tag
stringclasses 9
values | question_body
stringlengths 61
12.9k
| accepted_answer
stringlengths 38
36.4k
| second_answer
stringlengths 63
33k
|
|---|---|---|---|
geometry | <p>I am trying to understand the notion of an orientable manifold.<br>
Let M be a smooth n-manifold. We say that M is orientable if and only if there exists an atlas $A = \{(U_{\alpha}, \phi_{\alpha})\}$ such that $\textrm{det}(J(\phi_{\alpha} \circ \phi_{\beta}^{-1}))> 0$ (where defined). My question is:<br>
Using this definition of orientation, how can one prove that the Möbius strip is not orientable? </p>
<p>Thank you!</p>
| <p>If you had an orientation, you'd be able to define at each point $p$ a unit vector $n_p$ normal to the strip at $p$, in a way that the map $p\mapsto n_p$ is continuous. Moreover, this map is completely determined once you fix the value of $n_p$ for some specific $p$. (You have two possibilities, this uses a tangent plane at $p$, which is definable using a $(U_\alpha,\phi_\alpha)$ that covers $p$.)</p>
<p>The point is that the positivity condition you wrote gives you that the normal at any $p'$ is independent of the specific $(U_{\alpha'},\phi_{\alpha'})$ you may choose to use, and path connectedness gives you the uniqueness of the map. Now you simply check that if you follow a loop around the strip, the value of $n_p$ changes sign when you return to $p$, which of course is a contradiction.</p>
<p>(This is just a formalization of the intuitive argument.)</p>
| <p>Let $M:=\{(x,y)|x\in\mathbb R, -1<y<1\}$ be an infinite strip and choose an $L>0$. The equivalence relation $(x+L,-y)\sim(x,y)$ defines a Möbius strip $\hat M$. Let $\pi: M \to \hat M$ be the projection map. The Möbius strip $\hat M$ inherits the differentiable structure from ${\mathbb R}^2$. We have to prove that $\hat M$ does not admit an atlas of the described kind which is compatible with the differentiable structure on $\hat M$. Assume that there is such an atlas $(U_\alpha,\phi_\alpha)_{\alpha\in I}$. We then define a function $\sigma:{\mathbb R}\to\{-1,1\}$ as follows: For given $x\in{\mathbb R}$ the point $\pi(x,0)$ is in $\hat M$, so there is an $\alpha\in I$ with $\pi(x,0)\in U_\alpha$. The map $f:=\phi_\alpha^{-1}\circ\pi$ is a diffeomorphism in a neighbourhood $V$ of $(x,0)$. Put $\sigma(x):=\mathrm{sgn}\thinspace J_f(x,0)$, where $J_f$ denotes the Jacobian of $f$. One easily checks that $\sigma(\cdot)$ is well defined and is locally constant, whence it is constant on ${\mathbb R}$. On the other hand we have $f(x+L,y)\equiv f(x,-y)$ in $V$ which implies $\sigma(L)=-\sigma(0)$ -- a contradiction.</p>
|
linear-algebra | <p>Given two square matrices <span class="math-container">$A$</span> and <span class="math-container">$B$</span>, how do you show that <span class="math-container">$$\det(AB) = \det(A) \det(B)$$</span> where <span class="math-container">$\det(\cdot)$</span> is the determinant of the matrix?</p>
| <p>Let's consider the function <span class="math-container">$B\mapsto \det(AB)$</span> as a function of the columns of <span class="math-container">$B=\left(v_1|\cdots |v_i| \cdots | v_n\right)$</span>. It is straight forward to verify that this map is multilinear, in the sense that
<span class="math-container">$$\det\left(A\left(v_1|\cdots |v_i+av_i'| \cdots | v_n\right)\right)=\det\left(A\left(v_1|\cdots |v_i| \cdots | v_n\right)\right)+a\det\left(A\left(v_1|\cdots |v_i'| \cdots | v_n\right)\right).$$</span> It is also alternating, in the sense that if you swap two columns of <span class="math-container">$B$</span>, you multiply your overall result by <span class="math-container">$-1$</span>. These properties both follow directly from the corresponding properties for the function <span class="math-container">$A\mapsto \det(A)$</span>.</p>
<p>The determinant is completely characterized by these two properties, and the fact that <span class="math-container">$\det(I)=1$</span>. Moreover, any function that satisfies these two properties must be a multiple of the determinant. If you have not seen this fact, you should try to prove it. I don't know of a reference online, but I know it is contained in Bretscher's linear algebra book.</p>
<p>In any case, because of this fact, we must have that <span class="math-container">$\det(AB)=c\det(B)$</span> for some constant <span class="math-container">$c$</span>, and setting <span class="math-container">$B=I$</span>, we see that <span class="math-container">$c=\det(A)$</span>.</p>
<hr />
<p>For completeness, here is a proof of the necessary lemma that any a multilinear, alternating function is a multiple of determinant.</p>
<p>We will let <span class="math-container">$f:\mathbb (F^n)^n\to \mathbb F$</span> be a multilinear, alternating function, where, to allow for this proof to work in characteristic 2, we will say that a multilinear function is alternating if it is zero when two of its inputs are equal (this is equivalent to getting a sign when you swap two inputs everywhere except characteristic 2). Let <span class="math-container">$e_1, \ldots, e_n$</span> be the standard basis vectors. Then <span class="math-container">$f(e_{i_1},e_{i_2}, \ldots, e_{i_n})=0$</span> if any index occurs twice, and otherwise, if <span class="math-container">$\sigma\in S_n$</span> is a permutation, then <span class="math-container">$f(e_{\sigma(1)}, e_{\sigma(2)},\ldots, e_{\sigma(n)})=(-1)^\sigma$</span>, the sign of the permutation <span class="math-container">$\sigma$</span>.</p>
<p>Using multilinearity, one can expand out evaluating <span class="math-container">$f$</span> on a collection of vectors written in terms of the basis:</p>
<p><span class="math-container">$$f\left(\sum_{j_1=1}^n a_{1j_1}e_{j_1}, \sum_{j_2=1}^n a_{2j_2}e_{j_2},\ldots, \sum_{j_n=1}^n a_{nj_n}e_{j_n}\right) = \sum_{j_1=1}^n\sum_{j_2=1}^n\cdots \sum_{j_n=1}^n \left(\prod_{k=1}^n a_{kj_k}\right)f(e_{j_1},e_{j_2},\ldots, e_{j_n}).$$</span></p>
<p>All the terms with <span class="math-container">$j_{\ell}=j_{\ell'}$</span> for some <span class="math-container">$\ell\neq \ell'$</span> will vanish before the <span class="math-container">$f$</span> term is zero, and the other terms can be written in terms of permutations. If <span class="math-container">$j_{\ell}\neq j_{\ell'}$</span> for any <span class="math-container">$\ell\neq \ell'$</span>, then there is a unique permutation <span class="math-container">$\sigma$</span> with <span class="math-container">$j_k=\sigma(k)$</span> for every <span class="math-container">$k$</span>. This yields:</p>
<p><span class="math-container">$$\begin{align}\sum_{j_1=1}^n\sum_{j_2=1}^n\cdots \sum_{j_n=1}^n \left(\prod_{k=1}^n a_{kj_k}\right)f(e_{j_1},e_{j_2},\ldots, e_{j_n}) &= \sum_{\sigma\in S_n} \left(\prod_{k=1}^n a_{k\sigma(k)}\right)f(e_{\sigma(1)},e_{\sigma(2)},\ldots, e_{\sigma(n)}) \\ &= \sum_{\sigma\in S_n} (-1)^{\sigma}\left(\prod_{k=1}^n a_{k\sigma(k)}\right)f(e_{1},e_{2},\ldots, e_{n}) \\ &= f(e_{1},e_{2},\ldots, e_{n}) \sum_{\sigma\in S_n} (-1)^{\sigma}\left(\prod_{k=1}^n a_{k\sigma(k)}\right). \end{align}
$$</span></p>
<p>In the last line, the thing still in the sum is the determinant, although one does not need to realize this fact, as we have shown that <span class="math-container">$f$</span> is completely determined by <span class="math-container">$f(e_1,\ldots, e_n)$</span>, and we simply define <span class="math-container">$\det$</span> to be such a function with <span class="math-container">$\det(e_1,\ldots, e_n)=1$</span>.</p>
| <p>The proof using elementary matrices can be found e.g. on <a href="http://www.proofwiki.org/wiki/Determinant_of_Matrix_Product" rel="noreferrer">proofwiki</a>. It's basically the same proof as given in Jyrki Lahtonen 's comment and Chandrasekhar's link.</p>
<p>There is also a proof using block matrices, I googled a bit and I was only able to find it in <a href="http://books.google.com/books?id=N871f_bp810C&pg=PA112&dq=determinant+product+%22alternative+proof%22&hl=en&ei=EPlZToKHOo6aOuyapJIM&sa=X&oi=book_result&ct=result&resnum=2&ved=0CC8Q6AEwAQ#v=onepage&q=determinant%20product%20%22alternative%20proof%22&f=false" rel="noreferrer">this book</a> and <a href="http://www.mth.kcl.ac.uk/%7Ejrs/gazette/blocks.pdf" rel="noreferrer">this paper</a>.</p>
<hr />
<p>I like the approach which I learned from Sheldon Axler's Linear Algebra Done Right, <a href="http://books.google.com/books?id=ovIYVIlithQC&printsec=frontcover&dq=linear+algebra+done+right&hl=en&ei=H-xZTuutJoLrOdvrwaMM&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCkQ6AEwAA#v=onepage&q=%2210.31%22&f=false" rel="noreferrer">Theorem 10.31</a>.
Let me try to reproduce the proof here.</p>
<p>We will use several results in the proof, one of them is - as far as I can say - a little less known. It is the <a href="http://www.proofwiki.org/wiki/Determinant_as_Sum_of_Determinants" rel="noreferrer">theorem</a> which says, that if I have two matrices <span class="math-container">$A$</span> and <span class="math-container">$B$</span>, which only differ in <span class="math-container">$k$</span>-th row and other rows are the same, and the matrix <span class="math-container">$C$</span> has as the <span class="math-container">$k$</span>-th row the sum of <span class="math-container">$k$</span>-th rows of <span class="math-container">$A$</span> and <span class="math-container">$B$</span> and other rows are the same as in <span class="math-container">$A$</span> and <span class="math-container">$B$</span>, then <span class="math-container">$|C|=|B|+|A|$</span>.</p>
<p><a href="https://math.stackexchange.com/questions/668/whats-an-intuitive-way-to-think-about-the-determinant">Geometrically</a>, this corresponds to adding two parallelepipeds with the same base.</p>
<hr />
<p><strong>Proof.</strong>
Let us denote the rows of <span class="math-container">$A$</span> by <span class="math-container">$\vec\alpha_1,\ldots,\vec\alpha_n$</span>. Thus
<span class="math-container">$$A=
\begin{pmatrix}
a_{11} & a_{12}& \ldots & a_{1n}\\
a_{21} & a_{22}& \ldots & a_{2n}\\
\vdots & \vdots& \ddots & \vdots \\
a_{n1} & a_{n2}& \ldots & a_{nn}
\end{pmatrix}=
\begin{pmatrix}
\vec\alpha_1 \\ \vec\alpha_2 \\ \vdots \\ \vec\alpha_n
\end{pmatrix}$$</span></p>
<p>Directly from the definition of matrix product we can see
that the rows of <span class="math-container">$A\cdot B$</span> are of the form <span class="math-container">$\vec\alpha_kB$</span>, i.e.,
<span class="math-container">$$A\cdot B=\begin{pmatrix}
\vec\alpha_1B \\ \vec\alpha_2B \\ \vdots \\ \vec\alpha_nB
\end{pmatrix}$$</span>
Since <span class="math-container">$\vec\alpha_k=\sum_{i=1}^n a_{ki}\vec e_i$</span>, we can
rewrite this equality as
<span class="math-container">$$A\cdot B=\begin{pmatrix}
\sum_{i_1=1}^n a_{1i_1}\vec e_{i_1} B\\
\vdots\\
\sum_{i_n=1}^n a_{ni_n}\vec e_{i_n} B
\end{pmatrix}$$</span>
Using the theorem on the sum of determinants multiple times we get
<span class="math-container">$$
|{A\cdot B}|= \sum_{i_1=1}^n a_{1i_1}
\begin{vmatrix}
\vec e_{i_1}B\\
\sum_{i_2=1}^n a_{2i_2}\vec e_{i_2} B\\
\vdots\\
\sum_{i_n=1}^n a_{ni_n}\vec e_{i_n} B
\end{vmatrix}= \ldots =
\sum_{i_1=1}^n \ldots \sum_{i_n=1}^n a_{1i_1} a_{2i_2} \dots
a_{ni_n}
\begin{vmatrix}
\vec e_{i_1} B \\ \vec e_{i_2} B \\ \vdots \\ \vec e_{i_n} B
\end{vmatrix}
$$</span></p>
<p>Now notice that if <span class="math-container">$i_j=i_k$</span> for some <span class="math-container">$j\ne k$</span>, then
the corresponding determinant in the above sum is zero (it has two <a href="http://www.proofwiki.org/wiki/Square_Matrix_with_Duplicate_Rows_has_Zero_Determinant" rel="noreferrer">identical rows</a>).
Thus the only nonzero summands are those one, for which the <span class="math-container">$n$</span>-tuple
<span class="math-container">$(i_1,i_2,\dots,i_n)$</span> represents a permutation of the numbers <span class="math-container">$1,\ldots,n$</span>.
Thus we get
<span class="math-container">$$|{A\cdot B}|=\sum_{\varphi\in S_n} a_{1\varphi(1)} a_{2\varphi(2)} \dots a_{n\varphi(n)}
\begin{vmatrix}
\vec e_{\varphi(1)} B \\ \vec e_{\varphi(2)} B \\ \vdots \\ \vec
e_{\varphi(n)} B
\end{vmatrix}$$</span>
(Here <span class="math-container">$S_n$</span> denotes the set of all permutations of <span class="math-container">$\{1,2,\dots,n\}$</span>.)
The matrix on the RHS of the above equality is the matrix
<span class="math-container">$B$</span> with permuted rows. Using several transpositions of rows we can get the matrix
<span class="math-container">$B$</span>. We will show that this can be done using
<span class="math-container">$i(\varphi)$</span> transpositions, where <span class="math-container">$i(\varphi)$</span> denotes the number of <a href="http://en.wikipedia.org/wiki/Inversion_%28discrete_mathematics%29" rel="noreferrer">inversions</a> of <span class="math-container">$\varphi$</span>.
Using this fact we get
<span class="math-container">$$|{A\cdot B}|=\sum_{\varphi\in S_n} a_{1\varphi(1)} a_{2\varphi(2)} \dots a_{n\varphi(n)} (-1)^{i(\varphi)} |{B}| =|A|\cdot |B|.$$</span></p>
<p>It remains to show that we need <span class="math-container">$i(\varphi)$</span> transpositions.
We can transform the "permuted matrix" to matrix <span class="math-container">$B$</span>
as follows: we first move the first row of <span class="math-container">$B$</span> on the first place
by exchanging it with the preceding row until it is on the correct position.
(If it already is in the first position, we make no exchanges at all.)
The number of transpositions we have used is exactly the number
of inversions of <span class="math-container">$\varphi$</span> that contains the number 1.
Now we can move the second row to the second place in the same way.
We will use the same number of transposition as the number of inversions of <span class="math-container">$\varphi$</span>
containing 2 but not containing 1. (Since the first row is already in place.)
We continue in the same way.
We see that by using this procedure we obtain the matrix
<span class="math-container">$B$</span> after <span class="math-container">$i(\varphi)$</span> row transpositions.</p>
|
probability | <p>This sounds more like a brain teaser, but I had some kink to think it through :( Suppose you're parking at a non-parking zone, the probability to get a parking ticket is 80% in 1 hour, what is the probability to get a ticket in half an hour? Please show how you deduce the answer. Thanks!</p>
| <p>It really depends on what model is assumed. However, if the idea is that no matter how long you leave your car there, you have a $20$% chance of getting through any given hour unscathed, you can treat it as an exponential decay problem. Let $p(t)$ be the probability that you do <strong>not</strong> get a ticket in the first $t$ hours. Then $p(1)=0.2$, $p(2)=0.2^2$ (a $20$% chance of making it through the first hour times a $20$% chance of making it through the second), and in general $p(t)=0.2^t$. The probability of <strong>not</strong> getting a ticket in the first half hour is then $p(1/2)=0.2^{1/2}=\sqrt{0.2}\approx 0.4472$, and the probability that you <strong>do</strong> get a ticket in the first half hour is about $1-0.4472=0.5528$.</p>
| <p>If my probability of getting a parking ticket in a half hour is T, then</p>
<p>$$T+(1-T)T$$</p>
<p>are the odds I got one in 1 hour. The first term says that I got it in the 1st half hour (so it doesn't matter what happens after that), and the second term says that I got it in the 2nd half hour (so not in the first). Solving and taking the sensible solution:</p>
<p>$$2T-T^2=.8\implies T^2-2T+.8$$</p>
<p>$$T=1-\sqrt{.2}$$</p>
<p>Which are roughly $55.3\%$ odds in a half hour.</p>
|
game-theory | <p>Players A and B alternate writing one digit to make a six-figure number. That means A writes digit $a$, B writes digit $b$, ... to make a number $\overline{abcdef}$. </p>
<p>$a,b,c,d,e,f$ are distinct, $a\neq 0$.</p>
<p>A is the winner if this number is composite, otherwise B is. Is there any way to help A or B always win?</p>
| <p>Player A has a winning strategy. First, Player A picks $a = 3$. </p>
<p>Case I: Player B picks $b = 9$. </p>
<p>Then, Player A picks $c = 1$. If Player B picks $d = 7$, Player A can pick $e$ arbitrarily. If Player B picks $d \neq 7$, Player A picks $e = 7$. Following this strategy, $\{1,3,7,9\} \subseteq \{a,b,c,d,e\}$. So Player B is forced to pick $f \in \{0,2,4,5,6,8\}$, which makes $\overline{abcdef}$ composite. </p>
<p>Case II: Player B picks $b \neq 9$.</p>
<p>Then, Player A picks $c = 9$ and waits for Player B to pick $d$.</p>
<p>If $\{b,d\} = \{0,6\}, \{1,5\}, \{1,8\}, \{4,5\}, \{4,8\}, \{7,5\}, \{7,8\}$, then Player A picks $e = 2$. </p>
<p>If $\{b,d\} = \{1,2\}, \{4,2\}, \{7,2\}$, then Player A picks $e = 5$. </p>
<p>If $\{b,d\} = \{0,4\}, \{0,7\}, \{6,4\}, \{6,7\}, \{2,5\}, \{2,8\}, \{5,8\}$, then Player A picks $e = 1$.</p>
<p>If $\{b,d\} = \{0,1\}, \{6,1\}$, then Player A picks $e = 4$. </p>
<p>If $\{b,d\} = \{6,2\}, \{6,5\}, \{6,8\}, \{1,4\}, \{1,7\}, \{4,7\}$, then Player A picks $e = 0$.</p>
<p>If $\{b,d\} = \{0,2\}, \{0,5\}, \{0,8\}$, then Player A picks $e = 6$. </p>
<p>In all of these cases, $e$ was chosen such that $b+d+e \equiv 2\pmod{3}$. Hence, $a+b+c+d+e \equiv 2 \pmod{3}$. If Player B picks $f \in \{1,7\}$, then $a+b+c+d+e+f$ will be a multiple of $3$, and thus, $\overline{abcdef}$ is a multiple of $3$, and hence, composite. Otherwise, $f \in \{0,2,4,5,6,8\}$, and $\overline{abcdef}$ is composite. </p>
<p>Therefore, Player A can follow this strategy to guarantee that $\overline{abcdef}$ is composite, and thus, ensure that Player A wins the game.</p>
| <p>This is community wiki because it is JimmyK4542's answer with different reasoning.</p>
<p>First note that if the final digit is 0, 2, 4, 5, 6, or 8 then the result is composite and A wins. In other words B can only win if the final digit is 1, 3, 7, or 9 and even then it is not guaranteed.</p>
<p>A gets to choose three digits. As in JimmyK4542's answer A chooses 3 first (9 works as well). If B chooses any of 1, 3, or 7 then A can choose the other two and guarantee a win. Thus, we only need to consider the cases where B's choice lies outside this set.</p>
<p>A chooses 9 as his second digit (or 3 if he chose 9 first). Again, if B selects either 1 or 7 then A selects the other and wins. Thus when A is going to select his third digit we need only consider the cases where B has selected two digits from the set $\{0, 2, 4, 5, 6, 8\}$.</p>
<p>A's strategy now is to select a number such that even if B chooses either of 1 or 7 that the number will be composite. Note that 1 and 7 are both equal to 1 mod 3 and 3 and 9 are both 0 mod 3. Therefore, A must ensure that his choice added to B's two previous choices equals 2 mod 3. Let us examine the three cases:</p>
<p>1) The sum of B's two numbers is 0 mod 3. We know that he cannot have chosen both 2 and 8, so choose one of them.</p>
<p>2) The sum of B's two numbers is 1 mod 3. Choose 1 or 7.</p>
<p>3) The sum of B's two numbers is 2. We know that he cannot have chosen both 0 and 6 so choose one of them.</p>
<p>And we are done.</p>
<p>I have added this answer because the reasoning may appear more intuitive to some and the strategy perhaps more direct.</p>
|
game-theory | <p>Yesterday at work we had a staff day, where we were asked to play an interesting game as an icebreaker.</p>
<p>We (50 or so people) were told to stand in a circle and choose 2 people at random out of the group. We were then asked to walk to a point so that we would be equidistant to those two people. After a few eyerolls, we set off, but contrary to most people's expectations, the outcome was quite fascinating! We found that the group as a whole appeared to be in constant movement, although finally we did reach some sort of equillibrium.</p>
<p>This got me thinking, would a collection of points set up in this way ever reach a position of stasis?</p>
<p>Of course, there are obvious conditions under which there might be no movement at all (eg if each person chose the 2 people adjacent to him/her to start with), but apart from these unlikely conditions, I have no idea whether, given unlimited runtime, stasis would ever be achieved.</p>
<p>In the "game" we enacted as a group, participants were allowed to triangulate their positions to achieve equidistance. However, I would like to add the extra constraint that equidistance in this case should mean the shortest possible equidistance, since triangulation adds all sort of complications.</p>
<p>I am also assuming that the points are infinitesimally small, so collision is not an issue.</p>
<p>Ok. So I was a bit geeky, and went home and made a little simulation of it (imagine the people are holding black umbrellas and an aerial film was shot of the first 300 frames!):</p>
<p><a href="https://i.sstatic.net/ctWPV.gif" rel="noreferrer"><img src="https://i.sstatic.net/ctWPV.gif" alt="enter image description here"></a></p>
<p><a href="https://raw.githubusercontent.com/martinq321/pursuit/master/improved_code" rel="noreferrer">code</a> <em>(improvements in scaling)</em></p>
<p>Looking at the gif, other questions arise of course, like what on earth is going on with this order$\rightarrow$chaos$\rightarrow$ strange-attractor-type shapes? And why do they seem to exhibit this <a href="https://www.youtube.com/watch?v=wHfpacPLHIA" rel="noreferrer">mean curvature flow</a>-type behaviour?</p>
<h1>Added</h1>
<p>As requested by @AlexR. here is a second gif of $k=100$ points slowed down to show what happens between frames 1 and 100 <em>(as process appears to start converging to the loops just before $k$ frames)</em>.</p>
<p><a href="https://i.sstatic.net/um6vH.gif" rel="noreferrer"><img src="https://i.sstatic.net/um6vH.gif" alt="enter image description here"></a></p>
<h1><em>Note</em></h1>
<p><em>Original code was faulty (whole thing was wrapped in <code>manipulate</code> which meant random seed was constantly changing). Code and gifs now updated.</em> The images certainly seem to make more sense - it appears that the group tend towards eventually reforming back into a circle, but this is only speculation. The same questions still apply though.</p>
<p><strong><em>Point mechanics</em></strong></p>
<p>Just to clarify (to reiterate comment below in answer to question by @Anaedonist), given a step size of 1 unit (the unit circle around the black point), the black point will move 1 unit in the direction of the midpoint (dark blue) of its corresponding pair (red), so its new position is the light blue point <em>(image on left)</em>, unless the midpoint is inside the unit circle, in which case, the step size is smaller, and the light blue point reaches the midpoint <em>(image on right)</em>:</p>
<p><img src="https://i.sstatic.net/7ApE6.png" width="300" height="300">
<img src="https://i.sstatic.net/AU1yN.png" width="300" height="300"></p>
<p>I should probably add the requirement that each point must be selected <em>(ie there are no points that are unselected /unpaired)</em>. This simplifies it somewhat, and probably accounts for the regularity seen in the above images. It also avoids the problem of potential breaking off into smaller sub-groups. (In a real-life enactment, this might be remedied by selecting participant names out of a hat, or similar.)</p>
<p><strong><em>A note on scaling</em></strong></p>
<p>Since the code update, it appears that scaling has an impact on point behaviour, issofar as if the points are small in comparison to the step size <em>(ie the unit circle within which movement is permitted per frame)</em>, the points enter into the strange looping structures as seen in the gifs. The further apart the points are spread to begin with, the more quickly convergence to elliptical uniformity appears to occur:</p>
<p><img src="https://i.sstatic.net/FLlQ3.gif" width="300" height="300">
<img src="https://i.sstatic.net/scJqC.gif" width="300" height="300"></p>
<p><em>Left image shows $k=100$ points, starting positions within unit circle produced with</em> <code>pursuit[#, Floor[Sqrt[#] #], 1] &[100]</code>; <em>right-hand image shows $k=100$ points, starting positions within unit circle $\times k \log k$ produced with</em> <code>pursuit[#, Floor[Sqrt[#] #], # Log[#]] &[100]</code>.</p>
<p><strong><em>Likely convergence conditions</em></strong></p>
<p><a href="https://i.sstatic.net/J0O0n.gif" rel="noreferrer"><img src="https://i.sstatic.net/J0O0n.gif" alt="enter image description here"></a></p>
<p>Its fairly clear from this that the closed curves start when the points converge to within the step size of the unit circle.</p>
<p><a href="https://raw.githubusercontent.com/martinq321/pursuit/master/manual_code" rel="noreferrer">code</a> <em>(lengthy, but far more transparent, added for straightforward manual edits)</em>
or less transparent <a href="https://raw.githubusercontent.com/martinq321/pursuit/master/string_based_code" rel="noreferrer">code for any k</a>.</p>
<hr>
<p><a href="https://raw.githubusercontent.com/martinq321/pursuit/master/closer" rel="noreferrer">This code</a> is more faithful to the true <a href="http://mathworld.wolfram.com/PursuitCurve.html" rel="noreferrer">pursuit curve</a>. Though not a true pursuit curve, it mimics it, and the difference can be seen in the smoother transitions:</p>
<p><a href="https://i.sstatic.net/DGgTq.gif" rel="noreferrer"><img src="https://i.sstatic.net/DGgTq.gif" alt="enter image description here"></a></p>
<p><em>Comparison with true pursuit curve:</em></p>
<p><a href="https://i.sstatic.net/zsqm8.gif" rel="noreferrer"><img src="https://i.sstatic.net/zsqm8.gif" alt="enter image description here"></a></p>
<p>where red $=$ true pursuit curve and blue $=$ mimicking function.</p>
<p><a href="https://raw.githubusercontent.com/martinq321/pursuit/master/comparison" rel="noreferrer">code</a></p>
| <p>Doesn't this game converge to a point, assuming smallest possible equidistance? Assume that it doesn't. Instead it converges to some equilibrium position L. Then there is a unique convex polygon joining the points on the outside of L. But on the next step , each of these points on the outside must move inside L, hence L cannot be an equilibrium. </p>
<p>I would add that I think the step size rule should be that a point always moves a percentage of the distance to the target midpoint, to make the process more continuous. </p>
| <p>Following the OP's explanation in the comments, I would like to add some considerations, leaving for clarity's sake my initial clarification requests at the end.</p>
<p>Firsty, I believe the algorithm used for the simulations might introduce some complications compared to the description given, as the distance step taken by each point at each time step is bounded by $1$.</p>
<p>I tried to get a simplified, more tractable description, by modelling the problem as a system of differential equations.
Denoting the coordinates $(x,y)$ of the $i$-th player as $p_i$, and its chosen pair $p_k$ and $p_j$, an ODE such $$ \dot{p_i} = \gamma \bigg(p_k - p_i + \frac{p_j - p_k}{2} \bigg)$$
could be enforced for each $i = 1,2 \dots n$ (where $n$ is the total number of players) translating the condition that one player instaneously directs itself to the middle point between the two chosen neighbours.</p>
<p>Setting time derivatives to zero one verifies that the trivial solution whereby all the players collapse to a point exists.</p>
<p>As a matter of fact, it seems intuitive to conjecture even that $\max_{i,j} {\vert p_i - p_j \vert}$ is exponentially vanishing: so the "collapsed" configuration will be reached in exponential time, and the players will progressively occupy an ever smaller region of the playing ground. </p>
<p>For sure a mathematical description better than the one hereinattempted is needed, allowing the solution whereby the $n$ players sit on the vertices of an $n$-gon (compatibl with the choices made for the neighbours): the main problem I have is how to define distances to allow for such case.</p>
<p><strong>Previous clarification request</strong>: I just have some questions to ensure I understood this interesting setting well.</p>
<p>The main problem is, will the equilibrium position attained, starting from any configuration of players around a circle and any choice of equidistant neighbours made by any player?</p>
<p>Firstly, I believe there are some choices the players could make, which cannot yield an equilibrium configuration.
For example in the case say $n = 5$, if the player $p_4$ wants to stay between players $p_3$ and $p_5$, and the player $p_3$ wants to stay between $p_4$ and $p_5$, no equilibrium configuration can be attained (if I understood your comment on shortest possible equidistance).</p>
<p>Let us assume for a moment that we can characterize all “compatible” choices, i.e. initial choices of equidistant neighbours made by each player such that an equilibrium configuration exists.</p>
<p>Then I thought, we could see if such configuration is attained by reversing the game, i.e. starting from the equilibrium configuration and going backwards in time: if we verify that any starting configuration is attainable (something akin “ergodicity”), some progress would be made.</p>
<p>At this point I notice I am unsure on what rules you implemented for your game and your (nice) plots.</p>
<p>At each time step, how is the motion of the $p_i$ player defined?
Let me clarify with an example.</p>
<p>At the $k$-th time step, the configuration reads $ 1, 3, 4, 2, 5$ (clockwise sense). Let us assume the $p_5$ player has chosen $p_3$ and $p_2$ as desired neighbours.</p>
<p>In your simulation, where will $p_5$ go? He can move between $p_3$ and $p_4$, or between $p_4$ and $p_2$.
Also, do you decide the movements of each $p_i$ player at the beginning of the $k$-th time step, given the configuration attained at the $k-1$-th time step, or do you move $p_1$, and then decide the moves of the other players based on the updated configuration (after $p_1$ has moved)?</p>
<p>I am far from confident I can contribute much to this problem, but it is fun to think about. Once the dynamics you use is clear, it would be interesting to check if the reverse dynamics can be somehow characterized, starting from the equilibrium position.</p>
|
combinatorics | <p>I want to know if the popular Sudoku puzzle is a Cayley table for a group. </p>
<p>Methods I've looked at: Someone I've spoken to told me they're not because counting the number of puzzle solutions against the number of tables with certain permutations of elements, rows and columns, the solutions are bigger than the tables, but I can't see why because I don't know how to count the different tables for a group of order 9, and then permute the rows, columns and elements in different ways. Also I believe the rotations/reflections will matter in comparing these numbers too. It would also be nice if there was a way to know if the operation is associative just from the table.</p>
| <p>A Cayley table for a group can never be a sudoku. Assume you have a $9 \times 9$ Cayley table for a group of order $9$, and say your identity element is at index $1 \le i \le 9$. Then row $i$ and column $i$ are symmetric to each other because they correspond to multiplication with the identity. In particular, if you look at the $3\times3$ sub-square containing the element of coordinates $(i,i)$, this square has duplicates (because it contains symmetric elements of row $i$ and column $i$). So the table isn't a Sudoku.</p>
<p>If you allow to swap the rows or columns, this is possible. Take the table of $G = \mathbb{Z}/9\mathbb{Z}$ (I wrote $0$ instead of $9$ for convenience):
$$
\begin{array}{r|lllllllll}
+&0&1&2&3&4&5&6&7&8\\
\hline
0&0&1&2&3&4&5&6&7&8\\
1&1&2&3&4&5&6&7&8&0\\
2&2&3&4&5&6&7&8&0&1\\
3&3&4&5&6&7&8&0&1&2\\
4&4&5&6&7&8&0&1&2&3\\
5&5&6&7&8&0&1&2&3&4\\
6&6&7&8&0&1&2&3&4&5\\
7&7&8&0&1&2&3&4&5&6\\
8&8&0&1&2&3&4&5&6&7\\
\end{array}$$</p>
<p>And swap the rows in order $0,3,6,1,4,7,2,5,8$, to obtain the Sudoku</p>
<p>$$
\begin{array}{r|lll|lll|lll}
+&0&1&2&3&4&5&6&7&8\\
\hline
0&0&1&2&3&4&5&6&7&8\\
3&3&4&5&6&7&8&0&1&2\\
6&6&7&8&0&1&2&3&4&5\\
\hline
1&1&2&3&4&5&6&7&8&0\\
4&4&5&6&7&8&0&1&2&3\\
7&7&8&0&1&2&3&4&5&6\\
\hline
2&2&3&4&5&6&7&8&0&1\\
5&5&6&7&8&0&1&2&3&4\\
8&8&0&1&2&3&4&5&6&7\\
\end{array}$$</p>
| <p>It is easy to construct a Sudoku that cannot be a Cayley table. <strong>Not even if you label the rows and columns independently from each other.</strong> Consider the following.
$$
\begin{array}{ccc|ccc|ccc}
1&2&3&4&5&6&7&8&9\\
4&5&6&7&8&9&1&3&2\\
7&8&9&1&2&3&4&5&6\\
\hline
2&3&1&5&6&4&9&7&8\\
5&6&4&8&9&7&2&1&3\\
8&9&7&2&3&1&6&4&5\\
\hline
3&1&2&6&4&5&8&9&7\\
6&4&5&9&7&8&3&2&1\\
9&7&8&3&1&2&5&6&4
\end{array}
$$
My argument only needs the two first rows. The rest were filled in just to leave no doubt about the fact that those two rows form a part of a complete Sudoku.</p>
<p>We see that we get the second row from the first by applying the permutation
$\sigma=(147)(258369)$ to it. This is all we need to prove that this is not a Cayley table. If in a Cayley table the first row has $a$ as the common left factor and the second row has $b$ as the common left factor, we get the second row from the first by multiplying everything from the left by $c:=ba^{-1}$.
If $n=\operatorname{ord}(c)$ then the permutation $\sigma$ bringing the first row to the second will then only have cycles of length $n$ as the cycles
act transitively on right cosets $Hx$, $H=\langle c\rangle, x\in G$.</p>
<p>But our $\sigma$ has cycle type $(3,6)$. This already gives two obvious contradictions:</p>
<ul>
<li>the lengths of the cycles vary, and</li>
<li>$\sigma$ has order six, so cannot be an element of a group of order nine.</li>
</ul>
|
matrices | <p>I found out that there exist positive definite matrices that are non-symmetric, and I know that symmetric positive definite matrices have positive eigenvalues.</p>
<p>Does this hold for non-symmetric matrices as well?</p>
| <p>Let <span class="math-container">$A \in M_{n}(\mathbb{R})$</span> be any non-symmetric <span class="math-container">$n\times n$</span> matrix but "positive definite" in the sense that: </p>
<p><span class="math-container">$$\forall x \in \mathbb{R}^n, x \ne 0 \implies x^T A x > 0$$</span>
The eigenvalues of <span class="math-container">$A$</span> need not be positive. For an example, the matrix in David's comment:</p>
<p><span class="math-container">$$\begin{pmatrix}1&1\\-1&1\end{pmatrix}$$</span></p>
<p>has eigenvalue <span class="math-container">$1 \pm i$</span>. However, the real part of any eigenvalue <span class="math-container">$\lambda$</span> of <span class="math-container">$A$</span> is always positive.</p>
<p>Let <span class="math-container">$\lambda = \mu + i\nu\in\mathbb C $</span> where <span class="math-container">$\mu, \nu \in \mathbb{R}$</span> be an eigenvalue of <span class="math-container">$A$</span>. Let <span class="math-container">$z \in \mathbb{C}^n$</span> be a right eigenvector associated with <span class="math-container">$\lambda$</span>. Decompose <span class="math-container">$z$</span> as <span class="math-container">$x + iy$</span> where <span class="math-container">$x, y \in \mathbb{R}^n$</span>.</p>
<p><span class="math-container">$$(A - \lambda) z = 0 \implies \left((A - \mu) - i\nu\right)(x + iy) = 0
\implies \begin{cases}(A-\mu) x + \nu y = 0\\(A - \mu) y - \nu x = 0\end{cases}$$</span>
This implies</p>
<p><span class="math-container">$$x^T(A-\mu)x + y^T(A-\mu)y = \nu (y^T x - x^T y) = 0$$</span></p>
<p>and hence
<span class="math-container">$$\mu = \frac{x^TA x + y^TAy}{x^Tx + y^Ty} > 0$$</span></p>
<p>In particular, this means any real eigenvalue <span class="math-container">$\lambda$</span> of <span class="math-container">$A$</span> is positive.</p>
| <p>I am answering the first part of @nukeguy's comment, who asked:</p>
<blockquote>
<p>Is the converse true? If all of the eigenvalues of a matrix <span class="math-container">$𝐴$</span> have
positive real parts, does this mean that <span class="math-container">$𝑥^𝑇𝐴𝑥>0$</span> for any
<span class="math-container">$𝑥≠0∈ℝ^𝑛$</span>? What if we assume <span class="math-container">$𝐴$</span> is diagonalizable?</p>
</blockquote>
<p>I have a counterexample, where <span class="math-container">$A$</span> has positive eigenvalues, but it is not positive definite:
<span class="math-container">$ A=
\begin{bmatrix}
7 & -2 & -4 \\
-17 & 40 & -19 \\
-21 & -9 & 31
\end{bmatrix}
$</span>. Eigenvalues of this matrix are <span class="math-container">$1.2253$</span>, <span class="math-container">$27.4483$</span>, and <span class="math-container">$49.3263$</span>, but it indefinite because if <span class="math-container">$x_1 = \begin{bmatrix}-48 & -10& -37\end{bmatrix}$</span> and <span class="math-container">$x_2 = \begin{bmatrix}-48 &10 &-37\end{bmatrix}$</span>, then we have <span class="math-container">$𝑥_1𝐴𝑥_1^T = -1313$</span> and <span class="math-container">$𝑥_2𝐴𝑥_2^T = 37647.$</span></p>
|
linear-algebra | <p>I am relatively new to the world of academical mathematics, but I have noticed that most, if not all, mathematical textbooks that I've had the chance to come across, seem completely oblivious to the existence of lambda notation. </p>
<p>More specifically, in a linear algebra course I'm taking, I found it a lot easier to understand "higher order functionals" from the second dual space, by putting them in lambda expressions. It makes a lot more sense to me to put them in the neat, clear notation of lambda expressions, rather than in multiple variable functions where not all the arguments are of the same "class" as some are linear functionals and others are vectors. For example, consider the canonical isomorphism -
$$A:V \rightarrow V^{**}$$</p>
<p>It would usually be expressed by $$Av(f) = f(v)$$
This was a notation I found particularly difficult to understand at first as there are several processes taking place "under the hood", that can be put a lot more clearly, in my opinion, this way:</p>
<p>$$A = \lambda v \in V. \lambda f \in V^{*}. f(v)$$</p>
<p>I agree that this notation may become tedious and over-explanatory over time, but as a first introduction of the concept I find it a lot easier as it makes it very clear what goes where.</p>
<p>My question is, basically, why isn't this widespread, super popular notation in the world of computer science, not as popular in the field of mathematics? Or is it, and I'm just not aware? </p>
| <p>As Derek already said, there is no essential difference between functions $A\times B \to C$ and functions $A\to (B \to C)$ via Currying (this is also more abstractly expressed by the universal property of an <a href="https://en.wikipedia.org/wiki/Exponential_object" rel="noreferrer">exponential</a> which unifies the set-theoretical currying and currying in a typed lambda calculus).</p>
<p>On the notational side of things, I <em>personally</em> prefer $x\mapsto f(x)$ to $\lambda x. f(x)$ and I suspect many other mathematicians feel the same (especially since $\lambda$ is such a commonly used letter).</p>
<hr>
<p>EDIT: (now that my answer stopped being one, let me add some rambling that the 29 people so far have <em>not</em> upvoted for):</p>
<p>I'm guessing many mathematicians are less "comfortable" with nested expressions like $v\mapsto (f \mapsto f(v))$. That would be nothing extraordinary, since there are various concepts that some mathematicians feel less comfortable about. Here are two (unrelated) things that I have encountered:</p>
<ul>
<li>empty metric spaces: Some people deliberately require metric spaces to be non-empty which is a nuisance: given a metric space $(X,d)$ and $Y\subseteq X$, $(Y,d|_{Y^2})$ is a metric space again... unless of course $Y=\emptyset$; apparently it doesn't feel "right" for metric spaces to be empty</li>
<li>$f(x)$ instead of $f$: Some people refer to a function $f$ as $f(x)$; this is (unfortunately) what I learned in high school and is (rein)forced by notation like $\frac{d f(x)}{d x}$ and $\int f(x) \,dx$</li>
</ul>
<p>Although, your example:</p>
<blockquote>
<p>Let $A : V\to V^{**}$ such that $Av(f) = f(v)$ for all $v\in V$ and $f\in V^*$</p>
</blockquote>
<p>is fine and not hard to understand, in my opinion. For every $v\in V$ we have $Av\in V^{**}$, i.e. $Av : V^* \to \mathbb K$. Hence we can plug in an $f\in V^*$ to get $f(v) \in \mathbb{K}$. If the author thinks it is easy to understand <em>and</em> is more used to it than $v\mapsto (f \mapsto f(v))$ then they would obviously have no reason to change the notation.</p>
<p>So the reason why $v\mapsto (f\mapsto f(v))$ (or a variant thereof) is not used as much is probably: "I'm not used to this notation and I'm perfectly happy with mine."</p>
<p>By the way, <em>my personal</em> favourite is <em>also</em> not:
$$A : V \to V^{**}, v \mapsto (f\mapsto f(v))$$ but
$$A : V\to V^{**}, v\mapsto \_(v)$$ where it is implied that $\_$ is a placeholder, i.e. $\_(v) : V^* \to \mathbb{K}, f\to f(v)$.</p>
| <p>Lambda calculus is related with computer science through and through. To quote Wikipedia:</p>
<blockquote>
<p>Lambda calculus (also written as λ-calculus) is a formal system in mathematical logic for expressing <em>computation</em> based on function abstraction and <em>application</em> using variable binding and <em>substitution</em>.</p>
</blockquote>
<p>Highlights mine. Here, "computation", "application" and "substitution" are very well defined operations on symbols as understood in CS. That is literally what lambda calculus is all about, to start out with: to reason about substituting symbols in formal languages.</p>
<p>Processes like Currying are there because they have relatively practical applications - for example, they make abstract reasoning easier (by reducing all lambdas with multiple arguments to ones with single arguments). "Meta" topics like lazy evalation, typing, strictness etc. can all be explored in the context of lambda calculus and have little impact on general mathematic formulae. For CS, it is important to be super exact with these things, as computers, basically, are machines for manipulating symbols.</p>
<p>So, lambdas have use for the theoretical computer linguist / computer scientist / logician; on the surface you could probably use the notation for general mathematics, but many of the advanced "benefits" do not transfer (or at least not in a helpful manner). In most parts of mathematics, especially applied mathematics (physics...), the question of how exactly to "apply" and "substitute" variables is crystal clear and of little interest to anybody - it is often quite usual to skip writing bound variables completely. </p>
<p>Oh, and the other answer: people are just used to the usual representation. Plenty of mathematical areas tend to have their own notations for quite similar things. It's just how it is.</p>
|
combinatorics | <p>So, the full problem goes like this:</p>
<blockquote>
<p>There are $25$ people at a party. Assuming that among any three people, at least two of them know each other, prove that there exists one person who must know at least twelve people.</p>
</blockquote>
<p>I've been stuck on this problem for a while and haven't really figured out how to proceed. I'm pretty sure that there is an answer that can be found via the <a href="https://en.m.wikipedia.org/wiki/Pigeonhole_principle" rel="noreferrer">pigeonhole principle</a> or some graph theory, but I'm not really sure how to get started. Any help would be appreciated.</p>
| <p>If everyone knows everyone, then you are done.</p>
<p>Otherwise choose two people, A and B say, who don't know each other. These two people are part of $23$ triples. In each of these triples, either A knows the third person, or B knows the third person.</p>
<p>Thus one of A or B knows (at least) $12$ people.</p>
| <p>Pick a vertex <span class="math-container">$v$</span>. If <span class="math-container">$\deg(v) \geq 12$</span> you are done.</p>
<p>Otherwise <span class="math-container">$v$</span> is connected with at most 11 vertices. Let <span class="math-container">$C$</span> be the vertices connected to <span class="math-container">$v$</span> and <span class="math-container">$N$</span> be the vertices not connected to <span class="math-container">$v$</span>. Note that <span class="math-container">$N$</span> has at least <span class="math-container">$13$</span> vertices.</p>
<p>Fix one vertex <span class="math-container">$u \in N$</span>.</p>
<p>Now, for each <span class="math-container">$w \in N$</span> look at the group <span class="math-container">$\{ u, v, w\}$</span>. The only possible edge in this group is <span class="math-container">$uw$</span>. Therefore, <span class="math-container">$uw \in E(G)$</span>.</p>
<p>This shows that <span class="math-container">$u$</span> is connected to all the other vertices in <span class="math-container">$N$</span>.</p>
<p><strong>Note</strong> The proof is basically the following:</p>
<p>The given condition shows that if you fix one vertex <span class="math-container">$v$</span>, and you look to all the vertices <span class="math-container">$N$</span> which are not connected to <span class="math-container">$v$</span>, then the induced graph on <span class="math-container">$N$</span> is the complete graph.</p>
<p>So if <span class="math-container">$|N| \geq 13$</span> you are done, otherwise <span class="math-container">$|N| \leq 12$</span> which means <span class="math-container">$\deg(v) \geq 12$</span>.</p>
|
geometry | <p>I'm a fifteen year old who is currently studying circle geometry (if that is the appropriate term) and our teacher stated that concentric circles are similar. I thought about this, and it doesn't make sense to me. The reason is because of proportionality. For example, similar triangles are similar because they have the same angles and they have proportional sides. However, circles can not be compared for angles, so that's out (as they all have the same 360 degree angle at the center) and the only factor is their size, which is directly influenced by their radius. If the radius is the only variable involved in a triangle like this, how can a circle be NOT proportional to another circle? If a case of that existed, there would be meaning (at least from my current perspective) to the term "similar circle."</p>
<p>Help and critique on my logic is requested, and an explanation as to the term "similar circle."</p>
| <p>You're right: any two circles are similar (and so there's not much point of talking about "similar circles")! In general, two shapes are "congruent" if you can turn one into the other by translations (moving around in the plane), rotations, and reflections. Two shapes are "similar" if you can rescale ("zoom in or out" on the picture) one of them to turn it into a shape that is congruent to the other. Given two circles, you can rescale one so that it has the same radius as the other, and then any two circles with the same radius are congruent since you can just translate the center of one to the center of the other.</p>
| <p>Yes indeed. Every circle is similar. You can always scale one of them to match the other. Actually, this is the definition of similarity. In case of triangles, this definition yields the result that the sides are proportional. "The sides of one triangle are proportional to the other" is not the actual definition of similarity.
You may have a look <a href="https://learnzillion.com/resources/73031-prove-that-all-circles-are-similar">here</a></p>
|
combinatorics | <p>Yesterday the aliens took 100 scientists from Earth as prisoners. They want to test how smart the humans are.</p>
<p>The aliens made 101 headbands, numbered from 1 to 101. On the contest day, they throw away one of the headbands, and then from the remaining headbands randomly put one on each of the scientists. Then they line up the scientists in a queue in such a way that each person can see the numbers written on the headbands of the people who stand in front of him, but can't see his/her own number, nor the number of the scientists behind him.</p>
<p>Then the aliens will force the scientists to guess their numbers, and after everybody finished saying a number, they will kill those who said a number different from what is written on their headbands. Note that each scientist can only say one number in the range 1-101, and no one can say a number that has already been said. Each scientist can independently choose when to speak their guess; there is no pre-set order on when they have to speak. As you are very smart, the scientists asked you to find a way to save the most of them. Find this way, and then prove it.</p>
| <p>You can save 99.5 scientists.</p>
<p>Imagine that the 101st headband was left on the ground behind the first scientist, making a line of 101 headbands.</p>
<p>The strategy is that everyone should assume that the headbands are laid out in an even permutation of the numbers {1, 2, ..., 101}. Each scientist will always have a choice of two headbands: one results in an even permutation and one results in an odd permutation. They always make the choice which results in an even permutation.</p>
<p>If the headbands are actually in an even permutation (half the time), everyone guesses correctly. (More formal inductive proof below.)</p>
<p>If the headbands are actually in an odd permutation then the first scientist is dead. But the other scientists are still saved: they will end up naming a permutation which is correct except for the switch of the first scientist's headband with the one on the ground.</p>
<p>Proof that everyone guesses correctly if the headbands are actually in an even permutation:</p>
<p>Consider the first scientist. He can see 99 headbands, so he knows there are only two possible arrangements. These arrangements are related by a single swap (swapping his headband for the one left on the ground), so one of them is an even permutation and the other is odd. His strategy is to say the number which corresponds to the even permutation, and we're assuming here that it actually <em>is</em> an even permutation, so his guess is correct.</p>
<p>Now consider the nth scientist. We can assume for induction that the first (n-1) scientists have guessed correctly, and he can see a further (100-n) headbands. So he knows a total of 99 numbers. This means that, again, there are only two possible arrangements, and the two arrangements are related by a single swap (swapping his headband for the one on the ground). So precisely one of these arrangements is an even permutation, and this is the one that the nth scientist will guess. This means that the nth scientist has also guessed correctly, since we know that the actual arrangement is the even permutation.</p>
| <p>The scientist in the rear states the sum of all 99 visible numbers, $S_{99}$. He or she will die. The second to last scientist is hopefully clever enough to understand why the prior scientist clearly picked a faulty number. This scientist can now can sum the 98 visible numbers, $S_{98}$, to uniquely determine his or her own number, $S_{99} - S_{98}$.</p>
<p>This idea is repeated, saving all the remaining scientists.</p>
<p>All but one survive.</p>
|
matrices | <p>In one of my exams I'm asked to prove the following </p>
<blockquote>
<p>Suppose <span class="math-container">$A,B\in \mathbb R^{n\times n}$</span>, and <span class="math-container">$AB=BA$</span>, then <span class="math-container">$A,B$</span> share the same eigenvectors. </p>
</blockquote>
<p>My attempt is let <span class="math-container">$\xi$</span> be an eigenvector corresponding to <span class="math-container">$\lambda$</span> of <span class="math-container">$A$</span>, then <span class="math-container">$A\xi=\lambda\xi$</span>, then I want to show <span class="math-container">$\xi$</span> is also some eigenvector of <span class="math-container">$B$</span> but I get stuck.</p>
| <p>The answer is in the book <em>Linear Algebra and its Application</em> by Gilbert Strang. I'll just write down what he said in the book.</p>
<blockquote>
<p>Starting from <span class="math-container">$Ax=\lambda x$</span>, we have</p>
<p><span class="math-container">$$ABx = BAx = B \lambda x = \lambda Bx$$</span></p>
<p>Thus <span class="math-container">$x$</span> and <span class="math-container">$Bx$</span> are both eigenvectors of <span class="math-container">$A$</span>, sharing the same <span class="math-container">$\lambda$</span> (or else <span class="math-container">$Bx = 0$</span>). If we assume for convenience that the eigenvalues of <span class="math-container">$A$</span> are distinct – the eigenspaces are one dimensional – then <span class="math-container">$Bx$</span> must be a multiple of <span class="math-container">$x$</span>. In other words <span class="math-container">$x$</span> is an eigenvector of <span class="math-container">$B$</span> as well as <span class="math-container">$A$</span>.</p>
</blockquote>
<p>There's another proof using diagonalization in the book.</p>
| <p>Commuting matrices do not necessarily share <em>all</em> eigenvector, but generally do share <em>a</em> common eigenvector.</p>
<p>Let <span class="math-container">$A,B\in\mathbb{C}^{n\times n}$</span> such that <span class="math-container">$AB=BA$</span>. There is always a nonzero subspace of <span class="math-container">$\mathbb{C}^n$</span> which is both <span class="math-container">$A$</span>-invariant and <span class="math-container">$B$</span>-invariant (namely <span class="math-container">$\mathbb{C}^n$</span> itself). Among all these subspaces, there exists hence an invariant subspace <span class="math-container">$\mathcal{S}$</span> of the minimal (nonzero) dimension.</p>
<p>We show that <span class="math-container">$\mathcal{S}$</span> is spanned by some common eigenvectors of <span class="math-container">$A$</span> and <span class="math-container">$B$</span>.
Assume that, say, for <span class="math-container">$A$</span>, there is a nonzero <span class="math-container">$y\in \mathcal{S}$</span> such that <span class="math-container">$y$</span> is not an eigenvector of <span class="math-container">$A$</span>. Since <span class="math-container">$\mathcal{S}$</span> is <span class="math-container">$A$</span>-invariant, it contains some eigenvector <span class="math-container">$x$</span> of <span class="math-container">$A$</span>; say, <span class="math-container">$Ax=\lambda x$</span> for some <span class="math-container">$\lambda\in\mathbb{C}$</span>. Let <span class="math-container">$\mathcal{S}_{A,\lambda}:=\{z\in \mathcal{S}:Az=\lambda z\}$</span>. By the assumption, <span class="math-container">$\mathcal{S}_{A,\lambda}$</span> is a proper (but nonzero) subspace of <span class="math-container">$\mathcal{S}$</span> (since <span class="math-container">$y\not\in\mathcal{S}_{A,\lambda}$</span>).</p>
<p>We know that for any <span class="math-container">$z\in \mathcal{S}_{A,\lambda}$</span>, <span class="math-container">$Bz\in \mathcal{S}$</span> since <span class="math-container">$\mathcal{S}_{A,\lambda}\subset\mathcal{S}$</span> and <span class="math-container">$\mathcal{S}$</span> is <span class="math-container">$B$</span>-invariant. However, <span class="math-container">$A$</span> and <span class="math-container">$B$</span> commute so
<span class="math-container">$$
ABz=BAz=\lambda Bz \quad \Rightarrow\quad Bz\in \mathcal{S}_{A,\lambda}.
$$</span>
This means that <span class="math-container">$\mathcal{S}_{A,\lambda}$</span> is <span class="math-container">$B$</span>-invariant. Since <span class="math-container">$\mathcal{S}_{A,\lambda}$</span> is both <span class="math-container">$A$</span>- and <span class="math-container">$B$</span>-invariant and is a proper (nonzero) subspace of <span class="math-container">$\mathcal{S}$</span>, we have a contradiction. Hence every nonzero vector in <span class="math-container">$\mathcal{S}$</span> is an eigenvector of both <span class="math-container">$A$</span> and <span class="math-container">$B$</span>.</p>
<hr>
<p><strong>EDIT:</strong> A nonzero <span class="math-container">$A$</span>-invariant subspace <span class="math-container">$\mathcal{S}$</span> of <span class="math-container">$\mathbb{C}^n$</span> contains an eigenvector of <span class="math-container">$A$</span>.</p>
<p>Let <span class="math-container">$S=[s_1,\ldots,s_k]\in\mathbb{C}^{n\times k}$</span> be such that <span class="math-container">$s_1,\ldots,s_k$</span> form a basis of <span class="math-container">$\mathcal{S}$</span>. Since <span class="math-container">$A\mathcal{S}\subset\mathcal{S}$</span>, we have <span class="math-container">$AS=SG$</span> for some <span class="math-container">$G\in\mathbb{C}^{k\times k}$</span>. Since <span class="math-container">$k\geq 1$</span>, <span class="math-container">$G$</span> has at least one eigenpair <span class="math-container">$(\lambda,x)$</span>. From <span class="math-container">$Gx=\lambda x$</span>, we get <span class="math-container">$A(Sx)=SGx=\lambda(Sx)$</span> (<span class="math-container">$Sx\neq 0$</span> because <span class="math-container">$x\neq 0$</span> and <span class="math-container">$S$</span> has full column rank). The vector <span class="math-container">$Sx\in\mathcal{S}$</span> is an eigenvector of <span class="math-container">$A$</span> and, consequently, <span class="math-container">$\mathcal{S}$</span> contains at least one eigenvector of <span class="math-container">$A$</span>.</p>
<hr>
<p><strong>EDIT:</strong> There is a nonzero <span class="math-container">$A$</span>- and <span class="math-container">$B$</span>-invariant subspace of <span class="math-container">$\mathbb{C}^n$</span> of the least dimension.</p>
<p>Let <span class="math-container">$\mathcal{I}$</span> be the set of all nonzero <span class="math-container">$A$</span>- and <span class="math-container">$B$</span>-invariant subspaces of <span class="math-container">$\mathbb{C}^n$</span>. The set is nonempty since <span class="math-container">$\mathbb{C}^n$</span> is its own (nonzero) subspace which is both <span class="math-container">$A$</span>- and <span class="math-container">$B$</span>-invariant (<span class="math-container">$A\mathbb{C}^n\subset\mathbb{C}^n$</span> and <span class="math-container">$B\mathbb{C}^n\subset\mathbb{C}^n$</span>). Hence the set <span class="math-container">$\mathcal{D}:=\{\dim \mathcal{S}:\mathcal{S}\in\mathcal I\}$</span> is a nonempty subset of <span class="math-container">$\{1,\ldots,n\}$</span>. By the <a href="http://en.wikipedia.org/wiki/Well-ordering_principle" rel="noreferrer">well-ordering principle</a>, <span class="math-container">$\mathcal{D}$</span> has the least element and hence there is a nonzero <span class="math-container">$\mathcal{S}\in\mathcal{I}$</span> of the least dimension.</p>
|
combinatorics | <p>Aside from <span class="math-container">$1!\cdot n!=n!$</span> and <span class="math-container">$(n!-1)!\cdot n! = (n!)!$</span>, the only product of factorials known is <span class="math-container">$6!\cdot 7!=10!$</span>.</p>
<p>One might naturally associate these numbers with the permutations on <span class="math-container">$6, 7,$</span> and <span class="math-container">$10$</span> objects, respectively, and hope that this result has some kind of connection to a sporadic relation between such permutations - numerical "coincidences" often have deep math behind them, like how <span class="math-container">$1^2+2^2+\ldots+24^2=70^2$</span> can be viewed as an ingredient that makes the Leech lattice work.</p>
<p>The most natural thing to hope for would be a product structure on the groups <span class="math-container">$S_6$</span> and <span class="math-container">$S_7$</span> mapping to <span class="math-container">$S_{10}$</span>, but as <a href="https://mathoverflow.net/questions/324436/does-the-symmetric-group-s-10-factor-as-a-knit-product-of-symmetric-subgroup">this MathOverflow thread</a> shows, one cannot find disjoint copies of <span class="math-container">$S_6$</span> and <span class="math-container">$S_7$</span> living in <span class="math-container">$S_{10}$</span>, so a product structure seems unlikely.</p>
<p>However, I'm holding out hope that some weaker kind of bijection can be found in a "natural" way. Obviously one <em>can</em> exhibit a bijection. For instance, identify the relative ordering of <span class="math-container">$1,2,\ldots 7$</span> in a permutation of size <span class="math-container">$10$</span>, and then biject <span class="math-container">$_{10}P_{3}=720$</span> with <span class="math-container">$S_6$</span> in some way. But I'd like to know if there is a way to define such a bijection which arises naturally from the permutation structures on these sets, and makes it clear why the construction does not extend to other orders.</p>
<p>I tried doing something with orderings on polar axes of the dodecahedron (<span class="math-container">$10!$</span>) and orderings on polar axes of the icosahedron (<span class="math-container">$6!$</span>), in the hopes that the sporadic structure and symmetry of these Platonic solids would allow for interesting constructions that don't generalize, but ran into issues with the dodecahedron (sequences of dodecahedral axes aren't particularly nice objects) and the question of how to extract a permutation of length <span class="math-container">$7$</span>.</p>
<p>I'm curious if someone can either devise a natural bijection between these sets or link to previous work on this question.</p>
| <p>This family of bijections (of sets) <span class="math-container">$S_6\times S_7 \to S_{10}$</span> has already been suggested in comments and linked threads, but it is so pretty I wanted to spell it out:</p>
<p>There are <span class="math-container">$10$</span> ways of partitioning the numbers <span class="math-container">$1,2,3,4,5,6$</span> into two (unordered) pieces of equal size: <span class="math-container">$P_1,P_2,\cdots,P_{10}$</span>. Thus we have a canonical embedding <span class="math-container">$S_6\hookrightarrow S_{10}$</span>, coming from the induced action on the <span class="math-container">$P_i$</span>.</p>
<p>Any distinct pair <span class="math-container">$P_i,P_j$</span> will be related by a unique transposition. For example <span class="math-container">$\{\{1,2,3\},\{4,5,6\}\}$</span> (denoted hereafter <span class="math-container">$\left(\frac{123}{456}\right)$</span>) is related to <span class="math-container">$\left(\frac{126}{453}\right)$</span> via the transposition <span class="math-container">$(36)$</span>.</p>
<p>There are two types of ordered (distinct) triples <span class="math-container">$P_i, P_j,P_k$</span>:</p>
<ol>
<li><p>They may be related pairwise via transpositions <span class="math-container">$(ab),(cd),(ef)$</span> with <span class="math-container">$a,b,c,d,e,f$</span> distinct and each of <span class="math-container">$\{a,b\}, \{c,d\},\{e,f\}$</span> not on the same side of any of <span class="math-container">$P_i, P_j,P_k$</span>:<span class="math-container">$$
\left(\frac{ace}{bdf}\right), \left(\frac{bce}{adf}\right), \left(\frac{ade}{bcf}\right).$$</span><br />
Here, there are <span class="math-container">$10$</span> choices for <span class="math-container">$P_i$</span>, <span class="math-container">$9$</span> choices for <span class="math-container">$P_j$</span> and <span class="math-container">$4$</span> choices for <span class="math-container">$P_k$</span>, giving <span class="math-container">$360$</span> triples in total.</p>
</li>
<li><p>They may be related pairwise via transpositions <span class="math-container">$(ab),(ca),(bc)$</span> with <span class="math-container">$a,b,c$</span> distinct:
<span class="math-container">$$
\left(\frac{ace}{bdf}\right), \left(\frac{bce}{adf}\right), \left(\frac{abe}{cdf}\right).$$</span><br />
Again, there are <span class="math-container">$10$</span> choices for <span class="math-container">$P_i$</span>, <span class="math-container">$9$</span> choices for <span class="math-container">$P_j$</span> and <span class="math-container">$4$</span> choices for <span class="math-container">$P_k$</span>, giving <span class="math-container">$360$</span> triples in total.</p>
</li>
</ol>
<p>An element of the stabiliser (in <span class="math-container">$S_6$</span>) of a type 1 ordered triple (written as above) must preserve the pairs <span class="math-container">$\{a,b\}, \{c,d\},\{e,f\}$</span>. Further if it swaps any of these pairs it must swap all of them, so the only non-trivial element of the stabiliser is an odd permutation: <span class="math-container">$(ab)(cd)(ef)$</span>.</p>
<p>An element of the stabiliser (in <span class="math-container">$S_6$</span>) of a type 2 ordered triple (written as above) must preserve the sets <span class="math-container">$\{d,f\}, \{e\},\{a,c,b\}$</span>. Further it must fix each of <span class="math-container">$a,b,c$</span>. Thus the only non-trivial element of the stabiliser is an odd permutation: <span class="math-container">$(df)$</span>.</p>
<p>As <span class="math-container">$|A_6|=360$</span>, in particular this means there is a unique element of <span class="math-container">$A_6$</span> taking the ordered triple <span class="math-container">$P_1,P_2,P_3$</span> to a specified ordered triple <span class="math-container">$P_i,P_j,P_k$</span> of the same type as <span class="math-container">$P_1,P_2,P_3$</span>.</p>
<p>Fix <span class="math-container">$t\in S_{10}$</span> a permutation taking <span class="math-container">$P_1,P_2,P_3$</span> to an ordered triple of the other type. Then there is a unique element in <span class="math-container">$A_6$</span> which composed with <span class="math-container">$t$</span> takes the ordered triple <span class="math-container">$P_1,P_2,P_3$</span> to a specified ordered triple <span class="math-container">$P_i,P_j,P_k$</span> of the other type to <span class="math-container">$P_1,P_2,P_3$</span>.</p>
<p>Let <span class="math-container">$S_7$</span> denote the group of permutations of <span class="math-container">$P_4,P_5,\cdots,P_{10}$</span>. Then any permutation in <span class="math-container">$S_{10}$</span> may be written uniquely as an element of <span class="math-container">$S_7$</span> followed by an element of <span class="math-container">$(A_6\sqcup tA_6)$</span>, where the latter is determined by where <span class="math-container">$P_1,P_2,P_3$</span> are mapped to.</p>
<p>Thus we have established a bijection of sets <span class="math-container">$$S_{10}\to (A_6\sqcup tA_6)\times S_7.$$</span>
Once we fix an odd permutation <span class="math-container">$t'\in S_6$</span>, we may identify the sets <span class="math-container">$$(A_6\sqcup t'A_6)\to S_6.$$</span>
Composing we get: <span class="math-container">$$S_{10}\to (A_6\sqcup tA_6)\times S_7\to (A_6\sqcup t'A_6)\times S_7\to S_6\times S_7.$$</span></p>
<p>That is for any choice of the permutations <span class="math-container">$t,t'$</span> we have the required bijection of sets.</p>
| <p>It may be connected with, of all things, the <span class="math-container">$3-4-5$</span> right triangle! This triangle and its multiples stand out as having the sides in arithmetic progression. Such an arithmetic progression leads to factorial expressions when the sides are multiplied together.</p>
<p>As a preliminary step, consider a relatively unheralded property of right triangles: the diameter of the incircle plus the hypotenuse equals the sum of the other two sides. Suppose that the legs are <span class="math-container">$a$</span> and <span class="math-container">$b$</span>, and the hypoteneuse is <span class="math-container">$c$</span> where <span class="math-container">$c^2=a^2+b^2$</span>. The diameter of the incircle is then <span class="math-container">$2ab/(a+b+c)$</span> while the Pythagorean relation implies <span class="math-container">$$(a+b+c)(a+b-c)=(a^2+2ab+b^2)-(a^2+b^2)=2ab$$</span>
Thereby the diameter of the incircle reduces to <span class="math-container">$a+b-c$</span>. Should there be a right triangle whose sides are in arithmetic progression, then, the diameter of the incircle will join this progression, making it longer and thus perhaps generating a bigger factorial upon multiplication.</p>
<p>In <a href="https://math.stackexchange.com/questions/3626718/triangle-construction-given-semiperimeter-and-radii-of-inscribed-and-circumscrib/3627358?r=SearchResults#3627358">this question</a> it is shown that the product of the sides of any triangle is half the product of the diameter of the circumcircle (circumdiameter), the diameter of the incircle (indiameter), and the perimeter. Let us see where that leads if we apply it to a right triangle having sides <span class="math-container">$3,4,5$</span>.
Multiplying the sides together then gives</p>
<p><span class="math-container">$3×4×5=\text{circumdiameter}×\text{indiameter}×\text{perimeter}/2$</span></p>
<p>We double the sides of the triangle to clear the fraction on the right side:</p>
<p><span class="math-container">$6×8×10=\text{circumdiameter}×\text{indiameter}×\text{perimeter}×4$</span></p>
<p>The circumdiameter is the hypoteneuse of the <span class="math-container">$3-4-5$</span> triangle, thus <span class="math-container">$5$</span> -- which is in the aforementioned arithmetic progression. The indiameter is <span class="math-container">$2$</span> from the above lemma, which precedes <span class="math-container">$3,4,5$</span> in the arithmetic progression. And the perimeter of the triangle is three times the longer leg, again due to the arithmetic progression, thus <span class="math-container">$4×3$</span>. Substituting these results into the above product equality then gives</p>
<p><span class="math-container">$6×8×10=5×2×(3×4)×4=5!×4$</span></p>
<p>And there is our factorial. To make it cleaner we should multiply by <span class="math-container">$3/2$</span>, absorbing the dangling factor <span class="math-container">$4$</span> into the factorial. We then get three different three-term products on the left side, depending on which of the factors <span class="math-container">$6,8,10$</span> we increment:</p>
<p><span class="math-container">$\color{blue}{8×9×10}=6×10×12=6×8×15=5×2×(3×4)×6=6!$</span></p>
<p>And from the three-term product shown in blue, we have</p>
<p><span class="math-container">$6!=10!/7!$</span></p>
<p>Why is this uniquely chosen? We see that the sides of a right triangle being in arithmetic progression lead to the factorial on the right in two ways, by making the perimeter a simple multiple of one leg and by incorporating the circum diameter into the arithmetic progression. Only the <span class="math-container">$3-4-5$</span> right triangle has these properties, and it leads specifically to <span class="math-container">$6!$</span> also being a factorial ratio.</p>
|
matrices | <p>I can follow the definition of the transpose algebraically, i.e. as a reflection of a matrix across its diagonal, or in terms of dual spaces, but I lack any sort of geometric understanding of the transpose, or even symmetric matrices.</p>
<p>For example, if I have a linear transformation, say on the plane, my intuition is to visualize it as some linear distortion of the plane via scaling and rotation. I do not know how this distortion compares to the distortion that results from applying the transpose, or what one can say if the linear transformation is symmetric. Geometrically, why might we expect orthogonal matrices to be combinations of rotations and reflections?</p>
| <p>To answer your second question first: an orthogonal matrix $O$ satisfies $O^TO=I$, so $\det(O^TO)=(\det O)^2=1$, and hence $\det O = \pm 1$. The determinant of a matrix tells you by what factor the (signed) volume of a parallelipiped is multipled when you apply the matrix to its edges; therefore hitting a volume in $\mathbb{R}^n$ with an orthogonal matrix either leaves the volume unchanged (so it is a rotation) or multiplies it by $-1$ (so it is a reflection).</p>
<p>To answer your first question: the action of a matrix $A$ can be neatly expressed via its singular value decomposition, $A=U\Lambda V^T$, where $U$, $V$ are orthogonal matrices and $\Lambda$ is a matrix with non-negative values along the diagonal (nb. this makes sense even if $A$ is not square!) The values on the diagonal of $\Lambda$ are called the singular values of $A$, and if $A$ is square and symmetric they will be the absolute values of the eigenvalues.</p>
<p>The way to think about this is that the action of $A$ is first to rotate/reflect to a new basis, then scale along the directions of your new (intermediate) basis, before a final rotation/reflection.</p>
<p>With this in mind, notice that $A^T=V\Lambda^T U^T$, so the action of $A^T$ is to perform the inverse of the final rotation, then scale the new shape along the canonical unit directions, and then apply the inverse of the original rotation.</p>
<p>Furthermore, when $A$ is symmetric, $A=A^T\implies V\Lambda^T U^T = U\Lambda V^T \implies U = V $, therefore the action of a symmetric matrix can be regarded as a rotation to a new basis, then scaling in this new basis, and finally rotating back to the first basis. </p>
| <p>yoyo has succinctly described my intuition for orthogonal transformations in the comments: from <a href="http://en.wikipedia.org/wiki/Polarization_identity">polarization</a> you know that you can recover the inner product from the norm and vice versa, so knowing that a linear transformation preserves the inner product ($\langle x, y \rangle = \langle Ax, Ay \rangle$) is equivalent to knowing that it preserves the norm, hence the orthogonal transformations are precisely the linear <a href="http://en.wikipedia.org/wiki/Isometry">isometries</a>. </p>
<p>I'm a little puzzled by your comment about rotations and reflections because for me a rotation is, <em>by definition</em>, an orthogonal transformation of determinant $1$. (I say this not because I like to dogmatically stick to definitions over intuition but because this definition is elegant, succinct, and agrees with my intuition.) So what intuitive definition of a rotation are you working with here?</p>
<p>As for the transpose and symmetric matrices in general, my intuition here is not geometric. First, here is a comment which may or may not help you. If $A$ is, say, a stochastic matrix describing the transitions in some <a href="http://en.wikipedia.org/wiki/Markov_chain">Markov chain</a>, then $A^T$ is the matrix describing what happens if you run all of those transitions backwards. Note that this is not at all the same thing as inverting the matrix in general. </p>
<p>A slightly less naive comment is that the transpose is a special case of a structure called a <a href="http://en.wikipedia.org/wiki/Dagger_category">dagger category</a>, which is a category in which every morphism $f : A \to B$ has a dagger $f^{\dagger} : B \to A$ (here the adjoint). The example we're dealing with here is implicitly the dagger category of Hilbert spaces, which is relevant to quantum mechanics, but there's another dagger category relevant to a different part of physics: the $3$-<a href="http://en.wikipedia.org/wiki/Cobordism">cobordism category</a> describes how space can change with time in relativity, and here the dagger corresponds to just flipping a cobordism upside-down. (Note the similarity to the Markov chain example.) Since relativity and quantum mechanics are both supposed to describe the time evolution of physical systems, it's natural to ask for ways to relate the two dagger categories I just described, and this is (roughly) part of <a href="http://en.wikipedia.org/wiki/Topological_quantum_field_theory">topological quantum field theory</a>.</p>
<p>The punchline is that for me, "adjoint" is intuitively "time reversal." (Unfortunately, what this has to do with self-adjoint operators as observables in quantum mechanics I'm not sure.)</p>
|
logic | <p>Consider propositional logic with primitive connectives $\{{\to},{\land},{\lor},{\bot}\}$. We view $\neg \varphi$ as an abbreviation of $\varphi\to\bot$ and $\varphi\leftrightarrow\psi$ as an abbreviation of $(\varphi\to\psi)\land(\psi\to\varphi)$, etc.</p>
<p>The <em>classical sequent calculus</em> LK has rules such as</p>
<p>$$\frac{\Gamma, \varphi\vdash \Pi \qquad \Sigma, \psi\vdash \Pi}
{\Gamma, \Sigma, \varphi\lor\psi \vdash \Pi}{\lor}L \qquad
\frac{\Gamma\vdash\varphi,\Delta}{\Gamma\vdash\varphi\lor\psi,\Delta}{\lor}R_1 \qquad
\frac{\Gamma \vdash \psi, \Delta}{\Gamma\vdash\varphi\lor\psi,\Delta}{\lor}R_2
$$
$$
\frac{\Gamma\vdash\varphi,\Delta \qquad \Sigma,\psi\vdash\Pi}
{\Gamma, \Sigma, \varphi\to\psi \vdash \Delta,\Pi}{\to}L \qquad
\frac{\Gamma,\varphi\vdash \psi,\Delta}{\Gamma\vdash\varphi\to\psi,\Delta}{\to}R
$$
and so forth, where $\Gamma$, $\Sigma$, $\Pi$, and $\Delta$ are finite multisets of formulae.</p>
<p>It is well known that if we restrict the shape of all sequents to have exactly one formula to the right of the $\vdash$, we get a proof system LJ for <em>intuitionistic</em> propositional logic. This corresponds to requiring that every $\Delta$ is empty and every $\Pi$ is a singleton in the formulations of the rules above.</p>
<p><strong>Do we still get intuitionistic logic if the only rule we make this change to is ${\to}R$?</strong> In other words, we have
$$\frac{\Gamma,\varphi\vdash\psi}{\Gamma\vdash\varphi\to\psi}{\to}R'$$
together with all of the other rules in their classical formulation, including structural rules with long $\Delta$s and $\Pi$s.</p>
<hr>
<p>(Motivation: I'm trying, yet again, to wrap my head around what the <em>essential</em> difference between intuitionistic and classical logic is. It's often said that intuitionistic logic grew out of Brouwer's stricter demands on how a <em>disjunction</em> can be proved, but that can't be the whole story because there's a difference even for the implicational fragment (with no disjunctions). Now I'm wondering whether disjunction is actually part of the story at all. Kripke semantics treats it completely classically, and intuitionistic logic becomes fully classical if we add Peirce's law as an axiom (which again does not mention disjunction). The above conjecture is inspired by the Curry-Howard isomorphism where Peirce's law maps to <code>call/cc</code>, more or less. Therefore it would make sense that we can get intuitionistic logic simply by forbidding lambda abstractions from capturing continuation variables.)</p>
| <p>This rings a bell. And ah yes, on p. 48 of Sara Negri and Jan van Plato's admirable book <em>Structural Proof Theory</em> (CUP, 2001), they write</p>
<blockquote>
<p>It is not the [general] feature of having a multiset as a succedent that leads
to classical logic, but the unrestricted $R\!\supset$ rule,</p>
</blockquote>
<p>where by the unrestricted $R\!\supset$ they mean your classical $\to\! R$ rule, and the restricted rule would be your $\to\! R'$ rule. </p>
<p>Then on p. 108, they introduce an intuitionistic multisuccedent calculus they call <strong>G3im</strong>, which is exactly like a classical multisuccedent calculus except for the $\supset$ rules (though both left and right rules get tinkered with). I guess that the ensuing discussion, and the 1988 book by Dragalin <em>Mathematical Intuitionism</em> to which the calculus is due, would seem to be good starting points for further investigation (and I'd be interested to hear more about how things go!).</p>
| <p>Peter Smith found a nice reference which more or less settles the question. Here are some concluding remarks that won't fit as comments:</p>
<p>Knowing that it's true gave me the courage to try proving it, which turned out to be easier than expected. In fact, each of the classical rules <em>except</em> ${\to}R$ (but including ${\to}L$) is actually <em>derivable</em> in the single-succedent calculus (with Cut) if $\Gamma\vdash\varphi,\psi,\ldots,\sigma$ is taken to abbreviate $\Gamma\vdash \varphi\lor\psi\lor\cdots\lor\sigma$. This verification turns out to be completely routine.</p>
<p>This supports my hypothesis that the essential difference between classical and intuitionistic is that the intuitionistic $\varphi\to\psi$ is a <em>stronger claim</em> in the sense of being more difficult to prove (but <em>not</em> easier to conclude something from). Since $\neg$ abbreviates an implication, this also makes negations more difficult to prove in intuitionistic logic.</p>
<p>What prevents the unrestricted ${\to}R$ from being derivable is that deriving it in the same way as the others would require reasoning from $\varphi\to(\psi\lor \sigma)$ to $(\varphi\to\psi)\lor\sigma$, which is not intuitionistically possible.</p>
<p>In the particular case $\psi=\bot$, what this says is that we can't reason from $\varphi\to(\bot\lor\sigma)$ (which is obviously equivalent to $\varphi\to\sigma$) to $\neg\varphi\lor \sigma$. This can look like $\lor$ is harder to prove in intuitionistic logic, but it's really the $\neg$ that makes $\neg\varphi\lor\sigma$ harder to conclude.</p>
|
logic | <p>In philosophical contexts, the <em>Principia Mathematica</em> is sometimes held in high regard as a demonstration of a logical system.</p>
<p>But what did Whitehead and Russell's <em>Principia Mathematica</em> achieve for mathematics?</p>
| <p>I'll try to answer referring to the <em>Introduction</em> to the 1st edition of W&R's <a href="http://plato.stanford.edu/entries/principia-mathematica/" rel="nofollow noreferrer">Principia</a> (3 vols, 1910-13); see :</p>
<ul>
<li>Alfred North Whitehead & Bertrand Russell, <a href="https://books.google.it/books?id=ke9yGmFy24sC&pg=PA1" rel="nofollow noreferrer">Principia Mathematica to *56</a> (2nd ed,1927), page 1:</li>
</ul>
<blockquote>
<p>THE mathematical logic which occupies Part I of the present work has
been constructed under the guidance of <em>three different purposes</em>. In the <em>first
place</em>, it aims at effecting the greatest possible <em>analysis of the ideas with
which it deals and of the processes by which it conducts demonstrations</em>,
and at diminishing to the utmost the number of the undefined ideas and
undemonstrated propositions (called respectively primitive ideas and primitive
propositions) from which it starts. In the <em>second place</em>, it is framed with a
view to the <em>perfectly precise expression, in its symbols, of mathematical
propositions</em>: to secure such expression, and to secure it in the simplest and
most convenient notation possible, is the chief motive in the choice of topics.
In the <em>third place</em>, the system is specially framed <em>to solve the paradoxes</em>
which, in recent years, have troubled students of symbolic logic and the
theory of aggregates; it is believed that the theory of types, as set forth in
what follows, leads both to the avoidance of contradictions, and to the
detection of the precise fallacy which has given rise to them. [<em>emphasis added</em>.]</p>
</blockquote>
<p>Simplifying a little bit, the three purposes of the work are :</p>
<ul>
<li><p>the foundation of <em>mathematical logic</em></p>
</li>
<li><p>the <em>formalization</em> of mathematics</p>
</li>
<li><p>the development of the <em>philosophical project</em> called <a href="http://plato.stanford.edu/entries/logicism/" rel="nofollow noreferrer">Logicism</a>.</p>
</li>
</ul>
<p>I'll not touch the third point.</p>
<hr />
<p>Regarding the first one, <em>PM</em> are unquestionably the basic building block of modern mathemtical logic.</p>
<p>Unfortunately, its cumbersome <a href="http://plato.stanford.edu/entries/pm-notation/" rel="nofollow noreferrer">notation</a> and the intermingling of technical aspects and philosophical ones prevent for using it (at least the initial chapters) as a textbook.</p>
<p>Compare with the first fully modern textbook of mathematical logic :</p>
<ul>
<li>David Hilbert & Wilhelm Ackermann <a href="https://books.google.it/books?id=45ZGMjV9vfcC&pg=PA2" rel="nofollow noreferrer">Principles of Mathematical Logic</a>, the 1950 translation of the 1938 second edition of text <a href="https://en.wikipedia.org/wiki/Principles_of_Mathematical_Logic" rel="nofollow noreferrer">Grundzüge der theoretischen Logik</a>, firstly published in 1928.</li>
</ul>
<p>See the <em>Introduction</em> [page 2] :</p>
<blockquote>
<p>symbolic logic received a new impetus from the need of mathematics for an exact foundation and strict axiomatic treatment. <a href="http://plato.stanford.edu/entries/frege/" rel="nofollow noreferrer">G.Frege</a> published his <a href="https://en.wikipedia.org/wiki/Begriffsschrift" rel="nofollow noreferrer">Begriffsschrift</a> in 1879
and his <a href="http://plato.stanford.edu/entries/frege-theorem/" rel="nofollow noreferrer">Grundgesetze der Arithmetik</a> in 1893-1903. <a href="https://en.wikipedia.org/wiki/Giuseppe_Peano" rel="nofollow noreferrer">G.Peano</a> and his co-workers began in 1894 the publication of the <a href="https://en.wikipedia.org/wiki/Formulario_mathematico" rel="nofollow noreferrer">Formulaire
de Mathématiques</a>, in which all the mathematical disciplines were to be presented in terms of the logical calculus. A high point of this development is the appearance of the <em>Principia Mathematica</em> (1910-1913) by A.N. Whitehead and B. Russell.</p>
</blockquote>
<p>We may consider H&A's work as a textbook because - in spite of Hilbert's deep involvement with is <a href="http://plato.stanford.edu/entries/hilbert-program/" rel="nofollow noreferrer">foundational project</a> - it is devoted to a plain exposition of technical issues, without philosophical discussions.</p>
<hr />
<p>Now for my tentative answer to the question :</p>
<blockquote>
<p>what did Whitehead and Russell's Principia Mathematica achieve for mathematics?</p>
</blockquote>
<p>The first (and unprecedented) fully-flegded <em>formalization</em> of a huge part of mathematics, mainly the Cantorian mathematics of the infinite.</p>
<p>Unfortunately again, we have a cumbersome symbolism, as well as an axiomatization based on the theory of <em>classes</em> (and not : <em>sets</em>) that has been subsequently surpassed by <a href="http://plato.stanford.edu/entries/zermelo-set-theory/" rel="nofollow noreferrer">Zermelo</a>'s axiomatization.</p>
<p>But we can find there "perfectly precise expression of mathematical propositions [and concepts]", starting from the elementary ones.</p>
<p>Some examples regarding <a href="https://books.google.it/books?id=ke9yGmFy24sC&pg=PA205" rel="nofollow noreferrer">operations on classes</a>:</p>
<blockquote>
<p>*22.01. <span class="math-container">$\alpha \subset \beta \ . =_{\text {Df}} . \ x \in \alpha \supset_x x \in \beta$</span></p>
<p>[in modern notation : <span class="math-container">$\forall x \ (x \in \alpha \to x \in \beta)$</span>]</p>
<p>This defines "the class <span class="math-container">$\alpha$</span> is contained in the class <span class="math-container">$\beta$</span>," or "all <span class="math-container">$\alpha$</span>'s are <span class="math-container">$\beta$</span>'s."</p>
</blockquote>
<p>and the definition of <a href="https://books.google.it/books?id=ke9yGmFy24sC&pg=PA338" rel="nofollow noreferrer">singleton</a>:</p>
<blockquote>
<p>[the] function <span class="math-container">$\iota 'x$</span>, meaning "the class of terms which are identical with <span class="math-container">$x$</span>" which is the same thing as "the class whose only member is <span class="math-container">$x$</span>." We are thus to have</p>
<p><span class="math-container">$$\iota'x = \hat y(y = x).$$</span></p>
</blockquote>
<p>[in modern notation : <span class="math-container">$\{ x \} = \{ y \mid y=x \}]$</span></p>
<blockquote>
<p>[...] The distinction between <span class="math-container">$x$</span> and <span class="math-container">$\iota'x$</span> is one of the merits of Peano's symbolic logic, as well as of Frege's. [....] Let <span class="math-container">$\alpha$</span> be a class; then the class whose only member is <span class="math-container">$\alpha$</span> has only one member, namely <span class="math-container">$\alpha$</span>, while <span class="math-container">$\alpha$</span> may have many members. Hence the class whose only member is a cannot be identical with <span class="math-container">$\alpha$</span>*. [...]</p>
<p>*51.15. <span class="math-container">$y \in \iota'x \ . \equiv . \ y = x$</span></p>
</blockquote>
<p>[in modern notation : <span class="math-container">$y \in \{ x \} \leftrightarrow y=x$</span>].</p>
| <p>Whilst Whitehead and Russell failed in their main aim of deriving mathematics from logic, the logic that they developed is the mathematical logic that is used universally today.</p>
|
linear-algebra | <p>I’m learning multivariate analysis and I have learnt linear algebra for two semesters when I was a freshman.</p>
<p>Eigenvalues and eigenvectors are easy to calculate and the concept is not difficult to understand. I found that there are many applications of eigenvalues and eigenvectors in multivariate analysis. For example:</p>
<blockquote>
<p>In principal components, proportion of total population variance due to <span class="math-container">$k$</span>th principal component equals <span class="math-container">$$\frac{\lambda_k}{\lambda_1+\lambda_2+\ldots+\lambda_k}$$</span></p>
</blockquote>
<p>I think eigenvalue product corresponding eigenvector has same effect as the matrix product eigenvector geometrically.</p>
<p>I think my former understanding may be too naive so that I cannot find the link between eigenvalue and its application in principal components and others.</p>
<p>I know how to induce almost every step form the assumption to the result mathematically. I’d like to know how to <em><strong>intuitively</strong></em> or <em><strong>geometrically</strong></em> understand eigenvalue and eigenvector in the context of <strong>multivariate analysis</strong> (in linear algebra is also good).</p>
<p>Thank you!</p>
| <p>Personally, I feel that intuition isn't something which is easily explained. Intuition in mathematics is synonymous with experience and you gain intuition by working numerous examples. With my disclaimer out of the way, let me try to present a very informal way of looking at eigenvalues and eigenvectors.</p>
<p>First, let us forget about principal component analysis for a little bit and ask ourselves exactly what eigenvectors and eigenvalues are. A typical introduction to spectral theory presents eigenvectors as vectors which are fixed in direction under a given linear transformation. The scaling factor of these eigenvectors is then called the eigenvalue. Under such a definition, I imagine that many students regard this as a minor curiosity, convince themselves that it must be a useful concept and then move on. It is not immediately clear, at least to me, why this should serve as such a central subject in linear algebra.</p>
<p>Eigenpairs are a lot like the roots of a polynomial. It is difficult to describe why the concept of a root is useful, not because there are few applications but because there are too many. If you tell me all the roots of a polynomial, then mentally I have an image of how the polynomial must look. For example, all monic cubics with three real roots look more or less the same. So one of the most central facts about the roots of a polynomial is that they <em>ground</em> the polynomial. A root literally <em>roots</em> the polynomial, limiting it's shape.</p>
<p>Eigenvectors are much the same. If you have a line or plane which is invariant then there is only so much you can do to the surrounding space without breaking the limitations. So in a sense eigenvectors are not important because they themselves are fixed but rather they limit the behavior of the linear transformation. Each eigenvector is like a skewer which helps to hold the linear transformation into place.</p>
<p>Very (very, very) roughly then, the eigenvalues of a linear mapping is a measure of the distortion induced by the transformation and the eigenvectors tell you about how the distortion is oriented. It is precisely this rough picture which makes PCA very useful. </p>
<p>Suppose you have a set of data which is distributed as an ellipsoid oriented in $3$-space. If this ellipsoid was very flat in some direction, then in a sense we can recover much of the information that we want even if we ignore the thickness of the ellipse. This what PCA aims to do. The eigenvectors tell you about how the ellipse is oriented and the eigenvalues tell you where the ellipse is distorted (where it's flat). If you choose to ignore the "thickness" of the ellipse then you are effectively compressing the eigenvector in that direction; you are projecting the ellipsoid into the most optimal direction to look at. To quote wiki:</p>
<blockquote>
<p>PCA can supply the user with a lower-dimensional picture, a "shadow" of this object when viewed from its (in some sense) most informative viewpoint</p>
</blockquote>
| <p>First let us think what a square matrix does to a vector. Consider a matrix <span class="math-container">$A \in \mathbb{R}^{n \times n}$</span>. Let us see what the matrix <span class="math-container">$A$</span> acting on a vector <span class="math-container">$x$</span> does to this vector. By action, we mean multiplication i.e. we get a new vector <span class="math-container">$y = Ax$</span>.</p>
<p>The matrix acting on a vector <span class="math-container">$x$</span> does two things to the vector <span class="math-container">$x$</span>.</p>
<ol>
<li>It scales the vector.</li>
<li>It rotates the vector.</li>
</ol>
<p>However, for any matrix <span class="math-container">$A$</span>, there are some <em>favored vectors/directions</em>. When the matrix acts on these favored vectors, the action essentially results in just scaling the vector. There is no rotation. These favored vectors are precisely the eigenvectors and the amount by which each of these favored vectors stretches or compresses is the eigenvalue.</p>
<p>So why are these eigenvectors and eigenvalues important? Consider the eigenvector corresponding to the maximum (absolute) eigenvalue. If we take a vector along this eigenvector, then the action of the matrix is maximum. <strong>No other vector when acted by this matrix will get stretched as much as this eigenvector</strong>.</p>
<p>Hence, if a vector were to lie "close" to this eigen direction, then the "effect" of action by this matrix will be "large" i.e. the action by this matrix results in "large" response for this vector. The effect of the action by this matrix is high for large (absolute) eigenvalues and less for small (absolute) eigenvalues. Hence, the directions/vectors along which this action is high are called the principal directions or principal eigenvectors. The corresponding eigenvalues are called the principal values.</p>
|
linear-algebra | <p>It seems, at times, that physicists and mathematicians mean different things when they say the word "tensor." From my perspective, when I say tensor, I mean "an element of a tensor product of vector spaces." </p>
<p>For instance, here is a segment about tensors from Zee's book <em>Einstein Gravity in a Nutshell</em>:</p>
<blockquote>
<p>We already saw in the preceding chapter
that a vector is defined by how it transforms: $V^{'i}
= R^{ij}V^j$ . Consider a collection of “mathematical
entities” $T^{ij}$ with $i , j = 1, 2, . . . , D$ in $D$-dimensional space. If they transform
under rotations according to
$T^{ij} \to T^{'ij} = R^{ik}R^{jl}T^{kl}$ then we say that $T$ transforms like a tensor.</p>
</blockquote>
<p>This does not really make any sense to me. Even for "vectors," and before we get to "tensors," it seems like we'd have to be given a sense of what it means for an object to "transform." How do they divine these transformation rules?</p>
<p>I am not completely formalism bound, but I have no idea how they would infer these transformation rules without a notion of what the object is <em>first</em>. For me, if I am given, say, $v \in \mathbb{R}^3$ endowed with whatever basis, I can <em>derive</em> that any linear map is given by matrix multiplication as it seems the physicists mean. But, I am having trouble even interpreting their statement. </p>
<p>How do you derive how something "transforms" without having a notion of what it is? If you want to convince me that the moon is made of green cheese, I need to at least have a notion of what the moon is first. The same is true of tensors. </p>
<p>My questions are:</p>
<ul>
<li>What exactly are the physicists saying, and can someone translate what they're saying into something more intelligible? How can they get these "transformation rules" without having a notion of what the thing is that they are transforming?</li>
<li>What is the relationship between what physicists are expressing versus mathematicians? </li>
<li>How can I talk about this with physicists without being accused of being a stickler for formalism and some kind of plague? </li>
</ul>
| <p>What a physicist probably means when they say "tensor" is "a global section of a tensor bundle." I'll try and break it down to show the connection to what mathematicians mean by tensor.</p>
<p>Physicists always have a manifold <span class="math-container">$M$</span> lying around. In classical mechanics or quantum mechanics, this manifold <span class="math-container">$M$</span> is usually flat spacetime, mathematically <span class="math-container">$\mathbb{R}^4$</span>. In general relativity, <span class="math-container">$M$</span> is the spacetime manifold whose geometry is governed by Einstein's equations.</p>
<p>Now, with this underlying manifold <span class="math-container">$M$</span> we can discuss what it means to have a vector field on <span class="math-container">$M$</span>. Manifolds are locally euclidean, so we know what tangent vector means locally on <span class="math-container">$M$</span>. The question is, how do you make sense of a vector field globally? The answer is, you specify an open cover of <span class="math-container">$M$</span> by coordinate patches, say <span class="math-container">$\{U_\alpha\}$</span>, and you specify vector fields <span class="math-container">$V_\alpha=(V_\alpha)^i\frac{\partial}{\partial x^i}$</span> defined locally on each <span class="math-container">$U_\alpha$</span>. Finally, you need to ensure that on the overlaps <span class="math-container">$U_\alpha \cap U_\beta$</span> that <span class="math-container">$V_\alpha$</span> "agrees" with <span class="math-container">$V_\beta$</span>. When you take a course in differential geometry, you study vector fields and you show that the proper way to patch them together is via the following relation on their components:
<span class="math-container">$$
(V_\alpha)^i = \frac{\partial x^i}{\partial y^j} (V_\beta)^j
$$</span>
(here, Einstein summation notation is used, and <span class="math-container">$y^j$</span> are coordinates on <span class="math-container">$U_\beta$</span>). With this definition, one can define a vector bundle <span class="math-container">$TM$</span> over <span class="math-container">$M$</span>, which should be thought of as the union of tangent spaces at each point. The compatibility relation above translates to saying that there is a well-defined global section <span class="math-container">$V$</span> of <span class="math-container">$TM$</span>. So, when a physicist says "this transforms like this" they're implicitly saying "there is some well-defined global section of this bundle, and I'm making use of its compatibility with respect to different choices of coordinate charts for the manifold."</p>
<p>So what does this have to do with mathematical tensors? Well, given vector bundles <span class="math-container">$E$</span> and <span class="math-container">$F$</span> over <span class="math-container">$M$</span>, one can form their tensor product bundle <span class="math-container">$E\otimes F$</span>, which essentially is defined by
<span class="math-container">$$
E\otimes F = \bigcup_{p\in M} E_p\otimes F_p
$$</span>
where the subscript <span class="math-container">$p$</span> indicates "take the fiber at <span class="math-container">$p$</span>." Physicists in particular are interested in iterated tensor powers of <span class="math-container">$TM$</span> and its dual, <span class="math-container">$T^*M$</span>. Whenever they write "the tensor <span class="math-container">$T^{ij...}_{k\ell...}$</span> transforms like so and so" they are talking about a global section <span class="math-container">$T$</span> of a tensor bundle <span class="math-container">$(TM)^{\otimes n} \otimes (T^*M)^{\otimes m}$</span> (where <span class="math-container">$n$</span> is the number of upper indices and <span class="math-container">$m$</span> is the number of lower indices) and they're making use of the well-definedness of the global section, just like for the vector field.</p>
<p>Edit: to directly answer your question about how they get their transformation rules, when studying differential geometry one learns how to take compatibility conditions from <span class="math-container">$TM$</span> and <span class="math-container">$T^*M$</span> and turn them into compatibility relations for tensor powers of these bundles, thus eliminating any guesswork as to how some tensor should "transform."</p>
<p>For more on this point of view, Lee's book on Smooth Manifolds would be a good place to start.</p>
| <p>Being a physicist by training maybe I can help.</p>
<p>The "physicist" definition of a vector you allude to, in more mathematicians-friendly terms would become something like</p>
<blockquote>
<p>Let $V$ be a vector space and fix a reference frame $\mathcal{F}$ (mathematicians lingo: a basis.) A collection $\{v^1, \ldots, v^n\}$ of real numbers is called a vector if upon a change of reference frame $\mathcal{F}^\prime = R ^{-1} \mathcal{F}$ it becomes the collection $\{ v^{\prime 1}, \dots, v^{\prime n}\}$ where $v^{\prime i} =R^i_{\ j} v^j$.</p>
</blockquote>
<p>If you like, you are defining a vector as an equivalence class of $n$-tuples of real numbers.</p>
<p>Yes, in many physics books most of what I wrote is tacitly implied/shrugged off as non-important. Anyway, the definition of tensors as collections of numbers transforming according to certain rules is not so esoteric/rare as far as I am aware, and as others have pointed out it's also how mathematicians thought about them back in the past.</p>
<p>Physicists often prefer to describe objects in what they find to be more intuitive and less abstract terms, and one of their strength is the ability to work with vaguely defined objects! (Yes, I consider it a strength and yes, it has its drawback and pitfalls, no need to start arguing about that).</p>
<p>The case of tensors is similar, just think of the collection of numbers with indices as the components with respect to some basis. Be warned that sometimes what a physicist calls a tensor is actually a tensor field.</p>
<p>As to why one would use the definition in terms of components rather than more elegant invariant ones: it takes less technology and is more down-to-earth then introducing a free module over a set and quotienting by an ideal.</p>
<p>Finally, regarding how to communicate with physicists: this has always been a struggle on both sides but</p>
<ol>
<li><p>Many physicists, at least in the general relativity area, are familiar with the definition of a tensor in terms of multilinear maps. In fact, that is how they are defined in all GR books I have looked at (Carroll, Misner-Thorne-Wheeler, Hawking-Ellis, Wald).</p></li>
<li><p>It wouldn't hurt for you to get acquainted, if not proficient, with the index notation. It has its own strengths <em>and</em> is still intrinsic. See Wald or the first volume of Penrose-Rindler "Spinors and space-time" under abstract index notation for more on that.</p></li>
</ol>
|
matrices | <p>I happened to stumble upon the following matrix:
$$ A = \begin{bmatrix}
a & 1 \\
0 & a
\end{bmatrix}
$$</p>
<p>And after trying a bunch of different examples, I noticed the following remarkable pattern. If $P$ is a polynomial, then:
$$ P(A)=\begin{bmatrix}
P(a) & P'(a) \\
0 & P(a)
\end{bmatrix}$$</p>
<p>Where $P'(a)$ is the derivative evaluated at $a$.</p>
<p>Futhermore, I tried extending this to other matrix functions, for example the matrix exponential, and wolfram alpha tells me:
$$ \exp(A)=\begin{bmatrix}
e^a & e^a \\
0 & e^a
\end{bmatrix}$$
and this does in fact follow the pattern since the derivative of $e^x$ is itself!</p>
<p>Furthermore, I decided to look at the function $P(x)=\frac{1}{x}$. If we interpret the reciprocal of a matrix to be its inverse, then we get:
$$ P(A)=\begin{bmatrix}
\frac{1}{a} & -\frac{1}{a^2} \\
0 & \frac{1}{a}
\end{bmatrix}$$
And since $f'(a)=-\frac{1}{a^2}$, the pattern still holds!</p>
<p>After trying a couple more examples, it seems that this pattern holds whenever $P$ is any rational function.</p>
<p>I have two questions:</p>
<ol>
<li><p>Why is this happening?</p></li>
<li><p>Are there any other known matrix functions (which can also be applied to real numbers) for which this property holds?</p></li>
</ol>
| <p>If $$ A = \begin{bmatrix}
a & 1 \\
0 & a
\end{bmatrix}
$$
then by induction you can prove that
$$ A^n = \begin{bmatrix}
a^n & n a^{n-1} \\
0 & a^n
\end{bmatrix} \tag 1
$$
for $n \ge 1 $. If $f$ can be developed into a power series
$$
f(z) = \sum_{n=0}^\infty c_n z^n
$$
then
$$
f'(z) = \sum_{n=1}^\infty n c_n z^{n-1}
$$
and it follows that
$$
f(A) = \sum_{n=0}^\infty c_n A^n = I + \sum_{n=1}^\infty c_n
\begin{bmatrix}
a^n & n a^{n-1} \\
0 & a^n
\end{bmatrix} = \begin{bmatrix}
f(a) & f'(a) \\
0 & f(a)
\end{bmatrix} \tag 2
$$
From $(1)$ and
$$
A^{-1} = \begin{bmatrix}
a^{-1} & -a^{-2} \\
0 & a^{-1}
\end{bmatrix}
$$
one gets
$$
A^{-n} = \begin{bmatrix}
a^{-1} & -a^{-2} \\
0 & a^{-1}
\end{bmatrix}^n =
(-a^{-2})^{n} \begin{bmatrix}
-a & 1 \\
0 & -a
\end{bmatrix}^n \\ =
(-1)^n a^{-2n} \begin{bmatrix}
(-a)^n & n (-a)^{n-1} \\
0 & (-a)^n
\end{bmatrix} =
\begin{bmatrix}
a^{-n} & -n a^{-n-1} \\
0 & a^{-n}
\end{bmatrix}
$$
which means that $(1)$ holds for negative exponents as well.
As a consequence, $(2)$ can be generalized to functions
admitting a Laurent series representation:
$$
f(z) = \sum_{n=-\infty}^\infty c_n z^n
$$</p>
| <p>It's a general statement if <span class="math-container">$J_{k}$</span> is a Jordan block and <span class="math-container">$f$</span> a function matrix then
<span class="math-container">\begin{equation}
f(J)=\left(\begin{array}{ccccc}
f(\lambda_{0}) & \frac{f'(\lambda_{0})}{1!} & \frac{f''(\lambda_{0})}{2!} & \ldots & \frac{f^{(n-1)}(\lambda_{0})}{(n-1)!}\\
0 & f(\lambda_{0}) & \frac{f'(\lambda_{0})}{1!} & & \vdots\\
0 & 0 & f(\lambda_{0}) & \ddots & \frac{f''(\lambda_{0})}{2!}\\
\vdots & \vdots & \vdots & \ddots & \frac{f'(\lambda_{0})}{1!}\\
0 & 0 & 0 & \ldots & f(\lambda_{0})
\end{array}\right)
\end{equation}</span>
where
<span class="math-container">\begin{equation}
J=\left(\begin{array}{ccccc}
\lambda_{0} & 1 & 0 & 0\\
0 & \lambda_{0} & 1& 0\\
0 & 0 & \ddots & 1\\
0 & 0 & 0 & \lambda_{0}
\end{array}\right)
\end{equation}</span>
This statement can be demonstrated in various ways (none of them short), but it's a quite known formula. I think you can find it in various books, like in Horn and Johnson's <em>Matrix Analysis</em>.</p>
|
number-theory | <p><strong>Conjecture:</strong> <span class="math-container">$100$</span> is the only square number of the form <span class="math-container">$a^b+b^a$</span> for integers <span class="math-container">$b>a>1$</span>.</p>
<p>In other words, <span class="math-container">$(a,b)=(2,6)$</span> is the only solution. Can we prove/disprove this?</p>
<p><strong>Observations:</strong></p>
<ul>
<li><p>The solution mentioned should not come as a surprise, since <span class="math-container">$2^6+6^2=8^2+6^2=10^2$</span> is a (non-primitive) Pythagorean triple. It is possible to show that <span class="math-container">$2^b+b^2$</span> has no other solutions. See Remark 1.</p>
</li>
<li><p>In the general case where <span class="math-container">$a$</span> is a power of <span class="math-container">$2$</span>; that is, <span class="math-container">$a=2^d$</span> for some positive integer <span class="math-container">$d$</span>, a similar approach can be followed. See Remark 2.</p>
</li>
<li><p>We can eliminate some values of <span class="math-container">$b$</span> when <span class="math-container">$a=5^r,6^r$</span>, since no matter the value of <span class="math-container">$r$</span>, we have <span class="math-container">$a\equiv5,6\pmod{10}$</span> respectively.</p>
</li>
<li><p>PARI/GP code: If the conjecture is true it should only ever print <code>2 6</code>.</p>
<pre><code>sqfun(a,b)={for(i=2,a,for(j=2,b,if(issquare(i^j+j^i)==1,print(i," ",j))));}
</code></pre>
</li>
</ul>
<hr />
<p><strong>Remark 1:</strong> Suppose that there is a positive integer <span class="math-container">$b$</span> that admits <span class="math-container">$2^b+b^2=t^2$</span> for some integer <span class="math-container">$t$</span>. Then we can write the equation as <span class="math-container">$$2^b=(t+b)(t-b)\implies\begin{cases}t+b=2^c\\t-b=2^{b-c}\end{cases}$$</span> for some positive integer <span class="math-container">$c>\dfrac b2$</span>. Subtracting the two equations yields <span class="math-container">$$2b=2^c-2^{b-c}\implies b=2^{b-c-1}(2^{2c-b}-1).$$</span> If <span class="math-container">$b$</span> is odd, it cannot have a factor of <span class="math-container">$2$</span>, forcing <span class="math-container">$b-c-1=0\implies t=b+2$</span> and substituting gives <span class="math-container">$2^b+b^2=(b+2)^2$</span>, or <span class="math-container">$2^b=4(b+1)$</span>. No solutions exist.</p>
<p>If <span class="math-container">$b$</span> is even, then <span class="math-container">$b=2k$</span> for some positive integer <span class="math-container">$k$</span>, so we must have <span class="math-container">$$\begin{cases}2^k=s(m^2-n^2)\\2k=2mns\end{cases}$$</span> for some integers <span class="math-container">$m,n,s$</span>, so that <span class="math-container">$2^{mns}=s(m^2-n^2)$</span>. Without loss of generality, let <span class="math-container">$m>n>0$</span> and <span class="math-container">$s>0$</span>. [Servaes: If <span class="math-container">$mns\ge4$</span> then <span class="math-container">$2^{mns}\geq(mns)^2\geq sm^2> s(m^2-n^2)$</span>, so the only solutions with even <span class="math-container">$b$</span> are <span class="math-container">$b=4,6$</span>, and the first case does not yield a square.]</p>
<p><strong>Remark 2:</strong> If <span class="math-container">$b$</span> is odd, it boils down to the equation <span class="math-container">$$2^{db}=4\left(b^{2^{d-1}}+1\right)\implies 2^{db-2}-1=b^{2^{d-1}}.$$</span> [Haran: For <span class="math-container">$db-2>1$</span>, the LHS is congruent to <span class="math-container">$3\pmod4$</span>, and since the RHS is a square for <span class="math-container">$d>1$</span>, we reach a contradiction unless <span class="math-container">\begin{cases}d=1\implies a=2\quad\text{case covered above}\\db-2=1\implies1=b^{2^{d-1}}\implies 2^{d-1}=0\end{cases}</span> which is impossible.]</p>
<p>If <span class="math-container">$b$</span> is even, then <span class="math-container">$b=2k$</span> for some positive integer <span class="math-container">$k$</span>, and the Pythagorean triplet forces <span class="math-container">$$\begin{cases}2^{dk}=s(m^2-n^2)\\(2k)^{2^{d-1}}=2mns.\end{cases}$$</span> [Servaes: From the first equation, all three factors on the RHS are powers of two, so <span class="math-container">$$\begin{cases}m+n=2^u\\m-n=2^v\end{cases}\implies\begin{cases}m=2^{v-1}(2^{u-v}+1)\\n=2^{v-1}(2^{u-v}-1)\end{cases}$$</span> with <span class="math-container">$u>v>0$</span>. Since <span class="math-container">$m$</span> and <span class="math-container">$n$</span> are coprime, we have <span class="math-container">$v=1$</span>. Plugging this into the first equation yields <span class="math-container">$$2^{dk}=s(m-n)(m+n)=2^{u+1}s\implies s=2^{dk-u-1}.$$</span> Substituting this into the second equation yields <span class="math-container">$$(2k)^{2^{d-1}}=2mns=2(2^{u-1}+1)(2^{u-1}-1)s=(2^{2u-2}-1)2^{dk-u}$$</span> which is impossible; if we let <span class="math-container">$k=2^w\ell$</span> with <span class="math-container">$\ell$</span> odd then this implies <span class="math-container">$\ell^{2^{d-1}}=2^{2u-2}-1$</span> which by Catalan's conjecture/Mihailescu's theorem is impossible if <span class="math-container">$d>1$</span>. Note that <span class="math-container">$u>v$</span> hence <span class="math-container">$u\geq2$</span>.]</p>
| <p><span class="math-container">$\newcommand{\eps}{\varepsilon}$</span>
<span class="math-container">$\newcommand{\rad}{\mathrm{rad}}$</span></p>
<p>At least, under the <a href="https://en.wikipedia.org/wiki/Abc_conjecture" rel="noreferrer">abc conjecture</a>, there can be only finitely many pairs <span class="math-container">$(a,b)$</span> <em>with <span class="math-container">$b>a>1$</span> coprime</em> such that <span class="math-container">$a^b+b^a$</span> is a square.</p>
<p>As a reminder, the conjecture says that to any <span class="math-container">$\eps>0$</span> there corresponds some <span class="math-container">$K_\eps>0$</span> such that whenever <span class="math-container">$u,v$</span>, and <span class="math-container">$w$</span> are coprime positive integers with <span class="math-container">$u+v=w$</span>, one has <span class="math-container">$\rad(uvw)>K_\eps w^{1-\eps}$</span>. Here <span class="math-container">$\rad(z)$</span> is the product of all primes dividing <span class="math-container">$z$</span> (thus, for instance, <span class="math-container">$\rad(8)=2$</span>, <span class="math-container">$\rad(9)=3$</span>, <span class="math-container">$\rad(10)=10$</span>, <span class="math-container">$\rad(11)=11$</span>, and <span class="math-container">$\rad(12)=6$</span>).</p>
<p>Suppose now that <span class="math-container">$a^b+b^a=c^2$</span> with coprime integers <span class="math-container">$b>a\ge 3$</span> and <span class="math-container">$c>0$</span> (the case <span class="math-container">$a=2$</span> is resolved above). Applying the abc conjecture with <span class="math-container">$u=a^b$</span>, <span class="math-container">$v=b^a$</span>, <span class="math-container">$w=c^2$</span>, and <span class="math-container">$\eps=0.05$</span>, and making the key observation <span class="math-container">$\rad(a^bb^ac^2)\le abc$</span>, we conclude that
<span class="math-container">$$ Kc^{2\cdot 0.95} < abc $$</span>
with an absolute constant <span class="math-container">$K>0$</span>. At the same time, we have <span class="math-container">$c^2>a^b$</span> and <span class="math-container">$c^2>b^a$</span>, implying <span class="math-container">$a<c^{2/b}$</span> and <span class="math-container">$b<c^{2/a}$</span>, respectively. Consequently,
<span class="math-container">$$ Kc^{1.9} < c^{(2/b)+(2/a)+1}, $$</span>
showing that either <span class="math-container">$\frac1b+\frac1a>0.4$</span>, or <span class="math-container">$Kc^{0.1}<1$</span>. Clearly, there are only finitely many values of <span class="math-container">$c$</span> satisfying the latter condition, and to each value corresponds finitely many pairs <span class="math-container">$(a,b)$</span>. On the other hand, since <span class="math-container">$\frac1b+\frac13\ge\frac1b+\frac1a>0.4$</span> implies <span class="math-container">$b<15$</span>, there are only finitely many pairs <span class="math-container">$(a,b)$</span> satisfying the former condition. Thus, the total number of exceptional pairs <span class="math-container">$(a,b)$</span> is also finite.</p>
| <p><strong>(21-03-2020) Update:</strong> As no unconditional answer has yet been given, I will include a few more conditions that any solution must satisfy. The results used are rather advanced, so I will only include references, not proofs.</p>
<p><strong>Lemma 5:</strong> Let <span class="math-container">$a$</span> and <span class="math-container">$b$</span> be positive integers such that <span class="math-container">$a^b+b^a$</span> is a perfect square. Then <span class="math-container">$\gcd(a,b)\leq3$</span>.</p>
<p><em>Proof.</em> Such a pair of positive integers yields a nontrivial integral solution to
<span class="math-container">$$x^d+y^d=z^2,\tag{3}$$</span>
where <span class="math-container">$d:=\gcd(a,b)$</span>. This implies <span class="math-container">$d\leq3$</span> by <a href="http://www.math.mcgill.ca/darmon/pub/Articles/Research/18.Merel/paper.pdf" rel="noreferrer">this paper</a><span class="math-container">${}^1$</span>.<span class="math-container">$\qquad\square$</span></p>
<p><strong>Proposition 6:</strong> Let <span class="math-container">$a$</span> and <span class="math-container">$b$</span> be positive integers such that <span class="math-container">$a^b+b^a$</span> is a perfect square. Then <span class="math-container">$\gcd(a,b)\leq2$</span>.</p>
<p><em>Proof.</em> By the preceding lemma it suffices to show that <span class="math-container">$\gcd(a,b)=3$</span> is impossible. If <span class="math-container">$\gcd(a,b)=3$</span> then <span class="math-container">$a^{\tfrac b3}$</span> and <span class="math-container">$b^{\tfrac a3}$</span> are integers satisfying
<span class="math-container">$$\Big(a^{\tfrac b3}\Big)^3+\Big(b^{\tfrac a3}\Big)^3=z^2,$$</span>
for some integer <span class="math-container">$z$</span>, which means that, after swapping <span class="math-container">$a$</span> and <span class="math-container">$b$</span> if necessary, <strong>either</strong>
<span class="math-container">$$a^{\tfrac b3}=\frac{x(x^3-8y^3)}{w^2}z^2
\qquad\text{ and }\qquad
b^{\tfrac a3}=\frac{4y(x^3+y^3)}{w^2}z^2,$$</span>
for integers <span class="math-container">$x$</span>, <span class="math-container">$y$</span> and <span class="math-container">$z$</span> with <span class="math-container">$x$</span> odd and <span class="math-container">$x$</span> and <span class="math-container">$y$</span> coprime, and <span class="math-container">$w:=\gcd(3,x+y)$</span>, <strong>or</strong>
<span class="math-container">$$a^{\tfrac b3}=\frac{x^4+6x^2y^2-3y^4}{w^2}z^2
\qquad\text{ and }\qquad
b^{\tfrac a3}=\frac{3y^4+6x^2y^2-x^4}{w^2}z^2,$$</span>
for <span class="math-container">$x$</span>, <span class="math-container">$y$</span> and <span class="math-container">$z$</span> with <span class="math-container">$x$</span> and <span class="math-container">$y$</span> coprime and <span class="math-container">$x$</span> coprime to <span class="math-container">$3$</span>, and <span class="math-container">$w=\gcd(2,x+1,y+1)$</span>. These complete parametrizations of the integral solutions to <span class="math-container">$(3)$</span> when <span class="math-container">$d=3$</span> are taken from section 7.2 of <a href="https://londmathsoc.onlinelibrary.wiley.com/doi/10.1112/blms/27.6.513" rel="noreferrer">this article</a><span class="math-container">${}^2$</span>.</p>
<p>For the second parametrization, because <span class="math-container">$x$</span> is coprime to <span class="math-container">$3$</span> we see that
<span class="math-container">$$\nu_3\Big(a^{\tfrac b3}\Big)=\nu_3(z^2)
\qquad\text{ and }\qquad
\nu_3\Big(b^{\tfrac a3}\Big)=\nu_3(z^2),$$</span>
where <span class="math-container">$\nu_p(t)$</span> denotes largest integer <span class="math-container">$k$</span> such that <span class="math-container">$p^k$</span> divides <span class="math-container">$t$</span>. It follows that <span class="math-container">$a\nu_3(b)=b\nu_3(a)$</span>, and from <span class="math-container">$\gcd(a,b)=3$</span> it follows that either <span class="math-container">$\nu_3(a)=1$</span> or <span class="math-container">$\nu_3(b)=1$</span>, so either <span class="math-container">$a$</span> divides <span class="math-container">$b$</span> or <span class="math-container">$b$</span> divides <span class="math-container">$a$</span>, respectively. This means either <span class="math-container">$a=3$</span> or <span class="math-container">$b=3$</span>.</p>
<p>If <span class="math-container">$a=3$</span> then the identity
<span class="math-container">$$3^{\tfrac b3}=a^{\tfrac b3}=\frac{x^4+6x^2y^2-3y^4}{w^2}z^2,$$</span>
shows that <span class="math-container">$z^2=3^{\tfrac b3}$</span>. The parametrization for <span class="math-container">$b^{\tfrac a3}$</span> then shows that <span class="math-container">$3^{\tfrac b3}$</span> divides <span class="math-container">$b^{\tfrac a3}=b$</span>, which quickly implies <span class="math-container">$b=3$</span>. But <span class="math-container">$3^3+3^3=54$</span> is not a perfect square; a contradiction. If <span class="math-container">$b=3$</span> a similar argument shows that then <span class="math-container">$a=3$</span>, and this shows that the second parametrization yields no solutions to our original problem.</p>
<p>For the first parametrization, note that <span class="math-container">$b$</span> is even and hence <span class="math-container">$a^{\tfrac b3}$</span> is a perfect square, hence so is
<span class="math-container">$$w^2z^{-2}a^{\tfrac b3}=x(x^3-8y^3)=x^4-8xy^3.$$</span>
This means there is some integer <span class="math-container">$c$</span> such that <span class="math-container">$x^4-8xy^3=(x^2-2c)^2$</span> and so
<span class="math-container">$$cx^2+2y^3x-c^2=0.$$</span>
In particular the discriminant <span class="math-container">$\Delta$</span> of this quadratic polynomial in <span class="math-container">$x$</span> is a perfect square, say <span class="math-container">$\Delta=(2e)^2$</span>, where
<span class="math-container">$$4e^2=\Delta=(2y^3)^2-4c(-c)^2=4(y^6+c^3).$$</span>
It is a classical result that then for some nonnegative integer <span class="math-container">$k$</span> we have
<span class="math-container">$$(|e|,c,|y|)=(3k^3,2k^2,k).$$</span>
Plugging this in and solving the quadratic equation for <span class="math-container">$x$</span> yields <span class="math-container">$x\in\{\pm k,\pm2k\}$</span> where <span class="math-container">$y=\pm k$</span>. Because <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are coprime it follows that <span class="math-container">$k=1$</span>, so <span class="math-container">$y=\pm1$</span> and <span class="math-container">$x\in\{\pm1,\pm2\}$</span>. Then for <span class="math-container">$a^{\tfrac b3}$</span> to be a perfect square we must have <span class="math-container">$\{x,y\}=\{1,-1\}$</span>, but then <span class="math-container">$b=0$</span>, a contradiction. This shows that the second parametrization also yields no solutions to our original problem. In conclusion <span class="math-container">$\gcd(a,b)=3$</span> is impossible. <span class="math-container">$\quad\square$</span>.</p>
<p>A small step towards a complete proof of the original problem would be to show that any solution other than <span class="math-container">$\{a,b\}=\{2,6\}$</span> must have <span class="math-container">$a$</span> and <span class="math-container">$b$</span> coprime, which seems likely. So for now, I propose the following conjecture:</p>
<p><strong>Conjecture 7:</strong> Let <span class="math-container">$a$</span> and <span class="math-container">$b$</span> be positive integers such that <span class="math-container">$a^b+b^a$</span> is a perfect square and <span class="math-container">$\gcd(a,b)=2$</span>. Then <span class="math-container">$\{a,b\}=\{2,6\}$</span>.</p>
<p>Of course such solutions yield Pythagorean triples, for which the parametrization is well known. Perhaps arguments similar to those of Proposition 6 can be used here. I hope to give another update resolving this conjecture soon.</p>
<hr>
<h2>References</h2>
<ol>
<li>H. Darmon and L. Merel, <em>Winding quotients and some variants of Fermat’s last theorem</em>, J.Reine Angew. Math. <strong>490</strong> (1997), 81–100.</li>
<li>H. Darmon, A. Granville, <em>On the equations <span class="math-container">$x^p+y^q=z^r$</span> and <span class="math-container">$z^m=f(x, y)$</span></em>, Bulletin of the London Math. Society, no 129, <strong>27</strong> part 6, November 1995, pp. 513–544.</li>
</ol>
<hr>
<p><strong>Original answer:</strong></p>
<p>I'll collect a few partial results here. Let <span class="math-container">$a$</span>, <span class="math-container">$b$</span> and <span class="math-container">$c$</span> be positive integers with <span class="math-container">$a,b>1$</span> such that
<span class="math-container">$$a^b+b^a=c^2,$$</span>
and let <span class="math-container">$d=\gcd(a,b)$</span>. First two lemmas that are repeatedly useful.</p>
<p><strong>Lemma 1:</strong> If <span class="math-container">$m$</span> and <span class="math-container">$n$</span> are positive integers with <span class="math-container">$m>n$</span> and not both even, such that <span class="math-container">$m+n$</span> and <span class="math-container">$m-n$</span> are both powers of <span class="math-container">$2$</span>, then <span class="math-container">$m=2^k+1$</span> and <span class="math-container">$n=2^k-1$</span> for some positive integer <span class="math-container">$k$</span>.</p>
<p><em>Proof.</em> If <span class="math-container">$m+n=2^u$</span> and <span class="math-container">$m-n=2^v$</span> then
<span class="math-container">$$m=\frac{(m+n)+(m-n)}2=\frac{2^u+2^v}2=2^{v-1}(2^{u-v}+1),$$</span>
<span class="math-container">$$n=\frac{(m+n)-(m-n)}2=\frac{2^u-2^v}2=2^{v-1}(2^{u-v}-1),$$</span>
and hence <span class="math-container">$v=1$</span> because one of <span class="math-container">$m$</span> and <span class="math-container">$n$</span> is odd. Then <span class="math-container">$k=u-v$</span>.<span class="math-container">$\qquad\square$</span></p>
<p><strong>Lemma 2:</strong> The only perfect power that is one less than a square is <span class="math-container">$8$</span>.</p>
<p><em>Proof.</em> There are fairly elementary proofs, but it also follows from <a href="https://en.wikipedia.org/wiki/Catalan%27s_conjecture" rel="noreferrer">Mihailescu’s theorem</a>.<span class="math-container">$\qquad\square$</span></p>
<p><strong>Proposition 3:</strong> If <span class="math-container">$a$</span> is a power of <span class="math-container">$2$</span> then <span class="math-container">$(a,b)=(2,6)$</span>.</p>
<p>Most of this was proved in the original question by TheSimpliFire and Haran.</p>
<p><em>Proof.</em> Let <span class="math-container">$a=2^e$</span>. If <span class="math-container">$b$</span> is odd then writing
<span class="math-container">$$(c-b^{2^{e-1}})(c+b^{2^{e-1}})=c^2-b^a=a^b=2^{be},$$</span>
shows that both factors on the left hand side are powers of <span class="math-container">$2$</span>. Then by Lemma 1 we have <span class="math-container">$c=2^v+1$</span> and
<span class="math-container">$$b^{2^{e-1}}=2^v-1,$$</span>
for some positive integer <span class="math-container">$v$</span> because <span class="math-container">$b$</span> is odd. Hence by Lemma 2 either <span class="math-container">$v=1$</span> or <span class="math-container">$2^{e-1}=1$</span>. Clearly <span class="math-container">$v=1$</span> is impossible, so <span class="math-container">$2^{e-1}=1$</span> and so <span class="math-container">$e=1$</span>. Then comparing exponents shows that <span class="math-container">$b=v+2$</span> and so
<span class="math-container">$$v+2=b=2^v-1,$$</span>
which is easily seen to have no integral solutions. Hence <span class="math-container">$b$</span> is even, say <span class="math-container">$b=2f$</span>. Then we have the following Pythagorean triple:
<span class="math-container">$$c^2=a^b+b^a=(2^e)^{2f}+(2f)^{2^e}=(2^{ef})^2+((2f)^{2^{e-1}})^2.$$</span>
Then there exist positive integers <span class="math-container">$k$</span>, <span class="math-container">$m$</span> and <span class="math-container">$n$</span> with <span class="math-container">$m>n$</span> and <span class="math-container">$\gcd(m,n)=1$</span> such that either
<span class="math-container">$$c=k(m^2+n^2),\qquad2^{ef}=k(m^2-n^2),\qquad (2f)^{2^{e-1}}=2kmn,\tag{1}$$</span>
<span class="math-container">$$\text{or}$$</span>
<span class="math-container">$$c=k(m^2+n^2),\qquad2^{ef}=2kmn,\qquad (2f)^{2^{e-1}}=k(m^2-n^2).\tag{2}$$</span>
In case the triple is of the form <span class="math-container">$(2)$</span>, the middle identity shows that <span class="math-container">$k$</span>, <span class="math-container">$m$</span> and <span class="math-container">$n$</span> are all powers of <span class="math-container">$2$</span>, so in particular <span class="math-container">$n=1$</span> because <span class="math-container">$m$</span> and <span class="math-container">$n$</span> are coprime and <span class="math-container">$m>n$</span>. Then the latter identity shows that
<span class="math-container">$$(2f)^{2^{e-1}}=k(m^2-1)=k(m-1)(m+1),$$</span>
where the factors <span class="math-container">$m-1$</span> and <span class="math-container">$m+1$</span> are odd and <span class="math-container">$k$</span> is a power of <span class="math-container">$2$</span>, so both <span class="math-container">$m-1$</span> and <span class="math-container">$m+1$</span> are <span class="math-container">$2^{e-1}$</span>-th powers. But for <span class="math-container">$e>1$</span> no two <span class="math-container">$2^{e-1}$</span>-th powers of positive numbers differ by <span class="math-container">$2$</span>, so <span class="math-container">$e=1$</span>. Writing <span class="math-container">$k=2^u$</span> and <span class="math-container">$m=2^v$</span> we see that <span class="math-container">$u+v+1=f$</span>, where <span class="math-container">$v\geq1$</span> because <span class="math-container">$m>n$</span>.
By comparing powers in the above we find that
<span class="math-container">$$u+v+1=2^{u-1}(2^{2v}-1)=2^{u-1}(2^v-1)(2^v+1).$$</span>
Of course <span class="math-container">$2^v+1>2$</span>, so <span class="math-container">$2^{u-1}=1$</span> as otherwise
<span class="math-container">$$2^{u-1}(2^v+1)>2^{u-1}+2^v+1\geq u+v+1,$$</span>
a contradiction. Hence <span class="math-container">$u=1$</span> and <span class="math-container">$2^{2v}-1=v+2$</span>, so also <span class="math-container">$v=1$</span>. This yields the solution <span class="math-container">$(a,b)=(2,6)$</span>.</p>
<p>On the other hand, if the Pythagorean triple is of the form <span class="math-container">$(1)$</span> then <span class="math-container">$k$</span>, <span class="math-container">$m-n$</span> and <span class="math-container">$m+n$</span> are powers of <span class="math-container">$2$</span> because
<span class="math-container">$$2^{ef}=k(m^2-n^2)=k(m-n)(m+n).$$</span>
Because <span class="math-container">$m$</span> and <span class="math-container">$n$</span> are not both even, by Lemma 1 there exists a positive integer <span class="math-container">$v$</span> such that <span class="math-container">$m=2^v+1$</span> and <span class="math-container">$n=2^v-1$</span>, and so the above shows that <span class="math-container">$k=2^{ef-v-2}$</span>. Plugging this into the other equation yields
<span class="math-container">$$(2f)^{2^{e-1}}=2kmn=2^{ef-v-1}(2^{2v}-1),$$</span>
and writing <span class="math-container">$f=2^wg$</span> with <span class="math-container">$g$</span> odd then implies
<span class="math-container">$$g^{2^{e-1}}=2^{2v}-1,$$</span>
which by Lemma 2 implies that <span class="math-container">$2^{e-1}=1$</span>, and hence <span class="math-container">$e=1$</span>. Then <span class="math-container">$f=kmn$</span> and if <span class="math-container">$kmn\geq4$</span> then
<span class="math-container">$$2^{kmn}\geq(kmn)^2\geq km^2>k(m^2-n^2),$$</span>
so <span class="math-container">$f=kmn\leq3$</span>. Then <span class="math-container">$b\leq6$</span> and clearly <span class="math-container">$b=2$</span> and <span class="math-container">$b=4$</span> do not yield solutions.<span class="math-container">$\qquad\square$</span></p>
<p><strong>Proposition 4:</strong> If <span class="math-container">$d$</span> is even then <span class="math-container">$d=2$</span>.</p>
<p><em>Proof.</em> Suppose <span class="math-container">$d=2e$</span> and let <span class="math-container">$a=2eA$</span> and <span class="math-container">$b=2eB$</span>. Then <span class="math-container">$e$</span> is odd as otherwise <span class="math-container">$c^2$</span> is the sum of two fourth powers, which is well known to be impossible by a classical result by Fermat. Now
<span class="math-container">$$c^2=a^b+b^a=(a^{eB})^2+(b^{eA})^2,$$</span>
is a pythagorean triple and hence there exist positive integers <span class="math-container">$k$</span>, <span class="math-container">$m$</span> and <span class="math-container">$n$</span> with <span class="math-container">$m>n$</span> and <span class="math-container">$\gcd(m,n)=1$</span> such that
<span class="math-container">$$a^{eB}=k(m^2-n^2),\qquad b^{eA}=2kmn,\qquad c=k(m^2+n^2),$$</span>
where we can exchange the roles of <span class="math-container">$a$</span> and <span class="math-container">$b$</span> if necessary. It is clear that
<span class="math-container">$$k=\gcd(a^{eB},b^{eA})=\gcd((dA)^B,(dB)^A)^e=d^{e\ell},$$</span>
where <span class="math-container">$\ell\geq\min\{A,B\}$</span>. In particular <span class="math-container">$k$</span> is an <span class="math-container">$e$</span>-th power, and hence the factorizations
<span class="math-container">$$(a^B)^e=k(m^2-n^2)=k(m-n)(m+n)
\qquad\text{ and }\qquad
(b^A)^e=2kmn,$$</span>
show that, up to powers of <span class="math-container">$2$</span>, the factors <span class="math-container">$m$</span>, <span class="math-container">$n$</span>, <span class="math-container">$m+n$</span> and <span class="math-container">$m-n$</span> are also <span class="math-container">$e$</span>-th powers. That is to say,
<span class="math-container">$$m=2^tp^e,\qquad n=2^uq^e,\qquad m+n=2^vr^e,\qquad m-n=2^ws^e,$$</span>
for odd positive integers <span class="math-container">$p$</span>, <span class="math-container">$q$</span>, <span class="math-container">$r$</span> and <span class="math-container">$s$</span>, and nonnegative integers <span class="math-container">$t$</span>, <span class="math-container">$u$</span>, <span class="math-container">$v$</span> and <span class="math-container">$w$</span>, and <span class="math-container">$t+u+1\equiv v+w\equiv0\pmod{e}$</span>. Then
<span class="math-container">$$m=\frac{(m+n)+(m-n)}{2}=\frac{2^vr^e+2^ws^e}{2}=2^{v-1}r^e+2^{w-1}s^e,$$</span>
<span class="math-container">$$n=\frac{(m+n)-(m-n)}{2}=\frac{2^vr^e-2^ws^e}{2}=2^{v-1}r^e-2^{w-1}s^e,$$</span>
and at least one of <span class="math-container">$m$</span> and <span class="math-container">$n$</span> is odd, so either <span class="math-container">$v=1$</span> or <span class="math-container">$w=1$</span> (but not both) or <span class="math-container">$v=w=0$</span>.</p>
<p>If either <span class="math-container">$v=1$</span> or <span class="math-container">$w=1$</span> (but not both) then <span class="math-container">$m$</span> and <span class="math-container">$n$</span> are both odd, so <span class="math-container">$t=u=0$</span> and hence <span class="math-container">$e=1$</span>.</p>
<p>If <span class="math-container">$v=w=0$</span> then still either <span class="math-container">$t=0$</span> or <span class="math-container">$u=0$</span> because <span class="math-container">$m$</span> or <span class="math-container">$n$</span> is odd. If <span class="math-container">$u=0$</span> then <span class="math-container">$e\mid t+1$</span> and
<span class="math-container">$$2^{t+1}p^e=2m=(m+n)+(m-n)=r^e+s^e,$$</span>
and so it follows from Fermats last theorem that <span class="math-container">$e=1$</span>. The same holds if <span class="math-container">$t=0$</span>.<span class="math-container">$\qquad\square$</span></p>
|
logic | <p>I am trying to understand what mathematics is really built up of. I thought mathematical logic was the foundation of everything. But from reading a book in mathematical logic, they use "="(equals-sign), functions and relations.</p>
<p>Now is the "=" taken as undefined? I have seen it been defined in terms of the identity relation.</p>
<p>But in order to talk about functions and relations you need set theory.
However, <a href="https://en.wikipedia.org/wiki/Set_theory">set theory</a> seems to be a part of mathematical logic.</p>
<p>Does this mean that (naive) set theory comes before sentential and predicate logic? Is (naive)set-theory at the absolute bottom, where we can define relations and functions and the eqality relation. And then comes sentential logic, and then predicate logic?</p>
<p>I am a little confused because when I took an introductory course, we had a little logic before set-theory. But now I see in another book on introduction to proofs that set-theory is in a chapter before logic. <strong>So what is at the bottom/start of mathematics, logic or set theory?, or is it circular at the bottom?</strong></p>
<p>Can this be how it is at the bottom?</p>
<p>naive set-theory $\rightarrow$ sentential logic $\rightarrow $ predicate logic $\rightarrow$ axiomatic set-theory(ZFC) $\rightarrow$ mathematics</p>
<p>(But the problem with this explanation is that it seems that some naive-set theory proofs use logic...)</p>
<p>(The arrows are of course not "logical" arrows.)</p>
<p><strong>simple explanation of the problem:</strong></p>
<p><strong>a book on logic uses at the start</strong>: functions, relations, sets, ordered pairs, "="</p>
<p><strong>a book on set theory uses at the start:</strong> logical deductions like this: "$B \subseteq A$", means every element in B is in A, so if $C \subseteq B, B \subseteq A$, a proof can be "since every element in C is in B, and every element in B is in A, every element of C is in A: $C \subseteq A$". But this is first order logic? ($(c \rightarrow b \wedge b \rightarrow a)\rightarrow (c\rightarrow a)$).</p>
<p><strong>Hence, both started from each other?</strong></p>
| <p>Most set theories, such as ZFC, require an underlying knowledge of first-order logic formulas (as strings of symbols). This means that they require acceptance of facts of string manipulations (which is essentially equivalent to accepting arithmetic on natural numbers!) First-order logic does not require set theory, but if you want to prove something <strong>about</strong> first-order logic, you need some stronger framework, often called a meta theory/system. Set theory is one such stronger framework, but it is not the only possible one. One could also use a higher-order logic, or some form of type theory, both of which need not have anything to do with sets.</p>
<p>The circularity comes only if you say that you can <strong>justify</strong> the use of first-order logic or set theory or whatever other formal system by proving certain properties about them, because in most cases you would be using a stronger meta system to prove such meta theorems, which <strong>begs the question</strong>. However, if you use a <strong>weaker</strong> meta system to prove some meta theorems about stronger systems, then you might consider that justification more reasonable, and this is indeed done in the field called Reverse Mathematics.</p>
<p>Consistency of a formal system has always been the worry. If a formal system is inconsistent, then anything can be proven in it and so it becomes useless. One might hope that we can use a weaker system to prove that a stronger system is consistent, so that if we are convinced of the consistency of the weaker system, we can be convinced of the consistency of the stronger one. However, as Godel's incompleteness theorems show, this is impossible if we have arithmetic on the naturals.</p>
<p>So the issue dives straight into philosophy, because any proof in any formal system will already be a finite sequence of symbols from a finite alphabet of size at least two, so simply <strong>talking about</strong> a proof requires understanding finite sequences, which (almost) requires natural numbers to model. This means that any meta system powerful enough to talk about proofs and 'useful' enough for us to prove meta theorems in it (If you are a Platonist, you could have a formal system that simply has all truths as axioms. It is completely useless.) will be able to do something equivalent to arithmetic on the naturals and hence suffer from incompleteness.</p>
<p>There are two main parts to the 'circularity' in Mathematics (which is in fact a sociohistorical construct). The first is the understanding of logic, including the conditional and equality. If you do not understand what "if" means, no one can explain it to you because any purported explanation will be circular. Likewise for "same". (There are many types of equality that philosophy talks about.) The second is the understanding of the arithmetic on the natural numbers including induction. This boils down to the understanding of "repeat". If you do not know the meaning of "repeat" or "again" or other forms, no explanation can pin it down.</p>
<p>Now there arises the interesting question of how we could learn these basic undefinable concepts in the first place. We do so because we have an innate ability to recognize similarity in function. When people use words in some ways consistently, we can (unconsciously) learn the functions of those words by seeing how they are used and abstracting out the similarities in the contexts, word order, grammatical structure and so on. So we learn the meaning of "same" and things like that automatically.</p>
<p>I want to add a bit about the term "mathematics" itself. What we today call "mathematics" is a product of not just our observations of the world we live in, but also historical and social factors. If the world were different, we will not develop the same mathematics. But in the world we do live in, we cannot avoid the fact that there is no non-circular way to explain some fundamental aspects of the mathematics that we have developed, including equality and repetition and conditionals as I mentioned above, even though these are based on the real world. We can only explain them to another person via a shared experiential understanding of the real world.</p>
| <p>What you are butting your head against here IMO is the fact that you need a meta-language at the beginning. Essentially at some point you have to agree with other people what your axioms and methods of derivation are and these concepts cannot be intrinsic to your model. </p>
<p>Usually I think we take axioms in propositional logic as understood, with the idea that they apply to purely abstract notions of sentences and symbols. You might somethimes see proofs of the basic axioms such as Modus Ponens in terms of a meta-language i.e. not inside the system of logic but rather outside of it.</p>
<p>There is a lot of philosophical fodder at this level since really you need some sort of understanding between different people (real language perhaps or possibly just shared brain structures which allow for some sort of inherent meta-deduction) to communicate the basic axioms.</p>
<p>There is some extra confusion in the way these subjects are usually taught since propositional logic will often be explained in terms of, for example, truth tables, which seem to already require having some methods for modeling in place. The actual fact IMO is that at the bottom there is a shared turtle of interhuman understanding which allows you to grasp what the axioms you define are supposed to mean and how to operate with them.</p>
<p>Anyhow that's my take on the matter.</p>
|
geometry | <p>A hypothetical (and maybe practical) question has been nagging at me.</p>
<p>If you had a piece of paper with dimensions 4 and 3 (4:3), folding it in half along the long side (<strong><em>once</em></strong>) would result in 2 inches and 3 inches (2:3), which wouldn't retain its ratio. For example, here is a piece of paper that doesn't retain its ratio when folded:
<img src="https://i.sstatic.net/l14ls.jpg" alt="enter image description here"></p>
<p>Is retaining the ratio technically possible? If so, what is the side length and ratio that fulfills this requirement? Any help would be appreciated.</p>
<p><strong><em>Update:</em></strong></p>
<p>I added "<strong><em>once</em></strong>" because I got an answer saying that any recectangle would work, as any rectangle folded twice has the original ratio. Nice answer, but not quite what I was looking for. As for the other answers, I got 3x as much information as I needed! Thanks!</p>
| <p>The $1:\sqrt{2}$ ratio ensures exactly that. That is the idea behind the <a href="http://en.wikipedia.org/wiki/ISO_216">ISO 216</a> standard for paper sizes, which was adopted from the German <a href="https://en.wikipedia.org/wiki/Deutsches_Institut_f%C3%BCr_Normung">DIN</a> 476 standard.</p>
<p>Its most common usage is the <a href="http://en.wikipedia.org/wiki/ISO_216#A_series">A series</a> which especially in Europe is a collection of very common paper sizes. The base size, A0, has an area of a square meter, and every next smaller paper size is constructed by folding it in half.</p>
| <p>A ratio of $1:\sqrt{2}$ will do the trick!</p>
<p>The original rectangle will have a ratio of $x:y$, where $y$ is the longer side and $x$ the shorter side. Then the folded rectangle will have a ratio of $\frac{1}{2}y:x$ and we want</p>
<p>$$\begin{align}
\frac{x}{y} & =\frac{\frac{1}{2}y}{x} \\
x^2 & = \tfrac{1}{2}y^2 \\
x & = \tfrac{1}{\sqrt{2}} \cdot y \\
y & = \sqrt{2}\cdot x
\end{align}$$</p>
|
logic | <p>I've been reading a book on Gödel's incompleteness theorems and it makes the following claim regarding provability of statements in Peano arithmetic (paraphrased):</p>
<blockquote>
<p>There exists a formula $A(x)$ such that the statements $A(0), A(1), A(2), \dots$ are all provable, but $\forall x\, A(x)$ is not provable.</p>
</blockquote>
<p>It goes on to say that while Gödel's first incompleteness theorem guarantees its existence, it is not easy to find such a property for a theory as strong as PA.</p>
<p>Is there a specific example of such a formula or has none been found yet?</p>
| <p>Great question! Yes, there are specific examples. One of the most famous is <a href="https://en.wikipedia.org/wiki/Goodstein%27s_theorem">Goodstein's theorem</a>. If $A(n)$ is the statement that Goodstein's sequence starting at $n$ terminates, then it is known (via stronger theories than PA, namely use of a clever ordinal argument) that $\forall n \; A(n)$ is true.</p>
<p>Moreover, for a <em>specific</em> $n$, given that $A(n)$ is true, it must be provable: just write out the finite sequence, and that gives a proof that it terminates. (Once you get to $n = 4$ or larger, it will be a very very long proof -- one that we humans cannot actually have the space or memory to write down.)</p>
<p>However, $\forall n \; A(n)$ is known to be unprovable in PA.</p>
| <p>Possibly the easiest example is to let $A(x)$ say that $x$ is not the Gödel number of a proof of $0=1$ from the axioms of PA. </p>
<p>Because there is no such proof, $A(0)$ is true, $A(1)$ is true, etc., and because of their syntactic form each of these is provable in PA.</p>
<p>However, $(\forall x)A(x)$ is the statement $\text{Con}(\text{PA})$ which the second incompleteness theorem shows is not provable in PA. </p>
<p>There are other examples, as well. Some are related to the Paris-Harrington theorem and other independence results; others are related more directly to the incompleteness theorems. </p>
|
differentiation | <p>In calculus of one variable I read:</p>
<p>A function <span class="math-container">$f$</span> is differentiable on an <strong>open interval</strong> if it is differentiable at every number of the interval.</p>
<p>I wonder why in the definition we suppose that the interval is open. This is the case in Rolle theorem, Mean value theorem,etc.</p>
<blockquote>
<p>Do we have a notion of a function being differentiable on a closed interval <span class="math-container">$[a,b]$</span>?</p>
</blockquote>
| <p>The problem is one of <em>consistent definitions</em>. Intuitively we can make sense of differentiable on a closed interval: but it requires a slightly more careful phrasing of the definition of "differentiable at a point". I don't know which book you are using, but I am betting that it contains (some version of the) following (naive) definition:</p>
<blockquote>
<p><strong>Definition</strong> A function <span class="math-container">$f$</span> is differentiable at <span class="math-container">$x$</span> if <span class="math-container">$\lim_{y\to x} \frac{f(y) - f(x)}{y-x}$</span> exists and is finite. </p>
</blockquote>
<p>To make sense of the limit, often times the textbook will explicitly require that <span class="math-container">$f$</span> be <em>defined on an open interval containing <span class="math-container">$x$</span></em>. And if the definition of differentiability at a point requires <span class="math-container">$f$</span> to be defined on an open interval of the point, the definition of differentiability on a set can only be stated for sets for which every point is contained in an open interval. To illustrate, consider a function <span class="math-container">$f$</span> defined <em>only</em> on <span class="math-container">$[0,1]$</span>. Now you try to determine whether <span class="math-container">$f$</span> is differentiable at <span class="math-container">$0$</span> by naively applying the above definition. But since <span class="math-container">$f(y)$</span> is undefined if <span class="math-container">$y<0$</span>, the limit</p>
<p><span class="math-container">$$ \lim_{y\to 0^-} \frac{f(y) - f(0)}{y} $$</span></p>
<p>is <em>undefined</em>, and hence the derivative cannot exist at <span class="math-container">$0$</span> using one particular reading of the above definition. </p>
<p>For this purposes some people use the notion of <a href="http://en.wikipedia.org/wiki/Semi-differentiability" rel="noreferrer">semi-derivatives or one-sided derivatives</a> when dealing with boundary points. Other people just make the convention that when speaking of <em>closed</em> intervals, on the boundary the derivative is necessarily defined using a one-sided limit. </p>
<hr>
<p>Your textbook is not just being pedantic, however. If one wishes to study multivariable calculus, the definition of differentiability which requires taking limits in all directions is much more robust, compared to one-sided limits: the main problem being that in one dimension, given a boundary point, there is clearly a "left" and a "right", and each occupies "half" the available directions. This is no longer the case for domains in higher dimensions. Consider the domain</p>
<p><span class="math-container">$$ \Omega = \{ y \leq \sqrt{|x|} \} \subsetneq \mathbb{R}^2$$</span></p>
<p><span class="math-container">$\hspace{5cm}$</span><a href="https://i.sstatic.net/3gAKo.png" rel="noreferrer"><img src="https://i.sstatic.net/3gAKo.png" alt="enter image description here"></a></p>
<p>A particular boundary point of <span class="math-container">$\Omega$</span> is the origin. However, from the origin, almost all directions point to inside <span class="math-container">$\Omega$</span> (the only one that doesn't is the one that points straight up, in the positive <span class="math-container">$y$</span> direction). So the <a href="http://en.wikipedia.org/wiki/Total_derivative" rel="noreferrer">total derivative</a> cannot be defined at the origin if a function <span class="math-container">$f$</span> is only defined on <span class="math-container">$\Omega$</span>. But if you try to loosen the definitions and allow to consider only those "defined" directional derivatives, they may not patch together nicely at all. (A canonical example is the function
<span class="math-container">$$f(x,y) = \begin{cases} 0 & y \leq 0 \\ \text{sgn}(x) y^{3/2} & y > 0\end{cases}$$</span>
where <span class="math-container">$\text{sgn}$</span> return <span class="math-container">$+1$</span> if <span class="math-container">$x > 0$</span>, <span class="math-container">$-1$</span> if <span class="math-container">$x < 0$</span>, and <span class="math-container">$0$</span> if <span class="math-container">$x = 0$</span>. Its graph looks like what happens when you tear a piece of paper.) </p>
<p><span class="math-container">$\hspace{4cm}$</span><a href="https://i.sstatic.net/kEqHq.png" rel="noreferrer"><img src="https://i.sstatic.net/kEqHq.png" alt="enter image description here"></a></p>
<hr>
<p>But note that this is mainly a <em>failure</em> of the original naive definition of differentiability (which, however, may be pedagogically more convenient). A much more general notion of differentiability can be defined:</p>
<blockquote>
<p><strong>Definition</strong> Let <span class="math-container">$S\subseteq \mathbb{R}$</span>, and <span class="math-container">$f$</span> a <span class="math-container">$\mathbb{R}$</span>-valued function defined over <span class="math-container">$S$</span>. Let <span class="math-container">$x\in S$</span> be a <a href="http://en.wikipedia.org/wiki/Limit_point" rel="noreferrer">limit point</a> of <span class="math-container">$S$</span>. Then we say that <span class="math-container">$f$</span> is differentiable at <span class="math-container">$x$</span> if there exists a linear function <span class="math-container">$L$</span> such that for every sequence of points <span class="math-container">$x_n\in S$</span> different from <span class="math-container">$x$</span> but converging to <span class="math-container">$x$</span>, we have that
<span class="math-container">$$ \lim_{n\to\infty} \frac{f(x_n) - f(x) - L(x_n-x)}{|x_n - x|} = 0 $$</span></p>
</blockquote>
<p>This definition is a mouthful (and rather hard to teach in an introductory calculus course), but it has several advantages:</p>
<ol>
<li>It readily includes the case of the closed intervals.</li>
<li>It doesn't even need intervals. For example, you can let <span class="math-container">$S$</span> be the set <span class="math-container">$\{0\} \cup \{1/n\}$</span> where <span class="math-container">$n$</span> range over all positive integers. Then <span class="math-container">$0$</span> is a limit point, and so you can consider whether a function defined on this set is differentiable at the origin.</li>
<li>It easily generalises to higher dimensions, and vector valued functions. Just let <span class="math-container">$f$</span> take values in <span class="math-container">$\mathbb{R}^n$</span>, and let the domain <span class="math-container">$S\subseteq \mathbb{R}^d$</span>. The rest of the definition remains unchanged. </li>
<li>It captures, geometrically, the essence of the differentiation, which is "approximation by tangent planes". </li>
</ol>
<p>For this definition, you can easily add</p>
<blockquote>
<p><strong>Definition</strong> If <span class="math-container">$S\subseteq \mathbb{R}$</span> is such that every point <span class="math-container">$x\in S$</span> is a limit point of <span class="math-container">$S$</span>, and <span class="math-container">$f$</span> is a real valued function on <span class="math-container">$S$</span>, we say that <span class="math-container">$f$</span> is differentiable on <span class="math-container">$S$</span> if <span class="math-container">$f$</span> is differentiable at all points <span class="math-container">$x\in S$</span>. </p>
</blockquote>
<p>Note how this looks very much like the statement you quoted in your question. In the definition of pointwise differentiability we replaced the condition "<span class="math-container">$x$</span> is contained in an open neighborhood" by "<span class="math-container">$x$</span> is a limit point". And in the definition of differentiability on a set we just replaced the condition "every point has an open neighborhood" by "every point is a limit point". (This is what I meant by consistency: however you define pointwise differentiability necessarily effect how you define set differentiability.)</p>
<hr>
<p>If you go on to study differential geometry, this issue manifests behind the definitions for "manifolds", "manifolds with boundaries", and "manifolds with corners". </p>
| <p>A function is differentiable on a set $S$, if it is differentiable at every point of $S$. This is the definition that I seen in the beginning/classic calculus texts, and this mirrors the definition of continuity on a set. </p>
<p>So $S$ could be an open interval, closed interval, a finite set, in fact, it could be any set you want.</p>
<p>So yes, we do have a notion of a function being differentiable on a closed interval.</p>
<p>The reason Rolle's theorem talks about differentiabilty on the open interval $(a,b)$ is that it is a <em>weaker</em> assumption than requiring differentiability on $[a,b]$.</p>
<p>Normally, theorems might try to make the assumptions as weak as possible, to be more generally applicable.</p>
<p>For instance, the function:</p>
<p>$$f(x) = x \sin \frac{1}{x}, x \gt 0$$
$$f(0) = 0$$</p>
<p>is continuous at $0$, and differentiable everywhere except at $0$.</p>
<p>You can still apply Rolle's theorem to this function on say the interval $(0,\frac{1}{\pi})$. If the statement of Rolle's theorem required the use of the closed interval, then you could not apply it to this function.</p>
|
probability | <h1>Problem</h1>
<p>The premise is <em>almost</em> the same as in <a href="https://math.stackexchange.com/questions/29242/probability-that-a-quadratic-polynomial-with-random-coefficients-has-real-roots?rq=1">this question</a>. I'll restate for convenience.</p>
<blockquote>
<p>Let <span class="math-container">$A$</span>, <span class="math-container">$B$</span>, <span class="math-container">$C$</span> be independent random variables uniformly distributed between <span class="math-container">$(-1,+1)$</span>. What is the probability that the polynomial <span class="math-container">$Ax^2+Bx+C$</span> has real roots?</p>
</blockquote>
<p><strong>Note:</strong> The distribution is now <span class="math-container">$-1$</span> to <span class="math-container">$+1$</span> instead of <span class="math-container">$0$</span> to <span class="math-container">$1$</span>.</p>
<h1>My Attempt</h1>
<h2>Preparation</h2>
<p>When the coefficients are sampled from <span class="math-container">$\mathcal{U}(0,1)$</span>, the probability for the discriminant to be non-negative that is, <span class="math-container">$P(B^2-4AC\geq0) \approx 25.4\% $</span>. This value can be obtained theoretically as well as experimentally. The link I shared above to the older question has several good answers discussing both approaches.</p>
<p>Changing the sampling interval to <span class="math-container">$(-1, +1)$</span> makes things a bit difficult from the theoretical perspective. Experimentally, it is rather simple. <a href="https://github.com/hungrybluedev/root-from-parameters/blob/master/Empirical%20Probability%20of%20obtaining%20a%20real%20root.ipynb" rel="noreferrer">This is the code</a> I wrote to simulate the experiment for <span class="math-container">$\mathcal{U}(0,1)$</span>. Changing it from <code>(0, theta)</code> to <code>(-1, +1)</code> gives me an average probability of <span class="math-container">$62.7\%$</span> with a standard deviation of <span class="math-container">$0.3\%$</span></p>
<p>I plotted the simulated PDF and CDF. In that order, they are:</p>
<p><a href="https://i.sstatic.net/70f4U.png" rel="noreferrer"><img src="https://i.sstatic.net/70f4U.png" alt="PDF" /></a><a href="https://i.sstatic.net/yAIu6.png" rel="noreferrer"><img src="https://i.sstatic.net/yAIu6.png" alt="CDF" /></a></p>
<p>So I'm aiming to find a CDF that looks like the second image.</p>
<h2>Theoretical Approach</h2>
<p>The approach that I find easy to understand is outlined in <a href="https://math.stackexchange.com/a/2874497/203386">this answer</a>. Proceeding in a similar manner, we have</p>
<p><span class="math-container">$$
f_A(a) = \begin{cases}
\frac{1}{2}, &-1\leq a\leq+1\\
0, &\text{ otherwise}
\end{cases}
$$</span></p>
<p>The PDFs are similar for <span class="math-container">$B$</span> and <span class="math-container">$C$</span>.</p>
<p>The CDF for <span class="math-container">$A$</span> is</p>
<p><span class="math-container">$$
F_A(a) = \begin{cases}
\frac{a + 1}{2}, &-1\leq a\geq +1\\
0,&a<-1\\
1,&a>+1
\end{cases}
$$</span></p>
<p>Let us assume <span class="math-container">$X=AC$</span>. I proceed to calculate the CDF for <span class="math-container">$X$</span> (for <span class="math-container">$x>0$</span>) as:</p>
<p><span class="math-container">$$
\begin{align}
F_X(x) &= P(X\leq x)\\
&= P(AC\leq x)\\
&= \int_{c=-1}^{+1}P(Ac\leq x)f_C(c)dc\\
&= \frac{1}{2}\left(\int_{c=-1}^{+1}P(Ac\leq x)dc\right)\\
&= \frac{1}{2}\left(\int_{c=-1}^{+1}P\left(A\leq \frac{x}{c}\right)dc\right)\\
\end{align}
$$</span></p>
<p>We take a quick detour to make some observations. First, when <span class="math-container">$0<c<x$</span>, we have <span class="math-container">$\frac{x}{c}>1$</span>. Similarly, <span class="math-container">$-x<c<0$</span> implies <span class="math-container">$\frac{x}{c}<-1$</span>. Also, <span class="math-container">$A$</span> is constrained to the interval <span class="math-container">$[-1, +1]$</span>. Also, we're only interested when <span class="math-container">$x\geq 0$</span> because <span class="math-container">$B^2\geq 0$</span>.</p>
<p>Continuing, the calculation</p>
<p><span class="math-container">$$
\begin{align}
F_X(x) &= \frac{1}{2}\left(\int_{c=-1}^{+1}P\left(A\leq \frac{x}{c}\right)dc\right)\\
&= \frac{1}{2}\left(\int_{c=-1}^{-x}P\left(A\leq \frac{x}{c}\right)dc + \int_{c=-x}^{0}P\left(A\leq \frac{x}{c}\right)dc + \int_{c=0}^{x}P\left(A\leq \frac{x}{c}\right)dc + \int_{c=x}^{+1}P\left(A\leq \frac{x}{c}\right)dc\right)\\
&= \frac{1}{2}\left(\int_{c=-1}^{-x}P\left(A\leq \frac{x}{c}\right)dc + 0 + 1 + \int_{c=x}^{+1}P\left(A\leq \frac{x}{c}\right)dc\right)\\
&= \frac{1}{2}\left(\int_{c=-1}^{-x}\frac{x+c}{2c}dc + 0 + 1 + \int_{c=x}^{+1}\frac{x+c}{2c}dc\right)\\
&= \frac{1}{2}\left(\frac{1}{2}(-x+x(\log(-x)-\log(-1)+1) + 0 + 1 + \frac{1}{2}(-x+x(-\log(x)-\log(1)+1)\right)\\
&= \frac{1}{2}\left(2 + \frac{1}{2}(-x+x(\log(x)) -x + x(-\log(x))\right)\\
&= 1 - x
\end{align}
$$</span></p>
<p>I don't think this is correct.</p>
<h1>My Specific Questions</h1>
<ol>
<li>What mistake am I making? Can I even obtain the CDF through integration?</li>
<li>Is there an easier way? I used this approach because I was able to understand it well. There are shorter approaches possible (as is evident with the <span class="math-container">$\mathcal{U}(0,1)$</span> case) but perhaps I need to read more before I can comprehend them. Any pointers in the right direction would be helpful.</li>
</ol>
| <p>I would probably start by breaking into cases based on <span class="math-container">$A$</span> and <span class="math-container">$C$</span>.</p>
<p>Conditioned on <span class="math-container">$A$</span> and <span class="math-container">$C$</span> having different signs, there are always real roots (because <span class="math-container">$4AC\leq 0$</span>, so that <span class="math-container">$B^2-4AC\geq0$</span>). The probability that <span class="math-container">$A$</span> and <span class="math-container">$C$</span> have different signs is <span class="math-container">$\frac{1}{2}$</span>.</p>
<p>Conditioned on <span class="math-container">$A\geq0$</span> and <span class="math-container">$C\geq 0$</span>, you return to the problem solved in the link above. Why? Because <span class="math-container">$B^2$</span> has the same distribution whether you have <span class="math-container">$B$</span> uniformly distributed on <span class="math-container">$(0,1)$</span> or on <span class="math-container">$(-1,1)$</span>. At the link, they computed this probability as <span class="math-container">$\frac{5+3\log4}{36}\approx0.2544134$</span>. The conditioning event here has probability <span class="math-container">$\frac{1}{4}$</span>.</p>
<p>Finally, if we condition on <span class="math-container">$A<0$</span> and <span class="math-container">$C<0$</span>, we actually end up with the same probability, as <span class="math-container">$4AC$</span> has the same distribution in this case as in the case where <span class="math-container">$A\geq0$</span> and <span class="math-container">$C\geq 0$</span>. So, this is an additional <span class="math-container">$\frac{5+3\log 4}{36}\approx0.2544134$</span> conditional probability, and the conditioning event has probability <span class="math-container">$\frac{1}{4}$</span>.</p>
<p>So, all told, the probability should be
<span class="math-container">$$
\begin{align*}
P(B^2-4AC\geq0)&=1\cdot\frac{1}{2}+\frac{1}{4}\cdot\frac{5+3\log4}{36}+\frac{1}{4}\cdot\frac{5+3\log 4}{36}\\
&=\frac{1}{2}+\frac{5+3\log4}{72}\\
&\approx0.6272...
\end{align*}
$$</span></p>
| <p><span class="math-container">$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$</span>
Hereafter, <span class="math-container">$\ds{\bracks{P}}$</span> is an
<a href="https://en.wikipedia.org/wiki/Iverson_bracket" rel="noreferrer">Iverson Bracket</a>. Namely,
<span class="math-container">$\ds{\bracks{P} = \color{red}{1}}$</span> whenever <span class="math-container">$\ds{P}$</span> is <span class="math-container">$\ds{\tt true}$</span> and <span class="math-container">$\ds{\color{red}{0}}$</span>
<span class="math-container">$\ds{\tt otherwise}$</span>. They are very convenient whenever we have to <em>manipulate constraints</em>.</p>
<hr>
<span class="math-container">\begin{align}
&\bbox[5px,#ffd]{\int_{-1}^{1}{1 \over 2}\int_{-1}^{1}
{1 \over 2}\int_{-1}^{1}{1 \over 2}\bracks{b^{2} - 4ac > 0}
\dd c\,\dd a\,\dd b}
\\[5mm] = &\
{1 \over 4}\int_{0}^{1}\int_{-1}^{1}
\int_{-1}^{1}\bracks{b^{2} - 4ac > 0}
\dd c\,\dd a\,\dd b
\\[5mm] = &\
{1 \over 4}\int_{0}^{1}\int_{-1}^{1}
\int_{0}^{1}\braces{\bracks{b^{2} - 4ac > 0} +
\bracks{b^{2} + 4ac > 0}}
\dd c\,\dd a\,\dd b
\\[5mm] = &\
{1 \over 4}\int_{0}^{1}\int_{0}^{1}
\int_{0}^{1}\left\{\bracks{b^{2} - 4ac > 0} +
\bracks{b^{2} + 4ac > 0}\right.
\\[2mm] &\ \phantom{{1 \over 4}\int_{0}^{1}\int_{-1}^{1}
\int_{0}^{1}}
\left. + \bracks{b^{2} + 4ac > 0} +
\bracks{b^{2} - 4ac > 0}\right\}\dd c\,\dd a\,\dd b
\\[5mm] = &\
{1 \over 2} +
{1 \over 2}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}
\bracks{b^{2} - 4ac > 0}\dd c\,\dd a\,\dd b
\\[5mm] = &\
{1 \over 2} +
{1 \over 2}\int_{0}^{1}\int_{0}^{1}{1 \over a}\int_{0}^{a}
\bracks{b^{2} - 4c > 0}\dd c\,\dd a\,\dd b
\\[5mm] = &\
{1 \over 2} +
{1 \over 2}\int_{0}^{1}\int_{0}^{1}\bracks{b^{2} - 4c > 0}
\int_{c}^{1}{1 \over a}\,\dd a\,\dd c\,\dd b
\\[5mm] = &\
{1 \over 2} -
{1 \over 2}\int_{0}^{1}\int_{0}^{1}
\bracks{c < {b^{2} \over 4}}\ln\pars{c}\,\dd c\,\dd b
\\[5mm] = &\
{1 \over 2} -
{1 \over 2}\int_{0}^{1}\int_{0}^{b^{2}/4}
\ln\pars{c}\,\dd c\,\dd b
\\[5mm] = &\
{1 \over 2} -
{1 \over 2}\int_{0}^{1}\bracks{%
-\,{1 + 2\ln\pars{2} \over 4}\,b^{2} +
{1 \over 2}\,b^{2}\ln\pars{b}}\,\dd b
\\[5mm] = &
\bbx{{\ln\pars{2} \over 12} + {41 \over 72}}
\approx 0.6272 \\ &
\end{align}</span>
|
matrices | <p>I am taking a proof-based introductory course to Linear Algebra as an undergrad student of Mathematics and Computer Science. The author of my textbook (Friedberg's <em>Linear Algebra</em>, 4th Edition) says in the introduction to Chapter 4:</p>
<blockquote>
<p>The determinant, which has played a prominent role in the theory of linear algebra, is a special scalar-valued function defined on the set of square matrices. <strong>Although it still has a place in the study of linear algebra and its applications, its role is less central than in former times.</strong> </p>
</blockquote>
<p>He even sets up the chapter in such a way that you can skip going into detail and move on:</p>
<blockquote>
<p>For the reader who prefers to treat determinants lightly, Section 4.4 contains the essential properties that are needed in later chapters.</p>
</blockquote>
<p>Could anyone offer a didactic and simple explanation that refutes or asserts the author's statement?</p>
| <p>Friedberg is not wrong, at least on a historical standpoint, as I am going to try to show it.</p>
<p>Determinants were discovered "as such" in the second half of the 18th century by Cramer who used them in his celebrated rule for the solution of a linear system (in terms of quotients of determinants). Their spread was rather rapid among mathematicians of the next two generations ; they discovered properties of determinants that now, with our vision, we mostly express in terms of matrices.</p>
<p>Cauchy has given two important results about determinants as explained in the very nice article by Hawkins referenced below :</p>
<ul>
<li><p>around 1815, Cauchy discovered the multiplication rule (rows times columns) of two determinants. This is typical of a result that has been completely revamped : nowadays, this rule is for the multiplication of matrices, and determinants' multiplication is restated as the homomorphism rule <span class="math-container">$\det(A \times B)= \det(A)\det(B)$</span>.</p>
</li>
<li><p>around 1825, he discovered eigenvalues "associated with a symmetric <em>determinant</em>" and established the important result that these eigenvalues are real ; this discovery has its roots in astronomy, in connection with Sturm, explaining the word "secular values" he attached to them: see for example <a href="http://www2.cs.cas.cz/harrachov/slides/Golub.pdf" rel="noreferrer">this</a>.</p>
</li>
</ul>
<p>Matrices made a shy apparition in the mid-19th century (in England) ; "matrix" is a term coined by Sylvester <a href="http://mathworld.wolfram.com/Matrix.html" rel="noreferrer">see here</a>. I strongly advise to take a look at his elegant style in his <a href="https://archive.org/stream/collectedmathema04sylvuoft#page/n7/mode/2up" rel="noreferrer">Collected Papers</a>.</p>
<p>Together with his friend Cayley, they can rightly be named the founding fathers of linear algebra, with determinants as permanent reference. Here is a major quote of Sylvester:</p>
<p><em>"I have in previous papers defined a "Matrix" as a rectangular array of terms, out of which different systems of determinants may be engendered as from the womb of a common parent".</em></p>
<p>A lot of important polynomials are either generated or advantageously expressed as determinants:</p>
<ul>
<li><p>the characteristic polynomial (of a matrix) is expressed as the famous <span class="math-container">$\det(A-\lambda I)$</span>,</p>
</li>
<li><p>in particular, the theory of orthogonal polynomials mainly developed at the end of 19th century, can be expressed in great part with determinants,</p>
</li>
<li><p>the "resultant" of two polynomials, invented by Sylvester (giving a condition for these polynomials to have a common root), etc.</p>
</li>
</ul>
<p>Let us repeat it : for a mid-19th century mathematician, a <em>square</em> array of numbers has necessarily a <strong>value</strong> (its determinant): it cannot have any other meaning. If it is a <em>rectangular</em> array, the numbers attached to it are the determinants of submatrices that can be "extracted" from the array.</p>
<p>The identification of "Linear Algebra" as an integral (and new) part of Mathematics is mainly due to the German School (say from 1870 till the 1930's). I don't cite the names, there are too many of them. An example among many others of this german domination: the germenglish word "eigenvalue". The word "kernel" could have remained the german word "kern" that appears around 1900 (see <a href="http://jeff560.tripod.com/mathword.html" rel="noreferrer">this site</a>).</p>
<p>The triumph of Linear Algebra is rather recent (mid-20th century). "Triumph" meaning that now Linear Algebra has found a very central place. Determinants in all that ? Maybe the biggest blade in this swissknife, but not more ; another invariant (this term would deserve a long paragraph by itself), the <strong>trace</strong>, would be another blade, not the smallest.</p>
<p>In 19th century, Geometry was still at the heart of mathematical education; therefore, the connection between geometry and determinants has been essential in the development of linear algebra. Some cornerstones:</p>
<ul>
<li>the development of projective geometry, <em>in its analytical form,</em> in the 1850s. This development has led in particular to place homographies at the heart of projective geometry, with their associated matricial expression. Besides, conic curves, described by a quadratic form, can as well be written under an all-matricial expression <span class="math-container">$X^TMX=0$</span> where <span class="math-container">$M$</span> is a symmetrical <span class="math-container">$3 \times 3$</span> matrix. This convergence to a unique and new "algebra" has taken time to be recognized.</li>
</ul>
<p>A side remark: this kind of reflexions has been capital in the decision of Bourbaki team to avoid all figures and adopt the extreme view of reducing geometry to linear algebra (see the <a href="https://hsm.stackexchange.com/q/2578/3730">"Down with Euclid"</a> of J. Dieudonné in the sixties).</p>
<p>Different examples of the emergence of new trends :</p>
<p>a) the concept of <strong>rank</strong>: for example, a pair of straight lines is a conic section whose matrix has rank 1. The "rank" of a matrix used to be defined in an indirect way as the "dimension of the largest nonzero determinant that can be extracted from the matrix". Nowadays, the rank is defined in a more straightforward way as the dimension of the range space... at the cost of a little more abstraction.</p>
<p>b) the concept of <strong>linear transformations</strong> and <strong>duality</strong> arising from geometry: <span class="math-container">$X=(x,y,t)^T\rightarrow U=MX=(u,v,w)$</span> between points <span class="math-container">$(x,y)$</span> and straight lines with equations <span class="math-container">$ux+vy+w=0$</span>. More precisely, the tangential description, i.e., the constraint on the coefficients <span class="math-container">$U^T=(u,v,w)$</span> of the tangent lines to the conical curve has been recognized as associated with <span class="math-container">$M^{-1}$</span> (assuming <span class="math-container">$\det(M) \neq 0$</span>!), due to relationship</p>
<p><span class="math-container">$$X^TMX=X^TMM^{-1}MX=(MX)^T(M^{-1})(MX)=U^TM^{-1}U=0$$</span>
<span class="math-container">$$=\begin{pmatrix}u&v&w\end{pmatrix}\begin{pmatrix}A & B & D \\ B & C & E \\ D & E & F \end{pmatrix}\begin{pmatrix}u \\ v \\ w \end{pmatrix}=0$$</span></p>
<p>whereas, in 19th century, it was usual to write the previous quadratic form as :</p>
<p><span class="math-container">$$\det \begin{pmatrix}M^{-1}&U\\U^T&0\end{pmatrix}=\begin{vmatrix}a&b&d&u\\b&c&e&v\\d&e&f&w\\u&v&w&0\end{vmatrix}=0$$</span></p>
<p>as the determinant of a matrix obtained by "bordering" <span class="math-container">$M^{-1}$</span> precisely by <span class="math-container">$U$</span></p>
<p>(see the excellent lecture notes (<a href="http://www.maths.gla.ac.uk/wws/cabripages/conics/conics0.html" rel="noreferrer">http://www.maths.gla.ac.uk/wws/cabripages/conics/conics0.html</a>)). It is to be said that the idea of linear transformations, especially orthogonal transformations, arose even earlier in the framework of the theory of numbers (quadratic representations).</p>
<p>Remark: the way the former identities have been written use matrix algebra notations and rules that were unknown in the 19th century, with the notable exception of Grassmann's "Ausdehnungslehre", whose ideas were too ahead of his time (1844) to have a real influence.</p>
<p>c) the concept of <strong>eigenvector/eigenvalue</strong>, initially motivated by the determination of "principal axes" of conics and quadrics.</p>
<ul>
<li>the very idea of "geometric transformation" (more or less born with Klein circa 1870) associated with an array of numbers (when linear or projective). A matrix, of course, is much more that an array of numbers... But think for example to the persistence of expression "table of direction cosines" (instead of "orthogonal matrix") as can be found for example still in the 2002 edition of Analytical Mechanics by A.I. Lorrie.</li>
</ul>
<p>d) The concept of "companion matrix" of a polynomial <span class="math-container">$P$</span>, that could be considered as a tool but is more fundamental than that (<a href="https://en.wikipedia.org/wiki/Companion_matrix" rel="noreferrer">https://en.wikipedia.org/wiki/Companion_matrix</a>). It can be presented and "justified" as a "nice determinant" :
In fact, it has much more to say, with the natural interpretation for example in the framework of <span class="math-container">$\mathbb{F}_p[X]$</span> (polynomials with coefficients in a finite field) as the matrix of multiplication by <span class="math-container">$P(X)$</span>. (<a href="https://glassnotes.github.io/OliviaDiMatteo_FiniteFieldsPrimer.pdf" rel="noreferrer">https://glassnotes.github.io/OliviaDiMatteo_FiniteFieldsPrimer.pdf</a>), giving rise to matrix representations of such fields. Another remarkable application of companion matrices : the main numerical method for obtaining the roots of a polynomial is by computing the eigenvalues of its companion matrix using a Francis "QR" iteration (see (<a href="https://math.stackexchange.com/q/68433">https://math.stackexchange.com/q/68433</a>)).</p>
<p>References:</p>
<p>I discovered recently a rather similar question with a very complete answer by Denis Serre, a specialist in the domain of matrices :
<a href="https://mathoverflow.net/q/35988/88984">https://mathoverflow.net/q/35988/88984</a></p>
<p>The article by Thomas Hawkins : "Cauchy and the spectral theory of matrices", Historia Mathematica 2, 1975, 1-29.</p>
<p>See also (<a href="http://www.mathunion.org/ICM/ICM1974.2/Main/icm1974.2.0561.0570.ocr.pdf" rel="noreferrer">http://www.mathunion.org/ICM/ICM1974.2/Main/icm1974.2.0561.0570.ocr.pdf</a>)</p>
<p>An important bibliography is to be found in (<a href="http://www-groups.dcs.st-and.ac.uk/history/HistTopics/References/Matrices_and_determinants.html" rel="noreferrer">http://www-groups.dcs.st-and.ac.uk/history/HistTopics/References/Matrices_and_determinants.html</a>).</p>
<p>See also a good paper by Nicholas Higham : (<a href="http://eprints.ma.man.ac.uk/954/01/cay_syl_07.pdf" rel="noreferrer">http://eprints.ma.man.ac.uk/954/01/cay_syl_07.pdf</a>)</p>
<p>For conic sections and projective geometry, see a) this excellent chapter of lectures of the University of Vienna (see the other chapters as well) : (<a href="https://www-m10.ma.tum.de/foswiki/pub/Lehre/WS0809/GeometrieKalkueleWS0809/ch10.pdf" rel="noreferrer">https://www-m10.ma.tum.de/foswiki/pub/Lehre/WS0809/GeometrieKalkueleWS0809/ch10.pdf</a>). See as well : (maths.gla.ac.uk/wws/cabripages/conics/conics0.html).</p>
<p>Don't miss the following very interesting paper about various kinds of useful determinants : <a href="https://arxiv.org/pdf/math/9902004.pdf" rel="noreferrer">https://arxiv.org/pdf/math/9902004.pdf</a></p>
<p>See also <a href="https://ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/wiki/Matrix_(mathematics).html" rel="noreferrer">this</a></p>
<p>Very interesting precisions on determinants in <a href="https://arxiv.org/pdf/math/9902004.pdf" rel="noreferrer">this text</a> and in these <a href="https://math.stackexchange.com/q/194579">answers</a>.</p>
<p>A fundamental work on "The Theory of Determinants" in 4 volumes has been written by Thomas Muir : <a href="http://igm.univ-mlv.fr/%7Eal/Classiques/Muir/History_5/VOLUME5_TEXT.PDF" rel="noreferrer">http://igm.univ-mlv.fr/~al/Classiques/Muir/History_5/VOLUME5_TEXT.PDF</a> (years 1906, 1911, 1922, 1923) for the last volumes or, for all of them <a href="https://ia800201.us.archive.org/17/items/theoryofdetermin01muiruoft/theoryofdetermin01muiruoft.pdf" rel="noreferrer">https://ia800201.us.archive.org/17/items/theoryofdetermin01muiruoft/theoryofdetermin01muiruoft.pdf</a>. It is very interesting to take random pages and see how the determinant-mania has been important, especially in the second half of the 19th century. Matrices appear at some places with the double bar convention that lasted a very long time. Matrices are mentionned here and there, rarely to their advantage...</p>
<p>Many historical details about determinants and matrices can be found <a href="https://mathshistory.st-andrews.ac.uk/HistTopics/Matrices_and_determinants/" rel="noreferrer">here</a>.</p>
| <p><strong>It depends who you speak to.</strong></p>
<ul>
<li>In <strong>numerical mathematics</strong>, where people actually have to compute things on a computer, it is largely recognized that <strong>determinants are useless</strong>. Indeed, in order to compute determinants, either you use the Laplace recursive rule ("violence on minors"), which costs <span class="math-container">$O(n!)$</span> and is infeasible already for very small values of <span class="math-container">$n$</span>, or you go through a triangular decomposition (Gaussian elimination), which by itself already tells you everything you needed to know in the first place. Moreover, for most reasonably-sized matrices containing floating-point numbers, determinants overflow or underflow (try <span class="math-container">$\det \frac{1}{10} I_{350\times 350}$</span>, for instance). To put another nail on the coffin, computing eigenvalues by finding the roots of <span class="math-container">$\det(A-xI)$</span> is hopelessly unstable. In short: in numerical computing, whatever you want to do with determinants, there is a better way to do it without using them.</li>
<li>In <strong>pure mathematics</strong>, where people are perfectly fine knowing that an explicit formula exists, all the examples are <span class="math-container">$3\times 3$</span> anyway and people make computations by hand, <strong>determinants are invaluable</strong>. If one uses Gaussian elimination instead, all those divisions complicate computations horribly: one needs to take different paths whether things are zero or not, so when computing symbolically one gets lost in a myriad of cases. The great thing about determinants is that they give you an explicit polynomial formula to tell when a matrix is invertible or not: this is extremely useful in proofs, and allows for lots of elegant arguments. For instance, try proving this fact without determinants: given <span class="math-container">$A,B\in\mathbb{R}^{n\times n}$</span>, if <span class="math-container">$A+Bx$</span> is singular for <span class="math-container">$n+1$</span> distinct real values of <span class="math-container">$x$</span>, then it is singular for all values of <span class="math-container">$x$</span>. This is the kind of things you need in proofs, and determinants are a priceless tool. Who cares if the explicit formula has a exponential number of terms: they have a very nice structure, with lots of neat combinatorial interpretations. </li>
</ul>
|
geometry | <p>I was just wondering why we have 90° degrees for a perpendicular angle. Why not 100° or any other number?</p>
<p>What is the significance of 90° for the perpendicular or 360° for a circle?</p>
<p>I didn't ever think about this during my school time.</p>
<p>Can someone please explain it mathematically? Is it due to some historical reason?</p>
| <p>360 is an incredibly <a href="http://en.wikipedia.org/wiki/Abundant_number">abundant</a> number, which means that there are many factors. So it makes it easy to divide the circle into $2, 3, 4, 5, 6, 8, 9, 10, 12,\ldots$ parts. By contrast, 400 gradians cannot even be divided into 3 equal whole-number parts. While this may not necessarily be why 360 was chosen in the first place, it could be one of the reasons we've stuck with the convention.</p>
<p>By the way, when working in radians, we just "live with" the fact that most common angles are fractions involving $\pi$. There's a small group of people who prefer to use a constant called $\tau$, which is just $2\pi$. Then angles seem naturally to be divisions of the circle: The angle that divides a circle into $n$ equal parts is $\tau/n$ (radians).</p>
<p>Hope this helps!</p>
| <p>I have heard that the ancient Babylonians used a base-$60$ numeral system with sub-base $10$.</p>
<p>Certainly such a system was used by Ptolemy in the second century AD. See Gerald Toomer's translation of Ptolemy's <em>Almagest</em>. In particular Ptolemy divided the circle into $360$ degrees. See <a href="http://en.wikipedia.org/wiki/Ptolemy%27s_table_of_chords">http://en.wikipedia.org/wiki/Ptolemy%27s_table_of_chords</a>, <a href="http://en.wikipedia.org/wiki/Almagest">http://en.wikipedia.org/wiki/Almagest</a>, and <a href="http://hypertextbook.com/eworld/chords.shtml">http://hypertextbook.com/eworld/chords.shtml</a> .</p>
|
probability | <p>This morning, I wanted to flip a coin to make a decision but only had an SD card:</p>
<p><img src="https://i.sstatic.net/T7fUT.png" alt="enter image description here"></p>
<p>Given that <em>I don't know</em> the bias of this SD card, would flipping it be considered a "fair toss"?</p>
<p>I thought if I'm just as likely to assign an outcome to one side as to the other, then it must be a fair. But this also seems like a recasting of the original question; instead of asking whether the <em>unknowing of the SD card's construction</em> defines fairness, I'm asking if the <em>unknowing of my own psychology</em> (<em>e.g.</em> which side I'd choose for which outcome) defines fairness. Either way, I think I'm asking: What's the exact relationship between <em>not knowing</em> and "fairness"?</p>
<p>Additional thought: An SD card might be "fair" <em>to me</em>, but not at all fair to, say, a design engineer looking at the SD card's blueprint, who immediately sees that the chip is off-center from the flat plane. So it seems <em>fairness</em> even depends on the subjects to whom fairness <em>matters</em>. In a football game then, does an SD card remain "fair" as long as no design engineer is there to discern the object being tossed?</p>
| <p>Here's a pragmatic answer from an engineer. You can always get a fair 50/50 outcome with <strong>any</strong> "coin" (or SD card, or what have you), <em>without having to know whether it is biased, or how biased it is</em>:</p>
<ul>
<li>Flip the coin twice. </li>
<li>If you get $HH$ or $TT$, discard the trial and
repeat. </li>
<li>If you get $HT$, decide $H$. </li>
<li>If you get $TH$, decide $T$.</li>
</ul>
<p>The only conditions are that (i) the coin is not completely biased (i.e., $\Pr(H)\neq 0, \Pr(T)\neq 0$), and (ii) the bias does not change from trial to trial.</p>
<p>The procedure works because whatever the bias is (say $\Pr(H)=p$, $\Pr(T)=1-p$), the probabilties of getting $HT$ and $TH$ are the same: $p(1-p)$. Since the other outcomes are discarded, $HT$ and $TH$ each occur with probability $\frac{1}{2}$.</p>
| <p>That is a very good question!</p>
<p>There are (at least) two different ways to define probability: as a measure of frequencies, and as a measure of (subjective) knowlegde of the result.</p>
<p>The frequentist definition would be: the probability of the sd card landing "heads" is the proportion of times it lands "heads", if you toss it many times
(details ommited partially because of ignorance: what do we mean by 'many'?)</p>
<p>The "knowledge" approach (usually called bayesian) is harder to define. It asks how likely you (given your information) think an outcome is. As you have no information about the construction of the sd card, you might think both sides are equally likely to appear. </p>
<p>In more concrete terms, say I offer you a bet: I give you one dollar if 'heads', and you give me one if 'tails'. If we both are ignorant about the sd card, then, for us both, the bet sounds neither good nor bad. In a sense, it is a fair bet.</p>
<p>Notice that the bayesian approach defines more probabilities that the frequentist. I can, say, talk about the probability that black holes exist. Well, either they do, or they don't, but that does not mean there are no bets I would consider advantageous on the matter: If you offer me a million dollars versus one dollar, saying that they exist, I might take that bet (and that would 'imply' that I consider the probability that they don't exist to be bigger than 1 millionth).</p>
<p>Now, the question of fairness: if no one knows anything about the sd card, I would call your sd card toss fair. In a very meaningfull way: neither of the teams, given a side, would have reason to prefer the other side. However, obviously, it has practical drawbacks: a team might figure something out latter on, and come to complain about it. (that is: back when they chose a side, their knowledge did not allow them to distinguish the sides. Now, it does)</p>
<p>It the end: there is not one definition of probability that is 100% accepted. Hence, there is no definition of fair that is 100% accepted.</p>
<p><a href="http://en.wikipedia.org/wiki/Probability_interpretations">http://en.wikipedia.org/wiki/Probability_interpretations</a></p>
|
linear-algebra | <p>In my linear algebra class, we just talked about determinants. So far I’ve been understanding the material okay, but now I’m very confused. I get that when the determinant is zero, the matrix doesn’t have an inverse. I can find the determinant of a <span class="math-container">$2\times 2$</span> matrix by the formula. Our teacher showed us how to compute the determinant of an <span class="math-container">$n \times n$</span> matrix by breaking it up into the determinants of smaller matrices. Apparently there is a way by summing over a bunch of permutations. But the notation is really hard for me and I don’t really know what’s going on with them anymore. Can someone help me figure out what a determinant is, intuitively, and how all those definitions of it are related?</p>
| <p>Your trouble with determinants is pretty common. They’re a hard thing to teach well, too, for two main reasons that I can see: the formulas you learn for computing them are messy and complicated, and there’s no “natural” way to interpret the value of the determinant, the way it’s easy to interpret the derivatives you do in calculus at first as the slope of the tangent line. It’s hard to believe things like the invertibility condition you’ve stated when it’s not even clear what the numbers mean and where they come from.</p>
<p>Rather than show that the many usual definitions are all the same by comparing them to each other, I’m going to state some general properties of the determinant that I claim are enough to specify uniquely what number you should get when you put in a given matrix. Then it’s not too bad to check that all of the definitions for determinant that you’ve seen satisfy those properties I’ll state.</p>
<p>The first thing to think about if you want an “abstract” definition of the determinant to unify all those others is that it’s not an array of numbers with bars on the side. What we’re really looking for is a function that takes N vectors (the N columns of the matrix) and returns a number. Let’s assume we’re working with real numbers for now.</p>
<p>Remember how those operations you mentioned change the value of the determinant?</p>
<ol>
<li><p>Switching two rows or columns changes the sign.</p>
</li>
<li><p>Multiplying one row by a constant multiplies the whole determinant by that constant.</p>
</li>
<li><p>The general fact that number two draws from: the determinant is <em>linear in each row</em>. That is, if you think of it as a function <span class="math-container">$\det: \mathbb{R}^{n^2} \rightarrow \mathbb{R}$</span>, then <span class="math-container">$$ \det(a \vec v_1 +b \vec w_1 , \vec v_2 ,\ldots,\vec v_n ) = a \det(\vec v_1,\vec v_2,\ldots,\vec v_n) + b \det(\vec w_1, \vec v_2, \ldots,\vec v_n),$$</span> and the corresponding condition in each other slot.</p>
</li>
<li><p>The determinant of the identity matrix <span class="math-container">$I$</span> is <span class="math-container">$1$</span>.</p>
</li>
</ol>
<p>I claim that these facts are enough to define a <em>unique function</em> that takes in N vectors (each of length N) and returns a real number, the determinant of the matrix given by those vectors. I won’t prove that, but I’ll show you how it helps with some other interpretations of the determinant.</p>
<p>In particular, there’s a nice geometric way to think of a determinant. Consider the unit cube in N dimensional space: the set of N vectors of length 1 with coordinates 0 or 1 in each spot. The determinant of the linear transformation (matrix) T is the <em>signed volume of the region gotten by applying T to the unit cube</em>. (Don’t worry too much if you don’t know what the “signed” part means, for now).</p>
<p>How does that follow from our abstract definition?</p>
<p>Well, if you apply the identity to the unit cube, you get back the unit cube. And the volume of the unit cube is 1.</p>
<p>If you stretch the cube by a constant factor in one direction only, the new volume is that constant. And if you stack two blocks together aligned on the same direction, their combined volume is the sum of their volumes: this all shows that the signed volume we have is linear in each coordinate when considered as a function of the input vectors.</p>
<p>Finally, when you switch two of the vectors that define the unit cube, you flip the orientation. (Again, this is something to come back to later if you don’t know what that means).</p>
<p>So there are ways to think about the determinant that aren’t symbol-pushing. If you’ve studied multivariable calculus, you could think about, with this geometric definition of determinant, why determinants (the Jacobian) pop up when we change coordinates doing integration. Hint: a derivative is a linear approximation of the associated function, and consider a “differential volume element” in your starting coordinate system.</p>
<p>It’s not too much work to check that the area of the parallelogram formed by vectors <span class="math-container">$(a,b)$</span> and <span class="math-container">$(c,d)$</span> is <span class="math-container">$\Big|{}^{a\;b}_{c\;d}\Big|$</span>
either: you might try that to get a sense for things.</p>
| <p>You could think of a determinant as a volume. Think of the columns of the matrix as vectors at the origin forming the edges of a skewed box. The determinant gives the volume of that box. For example, in 2 dimensions, the columns of the matrix are the edges of a rhombus.</p>
<p>You can derive the algebraic properties from this geometrical interpretation. For example, if two of the columns are linearly dependent, your box is missing a dimension and so it's been flattened to have zero volume.</p>
|
differentiation | <p>I am a senior in high school so I know I am simply misunderstanding something but I don't know what, please have patience.</p>
<p>I was tasked to find the derivative for the following function:</p>
<p><span class="math-container">$$ y = \frac{ (4x)^{1/5} }{5} + { \left( \frac{1}{x^3} \right) } ^ {1/4} $$</span></p>
<p>Simplifying:</p>
<p><span class="math-container">$$ y = \frac{ 4^{1/5} }{5} x^{1/5} + { \frac{1 ^ {1/4}}{x ^ {3/4}} } $$</span></p>
<p><span class="math-container">$$ y = \frac{ 4^{1/5} }{5} x^{1/5} + { \frac{\pm 1}{x ^ {3/4}} } $$</span></p>
<p>Because <span class="math-container">$ 1 ^ {1/n} = \pm 1 $</span>, given <span class="math-container">$n$</span> is even</p>
<p><span class="math-container">$$ y = \frac{ 4^{1/5} }{5} x^{1/5} \pm { x ^ {-3/4} } $$</span></p>
<p>Taking the derivative using power rule:</p>
<p><span class="math-container">$$ \frac{dy}{dx} = \frac{ 4^{1/5} }{25} x^{-4/5} \pm \frac{-3}{4} { x ^ {-7/4} } $$</span></p>
<p>which is the same as</p>
<p><span class="math-container">$$ \frac{dy}{dx} = \frac{ 4^{1/5} }{25} x^{-4/5} \pm \frac{3}{4} { x ^ {-7/4} } $$</span></p>
<p>And that is the part that I find difficult to understand. I know that I should be adding the second term(I graphed it multiple times to make sure), but I cannot catch my error and my teacher did't want to discuss it.</p>
<p>So I know I am doing something wrong because one function cannot have more than one derivative.</p>
| <p>The <span class="math-container">$(\cdot)^{\frac{1}{4}}$</span> operation has to be understood as a function. A function can only have one image for any argument. Depending upon how you interpret the fourth root, the image could be positive or negative. But once you set how you interpret your function (positive or negative valued), you have to stick with that interpretation throughout. </p>
<p>When you write <span class="math-container">$y = \frac{ 4^{1/5} }{5} x^{1/5} \pm { \frac{1}{x ^ {3/4}} }$</span>, you are working with both interpretations simultaneously. In other words, when you differentiate, you don't get two derivatives for one function, rather two derivatives corresponding to two different functions, one <span class="math-container">$y = \frac{ 4^{1/5} }{5} x^{1/5} + { \frac{1}{x ^ {3/4}} }$</span>, and the other, <span class="math-container">$y = \frac{ 4^{1/5} }{5} x^{1/5} - { \frac{1}{x ^ {3/4}} }$</span>.</p>
| <p>You are confused about what <span class="math-container">$y^{1/4}$</span> actually means.</p>
<p>Suppose that <span class="math-container">$x^4=1$</span>. We could raise both sides to the <span class="math-container">$1/4$</span> power: <span class="math-container">$$\left(x^4\right)^{1/4}=1^{1/4}$$</span></p>
<p>The right side is <em>unambiguously</em> <span class="math-container">$1$</span>. It is not <span class="math-container">$\pm1$</span>. But read on. <span class="math-container">$$\left(x^4\right)^{1/4}=1$$</span> The left side does <em>not</em> simplify to <span class="math-container">$x$</span> unless you somehow know ahead of time that <span class="math-container">$x$</span> is positive. Otherwise, all you can say is the left side simplifies to <span class="math-container">$\lvert x\rvert$</span>. So you have <span class="math-container">$$\lvert x\rvert = 1$$</span> That implies that "either <span class="math-container">$x=1$</span> or <span class="math-container">$x=-1$</span>". Out of laziness (or a minor efficiency boost) people write <span class="math-container">$x=\pm1$</span>.</p>
<p>Now we started with <span class="math-container">$x^4=1$</span> and ended with <span class="math-container">$x=\pm1$</span>. And because of this and applying the <span class="math-container">$1/4$</span> power in the middle of that process, you have inferred that <span class="math-container">$1^{1/4}=\pm1$</span>. But that is a misunderstanding of the process in its entirety. <span class="math-container">$1^{1/4}$</span> is unambiguously equal to <span class="math-container">$1$</span> when working with arithmetic and real numbers.</p>
|
probability | <p>I found this math "problem" on the internet, and I'm wondering if it has an answer:</p>
<blockquote>
<p><strong>Question:</strong> If you choose an answer to this question at random, what is the probability that you will be correct?</p>
<p><strong>a.</strong> <span class="math-container">$25\%$</span></p>
<p><strong>b.</strong> <span class="math-container">$50\%$</span></p>
<p><strong>c.</strong> <span class="math-container">$0\%$</span></p>
<p><strong>d.</strong> <span class="math-container">$25\%$</span></p>
</blockquote>
<p>Does this question have a correct answer?</p>
| <p>No, it is not meaningful. 25% is correct iff 50% is correct, and 50% is correct iff 25% is correct, so it can be neither of those two (because if <em>both</em> are correct, the only correct answer could be 75% which is not even an option). But it cannot be 0% either, because then the correct answer would be 25%. So none of the answers are correct, so the answer must be 0%. But then it is 25%. And so forth.</p>
<p>It's a multiple-choice variant (with bells and whistles) of the classical <em>liar paradox</em>, which asks whether the statement</p>
<blockquote>
<p>This statement is false.</p>
</blockquote>
<p>is true or false. There are various more or less contrived "philosophical" attempts to resolve it, but by far the most common resolution is to deny that the statement means anything in the first place; therefore it is also meaningless to ask for its truth value.</p>
<hr>
<p><strong>Edited much later to add:</strong> There's a variant of this puzzle that's very popular on the internet at the moment, in which answer option (c) is 60% rather than 0%. In this variant it is at least internally consistent to claim that all of the answers are wrong, and so the possibility of getting a right one by choosing randomly is 0%.</p>
<p>Whether this actually resolves the variant puzzle is more a matter of taste and temperament than an objective mathematical question. It is not in general true for self-referencing questions that simply being internally consistent is enough for an answer to be unambiguously right; otherwise the question</p>
<blockquote>
<p>Is the correct answer to this question "yes"?</p>
</blockquote>
<p>would have two different "right" answers, because "yes" and "no" are both internally consistent. In the 60% variant of the puzzle it is happens that the <em>only</em> internally consistent answer is "0%", but even so one might, as a matter of caution, still deny that such reasoning by elimination is valid for self-referential statements at all. If one adopts this stance, one would still consider the 60% variant meaningless.</p>
<p>One rationale for taking this strict position would be that we don't want to accept reasoning by elimination on</p>
<blockquote>
<p>True or false?</p>
<ul>
<li>The Great Pumpkin exists.</li>
<li>Both of these statements are false.</li>
</ul>
</blockquote>
<p>where the only internally consistent resolution is that the first statement is true and the second one is false. However, it appears to be unsound to conclude that the Great Pumpkin exists on the basis simply that the puzzle was posed.</p>
<p>On the other hand, it is difficult to argue that there is <em>no</em> possible principle that will cordon off the Great Pumpkin example as meaningless while still allowing the 60% variant to be meaningful.</p>
<p>In the end, though, these things are more matters of taste and philosophy than they are mathematics. <em>In mathematics</em> we generally prefer to play it safe and completely refuse to work with explicitly self-referential statements. This avoids the risk of paradox, and does not seem to hinder mathematical arguments about the things mathematicians are ordinarily interested in. So whatever one decides to do with the question-about-itself, what one does is not really mathematics.</p>
| <p>The question is underspecified since it doesn't say which distribution is used in choosing an answer at random. Any of the answers could be correct:</p>
<p>If I choose a. with probability 25% and b. with probability 75%, a and d are correct.</p>
<p>If I choose a. with probability 50% and b. with probability 50%, b is correct.</p>
<p>If I choose a. with probability 75% and b. with probability 25%, c is correct.</p>
<p>From the design of the question, it seems that whoever wrote it had in mind a uniform distribution over all four answers, but forgot to specify that. In that case Henning's answer applies.</p>
|
probability | <blockquote>
<p>If a $1$ meter rope is cut at two uniformly randomly chosen points (to give three pieces), what is the average length of the smallest piece?</p>
</blockquote>
<p>I got this question as a mathematical puzzle from a friend. It looks similar to MathOverflow question <a href="https://mathoverflow.net/q/2014/">If you break a stick at two points chosen uniformly, the probability the three resulting sticks form a triangle is 1/4.</a></p>
<p>However, in this case, I have to find the expected length of the smallest segment. The two points where the rope is cut are selected uniformly at random. </p>
<p>I tried simulating it and I got an average value of $0.1114$. I suspect the answer is $1/9$ but I don't have any rigorous math to back it up.</p>
<p>How do I solve this problem?</p>
| <p>Update: The current version of this answer is more intuitive (IMHO) than the previous one. See also <a href="https://math.stackexchange.com/questions/14190/average-length-of-the-longest-segment/14194#14194">this similar answer</a> to a similar question.</p>
<p>The generalization of this problem to <span class="math-container">$n$</span> pieces is discussed extensively in David and Nagaraja's <em><a href="http://books.google.com/books?id=bdhzFXg6xFkC&printsec=frontcover&dq=david+and+nagaraja+order+statistics&source=bl&ots=OaK_k-BbTf&sig=vAtEAURNfZTe3NDV9WodQZ-3ip4&hl=en&ei=DfsDTdiXPML88AafsKnqAg&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBgQ6AEwAA#v=onepage&q&f=false" rel="noreferrer">Order Statistics</a></em> (pp. 133-135, and p. 153). </p>
<p>If <span class="math-container">$X_1, X_2, \ldots, X_{n-1}$</span> denote the positions on the rope where the cuts are made, let <span class="math-container">$V_i = X_i - X_{i-1}$</span>, where <span class="math-container">$X_0 = 0$</span> and <span class="math-container">$X_n = 1$</span>. So the <span class="math-container">$V_i$</span>'s are the lengths of the pieces of rope.
<BR><BR>
The key idea is that the probability that any particular <span class="math-container">$k$</span> of the <span class="math-container">$V_i$</span>'s simultaneously have lengths longer than <span class="math-container">$c_1, c_2, \ldots, c_k$</span>, respectively (where <span class="math-container">$\sum_{i=1}^k c_i \leq 1$</span>), is <span class="math-container">$$(1-c_1-c_2-\ldots-c_k)^{n-1}.$$</span> This is proved formally in David and Nagaraja's <em><a href="http://books.google.com/books?id=bdhzFXg6xFkC&pg=PA133&lpg=PA133&dq=david+and+nagaraja+order+statistics+division+of+random+interval&source=bl&ots=OaK_l7B9-j&sig=por_GsntbxBDII72xTWISQ9Keas&hl=en&ei=urkGTevVIZTmsQOM_7HsBw&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBMQ6AEwAA#v=onepage&q&f=false" rel="noreferrer">Order Statistics</a></em>, p. 135. Intuitively, the idea is that in order to have pieces of size at least <span class="math-container">$c_1, c_2, \ldots, c_k$</span>, all <span class="math-container">$n-1$</span> of the cuts have to occur in intervals of the rope of total length <span class="math-container">$1 - c_1 - c_2 - \ldots - c_k$</span>. For example, <span class="math-container">$P(V_1 > c_1)$</span> is the probability that all <span class="math-container">$n-1$</span> cuts occur in the interval <span class="math-container">$(c_1, 1]$</span>, which, since the cuts are randomly distributed in <span class="math-container">$[0,1]$</span>, is <span class="math-container">$(1-c_1)^{n-1}$</span>.
<BR><BR>
If <span class="math-container">$V_{(1)}$</span> denotes the shortest piece of rope, then for <span class="math-container">$x \leq \frac{1}{n}$</span>, (following Raskolnikov's comment) <span class="math-container">$$P(V_{(1)} > x) = P(V_1 > x, V_2 > x, \ldots, V_n > x) = (1 - nx)^{n-1}.$$</span></p>
<p>Therefore,
<span class="math-container">$$E[V_{(1)}] = \int_0^{\infty} P(V_{(1)} > x) dx = \int_0^{1/n} (1-nx)^{n-1} dx = \frac{1}{n^2}.$$</span></p>
<p>David and Nagaraja also give the formula Yuval Filmus mentions (as Problem 6.4.2):</p>
<p><span class="math-container">$$E[V_{(r)}] = \frac{1}{n} \sum_{j=1}^r \frac{1}{n-j+1}.$$</span></p>
| <p>My approach is maybe more naive than the others posted.</p>
<p>Break the unit interval at $x$ and $y$ where $x < y$. Our lengths are then $x$, $y - x$, and $1 - y$. It's not hard to show that they all have probability $1/3$ of being the shortest. In any case, our joint PDF is given by $f(x,y) = 6$ (since $x$ and $y$ remain uniform random variables on $1/6$th of the square $[0,1] \times [0,1]$). Each triangle in the diagram below corresponds to the domain of the PDF for one of the three cases.</p>
<p><img src="https://i.sstatic.net/obVJs.png" alt="Each triangle corresponds to which length is shortest"></p>
<p>I'll take care of the case when $x$ is shortest, that is, $x \leq y - x$ and $x \leq 1 - y$. This is the leftmost triangle. Since we're assuming $x$ is least, we are looking for
$$E[x] = \int_0^{1/3} \int_{2x}^{1 - x} 6x \;dy \;dx = 1/9$$</p>
<p>The cases when $y - x$ and $1 - y$ are shortest are similar.</p>
|
probability | <p>Choose a random number between $0$ and $1$ and record its value. Do this again and add the second number to the first number. Keep doing this until the sum of the numbers exceeds $1$. What's the expected value of the number of random numbers needed to accomplish this?</p>
| <p>Here is a (perhaps) more elementary method. Let $X$ be the amount of numbers you need to add until the sum exceeds $1$. Then (by linearity of expectation):</p>
<p>$$ \mathbb{E}[X] = 1 + \sum_{k \geq 1} \Pr[X > k] $$</p>
<p>Now $X > k$ if the sum of the first $k$ numbers $x_1,\ldots,x_k$ is smaller than $1$. This is exactly equal to the volume of the $k$-dimensional set:</p>
<p>$$ \left\{(x_1,\ldots,x_k) : \sum_{i=1}^k x_i \leq 1, \, x_1,\ldots,x_k \geq 0\right\}$$</p>
<p>This is known as the $k$-dimensional <em>simplex</em>. When $k = 1$, we get a line segment of length $1$. When $k = 2$, we get a right isosceles triangle with sides of length $1$, so the area is $1/2$. When $k=3$, we get a triangular pyramid (tetrahedron) with unit sides, so the volume is $1/6$. In general, the volume is $1/k!$, and so</p>
<p>$$ \mathbb{E}[X] = 1 + \sum_{k \geq 1} \frac{1}{k!} = e. $$</p>
| <p>Assuming the numbers come from a uniform distribution over $[0,1]$ and that the trials are independent, here is an outline (this is example 7.4 4h. in Sheldon Ross' <i>A First Course in Probability</i>, sixth edition): </p>
<ol>
<li><p>Let $X_i$ be the number obtained on the $i$'th trial. </p></li>
<li><p>For $0\le x\le1$, let $Y(x)$ be the minimum number of trials needed so that the sum of the $X_i$ exceeds $x$.
Set $e(x)=\Bbb E [Y(x)]$.</p></li>
<li><p>Compute $e(x)$ by conditioning on the value of $X_1$:
$$\tag{1}
e(x)=\int_0^1 \Bbb E [ Y(x) | X_1=y]\, dy.
$$</p></li>
</ol>
<p>Here, use the fact that
$$\tag{2}\Bbb E [ Y(x) | X_1=y] = \cases{1,& $y>x$\cr 1+e(x-y),& $y\le x $}.$$</p>
<p>Substitution of $(2)$ into $(1)$ will give
$$\tag{3}
e(x)=1+\int_0^x e(u)\,du.
$$</p>
<ol start="4">
<li><p>Solve equation $(3)$ (by differentiating both sides with respect to $x$ first) for $e(x)$.</p></li>
<li><p>You wish to find $e(1)$.</p></li>
</ol>
|
combinatorics | <p>A friend of mine was doodling with numbers arranged somewhat reminiscent of Pascal's Triangle, where the first row was $ 1^{n-1} \ \ 2^{n-1} \ \cdots \ n^{n-1} $ and subsequent rows were computed by taking the difference of adjacent terms. He conjectured that the number we get at the end is $ n! $ but I've not been able to prove or disprove this. The first few computations are given below:
$$
\begin{pmatrix}
1 \\
\end{pmatrix}
$$</p>
<p>$$
\begin{pmatrix}
1 & & 2 \\
& 1 & \\
\end{pmatrix}
$$</p>
<p>$$
\begin{pmatrix}
1 & & 4 & & 9 \\
& 3 & & 5 & \\
& & 2 & & \\
\end{pmatrix}
$$</p>
<p>$$
\begin{pmatrix}
1 & & 8 & & 27 & & 64 \\
& 7 & & 19 & & 37 & \\
& & 12 & & 18 & & \\
& & & 6 & & & \\
\end{pmatrix}
$$</p>
<p>$$
\newcommand\pad[1]{\rlap{#1}\phantom{625}}
\begin{pmatrix}
1 & & 16 & & 81 & & 256 & & 625 \\
& 15 & & 65 & & 175 & & 369 & \\
& & 50 & & 110 & & 194 & & \\
& & & 60 & & 84 & & & \\
& & & & 24 & & & & \\
\end{pmatrix}
$$</p>
<p>I attempted to write down the general term and tried to reduce that to the required form. The general term worked out as
$$
\sum_{i=0}^n (-1)^{n-i} \binom{n}{i} (i+1)^{n}.
$$
I tried applying various identities of the binomial coefficients but I'm barely making any progress. Any help would be appreciated.</p>
<p><strong>Small note</strong>: If I instead start with the first row as $ 0^{n} \ \ 1^{n} \ \cdots \ n^{n} $ then I still get $n!$ at the end of the computation, and the general formula in this case works out as
$$
\sum_{i=0}^n (-1)^{n-i} \binom{n}{i} i^{n}.
$$
In fact, we can start with any $n$ consecutive natural numbers, each raised to the $(n-1)$th power, and we still get $n!$ at the end of the computation.</p>
| <p>The top rows are indeed made of the powers $i^n=P_n(i)$, which are polynomials of degree $n$, with the leading coefficient $1$.</p>
<p>On the next row you take the first order difference. By the binomial formula, we have</p>
<p>$$P_{n-1}(i)=P_n(i+1)-P_n(i)=i^n+ni^{n-1}+\cdots-i^n=ni^{n-1}+\cdots$$</p>
<p>which is a polynomial of degree $n-1$ with the leading coefficient $n$.</p>
<p>For the next row, $$P_{n-2}(i)=P_{n-1}(i+1)-P_{n-1}(i)=n(n-1)i^{n-2}+\cdots$$ and so on.</p>
<p>On the last row, we have a polynomial of degree $0$ with the leading coefficient $n!$, and all the rest has vanished.</p>
<hr>
<p>Actually you will make the same observation starting with any polynomial in $i$: the final value is $p_nn!$, where $p_n$ is the initial leading coefficient. And if you enlarge the table to the right, the bottom row remains constant.</p>
<p>E.g.</p>
<p>$$2i^3+i\\\Delta_1=6i^2+6i+3\to2\cdot3\,i^2\\\Delta_2=12i+4\to2\cdot3\cdot2\,i^1\\\Delta_3=12\to2\cdot3\cdot2\cdot1\,i^0.$$</p>
| <p>Here is another way to show that
$$ \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} k^n = n!$$</p>
<p>We consider the set $S$ consisting of all strings of length $n$ consisting of the symbols $a_1, a_2, \cdots, a_n$, with repetition allowed.</p>
<p>Then we clearly have that $|S|=n^n$ because there are $n$ choices for each symbol in the string.</p>
<p>Now let $A_k$ be the set of all such strings which does not contain the symbol $a_k$.</p>
<p>For any natural numbers $i_1, i_2, \cdots, i_k$ (where $k\leq n$ is some natural number) such that
$$ 0 < i_1 < i_2 < i_3 \cdots < i_k \leq n$$
we can calculate the cardinality of the set
$$ A_{i_1}\cap A_{i_2}\cap A_{i_3} \cap\cdots\cap A_{i_k} $$</p>
<p>There are $(n-k)$ options for each symbol in some string in the intersection above, since such a string can consist of (and can only consist of) any of the symbols which are not $a_{i_1}, a_{i_2}, \cdots, a_{i_k}$.</p>
<p>We thus see that
$$ \left|A_{i_1}\cap A_{i_2}\cap A_{i_3} \cap\cdots\cap A_{i_k} \right| = (n-k)^n $$</p>
<p>We can now apply the inclusion-exclusion principle to find the cardinality of the set
$$ A_1\cup A_2\cup A_3 \cup\cdots\cup A_n $$</p>
<p>We have that
$$ \left|A_1\cup A_2\cup A_3 \cup\cdots\cup A_n \right| = \sum_{k=1}^{n} (-1)^{k+1} \left(\sum_{0<i_1<i_2<\cdots<i_k\leq n} \left|A_{i_1}\cap A_{i_2}\cap A_{i_3} \cap\cdots\cap A_{i_k} \right|\right) $$</p>
<p>For each $k$, there are $\binom{n}{k}$ ways to choose the numbers $i_1, i_2, i_3, \cdots, i_k$, and so we see that the above sum is equal to
$$ \sum_{k=1}^{n} (-1)^{k+1}\binom{n}{k} (n-k)^n = \sum_{k=0}^{n-1} (-1)^{n-k+1} \binom{n}{k} k^n $$</p>
<p>Finally, we consider the set
$$ S \setminus \left(A_1\cup A_2\cup A_3\cup\cdots\cup A_n\right) $$</p>
<p>From our work above, we can see that its cardinality is
$$\begin{gather}
|S| - \left|A_1\cup A_2\cup A_3 \cup\cdots\cup A_n \right| = n^n - \sum_{k=0}^{n-1} (-1)^{n-k+1} \binom{n}{k} k^n \\ = n^n + \sum_{k=0}^{n-1} (-1)^{n-k} \binom{n}{k} k^n = \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} k^n
\end{gather}$$
which is the sum which we set out to evaluate. We wish to show that this is equal to $n!$.</p>
<p>Now any element of the set
$$ S \setminus \left(A_1\cup A_2\cup A_3\cup\cdots\cup A_n\right) $$
must contain all of the symbols $a_1, a_2, a_3, \cdots, a_n$, since if it did not contain $a_k$ for some $k$, then it would be an element of $A_k$, and hence of
$$ A_1\cup A_2\cup A_3\cup\cdots\cup A_n $$</p>
<p>Conversely, any string which contains all of the symbols $a_1, a_2, a_3, \cdots, a_n$ is an element of
$$ S \setminus \left(A_1\cup A_2\cup A_3\cup\cdots\cup A_n\right) $$
since such a string is in $S$, but not in any of the $A_k$'s.</p>
<p>We see that
$$ S \setminus \left(A_1\cup A_2\cup A_3\cup\cdots\cup A_n\right) $$
consists precisely of the permutations of the symbols $a_1, a_2, a_3, \cdots, a_n$ and so its cardinality is $n!$. We have thus shown that
$$ \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} k^n = n!$$
as desired.</p>
|
linear-algebra | <p>I have a couple of questions about tensor products:</p>
<p>Why is $\text{Hom}(V,W)$ the same thing as $V^* \otimes W$? </p>
<p>Why is an element of $V^{*\otimes m}\otimes V^{\otimes n}$ the same thing as a multilinear map $V^m \to V^{\otimes n}$? </p>
<p>What is the general formulation of this principle?</p>
| <p>The result is generally wrong for infinite-dimensional spaces: see <a href="https://math.stackexchange.com/questions/573378/u-otimes-v-versus-lu-v-for-infinite-dimensional-spaces/573416#573416">this question</a>.</p>
<p>For finite dimensional space $V$, let's build an isomorphism $f : V^* \otimes W \to \hom(V,W)$ by defining</p>
<p>$$f(\phi \otimes w)(v) = \phi(v) w$$</p>
<p>This clearly defines a linear map $V^* \otimes W \to \hom(V,W)$ (it's bilinear in $V^* \times W$). Reciprocally, take a basis $(e_i)$ of $V$, then define $g : \hom(V,W) \to V^* \otimes W$ by:</p>
<p>$$g(u) = \sum_i e_i^* \otimes u(e_i)$$</p>
<p>Where $(e_i^*)$ is the dual basis to $(e_i)$ (I will use a few of its properties in $\color{red}{red}$ below). This is well-defined because $V$ is finite-dimensional (the sum is finite). Let's check that $f$ and $g$ are inverse to each other:</p>
<ul>
<li><p>For $u : V \to W$, $$f(g(u))(v) = \sum e_i^*(v) u(e_i) = u \left( \sum e_i^*(v) e_i \right) \color{red}{=} u(v)$$ and so $f(g(u)) = u$.</p></li>
<li><p>For $\phi \otimes w \in V^* \otimes W$, $$g(f(\phi \otimes w)) = \sum e_i^* \otimes f(\phi \otimes w)(e_i) = \sum e_i^* \otimes \phi(e_i) w = \sum \phi(e_i) e_i^* \otimes w \color{red}{=} \phi \otimes w$$</p></li>
</ul>
<p>And so $f$ and $g$ are isomorphisms, inverse to each other.</p>
<hr>
<p>It is known that for finite dimensional $V$, then $(V^*)^{\otimes m} = (V^{\otimes m})^*$. Then an element of $V^{* \otimes m} \otimes V^{\otimes n}$ is an element of $(V^{\otimes m})^* \otimes V^{\otimes n} = \hom(V^{\otimes m}, V^{\otimes n})$. So by definition / universal property of the tensor product, it's a multilinear map $V^m \to V^{\otimes n}$.</p>
| <p>One general form of that assertion (noting that it cannot be quite as simple as one might imagine) is the Cartan-Eilenberg adjunction
<span class="math-container">$$
\mathrm{Hom}(X\otimes Y,Z)\;\approx\;\mathrm{Hom}(X,\mathrm{Hom}(Y,Z))
$$</span>
in some reasonable additive (or whatever) category, where, significantly, the tensor product must be a <em>genuine</em> categorical tensor product, as opposed to a "projective" or "injective" tensor product, which have only half the properties of a genuine tensor product. So, for example, there is <em>no</em> genuine tensor product of (infinite-dimensional) Hilbert spaces in <em>any</em> reasonable category of topological vector spaces. (The thing often called the "Hilbert-space tensor product" has only half the requisite properties.</p>
|
differentiation | <p>I thought I had a good idea on why/how implicit differentiation works until I read the following passage in my Calculus book:</p>
<blockquote>
<p>Furthermore, implicit differentiation works just as easily for equations such as
$$x^5+5x^4y^2+3xy^3+y^5=1$$
which are actually <i>impossible</i> to solve for $y$ in terms of $x$</p>
</blockquote>
<p>My problem with it is the following:</p>
<p>The way we go about differentiating, for instance, $xy=1$ is by differentiating the whole equation through with respect to $x$ and treating $y$ as a function of $x$. But (and at least that's how I see it) we can only treat $y$ as $f(x)$ because the equation <i>determines</i> $y$ as a function of $x$ in a relation that <i>can</i> ben written expliclity (in this case, $y=\frac{1}{x}$). If we have an equation such as the quoted one, in which we just can't solve for $y$, doesn't that mean that $y$ is <i>not</i> a function of $x$? In such a case, wouldn't treating it as such be an invalid move?</p>
<p>I hope I have made myself understood. Any clarification will be appreciated. Thanks</p>
| <p>At a basic level, I think this question is really about the difference between saying that something <em>exists</em>, on the one hand, and being able to <em>write a formula for it</em> on the other. It's important to distinguish between three different (but closely related) ideas:</p>
<ol>
<li>It may be that the relationship between $x$ and $y$ does not define $y$ as a function of $x$ because there is more than one $y$-value associated to a given $x$-value. To take a simple example, the equation of a unit circle ($x^2+y^2=1$) does not define $y$ as a function of $x$.</li>
<li>However, even if $y$ is not <em>globally</em> a function of $x$, it is nevertheless possible that <em>locally</em> (i.e. in the vicinity of some point) there may be a function of $x$ that "matches" the graph of the relationship, in a precise sense. For example, the point $(0.6, -0.8)$ lies on the bottom half of the unit circle. and the function $f(x) = -\sqrt{1-x^2}$ is a local solution for $y$ in terms of $x$ that includes that point. The <em><a href="https://en.wikipedia.org/wiki/Implicit_function_theorem" rel="noreferrer">Implicit Function Theorem</a></em> provides conditions under which such a function exists.</li>
<li>On the other hand, even when such a local function <em>exists</em>, it may be impossible to write down an <em>explicit formula</em> for it. That, I think, is what your textbook means by "impossible to solve". It's not that the function doesn't exist, but rather that there is no way to write down a formula for the function.</li>
</ol>
<p>To elaborate on this last point, consider the implicitly defined relation
$$ y^3 + 2^y = \cos(2\pi x^2) + x $$
This relationship implicitly defines $y$ as a function of $x$: choose any specific value of $x$, say $x=2$. Then the right-hand side of the equation is $\cos(2\pi\cdot4) + 2 = 3$. The equation then asks us to find a value of $y$ such that $y^3 + 2^y = 3$. Such a $y$ is guaranteed to exist, and is in fact unique, as you can convince yourself of by looking at the graph of the function $h(t) = t^3 + 2^t$ (it's strictly increasing because its derivative is always positive, and its range is $(-\infty ,\infty)$ . And there's nothing special about the choice $x=2$ in this example; choose <em>any</em> value of $x$, and there is a unique $y$ value associated to that value of $x$ by the relation $ y^3 + 2^y = \cos(2\pi x^2) + x $. In fact you can see the graph of this implicitly-defined function below.
<a href="https://i.sstatic.net/VzVgi.png" rel="noreferrer"><img src="https://i.sstatic.net/VzVgi.png" alt="enter image description here"></a></p>
<p>But go ahead, try to find a <em>formula</em> for explicitly calculating $y$ in terms of $x$. I'll wait.</p>
<p>(Okay, this is the point where someone jumps into the comments and says "Well, actually..." and goes on to explain that you <em>can</em> explicitly calculate $y$ in terms of $x$ by introducing a Lambert function or something. Let me try to pre-empt that by arguing that such a"solution" just sweeps the implicitness under the rug. In any case it misses the point of the example, which is that a relationship may implicitly define a function even if you lack an explicit formula for computing one variable in terms of the other.)</p>
<p>On the other hand, consider this closely-related example:
$$ y^2 + 2^y = \cos(2\pi x^2) + x $$
(The only change is the exponent on the $y$ on the left-hand side.) This relationship most definitely does <em>not</em> define $y$ as a function of $x$, as can be seen in the graph below:</p>
<p><a href="https://i.sstatic.net/YiT37.png" rel="noreferrer"><img src="https://i.sstatic.net/YiT37.png" alt="enter image description here"></a></p>
<p>We can see that for many values of $x$ there are two different $y$ values that both satisfy $ y^2 + 2^y = \cos(2\pi x^2) + x $, so this is not a function. Nevertheless if we choose a point on the graph — $(2,1)$ is a convenient one — and zoom in on a neighborhood of that point, it <em>locally</em> looks like a function:</p>
<p><a href="https://i.sstatic.net/ZNxoD.png" rel="noreferrer"><img src="https://i.sstatic.net/ZNxoD.png" alt="enter image description here"></a></p>
<p>The power of implicit differentiation as a technique is precisely that it allows us to find an <em>explicit</em> formula for the slope of the tangent line at $(x,y)$, even when we can't find an explicit formula for $y$ in terms of $x$, <em>and even when the "function" isn't really a function at all</em>. (The trade-off is that the "explicit" formula for the slope is expressed in terms of both variables, so there is still some lurking implicitness in the problem.)</p>
| <p>That's an excellent question. </p>
<p>Part of the <em>conclusion</em> of the Implicit Differentiation Theorem (perhaps better known as the Implicit Function Theorem) is that the equation <strong>locally</strong> defines $y$ as a function of $x$.</p>
<p>The theorem says, roughly, this. Suppose that $(x,y)=(a,b)$ lies in the solution set, and suppose furthermore that the partial derivative $\frac{\partial F}{\partial y}$ of the left hand side is nonzero at $(a,b)$. Then there exists an open ball $B \subset \mathbb{R}^2$ of some positive radius $\epsilon>0$ centered on the point $(a,b)$, such that the intersection of the solution set with the ball $B$ is, indeed, the graph of a differentiable function $y=f(x)$, and furthermore its derivative $\frac{dy}{dx}$ can be calculated using the method you learned in calculus.</p>
<p>Now you may ask: how do we find a formula for $y=f(x)$? </p>
<p>There's not really an answer there. In general you cannot find the formula, although if you follow through the proof of the theorem then you can find numerical approximation methods. That's the real power of the Implicit Differentiation Theorem, proving the existence of a differentiable function without even being able to write down a formula for it.</p>
|
combinatorics | <p>I need to calculate the possible combinations for 8 characters password. The password must contain at least one of the following: (lower case letters, upper case letters, digits, punctuations, special characters).</p>
<p>Assume I have 95 ascii characters (lower case letters, upper case letters, digits, punctuations, special characters).</p>
<ul>
<li>lower case letters = $26$</li>
<li>upper case letters = $26$</li>
<li>digits = $10$</li>
<li>punctuations & special characters = $33$</li>
</ul>
<p>The general formula for the possible passwords that I can from from these 95 characters is: $95^8$.</p>
<p>But, accurately, I feel the above formula is incorrect. Please, correct me.
The password policy requires at least one of the listed above ascii characters. Therefore, the password possible combinations = $(26)*(26)*(10)*(33)*(95)*(95)*(95)*(95)$</p>
<p>Which calculation is correct?</p>
<p>EDIT: Please, note that I mean 8 characters password and exactly 8. Also, There is no order specified (i.e. it could start with small letter, symbol, etc.). But it should contain at least one of the specified characters set (upper case, lower case, symbol, no., etc.).</p>
| <p>Start with all $8$-character strings: $95^8$</p>
<p>Then remove all passwords with no lowercase ($69^8$), all passwords with no uppercase ($69^8$), all passwords with no digit ($85^8$) and all passwords with no special character ($62^8$).</p>
<p>But then you removed some passwords twice. You must add back all passwords with:</p>
<ul>
<li>no lowercase AND no uppercase: $43^8$</li>
<li>no lowercase AND no digit: $59^8$</li>
<li>no lowercase AND no special: $36^8$</li>
<li>no uppercase AND no digit: $59^8$</li>
<li>no uppercase AND no special: $36^8$</li>
<li>no digit AND no special: $52^8$</li>
</ul>
<p>But then you added back a few passwords too many times. For instance, an all-digit password was remove three times in the first step, then put back three times in the second step, so it must be removed again:</p>
<ul>
<li>only lowercase: $26^8$</li>
<li>only uppercase: $26^8$</li>
<li>only digits: $10^8$</li>
<li>only special: $33^8$</li>
</ul>
<p>Grand total: $95^8 - 69^8 - 69^8 - 85^8 - 62^8 + 43^8 + 59^8 + 36^8 + 59^8 + 36^8 + 52^8 - 26^8 - 26^8 - 10^8 - 33^8 = 3025989069143040 \approx 3.026\times10^{15}$</p>
| <p>The answer I get is 0.3051925477389360000E+16 = 3,051,925,477,389,360. This answer is taken from the answer at <a href="https://math.stackexchange.com/questions/2103361/derive-an-algorithm-for-computing-the-number-of-restricted-passwords-for-the-gen#2103361">Derive an algorithm for computing the number of restricted passwords for the general case?</a> . This problem is more difficult than it appears. $95^8$ is not the right answer. (26)∗(26)∗(10)∗(33)∗(95)∗(95)∗(95)∗(95) is not the right answer and you can't just multiply this product by the number of permutations. This is because if for example the punctuation special character first appears in the fifth position in the password with the first lower case letter, first upper case letter, and first digit appearing in the first 3 positions of the password then there are only $(95-33) = 62$ characters that can appear in the fourth position of the password. Because of this problem a different approach is necessary. An approach that gives the right answer is done by summing as follows:</p>
<p>The number of passwords for permutation $I_k$ with password positions j1, j2, j3, and j4 is </p>
<p>$$S(I_k,j1,j2,j3,j4)=f1*g_{I(1)}*(n_t-(g_{I(2)}+g_{I(3)}+g_{I(4)}))^{(j2-1-j1)}*g_{I(2)}*(n_t-(g_{I(3)}+g_{I(4)}))^{(j3-1-j2)}*gI_{(3)}*(n_t-g_{I(4)})^{(j4-1-j3)}*g_{I(4)}*n_t^{(n-j4)}$$ .</p>
<p>$$S(I_k)=\sum_{j4=4}^n \sum_{j3=3}^{j4-1} \sum_{j2=2}^{j3-1} \sum_{j1=1}^{j2-1} S(I_k,j1,j2,j3,j4)$$</p>
<p>gives the number of passwords corresponding to permutation, $I_k$. </p>
<p>$$Total=\sum_{k=1}^{24} S(I_k)$$
gives the total number of passwords satisfying the requirement.
The algorithm for computing the number of passwords meeting the requirement is more completely described at the web site indicated above.</p>
|
probability | <p>What is the average number of times it would it take to roll a fair <span class="math-container">$6$</span>-sided die and get all numbers on the die? The order in which the numbers appear does not matter.</p>
<p>I had this questions explained to me by a professor (not math professor), but it was not clear in the explanation. We were given the answer <span class="math-container">$(1-(\frac56)^n)^6 = .5$</span> or <span class="math-container">$n = 12.152$</span></p>
<p>Can someone please explain this to me, possibly with a link to a general topic?</p>
| <p>The time until the first result appears is $1$. After that, the random time until a second (different) result appears is geometrically distributed with parameter of success $5/6$, hence with mean $6/5$ (recall that the mean of a geometrically distributed random variable is the inverse of its parameter). After that, the random time until a third (different) result appears is geometrically distributed with parameter of success $4/6$, hence with mean $6/4$. And so on, until the random time of appearance of the last and sixth result, which is geometrically distributed with parameter of success $1/6$, hence with mean $6/1$. This shows that the mean total time to get all six results is
$$\sum_{k=1}^6\frac6k=\frac{147}{10}=14.7.$$</p>
<hr>
<p><em>Edit:</em> This is called the <a href="http://en.wikipedia.org/wiki/Coupon_collector%27s_problem">coupon collector problem</a>. For a fair $n$-sided die, the expected number of attempts needed to get all $n$ values is
$$n\sum_{k=1}^n\frac1k,$$ which, for large $n$, is approximately $n\log n$. This stands in contrast with the mean time needed to complete only some proportion $cn$ of the whole collection, for some fixed $c$ in $(0,1)$, which, for large $n$, is approximately $-\log(1-c)n$. One sees that most of the $n\log n$ steps needed to complete the full collection are actually spent completing the last one per cent, or the last one per thousand, or the last whatever percentage of the collection.</p>
| <p>Here's the logic:</p>
<p>The chance of rolling a number you haven't yet rolled when you start off is $1$, as any number would work. Once you've rolled this number, your chance of rolling a number you haven't yet rolled is $5/6$. Continuing in this manner, after you've rolled $n$ different numbers the chance of rolling one you haven't yet rolled is $(6-n)/6$.</p>
<p>You can figure out the mean time it takes for a result of probability $p$ to appear with a simple formula: $1/p$. Furthermore, the mean time it takes for multiple results to appear is the sum of the mean times for each individual result to occur.</p>
<p>This allows us to calculate the mean time required to roll every number:
$t = 1/1 + 6/5 + 6/4 + 6/3 + 6/2 + 6/1 = 1 + 12/10 + 15/10 + 2 + 3 + 6 = 12 + 27/10 = 14.7$</p>
|
linear-algebra | <p>What is an intuitive explanation of a positive-semidefinite matrix? Or a simple example which gives more intuition for it rather than the bare definition. Say $x$ is some vector in space and $M$ is some operation on vectors.</p>
<p>The definition is:</p>
<p>A $n$ × $n$ Hermitian matrix M is called <em>positive-semidefinite</em> if</p>
<p>$$x^{*} M x \geq 0$$</p>
<p>for all $x \in \mathbb{C}^n$ (or, all $x \in \mathbb{R}^n$ for the real matrix), where $x^*$ is the conjugate transpose of $x$.</p>
| <p>One intuitive definition is as follows. Multiply any vector with a positive semi-definite matrix. The angle between the original vector and the resultant vector will always be less than or equal $\frac{\pi}{2}$. The positive definite matrix tries to keep the vector within a certain half space containing the vector. This is analogous to what a positive number does to a real variable. Multiply it and it only stretches or contracts the number but never reflects it about the origin.</p>
| <p>First I'll tell you how I think about Hermitian positive-definite matrices. A Hermitian positive-definite matrix $M$ defines a sesquilinear inner product $\langle Mv, w \rangle = \langle v, Mw \rangle$, and in fact every inner product on a finite-dimensional inner product space $V$ has this form. In other words it is a way of computing angles between vectors, or a way of projecting vectors onto other vectors; over the real numbers it is the key ingredient to doing Euclidean geometry. An inner product can be recovered from the norm $\langle Mv, v \rangle = \langle v, Mv \rangle$ it induces, and a norm in turn can be recovered from its unit sphere $\{ v : \langle Mv, v \rangle = 1 \}$. This unit sphere is a distorted version of the usual unit sphere; the distortions will occur along axes corresponding to the eigenvectors of $M$, and the amount of distortion corresponds to the (inverses of the) corresponding eigenvalues. For example when $\dim V = 2$ it is an ellipse and when $\dim V = 3$ it is an ellipsoid.</p>
<p>A Hermitian positive-semidefinite matrix $M$ no longer describes an inner product because it is not necessarily positive-definite, but it still defines a sesquilinear form. It also defines a function $\langle Mv, v \rangle$ which is no longer a norm because it is not necessarily positive-definite; some people call these "pseudonorms," I think. The corresponding unit sphere $\{ v : \langle Mv, v \rangle = 1 \}$ might now be lower-dimensional than the usual unit sphere, depending on how many eigenvalues are equal to zero; for example if $\dim V = 3$ it might be an ellipsoid, or an ellipse, or two points. </p>
|
matrices | <p>A Hermitian matrix always has real eigenvalues and real or complex orthogonal eigenvectors. A real symmetric matrix is a special case of Hermitian matrices, so it too has orthogonal eigenvectors and real eigenvalues, but could it ever have complex eigenvectors?</p>
<p>My intuition is that the eigenvectors are always real, but I can't quite nail it down.</p>
| <p>Always try out examples, starting out with the simplest possible examples (it may take some thought as to which examples are the simplest). Does for instance the identity matrix have complex eigenvectors? This is pretty easy to answer, right? </p>
<p>Now for the general case: if $A$ is any real matrix with real eigenvalue $\lambda$, then we have a choice of looking for real eigenvectors or complex eigenvectors. The theorem here is that the $\mathbb{R}$-dimension of the space of real eigenvectors for $\lambda$ is equal to the $\mathbb{C}$-dimension of the space of complex eigenvectors for $\lambda$. It follows that (i) we will always have non-real eigenvectors (this is easy: if $v$ is a real eigenvector, then $iv$ is a non-real eigenvector) and (ii) there will always be a $\mathbb{C}$-basis for the space of complex eigenvectors consisting entirely of real eigenvectors.</p>
<p>As for the proof: the $\lambda$-eigenspace is the kernel of the (linear transformation given by the) matrix $\lambda I_n - A$. By the rank-nullity theorem, the dimension of this kernel is equal to $n$ minus the rank of the matrix. Since the rank of a real matrix doesn't change when we view it as a complex matrix (e.g. the reduced row echelon form is unique so must stay the same upon passage from $\mathbb{R}$ to $\mathbb{C}$), the dimension of the kernel doesn't change either. Moreover, if $v_1,\ldots,v_k$ are a set of real vectors which are linearly independent over $\mathbb{R}$, then they are also linearly independent over $\mathbb{C}$ (to see this, just write out a linear dependence relation over $\mathbb{C}$ and decompose it into real and imaginary parts), so any given $\mathbb{R}$-basis for the eigenspace over $\mathbb{R}$ is also a $\mathbb{C}$-basis for the eigenspace over $\mathbb{C}$.</p>
| <p>If $A$ is a symmetric $n\times n$ matrix with real entries, then viewed as an element of $M_n(\mathbb{C})$, its eigenvectors always include vectors with non-real entries: if $v$ is any eigenvector then at least one of $v$ and $iv$ has a non-real entry.</p>
<p>On the other hand, if $v$ is any eigenvector then at least one of $\Re v$ and $\Im v$ (take the real or imaginary parts entrywise) is non-zero and will be an eigenvector of $A$ with the same eigenvalue. So you can always pass to eigenvectors with real entries.</p>
|
matrices | <p>How can I understand that $A^TA$ is invertible if $A$ has independent columns? I found a similar <a href="https://math.stackexchange.com/questions/1181271/if-ata-is-invertible-then-a-has-linearly-independent-column-vectors">question</a>, phrased the other way around, so I tried to use the theorem</p>
<p>$$
rank(A^TA) \le min(rank(A^T),rank(A))
$$</p>
<p>Given $rank(A) = rank(A^T) = n$ and $A^TA$ produces an $n\times n$ matrix, I can't seem to prove that $rank(A^TA)$ is actually $n$.</p>
<p>I also tried to look at the question another way with the matrices</p>
<p>$$
A^TA
= \begin{bmatrix}a_1^T \\ a_2^T \\ \ldots \\ a_n^T \end{bmatrix}
\begin{bmatrix}a_1 a_2 \ldots a_n \end{bmatrix}
= \begin{bmatrix}A^Ta_1 A^Ta^2 \ldots A^Ta_n\end{bmatrix}
$$</p>
<p>But I still can't seem to show that $A^TA$ is invertible. So, how should I get a better understanding of why $A^TA$ is invertible if $A$ has independent columns?</p>
| <p>Consider the following:
$$A^TAx=\mathbf 0$$
Here, $Ax$, an element in the range of $A$, is in the null space of $A^T$. However, the null space of $A^T$ and the range of $A$ are orthogonal complements, so $Ax=\mathbf 0$.</p>
<p>If $A$ has linearly independent columns, then $Ax=\mathbf 0 \implies x=\mathbf 0$, so the null space of $A^TA=\{\mathbf 0\}$. Since $A^TA$ is a square matrix, this means $A^TA$ is invertible.</p>
| <p>If $A $ is a real $m \times n $ matrix then $A $ and $A^T A $ have the same null space. Proof: $A^TA x =0\implies x^T A^T Ax =0 \implies (Ax)^TAx=0 \implies \|Ax\|^2 = 0 \implies Ax = 0 $. </p>
|
logic | <p>Assuming ZFC has a model, is there a model of ZFC such that in that model, ZFC has no model?</p>
<p>Also, assuming ZFC has a model, is there a model of ZFC such that in that model, ZFC is inconsistent?</p>
| <p>Yes.</p>
<p>Recall that by the Completeness Theorem, having a model and being consistent are the same thing. Also, by Incompleteness, ZFC doesn't prove its own consistency. Finally, ZFC can prove the Soundness Theorem - that an inconsistent theory has no models!</p>
<p>So - assuming ZFC has a model - ZFC is consistent. If ZFC is consistent, then ZFC can't prove "ZFC is consistent." By completeness, this means there's a model of ZFC satisfying "ZFC is inconsistent." Since ZFC proves the Soundness Theorem, this model must think that ZFC has no model!</p>
<hr>
<p>Note that there's a subtlety to the claim "There is a model $M$ of ZFC not containing a model of ZFC." What that means is that there is an $M$ which satisfies ZFC+"There is no model of ZFC." However, since ZFC is not finitely axiomatizable, what $M$ thinks ZFC is will be different from what ZFC actually is! And in fact every model $M$ of ZFC contains a structure $N$ which <em>classically</em> is a model of ZFC, but <em>within $M$</em> appears to not be a model of ZFC! Neat, huh? See Joel David Hamkins' answer to <a href="https://mathoverflow.net/questions/51754/clearing-misconceptions-defining-is-a-model-of-zfc-in-zfc">https://mathoverflow.net/questions/51754/clearing-misconceptions-defining-is-a-model-of-zfc-in-zfc</a>.</p>
| <p>Noah gave the correct answer. It is true, that if $\sf ZFC$ has any models, then there are models of $\sf ZFC$ in which there are no models of $\sf ZFC$.</p>
<p>But let me point out an important piece of information here.</p>
<p>We formulate statements about consistency in terms of natural numbers. This is because the language of set theory is finite, so we can encode formulas as natural numbers in a very robust way. How robust? Well, like a computer-program robust. It's a mechanical algorithm to move on thing into the other.</p>
<p>Why am I bringing this up? Because it serves the point that natural numbers are important to the statement that something is consistent or inconsistent. </p>
<p>More specifically, we work in a universe $V$. In $V$, it is true that $\sf ZFC$ is consistent. Now, it is important to point out that by this we mean "what $V$ interprets as the predicate which defines '$\sf ZFC$' in its own natural numbers". But we can show that if $M$ is a model of $\sf ZFC$ inside $V$, and $M$ and $V$ agree on the notion of "natural number", then both have the same first-order theory for the natural numbers.</p>
<p>In particular, the predicate '$\sf ZFC$' and the statement of its consistency are absolute between $V$ and $M$. And since $V$ thinks that $\sf ZFC$ is consistent, so must $M$.</p>
<p>This means that if $M$ is a model of $\sf ZFC$ which thinks that $\sf ZFC$ is inconsistent two things must be true:</p>
<ol>
<li>$M$ has non-standard integers (compared to $V$).</li>
<li>The interpretation of $\sf ZFC$ inside $M$ is different the one in $V$. </li>
</ol>
<p>Namely, '$\sf ZFC$' can be seen as a predicate in the language of arithmetic, and since in $M$ we have <em>more</em> natural numbers, this predicate is interpreted to have <em>more</em> axioms than it should really have; and first-order logic have <em>more</em> inference rules; and proofs can suddenly be "too long" as far as $V$ is concerned. All these things allow $M$ to find out a proof of contradiction from $\sf ZFC$.</p>
<p>This is the reason why we can find such models. And models whose natural numbers agree with that of the universe are called $\omega$-models, and in such models it is necessarily the case that $\sf ZFC$ is consistent, simply because the natural numbers agree with the universe, so the interpretation of '$\sf ZFC$' and the rules of first-order logic, are the same as in $V$. Since in $V$ we know that $\sf ZFC$ is consistent, it must be consistent in any $\omega$-model. </p>
<p>So while $\sf ZFC$ proves that if there are models of $\sf ZFC$, then there are models of $\sf ZFC+\lnot\operatorname{Con}(ZFC)$; it is not the case that $\sf ZFC$ proves that if there are models of $\sf ZFC$, then there are $\omega$-models of $\sf ZFC$, as this would be a violation of the incompleteness theorem.</p>
|
probability | <p>Of course, we've all heard the colloquialism "If a bunch of monkeys pound on a typewriter, eventually one of them will write Hamlet."</p>
<p>I have a (not very mathematically intelligent) friend who presented it as if it were a mathematical fact, which got me thinking... Is this really true? Of course, I've learned that dealing with infinity can be tricky, but my intuition says that time is countably infinite while the number of works the monkeys could produce is uncountably infinite. Therefore, it isn't necessarily given that the monkeys would write Hamlet.</p>
<p>Could someone who's better at this kind of math than me tell me if this is correct? Or is there more to it than I'm thinking?</p>
| <p>I found online the claim (which we may as well accept for this purpose) that there are $32241$ words in Hamlet. Figuring $5$ characters and one space per word, this is $193446$ characters. If the character set is $60$ including capitals and punctuation, a random string of $193446$ characters has a chance of $1$ in $60^{193446}$ (roughly $1$ in $10^{344000}$) of being Hamlet. While very small, this is greater than zero. So if you try enough times, and infinity times is certainly enough, you will probably produce Hamlet. But don't hold your breath. It doesn't even take an infinite number of monkeys or an infinite number of tries. Only a product of $10^{344001}$ makes it very likely. True, this is a very large number, but most numbers are larger.</p>
| <p>Some references (I am mildly surprised that no one has done this yet). This is called the <a href="http://en.wikipedia.org/wiki/Infinite_monkey_theorem">infinite monkey theorem</a> in the literature. It follows from the second <a href="http://en.wikipedia.org/wiki/Borel%E2%80%93Cantelli_lemma">Borel-Cantelli lemma</a> and is related to <a href="http://en.wikipedia.org/wiki/Kolmogorov%27s_zero-one_law">Kolmogorov's zero-one law</a>, which is the result that provides the intuition behind general statements like this. (The zero-one law tells you that the probability of getting Hamlet is either zero or one, but doesn't tell you which. This is usually the hard part of applying the zero-one law.) Since others have addressed the practical side, I am telling you what the mathematical idealization looks like.</p>
<blockquote>
<p>my intuition says that time is countably infinite while the number of works the monkeys could produce is uncountably infinite.</p>
</blockquote>
<p>This is a good idea! Unfortunately, the number of finite strings from a finite alphabet is countable. This is a good exercise and worth working out yourself.</p>
<p>Edit: also, regarding some ideas which have come up in the discussions on other answers, Jorge Luis Borges' short story <a href="http://en.wikipedia.org/wiki/The_Library_of_Babel">The Library of Babel</a> is an interesting read.</p>
|
logic | <blockquote>
<p><em><strong>Theorem 1</strong> [ZFC, classical logic]:</em> If $A,B$ are sets such that $\textbf{2}\times A\cong \textbf{2}\times B$, then $A\cong B$.</p>
</blockquote>
<p>That's because the axiom of choice allows for the definition of cardinality $|A|$ of any set $A$, and for $|A|\geq\aleph_0$ we have $|\textbf{2}\times A|=|A|$.</p>
<blockquote>
<p><em><strong>Theorem 2</strong>:</em> Theorem 1 still holds in ZF with classical logic.</p>
</blockquote>
<p>This is less trivial and explained in Section 5 of <a href="https://math.dartmouth.edu/~doyle/docs/three/three.pdf" rel="noreferrer">Division by Three</a> - however, though the construction does not involve any choices, it <em>does</em> involve the law of excluded middle.</p>
<blockquote>
<p><strong><em>Question:</em></strong> Are there <em>intuitionistic</em> set theories in which one can prove $$\textbf{2}\times A\cong \textbf{2}\times B\quad\Rightarrow\quad A\cong B\quad\text{?}$$ </p>
</blockquote>
<p>For example, is this statement true in elementary topoi or can it be proved in some intuitionistic type theory?</p>
<blockquote>
<p>In his comment below Kyle indicated that the statement is unprovable in some type theory - does somebody know the argument or a reference for that?</p>
</blockquote>
<p><em>Edit</em> See also the related question <a href="https://math.stackexchange.com/questions/1114752/does-a-times-a-cong-b-times-b-imply-a-cong-b">Does $A\times A\cong B\times B$ imply $A\cong B$?</a> about 'square roots'</p>
| <p>There was a paper recently posted to arXiv about this question: Swan, <a href="https://arxiv.org/abs/1804.04490" rel="noreferrer"><em>On Dividing by Two in Constructive Mathematics</em></a>.</p>
<p>It turns out that there are examples of toposes where you can't divide by two.</p>
| <p>(The following is not really an answer, or just a very partial one, but it's definitely relevant and too long for a comment.)</p>
<p>There is a theorem of Richard Friedberg ("<a href="https://eudml.org/doc/169898" rel="nofollow noreferrer">The uniqueness of finite division for recursive equivalence types</a>", <em>Math. Z.</em> <strong>75</strong> (1961), 3–7) which goes as follows (all of this is in classical logic):</p>
<p>For $A$ and $B$ subsets of $\mathbb{N}$, define $A \sim B$ when there exists a partial computable function $f:\mathbb{N}\rightharpoonup\mathbb{N}$ that is one-to-one on its domain and defined at least on all of $A$ such that $f(A) = B$. (One also says that $A$ and $B$ are <em>computably equivalent</em> or <em>recursively equivalent</em>, and it is indeed an equivalence relation, not to be confused with "computably/recursively isomorphic", <a href="https://mathoverflow.net/questions/228599/partial-computably-isomorphic-sets">see here</a>.) Then [Friedberg's theorem states]: if $n$ is a positive integer then $(n\times A) \sim (n \times B)$ implies $A\sim B$ (here, $n\times A$ is the set of natural numbers coding pairs $(i,k)$ where $0\leq i<n$ and $k\in A$ for some standard coding of pairs of natural numbers by natural numbers).</p>
<p>To make this assertion closer to the question asked here, subsets of $\mathbb{N}$ can be considered as objects, indeed subobjects of $\mathcal{N}$, in the <a href="https://en.wikipedia.org/wiki/Effective_topos" rel="nofollow noreferrer">effective topos</a> (an elementary topos with n.n.o. $\mathcal{N}$ such that all functions $\mathcal{N}\to\mathcal{N}$ are computable), in fact, these subobjects are exactly those classified by maps $\mathcal{N} \to \Omega_{\neg\neg}$ where $\Omega_{\neg\neg} = \nabla 2$ is the subobject of the truth values $p\in\Omega$ such that $\neg\neg p = p$; moreover, to say that two such objects are isomorphic, or internally isomorphic, in the effective topos, is equivalent to saying that $A$ and $B$ are computably isomorphic as above. So Friedberg's result can be reinterpreted by saying that if $A$ and $B$ are such objects of the effective topos and if $n\times A$ and $n\times B$ are isomorphic then $A$ and $B$ are.</p>
<p>I'm not sure how much this can be internalized (e.g., does the effective topos validate "if $A$ and $B$ are $\neg\neg$-stable sets of natural numbers and $n\times A$ is isomorphic to $n\times B$ then $A$ is isomorphic to $B$" for explicit $n$? and how about for $n$ quantified inside the topos?) or generalized (do we really need $\neg\neg$-stability?). But this may be worth looking into, and provides at least a positive kind-of-answer to the original question.</p>
|
logic | <p>Today I had an argument with my math teacher at school. We were answering some simple True/False questions and one of the questions was the following:</p>
<p><span class="math-container">$$x^2\ne x\implies x\ne 1$$</span></p>
<p>I immediately answered true, but for some reason, everyone (including my classmates and math teacher) is disagreeing with me. According to them, when <span class="math-container">$x^2$</span> is not equal to <span class="math-container">$x$</span>, <span class="math-container">$x$</span> also can't be <span class="math-container">$0$</span> and because <span class="math-container">$0$</span> isn't excluded as a possible value of <span class="math-container">$x$</span>, the sentence is false. After hours, I am still unable to understand this ridiculously simple implication. I can't believe I'm stuck with something so simple.<br><br>
<strong>Why I think the logical sentence above is true:</strong><br>
My understanding of the implication symbol <span class="math-container">$\implies$</span> is the following:
If the left part is true, then the right part must be also true. If the left part is false, then nothing is said about the right part. In the right part of this specific implication nothing is said about whether <span class="math-container">$x$</span> can be <span class="math-container">$0$</span>. Maybe <span class="math-container">$x$</span> can't be <span class="math-container">$-\pi i$</span> too, but as I see it, it doesn't really matter, as long as <span class="math-container">$x \ne 1$</span> holds. And it always holds when <span class="math-container">$x^2 \ne x$</span>, therefore the sentence is true.</p>
<h3>TL;DR:</h3>
<p><strong><span class="math-container">$x^2 \ne x \implies x \ne 1$</span>: Is this sentence true or false, and why?</strong></p>
<p>Sorry for bothering such an amazing community with such a simple question, but I had to ask someone.</p>
| <p>The short answer is: Yes, it is true, because the contrapositive just expresses the fact that $1^2=1$.</p>
<p>But in controversial discussions of these issues, it is often (but not always) a good idea to try out non-mathematical examples:</p>
<hr>
<p>"If a nuclear bomb drops on the school building, you die."</p>
<p>"Hey, but you die, too."</p>
<p>"That doesn't help you much, though, so it is still true that you die." </p>
<hr>
<p>"Oh no, if the supermarket is not open, I cannot buy chocolate chips cookies."</p>
<p>"You cannot buy milk and bread, either!"</p>
<p>"Yes, but I prefer to concentrate on the major consequences."</p>
<hr>
<p>"If you sign this contract, you get a free pen."</p>
<p>"Hey, you didn't tell me that you get all my money."</p>
<p>"You didn't ask."</p>
<hr>
<p>Non-mathematical examples also explain the psychology behind your teacher's and classmates' thinking. In real-life, the choice of consequences is usually a loaded message and can amount to a lie by omission. So, there is this lingering suspicion that the original statement suppresses information on 0 on purpose. </p>
<p>I suggest that you learn about some nonintuitive probability results and make bets with your teacher.</p>
| <p>First, some general remarks about logical implications/conditional statements. </p>
<ol>
<li><p>As you know, $P \rightarrow Q$ is true when $P$ is false, or when $Q$ is true. </p></li>
<li><p>As mentioned in the comments, the <em>contrapositive</em> of the implication $P \rightarrow Q$, written $\lnot Q \rightarrow \lnot P$, is logically equivalent to the implication. </p></li>
<li><p>It is possible to write implications with merely the "or" operator. Namely, $P \rightarrow Q$ is equivalent to $\lnot P\text{ or }Q$, or in symbols, $\lnot P\lor Q$.</p></li>
</ol>
<p>Now we can look at your specific case, using the above approaches. </p>
<ol>
<li>If $P$ is false, ie if $x^2 \neq x$ is false (so $x^2 = x$ ), then the statement is true, so we assume that $P$ is true. So, as a statement, $x^2 = x$ is false. Your teacher and classmates are rightly convinced that $x^2 = x$ is equivalent to ($x = 1$ or $x =0\;$), and we will use this here.
If $P$ is true, then ($x=1\text{ or }x =0\;$) is false. In other words, ($x=1$) AND ($x=0\;$) are both false. I.e., ($x \neq 1$) and ($x \neq 0\;$) are true.
I.e., if $P$, then $Q$. <br></li>
<li>The contrapositive is $x = 1 \rightarrow x^2 = x$. True.</li>
<li>We use the "sufficiency of or" to write our conditional as: $$\lnot(x^2 \neq x)\lor x \neq 1\;.$$ That is, $x^2 = x$ or $x \neq 1$,
which is $$(x = 1\text{ or }x =0)\text{ or }x \neq 1,$$ which is
$$(x = 1\text{ or }x \neq 1)\text{ or }x = 0\;,$$ which is
$$(\text{TRUE})\text{ or }x = 0\;,$$ which is true. </li>
</ol>
|
linear-algebra | <p>Let $A$ be a symmetric $n\times n$ matrix.</p>
<p>I found <a href="https://ece.uwaterloo.ca/~dwharder/NumericalAnalysis/04LinearAlgebra/posdef/" rel="noreferrer">a method on the web to check if $A$ is <strong>positive definite</strong></a>:</p>
<blockquote>
<p>$A$ is positive-definite if all the diagonal entries are positive, and
each diagonal entry is greater than the sum of the absolute values of all other entries in the corresponding row/column.</p>
</blockquote>
<p>I couldn't find a proof for this statement. I also couldn't find a reference in my linear algebra books.</p>
<p>I've a few questions.</p>
<ol>
<li><p>How do we prove the above statement?</p></li>
<li><p>Is the following <strong>slightly weaker</strong> statement true? </p></li>
</ol>
<blockquote>
<p>A symmetric matrix $A$ is positive-definite if all the diagonal entries are positive, each diagonal entry is greater than <strong>or equal to</strong> the sum of the absolute values of all other entries in the corresponding row/column, and there exists one diagonal entry which is strictly greater than the sum of the absolute values of all other entries in the corresponding row/column.</p>
</blockquote>
| <p>These matrices are called (strictly) <a href="http://en.wikipedia.org/wiki/Diagonally_dominant_matrix">diagonally dominant</a>. The standard way to show they are positive definite is with the <a href="http://en.wikipedia.org/wiki/Gershgorin_circle_theorem">Gershgorin Circle Theorem</a>. Your weaker condition does not give positive definiteness; a counterexample is $
\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{matrix} \right]
$.</p>
| <p>Before continuing, let me add the caution that a symmetric matrix can violate your rules and still be positive definite, give me a minute to check the eigenvalues</p>
<p><span class="math-container">$$ H \; = \;
\left( \begin{array}{rrr}
3 & 2 & 0 \\
2 & 3 & 2 \\
0 & 2 & 3
\end{array}
\right) .
$$</span>
This is positive by Sylvester's Law of Inertia,
<a href="https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia" rel="noreferrer">https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia</a></p>
<p>A proof is given <a href="https://planetmath.org/propertiesofdiagonallydominantmatrix" rel="noreferrer">here</a> as a consequence of Gershgorin's circle theorem. For additional information, see
<a href="http://en.wikipedia.org/wiki/Diagonally_dominant_matrix" rel="noreferrer">http://en.wikipedia.org/wiki/Diagonally_dominant_matrix</a> and <a href="http://mathworld.wolfram.com/DiagonallyDominantMatrix.html" rel="noreferrer">http://mathworld.wolfram.com/DiagonallyDominantMatrix.html</a>
or just Google "diagonally dominant symmetric"</p>
<p>Later methodology, amounting to repeated completing the square:</p>
<p><span class="math-container">$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$</span></p>
<p><span class="math-container">$$ P^T H P = D $$</span></p>
<p><span class="math-container">$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
- \frac{ 2 }{ 3 } & 1 & 0 \\
\frac{ 4 }{ 5 } & - \frac{ 6 }{ 5 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
3 & 2 & 0 \\
2 & 3 & 2 \\
0 & 2 & 3 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & - \frac{ 2 }{ 3 } & \frac{ 4 }{ 5 } \\
0 & 1 & - \frac{ 6 }{ 5 } \\
0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
3 & 0 & 0 \\
0 & \frac{ 5 }{ 3 } & 0 \\
0 & 0 & \frac{ 3 }{ 5 } \\
\end{array}
\right)
$$</span>
<span class="math-container">$$ $$</span></p>
<p><span class="math-container">$$ Q^T D Q = H $$</span></p>
<p><span class="math-container">$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
\frac{ 2 }{ 3 } & 1 & 0 \\
0 & \frac{ 6 }{ 5 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
3 & 0 & 0 \\
0 & \frac{ 5 }{ 3 } & 0 \\
0 & 0 & \frac{ 3 }{ 5 } \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & \frac{ 2 }{ 3 } & 0 \\
0 & 1 & \frac{ 6 }{ 5 } \\
0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
3 & 2 & 0 \\
2 & 3 & 2 \\
0 & 2 & 3 \\
\end{array}
\right)
$$</span></p>
<p><span class="math-container">$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$</span></p>
<p>Algorithm discussed at <a href="http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr">reference for linear algebra books that teach reverse Hermite method for symmetric matrices</a><br />
<a href="https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia" rel="noreferrer">https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia</a><br />
<span class="math-container">$$ H = \left(
\begin{array}{rrr}
3 & 2 & 0 \\
2 & 3 & 2 \\
0 & 2 & 3 \\
\end{array}
\right)
$$</span>
<span class="math-container">$$ D_0 = H $$</span>
<span class="math-container">$$ E_j^T D_{j-1} E_j = D_j $$</span>
<span class="math-container">$$ P_{j-1} E_j = P_j $$</span>
<span class="math-container">$$ E_j^{-1} Q_{j-1} = Q_j $$</span>
<span class="math-container">$$ P_j Q_j = Q_j P_j = I $$</span>
<span class="math-container">$$ P_j^T H P_j = D_j $$</span>
<span class="math-container">$$ Q_j^T D_j Q_j = H $$</span></p>
<p><span class="math-container">$$ H = \left(
\begin{array}{rrr}
3 & 2 & 0 \\
2 & 3 & 2 \\
0 & 2 & 3 \\
\end{array}
\right)
$$</span></p>
<p>==============================================</p>
<p><span class="math-container">$$ E_{1} = \left(
\begin{array}{rrr}
1 & - \frac{ 2 }{ 3 } & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
$$</span>
<span class="math-container">$$ P_{1} = \left(
\begin{array}{rrr}
1 & - \frac{ 2 }{ 3 } & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
, \; \; \; Q_{1} = \left(
\begin{array}{rrr}
1 & \frac{ 2 }{ 3 } & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
, \; \; \; D_{1} = \left(
\begin{array}{rrr}
3 & 0 & 0 \\
0 & \frac{ 5 }{ 3 } & 2 \\
0 & 2 & 3 \\
\end{array}
\right)
$$</span></p>
<p>==============================================</p>
<p><span class="math-container">$$ E_{2} = \left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & - \frac{ 6 }{ 5 } \\
0 & 0 & 1 \\
\end{array}
\right)
$$</span>
<span class="math-container">$$ P_{2} = \left(
\begin{array}{rrr}
1 & - \frac{ 2 }{ 3 } & \frac{ 4 }{ 5 } \\
0 & 1 & - \frac{ 6 }{ 5 } \\
0 & 0 & 1 \\
\end{array}
\right)
, \; \; \; Q_{2} = \left(
\begin{array}{rrr}
1 & \frac{ 2 }{ 3 } & 0 \\
0 & 1 & \frac{ 6 }{ 5 } \\
0 & 0 & 1 \\
\end{array}
\right)
, \; \; \; D_{2} = \left(
\begin{array}{rrr}
3 & 0 & 0 \\
0 & \frac{ 5 }{ 3 } & 0 \\
0 & 0 & \frac{ 3 }{ 5 } \\
\end{array}
\right)
$$</span></p>
<p>==============================================</p>
<p><span class="math-container">$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$</span></p>
<p><span class="math-container">$$ P^T H P = D $$</span></p>
<p><span class="math-container">$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
- \frac{ 2 }{ 3 } & 1 & 0 \\
\frac{ 4 }{ 5 } & - \frac{ 6 }{ 5 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
3 & 2 & 0 \\
2 & 3 & 2 \\
0 & 2 & 3 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & - \frac{ 2 }{ 3 } & \frac{ 4 }{ 5 } \\
0 & 1 & - \frac{ 6 }{ 5 } \\
0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
3 & 0 & 0 \\
0 & \frac{ 5 }{ 3 } & 0 \\
0 & 0 & \frac{ 3 }{ 5 } \\
\end{array}
\right)
$$</span>
<span class="math-container">$$ $$</span></p>
<p><span class="math-container">$$ Q^T D Q = H $$</span></p>
<p><span class="math-container">$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
\frac{ 2 }{ 3 } & 1 & 0 \\
0 & \frac{ 6 }{ 5 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
3 & 0 & 0 \\
0 & \frac{ 5 }{ 3 } & 0 \\
0 & 0 & \frac{ 3 }{ 5 } \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & \frac{ 2 }{ 3 } & 0 \\
0 & 1 & \frac{ 6 }{ 5 } \\
0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
3 & 2 & 0 \\
2 & 3 & 2 \\
0 & 2 & 3 \\
\end{array}
\right)
$$</span></p>
<p><span class="math-container">$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$</span></p>
|
differentiation | <p>How to calculate the gradient with respect to $X$ of:
$$
\log \mathrm{det}\, X^{-1}
$$
here $X$ is a positive definite matrix, and det is the determinant of a matrix.</p>
<p>How to calculate this? Or what's the result? Thanks!</p>
| <p>I assume that you are asking for the derivative with respect to the elements of the matrix. In this cases first notice that</p>
<p>$$\log \det X^{-1} = \log (\det X)^{-1} = -\log \det X$$</p>
<p>and thus</p>
<p>$$\frac{\partial}{\partial X_{ij}} \log \det X^{-1} = -\frac{\partial}{\partial X_{ij}} \log \det X = - \frac{1}{\det X} \frac{\partial \det X}{\partial X_{ij}} = - \frac{1}{\det X} \mathrm{adj}(X)_{ji} = - (X^{-1})_{ji}$$</p>
<p>since $\mathrm{adj}(X) = \det(X) X^{-1}$ for invertible matrices (where $\mathrm{adj}(X)$ is the adjugate of $X$, see <a href="http://en.wikipedia.org/wiki/Adjugate">http://en.wikipedia.org/wiki/Adjugate</a>).</p>
| <p>The simplest is probably to observe that
$$-\log\det (X+tH) = -\log\det X -\log\det(I+tX^{-1}H)
\\= -\log\det X - t \textrm{Tr}(X^{-1}H) + o(t),$$</p>
<p>where is used the "obvious" fact that $\det(I+A) = 1+\textrm{Tr}(A)+o(|A|)$ (all the other terms are quadratic expressions of the coefficients of $A$).</p>
<p>Notice that $\textrm{Tr}(X^{-1}H)=(X^{-T},H)$ in the Frobenius scalar product, hence $\nabla [-\log\det(X)] = -X^{-T}$ in this scalar product. (This gives another proof that $\nabla\det (X) = cof(X)$.)</p>
<p>Of course if $X$ is symmetric positive definite then $-X^{-1}$ is also a valid expression. Moreover, one has in this case, for $X,Y$ positive definite, $(-X^{-1}+Y^{-1},X-Y)\ge 0$.</p>
|
matrices | <p>I gave the following problem to students:</p>
<blockquote>
<p>Two $n\times n$ matrices $A$ and $B$ are <em>similar</em> if there exists a nonsingular matrix $P$ such that $A=P^{-1}BP$.</p>
<ol>
<li><p>Prove that if $A$ and $B$ are two similar $n\times n$ matrices, then they have the same determinant and the same trace.</p></li>
<li><p>Give an example of two $2\times 2$ matrices $A$ and $B$ with same determinant, same trace but that are not similar.</p></li>
</ol>
</blockquote>
<p>Most of the ~20 students got the first question right. However, almost none of them found a correct example to the second question. Most of them gave examples of matrices that have same determinant and same trace. </p>
<p>But computations show that their examples are similar matrices. They didn't bother to check that though, so they just tried <em>random</em> matrices with same trace and same determinant, hoping it would be a correct example.</p>
<p><strong>Question</strong>: how to explain that none of the random trial gave non similar matrices?</p>
<p>Any answer based on density or measure theory is fine. In particular, you can assume any reasonable distribution on the entries of the matrix. If it matters, the course is about matrices with real coefficients, but you can assume integer coefficients, since when choosing numbers <em>at random</em>, most people will choose integers.</p>
| <p>If <span class="math-container">$A$</span> is a <span class="math-container">$2\times 2$</span> matrix with determinant <span class="math-container">$d$</span> and trace <span class="math-container">$t$</span>, then the characteristic polynomial of <span class="math-container">$A$</span> is <span class="math-container">$x^2-tx+d$</span>. If this polynomial has distinct roots (over <span class="math-container">$\mathbb{C}$</span>), then <span class="math-container">$A$</span> has distinct eigenvalues and hence is diagonalizable (over <span class="math-container">$\mathbb{C}$</span>). In particular, if <span class="math-container">$d$</span> and <span class="math-container">$t$</span> are such that the characteristic polynomial has distinct roots, then any other <span class="math-container">$B$</span> with the same determinant and trace is similar to <span class="math-container">$A$</span>, since they are diagonalizable with the same eigenvalues.</p>
<p>So to give a correct example in part (2), you need <span class="math-container">$x^2-tx+d$</span> to have a double root, which happens only when the discriminant <span class="math-container">$t^2-4d$</span> is <span class="math-container">$0$</span>. If you choose the matrix <span class="math-container">$A$</span> (or the values of <span class="math-container">$t$</span> and <span class="math-container">$d$</span>) "at random" in any reasonable way, then <span class="math-container">$t^2-4d$</span> will usually not be <span class="math-container">$0$</span>. (For instance, if you choose <span class="math-container">$A$</span>'s entries uniformly from some interval, then <span class="math-container">$t^2-4d$</span> will be nonzero with probability <span class="math-container">$1$</span>, since the vanishing set in <span class="math-container">$\mathbb{R}^n$</span> of any nonzero polynomial in <span class="math-container">$n$</span> variables has Lebesgue measure <span class="math-container">$0$</span>.) Assuming that students did something like pick <span class="math-container">$A$</span> "at random" and then built <span class="math-container">$B$</span> to have the same trace and determinant, this would explain why none of them found a correct example.</p>
<p>Note that this is very much special to <span class="math-container">$2\times 2$</span> matrices. In higher dimensions, the determinant and trace do not determine the characteristic polynomial (they just give two of the coefficients), and so if you pick two matrices with the same determinant and trace they will typically have different characteristic polynomials and not be similar.</p>
| <p>As Eric points out, such $2\times2$ matrices are special.
In fact, there are only two such pairs of matrices.
The number depends on how you count, but the point is that such matrices have a <em>very</em> special form.</p>
<p>Eric proved that the two matrices must have a double eigenvalue.
Let the eigenvalue be $\lambda$.
It is a little exercise<sup>1</sup> to show that $2\times2$ matrices with double eigenvalue $\lambda$ are similar to a matrix of the form
$$
C_{\lambda,\mu}
=
\begin{pmatrix}
\lambda&\mu\\
0&\lambda
\end{pmatrix}.
$$
Using suitable diagonal matrices shows that $C_{\lambda,\mu}$ is similar to $C_{\lambda,1}$ if $\mu\neq0$.
On the other hand, $C_{\lambda,0}$ and $C_{\lambda,1}$ are not similar; one is a scaling and the other one is not.</p>
<p>Therefore, up to similarity transformations, the only possible example is $A=C_{\lambda,0}$ and $B=C_{\lambda,1}$ (or vice versa).
Since scaling doesn't really change anything, <strong>the only examples</strong> (up to similarity, scaling, and swapping the two matrices) are
$$
A
=
\begin{pmatrix}
1&0\\
0&1
\end{pmatrix},
\quad
B
=
\begin{pmatrix}
1&1\\
0&1
\end{pmatrix}
$$
and
$$
A
=
\begin{pmatrix}
0&0\\
0&0
\end{pmatrix},
\quad
B
=
\begin{pmatrix}
0&1\\
0&0
\end{pmatrix}.
$$
If adding multiples of the identity is added to the list of symmetries (then scaling can be removed), then there is only one matrix pair up to the symmetries.</p>
<p>If you are familiar with the <a href="https://en.wikipedia.org/wiki/Jordan_normal_form" rel="noreferrer">Jordan normal form</a>, it gives a different way to see it.
Once the eigenvalues are fixed to be equal, the only free property (up to similarity) is whether there are one or two blocks in the normal form.
The Jordan normal form is invariant under similarity transformations, so it gives a very quick way to solve problems like this.</p>
<hr>
<p><sup>1</sup>
You only need to show that any matrix is similar to an upper triangular matrix.
The eigenvalues (which now coincide) are on the diagonal.
You can skip this exercise if you have Jordan normal forms at your disposal.</p>
|
linear-algebra | <p>I am currently trying to self-study linear algebra. I've noticed that a lot of the definitions for terms (like eigenvectors, characteristic polynomials, determinants, and so on) require a <strong>square</strong> matrix instead of just any real-valued matrix. For example, <a href="http://mathworld.wolfram.com" rel="noreferrer">Wolfram</a> has this in its <a href="http://mathworld.wolfram.com/CharacteristicPolynomial.html" rel="noreferrer">definition</a> of the characteristic polynomial:</p>
<blockquote>
<p>The characteristic polynomial is the polynomial left-hand side of the characteristic equation $\det(A - I\lambda) = 0$, where $A$ is a square matrix.</p>
</blockquote>
<p>Why must the matrix be square? What happens if the matrix is not square? And why do square matrices come up so frequently in these definitions? Sorry if this is a really simple question, but I feel like I'm missing something fundamental.</p>
| <p>Remember that an $n$-by-$m$ matrix with real-number entries represents a linear map from $\mathbb{R}^m$ to $\mathbb{R}^n$ (or more generally, an $n$-by-$m$ matrix with entries from some field $k$ represents a linear map from $k^m$ to $k^n$). When $m=n$ - that is, when the matrix is square - we're talking about a map from a space to itself.</p>
<p>So really your question amounts to:</p>
<blockquote>
<p>Why are maps from a space to <em>itself</em> - as opposed to maps from a space to <em>something else</em> - particularly interesting?</p>
</blockquote>
<p>Well, the point is that when I'm looking at a map from a space to itself inputs to and outputs from that map are the same "type" of thing, <em>and so I can meaningfully compare them</em>. So, for example, if $f:\mathbb{R}^4\rightarrow\mathbb{R}^4$ it makes sense to ask when $f(v)$ is parallel to $v$, since $f(v)$ and $v$ lie in the same space; but asking when $g(v)$ is parallel to $v$ for $g:\mathbb{R}^4\rightarrow\mathbb{R}^3$ doesn't make any sense, since $g(v)$ and $v$ are just different types of objects. (This example, by the way, is just saying that <em>eigenvectors/values</em> make sense when the matrix is square, but not when it's not square.)</p>
<hr>
<p>As another example, let's consider the determinant. The geometric meaning of the determinant is that it measures how much a linear map "expands/shrinks" a unit of (signed) volume - e.g. the map $(x,y,z)\mapsto(-2x,2y,2z)$ takes a unit of volume to $-8$ units of volume, so has determinant $-8$. What's interesting is that this applies to <em>every</em> blob of volume: it doesn't matter whether we look at how the map distorts the usual 1-1-1 cube, or some other random cube.</p>
<p>But what if we try to go from $3$D to $2$D (so we're considering a $2$-by-$3$ matrix) or vice versa? Well, we can try to use the same idea: (proportionally) how much <em>area</em> does a given <em>volume</em> wind up producing? However, we now run into problems:</p>
<ul>
<li><p>If we go from $3$ to $2$, the "stretching factor" is no longer invariant. Consider the projection map $(x,y,z)\mapsto (x,y)$, and think about what happens when I stretch a bit of volume vertically ...</p></li>
<li><p>If we go from $2$ to $3$, we're never going to get any volume at all - the starting dimension is just too small! So regardless of what map we're looking at, our "stretching factor" seems to be $0$.</p></li>
</ul>
<p>The point is, in the non-square case the "determinant" as naively construed either is ill-defined or is $0$ for stupid reasons.</p>
| <p>Lots of good answers already as to why square matrices are so important. But just so you don't think that other matrices are not interesting, they have analogues of the inverse (e.g., the <a href="https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse" rel="noreferrer">Moore-Penrose inverse</a>) and non-square matrices have a <a href="https://en.wikipedia.org/wiki/Singular-value_decomposition" rel="noreferrer">singular-value decompition</a>, where the singular values play a role loosely analogous to the eigenvalues of a square matrix. These topics are often left out of linear algebra courses, but they can be important in numerical methods for statistics and machine learning. But learn the square matrix results before the fancy non-square matrix results, since the former provide a context for the latter.</p>
|
logic | <p>The Continuum Hypothesis says that there is no set with cardinality between that of the reals and the natural numbers. Apparently, the Continuum Hypothesis can't be proved or disproved using the standard axioms of set theory.</p>
<p>In order to disprove it, one would only have to construct one counterexample of a set with cardinality between the naturals and the reals. It was proven that the CH can't be disproven. Equivalently, it was proven that one cannot construct a counterexample for the CH. Doesn't this prove it?</p>
<p>Of course, the issue is that it was also proven that it can't be proved. I don't know the details of this unprovability proof, but how can it avoid a contradiction? I understand the idea of something being independent of the axioms, I just don't see how if there is provably no counterexample the hypothesis isn't immediately true, since it basically just says a counterexample doesn't exist.</p>
<p>I'm sure I'm making some horrible logical error here, but I'm not sure what it is.</p>
<p>So my question is this: what is the flaw in my argument? Is it a logical error, or a gross misunderstanding of the unprovability proof in question? Or something else entirely?</p>
| <p>Here's an example axiomatic system:</p>
<ol>
<li>There exist exactly three objects $A, B, C$.</li>
<li>Each of these objects is either a banana, a strawberry or an orange.</li>
<li>There exists at least one strawberry.</li>
</ol>
<p>Let's name the system $X$.</p>
<p><strong>Vincent's Continuum Hypothesis (VCH)</strong>: Every object is either a banana or a strawberry (i.e., there are no oranges).</p>
<p>Now, to disprove this in $X$, you would have to show that one of $A, B, C$ is an orange ("construct a counterexample"). But this does not follow from $X$, because the following model is consistent with $X$: A and B are bananas, C is a strawberry.</p>
<p>On the other hand, VCH does not follow from $X$ either, because the following model is consistent with $X$: A is a banana, B is a strawberry, C is an orange.</p>
<p>As you can see, there is no contradiction, because you have to take into account different models of the axiomatic system.</p>
| <p>I think the basic problem is in your statement that "In order to disprove it, one would only have to construct one counterexample of a set with cardinality between the naturals and the reals." Actually, to disprove CH by this strategy, one would have to produce a counterexample <strong>and prove</strong> that it actually has cardinality between those of $\mathbb N$ and $\mathbb R$. </p>
<p>So, from the fact that CH can't be disproved in ZFC, you can't infer that there is no counterexample but only that no set can be proved in ZFC to be a counterexample.</p>
|
matrices | <p>I have one triangle in <span class="math-container">$3D$</span> space that I am tracking in a simulation. Between time steps I have the previous normal of the triangle and the current normal of the triangle along with both the current and previous <span class="math-container">$3D$</span> vertex positions of the triangles.</p>
<p>Using the normals of the triangular plane I would like to determine a rotation matrix that would align the normals of the triangles thereby setting the two triangles parallel to each other. I would then like to use a translation matrix to map the previous onto the current, however this is not my main concern right now.</p>
<p>I have found <a href="http://forums.cgsociety.org/archive/index.php/t-741227.html" rel="noreferrer">this</a> website that says I must</p>
<ul>
<li>determine the cross product of these two vectors (to determine a rotation axis)</li>
<li>determine the dot product ( to find rotation angle)</li>
<li>build quaternion (not sure what this means)</li>
<li>the transformation matrix is the quaternion as a <span class="math-container">$3 \times 3$</span> (not sure)</li>
</ul>
<p>Any help on how I can solve this problem would be appreciated.</p>
| <p>Suppose you want to find a rotation matrix $R$ that rotates unit vector $a$ onto unit vector $b$.</p>
<p>Proceed as follows:</p>
<p>Let $v = a \times b$</p>
<p>Let $s = \|v\|$ (sine of angle)</p>
<p>Let $c = a \cdot b$ (cosine of angle)</p>
<p>Then the rotation matrix R is given by:
$$R = I + [v]_{\times} + [v]_{\times}^2\frac{1-c}{s^2},$$</p>
<p>where $[v]_{\times}$ is the skew-symmetric cross-product matrix of $v$,
$$[v]_{\times} \stackrel{\rm def}{=} \begin{bmatrix}
\,\,0 & \!-v_3 & \,\,\,v_2\\
\,\,\,v_3 & 0 & \!-v_1\\
\!-v_2 & \,\,v_1 &\,\,0
\end{bmatrix}.$$</p>
<p>The last part of the formula can be simplified to
$$
\frac{1-c}{s^2} = \frac{1-c}{1-c^2} = \frac{1}{1+c},
$$
revealing that it is <em>not</em> applicable only for $\cos(\angle(a, b)) = -1$, i.e., if $a$ and $b$ point into exactly opposite directions.</p>
| <p>Using <a href="https://math.stackexchange.com/a/180436/76513">Kjetil's answer</a> answer, with <a href="https://math.stackexchange.com/users/17349/process91">process91</a>'s comment, we arrive at the following procedure.</p>
<h2>Derivation</h2>
<p>We are given two unit column vectors, $A$ and $B$ ($\|A\|=1$ and $\|B\|=1$). The $\|\circ\|$ denotes the L-2 norm of $\circ$.</p>
<p>First, note that the rotation from $A$ to $B$ is just a 2D rotation on a plane with the normal $A \times B$. A 2D rotation by an angle $\theta$ is given by the following augmented matrix:
$$G=\begin{pmatrix}
\cos\theta & -\sin\theta & 0 \\
\sin\theta & \cos\theta & 0 \\
0 & 0 & 1
\end{pmatrix}.$$</p>
<p>Of course we don't want to actually <em>compute</em> any trig functions. Given our unit vectors, we note that $\cos\theta=A\cdot B$, and $\sin\theta=||A\times B||$. Thus
$$G=\begin{pmatrix}
A\cdot B & -\|A\times B\| & 0 \\
\|A\times B\| & A\cdot B & 0 \\
0 & 0 & 1\end{pmatrix}.$$</p>
<p>This matrix represents the rotation from $A$ to $B$ in the base consisting of the following column vectors:</p>
<ol>
<li><p>normalized <a href="http://en.wikipedia.org/wiki/Vector_projection#Vector_projection" rel="noreferrer">vector projection</a> of $B$ onto $A$: $$u={(A\cdot B)A \over \|(A\cdot B)A\|}=A$$</p></li>
<li><p>normalized <a href="http://en.wikipedia.org/wiki/Vector_projection#Vector_rejection" rel="noreferrer">vector rejection</a> of $B$ onto $A$: $$v={B-(A\cdot B)A \over \|B- (A\cdot B)A\|}$$</p></li>
<li><p>the cross product of $B$ and $A$: $$w=B \times A$$</p></li>
</ol>
<p>Those vectors are all orthogonal, and form an orthogonal basis. This is the detail that Kjetil had missed in his <a href="https://math.stackexchange.com/a/180436/76513">answer</a>. You could also normalize $w$ and get an orthonormal basis, if you needed one, but it doesn't seem necessary.</p>
<p>The basis change matrix for this basis is:
$$F=\begin{pmatrix}u & v & w \end{pmatrix}^{-1}=\begin{pmatrix} A & {B-(A\cdot B)A \over \|B- (A\cdot B)A\|} & B \times A\end{pmatrix}^{-1}$$</p>
<p>Thus, in the original base, the rotation from $A$ to $B$ can be expressed as right-multiplication of a vector by the following matrix: $$U=F^{-1}G F.$$</p>
<p>One can easily show that $U A = B$, and that $\|U\|_2=1$. Also, $U$ is the same as the $R$ matrix from <a href="https://math.stackexchange.com/a/476311/76513">Rik's answer</a>.</p>
<h2>2D Case</h2>
<p>For the 2D case, given $A=\left(x_1,y_1,0\right)$ and $B=\left(x_2,y_2,0\right)$, the matrix $G$ is the forward transformation matrix itself, and we can simplify it further. We note
$$\begin{aligned}
\cos\theta &= A\cdot B = x_1x_2+y_1y_2 \\
\sin\theta &= \| A\times B\| = x_1y_2-x_2y_1
\end{aligned}$$</p>
<p>Finally,
$$U\equiv G=\begin{pmatrix}
x_1x_2+y_1y_2 & -(x_1y_2-x_2y_1) \\
x_1y_2-x_2y_1 & x_1x_2+y_1y_2
\end{pmatrix}$$
and
$$U^{-1}\equiv G^{-1}=\begin{pmatrix}
x_1x_2+y_1y_2 & x_1y_2-x_2y_1 \\
-(x_1y_2-x_2y_1) & x_1x_2+y_1y_2
\end{pmatrix}$$</p>
<h2>Octave/Matlab Implementation</h2>
<p>The basic implementation is very simple. You could improve it by factoring out the common expressions of <code>dot(A,B)</code> and <code>cross(B,A)</code>. Also note that $||A\times B||=||B\times A||$.</p>
<pre><code>GG = @(A,B) [ dot(A,B) -norm(cross(A,B)) 0;\
norm(cross(A,B)) dot(A,B) 0;\
0 0 1];
FFi = @(A,B) [ A (B-dot(A,B)*A)/norm(B-dot(A,B)*A) cross(B,A) ];
UU = @(Fi,G) Fi*G*inv(Fi);
</code></pre>
<p>Testing:</p>
<pre><code>> a=[1 0 0]'; b=[0 1 0]';
> U = UU(FFi(a,b), GG(a,b));
> norm(U) % is it length-preserving?
ans = 1
> norm(b-U*a) % does it rotate a onto b?
ans = 0
> U
U =
0 -1 0
1 0 0
0 0 1
</code></pre>
<p>Now with random vectors:</p>
<pre><code>> vu = @(v) v/norm(v);
> ru = @() vu(rand(3,1));
> a = ru()
a =
0.043477
0.036412
0.998391
> b = ru()
b =
0.60958
0.73540
0.29597
> U = UU(FFi(a,b), GG(a,b));
> norm(U)
ans = 1
> norm(b-U*a)
ans = 2.2888e-16
> U
U =
0.73680 -0.32931 0.59049
-0.30976 0.61190 0.72776
-0.60098 -0.71912 0.34884
</code></pre>
<h2>Implementation of Rik's Answer</h2>
<p>It is computationally a bit more efficient to use <a href="https://math.stackexchange.com/a/476311/76513">Rik's</a> answer. This is also an Octave/MatLab implementation.</p>
<pre><code>ssc = @(v) [0 -v(3) v(2); v(3) 0 -v(1); -v(2) v(1) 0]
RU = @(A,B) eye(3) + ssc(cross(A,B)) + \
ssc(cross(A,B))^2*(1-dot(A,B))/(norm(cross(A,B))^2)
</code></pre>
<p>The results produced are same as above, with slightly smaller numerical errors since there are less operations being done.</p>
|
linear-algebra | <p>I have looked extensively for a proof on the internet but all of them were too obscure. I would appreciate if someone could lay out a simple proof for this important result. Thank you.</p>
| <p>These answers require way too much machinery. By definition, the characteristic polynomial of an $n\times n$ matrix $A$ is given by
$$p(t) = \det(A-tI) = (-1)^n \big(t^n - (\text{tr} A) \,t^{n-1} + \dots + (-1)^n \det A\big)\,.$$
On the other hand, $p(t) = (-1)^n(t-\lambda_1)\dots (t-\lambda_n)$, where the $\lambda_j$ are the eigenvalues of $A$. So, comparing coefficients, we have $\text{tr}A = \lambda_1 + \dots + \lambda_n$.</p>
| <p>Let $A$ be a matrix. It has a <a href="http://en.wikipedia.org/wiki/Jordan_normal_form">Jordan Canonical Form</a>, i.e. there is matrix $P$ such that $PAP^{-1}$ is in Jordan form. Among other things, Jordan form is upper triangular, hence it has its eigenvalues on its diagonal. It is therefore clear for a matrix in Jordan form that its trace equals the sum of its eigenvalues. All that remains is to prove that if $B,C$ are <a href="http://en.wikipedia.org/wiki/Similar_%28linear_algebra%29">similar</a> then they have the same eigenvalues.</p>
|
combinatorics | <p>It is quite easy to calculate the total number of functions from a set $X$ with $m$ elements to a set $Y$ with $n$ elements ($n^{m}$), and also the total number of injective functions ($n^{\underline{m}}$, denoting the falling factorial). But I am thinking about how to calculate the total number of surjective functions $f\colon X \twoheadrightarrow Y $.</p>
<p>The way I thought of doing this is as follows: firstly, since all $n$ elements of the codomain $Y$ need to be mapped to, you choose any $n$ elements from the $m$ elements of the set $X$ to be mapped one-to-one with the $n$ elements of $Y$. This results in $n!$ possible pairings. But the number of ways of choosing $n$ elements from $m$ elements is $\frac{m!}{(m-n)!\,n!}$, so the total number of ways of matching $n$ elements in $X$ to be one-to-one with the $n$ elements of $Y$ is $\frac{m!}{(m-n)!\,n!} \times n! = \frac{m!}{(m-n)!}$.</p>
<p>Now we have 'covered' the codomain $Y$ with $n$ elements from $X$, the remaining unpaired $m-n$ elements from $X$ can be mapped to any of the elements of $Y$, so there are $n^{m-n}$ ways of doing this. Therefore I think that the total number of surjective functions should be $\frac{m!}{(m-n)!} \, n^{m-n}$.</p>
<p>Is this anything like correct or have I made a major mistake here?</p>
| <p>Consider $f^{-1}(y)$, $y \in Y$. This set must be non-empty, regardless of $y$. What you're asking for is the number of ways to distribute the elements of $X$ into these sets.</p>
<p>The number of ways to distribute m elements into n non-empty sets is given by the <a href="http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind">Stirling numbers of the second kind</a>, $S(m,n)$. However, each element of $Y$ can be associated with any of these sets, so you pick up an extra factor of $n!$: the total number should be $S(m,n) n!$</p>
<p>The Stirling numbers have interesting properties. They're worth checking out for their own sake.</p>
| <p>Here is a solution that does not involve the Stirling numbers of the second kind, $S(n,m)$. The number of surjective functions from a set $X$ with $m$ elements to a set $Y$ with $n$ elements is</p>
<p>$$
\sum_{i=0}^{n-1} (-1)^i{n \choose i}(n-i)^m
$$</p>
<p>or more explicitly
$$
{n \choose 0}n^m - {n \choose 1}(n-1)^m + {n \choose 2}(n-2)^m - \cdots \pm {n \choose n-2}2^m \mp {n \choose n-1}1^m
$$</p>
<p>We begin by counting the number of functions from $X$ to $Y$, which is already mentioned to be $n^m$. Next we subtract off the number $n(n-1)^m$ (roughly the number of functions that miss one or more elements). Of course this subtraction is too large so we add back in ${n \choose 2}(n-2)^m$ (roughly the number of functions that miss 2 or more elements). But again, this addition is too large, so we subtract off the next term and so on. This is a rough sketch of a proof, it could be made more formal by using induction on $n$.</p>
|
game-theory | <p>I am from programming background but with very limited knowledge of maths.</p>
<p>I am very much eager to learn and apply <a href="http://www.britannica.com/EBchecked/topic/224893/game-theory" rel="noreferrer">game theory</a> to understand dynamics of International Politics and economics. But I am facing difficulties to understand the maths involved in it. </p>
<p>So, </p>
<ol>
<li>What are <strong>prerequisites for understanding the math in the game theory</strong>?</li>
<li>What are the authentic sources to understand game theory mathematically and
and <strong>its application in the international politics and economics</strong>? </li>
</ol>
| <p>As @JordanMahar mentions, Fudenberg and Tirole is the standard graduate-level text. But I would start with <strong><em>Game Theory for Applied Economists</em></strong> by <strong>Gibbons</strong>. It is very readable. </p>
<p>Prerequisites for Gibbons are minimal. A little algebra and probability will do just fine. </p>
| <p>In economics, the classic source for game theory is:</p>
<p>Fudenberg, D. and Tirole, J. (1991). <a href="http://www.amazon.ca/Game-Theory-Drew-Fudenberg/dp/0262061414/ref=sr_1_1?ie=UTF8&qid=1369889236&sr=8-1&keywords=game+theory+tirole" rel="nofollow">Game Theory</a></p>
<p>At least with Game Theory applied to economics, you can begin with a minimal knowledge of mathematics (applied calculus and some set theory will usually suffice). </p>
|
logic | <p><em>Background: I'm a logic student with very little background in cohomology etc., so this question is fairly naive.</em></p>
<hr>
<p>Although mathematical logic is generally perceived as sitting off on its own, there are some striking applications of algebraic/geometric/combinatorial ideas to logic. In general, I'm very interested in the following broad question: </p>
<blockquote>
<p>"How should I go about looking for pieces of mathematics far from mathematical logic, which have bearing on some piece of mathematical logic?"</p>
</blockquote>
<p>Right now, I'm specifically interested in the following: </p>
<blockquote>
<p>"When should I think 'cohomology!'?"</p>
</blockquote>
<p>The specific example I'm motivated by is a pair of papers by Dan Talayco (<a href="http://arxiv.org/pdf/math/9311205.pdf">http://arxiv.org/pdf/math/9311205.pdf</a>, <a href="http://www.sciencedirect.com/science/article/pii/0168007295000240">http://www.sciencedirect.com/science/article/pii/0168007295000240</a>) in which he develops cohomology theories for two purely set-theoretic objects: Hausdorff gaps, and particularly weird infinite trees ("Todorcevic trees").</p>
<p>At the beginning of his paper on Hausdorff gaps, Talayco mentions that</p>
<blockquote>
<p>"the original observation that gaps are cohomological in nature is due to Blass."</p>
</blockquote>
<p>This is something I want to be able to do! I can tell that e.g. Hausdorff gaps are all about "not being able to fill something in," but it's a long way between that vague statement and the intuition that there should be a cohomology theory around it, let alone coming up with the specifics. So my question is:</p>
<blockquote>
<p><strong>Question.</strong> When should I suspect that some piece of mathematics (ideally far from algebra/geometry) has a cohomological interpretation, and how should I go about figuring out what the specifics should be?</p>
</blockquote>
<p>To clarify: although 'useful' is always good, I'm just asking how I can tell that cohomology <em>can</em> be attached to some piece of mathematics (especially logic), regardless of whether it yields new results.</p>
| <p>I guess I will take a crack at this.</p>
<p>First of all, it is probably worthwhile for you to learn some cohomology in its original home so that you have some intuition for it, and some knowledge of what theorems there are, how to compute it, etc. You do not give much indication in your answer as to how much knowledge of cohomology you have currently.</p>
<p>Generally the intuition behind cohomology groups is that they measure the failure of "locally consistent" things to be "globally consistent".</p>
<p>Examples:</p>
<p>The first de Rham cohomology group of the punctured plane is 1 dimensional since there is (up to the gradient of a global function) only 1 vector field on this space which is locally a gradient of a function but not globally the gradient of a function.</p>
<p>The <a href="http://en.wikipedia.org/wiki/Penrose_triangle" rel="noreferrer">Penrose triangle</a> represents a nontrivial cohomology class over the multiplicative group of positive reals, since it "locally" looks like a perspective drawing, but there is no "global" object realizing that.</p>
<p>If local exchange rates between countries allow arbitrage, then there is no globally consistent exchange rate, so current exchange rates give a nontrivial cohomology class.</p>
<p>The axiom of choice says every surjection splits. In fact, even without the axiom of choice, every surjection splits locally (for every point in the codomain, I can find an inverse image), and so the axiom of choice is a a local to global statement: these local inverses can be assembled into a global section. Blass has written a bit about this in <a href="http://www.ams.org/journals/tran/1983-279-01/S0002-9947-1983-0704615-7" rel="noreferrer">Blass - Cohomology detects failure of the axiom of choice</a>, but there is still a lot more work to be done with this concept.</p>
<p>The moral is just to be on the lookout for situations where things seems to fit together in small bits, but somehow the whole does not work out. There is, more than likely, cohomology playing into this somehow. </p>
<p>I will mention that (from my perspective) sheaf cohomology probably formalizes this intuitive perspective the best, since you do not have to start with a sequence of maps with differentials (where do those come from?) just a notion of local objects and how to patch them. So I would recommend learning some sheaf cohomology if you are planning on looking for cohomology far from algebraic topology.</p>
| <p>As Steven Gubkin and Ryan Budney have already pointed out, cohomology is often used to measure how far a "locally consistent" object is from being "globally consistent". I thought I might describe another example of this in set theory, which it turns out contains a lot of open problems. As far as I know, this example hasn't been written down anywhere, so I'll have to be a bit verbose. I won't attempt an actual answer to the OP's question, but I hope that what I write down might be helpful.</p>
<p>Disclaimer: Justin Moore told me about this problem many years ago, when I was just beginning grad school. What I'm writing down is what I've been able to reconstruct from my memory of that conversation; it might not be the most up-to-date information on the problem, or the best description of it.</p>
<p>Given a subset $A$ of $\omega$, let $\Gamma(A) = \prod_{n\in A} \mathbb{Z} / \bigoplus_{n\in A} \mathbb{Z}$. Thus, an element of $\Gamma(A)$ can be described as the equivalence class of a function $f : A\to\mathbb{Z}$, where the equivalence is "$f$ and $g$ differ on at most finitely-many coordinates."</p>
<p>Now given a family $\mathcal{A}\subseteq\mathcal{P}(\omega)$, and an $n < \omega$, we define the groups $$ C_n(\mathcal{A}) = \prod_{A_1,\ldots,A_n} \Gamma(A_1\cap\cdots \cap A_n)$$</p>
<p>When $n = 0$, this definition is a little ambiguous, so we take the opportunity to identify $C_0(\mathcal{A})$ with $\Gamma(\omega)$. (One can argue that that's the correct way of doing things, but I'll leave it to the reader.) Now we can define coboundary maps $\delta_n : C_n\to C_{n+1}$ (for all $n$, <em>including</em> $n = 0$) by $$\delta_n(F)(A_1,\ldots,A_{n+1}) = \sum_{k=0}^n (-1)^k F(A_1,\ldots,\widehat{A_{k+1}},\ldots,A_{n+1})$$ where as usual $\widehat{A_{k+1}}$ means we drop $A_{k+1}$ from the list. One can prove as usual that $\delta_{n+1}\circ \delta_n = 0$, so $H_n(\mathcal{A}) = \ker{\delta_{n+1}} / \textrm{im}\;\delta_n$ makes sense.</p>
<p>If you work through the definitions, you can see that $\delta_0$ maps a function (or more accurately, its equivalence class) to its restrictions to elements of $\mathcal{A}$. $\delta_1$ takes a collection of functions defined on members of $\mathcal{A}$ to their differences (on the pairwise intersections).</p>
<p>Hence, a member of $\ker{\delta_1}$ is a family of functions $f_A : A\to \mathbb{Z}$ ($A\in\mathcal{A}$) such that $f_A\upharpoonright A\cap B$ and $f_B\upharpoonright A\cap B$ agree mod-finite for every $A,B$. Such families have been studied before by set theorists, and are called <em>coherent families</em>. The question of whether such a coherent family is in $\textrm{im}\;\delta_0$ is exactly what comes up in Dow, Simon and Vaughan's paper "Strong homology and the proper forcing axiom". They prove there the following (I'm paraphrasing a little bit):</p>
<p><strong>Theorem 1</strong>: Assume $\mathfrak{d} = \omega_1$. Then there is a $P$-ideal $\mathcal{I}\subseteq\omega$ (in fact, $\mathcal{I}$ is just $\emptyset\times\textrm{fin}$) such that $H_0(\mathcal{I}) \neq 0$.</p>
<p><strong>Theorem 2</strong>: Assume the Proper Forcing Axiom. Then $H_0(\mathcal{I}) = 0$ for every $P_{\aleph_1}$-ideal $\mathcal{I}$.</p>
<p>Actually, it's not hard to prove that $2^\omega < 2^{\omega_1}$ implies that $H_0(\mathcal{A})$ has size $2^{\omega_1}$, whenever $\mathcal{A}$ is a $\subset^*$-increasing $\omega_1$-sequence in $\mathcal{P}(\omega)$. Moreover, Velickovic proves Theorem 2 from just OCA in his paper "OCA and automorphisms of $\mathcal{P}(\omega)/\mathrm{fin}$".</p>
<p>Okay, so we have a lot of natural questions!</p>
<p><strong>Question</strong>: What is the possible behavior of $H_n(\mathcal{A})$ for various sets $\mathcal{A}$, and $n\ge 1$? Does PFA (or MM, or whatever) imply that they're all trivial, whenever (say) $\mathcal{A}$ is a $P$-ideal? Can we consistently get $H_n(\mathcal{A}) \neq 0$ for all $n\ge 1$? All at the same time? Etc.</p>
<p>Here's another, entirely unrelated problem. It's known that for every $n$, there is a $\sigma$-$n$-linked poset of size $\mathfrak{b}$, which has no $n+1$-linked subset of size $\mathfrak{b}$. (See Todorcevic, "Remarks on cellularity in products.") Can you express this using cohomology (or maybe homology)?</p>
|
logic | <p>I am teaching a first-semester course in abstract algebra, and we are discussing group isomorphisms. In order to prove that two group are not isomorphic, I encourage the students to look for a group-theoretic property satisfied by one group but not by the other. I did not give a precise meaning to the phrase "group-theoretic property", but some examples of the sort of properties I have in mind are
<span class="math-container">$$
\forall g,h\in G:\exists n,m\in\mathbb{Z}:(n,m)\neq (0,0)\wedge g^n=h^m,\\
\forall H\leq G:\exists g,h\in G:H=\langle g,h\rangle,\\
\forall g,h\in G:\exists i\in G: \langle g,h\rangle = \langle i\rangle
$$</span>
One of my students asked if, give two non-isomorphic groups, there is always a group-theoretic property satisfied by one group but not the other. In a sense, "being isomorphic to that group over there" is a group-theoretic property. But this is not really what I have in mind.</p>
<p>To pin down the class of properties I have in mind, let's say we allow expressions involving</p>
<ul>
<li>quantification over <span class="math-container">$G$</span>, subgroups of <span class="math-container">$G$</span>, and <span class="math-container">$\mathbb{Z}$</span>,</li>
<li>group multiplication, inversion, and subgroups generated by a finite list of elements</li>
<li>the symbol <span class="math-container">$1_G$</span> (the group identity element),</li>
<li>addition, subtraction, multiplication, exponentiation (provided the exponent is non-negative), and inequalities of integers ,</li>
<li>the integer symbols <span class="math-container">$0$</span> and <span class="math-container">$1$</span>,</li>
<li>raising a group element to an integer power, and</li>
<li>equality, elementhood, and logical connectives.</li>
</ul>
<p>I do not know much about model theory or logic, but my understanding is that this is not the first-order theory of groups. In particular, <a href="https://math.stackexchange.com/questions/2137508/can-a-torsion-group-and-a-nontorsion-group-be-elementarily-equivalent">this MSE question</a> indicates that there exist a torsion and a non-torsion group which are elementarily equivalent (meaning they cannot be distinguished by a first-order statement in the language of groups), but these groups can be distinguished by a property of the above form. I have also heard that free groups of different rank are elementarily equivalent, but these can also be distinguished by a property of the above form.</p>
<p>My questions are:</p>
<p>(1) Is there a name for the theory I am considering? Or something closely (or distantly) related?</p>
<p>(2) Are there examples of non-isomorphic groups that cannot be distinguished by a property of the above form? Are there examples where the groups involved could be understood by an average first-semester algebra student?</p>
| <p>First, let's start with the silly answer. Your language only has countably many different expressions, so can only divvy groups up into continuum-many classes - so there are definitely non-isomorphic groups it can't distinguish! In general this will happen as long as your language has <em>only set-many</em> expressions: you need a proper class sized logic like <span class="math-container">$\mathcal{L}_{\infty,\infty}$</span> to distinguish between all pairs of non-isomorphic structures.</p>
<p>That said, you're right that you're looking at something much stronger than first-order logic. Specifically, you're describing a sublogic of <strong>second-order logic</strong>, the key difference being that second-order logic lets you quantify over arbitrary subsets of the domain, and indeed <em>functions and relations of arbitrary arity</em> over the domain, and not just subgroups. Second-order logic doesn't have an explicit ability to refer to (say) integers built in, but it can do so via tricks of quantifying over finite configurations.</p>
<p>While the exact strength of the system you describe isn't clear to me, second-order logic is known to be extremely powerful. In particular, I believe there are no known natural examples of non-isomorphic second-order-elementarily-equivalent structures <strong>at all</strong>, although as per the first paragraph of this answer such structures certainly have to exist! So second-order-equivalence is a pretty strong equivalence relation, and in practice will suffice to distinguish all the groups your students run into.</p>
| <p>Here are some simple examples where you at least need to make some decisions about what you believe about set theory to determine whether two groups are isomorphic. Assuming the axiom of choice every vector space has a basis, so <span class="math-container">$\mathbb{R}$</span> is isomorphic (as a group) to some direct sum of copies of <span class="math-container">$\mathbb{Q}$</span> (in fact necessarily to a direct sum of <span class="math-container">$|\mathbb{R}|$</span> copies of <span class="math-container">$\mathbb{Q}$</span>). The existence of such a basis for <span class="math-container">$\mathbb{R}$</span> over <span class="math-container">$\mathbb{Q}$</span> allows you to construct <a href="https://en.wikipedia.org/wiki/Vitali_set" rel="noreferrer">Vitali sets</a>, which are non-measurable, and there are models of <span class="math-container">$ZF \neg C$</span> in which every subset of <span class="math-container">$\mathbb{R}$</span> is measurable, so <span class="math-container">$\mathbb{R}$</span> fails to have a basis in such models.</p>
<p>Another example along the same lines is <span class="math-container">$\left( \prod_{\mathbb{N}} \mathbb{Q} / \bigoplus_{\mathbb{N}} \mathbb{Q} \right)^{\ast}$</span>, taking the dual as a <span class="math-container">$\mathbb{Q}$</span>-vector space. Assuming the axiom of choice this is a direct sum of <span class="math-container">$|\mathbb{R}|$</span> copies of <span class="math-container">$\mathbb{Q}$</span> again, but without at least enough choice to construct something like non-principal ultrafilters on <span class="math-container">$\mathbb{N}$</span> it's not clear how to write down a single nonzero element of this group!</p>
|
logic | <p>I want to get this as a tattoo as I love the role maths plays in the universe and the idea that the farthest reaches of what we can ever know, fall short of the limits of what is true, even in mathematics. I've done a lot of research but have yet to find a mathematical symbol version of this theorem.</p>
<p>Is there a way to show even a short part of/ the essence of it/ an example of it using symbols?</p>
<p>Sorry if it's inappropriate to ask this here - you just seem like the corner of the internet with my best hope. Any help would be greatly appreciated - I've been trying to find an answer to this for almost a year. Thank you!</p>
| <p>If you have enough skin and endurance, the following is not exactly Gödel's theorem, but an instance of a statement as predicted by Gödel's theorem, that is - with a suitable interpretation - it states that it is not provable (which means that it must be true, but unprovable - or PA is inconsistent). But don't complain if it turns out there is a typo buried inside ... </p>
<p>¬∃b:∃c:∃d:〈∃e:∃f:∃g:〈a=((((((e+ f)+ g)· ((e+ f)+ g))· ((e+ f)+ g))+ ((e+ f)· (e+ f)))+ e) ∧ ∃h:∃i:〈〈∃j:h=(j· Si) ∧ 〈∃j:h=(d+ (j· S(i· Sg))) ∧ ∃j:(d+ j)=(i· Sg)〉〉 ∧ ∀j:〈∃k:S(j+ k)=g ⇒ ∀k:∀l:〈〈〈∃m:h=(k+ (m· S(i· Sj))) ∧ ∃m:(k+ m)=(i· Sj)〉 ∧ 〈∃m:e=(l+ (m· S(f· Sj))) ∧ ∃m:(l+ m)=(f· Sj)〉〉 ⇒ 〈〈¬l=SSSSSSSSSS0 ⇒ 〈〈∃m:h=S(k+ (m· S(i· SSj))) ∧ ∃m:S(k+ m)=(i· SSj)〉 ∧ 〈∃m:b=(l+ (m· S(c· Sj))) ∧ ∃m:(l+ m)=(c· Sj)〉〉〉 ∧ 〈l=SSSSSSSSSS0 ⇒ 〈〈〈∃m:h=S((a+ k)+ (m· S(i· SSj))) ∧ ∃m:S((a+ k)+ m)=(i· SSj)〉 ∧ 〈∃m:b=SSSSSSSSS(m· S(c· S(a+ k))) ∧ ∃m:SSSSSSSSSm=(c· S(a+ k))〉〉 ∧ ∀m:〈∃n:S(m+ n)=a ⇒ 〈∃n:b=SSSSSSSS(n· S(c· S(m+ k))) ∧ ∃n:SSSSSSSSn=(c· S(m+ k))〉〉〉〉〉〉〉〉〉 ∧ ∃e:∃f:∃g:∃h:∃i:〈〈〈〈〈∃j:i=(j· Sf) ∧ ∃j:i=(j· S(f· Sg))〉 ∧ ∀j:∀k:∀l:〈〈〈∃m:S(j+ m)=g ∧ 〈∃m:i=(k+ (m· S(f· Sj))) ∧ ∃m:(k+ m)=(f· Sj)〉〉 ∧ 〈∃m:i=(l+ (m· S(f· SSj))) ∧ ∃m:(l+ m)=(f· SSj)〉〉 ⇒ 〈k=l ∨ 〈∃m:e=(m· S(f· Sj)) ∧ 〈k=Sl ∨ l=Sk〉〉〉〉〉 ∧ ∀j:〈∃k:S(j+ k)=g ⇒ ∀k:∀l:∀m:∀n:〈〈〈〈〈∃o:e=(k+ (o· S(f· Sj))) ∧ ∃o:(k+ o)=(f· Sj)〉 ∧ 〈∃o:h=(l+ (o· S(f· Sj))) ∧ ∃o:(l+ o)=(f· Sj)〉〉 ∧ 〈∃o:h=(m+ (o· S(f· SSj))) ∧ ∃o:(m+ o)=(f· SSj)〉〉 ∧ 〈∃o:h=(n+ (o· S(f· SS(m+ j)))) ∧ ∃o:(n+ o)=(f· SS(m+ j))〉〉 ⇒ 〈〈〈〈〈〈∃o:((j+ l)+ o)=g ∧ 〈∃o:SSSSSSSSSo=k ⇒ l=S0〉〉 ∧ 〈〈〈k=0 ∨ k=SSSSSSS0〉 ∨ k=SSSSSSSS0〉 ⇒ l=Sm〉〉 ∧ 〈〈〈〈〈〈k=S0 ∨ k=SS0〉 ∨ k=SSS0〉 ∨ k=SSSS0〉 ∨ k=SSSSS0〉 ∨ k=SSSSSS0〉 ⇒ 〈l=S(m+ n) ∧ ∀o:〈〈∃p:〈∃q:S(p+ q)=m ∧ 〈∃q:e=(o+ (q· S(f· SS(j+ p)))) ∧ ∃q:(o+ q)=(f· SS(j+ p))〉〉 ∧ ∃p:〈∃q:S(p+ q)=n ∧ 〈∃q:e=(o+ (q· S(f· SS((m+ j)+ p)))) ∧ ∃q:(o+ q)=(f· SS((m+ j)+ p))〉〉〉 ⇒ 〈〈∀p:¬〈〈∃q:S(p+ q)=m ∧ 〈∃q:e=SSSSSS(q· S(f· SS(j+ p))) ∧ ∃q:SSSSSSq=(f· SS(j+ p))〉〉 ∧ 〈∃q:e=(o+ (q· S(f· SSS(j+ p)))) ∧ ∃q:(o+ q)=(f· SSS(j+ p))〉〉 ∨ ∀p:¬〈〈∃q:S(p+ q)=n ∧ 〈∃q:e=SSSSSS(q· S(f· SS((m+ j)+ p))) ∧ ∃q:SSSSSSq=(f· SS((m+ j)+ p))〉〉 ∧ 〈∃q:e=(o+ (q· S(f· SSS((m+ j)+ p)))) ∧ ∃q:(o+ q)=(f· SSS((m+ j)+ p))〉〉〉 ⇒ ∀p:¬〈〈∃q:S(p+ q)=(m+ n) ∧ 〈∃q:e=SSSSSS(q· S(f· SS(j+ p))) ∧ ∃q:SSSSSSq=(f· SS(j+ p))〉〉 ∧ 〈∃q:e=(o+ (q· S(f· SSS(j+ p)))) ∧ ∃q:(o+ q)=(f· SSS(j+ p))〉〉〉〉〉〉〉 ∧ ∀o:〈〈∃p:e=(o+ (p· S(f· SSj))) ∧ ∃p:(o+ p)=(f· SSj)〉 ⇒ 〈〈〈〈〈〈k=0 ∨ k=S0〉 ∨ k=SS0〉 ∨ k=SSSSSSS0〉 ⇒ 〈〈〈〈o=S0 ∨ o=SS0〉 ∨ o=SSS0〉 ∨ o=SSSSSS0〉 ∨ o=SSSSSSS0〉〉 ∧ 〈〈〈〈k=SSS0 ∨ k=SSSS0〉 ∨ k=SSSSS0〉 ∨ k=SSSSSSSS0〉 ⇒ 〈〈o=SSSS0 ∨ o=SSSSS0〉 ∨ ∃p:SSSSSSSSp=o〉〉〉 ∧ 〈k=SSSSSS0 ⇒ ∃p:SSSSSSSSSSp=o〉〉〉〉 ∧ ∀o:〈〈∃p:e=(o+ (p· S(f· SS(m+ j)))) ∧ ∃p:(o+ p)=(f· SS(m+ j))〉 ⇒ 〈〈〈k=0 ⇒ k=o〉 ∧ 〈〈〈k=S0 ∨ k=SS0〉 ∨ k=SSSSSS0〉 ⇒ 〈〈〈〈o=S0 ∨ o=SS0〉 ∨ o=SSS0〉 ∨ o=SSSSSS0〉 ∨ o=SSSSSSS0〉〉〉 ∧ 〈〈〈k=SSS0 ∨ k=SSSS0〉 ∨ k=SSSSS0〉 ⇒ 〈〈o=SSSS0 ∨ o=SSSSS0〉 ∨ ∃p:SSSSSSSSp=o〉〉〉〉〉 ∧ 〈j=0 ⇒ k=0〉〉〉〉〉 ∧ ∃j:〈〈〈g=S(j+ d) ∧ ∃k:e=(k· S(f· Sj))〉 ∧ ∀k:¬〈〈∃l:S(k+ l)=d ∧ 〈∃l:b=SSSSSS(l· S(c· Sk)) ∧ ∃l:SSSSSSl=(c· Sk)〉〉 ∧ ∃l:b=(l· S(c· SSk))〉〉 ∧ ∀k:∀l:〈〈∃m:S(k+ m)=d ∧ 〈∃m:b=(l+ (m· S(c· Sk))) ∧ ∃m:(l+ m)=(c· Sk)〉〉 ⇒ 〈∃m:e=(l+ (m· S(f· SS(j+ k)))) ∧ ∃m:(l+ m)=(f· SS(j+ k))〉〉〉〉 ∧ ∀j:∀k:〈〈∃l:e=(l· S(f· Sj)) ∧ 〈∃l:h=(k+ (l· S(f· SSj))) ∧ ∃l:(k+ l)=(f· SSj)〉〉 ⇒ 〈〈〈〈∃l:e=SSSSSS(l· S(f· SSj)) ∧ ∃l:SSSSSSl=(f· SSj)〉 ∧ 〈∃l:e=SSSSSSSSSS(l· S(f· SSSj)) ∧ ∃l:SSSSSSSSSSl=(f· SSSj)〉〉 ∧ 〈〈〈〈〈〈∃l:e=SSSSSSS(l· S(f· SSSSj)) ∧ ∃l:e=SSS(l· S(f· SSSSSj))〉 ∧ ∃l:e=SSSSSSSS(l· S(f· SSSSSSj))〉 ∧ ∃l:e=SSSSSSSSSS(l· S(f· SSSSSSSj))〉 ∧ ∃l:e=SSSSSSSSS(l· S(f· SSSSSSSSj))〉 ∨ 〈〈〈∃l:e=SSS(l· S(f· SSSSj)) ∧ ∃l:e=SSSSSSSSSS(l· S(f· SSSSSSj))〉 ∧ ∃l:e=SSSSSSSSS(l· S(f· SSSSSSSj))〉 ∧ 〈〈∃l:e=SSSS(l· S(f· SSSSSj)) ∧ ∃l:e=SSSSSSSSSS(l· S(f· SSSSSSSSj))〉 ∨ 〈∃l:e=SSSSS(l· S(f· SSSSSj)) ∧ ∃l:e=SSSSSSSSS(l· S(f· SSSSSSSSj))〉〉〉〉 ∨ 〈〈〈〈〈〈〈〈∃l:e=SSSSSS(l· S(f· SSSSj)) ∧ ∃l:e=SSSSSSSSSSS(l· S(f· SSSSSj))〉 ∧ ∃l:e=SSS(l· S(f· SSSSSSj))〉 ∧ ∃l:e=SSSSSSSSSS(l· S(f· SSSSSSSSj))〉 ∧ ∃l:e=SSSSSSSS(l· S(f· SSSSSSSSSj))〉 ∧ ∃l:e=SSSSSSSSSSS(l· S(f· SSSSSSSSSSj))〉 ∧ ∃l:e=SSSSSSSSSS(l· S(f· SSSSSSSSSSSSSj))〉 ∧ ∃l:e=SSSSSSSSSSS(l· S(f· SSSSSSSSSSSSSSj))〉 ∧ 〈〈〈∃l:e=SSSS(l· S(f· SSSSSSSj)) ∧ ∃l:e=SSSSSSSS(l· S(f· SSSSSSSSSSSj))〉 ∧ ∃l:e=SSSS(l· S(f· SSSSSSSSSSSSj))〉 ∨ 〈〈〈∃l:e=SSSSS(l· S(f· SSSSSSSj)) ∧ ∃l:e=SSSS(l· S(f· SSSSSSSSSSSj))〉 ∧ 〈∃l:e=SSSSS(l· S(f· SSSSSSSSSSSSj)) ∧ ∃l:SSSSSl=(f· SSSSSSSSSSSSj)〉〉 ∧ ∃l:e=SSSSSSSSSS(l· S(f· SSSSSSSSSSSSSSSj))〉〉〉〉〉 ∨ ∃l:∃m:〈〈〈〈∃n:S(l+ n)=j ∧ ∃n:e=(n· S(f· Sl))〉 ∧ 〈∃n:h=(m+ (n· S(f· SSl))) ∧ ∃n:(m+ n)=(f· SSl)〉〉 ∧ ∀n:∀o:∀p:〈〈〈〈∃q:i=(n+ (q· S(f· SSl))) ∧ ∃q:(n+ q)=(f· SSl)〉 ∧ ∃q:((l+ o)+ q)=j〉 ∧ 〈∃q:i=(p+ (q· S(f· SS(l+ o)))) ∧ ∃q:(p+ q)=(f· SS(l+ o))〉〉 ⇒ ∃q:(p+ q)=n〉〉 ∧ 〈〈〈〈〈〈〈〈〈〈〈∀n:∀o:〈〈∃p:S(n+ p)=m ∧ 〈∃p:e=(o+ (p· S(f· SS(n+ j)))) ∧ ∃p:(o+ p)=(f· SS(n+ j))〉〉 ⇒ ∃p:e=(o+ (p· S(f· SS(n+ l))))〉 ∨ 〈〈〈∃n:e=SSS(n· S(f· SSj)) ∧ ∃n:SSSn=(f· SSj)〉 ∧ k=m〉 ∧ ∃n:∃o:〈〈S(n+ o)=k ∧ ∀p:∀q:〈〈∃r:S(p+ r)=o ∧ 〈∃r:e=(q+ (r· S(f· SSS(p+ j)))) ∧ ∃r:(q+ r)=(f· SSS(p+ j))〉〉 ⇒ ∃r:e=(q+ (r· S(f· SSS(p+ (l+ n)))))〉〉 ∧ ∀p:∀q:〈〈∃r:S(p+ r)=n ∧ 〈∃r:e=(q+ (r· S(f· SSS(p+ (j+ o))))) ∧ ∃r:(q+ r)=(f· SSS(p+ (j+ o)))〉〉 ⇒ ∃r:e=(q+ (r· S(f· SSS(p+ l))))〉〉〉〉 ∨ 〈〈〈∃n:e=SSSSSSS(n· S(f· SSj)) ∧ ∃n:SSSSSSSn=(f· SSj)〉 ∧ k=SSm〉 ∧ ∀n:∀o:〈〈∃p:S(n+ p)=m ∧ 〈∃p:e=(o+ (p· S(f· SSSS(n+ j)))) ∧ ∃p:(o+ p)=(f· SSSS(n+ j))〉〉 ⇒ 〈∃p:e=(o+ (p· S(f· SS(n+ l)))) ∧ ∃p:(o+ p)=(f· SS(n+ l))〉〉〉〉 ∨ 〈〈〈∃n:e=SSSSSSS(n· S(f· SSl)) ∧ ∃n:SSSSSSSn=(f· SSl)〉 ∧ m=SSk〉 ∧ ∀n:∀o:〈〈∃p:S(n+ p)=k ∧ 〈∃p:e=(o+ (p· S(f· SSSS(n+ l)))) ∧ ∃p:(o+ p)=(f· SSSS(n+ l))〉〉 ⇒ 〈∃p:e=(o+ (p· S(f· SS(n+ j)))) ∧ ∃p:(o+ p)=(f· SS(n+ j))〉〉〉〉 ∨ ∃n:∃o:〈〈〈〈〈〈∃p:h=(n+ (p· S(f· SSSl))) ∧ ∃p:(n+ p)=(f· SSSl)〉 ∧ S(n+ o)=m〉 ∧ 〈∃p:e=SSSSSSSS(p· S(f· SSSj)) ∧ ∃p:SSSSSSSSp=(f· SSSj)〉〉 ∧ 〈∃p:e=SSSSSSSS(p· S(f· SSSS(j+ n))) ∧ ∃p:SSSSSSSSp=(f· SSSS(j+ n))〉〉 ∧ ∀p:∀q:〈〈∃r:S(p+ r)=n ∧ 〈∃r:e=(q+ (r· S(f· SSSS(p+ j)))) ∧ ∃r:(q+ r)=(f· SSSS(p+ j))〉〉 ⇒ 〈∃r:e=(q+ (r· S(f· SSS(p+ l)))) ∧ ∃r:(q+ r)=(f· SSS(p+ l))〉〉〉 ∧ ∀p:∀q:〈〈∃r:S(p+ r)=o ∧ 〈∃r:e=(q+ (r· S(f· SSSSS(p+ (j+ n))))) ∧ ∃r:(q+ r)=(f· SSSSS(p+ (j+ n)))〉〉 ⇒ 〈∃r:e=(q+ (r· S(f· SSS(p+ (l+ n))))) ∧ ∃r:(q+ r)=(f· SSS(p+ (l+ n)))〉〉〉〉 ∨ ∃n:∃o:〈〈〈〈〈〈∃p:h=(n+ (p· S(f· SSSj))) ∧ ∃p:(n+ p)=(f· SSSj)〉 ∧ S(n+ o)=k〉 ∧ 〈∃p:e=SSSSSSSS(p· S(f· SSSl)) ∧ ∃p:SSSSSSSSp=(f· SSSl)〉〉 ∧ 〈∃p:e=SSSSSSSS(p· S(f· SSSS(l+ n))) ∧ ∃p:SSSSSSSSp=(f· SSSS(l+ n))〉〉 ∧ ∀p:∀q:〈〈∃r:S(p+ r)=n ∧ 〈∃r:e=(q+ (r· S(f· SSSS(p+ l)))) ∧ ∃r:(q+ r)=(f· SSSS(p+ l))〉〉 ⇒ 〈∃r:e=(q+ (r· S(f· SSS(p+ j)))) ∧ ∃r:(q+ r)=(f· SSS(p+ j))〉〉〉 ∧ ∀p:∀q:〈〈∃r:S(p+ r)=o ∧ 〈∃r:e=(q+ (r· S(f· SSSSS(p+ (l+ n))))) ∧ ∃r:(q+ r)=(f· SSSSS(p+ (l+ n)))〉〉 ⇒ 〈∃r:e=(q+ (r· S(f· SSS(p+ (j+ n))))) ∧ ∃r:(q+ r)=(f· SSS(p+ (j+ n)))〉〉〉〉 ∨ 〈〈∃n:e=SS(n· S(f· SSl)) ∧ ∃n:SSn=(f· SSl)〉 ∧ ∀n:∀o:〈〈∃p:S(n+ p)=k ∧ 〈∃p:e=(o+ (p· S(f· SS(n+ j)))) ∧ ∃p:(o+ p)=(f· SS(n+ j))〉〉 ⇒ ∃p:e=(o+ (p· S(f· SSS(n+ l))))〉〉〉 ∨ ∃n:〈〈〈∃o:h=(n+ (o· S(f· SSSl))) ∧ ∃o:(n+ o)=(f· SSSl)〉 ∧ 〈∃o:e=SS(o· S(f· SSl)) ∧ ∃o:SSo=(f· SSl)〉〉 ∧ ∀o:∀p:〈〈∃q:S(o+ q)=k ∧ 〈∃q:e=(p+ (q· S(f· SS(o+ j)))) ∧ ∃q:(p+ q)=(f· SS(o+ j))〉〉 ⇒ ∃q:e=(p+ (q· S(f· SSS(o+ (l+ n)))))〉〉〉 ∨ 〈k=SSm ∧ ∃n:∃o:〈〈〈〈S(n+ o)=m ∧ 〈∃p:e=S(p· S(f· SSj)) ∧ ∃p:Sp=(f· SSj)〉〉 ∧ 〈∃p:e=S(p· S(f· SSl)) ∧ ∃p:Sp=(f· SSl)〉〉 ∧ ∀p:∀q:〈〈∃r:S(p+ r)=o ∧ 〈∃r:e=(q+ (r· S(f· SSSSS(p+ (j+ n))))) ∧ ∃r:(q+ r)=(f· SSSSS(p+ (j+ n)))〉〉 ⇒ ∃r:e=(q+ (r· S(f· SSS(p+ l))))〉〉 ∧ ∀p:∀q:〈〈∃r:S(p+ r)=n ∧ 〈∃r:e=(q+ (r· S(f· SSSS(p+ j)))) ∧ ∃r:(q+ r)=(f· SSSS(p+ j))〉〉 ⇒ ∃r:e=(q+ (r· S(f· SSS(p+ (l+ o)))))〉〉〉〉 ∨ ∃n:∃o:∃p:∃q:〈〈〈m=SSq ∧ 〈∃r:e=SSSSSS(r· S(f· SSl)) ∧ ∃r:SSSSSSr=(f· SSl)〉〉 ∧ 〈∃r:e=(n+ (r· S(f· SSSl))) ∧ ∃r:(n+ r)=(f· SSSl)〉〉 ∧ 〈∀r:〈∃s:〈∃t:S(s+ t)=p ∧ 〈∃t:e=(r+ (t· S(f· S(o+ s)))) ∧ ∃t:(r+ t)=(f· S(o+ s))〉〉 ⇒ ∀s:¬〈〈∃t:S(s+ t)=q ∧ 〈∃t:e=SSSSSS(t· S(f· SSSS(l+ s))) ∧ ∃t:SSSSSSt=(f· SSSS(l+ s))〉〉 ∧ 〈∃t:e=(r+ (t· S(f· SSSSS(l+ s)))) ∧ ∃t:(r+ t)=(f· SSSSS(l+ s))〉〉〉 ∧ ∃r:∃s:〈〈∃t:r=(t· Ss) ∧ 〈∃t:r=(k+ (t· S(s· Sq))) ∧ ∃t:(k+ t)=(s· Sq)〉〉 ∧ ∀t:∀u:∀v:∀w:〈〈〈〈∃x:S(t+ x)=q ∧ 〈∃x:r=(u+ (x· S(s· St))) ∧ ∃x:(u+ x)=(s· St)〉〉 ∧ 〈∃x:r=(v+ (x· S(s· SSt))) ∧ ∃x:(v+ x)=(s· SSt)〉〉 ∧ 〈∃x:e=(w+ (x· S(f· SSSS(l+ t)))) ∧ ∃x:(w+ x)=(f· SSSS(l+ t))〉〉 ⇒ 〈〈¬w=n ⇒ 〈〈∃x:e=(w+ (x· S(f· SS(j+ u)))) ∧ ∃x:(w+ x)=(f· SS(j+ u))〉 ∧ v=Su〉〉 ∧ 〈w=n ⇒ 〈∀x:∀y:〈〈∃z:S(x+ z)=p ∧ 〈∃z:e=(y+ (z· S(f· S(o+ x)))) ∧ ∃z:(y+ z)=(f· S(o+ x))〉〉 ⇒ 〈∃z:e=(y+ (z· S(f· SS((x+ u)+ j)))) ∧ ∃z:(y+ z)=(f· SS((x+ u)+ j))〉〉 ∧ v=(p+ u)〉〉〉〉〉〉〉〉 ∨ ∃n:〈∀o:〈〈〈∃p:S(o+ p)=j ∧ ∃p:e=(p· S(f· So))〉 ∧ ∀p:∀q:∀r:〈〈〈〈∃s:i=(q+ (s· S(f· So))) ∧ ∃s:(q+ s)=(f· So)〉 ∧ 〈∃s:i=(r+ (s· S(f· SS(p+ o)))) ∧ ∃s:(r+ s)=(f· SS(p+ o))〉〉 ∧ ∃s:S((o+ p)+ s)=j〉 ⇒ ∃s:S(q+ s)=r〉〉 ⇒ ∀p:¬〈〈〈∃q:h=(p+ (q· S(f· So))) ∧ ∃q:(p+ q)=(f· So)〉 ∧ ∃q:〈∃r:S(q+ r)=p ∧ 〈∃r:e=(n+ (r· S(f· S(o+ q)))) ∧ ∃r:(n+ r)=(f· S(o+ q))〉〉〉 ∧ ∀q:¬〈〈∃r:S(q+ r)=p ∧ 〈∃r:e=SSSSSS(r· S(f· S(o+ q))) ∧ ∃r:SSSSSSr=(f· S(o+ q))〉〉 ∧ 〈∃r:e=(n+ (r· S(f· SS(o+ q)))) ∧ ∃r:(n+ r)=(f· SS(o+ q))〉〉〉〉 ∧ 〈〈〈∃o:e=SSSSSS(o· S(f· SSj)) ∧ ∃o:SSSSSSo=(f· SSj)〉 ∧ 〈∃o:e=(n+ (o· S(f· SSSj))) ∧ ∃o:(n+ o)=(f· SSSj)〉〉 ∧ ∀o:∀p:〈〈∃q:S(o+ q)=m ∧ 〈∃q:e=(p+ (q· S(f· SSSS(o+ j)))) ∧ ∃q:(p+ q)=(f· SSSS(o+ j))〉〉 ⇒ 〈∃q:e=(p+ (q· S(f· SS(o+ l)))) ∧ ∃q:(p+ q)=(f· SS(o+ l))〉〉〉〉〉 ∨ ∃n:∃o:〈〈〈〈∃p:S(n+ p)=j ∧ ∃p:e=(p· S(f· Sn))〉 ∧ 〈∃p:h=(o+ (p· S(f· SSn))) ∧ ∃p:(o+ p)=(f· SSn)〉〉 ∧ ∀p:∀q:∀r:〈〈〈〈∃s:i=(p+ (s· S(f· SSn))) ∧ ∃s:(p+ s)=(f· SSn)〉 ∧ ∃s:((n+ q)+ s)=j〉 ∧ 〈∃s:i=(r+ (s· S(f· SS(n+ q)))) ∧ ∃s:(r+ s)=(f· SS(n+ q))〉〉 ⇒ ∃s:(r+ s)=p〉〉 ∧ 〈〈〈〈〈〈〈∃p:e=SS(p· S(f· SSj)) ∧ ∃p:SSp=(f· SSj)〉 ∧ ∀p:∀q:〈〈∃r:S(p+ r)=m ∧ 〈∃r:e=(q+ (r· S(f· SSS(p+ j)))) ∧ ∃r:(q+ r)=(f· SSS(p+ j))〉〉 ⇒ ∃r:e=(q+ (r· S(f· SS(p+ l))))〉〉 ∧ ∀p:∀q:〈〈∃r:S(p+ r)=o ∧ 〈∃r:e=(q+ (r· S(f· SSS(p+ (j+ m))))) ∧ ∃r:(q+ r)=(f· SSS(p+ (j+ m)))〉〉 ⇒ ∃r:e=(q+ (r· S(f· SS(p+ n))))〉〉 ∨ 〈〈〈∃p:e=S(p· S(f· SSl)) ∧ ∃p:Sp=(f· SSl)〉 ∧ ∀p:∀q:〈〈∃r:S(p+ r)=o ∧ 〈∃r:e=(q+ (r· S(f· SSS(p+ l)))) ∧ ∃r:(q+ r)=(f· SSS(p+ l))〉〉 ⇒ 〈∃r:e=(q+ (r· S(f· SS(p+ n)))) ∧ ∃r:(q+ r)=(f· SS(p+ n))〉〉〉 ∧ ∀p:∀q:〈〈∃r:S(p+ r)=k ∧ 〈∃r:e=(q+ (r· S(f· SS(p+ j)))) ∧ ∃r:(q+ r)=(f· SS(p+ j))〉〉 ⇒ 〈∃r:e=(q+ (r· S(f· SSS(p+ (l+ o))))) ∧ ∃r:(q+ r)=(f· SSS(p+ (l+ o)))〉〉〉〉 ∨ 〈〈∃p:e=SSS(p· S(f· SSj)) ∧ ∃p:SSSp=(f· SSj)〉 ∧ ∃p:∃q:∃r:〈〈〈〈〈S(p+ q)=k ∧ S(p+ r)=m〉 ∧ S(r+ q)=o〉 ∧ ∀s:∀t:〈〈∃u:S(s+ u)=p ∧ 〈∃u:e=(t+ (u· S(f· SSS(s+ j)))) ∧ ∃u:(t+ u)=(f· SSS(s+ j))〉〉 ⇒ ∃u:e=(t+ (u· S(f· SSS(s+ l))))〉〉 ∧ ∀s:∀t:〈〈∃u:S(s+ u)=r ∧ 〈∃u:e=(t+ (u· S(f· SSS(s+ n)))) ∧ ∃u:(t+ u)=(f· SSS(s+ n))〉〉 ⇒ 〈∃u:e=(t+ (u· S(f· SSS(s+ (l+ p))))) ∧ ∃u:(t+ u)=(f· SSS(s+ (l+ p)))〉〉〉 ∧ ∀s:∀t:〈〈∃u:S(s+ u)=q ∧ 〈∃u:e=(t+ (u· S(f· SSS(s+ (j+ p))))) ∧ ∃u:(t+ u)=(f· SSS(s+ (j+ p)))〉〉 ⇒ ∃u:e=(t+ (u· S(f· SSS(s+ (n+ r)))))〉〉〉〉 ∨ ∃p:〈〈〈〈〈〈〈〈∃q:e=SSSSSS(q· S(f· SSj)) ∧ ∃q:SSSSSSq=(f· SSj)〉 ∧ 〈∃q:e=(p+ (q· S(f· SSSj))) ∧ ∃q:(p+ q)=(f· SSSj)〉〉 ∧ 〈∃q:e=SSSSSS(q· S(f· SSn)) ∧ ∃q:SSSSSSq=(f· SSn)〉〉 ∧ 〈∃q:e=(p+ (q· S(f· SSSn))) ∧ ∃q:(p+ q)=(f· SSSn)〉〉 ∧ 〈∃q:e=S(q· S(f· SSSSn)) ∧ ∃q:Sq=(f· SSSSn)〉〉 ∧ ∀q:∀r:〈〈∃s:S(q+ s)=m ∧ 〈∃s:e=(r+ (s· S(f· SSSS(q+ j)))) ∧ ∃s:(r+ s)=(f· SSSS(q+ j))〉〉 ⇒ 〈∃s:e=(r+ (s· S(f· SSSSS(q+ n)))) ∧ ∃s:(r+ s)=(f· SSSSS(q+ n))〉〉〉 ∧ ∀q:∀r:∀s:〈〈〈∃t:S(q+ t)=m ∧ 〈∃t:e=(r+ (t· S(f· SS(l+ q)))) ∧ ∃t:(r+ t)=(f· SS(l+ q))〉〉 ∧ 〈∃t:e=(s+ (t· S(f· SSSS(j+ q)))) ∧ ∃t:(s+ t)=(f· SSSS(j+ q))〉〉 ⇒ 〈〈s=p ⇒ r=SSSSSSSSS0〉 ∧ 〈¬s=p ⇒ r=s〉〉〉〉 ∧ ∃q:∃r:〈〈∃s:q=SSSS((m+ l)+ (s· S(r· SSSSj))) ∧ ∃s:SSSS((m+ l)+ s)=(r· SSSSj)〉 ∧ ∀s:∀t:∀u:∀v:∀w:〈〈〈〈〈∃x:S(s+ x)=m ∧ 〈∃x:q=(t+ (x· S(r· SSSS(j+ s)))) ∧ ∃x:(t+ x)=(r· SSSS(j+ s))〉〉 ∧ 〈∃x:q=(u+ (x· S(r· SSSSS(j+ s)))) ∧ ∃x:(u+ x)=(r· SSSSS(j+ s))〉〉 ∧ 〈∃x:e=(v+ (x· S(f· SSSS(j+ s)))) ∧ ∃x:(v+ x)=(f· SSSS(j+ s))〉〉 ∧ 〈∃x:e=(w+ (x· S(f· St))) ∧ ∃x:(w+ x)=(f· St)〉〉 ⇒ 〈〈v=p ⇒ 〈〈u=SSt ∧ 〈∃x:e=(p+ (x· S(f· SSt))) ∧ ∃x:(p+ x)=(f· SSt)〉〉 ∧ w=SSSSSSSS0〉〉 ∧ 〈¬v=p ⇒ 〈u=St ∧ w=v〉〉〉〉〉〉〉 ∨ 〈∃p:〈〈〈S(n+ o)=j ∧ 〈∃q:i=(p+ (q· S(f· Sl))) ∧ ∃q:(p+ q)=(f· Sl)〉〉 ∧ 〈∃q:i=(p+ (q· S(f· SSj))) ∧ ∃q:(p+ q)=(f· SSj)〉〉 ∧ ∀q:∀r:〈〈∃s:S((l+ q)+ s)=j ∧ 〈∃s:i=(r+ (s· S(f· SS(l+ q)))) ∧ ∃s:(r+ s)=(f· SS(l+ q))〉〉 ⇒ ∃s:S(p+ s)=r〉〉 ∧ 〈〈〈∃p:e=S(p· S(f· SSj)) ∧ ∃p:Sp=(f· SSj)〉 ∧ ∀p:∀q:〈〈∃r:S(p+ r)=m ∧ 〈∃r:e=(q+ (r· S(f· SSS(p+ j)))) ∧ ∃r:(q+ r)=(f· SSS(p+ j))〉〉 ⇒ 〈∃r:e=(q+ (r· S(f· SS(p+ l)))) ∧ ∃r:(q+ r)=(f· SS(p+ l))〉〉〉 ∧ ∀p:∀q:〈〈∃r:S(p+ r)=o ∧ 〈∃r:e=(q+ (r· S(f· SS(p+ n)))) ∧ ∃r:(q+ r)=(f· SS(p+ n))〉〉 ⇒ 〈∃r:e=(q+ (r· S(f· SSS(p+ (j+ m))))) ∧ ∃r:(q+ r)=(f· SSS(p+ (j+ m)))〉〉〉〉〉〉〉〉〉〉〉〉</p>
| <p>Gödel's second incompleteness theorem has a succinct description:</p>
<p>$$ PA \not{\vdash} Con(PA). $$</p>
|
probability | <p><strong>Question :</strong> What is the difference between Average and Expected value?</p>
<hr />
<p>I have been going through the definition of expected value on <a href="http://en.wikipedia.org/wiki/Expected_value" rel="noreferrer">Wikipedia</a> beneath all that jargon it seems that the expected value of a distribution is the average value of the distribution. Did I get it right ?</p>
<p>If yes, then what is the point of introducing a new term ? Why not just stick with the average value of the distribution ?</p>
| <p>The concept of expectation value or expected value may be understood from the following example. Let $X$ represent the outcome of a roll of an unbiased six-sided die. The possible values for $X$ are 1, 2, 3, 4, 5, and 6, each having the probability of occurrence of 1/6. The expectation value (or expected value) of $X$ is then given by </p>
<p>$(X)\text{expected} = 1(1/6)+2\cdot(1/6)+3\cdot(1/6)+4\cdot(1/6)+5\cdot(1/6)+6\cdot(1/6) = 21/6 = 3.5$</p>
<p>Suppose that in a sequence of ten rolls of the die, if the outcomes are 5, 2, 6, 2, 2, 1, 2, 3, 6, 1, then the average (arithmetic mean) of the results is given by</p>
<p>$(X)\text{average} = (5+2+6+2+2+1+2+3+6+1)/10 = 3.0$</p>
<p>We say that the average value is 3.0, with the distance of 0.5 from the expectation value of 3.5. If we roll the die $N$ times, where $N$ is very large, then the average will converge to the expected value, i.e.,$(X)\text{average}=(X)\text{expected}$. This is evidently because, when $N$ is very large each possible value of $X$ (i.e. 1 to 6) will occur with equal probability of 1/6, turning the average to the expectation value.</p>
| <p>The expected value, or mean $\mu_X =E_X[X]$, is a parameter associated with the distribution of a random variable $X$.</p>
<p>The average $\overline X_n$ is a computation performed on a sample of size $n$ from that distribution. It can also be regarded as an unbiased estimator of the mean, meaning that if each $X_i\sim X$, then $E_X[\overline X_n] = \mu_X$.</p>
|
number-theory | <blockquote>
<p>Let $a,b,c \in \mathbb N$ find integer in the form: $$I=\frac{a}{b+c}
+ \frac{b}{c+a} + \frac{c} {a+b}$$</p>
</blockquote>
<p>Using Nesbitt's inequality: $I \ge \frac 32$</p>
<p>I am trying to prove $I \le 2$ to implies there $\nexists \ a,b,c$ such that $I\in \mathbb Z$: $$I\le 2 \\ \iff c{a}^{2}+3\,acb+{a}^{3}+{a}^{2}b+a{b}^{2}+{b}^{3}+c{b}^{2}+{c}^{2}b+{c
}^{3}+{c}^{2}a-2\, \left( b+c \right) \left( c+a \right) \left( a+b
\right) \le 0 \\ \iff {a}^{3}+{b}^{3}+{c}^{3}\leq c{a}^{2}+{a}^{2}b+a{b}^{2}+c{b}^{2}+{c}^{2
}a+{c}^{2}b+abc
$$
and stuck.</p>
<p>EDIT: Look like prove $I \le 2$ not a good thinking :P</p>
| <p>I find a solution to $n=4,$when</p>
<blockquote>
<p>a=2332797891204725453580403814216955612718693675609518139813675622446336
8530351921955206357565424226029748329737767516130520072674084336131550
2597616224970927979227396663481447506019173462295157784788781420305046
5201815993661680059006448575315523206103260762210944137954571975497854
9786027663601160534574317253280344812956727894696796553762212813889660
9065956718516224446015577143267128739011935697434909021669583635379832
35022557869209259</p>
<p>b=1161640217306132458900911441651415023972393417197892812143262449233898
8034221463466278254018560734492913221738943224762433374574861704275058
0062902808034990817009121975186967451351814311101112040391014295321972
8784138582766210837461563508481437266175417187186208008663435889653439
7066554486263784443013141020886435995672339322997499528376940620045001
1919735272479457688230567501843839892799164246003766614214017398378635
0444307965016411</p>
<p>c=5054729227475450427274369484803239479825091305751388135572448603576037
6549781961422098862259430557133842304461180359698183208339647924784255
6816542651386138853492649101592171641096957016404851774814750638840260
3496289958758089911825477669004739864966841494437579004665357462952425
4130327474390635537868978719887059697148297723373566417781389238382736
3204638301684342182024187145267526992579708085994452308601529371953916
7125415529515145</p>
</blockquote>
<p>This may be not the smallest solution to $n=4$,but I think the smallest one is also very huge.</p>
<p>I'd like to tell something about how I get this answer.</p>
<p>I estimate that there will be a solution to $n=4,$(I also tried $n=3$,but failed.)</p>
<p>Let $a,b,c$ be rational numbers,and $a+b+c=1,a,b,c>0,$we get
$$\frac{a}{1-a}+\frac{b}{1-b}+\frac{1-a-b}{a+b}=4.\tag{eq.1}$$
denote $\dfrac{a}{1-a}=s,\dfrac{b}{1-b}=t,$then $a=\dfrac{s}{1+s},b=\dfrac{t}{1+t},$
$eq.1$ is equivalent to $$s+t+\dfrac{1-st}{s+t+2st}=4,\tag{eq.2}$$
denote $s=x+y,t=x-y,$then$$\frac{4 x^3+3 x^2-4 x y^2+y^2+1}{2 \left(x^2+x-y^2\right)}=4$$
$$y^2=\frac{4 x^3-5 x^2-8 x+1}{4 x-9}$$
denote $x=\dfrac{9k+1}{4k},$then $$y^2=\frac{(13 k+1) \left(4 k^2+9 k+1\right)}{16 k^2}$$
denote $r=4ky,$then $$r^2=(13 k+1) \left(4 k^2+9 k+1\right)=1 + 22 k + 121 k^2 + 52 k^3$$
denote $X=52k,Y=52r,$then $$Y^2 = X^3 + 121 X^2 + 1144 X + 2704.\tag{eq.3}$$
So we just need to find a rational solution of $eq.3$ which makes $a,b,c>0.$</p>
<p>I find two basic solutions of $eq.3,(X,Y)=(-104,260)(52,728),$we denote $p=(-104,260),q=(52,728),$</p>
<p>At first,I find that $6q=O,$when I calculate $p,2p,3p,4p,\cdots$ ,I find that $21p$ makes $-52<X<-33.8564,$in fact,$X\approx -36.8062,$this is what we need,because when $-52<X<-33.8564,a,b,c$ will all be positive(it need some calculation).</p>
<p>Finally,we can get $a,b,c>0$ from $X,Y,$ and then multiply by the LCM of the denominators of $a,b,c,$ we are done.</p>
<p>You can see that I only use $p,$ but no $q$ or other points on $eq.3,$so I have to plus $21$ times by $p$,which makes the digits so huge.</p>
| <p>I found a few solutions $(a,b,c)$ of equatoin
$$
\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=n \qquad\qquad (a,b,c,n\in\mathbb{N})\tag{1}
$$
for even $n$.
<br/>
And I have a doubt whether there are smaller solutions.</p>
<hr>
<p>$\color{#FF4400}{n=4:}$ </p>
<p>$\small
a=4373612677928697257861252602371390152816537558161613618621437993378423467772036$;</p>
<p>$\small b=36875131794129999827197811565225474825492979968971970996283137471637224634055579$;</p>
<p>$\small c=154476802108746166441951315019919837485664325669565431700026634898253202035277999$.</p>
<p>This solution is much smaller, but is still big: <br/>
$\log_2 (a+b+c) \approx 266.723$,
$\log_{10} (a+b+c) \approx 80.2916$.</p>
<p>Other solutions for $n=4$:</p>
<p>a=16666476865438449865846131095313531540647604679654766832109616387367203990642764342248100534807579493874453954854925352739900051220936419971671875594417036870073291371;</p>
<p>b=184386514670723295219914666691038096275031765336404340516686430257803895506237580602582859039981257570380161221662398153794290821569045182385603418867509209632768359835;</p>
<p>c=32343421153825592353880655285224263330451946573450847101645239147091638517651250940206853612606768544181415355352136077327300271806129063833025389772729796460799697289;</p>
<p>$\log_2(a+b+c)\approx 555.985$;
$\log_{10}(a+b+c) \approx 167.368$.</p>
<p>a=1054210182683112310528012408530531909717229064191793536540847847817849001214642792626066010344383473173101972948978951703027097154519698536728956323881063669558925110120619283730835864056709609662983759100063333396875182094245046315497525532634764115913236450532733839386139526489824351;</p>
<p>b=1440354387400113353318275132419054375891245413681864837390427511212805748408072838847944629793120889446685643108530381465382074956451566809039119353657601240377236701038904980199109550001860607309184336719930229935342817546146083848277758428344831968440238907935894338978800768226766379;</p>
<p>c=9391500403903773267688655787670246245493629218171544262747638036518222364768797479813561509116827252710188014736501391120827705790025300419608858224262849244058466770043809014864245428958116544162335497194996709759345801074510016208346248254582570123358164225821298549533282498545808644;</p>
<p>$\log_2(a+b+c)\approx 950.321$;
$\log_{10}(a+b+c)\approx 286.075$.</p>
<hr>
<p>$\color{#FF4400}{n=6:}$ </p>
<p>a=1218343242702905855792264237868803223073090298310121297526752830558323845503910071851999217959704024280699759290559009162035102974023;</p>
<p>b=2250324022012683866886426461942494811141200084921223218461967377588564477616220767789632257358521952443049813799712386367623925971447;</p>
<p>c=20260869859883222379931520298326390700152988332214525711323500132179943287700005601210288797153868533207131302477269470450828233936557.</p>
<p>$\log_2 (a+b+c) \approx 443.063$;
$\log_{10} (a+b+c) \approx 133.375$.</p>
<hr>
<p>$\color{#FF4400}{n=10:}$ </p>
<p>a=221855981602380704196804518854316541759883857932028285581812549404634844243737502744011549757448453135493556098964216532950604590733853450272184987603430882682754171300742698179931849310347;</p>
<p>b=269103113846520710198086599018316928810831097261381335767926880507079911347095440987749703663156874995907158014866846058485318408629957749519665987782327830143454337518378955846463785600977;</p>
<p>c=4862378745380642626737318101484977637219057323564658907686653339599714454790559130946320953938197181210525554039710122136086190642013402927952831079021210585653078786813279351784906397934209.</p>
<p>$\log_2(a+b+c) \approx 630.265$;
$\log_{10}(a+b+c) \approx 189.729$.</p>
<hr>
<p>Instead of $a,b,c \in \mathbb{N}$, we can consider positive rational numbers $p,q,r$:
$$
p = \dfrac{a}{a+b+c}, \qquad q = \dfrac{b}{a+b+c}, \qquad r = \dfrac{c}{a+b+c} = 1-p-q.
$$</p>
<p>So, we'll search rational solutions of equation
$$
\dfrac{p}{1-p}+\dfrac{q}{1-q}+\dfrac{1-p-q}{2p} = n, \qquad (n\in \mathbb{N}), \tag{2}
$$
like (eq.1) in <strong>Hecke</strong>'s answer.</p>
<p>Denote $p=m-d$, $\quad$ $q=m+d$. $\quad$ $\quad$ $(2) \rightarrow (3)$:
$$
\dfrac{m-d}{(1-m)-d} + \dfrac{m+d}{(1-m)+d} + \dfrac{1-2m}{2m} = n, \tag{3}
$$
then $(3) \rightarrow (4)$:</p>
<p>$$
\Bigl((2n+6)m-1\Bigr)d^2 = (2n+6)m^3 - (4n+9)m^2+(2n+4)m-1. \tag{4}
$$</p>
<p>Denote
$m = \dfrac{x-1}{(2n+6)x}$, $d = \dfrac{y}{(2n+6)x}$. Then we get equation:
$$
y^2 = 1 + (4n+6)x + (2n+3)^2x^2 + (8n+20)x^3, \qquad n \in \mathbb{N}; \tag{5}
$$
$$
y^2 = \Bigl( 1 + (2n+5)x \Bigr) \Bigl(1 + (2n+1)x + 4x^2 \Bigr), \qquad n \in \mathbb{N}.
$$</p>
<p>We need to find rational solutions $(x,y)$ of equation $(5)$, which give $ |d| <m<0.5$.</p>
<p>If $n=4$, then we consider rational points on elliptic curve $y^2=1 + 22 x + 121 x^2 + 52 x^3$.
<br/>
Starting rational points on this EC: <br/>
points, where $x$ is $-2, 0, 1, \dfrac{-1}{4}, \dfrac{-5}{9}, \dfrac{-15}{13}, \dfrac{-2}{13}, \dfrac{-1}{13}, \dfrac{-3}{25}, \dfrac{168}{25}, \dfrac{611}{121}, \ldots$ .
<br/>
We need to find/construct one with $x \in (-2.13278;-2.08649) \bigcup (-0.22120;-0.16667)$.</p>
<p>If $n=6$, then we consider rational points on elliptic curve $y^2=1 + 30 x + 225 x^2 + 68 x^3$, <br/> where
$x \in (-3.17116;-3.14447) \bigcup (-0.14965;-0.125)$.</p>
<p>If we'll find appropriate $(x,y)$, then
$$
p = \dfrac{x-y-1}{(2n+6)x}, \qquad
q = \dfrac{x+y-1}{(2n+6)x}, \qquad
r = \dfrac{(2n+4)x+2}{(2n+6)x}.
$$</p>
<hr>
<p>It seems, that if $n$ is odd, then there are only 5 rational points on corresponding EC, and there is no integer solution $(a,b,c)$ of equation $(1)$.</p>
<p>In the most cases of even $n$ $(n=2,4,6,10,12,14,...)$ there are many rational points on corresponding EC.</p>
|
geometry | <blockquote>
<p><strong>Problem:</strong></p>
<p><em>A vertex of one square is pegged to the centre of an identical square, and the overlapping area is blue. One of the squares is then rotated about the vertex and the resulting overlap is red.</em></p>
<p><em>Which area is greater?</em></p>
</blockquote>
<p><img src="https://i.sstatic.net/jmjxu.png" alt="a diagram showing overlapped squares, one forming a smaller blue square and the other an irregular red quadrilateral" /></p>
<p>Let the area of each large square be exactly <span class="math-container">$1$</span> unit squared. Then, the area of the blue square is exactly <span class="math-container">$1/4$</span> units squared. The same would apply to the red area if you were to rotate the square <span class="math-container">$k\cdot 45$</span> degrees for a natural number <span class="math-container">$k$</span>.</p>
<p>Thus, I am assuming that no area is greater, and that it is a trick question <span class="math-container">$-$</span> although the red area might appear to be greater than the blue area, they are still the same: <span class="math-container">$1/4$</span>.</p>
<p>But how can it be <em>proven</em>?</p>
<p>I know the area of a triangle with a base <span class="math-container">$b$</span> and a height <span class="math-container">$h\perp b$</span> is <span class="math-container">$bh\div 2$</span>. Since the area of each square is exactly <span class="math-container">$1$</span> unit squared, then each side would also have a length of <span class="math-container">$1$</span>.</p>
<p>Therefore, the height of the red triangle area is <span class="math-container">$1/2$</span>, and so <span class="math-container">$$\text{Red Area} = \frac{b\left(\frac 12\right)}{2} = \frac{b}{4}.$$</span></p>
<p>According to the diagram, the square has not rotated a complete <span class="math-container">$45$</span> degrees, so <span class="math-container">$b < 1$</span>. It follows, then, that <span class="math-container">$$\begin{align} \text{Red Area} &< \frac 14 \\ \Leftrightarrow \text{Red Area} &< \text{Blue Area}.\end{align}$$</span></p>
<blockquote>
<p><strong>Assertion:</strong></p>
<p>To conclude, the <span class="math-container">$\color{blue}{\text{blue}}$</span> area is greater than the <span class="math-container">$\color{red}{\text{red}}$</span> area.</p>
</blockquote>
<p>Is this true? If so, is there another way of proving the assertion?</p>
<hr />
<p>Thanks to users who commented below, I did not take account of the fact that the red area is <em>not</em> a triangle <span class="math-container">$-$</span> it does not have three sides! This now leads back to my original question on whether my hypothesis was correct.</p>
<p>This question is very similar to <a href="https://math.stackexchange.com/questions/2369403/which-area-is-larger-the-blue-area-or-the-white-area?rq=1">this post</a>.</p>
<hr />
<p><strong>Source:</strong></p>
<p><a href="https://www.youtube.com/watch?v=sj8Sg8qnjOg" rel="noreferrer">The Golden Ratio (why it is so irrational) <span class="math-container">$-$</span> Numberphile</a> from <span class="math-container">$14$</span>:<span class="math-container">$02$</span>.</p>
| <p><a href="https://i.sstatic.net/mXMhQ.jpg" rel="noreferrer"><img src="https://i.sstatic.net/mXMhQ.jpg" alt="some text"></a> </p>
<p>The four numbered areas are congruent.</p>
<hr>
<p>[Added later] The figure below is from a suggested edit by @TomZych, and it shows the congruent parts more clearly. Given all the upvotes to the (probably tongue-in-cheek) comment “This answer also deserves the tick for artistic reasons,” I’m leaving my original “artistic” figure but also adding Tom’s improved version to my answer.</p>
<p><a href="https://i.sstatic.net/RgrTs.png" rel="noreferrer"><img src="https://i.sstatic.net/RgrTs.png" alt="enter image description here"></a></p>
| <p>I think sketching the two identical triangles marked with green below makes this rather intuitive. This could also be turned into a formal proof quite easily.</p>
<p><a href="https://i.sstatic.net/zilQU.png" rel="noreferrer"><img src="https://i.sstatic.net/zilQU.png" alt="Identical triangles"></a></p>
|
linear-algebra | <p>Singular value decomposition (<a href="http://en.wikipedia.org/wiki/Singular_value_decomposition" rel="noreferrer">SVD</a>) and principal component analysis (<a href="https://en.wikipedia.org/wiki/Principal_component_analysis" rel="noreferrer">PCA</a>) are two eigenvalue methods used to reduce a high-dimensional data set into fewer dimensions while retaining important information. Online articles say that these methods are 'related' but never specify the exact relation.</p>
<p>What is the intuitive relationship between PCA and SVD? As PCA uses the SVD in its calculation, clearly there is some 'extra' analysis done. What does PCA 'pay attention' to differently than the SVD? What kinds of relationships do each method utilize more in their calculations? Is one method 'blind' to a certain type of data that the other is not?</p>
| <p>(I assume for the purposes of this answer that the data has been preprocessed to have zero mean.)</p>
<p>Simply put, the PCA viewpoint requires that one compute the eigenvalues and eigenvectors of the covariance matrix, which is the product <span class="math-container">$\frac{1}{n-1}\mathbf X\mathbf X^\top$</span>, where <span class="math-container">$\mathbf X$</span> is the data matrix. Since the covariance matrix is symmetric, the matrix is diagonalizable, and the eigenvectors can be normalized such that they are orthonormal:</p>
<p><span class="math-container">$\frac{1}{n-1}\mathbf X\mathbf X^\top=\frac{1}{n-1}\mathbf W\mathbf D\mathbf W^\top$</span></p>
<p>On the other hand, applying SVD to the data matrix <span class="math-container">$\mathbf X$</span> as follows:</p>
<p><span class="math-container">$\mathbf X=\mathbf U\mathbf \Sigma\mathbf V^\top$</span></p>
<p>and attempting to construct the covariance matrix from this decomposition gives
<span class="math-container">$$
\frac{1}{n-1}\mathbf X\mathbf X^\top
=\frac{1}{n-1}(\mathbf U\mathbf \Sigma\mathbf V^\top)(\mathbf U\mathbf \Sigma\mathbf V^\top)^\top
= \frac{1}{n-1}(\mathbf U\mathbf \Sigma\mathbf V^\top)(\mathbf V\mathbf \Sigma\mathbf U^\top)
$$</span></p>
<p>and since <span class="math-container">$\mathbf V$</span> is an orthogonal matrix (<span class="math-container">$\mathbf V^\top \mathbf V=\mathbf I$</span>),</p>
<p><span class="math-container">$\frac{1}{n-1}\mathbf X\mathbf X^\top=\frac{1}{n-1}\mathbf U\mathbf \Sigma^2 \mathbf U^\top$</span></p>
<p>and the correspondence is easily seen (the square roots of the eigenvalues of <span class="math-container">$\mathbf X\mathbf X^\top$</span> are the singular values of <span class="math-container">$\mathbf X$</span>, etc.)</p>
<p>In fact, using the SVD to perform PCA makes much better sense numerically than forming the covariance matrix to begin with, since the formation of <span class="math-container">$\mathbf X\mathbf X^\top$</span> can cause loss of precision. This is detailed in books on numerical linear algebra, but I'll leave you with an example of a matrix that can be stable SVD'd, but forming <span class="math-container">$\mathbf X\mathbf X^\top$</span> can be disastrous, the <a href="https://doi.org/10.1007/BF01386022" rel="noreferrer">Läuchli matrix</a>:</p>
<p><span class="math-container">$\begin{pmatrix}1&1&1\\ \epsilon&0&0\\0&\epsilon&0\\0&0&\epsilon\end{pmatrix}^\top,$</span></p>
<p>where <span class="math-container">$\epsilon$</span> is a tiny number.</p>
| <p><em><a href="https://arxiv.org/pdf/1404.1100.pdf" rel="nofollow noreferrer">A tutorial on Principal Component Analysis</a></em> by Jonathon Shlens is a good tutorial on PCA and its relation to SVD. Specifically, section VI: A More General Solution Using SVD.</p>
|
logic | <p>In <a href="https://math.stackexchange.com/a/440219/1543">this answer</a>, user18921 wrote that the $\iff$ operation is associative, in the sense that</p>
<ul>
<li>$(A\iff B)\iff C$</li>
<li>$A\iff (B\iff C)$ </li>
</ul>
<p>are equivalent statements. </p>
<p>One can brute-force a proof fairly easily (just write down the truth table). But is there a more intuitive reason why this associativity must hold? </p>
| <p>I would say that I personally don't feel that this equivalence is that intuitive. (Maybe it is because I have been thinking too constructively these days.) In classical logic, we have the isomorphism with the Boolean algebra, and that makes it possible to think of $A \leftrightarrow B$ as $A \oplus B \oplus 1$ where $\oplus$ is the exclusive OR. This makes quite a short proof of the associativity of the biconditional. To me, this is the most intuitive way I can think of to justify the associativity.</p>
<p>But as J Marcos mentioned, the equivalence is no longer true in intuitionistic logic. This may serve as an explanation why the equivalence is not supposed to be as <em>intuitive</em>.</p>
<p>(It is straightforward to find a counterexample using the topological model. I will just work out the tedious details for you. Assume $\mathbb R$ is the topological space of interest with the usual topology. Define the valuation $[\![\cdot]\!]$ by $[\![A]\!] = (-1, 1)$, $[\![B]\!] = (0, \infty)$ and $[\![C]\!] = (-\infty, -1) \cup (0, 1)$. It follows that
\begin{align*}
[\![A \leftrightarrow B]\!] & =
[\![A \rightarrow B]\!] \cap [\![B \rightarrow A]\!]
= \text{int}([\![A]\!]^c \cup [\![B]\!]) \cap \text{int}([\![B]\!]^c \cup [\![A]\!]) \\
& = (-\infty, -1) \cup (0, 1) = [\![C]\!] \\
[\![(A \leftrightarrow B) \leftrightarrow C]\!]
& = [\![(A \leftrightarrow B) \to C]\!] \cap [\![C \to (A \leftrightarrow B)]\!]\\
& = \text{int}([\![A \leftrightarrow B]\!]^c \cup [\![C]\!]) \cap
\text{int}([\![C]\!]^c \cup [\![A \leftrightarrow B]\!]) = \mathbb R \\
[\![B \leftrightarrow C]\!] & =
[\![B \rightarrow C]\!] \cap [\![C \rightarrow B]\!] = \text{int}([\![B]\!]^c \cup [\![C]\!]) \cap \text{int}([\![C]\!]^c \cup [\![B]\!]) \\
& = (-1, 0) \cup (0, 1) = A - \{0\}\\
[\![A \leftrightarrow (B \leftrightarrow C)]\!]
& = [\![(B \leftrightarrow C) \to A]\!] \cap [\![A \to (B \leftrightarrow C)]\!] \\
& = \text{int}([\![B \leftrightarrow C]\!]^c \cup [\![A]\!]) \cap
\text{int}([\![A]\!]^c \cup [\![B \leftrightarrow C]\!]) \\
& = \mathbb R - \{0\} \ne \mathbb R
\end{align*}
where int is the interior and $^c$ is the complement.)</p>
| <p>This might be one of the situations where a more general (and therefore stronger) result is more intuitive (depending on how your intuition works). The result I have in mind is that, if you combine any list $\mathcal L$ of formulas (not just three like the $A,B,C$ in the question) by $\iff$, then no matter how you put the parentheses to make it unambiguous, the resulting formula is true if and only if an even number of the formulas in the list $\mathcal L$ are false. In other words, iterated $\iff$ amounts to a parity connective, regardless of the placement of parentheses.</p>
|
game-theory | <p>A stack of silver coins is on the table. </p>
<p>For each step we can either remove a silver coin and write the number of gold coins on a piece of paper, or we can add a gold coin and write the number of silver coins on another piece of paper. </p>
<p>We stop when only gold coins are left.</p>
<p>Prove that the sums of numbers on these two papers are equal.</p>
<p>Tried playing the game and the problem seems right each time, but I can't proceed from there. Is it done by induction?</p>
| <p>The state of the game can be desribed by
$$
(g,s,G,S),
$$
where $g$ is the number of golden coins on the table, $s$ is the number of silver coins on the table, $G$ is the sum of the numbers in the first paper, and $S$ is the sum of the numbers in the second paper. The initial state is $(0,n,0,0)$, and we want to show that if the state of the game is $(g,0,G,S)$, then $G=S$.</p>
<p>If we are at $(g_i,s_i,G_i,S_i)$ and add a golden coin, the state changes to
$$
(g_{i+1},s_{i+1},G_{i+1},S_{i+1}) = (g_i+1,s_i,G_i,S_i+s_i),
$$
and if we remove a silver coin, the state changes to
$$
(g_{i+1},s_{i+1},G_{i+1},S_{i+1}) = (g_i,s_i-1,G_i+g_i,S_i).
$$</p>
<p>One plan to solve the problem is to find an <em>invariant</em>, for example, a function from $(g,s,G,S)$ to integers, such that these transformations do not change the value of that function. Looking at the equations for a while suggests something with $gs$ because that's how we would get changes of size $g$ and $s$. A bit more looking gives us
$$
f(g,s,G,S) = gs+G-S.
$$
Once we have found the above formula, it is easy to verify that a step does not affect the value of $gs+G-S$. </p>
<p>Thus if we start from $f(0,n,0,0)=0$ and end with $f(g,0,G,S) = G-S$, we can see that $G=S$. </p>
| <p>When you add a gold coin, you write $n$ for the number of silver coins left.</p>
<p>Every time you remove one of those $n$ silver coins, that gold coin gets counted once as part of the number of gold coins - a total of $n$ times, since all the silver coins are eventually removed.</p>
|
linear-algebra | <p>When I came across the Cauchy-Schwarz inequality the other day, I found it really weird that this was its own thing, and it had lines upon lines of <a href="https://www.khanacademy.org/math/linear-algebra/vectors_and_spaces/dot_cross_products/v/proof-of-the-cauchy-schwarz-inequality">proof</a>.</p>
<p>I've always thought the geometric definition of dot multiplication:
$$|{\bf a }||{\bf b }|\cos \theta$$ is equivalent to the other, algebraic definition: $$a_1\cdot b_1+a_2\cdot b_2+\cdots+a_n\cdot b_n$$
And since the inequality is directly implied by the geometric definition (the fact that $\cos(\theta)$ is $1$ only when $\bf a$ and $\bf b$ are collinear), then shouldn't the Cauchy-Schwarz inequality be the world's most obvious and almost-no-proof-needed thing?</p>
<p>Can someone correct me on where my thought process went wrong?</p>
| <p><em>Side note: it's actually the Cauchy-Schwarz-Bunyakovsky inequality, and don't let anyone tell you otherwise.</em></p>
<p>The problem with using the geometric definition is that you have to define what an angle is. Sure, in three dimensional space, you have pretty clear ideas about what an angle is, but what do you take as $\theta$ in your equation when $i$ and $j$ are $10$ dimensional vectors? Or infinitely-dimensional vectors? What if $i$ and $j$ are polynomials?</p>
<p>The Cauchy-Schwarz inequality tells you that <strong>anytime</strong> you have a vector space and an inner product defined on it, you can be sure that for any two vectors $u,v$ in your space, it is true that $\left|\langle u,v\rangle\right| \leq \|u\|\|v\|$.</p>
<p>Not all vector spaces are simple $\mathbb R^n$ businesses, either. You have the vector space of all continuous functions on $[0,1]$, for example. You can define the inner product as</p>
<p>$$\langle f,g\rangle=\int_0^1 f(x)g(x)dx$$</p>
<p>and use Cauchy-Schwarz to prove that for any pair $f,g$, you have
$$\left|\int_{0}^1f(x)g(x)dx\right| \leq \sqrt{\int_0^1 f^2(x)dx\int_0^1g^2(x)dx}$$</p>
<p>which is not a trivial inequality.</p>
| <ol>
<li><p>The inequality is ubiquitous, so some name is needed.</p></li>
<li><p>As there is no cosine in the statement of the inequality, it cannot be called "cosine inequality" or anything like that.</p></li>
<li><p>The geometric interpretation with cosines only works for finite-dimensional real Euclidean space, but the inequality holds and is used more generally than that. That is Schwarz' contribution.</p></li>
<li><p>Schwarz founded the field of <em>functional analysis</em> (infinite-dimensional metrized linear algebra) with his proof of the inequality. That is important enough to warrant a name. In terms of consequences per line of proof it is one of the greatest arguments of all time.</p></li>
<li><p>The Schwarz proof was part of the historical realization that Euclidean geometry, with its mysterious angle measure that seems to depend on notions of arc-length from calculus, <em>is</em> the theory of a vector space equipped with a quadratic form. That is a major shift in viewpoint. </p></li>
<li><p>Stating the inequality in terms of cosines assumes that the inner product restricts to the standard Euclidean one on the 2 (or fewer) dimensional subspace spanned by the two vectors, and that you have proved the inequality for holds for standard Euclidean space of 2 dimensions or less. How do you know those things are correct without a much longer argument? That argument will, probably, include somewhere a proof of the Cauchy-Schwarz inequality, maybe written for 2-dimensions but working for the whole $n$-dimensional space, so it might as well be stated as a direct proof for $n$ dimensions. Which is what Cauchy and Schwarz did.</p></li>
</ol>
|
combinatorics | <p>At the end of Probability class, our professor gave us the following puzzle:</p>
<blockquote>
<p>There are 100 bags each with 100 coins, but only one of these bags has gold coins in it. The gold coin has weight of 1.01 grams and the other coins has weight of 1 gram. We are given a digital scale, but we can only use it once. How can we identify the bag of gold coins?</p>
</blockquote>
<p>After about 5 minutes waiting, our professor gave us the solution (the class had ended and he didn't want to wait any longer):</p>
<blockquote>
<p>Give the bags numbers from 0 through 99, then take 0 coins from the bag number 0, 1 coin from the bag number 1, 2 coins from the bag number 2, and so on until we take 99 coins from the bag number 99. Gather all the coins we have taken together and put them on the scale. Denote the weight of these coins as $W$ and the number of bag with gold coins in it as $N$, then we can identify the bag of gold coins using formula $$N=100(W-4950)$$
For instance, if the weight of all coins gathered is $4950.25$ grams, then using the formula above the bag number 25 has the gold coins in it.</p>
</blockquote>
<p>My questions are:</p>
<ol>
<li>How does the formula work? Where does it come from?</li>
<li>Do we have other ways to solve this puzzle? If yes, how?</li>
<li>If the digital scale is replaced by a traditional scale, the scale like symbol of libra or the scale in Shakespeare's drama: The Merchant of Venice (I don't know what is the name in English), then how do we solve this puzzle?</li>
</ol>
| <p>To understand the formula, it would be easiest to explain how it works conceptually before we derive it.</p>
<p>Let's simplify the problem and say there are only 3 bags each with 2 coins in them. 2 of those bags have the 1 gram coins and one has the 1.01 gram gold coins. Let's denote the bags arbitrarily as $Bag_0$, $Bag_1$, and $Bag_2$. Similarly to your problem, let's take 0 coins from $Bag_0$, 1 coin from $Bag_1$, and 2 coins from $Bag_2$. We know that the gold coins must be in one of those bags, so there are three possibilities when we weigh the three coins we removed:</p>
<p>Gold Coins in $Bag_0$: So the weight of the 3 coins on the scale are all 1 gram. So the scale will read 3 grams.</p>
<p>Gold Coins in $Bag_1$: So the weight of 1 of the coins is 1.01 grams and 2 of the coins are 2 grams. So the scale will read 3.01 grams.</p>
<p>Gold Coins in $Bag_2$: So the weight of 2 of the coins is 2.02 grams and 1 of the coins is 1 gram. So the scale will read 3.02 grams.</p>
<p>So each possibility has a unique scenario. So if we determine the weight, we can determine from which bag those coins came from based on that weight.</p>
<p>We can generalize our results from this simplified example to your 100 bag example.</p>
<p>Now for deriving the formula. Say hypothetically, of our 100 bags, all 100 coins in each of the 100 bags weigh 1 gram each. In that case, when we remove 0 coins from $Bag_0$, 1 from $Bag_1$, up until 99 coins from $Bag_{99}$, we'll have a total of 4950 coins on the scale, which will equivalently be 4950 grams. Simply put, if $n$ is our Bag number (denoted $Bag_n$), we've placed $n$ coins from each $Bag_n$ onto the scale for $n = 0, 1, 2, ... 99 $.</p>
<p>So the weight of the coins will be $Weight = 1 + 2 + 3 + ... + 99 = 4950$</p>
<p>But we actually have one bag with gold coins weighing 1.01 grams. And we know that those 1.01 gram coins must be from some $Bag_n$. In our hypothetical example, all of our coins were 1 gram coins, so we must replace the $n$ coins weighed from $Bag_n$ with $n$ gold coins weighing 1.01 grams. Mathematically, we would have:
$Weight = 4950 - n + 1.01n = 4950 + .01n = 4950 + n/100$</p>
<p>Rearranging the formula to solve for n, we have:
$100(Weight-4950) = n$, where $Weight$ is $W$ and $n$ is $N$ in your example.</p>
<p>I have no knowledge of an alternative answer to this puzzle, but perhaps another member's answer may be enlightening if there is. Technically speaking, you could have denoted the bags from 1 to 100 and gone through a similar process as above, but the method is still the same, so I wouldn't treat it as a new answer.</p>
<p>If our electric scale is replaced by a scale of libra, I don't believe it would be possible to answer this puzzle with only one measurement of weight. But again, perhaps another answer may be enlightening on that.</p>
| <p>For #1: imagine for a moment that all the coins are fake. If we took 0 coins from bag 0, 1 coin from bag 1, 2 coins from bag 2... we'd have $99\times100/2=4,950$ coins, and those 4,950 coins would weigh a total of 4,950 grams. But now, say that bag 25 were the one with real coins that are slightly heavier: 0.01 grams heavier, in fact. So the total weight of the coins is $W=4950+0.01N$, where $N$ is the number of the bag with the real coins. But -- we have the weight, not the bag number. So let's <strong>invert</strong> the equation: we want to find N given W, not the other way around.</p>
<p>$$\begin{align}W&=4950+0.01N\\W-4950&=0.01N\\100(W-4950)&=N\end{align}$$</p>
<p>For #2, aside from renumbering the bags, there isn't a different way to do this; no matter what, we have to have a different number coming from the scale for each different possible result.</p>
<p>For #3, you need $\lceil\log_3k\rceil$ weighings to discover the odd coin out; for 100 coins, that's 5 weighings: the first splits the coins into groups of (up to) 34; the second into groups of (up to) 12; the third into groups of (up to) 4; the fourth into groups of (up to) 2; the fifth finds it guaranteed.</p>
<p>Why $\lceil\log_3k\rceil$? Each use of the balance scales actually compares three different groups of coins: the one on the left scale, the one on the right scale, and the one not on the scale at all. If one of the two groups on the scale is heavier, then the gold coin is in that group; if neither, then the gold coin is in the group not on the scale. Thus, each weighing can distinguish between 3 states, and $n$ weighings can distinguish between $3^n$ states. We need an integer solution to $3^n\ge k$, thus $n\ge\log_3k$, thus $n=\lceil\log_3k\rceil$.</p>
|
geometry | <p>I'm an 8th grader. After browsing aops.com, a math contest website, I've seen a lot of problems solved by Cauchy Schwarz. I'm only in geometry (have not started learning trigonometry yet). So can anyone explain Cauchy Schwarz in layman's terms, as if you are explaining it to someone who has just started geo in 8th grade?</p>
| <p>In geometry terms that you can understand, the Cauchy-Schwarz inequality says that:</p>
<blockquote>
<p>Among all the parallelograms with sides <em>a</em> and <em>b</em>, the rectangle is the one with the largest area.</p>
</blockquote>
<p>Usually you can use this inequality when you are looking for an upper (or lower) bound of an expression.</p>
<p>I wanted to give you an example, but my geometry studies are too far to remember an easy demonstration. Maybe you can ask somebody to give you a compass-and-straightedge demonstration of the equivalence between Cauchy-Schwarz and triangle inequalities.</p>
| <p>Since you said you browse AoPS, maybe you're interested in math contests. In math contests, the following forms of Cauchy-Schwarz (C-S) inequality are used:</p>
<p>For all $a_i,b_i\in\Bbb R$:</p>
<p>$$\left(a_1^2+a_2^2+\cdots+a_n^2\right)\left(b_1^2+b_2^2+\cdots+b_n^2\right)\ge (a_1b_1+a_2b_2+\cdots+a_nb_n)^2$$</p>
<p>Equality holds if and only if either $a_1=b_1k, a_2=b_2k,\ldots, a_n=b_nk$ for some $k\in\Bbb R$ or $a_1=a_2=\cdots=a_n=0$ or $b_1=b_2=\cdots=b_n=0$.</p>
<p>For all $a_i,b_i\in\Bbb R^+$:</p>
<p>$$\sqrt{a_1+a_2+\cdots+a_n}\sqrt{b_1+b_2+\cdots+b_n}\ge \sqrt{a_1b_1}+\sqrt{a_2b_2}+\cdots+\sqrt{a_nb_n}$$</p>
<p>Equality holds if and only if $\frac{a_1}{b_1}=\frac{a_2}{b_2}=\cdots=\frac{a_n}{b_n}$.</p>
<p>The following is also called Titu's lemma, or Engel's form of C-S inequality (to prove it, multiply both sides by $b_1+b_2+\cdots+b_n$ and apply the first form of C-S inequality):</p>
<p>For $b_i\in\Bbb R^+, a_i\in\Bbb R$:</p>
<p>$$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\cdots+\frac{a_n^2}{b_n}\ge \frac{(a_1+a_2+\cdots+a_n)^2}{b_1+b_2+\cdots+b_n}$$</p>
<p>with equality if and only if $\frac{a_1}{b_1}=\frac{a_2}{b_2}=\cdots=\frac{a_n}{b_n}$.</p>
<p>The more general is <a href="https://hcmop.wordpress.com/2012/04/19/holders-inequality/" rel="nofollow">Hölder's inequality</a>:</p>
<p>For all $a_i,b_i,c_i\in\Bbb R$:</p>
<p>$$\left(a_1^3+a_2^3+\cdots+a_n^3\right)\left(b_1^3+b_2^3+\cdots+b_n^3\right)\left(c_1^3+c_2^3+\cdots+c_n^3\right)$$</p>
<p>$$\ge (a_1b_1c_1+a_2b_2c_2+\cdots+a_nb_nc_n)^3$$</p>
<p>The same holds for all $a_i,b_i,c_i,d_i\in\Bbb R$, etc. I.e., any number of arrays $a_i,b_i,c_i,d_i,\ldots$. In general, all these inequalities are called Hölder's inequalities.</p>
|
probability | <p>I might just be slow (or too drunk), but I'm seeing a conflict in the equations for adding two normals and scaling a normal. According to <a href="http://www.cs.toronto.edu/~yuvalf/CLT.pdf" rel="noreferrer">page 2 of this</a>, if $X_1 \sim N(\mu_1,\sigma_1^2)$ and $X_2 \sim N(\mu_2,\sigma_2^2)$, then $X_1 + X_2 \sim N(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2)$, and for some $c \in \mathbb{R}$, $cX_1 = N(c\mu_1, c^2\sigma_1^2)$.</p>
<p>Then for $X \sim N(\mu,\sigma^2)$, we have $X + X = N(\mu + \mu,\sigma^2 + \sigma^2) = N(2\mu,2\sigma^2)$, but also $X + X = 2X = N(2\mu,2^2\sigma^2) = N(2\mu,4\sigma^2)$ ? Ie, the variances disagree.</p>
<p>edit: Oh, am I mistaken in saying that $2X = X + X$? Is the former "rolling" $X$ just once and doubling it while the latter "rolls" twice and adds them?</p>
| <p>On the first page of the cited document, $X_1$ and $X_2$ were previously defined to be two (distinct) <em>independent</em>, identically distributed random variables. For your purposes, the "identically distributed" part is not important, but the "independent" part <em>is</em>.</p>
<p>On the second page, where $X_1$ and $X_2$ are considered to be normal variables, there's still the assumption that they're independent. Possibly this could have been stated more clearly, but in context this assumption makes sense.</p>
<p>When you consider $2X = X + X$, you are not dealing with two independent variables.
The two "copies" of $X$ are correlated (in fact, as correlated as any two variables can be).
The formula for the sum of two independent normal variables therefore does not apply.</p>
| <p>Expectation is always linear. So for any two variables <span class="math-container">$X,Y$</span>, we have <span class="math-container">$E[X+Y]=E[X]+E[Y]$</span>. And <span class="math-container">$E[\underbrace{X+X+\ldots+X}_{k \text{ times}}]=E[kX]=kE[X]$</span></p>
<p>Variance is linear when the variables are <em>independent</em>. In this case, <span class="math-container">$V[X+Y]=V[X]+V[Y]$</span>. However when the variables are the same, i.e. when we scale, we have <span class="math-container">$V[kX]=k^2 V[X]$</span>.</p>
<p>These are true no matter what the distribution. Determining the distribution of the sum of random variables is, in general, <a href="https://stats.stackexchange.com/q/331973/89612">difficult</a>. However when <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> are independent normal variables, then the sum of <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> is also normally distributed (and the means and variances add as above).</p>
<p>In my opinion your question is a good one and it is very easy to become confused about what is adding and what is scaling. A very important example which involves adding independent distributions <em>and</em> scaling is when you compute the variance of the mean <span class="math-container">$\bar{X}$</span> of independent identically distributed (iid) samples from the same distribution.</p>
<p>To keep things simple we have <span class="math-container">$n$</span> samples and let's say each sample is normally distributed with variance <span class="math-container">$\sigma^2$</span>. So each sample is drawn from <span class="math-container">$X_i \sim N(\mu,\sigma^2)$</span>. Then <em>adding</em> independent random variables the variance of the sum is <span class="math-container">$V(S) = V\left(\sum_{i=1}^n X_i\right) = \sum_{i=1}^n V\left(X_i\right) = n\sigma^2$</span>. But the mean <span class="math-container">$\bar{X}=S/n$</span> and so by <em>scaling</em> <span class="math-container">$V(\bar{X}) = V(S/n) = n\sigma^2 / n^2 = \sigma^2/n$</span>. It is this combination of adding and scaling which leads to the famous relationship that standard deviation of the sum increases according to the square root of <span class="math-container">$n$</span>, and of the mean as <span class="math-container">$1/\sqrt{n}$</span>.</p>
|
logic | <p>A friend of mine asked me if I could explain this statement: "It's not logically possible to prove that something can't be done". The actual reason is the understanding of this strip: </p>
<p><img src="https://i.sstatic.net/YtHMp.gif" alt="enter image description here"></p>
<p>Since I'm not an expert on logic, but at least know the basics, I told him that the statement is false. To prove it I gave him the counterexample of the impossibility of trisecting an angle by using only straightedge and compass, which was proved by Pierre Wantzel. This shows clearly that it's possible to prove that something can't be done.</p>
<p>When I told him this, he explained to me that this is sort of naive because Asok, the guy who makes the statement, is a graduate from IIT and also I read that he's always a brilliant character in the strip. </p>
<p>Can you guys please tell me what you think about this? Also I'd like to know how you would interpret this dialog. </p>
| <p>What the strip most likely was (awkwardly) referring to is not that it's impossible to prove something can't be done, but rather that it's impossible to prove within a (sufficiently powerful) formal system that something can't be <em>proven</em> in that system — in other words, letting $P(\phi)$ be the statement 'there exists a proof of $\phi$', then it's impossible to prove any statement of the form $\neg P(\phi)$ (i.e., the statement $P(\neg P(\phi))$ can't be true).</p>
<p>The reason behind this has to do with Godel's second incompleteness theorem that no formal system can prove its own consistency - in other words, the system can't prove the statement $\neg P(\bot)$ (or equivalently, $P(\neg P(\bot))$ is false). The logical chain follows from the statement 'false implies anything': $\bot\implies\phi$ for any proposition $\phi$. By the rules of deduction, this gives that $P(\bot)\implies P(\phi)$, and then taking the contrapositive, $\neg P(\phi)\implies\neg P(\bot)$ for <em>any</em> proposition $\phi$; using deduction once more, this gives $P(\neg P(\phi))\implies P(\neg P(\bot))$ — if the system can prove the unprovability of any statement, then it can prove its own consistency, which would violate Godel's second incompleteness theorem.</p>
<p>Note the keyword 'sufficiently powerful' here - and also that we're referring to a formal system talking about proofs <em>within itself</em>. This is what allows statements like the impossibility of angle trisection off the hook: the axioms of ruler-and-compass geometry aren't sufficiently strong for Godel's theorem to apply to the system, so that geometry itself can't talk about the notion of <em>proof</em>, and our statements of proofs of impossibility within that system are proofs from outside of that system.</p>
| <p>Scott Adams may have intended something different, but in the given context and with the given wording, Dan is right and Asok is wrong.</p>
<p>Imagine Asok writing a piece of software that crucially depends on deciding the <a href="http://en.wikipedia.org/wiki/Halting_problem">halting problem</a> (or any other property covered by <a href="http://en.wikipedia.org/wiki/Rice%27s_theorem">Rice's theorem</a>), then Dan is right: The halting problem is <a href="http://en.wikipedia.org/wiki/Undecidable_problem">not decidable</a> and therefore the software will never work. And Dan can indeed prove it, because the undecidability of the halting problem is provable.</p>
|
logic | <p>An extreme form of constructivism is called <em>finitisim</em>. In this form, unlike the standard axiom system, infinite sets are not allowed. There are important mathematicians, such as Kronecker, who supported such a system. I can see that the natural numbers and rational numbers can easily defined in a finitist system, by easy adaptations of the standard definitions. But in order to do any significant mathematics, we need to have definitions for the irrational numbers that one is likely to encounter in practice, such as <span class="math-container">$e$</span> or <span class="math-container">$\sqrt{2}$</span>. In the standard constructions, real numbers are defined as Dedekind cuts or Cauchy sequences, which are actually sets of infinite cardinality, so they are of no use here. My question is, how would a real number like those be defined in a finitist axiom system (Of course we have no hope to construct the entire set of real numbers, since that set is uncountably infinite).</p>
<p>After doing a little research I found a constructivist definition in Wikipedia <a href="http://en.wikipedia.org/wiki/Constructivism_(mathematics)#Example_from_real_analysis" rel="noreferrer">http://en.wikipedia.org/wiki/Constructivism_(mathematics)#Example_from_real_analysis</a> , but we need a finitist definition of a function for this definition to work (Because in the standard system, a function over the set of natural numbers is actually an infinite set).</p>
<p>So my question boils down to this: How can we define a function f over the natural numbers in a finitist axiom system? </p>
<p><em>Original version of this question, <a href="http://meta.math.stackexchange.com/questions/172/why-did-you-close-my-question-if-all-sets-were-finite">which had been closed during private beta</a>, is as follows:</em></p>
<blockquote>
<p><strong>If all sets were finite, how would mathematics be like?</strong></p>
<p>If we replace the axiom that 'there
exists an infinite set' with 'all sets
are finite', how would mathematics be
like? My guess is that, all the theory
that has practical importance would
still show up, but everything would be
very very unreadable for humans. Is
that true?</p>
<p>We would have the natural numbers,
athough the class of all natural
numbers would not be a set. In the
same sense, we could have the rational
numbers. But could we have the real
numbers? Can the standard
constructions be adapted to this
setting?</p>
</blockquote>
| <p>Set theory with all sets finite has been studied, is a familiar theory in disguise, and is enough for most/all concrete real analysis.</p>
<p>Specifically, Zermelo-Fraenkel set theory with the Axiom of Infinity replaced by its negation (informally, "there is no infinite set") is equivalent to first-order Peano Arithmetic. Call this system <em>finite ZF</em>, the theory of hereditarily finite sets. Then under the Goedel arithmetic encoding of finite sets, Peano Arithmetic can prove all the theorems of Finite ZF, and under any of the standard constructions of integers from finite sets, Finite ZF proves all the theorems of Peano Arithmetic. </p>
<p>The implication is that theorems unprovable in PA involve intrinsically infinitary reasoning. Notably, finite ZF was used as an equivalent of PA in the Paris-Harrington paper "A Mathematical Incompleteness in Peano Arithmetic" which proved that their modification of the finite Ramsey theorem can't be proved in PA.</p>
<p>Real numbers and infinite sequences are not directly objects of the finite ZF universe, but there is a clear sense in which real (and complex, and functional) analysis can be performed in finite ZF or in PA. One can make statements about $\pi$ or any other explicitly defined real number, as theorems about a specific sequence of rational approximations ($\forall n P(n)$) and these can be formulated and proved using a theory of finite sets. PA can perform very complicated induction proofs, i.e., transfinite induction below $\epsilon_0$. In practice this means any concrete real number calculation in ordinary mathematics. For the example of the prime number theorem, using complex analysis and the Riemann zeta function, see Gaisi Takeuti's <em>Two Applications of Logic to Mathematics</em>. More discussion of this in a MO thread and my posting there:</p>
<p><a href="https://mathoverflow.net/questions/31846/is-the-riemann-hypothesis-equivalent-to-a-pi-1-sentence">https://mathoverflow.net/questions/31846/is-the-riemann-hypothesis-equivalent-to-a-pi-1-sentence</a></p>
<p><a href="https://mathoverflow.net/questions/31846/is-the-riemann-hypothesis-equivalent-to-a-pi-1-sentence/31942#31942">https://mathoverflow.net/questions/31846/is-the-riemann-hypothesis-equivalent-to-a-pi-1-sentence/31942#31942</a></p>
<p>Proof theory in general and reverse mathematics in particular contain analyses of the logical strength of various theorems in mathematics (when made suitably concrete as statements about sequences of integers), and from this point of view PA, and its avatar finite set theory, are very powerful systems. </p>
| <p>Disclaimer: I am not a finitist --- but as a theoretical computer scientist, I have a certain sympathy for finitism. The following is the result of me openly speculating what an "official" finitist response would be, based on grounds of computability.</p>
<p>The short version is this: <strong>(a)</strong> It depends on what you mean by a 'number', but there's a reasonable approach which makes it reasonable to talk about finitistic approaches to real numbers; <strong>(b)</strong> What you can do finitisitically with numbers, real, rational, or otherwise, depends on how you represent those numbers.</p>
<ol>
<li><p><strong>What is a number?</strong> Is −1 a number? Is sqrt(2) a number? Is <em>i</em> = sqrt(−1) a number? What about quaternions? --- I'm going to completely ignore this question and suggest a pragmatic, formalist approach: a "number" is an element of a "number system"; and a "number system" is a collection of expressions which you can transform or describe properties of in some given ways (<em>i.e.</em> certain given arithmetic operations) and test certain properties (<em>e.g.</em> tests for equality, ordering, <em>etc.</em>) These expressions don't have to have a meaningful interpretation in terms of quantities or magnitudes as far as I'm concerned; <em>you</em> get to choose which operations/tests you care about.<br><br> A finitist would demand that any operation or property be described by an algorithm which provably terminates. That is, it isn't sufficient to prove existence or universality <em>a la</em> classical logic; existence proofs must be finite constructions --- of a "number", that is a representation in some "number system" --- and univserality must be shown by a computable test. </p></li>
<li><p><strong>Representation of numbers:</strong> How we represent the numbers matters. A finitist should have no qualms about rational numbers: ratios which ultimately boil down to ordered pairs. Despite this, the decimal expansions of these numbers may be infinitely long: 1/3 = 0.33333... what's going on here?<br><br> Well, the issue is that we have two representations for the same number, one of which is finite in length (and allows us to perform computations) and another which is not finite in length. However, the decimal expansion can be easily expressed as a function: for all <em>k</em>, the <em>k</em><sup>th</sup> decimal place after the point is '3'; so you can still characterize it precisely in terms of a finite rule.<br><br> What's important is that there exists <strong>some</strong> finite way to express the number. But the way in which we choose to <em>define</em> the number (as a part of system or numbers, using some way of expressing numbers) will affect what we can do with it...
there is now a question about what operations we can perform.<br><br>--- For rationals-as-ratios, we can add/subtract, multiply/divide, and test order/equality. So this representation is a very good one for rationals. <br><br>--- For rationals-as-decimal-expansions, we can still add/subtract and multiply/divide, by defining a new digit-function which describes how to compute the result from the decimal expansions; these will be messier than the representations as ratios. Order comparisons are still possible for <em>distinct</em> rationals; but you cannot test equality for arbitrary decimal-expansion representations, because you cannot necessarily verify that all decimal places of the difference |<em>a</em>−<em>b</em>| are 0. The best you can do in general is testing "equality up to precision ε", wherein you show that |<em>a</em>−<em>b</em>| < ε, for some desired precision ε. This is a number system which informally we may say has certain amount of "vagueness"; but it is in principle completely specified --- there's nothing wrong with this in principle. It's just a matter of how you wish to define your system of arithmetic.</p></li>
<li><p><strong>What representation of reals?</strong> Obviously, because there are uncountably many real numbers, you cannot represent all real numbers even if you <em>aren't</em> a finitist. But we can still express some of them. The same is true if you're a finitist: you just don't have access to as many, and/or you're restricted in what you can do with them, according to what your representation can handle.<br><br>
--- Algebraic irrational numbers such as sqrt(2) can be expressed simply like that: "sqrt(2)". There's nothing wrong with the expressions "sqrt(2) − 1" or "[1 + sqrt(5)]/2" --- they express quantities perfectly well. You can perform arithmetic operations on them perfectly well; and you can also perform ordering/equality tests by transforming them into a normal form of the type "[sum of integers and roots of integers]/[positive integer]"; if the difference of two quantities is zero, the normal form of the difference will just end up being '0'. For order comparisons, we can compute enough decimal places of each term in the sum to determine whether the result is positive or negative, a process which is guaranteed to terminate.<br><br>
--- Numbers such as π and e can be represented by decimal expansions, and computed with in this form, as with the rational numbers. The decimal expansions can be gotten from classical equalities (<em>e.g.</em> "infinite" series, except computing only <em>partial</em> sums; a number such as e may be expressed by some finite representation of such an 'exact' formula, together with a computable function which describes how many terms of the series are required to get a correct evaluation of the first <em>k</em> decimal places.) Of course, what you can do finitistically with these representations is limited in the same way as described above with the rationals; specifically, you cannot always test equality.</p></li>
</ol>
|
game-theory | <p>Imagine Rock Paper Scissors, but where winning with a different hand gives a different reward.</p>
<ul>
<li><p>If you win with Rock, you get \$9. Your opponent loses the \$9.</p>
</li>
<li><p>If you win with Paper, you get \$3. Your opponent loses the \$3.</p>
</li>
<li><p>If you win with Scissors, you get \$5. Your opponent loses the \$5.</p>
</li>
<li><p>If you tie, you get $0</p>
<p>My first intuition would be that you should play Rock with a probability of <code>9/(9+3+5)</code>, Paper with <code>3/(9+3+5)</code> and Scissors with <code>5/(9+3+5)</code> however this seems wrong, as it doesn't take into consideration the risk you expose yourself to (if you play <code>Paper</code>, you have an upside of \$3 but a downside of \$5).</p>
</li>
</ul>
<p>So I put the question to you, in such a game -- what is the ideal strategy.</p>
<p>Edit: By "ideal" strategy, I mean playing against an adversarial player who knows your strategy.</p>
| <p>Let $(x_1,x_2,x_3)$ be the first player's strategy (i.e., his probabilities of playing rock, paper, and scissors respectively), and let $(y_1,y_2,y_3)$ be the second player's strategy. The expected payoff to the first player is
$$
P(x,y)=9(x_1y_3-x_3y_1)+3(x_2y_1-x_1y_2)+5(x_3y_2-x_2y_3).
$$
To constrain the probability sums to be $1$, we take $x_3=1-x_1-x_2$ and $y_3=1-y_1-y_2$. So
$$
P(x,y)=9\left(x_1(1-y_1-y_2)-(1-x_1-x_2)y_1\right)+3(x_2y_1-x_1y_2)+5\left((1-x_1-x_2)y_2-x_2(1-y_1-y_2)\right) \\
=9(x_1-y_1)+ 17(x_2 y_1 -x_1y_2) + 5(y_2-x_2).
$$
The first derivatives are zero when
$$
\frac{\partial}{\partial x_1}P(x,y)= 9 -17y_2=0 \\
\frac{\partial}{\partial x_2}P(x,y)= 17y_1 -5=0 \\
\frac{\partial}{\partial y_1}P(x,y)=-9+17x_2 = 0\\
\frac{\partial}{\partial y_2}P(x,y)=-17x_1+5=0,
$$
or at $(x_1,x_2,x_3)=(y_1,y_2,y_3)=(5/17, 9/17, 3/17)$. The Nash equilibrium is to play to beat each move with probability proportional to that move's reward.</p>
<p>To check that this is indeed a Nash equilibrium, suppose $y_1=5/17$ and $y_2=9/17$. Then
$$
P(x)=9(x_1-5/17)+17\left(x_2 (5/17)-x_1 (9/17)\right)-5(x_2-9/17)=0;
$$
that is, the first player's expected payoff is zero with any strategy. So the first player cannot improve his payoff by changing his strategy unilaterally, and by symmetry, neither can the second player; this is the definition of a Nash equilibrium.</p>
| <p>This table summarizes the possible outcomes in playing this game once:
$$
\begin{array}{c|c|c|c|c}
\text{Hero Plays} & \text{Villain Plays} & \text{Hero's Earnings} \\ \hline
R & R & +\$0 \\
R & P & -\$3 \\
R & S & +\$9 \\
P & R & +\$3 \\
P & P & +\$0 \\
P & S & -\$5 \\
S & R & -\$9 \\
S & P & +\$5 \\
S & S & +\$0
\end{array}
$$
Assume that all outcomes are equally likely.
Letting $X$ be our earnings we can compute our expected earnings given which choice we make
\begin{align*}
\Bbb E(X|R) &= \frac{1}{3}\cdot (\$0)+\frac{1}{3}\cdot (-\$3)+\frac{1}{3}\cdot(\$9) \\
&= -\$1+\$3 \\
&= \$2 \\
\Bbb E(X|P) &= \frac{1}{3}\cdot (\$3)+\frac{1}{3}\cdot (\$0)+\frac{1}{3}\cdot(-\$5) \\
&= \$1-\$\frac{5}{3} \\
&= -\$\frac{2}{3} \\
\Bbb E(X|S) &= \frac{1}{3}\cdot (-\$9)+\frac{1}{3}\cdot (\$5)+\frac{1}{3}\cdot(\$0) \\
&= -\$3+\$\frac{5}{3} \\
&= -\$\frac{4}{3} \\
\end{align*}
According to these computations, the only winning strategy is to play rock in which case the expected value is $\$2$.</p>
<p>Note: The above assumes that hero makes a choice against a villain that is playing randomly. </p>
<p>Edit: I missed the "Nash-Equilibrium" tag when I first read this question. You're probably looking for something a bit more sophisticated than this! I'll keep the answer posted in case others find these computations useful.</p>
|
matrices | <blockquote>
<p>How can one prove that <span class="math-container">$ \|A\|_2 \le \|A\|_F $</span> without using <span class="math-container">$ \|A\|_2^2 := \lambda_{\max}(A^TA) $</span>?</p>
</blockquote>
<hr />
<p>It makes sense that the <span class="math-container">$2$</span>-norm would be less than or equal to the Frobenius norm but I don't know how to prove it. I do know:</p>
<p><span class="math-container">$$\|A\|_2 = \max_{\|x\|_2 = 1} {\|Ax\|_2}$$</span></p>
<p>and I know I can define the Frobenius norm to be:</p>
<p><span class="math-container">$$\|A\|_F^2 = \sum_{j=1}^n {\|Ae_j\|_2^2}$$</span>
but I don't see how this could help. I don't know how else to compare the two norms though.</p>
| <p>Write $x=\sum_{j=1}^nc_je_j$, for coefficients $c_1,\ldots,c_n$. Suppose that $\|x\|_2=1$, i.e. $\sum_j |c_j|^2=1$.
Then
\begin{align}
\|Ax\|_2^2&=\left\|\sum_j c_j\,Ae_j\right\|_2^2\leq\left(\sum_j|c_j|\,\|Ae_j\|_2\right)^{2}\\ \ \\ &\leq\left(\sum_j|c_j|^2\right)\sum_j\|Ae_j\|_2^2=\sum_j\|Ae_j\|_2^2=\|A\|_F^2,
\end{align}
where the triangle inequality is used in the first $\leq$ and Cauchy-Schwarz in the second. </p>
<p>As $x$ was arbitrary, we get $\|A\|_2\leq\|A\|_F$.</p>
| <p>In fact, the proof from $\left\| \mathbf{A}\right\|_2 =\max_{\left\| \mathbf{x}\right\|_2=1} \left\| \mathbf{Ax} \right\|_2$ to $\left\| \mathbf{A}\right\|_2 = \sqrt{\lambda_{\max}(\mathbf{A}^H \mathbf{A})}$ is straight forward. We can first simply prove when $\mathbf{P}$ is Hermitian
$$
\lambda_{\max} = \max_{\| \mathbf{x} \|_2=1} \mathbf{x}^H \mathbf{Px}.
$$
That's because when $\mathbf{P}$ is Hermitian, there exists one and only one unitary matrix $\mathbf{U}$ that can diagonalize $\mathbf{P}$ as $\mathbf{U}^H \mathbf{PU}=\mathbf{D}$ (so $\mathbf{P}=\mathbf{UDU}^H$), where $\mathbf{D}$ is a diagonal matrix with eigenvalues of $\mathbf{P}$ on the diagonal, and the columns of $\mathbf{U}$ are the corresponding eigenvectors. Let $\mathbf{y}=\mathbf{U}^H \mathbf{x}$ and substitute $\mathbf{x} = \mathbf{Uy}$ to the optimization problem, we obtain</p>
<p>$$
\max_{\| \mathbf{x} \|_2=1} \mathbf{x}^H \mathbf{Px} = \max_{\| \mathbf{y} \|_2=1} \mathbf{y}^H \mathbf{Dy} = \max_{\| \mathbf{y} \|_2=1} \sum_{i=1}^n \lambda_i |y_i|^2 \le \lambda_{\max} \max_{\| \mathbf{y} \|_2=1} \sum_{i=1}^n |y_i|^2 = \lambda_{\max}
$$</p>
<p>Thus, just by choosing $\mathbf{x}$ as the corresponding eigenvector to the eigenvalue $\lambda_{\max}$, $\max_{\| \mathbf{x} \|_2=1} \mathbf{x}^H \mathbf{Px} = \lambda_{\max}$. This proves $\left\| \mathbf{A}\right\|_2 = \sqrt{\lambda_{\max}(\mathbf{A}^H \mathbf{A})}$.</p>
<p>And then, because the $n\times n$ matrix $\mathbf{A}^H \mathbf{A}$ is positive semidefinite, all of its eigenvalues are not less than zero. Assume $\text{rank}~\mathbf{A}^H \mathbf{A}=r$, we can put the eigenvalues into a decrease order:</p>
<p>$$
\lambda_1 \geq \lambda_2 \geq \lambda_r > \lambda_{r+1} = \cdots = \lambda_n = 0.
$$</p>
<p>Because for all $\mathbf{X}\in \mathbb{C}^{n\times n}$,
$$
\text{trace}~\mathbf{X} = \sum\limits_{i=1}^{n} \lambda_i,
$$
where $\lambda_i$, $i=1,2,\ldots,n$ are eigenvalues of $\mathbf{X}$; and besides, it's easy to verify
$$
\left\| \mathbf{A}\right\|_F = \sqrt{\text{trace}~ \mathbf{A}^H \mathbf{A}}.
$$</p>
<p>Thus, through
$$
\sqrt{\lambda_1} \leq \sqrt{\sum_{i=1}^{n} \lambda_i} \leq \sqrt{r \cdot \lambda_1}
$$
we have
$$
\left\| \mathbf{A}\right\|_2 \leq \left\| \mathbf{A}\right\|_F \leq \sqrt{r} \left\| \mathbf{A}\right\|_2
$$</p>
|
probability | <p>The PDF describes the probability of a random variable to take on a given value:</p>
<p>$f(x)=P(X=x)$</p>
<p>My question is whether this value can become greater than $1$?</p>
<p>Quote from wikipedia:</p>
<p>"Unlike a probability, a probability density function can take on values greater than one; for example, the uniform distribution on the interval $[0, \frac12]$ has probability density $f(x) = 2$ for $0 \leq x \leq \frac12$ and $f(x) = 0$ elsewhere."</p>
<p>This wasn't clear to me, unfortunately. The question has been asked/answered here before, yet used the same example. Would anyone be able to explain it in a simple manner (using a real-life example, etc)?</p>
<p>Original question:</p>
<p>"$X$ is a continuous random variable with probability density function $f$. Answer with either True or False.</p>
<ul>
<li>$f(x)$ can never exceed $1$."</li>
</ul>
<p>Thank you!</p>
<p>EDIT: Resolved.</p>
| <p>Discrete and continuous random variables are not defined the same way. Human mind is used to have discrete random variables (example: for a fair coin, -1 if it the coin shows tail, +1 if it's head, we have that $f(-1)=f(1)=\frac12$ and $f(x)=0$ elsewhere). As long as the probabilities of the results of a discrete random variable sums up to 1, it's ok, so they have to be at most 1.</p>
<p>For a continuous random variable, the necessary condition is that $\int_{\mathbb{R}} f(x)dx=1$. Since an integral behaves differently than a sum, it's possible that $f(x)>1$ on a small interval (but the length of this interval shall not exceed 1).</p>
<p>The definition of $\mathbb{P}(X=x)$is not $\mathbb{P}(X=x)=f(x)$ but more $\mathbb{P}(X=x)=\mathbb{P}(X\leq x)-\mathbb{P}(X<x)=F(x)-F(x^-)$. In a discrete random variable, $F(x^-)\not = F(x)$ so $\mathbb{P}(X=x)>0$. However, in the case of a continuous random variable, $F(x^-)=F(x)$ (by the definition of continuity) so $\mathbb{P}(X=x)=0$. This can be seen as the probability of choosing $\frac12$ while choosing a number between 0 and 1 is zero.</p>
<p>In summary, for continuous random variables $\mathbb{P}(X=x)\not= f(x)$.</p>
| <p>Your conception of <a href="https://en.wikipedia.org/wiki/Probability_density_function" rel="noreferrer">probability density function</a> is wrong.</p>
<p>You are mixing it up with <a href="https://en.wikipedia.org/wiki/Probability_mass_function" rel="noreferrer">probability mass function</a>.</p>
<p>If <span class="math-container">$f$</span> is a PDF then <span class="math-container">$f(x)$</span> is not a probability and does not have the restriction that it cannot exceed <span class="math-container">$1$</span>.</p>
|
probability | <p>what are some good books on probabilities and measure theory?
I already know basic probabalities, but I'm interested in sigma-algrebas, filtrations, stopping times etc, with possibly examples of "real life" situations where they would be used</p>
<p>thanks</p>
| <p>I'd recommend Klenke's <a href="http://books.google.ch/books?id=tcm3y5UJxDsC&printsec=frontcover&hl=de#v=onepage&q&f=false"><em>Probability Theory</em></a>.</p>
<p>It gives a good overview of the basic ideas in probability theory. In the beginning it builds up the basics of measure theory and set functions.</p>
<p>There are also some examples of applications of probability theory.</p>
| <p>I think Chung's <a href="http://rads.stackoverflow.com/amzn/click/0121741516" rel="noreferrer">A Course in Probability Theory</a> is a good one that is rigorous. Also Sid Resnick's <a href="http://rads.stackoverflow.com/amzn/click/081764055X" rel="noreferrer">A Probability Path</a> is advanced but easy to read.</p>
|
linear-algebra | <p>How can I understand that $A^TA$ is invertible if $A$ has independent columns? I found a similar <a href="https://math.stackexchange.com/questions/1181271/if-ata-is-invertible-then-a-has-linearly-independent-column-vectors">question</a>, phrased the other way around, so I tried to use the theorem</p>
<p>$$
rank(A^TA) \le min(rank(A^T),rank(A))
$$</p>
<p>Given $rank(A) = rank(A^T) = n$ and $A^TA$ produces an $n\times n$ matrix, I can't seem to prove that $rank(A^TA)$ is actually $n$.</p>
<p>I also tried to look at the question another way with the matrices</p>
<p>$$
A^TA
= \begin{bmatrix}a_1^T \\ a_2^T \\ \ldots \\ a_n^T \end{bmatrix}
\begin{bmatrix}a_1 a_2 \ldots a_n \end{bmatrix}
= \begin{bmatrix}A^Ta_1 A^Ta^2 \ldots A^Ta_n\end{bmatrix}
$$</p>
<p>But I still can't seem to show that $A^TA$ is invertible. So, how should I get a better understanding of why $A^TA$ is invertible if $A$ has independent columns?</p>
| <p>Consider the following:
$$A^TAx=\mathbf 0$$
Here, $Ax$, an element in the range of $A$, is in the null space of $A^T$. However, the null space of $A^T$ and the range of $A$ are orthogonal complements, so $Ax=\mathbf 0$.</p>
<p>If $A$ has linearly independent columns, then $Ax=\mathbf 0 \implies x=\mathbf 0$, so the null space of $A^TA=\{\mathbf 0\}$. Since $A^TA$ is a square matrix, this means $A^TA$ is invertible.</p>
| <p>If $A $ is a real $m \times n $ matrix then $A $ and $A^T A $ have the same null space. Proof: $A^TA x =0\implies x^T A^T Ax =0 \implies (Ax)^TAx=0 \implies \|Ax\|^2 = 0 \implies Ax = 0 $. </p>
|
geometry | <p>The derivative of the volume of a sphere, with respect to its radius, gives its surface area. I understand that this is because given an infinitesimal increase in radius, the change in volume will only occur at the surface. Similarly, the derivative of a circle's area with respect to its radius gives its circumference for similar reasons.</p>
<p>A similar logic can be applied to other simple geometric shapes and the pattern holds.
For a cylinder, it becomes more interesting. Its volume is <span class="math-container">$\pi r^2h$</span>, with the height <span class="math-container">$h$</span> and radius <span class="math-container">$r$</span> being independent.</p>
<p>Incremental increases in a cylinders height will "add a circle" to the bottom of the cylinder, so it makes sense to me that the derivative of volume with respect to height is the are of that circle <span class="math-container">$\pi r^2$</span>. Incremental changes to its radius, will add a "tube" around the cylinder, whose area is <span class="math-container">$2\pi rh$</span>, the derivative of volume with respect to radius.</p>
<p>I thought I had found a nice heuristic reason for why the derivatives of volumes give surface areas, but it breaks down for the cone. The volume is <span class="math-container">$\frac{\pi r^2 h}{3}$</span>. Assuming the angle <span class="math-container">$\theta$</span> of the cone remains constant, <span class="math-container">$h$</span> and <span class="math-container">$r$</span> are dependent (<span class="math-container">$r=h\tan\theta$</span>).</p>
<p>Now, the derivative of the cone's volume with respect to height gives the area of the base of the cone, this intuitively makes sense, but the derivative with respect to radius doesn't seem to match any geometric quantity.</p>
<p><span class="math-container">$V=\dfrac{\pi r^3}{3\tan\theta}~$</span> and <span class="math-container">$~\dfrac{\text{d}V}{\text{d}r}=\dfrac{\pi r^2}{\tan\theta}$</span></p>
<p>Similarly, I'd hoped to find some way to drive the surface area of the cone without the base <span class="math-container">$\pi rl$</span> where <span class="math-container">$l$</span> is the distance from the point of the cone to the edge of the circle at its base. However, the derivative with respect to <span class="math-container">$l$</span> also doesn't seem to have any meaningful geometric significance.</p>
<p>What's going on here? Is it an issue of not defining the shape rigorously enough? Do I need to parametrise the volume rigorously in some way?</p>
| <p>If you continuously enlarge a solid cone by adding material to the base, then every inch of added height corresponds to an inch-thick layer added to the circular base, so the rate of change of volume is equal to the area of the base times the rate of change of the height. That is, <span class="math-container">$dV/dh=A_\text{base}=\pi r^2$</span>.</p>
<p>However, if you enlarge the cone by adding material to the "cap", then adding an inch of height (or radius) does <em>not</em> correspond to exactly an inch-thick layer added to the cap. The added thickness is actually <span class="math-container">$\Delta t=\Delta r \cos\theta=\Delta h \sin\theta$</span>, which you can see by drawing a couple of triangles (with a segment of length <span class="math-container">$t$</span> perpendicular to the surface of the cone and with an endpoint at the center of the base). The rate of change of volume is the area of the cap times the rate at which we add thickness to the cap: <span class="math-container">$dV=A_\text{cap}dt$</span>, so we have <span class="math-container">$A_\text{cap}=\frac{dV}{dt}=\frac1{\cos\theta} \frac{dV}{dr}=\frac{\pi rh}{\cos\theta}=\pi rl$</span>, where <span class="math-container">$l$</span> is the slant height.</p>
| <p>My intuition tells me that you should be able to find such a formula by choosing the right parameterization. A couple of motivating examples:</p>
<h4>First example: a disc</h4>
<p>Consider a disk of radius <em>r</em>. The area in terms of the radius is <span class="math-container">$\pi r^2$</span> and in terms of the diameter <span class="math-container">$\pi (\frac{d}{2})^2$</span>. If we take the derivatives of these expressions with respect to each parameter we find:</p>
<ul>
<li><span class="math-container">$2 \pi r$</span> for the radius, which is the circumference, But</li>
<li><span class="math-container">$\frac{1}{2} \pi d$</span> for the diameter, which is not the circumference.</li>
</ul>
<p>The parameterization affects how "fast" the area grows, and varying the diameter makes the area of the disk "grow too slowly" for the change in area to be the circle that surrounds it.</p>
<h4>Second example: an equilateral triangle</h4>
<p>Consider an equilateral triangle of side <span class="math-container">$s$</span>. The area is <span class="math-container">$A(s)=\frac{\sqrt{3}}{4}s^2$</span> and the perimeter is <span class="math-container">$P(s)=3s$</span>. The derivative of the area with respect to <span class="math-container">$s$</span> is <span class="math-container">$\frac{dA}{ds} = \frac{\sqrt{3}}{2}s$</span>, which is not the perimeter.</p>
<p>So let's introduce a new parameter <span class="math-container">$q$</span>: <span class="math-container">$s = kq$</span>, for <span class="math-container">$k$</span> a constant. We have <span class="math-container">$A(q)=\frac{\sqrt{3}}{4}k^2q^2$</span> and <span class="math-container">$P(q)=3kq$</span>. If we set <span class="math-container">$\frac{dA}{dq} = P(q)$</span> we find <span class="math-container">$k=\frac{\sqrt{3}}{2}$</span>. This gives a parameter of <span class="math-container">$q = \frac{2}{\sqrt{3}}s$</span>, which happens to be the diameter of the circumscribed circle of our triangle!</p>
<p>So if we parametrize the area of an equilateral triangle by the diameter of its circumscribed circle, then the rate of change of the area is the circumference.</p>
<p>I suspect a similar approach could give you a parametrization with the same property for the cone. As a first step, you could try doing the same for a cone whose base is equal to its side length and see what you get. Who knows, maybe the parameter you'll find will have an interpretation in terms of the radius of the circumscribed sphere.</p>
|
differentiation | <p>I have <span class="math-container">$x= \exp(At)$</span> where <span class="math-container">$A$</span> is a matrix. I would like to find derivative of <span class="math-container">$x$</span> with respect to each element of <span class="math-container">$A$</span>. Could anyone help with this problem?</p>
| <p>Considering the expression <span class="math-container">$x = \exp(tA)$</span> I can think of two derivatives.</p>
<p>First, the derivative with respect to the real variable <span class="math-container">$t$</span> of the matrix-valued function <span class="math-container">$t \mapsto \exp(tA)$</span>. Here the result is easily derived from direct calculation of the series definition of the matrix exponential:
<span class="math-container">\begin{align} \frac{d}{dt} \exp(tA) &= \frac{d}{dt} \left[ I+tA+\frac{1}{2}t^2A^2+\frac{1}{3!}t^3A^3+ \cdots\right] \\
&= A+tA^2+\frac{1}{2}t^2A^3+ \cdots \\
&= A\exp(tA)
\end{align}</span>
Thus, <span class="math-container">$\frac{d}{dt} \exp(tA) = A\exp(tA)$</span>. (Edited to fix typo)</p>
<p>Second, we can differentiate with respect to the component <span class="math-container">$A_{ij}$</span> of <span class="math-container">$A = \sum A_{ij} E_{ij}$</span> where <span class="math-container">$E_{ij}=e_ie_j^T$</span> is the matrix which is zero except in the <span class="math-container">$ij$</span>-th spot where there is a <span class="math-container">$1$</span>. In other words, <span class="math-container">$(E_{ij})_{kl} = \delta_{ik}\delta_{jl}$</span>. I'll look at the derivative as a directional derivative essentially: calculate the difference along the <span class="math-container">$E_{ij}$</span> direction: considering <span class="math-container">$f(t,A)=\exp(tA)$</span>
<span class="math-container">$$ \frac{\partial f}{\partial A_{ij}}= \lim_{h \rightarrow 0}\frac{1}{h} \left[\exp(t(A+hE_{ij}))-\exp(tA)\right]$$</span>
I expect this can be simplified.</p>
<p>Ok, the matrix exponential satisfies the Baker-Campbell-Hausdorf relation:
<span class="math-container">$$ \exp(A)\exp(B) = \exp\left(A+B+ \frac{1}{2}[A,B] + \cdots\right)$$</span>
From this we derive the Zassenhaus formula,
<span class="math-container">$$ \exp(A+B) = \exp(A)\exp(B)\exp\left(-\frac{1}{2}[A,B] + \cdots\right) $$</span>
I'll use this to simplify <span class="math-container">$\exp( t(A+hE_{ij})) = \exp\left(tA+ thE_{ij}\right)$</span>
<span class="math-container">$$ \exp\left(tA+ thE_{ij}\right) = \exp(tA)\exp\left( thE_{ij}\right)
\exp\left( -\frac{1}{2}[tA,thE_{ij}]+ \cdots\right)$$</span>
hence
<span class="math-container">$$ \exp\left(tA+ thE_{ij}\right) = \exp(tA)\exp\left( thE_{ij} -\frac{1}{2}[tA,thE_{ij}]+ \cdots\right)$$</span>
where I am omitting terms with <span class="math-container">$h^2,h^3,\dots$</span> as those vanish in the limit and I am also omitting terms with nested commutators of <span class="math-container">$A$</span> so the answer below is just the first couple terms in an infinite series flowing from the BCH relation.
<span class="math-container">\begin{align} \frac{\partial f}{\partial A_{ij}}&= -\lim_{h \rightarrow 0}\frac{1}{h} \left[\exp(tA)-\exp(t(A+hE_{ij}))\right] \\
&=-\lim_{h \rightarrow 0}\frac{1}{h} \left[\exp(tA)-\exp(tA)\exp\left( thE_{ij} -\frac{1}{2}[tA,thE_{ij}]+ \cdots\right) \right] \\
&=-\exp(tA)\lim_{h \rightarrow 0}\frac{1}{h} \left[I-\exp\left( thE_{ij} -\frac{1}{2}[tA,thE_{ij}]+ \cdots\right) \right] \\
&=-\exp(tA)\lim_{h \rightarrow 0}\frac{1}{h} \left[I-I-thE_{ij} +\frac{1}{2}[tA,thE_{ij}]+ \cdots \right] \\
&=-\exp(tA)\left[-tE_{ij} +\frac{t^2}{2}[A,E_{ij}]+ \cdots \right].
\end{align}</span>
Note the terms linear in <span class="math-container">$h$</span> do survive the limit and there are such terms (indicated by the <span class="math-container">$+ \cdots$</span>) stemming from <span class="math-container">$[tA,[tA,hE_{ij}]]$</span> and <span class="math-container">$[tA,[tA,[tA,hE_{ij}]]]$</span> etc.
Now, you can calculate: <span class="math-container">$[A,E_{ij}] = \sum_{k=1}^n \left(A_{ki}E_{kj}-A_{jk}E_{ik} \right)$</span> so,
<span class="math-container">$$ \frac{\partial f}{\partial A_{ij}} = -\exp(tA) \left[-tE_{ij}+ \frac{t^2}{2}\left(A_{ki}E_{kj}-A_{jk}E_{ik} +\cdots \right)\right]$$</span>
For what it's worth, you can simplify the nested commutator:
<span class="math-container">$$ [A,[A,E_{ij}]] = \sum_{k,l=1}^n \left( A_{lk}A_{ki}E_{lj}-2A_{ki}A_{jl}E_{kl}+A_{jl}A_{lk}E_{ik} \right)$$</span>
Or, in Einstein notation,
<span class="math-container">$$ [A,[A,E_{ij}]] = (A^2)_{li}E_{lj}-2A_{ki}A_{jl}E_{kl}+(A^2)_{jk}E_{ik}.$$</span>
Anyway, I hope this helps. Notice if <span class="math-container">$i=j$</span> and <span class="math-container">$A$</span> is diagonal or if simply <span class="math-container">$[A,E_{ij}]=0$</span> then we obtain:
<span class="math-container">$$ \frac{\partial }{\partial A_{ij}} \exp(tA) = t\exp(tA)E_{ij}.$$</span>
(I fixed the sign-errors, sorry for all weridly placed minus signs: 9-18-21).</p>
| <p>If $A\in{\mathbb R}^{n\times n}$, then you can use Higham's "Complex Step Approximation" to calculate each component
$$
\frac {\partial f} {\partial A_{jk}} = {\rm Im}\bigg(\frac{f(A+ihE_{jk})}{h}\bigg)
$$
where $f(A)={\rm exp}(tA)$ and $h=10^{-20}$. </p>
|
differentiation | <p>Consider a function <span class="math-container">$g: \mathbb{R^3} \to \mathbb{R}$</span> defined by <span class="math-container">$g(x,y,z) = z^2 -x^3 + 2z + 3y^3$</span> </p>
<p>Find the gradient of <span class="math-container">$g$</span> at the point <span class="math-container">$(2,1,-1)$</span>.</p>
<p>Is the gradient <span class="math-container">$\nabla g(2,1,-1)$</span> given by a vector, that is, <span class="math-container">$\nabla g(2,1,-1) = -12i + 9j$</span>? If so, then what does the directional derivative mean?</p>
| <p>The gradient is a vector; it points in the direction of steepest ascent.</p>
<p>The directional derivative is a number; it is the rate of change when your point in $\Bbb R^3$ moves in that direction. (You can imagine "reducing" your function to a function of a single variable, say $t$, by "slicing" the curve in that direction; the directional derivative is just then the 1-D derivative of that "sliced" function.)</p>
| <p>Be careful that directional derivative of a function is a scalar while gradient is a vector.</p>
<p>The only difference between <em>derivative</em> and <em>directional derivative</em> is the definition of those terms. Remember:</p>
<ul>
<li>Directional derivative is the instantaneous rate of change (which is a scalar) of $f(x,y)$ in the direction of the unit vector $u$.</li>
<li>Derivative is the rate of change of $f(x,y)$, which can be thought of the slope of the function at a point $(x_0,y_0)$.</li>
</ul>
|
differentiation | <p>I have a clarifying question about this question:</p>
<p><a href="https://math.stackexchange.com/questions/183325/what-is-the-difference-between-d-and-partial">What is the difference between $d$ and $\partial$?</a></p>
<p>I understand the idea that $\frac{d}{dx}$ is the derivative where all variables are assumed to be functions of other variables, while with $\frac{\partial}{\partial x}$ one assumes that $x$ is the only variable and every thing else is a constant (as stated in one of the answers).</p>
<p><strong>Example 1</strong>: If $z = xa + x$, then I would guess that
$$
\frac{\partial z}{\partial x} = a + 1
$$
and
$$
\frac{d z}{d x} = a + x\frac{da}{dx} + 1.
$$
since now $a$ should be considered a function.</p>
<p>When we in calculus 1 have $y = ax^2 + bx + c$, then technically we should use $\partial$ as we are assuming $a, b$, and $c$ are constants?</p>
<p><strong>Is this correct?</strong></p>
<p><strong>Example 2</strong>: Maybe the thing that is confusing me is that when we do implicit differentiation we use $d$. So if
$$
x^2 + y^2 = 1
$$
then taking $\frac{d}{dx}$ gives
$$
2x + 2y\frac{dy}{dx} = 0
$$
again because $y$ is considered a function.</p>
<p>How would taking $\frac{\partial}{\partial x}$ of an equation like $x^2 + y^2 =1$ work? Does that even make sense?</p>
<p><strong>Example 3</strong>: Is it ever possible that using $\partial$ and $d$ can give the same? If, for example $y = x^2$, does it make sense to say that
$$
\frac{\partial}{\partial x} y = 2x?
$$</p>
<p>Edit: My overall question, I guess, is how the notations of partial derivatives vs. ordinary derivatives are <em>formally</em> defined. I am looking for a bit more background.</p>
| <p>Some key things to remember about partial derivatives are:</p>
<ul>
<li>You need to have a function of one or more variables.</li>
<li>You need to be very clear about what that function is.</li>
<li>You can only take partial derivatives of that function with respect to each of the variables it is a function of.</li>
</ul>
<p>So for your Example 1, $z = xa + x$, if what you mean by this to define $z$
as a function of two variables,
$$z = f(x, a) = xa + x,$$
then $\frac{\partial z}{\partial x} = a + 1$ and
$\frac{dz}{dx} = a + 1 + x\frac{da}{dx},$ as you surmised,
though you could also have gotten that last result by considering $a$ as a
function of $x$ and applying the Chain Rule.</p>
<p>But when we write something like
$y = ax^2 + bx + c,$ and we say explicitly that $a$, $b$, and $c$ are
(possibly arbitrary) constants, $y$ is really only a function of one variable:
$$y = g(x) = ax^2 + bx + c.$$
Sure, you can say that $\frac{\partial y}{\partial x}$ is what happens
when you vary $x$ while holding $a$, $b$, and $c$ constant, but that's
about as meaningful as saying you vary $x$ while holding the number $3$ constant. </p>
<p>I suppose technically $\frac{\partial y}{\partial x}$
is defined even if $y$ is a single-variable function of $x$,
but it would then just be $\frac{dy}{dx}$ (the ordinary derivative),
and I can't remember seeing such a thing ever written as a partial derivative.
It would not make it possible to do anything you cannot do with
the ordinary derivative, and it might confuse people (who might try to
guess what other variables $y$ is a function of).</p>
<p>The previous paragraph implies that the answer to your Example 3 is "yes."
It also hints at why I almost wrote "a function of two or more variables"
as part of the first requirement for using partial derivatives.
Technically I think you only <em>need</em> a function of one or more variables,
but you should <em>want</em> a function of at least two variables before you
think about taking partial derivatives.</p>
<p>For Example 2, where we have $x^2 + y^2 = 1$, it is not obvious
what the function is that we would be taking partial derivatives of.
Either $x$ or $y$ could be a function of the other.
(The function would be defined only over a limited domain,
and would produce only some of the points that satisfy the equation, but
it can still be useful to do some analysis under those conditions.)
If you write something besides the equation to make it clear that
(say) $y$ is a function of $x$, giving a sufficiently clear idea <em>which</em>
of the possible functions of $x$ you mean, then I think technically you
could write $\frac{\partial y}{\partial x}$, and you might even find that
$\frac{\partial y}{\partial x} = 2x$, but again this is a lot of trouble
and confusion to get a result you could get simply by using
ordinary derivatives.</p>
<p>On the other hand, suppose we say that
$$h(x,y) = x^2 + y^2 - 1,$$
and we are interested in the points that satisfy $x^2 + y^2 = 1$,
that is, where $h(x,y) = 0$.
<em>Now</em> we have a function of multiple variables, so we can do interesting
things with partial derivatives,
such as compute $\frac{\partial h}{\partial x}$ and
$\frac{\partial h}{\partial y}$ and perhaps use these to look for trajectories
in the $x,y$ plane along which $h$ is constant.
OK, we don't really need partial derivatives to figure out that
those trajectories will run along circular arcs, but we could have
some other two-variable function where the answer is not so obvious.</p>
| <p>I hope this answers your question.</p>
<p>The partial derivative notation is used to specify the derivative of a function of more than one variable with respect to one of its variables.</p>
<p>e.g. Let y be a function of 3 variables such that $y(s, t, r) = r^2 - srt$</p>
<p>$$\frac{\partial y}{\partial r} = 2r-st$$</p>
<p>$\frac{d}{dx}$ notation is used when the function to be differentiated is only of one variable e.g. $y(x) = x^2 \ \implies \frac{dy}{dx} = 2x$</p>
<p>I hope this clarifies it a bit for you.</p>
<p>So really, they both mean the same thing but one is used within the context of multivariable calculus whilst the other is reserved for univariate calculus.</p>
|
number-theory | <blockquote>
<p>Let <span class="math-container">$k$</span> be a natural number. Then <span class="math-container">$3k+1$</span> , <span class="math-container">$4k+1$</span> and <span class="math-container">$6k+1$</span> cannot all be square numbers.</p>
</blockquote>
<p>I tried to prove this by supposing one of them is a square number and by substituting the corresponding <span class="math-container">$k$</span> value. But I failed to prove it.</p>
<p>If we ignore one term, we can make the remaining terms squares. For example, <span class="math-container">$3k+1$</span> and <span class="math-container">$4k+1$</span> are both squares if <span class="math-container">$k=56$</span> (they are <span class="math-container">$13^2$</span> and <span class="math-container">$15^2$</span>); <span class="math-container">$3k+1$</span> and <span class="math-container">$6k+1$</span> are both squares if <span class="math-container">$k=8$</span> (they are <span class="math-container">$5^2$</span> and <span class="math-container">$7^2$</span>); <span class="math-container">$4k+1$</span> and <span class="math-container">$6k+1$</span> are both squares if <span class="math-container">$k=20$</span> (they are <span class="math-container">$9^2$</span> and <span class="math-container">$11^2$</span>).</p>
| <p><strong>Complete answer from the literature:</strong></p>
<p>I found a list of solved Pell systems in <a href="http://www.kurims.kyoto-u.ac.jp/EMIS/journals/AMI/2007/ami2007-szalay.pdf">Szalay</a>, Appendix 4, page 84. </p>
<p><a href="http://www.ams.org/journals/mcom/1998-67-222/S0025-5718-98-00918-1/S0025-5718-98-00918-1.pdf">Kiran Kedlaya published <em>Solving Constrained Pell Equations</em> in Mathematics of Computation, Volume 67, April 1998, pages 833-842</a>.</p>
<p>As an example of his method, on page 840 he does the harder part of the Lucas problem (bottom page 238), namely</p>
<blockquote>
<p>Solving $n+1 = 2 x^2,$ $2n+1 = 3 y^2,$ $n=z^2$</p>
</blockquote>
<p>with the conclusion </p>
<blockquote>
<p>Possible values of $n:$ $\{1\}$</p>
</blockquote>
<p>This fits the problem above by taking $x^2 = 3k+1,$ $y^2 = 4k+1,$ $z^2 = 6k+1$ which gives the system
$$ 2 x^2 - z^2 = 1,$$
$$ 3 y^2 - 2 z^2 = 1, $$
with the conclusion that we can only have $z^2 = 1,$ so $k=0.$</p>
<p>Kedlaya refers to his own solution as elementary, references bottom of page 838 to top of page 839. He mentions other articles with elementary methods; I think this is a case where "elementary" is in the eye of the beholder.</p>
| <p>$\def\Q{\mathbf{Q}}$
$\def\Qbar{\overline{\Q}}$
$\def\Z{\mathbf{Z}}$
$\def\F{\mathbf{F}}$</p>
<p>I want to try to explain why there won't be any slick elementary argument in this case. The usual competition techniques involve a tricky use of congruences. A more advanced trick is to use infinite descent. However, in the end, this problem boils down to finding all integral points on an elliptic curve with positive rank (the curve $Y^2 = X^3 - 36 X$). There are some pretty standard techniques to do this - but none of them are elementary. The reason why elementary arguments (such as congruences) are doomed to fail is related to the fact that there are many solutions to this problem with $k \in \Q$. </p>
<p>Any odd square is $1 \mod 8$. Hence $k$ is even, and so we need to find $k$ such that
$6k+1$, $8k+1$, and $12k + 1$ are squares.</p>
<p>As others have observed, can write down equations for this problem as $a^2 = 6k+1, b^2 = 8k+1$, and $c^2 = 12k+1$. This leads to the sytem equations
$$C: 3 b^2 - 4 a^2 = -d^2, \quad c^2 - 2 a^2 = -d^2,$$
where $d = 1$. Here I introduced the variable $d$ to turn the affine equation into a projective one.
The intersection of two quadrics in $\mathbf{P}^3$ has genus one, and so this equation is an elliptic curve (for some choice of base point, note that $[1;1;1;1] \in C(\Q)$.)
It is a general theorem (of Siegel) that, given an elliptic curve has only finitely
many integral points. On the other hand, most elementary solutions to these sort of problems
typically prove
the stronger claim that $E(\mathbf{Q})$, the set of rational points, is also finite. Proofs of this kind of
statement were first found by Fermat, and use the method of infinite descent.
For example, Fermat proved that there are no integral solutions to $x^4 + y^4 = z^4$ by considering
the equation $x^4 + y^4 = z^2$ in rational numbers. After scaling, one can let $y = 1$, and then $z^2 = x^4 + 1$ is an affine open inside an elliptic curve.
Unfortunately, as we shall see, this is not one of those examples.</p>
<p>For those who like to think of elliptic curves in terms of Weierstrass equations, it's easy to write down an elliptic curve $E$ isogenous to $C$ of
the form $y^2$ is a cubic. Namely, if $6k + 1$, $8k+1$, and $12k+1$ are all squares, then so is their product, hence we also get a solution
in integers (and hence in rational numbers) to the equation</p>
<p>$$E: z^2 = (6x+1)(8x+1)(12x+1),$$</p>
<p>Note that by replacing $k$ with $2k$ (which makes no difference to rational solutions), it makes the coefficient of $x^3$ equal to a square ($576 = 24^2$). And now, letting $z = 24y$, we can also write as</p>
<p>$$E: y^2 = (x + 1/6)(x + 1/8)(x + 1/12),$$</p>
<p>and the point $\infty$ at infinity is a rational point.
There is a degree $2$ isogeny $C \rightarrow E$ given by
$$(a,b,c) \mapsto \left(\frac{a - 1}{6},\frac{abc}{24} \right).$$
There will also be a dual isogeny $E \rightarrow C$ of degree $2$, but it will be a little messy to write down.</p>
<p><b>Descent:</b> The main elementary way to find rational points on an elliptic curve $E$ is
to do a $2$-descent. In the case
when all the $2$-torsion is rational, this is pretty concrete. Explicitly, there is an injective map:
$$E(\Q) \rightarrow \Q^{\times}/\Q^{\times 2} \times \Q^{\times}/\Q^{\times 2} \times \Q^{\times}/\Q^{\times 2}$$
which sends a point $P = [x,y]$ to the class:
$$(x + 1/6, x + 1/8,x + 1/12).$$
Note that this class lies inside the subgroup $(a,b,c) \subset (\Q^{\times}/\Q^{\times 2})^3$
with $abc = 1$, so we can really just take the first two entries in
$(\Q^{\times}/\Q^{\times 2})^2$.
The key fact about this map is that the kernel is precisely $2 E(\Q)$, so the induced
map from $E(\Q)/2 E(\Q)$ is injective.
A slight subtlety --- this doesn't quite make sense
for $2$-torsion points, because one of the entries is zero. But you can work out the
correct image by noting that it should land inside the diagonal. Hence
$$\infty \mapsto (1,1,1),$$
$$e_1 = [-1/6,0] \mapsto (*, -1/24,-1/12) \equiv (2,-6,-3) \in (\Q^{\times}/\Q^{\times 2})^3,$$
$$e_2 = [-1/8,0] \mapsto (1/24,*,1/24) \equiv (6,-1,-6) \in (\Q^{\times}/\Q^{\times 2})^3,$$
$$e_1 + e_2 = [-1/12,0] \mapsto (1/12,1/24,*) \equiv (3,6,2) \in (\Q^{\times}/\Q^{\times 2})^3.$$</p>
<p>On the other hand, $E$ also has an obvious rational point $P = [0,1]$. We find that
$$P \mapsto (6,2,3) \in (\Q^{\times}/\Q^{\times 2})^3,$$
$$P + e_1 \mapsto (3,-3,1) \in (\Q^{\times}/\Q^{\times 2})^3, $$
$$P + e_2 \mapsto (1,-2,-2) \in (\Q^{\times}/\Q^{\times 2})^3,$$
$$P + e_3 \mapsto (2,3,6) \in (\Q^{\times}/\Q^{\times 2})^3.$$</p>
<p>Let's now make the following observation, using the fact that $\mathbf{Q}$ has unique factorization.
If $(x + 1/6)$, $(x + 1/8)$, and $(x + 1/12)$ have a square product, then they are all squares up to possible common factors. (Any power of $p > 3$ in the numerator will have to be a square as well by an elementary computation). However, any common factor must divide
$$\Delta = (1/6 - 1/8)^2(1/6 - 1/12)^2 (1/8 - 1/12)^2 = 2^{-16} \cdot 3^{-6}.$$
This implies that the image of $E(\mathbf{Q})/2 E(\mathbf{Q})$ lands in the smaller group
$$(S^{\times}/S^{\times 2 } )^2,$$
Where $S \subset \mathbf{Q}^{\times}$ is generated by $2$, $3$, and $-1$.
This is a group of order $8 \times 8 = 64$. We have computed a subgroup of order $8$ that lies in the image.
If we are lucky, we can show that none of the other $56$ points lie in the image for local reasons, and this would imply that $E(\mathbf{Q})/2 E(\mathbf{Q}) = (\mathbf{Z}/2 \mathbf{Z})^3$.
We don't need to check all $56$ cases, because the image is a group. And most of them are easy. Namely, we write:
$$x + \frac{1}{6} = a u^2, x + \frac{1}{8} = b v^2, x + \frac{1}{12} = a b w^2,$$
or
$$a u^2 - b v^2 = \frac{1}{24}, \qquad a u^2 - a b w^2 = \frac{1}{12},$$
where $(a,b) \in (S^{\times}/S^{\times 2 } )^2$.
We can immediately rule out many examples: If $b > 0$ and $a < 0$ the first equation has
no solutions. If $a < 0$ and $b < 0$, the second equation has no solutions. Hence $a > 0$.
This leaves us with $32$ possibilities. Considering the equation $2$-adically rules out half of the remaining possibilities, and $3$-adic considerations reduce us by half again to end up with $8$.
Thus we have shown:</p>
<p>$$E(\mathbf{Q}) \simeq \mathbf{Z} \oplus (\mathbf{Z}/2 \mathbf{Z})^2,$$
and $P = [0,1]$ is a generator. (Computing the torsion subgroup is easy:
note that $|E(\F_{11})| = 12$ and $E(\F_{5}) = 8$, so the torsion subgroup has order dividing $4$.)</p>
<p>So we have now found (in some sense) all rational points on $E$. However, the answer is a little unexpected, because we had hoped to show that the original equation had no solutions by showing
that there were no rational solutions. But there are many rational solutions!</p>
<p>Let's think of solutions to $E(\mathbf{Q})$ that actually come from $C(\mathbf{Q})$.
We want $6x+1$, $8x +1$, and $12x+1$ to all be squares. Hence, we want the image
of the point in $E$ in $(\mathbf{Q}^{\times}/\mathbf{Q}^{\times 2}$ to be
$$(1/6,1/8,1/12) = (6,2,3).$$
Clearly $P$ is such a point, and this corresponds to $x = 0$ or $k = 0$.
Any other point which maps to this must be equal to $P$ in $E(\mathbf{Q})/2 E(\mathbf{Q})$,
and thus $P \mod 2 E(\mathbf{Q})$. Hence we deduce:</p>
<p><b>Rational points on $C(\mathbf{Q})$ correspond to points on $E(\mathbf{Q})$ of the form
$2Q + P$, or equivalently, $n P$ for $n$ odd. </b></p>
<p>For example,
$$3P = \left[ \frac{-140}{2209},\frac{-28083}{103823} \right],$$
and we correspondingly see that
$$3 \left(\frac{-280}{2209}\right) + 1 = \left(\frac{37}{47}\right)^2, 4 \left(\frac{-280}{2209}\right) + 1 = \left(\frac{33}{47}\right)^2, 6 \left(\frac{-280}{2209}\right) + 1 = \left(\frac{23}{47}\right)^2$$
Taking $5P$ gives
$$3 \left(\frac{369572280}{537451489}\right) + 1 = \left(\frac{40573}{23183} \right)^2,
4 \left(\frac{369572280}{537451489}\right) + 1 = \left(\frac{44897}{23183}\right)^2, 6 \left(\frac{369572280}{537451489}\right) + 1 = \left(\frac{52487}{23183}\right)^2,$$
and so on.</p>
<p>But what does this say about our original question? A weaker question than asking
for solutions with integral $x$ is to ask for solutions with $72x \in \mathbf{Z}$. Then we can
make the transformation
$$X = \frac{x}{144} - \frac{1}{8}, \quad Y = \frac{y}{1728},$$
to get the equation
$$D: Y^2 = X^3 - 6^2 X.$$
This is isomorphic to $E$ over $\Q$, but it is a different integral model. This is a familiar curve; it is the congruent
number curve for $d = 6$. There is an integral solution related to the fact that $(3,4,5)$ is a
right angled triangle with area $6$.
But there are other integral solutions as well, namely:
$$[0,0], [\pm 6, 0], [-3, \pm 9], [-2, \pm 8], [12, \pm 36],
[18, \pm 72], [294, \pm 5040].$$
This is, in fact, the complete list. If you like, you can confirm this in magma as follows:</p>
<p>$$\begin{aligned}
& \ \texttt{> E := EllipticCurve([0,0,0,-36,0]);} \\
& \
\texttt{> IntegralPoints(E);} \\
& \ \texttt{
[ (-6 : 0 : 1), (-3 : -9 : 1), (-2 : 8 : 1), (0 : 0 : 1),} \\
& \ \texttt{ (6 : 0 : 1), (12 : -36
: 1), (18 : 72 : 1), (294 : -5040 : 1) ]
} \end{aligned}$$</p>
<p>The observatio that $E$ has many rational points and lots of $S$-integral
points for $S \in \{2,3\}$ suggests that there won't be a quick
solution to the original problem.
There are fairly standard methods to (certifiably) compute all the integral
points of an elliptic curve $D$ given the Mordell-Weil group (which
$\texttt{magma}$ uses), but they are not particularly
elementary. </p>
|
linear-algebra | <p>What is the "standard basis" for fields of complex numbers?</p>
<p>For example, what is the standard basis for $\Bbb C^2$ (two-tuples of the form: $(a + bi, c + di)$)? I know the standard for $\Bbb R^2$ is $((1, 0), (0, 1))$. Is the standard basis exactly the same for complex numbers?</p>
<p><strong>P.S.</strong> - I realize this question is very simplistic, but I couldn't find an authoritative answer online.</p>
| <p>Just to be clear, by definition, a vector space always comes along with a field of scalars $F$. It's common just to talk about a "vector space" and a "basis"; but if there is possible doubt about the field of scalars, it's better to talk about a "vector space over $F$" and a "basis over $F$" (or an "$F$-vector space" and an "$F$-basis").</p>
<p>Your example, $\mathbb{C}^2$, is a 2-dimensional vector space over $\mathbb{C}$, and the simplest choice of a $\mathbb{C}$-basis is $\{ (1,0), (0,1) \}$.</p>
<p>However, $\mathbb{C}^2$ is also a vector space over $\mathbb{R}$. When we view $\mathbb{C}^2$ as an $\mathbb{R}$-vector space, it has dimension 4, and the simplest choice of an $\mathbb{R}$-basis is $\{(1,0), (i,0), (0,1), (0,i)\}$.</p>
<p>Here's another intersting example, though I'm pretty sure it's not what you were asking about:</p>
<p>We can view $\mathbb{C}^2$ as a vector space over $\mathbb{Q}$. (You can work through the definition of a vector space to prove this is true.) As a $\mathbb{Q}$-vector space, $\mathbb{C}^2$ is infinite-dimensional, and you can't write down any nice basis. (The existence of the $\mathbb{Q}$-basis depends on the axiom of choice.)</p>
| <p>The "most standard" basis is also $\left\lbrace(1,0),\, (0,1)\right\rbrace$. You just take complex combinations of these vectors. Simple :)</p>
|
geometry | <p>I'm searching to develop the intuition (rather than memorization) in relating the two forms of a dot product (by an angle theta between the vectors and by the components of the vector ). </p>
<p>For example, suppose I have vector $\mathbf{a} = (a_1,a_2)$ and vector $\mathbf{b}=(b_1,b_2)$. What's the physical or geometrical meaning that </p>
<p>$$a_1b_1 + a_2b_2 = |\mathbf{a}||\mathbf{b}|\cos(\theta)\;?$$</p>
<p>Why is multiplying $|\mathbf{b}|$ times $|\mathbf{a}|$ in direction of $\mathbf{b}$ the same as multiplying the first and second components of $\mathbf{a}$ and $\mathbf{b}$ and summing ? </p>
<p>I know this relationship comes out when we use the law of cosines to prove, but even then i cant get a intuition in this relationship.</p>
<p>This image clarifies my doubt:</p>
<p><img src="https://i.sstatic.net/Bpyfg.jpg" alt="enter image description here"></p>
<p>Thanks</p>
| <p>Start with the following definition:</p>
<p><a href="https://i.sstatic.net/o2kgw.jpg" rel="noreferrer"><img src="https://i.sstatic.net/o2kgw.jpg" alt="enter image description here"></a></p>
<p>(with a negative dot product when the projection is onto $-\mathbf{b}$)</p>
<p>This implies that the dot product of perpendicular vectors is zero and the dot product of parallel vectors is the product of their lengths.</p>
<p>Now take any two vectors $\mathbf{a}$ and $\mathbf{b}$. </p>
<p><a href="https://i.sstatic.net/sPoEK.jpg" rel="noreferrer"><img src="https://i.sstatic.net/sPoEK.jpg" alt="enter image description here"></a></p>
<p>They can be decomposed into horizontal and vertical components $\mathbf{a}=a_x\mathbf{i}+a_y\mathbf{j}$ and $\mathbf{b}=b_x\mathbf{i}+b_y\mathbf{j}$:</p>
<p><a href="https://i.sstatic.net/HigB9.jpg" rel="noreferrer"><img src="https://i.sstatic.net/HigB9.jpg" alt="enter image description here"></a></p>
<p>and so </p>
<p>$$\begin{align*}
\mathbf{a}\cdot \mathbf{b}&=(a_x\mathbf{i}+a_y\mathbf{j})\cdot(b_x\mathbf{i}+b_y\mathbf{j}),
\end{align*}$$</p>
<p>but the perpendicular components have a dot product of zero while the parallel components have a dot product equal to the product of their lengths.</p>
<p>Therefore</p>
<p>$$\mathbf{a}\cdot\mathbf{b}=a_xb_x+a_yb_y.$$</p>
| <p>I found a reasonable proof using polar coordinates. Lets suppose the point "$a$" points is $(|a|\cos(r)$ , $|a|\sin(r) )$ and the point vector "$b$" points is ($|b|\cos(s),|b|\sin(s) $).
Then doing the definition of the scalar product we get : </p>
<p>$a\cdot b = |a||b|\cos(r)\cos(s) + |b||a|\sin(r)\sin(s) = |a||b|\cos(r - s)$. But $\cos(r-s) = \cos(\theta)$ where theta is the angle between the vectors.</p>
<p>So, $a\cdot b = |a||b|\cos(\theta)$.</p>
|
geometry | <p>For <span class="math-container">$n = 2$</span>, I can visualize that the determinant <span class="math-container">$n \times n$</span> matrix is the area of the parallelograms by actually calculating the area by coordinates. But how can one easily realize that it is true for any dimensions?</p>
| <p>If the column vectors are linearly dependent, both the determinant and the volume are zero.
So assume linear independence.
The determinant remains unchanged when adding multiples of one column to another. This corresponds to a skew translation of the parallelepiped, which does not affect its volume.
By a finite sequence of such operations, you can transform your matrix to diagonal form, where the relation between determinant (=product of diagonal entries) and volume of a "rectangle" (=product of side lengths) is apparent.</p>
| <p>Here is the same argument as Muphrid's, perhaps written in an elementary way.<br /><br /></p>
<p>Apply Gram-Schmidt orthogonalization to $\{v_{1},\ldots,v_{n}\}$, so that
\begin{eqnarray*}
v_{1} & = & v_{1}\\
v_{2} & = & c_{12}v_{1}+v_{2}^{\perp}\\
v_{3} & = & c_{13}v_{1}+c_{23}v_{2}+v_{3}^{\perp}\\
& \vdots
\end{eqnarray*}
where $v_{2}^{\perp}$ is orthogonal to $v_{1}$; and $v_{3}^{\perp}$
is orthogonal to $span\left\{ v_{1},v_{2}\right\} $, etc. </p>
<p>Since determinant is multilinear, anti-symmetric, then
\begin{eqnarray*}
\det\left(v_{1},v_{2},v_{3},\ldots,v_{n}\right) & = & \det\left(v_{1},c_{12}v_{1}+v_{2}^{\perp},c_{13}v_{1}+c_{23}v_{2}+v_{3}^{\perp},\ldots\right)\\
& = & \det\left(v_{1},v_{2}^{\perp},v_{3}^{\perp},\ldots,v_{n}^{\perp}\right)\\
& = & \mbox{signed volume}\left(v_{1},\ldots,v_{n}\right)
\end{eqnarray*}</p>
|
combinatorics | <p>Popular mathematics folklore provides some simple tools
enabling us compactly to describe some truly enormous
numbers. For example, the number $10^{100}$ is commonly
known as a <a href="http://en.wikipedia.org/wiki/Googol">googol</a>,
and a <a href="http://en.wikipedia.org/wiki/Googolplex">googol
plex</a> is
$10^{10^{100}}$. For any number $x$, we have the common
vernacular:</p>
<ul>
<li>$x$ <em>bang</em> is the factorial number $x!$</li>
<li>$x$ <em>plex</em> is the exponential number $10^x$</li>
<li>$x$ <em>stack</em> is the number obtained by iterated exponentiation
(associated upwards) in a tower of height $x$, also denoted $10\uparrow\uparrow x$,
$$10\uparrow\uparrow x = 10^{10^{10^{\cdot^{\cdot^{10}}}}}{\large\rbrace} x\text{ times}.$$</li>
</ul>
<p>Thus, a googol bang is $(10^{100})!$, and a googol stack is
$10\uparrow\uparrow 10^{100}$. The vocabulary enables us to
name larger numbers with ease:</p>
<ul>
<li>googol bang plex stack. (This is the exponential tower $10^{10^{\cdot^{\cdot^{^{10}}}}}$ of height $10^{(10^{100})!}$)</li>
<li>googol stack bang stack bang</li>
<li>googol bang bang stack plex stack</li>
<li>and so on…</li>
</ul>
<p>Consider the collection of all numbers that can be named in
this scheme, by a term starting with googol and having
finitely many adjectival operands: bang, stack, plex, in
any finite pattern, repetitions allowed. (For the purposes
of this question, let us limit ourselves to these three
operations and please accept the base 10 presumption of the
stack and plex terminology simply as an artifact of its
origin in popular mathematics.)</p>
<p>My goal is to sort all such numbers nameable in this
vocabulary by size.</p>
<p>A few simple observations get us started. Once $x$ is large
enough (about 20), then the factors of $x!$ above $10$
compensate for the few below $10$, and so we see that
$10^x\lt x!$, or in other words, $x$ plex is less than $x$
bang. Similarly, $10^{10^{:^{10}}}x$ times is much larger
than $x!$, since $10^y\gt (y+1)y$ for large $y$, and so for
large values we have</p>
<ul>
<li>$x$ plex $\lt$ $x$ bang $\lt$ $x$ stack.</li>
</ul>
<p>In particular, the order for names having at most one
adjective is:</p>
<pre><code> googol
googol plex
googol bang
googol stack
</code></pre>
<p>And more generally, replacing plex with bang or bang with
stack in any of our names results in a strictly (and much)
larger number.</p>
<p>Continuing, since $x$ stack plex $= (x+1)$ stack, it
follows that</p>
<ul>
<li>$x$ stack plex $\lt x$ plex stack.</li>
</ul>
<p>Similarly, for large values,</p>
<ul>
<li>$x$ plex bang $\lt x$ bang plex,</li>
</ul>
<p>because $(10^x)!\lt (10^x)^{10^x}=10^{x10^x}\lt 10^{x!}$.
Also,</p>
<ul>
<li>$x$ stack bang $\lt x$ plex stack $\lt x$ bang stack,</li>
</ul>
<p>because $(10\uparrow\uparrow x)!\lt (10\uparrow\uparrow
x)^{10\uparrow\uparrow x}\lt 10\uparrow\uparrow 2x\lt
10\uparrow\uparrow 10^x\lt 10\uparrow\uparrow x!$. It also
appears to be true for large values that</p>
<ul>
<li>$x$ bang bang $\lt x$ stack.</li>
</ul>
<p>Indeed, one may subsume many more iterations of plex and
bang into a single stack. Note also for large values that</p>
<ul>
<li>$x$ bang $\lt x$ plex plex</li>
</ul>
<p>since $x!\lt x^x$, and this is seen to be less than
$10^{10^x}$ by taking logarithms.</p>
<p>The observations above enable us to form the following
order of all names using at most two adjectives.</p>
<pre><code> googol
googol plex
googol bang
googol plex plex
googol plex bang
googol bang plex
googol bang bang
googol stack
googol stack plex
googol stack bang
googol plex stack
googol bang stack
googol stack stack
</code></pre>
<p>My request is for any or all of the following:</p>
<ol>
<li><p>Expand the list above to include numbers named using
more than two adjectives. (This will not be an
end-extension of the current list, since googol plex plex
plex and googol bang bang bang will still appear before
googol stack.) If people post partial progress, we can
assemble them into a master list later.</p></li>
<li><p>Provide general comparison criteria that will assist
such an on-going effort.</p></li>
<li><p>Provide a complete comparison algorithm that works for
any two expressions having the same number of adjectives.</p></li>
<li><p>Provide a complete comparison algorithm that compares
any two expressions.</p></li>
</ol>
<p>Of course, there is in principle a computable comparison
procedure, since we may program a Turing machine to
actually compute the two values and compare their size.
What is desired, however, is a simple, feasible algorithm.
For example, it would seem that we could hope for an
algorithm that would compare any two names in polynomial
time of the length of the names.</p>
| <p>OK, let's attempt a sorting of the names having at most
three operands. I'll make several observations, and then
use them to assemble the order section by section,
beginning with the part below googol stack.</p>
<ul>
<li><p>googol bang bang bang $\lt$ googol
stack. It seems clear that we shall be able to iterated bangs many
times before exceeding googol stack. Since googol bang bang
bang is the largest three-operand name
using only plex and bang, this means that all such names will interact
only with each below googol stack.</p></li>
<li><p>plex $\lt$ bang. This was established in the question.</p></li>
<li><p>plex bang $\lt$ bang plex. This was established in the
question, and it allows us to make many comparisons in
terms involving only plex and bang, but not quite all of
them.</p></li>
<li><p>googol bang bang $\lt$ googol plex plex plex. This is
because $g!!\lt (g^g)^{g^g}=g^{gg^g}=10^{100\cdot gg^g}$, which is less than
$10^{10^{10^g}}$, since $100\cdot gg^g=10^{102\cdot
10^{100}}\lt 10^{10^g}$. Since googol bang bang is the largest two-operand name using only
plex and bang and googol plex plex plex is the smallest three-operand name, this means that
the two-operand names using only plex and bang will all come
before all the three-operand names.</p></li>
<li><p>googol plex bang bang $\lt$ googol bang plex plex. This
is because $(10^g)!!\lt
((10^g)^{10^g})!=(10^{g10^g})!=(10^{10^{g+100}})!\lt
(10^{10^{g+100}})^{10^{10^{g+100}}}=10^{10^{g+100}10^{10^{g+100}}}=
10^{10^{(g+100)10^{g+100}}}\lt
10^{10^{g!}}$.</p></li>
</ul>
<p>Combining the previous observations leads to the following
order of the three-operand names below googol stack:</p>
<pre><code> googol
googol plex
googol bang
googol plex plex
googol plex bang
googol bang plex
googol bang bang
googol bang bang
googol plex plex plex
googol plex plex bang
googol plex bang plex
googol plex bang bang
googol bang plex plex
googol bang plex bang
googol bang bang plex
googol bang bang bang
googol stack
</code></pre>
<p>Perhaps someone can generalize the methods into a general
comparison algorithm for larger smallish terms using only
plex and bang? This is related to the topic of the Velleman
article linked to by J. M. in the comments.</p>
<p>Meanwhile, let us now turn to the interaction with stack.
Using the observations of the two-operand case in the
question, we may continue as follows:</p>
<pre><code> googol stack plex
googol stack bang
googol stack plex plex
googol stack plex bang
googol stack bang plex
googol stack bang bang
</code></pre>
<p>Now we use the following fact:</p>
<ul>
<li>stack bang bang $\lt$ plex stack. This is established as
in the question, since $(10\uparrow\uparrow x)!!\lt
(10\uparrow\uparrow x)^{10\uparrow\uparrow x}!\lt$
$(10\uparrow\uparrow x)^{(10\uparrow\uparrow
x)(10\uparrow\uparrow x)^{10\uparrow\uparrow x}}=$
$(10\uparrow\uparrow x)^{(10\uparrow\uparrow
x)^{1+10\uparrow\uparrow x}} 10\uparrow\uparrow 4x\lt
10\uparrow\uparrow 10^x$. In fact, it seems that we will be
able to absorb many more iterated bangs after stack into
plex stack.</li>
</ul>
<p>The order therefore continues with:</p>
<pre><code> googol plex stack
googol plex stack plex
googol plex stack bang
</code></pre>
<ul>
<li>plex stack bang $\lt$ bang stack. To see this, observe
that $(10\uparrow\uparrow 10^x)!\lt (10\uparrow\uparrow
10^x)^{10\uparrow\uparrow 10^x}\lt 10\uparrow\uparrow
2\cdot10^x$, since associating upwards is greater, and this
is less than $10\uparrow\uparrow x!$. Again, we will be
able to absorb many operands after plex stack into bang
stack.</li>
</ul>
<p>The order therefore continues with:</p>
<pre><code> googol bang stack
googol bang stack plex
googol bang stack bang
</code></pre>
<ul>
<li>bang stack bang $\lt$ plex plex stack.
This is because $(10\uparrow\uparrow x!)!\lt
(10\uparrow\uparrow x!)^{10\uparrow\uparrow x!}\lt
10\uparrow\uparrow 2x!\lt 10\uparrow 10^{10^x}$.</li>
</ul>
<p>Thus, the order continues with:</p>
<pre><code> googol plex plex stack
googol plex bang stack
googol bang plex stack
googol bang bang stack
</code></pre>
<p>This last item is clearly less than googol stack stack, and
so, using all the pairwise operations we already know, we
continue with:</p>
<pre><code> googol stack stack
googol stack stack plex
googol stack stack bang
googol stack plex stack
googol stack bang stack
googol plex stack stack
googol bang stack stack
googol stack stack stack
</code></pre>
<p>Which seems to complete the list for three-operand names.
If I have made any mistakes, please comment below.</p>
<p>Meanwhile, this answer is just partial progress, since we
have the four-operand names, which will fit into the
hierarchy, and I don't think the observations above are
fully sufficient for the four-operand comparisons, although
many of them will now be settled by these criteria. And of course, I am nowhere near a general comparison algorithm.</p>
<p>Sorry for the length of this answer. Please post comments if I've made any errors.</p>
| <p>The following describes a comparison algorithm that will work for expressions where the number of terms is less than googol - 2.</p>
<p>First, consider the situation with only bangs and plexes. To compare two numbers, first count the total number of bangs and plexes in each. If one has more than the other, that number is bigger. If the two numbers have the same number of bangs and plexes, compare the terms lexicographically, setting bang > plex. So googol plex bang plex plex > googol plex plex bang bang, since the first terms are equal, and the second term favors the first.</p>
<p>To prove this, first note that x bang > x plex for $x \ge 25$. To show that the higher number of terms always wins, it suffices to show that googol plex$^{k+1} >$ googol bang$^k$. We will instead show that $x$ plex$^{k+1} > x$ bang $^k$ for $x \ge 100$. Set $x = 10^y$.</p>
<p>$10^y$ bang $< (10^y)^{10^y} = 10^{y*10^y} < 10^{10^{10^y}} = 10^y$ plex plex</p>
<p>$10^y$ bang bang $< (10^{y*10^y})^{10^{y*10^y}} $</p>
<p>$= 10^{10^{y*10^y + y + \log_{10} y}}$ </p>
<p>$= 10^{10^{10^{y + \log_{10} y} (1 + \frac{y + \log_{10} y}{10^{y + \log_{10} y}})}}$ </p>
<p>$< 10^{10^{10^{y + \log_{10} y} (1 + \frac{y}{10^{y}})}}$</p>
<p>(We use the fact that x/10^x is decreasing for large x.) </p>
<p>$= 10^{10^{10^{y + \log_{10} y + \log_{10}(1 + \frac{y}{10^{y}})}}}$ </p>
<p>$< 10^{10^{10^{y + \log_{10} y + \frac{y}{10^{y}}}}}$ </p>
<p>(We use the fact that ln(1+x) < x, so log_10 (1+x) < x)</p>
<p>$< 10^{10^{10^{2y}}} < 10^{10^{10^{10^y}}} = 10^y$ plex plex plex</p>
<p>$10^y$ bang bang bang < $(10^{10^{10^{y + \log_{10} y + \frac{y}{10^y}}}})^{10^{10^{10^{y + \log_{10} y + \frac{y}{10^y}}}}} $</p>
<p>$= 10^{10^{(10^{10^{y + \log_{10} y + \frac{y}{10^y}}} + 10^{y + \log_{10} y + \frac{y}{10^y}})}}$ </p>
<p>$= 10^{10^{(10^{10^{y + \log_{10} y + \frac{y}{10^y}}}(1 + \frac{10^{y + \log_{10} y + \frac{y}{10^y}}}{10^{10^{y + \log_{10} y + \frac{y}{10^y}}}})}}$ </p>
<p>$< 10^{10^{(10^{10^{y + \log_{10} y + \frac{y}{10^y}}}(1 + \frac{10^{y }}{10^{10^{y}}})}}$ </p>
<p>$= 10^{10^{10^{(10^{y + \log_{10} y + \frac{y}{10^y}} + \log_{10}(1+\frac{10^{y }}{10^{10^{y}}}))}}}$ </p>
<p>$< 10^{10^{10^{(10^{y + \log_{10} y + \frac{y}{10^y}} + \frac{10^{y }}{10^{10^{y}}})}}}$ </p>
<p>$= 10^{10^{10^{(10^{y + \log_{10} y + \frac{y}{10^y}} (1 + \frac{10^{y }}{10^{10^{y}} * (10^{y + \log_{10} y + \frac{y}{10^y}})}))}}}$ </p>
<p>$< 10^{10^{10^{(10^{y + \log_{10} y + \frac{y}{10^y}} (1 + \frac{1}{10^{10^{y}} }))}}}$</p>
<p>$= 10^{10^{10^{10^{y + \log_{10} y + \frac{y}{10^y} + \frac{1}{10^{10^{y}} }} }}}$ </p>
<p>$< 10^{10^{10^{10^{2y}}}} < 10^{10^{10^{10^{10^y}}}} = 10^y$ plex plex plex plex</p>
<p>We can see that the third bang added less than $\frac{1}{10^{10^y}}$ to the top exponent. Similarly, adding a fourth bang will add less than $\frac{1}{10^{10^{10^y}}}$, adding a fifth bang will add less than $\frac{1}{10^{10^{10^{10^y}}}}$, and so on. It's clear that all the fractions will add up to less than 1, so in general,</p>
<p>$10^y$ bang$^{k} < 10^{10^{10^{\cdot^{\cdot^{10^{y + \log_{10} y + 1}}}}}}{\large\rbrace} k+1\text{ 10's} < 10^{10^{10^{\cdot^{\cdot^{10^{10^y}}}}}}{\large\rbrace} k+2\text{ 10's} = 10^y$ plex$^{k+1}$.</p>
<p>Next, we have to show that the lexicographic order works. We will show that it works for all $x \ge 100$. Suppose our procedure failed; take two numbers with the fewest number of terms for which it fails, e.g. $x s_1 ... s_n$ and $x t_1 ... t_n$. It cannot be that $s_1$ and $t_1$ are both plex or both bang, since then $(x s_1) s_2 ... s_n$ and $(x s_1) t_2 ... t_n$ would be a failure of the procedure with one fewer term. So set $s_1 =$ bang and $t_1 =$ plex. Since our procedure tells us that $x s_1 ... s_n$ > $x t_1 ... t_n$, and our procedure fails, it must be that $x s_1 ... s_n$ < $x t_1 ... t_n$. Then</p>
<p>$x$ bang plex ... plex $< x$ bang $s_2 ... s_n < x$ plex $t_2 ... t_n < x$ plex bang ... bang.</p>
<p>So to show our procedure works, it suffices to show that x bang plex$^k$ > x plex bang$^k$. Set x = 10^y.</p>
<p>$10^y$ bang > $(\frac{10^y}{e})^{10^y} > (10^{y - \frac{1}{2}})^{10^y} = 10^{(y-\frac{1}{2})10^y}$</p>
<p>$10^y$ bang plex$^k > 10^{10^{10^{\cdot^{\cdot^{10^{(y-\frac{1}{2})10^y}}}}}}{\large\rbrace} k+1\text{ 10's}$</p>
<p>To determine $10^y$ plex bang$^k$, we can use our previous inequality for $10^y$ bang$^k$ and set $x = 10^y$ plex $= 10^{10^y}$, i.e. substitute $10^y$ for $y$. We get</p>
<p>$10^y$ plex bang$^k < 10^{10^{10^{\cdot^{\cdot^{10^{(10^y + \log_{10}(10^y) + 1}}}}}}{\large\rbrace} k+1\text{ 10's} = 10^{10^{10^{\cdot^{\cdot^{10^{10^y + y + 1}}}}}}{\large\rbrace} k+1\text{ 10's}$</p>
<p>$< 10^{10^{10^{\cdot^{\cdot^{10^{(y-\frac{1}{2})10^y}}}}}}{\large\rbrace} k+1\text{ 10's} < 10^y$ bang plex$^k$.</p>
<p>Okay, now for terms with stack. Given two expressions, first compare the number of times stack appears; the number in which stack appears more often is the winner. If stack appears n times for both expressions, then in each expression consider the n+1 groups of plexes and bangs separated by the n stacks. Compare the n+1 groups lexicographically, using the ordering we defined above for plexes and bangs. Whichever expression is greater denotes the larger number.</p>
<p>Now, this procedure clearly does not work all the time, since a googol followed be a googol-2 plexes is greater than googol stack. However, I believe that if the number of terms in the expressions are less than googol-2, then the procedure is correct.</p>
<p>First, observe that $x$ plex stack > $x$ stack plex and $x$ bang stack > $x$ stack bang, since </p>
<p>$x$ stack plex $< x$ stack bang $< (10\uparrow\uparrow x)^{10\uparrow\uparrow x} < 10\uparrow\uparrow (2x) < x$ plex stack $< x$ bang stack.</p>
<p>Thus if googol $s_1 ... s_n$ is some expression with fewer stacks than googol $t_1 ... t_m$, we can move all the plexes and bangs in $s_1 ... s_n$ to the beginning. Let $s_1 ... s_i$ and $t_1 ... t_j$ be the initial bangs and plexes before the first stack. There will be less than googol-2 bangs and plexes, and </p>
<p>googol bang$^{\text{googol}-3} < 10^{10^{10^{\cdot^{\cdot^{10^{100 + \log_{10} 100 + 1}}}}}}{\large\rbrace} \text{googol-2 10's} < 10^{10^{10^{\cdot^{\cdot^{10^{103}}}}}}{\large\rbrace} \text{googol-2 10's}$</p>
<p>$ < 10 \uparrow\uparrow $googol = googol stack</p>
<p>and so googol $s_1 ... s_i$ will be less than googol $t_1 ... t_{j+1}$ ($t_{j+1}$ is a stack). $s_{i+1} ... s_n$ consists of $k$ stacks, and $t_{j+2} ... t_m$ consists of at least $k$ stacks and possibly some plexes and bangs. Thus googol $s_1 ... s_n$ will be less than googol $t_1 ... t_m$.</p>
<p>Now consider $x S_1$ stack $S_2$ stack ... stack $S_n$ versus $x T_1$ stack $T_2$ stack ... stack $T_n$, where the $S_i$ and $T_i$ are sequences of plexes and bangs. Without loss of generality, we can assume that $S_1 > T_1$ in our order. (If $S_1 = T_1$, we can consider ($x S_1$ stack) $S_2$ stack ... stack $S_n$ versus ($x T_1$ stack) $T_2$ stack ... stack $T_n$, and compare $S_2$ versus $T_2$, etc., until we get to an $S_i$ and $T_i$ that are different.) $x S_1$ stack $S_2$ stack ... stack $S_n$ is, at the minimum, $x S_1$ stack ... stack, while $x T_1$ stack $T_2$ stack ... stack $T_n$, is, at the maximum, $x T_1$ stack bang$^{\text{googol}-3}$ stack .... stack. So it is enough to show</p>
<p>x S_1 stack > x T_1 stack bang$^{\text{googol}-3}$</p>
<p>We have seen that $x$ bang$^k < 10^{10^{10^{\cdot^{\cdot^{10^x}}}}}{\large\rbrace} k+1\text{ times}$ so $x$ bang$^{\text{googol}-3} < 10^{10^{10^{\cdot^{\cdot^{10^x}}}}}{\large\rbrace} \text{googol-2 times}$, and $x T_1$ stack bang$^{\text{googol}-3} < ((x T_1) +$ googol) stack. Thus we must show $x S_1 > (x T_1) +$ googol.</p>
<p>We can assume without loss of generality that the first term of $S_1$ and the first term of $T_1$ are different. (Otherwise set $x = x s_1 ... s_{i-1}$ where i is the smallest number such that s_i and t_i are different.) We have seen above that it is enough to consider </p>
<p>x (plex)^(k+1) versus x (bang)^k
x bang (plex)^k versus x plex (bang)^k</p>
<p>We have previously examined these two cases. In both cases, adding a googol to the smaller leads to the same inequality.</p>
<p>And with that, we are done.</p>
<hr>
<p>What are the prospects for a general comparison algorithm, when the number of terms exceeds googol-3? The difficulty can be illustrated by considering the following two expressions:</p>
<p>$x$ $S_1$ stack plex$^k$</p>
<p>$x$ $T_1$ stack</p>
<p>The two expressions are equal precisely when k = $x$ $T_1$ - $x$ $S_1$. So a general comparison algorthm must allow for the calculation of arbitrary expressions, which perhaps makes our endeavor pointless.</p>
<p>In light of this, I believe the following general comparison algorithm is the best that can be done.</p>
<p>We already have a general comparison algorithm for expressions with no appearances of stack. If stack appears in both expressions, let them be $x$ $S_1$ stack $S_2$ and $x$ $T_1$ stack $T_2$, where $S_2$ and $T_2$ have no appearances of stack. Replace $x$ $S_1$ stack with plex$^{(x S_1)}$, and $x$ $T_1$ stack with plex$^{(x T_1)}$, and do our previous comparison algorithm on the two new expressions. This clearly works because $x$ $S_1$ stack = $10^{10}$ plex$^{(x S_1 -2)}$ and
$x$ $T_1$ stack = $10^{10}$ plex$^{(x T_1-2)}$.</p>
<p>The remaining case is where one expression has stack and the other does not, i.e. googol $S_1$ stack $S_2$ versus googol $T$, where $S_2$ and $T$ have no appearances of stack. Let $s$ and $t$ be the number of terms in $S_2$ and $T$ respectively. Then googol $T$ is greater than googol $S_1$ stack $S_2$ iff $t \ge $ googol $S_1 + s - 2$.</p>
<p>Indeed, if $t \ge $ googol $S_1 + s - 2$,</p>
<p>googol $T \ge$ googol plex$^{\text{googol} S_1 + s - 2} = 10^{10^{10^{\cdot^{\cdot^{10^{100}}}}}}{\large\rbrace} $ googol $S_1$ $+s-1$ 10's $ > 10^{10^{10^{\cdot^{\cdot^{10^{10} + 10 + 1}}}}}{\large\rbrace} $ googol $S_1 +s-1$ 10's > googol $S_1$ stack bang$^s$ </p>
<p>$\ge$ googol $S_1$ stack $S_2$.</p>
<p>If $t \le $ googol $S_1 + s - 3$,</p>
<p>googol $T \le$ googol bang$^{\text{googol} S_1 + s - 3} < 10^{10^{10^{\cdot^{\cdot^{10^{103}}}}}}{\large\rbrace} $ googol $S_1$ $+s-2$ 10's $ < 10^{10^{10^{\cdot^{\cdot^{10^{10^{10}}}}}}}{\large\rbrace} $ googol $S_1 +s$ 10's = googol $S_1$ stack plex$^s$ </p>
<p>$\le$ googol $S_1$ stack $S_2$.</p>
<p>So the comparison algorithm, while not particularly clever, works. </p>
<p>In one of the comments, someone raised the question of a polynomial algorithm (presumably as a function of the maximum number of terms). We can implement one as follows. Let n be the maximum number of terms. We use the following lemma.</p>
<p>Lemma. For any two expressions googol $S$ and googol $T$, if googol $S$ > googol $T$, then googol $S$ > 2 googol $T$.</p>
<p>This lemma is not too hard to verify, but for reasons of space I will not do so here.</p>
<p>As before, we have a simple algorithm in O(n) time when both expressions do not contain stack. If exactly one of the expressions has a stack, we compute $x$ $S_1$ as above, but we stop if the calculation exceeds n. If the calculation finishes, then we can do the previous comparison in O(log n) time; if the calculation stops, then we know that $x$ $S_1$ stack $S_2$ is larger, again in O(log n) time.</p>
<p>If both expressions have stack, then from our previous precedure we calculate both $x$ $S_1$ and $x$ $T_1$. Now we stop if either calculation exceeds $2m$, where $m = $the maximum length of $S_2$ and $T_2$ (Clearly, $2m < 2n$). If the caculation finishes, then we can do our previous procedure in O(n) time. If the calculation stops prematurely, than the larger of $x$ $S_1$ or $x$ $T_1$ will determine the larger original expression. indeed, if $y = x$ $S_1$ and $z = x$ $T_1$, and $y > z$, then by the Lemma $y > 2z$, so since $y > 2m$, we have $y > z+m$. In our procedure we replace $x$ $S_1$ stack by plex$^y$ and $x$ $T_1$ stack by plex$^z$; since $y$ is more than $m$ more than $z$, plex$^y$ $S_2$ will be longer than plex$^z$ $T_2$, so the first expression will be larger. So we apply our procedure to $x$ $S_1$ and $x$ $S_2$; this will reduce the sum of the lengths by at least $m+2$, having used O(log m) operations.</p>
<p>So we wind up with an algorithm that takes O(n) operations.</p>
<hr>
<p>We could extend the notation to have suffixes that apply k -> 10^^^k (Pent), k -> 10^^^^k (Hex), etc. I believe the obvious extension of the above procedure will work, e.g. for expressions with plex, bang, stack, and pent, first count the number of pents; the expression with more pents will be the larger. Otherwise, compare the n+1 groups of plex bang and stack lexicographically by our previously defined procedure. So long as the number of terms is less than googol-2, this procedure should work.</p>
|
linear-algebra | <p>In my opinion both are almost same. However there should be some differenes like any two elements can be multiplied in a field but it is not allowed in vector space as only scalar multiplication is allowed where scalars are from the field.</p>
<p>Could anyone give me atleast one counter- example where field and vector space are both same.
Every field is a vector space but not every vectorspace is a field.
I need an example for which a vector space is also a field.</p>
<p>Thanks in advance. (I'm not from mathematical background.)</p>
| <p>It is true that vector spaces and fields both have operations we often call multiplication, but these operations are fundamentally different, and, like you say, we sometimes call the operation on vector spaces <em>scalar multiplication</em> for emphasis.</p>
<p>The operations on a field <span class="math-container">$\mathbb{F}$</span> are</p>
<ul>
<li><span class="math-container">$+$</span>: <span class="math-container">$\mathbb{F} \times \mathbb{F} \to \mathbb{F}$</span></li>
<li><span class="math-container">$\times$</span>: <span class="math-container">$\mathbb{F} \times \mathbb{F} \to \mathbb{F}$</span></li>
</ul>
<p>The operations on a vector space <span class="math-container">$\mathbb{V}$</span> over a field <span class="math-container">$\mathbb{F}$</span> are</p>
<ul>
<li><span class="math-container">$+$</span>: <span class="math-container">$\mathbb{V} \times \mathbb{V} \to \mathbb{V}$</span></li>
<li><span class="math-container">$\,\cdot\,$</span>: <span class="math-container">$\mathbb{F} \times \mathbb{V} \to \mathbb{V}$</span></li>
</ul>
<p>One of the field axioms says that any nonzero element <span class="math-container">$c \in \mathbb{F}$</span> has a multiplicative inverse, namely an element <span class="math-container">$c^{-1} \in \mathbb{F}$</span> such that <span class="math-container">$c \times c^{-1} = 1 = c^{-1} \times c$</span>. There is no corresponding property among the vector space axioms.</p>
<p>It's an important example---and possibly the source of the confusion between these objects---that any field <span class="math-container">$\mathbb{F}$</span> is a vector space over itself, and in this special case the operations <span class="math-container">$\cdot$</span> and <span class="math-container">$\times$</span> coincide.</p>
<p>On the other hand, for any field <span class="math-container">$\mathbb{F}$</span>, the Cartesian product <span class="math-container">$\mathbb{F}^n := \mathbb{F} \times \cdots \times \mathbb{F}$</span> has a natural vector space structure over <span class="math-container">$\mathbb{F}$</span>, but for <span class="math-container">$n > 1$</span> it does not in general have a <em>natural</em> multiplication rule satisfying the field axioms, and hence does not have a natural field structure.</p>
<p><strong>Remark</strong> As @hardmath points out in the below comments, one can often realize a finite-dimensional vector space <span class="math-container">$\mathbb{F}^n$</span> over a field <span class="math-container">$\mathbb{F}$</span> as a field in its own right <em>if</em> one makes additional choices. If <span class="math-container">$f$</span> is a polynomial irreducible over <span class="math-container">$\mathbb{F}$</span>, say with <span class="math-container">$n := \deg f$</span>, then we can form the set
<span class="math-container">$$\mathbb{F}[x] / \langle f(x) \rangle$$</span>
over <span class="math-container">$\mathbb{F}$</span>: This just means that we consider the vector space of polynomials with coefficients in <span class="math-container">$\mathbb{F}$</span> and declare two polynomials to be equivalent if their difference is some multiple of <span class="math-container">$f$</span>. Now, polynomial addition and multiplication determine operations <span class="math-container">$+$</span> and <span class="math-container">$\times$</span> on this set, and it turns out that because <span class="math-container">$f$</span> is irreducible, these operations give the set the structure of a field. If we denote by <span class="math-container">$\alpha$</span> the image of <span class="math-container">$x$</span> under the map <span class="math-container">$\mathbb{F}[x] \to \mathbb{F}[x] / \langle f(x) \rangle$</span> (since we identify <span class="math-container">$f$</span> with <span class="math-container">$0$</span>, we can think of <span class="math-container">$\alpha$</span> as a root of <span class="math-container">$f$</span>), then by construction <span class="math-container">$\{1, \alpha, \alpha^2, \ldots, \alpha^{n - 1}\}$</span> is a basis of (the underlying vector space of) <span class="math-container">$\mathbb{F}[x] / \langle f \rangle$</span>; in particular, we can identify the span of <span class="math-container">$1$</span> with <span class="math-container">$\Bbb F$</span>, which we may hence regard as a subfield of <span class="math-container">$\mathbb{F}[x] / \langle f(x) \rangle$</span>; we thus call the latter a <em>field extension</em> of <span class="math-container">$\Bbb F$</span>. In particular, this basis defines a vector space isomorphism
<span class="math-container">$$\mathbb{F}^n \to \mathbb{F}[x] / \langle f(x) \rangle, \qquad (p_0, \ldots, p_{n - 1}) \mapsto p_0 + p_1 \alpha + \ldots + p_{n - 1} \alpha^{n - 1}.$$</span> Since <span class="math-container">$\alpha$</span> depends on <span class="math-container">$f$</span>, this isomorphism <em>does</em> depend on a choice of irreducible polynomial <span class="math-container">$f$</span> of degree <span class="math-container">$n$</span>, so the field structure defined on <span class="math-container">$\mathbb{F}^n$</span> by declaring the vector space isomorphism to be a field isomorphism is not natural.</p>
<p><strong>Example</strong> Taking <span class="math-container">$\Bbb F := \mathbb{R}$</span> and <span class="math-container">$f(x) := x^2 + 1 \in \mathbb{R}[x]$</span> gives a field
<span class="math-container">$$\mathbb{C} := \mathbb{R}[x] / \langle x^2 + 1 \rangle.$$</span>
In this case, the image of <span class="math-container">$x$</span> under the canonical quotient map <span class="math-container">$\mathbb{R}[x] \to \mathbb{R}[x] / \langle x^2 + 1 \rangle$</span> is usually denoted <span class="math-container">$i$</span>, and this field is exactly the complex numbers, which we have realized as a (real) vector space of dimension <span class="math-container">$2$</span> over <span class="math-container">$\mathbb{R}$</span> with basis <span class="math-container">$\{1, i\}$</span>.</p>
| <p>A <em>field</em> is an algebraic structure allowing the four basic operations $+$, $-$, $\cdot$, and $:\,$, such that the usual rules of algebra hold, e.g., $(x+y)\cdot z=(x\cdot z) +(y\cdot z)$, etcetera, and division by $0$ is forbidden. The elements of a given field should be considered as "numbers". The systems ${\mathbb Q}$, ${\mathbb R}$, and ${\mathbb C}$ are fields, but there are many others, e.g., the field ${\mathbb F}_2$ consisting only of the two elements $0$, $1$ and satisfying (apart from the obvious relations) $1+1=0$.</p>
<p>A <em>vector space</em> $X$ is in the first place an "additive structure" satisfying the rules we associate with such structures, e.g., $a+({-a})=0$, etc. In addition any vector space has associated with it a certain field $F$, the <em>field of scalars</em> for that vector space. The elements $x$, $y\in X$ cannot only be added and subtracted, but they can be as well <em>scaled</em> by "numbers" $\lambda\in F$. The vector $x$ scaled by the factor $\lambda$ is denoted by $\lambda x$. This scaling satisfies the laws we are accustomed to from the scaling of vectors in ${\mathbb R}^3$:
$$\lambda(x+y)=\lambda x+\lambda y,\qquad (\lambda+\mu)x=\lambda x+\mu x\ .$$</p>
<p>Asking "What is the difference between a vector space and a field" is similar to asking "What is the difference between tension and charge" in electrodynamics. In both cases the simple answer would be: "They are different notions making sense in the same discipline".</p>
|
differentiation | <p>Considering many elementary functions have an antiderivative which is not elementary, why does this type of thing not also happen in differential calculus?</p>
| <p>Just think of how we find those elementary functions:</p>
<ul>
<li><p>We start with the constant functions, which have derivative $0$, and the identity function $f(x)=x$ which has derivative $1$.</p></li>
<li><p>We combine functions by means of addition, subtraction, multiplication, division, composition. For all of those cases we have explicit rules for the derivative.</p></li>
<li><p>We define new functions as the integral of other functions (e.g. $\ln x$ as integral of $1/x$). Obviously when deriving those we get back the function we started with.</p></li>
<li><p>We define functions as the inverse of another function. Again, we've got an explicit formula for derivatives of inverse functions.</p></li>
</ul>
<p>Any function that cannot be defined by a chain of such operations (and also some which can, using the integration rule) we don't consider elementary.</p>
<p>So basically the reason is in the way we construct elementary functions. In some sense, one could say it is because of what functions we consider elementary.</p>
<p>Indeed, this hold not only for elementary functions; even most non-elementary functions we use are defined through such operations (in particular by integrals).</p>
| <p>The short answer is that we have differentiation rules for all the elementary functions, and we have differentiation rules for every way we can combine elementary functions (addition, multiplication, composition), where the derivative of a combination of two functions may be expressed using the functions, their derivatives and the different forms of combination.</p>
<p>Integration, on the other hand, neither has a direct rule for multiplication of two functions nor for composition of two functions. We can integrate the corresponding rules for differentiation and get something that <em>looks</em> like it (integration by parts and substitution), but it only works if you're lucky with what elementary functions are combined in what way.</p>
<p>You might say that there is a hope that there are rules out there, that we just haven't found them yet. This is not true; it's been proven that there are <em>always</em> integrals of elementary functions that are not elementary themselves (under most reasonable definitions of "elementary functions"). It's a deep result known as <a href="https://en.wikipedia.org/wiki/Liouville%27s_theorem_(differential_algebra)" rel="noreferrer">Liouville's theorem</a>.</p>
|
game-theory | <p>You select 10 numbers from the set $\{2,3,\dots,12\}$.
You then continually roll 2 fair dice and sum them up, until your selection of 10 numbers come up. </p>
<p>For example if your selection was 7,7,7,7,8,8,8,6,6,6 (4 7's, 3 8's and 3 6's), and you roll the dice repeatedly and get 7,7,6,5,8,7,7,9,8,3,5,10,12,6,6,3,2,5,7,9,8, you may stop now because 4 7's have come up, 3 8's and 3 6's.</p>
<p>What is the best choice of 10 numbers so as to minimise the number of rolls ?</p>
| <p>Using a brute-force exhaustive search, and a recursion for each case, I get the following answer:
<p>
There are two equivalent optimal strategies, namely
$$4,5,6,6,7,7,7,8,8,9$$
$$5,6,6,7,7,7,8,8,9,10$$
yielding, for the expected number of rounds,
$$
e=\frac{a}{b} \approx 28.26676327
$$
where
$$
\begin{align*}
a&=71526610479792733682076713232552201067\\[4pt]
b&=2530413892848114144358747803518976000\\[4pt]
\end{align*}
$$
<strong>Remarks:</strong>
<p>
Intuitively, the multiplicities of the selected numbers should be in proportion to their probability of occurring in a single round.
<p>
Thus, for example, for the simpler game where you roll a single die, and get to guess $6$ numbers, the optimal selection is $1,2,3,4,5,6$.
<p>
If in each round you roll two dice, and get to guess $36$ numbers, I suspect the optimal selection is
$$2,
3,3,
4,4,4,
5,5,5,5,
6,6,6,6,6,
7,7,7,7,7,7,
8,8,8,8,8,
9,9,9,9,
10,10,10,
11,11,
12
$$
For the game in question, where in each round you roll two dice, and get to guess $10$ numbers, then, since the multiplicities for the selected values must be nonnegative integers, it's not possible for the multiplicities to be in proportion to the associated probabilities. Yet, it's intuitive that they should be <em>approximately</em> in those proportions.</p>
| <p>(too long for a comment, but not a full answer). I'll work it out for $n=2$, $n$ being the number of values you select.</p>
<p>Let $E[a,b]$ be the expected number of throws it take assuming you chose $a,b$. Similarly, let $E[a]$ be the expected number of throws it takes just to see $a$. Of course $E[a]=\frac 1{P(a)}$.</p>
<p>For $a=b$ clearly choosing $7$ is best. In that case one sees at once that $$\boxed{E[7,7]=\frac 2{P(7)}=12}$$</p>
<p>Now assume that $a\neq b$.</p>
<p>Considering the possible outcomes of the first toss (i.e. "$a$", "$b$", or neither) it is easy to see that $$E[a,b]=P(a)\times \left(E[b]+1\right)+P(b)\times \left(E[a]+1\right)+(1-P(a)-P(b))\times \left(E[a,b]+1\right)$$</p>
<p>This implies that $$\boxed {E[a,b]=\frac {P(a)P(b)+P(a)+P(b)}{P(a)P(b)(P(a)+P(b))}}$$</p>
<p>It is easy to compute that $$E[7,8]=E[7,6]=9.92727\cdots<12$$ and that $$E[6,8]=10.8$$ so as the OP expected, choosing $\{7,6\}$ or $\{7,8\}$ is optimal. (remark: it is not difficult to rule out the other possibilities, by direct computation if nothing else.)</p>
<p>Note I: this method generalizes to more choices but the cases start to multiply badly. My guess for $n=3$ would be $\{6,7,8\}$ but I have not verified this and it could easily be incorrect.</p>
<p>Note II: this method certainly lends itself to automation. If you have computed all the expectations for $n$ choices then you can get them for $n+1$ by a recursion as was done above. This is hardly a pencil and paper method, however. there are $\binom {19}9=92378$ possible selections when $n=9$ and $\binom {20}{10}=184756$ when $n=10$.</p>
|
differentiation | <p>Here's a statement in Zygmund's Measure and Integral on page 17:</p>
<blockquote>
<p>If $f$ has a continuous derivative on $[a,b]$, then (by the mean-value theorem) $f$ satisfies a Lipschitz condition on $[a,b]$.</p>
</blockquote>
<p>This does not seem obvious to me. How can I show it?</p>
<p>Also, what does a continuous derivative imply? Can we conclude the function is differentiable? If so, how can I prove it?</p>
| <p>By the mean value theorem, </p>
<p><span class="math-container">$$f(x) - f(y) = f'(\xi)(x-y)$$</span>
for some <span class="math-container">$\xi \in (y,x)$</span>. But since <span class="math-container">$f'$</span> is continuous and <span class="math-container">$[a,b]$</span> is compact, then <span class="math-container">$f'$</span> is bounded in that interval, say by <span class="math-container">$C$</span>. Thus taking absolute values yields </p>
<p><span class="math-container">$$\lvert f(x) - f(y)\rvert \le C \lvert x-y\rvert$$</span> </p>
| <p>Since $f'$ is continuous it is bounded on $[a,b]$. Choose $M\in \Bbb R$ such that $|f'(x)|\le M$ for all $x\in [a,b]$. If $x,y\in [a,b]$, then by the MVT there is $\xi\in [a,b]$ with $f(x)-f(y)=f'(\xi)(x-y)$, thus $|f(x)-f(y)|\le M|x-y|$. In fact you can show that a differentiable function on an open interval (not necessarily a bounded interval) is Lipschitz continuous if and only if it has a bounded derivative. This is because any Lipschitz constant gives a bound on the derivative and conversely any bound on the derivative gives a Lipschitz constant.<br><br>
To your other question: Continuous derivative does not imply twice differentiable. In fact, we know that there are functions $f$ (such as the Weierstrass-function) which are continuous but not differentiable. Taking the antiderivative $F$ gives us a differentiable function with derivative $f$ (by the Fundamental Theorem of Calculus), so it has a continuous derivative, but as $f$ is not differentiable, $F$ is not twice differentiable.</p>
|
logic | <p>There are many classic textbooks in <strong>set</strong> and <strong>category theory</strong> (as possible foundations of mathematics), among many others Jech's, Kunen's, and Awodey's.</p>
<blockquote>
<p>Are there comparable classic textbooks in <strong>type theory</strong>, introducing and motivating their matter in a generally agreed upon manner from the ground up and covering the whole field, essentially?</p>
</blockquote>
<p>If not so: why?</p>
| <p>Although not as comprehensive a textbook as, say, Jech's classic book on set theory, Jean-Yves Girard's <a href="http://www.paultaylor.eu/stable/Proofs+Types"><em>Proofs and Types</em></a> is an excellent starting point for reading about type theory. It's freely available from translator Paul Taylor's website as a PDF. Girard does assume some knowledge of the lambda calculus; if you need to learn this too, I recommend Hindley and Seldin's <a href="http://www.cambridge.org/gb/knowledge/isbn/item1175709/?site_locale=en_GB"><em>Lambda-Calculus and Combinators: An Introduction</em></a>.</p>
<p>As others have mentioned, Martin-Löf's <em>Intuitionistic Type Theory</em> would then be a good next step.</p>
<p>A different approach would be to read Benjamin Pierce's wonderful textbook, <a href="http://www.cis.upenn.edu/~bcpierce/tapl/"><em>Types and Programming Languages</em></a>. This is oriented towards the practical aspects of understanding types in the context of writing programming languages, rather than purely its mathematical characteristics or foundational promise, but nonetheless it's a very clear and well-written book, with numerous exercises.</p>
<p>The bibliography provided by the <a href="http://plato.stanford.edu/entries/type-theory/">Stanford Encyclopedia of Philosophy entry on type theory</a> is quite extensive, and might provide alternative avenues for your research.</p>
| <p>There are two main settings in which I see type theory as a foundational system.</p>
<p>The first is intuitionistic type theory, particularly the system developed by Martin-Löf. The book <em>Intuitionistic Type Theory</em> (1980) seems to be floating around the internet. </p>
<p>The other setting is second-order (and higher-order) arithmetic. Two main books on this are <em>Foundations without foundationalism</em> by Stewart Shapiro (1991) and <em>Subsystems of second order arithmetic</em> by Stephen Simpson (1999). A decent amount of constructive mathematics, for example the material in <em>Constructive Analysis</em> by Bishop and Bridges (1985), can also be formalized directly in constructive higher-order arithmetic, however the taste of many constructivists is to avoid doing this. </p>
|
combinatorics | <blockquote>
<p>For which <span class="math-container">$n\in \mathbb{N}$</span> can we divide the set <span class="math-container">$\{1,2,3,\ldots,3n\}$</span> into <span class="math-container">$n$</span> subsets each with <span class="math-container">$3$</span> elements such that in each subset <span class="math-container">$\{x,y,z\}$</span> we have <span class="math-container">$x+y=3z$</span>?</p>
</blockquote>
<p>Since <span class="math-container">$x_i+y_i=3z_i$</span> for each subset <span class="math-container">$A_i=\{x_i,y_i,z_i\}$</span>, we have <span class="math-container">$$4\sum _{i=1}^n z_i=\sum _{i=1}^{3n}i = {3n(3n+1)\over 2} \implies 8\mid n(3n+1) $$</span>
so <span class="math-container">$n=8k$</span> or <span class="math-container">$n=8k-3$</span>. Now it is not difficult to see that if <span class="math-container">$k=1$</span> we have such partition.</p>
<ul>
<li>For <span class="math-container">$n=5$</span> we have:
<span class="math-container">$$A_1= \{9,12,15\}, A_2= \{4,6,14\}, A_3= \{2,5,13\}, \\A_4= \{10,7,11\}, A_5= \{1,3,8\}$$</span></li>
<li>For <span class="math-container">$n=8$</span> we have:
<span class="math-container">$$A_1= \{24,21,15\}, A_2= \{23,19,14\}, A_3= \{22,2,8\}, A_4= \{20,1,7\}, \\A_5= \{17,16,11\}, A_6= \{18,12,10\}, A_7= \{13,5,6\}, A_8= \{9,3,4\}$$</span></li>
</ul>
<p>What about for <span class="math-container">$k\geq 2$</span>? Some clever induction step? Or some ''well'' known configuration?</p>
<p>Source: Serbia 1983, municipal round, 3. grade</p>
| <p>If there is a solution for <span class="math-container">$N$</span>, then there is a solution for <span class="math-container">$7N+5$</span>.<br />
The solution for <span class="math-container">$N$</span> uses up numbers from <span class="math-container">$1$</span> to <span class="math-container">$3N$</span>. Then
<span class="math-container">$$(3N+k, 15N+9+2k, 6N+3+k), k=1..3N+3\\
(12N+8+k,15N+10+2k,9N+6+k), k=1..3N+2$$</span>
sits the numbers from <span class="math-container">$3N+1$</span> to <span class="math-container">$21N+15$</span> on top of them.</p>
<p>A similar method gives a solution for <span class="math-container">$25N+8Q$</span>, for all <span class="math-container">$-13\le Q\le11$</span>, whenever there is a solution for <span class="math-container">$N\ge 13$</span>. Together with @RobPratt's solution, that covers all <span class="math-container">$N=8M$</span> and all <span class="math-container">$N=8M-3$</span>.</p>
<p>I have started a new question for a different version at <a href="https://math.stackexchange.com/questions/4190163/split-1-2-3n-into-triples-with-xy-4z">Split $\{1,2,...,3n\}$ into triples with $x+y=4z$</a> and also <a href="https://math.stackexchange.com/questions/4195206/split-1-3n-into-triples-with-xy-5z-no-solutions">Split $\{1,...,3n\}$ into triples with $x+y=5z$ - no solutions?</a></p>
| <p>Here is the integer linear programming approach I used to find partitions for all such <span class="math-container">$n\le 496$</span> with <span class="math-container">$n \equiv 0,5 \pmod 8$</span>. First enumerate all triples <span class="math-container">$\{x,y,z\}$</span> with <span class="math-container">$x+y=3z$</span> and <span class="math-container">$x,y,z$</span> distinct elements of <span class="math-container">$[3n]:=\{1,\dots,3n\}$</span>. For each such triple <span class="math-container">$T$</span>, let binary decision variable <span class="math-container">$u_T$</span> indicate whether <span class="math-container">$T$</span> appears in the partition. The constraints
<span class="math-container">$$\sum_{T:\ i\in T} u_T = 1 \quad \text{for $i\in[3n]$} \tag1$$</span>
enforce that each element appears exactly once in the partition.</p>
<p>An alternative approach is to introduce nonnegative slack variables <span class="math-container">$s_i$</span>, replace the set partitioning constraints <span class="math-container">$(1)$</span> with (set covering and cardinality) constraints
<span class="math-container">\begin{align}
\sum_{T:\ i\in T} u_T + s_i &\ge 1 &&\text{for $i\in[3n]$} \tag2 \\
\sum_T u_T &= n \tag3
\end{align}</span>
and minimize <span class="math-container">$\sum_{i=1}^{3n} s_i$</span>. A partition of <span class="math-container">$[3n]$</span> into <span class="math-container">$n$</span> triples with <span class="math-container">$x+y=3z$</span> exists if and only if the optimal objective value is <span class="math-container">$0$</span>.</p>
|
differentiation | <p>I have just started learning about differential equations, as a result I started to think about this question but couldn't get anywhere. So I googled and wasn't able to find any particularly helpful results. I am more interested in the reason or method rather than the actual answer. Also I do not know if there even is a solution to this but if there isn't I am just as interested to hear why not.</p>
<p>Is there a solution to the differential equation:</p>
<p>$$f(x)=\sum_{n=1}^\infty f^{(n)}(x)$$</p>
| <p>$f(x)=\exp(\frac{1}{2}x)$ is such a function, since $f^{(n)}=2^{-n} f(x)$, you have</p>
<p>$$\sum_{n=1}^\infty f^{(n)}(x)=\sum_{n=1}^\infty 2^{-n}f(x)=(2-1)f(x)=f(x)$$</p>
<p>This is the only function (up to a constant prefactor) for which $\sum_{n}f^{(n)}$ and its derivatives converge uniformly (on compacta), as $$f'=\sum_{n=1}^\infty f^{(n+1)}=f-f'$$ follows from this assumption. But this is the same as $f-2 f'=0$, of which the only (real) solutions are $f(x)= C \exp{\frac{x}{2}}$ for some $C \in \mathbb R$.</p>
| <p>Differentiate both sides to get $f'(x)=f''(x)+f^{(3)}(x)+...$ </p>
<p>So the starting equation becomes $f(x)=f'(x)+f'(x)\Rightarrow f(x)=2f'(x)$ </p>
<p>Multiply now both sides by $e^{-\frac{x}{2}}$ and this becomes<br>
$$[e^{-\frac{x}{2}}f(x)]'=0$$<br>
So $f(x)=ce^{\frac{x}{2}}$<br>
Done.</p>
|
combinatorics | <p>I checked several thousand natural numbers and observed that $\lfloor n!/e\rfloor$ seems to always be an even number. Is it indeed true for all $n\in\mathbb N$? How can we prove it?</p>
<p>Are there any positive irrational numbers $a\ne e$ such that $\lfloor n!/a\rfloor$ is even for all $n\in\mathbb N$?</p>
| <p>Note that:</p>
<p>$$e^{-1}=\sum_{k=0}^\infty \frac{(-1)^k}{k!}$$</p>
<p>Then:</p>
<p>$$\frac{n!}e=n!e^{-1} = \left(\sum_{k=0}^{n} (-1)^k\frac{n!}{k!}\right) + \sum_{k=n+1}^{\infty} (-1)^{k}\frac{n!}{k!}$$</p>
<p>Show that if $a_n=\sum_{k=n+1}^{\infty} (-1)^{k}\frac{n!}{k!}$ then $0<|a_{n}|<1$ and $a_n>0$ if and only if $n$ is odd.</p>
<p>So the when $n$ is odd, the value is:
$$\left\lfloor\frac{n!}{e}\right\rfloor=\sum_{k=0}^{n} (-1)^k\frac{n!}{k!}\tag{1}$$
When $n$ is even it is one less:
$$\left\lfloor\frac{n!}{e}\right\rfloor=-1+\sum_{k=0}^{n} (-1)^k\frac{n!}{k!}\tag{2}$$</p>
<p>Now, almost all of these terms are even. The last term $n!/n!=1$ is odd. When $n$ is odd, the second-to-last term $n!/(n-1)!$ is odd, also. But all other terms are even.</p>
<p>So for $n$ odd, there are two odd terms in the sum, $k=n,n-1$.</p>
<p>For $n$ even, there are two odd terms in the sum, $-1$ and $k=n.$</p>
<hr>
<p>The trick, then, is to show that the $a_n$ has these properties:
$$\begin{align}
&0<|a_n|<1\\
&a_n>0\iff n\text{ is odd}
\end{align}$$</p>
<p>To show these, we note that $\frac{n!}{k!}$ is strictly decreasing for $k>n$ and $(-1)^k\frac{n!}{k!}$ is alternating. In general, any alternating sum of a decreasing series converges to a value strictly between $0$ and the first term of the sequence, which in this case is $\frac{(-1)^{n+1}}{n+1}.$</p>
| <p>The number of <a href="https://en.wikipedia.org/wiki/Derangement">derangements</a> of $[n]=\{1,\ldots,n\}$ is </p>
<p>$$d_n=n!\sum_{k=0}^n\frac{(-1)^k}{k!}\;,$$</p>
<p>so</p>
<p>$$\frac{n!}e-d_n=n!\sum_{k>n}\frac{(-1)^k}{k!}\;,$$</p>
<p>which is less than $\frac1{n+1}$ in absolute value. Thus for $n\ge 1$, $d_n$ is the integer nearest $\frac{n!}e$, and</p>
<p>$$d_n=\begin{cases}
\left\lfloor\frac{n!}e\right\rfloor,&\text{if }n\text{ is odd}\\\\
\left\lceil\frac{n!}e\right\rceil,&\text{if }n\text{ is even}\;.
\end{cases}$$</p>
<p>The recurrence $d_n=nd_{n-1}+(-1)^n$ is also well-known. We have $d_0=1$, so an easy induction shows that $d_n$ is odd when $n$ is even, and even when $n$ is odd. Thus, for odd $n$ we have $\left\lfloor\frac{n!}e\right\rfloor=d_n$ is even, and for even $n$ we have $\left\lfloor\frac{n!}e\right\rfloor=\left\lceil\frac{n!}e\right\rceil-1$ is again even.</p>
|
probability | <p>Suppose I have a line segment of length $L$. I now select two points at random along the segment. What is the expected value of the distance between the two points, and why?</p>
| <p>Byron has already answered your question, but I will attempt to provide a detailed solution...</p>
<p>Let $X$ be a random variable uniformly distributed over $[0,L]$, i.e., the probability density function of $X$ is the following</p>
<p>$$f_X (x) = \begin{cases} \frac{1}{L} & \textrm{if} \quad{} x \in [0,L]\\ 0 & \textrm{otherwise}\end{cases}$$</p>
<p>Let us randomly pick two points in $[0,L]$ <em>independently</em>. Let us denote those by $X_1$ and $X_2$, which are random variables distributed according to $f_X$. The distance between the two points is a new random variable</p>
<p>$$Y = |X_1 - X_2|$$</p>
<p>Hence, we would like to find the expected value $\mathbb{E}(Y) = \mathbb{E}( |X_1 - X_2| )$. Let us introduce function $g$</p>
<p>$$g (x_1,x_2) = |x_1 - x_2| = \begin{cases} x_1 - x_2 & \textrm{if} \quad{} x_1 \geq x_2\\ x_2 - x_1 & \textrm{if} \quad{} x_2 \geq x_1\end{cases}$$</p>
<p>Since the two points are picked independently, the joint probability density function is the product of the pdf's of $X_1$ and $X_2$, i.e., $f_{X_1 X_2} (x_1, x_2) = f_{X_1} (x_1) f_{X_2} (x_2) = 1 / L^2$ in $[0,L] \times [0,L]$. Therefore, the expected value $\mathbb{E}(Y) = \mathbb{E}(g(X_1,X_2))$ is given by</p>
<p>$$\begin{align} \mathbb{E}(Y) &= \displaystyle\int_{0}^L\int_{0}^L g(x_1,x_2) \, f_{X_1 X_2} (x_1, x_2) \,d x_1 \, d x_2\\[6pt]
&= \frac{1}{L^2} \int_0^L\int_0^L |x_1 - x_2| \,d x_1 \, d x_2\\[6pt]
&= \frac{1}{L^2} \int_0^L\int_0^{x_1} (x_1 - x_2) \,d x_2 \, d x_1 + \frac{1}{L^2} \int_0^L\int_{x_1}^L (x_2 - x_1) \,d x_2 \, d x_1\\[6pt]
&= \frac{L^3}{6 L^2} + \frac{L^3}{6 L^2} = \frac{L}{3}\end{align}$$</p>
| <p>Sorry. I posted a cryptic comment just before running off to class. What I meant was that if $X,Y$ are independent uniform $(0,1)$ random variables, then the triple
$$(A,B,C):=(\min(X,Y),\ \max(X,Y)-\min(X,Y),\ 1-\max(X,Y))$$
is an exchangeable sequence. In particular, $\mathbb{E}(A)=\mathbb{E}(B)=\mathbb{E}(C),$ and since $A+B+C=1$ identically we must have $\mathbb{E}(B)=\mathbb{E}(\mbox{distance})={1\over 3}.$ </p>
<p>Intuitively, the "average" configuration of two random points on a interval
looks like this: <img src="https://i.sstatic.net/NTD7P.jpg" alt="enter image description here"></p>
|
number-theory | <p>Some time ago when decomponsing the natural numbers, $\mathbb{N}$, in prime powes I noticed a pattern in their powers. Taking, for example, the numbers $\lbrace 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16 \rbrace$ and factorize them, we will get
$$\begin{align} 1&=2^0\times 3^0\times 5^0\times 7^0\times 11^0\times 13^0\times\ldots \\ 2&=2^1\times 3^0\times 5^0\times 7^0\times 11^0\times 13^0\times\ldots \\3&=2^0\times 3^1\times 5^0\times 7^0\times 11^0\times 13^0\times\ldots \\ 4&=2^2\times 3^0\times 5^0\times 7^0\times 11^0\times 13^0\times\ldots \\ 5&=2^0\times 3^0\times 5^1\times 7^0\times 11^0\times 13^0\times\ldots \\ 6&=2^1\times 3^1\times 5^0\times 7^0\times 11^0\times 13^0\times\ldots \\ 7&=2^0\times 3^0\times 5^0\times 7^1\times 11^0\times 13^0\times\ldots \\ 8&=2^3\times 3^0\times 5^0\times 7^0\times 11^0\times 13^0\times\ldots \\ 9&=2^0\times 3^2\times 5^0\times 7^0\times 11^0\times 13^0\times\ldots \\ 10&=2^1\times 3^0\times 5^1\times 7^0\times 11^0\times 13^0\times\ldots \\ 11&=2^0\times 3^0\times 5^0\times 7^0\times 11^1\times 13^0\times\ldots \\ 12&=2^2\times 3^1\times 5^0\times 7^0\times 11^0\times 13^0\times\ldots \\ 13&=2^0\times 3^0\times 5^0\times 7^0\times 11^0\times 13^1\times\ldots \\ 14&=2^1\times 3^0\times 5^0\times 7^1\times 11^0\times 13^0\times\ldots \\ 15&=2^0\times 3^1\times 5^1\times 7^0\times 11^0\times 13^0\times\ldots \\ 16&=2^4\times 3^0\times 5^0\times 7^0\times 11^0\times 13^0\times\ldots \\\end{align}$$
Now if we look at the powers of $2$ we will notice that they are $$\lbrace f_2(n)\rbrace=\lbrace 0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,4\rbrace$$ and for the powers of $3$ we have $$\lbrace f_3(n)\rbrace=\lbrace 0,0,1,0,0,1,0,0,2,0,0,1,0,0,1,0\rbrace$$
This, of course, is a well known fact.</p>
<p>Since then I wondered if there was a formula for $f_2(n)$ or $f_3(n)$ or $f_p(n)$, with $p\in \mathbb{P}$. It seemed impossible but I was able to devise the suitable formulas. They are
$$\displaystyle\begin{align} f_2(n)=\sum_{r=1}^{\infty}\frac{r}{{2^{r+1}}}\sum_{k=0}^{2^{r+1}-1}\cos\left( \frac{2k\pi(n+2^{r})}{2^{r+1}} \right)\end{align}$$
and for the general case we have
$$\displaystyle f_p(n)=\sum_{r=1}^{\infty}\frac{r}{p^{r+1}}\sum_{j=1}^{p-1}\left(\sum_{k=0}^{p^{r+1}-1}\cos\left( \frac{2k\pi(n+(p-j)p^{r})}{p^{r+1}} \right)\right)$$
If one cares to analyse the formula for $f_p(n)$ it can be concluded that it needs not to be restricted to the prime numbers, so that we have $f_m(n), m \in \mathbb{N}$ and similar patterns for $\lbrace f_m(n)\rbrace$ will result. Now, the wonderfull thing is that we can express the <em>arithmetical divisor functions</em> $\sigma_k(n)$ in terms of $f_m(n)$ as follows
$$\displaystyle \sigma_a(n)=1+\sum_{m=2}^{\infty}\sum_{r=1}^{\infty}\frac{m^{a}}{m^{r+1}}\sum_{j=1}^{m-1}\left(\sum_{k=0}^{m^{r+1}-1}\cos\left( \frac{2k\pi(n+(m-j)m^{r})}{m^{r+1}} \right)\right)$$
And, if we consider the <em>divisor summatory function</em>, $D(n)$, as
$$D(n)=\sum_{m \leq n}d(m)$$
with $$d(n)=\sigma_{0}(n)=\sum_{d|n}1$$
we can express $D(n)$ as
$$D(n)=\sum_{m=2}^{\infty}\sum_{r=1}^{\infty}\frac{r}{m^{r+1}}\sum_{j=1}^{m-1}\left(\sum_{k=0}^{m^{r+1}-1}\cos\left( \frac{2k\pi(2^{n}+(m-j)m^{r})}{m^{r+1}} \right)\right)$$
Now, we know that, $d(n)$ and $D(n)$ are related to the Riemann zeta-function by
$$\zeta^{2}(z)=\sum_{n=1}^{\infty}\frac{d(n)}{n^{z}}$$
and
$$\zeta^{2}(z)=z\int_{1}^{\infty}\frac{D(x)}{x^{z+1}}dx$$</p>
<p>Now, my questions</p>
<ul>
<li>What can we say about the convergence of $f_m(z)$, $\sigma_a(z)$ and $D(z)$ with $z \in \mathbb{C}$? We can see that they converge for $z \in \mathbb{N}$.</li>
<li>I think that $\sigma_a(z)$ and $D(z)$ are only <a href="https://mathoverflow.net/questions/54625/how-important-are-the-formulas-at-number-theory-closed">curiosities</a> and aren't interesting in the context of the Riemann zeta-function because they are hard to compute. What do you think? </li>
<li>Are formulas $f_m(z)$, $\sigma_a(z)$ and $D(z)$ original. I think they are. I'd like to know if anyone has found something like this before. I've posted this as an answer to <a href="https://mathoverflow.net/questions/29828/greatest-power-of-two-dividing-an-integer">this</a> post sometime ago.</li>
<li>Finally, is this intereting enough to publish somewhere? I'm just an amateur...</li>
</ul>
<p>To conclude I'd like to apologise for presenting all this formulas without showing how I got them but you can consider <a href="https://math.stackexchange.com/questions/5574/a-triangular-representation-for-the-divisor-summatory-function-dx">this</a> previous post of mine and the question <a href="https://mathoverflow.net/questions/29828/greatest-power-of-two-dividing-an-integer">Greatest power of two dividing an integer</a>, <a href="https://mathoverflow.net/questions/40427/difficult-infinite-sum">Difficult Infinite Sum</a> and <a href="http://fs.gallup.unm.edu/Atanassov-SomeProblems.pdf" rel="noreferrer">On the 61-st, the 62-nd, and the 63-rd Smarandache's problem</a> page 38.</p>
<p>And now a challenge, can you present a formula for the <em>characteristic function of the prime numbers</em>?</p>
<p><strong>EDIT:</strong>
I'm answering my challenge and leaving another. Considering that the <em>characteristic function of the primes</em>, $u(n)$, is given by </p>
<p>$$
\begin{equation}
u(n)=\begin{cases}
&1\;\;\;\text{ if } n \in \mathbb{P} \\
&\\
&\\
&0\;\;\;\text{ if } n \notin \mathbb{P}
\end{cases}
\end{equation}
$$</p>
<p>I have found that $u(n)$ is given by the following formula</p>
<p>$$
\begin{equation}
u(n)=\prod_{m=2}^{\infty}\;\;\prod _{r=1}^{\infty} \left\{1-\frac{1}{m^{r+1}} \sum _{j=1}^{m-1}\;\;\;\sum _{k=0}^{m^{r+1}-1} \cos\left(2 k \pi \cdot\frac{n-m+(m-j) m^r }{m^{r+1}}\right)\right\}
\end{equation}
$$
Now, in the same spirit, what is formula for the <em>prime counting function</em>, $\pi(x)$?</p>
| <p>You should take a look at this <a href="https://math.stackexchange.com/questions/44290/can-insight-be-derived-from-direct-formulae-for-prime-number-functions">math stack exchange post</a>, and the answer there, as it applies directly here.</p>
<p>For a formula for $\pi (x)$ you can take a look at <a href="http://en.wikipedia.org/wiki/Formula_for_primes" rel="nofollow noreferrer">this wikipedia page</a>.</p>
<p>Such elementary formulas tend to encode some complex algorithm which would check the result, and evaluate the function. Usually they are not of too much interest because: </p>
<p>(1) Any analysis is very difficult, and people do not tend to be able to use such formulas to prove theorems regarding the primes or $\sigma(n)$
(2) Evaluating the series takes longer then other known computational methods for $\sigma(n)$ or $\pi(x)$.
(3) They do not tell us about the nature of the functions at hand, as they simply encode an algorithm which calculates those functions. Studying the algorithm would be more fruitful then the series for a greater understanding.
(4) There seems to be a lot of these series.</p>
<p>For example, the formula for $\pi(x)$,</p>
<p>$$\pi(k) = k - 1 + \sum_{j=1}^k \left\lfloor {2 \over j} \left(1 + \sum_{s=1}^{\left\lfloor\sqrt{j}\right\rfloor} \left(\left\lfloor{ j-1 \over s}\right\rfloor - \left\lfloor{j \over s}\right\rfloor\right) \right)\right\rfloor$$</p>
<p>is an encoding of the Sieve of Eratosthenes, which is more interesting and easier to understand when just looked at by itself. Indeed, the idea behind the sieve can be explained to a highschool student, but trying to understand the formula above without having seen it become could intimidate many strong undergraduates or graduate students.</p>
<p>We also have the formula for the $n^{th}$ prime number:</p>
<p>$$p_n = 1 + \sum_{k=1}^{2(\lfloor n \ln(n)\rfloor+1)} \left(1 - \left\lfloor{\frac{1}{n}\left(k - 1 + \sum_{j=1}^k \left\lfloor {2 \over j} \left(1 + \sum_{s=1}^{\left\lfloor\sqrt{j}\right\rfloor} \left(\left\lfloor{ j-1 \over s}\right\rfloor - \left\lfloor{j \over s}\right\rfloor\right) \right)\right\rfloor\right)} \right\rfloor\right)$$</p>
<p>Having said all of this, I still think it is quite interesting that you managed to discover these formulas. I would encourage you to write it up, and explain the ideas behind why they are correct. (Also perhaps have some computations verifying the formulas to comfort the reader.) Someone might have been looking for exactly these formulas, and it helps others to know that they even exists. Also, perhaps the ideas that go into your formulas are deeper then what I described above.</p>
| <p>These formulas are convoluted and what looks to me as probably useless representations of the p-adic order of an integer. All you seem to have done is taken advantage of the fact that: $$1_{p^j\mid n}=\frac{1}{p^j}\sum_{t=0}^{p^j-1}e^{2\pi i t\frac{n}{p^j}}$$ And then re-wrote it it without the imaginary part of each root of unity, so you got a sum of cosines, and then summed over the powers in $j$, so that it counted the multiplitices $p$ of a given integer, which you gave as a complex double sum. In addition I doubt this is of any use in a computational context, since: $$v_p(n)=\gcd(p^{\lfloor \log_p(n) \rfloor},n)$$Which can be calculated very fast using the euclidean algorithm.</p>
<p>Also, just to show you that it is not hard to obtain results about this function: $$\frac{\zeta(s)}{p^s-1}=\sum_{n=1}^\infty\frac{v_p(n)}{n^s}=\sum_{n=1}^\infty\frac{f_p(n)}{n^s}$$
$$\sum_{k=1}^n v_p(k)=\sum_{k=1}^nf_p(k)=\sum_{j=1}^{\lfloor \log_p(n) \rfloor}\lfloor \frac{n}{p^j} \rfloor$$
$$\sum_{k=1}^np^{v_p(k)}=\sum_{k=1}^np^{f_p(k)}=n+(1-\frac{1}{p})\sum_{j=1}^{\lfloor \log_p(n) \rfloor}p^j\lfloor \frac{n}{p^j} \rfloor$$
And in general if we define for a fixed prime $p$ and an arbitrary function $g$ the function $\delta_p$: $$
\delta_p(k) \stackrel{}{=}
\begin{cases}
g(j)-g(j-1) & \text{if } k = p^j \text{ with } j\ge 1, \\
g(0) & \text{if } k=1 \\
0 & \text{otherwise}
\end{cases}
$$</p>
<p>So that $\sum_{d\mid k}\delta_p(d)=g(v_p(k))=g(f_p(k))$</p>
<p>Then we have for arbitrary functions $g$ and $h$:</p>
<p>$$\sum_{k=1}^ng(f_p(k))h(k)=\sum_{k=1}^n(\sum_{d\mid k}\delta_p(d))h(k)=\sum_{k=1}^n\delta_p(k)\sum_{m=1}^{\lfloor n/k\rfloor}h(mk)$$</p>
<p>$$=g(0)\sum_{m=1}^nh(m)+\sum_{j=1}^{\lfloor \log_p(n) \rfloor}(g(j)-g(j-1))\sum_{m=1}^{\lfloor n/p^j \rfloor}h(mp^j)$$ Where the upper index $n$ in this sum has been simplified to a sum with an upper index of $O(\ln(n))$ thus allowing any sort of sum evolving the p-adic order or as you are refering to it with the function $f_p$ to be calculatetable in an exponentially faster time then the sums you use just to represent one value of $f_p$.</p>
<p>So I wouldn't say your formulas are original, nor would I say they are very practical. Though for whatever my opinion is worth (probably not much) I would say that despite this, it is still great you are exploring and finding new things that interest you.</p>
|
linear-algebra | <p>This may be a trivial question yet I was unable to find an answer:</p>
<p>$$\left \| A \right \| _2=\sqrt{\lambda_{\text{max}}(A^{^*}A)}=\sigma_{\text{max}}(A)$$</p>
<p>where the spectral norm $\left \| A \right \| _2$ of a complex matrix $A$ is defined as $$\text{max} \left\{ \|Ax\|_2 : \|x\| = 1 \right\}$$</p>
<p>How does one prove the first and the second equality?</p>
| <p>Put <span class="math-container">$B=A^*A$</span> which is a Hermitian matrix. A linear transformation of the Euclidean vector space <span class="math-container">$E$</span> is Hermite iff there exists an orthonormal basis of E consisting of all the eigenvectors of <span class="math-container">$B$</span>. Let <span class="math-container">$\lambda_1,...,\lambda_n$</span> be the eigenvalues of <span class="math-container">$B$</span> and <span class="math-container">$\left \{ e_1,...e_n \right \}$</span> be an orthonormal basis of <span class="math-container">$E$</span>. Denote by <span class="math-container">$\lambda_{j_{0}}$</span> to be the largest eigenvalue of <span class="math-container">$B$</span>.</p>
<p>For <span class="math-container">$x=a_1e_1+...+a_ne_n$</span>, we have <span class="math-container">$\left \| x \right \|=\left \langle \sum_{i=1}^{n}a_ie_i,\sum_{i=1}^{n}a_ie_i \right \rangle^{1/2} =\sqrt{\sum_{i=1}^{n}a_i^{2}}$</span> and
<span class="math-container">$Bx=B\left ( \sum_{i=1}^{n}a_ie_i \right )=\sum_{i=1}^{n}a_iB(e_i)=\sum_{i=1}^{n}\lambda_ia_ie_i$</span>. Therefore:</p>
<p><span class="math-container">$\left \| Ax \right \|=\sqrt{\left \langle Ax,Ax \right \rangle}=\sqrt{\left \langle x,A^*Ax \right \rangle}=\sqrt{\left \langle x,Bx \right \rangle}=\sqrt{\left \langle \sum_{i=1}^{n}a_ie_i,\sum_{i=1}^{n}\lambda_ia_ie_i \right \rangle}=\sqrt{\sum_{i=1}^{n}a_i\overline{\lambda_ia_i}} \leq \underset{1\leq j\leq n}{\max}\sqrt{\left |\lambda_j \right |} \times (\left \| x \right \|)$</span></p>
<p>So, if <span class="math-container">$\left \| A \right \|$</span> = <span class="math-container">$\max \left\{ \|Ax\| : \|x\| = 1 \right\}$</span> then <span class="math-container">$\left \| A \right \|\leq \underset{1\leq j\leq n}\max\sqrt{\left |\lambda_j \right |}$</span>. (1)</p>
<p>Consider: <span class="math-container">$x_0=e_{j_{0}}$</span> <span class="math-container">$\Rightarrow \left \| x_0 \right \|=1$</span> so that <span class="math-container">$\left \| A \right \|^2 \geq \left \langle x_0,Bx_0 \right \rangle=\left \langle e_{j_0},B(e_{j_0}) \right \rangle=\left \langle e_{j_0},\lambda_{j_0} e_{j_0} \right \rangle = \lambda_{j_0}$</span>. (2)</p>
<p>Combining (1) and (2) gives us <span class="math-container">$\left \| A \right \|= \underset{1\leq j\leq n}{\max}\sqrt{\left | \lambda_{j} \right |}$</span> where <span class="math-container">$\lambda_j$</span> is the eigenvalue of <span class="math-container">$B=A^*A$</span></p>
<p>Conclusion: <span class="math-container">$$\left \| A \right \| _2=\sqrt{\lambda_{\text{max}}(A^{^*}A)}=\sigma_{\text{max}}(A)$$</span></p>
| <p>First of all,
<span class="math-container">$$\begin{align*}\sup_{\|x\|_2 =1}\|Ax\|_2 & = \sup_{\|x\|_2 =1}\|U\Sigma V^Tx\|_2 = \sup_{\|x\|_2 =1}\|\Sigma V^Tx\|_2\end{align*}$$</span>
since <span class="math-container">$U$</span> is unitary, that is, <span class="math-container">$\|Ux_0\|_2^2 = x_0^TU^TUx_0 = x_0^Tx_0 = \|x_0\|_2^2$</span>, for some vector <span class="math-container">$x_0$</span>.</p>
<p>Then let <span class="math-container">$y = V^Tx$</span>. By the same argument above, <span class="math-container">$\|y\|_2 = \|V^Tx\|_2 = \|x\|_2 = 1$</span> since <span class="math-container">$V$</span> is unitary.
<span class="math-container">$$\sup_{\|x\|_2 =1}\|\Sigma V^Tx\|_2 = \sup_{\|y\|_2 =1}\|\Sigma y\|_2$$</span>
Since <span class="math-container">$\Sigma = \mbox{diag}(\sigma_1, \cdots, \sigma_n)$</span>, where <span class="math-container">$\sigma_1$</span> is the largest singular value. The max for the above, <span class="math-container">$\sigma_1$</span>, is attained when <span class="math-container">$y = (1,\cdots,0)^T$</span>. You can find the max by, for example, solving the above using a Lagrange Multiplier.</p>
|
geometry | <p>When I came across the Cauchy-Schwarz inequality the other day, I found it really weird that this was its own thing, and it had lines upon lines of <a href="https://www.khanacademy.org/math/linear-algebra/vectors_and_spaces/dot_cross_products/v/proof-of-the-cauchy-schwarz-inequality">proof</a>.</p>
<p>I've always thought the geometric definition of dot multiplication:
$$|{\bf a }||{\bf b }|\cos \theta$$ is equivalent to the other, algebraic definition: $$a_1\cdot b_1+a_2\cdot b_2+\cdots+a_n\cdot b_n$$
And since the inequality is directly implied by the geometric definition (the fact that $\cos(\theta)$ is $1$ only when $\bf a$ and $\bf b$ are collinear), then shouldn't the Cauchy-Schwarz inequality be the world's most obvious and almost-no-proof-needed thing?</p>
<p>Can someone correct me on where my thought process went wrong?</p>
| <p><em>Side note: it's actually the Cauchy-Schwarz-Bunyakovsky inequality, and don't let anyone tell you otherwise.</em></p>
<p>The problem with using the geometric definition is that you have to define what an angle is. Sure, in three dimensional space, you have pretty clear ideas about what an angle is, but what do you take as $\theta$ in your equation when $i$ and $j$ are $10$ dimensional vectors? Or infinitely-dimensional vectors? What if $i$ and $j$ are polynomials?</p>
<p>The Cauchy-Schwarz inequality tells you that <strong>anytime</strong> you have a vector space and an inner product defined on it, you can be sure that for any two vectors $u,v$ in your space, it is true that $\left|\langle u,v\rangle\right| \leq \|u\|\|v\|$.</p>
<p>Not all vector spaces are simple $\mathbb R^n$ businesses, either. You have the vector space of all continuous functions on $[0,1]$, for example. You can define the inner product as</p>
<p>$$\langle f,g\rangle=\int_0^1 f(x)g(x)dx$$</p>
<p>and use Cauchy-Schwarz to prove that for any pair $f,g$, you have
$$\left|\int_{0}^1f(x)g(x)dx\right| \leq \sqrt{\int_0^1 f^2(x)dx\int_0^1g^2(x)dx}$$</p>
<p>which is not a trivial inequality.</p>
| <ol>
<li><p>The inequality is ubiquitous, so some name is needed.</p></li>
<li><p>As there is no cosine in the statement of the inequality, it cannot be called "cosine inequality" or anything like that.</p></li>
<li><p>The geometric interpretation with cosines only works for finite-dimensional real Euclidean space, but the inequality holds and is used more generally than that. That is Schwarz' contribution.</p></li>
<li><p>Schwarz founded the field of <em>functional analysis</em> (infinite-dimensional metrized linear algebra) with his proof of the inequality. That is important enough to warrant a name. In terms of consequences per line of proof it is one of the greatest arguments of all time.</p></li>
<li><p>The Schwarz proof was part of the historical realization that Euclidean geometry, with its mysterious angle measure that seems to depend on notions of arc-length from calculus, <em>is</em> the theory of a vector space equipped with a quadratic form. That is a major shift in viewpoint. </p></li>
<li><p>Stating the inequality in terms of cosines assumes that the inner product restricts to the standard Euclidean one on the 2 (or fewer) dimensional subspace spanned by the two vectors, and that you have proved the inequality for holds for standard Euclidean space of 2 dimensions or less. How do you know those things are correct without a much longer argument? That argument will, probably, include somewhere a proof of the Cauchy-Schwarz inequality, maybe written for 2-dimensions but working for the whole $n$-dimensional space, so it might as well be stated as a direct proof for $n$ dimensions. Which is what Cauchy and Schwarz did.</p></li>
</ol>
|
differentiation | <p>Let $f$ be a real-valued function continuous on $[a,b]$ and differentiable on $(a,b)$.<br>
Suppose that $\lim_{x\rightarrow a}f'(x)$ exists.<br>
Then, prove that $f$ is differentiable at $a$ and $f'(a)=\lim_{x\rightarrow a}f'(x)$. </p>
<p>It seems like an easy example, but a little bit tricky.<br>
I'm not sure which theorems should be used in here. </p>
<p>==============================================================</p>
<p>Using @David Mitra's advice and @Pete L. Clark's notes<br>
I tried to solve this proof.
I want to know my proof is correct or not.</p>
<p>By MVT, for $h>0$ and $c_h \in (a,a+h)$
$$\frac{f(a+h)-f(a)}{h}=f'(c_h)$$<br>
and $\lim_{h \rightarrow 0^+}c_h=a$. </p>
<p>Then $$\lim_{h \rightarrow 0^+}\frac{f(a+h)-f(a)}{h}=\lim_{h \rightarrow 0^+}f'(c_h)=\lim_{h \rightarrow 0^+}f'(a)$$ </p>
<p>But that's enough? I think I should show something more, but don't know what it is. </p>
| <p>Some hints:</p>
<p>Using the definition of derivative, you need to show that
$$
\lim_{h\rightarrow 0^+} {f(a+h)-f(a)\over h }
$$
exists and is equal to $\lim\limits_{x\rightarrow a^+} f'(x)$.</p>
<p>Note that for $h>0$ the Mean Value Theorem provides a point $c_h$ with $a<c_h<a+h$ such that
$$
{f(a+h)-f(a)\over h } =f'(c_h).
$$</p>
<p>Finally, note that $c_h\rightarrow a^+$ as $h\rightarrow0^+$.</p>
| <p>The result is essentially Theorem 5.29 from <a href="http://alpha.math.uga.edu/%7Epete/2400full.pdf" rel="nofollow noreferrer">my honors calculus notes</a>. As I mention, I learned this result from Spivak's <em>Calculus</em>. I say "essentially" because the version discussed in my notes is for an interior point of an interval whereas your version is at an endpoint, but to prove the two-sided version you just make a one-sided argument twice, so it's really the same thing.</p>
<p>[And David Mitra is right: the proof uses the Mean Value Theorem and not much else.]</p>
<p><b>Added</b>: Since we are talking about this result anyway: although I call it a "theorem of Spivak", this is not entirely serious -- the result is presumably much older than Michael Spivak. I am just identifying my (probably secondary) source. If someone knows a primary source, I'd be very happy to hear it.</p>
<p>Also it is of interest to ask what this result is used for. In my notes it isn't used for anything but is only a curiosity. I think Spivak does use it for something, though I forget what at the moment. Moreover a colleague of mine called my attention to this result in the context of, IIRC, Taylor's Theorem. Does anyone know of further applications?</p>
|
logic | <p>Here's the picture I have in my head of Model Theory:</p>
<ul>
<li>a <em>theory</em> is an axiomatic system, so it allows proving some statements that apply to <em>all</em> models consistent with the theory</li>
<li>a <em>model</em> is a particular -- consistent! -- function that assigns every statement to its truth value, it is to be thought of as a "concrete" object, the kind of thing we actually usually think about. It's only when it comes to <em>models</em> that we have the law of the excluded middle.</li>
</ul>
<p>My understanding of Gödel's first incompleteness theorem is that <strong>no theory that satisfies some finiteness condition can uniquely pin down a model</strong>.</p>
<p>So I am not really surprised by it. The idea of theories being incomplete -- of not completely pinning down a particular model -- is quite normal. The fact that no theory is complete seems analogous to how no Turing machine can compute every function.</p>
<p>But then I read <a href="https://math.stackexchange.com/questions/1052299/what-is-a-simple-example-of-an-unprovable-statement">this thread</a> and there were two claims there in the answers which <em>made no sense to me</em>:</p>
<ol>
<li><strong>Self-referential statements as examples of unprovable statements</strong> -- Like "<a href="https://math.stackexchange.com/a/1057014/78451">there is no number whose ASCII representation proves this statement</a>".</li>
</ol>
<p>A statement like this <em>cannot be constructed in propositional logic</em>. I'm guessing this has to do with the concept of a "language", but why would anyone use a language that permits self-reference?</p>
<p>Wouldn't that be completely defeat the purpose of using classical logic as the system for syntactic implications?</p>
<p>If we permit this as a valid sentence, wouldn't we also have to permit the liar paradox (and then the system would be inconsistent)?</p>
<ol start="2">
<li><strong>Unprovable statements being "intuitively true/false"</strong> -- According to <a href="https://math.stackexchange.com/a/1054255/78451">this answer</a>, if we found that the Goldbach conjecture was unprovable, then in particular that means we cannot produce a counter-example, so we'd "intuitively" know that the conjecture is true.</li>
</ol>
<p>How is this only <em>intuitive</em>? If there exist <span class="math-container">$\sf PA$</span>-compatible models <span class="math-container">$M_1$</span>, <span class="math-container">$M_2$</span> where Goldbach is true in <span class="math-container">$M_1$</span> but not <span class="math-container">$M_2$</span>, then <span class="math-container">$\exists n, p, q$</span> such that <span class="math-container">$n= p+q$</span> in <span class="math-container">$M_1$</span> but not in <span class="math-container">$M_2$</span>. But whether <span class="math-container">$n=p+q$</span> is decidable from <span class="math-container">$\sf PA$</span>, so either "<span class="math-container">$\sf{PA}+\sf{Goldbach}$</span>" or "<span class="math-container">$\sf{PA}+\lnot\sf{Goldbach}$</span>" must be inconsistent, and Goldbach cannot be unprovable. Right?</p>
<p>In any case, I don't know what it means for the extension to be "intuitively correct". Do we know something about the consistency of each extension or do we not?</p>
<p>Further adding to my confusion, the answer claims that the irrationality of <span class="math-container">$e+\pi$</span> is <em>not</em> such a statement, that it can truly be unprovable. I don't see how this can be -- surely the same argument applies; if <span class="math-container">$e+\pi$</span>'s rationality is unprovable, there does not exist <span class="math-container">$p/q$</span> that it equals, thus it is irrational. Right?</p>
| <p>This answer only addresses the second part of your question, but you asked many questions so hopefully it's okay.</p>
<p>First, there is in the comments a statement: "If Goldbach is unprovable in PA then it is necessarily true in all models." This is incorrect. If Goldbach were true in all models of PA then PA would prove Goldbach by Godel's <em>Completeness</em> Theorem (less popular, still important).</p>
<p>What is true is:</p>
<p><strong>Lemma 1:</strong> Any <span class="math-container">$\Sigma_1$</span> statement true in <span class="math-container">$\mathbb{N}$</span> (the "standard model" of PA) is provable from PA.</p>
<p>These notes (see Lemma 3) have some explanation: <a href="http://journalpsyche.org/files/0xaa23.pdf" rel="noreferrer">http://journalpsyche.org/files/0xaa23.pdf</a></p>
<p>So the correct statement is:</p>
<p><strong>Corollary 2:</strong> If PA does not decide Goldbach's conjecture then it is true in <span class="math-container">$\mathbb{N}$</span>.</p>
<p><strong>Proof:</strong> The negation of Goldbach's conjecture is <span class="math-container">$\Sigma_1$</span>. So if PA does not prove the negation, then the negation of Goldbach is not true in <span class="math-container">$\mathbb{N}$</span> by Lemma 1.</p>
<p>Remember that <span class="math-container">$\mathbb{N}$</span> is a model so any statement is either true or false in it (in our logic). But PA is an incomplete theory (assuming it's consistent), so we don't get the same dichotomy for things it can prove.</p>
<p>Now, it could be the case that PA does prove Goldbach (so its true in all models of PA including <span class="math-container">$\mathbb{N}$</span>). But if we are in the situation of Corollary 2 (PA does not prove Goldbach or its negation) then Goldbach is true in <span class="math-container">$\mathbb{N}$</span> but false in some other model of PA. (This would be good enough for the number theorists I imagine.) This is also where the problem in your reasoning is. It is NOT true that if Goldbach fails in some model <span class="math-container">$M$</span> of PA then there is a <em>standard</em> <span class="math-container">$n$</span> in <span class="math-container">$\mathbb{N}$</span> that is not the sum of two primes. Rather the witness to the failure of Goldbach is just some element that <span class="math-container">$M$</span> believes is a natural number. In some random model, this element need not be in the successor chain of <span class="math-container">$0$</span>.</p>
<p>On the other hand, the rationality of <span class="math-container">$\pi+e$</span> is not known to be expressible by a <span class="math-container">$\Sigma_1$</span> statement. So we can't use Lemma 1 in the same way.</p>
<p><strong>Edited later:</strong> I don't have much to say about the question on self-referential statements beyond what others have said. But I'll just say that one should be careful to distinguish propositional logic and predicate logic. This also goes for your "general picture of Model Theory". Part of the interesting thing with the incompleteness theorems is that they permit self-reference without being so obvious about it. In PA there is enough expressive power to code statements and formal proofs, and so the self-referential statements about proofs and so forth are fully rigorous and uncontroversial.</p>
| <p>Let me try to get at the heart of your misunderstanding as concise as possible:</p>
<p><strong>1. We are not deliberately choosing to use a language that permits self-reference, we are forced to do so.</strong></p>
<p>The only choice we made is that of a logic that is sufficiently strong to include integer arithmetic. What Gödel then proves is that access to the integers automatically allows us to construct somewhat self-referential statements. If we want integers, then we have to accept self-referentiality. The same is true in the theory of computability. Turing machines are not chosen because they can emulate themselves, they are chosen because they allow all operations we expect a general computer to do, which just happens to include emulating turing machines.</p>
<p><strong>2. We are self-referential with respect to the theory, not the model.</strong></p>
<p>The kind of sentences that Gödels procedure allows us to construct are of the form "X can not be infered from Y", as the integers are only used to build a copy of logical reasoning. If we pick the set of axioms of a given theory as Y, then we can construct sentences like "X is not provable in the theory" which is what leads to the incompleteness theorem if X is the sentence itself. There is no way to access a specific model of the theory and thus no way of constructing sentences like "X is false", which would be needed for the liar's paradoxon.</p>
|
geometry | <p>I have seen some examples of duality. Sometimes applied to theorems, as for example Desargues theorem and Pappus theorem. Sometimes applied to spaces, for example the dual space of a vector space. Sometimes even applied to a method like simplex and dual simplex methods in linear programming.</p>
<p>My question is what is the general meaning behind the term duality and what is its relevance to mathematics. Do we mean the same always when we use the term? Or the examples I have put have no connection whatsoever?</p>
<p>Thanks a lot</p>
| <p>Duality is a very general and broad concept, without a strict definition that captures all those uses. When applied to specific concepts, there usually is a precise definition for just that context. The common idea is that there are two things which basically are just <strong>two sides of the same coin</strong>.</p>
<p>Common themes in this topic include:</p>
<ul>
<li>Two different interpretations or descriptions of fundamentally the same structure or object<br>
(e.g. roles of points and lines interchanged, roles of variables in LP changed)</li>
<li>Primal and dual often are the same kind of object<br>
(e.g. incidence configuration, vector space, linear program, planar graph, …)</li>
<li>The dual of the dual is again the primal</li>
</ul>
<p>Not every use of the word strictly satisfies all of these aspects, but the general idea usually is still the same.</p>
| <p>It is true that "duality" is a quite broad concept in many fields and the field of mathematics is not the exception. For example the <a href="https://en.wikipedia.org/wiki/Duality_(mathematics)" rel="noreferrer">Wikipedia website</a> shows many forms of duality in Mathematics. As mathematicians we want to abstract the meaning of a word into a unique concept that ecompasses all situations.</p>
<p>For duality we need to define two spaces of objects and an attribute (property) of those objects. Then establish a relation between objects of one space and the other thru this attribute. If this relation is unique we say that objects of one space are duals of objects in the other space. This is not anything else that the definition of a bijective function. Any duality in mathematics can be expressed as a bijective function between two spaces of objects. So $a \in A$ is dual of $b \in B$ if there is some relation $f$ such that $b=f(a)$ and $a=f^{-1}(b)$ in a unique way.</p>
<p>Two properties should be always present in a duality:</p>
<ol>
<li>Symmetry: If $a$ is dual of $b$, $b$ is dual of $a$.</li>
<li>Idempotence: If $d$ is dual operation then $d^2 = I$.In words, the dual of the dual is the original object.</li>
</ol>
<p>If the two spaces of objects are the same, the function $f$ described in the first paragraph, at the top ot this post, is idemponent. That is $f^2=I$. Otherwise we need two different functions $f$ and $f^{-1}$ to get back to the starting point.
For example in linear algebra, in the space of square matrices, the transpose is an idempotent operator and so it is a dual operation which does not exit the space (it is closed). If a matrix is not square, its transpose leaves in a different space and the trasposition function it is not idempotent anymore. This generalizes to functional analysis with the concept of adjoint. In spherical geometry a north pole is a dual of its equator but the objects do not live in the same space: the first are points and the second "lines" in the sphere. But in spherical geometry triangles are dual of polar triangles and they leave in the same space so the duality here is an idempotence.</p>
<p>Please observe that the dual in this discussion can be confused with the concept of inverse and a claritication is needed. In the language of the first paragraph it is, since we establish a bijection between objects and, by definition of inverse, duality and inversion are linked together. However it is not an inverse in many respects. In the space of matrices, the transpose is a dual but not an inverse matrix and in general the adjoint is not the inverse for the more general context of functional analysis. Now, if we define a function in the space of square matrices as sending a matrix into its transpose, then the inverse of this function coincides with the concept of dual, but it is not the inverse of the matrix, so by "inverse" we need to be precise of what type of inverse we are looking for.</p>
<p>We then say that duality is a word attached to objects and spaces where those objects leave. We could say that two spaces are dual of each other if there is a bijective function between them. In this sense duality is an equivalence relation:</p>
<ol>
<li>Reflexive: Every space is dual to itself. The identity function is always a duality.</li>
<li>Symmetric: If a space $A$ is dual to a space $B$, then a space $B$ is dual to a space $A$. This is by definition since it exist a bijective function between the two spaces.</li>
<li>Transitive: Function composition of bijective functions.</li>
</ol>
|
matrices | <p>Are there any open source math programs out there that have an interactive terminal and that work on linux?</p>
<p>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.</p>
| <p><a href="https://en.wikipedia.org/wiki/Sage_%28mathematics_software%29">Sage</a> is basically a Python program/interpreter that aims to be an open-source mathematical suite (ala Mathematica and Magma etc.). There are many algorithms implemented as a direct part of Sage, as well as wrapping many other open-source mathematics packages, all into a single interface (the user never has to tell which package or algorithm to use for a given computation: it makes the decisions itself). It includes GP/Pari and maxima, and so does symbolic manipulations and number theory at least as well as them.</p>
<p>It has a command-line mode, as well as a web notebook interface (as an example, <a href="http://sagenb.org/">a public server run by the main developers</a>).</p>
<p>And, although this might not be relevant since your use-case sounds very simple, the syntax is just Python with a small preprocessing step to facilitate some technical details and allow some extra notation (like <code>[1..4]</code> which expands to <code>[1,2,3,4]</code>), so many people already know it and, if not, learning it is very easy.</p>
<p>As a slight tangent, Sage is actually the origin of the increasingly popular <a href="http://cython.org/">Cython</a> language for writing fast "Python", and even offers an easy and transparent method of using Cython for sections of code in the notebook.</p>
| <p>I am aware of two packets that might help You:</p>
<ol>
<li><a href="http://www.gnu.org/software/octave/"><strong>Octave</strong></a> - free clone of Matlab, </li>
<li><a href="http://maxima.sourceforge.net/"><strong>maxima</strong></a> - a packet for symbolic computations.</li>
</ol>
<p>Octave is good when you want to do numerical calculations and you can accept numerical errors. Especially good for matrix/vector operations. I think this is what you want. Maxima is rather for symbolic computations. Both of them have text interface. If you want more user-friendly interface for the latter one, try wxMaxima.</p>
|
matrices | <p>I am trying to write a program that will perform <a href="http://en.wikipedia.org/wiki/Optical_character_recognition" rel="noreferrer">OCR</a> on a mobile phone, and I recently encountered this article : </p>
<hr>
<p><img src="https://i.sstatic.net/2jenb.png" alt="article excerpt"></p>
<hr>
<p>Can someone explain this to me ? </p>
| <p><a href="https://math.stackexchange.com/a/92176/4423">J.M. has given a very good answer</a> explaining singular values and how they're used in low rank approximations of images. However, a few pictures always go a long way in appreciating and understanding these concepts. Here is an example from one of my presentations from I don't know when, but is exactly what you need. </p>
<p>Consider the following grayscale image of a hummingbird (left). The resulting image is a $648\times 600$ image of MATLAB <code>double</code>s, which takes $648\times 600\times 8=3110400$ bytes. </p>
<p><img src="https://i.sstatic.net/twFL0m.png" alt="enter image description here">
<img src="https://i.sstatic.net/LzRaNm.png" alt="enter image description here"></p>
<p>Now taking an SVD of the above image gives us $600$ singular values that, when plotted, look like the curve on the right. Note that the $y$-axis is in decibels (i.e., $10\log_{10}(s.v.)$). </p>
<p>You can clearly see that after about the first $20-25$ singular values, it falls off and the bulk of it is so low, that any information it contains is negligible (and most likely noise). So the question is, <em>why store all this information if it is useless?</em></p>
<p>Let's look at what information is actually contained in the different singular values. The figure on the left below shows the image recreated from the first 10 singular values ($l=10$ in J.M.'s answer). We see that the <em>essence</em> of the picture is basically captured in just 10 singular values out of a total of 600. Increasing this to the first 50 singular values shows that the picture is almost exactly reproduced (to the human eye). </p>
<p>So if you were to save just the first 50 singular values and the associated left/right singular vectors, you'd need to store only $(648\times 50 + 50 + 50\times 600)\times 8=499600$ bytes, which is only about 16% of the original! (I'm sure you could've gotten a good representation with about 30, but I chose 50 arbitrarily for some reason back then, and we'll go with that.) </p>
<p><img src="https://i.sstatic.net/bMUWbm.png" alt="enter image description here">
<img src="https://i.sstatic.net/1BJZ5m.png" alt="enter image description here"></p>
<p>So what exactly do the smaller singular values contain? Looking at the next 100 singular values (figure on the left), we actually see some fine structure, especially the fine details around the feathers, etc., which are generally indistinguishable to the naked eye. It's probably very hard to see from the figure below, but you certainly can in <a href="https://i.sstatic.net/eibGG.png" rel="noreferrer">this larger image</a>. </p>
<p><img src="https://i.sstatic.net/eibGGm.png" alt="enter image description here">
<img src="https://i.sstatic.net/hPktum.png" alt="enter image description here"></p>
<p>The smallest 300 singular values (figure on the right) are complete junk and convey no information. These are most likely due to sensor noise from the camera's CMOS. </p>
| <p>It is rather unfortunate that they used "eigenvalues" for the description here (although it's correct); it is better to look at principal component analysis from the viewpoint of singular value decomposition (SVD).</p>
<p>Recall that any $m\times n$ matrix $\mathbf A$ possesses the singular value decomposition $\mathbf A=\mathbf U\mathbf \Sigma\mathbf V^\top$, where $\mathbf U$ and $\mathbf V$ are orthogonal matrices, and $\mathbf \Sigma$ is a diagonal matrix whose entries $\sigma_k$ are called <em>singular values</em>. Things are usually set up such that the singular values are arranged in nonincreasing order: $\sigma_1 \geq \sigma_2 \geq \cdots \geq \sigma_{\min(m,n)}$.</p>
<p>By analogy with the eigendecomposition (eigenvalues and eigenvectors), the $k$-th column of $\mathbf U$, $\mathbf u_k$, corresponding to the $k$-th singular value is called the <em>left singular vector</em>, and the $k$-th column of $\mathbf V$, $\mathbf v_k$, is the <em>right singular vector</em>. With this in mind, you can treat $\mathbf A$ as a sum of outer products of vectors, with the singular values as "weights":</p>
<p>$$\mathbf A=\sum_{k=1}^{\min(m,n)} \sigma_k \mathbf u_k\mathbf v_k^\top$$</p>
<p>Okay, at this point, some people might be grumbling "blablabla, linear algebra nonsense". The key here is that if you treat images as matrices (gray levels, or RGB color values sometimes), and then subject those matrices to SVD, it will turn out that some of the singular values $\sigma_k$ are <em>tiny</em>. The key to "approximating" these images then, is to treat those tiny $\sigma_k$ as zero, resulting in what is called a "low-rank approximation":</p>
<p>$$\mathbf A\approx\sum_{k=1}^\ell \sigma_k \mathbf u_k\mathbf v_k^\top,\qquad \ell \ll \min(m,n)$$</p>
<p>The matrices corresponding to images can be big, so it helps if you don't have to keep all those singular values and singular vectors around. (The "two", "ten", and "thirty" images mean what they say: $n=256$, and you have chosen $\ell$ to be $2$, $10$, and $30$ respectively.) The criterion of when to zero out a singular value depends on the application, of course.</p>
<p>I guess that was a bit long. Maybe I should have linked you to <a href="http://www.mathworks.com/moler/eigs.pdf" rel="noreferrer">Cleve Moler's book</a> instead for this (see page 21 onwards, in particular).</p>
<hr>
<p><strong>Edit 12/21/2011:</strong></p>
<p>I've decided to write a short <em>Mathematica</em> notebook demonstrating low-rank approximations to images, using the famous <a href="http://www.cs.cmu.edu/~chuck/lennapg/" rel="noreferrer">Lenna test image</a>. The notebook can be obtained from me upon request.</p>
<p>In brief, the $512\times 512$ example image can be obtained through <code>ImageData[ColorConvert[ExampleData[{"TestImage", "Lena"}], "Grayscale"]]</code>. A plot of $\log_{10} \sigma_k$ looks like this:</p>
<p><img src="https://i.sstatic.net/XD4wd.png" alt="log singular values of Lenna"></p>
<p>Here is a comparison of the original Lenna image with a few low-rank approximations:</p>
<p><img src="https://i.sstatic.net/FAC0i.png" alt="Lenna and low-rank approximations"></p>
<p>At least to my eye, taking $120$ out of $512$ singular values (only a bit more than $\frac15$ of the singular values) makes for a pretty good approximation.</p>
<p>Probably the only caveat of SVD is that it is a rather slow algorithm. For images that are quite large, taking the SVD might take a long time. At least one only deals with a single matrix for grayscale pictures; for RGB color pictures, one must take the SVD of three matrices, one for each color component.</p>
|