instruction
stringlengths 12
30k
|
|---|
Suppose that $M$ is simply connected, compact manifold.
Then I want to show that every closed $1-$ form is exact on $M$.
This is same showing that the first de-Rham cohomology space of $M$ is $0$.
**Question**: Is it possible to answer this without using universal coefficients theorem?
(The reason for not using the universal coefficients theorem is that it contains a mystery term ext which I'm not comfortable using. Digging deeper into ext, it turns out that it is a core algebraic term which I'm not good enough at at this stage.)
Since $M$ is simply connected, its fundamental group $\pi_1$ is $0$.
First homology group $H_1$ is abeliazation of $\pi_1$ so $H_1 =0$.
To conclude $H^1=0$, I would need universal coefficients theorem, which is what I want to avoid.
Does the following work?
Suppose that $\omega$ is a closed $1$ form on $M$. Consider any $p$ in M. Take a chart at $p$ and call it $h$. Then $\tau=(h^{-1})^\ast \omega$ is a $1$ form on $\mathbb R^n$ hence equal to $df$ for some smooth $f$ on $\mathbb R^n$.
Pulling $\tau$ back to $M$ and using that fact that $d$ and $\ast$ commute, I have $\omega= d(h^\ast f)$. This is true for every point $p$. Locally this is fine.
But here $h$ is not the same function throughout. $h$ depends upon $p$. So how to create a $0$ form $g$ on $M$ such that $\omega =dg$?
|
Let $X$ be a topological space and $A\subseteq X$ a subset. Suppose I know that $V\subseteq X$ is such that $V\subseteq X\setminus A$ and $V$ is open
Can I conclude $V\subseteq X\setminus\overline{A}$? Why?
|
> What numbers can be written uniquely as a sum of two squares?
I was looking at sequence [A125022](https://oeis.org/A125022), which shows the numbers that can be uniquely written as a sum of two squares. Here are a few things that I noticed from the first numbers. We have $1$, $2$, $4$, $8$, $16$, $32$, $64$, $128$. It is then safe to assume that all numbers of the form $2^{s}$ can be written uniquely, where $s \in \mathbb{Z}_{+} \cup \{0\}$. Moreover, primes of the form $4k+1$, for example, $5$, $13$, $17$, $29$, also appear and, interestingly enough, $5^2$ and $13^{2}$ do not. So, I could also say that $p^{s}$ has a unique representation only when $s = 0$ or $s = 1$. If we analyze $A125022$ a bit more, we notice that $3^{2}$, $7^{2}$, $11^{2}$ are there, so we can also conjecture that numbers of the form $q^{2}$ have a unique representation, where $q$ is a prime of the form $4k+3$. Furthermore, for reasons I will say later, I believe $d^{2}$, where $d$ has all of its prime factors of the form $4k+3$, can be uniquely represented as a sum of two squares. It is also possible to see that products between these three are in the sequence, for example $2^{2}\cdot 5$, $2 \cdot 5 \cdot 3^{2}$ and $2 \cdot 7^{2}$.
*Conjecture.* A number $n \in \mathbb{Z}_{+}$ can be written uniquely as a sum of two squares if, and only if, $n = 2^{s} d^{2} p^{e_1}$, where $s \in \mathbb{Z}_{+} \cup \{0\}$, $d$ has all of its prime divisors of the form $4k+3$, $p$ is a prime of the form $4k+1$ and $e_{1} \in \{0,1\}$.
It is known that a number can be written as a sum of two squares if, and only if, it can be written as $2^{s} t^{2} l$, where $s \in \mathbb{Z}_{+} \cup \{0\}$ and $l$ is a square-free positive integer with all of its prime factors of the form $4k+1$. Thus, we know the number $n$ we conjectured above can in fact be written as a sum of two squares. We only need to understand uniqueness. It is more natural to study these questions with the Gaussian integers, $\mathbb{Z}[i]$. If, for example, we have
$$n = a^{2} + b^{2} = (a+ib)(a-ib) = (\pi_1 \cdots \pi_k) (\overline{\pi_1} \cdots \overline{\pi_k}),$$
where the last expression is the factorization of $n$ in primes of $\mathbb{Z}[i]$, then we might be able to get different representations of $n$ by exchanging, say, $\pi_j$ for $\overline{\pi_j}$. That is,
$$(\pi_1 \cdots \overline{\pi_j} \cdots \pi_n)(\overline{\pi_1} \cdots \pi_j \cdots \overline{\pi_n})$$
should yield a different sum when $\pi_j \neq \overline{\pi_j}$ and at least one of the other primes, say $\pi_i$, also satisfies $\pi_i \neq \overline{\pi_i}$. This does not seem to occur precisely for the numbers conjectured above, which makes me think those are the only numbers that can be uniquely represented.
*Question.* Is my guess correct or am I missing other numbers?
|
What numbers can be wrriten uniquely as a sum of two squares?
|
And what if $f > g$ ?
I tried to find some counter-example but it's less obvious !
|
What numbers can be written uniquely as a sum of two squares?
|
I am confused about what can and cannot be done in a proof in order to avoid having an infinite number of "steps", thus making the argument invalid.
For instance, let $A_{0}, A_{1}, \dotsc$ be subsets of $\mathbb{R}^{n}$, and suppose that I want to show that $A_{0}$ satisfies some desired property *after shrinking $A_{1}, A_{2}, \dotsc$*.
Would this be OK? Or would it involve infinitely many "steps"?
In general, how can I make sure my proofs are free from this kind of problems?
|
How do I know if I am using infinitely many "steps" in a proof?
|
And what if $f > g$ ?
I tried to find some counter-example but it's less obvious !
Feels like we can prove it on segments, no ?
|
I have the following proof in my notes. It is a particular case of the more general rank theorem.
Theorem (Rank theorem for injective differential). Suppose $M$ is a smooth manifold of dimension $m$, and that $N$ is a smooth manifold of dimension $n$. Suppose $F : M \to N$ is smooth. Let $p \in M$. If $dF_p$ is injective, then there are charts $(U, \varphi)$ of M around p and $(V,\psi )$ of N around $F(p)$ such that $F(U) \subseteq V$
and for all $x \in\varphi(U)$, and
$\psi\circ F \circ \varphi^{−1}(x) = (x, 0_{n−m})$
**Proof**
We prove the theorem **in the case** that $M$ is an open subset of $\Bbb R^m$, $N$ open subset of $\Bbb R^n$, $p = 0_m$ and $F(p) = 0_n$ By using appropriate charts around p and F(p), we can prove the general case.
Suppose $dF_p$ is injective. Then $m \le n$. Define $Q: M \to \Bbb R^m$ and
$R: M \to \Bbb R^{n−m}$ by $F(x) = (Q(x), R(x))$ for $x \in M$.
Since $DF(p)$ is injective, its matrix has an $m × m$ invertible submatrix. We can do this by possibly exchanging coordinates ensuring that $DQ(p)$ is such a matrix; we will assume this from now on. Because $DQ(p)$ is invertible, then due to the inverse function theorem there are nbhs $U$ of $p$ and $\tilde V$ of $Q(p)$ such that $\varphi := Q|_U : U \to\tilde V$ is a diffeomorphism
Define $V := \tilde V \times \Bbb R^{n−m}$.This is an open nbhd of $F(p) = 0_n$, because $Q(p) = 0_m$.
Define :$ \psi:V \to \Bbb R^n: \psi(v; w) = (v, w − R \circ \varphi^{−1}(v))$, for $v \in \tilde V$ and $w \in \Bbb R^{n−m}$.
This is a diffeomorphism onto its image, with inverse given by
$\psi^{−1}(s, t) = (s, t + \varphi^{−1}(s))$, for $s \in \Bbb R^m $ and $t \in \Bbb R^{n−m}$ such that $(s, t) \in \psi(V )$.
And for all $x \in \varphi(U)$,
$\psi\circ F \circ \varphi^{−1}(x) =\psi(x,R\circ\varphi^{-1}(x))= (x, 0_{n−m}) \tag{*}$
**I have a couple of doubts**
1) How does one come up a priori with the idea to define $\psi$ like that? It is supposed to be a chart of $ N$ around $F(p)$, so how does one know that it has that form?
2) How do I prove $F(U)\subset V$? It seems to be missing or maybe I just can't see it in the middle of this proof.
Thanks for your help.
[![enter image description here][1]][1]
[1]: https://i.stack.imgur.com/gjz6v.png
|
> What numbers can be written uniquely as a sum of two squares?
I was looking at sequence [A125022](https://oeis.org/A125022), which shows the numbers that can be uniquely written as a sum of two squares. Here are a few things that I noticed from the first numbers. We have $1$, $2$, $4$, $8$, $16$, $32$, $64$, $128$. It is then safe to assume that all numbers of the form $2^{s}$ can be written uniquely, where $s \in \mathbb{Z}_{+} \cup \{0\}$. Moreover, primes of the form $4k+1$, for example, $5$, $13$, $17$, $29$, also appear and, interestingly enough, $5^2$ and $13^{2}$ do not. So, I could also say that $p^{s}$ has a unique representation only when $s = 0$ or $s = 1$. If we analyze $A125022$ a bit more, we notice that $3^{2}$, $7^{2}$, $11^{2}$ are there, so we can also conjecture that numbers of the form $q^{2}$ have a unique representation, where $q$ is a prime of the form $4k+3$. Furthermore, for reasons I will say later, I believe $d^{2}$, where $d$ has all of its prime factors of the form $4k+3$, can be uniquely represented as a sum of two squares. It is also possible to see that products between these three are in the sequence, for example $2^{2}\cdot 5$, $2 \cdot 5 \cdot 3^{2}$ and $2 \cdot 7^{2}$.
**Conjecture.** A number $n \in \mathbb{Z}_{+}$ can be written uniquely as a sum of two squares if, and only if, $n = 2^{s} d^{2} p^{e_1}$, where $s \in \mathbb{Z}_{+} \cup \{0\}$, $d$ has all of its prime divisors of the form $4k+3$, $p$ is a prime of the form $4k+1$ and $e_{1} \in \{0,1\}$.
It is known that a number can be written as a sum of two squares if, and only if, it can be written as $2^{s} t^{2} l$, where $s \in \mathbb{Z}_{+} \cup \{0\}$ and $l$ is a square-free positive integer with all of its prime factors of the form $4k+1$. Thus, we know the number $n$ we conjectured above can in fact be written as a sum of two squares. We only need to understand uniqueness. It is more natural to study these questions with the Gaussian integers, $\mathbb{Z}[i]$. If, for example, we have
$$n = a^{2} + b^{2} = (a+ib)(a-ib) = (\pi_1 \cdots \pi_k) (\overline{\pi_1} \cdots \overline{\pi_k}),$$
where the last expression is the factorization of $n$ in primes of $\mathbb{Z}[i]$, then we might be able to get different representations of $n$ by exchanging, say, $\pi_j$ for $\overline{\pi_j}$. That is,
$$(\pi_1 \cdots \overline{\pi_j} \cdots \pi_n)(\overline{\pi_1} \cdots \pi_j \cdots \overline{\pi_n})$$
should yield a different sum when $\pi_j \neq \overline{\pi_j}$ and at least one of the other primes, say $\pi_i$, also satisfies $\pi_i \neq \overline{\pi_i}$. This does not seem to occur precisely for the numbers conjectured above, which makes me think those are the only numbers that can be uniquely represented.
*Question.* Is my guess correct or am I missing other numbers?
|
>Surely if $X_t$ is nice and smooth, then $\alpha > \frac{1}{2}$, but then $2\alpha > 1$, and wouldn't this degenerate into saying that $\mathbb{X}$ is constant? From the known result on Holder continuous greater than 1 $\implies$ constant?
As mentioned in the comments, the rough path lift $\mathbb{X}_{s,t}$ is not an increment. For example, in the case of Brownian motion we can take
$$\mathbb{B}_{s,t}^{Itô}:=\int_{s}^{t}(B_{r}-B_{s})\otimes dB_{r}.$$
This definition originates from the "Taylor"-like-expansion argument by repeatedly using FTC
$$\int^t_sF(X_u)dX_u = F(X_s)\int^t_sdX_u + \int^t_s(F(X_u)-F(X_s))\otimes dX_u$$
$$\stackrel{FTC}{=} F(X_s)\int^t_sdX_u + \int^t_s\int_{s}^{u}DF(X_v)dX_{v}\otimes dX_u$$
$$= F(X_s)\int^t_sdX_u + DF(X_s)\int^t_s\int_{s}^{u}dX_{v}\otimes dX_u+\int^t_s\int_{s}^{u}(DF(X_v)-DF(X_s))dX_{v}\otimes dX_u.$$
>How do I show that a smooth path is a particular case of rough paths with the obvious choice of signature?
Indeed as mentioned in the comments, when we have two functions $Y\in C^{\alpha},X\in C^{\beta}$ with $\alpha+\beta>1$, then we can define their Young integral
$$\int YdX\approx \sum Y_{t_{i}} (X_{t_{i+1}}-X_{t_{i}})$$
and use Young's inequality as mentioned in "a course in rough paths" eq. (4.3)
$$|\int_{s}^{t} Y_{r}dX_{r}-Y_{s}X_{s,t}|\leq c \|Y\|_{\beta} \|X\|_{\alpha}|t-s|^{\alpha+\beta}.$$
So in the case of differentiable $X$ i.e. $\alpha\geq 1$,we can just take rough path lift $\mathbb{X}$ to be
$$\mathbb{X}_{s,t}:=\int_{s}^{t} (X_{r}-X_{s})dX_{r}$$
and from above this indeed satisfies $\frac{|\mathbb{X}_{s,t}|}{|t-s|^{2\alpha}}<c\|X\|_{\alpha}^{2}$.
|
For any two finite subsets $A,B$, of an abelian group, is
$$|A+B|^2 |A-B|^2 \geq |A+A||A-A||B+B||B-B| \ ?$$
I’m interested in finding out if there are sumset inequalities that are sharper than the triangle inequalities and that show that some quantity related to sums and differences is smaller when the sums and differences are over the same sets.
Smaller examples don’t seem to work:
$|A-B|^2 \geq |A-A||B-B|$ fails when $B = -A$ and $|A+A| < |A-A|$.
$|A+B|^2 \geq |A+A||B+B|$ doesn’t work when $B = -A$ and $|A-A| < |A+A|$.
The candidate given in the title combines the two and avoids these difficulties because of its symmetry. It also holds when A is an arithmetic progression or a subspace (if such finite subspaces exist) and B is a random set.
|
Let $X$ be a topological space and $A\subseteq X$ a subset. Suppose I know that $V\subseteq X$ is such that $V\subseteq X\setminus A$ and $V$ is open
Can I conclude $V\subseteq X\setminus\overline{A}$? Why?
Equivalently, I tried to prove $\overline{A}\subseteq X\setminus V$.
If $x\in A$ then $x\in X\setminus V$ by hypothesis. But I can't really conclude the case $x\in \overline{A}\setminus A$, i guess I should use the fact that $V$ is open. Where precisely? Is this statement even true?
|
If $|z - i| \le 2$, then $z = 3i$ certainly satisfies this inequality, since
$|3i - i| = |2i| = 2$, yet $|3i| = 3 > 1$. So you clearly have an error. Where is it?
Well, the triangle inequality does say that $$2 = |z-i| = |z+(-i)| \color{red}{\le} |z| + |-i| = |z| + 1. \tag{1}$$ But the problem here is that you write
$$|z - i| = 2 \implies |z| + 1 = 2,$$ when instead the inequality $(1)$ clearly is saying
$$2 \color{red}{\le} |z| + 1,$$ or
$$|z| \color{red}{\ge} 1.$$
A better approach would be to let $w = iz+z_0$ and note that if $|z-i| \le 2$, then
$$\begin{align}
2 &\ge |z-i| \\
&= |i||z-i| \\
&= |iz-i^2| \\
&= |iz+1| \\
&= |iz+(5+3i)-(4+3i)| \\
&= |iz+z_0 - (4+3i)| \\
&= |w - (4+3i)|. \tag{2}
\end{align}$$
We want to find, for all $w$ satisfying $(2)$, the one with maxmimum $|w|$. Geometrically, what is $(2)$? It is the disk in the complex plane with radius $2$ and center $4+3i$. The point in this disk that is furthest away from the origin, and thus has largest magnitude, is the one that lies on the line joining $0$ and $4+3i$. Since $|4+3i| = 5$ and the disk has radius $2$, this means $|w|_{\text{max}} = 5 + 2 = 7$.
|
I have the following proof in my notes. It is a particular case of the more general rank theorem.
Theorem (Rank theorem for injective differential). Suppose $M$ is a smooth manifold of dimension $m$, and that $N$ is a smooth manifold of dimension $n$. Suppose $F : M \to N$ is smooth. Let $p \in M$. If $dF_p$ is injective, then there are charts $(U, \varphi)$ of M around p and $(V,\psi )$ of N around $F(p)$ such that $F(U) \subseteq V$
and for all $x \in\varphi(U)$, and
$\psi\circ F \circ \varphi^{−1}(x) = (x, 0_{n−m})$
**Proof**
We prove the theorem **in the case** that $M$ is an open subset of $\Bbb R^m$, $N$ open subset of $\Bbb R^n$, $p = 0_m$ and $F(p) = 0_n$ By using appropriate charts around p and F(p), we can prove the general case.
Suppose $dF_p$ is injective. Then $m \le n$. Define $Q: M \to \Bbb R^m$ and
$R: M \to \Bbb R^{n−m}$ by $F(x) = (Q(x), R(x))$ for $x \in M$.
Since $DF(p)$ is injective, its matrix has an $m × m$ invertible submatrix. We can do this by possibly exchanging coordinates ensuring that $DQ(p)$ is such a matrix; we will assume this from now on. Because $DQ(p)$ is invertible, then due to the inverse function theorem there are nbhs $U$ of $p$ and $\tilde V$ of $Q(p)$ such that $\varphi := Q|_U : U \to\tilde V$ is a diffeomorphism
Define $V := \tilde V \times \Bbb R^{n−m}$.This is an open nbhd of $F(p) = 0_n$, because $Q(p) = 0_m$.
Define :$ \psi:V \to \Bbb R^n: \psi(v; w) = (v, w − R \circ \varphi^{−1}(v))$, for $v \in \tilde V$ and $w \in \Bbb R^{n−m}$.
This is a diffeomorphism onto its image, with inverse given by
$\psi^{−1}(s, t) = (s, t + \varphi^{−1}(s))$, for $s \in \Bbb R^m $ and $t \in \Bbb R^{n−m}$ such that $(s, t) \in \psi(V )$.
And for all $x \in \varphi(U)$,
$\psi\circ F \circ \varphi^{−1}(x) =\psi(x,R\circ\varphi^{-1}(x))= (x, 0_{n−m}) \tag{*}$
**I have a couple of doubts**
1) How does one come up a priori with the idea to define $\psi$ like that? It is supposed to be a chart of $ N$ around $F(p)$, so how does one know that it has that form?
2) How do I prove the general case ? (that is removing the assumptions: $M$ is an open subset of $\Bbb R^m$, $N$ open subset of $\Bbb R^n$, $p = 0_m$ and $F(p) = 0_n$ but still considering injective derivative)I am having trouble writting it down. Moreover why are they using charts in the proof above if they are not working with general manifolds?
Thanks for your help.
[![enter image description here][1]][1]
[1]: https://i.stack.imgur.com/gjz6v.png
|
How do I remove the specific assumptions in the proof of the rank theorem in the special case of injective differential?
|
Let $\{X_i\}_{i\ge 1}$ be a sequence of random variables defined on $(\mathbb{R}, \mathcal{B})$. Assume that $E[X_i]=0$, $E[X_i^2]=1$, and $E[X_iX_j]=0$ for all $i\neq j$ and $i\ge 1$. Let $S_n=\sum_{j=1}^n X_j$. Show that
1. for $\alpha>1$, $\frac{S_n}{n^\alpha}\to 0$ a.s.
2. for $\alpha>1/2$, $\frac{S_n}{n^\alpha}\to 0$ in probability.
---
My proof is that
1. Since we have $E[S_n^2]=n$, then by Markov's inequality
$$
P\left(\Big|\frac{S_n}{n^\alpha}\Big|\ge n^{-0.5(\alpha-1)}\right)\le n^{-(\alpha-1)} E|n^{-\alpha}S_n|^2=n^{-\alpha}.
$$
Then for $\alpha>1$,
$$
\sum _{n=1}^\infty P\left(\Big|\frac{S_n}{n^\alpha}\Big|\ge n^{-0.5(\alpha-1)}\right)<\infty
$$
By Borel-Cantelli Lemma, we proved our result in 1.
2. By Markov's inequality, for $\epsilon>0$,
$$
P\left(\Big|\frac{S_n}{n^\alpha}\Big|\ge \epsilon\right)\le \epsilon^{-2} E\left[\Big|\frac{S_n}{n^\alpha}\Big|^2\right]=\epsilon^{-2} n^{2\alpha-1}
$$
So as $n\to \infty$, $\frac{S_n}{n^\alpha}\to 0$ in probability.
Does this proof look good?
|
Given $f(x) = \frac{x+1}{x^2+2 \sqrt{x}}$, How to prove $\lim_{x \rightarrow 4} f(x)$ using $\varepsilon$-$\delta$ definition?
|
Is it possible to simplify the following expression, where $k$ is a given constant? I want to simplify it to something in terms of only $k$.
$$ \sum_{i=0}^{\lfloor \frac{k}{2} \rfloor} 2^{k} \binom{k-i}{i}$$
I was confused on how to deal with the $i$ in the combination.
|
Let $e_1,\ldots,e_n$ be the standard basis in $\mathbb{R}^n$. Suppose the scalars $\lambda_1,\ldots,\lambda_n$ satisfy $0< \lambda_1,\ldots,\lambda_n\leq1$ and $\lambda_1^2+\ldots+\lambda_n^2 = m$, where $m$ is a positive integer less than or equal to $n$. Does there always exist a $m$-dimensional subspace $W$ of $\mathbb{R}^n$, such that the length of the orthogonal projection of $e_i$ onto $W$ equals $\lambda_i$ for each $i=1,\ldots,n$?
I tried to reformulate the question by considering an orthonormal basis $v_1,\ldots,v_m$ of $W$, then the orthogonal projection of $e_i$ onto $W$ is $\sum_{j=1}^{m}\langle e_i,v_j \rangle\cdot v_j$, and the length of the projection will be
$$\sqrt{\sum\limits_{j=1}^{m}\langle e_i,v_j \rangle^2} = \sqrt{\sum\limits_{j=1}^{m}A_{ij}^2}$$
where $A$ denotes the $n\times m$ matrix with column vectors $v_1,\ldots,v_m$.
I also solved the case $m=1$, where one can just take $v_1 = (\lambda_1,\ldots,\lambda_n)$, and from this I can also solve the case $m=n-1$, where one can take $W$ to be the orthogonal complement of $\text{span}\{(\sqrt{1-\lambda_1^2},\ldots,\sqrt{1-\lambda_n^2})\}$.
This question originated from the observation that if $l_1,\ldots,l_n$ are the lengths of orthogonal projections of $e_1,\ldots,e_n$ onto a $m$-dimensional subspace, respectively, then the equality $l_1^2+\cdots+l_n^2 = m$ would always hold, so I wondered if the converse would also be true.
|
Let $a\neq -1,0,1$ be an integer. Write $a=(b^2c)^k$, where $b^2c$ is not a perfect power, and $c$ is squarefree. Artin's conjecture on primitive roots states that the asymptotic density of the set of prime numbers $p$ such that $a$ is a primitive root modulo $p$ is
$$
\tag{$\star$}
A(a)=\begin{cases}
\left(\displaystyle\prod_{q|k,q\text{ prime}}\dfrac{q(q-2)}{q^2-q-1}\right)C,&c\not\equiv 1\pmod 4;\\
\left(\displaystyle\prod_{q|k,q\text{ prime}}\dfrac{q(q-2)}{q^2-q-1}\right)\left(1-\displaystyle\prod_{p|c,p\text{ prime}}\dfrac{-1}{p^2-p-1}\right)C,&c\equiv 1\pmod 4.
\end{cases}
$$
The formula is copied from [here](https://www.numericana.com/answer/constants.htm#artin).
Everything looks well, except that at certain inputs the formula can't work: Let $q\neq 2$ be a prime and $m$ be a positive odd integer. Write $qm=q^rm'$ with $m'$ not divisible by $q$, and $r\in\mathbb{N}^*$, then we have
$$
\tilde{q}^{qm}\text{ is a primitive root modulo }p\Leftrightarrow \tilde{q}^{m'}\text{ is a primitive root modulo }p,
$$
where
$$
\tilde{q}:=(-1)^{\frac{q-1}{2}}q=\begin{cases}q,&q\equiv 1\pmod 4;\\ -q,&q\equiv 3\pmod 4.\end{cases}
$$
Here "$\Rightarrow$" is obvious; for "$\Leftarrow$", note that $\tilde{q}^{m'}\equiv 1\pmod 4$, so $\tilde{q}^{m'}$ being a quadratic nonresidue modulo $p$ implies that $p\not\equiv 1\pmod q$, then
$$
\operatorname{ord}_{p}(\tilde{q}^{qm}) = \dfrac{\operatorname{ord}_{p}(\tilde{q}^{m'})}{\gcd(\operatorname{ord}_{p}(\tilde{q}^{m'}),q^r)} = \dfrac{p-1}{\gcd(p-1,q^r)} = p-1.
$$
(As concrete examples, we have
\begin{align*}
&q=3,m=1:-27=(-3)^3\text{ is a primitive root modulo }p\Leftrightarrow -3\text{ is a primitive root modulo }p;\\
&q=5,m=1:3125=5^5\text{ is a primitive root modulo }p\Leftrightarrow 5\text{ is a primitive root modulo }p.
\end{align*}
However, the formula ($\star$) gives
$$
\dfrac{A(\tilde{q}^{qm})}{A(\tilde{q}^{m'})} = \dfrac{q(q-2)}{q^2-q-1}\neq 1.
$$
So my question is: **Are numbers of the form
$$
((-1)^{\frac{q-1}{2}}q)^{qm},\quad q\neq 2\text{ prime},\quad m\text{ odd}
$$
the only inputs for which the formula ($\star$) is false?** (Assuming that Artin's conjecture is true, of course.) If not, could you please give a modified formula that would work for all $a$? Any hint/reference appreciated.
|
An upper bound for the [primorial](https://en.wikipedia.org/wiki/Primorial) can be found based on the first [chebyshev function](https://en.wikipedia.org/wiki/Chebyshev_function).
From $\vartheta(x) < 1.00028x$, it is clear that:
$$\prod\limits_{p \le x}p \le e^{1.00028x} < (2.72)^x$$
I am looking at the following and am having trouble figuring out the upper bound.
$$\prod\limits_{p \le x}p^{\frac{1}{p}}$$
Is there a function similiar to the first chebyshev function that can be used? The second chebyshev function is not helpful since:
$$\psi(x) = \sum\limits_{n=1}^{\infty}{\vartheta(x^{\frac{1}{n}}})$$
Is there a better way to estimate this upper bound? I am wondering if there is an upper bound better than a very simple approach such as this:
$$\prod\limits_{p \le x}p^{\frac{1}{p}} < \prod\limits_{p \le x}p^{\frac{1}{2}} < \sqrt{(2.72)^x} < (1.65)^x$$
|
From the ["Finite affine planes" section](https://en.wikipedia.org/wiki/Affine_plane_(incidence_geometry)#Finite_affine_planes) of the Wikipedia article "Affine plane (incidence geometry)", in a (finite) affine plane of order $n$:
- each line contains $n$ points,
- each point is contained in $n+1$ lines,
- there are $n^2$ points in all, and
- there is a total of $n^2+n$ lines.
I've read enough to know that a common notation of an affine plane of order $n$ (or at least one obtained from (removing a line and all points on that line on) a Desarguesian projective plane) is $AG(2,n)$, just as a common notation of a Desarguesian projective plane of order $n$ is $PG(2,n)$. For orders (such as order $2$) in which there is only one Desarguesian projective plane up to isomorphism, there is also only one affine plane under isomorphism. From the above bulleted properties of an affine plane of order $n$, it seems clear to me that in the affine plane of order $2$, $AG(2,2)$:
- each line contains $2$ points,
- each point is contained in $2+1=3$ lines,
- there are $2^2=4$ points in all, and
- there is a total of $2^2+2=6$ lines.
$AG(2,2)$ is clearly... (I'm not sure exactly what the correct word is but equivalent or isomorphic or something like that) to the complete graph $K_4$. With the nodes of that graph displayed as the three vertices and center of an equilateral triangle, $AG(2,2)$ looks like the common representation of the Fano plane, $PG(2,2)$, without the line shown as a circle and the points on that circle, and with the three medians of the triangle, going from the vertices, truncating at the triangle center.
I feel pretty confident in what I've written so far (although confirmation would be appreciated), but my main interest in asking this question here is in the (finite) affine 3-space of order $2$, $AG(3,2)$, and its tetrahedral representation (akin to the tetrahedral representation of $PG(3,2)$). From the [Wikipedia article on $PG(3,2)$](https://en.wikipedia.org/wiki/PG(3,2)), that projective 3-space has the following properties:
- It has $15$ points, $35$ lines and $15$ planes,
- Each point is contained in $7$ lines and $7$ planes,
- Each line is contained in $3$ planes and contains $3$ points,
- Each plane contains $7$ points and $7$ lines,
- Each plane is isomorphic to the Fano plane, $PG(2,2)$,
- Every pair of distinct planes intersect in a line,
- A line and a plane not containing the line intersect in exactly $1$ point.
The article goes on to describe how $PG(3,2)$ can be represented as a tetrahedron. "The $15$ points correspond to the $4$ vertices + $6$ edge-midpoints + $4$ face-centers + $1$ body-center. The $35$ lines correspond to the $6$ edges + $12$ face-medians + $4$ face-incircles + $4$ altitudes from a face to the opposite vertex + $3$ lines connecting the midpoints of opposite edges + $6$ ellipses connecting each edge midpoint with its two non-neighboring face centers. The $15$ planes consist of the $4$ faces + the $6$ "medial" planes connecting each edge to the midpoint of the opposite edge + $4$ "cones" connecting each vertex to the incircle of the opposite face + $1$ "sphere" with the 6 edge centers and the body center."
It seemed clear to me that each plane in $AG(3,2)$ would be isomorphic to $AG(2,2)$ and $K_4$, and so I started with assuming there would be, for starters, on the outside of the tetrahedron, $8$ points ($4$ vertices + $4$ face centers), $18$ lines ($6$ tetrahedral edges + $12$ truncated triangle medians ($3$ for each external tetrahedral face) that go from the vertex to the center of that face. That would make each tetrahedral vertex be on $6$ lines and $4$ planes, while the face centers would each be on three lines. Trying to balance the number of lines for each point, while keeping each plane isomorphic to $K_4$, I at first just added a new plane (a hybrid of sorts of the "cones" and "sphere" in $PG(3,2)$) consisting of the $4$ face centers and the $6$ lines connecting them (each of those lines would be the ellipses in $PG(3,2)$ without the edge midpoints). That would mean that each point was on $6$ lines and that each of the $5$ planes would be isomorphic to $K_4$. I thought I had it at that point. $8$ points, $18$ lines and $5$ planes, with each plane having $4$ points and $6$ lines and each point being on... wait a minute! The tetrahedral vertices are on $3$ planes (the $K_4$s on the triangular faces including that vertex) while the face centers are only on $2$ of them (one of the external planes plus the "final" plane of the face centers and lines connecting them). The tetrahedral edges would each be on $2$ planes (both external) while the truncated edges would only be on $1$ of them (the one in the interior of the tetrahedron). Something wasn't adding up.
It was actually in the course of writing this question that I both realized that what I thought I had come up with didn't work and that I came up with (I believe) a solution: "pinch" the four triangular sides of the tetrahedron so that their centers are all at the same point, the center of the tetrahedron. That reduces the total number of points from $8$ down to $5$ and the number of truncated medians from $12$ down to $4$ (so the total number of points is reduced from $18$ down to $10$. The fifth plane is no longer on the inside of the tetrahedron but on the outside, consisting of the entire exterior of the tetrahedron: all four tetrahedral vertices (each of which will also be on three of the "pinched" triangles) and all six tetrahedral edges (each of which will also be on two of the "pinched" triangles). The lines from the vertices to the tetrahedral center will be on three of the "pinched" triangles, and the tetrahedral center itself will be on all four of the pinched triangles. This all leaves $AG(3,2)$ with the following properties:
- It has $5$ points, $10$ lines and $5$ planes,
- Each point is contained in $4$ lines and $4$ planes,
- Each line is contained in $3$ planes and contains $2$ points,
- Each plane contains $4$ points and $6$ lines,
- Each plane is isomorphic to $AG(2,2)$ or $K_4$,
- Every pair of distinct planes intersect in $3$ lines and $3$ points (a $K_3$),
- A line and a plane not containing the line intersect in exactly $1$ point.
Leaving the tetrahedral representation aside for a moment, $AG(3,2)$ is basically the complete graph $K_5$ with all $5$ of the component $K_4$s being a plane. My "pinched" tetrahedron idea with one affine plane covering the entire exterior is one way to visualize it. You could also think of it as the vertices and edges of a tetrahedron (still being all on the same affine plane) and a "point at infinity" with lines connecting that point to all four vertices and four planes each consisting of that "point of infinity" and the points on one of the four faces of the tetrahedron.
More generally, $AG(n,2)$, $n\ge2$, is basically the complete graph $K_{n+2}$, with smaller-rank constituent affine spaces being smaller complete graphs down to $AG(2,2)$ and $K_4$. This can potentially be visualized in $n$-space by an $n$-simplex representation along the lines of the tetrahedral representation I described for $AG(3,2)$.
Am I correct?
Thank you for reading this question and for any attempts to answer it.
[Edited to correct the false statement that in my designed 3-space (what I believe is $AG(3,2)$, isomorphic to $K_5$) every pair of distinct planes intersect in a single line; they actually intersect in $3$ lines, but also in $3$ points (the same number of points as any two distinct planes in $PG(3,2)$).]
|
An upper bound for the [primorial](https://en.wikipedia.org/wiki/Primorial) can be found based on the first [chebyshev function](https://en.wikipedia.org/wiki/Chebyshev_function).
From $\vartheta(x) < 1.00028x$, it is clear that:
$$\prod\limits_{p \le x}p \le e^{1.00028x} < (2.72)^x$$
I am looking at the following and am having trouble figuring out the upper bound.
$$\prod\limits_{p \le x}p^{\frac{1}{p}}$$
Is there a function similiar to the first chebyshev function that can be used? The second chebyshev function is not helpful since:
$$\psi(x) = \sum\limits_{n=1}^{\infty}{\vartheta(x^{\frac{1}{n}}})$$
I am wondering if there is an upper bound better than a very simple approach such as this:
$$\prod\limits_{p \le x}p^{\frac{1}{p}} < \prod\limits_{p \le x}p^{\frac{1}{2}} < \sqrt{(2.72)^x} < (1.65)^x$$
|
>Surely if $X_t$ is nice and smooth, then $\alpha > \frac{1}{2}$, but then $2\alpha > 1$, and wouldn't this degenerate into saying that $\mathbb{X}$ is constant? From the known result on Holder continuous greater than 1 $\implies$ constant?
As mentioned in the comments, the rough path lift $\mathbb{X}_{s,t}$ is not an increment. For example, in the case of Brownian motion we can take
$$\mathbb{B}_{s,t}^{Itô}:=\int_{s}^{t}(B_{r}-B_{s})\otimes dB_{r}.$$
This definition originates from the "Taylor"-like-expansion argument by repeatedly using FTC
$$\int^t_sF(X_u)dX_u = F(X_s)\int^t_sdX_u + \int^t_s(F(X_u)-F(X_s))\otimes dX_u$$
$$\stackrel{FTC}{=} F(X_s)\int^t_sdX_u + \int^t_s\int_{s}^{u}DF(X_v)dX_{v}\otimes dX_u$$
$$= F(X_s)\int^t_sdX_u + DF(X_s)\int^t_s\int_{s}^{u}dX_{v}\otimes dX_u+\int^t_s\int_{s}^{u}(DF(X_v)-DF(X_s))dX_{v}\otimes dX_u.$$
>How do I show that a smooth path is a particular case of rough paths with the obvious choice of signature?
Indeed as mentioned in the comments, when we have two functions $Y\in C^{\alpha},X\in C^{\beta}$ with $\alpha+\beta>1$, then we can define their Young integral
$$\int YdX\approx \sum Y_{t_{i}} (X_{t_{i+1}}-X_{t_{i}})$$
and use Young's inequality as mentioned in "a course in rough paths" eq. (4.3)
$$|\int_{s}^{t} Y_{r}dX_{r}-Y_{s}X_{s,t}|\leq c \|Y\|_{\beta} \|X\|_{\alpha}|t-s|^{\alpha+\beta}.$$
So in the case of differentiable $X$ i.e. $\alpha\geq 1$,we can just take a rough path lift $\mathbb{X}$ to be
$$\mathbb{X}_{s,t}:=\int_{s}^{t} (X_{r}-X_{s})dX_{r}$$
and from above this indeed satisfies $\frac{|\mathbb{X}_{s,t}|}{|t-s|^{2\alpha}}<c\|X\|_{\alpha}^{2}$.
|
The definition of a Pentagon says (https://en.wikipedia.org/wiki/Pentagon) "five-sided polygon" but let ABCD be a square and E the middle of AB. Is AEBCD a pentagon?
What in the definition of a pentagon "prevents" this from happening? And more generally hexagons, etc ...
Does "five-sided" does not work if the side defined by AE is the same as the one by EB?
|
I have to test the following Nullspace property: $$ Null(A) \subset Null(BA) $$
where $$ Null(A) = {x\in \mathbb{R}^n/ A x=0} $$
$$ Null(BA) = {y\in \mathbb{R}^n/ AB y=0} $$
For any matrix $A$,$B$ such that they have the correct dimensions to make the product.\
The problem is that when I write the linear combination of the vectors $y$, to show that every vector x can be written as a linear combination of the vectors $y$, the vector $x$ that I am looking for does not appear, therefore I cannot prove that all $x$ can be written from the vectors $y$.\
How can I test this property? any ideas?\
Refering to the dimension of each nullspace, are they the same? that is, the symbol would be $\subseteq$ instead of $\subset$?\
thanks!
|
How can I test this property of the nullspace subspace?
|
From this [answer][1], it implicitly becomes evident that
$$\sum_{k=1}^\infty \frac{kn^k}{(k+n)!} =\frac{1}{(n-1)!}$$
holds for any integer $n$.
I **guess** that the following general formula holds for any $x>0$:
$$\color{blue}{\sum_{k=1}^\infty \frac{kx^k}{\Gamma(k+1+x)}=\frac{1}{\Gamma(x)}}.$$
I could not find the above identity in the literature of the Gamma function. In fact, known expansions of $\frac{1}{\Gamma(x)}$ are complex.
For $x=1$, see this related [question][2].
Could you find a solution or approve the guess for the above series?
[1]: https://math.stackexchange.com/a/4890225/1231520
[2]: https://math.stackexchange.com/a/1917115/1231520
|
Consider the numbers $ax^2$, $c-by^2$, where $x,y=0,1,…,p-1$. There are $2p$ such numbers. Any residue class, except for $0$, can have at most two elements of the form $ax^2$ and at most two elements of the form $c-by^2$ from the ones listed above.
I’m trying to understand the logical deductions happening here. Any residue class refers to residue classes of $1$ (mod p), $2$ (mod p),…,$p-1$ (mod p). Then, for residue class $1$ (mod p), there are at most two elements of the form $ax^2$ because both $a(-1)^2$ and $a(1)^2$ are elements of the residue class $1$ (mod p). Is my understanding correct? If my understanding is incorrect, then please tell me what’s happening here. Oh, by the way, $a$ and $b$ are not divisible by $p$.
|
Let $A$ be a compact topological space equipped with the Borel $\sigma$-algebra, and $X=B_b(A)$ be the vector space of bounded measurable functions. Let $Y=\mathcal M(A)$ be the vector space of finite signed measure on $A$. Define the dual pair $<\cdot, \cdot>$ between $(X,Y)$ such that $
<f, \mu >=\int_A f(a)\mu(da)
$.
Let $\sigma(X,Y)$ be the weakest topology such that for all $\mu\in Y$, the linear map $X\ni f\mapsto <f,\mu>\in \mathbb R$ is continuous.
Define the set $U=\{f\in X\mid \sup_{a\in A}|f(x)|\ge 1\}$. Is the set $U$ closed in the $\sigma(X,Y)$ topology?
I am not sure how to proceed to prove or disprove the claim.
|
Let $\mathcal{O}_{\mathbb{P}^m}(1)$ be the tautological bundle on $\mathbb{P}^m$ and let $T := (\mathbb{C}^\times)^{m+1}$ act on $\mathbb{P}^m$ diagonally, i.e.,
$$
(t_0,\ldots,t_m) \cdot [x_0:\cdots:x_m] := [t_0x_0:\cdots:t_mx_m].
$$
The total space of $\mathcal{O}_{\mathbb{P}^m}(1)$ is given by $\mathbb{C} \times \mathbb{P}^m$.
What is the induced action on the total space of the tautological bundle?
My guess: the action is trivial on $\mathbb{C}$ and the same on $\mathbb{P}^m$.
Furthermore, what is the action on the (other) line bundles $\mathcal{O}_{\mathbb{P}^m} (k)$ for $k\in \mathbb{Z}$?
Any help is appreciated.
|
We consider the sequence $(f_n)_{n \in \mathbb{N}}$ defined by $f_n : [0, 2] \rightarrow \mathbb{R}$ as follows:
$f_n(x) =\begin{cases}
x^n, & \text{if } 0 \leq x \leq 1 \\
1, & \text{if } 1 < x \leq 2
\end{cases}$
for $n \in \mathbb{N}$.
Show that the sequence $(f_n)_{n \in \mathbb{N}} \in (C^0([0,2]),||\cdot||_1)$, is a Cauchy-sequence, but does not converge.
Note: $||f_n(x)||_1 = \int_0^2 |f_n(x)|dx$.
(a) Show that $(f_n)_{n \in \mathbb{N}}$ is a Cauchy-Sequence:
We have to show $\forall \epsilon >0 \exists N \in \mathbb{N} : \forall m,n > N: ||f_n(x)-f_m(x)||_1 < \epsilon $. We assume w.l.o.g. $n<m$. Choose $\epsilon >0$ arbitrary, then by choosing a corresponding $N(\epsilon) = \frac{1}{\epsilon}$ we can show that:
$$||f_n(x)-f_m(x)||_1 = \int_0^2 |f_n(x)-f_m(x)|dx = \int_0^1 |f_n(x)-f_m(x)|dx + \int_1^2|f_n(x)-f_m(x)|dx$$
Since $|f_n(x)-f_m(x)|$ for all $x \in (1,2]$ is zero we can drop that part. Also, we can neglect the absolute value, due to $n<m$.
$$ ´\int_0^1 f_n(x)-f_m(x)dx = \int_0^1 f_n(x)dx - \int_0^1 f_m(x)dx = \frac{x^{n+1}}{n+1} \bigg|_0^1 - \frac{x^{m+1}}{m+1} \bigg|_0^1 $$
$$\frac{1}{n+1} - \frac{1}{m+1} < \frac{1}{n+1} < \frac{1}{N} = \epsilon \quad \square $$
(b) Show the sequence $(f_n)_{n \in \mathbb{N}}$ does not converge:
I'm currently working on an exercise with two parts. In part (a), I need to demonstrate that the sequence forms a Cauchy sequence. I would greatly appreciate it if someone could review my work to ensure its accuracy. For part (b), I've hit a roadblock. I'm familiar with two different types of convergence: pointwise and uniform. Unfortunately, the problem doesn't specify which one to use. My intuition suggests that the sequence converges uniformly, because I can make the difference $||f_n(x)-f(x)||$ as small as I like. Could someone tell me why I am wrong? If so, I'd appreciate a hint on how to proceed with proving it. Thank you very much for your assistance.
|
Is it possible to simplify the following expression, where $k$ is a given constant? I want to simplify it to something in terms of only $k$.
$ \sum_{i=0}^{\lfloor \frac{k}{2} \rfloor} 2^{k} \binom{k-i}{i}$.
I was confused on how to deal with the $i$ in the combination.
|
> What numbers can be written uniquely as a sum of two squares?
I was looking at sequence [A125022](https://oeis.org/A125022), which shows the numbers that can be uniquely written as a sum of two squares. Here are a few things that I noticed from the first numbers. We have $1$, $2$, $4$, $8$, $16$, $32$, $64$, $128$. It is then safe to assume that all numbers of the form $2^{s}$ can be written uniquely, where $s \in \mathbb{Z}_{+} \cup \{0\}$. Moreover, primes of the form $4k+1$, for example, $5$, $13$, $17$, $29$, also appear and, interestingly enough, $5^2$ and $13^{2}$ do not. So, we could also say that $p^{s}$ has a unique representation only when $s = 0$ or $s = 1$. If we analyze $A125022$ a bit more, we notice that $3^{2}$, $7^{2}$, $11^{2}$ are there, so we can also conjecture that numbers of the form $q^{2}$ have a unique representation, where $q$ is a prime of the form $4k+3$. Furthermore, for reasons I will say later, I believe $d^{2}$, where $d$ has all of its prime factors of the form $4k+3$, can be uniquely represented as a sum of two squares. It is also possible to see that products between these three are in the sequence, for example $2^{2}\cdot 5$, $2 \cdot 5 \cdot 3^{2}$ and $2 \cdot 7^{2}$.
**Conjecture.** A number $n \in \mathbb{Z}_{+}$ can be written uniquely as a sum of two squares if, and only if, $n = 2^{s} d^{2} p^{e_1}$, where $s \in \mathbb{Z}_{+} \cup \{0\}$, $d$ has all of its prime divisors of the form $4k+3$, $p$ is a prime of the form $4k+1$ and $e_{1} \in \{0,1\}$.
It is known that a number can be written as a sum of two squares if, and only if, it can be written as $2^{s} t^{2} l$, where $s \in \mathbb{Z}_{+} \cup \{0\}$ and $l$ is a square-free positive integer with all of its prime factors of the form $4k+1$. Thus, we know the number $n$ we conjectured above can in fact be written as a sum of two squares. We only need to understand uniqueness. It is more natural to study these questions with the Gaussian integers, $\mathbb{Z}[i]$. If, for example, we have
$$n = a^{2} + b^{2} = (a+ib)(a-ib) = (\pi_1 \cdots \pi_k) (\overline{\pi_1} \cdots \overline{\pi_k}),$$
where the last expression is the factorization of $n$ in primes of $\mathbb{Z}[i]$, then we might be able to get different representations of $n$ by exchanging, say, $\pi_j$ for $\overline{\pi_j}$. That is,
$$(\pi_1 \cdots \overline{\pi_j} \cdots \pi_n)(\overline{\pi_1} \cdots \pi_j \cdots \overline{\pi_n})$$
should yield a different sum when $\pi_j \neq \overline{\pi_j}$ and at least one of the other primes, say $\pi_i$, also satisfies $\pi_i \neq \overline{\pi_i}$. This does not seem to occur precisely for the numbers conjectured above, which makes me think those are the only numbers that can be uniquely represented.
*Question.* Is my guess correct or am I missing other numbers?
|
We consider the sequence $(f_n)_{n \in \mathbb{N}}$ defined by $f_n : [0, 2] \rightarrow \mathbb{R}$ as follows:
$f_n(x) =\begin{cases}
x^n, & \text{if } 0 \leq x \leq 1 \\
1, & \text{if } 1 < x \leq 2
\end{cases}$
for $n \in \mathbb{N}$.
Show that the sequence $(f_n)_{n \in \mathbb{N}} \in (C^0([0,2]),||\cdot||_1)$, is a Cauchy-sequence, but does not converge.
Note: $||f_n(x)||_1 = \int_0^2 |f_n(x)|dx$.
(a) Show that $(f_n)_{n \in \mathbb{N}}$ is a Cauchy-Sequence:
We have to show $\forall \epsilon >0 \exists N \in \mathbb{N} : \forall m,n > N: ||f_n(x)-f_m(x)||_1 < \epsilon $. We assume w.l.o.g. $n<m$. Choose $\epsilon >0$ arbitrary, then by choosing a corresponding $N(\epsilon) = \frac{1}{\epsilon}$ we can show that:
$$||f_n(x)-f_m(x)||_1 = \int_0^2 |f_n(x)-f_m(x)|dx$$
$$\int_0^1 |f_n(x)-f_m(x)|dx + \int_1^2|f_n(x)-f_m(x)|dx$$
Since $|f_n(x)-f_m(x)|$ for all $x \in (1,2]$ is zero we can drop that part. Also, we can neglect the absolute value, due to $n<m$.
$$ ´\int_0^1 f_n(x)-f_m(x)dx = \int_0^1 f_n(x)dx - \int_0^1 f_m(x)dx = \frac{x^{n+1}}{n+1} \bigg|_0^1 - \frac{x^{m+1}}{m+1} \bigg|_0^1 $$
$$\frac{1}{n+1} - \frac{1}{m+1} < \frac{1}{n+1} < \frac{1}{N} = \epsilon \quad \square $$
(b) Show the sequence $(f_n)_{n \in \mathbb{N}}$ does not converge:
I'm currently working on an exercise with two parts. In part (a), I need to demonstrate that the sequence forms a Cauchy sequence. I would greatly appreciate it if someone could review my work to ensure its accuracy. For part (b), I've hit a roadblock. I'm familiar with two different types of convergence: pointwise and uniform. Unfortunately, the problem doesn't specify which one to use. My intuition suggests that the sequence converges uniformly, because I can make the difference $||f_n(x)-f(x)||$ as small as I like. Could someone tell me why I am wrong? If so, I'd appreciate a hint on how to proceed with proving it. Thank you very much for your assistance.
|
I'd like to know if the following statement is true:
For any $f \in \mathcal{C}^\infty(\mathbb{R})$ such that $f(0) = 0$, then $f(x)/x$ is smooth at $0$.
- if $f$ is analytic, then this is clear by simply Taylor-expanding it.
- hence, my question is for when $f$ is non-analytic. I am hoping that, maybe the fact that derivatives of $f$ look bizarre means that there won't be a singularity at $0$, even after dividing by $x$. (Which seems to be the case for the prototypical example of smooth but non-analytic function, the $\exp(-1/x)$ example).
I wonder if this can be done through some sort of polynomial approximation scheme, but am not at all versed in the topic.
----------
It is a very bizarre question, and I am not *a priori* convinced it is true. However, this result seems to be used in a paper I am reading. Indeed:
I am working in the space of germs $\mathcal{C}_0^\infty(\mathbb{R})$ at $0$, which I treat as an $\mathbb{R}$-vector space. I consider the subspace:
$V =$ {$ax + K | a \in \mathcal{C}_0^\infty(\mathbb{R}), K \in \mathbb{R}$} $\subset \mathcal{C}_0^\infty(\mathbb{R})$
They claim, in Example 2 of §1 of [this paper][1] (whose DOI you might find useful if you want to download it; but they do not say much more), that $\dim_\mathbb{R}(\mathcal{C}_0^\infty(\mathbb{R})/V) = 0$, which seems to imply that $V = \mathcal{C}_0^\infty(\mathbb{R})$. Which in turn seems to indicate that any germ in $\mathcal{C}_0^\infty(\mathbb{R})$ can be written in the form of an element of $V$.
It is clearly true that any *analytic* germ can be put in this form. But since there definitely exist non-analytic germs, I don't see why that statement would hold; unless the question I asked holds.
Indeed, if for every non-analytic function $f$ such that $f(0) = 0$, we have that $f(x)/x$ is smooth at $0$, then this allows us to put any germ $f \in \mathcal{C}_0^\infty(\mathbb{R})$ in the form $f(x) = x \dfrac{f(x)-f(0)}{x} + f(0) \in V$, as required.
Hence: is this true?
[1]: https://www.researchgate.net/publication/243097390_An_Introduction_to_Catastrophe_Theory_and_Its_Applications
|
$f$ smooth non-analytic, $f(0) = 0$. Is $f(x)/x$ smooth at $0$?
|
In most of the cases I have studied, I haven't seen what it means for a form "to be" on the manifold. We usually let $M$ to be a manifold, and we let $\omega\in \bigwedge ^kT^*_pM$ in the exterior algebra of the dual tangent space, we then say $\omega$ is of the form $\sum_{I}f_Idx^I$ (I is a multiindex) and we don't give too much meaning. But all this is abstract and much general.
How we can make sense of a form been on a manifold. Apart from the obvious, to say that $\omega$ "takes values" from the tangent space of the manifold.
Can we say anything about the coefficients (functions) of the form ?
I have been trying to answer this by looking the example of the sphere $S^2$, trying to deduce some information about the functions of the form by looking some evaluations of vector fields on the sphere.
So I think a nice answer to this question would be, to give an example of a $2$-form on the sphere.
Any example of your choice that answers this will do. But I always take the sphere as a "natural choice".
|
Given the following relation, which properties does it fulfill?
Is my reasoning correct or am I missing something?
$(m, n) ∈ R \iff m > n$
***1.) reflexivity***
no, $1 > 1$ (trivially false)
***2.) symmetry***
no, since $x > y\not\rightarrow y > x$, since $2 > 3 \not\rightarrow 3 > 2$
***3.) antisymmetry***
no, since $(2 > 2)\wedge(2>2) \not\rightarrow 2 = 2 $ (trivially false)
***4.) transitivity***
yes, since one can prove that $\forall m,n,r\in R:((m> n ) \wedge (n>r)) \rightarrow m > r$
|
I'd like to know if the following statement is true:
For any $f \in \mathcal{C}^\infty(\mathbb{R})$ such that $f(0) = 0$, then $f(x)/x$ is smooth at $0$.
- if $f$ is analytic, then this is clear by simply Taylor-expanding it.
- hence, my question is for when $f$ is non-analytic. I am hoping that, maybe the fact that derivatives of $f$ look bizarre means that there won't be a singularity at $0$, even after dividing by $x$. (Which seems to be the case for the prototypical example of smooth but non-analytic function, the $\exp(-1/x)$ example).
I wonder if this can be done through some sort of polynomial approximation scheme, but am not at all versed in the topic.
----------
It is a very bizarre question, and I am not *a priori* convinced it is true. However, this result seems to be used in a paper I am reading. Indeed:
I am working in the space of germs $\mathcal{C}_0^\infty(\mathbb{R})$ at $0$, which I treat as an $\mathbb{R}$-vector space. I consider the subspace:
$V =$ {$ax + K | a \in \mathcal{C}_0^\infty(\mathbb{R}), K \in \mathbb{R}$} $\subset \mathcal{C}_0^\infty(\mathbb{R})$
They claim, in Example 2 of §1 of [this paper][1] (whose DOI you might find useful if you want to download it; but they do not say much more), that $\dim_\mathbb{R}(\mathcal{C}_0^\infty(\mathbb{R})/V) = 0$, which seems to imply that $V = \mathcal{C}_0^\infty(\mathbb{R})$. Which in turn seems to indicate that any germ in $\mathcal{C}_0^\infty(\mathbb{R})$ can be written in the form of an element of $V$.
It is clearly true that any *analytic* germ can be put in this form. But since there definitely exist non-analytic germs, I don't see why that statement would hold; unless the question I asked holds.
Indeed, if for every non-analytic function $f$ such that $f(0) = 0$, we have that $f(x)/x$ is smooth at $0$, then this allows us to put any germ $f \in \mathcal{C}_0^\infty(\mathbb{R})$ in the form $f(x) = x \dfrac{f(x)-f(0)}{x} + f(0)$, so that $f \in V$ as required.
Hence: is this true?
[1]: https://www.researchgate.net/publication/243097390_An_Introduction_to_Catastrophe_Theory_and_Its_Applications
|
From this [answer][1], it implicitly becomes evident that
$$\sum_{k=1}^\infty \frac{kn^k}{(k+n)!} =\frac{1}{(n-1)!}$$
holds for any integer $n$.
I **guess** that the following general formula holds for any $x>0$:
$$\color{blue}{\sum_{k=1}^\infty \frac{kx^k}{\Gamma(k+1+x)}=\frac{1}{\Gamma(x)}}.$$
I could not find the above identity in the literature of the Gamma function. In fact, known expansions of $\frac{1}{\Gamma(x)}$ are complex.
For $x=1$, see this related [question][2].
Could you prove the guess?
[1]: https://math.stackexchange.com/a/4890225/1231520
[2]: https://math.stackexchange.com/a/1917115/1231520
|
Suppose $S$ is an invertible symmetric matrix with the following property:
For the entry in the $i$th row and $j$th column, if $|i-j|$ is an odd number then $S_{ij} = 0$; if $|i-j|$ is an even number then $S_{ij} \neq 0$.
Is it possible to prove that the inverse matrix $S^{-1}$ also has the above property?
|
I was working on another problem, and the last step to prove it would be finding a function like this. So to be clear, I am looking for a function $f(x)=\sum_{n=0}^\infty a_nx^n$ such that $0=f(1)=f'(1)=f''(1)=\ldots$ , and $a_n\to 0$. I might be just too tired, but I haven't been able to think of one. Does a function like this exist?
|
I am currently working through [this][1] paper by Denef, and I have run into a little bit of a snafu with Lemma 3.1.
The setup is as follows: let $K$ be any field of characteristic zero, and let $E_0$ be the elliptic curve defined by the equation
$$Y^2 = X^3 + aX + b$$
Identify $T$ with the rational function $(x,y) \to x$ on $E_0$ and identify $U$ with the rational function $(x,y) \to y$ on $E_0$. Consider the field of rational functions $F = K(T,U)$, where $U^2 = T^3 + aT + b$. Denote the space of $K$-rational maps from $E_0$ to $E_0$ by $Rat_K(E_0,E_0)$ and let $\psi_2$ be the map
$$\psi_2: E_0(F) \to Rat_K(E_0,E_0)$$
which sends the point $(V,W)$ on $E_0(F)$ (the set of rational points of the elliptic curve $Y^2 = X^3 + aX + b$ where, in this case, $(x,y)\in F$) to the $K$-rational map
$$\psi_2(V,W): E_0(K) \to E_0(K): (x,y) \mapsto (V(x,y), W(x,y))$$
The claim is then that $\psi_2$ is a homomorphism. This seems clear to me since if we were to try and construct a rational map on $E_0$, this amount to sending a rational point $(X,Y)$ satisfying
$$Y^2 = X^3 + aX + b$$
to some other point satisfying the same elliptic curve equation, and we know that
$$W(x,y)^2 = V(x,y)^3 + aV(x,y) + b$$
since we made the assumption that these rational maps were members of $F$ (well, technically, they are identified with members of $F$ which are rational functions of $T$ and $U$, but I think that the point is clear).
My question is then how one would explicitly write down the homomorphism described by $\psi_2$? The first place that I am having trouble is trying to figure out how to write down the image of the identity of the elliptic curve would map to the identity map? I tried to write this down explicilty in terms of projective coordinates so that $[T:U:Z] = [0:1:0] \mapsto Id_{E_0}$, but I think that I am either missing or misunderstanding something since this is seeming to be a lot more trouble than I feel like it should be. Any help is greatly appreciated; thank you!
[1]: https://www.ams.org/journals/tran/1978-242-00/S0002-9947-1978-0491583-7/S0002-9947-1978-0491583-7.pdf
|
I am working on this question:
> Consider $\mathcal{P}_2(\mathbb{R})$ be all polynomials of degree at most 2 with real coefficients. Let $\mathcal{S}$ be a subspace defined by $\mathcal{S} = \{p \in \mathcal{P}_2(\mathbb{R}): p(0) = p(1) \}$. Let T: $\mathcal{S} \rightarrow \mathcal{S}$ be a linear operator defined by T(p) = p+p'. Prove or disprove T is diagonalizable.
My approach is to use the theorem "T is diagonalizable iff its algebraic and geometric multiplicity is the same". So, I started by finding a basis of T. Since every element in $\mathcal{S}$ is in the form $ax^2-ax+b$, we have $T(ax^2-ax+b) = ax^2+(a-1)x+b$. I feel like the basis should only contain 2 elements (since it is governed by 2 variables), but besides 1, what could be the other basis?
AI give me the opinion of $\{1, x, x^2-x\}$ as a basis, but I do not know if I should trust this.
Any help is appreciated!
|
Consider two random variables $X$ and $Y$ such that $P(X \le Y) = 1$ but $P(X < Y) < 1$, that is $P(X < Y) > 0$. Then, how do we prove that $E[X] < E[Y]$?
|
How to prove that E[X] < E[Y]?
|
> What numbers can be written uniquely as a sum of two squares?
I was looking at sequence [A125022](https://oeis.org/A125022), which shows the numbers that can be uniquely written as a sum of two squares. Here are a few things that I noticed from the first numbers. We have $1$, $2$, $4$, $8$, $16$, $32$, $64$, $128$. It is then safe to assume that all numbers of the form $2^{s}$ can be written uniquely, where $s \in \mathbb{Z}_{+} \cup \{0\}$. Moreover, primes of the form $4k+1$, for example, $5$, $13$, $17$, $29$, also appear and, interestingly enough, $5^2$ and $13^{2}$ do not. So, we could also say that $p^{s}$ has a unique representation only when $s = 0$ or $s = 1$. If we analyze $A125022$ a bit more, we notice that $3^{2}$, $7^{2}$, $11^{2}$ are there, so we can also conjecture that numbers of the form $q^{2}$ have a unique representation, where $q$ is a prime of the form $4k+3$. Furthermore, for reasons I will say later, I believe $d^{2}$, where $d$ has all of its prime factors of the form $4k+3$, can be uniquely represented as a sum of two squares. It is also possible to see that products between these three cases are in the sequence, for example $2^{2}\cdot 5$, $2 \cdot 5 \cdot 3^{2}$ and $2 \cdot 7^{2}$.
**Conjecture.** A number $n \in \mathbb{Z}_{+}$ can be written uniquely as a sum of two squares if, and only if, $n = 2^{s} d^{2} p^{e_1}$, where $s \in \mathbb{Z}_{+} \cup \{0\}$, $d$ has all of its prime divisors of the form $4k+3$, $p$ is a prime of the form $4k+1$ and $e_{1} \in \{0,1\}$.
It is known that a number can be written as a sum of two squares if, and only if, it can be written as $2^{s} t^{2} l$, where $s \in \mathbb{Z}_{+} \cup \{0\}$ and $l$ is a square-free positive integer with all of its prime factors of the form $4k+1$. Thus, we know the number $n$ we conjectured above can in fact be written as a sum of two squares. We only need to understand uniqueness. It is more natural to study these questions with the Gaussian integers, $\mathbb{Z}[i]$. If, for example, we have
$$n = a^{2} + b^{2} = (a+ib)(a-ib) = (\pi_1 \cdots \pi_k) (\overline{\pi_1} \cdots \overline{\pi_k}),$$
where the last expression is the factorization of $n$ in primes of $\mathbb{Z}[i]$, then we may get different representations of $n$ by exchanging, say, $\pi_j$ for $\overline{\pi_j}$. That is,
$$(\pi_1 \cdots \overline{\pi_j} \cdots \pi_n)(\overline{\pi_1} \cdots \pi_j \cdots \overline{\pi_n})$$
should yield a different sum when $\pi_j \neq \overline{\pi_j}$ and at least one of the other primes, say $\pi_i$, also satisfies $\pi_i \neq \overline{\pi_i}$. This does not seem to occur precisely for the numbers conjectured above, which makes me think those are the only numbers that can be uniquely represented.
*Question.* Is my guess correct or am I missing other numbers?
|
Consider two random variables $X$ and $Y$ such that $P(X \le Y) = 1$ but $P(X < Y) < 1$, that is $P(X < Y) > 0$. Then, how do we prove that $E[X] < E[Y]$?
My attempt is as below:
Let $A = \{x \in \mathbb{R}: x = X(w) < Y(w) = y \}$,
then
$$
E[Y]
=
\int_{\mathbb{R}} x \, dF_{Y}(x)
=
\int_{\mathbb{R}/A} x \, dF_{Y}(x)
+
\int_{A} x \, dF_{Y}(x)
$$
but not able to get it to the end.
And, I am not familiar with the measure theoretic approach.
|
In the paper [Magnetic Bloch analysis and Bochner Laplacians](https://www.academia.edu/65733916/Magnetic_Bloch_analysis_and_Bochner_Laplacians) of Ruedi Seiler, this expression appears in this context:
[![enter image description here][1]][1]
The conventional notation for a fiber bundle would be $\pi: b\ \to\ M$.
where $\pi$ is a continuous surjection from $b$ to $M$. Does $\varnothing$ mean a continuous surjection? Is this a conventional notation for this? Or does it mean something else here?
[1]: https://i.stack.imgur.com/3E4z4.png
|
What kind of notation is $\pi: b\ \varnothing\ M$?
|
So let's say we have two vectors $a \in \mathbb R^n$ and $b \in \mathbb R^m$. Let $L=\text{lcm}(n,m)$.
Consider the natural inclusions $a,b \mapsto \mathbb R^L$, which simply continually concatenate $a$ or $b$ to itself until the dimension matches $L$, which of course is possible due to $L$ being the least common multiple of the dimensions of $a$ and $b$ themselves.
Now, take the fourier transform of the inclusions. For $a$, you end up with every $L/n$ frequencies being identical.
It seems like it would make more sense to "spread out" the vector being included, so that the amplitude of the wave relative to how far percentage-wise you are into the vector should be identical? So that for example $(1\ 0\ 1) \mapsto (1\ 0\ 0\ 0\ 1\ 0)$ would make sense. If you expand in this way, what you will find is that the fourier coefficients themselves will begin to repeat across the period $L$. That is, if you go from $n \mapsto L$ by multiplying by $k$, then, the fourier expansion of $a$ will be $k$-periodic, again, only if you're putting 0's in between. i.e. $111$ maps to $100010001000$ if $k=4$. Trying to make the cases clear here.
So either way you take it, you can expand a or b in terms of a L dimensional repeated, spaced out, vector. However, both of those lead to artifacts being present in the fourier transform coefficients. In the former case, you have spaced out fourier coefficients, and in the latter case, you have periodic fourier coefficients. So there doesn't seem to be a natural way to take the dot in fourier space of two vectors of different lengths, since if you try to include upwards into the LCM dimension space, it is not clear how meaningful it would be to at that point dot together these two vectors. Because, you end up recording artifacts of the dimensions $n$ and $m$, rather than simply elucidating structure within $a$ and $b$. Perhaps, you can modulo out this lcm dimensional structure and be left with a lower dimensional fourier object you can then compute a dot product for, but it doesn't seem likely. My question is essentially, how can we compute a meaningful dot product between $a$ and $b$ in the fourier space, so that, let's say a major chord $a$ and a major seven chord $b$ (I'm referencing the fourier coefficients of vectors $a$ and $b$ of different lengths, meaning 4,5,6, but not 7, is highly present in $a$ and 4,5,6,7 is highly present in $b$), could dot together and get a higher number than let's say $a$ dotted in the fourier space with a vector $b$ whose fourier coefficients (in $m$ dimensions, by the way) is like a half diminished seventh chord. Seems like it would be natural that $a$ dot $b$ in fourier space (whatever way you define it, which I don't know how to yet because of what I wrote above, and that's where I need your help) should be larger in magnitude than $a$ dot $b'$ where $b'$.
You may assume the norms $||a||=||b||=1$ and $a_i,b_i \in \mathbb C$.
|
Is this summation over Fourier coefficients a meaningful way to compute dot product despite its weaknesses? Or what is the better way to define it?
|
From the ["Finite affine planes" section](https://en.wikipedia.org/wiki/Affine_plane_(incidence_geometry)#Finite_affine_planes) of the Wikipedia article "Affine plane (incidence geometry)", in a (finite) affine plane of order $n$:
- each line contains $n$ points,
- each point is contained in $n+1$ lines,
- there are $n^2$ points in all, and
- there is a total of $n^2+n$ lines.
I've read enough to know that a common notation of an affine plane of order $n$ (or at least one obtained from (removing a line and all points on that line on) a Desarguesian projective plane) is $AG(2,n)$, just as a common notation of a Desarguesian projective plane of order $n$ is $PG(2,n)$. For orders (such as order $2$) in which there is only one Desarguesian projective plane up to isomorphism, there is also only one affine plane under isomorphism. From the above bulleted properties of an affine plane of order $n$, it seems clear to me that in the affine plane of order $2$, $AG(2,2)$:
- each line contains $2$ points,
- each point is contained in $2+1=3$ lines,
- there are $2^2=4$ points in all, and
- there is a total of $2^2+2=6$ lines.
$AG(2,2)$ is clearly... (I'm not sure exactly what the correct word is but equivalent or isomorphic or something like that) to the complete graph $K_4$. With the nodes of that graph displayed as the three vertices and center of an equilateral triangle, $AG(2,2)$ looks like the common representation of the Fano plane, $PG(2,2)$, without the line shown as a circle and the points on that circle, and with the three medians of the triangle, going from the vertices, truncating at the triangle center.
I feel pretty confident in what I've written so far (although confirmation would be appreciated), but my main interest in asking this question here is in the (finite) affine $3$-space of order $2$, $AG(3,2)$, and its tetrahedral representation (akin to the tetrahedral representation of $PG(3,2)$). From the [Wikipedia article on $PG(3,2)$](https://en.wikipedia.org/wiki/PG(3,2)), that projective $3$-space has the following properties:
- It has $15$ points, $35$ lines and $15$ planes,
- Each point is contained in $7$ lines and $7$ planes,
- Each line is contained in $3$ planes and contains $3$ points,
- Each plane contains $7$ points and $7$ lines,
- Each plane is isomorphic to the Fano plane, $PG(2,2)$,
- Every pair of distinct planes intersect in $1$ line,
- A line and a plane not containing the line intersect in exactly $1$ point.
The article goes on to describe how $PG(3,2)$ can be represented as a tetrahedron. "The $15$ points correspond to the $4$ vertices + $6$ edge-midpoints + $4$ face-centers + $1$ body-center. The $35$ lines correspond to the $6$ edges + $12$ face-medians + $4$ face-incircles + $4$ altitudes from a face to the opposite vertex + $3$ lines connecting the midpoints of opposite edges + $6$ ellipses connecting each edge midpoint with its two non-neighboring face centers. The $15$ planes consist of the $4$ faces + the $6$ "medial" planes connecting each edge to the midpoint of the opposite edge + $4$ "cones" connecting each vertex to the incircle of the opposite face + $1$ "sphere" with the 6 edge centers and the body center."
Removing the "sphere" from $PG(3,2)$ and all the points (the $6$ edge-midpoints and $1$ body center) and lines (the $4$ face-incircles and $4$ lines connecting the midpoints of opposite edges), and then doing the associated truncating of the face-medians and ellipses to just being lines from the vertices to the body center and lines between face-centers, respectively, the resulting $3$-space would have the following properties:
- It has $8$ points, $28$ lines and $14$ planes,
- Each point is contained in $7$ lines and $7$ planes,
- Each line is contained in $3$ planes and contains $2$ points,
- Each plane contains $4$ points and $6$ lines,
- Each plane is isomorphic to $AG(2,2)$ or $K_4$,
- Every pair of distinct planes intersect in $1$ line,
- A line and a plane not containing the line intersect in $\le1$ point.
The $8$ points correspond to the $4$ vertices + $4$ face-centers. The $28$ lines correspond to the $6$ edges + $12$ truncated face-medians (lines from the face-centers to the vertices on that race) + $4$ altitudes from a face to the opposite vertex (these altitudes would cross each other in the center of the tetrahedron if not bowed somehow, but there wouldn't be an actual point in the $3$-space there) + $6$ lines ("truncated ellipses") connecting each pair of face-centers. The $14$ planes consist of the $4$ faces + the $6$ "medial" planes bisecting the dihedral angle at each edge (but no longer extending to the midpoint of the opposite edge, which is no longer a point in the $3$-space) + $4$ "mini-cones" connecting each vertex to centers of the adjacent faces.
The vertices are each on the $3$ adjacent faces + $3$ "medial" planes + $1$ "mini-cone" including that vertex, while the face-centers are each on the $1$ face that it's the center of + the $3$ "medial" planes that include the vertex opposite that face + the $3$ "mini-cones" that don't include that vertex. The edges are each on the $2$ adjacent faces + $1$ "medial" plane extending from that edge. The truncated face-medians are each on the $1$ face it's a truncated median of + $1$ "medial" plane + $1$ "mini-cone". The altitudes are each on the $3$ "medial" planes extend from an edge of the face that is the base of that altitude. And the lines connecting pairs of face-centers ("truncated ellipses") are each in $1$ "medial" plane and $2$ "mini-cones".
The shape described in the last two paragraphs and the group of bullet points immediately before them is, I believe now, one way to represent $AG(3,2)$.
Am I correct?
When I first posted this question, I thought $AG(3,2)$ is basically the complete graph $K_5$ with all $5$ of the component $K_4$s being a plane, with its tetrahedral representation having one "plane" including the entire exterior of the tetrahedron and having either "pinched" triangles or a point at infinity for the other four "planes". But I remembered reading that affine planes were just projective planes minus a single line and all the points on it, and that the relationship was similar for higher-dimensional affine spaces.
Thank you for reading this question and for any attempts to answer it.
[Basically rewritten (apart from the $2$-dimensional part and the description of $PG(3,2)$) to reflect a new hypothesis for what $AG(3,2)$ is and looks like.]
|
Consider two random variables $X$ and $Y$ such that $P(X \le Y) = 1$ but $P(X = Y) < 1$, that is $P(X < Y) > 0$. Then, how do we prove that $E[X] < E[Y]$?
My attempt is as below:
Let $A = \{x \in \mathbb{R}: x = X(w) < Y(w) = y \}$,
then
$$
E[Y]
=
\int_{\mathbb{R}} x \, dF_{Y}(x)
=
\int_{\mathbb{R}/A} x \, dF_{Y}(x)
+
\int_{A} x \, dF_{Y}(x)
$$
but not able to get it to the end.
And, I am not familiar with the measure theoretic approach.
|
From the ["Finite affine planes" section](https://en.wikipedia.org/wiki/Affine_plane_(incidence_geometry)#Finite_affine_planes) of the Wikipedia article "Affine plane (incidence geometry)", in a (finite) affine plane of order $n$:
- each line contains $n$ points,
- each point is contained in $n+1$ lines,
- there are $n^2$ points in all, and
- there is a total of $n^2+n$ lines.
I've read enough to know that a common notation of an affine plane of order $n$ (or at least one obtained from (removing a line and all points on that line on) a Desarguesian projective plane) is $AG(2,n)$, just as a common notation of a Desarguesian projective plane of order $n$ is $PG(2,n)$. For orders (such as order $2$) in which there is only one Desarguesian projective plane up to isomorphism, there is also only one affine plane under isomorphism. From the above bulleted properties of an affine plane of order $n$, it seems clear to me that in the affine plane of order $2$, $AG(2,2)$:
- each line contains $2$ points,
- each point is contained in $2+1=3$ lines,
- there are $2^2=4$ points in all, and
- there is a total of $2^2+2=6$ lines.
$AG(2,2)$ is clearly... (I'm not sure exactly what the correct word is but equivalent or isomorphic or something like that) to the complete graph $K_4$. With the nodes of that graph displayed as the three vertices and center of an equilateral triangle, $AG(2,2)$ looks like the common representation of the Fano plane, $PG(2,2)$, without the line shown as a circle and the points on that circle, and with the three medians of the triangle, going from the vertices, truncating at the triangle center.
I feel pretty confident in what I've written so far (although confirmation would be appreciated), but my main interest in asking this question here is in the (finite) affine $3$-space of order $2$, $AG(3,2)$, and its tetrahedral representation (akin to the tetrahedral representation of $PG(3,2)$). From the [Wikipedia article on $PG(3,2)$](https://en.wikipedia.org/wiki/PG(3,2)), that projective $3$-space has the following properties:
- It has $15$ points, $35$ lines and $15$ planes,
- Each point is contained in $7$ lines and $7$ planes,
- Each line is contained in $3$ planes and contains $3$ points,
- Each plane contains $7$ points and $7$ lines,
- Each plane is isomorphic to the Fano plane, $PG(2,2)$,
- Every pair of distinct planes intersect in $1$ line,
- A line and a plane not containing the line intersect in exactly $1$ point.
The article goes on to describe how $PG(3,2)$ can be represented as a tetrahedron. "The $15$ points correspond to the $4$ vertices + $6$ edge-midpoints + $4$ face-centers + $1$ body-center. The $35$ lines correspond to the $6$ edges + $12$ face-medians + $4$ face-incircles + $4$ altitudes from a face to the opposite vertex + $3$ lines connecting the midpoints of opposite edges + $6$ ellipses connecting each edge midpoint with its two non-neighboring face centers. The $15$ planes consist of the $4$ faces + the $6$ "medial" planes connecting each edge to the midpoint of the opposite edge + $4$ "cones" connecting each vertex to the incircle of the opposite face + $1$ "sphere" with the 6 edge centers and the body center."
Removing the "sphere" from $PG(3,2)$ and all the points (the $6$ edge-midpoints and $1$ body center) and lines (the $4$ face-incircles and $4$ lines connecting the midpoints of opposite edges), and then doing the associated truncating of the face-medians and ellipses to just being lines from the vertices to the body center and lines between face-centers, respectively, the resulting $3$-space would have the following properties:
- It has $8$ points, $28$ lines and $14$ planes,
- Each point is contained in $7$ lines and $7$ planes,
- Each line is contained in $3$ planes and contains $2$ points,
- Each plane contains $4$ points and $6$ lines,
- Each plane is isomorphic to $AG(2,2)$ or $K_4$,
- Every pair of distinct planes intersect in $1$ line,
- A line and a plane not containing the line intersect in $\le1$ point.
In a tetrahedral representation of this $3$-space (akin to the tetrahedral representation of $PG(3,2)$), the $8$ points correspond to the $4$ vertices + $4$ face-centers. The $28$ lines correspond to the $6$ edges + $12$ truncated face-medians (lines from the face-centers to the vertices on that race) + $4$ altitudes from a face to the opposite vertex (these altitudes would cross each other in the center of the tetrahedron if not bowed somehow, but there wouldn't be an actual point in the $3$-space there) + $6$ lines ("truncated ellipses") connecting each pair of face-centers. The $14$ planes consist of the $4$ faces + the $6$ "medial" planes bisecting the dihedral angle at each edge (but no longer extending to the midpoint of the opposite edge, which is no longer a point in the $3$-space) + $4$ "mini-pyramids" connecting each vertex to centers of the adjacent faces.
The vertices are each on the $3$ adjacent faces + $3$ "medial" planes + the $1$ "mini-pyramid" including that vertex, while the face-centers are each on the $1$ face that it's the center of + the $3$ "medial" planes that include the vertex opposite that face + the $3$ "mini-pyramids" that don't include that vertex. The edges are each on the $2$ adjacent faces + $1$ "medial" plane extending from that edge. The truncated face-medians are each on the $1$ face it's a truncated median of + $1$ "medial" plane + $1$ "mini-pyramid". The altitudes are each on the $3$ "medial" planes extend from an edge of the face that is the base of that altitude. And the lines connecting pairs of face-centers ("truncated ellipses") are each in $1$ "medial" plane and $2$ "mini-pyramids".
The shape described in the last two paragraphs and the group of bullet points immediately before them is, I believe now, one way to represent $AG(3,2)$.
Am I correct?
When I first posted this question, I thought $AG(3,2)$ is basically the complete graph $K_5$ with all $5$ of the component $K_4$s being a plane, with its tetrahedral representation having one "plane" including the entire exterior of the tetrahedron and having either "pinched" triangles or a point at infinity for the other four "planes". But I remembered reading that affine planes were just projective planes minus a single line and all the points on it, and that the relationship was similar for higher-dimensional affine spaces.
Thank you for reading this question and for any attempts to answer it.
[Basically rewritten (apart from the $2$-dimensional part and the description of $PG(3,2)$) to reflect a new hypothesis for what $AG(3,2)$ is and looks like.]
|
I am currently reading Range's *Holomorphic functions and integral representations in several complex variables* and would like some clarification on the following definition.
> Let $M$ be a complex submanifold of $\mathbb{C}^n$ and let $p \in M$. Then a function $f : M \to \mathbb{C}$ is *holomorphic* at $p$ if $f \circ H$ is holomorphic at $H^{-1}(p)$ for a local parametrisation $H$ of $M$ at $p$.
There is a theorem in the book that says that a subset $M$ of $\mathbb{C}^n$ is a complex submanifold of dimension $k$ if and only if for every $p \in M$, there exist an open neighbourhood $U$ of $p$ in $\mathbb{C}^n$, an open ball $B^{(k)}(a, \varepsilon) \subset \mathbb{C}^k$, and a non-singular holomorphic map $H : B^{(k)}(a, \varepsilon) \to \mathbb{C}^n$ such that $H(B^{(k)}(a, \varepsilon)) = M \cap U$. Here, Range says that a map $H$ which satisfies these conditions is called a *local parametrisation of $M$ at $p$*.
My uncertainty is how this definition of local parametrisation is compatible with the definition of a holomorphic function on a complex submanifold. With the conditions as stated, $H$ being non-singular says that it is only locally injective. But it seems that in the definition of a holomorphic function on a complex submanifold, we are able to take the inverse of $H$ not only on a smaller subset of $M \cap U$, but on all of $M \cap U$? Can this really be done?
Another reason why I am uncertain with these definitions is that it is left as an exercise for the reader to prove that if $H$ is a local parametrisation of $M$ at $p$ with image $M \cap U$, then $H^{-1} : M \cap U \to \mathbb{C}^{\dim M_p}$ is holomorphic. So it does seem like somehow it is possible to obtain not only a non-singular map, but a bijective one with all the properties of a general local parametrisation. Being able to do this is very unclear to me, so I'd greatly appreciate it if I could get some help. Thanks in advance!
|
The other solution uses class field theory. I will avoid that here.
Your question is the same as asking when $\mathbf Q(\sqrt{n}) \subset \mathbf Q(\omega_m)$.
By replacing $n$ by is biggest squarefree factor, we can assume $n$ is *squarefree*. And surely you are not interested in $n$ being a square, so we take $n$ to be a squarefree integer other than $1$. When $\mathbf Q(\sqrt{n}) \subset \mathbf Q(\omega_m)$, primes that ramify in the quadratic field also ramify in the cyclotomic field.
When $n$ is squarefree and not $1$, which primes ramify in $\mathbf Q(\sqrt{n})$? An odd prime ramifies in $\mathbf Q(\sqrt{n})$ if and only if it divides $n$, while $2$ ramifies in $\mathbf Q(\sqrt{n})$ if and only if $n \equiv 2, 3 \bmod 4$.
Which primes ramify in $\mathbf Q(\omega_m)$? When $p \nmid m$, $p$ is unramified in $\mathbf Q(\omega_m)$ since $x^m - 1 \bmod p$ is separable. When $p \mid m$ and $p > 2$, $p$ is ramified in $\mathbf Q(\omega_p)$, so $p$ is ramified in the larger field $\mathbf Q(\omega_m)$. When $m$ is divisible by $2$ just once, $2$ is unramified in $\mathbf Q(\omega_m)$ since this field equals $\mathbf Q(\omega_{m/2})$ and $m/2$ is odd. When $4 \mid m$, $2$ is ramified in $\mathbf Q(i)$ and thus also in the larger field $\mathbf Q(\omega_m)$.
By looking at ramified primes, if $\mathbf Q(\sqrt{n}) \subset \mathbf Q(\omega_m)$ then every *odd* prime factor of $n$ is a factor of $m$, so when $n$ is odd we have $n \mid m$ (because $n$ is squarefree). When $n \equiv 3 \bmod 4$, also $2$ ramifies in $\mathbf Q(\sqrt{n})$, so $4 \mid m$, and thus $4n \mid m$.
I am going to look only at odd $n$ for simplicity. A necessary condition to have $\mathbf Q(\sqrt{n}) \subset\mathbf Q(\omega_m)$ is then
$n \mid m$ when $n \equiv 1 \bmod 4$ and $4n \mid m$ when $n \equiv 3 \bmod 4$.
Now we want to prove the converse: if $n$ is odd with $n \mid m$ when $n \equiv 1 \bmod 4$ and $4n \mid m$ when $n \equiv 3 \bmod 4$, then
$\mathbf Q(\sqrt{n}) \subset \mathbf Q(\omega_m)$.
When $n = -1$ this is immediate: $4 \mid m$ and $\mathbf Q(\sqrt{n}) = \mathbf Q(i)\subset \mathbf Q(\omega_m)$. Now let $n$ be odd and not be $\pm 1$, so $n$ has prime factors.
Write $n = \pm p_1\cdots p_r$ where the $p_i$'s are distinct odd primes.
In order to show $\sqrt{n} \in \mathbf Q(\omega_m)$, we're going to show $\sqrt{\pm p_i} \in \mathbf Q(\omega_m)$ for a *careful* choice of signs on each prime and then multiply together those square roots to show $\sqrt{n} \in \mathbf Q(\omega_m)$.
To handle signs nicely, it is convenient to consider, when $n$ is odd, the odd number $n^* = (-1)^{(n-1)/2}n$. So $n^* = \pm n$ with $n^* \equiv 1 \bmod 4$. Check $(-n)^* = -(n^*)$. Thus $(\varepsilon n)^* = \varepsilon (n^*)$ when $\varepsilon = \pm 1$. Also check $(n_1n_2)^* = n_1^*n_2^*$ when $n_1$ and $n_2$ are odd.
Let the prime factorization of $n$ be $\varepsilon p_1\cdots p_r$ where $\varepsilon = \pm 1$ and the $p_i$'s are distinct odd primes. Then $\varepsilon n = p_1\cdots p_r$, so
$(\varepsilon n)^* = p_1^*\cdots p_r^*$. Since $(\varepsilon n)^* = \varepsilon n^*$,
$$
n^* = \varepsilon p_1^*\cdots p_r^*.
$$
Case 1: $n \equiv 1 \bmod 4$, so $n \mid m$. Then $n^* = n$, so $n = p_1^*\cdots p_r^*$.
Thus $\sqrt{p_1^*}\cdots\sqrt{p_r^*}$ is a square root of $n$. Using Gauss sums, $\sqrt{p^*} \in \mathbf Q(\omega_p)$ when $p$ is an odd prime. (Explicitly, $(\sum_{k=1}^{p-1} (\frac{k}{p})\omega^k)^2 = p^*$, so the Gauss sum $\sum_{k=1}^{p-1} (\frac{k}{p})\omega^k$ is a square root of $p^*$ and its very formula shows it is in $\mathbf Q(\omega_p)$.) When $n \equiv 1 \bmod 4$, each $\sqrt{p_i^*}$ is in $\mathbf Q(\omega_{p_i})$, which is in $\mathbf Q(\omega_m)$, so their product $\sqrt{n}$ is in $\mathbf Q(\omega_m)$.
Case 2: $n \equiv 3 \bmod 4$, so $4n \mid m$. Then $i \in \mathbf Q(\omega_m)$. We have $n^* = -n$, so $-n = p_1^*\cdots p_r^*$ and
$\sqrt{p_1^*}\cdots\sqrt{p_r^*}$ is a square root of $-n$. Using Gauss sums, $\sqrt{p_i^*} \in \mathbf Q(\omega_{p_i}) \subset \mathbf Q(\omega_m)$.
so $\sqrt{-n} \in \mathbf Q(\omega_m)$. Since $i \in \mathbf Q(\omega_m)$, we also get $\sqrt{n} \in \mathbf Q(\omega_m)$.
**Remark**. When $n$ is squarefree, the least $m$ such that $\mathbf Q(\sqrt{n}) \subset \mathbf Q(\omega_m)$ is $|{\rm disc}(\mathbf Q(\sqrt{n})|$, which is $|n|$ when $n \equiv 1 \bmod 4$ and is $4|n|$ when $n \not\equiv 1 \bmod 4$.
|
I was working on another problem, and the last step to prove it would be finding a function like this. So to be clear, I am looking for a function $f(x)=\sum_{n=0}^\infty a_nx^n$ such that $0=f(1)=f'(1)=f''(1)=\ldots$ , and $a_n\to 0$, and there exists a nonzero coefficient. I might be just too tired, but I haven't been able to think of one. Does a function like this exist?
|
Definition 3.1.1 in page 29 of [this book][1] is the definition of quasiperod and Proposition 3.1.3. shows that gcd of two quasiperiods is a quasiperiod. The whole proof is clear except for the part about CRT.
I would appreciate a simple explanation of the following claim from the proof of Proposition 3.1.3. :
Now choose an integer $w_1$ such that it is from the prescribed residue class modulo $d_2/ gcd(d_1, d_2)$, and that for any prime divisor $p$ of $q$ not dividing $d_1d_2, w_1 \notequiv −m/d_1$ mod $p$. The existence of such integers is guaranteed by the Chinese Remainder Theorem. How holds $w_1 \notequiv −m/d_1$ mod $p$ and how it comes from CRT?
[1]: https://users.renyi.hu/~magap/classes/ceu/19_fall_analytic_number_theory/classical_analytic_number_theory.pdf
|
Does anyone have an intuitive explanation for this result? My textbook uses a proof by contradiction. But it just doesn't come logically to me why this holds. What would go wrong if, for example, |a|=3 and |b|=4, where b is the largest order element?
|
I'm looking for a pairing function $f(x,y)$ which gives unique values for every combination of integers $x,y>0$ and and as special property 0 for $x=0$ $\forall y$.<br>
Or to be more general we can also replace $0$ with constants. <br>
- We do know the max values of $x$ and $y$ can achieve
- We are allowed to shift $x, y$ values. So we also know their min values. Which don't need to be 0 <br>
So the more general definitions with constants $c_x, c_f$ (instead of $0$):
$$ x_{min} \le x \le x_{max}$$
$$ y_{min} \le y \le y_{max}$$
$$\forall y: f(c_x,y) = c_f $$
$$\forall y, \forall x\not= c_x: f(x,y)\not = f(y,x) \not = c_f$$
- It will be used in a computer program. The goal is finding a function with a max result as small as possible (bit's needed for representation) while still being easy and fast to compute by a machine (not using too much storage). Therefore integer calculations are appreciated. Floating point can lead to inaccuracies.<br>
- The inverse can be anything. We only need to check if two variables $x,y$ lead to the target number.<br>
- No case selections are allowed (e.g. for $x=0$). Something like Kronecker Delta function which can only be $0$ for input $0$ and $1$ for anything else is also not allowed.
Target value size, the number of different values $|${$x$}$|$ $= 2^{16} = 65536$ and
$|${$y$}$|$ at least $2^{26} = 67108864$. <br>
Size $|${$y$}$|$ can be bigger but $\max f(x,y) < 2^{92}$
Can we find any such function?
----
Here some **Examples:**
**I**) Pairing functions which **do not work** ( not $0$ or a constant for any $x = c_x$ ) :<br>
Cantor's pairing function :
$$f(x,y)=\frac{(x+y)\cdot (x+y+1)}{2} + a$$
Szudzik pairing function:
$$
\begin{eqnarray*}
f(x,y) =
\begin{cases}
y^2 + x, &\text{if }x < y,
\\ y^2 + x + y, &\text{if }x \ge y
\end{cases}
\end{eqnarray*}
$$
This has some cases selection we do not want. We can shift $y$ to be always larger than $x$ and with this reduce it to:
$$f(x,y) = (y+x_{max} +1 )^2 + x$$
Can we modify them to be a constant for a certain $x = c_x$?
----------
**II**)
Pairing function which **does work** but needs extra memory/time : <br>
If we use $0 \le x \le 2^{16}-1$ it can not surpass value $2^{16} = 65536$. Next prime above it is 65537. This is the 6543'th prime number. <br>
Let $p(i)$ return the $i'th$ prime number.<br>
Let $0 \le y \le 2^{26}-1$ <br>
With this we can do the pairing function:
$$f(x,y) = x \cdot p(y + 6543)$$
This returns a unique value for every combination $x,y$ except for $x = c_x = 0$ it will always return the constant $c_f = 0$.<br>
However this works good in theory but calculating $p$ takes quite a long time. Storing all values would take a lot of memory ($\approx 268$ MB) - not a nice option. <br>
Max value of $f(x,y)$ would be $2^{46.31}$ needing $47$ bits which is not too far from optimum $42$ bits. (Cantor and Szudzik are much bigger)
To reduce memory we can split $y$ into half bit-size parts. For $2^{26}$ different values this would be $2^{13}$. First part gets represented by the first $2^{13}$ primes after $2^{16}$, second part by the $2^{13}$ primes after that.
$$f(x,y) = x \cdot p(y_{\text{bits } 1..13} + 6543) \cdot p(y_{\text{bits } 14..26} + 6543 + 2^{13}) $$
This would reduce memory but scales the max value of $f(x,y)$ to $2^{51.3}$, so we need $52$ bits.
------
**Question:** Can we find any such pairing function which does not need to replace parameter $x,y$ with their primes (or similar) while the resulting value $f(x,y)$ won't get much bigger than $2^{52}$ (at most $2^{92}-1$ ) ?
----
**III**) Pairing function which **might work** but is **too big**:<br>
While testing around I came up with <br>
$$f(x,y) = ((x+1)^2+y^2) \cdot (x^3+(y+1)^2) \cdot x$$
I have **no proof** that this is a valid pairing function for all combinations $x,y$ but in tests it did work for all combinations $x<2^{8}, y<2^{20}$<br>
However the results can get too big. It needs up to $121$ bits. <br>
Can we find a pairing function with a smaller max value?
--------
Edit:<br>
**IV**) For $0\ge x < 2^{m}$, $0\ge y < 2^{n}$ (and $n\ge m$) we could do
$$ f(x,y) = 2^{m+n}\cdot x \cdot y + 2^{m} \cdot x$$
With target $m=16, n=24$ this will give us $84$ bit numbers.<br>
Pretty big but valid numbers. Can we go smaller?
|
How to show that gcd of two quasiperiods is a quasiperiod?
|
Definition 3.1.1 in page 29 of [this book][1] is the definition of quasiperod and Proposition 3.1.3. shows that gcd of two quasiperiods is a quasiperiod. The whole proof is clear except for the part about CRT.
I would appreciate a simple explanation of the following claim from the proof of Proposition 3.1.3. :
Now choose an integer $w_1$ such that it is from the prescribed residue class modulo $d_2/ gcd(d_1, d_2)$, and that for any prime divisor $p$ of $q$ not dividing $d_1d_2, w_1 \notequiv −m/d_1$ mod $p$. The existence of such integers is guaranteed by the Chinese Remainder Theorem. How holds $w_1 \notequiv −m/d_1$ mod $p$ and how it comes from CRT?
PS this is an exercise in Apostol's book Ch8 and aslo Montgomery's book Ch9. In Apostol "quasiperiod" is named "induced modulus"
[1]: https://users.renyi.hu/~magap/classes/ceu/19_fall_analytic_number_theory/classical_analytic_number_theory.pdf
|
Definition 3.1.1 in page 29 of [this book][1] is the definition of quasiperod and Proposition 3.1.3. shows that gcd of two quasiperiods is a quasiperiod. The whole proof is clear except for the part about CRT.
I would appreciate a simple explanation of the following claim from the proof of Proposition 3.1.3:
Now choose an integer $w_1$ such that it is from the prescribed residue class modulo $d_2/ \gcd(d_1, d_2)$, and that for any prime divisor $p$ of $q$ not dividing $d_1d_2$, we have $w_1 \not\equiv −m/d_1\pmod p$. The existence of such integers is guaranteed by the Chinese Remainder Theorem. How holds $w_1 \not\equiv −m/d_1\pmod p$ and how it comes from CRT?
PS this is an exercise in Apostol's book Ch8 and also Montgomery's book Ch9. In Apostol "quasiperiod" is named "induced modulus".
[1]: https://users.renyi.hu/~magap/classes/ceu/19_fall_analytic_number_theory/classical_analytic_number_theory.pdf
|
Definition $3.1.1$ in page $29$ of [this book][1] is the definition of quasiperod and Proposition $3.1.3.$ shows that gcd of two quasiperiods is a quasiperiod. The whole proof is clear except for the part about CRT.
I would appreciate a simple explanation of the following claim from the proof of Proposition $3.1.3.$ :
Now choose an integer $w_1$ such that it is from the prescribed residue class modulo $d_2/ gcd(d_1, d_2)$, and that for any prime divisor $p$ of $q$ not dividing $d_1d_2, w_1 \not\equiv −m/d_1$ mod $p$. The existence of such integers is guaranteed by the Chinese Remainder Theorem. How holds $w_1 \not\equiv −m/d_1$ mod $p$ and how it comes from CRT?
PS this is an exercise in Apostol's book Ch8 and aslo Montgomery's book Ch9. In Apostol "quasiperiod" is named "induced modulus"
[1]: https://users.renyi.hu/~magap/classes/ceu/19_fall_analytic_number_theory/classical_analytic_number_theory.pdf
|
Definition 3.1.1 in page 29 of [this book][1] is the definition of quasiperod and Proposition 3.1.3. shows that gcd of two quasiperiods is a quasiperiod. The whole proof is clear except for the part about CRT.
I would appreciate a simple explanation of the following claim from the proof of Proposition 3.1.3:
Now choose an integer $w_1$ such that it is from the prescribed residue class modulo $d_2/ \gcd(d_1, d_2)$, and that for any prime divisor $p$ of $q$ not dividing $d_1d_2$, we have $w_1 \not\equiv −m/d_1\pmod p$. The existence of such integers is guaranteed by the Chinese Remainder Theorem. How holds $w_1 \not\equiv −m/d_1\pmod p$ and how it comes from CRT?
PS this is an exercise in Apostol's book Ch8 and also Montgomery's book Ch9. In Apostol "quasiperiod" is named "induced modulus".
[1]: https://users.renyi.hu/~magap/classes/ceu/19_fall_analytic_number_theory/classical_analytic_number_theory.pdf
|
Definition $3.1.1$ in page $29$ of [this book][1] is the definition of quasiperod and Proposition $3.1.3.$ shows that gcd of two quasiperiods is a quasiperiod. The whole proof is clear except for the part about CRT.
I would appreciate a simple explanation of the following claim from the proof of Proposition $3.1.3.$ :
Now choose an integer $w_1$ such that it is from the prescribed residue class modulo $d_2/ gcd(d_1, d_2)$, and that for any prime divisor $p$ of $q$ not dividing $d_1d_2, w_1 \not\equiv −m/d_1$ mod $p$. The existence of such integers is guaranteed by the Chinese Remainder Theorem. How holds $w_1 \not\equiv −m/d_1$ mod $p$ and how it comes from CRT?
PS this is an exercise in Apostol's book Ch8 and aslo Montgomery's book Ch9. In Apostol "quasiperiod" is named "induced modulus"
[1]: https://users.renyi.hu/~magap/classes/ceu/19_fall_analytic_number_theory/classical_analytic_number_theory.pdf
|
Definition $3.1.1$ in page $29$ of [this book][1] is the definition of quasiperod and Proposition $3.1.3.$ shows that gcd of two quasiperiods is a quasiperiod. The whole proof is clear except for the part about CRT.
I would appreciate a simple explanation of the following claim from the proof of Proposition $3.1.3.$ :
Now choose an integer $w_1$ such that it is from the prescribed residue class modulo $d_2/ gcd(d_1, d_2)$, and that for any prime divisor $p$ of $q$ not dividing $d_1d_2, w_1 \not\equiv −m/d_1$ mod $p$. The existence of such integers is guaranteed by the Chinese Remainder Theorem. How holds $w_1 \not\equiv −m/d_1$ mod $p$ and how it comes from CRT?
PS this is an exercise in Apostol's book Ch8 and aslo Montgomery's book Ch9. In Apostol "quasiperiod" is named "induced modulus".
[1]: https://users.renyi.hu/~magap/classes/ceu/19_fall_analytic_number_theory/classical_analytic_number_theory.pdf
|
I'm looking for a pairing function $f(x,y)$ which gives unique values for every combination of integers $x,y>0$ and and as special property 0 for $x=0$ $\forall y$.<br>
Or to be more general we can also replace $0$ with constants. <br>
- We do know the max values of $x$ and $y$ can achieve
- We are allowed to shift $x, y$ values. So we also know their min values. Which don't need to be 0 <br>
So the more general definitions with constants $c_x, c_f$ (instead of $0$):
$$ x_{min} \le x \le x_{max}$$
$$ y_{min} \le y \le y_{max}$$
$$\forall y: f(c_x,y) = c_f $$
$$\forall y, \forall x\not= c_x: f(x,y)\not = f(y,x) \not = c_f$$
- It will be used in a computer program. The goal is finding a function with a max result as small as possible (bit's needed for representation) while still being easy and fast to compute by a machine (not using too much storage). Therefore integer calculations are appreciated. Floating point can lead to inaccuracies.<br>
- The inverse can be anything. We only need to check if two variables $x,y$ lead to the target number.<br>
- No case selections are allowed (e.g. for $x=0$). Something like Kronecker Delta function which can only be $0$ for input $0$ and $1$ for anything else is also not allowed.
Target value size, the number of different values $|${$x$}$|$ $= 2^{16} = 65536$ and
$|${$y$}$|$ at least $2^{26} = 67108864$. <br>
Size $|${$y$}$|$ can be bigger but $\max f(x,y) < 2^{92}$
Can we find any such function?
----
Here some **Examples:**
**I**) Pairing functions which **do not work** ( not $0$ or a constant for any $x = c_x$ ) :<br>
Cantor's pairing function :
$$f(x,y)=\frac{(x+y)\cdot (x+y+1)}{2} + a$$
Szudzik pairing function:
$$
\begin{eqnarray*}
f(x,y) =
\begin{cases}
y^2 + x, &\text{if }x < y,
\\ y^2 + x + y, &\text{if }x \ge y
\end{cases}
\end{eqnarray*}
$$
This has some cases selection we do not want. We can shift $y$ to be always larger than $x$ and with this reduce it to:
$$f(x,y) = (y+x_{max} +1 )^2 + x$$
Can we modify them to be a constant for a certain $x = c_x$?
----------
**II**)
Pairing function which **does work** but needs extra memory/time : <br>
If we use $0 \le x \le 2^{16}-1$ it can not surpass value $2^{16} = 65536$. Next prime above it is 65537. This is the 6543'th prime number. <br>
Let $p(i)$ return the $i'th$ prime number.<br>
Let $0 \le y \le 2^{26}-1$ <br>
With this we can do the pairing function:
$$f(x,y) = x \cdot p(y + 6543)$$
This returns a unique value for every combination $x,y$ except for $x = c_x = 0$ it will always return the constant $c_f = 0$.<br>
However this works good in theory but calculating $p$ takes quite a long time. Storing all values would take a lot of memory ($\approx 268$ MB) - not a nice option. <br>
Max value of $f(x,y)$ would be $2^{46.31}$ needing $47$ bits which is not too far from optimum $42$ bits. (Cantor and Szudzik are much bigger)
To reduce memory we can split $y$ into half bit-size parts. For $2^{26}$ different values this would be $2^{13}$. First part gets represented by the first $2^{13}$ primes after $2^{16}$, second part by the $2^{13}$ primes after that.
$$f(x,y) = x \cdot p(y_{\text{bits } 1..13} + 6543) \cdot p(y_{\text{bits } 14..26} + 6543 + 2^{13}) $$
This would reduce memory but scales the max value of $f(x,y)$ to $2^{51.3}$, so we need $52$ bits.
------
**Question:** Can we find any such pairing function which does not need to replace parameter $x,y$ with their primes (or similar) while the resulting value $f(x,y)$ won't get much bigger than $2^{52}$ (at most $2^{92}-1$ ) ?
----
**III**) Pairing function which **might work** but is **too big**:<br>
While testing around I came up with <br>
$$f(x,y) = ((x+1)^2+y^2) \cdot (x^3+(y+1)^2) \cdot x$$
I have **no proof** that this is a valid pairing function for all combinations $x,y$ but in tests it did work for all combinations $x<2^{8}, y<2^{20}$<br>
However the results can get too big. It needs up to $121$ bits. <br>
Can we find a pairing function with a smaller max value?
--------
Edit:<br>
**IV**) For $0\ge x < 2^{m}$, $0\ge y < 2^{n}$ (and $n\ge m$) we could do
$$ f(x,y) = 2^{m+n}\cdot x \cdot y + x$$
With target $m=16, n=24$ this will give us $84$ bit numbers.<br>
Pretty big but valid numbers. Can we go smaller?
|
Suppose we have a vector field $\vec{A}$ in $\mathbb{R}^n.$ The following “vector quantity”:
$$(d\vec{r} \cdot \nabla)\vec{A}$$
looks strange perhaps, but when expanded is seen to be a simple quantity:
$$(d\vec{r} \cdot \nabla)\vec{A} = (dx \frac{\partial}{\partial x} + dy \frac{\partial}{\partial y} )\vec{A}= \begin{pmatrix}\frac{\partial A_x}{\partial x} & \frac{\partial A_x}{\partial y} \\ \frac{\partial A_y}{\partial x} & \frac{\partial A_y}{\partial y}\end{pmatrix}\begin{pmatrix} dx \\ dy \end{pmatrix} = \begin{pmatrix}\frac{\partial A_x}{\partial x}dx + \frac{\partial A_x}{\partial y}dy \\ \frac{\partial A_y}{\partial x}dx + \frac{\partial A_y}{\partial y}dy \end{pmatrix} = \Delta \vec{A}$$
so it is just the infinitesimal change in the direction $d\vec{r}.$ Or, to put it differently for a velocity vector $\vec{v}$ we have
$$(\vec{v} \cdot \nabla)\vec{A} = \frac{d \vec{A}}{dt}\Big|_\vec{v}$$
as a "directional derivative." (This generalizes beyond two dimensions).
My question is:
> Is there an analogous natural geometric interpretation for the quantity $\nabla(d\vec{r} \cdot \vec{A}),$ or equivalently, $\nabla(\vec{v} \cdot \vec{A})$?
Playing the same game as above shows
$$ \nabla(d\vec{r} \cdot \vec{A}) = \begin{pmatrix}\frac{\partial A_x}{\partial x} & \frac{\partial A_y}{\partial x} \\ \frac{\partial A_x}{\partial y} & \frac{\partial A_y}{\partial y}\end{pmatrix}\begin{pmatrix} dx \\ dy \end{pmatrix}$$
which, interestingly, is the transpose of the above. Interpreting though, for example, the quantity $\frac{\partial A_y}{\partial x} dy$ is difficult, let alone $\frac{\partial A_x}{\partial x} dx + \frac{\partial A_y}{\partial x} dy$ .
|
Let $f \colon [0, 1] \to \mathbb{R}$ be defined by
$$
f(x) = \sum_{n=1}^\infty \frac{a_n}{n} \sin(n\pi x),
$$
for some square-summable real sequence $a_n$.
Is it true that $f$ is absolutely continuous? I tried a basic estimate,
$$
\sum_{k=1}^K |f(u_k) - f(v_k)| \leq \sum_{n \geq 1} \sum_{k \leq K} \frac{|a_n|}{n} |\sin(n \pi u_k) - \sin(n\pi v_k)|
\lesssim
\delta \sum_{n \geq 1} |a_n|,
$$
provided that $\sum_k |u_k - v_k| \leq \delta$, however this does not seem to be strong enough.
Is there another argument? Or is this not true?
|
Is the function $f(x) = \sum_{n \geq 1} \tfrac{a_n}{n} \sin(n\pi x)$ absolutely continuous on $[0, 1]$?
|
Definition 3.1.1 in page 29 of [this book][1] is the definition of quasiperod and Proposition 3.1.3. shows that gcd of two quasiperiods is a quasiperiod. The whole proof is clear except for the part about CRT.
I would appreciate a simple explanation of the following claim from the proof of Proposition 3.1.3:
Now choose an integer $w_1$ such that it is from the prescribed residue class modulo $d_2/ \gcd(d_1, d_2)$, and that for any prime divisor $p$ of $q$ not dividing $d_1d_2$, we have $w_1 \not\equiv −m/d_1\pmod p$. The existence of such integers is guaranteed by the Chinese Remainder Theorem. How holds $w_1 \not\equiv −m/d_1\pmod p$ and how it comes from CRT?
PS this is an exercise in Apostol's book Ch8 and also Montgomery's book Ch9. In Apostol "quasiperiod" is named "induced modulus".
[1]: https://users.renyi.hu/~magap/classes/ceu/19_fall_analytic_number_theory/classical_analytic_number_theory.pdf
|
Definition $3.1.1$ in page $29$ of [this book][1] is the definition of quasiperod and Proposition $3.1.3.$ shows that gcd of two quasiperiods is a quasiperiod. The whole proof is clear except for the part about CRT.
I would appreciate a simple explanation of the following claim from the proof of Proposition $3.1.3.$ :
Now choose an integer $w_1$ such that it is from the prescribed residue class modulo $d_2/ gcd(d_1, d_2)$, and that for any prime divisor $p$ of $q$ not dividing $d_1d_2, w_1 \not\equiv −m/d_1$ mod $p$. The existence of such integers is guaranteed by the Chinese Remainder Theorem. How holds $w_1 \not\equiv −m/d_1$ mod $p$ and how it comes from CRT?
PS this is an exercise in Apostol's book Ch8 and aslo Montgomery's book Ch9. In Apostol "quasiperiod" is named "induced modulus"
[1]: https://users.renyi.hu/~magap/classes/ceu/19_fall_analytic_number_theory/classical_analytic_number_theory.pdf
|
Definition 3.1.1 in page 29 of [this book][1] is the definition of quasiperod and Proposition 3.1.3. shows that gcd of two quasiperiods is a quasiperiod. The whole proof is clear except for the part about CRT.
I would appreciate a simple explanation of the following claim from the proof of Proposition 3.1.3:
Now choose an integer $w_1$ such that it is from the prescribed residue class modulo $d_2/ \gcd(d_1, d_2)$, and that for any prime divisor $p$ of $q$ not dividing $d_1d_2$, we have $w_1 \not\equiv −m/d_1\pmod p$. The existence of such integers is guaranteed by the Chinese Remainder Theorem. How holds $w_1 \not\equiv −m/d_1\pmod p$ and how it comes from CRT?
PS this is an exercise in Apostol's book Ch8 and also Montgomery's book Ch9. In Apostol "quasiperiod" is named "induced modulus".
[1]: https://users.renyi.hu/~magap/classes/ceu/19_fall_analytic_number_theory/classical_analytic_number_theory.pdf
|
This question was inspired by the solution to this problem: https://math.stackexchange.com/q/913780.
**Question:**
Suppose you have the real-valued piecewise function:
\begin{equation*}
f(x) =
\begin{cases}
f_1(x), & \text{if } x < a \\
f_2(x), & \text{if } x \geq a
\end{cases},
\end{equation*}
where $f_1$ is continuous and differentiable on the left side of $a$ and $f_2$ is continuous and differentiable on the right side of $a$. Is there a systematic way to write $f(x)$ without making it piecewise? I suspect the answer will have something like the form:
$f(x)=f_1(x) \space + \space $(something involving $f_1'$ and $f_2'$) $ \space \cdot \space$ (some mix of $f_1$ and $f_2$ along with $a$ involving absolute values),
but the only reason I suspect this is because of the special case where the $f_i$s are linear.
**Note:**
The case where $f$ is continuous and piecewise-linear is straightforward:
\begin{equation*}
f(x) =
\begin{cases}
\alpha_1 x+\beta_1, & \text{if } x < a \\
\alpha_2 x+\beta_2, & \text{if } x \geq a
\end{cases},
\end{equation*}
\begin{equation}
= (\alpha_1 x+\beta_1)+\frac{\alpha_2-\alpha_1}{2}\cdot(|x-a|+x-a)
\end{equation}
which can be reasoned to graphically and algebraically through examples, but I don't see how to generalize it from there.
|
Is this Proof on 1-Form with Compact Support Correct?
|
Let $\mathcal{O}_{\mathbb{P}^m}(1)$ be the tautological bundle on $\mathbb{P}^m$ and let $T := (\mathbb{C}^\times)^{m+1}$ act on $\mathbb{P}^m$ diagonally, i.e.,
$$
(t_0,\ldots,t_m) \cdot [x_0:\cdots:x_m] := [t_0x_0:\cdots:t_mx_m].
$$
The total space of $\mathcal{O}_{\mathbb{P}^m}(-1)$ is given by $\mathbb{C} \times \mathbb{P}^m$.
What is the induced action on the total space of the tautological bundle?
My guess: the action is trivial on $\mathbb{C}$ and the same on $\mathbb{P}^m$.
Furthermore, what is the action on the (other) line bundles $\mathcal{O}_{\mathbb{P}^m} (k)$ for $k\in \mathbb{Z}$?
Any help is appreciated.
|
I'm learning about types of function in discrete maths. I like to learn by exploring concepts through computer code. I came across the following code purporting to determine whether a function is injective:
```
def is_injective(function):
# Check if the function is injective (one-to-one)
codomain_set = set(function.values())
return len(codomain_set) == len(function)
# Example usage
domain = ["A","B","C"] # Replace with your domain
codomain = [1,2,3,4] # Replace with your codomain
function = {
"A": 1, # Replace with your function mapping
"B": 2, # Replace with your function mapping
"C": 4, # Replace with your function mapping
}
if is_injective(function):
print("The function is injective (one-to-one).")
```
I'm having a bit of trouble understanding how this works or whether it is correct. In `is_injective()`, `codomain_set` seems to create a set from the function values (which I think represent the range?). I assume creating a set is to remove duplicates.
However, I'm not seeing how comparing a set comprising the range values to the number of values from the domain in the mapping is equivalent to the standard mathematical definition of an injective function.
Can anyone please provide some clarification?
|
I'm looking for a pairing function $f(x,y)$ which gives unique values for every combination of integers $x,y>0$ and and as special property 0 for $x=0$ $\forall y$.<br>
Or to be more general we can also replace $0$ with constants. <br>
- We do know the max values of $x$ and $y$ can achieve
- We are allowed to shift $x, y$ values. So we also know their min values. Which don't need to be 0 <br>
So the more general definitions with constants $c_x, c_f$ (instead of $0$):
$$ x_{min} \le x \le x_{max}$$
$$ y_{min} \le y \le y_{max}$$
$$\forall y: f(c_x,y) = c_f $$
$$\forall y, \forall x\not= c_x: f(x,y)\not = f(y,x) \not = c_f$$
- It will be used in a computer program. The goal is finding a function with a max result as small as possible (bit's needed for representation) while still being easy and fast to compute by a machine (not using too much storage). Therefore integer calculations are appreciated. Floating point can lead to inaccuracies.<br>
- The inverse can be anything. We only need to check if two variables $x,y$ lead to the target number.<br>
- No case selections are allowed (e.g. for $x=0$). Something like Kronecker Delta function which can only be $0$ for input $0$ and $1$ for anything else is also not allowed.
Target value size, the number of different values $|${$x$}$|$ $= 2^{16} = 65536$ and
$|${$y$}$|$ at least $2^{26} = 67108864$. <br>
Size $|${$y$}$|$ can be bigger but $\max f(x,y) < 2^{92}$
Can we find any such function?
----
Here some **Examples:**
**I**) Pairing functions which **do not work** ( not $0$ or a constant for any $x = c_x$ ) :<br>
Cantor's pairing function :
$$f(x,y)=\frac{(x+y)\cdot (x+y+1)}{2} + a$$
Szudzik pairing function:
$$
\begin{eqnarray*}
f(x,y) =
\begin{cases}
y^2 + x, &\text{if }x < y,
\\ y^2 + x + y, &\text{if }x \ge y
\end{cases}
\end{eqnarray*}
$$
This has some cases selection we do not want. We can shift $y$ to be always larger than $x$ and with this reduce it to:
$$f(x,y) = (y+x_{max} +1 )^2 + x$$
Can we modify them to be a constant for a certain $x = c_x$?
----------
**II**)
Pairing function which **does work** but needs extra memory/time : <br>
If we use $0 \le x \le 2^{16}-1$ it can not surpass value $2^{16} = 65536$. Next prime above it is 65537. This is the 6543'th prime number. <br>
Let $p(i)$ return the $i'th$ prime number.<br>
Let $0 \le y \le 2^{26}-1$ <br>
With this we can do the pairing function:
$$f(x,y) = x \cdot p(y + 6543)$$
This returns a unique value for every combination $x,y$ except for $x = c_x = 0$ it will always return the constant $c_f = 0$.<br>
However this works good in theory but calculating $p$ takes quite a long time. Storing all values would take a lot of memory ($\approx 268$ MB) - not a nice option. <br>
Max value of $f(x,y)$ would be $2^{46.31}$ needing $47$ bits which is not too far from optimum $42$ bits. (Cantor and Szudzik are much bigger)
To reduce memory we can split $y$ into half bit-size parts. For $2^{26}$ different values this would be $2^{13}$. First part gets represented by the first $2^{13}$ primes after $2^{16}$, second part by the $2^{13}$ primes after that.
$$f(x,y) = x \cdot p(y_{\text{bits } 1..13} + 6543) \cdot p(y_{\text{bits } 14..26} + 6543 + 2^{13}) $$
This would reduce memory but scales the max value of $f(x,y)$ to $2^{51.3}$, so we need $52$ bits.
------
**Question:** Can we find any such pairing function which does not need to replace parameter $x,y$ with their primes (or similar) while the resulting value $f(x,y)$ won't get much bigger than $2^{52}$ (at most $2^{92}-1$ ) ?
----
**III**) Pairing function which **might work** but is **too big**:<br>
While testing around I came up with <br>
$$f(x,y) = ((x+1)^2+y^2) \cdot (x^3+(y+1)^2) \cdot x$$
I have **no proof** that this is a valid pairing function for all combinations $x,y$ but in tests it did work for all combinations $x<2^{8}, y<2^{20}$<br>
However the results can get too big. It needs up to $121$ bits. <br>
Can we find a pairing function with a smaller max value?
--------
Edit:<br>
**IV**) For $0\ge x < 2^{m}$, $0\ge y < 2^{n}$ we could do
$$ f(x,y) = 2^{m+m}\cdot x \cdot y + x$$
With target $m=16, n=24$ this will give us $74$ bit numbers.<br>
Pretty big but valid numbers. Can we go smaller?
|
We know that
$$\sin x = x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \cdots = \sum_{n\ge 0} \frac{(-1)^n}{(2n+1)!}x^{2n+1}$$
But how could we derive this without calculus?
There were some approach using $e^{ix} = \cos x + i \sin x$, pls notice I also would like to avoid such definition as to prove $e^{ix} = \cos x + i \sin x$ again we need the expansion of $\sin x$ and $\cos x$.
One approach I tried is to start with $\sin^2 x$: let
$$\sin^2 x := \sum_{n\ge 1} a_n x^{2n}$$
Note: I guess such an ansatz as $\sin x$ is an odd function -- so that $\sin x$ has only $x$'s power of odd numbers, so $\sin^2 x$ only has $x$'s power of even numbers.
Now if I can arrive $$\sin^2 x = \sum_{n\ge 1} \frac{(-1)^{n+1} 2^{2n-1} }{(2n)!} x^{2n}$$, then via $\cos^2x = 1-2\sin^2 x$ I can get the expansion of $\cos x$ then $\sin x$.
To derive $a_n$, first I use $\lim_{x\rightarrow 0} \frac{\sin x}{x} = 1$ from geometric interpretation (the arc is almost the opponent edge for small angle $x$), to get $$a_1=1$$
Then from $$\sin^2 2x = 4 \sin^2 x ( 1 - \sin^2 x) \Rightarrow \sin^2 x - \frac14 \sin^2 2x = \sin^4 x$$
, I get
$$\sum_{n\ge1} (1-2^{2n-2}) a_n x^{2n} \equiv \left(\sum_n a_n x^{2n}\right)^2$$
This leads to the recursive formula
$$ (1-2^{2n}) a_{n+1} = \sum_{k=1}^{n} a_k a_{n-k} $$
, with $a_1=1$, I do get $$a_2=-\frac13, a_3=\frac{2}{45}, a_4 = -\frac{1}{315}, a_5 = \frac{2}{14175}$$ etc, however, I can only get such results via manual calculation -- there is convolution involved, I could not derive a formula for $a_n$.
Is there a way out pls?
|
I have to test the following Nullspace property: $$ Null(A) \subset Null(BA) $$
where $$ Null(A) = {x\in \mathbb{R}^n/ A x=0} $$
$$ Null(BA) = {y\in \mathbb{R}^n/ BA y=0} $$
For any matrix $A$,$B$ such that they have the correct dimensions to make the product.\
The problem is that when I write the linear combination of the vectors $y$, to show that every vector x can be written as a linear combination of the vectors $y$, the vector $x$ that I am looking for does not appear, therefore I cannot prove that all $x$ can be written from the vectors $y$.\
How can I test this property? any ideas?\
Refering to the dimension of each nullspace, are they the same? that is, the symbol would be $\subseteq$ instead of $\subset$?\
thanks!
|
I'm looking for a pairing function $f(x,y)$ which gives unique values for every combination of integers $x,y>0$ and and as special property 0 for $x=0$ $\forall y$.<br>
Or to be more general we can also replace $0$ with constants. <br>
- We do know the max values of $x$ and $y$ can achieve
- We are allowed to shift $x, y$ values. So we also know their min values. Which don't need to be 0 <br>
So the more general definitions with constants $c_x, c_f$ (instead of $0$):
$$ x_{min} \le x \le x_{max}$$
$$ y_{min} \le y \le y_{max}$$
$$\forall y: f(c_x,y) = c_f $$
$$\forall y, \forall x\not= c_x: f(x,y)\not = f(y,x) \not = c_f$$
- It will be used in a computer program. The goal is finding a function with a max result as small as possible (bit's needed for representation) while still being easy and fast to compute by a machine (not using too much storage). Therefore integer calculations are appreciated. Floating point can lead to inaccuracies.<br>
- The inverse can be anything. We only need to check if two variables $x,y$ lead to the target number.<br>
- No case selections are allowed (e.g. for $x=0$). Something like Kronecker Delta function which can only be $0$ for input $0$ and $1$ for anything else is also not allowed.
Target value size, the number of different values $|${$x$}$|$ $= 2^{16} = 65536$ and
$|${$y$}$|$ at least $2^{26} = 67108864$. <br>
Size $|${$y$}$|$ can be bigger but $\max f(x,y) < 2^{92}$
Can we find any such function?
----
Here some **Examples:**
**I**) Pairing functions which **do not work** ( not $0$ or a constant for any $x = c_x$ ) :<br>
Cantor's pairing function :
$$f(x,y)=\frac{(x+y)\cdot (x+y+1)}{2} + a$$
Szudzik pairing function:
$$
\begin{eqnarray*}
f(x,y) =
\begin{cases}
y^2 + x, &\text{if }x < y,
\\ y^2 + x + y, &\text{if }x \ge y
\end{cases}
\end{eqnarray*}
$$
This has some cases selection we do not want. We can shift $y$ to be always larger than $x$ and with this reduce it to:
$$f(x,y) = (y+x_{max} +1 )^2 + x$$
Can we modify them to be a constant for a certain $x = c_x$?
----------
**II**)
Pairing function which **does work** but needs extra memory/time : <br>
If we use $0 \le x \le 2^{16}-1$ it can not surpass value $2^{16} = 65536$. Next prime above it is 65537. This is the 6543'th prime number. <br>
Let $p(i)$ return the $i'th$ prime number.<br>
Let $0 \le y \le 2^{26}-1$ <br>
With this we can do the pairing function:
$$f(x,y) = x \cdot p(y + 6543)$$
This returns a unique value for every combination $x,y$ except for $x = c_x = 0$ it will always return the constant $c_f = 0$.<br>
However this works good in theory but calculating $p$ takes quite a long time. Storing all values would take a lot of memory ($\approx 268$ MB) - not a nice option. <br>
Max value of $f(x,y)$ would be $2^{46.31}$ needing $47$ bits which is not too far from optimum $42$ bits. (Cantor and Szudzik are much bigger)
To reduce memory we can split $y$ into half bit-size parts. For $2^{26}$ different values this would be $2^{13}$. First part gets represented by the first $2^{13}$ primes after $2^{16}$, second part by the $2^{13}$ primes after that.
$$f(x,y) = x \cdot p(y_{\text{bits } 1..13} + 6543) \cdot p(y_{\text{bits } 14..26} + 6543 + 2^{13}) $$
This would reduce memory but scales the max value of $f(x,y)$ to $2^{51.3}$, so we need $52$ bits.
------
**Question:** Can we find any such pairing function which does not need to replace parameter $x,y$ with their primes (or similar) while the resulting value $f(x,y)$ won't get much bigger than $2^{52}$ (at most $2^{92}-1$ ) ?
----
**III**) Pairing function which **might work** but is **too big**:<br>
While testing around I came up with <br>
$$f(x,y) = ((x+1)^2+y^2) \cdot (x^3+(y+1)^2) \cdot x$$
I have **no proof** that this is a valid pairing function for all combinations $x,y$ but in tests it did work for all combinations $x<2^{8}, y<2^{20}$<br>
However the results can get too big. It needs up to $121$ bits. <br>
Can we find a pairing function with a smaller max value?
--------
Edit:<br>
**IV**) For $0\ge x < 2^{m}$, $0\ge y < 2^{n}$ we could do
$$ f(x,y) = 2^{m}\cdot x \cdot y + x$$
With target $m=16, n=24$ this will give us $58$ bit numbers.<br>
Can we go smaller?
|
Give an example of a sequence of function in $C^1(\mathbb{R})$ to show that the Poincaré inequality does not hold on a general non-compact domain.
I tried to find the sequence in $C^1(\mathbb{R})$ such that it is not integrable on unbounded domain but its dirivative is integrable in that unbounded domain.
Consider
$f(x)= x^{-1/3}$ and $f'(x)=-\dfrac{1}{3}x^{-4/3}$ on the domain $[1, \infty),$ we have
$$\int_1^{\infty}x^{-2/3}dx=+\infty$$
but
$$\int_1^{\infty} -\dfrac{1}{3}[x^{-4/3}]^2dx=\dfrac{5}{27}.$$
It is easy to find an example that it is not integrable on unbounded domain but its dirivative is integrable in that unbounded domain. But the problem is $f$ is not continuously differentiable at $0.$
So I have trouble to find the function belongs to $C^1(\mathbb{R}),$ counte-examples that I found only countinuously differentiable on limit domain not for $\mathbb{R}.$
I also saw a lot of the examples with $u(x) \in H^1_0(\Omega)$ with $u$ is even not continuous.
Could you provide me any idea?
|
Let $u,v \in \mathbb{C}[x]$.
Assume that $\mathbb{C}(u,v)=\mathbb{C}(x)$.
Then by [this answer][1], there exists $\alpha,\beta,\gamma \in \mathbb{C}$
such that
$\langle u-\alpha,v-\beta \rangle = \langle x-\gamma \rangle$,
where $\langle * \rangle$ is the ideal generated by $*$.
> Is this result can be generalized to $\mathbb{C}[x,y]$?
**First attempt**
Let $s,t,u,v \in \mathbb{C}[x,y]$.
Assume that $\mathbb{C}(s,t,u,v)=\mathbb{C}(x,y)$.
Is it true that there exist $\alpha,\beta,\gamma,\delta,\epsilon_1,\epsilon_2 \in \mathbb{C}$ such that
$\langle s-\alpha,t-\beta,u-\gamma,v-\delta \rangle = \langle x-\epsilon_1,y-\epsilon_2 \rangle$?
**Second attempt:**
Let $u,v \in \mathbb{C}[x,y]$.
Assume that $\mathbb{C}(u,v)=\mathbb{C}(x,y)$.
Is it true that there exist $\alpha,\beta,\gamma,\delta \in \mathbb{C}$ such that
$\langle u-\alpha,v-\beta \rangle = \langle x-\gamma,y-\delta \rangle$?
Are there counterexamples for both attempts?
Thank you very much!
**Edit:** What happens if we replace $y$ by $y^2$, namely:
Let $u,v \in \mathbb{C}[x,y]$.
Assume that $\mathbb{C}(u,v)=\mathbb{C}(x,y^2)$.
Is it true that there exist $\alpha,\beta,\gamma,\delta \in \mathbb{C}$ such that
$\langle u-\alpha,v-\beta \rangle = \langle x-\gamma,y^2-\delta \rangle$?
It seems that this result is also true or am I missing something?
[1]: https://math.stackexchange.com/questions/3042016/why-mathbbcft-gt-mathbbct-implies-that-gcdft-a-gt-b-t-c?noredirect=1&lq=1
[2]: https://math.stackexchange.com/questions/2803632/a-sufficient-and-necessary-condition-for-mathbbcfx-gx-mathbbcx?noredirect=1&lq=1
|
I am trying to figure out how to produce this exact integral sign (in LaTeX, if possible; if not, in Microsoft Word?). Would anyone know how to? I know that it's really specific, but it could really make the difference for me in a class. I've also attached an image of this integral with an integrand. Thank you!
[The integral][1] [The integral with an integrand][2]
[1]: https://i.stack.imgur.com/iy9Ac.png
[2]: https://i.stack.imgur.com/tV7G4.png
|
Does anyone know what font this is?
|
Give an example of a sequence of function in $C^1(\mathbb{R})$ to show that the Poincaré inequality does not hold on a general non-compact domain.
I tried to find the sequence in $C^1(\mathbb{R})$ such that it is not belong to $L^2$ on unbounded domain in but its dirivative is in $L^2$ in that unbounded domain.
Consider
$f(x)= x^{-1/3}$ and $f'(x)=-\dfrac{1}{3}x^{-4/3}$ on the domain $[1, \infty),$ we have
$$\int_1^{\infty}x^{-2/3}dx=+\infty$$
but
$$\int_1^{\infty} -\dfrac{1}{3}[x^{-4/3}]^2dx=\dfrac{5}{27}.$$
It is easy to find an example that it is not integrable on unbounded domain but its dirivative is integrable in that unbounded domain. But the problem is $f$ is not continuously differentiable at $0.$
So I have trouble to find the function belongs to $C^1(\mathbb{R}),$ counte-examples that I found only countinuously differentiable on limit domain not for $\mathbb{R}.$
I also saw a lot of the examples with $u(x) \in H^1_0(\Omega)$ with $u$ is even not continuous.
Could you provide me any idea?
|
I am trying to figure out how to produce this exact integral sign (in pdfLaTeX, if possible; if not, in Microsoft Word?). Would anyone know how to? I know that it's really specific, but it could really make the difference for me in a class. I've also attached an image of this integral with an integrand. Thank you!
[The integral][1] [The integral with an integrand][2]
[1]: https://i.stack.imgur.com/iy9Ac.png
[2]: https://i.stack.imgur.com/tV7G4.png
|
I am new to measure theory. I am confused by the following claim from *Measure Theory* by Donald Cohn (Section 1.5 Completeness and Regularity):
> **Claim**$\quad$ *The restriction of Lebesgue measure to the $\sigma$-algebra of Borel subsets of $\mathbb{R}$ is not complete.*
In this post, I will denote the Lebesgue outer measure by $\lambda^*$.
According to the textbook,
> **Definition**$\quad$ The restriction of Lebesgue outer measure on $\mathbb{R}$ (or on $\mathbb{R}^d$) to $\mathcal{B}(\mathbb{R})$ or to $\mathcal{B}(\mathbb{R}^d)$ is called Lebesgue measure and will be denoted by $\lambda$.
>
> The restriction of Lebesgue outer measure on $\mathbb{R}$ (or on $\mathbb{R}^d$) to the collection of Lebesgue measurable subsets of $\mathbb{R}$ (or of $\mathbb{R}^d$) is also called Lebesgue measure and will be denoted by $\lambda$ as well.
> **Definition**$\quad$ Let $(X,\mathcal{A},\mu)$ be a measure space. The measure $\mu$ (or the measure space $(X,\mathcal{A},\mu)$) is *complete* if the relations $A\in\mathcal{A}$, $\mu(A)=0$, and $B \subseteq A$ together imply that $B\in\mathcal{A}$.
So, my understanding of this claim is the following:
> Let $(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)$ be a measure space. Let $A$ be a Borel subset of $\mathbb{R}$ with $\lambda(A)=0$. There exists a subset $B$ of $A$ such that $B \notin \mathcal{B}(\mathbb{R})$.
I saw that this [post][1] used a Vitali set as an example to show that such a set $B$ does exist. However, I am very confused by its explanation. Let $V$ be a Vitali set in $[0,1]$. I know that, since $V$ is not measurable with respect to the Lebesgue outer measure $\lambda^*$, it follows that $V \notin \mathcal{B}(\mathbb{R})$. I think we would want to find a set $A$ such that $A \in \mathcal{B}(\mathbb{R})$, $\lambda(A)=0$, and $V \subseteq A$. But what exactly is this $A$ in the example?
In that post I mentioned above, it seems to me that it wanted to say that the Vitali set $V$ is of Lebesgue measure zero in $\mathbb{R}^2$ and itself is not a Borel subset of $\mathbb{R}^2$. However, if $V$ is not even Lebesgue measurable (i.e., $V$ is not measurable with respect to the Lebesgue outer measure), wouldn't $\lambda(V)$ be not defined? In addition, I couldn't see how his example showed the claim is true, because there was no such set $A \subseteq \mathcal{B}(\mathbb{R})$.
Basically, I am completely lost. I would really appreciate it if someone could help me clarify it or present another example!
----
As @LeeMosher pointed out. The claim should have been understood in the following way:
> Let $(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)$ be a measure space. For some Borel subset $A$ of $\mathbb{R}$ with $\lambda(A) = 0$, there exists a subset $B$ of $A$ where $B \notin \mathcal{B}(\mathbb{R})$.
[1]: https://math.stackexchange.com/questions/575396/borel-not-complete
|
Is it necessary to go back to the definition with epsilons or can you help me find two sub-suites that don't have the same limit?
|
How can you show that the absolute value of sinx diverges?
|
I have prices $p_i$, sizes $s_i$, with average weighted price $A=\frac{\sum(p_i s_i)}{\sum(s_i)}$
I want to calculate $s'_i$, such that
$\sum(s'_i) = \sum(s_i)$
to give a desired new average weighted price
$B = \frac{\sum(p_i s'_i)} { \sum(s'_i)}$
What is the best way to do this?
Would it be possible to solve this with a linear relationship:
$s'_i = ms_i+c$.
(EDITING IN RESPONSE TO ISCO'S ACCEPTED ANSWER)
For people lacking algebra skills, this can be solved using Sympy, which confirms the accepted answer.
from sympy import symbols, Sum
m = symbols("m")
n = symbols("n")
c = symbols("c")
sum_p = symbols("sum_p")
sum_ps = symbols("sum_ps")
sum_s = symbols("sum_s")
A = symbols("A")
B = symbols("B")
eqn1 = 1-n*c/sum_s - m
eqn2 = (1/A)*(B - c*sum_p/sum_s) - m
# Solve the equations
solution = solve((eqn1, eqn2), (m, c))
# Print the solutions
print("Solutions:")
print("m =", solution[m])
print("c =", solution[c])
|
From observations, for quadratic functions, the average rate of change in $[a,b]$ is the **arithmetic** average of the derivative evaluated at $a$ and $b$.
$${\rm{Avg. ROC}_{[a,b]}}=\frac{f'(a)+f'(b)}{2}\qquad f(x)=c_0+c_1x+c_2x^2$$
So, motivated by [this video](https://www.youtube.com/watch?v=3r1t9Pf1Ffk) by Michael Penn, I set out to find "which" average works to find the average ROC for the exponential function and others.
>**Video Overview:**\
>Basically, all averages can be thought of as special cases of
>$$A(t)=\frac{\int_a^bx^{t}dx}{\int_a^bx^{t-1}dx}\tag{1}$$
>For example, $A(1)$ is the arithmetic mean, $A(-0.5)$ is the geometric mean, $A(2)$ is the centroidal mean, etc.
I experimented with values of $t$ in $\rm Eq. (1)$ until a satisfying result was found for respective kinds of functions. Here are my results:
- Exponentials: $0$ ("Logarithmic Mean")
- Quadratic: $1$ ("Arithmetic Mean")
- Cubics: $\frac12$ ("Heronian Mean")
- Quartic: $\frac13$ (?)
- Quintic: $\frac14$ (?)
- (Hypothesis) $n$-degree polynomial: $\frac{1}{n-1}$
- Logarithms: $-1$ (?)
- $x^x$: $-1.0326246...$ (?)
- $x!=\Gamma(x+1)$ : $-1.0364826...$ (?)
The last two numbers were tested for small $a$ and $b$, and are very very crude approximations using Desmos. Here is my [Desmos](https://www.desmos.com/calculator/qtbzznem38) (edited and cleaned for others to use!).
I want an explanation of these various phenomena and the pattern behind these numbers.
---
**Edit.** I deleted this question because I thought I knew the answer. As it turns out, I didn't. After almost a month, I returned to this question to try it again. Unfortunately, I was stuck.
$$\frac{\int_a^bx^{t}dx}{\int_a^bx^{t-1}dx}=\frac{n}{n+1}\frac{[f'(b)]^{n+1}-[f'(a)]^{n+1}}{[f'(b)]^n-[f'(a)]^n}$$
By our condition that it should equal the average rate of change,
$$\frac{n}{n+1}\frac{[f'(b)]^{n+1}-[f'(a)]^{n+1}}{[f'(b)]^n-[f'(a)]^n}=\frac{f(b)-f(a)}{b-a}\tag{2}$$
I need to somehow find an equation that relates $n$ and a function $f(x)$ with $\rm Eq. (2)$. I didn't know where to go from here on. I spent almost an hour fiddling around with Laplace transforms and other algebraic manipulations, but I got nowhere.
|
Use tensor product to show that $ \mathscr{T}^n=\sum_{j \geq 1} \lambda_j^n e_j \otimes e_j $?
|
From observations, for quadratic functions, the average rate of change in $[a,b]$ is the **arithmetic** average of the derivative evaluated at $a$ and $b$.
$${\rm{Avg. ROC}_{[a,b]}}=\frac{f'(a)+f'(b)}{2}\qquad f(x)=c_0+c_1x+c_2x^2$$
So, motivated by [this video](https://www.youtube.com/watch?v=3r1t9Pf1Ffk) by Michael Penn, I set out to find "which" average works to find the average ROC for the exponential function and others.
>**Video Overview:**\
>Basically, all averages can be thought of as special cases of
>$$A(t)=\frac{\int_a^bx^{t}dx}{\int_a^bx^{t-1}dx}\tag{1}$$
>For example, $A(1)$ is the arithmetic mean, $A(-0.5)$ is the geometric mean, $A(2)$ is the centroidal mean, etc.
I experimented with values of $t$ in $\rm Eq. (1)$ until a satisfying result was found for respective kinds of functions. Here are my results:
- Exponentials: $0$ ("Logarithmic Mean")
- Quadratic: $1$ ("Arithmetic Mean")
- Cubics: $\frac12$ ("Heronian Mean")
- Quartic: $\frac13$ (?)
- Quintic: $\frac14$ (?)
- (Speculation) $n$-degree polynomial: $\frac{1}{n-1}$
- Logarithms: $-1$ (?)
- (Speculation) $\operatorname{Li}_n$ : $-n$
The last two numbers were tested for small $a$ and $b$, and are very very crude approximations using Desmos. Here is my [Desmos](https://www.desmos.com/calculator/qtbzznem38) (edited and cleaned for others to use!).
I want an explanation of these various phenomena and the pattern behind these numbers.
---
**Edit.** I deleted this question because I thought I knew the answer. As it turns out, I didn't. After almost a month, I returned to this question to try it again. Unfortunately, I was stuck.
$$\frac{\int_a^bx^{t}dx}{\int_a^bx^{t-1}dx}=\frac{n}{n+1}\frac{[f'(b)]^{n+1}-[f'(a)]^{n+1}}{[f'(b)]^n-[f'(a)]^n}$$
By our condition that it should equal the average rate of change,
$$\frac{n}{n+1}\frac{[f'(b)]^{n+1}-[f'(a)]^{n+1}}{[f'(b)]^n-[f'(a)]^n}=\frac{f(b)-f(a)}{b-a}\tag{2}$$
I need to somehow find an equation that relates $n$ and a function $f(x)$ with $\rm Eq. (2)$. I didn't know where to go from here on. I spent almost an hour fiddling around with Laplace transforms and other algebraic manipulations, but I got nowhere.
|
Give an example of a sequence of function in $C^1(\mathbb{R})$ to show that the Poincaré inequality does not hold on a general non-compact domain.
I tried to find the sequence in $C^1(\mathbb{R})$ such that it is not belong to $L^2$ on unbounded domain in but its dirivative is in $L^2$ in that unbounded domain.
Consider
$f(x)= x^{-1/3}$ and $f'(x)=-\dfrac{1}{3}x^{-4/3}$ on the domain $[1, \infty),$ we have
$$\int_1^{\infty}x^{-2/3}dx=+\infty$$
but
$$\int_1^{\infty} -\dfrac{1}{3}[x^{-4/3}]^2dx=\dfrac{5}{27}.$$
It is easy to find an example that it is not integrable on unbounded domain but its dirivative is integrable in that unbounded domain. But the problem is $f$ is not continuously differentiable at $0.$
So I have trouble to find the function belongs to $C^1(\mathbb{R}),$ counte-examples that I found only countinuously differentiable on limit domains not for $\mathbb{R}.$
I also saw a lot of the examples with $u(x) \in H^1_0(\Omega)$ with $u$ is even not continuous.
Could you provide me any idea?
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.