Saturday, July 17, 2021

Recent Questions - Mathematics Stack Exchange

Recent Questions - Mathematics Stack Exchange

Divergence of vector product

Posted: 17 Jul 2021 07:54 PM PDT

I came across this identity in a Fluid Mechanics book where while deriving Cauchy's equation

∇.(VV)= V∇.V + (V.∇)V where, V is the velocity vector field. I tried to look for the identity but couldn't find any.I have attached an image for reference

Does $\star(\text{d}\omega)=0$ implies $\text{d}\omega=0$?

Posted: 17 Jul 2021 07:54 PM PDT

Yes, $\star(\text{d}\omega)$ is one definition to $\text{rot} \ v$. I need to prove that if $\text{rot} \ v=0$ then $v$ is locally exact (derives locally from a potential, more context can be provided here). And yes, I've checked others answers, like this one (and I know how to solve this problem). But this came to my mind, because in this case, as a consequence, $\text{d}\omega = 0$. Is this true in general?

The definition I have for the Hodge Star Operation on a $k-$form of $\Bbb{R}^{n}$ is that \begin{align*} \star(d_{x_{i_{1}}} \wedge \dots \wedge d_{x_{i_{k}}}) = (-1)^{\text{sgn}\ \sigma} (d_{x_{j_{1}}} \wedge \dots \wedge d_{x_{j_{n-k}}}),\ \text{where} \ (i_{1}\dots i_{k}j_{i}\dots j_{n-k}) \in S_{n}. \end{align*} If the latter is $0$, then $d_{x_{j_{a}}}=d_{x_{j_{b}}}$ for some $a,b$. But all of $d_{x}$'s form a base for the dual of $k-$form of $\Bbb{R}^{n}$, so this can't happen. This is enough to conclude that? Thanks in advance!

Condition that CR equation implies holomorphic

Posted: 17 Jul 2021 07:50 PM PDT

There are many version of criterion for a function to be holomoprhic when CR equation holds ,for example in Stein and Shakarchi's complex analysis book has a version that :

Theorem $2.4$ Suppose $f=u+i v$ is a complex-valued function defined on an open set $\Omega .$ If $u$ and $v$ are continuously differentiable and satisfy the Cauchy-Riemann equations on $\Omega$, then $f$ is holomorphic on $\Omega$ and $f^{\prime}(z)=\partial f / \partial z$

I paste the proof below:

Proof. Write $$ u\left(x+h_{1}, y+h_{2}\right)-u(x, y)=\frac{\partial u}{\partial x} h_{1}+\frac{\partial u}{\partial y} h_{2}+|h| \psi_{1}(h) $$ and $$ v\left(x+h_{1}, y+h_{2}\right)-v(x, y)=\frac{\partial v}{\partial x} h_{1}+\frac{\partial v}{\partial y} h_{2}+|h| \psi_{2}(h) $$ where $\psi_{j}(h) \rightarrow 0$ (for $j=1,2$ ) as $|h|$ tends to 0, and $h=h_{1}+i h_{2}$. Using the Cauchy-Riemann equations we find that $$ f(z+h)-f(z)=\left(\frac{\partial u}{\partial x}-i \frac{\partial u}{\partial y}\right)\left(h_{1}+i h_{2}\right)+|h| \psi(h) $$ where $\psi(h)=\psi_{1}(h)+\psi_{2}(h) \rightarrow 0$, as $|h| \rightarrow 0$. Therefore $f$ is holomorphic and $$ f^{\prime}(z)=2 \frac{\partial u}{\partial z}=\frac{\partial f}{\partial z} $$

There are some observation here,First we can find that to check $f$ holomorphic at some point,we only need to check that:

(1)$f$ has $u,v $ is $C^1$ at that point

(2) CR equation holds at that point.

I found in the proof,we only use the fact that $u,v$ is differentiable ,do not need $u,v$ be continuous differentialbe at that point correct?

See relavent question here and here

When you change the sign of the exponent in an exponential function like e^x into e^-x, what is it called?

Posted: 17 Jul 2021 07:43 PM PDT

When you change the sign of x in e^x to e^-x, is it called getting the conjugate even though its not a complex number in the form of e^jx? I'm just confused some people I talk to refer to it as conjugate because I thought the term was only applicable to complex numbers and I guess binomials where you change the sign of the middle term.

Complicated first order non linear differential equation

Posted: 17 Jul 2021 07:40 PM PDT

How do I solve the following equation.

$$a'(t)-\frac{\sqrt{r+a(t)(da(t)^3+m)}}{a(t)}=0$$

Where r, m and d are positive constants of dimension $\frac{1}{t^2}$. a(t) is dimensionless. The only initial condition given is a(0)=1.

I tried rewritting it in a more intuitive form for ODE's: $$a'(t)^2-da(t)^2-\frac{r}{a(t)^2}-\frac{m}{a(t)}=0$$ But I don't know what substitution is the most convenient.

Find the rules of the sequences below

Posted: 17 Jul 2021 07:36 PM PDT

Help me with this problem enter image description here

I think these sequences have something special because each only have limit numbers.

Points above a $k$-rational point have trivial stabilizer in algebraic fundamental group of the base change to $\bar{k}$

Posted: 17 Jul 2021 07:33 PM PDT

Let $k$ be a perfect field with a fixed algebraic closure $\bar{k}$, $X$ an integral proper normal $k$-curve, $U \subset X$ open and non-empty. Assume that $X$ is geometrically integral and that $p \in U$ is $k$-rational. Take some pro-point $\tilde{q}$ of $\tilde{X}$, the profinite branched cover of $X$ pro-étale over $U$. Let $\bar{q}$ be the image of $\tilde{q}$ in $U_\bar{k} \subset X_\bar{k}$, the base change of $X$ to $\bar{k}$. My goal is to show that it has trivial stabilizer in $\pi_1(U_\bar{k})$, the étale fundamental group of $U_\bar{k}$.

Honestly I am unsure where to start. I know that an element of the stabilizer is going to be an automorphism of $K_U$, the compositum of all finite extensions corresponding to étale morphisms onto $U$, that fixes $K \bar{k}$. My first thought was that an element of the stabilizer of $\bar{q}$ is going to induce an automorphism in $\operatorname{Gal}(\kappa(q) \mid \kappa(p)= k)$ just like an element of the stabilizer of $\tilde{q}$ in $\pi_1(U)$ will. I guess the hope would be that somehow we can look at this induced automorphism in the $\pi_1(U_\bar{k})$ case to show that any such $g$ in the stabilizer will be trivial, unlike in the $\pi_1(U)$ case.

Apologies if what I have so far is unhelpful or wrong: I am pretty stuck on this claim.

If $0 < x < \pi$ and $\cos x + \sin x = 1/2,$ then value of $\tan x$ is? [closed]

Posted: 17 Jul 2021 07:52 PM PDT

Options: A) $\frac{1-\sqrt{7}}{4}$ B) $\frac{4-\sqrt{7}}{3}$ C) $-\frac{4+\sqrt{7}}{3}$ D) $\frac{1+\sqrt{7}}{4}$

Recursion formula for volume of an unit ball in $\mathbb{R^n}$

Posted: 17 Jul 2021 07:57 PM PDT

Let $B^n\subset \mathbb{R^n}$ be an unit ball in $\mathbb{R^n}$ and $V_n(A)$ be the volume of $A\subset \mathbb{R^n}$.Then, prove

\begin{equation} V_n(B^n)=2 V_{n-1} (B^{n-1}) \displaystyle\int_0^1 > (1-u^2)^{\frac{n-1}{2}} du. \end{equation}

using Fubini's theorem.

My idea is here: Let $m_n$ be $n$- dimentional Lebesgue measure. Then, $$V_n(B^n)=m_n (B^n)=\displaystyle \int_{\mathbb{R^n}} \chi_{B^n} d\mu .$$

Since the claim says "using the Fubini", I think I have to separate $\int_{R^n}$ as $\int_{R^n}=\int_{R^{n-1}}\int_{R} $. So I have $$\int_{\mathbb{R^n}} \chi_{B^n} d\mu=\int_{R^{n-1}} \left(\int_{R} \chi_{B^n} \ dy \right)dx.$$

I don't know how I have to proceed.

determining whether a set is dense in a metric space of continuous functions

Posted: 17 Jul 2021 07:12 PM PDT

Let $B$ be the closed unit disc $B = \overline{B}(0,1) = \{z\in \mathbb{Z} |\lVert z\rVert \leq 1\}$ in $\mathbb{C}.$ Determine whether the algebra $\mathcal{P} := \mathcal{P}(B,\mathbb{C})$ of polynomial functions $p:B\to\mathbb{C},$ where $p$ has coefficients in $\mathbb{C},$ is dense in $\mathcal{C}(D,\mathbb{C})$ under the supremum metric.

If the Stone-Weierstrass theorem is useful, I'd need to verify that $\mathcal{P}(B,\mathbb{C})$ separates points and vanishes nowhere and is closed under conjugation (since $B$ is compact in $\mathbb{C}$). The identity function $x\mapsto x$ is in $\mathcal{P}(B,\mathbb{C}),$ so it separates points. Also, $\mathcal{P}$ vanishes nowhere as the constant functions are in it. I just need to verify that $\mathcal{P}$ is closed under conjugation. But this is also clear as if $p:B\to \mathbb{C}$ has coefficients in $\mathbb{C},$ so does its conjugate.

Am I doing something wrong? Would it be useful to consider the map $G : \mathcal{C}(B,\mathbb{C}) \to \mathbb{C} $ given by $G(f) = \int_{t=0}^{2\pi} f(e^{it})e^{it} dt$?

If $n$ is a composite integer larger than $8$, then $n \mid(n-3) !$

Posted: 17 Jul 2021 07:54 PM PDT

So far my understanding is that let $n$ be the product of some integer $(a,b,c...)$, and try to prove $a,b,c...$ $\in$ $[1,n-3]$. Then I stuck at here and wonder if I am on the right truck? Could anyone give me some hints? Thank you!

existence of solution to a differential equation

Posted: 17 Jul 2021 06:59 PM PDT

Define $G:\mathbb{R}^2\to \mathbb{R}$ by $G(x,y) = 3y^{2/3}.$ Determine whether $G$ satisfies the hypothesis of the Picard-Lindelof theorem, whether $G$ satisfies the hypothesis of the Peano existence theorem, and whether there exists $\delta > 0$ so that the differential equation $\frac{dy}{dx} = G(x,y)$ has a unique solution $y=g(x)$ with $g(0) = 0,$ defined for all $x\in (-\delta, \delta).$

The version of the Picard-Lindelof theorem being used is as follows: Let $O$ be an open set in $\mathbb{R}^2$, let $(a,b)\in O,$ and let $G:O\to \mathbb{R}$ satisfy a Lipschitz condition in the second coordinate $y$ on $U$. Then there exists $\delta > 0$ so that the differential equation $\frac{dy}{dx}=G(x,y)$ has a unique solution $y=g(x)$ with $f(a)=b,$ defined for all $x\in [x-\delta, a+\delta].$ The version of the Peano existence theorem being used is as follows. Let $O\subseteq \mathbb{R}^2$ be open, let $(a,b) \in O,$ and let $G:O\to\mathbb{R}$ be continuous. Then there exists $k > 0$ so that the differential equation $\frac{dy}{dx} = G(x,y)$ has a solution $y=g(x)$ which is defined for all $x\in [a-k,a+k].$

I think $G$ is not Lipschitz in $y$. Fix $x\in \mathbb{R}.$ For $n\geq 1,$let $y_1, y_2 \in (0,\frac{1}n).$ Then $|G(x,y_1) - G(x,y_2)| = |y_1^{2/3}-y_2^{2/3}| = |\frac{2}3 c^{-1/3}||y_1-y_2|$ where $c$ is between $y_1$ and $y_2.$ But this is at least $|\frac{2}3 n^{1/3}||y_1-y_2|,$ so $G$ cannot be Lipschitz in $y$ and hence $G$ does not satisfy the hypothesis of the Picard-Lindelof theorem (so it's possible that the differential equation may not have a unique solution). $G$ is clearly continuous, being an elementary function (indeed the projection on the second coordinate/component is continuous and $G$ is the composition of this map with the continuous map $y\mapsto 3y^{2/3}$). So the hypothesis of the Peano existence theorem is satisfied. The differential equation simplifies to solving $\frac{dy}{dx} = 3y^{2/3}\Rightarrow \int \frac{1}3 y^{-2/3} dy = \int dx\Rightarrow y^{1/3} = x+C\Rightarrow y = (x+C)^3,$ where $C$ is some constant. The only way $y(0) = 0$ is if $0=C^3\Rightarrow C=0.$ So the unique solution is $y = x^3,$ which is defined for all $x\in \mathbb{R}$ (any $\delta > 0$ in the problem statement will work).

Have I made a mistake somewhere?

Prove the following statement by mathematical induction

Posted: 17 Jul 2021 07:08 PM PDT

Prove the following below by mathematical induction:

$$\forall n\in\mathbb N, \left(\frac{1^2}{1+1}\right) \left(\frac{2^2}{2+1}\right)\cdots \left(\frac{n^2}{n+1}\right)=\frac{n!}{n+1}$$

Let $f = 0$ s.t. $x_n$ is implicitly function $\phi$ of $x_1, ..., x_{n-1}$. Prove a partial derivative of $\phi$ is a quotient of partials of $f$.

Posted: 17 Jul 2021 07:56 PM PDT

Let $f : \mathbb{R}^n \rightarrow \mathbb{R}$. Suppose $f(\vec a) = 0$ and $ \frac{\partial f (\vec a)}{\partial x_n} \neq 0$ and using the implicit function theorem let $x_n$ be implicitly defined from $f= 0$ by $x_1,...,x_{n-1} = \underline x$ such that $\phi(\underline x) = x_n$. Notice therefore that $\vec x = \langle \underline x, \phi(\underline x) \rangle$

I'm trying to figure out a geometrical interpretation in terms of the function $\phi$ in order to prove $\frac{\partial \phi( {\underline a)}}{\partial x_j}= -\frac{\frac{\partial f(\vec a)}{\partial x_j}}{{\frac{\partial f(\vec a)}{\partial x_n}}}$. I have drawn the following diagram, with two level curves $z = f = 0$ and $z=f=1$ where subscripted $e$ designates a standard vector and $b = \langle \underline a + t\vec e_j, \phi(\underline a + t\vec e_j)\rangle$ (notice it is part of the definition for $\frac{\partial \phi(\underline a)}{x_j}$), and where $f(\vec a) = f(\vec b) = 0$: enter image description here

My question is: is there any geometrical interpretation for $\frac{\delta f}{\delta x_j}$ and $\frac{\delta f}{\delta x_n}$ in terms of the function $\phi$ that would make $\frac{\partial \phi( {\underline a)}}{\partial x_j}= -\frac{\frac{\partial f(\vec a)}{\partial x_j}}{{\frac{\partial f(\vec a)}{\partial x_n}}}$ easier to prove? More specifically, how could we geometrically interpret change in $\phi$ with respect to $x_j$ in terms of change in $f$ with respect to $x_j, x_n$?

Thank you very much!

Simply connected subset of $\mathbb{S}^1$

Posted: 17 Jul 2021 07:24 PM PDT

I was trying. to think of a generalization for theorem stating that if $X=U\cup V$ where $U,V$ are simply connected with path-connected intersection, then $X$ is simply connected.

But I was given the counter-example of covering $\mathbb{S}^1$ by 3 arcs each of length $=\left(\frac{1}{3}+\epsilon\right)*$(circumference of the circle), and I cannot understand how this counter example demonstrates simply connected subsets, or how the same argument couldn't be used for the case of the initial theorem? i.e. Having two arcs of length$=\left(\frac{1}{2}+\epsilon\right)*$(circumference of the circle)?

Pigeon hole problem for custom deck of cards

Posted: 17 Jul 2021 07:50 PM PDT

Suppose you have a custom deck of cards. In a traditional deck of cards, each card has 1 of 4 suits and 1 of 13 numbers (ignoring jokers). "Suits" and "numbers" are what I will refer to as attributes and the respective suit and number of each card is what I will refer to as "value."

In this custom deck of cards, we have 4 attributes, each with 3 possible values. So there are $3^4$ different types of cards. You draw cards from an infinite shuffled deck of these custom cards. Without looking at the cards you drew, what is the minimum number of draws that guarantee that you've got 3 cards (from the ones you've drawn, not necessarily consecutively drawn) that satisfy the following criteria:

(1) For each attribute of the 3 cards, they all have different values.

or

(2) For each attribute of the 3 cards, they have all of the same values.

Here is an example. Suppose the 4 attributes and corresponding values are:

(a) Color: red, green, blue

(b) Number: 1, 2, 3

(c) Suit: circle, square, triangle

(d) Roman numeral: I, II, III

Say we have 3 cards. The first card is "red 2 circle I." Second card is "green 2 square II." Third card is "blue 2 triangle III." This is a set of 3 cards that satisfies all the criteria because each attribute of the 3 cards either has all different values (Color, suit, roman numeral) or have the same value (Number).

I want to know the minimum number of draws that guarantees I have 3 cards that satisfy these conditions. Obviously, if you draw 163 cards, you'll definitely have 3 cards that satisfy these conditions because 163 cards guarantees that you have 3 identical cards since there are only 81 different cards, and by the pigeon whole principle with each of the 81 different card types as a hole, one of the holes will certainly have 3 pigeons after 163 draws.

But I don't think 163 is the minimum. I think there should be a smaller number of draws. With 163, we are guaranteed there are 3 cards such that all 4 attributes satisfies criteria (2). So I think I need to apply the pigeon pigeon principle to such that all 4 attributes satisfies criteria (1), and also for the case where the 4 attributes satisfies a mixture of criteria (1) and (2), and perhaps one of these will give a lower bound than 163. How can I go about this?

I tried to think about using combinations, but there too many variables to keep track of that I'm getting lost.

$|x| ^\alpha + |y-x|^\alpha \geq |y|^\alpha$ for $1 \geq \alpha \geq \frac12$?

Posted: 17 Jul 2021 07:50 PM PDT

We want to prove that $\forall x,y$ and $ 1 \geq \alpha \geq \frac12$ $$|x| ^\alpha + |y-x|^\alpha \geq |y|^\alpha$$

Is there an alternative proof ?

  • We consider for $\beta \geq 1$, $\Phi(t)= (a+t)^\beta - a^\beta - t^\beta$

  • $\Phi$ is increasing (its derivative is positive) and $\Phi(0)=0$

  • Therefore $(a+t)^\beta - a^\beta - t^\beta \geq 0$

  • then we choose $a=x-y, t=y, \beta= \frac{1}{\alpha}$ so $\alpha \leq 1$

  • We obtain $ | (x-y)^{\alpha} + y^{\alpha}|^{\frac{1}{\alpha}}\geq |x-y| +|y| \geq |x|$

  • $|x-y|^{\alpha} + |y|^{\alpha} \geq |x|^{ \alpha}$

  • By exchanging $x$ and $y$, we have $||x|^{\alpha} -|y|^{\alpha} | \leq |x-y| ^ { \alpha}$

What is the correct way to obtain equations for the intersection of two cylinders and how does one plot the intersection in Maple?

Posted: 17 Jul 2021 07:52 PM PDT

Given two cylinders

$$x^2+z^2=4$$

$$x^2+y^2=4$$

What is the correct way to parametrize their intersection?

In spherical coordinates:

$$\rho^2 \sin^2{\phi} \cos^2{\theta}+\rho^2 \cos^2{\phi}=4$$ $$\rho^2 \sin^2{\phi} \cos^2{\theta} + \rho^2 \sin^2{\phi} \sin^2{\theta}=4 \implies \rho^2 \sin^2{\phi}=4$$ $$\implies 4 \cos^2{\theta}+\rho^2 \cos^2{\phi}=4\tag{1}$$

I believe $(1)$ gives the spherical coordinates for the intersection.

If we equate the first two equations, we are able to get rid of $\rho$ however.

$$\rho^2 \sin^2{\phi} \cos^2{\theta} +\rho^2 \cos^2{\phi}=\rho^2 \sin^2{\phi}$$ $$\implies \rho^2 \sin^2{\phi}(1-\cos^2{\theta} = \rho^2 \cos^2{\phi}$$ $$\implies \rho^2 \sin^2{\phi} \sin^2{\theta}= \rho^2 \cos^2{\phi}$$ $$\implies \sin^2{\phi} \sin^2{\theta}=\cos^2{\phi}\tag{2}$$

What is the difference between $(1)$ and $(2)$? It seems information has been lost in $(2)$.

Finally, what is the correct way to plot the intersection in Maple?

with(plots);  implicitplot3d(sin(p)^2*sin(t)^2 = cos(p)^2, r = 1 .. 1.1, t = 0 .. 2*Pi, p = 0 .. Pi, coords = spherical);

Attempt at plotting intersection of two cylinders

The above command produces a result which looks roughly correct, but something seems off. In particular, I had to specify a range for $\rho$, and that is making the graph a surface and not a curve. At least that is what I think. How do we get the correct curves in Maple?

If a function is measurable with respect to Prog then it is progressively measurable.

Posted: 17 Jul 2021 07:53 PM PDT

I have come across a statement I am not sure how to prove.

Assume you have a filtered probability space $(\Omega, \mathcal{F},\{\mathcal{F}_t\},P)$. A stochastic process $X_t$ is progressively measurable if for every $t\ge 0$, the function $f(\omega,s) $from $\Omega\times[0,t]$ to $\mathbb{R}$ given by $f(\omega,s)=X_s(\omega)$, is $\mathcal{F}_t\otimes\mathcal{B}([0,t])$ measurable.

Prog is the smallest sigma-algebra on $\Omega\times[0,\infty)$ such that all progressively measurable processes are Prog-measurable.

Obviously we have that every progressively measurable process is Prog-measurable. However, assume you have a function $g: \Omega\times[0,\infty)\rightarrow \mathbb{R}$, and assume that it is Prog-measurable, then it is stated that the stochastic process $Y_t$ defined by $Y_t(\omega)=g(\omega,t)$ is progressively measurable.

Do you know how to show this?

Attempt:

My only idea is using something like the Doob-Dynkin lemma. The Doob-Dynkin lemma states that if a function $f$ is $\sigma(h)$ measurable then there exists a borel function k, such that $f=k(h).$ The problem I have is that Prog is not generated by a single progressively measurable process it is generated by an infinite amount, so I don't think I can use the Doob-Dynkin lemma directly? But if I can prove that if $g$ is Prog-measurable, then it is some kind of limit or some kind of Borel function of the progressively measurable processes, then it will also be progressively measurable. Is there a way to do this?

One question on boundedness

Posted: 17 Jul 2021 07:11 PM PDT

Let $T$ closed unbounded operator, and $S\neq 0$, nor multiple of $I$ is bounded operator. If $TS=ST$, then is it true $TS$ is unbounded? If not, please give counterexample. Furthermore, under what condition on $S$, we always have $TS$ is bounded. More precisely, what is the necessary condition to have $TS$ is bounded? For instance, what can happen for $S$ unitary?

Comparison of critical regions of Multiple univariates and multivariate tests.

Posted: 17 Jul 2021 08:02 PM PDT

Let $X_i\sim\mathcal{N}(\mu_i,\sigma_i^2)$ for $i = 1,\ldots,n$. Also let $\boldsymbol{X} = (X_1,\ldots,X_n)^{\rm T}\sim MVN(\boldsymbol{\mu},\mathrm{\Sigma})$, where $\boldsymbol{\mu} = (\mu_1,\ldots,\mu_n)^{\rm T}$. I am expecting the following inequality to be true: $\mathbb{P}(|\sqrt{n}(\overline{X}_i-\mu_i)/\sigma_i|\leq \alpha:\;\mbox{for all}\;i=1,\ldots,n)\geq \mathbb{P}(n(\boldsymbol{X}-\boldsymbol{\mu})^{\rm T}\mathrm{\Sigma}^{-1}(\boldsymbol{X}-\boldsymbol{\mu})\leq \alpha^2)$. It can be proved for $n=2$ by comparing the regions inside the probabilities (first one is rectangle and the second is ellipse). Any hint to to prove it for any $n$?

Question about Legendre Symbol and Division Q12(a)

Posted: 17 Jul 2021 07:57 PM PDT

Let $q>3$ be an odd prime and let $q=2Q+1$. Prove that:

(a) $q\,|\,(3^Q-1) \iff q\equiv ±1 \pmod {12}$

adpated from Number Theory with Applications, James A. Anderson, James M. Bell. Ch 3.9 Q12 (a)

Any hint from the question? Thanks.

Edit 1: (DELETED BECAUSE IT IS WRONG) Edit 2: I have proved for the one-sided statement. Let me put it here first and wait for me to prove the another side of statement

Lemma 1: Iff $\exists x \in \mathbb{Z}^+, x>3: x^2\equiv 3 \pmod {p}$, then $p\equiv\pm1 \pmod{12}$

Proof:

Forward Statement:

By quadratic reciprocity, $(\frac{p}{3})=(\frac{3}{p})(-1)^{\frac{p-1}{2} \frac{3-1}{2}}=(-1)^{\frac{p-1}{2}}$

For $y^2\equiv p \pmod{3}$,

Case 1: If $(\frac{p}{3})=0$, then $3 \mid p \Rightarrow p \equiv 0 \pmod{3}$

Case 2: If $(\frac{p}{3})=1$, then by Fermat little thm, $y^2 \equiv 1 \pmod{3}$, $\therefore p \equiv 1 \pmod{3}$

On the other hand, $1=(-1)^{\frac{p-1}{2}} \Rightarrow p$ is even.

$\therefore$ By Chinese Remainder theorem, $p \equiv 1 \pmod{12}$

Case 3: If $(\frac{p}{3})=-1$, then $y^2 \equiv 2 \pmod{3}$, $\therefore p \equiv 2 \pmod{3}$

On the other hand, $-1=(-1)^{\frac{p-1}{2}}$

$\therefore \exists t \in \mathbb{Z}^+: \frac{p-1}{2} = 2t+1 \Rightarrow p-1=4t+2\Rightarrow p=4t+3\Rightarrow p\equiv 3 \pmod{4}$

By Chinese Remainder theorem, $p \equiv 11 \pmod{12}$

Combining the latter two cases, $p \equiv \pm 1 \pmod{12}$

Edit 3: Here is the proof of backward statement:

If $p$ is an odd prime, then $(\frac{p}{3}) \equiv p \pmod{3}$

Case 1: $p\equiv 1 \pmod{12} \Rightarrow p\equiv 1 \pmod{3}, \therefore (\frac{p}{3})=1$

Also, $\exists \alpha \in \mathbb{Z}^+: p-11=12 \beta \Rightarrow p=12\beta +11$

$(\frac{3}{p})=(\frac{p}{3}) (-1)^{\frac{p-1}{2}}=1(-1)^{\frac{12t}{2}}=1$

Case 2: $p\equiv 11 \pmod{12} \Rightarrow p\equiv -1 \pmod{3}, \therefore (\frac{p}{3})=-1$

$\therefore \exists \beta \in \mathbb{Z}^+ : p-11=12 \beta \Rightarrow p=12\beta +11$

$(\frac{3}{p})=(\frac{p}{3})(-1)^{\frac{12\beta+11-1}{2}}=(-1)(-1)=1$

Combining the above two cases, $(\frac{p}{3}) = 1 \pmod{3}$

Edit 4:

There are two cases: $q \equiv 1 \pmod{4}$ and $q\equiv 3 \pmod{4}$:

Case 1: $q \equiv 1 \pmod{4} \Rightarrow (\frac{q}{3}) = (\frac{3}{q})(-1)^{\frac{q-1}{2}} = 1, \therefore q\equiv 1 \pmod{3}$

$\therefore$ By Chinese Remainder thm, $q \equiv 1 \pmod{12}$

Case 2: $q \equiv 3 \pmod{4} \Rightarrow (\frac{q}{3}) = -1, \therefore q \equiv 2 \pmod{3}$

$\therefore$ By Chinese Remainder thm, $q \equiv 11 \pmod{12}$

(The converse way is similar to that in edit 3).

(Edit 5: I changed the above from "statement" to "lemma 1")

Edit 5:

By using Lemma 1,

$(\frac{3}{q}) = 1 \Leftrightarrow (\frac{3}{q}) \equiv 3^{\frac{p-1}{2}} \pmod{q} \Leftrightarrow 1 \equiv 3^{\frac{p-1}{2}} \pmod{p}$

$\therefore$ The thm is proved.

All conjugacy classes of $\operatorname{SL}_2 \mathbb{F}_p$

Posted: 17 Jul 2021 07:46 PM PDT

For an odd prime $p$, is the number of the conjugacy classes of $\operatorname{SL}_2 \mathbb{F}_p$ p+4 ?

I showed a partial result:

Let $A$ be a matrix. Consider its characteristic polynomial $p$. Then over $\mathbb{F}_{p^2}$, $p$ has roots. Write it $\xi, \zeta$. Then since the determinant is $1$, we have that $\zeta = \xi ^{-1}$. And since the trace is in $\mathbb{F}_p$, we have that $\xi + \xi ^{-1} \in \mathbb{F}_p$. Thus $(\xi^{p-1} - 1)(\xi ^{p+1} - 1)=0$

  1. The case that $\xi = \pm 1$. In this case we have that $A = \pm \left( \begin{matrix}1 & * \\ 0 & 1 \end{matrix} \right)$. (6 matrices are in this case.)

  2. The case that $\xi \not= \pm 1$ and $\xi^{p-1} = 1$. In this case $\xi \in \mathbb{F}_p$. So $A$ is conjugate to $\left( \begin{matrix}\xi & * \\ 0 & \xi \end{matrix} \right)$...? ( $(p-3) / 2$ matrices are in this case?)

  3. The case that $\xi \not= \pm 1$ and $\xi^{p+1} = 1$. ...?

From them how can I show the result?

Why does $\Omega_{ij} = \overline{\Omega}_{ji}$ only imply $\Omega$ is Hermitian in the finite dimensional case?

Posted: 17 Jul 2021 07:42 PM PDT

Using bra-ket notation:

Hermitian definition. Let us use the definition that $\Omega$ is Hermitian iff

$$ \langle v | \Omega | w \rangle = \overline{\langle w | \Omega | v \rangle} $$

instead of merely just

$$ \langle v | \Omega | w \rangle = \overline{\langle w | \Omega^{\dagger} | v \rangle} $$

for arbitrary vectors $|v\rangle$ and $|w\rangle$.

The finite case. Let $\Omega$ be a finite-dimensional matrix satisfying

$$ \Omega_{ij} = \overline{\Omega}_{ji} $$

then apparently this is enough for a $\Omega$ to be established as Hermitian. It seems to me that the reason this is true is that if we assume, by hypothesis, that

$$ \langle i|\Omega| j\rangle =\overline{\langle j|\Omega| i\rangle} $$

for $|i\rangle$ and $|j\rangle$ basis elements, then

\begin{aligned} \langle v|\Omega| \omega\rangle &=\sum_i v_i^{*}\langle i|\Omega \sum_j w_j|j\rangle \text{ (via basis expansion)}\\ &=\sum_{i} \sum_j v_{i}^{*}\left\langle i\left|\Omega\right| j\right\rangle w_j \text{ (via linearity)} \\ &=\sum_i \sum_j v_i^{*}\overline{\left\langle\left.\dot{j \mid} \Omega^{\dagger}\right|{i}\right\rangle} w_{j} \text{ (via hypothesis)} \\ &=\overline{\langle\omega|\Omega^\dagger| v\rangle} \end{aligned}

and thus $\Omega$ must be Hermitian.

The infinite case. Something analogous apparently does not hold in the infinite case. Page 65 of Principles of Quantum Mechanics presents a counter-example using the differential operator

$$ K = -iD $$

where $D$ is the differential operator:

enter image description here

Counter-proof. I understand the example presented in the textbook, but don't understand why similar reasoning to the finite case doesn't generate a contradictory proof that the theorem holds in the infinite case as well.

Suppose we assume by hypothesis that $K_{x x'} = \overline{K_{x'x}}$, or in the notation of the textbook that

$$ \left\langle x'\left|K^\dagger\right| x\right\rangle = \left\langle x'|K| x\right\rangle. $$

Then why can't we just say:

\begin{aligned} &\langle g|K| f\rangle \\ =& \int_a^b \int_a^b\langle g \mid x\rangle\left\langle x|K| x^{\prime}\right\rangle\left\langle x' \mid f\right\rangle \, d x \, d x' \\ =&\int_a^b \int_a^b \overline{\left\langle x| g\right\rangle}\overline{\left\langle x' \left|K^\dagger\right| x\right\rangle} \overline{\left\langle f \mid x^{\prime}\right\rangle} \, d x \, d x' \quad(\text {conjugate symmetry}) \\ =& \int_a^b \int_a^b \overline{\left\langle f \mid x^{\prime}\right\rangle\left\langle x'|K| x\right\rangle\langle x \mid g\rangle} \, d x \, d x' \quad\left(\text{hypothesis}\right) \\ =&\left(\int_{a}^{b} \int_a^b \left\langle f \mid x^{\prime}\right\rangle\left\langle x^{\prime}|K| x\right\rangle\langle x \mid g\rangle \, d x \, d x' \right)^{*} \\ =&\langle f|K| g\rangle^{*} \end{aligned}

This supposed proof (which I know must be wrong) basically mirrors the reasoning of the finite case. So what's wrong with it?

EDIT: There's been some discussion in the comments about complications that arise when looking at $\int_{-\infty}^{\infty}$ instead of $\int_a^b$, but I don't see how this supposed "counter-proof" would change if we replaced $a$ with $-\infty$ and $b$ with $\infty$. What step would suddenly become invalid?

Convergence of winning probability in a one-player dice-throwing game

Posted: 17 Jul 2021 07:24 PM PDT

In this (one-player) game, the player starts with a total of $n$ points. On each turn, they choose to throw either a four-, six-, or eight-sided die, and then subtract the number thrown from their point total. The game continues until the player's point total reaches zero exactly, and they win, or falls below zero, in which case they lose.

The strategy does not seem obvious. (Even for the simple case of $n=5$, one must do a calculation.) And dynamic programming techniques show that the strategy is not straightforward. For $n≤20$, the four-sided die is optimal, except for $n=5, 6, 13, 20$, where the six-sided die is optimal, and $n=8, 14, 16$, where the eight-sided die is optimal.

For large $n$, the probability of winning approaches $0.456615178766744$, which is not a number that I recognize. (The ISC does not recognize it either.)

It's not particularly surprising that the probability of winning converges as $n$ increases, because the probability of winning with $n$ points is the mean of probability of winning with $n-i$ points, taken over a small range of $i$. Considering the probabilities as a sequence, each element lies inside the range of the previous few elements, and tends to be in the middle of that range. As one goes farther out in the sequence, variations away from the mean tend to be damped out.

My questions are:

  • What's this $0.456615178766744$? (For a single $d$-sided die the probability of winning approaches $\frac 2{d+1}$, but for more dice the problem seems harder.)
  • Is there any regularity to the optimal move, as $n$ increases?
  • Is there any good way estimate a good strategy, short of exhaustive computer calculations? For example, in the $n=7$ situation, the four-sided die is significantly better than the six-sided die. Is there some way to see this, or at least to guess that it is so?
  • Someone must have studied this before. Does the problem have a name? Can someone give me a reference to the literature?

Evaluate integral: $\int_0^1 \frac{\log^3(1-x)}{x}dx$

Posted: 17 Jul 2021 08:03 PM PDT

I need your help to evaluate the following integral:

$$I=\int_0^1 \frac{\log^3(1-x)}{x}dx$$

Using the fact that for $|x|<1$ I get

$$\log(1-x)=-\sum_{n=1}^{\infty}\frac{x^k}{k}$$

One can rewrite $I$ as

$$I=-\sum_{k=0}^{\infty}\frac{1}{k+1}\left(\int_0^1 x^k\log^2(1-x)dx\right)$$

I tried to rewrite $\log(1-x)$ as a sum but I get some "monstrous" summation to calculate.

Poisson Distribution Problem Exam P

Posted: 17 Jul 2021 07:03 PM PDT

I am prepping for Exam P and getting a little confused on how to solve/set up part B?:

Claims filed in a year by a policyholder of an insurance company have a Poisson distribution with $\lambda=.40$. The number of claims filed by two different policyholders are independent events.

(a) If two policyholders are selected at random, what is the probability that each of them will file one claim during the year? (b) What is the probability that at least one of them will file no claims?

a). $[\exp(-0.4) * 0.4^1]/1! = \text{Probability 1 claim is filed} \implies$ $P(A \cap B) = 0.268128 * 0.268128 = 0.0719$

b). I'm thinking the answer is $[\exp(-0.4) * 0.4^0]/0! + [\exp(-0.4) * 0.4^1]/1! = 0.938448064$

But the book says it's $.8913$. What am I doing wrong?

Josephine problem

Posted: 17 Jul 2021 08:01 PM PDT

So the problem is Suppose there are 2n people in a circle; the first n are "good guys" and the last n are "bad guys." Show that there is always an integer m (depending on n) such that, if we go around the circle executing every mth person, all the bad guys are first to go.

I can see its true by simply visualizing n=8 k=2, so that if prisoners were at positions 2,4,6,8 then all the bad guys would be dead. but how can i generalize and make proof work for everthing.

Huffman codes: does less entropy imply less weighted average codeword length?

Posted: 17 Jul 2021 07:52 PM PDT

Let $\Sigma$ be a source alphabet with a probability distribution over its symbols $P$. Then, the Shannon entropy of $\Sigma$ is $$-\sum p_j \times \mbox{log}_2(p_j)$$ where $p_j$ is the probability of the j'th symbol. We know, thanks to the Shannon's source coding theorem, that this entropy is a (very close) lower bound of the minimum weighted average codeword length, i.e. the weighted average length of the codewords obtained with the Huffman algorithm.

The question: Let $P_1$ and $P_2$ be different probability distributions of the symbols of a given alphabet such that the shannon entropy of $P_1$ is smaller than the shannon entropy of $P_2$. Does this imply that the mininmum weighted average codeword length that we will be able to find for $P_1$ will be at most as large as the minimum weighted average codeword length that we will be able to find for $P_2$?

I am looking for either a counter-example of this or a proof that this is always the case.

Note: this is indeed true when the shannon entropy of $P_1$ is $\leq$ than the shannon entropy of $P_2 + 1$, as the Shannon's source coding theorem tells us that the entropy + 1 is an upper bound of the minimum weighted average codeword length. But when this is not the case (i.e. the difference is less than 1), it is not clear.

Find all cycles (faces) in a graph

Posted: 17 Jul 2021 07:58 PM PDT

I have a graph $G$ with a list of edges $E$, all the edges can form cycles, as shown below:

All the coordinates for the vertex $V$ are known.

Wat is the algorithm to find all the faces $F$? For the example above, there are 4 such faces, as indicated by the looping arrow.

Update: Why is it that the algorithm to find the face in the case without coordinate is different from the one with coordinates? Simply because by specifying the coordinates there is a unique set of faces, and without coordinates there could be sets of faces that are topologically equal. See the diagram below:

Update 2: Clear up the terminology

UPdate 3: It is suggested that I construct a cyclically linked list at each vertex, sorted according to the angles made between the incoming edge and the outgoing edge. Then one would just have to continue along the edge, but I afraid this algorithm doesn't work with the so called butterfly graph. Consider this graph:

Assume that we start at $e_1$, and the arrow denotes the direction of the movement. If we take the left turn as our orientation then we would be visiting $v_1$ twice before completing a cycle-- and the cycle is not really a cycle.

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