $X$ into $\mathbb{P}(H^0(X, \mathcal{O}(D))^{\vee})$ by the standard construction. - -Is the degree of $X$ inside this projective space equal to the - self-intersection $D^d$? - -This is what I have always understood the phrase "the degree of $D$", without further context, to mean. -As Georges points out, another reasonable interpretation is that we already have some embedding $\phi: X \to \mathbb{P}^N$ and we ask about the degree of $\phi(D)$; that has no relation to the self intersection of $D$. -Now, even with the interpretation above, one needs some caveats. An arbitrary divisor $D$ may not give an embedding to projective space. (Indeed, for an arbitrary $D$, $H^0(X, \mathcal{O}(D))$ may be zero.) But, if $X$ does embed in $\mathbb{P}(H^0(X, \mathcal{O}(D))^{\vee})$ (i.e. if $D$ is very ample), the answer is yes. Correctly formulated, one could probably weaken "very ample" to "ample", or maybe to one of the weaker adjectives like "big" or "nef", but I'll just stick to the very ample case, where there is literally an embedding. -Proof: Let $d= \dim X$. The degree of $X$ in $\mathbb{P}(H^0(X, \mathcal{O}(D))^{\vee})$ is the intersection of $X$ with $d$ generic hyperplanes; $X \cap H_1 \cap H_2 \cap \cdots \cap H_d$. Each those hyperplanes pulls back to a divisor on $X$ which is linearly equivalent to $D$. So $X \cap \bigcap H_i = \bigcap (X \cap H_i)$ and the latter has size $D^d$.<|endoftext|> -TITLE: non-residually finite group -QUESTION [6 upvotes]: Let $G$ be the subgroup of $\text{Bij}(\mathbb{Z})$ generated by $\sigma : n \mapsto n+1$ and $\tau$ which switches $0$ and $1$. -How can we prove that $G$ is not residually finite? Is it hopfian? - -REPLY [7 votes]: $G$ is not residually finite. Specifically, any map $f:G\to H$ with $H$ finite sends $g=(123)$ to the identity. To see this, pick $n$ with $n!>\max(5,|H|)$ and note that $f$ restricts to a map $S_n\to H$, which must have a non-trivial kernel, but the only normal subgroups of $S_n$ are $1$, $A_n$ and $S_n$. -$G$ is hopfian. Consider an epimorphism $f:G\to G$. If $\ker(f)$ contains an element $g$ of infinite order, write $g=g'\tau^k$ where $g'$ is a finitary permutation. There exists $h$ with $f(h)=\tau$; write $h=h'\tau^\ell$ similarly. Then $h^kg^{-\ell}$ is a finitary permutation, so $f(h^kg^{-\ell})=\tau^k$ has finite order, but this is a contradiction. If $\ker(f)$ contains a non-trivial element $g$ of finite order then the restriction $f:S_X\to G$ has a non-trivial kernel for some finite set $X=\{-n,\dots,n\}$. But we can choose $|X|\geq 5$ and derive a contradiction as before unless the kernel is the group of even permutations. But in the last case, since $n$ can be chosen arbitrarily large, $\ker(f)$ is the group of even permutations, so $G/\ker(f)$ is isomorphic to $\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$ (I think), contradicting the surjectivity of $f$. -So $f$ must be an isomorphism.<|endoftext|> -TITLE: Why do we distinguish the continuous spectrum and the residual spectrum? -QUESTION [23 upvotes]: As we know, continuous spectrum and residual spectrum are two cases in the spectrum of an operator, which only appear in infinite dimension. -If $T$ is a operator from Banach space $X$ to $X$, $aI-T$ is injective, and $R(aI-T)$ is not $X$. If $R(aI-T)$ is dense in $X$, then $a$ belongs to the continuous specturm, it not, $a$ belongs to the residual spectrum. -I want to know why do we care about whether $R(aI-T)$ is dense, thanks. - -REPLY [9 votes]: As pointed out by M. Reed, B. Simon in the section VI.3 of their "Methods of Modern Mathematical Physics: Functional Analysis": - -The reason that we single out the residual spectrum is that it does not occur for a large class of operators, for example, for self-adjoint operators.<|endoftext|> -TITLE: Harmonic coordinates for Ricci flow -QUESTION [6 upvotes]: It is customary to use DeTurck's argument (or Hamilton's original one involving the Nash-Moser iteration) for proving local existence of the Ricci flow. I am wondering why one cannot use harmonic coordinates for this purpose, as can be done for the Einstein equations. - -REPLY [7 votes]: There are at least two problems if you naively approach the problem from that point of view: - -How do you propose to time-evolve the harmonic coordinate system? For the Einstein's equation the coordinate system is more properly called "wave coordinates" as the coordinate functions solve the wave equation (compare to the Riemannian case when the coordinate functions solve the Laplace equation). -Harmonic coordinate systems are not necessarily global. For hyperbolic equations like Einstein's equation, finite speed of propagation implies that you can cut up into small neighborhoods, and solve in the domain of dependence o that neighborhood in the (local) wave coordinate system. For parabolic equation one cannot use the same method. - -That said, the DeTurck trick can be interpreted as the "harmonic coordinate" version of the proof. In the DeTurck trick you solve the modified Ricci flow with a vector field $X^a = g^{bc}\Gamma_{bc}^a$ where the $\Gamma$ are the analogues of the Christoffel symbol "relative to a fixed background metric". Now, the choice of harmonic coordinate system is one in which $0 = g^{bc}\Gamma_{bc}^a$ (the real Christoffel symbol relative to the coordinates now). So we can see DeTurck's trick as circumventing the difficulty in choosing a global, truly harmonic coordinate system, by compensating it with a time-dependent diffeomorphism that "gets rid of" that extra non-harmonicity. -The correct analogue for the DeTurck trick in Einstein's equations is not the simple o' wave coordinate system. Instead, it is what Choquet-Bruhat calls the "$\hat{e}$ wave gauge" and what is sometimes known as the "wave-map gauge"; for more details you should consult chapter VI.7 of her recent monograph General Relativity and the Einstein Equations.<|endoftext|> -TITLE: Is the axiom of choice needed to show that $a^2=a$? -QUESTION [10 upvotes]: A comment on this answer states that choice is needed for the statement that $a^2=a$ for all infinite cardinals $a$. In Thomas Jech's Set Theory (3rd edition), his theorem 3.5 proves this statement when $a = \aleph_b$ for some $b$. -It's not clear to me in the proof where the axiom of choice is used - can someone clarify that? -I would also like to know if this statement is true without choice when we are working with a set that is already well-ordered. Specifically: suppose we have a well-ordered set $A$, then is it true that for any infinite $B \subset A$, we have -$$ \text{card}(B)^2 = \text{card}(B) ? $$ -I believe that Jech's proof of 3.5 would hold in that case without the need for AC, but I would like to check this. - -REPLY [15 votes]: For [infinite] every well-ordered set $A$ it holds that $A^2\sim A$. Furthermore if $B\subseteq A$ and $A$ can be well-ordered then $B$ can be well-ordered as well, and therefore this property is true for $B$ as well (if it is infinite). -However it is possible that for a non well-ordered set it might not be true anymore, in fact we can easily generate such set from every non well-orderable set. If $A$ is a non well-ordered set and $\aleph(A)$ is an ordinal such that $\aleph(A)\nleq A$ (e.g. the Hartogs number of $A$) then $A+\aleph(A)$ has the property: $$\big(A+\aleph(A)\big)^2>A+\aleph(A)$$ -One should remark that it is not true for all non well-ordered sets. Even if $\mathbb R$ cannot be well-ordered it still holds that $\mathbb R^2\sim\mathbb R$. - -Some proofs: - -To see the proof that $A^2\sim A$ implies choice, see my answer: For every infinite $S$, $|S|=|S\times S|$ implies the Axiom of choice. -To see that for well-ordered sets it holds, see my answer: About a paper of Zermelo. - - -Let me add a bit of history, and refer to the Jech citation. Zermelo formulated the axiom of choice and proved (without it) that every [infinite] well-orderable set has this property in $1904$. The proof appears in the second link above. When one proves that the axiom of choice is equivalent to the fact that every set can be well-ordered, one immediately sees that the axiom of choice implies that every infinite set is bijectible with its square. -On the other hand, Tarski proved in $1923$ that the opposite holds, in particular the proof relies on an interesting lemma which abuses the example I gave for a set without this squaring property. The details appear in the first answer I have linked. -Interestingly enough, in the early $1970$'s two proofs were announced that $A+A\sim A$ does not imply the axiom of choice. I only know that one was published, it was in the Ph.D. dissertation of Sageev, and my advisor told me that he (an undergrad at the time) remembers Sageev sitting on the bench and working on this proof, and that it took him a long time to finish it. (Every footnote referring to this model suggests that the proof is very hard.)<|endoftext|> -TITLE: are there any bounds on the eigenvalues of products of positive-semidefinite matrices? -QUESTION [8 upvotes]: I have real positive semidefinite matrices (symmetric) $A$ and $B$, both are $n \times n$. -I am looking for upper bounds and lower bounds on the $m$th largest eigenvalue of $AB$, in terms of the eigenvalues of $A$ and $B$. -I know, for example, that the $\det(AB) = \det(A)\det(B)$ and that the determinant of a square matrix equals the product of its eigenvalues, but that wouldn't give me what I am looking for, because I want to focus only on the top $m$ largest eigenvalues of $AB$. - -REPLY [6 votes]: Let the eigenvalues of $A$ be $a_1\ge a_2\ge \dots \ge a_n$, and the eigenvalues of $B$ be $b_1\ge b_2\ge \dots \ge b_n$. According to the discussion in comments, we are looking for bounds on the singular values of $AB$, denoted $s_1\ge s_2\ge \dots \ge s_n$. The min-max principle is the natural tool to use. Actually, the Wikipedia article does not state the form I'd like to use here: the $k$th largest singular value of $M$ is $\inf\{\|M-T\|:\mathrm{rank}\, T\le k-1\}$. This formula appears in another article with attribution to Allakhverdiev (and no reference). I think the result of Allakhverdiev was the extension to compact operators in Hilbert spaces, but don't know for sure. In general, Wikipedia articles on singular values are a toxic mess. -Claim 1: $s_k\le \min\{ a_i b_{j} : i+j=k+1\}$. -Proof: Let $T$ be an operator of rank $\le i-1$ such that $a_i=\|A-T\|$. Similarly, let $S$ be an operator of rank $\le j-1$ such that $b_{j}=\|B-S\|$. Since $\|(A-T)(B-S)\|\le a_i b_{j}$, it remains to estimate the rank of $(A-T)(B-S)-AB$. Writing $(A-T)(B-S)-AB=-T(B-S)-AS$, we see that the rank is at most $j-1+i-1=k-1$, as desired. -Claim 2: $s_k\ge \max\{ a_i b_{j} : i+j=k+n\}$. -If $A$ and $B$ are invertible, Claim 2 follows by applying Claim 1 to $(AB)^{-1}$. Indeed, the $k$th largest singular value of $AB$ is the reciprocal of the $(n+1-k)$th largest singular value of $(AB)^{-1}$. The $(n+1-k)$th largest singular value of $(AB)^{-1}$ does not exceed $\min \{a_{n+1-i}^{-1}b_{n+1-j}^{-1} : i+j=n+2-k\}$. Decoding these sums of indices, we get Claim 2. The general case follows by considering $A+\epsilon I$ and $B+\epsilon I$, and letting $\epsilon\to 0$.<|endoftext|> -TITLE: Elementary Geometry Nomenclature: why so bad? -QUESTION [16 upvotes]: A long-ish wall of text, and I apologize. -Some background: when I was a first-year university student, my chemistry professor was lecturing and was trying to find the word to describe a shape. A student piped up and said, "that's a rhombus." The professor stopped mid-stride, looked at him squarely, and said, "rhombus? That's a stupid word. What's a rhombus? I don't even think that's a word. The word I was thinking of was 'parallelogram'." This was shocking, because this was an American professor, at an American university, and in my American public education, I was taught what a rhombus was in the second or third grade. -Recently, however, I was thinking that maybe my professor wasn't wrong. Consider the naming system for quadrilaterals. The term "quadrilateral" makes some sense: "quad" from Latin for "four", and "lateral" meaning side. And then you get parallelogram, with "parallel" meaning "parallel" and "gram" from Greek meaning "drawn". But then a rectangle is a special case of a parallelogram where the angles are all right angles, which follows clearly enough, and a square is a special case of a rectangle, and important enough to merit its own term. -But then a quadrilateral with only two parallel sides is a trapezoid, which derives from Greek for "table shaped". And then a rhombus is the complement to the square in the special cases of parallelograms -- its angles are anything but right angles! -Confusing yet? We've got the following suffixes describing shapes: -lateral, -gram, -zoid. -We also have triangles, which makes sense because it's "three angles." Yet a "quadrangle" is a region in a university campus. -Increasing the number of sides in the shape, we go from "quadrilaterals" to "pentagons". Ok, now we've gone from the Latin prefix for "four" and a suffix meaning "side" to the Greek for "five" and a totally different suffix. Sometimes we describe the word using a root that means "drawn", and sometimes we describe it by the way that it looks. -And still "rhombus" fits in nowhere in this crazy, convoluted scheme! -To bring this all back to mathematics, and to ask my original question: -Individually, I can find the etymology of each of these terms. But why did the mathematics community adhere to these terms, particularly in elementary education? Did these terms get translated haphazardly from Elements? Is this one of those consequences of the somewhat insular nature of the mathematical community during the Renaissance era? The mathematics community has evolved to be fairly precise in its use of terminology. Why is the terminology surrounding elementary geometry so fragmented? - -REPLY [4 votes]: We need to get better at naming things and it should follow an understandable pattern for easy comprehension. When we're inconsistent with our nomenclature, it takes the fun out of learning. -Unfortunately, this problem goes all the way back to the numbers we use. Eighty-one is 8 tens and one, which makes sense, but explain "eleven" and "twelve". Then try explaining "thirteen" which is 3 ones and ten (which is backwards). -Where English has unique names for 11 and 12, Spanish has unique names for 11-15, then 16 translates to "ten and six". -Adding these poorly named numbers makes you do an extra step in your head. "Eleven plus seventeen plus eighty-one"? First you have to think about what these words mean, then rearrange their orders in your head by tens and ones, then add them. -Science makes you memorize the names of long dead folks (Ohms, Plancks, Coulombs,...) and computer programming is awful as well. (Did you know that in Unix, the command used to OPEN a file is "LESS"?) -Every subject has things which need new names.<|endoftext|> -TITLE: If $\phi \in C^1_c(\mathbb R)$ then $ \lim_n \int_\mathbb R \frac{\sin(nx)}{x}\phi(x)\,dx = \pi\phi(0)$. -QUESTION [10 upvotes]: Let $\phi \in C^1_c(\mathbb R)$. Prove that $$ -\lim_{n \to +\infty} \int_\mathbb R \frac{\sin(nx)}{x}\phi(x) \, dx = \pi\phi(0). -$$ - -Unfortunately, I didn't manage to give a complete proof. First of all, I fixed $\varepsilon>0$. Then there exists a $\delta >0$ s.t. -$$ -\vert x \vert < \delta \Rightarrow \vert \phi(x)-\phi(0) \vert < \frac{\varepsilon}{\pi}. -$$ -Now, I would use the well-known fact that -$$ -\int_\mathbb R \frac{\sin x}{x} \, dx = \pi. -$$ -On the other hand, by substitution rule, we have also -$$ -\int_\mathbb R \frac{\sin(nx)}{x} \, dx = \int_\mathbb R \frac{\sin x}{x} \, dx = \pi. -$$ -Indeed, I would like to estimate the quantity -$$ -\begin{split} -& \left\vert \int_\mathbb R \frac{\sin(nx)}{x}\phi(x) \, dx - \pi \phi(0) \right\vert = \\ -& = \left\vert \int_\mathbb R \frac{\sin(nx)}{x}\phi(x) \, dx - \phi(0)\int_\mathbb R \frac{\sin{(nx)}}{x}dx \right\vert \le \\ -& \le \int_\mathbb R \left\vert \frac{\sin(nx)}{x}\right\vert \cdot \left\vert \phi(x)-\phi(0) \right\vert dx -\end{split} -$$ -but the problem is that $x \mapsto \frac{\sin(nx)}{x}$ is not absolutely integrable over $\mathbb R$. Another big problem is that I don't see how to use the hypothesis $\phi$ has compact support. -I think that I should use dominated convergence theorem, but I've never done exercises about this theorem. Would you please help me? Thank you very much indeed. - -REPLY [4 votes]: Assume that $\phi(x)$ is supported in $|x|< L$. Since $\phi$ is differentiable, $\frac{\phi(x)-\phi(0)}{x}$ is bounded and therefore integrable on $|x| -TITLE: Sum of truncated normals -QUESTION [5 upvotes]: Suppose $X_1, \dots, X_n$ are truncated standard normal variables, truncated so that $X_i \geq 0$ (that is, $X_i$ is drawn as a standard normal, conditional on $X_i \geq 0$) -Let $c_1, \dots, c_n$ be non-negative coefficients. -What does the distribution of $\sum_i c_i Y_i$ look like? Does it have, or approximately have, a standard distribution, such as a truncated normal distribution? - -Original question: -Suppose $X_1, \dots, X_n$ are iid Normal random variables, with mean 0 and variances $\sigma_1, \dots, \sigma_n$. -Let $Y_i = \max(0,X_i)$. (So $Y_i$ is a truncated normal random variable). -What does the distribution of $\sum_i Y_i$ look like? Does it have, or approximately have, a standard distribution? - -REPLY [5 votes]: This post answers the original question. -Please be careful, $Y_k$ is not a truncated normal random variable, it is censored normal random variable. In particular: -$$ - \mathbb{P}(Y_k = 0) = \mathbb{P}\left(X_k \leqslant 0 \right) = \frac{1}{2} \not= 0 -$$ -meaning that $Y_k$ is not an absolutely continuous random variable. Rather, $Y_k$ can be thought of as the mixture of a degenerate random variable, concentrated at $x=0$, and a normal random variable, truncated to the positive semi-axis. -With this said, $Z = \sum\limits_{k=1}^n Y_k$ is not absolutely continuous either, since -$$ - \mathbb{P}\left(Z=0\right) = \mathbb{P}\left(X_1 \leqslant 0, \ldots, X_n \leqslant 0\right) = \mathbb{P}\left(X_1 \leqslant 0\right) \cdots \mathbb{P}\left(X_n \leqslant 0\right) = \frac{1}{2^n} -$$ -The absolutely continuous part of $Z$ is not equal in distribution to any standard distribution. One can compute the characteristic function of $Z$ rather easily. -$$ \begin{eqnarray} - \phi_{Y_k}(t) &=& \mathbb{E}\left(\exp\left(i \max(0,X_k)t\right)\right) = \mathbb{P}\left(X_k \leqslant 0\right) + \mathbb{E}\left(\exp\left(i X_k t\right): X_k > 0\right)\\ &=& \frac{1}{2} + \exp\left(-\frac{t^2}{2} \sigma_k^2\right) + \frac{2 i}{\sqrt{\pi}} D_F\left( \frac{t \sigma_k}{\sqrt{2}} \right) -\end{eqnarray}$$ -where $D_F(x)$ denotes Dawson's F-function. -Since the random variables $Y_k$ are independent: -$$ - \phi_Z(t) = \phi_{Y_1}(t) \cdots \phi_{Y_n}(t) -$$ -Here is a histogram for several small values of $n$, assuming equal unit variance: - -You can explicitly see how the cumulative distribution function is not continuous at $x=0$, and how the size of the discontinuity jump decreases as $n$ grows.<|endoftext|> -TITLE: Why can't we construct a counter-example to the Fundamental Lemma of the Calculus of Variations? -QUESTION [6 upvotes]: "The fundamental lemma of the calculus of variations states that if the definite integral of the product of a continuous function $f(x)$ and $h(x)$ is zero, for all continuous functions $h(x)$ that vanish at the endpoints of the range of integration and have their first two derivatives continuous, then $f(x)=0$." -Why can't we construct some $h(x)$ that starts at $(a,0)$, has a positive trajectory - imagine, say, an upside down parabola - and then comes back down to end at $(b,0)$, and then some $f(x)$ that is positive for the first half of the interval and negative for the second half (think of some sinusoidal curve)? Then when we integrate, we would basically be adding up a series of positive quantities ($h(x)f(x)$ would be positive), and then a series of negative quantities ($f(x)h(x)$ would be $\text{negative}\times\text{positive} = \text{negative}$). We could then fine-tune to get this to equal zero. -What precisely am I not understanding here? - -REPLY [16 votes]: This seems to be a problem with quantifiers. -If I'm understanding you correctly, the fundamental lemma states that: - -If $f \in C[a,b]$ is such that $\int_a^b f(x)h(x)\,dx = 0$ for all $h\in C^2[a,b]$ with $h(a) = h(b) = 0$, then $f \equiv 0$. - -A counter-example to this claim would mean finding a function $f \in C[a,b]$ with $f \not \equiv 0$ such that $\int_a^b f(x) h(x)\,dx = 0$ for all $h \in C^2[a,b]$ with $h(a) = h(b) = 0$. -In other words: You found a function $f\not \equiv 0$ such that $\int_a^b f(x)h(x)\,dx = 0$ for some function $h \in C^2[a,b]$ with $h(a) = h(b) = 0$. Your function $f$, however, will not satisfy $\int_a^b f(x)h(x)\,dx = 0$ for all $h \in C^2[a,b]$ with $h(a) = h(b) = 0$.<|endoftext|> -TITLE: A trigonometric identity -QUESTION [5 upvotes]: If one sees the simplification done in equation $5.3$ (bottom of page 29) of this paper it seems that a trigonometric identity has been invoked of the kind, -$$\ln(2) + \sum _ {n=1} ^{\infty} \frac{\cos(n\theta)}{n} = - \ln \left\vert \sin\left(\frac{\theta}{2}\right)\right\vert $$ -Is the above true and if yes then can someone help me prove it? - -REPLY [7 votes]: Hint 1: Use that -$$ -\log(1-z)=-\sum\limits_{n=1}^\infty\frac{z^n}{n} -$$ -Hint 2: Set -$$ -z=r e^{i\theta} -$$ -Hint 3: Take a real part -Hint 4: Take a limit $r\to 1-0$ and use Abel's summation formula<|endoftext|> -TITLE: Are all $p$-adic number systems the same? -QUESTION [24 upvotes]: After just having learned about $p$-adic numbers I've now got another question which I can't figure out from the Wikipedia page. -As far as I understand, the $p$-adic numbers are basically completing the rational numbers in the same way the real numbers do, except with a different notion of distance where differences in more-significant digits correspond to small distances, instead of differences in less-significant digits. So if I understand correctly, the $p$-adic numbers contain the rational numbers, but not the irrational numbers, while the non-rational $p$-adic numbers are not in $\mathbb{R}$ (someone please correct me if I'm wrong). -Now the real numbers do not depend on the base you write the numbers in. However the construction of the $p$-adic numbers seems to depend on the $p$ chosen. On the other hand I am sure that the construction of the real numbers can be written in a way that it apparently depends on the base, so the appearance might be misleading. -Therefore my question: Are the $p$-adic numbers the same for each $p$ (that is, are e.g. $2$-adic and $3$-adic numbers the same numbers, only written in different bases), or are they different (except for the rational numbers, of course). For example, take the $2$-adic number $x := ...1000001000010001001011$ (i.e. $\sum_{n=0}^\infty 2^{n(n+1)/2}$), which IIUC isn't rational (because it's not periodic). Can $x$ also be written as $3$-adic number, or is there no $3$-adic number corresponding to this series? -In case they are different, is there some larger field which contains all $p$-adic numbers for arbitrary $p$? - -REPLY [40 votes]: No, the different $p$-adic number systems are not in any way compatible with one another. -A $p$-adic number is a not a number that is $p$-adic; it is a $p$-adic number. Similarly, a real number is not a number that is real, it is a real number. There is not some unified notion of "number" that these are all subsets of; they are entirely separate things, though there may be ways of identifying bits of them in some cases (e.g., all of them contain a copy of the rational numbers). -Now, someone here is bound to point out that if we take the algebraic closure of some $\mathbb{Q}_p$, the result will be algebraically isomorphic to $\mathbb{C}$. But when we talk about $p$-adic numbers we are not just talking about their algebra, but also their absolute value, or at least their topology; and once you account for this they are truly different. (And even if you just want algebraic isomorphism, this requires the axiom of choice; you can't actually identify a specific isomorphism, and there's certainly not any natural way to do so.) -How can we see that they are truly different? Well, first let's look at the algebra. The $5$-adics, for instance, contain a square root of $-1$, while the $3$-adics do not. So if you write down a $5$-adic number which squares to $-1$, there cannot be any corresponding $3$-adic number. -But above I claimed something stronger -- that once you account for the topology, there is no way to piece the various $p$-adic number systems together, which the above does not rule out. How can we see this? Well, let's look at the topology when we look at the rational numbers, the various $p$-adic topologies on $\mathbb{Q}$. These topologies are not only distinct -- any finite set of them is independent, meaning that if we let $\mathbb{Q}_i$ be $\mathbb{Q}$ with the $i$'th topology we're considering, then the diagonal is dense in $\mathbb{Q}_1 \times \ldots \times \mathbb{Q}_n$. -Put another way -- since these topologies all come from metrics -- this means that for any $c_1,\ldots,c_n\in\mathbb{Q}$, there exists a sequence of rational numbers $a_1,a_2,\ldots$ such that in topology number 1, this converges to $c_1$, but in topology number two, it converges to $c_2$, and so forth. (In fact, more generally, given any finite set of inequivalent absolute values on a field, the resulting topologies will be independent.) -So even on $\mathbb{Q}$, the different topologies utterly fail to match up, so there is no way they can be pieced together by passing to some larger setting.<|endoftext|> -TITLE: Selfadjoint compact operator with finite trace -QUESTION [16 upvotes]: I have a compact selfadjoint operator $T$ on a separable Hilbert space. For some fixed orthonormal basis, the operator's diagonal is in $\ell^1(\mathbb{N})$. - -Can we conclude that $T$ is trace class? - -REPLY [2 votes]: Disclaimer: Non-Compact Operators! -Given the Hilbert space $\ell^2(\mathbb{N})$. -Consider sum of shifts:* -$$A_\pm:\ell^2(\mathbb{N})\to\ell^2(\mathbb{N}):\quad A_\pm:=R\pm L$$ -They have finite trace: -$$\sum_n\langle A_\pm\delta_n,\delta_n\rangle=\sum_n0=0$$ -But for the shifts: -$$\sum_n\langle|R|\delta_n,\delta_n\rangle=\sum_n1=\infty$$ -$$\sum_n\langle|L|\delta_n,\delta_n\rangle=\sum_n1=\infty$$ -Thus for the sum: -$$\operatorname{Tr}A_\pm<\infty\implies\operatorname{Tr}A_\mp<\infty$$ -Concluding counterexample. -*Shifts: Right & Left<|endoftext|> -TITLE: A (probably trivial) induction problem: $\sum_2^nk^{-2}\lt1$ -QUESTION [23 upvotes]: So I'm a bit stuck on the following problem I'm attempting to solve. Essentially, I'm required to prove that $\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2} < 1$ for all $n$. I've been toiling with some algebraic gymnastics for a while now, but I can't seem to get the proof right. Proving it using calculus isn't a problem, but I'm struggling hither. - -REPLY [3 votes]: Yet another approach : -Let us first analyze the sum till infinity. Let $$ S= \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+ \cdots \infty$$ $$\Rightarrow S=(\frac{1}{2^2}+\frac{1}{4^2}+ \cdots\infty) +(\frac{1}{3^2}+\frac{1}{5^2}+ \cdots\infty ) $$$$\Rightarrow S= \frac{1}{4}(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots\infty)+ S' $$ Where $$S'=\frac{1}{3^2}+\frac{1}{5^2}+ \cdots\infty $$ $$\Rightarrow S=\frac{1}{4}(1+S)+S'$$ $$\Rightarrow 3S=4S'+1........ Eqn(1)$$ Now examine the following inequality $$ (\frac{1}{2^2}-\frac{1}{3^2})+(\frac{1}{4^2}-\frac{1}{5^2})+ \cdots \infty > 0$$ $$\Rightarrow \frac{1}{2^2}+\frac{1}{4^2}+\cdots > \frac{1}{3^2}+\frac{1}{5^2}+ \cdots$$ $$\Rightarrow \frac{1}{4}(1+\frac{1}{2^2}+\cdots) >\frac{1}{3^2}+\frac{1}{5^2}+ \cdots $$ $$\Rightarrow \frac{1}{4}(1+S)> S'$$ $$\Rightarrow (1+S)> 4S'......Eqn(2)$$ From Equation 1 and 2 we get $$ 1+S> 3S-1$$ $$\Rightarrow 1> S$$ Which shows $$ 1> \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+ \cdots \infty$$<|endoftext|> -TITLE: Zeros of a complex polynomial -QUESTION [7 upvotes]: The question is: -Show that $$ P(z) = z^4 + 2z^3 + 3z^2 + z +2$$ has exactly one root in each quadrant of the complex plane. -My initial thought was to use Rouche's Theorem (since that's generally what I use to find how many roots a complex polynomial has), but the more I think about it, the more I'm not sure how to make it work. Here is my attempt: -First, pick a radius for a circle that can encompass all four roots of the polynomial. For simplicity sake (in my opinion), I went with |z| = 5. -Setting $$f(z) = z^4$$ and $$g(z) = 2z^3 + 3z^2 + z + 2$$ -I get |f(z)| = 625 and |g(z)| = 332, so by Rouche's Theorem we have four roots in the disc. -Now, my thought was that I could somehow seperate the quadrants by breaking up my circle into quarter-circles (like four slices of pie) and applying Rouche's Theorem again on each of these new domains. However, finding the place on the boundary where the value hits its max for some f(z) or g(z) would be messy (at best), since if I juse use |z| = 5, I'm right back where I started. There's also the issue that these zeroes may occur on the boundary, so in the real/imaginary axis, which wouldn't be what I'm trying to show. So now, I'm just stuck, so if anyone can see how to tackle this, it would be greatly appreciated. - -REPLY [4 votes]: First dispose of real roots: e.g. $P(z) = z^2 (z+1)^2 + 2 (z+1/4)^2 + 15/8$. -Let's look at what happens to $P(z)$ as $z$ goes around a contour around part of the -first quadrant. As $z$ goes from $0$ to some large positive $R$ on the real axis, -$P(z)$ increases from $2$ to $P(R) >> 0$. Then go on the quarter-arc of the circle $|z| = R$ from $R$ to $iR$: $P(z)$ goes almost in a circle, ending at $P(iR)$ which is in the fourth quadrant. Now come back in to the origin on the imaginary axis. Note that $\text{Re}(P(it)) = t^4 - 3 t^2 + 2 = 0$ at $t = 1$ and $t=\sqrt{2}$, while $\text{Im}(P(it)) = - 2 t^3 + t = 0$ at $t=0$ and $t = \sqrt{2}/2$. So you hit the negative imaginary axis at $t=\sqrt{2}$ and again at $t=1$, then the positive real axis at $t=\sqrt{2}/2$ and $t=0$, but not the negative real or positive imaginary axis. Thus as $z$ goes around this contour, the winding number of $P(z)$ around $0$ is $1$, indicating that there is exactly one zero of $P(z)$ inside the contour. -Here's a plot of the case $R=1.6$:<|endoftext|> -TITLE: Complete undergraduate bundle-pack -QUESTION [38 upvotes]: First of all I'm sorry if this is not the right place to post this. I like math a lot. But I'm not sure if i want to do a math major in college. My question is: Can you give me a list of books that will give me the knowledge of the subjects a person doing a math major would have? I think I know all the stuff a good high school student knows. Thanks. - -REPLY [5 votes]: Chicago undergraduate mathematics bibliography - - -ELEMENTARY -This includes “high school topics” and first-year calculus. Contents - - -Algebra $(4)$ - - -Gelfand/Shen, Algebra -Gelfand/Glagoleva/Shnol, Functions and graphs -Gelfand/Glagoleva/Kirillov, The method of coordinates -Cohen, Precalculus with unit circle trigonometry, and its content is here. - -Geometry $(2)$ - - -Euclid, The elements -Coxeter, Geometry revisited - -Foundations $(1)$ - - -Rucker, Infinity and the mind - -Problem solving $(4)$ - - -New Mathematical Library problem books -Larson, Problem solving through problems -Pólya, How to solve it -Pólya, Mathematics and plausible reasoning, I and II - -Calculus $(6)$ - - -Spivak, Calculus -Spivak, The hitchhiker's guide to calculus -Hardy, A course of pure mathematics -Courant, Differential and integral calculus -Apostol, Calculus -Janusz, Calculus - -Bridges to intermediate topics $(2) $ - - -Smith, Introduction to mathematics: algebra and analysis -Johnson, Introduction to logic via numbers and sets - - - - -INTERMEDIATE -Roughly, general rather than specialized texts in higher mathematics. I would not hesitate to recommend any book here to honors second-years, but they might not find easy going in some of them. - - -Foundations $(5)$ - - -Halmos, Naive set theory -Fraenkel, Abstract set theory -Ebbinghaus/Flum/Thomas, Mathematical logic -Enderton, A mathematical introduction to logic -Landau, Foundations of analysis - -General abstract algebra $(7)$ (difficulty: $\color{orange}{\mathscr{m}}$oderate-$\color{red}{\mathscr{h}}$igher) - - -$\color{orange}{\mathscr{m}}$ - Dummit/Foote, Abstract algebra -$\color{orange}{\mathscr{m}}$ - Herstein, Topics in algebra -$\color{orange}{\mathscr{m}}$ - Artin, Algebra -$\color{red}{\mathscr{h}}$ - Jacobson, Basic algebra I -$\color{red}{\mathscr{h}}$ - Hungerford, Algebra -$\color{red}{\mathscr{h}}$ - Lang, Algebra -$\color{red}{\mathscr{h}}$ - Mac Lane/Birkhoff, Algebra - -Linear algebra $(3)$ - - -Halmos, Finite dimensional vector spaces -Curtis, Abstract linear algebra -Greub, Linear algebra and Multilinear algebra - -Number theory $(5)$ - - -Ireland/Rosen, A classical introduction to modern number theory -Burn, A pathway into number theory -- Hardy/Wright, Introduction to number theory -- - -Combinatorics and discrete mathematics $(1)$ -Real analysis $(10)$ -- -- -- -- - -- - -Multivariable calculus $(2)$ - -- - -Complex analysis $(5)$ -- -- -Differential equations $(2)$ - -- - -Point-set topology $(5)$ -- -- -Differential geometry $(4)$ -- - -- - -Classical geometry $(3) $ -- - - -TO BE CONTINUED<|endoftext|> -TITLE: Evaluation of $\sum_{x=0}^\infty e^{-x^2}$ -QUESTION [21 upvotes]: Most of us are aware of the classic Gaussian Integral -$$\int_0^\infty e^{-x^2}\, dx=\frac{\sqrt{\pi}}{2}$$ -I would be interested in evaluating the similar sum -$$\sum_{x=0}^\infty e^{-x^2}$$ -Now, because $\exp(-\lfloor x \rfloor^2) \ge \exp(-x)$, we find -$$\sum_{x=0}^\infty e^{-x^2}= \int_0^\infty e^{-\lfloor x \rfloor^2}\, dx \ge \int_0^\infty e^{-x^2}\, dx=\frac{\sqrt{\pi}}{2}$$ -Does a closed form for this sum exist? If so, what would it be? I would be very interested in how a closed form would be found for this function. - -REPLY [14 votes]: As mentioned in the comments, this sum is related to one of the Jacobi theta functions. -$$\begin{eqnarray*} -\sum_{n=0}^\infty e^{-n^2} -&=& \frac{1}{2}\left(1 + \sum_{n=-\infty}^\infty \left(\frac{1}{e}\right)^{n^2}\right) \\ -&=& \frac{1}{2}\left[1+\vartheta_3\left(0,\frac{1}{e}\right)\right] \\ -&\simeq& 1.386 -\end{eqnarray*}$$<|endoftext|> -TITLE: Maximal real subfield of $\mathbb{Q}(\zeta )$ -QUESTION [8 upvotes]: Let $l$ be an odd prime number and $\zeta$ be a primitive $l$-th root of unity in $\mathbb{C}$. -Let $K = \mathbb{Q}(\zeta)$. -My question: Is the following proposition true? If yes, how would you prove this? -Proposition -(1) $K_0 = \mathbb{Q}(\zeta + \zeta^{-1})$ is the maximal real subfield of $K$. -(2) $[K_0 : \mathbb{Q}] = (l - 1)/2$ -(3) The ring of algebraic integers $A_0$ in $K_0$ is $\mathbb{Z}[\zeta + \zeta^{-1}]$. -(4) $\zeta + \zeta^{-1}$ and its conjugates constitute an integral basis of $A_0$. -Related questions -This and this. - -REPLY [8 votes]: As Dylan points out, parts (1) and (2) are clear. Moreover, $\mathbb{Z}[\zeta + \zeta^{-1}]$ contains $\zeta^j + \zeta^{-j}$ for all $j \ge 1$ (by induction using the binomial theorem); these include all the conjugates of $\zeta + \zeta^{-1}$, so (4) implies (3). Thus it suffices to prove (4), which follows from the corresponding fact for the full cyclotomic field $\mathbb{Q}(\mu_\ell)$ (which is well known), as follows: -Let $u \in A_0$. Because $u$ is an algebraic integer in $\mathbb{Q}(\zeta)$, we can write $u = \sum_{i = 0}^{\ell - 1} u_i \zeta^i$ for some $u_i \in \mathbb{Z}$. But since $u = \overline{u}$, we have $u = \sum u_i \zeta^{-i} = \sum u_i \zeta^{\ell - i}$. Hence $u_i = u_{\ell - i}$. Thus we have $u = \sum_{i = 0}^{(\ell - 1)/2} u_i (\zeta_i + \zeta^{-i})$, and (4) is proved.<|endoftext|> -TITLE: Why are $\sin$ and $\cos$ (and perhaps $\tan$) "more important" than their reciprocals? -QUESTION [13 upvotes]: (My personal "feel" is that $\sin$ and $\cos$ are first-class citizens, $\tan$ is "1.5th-class," and the rest are second-class; I'm sure there are others who feel the same.) -Main question(s): From a purely high-school-geometric/"ninth-century-geometer's" standpoint, is there any reason why this should be so? Given the usual elementary knowledge of triangles when one is first introduced to these functions, I think it appears pretty arbitrary. How should I convince a high school student that $\sin$, $\cos$, and $\tan$, instead of their reciprocals, should be our main objects of study? How did history decide on their superiority? -Of course, with real analysis goggles, things look quite a bit different: $\sin$ and $\cos$ are the only ones that are continuous everywhere; $f'' = -f$ characterizes all their linear combinations; they have much nicer series representations; etc. But I suspect this is all hindsight. -(I don't pretend to know enough about complex analysis, but I suspect even more nice things happen there with $\sin$ and $\cos$, and even more ugly things happen with the other four. In any case, I doubt history chose $\sin$ and $\cos$ to be first-class citizens because of their complex properties.) -Secondary questions: Is there any reason why $\sin$ is the "main" function and $\cos$ is "only" its complement, or is this arbitrary as well? Is there any reason why $\tan$ is preferred to $\cot$? - -REPLY [11 votes]: Historically, trigonometric functions were originally computed in terms of "chords" (the angles subtended by a circular arc), and the sine was computed from a bisected chord (so a "half-chord"). In fact, the word "sine" originates from a mis-translation (from Arabic) of the Sanskrit word "jyā" which literally means "bow-string". -(The word "jyā" became "jība" in Arabic, and then subsequently "jaib". A cognate of "jaib" in Arabic has the meaning of "bosom", and jaib was mistakenly rendered into Latin as "sinus". So yes, the word "sine" was originally not safe for work). -The importance of the cosine seems also to have been first recognized by Indian mathematicans, who called it "koṭi-jyā" or "kojyā" meaning (roughly) "sine of the extreme angle" ("koṭi" means "the extreme end of a bow" or "extremity" in general). -So the sine as the "main" function, and cosine as the "adjunct" function goes back at least 16 centuries (in the Surya Siddhanta, written some time in the 4th century).<|endoftext|> -TITLE: Intersection of all neighborhoods of zero is a subgroup -QUESTION [14 upvotes]: Let $G$ be a topological abelian group. Let $H$ be the intersection of all neighborhoods of zero. -How is $H = \mathrm{cl}(\{0\})$? Isn't the closure of a set $A$ the smallest closed set containing $A$ which is the same as the intersection of all closed sets containing $A$? But neighborhoods in $G$ are not necessarily closed. Thanks. - -(Edit) -To see that $H$ is a subgroup: -First note that by construction $H$ contains $0$. -Furthermore, $f: x \mapsto -x$ is continuous and its own inverse so that $f$ is also open. Hence $U$ is a neighborhood of $0$ if and only if $-U$ is. Now let $x \in H$. Then $x$ is in every neighborhood $U$ of $0$. Hence $x$ is in every neighborhood $-U$ of $0$. Hence $-x$ is in $U$ and hence in $H$. -Alternatively one can verify it as follows: $$x \in H \iff x \in \bigcap_{U \text{ nbhd of } 0} U \iff x \in \bigcap_{U \text{ nbhd of } 0} -U \iff -x \in \bigcap_{U \text{ nbhd of } 0} U \iff -x \in H$$ -To see that $x+y$ is in $H$ if $x,y \in H$, note that $g: (x,y) \mapsto x+y$ is continuous. Now let $V$ be an arbitrary neighborhood of $0$. Then since $g$ is continuous there exists a neighborhood $N \times M$ of $(0,0)$ such that $g(N \times M) \subset V$. Since $G \times G$ has the product topology, $N \times M$ is a neighborhood of $0$ if and only if $N$ and $M$ are neighborhoods of $0$. Hence $x,y \in N$ and $x,y \in M$ and hence $g((x,y)) = x + y \in V$ since $g(N \times M) \subset V$. - -REPLY [12 votes]: $\def\cl{\mathop{\mathrm{cl}}}$ For $x \in G$, let $\mathcal U_x$ denote the set of all neighbourhoods of $x$. Then we have that $x - \mathcal U_0 = \mathcal U_x$ for each $x \in G$. It follows -\begin{align*} - x \in \cl\{0\} &\iff \forall U \in \mathcal U_x : U \cap \{0\} \ne \emptyset\\ - &\iff \forall V \in \mathcal U_0: (x - V) \cap \{0\} \ne \emptyset\\ - &\iff \forall V \in \mathcal U_0: 0 \in x - V\\ - &\iff \forall V \in \mathcal U_0 : x \in V\\ - &\iff x \in \bigcap \mathcal U_0. -\end{align*}<|endoftext|> -TITLE: Compute $ \sum\limits_{m=1}^{\infty} \sum\limits_{n=1}^{\infty} \sum\limits_{p=1}^{\infty}\frac{(-1)^{m+n+p}}{m+n+p}$ -QUESTION [14 upvotes]: How would you compute this sum? It's not a problem I need to immediately solve, but a problem that came to my mind today. I think that the generalization to more than three nested sums would be interesting as well. -$$ \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \sum_{p=1}^{\infty}\frac{(-1)^{m+n+p}}{m+n+p}$$ - -REPLY [26 votes]: Here is a simple lemma: - -Let $(u_n)_{n\geqslant1}$ denote a decreasing sequence of positive functions defined on $(0,1)$, which converges pointwise to zero and such that $u_1$ is integrable on $(0,1)$. Then, - $$ -\sum\limits_{n=1}^{+\infty}(-1)^n\int_0^1u_n(s)\,\mathrm ds=\int_0^1u(s)\,\mathrm ds,\qquad u(s)=\sum\limits_{n=1}^{+\infty}(-1)^nu_n(s). -$$ - -Now, let us consider the multiple series the OP is interested in. One sees readily that it does not converge absolutely hence the idea is to apply the lemma three times. - -First, fix $n$ and $m$ and, for every $p\geqslant1$, consider $u_p(s)=s^{m+n+p-1}$. Then $u(s)=-\dfrac{s^{m+n}}{1+s}$ hence the lemma yields -$$ -\sum\limits_{p=1}^{+\infty}\frac{(-1)^{m+n+p}}{m+n+p}=(-1)^{m+n}\sum\limits_{p=1}^{+\infty}(-1)^{p}\int_0^1u_p(s)\,\mathrm ds=(-1)^{m+n+1}\int_0^1\frac{s^{m+n}}{1+s}\,\mathrm ds. -$$ -Second, fix $m$ and, for every $n\geqslant1$, consider $u_n(s)=\dfrac{s^{m+n}}{1+s}$. Then $u(s)=-\dfrac{s^{m+1}}{(1+s)^2}$ hence the lemma yields -$$ -\sum\limits_{n=1}^{+\infty}(-1)^{m+n+1}\int_0^1\frac{s^{m+n}}{1+s}\,\mathrm ds=(-1)^m\int_0^1\frac{s^{m+1}}{(1+s)^2}\,\mathrm ds -$$ -Third and finally, for every $m\geqslant1$, consider $u_m(s)=\dfrac{s^{m+1}}{(1+s)^2}$. Then $u(s)=-\dfrac{s^{2}}{(1+s)^3}$ hence the lemma yields -$$ -\sum\limits_{m=1}^{+\infty}(-1)^m\int_0^1\frac{s^{m+1}}{(1+s)^2}\,\mathrm ds=-\int_0^1\frac{s^{2}}{(1+s)^3}\,\mathrm ds. -$$ - -Thus, the triple series the OP is interested in converges and the value $S_3$ of the sum is -$$ -\color{red}{S_3=-\int_0^1\frac{s^{2}}{(1+s)^3}\,\mathrm ds}=-\int_1^2\frac{s^{2}-2s+1}{s^3}\,\mathrm ds=-\left[\log(s)+\frac2s-\frac1{2s^2}\right]_1^2, -$$ -that is, $\color{red}{S_3=-\log(2)+\frac58}=-0.06814718\ldots$ -The technique above shows more generally that, for every $k\geqslant1$, the analogous series over $k$ indices converges and that the value of its sum is -$$ -S_k=(-1)^k\int_0^1\frac{s^{k-1}}{(1+s)^k}\,\mathrm ds=(-1)^k\left(\log(2)+\sum_{i=1}^{k-1}(-1)^i{k-1\choose i}\frac1i(1-2^{-i})\right). -$$<|endoftext|> -TITLE: Cancellation law for Minkowski sums -QUESTION [5 upvotes]: Let $(X,\|\cdot\|)$ be a Banach space and $A,B,C\subset X$ closed bounded non-empty convex subsets. Let $+$ denote the Minkowski symbol for addition. -Does the $+$ satisfy: $$A+C\subset B+C\implies A\subset B$$ - -REPLY [4 votes]: It does. Suppose $a\in A$ with $a\notin B$. Since $B$ is closed and not empty, there is $b\in B$ with minimal distance from $A$. Project everything onto $a-b$, and denote the projected objects by primes. Since the sets are closed and convex, their projections are closed intervals. $B'$ is all on one side of $a'$, since the line segment from $b$ to a point whose projection is on the other side of $a'$ would contain points closer to $a$ than $b$. Thus $a'+C'\not\subset B'+C'$, hence $a+C\not\subset B+C$ and thus $A+C\not\subset B+C$. -[Edit:] -As t.b. pointed out, this doesn't work in general, but as D. Thomine pointed out, it can be fixed using the Hahn–Banach theorem. Again, suppose $a\in A$ with $a\notin B$. Since $\{a\}$ is convex and compact and $B$ is convex and closed, there is a continuous linear map $\lambda:X\to\mathbb R$ strictly separating the two, so there is $s\in\mathbb R$ such that $\lambda(a)\lt s\lt\lambda(b)$ for all $b\in B$. Since $C$ is bounded and $\lambda$ is continuous, $\lambda(C)$ is bounded. Thus $\lambda(a)+\lambda(C)\not\subset\lambda(B)+\lambda(C)$, hence $a+C\not\subset B+C$ and thus $A+C\not\subset B+C$. -Note that only the boundedness of $C$, not that of $A$ or $B$ has been used.<|endoftext|> -TITLE: When is the integral of a periodic function periodic? -QUESTION [8 upvotes]: I'm attempting some questions from Zwiebach - A First Course in String Theory, and have got stuck. I've proved that a function $h'(u)$ is periodic. The question then asks me to show that $h(u)=au+f(u)$ where $a$ is a constant and $f(u)$ a periodic function. I can't see how to do this directly from the periodicity of $h'$. Is this possible, or true? -Many thanks! - -REPLY [3 votes]: Since $h'$ is periodic, integrating $h'$ over an interval the size of its period will always give the same value, let's call it $I$. -Integrating e.g. from $0$ to $T$ will give $I$, from $0$ to $2T$ will give $2I$, etc. In general, $\forall x$, $\int_x^{x+T} h'(u) du =I$. You'll see then that $h(x+T)-h(x)=I$. (take any point $x_0$ to define $h(x)=\int_{x_0}^x h' +h(x_0)$ -Case 1: $I=0$. Then $h$ is a periodic function of period $T$. -Case 2: $I\neq 0$. Then consider the function $g=h'-I/T$. $g$ is also periodic of period $T$ and the integral of $g$ over it's period is $0$. We are back to case 1. You'll then find that $a=I/T$ and any integration constant can go in $f(u)$.<|endoftext|> -TITLE: What is the unit of the FFT output? -QUESTION [6 upvotes]: Consider a signal, f(t), with impulse samples taken N times, i.e f[0],f[1],f[2],...f[N-1] Let us perform FFT on it. Now, we have the amplitude on the y-axis and the frequency on the x-axis. I want to know if the unit of the quantity on the y-axis remains the same. If yes, why? If no, what happens to it? -Example: If we consider a voltage signal. What will be the unit of the quantity on y-axis after FFT of f(t)? - -REPLY [6 votes]: It's still a voltage. If you do a continuous Fourier transform, you go from signal to signal integrated over time, which is signal per frequency, but in a discrete Fourier transform you're just summing discrete voltages with coefficients, and the result is still a voltage. Of course if you want you can multiply it by the time interval between sample points to get a voltage per frequency unit.<|endoftext|> -TITLE: A compact operator in $L^2(\mathbb R)$ -QUESTION [9 upvotes]: Let $g \in L^{\infty}(\mathbb R)$. Consider the operator - $$ -\begin{split} -T_g\colon & L^2(\mathbb R)\to L^2(\mathbb R) \\ -& f \mapsto gf -\end{split} -$$ - Prove that $T_g$ is compact (i.e., the image under $T_g$ of bounded closed sets is compact) if and only if $g=0$ a.e. - -I do not know how to start and I'm very puzzled. I know very little about compactness in $L^p$: of course they are complete metric spaces, therefore a subspace is compact if and only if it is closed (complete) and totally bounded. A singleton is of course totally bounded and I think it is closed: therefore I can say that if $g=0$ a.e. then the image of every subspace is $\{0\}$ which is compact, so the operator is compact. What about the inverse direction? It seems hard to prove. -Would you help me, please? Thanks. - -REPLY [3 votes]: If $\lVert g\rVert_{\infty}\neq 0$, fix, $\Delta_n$ a sequence of measurable subsets of $\Bbb R$ such that $\lambda(\Delta_n)\in (0,+\infty)$ and for all $n\in \Bbb N$, $x\in\Delta_n$, we have $|g(x)|\geq\lVert g\rVert_{\infty}-\frac 1n$. - -We can assume that $\lambda(\Delta_n)\to 0$, since the sequence $\{\Delta_n\}$ can wbe chosen decreasing, and if the measure of the intersection is non-negative, $g$ would be constant equal to $C$ over a set of positive measure $A_0$. If we take a sequence of the form $\chi_{A_0}f_n$ which is weakly convergent but non fo the norm. Then $T_g(\chi_{A_0}f_n)=C\chi_{A_0}f_n$, which is not strongly convergent. Hence in this case we already have that $T_g$ is not compact. -We would have, if $T_g$ were compact, that -$$\tag{*} \lim_{N\to +\infty}\sup_{n\in\Bbb N}\int_{\{1\geq N\lambda(\Delta_n)\}}g^2\frac{\chi_{\Delta_n}}{\lambda(\Delta_n)}dx=0.$$ -To see this, use precompactness and a "$2\varepsilon$" argument. -With the assumption made in the first point, we can, for each integer $N$, pick an integer $k(N)$ with $1\geq N\lambda(\Delta_{k(N)})$. We can choose the sequence $\{k(N)\}$ strictly increasing, hence $1/k(N)< \lVert g\rVert_{\infty}$ for $N$ large enough. We would have -$$\lim_{N\to +\infty}\left(\lVert g\rVert_{\infty}-\frac 1{k(N)}\right)^2.$$ -But this limit is $\lVert g\rVert_{\infty}$. - -An alternative method is the following: we use the fact that the spectrum of a multiplication operator is the essential range of $g$. The spectrum of a compact self-adjoint operator is a sequence which converge to $0$, hence the essential image of $g$ is $\{0\}$.<|endoftext|> -TITLE: Simple simple Euler Lagrange Equation -QUESTION [5 upvotes]: Just starting a course on Lagrangian Mechanics and I'm just wondering what about the Euler-Lagrange equation, and more specifically what I'm meant to be trying to do .. -One of the questions from my textbook reads : -Solve the Euler-Lagrange equation for the following function -\begin{align*} -f(y,y') = y^2+y'^2 -\end{align*} -Looks simple enough... But where should I be headed to start? -The Euler-Lagrange Equation as we have it is -\begin{align*} -\frac{\partial f}{\partial y} = \frac{d}{dx} \frac{\partial f}{\partial y'} -\end{align*} -Do I just find the partial derivatives of f treating y and y' as independent variables as follows: -$$\frac{\partial f}{\partial y} = 2y,$$ -$$\frac{\partial f}{\partial y'} = 2y$$ -so then we have : -$$2y = \frac{d}{dx}(2y')$$ -$$y = \frac{d}{dx}(y')$$ -$$yx+c = y'$$ -$$\frac{yx^2}{2}+cx+d = y$$ -And then simplify with the y isolated on the right so it looks nice .. or at least thats what I thought.. -Any comments ? right track, wrong track? -Also, what is this euler equation actually solving for? Is it just a nice way of solving differential equations in this form ? -When we did this in class, we did a monster proof showing it minimizes the path between 2 points or finds the extrema for the function $I(x) = \int^a_b \ F(y(x),y'(x),x) dx,$ which is all great but I'm not too sure how this relates to the equation we solve in the question .. Thanks for all - -REPLY [6 votes]: The function $f(y,y') = {y'}^2+y^2$ can be interpreted as a Lagrangian $L = T-V$, where $T={y'}^2$ is the kinetic energy and $V = -y^2$ the potential energy of some object.${}^\dagger$ -Among other things, this is the Lagrangian of pencil that is balanced vertically on its tip. -$y$ is then interpreted as the angle the pencil makes with the vertical, and must be assumed to be small. -$x$ should be interpreted as time. -We expect at least one unstable solution since the pencil can fall over. -The action is extremized by solutions to the Euler-Lagrange equation. -The equations of motion found by the principle of least action are equivalent to Newton's laws of motion---we are going to find out how the pencil drops. -First, note that -$$\begin{eqnarray*} -\frac{d}{dx} \frac{\partial f}{\partial y'} -&=& \frac{d}{dx} 2 y' \\ -&=& 2y'' \\ -\frac{\partial f}{\partial y} -&=& 2 y. -\end{eqnarray*}$$ -The Euler-Lagrange equation gives -$$y'' = y.$$ -The solutions are -$$y(x) = A e^x + B e^{-x},$$ -where $A$ and $B$ are some constants determined by the initial conditions. -For example, if $y(0)=0$ and $y'(0)=1$ we find -$$y(x) = \sinh x.$$ -This solution is indeed unstable, the angle increases without bound. -Of course, this can only be taken seriously for small $y$. -The principal advantage of finding equations of motion in this way is it allows the simple use of generalized coordinates. -With practice it becomes automatic to write the total kinetic and potential energy of a system in terms of some convenient coordinates and to find the equations of motion via the Euler-Lagrange equations, which have the same form for any system of generalized coordinates. -As a cultural side note, be aware that variational methods extend far beyond their application to the Lagrangian formulation of classical mechanics. - - -${}^\dagger$The force corresponding to this potential is $F = -\frac{d}{dy}V = 2y$. This is Hooke's law with the wrong sign. -What happens if $f(y,y') = {y'}^2 - y^2$?<|endoftext|> -TITLE: Is there a known closed form number for $\prod\limits_{k=2}^{ \infty } \sqrt[k^2]{k}$ -QUESTION [5 upvotes]: $f(x)=\sum\limits_{k = 2 }^ \infty e^{-kx} \ln(k) $ -$\int\limits_0^{\infty}\int\limits_x^{\infty}\, f(\gamma)\, d\gamma dx=\sum\limits_{k = 2 }^ \infty \frac{1}{k^2} \ln(k) $ -$\int\limits_0^{\infty}\int\limits_x^{\infty} f(\gamma)\, d\gamma dx=\sum\limits_{k=2}^ \infty\ln(k^{\frac{1}{k^2}})=\ln(\prod\limits_{k=2}^{\infty}k^{\frac{1}{k^2}}) $ -$\prod\limits_{k=2}^{ \infty }k^{\frac{1}{k^2}}=\prod\limits_{k=2}^{ \infty } \sqrt[k^2]{k}=e^{\int\limits_0^{\infty}\int\limits_x^{\infty} f(\gamma) \,d\gamma dx}$ -$f(x)=\sum\limits_{k = 2 }^ \infty e^{-kx} \ln(k) $ -$f(x)=\sum\limits_{k = 1 }^ \infty e^{-(k+1)x} \ln(k+1) $ -$f(x)=e^{-x}\sum\limits_{k = 1 }^ \infty e^{-kx} \ln(k+1) $ -$f(x)=e^{-x}\sum\limits_{n = 1 }^ \infty \frac{(-1)^{n+1}}{n} \sum\limits_{k = 1 }^ \infty k^n e^{-kx}$ -We know that -$\sum\limits_{k = 1 }^ \infty e^{-kx}= \frac{1}{e^{x}-1} $ -$\sum\limits_{k = 1 }^ \infty k^n e^{-kx}= (-1)^n\frac{d^n}{dx^n}(\frac{1}{e^{x}-1}) $ -$f(x)=e^{-x}\sum\limits_{n = 1 }^ \infty \frac{(-1)^{n+1}}{n} \sum\limits_{k = 1 }^ \infty k^n e^{-kx} = e^{-x}\sum\limits_{n = 1 }^ \infty \frac{(-1)^{n+1}}{n} (-1)^n\frac{d^n}{dx^n}(\frac{1}{e^{x}-1})$ -$f(x)=-e^{-x}\sum\limits_{n = 1 }^ \infty \frac{1}{n} \frac{d^n}{dx^n}(\frac{1}{e^x-1})$ -$\int\limits_0^{\infty}\int\limits_x^{\infty} f(\gamma) \,d\gamma dx= -\int\limits_0^{\infty}\int\limits_x^{\infty} e^{-\gamma}\sum\limits_{n = 1 }^ \infty \frac{1}{n} \frac{d^n}{d\gamma^n}(\frac{1}{e^{\gamma}-1})\, d\gamma dx$ -I have lost my way after that. -Is it possible to find a closed form in my way? or I need to follow a different way. -I need your mathematical sense. -Thanks a lot for answers and advice. - -REPLY [7 votes]: Yes $\boxed{\displaystyle e^{-\zeta'(2)}}$ I think. -To prove it start with : -$$\zeta(2-x)=\sum_{k=1}^\infty \frac {k^x}{k^2}$$ -and compute the derivative! -The trick is that the derivation will create a $\ln(k)$ - term at the numerator. At the end take the limit as $x\to 0$. - -REPLY [2 votes]: Your infinite product equals to : $e^{-{\zeta}'(2)}$.<|endoftext|> -TITLE: Question about $p$-adic numbers and $p$-adic integers -QUESTION [11 upvotes]: I've been trying to understand what $p$-adic numbers and $p$-adic integers are today. Can you tell me if I have it right? Thanks. -Let $p$ be a prime. Then we define the ring of $p$-adic integers to be -$$ \mathbb Z_p = \{ \sum_{k=m}^\infty a_k p^k \mid m \in \mathbb Z, a_k \in \{0, \dots, p-1\} \} $$ -That is, the $p$-adic integers are a bit like formal power series with the indeterminate $x$ replaced with $p$ and coefficients in $\mathbb Z / p \mathbb Z$. So for example, a $3$-adic integers could look like this: $1\cdot 1 + 2 \cdot 3 + 1 \cdot 9 = 16$ or $\frac{1}{9} + 1 $ and so on. Basically, we get all natural numbers, fractions of powers of $p$ and sums of those two. -This is a ring (just like formal power series). Now we want to turn it into a field. To this end we take the field of fractions with elements of the form -$$ \frac{\sum_{k=m}^\infty a_k p^k}{\sum_{k=r}^\infty b_k p^k}$$ -for $\sum_{k=r}^\infty b_k p^k \neq 0$. We denote this field by $\mathbb Q_p$. -Now as it turns out, $\mathbb Q_p$ is the same as what we get if we take the ring of fractions of $\mathbb Z_p$ for the set $S=\{p^k \mid k \in \mathbb Z \}$. This I don't see. Because then this would mean that every number $$ \frac{\sum_{k=m}^\infty a_k p^k}{\sum_{k=r}^\infty b_k p^k}$$ can also be written as $$ \frac{\sum_{k=m}^\infty a_k p^k}{p^r}$$ -and I somehow don't believe that. So where's my mistake? Thanks for your help. - -REPLY [15 votes]: I want to emphasize that $\mathbb Z_p$ is not just $\mathbb F_p[[X]]$ in disguise, though the two rings share many properties. For example, in the $3$-adics one has -\[ -(2 \cdot 1) + (2 \cdot 1) = 1 \cdot 1 + 1 \cdot 3 \neq 1 \cdot 1. -\] -I know three ways of constructing $\mathbb Z_p$ and they're all pretty useful. It sounds like you might enjoy the following description: -\[ -\mathbb Z_p = \mathbb Z[[X]]/(X - p). -\] -This makes it clear that you can add and multiply elements of $\mathbb Z_p$ just like power series with coefficients in $\mathbb Z$. The twist is that you can always exchange $pX^n$ for $X^{n + 1}$. This is the “carrying over” that Thomas mentions in his helpful series of comments.<|endoftext|> -TITLE: Definition of a Functor of Abelian Categories -QUESTION [5 upvotes]: What is the precise definition of a functor of abelian categeries. I've looked on the internet but can't find one. From the Wikipedia definition of an abelian category, I'm guessing that, for two abelian categories ${\cal C,D}$, a functor $F:{\cal C} \to {\cal D}$, it must satisfy: -(i) the function Hom$(A,B) \to$ Hom($F(A),F(B))$ induce by $F$ must be a group homomorphism, for any two objects $A,B \in {\cal C}$, -(ii) $F$ must preserve kernels and cokernels, -(iii) $F$ must preserve direct sums. - -REPLY [8 votes]: When one has abelian categories, one is usually interested in additive functors. By definition, these are functors $F : \mathcal{C} \to \mathcal{D}$ whose action on morphisms is an abelian group homomorphism $\mathcal{C}(A, B) \to \mathcal{D}(F A, F B)$. -Proposition. If $\mathcal{C}$ and $\mathcal{D}$ are additive categories (i.e. $\textbf{Ab}$-enriched categories with finite direct sums) and $F : \mathcal{C} \to \mathcal{D}$ is an ordinary functor, then the following are equivalent: - -$F$ preserves finite coproducts (including the initial object) -$F$ preserves finite products (including the terminal object) -$F$ preserves the zero object and binary direct sums -$F$ is additive - -Proof. (1), (2), and (3) are equivalent because coproducts, products, and direct sums all coincide in an abelian category. One shows that (4) implies (3) by observing that being a direct sum in an $\textbf{Ab}$-enriched category is a purely equational condition: given objects $A$ and $B$, $(A \oplus B, \iota_1, \iota_2, \pi_1, \pi_2)$ is a direct sum of $A$ and $B$ if and only if -\begin{align} -\pi_1 \circ \iota_1 & = \textrm{id} & -\pi_1 \circ \iota_2 & = 0 \\ -\pi_2 \circ \iota_1 & = 0 & -\pi_2 \circ \iota_2 & = \textrm{id} -\end{align} -$$\iota_1 \circ \pi_1 + \iota_2 \circ \pi_2 = \textrm{id}$$ -where $\iota_1 : A \to A \oplus B$ and $\iota_2 : B \to A \oplus B$ are the coproduct insertions and $\pi_1 : A \oplus B \to A$ and $\pi_2 : A \oplus B \to B$ are the product projections. -On the other hand, (3) implies (4) by the following trick: given $f, g : A \to B$ in an abelian category $\mathcal{C}$, we have -$$f + g = \nabla_B \circ (f \oplus g) \circ \Delta_A$$ -where $\Delta_A : A \to A \oplus A$ is the diagonal map and $\nabla_B : B \oplus B \to B$ is the fold map; this can be verified by using the last equation in the above paragraph: -\begin{align} -\textrm{id} \circ (f \oplus g) \circ \Delta_A -& = \textrm{id} \circ \langle f, g \rangle \\ -& = (\iota_1 \circ \pi_1 + \iota_2 \circ \pi_2) \circ \langle f, g \rangle \\ -& = \iota_1 \circ f + \iota_2 \circ g -\end{align} -and so $\nabla_B \circ (f \oplus g) \circ \Delta_A = \nabla_B \circ (\iota_1 \circ f + \iota_2 \circ g) = f + g$. Hence, if $F$ preserves the zero object and direct sums, it must also preserve addition of morphisms. 　◼ - -One often also considers left/right exact functors between abelian categories. Officially, these are functors that preserve all finite limits/colimits (resp.), but in the case of abelian categories, it is enough that they be additive and preserve all kernels/cokernels (resp.). An exact functor is one that is both left and right exact. -These are all non-trivial conditions: the subject of homological algebra is essentially the study of the difference between left/right exact functors and exact functors! For example, $\textrm{Hom}(A, -)$ and $\textrm{Hom}(-, B)$ are both left exact functors; $\textrm{Hom}(A, -)$ is exact if and only if $A$ is a projective object, and $\textrm{Hom}(-, B)$ is exact if and only if $B$ is an injective object.<|endoftext|> -TITLE: Definition of a *Monoidal Abelian Category* -QUESTION [9 upvotes]: Does the definition of a monoidal abelian category require any coherence between the abelian structure and the monoidal structure? - -REPLY [2 votes]: At a minimum, you should require that the monoidal structure is additive in each variable, as Zhen Lin says in the comments.<|endoftext|> -TITLE: A curve that intersects every plane in finitely but arbitrarily many points -QUESTION [9 upvotes]: Does there exist a piecewise smooth curve in $\mathbb{R}^3$ such that every plane intersects the curve at finitely many points and the number of intersection points can be arbitrary large? -If the number of intersection points for each plane is bounded then there is an example: -$\gamma(t)=(t,t^3,t^5)$ intersects every plane at most five points. - -REPLY [4 votes]: Try $(x,y,z) = (t + \sin(t^2), t^2, t^3)$. For any $a,b,c$ (not all $0$) and $d$, -$|a (t+\sin(t^2)) + b t^2 + c t^3 - d| \to \infty$ as $t \to \pm \infty$ so the intersections are in a finite interval. And since $a (t+\sin(t^2)) + b t^2 + c t^3 - d$ is analytic and not constant, it has finitely many zeros in a compact set. So any plane has only finitely many intersections with the curve. -The curve intersects the plane $x = \sqrt{2 m \pi}$ (where $m$ is a positive integer) when $t + \sin(t^2) = \sqrt{2 m \pi}$. For $t = \sqrt{2m \pi}+s$ that says -$$s + \sin((\sqrt{2m\pi}+s)^2) = s + \sin(2 \sqrt{2m\pi} s + s^2) = 0$$ -In the interval $-1/2 < s < 1/2$, $2 \sqrt{2m\pi}s +s^2$ runs from $-\sqrt{2m\pi} + 1/4$ -to $+\sqrt{2m\pi} + 1/4$, and thus passes through approximately $\sqrt{2m/\pi}$ odd multiples -of $\pi/2$, at which the sine is alternately $\pm 1$, and thus $ s + \sin(2 \sqrt{2m\pi} s + s^2)$ has approximately $\sqrt{2m/\pi}$ sign changes. Thus the number of intersection points is unbounded.<|endoftext|> -TITLE: Complexity of the set of surjective continuous functions -QUESTION [8 upvotes]: Let $X,Y$ be complete separable metric spaces, with $X$ locally compact, and $C(X,Y)$ the space of continuous functions from $X$ to $Y$, equipped with the topology of uniform convergence on compact sets. If I am not mistaken, $C(X,Y)$ is Polish. -Let $S \subset C(X,Y)$ be the set of functions in $C(X,Y)$ which are surjective. - -Is $S$ Borel? If not, what can we say about its complexity? - -If $X$ is compact, it is easy to show that $S$ is closed. But the locally compact case seems harder. -In particular, I have in mind something like $X = \mathbb{R} \times [0,1]^2$ and $Y = \mathbb{R}^d$. -Thanks! - -REPLY [3 votes]: Thanks to a suggestion from Clinton Conley, we can show that this set is $\Pi_1^1$, i.e. co-analytic. -$X$ is $\sigma$-compact, so we can write $X = \bigcup K_n$ for compact $K_n$. Then it is easy to check that -$$B_n := \{(f,y) \in C(X,Y) \times Y : y \notin f(K_n)\}$$ -is open in $C(X,Y) \times Y$. Thus $$B := \bigcap_n B_n = \{(f,y) \in C(X,Y) \times Y: y \notin f(X)\}$$ -is $G_\delta$ and in particular Polish. The projection of $B$ onto $C(X,Y)$ is thus analytic, but this projection is exactly the complement of $S$. -This leaves open the question of whether $S$ is necessarily Borel. However, it does guarantee that $S$ is universally measurable. Moreover, and this is what I really wanted, for any Polish $Z$ and any continuous $g : Z \to C(X,Y)$, $g^{-1}(S)$ is again $\Pi_1^1$ and hence universally measurable.<|endoftext|> -TITLE: Can a meromorphic function be written as ratio of holomorphic function? -QUESTION [16 upvotes]: Well, I want to know whether a meromorphic function can be written as ratio of two holomorphic function on $\mathbb{C}$ or on a Riemann surface. -Thank you for help. - -REPLY [20 votes]: a) On a compact Riemann surface $X$ holomorphic functions are constant so that the the quotients of holomorphic functions are just the constants too. In formulas: $$\mathcal O(X)=\mathbb C \; ,\quad \text{Frac} (\mathcal O(X))=\mathbb C$$ However a deep theorem (Riemann's Existence Theorem) assures us that there exists a non-constant meromorphic function on $X$ and that these meromorphic functions form a finitely generated field of transcendence degree one over $\mathbb C$ ($\;trdeg_ \mathbb C\mathcal M(X)=1$), so that the answer to your question is negative for compact Riemann surfaces: $$\text{Frac} (\mathcal O(X))=\mathbb C \subsetneq \mathcal M(X)$$ -b) On a non-compact Riemann surface $Y$ however another difficult theorem, first proved only in 1948 by Behnke and Stein, says that indeed every meromorphic function is the quotient of two holomorphic functions . In formula: $$ \text{Frac} (\mathcal O(Y))= \mathcal M(Y) $$ -The modern point of view is that this is an easy consequence of the difficult result that $Y$ is a Stein manifold, the analogue in complex-analytic geometry of an affine algebraic variety. -Bibliography As usual, Forster's Lectures on Riemann Surfaces is the best reference for these questions.<|endoftext|> -TITLE: Normal to Ellipse and Angle at Major Axis -QUESTION [9 upvotes]: I've tried to detail my question using the image shown in this post. - -Consider an ellipse with 5 parameters $(x_C, y_C, a, b, \psi)$ where $(x_C, y_C)$ is the center of the ellipse, $a$ and $b$ are the semi-major and semi-minor radii and $\psi$ is the orientation of the ellipse. -Now consider a point $(x,y)$ on the circumference of the ellipse. The normal at this point on the circumference of the ellipse intersects the major axis at a point $(x_D, y_D)$. This normal makes an angle $/phi$ with the major axis. However, the angle subtended by this point at the center of the ellipse is $\theta$. For a circle, $\theta = \phi$ for all points on its circumference because the normal at the circle is the radial angle subtended by the point on the circumference. -Is there a relationship between the angles $\theta$ and $\phi$ for an ellipse. -For some context, I am trying to "extract" points from the circumference of an ellipse given its parameters $(x_C, y_C, a, b, \psi)$. For such an ellipse, I start from $(x_C, y_C$) and with angle $\theta = 0^\circ$ and I start sweeping until $360^\circ$. Using the equation $\left[\begin{array}{c} x \\ y\end{array}\right] = \left[\begin{array}{c c} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{array}\right] \left[\begin{array}{c} a\cos(\psi) \\ b\sin(\psi) \end{array}\right]$, I get the $(x,y)$ location of the point that is supposed to be on the ellipse circumference. I then look up this location in a list of "edge" points. Along with this list of edge points, I also have gradient angle information for each edge point. This corresponds to the angle $\phi$. -Here is the crux of the question, for a circle, I am confident that the edge point lies on the circumference of the circle if $|\theta - \phi| < \text{threshold}$. But, for an ellipse, how do I get a similar relationship ? - -REPLY [8 votes]: With the center at $(0,0)$ and $\psi=0$, the equation of the ellipse is $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{.}$$ -So $$\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0$$ and therefore $$\frac{dy}{dx}= -{\frac{xb^2}{ya^2}}\text{.}$$ -The slope of the normal line is the negative reciprocal of this, so $$\tan(\phi)=\frac{ya^2}{xb^2}\text{.}$$ -Meanwhile, $$\tan(\theta)=\frac{y}{x}\text{.}$$ So, eliminating $\frac{y}{x}$ from these two equations and clearing denominators, the relationship between $\phi$ and $\theta$ is: $$b^2\tan(\phi)=a^2\tan(\theta)$$<|endoftext|> -TITLE: How to prove that $\frac{10^{\frac{2}{3}}-1}{\sqrt{-3}}$ is an algebraic integer -QUESTION [7 upvotes]: As the title says, I'm trying to show that $\frac{10^{\frac{2}{3}}-1}{\sqrt{-3}}$ is an algebraic integer. -I suppose there's probably some heavy duty classification theorems that give one line proofs to this but I don't have any of that at my disposable so basically I'm trying to construct a polynomial over $\mathbb{Z}$ which has this complex number as a root. -My general strategy is to raise both sides of the equation -$$x = \frac{10^{\frac{2}{3}}-1}{\sqrt{-3}}$$ -to the $n^{th}$ power and then break up the resulting sum in such a way as to resubstitute back in smaller powers of $x$. Also since this root is complex I know it must come in a conjugate pair for the coefficients of my polynomial to be real, thus I know that -$$x = -\frac{10^{\frac{2}{3}}-1}{\sqrt{-3}}$$ -must also be a root of my polynomial. Hence from this I obtain: -$$3x^2 = (10^{\frac{2}{3}} -1 )^2$$ -However since my root is pure imaginary I don't really get any more information from this, so I'm a bit stumped, I tried raising both sides of $3x^2 -1 = 10^{\frac{4}{3}} - (2)10^{\frac{2}{3}}$ to the third power but it doesn't look like it's going to break up correctly, can anyone help me with this? Thanks. - -REPLY [5 votes]: A somewhat less computational approach can be based on the idea that an algebraic number is an algebraic integer if and only if it is everywhere locally integral. Certainly $\sqrt{-3}$ is a $p$-adic unit (possibly in a quadratic extension) everywhere except $3$, so the only issue is at $3$. -Second, observe/speculate that there is a cube root $\alpha$ of $10$ in $\mathbb Q_3$, using Hensel's Lemma: $10$ is pretty close to $1$ modulo powers of $3$, that is, $3$-adically. A useful form of Hensel's lemma here is that a monic polynomial $f$ with coefficients in $\mathbb Z_3$ and $x_o\in \mathbb Z_3$ such that $|f(x_o)/f'(x_o)^2|_3<1$ produces a root of $f(x)=0$ in $\mathbb Z_3$. -From this, we see that if we can find a cube root of $10$ mod $3^3=27$, we have a $3$-adic cube root. Indeed, $4^3=64=10+2\cdot 3^3$. -So there is $\alpha\in \mathbb Z_3$ with $\alpha-4=0 \mod 3$, so ${\alpha-1\over 3}$ is a $3$-adic integer, and certainly ${\alpha-1\over \sqrt{-3}}$ is integral over $\mathbb Z_3$, although in a quadratic extension. -A similar pattern occurs for all odd primes: $(1+p)^p=1+p^2 \mod p^3$, so there is a $p$-th root of $1+p^2$ in $\mathbb Z_p$.<|endoftext|> -TITLE: An inequality in a proof of Kunen's Inconsistency -QUESTION [5 upvotes]: This may be a silly question, but I keep coming back to it. -Let $j:V\prec M$ be a non-trivial elementary embedding with $M$ a transitive class and $\kappa$ the critical point of $j$. Define the critical sequence for $j$ as usual, setting $\kappa_0=\kappa$ and $\kappa_n=j^n(\kappa)$ (i.e., the $n^{th}$ iterate of $j$ evaluated at $\kappa$). Finally, let $\lambda=\sup_{n<\omega}(\kappa_n)$. Then $j(\lambda)=\lambda$ and $\lambda$ is a strong limit of cofinality $\omega$. -The trouble I'm having is the claim that $j(\lambda^+)=\lambda^+$. The claim follows from the string of inequalities: $\lambda^+\leq j(\lambda^+)=(\lambda^+)^M\leq\lambda^+$ However, I don't understand why the final inequality is true. I feel as if I'm just overlooking something obvious. -For reference, the string of inequlaities is from proof 2 (due to Woodin) of Kunen's Inconsistency on page 320 in Kanamori's The Higher Infinite. - -REPLY [7 votes]: We have indeed that $j(\lambda)=\lambda$. Since $j$ is elementary $j(\lambda^+)$ is the successor of $j(\lambda)=\lambda$ in $M$, so indeed $j(\lambda^+)=(\lambda^+)^M$. -However if if $M\subseteq V$ is an inner model then $(\alpha^+)^M\leq(\alpha^+)^V$ for trivial reasons. -Therefore we have the equality $\lambda^+=j(\lambda^+)$.<|endoftext|> -TITLE: Inverse limit by example -QUESTION [8 upvotes]: I'm trying to understand inverse limits. For this I am looking at the example (mentioned in Atiyah-Macdonald, page 102): We start with the topological abelian group $G = \mathbb Z$ (endowed with the topology induced by $|\cdot|_p$) and observes that we have an inverse system, that is, a sequence of groups $G_n$ and group homomorphisms $f_{ji}: G_i \to G_j$ ($j \leq i$) such that -(i) $f_{ii} = id$ -(ii) $f_{kj} \circ f_{ji} = f_{ki}$ -given by $G_n = \mathbb Z / p^n \mathbb Z$ and homomorphisms $f_{ji}: \mathbb Z / p^i \mathbb Z \hookrightarrow \mathbb Z / p^j \mathbb Z, \bar{x} \mapsto x \mod p^j$. Then, by definition, the inverse limit is -$$\lim_{\longleftarrow_{n \in \mathbb N}} G_n = \{(x_n) \in \prod_{n \in \mathbb N} G_n \mid f_{ji} (x_i) = x_j , \text{ for all } j \leq i \}$$ -Now since $f_{ji}$ are the "projections" (they are $\mod p^j$ really) we see that the requirement $f_{ji} (x_i) = x_j$ is nicely fulfilled if $x_{i+1}$ only adds stuff times $p^{i+1}$ since then the added stuff gets deleted when projecting down. -Hence we "see" (we don't really but we already know what it should be) that sequences in this inverse limit look like $p$-adic numbers, -since $x_i = \sum_{k=0}^i a_k p^k$. -Question 1: Do I have it right? -Question 2: Why do we need a topology on the group? I don't see where we used the topology. -And question 3: Would someone give me another (very easy!) example of an inverse limit, please? - -REPLY [7 votes]: The topology is not, strictly speaking, necessary. But there many reasons why it is desirable. -To begin, many subgroup filtrations on $G$ will induce the same topology. From the topological viewpoint it is clear that the completions of $G$ for these filtrations will be isomorphic. This makes clear the value of the Artin–Rees lemma, for example. -But beyond that there is some psychological value in viewing the higher powers of $p$ and $X$ in $\mathbb Z_p$ and $k[[X]]$, respectively, as being small. Now everything in $\mathbb Z_p$ is truly a limit of something in $\mathbb Z$, and certain infinite sums make sense. You might start to wonder if there is a useful notion of analysis in the $p$-adic world. -We also get a useful universal property which makes no reference to specific filtrations or connecting homomorphisms: if $H$ is complete and separated for such a topology and $f\colon G \to H$ is continuous, then $f$ factors through a unique map $\hat G \to H$. -A good (not so easy) example to look at is a nodal cubic. Let $k$ be a field of characteristic $\neq 2$. Below is a drawing that I ripped off from Silverman‘s book of the curve in the plane corresponding to $A = k[x, y]/(y^2 - x^3 - x^2)$. - -The completion of $A$ at the ideal $(x, y)$ is isomorphic to $k[[u, v]]/(uv)$. So locally, at the origin, our curve is just two lines crossing.<|endoftext|> -TITLE: Simple dice game: Optimal strategy? -QUESTION [13 upvotes]: Here's the description of a dice game which puzzles me since quite some time (the game comes from a book which offered a quite unsatisfactory solution — but then, its focus was on programming, so this is probably excusable). -The game goes as follows: -Two players play against each other, starting with score 0 each. Winner is the player to first reach a score of 100 or more. The players play in turn. The score added in each round is determined as follows: The player throws a die. If the die does not show an 1, he has the option to stop and have the points added to his score, or to continue throwing until either he stops or gets an 1. As soon as he gets an 1, his turn ends and no points are added to his score, any points he has accumulated in this round are lost. Afterward it is the second player's turn. -The question is now what is the best strategy for that game. The book suggested to try out which of the following two strategies gives better result: - -Throw 5 times (if possible), then stop. -If the accumulated points in this round add up to 20 or more, stop, otherwise continue. - -The rationale is that you want the next throw to increase the expected score. Of course it doesn't need testing to see that the second strategy is better: If you've accumulated e.g. 10 points, it doesn't matter whether you accumulated them with 5 times throwing a 2, or with 2 times throwing a 5. -However it is also easy to see that this second strategy isn't the best one either: After all, the ultimate goal is not to maximize the increase per round, but to maximize the probability to win.; both are related, but not the same. For example, imagine you have been very unlucky and are still at a very low score, but your opponent has already 99 points. It's your turn, and you've already accumulated some points (but those points don't get you above 100) and have to decide whether to stop, or to continue. If you stop, you secure the points, but your opponent has a 5/6 chance to win in the next move. Let's say that if you stop, the optimal strategy in the next move will be to try to get 100 points in one run, and that the probability to reach that is $p$. Then if you stop, since your opponent then has his chance to win first, your total probability to win is just $1/6(p + (1-p)/6 (p + (1-p)/6 (p + ...))) = p/(p+5)$. On the other hand, if you continue to 100 points right now, you have the chance $p$ to win this round before the other has a chance to try, but a lower probability $p'$ to win in later rounds, giving a probability $p + p'/(5+p')$. It is obvious that even if we had $p'=0$ (i.e. if you don't succeed now, you'll lose), you'd still have the probability $p>p/(p+5)$ to win by continuing, so you should continue no matter how slim your chances, and even if your accumulated points this round are above 20, because if you stop, you chances will be worse for sure. Since at some time, the optimal strategy will have a step where you try to go beyond 100 (because that's where you win), by induction you can say that if your opponent has already 99 points, your best strategy is, unconditionally, to try to get 100 points in one run. -Of course this "brute force rule" is for that specific situation (it also applies if the opponent has 98 points, for obvious reasons). If you'd play that brute-force rule from the beginning, you'd lose even against someone who just throws once each round. Indeed, if both are about equal, and far enough from the final 100 points, intuitively I think the 20 points rule is quite good. Also, intuitively I think if you are far advanced against your opponent, you should even play more safe and stop earlier. -As the current game situation is described by the three numbers your score ($Y$), your opponent's score ($O$) and the points already collected in this round ($P$), and your decision is to either continue ($C$) or to stop ($S$), a strategy is completely given by a function -$$s:\{(Y, O, P)\}\to \{C,S\}$$ -where the following rules are obvious: - -If $Y+S\ge 100$ then $s(Y,O,P)=S$ (if you already have collected 100 points, the only reasonable move is to stop). -$s(Y, O, 0)=C$ (it doesn't make sense to stop before you threw at least once). - -Also, I just above derived the following rule: - -$f(Y,98,P)=f(Y,99,P)=C$ unless the first rule kicks in. - -I believe the following rule should also hold (but have no idea how to prove it): - -If $f(Y,98,P)=S$ then also $f(Y,98,P+1)=S$ - -If that believe is true, then the description of a strategy can be simplified to a function $g(Y,O)$ which gives the smallest $P$ at which you should stop. -However, that's all I've figured out. What I'd really like to know is: What is the optimal strategy for this game? - -REPLY [3 votes]: There is no simplified description of the Nash equilibrium of this game. -You can compute the best strategy starting from positions where both players are about to win and going backwards from there. -Let $p(Y,O,P)$ the probability that you win if you are at the situation $(Y,O,P)$ and if you make the best choices. The difficulty is that to compute the strategy and probability to win at some situation $(Y,O,P)$, you make your choice depending on the probability $p(O,Y,0)$. So you have a (piecewise affine and contracting) decreasing function $F_{(Y,O,P)}$ such that $p(Y,O,P) = F_{(Y,O,P)}(p(O,Y,0))$, and in particular, you need to find the fixpoint of the composition $F_{(Y,O,0)} \circ F_{(O,Y,0)}$ in order to find the real $p(O,Y,0)$, and deduce everything from there. -After computing this for a 100 points game and some inspecting, there is no function $g(Y,O)$ such that the strategy simplifies to "stop if you accumulated $g(Y,O)$ points or more". For example, at $Y=61,O=62$,you should stop when you have exactly $20$ or $21$ points, and continue otherwise. -If you let $g(Y,O)$ be the smallest number of points $P$ such that you should stop at $(Y,O,P)$, then $g$ does not look very nice at all. It is not monotonous and does strange things, except in the region where you should just keep playing until you lose or win in $1$ move.<|endoftext|> -TITLE: How to show that this set is compact in $\ell^2$ -QUESTION [15 upvotes]: Let $(a_n)_{n}\in\ell^2:=\ell^2(\mathbb{R})$ be a fixed sequence. Consider the subspace $$C=\{(x_n)_{n}\in\ell^2 : |x_n|\le a_n\text{ for all }n\in\mathbb{N}\}.$$ -According to the book [Dunford and Schwartz, Linear operators part I, page 453] $C$ is compact in the $\ell^2$-norm, but there is no proof. How can I show that $C$ is indeed compact in $\ell^2$ ? - -REPLY [2 votes]: I just want to mention that the "diagonal argument" mentioned in comments is nothing other than the compactness of $\prod_{i \in \mathbb{N}} [a_i, b_i]$ (or equivalently $[0,1]^\mathbb{N}$), an important (and easier to prove) special case of Tychonoff's theorem. Essentially all compactness results in infinite-dimensional spaces flow from this fact, so it would be well to become comfortable with it. -This guarantees there is a subsequence $x_{n_k}$ which converges pointwise (i.e. coordinate-wise) to some $x$. But $|(x_{n_k})_i| \le |a_i|$ with $a_i \in \ell^2$, so by dominated convergence, the convergence also takes place in $\ell^2$.<|endoftext|> -TITLE: Gradient And Hessian Of General 2-Norm -QUESTION [9 upvotes]: Given $f(\mathbf{x}) = \|\mathbf{Ax}\|_2 = (\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{Ax} )^{1/2}$, -$\nabla f(\mathbf{x}) = \frac {\mathbf{A}^\mathrm{T} \mathbf{Ax}} {\|\mathbf{Ax}\|_2} = \frac {\mathbf{A}^\mathrm{T} \mathbf{Ax}} {(\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{Ax} )^{1/2}}$ -$\nabla^2 f(\mathbf{x}) = \frac { (\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{Ax} )^{1/2} \cdot \mathbf{A}^\mathrm{T} \mathbf{A} - (\mathbf{A}^\mathrm{T} \mathbf{Ax})^\mathrm{T} (\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{Ax} )^{-1/2} \mathbf{A}^\mathrm{T} \mathbf{Ax} } {(\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{Ax} ) } = \frac { \mathbf{A}^\mathrm{T} \mathbf{A} } { (\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{Ax} )^{1/2}} - \frac {\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{A} \mathbf{A}^\mathrm{T} \mathbf{Ax} } { (\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{Ax} )^{3/2} }$ -I guess I am looking for confirmation that I have done the above correctly. The dimensions match up except for the second term of the Hessian is a scalar, which makes me think that something is missing. -Edit: Also, the last equality reduces to -$\nabla^2 f(\mathbf{x}) = \frac {\mathbf{A}^\mathrm{T} \mathbf{A} - \nabla f(\mathbf{x})^\mathrm{T} \nabla f(\mathbf{x})} {\|\mathbf{Ax}\|_2}$ - -REPLY [9 votes]: It is easier to work with $\phi(x) = \frac{1}{2} f^2(x)$. Just expand $\phi$ around $x$. -$\phi(x+\delta) = \frac{1}{2} (x + \delta)^T A^T A (x + \delta) = \phi(x) + x^TA^TA \delta + \frac{1}{2} \delta^T A^T A \delta$. It follows from this that the gradient $\nabla \phi(x) = A^T A x$, and the Hessian is $H = A^TA$. -To finish, let $g(x) = \sqrt{2x}$, and note that $f = g \circ \phi$. To get the first derivative, use the composition rule to get -$D f(x) = Dg(\phi(x)) D \phi(x)$, which gives $Df(x) = \frac{1}{\sqrt{2 \phi(x)}} x^T A^T A = \frac{1}{\|Ax\|} x^T A^T A$. -Let $\eta(x) = \frac{1}{\|Ax\|}$, and $\gamma(x) = x^T A^T A$, and note that $D f(x) = \eta(x) \cdot \gamma(x)$, so we can use the product rule. Let $h(x) = Df(x)$ then the product rule gives $D h(x) (\delta) = (D \eta(x) (\delta)) \gamma(x) + \eta(x) D \gamma(x) (\delta)$. -Expanding this yields: $Dh(x)(\delta) = (- \frac{1}{\|Ax\|^2} \frac{1}{\|Ax\|} x^T A^T A \delta) x^T A^T A + \frac{1}{\|Ax\|} \delta^T A^T A $. Noting that $x^T A^T A \delta = \delta^T A^T A x$, we can write this as: -$$Dh(x)(\delta) = \delta^T(\frac{1}{\|Ax\|} A^T A - \frac{1}{\|Ax\|^3 } A^T A x x^T A^T A),$$ -or alternatively: -$$D^2 f(x) = \frac{1}{\|Ax\|} A^T A - \frac{1}{\|Ax\|^3 } A^T A x x^T A^T A .$$ -The only difference with the formula given in the question is that the latter dyad was written incorrectly (instead of the dyad $g g^T$, you have the scalar $g^T g$).<|endoftext|> -TITLE: The group of roots of unity in an algebraic number field -QUESTION [10 upvotes]: Is the following proposition true? If yes, how would you prove this? -Proposition. -Let $K$ be an algebraic number field. -The group of roots of unity in $K$ is finite. -In other words, the torsion subgroup of $K^*$ is finite. -Motivation. -Let $A$ be the ring of algebraic integers in $K$. -A root of unity in $K$ is a unit (i.e. an invertible element of $A$). It is important to determine the structure of the group of units in $K$ to investigate the arithmetic properties of $K$. -Remark. -Perhaps, the following fact can be used in the proof. -Every conjugate of a root of unity in $K$ has absolute value 1. -Related question: -The group of roots of unity in the cyclotomic number field of an odd prime order -Is an algebraic integer all of whose conjugates have absolute value 1 a root of unity? - -REPLY [4 votes]: Lemma. Let $f(x)$ be a degree $n$ polynomial with rational coefficients all of whose roots have absolute value 1. Then the coefficient of $x^k$ in $f$ has absolute value $\leq \binom{n}{k}$. -Proof: Apply Vieta's formulas and the triangle inequality to conclude. $\square$ -Proposition. There are only finitely many $n$-th roots of unity in an algebraic number field $K$. -Proof. It will be enough to prove that there are only finitely many algebraic integers $\alpha$ of fixed degree $n$, all of whose Galois conjugates (including $\alpha$) have absolute value $1$. Let $f_\alpha(x)$ be the minimal polynomial of $\alpha$ with coefficients in $\Bbb{Z}$. Now for each $k$ with $1 \leq k\leq n$ there are only finite many integers of absolute value $\leq \binom{n}{k}$. Since given any $\alpha$ every Galois conjugate of $\alpha$ has absolute value $1$ we apply the Lemma above to conclude that only finitely many such $f_\alpha$ can exist. Thus only finitely many such $\alpha$ can exist which completes the proof of the proposition. $\square$<|endoftext|> -TITLE: The Star Trek Problem in Williams's Book -QUESTION [11 upvotes]: This problem is from the book Probability with martingales by Williams. It's numbered as Exercise 12.3 on page 236. It can be stated as follows: - -The control system on the starship has gone wonky. All that one can do is to set a distance to be traveled. The spaceship will then move that distance in a randomly chosen direction and then stop. The object is to get into the Solar System, a ball of radius $r$. Initially, the starship is at a distance $R>r$ from the sun. -If the next hop-length is automatically set to be the current distance to the Sun.("next" and "current" being updated in the obvious way). Let $R_n$ be the distance from Sun to the starship after $n$ space-hops. Prove $$\sum_{n=1}^\infty \frac{1}{R^2_n}<\infty$$ -holds almost everywhere. - -It has puzzled me a long time. I tried to prove the series has a finite expectation, but in fact it's expectation is infinite. Does anyone has a solution? - -REPLY [3 votes]: I think this way works basically, but there are some details to be completed. -Let $A_k$ be the event that the starship goes back to the Solar System just after the n th space jump. Then $\{A_k\}_{k=1}^\infty$ is a sequence of disjoint events. Obviously $A_k$ is adapt to $\mathcal{F}_n$, here $\mathcal{F}_n$ is the $\sigma$- field generated by the first $n$ jumps. -Let $p_{i+1}=P(A_{i+1}|\mathcal{F}_{i})$, then$$p_{i+1}= \frac{R_n-\sqrt{R^2_n-r^2}}{2R_n}$$ -this follows from the area formula of a sphere cap. -(Here I am not sure because $ \frac{R_n-\sqrt{R^2_n-r^2}}{2R_n}$ is the conditional probability that given the ship is outside the solar system after n jumps, it will go back home after the next (n+1 th) jump, is it the same with $P(A_{i+1}|\mathcal{F}_{i})$?) -suppose it is, then we use the inequality -$$1-\sqrt{1-x}\geq \frac{x}{2}\quad (0 -TITLE: $\sum a_n$ converges absolutely, does $\sum (a_n + \cdots + a_n^n)$ converge -QUESTION [15 upvotes]: Suppose $\sum_{n=1}^\infty a_n$ converges absolutely. Does this imply that the series $$\sum_{n=1}^\infty (a_n + \cdots + a_n^n)$$ converges? - -I believe the answer is yes, but I can't figure out how to prove it. Any help would be appreciated. -Thanks. - -REPLY [7 votes]: Another way to see this is to notice that each term is less than $a_n + a_n^2 + a_n^3 + ...$ which, when $a_n < 1$ is $\frac{a_n}{1-a_n}$. When $a_n < \frac{1}{2}$ then $\frac{a_n}{1-a_n} < 2a_n$.<|endoftext|> -TITLE: $p$-adic completion of integers -QUESTION [8 upvotes]: I'm trying to do the following exercise: -Let $p$ be a prime and for $n\geq 1$ let $\alpha_n :\mathbb Z/p \mathbb Z \to \mathbb Z/p^n \mathbb Z$ be the injection of abelian groups given by $1 \mapsto p^{n−1}$. Consider the direct sum $\alpha : A \to B$ of these maps where $A$ is a countable direct sum of copies of $\mathbb Z/p \mathbb Z$ and $B$ is the direct sum of the groups $\mathbb Z/p^n \mathbb Z$. Show that the $p$-adic completion of $A$ is just $A$ but that the completion of $A$ for the topology induced from the $p$-adic topology on $B$ is the direct product of the $\mathbb Z/p \mathbb Z$. Deduce that $p$-adic completion is not a right exact functor on the category of all $\mathbb Z$-modules. -At first I thought $A$ was just the normal integers but it's not since for example for $p=2$, $-1 = 01111\dots$ is not in the space. The direct sum are all things with only finitely many non-zero terms, so for example the sequence $a_0 = 10000\dots a_1 = 110000\dots, a_2 = 111000\dots$ is a sequence in $A$ with a limit not in $A$. -I guess I am confused about what "$p$-adic completion" means: I assumed it meant that I take the equivalence classes of Cauchy sequences (Cauchy with respect to $|\cdot|_p$) where two sequences are equivalent if their difference tends to zero. But if that was what "$p$-adic completion" really meant then the sequence $a_k$ I gave above would be Cauchy and didn't have a limit in $A$ which is a counter example to what the exercise asks me to show. -Would someone explain to me what "$p$-adic completion" means? Thanks. -Edit -I'm bumping this question because the answerer is on holiday and I still have a bunch of questions. Thanks for your help. - -REPLY [6 votes]: I thought it would be a good exercise to post an answer of my own: -(i) We want to know the completion of the topological ring $A = \bigoplus_{n \in \mathbb N} \mathbb Z / p \mathbb Z$ with respect to the $p$-adic topology, i.e., the topology induced by neighbourhoods of zero of the form $A_n = p^n A$ so that $A=A_0 \supset A_1 \supset A_2\supset \dots$. -From chapter 10 in Atiyah-MacDonald we know that since this is a topological Abelian group with a countable neighbourhood basis of zero such that $A_n \supset A_{n+1}$, the completion is isomorphic to the inverse limit of the inverse system $X_n = A/A_n$ and the transition maps $f_n : A/A_n \to A/A_{n-1}$, $(x \mod p^{n}) \mapsto (x \mod p^{n-1})$. But since $pA = 0$ we get $X_n = A$ and $f_n = id_A$. Now the inverse limit are sequences $\vec{a} \in \bigoplus_n A/A_n = \bigoplus_n A$ such that $id(a_n) = a_{n-1}$, that is, constant sequences. Now clearly, $\varprojlim_n A \cong A$ via the map $(a,a,a, \dots ) \mapsto a$. - -(ii) Next we would like to compute the completion of $A$ with respect to the topology induced by the $p$-adic topology on $B = \bigoplus_n \mathbb Z / p^n \mathbb Z$. The map $\alpha : A \to B$ is the inclusion map so that a set $U$ is a neighbourhood in $A$ if and only if $\alpha^{-1}(V) = V \cap A = U$ for $V$ open in $B$. Open sets in $B$ are of the form $p^k B$. Since $B = \bigoplus_n \mathbb Z / p^n \mathbb Z$, $p^k B = \bigoplus_{n=0}^{\infty} p^k \mathbb Z / p^n \mathbb Z$ where for $n \leq k$ the component is zero. We compute the inverse image $\alpha^{-1}(p^kB)$ as follows: Let $(A)_n = \mathbb Z / p \mathbb Z$ be the $n$-th component of $A$. For the $n$-th component of $p^k B$ we get -$$ (p^kB)_n = -\begin{cases} -0 & n \leq k \\ -p^k \mathbb Z / p^n \mathbb Z & n > k\\ -\end{cases} -$$ -We have $\alpha(A)_n = \alpha_n(\mathbb Z / p \mathbb Z) = p^{n-1} \mathbb Z / p^n \mathbb Z$ so that -$$ \alpha^{-1}(p^k B) = -\begin{cases} -0 & n \leq k \hspace{0.2cm} (\text{since } \alpha \text{ is injective}) \\ -\mathbb Z / p \mathbb Z & n > k \hspace{0.2cm} (\text{since } \mathrm{im}(\alpha ) = p^{n-1} \mathbb Z / p^n \mathbb Z \subset p^{k} \mathbb Z / p^n \mathbb Z)\\ -\end{cases}$$ -Hence open sets in this topology on $A$ look like $O_k = 0 \oplus \dots \oplus 0 \oplus \mathbb Z / p \mathbb Z \oplus \mathbb Z / p \mathbb Z \dots $ where the first $k$ entries are zero. -For our inverse system this gives us $X_n = A / O_k = \mathbb Z / p \mathbb Z \oplus \dots \oplus \mathbb Z / p \mathbb Z \oplus 0 \oplus \dots $ where the first $k$ entries are non-zero. For the transition maps $t_n: X_n \to X_{n-1}$ this means $(x_1, x_2, \dots,x_{k-1}, x_k, 0 , 0 , \dots) \mapsto (x_1, x_2, \dots,x_{k-1}, 0 , 0 , \dots)$. -For the inverse limit of this system this means that it is all sequences with $x_n \in X_n$, which gives us $\varprojlim_n X_n = \prod_n X_n = \prod_n \mathbb Z / p \mathbb Z$. - -Now to conclude that $\varprojlim_n$ is not a right-exact functor on $\mathbb Z - \mathrm{\mathbf{Mod}}$, let $A_k = \alpha^{-1}(p^k B)$. Then the following sequence is exact: -$$ 0 \to A_k \hookrightarrow A \xrightarrow{\pi_k} A/A_k \to 0$$ -But for the inverse limit we get -$$ 0 \to 0 \hookrightarrow A = \bigoplus_n \mathbb Z / p \mathbb Z \xrightarrow{\pi} \prod_n \mathbb Z / p \mathbb Z $$ -where $\pi$ cannot be surjective since it is a group homomorphism mapping zero in the $n$-th component to zero.<|endoftext|> -TITLE: How follows the Strong Law of Large Numbers from Birkhoff's Ergodic Theorem? -QUESTION [5 upvotes]: We want to prove the strong law of large numbers with Birkhoff's ergodic theorem. -Let $X_k$ be an i.i.d. sequence of $\mathcal{L}^1$ random variables. This is a stochastic process with measure-preserving operation $\theta$ (the shift operator). From Birkhoff's ergodic theorem, we obtain $\frac{X_0 + \dotsb + X_{n-1}}{n} \to Y$ a.s., with $Y=\mathbb{E}[X_1 \mid \mathcal{J}_{\theta}]$ a.s. -Now, if $Y$ constant a.s., $Y= \mathbb{E}[X_1]$ a.s., and we would have the desired result. But why is $Y$ constant a.s.? - -REPLY [4 votes]: The transformation $\theta$ on $\Omega^{\Bbb N}$ is ergodic. Indeed, it's enough to show that for each cylinder $A$ and $B$, we have -$$\frac 1n\sum_{k=0}^{n-1}\mu(\theta^{-k}A\cap B)\to \mu(A)\mu(B),$$ -where $\mu$ is the measure on the product $\sigma$-algebra. -If $A=\prod_{j=0}^NA_j\times \Omega\times\dots$ and $B=\prod_{j=0}^NB_j\times \Omega\times\dots$, we have for $k>N$ -\begin{align} -\theta^{-k}A\cap B&=\{(x_j)_{j\geq 0}, (x_{j+k})_{j\geq 0}\in A, (x_j)_{j\geq 0}\in B\}\\ -&=\{(x_j)_{j\geq 0},x_{j+k}\in A_j, 0\leq j\leq N, x_j\in B_j,0\leq j\leq N\}\\ -&=B_0\times \dots\times B_N\times \Omega\times\dots\times \Omega\times A_0\times\dots\times A_n\times \Omega\times\dots, -\end{align} -and we use the definition of product measure $\mu$ on cylinders (the $N$ first terms doesn't matter). -Since $\theta$ is ergodic, $\mathcal J_{\theta}$ consists only of events of measure $0$ or $1$. The conditional expectation with respect such a $\sigma$-algebra is necessarily constant.<|endoftext|> -TITLE: Uniform convergence of functions, Spring 2002 -QUESTION [6 upvotes]: The question I have in mind is (see here, page 60, the solution is at page 297): -Assume $f_{n}$ is a sequence of functions from a metric space $X$ to $Y$. Suppose $f_{n}\rightarrow f$ uniformly and has inverse $g_{n}$. Now assume $f$'s inverse $g$ is uniformly continuous on $Y$. Prove that $g_{n}\rightarrow g$ uniformly. -I could not prove it using standard techniques as I do not know how to bound $|g_{n}(y)-g(y)|$ when $n$ becomes very large. The authors argue that the convergence of $g_{n}(y)\rightarrow g(y)$ is similar to $f(g_{n}(y))\rightarrow f(g(y))$ because the mapping by a uniformly convergent function series keeps uniform convergence. Thus they give the following argument that $$d(f(g(y)),f(g_{n}(y)))=d(y,f(g_{n}(y)))\le d(y,f_{n}(g_{n}(y)))+d(f_{n}(g_{n}(y)),f(g_{n}(y)))=d(f_{n}(g_{n}(y)),f(g_{n}(y)))$$ -So since $f_{n}\rightarrow f$ uniformly by hypothesis the statement is proved. My question is: Is the step of substituting $|g_{n}(y)-g(y)|$ by $|f(g(y))-f(g_{n}(y))|$ really justified? I could not get the "keep uniform convergence" thing the author is talking about. But I also could not come up with a better proof. - -REPLY [4 votes]: If $h_n\colon S\to S'$ converges uniformly to $h$ on $S$ and $\varphi\colon S'\to S''$ is uniformly continuous on $S$, fix $\varepsilon>0$. There is a $\delta>0$ such that if $d_{S'}(x,y)\leq \delta$ then $d_{S''}(\varphi(x),\varphi(y))\leq\varepsilon)$. Now, we use the fact that there is an integer $n_0$ such that for all $n\geq n_0$, we have -$$\sup_{x\in S}d_S(h_n(x),h(x))\leq \delta.$$ -Then -$$\sup_{x\in S}d_S(\varphi(h_n(x)),\varphi(h(x)))\leq\varepsilon.$$ -Now, we apply this result with $\varphi=g$ and $h_n:=f\circ g_n$.<|endoftext|> -TITLE: Is there a Definite Integral Representation for $n^n$? -QUESTION [9 upvotes]: The factorial $n!$ has a nice representation as definite integral: -$$ - n!=\Gamma(n+1)=\int_0^\infty t^{n} e^{-t}\, \mathrm{d}t. \! -$$ -Is it possible to write down such an integral for $n^n$ as well? -I tried to come up with an integral that reproduces a $n$ factor, $n$-times, but without success. I don't see a way to stop the partial integration process like in the $n!$ case. So this might not work here and I currently can't think of another way. If it helps to restrict $n$, feel free to do so. -The only thing a found online so far is the Lambert's $W$ function, which is involved when solving $x^x=z$, but I'm not sure if this helps. -EDIT: Answers with integrals of the form $\displaystyle n^n=\int_0^\infty \cdots dt$ are preferred. - -REPLY [2 votes]: The correct question (which someone has just asked me) may be as follows: -Find $f(t)$ (positive and independent of $n$) such that -$$n^n = \int_0^\infty t^n f(t) \,dt,\quad \forall\,n$$ -That is, find $f(t)$ such that $\{n^n\}$ is the sequence of moments of f(t)dt. -This is a particular instance of the Stieltjes moment problem. -There is criterion (due to Carleman) which guarantees that such -a solution is unique, but I don't know an explicit form for $f(t)$.<|endoftext|> -TITLE: Solution af a system of 2 quadratic equations -QUESTION [5 upvotes]: I have a system of two quadratic equations with unknowns $x$ and $y$: -$$a_{1 1} x y + a_{1 2} x^2 + a_{1 3} y^2 + a_{1 4} x + a_{1 5} y + a_{1 6} = 0,\\ -a_{2 1} x y + a_{2 2} x^2 + a_{2 3} y^2 + a_{2 4} x + a_{2 5} y + a_{2 6} = 0,$$ -where $a_{i j}$ are arbitrary scalars. -Is there an algebraic solution of the above system? - -REPLY [8 votes]: From Intersecting two conics: - -The solutions to a two second degree equations system in two variables may be seen as the coordinates of the intersections of two generic conic sections. In particular two conics may possess none, two or four possibly coincident intersection points. The best method of locating these solutions exploits the homogeneous matrix representation of conic sections, i.e. a 3x3 symmetric matrix which depends on six parameters. - The procedure to locate the intersection points follows these steps: - -given the two conics $C_1$ and $ C_2$ consider the pencil of conics given by their linear combination $\lambda C_1 + \mu C_2$ -identify the homogeneous parameters $(\lambda,\mu)$ which corresponds to the degenerate conic of the pencil. This can be done by imposing that $\det(\lambda C_1 + \mu C_2) = 0$, which turns out to be the solution to a third degree equation. -given the degenerate conic $C_0$, identify the two, possibly coincident, lines constituting it -intersects each identified line with one of the two original conic; this step can be done efficiently using the dual conic representation of $C_0$ -the points of intersection will represent the solution to the initial equation system<|endoftext|> -TITLE: P[random x is composite | $2^{x-1}$ mod $x = 1$ ]? -QUESTION [5 upvotes]: Select a uniformly random integer $n$ between $2^{1024}$ and $2^{1025}$ -(Q) What is the probability that n is composite given that $2^{n-1}$ mod $n = 1$ ? -How did you calculate this? -More info: -One way to calculate this would be if you had the following two variables: -$$P_Q(n) = 1 - { P_{prime}(n) \over P_{cong}(n) }$$ -Where: - -$P_{prime}(n)$ is the probability n is prime -$P_{cong}(n)$ is the probability that $2^{n-1}$ mod $n = 1$ - -So answering the following two questions would be sufficient to answer the main one: -What is $P_{prime}(n)$ equal to? -What is $P_{cong}(n)$ equal to ? -(This holds because the probability that n is prime if the congruence is false is 0.) -Based on the PouletNumber forumale given below: -exp((ln(2^1025))^(5/14))-exp((ln(2^1024))^(5/14)) - -= 123 - -and -(2^1025)*exp(-ln(2^1025)*ln(ln(ln(2^1025))) / (2*ln(ln(2^1025)))) - -(2^1024)*exp(-ln(2^1024)*ln(ln(ln(2^1024))) / (2*ln(ln(2^1024)))) - -= 9.82e263 - -So its between 123 < x < 9.82e263 -?? -And so $P_Q$ is: -3.29e-306 < P_Q < 2.63e-44 - -REPLY [4 votes]: Due to Fermat's Little Theorem, -$$ -a^{p-1} \bmod p=1 -$$ -if $a$ and $p$ are coprime. In your case $a=2$ and $p$ are all other primes, as pointed out by tomasz in the comment. -But as I recognized right now you are looking for composite $x$, so you are looking for Fermat Pseudo primes. Their distribution can be found here. A file with pseudo primes up to $10^{15}$ can be found here. -They are also called Poulet numbers, and according to the MathWorld page, for large $x$ their number below $x$ is given by -$$ -\exp((\ln(x))^{5/14})19$, there exists a Poulet number between $n$ and $n^2$. The theorem was proved in 1965. -Let $p$, $q$, $\ell$ be odd primes, and let $P_2(x)$ be the counting function for odd pseudoprimes with two -distinct prime factors, $P_2(x) := \#\{n \leq x : n = pq, p < q, psp(n)\}$ (where $psp(n)$ means $n$ is pseudoprime). Then $P_2(x)\sim C\sqrt{x} / \ln^2(x)$ as $x\to \infty$. -For more read here. -The notation $P_2(x)$ (the counting function for odd pseudoprimes with $2$ -distinct prime factors) seems to be a little confusing. In the last link they also state a result by Erdős, saying $c_1\log x\leq P_k(x)\le c_2\frac x{\log^k x}$. Maybe you need to sum the $P_k(x)$ to get to what you need.<|endoftext|> -TITLE: Question about $L^1$-$L^2$ integrable functions -QUESTION [5 upvotes]: Can somebody tell me what's wrong with the following argument? -If $f$ is $L^1$ Lebesgue-integrable, say $f$ positive, then it is bounded almost everywhere by some bound $M$. Then $f^2 < M\cdot f$ which is in $L^1$, then $f$ is in $L^2$ and $L^1$ lies in $L^2$. -It seems to me that the map $x^{-1/2}$ is $L^1$ but not $L^2$ on $(0,1)$, hence a counterexample... -So I'm a bit confused. - -REPLY [6 votes]: Your argument shows that if $f\in L^1$ and it is bounded a.e., then it is in $L^2$. But, as your own example shows, not every function in $L^1$ is bounded.<|endoftext|> -TITLE: Structures on torus -QUESTION [10 upvotes]: Quotienting $\mathbb R^2$ by different lattices isomorphic to $\mathbb Z^2$, we get different tori. -Somehow I think of the tori as having different "structures", but thinking more about it, I am not quite sure what different structures I am really thinking of. Two structures I am guessing at are complex structures and metrics. -Could someone explain how these differ? Also, am I thinking of the right kind of structure? Are there other structures which vary with the lattice chosen? - -REPLY [8 votes]: You're mostly right. -The relation between complex structures and metrics comes from their common passion about angles. -Basically, a complex structure on a Riemann surface is just a procedure for turning tangent vectors 90° counterclockwise. (Well, actually, that is an almost complex structure but they are the same thing in (complex) dimension 1). But that is one of the things a metric (with an orientation) allows you to do! So a metric on a surface defines canonically a complex structure. Of course, many metrics give the same structure (example 1: you can rescale the metric; example 2: the sphere $S^2$ has a lot of metrics but only one complex structure.) -The good notion is the notion of conformally equivalent metrics. Roughly speaking, two metrics are conformally equivalent if they define the same notions of angles between two tangent vectors. That means that they are proportional to each other (the ratio being a positive smooth function on the manifold.) So you get an important fact about surfaces: complex structures and conformal classes of metrics are basically the same thing. So, most of the Riemann surface theory can be stated equivalently in the holomorphic world or in the conformal world. (Example: the uniformisation theorem says either “Any Riemann surface is a quotient of $S^2$, $\mathbb C$ or $\mathbb H$” or “Any Riemannian metric on a surface is conformally equivalent to a metric of constant curvature.”) This polyvalence clearly is one of the riches of Riemann surface theory and explains partially the huge number of dedicated textbooks, as there is room for lots of different approaches. -So, here are two possible answers to your question: the relevant structures are “complex structures” or “conformal Riemannian structures”. This gives the same notion of “equivalent” lattices: two of them are equivalent if there is an affine map sending one to the other. -(One can imagine variants: for example, one could choose to call two lattices equivalent if there is a volume-preserving map sending one to the other. In that case, one has to enrich the relevant structures on the quotient. Of course, the finer the equivalence relation on the lattices, the richer the structure, but affine equivalence is a popular choice, because the two structures I mentioned are very important and very natural.)<|endoftext|> -TITLE: What are the best methods for solving cubic and quartic equations by computer programs? -QUESTION [8 upvotes]: We know that there are closed form formulas for real roots of degree 4 and 3 polynomials, but people sometimes advise to use numerical (e.g. Newton) methods anyway. They claim that closed formulas (Cardano or Viéte...) lack robustness. -What is the best approach to this problem, for a programmer? - -Which library (possibly open source) do you recommend? -Iterative or algebraic approach? -What are the problems with algebraic methods? -How about "speed" concerns? - -REPLY [7 votes]: I might as well expand on the comments I gave. -Firstly, note that square roots and cube roots, and even division in itself are computed through Newton-Raphson or some other iterative method, so the dichotomy of "iterative" versus "algebraic" methods in implementing these on a computer are more or less moot; they're all iterative. -Secondly, by virtue of any of these things being iterative, there is no surefire way to say whether directly applying the quadratic formula (to use the simplest case) or using Newton-Raphson on $ax^2+bx+c=0$ would be quicker. (Also, compilers do the darndest things, so always do some testing.) The little snag that these people who keep pushing "iterative" over "algebraic" approaches apparently forget to talk about is that Newton-Raphson only works out nicely if you have a good starting point. If you're very much concerned with efficiency, you can't afford to be a slouch in choosing the starting point, and that's not always easy to do automatically. -It must be noted, however, that there are certain modifications of Newton-Raphson that do not require accurate starting values. These usually fall under the rubric of "simultaneous iteration methods", the prototypical example of which is the (Weierstrass-)Durand-Kerner method. This is effectively the application of Newton-Raphson to the Vieta formulae relating the roots and coefficients of a polynomial. (See this for more details.) -But, getting back to the topic at hand, even the numerically sound implementation of the classical algebraic formulae is not completely trivial, and most people are careless, writing lousy code that does not take into account things like subtractive cancellation. To again use the quadratic formula as an example: -$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ -consider the case where $b^2\gg4ac$, with which the square root of the discriminant is nearly the same in magnitude as $b$. Then, the computation corresponding to the plus sign involves what is called subtractive cancellation, where one is subtracting nearly equal quantities to obtain a tiny result. This is a good way to lose significant figures in your results. -The right way starts by first considering the "Citardauq" formula: -$$x=\frac{2c}{-b\pm\sqrt{b^2-4ac}}$$ -In that case, one of the roots is accurately computed, and the other one is the one prone to subtractive cancellation. The good news is that the two formulae are complementary: one is able to accurately compute the root that the other method is not too good at. Thus, if one wants to stably implement the "algebraic approach", one should first compute -$$q=-\frac12\left(b+\mathrm{sign}(b)\sqrt{b^2-4ac}\right)$$ -(assuming that the coefficients are all real; if the coefficients are complex, things are slightly more intricate), and after which, the two roots of $ax^2+bx+c=0$ are computed as $q/a$ and $c/q$. Similar stabilized rearrangements are possible for the cubic formula (both the trigonometric form that is needed for casus irreducibilis and the Cardano form that you can use when only one of the roots is real), and then one has to note that the solution of a quartic hinges on the solution of a (resolvent) cubic and a quadratic. -In short: you still have to do the testing for determining which is "best" for your application.<|endoftext|> -TITLE: How to properly use GAGA correspondence -QUESTION [28 upvotes]: currently studying algebraic surfaces over the complex numbers. Before i did some algebraic geometry (I,II,start of III of Hartshorne) and a course on Riemann surfaces. -Now i understood that by GAGA, a lot of results transfer from complex analytic geometry to algebraic geometry over $\mathbb{C}$. -Question: in my RS surface course the genus for a compact RS was defined as the number of holes, formally half of the dimension of the first de rham cohomology group (or singular with coefficients in some field of characteristic zero). -Now i encountered the algebraic definition: the dimension of the first cohomology group of the structure sheaf. -I expect them to be the same, but did not find a quick proof. Can anyone show me how this works? I learned that de rham cohomology with coefficients in $\mathbb{C}$ corresponds to sheaf cohomology of the constant sheaf $\mathbb{C}$. However i'm not sure, can someone either confirm this or tell me why it's not true? If it is true, i find it hard to work with, since up till now i've been basically using Serre duality and Riemann Roch for bundles to reduce computing homology to computing spaces of global sections, however since the constant sheaf is not locally free i cannot apply this trick (or can i?). -Also, the constant sheaf makes sense both algebraically and analytically, so which one should i take in the de rham <-> constant sheaf correspondence? Or do both work? (with GAGA in mind, i expect them to have the same cohomology, but this might not be true) -So summarizing - -Do both definitions of genus agree? (of course assuming the algebraic curve to be smooth, so it is a RS) -If they do, can you show me a proof? (preferably using some sheaf cohomology) -Is it true that de rham cohomology corresponds to sheaf cohomology of the constant sheaf, if so is it the the analytic one, the algebraic one, or both? -Am i limiting myself unnecessarily by using Riemann Roch and Serre duality just for bundles, i.e. can i use them for all sheaves? - -Lastly, when answering my questions, it would be immensely appreciated if you could elaborate a little on how to use GAGA in general. -Hope this makes sense, i suspect them to be silly questions once i understand them, but right now it's a fuzz to me.. -Joachim -Edit: i just found a result from Hodge theory for surfaces in Beauville, stating $h^{0}(S,\Omega^1_S) = \frac{1}{2}h^{1}(S,\mathbb{R})$, all cohomology being analytic. So assuming (1) the same thing holds for curves and (2) GAGA identifies the algebraic and analytic cotangent bundle, Serre duality states that the left hand side is equal to $h^{1}(C,\mathcal{O}_C)$, i.e. the genus found in my AG textbook and i proved one thingwhat i wanted to proof. I was trying to find a reference on Hodge theory so i could verify (1) but a quick scan yielded no results. Does anyone have good idea on this? -Also i was wondering about (2) for a longer time, i'd be really happy with any comments on this. - -REPLY [26 votes]: First let me emphasize what Serre did : in 1956 he wrote an article Géométrie algébrique et géométrie analytique, which everybody calls GAGA (I heard him explain that he deliberately chose the title so that this would happen!). -This article proves many results, the gist of which is that calculations for coherent sheaves on a projective complex variety $X$ give the same (or isomorphic) results as those done on the underlying holomorphic variety $X^h$. -However he did not consider non coherent sheaves and for good reason: if for example you consider the constant sheaf $\mathbb C_X$ on $X$, it is flabby ("flasque") so that all its cohomology groups vanish: $H^i(X,\mathbb C_X)=0$, whereas of course the cohomology vector spaces $H^i(X^h,\mathbb C_{X^h})$ are not zero in general: -For example if $X$ is a smooth projective curve of genus $g$, then $X^h$ is a Riemann surface and $dim _\mathbb C H^1(X^h,\mathbb C_{X^h})=2g $. -Fortunately thanks to the rich techniques known for holomorphic manifolds many apparently transcendental invariants of a projective smooth manifold can be computed using only coherent sheaves. -A splendid example is Dolbeault's 1953 result for Betti numbers $$b_r(X^h)=\sum_{p+q=r} dim_\mathbb C H^q (X^h,\Omega_{X^h}^p)$$ Since the right hand size is cohomology of coherent sheaves,it can be calculated algebraically thanks to GAGA: $$ dim_\mathbb C H^q (X^h,\Omega_{X^h}^p)=dim_\mathbb C H^q (X,\Omega_{X}^p) $$ -In particular, for a smooth projective curve $X$, the genus of the associated Riemann surface $X^h$ is given algebraically by $$g(X^h)=\frac {1}{2}(b_1(X^h))=\frac {1}{2}(dim_\mathbb C H^1 (X,\mathcal O_X)+dim_\mathbb C H^0 (X,\Omega_{X}^1))$$ where the two summands on the right are equal by Serre duality (him again!). -Warning This answer is meant to address your request "it would be immensely appreciated if you could elaborate a little on how to use GAGA in general ", not to show the shortest or most elementary road to technically answer your questions on Riemann surfaces.<|endoftext|> -TITLE: Proving that the general linear group is a differentiable manifold -QUESTION [21 upvotes]: We know that the the general linear group is defined as the set $\{A\in M_n(R): \det A \neq 0\}$. I have a homework on how to prove that it is a smooth manifold. So far my only idea is that we can think of each matrix, say $A$, in that group as an $n^2-$dimensional vector. So i guess that every neighborhood of $A$ is homeomorphic to an open ball in $\mathbb{R}^{n^2}$ (However, i don't know how to prove this.) -Now, I'm asking for help if anyone could give me a hint on how to prove that the general linear group is a smooth manifold since I really don't have an idea on how to do this. (By the way, honestly, I don't really understand what a $C^{\infty}-$smooth structure means which is essential to the definition of a smooth manifold.) Your help will be greatly appreciated. :) - -REPLY [21 votes]: Construct a map $f:M_n(\mathbb{R}) \rightarrow \mathbb{R}$ by taking each matrix to its determinant, where $M_n(\mathbb{R})$ is the set of all $n \times n$ matrices. $f^{-1}(\mathbb{R}\backslash\{0\})=GL_n(\mathbb{R})$, and $\mathbb{R}\backslash\{0\}$ is an open subset of $\mathbb{R}$. Therefore, $GL_n(\mathbb{R})$ is an open subset of $M_n(\mathbb{R})$. I'll leave the rest to you.<|endoftext|> -TITLE: What does "harmonic" mean in "harmonic mean"? -QUESTION [9 upvotes]: Pun aside, what is harmonic about the harmonic series or the harmonic average? I assume it has a direct connection to music, but I cannot see it. - -REPLY [2 votes]: Just to indicate the relation between a harmonic average and a harmonic progression: if the harmonic mean of some subset of terms of a harmonic progression happens to be another term of that sequence, then the position of the "mean" term is the (ordinary arithmetic) mean of the positions of that subset. More generally, if one imagines that terms of a harmonic progression can be interpolated at arbitrary (non-integral) positions, then the harmonnic mean can be computed by locating its operands in a harmonic progression, taking the average of their positions and looking up the term at that point in the progression.<|endoftext|> -TITLE: Principles of Mathematical Analysis, Dedekind Cuts, Multiplicative Inverse -QUESTION [12 upvotes]: At the top of the page 20 of Rudin's book ''Principles of Mathematical Analysis'' he writes: -''The proofs (of the multiplication axioms) are so similar to the ones given in detail in Step 4 (proof of the addition axioms) that we omit them''. I tried to prove them but I got stuck in the proof of -\begin{equation}\alpha \cdot {\alpha }^{-1}=1^*\end{equation} - where $\alpha$ is positive cut and ${\alpha }^{-1}=\mathbb{Q}_{-}\bigcup\left\{0\right\}\bigcup\left\{t\in \mathbb{Q}:00$ with the following property: There exists $r>1$ such that $(1/p)/r \notin \alpha$. -We show that $\beta \in \mathbb R^+$ and $\alpha\beta = 1^*$. -If $s \notin \alpha$ [so that $s > 0$] and $p=(1/s)/2$, then $(1/p)/2=s \notin \alpha$, hence $p \in \beta$. So $\beta^+$ is not empty [because $p > 0$]. If $0 < q \in \alpha$ [which exists because $0^* < \alpha$], then $1/q \notin \beta$. So $\beta \ne \mathbb Q$. Hence $\beta$ satisfies (I). -Pick $0 < p \in \beta$ [which exists because $\beta^+$ is nonempty], and pick $r > 1$, so that $(1/p)/r \notin \alpha$. If $0 < q < p$, then $0 < (1/p)/r < (1/q)/r$, hence $(1/q)/r \notin \alpha$. Thus $q \in \beta$, and (II) holds. -[For the proof of (III), Rudin uses the expression '$r/2$', which in multiplicative language would be '$\sqrt r$', which may not be rational. We can do just as well with rational $j,k$ satisfying $1 p$, and $(1/t)/k = (1/p)/r \notin\alpha$, so that $t \in \beta$. Hence $\beta$ satisfies (III). -We have proved that $\beta \in \mathbb R$ [and, since $\beta^+$ is nonempty, that $\beta \in \mathbb R^+$]. -If $0 < r \in \alpha$ and $0 < s \in \beta$, then $1/s \notin \alpha$, hence $r < 1/s$, $rs < 1$. Thus $\alpha\beta \subseteq 1^*$. -To prove the opposite inclusion, pick $v \in 1^*$, $v > 0$. [Once again, we would like to follow Rudin and set $w = 1/\sqrt v$, but $\sqrt v$ may not be rational. It suffices to take rational $j,k$ satisfying $0 1$, and there is a nonnegative integer $n$ such that $w^n \in \alpha$ but $w^{n+1} \notin \alpha$. (See Comment below.) [This follows from $w^n = ((w-1)+1)^n > n(w-1)$ by binomial expansion, then using the archimedean property of $\mathbb Q$, since $w-1>0$.] Put $p = k/w^{n+1}$. Then $p \in \beta$, since $(1/p)/(1/k) \notin \alpha$, and $$v = w^np \in \alpha\beta.$$ Thus $1^* \subseteq \alpha\beta$. -[Comment: This argument does not work if $w^0 = 1$ is not in $\alpha$, for in that case there is no nonnegative integer $n$ with the required property. However, in that case we can simply switch the roles of $\alpha$ and $\beta$: Indeed, if $1 \notin \alpha$, then $\alpha < 1^*$, from which it follows that $\beta > 1^*$. And there is a symmetry between $\alpha$ and $\beta$, in that $\alpha$ is the union of $(-\infty,0]$ with $\alpha^+$, the set of all $p > 0$ such that there exists $r>1$ which satisfies $(1/p)/r \notin \beta$. Hence we can switch $\alpha$ and $\beta$ in the preceding paragraph and draw the same conclusion, as claimed.] -We conclude that $\alpha\beta = 1^*$.<|endoftext|> -TITLE: 3xy + 14x + 17y + 71 = 0 need some advice -QUESTION [6 upvotes]: $$3xy + 14x + 17y + 71 = 0$$ -Need to find both $x$ and $y$. If there was only one variable then this is easy problem. -Have tried: -$$\begin{align}3xy &= -14x - 17y - 71 \\ -x &= \frac{-14x - 17y - 71}{3y}\end{align}$$ -Then tried to put this expression everywhere instead of $x$ but it tooks forever to find both $x$ and $y$. -I don't even know how to get on right track. -Please give any advice. Thanks. - -REPLY [4 votes]: This is another way to find the integer solutions. -Define $d=y-x$ and the equation becomes $3x(x+d)+14x+17(x+d)+71=0$ -which boils down to $$3x^2+(3d+31)x+17d+71=0.$$ -The quadratic formula gives the roots:$$6x=-(3d+31)\pm\sqrt{9d^2-18d+109},$$ -and we want the discriminant, $9(d-1)^2+100 = (3(d-1))^2+10^2$ to be a square. -$d=1$ is obviously one candidate, leading to final values $x=-4, y=-3$. -Otherwise we are just looking for Pythagorean triples. A quick look at a list of Pythagorean triples shows that there is just one triple with $10$ as one of the lower values: (10, 24, 26). -Thus $9(d-1)^2=24^2$, giving $d=9$ or $d=-7$, from which the appropriate values of $x$ and $y$ can be found.<|endoftext|> -TITLE: The longest sum of consecutive primes that add to a prime less than 1,000,000 -QUESTION [9 upvotes]: In Project Euler problem $50,$ the goal is to find the longest sum of consecutive primes that add to a prime less than $1,000,000. $ - -I have an efficient algorithm to generate a set of primes between $0$ and $N.$ -My first algorithm to try this was to brute force it, but that was impossible. -I tried creating a sliding window, but it took much too long to even begin to cover the problem space. -I got to some primes that were summed by $100$+ consecutive primes, but had only run about $5$-$10$% of the problem space. -I'm self-taught, with very little post-secondary education. - -Where can I read about or find about an algorithm for efficiently calculating these consecutive primes? -I'm not looking for an answer, but indeed, more for pointers as to what I should be looking for and learning about in order to solve this myself. - -REPLY [4 votes]: To flesh out a comment: the key insight for speeding up this problem is that by using $O(n)$ space, the intermediate sums can be calculated in constant time each: $S_{m,n} = S_{1,n}-S_{1,(m-1)}$, so storing partial sums from $1$ to $i$ lets all other partial sums be easily computed. This removes an $O(n)$ factor from the running time and makes the problem much more tractable.<|endoftext|> -TITLE: Every multiplicative linear functional on $\ell^{\infty}$ is the limit along an ultrafilter. -QUESTION [15 upvotes]: It is well-known that for any ultrafilter $\mathscr{u}$ in $\mathbb{N}$, the map\begin{equation}a\mapsto \lim_{\mathscr{u}}a\end{equation} is a multiplicative linear functional, where $\lim_{\mathscr{u}}a$ is the limit of the sequence $a$ along $\mathscr{u}$. -I vaguely remember someone once told me that every multiplicative linear functional on $\ell^{\infty}$ is of this form. That is, given a multiplicative linear functional $h$ on $\ell^{\infty}$, there is an ultrafilter $\mathscr{u}$ such that \begin{equation} -h(a)=\lim_{\mathscr{u}}a -\end{equation} for all $a\in\ell^{\infty}$. -However, I cannot find a proof to this. I can show that if $h$ is the evaluation at $n$, then $h$ corresponds to the principal ultrafilter centered at $n$, but there are other kinds of multiplicative functionals (all these must vanish on any linear combinations of point masses though). -Can somebody give a hint on how to do this latter case? -Thanks! - -REPLY [3 votes]: Let me give a slightly different approach to the second half of Martin Argerami's answer. Suppose $\varphi:\ell^\infty\to\mathbb{C}$ is a multiplicative linear functional, and as in Martin Argerami's answer define $\mathcal{U}=\{A:\varphi(1_A)=1\}$ and prove that $\mathcal{U}$ is an ultrafilter. Now we just need to prove that $$\varphi(c)=\lim\limits_\mathcal{U} c$$ for any $c\in\ell^\infty$. -To prove this, let $L=\lim\limits_\mathcal{U} c$ and fix $\epsilon>0$. Since $L$ is the limit of $c$ along $\mathcal{U}$, the set $$A=\{n:|c(n)-L|<\epsilon\}$$ is in $\mathcal{U}$. Let $d=c\cdot 1_A$. Since $\varphi$ is multiplicative, $\varphi(d)=\varphi(c)\varphi(1_A)=\varphi(c)\cdot 1=\varphi(c)$. But by definition of $A$, $\|d-L1_A\|\leq\epsilon$. Thus $$|\varphi(c)-L|=|\varphi(c)-\varphi(L1_A)|=|\varphi(d)-\varphi(L1_A)|=|\varphi(d-L1_A)|\leq \|\varphi\|\epsilon.$$ Since $\epsilon$ was arbitrary, we conclude that $\varphi(c)=L$. -(Note that this argument, like Martin Argerami's, assumes $\varphi$ is bounded. You don't actually need to assume this; you can prove it. Indeed, $\|c\|$ can be described as the least $r\in [0,\infty)$ such that for all $\lambda\in\mathbb{C}$ such that $|\lambda|> r$, $\lambda-c$ has a multiplicative inverse. Since $\varphi$ preserves multiplicative inverses, $\lambda-\varphi(c)\neq 0$ whenever $|\lambda|> \|c\|$, so $|\varphi(c)|\leq \|c\|$.)<|endoftext|> -TITLE: Is the following matrix invertible? -QUESTION [150 upvotes]: $$\begin{bmatrix} 1235 &2344 &1234 &1990\\ -2124 & 4123& 1990& 3026 \\ -1230 &1234 &9095 &1230\\ -1262 &2312& 2324 &3907 -\end{bmatrix}$$ -Clearly, its determinant is not zero and, hence, the matrix is invertible. -Is there a more elegant way to do this? -Is there a pattern among these entries? - -REPLY [500 votes]: Find the determinant. To make calculations easier, work modulo $2$! The diagonal is $1$'s, the rest are $0$'s. The determinant is odd, and therefore non-zero.<|endoftext|> -TITLE: Finding shortest distance between a point and a surface -QUESTION [15 upvotes]: Consider the surface $S$ (in $\mathbb R^3$) given by the equation $z=f(x,y)=\frac32(x^2+y^2)$. How can I find the shortest distance from a point $p=(a,b,c)$ on $S$ to the point $(0,0,1)$. -This is what I have done: Define $d(a,b,c)=a^2+b^2+(c-1)^2$, for all points $p=(a,b,c)\in S$. Then $\sqrt d$ is the distance from $S$ to $(0,0,1)$. I think that the method of Lagrange multipliers is the easiest way to solve my question, but how can I find the Lagrangian function? Or is there an easier way to find the shortest distance? - -REPLY [8 votes]: I think that the method of Lagrange multipliers is the easiest way to -solve my question, but how can I find the Lagrangian function? - -As shown by other answers and in note 1 there are easier ways to find the shortest distance, but here is a detailed solution using the method of Lagrange multipliers. You need to find the minimum of the distance function -$$\begin{equation} -d(x,y,z)=\sqrt{x^{2}+y^{2}+(z-1)^{2}} \tag{1a} -\end{equation}$$ -subject to the constraint given by the surface equation $z=f(x,y)=\frac32(x^2+y^2)$ -$$\begin{equation} -g(x,y,z)=z-\frac{3}{2}\left( x^{2}+y^{2}\right) =0. \tag{2} -\end{equation}$$ -Since $\sqrt{x^{2}+y^{2}+(z-1)^{2}}$ increases with $x^{2}+y^{2}+(z-1)^{2}$ you can simplify the -computations if you find the minimum of -$$\begin{equation} -[d(x,y,z)]^2=x^{2}+y^{2}+(z-1)^{2} \tag{1b} -\end{equation}$$ -subject to the same constraint $(2)$. The Lagrangian function is then -defined by -$$\begin{eqnarray} -L\left( x,y,z,\lambda \right) &=&[d(x,y,z)]^2+\lambda g(x,y,z) \\ -L\left( x,y,z,\lambda \right) &=&x^{2}+y^{2}+(z-1)^{2}+\lambda \left( z- -\frac{3}{2}\left( x^{2}+y^{2}\right) \right), \tag{3} -\end{eqnarray}$$ -where $\lambda $ is the Lagrange multiplier. By this method you need to -solve the following system -$$\begin{equation} -\left\{ \frac{\partial L}{\partial x}=0,\frac{\partial L}{\partial y}=0, -\frac{\partial L}{\partial z}=0,\frac{\partial L}{\partial \lambda } -=0,\right. \tag{4} -\end{equation}$$ -which results in -$$\begin{eqnarray} -\left\{ -\begin{array}{c} -2x+3\lambda x=0 \\ -2y+3\lambda y=0 \\ -2z-2-\lambda =0 \\ --z+\frac{3}{2}\left( x^{2}+y^{2}\right) =0 -\end{array} -\right. &\Leftrightarrow &\left\{ -\begin{array}{c} -x=0\vee 2+3\lambda =0 \\ -y=0\vee 2+3\lambda =0 \\ -2z-2-\lambda =0 \\ --z+\frac{3}{2}\left( x^{2}+y^{2}\right) =0 -\end{array} -\right. \\ -&\Leftrightarrow &\left\{ -\begin{array}{c} -x=0 \\ -y=0 \\ -\lambda =2 \\ -z=0 -\end{array} -\right. \vee \left\{ -\begin{array}{c} -\lambda =-2/3 \\ -z=2/3 \\ -x^{2}+y^{2}=4/9 -\end{array} -\right. \tag{5} -\end{eqnarray}$$ -For $x=y=x=0$ we get $d(0,0,0)=1$. And for $x^2+y^2=4/9,z=2/3$ we get the minimum distance subject to the given conditions -$$\begin{equation} -\underset{g(x,y,z)=0}\min d(x,y,z)=\sqrt{\frac{4}{9}+(\frac{2}{3}-1)^{2}}=\frac{1}{3}\sqrt{5}. \tag{6} -\end{equation}$$ -It is attained on the intersection of the surface $z=\frac{3}{2}\left( -x^{2}+y^{2}\right) $ with the vertical cylinder $x^{2}+y^{2}=\frac{4}{9}$ or equivalently with the horizontal plane $z=\frac{2}{3}$. - -$$\text{Plane }z=\frac{2}{3} \text{(blue) and surface }z=\frac{3}{2}\left( -x^{2}+y^{2}\right) $$ -Notes. - -As the solution depends only on the sum $r^{2}=x^{2}+y^{2}$ we could just find -$$\begin{equation} -\min [d(r)]^2=r^{2}+(\frac{3}{2}r^{2}-1)^{2} \tag{7} -\end{equation}$$ -and then find $d(r)=\sqrt{[d(r)]^2}$ at the minimum. -The surface $z=\frac{3}{2}\left( -x^{2}+y^{2}\right) $ is a surface of revolution around the $z$ axis.<|endoftext|> -TITLE: What function satisfies $x^2 f(x) + f(1-x) = 2x-x^4$? -QUESTION [6 upvotes]: What function satisfies $x^2 f(x) + f(1-x) = 2x-x^4$? I'm especially curious if there is both an algebraic and calculus-based derivation of the solution. - -REPLY [2 votes]: Given, $x^2f(x)+f(1−x)=2x−x^4$ -$f(-1) = 0$ -$f(0) = 1$ -$f(1) = 0$ -$f(2) = -3$ -$f(3) = - 8$ -$f(4) = -15$ -By careful observation one can see that the difference between the terms vanishes after the second stage -i.e. $f(x)$ is of the form $a{x^2} + b{x} +c$ -i.e. $f(x) = a{x^2} + b{x} + c$ -------(1) -i.e. $f(0) = c = 1$ -i.e. $c = 1$ -------(2) -$f(1) = a + b + c = 0$ -i.e. $f(1) = a + b + 1= 0$ -i.e. $a + b = -1$ -------(3) -$f(2) = 4a + 2b + c = -3$ -i.e. $4a + 2b + 1 = -3$ -i.e. $4a + 2b = -2$ -i.e. $2a + b = -2$ -------(4) -Solve equations $(3)$ and $(4)$ to get $a = -1$ and $b = 0$ -i.e. $f(x) = {-1}{x^2} + {0}{x} + 1$ -i.e. $f(x) = {-x^2} + 1$<|endoftext|> -TITLE: Cauchy-Schwarz matrix inequality for random vectors -QUESTION [6 upvotes]: If $X$ and $Y$ are random scalars, then Cauchy-Schwarz says that -$$| \mathrm{Cov}(X,Y) | \le \mathrm{Var}(X)^{1/2}\mathrm{Var}(Y)^{1/2}.$$ -If $X$ and $Y$ are random vectors, is there a way to bound the covariance matrix $\mathrm{Cov}(X,Y)$ in terms of the matrices $\mathrm{Var}(X)$ and $\mathrm{Var}(Y)$? -In particular, is it true that -$$\mathrm{Cov}(X,Y) \le \mathrm{Var}(X)^{1/2}\mathrm{Var}(Y)^{1/2},$$ -where the square roots are Cholesky decompositions, and the inequality is read as meaning that the right hand side minus the left hand side is positive semidefinite? - -REPLY [6 votes]: There is a generalization of Cauchy Schwarz inequality from Tripathi [1] that says that: -\begin{equation} -\mathrm{Var}(Y) \ge \mathrm{Cov}(Y,X)\mathrm{Var}(X)^{-1}\mathrm{Cov}(X,Y) -\end{equation} -in the sense that the diference is semidefinite positive. -He actually says that a student asked about it and couldn't find any other reference (1998!). -[1]: G. Tripathi, ”A matrix extension of the Cauchy-Schwarz inequality”, Economics Letters, vol. 63, nr 1, s. 1–3, apr. 1999, doi: 10.1016/S0165-1765(99)00014-2.<|endoftext|> -TITLE: Intuitionistic Banach-Tarski Paradox -QUESTION [11 upvotes]: While the Banach-Tarski paradox is a counter-intuitive result which requires the Axiom of Choice, leading some people to argue specifically against Choice, and others to argue for constructive mathematics, as the use of Choice is the only non-constructive step in the proof, and traditional accounts of constructive logic do not contain choice. -However, newer frameworks of constructive logic such as intensional type theory (ITT) do admit Choice as a rather trivial theorem. Does this mean that the Banach Tarski theorem is also a theorem of ITT? - -REPLY [13 votes]: As Kaveh says in a comment, the issue is not going to be choice. Usually, when we examine what goes wrong when we try to formalize a "nonconstructive" classical proof in a constructive system, the problem that we discover is that the proof is already nonconstructive before the axiom of choice is invoked. -In this case, the issue is that the entire concept of partitions and equivalence classes, which underlies the usual proof of the Banach-Tarski paradox, behaves very differently in constructive systems. -Consider the simpler case of a Vitali set in the real line. Two reals are defined to be equivalent if their difference is rational. That is a perfectly good definition in constructive systems, but it is easy to miss the fact that it involves an implicit universal quantifier over the rationals (or the ability to tell whether an arbitrary real is rational). In this particular case, standard methods show that it will not be possible to prove constructively[1] that "for all reals $x$ and $y$, either $x$ is equivalent to $y$ or $x$ is not equivalent to $y$". Thus it is no longer possible to prove that two equivalence classes $[x]$ and $[y]$ are disjoint or identical, that is, impossible to prove that the equivalence classes form a partition of the real line. But that fact is crucial for the usual proof that the Vitali set is nonmeasurable. The same sort of problem is easy to spot in the usual proof of the Banach-Tarski paradox. - -More precisely, this will be impossible in any of the usual formalized systems for constructive mathematics, including intuitionistic type theory.<|endoftext|> -TITLE: How do I get insight into equations with the $\sum$ notation without actually expanding it for a specific n every time? -QUESTION [5 upvotes]: I am reading Goldstein's Classical Mechanics and I've noticed there is copious use of the $\sum$ notation. He even writes the chain rule as a sum! I am having a real hard time following his arguments where this notation is used, often with differentiation and multiple indices thrown in for good measure. How do I get some working insight into how sums behave without actually saying "Now imagine n=2. What does the sum become in this case?" Is there an easier way to do this? Is there an "algebra" or "calculus" of sums, like a set of rules for manipulating them? I've seen some documents on the web but none of them seem to come close to Goldstein's usage in terms of sophistication. Where can I get my hands on practice material for this notation? - -REPLY [2 votes]: I remember that I completely lost my uneasines with sums after reading first several chapters of this book. Apart from being very educative, having lots of various excercises, and $\sum$ letter on its cover -- it is also very fun to read. - -REPLY [2 votes]: Just look at the formulas in the various derivations, abstract from them the operations that apparently have been going on to reaching the rhs from the lhs, keep a list of these rules, and take the recurring ones as permissions to do this sort of rearrangement or substitution. -If a text isn't too short of intermediate steps and doesn't have too many misprints, this reveals the hidden secrets in most calculations, not only sums. - -REPLY [2 votes]: Donald Knuth's book "The Art of Computer Programming" Volume 1 contains probably all the tricks of manipulating sums, you will ever need. If you just want to learn the material in Goldstein and find it hard, you should consider an alternative book with the same material.<|endoftext|> -TITLE: On the element orders of finite group -QUESTION [5 upvotes]: Let $G$ be a finite group. Suppose that $G$ has a normal Sylow $p-$subgroup $P$ such that $|P|=p^2$ where $p\neq 2$, but $P$ does not contain an element of order $p^2$ or equivalently $P$ is not cyclic. -Is it the case that if $g \in G$ has order $k$ such that $\gcd(k,p)=1$ then there is an element of $G$ of order $kp$? Thanks in advance. - -REPLY [2 votes]: The alternating group of degree 4 provides a model for counterexamples for all primes $p$ (and all powers $p^f$, not just $p^2$). Rather than viewing the alternating group acting on 4 unstructured points, think of it acting on the underlying set of a finite field. Rather than viewing it as all possible “even” permutations of the set, view it as all possible “affine” permutations of the set. -$$\newcommand{\AGL}{\operatorname{AGL}} G=\AGL(1,K) = \{ f : K \to K : x \mapsto \alpha x + \beta ~\mid~ \alpha,\beta \in K, \alpha \neq 0 \}$$ -In the case of $A_4$, we take $K$ of order $4$. The subgroup $$K_+ = \{ f : K \to K : x \mapsto x + \beta ~\mid~ \beta \in K \}$$is a normal, elementary abelian, Sylow $p$-subgroup of $\AGL(1,K)$ whenever $K$ is a a finite field of characteristic $p$. The subgroup $$K^\times = \{ f : K \to K : x \mapsto \alpha x ~\mid~ \alpha \in K, \alpha \neq 0\}$$ -is a cyclic subgroup of order $k_0=|K|-1$, relatively prime to the order of $K$. -If $K$ is chosen to be a field of size $p^2$, then $K_+$ is a normal, noncyclic Sylow $p$-subgroup of $G$ of order $p^2$, and $k$ divides $|G|=kp^2$, but of course $G$ has no subgroup of order $k_0p$, much less an element of order $k_0p$. -Proposition: The orders of elements of $G$ are exactly $\{p\} \cup \{ k : k \text{ divides } k_0 \}$. -Proof: Clearly $K_+$ has elements of order $p$, and the cyclic group $K^\times$ has elements of the other orders. If $g$ has order $kp$ for $k$ dividing $k_0$, then because $\gcd(k,p)=1$ we can write $1=uk+vp$, and so $g=g^1 = g^{(uk)} g^{(vp)}$ and $b=g^{(uk)}$ has order $p$ while $a=g^{(vp)}$ has order $k$. Hence we have an element $a$ of order $k$ commuting with an element $b$ of order $p$. All such elements $b$ lie in $K_+$. Write $a = (x\mapsto \alpha x + \beta)$ and $b=(x\mapsto x+\gamma)$. Then one product of $a$ and $b$ is $x\mapsto \alpha x + (\beta+\gamma)$ and the other is $x\mapsto \alpha x + (\beta+\alpha \gamma)$. For these two maps to be the same, they must act the same on $0 \in K$, and in particular $\beta+\gamma = \beta+\alpha\gamma$ so that $\alpha=1$ or $\gamma=0$. In the first case, this means $\alpha =1$ and $a \in K_+$ has order dividing both $p$ and $k_0$, so $k=1$. In the second case, this means $b$ is the identity, contradicting $g$ having order a multiple of $p$. $\square$<|endoftext|> -TITLE: Show $1 + 2 \sum_{n=1}^N \cos n x = \frac{ \sin (N + 1/2) x }{\sin \frac{x}{2}}$ for $x \neq 0$ -QUESTION [6 upvotes]: For $x \neq 0$, $$ 1 + 2 \sum_{n=1}^N \cos n x = \frac{ \sin (N + 1/2) x }{\sin \frac{x}{2}} $$ - -REPLY [12 votes]: Here is a well known trigonometric trick -$$ -1+2\sum\limits_{n=1}^N\cos (nx)= -1+\frac{1}{\sin(x/2)}\sum\limits_{n=1}^N 2\cos (nx)\sin (x/2)=\\ -1+\frac{1}{\sin (x/2)}\sum\limits_{n=1}^N(\sin (nx+x/2)-\sin (nx-x/2))=\\ -1+\frac{1}{\sin (x/2)}(\sin (Nx+x/2)-\sin (x/2))=\\ -1+\frac{\sin (Nx+x/2)}{\sin (x/2)}-1= -\frac{\sin (N+1/2)x}{\sin (x/2)} -$$ -And this is a complex analysis approach -$$ -1+2\sum\limits_{n=1}^N\cos(nx)= -e^{i0x}+\sum\limits_{n=1}^N(e^{inx}+e^{-inx})= -$$ -$$ -\sum\limits_{n=-N}^N e^{inx}= -\frac{e^{-iNx}(e^{i(2N+1)x}-1)}{e^{ix}-1}= -\frac{e^{i(N+1)x}-e^{-iNx}}{e^{ix}-1}= -$$ -$$ -\frac{e^{i(N+1/2)x}-e^{-i(N+1/2)x}}{e^{ix/2}-e^{-ix/2}}= -\frac{2i\sin(N+1/2)x}{2i\sin(x/2)}= -\frac{\sin(N+1/2)x}{\sin(x/2)} -$$<|endoftext|> -TITLE: Good ways to help people learn math. -QUESTION [32 upvotes]: I have a student position at my university, which involves helping people who are struggling with basic calculus, precalculus, and discrete math. In general, the students that I help are hostile towards the subject, and are easily frustrated. For instance, many people have told me that "we don't need anyone that good in math"--in earnest--after I answered their question about what I'm going to do after I graduate (go to grad school). For anouther example, and I regretfully say that this is common, students frequently have trouble with online homework systems. They do no enter their, say equation of a line, in the format required by the system. One time, I helped someone who was using this software to pratice for a test they had in one hour. I solved the problem for them (it said not graded on the page), to show them how to go about solving the types of questions that were being asked. I told her the answer was $2y+3x=4$ , 4and the machine marked it wrong. What was going on was that it wanted the equation to look like $3x+2y=4$... She was angry at me, and asked me if I knew "even know wtf this even means." -All in all, I am asking a few questions: - -How do you deal with students who treat you like garbage? -How do you help a student who is learning extremly basic math without frustrating them, or somehow making them feel stupid. -How do you deal with people who think that your disipline is pointless, but want your help anyways? - -Obviously, these 3 questions are subjective, and this post should be in as a community wiki, rather than on the Q and A part of the site. However, I do not know how to do this. If anyone knows how do that, feel free to make it a community wiki. - -REPLY [5 votes]: In my opinion mathematics/science should be taught with some history. Showing the original work of the author of a mathematical concept, for example, will help the student see, hopefuly understand, the thought process behind it and a question like "Why was this concept was introduced?". Study the discovery step by step and then see how it evolved and how it assumed its current form.<|endoftext|> -TITLE: How to prove that $R/I \otimes_R M \cong M / IM$ -QUESTION [7 upvotes]: Possible Duplicate: -Showing that if $R$ is local and $M$ an $R$-module, then $M \otimes_R (R/\mathfrak m) \cong M / \mathfrak m M$. - -In one of the answers to one of my previous questions the following claim was mentioned: -$R/I \otimes_R M \cong M / IM$ -So I tried to prove it. Can you help me finish my proof? Thanks! - -We recall that $M \otimes_R -$ is a covariant right-exact functor that it is exact if $M$ is flat and we observe that the following is an exact sequence: -$$ 0 \to IM \xrightarrow{i} M \xrightarrow{\pi} M / IM \to 0$$ -$R/I$ is an $R$ module since $R/I$ is a subring of $R$ closed with respect to multiplication from $R$ but it is not necessarily flat hence we only get exactness on one side: -$$ (R/I) \otimes_R IM \xrightarrow{id \otimes i} (R/I) \otimes_R M \xrightarrow{id \otimes \pi} (R/I) \otimes_R M / IM \to 0$$ -Then $$ \mathrm{Im(id \otimes \pi)} = (R/I) \otimes_R M / IM \cong (R/I) \otimes_R M / \mathrm{Ker} (id \otimes i) $$ -so we want to show two things: -(i) that $\mathrm{Ker} (id \otimes i) = \{0\}$. To this end let $r + I \otimes im \in (R/I) \otimes_R IM $ and assume $ id \otimes i(r + I \otimes im ) = r + I \otimes im = 0 + I \otimes 0$. I'm not sure how to proceed from here. -(ii) and that $M /IM \cong (R/I) \otimes_R M / IM $ -Can you help me? Thanks. - -REPLY [14 votes]: Consider -$$ - 0 \to I \to R \to R/I \to 0 -$$ -Apply the right exact functor $-\otimes_R M$, and you get -$$ - I\otimes_R M \to R\otimes_R M \to (R/I)\otimes_R M \to 0 -$$ -But $R\otimes_R M$ is canonically identified with $M$ by $a\otimes m \mapsto am$. Then $I\otimes_R M \to R\otimes_R M = M$ is $a\otimes m \mapsto am$ so, by definition, its image is $IM$. So the exactness of the sequence tells you that -$$ - M/IM \cong (R/I)\otimes_R M -$$ - -REPLY [4 votes]: In general, playing around with elements of a tensor product is something that should be avoided if you can help it (defining maps from a tensor product is also usually better done by defining them from the product and then checking that they are in fact $R$-balanced and the using the universal property). A proof which avoids dealing with explicit elements of the tensor product can be done in the following way: -Consider the map $R/I\times M\to M/IM$ given by $(r+I,m)\mapsto rm+IM$. This is well-defined, as the image of any two representatives of $r+I$ differ by an element of $IM$. Since it is $R$-balanced, it descends to a map from the tensor product. Now consider the map $M \to R/I\otimes M$ given by sending $m\mapsto (1+I)\otimes m$. This descends to the quotient $M/IM$ by the 1st isomorphism theorem, and then it is easy to see that these maps are mutual inverses.<|endoftext|> -TITLE: Show $ \int_{-\pi}^{\pi} \frac{1 - \cos (n+1) x}{1- \cos x} dx = (n+1) 2 \pi$ for $n \in \mathbb N$ -QUESTION [6 upvotes]: Possible Duplicate: -Compute the trigonometric integrals - -For $n \in \mathbb N$, $$ \int_{-\pi}^{\pi} \frac{1 - \cos (n+1) x}{1- \cos x} dx = (n+1) 2 \pi$$ - -REPLY [3 votes]: Setting $z=e^{ix}$ and denoting by $C$ the unit circle, one has -$$ -I_n:=\int_{-\pi}^\pi\frac{1-\cos(n+1)x}{1-\cos x}dx=-i\oint_Cf(z)dz, -$$ -where -$$ -f(z)=\frac{z^{n+1}+z^{-n-1}-2}{z(z+z^{-1}-2)}=\frac{(1+z+\ldots+z^n)^2}{z^{n+1}}=\frac{(P_n(z))^2}{z^{n+1}}. -$$ -Thanks to the Residue Theorem, one gets -\begin{eqnarray} -I_n&=&2\pi\text{Res}(f,0)=\frac{2\pi}{n!}\lim_{z \to 0}\frac{d^n}{dz^n}\left[z^{n+1}f(z)\right]=\frac{2\pi}{n!}\lim_{z \to 0}\frac{d^n}{dz^n}(P_n(z))^2\cr -&=&\frac{2\pi}{n!}\lim_{z \to 0}\sum_{k=0}^n{n\choose k}P_n^{(k)}(z)P_n^{(n-k)}(z) -=\frac{2\pi}{n!}\sum_{k=0}^n{n\choose k}P_n^{(k)}(0)P_n^{(n-k)}(0). -\end{eqnarray} -Since $P_n(z)=1+z+\ldots+z^n$, one has $P_n^{(k)}(0)=k!$ for every $k \in \{0,\ldots, n\}$. Hence -$$ -I_n=\frac{2\pi}{n!}\sum_{k=0}^n{n\choose k}k!(n-k)!=2\pi\sum_{k=0}^n1=2(n+1)\pi. -$$<|endoftext|> -TITLE: How to prove every non-compact, connected 2 dimensional surface is homotopical to a bouquet of flowers? -QUESTION [5 upvotes]: This is one of my old unsolved questions when I reading Novikov's book on homology theory. I do not know how to prove it because standard triangulation, fundamental diagram, etc does not help and it should be easy to prove. - -REPLY [2 votes]: This question just floated up again, so let me put some references here. -This MO question gives several different proofs that for your surface $S$, $\pi_1(S)$ is free. In fact, Lee Mosher's answer gives a direct proof that $S$ is homotopy equivalent to a graph, and hence to a bouquet of circles. You can also proceed by noting the universal cover of $S$ is contractible, and hence $S$ is homotopy equivlanent to any $K(\pi_1(S),1)$, of which the appropriate bouquet of circles is one.<|endoftext|> -TITLE: Do such sequences exist? -QUESTION [5 upvotes]: I wish to know if there are real sequences $(a_k)$, $(b_k)$ (and if there are, how to construct such sequences) such that: -$b_k<0$ for each $k \in \mathbb{N}$ with $\lim\limits_{k \rightarrow \infty} b_k=-\infty,$ -$$\sum_{k=1}^\infty |a_k| |b_k|^n< \infty, \space\forall n\in\mathbb{N}\cup\{0\}$$ -$$\sum_{k=1}^\infty a_k b_k^n=1, \space \forall n\in\mathbb{N}\cup\{0\}$$ - -REPLY [3 votes]: For any unbounded sequence $(b_k)$ and any sequence $(\lambda_n)$, you can find a sequence $(a_k)$ such that for any $n$, $\sum_{k=1}^\infty a_k b_k^n$ converges absolutely to $\lambda_n$ : -Suppose you have constructed $a_k$ up to some index $k_0$, such that for all $m < n$, $\sum_{k=1}^{k_0} a_k b_k^m = \lambda_m$. The goal now is to extend this up to some index $k_1$ such that for all $m \le n$, $\sum_{k=1}^{k_1} a_k b_k^m = \lambda_m$, with "arbitrarily small coefficients". -This means we have to add $y = \lambda_n - \sum_{k=1}^{k_0} a_k b_k^n$ to the partial sum of the $n$th series, and $0$ to the first $n-1$ series. -In order to do that, look at the next $n-1$ distincts values of $b_k$ (for $k > k_0$), say they are $c_1,c_2, \ldots c_{n-1}$, look at the Vandermonde matrix -$$ M(X) = \begin{pmatrix} c_1^1 & \ldots & c_{n-1}^1 & X^1 \\ \vdots & & \vdots & \vdots \\ c_1^n & \ldots & c_{n-1}^n & X^n \end{pmatrix}$$ -and solve for the vector $A(X)$ in the equation $M(X) A(X) = (0,0,\ldots,0,y)$. -We get that $A(X)$ has to be $y$ times the $n$th column of $M(X)^{-1}$. -Thus $A(X)$ is $y/det(M(X))$ times the transpose of the $n$th row of the comatrix of $M(X)$. -If you look carefully at this row, each entry there is a polynomial in $X$ of degree $n-1$, except the last one which is a constant. Furthermore, the determinant of $M(X)$ is of degree $n$, so the entries of the vector $A(X)$ are rational fractions of degree $-1$ and $-n$. -In particular, for $X$ large enough, $\sum_{1 \le i,j \le n-1} |A_i(X)c_i^j| < 2^{-n}$ -Thus, if you pick $k_1$ such that $b_{k_1}$ is large enough, you will get from $A(b_{k_1})$ the coefficients to put in front of $c_1^n, c_2^n, \ldots, c_{n-1}^n, b_{k_1}^n$ such that all the partial sums are what we want, and the absolute contribution to what we added to each partial sum is less than $2^{-n-1}$. -Now, repeat this procedure for all $n$, and you get a suitable sequence $(a_k)$.<|endoftext|> -TITLE: Question about proof of $A[X] \otimes_A A[Y] \cong A[X, Y] $ -QUESTION [7 upvotes]: As far as I understand universal properties, one can prove $A[X] \otimes_A A[Y] \cong A[X, Y] $ where $A$ is a commutative unital ring in two ways: -(i) by showing that $A[X,Y]$ satisfies the universal property of $A[X] \otimes_A A[Y] $ -(ii) by using the universal property of $A[X] \otimes_A A[Y] $ to obtain an isomorphism $\ell: A[X] \otimes_A A[Y] \to A[X,Y]$ -Now surely these two must be interchangeable, meaning I can use either of the two to prove it. So I tried to do (i) as follows: -Define $b: A[X] \times A[Y] \to A[X,Y]$ as $(p(X), q(Y)) \mapsto p(X)q(Y)$. Then $b$ is bilinear. Now let $N$ be any $R$-module and $b^\prime: A[X] \times A[Y] \to N$ any bilinear map. -I can't seem to define $\ell: A[X,Y] \to N$ suitably. The "usual" way to define it would've been $\ell: p(x,y) \mapsto b^\prime(1,p(x,y)) $ but that's not allowed in this case. -Question: is it really not possible to prove the claim using (i) in this case? - -REPLY [3 votes]: For a commutative unital ring $R$ we can show the following $R$-module isomorphism in two ways: $R[x,y] \cong R[x] \otimes R[y]$. - -(i) Obtain an isomorphism by using the universal property of $(R[x] \otimes R[y], b)$ where $b: R[x] \times R[y] \to R[x] \otimes R[y]$ is the bilinear map $(p(x), q(y)) \mapsto p(x) \otimes q(y)$ (it looks like this by construction of the tensor product). For $R[x,y]$ we observe that $b^\prime: R[x]\times R[y] \to R[x,y]$, $(p(x),q(y)) \mapsto p(x)q(y)$ is bilinear hence there exists a unique linear map $l: R[x] \otimes R[y] \to R[x,y]$ such that $l \circ b = b^\prime$. We claim that $l$ is an isomorphism, that is, has a two sided inverse. To see this, define $f: R[x,y] \to R[x] \otimes R[y]$ as $\sum a_{ij} x^i y^j \mapsto \sum a_{ij} (x^i \otimes y^j)$. Then -$$ l (f (\sum a_{ij} x^i y^j)) = \sum a_{ij} x^i y^j$$ and -$$ f (l (\sum a_{ij} (x^i \otimes y^j))) = \sum a_{ij} (x^i \otimes y^j)$$ - -(ii) Show that $R[x,y]$ together with the map $b: R[x] \times R[y] \to R[x,y]$, $(p(x) , q(y)) \mapsto p(x) q(y)$ satisfies the universal property of $R[x] \otimes R[y]$. That is, for an $R$-module $M$ and a a bilinear map $b^\prime : R[x] \times R[y] \to M$ there is a unique linear map $l: R[x,y] \to M$ such that $l \circ b = b^\prime$. Define $l: \sum a_{ij} x^i y^j \mapsto \sum a_{ij}b^\prime (x^i, y^j)$. Then $l$ is linear, $l \circ b = b^\prime$ and $l$ is unique: let $l^\prime$ be such that $l^\prime \circ b = b^\prime$. Then $l^\prime \circ b (x^i, y^j) = b^\prime(x^i, y^j) = l \circ b (x^i, y^j)$ and hence $l(x^i y^j) = l^\prime(x^i y^j)$ and hence $l = l^\prime$.<|endoftext|> -TITLE: Is product of Banach spaces a Banach space? -QUESTION [12 upvotes]: If $X$ is a Banach space, then I want to know if $X\times X$ is also Banach. What is the norm of that space? -So for example, we know $C^k(\Omega)$ is Banach and I have a vector $v = (u_1, u_2)$ where $u_i \in C^k(\Omega)$ and I want to use properties on this vector. -Thanks - -REPLY [15 votes]: Yes, though the product is more commonly called a direct sum. Many equivalent norms are possible; common choices are $\lVert(u_1,u_2)\rVert=\lVert u_1\rVert+\lVert u_2\rVert$ and $\lVert(u_1,u_2)\rVert=\max(\lVert u_1\rVert,\lVert u_2\rVert)$, or more generally $\lVert(u_1,u_2)\rVert=(\lVert u_1\rVert^p+\lVert u_2\rVert^p)^{1/p}$ for $1\le p<\infty$. - -REPLY [6 votes]: Yes. -Let $(X, \lVert - \rVert)$ be a Banach space. Then $X \times X$ is a Banach space under the norm $\lVert (x,y) \rVert = \lVert x \rVert + \lVert y \rVert$. -Proof: It's easy to check that this defines a norm, so we just need completeness. Let $(x_n, y_n)$ be a Cauchy sequence in this norm. Let $\varepsilon > 0$ and let $N \in \mathbb{N}$ be such that for all $m,n \ge N$ we have -$$\lVert (x_n, y_n) - (x_m, y_m) \rVert < \varepsilon$$ -Then by the definition of our norm we must also have $\lVert x_n - x_m \rVert + \lVert y_n - y_m \rVert < \varepsilon$, and hence $(x_n)$ and $(y_n)$ are Cauchy in $X$, hence convergent, and it's easy to check that if $x_n \to x$ and $y_n \to y$ then $(x_n,y_n) \to (x,y)$. $\square$ -(In fact we may choose other norms for $X \times X$.) - -REPLY [3 votes]: The product of a finite number of Banach spaces can easily made into a Banach space by, e.g., adding the norms or by taking their maximum. There are more choices, but none of them is natural, to my knowledge, or preferred. -A better way to put it is, in my opinion, to say that such a product is, in a natural way, a topological vector space, which admits a complete norm.<|endoftext|> -TITLE: Compute $ I_{n}=\int_{-\infty}^\infty \frac{1-\cos x \cos 2x \cdots \cos nx}{x^2}\,dx$ -QUESTION [25 upvotes]: I'm very curious about the ways I may compute the following integral. I'd be very glad to know your approaching ways for this integral: -$$ -I_{n} \equiv -\int_{-\infty}^\infty -{1-\cos\left(x\right)\cos\left(2x\right)\ldots\cos\left(nx\right) \over x^{2}} -\,{\rm d}x -$$ -According to W|A, $I_1=\pi$, $I_2=2\pi$, $I_3=3\pi$, and one may be tempted to think that it's about an arithmetical progression here, but things change (unfortunately) from $I_4$ that is $\frac{9 \pi}{2}$. This problem -came to my mind when I was working on a different problem. - -REPLY [4 votes]: I thought it would be worth mentioning that $$\int_{-\infty}^{\infty} \frac{1-\prod_{k=1}^{n}\cos (a_{k}x)}{x^{2}} = \pi a_{n} \tag{1}$$ if the $a_{k}$'s are positive parameters and $a_{n} \ge \sum_{k=1}^{n-1} a_{k}. $ -This integral is somewhat similar to the Borwein integral. -We can show $(1)$ by integrating the function $$f(z) = \frac{1-e^{ia_{n}z}\prod_{k=1}^{n-1}\cos(a_{k}z)}{z^{2}}$$ around an indented contour that consists of the real axis and the semicircle above it. -If $a_{n} \ge \sum_{k=1}^{n-1} a_{k}$, then $$1-e^{ia_{n}z}\prod_{k=1}^{n-1}\cos(a_{k}z) = 1- e^{ia_{n}x}\prod_{k=1}^{n-1}\frac{e^{ia_{k}z}-e^{-ia_{k}z}}{2} $$ is bounded in the upper half-plane. (If you expand, all of the exponentials are of the form $e^{ibz}$, where $b\ge 0$.) -So integrating around the contour, we get $$\operatorname{PV} \int_{-\infty}^{\infty} f(x) \, dx - i \pi \operatorname{Res} [f(z), 0] = 0, $$ where $$ \begin{align}\operatorname{Res}[f(z),0] &= \lim_{z \to 0} \frac{1-e^{ia_{n}z}\prod_{k=1}^{n-1}\cos(a_{k}z)}{z} \\ &= \lim_{z \to 0} \left(-ia_{n}e^{ia_{n}z}\prod_{k=1}^{n-1}\cos(a_{k}z)+e^{ia_{n}z} \sum_{k=1}^{n-1} a_{k} \sin(a_{k}z) \prod_{\underset{j \ne k}{j =1}}^{n-1} \cos(a_{j}z)\right) \\ &= -i a_{n}. \end{align}$$ -The result follows by equating the real parts on both sides of the equation.<|endoftext|> -TITLE: Partial Fractions Expansion of $\tanh(z)/z$ -QUESTION [10 upvotes]: I have seen the following formula in papers (without citations) and in Mathematica's documentation about Tanh[]: -$$ -\frac{\tanh(z)}{8z}=\sum_{k=1}^{\infty} \frac{1}{(2k-1)^2 \pi^2+4z^2} -$$ -I have no idea how to prove it and I have also encountered in my research similar sums involving, for instance, $\mathrm{coth}$. It would be nice to have a general method for working with these problems; any suggestions? - -REPLY [5 votes]: In this answer, it is shown that for all $z\in\mathbb{C}\setminus\mathbb{Z}$, -$$ -\pi\cot(\pi z)=\frac1z+\sum_{k=1}^\infty\frac{2z}{z^2-k^2}\tag{1} -$$ -Applying the identity $\tan(x)=\cot(x)-2\cot(2x)$ to $(1)$ gives -$$ -\begin{align} -\pi\tan(\pi z) -&=\sum_{k=1}^\infty\frac{2z}{z^2-k^2}-\sum_{k=1}^\infty\frac{2z}{z^2-\frac{k^2}{4}}\\ -&=\sum_{k=1}^\infty\frac{8z}{4z^2-(2k)^2}-\sum_{k=1}^\infty\frac{8z}{4z^2-k^2}\\ -&=-\sum_{k=1}^\infty\frac{8z}{4z^2-(2k-1)^2}\\ -&=\sum_{k=1}^\infty\frac{8z}{(2k-1)^2-4z^2}\tag{2} -\end{align} -$$ -Applying the identity $\tanh(x)=-i\tan(ix)$ to $(2)$ yields -$$ -\pi\tanh(\pi z)=\sum_{k=1}^\infty\frac{8z}{(2k-1)^2+4z^2}\tag{3} -$$ -Finally, applying the change variables $z\mapsto z/\pi$ to $(3)$ reveals -$$ -\frac{\tanh(z)}{8z}=\sum_{k=1}^\infty\frac{1}{(2k-1)^2\pi^2+4z^2}\tag{4} -$$<|endoftext|> -TITLE: An infinite series of a product of three logarithms -QUESTION [5 upvotes]: I was told this interesting question today, but I haven't managed to get very far: - -Evaluate $$\sum_{n=1}^\infty \log \left(1+\frac{1}{n}\right)\log \left(1+\frac{1}{2n}\right)\log \left(1+\frac{1}{2n+1}\right).$$ - -I am interested in seeing at least a few solutions. - -REPLY [7 votes]: Here is a solution I just found. Notice that $$\log\left(1+\frac{1}{2n+1}\right)=\log\left(1+\frac{1}{n}\right)-\log\left(1+\frac{1}{2n}\right)$$ so that our series becomes $$\sum_{n=1}^{\infty}\left(\log\left(1+\frac{1}{n}\right)^{2}\log\left(1+\frac{1}{2n}\right)-\log\left(1+\frac{1}{n}\right)\log\left(1+\frac{1}{2n}\right)^{2}\right).$$ Since $$\log\left(1+\frac{1}{2n+1}\right)^{3}=\log\left(1+\frac{1}{n}\right)^{3}-3\log\left(1+\frac{1}{n}\right)^{2}\log\left(1+\frac{1}{2n}\right)+3\log\left(1+\frac{1}{n}\right)\log\left(1+\frac{1}{2n}\right)^{2}-\log\left(1+\frac{1}{2n}\right)^{3},$$ we see that our series equals $$\frac{1}{3}\left(\sum_{n=1}^{\infty}\log\left(1+\frac{1}{n}\right)^{3}-\log\left(1+\frac{1}{2n}\right)^{3}-\log\left(1+\frac{1}{2n+1}\right)^{3}\right),$$ and the above telescopes and equals $$\frac{\left(\log2\right)^{3}}{3}.$$<|endoftext|> -TITLE: Explain this code to compute $\log(1+x)$ -QUESTION [8 upvotes]: It's well known that you need to take care when writing a function to compute $\log(1+x)$ when $x$ is small. Because of floating point roundoff, $1+x$ may have less precision than $x$, which can translate to large relative error in computing $\log(1+x)$. In numerical libraries you often see a function log1p(), which computes the log of one plus its argument, for this reason. -However, I was recently puzzled by this implementation of log1p (commented source): -def log1p(x): - y = 1 + x - z = y - 1 - if z == 0: - return x - else: - return x * (log(y) / z) - -Why is x * (log(y) / z) a better approximation to $\log(1+x)$ than simply log(y)? - -REPLY [8 votes]: The answer involves a subtlety of floating point numbers. I'll use decimal floating point in this answer, but it applies equally well to binary floating point. -Consider decimal floating point arithmetic with four places of precision, and let -$$x = 0.0001234$$ -Then under floating point addition $\oplus$, we have -$$y = 1 \oplus x = 1.0001$$ -and -$$z = y \ominus 1 = 0.0001000$$ -If we now denote the lost precision in $x$ by $s = 0.0000234$, and the remaining part by $\bar{x}$, then we can write -$$x = \bar{x} + s$$ -$$y = 1 + \bar{x}$$ -$$z = \bar{x}$$ -Now, the exact value of $\log(1+x)$ is -$$\log(1+x) = \log(1+\bar{x}+s) = \bar{x}+s + O(\bar{x}^2) = 0.0001234$$ -If we compute $\log(y)$ then we get -$$\log(1+\bar{x}) = \bar{x} + O(\bar{x}^2) = .0001000$$ -On the other hand, if we compute $x \times (\log(y)/z)$ then we get -$$(\bar{x}+s) \otimes (\log(1+ \bar{x}) \div \bar{x}) = (\bar{x}+s)(\bar{x}\div \bar{x}) = \bar{x}+s = 0.0001234$$ -so we keep the digits of precision that would have been lost without this correction.<|endoftext|> -TITLE: A Dirichlet Convolution involving $\mu(n)$ and $\log n$ -QUESTION [8 upvotes]: The following arithmetic identity holds: -\begin{align} -\Lambda(n) = \sum_{d \mid n} \mu(d) \log \frac{n}{d} -\end{align} -where $\mu(n)$ is the Moebius function and $\Lambda(n)$ is the von Mangoldt function. Does the following related Dirichlet convolution simplify to known (or simpler) functions? -\begin{align} -n \sum_{d \mid n} \frac{\mu(d)}{d} \log \frac{n}{d} -\end{align} - -REPLY [4 votes]: Yes,$$n \sum_{d \mid n} \frac{\mu(d)}{d} \log \frac{n}{d}=n \sum_{d \mid n} \frac{\mu(d)}{d} (\log(n)-\ln(d))$$ -$$= \sum_{d \mid n} \frac{\mu(d)}{d} n\log(n)-n\frac{\mu(d)}{d}\ln(d)=n\ln(n) \sum_{d \mid n} \frac{\mu(d)}{d} -n\sum_{d\mid n}\frac{\mu(d)}{d}\ln(d)$$ -$$=\ln(n)\phi(n)-n\sum_{d\mid n}\frac{\mu(d)}{d}\ln(d)=\ln(n)\phi(n)+\phi(n)\sum_{p\mid n}\frac{\ln(p)}{p-1}$$ -Thus, -$$n \sum_{d \mid n} \frac{\mu(d)}{d} \log \frac{n}{d}=\phi(n)( \ln(n)+\sum_{p\mid n}\frac{\ln(p)}{p-1})$$<|endoftext|> -TITLE: Distance in the Poincare Disk model of hyperbolic geometry -QUESTION [12 upvotes]: I am trying to understand the Poincare Disk model of a hyperbolic geometry and how to measure distances. I found the equation for the distance between two points on the disk as: -$d^2 = (dx^2 + dy^2) / (1-x^2-y^2)^2$ -Given two points on the disk, I am assuming that $dx$ is the difference in the euclidean $x$ coordinate of the points, and similar for $dy$. So, what are $x$ and $y$ in the formula? Also, I see some formulas online as: -$d^2 = 4(dx^2 + dy^2) / (1-x^2-y^2)^2$ -I am not sure which one is correct. -Assuming that $X$ and $Y$ are coordinates of one of the points, I have tried a few examples, but can't get the math to work out. For instance, if $A = (0,0)$, $B = (0,.5)$, $C = (0,.75)$, then what are $d(A,B)$, $d(B,C)$, and $d(A,C)$? The value $d(A,B) + d(B,C)$ should equal $d(A,C)$, since they are on the same line, but I can't get this to work out. I get distances of $.666$, $.571$, and $1.714$ respectively. -MathWorld - Hyperbolic Metric - -REPLY [11 votes]: As others have indicated one has to distinguish a metric $(x,y)\mapsto d(x,y)$ which measures distances between points $x$, $y$ in a space $X$ and a Riemannian metric which tells us how lengths of curves $\gamma$ in a manifold $X$ should be computed. -In the complex plane we have the usual euclidean metric $d(z_1,z_2):=|z_2-z_1|$ and at the infinitesimal scale the Riemannian metric -$$ds^2:=|dz|^2=dx^2+dy^2\ .$$ -The latter formula says that the length of an arbitrary curve -$$\gamma: \quad t\mapsto\bigl(x(t),y(t)\bigr)\qquad(a\leq t\leq b)$$ -should be computed as -$$L(\gamma)=\int_\gamma ds=\int_a^b\sqrt{x'^2(t)+y'^2(t)}\ dt = \int_a^b|z'(t)|\ dt\ .$$ -This formula implies that the length of a segment $\sigma$ connecting two points $z_1$ and $z_2$ is just $|z_2-z_1|$. -On the "Poincare disc" $P$ we are given a priori only a Riemannian metric -$$ds:= {|dz|\over 1-|z|^2}$$ -(resp., $ds^2=\ldots$). This metric allows us to compute the lengths of arbitrary curves in $P$: -$$L(\gamma)=\int_a^b{|z'(t)|\over1-|z(t)|^2}\ dt\ .$$ The particular definition of $ds$ is chosen such that this hyperbolic length is invariant under arbitrary conformal movements of $P$ and the curves therein. -A posteriori one can define a metric $d(\cdot,\cdot)$ on $P$ by letting the distance $d(z_1,z_2)$ between two points $z_1$, $z_2\in P$ be the hyperbolic length of the shortest curve $\gamma$ connecting $z_1$ and $z_2$. The actual carrying out of this idea shows that $d(z_1,z_2)$ can be written as an elementary function (using ${\rm artanh}$, etc.) in terms of $z_1$ and $z_2$.<|endoftext|> -TITLE: Combinatorial Interpretation of a Certain Product of Factorials -QUESTION [21 upvotes]: Let $\mu$ denote the Moebius function. What is a combinatorial interpretation of the following integer, -\begin{align} -\prod_{d \mid n} d!^{\,\mu(n/d)}, -\end{align} -where the product is taken over divisors of $n$? Does it have a simpler representation in terms of known functions? Note: The Online Encyclopedia of Integer Sequences does not have an entry containing the corresponding sequence. - -REPLY [4 votes]: This is an idea of approach, and it does not completely answer to the question. Define -$$ \mathcal{R}_n(x) = \{r \le x: (r,n) = 1\}$$ -and note that, by Inclusion-Exclusion principle -$$ | \mathcal{R}_n(x) | = \sum_{d \mid n} \mu(d) \left \lfloor \frac{x}{d} \right \rfloor $$ -Let $S_n$ be the sum in the text. Now, for every prime $p \le n$, consider $ V_p(S_n) $. -Recalling Polignac identity on $V_p(k!)$, we have -$$ V_p(S_n) = \sum_{d \mid n} \mu(n/d) V_p(d!) = \sum_{d \mid n} \mu(n/d) \sum_{k=1}^{\infty} \left \lfloor \frac{d}{p^k} \right \rfloor = \sum_{k=1}^{\infty} \sum_{d \mid n} \mu(d) \left \lfloor \frac{n}{dp^k} \right \rfloor = \sum_{k=1}^{\infty} |\mathcal{R}_n(n/p^k) | $$ -Take $r \in \mathcal{R}_n(n)$: the number of times it is counted in the last infinite sum is the number of positive integer solutions $k$ of $n/p^k \ge r$, i.e. $ \lfloor \log_p n/r \rfloor $. So -$$ \sum_{k=1}^{\infty} |\mathcal{R}_n(n/p^k)| = \sum_{r \in \mathcal{R}_n(n)} \lfloor \log_p (n/r) \rfloor $$ Define $q_p(x) := \max \{ p^n \ | \ p^n \le x\}$, and note that $ q_p(x) =p^{ \lfloor \log_p x \rfloor} $. -So, in conclusion: -$$ S_n = \prod_{p \le n} p^{V_p(S_n)} = \prod_{\substack{p \le n \\ r \in \mathcal{R}_n(n) }} p^{ \lfloor \log_p (n/r) \rfloor } = \prod_{\substack{r,p \le n \\(r,n)=1 }} q_p(n/r) $$ -I don't know, actually, if this can be useful; I think that it gives a combinatorial meaning to involved quantities, even if not much to the formula itself. -Hope it helps, Andrea -EDIT: I found an error in an inequality I used; so I can't actually state two bounds (upper and lower) of the same order. I would like to know more about this: Second-order asymptotics for $\pi(n), \theta(n)$<|endoftext|> -TITLE: Idempotents in $\mathbb Z_n$ -QUESTION [30 upvotes]: An element $a$ of the ring $(P,+,\cdot)$ is called idempotent if $a^2=a$. An idempotent $a$ is called nontrivial if $a \neq 0$ and $a \neq 1$. -My question concerns idempotents in rings $\mathbb Z_n$, with addition and multiplication modulo $n$, where $n$ is natural number. Obviously when $n$ is a prime number then there is no nontrivial idempotent. If $n$ is nonprime it may happen, for example $n=4, n=9$, that also there is no. - -Is it known, in general, for what $n$ there are nontrivial idempotents and what is a form of such idempotents? - -REPLY [2 votes]: Let $m=p^{c_{1}}_{1}...p^{c_{n}}_{n}$ be a prime factorization of an integer $m$ with $c_{i}\geq1$ and $p_{i}$ are distinct prime numbers. Then the ring $\mathbb{Z}/m\mathbb{Z}$ has $2^{n}$ idempotents and (modulo $m$) these are precisely of the form $\sum\limits_{k=1}^{n}h_{k}\epsilon_{k}$ where $\epsilon_{k}\in\{0,1\}$ and $h_{k}\in\left(\prod\limits_{\substack{i=1,\\ -i\neq k}}^{n}p^{c_{i}}_{i}\right)\mathbb{Z}$ such that $h_{k}-1\in p^{c_{k}}_{k}\mathbb{Z}$.<|endoftext|> -TITLE: Formal notation for a required statement -QUESTION [5 upvotes]: Is there a formal notation to distinguish an equality that is a true statement, e.g., - -[...] and hence, - $$ - x = x^2 - 1 - $$ - [...] - -from a demand, e.g., - -[...] so we require - $$ - x \stackrel{!}{=} x^2 - 1 -$$ - [...] - -? -The same thing could apply to membership to sets $x\stackrel{!}{\in}\mathbb{R}$ and more. -I've seen the exclamation mark syntax once, and I faintly remember having seen some other notation, but I'm not sure if any of this is commonly used. - -REPLY [3 votes]: Basically if I were you I would write the exclamation mark. If you fear that it is not understandable then remark in your text that this should specify that it is a demand and not a statement. -I think a lot of mathematicians use the exclamation mark in the sense you think of it. However, I don't think that it is "official" notation (like $e$ for the Euler number or so...). - -(Although it doesn't exactly go for the question itself as the following deals with statements rather than demands I want to shortly mention here an) -Interesting note: Frege (the founder of modern logic) introduced a special sign to indicate that what follows it is an assertion rather than a truth. This sign is "$\vdash$". -So he would write at the beginning of a proof: -$\vdash 1+1=2$ in $\mathbb{R}$. -To say that he states that $1+1=2$ in $\mathbb{R}$ is true. -Contrariwise -$1+1=2$ in $\mathbb{R}$, -is for Frege a truth value (or to be more accuarte: the Truth itself.) -Actually people in mathematical logic use this very sign to indicate tautologies. However, I don't really know if there is any connection between this use and Frege...<|endoftext|> -TITLE: Spectrum of a field -QUESTION [13 upvotes]: Let's $F$ be a field. What is $\operatorname{Spec}(F)$? I know that $\operatorname{Spec}(R)$ for ring $R$ is the set of prime ideals of $R$. But field doesn't have any non-trivial ideals. -Thanks a lot! - -REPLY [2 votes]: When people in modern algebraic geometry talk about $\operatorname{Spec}(k)$, the prime spectrum of a field $k$, they mean it in the sense of schemes. -Every scheme has an underlying topological space; in the case of $\operatorname{Spec}(k)$ this is a space with a single point, namely the zero ideal in $k$ (the only prime ideal in $k$). -But a scheme is more than just a topological space: it is a "ringed space" satisfying certain properties. In particular, there's a commutative ring assigned to every open set of the space. In our case of $\operatorname{Spec}(k)$, there are only two open sets: the empty set and the entire space. The ring assigned to the empty set is the trivial ring $\{0\}$, and the ring assigned to the entire space is our field $k$. -So in a sense, $\operatorname{Spec}(k)$ is a "single point that remembers the field it's defined over, namely $k$". This is useful in order to define more general schemes defined over $k$.<|endoftext|> -TITLE: Almost a perfect cuboid -QUESTION [7 upvotes]: While reading a very old book on diophantine equations, I came across this exercise: -Find an infinite number of positive integer solutions of the equations -$$x^2 + y^2 = u^2$$ -$$y^2 + z^2 = v^2$$ -$$z^2 + x^2 = w^2$$ -I have found a few solutions by hand, for example $x=240$, $y = 117$, $z = 44$, and trivially multiples will also produce solutions, but I assume that the book is really asking for solutions where there is no common factor of $x$, $y$, $z$. -I have spent a few hours trying to get something from the standard parametric solutions of $x^2 + y^2 = z^2$ without success, and wondered if anyone has any insight they can offer. -Clearly this is (slightly) connected with the problem of finding an integer cuboid with all the face diagonals integral, and the main diagonal also integral, which I assume is still an open problem. - -REPLY [2 votes]: There's an interesting aspect to Euler bricks that can be pointed out. The four smallest Euler bricks $1 -TITLE: A formula for the roots of a solvable polynomial -QUESTION [7 upvotes]: Let $F$ be a field and $p(x)\in F[x]$ a separable polynomial, denote -$K$ as the splitting field of $p$ and assume that $K/F$ is Galois -with a solvable Galois group. -I don't understand if this imply of any formula (in radicals) for -the roots of $p$ (however, I do understand how a formula would imply -that $p$ is solvable by roots). -Is there some kind of a way to obtain the roots of a solvable polynomial -? - -REPLY [8 votes]: Lagrange and Vandermonde (and others) knew how to treat various cases before "Galois theory", by "Lagrange resolvents". Some of my algebra notes show how this idea recovers the formula for the solution of cubics: http://www.math.umn.edu/~garrett/m/algebra/notes/23.pdf -By the late 19th century such ideas were well known, and in many cases of solvable Galois groups (though people were not quite able to say things so simply) this device produce solutions in radicals. -Expressing roots of unity in radicals is another example where Lagrange resolvents gives expressions in radicals. This quickly becomes computation-heavy, however. A slightly more sophisticated implementation of "Lagrange resolvents" that does bigger cyclotomic examples before getting bogged down is worked out at http://www.math.umn.edu/~garrett/m/v/kummer_eis.pdf In that case, some additional ideas of algebraic number theory are used. -Van der Waarden's "Algebra" (based on notes from E. Noether) was the place I saw Lagrange resolvents, and it was a revelation. Indeed, many treatments of "Galois theory" give no inkling of how to do any computations.<|endoftext|> -TITLE: Good book on evaluating difficult definite integrals (without elementary antiderivatives)? -QUESTION [39 upvotes]: I am very interested in evaluating difficult definite integrals without elementary antiderivatives by manipulating the integral somehow (e.g. contour integration, interchanging order of integration/summation, differentiation under the integral sign, etc.), especially if they have elegant solutions. However, I simply cannot seem to find a good book that covers many ways to evaluate these. The single book that I know that covers some good techniques is the Schaum's Advanced Calculus. -What is another good book that explains methods and techniques of integration of these fun integrals? - -REPLY [19 votes]: My favorite is a book of 1992 from Daniel Zwillinger : "Handbook of Integration" it is a "compilation of the most important methods" in 360 pages. Recommanded!<|endoftext|> -TITLE: Are minimal prime ideals in a graded ring graded? -QUESTION [10 upvotes]: Let $A=\oplus A_i$ be a graded ring. Let $\mathfrak p$ be a minimal prime in $A$. Is $\mathfrak p$ a graded ideal? - -Intuitively, this means the irreducible components of a projective variety are also projective varieties. When $A$ is Noetherian, I can give a proof, as follows. -There is some filtration of $A$, as an $A$ module, -$$0=M_0\subset M_1\subset\cdots\subset M_n=A$$ -such that $M_i/M_{i-1}\cong A/\mathfrak p_i$, for some graded prime ideal $\mathfrak p_i$. -Then I claim that the nilradical is $\cap\mathfrak p_i$. This is because -$$x^n=0 \Rightarrow x^nA=0 \Rightarrow x^nM_i\subset M_{i-1}\forall i \Rightarrow x^n\in \cap \mathfrak p_i \Leftrightarrow x\in \cap \mathfrak p_i $$ and -$$ x\in \cap \mathfrak p_i \Rightarrow xM_i\subset M_{i+1},\forall i \Rightarrow x^nA=0 \Leftrightarrow x=0.$$ -Hence the mininal primdes are just the minimal elements in $\{\mathfrak p_i\}$. -I would like to know if this assertion is still true if we drop the Noetherian condition, or if anyone has some more direct proofs. -Thanks! - -REPLY [9 votes]: Yes, the minimal primes of a graded ring are graded. If $\mathfrak{p}$ is any prime, then the ideal $\mathfrak{p}^h$ generated by the homogeneous elements of $\mathfrak{p}$ is also prime, and certainly graded. So if $\mathfrak{p}$ is minimal, $\mathfrak{p}=\mathfrak{p}^h$ is graded.<|endoftext|> -TITLE: Count the number of n-bit strings with an even number of zeros. -QUESTION [8 upvotes]: I am currently self-studying introductory combinatorics by reading Introduction to combinatorial mathematics. I am currently in the first chapter, and I have a question regarding one of the examples. The question was asking to count the number of n-bit strings with an even number of zeros. The answer is of course $2^{n-1}$. The author gave 2 solutions. I however didn't completely understand what I think is the straightforward one. The solution I got was that he took out 1 bit, leaving $(n-1)$ bits, if the number of zeros is even in the $(n-1)$-bit number, then he will just append a 1, if not then he will append a zero. So in the end we just needed to count the number of $(n-1)$-bit strings. The other solution (the straightforward one) that I didn't understand examined the symmetry that half of the $2^n$ must have an even number of zeros, and the other half will have an odd number of zeros. I just don't get why this property must hold. I can understand that half of the $2^n$ numbers will have even parity, but I can't see how it holds for the parity of the number of zero or one bits. If anyone can show me how that property holds, I'd be very grateful. I'd also be interested to see different explanations and proofs if possible. -Thank you. - -REPLY [8 votes]: One standard approach is that the number of strings with an even number of $0$'s is -$$\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots.$$ -The number with an odd number of $0$'s is -$$\binom{n}{1}+\binom{n}{3}+\binom{n}{5}+\cdots.$$ -Recall that by the Binomial Theorem, -$$(1+x)^n=\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+\binom{n}{3}x^3+\binom{n}{4}x^4+\cdots.$$ -Put $x=-1$. We get -$$0=\binom{n}{0}-\binom{n}{1}+\binom{n}{2}-\binom{n}{3}+\binom{n}{4}-\cdots.$$ -So the number with an even number of $0$'s is the same as the number with an odd number of $0$'s. -Overkill, but this generating functions approach can be used for other results. Not first chapter stuff, maybe second.<|endoftext|> -TITLE: root-finding methods to invert numerically a function -QUESTION [8 upvotes]: let be the equation $ y-f(x)=0 $ the idea is to get $ s=g(y) $ that is x as a function of 'y' -can this be made by a root finding algorithm ?? i mean you treat $ y $ as a numerical free parameter and find the roots of $ y-f(x)=0 $ in general these roots will depend on 'y' so we can represent every solution of $ y-f(x) $ for different parameters of 'y' -for example in Newton's method -$ x_{n+1} (y)=x_{n} (y)- \frac{y-f(x)}{-f'(x)} $ - -REPLY [4 votes]: The inverse of a single-variable function is a reflection over the line $y = x$. If you want the numerical inverse, just flip the coordinates. For example, -Function: -y = f(x); -plot(x,y); - -Inverse -y = f(x); -plot(y,x);<|endoftext|> -TITLE: Minimize sum of smallest and largest among integers on the real line. -QUESTION [5 upvotes]: Suppose there are 3 non-negative integers $x$, $y$ and $z$ on the real line. -We are told that $x + y + z = 300$. Without loss of generality, assume - $x$ to be the smallest integer, and $z$ to be the largest. -How do I minimize $(x + z)$? -Attempt: $x + z = 300 - y$, so for a start I should maximize y. This occurs at $y = z - 1$. So, we have $x + 2z = 301$. Now, $z = \dfrac {301}2 - \dfrac x2$. $\dfrac {dz}{dx} = -\dfrac 12$. Increasing $x$ by $1$ decreases $z$ only by $\dfrac12$. So, I should pick the smallest possible $x$, which is $1$. Then, $z = 150$. $\min (x+z) = 151$. -Questions - -Is my logic correct? -Is there a systematic way to solve questions of this kind? i.e. given non-negative numbers on the real line that sum up to a fixed value, how to minimize the sum of the largest and smallest of them? - -REPLY [2 votes]: Your logic is fine. Working with integers there are sometimes "end effects". You seem to be requiring that $y \ne z$ and an alternate solution is $(0,149,151)$ but that has the same sum of $x+z$. Without the restriction that $y \ne z$ you could have $(0,150,150)$ for a sum of $150$ -Your approach is quite systematic. If you had to have $7$ different non-negative integers sum to $300$ and wanted to minimize the sum of smallest plus largest, you would argue the same way-you want the middle ones to be as large as possible, so you have six numbers that add to $300$ (or a little less), so the average one is $50$, so they are $(47,48,49,50,51,52)$ and you need to add a $3$ to make $300$ and the sum is $3+52=55$. You don't really need the derivative here.<|endoftext|> -TITLE: Solving for unknown functions -QUESTION [9 upvotes]: I am not a mathematician, so excuse if my question is silly or badly stated. I have the following problem. I have 2 conditions on two unknown continuously differentiable functions: -$$A(t)=\frac{1}{B(t)}+C -\\ -B(t)=D-A(t)-\int_0^t A(\tau) d\tau.$$ -C and D are constants. I also know $A(0)$ and $B(0)$. I am looking for a way to get the value of $A(t)$ and $B(t)$ for small $t>0$. So far I have a numerical solution, but that involves a lot of interpolation and I don't think it is very good. -I was wondering if there is some way to get an analytic solution for this problem. I don't expect you to solve the problem for me, I'm willing to learn and I'd be very grateful if you could point me towards possible readings. -Thanks in advance. - -REPLY [3 votes]: UPDATED: Let's write everything in function of $E(t):=A(t)-C$ : -$$B(t)=\frac 1{E(t)}$$ $$\frac 1{E(t)}+E(t)=D-C-\int_0^t E(t)+C\ dt$$ -after derivation the second equation becomes : -$$E'(t)\left(1-\frac 1{E(t)^2}\right)=-E(t)-C$$ -$$\frac {E'(t)}{E(t)+C}-\frac {E'(t)}{(E(t)+C)E(t)^2}=-1$$ -$$\frac {E'(t)}{E(t)+C}-\frac {E'(t)}{C^2}\left(\frac {1}{E(t)+C}-\frac {E(t)-C}{E(t)^2}\right)=-1$$ -$$\frac {E'(t)}{E(t)+C}-\frac {E'(t)}{C^2}\left(\frac {1}{E(t)+C}-\frac 1{E(t)}+\frac C{E(t)^2}\right)=-1$$ -Integrating this we get : -$$\log(E(t)+C)-\frac 1{C^2}\log(E(t)+C)+\frac 1{C^2}\log(E(t))+ -\frac 1{CE(t)}=-t+F$$ -that we may rewrite as (changing sign) : -$$t-F=\frac {1-C^2}{C^2}\log(E(t)+C)-\frac 1{C^2}\log(E(t))- -\frac 1{CE(t)}$$ -with $F$ a constant and $E(t)=A(t)-C$ -This is the same result as countinghaus' except that I expressed $t$ in function of $E(t)=A(t)-C$ while his result was in function of $B(t)=\dfrac 1{E(t)}$. Let's rewrite our result with $B(t)$ : -$$t-F=\frac {1-C^2}{C^2}\log\left(\frac 1{B(t)}+C\right)+\frac 1{C^2}\log(B(t))- -\frac {B(t)}C$$ -$$\boxed{\displaystyle t=- -\frac {B(t)}C-\frac {C^2-1}{C^2}\log(1+C B(t))+\log(B(t))+F}$$ -Obtaining $A(t)$ (or $B(t)$) in function of $t$ in 'closed form' is probably not possible except for specific values of ($C$) : - -$C=1$ : solutions $B(t)=-1$ and $\displaystyle B(t)=-W\left(-e^{F-t}\right)$ -$C=-1$ : solutions $B(t)=1$ and $\displaystyle B(t)=W\left(e^{F-t}\right)$ -$C=0$ : solutions $\displaystyle B(t)=\pm\frac {\sqrt{W\left(-e^{F-2t}\right)}}i$ -(with $W$ the Lambert W function and $F$ a constant ; you may search other solutions using Alpha by changing the ($C=1$ in the example) constant before $x^2$ in the query and get the plot of the result)<|endoftext|> -TITLE: Calculate the expected value of $Y=e^X$ where $X \sim N(\mu, \sigma^2)$ -QUESTION [30 upvotes]: I got a problem of calculating $E[e^X]$, where X follows a normal distribution $N(\mu, \sigma^2)$ of mean $\mu$ and standard deviation $\sigma$. -I still got no clue how to solve it. Assume $Y=e^X$. Trying to calculate this value directly by substitution $f(x) = \frac{1}{\sqrt{2\pi\sigma^2}}\, e^{\frac{-(x-\mu)^2}{2\sigma^2}}$ then find $g(y)$ of $Y$ is a nightmare (and I don't know how to calculate this integral to be honest). -Another way is to find the inverse function. Assume $Y=\phi(X)$, if $\phi$ is differentiable, monotonic, and have inverse function $X=\psi(Y)$ then $g(y)$ (PDF of random variable $Y$) is as follows: $g(y)=f[\psi(y)]|\psi'(y)|$. -I think we don't need to find PDF of $Y$ explicitly to find $E[Y]$. This seems to be a classic problem. Anyone can help? - -REPLY [3 votes]: Since mgf of $X$ is $\;M_x(t)=e^{\large\mu t + \sigma^2 t^2/2}$ -$$E[Y]=E[e^X]=M_x(1)=e^{\large\mu + \sigma^2/2}$$<|endoftext|> -TITLE: recovering a representation from its character -QUESTION [5 upvotes]: The character of a representation of a finite group or a finite-dimensional Lie group determines the representation up to isomorphism. -Is there an algorithmic way of recovering the representation given the character and the conjugacy structure of the group? - -REPLY [3 votes]: Below by "representation" I mean "finite-dimensional complex continuous representation." -This is false for Lie groups in general; for example, the character of a representation of $\mathbb{R}$ does not distinguish the $2$-dimensional trivial representation from the representation -$$r \mapsto \left[ \begin{array}{cc} 1 & r \\ 0 & 1 \end{array} \right].$$ -(Technically speaking, any countable discrete group is also a Lie group as it is a $0$-dimensional manifold for any reasonable definition of manifold, but now I'm being far too picky.) -The correct statement is for compact groups (not necessarily Lie). It suffices to address the problem for irreducible representations. If $G$ is such a group, $\mu$ is normalized Haar measure, and $\chi_V$ is the character of an irreducible representation $V$, let $L_g : L^2(G) \to L^2(G)$ be the map which translates a function by $g$ (that is take $L_g(f(h)) = f(g^{-1}h)$). Then -$$\dim(V) \int_G \overline{ \chi_V(g) } L_g \, d \mu$$ -is a projection $L^2(G) \to L^2(G)$ whose image is the $V$-isotypic component of $L^2(G)$ (exercise). Taking any nonzero vector in this image and applying suitably many elements of $G$ to it will give you explicit vectors spanning a subspace of $L^2(G)$ isomorphic to $V$; turn these into a basis, and then you get explicit matrices for the elements of $G$. -Since $L^2(G)$ is a bit large, an alternate method (which only works for Lie groups) is to start with a faithful representation $W$ and take a tensor power $W^{\otimes n} (W^{\ast})^{\otimes m}$ containing $V$ (this is always possible, see this MO question), then apply the projection -$$\dim(V) \int_G \overline{\chi_V(g)} g \, d \mu.$$<|endoftext|> -TITLE: When does a normal subgroup contain precisely one non-identity conjugacy class? -QUESTION [9 upvotes]: Every normal subgroup $N$ of a group $G$ is a union of conjugacy classes. Since every subgroup contains the identity, and the identity is in a class by itself, every normal subgroup already contains the conjugacy class of the identity. -So when is a normal subgroup comprised of exactly two conjugacy classes? -$N = \{1\} \cup \mathcal K$ -Here is what I see so far: - -Unless $|N|=2$ and $N \leq Z(G)$, the subgroup must have trivial intersection with the center, since each element in the center is contained in its own conjugacy class. -Since $|\mathcal K|$ is the index of the centralizer $C_G(k)$ of any $k\in\mathcal K$, and $|G:C_G(k)| = |N|-1$ divides $|G|$, we must have that G is divisible by the product $|N|(|N|-1)$ of two consecutive numbers. This also implies $|G|$ is even. - -Any inner automorphism fixes $N$, but I don't know about outer automorphisms, so $N$ may not have to be a characteristic subgroup. -What is the full characterization of these types of normal subgroups? Do they have any important properties? -Edit: Ted is correct. - -REPLY [6 votes]: All groups are finite. The following is an old result of Wielandt, I believe. -Proposition: Such a subgroup $N$ must be an elementary abelian $p$-group, and every elementary abelian $p$-group is such a subgroup in a certain group $G$. -Proof: If two elements of $G$ are conjugate, then they have the same order. Hence every non-identity element of $N$ has the same order. The order cannot be composite since $g^a$ has order $b$ if $g$ has order $ab$. Hence $N$ is a $p$-group. The commutator subgroup of $N$ is characteristic in $N$ and so normal in $G$. Hence it is either all of $N$ or just $1$; however, in a non-identity $p$-group the commutator subgroup is always a proper subgroup. In particular, $N$ is abelian and every element has order $p$. -Now suppose such an elementary abelian group $N$ is given. Let $G=\operatorname{AGL}(1,p^n)$ be the set of affine transformations of the one-dimensional vector space $K$ over the field $K$ of $p^n$ elements. That is $G$ consists of all $\{ f : K \to K :x \mapsto \alpha x + \beta ~\mid~ \alpha,\beta \in K, \alpha \neq 0 \}$. Then every non-identity element of $N=K_+=\{ f : K \to K : x \mapsto x + \beta ~\mid ~ \beta \in K \}$ is conjugate under $K^\times = \{ f : K \to K : x \mapsto \alpha x ~\mid~ \alpha \in K, \alpha \neq 0 \}$. Thus the proof is complete. $\square$.<|endoftext|> -TITLE: Convergence of harmonic functions. -QUESTION [7 upvotes]: I am looking for the proof of the following Theorem : Does anyone know where i can find out ? -If $\Omega$ is open and connected and $u_k$ be uniformly bounded sequence of harmonic functions . There exists a subsequence that converges uniformly to a harmonic function $u:\Omega \to \mathbb R$ on any compact subset of $\Omega$. -In case you think that its not hard, i look forward to hints as well. -I hope the statement is true . -Thanks. - -REPLY [2 votes]: You can see this book -D.H. Armitage, S.J. Gardiner, Classical Potential Theory, Springer, London, 2000. -proposition 1.5.11. This will exactly answer your question. -This is a link of this book in booksgoogle -Enjoy.<|endoftext|> -TITLE: Is First Order Logic (FOL) the only fundamental logic? -QUESTION [33 upvotes]: I'm far from being an expert in the field of mathematical logic, but I've been reading about the academic work invested in the foundations of mathematics, both in a historical and objective sense; and I learned that it all seems to reduce to a proper -axiomatic- formulation of set theory. -It also seems that all set theories (even if those come in ontologically different flavours, such as the ones which pursue the "iterative approach" like ZFC, versus the "stratified approach" -inspired by Russell's and Whitehead's type theory first formulated in their Principia- such as Quine's NFU or Mendelson's ST) are built as collections of axioms expressed in a common language, which invariably involves an underlying first order predicate logic augmented with the set-membership binary relation symbol. From this follows that FOL makes up the (necessary) "formal template" in mathematics, at least from a foundational perspective. -The justification of this very fact, is the reason behind this question. All the stuff I've read about the metalogical virtues of FOL and the properties of its "extensions" could be summarized as the statements below: - -FOL is complete (Gödel, 1929), compact and sound, and all its particular formalizations as deductive systems are equivalent (Lindström, 1969). That means that, given a (consistent) collection of axioms on top of a FOL deductive system, the set of all theorems which are syntactically provable, are semantically satisfied by a model of the axioms. The specification of the axioms absolutely entails all its consequences; and the fact that every first order deductive system is equivalent, suggests that FOL is a context-independent (i.e. objective), formal structure. -On the other hand, the Löwenheim–Skolem theorem implies that FOL cannot categorically characterize infinite structures, and so every first order theory satisfied by a model of a particular infinite cardinality, is also satisfied by multiple additional models of every other infinite cardinality. This non-categoricity feature is explained to be caused by the lack of expressive power of FOL. -The categoricity results that FOL-based theories cannot achieve, can be obtained in a Second Order Logic (SOL) framework. Examples abound in ordinary mathematics, such as the Least Upper Bound axiom, which allows the definition of the real number system up to isomorphism. Nevertheless, SOL fails to verify an analog to the completeness results of FOL, and so there is no general match between syntactic provability and semantic satisfiability (in other words, it doesn't admit a complete proof calculus). That means that, even if a chosen collection of axioms is able to categorically characterize an infinite mathematical structure, there is an infinite set of wff's satisfied by the unique model of the axioms which cannot be derived through deduction. -The syntactic-semantic schism in SOL also implies that there is no such a thing as an equivalent formulation of potential deductive systems, as is the case in FOL and stated by Lindström's theorem. One of the results of this fact is that the domain over which second order variables range must be specified, otherwise being ill-defined. If the domain is allowed to be the full set of subsets of the domain of first order variables, the corresponding standard semantics involve the formal properties stated above (enough expressive power to establish categoricity results, and incompleteness of potential, non-equivalent deductive systems). On the other hand, through an appropriate definition of second order domains for second order variables to range over, the resultant logic exhibits nonstandard semantics (or Henkin semantics) which can be shown to be equivalent to many-sorted FOL; and as single-sorted FOL, it verifies the same metalogical properties stated at the beginning (and of course, its lack of expressive power). -The quantification extension over variables of successive superior orders can be formalized, or even eliminate the distinction between individual (first order) variables and predicates; in each case, is obtained -for every N- an Nth Order Logic (NOL), and Higher Order Logic (HOL), respectively. Nevertheless, it can be shown (Hintikka, 1955) that any sentence in any logic over FOL with standard semantics to be equivalent (in an effective manner) to a sentence in full SOL, using many-sorting. -All of this points to the fact that the fundamental distinction, in logical terms, lies between FOL (be it single-sorted or many-sorted) and SOL (with standard semantics). Or what seems to be the case, the logical foundations of every mathematical theory must be either non-categorical or lack a complete proof calculus, with nothing in between that trade-off. - -Why, so, is FOL invariably chosen as the underlying logic on top of which the set theoretical axioms are established, in any potentially foundational formalization of mathematics? -As I've said, I'm not an expert in this topic, and I just happen to be interested in these themes. What I wrote here is a summary of what I assume I understood of what I read (even though I'm personally inclined against the people who speaks about what they don't fully understand). In this light, I'd be very pleased if any answer to this question involves a rectification of any assertion which happened to be wrong. -P.S. : this is an exact repost of the question I originally asked at Philosophy .SE, because I assumed this to be an overly philosophical matter, and so it wouldn't be well received by the mathematics community. The lack of response there (be it because I wrong, and this actually makes up a question which can only be answered with a technical background on the subject, or because it's of little philosophical interest) is the reason why I decided to ask it here. Feel free to point out if my original criteria was actually correct, and of course, I'll take no offense if any moderator takes actions because of the probable unsuitability of the question in this site. - -REPLY [5 votes]: I have not an answer, just an additional note about this. -In 2000, John Alan Robinson (known for joining the "cut" logic inference rule and unification into the Resolution logic inference rule, thus giving logic programming its practical and unifying processing principle) authored an eminently readable overview of the "computational logic" research domain: "Computational Logic: Memories of the Past and Challenges for the Future". On page 2 of that overview, he wrote the following: - -First Order Predicate Calculus: All the Logic We Have and All the Logic We Need. -By logic I mean the ideas and notations comprising the classical first order -predicate calculus with equality (FOL for short). FOL is all the logic we have and all the logic we need. (...) -Within FOL we are completely free to postulate, by formulating suitably axiomatized first order theories, whatever more exotic constructions we may wish to contemplate in our ontology, or to limit ourselves to more parsimonious means of inference than the full classical repertoire. -The first order theory of combinators, for example, provides the semantics of the lambda abstraction notation, which is thus available as syntactic sugar for a deeper, first-order definable, conceptual device. -Thus FOL can be used to set up, as first order theories, the many “other logics” such as modal logic, higher order logic, temporal logic, dynamic logic, concurrency logic, epistemic logic, nonmonotonic logic, relevance logic, linear logic, fuzzy logic, intuitionistic logic, causal logic, quantum logic; and so on and so on. -The idea that FOL is just one among many "other logics" is an unfortunate source of confusion and apparent complexity. The "other logics" are simply notations reflecting syntactically sugared definitions of -notions or limitations which can be formalized within FOL. (...) All those “other logics”, including higher-order logic, are thus theories formulated, like general set theory and indeed all of mathematics, within FOL. - -So I wanted to know what he meant by this and whether this statement can be defended. This being Robinson, I would assume that yes. I am sure there would be domains where expressing the mathematical constraints using FOL instead of something more practical would be possible in principle, but utterly impractical in fact. -The question as stated is better than I could have ever have formulated (because I don't know half of what is being mentioned). Thank you! -Addendum (very much later) -In The Emergence of First-Order Logic, Gregory H. Moore explains (a bit confusingly, these papers on history need timelines and diagrams) how mathematicians converged on what is today called (untyped) FOL from initially richer logics with Second-Order features. In particular, to formalize Set Theory. -The impression arises that FOL is not only non-fundamental in any particular way but objectively reduced in expressiveness relative to a Second-Order Logic. It has just been studied more. -Thus, Robinson's "FOL as all of logic with anything beyond reducible to FOL" sounds like a grumpy and ill-advised statement (even more so as the reductions, if they even exist, will be exponentially large or worse). Robinson may be referring to Willard Van Orman Quine's works. Quine dismisses anything outside of FOL as "not a logic", which to me is frankly incomprehensible. Quine's attack on Second-Order Logic is criticized by George Boolos in his 1975 paper "On Second-Order Logic" (found in "Logic, Logic and Logic"), which is not available on the Internet, but searching for that paper brings up other hits of interest, like "Second-Order Logic Revisited" by Otávio Bueno. In any case, I don't understand half of the rather fine points made. The fight goes on. -Let me quote liberally from the conclusion of Gregory Moore's fine paper: - -As we have seen, the logics considered from 1879 to 1923 — such as -those of Frege, Peirce, Schröder, Löwenheim, Skolem, Peano, and -Russell — were generally richer than first-order logic. This richness -took one of two forms: the use of infinitely long expressions (by -Peirce, Schröder, Hilbert, Löwenheim, and Skolem) and the use of a -logic at least as rich as second-order logic (by Frege, Peirce, -Schröder, Löwenheim, Peano, Russell, and Hilbert). The fact that no -system of logic predominated — although the Peirce-Schröder tradition -was strong until about 1920 and Principia Mathematica exerted a -substantial influence during the 1920s and 1930s — encouraged both -variety and richness in logic. -First-order logic emerged as a distinct subsystem of logic in -Hilbert's lectures (1917) and, in print, in (Hilbert and Ackermann -1928). Nevertheless, Hilbert did not at any point regard first-order -logic as the proper basis for mathematics. From 1917 on, he opted for -the theory of types — at first the ramified theory with the -Axiom of Reducibility and later a version of the simple theory of -types ($\omega$ - order logic). Likewise, it is inaccurate to regard -what Löwenheim did in (1915) as first-order logic. Not only did he -consider second-order propositions, but even his first-order subsystem -included infinitely long expressions. -It was in Skolem's work on set theory (1923) that first-order logic -was first proposed as all of logic and that set theory was first -formulated within first-order logic. (Beginning in [1928], Herbrand -treated the theory of types as merely a mathematical system with an -underlying first-order logic.) Over the next four decades Skolem -attempted to convince the mathematical community that both of his -proposals were correct. The first claim, that first-order logic is all -of logic, was taken up (perhaps independently) by Quine, who argued -that second-order logic is really set theory in disguise (1941). This -claim fared well for a while. After the emergence of a distinct -infinitary logic in the 1950s (thanks in good part to Tarski) and -after the introduction of generalized quantifiers (thanks to -Mostowski [1957]), first-order logic is clearly not all of logic. -Skolem's second claim, that set theory should be formulated in -first-order logic, was much more successful, and today this is how -almost all set theory is done. -When Gödel proved the completeness of first-order logic (1929, 1930) -and then the incompleteness of both second-order and co-order logic -(1931), he both stimulated first-order logic and inhibited the growth -of second-order logic. On the other hand, his incompleteness results -encouraged the search for an appropriate infinitary logic—by Carnap -(1935) and Zermelo (1935). The acceptance of first-order logic as one -basis on which to formulate all of mathematics came about gradually -during the 1930s and 1940s, aided by Bernays's and Gödel's first-order -formulations of set theory. -Yet Maltsev (1936), through the use of uncountable first-order -languages, and Tarski, through the Upward Löwenheim-Skolem Theorem and -the definition of truth, rejected the attempt by Skolem to restrict -logic to countable first-order languages. In time, uncountable -first-order languages and uncountable models became a standard part of -the repertoire of first-order logic. Thus set theory entered logic -through the back door, both syntactically and semantically, though it -failed to enter through the front door of second-order logic. - -There sure is space for a few updated books in all of this.<|endoftext|> -TITLE: 4 by 4 Matrix Puzzle -QUESTION [11 upvotes]: I was solving the puzzle for the Company interview exam. I found this puzzle, I cannot come up with the solution. How to solve it and what is the correct answer? - -Determine the number of $4\times 4$ matrices having all entries 0 or 1 that have an odd number of $1$s in each row and each column. - -REPLY [11 votes]: Fill the upper-left hand $3\times 3\,$ arbitrarily with $0$'s and/or $1$'s. This can be done in $2^9$ ways. -For any such choice of $0$'s and/or $1$'s, fill in the first three entries in the fourth row, and the first three entries in the fourth column, so that the number of $1$'s in each of the first three columns, and in each of the first three rows, is odd. This can be done in precisely one way. -Now put a $0$ or a $1$ in the lower right-hand corner, to make the number of $1$'s in the bottom row odd. It turns out that this makes the number of $1$'s in the rightmost column odd. To check this, work modulo $2$.<|endoftext|> -TITLE: Conditional and joint probability manipulations when there are 3 variables -QUESTION [46 upvotes]: I'm having trouble verifying why the following is correct. -$$p(x, y \mid z)= p(x \mid y, z) p(y \mid z)$$ -I tried grouping the $(x, y)$ together and split by the conditional, which gives me -$$p(x, y \mid z) = p(z\mid x, y) p(x, y)/p(z)$$ -However, this did not bring me any closer. I'm uncertain about what kind of manipulations are allowed given more than 2 variables. -Say an expression like: -$$p(a, b, c)$$ -Then I know from the chain rule that I can break it down to: -$$p(a, b, c)=p(a \mid b, c) p(b, c) = p(a \mid b, c) p(b \mid c) p(c)$$ -Is it allowed to split by the second comma: -$$p(a, b, c) = p(a, b \mid c) p(c) ?$$ -And even more complicated and expression like: -$$p(a|b,c)$$ -Am I allowed to rewrite this expression by grouping (a|b) together to give me something like -$$p(a|b,c)=p((a|b)|c)p(c)$$ -And does this expression even make sense? - -REPLY [44 votes]: $\Pr(a,b,c)=\Pr(a,b\mid c)\Pr(c)$ is allowed. -You are simply saying $\Pr(d,c)=\Pr(d\mid c)\Pr(c)$ where $d = a \cap b$. -Combine this with $\Pr(a,b,c)=\Pr(a\mid b,c)\Pr(b,c)=\Pr(a\mid b,c)\Pr(b\mid c)\Pr(c)$ and divide through by nonzero $\Pr(c)$ to get $\Pr(a,b\mid c)=\Pr(a\mid b,c)\Pr(b\mid c)$.<|endoftext|> -TITLE: Two questions on product measures -QUESTION [6 upvotes]: Do you know where to find references on the following questions (I suspect they are well-known)? -1) The product $\sigma$-algebra is defined to be the smallest $\sigma$-algebra containing all the "measurable rectangles." Is it possible to write every element in the product $\sigma$-algebra as a sequence of countable unions and intersections of the measurable rectangles? (I suspect the answer is no, but I can't think of a good reason why.) -(More precisely, given two $\sigma$-algebras $\mathcal{A}$ and $\mathcal{B}$, is it possible to an arbitrarily element of $\mathcal{A}\times \mathcal{B}$ as $\bigcup_{i_{1}=1}^{\infty}\bigcap_{i_{2}=1}^{\infty}\cdots\bigcap_{i_{n}}^{\infty}A_{i_{1},\ldots,i_{n}}\times B_{i_{1},\ldots,i_{n}}$ for some $n$ and all $A_{i_{1},\ldots,i_{n}}\in\mathcal{A},\ B_{i_{1},\ldots,i_{n}}\in\mathcal{B}$?) -2) Given two independent $\sigma$-algebras $\mathcal{A}$ and $\mathcal{B}$ (with respect to a measure $\mu$), why is not possible describe the measure $\mu$ on $\sigma(\mathcal{A}\cup\mathcal{B})$ as a product measure $\mathcal{A}\times \mathcal{B}$ (where we would define the measure of a rectangle $A\times B$ for $A\in\mathcal{A}$ and $B\in\mathcal{B}$ as $\mu(A\cap B)=\mu(A)\mu(B)$)? -Again, it makes sense that there are lot more ways to construct independent $\sigma$-algebras than taking products, but I feel like I don't understand a lot if I can't distinguish between the two. - -REPLY [2 votes]: Consider the Cantor Set. It is compact, and therefore measurable. Let $\mathcal{A}$ denote the half-open (right closed, left open) intervals, along with the null set. Then $\mathcal{A}$ is an algebra of sets. $\sigma(\mathcal{A})$ is the Borel sets. -However, the Cantor set is totally disconnected and uncountable. It is not a countable intersection or union of elements of $\mathcal{A}$. The appearance of $\sigma(\mathcal{A})$ in terms of $\mathcal{A}$ is subtle. -Consider a Cantor Set constructed along the line $y=x$ in $[0,1]^2$.<|endoftext|> -TITLE: When is a quasi-projective variety affine? -QUESTION [7 upvotes]: By an affine variety I mean a variety that is isomorphic to some irreducible algebraic set in $\mathbb A^n$ and by a quasi-projective variety I mean a locally closed subset of $\mathbb P^n$, with the usual Zariski topology and structure sheaf. (I am not quite familiar with the language of schemes.) -Assume that the underlying field is algebraically closed. -Let $X$ be a quasi-projective variety. Are there any effective ways to tell whether $X$ is affine? -Let me be more specific. If $X$ is affine, then $\mathcal O_X(X)$ is large enough so that it completely determines the variety. In particular, the Nullstellensatz holds, i.e. the map -$$X\to \mathrm {spm}(\mathcal O_X(X))$$ -$$x\mapsto \{f\in\mathcal O_X(X)|f(x)=0\}$$ -is a 1-1 correspondence, where $\mathrm{spm}(\mathcal O_X(X))$ denotes the maximal ideals in $\mathcal O_X(X)$. -Now I'd like to pose the question: If $X$ is quasi-projective and the above map is a bijection, is $X$ necessarily affine? -Thanks! - -REPLY [9 votes]: If the map from $X$ to the maxspec of $\mathcal O(X)$ is a bijection, then $X$ is indeed affine. -Here is an argument: -By assumption $X \to $ maxspec $\mathcal O(X)$ is bijective, thus quasi-finite, -and so by (Grothendieck's form of) Zariski's main theorem, this map factors as an open embedding of $X$ into a variety that is finite over maxspec $\mathcal O(X)$. -Any variety finite over an affine variety is again affine, and hence $X$ is an open subset of an affine variety, i.e. quasi-affine. So we are reduced to considering the case when $X$ is quasi-affine, which is well-known and straightforward. -(I'm not sure that the full strength of ZMT is needed, but it is a natural tool -to exploit to get mileage out of the assumption of a morphism having finite fibres, which is what your bijectivity hypothesis gives.) - -In fact, the argument shows something stronger: suppose that we just assume -that the morphism $X \to $ maxspec $\mathcal O(X)$ has finite non-empty fibres, -i.e. is quasi-finite and surjective. -Then the same argument with ZMT shows that $X$ is quasi-affine. But it is standard that the map $X \to $ maxspec $\mathcal O(X)$ is an open immersion when $X$ is quasi-affine, -and since by assumption it is surjecive, it is an isomorphism. -Note that if we omit one of the hypotheses of surjectivity or quasi-finiteness, we can find a non-affine $X$ satisfying the other hypothesis. -E.g. if $X = \mathbb A^2 \setminus \{0\}$ (the basic example of a quasi-affine, -but non-affine, variety), then maxspec $\mathcal O(X) = \mathbb A^2$, and the open immersion $X \to \mathbb A^2$ is evidently not surjective. -E.g. if $X = \mathbb A^2$ blown up at $0$, then maxspec $\mathcal O(X) = -\mathbb A^2$, and $X \to \mathbb A^2$ is surjective, but has an infinite fibre -over $0$. - -Caveat/correction: I should add the following caveat, namely that it is not always true, for a variety $X$ over a field $k$, that $\mathcal O(X)$ is finitely generated over $k$, in which case maxspec may not be such a good construction to apply, and the above argument may not go through. So in order to conclude that $X$ is affine, one should first insist that $\mathcal O(X)$ is finitely generated over $k$, and then that futhermore the natural map $X \to $ maxspec $\mathcal O(X)$ is quasi-finite and surjective. -(Of course, one could work more generally with arbitrary schemes and Spec rather than -maxspec, but I haven't thought about this general setting: in particular, ZMT requires some finiteness hypotheses, and I haven't thought about what conditions might guarantee that the map $X \to $ Spec $\mathcal O(X)$ satisfies them.) -Incidentally, for an example of a quasi-projective variety with non-finitely generated ring of regular functions, see this note of Ravi Vakil's<|endoftext|> -TITLE: Keep getting generating function wrong (making change for a dollar) -QUESTION [5 upvotes]: Possible Duplicate: -Making Change for a Dollar (and other number partitioning problems) - -I am working on the classic coin problem where I would like to calculate the number of ways to make change for a dollar with a given number of denominations. From here, I am also going to be working on how to partition the number $100$ with at least two positive integers below $100$. -I have read over all the same posts on here and other sites and still I am unsure of what I am doing wrong. - -The answer to our problem ($293$) is the coefficient of $x^{100}$ in the - reciprocal of the following: -$$(1-x)(1-x^5)(1-x^{10})(1-x^{25})(1-x^{50})(1-x^{100}).$$ - -I do out this equation with $x = 100$ and get a really large return. My skills are very limited and most of the sites I have been reading over use terminology and operators/symbols I am unfamiliar with. -I look at this and think it seems very straight forward but, I get answers that way off. Are there any tips or steps that I could be overlooking? - -REPLY [12 votes]: Here is how you can compute the coefficient you are after without using symbolic algebra software, just any programming language where you can handle an array of $101$ integers; I'll assume for convenience that the array is called $c$ and is indexed as $c[i]$ where $i$ ranges from $0$ to $100$, because then $c[i]$ records the coefficient of $x^i$ in a power series. -To multiply such a power series by a binomial of the form $(1-ax^k)$ (with $k>0$) is easy: the coefficient $c[i]$ must be subtracted $a$ times from $c[i+k]$ for every $i$ for which this is possible (so $i+k\leq100$). One must make sure that the old value of $c[i]$ is being subtracted from $c[i+k]$, which can be achieved (in sequential operation) by traversing the values of $i$ in decreasing order. -To divide such a power series by a binomial of the form $(1-ax^k)$ is the inverse operation, which turns out to be even easier. The inverse of subtracting $a*c[i]$ from $c[i+k]$ is adding $a*c[i]$ to $c[i+k]$, and this must be performed for all $i$ in reverse order to what was done for multiplication, namely in increasing order. -So here is schematically the computation you need to perform: - -Initialise your array so that $c[0]=1$ and $c[i]=0$ for all $i>0$. -For $k=1,5,10,25,50,100$ (in any order; multiplication is commutatitve) do: - -for $i=0,1,\ldots,100-k$ (in this order) do: - -add $c[i]$ to $c[i+k]$. - - -Now $c[100]$ gives your answer. - -This computation gives you the coefficient of $x^{100}$ in the power series for $1/((1-x)(1-x^5)(1-x^{10})(1-x^{25})(1-x^{50})(1-x^{100}))$. -Note however that your question asks to "partition the number $100$ with at least two positive integers below $100$", and to find that, one should omit the case $k=100$ from the outer loop, as this contributes (only) the partition of $100$ into a single part equal to $100$. The result with this modification is $292$.<|endoftext|> -TITLE: Approximation of bounded measurable functions with continuous functions -QUESTION [8 upvotes]: This is not homework. I was reading a paper where the authors showed a result for all continuous functions and then just proceeded to write "the usual limiting Argument gives the result for all bounded functions" - so I am asking myself what this "usual limiting argument" might be. I do not know whether they mean uniform or pointwise convergence. As I see it pointwise convergence should suffice :D -Thus I was am wondering whether there is a theorem having or leading to the following statement: - -Let $K\subset\mathbb{R}^2$ be compact. Any bounded measurable function $f:K\to\mathbb{R} $ can be approximated by a sequence of continuous functions $(g_m)$ on $K$. - -Nate Eldredge suggested that I post some excerpt from the original to provide more context for the problem. Here I go: -The goal is to proof the existence of a weak limit for a tight sequence of probability measures on $\mathcal{C}^0([0,1]^2,\mathbb{R})$ associated with reflecting Brownian Motions on the compact the set $[0,1]^2$ which is a Lipshitz Domain. Thus we already, know that some weak limit must exist and it remains to show that two limit-Points agree. Weak-Convergence is generally defined via bounded measurable functions. Now let $P'$ and $P''$ be two subsequential limit points. The authors show that $f \in \mathcal{C}^0([0,1]^2,\mathbb{R})$ the following holds -(here $X_s$ the canonical process) - -$E'f(X_s)=E''f(X_s)$ - -And now comes the actual source of my question: -"The usual limiting argument gives the result for bounded $f$ and hence the one-dimensional distributions agree." (the second part I understand only the "standard limiting argument thing" is somewhat confusing) -Any Help is much appreciated and Thanks in Advance :D - -REPLY [4 votes]: To respond to your comment on Byron's answer: -The functional monotone class theorem is a very useful result and well worth knowing. However, you can also get this result with arguments that may be more familiar. To recap, we want to show: - -Suppose $\mu', \mu''$ are two probability measures on $\mathbb{R}$, and we have $\int f\,d\mu' = \int f\,d\mu''$ for all bounded continuous $f$. Then $\mu' = \mu''$. - -One could proceed as follows: - -Exercise. For any open interval $(a,b)$, there is a sequence of nonnegative bounded continuous functions $f_n$ such that $f_n \uparrow 1_{(a,b)}$ pointwise. - -(For example, some trapezoidal-shaped functions would work.) -If $f_n$ is such a sequence, we have $\int f_n \,d\mu' = \int f_n \,d\mu''$ for each $n$. By monotone convergence, the left side converges to $\int 1_{(a,b)}\,d\mu' = \mu'((a,b))$ and the right side converges to $\mu''((a,b))$. So $\mu'((a,b)) = \mu''((a,b))$, and this holds for any interval $(a,b)$. -Now you can use Dynkin's $\pi$-$\lambda$ lemma, once you show: - -Exercise. The collection - $$\mathcal{L} := \{B \in \mathcal{B}_\mathbb{R} : \mu'(B) = \mu''(B)\}$$ - is a $\lambda$-system. (Here $\mathcal{B}_{\mathbb{R}}$ is the Borel $\sigma$-algebra on $\mathbb{R}$.) - -We just showed that the open intervals are contained in $\mathcal{L}$. But the open intervals are a $\pi$-system which generates $\mathcal{B}_{\mathbb{R}}$. So by Dynkin's lemma, $\mathcal{B}_\mathbb{R} \subset \mathcal{L}$, which is to say $\mu' = \mu''$.<|endoftext|> -TITLE: Rain droplets falling on a table -QUESTION [41 upvotes]: Suppose you have a circular table of radius $R$. This table has been left outside, and it begins to rain at a constant rate of one droplet per second. The drops, which can be considered points as they fall, can only land in such a way such that they impact the surface of the table. Once they strike the table, they form a puddle of radius $r$, centered at their point of impact. What is the expected number of droplets it takes to cover the table in water? -The answer should be left in terms of $R$ and $r$. However, if you can simulate this, while changing both $r$ and $R$ so that some regression might be applied to find the approximate relation, that would be nice as well. I'd really like some intuition as to how the two are related. -I have tried decomposing the problem by considering only the 1-dimensional case with line segments, but even its solution has eluded me. A potential starting point could be the discrete case of marbles falling into buckets. -The following edit was suggested by Jbeuh: -A more formal restatement of the problem : -Consider a sequence $X_1,\ldots,X_n$ of independent random variables following a uniform distribution on a disk $D$ of radius $R$. For each $i$, let $C_i$ be the disk centered at $X_i$ with radius $r$. Let $Y$ be the first $i$ such that the disks $C_1,\ldots,C_i$ forms a cover of $D$. What is the expected value of $Y$? - -REPLY [9 votes]: For any covering of the target region with shapes of diameter $\le r$, if one droplet center lies within each shape, this is sufficient to cover the entire region with droplets. This can be used to calculate an upper bound on the expected time to cover the table. Note that this upper bound will be tighter when the number of shapes is smaller; i.e., we want the area of each shape to be as large as possible. Suppose the covering uses regular hexagons of diameter $r$: these have area $3r^2\sqrt{3}/8$, so there will be about -$$ -N_{\rm hex} = \frac{8\pi}{3\sqrt{3}}\left(\frac{R}{r}\right)^2 -$$ -of them. The coupon collector's problem then states that the expected time to hit all the hexagons is $E[T_{\rm hex}] \sim N_{\rm hex}\ln N_{\rm hex}$. -On the other hand, for any packing of the target region with disks of radius $r$, it is necessary for at least one droplet center to fall within each disk in order for the target region to be covered (in particular, the disk centers cannot be covered otherwise). This can be used to provide a lower bound on the expected time. Here we again use a hexagonal packing, which covers a fraction $\pi/\sqrt{12}$ of the table's surface, but now the hexagon area is $2r^2\sqrt{3}$. The number of disks is then -$$ -N_{\rm disc}=\frac{\pi}{2\sqrt{3}}\left(\frac{R}{r}\right)^2=\frac{3}{16}N_{\rm hex}. -$$ -The coupon collector's problem in this case states that the expected time to hit all the disks is $E[T_{\rm disc}] \sim \frac{\sqrt{12}}{\pi}N_{\rm disc}\ln N_{\rm disc}=\left(\frac{R}{r}\right)^{2}\ln N_{\rm disc}$. Note the extra factor of $\sqrt{12}/\pi$, which comes from the fact that each droplet only "collects a coupon" with probability $\pi/\sqrt{12}$. -Putting these bounds together, we have that the expected time to cover the table satisfies -$$ -2 \mu^2 \ln \mu + O(\mu^2) = E[T_{\rm disc}] \le E[T] \le E[T_{\rm hex}] = \frac{16\pi}{3\sqrt{3}} \mu^2 \ln \mu + O(\mu^2) -$$ -for $\mu \equiv (R/r) \gg 1$. In particular, the expected time is proven to be $\Theta(\mu^2 \ln \mu)$, with a multiplicative factor bounded between $2$ and about $9.68$.<|endoftext|> -TITLE: Why the number e(=2.71828) was chosen as the natural base for logarithm functions? -QUESTION [10 upvotes]: Possible Duplicate: -What's so “natural” about the base of natural logarithms? - -Why the number e(=2.71828) was chosen as the natural base for logarithm functions ? Mainly I am interested in knowing why is it called "natural " . The number "2" could instead have been chosen as the most natural base. - -REPLY [18 votes]: The simplest answer is this: -If you draw the graphs of $y=a^x$ for varying values of $a$, you find that they all pass through the point$(0,1)$ on the $y$-axis. There is exactly one of these curves that passes through that point with a gradient of exactly 1, and that value is obtained by taking $a=2.718281828459 \dots$. -In more analytical terms, this means that this is the value of $a$ which makes the derivative of $a^x$ equal to $a^x$, rather than a constant multiple of $a^x$.<|endoftext|> -TITLE: Statement about the order of $2$ modulo prime powers -QUESTION [6 upvotes]: I computed the factorization of $2^n-1$ for many $n$'s and came up with the following conjecture that for any odd prime $p$, -$$ -p^k || 2^n-1 \quad \Longleftrightarrow \quad O_p(2) p^{k-1} \, | \, n \quad with \quad p^{k-1} \, || \, n. -$$ -where $O_p(2)$ is the order of $2 \pmod p$. Here are my thoughts on the question. -We clearly need to assume $O_p(2) \, | \, n$ if $k \ge 1$, so the statement can be made easier to prove because we only need to show that by assuming this, $p^k \, | \, 2^n-1$ iff $p^{k-1} \, | \, n$. Let's prove $(\Longleftarrow)$ by induction. For $k = 1$ we have nothing to show. Say we take a general $k$, then by induction on $k$, -$$ -p^{k-1} \, | \, n \quad \Longrightarrow \quad 2^{p^{k-1} O_p(2)} - 1 = (2^{p^{k-2}O_p(2)})^p-1^p = (2^{p^{k-2}O_p(2)} - 1)(1 + 2^{p^{k-2}O_p(2)} + \dots + (2^{p^{k-2}O_p(2)})^{p-1} ) \equiv (p^{k-1}m)(\underset{p \text{ times}}{\underbrace{1 + \dots + 1}})\equiv 0 \pmod {p^k}. -$$ -So I've proved one direction. (Note that when $2$ is a primitive root $\pmod p$, this is actually quite trivial because my statement in the direction I proved it says 'since $\varphi(p^k) \, | \, n$, then $2^n \equiv 1 \pmod {p^k}$'.) -In the other direction the question seems quite hard though. For instance, if I just want to show this for $k = 2$, I need to prove that if $p^2 \, | \, 2^n-1$ then $p \, | \, n$. If I try to show that, then working modulo $p^2$ I get -$$ -0 \equiv 2^n-1 = 2^{O_p(2) m} -1 = (2^{O_p(2)}-1)(1 + 2^{O_p(2)} + \dots + 2^{O_p(2)(m-1)}) \\\ - \equiv (kp)(\underset{m \text{ times}}{\underbrace{1 + 1 + \dots + 1}} ) \equiv kpm \pmod {p^2} \\\ -$$ -and what I would need is to assume $p \, || \, 2^{O_p(2)}-1$, so that $p \, | \, m \, | \, n$... I think that if I have this the claim follows by induction, but I have absolutely no idea how to show this. I believe it is true though (computations). -Any ideas on this problem? - -REPLY [3 votes]: This happens to be false for $p=1093$, $k=1$ and $n=364 = O(2,1093)$. The conjecture breaks down because $2^{364} \equiv 1 \pmod {1093^2}$, i.e. the division on the LHS is not exact. I was stuck at proving the thing so I pushed Mathematica a little further to try some more number crunching and he popped out this counterexample...<|endoftext|> -TITLE: A prime number pattern -QUESTION [31 upvotes]: The algorithm -Given a natural number $n$ define a procedure as follows: - -Generate a list of primes upto and possibly including, $n$ -Assign $Z = n$ -If $Z > 0$, subtract the largest prime from list which we haven't considered yet. Otherwise, add it to $Z$. If $n$ is prime, it is assumed accounted for by the first step. -Repeat until all primes have been considered. - -For example, take $25$. The list of primes would be $2, 3, 5, 7, 11, 13, 17, 19, 23$. Subtracting $23$ from $Z=25$, we get $Z=2$. Next, we get $Z=2-19= -17$. And so on. Consequently, $Z$ assumes the values $25, 2, -17, 0, 13, 2, -5, 0, 3, 1$. -Note: Only an example. The conjecture as stated deals with applying the algorithms on primes. However, other numbers also seem to exhibit interesting patterns. -The Pattern - -Beginning at $3$ and every other prime thereafter, following the algorithm always (seems to) land us at $1$. -For the rest of the primes, $Z$ has a final value of either $0$ or $2$. - -The problem -Please read @alex.jordan's answer as he cleverly limits the range of values $Z$ can reach (say, terminal Z, or $Z_t$) to $\{-1,0,1,2\}$. As a result, the problem is now reduced to: - -For any prime number, prove that $-1$ cannot be a terminal. - -Also being discussed: here - -REPLY [9 votes]: Here is an outline: - -No matter what $N$ you start with, it is impossible to end with $3$ or higher. The penultimate $N$-value would have either been $N=1$ (impossible, since your next step would need to have been subtracting 2) or $N=5$. Since $N$ was $5$ at some point, the prime $3$ was in the original set of primes, and this necessitates an even earlier step in the chain. The step before that would have either been with $N=2$ (again impossible, since you would have subtracted 3) or with $N=8$. And again, we have a number so large that there had to have been another prime in the original list ($5$); there had to be an earlier step; and there had to have been a point when $N$ was much larger ($13$). Since the value of $N$ is getting larger at a more than quadratic rate (summing primes grows almost like like $n^2\log(n)$) while prime values are getting larger much less quickly, this process will never have a chance to end. -No matter what $N$ you start with, it is impossible to end with $-2$ or lower. A similar, but negative, argument applies. If we end with $-2$, then the penultimate value was either $N=0$ (not possible, since our last step would then have been adding $2$) or $N=-5$. Since $N$ is negative at this step, there must have been a previous step, where either $N=-2$ (again not possible) or $N=-8$. And this continues; we never return to an initial $N$-value that is positive, since $N$ is growing negative so quickly relative to the next largest prime. -Since you are subtracting and adding primes (which are mostly odd) and you are starting with $N$ prime (and odd), the parity of the terminal number only depends on how many steps there are in your sequence - how many primes there are up to $N$. For example, since $N=11$ yields a prime collection $\{2,3,5,7,11\}$ with 5 elements, you will add/subtract 4 odd numbers, and end up with an odd terminal $N$-value. But with $N=13$, you will add/subtract 5 odd numbers, and end up with an even terminal $N$-value. - -This explains why the terminal values of $N$ would have to alternate between something in $\{-1,1\}$ and something in $\{0,2\}$ as $N$ increments through prime values. -Your conjecture would be proved if we could rule out $-1$ as a terminal value when $N$ is prime. I've tried assuming that $-1$ is a terminal value, hoping that this led to only finitely many possible initial $N$, none of which are prime. However it appears that many $N$ lead to $-1$. The smallest are: $4, 10, 16, 22$. Perhaps it can be show that if $-1$ is terminal for $N$, then $N$ was even. But maybe someone else can take it from here. -Hope this much helps! - -EDIT added MUCH later -We can indeed show that if the terminal value is $-1$, then $N$ must have been even. -Working again from the end of the process backward, let's denote $p_i$ to be the $i$th prime in the list, and $Z_i$ the $i$th value for $Z$. Again, these indices are from the end of the process. So we take $Z_0=-1$ and $p_0=2$. -$p_1$ is $3$. And $Z_1$ must have either been $-1+p_0$ or $-1-p_0$; that is, either $1$ or $-3$. Note that the only possibilities for $Z_1$ are odd numbers. -Next, $p_2$ is 5, and $Z_2$ must be one of the preceding possibilities for $Z_1$ plus or minus $3$. So $Z_2$ must be even. -Since all further primes are odd, if we continue in this way we see that for $n>0$, whatever $Z_n$ is, it's odd if $n$ is odd and even if $n$ is even. -Now only for some of the potential $Z_n$ that we can reverse engineer this way, is it possible for them to have been initial $N$ values. Still, we know now that if we start with an $N$ that has $-1$ as its terminal $Z$-value, and if it takes an even number of steps to get there, then $N$ must have been even, and therefore not prime. -All told, this process has terminating $Z$-values that alternate between $1$ and either $0$ or $2$ as $N$ increments through prime values.<|endoftext|> -TITLE: Continuous maps between compact manifolds are homotopic to smooth ones -QUESTION [13 upvotes]: If $M_1$ and $M_2$ are compact connected manifolds of dimension $n$, and $f$ is a continuous map from $M_1$ to $M_2$, f is homotopic to a smooth map from $M_1$ to $M_2$. - -Seems to be fairly basic, but I can't find a proof. It might be necessary to assume that the manifolds are Riemannian. -It should be possible to locally solve the problem in Euclidean space by possibly using polynomial approximations and then patching them up, where compactness would tell us that approximating the function in finitely many open sets is enough. I don't see how to use the compactness of the target space though. - -REPLY [11 votes]: It is proved as Proposition 17.8 on page 213 in Bott, Tu, Differential Forms in Algebraic Topology. For the necessary Whitney embedding theorem, they refer to deRham, Differential Manifolds. -This Whitney Approximation on Manifolds is proved as Proposition 10.21 on page 257 in Lee, Introduction to Smooth Manifolds. -There you can find even the proof of Whitney embedding Theorem.<|endoftext|> -TITLE: A question of the norm calculation of Hermite function. -QUESTION [7 upvotes]: Define the Hermite function $H_n (x)$ by $$H_n (x) = (-1)^n e^{x^2} \frac{d^n}{dx^n} e^{-x^2} $$ then prove that $$ \int_{\mathbb R} |H_n (x) |^2 e^{-x^2} dx = 2^n n! \sqrt{\pi}$$ - -REPLY [4 votes]: We have the recursive relation -$$H'_n(x)=(-1)^ne^{x^2}\left(2x\frac{d^n}{dx^n}e^{—x^2}+\frac{d^{n+1}}{dx^{n+1}}e^{-x^2}\right)=-H_{n+1}(x)+2xH_n(x).$$ -We use this in the integral: -\begin{align} -\int_{\Bbb R}H_n(x)^2e^{-x^2}dx&=\int_{\Bbb R}(2xH_{n-1}(x)-H'_{n-1}(x))H_n(x)e^{-x^2}dx\\ -&=\int_{\Bbb R}2xe^{-x^2}H_{n-1}(x)H_n(x)dx-\int_{\Bbb R}H'_{n-1}(x)H_n(x)e^{—x^2}dx\\ -&=\left[-e^{-x^2}H_{n-1}(x)H_n(x)\right]_{—\infty}^{+\infty}+\int_{\Bbb R}e^{-x^2}(H'_{n-1}(x)H_n(x)+H_{n-1}(x))H'_n(x))\\ -&-\int_{\Bbb R}H'_{n-1}(x)H_n(x)e^{—x^2}dx\\ -&=\int_{\Bbb R}H_{n-1}H'_n(x)e^{-x^2}dx. -\end{align} -By induction, we get -$$\int_{\Bbb R}H_n(x)^2e^{-x^2}dx=\int_{\Bbb R}H_0(x)H^{(n)}_n(x)e^{-x^2}dx=\int_{\Bbb R}H^{(n)}_n(x)e^{-x^2}dx.$$ -Indeed, denote $\langle P,Q\rangle:=\int_{\Bbb R}P(x)Q(x)e^{-x^2}dx$ for two polynomials $P$ and $Q$. We can show by an analogous way that $\langle H_n,H_{n+k}\rangle=\langle H_{n-1},H'_{n+k+}\rangle$ for integers $k$ and $n$, then the induction relation. -Now, to get the result, we need the following facts: - -$H_n$ is a polynomial of degree $n$, whose leading terms is $2^n$ (show it by induction); -$\int_{\Bbb R}e^{-x^2}=\sqrt \pi$. - -By the way, this shows that the Hermite polynomials are orthogonal. - -REPLY [3 votes]: Due to orthogonality, -$$\int_{-\infty}^\infty x^k H_n(x) \exp(-x^2)\mathrm dx=0\qquad k -TITLE: Example of application of the Uniform Boundedness Principle -QUESTION [7 upvotes]: I've been trying to come up with an easy example of an application of the uniform boundedness principle (or Banach-Steinhaus theorem). I was thinking of something like the following, which is unfortunately a non-example: - -Non-example: -Consider the Banach space $(B(X), \|\cdot\|_\infty)$, i.e. the space of bounded functions with the sup norm. Let's choose $X=\mathbb R$ and let $\mu$ be a measure that is finite on compact sets, like for example the Lebesgue measure. Define -$$ T_t : (B(X), \|\cdot\|_\infty) \to \mathbb R$$ as -$$ f \mapsto \int_{[-t,t]} f d \mu$$ -Then $T_t$ is linear and bounded: If $\|f\|_\infty = 1$ then $\|T_t f\| = |T_t f| = \int_{[-t,t]} f d \mu \leq 2t$ so that the operator norm $\|T_t\| \leq 2t < \infty$. The condition that fails that prevents me from applying Banach-Steinhaus is that the family $\{T_t\}_{t \in \mathbb R}$ is not pointwise bounded: For $f$ with $\|f\|_\infty$ we have $\sup_{t \in \mathbb R} \|T_t\| \geq \sup_{t \in \mathbb R} 2 t = \infty$. - -My non-example was supposed to conclude that the integral operator is continuous on the whole space. Of course it's not on $\mathbb R$ for bounded functions but you see what I'm trying to do. Although one might consider showing that the integral is a continuous operator by applying Banach-Steinhaus "cracking nuts with a sledge hammer", this one is for educational purposes so it's acceptable. -Can someone either modify my example so that it works or show me an equally easy example? Thanks a lot for your help. - -REPLY [5 votes]: Here's another slightly more important example: -We can use Banach-Steinhaus to show that even though we have convergence in $L^2$ for Fourier series of $f \in C(\mathbb T)$, convergence does not hold pointwise. This we can show by showing that there exists an $f \in C(\mathbb T)$ such that the Fourier series of $f$ diverges at $0$. -For this we need the following three facts (proofs omitted): -Fact 1: If $D_n = \sum_{k=-n}^n e^{2 \pi i k}$ is the $n$-th Dirichlet kernel then $$ D_n \ast f(x) = \sum_{k=-n}^n \hat{f}_k e^{ik 2 \pi x}$$ -that is, the convolution with the Dirichlet kernel gives us the $n$-th partial sum of the Fourier series of $f$. -Fact 2: $$ \int_0^1 \left | D_n(x) \right | dx \xrightarrow{n \to \infty} \infty$$ -Fact 3: The map $T_n: f \mapsto D_n \ast f (0)$ is a bounded linear operator with norm $\|T_n\| = \int_0^1 \left | D_n(x) \right | dx$. -Now all the prerequisites for Banach-Steinhaus are fulfilled: -(i) $(C(\mathbb T), \|\cdot\|_{L^2})$ is a Banach space, $\mathbb R$ is a normed space -(ii) $\{T_n\}$ is a family of bounded linear operators: for each $n$, $T_n$ is bounded by $\|T_n\|$ (Fact 3) -Now if $\{T_n\}$ was pointwise bounded, i.e., for a given $f$, $\sup_{n \in \mathbb N} \|T_n f \| < \infty $ then we could apply Banach-Steinhaus to get $\sup_{n \in \mathbb N} \| T_n \| < \infty$ but that would be a contradiction to $\| T_n \| = \int_0^1 |D_n(x) | dx$ and $ \int_0^1 \left | D_n(x) \right | dx \xrightarrow{n \to \infty} \infty$. Hence $\{T_n\}$ cannot be pointwise bounded that is, there exist an $f$ such that $T_nf$ diverges for $n \to \infty$.<|endoftext|> -TITLE: Stronger than ZF, weaker than ZFC -QUESTION [7 upvotes]: Can you please name axiom system that is strictly weaker than ZFC and strictly stronger than ZF? (such as DC, AC$_\omega$) -I searched for it but i could only find these two. -If there are statements equivalent to DC or AC$_\omega$ in ZF, please tell me or give me a link introducing those.. -(I want some equivalent statements, since I don't know where these two can be used usefully.) - -REPLY [7 votes]: There are infinitely many assertions which are between ZF and ZFC, these are often called choice principles. -Slightly more formally (but naively, as Carl indicates in the comments), we say that a sentence in the language of set theory $\varphi$ is a choice principle if it is provable from ZFC but not from ZF, we say that $\varphi$ is a weak choice principle if ZF+$\varphi$ does not prove AC. -Examples for choice principles: - -the axiom of choice; -the axiom of choice for countable families; -countable unions of countable sets are countable; -the axiom of choice for countable families of finite sets; -the axiom of choice for countable families of pairs. - -The first one is not a weak choice principle, it is in fact the axiom of choice in full. It can be shown that each choice principle implies the following, and that all those implications are strict. The list can be expanded much further, even if we require it will remain "linear" (that for every two statements one of them implies the other). -However there are other choice principles, for example: - -The ultrafilter lemma: Every filter can be extended to an ultrafilter. - -This is a particularly important choice principle with many equivalents and uses throughout mathematics. For example, it is equivalent to the compactness theorem in logic; or to the Tychonoff theorem for Hausdorff compact spaces. This choice principle implies that every set can be linearly ordered, as well the Hahn-Banach theorem and in turn - the Banach-Tarski paradox. From this family of choice principles also follow the fact that every field has an algebraic closure, and that the closure is unique up to isomorphism. -There are many many other choice principles, some of them are equivalent to the ultrafilter lemma; others to the axiom of choice for families of some sort (countable; etc.); and other principles are just "out there" and are not equivalent to the variants of these two above, for example KWP($n$) and SVC are two important examples. -Specifically for DC, we have an interesting equivalent: - -The Baire category theorem for complete metric spaces. - -As for the axiom of countable choice, it too has several equivalents and one of the interesting ones, for example: - -$\sigma$-compact spaces are Lindelof. - - -To learn more about such principles you may find yourself interested in reading these books, all contain proofs, lists and diagrams of various choice principles and the known implications between them: - -Herrlich, H. Axiom of Choice. Lecture Notes in Mathematics, Springer, 2006. -Jech, T. The Axiom of Choice. North-Holland (1973). -Howard, P. and Rubin, J.E. Consequences of the Axiom of Choice. American Mathematical Soc. (1998). Also see the online database for the book. -Moore, G. H. Zermelo's Axiom of Choice. Springer-Verlag (1982).<|endoftext|> -TITLE: Every compact subset must be closed? -QUESTION [5 upvotes]: This is an exercise from a topological book. -In $T_1$ space, every compact subset must be closed? For any two compact subset, their intersection must be compact? -Thanks for any help:) - -REPLY [4 votes]: I would like to advertise the following result, and some consequences, just in case it is not so well reported: -Theorem: Let $B,C$ be compact subsets of $X,Y$ respectively and let $\mathcal W$ be a cover of $B \times C$ by sets open in $X \times Y$. Then $B,C$ have open neighbourhoods $U,V$ respectively such that $U \times V$ is covered by a finite number of sets of $\mathcal W$. -This has a number of consequences from "the product of two compact spaces is compact" to "a compact subset of a Hausdorff space is closed". (Two others are listed in my book "Topology and groupoids", p. 86, but I forget where I first learned of the theorem, back in the 1960s!)<|endoftext|> -TITLE: What is the standard proof that $\dim(k^{\mathbb N})$ is uncountable? -QUESTION [8 upvotes]: This is my (silly) proof to a claim on top of p. 54 of Rotman's "Homological algebra". -For $k$ an infinite field (the finite case is trivial) prove that $k^\mathbb{N}$, the $k$-space of functions from the positive integers $\mathbb{N}$ to $k$, has uncountable dimension. -Lemma. There is an uncountable family $(A_r)_{r \in \mathbb{R}}$ of almost disjoint infinite subsets of $\mathbb{N}$, i.e., $|A_r \cap A_s| < \infty$ for $r \neq s$. -The proof is standard, let $f : \mathbb{Q} \stackrel{\sim}{\to} \mathbb{N}$ and $A_r := \{f(r_1), f(r_2), \ldots\}$ for $(r_j)_{j \in \mathbb{N}}$ a sequence of distinct rationals whose limit is $r$. Of course, these must be chosen simultaneously for all $r$, but any choice will work. -Now the family $f_r : \mathbb{N} \to k$, $f_r(x) = 1$ for $x \in A_r$ and $0$ elsewhere is linearly independent, since $a_1f_{r_1} + \cdots + a_kf_{r_k} = 0$ yields $a_1 = 0$ if applied to $x \in A_{r_1} \backslash (A_{r_2} \cup \cdots \cup A_{r_k})$, etc. -Curiously, this shows that $\dim(k^\mathbb{N}) \ge |\mathbb{R}|$, which is "a bit more". I'd like to see the folklore trivial "one-line argument", since I don't remember to have learned about it. -Thanks in advance. Also, greetings to stackexchange (this being my first topic here). - -REPLY [3 votes]: One liner argument which uses a much more difficult theorems (swatting gnats with cluster bombs kind of proof): -$k^\mathbb N$ is the algebraic dual of the polynomials in one variable, $k[x]$ which has a countable dimension. If $k^\mathbb N$ had a countable basis then $k[x]$ would be isomorphic to its dual, and since this cannot be we conclude that $k^\mathbb N$ has a basis of uncountable size. -The arguments given in Arturo's answer show that the above is indeed a proof (in particular Lemma 2 with $\kappa=\aleph_0$).<|endoftext|> -TITLE: Cauchy+pointwise convergence $\Rightarrow$ uniform converges (for an operator in a Hilbert space) -QUESTION [5 upvotes]: Suppose that the sequence of operators in a Hilbert space $H$, $\left(T_{n}\right)_{n}$, -is Cauchy (with respect to the operator norm) and that there is an -operator $L$, such that $Lx=\lim_{n\rightarrow\infty}T_{n}x$, for -all $x\in H$ (i.e. the $T_{n}$ converge pointwise to $L$). -How -can I prove then, that $\left(T_{n}\right)_{n}$ converges with respect -to the operator norm to $L$ (i.e. $\left(T_{n}\right)_{n}$ converges -uniformly to $L$)? - -REPLY [2 votes]: The space of bounded linear operators under the operator norm is complete. Hence, since $\{T_n\}$ is Cauchy, there exists an operator $T$ such that $T_n \to T$ in operator norm. In particular, $T_n \to T$ pointwise, since $\|T_n x - T x\| \le \|T_n -T\| \|x\|$. But $T_n \to L$ pointwise also, so we must have $T=L$.<|endoftext|> -TITLE: Unramified extension is normal if it has normal residue class extension -QUESTION [5 upvotes]: Let $K/F$ be an unramified extension such that $\rho_K / \rho_F$ (the corresponding extension of residue classes) is normal. Prove $K/F$ is normal. -I guess I need to do some polynomial lifting, but if we take $f(x)$ the minimal polynomial of $\alpha \in K$, we can't lift it to $\rho_F$ unless his coefficients are in $O_{F} = \{x\in F : |x| \leq 1\}$. - -REPLY [5 votes]: I'm assuming $K$ and $F$ are complete, discretely valued fields and $K/F$ is finite separable. The term ``unramified" means the residue extension $\rho_K/\rho_F$ is separable and if $\pi$ is a uniformizer of $\mathscr{O}_F$, it is also a uniformizer of $\mathscr{O}_K$. So $[K:F]=[\rho_K:\rho_F]$. Let $f$ be a monic lift in $\mathscr{O}_F$ of the minimal polynomial for a generator $\bar{b}$ of $\rho_K$ over $\rho_F$ (which exists because the residue extension is finite separable). Then $f$ is irreducible because its reduction mod $\pi$ is. Because $\bar{f}$ is separable, we can uniquely lift $\bar{b}$ to a root $b$ in $\mathscr{O}_K$ of $f$ by Hensel's lemma. Thus we have an $F$-algebra injection $F[X]/(f(X))\hookrightarrow K$ sending $X$ to $b$. This must be an isomorphism by dimension counting. By Hensel's lemma, each root of $\bar{f}$ in $\rho_K$ lifts to a (unique) root of $f$ in $\mathscr{O}_K$. So, since $\bar{f}$ splits over $\rho_K$, $f$ splits over $K$. Thus $K$ is normal over $F$.<|endoftext|> -TITLE: Torsion in two dimensions? -QUESTION [16 upvotes]: This question is about the notion of a connection with torsion in differential geometry, i.e., a connection that is not Levi-Civita. (It's not about the torsion of a curve in three dimensions.) -Torsion has various physical applications in $\ge 4$ dimensions (Einstein-Cartan theory, string theory, experimental searches for coupling of spin to torsion) and in 3 dimensions (crystallography). -The simplest examples of any given concept are always good to study, and lower-dimensional spaces are simpler. It seems that torsion should be a reasonable thing to study in two dimensions, although you can't have torsion that preserves tangent vectors, since there can't be a rotation around a fixed axis in two dimensions. This is equivalent to the fact that you can't have a totally antisymmetric torsion tensor $\tau_{abc}$ in two dimensions. -Is there any mathematical reason why torsion in two dimensions is impossible, dull, or trivial? -It would seem that if $\tau_{abc}$ is only required by definition to be antisymmetric on the two final indices, not all three, then you have 2 independent quantities, $\tau_{112}$ and $\tau_{212}$. Geometrically, I'm not clear on why there are two independent quantities and not just one. It seems to me that the only intrinsic quantity should be the failure of a parallelogram to close, and this should only give one degree of freedom, since I don't think the orientation of the parallelogram can make a difference in two dimensions. (A 90-degree rotation is the same as interchanging the two axes, which won't give an independent result.) -If there is no mathematical reason why the 2-dimensional case doesn't work, are there any interesting real-world applications of torsion in two dimensions? Failing that, I would still be interested in any discussion of mathematically interesting, simple examples in two dimensions, e.g., a discussion of what phenomena occur in a flat space with constant torsion, or a space of constant curvature with constant torsion. For example, I suppose that in the flat case with constant torsion, curves of extremal length are lines, whereas curves that parallel-transport their own tangent vectors are circles. In the positive-curvature case, I haven't thought this through carefully, but it seems that constant torsion on a sphere would be impossible due to something like the hairy ball theorem. -Are there simple models of such geometries, in roughly the same sense that elliptic geometry can be modeled by identifying antipodal points on a sphere? (I know that you can't model a two-dimensional space with torsion in exactly this way, since all you naturally induce is the torsion-free structure.) - -REPLY [7 votes]: Here is a 2D example with a picture: -\begin{align*} - &\nabla_y {\bf e}_x = -{\bf e}_y; \quad \nabla_y {\bf e}_y = {\bf e}_x\\ - &\nabla_x {\bf e}_x = \nabla_x {\bf e}_y =0 -\end{align*} -These equations say that the standard xy frame rotates clockwise (according to the connection) as it's transported upwards, and so if we parallel transport a vector upwards, it rotates counterclockwise according to the standard xy frame. Picture: - -On the left hand side, I've parallel transported the unit frame from the origin to a bunch of lattice points. On the right hand side, I've shown (as you say) the failure of a parallelogram to close. -This connection has zero curvature, as you can see by parallel transporting a vector around a coordinate rectangle: the rotation up one vertical side is exactly canceled by the counter-rotation down the opposite side. -As for the number of independent coefficients: we have $\tau^x_{\;\;xy}=0$, $\tau^y_{\;\;xy}=-1$. (I am using the Misner, Thorne, Wheeler convention where $\nabla_i{\bf e}_j=\Gamma^k_{\;\;ji}{\bf e}_k$, i.e., the differentiating index comes last.) You can also see that the non-closed "failed parallelogram" is not invariant under a 90 degree rotation. -I wrote the torsion coefficients with one index up, since I don't know what metric you'd want to use. This is not, as you say, a Levi-Civita connection. Since the connection is flat, I suppose the most natural metric would be the standard Euclidean $g_{ij}=\delta_{ij}$. Curves of extremal length would be lines, as you say. However, circles do not parallel transport their tangent vectors. A circle of the proper radius would start doing that if you begin at the rightmost point and go upward. But at the top and bottom, the tangent vector is still turning while parallel transport does not induce any rotation w.r.t. xy coordinates. -I don't know of any practical applications of this example, but Frankel, in The Geometry of Physics (3rd ed., p.252) mentions something similar in geodesy: - -Surveyors could introduce a frame of 3 orthonormal vectors in a small 3-dimensional neighborhood of a point on the irregular Earth’s surface as follows: ${\bf e}_3$ is an upward pointing unit vector defined by a plumb line, ${\bf e}_1$ is a horizontal unit vector pointing magnetic north, and ${\bf e}_2 = {\bf e}_3\times{\bf e}_1$ points “west.” It is thus natural for surveyors to - introduce (locally) a distinguished frame of vectors defining a distant parallelism with curvature 0, and this frame is not associated with any coordinate system; the torsion does not vanish! - -Frankel gives a reference to an article I haven't read: - -Grossman, N. Holonomic measurables in geodesy. J. Geophysical Res. 79. (1974), - pp. 689–94. - -I hope this all helps.<|endoftext|> -TITLE: Is there a simple explanation why degree 5 polynomials (and up) are unsolvable? -QUESTION [34 upvotes]: We can solve (get some kind of answer) equations like: -$$ ax^2 + bx + c=0$$ -$$ax^3 + bx^2 + cx + d=0$$ -$$ax^4 + bx^3 + cx^2 + dx + e=0$$ -But why is there no formula for an equation like $$ax^5 + bx^4 + cx^3 + dx^2 + ex + f=0$$ -I'm not sure if this has anything to do with the Galois theory, but is there a dumbed-down simple explanation as to why degree 5 polynomials (and up) are unsolvable? - -REPLY [2 votes]: There is already one "dumbed" answer above, so forgive me for adding an even "dumber". -What it means by solvable is to be able to write down a formula for a solution of a polynomial equation like $ax^2 + bx + c = 0$ such this one: -${x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}}$ -In other words, we need the solution (whatever it would be in $x$) could be expressed in some algebraic expressions. In the example above: $-, b, \pm, \sqrt, \: ^2, 4, a, c, /,$ and $2$ are all algebraic expressions. -For most of all numbers we know, they could be expressed in algebraic expressions. For example: $2, 6/3, \sqrt{2}, \sqrt{-1},$ etc and even for a non-repeating irrational number like $\pi$ could be expressed in $\frac{C}{D}$ where $C$ is the circumference of a circle and $D$ is the diameter. -Amazingly there are numbers that could not be expressed in any algebraic expressions we know. -Historically, it was discovered when observing that polynomial equations of degree higher than 4 may not necessarily have a solution that could be expressed in algebraic expressions. It was first observed by Joseph-Louis Lagrange in 1770, partly proven by Paolo Ruffini in 1799, and then completed by Niels_Henrik_Abel in 1824, establishing Abel–Ruffini theorem. But not until 1830 Évariste Galois at the very young age of 18 showed the necessary condition of a polynomial equation to have a solution hence called solvable. -The theory is called Galois theory, but unfortunately it is too complex to be explained satisfactorily here. If we have enough background in fundamental algebra, it would still need at least 200 hundreds pages of hard math to understand it. Nevertheless the necessary condition showed by Galois theory also shows that there are polynomial equations that their solutions could not be expressed in any algebraic expressions. -An example of a such unsolvable polynomial equation is $x^5 - x - 1 = 0$. Note to not be confused with the algebraic expression mentioned above. In the equation $x^5 - x - 1 = 0$, we cannot transform it into a solution expression where a single $x$ is put on the left side of the equation such $x = \: ...$ because, well, we don't have the algebraic expression to be written down in the right side of the equation. -Of course we could approximate the solution with some number close enough, but it is still NOT the solution to the equation because simply it is just close enough BUT not the exact solution. The exact solution is not expressible yet. Amazing right!<|endoftext|> -TITLE: Alternating sum of binomial coefficients -QUESTION [5 upvotes]: Calculate the sum: -$$ \sum_{k=0}^n (-1)^k {n+1\choose k+1} $$ - -I don't know if I'm so tired or what, but I can't calculate this sum. The result is supposed to be $1$ but I always get something else... - -REPLY [2 votes]: Another way to see it: prove that -$$\binom{n}{k}+\binom{n}{k+1}=\binom{n+1}{k+1}\,\,\,,\,\,\text{so}$$ -$$\sum_{k=0}^n(-1)^k\binom{n+1}{k+1}=\sum_{k=0}^n(-1)^k\cdot 1^{n-k}\binom{n}{k}+\sum_{k=0}^n(-1)^k\cdot 1^{n-k}\binom{n}{k+1}=$$ -$$=(1+(-1))^n-\sum_{k=0}^n(-1)^{k+1}\cdot 1^{n-k-1}\binom{n}{k+1}=0-(1-1)^n+1=1$$<|endoftext|> -TITLE: If $f(x)$ is irreducible in $k[x]$, why is it also irreducible in $k(t)[x]$, for $t$ an indeterminate? -QUESTION [5 upvotes]: I've been thinking on this a few days, but I'm stuck. - -Let $k$ be a field, $f(x)$ irreducible in $k[x]$. Why is $f(x)$ also irreducible in $k(t)[x]$, for $t$ an indeterminate? - - -I write $f(x)=c_0+c_1x+\cdots+c_nx^n$ for $c_i\in k$. Suppose that $f(x)$ is reducible in $k(t)[x]$, so -$$ -f(x)=(a_0(t)+a_1(t)x+\cdots+a_r(t)x^r)(b_0(t)+b_1(t)x+\cdots+b_s(t)x^s) -$$ -for $r,s>0$. Since $c_0=a_0(t)b_0(t)$, $a_0(t)$ and $b_0(t)\in k$. Same goes for $a_r(t)$ and $b_s(t)$. -I tried something like letting $b_j(t)$ be the least index such that $b_j(t)\in k$. Then -$$ -c_j=a_0b_j+a_1b_{j-1}+\cdots -\text{ and } -c_j-a_0b_j=a_1b_{j-1}+\cdots -$$ -where $c_j-a_ob_j\in k$, and the right hand side is the sum of polynomials in $k(t)$ of degree at least $1$. I don't see if there is a contradiction to be had. What's the right approach? Thanks. - -REPLY [9 votes]: If $k$ is infinite you do not need Gauss's lemma (in fact you hardly need any tools at all). Let us start the same way as you started: suppose $f$ admitted a nontrivial factorization in $k(t)[x]$: -$$f(x) = (a_0(t) + ... + a_n(t) x^n)(b_0(t) + ... + b_m(t) x^m).$$ -Pick $c \in k$ such that the denominators of each $a_i(t)$ and $b_i(t)$ are nonzero when $t = c$ and such that the numerators of $a_n(t), b_n(t)$ are also nonzero (this is always possible iff $k$ is infinite). Then -$$f(x) = (a_0(c) + ... + a_n(c) x^n)(b_0(c) + ... + b_m(c) x^m)$$ -is a nontrivial factorization of $f$ in $k[x]$. -(The takeaway lesson here is that a polynomial in $k(t)[x]$ really lies in $k[t][x]$ localized away from finitely many elements of $k[t]$.) -If $k$ is finite you can do something similar but sneakier: pass to the algebraic closure $\bar{k}$ to find $c$ as above. By choosing $c$ more carefully you can deduce that either the $a_i$ are all constant multiples of each other (resp. the $b_i$) or $f$ admits a nontrivial factorization in $\bar{k}[x]$ with an irreducible factor having a Galois orbit larger than the degree of $f$, which is a contradiction.<|endoftext|> -TITLE: Understanding the concept behind the Lagrangian multiplier -QUESTION [17 upvotes]: I've been trying to understand the principles behind the Lagrangian multipliers and I think I've got a rough understanding of it. Would appreciate it if you guys could help me answer a few questions! -I've pretty much self-studied this from here and here. -As far as I understand it, the Lagrangian multiplier essentially works by ensuring the gradient of the function is equal to the gradient of the restraint. Assuming that $g(x,y) = c$ is the restraint, it also ensures the point satisfies the restraint. -However, I don't understand the reasoning behind some of the statements in the ideashop link: -1) "The most important thing to know about gradients is that they always point in the direction of a function's steepest slope at a given point." . Apparently a gradient is a collection of partial first derivatives but I don't understand how one gets from a collection of partial first derivatives to 'Always pointing in the direction of the steepest slope'. Wouldn't it be changing direction as one went around a 3D surface? -2) Why is the gradient always perpendicular to the level curve? -3) How does one get from $$\nabla L=\begin{bmatrix}\frac{\partial f}{\partial x_1}-\lambda\frac{\partial g}{\partial x_1}\\\frac{\partial f}{\partial x_2}-\lambda\frac{\partial g}{\partial x_2}\\g(x_1,x_2)-c\end{bmatrix}=0$$ to $$L(x_1,x_2,\lambda)=f(x_1,x_2)-\lambda(g(x_1,x_2)-c)$$ (Both sourced from ideashop) ? -Thank you for your help in advance! - -REPLY [6 votes]: I'll explain it in a three-dimensional setting. This means a priori you are given a "temperature" function $f:\,{\mathbb R}^3\to{\mathbb R}$. -Consider a fixed point ${\bf p}\in{\mathbb R}^3$ and a point $t\mapsto{\bf x}(t)$ $\ (t\geq0)$ moving away from ${\bf p}$ in direction ${\bf u}\,$; so ${\bf x}(0)={\bf p}$, $\,{\bf x}'(0)={\bf u}$. The temperature at the moving point is a function of $t$ and is given by $\phi(t):=f\bigl({\bf x}(t)\bigr)$. According to the chain rule we have -$$\phi'(0)=\nabla f\bigl({\bf x}(0)\bigr)\cdot{\bf x}'(0)=\nabla f({\bf p})\cdot{\bf u}\ .\qquad(1)$$ -If $\nabla f({\bf p})\ne{\bf 0}$ then there will always be direction vectors ${\bf u}$ such that the scalar product on the right is positive and other such vectors where it is negative. In this case the temperature $f$ can neither be locally minimal nor locally maximal at ${\bf p}$. In particular, for $|{\bf u}|=1$ the increase $\phi'(0)$ is maximal ($=|\nabla f({\bf p})|$) if ${\bf u}$ points in the direction of $\nabla f({\bf p})$, the temperature remains stationary, i.e., $\phi'(0)=0$, for directions ${\bf u}\perp\nabla f({\bf p})$, and decreases fastest in the direction $-\nabla f({\bf p})$. -When the moving point stays on the isothermal surface $S_f$ through ${\bf p}$ then $\phi'(0)=0$. Looking at $(1)$ we see that all "$f$-isothermal tangent directions" are orthogonal to $\nabla f({\bf p})$. This implies that $\nabla f({\bf p})$ spans the orthogonal complement of the tangent plane $T_{f,{\bf p}}$ of $S_f$. -Assume now that in addition to $f$ we have a constraint $g({\bf x})=0$ defining an "admissible" surface $S_g\subset{\mathbb R}^3$, and assume that our point ${\bf p}$ satisfies the constraint. In this case only "$g$-isothermal tangent directions" ${\bf u}$ are allowed for the moving point. -If ${\bf p}$ aspires to be a conditionally extremal point, whence a conditionally stationary point, of $f$ then we should have $\phi'(0)=0$ for all allowed directions. This means by $(1)$ that $\nabla f({\bf p})$ should be orthogonal to all "$g$-isothermal tangent directions", or that $\nabla f({\bf p})$ is in the orthogonal complement of the tangent plane $T_{g,{\bf p}}$. As the latter is spanned by $\nabla g({\bf p})$ (assumed $\ne{\bf 0}$) we would have an equality of the form $\nabla f({\bf p})=\lambda\, \nabla g({\bf p})$ for some $\lambda\in{\mathbb R}$. This in term implies that the point ${\bf p}$ will come to the fore when we solve the equations -$$\nabla f({\bf x})=\lambda\, \nabla g({\bf x}),\quad g({\bf x})=0\qquad(2)$$ -for ${\bf x}$ (and $\lambda$). -Applying the "Lagrange method" means no more and no less than solving the equations $(2)$.<|endoftext|> -TITLE: Are non-circular swords possible? -QUESTION [17 upvotes]: I was reminded of this by our recent discussion of the old chestnut about possible shapes for utility hole covers. Perhaps this question is less familiar. -A sword can be made in any shape at all, but if you want to be able to put it into a scabbard, only certain shapes will do. For simplicity, let us neglect the width of the sword, and take it to be the image of a continuous mapping $[0,1]\to {\Bbb R}^3$. Clearly, a sword in the shape of a circular arc will fit into a similar scabbard, as long as the arc is less than about half of a full circle: - -(For non-mathematical reasons, such swords are rarely more than about $\frac1{36}$ of circle.) -If we let the curvature go to zero, we get the special case of a straight segment, which can be of any length: - -It seems to me that if one had a sword which was a helical segment, it would go into a matching scabbard. The circular and straight swords are special cases of this, where the helical pitch is zero. (A cursory search for helical swords or scabbards turned up nothing—unsurprisingly, since a sword which has to be worked into the enemy like a corkscrew is a really stupid idea.) -Are there any other shapes of swords which will go into scabbards? - -REPLY [14 votes]: Clearly the curvature and torsion of the sword must be constant. A curve of nonzero constant curvature and torsion is a helix.<|endoftext|> -TITLE: Complex and real forms of the Poisson integral formula -QUESTION [7 upvotes]: In my complex analysis book there is the expression -$$\frac{1 - |z|^2}{|1 - \bar z e^{it}|^2}$$ -and it says that when $z = re^{it}$, we can write the above expression as -$$P_r(t) = \frac{1 - r^2}{1 - 2r\cos t + r^2} = \text{Re}\left( \frac{1 + z}{1 - z} \right)$$ -I do not see where the $\cos t$ comes from though. -Isn't $\bar z = re^{-it}$, so the top is $1 - r^2$ and the bottom is $|1 - r|^2 = 1 - 2r + r^2$. I have not really figured out where the $\text{Re}(1 + z)/(1 - z)$ comes from either. - -REPLY [4 votes]: Here is a link a of the book. $P_r(t)$ is defined at equation (5) as $\frac{1-r^2}{1-2r\cos t+r^2}$. If $z=re^{it}$, then -$$\frac{1+z}{1-z}=\frac{1+re^{it}}{1-re^{it}}=\frac{(1+re^{it})(1-re^{-it})}{(1-re^{it})(1-re^{-it})}=\frac{1-re^{-it}+re^{it}-r^2}{1-2r\cos t+r^2},$$ -and the real part of $re^{-it}+re^{it}$ is $0$. -It helps us to write a formula for $f(re^{it})$, involving $P_r$.<|endoftext|> -TITLE: For what values for m does $\sum \limits_{k=2}^{\infty}\frac{1}{(\ln{k})^m}$ converges? -QUESTION [7 upvotes]: For what values for m does $$\sum \limits_{k=2}^{\infty}\frac{1}{(\ln{k})^m}$$ converge? -What about -$$\sum_{k=2}^{\infty}\frac{1}{(\ln(\ln{k}))^m}$$ or more generally -$$\sum_{k=2}^{\infty}\frac{1}{(\ln(\cdots (\ln{k}))\cdots)^m}$$ ? - -REPLY [4 votes]: For the first series,we can use Cauchy's condensation criterion when $p>0$ (when $p\leq 0$ the series is divergent): - -If $\{a_n\}$ is a decreasing sequence of positive numbers, the series $\sum_ka_k$ is convergent if and only if $\sum_k2^ka_{2^k}$ is convergent. - -We have with $a_k=\frac 1{(\ln k)^m}$ that $2^ka_k=\frac{2^k}{k^m(\ln 2)^m}$ and it diverges. -An alternative way is to note that $\ln k\leq k^{1/m}$ for a $m$ large enough (which depend on $m$), and use the divergence of harmonic series. -We have for $x\geq 1$ that $\log x\leq x$. Denote $f_N$ the $N$-th iterate of the logarithm. For a fixed $N$, we can find $n(N)$ such that for $k\geq n(N)$, $f_{N-1}(k)\geq 1$. Hence $f_N(k)\leq \log k$ for $k$ large enough, which shows that the series is divergent for all $m$ and all $N$. -The problem here is that the logarithm grows too slowly. Taking iterates doesn't make the thing better, and the exponent doesn't change anything.<|endoftext|> -TITLE: What does the letter epsilon signify in mathematics? -QUESTION [27 upvotes]: This letter "$\varepsilon$" is called epsilon right ? What does it signify in mathematics ? - -REPLY [5 votes]: Hilbert's epsilon-calculus used the letter $\varepsilon$ to denote a value satisfying a predicate. If $\phi(x)$ is any property, then $\varepsilon x. \phi(x)$ is a term $t$ such that $\phi(t)$ is true, if such $t$ exists. One can define the usual existential and universal quantifiers $\exists$ and $\forall$ in terms of the $\varepsilon$ quantifier: -$$\begin{eqnarray} -\def\hil#1{#1(\varepsilon x. #1(x))} -\exists x.\phi(x) & \equiv & \hil{\phi}\\ -\forall x.\phi(x) & \equiv & \phi(\varepsilon x.\lnot\phi(x)) -\end{eqnarray} -$$<|endoftext|> -TITLE: Domain of an operator in functional analysis -QUESTION [12 upvotes]: I used to think that if we say $f$ is a function from a set $X$ to $Y$ then this implied that $f$ was defined on all of $X$. Because the definition of function is that it's a set $\{(x,y) \mid \text{ for every } x \in X \text{ there is exactly one } (x,y) \text{ where } y \in Y \}$. -Now I'm reading the definition of closed graph of a linear operator: - -So a linear operator is somehow not a function. But then what is it? Why does it make sense to say that $T$ is a map from $X$ to $Y$ when it's not even defined on all of $X$? - -In response to comment: - - -Edit -Perhaps my question is: why and when does it make sense to say $T:X \to Y$ is a linear operator when $T$ is not even defined on all of $X$? - -REPLY [17 votes]: This is an admittedly confusing abuse of terminology which your book appears to make even more confusing by using somewhat nonstandard terminology. The following is more standard but also confusing in its own way. For what follows let $X$ and $Y$ be Banach spaces. -A bounded linear operator from $X$ to $Y$ is a function $T: X \to Y$ which is linear and norm continuous. -An unbounded linear operator from $X$ to $Y$ is a pair $(T,D_T)$ where $D_T$ is a linear subspace of $X$ and $T: D_T \to Y$ is a linear map. -Comments: - -An unbounded linear operator from $X$ to $Y$ is not in general a function from $X$ to $Y$, and the definition does not claim it is. The phrase "from $X$ to $Y$" is part of what is being defined. -Because unbounded operators are not functions, extra care must be taken: two unbounded linear operators from $X$ to $Y$ cannot necessarily be added, and given unbounded linear operators from $X$ to $Y$ and from $Y$ to $Z$ they do not necessarily have a composition. -"Unbounded" really means "not necessarily bounded"; every bounded linear operator is an unbounded linear operator. -Some authors require that the domain $D_T$ of $T$ is a dense subspace of $X$. With this convention one can conceive of operators which are neither bounded nor unbounded. When this is not part of the definition, the assumption is added via the phrase "densely defined". -Adjoints can be very confusing: there are densely defined unbounded operators such that the largest possible domain of definition for the adjoint is $\{0\}$. - -The only real way to sort through all of this language is to look at some examples and results. The theory was invented by Von Neumann in the 20's to make precise sense of what physicists were doing in quantum mechanics. They had stumbled onto the idea that classical notions like position and momentum can be profitably viewed as linear operators on Hilbert space which satisfy certain relations. Mathematicians noticed, however, that there are no bounded operators on Hilbert space which satisfy these relations, and in particular the physicists' use of spectral theory had no mathematical basis. So Von Neumann invented the language of unbounded operators to make sense of the operators that physicists were doing. In particular he generalized the spectral theorem to certain classes of unbounded operators. -Maybe it would help to look at a quick example. Consider the operator $D = i \frac{d}{dx}$, a linear map $C_{per}^\infty[0,1] \to C_{per}^\infty[0,1]$, the space of smooth periodic functions on the interval $[0,1]$. Notice that $D$ has a rich supply of eigenfunctions: setting $e_n(x) = e^{-2 \pi i n x}$, we find that $e_n$ is an eigenfunction of $D$ with eigenvalue $2 \pi n$. The set of functions $\{e_n\}$ forms an orthonormal basis for the Hilbert space $L^2[0,1]$, so even though $D$ isn't an operator on $L^2[0,1]$ we can in a sense diagonalize it and use Hilbert space techniques to study it. -By contrast, consider the same operator $D = i \frac{d}{dx}$, this time viewed as a linear map $C_0^\infty(0,1) \to C_0^\infty(0,1)$, the space of smooth functions on the interval $[0,1]$ which vanish at $0$ and $1$. Now the only eigenfunctions for $D$ are constant functions, so it is not clear how one would diagonalize $D$. Secretly we know that we should just enlarge the domain of $D$ to include all periodic functions, but this isn't immediately obvious. Part of Von Neumann's accomplishment was to understand abstractly why $D$ with the domain $C_{per}^\infty[0,1]$ is "good" but $D$ with the domain $C_0^\infty(0,1)$ is not.<|endoftext|> -TITLE: Asymptotic behavior of iterative sequences -QUESTION [12 upvotes]: Suppose you have a sequence -$$ -a_10$. Can we conclude that $a_n\sim Cn\log n$ for some $C$? - -REPLY [2 votes]: For the moment I will only provide a solution for this problem : -Exercise. Given $x_0 >e^{-C}$ and $x_{n+1}=x_n+f(x_n)$ where $f(t):=\log t + C +o(1)$. Show that $x_n \sim n \log n$. -The following proof has not been proofread yet so there might be some errors (and I hope no real errors), I will proofread when I have time. -Proof. First $(x_n)$ is an increasing sequence and thus have a limit $\ell > e^{-C}$, but if it is finite then $\log \ell + C =0$ but it is not the case. -So for the moment we only have $x_n \to + \infty$. To be more precise one powerful method is the following : $x_{n+1}-x_n \simeq \log x_n$. The "continuous" problem associated is $y'(t) = \log y(t)$ that writes : $$\int_2^{y(t)} \frac{\mathrm{d}u}{\log u} = t- 2$$ hence we get $\frac{\mathrm{d}}{\mathrm{d}t} F(y(t)) = 1$ with : $$F(z)= \int_2^z \frac{\mathrm{d}u}{\log u}.$$ One can see (using by part integration) that : $$F(z) \sim \frac{z}{\log z}$$ as $z \to \infty$, we will use it shortly. -Then it might be interesting to look at the discrete equivalent for $\frac{\mathrm{d}}{\mathrm{d}t} F(y(t)) = 1$ which is : $$\int_{x_n}^{x_{n+1}} \frac{\mathrm{d}u}{\log u} \to 1.$$ In fact we can achieve just integrating the comparaison : $$\log x_n \leqslant \log u \leqslant \log x_{n+1} = \log x_n +o(\log x_n).$$ -Thus we have $\displaystyle \int_2^{x_n} \frac{\mathrm{d}u}{\log u} \sim n$ thus we get $\frac{x_n}{\log x_n} \sim n$ and so $\log x_n \sim \log n$ and finally $x_n \sim n \log n$.<|endoftext|> -TITLE: Variance of a max function -QUESTION [8 upvotes]: Say $x_1$ and $x_2$ are normal random variables with known means and standard deviations and $C$ is a constant. If $y = \max(x_1,x_2,C)$, what is $\mathrm{Var}(y)$? -Well, I forgot to tell that $x_1$ and $x_2$ are independent. - -REPLY [2 votes]: What I've got isn't pretty, but here goes (note: I use $\varphi(\cdot)$ and $\Phi(\cdot)$ to denote the standard normal density and CDF, resp.): -First, let's put everything in terms of standard normals: -$y_1=\frac{x_1-\mu_1}{\sigma_1},y_2=\frac{x_2-\mu_2}{\sigma_2},C_1=\frac{C-\mu_1}{\sigma_1},C_2=\frac{C-\mu_2}{\sigma_2}$ -The first key is drawing proper regions for the evaluation of the Max function: - -With this we can see that: -$$\begin{eqnarray*} -\mathbb{E}[Max\{X_{1},X_{2},C\}] & = & \intop\limits _{-\infty}^{C_{2}}\intop\limits _{-\infty}^{C_{1}}Cd\Phi(y_{1})d\Phi(y_{2})\\ - & & +\intop\limits_{C_1}^{\infty} \intop\limits_{-\infty}^{y_1} y_1 d\Phi(y_2)d\Phi(y_1) \\ - & & +\intop\limits_{C_2}^{\infty} \intop\limits_{-\infty}^{y_2} y_2 d\Phi(y_1)d\Phi(y_2) \\ -\end{eqnarray*}$$ -Simplifying what we can: -$\mathbb{E}[Max\{X_{1},X_{2},C\}] = C\Phi(C_1)\Phi(C_2)+(1-\Phi(C_1))\mathbb{E}[z\Phi(z) | Z>C_1]+(1-\Phi(C_2))\mathbb{E}[z\Phi(z) | Z>C_2]$ -($Z$ distributed $N(0,1)$) -Similarly, we get: -$$\begin{eqnarray*} -\mathbb{E}[(Max\{X_{1},X_{2},C\})^2] & = & \intop\limits _{-\infty}^{C_{2}}\intop\limits _{-\infty}^{C_{1}}C^2d\Phi(y_{1})d\Phi(y_{2})\\ - & & +\intop\limits_{C_1}^{\infty} \intop\limits_{-\infty}^{y_1} y_1^2 d\Phi(y_2)d\Phi(y_1) \\ - & & +\intop\limits_{C_2}^{\infty} \intop\limits_{-\infty}^{y_2} y_2^2 d\Phi(y_1)d\Phi(y_2) \\ -\end{eqnarray*}$$ -Which simplifies slightly: -$\mathbb{E}[(Max\{X_{1},X_{2},C\})^2] = C^2\Phi(C_1)\Phi(C_2)+(1-\Phi(C_1))\mathbb{E}[z^2\Phi(z)|Z>C_1]+(1-\Phi(C_2))\mathbb{E}[z^2\Phi(z)|Z>C_2]$ -That's about as far as I've gotten.<|endoftext|> -TITLE: Prove trigonometry identity for $\cos A+\cos B+\cos C$ -QUESTION [9 upvotes]: I humbly ask for help in the following problem. -If -\begin{equation} -A+B+C=180 -\end{equation} -Then prove -\begin{equation} -\cos A+\cos B+\cos C=1+4\sin(A/2)\sin(B/2)\sin(C/2) -\end{equation} -How would I begin the problem I mean I think $\cos C $ can be $\cos(180-A+B)$. But I am unsure what to do next. - -REPLY [3 votes]: $\cos A+\cos B+\cos C$ -$=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}+1-2\sin^2\frac{C}{2}$ as $\cos2x=1-2\sin^2x$ -Now $\cos\frac{A+B}{2}=\cos\frac{180^\circ - C}{2}=\cos(90^\circ-\frac{C}{2})=\sin\frac{C}{2}$ -So, $\cos A+\cos B+\cos C$ becomes $2\sin\frac{C}{2}\cos\frac{A-B}{2}+1-2\sin^2\frac{C}{2}$ -$=1+2\sin\frac{C}{2}(\cos\frac{A-B}{2}-\sin\frac{C}{2})$ -$=1+2\sin\frac{C}{2}(\cos\frac{A-B}{2}-\cos\frac{A+B}{2})$ replacing $\sin\frac{C}{2}$ with $\cos\frac{A+B}{2}$ -$=1+2\sin\frac{C}{2}(2\sin\frac{A}{2}\sin\frac{B}{2})$ applying $\cos(x-y)-\cos(x+y)= 2 \sin x \sin y$ -$=1+4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$<|endoftext|> -TITLE: A trigonometry equation problem -QUESTION [5 upvotes]: Let $a,b \in \left(0,\frac{\pi}{2}\right)$, satisfying -$$ -\frac{1-\cos{2a}}{1+\sin{a}}+\frac{1+\cos{2a}}{1+\cos{b}}=\frac{1-\cos{(2a+2b)}}{1+\sin{(a+2b)}} -$$ -Prove that: -$$ -a+b=\frac{\pi}{2} -$$ - -REPLY [4 votes]: Let, without loss of generality, $a=\frac{\pi}{2}-(b-c)$, -then by using the fact that you can suck up $\frac{\pi}{2}$ (and thereby change $\sin$ to $\cos$) as well as $\pi$ (and thereby change a sign of the trigonometric function) you can get rid of all $\pi$'s and $a$'s: -$$ -\frac{1-\cos{2a}}{1+\sin{a}}+\frac{1+\cos{2a}}{1+\cos{b}}=\frac{1-\cos{(2a+2b)}}{1+\sin{(a+2b)}} -$$ -$$\Longrightarrow$$ -$$ -\frac{1+\cos{(2(b-c))}}{1+\cos{(b-c)}}+\frac{1-\cos{(2(b-c))}}{1+\cos{b}}=\frac{1+\cos{(2c)}}{1+\cos{(b+c)}} -$$ -Now it is acutally clear that $a+b=\frac{\pi}{2}$ is a solution, because if you plug in $c=0$, all three denomiators become $1+\cos{(b)}$ and the equation reads $2=2$. -If you want to go futher you can use $$\cos ( b \pm c ) = \cos (b) \cos (c) \mp \sin (b) \sin (c)$$ to isolate the two trigonometric functions of $c$, substitue $t=\tan{\frac{c}{2}}$ to get polynomial expression in $t$, put everything on one equal denominator and solve the thing. This will amount to $t=\tan{(0)}=0$.<|endoftext|> -TITLE: Are bounded open regions in $\mathbb{R}^n$ determined by their boundary? -QUESTION [5 upvotes]: Let $U$ and $V$ be two bounded open regions in $\mathbb{R}^n$, and let us further assume that their topological boundaries are nice enough that they are homeomorphic to finite simplicial complexes. -Assume $\partial U$ is homeomorphic to $\partial V$. - -Is $U$ be homeomorphic to $V$? -A weaker question: if $U$ is contractible, is $V$ contractible? - -For $n=1$, the answer to both is yes, since $\partial U\simeq \partial V$ are sets of $2i$-many points, which both divide $\mathbb{R}^1$ into $i$-many open intervals. -For $n=2$, the answer to both is yes, since $\partial U\simeq \partial V$ are sets of $i$-many topological circles, which both divide $\mathbb{R}^2$ into $i$-many open discs. - -REPLY [3 votes]: For $n=3$ you could consider a ball minus a knot. In all cases the boundary is $S^2 \coprod S^1$, but inequivalent knots will yield non-homeomorphic sets. These have the added advantage of being connected.<|endoftext|> -TITLE: Prerequisite reading before studying the Collatz $3x+1$ Problem -QUESTION [5 upvotes]: Let's assume I am starting college and have just finished calculus. I've been reading a bit online about the Collatz $3x+1$ Problem and find it to be very intriguing. However, a lot of what I'm reading uses terms and techniques that I have not seen before. I'm wondering what prerequisite (text book) reading is required before starting to study this problem? -Put another way: I'm thinking about reading The Ultimate Challenge: The 3x+1 Problem by Jeffrey C. Lagarias. What areas of mathematics will I need to understand first before being able to fully understand this book? - -REPLY [7 votes]: The Lagarias book is a compilation of papers by various authors about various aspects of the problem. Different papers have different prerequisites. Some of the more expository papers have essentially no prerequisites at all; for others, you'll want to know about dynamical systems, Markov chains, ergodic theory, $p$-adic numbers, Turing machines and undecideability, and, of course, elementary Number Theory. And each of these has prerequisites, e.g., ergodic theory is based on measure theory, Markov chains involve Linear Algebra, etc., etc., etc. But don't be disheartened! You don't need all these for every paper, not by any means, and a well-written paper will teach you something useful in its introductory paragraphs even if the rest of the paper is beyond you. -I think the best thing is to jump in, start reading something you find interesting, and then, if you get stuck, come back here to ask something like, "What do I need to know to understand the proof that all furbles are craginacs, as given on page 977 of Peeble and Zimp, The Elephant and the $3x+1$ Problem?" It's much easier to give prerequisites when you have a narrowly-focussed problem in mind, than when it's as broad as "I want to learn about the $3x+1$ problem".<|endoftext|> -TITLE: spherical triangle inequality for multiple edges -QUESTION [6 upvotes]: Consider a spherical triangle, with side lengths $0 \leq a,b,c \leq \pi$. In texts on spherical geometry we see proofs that $a \leq b + c$ but this inequality is unnecessarily weak, in that some sets of lengths of sides satisfy the triangle inequality (and its cyclic permutations) but no triangle can be constructed with the given side lengths; for example $a=\pi, b=\pi, c=\frac{\pi}{2}$. -Stronger is -$$a \leq \begin{cases}b + c, &b+c\leq \pi\\2\pi-b-c, &b+c > \pi\end{cases},$$ -or more concisely, -$$\cos a \geq \cos(b+c),$$ -which does have the property that a triangle can be constructed from edges of length $a,b,c$ if and only if the lengths satisfy the triangle inequality. -Does this inequality generalize in a clean way to general polygons of side lengths $0 \leq a, b, c, d, \ldots \leq \pi$? - -REPLY [3 votes]: "There exists a spherical $n$-gon with sidelengths $2\pi \nu_1,\dots, 2\pi \nu_n$ if and only for every subset $P\subset \{1,\dots,n\}$ of odd cardinality $|P|$ the inequality -$$\sum_{i\in P}\nu_i -\sum_{i\notin P}\nu_i - \frac{|P|-1}{2}\le 0$$ -holds." -Source: Spherical polygons and unitarization by R. Buckman and N. Schmidt. Apparently, this REU paper was not formally published. See also the UMass REU site.<|endoftext|> -TITLE: Why need two directions to make $\sim_{\rm wa}$ an equivalence relation? -QUESTION [7 upvotes]: Let $\pi$ and $\sigma$ be representations of a $C^*$-algebra $\mathcal{A}$. They are weak approximately equivalent ($\pi\mathbin{\sim_{\rm wa}}\sigma$) if there are sequences of unitary operators $\{U_n\}$ and $\{V_n\}$ such that \begin{equation} -\sigma(A)=\operatorname{WOT-lim} U_n\pi(A) U_n^*, -\end{equation} \begin{equation} -\pi(A)=\operatorname{WOT-lim} V_n\sigma(A) V_n^* -\end{equation} for all $A\in\mathcal{A}$. -Many books point out that both directions are needed to obtain an equivalence relation but no clue is given why. Since for approximate equivalence ($\mathbin{\sim_{\rm a}}$), only one direction is needed, I wonder why for $\mathbin{\sim_{\rm wa}}$ we need two. -Thanks! - -REPLY [2 votes]: I think I have an example. The representations are degenerate, but I don't see any assumption in Davidson that they shouldn't be. -Let $A=B(\ell^2(\mathbb Z_{\geq 0}))$ (or any nonzero $C^*$-subalgebra). Let $P:\ell^2(\mathbb Z)\to\ell^2(\mathbb Z_{\geq 0})$ be orthogonal projection, and define $\pi:A\to B(\ell^2(\mathbb Z))$ by $\pi(a)=P^*aP$ (essentially embedding $A$ in the lower right corner of $B(\ell^2(\mathbb Z))$). Let $U$ be the right shift on $\ell^2(\mathbb Z)$, and define the sequence $(U_n)_{n\geq 1}$ of unitary operators on $\ell^2(\mathbb Z)$ by $U_n=U^n$. Then for all $a\in A$, $\text{WOT-}\lim_n U_n \pi(a) U_n^*=0$. Thus $\pi$ is "halfway" weak approximately equivalent to the $0$ representation $\sigma(a)=0$ on $\ell^2(\mathbb Z)$, but not weak approximately equivalent to the $0$ representation.<|endoftext|> -TITLE: What function has a graph that looks like this? -QUESTION [5 upvotes]: I delete my file which I used to produce this graph. Does anybody have some idea how to produce it again? - -Thanks for a while. - -REPLY [11 votes]: A few key things that jump out looking at this graph: - -It's got a sort of 'radial symmetry' about it: if you look at a circular 'cross-section' centered around the origin then the shape looks roughly the same, just scaled. Similarly all of the 'radial' cross-sections along lines through the origin look roughly the same. This means that it's going to be best expressed as the plot of a function $z=f(r,\theta)$ where the 'base plane' is represented in polar coordinates. Moreover, the structure of it (a sequence of radial spires, of sorts) suggests that it's 'separable' in the sense that it can be written as $f(r,\theta) = g(r)\cdot h(\theta)$ for two individual functions $g$ and $h$. -Going around the origin once seems (though it's unclear from the diagram) to encounter eight distinct peaks (possibly 7, but I think that's an artifact of the projection and there are actually 8), so the 'angular' part of the function, $h$, can be written as $h(\theta) = \sin(8\theta)$. -Going out along any of the radial lines appears to be a straight line, so I'd say the radial portion of the function, $g$, can be written as $g(r) = c\cdot r$ for some small $c$, say $c=0.3$ or something on that order (but that's a function of the scale of the grap). - -So, putting this together, the plot looks to be of an equation roughly like $z=.3r\sin(8\theta)$. - -REPLY [9 votes]: For example : -$$f(x,y)=\sqrt{x^2+y^2}\sin(8 \arctan(y/x))$$<|endoftext|> -TITLE: Is there a Cantor-Schroder-Bernstein statement about surjective maps? -QUESTION [36 upvotes]: Let $A,B$ be two sets. The Cantor-Schroder-Bernstein states that if there is an injection $f\colon A\to B$ and an injection $g\colon B\to A$, then there exists a bijection $h\colon A\to B$. -I was wondering whether the following statements are true (maybe by using the AC if necessary): - -Suppose $f \colon A\to B$ and $g\colon B\to A$ are both surjective, does this imply that there is a bijection between $A$ and $B$. -Suppose either $f\colon A\to B$ or $g\colon A\to B$ is surjective and the other one injective, does this imply that there is a bijection between $A$ and $B$. - -REPLY [27 votes]: For the first one the need of the axiom of choice is essential. There are models of ZF such that $A,B$ are sets for which exists surjections from $A$ onto $B$ and vice versa, however there is no bijection between the sets. -Using the axiom of choice we can simply inverse the two surjections and have injections from $A$ into $B$ and vice versa, then we can use Cantor-Bernstein to ensure a bijection exists. -The second one, I suppose should be $f\colon A\to B$ injective and $g\colon A\to B$ surjective, again we need the axiom of choice to ensure that there is a bijection, indeed there are several models without it where such sets exist but there is no bijection between them. Using the axiom of choice we reverse the surjection and use Cantor-Bernstein again. -It should be noted that without the axiom of choice it is true that if $f\colon A\to B$ is injective then there is $g\colon B\to A$ surjective. Therefore if the first statement is true, so is the second, and if the second is false then so is the first. -Another interesting point on this topic is this: The Partition Principle says that if there is $f\colon A\to B$ surjective then there exists an injective $g\colon B\to A$. Note that we do not require that $f\circ g=\mathrm{id}_B$, but simply that such injection exists. -This principle implies both the statements, and is clearly implied by the axiom of choice. It is open for over a century now whether or not this principle is equivalent to the axiom of choice or not. -Lastly, as stated $f\colon A\to B$ injective and $g\colon B\to A$ surjective cannot guarantee a bijection between $A$ and $B$ with or without the axiom of choice. Indeed the identity map is injective from $\mathbb Z$ into $\mathbb R$, as well the floor function, $x\mapsto\lfloor x\rfloor$ is surjective from $\mathbb R$ to $\mathbb Z$ but there is no bijection between $\mathbb Z$ and $\mathbb R$. - -REPLY [5 votes]: As stated, 2. is clearly false (just take $A=\{ 0,1\} ,B=\{ 0\}$ with $f$ identically zero, and $g$ likewise). I will assume that it's actually $f:A\to B$ and $g:A\to B$. -Using axiom of choice, both statements can be shown to be true, simply because when we have a surjection $f:A\to B$, then by axiom of choice we can choose a right inverse $f^{-1}:B\to A$ which will be injective, so we can reduce both statements to the usual C-B-S. -Without choice, neither statement can be proved. -For the first one, see https://mathoverflow.net/questions/38771 (apparently, it would imply countable choice). -For the second one, see https://mathoverflow.net/questions/65369/half-cantor-bernstein-without-choice.<|endoftext|> -TITLE: On the relationship between Fermats Last Theorem and Elliptic Curves -QUESTION [5 upvotes]: I have to give a presentation on elliptic curves in general. It does not have to be very in depth. I have a very basic understanding of elliptic curves (The most I understand is the concept of ranks). I was wondering if anyone could explain to me simply what is the connection between the equation $x^n +y^n=z^n$ and elliptic curves. - -REPLY [18 votes]: Given non-zero integers $A,$ $B$, and $C$, such that $A + B = C$, we can form the -so-called Frey curve (named after the mathematician Frey, who first considered -elliptic curves in the context of FLT) -$$E: y^2 = x(x-A)(x+B),$$ -which has discriminant (up to some power of $2$ which one can compute precisely, but which I will ignore here) equal to $ABC$. -Suppose now that $A = a^p$, $B = b^p$, and $C = c^p$ (so that we have a solution -to the Fermat equation of exponent $p$). Then the elliptic curve $E$ has -a discriminant which (up to the power of $2$ that I am ignoring) is a perfect $p$th power. -This means that the group of $p$-torsion points $E[p]$ on $E$ (which is a two-dimensional vector space over the field $\mathbb F_p$ of $p$-elements, equipped with an action of the Galois group of $\overline{\mathbb Q}$ over $\mathbb Q$) -has very special properties --- in algebraic number theory terms, it is very close to being unramified. (More specifically, but more technically, it is unramified except possibly at $2$ and $p$, and at $p$ the ramification is very mild --- it is finite flat.) -Now the Shimura--Taniyama conjecture, which is what Wiles (together with Taylor) -proved, shows that $E$, and so $E[p]$, arises from a weight two modular form. Ribet's earlier results on Serre's epsilon conjecture imply that this modular -form must actually be of level $2$. (This is where we use the above information about the ramification.) But there are no non-zero cuspforms of weight $2$ and level $2$, and we get a contradiction. -Although it is much harder (in that the only way we know to rule out the existence of $E[p]$ is by the --- quite difficult --- Shimura--Taniyama -conjecture, or else by related more recent results such as Khare and Wintenberger's work on Serre's conjecture), one can think of the non-existence -of $E[p]$ as being analogous to Minkowski's theorem in algebraic number theory, which says that an everywhere-unramified extension of $\mathbb Q$ cannot exist.<|endoftext|> -TITLE: L'hospital rule for two variable. -QUESTION [25 upvotes]: How to use L'hospital rule to compute the limit of the given function - -$$\lim_{(x,y)\to (0,0)} \frac{x^{2}+y^{2}}{x+y}?$$ - -REPLY [14 votes]: The answer by Christian Blatter is correct. This particular limit is undefined. -I have recently written a paper giving a l'Hospital's rule for many types of multivariable limits. You can find it at -http://arxiv.org/abs/1209.0363<|endoftext|> -TITLE: A question on the compact subset -QUESTION [5 upvotes]: This is an exercise from a topological book. - -Let $X$ is Hausdorff and $K$ is a compact subset of $X$. $\{U_i:i=1,2,...,k\}$ is the open sets of $X$ which covers $K$. How to prove that there exist compact subsets of $X$: $\{K_i:i=1,2,...,k\}$ such that $K=\cup^k_{i=1}K_i$ and for any $i\le k$, $K_i \subset U_i$? - -What I've tried: I try to let $K_i = K\cap U_i$, then it is obvious $K=\cup^k_{i=1}K_i$, however, I'm not sure such $K_i$ is still compact in $X$. I don't know how to go on. -Could anybody help me? Thanks ahead:) - -REPLY [3 votes]: If $x \in U_i$, since $K \backslash U_i$ is compact we can take disjoint open neighbourhoods $V$ and $W$ of $x$ and $K \backslash U_i$ respectively. Then the closure of $V$ is contained in $U_i$. And so each $x \in K$ has an open neighbourhood $V_x$ whose closure is contained in some $U_i$. These form an open cover of $K$, so we can take a finite subcover $V_1, \ldots, V_m$. Let $K_i$ be the union of the closures of those $V_j$ whose closures are contained in $U_i$.<|endoftext|> -TITLE: $\int_{-\infty}^\infty e^{ikx}dx$ equals what? -QUESTION [15 upvotes]: What would $\int\limits_{-\infty}^\infty e^{ikx}dx$ be equal to where $i$ refers to imaginary unit? What steps should I go over to solve this integral? -I saw this in the Fourier transform, and am unsure how to solve this. - -REPLY [3 votes]: I just want to clarify what is meant by saying that "as a distribution", -$$ -2\pi\delta(k) = \int_{-\infty}^\infty e^{ikx}\,dx. -$$ -The right hand side certainly doesn't make sense, and the left hand side can only make sense for $k\not=0$. What we mean though is that for any smooth function $\phi$ with rapid decay, -$$ -2\pi \phi(0) = \int_{-\infty}^\infty \int_{-\infty}^\infty e^{ikx} \phi(k) \,dk\,dx. -$$ -Now both sides make sense. This isn't trivial to prove, and in fact it implies the Fourier inversion formula (using the relationship between modulation and translation under the Fourier transform). The proof often goes through a calculation similar to what @robjohn's answer describes.<|endoftext|> -TITLE: How can I generate the binary representation of any real number? -QUESTION [9 upvotes]: In p. 30 of Baby Rudin, I find a reference to the fact that the binary representation of a real number implies the uncountablity of the set of real numbers. But I have two questions: - -Does every real number have a binary representation? If yes, how do I prove it? -How can I generate the binary representation of a given real number $a$? - -I'm aware of the binary representation of integers, but had never thought of a binary representation of real numbers earlier. - -REPLY [7 votes]: Every real number $\alpha>0$ (to make things simpler) can be developed into an infinite binary expansion of the form $$110\ldots10\,.\,1011101\ldots\ ,$$ -where to the left of the "decimal" point we only have finitely many digits and to the right an unending sequence of digits. Numbers of the form $n/2^N$ $\ (n, N\in{\mathbb N}_{\geq0})$ have exactly two such representations (this is the $0.999\ldots=1.000\ldots$ phenomenon in decimal), all other numbers have exactly one representation. In fact it is possible to construct the full real number system (with addition, multiplication and order) as such an "uncountable list" of binary fractions. This has been sketched by Gowers here: http://www.dpmms.cam.ac.uk/~wtg10/decimals.html -A real number $\alpha>0$ can be considered as "given" when for each $N$ it is possible to name an integer $a_N$ such that -$${a_N\over 2^N}\leq \alpha<{a_N+1\over 2^N}\ .$$ -Using induction one easily proves that -$$2 a_N\leq a_{N+1}\leq 2a_N+1\ ,$$ -and this implies that the finite binary representation of $a_{N+1}$ is obtained from the representation of $a_N$ by appending a $0$ or a $1$. Now the quotients $a_N/2^N$ approximate the given number $\alpha$. Writing $a_N$ in binary -and separating the last $N$ digits by a "decimal" point we therefore get a finite binary approximation of $\alpha$, and things work out such that for $N'>N$ the first $N$ places (after the "decimal" point) of $a_N$ and $a_{N'}$ coincide. In this way we get an ever better approximation of $\alpha$ just by adding new "binary places" to the right.<|endoftext|> -TITLE: Addition and multiplication are continuous in the $I$-adic topology -QUESTION [8 upvotes]: Can you tell me if this is correct? -Let $R$ be a ring and let it have the $I$-adic topology for some ideal $I$ in $R$. I want to show that $+: R \times R \to R$ is continuous at $(x_0, y_0)$. -Proof: Neighbourhoods of zero look like $I^nR$ so that neighbourhoods of $x$ in $R$ look like $x + I^nR$. What I want to show is that for a neighbourhood $x_0 + y_0 + I^{n_0}R$ of $x_0 + y_0$ there exists a neighbourhood $N = I^nR \times I^mR$ of $(x_0, y_0)$ such that for all $(x,y) \in I^nR \times I^mR$ we have $x+y \in x_0 + y_0 + I^{n_0}R$. Pick $n = m = n_0$. Now $+(x_0 + I^{n_0}R \times y_0 + I^{n_0}R) = x_0 + y_0 + I^{n_0}R$ which proves the claim. - -Similarly, $\cdot : R \times R \to R$ is continuous at $(x_0, y_0)$. Let $N=I^{n_0}R$ be a neighbourhood of zero so that $x_0 y_0 + I^{n_0} R$ is a neighbourhood of $x_0 y_0$. Now we want to find $n,m$ such that $\cdot(x_0 + I^n R \times y_0 + I^m R ) \subset x_0y_0 + I^{n_0}R$. Again let $n=m=n_0$. Then $\cdot(x_0 + I^{n_0} R \times y_0 + I^{n_0} R ) = x_0 y_0 + I^{2n_0}R \subset x_0 y_0 + I^{n_0}R $. Which proves the claim. - -Thanks for your help. - -REPLY [2 votes]: As Andrea Mori remarks, the proofs appear correct. -One minor technicality: Due to the appearance of cross terms in $$(x_0+i)(y_0+j) = x_0y_0 + x_0j+iy_0+ij$$ we can only infer that $$(x_0+I^{n_0}R) \cdot (y_0+I^{n_0}R) \subseteq x_0y_0+I^{n_0}R$$ -Of course, this does not change the essence of the proof.<|endoftext|> -TITLE: About the asymptotic formula of Bessel function -QUESTION [16 upvotes]: For $ \nu \in \Bbb R$, I want to prove the well-known formula $$ J_\nu (x) \sim \sqrt{\frac{2}{\pi x}} \cos \left( x - \frac{2 \nu +1}{4} \pi \right) + O \left( \frac{1}{x^{3/2}} \right) \;\;\;\;(x \to \infty)$$ where $J_\nu$ denotes the Bessel function. How can I show this? Or would you tell me the Internet site which proves this formula? I could not find the proof of this. - -REPLY [13 votes]: First, some preliminary series expansions. Consider the substitution -$$ -\cos(t)=1-u^2/2\tag{1} -$$ -We get the power series for $u=2\sin(t/2)$: -$$ -u=t-t^3/24+t^5/1920-t^7/322560+t^9/92897280+O(t^{11})\tag{2} -$$ -and the inverse series for $t$; -$$ -t=u+u^3/24+3u^5/640+5u^7/7168+35u^9/294912+O(u^{11})\tag{3} -$$ - -We need to concentrate on the stationary points at $t=\pi/2$ and $t=-\pi/2$, away from which the integral decays exponentially in $x$. The contribution at $-\pi/2$ is the conjugate of the contribution at $\pi/2$, so the whole contribution is twice the real part of the contribution at $\pi/2$. -$$ -\begin{align} -J_\nu(x) -&=\frac1{2\pi}\int_{-\pi}^\pi e^{-i(\nu t-x\sin(t))}\,\mathrm{d}t\\ -&=2\mathrm{Re}\left(\frac1{2\pi}\int_0^\pi e^{-i(\nu t-x\sin(t))}\,\mathrm{d}t\right)\\ -&=2\mathrm{Re}\left(e^{-i\nu\pi/2}\frac1{2\pi}\int_{-\pi/2}^{\pi/2} e^{-i\nu t}e^{ix\cos(t)}\,\mathrm{d}t\right)\\ -&\sim2\mathrm{Re}\left(e^{-i\nu\pi/2}\frac1{2\pi}\int_{-\infty}^\infty\left(1-\nu^2u^2/2+O(u^4)\right)e^{ix(1-u^2/2)}\left(1+u^2/8+O(u^4)\right)\,\mathrm{d}u\right)\\ -&=2\mathrm{Re}\left(e^{i(x-\nu\pi/2)}\frac1{2\pi}\int_{-\infty}^\infty\left(1-(4\nu^2-1)u^2/8+O(u^4)\right)e^{-ixu^2/2}\,\mathrm{d}u\right)\\ -&=2\mathrm{Re}\left(e^{i(x-(2\nu+1)\pi/4)}\frac1{2\pi}\int_{-\infty}^\infty\left(1+i(4\nu^2-1)v^2/8+O(v^4)\right)e^{-xv^2/2}\,\mathrm{d}v\right)\\ -&=2\mathrm{Re}\left(e^{i(x-(2\nu+1)\pi/4)}\frac1{2\pi}\left(\sqrt{\frac{2\pi}{x}}+i\frac{4\nu^2-1}{8}\frac1{2\pi}\sqrt{\frac{2\pi}{x}}^3+O\left(x^{-5/2}\right)\right)\right)\\ -&=\cos\left(x-\frac{2\nu+1}{4}\pi\right)\sqrt{\frac{2}{\pi x}}-\sin\left(x-\frac{2\nu+1}{4}\pi\right)\frac{4\nu^2-1}{8}\sqrt{\frac{2}{\pi x^3}}+O\left(x^{-5/2}\right) -\end{align} -$$ -In the $\sim$ step, we make the $u$ substitution whose series is given above, and integrate over $(-\infty,\infty)$ instead of $[-\sqrt{2},\sqrt{2}]$ since the part outside the compact interval decays exponentially. -We also use the substitution $u=e^{-i\pi/4}v$, so that $u^2=-iv^2$, and change the path of integration, which is allowed since there are no singularities. -In the end, we get the first two terms of the asymptotic expansion of $J_\nu(x)$.<|endoftext|> -TITLE: How to deal with $A^{26}=I$? -QUESTION [6 upvotes]: I got stuck in this problem: -Let $A:\mathbb{R}^{6}\rightarrow \mathbb{R}^{6}$ be a linear transformation. Assume $A^{26}=I$, prove that $R^{6}=\oplus_{i=1}^{3} V_{i}$, with $AV_{i}\subset V_{i}$(the explicit condition is $V_{i}$ are 2-dimensional invariant subspaces of $\mathbb{R}^{6}$ under $A$). -My thought is $A$ must have a minimal polynomial of degree less or equal to 6. Thus since it divides $x^{26}-1$, the only choices are: $$x-1,x+1,x^{2}-1$$ since the rest term $$(x^{13}-1)/(x-1)*(x^{13}+1)/(x+1)$$ has factors irreducible and degree higher than 6. And the claim is trivial in the case $A=\pm I$. But I do not know how to deal with the case $A^{2}-I=0$ - $A$ can only have eigenvalues $1$ and $-1$, but how this helps to solve the problem? - -Edit: -In the light of did's comments $\sum^{12}_{i=0}x^{i}$ and $\sum^{12}_{i=0}(-1)^{i}x^{i}$ can be reducible over the reals in pairs of 6 quadratics, and the corresponding $A$'s are rotations. But I still feel rather confused as if problem is solved at this stage by suggesting $A$'s minimal polynomial must be a product of $$x-1,x+1,x^{2}-1, x^{2}-\cos[\theta]x+1$$ -which are dealt with respectively by $I,-I$, selecting linearly independent vectors and run with $A$, and selecting the rotational invariant subspace. Since obviously cases like $$(x\pm 1)(x^{2}-\cos[\theta]x+1)$$ or even $$(x+1)(x-1)^{2}$$ could happen. - -REPLY [3 votes]: The decomposition of $x^{26}-1$ into irreducible factors over $\mathbb R$ is $$ -(x-1)\cdot(x+1)\cdot\prod\limits_{k=1}^{12}p_k(x),\qquad p_k(x)=x^2-2\cos(k\pi/13)x+1. -$$ -This almost determines the minimal polynomial $\mu_A(x)$ of $A$, since $\mu_A(x)$ divides $x^{26}-1$. Hence $\mu_A(x)$ is the product of at most three factors $p_k(x)$ and possibly a factor $x-1$ and possibly a factor $x+1$. In any case, $\mu_A(x)$ has no repeated irreducible factor hence $A$ is equivalent over $\mathbb R$ to a matrix diagonal by blocks, with (1.) possibly an even number of blocks of size $1\times1$ equal to $+1$ or to $-1$ and (2.) possibly some blocks of size $2\times2$ equal to rotation matrices -$$ -\begin{pmatrix}\cos(k\pi/13)&\sin(k\pi/13)\\ -\sin(k\pi/13)&\cos(k\pi/13)\end{pmatrix}. -$$ -This yields the desired decomposition as follows: choose each plane left invariant by one of the rotation matrices, if there are some, and complete by any planes made of eigenvectors with eigenvalues $\pm1$, if necessary. (Note that the result is such that $AV_i=V_i$ for every $i$.)<|endoftext|> -TITLE: How to show $A$ cannot have more than one Jordan block for any eigenvalue? -QUESTION [5 upvotes]: I got stuck in this problem from Spring 99, Berkeley Problems in Mathematics: -Let $A$ be a $n\times n$ matrix such that $a_{ij}\not=0$ if $i=j+1$ but $a_{ij}=0$ if $i\ge j+2$. Prove that $A$ cannot have more than one Jordan block for any eigenvalue. -I thought the matrix would satisfy some obvious relationship like $A^{2}=0$, but I realized the entries not listed are not even specified; thus such a gross simplification cannot hold. Working on toy examples does not tell me much, so I decided to ask in here. - -REPLY [3 votes]: So, let $\lambda$ be an eigenvalue. The number of Jordan Blocks for $\lambda$ is well known to be the geometric multiplicity of $\lambda$. As the geometric multiplicity of an eigenvalue is the dimension of the nullspace of $A - \lambda I$, therefore, we need to show that this dimension is $1$. Hence, we only need to show that the nullspace of $B = A - \lambda I$ has only one vector in it (i.e. all the vectors in the null space are multiples of that vector). -To show this, let $v_1, v_2, \cdots, v_n$ be the column vectors of $B$. Note that if $v = (a_1, a_2, \cdots, a_n)$ is a vector in the nullspace, then $a_1 v_1 + a_2 v_2 + \cdots + a_nv_n = 0$. Let's fix $a_1 =1$. Comparing the first component of the columns, gives a non trivial relationship between $a_1,a_2$ as the rest of the components are $0$. Hence, $a_2$ is fixed. Similarly, comparing second component fixes $a_3$. Continuing this procedure, we get the assertion.<|endoftext|> -TITLE: Real life applications of general vector spaces -QUESTION [20 upvotes]: Students familiar with Euclidean space find the introduction of general vectors spaces pretty boring and abstract particularly when describing vector spaces such as set of polynomials or set of continuous functions. Is there a tangible way to introduce this? Are there examples which will have a real impact? I would like to introduce this in an engaging manner to introductory students. Are there any real life applications of general vector spaces? - -REPLY [2 votes]: Well you could talk about the word vectors? Or even thought vectors, really any time you want a categorical piece of data to be represented in a unique relational database, you can give that piece of data numerical "features", or in other words, dimensions that allow one to classify people, places, words, things, literally anything you want. -For example Netflix vectorizes movies, and they actually then insert the user as a vector into the same vector space as the movies to get an idea of what other movies to suggest to the user. Vectors are heavily used in machine learning and have so many cool use cases. -What's really awesome about those kinds of examples is you don't really have to understand the linear algebra to even really understand what is going on, at least from a conceptual point of view. Geoff Hinton (one of the inventors of back-propagation) does work with thought vectors, I really suggest reading up on that because the applications are literally endless. Like Google is developing an algorithm that can flirt with thought vectors. They already have a program that can answer email for you based on the same technology. -Hope this helps!<|endoftext|> -TITLE: Integral:$ \int^\infty_{-\infty}\frac{\ln(x^{2}+1)}{x^{2}+1}dx $ -QUESTION [12 upvotes]: How to evaluate: -$$ \int^\infty_{-\infty}\frac{\ln(x^{2}+1)}{x^{2}+1}dx $$ -Maybe we can evaluate it using the well-known result:$\int_{0}^{\frac{\pi}{2}} \ln{\sin t} \text{d}t=\int_{0}^{\frac{\pi}{2}} \ln{\cos t} \text{d}t=-\frac{\pi}{2}\ln{2}$ -But how do I evaluate it, using that ? - -REPLY [8 votes]: Let -$$I(\alpha)=\int_{-\infty}^{\infty}\frac{\ln(\alpha x^2+1)}{x^2+1}dx.$$ -Then -\begin{eqnarray*} -I'(\alpha)&=&\int_{-\infty}^{\infty}\frac{x^2}{(\alpha x^2+1)(x^2+1)}dx\\ -&=&\frac{1}{\alpha-1}\int_{-\infty}^{\infty}\left[\frac{1}{x^2+1}-\frac{1}{\alpha x^2+1}\right]dx\\ -&=&\frac{\pi}{\sqrt{\alpha}+\alpha} -\end{eqnarray*} -and hence -\begin{eqnarray*} -I(\alpha)&=&\int\frac{\pi}{\sqrt{\alpha}+\alpha}d\alpha\\ -&=&2\pi\ln(1+\sqrt{\alpha})+C. -\end{eqnarray*} -Clearly $I(0)=0$ implies $C=0$. Thus -$$\int_{-\infty}^{\infty}\frac{\ln(x^2+1)}{x^2+1}dx=I(1)=2\pi\ln 2.$$<|endoftext|> -TITLE: Tips for an adult to learn math -- from the beginning. -QUESTION [16 upvotes]: First let me start with I am an adult and I can't do simple maths. I some how got through all of my math courses in University (after several attempts) but I honestly couldn't tell you how... -I cannot do these: - -Add/Subtract with decimals -Add/Subtract fractions -Basic multiplication -Basic division -More advanced math that uses these principles - -How can I go about learning these things now? Is there a particular book that I can study from (with worksheets). I'd really rather not do exercises that involve connecting dots to create pictures, or coloring things... -As a side note, I believe I suffer from dyscalculia. - -REPLY [2 votes]: Thinking Blocks is a great resource for understanding ratios and proportions.(And it also has addition and subtraction. It's very interactive. It guides you one step at a time, and gives you feedback every step, so you know if your thinking is correct. You don't have to wait until the end of the problem to find out.<|endoftext|> -TITLE: Complex polynomial and the unit circle -QUESTION [10 upvotes]: Given a polynomial $ P(z) = z^n + a_{n-1}z^{n-1} + \cdots + a_0 $, such that - $\max_{|z|=1} |P(z)| = 1 $ -Prove: $ P(z) = z^n $ -Hint: Use cauchy derivative estimation - $$ |f^{(n)} (z_0)| \leq \frac{n!}{r^n} \max_{|z-z_0|\leq r} |f(z)| $$ and look at the function $ \frac{P(z)}{z^n} $ - -It seems to be maximum principle related, but I can't see how to use it and I can't understand how to use the hint. - -REPLY [4 votes]: Another way to do it is to use the orthogonality of the exponentials to see -$$\frac{1}{2\pi}\int_0^{2\pi}|P(e^{it})|^2\,dt = 1^2 +|a_{n-1}|^2 + \cdots + |a_0|^2.$$ -Since $|P|\le 1$ on the circle, the expression on the left is $\le 1.$ It follows that $a_k=0$ for $k=0,\dots ,n-1.$<|endoftext|> -TITLE: How to measure the volume of rock? -QUESTION [6 upvotes]: I have a object which is similar to the shape of irregular rock like this -I would like to find the volume of this. How to do it? -If I have to find the volume, what are the things I would need. eg., If it is cylindrical, I would measure length and diameter. But, it is irregularly shaped. Like the above rock. -Where should I start? Couple of google search says something related to integration and contours. Somebody pls give me some handle :) I would say i'm very beginner level in math. -Many Thanks :) -Edit: -60 to 70% accuracy would be helpful. - -REPLY [3 votes]: There is a very simple and elegant algorithm for calculating the volume of an arbitrary, closed mesh described in this paper (Zhang, Cha, and Tsuhan Chen. "Efficient feature extraction for 2D/3D objects in mesh representation." Image Processing, 2001. Proceedings. 2001 International Conference on. Vol. 3. IEEE, 2001. Page 2). The trick is to base a tetrahedron on each triangle and top it off at the origin and then sum up all volumes, signed by whether the triangle faces the origin.<|endoftext|> -TITLE: Is the product of two sets well-defined if one is empty -QUESTION [8 upvotes]: Let $X$ be a set. What is $X\times \emptyset$ supposed to mean? Is it just the empty set? - -REPLY [9 votes]: Recall the definition of $A\times B$: $z\in A\times B$ if and only if $z=\langle a,b\rangle$ where $a\in A$ and $b\in B$. That is $A\times B$ is the set of all ordered pairs whose first coordinate is in $A$ and second in $B$. -If $B$ is empty then there are no ordered pairs $\langle a,b\rangle$ such that $b\in\varnothing$, therefore $A\times\varnothing=\varnothing$. Similarly $\varnothing\times B=\varnothing$.<|endoftext|> -TITLE: Prove $\sin^2(A)+\sin^2(B)-\sin^2(C)=2\sin(A)\sin(B) \cos(C)$ if $A+B+C=180$ degrees -QUESTION [9 upvotes]: I most humbly beseech help for this question. -If $A+B+C=180$ degrees, then prove -$$ -\sin^2(A)+\sin^2(B)-\sin^2(C)=2\sin(A)\sin(B) \cos(C) -$$ -I am not sure what trig identity I should use to begin this problem. - -REPLY [3 votes]: $\sin^2A+\sin^2B-\sin^2C=\sin^2A+\sin(B+C)\sin(B-C)$ - either using the identity $\sin^2B-\sin^2C=\sin(B+C)\sin(B-C)$ -or $\sin^2B-\sin^2C=\frac{1}{2}(2\sin^2B-2\sin^2C)=\frac{1}{2}(1-\cos2B-(1-\cos2C))=\frac{1}{2}(\cos2C-\cos2B)=-\frac{1}{2}2\sin(B+C)\sin(C-B)=\sin(B+C)\sin(B-C)$ -as $\sin(-x)=-\sin(x)$ -Now, $\sin(B+C)=\sin(180^\circ-A)=\sin{A}$ -So, $\sin^2A+\sin^2B-\sin^2C=\sin^2A+\sin{A} \sin(B-C)$ -$=\sin{A}(\sin{A}+\sin(B-C))$ -$=\sin{A}(\sin(B+C)+\sin(B-C))$ replacing $\sin{A}$ with $\sin(B+C)$ -$=\sin{A}(2\sin{B}\cos{C})$ -$=2\sin{A}\sin{B}\cos{C}$<|endoftext|> -TITLE: Maximum area of a triangle -QUESTION [14 upvotes]: [Edit:I had a couple of links to the original probelm but they have gone the way of all things.] -I have been attempting to solve this problem: - -Given three concentric circles of radii 1, 2, and 3, respectively, find the maximum area of a triangle that has one vertex on each of the three circles. - - -Here is a partial solution (not my own) which I have edited it a little for clarity. Note that $A=1$, $B=2$ and $C=3$: - -Let radii A,B, and C be at angles a,b, and c respectively. Position radius A on the positive x-axis at angle $a=0$ (no loss in generality). From the equation for triangle area -(1) area = $\frac12 BC \sin(b-c) + \frac12 CA \sin(c) + \frac12 AB \sin (2\pi -b)$. -Take the total partial of -area w.r.t $b$ and $c$ and set equal to $0$. This gives -(2) $C \cos c = B \cos b$. -Also, from condition (2) extended radii are perpendicular to the triangle side. Next, -the value of angle $b$ is determined. A little fancy geometry shows that $b$ is $225^o$ from -$A$ ($-45^o$ in the third quadrant). From (2) angle $c$ is obtained. - -I am happy with the expression for the triangle's area, and also with the differentiation and derivation of $C\cos c = B\cos b$. -But I don't see why the extended radii are perpendicular to the triangle's sides, which makes the centre of the concentric circles the orthocentre of the triangle. And I'm also not seeing the "fancy geometry" that gives the angle $b$, nor, indeed, why angle $b$ is constant. -Could someone please explain what's happening here? - -REPLY [4 votes]: Some algebraic computation -The accepted answer by robjohn explains nicely why the origin has to be the orthocenter. I want to suggest an alternative approach to the computation that follows. To do so, I want to use coordinates for the points in question: -$$O=(0,0)\qquad A=(1,0)\qquad B=(x,b)\qquad C=(x,c)$$ -The use of the same $x$ coordinate for both $B$ and $C$ already ensures that $OA$ is perpendicular to $BC$. You have three conditions to determine these three variables: -\begin{align*} -\lVert B\rVert&=2 & x^2+b^2 &= 2^2 \\ -\lVert C\rVert&=3 & x^2+c^2 &= 3^2 \\ -\langle C-A,B\rangle&=0 & x^2+bc-x &= 0 -\end{align*} -The scalar product expresses the condition $(C-A)\perp B$, while the third condition $(C-B)\perp A$ is already implied by these conditions (since the altitudes of any triangle meet in a single point). -You can feed this to Wolfram Alpha to obtain numeric approximations of the relevant coordinates. But you can also eliminate variables, e.g. using resultants, to obtain polynomials which describe your solutions: -\begin{align*} -x^3 - 7x^2 + 18 &= 0 \\ -b^6 + 37b^4 - 92b^2 + 36 &= 0 \\ -c^6 + 22c^4 - 387c^2 + 1296 &= 0 -\end{align*} -No ruler and compass solution possible -These are irreducible cubic polynomials in $x,b^2,c^2$ resp. so they cannot be the result of a ruler and compass construction. Every ruler and compass construction which starts from integer coordinates (like the points $O,A$ and integer radii) can only result in coordinates which are sums, differences, products, quotients or square roots of other constructible coordinates. Cubic roots are not possible. -So the main result of this investigation is this: -The triangle of maximal area cannot be constructed using ruler and compass. -Anyone looking for a “fancy geometry” construction to achieve this can stop now, unless he is knowingly looking for a good approximation only. -The value of the maximal area -Using a little bit of computation on the above coorinates using algebraic numbers in sage shows that the area $a$ satisfies the equation -$$16a^6-392a^4+133a^2+900=0$$ -so that again is a cubic polynomial in $a^2$. Feeding that polynomial into WA (surprisingly only when omitting commands like “solve”) I get an ugly but exact expression for the solution in question: -$$a=\frac12\sqrt{\frac13\left(98+ -\frac{9205}{\sqrt[3]{833939+7974i\sqrt{1329}}}+ -\sqrt[3]{833939+7974i\sqrt{1329}}\right)}$$ -This is interesting because the post you referred to claims a much simpler formula, but that was based on the incorrect $45°$ assumption robjohn already disproved. The last post by ferret, however, results in the same area I computed myself. So they, too, eventually found an exact solution.<|endoftext|> -TITLE: Is there a version of the Gershgorin circle theorem that is suitable for nearly triangular matricies? -QUESTION [6 upvotes]: The Gershgorin circle theorem, http://en.wikipedia.org/wiki/Gershgorin_circle_theorem, gives bounds on the eigenvalues of a square matrix, and works well for nearly diagonal matrices. -For a triangular matrix, however, the bounds are not useful in general, despite the fact that the eigenvalues are known to be the diagonal elements. Is there a version (or can some helpful person develop a version) of the Gershgorin circle theorem that gives more useful bounds in the nearly triangular case? - -REPLY [2 votes]: There is a difference between diagonal and triangular matrices. Diagonal matrices are normal, so its eigenvalues are well-conditioned. Gershgorin's theorem makes this explicit. Triangular matrices are not necessarily normal, so its eigenvalues can be ill-conditioned. Explicit example: the eigenvalues of $\bigl[ \begin{smallmatrix} 0 & 1 \\ \epsilon & 0 \end{smallmatrix} \bigr]$ are $\pm\sqrt\epsilon$. -This means that perturbation results for triangular matrices cannot be as strong as for diagonal matrices. Gershgorin's theorem can be formulated as follows: If $D$ is a diagonal matrix, then the eigenvalues of $D+E$ are within $|E|_1$ of the eigenvalues of $D$, where the norm is defined by $|E|_1 = \sum_{i,j} |e_{ij}|$. However, the example shows that there cannot be a theorem of the form: If $T$ is a triangular matrix, then the eigenvalues of $T+E$ are within $C|E|_*$ of the eigenvalues of $T$ (for some norm $|\cdot|_*$ and some constant $C$).<|endoftext|> -TITLE: Derivative of convolution -QUESTION [38 upvotes]: Assume that $f(x),g(x)$ are positive and are in $L^1$. Moreover, they are differentiable and their derivative is integrable. Let $h(x)=f(x)*g(x)$, the convolution of $f$ and $g$. Does the derivative of $h(x)$ exist? If yes, how can we prove that -$$ \frac{d}{dx}(f(x)*g(x)) = \left(\frac{d}{dx}f(x)\right)*g(x)$$ -Thanks - -REPLY [24 votes]: Definition: $$h(x)=f*g(x)=\int_A f(x-t)g(t)dt$$ where A is a support of function $q()$, i.e. $A=\{t:q(t)\ne 0\}$ -Let's calculate derivative: -$$\frac {dh}{dx}=\underset{dx\rightarrow0}{\lim} \frac {(\int_A f(x+dx-t)g(t)dt-\int_A f(x-t)g(t)dt)}{dx}=\underset{dx\rightarrow0}{\lim}(\int_A \frac{(f(x+dx-t)-f(x-t))}{dx}g(t)dt)$$ -If we assume that there exists some integrable function $q(t)$, such that for $t$ almost everywhere -$$ -\left| \frac{(f(x+dx-t)-f(x-t))}{dx} \right| < q(t), \forall dx>0 -$$ -I.e. -$$ -\mu\{t: \left| \frac{(f(x+dx-t)-f(x-t))}{dx} \right| \ge q(t)\}=0,\forall dx >0 -$$ -then by the Lebesgue dominated convergence theorem we can push the limit inside integral. -$$\frac {dh}{dx}=\frac{d}{dx}(f*g(x))=\int_A f'(x-t)g(t)dt=f'*g$$ -Under assumption that: $\int_A q(t)dt$ is bounded above. -One situation is when A is a compact set and $f,g$ are continuous function in the set A with a finite number of dicontinuities.<|endoftext|> -TITLE: Dependence of the Sobolev embedding constants on the domain -QUESTION [6 upvotes]: Let $\Omega$ be a sufficiently nice domain in $\mathbb{R}^n$. -If $ 1 \leq p < n $ and $ p^* = \frac{np}{n-p} $ then there exists a constant $C_1$ such that for all $ u \in W^{1,p}(\Omega) $ we have -$$ (I)~~~~||u||_{L^{p^*}(\Omega)} \leq C_1||u||_{W^{1,p}(\Omega)}. $$ -If $ p > n $ then there exists a constant $C_2$ such that for all $ u \in W^{1,p}(\Omega) $ we have -$$ (II)~~~~||u||_{L^{\infty}(\Omega)} \leq C_2||u||_{W^{1,p}(\Omega)}. $$ -In general, the constants $C_i$ depend on the domain $\Omega$. Can someone point me to some references that discuss the dependence between the embedding constants and the domain? -I'm interested in conditions under which, given some family of domains $ \Omega_\alpha $, I can get Sobolev embedding inequalities as above with a constant that doesn't depend on $\alpha$. To be even more specific, I'm interested in the case $ n = 2 $ and when the domains are families of balls or annuli. -For example, if I consider inequality (II) and a family of balls, then an obvious sufficient condition is to have both an upper and a lower bound on the radii of the balls. One can't get away without a lower bound (consider the function $u \equiv 1$) but can get away without an upper bound by using a translation argument. What about inequality (I)? What can I use to answer such questions? -Since one way to prove the inequalities above is to use an extension operator, and then "steal" the inequality from $\mathbb{R}^n$, this question is related to dependence of the minimal norm of an extension operator $W^{1,p}(\Omega) \rightarrow W^{1,p}(\mathbb{R}^n)$ on the domain - -REPLY [2 votes]: By a scaling argument, one does not need upper bound on the radii in (I) as well. The lower bound is needed because of the inhomogeneous term in the Sobolev norm in the right hand side. If you use the Sobolev seminorm $\|Du\|_{L^p}$ instead of the full norm you will have a scale independent inequality.<|endoftext|> -TITLE: Who came up with the Euler-Lagrange equation first? -QUESTION [5 upvotes]: Could someone explain who came up with the specific equation first? -http://en.wikipedia.org/wiki/Euler-Lagrange -makes it sound like Lagrange got it first, in 1755, then sent it to Euler. -but: -http://en.wikipedia.org/wiki/Calculus_of_variations -sort of makes it sound like Euler got it first in the 1730s. -It seems like a straightforward question, but I can't find an answer anywhere. Who came up with the equation, Euler or Lagrange? And what precisely did the other man contribute to get his name on there? - -REPLY [3 votes]: I'm writing this partially not to let this question go unanswered and partially to include some details that I didn't find at the MO post. -There is a book by Herman Goldstine titled "A History of the Calculus of Variations from the 17th through the 19th Century" that covers the history and development of the Euler-Lagrange equation (and much more, naturally), and it covers this topic well. In it, Goldstine writes that Euler first discovered what we now call the Euler-Lagrange equation prior to April 15, 1743, which we know as a result of a letter from that date sent by Euler to Daniel Bernoulli containing his discovery. Euler then published this finding to a broader audience in his 1744 Methodus Inveniendi. -Euler's derivation approximated a curve by $N$ points and then let $N$ go to infinity to find extremals. This method was somewhat tedious in its implementation and Euler himself was interested in finding a method that did not rely on any geometry as his method did. -Eleven years later, in a letter dated August 12, 1755, Lagrange (at just 19 years old) sent Euler a letter in which he re-derived Euler's result using purely analytical methods. Lagrange's derivation was powerful enough to handle other types of problems and had Euler's earlier result as a nearly automatic consequence. Euler himself much preferred Lagrange's derivation and gave it the name "calculus of variations," and it is essentially Lagrange's technique that is used today. -The name of the equation, then, is very reasonable given that Euler found it first and Lagrange refined his approach.<|endoftext|> -TITLE: Equivalents norms in Sobolev Spaces -QUESTION [6 upvotes]: I know that this is classical but I have never do the calculations to show that the norms in the sobolev space $W^{k,p}(\Omega)$ -\begin{equation} -\|u\|_{k,p,\Omega}= \Bigl(\int_{\Omega} \sum_{|\alpha| \le k}|D^{\alpha} u |^{p} dx \Bigr)^{1/p} -\end{equation} -and -\begin{equation} -\|u\|_{W^{k.p}(\Omega)} =\sum_{|\alpha|\le k}\|D^{\alpha}u\|_{p} -\end{equation} -are equivalents. - -REPLY [6 votes]: Let $N$ the number of $\alpha$ such that $|\alpha|\leq k$, and enumerate them as $\alpha^{(1)},\dots,\alpha^{(N)}$. Since all the norms are equivalent on $\Bbb R^N$, we can find two constants $C_1,C_2>0$ such that for all $(a_1,\dots,a_N)\in\Bbb R^N$, -$$C_1\sum_{j=1}^N|a_j|\leq \left(\sum_{j=1}^N|a_j|^p\right)^{\frac 1p}\leq C_2\sum_{j=1}^N|a_j|.$$ -Then apply this inequality to $a_j:=\left(\int_{\Omega}|D^{\alpha^{(j)}}u(x)|^pdx\right)^{\frac 1p}$.<|endoftext|> -TITLE: Linear isometry and operator norm $=1$ -QUESTION [10 upvotes]: For some reason I used to think that if $T$ is a linear operator on normed spaces $V \to W$ then saying $T$ is an isometry is the same as saying $\|T\|_{op} = 1$. -Well, I got stuck on a proof and subsequently looked up the definition and realised that the definition of isometry is that a linear operator $T$ is an isometry if $\|Tx\| = \|x\|$. -Now I've been wondering whether we have that $\|T\|_{op} = 1$ implies $T$ is an isometry? -The other direction holds: if $\|Tx\| = \|x\|$ for all $x$ then $\frac{\|Tx\|}{\|x\|} = 1$ for all $x \neq 0$ and hence $\sup_{\|x\|=1}\|Tx\| = \sup_{\|x\|=1} \frac{\|Tx\|}{\|x\|} = \|T\| = 1$. -Thanks for your help. - -REPLY [9 votes]: For example, on a finite dimensional space, $\|T\|=1$ if and only if for all $x$, $\|Tx\|\leq \|x\|$, and there exists $x_0\neq 0$ such that $\|Tx_0\|=\|x_0\|$. On the other hand, for $T$ to be an isometry requires for all $x$, $\|Tx\|=\|x\|$. You can give many examples of the former that do not satisfy the latter if the dimension is greater than $1$, as paul garrett commented, even diagonal matrices as Qiaochu Yuan commented. The former condition even allows $T$ to have nontrivial kernel as Martin Sleziak commented. -As David Mitra commented, if $T\neq 0$ then $\dfrac{T}{\|T\|}$ has norm $1$, and note that if there exist nonzero vectors $x$ and $y$ such $\dfrac{\|Tx\|}{\|x\|}\neq \dfrac{\|Ty\|}{\|y\|}$, then $T$ is not a multiple of an isometry. For example, if $T$ is nonzero and noninjective, take $x\in\ker T$ and $y\not\in\ker T$.<|endoftext|> -TITLE: On linearly independent matrices -QUESTION [12 upvotes]: I have been reading J.S. Milne's lecture notes on fields and Galois theory and came across the normal basis theorem (Thm 5.18 on page 66 of the notes). Trying to find my own proof, the following problem in linear algebra quickly arose: - -Question: Let $F$ be a field. Given linearly independent matrices $A_1, \dots, A_n \in \operatorname{GL}_n(F)$, does there necessarily exist some $b\in F^n$ such that $A_1b, \dots, A_nb$ are linearly independent over $F$? - -This is clearly not true if the matrices are not linearly independent. Also, if they are not invertible, the claim is false in general: e.g. set $n=2$ and consider -$$A_1 = \begin{pmatrix} 1 & 0 \\ 0& 0\end{pmatrix},\quad A_2 = \begin{pmatrix} 0 & 1 \\ 0& 0\end{pmatrix}$$ -Going through a case-by-case analysis, I think I can prove the claim for $n=2$, so there seems to be some hope... -Any help would be appreciated. Thanks! - -REPLY [10 votes]: The answer is no. -Consider, for $F=\mathbb{R}$, $n=3$, -$$ -A_1=\begin{bmatrix}1&0&0\\0&1&0\\ 1&2&3\end{bmatrix}, \ -A_2=\begin{bmatrix}1&0&0\\0&1&0\\ 2&3&1\end{bmatrix}, \ -A_3=\begin{bmatrix}1&0&0\\0&1&0\\ 3&1&2\end{bmatrix}. -$$ -These three matrices are linearly independent, because the last three rows are. They are invertible, since the are triangular with nonzero diagonal. -For any $b=\begin{bmatrix}x\\ y\\ z\end{bmatrix}\in\mathbb{R}^3$, the three vectors we obtain are -$$ -A_1b=\begin{bmatrix}x\\ y\\ x+2y+3z\end{bmatrix}, \ -A_2b=\begin{bmatrix}x\\ y\\ 2x+3y+z\end{bmatrix}, \ -A_3b=\begin{bmatrix}x\\ y\\ 3x+y+2z\end{bmatrix}. \ -$$ -And for any choice of $x,y,z$, these vectors are linearly dependent. To see this, we need to find coefficients, not all zero, such that $\alpha\,A_1b+\beta\,A_2b+\gamma\,A_3b=0$. The first two rows require $\alpha+\beta+\gamma=0$. And the third row requires $(x+2y+3x)\alpha+(2x+3y+z)\beta+(2x+y+2x)\gamma=0$. As it is a homogeneous system of two equations in three variables, it will always have nonzero solutions.<|endoftext|> -TITLE: Fast algorithms for calculating the Möbius inversion -QUESTION [10 upvotes]: Recall the Möbius inversion formula: if we have two functions $f,g \colon \mathbf{N} \to \mathbf{N}$ such that -$$g(n) = \sum_{k=1}^n f\left(\left\lfloor \frac{n}{k} \right\rfloor\right)$$ -holds for every $n \in \mathbf{N}$, then the values of $f$ can be recovered as -$$f(n) = \sum_{k=1}^n \mu(k) g\left(\left\lfloor \frac{n}{k} \right\rfloor\right),$$ -where $\mu(k)$ is the Möbius function. -I am interested in fast algorithms for calculating a single value $f(n)$, assuming that $g$ can be calculated in $O(1)$ time. -The best algorithm I know of is as follows: There are $O(\sqrt{n})$ different values $\left\lfloor \frac{n}{k} \right\rfloor$, $1 \le k \le n$. If we have already calculated $f\left(\left\lfloor \frac{n}{k} \right\rfloor\right)$ for $2 \le k \le n$, then -$$f(n) = g(n) - \sum_{k=2}^n f\left(\left\lfloor \frac{n}{k} \right\rfloor\right)$$ -can be calculated in $O(\sqrt{n})$ time by counting the multiplicities of the terms in the sum. By calculating the $f\left(\left\lfloor \frac{n}{k} \right\rfloor\right)$ from bigger to lower $k$ the total time required will then be $O(n^{3/4})$ while using $O(\sqrt{n})$ space. -I'm also interested in possible improvements to the above algorithm, even if they reduce the required time just by a constant factor. - -REPLY [8 votes]: You can improve the $O(n^{3/4})$ algorithm by a constant factor of roughly three by computing only the odd summands of $k$ by using the identity: -$$\begin{array}{lll} - f \left( n \right) & = & g \left( n \right) - g \left( \frac{n}{2} \right) - - \sum_{3 \leq k \leq n, k \:\mathrm{odd}} f \left( \left\lfloor \frac{n}{k} - \right\rfloor \right) -\end{array}$$ -which can be derived by subtracting $g(\frac{n}{2})$ from $g(n)$. The odd-only approach is hinted at in [Lehman 1960] (search for "in practice"). -You can further improve the time complexity to $O(n^{2/3+\epsilon})$ if you can compute $f(k)$ for each $k\in \left\{1,..,u \right\}$ in $O(u^{1+\epsilon})$, as is typical for arithmetic functions by using a (possibly segmented) sieve. -Two references for the $O(n^{2/3})$ approach are [Pawlewicz 2008] and [Deleglise 1996].<|endoftext|> -TITLE: How to disprove there exists a real number $x$ with $x^2 < x < x^3$ -QUESTION [5 upvotes]: I realize that the only method is to show various cases: -I must test for $x > 1$, $x < -1$, $0 \leq x \leq 1$, and $-1\leq x \leq0$. -But even with this, I don't understand how to inject the properties of these four distinct possible $x$'s into the inequality (from the title) in order to show that none of these work. -Thanks for any help - -REPLY [2 votes]: For the full inequality to be true, both halves must be true. Let's take $x^2 < x$ first. For any $x<0$, $x^2 > x$ because $x^2 = |x|^2$; all squares are nonnegative and so are greater than any negative root. For any $x>1$, $x^2>x$. That means that the only values for which the first inequality is true is where $0 < x < 1$. -So, for the entire inequality to be true, $x < x^3$ for some $0 -TITLE: Cartesian products of families in Halmos' book. -QUESTION [11 upvotes]: I'm studying some set theory from Halmos' book. He introduces the generalization of cartesian products by means of families. However, I can't understand what is going on. I get the first introduction "The notation..." to "... one-to-one correspondence". What I'm having trouble is with - -If $\{X_i\}$ is a family of sets $(i\in I)$, the Cartesian product of the family is, by definition, the set of all families $\{x_i\}$ with $x_i\in X_i$ for each $i$ in $I$. - -Could you explain to me the motivation of this definition? I know families are itself functions $f:I\to X$ such that to each $i$ there corresponds a subset of $X$, $x_i$. Instead of this we write them succintly as $\{x_i\}_{i\in I}$ to put emphasis on the range (indexed sets) of the function and the domain (indexing set) in question. -For example, in my case, the family is $f:I\to X$ with $f(i)=A_i$ with ${\rm dom} f=\{0,1,2,3\}$ and ${\rm ran} f =\left\{ {{A_0},{A_1},{A_2},{A_3}} \right\}$. -I'm thinking that we can talk about the cartesian product of sets as a set of tuples. However, I can't understand the definition for families of sets. -I leave the page in question: -$\hspace{1 cm} $ - -REPLY [12 votes]: I think this will help other readers that have this same question (Mendelson, Introduction to Topology): - -Let $X_1,X_2,\dots,X_n$ be sets. We have defined a point $$x=(x_1,\dots,x_n)\in \prod_{i=1}^n X_i$$ -as an ordered sequence such that $x_i\in X_i$. Given such a point, by setting $x(i)=x_i$ we obtain a function $x$ which associates to each integer $i$,$1\leq i \leq n$, the element $x(i)\in X_i$. Conversely, given a function $x$ which associates to each integer $i$,$1\leq i \leq n$, an element $x(i)\in X_i$ we obtain the point -$$(x(1),\dots,x(n))\in \prod_{i=1}^n X_i$$ -It is easily seen that this correspondence between points of $\displaystyle\prod_{i=1}^n X_i$ and functions of the above type is one-one and onto, so that a point of $\displaystyle\prod_{i=1}^n X_i$ may also be defined as a function $x$ which associates to each integer $i$,$1\leq i \leq n$, a point $x(i)\in X_i$. The advantage of this second approach is that it allows us to define the product of an arbitrary family of sets. -DEFINITION Let $\{X_\alpha\}_{\alpha \in I}$ be an indexed family of sets. The product of the sets $\{X_\alpha\}_{\alpha \in I}$, - written $\prod_{x\in I}X_\alpha$ consists of all functions $x$ with domain the indexing set $I$ having the property that for each $\alpha \in I$, $x(\alpha)\in X_\alpha$. -Given a point $x\in \prod_{x\in I}X_\alpha$, one may refer to $x(\alpha)$ as the $\alpha$-th coordinate of $x$. However, unless the indexing set has been ordered in some fashion (as is the case with finite products in our earlier discussion), there is no first coordinate, second coordinate, and so on.<|endoftext|> -TITLE: Computing $\frac{d^k}{dx^k}\left(f(x)^k\right)$ where $k$ is a positive integer -QUESTION [10 upvotes]: Does anyone know a formula for the derivative $$\frac{d^k}{dx^k}\left(f(x)^k\right)$$ where $k$ is some positive integer? -I started trying to work it out but it got messy. - -REPLY [2 votes]: After Raymond's explication of Davide's Faa di Bruno-formula I decomposed it in an even more memorizable form. Example k=4 -$$ \begin{eqnarray} {(f^4)^{(4)}\over 4! \cdot 4! } &=& - {{ f^{(0)}\over 0!} { f^{(0)}\over 0!}{ f^{(0)}\over 0!} \over 3! }{{ f^{(4)}\over 4!} \over 1! } \\ \\ - &+ & {{ f^{(0)}\over 0!} { f^{(0)}\over 0!} \over 2! }{{ f^{(1)}\over 1!} \over 1! } {{ f^{(3)}\over 3!} \over 1! } \\ \\ - &+ & {{ f^{(0)}\over 0!} { f^{(0)}\over 0!} \over 2! }{{ f^{(2)}\over 2!}{ f^{(2)}\over 2!} \over 2! } - &+ & {{ f^{(0)}\over 0!} \over 1! } {{ f^{(1)}\over 1!} { f^{(1)}\over 1!} \over 2! }{{ f^{(2)}\over 2!} \over 1! } \\ \\ - &+ & {{ f^{(1)}\over 1!} { f^{(1)}\over 1!} { f^{(1)}\over 1!} { f^{(1)}\over 1!}\over 4! } \\ \\ - \end{eqnarray} $$ -The key for that decomposition is, that for some k we need k factors in each summand. Then the sum of all derivation-indices (with their multiplicity!) must also equal k. A j-fold multiplicity of a factor must be compensated by a j! in the denominator. -If we write -$\displaystyle [k]= { f^{(k)}\over k! } $ and $ \displaystyle [k \times j] = { [k]^j \over j!} $ we can denote it even more compact: -$$ \begin{eqnarray}{(f^4)^{(4)} \over 4!4! }&=& [0 \times 3][4 \times 1] \\ &+& [0 \times 2][1 \times 1][3 \times 1] \\ &+& [0 \times 2][2 \times 2] + [0 \times 1][1 \times 2][2 \times 1] \\ -&+&[1 \times 4] \end{eqnarray} -$$ where, if we take the brackets as products and the terms as sums then each term has the sum k. It is a simple mechanic to write down the terms by hand for some small k, even for $k=10$ or so this should be doable with paper and pen.<|endoftext|> -TITLE: Prove $\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots$ converges to $\frac 1 2 $ -QUESTION [10 upvotes]: Show that -$$\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots = \frac{1}{2}.$$ -I'm not exactly sure what to do here, it seems awfully similar to Zeno's paradox. -If the series continues infinitely then each term is just going to get smaller and smaller. -Is this an example where I should be making a Riemann sum and then taking the limit which would end up being $1/2$? - -REPLY [2 votes]: You can prove it with partial sums: -$$ -S_n=\sum_{k=1}^n\frac{1}{(2k-1)(2k+1)}=\sum_{k=1}^n\left(\frac{1}{2(2k-1)}-\frac{1}{2(2k+1)}\right)=\frac{1}{2}\left(\sum_{k=1}^n\frac{1}{2k-1}-\sum_{k=2}^{n+1}\frac{1}{2k-1}\right) -$$ $$ -=\frac{1}{2}\left(\frac{1}{2(1)-1}-\frac{1}{2(n+1)-1}\right)=\frac{1}{2}-\frac{1}{2(2n+1)} -$$ -Hence, -$$ -\sum_{k=1}^\infty\frac{1}{(2k-1)(2k+1)}=\lim_{n\to\infty}S_n=\lim_{n\to\infty}\left(\frac{1}{2}-\frac{1}{2(2n+1)}\right)=\frac{1}{2} -$$<|endoftext|> -TITLE: a question about an infinite series calculation. -QUESTION [6 upvotes]: I want to prove that for $y >0$, $ x \in \mathbb R$, -$$ \sum_{n=-\infty}^\infty \frac{y}{(x+n)^2 + y^2} = \frac{1}{2} \frac{1 - e^{-4 \pi y }}{1 - 2 e^{-2 \pi y} \cos ( 2 \pi x ) + e^{-4 \pi y}}$$ - -REPLY [4 votes]: Let's use J.M.'s excellent hint : -$$\sum_{n=-\infty}^\infty \frac{y}{(x+n)^2 + y^2}=-\ \Im{\sum_{n=-\infty}^\infty \frac 1{x+iy+n}}$$ -Setting $z:=x+iy\ $ we will evaluate : -$$\sum_{n=-\infty}^\infty \frac 1{z+n}=\frac 1z+\sum_{n=1}^\infty \frac 1{z+n}+\frac 1{z-n}=\frac 1z+\sum_{n=1}^\infty \frac {2z}{z^2-n^2}$$ -The series at the right may be simplified (this was proved many times at S.E. for example here or here or here) getting : -$$\pi\cot(\pi z)=\frac1{z}+\sum_{n=1}^{\infty}\frac{2z}{z^2-n^2}$$ -and the simple result : -$$\sum_{n=-\infty}^\infty \frac{y}{(x+n)^2 + y^2}= -\pi\;\Im{(\cot\pi (x+iy))}$$ -Your remaining problem is that this simple answer doesn't appear related to your more complicated result. Worse your L.H.S and R.H.S. terms don't numerically correspond so that it seems that you have a problem in your question ! -Let's add that your R.H.S. term may be rewritten as half : -$$\frac{1 - e^{-4 \pi y }}{1 - 2 e^{-2 \pi y} \cos ( 2 \pi x ) + e^{-4 \pi y}}=\frac{\sinh(2\pi y)}{\cosh(2 \pi y) - \cos ( 2 \pi x )}$$ - -UPDATE: In fact your equation should read (simply multiplying your R.H.S. by $2\pi$) : -$$\boxed{\displaystyle \sum_{n=-\infty}^\infty \frac{y}{(x+n)^2 + y^2} = \pi\frac{1 - e^{-4 \pi y }}{1 - 2 e^{-2 \pi y} \cos ( 2 \pi x ) + e^{-4 \pi y}}}$$ -We want : -$$\sum_{n=-\infty}^\infty \frac{y}{(x+n)^2 + y^2}= -\pi\Im{(\cot \pi (x+iy))}$$ -Let's observe that $\ \sin a-\sin b=2\sin\frac{a-b}2\cos\frac{a+b}2$ and $\ \cos a-\cos b=-2\sin\frac{a+b}2\sin\frac{a-b}2$ -implies (dividing these expressions) : -$$\frac {\sin a-\sin b}{\cos a-\cos b}=-\cot \frac{a+b}2$$ -Set $a:=2\pi x,\ b:=2\pi i y\ $ to get : -$$-\cot\pi (x+iy)=\frac {\sin 2\pi x-\sin 2\pi i y}{\cos 2\pi x-\cos 2\pi i y}=\frac {\sin 2\pi x-i\sinh 2\pi y}{\cos 2\pi x-\cosh 2\pi y}$$ -so that : -$$\sum_{n=-\infty}^\infty \frac{y}{(x+n)^2 + y^2}=-\pi\frac {\sinh 2\pi y}{\cos 2\pi x-\cosh 2\pi y}$$ -$$= \pi\frac{1 - e^{-4 \pi y }}{1 - 2 e^{-2 \pi y} \cos ( 2 \pi x ) + e^{-4 \pi y}}$$<|endoftext|> -TITLE: Universally measurable sets of $\mathbb{R}^2$ -QUESTION [25 upvotes]: $$\text{Is }{{\cal B}(\mathbb{R}^2})^u={{\cal B}(\mathbb{R}})^u\times {{\cal B}(\mathbb{R}})^u\,?\tag1$$ -Is the $\sigma$-algebra of universally measurable sets on $\mathbb{R}^2$ equal to the -product $\sigma$-algebra of two copies of the universally measurable sets on $\mathbb{R}$? It is not hard -to see that (1) is true with $\supseteq$ instead of $=$, -and I would be astonished if (1) were true, but I'm not sure. -Has anyone encountered this problem, or know a reference that might help? - -REPLY [12 votes]: As you expected, the answer to your question is negative. -Recall that a set $U \subset X \times Y$ is universal for a class $\Gamma \subset \mathcal{P}(Y)$ if for every $G \in \Gamma$ there is $x \in X$ such that $G$ is the vertical section $U_x$ of $U$, that is: $G = U_x = \{ y \in Y : (x,y) \in U\}$. Recall also that analytic sets are universally measurable. -One standard way to prove that there are analytic sets in $X$ that aren't Borel sets is to exhibit an analytic set $U \subset X \times X$ which is universal for the analytic sets in $X$. The set of diagonal points $D = \{x \in X: (x,x) \in U\}$ of $U$ turns out to be an analytic set which is not Borel. The proof is a diagonal argument reminiscent of the Russell paradox and of Cantor's $\kappa \lt 2^\kappa$-theorem. See for example Srivastava, A course on Borel sets, Theorem 4.1.5, p.130. -A clever refinement of this argument yields the answer to your question: - -Let $U \subset [0,1] \times [0,1]$ be an analytic set which is universal for the analytic sets in $[0,1]$. Then $U$ is not an element of the product $\sigma$-algebra $\mathcal{P}[0,1] \times \mathcal{L}[0,1]$ of the power set $\sigma$-algebra $\mathcal{P}[0,1]$ and the Lebesgue $\sigma$-algebra $\mathcal{L}[0,1]$. -In particular, being analytic, $U$ is universally measurable in $[0,1]^2$, but it does not even belong to the product $\sigma$-algebra $\mathcal{L}[0,1] \times \mathcal{L}[0,1] \supset \mathcal{B}([0,1])^u \times \mathcal{B}([0,1])^u$. - -This is proved in B.V. Rao, Remarks on analytic sets, Fund. Math., 66 (1970), 237-239. A variant was independently found by R. Mansfield in the first part of The solution to one of Ulam's problems concerning analytic sets, I and II by different and more complicated methods (the argument in the second part is essentially the same as the one given by Rao). Srivastava's book, Theorem 4.3.4, page 141 also contains a proof of this. Of course, one can replace $[0,1]$ by $\mathbb{R}$ or any other uncountable Polish space. -Further elaborations and refinements can be found in Arnold W. Miller's Measurable rectangles. See in particular Theorem 1 where he proves that $\mathcal{L}[0,1]$ can be replaced by the Baire property algebra $\mathcal{BP}[0,1]$ in the Mansfield-Rao theorem. -Moreover, Miller showed in section 4 of On the length of Borel hierarchies that it is relatively consistent with ZFC that no universal analytic set belongs to the product $\sigma$-algebra $\mathcal{P}[0,1] \times \mathcal{P}[0,1]$.<|endoftext|> -TITLE: Definition of Cantor Set without AC -QUESTION [5 upvotes]: You can see the original text that i thought AC is used here; -From Walter Rudin: Principles of Mathematical Analysis, 3rd ed., ISBN 0-07-054235-X, p.41-42. - -2.44 The Cantor set The set which we are now going to construct shows - that there exist perfect sets in $R^1$ which contain no segment. -Let $E_0$ be the interval $[0, 1]$. Remove the segment $(\frac13,\frac23)$, and let $E_1$ be - the union of the intervals - $$[0,\frac13], [\frac23,1].$$ - Remove the middle thirds of these intervals, and let $E_2$ be the union of the - intervals - $$[0,\frac19], [\frac29,\frac39], [\frac69,\frac79],[\frac89,1]$$ - Continuing in this way, we obtain a sequence of compact sets $E_n$, such that - (a) $E_1\supset E_2 \supset E_3 \dots $; - (b) $E_n$ is the union of $2^n$ intervals, each of length $3^{-n}$. -The set - $$P=\bigcap_{n=1}^\infty E_n$$ - is called the Cantor set. $P$ is clearly compact, and Theorem 2.36 shows that $P$ - is not empty. -No segment of the form - $$\left(\frac{3k+1}{3^m},\frac{3k+2}{3^m}\right)\tag{24},$$ - where $k$ and $m$ are positive integers, has a point in common with $P$. - Since every segment $(\alpha,\beta)$ contains a segment of the form (24), if - $$3^{-m}<\frac{\beta-\alpha}6,$$ - $P$ contains no segment. -To show that $P$ is perfect, it is enough to show that $P$ contains no isolated - point. Let $x \in P$, and let $S$ be any segment containing $x$. Let $I_n$ be that interval - of $E_n$ which contains $x$. Choose $n$ large enough, so that $I_n\subset S$. Let $x_n$ be an - endpoint of $I_n$, such that $x_n\ne x$. -It follows from the construction of $P$ that $x_n\in P$. Hence $x$ is a limit point - of $P$, and $P$ is perfect. -One of the most interesting properties of the Cantor set is that it provides - us with an example of an uncountable set of measure zero (the concept of - measure will be discussed in Chap. 11). - - -I really didn't like the definition of Cantor set in my book (It defines cantor set by using AC$_\omega$), so i tried to construct it in more constructive way. (That is, without AC) -Let $E_n = [0,1]\setminus \bigcup_{k\in 3^n}(3k+1/3^{n+1}, 3k+2/3^{n+1})$. -(Exponentiation here is an ordinal exponentiation) -Let $C=\bigcap_{n\in \omega} E_n$. -I proved that $C$ is nonempty and compact, and $C$ contains no segment. -Now, I'm trying to prove that $C$ is perfect but there's a problem. -This is a lemma I made to prove this; -Let $T_n = \{3k/3^{n+1}, 3k+1/3^{n+1}, 3k+2/3^{n+1}, 3k+3/3^{n+1} \in [0,1] | k\in 3^n \}$. -'For every $n,m\in \omega$, if $nn$. -I tried it and proved that 'If $m>n$ and intersection of an interval $I_n$ in $E_n$ and an interval $I_m$ in $E_m$ is nonempty, the intersection is a set of exactly one endpoint of $I_m$ or $I_m \subset I_n$. (If one holds, the other does not hold)' too. -Here, I have no idea how to derive a contradiction that 'if $x\in I_m \cap I_n$, then $I_m \cap I_n$ is NOT a singleton' ($x\in C$) -Help - -REPLY [9 votes]: Converting my above comment into an answer: - -While Rudin may be accused on hand-waving, I'm pretty sure that his basic idea uses no Choice principles. Note that once $E_n$ is given, then the next set $E_{n+1}$ is determined, and not chosen out of some collection of possibilities (or, to hedge, it is chosen out of a collection of exactly one possibility, but no Choice is required to do this). This is a recursive construction (much like the definition of ordinal arithmetic is recursive).<|endoftext|> -TITLE: Some exact sequence of ideals and quotients -QUESTION [11 upvotes]: I saw an exact sequence of ideals -$$0 \rightarrow I \cap J\rightarrow I \oplus J \rightarrow I + J \rightarrow 0$$In this sequence, maps are ring homomorphisms or module homomorphisms? -And how the above sequence yield the exact seqeunce -$$0 \rightarrow R/I \cap J\rightarrow R/I \oplus R/J \rightarrow R/(I + J) \rightarrow 0$$ - -REPLY [6 votes]: Matt E. answers the question, but I just wanted to add a couple of things. -First note a proper ideals of $R$ does not inherit a ring structure from $R$ because it does not contain $1$. Also, there is no notion of exact sequences for rings, since the kernel of a ring homomorphism is not a ring (but an ideal). -The maps are $R$-module homomorphisms, so indeed the first and last term of the first sequence are ideals of $R$. The middle term $I \oplus J$ is just the direct sum of $I$ and $J$ as $R$-modules; it is not an ideal of $R$. -The second exact sequence comes from applying the snake lemma to the morphism of short exact sequences which Matt points to; its "cokernel" part is your sequence.<|endoftext|> -TITLE: Poincaré Duality Example -QUESTION [9 upvotes]: It is intuitively clear that the Poincaré dual of a ray $\{(r,0);r>0\}$ in $\mathbb{R}^2-\{0\}$ is the form $\frac{1}{2\pi} d\theta$. -For some reason I do not understand, I failed to prove this rigorously. Can somebody help me? -(this example comes from the book of Bott & Tu, Differential Forms in Algebraic Topology) -More precisely, I failed to prove the equality of the integral of a closed form $\omega$ on the ray with the integral of the same form wedge $d\theta$ on the whole punctured plane. - -REPLY [9 votes]: Write $\omega = f(r,\theta) \, dr + g(r,\theta) \, d\theta$ in polar coordinates. The crucial observation is the following: - -The integral of $\omega$ along a ray emanating from $0$ does not depend on the particular choice of the ray! - -For the ray $R_\theta = \{(r\cos\theta,r\sin \theta)\mid r>0\}$ we have $\int_{R_\theta} \omega = \int_0^\infty f(r,\theta) \, dr$. By closedness of $\omega$, we have $\frac{\partial f}{\partial \theta} = \frac{\partial g}{\partial r}$. Now this implies -$$\frac{d}{d\theta} \int_{R_\theta} \omega = \frac{d}{d\theta}\int_0^\infty f(r, \theta) \, dr = \int_0^\infty \frac{\partial f(r, \theta)}{\partial \theta} \, dr = \int_0^\infty \frac{\partial g(r, \theta)}{\partial r} \, dr = 0$$ -where the compact support of $\omega$ was used in the last equality. So indeed, $\int_{R_\theta} \omega$ is independent of $\theta$. Therefore we conclude -$$\int_{\mathbb R^2\setminus \{0\}} \omega \wedge \frac{d\theta}{2\pi} = \frac{1}{2\pi} \int_0^{2\pi} \int_0^\infty f(r,\theta) \, dr \, d\theta = \int_0^\infty f(r,0)\, dr = \int_{\{(r,0)\mid r>0\}} \omega$$ -i.e. $\frac{d\theta}{2\pi}$ is the Poincaré dual of $\{(r,0)\mid r>0\}$ (and of any other ray emanating from $0$).<|endoftext|> -TITLE: Supermartingale with constant Expectation is a martingale -QUESTION [14 upvotes]: In my lecture notes they use the fact, that every supermartingale $(M_t)$ for which the map $t\mapsto E[M_t]$ is constant is already a martingale. Unfortunately I can't prove it. Some help would be appreciated. -By definition of a supermartingale we have: $E[M_0]\ge E[M_s]\ge E[M_t]$ for $t\ge s\ge 0$. -I also know that $M_s\ge E[M_t|\mathcal{F}_s]$. If I would take expectation I would get an equality by assumption. However I do not see how this helps here to prove $M_s= E[M_t|\mathcal{F}_s]$ - -REPLY [19 votes]: Fix $t\geq s$. Then $M_s-E[M_t\mid M_s]$ is a non-negative random variable. Its expectation is -$$E[M_s]-E[E[M_t\mid M_s]]=E[M_s]-E[M_t]=0$$ -by assumption. -A non-negative random variable $X$ with $0$ expectation is $0$ almost everywhere, since -$$P(X\geq 2^{-n})\leq 2^nE X=0.$$<|endoftext|> -TITLE: What is a Lawvere-Tierney topology? -QUESTION [5 upvotes]: I've read some articles and books for the definition and use of Lawvere-Tierney topologies, but I still don't understand their role. -Some people introduce these topologies as modal operators for logic. Others speak about them as a way to define "locally true" propositions, although I don't see what it means. -When others introduce them as topologies on a category, I don't understand if it allows to define open sets on the category or on the image sets by any presheaf. -Could someone give me some clarifications about this ? In particular I'd like to see some examples of them in simple cases (I've been working with the Sierpinsky topos for example) and their use and advantage. - -REPLY [3 votes]: I'm going to give an example, which works on any topos, and therefore also on toposes of sheaves on topological spaces. -Double negation $\neg\neg$ is a local operator or Lawvere-Tierney topology. It acts as follows on the lattice of open subsets of a topological space $X$: for each open set $U$, $\neg\neg U$ is the interior of the closure of $U$. To every open set $U$ the operator joins all isolated subsets of the closed complement of $U$. Open sets that are stable for this topology (this means $\neg\neg U=U$) are known as regular open subsets of the space. -Lawvere-Tierney topologies in general redefine what an open cover is, without changing much about open sets themselves. In our example a family of open sets $\{U_i|i\in\kappa\}$ covers $X$ if $X=\neg\neg (\bigcup_{i\in\kappa} U)$, i.e., we ignore it when covers miss isolated closed subsets of $X$. -The lattice of regular open subsets is always a Boolean algebra, and one of the applications of $\neg\neg$ is in constructing models for classical theories. In fact, $\neg\neg$ is part of the construction of Cohen's counterexample to the continuum hypothesis (see MacLane and Moerdijk's "Sheaves in Geometry and Logic" for this).<|endoftext|> -TITLE: What is the $\tau$ symbol in the Bourbaki text? -QUESTION [8 upvotes]: I'm reading the first book by Bourbaki (set theory) and he introduces this logical symbol $\tau$ to later define the quantifiers with it. It is such that if $A$ is an assembly possibly contianing $x$ (term, variable?), then $\tau_xA$ does not contain it. - -Is there a reference to this symbol $\tau$? Is it not useful/necessary? Did it die out of fashion? -How to read it? - -At the end of the chapter on logic, they make use of it by... I don't know treating it like $\tau_X(R)$ would represent the solution of an equation $R$, this also confuses me. - -REPLY [3 votes]: The Hilbert operator was introduced by Hilbert in 1923. Hilbert was -convinced of the axiom of choice as an indispensable principle but refrained to introduce it in its logical system. Instead he postulated another axiom, -similar in contents to the axiom of choice, introducing new kind of terms: the $\tau$-terms (in fact Hilbert used the letter $\epsilon$ instead of $\tau$ but Bourbaki preferred the latter to avoid any confusion with the set-theoretical $\in$ symbol). -Hilbert's operator offers a systematic mechanism of attaching to a first-order predicate $R[x]$ a $\tau$-term by the procedure you described above, i.e. $\tau_x(R)$. Formally this is a string where all the occurrences of $x$, if any, are muted and tightly bound to the initial $\tau$ symbol by an upper link (the largest link indicates the scope of the binding). -The intended intuitive meaning is according to Bourbaki the following: -"If there exists at least one object for which the predicate $R[x]$ is true, then $\tau_x(R)$ represents some kind of an "ideal" (Hilbert's terminology) "preferred" (Bourbaki's terminology) object of the theory for which the predicate $R[x]$ is certainly true. Otherwise, strictly nothing can be said about it". -According to this naïve interpretation $(\exists x)R$ can be considered as an alias for the relation $R[\tau_x(R)]$ where the $x$ variable is in fact muted and bound in all of its occurrences in $R$. Similarly $(\forall x)R$ can be considered as an alias for the string $\neg(\exists x)(\neg R[x])$, which becomes now expandable into an explicit formula based on Hilbert's $\tau$-term. -Two logically equivalent 1st-order predicates in variable $x$ should lead to the same ideal $\tau$-term. The axiom schema about $\tau$-terms and equality ensures that: $(R\iff S)\implies \tau_x(R)=\tau_x(S)$. -The net result of that schema is that when a predicate is one-to-one -in $x$: -$$R[x] \wedge R[x'] \implies x=x'$$ -then $R[x]\implies x=\tau_x(R)$. The $\tau$-term attached to predicate $R$ is the "obligated" logical solution of $R[x]$, if such a solution does exist. This was the axiom schema postulated by Hilbert. -Conversely if such a solution does really exists, i.e. if $(\exists x) R[x]$ literally $R[\tau_x(R)]$ is true, then $R[x]$ logical equation is equivalent to $x=\tau_x(R)$: the converse implication in Hilbert's axiom schema is also true. -$\tau$-term $\tau_x(R)$ can thus be regarded as the result of some particular choice (an ideal one) among all possible logical solutions of predicate $R[x]$ and can thus be constructed from a choice function which usage is set-theoretically justified by the axiom of choice.<|endoftext|> -TITLE: Massey products in the Adams Spectral Sequence -QUESTION [6 upvotes]: I've never quite 'got' Massey products - this question, I guess, is to work out a small example that might shed some light for me. -So following Wikipedia, let $\Gamma$ be a differential graded algebra with differential $d$, let $\bar{u} = (-1)^{|u|+1}u$ and let $[u]$ denote the cohomology class of an element $u \in \Gamma$. Then suppose that $uv = 0$ and $vw=0$. Then there are $s,t$ such that $d(s)=\bar{u}v$ and $d(t)=\bar{v}w$, and then the cycle $\bar{s}w+\bar{u}t$ represents an element of the Massey product $\langle u,v,w\rangle$. -The particular example I will be interested in is applying this to the cobar complex in the Adams spectral sequence for the sphere spectrum (at $p=2$). This has $E_2$ term $\text{Ext}_{\mathcal{A}}^s(\mathbb{F}_2,\Sigma^t \mathbb{F}_2)$. We have the following theorem - -The 2-line $\text{Ext}^2_{\mathcal{A}}(\mathbb{F}_2,\Sigma^\ast \mathbb{F}_2)$ is the graded vector space generated by $h_ih_j$ subject to relations $h_ih_j = h_jh_i$ and $h_ih_{i+1}=0$. - -The notes of Paul Goerss then point out that this is the defining relation for a Massey product and that it is a 'fun exercise' to show that -$\langle h_0,h_1h_0 \rangle = h_1^2$ and $\langle h_1,h_0,h_1 \rangle = h_0h_2$ -I'm interested in doing these calculations, but I can't seem to work out how! Firstly, I presume the first is a typo and it should be $\langle h_0,h_1,h_0 \rangle $? Secondly I feel like, no matter what classes I chose for $s$ and $t$ that the answer should involve $h_1$, since it is taking the place of $u$ and $w$ in the above. -If anyone can shed any light on how to do these calculations, it would be much appreciated. - -REPLY [4 votes]: I have worked through two rather explicit approaches to computing the -Massey product. One way is to follow the notes of Bruner, using -theorem 2.6.5 and the paragraphs which follow. The notes are very good, and there isn't much I could possibly add to them. The second is to use -the cobar complex and work out the Massey product explicitly. For this -approach, the computation is much simpler if we use the results in -McCleary's book "A User's Guide to Spectral Sequences," see pages -378--379, and 128--130. -Let $\mathcal{A}$ denote the Steenrod algebra at the prime 2. Recall -$\mathcal{A}$ is generated by the Steenrod squares $Sq^i$ for $i\geq -0$ and the admissible monomials form an $\mathbb{F}_2$-basis for -$\mathcal{A}$. I will use $-^{\vee}$ for the functor of graded -$\mathcal{A}$-modules $Hom^{*}_{\mathcal{A}}(-, \mathbb{F}_2)$. Recall -that $\mathcal{A}$ has an augmentation $\epsilon : \mathcal{A} \to -\mathbb{F}_2$. Denote $\ker(\epsilon)$ by $I(\mathcal{A})$. -We first need the (reduced) bar resolution of $\mathbb{F}_2$. This -takes the form -\begin{equation*} - \mathbb{F}_2 \leftarrow \mathcal{A} \leftarrow \mathcal{A} \otimes I(\mathcal{A}) \leftarrow \mathcal{A} \otimes I(\mathcal{A}) \otimes I(\mathcal{A}) \leftarrow \cdots -\end{equation*} -Let $B_n = \mathcal{A} \otimes I(\mathcal{A})^{\otimes n}$ to simplify -notation. The tensor products are over $\mathbb{F}_2$, and the -traditional notation for an element in $B_n$ is $\gamma[\gamma_1 \vert -\gamma_2 \vert \cdots \vert \gamma_n]\equiv \gamma \otimes \gamma_1 -\otimes \cdots \otimes \gamma_n$. A reason for this is that we -consider $B_n$ to be an $\mathcal{A}$-module with action $x \cdot -\gamma[\gamma_1 \vert \cdots \vert \gamma_n] = (x\gamma)[\gamma_1 -\vert \cdots \vert \gamma_n]$. The maps $d_n : B_n \to B_{n-1}$ in the -resolution are defined by -\begin{equation*} - d_n(\gamma[\gamma_1 \vert \cdots \vert \gamma_n]) = \gamma\gamma_1[\gamma_2 \vert \cdots \vert \gamma_n] + \gamma[\gamma_1\gamma_2 \vert \cdots \vert \gamma_n] + \cdots + \gamma[\gamma_1 \vert \cdots \vert \gamma_{n-2} \vert \gamma_{n-1}\gamma_{n}]. -\end{equation*} -This is a resolution of $\mathbb{F}_2$ by free $\mathcal{A}$-modules, -as one can check. See pages 242--248 in McCleary for more -details. With this in hand, if we apply $-^{\vee} = -Hom^*_{\mathcal{A}}(-,\mathbb{F}_2)$ and take cohomology of the -resulting chain complex, we will get -$Ext^{**}_{\mathcal{A}}(\mathbb{F}_2,\mathbb{F}_2)$. -Let us now dualize the bar construction above, i.e., apply our functor -$-^{\vee}$. We will get -\begin{equation*} - \mathcal{A}^{\vee} \rightarrow (\mathcal{A} \otimes - I(\mathcal{A}))^{\vee} \rightarrow (\mathcal{A} \otimes I(\mathcal{A}) \otimes - I(\mathcal{A}))^{\vee} \rightarrow \cdots -\end{equation*} -which is -\begin{equation*} - \mathbb{F}_2 \rightarrow Hom^*_{\mathbb{F}_2}(I(\mathcal{A}),\mathbb{F}_2) \rightarrow Hom^*_{\mathbb{F}_2}(I(\mathcal{A}),\mathbb{F}_2) \otimes Hom^*_{\mathbb{F}_2}(I(\mathcal{A}),\mathbb{F}_2) \rightarrow \cdots. -\end{equation*} -An $\mathcal{A}$-module map $B_n \to \mathbb{F_2}$ is determined by its -values on an $\mathcal{A}$-basis, and the elements of the form -$1[\gamma_1 \vert \cdots \gamma_n]$ is one such basis, where all of -the $\gamma_i$'s are admissible monomials of positive degree. Thus we -see $B_n^{\vee} \cong Hom^*_{\mathbb{F}_2}(I(\mathcal{A}), -\mathbb{F}_2) = I(\mathcal{A})^{dual}$. Here we define $X^{dual} = Hom_{\mathbb{F}_2}(X, \mathbb{F}_2)$. -The dualized maps $d_n^{\vee} : B_{n-1}^{\vee} \to B_{}^{\vee}$ have a -very handy description if we use the description of the dual Steenrod -algebra $\mathcal{A}^{dual} = Hom_{\mathbb{F}_2}^*(\mathcal{A}, -\mathbb{F}_2)$ determined by Milnor in his 1958 paper ``The Steenrod -algebra and its dual.'' Namely, $\mathcal{A}^{dual} = -\mathbb{F}_2[\zeta_1, \zeta_2, ... ]$ where $\mathrm{deg} \zeta_i = 2^i - -1$. Furthermore, the coproduct $\phi^{dual}$ for $\mathcal{A}^{dual}$ -is determined by $\phi^{dual}(\zeta_i) = \sum_{j+k = i} -\zeta_j^{2^k}\otimes \zeta_k$. As $\phi^{dual}$ is an algebra -homomorphism, this determines the coproduct. To start off, $d_1^{\vee} -: \mathbb{F}_2 \to I(\mathcal{A})^{dual}$ is the zero map $1 -\mapsto 0$. Furthermore, the map $d_2^{\vee} : -I(\mathcal{A})^{dual} \to I(\mathcal{A})^{dual}\otimes -I(\mathcal{A})^{dual}$ is given by $[x] \mapsto \sum [x' \vert x'']$ where -$\phi^{dual}(x) = 1\otimes x + x \otimes 1 + \sum x' \otimes x''$. -Here are some concrete computations. As $\phi^{dual}(\zeta_1) = -1\otimes \zeta_1 + \zeta_1 \otimes 1$, we have $\phi^{dual}(\zeta_1^2) -= \zeta_1^2 \otimes 1 + 1 \otimes \zeta_1^2$ and -\begin{equation*} -\phi^{dual}(\zeta_1^3) = 1\otimes \zeta_1^3 + \zeta_1 \otimes -\zeta_1^2 + \zeta_1^2 \otimes \zeta_1 + \zeta_1^3 \otimes 1 -\end{equation*} -Therefore $d_2^{\vee}(\zeta_1) = 0$, $d_2^{\vee}(\zeta_1^2) = 0$ and -\begin{equation*} -d_2^{\vee}([\zeta_1^3]) = [\zeta_1 \vert \zeta_1^2] + [\zeta_1^2 \vert \zeta_1]. -\end{equation*} -Another computation we will need is for $\zeta_2$. Recall $\zeta_2$ is -of degree $3$, and is the dual of $Sq^2Sq^1$. The coproduct of -$\zeta_2$ is $\phi^{dual}(\zeta_2) = 1\otimes \zeta_2 + \zeta_1^2 -\otimes \zeta_1 + \zeta_2 \otimes 1$. Therefore -\begin{equation*} - d_2^{\vee}([\zeta_2]) = [\zeta_1^2 \vert \zeta_1]. -\end{equation*} -Now what makes it possible to compute (and define I suppose!) the -Massey products in $Ext_{\mathcal{A}}(\mathbb{F}_2, \mathbb{F}_2)$ is -that the dual of the bar resolution---which is called the cobar -construction---is a DG algebra whose cohomology is -$Ext_{\mathcal{A}}(\mathbb{F}_2, \mathbb{F}_2)$. One product in the -bar resolution is the juxtaposition product (or concatenation -product). That is, define -\begin{equation*} [\gamma_1 \vert \cdots \vert \gamma_n] * [\beta_1 - \vert \cdots \vert \beta_m] = [\gamma_1 \vert \cdots \vert \gamma_n - \vert \beta_1 \vert \cdots \vert \beta_m]. -\end{equation*} -This turns out to agree with the composition product, which is a good -thing, see McCleary pg. 383. So with respect to the juxtaposition -product, the differentials $d_*^{\vee}$ satisfy the Leibniz rule. So -we can define the Massey products! The juxtaposition product is nice -because it allows us to do computations easily. -Let's now finally compute $\langle h_0, h_1, h_0 \rangle = \left\{[ a - *h_0 + h_0 *b] \, \vert \, d_2^{\vee}a = h_0*h_1 \, d_2^{\vee} b = - h_1 * h_0 \right\}$. In this definition, the brackets mean -cohomology class. We must first figure out what we mean by $h_0$ and -$h_1$. Let us compute $Ext_{\mathcal{A}}^{1,1}$ and -$Ext_{\mathcal{A}}^{1,2}$. For the first, the image is 0, and the -kernel of $d_2^{\vee}$ in degree 1 is $[\zeta_1]$. So let us define -$h_0 = [\zeta_1]$. Next, we see that the kernel of $d_2^{\vee}$ in -degree 2 is $[\zeta_1^2]$ and again the image is 0. So let us define -$h_1 = [\zeta_1^2]$. The product $h_0 * h_1$ is vanishes in -$Ext^{1,3}$, so it must be the boundary of something. By our above -calculations, we see that $d_2^{\vee}([\zeta_1^3] + [\zeta_2]) = -[\zeta_1 \vert \zeta_1^2]$. Also, the product $h_1 * h_0 = [\zeta_1^2 -\vert \zeta_1]$ vanishes in $Ext^{1,3}$, so it must also be the -boundary of something. Our above calculations show that -$d_2^{\vee}([\zeta_2]) = [\zeta_1^2 \vert \zeta_1]$. We now can -identify one element in the Massey product. Namely: -\begin{equation*} - [\zeta_1^3 + \zeta_2 \vert \zeta_1] + [ \zeta_1 \vert \zeta_2] = - [\zeta_1^3 \vert \zeta_1] + [\zeta_2 \vert \zeta_1 ] + [\zeta_1 \vert \zeta_2]. -\end{equation*} -Which element is this in $Ext^{2,4}$? Well, let us calculate some -more. First off, -\begin{align*} - \phi^{dual}(\zeta_2\zeta_1)& = (1\otimes \zeta_2 + \zeta_1^2 \otimes \zeta_1 + \zeta_2 \otimes 1)(1\otimes \zeta_1 + \zeta_1\otimes1)\\ - & = 1\otimes \zeta_2\zeta_1 + \zeta_2\zeta_1 \otimes 1 + \zeta_1^3 \otimes \zeta_1 - + \zeta_2 \otimes \zeta_1 + \zeta_1^2 \otimes \zeta_2^2 + \zeta_2 \otimes \zeta_2, -\end{align*} -hence -\begin{equation*} - d_2^{\vee}([\zeta_2\zeta_1]) = [\zeta_1^3 \vert \zeta_1 ] + [\zeta_2 \vert \zeta_1] + [\zeta_1^2 \vert \zeta_1^2] + [\zeta_1\vert \zeta_2]. -\end{equation*} -Thus we see that our class $[\zeta_1^3 \vert \zeta_1] + [\zeta_2 \vert -\zeta_1 ] + [\zeta_1 \vert \zeta_2]$ in $\langle h_0 , h_1, -h_0\rangle$ is cohomologous to $[\zeta_1^2 \vert \zeta_1^2 ] = h_1 * -h_1 = h_1^2$ which is a non-zero class in $Ext^{1,3}$. By further direct -computation, one can determine that the indeterminacy, which is -$h_0Ext^{2,3} + Ext^{2,3}h_0$, of the Massey product $\langle h_0, h_1, -h_0\rangle$ is $0$, so that we have completely determined it.<|endoftext|> -TITLE: $I\cdot J$ principal implies $I$ and $J$ principal? -QUESTION [6 upvotes]: Let $R$ be a Noetherian domain, and let $I$ and $J$ be two ideals of $R$ such that their product $I\cdot J$ is a non-zero principal ideal. Is it true that $I$ and $J$ are principal ideals ? This seems an easy question to settle, but I can't find an answer. -Any idea is welcome, thanks ! - -Thanks a lot for your enlightening answers. I admit I'm more interested in a geometric setting (i.e. when $R$ is an algebra, finitely generated, or a localization of that). I fail to adapt examples coming from number theory to this setting. What do you think ? - -REPLY [8 votes]: At Lierre's request , here is a geometric example. -Consider an elliptic curve $\bar E$ (say over $\mathbb C$), a point $P\in \bar E$ of order $2$ in the group $\bar E(\mathbb C)$ (there are 3 such) and the complement $E=\bar E \setminus \lbrace O\rbrace $ of the origin in $\bar E$. -Like all non complete integral curves $E$ is affine, with ring $R=\Gamma (E, \mathcal O_E)$. -The ideal $I=\mathfrak m_P\subset R$ of functions vanishing at $P$ is not principal because $\mathcal O_E(-P)$ is a non-trivial line bundle (use Abel-Jacobi's theorem). -However $I^2=\mathfrak m_P^2$ is principal because the line bundle $\mathcal O_E(-P)$ has as its square $\mathcal O_E(-2P)=\mathcal O_E(0)=\mathcal O_E= $ the trivial line bundle, so that $I$ is an example of non-principal ideal with $I\cdot I$ principal.<|endoftext|> -TITLE: Retracts are Submanifolds -QUESTION [9 upvotes]: Looking over some old qualifying exams, we found this: -Let $A\subseteq M$ be a connected subset of a manifold $M$. If there exists a smooth retraction $r:M\longrightarrow A$, then $A$ is a submanifold. -Our thought to prove this statement was that since $r$ is smooth and the identity on $A$, then the inclusion $i:A\longrightarrow M$ is smooth. Also, since $i\circ r=\operatorname{Id}_A$, then $i_*:TA\longrightarrow TM$ is injective. Thus $i$ is a smooth immersion. Therefore $A$ is a submanifold. But, nowhere did we use that $A$ is connected. What is wrong with the argument? And, what is the correct proof? - -REPLY [6 votes]: As pointed out in the comments, since you don't know $A$ is a manifold, you can't speak about smooth immersion of $A$ into $M$. -To prove the statement, you have to show there exists an open neighborhood $U$ of $A$ in $M$ such that the rank of $T_y r$ is constant for $y\in U$. Then applying the constant rank theorem, the result follows. -If $A$ was not connected, in general, the rank of $T_y r$ would have a different value in each connected component and $A$ would not be a pure manifold. -For the proof details you can look at P. W. Michor, Topics in Differential Geometry, section 1.15.<|endoftext|> -TITLE: Learning how to prove that a function can't proved total? -QUESTION [7 upvotes]: In proof-theory one can prove that in, say, Peano Arithmetic one can't prove a function $f$ total. Often this seems to mean $f$ is growing too fast to be provably total. -I have some background in logic and know the very basics about formal proofs in FO, but overall I'm more familiar with finite model theory than proof-theory. Now I want to learn how to prove a function total in PA and how to prove that a function can't be proved total in PA. -Are there any books that would cover the topic with my background? Also, if there are any lecture notes near the area, I would be very happy receive pointers to them (books do cost a lot and I think I can only afford one). - -REPLY [6 votes]: The proof theoretic terminology for this concept is "Provable Total Functions"/"Provably Recursive Functions" of a theory. -You will find quite a number of articles on the topic if you search for these terms. -The third chapter of Samuel Buss, "Handbook of Proof Theory", 1998, is devoted to the topic: - -Matt Fairtlough and Stanley S. Wainer, "Hierarchies of Provably Recursive Functions". - -Part 5 of the chapter is "Independence results for PA". -The totalness of (a $\Sigma^0_1$ representation of) a function corresponds to a $\Sigma^0_1$ formula, or equivalently $\Pi^0_2$ sentence. Therefore proving that a function is not total in a theory corresponds to an independence result for a $\Pi^0_2$ sentence, and there are not that many core techniques to do this. One technique as you mentioned is to show that the function is growing too fast, so fast that its totality would imply the consistency of the theory at hand. (There are some tricky issues here about the representation chosen for the function.)<|endoftext|> -TITLE: Finding the convergence interval of $\sum_{n=0}^\infty\frac{n!x^n}{n^n}$. -QUESTION [20 upvotes]: I want to find the convergence interval of the infinite series $\sum\limits_{n=0}^\infty \dfrac{n!x^n}{n^n}$. -I will use the ratio test: if I call $u_n = \dfrac{n!x^n}{n^n}$, the ratio test says that, if the following is true for some values of $x$, the series will be convergent for these values of $x$: -$$\lim_{n\to+\infty}\left|\frac{u_{n+1}}{u_n}\right|<1$$ -So, I will first calculate the value of $\left|\dfrac{u_{n+1}}{u_n}\right|$: -$$\left|\dfrac{\dfrac{(n+1)!x^{n+1}}{(n+1)^{n+1}}}{\dfrac{n!x^n}{n^n}}\right|=\dfrac{(n+1)!|x|^{n+1}}{(n+1)^{n+1}}\times\dfrac{n^n}{n!|x|^n}=\frac{(n+1)n^n|x|}{(n+1)^{n+1}}=|x|\left(\frac{n}{n+1}\right)^n$$ -So, $\lim\limits_{n\to+\infty}\left|\dfrac{u_{n+1}}{u_n}\right|$ becomes: -$$\lim_{n\to+\infty}\left|\frac{u_{n+1}}{u_n}\right|=\lim_{n\to+\infty}|x|\left(\frac{n}{n+1}\right)^n=|x|\lim_{n\to+\infty}\left(\frac{n}{n+1}\right)^n$$ -Now I must evaluate the value of $\lim\limits_{n\to+\infty}\left(\dfrac{n}{n+1}\right)^n$. For this, let $y = \left(\dfrac{n}{n+1}\right)^n$; so, instead of calculating $\lim\limits_{n\to+\infty}y$, I will first calculate $\lim\limits_{n\to+\infty}\ln y$: -$$\lim_{n\to+\infty}\ln y=\lim_{n\to+\infty}\ln \left(\dfrac{n}{n+1}\right)^n=\lim_{n\to+\infty}n\ln\left(\frac{n}{n+1}\right) -=\lim_{n\to+\infty}\frac{\ln\left(\frac{n}{n+1}\right)}{\frac{1}{n}}$$ -Applying L'Hôpital's rule: -$$\lim_{n\to+\infty}\frac{\ln\left(\frac{n}{n+1}\right)}{\frac{1}{n}} -=\lim_{n\to+\infty}\frac{\frac{1}{n(n+1)}}{-\frac{1}{n^2}} -=\lim_{n\to+\infty}\left(-\frac{n}{n+1}\right)=-1$$ -Now, since we know that $\lim\limits_{n\to+\infty}\ln y = -1$, we have that: -$$\lim_{n\to+\infty}y=\lim_{n\to+\infty}e^{\ln y} = e^{-1} = \frac{1}{e}$$. -Substituting this back into the expression $\lim\limits_{n\to+\infty}\left|\frac{u_{n+1}}{u_n}\right| = |x|\lim\limits_{n\to+\infty}\left(\frac{n}{n+1}\right)^n$, we have that the limit of $\left|\dfrac{u_{n+1}}{u_n}\right|$ as $n\to+\infty$ is: -$$\lim_{n\to+\infty}\left|\frac{u_{n+1}}{u_n}\right|=\frac{|x|}{e}$$ -Therefore, the series will certainly be convergent for the values of $x$ for which $\dfrac{|x|}{e}<1$, that is, $|x|\dfrac{n^n}{n!}$; therefore, $\dfrac{n!e^n}{n^n} > 1$, and, thus, we show that $\sum\limits_{n=0}^{\infty} \dfrac{n!e^n}{n^n}$ is divergent, because its term doesn't approach zero. -$\sum\limits_{n=0}^{\infty} \dfrac{(-1)^nn!e^n}{n^n}$ is also divergent, because, although the absolute value of the ratio between two successive terms, $|e|\left(\frac{n}{n+1}\right)^n$, approaches 1 as $n\to\infty$, it approaches 1 from values bigger than 1; therefore, the absolute value of a term is always greater than the absolute value of the previous term. - -REPLY [2 votes]: This may be overkill but we have $$n!\approx n^{n+1/2}e^{-n}$$ by Stiriling's formula when $n$ goes very large. Thus the original sum becomes approximately $$n^{1/2}\frac{x^{n}}{e^{n}}$$ So if $x\ge e$ this series obviously will not converge. When $0 -TITLE: When was $\pi$ first suggested to be irrational? -QUESTION [24 upvotes]: When was $\pi$ first suggested to be irrational? -According to Wikipedia, this was proved in the 18th century. -Who first claimed / suggested (but not necessarily proved) that $\pi$ is irrational? -I found a passage in Maimonides's Mishna commentary (written circa 1168, Eiruvin 1:5) in which he seems to claim that $\pi$ is irrational. Is this the first mention? - -REPLY [6 votes]: Bhaskara I (the less famous of the two Bhaskara) wrote a comment on Aryabhata in 629, where he gives Aryabhata's approximation $\pi\approx {62832\over 20000}=3.1416$, and states that a nonapproximate value for this ratio is impossible. -See Kim Plofker: Mathematics in India, p. 140. -On the surface this would seem like a clear statement equivalent to the irrationality of $\pi$, but perhaps it is not quite so easy. The "reason" Bhaskara gives is that "surds (square roots of nonsquare numbers) do not have a statable size". It is believed that Aryabhata's method of approximating $\pi$ is essentially the same as Archimedes', computing the circumference of an inscribed regular polygon in a circle with 384 sides. This requires the computation of many surds, so perhaps Bhaskara only meant that he could not get an exact value of the circumference of the inscribed polygon.<|endoftext|> -TITLE: Proving determinant product rule combinatorially -QUESTION [11 upvotes]: One of definitions of the determinant is: -$\det ({\mathbf C}) - =\sum_{\lambda \in S_n} ({\operatorname {sgn} ({\lambda}) \prod_{k=1}^n C_{k \lambda ({k})}})$ -I want to prove from this that -$\det \left({\mathbf {AB}}\right) = \det({\mathbf A})\det({\mathbf B})$ -What I have so far: -$(AB)_{k\lambda ({k})} = \sum_{j=1}^n A_{kj}B_{j\lambda(k)}$ -so we have for the determinant of $\mathbf {AB}$ -$\det ({\mathbf {AB}}) - =\sum_{\lambda \in S_n} ({\operatorname {sgn} ({\lambda}) \prod_{k=1}^n \sum_{j=1}^n A_{kj}B_{j\lambda(k)}})$ -Now I'm not sure how to denote this, but the product of the sum I think is the sum over all combinations of n terms, each ranging from 1 to n, so I'll denote -this set of all combinations $C_n(n)$ for n terms each ranging from 1 to n, -analogous to the permutation set, but all combinations instead of permutations. -then I get -$\det ({\mathbf {AB}}) - =\sum_{\lambda \in S_n} ({\operatorname {sgn} ({\lambda}) \sum_{\gamma \in C_n(n)} \prod_{k=1}^n A_{k\gamma(k)}B_{\gamma(k)\lambda(k)}} )$ -then I can at least seperate the product: -$\det ({\mathbf {AB}}) - =\sum_{\lambda \in S_n} ({\operatorname {sgn} ({\lambda}) \sum_{\gamma \in C_n(n)} \prod_{k=1}^n A_{k\gamma(k)} \prod_{r=1}^n B_{\gamma(r)\lambda(r)}} )$ -I changed the k to an r in one product because it's a dummy variable so I think it doesn't matter, I don't really know if this thing is helpful but this is my attempt at a solution so far. -Thanks to anyone who helps! - -REPLY [13 votes]: From -$$\det ({\mathbf {AB}}) - =\sum_{\lambda \in S_n} \operatorname {sgn} (\lambda) \sum_{\gamma \in C_n(n)} \left(\prod_{k=1}^n A_{k\gamma(k)} \right) \left(\prod_{r=1}^n B_{\gamma(r)\lambda(r)} \right)$$ -reorder the sums and factor out the first product : -$$\det (\mathbf {AB}) - =\sum_{\gamma \in C_n(n)} \left(\prod_{k=1}^n A_{k\gamma(k)} \right) \left( \sum_{\lambda \in S_n} \operatorname {sgn} (\lambda) \prod_{r=1}^n B_{\gamma(r)\lambda(r)} \right)$$ -For $\gamma \in C_n(n)$, suppose that $\gamma$ is not a permutation : there are two indices $i,j$ such that $\gamma(i) = \gamma(j)$. -Let $\tau$ be the transposition swapping $i$ and $j$. In particular, $\gamma \circ \tau = \gamma$. -For any $\lambda \in S_n$, by reordering the product, we get : -$$\operatorname {sgn} (\lambda)\prod_{r=1}^n B_{\gamma(r)\lambda(r)} = \operatorname {sgn} (\lambda)\prod_{r=1}^n B_{\gamma(\tau(r))\lambda(\tau(r))} = - \operatorname {sgn} (\lambda \circ \tau)\prod_{r=1}^n B_{\gamma(r)\lambda \circ \tau(r)} $$ -Thus, by duplicating and reorganizing the sum, -$$ \sum_{\lambda \in S_n} \operatorname {sgn} (\lambda) \prod_{r=1}^n B_{\gamma(r)\lambda(r)} = \frac 12 \sum_{\lambda \in S_n} \left( \operatorname {sgn} (\lambda) \prod_{r=1}^n B_{\gamma(r)\lambda(r)} + \operatorname {sgn} (\lambda \circ \tau) \prod_{r=1}^n B_{\gamma(r)\lambda \circ \tau(r)(r)}\right) = 0$$ -Therefore we may only keep the $\gamma$ that are permutations, and by reordering the second product, we have : -$$\det (\mathbf {AB}) - =\sum_{\gamma \in S_n} \left(\prod_{k=1}^n A_{k\gamma(k)} \right) \left( \sum_{\lambda \in S_n} \operatorname {sgn} (\lambda) \prod_{r=1}^n B_{r\lambda\circ\gamma^{-1}(r)} \right)$$ -By reorganizing the second sum, we get : -$$\det (\mathbf {AB}) - =\sum_{\gamma \in S_n} \left(\prod_{k=1}^n A_{k\gamma(k)} \right) \left( \sum_{\lambda \in S_n} \operatorname {sgn} (\lambda) \operatorname {sgn} (\gamma)\prod_{r=1}^n B_{r\lambda(r)} \right) $$ -We can factor out $\operatorname{sgn}(\gamma)$ from the second sum, then factor out the whole second sum from the first sum, to get -$$\det (\mathbf {AB}) = \left( \sum_{\gamma \in S_n} \operatorname {sgn} (\gamma) \prod_{k=1}^n A_{k\gamma(k)} \right) \left( \sum_{\lambda \in S_n} \operatorname {sgn} (\lambda) \prod_{r=1}^n B_{r\lambda(r)} \right) = \det (\mathbf {A})\det (\mathbf {B})$$<|endoftext|> -TITLE: Smallest possible rank of an $n×n$ matrix that has zeros along the main diagonal and strictly positive real numbers off the main diagonal -QUESTION [6 upvotes]: A problem in IMC 2012 in which i'm interested but I have no answer. Can you help me? Many thanks. -Problem : Let $n$ be a fixed positive integer. Determine the smallest possible rank of an $n\times n$ matrix that has zeros along the main diagonal and strictly positive real numbers off the main diagonal. - -REPLY [3 votes]: The problem has been solved at the website Art Of Problem Solving. The minimal rank is $2$ for two times two matrices, and $3$ for $n\geq 3$. A matrix of minimal rank is given: its entries are $a_{i,j}:=(i-j)^2$.<|endoftext|> -TITLE: How to tell $i$ from $-i$? -QUESTION [40 upvotes]: Suppose now we are trying to explain to students who do not know complex numbers, how do we distinguish $i$ and $-i$ to them? They will object that they both squared to $-1$ and thus they are indistinguishable. Is there a way of explaining this in an elementary way without go into introductory things in complex analysis? - -REPLY [9 votes]: This question relates directly to issues that are often discussed in the philosophy of mathematics. -According to one of the standard accounts of structuralism, what mathematical objects are at bottom are the structural roles that they play within a mathematical system. One might define the equivalence relation whereby element $a$ in structure $A$ is equivalent to element $b$ in structure $B$ exactly when there is an isomorphism of $A$ to $B$ taking $a$ to $b$. This results in what I have called the isomorphism orbit of the object $a$ in $A$. - -According to some accounts of the philosophy of abstract structuralism, what a mathematical object is, is the structural role that it plays in abstract mathematical structures. -The problem for this view arises with nonrigid mathematical structures such as the complex field $\newcommand\C{\mathbb{C}}\C$. -Here is how I describe the situation is my book, Lectures on the Philosopy of Mathematics (MIT Press) - -Although one conventionally describes $i$ as "the square root of negative one," nevertheless one might reply to this, ``Which one?'' in light of the fact that $\newcommand\unaryminus{-}\unaryminus i$ also is such a root: -$$(\unaryminus i)^2=(\unaryminus 1\cdot i)^2=(\unaryminus 1)^2i^2=i^2=\unaryminus 1.$$ -Indeed, the complex numbers admit an automorphism, an isomorphism of themselves with themselves, induced by swapping $i$ with $\unaryminus i$---namely, complex conjugation: -$$z=a+bi\qquad\mapsto\qquad\bar z= a-bi.$$ -The conjugation map preserves the field structure, since $\overline{y+z}=\bar y+\bar z$ and $\overline{y\cdot z}=\bar y\cdot\bar z$, and therefore the complex field is not a rigid mathematical structure. Since conjugation swaps $i$ and $\unaryminus i$, it follows that $i$ can have no structural property in the complex numbers that $\unaryminus i$ does not also have. So there can be no principled, structuralist reason to pick one of them over the other. Is this a problem for structuralism? It does seem to be a problem for singular terms, since how do we know that the $i$ appearing in my calculations this week is the same number as what will appear in your calculations next week? Perhaps my $i$ is your $\unaryminus i$, and we do not even realize it. - -If one wants to understand mathematical objects as abstract positions within a structure, as in abstract structuralism, then one must grapple with the fact that in light of the conjugation automorphism, the numbers $i$ and $\unaryminus i$ play exactly the same roles in this structure (see Shapiro, 2012). The numbers $i$ and $\unaryminus i$ have the same isomorphism orbit with respect to the complex field, and so in this sense, although distinct, they each play exactly the same structural role in $\C$. This would seem to undermine the idea that mathematical objects are abstract positions in a structure, since we want to regard these as distinct complex numbers. -The point is that one cannot understand the mathematical objects as identical to the structural roles that they play within a system, since $i$ and $\unaryminus i$ play exactly the same role in $\C$ as each other, yet are to be regarded as distinct complex numbers. -In fact, there is nothing special about the numbers $i$ and $\unaryminus i$ in this argument. For example, the numbers $\sqrt{2}$ and $\unaryminus\!\sqrt{2}$ also happen to play the same structural role in the complex field $\C$, because there is an automorphism of $\C$ that swaps them (although one uses the axiom of choice to prove this). Contrast this with the real field $\newcommand\R{\mathbb{R}}\R$, where $\sqrt{2}$ and $\unaryminus\!\sqrt{2}$ are of course discernible, since one is positive and the other is negative, and the order is definable from the field operations in $\R$ via $x\leq y\iff\exists u\ x+u^2=y$. It follows that the real number field is not definable in the complex field by any assertion in the language of fields. In fact, there is an enormous diversity of automorphisms of the complex field; one may move $\sqrt[3]{2}$, for example, to one of the nonreal cube roots of $2$, such as $\sqrt[3]{2}(\sqrt{3}i-1)/2$. Therefore, the numbers $\sqrt[3]{2}$ and $\sqrt[3]{2}(\sqrt{3}i-1)/2$ are indiscernible in the complex field---there is no property expressible in the language of fields that will distinguish them. Indeed, except for the rational numbers, every single complex number is part of a nontrivial orbit of automorphic copies, from which it cannot be distinguished in the field structure. So the same issue as with $i$ and $\unaryminus i$ occurs with every irrational complex number. For this reason, it is problematic to try to identify complex numbers with the abstract positions or roles that the numbers play in the complex field. -Meanwhile, one may recovers the uniqueness of the structural roles simply by augmenting the complex numbers with additional natural structure. Specifically, once we augment the complex field $\C$ with the standard operators for the real and imaginary parts: -$$\text{Re}(a+bi)=a\qquad\qquad\text{Im}(a+bi)=b,$$ -then the expanded structure $\langle\C,+,\cdot,\text{Re},\text{Im}\rangle$ becomes rigid, meaning that it has no nontrivial automorphisms. Thus, every complex number plays a unique structural role in this new structure, which is Leibnizian. This additional structure is implicit in the complex plane conception of the complex numbers, which is part of why the number $i$ appears fine as a singular term---it refers to the point $(0,1)$ in the complex plane---whereas $\unaryminus i$ refers to $(0,-1)$. The complex plane is not merely a field, for it carries along its coordinate information by means of the real-part and imaginary-part operators, making it rigid. In the complex plane, every complex number plays a different role. -This answer was copied from a related post at meta.mathoverflow.<|endoftext|> -TITLE: An interesting pattern in solutions to differential equations -QUESTION [6 upvotes]: OK, watch this: -Suppose I have a weight on the end of a spring. Assuming the spring obeys Hooke's law, as the weight is displaced from its rest position, the spring exerts a restoring force in the opposite direction who's magnitude is equal to the displacement multiplied by the spring constant. -Suppose that $f(t)$ represents the displacement of the weight at time $t$. If we assume that the spring constant and the mass of the weight are both unity, we have -$$f''(t) = -f(t)$$ -This is an equation involving both $f$ itself and its derivative $f''$, so this is presumably a "differential equation". I don't know how to deal with such a thing. But it is clear that this does not yet tell me what $f$ is, only that it must satisfy a specific property. -Thinking about this for a moment, it is clear that $f(x) = 0$ has the requested property. This corresponds to the weight remaining stationary for all eternity - a physically valid, but rather "boring" result. -Contemplating this further, it occurs to me that the derivative of $\sin$ is $\cos$, and the derivative of $\cos$ is $-\sin$. So if $f(t)=\sin(t)$ then $f''(t)=-\sin(t)$, which satisfies the equation. By nearly identical reasoning, $f(t)=\cos(t)$ would also work. In short, if you ping the weight, it oscillates around zero. -Now suppose that, by some bizarre mechanism, the restoring force were to somehow be in the same direction as the displacement. Impossible, I know. But imagine. Now our equation becomes -$$f''(t)=f(t)$$ -Again $f(t)=0$ would work. But what else? Well, there is exactly one function who's derivative equals itself: $\exp$. This is a stronger property than we need, but still, if $f(t)=\exp(t)$ then every derivative of $f$ (including $f''$) would equal $f$. This describes the weight accelerating away exponentially - rather as you might expect. -So far, we have two equations. The solution to one is $\sin$. The solution to the other is $\exp$. Two similar equations, two totally different solutions. Or at least, they look different. But now I'm thinking about something Euler once wrote: -$$\exp(ix) = \cos(x) + i \sin(x)$$ -Say that, and suddenly these solutions don't look so dissimilar at all. Now they suddenly look suspiciously similar! -My question: Is this the result of some deep and meaningful connection? Or is it merely a coincidence? - -Holy smokes, you guys are right! -I know, of course, of $\sinh$ and $\cosh$. (For real numbers, they look utterly unrelated. But in the complex plane, one is a trivially rotated version of the other.) What I didn't know, until I looked it up, was the derivatives of these functions. -Since they're defined in terms of $\exp$, I was expecting some really complicated derivative. However, what I actually found (as you presumably all know) is that $\sinh'=\cosh$ and, unlike in the circular case, $\cosh'=\sinh$! -So yes, for $f''=-f$ we have $f=\sin$ or $f=\cos$, and for $f''=f$ we have $f=\sinh$ or $f=\cosh$. So flipping the sign of the differential equation rotates the function in the complex plane. Physically, it doesn't look very meaningful to talk about complex-valued seconds, but mathematically it all looks vastly too perfect to be mere coincidence. - -REPLY [4 votes]: Continuing on Qiaochu Yuan's path, from $p(D)f=0$ you might think that there are as many solutions as the degree of $p$, just like there are the same number of roots of a polynomial as its degree. This is correct. -Your equations are linear, which means that no powers of $f$ appear (though powers of $D$ and the independent variable can occur) and homogeneous (which means there is no constant term). You can then prove that any linear combination of solutions is again a solution. So having found that $\exp(t)$ and $\exp(-t)$ are solutions of $f''(t)=f(t)$ you know that $A\exp(t)+B\exp(-t)$ is again a solution for any real $A,B$. The fact that you have found two and the degree of $D^2-1$ is $2$ means you have found them all. For $(D^2+1)f=0$ you noted that $\sin t$ and $\cos t$ are solutions and Qiaochu Yuan noted that $exp (it)$ and $\exp (-it)$ are solutions, but $\cos t=\frac {exp (it) +\exp (-it)}2$ and $\sin t=\frac {exp (it) -\exp (-it)}{2i}$ so these are linear combinations of each other.<|endoftext|> -TITLE: Why we need to know how to solve a quadratic? -QUESTION [19 upvotes]: Five years ago I was tutoring orphans in a local hospital. One of them asked me the following question when I tried to ask him to solve a quadratic: - -Why do I need how to solve a quadratic? I am not going to use it for - my future job! - -This question is, largely not mathematical. Substituting 'quadratic' with 'linear forms' or 'calculus' or 'Hamlet' would not make much difference since the specific knowledge is not used on a day to day basis in most occupations except academia. But I feel puzzled as how to justify myself that 'learning quadratics is important enough that you must learn it'. At that day, I used a pragmatic argument that he need to pass various qualification exams to get to college, and after college he can find a job he wants. But this feels self-defeating - we are not learning for the sake of passing tests or getting high grades. I do not know how to make the kid understand that "knowing how to solve a quadratic is interesting and knowing how to solve higher degree ones can be awesome" - because knowing $(x-p)(x-q)=x^2-(p+q)x+pq$ is not very interesting to him. -Since I am still puzzled over it I decided to ask others who may had similar experience. What do you say when others ask you "what is the benefit of knowing $xxx$ theorem? Will you respond that "knowing $xxx$ is helpful/interesting because of $a,b,c,d$ reasons?"(thus refute the utilitarian argument), or arguing as this post that some knowledge is essential to know for anyone? -My father asked me "What is the importance of proving $1+1$ (the Goldbach Conjecture)" when I returned from college. I do not know how to answer as well even though I know the history behind the conjecture. Now I am going to become a teaching assistant, I think I should be able to answer such questions before I am at the stage and someone ask me questions like "Why do I need to know calculus"? again. So I post this at here. - -REPLY [3 votes]: I'll try to answer your question from another point of view. To me, it's important not to know how to solve quadratic equations, but to show your students that they already know how to do that. -Consider the following rectangle as an easy way to show that $(x+y)(a+b+c+d)$ is indeed a sum of smaller rectangles formed by those lines and segments: - -(source: popmath.ru) -The same picture show that any expressions of that sort will consist of $2 \cdot 4$ members, which can come quite handy while exlaining the intuition behind binomial formula. -Did your student have to know anything to come up with this conclusion? No, he already knew that. -Going on, the same can be said about quadratics, consider that $(a+b)^2$ expression is beautifully summed up in this picture: - -(source: popmath.ru) -So to come up with a way of solving it your student only needs to understand that any equation with $a^2$ is a crippled square of sum formula, thus by completing the square he can solve it (at least in trivial case without any fancy numbers) by not memorizing anything. -And that's the point of math — it can be treated as an art of arriving at unexpected conclusions by reflecting on the things you already know. Quadratics just serve as a simple example of that approach to make sure your students get the general idea. -Hope that helped to answer your question.<|endoftext|> -TITLE: Is there a Hamiltonian path for the graph of English counties? -QUESTION [10 upvotes]: The mainland counties of England form a graph with counties as vertices and edges as touching borders. Is there a Hamiltonian path one can take? This is not homework, I just have an idea for a holiday around England where I visit every county only once! - -REPLY [17 votes]: Let's assume we can access the Isle of Wight through Hampshire. Then the answer is yes: - -N.B.: This uses ceremonial counties instead of administrative counties; see comments for discussion. -Edit (after a comment below): The background image from Wikipedia. I found the path mostly by luck, with the knowledge that I had to start in the Isle of Wight and finish in Cornwall.<|endoftext|> -TITLE: Positive part of the kernel -QUESTION [5 upvotes]: Let $(E,\mathscr E)$ be a measurable space and $Q:E\times \mathscr E\to\Bbb [-1,1]$ be a signed bounded kernel, i.e. $Q_x(\cdot)$ is a finite measure on $(E,\mathscr E)$ for any $x\in E$ and $x\mapsto Q_x(A)$ is a measurable function for any set $A\in \mathscr E$. -For any fixed $x$, let the measure $Q^+_x$ be a positive part of the signed measure $Q_x$ as in Hahn-Jordan decomposition. Is it true that $Q^+$ is a kernel, i.e. is the function $x\mapsto Q_x^+(A)$ measurable for any $A\in \mathscr E$? It clearly holds if $Q$ is an integral kernel, but I am interested in the general case. - -Update: after three weeks and 1 bounty I decided to post this question also on MO - -REPLY [2 votes]: Here is another proof, taken from "Markov Chains" by D. Revuz (lemma 1.5 page 190), always under the assumption that $(E, \mathscr{E})$ is countably generated. -The following answer is almost a paraphrase of the proof of Revuz. -By assumption, there is a sequence of finite partitions $\mathscr{P}_n$ of $E$, such that $\mathscr{P}_{n+1}$ is a refinement of $\mathscr{P}_n$, and $\mathscr{E}$ is generated by $\cup_{n \ge 0} \mathscr{P}_n$. For every $x \in E$, there exists a unique $E_n^x \in \mathscr{P}_n$ with $x \in E_n^x$. -Let $x \in E$ be fixed for the moment. Define $\lambda_x$ as the probability measure which is a multiple of $|Q_x|$. (if $Q_x = 0$, choose it as you want) -Then define a function $f_n$ on $E$ by $\displaystyle f_n(y) = \frac{Q_x(E_n^y)}{\lambda_x(E_n^y)}$ if $\lambda_x(E_n^y) > 0$ and $f_n(y) = 0$ otherwise. -By martingale convergence theorem, we have that $f_n$ converges $\lambda_x$-a.s. to the density of $Q_x$ with respect to $\lambda_x$. Hence $f_n^+$ converges $\lambda_x$-a.s. to the density of $Q_x^+$, and since $f_n$ are uniformly bounded, we have for all $A \in \mathscr{E}$ : $$Q_x^+(A) = \lim_n \int_A f_n^+ \, d \lambda_x$$ -Now, if $A \in \mathscr{P}_k$, then for all $n > k$, $\int_A f_n^+ \, d \lambda_x = Q_{x,n}^+(A)$, where $Q_{x,n}^+$ is the positive part of the restriction of $Q_x$ to the $\sigma$-algebra generated by $\mathscr{P}_n$. It's easy to see that the map $x \to Q_{x,n}^+(A)$ is measurable, and so is the map $x \to Q_x^+(A)$. -We have hence proven that $x \to Q_x^+(A)$ is measurable for every $A \in \cup_{n \ge 0} \mathscr{P}_n$, and then a Dynkin class argument finishes the proof.<|endoftext|> -TITLE: Quadratic field such that a certain finite set of primes split -QUESTION [5 upvotes]: Given a finite set $S$ of primes, is it possible to find an imaginary quadratic field $K$ such that all primes in $S$ are split completely in $K$? - -REPLY [3 votes]: Sure. Let me just assume WLOG that $S$ contains $2$. Let $D$ be squarefree. If $D \equiv 1 \bmod 4$, then $\mathbb{Q}(\sqrt{D})$ has ring of integers $\mathbb{Z}[x]/(x^2 - x - \frac{D-1}{4})$. Then $2$ splits if and only if $D \equiv 1 \bmod 8$. An odd prime $p$ splits if $D \equiv 1 \bmod p$, so it suffices to find a (negative) squarefree $D \equiv 1 \bmod 8$ such that $D \equiv 1 \bmod p$ for all odd primes $p \in S$. In fact a prime with this property exists by Dirichlet's theorem.<|endoftext|> -TITLE: Indefinite Integral of $\sqrt{\sin x}$ -QUESTION [18 upvotes]: $$\int \sqrt{\sin x} ~dx.$$ -Does there exist a simple antiderivative of $\sqrt{\sin x}$? How do I integrate it? - -REPLY [25 votes]: Since $\sqrt{\sin(x)} = \sqrt{1 - 2 \sin^2\left(\frac{\pi}{4} -\frac{x}{2}\right)}$, this matches with the elliptic integral of the second kind: -$$\begin{align*} - \int \sqrt{\sin(x)} \, \mathrm{d} x &\stackrel{u = \frac{\pi}{4}-\frac{x}{2}}{=} -2 \int \sqrt{1-2 \sin^2(u)} \,\mathrm{d} u\\ &= -2 E\left(u\mid 2\right) + c = -2 E\left(\frac{\pi}{4}-\frac{x}{2}\middle|\, 2\right) + c -\end{align*}$$ -where $c$ is an integration constant.<|endoftext|> -TITLE: Returning Paths on Cubic Graphs Without Backtracking -QUESTION [12 upvotes]: I was once interested in the returning paths on cubic graphs . But I'm even more curious to have the number of ways without backtracking, which means doing one step forward and than one back (which might be good for dancing), e.g. $1\to 2\to 1$ . -The solution with the powers of the adjacency matrix doesn't seem to work here. Does anybody know a solution? - -REPLY [5 votes]: Chris Godsil's beautiful answer is marred, I believe, by errors in the last two lines. Since numerous other posts refer to that answer and since there seems to be some reluctance to modify it, while at the same time there still seems to be some confusion/disagreement about what the correct result is, I have written this community wiki answer in order to place what I believe to be a corrected version of Chris's answer on the record. If this correction is in error, I hope that people will downvote it mercilessly; if it is valid, I hope that some form of the corrections will be incorporated into the original post so that this answer can be deleted. -Here is Chris's answer, with correction appended: - -Call a walk reduced if it does not backtrack. - If $A=A(X)$ for a graph $X$, define $p_r(A)$ to be the matrix (of the same order as $A$) such - that $(p_r(A)_{u,v})$ is the number of reduced walks in $X$ from $u$ to $v$. - Observe that - $$ - p_0(A)=I,\quad p_1(A) =A,\quad p_2(A) = A^2-\Delta, -$$ - where $\Delta$ is the diagonal matrix of valencies of $X$. If $r\ge3$ we have the recurrence - $$ - Ap_r(A) = p_{r+1}(A) +(\Delta-I) p_{r-1}(A). -$$ - These calculations were first carried out by Norman Biggs, who observed the implication that $p_r(A)$ is a polynomial in $A$ and $\Delta$, of degree $r$ in $A$. -If $X$ is cubic, $\Delta=3I$ and we want the polynomials $p_r(t)$ satisfying the recurrence - $$ - p_{r+1}(t) = tp_r(t)-2p_{r-1}(t) -$$ - -with initial conditions $p_1=t$ and $p_2=t^2-3$. Note that the recurrence does not hold when $r=1$ since $tp_1(t)-2p_0(t)=t^2-2$ is not equal to $p_2(t)=t^2-3$. The function $q_r(t)=2^{-r/2}p_r(2^{3/2}t)$ satisfies the recurrence of the Chebyshev polynomials, -$$ -q_{r+1}(t)=2tq_r(t)-q_{r-1}(t) -$$ -with initial conditions -$$\begin{aligned} -q_1(t)&=2t=U_1(t)=U_1(t)-\frac{1}{2}U_{-1}(t),\\ -q_2(t)&=4t^2-\frac{3}{2}=4t^2-1-\frac{1}{2}=U_2(t)-\frac{1}{2}U_0(t). -\end{aligned} -$$ -Here $U_r(t)$ are the Chebyshev polynomials of the second kind, which satisfy the initial conditions -$$ -\begin{aligned} -U_0(t)&=1,\\ -U_1(t)&=2t, -\end{aligned} -$$ -and the further relations -$$ -\begin{aligned} -U_{-1}(t)&=0,\\ -U_2(t)&=4t^2-1, -\end{aligned} -$$ -as implied by the recurrence. Since the recurrence is linear, we conclude that -$$ -q_r(t)=\begin{cases}1 & \text{if $r=0$,}\\ U_r(t)-\frac{1}{2}U_{r-2}(t) & \text{if $r\ge1$.}\end{cases} -$$ -From this it follows that -$$ -p_r(t)=\begin{cases}1 & \text{if $r=0$,}\\ 2^{r/2}U_r(t/2^{3/2})-2^{(r-2)/2}U_{r-2}(t/2^{3/2}) & \text{if $r\ge1$.}\end{cases} -$$<|endoftext|> -TITLE: Computer Algebra Systems which implement Cylindrical Algebraic Decomposition -QUESTION [8 upvotes]: My understanding is that Mathematica's Reduce function is based on Cylindrical Algebraic Decomposition (CAD). -The only other system I've seen which implements CAD is QEPCAD. QEPCAD isn't a general CAS like Mathematica or Maple. -Do any mainstream CASes feature a solver as powerful as Mathematica's Reduce? Are they known to be based on CAD as well? Are any of the open source CASes working on adding a CAD based solver? - -REPLY [5 votes]: Open source: In Axiom (also FriCAS and OpenAxiom) see the axiom-developer email list -Re: [Axiom-developer] installing Axiom on Fedora 16 -From: Renaud . Rioboo -Date: Fri, 13 Jul 2012 14:15:46 -0700 - -Dear Axiom Gurus, -there seems to be again some interests on cylindrical algebraic -decomposition and I wanted to recompile my CAD package which is -available at -http://rioboo.free.fr/CadPub/ -The package compiles and runs under open-axiom ... -http://www.ensiie.fr/~renaud.rioboo/<|endoftext|> -TITLE: Derivative of $x^x$ at $x=1$ from first principles -QUESTION [6 upvotes]: Find the derivative of $x^x$ at $x=1$ by definition (i.e. using the limit of the incremental ratio). - -The only trick I know is $x^x = e^{x \ln x}$ but it doesn't work. - -REPLY [13 votes]: Using the definition: -$$ -\begin{align} -f'(1)&=\lim_{x\rightarrow1}\frac{x^x-1}{x-1}\\ -&=\lim_{x\rightarrow1}\frac{e^{x\log{x}}-1}{x-1}\\ -&=\lim_{y\rightarrow0}\frac{e^{(1+y)\log(1+y)}-1}{y}\\ -&=\lim_{y\rightarrow0}\frac{e^{(1+y)\log(1+y)}-1}{(1+y)\log(1+y)}\frac{(1+y)\log(1+y)}{y}\\ -&=\lim_{t\rightarrow0}\frac{e^{t}-1}{t}\lim_{y\rightarrow0}(1+y)\frac{\log(1+y)}{y}=1 -\end{align} -$$ -where $y=x-1$, and $t=(1+y)\log(1+y).$<|endoftext|> -TITLE: Hatcher Algebraic Topology 0.24 -QUESTION [8 upvotes]: This is my second question from Hatcher chapter 0 (and final I think). For $X$, $Y$ CW complexes, it asks one to show that -$$X \ast Y = S(X \wedge Y)$$ -by showing -$$X \ast Y/(X \ast y_0 \cup x_0 \ast Y) = S(X \wedge Y) / S(x_0 \wedge y_0),$$ -where $\ast$ is topological join, $\wedge$ is smash product, $S$ is suspension and $=$ is homeomorphism. -Unfortunately I am quite at loss on this question besides from knowing some definitions of these spaces. $S(x_0\wedge y_0)$ is suspension of a point which is an interval I. $X \wedge Y$ is the space obtained by the quotient $X \times Y / X \vee Y$ where $X \vee Y$ is wedge sum of $X$ and $Y$ at the point $x_0$ and $y_0$. Suspension of this is its product with interval $I$ and collapsing the end points. Overall we have -$$X \times Y \rightarrow X \times Y / X \vee Y \rightarrow (X \times Y / X \vee Y) \times I \rightarrow Z = (X \times Y / X \vee Y) \times I/(\text{contract $t=0$ and $t=1$ to a point}) \rightarrow Z / S(x_0\wedge y_0).$$ -$(X \ast {y_0} \cup {x_0} \ast Y)$ is the union of two cones over $X$ and $Y$ and finally $X \ast Y$ could be seen as union of cones $C(X,y)$ for all $y$ or the other way around. Then this space is -$$X \times Y \times I \rightarrow A= X \times Y \times I/(\text{contract $Y$ at $t=0$ and $X$ at $t=1$}) \rightarrow A/(X \ast {y_0} \cup {x_0} \ast Y).$$ -However I cant see how to relate these operations. I know some theorems on these space making homotopy relations but this questions requires this to be a homeomorphism. -Thanks a lot. - -REPLY [3 votes]: Some pictures from "Topology and Groupoids", Chapter 5, which you may find helpful are the join $X * Y$ as -and the subspace to be collapsed to a point to give the suspension of the smash product is - where the two vertices on the mid line are the base points.<|endoftext|> -TITLE: PDEs of higher order than three? -QUESTION [7 upvotes]: Motivation for question: I know just a little about the general theory of PDEs. I'm working on a project which happens to need examples of PDEs like Laplace's equation. The next step is to look at higher order PDEs in my investigation. I own a few PDE texts and I had hoped to find further examples, but I've noticed that the bulk of my PDE books concern mainly second order PDEs over various dimensional spaces (for example, $n=2$: $u_{xx}+u_{yy}=0$, or $n=3$: $u_{xx}+u_{yy}=u_{tt}$). -My question is this: do higher order equations (like $u_{xxx}=u_{tt}$, to give a silly example) have a well-developed theory? -In the case of ODEs I'm aware that we can reduce to a system of first order equations thus solutions to the $n^{th}$-order problem are furnished by the first-order theory. -Is there a similar story for PDEs? Or, is it just that higher-order PDEs are less interesting? -Thanks in advance for any insights! - -REPLY [5 votes]: The first example that comes to mind is the clamped plate equation, which is of 4th order. To see how engineers approach it, click here. More generally, any time your model calls for prescribed values and prescribed derivative on the boundary, you'll probably be dealing with a PDE of order above $2$. For second-order equations such boundary value problems are overdetermined. -There is also a general method for the Dirichlet problem for elliptic operators of any order, namely one can use Gårding's inequality to verify the coercitivity assumption in the Lax-Milgram theorem. -On the other hand, a recent paper by one of my colleagues opens with a sobering statement - -There is nothing in the theory of linear strongly elliptic differential operators [of higher orders] that can be called a general existence theorem for solutions to the Neumann problem. - -On the positive side, the Fourier transform goes a long way for linear PDE of all orders, especially as far as the regularity of solutions is concerned. The very size of Hörmander's 4-volume treatise "The analysis of linear partial differential operators" suggests that there is something well-developed in there. -One of the things that make second order equations easier to deal with is the fact that the calculus test for maxima/minima is the second derivative test. The maximum principle is a workhorse in the theory of elliptic equations of 2nd order. In contrast, the Hadamard maximum principle for biharmonic equation does not go very far (see page 2 of the paper).<|endoftext|> -TITLE: showing almost equal function are actually equal -QUESTION [5 upvotes]: I am trying to show that if $f$ and $g$ are continuous functions on $[a, b]$ and if $f=g$ a.e. on $[a, b]$, then, in fact, $f=g$ on $[a, b]$. Also would a similar assertion be true if $[a, b]$ was replaced by a general measurable set $E$ ? -Some thoughts towards the proof - -Since $f$ and $g$ are continuous functions, so for all open sets $O$ and $P$ in $f$ and $g$'s ranges respectfully the sets $f^{-1}\left(O\right) $ and $g^{-1}\left(P\right) $ are open. -Also since $f=g$ a.e. on $[a, b]$ I am guessing here implies their domains and ranges are equal almost everywhere(or except in the set with measure zero). -$$m(f^{-1}\left(O\right) - g^{-1}\left(P\right)) = 0$$ - -I am not so sure if i can think of clear strategy to pursue here. Any help would be much appreciated. -Also i would be great full you could point out any other general assertions which if established would prove two functions are the same under any domain or range specification conditions. -Cheers. - -REPLY [2 votes]: A subset of an interval of full measure is dense in that interval. If two functions agree on a dense set, they are equal everywhere.<|endoftext|> -TITLE: About differentiation under the product integral sign -QUESTION [5 upvotes]: It is known that -$$\dfrac{d}{dx}\int_{a(x)}^{b(x)}f(x,t)~dt=\dfrac{db(x)}{dx}f(x,b(x))-\dfrac{da(x)}{dx}f(x,a(x))+\int_{a(x)}^{b(x)}\dfrac{\partial}{\partial x}f(x,t)~dt.$$ -How about -$$\dfrac{d}{dx}\prod_{a(x)}^{b(x)}f(x,t)^{dt}?$$ - -REPLY [7 votes]: Because we are in a commutative setting, the product integral satisfies -$$\prod_{a(x)}^{b(x)} f(x,t)^{dt}=\exp\left(\int_{a(x)}^{b(x)}\log f(x,t)\,dt\right).$$ -Therefore with the chain rule and the rule you state, we derive -$$\frac{d}{dx}\log \left(\prod_{a(x)}^{b(x)} f(x,t)^{dt}\right) = \frac{d}{dx}\int_{a(x)}^{b(x)}\log f(x,t)dt$$ $$ = b'(x)\log f(x,b(x))-a'(x)\log f(x,a(x))+\int_{a(x)}^{b(x)}\frac{\partial}{\partial x}\log f(x,t)dt$$ -hence -$$\frac{d}{dx}\prod_{a(x)}^{b(x)} f(x,t)^{dt}=\left(\prod_{a(x)}^{b(x)} f(x,t)^{dt}\right)\left(\log \frac{f(x,b(x))^{b'(x)}}{f(x,a(x))^{a'(x)}}+\int_{a(x)}^{b(x)}\frac{\frac{\partial}{\partial x}f(x,t)}{f(x,t)}dt\right).$$<|endoftext|> -TITLE: Sums of sums of sums of...of numbers -QUESTION [6 upvotes]: If we introduce the following notation -$$S_r^q=\overbrace{\sum_{a_{r-1}=1}^q\sum_{a_{r-2}=1}^{a_{r-1}}\cdots\sum_{a_1=1}^{a_2}\sum^{a_1}}^{\mbox{a total of $r$ sums}}1$$ -for example, $S^q_1=q$, $S^q_2=q(q+1)/2$ and so on, then one can show that -$$S^p_{q-1}=S^q_{p-1},$$ -where $p$ and $q$ are positive integers. What is the simplest proof of this? I know of one but suspect that there exists simpler ones. Is there any generalisation of this statement. Can somebody also direct me to some references on related material. Thanks a lot in advance! - -REPLY [11 votes]: Using the formula -$$ -\sum_{k=0}^n\binom{k}{j}=\binom{n+1}{j+1}\tag{1} -$$ -we get inductively that -$$ -S_q^p=\binom{p+q-1}{q}\tag{2} -$$ -Thus, we have -$$ -S_{q-1}^p=\binom{p+q-2}{q-1} -$$ -and -$$ -S_{p-1}^q=\binom{p+q-2}{p-1} -$$ -and your identity is just the common symmetry identity $\displaystyle\binom{n}{k}=\binom{n}{n-k}$. - -Inductive Reasoning - -$(2)$ holds for $q=\color{#C00000}{1}$ because $\displaystyle S_{\color{#C00000}{1}}^p=p=\binom{p+\color{#C00000}{1}-1}{\color{#C00000}{1}}$ -Assume $(2)$ holds for some $q$. By definition, assumption, and $(1)$, -$$ -\begin{align} -S_{q+1}^p -&=\sum_{k=1}^pS_q^k\\ -&=\sum_{k=1}^p\binom{k+q-1}{q}\\ -&=\binom{p+q}{q+1}\\ -&=\binom{p+(q+1)-1}{q+1} -\end{align} -$$ -Therefore, $(2)$ holds for $q+1$.<|endoftext|> -TITLE: Is the intersection of two f.g. projective submodules f.g.? -QUESTION [5 upvotes]: Let $R$ be a commutative unital ring and $M$ a finitely generated projective $R$-module. My question is: if $N_1$ and $N_2$ are f.g. projective submodules of $M$, is $N_1 \cap N_2$ f.g.? Is it projective? -(Surely the answer is no but I haven't been able to find a counterexample. Also, sorry for the lack of motivation, but my reason to ask is so convoluted I think including it would seriously lower the interest/size ratio of this question.) - -REPLY [7 votes]: Thinking of a f.g. projective module as a vector bundle, it seems very likely the answer is no: consider the trivial bundle of rank 2 and two "twisted" subbundles of rank 1 whose intersection is 0-dimensional everywhere except over a closed subset with non-empty interior – so not a vector bundle, in particular. -Let's see where this thinking leads. Let $R$ be the ring of real-valued continuous functions on the real line $\mathbb{R}$. Let $M = R \oplus R$. Let $f : \mathbb{R} \to \mathbb{R}$ be the following continuous function: -$$f (x) = \begin{cases} -\exp (-(x+1)^{-2}) & x < 1 \\ -0 & -1 \le x \le 1 \\ -\exp (-(x-1)^{-2}) & x > 1 -\end{cases}$$ -Consider the submodules below: -\begin{align} -N_1 & = \{ (s_1, s_2) \in M : f s_1 - s_2 = 0 \} & -N_2 & = \{ (s_1, s_2) \in M : f s_1 + s_2 = 0 \} -\end{align} -$M$, $N_1$ and $N_2$ are all clearly f.g. free $R$-modules. Yet their intersection is not f.g. projective: -$$N = N_1 \cap N_2 = \{ (s_1, s_2) \in M : f s_1 = 0, s_2 = 0 \}$$ -Let $\mathfrak{m}$ be the maximal ideal of $R$ consisting of functions vanishing at $1$, and let $R_\mathfrak{m}$ be the localisation of $R$ at $\mathfrak{m}$. If $N$ were f.g. projective, then $N_\mathfrak{m}$ would be f.g. free over $R_\mathfrak{m}$. On the other hand, by construction, $N_\mathfrak{m}$ is annihilated by the germ of $f$, which is not zero, and $N_\mathfrak{m}$ is itself non-zero (take, for example, any bump function supported on $[-1, 1]$) – so $N_\mathfrak{m}$ cannot be free. Therefore $N$ is not f.g. projective. -Here is a picture of one of the monstrosities we constructed:<|endoftext|> -TITLE: Limit point and interior point -QUESTION [6 upvotes]: Is any interior point also a limit point? -Judging from the definition, I believe every interior point is a limit point, but I'm not sure about it. If this is wrong, could you give me a counterexample? -(Since an interior point $p$ of a set $E$ has a neighborhood $N$ with radius $r$ such that $N$ is a subset of $E$. Obviously any neighborhood of $p$ with radius less than $r$ is a subset of $E$. Also, any neighborhood with radius greater than $r$ contains $N$ as a subset, so (I think) it is a limit point.) - -REPLY [12 votes]: A discrete space has no limit points, but every point is an interior point.<|endoftext|> -TITLE: Proving that a metric space is compact -QUESTION [5 upvotes]: Let $H^\infty$ be the set of real sequences such that each element in each sequence has $|a_n|\leq 1$. The metric is defined as -$$d(\{a_n\}, \{b_n\}) = \sum_{n=1}^\infty \frac{|a_n - b_n|}{2^n}.$$ -Prove that $H^\infty$ is a compact metric space. -To prove this, I want to show that every sequence in $H^\infty$ has a convergent subsequence. I know that if we have a sequence $\{\{a_n\}^{(k)}\}$ in $H^\infty$, then for all $i$, the real sequence $\{a_i^{(k)}\}$ has a convergent subsequence, since it is bounded by 1. So we can get a convergent subsequence $\{a_1^{(k_j)}\}$, and then a convergent subsequence of $\{a_2^{(k_j)}\}$, and continue taking subsequences of subsequences until we have a convergent subsequence of $(a_1, a_2, a_3, ..., a_n)^{(k)}$ with $n$ some positive integer if we stop taking subsequences at the $nth$ subsequence; this gives a sequence $\{x_n\}$ where $x_n$ is the limit of the $n$th convergent subsequence of $\{a_n^{(k)}\}$. Ideally, we could show that the sequence in $H^\infty$ converges to $\{x_n\}$. -I know that if we have the $nth$ subsequence of $\{\{a_i\}^{(k)}\}$ defined in the way described above, then for any $\epsilon > 0$ there exist $N_1, ..., N_n$ such that if $k\geq \max_{i\leq n}\{N_i\}$, then for $1\leq i\leq n$, $|a_i^{(k)} - x_i| < \epsilon/2n$. By choosing $n$ sufficiently large that $\sum_{i=n+1}^\infty |a_i^{(k)} - x_i|/2^i\leq \sum_{i=n+1}^\infty 1/2^{i-1} < \epsilon/2$, we can ensure that $d(\{a_n\}^{(k)}, \{x_n\}) < \epsilon$. But the problem here is that for each $\epsilon$, we end up choosing a different convergent subsequence of the first $n$ terms (since we need to choose $n$, which determines how many subsequences of subsequences we take). Any idea on how to proceed? - -REPLY [8 votes]: One way is to take first a subsequence $(a_1^i)_i$ such that the first coordinate converges. Now take a subsequence of this subsequence $(a_2^i)_i$ such that the second coordinate converges. Proceeding this way you obtain a sequence of nested subsequences $(a_{k+1}^i)_i \subset (a_k^i)_i$. Now take the sequence $(a_k^k)_k$. By the last property every coordinate of this sequence converges, and this is equivalent to convergence in your metric. (Note that for each $k,i$ that $a_k^i$ is a sequence of real numbers so we have three layers of sequences...). By the way this is called a diagonal argument, and it's a really useful trick to have available. -Another way to do this would be to note that your space is basically $\prod_{n=1}^{\infty} [0,1]$ with the product topology, so an appeal to Tychonoff's theorem gives the conclusion.<|endoftext|> -TITLE: Olympic Badminton, or How to Design a Tournament -QUESTION [6 upvotes]: Hearing the recent news about disqualified Badminton players in the ongoing 2012 London Olympics got me wondering about how best to design tournaments to avoid situations where players are incentivized to throw matches. I have no doubt that much has been written about this but I have no idea where to start. - -Are there any Arrow-like theorems saying that "ideal tournament design" is impossible, i.e. given some short-ish list of generally agreeable desirable features of a tournament, one proves that they are contradictory? - -I'm a novice in this sort of mathematics so feel free to recommend introductory surveys or books as well. - -REPLY [2 votes]: There are no impossibility results "Arrow like" regarding tournament design but lots of open problems especially in the incomplete information case. -Kay Konrad's book on contests is a good general reference. -As tournaments can be modeled as all-pay auctions, the literature on optimal auction design may also be relevant, see Vijay Krishna's book. -I know the above economics literature well but I am not familiar with the computer science one that may be also relevant: see this thesis https://stacks.stanford.edu/file/druid:qk299yx6689/TV-thesis-final-augmented.pdf<|endoftext|> -TITLE: Localization of a non-commutative ring -QUESTION [5 upvotes]: Let $A$ be the non-commutative ring given by -$$ -A=\mathbb{C}\langle x,y,z \rangle /(xy=ayx,yz=bzy,zx=cxz) -$$ -for some $a,b,c\in \mathbb{C}$. What is the localization $A_{(x)}$ of A with respect to the (two-sided) ideal $(x)$? If it can be defined, is it graded ring? In general what condition is required to localize a ring? -I think of $A$ as a non-commutative $\mathbb{P}^2$ and wonder whether or not we can study it by local patch. -Thank you in advance. - -I should mention this; my main problem is I don't know about the definition of localization. Moreover even if it is defined I am not sure whether this technique is useful or not. I would like to conclude for example smoothness of the non-commutative $\mathbb{P}^2$ or a hypersurface in it. I hope to confirm this by checking it locally. - -REPLY [8 votes]: This isn't really an answer to your question (except perhaps the last part), but: Wikipedia claims that the localization of a noncommutative ring $R$ with respect to some subset $S$ does not always exist. This, I think, comes from using the wrong definition of localization. The definition that seems natural to me is the following. -Definition: The localization $S^{-1} R$ of a ring $R$ with respect to a subset $S$ is the universal ring equipped with a morphism $\phi : R \to S^{-1} R$ such that $\phi(s)$ is invertible for every $s \in S$. -The localization in this sense always exists. It can be constructed explicitly as consisting of sums of formal symbols $r_1 s_1^{-1} r_2 s_2^{-1} ... $ where $r_i \in R, s_i \in S$ quotiented by suitable relations. The real question is whether one can say anything reasonable about the localization in general (in particular determine whether it is nonzero)... -The concern that Wikipedia actually has seems to be with whether every element of the localization can be written $r s_1^{-1} s_2^{-1} ...$ (that is, whether we need to alternate elements of $R$). My understanding is that this is addressed by the Ore condition. For more general notions of noncommutative localization see for example the nLab.<|endoftext|> -TITLE: Characteristic functions based proof problem. -QUESTION [5 upvotes]: I am trying to show that if $T$ be a closed bounded interval and $E$ a measurable subset of $T$. Let $\epsilon >0$, then there is a step function $h$ on $T$ and a measurable subset $F$ of $T$ for which $h=\chi_E$ on $F$ and $m(T - F)<\epsilon$. -Some Thoughts towards the proof. -$\forall \epsilon >0$, there is a finite disjoint collection of open intervals $\left\{ I_{k}\right\} _{k=1}^{n}$ for which if $O=U_{k=1}^{n}I_{k}$ then $$m^{*}(E-O) + m^{*}(O-E) < \epsilon.$$ -Also $$\chi_{E}=\begin{cases}1, & x\in E \\ 0, & x\notin E. \end{cases} $$ -So $\forall \epsilon >0$ we have $\chi_{E} = \chi_{O}$. -Now $h$ being a step function looks like $$h =\sum _{i=0}^{n}\alpha _{i}\chi _{T_i}\left( x\right).$$ -I am unsure how to proceed from here and show the two results. -Any help would be much appreciated. - -REPLY [2 votes]: As David Mitra suggests in the comments, - -Apply Littlewood's first principle to find a finite collection of open intervals whose union $U$ satisfies $\mu(U\Delta E)<\varepsilon$. Take $F$ to be $(U\Delta E)^C$ and define the step function to be $1$ on $U$ and $0$ off $U$.<|endoftext|> -TITLE: Iterated prisoners dilemma with discount rate and infinite game averages -QUESTION [5 upvotes]: Suppose we have two players who are perfectly rational (with their perfect rationality common knowledge) playing a game. On round one both players play in a prisoners dilemma type game. With payoffs (1,1) for mutual cooperation, (.01,.01) for mutual defection and (1.01,0) and (0,1.01) for the situations where player 1 and player 2 defect respectively. They may talk before all rounds. -After playing a round they flip a coin, if coin is the result is tails, we end the game. If the result is heads, they play another round with payoffs multiplies by $.9^n$ where n is the number of rounds that have been played. If the result is tails, the game is over. -I'm confused by the two following, both seemingly plausible arguments. -Argument 1, The game is clearly equivalent to the following game. flip a coin until you get tails. call the number of coin flips n. Keep n hidden from the players. For n rounds the players will play the prisoners dilemma with payoffs as before (That is, as before, multiplied by $.9^n$ where n is the round number). Both players reason as follows "this game is nothing more than an iterated prisoner dilemma where I don't know the number of rounds, given that for any finite iterated prisoners dilemma (which this is with probability 1, where, because payoff is bounded, we can ignore the cases where it is not). Thus Because the iterated prisoners dilemma for any fixed finite number of rounds has unique nash equilibrium "defect" my optimal strategy, regardless of n, is to defect, thus I should defect. -Argument 2 I will tell my opponent that I am going to play a grim trigger strategy. That is, I will collaborate until he defects, after which, I will always defect. He will reply that he is also playing a grim trigger strategy. Given that there is always a 50% chance of a next round with .9 the payoff of the current one. Defecting on any round will cost me, on average at least .99*.45>.01 thus I will never have any incentive to defect. Neither will he. Thus we will always cooperate. -There is clearly a contradiction between these two arguments. I'm wary of both the first strategy's claim that if we use a hidden, but well defined, random variable to choose which of infinitely many games to play, all of which have the same move in their unique nash equilibria that same move is the nash equilbrium move of the averaged game. (Which, though seductive, seems close enough to infinity to not be inherently trust worthy). The second argument seems to me to rely on some symmetries which I'm not sure are valid assumptions. -I'm curious about two questions. Which of the above arguments is fallacious and where is the fallacy? And what is the (are the) nash equilbria and are they unique? - -REPLY [5 votes]: It is a well known fact that a repeated game in which the number of rounds follows a geometric distribution and payoffs are not discounted, but is unknown, is equivalent to a repeated game that goes on forever and in which payoffs are geometrically discounted. So this is simply a question of the repeated prisoners dilemma with geometric discounting. Always defect is always an equilibrium, and for sufficiently patient players grim trigger is too. -But in any finitely repeated prisoners dilemma, there is a unique Nash equilibrium in which both players always defect. So what is the difference. Here is the proof that in the finitely repeated prisoners dilemma, always defect is the unique Nash equilibrium outcome. Say Ann and Bob play the game for a finite number of rounds. Let $m$ be the last period in which any player cooperates in any Nash equilibrium. Since after $m$, players will defect anyways, and cooperating is costless, the player who cooperated in period $m$ will prefer to defect instead, and we do not have a Nash equilibrium. If the game is infinite or of unknown, but arbitrary long finite lenght, the round $m$ may not exists. -Indeed, what might keep the players cooperating, is that there is always a chance that cooperation will be rewarded in the future, at least with positive probability. This makes, quite generally, infinitely repeated games quite different from finitely repeated ones. Equilibria are usually not unique. As a matter of fact, there are certain Folk theorems that say pretty much anything is possible, within certain restricitons, if players are sufficiently patient. -Interestingly, for long, finitely repeated games something similar can work when the games has multiple equilibria, for then players can reward and punish by coordinating on a certain equilibrium. A folk theorem of this kind was proven by Benoît and Krishna.<|endoftext|> -TITLE: Riemannian Volume Form -QUESTION [9 upvotes]: There is a following exercise in my text: - -Let $S^n$ be $n-$ dim sphere in $R^{n+1}$ with inclusion function $i:S^{n}\to R^{n+1}$. Let -$$\omega=\sum_{i=1}^{n+1}(-1)^{i-1} x_i dx_1 \wedge... dx_{i-1}\wedge dx_{i+1}\wedge ... \wedge dx_{n+1}.$$ - -Prove that $i^*\omega \in \Omega^n(S^n)$ is Riemannian volume form on $S^n$. - -I treied to manually compute this expression and the one which uses definition of Riemannian volume form when they act on some vectors in $T_xS^n$ but things gets complicated when $n$ is large and involves sum of matrix determinants which I don't know how to resolve. I managed to prove the result for small values of $n$. -How would you go with the general case? - -REPLY [10 votes]: It's not so hard once you show that the volume form of a submanifold $N^{n-1}\subset M^n$ of codimension 1 is given by $\mathrm d vol_N(x) = (\iota_Z \mathrm{d}vol_{M})(x)$ for $x\in N$, where $Z$ is a normal vector field to $N$, $\mathrm d vol_M$ is the Riemannian volume form of $M$ and $\iota_Z \omega$ denotes the interior product. -In your case $Z(x) = x$ gives you a normal vector field when restricted to $N = S^{n-1}$ and for $M = \mathbb R^n$ we have $\mathrm d vol_{\mathbb R^n} = dx^1 \wedge \dots \wedge dx^n$. So -\begin{align} -\mathrm d vol_{S^{n-1}}(x) &= (\iota_Z dx^1 \wedge \dots \wedge dx^n)(x) \\ -&= \sum_{i=1}^{n-1} (-1)^{i-1} x_i \; dx^1 \wedge \dots \wedge dx^{i-1}\wedge dx^{i+1}\wedge \dots\wedge dx^{n} -\end{align} -for $x\in S^{n-1}$.<|endoftext|> -TITLE: Are right continuous functions measurable? -QUESTION [12 upvotes]: Are right-continuous function from $\mathbb{R}^n$ to $\mathbb{R}$ necessarily semi-continuous? -If not, are they necessarily Borel measurable? -Is there a topological characterization of right-continuous functions (as there is of continuous ones)? -Are CDFs of $n$-dimensional random vectors measurable? -Note: A function $f: \mathbb{R}^n \longrightarrow \mathbb{R}$ is right-continuous iff it is right-continuous at every point $x \in \mathbb{R}^n$. A function $f: \mathbb{R}^n \longrightarrow \mathbb{R}$ is right-continuous at $x \in \mathbb{R}^n$ iff given any infinite sequence of points in $\mathbb{R}^n$ $(y_0,y_1,\dots)$ that converges to $x$ from above (i.e. the sequence converges to $x$ in the usual, Euclidean sense and in addition every sequence element is greater than or equal to $x$ component-wise), the sequence $(f(y_0), f(y_1), \dots)$ converges to $f(x)$ in the usual sense. - -REPLY [9 votes]: Here is a proof that any function $F: \mathbb{R}^n \to \mathbb{R}$ which is right continuous is Borel measurable. -Define $F_n(x) = F\left(\frac{[nx]+1}{n}\right)$, where $[x]$ is the greatest integer smaller than $x$. Clearly $F_n$ are measurable since they are step functions. -From the right continuity of $F$ we get that $F_n\to F$, since $\frac{[nx]+1}{n} \downarrow x$. -Hence since $F$ is the pointwise limit of measurable functions, it is measurable too.<|endoftext|> -TITLE: When is $CaCl(X) \to Pic(X)$ surjective? -QUESTION [9 upvotes]: I am curious about following similar statements in algebraic geometry and complex geometry: - - -Algebraic Geometry Version: - If $X$ is an integral scheme, the map from Cartier divisor group to Picard group (i.e. group of invertible sheaves) $CaCl(X) \to Pic(X)$ is surjective. -Complex Geometry Version: If $X$ is a complex manifold, then the image of $CaCl(X) \to Pic(X)$ is generated by those line bundles $L \in Pic(X)$ with $H^0(X,L) \neq 0$. - - -More interesingly, there is a remark for the complex version: - - -Remark: $CaCl(X) \to Pic(X)$ may not be surjective even for very easy manifolds, e.g. a generic complex torus of dimension two, this is no longer the case. - - -My questions is twofold: -Explicit one: How to justify the above remark, i.e. $CaCl(X) \to Pic(X)$ may not be surjective for a generic complex torus of dimension two. -Inexplicit one: Though I don't know the exact meaning of "integral manifold" (or the proper prototype of integral scheme ), it seems for me that "integrality" is natrual endowed by a complex manifold(suppose it is connected). And thus, the difference is quite unexpected. How could this come out, and what are reasonable category to guarantee subjectivity. - -REPLY [10 votes]: For any complex torus $X$ of dimension $n$ the set of isomorphism classes of holomorphic line bundles $Pic(X)$ is a huge group: already the group $Pic^0(X)$ of line bundles $L$ with zero Chern class ($c_1(L)=0$) is itself a complex torus of dimension $n$ , called the dual torus $\hat X=Pic^0(X)$. -However there exist two-dimensional complex tori $X$ having no curves at all , so that its group of divisors is zero (see here for a detailed analysis). -So for such tori no non-trivial line bundle comes from a divisor. -A "philosophical" explanation -For integral (= reduced and irreducible) schemes Hartshorne proves that every line bundle comes from a divisor in Proposition II,6.15 -The proof uses that sheaves that are constant on some open covering of $X$ are constant on the whole of $X$. -This is due to irreducibility: two non-empty open subsets of $X$ have non-empty intersection. -But this is completely false for complex manifolds since a Hausdorff space containing at least two points is never irreducible. -A variant of this explanation is that on an integral scheme a rational function defined on an open subset of the scheme extends to a rational function on the whole scheme, whereas the analogous result for meromorphic functions on a complex manifold is completely false .<|endoftext|> -TITLE: When is an infinite product of natural numbers regularizable? -QUESTION [65 upvotes]: I only recently heard about the concept of $\zeta$-regularization, which allows the evaluation of things like -$$\infty != \prod_{k=1}^\infty k = \sqrt{2\pi}$$ -and -$$\infty \# = \prod_{k=1}^\infty p_k = 4\pi^2$$ -where $n\#$ is a primorial, and $p_k$ is the $k$-th prime. (The expression for the infinite product of primes is proven here.) That got me wondering if, given a sequence of positive integers $m_k$ (e.g. the Fibonacci numbers or the central binomial coefficients), it is always possible to evaluate the infinite product -$$\prod_{k=1}^\infty m_k$$ -in the $\zeta$-regularized sense. It would seem that this would require studying the convergence and the possibility of analytically continuing the corresponding Dirichlet series, but I am not too well-versed at these things. If such a regularization is not always possible, what restrictions should be imposed on the $m_k$ for a regularized product to exist? -I'd love to read up on references for this subject. Thank you! - -REPLY [4 votes]: Given an increasing sequence $0<\lambda_1<\lambda_2<\lambda_3<\ldots$ one defines the regularized infinite product -$$ -\prod^{\infty}_{n=1}\lambda_n=\exp\left(-\zeta'_{\lambda}(0)\right), -$$ -where $\zeta_{\lambda}$ is the zeta function associated to the sequence $(\lambda_n)$, -$$ -\zeta_{\lambda}(s)=\sum^{\infty}_{n=1}\lambda_n^{-s}. -$$ -(See the paper: E.Munoz Garcia and R.Perez-Marco."The Product over all Primes is $4\pi^2$". http://cds.cern.ch/record/630829/files/sis-2003-264.pdf ) -In the paper( https://arxiv.org/ftp/arxiv/papers/0903/0903.4883.pdf ) I have evaluated the -$$ -\prod_{p-primes}p^{\log p}=\prod_{p-primes}e^{\log^2p}=\exp\left(24\zeta''(0)+12\log^2(2\pi)\right) -$$ -where $\zeta''(0)$ is the second derivative of Riemann's Zeta function in $0$.<|endoftext|> -TITLE: An exercise in infinite combinatorics from Burris and Sankappanavar -QUESTION [7 upvotes]: Exercise 6.7 in chapter IV of Burris and Sankappanavar's A Course in Universal Algebra starts as follows: - -Show that for $I$ countably infinite there is a subset $S$ of the set of functions from $I$ to $2$ which has cardinality equal to that of the continuum such that for $f \not= g$ with $f,g \in S$, $\{i \in I : f(i) = g(i)\}$ is finite. - -I don't see how $S$ can have cardinality more than 2: let $f, g, h \in S$, with $f \not= g$ and $f \not= h$ and let $A = \{i \in I : f(i) = g(i)\}$ and $B = \{i \in I : f(i) = h(i)\}$. For $i$ in $I\backslash (A \cup B)$ we have $g(i) \not= f(i) \not= h(i)$, whence $g(i) = h(i)$, since $f(i), g(i), h(i) \in 2 = \{0, 1\}$. But $A$ and $B$ are finite, $I \backslash(A\cup B)$ is infinite. So $g(i) = h(i)$ holds for infinitely many $i$ and we must have $g = h$. -What have I missed? (A redefinition of 2, perhaps?) -I would also be grateful for a reference for what appears to be the goal of this exercise, namely to use an ultraproduct construction to obtain uncountable models from countable ones. - -REPLY [3 votes]: On the website of the book there is a newer version and - as the OP already pointed out in their comment the exercise is already corrected there (see page 150). It says: - -Show that, for $I$ countably infinite and $A$ infinite, there is a subset $S$ of the set of - functions from $I$ to $A$ which has cardinality equal to that of the continuum such that for - $f\ne g$ with $f,g\in S$, $\{i\in I; f(i)=g(i)\}$ is finite. Conclude that $|A^I/U|\ge 2^\omega$ if $U$ - is a nonprincipal ultrafilter over $I$. - - -The question basically boils down to finding such family for functions from an infinite countable set to an infinite countable set. (Since $A$ is infinite, it contains an infinite countable set.)$\newcommand{\N}{\mathbb{N}}\newcommand{\Q}{\mathbb{Q}}\newcommand{\R}{\mathbb{R}}$ -So we can take $I=\N$ and $A$ can be an arbitrary infinite countable set. - -For $A=\Q$ we can repeat basically the same trick as the trick which is used to show the existence of an almost disjoint family of cardinality $\mathfrak c$. See, for example, here or here. -For any real number $r\in\R$ we take any injective sequence $f_r \colon \N \to \Q$ of rational numbers, which converges to $r$. The system $\{f_r; r\in\R\}$ has the same cardinality as $r$. And for any $r\ne r'$ the set $\{i\in\N; f_r(i)=f_{r'}(i)\}$ is finite, since these two sequences have different limits. - -A different approach, which only shows that there is uncountably many such functions. -Lemma: If $\{f_n; n\in\N\}$ is a sequence of functions from $\N$ to $\N$ then there is a function $f$ such that each set $A_n=\{i\in\N; f(i)=f_n(i)\}$ is finite. -Proof. We can define -$$f(n)=\max\{f_i(j); i=0,\dots,n-1; j=0,\dots,n-1\}+1.$$ -Then we have $\{i\in\N; f_n(i)=f(i)\}\subseteq\{0,1,\dots,n\}$. -$\qquad\square$ -Using the above fact and transfinite induction we can get a family of functions with cardinality $\aleph_1$, which has the required properties.<|endoftext|> -TITLE: Eigenvalues of tridiagonal symmetric matrix with diagonal entries 2 and subdiagonal entries 1 -QUESTION [12 upvotes]: Problem: -Let A be a square matrix with all diagonal entries equal to 2, all entries directly above or below the main diagonal equal to 1, and all other entries equal to 0. Show that every eigenvalue of A is a real number strictly between 0 and 4. - -Attempt at solution: - -Since A is real and symmetric, we already know that its eigenvalues are real numbers. - -Since the entries in the diagonal of A are all positive (all 2), A is positive definite iff the determinants of all the upper left-hand corners of A are positive. I think this can be proven via induction (showing that each time the dimension goes up, the determinant goes up too) - -Since A is symmetric and positive definite, eigenvalues are positive, i.e. greater than 0. - - -But I can't get the upper bound of 4. Any help would be appreciated. Thank you. - -REPLY [12 votes]: The characteristic polynomial of $A-2I$ is the $n\times n$ determinant $D_n(X)$ of the matrix with entries $-1$ directly above or below the main diagonal, entries $X$ on the main diagonal, and entries $0$ everywhere else, hence -$$ -D_{n+2}(X)=XD_{n+1}(X)-D_n(X), -$$ -for every $n\geqslant0$ with $D_1(X)=X$ and the convention $D_0(X)=1$. -This recursion is obviously related to Chebyshev polynomials and one can prove: - -For every $u$ in $(0,\pi)$ and $n\geqslant0$, $D_{n}(2\cos(u))=\dfrac{\sin((n+1)u)}{\sin(u)}$. - -Assume that this holds for $n$ and $n-1$ for some $n\geqslant1$, then -$$ -D_{n+1}(2\cos(u))\sin(u)=2\cos(u)\sin(nu)-\sin((n-1)u)=\sin((n+1)u). -$$ -Since $D_1(2\cos(u))=2\cos(u)=\sin(2u)/\sin(u)$ and $D_0(2\cos(u))=1=\sin(u)/\sin(u)$, this proves the claim. Hence $x=2\cos(k\pi/(n+1))$ solves $D_n(x)=0$ for every $1\leqslant k\leqslant n$. These $n$ different values are the eigenvalues of $A-2I$. -Since the eigenvalues of $A-2I$ are all in the interval $[-2\cos(\pi/(n+1)),+2\cos(\pi/(n+1))]$, the eigenvalues of $A$ are all in the interval $]0,4[$.<|endoftext|> -TITLE: Prove the limit for a series of products. -QUESTION [6 upvotes]: Let $\beta > 0$, $\lambda > 1$. Show the identity -$$\sum_{n=0}^\infty\prod_{k=0}^{n} \frac{k+\beta}{\lambda + k + \beta} = \frac{\beta}{\lambda - 1}$$ -I have checked the statement numerically. -The special case $\beta = 1$, $\lambda = 2$ looks like this -$$\sum_{n=1}^\infty\prod_{k=1}^{n} \frac{k}{2 + k} = 1$$ - -This series arises in a probabilistic setting. Let $(Y_k)_{i\ge0}$ be independent exponentially distributed variables with parameters $k+\beta$ respectively and set $S_n := \sum_{k=0}^n Y_i$. For $t \ge 0$ let $X(t) := \#\{n \ge 0: S_n < t\}$. Let $Z_\lambda$ be exponentially distributed with parameter $\lambda$ and independent of $X(t)$. Then -\begin{align*} -EX(Z_\lambda) &= E\#\{n \ge 0: S_n < Z_\lambda\} \\ -& = \sum_{n=0}^\infty P(S_n < Z_\lambda) \\ -& = \sum_{n=0}^\infty Ee^{-\lambda S_n} \\ -& = \sum_{n=0}^\infty \prod_{k=0}^n Ee^{-\lambda Y_k} \\ -& = \sum_{n=0}^\infty \prod_{k=0}^n \frac{k+\beta}{\lambda + k + \beta} -\end{align*} - -REPLY [2 votes]: We want to evaluate : -$$f(\beta,\lambda)=\frac {\beta}{\lambda+\beta}+\frac {\beta(\beta+1)}{(\lambda+\beta)(\lambda+\beta+1)}+\frac {\beta(\beta+1)(\beta+2)}{(\lambda+\beta)(\lambda+\beta+1)(\lambda+\beta+2)}+\cdots$$ -Setting $a:=\lambda+\beta$ this becomes : -$$F(\beta,a)=\frac {\beta}{a}+\frac {\beta(\beta+1)}{a(a+1)}+\frac {\beta(\beta+1)(\beta+2)}{a(a+1)(a+2)}+\cdots$$ -This may be represented as a $\;_2F_1$ hypergeometric series since : -$$_2F_1(\beta,1;a;1)=1+\frac {\beta\cdot 1}{a\cdot 1!}+\frac {\beta(\beta+1)\cdot 1\cdot 2}{a(a+1)\cdot 2!}+\frac {\beta(\beta+1)(\beta+2)\cdot 1\cdot 2\cdot 3}{a(a+1)(a+2)\cdot 3!}+\cdots$$ -but this is simply (applying Gauss' theorem) : -$$_2F_1(a,b;c;1)=\frac {\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}$$ -so that : -$$_2F_1(\beta,1;a;1)=\frac {\Gamma(a)\Gamma(a-\beta-1)}{\Gamma(a-\beta)\Gamma(a-1)}=\frac {a-1}{a-\beta-1}=\frac {\lambda+\beta-1}{\lambda-1}$$ -minus $1$ gives $\frac {\beta}{\lambda-1}$ as wished.<|endoftext|> -TITLE: Why is it that a linear transformation can only preserve or reduce a vector space's dimension? -QUESTION [7 upvotes]: I am able to prove the above-asked using the fact that a linear map preserves linear dependence. I also vaguely suspect a connection from group theory (homomorphisms?). But I'm having trouble getting an intuition for that fact that a linear transformation cannot increase the dimension of a vector space. Help? - -REPLY [8 votes]: My intuition would be this: linear maps preserve lines (that's the intuition behind them, anyway). If they are degenerate, the lines might collapse to points, but points which are colinear remain such after the transformation. -Dimension of a space is, intuitively, the number of independent lines you can draw in a space. Because linear transformation preserves not just lines, but also linear subspaces of higher dimensions (so coplanar points remain coplanar etc.), it can't "split" a line into more of them, even if it can join some of them, and it can't turn lines which weren't independent into ones which are (because that would be "splitting" some higher-dimensional subspace; it's a little circular, but we're talking about the intuition here), so the dimension can't go up.<|endoftext|> -TITLE: Extending a subset of $\mathbb{Z}^n$ to a basis -QUESTION [5 upvotes]: Let $v_1,\dots,v_k\in \mathbb{Z}^n$. Is there a nice criterion for the existence of $v_{k+1},\ldots,v_n \in\mathbb{Z}^n$ such that $v_1,\ldots,v_n$ form a basis of $\mathbb{Z}^n$? - -For $k=1$, if $v_1 = (a_1,\ldots,a_n)$, it is not hard to see that it is iff $\gcd(a_1,\ldots,a_n)=1$. For bigger $k$ I can find out the answer for specific $v_1,\ldots,v_k$ algorithmically, but I would like to know a general criterion if there is one. - -REPLY [10 votes]: Stacking the vectors as a $k$-by-$n$ matrix $M$ with $k\le n$, the condition that the rows extend to a basis is that the gcd of all determinants of $k$-by-$k$ minors be $1$. -Proof: by the structure theorem for finitely-generated modules over a PID, such as $\mathbb Z$, there are $A,B$ integer matrices with integer inverses such that $AMB$ is diagonal. The left-and-right multiplication respects the gcd of minors. -For example, given $(1,0,a,b)$ and $(0,1,c,d)$ with arbitrary $a,b,c,d$ integers, there are $4$-choose-$2$ two-by-two minors, and the very first one is the two-by-two-identity, which has determinant $1$, so this will extend, as we know it does. True, there are other dets of minors, but the very first one gives $1$ already.<|endoftext|> -TITLE: fixed point of a holomorphic function on a disk -QUESTION [10 upvotes]: Let $ \mathbb{D} = \{ z : |z|<1 \} $ and $ f $ an holomorphic function on $ \mathbb{D} $ and continuous on $ \overline{\mathbb{D}} $ such that $ f(\overline{\mathbb{D}}) \subset \mathbb{D} $. -Prove the following: - -There exists single point $ z^* \in \mathbb{D} $ such that $ f(z^*)=z^* $ (obvious by Rouche theorem). -Let $ f_1=f,...,f_{n+1}=f(f_n) $ show that $ f_n(z) \longrightarrow z^* $ uniformly. - -REPLY [10 votes]: The key to all this is that $f(\bar{D}) \subset D$: - -Since $f(\bar{D})$ is compact, there exists $r_0>0$ such that $f(\bar{D})\subset D_{r_0}$. So for any $r_0r_0$ the uniqueness result follows. -Since $|f(z)/z|=|f(z)|$ is continuous on $\partial D$ it has a maximum $M$, and by hypothesis $M<1$. So, assuming for the moment that $f(0)=0$, by the maximum principle we get $|f(z)|\leq M|z|$ for $z\in D$. This gives that $|f_n(z)|\leq M|f_{n-1}(z)|$ in $D$, and so $|f_n(z)|\leq M^n|z|$. Taking supremums over $\bar{D}$ and then the limit as $n\to \infty$ the result follows. - -Assume now that $f(0)\neq 0$ then everything just said applies to $g=h\circ f\circ h^{-1}$ (with $h$ an appropiate automorphism of the disk), and the result follows in general.<|endoftext|> -TITLE: Integration of sawtooth, square and triangle wave functions -QUESTION [10 upvotes]: Context -After a discussion about how to plot the results of a frequency modulation between two signals on Stack Overflow, I understood that I need to find the time-integral of the following wave functions before using them in the general FM formula (as illustrated in the first answer). - -Research -Integrating a sine wave function is indeed easy, but things gets a lot complicated when it comes to other waveforms. Here follow the equations I'm using to display the waveforms: - -Sawtooth wave: -$ f(x) = \bmod(f_c x, 1.0); $ -Square wave: -$ f(x) = \operatorname{sign}(\cos(f_c x)); $ -Triangle wave: -$ f(x) = \frac{1}{f_c}|\bmod(x, f_c) - \frac{1}{2}f_c|$ - -These functions looks right, but as I don't have any particular background in mathematics or calculus I won't be surprised if I made some bad mistakes. Please be patient. - -Questions - -Is there a better way to describe mathematically the wave functions above? -If these are right, what is the correct time-integral? - - -Updates -Thanks to the the functions with period $T$ in the form Rahul suggested I get: -$$\begin{align}\operatorname{sawtooth}(x) = \int_0^x \frac{2x}T-1 \ \mathrm dx &= \frac{x(x - T)}T \end{align}$$ -$$\begin{align} -\operatorname{square}(x) &= \int_0^x \begin{cases}1&\text{if } x -TITLE: Question about example of non-separable Hilbert space -QUESTION [6 upvotes]: I have come across the following example of a non-separable Hilbert space: - -Example 2.84. Let $I$ be a set, equipped with the discrete topology and the counting measure $\lambda_{\text{ count}}$ defined on the $\sigma$-algebra $\Bbb P(I)$ of all subsets of $I$. Then - $$\ell^2(I)=L^2\big(I,\Bbb P(I),\lambda_{\text{ count}}\big)$$ - is a Hilbert space, and it comprises all functions $a:I\to\Bbb R$ (or $\Bbb C$) for which the support - $$F=\{i\in I : a(i)\ne0\},$$ - is finite or countable, and for which $\sum_{i\in I}|a_i|^2=\sum_{i\in F}|a_i|^2\lt\infty$. - -Why do I need the discrete topology on $I$? Or more generally: why do I need a topology? If we talk about $L^p$ spaces in general, we only want a measure space and we don't mention a topology because $f \in L^p$ doesn't have to be continuous. Thanks for your help. - -REPLY [2 votes]: To summarize the comments in an answer: -You are correct, the topology on $I$ is irrelevant and was probably mentioned by mistake. -(If someone will upvote this answer, the question will stop being bumped.)<|endoftext|> -TITLE: Proof that a periodic function is bounded and uniformly continuous. -QUESTION [5 upvotes]: I need to show that if $f:\mathbb{R}\to \mathbb{R}$ is continuous and $\forall x \in \mathbb R, f(x+1)=f(x)$, then: - -$f$ is bounded, -$f$ is uniformly continuous, -there exists $c\in \mathbb{R}$ such that $f(c)=f(c+\pi)$. - -REPLY [7 votes]: Let $$g : \begin{array}{ccc} \mathbb R & \to& \mathbb R\\ c &\mapsto &f(c+\pi) - f(c).\end{array}$$ That's a continuous function. -Now, let $c_0$ be a point where $f_{|[0,1]}$ has a minimum. Because of the periodicity, $c_0$ is a minimum for the whole of $f$. So $\forall x \in \mathbb R, f(x) \geq f(c_0)$. In particular $(x = c_0 + \pi)$, $g(c_0) \geq 0$. -The same argument with “maximum” instead of minimum gives a $c_1$ where $g(c_1) \leq0$. -The intermediate value theorem now gives a point $c$ on which $g$ vanishes, QED.<|endoftext|> -TITLE: Intuition for étale morphisms -QUESTION [44 upvotes]: Currently working on algebraic surfaces over the complex numbers. I did a course on schemes but at the moment just work in the language of varieties. -Now i encounter the term "étale morphism" every now and then (in the book by Beauville). I know Hartshorne's definition as a smooth morphism of relative dimension zero, and wikipedia states a bunch of equivalent ones. I can work with this, so no problem there. However, some more intuition about the concept would also be nice. -So basically, if you have worked with étale morphisms, could you explain what's your personal intuition for such things, in the case of varieties? If in your answer you could also mention smooth and flat morphisms, that would be really appreciated. -Thanks in advance! -Joachim - -REPLY [12 votes]: For four of Georges examples I can give a general hint. If $X\to Spec \ k$ is étale where $k$ is a field, then $X$ must just be a disjoint union $\coprod Spec \ L_i$ where each $L_i$ is a finite separable field extension of $k$. -Proving this fact is a good exercise and it goes back to the analogy people have been making. A "covering space" of a point should just topologically be a discrete set of points, but this is algebraic geometry so there should be algebraic information as well. The algebraic information is that the field extensions are all separable. This has to do with the fact that $X$ being smooth implies it is "geometrically reduced" and hence you see that we can't pick up nilpotents when base changing. -Now it should be pretty easy to do (e)-(h). Unfortunately, all the base fields there are perfect, so we don't have weird unnecessary complication. I'll throw in -i) $\displaystyle Spec \left(\frac{\mathbb{F}_p(t)[x]}{(x^p-t)}\right)\to Spec \ \mathbb{F}_p(t)$<|endoftext|> -TITLE: Prove that for any $x \in \mathbb N$ such that $x -TITLE: Ergodic theory in mathematics and physics -QUESTION [8 upvotes]: How is the theory of ergodic measure-preserving transformations related to ergodicity in the physical sense (which I understood as, very very roughly speaking, that a physical system is called ergodic if "averaging" over "states" of the physical system equals the "average" over time)? -I am sorry maybe the question is a bit unspecific for now, but I guess it's still of interest also for others who are about to dive into the subject. - -REPLY [3 votes]: They couldn't be related more closely: they're exactly the transformations for which the physical average equals the time average almost everywhere. -Specifically, they observe Birkhoff's ergodic theorem, that -$\int_X fd\mu=\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{k=0}^{n-1}f(T^kx), \mu$-almost everywhere, for any integrable $f$. A nice special case is if $f$ is the characteristic function of some subspace of $A$, in which case this says that the sequence $T^kx$ gets into $A$ proportionally often to the measure of $A$ in $X$ for just about every $x$.<|endoftext|> -TITLE: $SU(2)$ subgroups of $SU(3)$: Is my reasoning correct? -QUESTION [6 upvotes]: When looking at the standard $3\times3$ representation of $SU(3)$, one immediately recognizes some subgroups isomorphic to $SU(2)$: - -There is the subgroup acting on the first two elements of the vector, keeping the third one unchanged (and of course, the same for leaving any of the other elements intact). Indeed, it is not hard to see that for any unit vector, there's an $SU(2)$ subgroup acting only on the two vectors othogonal on it. It is also not hard to see that all those subgroups are conjugate to each other. -Since the $3\times3$ representation of $SU(2)$ also consists of special unitary matrices, it of course also corresponds a subgroup of $SU(3)$. The elements of this representation can e.g. be written as $\exp(\mathrm i\alpha L_z)\exp(\mathrm i\beta L_x)\exp(\mathrm i\gamma L_z)$ where $L_x=\frac{1}{\sqrt{2}}\begin{pmatrix}0&1&0\\1&0&1\\0&1&0\end{pmatrix}$ and $L_z=\operatorname{diag}(1,0,-1)$. - -Now a natural question is whether this last subgroup is conjugate to the previous ones. I've come to the conclusion that it is not, with the following argumentation: -It of course suffices to check conjugacy with the subgroup acting on the first two elements (because conjugacy is an equivalence relation). Now the elements of that subgroup all commute with the subgroup of transformations of the form $\operatorname{diag}(\mathrm e^{\mathrm i\phi}, \mathrm e^{\mathrm i\phi}, \mathrm e^{-2\mathrm i\phi})$. However, the members of the $3\times3$ representation of $SU(2)$ only commute with multiples of the unit matrix (three of which are members of $SU(3)$). Therefore the two subgroups cannot be conjugate to each other. -Now my question is: Is my reasoning correct? - -REPLY [5 votes]: Here's another way to see they're not conjugate. First, convince your self that conjugate subgroups are diffeomorphic. -In the $3\times 3$ representation of $SU(2)$, $-I$ acts trivially, so the induced map $SU(2)\rightarrow SU(3)$ actually factors through $SU(2)/-I = SO(3)$. This implies that the image of $SU(2)$ in $SU(3)$ is actually $SO(3)$, which is not even homotopy equivalent to $SU(2)$ (look at fundamental groups).<|endoftext|> -TITLE: What do you call a group that doesn't have a unique identity? -QUESTION [6 upvotes]: I have a set $M$ and an associative binary relation $+ : M \times M \to M$. -There exists an inversion operator $-$ , such that if $ m \in M$, then $m+(-m) \in Z$ where $Z$ is the set of all zeros ($Z \subset M$). -Additionally, if $z \in Z$ and $m \in M$, then $m+z=m$ -If $|Z| = 1,$ then this would form a group, right? But since $|Z|>1$, what would I call it? Is there any theory about categories like this? - -Also, what if there was no identity at all? That is, $m_1 + m_2 + (-m_2) = m_1$, but $m_2 + (-m_2)$ is undefined. In this case, associativity would not hold. - -REPLY [6 votes]: Suppose $M$ is a set with an associative binary operator $+$ such that for all $m \in M$ there is a unique $n$ such that $m+n+m = m$ and $n+m+n = n$ (thinking of $z_r=m+n$ we get $z_r + m = m$ and $n+z_r = n$; thinking of $z_l =n+m$ we get $m+z_l = m$ and $z_l+n=n$), then you have a well studied object called an “inverse semigroup”. -The set $Z$ you want is called the set of idempotents, $Z = \{ m \in M : m + m = m \}$. It also consists exactly of all $m + (-m)$ for $m \in M$. I also call them "partial identities", because: -There is a partial order $\leq$ on every inverse semigroup: $x \leq y$ iff $x = x + (-x) + y$. This let's you express how close to an identity each $z \in Z$ is: $z + x = x$ whenever $x \leq z$. -For every partial identity $z=g+(-g)$ there is an honest to goodness group called the $H$-class (or more accurately its Schutzenberger group) associated to $z$ consisting of all invertible $z+m+z$ for $m \in M$. -Inverse semigroups as semigroups of bijections -Every inverse semigroup is isomorphic to a semigroup consisting of bijections between subsets of a set $X$, where multiplication is the best version of composition you can manage: if $f:A\to B$ and $g:C \to D$ then $f\cdot g$ takes $f^{-1}(x)$ to $g(x)$ whenever $x \in B \cap C$. -In terms of bijections, the set $Z$ consists of all identity bijections $f:A \to A :a \mapsto a$ where $A$ is the domain of some $m \in M$. In terms of bijections, $-g$ is the inverse function $g^{-1} : D \to C : g(c) \mapsto c$. The partial identities $g+( -g)$ and $-g+g$ are the identity bijections on the domain and range of $g$. -The natural partial order is actually just “subset” if you consider a bijection to be a set of ordered pairs $(a,f(a))$, so that $f \leq g$ if and only if $f$ is the restriction of $g$ to the domain of $f$. -The H-class of an idempotent $f:A\to A:a\mapsto a$ is all the permutations of $A$, that is, bijections from $A$ to $A$. -Bibliography -I recommend both of these books as very clear and motivated introductions to these structures. Lawson's book is particularly readable and draws connections to topology, symmetry, groupoids, category theory, and computer science. - -Lawson, Mark V. -Inverse semigroups: The theory of partial symmetries. -World Scientific Publishing Co., Inc., River Edge, NJ, 1998. xiv+411 pp. ISBN: 981-02-3316-7 -MR1694900 -DOI10.1142/9789812816689 -Petrich, Mario. -Inverse semigroups. -Pure and Applied Mathematics (New York). A Wiley-Interscience Publication. John Wiley & Sons, Inc., New York, 1984. x+674 pp. ISBN: 0-471-87545-7 -MR752899<|endoftext|> -TITLE: Are two spaces manifolds if their product is a manifold? -QUESTION [11 upvotes]: Possible Duplicate: -Decomposition of a manifold - -For topological spaces $X,Y$, if their product space $X \times Y$ is a manifold, is it necessarily that $X,Y$ are manifolds? - -REPLY [11 votes]: No. The dogbone space $D$ is a topological space that is not a manifold but $D \times \mathbb{R} \cong \mathbb{R}^4$.<|endoftext|> -TITLE: Relation between left and right coset representatives of a subgroup -QUESTION [15 upvotes]: Let $G$ be a finite group and $H$ a subgroup. Is it true that -a set of right coset representatives of $H$ is also set of left coset -representatives of $H$? - -REPLY [9 votes]: This answer is mostly to give a nice bibliography for this problem of finding common transversals. I was going to include a proof, but all the proofs were combinatorial and my attempts at making them sound interesting failed. -Exists: yes, All: no -If $H \leq G$ and $K \leq G$ are two subgroups of the same finite index $[G:H]=[G:K]$, then they have a “common transversal”: a set $t_i \in G$ such that $H t_i \neq H t_j$ and $t_i K \neq t_j K$ if $i \neq j$ and such that $G = \cup H t_i = \cup t_i K$. Taking $H=K$ gives an answer to the original question, interpreting "a" as "there exists" rather than "for all". I discuss the publication history of this theorem in the next section. -One can easily check that $\{ (), (2,3), (1,2,3) \}$ is only a one-sided set of coset representatives of $\{(),(1,2)\}$ in the symmetric group $\{ (), (1,2), (2,3), (1,3), (1,2,3), (1,3,2) \}$. -History -The earliest proof is due to Miller (1910) with another similar proof in Chapman (1913). The same proof was used for a stronger result in Scorza (1927). Van der Waerden (1927) proved a combinatorial precursor to the Marriage theorem of Hall (1935) and explicitly cites Miller (1910) as his motivation. A proof is presented in textbook form in Zassenhaus (1937), theorem 3, page 12; I believe this is van der Waerden's proof, but Zassenhaus credits it to Willi Maak. Shü (1941) generalizes Scorza's result to finite index subgroups and gives an example to show the result is not true for a general subgroup. Ore (1958) continues this investigation. Hall (1967) uses the marriage theorem to prove the theorem in the finite index case as theorem 5.1.7 on page 55. Alonso (1972) proves less, but gives more examples and uses more primitive versions, somewhat similar to Hall (1967). Applegate–Onishi (1977) proves the result for continuous sections of profinite groups. -Bibliography - -Miller, G. A. -“On a method due to Galois.” -Quarterly Journal of Mathematics 41, (1910), 382-384. -JFM41.174.1 -Chapman, H. W. -“A note on the elementary theory of groups of finite order.” -Messenger of Mathematics (2) 42 (1913), 132-134; correction 43 (1914), 85. -JFM44.168.3 -(Correction in next volume) -Scorza, G. -“A proposito di un teorema del Chapman.” -Bollettino della Unione Matemàtica Italiana 6, 1-6 (1927). -JFM53.105.5 -van der Waerden, B. L. -“Ein Satz über Klasseneinteilungen von endlichen Mengen.” -Abhandlungen Hamburg 5, 185-187 (1927). -JFM53.171.2 -DOI10.1007/BF02952519 -Hall, P. -“On representatives of subsets.” -Journ. London Math. Soc. 10, 26-30 (1935). -JFM61.67.1 -ZBL10.345.3 -DOI:10.1112/jlms/s1-10.37.26 -Zassenhaus, Hans. -Lehrbuch der Gruppentheorie. -(Hamburg. Math. Einzelschriften 21) Leipzig, Berlin: B. G. Teubner. VI, 152 S. (1937). -JFM63.58.3 -ZBL18.9.1 -MR30947 -MR91275 -Shü, Shien-siu. -“On the common representative system of residue classes of infinite groups.” -J. London Math. Soc. 16, (1941). 101–104. -MR4624 -DOI:10.1112/jlms/s1-16.2.101 -Ore, Oystein. -“On coset representatives in groups.” -Proc. Amer. Math. Soc. 9 (1958) 665–670. -MR100639 -DOI:10.2307/2033229 -Hall, Marshall, Jr. -Combinatorial theory. -Blaisdell Publishing Co. Ginn and Co., Waltham, Mass.-Toronto, Ont.-London (1967) x+310 pp. -MR224481 -MR840216 -Alonso, James. -“Representatives for cosets.” -Amer. Math. Monthly 79 (1972), 886–890. -MR315004 -DOI:10.2307/2317669 -Appelgate, H.; Onishi, H. -“Coincident pairs of continuous sections in profinite groups.” -Proc. Amer. Math. Soc. 63 (1977), no. 2, 217–220. -MR442099 -DOI:10.2307/2041792<|endoftext|> -TITLE: Zeta-like function with offset -QUESTION [5 upvotes]: Is there a known function of the form: -$$\zeta(s,a) = \displaystyle\sum_{n=1}^\infty\frac{1}{n^s+a},$$ -and if so what is its name? - -REPLY [3 votes]: Let's assume $a$ is small, i.e. $0 -TITLE: Is there a theory that combines category theory/abstract algebra and computational complexity? -QUESTION [12 upvotes]: Category theory and abstract algebra deal with the way functions can be combined with other functions. Complexity theory deals with how hard a function is to compute. It's weird to me that I haven't seen anyone merge these fields of study, since they seem like such natural pairs. Has anyone done this before? - -As a motivating example, let's take a look at semigroups. Semigroups have an associative binary operation. It's well known that operations in semigroups can be parallelized due to the associativity. For example, if there's a billion numbers we want added together, we could break the problem into 10 problems of adding 100 million numbers together, then add the results. -But parallelizing the addition operator makes sense only because it can be computed in constant time and space. What if this weren't the case? For example, lists and the append operation form a common semigroup in functional programming, but the append operator takes time and space $O(n)$. -It seems like there should be a convenient way to describe the differences between these two semigroups. In general, it seems like algebras that took into account the complexity of their operations would be very useful to computer scientists like myself. - -REPLY [2 votes]: It seems that the idea of growth of groups is extremely relevant here. In the example in the question the operation may be viewed as concatenation of words, and so is very similar to the standard way in which we view the free group on $m$ generators $F(\mathbf{x}_m)$. The operation $V\circ W$ takes $O(|V|+|W|)$-time* to evaluate in $F(\mathbf{x}_m)$. -Every finitely-generated group can be constructed by taking such a (finite rank) free group $F(\mathbf{x}_m)$ and adding relators. These relators correspond to "short cuts" in our computation. For example, consider the free group $F(a)$ and add the relator $a^3=\epsilon$ (here, $\epsilon$ denotes the empty word) to obtain the group with presentation $\langle a\mid a^3\rangle$. (This group is actually cyclic of order $3$.) Now, I can evaluate $a^{100}\circ a^{34}$ to get $a^{134}$, but why would I? As I am in my group I know that $a^{100}=a$ and $a^{34}=a$ so $a^{100}\circ a^{34}=a\circ a=a^2$. The idea of growth of groups is to make sense of these "shortcuts" computationally. Roughtly, the "growth" of a group corresponds to how many group elements there are at distance $n$ from the identity element. Crucially, growth rate does not depend on the choice of generating set. Please see the above Wikipedia link for more details. However, I will leave you with some examples of groups and their growth rate (taken and amended from Wikipedia, of course!): - -A free group with a finite rank $k > 1$ has an exponential growth rate. -A finite group has constant growth – polynomial growth of order $0$. -A group has polynomial growth if and only if it has a nilpotent subgroup of finite index. This is a massive result due to Gromov see wiki. It was one of the first results to "link" analytic behaviour of a group with its algebraic structure. -If M is a closed negatively curved Riemannian manifold then its fundamental group $\pi _{1}(M)$ has exponential growth rate. Milnor proved this using the fact that the word metric on $\pi_{1}(M)$ is quasi-isometric to the universal cover of $M$. -$\mathbb{Z}^d$ has a polynomial growth rate of order $d$. -The discrete Heisenberg group $H^3$ has a polynomial growth rate of order $4$. This fact is a special case of the general theorem of Bass and Guivarch that is discussed in the article on Gromov's theorem on groups of polynomial growth. -The lamplighter group has an exponential growth. -The existence of groups with intermediate growth, i.e. subexponential but not polynomial was open for many years. It was asked by Milnor in 1968 and was finally answered in the positive by Grigorchuk in 1984 (see Girgorchuk's group). This is still an area of active research. -The class of triangle groups include infinitely many groups of constant growth, three groups of quadratic growth, and infinitely many groups of exponential growth. - -*This is an easily-computed upper bound. You can probably do something clever to reduce this though.<|endoftext|> -TITLE: A group of order $108$ has a proper normal subgroup of order $\geq 6$. -QUESTION [8 upvotes]: Problem: Let $G$ be a group of order $108 = 2^23^3$. Prove that $G$ has a proper normal subgroup of order $n \geq 6$. -My attempt: From the Sylow theorems, if $n_3$ and $n_2$ denote the number of subgroups of order $27$ and $4$, respectively, in $G$, then $n_3 = 1$ or $4$, since $n_3 \equiv 1$ (mod $3$) and $n_3~|~2^2$, and $n_2 = 1, 3, 9$ or $27$, because $n_2~|~3^3$. -Now, I don't know what else to do. I tried assuming $n_3 = 4$ and seeing if this leads to a contradiction, but I'm not even sure that this can't happen. I'm allowed to use only the basic results of group theory (the Sylow theorems being the most sophisticated tools). -Any ideas are welcome; thanks! - -REPLY [9 votes]: Let $\,P\,$ be a Sylow $3$-subgroup. of $\,G\,$ and let the group act on the left cosets of $\,P\,$ by left (or right) shift. This action determines a homomorphism of $\,G\,$ on $\,S_4\,$ whose kernel has to be non-trivial (why? Compare orders!) and either of order $27$ or of order $9$ (a subgroup of $ \, P \, $, say) , so in any case the claim's proved.<|endoftext|> -TITLE: About fractional differentiation under the integral sign -QUESTION [5 upvotes]: $1.$ Does $\dfrac{d^n}{dx^n}\int_a^bf(x,t)~dt=\int_a^b\dfrac{\partial^n}{\partial x^n}f(x,t)~dt$ correct when $n$ is a positive real number? -$2.$ How about $\dfrac{d^n}{dx^n}\int_{a(x)}^{b(x)}f(x,t)~dt$ when $n$ is a positive real number? - -REPLY [5 votes]: The first rule of Leibniz you mention will still hold for the fractional case, as the definitions of fractional derivatives are equivalent to whole-order differentiation of an integral which is operating orthogonal to the integral in another variable. Symbolically the proof for 1 using the traditional differintegral: -$$ -D^n_x = \frac{d^{\lceil n \rceil}}{dx^{\lceil n \rceil}}\frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^xf(X,t)(X-x)^{\lceil n \rceil - n -1}dX -$$ -so: -$$ -D^n_x\int_a^bf(x,t)dt = \frac{d^{\lceil n \rceil}}{dx^{\lceil n \rceil}}\frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^x \int_a^b f(X,t)dt(X-x)^{\lceil n \rceil - n -1}dX -$$ -by Fubini's theorem: -$$ -= \frac{d^{\lceil n \rceil}}{dx^{\lceil n \rceil}}\frac{1}{\Gamma(\lceil n \rceil - n)}\int_a^b \int_\alpha^x f(X,t) (X-x)^{\lceil n \rceil - n -1}dXdt -$$ -By Leibniz theorem (for integer integrals and derivatives): -$$ -= \int_a^b\frac{d^{\lceil n \rceil}}{dx^{\lceil n \rceil}}\frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^x (X-x)^{\lceil n \rceil - n -1} f(X,t)dXdt -$$ -$$ = \int_a^bD_x^nf(x,t)dt $$ -For Caputo's fractional differential the differential is inside of the integral with respect to x, so the use of Leibniz for integers is used before Fubini's theorem (the steps are just reversed). -For part 2: -$$ -D^n_x\int_{a(x)}^{b(x)}f(x,t)dt = \frac{d^{\lceil n \rceil}}{dx^{\lceil n \rceil}}\frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^x \int_{a(X)}^{b(X)} f(X,t)dt(X-x)^{\lceil n \rceil - n -1}dX -$$ -Fubini's theorem applies here but it must be noted that you're operating over an entire measure space simultaneously (the symbolic idea of reversing the integrals just doesn't make sense here): -$$ -=\frac{d^{\lceil n \rceil}}{dx^{\lceil n \rceil}}\frac{1}{\Gamma(\lceil n \rceil - n)}\int_{[a(X),b(X)]\times [\alpha,x]} f(X,t)(X-x)^{\lceil n \rceil - n -1}d(t,X) -$$ -This is where things get a bit dicey for me, as the integral in this case involves a bit more measure theory than I'm entirely comfortable with (due to the interdependence of the product that forms the integral). My inclination is that this fails to make sense, as you wind up with differentiation of a variable which is caught in the bounds of both directions in the integral. -if, however the Caputo differ-integral were used: -$$ -D^n_x\int_{a(x)}^{b(x)}f(x,t)dt = \frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^x \frac{d^{\lceil n \rceil}}{dX^{\lceil n \rceil}}\int_{a(X)}^{b(X)} f(X,t)dt(X-x)^{\lceil n \rceil - n -1}dX -$$ -applying Leibniz here makes more sense since the derivative variable matches the variables in the limits of integration (for 0 < n < 1): -$$ -D^n_x\int_{a(x)}^{b(x)}f(x,t)dt = -$$ -$$ -\frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^x \frac{d^{\lceil n \rceil}b(X)}{dX^{\lceil n \rceil}}f(X,b(X))(X-x)^{\lceil n \rceil - n -1} -\frac{d^{\lceil n \rceil}a(X)}{dX^{\lceil n \rceil}}f(X,a(X))(X-x)^{\lceil n \rceil - n -1} + \int_{a(X)}^{b(X)} \frac{d^{\lceil n \rceil} f(X,t)(X-x)^{\lceil n \rceil - n -1}}{dX}dtdX -$$ -At this point some breakdown can occur into three integrals: -$$ - = B(x)- A(x) + C(x) -$$ -By substituting the appropriate integral values for f(x,a(x)) and f(x,b(x)) so that they would cancel in the product of derivatives: -$$ - A(x) = \frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^x \frac{d^{\lceil n \rceil}a(X)}{dX^{\lceil n \rceil}}f(X,a(X))(X-x)^{\lceil n \rceil - n -1} dX -$$ -$$ - B(x) = \frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^x \frac{d^{\lceil n \rceil}b(X)}{dX^{\lceil n \rceil}}f(X,b(X))(X-x)^{\lceil n \rceil - n -1} dX -$$ -$$ - C(x) = \frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^x\int_{a(X)}^{b(X)} \frac{d^{\lceil n \rceil} f(X,t)(X-x)^{\lceil n \rceil - n -1}}{dX}dtdX -$$ -By the product rule for fractional differentiation we can then say: -$$ -A(x) = \sum_{i=1}^\infty \binom {n}{i} D_x^j(a(x))D_x^{n-j}(\int_\alpha^xf(X,b(X))(X-x)^{|n-\lceil n \rceil| -1} dX) -$$ -$$ -B(x) = \sum_{i=1}^\infty \binom {n}{i} D_x^j(b(x))D_x^{n-j}(\int_\alpha^xf(X,b(X))(X-x)^{|n-\lceil n \rceil| -1} dX) -$$ -The term for C(x) then solvable by substitution and the fundamental theorem: -$$ -C(x) = \frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^x f(X,b(X))(X-x)^{\lceil n \rceil - n -1}dX -$$ -$$ -- \frac{1}{\Gamma(\lceil n \rceil -- n)}\int_\alpha^x f(X,a(X))(X-x)^{\lceil n \rceil - n -1}dX -$$ -Which implies, by substituting the operators back in: -$$ -C(x) = D_x^nf(x,b(x)) - D^n_xf(x,a(x)) -$$ -So the total equation would then be: -$$ -D^n_x\int_{a(x)}^{b(x)}f(x,t)dt = \sum_{i=1}^\infty \binom {n}{i} D_x^j(b(x))D_x^{n-j}(\int_\alpha^xf(X,b(X))(X-x)^{|n-\lceil n \rceil| -1} dX) -$$ -$$ -- \sum_{i=1}^\infty \binom {n}{i} D_x^j(a(x))D_x^{n-j}(\int_\alpha^xf(X,b(X))(X-x)^{|n-\lceil n \rceil| -1} dX) -$$ -$$ -+ D_x^nf(x,b(x)) - D^n_xf(x,a(x)) -$$ -So I would be confident in saying that there is a general formula for the fractional derivative, but it doesn't necessarily match the Leibniz rule for normal integration and differentiation. I've already noted my issues trying to find an expression for the normal fractional derivative but after this I'd guess one exists. -It's also a bit late so there might be some typos in my work I can't see at the moment so I'll double check it tomorrow just to be sure.<|endoftext|> -TITLE: Product of quotient map a quotient map when domain is compact Hausdorff? -QUESTION [12 upvotes]: Suppose that $X$ is a compact Hausdorff space and that $q : X \to Y$ is a quotient map. Is it true that the product map $q \times q : X \times X \to Y \times Y$ is also a quotient map? Note I did not assume that the quotient space $Y$ was Hausdorff (which I know to be equivalent to closedness of the quotient map $q$ in this situation). Thanks! -Added: I ask for the following reason. Wikipedia claims that, for a compact Hausdorff space $X$ and a quotient map $q : X \to Y$, the following conditions are equivalent: - -$Y$ is Hausdorff. -$q$ is a closed map. -The equivalence relation $R = \{ (x,x') : q(x) = q(x') \}$ is closed in $X \times X$. - -I was able to prove that (1) and (2) are equivalent and also that (1) implies (3). However, I do not see how to deduce (1) or (2) from (3). Some googling yields the following ?proof? (the relevant paragraph being the last one) which relies on the claim that $q \times q$ is a quotient map. Thus the question. - -REPLY [5 votes]: It is possible to deduce ($2$) from ($3$) without $q\times q$ being a quotient map. -Let $A\subset X$ be closed. Assume $R=\{(x,x')\in X\times X\mid q(x)=q(x')\}$ is closed, thus compact. Then $A\times X\cap R$ is compact, and so is also its image under the projection $p_2$ onto the second factor. But $p_2(A\times X\cap R)=\{x\in X\mid\exists a\in A:q(a)\sim q(x)\}=q^{-1}(q(A))$. So the saturation of a closed set is compact, and hence closed, which means that $q$ is a closed map. -One could also omit the compactness of $X$ if one assumes directly that $R$ is compact, because the last step only uses the Hausdorff'ness to deduce that a compact set is closed. Note that compactness of $R$ also makes $\{x\}\times X\cap R$ a compact set, so fibers are compact and this makes $q$ a so-called perfect map. These maps preserve many properties of the domain, for example all the separation axioms (except $T_0$) -This doesn't answer the question in the title, which I would really like to know myself. I only know about the product of a quotient map with the Identity $q\times Id:X\times Z\to Y\times Z$, which is a quotient map if $Z$ is locally compact. -Edit: Actually, showing that $q$ is closed requires only the compactness of $X$. Indeed, if $X$ is compact, then the projection $p_2$ is a closed map, so if $R$ is closed, then for every closed $A\subseteq X$, the set $p_2(A\times X\cap R)$ is closed.<|endoftext|> -TITLE: Proving that the set of algebraic numbers is countable without AC -QUESTION [6 upvotes]: A complex number $z$ is said to be algebraic if there is a finite collection of integers $\{a_i\}_{i\in n+1}$, not all zero, such that $a_0z^n + … + a_n = 0$. -Then can I prove the set of all algebraic numbers is countable without AC? I can only prove this by assuming 'countable union of countable sets is countable'. I guess and hope there's a way to prove this in ZF. Help. - -REPLY [11 votes]: You do not need AC to prove the result because you can explicitly define an enumeration of the algebraic numbers within ZF. To do so, we start with any of the standard enumerations of the set $\bigcup_{n \in \omega} (\omega \times \mathbb{Z}^{n+1})$. This set consists of all sequences of the form $(k, a_0, \ldots, a_n)$ where $k \in \omega$ and $(a_0, \ldots, a_n)$ is a nonempty tuple of integers. We will use this as a starting point to define an enumeration of the algebraic numbers. -First, suppose that $w, z$ are two roots of the same nonzero polynomial $p(x) \in \mathbb{Z}[x]$. Say that $w <_{p(x)} z$ if either $|w| < |z|$ or else if $|w| = |z|$ and $\operatorname{arg}(w) < \operatorname{arg}(z)$. So $<_{p(x)}$ is a well ordering of the finite set of roots of $p(x)$ for every nonzero $p(x) \in \mathbb{Z}[x]$, and this ordering is uniformly definable with the tuple of coefficients of $p(x)$ as a parameter. -We form an enumeration of the algebraic numbers as follows. Say that a finite tuple of integers $(k, a_0, \ldots, a_n)$ represents a complex number $w$ if $w$ is a root of $p(x) = a_0 + a_1 x + \cdots + a_n x^n$, and $p(x)$ is not identically zero, and $k \in \omega$, and there are exactly $k$ other roots $z$ of $p(x)$ with $z <_{p(x)} w$. The relation "$(k, a_0, \ldots, a_{n})$ represents $w$" is definable in ZF, and ZF proves that for every algebraic $z$ there is at least one tuple that represents $z$, and every complex number that is represented is algebraic. Therefore, we can enumerate the algebraic numbers in the same order that they are represented by tuples, using the enumeration of the tuples from above. -Thus ZF proves that there is a surjection from $\omega$ to the set of algebraic numbers. Because $\omega$ is well ordered already, ZF can turn this into a bijection $f$ from $\omega$ to the set of algebraic numbers (namely, $f(n)$ is the $n$th distinct algebraic number to appear in the enumeration). Hence ZF can prove the set of algebraic numbers is countable.<|endoftext|> -TITLE: Extending continuous and uniformly continuous functions -QUESTION [12 upvotes]: What do you say about the following statement: - -Let $f\colon(a,b)\rightarrow \mathbb{R}$ be a uniformly continuous function. Then $f$ can be extended to a uniformly continuous function with domain $[a,b]$. -Let $f\colon(a,b)\rightarrow \mathbb{R}$ be a continuous function. Then $f$ can be extended to a continuous function with domain $[a,b]$. - -So, I think that 2) is true, will be like add the points $a,b$ to domain used the definition limit, but I'm not sure. And 1) is true, too. But I don't know a good explanation. - -REPLY [9 votes]: If $x_n$ is a Cauchy sequence, then uniform continuity of $f$ allows us to conclude that the sequence $f(x_n)$ is also Cauchy. -The sequences $a_n=a+\frac{1}{n}$ and $b_n =b-\frac{1}{n}$ are Cauchy, hence so are the sequences $f(a_n)$, $f(b_n)$. Since $\mathbb{R}$ is complete, these sequences converge to some numbers $f_a, f_b$ respectively. Define the function $\overline{f}:[a,b] \to \mathbb{R}$ by $\overline{f}(a) = f_a$, $\overline{f}(b) = f_b$ and $\overline{f}(x) = f(x)$ for $x \in (a,b)$. Clearly $\overline{f}$ is continuous in $(a,b)$, it only remains to show continuity at $a,b$. -Suppose $x_n\in [a,b]$, and $x_n \to a$. We have $|\overline{f}(x_n) - \overline{f}_a| \leq |\overline{f}(x_n) - \overline{f}(a_n)| + | \overline{f}(a_n) - \overline{f}_a|$. Let $\epsilon >0$, then by uniform continuity, there exists $\delta>0$ such that if $|x-y|< \delta$, with $x,y \in (a,b)$, then $|f(x)-f(y)| < \epsilon$. Choose $n$ large enough such that $|x_n-a_n| < \delta$, and $| f(a_n) - f_a| < \epsilon$. If $x_n = a$, then $|\overline{f}(x_n) - \overline{f}_a| = 0$, otherwise we have $|\overline{f}(x_n) - \overline{f}_a| \leq |f(x_n) - f(a_n)| + | f(a_n) - f_a| < 2 \epsilon$. Consequently $\overline{f}(x_n) \to \overline{f}_a$, hence $\overline{f}$ is continuous at $a$. Similarly for $b$. -For the second case, take $f(x) = \frac{1}{x-a}$. Then $f$ is continuous on $(a,b)$, but the domain cannot be extended to $[a,b]$ while keeping $f$ continuous, and $\mathbb{R}$ valued. To prove this, take the sequence $a_n$ above, then $f(a_n) = n$, and clearly $\lim_n f(a_n) = \infty$. If $f$ could be continuously extended to $\overline{f}$, then $\overline{f}(a) \in \mathbb{R}$, which would be a contradiction. - -REPLY [2 votes]: HINT - -For the second one, consider something like the function $f(x) = \dfrac{x^2}{(x+1)(x-1)}$ as a function $f:(-1,1) \to \mathbb{R}$ -For the first one, it might just work out. Alternatively, show that if $f$ is uniformly continuous and $x_n$ is a Cauchy sequence in the domain of $f$, then $f(x_n)$ is a Cauchy sequence as well. Then pick a Cauchy sequence in the domain of $f$ that goes to $a$, for example.<|endoftext|> -TITLE: Continuity equation on manifolds -QUESTION [5 upvotes]: Mass conservation is usually written as -$$\frac{\partial \rho}{\partial t} + \operatorname{div}(\rho \boldsymbol v) = 0$$ -$\rho$ is the density and $\boldsymbol v$ is the fluid velocity. My attempt to rewrite it in notions used in differential geometry: -$$\frac{\partial \rho}{\partial t} + \star\, \text{d} \star (\rho v^\flat) = 0$$ -But I have a doubt that it may have another form, probably like in classical mechanics Liouville equation. -Maybe there is a book or something about fluid dynamics on manifolds? (I've asked this last question on Physics.SE, but if somebody knows it here let me know). - -REPLY [8 votes]: Let $M$ be a smooth manifold with a given volume form $\mu$ on it. -Let $\mathbf v_t$ be the time-dependent vector-field on $M$ describing the fluid velocity, and $\rho_t$ the time-dependent function on $M$ describing the fluid density. - -The continuity equation $\partial_t\rho_t+\text{div}(\rho_t\mathbf v_t)$ states the invariance of the volume form $\rho_t\mu$ under $\text{Fl}^{\mathbf v}_{t,s},$ the time-dependent flow of $\mathbf{v},$ i.e., explicitly $$\frac{d}{dt}(\text{Fl}^{\mathbf v}_{t,t_0})^\ast(\rho_t\mu)=0.\tag{*}$$ - -This interpretation of the continuity equation is based on: - -Lie derivative theorem, i.e. $\frac{d}{dt}(\text{Fl}^{\mathbf v}_{t,t_0})^\ast T_t|_p=\left.\left[(\text{Fl}^{\mathbf v}_{t,t_0})^\ast\left(\mathcal{L}_{\mathbf v_t}T_t+\partial_t T_t\right)\right]\right|_p,$ -where $T_t$ is an arbitrary time-dependent tensor-field on $M.$ -the definition of divergence of a vector-field w.r.t. the volume form $\mu,$ by which $\mathcal{L}_{\mathbf v_t}(\rho_t\mu)=d(\mathbf v_t\rfloor\rho_t\mu)=\text{div}(\rho_t\mathbf v_t)\mu.$ - -Infact, putting 1. and 2. into (*), we get $(\text{Fl}^{\mathbf v}_{t,t_0})^\ast\left[\left(\text{div}(\rho_t\mathbf v_t)+\partial_t \rho_t\right)\mu\right]=0,$ i.e. $$\text{div}(\rho_t\mathbf v_t)+\partial_t \rho_t=0.$$ -Edit Note that the (*) can be rewritten in integral form as the law of conservation of mass: $$\tag{**}\frac{d}{dt}\int_{\text{Fl}_{t,t_0}^{\mathbf v}D}\rho_t\mu=0,\text{ for all compact }D\subseteq M.$$<|endoftext|> -TITLE: I want to start mathematics from scratch. What should I begin with? -QUESTION [16 upvotes]: I've just finished high school but I don't feel my knowledge of mathematics is good enough. I'd like to start again from scratch, possibly adding a bit (a lot) of problem solving to it. What is the order in which I should study the various branches of mathematics? What are some good books to do so? Keep in mind that for 'starting from scratch' I mean starting from (possibly at a university level) set theory and the four operations. Thanks in advance for the answers! - -REPLY [10 votes]: I feel like it depends on where you are headed. If you want to make mathematics you future profession, the way you take will be different from what say an engineer will take. -For example, in my case I am engineering student and i got to study plenty of calculus, probability and much of the fancy stuff but by the end of the day i still felt my knowledge of maths to be unsatisfactory(that's why i am on his site by the way). -So on the way to achieving your goal, here is what i can tell depending on my experience. -If engineering is your way: -You have to work very much on problem solving. A possible way to approach the task, here it is. -Start will "normal" calculus but now try to understand the concepts not just for computing answers but also try to understand what it means in real life. For example, say "limits". You must have studied those in high school. Understand carefully what it means. Try to find examples where this concept might fit. Here is an example: I am given a material whose 'flexibility' is modeled by a given function. And that function depends on temperature. Here limits may help you understand how the material behaves when the temperature tends towards a certain value. See ... Try to start thinking like that about concepts, not just solve some exercises - but don't get me wrong: exercises are of crucial importance in learning, but the difference between you and a maths software is that you must understand the why of every computation you are doing. -Now a possible road map: -I/ Calculus: - -Limits -Differentiation -Integration -Series -Gamma and Beta functions -Integral transforms - -Take a long pause after this be sure you really understand this stuff well - -Differential equations -Vector calculus -Complex analysis - -II/ Algebra - -Matrices and determinants -Linear equations -Vectors -Eigen vectors and eigen values. - - -From there you can go ahead and study other areas of interest mainly -(i) Engineering optimization and numerical analysis -(ii) Statistics and probability. -Those two because as an engineer the sooner you start producing results, the better off you are. -Starting with calculus is important because it has a lot of applications you can play with, it gives computational skills fast if you do exercises, has interesting concepts and forms the foundation of much mathematics engineers deal with. -Possible books: - -"Calculus" by Michael Spivak as already mentioned -"Differential and integral calculus" by Richard Courant -And some of the "(Applied) mathematics for scientists and engineers". I have no idea which one to recommend they are just so many and some are good. - -So basically, it boils down to - -Understand concepts -DO exercises -Find practical applications to related the math to real world things - -If mathematics is your way: -Now if you want to make mathematics your profession, you will need a different frame of mind. First i am neither a professional mathematician nor have i reached a level where i can say that i am thinking like one. Yet that is my goal too. So i will share with you what i have learnt so far. -First, mathematicians, from i can tell so far, work differently from say physicists and engineers. When a you hit a theorem, don't go ahead and read the proof, first try to prove it yourself. -That will form the basis of the mathematician in you. -Here is the books i can advice to start with. - -"How to prove it, A structured approach" by Daniel Velleman. Nice book for an introduction to proofs. I like the idea of givens and Goal. -"Book of proof" by Richard Hammack. Nice little book. You can either start with this one or Velleman. The thing i like with this one is that logic and set theory are separated in comparison with Velleman. - http://www.people.vcu.edu/~rhammack/BookOfProof/ - -Once you are grounded in Set theory ( not too much though, whatever is provided by the two previous will be enough ) and proofs, continue with these: - -Either "Principles of mathematical analysis" by Walter Rudin -Or "Topology without tears" by Sydney Morris - http://uob-community.ballarat.edu.au/~smorris/topbook.pdf -Or "Abstract Algebra: Theory and applications" by Thomas Judson - http://abstract.ups.edu/index.html - -Always try to prove theorems before reading the proof. Every time you read a mathematics book, usually graduate level ( don't be concerned about these for now ), and they tell you that a certain amount of mathematical maturity is expected from the reader, what that simply means is that they expect you to be able to prove the theorems or at least follow the logical arguments. -Mathematical literature -Also I highly advice like others that you try to read about mathematics in the general sense. Some books you may start with, here they are. - -"God created the integers - the mathematical breakthroughs that changed history" by Stephen Hawking. Interesting books, this is! -"What is mathematics" by Richard Courant -"The music of the primes - searching to solve the greatest mystery in mathematics" by Marcus du Sautoy - -You may not be able to follow, the proofs in the two first books but nonetheless, you will enjoy the ride!!! -So that's the best i can do for my level and I wish you good luck and success!<|endoftext|> -TITLE: The probability of one Gaussian larger than another. -QUESTION [9 upvotes]: For two Gaussian-distributed variables, $ Pr(X=x) = \frac{1}{\sqrt{2\pi}\sigma_0}e^{-\frac{(x-x_0)^2}{2\sigma_0^2}}$ and $ Pr(Y=y) = \frac{1}{\sqrt{2\pi}\sigma_1}e^{-\frac{(x-x_1)^2}{2\sigma_1^2}}$. What is probability of the case X > Y? - -REPLY [7 votes]: Suppose $X$ and $Y$ are jointly normal, i.e. no independence is needed. Define $Z = X - Y$. It is well known that $Z$ is Gaussian, and thus is determined by its mean $\mu$ and its variance $\sigma^2$. -$$ - \mu = \mathbb{E}(Z) = \mathbb{E}(X) - \mathbb{E}(Y) = \mu_1 - \mu_2 -$$ -$$ - \sigma^2 = \mathbb{Var}(Z) = \mathbb{Var}(X) + \mathbb{Var}(Y) - 2 \mathbb{Cov}(X,Y) = \sigma_1^2 + \sigma_2^2 - 2 \rho \sigma_1 \sigma_2 -$$ -where $\rho$ is the correlation coefficient. Now: -$$ - \mathbb{P}(X>Y) = \mathbb{P}(Z>0) = 1- \Phi\left(-\frac{\mu}{ \sigma}\right) = \Phi\left(\frac{\mu}{ \sigma}\right) = \frac{1}{2} \operatorname{erfc}\left(-\frac{\mu}{\sqrt{2}\sigma}\right) -$$<|endoftext|> -TITLE: 'Linux' math program with interactive terminal? -QUESTION [43 upvotes]: Are there any open source math programs out there that have an interactive terminal and that work on linux? -So for example you could enter two matrices and specify an operation such as multiply and it would then return the answer or a error message specifying why an answer can't be computed? I am just looking for something that can perform basic matrix operations and modular arithmetic. - -REPLY [2 votes]: freemat -FreeMat is a free environment for rapid engineering and scientific prototyping and data processing. It is similar to commercial systems<|endoftext|> -TITLE: Evaluating $\int_1^3\frac{\ln(x+2)}{x^2+2x+15} \ dx$ -QUESTION [11 upvotes]: Could you please give me a hint on how to compute: -$$ -\int_1^3\frac{\ln(x+2)}{x^2+2x+15}dx -$$ -Thank you for your help - -REPLY [7 votes]: This was supposed to be a comment on Raymond's answer, but it got too long. I started with trying to obtain Mayrand's fine expression from the dilogarithmic mess one might obtain through Mathematica or Raymond's route, but wound up with a satisfactorily simple expression. -We start from a version of Raymond's answer with the "elementary portion" already simplified: -$\begin{split} -\frac1{2\sqrt{14}}&\left(\log\frac53\;\arctan\sqrt{14}+\log\,15\;\arctan\frac{\sqrt{14}}{11}\right)+\\ -&\frac{i}{2\sqrt{14}}\left(\mathrm{Li}_2\left(\frac{2+i\sqrt{14}}{3}\right)+\mathrm{Li}_2\left(\frac{4-i\sqrt{14}}{5}\right)-\right.\\ -&\left.\left(\mathrm{Li}_2\left(\frac{2-i\sqrt{14}}{3}\right)+\mathrm{Li}_2\left(\frac{4+i\sqrt{14}}{5}\right)\right)\right)\end{split}$ -I grouped the terms in this way, since this allows the easy application of Landen's identity (see this paper for a survey of the various algebraic dilogarithm identities): -$$\mathrm{Li}_2(x)+\mathrm{Li}_2\left(\frac{x}{x-1}\right)=-\frac12\left(\log(1-x)\right)^2$$ -Now, we have the relations -$$\frac{\frac{2\pm i\sqrt{14}}{3}}{\frac{2\pm i\sqrt{14}}{3}-1}=\frac{4\mp i\sqrt{14}}{5}$$ -which when used with Landen's identity yields -$$\begin{split} -&\frac1{2\sqrt{14}}\left(\log\frac53\;\arctan\sqrt{14}+\log\,15\;\arctan\frac{\sqrt{14}}{11}\right)+\\ -&\frac{i}{2\sqrt{14}}\left(\left(-\frac12\left(\log\left(1-\frac{2+i\sqrt{14}}{3}\right)\right)^2\right)-\left(-\frac12\left(\log\left(1-\frac{2-i\sqrt{14}}{3}\right)\right)^2\right)\right) -\end{split}$$ -which, after a few more algebraic manipulations, finally yields -$$\frac1{2\sqrt{14}}\left(\log\frac53\;\arctan\sqrt{14}+\log\,15\;\arctan\frac{\sqrt{14}}{11}\right)-\frac{\arctan\sqrt{14}\log\frac53}{2\sqrt{14}}=\color{blue}{\frac1{2\sqrt{14}}\log\,15\;\arctan\frac{\sqrt{14}}{11}}$$ -which is even simpler than Mayrand's original result.<|endoftext|> -TITLE: How do I calculate the probability distribution of the percentage of a binary random variable? -QUESTION [7 upvotes]: I have an urn containing balls that are all either black or red. I'm interested in discovering the percentage of balls that are red. But I can only sample from the urn (without replacement), so the best I can do is calculate a probability distribution over possible percentages. -Obviously, if I've drawn no balls, I have no information, so the probability distribution is uniform from 0 to 1. But what is it once I start drawing balls? - -REPLY [5 votes]: Theory: if $reds$ is the number of red balls you've seen and $blacks$ is the number of black balls you've seen, then the distribution is: -$Beta(reds+1, blacks+1)$ -This starts out as $Beta(1,1)$ which is the uniform distribution we want. As we see blacks, it shifts toward zero; as we see reds, it shifts toward one. -Does anyone know if this is right?<|endoftext|> -TITLE: Proving an estimate for this integral -QUESTION [7 upvotes]: How can I show that$$\sqrt[3]6>\int_1^\infty\frac{(1+x)^{1/3}}{x^2}\mathrm dx?$$ - -REPLY [2 votes]: The integral can be written as -$$\int_1^{\infty} {1 \over x^{5 \over 3}} \bigg(1 + {1 \over x}\bigg)^{1 \over 3}\,dx$$ -By Taylor expanding, $(1 + {1 \over x})^{1 \over 3} = 1 + {1 \over 3x} +$ error term, where the error term is negative by the Lagrange form of the remainder. So the integral is less than -$$\int_1^{\infty} {1 \over x^{5 \over 3}} + {1 \over 3 x^{8 \over 3}}\,dx$$ -$$= {3 \over 2} + {1 \over 3}\cdot{3 \over 5}$$ -$$= 1.7$$ -$$ < \sqrt[3]6$$<|endoftext|> -TITLE: Quotienting a group by an equivalence relation -QUESTION [6 upvotes]: In Jacobson's book, Basic Algebra 1, he points out that you can form quotient groups by looking at the equivalence classes of a 'multiplication compatible' equivalence relation on the group. By multiplication compatible, he means that if $A,B$ are equivalence classes for $a, b$, and we also have that $c \in A, d \in B$, then the equivalence class of $ab$ is the equivalence class of $cd$. As it turns out, this forms a group. He then points out that the equivalence class of 1 is a normal subgroup and that the group that you obtained is equal to the original group quotiented by the equivalence class of 1. I'm curious as to if there are any situations where you would quotient a group by an equivalence relation, and not by a normal subgroup, given that these are equivalent. - -REPLY [5 votes]: Sure. Consider the quotient of a topological group $G$ by the relation "lies in the same connected component." You get a group $\pi_0(G)$ of connected components. This is equivalent to quotienting by the connected component of the identity (which incidentally proves that this subgroup is normal) but I think the equivalence relation is the more natural thing to think about, even if thinking about the normal subgroup is convenient.<|endoftext|> -TITLE: Why does Strassen's algorithm work for $2\times 2$ matrices only when the number of multiplications is $7$? -QUESTION [9 upvotes]: I have been reading Introduction to Algorithms by Cormen. Before explaining Strassen algorithm the book says this: - -Strassen’s algorithm is not at all obvious. (This might be the biggest understatement in this book.) - -The book just states the algorithm but don't explain why it works. So if we take the case of multiplication of two $2 \times 2$ matrices, it reduces the number of multiplications from $8$ to $7$ thereby reducing the complexity from $n^{3}$ to $n^{\lg 7}$ which is roughly $n^{2.8}$. -My question is why $7$ and say not $5$ or even lesser number? I mean they could have made the recursion tree even less "bushier". - -REPLY [16 votes]: Allow me to shamelessly copy an answer of mine to this question first. - -The idea behind Strassen multiplication is the same as the idea behind Karatsuba multiplication. -In this, if we have two numbers and a base $B$ (maybe $10^6$, say), and the numbers are written $a = x_0B + x_1$, $b = y_0B + y_1$, and we want to calculate $ab$, then naively we might say that it's -$ab = x_0 y_0 B^2 + (x_0 y_1 + x_1 y_0)B + x_1 y_1 \qquad$ (requiring 4 multiplications) -But we can be wittier, as $x_0y_1 + x_1y_0 = (x_0 + x_1)(y_0 + y_1) - x_0y_0 - x_1y_1 $ -So we can calculate $ab$ by knowing $x_0y_0, x_1y_1$ and $(x_0 + x_1)(y_0 + y_1)$, which only uses 3 multiplications. This is just a slick idea, akin to the many other slick arithmetic tricks out there that might be found by inspection, and it became popular in the 1960s. (There's a good story behind it - Karatsuba responded to a challenge of Kolmogorov, ultimately culminating with Kolmogorov writing a paper in Karatsuba's name). -For matrices, a direct port of Karatsuba doesn't quite work. But it's easy to verify that Strassen is true (just write it out). But it's very unclear how Strassen came across them. But it should be mentioned that Strassen was working on many numerical-methods style techniques to improve matrix calculations. It should also be mentioned that Strassen has an improved algorithm, better than this technique, using fast fourier transforms as well. -One might ask: well, we can do it with 7 multiplications. Can we do it with 6? Strassen multiplication has been around since 1969, so surely it's known? This was an open problem until just a few years ago, when Landsberg showed that 7 is optimal. It's nontrivial and a bit far afield for me, but his (corrected after it was published) article can be found on the arXiv. -One might also wonder about $3 \times 3$ matrices. How few multiplications are necessary? Naively, it takes 27 multiplications. But it can be done in 23. (See Julian D. Laderman, A noncommutative algorithm for multiplying 3×3 matrices using 23 muliplications, Bull. Amer. Math. Soc. 82 (1976) 126–128, MR0395320 (52 #16117).) Is that optimal? That's unknown - all that's known is that it takes at least 19. (But if you're just interested in computational savings, even 19 gives less savings when iterated than the $2 \times 2$ case). - -So in short, $7$ is optimal. It's not obvious that it's optimal. It was a very clever idea, even, to find the $7$.<|endoftext|> -TITLE: Finding all solutions to $y^3 = x^2 + x + 1$ with $x,y$ integers larger than $1$ -QUESTION [11 upvotes]: I am trying to find all solutions to -(1) $y^3 = x^2 + x + 1$, where $x,y$ are integers $> 1$ - -I have attempted to do this using...I think they are called 'quadratic integers'. It would be great if someone could verify the steps and suggest simplifications to this approach. I am also wondering whether my use of Mathematica invalidates this approach. -My exploration is based on a proof I read that $x^3 + y^3 = z^3$ has no non-trivial integer solutions. This proof uses the ring Z[W] where $W = \frac{(-1 + \sqrt{-3})}{2}$. I don't understand most of this proof, or what a ring is, but I get the general idea. The questions I have about my attempted approach are - - -Is it valid? -How could it be simplified? - - -Solution: -Let $w = (-1 + \sqrt{-3})/2$. (Somehow, this can be considered an "integer" even though it doesn't look anything like one!) -Now $x^3 - 1 = (x-1)(x-w)(x-w^2)$ so that, $(x^3 - 1)/(x-1) = x^2 + x + 1 = (x-w)(x-w^2)$. Hence -$y^3 = x^2 + x + 1 = (x-w)(x-w^2).$ -Since $x-w, x-w^2$ are coprime up to units (so I have read) both are "cubes". Letting $u$ be one of the 6 units in Z[w], we can say -$x-w = u(a+bw)^3 = u(c + dw)$ where -$c = a^3 + b^3 - 3ab^2, d = 3ab(a-b)$ -Unfortunately, the wretched units complicate matters. There are 6 units hence 6 cases, as follows: -1) $1(c+dw) = c + dw$ -2) $-1(c+dw) = -c + -dw$ -3) $w(c+dw) = -d + (c-d)w$ -4) $-w(c+dw) = d + (d-c)w$ -5) $-w^2(c+dw) = c-d + cw$ -6) $w^2(c+dw) = d-c + -cw$ -Fortunately, the first two cases can be eliminated. For example, if $u = 1$ then $x-w = c+dw$ so that $d = -1 = 3ab(a-b).$ But this is not possible for integers $a,b$. The same reasoning applies to $u = -1$. -For the rest I rely on a program called Mathematica, which perhaps invalidates my reasoning, as you will see. -We attack case 5. Here -$x = c-d = a^3 + b^3 - 3a^2b$, and $c = a^3 + b^3 - 3ab^2 = -1.$ -According to Mathematica the only integer solutions to $c = -1$ are -$(a,b) = (3,2), (1,1), (0,-1), (-1,0), (-1,-3), (-2,1).$ -Plugging these into $x = c-d$ we find that no value of x that is greater than 1. So case 5 is eliminated, as is 6 by similar reasoning. -Examining case 4 we see that $d-c = -(a^3 + b^3 - 3a^2*b) = -1$ with solutions -$(-2,-3), (-1,-1), (-1,2), (0,1), (1,0), (3,1).$ -Plugging these values into $x = d = 3ab(a-b)$ yields only one significant value, namely $x = 18$ (e.g. (a,b)=(3,1) . The same result is given by case 4. Hence the only solution to (1) is $7^3 = 18^2 + 18 + 1$ -However, I'm unsure this approach is valid because I don't know how Mathematica found solutions to expressions such as $a^3 + b^3 - 3ab^2=-1$. These seem more difficult than the original question of $y^3 = x^2 + x + 1$, although I note that Mathematica could not solve the latter. - -REPLY [2 votes]: In this answer to a related thread, I outline a very elementary solution to this problem. There are probably even simpler ones, but I thought you might appreciate seeing that.<|endoftext|> -TITLE: Closed set on Euclidean space that is not compact -QUESTION [21 upvotes]: I have read that a subset of Euclidean space may be called compact if it is both closed and bounded. I was wondering what a good example of a closed but unbounded set would be? -Would a closed ball inside a sphere with an infinite radius do the trick? If that example works are there any other examples people could think of? - -REPLY [6 votes]: A simple example of a closed but unbounded set is $[0,\infty)$.<|endoftext|> -TITLE: Question about math symbol $\bigsqcup$ -QUESTION [5 upvotes]: Can someone tell me what this symbol means? -$\bigsqcup$ - -REPLY [10 votes]: It is the disjoint union symbol- it is most commonly used informally to denote situations where you take the union of two disjoint sets. The actual definition though is more of a tagged union- intuitively, you index the sets to be unioned by some set $I$, and then the result is the collection of all the elements of each set, along with a "tag" that says which set it came from. -In your case, formally you have sets $A$ and $B$- let's re-label these $A_1$ and $A_2$. The disjoint union is $A_1 \bigsqcup A_2 =\{ (a,1) \vert a\in A_1\} \cup \{ (a,2) \vert a\in A_2\}$. So if they have some element $a$ in common, you end up with both $(a,1)$ and $(a,2)$ in your disjoint union. In the case that they have no common elements, the result is the same as the standard union.<|endoftext|> -TITLE: GAP semidirect product algorithm -QUESTION [8 upvotes]: Can anybody guide me towards, or possibly even explain here, the algorithm that GAP uses to compute the semidirectproduct of two permutation groups which outputs another permutation group? - -EXAMPLE: - -gap> C3:=CyclicGroup(IsPermGroup,3); Group([ (1,2,3) ]) - gap> C7:=CyclicGroup(IsPermGroup,7); Group([ (1,2,3,4,5,6,7) ]) - gap> A:=AutomorphismGroup(C7); < group with 1 generators > gap> elts := - Elements(A); [ IdentityMapping( Group([ (1,2,3,4,5,6,7) ]) ), - [ (1,2,3,4,5,6,7) ] -> [ (1,3,5,7,2,4,6) ], [ - (1,2,3,4,5,6,7) ] -> [ (1,4,7,3,6,2,5) ], [ (1,2,3,4,5,6,7) ] - -> [ (1,5,2,6,3,7,4) ], [ (1,2,3,4,5,6,7) ] -> [ (1,6,4,2,7,5,3) ], [ (1,2,3,4,5,6,7) ] -> [ (1,7,6,5,4,3,2) ] - ] gap> sigma := elts[2]; [ (1,2,3,4,5,6,7) - ] -> [ (1,3,5,7,2,4,6) ] gap> sigma^3; [ (1,2,3,4,5,6,7) ] - -> [ (1,2,3,4,5,6,7) ] gap> map := GroupHomomorphismByImages(C3, A, GeneratorsOfGroup(C3), [sigma]); [ (1,2,3) ] -> [ [ - (1,2,3,4,5,6,7) ] -> [ (1,3,5,7,2,4,6) ] ] gap> SDP := - SemidirectProduct(C3, map, C7); Group([ (2,3,5)(4,7,6), - (1,2,3,4,5,6,7) ]) - -REPLY [4 votes]: The code is on line 919 of lib/gprdperm.gi and is very easy to understand. -It rewrites the normal subgroup in its regular action (so $K$ acting on $|K|$ points), and then of course the complement subgroup acts on the normal subgroup giving the semidirect product of $H/C_H(K)$ with $K$. If $C_H(K) \neq 1$, then it does the subdirect product smooshy thing with $H$'s original rep: that is $H \ltimes K$ acts on $X \dot\cup K$ where $H$ acts on $X$ as usual, $K$ centralizes $X$, and $H,K$ have the previously described action on $K$. -In particular, this method is impractical with $S_{20} as the normal subgroup. -If $K$ acts on $Y$ and $|H|$ is small, then $H \ltimes K$ can act on $H \times Y$ as well, but this is not implemented. -There is a special method implemented on line 961 to catch a nice case: if $K$ acts on $Y$, and if the action of $H$ on $K$ lifts to a homomorphism from $H$ to $\operatorname{Sym}(Y)$, then one can view $H$ more compactly as a subgroup of $\operatorname{Sym}(X \dot\cup Y)$. This method is not guaranteed to succeed even if applicable if $C_{\operatorname{Sym}(Y)}(K)$ is large, as the method to check if the homomorphism lifts is just to choose a random section and see if it a homomorphism.<|endoftext|> -TITLE: The boundedness of an integral -QUESTION [11 upvotes]: Is there a constant $C$ which is independent of real numbers $a,b,N$, such that - -$$\left| {\int_{-N}^N \dfrac{e^{i(ax^2+bx)}-1}{x}dx} \right| \le C?$$ - -REPLY [8 votes]: These integrals are indeed uniformly bounded. As the only effect of changing the signs of $a,b,N$ on the integral is a sign change and/or complex conjugation, we can restrict to $a,b,N\gt0$. Setting $M=Nb$ and $\alpha=a/b^2$, we have -$$ -\int_{-N}^N\left(e^{i(ax^2+bx)}-1\right)\frac{dx}x=\int_0^M\left(e^{i(\alpha x^2+x)}-e^{i(\alpha x^2-x)}\right)\frac{dx}x. -$$ -However, the integrand can be written as $2ie^{i\alpha x^2}\frac{\sin x}x$, so is bounded by $2$ in absolute value. Therefore, fixing any constant $K\gt0$, the integral is bounded by -$$ -\begin{align} -2K+\int_K^{K\vee M}e^{i(\alpha x^2+x)}\frac{dx}x-\int_K^{K\vee M}e^{i(\alpha x^2-x)}\frac{dx}x.&&{\rm(1)} -\end{align} -$$ -I'll prove that this is bounded with the help of a lemma. - -Lemma: (van der Corput lemma) If $f\colon[A,B]\to\mathbb{R}$ is convex with $\lvert f^\prime(x)\rvert\ge\lambda\gt0$ and $g\colon[A,B]\to\mathbb{C}$ is differentiable then, - $$\left\lvert\int_A^Be^{if(x)}g(x)dx\right\rvert\le\frac2\lambda\left(\lvert g(B)\rvert+\int_A^B\lvert g^\prime(x)\rvert dx\right)$$ - -As I will only be interested in intervals $[A,B]\subset[K,\infty)$ with $g(x)=1/x$, the bound in the lemma can be written as -$$ -\begin{align} -\left\lvert\int_A^Be^{if(x)}\frac{dx}x\right\rvert\le\frac2{\lambda K}&&{\rm(2)} -\end{align} -$$ -Taking $f(x)=\alpha x^2+x$, this shows that the first integral in (1) is bounded by $2/K$. For the second integral, take $f(x)=\alpha x^2-x$, fix any $0\lt\epsilon\lt1/2$, and first look at value of the integral with integration range restricted to $[0,\epsilon/\alpha]$. In this range, we have $f^\prime(x)\le f^\prime(\epsilon/\alpha)=-(1-2\epsilon)$. So, by inequality (2), this part of the integral is bounded by $2K^{-1}(1-2\epsilon)^{-1}$. -Next, for any fixed $\gamma\gt1/2$, the value of the last integral in (1), restricted to the range $[\epsilon/\alpha,\gamma/\alpha]$ is bounded by -$$ -\int_{\epsilon/\alpha}^{\gamma/\alpha}\frac{dx}{x}=\log(\gamma/\epsilon). -$$ -Finally, look at the last integral in (1) restricted to the range $[\gamma/\alpha,\infty)$. As $f^\prime(x)\ge f^\prime(\gamma/\alpha)=(2\gamma-1)$ in this range, inequality (2) shows that this part of the integral is bounded by $2K^{-1}(2\gamma-1)^{-1}$. -Putting these together shows that the set of integrals in the question is bounded above by a constant. The upper bound obtained here is -$$ -2K+\frac2K+\frac2{K(1-2\epsilon)}+\frac2{K(2\gamma-1)}+\log(\gamma/\epsilon) -$$ -for arbitrary positive constants $K,\epsilon,\gamma$ with $\epsilon\lt1/2\lt\gamma$.<|endoftext|> -TITLE: Dyslexia in group actions of $SL_2$ on binary cubic forms -QUESTION [5 upvotes]: The group $SL_2$ (say, $SL_2(\mathbb{C})$, but we could take $SL_2$ of anything else, or probably regard $SL_2$ as a group scheme) acts on binary cubic forms. (Or binary $n$-ic forms in general.) What is the correct form of the action? -Let me give two different answers to this, both of which I have seen in the literature. -(1) Regard binary cubic forms as homomorphisms $\mathbb{A}^2 \rightarrow \mathbb{A}$, where $\mathbb{A}$ is affine space (i.e. just $\mathbb{C}$). We can define an action of $SL_2$ on $\mathbb{A}^2$ as follows: Write $g \in SL_2$ as a matrix, $v \in \mathbb{A}^2$ as a column vector, and then the action is just given by $g v$ (usual matrix multiplication). The action on $\mathbb{A}^1$ is trivial. With these conventions, for a binary cubic form $f$, write $(gf)(v) = f(g^{-1} v)$. This seems to be the representation-theoretic point of view; at any rate it is consistent with p. 4 of Fulton-Harris and this definition is given on p. 14 of Olver's Classical Invariant Theory. -(2) Use the definition $(gf)(v) = f(v g)$, where this time we write $v$ as a row vector, so that $v g$ is well-defined. This definition is quite common, appearing for example in Bhargava, Shankar, and Tsimerman's paper here among many other places (including papers I've co-authored!) -These definitions are equivalent (neither is "wrong") but they're not the same. Definition (1) feels "right" to me, where it seems that the point of (2) is to identify binary cubic forms with $\mathbb{A}^4$ instead of its dual. We would like to write $(gf)(v) = f(g(v))$, but this doesn't work for reasons that are more or less explained in (1). -A couple questions: First of all, is my explanation above accurate? -If it is, is there a good highbrow explanation of (2)? Somehow it feels like a hacky workaround to me. I think the notation (2) is much more natural than (1) in the paper I linked to (I am not trying to argue with their choice of notation) but it feels like cheating a bit, and it is confusing that the same group action is defined in different ways in the literature. -Is there a better perspective than the one I have offered, or do I just need to grin and bear it? -Thank you! - -REPLY [3 votes]: Perhaps some general remarks are in order. First, let $M$ be a monoid. Recall that a left action of $M$ is a set $S$ together with a map $M \times S \to S$ satisfying the usual associativity conditions. Equivalently, it is a set $S$ together with a homomorphism $M \to \text{End}(S)$ of monoids. A right action of $M$ is a set $S$ together with a map $S \times M \to S$ satisfying the usual associativity conditions. Equivalently, it is a set $S$ together with a homomorphism $M^{op} \to \text{End}(S)$, where $M^{op}$ is the opposite monoid (the monoid with the same underlying set as $M$ but whose multiplication is in the other direction) of $M$. -In general $M$ and $M^{op}$ are non-isomorphic, so left and right actions are genuinely different and must be carefully distinguished. Moreover, if $S$ is a left action and $T$ is another set, then the set of functions $\text{Hom}(S, T)$ inherits a right action via -$$m : f(s) \mapsto f(ms)$$ -and there is no way to turn this into a left action. (Category theorists should feel free to replace "monoid," "left action," and "right action" above with "category," "functor," and "contravariant functor.") -If $M$ is a group $G$ then it comes equipped with a canonical isomorphism to its opposite, namely $g \mapsto g^{-1}$. This gives a canonical way to turn all right actions into left actions, which is what we do when we pass from a representation $V$ of a group $G$ to the dual representation $V^{\ast}$ (which for a monoid is a right action if it acted from the left on $V$). - -Some specifics. Let $V$ be a vector space ("column vectors") and $G$ a group on which $V$ acts. Then $G$ acts on the space of polynomial functions on $V$, which can be identified with the symmetric algebra $S(V^{\ast})$ (since $V^{\ast}$ corresponds to linear polynomials on $V$). This gives the first action. -$G$ also acts on $V^{\ast}$ ("row vectors"), which induces an action on the space of polynomial functions on $V^{\ast}$, which can be identified with the symmetric algebra $S(V)$. This gives the second action. (You should write it as $v(g^{-1})^{-1}$ because you are implicitly taking the dual twice.) -The two actions are isomorphic in this case because $V$ is self-dual. Indeed, if $V$ is a $2$-dimensional vector space over a field $k$ then $G$ lies in $\text{SL}_2(k)$ if and only if it preserves the exterior product $V \times V \to \Lambda^2(V)$, and choosing an identification of $\Lambda^2(V)$ with the underlying field yields a $G$-equivariant dual pairing $V \times V \to k$.<|endoftext|> -TITLE: Complex matrix that commutes with another complex matrix. -QUESTION [6 upvotes]: I am trying to learn some linear algebra, and currently I am having a difficulty time grasping some of the concepts. I have this problem I found that I have no idea how to start. -Assume that $\bf A$ is an $n\times n$ complex matrix which has a cyclic vector. Prove that if $\bf B$ is an $n\times n$ complex matrix that commutes with $\bf A$, then ${\bf B}=p({\bf A})$ for some polynomial $p$. -All I know at this point is that ${\bf AB}={\bf BA}$. - -REPLY [10 votes]: Since $A^0v,A^1v,\dots,A^{n-1}v$ are linearly independent, they form a basis for ${\bf C}^n$. Thus, $$Bv=c_0A^0v+c_1A^1v+\cdots+c_{n-1}A^{n-1}v=p(A)v$$ where $$p(x)=c_0+c_1x+\cdots+c_{n-1}x^{n-1}$$ for some constants $c_0,c_1,\dots,c_{n-1}$. Since $B$ commutes with $A$, it commutes with all powers of $A$, so $$B(A^rv)=A^rBv=A^rp(A)v=p(A)(A^rv)$$ for $r=0,1,\dots,n-1$ (I've used $A^rp(A)=p(A)A^r$). But again the vectors $A^rv$ are a basis, so $B=p(A)$ ($Bx=Cx$ for all $x$ in a basis implies $Bx=Cx$ for all $x$ in the vector space, which implies $B=C$).<|endoftext|> -TITLE: Number of bit strings with 3 consecutive zeros or 4 consecutive 1s -QUESTION [10 upvotes]: I am trying to count the number of bit-strings of length 8 with 3 consecutive zeros or 4 consecutive ones. I was able to calculate it, but I am overcounting. The correct answer is $147$, I got $148$. -I calculated it as follows: -Number of strings with 3 consecutive zeros = $2^5+5\times2^4 = 112$, because the 3 zeros can start at bit number 1, 2, 3, .., 6 -Number of strings with 4 consecutive ones = $2^4+4\times2^3 = 48$, I used the same reasoning. -Now I am trying to count the number of bit-strings that contain both 3 consecutive zeros and 4 consecutive 1s. I reasoned as follows: -the strings can be of the following forms: 0001111x, 000x1111, x0001111..thus there are $2+2+2 = 6$ possibilities for bit-strings where the 3 consecutive zeros come first. Symmetrically there are $6$ bit-strings where the 4 consecutive ones come first. -Thus the answer should be = $112+48-12 = 148$. -clearly there's something wrong with my reasoning, if someone could point it out, that would be awesome. Thanks - -REPLY [3 votes]: $ \begin{array}{|l|r|l|} - \hline -format & N & exceptions \\ - \hline - 000***** & 32 & \\ -1000**** & 16 & \\ -*1000*** & 16 & \\ -**1000** & 16 & \\ -***1000* & 14 & 0001000* \\ -****1000 & 13 & 000*1000 , 10001000 \\ -1111**** & 13 & 1111000* , 11111000 \\ -01111*** & 7 & 01111000 \\ -*01111** & 8 & \\ -**01111* & 6 & 0001111* \\ -***01111 & 6 & *0001111 \\ - \hline - \end{array}$ -Total: $147$<|endoftext|> -TITLE: Proving Fermat's Last Theorem (easily) using "assumed" conjectures -QUESTION [5 upvotes]: It can easily be proven assuming Szpiro's conjecture that Fermat's Last Theorem is true for sufficiently large $n$. The proof consists of extremely straightforward computations. My question is, is there a refinement of that proof that prove's FLT for all $n$? Or maybe, are there any other unproven conjectures (ABC conjecture, etc.) that if proven would lead to a simple proof of FLT? If so, what is the statement of the conjecture and what is the proof of FLT assuming this conjecture? -Thanks! - -REPLY [5 votes]: I think that an effective form of ABC will give an effective upper bound on $n$ for which $x^n + y^n = z^n$ has non-trivial solutions. This would leave finitely many $n$ to check, which one could imagine (if the bound on $n$ is not too large) could be checked by other, more traditional means.<|endoftext|> -TITLE: Perfect set in $\mathbb{R}$ which contains no rational number -QUESTION [8 upvotes]: Possible Duplicate: -Perfect set without rationals - -Does there exist a nonempty perfect set in $\mathbb{R}$ which contains no rational number? -This problem is on p.44 PMA - Rudin -I found a proof of this on google but the proof is not 'suitable' for me, since the proof uses the concept of 'measure', which is in chapter 11, while this problem is on chapter 2. -Can one show this by direct construction? - -REPLY [6 votes]: Since the rational are countable, let $a_n$ enumerate all the rationals. Let $B_{2^{-n}}(a_n)$ be the open interval centered at $a_n$ of radius $2^{-n}$. Let $A = \bigcup_{n \in \mathbb{N}} B_{2^{-n}}(a_n)$. $A$ is open since it is it is a union of open sets. Note that $\mathbb{Q} \subset A$. Clearly $A$ is not all of $\mathbb{R}$ because "length" of $A$ is less than or equal to $2\sum_{n = 1}^\infty 2^{-n} < \infty$. (This is essentially the measure idea.) Because of this and fact that $\mathbb{R}$ is uncountable, you have that $C:= \mathbb{R} - A$ is an uncountable set. $C$ closed since it is the complement of an open set. Also $C \cap \mathbb{Q} = \emptyset$ since $\mathbb{Q} \subset A$. -Finally, by the Cantor Bendixson Theorem (Exercise 28 on page 45) which states that every uncountable closed set is (uniquely) the union of a perfec set and a countable. Hence there exists a perfect set $C'$ and a countable set $F$ such that $C = C' \cup F$. Since $C \cap \mathbb{Q} = \emptyset$, you also have $C' \cap \mathbb{Q} = \emptyset$. Hence $C'$ is a nonempty perfect set that does not contain any rational numbers. -By the way $C'$ can be constructed, look at the proof of the Cantor Bendixson Theorem to see exactly how it is made.<|endoftext|> -TITLE: How to compute the complex integral $\int_{\gamma} e^{\frac{1}{z^2 - 1}}\sin{\pi z} \, \mathrm dz$? -QUESTION [15 upvotes]: My Problem -I'm trying to solve problems from old qualifying exams in complex analysis and right now I'm stuck on the following exercise. - -Compute the complex integral -$$ -\int_{\gamma} e^{\frac{1}{z^2 - 1}}\sin{\pi z} \,\mathrm dz -$$ -where $\gamma$ is a closed curve in the right half plane that has index $N$ with respect to the point $1$. - - -My attempt -By the Residue theorem, since the function $f(z) := e^{\frac{1}{z^2 - 1}}\sin{\pi z}$ has only an isolated singularity at $z = 1$ in the right half plane, which in this case is an essential singularity, we can compute the integral by finding the residue at $1$, more precisely -$$ -\int_{\gamma} e^{\frac{1}{z^2 - 1}}\sin{\pi z} \, \mathrm dz = 2\pi i N \operatorname{Res}{(f(z); 1)} -$$ -where the $N$ comes from the assumption on the index of the curve. -Now my problem is that I can't compute this residue. The "obvious" things that I have tried are finding the Laurent expansions of the functions $e^{\frac{1}{z^2 - 1}}$ and $\sin{\pi z}$ around $z = 1$. It is easy to find that -$$ -\sin{\pi z} = -\frac{\pi}{1!}(z - 1) + \frac{\pi^3}{3!}(z - 1)^3 -\frac{\pi^5}{5!}(z - 1)^5 + \frac{\pi^7}{7!}(z - 1)^7 + \cdots -$$ -Then I tried doing the following with the exponential: -$$ -e^{\frac{1}{z^2 - 1}} = \sum_{n = 0}^{\infty}\frac{1}{n!}\frac{1}{(z^2 - 1)^n} -$$ -and I thought that maybe then expressing the fraction $\frac{1}{z^2 - 1}$ as -$$ -\frac{1}{z^2 - 1} = \frac{1}{2} \left ( \frac{1}{z-1} - \frac{1}{z+1} \right ) -$$ -and using the Laurent expansion -$$ -\frac{1}{z+ 1} = \frac{1}{z - 1 + 2} = \frac{1}{2}\frac{1}{1 + \frac{z - 1}{2}} = \frac{1}{2} \sum_{n = 0}^{\infty}\frac{(z - 1)^n}{2^n} -$$ -could be of help. -But now I don't see much hope of this working because I would have to put this last infinite series back into the series for the exponential and there's an $n$-th power there, and finally to top it all I would have to multiply by the Laurent series for the $\sin{\pi z}$ to try to get a hold of the coefficient of $\frac{1}{z - 1}$. - -I would really appreciate any help with this. Thanks. - -REPLY [3 votes]: Using your series expansions we can write: -$$ - e^{\frac1{z^2-1}} \sin \pi z = - \sum_{n,k \ge 0} (-1)^{k+1} \frac{\pi^{2k+1}}{(2k+1)!\,n!} - \frac1{(z+1)^n}\frac1{(z-1)^{n-2k-1}} -$$ -Now (for any $z$ inside the circle of radius 2 centered at $z=1$) -$$ - (z+1)^{-n} = - \sum_{l\ge0} \frac{(-n)(-n-1)\cdot\ldots\cdot(-n-l+1)}{l!}2^{-n-l} (z-1)^l -$$ -So in total we get -$$ - e^{\frac1{z^2-1}} \sin \pi z = - \sum_{n,k,l \ge 0} (-1)^{k+1} \frac{\pi^{2k+1}}{(2k+1)!\,n!} - {-n \choose l} \frac1{2^{n+l}} \frac1{(z-1)^{n-2k-l-1}} -$$ -The coefficient of $\frac1{z-1}$ is then -$$ - \sum_{n,k \ge 0} - \frac{(-1)^{k+1} \pi^{2k+1}}{(2k+1)!\,n!\,4^{n-k}} - {-n \choose n-2k} -$$ -There must be a nicer solution however....<|endoftext|> -TITLE: Connectedness of $\lbrace z\in\mathbb{C} : |z^2+az+b| -TITLE: What's the proof of correctness for Robert Floyd's algorithm for selecting a single, random combination of values? -QUESTION [10 upvotes]: I read about it in a SO answer: - -Algorithm to select a single, random combination of values? - -initialize set S to empty -for J := N-M + 1 to N do - T := RandInt(1, J) - if T is not in S then - insert T in S - else - insert J in S - -What is the proof of correctness for this? I've spent quite a bit of time on it and can prove to myself the $N^\text{th}$ value and the $1^\text{st}$ - $(N-M+1)^\text{th}$ values have $P$ (chosen) $= M/N$, but not the ones that remain. -e.g. -For $N$ choose $M$, each item enumerated from $1$ to $N$, I can prove to myself that using this algorithm, items $1$, $2$, $\ldots$, $N-M$, $N-M+1$, and $N$ each have the probability of $M/N$ to be chosen. I don't know for the other remaining items though. -e.g. 2 -For $8$ choose $4$, each item enumerated from $1$ to $8$, I can prove to myself that using this algorithm, items $1$, $2$, $3$, $4$, $5$, and $8$ each have the probability of $4/8$ to be chosen. - -REPLY [2 votes]: Base case -For $_{N}C_{M}$ where $M = 1$, T := RandInt(1, J) will execute with $J = N-M+1 = N$, randomly selecting one item $I$ of the $N$ items, as is expected for $_{N}C_{1}$. In other words, the trivial base case is $P_{chosen}(I_{i\in1..N}|_NC_1) = {1 \over N}$, which I'll denote using - -$P_{N,1}(I_i) = {1 \over N}$ - -Inductive hypothesis - -$P_{N,M}(I_i) = {M \over N}$ for choosing $M$ items from $N$ - -Inductive step -The inductive step is to solve for - -$P_{N',M'}(I_i)$ when $N'=N+1$ and $M'=M+1$ - -The inductive step is executed in the algorithm at every additional loop iteration. -For items $I_{i\in1..N}$ - -$P_{N',M'}(I_{i\in1..N}) = P_{N,M}(I_i) + \neg P_{N,M}(I_i) * {1 \over N'}$ -$P_{N', M'}(I_{i\in1..N})$ = "For items $I_i$ except $I_{N'}$ in the additional iteration" -$P_{N,M}(I_i)$ = "The chance of $I_i$ to have already been chosen earlier" -$\neg P_{N,M}(I_i)$ = "The chance of $I_i$ to have not yet been chosen earlier" -$1 \over N'$ = "The chance of $I_i$ to be chosen in the additional iteration by RandInt()" -Substitute in the hypothesis and simplify: -$P_{N',M'}(I_{i\in1..N}) = {M \over N} + (1-{M \over N}) * {1 \over N+1}$ -$P_{N',M'}(I_{i\in1..N}) = {M+1 \over N+1}$ - -For for the last item $I_{N'}$ added in the additional iteration - -$P_{N',M'}(I_{i=N+1}) = {1 \over N'} + {M \over N'} = {M+1 \over N+1}$ -${M \over N'}$ = "The chance that one of the $M$ items already chosen earlier is chosen again in the additional iteration" - -Now we know for all items $I_i$ in the additional iteration - -$P_{N',M'}(I_i) = {M+1 \over N+1} = {M' \over N'}$ - -which completes the inductive step. - -This proof works because for any $N'$ and $M'$, the inductive step reduces the problem to the sub-problem, $N=N'-1$ and $M=M'-1$. Eventually, $M$ will be reduced to $1$, giving us the base case, $_NC_1$. -Since the algorithm guarantees both - -$P(I)={M \over N}$ for all items $I$ -Exactly $M$ of $N$ items will be selected - -We know the algorithm correctly selects a random combination of $M$ items from a set of $N$ items.<|endoftext|> -TITLE: Evaluating $I(x)=\int_{0}^{\infty}\frac{e^{-xy}}{y^2+a^2}dy$ -QUESTION [5 upvotes]: I examine currently this integral: -$$I(x)=\int_{0}^{\infty}\frac{e^{-xy}}{y^2+a^2}dy;x\geqslant0$$ where $x$ and $a$ are real. -It seems that the integral has no closed form in terms of elementary functions. But perhaps it has a closed form in terms of special functions? -The end result should be series expansion of $I(x)$ at $x=0$ -Thanks! - -REPLY [10 votes]: Note that $$I''(x)=\int_0^\infty \frac{y^2 e^{-xy}}{y^2+a^2}dy$$ -so that $$\tag{1} I''(x)+a^2 I(x)= \int_0^\infty e^{-xy} dy=\frac 1x$$ -Solutions of the homogenous equation are $I(x)=\sin(ax)$ and $I(x)=\cos(ax)$ but specific solutions will require (as noticed by Norbert) sine and cosine integrals as we will show using variation of constants : -$$\tag{2} I(x)=c\;\sin(ax)$$ -$$I'(x)=c'\sin(ax)+c\;a\cos(ax)$$ -$$I''(x)=c''\sin(ax)+2\,c'a\cos(ax)-c\;a^2\sin(ax)$$ -$$I''(x)+a^2 I(x)=\frac 1x=c''\sin(ax)+2\,c'a\cos(ax)$$ -Let's set $\;b:=c'$ then we want $$b'\sin(ax)+2\,b\,a\cos(ax)=\frac 1x$$ -multiply this by $\,\sin(ax)$ to get : - $$b'\sin(ax)^2+b\,2\,a\sin(ax)\cos(ax)=\frac {\sin(ax)}x$$ - $$b'\sin(ax)^2+b \frac d{dx} \sin(ax)^2=\frac {\sin(ax)}x$$ - $$\frac d{dx} (b\;\sin(ax)^2)=\frac {\sin(ax)}x$$ -or $$c'=b=\frac 1{\sin(ax)^2} \left(C_0+\int \frac {\sin(ax)}{ax} d(ax)\right)=\frac {C_0+\rm{Si(ax)}}{\sin(ax)^2}$$ -and $c$ will be : -$$c=C_1+\int \frac {C_0+\rm{Si(ax)}}{\sin(ax)^2} dx$$ -but (int. by parts and since $\cot(x)'=-\frac 1{\sin(x)^2}$, $\rm{Si}'(x)=\frac {\sin(x)}x$ and $\rm{Ci}'(x)=\frac {\cos(x)}x$) : -$$\int \frac {\rm{Si}(ax)}{\sin(ax)^2} dx=\frac 1a\left[-\rm{Si}(ax)\cot(ax)\right]+\int \frac {\sin(ax)}{x} \cot(ax)dx$$ -$$\int \frac {\rm{Si}(ax)}{\sin(ax)^2} dx=-\frac 1a\left[\rm{Si}(ax)\cot(ax)\right]+\int \frac {\cos(ax)}{x}dx$$ -so that (for $C_0=C_1=0$) $c$ will simply be : -$$\tag{3} c=\frac{\rm{Ci}(ax)-\rm{Si}(ax)\cot(ax)}a$$ -multiplying $c$ by $\,\sin(ax)$ in $(2)$ we get the specific solution : -$$\tag{4} I(x)=\frac 1a\left(\rm{Ci}(ax)\sin(ax)-\rm{Si}(ax)\cos(ax)\right)$$ -with the general solution of the O.D.E. given by : -$$\tag{5} I(x)=C\,\sin(ax)+D\,\cos(ax)+\frac 1a\left(\rm{Ci}(ax)\sin(ax)-\rm{Si}(ax)\cos(ax)\right)$$ -You may use the specific case $I(0)=\left[\frac {\arctan(ax)}a\right]_0^\infty=\frac {\pi} -{2a}$ and the derivative $I'(0)=-\log(a)$ to find Norbert's expression : -$$\tag{6}\boxed{\displaystyle I(x)=\frac {\pi\cos(ax)}{2a}+\frac 1a\left(\rm{Ci}(ax)\sin(ax)-\rm{Si}(ax)\cos(ax)\right)}$$ -For numerical evaluation perhaps that the DLMF link or A&S' book will be helpful. -I got following series expansion of $\displaystyle (a\;I(x))$ (as explained by J.M. there is a logarithmic term coming from the $\rm{Ci}$ function that can't be expanded at $0$) : -$$\left[(\gamma+\ln(ax))\sin(ax)+\frac {\pi}2-(ax)-\frac{\pi}4 (ax)^2+\frac {11}{36}(ax)^3+\frac{\pi}{48}(ax)^4-\frac{137}{7200}(ax)^5+\rm{O}\left((ax)^6\right)\right]$$ -(with $\gamma\approx 0.5772156649$ the Euler constant) -To get more terms you may use following expansions : -$$\rm{Ci}(z)=\gamma+ \ln(z) +\sum_{n=1}^\infty \frac{(-1)^nz^{2n}}{(2n)(2n)!}$$ -$$\rm{Si}(z)=\sum_{n=0}^\infty \frac{(-1)^nz^{2n+1}}{(2n+1)(2n+1)!}$$ -P.S. It is quite interesting to notice that there is a reference to nearly the same integral in this other recent S.E. thread, the paper is Coffey 2012 'Certain logarithmic integrals, including solution of Monthly problem #tbd, zeta values, and expressions for the Stieltjes constants' where equation $(4.4)$ reads $\ \displaystyle I(k)\equiv \int_0^\infty \frac{e^{-(k-1)v}}{v^2+\pi^2}\,dv$.<|endoftext|> -TITLE: Can $(\Bbb{R}^2,+)$ be given the structure of a matrix Lie group? -QUESTION [10 upvotes]: I have an assignment problem that is coming from Brian Hall's book Lie Groups, Lie Algebras and Representations: An Elementary Introduction. - - -Suppose $G \subseteq GL(n_1;\Bbb{C})$ and $H \subset GL(n_2;\Bbb{C})$ are matrix Lie group and that $\Phi:G \to H$ is a Lie group homomorphism. Is the image of $G$ under $\Phi$ a matrix Lie group? - - -The definition of a matrix lie group that I have is a closed subgroup of $GL(n;\Bbb{C})$ for some $n$. I do not know about things like differentiable manifolds and the like. By a Lie group homomorphism, we mean a continuous group homomorphism from $G$ to $H$ (continuity is with respect to the usual topology coming from the euclidean metric). -Now an example I have in mind is the Heisenberg group $G$ that consists of all matrices of the form -$$\left\{\left(\begin{array}{ccc}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{array}\right) : a,b,c\in \Bbb{R} \right\}.$$ -I can get a surjective group homomorphism (which I believe is also continuous) from $G$ to $\Bbb{R}^2$ under addition simply by sending -$$\left(\begin{array}{ccc}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{array}\right) \to (a,c) \in \Bbb{R}^2.$$ -This would be a counterexample to the problem above if only I were to know how to show that $\Bbb{R}^2$ does not embed as a matrix lie group in $GL(n;\Bbb{C})$ for any $n \in \Bbb{N}$. The difficulty in showing this is it is not obvious if such an embedding exists or not. -For example $\Bbb{R}$ under addition surprisingly is a matrix lie group. I have a copy of $\Bbb{R}$ sitting inside of $GL(2;\Bbb{C})$, namely as the set of all matrices -$$\left\{ \left(\begin{array}{CC} 1 & a \\ 0 & 1 \end{array}\right) : a\in \Bbb{R} \right\}.$$ -How do I know if $(\Bbb{R}^2,+)$ can or cannot be given the structure of a matrix Lie group? If only I were to know representation theory, I believe this is asking if there exists a finite dimensional complex faithful representation of $\Bbb{R}^2$. -Thanks. - -REPLY [16 votes]: In fact $(\mathbb R^n,+)$ is a matrix Lie group for all $n$. Note that the map -$$\begin{pmatrix} -x_1\\ -\vdots\\ -x_n -\end{pmatrix}\mapsto \begin{pmatrix} -1 & 0 & \cdots & x_1\\ - & \ddots & & \vdots\\ -0 & \cdots & 1 & x_n\\ -0 & \cdots & 0 & 1 -\end{pmatrix}$$ -which sends a vector $x$ to the identity matrix with the zeroes of the last column replaced by coordinates of $x$, is an injective homomorphism from $(\mathbb R^n,+)$ to $GL(n+1,\mathbb C)$.<|endoftext|> -TITLE: Additive quotient group $\mathbb{Q}/\mathbb{Z}$ is isomorphic to the multiplicative group of roots of unity -QUESTION [11 upvotes]: I would like to prove that the additive quotient group $\mathbb{Q}/\mathbb{Z}$ is isomorphic to the multiplicative group of roots of unity. -Now every $X \in \mathbb{Q}/\mathbb{Z}$ is of the form $\frac{p}{q} + \mathbb{Z}$ for $0 \leq \frac{p}{q} < 1$ for a unique $\frac{p}{q} \in \mathbb{Q}.$ This suggest taking the map $f:\mathbb{Q}/\mathbb{Z} \mapsto C^{\times}$ defined with the rule $$f(\frac{p}{q} + \mathbb{Z}) = e^{\frac{2\pi i p}{q}}$$ where $\frac{p}{q}$ is the mentioned representative. -Somehow I have problems showing that this is a bijective function in a formal way. I suspect I do not know the properties of the complex roots of unity well enough. -Can someone point me out (perhaps with a hint) how to show that $f$ is injective and surjective? - -REPLY [5 votes]: To prove it is a bijection, one can use rather "primitive" methods. suppose that: -$f\left(\frac{p}{q} + \Bbb Z\right) = f\left(\frac{p'}{q'} + \Bbb Z\right)$, -then: $e^{2\pi ip/q} = e^{2\pi ip'/q'}$, so $e^{2\pi i(p/q - p'/q')} = 1$. -This, in turn, means that $\frac{p}{q} - \frac{p'}{q'} \in \Bbb Z$, so the cosets are equal. Hence $f$ is injective. -On the other hand, if $e^{2\pi i p/q}$ is any $q$-th root of unity, it clearly has the pre-image $\frac{p}{q} + \Bbb Z$ in $\Bbb Q/\Bbb Z$ (so $f$ is surjective). -One caveat, however. You haven't actually demonstrated $f$ is a function (i.e., that it is well-defined, although if you stare hard at the preceding, I'm sure it will come to you).<|endoftext|> -TITLE: How does a Class group measure the failure of Unique factorization? -QUESTION [16 upvotes]: I have been stuck with a severe problem from last few days. I have developed some intuition for my-self in understanding the class group, but I lost the track of it in my brain. So I am now facing a hell. -The Class group is given by $\rm{Cl}(F)=$ {Fractional Ideals of F} / {Principle fractional Ideals of F} , ($F$ is a quadratic number field) so that we are actually removing the Principal fractional ideals there (that's what I understood by quotient group). But how can that class group measure the failure of Unique Factorization ? -For example a common example that can be found in any text books is $\mathbb{Z[\sqrt{-5}]}$ -in which we can factorize $6=2\cdot3=(1+\sqrt{-5})(1-\sqrt{-5})$. So it fails to have unique factorization. Now can someone kindly clarify these points ? - -How can one construct $\rm{Cl}(\sqrt{-5})$ by using the quotient groups ? -What are the elements of $\rm{Cl}(\sqrt{-5})$ ? What do those elements indicate ? ( I think they must some-how indicate the residues that are preventing the $\mathbb{Z[\sqrt{-5}]}$ from having a unique factorization ) -What does $h(n)$ indicate ? ( Class number ). When $h(n)=1$ it implies that unique factorization exists . But what does the $1$ in $h(n)=1$ indicate. It means that there is one element in the class group , but doesn't that prevent Unique Factorization ? - -EDIT: - -I am interested in knowing whether are there any polynomial time running algorithms that list out all the numbers that fail to hold the unique factorization with in a number field ? - -I am expecting that may be Class group might have something to do with these things. By using the class group of a number field can we extract all such numbers ? For example, if we plug in $\mathbb{Z}[\sqrt{-5}]$ then we need to have $6$ and other numbers that don't admit to a unique factorization. -Please do answer the above points and save me from confusion . -Thank you. - -REPLY [5 votes]: First of all, I want to clarify one thing: taking the quotient by the subgroup of principal ideals is not the same thing as removing the subgroup. It means that two fractional ideals $\frak{A}$ and $\frak{B}$ are equivalent iff $\frak{A}^{-1}\frak{B}$ is a principal ideal. -In general, if one wants to compute the class group of a number field, the first thing you might want to do is compute the class number; a good way to tackle this is to first compute an upper bound for the class number. For this purpose, you can use Minkowski's bound: -$$M_F:=\sqrt{|D|}\left(\dfrac{4}{\pi}\right)^{r_2}\dfrac{n!}{n^n},$$ -where $F$ is a number field of degree $n$ over $\mathbb{Q}$, $D$ is the discriminant, $r_2$ is half the number of complex embeddings. In general, every class in the class group contains an ideal of norm at most $M_F$, and so the class group is generated by the prime ideals of norm at most $M_F$. By studying the splitting of the rational prime ideals in $F$, you can deduce a lot of information about the class group. To see example of computations, I highly recommend this note by Keith Conrad. - -Here are two ways of thinking about a Dedekind domain with class number equal to 1. First, for Dedekind domains, being a Unique factorisation domain (UFD) is equivalent to being a Principal ideal domain (PID) (this basically follows from the fact that Dedekind domains are Noetherian integral domains in which every nonzero prime ideal is maximal). Therefore, a Dedekind domain has a trivial class group if and only if every ideal is principal, if and only if it is a UFD. Second, we know that in Dedekind domains, every fractional ideal has a unique factorisation into prime ideals. Hence, you can think of the class group as being a comparison between unique factorisation of fractional ideals and unique factorisation per se.<|endoftext|> -TITLE: Evaluate $\lim_{n \to \infty }\frac{(n!)^{1/n}}{n}$. -QUESTION [7 upvotes]: Possible Duplicate: -Finding the limit of $\frac {n}{\sqrt[n]{n!}}$ - -Evaluate -$$\lim_{n \to \infty }\frac{(n!)^{1/n}}{n}.$$ -Can anyone help me with this? I have no idea how to start with. Thank you. - -REPLY [7 votes]: Let's work it out elementarily by wisely applying Cauchy-d'Alembert criterion: -$$\lim_{n\to\infty} \frac{n!^{\frac{1}{n}}}{n}=\lim_{n\to\infty}\left(\frac{n!}{n^n}\right)^{\frac{1}{n}} = \lim_{n\to\infty} \frac{(n+1)!}{(n+1)^{(n+1)}}\cdot \frac{n^{n}}{n!} = \lim_{n\to\infty} \frac{n^{n}}{(n+1)^{n}} =\lim_{n\to\infty} \frac{1}{\left(1+\frac{1}{n}\right)^{n}}=\frac{1}{e}. $$ -Also notice that by applying Stolz–Cesàro theorem you get the celebre limit: -$$\lim_{n\to\infty} (n+1)!^{\frac{1}{n+1}} - (n)!^{\frac{1}{n}} = \frac{1}{e}.$$ -The sequence $L_{n} = (n+1)!^{\frac{1}{n+1}} - (n)!^{\frac{1}{n}}$ is called Lalescu sequence, after the name of a great Romanian mathematician, Traian Lalescu. -Q.E.D.<|endoftext|> -TITLE: Show $\int_{0}^{\frac{\pi}{2}}\frac{x^{2}}{x^{2}+\ln^{2}(2\cos(x))}dx=\frac{\pi}{8}\left(1-\gamma+\ln(2\pi)\right)$ -QUESTION [40 upvotes]: Here is an interesting, albeit tough, integral I ran across. It has an interesting solution which leads me to think it is doable. But, what would be a good strategy?. -$$\int_{0}^{\frac{\pi}{2}}\frac{x^{2}}{x^{2}+\ln^{2}(2\cos(x))}dx=\frac{\pi}{8}\left(1-\gamma+\ln(2\pi)\right)$$ -This looks rough. What would be a good start?. I tried various subs in order to get it into some sort of shape to use series, LaPlace, something, but made no real progress. -I even tried writing a geometric series. But that didn't really result in anything encouraging. -$$\int_{0}^{\frac{\pi}{2}}\sum_{n=0}^{\infty}(-1)^{k}\left(\frac{\ln(2\cos(x))}{x}\right)^{2k}$$ -Thanks all. - -REPLY [26 votes]: In addition to the nice set of references by Raymond Manzoni, here is my proof of the identity. Frankly, I have not seen these references yet, thus I am not sure if this already appears in one of them. -Here I refer to the following identity -$$ \binom{\alpha}{\omega} = \frac{1}{2\pi} \int_{-\pi}^{\pi} \left(1 + e^{i\theta}\right)^{\alpha} e^{-i\omega \theta} \; d\theta, \ \cdots \ (1) $$ -whose proof can be found in my blog post. -Now let $x$ be a real number such that $|x| < \frac{\pi}{2}$. Then simple calculation shows that -$$ \log\left(1+e^{2ix}\right) = \log(2\cos x) + ix \quad \Longleftrightarrow \quad \Im \left( \frac{-x}{\log\left(1+e^{2ix}\right)} \right) = \frac{x^2}{x^2 + \log^2(2\cos x)},$$ -hence we have -$$ \begin{align*}I -&:= \int_{0}^{\frac{\pi}{2}} \frac{x^2}{x^2 + \log^2(2\cos x)} \; dx -= -\int_{0}^{\frac{\pi}{2}} \Im \left( \frac{x}{\log\left(1+e^{2ix}\right)} \right) \; dx \\ -&= -\frac{1}{8}\int_{-\pi}^{\pi} \Im \left( \frac{\theta}{\log\left(1+e^{i\theta}\right)} \right) \; d\theta -= \frac{1}{8}\Re \left( \int_{-\pi}^{\pi} \frac{i\theta}{\log\left(1+e^{i\theta}\right)} \; d\theta \right). -\end{align*}$$ -Differentiating both sides of $(1)$ with respect to $\omega$ and plugging $\omega = 1$, we have -$$ \frac{1}{2\pi} \int_{-\pi}^{\pi} (-i\theta) \left(1 + e^{i\theta}\right)^{\alpha} e^{-i\theta} \; d\theta = \alpha \left(\psi_0(\alpha) - \psi_0(2)\right). $$ -Now integrating both sides with respect to $\alpha$ on $[0, 1]$, -$$ \begin{align*} --\frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{i\theta}{\log \left(1 + e^{i\theta}\right)} \; d\theta -&= \int_{0}^{1} \alpha \left(\psi_0(\alpha) - \psi_0(2)\right) \; d\alpha \\ -&= \left[ \alpha \log \Gamma (\alpha) \right]_{0}^{1} - \int_{0}^{1} \log \Gamma (\alpha) \; d\alpha - \frac{1}{2}\psi_0(2) \\ -&= -\frac{1}{2}\left( 1 - \gamma + \log (2\pi) \right), -\end{align*}$$ -where we have used the fact that -$$ \psi_0 (1+n) = -\gamma + H_n, \quad n \in \mathbb{N}$$ -and -$$ \begin{align*} -\int_{0}^{1} \log \Gamma (\alpha) \; d\alpha -& = \frac{1}{2} \int_{0}^{1} \log \left[ \Gamma (\alpha) \Gamma (1-\alpha) \right] \; d\alpha \\ -&= \frac{1}{2} \int_{0}^{1} \log \left( \frac{\pi}{\sin \pi \alpha} \right) \; d\alpha \\ -&= \frac{1}{2} \left( \log \pi - \int_{0}^{1} \log \sin \pi \alpha \; d\alpha \right) \\ -&= \frac{1}{2} \log (2\pi). -\end{align*} $$ -Therefore we have the desired result.<|endoftext|> -TITLE: What is the practical benefit of a function being injective? surjective? -QUESTION [7 upvotes]: I have learned that an injective function is one such that no two elements of the domain map to the same value in the codomain. -Example: The function $x \mapsto x^2$ is injective when the domain is positive numbers but is not injective when the domain is all numbers because both $(-2)^2$ and $2^2$ map to the same value, $4$. -Is there an advantage of an injective function over a non-injective function? What is the practical benefit of injective functions? Or perhaps there is an advantage to a function not being injective? -I have also learned about surjective functions: a surjective function is one such that for each element in the codomain there is at least one element in the domain that maps to it. -What is the benefit of a function being surjective? Is there a danger to using functions that are not surjective? -I'd like to have a deeper understanding of injective and surjective than simply be able to parrot back their definitions. Please help. - -REPLY [4 votes]: If you are studying algebraic structures (let's say rings) an injective homomorphism $\phi: R \longrightarrow S$ provides you with a way to interpret one ring as a subring of the other. After all, this just means that $R$ will be isomorphic to $\mathrm{Im}(\phi)$. For example, let $\phi: \mathbb{Q} \longrightarrow \mathbb{Q}[X]$ be defined by $\phi(a) = a + 0x + 0x^{2} + \cdots$. One can then check (you check!) that $\phi$ is injective and a homomorphism. -And indeed, $\mathbb{Q}$ can be interpreted as a subsystem of the ring of polynomials with coefficients in $\mathbb{Q}$, just by viewing each rational as a constant polynomial.<|endoftext|> -TITLE: Understanding why the roots of homogeneous difference equation must be eigenvalues -QUESTION [6 upvotes]: There is some obvious relationship between the root solutions to a homogeneous difference equation (as a recurrence relation) and eigenvalues which I'm trying to see. I have read over the wiki article 3.2, 3.4 and the eigenvalues ($\lambda$ ) are hinted at as the roots, but I'm still not sure why these must be eigenvalues of some matrix, say $A_0$, and what the meaning of $A_0$ may be. -It seems that to solve a homogeneous linear difference equation we find the "characteristic polynomial" by simply factoring one difference equation. However, typically to find the "characteristic polynomial" I would solve the characteristic equation for some matrix, -$A_0 = \begin{pmatrix} 1 & 0 & 0\\ -0 & -2 & 0 \\ -0 & 0 & 3 \\ -\end{pmatrix}$ -$(A_0 - \lambda I)\mathbf x = \mathbf 0$, -then solve for the determinant equal to $0$, and then solve for each $\lambda$ e.g. -$ \det(A_0 - \lambda I) = 0$ -$(1 - \lambda)(2 + \lambda)(3 - \lambda) = 0$ -Now suppose this also happens to be a solution to some linear difference equation, and so here the characteristic polynomial is $\lambda^3 - 2\lambda^2 - 5\lambda + 6 = 0$, and the difference equation is. -$y_{k+3} - 2y_{k+2} - 5y_{k+1} + 6y_k = 0 $. Then, for example, $\lambda = 3$ is a solution for all k. -Now, given we have found this solution to this difference equation, how can we explain some special relationship to $A_0$, other than $\lambda = 3$ happens to be an eigenvalue of $A_0$? Is there any meaning to make of $A_0$? -(cf. 4.8, Linear Algebra 4th, D. Lay) - -REPLY [2 votes]: The roots are eigenvalues. What is the operator they are eigenvalues of? It is the shift operator. -Consider the vector space $V$ of sequences $a_0, a_1, a_2, ...$ (say of complex numbers). Define the left shift operator -$$S(a)_i = a_{i+1}.$$ -In other words, $S$ takes input a sequence $a_0, a_1, a_2, ...$ and returns the sequence $a_1, a_2, a_3, ...$. Now we need the following three fundamental observations. -Observation 1: The linear homogeneous recurrence with characteristic polynomial $p$ is precisely described by the equation $p(S) a = 0$. In other words, the sequences satisfying such a linear recurrence are precisely the sequences in the kernel of $p(S)$. -Observation 1.5: The shift operator $S$ acts on the space of solutions to any equation of the form $p(S)a = 0$. -Observation 2: Over the complex numbers, we can factor $p(S) = \prod (S - \lambda_i)$. Thus if $(S - \lambda_i) a = 0$, or equivalently if $a$ is an eigenvector of $S$ with eigenvalue $\lambda_i$, then $p(S) a = 0$. -Observation 3: $(S - \lambda_i) a = 0$ if and only if $a_n = c \lambda_i^n$ for some constant $c$. -$S$ resembles in some sense the differentiation operator, which it is sent to if one sends $V$ to the space of formal power series via -$$(a_0, a_1, a_2, ...) \mapsto \sum_{n=0}^{\infty} \frac{a_n}{n!} x^n.$$ -Thus everything above applies (at least formally) to linear homogeneous ODEs with constant coefficients and their characteristic polynomials as well. The corresponding eigenvectors are the functions $e^{\lambda_i x}$.<|endoftext|> -TITLE: What is the chance of an event happening a set number of times or more after a number of trials? -QUESTION [5 upvotes]: Assuming every trial is independent from all the others and the probability of a successful run is the same every trial, how can you determine the chance of a successful trial a set number of times or more? -For example, You run 20 independent trials and the chance of a "successful" independent trial each time is 60%. how would you determine the chance of 3 or more"successful" trials? - -REPLY [5 votes]: If the probability of success on any trial is $p$, then the probability of exactly $k$ successes in $n$ trials is -$$\binom{n}{k}p^k(1-p)^{n-k}.$$ -For details, look for the Binomial Distribution on Wikipedia. -So to calculate the probability of $3$ or more successes in your example, let $p=0.60$ and $n=20$. Then calculate the probabilities that $k=3$, $k=4$, and so on up to $k=20$ using the above formula, and add up. -A lot of work! It is much easier in this case to find the probability of $2$ or fewer successes by using the above formula, and subtracting the answer from $1$. So, with $p=0.60$, the probability of $3$ or more successes is -$$1-\left(\binom{20}{0}p^0(1-p)^{20}+\binom{20}{1}p(1-p)^{19}+\binom{20}{2}p^2(1-p)^{18} \right).$$ -For the calculations, note that $\binom{n}{k}=\frac{n!}{k!(n-k)!}$. In particular, $\binom{20}{0}=1$, $\binom{20}{1}=20$ and $\binom{20}{2}=\frac{(20)(19)}{2!}=190$.<|endoftext|> -TITLE: Eisenstein Series and cusps of $\Gamma_{1}(N)$ -QUESTION [11 upvotes]: Let $q = e^{2\pi i\tau}$, $\operatorname{Im}\tau > 0$ and let $$G(\tau) = 1 - 24\sum_{n = 1}^{\infty}\frac{nq^{n}}{1 - q^{n}}$$ be the weight 2 Eisenstein series for $\Gamma(1)$. Consider the inequivalent cusps of $\Gamma_{1}(N)$. For each cusp, there is an Eisenstein series of weight 2 associated to this cusp. How does one relate this Eisenstein series to a linear combination of $G(k\tau)$'s (for some integers $k$)? -In particular, the case I am considering is when $N = 7$. The inequivalent cusps are $0$, $2/7$, $1/3$, $3/7$, $1/2$, and $\infty$. Consider the cusp $2/7$ and denote by $E_{2/7}$ the associated Eisenstein series. I would like to write $E_{2/7}$ as a finite sum of $G(k\tau)$ for some integers $k$. One of the first issues I'm having is how would I know the $q$-expansions of Eisenstein series of $E_{2/7}$ (which is related to the issue that I can't seem to find a good reference for the definition of a weight 2 Eisenstein series associated to a cusp)? Second, would finding the desired linear combination of $G(k\tau)$'s be just an exercise in matching up finitely many coefficients in the $q$-expansion? - -REPLY [8 votes]: The question has some implicit hypotheses, possibly not clear to the questioner, and this implicitness and ambiguities about it complicate matters. First, the more natural descriptions of Eisenstein series for GL(2) of weights k>2 are not of the form in the question, but are $\sum_{c,d} 1/(cz+d)^k$. This does not converge for $k=2$, so $k=2$ has to be approached more delicately (via an analytic continuation, Hecke summation, producing _something_like_ the expression in the question). However, at level one, that is, for $SL_2(\mathbb Z)$, there is no truly-holomorphic Eisenstein series of weight $2$. The analytic continuation has an extra term, which one may discard, but then destroying the literal automorphy condition. Maybe that doesn't matter, but one should be careful about "understandings". -Thus, depending what one means, wants, or needs, while at higher levels the meromorphic continuation can produce holomorphic modular forms at level 2. (This positive outcome always occurs for Hilbert modular forms, that is, for ground fields totally real anything other than $\mathbb Q$.) First, whatever description one chooses for "Eisenstein series" (attached to cusps?), a suitable weighted average of the level-7 (for example) such should be level-one. A literal notion of holomorphy tells us there is no level-one, weight-two such. Thus, there are (at most) six linearly independent Eisenstein series at that level, so not quite possible to "attach" one to each cusp. It is not hard to say more. -Then there is the further issue of expressing various Eisenstein series in terms of each other, by the group action. At square-free level, the underlying (!) representation theory is simpler (Iwahori-fixed vectors in principal series are well understood, at least up to a very useful point.) -But, at this point, without knowing more precisely what the questioner wants, or may discover is wanted, there are too many things that can be said to know which to choose to say. :)<|endoftext|> -TITLE: Classifing groups of order 56: problems with the semidirect product -QUESTION [10 upvotes]: While I was doing an exercise about the classification of groups of order 56, I had some problems concerning the semidirect product. -Let $G$ a group of order 56 and let us suppose that the 7-Sylow is normal (let's call it $H$). Then we want to construct the non abelian group whose 2-Sylow $S$ is $S \cong \mathbb Z_8$. -First of all, we have to determine the homomorphism $\phi \colon \mathbb Z_8 \to \text{Aut}(\mathbb Z_7)$. -It is known that $\text{Aut}(\mathbb Z_7) \cong \mathbb Z_6$ and the isomorphism is given by -$$ -\begin{split} -& \mathbb Z_6 \to \text{Aut}(\mathbb Z_7) \\ -& a \mapsto \psi_a \colon \mathbb Z_7 \ni n \mapsto an \in \mathbb Z_7 -\end{split} -$$ -So we can start by finding the homomorphism $\mathbb Z_8 \to \mathbb Z_6$. There are exacly $(6,8)=2$ such homomorphism. Who are they? Simply the one who sends everything to $0$ and the multiplication by $3$ (which is the only element in $\mathbb Z_6$ whose order - 2 - divides 8). In multiplicative terms, they are the homomorphism which sends everything to $1$ and the homomorphism which sends $n \mapsto 6^n=(-1)^n$. -So, by composition, we have the two homomorphism -$$ -\begin{split} -\phi_1 \colon & \mathbb Z_8 \to \text{Aut}(\mathbb Z_7)\\ -& n \mapsto \text{id} -\end{split} -$$ -and -$$ -\begin{split} -\phi_2 \colon & \mathbb Z_8 \to \text{Aut}(\mathbb Z_7)\\ -& n \mapsto \psi_n \colon \mathbb Z_7 \ni x \mapsto 6^nx = (-1)^nx \in \mathbb Z_7 -\end{split} -$$ -Am I right? -Now, if we take $\phi_1$ we simply get the direct product. What if we take $\phi_2$? -For sake of simplicity, let's assume additive notation (this is stupid, I know but it has helped me somehow to understand). If I'm not wrong, we obtain that $H \rtimes_{\phi_2} \mathbb Z_8$ is the set $H \times \mathbb Z_8$ with the operation given by -$$ -(a,b) + (c,d) = (a+(-1)^bc,b+d) -$$ -Now if I do $(0,k)+(h,0)-(0,k) = ((-1)^k h, 0) = \phi_k(h)$ which is exactly what I want. -Now I must pass to the much more confortable multiplicative notation: so let's $C_7=\langle s \rangle$ and $C_8=\langle r \rangle$ be the cyclic groups of order 7 and 8. Then we define the automorphisms -$$ -\begin{split} -\phi_n \colon & C_7 \to C_7 \\ -& x \mapsto x^{(-1)^n} -\end{split} -$$ -and the homomorphism -$$ -\begin{split} -\psi \colon & C_8 \to \text{Aut}(C_7) \\ -& n \mapsto \phi_n -\end{split} -$$ -In other words, we can simply say that $\psi$ is the homomorphism which sends the generator $r$ to the inversion $x^{-1}$. Am I right so far? -Well, now $C_7 \rtimes_{\psi} C_8$ is the set $C_7 \times C_8$ with the operation given by -$$ -(a,b)(c,d) = (ac^{(-1)^b},bd) -$$ -I do again the calculation $(1,k)(h,1)(1,k)^{-1}=(h^{(-1)^k},1) = \phi_k(h)$ which is exactly what I want (also according to ineff's answer). -Are there any mistakes? -May I ask one more question? Who is this mysterious group I've built up? Is it isomorphic to some other (simpler) group? How can I do to write down a presentation? -I thank you in advance for your kind help. - -REPLY [4 votes]: I suppose your confusion is due to the double representation of the semidirect product: internal vs external. -The semidirect product's theorem states that if you have a group $G$ having two subgroups $H,K < G$ such that $H$ is normal in $G$, $H \cap K = \{1_G\}$ and $G=HK$ then there's an isomorphism $G \cong H \rtimes_\psi K$, for a certain $\psi \colon K \to \text{Aut}(H)$. -By the theorem we can represent every element of $G$ as a pair $(h,k) \in H \times K$ (which is the support of the group $H \rtimes K$). -Consider the two subgroups $\bar H = \{(h,1_K) | h \in H\} \leq H \rtimes K$ and $\bar K = \{(1_H,k)|k \in K\} \leq H \rtimes K$, these subgroups correspond, via the isomorphism, to the subgroups $H$ and $K$ of $G$. -In $H \rtimes K$ we have that for every $h \in H$ and $k \in K$ -$$(1_H,k) * (h,1_K) *(1_H,k)^{-1} = (\psi_k(h),1_K)$$ -if we identify every $h \in H$ with its corresponding element $(h,1_K)$ and every $k \in K$ with $(1_H,k)$ then this equality become (internally in $G$) -$$k*h*k^{-1}=\psi_k(h)$$ -The $\psi$ which determine the operation in the semidirect product is exactly the homomorphism sending every $k \in K$ in the (restriction to $H$ of the) automorphism $\psi_k \colon H \to H$ which send $h \in H$ in $khk^{-1}$ (this is clearly well defined because $H$ is normal in $G$. -Hope this help.<|endoftext|> -TITLE: Is every nilpotent linear group triangularizable? -QUESTION [5 upvotes]: Is it true that every finitely generated nilpotent group of matrices over $\mathbb C$ is conjugated to a subgroup of the upper triangular group? -If yes, what is a reference for that? - -REPLY [3 votes]: I realized that the answer to my question is NO. Take the quaternion group, realized by -$\pm I, \pm \left(\begin{matrix} i & 0 \\ 0 & -i \end{matrix}\right), \pm \left(\begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix}\right), \pm \left(\begin{matrix} 0 & i \\ i & 0 \end{matrix}\right).$ It is nilpotent but not triangularizable.<|endoftext|> -TITLE: Eigenvalues, singular values, and the angles between eigenvectors -QUESTION [5 upvotes]: Suppose the $n \times n$ matrix $A$ has eigenvalues $\lambda_1, \ldots, \lambda_n$ and singular values $\sigma_1, \ldots, \sigma_n$. It seems plausible that by comparing the singular values and eigenvalues we gets some sort of information about eigenvectors. Consider: -a. The singular values are equal to the absolute values of eigenvalues if and only if the matrix is normal, i.e., the eigenvectors are orthogonal (see http://en.wikipedia.org/wiki/Normal_matrix , item 11 of the "Equivalent definitions" section ). -b. Suppose we have two distinct eigenvalues $\lambda_1, \lambda_2$ with eigenvectors $v_1, v_2$. Suppose, hypothetically, we let $v_1$ approach $v_2$, while keeping all the other eigenvalues and eigenvectors the same. Then the largest singular value approaches infinity. This follows since $\sigma_{\rm max} = ||A||_2$ and $A$ maps the vector $v_1 - v_2$, which approaches $0$, to $\lambda_1 v_1 - \lambda_2 v_2$, which does not approach $0$. -It seems reasonable to guess that the ``more equal'' $|\lambda_1|, \ldots, |\lambda_n|$ and $\sigma_1, \ldots, \sigma_n$ are, the more the eigenvectors look like an orthogonal collection. So naturally my question is whether there is a formal statement to this effect. - -REPLY [2 votes]: A little help in understanding the relation or difference between eigenvectors and singular vectors is elaborated in detail in: SVD vs EVD - -The bookmark leads you to a comparison between evd and svd. But there's much more there if you browse through the pages. -EngleField and Farr show an elegant derivation for the angle between eigenvectors: -Englefield, M. J., & Farr, G. E. (2006). Eigencircles of 2 x 2 Matrices. Mathematics Magazine Vol. 79 Oct.,2006, 281-289. on page 285 -The angle between the singular vectors is the angle of the rotation $UV^T$ where $U$ and $V$ contain the singular vectors of $ A = U \Sigma V^T$ -Eigencircle representation leads you to a figure showing this angle on an eigencircle plot.<|endoftext|> -TITLE: Local solutions of a Diophantine equation -QUESTION [11 upvotes]: I am trying to prove that the equation -$$3x^3 + 4y^3 +5z^3 \equiv 0 \pmod{p}$$ -has a non-trivial solution for all primes $p$. -I am sure that this is a standard exercise, and I have done the easy parts: treating $p=2, 3, 5$ as special cases (very simple), and then for $p\geq 7$, those for which $p \equiv 2 \pmod{3}$ is also straightforward, as everything is a cubic residue $\pmod{p}$, but I am having a mental block about the remaining cases where $p \equiv 1 \pmod{3}$ and only $(p-1)/3$ of the integers $\pmod{p}$ are cubic residues. -I was hoping to be able to show that the original equation has non-trivial solutions in $\mathbb{Q}_p$, and that this might be an easy first step towards the $p$-adic case. -Any pointers, or references to a proof (I am sure there must be some in the literature) would be most gratefully received. - -REPLY [3 votes]: I would like to tell you about my approach to this problem; clearly enough, it can easily be adapted to yield an even more general result. -Let $p \equiv 1 \, \, (\mathrm{mod} \, \, 3)$ be a prime number and let us denote with $\mathbf{J}$ the number of solutions of the congruence $3x^{3}+4y^{3}+5y^{3} \equiv 0 \, \, (\mathrm{mod} \, \, p)$. Then , by basic properties of the complex exponential function it follows that -\begin{eqnarray*} -\mathbf{J} &=& \frac{1}{p} \sum_{x=0}^{p-1} \sum_{y=0}^{p-1} \sum_{z=0}^{p-1} \sum_{\lambda=0}^{p-1} e^{2\pi i \frac{3x^{3}+4y^{3}+5z^{3}}{p} \lambda} \\ -&=& \frac{1}{p} \sum_{\lambda=0}^{p-1} \left(\sum_{x=0}^{p-1}e^{2\pi i \frac{3\lambda}{p}x^{3}}\right) \left(\sum_{y=0}^{p-1}e^{2\pi i \frac{4 \lambda}{p}y^{3}}\right) -\left(\sum_{z=0}^{p-1}e^{2\pi i \frac{5 \lambda}{p}z^{3}}\right)\\ -&=& p^{2} + \frac{1}{p} \sum_{\lambda=1}^{p-1} \left(\sum_{x=0}^{p-1}e^{2\pi i \frac{3\lambda}{p}x^{3}}\right) \left(\sum_{y=0}^{p-1}e^{2\pi i \frac{4 \lambda}{p}y^{3}}\right) -\left(\sum_{z=0}^{p-1}e^{2\pi i \frac{5 \lambda}{p}z^{3}}\right). -\end{eqnarray*} -Appealing to the well-known estimate $ \left| \sum_{x=0}^{p-1} e^{2\pi i \frac{\omega}{p}x^{3}}\right| \leq 2 \sqrt{p}$ (which is valid for every $\omega$ coprime with $p$), we obtain that -$$ \mathbf{J} > p^{2}-8p^{1.5}.$$ -Since $p^{2}-8p^{1.5}>1$ for every $p \geq 65$, the problem has been reduced to verifying that the given congruence has a non-trivial solution for every $p \in E:=\{7, 13, 19, 31, 37, 43, 61\}$. I list below a non-trivial solution for the corresponding congruence for every $p \in E$ (even though at first sight they may strike you as having been chosen without rhyme or reason, they all were obtained by resorting to an idea of the Princeps Mathematicorum): -\begin{eqnarray*} -p&=&07: \quad (x=3, y=6, z=0)\\ -p&=&13: \quad (x=1, y=4, z=5)\\ -p&=&19: \quad (x=13,y=0, z=3)\\ -p&=&31: \quad (x=3, y=2, z=3)\\ -p&=&37: \quad (x=1, y=4, z=0)\\ -p&=&43: \quad (x=25,y=0, z=42)\\ -p&=&61: \quad (x=46,y=32, z=41)\\ -\end{eqnarray*} -QED.<|endoftext|> -TITLE: Combinations of characteristic functions: $\alpha\phi_1+(1-\alpha)\phi_2$ -QUESTION [8 upvotes]: Suppose we are given two characteristic functions: $\phi_1,\phi_2$ and I want to take a weighted average of them as below: -$\alpha\phi_1+(1-\alpha)\phi_2$ for any $\alpha\in [0,1]$ -Can it be proven that the result is also a characteristic function? If so, I am guessing this result could extend to any number of combinations $\alpha_i$ as long as $\sum_i\alpha_i=1$ - -Secondly, if $\phi$ is again a characteristic function, then $\mathfrak{R}e\phi(t)=\frac12(\phi(t)+\phi(-t))$ is also a characteristic function. I don't even know how to begin attempting this proof as I am not sure what the $\mathfrak{R}$ represents. - -Lastly, regarding the symmetry of characteristic functions, -$\phi$ is symmetric about zero iff it is real-valued iff the corresponding distribution is symmetric about zero. -Once again, my lack of familiarity with the complex plane leaves me in the dark here. Why can a complex-valued function not be symmetric about zero? - -REPLY [12 votes]: To prove that these are characteristic functions, using random variables yields simpler, more intuitive, proofs. -In the first case, assume that $\phi_1(t)=\mathrm E(\mathrm e^{itX_1})$ and $\phi_2(t)=\mathrm E(\mathrm e^{itX_2})$ for some random variables $X_1$ and $X_2$ defined on the same probability space and introduce a Bernoulli random variable $A$ such that $\mathrm P(A=1)=\alpha$ and $\mathrm P(A=0)=1-\alpha$, independent of $X_1$ and $X_2$. Then: - -The function $\alpha\phi_1+(1-\alpha)\phi_2$ is the characteristic function of the random variable $AX_1+(1-A)X_2$. - -The extension to more than two random variables is direct. Assume that $\phi_k(t)=\mathrm E(\mathrm e^{itX_k})$ for every $k$, for some random variables $X_k$ defined on the same probability space and introduce an integer valued random variable $A$ such that $\mathrm P(A=k)=\alpha_k$ for every $k$, independent of $(X_k)_k$. Then: - -The function $\sum\limits_k\alpha_k\phi_k$ is the characteristic function of the random variable $X_A=\sum\limits_kX_k\mathbf 1_{A=k}$. - -In the second case, assume that $\phi(t)=\mathrm E(\mathrm e^{itX})$ for some random variable $X$ and introduce a Bernoulli random variable $A$ such that $\mathrm P(A=1)=\mathrm P(A=-1)=\frac12$, independent of $X$. Then: - -The function $t\mapsto\frac12(\phi(t)+\phi(-t))$ is the characteristic function of the random variable $AX$.<|endoftext|> -TITLE: Newton's method - finding suitable starting point -QUESTION [6 upvotes]: I have some trouble solving a problem in my textbook: - -Given the following function: $$f(x) = x^{-1} - R$$ - Assume $R > 0$. Write a short algorithm to find $1/R$ by Newton's method applied to $f$. Do not use division or exponentiation in the algorithm. For positive $R$, what starting points are suitable? - -OK, so I've managed to solve the first part of the problem. Using Newton's method and rearranging terms I have gotten: -$$x_{n+1} = x_{n}(2 - Rx_{n})$$ -This is correct according to the book, and I can just use my standard algorithm for Newton's method with this expression. So far so good. -The problem is the second part, where I am supposed to find suitable starting points. I figured that if $x_{1} = -x_{0}$, then the iterations cycle. So then I get: -$$\begin{align*} --x_{0} &= x_{0}(2 - Rx_{0})\\ --3x_{0} &= - Rx_{0}^2\\ --3 &= -Rx_{0}\\ -x_{0} &= 3/R -\end{align*}$$ -Thus my answer would be that we must have $x_{0} < 3/R$. My book, however, says: -If $R > 1$ then the starting point $x_{0}$ should be close to $0$ but smaller than $1/R$. -So what is wrong with my reasoning here? If anyone can help me out, I would really appreciate it! - -REPLY [4 votes]: The points you can converge to are where -$$x=x(2-Rx)$$ -$$x=0\ \text{ or }\ x=\frac 1 R$$ -What you want are points where applying the function will get you closer to $\frac 1 R$ than you were previously. So what you want is where: -$$|f(x+\epsilon)-f(x)|<|\epsilon|$$ -So if you move $\epsilon$ away from $x$ (here $x=\frac 1 R$), applying the function takes you closer. For this to work in a stable way, it should work for arbitrarily small $\epsilon$. Applying this to your recurrence relation gives: -$$f(x)=x(2-Rx)$$ -$$(x+\epsilon)(2-Rx-R\epsilon)-x(2-Rx)<\epsilon$$ -$$2\epsilon-2Rx\epsilon-R\epsilon^2<\epsilon$$ -$$2-2Rx-R\epsilon<1,\ \ \epsilon>0$$ -$$2-2Rx-R\epsilon>1, \epsilon<0$$ -plugging in $x=\frac 1 R$, we get that -$$|\epsilon|<\frac 1 R$$ -So points within $(0,\frac 2 R)$ will converge to $\frac 1 R$ as intended. From what I can tell (I checked a few values of $R$ on a computer) this seems to work.<|endoftext|> -TITLE: Space of Complex Measures is Banach (proof?) -QUESTION [32 upvotes]: How can we prove that the space of Complex Measures is Complete? with the norm of Total Variation. -I have stuck on the last part of the proof where I have to prove that the limit function of a Cauchy sequence of measures has the properties of Complex measures. -We are using the norm of total variation :$$\lVert \mu\rVert = \lvert \mu \rvert(X)$$ - -REPLY [15 votes]: Another proof, using Radon-Nikodym. -Set -$$ -\lambda=\sum_{n=1}^\infty \frac{1}{n^2}|\mu_n|. -$$ -Then clearly, $\lambda$ is a bounded positive measure, and $\mu_n\ll\lambda$, for all $n\in\mathbb N$. Hence, due to Radon-Nikodym, there exist functions $\{f_n\}\in L^1(\lambda)$, such that $d\mu_n=f_n\,d\lambda$, for all $n\in\mathbb N$, and -$$ -\|\mu_m - \mu_n\|=\int_X|f_m(x)-f_n(x)|\,d\lambda(x)=\|f_m-f_n\|_{L^1(\lambda)} -$$ -Thus $\{f_n\}$ is a Cauchy sequence, and therefore, convergent to some $f\in L^1(\lambda)$, since $L^1(\lambda)$ is complete. If the measure $\mu$ is defined as -$d\mu=f\,d\lambda$, then -$$ -\|\mu_n-\mu\|=\int_X|f_n(x)-f(x)|\,d\lambda(x)=\|f_n-f\|_{L^1(\lambda)}\to 0. -$$<|endoftext|> -TITLE: Induction without integers (aka Structural Induction) -QUESTION [5 upvotes]: While composing the following question, I had an "ah-ha" moment. I still want to post the question along with my answer to show what I have learned. Any comments or additional answers will be greatly appreciated. -Recently I encountered a theorem in An Introcution to Mathematical Logic and Type Theory: To Truth through Proof by Peter B. Andrews where the author states that it can be used in inductive proofs without using integers. The theorem is stated as - -1000 Principle of Induction on the Construction of a Wff. Let $\mathscr{R}$ be a property of formulas, and let $\mathscr{R}(\mathbf{A})$ mean that $\mathbf{A}$ has property $\mathscr{R}$. Suppose -(1) $\mathscr{R}(\mathbf{q})$ for each propositional variable $\mathbf{q}$. -(2) Whenever $\mathscr{R}(\mathbf{A})$, then $\mathscr{R}(\mathord{\sim} \mathbf{A})$. -(3) Whenever $\mathscr{R}(\mathbf{A})$ and $\mathscr{R}(\mathbf{B})$, then $\mathscr{R}([\mathbf{A} \lor \mathbf{B}])$. -Then every wff has property $\mathscr{R}$. - -I am quite familiar with mathematical induction on integers. Mine typically take the following form: - -Blah, blah, blah. Therefore, the statement holds for $n=1$. -Now assume that the statement is true for some integer $n$. Now consider the statement for $n+1$. Yadda, yadda, yadda (eventually transforming the statement for $n+1$ into a statement involving $n$). Therefore, the statement holds for $n+1$. - -I'm also comfortable with using strong induction: - -Blah, blah, blah. Therefore the statement holds for $n=1$. -Now assume that the statement is true for all integers $k$ such that $1 \le k \le n$ for some integer $n$. Now consider the statement for $n+1$. Yadda, yadda, yadda (eventually transforming the statement for $n+1$ into a statement involving integers $k$ between $1$ and $n$). Therefore, the statement holds for $n+1$. - -One example where I would use this is in graph theory to prove a statement about trees. I proceed by induction on the number of vertices in a tree. This maps the object of interest (a tree) to the integers. -When I tried to apply the theorem about wffs, I got stuck how to proceed. How do I apply this theorem to a proof? - -REPLY [3 votes]: $\def\p{{\mathscr R}}$If you are uncomfortable with the idea of doing induction directly on the structure, or if you want to reduce the question to one of induction on the integers, you can proceed by induction on the size of the WFF, for some suitable definition of "size". One typical definition is that the size of the WFF is the number of logical operators it contains. -Then to prove that $\p(W)$ holds for every WFF $W$, you proceed as follows: - -Show that $\p(W)$ holds for each $W$ of the smallest possible size. (1, if you are counting atoms, or 0, if you are counting operators.) -Show that if $\p(W)$ holds for all $W$ of size less than $n$, then it must hold for all $W$ of size $n$, as follows: Let $W$ be a WFF of size $n$. Then it must have the form $\sim A$ for some wff $A$ of size $n-1$, or the form $A\oplus B$ for some operator $\oplus$ and some wffs $A$ and $B$ each of size at most $n-1$. Fill in the inductive argument for the two or more cases. - -Formulated in this way, the induction is just a regular strong induction on integers, where instead of proving the statement "$\p(W)$ holds for all wffs $W$", you reformulate it as "For all numbers $n$, $\p(W)$ holds for each wff of size $n$". -But this sort of transformation should not really be necessary. The induction principle can be stated more generally. Suppose $S$ is any set, and $\prec$ is a well-founded order on $S$; this means that every subset of $S$ is either empty, or contains a $\prec$-minimal element, which is an element $m\in S$ such that no other element $m'\in S$ has $m'\prec m$. -A typical example would be that wffs can be ordered by an ordering that says that $a\prec b$ whenever $a$ is a subformula of $b$. For example, $(x\land y)$ is a subformula of $\lnot(x\land y)\lor z$, so $(x\land y)\prec \lnot(x\land y)\lor z$. -The ordering $\prec$ need not be total, which means that it is possible that none of $a\prec b, a=b, $ and $b\prec a$ need be true. For the $\prec$ of the previous paragraph, we have neither $a\land b \prec a\lor b$ nor $a\lor b \prec a\land b$. That is quite all right. -Then you can use well-founded induction as follows: - -Let $F$ be the set of wffs for which $\p$ is false. Suppose $F$ is nonempty. Then by the well-foundedness of $\prec$, the set $F$ has a $\prec$-minimal element $m$. -Show that $m$ cannot have any subformulas, as follows: Using some properties of $\p$, show that $\p$ must fail for one of the subformulas of $m$. (This will depend on the details of $\p$ itself.) But then this subformula of $m$ would be an element of $F$, which would contradict the hypothesis that $m$ was $\prec$-minimal in $F$. -So $m$ has no subformulas. Rule out this case using some property of $\p$. - -This rules out the possibility that $F$ actually contains a $\prec$-minimal element $m$, and the only remaining possibility is that $F$ is empty, and so $\p$ holds for every wff. -Ordinary induction is a special case of well-founded induction which uses the well-founded relation $\lt$ on the natural numbers. The well-foundedness of $\lt$ is equivalent to the so-called well-ordering principle of the natural numbers, which states that every subset of $\Bbb N$ either contains a $\lt$-minimal element, or is empty.<|endoftext|> -TITLE: A Marcinkiewicz approach -QUESTION [6 upvotes]: The problem was to prove the following that the operator -$$Tf(x)=\int_{\mathbb{R}^N}\frac{f(y)}{|x-y|^\alpha}dy$$ -Is continuous from $$L^1 \to \ L_\mathrm{Weak}^{p}$$ where $0<\alpha1 $ and an intelligent use of Jensen's inequality.<|endoftext|> -TITLE: Proofs that every mathematician should know. -QUESTION [443 upvotes]: There are mathematical proofs that have that "wow" factor in being elegant, simplifying one's view of mathematics, lifting one's perception into the light of knowledge, etc. -So I'd like to know what mathematical proofs you've come across that you think other mathematicans should know, and why. - -REPLY [13 votes]: My all time favorite proof? -Furstenberg's proof of the infinity of primes by point-set topology. I know, a lot people think it's not a big deal. I think: -a) It's an immensely clever way to use point set topology to prove a result in a seemingly unrelated field, namely number theory. -b) I used it as the beginnings of my first research in additive number theory; looking to generalize this result to create similar proofs of results for sumsets and arithmetic progressions. Sadly, my health failed again,but I hope to return to this research soon.<|endoftext|> -TITLE: Showing that a set of trigonometric functions is linearly independent over $\mathbb{R}$ -QUESTION [5 upvotes]: I would like to determine under what conditions on $k$ the set $$ \begin{align} -A = &\{1,\cos(t),\sin(t), \\ -&\quad \cos(t(1+k)),\sin(t(1+k)),\cos(t(1−k)),\sin(t(1−k)), \\ -&\quad \cos(t(1+2k)),\sin(t(1+2k)),\cos(t(1−2k)),\sin(t(1−2k))\}, -\end{align}$$ -is linearly independent, where $k$ is some arbitrary real number. -As motivation, I know that the set defined by -$$ -\{1, \cos wt, \sin wt\}, \quad w = 1, \dots, n -$$ -is linearly independent on $\mathbb{R}$, which one generally proves by computing the Wronskian. I thought that I could extend this result to the set in question, but I haven't found a proper way to do so. My intuition tells me that $A$ will be linearly dependent when the arguments of the trig functions coincide, which will depend on the value of $k$. -Though, I'm at a loss for proving this is true. Computing the Wronskian for this set required an inordinate amount of time-- I stopped running the calculation after a day. Is there perhaps a way to reduce the set in question so that the Wronskian becomes manageable? -I'm interested in any suggestions/alternative methods for proving linear independence that could help my situation. Note that I'd like to have a result that holds for any $m = 0, \dots, n,$ where $n \in \mathbb{Z}$ if possible. -Thanks for your time. -EDIT: The set originally defined in the first instance of this post was incorrectly cited. My sincere apologies. - -REPLY [6 votes]: The answer is $k = 0, \pm 1, \pm \frac{1}{2}$. This follows from the following result. -Claim: The functions $\{ 1, \sin rt, \cos rt \}$ for $r$ a positive real are linearly independent over $\mathbb{R}$. -Proof 1. Suppose that $\sum s_r \sin rt + \sum c_r \cos rt = 0$ is a nontrivial linear dependence. Consider the largest positive real $r_0$ such that $c_{r_0} \neq 0$. Take a large even number of derivatives until the coefficient of $\cos r_0 t$ is substantially larger than the remaining coefficients of the other cosine terms and then substitute $t = 0$; we obtain a number which cannot be equal to zero, which is a contradiction. So no cosines appear. -Similarly, consider the largest positive real $r_1$ such that $s_{r_1} \neq 0$. Take a large odd number of derivatives until the coefficient of $\cos r_1 t$ is substantially larger than the remaining coefficients of the other cosine terms (which come from differentiating sine terms) and then substitute $t = 0$; we obtain a number which cannot be equal to zero, which is a contradiction. So no sines appear. -So $1$ is the only function which can appear in a nontrivial linear dependence, and so there are no such linear dependences. -Proof 2. It suffices to prove that the functions are all linearly independent over $\mathbb{C}$. Using the fact that -$$\cos rt = \frac{e^{irt} + e^{-irt}}{2}, \sin rt = \frac{e^{irt} - e^{-irt}}{2i}$$ -it suffices to prove that the functions $\{ e^{irt} \}$ for $r$ a real are linearly independent. This can be straightforwardly done by computing the Wronskian and in fact shows that in fact the functions $\{ e^{zt} \}$ for $z$ a complex number are linearly independent. -Proof 3. Begins the same as Proof 2, but we do not compute the Wronskian. Instead, let $\sum c_z e^{zt} = 0$ be a nontrivial linear dependence with a minimal number of terms and differentiate to obtain -$$\sum z c_z e^{zt} = 0.$$ -If $z_0$ is any complex number such that $z_0 \neq 0$ and $c_{z_0} \neq 0$ (such a number must exist in a nontrivial linear dependence), then -$$\sum (z - z_0) c_z e^{zt} = 0$$ -is a linear dependence with a fewer number of terms; contradiction. So there are no nontrivial linear dependences.<|endoftext|> -TITLE: Proof of infinitude of primes using the irrationality of π -QUESTION [15 upvotes]: According to the section Proof using the irrationality of $\pi$ of the Wikipedia article on Euclid's theorem, Euler proved that: -$$\frac{\pi}{4}=\frac34\cdot\frac54\cdot\frac78\cdot\frac{11}{12}\cdot\frac{13}{12}\cdots$$ -where "each denominator is the multiple of four nearest to the numerator". -Can someone please explain this formula? I see it, but cannot believe it. - -REPLY [10 votes]: The formula is described here (I am having a hard time finding a more authoritative reference); briefly, the OP's product in the usual product notation goes like -$$\frac{\pi}{4}=\prod_{k=2}^\infty \frac{p_k}{p_k-\chi(p_k)}$$ -where $p_k$ is the $k$-th prime and $\chi(n)$ is a character defined as -$$\chi(n)=\begin{cases}1&\text{if }n\equiv 1\pmod 4\\-1&\text{if }n\equiv 3\pmod 4\end{cases}$$ -As noted, the derivation is done by treating the usual Leibniz series -$$\frac{\pi}{4}=\sum_{k=1}^\infty \frac{(-1)^{k-1}}{2k-1}$$ -as a Dirichlet series, and then expanding that series as an Euler product. -Edit: Daniel has given another nice link in the comments.<|endoftext|> -TITLE: $f(x)$ absolutely continuous $\implies e^{f(x)}$ absolutely continuous, for $x \in [a,b]$? -QUESTION [5 upvotes]: If $f(x)$ is absolutely continuous (a.c.) on [a,b], is the function $e^{f(x)}$ also absolutely continuous on [a,b] ? -thanks - -REPLY [4 votes]: It is true. Suppose $g$ is Lipschitz with rank $L$ on $[a,b]$, and $f$ is AC. Then $g \circ f$ is also AC. -To see this, suppose $f$ is AC, and let $\epsilon>0$. Choose $\delta>0$ such that if $(y_k,x_k)$ are a finite collection of pairwise disjoint intervals in $[a,b]$ with $\sum |y_k-x_k| < \delta$, then $\sum |f(y_k)-f(x_k)| < \frac{\epsilon}{L}$. -Now consider $\sum |g \circ f(y_k)-g \circ f(x_k)| = \sum |g(f(y_k))-g ( f(x_k))| \leq L \sum |f(y_k)-f(x_k)| < \epsilon$. Hence $g \circ f$ is AC. -Since $x \mapsto e^x$ is smooth, it is Lipschitz on any compact interval, hence the function $ x \mapsto e^{f(x)}$ is AC.<|endoftext|> -TITLE: Which number fields can appear as subfields of a finite-dimensional division algebra over Q with center Q? -QUESTION [35 upvotes]: I have some idle questions about what's known about finite-dimensional division algebras over $\mathbb{Q}$ (thought of as "noncommutative number fields"). To keep the discussion focused, let's concentrate on these: - -Which number fields $K$ occur as subfields of a finite-dimensional division algebra over $\mathbb{Q}$ with center $\mathbb{Q}$? -Which pairs of number fields $K, L$ occur as subfields of the same finite-dimensional division algebra over $\mathbb{Q}$ with center $\mathbb{Q}$? - -There are some easy examples involving quaternions but I am curious how completely these kinds of questions are understood. Some preliminary Googling on my part was not successful. - -REPLY [8 votes]: $\def\Q{\mathbf{Q}}$ -$\def\Z{\mathbf{Z}}$ -$\def\Br{\mathrm{Br}}$ -$\def\inv{\mathrm{inv}}$ -$\def\Gal{\mathrm{Gal}}$ -The questions you ask are essentially straightforward enough, but require a little theory. In fact, I once asked a starting graduate student to answer your question as preparation for his qualifying exam. (Hint: if you ever tell your advisor "I basically understand the statements of class field theory but I don't really know anything about division algebras" be prepared for the answer "OK, your qual will center on explaining CFT through Brauer groups.") I guess he never came back to this website to actually post an answer to this question, and I'm bored now, so here's an answer. -I'm going to begin by stating a bunch of very standard results without either proofs or references. Any standard text on class field theory (including Milne's online notes) will have all the details. -Lemma: If $K/\Q$ can be centrally embedded into $D/\Q$ of dimension $n^2$, then there is a finite extension $L/K$ where $L$ can be centrally embedded into $D/\Q$ and $[L:\Q] = n$. -This is another way of saying that all maximal commutative subfields have degree $n$. -Lemma: $L/\Q$ of dimension $n$ can be centrally embedded into $D/\Q$ of dimension $n^2$ if and only if $L$ splits $D$. -Standard. -Theorem: There is an exact sequence: -$$0 \rightarrow \Br(\Q) \rightarrow \bigoplus_{p,\infty} \Br(\Q_p) \rightarrow \Q/\Z \rightarrow 0.$$ -This is a key result of global class field theory. -Lemma: $\Br(\Q_p) = \Q/\Z$, $\Br(\mathbf{R}) = \Z/2\Z$. -This follows from local class field theory. -Lemma: Let $L_v/\Q_p$ be a finite extension. Then $D_p/\Q_p$ splits over $L_v$ if and only if $[L_v:\Q_p]$ annihilates the class $D_p \in \Br(\Q_p)$, or equivalently -$$[L_v:\Q_p] \inv(D) = 0,$$ - where $\inv_p(D) \in \Q/\Z$ or $\Z/2\Z$ depending on whether $p$ is finite or not. -Standard. -Now, simply by connecting the dots, one obtains the following: - Lemma: A field $L/\Q$ of degree $n$ embeds centrally and maximally into $D/\Q$ of dimension $n^2$ if and only if, -for all primes $p$ and for all $v|p$ in $p$, -$$[L_v:\Q_p] \inv_p(D) = 0.$$ -There is a lot of freedom in constructing $D$. We can make $\inv_p(D)$ anything we want, as long as it is zero for all but finitely many primes $p$, has order dividing $2$ at infinity, and satisfies -$$\sum_p \inv_p(D) = 0.$$ -Hence we also deduce: - Lemma: A field $L/\Q$ of degree $n$ embeds centrally and maximally into $D/\Q$ of dimension $n^2$ if and only if, -for all primes $q | n$ with $q^m \| n$, we can find two distinct primes $p_i$ for $i = 1,2$ (one possibly infinite) such that, for any -prime $v|p_i$, we have -$$q^m | [L_v:\Q_{p_i}].$$ -Proof: Since $\sum \inv_p(D) = 0$ and $D$ has order $n$, there exist at least two (possibly one is infinite if $q^m = 2$) primes $p_i$ such that $\inv_p(D)$ has order divisible by $q^m$. This gives one direction. -For the converse, take the invariants to be the sum of the vector over $q^m | n$ whose contribution is $1/q^m$ at $p_1$ and $-1/q^m$ at $p_2$ and zero elsewhere. -Now we deduce: - Claim: A field $K/\Q$ of degree $n$ embeds into $D/\Q$ of dimension $N^2$ for some $N$ if and only if, -for all primes $q | n$ with $q^m \| n$, we can find two distinct primes $p_i$ (one possibly infinite) such that, for any -prime $v|p_i$, we have -$$q^m | [L_v:\Q_{p_i}].$$ -Proof: If there is such an embedding, there is a maximal embedding of an $L$ which contains $K$. -Suppose that $N = dn$, and let $q^M \| d$ so $q^{m+M} \| N$. We deduce from the previous Lemma that there is a prime $p_i$ such that -$$q^{m+M} | [L_w:\Q_{p_i}] = [L_w:K_v][K_v:\Q_{p_i}]$$ -for all $v|p_i$ and all $w|v$. Hence it follows that -$$q^{m+M} | \sum_{w|v} [L_w:K_v] [K_v:\Q_{p_i}] = [L:K][K_v:\Q_{p_i}] = d [K_v:\Q_{p_i}],$$ -and thus -$$q^m | [K_v:\Q_{p_i}],$$ -as required. For the converse, one can easily construct a totally ramified $L/K$ which has $L_w = K_v$ -for all $v|p_i$ for any finite set of $p_i$, and then use the previous Lemma. -The claim basically gives an answer to your first question, note that one also deduces that $K$ is included inside some $D$ if and only if it is included maximally for some $D$ of dimension $[K:\Q]^2$. The second question presents no real extra difficulties. -Let me give some examples. -Example: $K/\Q$ has prime degree $q$. Let $L$ be the Galois closure of $K$. We have $G = \Gal(L/\Q) \subset S_q$ has order divisible by $q$, and hence contains a $q$-cycle. Thus by Chebotarev there are infinitely many $p_i$ such that Frobenius at $p_i$ is a $q$-cycle. This implies that $[K_v:\Q_{p_i}] = q$ for all $v|q$, and so $K$ embeds. -Example: $K/\Q$ is Galois and cyclic. In this case, $G = \Gal(K/\Q) = \Z/n \Z \subset S_n$ also has an $n$ cycle, so as in the last question one is done by Cebotarev, and $K$ embeds. -Example: $K/\Q$ is bi-quadratic, and $[K_v:\Q_p] \le 2$ for all ramified primes. In this case, $K$ can never be embedded inside a $D/\Q$, because there are no primes $p_i$ such that $4 | [K_v:\Q_p]$. -Remark In the last example, one does require the hypothesis at the ramified primes, since it could happen that $[K_v:\Q_p] = 4$ for two primes dividing $\Delta_K$. Thus the answer depends not only on the Galois group but also at the inertia groups at primes of bad reduction. -Example: $K = \Q(\sqrt{-1},\sqrt{17})$ does not embed centrally into any $D/\Q$. This is a special case of the last example. -Example: $K = \Q(\sqrt{-1},\sqrt{3})$ does embed centrally into a $D/\Q$ (for example with invariants $\inv_2(D) = 1/4$ and $\inv_3(D) = -1/4$ and $\inv_p(D) = 0$ otherwise). This is a special non-case of the last example, because $[K_3:\Q_3] = 4$ and $[K_2:\Q_2] = 4$. - Example: $K/\Q$ is totally ramified at two primes $p_1$ and $p_2$. In this case, $[K_v:\Q_{p_i}] = [K:\Q]$, so we can aways embed $K$. -I think these examples are enough to get the point.<|endoftext|> -TITLE: Bézier approximation of archimedes spiral? -QUESTION [8 upvotes]: As part of an iOS app I’m making, I want to draw a decent approximation of an Archimedes spiral. The drawing library I’m using (CGPath in Quartz 2D, which is C-based) supports arcs as well as cubic and quadratic Bézier curves. What is a good method of approximating an Archimedes spiral using either of these path types? For example the wikipedia exemplar image says it was “drawn as a series of minimum-error Bézier segments.” How would one generate such segments? -My math background takes me through Calculus III plus some stuff I picked up from a classical mechanics class, but it’s been a couple of years so I’m rusty. What I have so far: -For a spiral r = a + b $\theta$, I used the information from this page to find that the cartesian slope at any point (r, $\theta$) is equal to -$$\frac{dy}{dx}=\frac{b\sin\theta\space+\space(a + b\theta)\cos\theta}{b\cos\theta\space-\space(a + b\theta)\sin\theta}$$ -From here, I could use point-slope to find the equation of a tangent line at any point, but how do I go about finding the proper lengths of the handles (i.e. the positions of the middle two points) for the curve? Or would an approximation with circular arc segments be better/easier/faster? -If I can’t figure it out, I’ll just use a static image in the app, but it occurs to me that I don’t even know of a way to generate a high-quality image of an Archimedes spiral! The Spiral tool in Illustrator, for example, does only logarithmic spirals. - -REPLY [3 votes]: So it looks like the Wikipedia reference image uses 45 degree sections of these curves. You can use the equation for the spiral to give you the tangent line at the beginning and end of these curve sections. Evaluate the derivative at these two points to get the tangent line slope and then shift your line appropriately to hit the point used. -The intersection Of these two lines should be your control point. -Once you have found your control point you can put it in the function 'CGPathAddQuadCurveToPoint' for the cx, cy (I think) along with the point you want to go to (also from the spiral equation). -For reference--check out the animation under 'quadratic curves' here -For extra speed, you only have to find 8 tangent lines max--just shift them out for the next cycle of the spiral and reuse them.<|endoftext|> -TITLE: Characterization of Rational Normal Curve -QUESTION [7 upvotes]: Given a curve $C\subset P^n$ in projective space such that any $n+1$ points on $C$ are linearly independent. I've heard from multiple people that this implies that $C$ is a rational normal curve, some people said that one needs that $C$ is rational for that. -I would like to know what precisely is true and how I can prove it, google is not my friend here... -Thanks for your help! - -REPLY [5 votes]: Let me elaborate on my hint, since there was some skepticism! ;) -Let $p$ be a point on $C$ and let $C' \subset \mathbb{P}^{n-1}$ denote the projection of $C$ from $p$. Think of $C'$ as a curve contained in a hyperplane in $\mathbb{P}^n$. Suppose that there are $n$ points of $C'$ not in general position; this means that there is a hyperplane of $\mathbb{P}^{n-1}$ containing them all. Then the hyperplane of $\mathbb{P}^n$ containing those $n$ points of $C'$ and the point $p$ is a hyperplane intersecting $C$ at $n+1$ points, contrary to the independence condition. Therefore, also the curve $C'$ has the independence property. -If we are very formal, then we reduce to the case $n=1$, where the assertion is a tautology. Otherwise, we stop at the case $n=2$ and the assumption of independence implies that no line in $\mathbb{P}^2$ intersects the curve in 3 points: $C$ is a conic. -Finally, to show that $C$ is a rational normal curve, you would need to specify what is the definition of rational normal curve you use: some people say it is a smooth connected curve of degree $n$ in $\mathbb{P}^n$, others say that it is the image of $\mathbb{P}^1$ under the complete linear system $\mathcal{O}(d)$, still others might even define it by the property you mention. What is your definition? -Comment. The argument given here is not especially different from the one given by Georges Elencwajg, except that he did not use the more geometric terminology: for instance, his parameterization is the result of projecting the curve away from the $n-1$ points $q_1,\ldots,q_{n-1}$.<|endoftext|> -TITLE: If $a^n-b^n$ is integer for all positive integral value of $n$, then $a$, $b$ must also be integers. -QUESTION [10 upvotes]: If $a^n-b^n$ is integer for all positive integral value of n with a≠b, then a,b must also be integers. -Source: Number Theory for Mathematical Contests, Problem 201, Page 34. -Let $a=A+c$ and $b=B+d$ where A,B are integers and c,d are non-negative fractions<1. -As a-b is integer, c=d. -$a^2-b^2=(A+c)^2-(B+c)^2=A^2-B^2+2(A-B)c=I_2(say),$ where $I_2$ is an integer -So, $c=\frac{I_2-(A^2-B^2)}{2(A-B)}$ i.e., a rational fraction $=\frac{p}{q}$(say) where (p,q)=1. -When I tried to proceed for the higher values of n, things became too complex for calculation. - -REPLY [3 votes]: Denote $I_n=a^n-b^n$. Note that if $(a,b)$ has the property, then so has $(ma,mb)$ for $m\in\mathbb N$ as this just replaces $I_n$ with $m^nI_n$. -We have $I_1\ne 0$ because $a\ne b$ and therefore find that $a=\frac{I_2+I_1^2}{2I_1}\in \mathbb Q$, say $a=\frac uv$ with $u\in\mathbb Z$, $v\in\mathbb N$. Then $w:=vb=u-vI_1$ is an integer, hence $b=\frac wv\in\mathbb Q$. If $v=1$, we are done. -And if $v>1$, let $p$ be a prime dividing $v$. By the observation above about multiples, we may assume wlog. that $v=p$. -Write $a=A+\frac rp$ with $0 -TITLE: What is the sum of only half the exponential terms that give the Dirac comb? -QUESTION [5 upvotes]: The following infinite sum of exponential terms gives a Dirac comb: -$$ \sum_{n=-\infty}^\infty e^{i n x} = 2 \pi \sum_{n=-\infty}^\infty \delta(x - 2 \pi n) $$ -Of course the sum doesn't strictly converge, but only in the same sense in which the Dirac delta-function is defined. -What is the result of a semi-infinite sum of such terms? -$$ \sum_{n=1}^\infty e^{i n x} =~? $$ - -REPLY [3 votes]: Hint: -(1) write the semi-infinite sum in the following way -$$\sum_{n=1}^\infty e^{i n x} = \frac{1}{2} \sum_{n=-\infty}^\infty e^{i n x} -+ \frac{1}{2} \sum_{n=-\infty}^\infty \mathop{\rm sgn}(n) e^{i n x} - \frac{1}{2}.$$ -The first summand is the Dirac-comb. So your question reduces to figuring out what the second term equals to... -(2) Lagrange's trigonometric identities might help there...<|endoftext|> -TITLE: Good exercises to do/examples to illustrate Seifert - Van Kampen Theorem -QUESTION [35 upvotes]: I have just learned about the Seifert-Van Kampen theorem and I find it hard to get my head around. The version of this theorem that I know is the following (given in Hatcher): - - -If $X$ is the union of path - connected open sets $A_\alpha$ each containing the basepoint $x_0 \in X$ and if each intersection $A_\alpha \cap A_\beta$ is path connected, then the homomorphism - $$\Phi:\ast_\alpha \pi_1(A_\alpha) \to \pi_1(X)$$ is surjective. If in addition each intersection triple intersection $A_\alpha \cap A_\beta \cap A_\gamma$ is path-connected, then $\ker \Phi$ is the normal subgroup $N$ generated by all elements of the form $i_{\alpha\beta}(\omega)i_{\beta\alpha}(\omega^{-1})$, and so $\Phi$ induces an isomorphism - $$\pi_1(X) \cong \ast_\alpha \pi_1(A_\alpha)/N.$$ - - -$i_{\alpha\beta}$ is the homomorphism $\pi_1(A_\alpha \cap A_\beta) \to \pi_1(A_\alpha)$ induced from the inclusion $A_\alpha \cap A_\beta \hookrightarrow A_\alpha$ and $\omega$ is an element of $\pi_1(A_\alpha \cap A_\beta)$. -Now I tried to get my head round this theorem by trying to understand the example in Hatcher on the computation of the fundamental group of a wedge sum. Suppose for the moment we look at $X = X_1 \vee X_2$. I cannot just apply the theorem blindly because $X_i$ is not open in $X$. So we need to look at -$$A_1 = X_1 \vee W_2, \hspace{3mm} A_2 = X_2 \vee W_1$$ -where $W_i$ is a neighbourhood about the basepoint $x_1$ of $X_1$ that deformation retracts onto $\{x_1\}$, similarly for $W_2$. I believe each of these is open in $X_1 \vee X_2$ because each $A_i$ is the union of equivalence classes that is open in $X_1 \sqcup X_2$. Now how do I see rigorously that $A_1 \cap A_2$ deformation retracts onto the point $p_0$ (that I got from identifying $x_1 \sim x_2$) in $X$? If I can see that, then I know by Proposition 1.17 (Hatcher) that -$$\pi_1(A_1 \cap A_2) \cong \pi_1(p_0) \cong 0$$ -from which it follows that $N= 0$ and the Seifert-Van Kampen Theorem tells me that -$$\pi_1(X_1\vee X_2) \cong \pi_1(X_1) \ast \pi_1(X_2).$$ - - -1) Is my understanding of this correct? -2) What other useful exercises/ examples/applications are there to illustrate the power of the Seifert-Van Kampen Theorem? I have also seen that you can use it to prove that $\pi_1(S^n) = 0$ for $n \geq 2$. - - -I have had a look at the examples section of Hatcher after the proof the theorem, but unfortunately I don't get much out of it. The only example I sort of got was the computation of $\pi_1(\Bbb{R}^3 - S^1)$. -I would appreciate it very much if I could see some other examples to illustrate this theorem. In particular, I heard that you can use it to compute group presentations for the fundamental group - it would be good if I could see examples like that. -Thanks. -Edit: Is there a way to rigorously prove that $A_1 \cap A_2$ deformation retracts onto the point $\{p_0\}$? - -REPLY [12 votes]: We use the Seifert Van-Kampen Theorem to calculate the fundamental group of a connected graph. This is Hatcher Problem 1.2.5: -It is a fact in graph theory that any connected graph $X$ contains a maximal tree $M$, namely a contractible graph that contains all the vertices of $X$. Now if the maximal tree $M = X$, then we are done because for any $x_0 \in M$, $\pi_1(M,x_0) = \pi_1(X,x_0) = 0$ that is trivially free. Now suppose $M \neq X$. Then there is an edge $e_i$ of $X$ not in $M$. Observe that for each edge $e_i$ we get a loop going in $M \cup e_i$ about some point $x_0 \in M$. Now fix out basepoint $x_0$ to be in $M$ and suppose that the edges not in $M$ are $e_1,\ldots,e_n$. Then it is clear that -$$X = \bigcup_{i=1}^n \left(M \cup e_i\right).$$ -The intersection of any two $M \cup e_i$ and $M \cup e_j$ contains at least $M$ and is path connected, so is the triple intersection of any 3 of these guys by the assumption that $X$ is a connected graph. So for any $x_0 \in M$, the Seifert-Van Kampen Theorem now tells us that -$$\pi_1(X,x_0) \cong \pi_1(M \cup e_1,x_0) \ast \ldots \ast \pi_1(M \cup e_n,x_0)/N$$ -where $N$ is the subgroup generated by words of the form $l_{ij}(w)l_{ji}(w)^{-1}$, where $l_{ij}$ is the inclusion map from $\pi_1((M\cup e_i) \cap (M \cup e_j),x_0) = \pi_1(M \cup (e_i \cap e_j),x_0)$. Now observe that if $i \neq j$ then $M \cup (e_i \cap e_j) = M$ and since $\pi_1(M,x_0) = 0$ we conclude that any loop $w \in \pi_1(M \cup (e_i \cap e_j),x_0)$ in here is trivial. If $i = j$, $l_{ij}$ is just the identity so that our generators for $N$ are just -$$l_{ij}(w)l_{ji}(w)^{-1} = ww^{-1} = 1$$ -completing our claim that $N$ was trivial. Now for each $i$, we have that $\pi_1(M\cup e_i,x_0)$ is generated by a loop that starts at $x_0$ and goes around the bounded complementary region formed by $M$ and $e_i$ and back to $x_0$ through the maximal tree. Such a path back to $M$ does not go through any other edge $e_j$ for $j$ different from $i$. It follows that $\pi_1(X,x_0)$ is a free group with basis elements consisting of loops about $x_0 \in M$ as described in the line before.<|endoftext|> -TITLE: Prove that: $\lim_{n\to\infty} f(n+1) - f(n) = \lim_{x\to\infty} (f(x))' $ -QUESTION [5 upvotes]: I conjecture that in some specific conditions a differentiating function gives the following equality: -$$\lim_{n\to\infty} f(n+1) - f(n) = \lim_{x\to\infty} (f(x))' $$ -However, I'm not sure yet what exactly those conditions are in order to precisely know where -I may apply this rule or not. If you wanna take a look over my posted problem here you'll immediately notice that this rule applies for that case. I really appreciate if you help me clarify this. - -REPLY [5 votes]: I believe that the minimal assumptions are: - -$f \colon [0,+\infty) \to \mathbb{R}$ is differentiable; -$\lim_{x \to +\infty} f'(x)$ exists. - -Then you easily check that $$\lim_{n \to +\infty} f(n+1)-f(n)=\lim_{x \to +\infty} f'(x),$$ -since you can apply Lagrange's theorem: $$f(n+1)-f(n)=f'(\xi_n)$$ for a suitable $\xi_n \in (n,n+1)$. -I think that the two limits are not equivalent to each other, since it may be impossible to control $f'$ by using the values of $f$ at discrete points.<|endoftext|> -TITLE: Intersection form on quotient manifold -QUESTION [6 upvotes]: I have a simple algebraic topology question. Let $M$ and $N$ be 2-dimensional oriented manifolds (say $H^{2}(M,\mathbb{Z})\cong \mathbb{Z}\alpha_{M}$ and $H^{2}(M,\mathbb{Z})\cong \mathbb{Z}\alpha_{N}$). Assume that a finite group $G$ acts on $M$ and $N$ in such a way that $G$ preserves the orientation of $M$ and $N$ and the induced action of $G$ on $M\times N$ has no fixed point. Then -$$ -X=(M\times N)/G -$$ is a 4-dimensional oriented manifold. -I would like to understand the intersection form on the middle cohomology $H^{2}(X,\mathbb{Z})$. There is a ono-to-one correspondence between -$$ -H^{2}(X,\mathbb{Z}) \ \longleftrightarrow H^{2}(M\times N,\mathbb{Z})^{G}, -$$ -i.e. any $G$-invariant element of $H^{2}(M\times N,\mathbb{Z})$ descends to $H^{2}(X,\mathbb{Z})$ and any element of $H^{2}(X,\mathbb{Z})$ can be pulled back to $H^{2}(M\times N,\mathbb{Z})^{G}$ by the quotient map. Since the $G$-action is free, the intersection form on $H^{2}(X,\mathbb{Z})$ is given by the intersection form on $H^{2}(M\times N,\mathbb{Z})^{G}$ divided by $|G|$. So, any intersection number on $H^{2}(M\times N,\mathbb{Z})^{G}$ must be a multiple of $|G|$. -On the other hand we have -$$ -p_{1}^{*}(\alpha_{M}), \ p_{2}^{*}(\alpha_{N})\in H^{2}(M\times N,\mathbb{Z})^{G} -$$ -(because $G$ preserves both $M$ and $N$) and -$$ -p_{1}^{*}(\alpha_{M})\cup \ p_{2}^{*}(\alpha_{N})=\alpha_{M\times N} -$$ -where $p_{i}$ is the $i$-th projection of $M\times N$ and $H^{2}(M\times N,\mathbb{Z})\cong \mathbb{Z}\alpha_{M\times N}$. This means that the intersection number $p_{1}^{*}(\alpha_{M})\cup \ p_{2}^{*}(\alpha_{N})$ is 1, not divisible by $|G|$ (unless $G$ is trivial). -Could anyone point out what is wrong with my argument? -Thank you in advance. - -REPLY [2 votes]: Why not try out both arguments on a specific example? -Take $M = N = S^1\times S^1$ and $G = \mathbb{Z}/2\mathbb{Z}$ acting by $-1 *(\theta,\phi) = (\theta +\pi,\phi +\pi)$. Then this action is free on each of $M$ and $N$, so the induced action on $M\times N$ is certainly free. -In this case, $(M\times N)/ G \cong (S^1)^4$, since, for example, $(M\times N)/ G$ is a compact abelian Lie group. (I'm sure there's an easy direct way to see it). -So, what do both computations give you? Which one must be wrong?<|endoftext|> -TITLE: How to integrate $\frac{1}{\sqrt{1+x^2}}$ using substitution? -QUESTION [9 upvotes]: How you integrate $\frac{1}{\sqrt{1+x^2}}$ using following substitution? $1+x^2=t$ $\Rightarrow$ $x=\sqrt{t-1} \Rightarrow dx = \frac{dt}{2\sqrt{t-1}}dt$... Now I'm stuck. I don't know how to proceed using substitution rule. - -REPLY [3 votes]: $$A=\int\frac{1}{\sqrt[]{1+x^2}}$$ -Let, $x = \tan\theta$ -$dx = \sec^{2}\theta{d\theta}$ -substitute, $x$, $dx$ -$$A=\int\left(\frac{1}{\sec\theta}\right){\sec^{2}\theta{d\theta}}$$ -$$A=\int{\sec\theta{d\theta}}$$ -$$A=\int{\sec\theta\left(\frac{\sec\theta + \tan\theta}{\sec\theta + \tan\theta}\right){d\theta}}$$ -$$A=\int{\left(\frac{\sec^2\theta + \sec\theta\tan\theta}{\sec\theta + \tan\theta}\right){d\theta}}$$ -Let, $(\sec\theta + \tan\theta) = u$ -$(\sec^2\theta + \sec\theta\tan\theta)d\theta = du$ -$$A=\int\frac{du}{u}$$ -$$A=\ln{u}+c$$ -$$A=\ln{\vert\sec\theta + \tan\theta\vert}+c$$ -$$A=\ln{\vert\sqrt[]{1+\tan^2\theta} + \tan\theta\vert}+c$$ -$A=\ln{\vert\sqrt[]{1+x^2} + x\vert}+c$, where $c$ is a constant<|endoftext|> -TITLE: How to introduce stress tensor on manifolds? -QUESTION [17 upvotes]: I want to understand the type of stress tensor $\mathbf{P}$ in classical physics. -Usually in physics it is said that the force $\text d \boldsymbol F$ (vector) acting on an infinitesimal area $\text d \boldsymbol s$ (vector) equals -$\text d \boldsymbol F = \mathbf{P} \cdot \text d \boldsymbol s$ -where $\cdot$ is a "scalar product". -How can it be rigourised? I guess directed area can be $\star s$ where $s$ is a 2-form, but can I avoid using $\star$ by employing the volume form for example? The force should be 1-form. -How is the power of surface forces is written? Usually it is given by -$$\frac{dA}{dt} = \int_S \boldsymbol v \cdot \text d \boldsymbol F$$ -$\boldsymbol v$ being the speed of the surface of the deformed body. -What would be the corresponding local form, that is the power density of surface forces? - -UPDATE 1 -If it helps, I found a whole appendix "The Classical Cauchy Stress Tensor and Equations of Motion" in the book "The Geometry of Physics: An Introduction" by Theodore Frankel. Particularly it says - -The Cauchy stress should be a vector-valued pseudo-$(n - 1)$-form. - -However currently I don't know what does it mean. Further development in the book is rather obscure and I'm afraid of that "pseudo". If a thing called "pseudo-something" I would prefer it stated as "actual another thing". - -UPDATE 2 -Stress tensor can also be viewed as a (molecular) flux of momentum. Then the equation for balance of momentum would be the Newton's second law. Probably this approach would be more fruitful, analogues can be made with the flux of density. - -REPLY [3 votes]: I found a paper supporting my comment that $P$ is a 1-form valued 2-form, that surface force f is also a 1-form valued 2-form, and power density is the 2-form that results from contracting f with the surface velocity. The paper is - -R. Segev and L. Falach. Velocities, stresses and vector bundle valued chains. J. Elast. 105:187-206, 2011.<|endoftext|> -TITLE: Applications of quadratic forms -QUESTION [7 upvotes]: It seems that a lot of great mathematicians spent quite a while of their time studying quadratic forms over $\mathbb{Z},\mathbb{Q},\mathbb{Q_p}$ etc. and there is indeed a vast and detailed theory of these. It usually qualifies as part of Number Theory as in Serre's book "A Course In Arithmetic" half of which is devoted to the topic, but it is also claimed to have applications in other areas such as differential topology, finite groups, modular forms (as stated in the preface of the aforementioned book). I would like to have examples of such applications just for general education in order to truly appreciate its universal importance. - -REPLY [3 votes]: You mention simple groups. Quadratic forms are a quiet element in the classification of Lie algebras, through Weyl chambers and so on. The short version is that any discussion involving Euclidean spaces that talks about reflections is using a quadratic form, typically the ordinary dot product. -So, finite simple groups were discovered as integral automorphism groups of integral lattices, meaning other positive quadratic forms over $\mathbb Z.$ This area has always used the language of Lie algebras, Dynkin diagrams, etc. However, the defining example is the Leech lattice, which is a peculiar object. Furthermore, the main calculation taking the Leech lattice to simple groups takes a detour through indefinite forms. -Three good references are ERROR CORRECTING, LATTICES AND CODES, and any edition of SPLAG. I am especially fond of section 4.5 in Lattices and Codes by Ebeling. In the preface to the second edition, he points out that section 4.5 is new in that edition. Amazing, as there was no other place that explained what I needed. There is something a little off about availability and price, I think I bought it for about $35.00 from the AMS, but now AMAZON. I wrote to Prof. Ebeling, he quite liked my application of the section. Known to Conway and Borcherds, of course, but worth publishing as a separate item. We will see.<|endoftext|> -TITLE: $\log_7 n$ is either an integer or an irrational number -QUESTION [6 upvotes]: Show that $\log_7 n$ is either an integer or an irrational number where n is a positive number. -I assumed that it is rational and tried to get a contradiction for $\log_7 n = a/b$, where b does not divide a, but how can I show that $7^{a/b}$ is not an integer to achieve a contradiction since n is an integer ? If I can exclude rational numbers from the range of log function then it is either integer or irrational. -Or do you suggest other methods ? - -REPLY [9 votes]: Interesting; usually one would assume not just that $b$ doesn't divide $a$ but that $a$ and $b$ are coprime, but in this case your assumption that $b$ doesn't divide $a$ is enough. -If $7^{a/b}=n$, then $7^a=n^b$. Thus $n$ must be a power of $7$, so we can write $n=7^k$ and thus $7^a=7^{kb}$, so $a=kb$, contradicting the assumption.<|endoftext|> -TITLE: Definition of Grothendieck group -QUESTION [5 upvotes]: I'm reading the Wiki article about the Grothendieck group. -What's the reason we define $[A] - [B] + [C] = 0 $ rather than $[A] + [B] - [C] = 0 $ (or something else) for every exact sequence $0 \to A \to B \to C \to 0$? What is the property we obtain if we define it this way? I suppose it has something to do with exactness at $B$ but what? - -REPLY [11 votes]: To get a feel for this kind of relation, consider a short exact sequence -$0 \rightarrow V_1 \rightarrow V_2 \rightarrow V_3 \rightarrow 0$ -of finite-dimensional vector spaces over a field. What is the relation between their dimensions?<|endoftext|> -TITLE: How to show x and y are equal? -QUESTION [5 upvotes]: I'm working on a proof to show that f: $\mathbb{R} \to \mathbb{R}$ for an $f$ defined as $f(x) = x^3 - 6x^2 + 12x - 7$ is injective. Here is the general outline of the proof as I have it right now: -Proof: For a function to be injective, whenever $x,y \in A$ and $x\neq y$, then $f(x) \neq f(y)$, i.e., where $A$, $B$ are finite sets, every two elements of $A$ must have distinct images in $B$, which also implies that there must be at least as many elements in $B$ as in $A$ such that the cardinality of $A$ is less than or equals the cardinality of $B$. -We shall prove the contra-positive: If $\exists$ $f(x) = f(y)$, then $x=y.$ -Let $x^3 - 6x^2 + 12x - 7 = y^3 - 6y^2 + 12y - 7$. -Then by addition and some algebra, we get $x(x^2 - 6x + 12) = y(y^2 - 6y + 12)$ - -This feels dumb to ask but how do I continue to finally get the result that $x = y$? - -REPLY [3 votes]: Hint $\rm\,\ 0 = f(x)\!-\!f(y) = (\color{#C00}{x\!-\!y})\,\left(\left(\color{blue}{y\!-\!2}\ +\dfrac{\color{#0A0}{x\!-\!2}}2\right)^2\! + 3\,\left(\dfrac{\color{#0A0}{x\!-\!2}}{2}\right)^2\right)\iff \begin{eqnarray}\rm \color{#C00}{x=y}\quad \ or\\\rm\ \color{#0A0}{x=2}\ \ and\ \ \color{blue}{y = 2}\end{eqnarray}$ -Remark $\ $ This approach doesn't require noticing that $\rm\:f(x) = (x-2)^3\!+1.\:$ Rather, we use only $\rm\:x\!-\!y\:|\:f(x)\!-f(y)\:$ (Factor Theorem), and we complete the square in the quadratic cofactor.<|endoftext|> -TITLE: Compute the limit of $\sqrt{1-a}\sum\limits_{n=0}^{+\infty} a^{n^2}$ when $a\to1^-$ -QUESTION [10 upvotes]: I need some suggestions, hints for the limit when $a \to 1^{-}$ of $$\sqrt{\,1 - a\,}\,\sum_{n = 0}^{\infty}a^{n^{2}}.$$ - -REPLY [8 votes]: This problem was asked in Mathematical Tripos 1932 and here is an answer based on a hint given in Hardy's Pure Mathematics. -Let $h$ be positive and $n$ be a positive integer. Since $e^{-x^{2}}$ is decreasing it is easy to see that $$\int_{h}^{(n + 1)h}e^{-x^{2}}\,dx < h \sum_{k = 1}^{n}e^{-k^{2}h^{2}} < \int_{0}^{nh}e^{-x^{2}}\,dx$$ If we let $n \to \infty$ we get $$\int_{h}^{\infty}e^{-x^{2}}\,dx \leq h\sum_{k = 1}^{\infty}e^{-k^{2}h^{2}} \leq \int_{0}^{\infty}e^{-x^{2}}\,dx$$ Now we put $t = e^{-h^{2}}$ so that as $h \to 0^{+}$ we have $t \to 1^{-}$ and also note that $h/\sqrt{1 - t} \to 1$ as $h \to 0^{+}$ and thus on taking limits as $h \to 0^{+}$ or as $t \to 1^{-}$ we get $$\lim_{t \to 1^{-}}\sqrt{1 - t}\sum_{k = 1}^{\infty}t^{k^{2}} = \int_{0}^{\infty}e^{-x^{2}}\,dx$$ Note that the question given by OP requires the lowest index in $\sum t^{k^{2}}$ to be $k = 0$ but above we have it as $k = 1$. This does not make any difference as the term corresponding to $k = 0$ is $1$ and hence a term $\sqrt{1 - t}$ gets added up which also tends to $0$. Hence we have $$\lim_{t \to 1^{-}}\sqrt{1 - t}\sum_{k = 0}^{\infty}t^{k^{2}} = \int_{0}^{\infty}e^{-x^{2}}\,dx$$<|endoftext|> -TITLE: Questions on fractional Laplacian graph spectra -QUESTION [8 upvotes]: Both the signed ($D-A$) and unsigned ($D+A$) Laplacian are of interest in spectral graph theory, see eg Cvetkovic: Bibliography on the signless Laplacian eigenvalues: first one hundred references. -Considering the spectral function $D+\rho A$ over the interval $\rho \in [-1,1]$ rather than just the extreme values $\rho={-1,1}$, results in the curves $\lambda_i(\rho)$. -For example, the following fractional spectra are are shown as point interpolations (as opposed to curve following, as in bifurcation theory) corresponding to some random Bernoulli graphs with increasing connectivity: - - - - - -These experimental results raise basic questions: - -If graphs are $M$-cospectral on $\rho = {-1,1}$ are they cospectral for $\rho \in [-1,1]$? -Are there no intersections of $\lambda_i(\rho)$ outside $\rho \in [-1,1]$? -Does $\lim_{\lambda_i \to \infty}\frac{d\lambda_i}{d\rho} = const?$ -What's the combinatorial interpretation of intersections in $\rho \in [-1,1]$? - -Feel free to add your own conjectures. - -REPLY [3 votes]: Let G0 be the graph with edge set -$$ - (0, 1), (0, 7), (0, 8), (1, 2), (2, 3), (2, 5), (2, 7), (2, 8), (3, 4), (3, 6), (4, 5), (5, 6), (6, 7) -$$ -and let G1 be the graph with edge set -$$ - (0, 1), (0, 3), (0, 5), (0, 8), (1, 2), (1, 4), (2, 3), (2, 8), (2, 9), (3, 4), (4, 5), (4, 8), (6, 7) -$$ -(Neither graph is connected and G0 has an isolated vertex.) -The Laplacian matrices of these two graphs are cospectral. As these graphs -are bipartite, their signless Laplacians are also cospectral. -However the respective degree sequences are -$$ - [0, 2, 2, 2, 3, 3, 3, 3, 3, 5], [1, 1, 1, 2, 3, 3, 3, 4, 4, 4] -$$ -and so the answer to your first question is no.<|endoftext|> -TITLE: Pell type equation: $x^2-py^2=a$ -QUESTION [5 upvotes]: Let $p=4k+1$ be a prime number such that $p=a^2+b^2$, where $a$ is an odd integer.Prove that the equation $$x^2-py^2=a$$ has at least a solution in $\mathbb{Z}$. - -REPLY [4 votes]: See e.g. Gary Walsh, On a question of Kaplansky".<|endoftext|> -TITLE: How $f^{2}=g^{6}-1$ implies $f$ and $g$ are constant? -QUESTION [7 upvotes]: From Harvard qualification exam, 1990. -Let $f,g$ be two entire holomorphic functions satisfy the property $$f(z)^{2}=g(z)^{6}-1,\forall z\in \mathbb{C}$$ Prove that $f,g$ are constant functions. Would this be the same if $f,g$ are allowed to be meromorphic functions? -The problem comes with a hint that I should think about the algebraic curve $$y^{2}=x^{6}-1$$but I do not see how they are related. I know this curve is hyperellipitic (from Riemann-Hurwitz or simply the wiki article). But how this help?(this curve should be of genus 2). Taking a short look at the entire function article also seems to be no help. Is the author implying $f,g$ is not best to be treated by classical Riemann Surface applications (as opposed to algebraic geometry ones)? - -REPLY [2 votes]: Enlightened by various hints, here is a 'proof' which is most likely to be wrong somewhere. I have not touched entire functions and covering maps for a long time. So suggestions on improvement is welcome. -If $f,g$ are meromorphic, then we can write them as quotients of rational functions. Thus we can extend $f,g$ to the Riemann Sphere by allowing $\infty$. Riemann-Hurwitz would imply we cannot map from low genus surface to a higher genus one: therefore there is no map from $S^{2}$ to $X$. And thus $f,g$ must be constants. -Assume $f,g$ are holomorphic over $\mathbb{C}$ with possible non-removable singularity at $\infty$. Since $f,g$ are both open maps if they are non-constant, together $(f,g)$ should map the open set $\overline{\mathbb{C}}-\{\infty\}$ to a connected open component to $X=\{(z,w),z^{2}=w^{6}-1\}$. $X$ is a Riemann surface that can be compactified to be homeomorphic to a two hole torus. Since $X$ is connected the map must be surjective. Further, by inverse mapping theorem since $f,g$ are assumed to non-constant, $(f',g')$ are not zero except in a discrete set of points. Ignore this for now (should be tractable by using local biholomorphic transformation to a locally ramified function) we may view the map $$F=(f,g):\mathbb{C}\rightarrow X$$ as a covering map since we have a discrete inverse image at every neighborhood of $X$ according to $F$'s degree at that point. -The universal cover of $X$ is a closed disk $D^{2}$ by the fundamental diagram. Therefore by the covering property we have a unique lift $p$ from $\mathbb{C}$ to $D^{2}$ that preserves $F$ such that $p:D^{2}\rightarrow \mathbb{C}$ satisfies $$p\circ F=q$$ where $q$ is the covering map from $D^{2}$ to $X$. Further $p$ is holomorphic. But this is contradictory since by mean value theorem (or maximal modulo principle) a non-constant holomorphic function attains its maximal absolute at the boundary. Thus $p$ must be bounded by some constant and cannot reach the whole complex plane. This showed at least one of $f,g$ must be a constant. And by definition this showed both $f,g$ are constants.<|endoftext|> -TITLE: Boundary of $L^1$ space -QUESTION [5 upvotes]: Is there any rigorous or heuristic notion of boundary of $L^1$ that is studied? I mean something loosely like the collection of functions or distributions defined by -$$\left\{f\notin L^1: f_n\to f\quad\text{a.e.}\quad \text{as} \quad n\to \infty \quad \text{where} \quad f_n\in L^1\right\}$$ -And what kind of characterizations or properties of this "surface" are known? -Edit: Changed to pointwise convergence. - -REPLY [7 votes]: I believe the set is just the set of all (non-integrable) measurable functions with $\sigma$-finite support. -In particular, if the space in question is $\sigma$-finite (like $\mathbf R$ with Lebesgue measure), then all measurable functions are pointwise limits of integrable functions. -Edit: cleaned up a bit. -In one direction: - -Clearly it is enough to show that for nonnegative functions; -Let $f$ be a nonnegative measurable function, and $S_n$ an increasing sequence of sets of finite measure such that $\mathrm{supp}f\subseteq \bigcup_nS_n$. -Let $A=\lbrace x\mid f(x)=\infty\rbrace$, $A_n=\lbrace x\mid f(x)1/m\rbrace$. -The support of the limit of $f_n$ is contained in $\bigcup_n B_n$ (because if at some point none of the functions is nonzero, neither is the limit), and hence $\sigma$-finite, so we're done.<|endoftext|> -TITLE: How to classify 3-sheeted covering space for $S_{1}\vee S_{1}$? -QUESTION [7 upvotes]: This might be a duplicate. This question also feels routine (it is also the execrise 10, page 88 in Hatcher). From Harvard qualification exam, 1990. -Let $X$ be figure eight. -1) How many 3-sheeted, connected covering space are there for $X$ up to isomorphism? -2) How many of these are normal (i.e Galois) covering spaces? -There are almost uncountably many covering spaces $Y$ for $X$ (one can check the corresponding page in Hatcher, page 58). The question is how to classify them nicely. I know that $p_{*}\pi_{1}(Y)$ has index 3 in $\pi_{1}(X)=\mathbb{Z}* \mathbb{Z}$(the free group generated by two generators). But I do not know how to find all index 3 subgroups of $\mathbb{Z}*\mathbb{Z}$. On the other hand if $H$ is normal in $\mathbb{Z}*\mathbb{Z}$, then the above question can be greatly simplified, but I still do not know how to solve it precisely. I tried to think in terms of deck transformations, etc but did not get anywhere. - -REPLY [9 votes]: Hint: Instead of thinking about index 3 subgroups of $\Bbb Z \star \Bbb Z$, consider what connected 3-fold covers of $S_1 \vee S_1$ look like. Any such cover is a connected graph on 3 vertices of valence 4 (why?), and there are only finitely many such graphs. Then use deck transformations to check if each cover is regular.<|endoftext|> -TITLE: What is the Birch and Swinnerton-Dyer Conjecture? -QUESTION [5 upvotes]: This is probably a really silly question, but I was wondering if someone could explain the Birch and Swinnerton-Dyer conjecture to me in a simple way. I've read a lot about it, but cannot understand it. Maybe posting here will help. -What I do understand is the concept of ranks on an elliptic curve; for ever point that generates an infinite number of points on the elliptic curve, it contributes 1 to the rank. Further, I know that Mordell proved that every elliptic curve must have a finite rank. -Finally, with regards to BSD, I understand that the conjecture says that the "analytic" way of computing the rank is the same as the "algebraic" way of computing the rank. But what I cannot seem to understand is what exactly is the analytic way of computing the rank, or the algebraic way. -If anyone wants to help me out, that would be much appreciated! Thanks! - -REPLY [18 votes]: Your phrasing of the problem, in terms of two ways to compute the rank (analytic and algebraic), is not quite correct. Rather the conjecture states that two numbers are equal. -On the one hand there is the rank of the elliptic curve, which is as you said: the number of independent rational points of infinite order. In the context of BSD, one calls this "the algebraic rank", $r_{alg}$, of the elliptic curve $E$. -Then, there is another quantity attached to $E$, also a non-negative integer, but of quite a different nature. To describe it, we begin by defining the $L$-function of $E$, -as a certain Euler product (a Dirichlet series given by taking a product -indexed by the primes), denoted $L(E,s)$, which is holomorphic in the half-plane $\Re s > 3/2$. Thanks to the modularity results of Wiles et. al. we know that this function has analytic continuation to the whole complex plane, and so in particular, we can consider its order of vanishing at $s = 1$. In the context of BSD, we call this "the analytic rank", $r_{an}$, of $E$. (Note though that this is just a name; $r_{an}$ is not defined in terms of the rank of any abelian group, but rather is the order of vanishing at $s = 1$ of an entire function; the only reason for calling it a "rank" comes from the BSD conjecture.) -The BSD conjecture is then that $r_{an} = r_{alg}$; in other words, we can determine the rank of $E$ by determining the order of vanishing of $L(E,s)$ at $s = 1$. To my mind the most striking consequence of this is that -$r_{an} > 0$ if and only if $r_{alg} > 0$; in other words, we (conjecturally) can determine whether or not $E$ has infinitely many rational points (i.e. has positive rank) by evaluating $L(E,1)$ and determining whether or not it equals zero. -In fact one direction of this weaker form of the conjecture is actually known: it is known that if $L(E,1) \neq 0,$ then $E$ has only finitely many rational points; this is due to Gross, Zagier, and Kolyvagin, with another proof more recently by Kato. - -(I won't give the definition of $L(E,s)$ here; you can find it in many places, -including in Wiles's write-up I'm sure, or just by googling "L-function of elliptic curve".)<|endoftext|> -TITLE: Show that $\overline{U\cap \overline{A}}=\overline{U\cap A}.$ -QUESTION [5 upvotes]: Show that: - -For every open set $U$ in a topological space $X$ and every $A\subset X$ we have $$\overline{U\cap \overline{A}}=\overline{U\cap A}.$$ - -The simple and new proof is welcome. Thanks for any help. - -REPLY [5 votes]: Clearly $\overline{ U \cap \overline{A} } \supseteq \overline{ U \cap A }$, so we need only show the opposite. -Suppose that $x \in \overline{ U \cap \overline{ A } }$, and let $V$ be any open neighbourhood of $x$. Then $V \cap ( U \cap \overline{ A } ) \neq \emptyset$. As $V \cap U$ is open, it then follows that $( V \cap U ) \cap A \neq \emptyset$ , and so $V \cap ( U \cap A ) \neq \emptyset$. Therefore $x \in \overline{ U \cap A }$. -(The only non-trivial step depends on the following fact, easily proved: If $U$ is an open subset of a topological space $X$ and $A \subseteq X$ is arbitrary, then $U \cap A = \emptyset$ implies $U \cap \overline{A} = \emptyset$.)<|endoftext|> -TITLE: Effective Upper Bound for the Number of Prime Divisors -QUESTION [18 upvotes]: Let $\omega(n) = \sum_{p \mid n} 1$. Robin proves for $n > 2$, -\begin{align} -\omega(n) < \frac{\log n}{\log \log n} + 1.4573 \frac{\log n}{(\log \log n)^{2}}. -\end{align} -Is there a similar tight effective upper bound for $\Omega(n) = \sum_{p \mid n} \text{ord}_{p}(n)$ or at least an upper bound in terms of $\omega(n)$? - -REPLY [15 votes]: The number of prime divisors counted with multiplicity is maximized for powers of $2$ and so -$$\Omega(n)\le\frac{\log n}{\log 2}=\log_2 n$$ -and since it is exactly equal for infinitely many $n$ it is also the tighest possible bound.<|endoftext|> -TITLE: Can you prove why consecutive diagonal intersection points show decreasing fractions inside a rectangle? -QUESTION [19 upvotes]: When I was in third grade, I was playing with rectangles and diagonal lines, and discovered something very interesting with fractions. I've shown several math teachers and professors over the years, and never got an answer. Just a few, "Wow, that's neat!" -Draw a rectangle. Draw a line from the top left corner to the bottom right corner. Then draw a line from the top right corner to the bottom left corner. The intersection obviously becomes 1/2 units of the rectangle's width. -Now draw a line from the last intersection to the bottom line of the rectangle, and then from that point to the top right corner of the rectangle. The new intersection becomes 1/3 units of the rectangle's width. -Keep doing this and the denominator of the fraction increases by one each time to infinite. Why does this happen? I don't know how to prove why this happens, but it would be interesting if someone could. Can you? I never became a mathematician to prove it, but if it's easy, please forgive my mathematical ignorance. I tried this several years ago with AutoCAD and it does in fact work out. - -REPLY [2 votes]: To address all those talking about projections, I can contribute a little here. This also constitutes a geometric proof which relies on projective geometry. Firstly, we can formalize the your method for drawing these lines using projections. Call the bottom of the rectangle M. Call the diagonal from the top left to bottom right L. Call the top right corner O. The projectivity you define is a product of two perspectivities. First is M to L with center of perspectivity O. Then, you go from L to M using the relevant point at infinity (where all the vertical lines meet). This tells us your method relates points on M projectively. -The heart of this proof relies on the fact that given four points, ABCD with distances $\frac{1}{n}, \frac{1}{n+1}, \frac{1}{n+2}, \frac{1}{n+3}$, from some fixed point (such as the bottom right corner), the cross ratio of (AC,BD) is constant: $\frac{-1}{3}$ in fact. You can check it if you want. -Note that I use two well known facts from projective geometry. The proofs for these are on Wikipedia. -1) Projectivites preserve cross ratios -2) Given three collinear points and a cross ratio, there exists a unique point satisfying the cross ratio equation. -Now, we induct. This is less painful than it sounds. You proved the case for $n = 1$. So base case is done. -Suppose that our our claim is true for some fixed n. Then take points points $x_1, x_2, x_3, x_4$ corresponding to $n$. We project all four points, giving us $x_2, x_3, x_4,$ and a new point $x_5$. By 1, cross ratios are invariant so the cross ratio of these four new points is $\frac{-1}{3}$. By 2, we know that this cross ratio corresponds only to points in the $\frac{1}{n}$ configuration described above. So $x_5$ has distance $\frac{1}{n+4}$ from the bottom right corner. -I'm open to critiques here. There is probably some way to condense this, or at least improve readability.<|endoftext|> -TITLE: Canonical Isomorphism Between $\mathbf{V}$ and $(\mathbf{V}^*)^*$ -QUESTION [12 upvotes]: For the finite-dimensional case, we have a canonical isomorphism between $\mathbf{V}$, a vector space with the usual addition and scalar multiplication, and $(\mathbf{V}^*)^*$, the "dual of the dual of $\mathbf{V}$." This canonical isomorphism means that the isomorphism is always the same, independent of additional choices. -We can define a map $I : \mathbf{V} \to (\mathbf{V}^*)^*$ by $$x \mapsto I(x) \in (\mathbf{V}^*)^* \ \text{ where } \ I(x)(f) = f(x) \ \text{for any } \ f \in \mathbf{V}^*$$ -My Question: what can go wrong in the infinite-dimensional case? The notes I am studying remark that if $\mathbf{V}$ is finite-dimensional, then $I$ is an isomorphism, but in the infinite-dimensional case we can go wrong? How? - -REPLY [4 votes]: (This is just a bit too longer to be a comment) -It is important to remark that the fact that for an infinite dimensional space $\bf V$ is not isomorphic to $\bf V^{**}$ requires the axiom of choice. Furthermore, the counterexample is not some pathological space. -In some models without the axiom of choice a peculiar thing happens: every linear operator from a Banach space into a normed space is continuous. One example of such model is Solovay's model in which all sets of reals are Lebesgue measurable. We remark that in this model the principle of Dependent Choice holds, which is enough to develop most classical analysis. -In such model $\ell_2$, a Banach space, has only continuous linear functionals, so the algebraic dual is the same as the topological dual. The fact that $\ell_2$ is a self-dual (in the topological sense) does not require much of the axiom of choice, not more than we have in Solovay's model anyway. -Now we need to verify that the evaluation map is a linear isomorphism, it is that and more. It is an isometry. This is not a very hard exercise in applying basic functional analysis theorems.<|endoftext|> -TITLE: Regular Polyhedrons -QUESTION [7 upvotes]: In $\mathbb{R}^3$, there are five regular polyhedrons (up to similarity), and can be parametrized by number of vertices, edges and faces. What is the number of regular polyhedrons in $\mathbb{R}^n$, and their parametrization? Please suggest the reference(s) also. (Thanks in advance.) - -REPLY [9 votes]: In short, what happens is the following. The $n$-dimensional analogue of a Platonic solid is called a regular polytope. In any dimension you are guaranteed three "boring" regular polytopes: the $n$-dimensional version of the tetrahedron (the $n$-simplex), the $n$-dimensional hypercube, and its dual, the $n$-dimensional version of the octahedron. In three dimensions, as you know, there are two others. In four dimensions there are others as well, called the 24-cell, 120-cell and 600-cell. In dimensions five and above the boring regular polytopes are the only ones that exist. -The wiki pages http://en.wikipedia.org/wiki/Regular_polytopes and http://en.wikipedia.org/wiki/Convex_regular_4-polytope are good places to start. Coxeter's book Regular Polytopes is very comprehensive. Another approach is to look at these things through their reflection symmetry groups: Coxeter's book is a good source for this too, see also http://en.wikipedia.org/wiki/Coxeter_group<|endoftext|> -TITLE: What is an example of a vector field that is not left-invariant? -QUESTION [5 upvotes]: Let $G$ be a Lie group, $L_g$ the left-translation on this group with differential $d L_g$. A vector field $X$ on $G$ is called left-invariant if -$$ X \circ L_g = d L_g \circ X \quad \forall g \in G$$ -i.e. -$$ X_{gh} = (d L_g)_h (X_h) \quad \forall g,h \in G. $$ -Now, this definition seems so natural to me that I cannot come up with a non-trivial counterexample for a vector field that is $\textit{not}$ left-invariant. In my mind, pushing forward on the tangent space is basically always the same as the group action... -Could you provide me with such a counterexample that helps understand the notion of left-invariance? - -REPLY [7 votes]: You could start taking the Lie group $(\mathbb R^n,+),$ and considering what does it means for a vector field $X$ on $\mathbb R^n$ to be left-invariant.<|endoftext|> -TITLE: Area Bounded by Polar Curves -QUESTION [7 upvotes]: I am answering sample exams for my Calculus class and my attention was caught by the following item. -Set-up the definite integral or sum of definite integrals equal to the area of the region above the polar axis, inside the limaçon $r = 3 + 2 \sin \theta$ and outside the lemniscate $r^2 = 32 \cos 2\theta$ given that the two curves intersect at $(4,\frac{\pi}{6})$. - -At first, I thought that the area is given by $$\dfrac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}{[(3 + 2\sin \theta)^2 - (32 \cos 2\theta)] \mathrm{d}\theta}$$ but I know that the area of the lemniscate is tricky so I may have given a smaller area. -My question is this: How do you know the limits of integration for lemniscates? (I know that the limits of integration for the area of the lemniscate alone is from $-\frac{\pi}{4}$ to $\frac{\pi}{4}$, but how about for small portions of the curve?) -I'll appreciate any help. Thank you so much. - -REPLY [5 votes]: There was lemniscate trouble, as you feared. Up to $\theta=\frac{\pi}{4}$, everything was fine. But from $\frac{\pi}{4}$ and for quite a while (up to $\frac{3\pi}{4}$), $\cos 2\theta$ is negative, so $r^2$ is negative and there is no curve. The integral doesn't know and doesn't care: it cheerfully "adds up" these negatives, giving the wrong answer. -So for the limaçon part, if you do things in your style, you will have to break up the limaçon integral into two integrals, $\frac{\pi}{6}$ to $\frac{\pi}{4}$ and then $\frac{3\pi}{4}$ to $\frac{5\pi}{6}$. -Since the integrals for the lemniscate and the limaçon are over different intervals, we cannot express their difference as a single integral. -What I would do is to use the symmetry, take the right-half of the region and multiply the resulting area by $2$. The limaçon part to be subtracted uses the integral from $\frac{\pi}{6}$ to $\frac{\pi}{4}$. -So there is $\frac{1}{2}$ as in your formula, but ultimately we multiply by $2$, so our area is -$$\int_{\pi/6}^{\pi/2}(3+2\sin \theta)^2\,d\theta-\int_{\pi/6}^{\pi/4}32\cos 2\theta\,d\theta.$$ -More pleasant, fewer minus signs to worry about!<|endoftext|> -TITLE: Does $\int_{0}^{\infty} \cos (x^2) dx$ diverge absolutely? -QUESTION [5 upvotes]: I believe it does, but i would like some help formulating a proof. - -REPLY [5 votes]: It's equivalent to the convergence of $\int_\pi^{\infty}\frac{|\cos t|}{\sqrt t}dt$, after having used the substitution $x^2=t$. -We have -$$ -\int_{\pi}^{N\pi}\frac{|\cos t|}{\sqrt t}dt=\sum_{n=1}^{N-1}\int_{n\pi}^{(n+1)\pi}\frac{|\cos t|}{\sqrt t}dt$$ -Use $\pi$ periodicity of $|\cos|$ and a substitution $s=t-n\pi$ to get bound which doesn't depend on $n$. -Find a good below bound will help to show the divergence. - -This argument can be applied for the divergence of $\int_0^{+\infty}|\cos(x^p)|dx$, $p>0$.<|endoftext|> -TITLE: What interpretations exists for the singleton of the singleton of the empty set $\{\{\varnothing\}\}$ and its singleton $\{\{\{\varnothing\}\}\}$? -QUESTION [5 upvotes]: What I know - -I know from Peano's axioms that the empty set is equivalent to the natural number $0$ and that the singleton of the empty set is equivalent to the natural number $1$. ( http://www.proofwiki.org/wiki/Peano%27s_Axioms_Uniquely_Define_Natural_Numbers , http://www.proofwiki.org/wiki/Axiom:Peano%27s_Axioms) -I know from Kuratowski that the set $\{\{A\}\}$ is equivalent to the ordered pair $\langle A,A\rangle$. - -What I want to know - -Are there any alternative interpretations of $\{\{A\}\}$ other than Kuratowski's? -How can we express an ordered homogeneous triple (e.g. Would $\langle A,A,A\rangle$ be equivalent to $\{\{\{A\}\}\}$?) -It seems like the set $\{\{\varnothing\}\}$ can be interpreted as either $\{1\}$ or $\langle\varnothing,\varnothing\rangle$. Should $\{\{\{\varnothing\}\}\}$ be interpreted as $\langle\varnothing,\varnothing,\varnothing\rangle$ or as $\langle 1,1\rangle$. Is it possible that these tuples are in some way equivalent? - -Note, re: Peano's Axioms -My understanding of Peano's axioms is consistent with the formulation at WolframAlpha, which is expressed as the 5 postulates: - -$0$ is a number. -If $a$ is a number, the successor of $a$ is a number. -$0$ is not the successor of a number -Two numbers of which the successors are equal are themselves equal. -If a set $S$ of numbers contains zero and also the successor of every number in $S$, then every number is in $S$. - -These axioms are expressed in set notation at http://www.proofwiki.org/wiki/Axiom:Peano%27s_Axioms / http://www.proofwiki.org/wiki/Peano%27s_Axioms_Uniquely_Define_Natural_Numbers . - -REPLY [10 votes]: You're taking the Kuratowski ordered pair definition too seriously. It doesn't have any intrinsic intuitive meaning, and in particular does not tell us anything deep about the sets it chooses to represent ordered pair. Its only role is as a technical detail in the proof of - -ORDERED-PAIR METATHEOREM. There exists a formula $\phi(x,y,p)$ with three free variables such that - -For all $x$ and $y$ there is exactly one $p$ such that $\phi(x,y,p)$. -If $\phi(x,y,p)$ and $\phi(x',y',p)$, then $x=x'$ and $y=y'$. - -are theorems of set theory. - -In the entire rest of the development of set theory (including the very formulation of several of the axioms), and of the development of the rest of mathematics as an application of set theory, we forget completely which formula this $\phi(x,y,p)$ actually is and just write it as $\langle x,y\rangle=p$ instead. -This involves taking care that we never depend on any properties of ordered pairs other than what the metatheorem guarantees -- in particular we never put ordered pairs and other things into the same set and expect to be able to tell them apart later, because a priori anything might be a pair. We could have proved the ordered-pair metatheorem for a different $\phi$ -- for example one that represents the particular pair $\langle \omega_3,\{42,19\}\rangle$ as $\varnothing$ and everything else as Kuratowski pairs in reverse order -- and that should not make any difference at all for the rest of the development. -So yes, it is true in a technical sense that $\langle 0,0\rangle -$ and $\{1\}$ are the same set. But that is just a coincidence -- an implementation detail -- and we deliberately don't depend on such details when we do mathematics. -It makes very good sense, as Asaf suggests, to view this as a type system. In that view, asking whether $A=B$ when we know $A$ as an ordered pair, and $B$ as a set with particular members, is a type error which produces an implementation-dependent (even if definite) truth value.<|endoftext|> -TITLE: Finite $G$ with involutory automorphism $\alpha$ with no nontrivial fixed points. Proving properties of $\alpha$. -QUESTION [6 upvotes]: The question at hand is: - -Let G be a finite group and $\alpha$ an involutory automorphism of G, which doesn't fixate any element aside from the trivial one. - 1) Prove that $ g \mapsto g^{-1}g^{\alpha} $ is an injection - 2) Prove that $\alpha$ maps every element to its inverse - 3) Prove that G is abelian - -I think I've found 1). -I assume $ g_1^{-1}g_1^{\alpha} = g_2^{-1}g_2^{\alpha} $ and from this I get $ (g_2g_1^{-1})^\alpha = g_2g_1^{-1} $, so $g_2 = g_1$ (is this correct?) -For 2) I'm sort of stumped though, not sure how to start proving that, so any help please? - -REPLY [3 votes]: 1) is correct as you've written. 3) is an easy consqeuence of 2), so I'll leave that part to you. -For 2) we need to use the fact that $\alpha$ is involutory... Note that $\varphi:g\mapsto g^{-1}g^\alpha$ is injective, so it is bijective (because $G$ is finite). Consider an arbitrary $g$, and $h:=\varphi^{-1}(g)$. What is $\varphi^2(h)$ (in terms of $h$ first, then in terms of $g$)?<|endoftext|> -TITLE: Showing the sum of orthogonal projections with orthogonal ranges is also an orthogonal projection -QUESTION [7 upvotes]: Show that if $P$ and $Q$ are two orthogonal projections with orthogonal ranges, then $P+Q$ is also an orthogonal projection. - -First I need to show $(P+Q)^\ast = P+Q$. I am thinking that since -\begin{align*} -((P+Q)^\ast f , g) & = (f,(P+Q)g) \\ - & = (f,Pg) + (f,Qg) \\ - & = (P^\ast f,g) + (Q^\ast f,g) \\ - & = (Pf,g) + (Qf,g) \\ - & = ((P+Q)f,g), -\end{align*} -we get $(P+Q)^\ast=P+Q$. -I am not sure if what I am thinking is right since I assumed that $(P+Q)f=Pf+Qf$ is true for any bounded linear operator $P$, $Q$. -For $(P+Q)^2=P+Q$, I use -$$(P+Q)^2= P^2 + Q^2 + PQ +QP,$$ -but I cant show $PQ=0$ and $QP=0$. -Anyone can help me? Thanks. - -REPLY [3 votes]: To complete your proof we need the following observations. -If $\langle f,g\rangle=0$ for all $g\in H$, then $f=0$. Indeed, take $g=f$, then you get $\langle f,g\rangle=0$. By definition of inner product this implies $f=0$. -Since $\mathrm{Im}(P)\perp\mathrm{Im}(Q)$, then for all $f,g\in H$ we have $\langle Pf,Qg\rangle=0$. which is equivalent to $\langle Q^*Pf,g\rangle=0$ for all $f,g\in H$. Using observation from previous section we see that $Q^*P(f)=0$ for all $f\in H$, i.e. $Q^*P=0$. Since $Q^*=Q$ and $P^*=P$ we conclude -$$ -QP=Q^*P=0 -$$ -$$ -PQ=P^*Q^*=(QP)^*=0^*=0 -$$ -In fact $R$ is an orthogonal projection iff $R=R^*=R^2$. In this case we can prove your result almost algebraically -$$ -(P+Q)^*=P^*+Q^*=P+Q -$$ -$$ -(P+Q)^2=P^2+PQ+QP+Q^2=P+0+0+Q=P+Q -$$ -I said almost, because prove of $PQ=QP=0$ requires some machinery with elements of $H$ and its inner product.<|endoftext|> -TITLE: Prerequisites for studying smooth manifold theory? -QUESTION [6 upvotes]: I am attending first year graduate school in about three weeks and one of the courses I am taking is an introduction to smooth manifolds. Unfortunately, my topology knowledge is minimal, limited to self study. Besides some basic topological definitions, are there specific areas where I should become acquainted with? I have heard multivariable calculus is a 'prerequisite' for the study of smooth manifolds -- is this from an intuition perspective, and if so, what parts and to what degree of rigor are they referring? -Thank you. (My apologies if this is a repeat question, by the way -- I could not find it) - -REPLY [14 votes]: You will need - -Calculus/real analysis of functions of one and several variables up to and including the implicit and inverse function theorem. This (the implicit function theorem) is the basic starting point for (smooth) manifold theory, you will not be able to get anywhere without it. -sound knowledge of linear algebra (at least in/for real vector spaces) -1-d integration (Riemann integral will suffice in the beginning), preferably basic knowledge of an n-dimensional integration theory. -basic theory of ODE will sooner or later become important, in order to be able to deal with, e.g, the flow of vector fields (actually one needs existence (Picard-Lindelöf) and rather soon smooth dependence of the solution on the initial values, but the latter may be stated and believed. It'll become difficult without existence of solutions). - -Depending on your book of choice everything else will probably be developed in the course of said book, see the other answers for some suggestion. The basics in topology you'll need will likely be known to you, once you really covered the above list. Spivak's calculus on manifolds comes to my mind.<|endoftext|> -TITLE: What are some examples of vector spaces that aren't graded? -QUESTION [12 upvotes]: From wikipedia: a vector space $V$ is graded if it decomposes into direct sum $ \oplus_{n \geq 0} V_n$ of vector spaces $V_n$. -So as far as I understand things, any vector space with a countable basis is graded: Let $V$ be a vector space over a field $k$ with basis $\{v_n\}_{n\in\mathbb{N}}$, then $V = \oplus_{n\geq 0} k\cdot v_n$. Then the only vector spaces that I can think of that aren't obviously graded are things like $C(X)$, the space of continuous functions on some manifold $X$ -Is this correct? are there any more? or do I not understand something? -Thanks - -REPLY [20 votes]: Grading isn't a property of a vector space: it's extra structure attached to a vector space, in the same way that a multiplication is an extra structure you attach to a set to make it a group. So this is a little like asking "what are some examples of sets that aren't groups?" (As it turns out, every set can be equipped with a group structure, and this is equivalent to the axiom of choice. But this is missing the point.) -Every vector space admits a trivial grading in which $V_0 = V$ and $V_n = 0$ for all $n \ge 1$. But we often encounter vector spaces (such as the space of polynomials) with natural gradings, they are usually nontrivial, and taking advantage of this extra structure is useful in various ways.<|endoftext|> -TITLE: The expected payoff of a dice game -QUESTION [51 upvotes]: There's a question in my Olympiad questions book which I can't seem to solve: - -You have the option to throw a die up to three times. You will earn - the face value of the die. You have the option to stop after each - throw and walk away with the money earned. The earnings are not additive. What is the expected payoff of this game? - -I found a solution here but I don't understand it. - -REPLY [3 votes]: Start from the end. -What happens if you do not care about the first two rolls? The expected payoff is $(1+2+3+4+5+6)/6 = 3.5$. Now you have a freedom to go step earlier and make a decision about your strategy after the second roll (you still do not care about the first roll). We can actually find the optimal scenario which maximizes payoff: -$$E[X] = \frac{n}{6}3.5 + \frac{1}{6}[6+5+ \ldots+(n+1)] = \frac{n}{6}3.5 + \frac{1}{6}\cdot\frac{7+n}{2}(6-n)$$ -It has a maximum for $n=3$ and then $E[X] = 4.25$. The meaning of the above expression is that if we roll $1,2,\ldots , n$ then we should roll once again. Otherwise we stop. -Ok, we move to the first roll and apply the same idea, but with a new expectation value $4.25$: -$$E[X] = \frac{n}{6}4.25 + \frac{1}{6}\cdot\frac{7+n}{2}(6-n)$$ -Maximum is for $n=4$ and we get $E[X] = 4.6666...$. -In the problem the optimal strategy is the following: - -If you roll $1,2,3$ or $4$ in the first try, then reroll the dice. Otherwise take what you got, because on average you will not get more than $4.25$. -In the second try. If you roll $1,2$ or $3$ you should roll once again. Otherwise stop, because on average you will not gain more than $3.5$.<|endoftext|> -TITLE: Four-parameter Beta distribution and Wikipedia -QUESTION [6 upvotes]: Sorry if it is not an appropriate place for such questions, but anyway can anybody please confirm that the formula for the density function of the four-parameter Beta distribution is correct in Wikipedia. It seems $(c - a)$ is missing in the denominator. Thank you. -Best regards, -Ivan - -REPLY [5 votes]: Yes, the factor is indeed missing. -Let $X$ be standard 2-parameter Beta random variable. The four-parameter one $Y$ is obtained by affine transformation $Y = (c-a) X + a$ for $c>a$. Then -$$ - f_Y(y) = \frac{1}{c-a} f_X\left(\frac{y-a}{c-a}\right) = \frac{1}{c-a} \left(\frac{y-a}{c-a} \right)^{\alpha-1} \left(\frac{c-y}{c-a} \right)^{\beta-1} \frac{\mathbf{1}(a < y -TITLE: The print problem: How to show it is not decidable? -QUESTION [5 upvotes]: I wonder the following reduction is correct. -I'm trying to show that the following problem "PRINT_BLANK" is not decidable. -Input: (a coding of) Turing machine M. -Question: Does the machine never types "blank" on the stripe when it runs on x? -An attempt for reduction: $AcceptProblem \leq PrintBlank: f(\langle M,x\rangle)= M'.$ -Given $M$ and $x$, we'll construct $M'$: For an input $y$ for $M'_x$, $M'_x$ simulates running of M on $w$. if $M$ accepts $w$, $M'_x$ writes "blank" and accepts y, otherwise it writes $x$ itself on the tape and rejects. -Any help? -Thanks! - -REPLY [4 votes]: HINT: to make your reduction work, you should alter $M$ slightly, can you figure out how? If not, let me know. -EDIT: What happens when M already writes a blank during it's execution.<|endoftext|> -TITLE: some elementary questions about cardinality -QUESTION [7 upvotes]: I know little about set theory and while reading some Algebra proof I had difficulty on some details. So my questions are : - -If $X$ is an infinite set and $Y$ is the set of all finite subsets of $X$, they have the same cardinality. How can I prove it ? -If $X$ is infinite, then it has the same cardinality of the product $X \times \mathbb{N}$, where $\mathbb{N}$ is the set of natural numbers. This seems pretty obvious, specially if you think separately when $X$ is countable and when it's not, but I cant prove this in a rigorous way. -Consider only infinite sets now. I know that if $X$ is countable, then it has the same cardinality as $X \times X$. Is this true in general (I mean, I know there is bijection between $\Bbb R$ and $\Bbb R^2$, but probably there is a counterexample that proves this is wrong for an arbitrary infinite set, or if not, I cannot prove it either) ? - -Thank you so much for the help ! - -REPLY [8 votes]: All these, and more, depend on an axiom called The Axiom of Choice. This axiom enforces infinite set to be very well-behaved, and the answer on all your questions become "yes". -If we do not assume the axiom of choice then the answers change, namely some of them become "sometimes" and others "no". -None of these propositions are trivial, especially without some background in set theory. It is even less trivial if you are not used to proving things from axiomatic systems like ZFC. -Let me answer your questions backwards. - -The axiom of choice is equivalent to the assertion "For every infinite $X$, $X\times X$ has the same cardinality as $X$". In particular this means that if we assume the axiom of choice then this is true for every infinite set, but if we assume that this axiom does not hold, then we are guaranteed a counterexample to exist. -The second question regarding $X\times\mathbb N$, again if we assume the axiom of choice the answer is "yes". One of the consequences of this axiom is that if $X$ is infinite then $X$ has a countably infinite subset. Combine with the above we have that $$|X|\leq|X\times\mathbb N|\leq|X\times X|=|X|$$ Therefore equality ensues. -This again amounts to the axiom of choice and the first answer. Assuming that $X\times X$ has the same cardinality as $X$, we can prove by induction that $|X^n|=|X|$ , and by the second answer we have that $|\bigcup X^n|=|X\times\mathbb N|=|X|$ again. -This is not exactly what we are looking for because this is the set of sequences rather than subsets, but now we can simply send every sequence $\langle a_i\mid i=0,\ldots,k\rangle$ to the set $\{a_i\mid i=0,\ldots,k\}$. Observe that in the sequences there might be repetitions which are removed when we take the set, but we still can get every finite subset. -Well, this is a surjection, so it does not quite yet prove that the set of all finite subsets of $X$ has the same cardinality as $X$, but another wonderful property following from the axiom of choice is that we can inverse surjective functions, namely there is an injection from the set of finite subsets into the set of finite sequences, and this is what we wanted. - -One last remark is that the counterexamples without the axiom of choice are not very natural, at least for a modern person. However they exist and have their place in mathematics. - -REPLY [6 votes]: The results described below freely use the Axiom of Choice. - -Theorem: If $X$ and $Y$ are infinite sets, then the cardinality of $X \times Y$ is the maximum of the cardinality of $X$ and the cardinality of $Y$. For a proof see e.g. Theorem 8 and the following discussion in these notes. - -This result answers your third question and also your second question, because every infinite set contains a countably infinite subset. -As for your first question: one way to do it is to write the set $Y$ of all finite subsets of your infinite set $X$ as a countable union of sets $Y_n$ for $n \in \mathbb{N}$, where each $Y_n$ is the set of subsets of $X$ having at most $n$ elements. There is a natural surjection $X^n \rightarrow Y_n$ which shows that $\# Y_n \leq \# X^n = \# X$. Thus $Y$ is a countable union of sets each having cardinality at most $X$, so $Y$ has cardinality at most $X$ (you can think of this in terms of $\# X \times \# \mathbb{N} = \# X$, for instance). On the other hand, the subset $Y_1$ of one element subsets of $X$ is naturally in bijection with $X$ itself, so also $\# X \leq \# Y$. It follows from Cantor-Bernstein that $\# X = \# Y$.<|endoftext|> -TITLE: If the unit sphere of a normed space is homogeneous is the space an inner product space? -QUESTION [7 upvotes]: Consider a normed vector space $V$. Suppose that for every pair of unit vectors $v,w$ there exists a linear isometry which sends $v$ to $w$ (and leaves the subspace spanned by $v$ and $w$ invariant). -Does it follow that $V$ is an inner product space? -By the formulation and the parallelogram law the problem immediately reduces to the two-dimensional case. I couldn't find a nice argument for that case. It would be nice to have a relatively elementary argument for this. -It seems to me that the answer for $\dim{V} = 2$ should be yes: the group $G$ of linear isometries of $V$ is a closed subgroup of $GL(2,\Bbb R)$ and since the unit sphere is compact, $G$ is compact. However, there aren't that many infinite compact subgroups of $GL(2,\Bbb R)$: up to conjugation, they are $O(2)$ and $SO(2)$. I would like this to tell me that I really have a circle as a unit sphere but I don't really know how to make this precise and how to conclude. -Could you please help me finishing this up or give me an alternative argument (or a counterexample)? -Bonus question: What happens if I drop the condition on leaving the subspace spanned by $v$ and $w$ invariant? -Thanks! - -REPLY [5 votes]: The answer to the bonus question in full generality is no. A Banach space is called transitive if for every unit vectors $u,v$ there is a surjective linear isometry mapping $u$ to $v$. There are nonseparable transitive Banach spaces which are not Hilbert spaces. The Banach-Mazur rotation problem asks whether every separable transitive Banach space is a Hilbert space. This remains unsolved. You can find a bunch of references by googling "transitive Banach space" or "Banach-Mazur problem". There is an MO thread with a proof for finite dimensions and a nonseparable counterexample.<|endoftext|> -TITLE: Zariski Open Sets are Dense? -QUESTION [21 upvotes]: Is it true that any nonempty open set is dense in the Zariski topology on $\mathbb{A}^n$? I'm pretty sure it is, but I can't think of a proof! Could someone possibly point me in the right direction? Many thanks! -Note: I am not asking about the Euclidean topology at all! - -REPLY [23 votes]: As an exercise, let's reduce everything to statements about polynomials. Every open set contains a basic open set $U$, which is the complement of the zero set of some nonzero polynomial $f$, so it suffices to show that these are Zariski dense. The Zariski closure of a set is the intersection of the zero sets of all polynomials vanishing on it. This is equal to $\mathbb{A}^n$ if and only if any polynomial vanishing on $U$ vanishes on $\mathbb{A}^n$. Thus the claim is equivalent to the following statement about polynomials: - -Suppose a polynomial $g$ has the property that if $f(x) \neq 0$, then $g(x) = 0$. Then $g(x) = 0$ for all $x$. - -But the condition is equivalent to the claim that $f(x) g(x) = 0$ for all $x$. Can you finish the problem from here? (Note that you need to assume $k$ infinite.)<|endoftext|> -TITLE: Generators of a cyclic group -QUESTION [10 upvotes]: In a paper there is a lemma: - -Let $G= \langle a,b \rangle$ be a finite cyclic group. Then $G=\langle ab^n \rangle$ for some integer $n$. - -The proof is omitted because it's "straightforward" but I'm not able to proof it. How does this work? - -REPLY [6 votes]: Yuval's solution, sans Dirichlet: let $n$ be the product of all the primes that divide the order of $G$ but don't divide $x$. Then $\gcd(x+ny,|G|)=1$. -Proof: Let $p$ be a prime dividing the order of $G$. If $p$ divides $x$, then it doesn't divide $y$ (since $\gcd(x,y,|G|)=1$), and it doesn't divide $n$ (by construction), so it doesn't divide $ny$, so it doesn't divide $x+ny$. -If $p$ doesn't divide $x$, then it divides $n$ (by construction), so it divides $ny$, so it doesn't divide $x+ny$. So no prime dividing the order of $G$ divides $x+ny$.<|endoftext|> -TITLE: Basic help with sigma algebras and borel sets -QUESTION [11 upvotes]: In non-rigorous, intuitive terms, can someone briefly define: -(i) a measurable set -(ii) a borel set -(iii) a sigma algebra -(iv) a borel sigma algebra -Im studying these concepts independently in preparation for a course in the fall and want to make sure I have a functional intuitive idea before learning them rigorously. Im not looking for references to textbooks, or textbook definitions, just a quick intuitive description from someone who is familiar. - -REPLY [11 votes]: A $\sigma$-algebra is, like a topology, a set of subsets of some space $X$. It's both bigger and smaller than a topology, though: smaller, because it's only required to be closed under countable unions, instead of all unions, but bigger, because it's also closed under complementation, and thus by de Morgan's law, countable intersections. So it's both the open and closed sets you'd get starting from the base of a topology if you only took countable unions but also allowed countable intersections. -A measurable set is just an element of some $\sigma$-algebra on $X$. The content comes in when you define measures, which are functions from the $\sigma$-algebra to $[0,\infty]$ that satisfy a few obvious properties of a generalization of length. -The Borel algebra on some topological space $X$ is the $\sigma$-algebra generated by its topology: take all the closed and open sets, countable unions and intersections of those, complements of those, countable unions and intersections of those, and so on. A Borel set is just an element of the Borel algebra. -Note in this case what I said about a $\sigma$-algebra being smaller than a topology does not hold at all! The Borel algebra is important specifically because it's the smallest $\sigma$-algebra containing the topology: we obviously would like to have open and closed sets of reals measurable, and to accomplish that we've got to let at least all the Borel sets be measurable as well. You can think of the Borel sets as every reasonable set; in particular the rationals are Borel in $\mathbb{R}$ with the standard topology, though they're neither closed nor open.<|endoftext|> -TITLE: Finding the sum of series; sum of squares -QUESTION [5 upvotes]: Is there any way to find the sum of the below series ? -$$\underbrace{10^2 + 14^2 + 18^2 +\cdots}_{41\text{ terms}}$$ -I got asked this question in a competitive exam. - -REPLY [5 votes]: Let $$A=10^2+14^2+18^2+\dots$$ to 41 terms. Then $A=4B$, where $$B=5^2+7^2+9^2+\dots$$ to 41 terms. Let $$C=6^2+8^2+10^2+\dots$$ to 41 terms. Then $$B+C=5^2+6^2+7^2+\dots=(1^2+2^2+3^2+\dots)-(1^2+2^2+3^2+4^2)$$ where the sums go to $86^2$. Also, $C=4D$, where $$D=3^2+4^2+5^2+\dots=(1^2+2^2+3^2+\dots)-(1^2+2^2)$$ where the sums go to $43^2$. So you can calculate $D$ from Argon's last formula, then from $D$ you can get $C$; you can get $B+C$ from Argon, then $B$, then, finally, $A$.<|endoftext|> -TITLE: Stone-Weierstrass theorem proof (Rudin) -QUESTION [7 upvotes]: I am reading Rudin's proof (3rd edition) and am wondering what substitution is made to make it true that $P_n(x)=$ the integral from $-x$ to $1-x$ is equal to the same function integrated from -1 to 1. He says there's a substitution but I haven't found the right one. Thanks a lot - -REPLY [3 votes]: (Edit after a comment by Kirk Boyer) The formula in question is -$$P_n(x):=\int_{-1}^1 f(x+t)Q_n(t)\ dt=\int_{-x}^{1-x}f(x+t)Q_n(t)\ dt=\int_0^1f(t)Q_n(t-x)\ dt\ .$$ -One obtains the second integral on account of the assumption that $f(t)$ is $\equiv0$ outside $[0,1]$, the third integral substituting $t:=t'-x$ $\,(0\leq t'\leq1)$, and finally writing again $t$ in place of $t'$.<|endoftext|> -TITLE: Compute the series : $\sum_{n=1}^{\infty} \frac{4^n n!}{(2n)!}$ -QUESTION [13 upvotes]: How would you compute the following series? I'm interested in some easy approaches that would allow me to work it out. -$$\sum_{n=1}^{\infty} \frac{4^n n!}{(2n)!}$$ - -REPLY [12 votes]: Here is a generalization. Consider the generating function -$$f(x) = \sum_{n \ge 1} \frac{x^{2n} n!}{(2n)!}.$$ -It's an easy check that $2$ is in the radius of convergence, so that we need to compute $f(2)$ to solve your problem. However, it's easy to see that -$$f''(x) = 1 + \sum_{n \ge 1} \frac{x^{2n} (n + 1)!}{(2n)!}$$ -Moreover, -$$xf'(x) = 2 \sum_{n \ge 1} \frac{x^{2n} n!}{(2n)!} \cdot n$$ -We get the second-order ODE -$$\frac{xf'(x)}{2} + f(x) = f''(x) - 1, f(0) = 0, f'(0) = 0.$$ -And the solution -$$f(x) = \frac{1}{2} \cdot e^{x^2/4} \cdot x \sqrt{\pi} \mathrm{erf}(x/2).$$ -Just plug in $x = 2$ to get your answer, $e\sqrt{\pi} \mathrm{erf}(1)$.<|endoftext|> -TITLE: Show that if $\kappa$ is an uncountable cardinal, then $\kappa$ is an epsilon number -QUESTION [5 upvotes]: Firstly, I give the definition of the epsilon number: - -$\alpha$ is called an epsilon number iff $\omega^\alpha=\alpha$. - -Show that if $\kappa$ is an uncountable cardinal, then $\kappa$ is an epsilon number and there are $\kappa$ epsilon numbers below $\kappa$; In particular, the first epsilon number, called $\in_0$, is countable. -I've tried, however I have not any idea for this. Could anybody help me? - -REPLY [4 votes]: Here is a slightly easier way: -Lemma: For every $\alpha,\beta$ the ordinal exponentiation $\alpha^\beta$ has cardinality of at most $\max\{|\alpha|,|\beta|\}$. -Now use the definition of $\omega^\kappa=\sup\{\xi^\omega\mid\xi<\kappa\}$, since $|\xi|<\kappa$ we have that $\omega^\kappa\leq\kappa$, but since $\xi\leq\omega^\xi$ for all $\xi$, $\omega^\kappa=\kappa$. - -Here is an alternative way (a variation on the above suggestion): -Lemma: If $\alpha$ is an infinite ordinal then there is some $\varepsilon_\gamma\geq\alpha$ such that $|\alpha|=|\varepsilon_\gamma|$ -Hint for the proof: Use the fact that you need to close under countably many operations, and by the above Lemma none changes the cardinality. -Now show that the limit of $\varepsilon$ numbers is itself an $\varepsilon$ number, this is quite simple: -If $\beta=\sup\{\alpha_\gamma\mid\alpha_\gamma=\omega^{\alpha_\gamma}\text{ for }\gamma<\tau\}$ (for some $\tau$ that is) then by definition of ordinal exponentiation $$\omega^\beta=\sup\{\omega^{\alpha_\gamma}\mid\gamma<\tau\}=\sup\{\alpha_\gamma\mid\gamma<\tau\}=\beta$$ -Now we have that below $\kappa$ there is a cofinal sequence of $\varepsilon$ numbers, therefore it is an $\varepsilon$ number itself; now by induction we show that there are $\kappa$ many of them: - -If $\kappa$ is regular then every cofinal subset has cardinality $\kappa$ and we are done; -if $\kappa$ is singular there is an increasing sequence of regular cardinals $\kappa_i$, such that $\kappa = \sup\{\kappa_i\mid i\in I\}$. Below each one there are $\kappa_i$ many $\varepsilon$ numbers, therefore below $\kappa$ there is $\sup\{\kappa_i\mid i\in I\}=\kappa$ many $\varepsilon$ numbers.<|endoftext|> -TITLE: Hatcher problem 1.2.3 - technicality in proof of simply connectedness -QUESTION [6 upvotes]: I am trying to prove that $\Bbb{R}^n$ minus finitely many points $x_1,\ldots,x_m$ is simply connected, where $n \geq 3$. For days now I have tried many different arguments but I have found flaws in all of them. I have finally come up with one, except that there is some small detail that I need to know how to prove. -I prove that $\Bbb{R}^n$ minus finitely many points is simply connected by inducting on the number of points that I remove. If I remove one point (the case $m=1$), I get that $\Bbb{R}^n - \{x_1\} \cong S^{n-1} \times \Bbb{R}$ which upon applying $\pi_1$ shows me that $\Bbb{R}^n - \{x_1\}$ is simply connected. - - -Inductive Hypothesis: Now suppose that $\Bbb{R}^n -\{x_1,\ldots,x_k\}$ is simply connected for all $ k 1$ and divide the points in $S = \{x_1, \ldots, x_m\}$ in two sets of smaller size (no matter how), say $A$ and $B$. -For convenience, let's assume that $A$ and $B$ are separated by the hyperplane $\mathcal{H}$, and that $N_{+}$ and $N_{-}$ are two open neighborhoods of the half-spaces that result. For an arbitrary base-point $x_0 \in \mathcal{H}$, Van Kampen theorem applies, giving a surjection from $\pi_1(N_{+} \backslash A) \ast \pi_1(N_{-} \backslash B)$ to $\pi_1(\mathbb{R}^n - S)$. Now just use the induction hypothesis to conclude that $\pi_1(N_{+} \backslash A) = \pi_1(N_{-} \backslash B) = \pi_1(\mathbb{R}^n - S) = 0$, as you wish. -Edit: Of course that $N_+ \backslash A$ and the other set are homeomorphic (or, if you want, homotopy equivalent) to $\mathbb{R^n} \backslash A$ in an obvious way. Such geometric observations are not a difficulty once you understood how to apply the theorem. - -REPLY [3 votes]: We may compactify $\mathbb{R}^{n}$ to be $\mathbb{S}^{n}$ by adding a point, and for $n\ge 3$ it should not influence the statement. Now we can proceed inductively since $S^{n}$ removed $k$ points should be homeomorphic to $\bigvee^{k-1}_{i=1} S^{n-1}_{i}$. For $n\ge 3$ we are left with spheres of dimension 2 or higher, so we can conclude the fundamental group must be trivial. It should not be too difficult to visualize.<|endoftext|> -TITLE: $C(X)$ with the pointwise convergence topology is not metrizable -QUESTION [11 upvotes]: I need to show that if $X$ is an uncountable Tychonoff space, then $C(X)$ is not metrizable. All I've been able to show so far is that that $F(X)$, the space of all functions with pointwise topology, is homeomorphic to $\mathbb{R}^X$ (the product) which is not metrizable, but I can't seem to get much further. Thanks. - -REPLY [8 votes]: Hint: If the pointwise convergence topology was metrizable, it would be first countable, so $0$ would have a countable neighbourhood base. Thus there would be a sequence of open neighbourhoods $U_i$ of $0$ such that every neighbourhood of $0$ contains some $U_i$. -By definition of the topology of pointwise convergence, for each $i$ there exist some finite set $J_i$ and some $\epsilon_i > 0$ such that $f \in U_i$ if $|f(x)| < \epsilon_i$ for all $x \in J_i$. Now define another neighbourhood $U$ of $0$, and use the fact that $X$ is completely regular to show that for any $i$ there is some $f \in C(X)$ such that $f \in U_i$ but $f \notin U$.<|endoftext|> -TITLE: Confusion on Cech cohomology -QUESTION [8 upvotes]: From Harvard math qualification exam, 1990. -Let $X$ be a smooth manifold with an open cover $N<\infty$ sets $\{B_{n}\}^{N}_{1}$ which are contractible. Assume that $$\pi_{0}(B_{n}\cap B_{m})\le k, \forall n,m$$ for all $n$ and $m$. Give an upper bound to the first betti number of $X$. -Recall that the first betti number is defined to be the rank of $H_{1}(X)$. The main trouble is how the conditon on $\pi_{0}$ relates to the bound. My thought (perhaps dumb) is to proceed this inductively. But an arbitrally large genus surface can be covered by a large enough open ball that is connected. So in this case $n=2g, k=0, N=1$. -I asked my professor today, and he gave the hint that this is nothing but Cech cohomology, in which the $B_{i}$s resemble the faces in simplicial homology. He suggest since $X$ has such a cover then it is the same as putting a simplicial structure on $X$. So the first betti number must be bounded. But how large is the bound given $N$ and $k$? Imagine a two hole torus covered by a million small open balls with one next to each other, then $k$ should be 2 if the balls are small enough. But this seems to carry over to a 3-hole torus or a torus of any genus. I suspect there is some fundamental misunderstanding on the statement of the problem or concepts behind, so I venture to ask. A closer look at the Cech cohomology article does not help me to come up with a bound. - -REPLY [7 votes]: If $\mathfrak{U}$ is any open cover of $X$, the Čech-to-derived functor spectral sequence assures us of an exact sequence of the form -$$0 \longrightarrow H^1(\check{C}^\bullet(\mathfrak{U})) \longrightarrow H^1(X, \mathbb{Z}) \longrightarrow H^0(\check{C}^\bullet(\mathfrak{U}, \mathscr{H}^1(\mathbb{Z}))) \longrightarrow \cdots$$ -where $\mathscr{H}^1(\mathbb{Z})$ is the presheaf $U \mapsto H^1(U, \mathbb{Z})$. (Here, $H^1$ refers to sheaf cohomology, but this is the same as singular cohomology when $X$ admits a triangulation.) Since we are assuming the open sets of $\mathfrak{U}$ are contractible, we have $\check{C}^0(\mathfrak{U}, \mathscr{H}^1(\mathbb{Z})) = 0$. So in fact we have an isomorphism -$$H^1(\check{C}^\bullet(\mathfrak{U})) \cong H^1(X, \mathbb{Z})$$ -and the universal coefficient theorem gives a short exact sequence -$$0 \longrightarrow \textrm{Ext}^1(H_0(X, \mathbb{Z}), \mathbb{Z}) \longrightarrow H^1(X, \mathbb{Z}) \longrightarrow \textrm{Hom}(H_1(X, \mathbb{Z}), \mathbb{Z}) \longrightarrow 0$$ -but if $X$ is path-connected then $H_0(X, \mathbb{Z})$ is a finitely-generated free abelian group, so $\textrm{Ext}^1(H_0(X, \mathbb{Z}), \mathbb{Z}) = 0$, so there is an isomorphism -$$H^1(X, \mathbb{Z}) \cong \textrm{Hom}(H_1(X, \mathbb{Z}), \mathbb{Z})$$ -and therefore the ranks of $H^1(X, \mathbb{Z})$ and $H_1(X, \mathbb{Z})$ are equal when finite. Thus, -$$b_1(X) = \operatorname{rank} H^1(\check{C}^\bullet(\mathfrak{U}))$$ -By simple linear algebra, $\operatorname{rank} H^1(\check{C}^\bullet(\mathfrak{U})) \le \operatorname{rank} \check{C}^1(\mathfrak{U})$, and -$$\operatorname{rank} \check{C}^1(\mathfrak{U}) = \sum_{\substack{\{U, V\} \\ U \ne V}} b_0 (U \cap V) \le \frac{1}{2} (N-1) N k$$ -by definition of $\check{C}^1(\mathfrak{U})$. -In summary, if $X$ admits a cover by $N$ contractible open subsets whose pairwise intersections have at most $k$ connected components, then: -$$b_1(X) \le \frac{1}{2} (N-1) N k$$<|endoftext|> -TITLE: Ring homomorphism with $\phi(1_R) \neq1_S$ -QUESTION [13 upvotes]: Let $R$ and $S$ be rings with unity $1_R$ and $1_S$ respectively. Let $\phi\colon R\to S$ be a ring homomorphism. Give an example of a non-zero $\phi$ such that $\phi(1_R)\neq 1_S$ -In trying to find a non-zero $\phi$ I've done the following observation: -Since for $\forall r\in R$ -$\phi(r) = \phi(r\times1_R) = \phi(r)\times\phi(1_R)$ we must have that $\phi(1_R)$ is an identity of $\phi(R)$ but not an identity of $S$. We must therefor construct a $\phi$ that is not onto and which have this property. I can't come up with any explicit example though, please help me. - -REPLY [7 votes]: Since for any homomorphism $\phi$, -$$\phi(1_R) = \phi(1_R \cdot 1_R) = \phi(1_R)\phi(1_R),$$ -any homomorphism must map $1_R$ to an idempotent element of $S$. If you map to an idempotent element other than $1_S$, the image of $\phi$ will be a subring $S'$ of $S$ and you will find that whatever element you mapped $\phi(1_R)$ to will end up being $1_{S'}$. -Note that not every idempotent element of $S$ is a valid candidate. Some concrete examples: -Valid: -Let $R = \mathbb{Z}/6\mathbb{Z}$ and $S = \mathbb{Z}/15\mathbb{Z}$. If we defined $\phi(1_R) = 10_S$, which is idempotent in $S$, we have defined a valid homomorphism. -Invalid: -Let $R = \mathbb{Z}/6\mathbb{Z}$ and $S = \mathbb{Z}/15\mathbb{Z}$. If we define $\phi(1_R) = 6_S$, both idempotent in $S$, the mapping is not well defined. For example, -$$\phi(0_R) = \phi(6_R)$$ -since $R \cong Z_6$, but -$$\phi(6_R) = 6_S \neq 0_S = \phi(0_R).$$ -Thus, such a mapping is not well defined. -A similar problem occurs if we define $\phi(1_R) = 1_S$. Thus when we only consider homomorphisms which map the identity of the domain to the identity of the codomain, we find the following theorem. -Theorem: A homomorphism $\phi: \mathbb{Z}/m\mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z}$ exists only if $n$ divides $m$.<|endoftext|> -TITLE: Irreducible polynomial over an algebraically closed field -QUESTION [8 upvotes]: Suppose $k$ is an algebraically closed field and $p(x,y)\in k[x,y]$ is an irreducible polynomial. - -Prove that there are only finite many $a\in k$ such that $p(x,y)+a$ is reducible, i.e. the set $\{a\mid p(x,y)+a $ is reducible over $k, a\in k\}$ is finite. - -More generally, suppose $p\in k[x_1,\ldots,x_n]$ is irreducible, is it true that there are only finite many $a\in k$ such that $p+a$ is reducible? -Thanks! - -REPLY [11 votes]: Yes, there are only finitely many values of $a$ for which $p(x_1,...,x_n)+a$ is reducible. -In order to use the mighty tools of projective algebraic geometry, let us consider the homogeneization $P(x_0,x_1,...,x_n)$ of $p(x_1,...,x_n)$ ($P$ is an irreducible homogeneous polynomial of degree $d$) and ask whether $P(x_0,x_1,...,x_n)+ax_0^n$ is irreducible with only finitely many exceptions (The factorizability of the homogeneized polynomial is equivalent to the factorizability of the original polynomial). -The answer is yes. -Indeed in the projective space $\mathbb P^{N_d}\: (N_d=\binom {n+d}{d}-1)$ of homogeneous polynomials of degree $d$ in $n+1$ variables , the reducible ones form a union of finitely many irreducible strict subvarieties: the images of the morphisms $\mathbb P^{N_e} \times \mathbb P^{N_f}\to \mathbb P^{N_d}\; (e+f=d)$ given by the multiplication of a polynomial of degree $e$ by one of degree $f$. -The rest is easy: the projective line in $\mathbb P^{N_d}$ given by the family $P(x_0,x_1,...,x_n)+ax_0^n$ is not included in the mentioned union of subvarieties because the polynomial corresponding to $a=0$ is irreducible. -Hence it cuts that variety for only finitely many values of $a\in k$ and for the other values of $a$ both $P(x_0,x_1,...,x_n)+ax_0^n$ and $p(x_1,...,x_n)+a$ are irreducible.<|endoftext|> -TITLE: Two sums with Fibonacci numbers -QUESTION [7 upvotes]: Find closed form formula for sum: $\displaystyle\sum_{n=0}^{+\infty}\sum_{k=0}^{n} \frac{F_{2k}F_{n-k}}{10^n}$ -Find closed form formula for sum: $\displaystyle\sum_{k=0}^{n}\frac{F_k}{2^k}$ and its limit with $n\to +\infty$. - - -First association with both problems: generating functions and convolution. But I have been thinking about solution for over a week and still can't manage. Can you help me? - -REPLY [2 votes]: The first one can be solved using the fact that the generating -function of the Fibonacci numbers is -$$\frac{z}{1-z-z^2}.$$ -Introduce the function -$$f(z) = \sum_{n\ge 0} z^n \sum_{k=0}^n \frac{F_{2k} F_{n-k}}{10^n}$$ -so that we are interested in $f(1).$ -Re-write $f(z)$ as follows: -$$f(z) = \sum_{k\ge 0} F_{2k} -\sum_{n\ge k} \frac{z^n}{10^n} F_{n-k} -= \sum_{k\ge 0} F_{2k} \frac{z^k}{10^k} -\sum_{n\ge 0} \frac{z^n}{10^n} F_n.$$ -Now we have -$$ \sum_{k\ge 0} F_{2k} z^{2k} = -\frac{1}{2} \frac{z}{1-z-z^2} -- \frac{1}{2} \frac{z}{1+z-z^2}$$ -and therefore $f(1)$ is -$$\left(\frac{1}{2} \frac{1/\sqrt{10}}{1-1/\sqrt{10}-1/10} -- \frac{1}{2} \frac{1/\sqrt{10}}{1+1/\sqrt{10}-1/10}\right) -\times -\frac{1/10}{1-1/10-1/100}$$ -which simplifies to -$$\frac{1}{2\sqrt{10}} -\frac{2/\sqrt{10}}{81/100-1/10} \times \frac{10}{89} -= \frac{1}{10} \times \frac{1}{71/100} \times \frac{10}{89} = -\frac{100}{89\times 71}.$$<|endoftext|> -TITLE: Best way to denote some trigonometric functions ("tg" vs "tan", "ctg" vs "cot") -QUESTION [9 upvotes]: What is the best way to denote tangent and other trigonometric functions: tg or tan, ctg or cot. What notation is commonly used and standardized? - -REPLY [8 votes]: Standard -ISO 80000-2 Quantities and units — Part 2: Mathematical signs and symbols to be -used in the natural sciences and technology -is perfectly clear: the right symbols are $\tan x$ (item 2-13.4) and $\cot x$ (item 2-13.5). -To cite the standard: "$\text{tg } x$, $\text{ctg } x$ should not be used."<|endoftext|> -TITLE: How to show that $\mathbb R^n$ with the $1$-norm is not isometric to $\mathbb R^n$ with the infinity norm for $n>2$? -QUESTION [17 upvotes]: Could you please give me a hint to prove that -$\mathbb{R}^n$ with the 1-norm $\lvert x\rvert_1=\lvert x_1\rvert+\cdots+\lvert x_n\rvert$ is not isometric to $\mathbb{R}^n$ with the infinity-norm $\lvert x\rvert_\infty = \max_{i=1,\ldots,n}(\lvert x_i\rvert)$ for $n\gt2$. -I do not see how the critical case $n=2$ enters the picture. -Thank you! - -REPLY [20 votes]: But one can do without the Mazur-Ulam theorem just as easily. I find it convenient to put the exponent $p$ in subscript, freeing the superscript for dimension: $\ell_\infty^n$ and $\ell_1^n$ -Suppose $F:\ell_\infty^n\to \ell_1^n$ is an isometry (which I do not assume to be surjective or linear). By translation we can achieve $F(0)=0$. Let $V\subset \ell_\infty^n$ be the set of vertices of the unit cube, i.e., the set of all vectors with entries $\pm1$. This set has three interesting properties: - -all points are at distance $1$ from the origin -all points are at distance $2$ from one another -the cardinality of the set is $2^n$. - -Let $W=F(V)$. The set $W\subset \ell_1^n$ also has the properties 1,2,3 listed above. For each vector $v\in W$ consider its positive and negative supports $\mathrm{supp}^+v=\{1\le j\le n : v_j> 0\}$ and $\mathrm{supp}^-v=\{1\le j\le n : v_j< 0\}$. If $n>2$, then $2^n>2n$. Therefore, we can assume that at least $n+1$ vectors in $W$ have nonempty positive supports. But then at least two of them, say $v'$ and $v''$, have overlapping positive supports. This creates cancellation when we compute distance between them: $\|v'-v''\|_1<\|v'\|_1+\|v''\|_1=2$ , a contradiction. QED -The proof actually shows that $\ell_{\infty}^n$ does not admit an isometric embedding (linear or not) into $\ell_1^N$ for $N<2^{n-1}$.<|endoftext|> -TITLE: How to invert this symmetric tridiagonal Toeplitz matrix? -QUESTION [9 upvotes]: What's the best way to invert a simple symmetric tridiagonal Toeplitz matrix of the following form? -$$ -A = \begin{bmatrix} 1 & a & 0 & \ldots & \ldots & 0 \\\ -a & 1 & a & \ddots & & \vdots \\\ -0 & a & 1 & \ddots & \ddots& \vdots \\\ -\vdots & \ddots & \ddots & \ddots & a & 0\\\ -\vdots & & \ddots & a & 1 & a\\\ -0 & \ldots & \ldots & 0 & a & 1 \end{bmatrix} -$$ - -REPLY [11 votes]: Usually, an eigendecomposition is the least efficient way to generate the inverse of a matrix, but in the case of the symmetric tridiagonal Toeplitz matrix, we have the nice eigendecomposition $\mathbf A=\mathbf V\mathbf D\mathbf V^\top$, where $$\mathbf D=\mathrm{diag}\left(1+2a\cos\frac{\pi}{n+1},\dots,1+2a\cos\frac{k\pi}{n+1},\dots,1+2a\cos\frac{n\pi}{n+1}\right)$$ and $\mathbf V$ is the symmetric and orthogonal matrix whose entries are $$v_{j,k}=\sqrt{\frac2{n+1}}\sin\frac{\pi jk}{n+1}$$ Thus, to generate the inverse, use $\mathbf A=\mathbf V\mathbf D^{-1}\mathbf V^\top$, and inverting a diagonal matrix is as easy as reciprocating the diagonal entries.<|endoftext|> -TITLE: Is there a difference between a model and a representation? -QUESTION [5 upvotes]: I'm thinking of models in logic here, vs. e.g. group representations. - -Is there a difference between a model and a representation? - -Could one not explain both at the same time? -A model gives an interpretation, but this might be viewed as a side effect to the work you do. You have some abstract axioms and you model/realize them with certain objects, which are part of another theory (e.g. the ordered pair concept is modeled via sets and $\in$). I don't see a real difference to e.g. a group representation, when you have some abstract multiplication laws and these come to live via a matrix representation of a specific dimension, say. - -Also, - -What kind of realizations do representable functors deal with? - -I mean beyond the realization of groups like above. - -Edit, from the comments: Like an example would be to consider integers $\mathbb{Z}$ and build the factor group $\mathbb{Z/2Z}$, this would be representing what is called $\mathbb{Z_2}$ (abstractly defined by the four relations between its two elements). The logic analog would be the sets and the abstract idea of a pair with its characterizing feature. - -REPLY [3 votes]: Well, I mean, they are representations of different kinds of things. But one can think of both as functors out of a given category. -For group representations, a group $G$ may be regarded as a category with one object and morphisms the elements of $G$. A group action of $G$ is then precisely a functor $G \to \text{Set}$; similarly, a linear representation of $G$ is precisely a functor $G \to \text{Vect}$. There are endless variations on this. -For models, a collection of axioms ought to describe category which is roughly speaking the free category containing a model of the axioms. This is easiest to explain for axioms nice enough that they can be described using a Lawvere theory, but I believe this formalism is more general than that. For example, the theory of groups can be described using a Lawvere theory which is described here. Given a Lawvere theory $C$, a model of $C$ is then precisely a product-preserving functor $C \to \text{Set}$. Again there are endless variations on this. See the Wikipedia article on categorical logic. -In the first example, the representable functor gives you the action of $G$ on itself (exercise), which is roughly speaking the free $G$-action on one element. In the second example, the representable functors for the Lawvere theory of groups give you free groups on finite sets (exercise). -Edit: Instead of reducing everything to category theory perhaps I should reduce everything to logic. I claim that representations are a special case of models. -For example, let $G$ be a group. Write down a first-order theory with one unary operation for every element of $G$ and one universally quantified axiom for every entry $g \times h = gh$ in the multiplication table of $G$. Then a model of this theory is precisely a set on which $G$ acts. (A similar but more complicated construction gives linear representations also as a special case of models.) Of course, much of this is redundant: it suffices to specify an operation for every element of a fixed set of generators of $G$ and an axiom for every relation in a fixed presentation relative to the generators.<|endoftext|> -TITLE: What do the $+,-$ mean in limit notation, like$\lim\limits_{t \to 0^+}$ and $\lim\limits_{t \to 0^-}$? -QUESTION [14 upvotes]: I'm working on Laplace Transforms and have got to a section where they are talking about zero to the power plus or minus and that they are different. I can't remember what this means though. -It's generally used in limits. -$\lim\limits_{t\to 0^-}$ or $\lim\limits_{t\to 0^+}$ -Any help would be much appreciated. - -REPLY [9 votes]: And a good, visual example could be something like -                             -$$\begin{align} -\lim_{x\to x_0^-}f&=y_1\\ -\lim_{x\to x_0^+}f&=y_2 -\end{align}$$<|endoftext|> -TITLE: How to integrate $\sec^3 x \, dx$? -QUESTION [6 upvotes]: Possible Duplicate: -Indefinite integral of secant cubed - -How to integrate $\sec^3 x \, dx$? -Can someone please give a method, I tried separating $\sec^3 x$ as $\sec x(\sec^2 x)$ then applying by-parts method but it didn't yield anything useful - -REPLY [6 votes]: There's a whole Wikipedia article about just this integral: Integral of secant cubed. -You're mistaken to think that integration by parts doesn't help.<|endoftext|> -TITLE: What are the points of discontinuity of $\tan x$? -QUESTION [10 upvotes]: $f(x) = \tan x$ is defined from $\mathbb R - \{\frac{\pi}{2} (2n+1) \mid n \in \mathbb Z\}$ to $\mathbb R$. For every $x$ in its domain, -$$f(x) = \frac{\sin x}{\cos x}$$ where $\cos x$ is never 0. Thus, (in short) $\tan x$ is defined for all points in its domain. -Now the question remains, is $\tan x$ discontinuous at $x = \pi/2$ (which is outside its domain)? -The question arises because the test for continuity in a textbook mentions that $f(x)$ is continuous at $x = c$ when: - -$f(c)$ exists. -$\lim_{x \to c} f(x)$ exists. -$f(c) = \lim_{x \to c} f(x)$. - -And my teacher says failure of any of the above results in $x = c$ being a point of discontinuity. Yet, according to me, first test above merely tests the point for its domain and should be the criteria for any point of discontinuity too. - -REPLY [3 votes]: There are no “right” or “wrong” definitions, but there are “standard” and “non-standard” definitions. In my opinion, the definition of continuity from your textbook is non-standard when applied to functions defined on sets with isolated points. For example, the function $f:{\mathbb Z} \to {\mathbb R}$ defined by $f(x) = x$ is a continuous function according to the standard definition but it is discontinuous according to the definition from your textbook. -Here is a standard definition. -Let $D \subset {\mathbb R}$ and $f:D \to {\mathbb R}$. We say that $f$ is continuous at point $c$ if one of the following condition holds: - -$c$ is a limit point of $D$ and $\lim_{x\to c} f(x) = c$, or -$c$ is an isolated point of $D$. - -Some textbooks define essential discontinuities even for points in $\bar D$ as follows. -Let $E \subset {\bar D}$. We say that $c\in E$ is an essential discontinuity of $f$ on $E$ if there is no function $\hat f: E \to {\mathbb R}$ such that - -${\hat f}(x) = f(x)$ for $x\in D\cap E$, -$\hat f(x)$ is continuous at point $c$. - -According to this definition, $\pi/2$ is an essential discontinuity of $\tan x$ on $\mathbb R$. Of course, it's important what the set $E$ is, in this definition. For example, consider the the function $g(x):{\mathbb R} \setminus {\mathbb Z}\to \mathbb R$ defined by $g(x) = x - \lfloor x \rfloor$. Then 0 is not an essential discontinuity of $g(x)$ on $[0, 1)$, nor on $(-1,0]$. But 0 is an essential discontinuity of $g(x)$ on $(-1,1)$. \ No newline at end of file