问:
Volume of region $\{ (x_1,\ldots,x_n)\in \mathbf{R}^n_{>0} \mid \prod_{i=1}^n\max(1,x_i)\leqslant d \}$
As part of a problem on algebraic number theory, I am faced with the following analysis/calculus exercise:

Show that the volume of the set
$$E_n:=\Big\{ (x_1,\ldots,x_n)\in \mathbf{R}^n_{>0} :\prod_{i=1}^n\max(1,x_i)\leqslant d \Big\}$$
is equal to $$d\sum_{i=0}^{n-1} \binom{n-1}{i} \frac{(\log d)^i}{i!}.$$

My attempt at solving this problem was as follows. Clearly, $E_n\subset [0,d]^n$. The volume of $E_n$ equals the following integral (abbreviating $m$ for $\max$):

$$\int_0^d \int_0^{\frac{d}{m(1,x_1)}}\cdots \int_0^{\frac{d}{m(1,x_1)\cdots m(1,x_{n-2})}} \int_0^{\frac{d}{m(1,x_1)\cdots m(1,x_{n-1})}} 1\, dx_n\,dx_{n-1}\cdots dx_2\,dx_1,$$
which can be simplified into
$$d\int_0^d \frac{1}{m(1,x_{1})} \int_0^{\frac{d}{m(1,x_1)}}\frac{1}{m(1,x_{2})}\cdots \int_0^{\frac{d}{m(1,x_1)\cdots m(1,x_{n-2})}} \frac{1}{m(1,x_{n-1})}\,dx_{n-1}\cdots dx_2\,dx_1.$$

I tried to prove the formula by induction. By splitting the middle integral, it evaluates to $1+\log d-\sum_{i=1}^{n-2}\log m(1,x_i)$. By linearity of the integral, we get $(1+\log d)E_{n-1}$ plus some other integral. However, this gets incredibly messy.

Is my approach correct? Is there any more clever way to attack this integral or an other method to prove this formula ? Is there maybe a probabilistic way of calculating its value ?

EDIT. In Theorem 6.5 of this article the integral is written as $\int_{y\in \mathbf{R}^n,\sum_i \max(0,y_i)\leq \delta} \exp \sum y_i\,dy$. Are my integral and this one equal ?

Any help is much appreciated.

(For the bounty: I would like to have a comprehensive, detailed answer.)

答: Assume $d>1$ (otherwise the volume is $0$). Call the volume $V_n(d)$ so$$V_n(d)=\int_0^1V_{n-1}(d)\mathrm{d}x+\int_1^dV_{n-1}(d/x)\mathrm{d}x=V_{n-1}(d)+\int_1^dV_{n-1}(x)\frac{d}{y^2}\mathrm{d}y,$$using $y=\frac{d}{x}$. Since $V_1(d)=d=d\sum_{i=0}^0\binom{0}{i}\frac{\ln^id}{i!}$, $n=1$ works. Now we use induction: if $V_k(d)=d\sum_{i=0}^{k-1}\binom{k-1}{i}\frac{\ln^id}{i!}$ then$$V_{k+1}(d)=d\sum_{i=0}^{k-1}\binom{k-1}{i}\frac{\ln^id}{i!}+d\int_1^d\frac1y\sum_{i=0}^{k-1}\binom{k-1}{i}\frac{\ln^iy}{i!}\mathrm{d}y.$$With $u=\ln y$, the integral is easy:$$\begin{align}V_{k+1}(d)&=d\sum_{i=0}^{k-1}\binom{k-1}{i}\frac{\ln^id}{i!}+d\sum_{i=0}^{k-1}\binom{k-1}{i}\frac{\ln^{i+1}d}{(i+1)!}\\&=d\sum_{i=0}^{k-1}\binom{k-1}{i}\frac{\ln^id}{i!}+d\sum_{i=1}^k\binom{k-1}{i-1}\frac{\ln^id}{i!}\\&=d\sum_{i=0}^k\binom{k-1}{i}\frac{\ln^id}{i!}+d\sum_{i=0}^k\binom{k-1}{i-1}\frac{\ln^id}{i!},\end{align}$$where binomial coefficients are defined combinatorially, so $\binom{r}{-1}=\binom{r}{r+1}=0$. Hence$$V_{k+1}(d)=d\sum_{i=0}^k\left(\binom{k-1}{i}+\binom{k-1}{i-1}\right)\frac{\ln^id}{i!}=d\sum_{i=0}^k\binom{k}{i}\frac{\ln^id}{i!}.$$


问:
Construct an entire function whose only zeroes are following
This question is from Ponnusamy and Silvermann Complex Variables pg 436, subsection wietestrauss product theorem

Question: Construct an entire function whose only zeroes are z=ln n.

Weierstrauss Product theorem guarentees that such a function exists, I thought of sin ($\pi z$) but it's zeroes will also be -1,-2,... . Similar problem is also for cos z and tan z.

Also, there are similar questions in same exercise wich ask for zeroes only at z=n of multiplicity n , zeroes only at z=$n^{3/4}$ .

Is there any general approach to these type of questions?

If yes kindly share. If no just answer 1 series asked.

Thanks!

Edit:there are similar questions in same exercise wich ask for zeroes only at z=n of multiplicity n.

There is an example before the exercise in which It is proved that an entire function woth zeroes only at z=+ve integers is $\prod_{n=1}^{\infty} (1-z/n) e^{z/n}$ and I adapted that proof to see if $\prod_{n=1}^{\infty} (1-z/n)^n e^{z/n}$ will be the polynomial asked in this but the mentioned product is not uniformly convergent product. So, I need an convergence producing factor which was $e^{z/n}$ in case of roots only at z=+ve integers. Can you please tell me how can I find such a convergence producing factor.

Also, I have solved the entire function with zeroes only at z=$n^{3/4}$ but not the above mentioned part in which I am unable to find the convergence producing factor.

Also ,this method used for entire function whose zeroes are only +ve integers can't be applied on ln (n) as the series $1\(ln n)^p $ diverges for all p .
So, No progress can be made on this part.

Kindly help.

答:
Since you changed your question.. Take $k_n$ such that  $$\sup_{|z|\le \frac12 \log n} |\log(1-z/\log n)+\sum_{m=1}^{k_n} \frac{(z/\log n)^m}{m}|\le 2^{-n}$$
(can you remind us why it exists ?)

Then $$\prod_{n\ge 2} (1-z/\log n) \exp(\sum_{m=1}^{k_n} \frac{(z/\log n)^m}{m})$$ is entire.

(can you remind us why ?)

ContentLicense: CC BY-SA 4.0


问:
Transform of a function of three random variables
Let $X$, $Y$, and $Z$ be independent random variables, where X is Bernoulli with parameter $1/3$, $Y$
is exponential with parameter $2$, and $Z$ is Poisson with parameter $3$.

(a) Consider the new random variable $U = XY + (1 - X)Z$ . Find the transform
associated with $U$.

(b) Find the transform associated with $2Z + 3$.

(c) Find the transform associated with $Y + Z$.

I know that I can get the transform of a sum of random variables by multiplying their transforms but I don't understand how I can get the complete transform of $U$.

答:
I don't understand how I can get the complete transform of U.

If $X=0$, the rv U is discrete and it is a poisson $Po(3)$. This happens with probabilty $\frac{2}{3}$

If $X=1$, the rv U is continuous and it is a $Exp(2)$. This happens with probabilty $\frac{1}{3}$

Thus

$$f_U(u)=\frac{2}{3}e^{-2u}\cdot\mathbb{1}_{[0;+\infty)}(u)+\frac{2}{3}\frac{e^{-3}3^u}{u!}\cdot\mathbb{1}_{\{0;1;2;3;...\}}(u)$$

Show your attempts for the rest of the questions (if you want that someone helps you)


问:
Derivative on parametric function. (Check my answer)

Let $x=t^2$ and $y=\sin 2t$. Determine $\dfrac{d^2y}{dx^2}$.

I have tried as below. Is this right or wrong answer?

\begin{align*}
  \dfrac{dy}{dx}&= \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{2\cos 2t}{2t}=
  \dfrac{\cos 2t}{t}.
\end{align*}
\begin{align*}
 \dfrac{d^2 y}{dx^2}&= \dfrac{d}{dx}\left(\dfrac{dy}{dx}\right) = \dfrac{dy'}{dx} =
 \dfrac{\dfrac{dy'}{dt}}{\dfrac{dx}{dt}} = \dfrac{\dfrac{-t \sin 2t-\cos 2t}{t^2}}{2t}=\dfrac{-t \sin 2t-\cos 2t}{2t^3}.
\end{align*}

答:
$$\frac{d^2y}{dx^2}= \dfrac{\dfrac{-(2)t \sin 2t-\cos 2t}{t^2}}{2t}=\dfrac{-(2)t \sin 2t-\cos 2t}{2t^3}.$$
Mind (2) in above which you have missed.


问:
How to integrate $\cos\left(\sqrt{x^2 + y^2}\right)$
Could you help me solve this?

$$\iint_{M}\!\cos\left(\sqrt{x^2+y^2}\right)\,dxdy;$$

$M: \frac{\pi^2}{4}\leq x^2+y^2\leq 4\pi^2$

I know that the region would look like this and I need to solve it as difference of two regions. Lets say $C=A\cup B$, then $A = C - B$

I was able to get the integrals, but don't know how to solve them.
$$\int_{0}^{2\pi}\!\int_{0}^{\sqrt{4\pi^2-x^2}}\!\cos\left(\sqrt{x^2+y^2}\right)\,dydx - \int_{0}^{\pi/2}\!\int_{0}^{\sqrt{\frac{\pi^2}{4}-x^2}}\!\cos\left(\sqrt{x^2+y^2}\right)\,dydx$$

答:
Change to polar coordinates. Thus $\sqrt{x^2+y^2}$ gets replaced by $r$. and $dx\,dy$ gets replaced by $r\,dr\,d\theta$.  So we end up needing to integrate $r\cos r\, dr\,d\theta$ over our region.

Note that the limits of integration for $r$ are $\frac{\pi}{2}$ and $2\pi$. If the region you are interested in is just the first quadrant region, as in the picture, then $\theta$ will run from $0$ to $\frac{\pi}{2}$.

The integration with respect to $r$ is done using integration by parts. Let $u=r$ and $dv=\cos r\,dr$. Then $du=dr$ and we can take $v=\sin r$. So
$$\int r\cos r\,dr=r\sin r-\int \sin r\, dr=r\sin r +\cos r+C.$$

Added: So for the first quadrant region, our integral is
$$\int_{\theta=0}^{\pi/2} \int_{r=\pi/2}^{2\pi} r\cos r\,dr\,d\theta.$$
The integration with respect to $\theta$ will be easy, since there is no dependence on $\theta$.

ContentLicense: CC BY-SA 3.0


问:
Finding the indefinite integral of $\int_0^\pi\sqrt{\sin^3x - \sin^5x}dx$
It's not hard to find that $\int\sqrt{\sin^3x - \sin^5x}dx = \frac{2\sin^{5/2}x}{5} + C$,

But the indefinite integral $\int_0^\pi\sqrt{\sin^3x - \sin^5x}dx = \frac{4}{5}$ does not conform to $\cfrac{2\sin^{5/2} \pi}{5} - \cfrac{2\sin^{5/2} 0}{5} = 0 - 0 = 0$

Why is that we can't simply use Newton-Leibniz axiom here and what is the way to find the value $\frac{4}{5}$ here ?

答: Your computation of the indefinite integral  is not correct. The square root sign in the integrand is the positve square root. You are assuming that $\cos x$ is positive in your computation. Note that $\cos x$ is positive for $x$ between $0$ and $\frac {\pi } 2$ but negative for $x$ between $\frac {\pi} 2$ and $\pi$.


答:
Use $\int_{0}^{2a} f(x) dx=2\int_{0}^{a} f(x) dx....(1)$ if $f(2a-x)=f(x)$, then
$$I=\int_{0}^{\pi} \sqrt{\sin^3 x-\sin ^5 x} dx= 2\int_{0}^{\pi/2} \sqrt{\sin^3 x-\sin ^5 x} dx \int_{0}^{\pi/2} (\sin x)^{3/2} \cos x dx$$
$$\implies I =2\int_{0}^{1} t^{3/2} dt=\frac{4}{5}.$$

Note: Always try to reduce the domain of integration if possible by using (1). Other wise $\sin x=t$ is not a good substitution in $(0, \pi)$ because both the limits on $t$ become zero. But it is a good substitution in $(0,\pi/2)$.

Title: Finding the indefinite integral of $\int_0^\pi\sqrt{\sin^3x - \sin^5x}dx$

Body:
It's not hard to find that $\int\sqrt{\sin^3x - \sin^5x}dx = \frac{2\sin^{5/2}x}{5} + C$,

But the indefinite integral $\int_0^\pi\sqrt{\sin^3x - \sin^5x}dx = \frac{4}{5}$ does not conform to $\cfrac{2\sin^{5/2} \pi}{5} - \cfrac{2\sin^{5/2} 0}{5} = 0 - 0 = 0$

Why is that we can't simply use Newton-Leibniz axiom here and what is the way to find the value $\frac{4}{5}$ here ?


问:
Show that $x^4-2$ is the minimal polynomial for $\sqrt[4]{2}$ over $\mathbb{Q}$
How do I prove that there is no other polynomial $p$ which is normed and of less or equal degree than $4$ which satisfies the equation $p(\sqrt[4]{2})=0$ ?

Is there some sort of algorithm one can use to solve those kind of questions?

I did not cover any galois theory. We just heared what algebraic and trancendent is

答:
Our problem is equivalent to showing that $1,\sqrt[4]{2}, \sqrt{2}$ and $\sqrt[4]8$ are linearly independent. Assume the contrary, then we have rational $a,b,c$ for which:
$$\sqrt[4] 8 = a\sqrt2+b\sqrt[4]2+c$$
Raising the whole equation to the power of $4$ gives a value in $\mathbb{Q}[\sqrt[4]2,\sqrt2]$. Hence, we would see that the values of $\sqrt[4]2, \sqrt2$ and $1$ are not linearly independent either. Putting $\sqrt[4]2$ in the LHS and repeating the process again will imply that $\sqrt2$ and $1$ are not linearly independent i.e. $\sqrt 2$ is rational, which is a contradiction.

Now, since $1,\sqrt[4]2,\sqrt2,\sqrt[4]8$ are linearly independent. There does not exist a polynomial of degree less than or equal to $3$ with $\sqrt[4]2$ as a root. Hence, the minimal polynomial of $\sqrt[4]2$ has degree $4$. Moreover, the polynomial $x^4-2$ is also unique upto scaling (else, you can use two minimal polynomials of degree $4$ and subtract them after scaling them appropriately and get a smaller minimal polynomial).


问:
Prove that the following lines so described are perpendicular.
If O be any point on the circumcircle of the triangle ABC, and OL be drawn parallel to BC to meet the circumcircle in L. OD and OF are drawn perpendicular to BC and AB respectively, then show that LA will be perpendicular to DF.

The above question is from my self-preparatory material for an exam(JEE). I tried a lot  thinking on this but I don't know from which theorem or property of triangles and circles to start with.
Any suggestion on how to tackle such problems will be appreciated.

答:
In quadrilateral AFGH we have:

$\angle AHG= \frac{Arc{GCA}}2$

$\angle AFG= \frac{Arc{AOG}}2$

Summing sides we get:

$\angle AHG+\angle AFG=\frac{Arc{GCA}}2+\frac{Arc{AOG}}2=180^0$

$\angle HGF= \frac{Arc{HOF}}2$

$\angle HAF= \frac{Arc{FCB}}2$

Summing both sides we get:

$\angle HGF+\angle HAF=\frac{Arc{HOF}}2+\frac{Arc{FCB}}2=180$

That is the sum of opposite angles of quadrilateral AFGH are equal which means it is an isosceles trapezoid which in turn means AH is parallel with FG. But $\angle HAL =90^o$(can you find why?) so $\angle DEL=90^o$ and this is what we had to show.

Additional observation: If B moves to H such that condition OL||BC reserved , B, D and H will be coincident. In this case while B moves on circle, D moves on extension of OH such that HD keeps being perpendicular on BC. Since HA is perpendicular on AL, it is reasonable to accept that AH keep being perpendicular on AL in its new positions including DF. In fact DF is transform of AH when B moves from H to it's own position.This shows that DE||AH.


问:
Distribution of ratio of sample variances
I have two independent zero mean populations (not necessarily normally distributed) and I am trying to estimate the ratio of variances of the two population. I have i.i.d. sample from each population of size $n$.
I do know that sample variance converges to population variance almost surely, so by Slutsky theorem, ratio of variances will converge to ratio population variances almost surely. Thus, ratio of sample variances is a consistent estimator.

But what can I say about it's asymptotic distribution, of ratio of sample variances without knowing the distribution of the population or assuming it to be normal?
I do know when population distribution is normal then it follows $F$ distribution with $(n-1,n-1)$ degrees of freedom. But what about the general case?

Thanks in Advance

答:
Let

$$ S_X^2 = \frac{1} {n} \sum_{i=1}^n X_i^2, S_Y^2 = \frac{1} {n} \sum_{i=1}^n Y_i^2$$

be the sample variance estimator of the variances $\sigma_X^2$ and $\sigma_Y^2$ respectively.

By the bivariate CLT,

$$ \sqrt{n} \left(\begin{bmatrix} S_X^2 \\ S_Y^2 \end{bmatrix} -
\begin{bmatrix} \sigma_X^2 \\ \sigma_Y^2 \end{bmatrix} \right)
\stackrel {d} {\to} \mathcal{N}
\begin{bmatrix} Var[X_1] & 0 \\ 0 & Var[Y_1] \end{bmatrix} $$

where

$$ Var[X_1] = E[X_1^4] - E[X_1^2]^2 = E[X_1^4] - \sigma_X^4$$
$$ Var[Y_1] = E[Y_1^4] - E[Y_1^2]^2 = E[Y_1^4] - \sigma_Y^4$$

Next we will need the bivariate Delta method to obtain the asymptotic distribution of $\displaystyle \frac {S_X^2} {S_Y^2}$

See, e.g. https://en.wikipedia.org/wiki/Taylor_expansions_for_the_moments_of_functions_of_random_variables

Consider $\displaystyle f(x, y) = \frac {x} {y}$. Then

$$ \frac {\partial f} {\partial x} = \frac {1} {y}, ~~
\frac {\partial f} {\partial y} = - \frac {x} {y^2}, ~~
\frac {\partial^2 f} {\partial x^2} = 0, ~~
\frac {\partial^2 f} {\partial y^2} = \frac {2x} {y^3}, ~~
\frac {\partial^2 f} {\partial x \partial y} = - \frac {1} {y^2}$$

We Taylor Expand $\displaystyle \frac {S_X^2} {S_Y^2} = f(S_X^2, S_Y^2)$ about the mean $(\sigma_X^2, \sigma_Y^2)$ up to the second order:

$$ \begin{align} f(S_X^2, S_Y^2) \approx &~ f(\sigma_X^2, \sigma_Y^2)
 + \left.\frac {\partial f} {\partial x}\right|_{(\sigma_X^2, \sigma_Y^2)}
(S_X^2 - \sigma_X^2)
 + \left.\frac {\partial f} {\partial y}\right|_{(\sigma_X^2, \sigma_Y^2)}
(S_Y^2 - \sigma_Y^2) \\
&~ + \frac {1} {2} \left.\frac {\partial^2 f} {\partial x^2}\right|_{(\sigma_X^2, \sigma_Y^2)}(S_X^2 - \sigma_X^2)^2
 + \frac {1} {2} \left.\frac {\partial^2 f} {\partial y^2}\right|_{(\sigma_X^2, \sigma_Y^2)}(S_Y^2 - \sigma_Y^2)^2 \\
&~ + \left.\frac {\partial^2 f} {\partial x \partial y}\right|_{(\sigma_X^2, \sigma_Y^2)}(S_X^2 - \sigma_X^2)(S_Y^2 - \sigma_Y^2) \\
\Rightarrow \frac {S_X^2} {S_Y^2} \approx &~ \frac {\sigma_X^2} {\sigma_Y^2}
 + \frac {1} {\sigma_Y^2} (S_X^2 - \sigma_X^2)
 - \frac {\sigma_X^2} {\sigma_Y^4} (S_Y^2 - \sigma_Y^2) \\
&~ + 0 + \frac {\sigma_X^2} {\sigma_Y^6} (S_Y^2 - \sigma_Y^2)^2
- \frac {1} {\sigma_Y^4} (S_X^2 - \sigma_X^2)(S_Y^2 - \sigma_Y^2)
\end{align} $$

Taking expectation,

$$ E\left[\frac {S_X^2} {S_Y^2}\right]
\approx \frac {\sigma_X^2} {\sigma_Y^2} + 0 + 0
 + \frac {\sigma_X^2} {\sigma_Y^6} Var[S_Y^2] - 0
= \frac {\sigma_X^2} {\sigma_Y^2}
 + \frac {\sigma_X^2(E[Y_1^4] - \sigma_Y^4)} {n\sigma_Y^6}  $$

and variance (with first order only)
$$ Var\left[\frac {S_X^2} {S_Y^2}\right]
\approx 0 + \frac {1} {\sigma_Y^4}Var[S_X^2] + \frac {\sigma_X^4} {\sigma_Y^8}Var[S_Y^2] = \frac {E[X_1^4] - \sigma_X^4} {n\sigma_Y^4}
 + \frac {\sigma_X^4(E[Y_1^4] - \sigma_Y^4)} {n\sigma_Y^8}$$

So by Delta method, collecting the terms,

$$ \sqrt{n} \left(\frac {S_X^2} {S_Y^2} - \frac {\sigma_X^2} {\sigma_Y^2}
\right) \stackrel {d} {\to} \mathcal{N} \left(0,
\frac {E[X_1^4] - \sigma_X^4} {\sigma_Y^4}
 + \frac {\sigma_X^4(E[Y_1^4] - \sigma_Y^4)} {\sigma_Y^8} \right)$$

The second order term here just serve as a correction term providing higher accuracy if needed.


问:
Determinant and inverse matrix calculation of a special matrix
Is there any smart way of calculating the determinant of this kind matrix?

\begin{pmatrix}
1 & 2 & 3 & \cdots &  n \\
2 & 1 & 2 & \cdots & n-1 \\
3 & 2 & 1 & \cdots & n-2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n & n-1 & n-2 & \cdots & 1 \end{pmatrix}

I encountered this in a problem for the case $n=4$. I need to find the inverse matrix.

I doubt the idea of this problem is to calculate all the cofactors and then the inverse matrix in the usual way.

I see the pattern here and some recursive relations... but I am not sure if this helps for calculating the determinant of the existing matrix.

答:
Call your matrix $A$. Let $\{e_1,e_2,\ldots,e_{n-1}\}$ be the standard basis of $\mathbb R^{n-1}$. Let also $e=\sum_ie_i=(1,1,\ldots,1)^T$ and
$$
L=\pmatrix{1\\ -1&1\\ &\ddots&\ddots\\ &&-1&1}.
$$
Then $B:=LAL^T=\pmatrix{1&e^T\\ e&-2I_{n-1}}$. Using Schur complement, we obtain
$$
\det(A)=\det(B)=\det(-2I_{n-1})\left(1-e^T(-2I_{n-1})^{-1}e\right)=(-2)^{n-1}\frac{n+1}{2}.
$$
To find $A^{-1}$, partition $L$ as $\pmatrix{1&0\\ -e_1&L'}$. Then
\begin{aligned}
A^{-1}
&=L^TB^{-1}L\\
&=\pmatrix{1&-e_1^T\\ 0&L'^T}
\pmatrix{\frac{2}{n+1}&\frac{1}{n+1}e^T\\ \frac{1}{n+1}e&\frac{1}{2(n+1)}ee^T-\frac12I_{n-1}}
\pmatrix{1&0\\ -e_1&L'}\\
&=\pmatrix{\frac{1}{n+1}&\frac{1}{2(n+1)}e^T+\frac12e_1^T\\ \ast&\frac{1}{2(n+1)}e_ne^T-\frac12L'^T}
\pmatrix{1&0\\ -e_1&L'}\\
&=\pmatrix{\frac{-n}{2(n+1)}&\frac{1}{2(n+1)}e_n^T+\frac12e_1^T\\ \ast&\frac{1}{2(n+1)}e_ne_n^T-\frac12L'^TL'}.
\end{aligned}
By direct calculation, $L'^TL'$ is the symmetric tridiagonal matrix whose main diagonal is $(2,\ldots,2,1)$ and whose superdiagonal is a vector of $-1$s. Since $A^{-1}$ is symmetric, we finally get
$$
A^{-1}=\pmatrix{
\frac{-n}{2(n+1)}&\frac12&&&\frac{1}{2(n+1)}\\
\frac12&-1&\frac12\\
&\frac12&-1&\ddots\\
&&\ddots&\ddots&\frac12\\
\frac{1}{2(n+1)}&&&\frac12&\frac{-n}{2(n+1)}
}.
$$


问:
$\sum_{\alpha<\omega_2}|\alpha|^{\aleph_0}=\aleph_2\cdot\aleph_1^{\aleph_0}$
I've seen this statement in multiple posts (e.g. here and here), but I can't seem to understand it. I can see why

$$\sum_{\alpha<\omega_2}|\alpha|^{\aleph_0}=\aleph_1^{\aleph_0},$$

by noting that every $\alpha<\omega_2$, so $|\alpha|\leq\aleph_1$, meaning

$$\sum_{\alpha<\omega_2}|\alpha|^{\aleph_0}=\max_{\alpha<\omega_2}|\alpha|^{\aleph_0}=\aleph_1^{\aleph_0}.$$

So how does $\aleph_2$ enter the equation?

答: The sum has $\aleph_2$ terms, each of which is at most $\aleph_1^{\aleph_0}$, so it’s bounded above by $\aleph_2\cdot\aleph_1^{\aleph_0}$. On the other hand, it’s clearly at least $\aleph_2$ and at least $\aleph_1^{\aleph_0}$, so its bounded below by their maximum, which is simply their product.</p>


问:
error function derivation
From contour integration, we get:
$$f(x,t) = \frac{1}{\pi}\int_{0}^{\infty} \exp(-ut) \sin(\sqrt\frac{u}{k} x) \frac{du}{u}$$
Then my instructor just wrote:

$$f(x,t) = \left(\text{erf} \left(\frac{x}{2\sqrt{kt}}\right)\right) $$

My instructor did a couple of confusing substitutions and then just wrote that the 1st equation is equal to the second equation.
Could someone clarify how my instructor got from the 1st equation to the 2nd equation?

答:
What I suppose your  instructor did is the following.

He probably considered
$$I=\Im \Bigg[ \int  e^{- u t} \,e^{i x \sqrt{\frac{u}{k}}}\,\frac {du}u\Bigg]$$ combined the two exponentials together, make $tu=v^2$ to arrive at
$$2\int \exp\left(-v^2+ \frac{ix}{\sqrt{k t}}v\right)\,\frac {dv}v$$ There is no antiderivative available (but the integral between $0$ and $\infty$ must exist.

In any manner, you see here all the elements required to arrive to the error function.

Sorry but I do not have the time to finish (may be later).

Title: error function derivation


问:
Natural deduction from non-first-order premises
Consider the following premises in a symbolic expression:
$$
A, B \subseteq \mathbb{R}\\
\forall x \in A: \exists y \in B: y \leq x\\
|A| < \infty
$$
Then, we can conclude that $\exists y \in B: \forall x \in A : y \leq x $. Note that the order of the bounded quantifiers is exchanged and this is not a first-order consequence.

One of the proofs can be informally stated that for each element of $A$, we can pick an element from $B$ that is not larger than $x$. Since $A$ is a finite set, the smallest element of $B$ is not larger than every element of $A$.

Can this proposition be proved by natural deduction? I tried to show this using rules of inference presented in a Fitch form in vain.

答:
As Mauro said, you did not express "$A$ is finite" in a logical form. So how would it be possible to do anything with it? Suppose you use this Fitch-style system. Then the simplest way to express it would be as follows.

Given any set $A$, define $A$ to be finite iff:

  $∃c{∈}ℕ\ ∃f{∈}FN(ℕ_{<c},A)\ ∀x{∈}A\ ∃k{∈}ℕ_{<c}\ ( \ f(k)=x \ )$.

In English, $A$ is finite iff there is some $c{∈}ℕ$ and some surjection from $ℕ_{<c}$ onto $A$. Here $ℕ_{<c}$ is defined as $\{ k : k{∈}ℕ ∧ k<c \}$. If you want to minimize set theory, you can use the following alternative (which is equivalent over set theory).

Given any set $A$, define $A$ to be finite iff:

  $∃c{∈}ℕ\ ∃f{∈}FN(ℕ,A)\ ∀x{∈}A\ ∃k{∈}ℕ\ ( \ k<c \land f(k)=x \ )$.

You are correct that you cannot automatically swap the quantifiers. To prove your desired theorem, you would need the induction rule (see under "Set Theory" in the linked post). See if you can find a proof!


问:
operator 2 norm and symmetric positive semidefinite matrix
Let $A$ be a symmetric positive semidefinite matrix and $u$ be a unit vector.

Then $u^TAu \leq \|A\|_2$ ?

Here $\Vert \cdot \Vert_2$ is the induced/operator 2-norm defined as

$\| A \|_2 = \sup \limits _{x \ne 0} \frac{\| A x\| _2}{\|x\|_2}$

答:
Yes, we have $u^TAu=\langle Au,\overline{u}\rangle\le\|Au\|_2\|\overline{u}\|_2\ \text{(Cauchy-Schwarz)}=\|Au\|_2\le\|A\|_2$.
This is true for every complex square matrix $A$. We don't need $A$ to be positive semidefinite.


问:
How to show the Hamiltonian is conserved
I have the langrangian $L = \frac 1 2m((1+4r^2)r’^2) + \frac 1 2 mr^2 θ’^2 -mgr^2  $

I worked out theHamiltonian and 4 Hamiltonian equations and got

$\frac {dpθ}{dt} = 0 $ for one of them $pθ$ is the momentum of the angle

Is that enough to show that the Hamiltonian is conserved.

答:
In the Euler-Lagrange equations with momentum variables, you get
$\newcommand{\pd}[2]{\frac{\partial#1}{\partial#2}}$
\begin{align}
p_r&=\pd{L}{\dot r}=m(1+r^2)\dot r\\
p_θ&=\pd{L}{\dot θ}=mr^2\dot θ
\end{align}
and then
\begin{align}
\dot p_r&=\pd{L}{r}=4mr\dot r^2+mr\dot θ^2-2mgr\\
\dot p_θ&=\pd{L}{θ}=0
\end{align}
For the Hamiltonian you should get
$$
H=p_r\dot r+p_θ\dot θ-L=\frac{p_r^2}{2m(1+4r^2)}+\frac{p_θ^2}{2mr^2}+mgr^2.
$$
It is now easy to check that $\dot q=H_p$ and $\dot p=-H_q$ gives the same equations as the Euler-Lagrange formalism, so that indeed $\dot H=H_p\dot p+H_q\dot q=0$.


问:
Interprete Banach spaces as subspace of the space of continuous functions on a compact space
At the end of the section of the Banach-Alaoglu theorem my professor wrote in his script:

"The B-A theorem can be used to prove that for every Banach space $X$ there exists a compact space $K$ and an isometric embedding $T: X \hookrightarrow C(K)$, i.e $T$ satiesfies $ ||Tx|| = ||x||$ but $T$ is not necessarily surjective. Therefore any Banach space can be realized as a subspace of the space of continuous functions on a compact space. Proving that is given as an exercise."

As it seems to follow directly from the Banach Alaoglu theorem I thought of using the closed unit ball in $X^{\star}$ with respect to the weak-$\star$ topology as $K$. When I am now looking at functions on this $K$ it seems to be related to elements in the bidual space, as we are mapping from $X^{\star}$. But I dont see why that should end up in an isometric embedding. What am I missing here and how could I prove the statement?

答:
Let $K$ be the closed unit ball of $X^{*}$ with its weak* topology. This is a compact Hausdorff space. Provide $C(K)$ with the usual sup norm.  Consider the map $\phi: X \to C(K)$ defined by $\phi (x) (x^{*})=x^{*}(x)$. This is a well-defined linear map and $\|\phi (x)\|_{C(K)} =\sup \{|x^{*}(x)|: \|x^{*}\| \leq 1\}=\|x\|$.

[Note that $|x^{*}(x)| \leq \|x^{*}\| \|x\| \leq \|x\|$ for all $x ^{*} \in K$. By Hahn Banach Theorem there exist $x^{*}$ such that $\|x^{*}\|=1$ and $|x^{*}(x)|=\|x\|$. Hence the supremum is exactly $\|x\|$.]


问:
Show that for $a,b>0$ and $a \ne b$ and $ n > 1$ we have that $a^n-b^n < n(a-b)a^{n-1}$
Show that for $a,b>0$ and $a \ne b$ and $ n > 1$ we have that:

$$a^n-b^n < n(a-b)a^{n-1}$$

For $1 \le k \le n-1$ : $$a^{n-1-k}b^k < a^{n-1}$$ $$\iff b^k <a^k$$ $$\iff a <b$$

And for $k=0$ the equality happens.
And so :

$$\sum_{k=0}^{n-1}a^{n-1-k}b^{k}<na^{n-1}$$

$$\left(a-b\right)\sum_{k=0}^{n-1}a^{n-1-k}b^{k}<n\left(a-b\right)a^{n-1}$$
$$a^{n}-b^{n}<n\left(a-b\right)a^{n-1}$$

I think the relation only holds if $a <b$ ,am I right?

Body:
Let $f(x)=x^n$ and $a>b$, $n>1$ Let us apply LMVT to f(x) in $(b,a)$, then

$$\frac{a^n-b^n}{a-b}=nc^{n-1}, c\in (b,a)~~~~(1)$$
next $$ b<c<a \implies c^{n-1}< a^{n-1}~~~~(2)$$
Combining (1) and (2), we get
$$a^n-b^n< n(a-b)a^{n-1}.$$


问:
Miller primality test bound
Good morning!

I'm on my way to implement a deterministic (though unproven due to GRH) Miller primality test.
On Wikipedia, it is said that it suffices to test all numbers in $[2, \lfloor\times2\ln^2n\rfloor]$

However, there is a problem that I only have bigint arithmetics and can afford neither a true natural logarithm function nor fractional numbers at all.

Ultimately, I only know the amount of bits in the tested number, which is actually $\lfloor\log_2n\rfloor+1$ (am I correct?).

Can anyone advise me what to do with this value to get as close as possible to the upper bound of Miller test, which is $\lfloor2\times\ln^2n\rfloor$, without getting away from the ring of integers?

答:
Frankly, the calculation of one logarithm takes so little time compared to doing so many tests that I'm not sure why you'd bother.  Probably the best approach in your case is to use a lookup table that tells you how many tests you need -- that way you can use A006945 which allows you to use 10 tests below 3825123056546413051 rather than the 3661 you'd need otherwise.

But if you'd really like to do this directly, $\ln x=\log_2x/\log_2e$ and you can use bignum multiplication with something like
$$\lceil2^n\ln2\rceil$$
for sufficiently large n, then shift back by n, then add 1.

Title: Miller primality test bound


问:
Exponential generating function with factorial.
Can't solve this example:

$$a_{n+1} = 2(n+1)a_n+(n+1)!$$

$$\sum \frac{a_{n+1}z^{n+1}}{(n+1)!} = \sum \frac{2(n+1)a_n z^{n+1}}{(n+1)!} + \sum \frac{(n+1)! z^{n+1}}{(n+1)!}$$

$$F(z)- a_0 = 2zF(z) + ...???...$$

I just don't guess how I can transform last term
($\sum (n+1)!(z^{n+1})/(n+1)!$) according list of generating functions. If you know how I can do it-tell me please!)

答:
As you've noticed, if we write

$$F(z) \triangleq \sum \frac{a_n}{n!}z^n$$

for the exponential generating function of $a_n$, then the recurrence becomes:

$$F(z) - a_0 = 2z F(z) + \sum \frac{(n+1)!}{(n+1)!} z^{n+1}$$

Of course, we can cancel the $(n+1)!$ terms to see

$$F(z) - a_0 = 2z F(z) + \sum z^{n+1}$$

It is at this point that one should recall $\frac{1}{1-z} = \sum z^n$, and so

$$F(z) - a_0 = 2z F(z) + \frac{z}{1-z}$$

Do you see how to finish it from here?

I hope this helps ^_^


问:
Elliptic integral $\int_0^1\int_b^c \frac{u}{\sqrt{1-u^2t^2}\sqrt{1-(u-a)^2}}dudt$
I am trying to find the solution to:

$$I=\int_0^1\int_b^c \frac{u}{\sqrt{1-u^2t^2}\sqrt{1-(u-a)^2}}dudt,$$
this definitely has the form of an elliptic integral, but for the life of me I can't figure out how to reduce this.

It's similar in form to $$F(\phi,t)=\int^{\sin\phi} \frac{du}{\sqrt{1-u^2t^2}\sqrt{1-u^2}},$$ but the factor of $u$ and the shifted argument make this difficult.

I have tried integration by parts ($[fg]_b^c-\int_b^cgdf=\int_b^cfdg$) where $$f=u$$ and $$dg=\frac{1}{\sqrt{1-u^2t^2}\sqrt{1-(u-a)^2}}.$$ Letting $$g=\mathcal{G}(u,t):=\int^u\frac{dx}{\sqrt{1-x^2t^2}\sqrt{1-(x-a)^2}},$$ and $df=du$ we have
$$\int_b^c \frac{u}{\sqrt{1-u^2t^2}\sqrt{1-(u-a)^2}}du=\left[u\mathcal{G}(u,t)\right]^c_b-\int_b^c\mathcal{G}(u,t)du,$$
thus $$I=\int_0^1\left[u\mathcal{G}(u,t)\right]^c_bdt-\int_0^1\int_b^c\mathcal{G}(u,t)dudt.$$
The closest I get to an expression for $\mathcal{G}(u,t)$ is
$$\mathcal{G}(u,t)=-\int^u\frac{d\phi}{\sqrt{1-(\cos\phi+a)^2t^2}}.$$ Mathematica gives a very complicated expression but I was wondering if there was anything simpler?

答:
This is not an elliptic integral. Perform the integration with respect to $t$ first; this is easy and leads to
$$I=\int_b^c\frac{\sin^{-1}u}{\sqrt{1-(u-a)^2}}\,du$$
If $a=0$ the indefinite integral in turn solves to $(\sin^{-1}u)^2/2$, but if $a\ne0$ there is no closed-form antiderivative.


答:
I'm going to elaborate on Kavi's answer:

The problem here is that the antiderivative $\frac{2}{5}\sin^{\frac{5}{2}}(x)+C$ is only valid when $x\in\left[0,\frac{\pi}{2}\right]$.

Looking at the Desmos plot I provided, it can be seen that the derivative of $\frac{2}{5}\sin^{\frac{5}{2}}(x)$ does not agree with $\sqrt{\sin^3(x)-\sin^5(x)}$ when $x\in\left(\frac{\pi}{2},\pi\right)$. This is because

\begin{align*}
\sqrt{\sin^3(x)-\sin^5(x)} &= \sqrt{\sin^3(x)\left[1-\sin^2(x)\right]}\\
&= \sin^{\frac{3}{2}}(x)\sqrt{\cos^2(x)}\\
&= \sin^{\frac{3}{2}}(x)|\cos(x)|
\end{align*}

and $\cos(x)$ is negative when $x\in\left(\frac{\pi}{2},\pi\right)$.

Don't lose hope, though! We can still find the correct antiderivative and apply $\text{FTC}2$. To begin, note that

$$\sqrt{\sin^3(x)-\sin^5(x)}$$

is continuous over $\left[0,\pi\right]$, so $\text{FTC}1$ asserts that the function

$$f(x)=\int_{0}^{x}\sqrt{\sin^3(t)-\sin^5(t)}\text{ }dt$$

is an antiderivative of $\sqrt{\sin^3(x)-\sin^5(x)}$. We can now boil down the problem of finding an antiderivative to finding a nice expression for this integral.

Using the identity $\sqrt{\sin^3(x)-\sin^5(x)}=\sin^{\frac{3}{2}}(x)|\cos(x)|$, we can write

$$f(x)=\int_{0}^{x}\sin^{\frac{3}{2}}(t)|\cos(t)|\text{ }dt$$

so for $x\in\left[0,\frac{\pi}{2}\right]$,

\begin{align*}
\int_{0}^{x}\sin^{\frac{3}{2}}(t)|\cos(t)|\text{ }dt &= \int_{0}^{x}\sin^{\frac{3}{2}}(t)\cos(t)\text{ }dt\\
&= \int_{0}^{\sin(x)}u^{\frac{3}{2}}\text{ }du\\
&= \frac{2}{5}\sin^{\frac{5}{2}}(x)
\end{align*}

For $x\in\left[\frac{\pi}{2},\pi\right]$,

\begin{align*}
\int_{0}^{x}\sin^{\frac{3}{2}}(t)|\cos(t)|\text{ }dt &= \int_{0}^{\frac{\pi}{2}}\sin^{\frac{3}{2}}(t)|\cos(t)|\text{ }dt+\int_{\frac{\pi}{2}}^{x}\sin^{\frac{3}{2}}(t)|\cos(t)|\text{ }dt\\
&= \frac{2}{5}\sin^{\frac{5}{2}}\left(\frac{\pi}{2}\right)-\int_{\frac{\pi}{2}}^{x}\sin^{\frac{3}{2}}(t)\cos(t)\text{ }dt\\
&= \frac{2}{5}-\int_{1}^{\sin(x)}u^{\frac{3}{2}}\text{ }du\\
&= \frac{2}{5}-\left(\frac{2}{5}\sin^{\frac{5}{2}}(x)-\frac{2}{5}\right)\\
&= \frac{4}{5}-\frac{2}{5}\sin^{\frac{5}{2}}(x)
\end{align*}

Thus,

$$f(x)=\begin{cases}
\frac{2}{5}\sin^{\frac{5}{2}}(x) & \text{if }x\in\left[0,\frac{\pi}{2}\right]\\
\frac{4}{5}-\frac{2}{5}\sin^{\frac{5}{2}}(x) & \text{if }x\in\left[\frac{\pi}{2},\pi\right]
\end{cases}$$

We can now (legally) evaluate the original integral with $\text{FTC}2$.

\begin{align*}
\int_{0}^{\pi}\sqrt{\sin^3(x)-\sin^5(x)}\text{ }dx &= f(\pi)-f(0)\\
&= \frac{4}{5}-\frac{2}{5}\sin^{\frac{5}{2}}(\pi)-\frac{2}{5}\sin^{\frac{5}{2}}(0)\\
&= \frac{4}{5}
\end{align*}


问:
$X$ topological space. $A$ open $A \cap Y = \emptyset \ \ \Longrightarrow A \cap \bar{Y} = \emptyset$?
I know this is an easy question, but I cannot demonstrate it properly.

Suppose by contradiction that $A \cap \bar{Y} \neq \emptyset$. Then $\exists \ x \in A \cap \bar{Y}$.

I need help formalizing this reasoning  (or correcting if it is wrong)

By the definition of $A$ open, $\exists \ U \subset A$ open neighborhood of $x$. Then my idea is to prove that $U \cap Y \neq \emptyset$ because $U$ in somewhat sense "touches" $Y$, because it's a neighborhood of $x$. But i can't write down a formal demonstration.
Any help is appreciated.
Thanks^^

答: Suppose that $A$ is open, and $A\cap Y=\varnothing$. Let $F=X\setminus A$; then $F$ is closed, and $Y\subseteq F$. Now take closures: $\operatorname{cl}Y\subseteq\operatorname{cl}F=F$, since $F$ is closed. But this immediately implies that $\operatorname{cl}Y\cap A=\varnothing$.</p>


问:
Elliptic Curve Discriminant
Here is my attempted approach to prove that the discriminant $\triangle = 4a^3+27b^2$ of an elliptic curve in the form of $y^2 = x^3 + ax +b$ is zero. I have a problem at the end which doesn't bring me to the expected conclusion. Below is the process, please let me know the error in the process. I try to approach this from the definition of non-singularity (part of the definition of an elliptic curve), which is equivalent to the statement: the equation $y^2 = x^3 + ax +b$ is differentiable everywhere on the graph. From that definition, I try to derive the derivative of $y$ in respect to $x$ through implicit differentiation:
\begin{align}
    y^2 &= x^3+ax+b\\ \label{ref1}
    2yy' &= 3x^2+a\\
    y' &= \frac{3x^2+a}{2y}
\end{align}
If the graph is singular, then $y'$ does not exist, in other words:
\begin{align}
    2y &= 0\\
    3x^2+a &\neq 0
\end{align}
I replace $y = 0$ in the equation of elliptic curve which yields
\begin{equation}
   0 \ = \ x^3 + ax + b
\end{equation}
I apply Cardano's method, who realize that the form could be represented as
\begin{equation}
    (\alpha-\beta)^3 + 3\alpha\beta(\alpha-\beta) = \alpha^3 - \beta^3
\end{equation}
in which
\begin{align}
    \alpha\beta &= \frac{a}{3} \\
    \alpha^3 - \beta^3 &= -b
\end{align}
By substituting $\alpha = \frac{a}{3\beta}$ in the second equation from above, I obtain
\begin{equation}
    (\frac{a}{3\beta})^3 - \beta^3 = -b
\end{equation}
I further simplify this by considering $\beta^3$ as a whole, i.e.
\begin{align}
    \frac{a^3}{27}-\beta^6 &= -b\beta^3\\
    (\beta^3)^2 - b\beta^3 -\frac{a^3}{27} &= 0
\end{align}
By quadratic formula,
\begin{equation}
    \beta^3 = \frac{b\pm \sqrt{b^2+\frac{4a^3}{27}}}{2} = \frac{b}{2} \pm \sqrt{\frac{b^2}{4}+\frac{a^3}{27}}
\end{equation}
From $\alpha^3 - \beta^3 = -b$, I get
\begin{equation}
    \alpha^3 = \beta^3 - b = -\frac{b}{2} \pm \sqrt{\frac{b^2}{4}+\frac{a^3}{27}}
\end{equation}
From $3x^2+a \neq 0$:
\begin{equation}
    3\left(\left(-\frac{b}{2} \pm \sqrt{\frac{b^2}{4}+\frac{a^3}{27}}\right)^\frac{1}{3} - \left(\frac{b}{2} \pm \sqrt{\frac{b^2}{4}+\frac{a^3}{27}}\right)^\frac{1}{3}\right)^2 + a \neq0
\end{equation}
The equation above is from $3(\alpha-\beta)^2+a \neq 0$. By De Moivre's Formula there are two other equations, $3\left(\alpha(\frac{-1}{2}+\frac{\sqrt{3}i}{2})-\beta(\frac{-1}{2}+\frac{\sqrt{3}i}{2})\right)^2+a \neq 0$ and $3\left(\alpha(\frac{-1}{2}+\frac{\sqrt{3}i}{2})^2-\beta(\frac{-1}{2}+\frac{\sqrt{3}i}{2})\right)^2+a \neq 0$. These result in several representations of b in terms of a and one integer solution, $a=0,b=0$ shouldn't exist at the same time, $b\neq\pm \frac{2ia^{3/2}}{3\sqrt{3}}$. These two meet my expectation, $\triangle = 0$. However, there are other four representations that do not meet my expectation $b \neq \pm \sqrt{\frac{a^3}{54}\pm\frac{5ia^3}{6\sqrt{3}}}$ and $b \neq \pm \sqrt{\frac{a^3}{54}\mp\frac{5ia^3}{6\sqrt{3}}}$. I also try to see what I could get from \begin{equation}
    3\left(\left(-\frac{b}{2}\right)^\frac{1}{3} - \left(\frac{b}{2}\right)^\frac{1}{3}\right)^2 + a \neq0
\end{equation}
which yields that $a=0,b=0$ shouldn't exist at the same time, or $b\neq\pm \frac{2ia^{3/2}}{3\sqrt{3}}$ these satisfy $\triangle = 0$ and also a weird pair of $b \neq \pm \frac{ia^{3/2}}{12\sqrt{3}}$, which doesn't even satisfy $\triangle = 0$. Could someone explain this?

答:
When  $3x^2 + a \ne0$ and $2y = 0$, the curve is still nonsingular, its just that the slope of the tangent line is infinite.
The singularity occurs when
$$3x^2 + a = 0 \text{ and } 2y =0$$
simultaneously i.e.
$a = - 3x^2$ so $0 = x^3 + (-3x^2)x + b$
giving
$$ b = 2x^3$$
this is what leads to the equation
$$27b^2 + 4a^3 = 0$$

Title: Elliptic Curve Discriminant


问:
Expectation of two random variables
Suppose I have $X_1,...X_{k} $ be random variables from the standard normal distribution. And I want to find the quantity $E[\overline X \overline X W\overline X + \overline X \overline XW^2]$, where $\overline X$ is the sample mean and $W$ is the sample variance, $W=S^2$.
My attempt is to use linearity to make it becomes $E[\overline X \overline X W\overline X] + E[\overline X \overline XW^2]$. We know sample mean and sample variance is independent, so = $E[\overline X^3]E[W] + E[\overline X \overline X]E[W^2] $. Here, I know the expectation of sample mean squared(3rd term) is just the mean of a chi-square distribution with $r=1$ and the second term is just $\sigma^2$.

However, I am struggling to find  $E[\overline X^3]$ because to my knowledge there is no distribution that is equal to $\overline X^3$ , same reason for $E[W^2]$. Is there a easy way to compute them? or do I have to actually take the derivative of the mgf from the normal distribution of $\overline X$ three times? Since $(\overline X=N(\mu,\sigma^2/k))$

答:
$\bar X$ is a normal random variable $\mathcal N(\mu,\sigma^2/k)$, it's third moment is $\mu^3+\frac{3}{k} \mu \sigma^2$ (See Moments of normal random variable).

For $W$, observe that $U=W/\sigma^2$ is a Chi-squared of degree $k$, we know it's mean is $k$ and variance is $2k$, such that $\mathbb E[U^2]=2k+k^2$ and $\mathbb E[W]=\sigma^4(2k+k^2)$.

Title: Expectation of two random variables


问:
Prove that a sequence is convergent using Cauchy's principle of convergence
I am facing difficulty in employing Cauchy's general principle of convergence to prove that a sequence is convergent.

Let's take this example:

Let $x_1 = 2$, and $x_{n+1} = \sqrt{x_n+20}$, for $n=1,2,3,\dots$

Prove that the sequence {$x_n$} converges.

I solved this using monotone convergence theorem. Using induction, it can be proved that the sequence is increasing and is bounded above by $5$.
Thus, it is convergent.

But how do I prove this using Cauchy's principle, which states that, for each $\epsilon > 0$ there exists a positive integer $m$ such that

$|S_{n+p}-S_n|<\epsilon$, $\forall$ $n\geq m $ and $p\geq1$,
$\{S_n\}$ being the sequence.

Body:
let $x_{n+1}>x_{n}$ then

$x_{n+1}+20 >x_n+20$ $\Rightarrow$ $\sqrt{x_{n+1}+20}=x_{n+2}>\sqrt{x_n+20}=x_{n+1}$

and we know that $x_2 > x_1$ so $x_n$ is increasing

also, note that $2 \leq x_n < 5$ $\Rightarrow$ $2\sqrt{22} \leq\sqrt{x_a+20}+\sqrt{x_b+20}<10$ hence $\frac{1}{\sqrt{x_a+20}+\sqrt{x_b+20}}>8$

$x_{n+p}-x_n = \sqrt{x_{n+p-1}+20}-\sqrt{x_{n-1}+20}=
\frac{x_{n+p-1}-x_{n-1}}{\sqrt{x_{n+p-1}+20}+\sqrt{x_{n-1}+20}}=\frac{\sqrt{x_{n+p-2}+20}-\sqrt{x_{n-2}+20}}{\sqrt{x_{n+p-1}+20}+\sqrt{x_{n-1}+20}}$

$\frac{\sqrt{x_{n+p-2}+20}-\sqrt{x_{n-2}+20}}{\sqrt{x_{n+p-1}+20}+\sqrt{x_{n-1}+20}}=\frac{x_{p+1}-\sqrt{22}}{(\sqrt{x_{n+p-1}+20}+\sqrt{x_{n-1}+20})...(\sqrt{x_{p+1}+20}+\sqrt{22})}< \frac{x_{p+1}-\sqrt{22}}{8^{n-2}}$

$\frac{x_{p+1}-\sqrt{22}}{8^{n-2}}<\frac{5-\sqrt{22}}{8^{n-2}}< \epsilon$ $\Rightarrow$ $\frac{1}{8^{n-2}}<\frac{\epsilon}{5-\sqrt{22}}$ and we know there exists and $m$ such that

$m > n \Rightarrow |\frac{1}{8^{n-2}}-0|=\frac{1}{8^{n-2}}<\frac{\epsilon}{5-\sqrt{22}}$

so $x_n$ is convergent


问:
Stokes Theorem on a Hemisphere
I have a question about Stokes Theorem. It's really confusing me.

So the problem asks,

Use Stokes theorem to set up the integral
$$\int_{a}^{b}g(t)dt$$
that you would use to compute
$$\iint_{S}^{}(\nabla\times \vec{F})\cdot d\vec{S}$$
where
$$\vec{F}(x,y,z)=\left \langle e^zsin(y),ze^x,2xcos(y) \right \rangle$$

and S is the hemisphere $y=\sqrt{9-x^2-z^2}$ oriented in the direction of the positive y-axis.

What I did was first calculate the curl:

$$\nabla\times \vec{F}=\left \langle -2xsin(y)-e^zcos(y),e^zsin(y)-2cos(y),ze^x-e^zcos(y) \right \rangle$$

Skipping steps,I have
$$\frac{\partial \vec{r}}{\partial x}\times \frac{\partial \vec{r}}{\partial z}
=\left \langle -\frac{x}{\sqrt{9-x^2-z^2}},-1,-\frac{z}{\sqrt{9-x^2-z^2}} \right \rangle=\vec{n}$$

$$\hat{n}=\left \langle -\frac{x}{3},-\frac{\sqrt{9-x^2-z^2}}{3},-\frac{z}{3} \right \rangle$$

I'm not sure where to go from here.

答:
As per Stokes' Theorem,

$\int_C \vec{F} \cdot d\vec{r} = \iint_S curl \vec{F} \cdot d\vec{S}$

which allows you to change the surface integral of the curl of the vector field to the line integral of the vector field around the boundary of the surface. The surface is hemisphere with $y = 0$ plane being the boundary, though the question should have been more clear on that. Also the question states we have orientation in positive $y$ direction.

By the way, there is a hint in the question to do line integral to find surface integral of the curl of the vector field.

Now we can parametrize the boundary as,

$\vec{r}(t) = (3 \cos t, 0, 3 \sin t) \,$ (as $y = 0$, it is a circle in $XZ$ plane with radius $3,$ $x^2 + z^2 = 9$)

$\vec{r'}(t) = (-3 \sin t, 0, 3 \cos t) \, $ ($0 \leq t \leq 2 \pi)$

Now substitute $x, y, z$ from above parametrization of the curve to evaluate $\vec{F}(r(t))$ and do the dot product with $\vec{r'}(t)$. Then do line integral wrt $dt$.

Title: Stokes Theorem on a Hemisphere


问:
Volume of region $\{ (x_1,\ldots,x_n)\in \mathbf{R}^n_{>0} \mid \prod_{i=1}^n\max(1,x_i)\leqslant d \}$
As part of a problem on algebraic number theory, I am faced with the following analysis/calculus exercise:

Show that the volume of the set
$$E_n:=\Big\{ (x_1,\ldots,x_n)\in \mathbf{R}^n_{>0} :\prod_{i=1}^n\max(1,x_i)\leqslant d \Big\}$$
is equal to $$d\sum_{i=0}^{n-1} \binom{n-1}{i} \frac{(\log d)^i}{i!}.$$

My attempt at solving this problem was as follows. Clearly, $E_n\subset [0,d]^n$. The volume of $E_n$ equals the following integral (abbreviating $m$ for $\max$):

$$\int_0^d \int_0^{\frac{d}{m(1,x_1)}}\cdots \int_0^{\frac{d}{m(1,x_1)\cdots m(1,x_{n-2})}} \int_0^{\frac{d}{m(1,x_1)\cdots m(1,x_{n-1})}} 1\, dx_n\,dx_{n-1}\cdots dx_2\,dx_1,$$
which can be simplified into
$$d\int_0^d \frac{1}{m(1,x_{1})} \int_0^{\frac{d}{m(1,x_1)}}\frac{1}{m(1,x_{2})}\cdots \int_0^{\frac{d}{m(1,x_1)\cdots m(1,x_{n-2})}} \frac{1}{m(1,x_{n-1})}\,dx_{n-1}\cdots dx_2\,dx_1.$$

I tried to prove the formula by induction. By splitting the middle integral, it evaluates to $1+\log d-\sum_{i=1}^{n-2}\log m(1,x_i)$. By linearity of the integral, we get $(1+\log d)E_{n-1}$ plus some other integral. However, this gets incredibly messy.

Is my approach correct? Is there any more clever way to attack this integral or an other method to prove this formula ? Is there maybe a probabilistic way of calculating its value ?

EDIT. In Theorem 6.5 of this article the integral is written as $\int_{y\in \mathbf{R}^n,\sum_i \max(0,y_i)\leq \delta} \exp \sum y_i\,dy$. Are my integral and this one equal ?

Any help is much appreciated.

(For the bounty: I would like to have a comprehensive, detailed answer.)

答:
Indeed the two representations are equivalent. Substituting $x_i = e^{y_i}$, we note that

$$ \prod_{i=1}^{n} \max\{1,x_i\} < d
\quad\Leftrightarrow\quad \prod_{i=1}^{n} \max\{1,e^{y_i}\} < d
\quad\Leftrightarrow\quad \sum_{i=1}^{n} \max\{0,y_i\} < \log d. $$

Also, since $\mathrm{d}x_i = e^{y_i} \, \mathrm{d} y_i$, we get

\begin{align*}
\operatorname{Vol}(E_n)
&= \int_{\mathbb{R}_{>0}^n} \mathbf{1}_{E_n} \, \mathrm{d}x_1\cdots\mathrm{d}x_n \\
&= \int_{\mathbb{R}^n} \mathbf{1}_{\{ \sum_{i=1}^{n} \max\{0,y_i\} < \log d\}} e^{y_1+\dots+y_n} \, \mathrm{d}y_1\cdots\mathrm{d}y_n
\end{align*}

From now on, we will assume $d > 1$ for an obvious reason. Write $[n] = \{1,\dots,n\}$ and set

$$ F(I) = \int_{A_I} \mathbf{1}_{\{ \sum_{i=1}^{n} \max\{0,y_i\} < \log d\}} e^{y_1+\dots+y_n} \, \mathrm{d}y_1\cdots\mathrm{d}y_n, $$

where $A_I = \bigcap_{i \in I} \{y_i \geq 0\} \cap \bigcap_{i \notin I} \{y_i < 0\} $ is the set of points $(y_1, \dots, y_n)$ such that $y_i \geq 0$ if and only if $i \in I$. Writing $k = \left| I \right|$ for the size of the set $I$, it is easy to find that the value of $F(I)$ depends only on $k$. In particular, if $k \geq 1$, then

\begin{align*}
F(I)
= F([k])
&= \left( \int_{-\infty}^{0} e^{y} \, \mathrm{d}y \right)^{n-k} \left( \int_{[0,\infty)^k} \mathbf{1}_{\{ \sum_{i=1}^{k} y_i \leq \log d \}} e^{y_1 + \dots + y_k} \, \mathrm{d}y_1 \dots \mathrm{d}y_k. \right) \\
&= \int_{[0,\infty)^k} \mathbf{1}_{\{ \sum_{i=1}^{k} y_i \leq \log d \}} e^{y_1 + \dots + y_k} \, \mathrm{d}y_1 \dots \mathrm{d}y_k.
\end{align*}

Substituting $(z_1, \dots, z_k) = (y_1, y_1+y_2, \dots, y_1+\dots+y_k)$, the region $[0, \infty)^k$ in $y$-domain transforms to $\{(z_1, \dots, z_k) : 0 < z_1 < z_2 < \dots < z_k \}$, and so,

\begin{align*}
F([k])
&= \int_{0}^{\log d}\int_{0}^{z_{k}} \dots \int_{0}^{z_2} e^{z_k} \, \mathrm{d}z_1 \dots \mathrm{d}z_k \\
&= \int_{0}^{\log d} \frac{z_k^{k-1}}{(k-1)!} e^{z_k} \, \mathrm{d}z_k.
\end{align*}

If in addition $k \geq 2$, then integration by parts shows that

\begin{align*}
F([k])
&= \frac{d (\log d)^{k-1}}{(k-1)!} - F([k-1]).
\end{align*}

Writing $[0] = \varnothing$ for convenience and noting that $F(\varnothing) = 1$, we find that this formula also extends to $ k = 1$. Therefore

\begin{align*}
\operatorname{Vol}(E_n)
&= \sum_{I \subseteq [n]} F(I) \\
&= \sum_{k=0}^{n} \binom{n}{k} F([k]) \\
&= \sum_{k=0}^{n} \left[ \binom{n-1}{k} + \binom{n-1}{k-1} \right] F([k]) \\
&= \sum_{k=0}^{n-1} \binom{n-1}{k} F([k]) + \sum_{k=0}^{n-1} \binom{n-1}{k} F([k+1]) \\
&= \sum_{k=0}^{n-1} \binom{n-1}{k} \frac{d (\log d)^{k}}{k!}.
\end{align*}


问:
spitting short exact sequence with projective module
I have a question about an argument from a proof in Hideyuki Matsumura's "Commutative Ring Theory" on page 9, Theorem 2.5:

Let $(A,\mathfrak{m})$ be a local ring; then a finitely generated projective module $M$ over $A$ is free.

The proof works as follows:

Choose a minimal $A$-basis $\omega_1,...,\omega_n$ of
$M$. Take into account that "minimal" means that there cannot exist another system of generators $b_1,..., b_m \in M$ with $m < n$ and $M = \sum_{i=1} ^m A b_i$.

Define a surjective map $\varphi:F \to M$ from the free module $F = Ae_1 \oplus \cdots \oplus Ae_n$ to $M$ by $\varphi(\sum a_i e_i) = \sum a_i\omega_i$. If we set $K = \operatorname{Ker}(\varphi)$ then, from the minimal basis property

$$\sum a_i \omega_i =0 \Rightarrow a_i \in \mathfrak{m} \text{ for all } i. $$

Thus $K \subset \mathfrak{m}F$. Because $M$ is projective, there exists $\psi: M \to F$ such that
$F = \psi(M)\oplus K$,

Why $F = \psi(M)\oplus K$? What I know is this exact sequence splits:

$$0\to \ker\varphi\to F\to M\to 0$$

Where the first non trivial homomorphism is the canonical injection
and the second one $\varphi$, since this sequence splits we have:
$F=\ker\varphi\oplus M$.But why $F = \psi(M)\oplus K$？
Can we show $M＝ \psi(M)$？

答:
It’s actually the same thing up to isomorphism.

You can take $\psi$ as a splitting morphism that goes from M to F.

By definition, a splitting morphism verifies $\phi \circ \psi =id$.
So $\psi$ is injective and $\psi(M)$ is isomorphic to M.

The main difference is that we work with internal direct sums (so it’s actually $F=\psi(M)\oplus K$ not just a isomorphism) when you use $\psi(M)$ which lives in F as well as K.


问:
Different definitions of Adjoint representation of SE(2)?
I have come across two different definitions of the Adjoint representation of SE(2) which similarly extends into SE(3):

$ \text{Ad}_{\mathbf{X}}= \begin{bmatrix} \mathbf{R} & 0_{3\times 3} \\ [\mathbf{p}_{1}]_{\times}\mathbf{R} & \mathbf{R}\end{bmatrix}$

and

$ \text{Ad}_{\mathbf{X}}= \begin{bmatrix} \mathbf{R} &  [\mathbf{p}_{1}]_{\times}\mathbf{R} \\ 0_{3\times 3} & \mathbf{R}\end{bmatrix}$

where $[\cdot]_{\times}$ denotes a skew-symmetrix.

What is the difference between the two?

Source for first representation: https://arxiv.org/abs/1510.06263

Source for second representation: http://ethaneade.com/lie.pdf

答:
The fundamental difference lies in the order of translation and rotation in se(3).

Specifically, for the first representation, se(3)=(w, t)^T, Rotate first and then translate;

for the second representation, se(3)=(t, w)^T, translate first and then Rotate.


问:
How to develop patience in mathematics?
How  to develop the patience in mathematics?

When  you read the books like Rudin, Conway, etc., it takes a lot of time for one problem; sometimes $10$ hours each.

I usually  devote one  problem for $30$ minutes. If I can't solve the problem within $30$ minutes, then I post the problem on this site for a solution.

Actually, I have no patience and can't wait for long hours for thinking about the problem.

How  can I improve myself and develop the patience?

答:
While developing patience is one thing, the counter-intuitive answer is that if you are taking too long to solve problems is that you are not solving enough problems. Mathematical problem solving is something that needs to be trained, both in general, but also in every subject.

If I, as someone without much training, were to run a marathon right now, I would probably take 10 hours to do so. But if I give up after 30 minutes, no one would tell me that I need more patience. A much more reasonable suggestion would be to first try to get into shape for a much shorter distance.

The same is true for textbook problems. There are of course outliers, but there is no reason for a normal textbook problem to need anywhere close to 10 hours of dedicated time to solve. If you are stuck at one problem, then you should try an easier related problem and come back later. This might come from the same, or a different textbook or can even be something you come up with on your own by modifying a given problem (e.g. dropping a condition and looking for a counterexample to something you're meant to prove).

Title: How to develop patience in mathematics?


问:
Given $dy/dt = 6y/39 -(y^2)/2002$ and the point $(0,14)$, what is the original equation in "$y=\dots$" form?
This is how I tried to solve it:
$$\frac{dy}{dt}=\frac{6y}{39}-\frac{y^2}{2002}$$
I multiplied and divided:
$$\frac{dy}{\frac{6y}{39}-\frac{y^2}{2002}}=dt$$
I simplified:
$$\frac{dy}{\frac{2y}{13}-\frac{y^2}{2002}}=dt$$
I found a common denominator:
$$\frac{dy}{\frac{308y-y^2}{2002}}=dt$$
I divided the fractions:
$$\frac{2002dy}{308y-y^2}=dt$$
I completed the square:
$$\frac{2002dy}{308y-y^2-154^2+154^2}=dt$$
$$\frac{2002dy}{-(y^2-308y+154^2)+154^2}=dt$$
$$\frac{2002dy}{-(y-154)^2+154^2}=dt$$
Now I am temporarily looking at only the left side of the equation to find the integral:
$$\int\frac{2002}{-(y-154)^2+154^2}dy$$
I substituted u:
$$2002\int\frac{1}{(-1)(u)^2+154^2}du$$
At this point I was very confused so I tried to use arctan when I probably wasn't allowed to. I used the following formula:
$$\int\frac{b}{b^2+a^2x^2}dx=\arctan(\frac{ax}{b})+c$$
My integral:
$$\frac{2002}{154}\int\frac{154}{(-1)(u)^2+154^2}du$$
$$\frac{2002}{154}\arctan(\frac{iu}{154})$$
$$13\arctan(\frac{i(y-154)}{154})$$
The integral of $1dt$ is $t$ so I got this equation:
$$13\arctan(\frac{i(y-154)}{154})=t+c$$
I then solved for y:
$$\frac{i(y-154)}{154}=\tan(\frac{t+c}{13})$$
$$y=\frac{154\tan(\frac{t+c}{13})}{i}+154$$
I then solved for c by substituting the initial point (0,14) into the equation:
$$14=\frac{154\tan(\frac{0+c}{13})}{i}+154$$
$$\frac{-140i}{154}=\tan(\frac{c}{13})$$
$$\arctan(\frac{-10i}{11})=\frac{c}{13}$$
$$c=13\arctan(\frac{-10i}{11})$$
And I substituted the found $c$ value into the equation:
$$y=\frac{154\tan(\frac{t+13\arctan(\frac{-10i}{11})}{13})}{i}+154$$
Is this technically correct or did do I something wrong?
If I did do something wrong, what did I do wrong?
What would be an alternative method of solving this?

答:
This is also a Bernoulli equation. So set
$$
u=\frac{2}{13y}+\frac1{2002},
$$
then
$$
u'=-\frac{2y'}{13y^2}=-\frac2{13}\left(\frac{2}{13y}+\frac1{2002}\right)=-\frac2{13}u
$$
so that
$$
u(t)=u(0)e^{-\frac2{13}t}.
$$
Now translate back to a formula for $y$.


问:
Compute $\iint _S (y^2+x^2)\, dA$ by change of variables.

Let $S$ be the region in the first quadrant bounded by the curves $xy=1$, $xy=3$, $x^2-y^2=1$, and $x^2-y^2=4$. Compute $$\iint _S (y^2+x^2)\, dA.$$

I have tried to use the change of variables $u=xy$ and $v=x+y$. Then $$\begin{align}x^2+y^2=v^2-2u \\ x^2-y^2=v^2-2u \end{align}$$

Since $v^2=x^2-y^2+2u$, $$0 \le v^2 \le 4-1=3$$, so $$0 \le v \le \sqrt{3}$$

The integral becomes
$$\begin{align} \iint _S (y^2+x^2)\, dA= \int_0^\sqrt{3} \int_1^3 (v^2-2u) \left|J \right| \,du\,dv \end{align}$$
but I can't solve $x,y$ in terms of $u,v$.

I have tried another one, $u=x^2+y^2$ and $v=x^2-y^2$.

I get $x=\sqrt{\frac{u+v}{2}}$ and $y=\sqrt{\frac{u-v}{2}}$

Then $1\le xy=\frac{\sqrt{u^2-v^2} }{2} \le 3$, and $$4\le u^2-v^2 \le 36$$

The integral becomes $$\begin{align} \iint _S (y^2+x^2)\, dA= \int_1^4 \int_{\sqrt{4+v^2}}^{\sqrt{36+v^2}} u \left|J \right| \,du\,dv \end{align}$$

Hope someone could give me some suggestion how to do it.

答:
EDIT

The substitution $u=xy$, $v=x^2-y^2$ does the job. The Jacobian with respect to $x$ and $y$ is

$$
J=\left(\begin{matrix}
y & x\\
2x & -2y
\end{matrix}\right)
$$

so $|\det(J)|=2(x^2+y^2)$, which means $dudv=2(x^2+y^2)dxdy$ and your integral becomes
$$\begin{align} \iint _S (y^2+x^2) dxdy= \int _1^3 \int_1^4 2 \;dudv \end{align}$$

As Ian said, this is no more than a special case of the "inverse method", used also in order to calculate one-dimensional integrals. E.g., if you have

$$
\int \tan(x)^4 dx
$$

and you apply the substitution $t=\tan(x)$, then it is easy to obtain $dt=(1+\tan(x)^2)dx=(1+t^2)dx$, and then $dx=\frac{dt}{1+t^2}$ (of course, in this case it is equally easy to start from $x=\arctan(t)$ and so on, but sometimes you can avoid some annoying calculation).


问:
Anyone encounter these Logic symbols?

These diagrams are equivalent representations of the 2-ary boolean functions. What are the symbols used in the top left diagram?

(Source: wikicommons user mate2code)

答:
Based on a comment by Henning Makholm and a short link trace I conducted, these are nonstandard symbols invented by Wikimedia user mate2code (formerly Lipedia).

mate2code calls them nibble shorthands, and they simply represent all possible nibbles (four digit binary strings, more commonly represented by the numbers $0$ to $15$ or by the hexadecimal digits 0 to F), in an order which can be seen here (by interpreting the red dots under each symbol as $1$'s and the white dots as $0$'s).

Note: The links may be volatile, as they refer to Wikimedia user pages.

Title: Anyone encounter these Logic symbols?


答:
Using the first method you have a standard situation for the use of the "inverse method": the Jacobian of $(u,v) \mapsto (x,y)$ is the inverse of the Jacobian of $(x,y) \mapsto (u,v)$ and so the determinants of these are reciprocals of each other. You should find the Jacobian is easy to find going the other way and then also easy to write in terms of only $u$ and $v$.

That said your bounds are not right in this method; since $x^2-y^2=v^2-2u$ is held between two constants, the bounds on $v$ as the inner variable will depend on $u$. This issue goes away if you select $v=x^2-y^2$ instead, and the Jacobian is still easily evaluated using the inverse method. This is what is detailed in user6530's answer.

Your second method at least in principle works without the use of the inverse method though I am not so confident that the integrals are easy to do in the end that way.

Title: Compute $\iint _S (y^2+x^2)\, dA$ by change of variables.

Body:
Let $S$ be the region in the first quadrant bounded by the curves $xy=1$, $xy=3$, $x^2-y^2=1$, and $x^2-y^2=4$. Compute $$\iint _S (y^2+x^2)\, dA.$$

I have tried to use the change of variables $u=xy$ and $v=x+y$. Then $$\begin{align}x^2+y^2=v^2-2u \\ x^2-y^2=v^2-2u \end{align}$$

Since $v^2=x^2-y^2+2u$, $$0 \le v^2 \le 4-1=3$$, so $$0 \le v \le \sqrt{3}$$

The integral becomes
$$\begin{align} \iint _S (y^2+x^2)\, dA= \int_0^\sqrt{3} \int_1^3 (v^2-2u) \left|J \right| \,du\,dv \end{align}$$
but I can't solve $x,y$ in terms of $u,v$.

I have tried another one, $u=x^2+y^2$ and $v=x^2-y^2$.

I get $x=\sqrt{\frac{u+v}{2}}$ and $y=\sqrt{\frac{u-v}{2}}$

Then $1\le xy=\frac{\sqrt{u^2-v^2} }{2} \le 3$, and $$4\le u^2-v^2 \le 36$$

The integral becomes $$\begin{align} \iint _S (y^2+x^2)\, dA= \int_1^4 \int_{\sqrt{4+v^2}}^{\sqrt{36+v^2}} u \left|J \right| \,du\,dv \end{align}$$

Hope someone could give me some suggestion how to do it.


问:
Eigenvalues of a 3x3 orthogonal matrix
If A is an arbitrary 3x3 orthogonal matrix with det(A)=1, then how do I show that the eigenvalues are 1, cos(x)+i sin(x), and cos(x)-i sin(X), where cos(x)=(tr(A)-1)/2.

I was thinking that maybe tr(A) equals det(A), and that gives a value for x, but I'm not sure if thats right

答:
I am assuming your matrix is real-valued (since it would not make sense to talk about orthogonality otherwise).

Since matrix is orthogonal, it is a normal operator -> it can be diagonalized ($\mathbf{M}=\mathbf{\Gamma}^\dagger.\mathbf{\Lambda}.\mathbf{\Gamma}$). Using the fact that for real-valued matrix transpose is the same as conjugate transpose:

$$
\mathbf{M}^T=\mathbf{M}^\dagger=\mathbf{M}^{-1}=\mathbf{\Gamma}^\dagger.\mathbf{\Lambda}^{\dagger}.\mathbf{\Gamma}=\mathbf{\Gamma}^\dagger.\mathbf{\Lambda}^{-1}.\mathbf{\Gamma}
$$

Since $\mathbf{\Gamma}$ is invertible, it follows that for matrix with eigenvalues in the diagonal:

$$
\mathbf{\Lambda}^\dagger=\left(
\begin{array}\\
\lambda_1^* & 0 & 0\\
0 & \lambda_2^*  & 0\\
0 & 0 & \lambda_3^*\\
\end{array}
\right)=
\mathbf{\Lambda}^{-1}=\left(
\begin{array}\\
\lambda_1^{-1} & 0 & 0\\
0 & \lambda_2^{-1}  & 0\\
0 & 0 & \lambda_3^{-1}\\
\end{array}
\right)
$$

So the eigenvalues of the matrix are all of the form $\lambda_s=\exp\left(i\phi_s\right),\,s=1,2,3$

Next $\mbox{det}\left[\mathbf{M}\right]=\Pi_s \exp\left(i\phi_s\right)=\exp\left(i\left[\phi_1+\phi_2+\phi_3\right]\right)=1$, so

$$
\phi_3=-\phi_1-\phi_2+2\pi n,\quad n=0,\pm1,\dots
$$

Next, if matrix is real-valued, the eigenvalues will be solutions to a cubic polynomial equation with real-valued coefficients. Therefore at least one of them will be real-valued, let it be the third one, so $\lambda_3=\pm1$. Also the other two eigenvalues will be complex conjugates of each other. So the only possibility is:

$$
\lambda_1=\exp\left(i\phi\right),\,\lambda_2=\exp\left(-i\phi\right),\,\lambda_3=1
$$

Therefore $\mbox{Tr}\left[\mathbf{M}\right]=2\cos\phi+1$

Title: Eigenvalues of a 3x3 orthogonal matrix


问:
l2 compact operator
Definined the mapping T:  ->  so that:  where
I'd like to show that if T is a compact operator then
I thought i could negate my thesis and then show that T would not be compact but i wasn't able to show it. Has anyone an idea?
Thanks!

答:
Hint:

If $\alpha$ is not in $c_0$, then you can find a number $\varepsilon > 0$ and a subsequence $(\alpha(n_k))_{k \in \mathbb{N}}$ such that $|\alpha(n_k)| \ge \varepsilon$ for each $k \in \mathbb{N}$.

Use this to prove that the sequence $(Te_{n_k})_{k \in \mathbb{N}}$ in $\ell^2$ has no convergent subsequence (where $e_n \in \ell^2$ denotes the $n$-the canonical unit vector).

Title: l2 compact operator


问:
Show that This infinite product is entire
This question is from Ponnusamy and silvermann complex analysis and I am making some mistake in this question and unable to find it.

Show that $\prod_{n=1}^{\infty} (1-z/n) e^{z/n +5z^2/n^2}$ is entire.

$1+ f_n (z)= (1-z/n) e^{z/n +5z^2/n^2}$ and finding a upper bound on $f_n (z)$ . $1+ f_n (z) = exp( Log(1-z/n) + z/n+ 5 z^2 /n^2)$ For $|z|\leq R$ then choose N large enough such that |z/n|<1 for all $n \geq N$ and using the identity$ Log(1-z/n) = -\sum_{k=1}^{\infty} 1/k (z/n)^k$ and the fact that|R/n|<1.

So, $|Log (-\sum_{k=1}^{\infty} 1/k (z/n)^k +z/n +5 z^2/n^2) | \leq  -5 -1/2 R^2/n^2 $, where I used Formula of infinite geometric series.

Now $|f_n(z)| \leq exp|e^{-5 -1/2 R^2 /n^2 -1}-1| exp( -5 -1/2 R^2 /n^2)$ -1 and using $e^x -1 \leq x e^x  for all x \geq 0 $ we get $|f_n(z)|\leq 5+ 1/2 R^2/ n^2 $ and if I sum $|f_n (z)| $ from N to $\infty$ i get Infinity as 5 is also there .

So, I think while approximating I am doing some mistake and I am unable to find it. Can you please point that out.

If there is some other method of proving it entire that is also welcome.

Thanks!

答:
Because $f(z)=\frac{\log(1-z)+z}{z^2}$ is analytic thus continuous for $|z|< 1$ then $$C= \sup_{|z|\le 1/2}|f(z)|<\infty$$
gives $|\log(1-z/n)+z/n|\le Cz^2/n^2$ for $|z|<n/2$.


问:
Exponential integral question
How would I solve the following problem?
$$f(x)=\int\!\frac{4}{\sqrt{e^x}}\,dx$$

Using $u$ substitution I have set $u=e^x$ andd $du=e^x dx$

so would I have $$4\int\!\frac{du}{\sqrt{u}}$$

What would I do from here?

答:
Stop!  Use law of exponents:

$$\sqrt{e^x} = e^{x/2}$$

Better yet:

$$\frac{1}{\sqrt{e^x}} = e^{-x/2}$$

By the way, using your method, you should have a factor of $1/u^{3/2}$ in the integral.

Title: Exponential integral question


问:
$E(|X_1+X_2|^p)<+\infty$ iff $E(|X_1|^p)<+\infty,E(|X_2|^p)<+\infty$
Let $X_1$ and $X_2$ be independent random variables. Show that for any $p>0$,
$$E(|X_1+X_2|^p)<+\infty \text{ iff } E(|X_1|^p)<+\infty,E(|X_2|^p)<+\infty$$
Show that this is false if $X_1$ and $X_2$ are not independent.

I tried to use the Hölder's inequality but it got nowhere. Any suggestions will be appreciated.

My attemp:

We remember that
$$ E (X) = \int_{\Omega} X d P $$
$(\Leftarrow)$
If $p \geqslant 1$,
\begin{eqnarray*}
    \| X_1 + X_2 \|_p & \leqslant & \| X_1 \|_p + \| X_2 \|_p\\
    \left( \int_{\Omega} | X_1 + X_2 |^p d P \right)^{\frac{1}{p}} & \leqslant
    & \left( \int_{\Omega} | X_1 |^p d P \right)^{\frac{1}{p}} + \left(
    \int_{\Omega} | X_2 |^p d P \right)^{\frac{1}{p}}\\
    E (| X_1 + X_2 |^p)^{\frac{1}{p}} & \leqslant & E (| X_1
    |^p)^{\frac{1}{p}} + E (| X_2 |^p)^{\frac{1}{p}}\\
    E (| X_1 + X_2 |^p)^{\frac{1}{p}} & < & + \infty\\
    E (| X_1 + X_2 |^p) & < & + \infty .
  \end{eqnarray*}

If $0 < p < 1,$
\begin{eqnarray*}
    (a + b)^p & \leqslant & a^p + b^p, \forall a, b \geqslant 0\\
    | X_1 +  X_2 |^p \leqslant (| X_1 | + | X_2 |)^p &
    \leqslant & | X_1 |^p + | X_2 |^p\\
    | X_1 + X_2 |^p & \leqslant & | X_1 |^p + | X_2
    |^p\\
    \int_{\Omega} | X_1 + X_2 |^p d P & \leqslant &
    \int_{\Omega} | X_1 |^p + | X_2 |^p d P\\
    \int_{\Omega} | X_1 + X_2 |^p d P & \leqslant &
    \int_{\Omega} | X_1 |^p d P + \int_{\Omega} | X_2 |^p d P\\
    E (| X_1 + X_2 |^p) & \leqslant & E (| X_1 |^p) + E
    (| X_2 |^p)\\
    E (| X_1 + X_2 |^p) & < & + \infty .
  \end{eqnarray*}
$(\Rightarrow)$ For the implication, I think I should use variable independence $X_1$,$X_2$, but I don't know how to use it.

答:
For the implication $\Rightarrow$: using Fubini's theorem and independence, we have
$$
\mathbb E\left[\lvert X_1+X_2\rvert^p \right]=\int_{\mathbb R}\mathbb E\left[\lvert X_1+y\rvert^p\right]d\mathbb P_{X_2}(y),
$$
where $\mathbb P_{X_2}$ is the law of $X_2$; this implies that $\mathbb E\left[\lvert X_1+y\rvert^p\right]$ is finite for at least one $y$ and we get that $\lvert X_1\rvert^p$ has a finite expectation.

This implication fails if we do not have independence: let $X_2=-X_1$, where $X_1$ is such that $\mathbb E\left[\lvert X_1\rvert^p\right]$ is infinite.


问:
Probability density function from known density function
Suppose I want to compute new density function $Z$, $Z= min(X_1,2X_1-X_2)$, where $X_1$ and $X_2$ are independent exponential distributions with $\lambda$. I have looked into Finding the density for $\min\{X, Y\}$ and in their case, $X, Y $ are independent, so they can split the cdf into a product. But here, I don't think  $X_1,2X_1-X_2$ are independent because of the $2X_1-X_2$ How would I tackle this? I also know that the sum of two exponential distributions is a gamma distribution but I don't know if they are independent so I cannot proceed. Any help is appreciated!

答:
Observing  that $3>2$ and $X_1,X_2 \geq 0$ it results to me that

$$3X_1\geq 2X_1-X_2$$

$\forall x_1,x_2 \in \mathbb{R}^+$

Thus

$$Z=2X_1-X_2$$

With usual transformation procedures the distribution of Z can be derived without particular problems

EDIT:

Let's use the definition:

$$ \bbox[yellow,5px,border:2px solid red]
{
\mathbb{P}[Z>z]=\mathbb{P}[min(X,2X-Y)>z]=\mathbb{P}[X>z;Y<2X-z]
\
}
$$

Do a drawing and derive the region to integrate $f_{XY}(x,y)$ to find $1-F_Z(z)$

sketch of solution:

FIrst observe that $Z \in \mathbb{R}$

If $Z\leq 0$ you can integrate the following region

$$\mathbb{P}[Z>z]=\int_0^{\infty}\lambda e^{-\lambda x}\Bigg[\int_0^{2x-z}\lambda e^{-\lambda y} dy \Bigg]dx=1-\frac{1}{3}e^{\lambda z}$$

Thus $F_Z(z)=\frac{1}{3}e^{\lambda z}$

If $Z>0$ you integrate the following region

$$\mathbb{P}[Z>z]=\int_z^{\infty}\lambda e^{-\lambda x}\Bigg[\int_0^{2x-z}\lambda e^{-\lambda y} dy \Bigg]dx=e^{-\lambda z}-\frac{1}{3}e^{-2\lambda z}$$

Putting the result together you find the CDF of Z

$$  F_Z(z) =
\begin{cases}
\frac{1}{3}e^{\lambda z},  & z\leq 0 \\
1-e^{-\lambda z}+\frac{1}{3}e^{-2\lambda z}, & z>0
\end{cases}$$

This is the drawing of $F_Z(z)$ for a fixed $\lambda=1$

As you can see, this is a nice CDF.

To find the desired pdf, just take the derivative of F


问:
Compilation of (exercises on) examples of forking, dividing, and independence
The resource I'm learning forking and dividing from (Tent and Ziegler) feels slightly light on examples to me. For instance, skimming through the chapter that introduces the concepts reveals only two concrete examples of the notion.

Example: For any $b_1<b_2\in\mathbb{Q}$, $b_1<v<b_2$ divides over $\emptyset$. On the other hand, the type $\{x>a:a\in\mathbb{Q}\}$ does not divide over $\emptyset$.

Exercise: Define the cyclical order on $\mathbb{Q}$ by $$\operatorname{cyc}(a,b,c)\iff(a<b<c)\vee(b<c<a)\vee(c<a<b).$$ Show that $(\mathbb{Q},\operatorname{cyc})$ has quantifier elimination, that $\operatorname{cyc}(a,x,b)$ divides over $\emptyset$ for any $a\neq b$, and that the (unique) type over $\emptyset$ forks over $\emptyset$.

Given that (as I understand it) one of the main utilities of these notions is in classifying classes of theories, it makes sense to me that there's an emphasis on general results rather than specific examples. However, I do find myself wishing for slightly more (exercises on) concrete examples; I want to be able to look at a formula or a type and immediately have some intuition as to whether it forks or divides over a particular parameter set, and (personally) I find that the best way of building such intuition is by working through many examples. I quite enjoyed the exercise above, for instance.

To this end, does anyone know of a worksheet or other resource that contains a large number of concrete examples of these ideas? I'm thinking of something vaguely along the lines of the book "Counterexamples in Topology", containing (for instance) a number of formulas and types in specific structures, and perhaps exercises on whether they fork or divide over a particular parameter set. If my skim-through of Tent and Ziegler was also uncharitable, and their later sections do contain more examples, I'd also be happy to hear this too. Hopefully this isn't too ill-posed or vague of a question; please feel free to close if it is!

答:
You could 'create' these exercises by yourself by looking at https://forkinganddividing.com/. It lists a whole lot of theories and places them in the classification picture. For a lot of theories there is a description of what forking (or dividing, but they coincide in the well-behaved cases, e.g. in simple theories) is in that theory. Just click on the theory and then click on characterization of forking in the bottom right box.

For example, in the theory of infinite sets we have that $\operatorname{tp}(A/BC)$ does not fork over $C$ precisely when $A \cap B \subseteq C$. You could try to prove this. In general: the more stable the examples are, the easier it will be to work things out. Strongly minimal theories are definitely the easiest, and the nonforking relation there is given by the pregeometry, something that Tent and Ziegler also covers.

Just a heads-up (although Tent and Ziegler also cover this): it is really the negation of dividing / forking that is the 'good' notion. Nonforking gives us a notion of independence, and we write this with an anchor symbol. See for example definition 7.2.8 in Tent and Ziegler. This notation is standard and the website I mentioned also uses this.


问:
Determine if subspace - complex numbers
$$ \mathbb{F} := \mathbb{C} \\ V := \mathbb{C}_2[x] $$

$V$ is the set of all polynomials with degree less than or equal to $2$ with complex scalars.

$$ U := \left \{ p(x) \in \mathbb{C}_2[x] | p(x) = p(2x) \right \} $$

Given $ U \subseteq V$ I need to determine if $U$ is a linear subspace of $V$

My try:

$$ p(x) = ax^2 + bx + c \\ p(2x) = a(2x)^2 +2bx + c \\ p(x) = p(2x) \Rightarrow ax^2 + bx + c = 4ax^2 + 2bx + c \\ 3ax^2 + bx = 0 $$

Now, I know that I need to check $3$ things:

$\vec{0} \in U $

If $\vec{v_1} \in U$ and $\vec{v_2} \in U$ then $\vec{v_1} + \vec{v_2} \in U$

If $t \in \mathbb{C}$ then $t \cdot \vec{v_1} \in U$

But I only got the rule of thumbs that " $ 3ax^2 + bx = 0$ "

Of course the zero vector is in $U$ because we can substitude $x=0$.

But how do I check closure over addition? what is really the rule of the subspace? I would appreciate you help.. I am lost. Thank you!

答:
Your justification for the fact that $0\in U$ is wrong. We have $0\in U$ because where $0$ is the null polynomial and if $n(x)$ is the null polynomial then it is indeed true that $n(x)=n(2x)$.

If $p(x),q(x)\in U$, then $p(x)=p(2x)$ and $q(x)=q(2x)$. So\begin{align}(p+q)(x)&=p(x)+q(x)\\&=p(2x)+q(2x)\\&=(p+q)(2x)\end{align}and therefore $p(x)+q(x)\in U$.

And if $p(x)\in U$ and $\alpha\in\Bbb C$, then $p(x)=p(2x)$ and \begin{align}(\alpha p)(x)&=\alpha p(x)\\&=\alpha p(2x)\\&=(\alpha p)(2x)\end{align}and therefore $(\alpha p)(x)\in U$.

Title: Determine if subspace - complex numbers


问:
Prove the lower bound of the function $f(x) = \sum_{i = 1}^{m} x_i\log{x_i}$ subject to $\sum_{i=1}^{m} x_i =1$
Can anyone prove the lower bound of the function $f(x) = \sum_{i = 1}^{m} x_i\log{x_i}$ subject to $\sum_{i=1}^{m} x_i =1$, thanks!

The lower bound is given as $f(x)\ge-\log{m}$, but why?

答:
Since $\log$ is concave, Jensen's inequality yields

$$\sum_{i=1}^m x_i \log\left({\frac{1}{x_i}}\right) ≤ \log\left(\sum_{i=1}^m x_i \cdot \frac{1}{x_i}\right) = \log m$$

This inequality is sharp because equality is attained when $x_1 = \cdots = x_m = 1/m$.


问:
Mean Square Error of using average to predict $X_{n+1}$

Let $X_1, X_2, \cdots, X_n$ be a sample from a population with mean $\mu$ and variance $\sigma^2$. Using $\bar{X_n} = \sum_{i=1}^n X_i / n$ as a predictor for $X_{n+1}$, determine the mean square error of this predictor.

I was able to solve this problem by directly calculating $E[(X_{n+1} - \bar{X_n})^2]$ and the answer is $(1+1/n)\sigma^2$.  However, if I try to calculate with $\text{MSE = Bias + Variance}$, I get a wrong answer. I think since $\bar{X_n}$ is unbiased, the MSE equals variance, which in this case is

$$V(\bar{X_n}) = \frac{1}{n^2} \sum V(X_i) = \frac{\sigma^2}{n}$$

Why is this wrong?

答:
Assuming that $X_1,X_2\ldots, X_{n+1}$ is a random sample,
$$
\mathsf{E}[X_{i+1}-\bar{X}_n]^2=\mathsf{E}[X_{i+1}-\mu-\bar{X}_n+\mu]^2=\operatorname{Var}(X_{i+1})+\operatorname{Var}(\bar{X}_n).
$$
Note that $X_{n+1}$ is random, and so the "Bias + Variance" formula is not appropriate here.


答:
Using Lagrange multipliers, consider
$$F= \sum_{i = 1}^{m} x_i\log(x_i)+\lambda\sum_{i = 1}^{m} x_i$$
$$\frac{\partial F}{\partial x_i}=1+\lambda+\log(x_i)=0$$ So, all $x_i$ are equal and then $x_i=\frac 1m$

Title: Prove the lower bound of the function $f(x) = \sum_{i = 1}^{m} x_i\log{x_i}$ subject to $\sum_{i=1}^{m} x_i =1$

Body:
Can anyone prove the lower bound of the function $f(x) = \sum_{i = 1}^{m} x_i\log{x_i}$ subject to $\sum_{i=1}^{m} x_i =1$, thanks!

The lower bound is given as $f(x)\ge-\log{m}$, but why?


问:
Convergence of Lax-Wendroff scheme for various initial conditions
I am considering the Lax-Wendroff discretisation
$$U_j^{(n+1)}=U_j^{(n)}-\frac{\Delta t}{2\Delta x}\left(U_{j+1}^{(n)}-U_{j-1}^{(n)}\right)+\frac{\Delta t^2}{2\Delta x^2}\left(U_{j+1}^{(n)}-2U_j^{(n)}+U_{j-1}^{(n)}\right)$$
of the simple advection equation $u_t+u_x=0$. I have proven that the truncation error is of $\mathcal{O}(\Delta t^2,\Delta x^2)$, and I have deduced by von Neumann stability analysis that we demand $\Delta t\leq\Delta x$ for $L_2$-stability. However, I am not sure how to interpret these results. In particular, I am not sure what these imply for convergence of the scheme dependent on the initial conditions such as for the cases i) $\mathrm{e}^{-x^2}$,  and ii) $\max\{1-|x|,0\}$. What should I be looking out for/arguing for? Thanks!

答:
The accuracy of the Lax-Wendroff scheme is presented in this post, and the stability analysis is performed here. Convergence results hold for any initial condition, but the notion of order of accuracy requires that the solution is sufficiently smooth (the Taylor series expansion used for the error analysis requires some smoothness). The main differences between (i) and (ii) are:

smoothness $C^\infty$ vs. smoothness $C^0$;

infinite support vs. bounded support.

In computational applications where the order of accuracy is measured 'experimentally', it is recommended to solve problems with very smooth solutions. Otherwise, the estimation of the order of accuracy might be inaccurate.


问:
Let $R$ be integral domain and $F$ a subring of $R$ and also a field then if $\dim_{F}(R)<\infty$ , then $R$ is a field
Because $F$ is a field we can describe $R$ as a verctorspace over $F$, we know that its dimension is $<\infty$ and we also got as a hint to look at the map $\phi:R\rightarrow R,r\mapsto rx$ where $x\notin 0\in R$. How does considering this maps help me to find an inverse for every $x\neq 0$ in $R$ ?

My approach is that if there exists an inverse for every element then those maps must be surjective because since there would be a $r$ such that $rx=1$, we could also describe every other element $z$ in $R$ with this map when we choose as input $zr$, so we have to pove that the map is surjective for any $x$, but how can we do it, and how do we use the dimension property?

We can interpretate the map as a linear map of the same vectorspaces. We could describe the map as matrix and look at the rank, but how do we show that the image of the basis vector is again a basis?

答: Hint: It is enough to show that $f_r(x)=rx$ is injective since an injective map between finite dimensional vector spaces is surjective.


答:
Let $x_i>0$, then $$f(x)=x \ln x \implies f''(x)=1/x>0$$ so according to Jensens inequality $$\frac{\sum_{i=1}^{n} x_i \ln x_i}{n} \ge \bar x \log \bar x=-\frac{\ln n}{n}$$ $$  \bar x=\frac{1}{n}\sum_{i=1}^{n} x_i=\frac{1}{n},~\text{given}~\sum_{i=1}^{n} x_i=1$$.
Finally, $$\sum_{i=1}^{n} x_i \ln x_i \ge -\ln n$$


问:
Normal line calculation from a given surface to the xy-plane
Question: Suppose you head toward the $xy$-plane from the surface $x^2+y^2-z^2=-1$ at the point $(1, 1, \sqrt{3})$ by following the normal line to the surface at that point. What are the $x$ and $y$ coordinates at which you will hit the $xy$-plane?$

My attempted Solution: I know that the normal line has the parametric equation

$$r(t) = <1+2t, 1+2t, \sqrt{3}+2\sqrt{3}t>$$
but I'm unsure of how to find the $xy$ coordinates at which this line hits the $xy$ plane. I tried setting $z=0$ and having $\sqrt{3}+2\sqrt{3}t = 0$. In this case, $t=-1/2$. Solving for $x$ and $y$, they are somehow also both zero.

Could someone please show me how this is done? Thanks!

答:
Normal line has equation
$$\vec{r}(t)=(1,1,\sqrt 3)+t(2,2,-2\sqrt 3)$$
Setting $z=0$ we get $\sqrt 3-2t\sqrt 3=0\to t=\frac12$

Thus the intersection point with $xy$-plane has coordinates $(2,2,0)$.


问:
Number of arbitrary constants in an ODE
Consider the general solution of a differential equation-
$$y=(C_1+C_2)\cos(x+C_3)+C_4\exp(x)+C_5$$
Without differentiating this equation and finding the differential equation for it, how can we say what the order of that DE is. Ofcourse, we can always count the number of arbitrary constants, which in this case I think are $4$ (since $C_1+ C_2$ is just one). But my textbook says that the DE has order $3$. Am I missing something? I think my mistake is in not absorbing one more constant, which I tried doing in the following way-
$$y=K (\cos C_3 \cos x-\sin C_3 \sin x)+C_4\exp(x)+C_5$$
and then we can also absorb the cosine and sine of $C_3$. But even here, it appears to me that we still have $4$ constants.
Any help is appreciated.

答:
As $C_1+C_2=C_0$, so there are effectively 4 parameters in the solution so its ODE must be of oder $4$.
Let us set up the ODE for
$$y=C_0 \cos(x+C_3)+C_4 e^x+C_5~~~~(1)$$
D. w. r. $x$, then
$$y'=-C_0 \sin(x+C_3)+C_4 e^x \implies e^{-x}y'=-C_0 e^{-x}\sin(x+C_3)+C_4~~~(2)$$
D. w.r. t. $x$
$$-e^{-x}y'+e^{-x}y''=C_0 e^{-x} \sin(x+C_3)-C_0 e^{-x} \cos (x+C_3)$$
$$\implies y''-y'=C_0 \sin(x+C_3)-C_0 \cos(x+C_3)~~~~(3).$$
D. W.R. t. $X$
$$\implies y'''-y''=C_0 \cos(x+C_3)+C_0 \sin (x+C_3)~~~(4)$$
Adding  (3) and (4), we get
$$y'''-y'=2C_0 \sin (x+C_3)~~~~(5)$$
Subtracting (3) and (4) we get
$$y''-y'-y'''+y''=-2C_0 \cos(x+C_3)~~~~(6)$$
D. w. r. t. $x$ in (5), we get
$$y''''-y''=2C_0 \cos (x+C_3)~~~~(7)$$
Adding (6)and  (7) , we successfully get the req\uired fouth degree ODE as:
$$y''''-y'''+y''-y'=0,$$
where all the parameters are eliminated.

Title: Number of arbitrary constants in an ODE


问:
Euler's formula for non-planar graph
Could anyone please provide an example of non-planar graph, that corresponds to Euler's formula:
$$v - e + f = 2$$

or show that there is no such a graph?

答: In general, Euler's formula can only be used to prove that a graph is non-planar. To prove planarity, one usually either finds an actual plane drawing or shows that the graph contains no $K_5$ or $K_{3,3}$ minors.

Title: Euler's formula for non-planar graph


问:
the ways completion of p-adic integers
I have heard that there are two ways of constructing completion of p-adic numbers:  One gives an equivalence relation on a Cauchy sequence. Unbeknownst to me , the another way  is using inverse limit.....(★)

To begin with, according to the following lecture, the author defines profinite completion.

$G$ be a group and $\mathcal{I}$ be a collection of normal subgroup
$N \triangleleft G $ of finite index. Let's efine $N \le N'$ iff $N'
 \subset N$. Then, this makes directed poset. If $N \subset N'$, then
we have a surjective homomorphism $G/N' \to G/N$ and the inverse limit
$\underset{\longleftarrow}{\lim}G/N =:\hat{G} $ is called profinite
completion.

Here is my question: Is the above definition really related to the completion as we know?
For example, (This is also an example within the following lecture )

Example )  p-adic integer ${\mathbb{Z}}_p$ is a profinite completion of $\mathbb{Z}$. i.e, ${\mathbb{Z}}_p=\underset{\longleftarrow}{\lim}\mathbb{Z}/p^n
\mathbb{Z}$

In the sense of completion, it looks nonsense for me because, unlike $\mathbb{Q}$, $\mathbb{Z}$ is already a complete set. Thus, when comparing such an Example  to constructing $\mathbb{Q}_p$, I think that profinite completion is not totally related to the concept of completion. In other words, the completion of a ring is just an algebraic concept that is not related to the completion. Though, my clumsy conclusion conflicts with the description that I have heard, (★).

答:
$\varprojlim \Bbb{Z/p^nZ}$ is the ring of sequences $a_n\in \Bbb{Z/p^nZ}$ such that $a_n=a_m\bmod p^m$,

isomorphic to the ring of limits of sequences of integers such that $a_n=a_m\bmod p^m$ (if $\forall n,a_n=0\bmod p^n$ then $\lim a_n=0$ in this ring)

isomorphic to the ring of limits of sequences of integers that converge $\bmod p^m$ for all $m$ (if $\forall m,\lim a_n=0\bmod p^m$ then $\lim a_n=0$ in this ring)

Clearly isomorphic to the completion of $\Bbb{Z}$ for $|.|_p$.

Title: the ways completion of p-adic integers


问:
Finding $\iint_D e^{x+y} \,\mathrm{d}x\,\mathrm{d}y$ without substitution
Is it possible to set this integral up without using substitution?

$$\iint_D e^{x+y} \,\mathrm{d}x\,\mathrm{d}y\,,$$ where

$$D = \left\{-1\le x+y \le 1, -1 \le -x + y \le 1\right\}$$

The answer is: $e-\frac{1}{e}$

答:
Sure, and indeed it is quite easy.
First you can divide $D$ in $D_1=\left\{0\leq x\leq 1, x-1 \le y \le -x+1\right\}$ and $D_2=\left\{-1\leq x\leq 0, -x-1 \le y \le x+1\right\}$, then

$$
\iint_{D_1} e^{x+y} dxdy=\int_0^1 \left(\int_{x-1}^{-x+1} e^{x+y} dy\right)dx= \int_0^1[e^{x+y}]_{x-1}^{-x+1} dx=\int_0^1 (e-e^{2x-1}) dx
$$

now yo can complete the whole exercise by yourself.


答:
You can indeed rewrite the solution as

$$y=a\cos x+b\sin x+ce^x+d$$ where the four constants are independent. As the functions $\cos x,\sin x,e^x$ and $1$ are linearly independent, the expression does have four DOFs.

The terms will be respectively eliminated by the operators $D^2+1,D-1$, and $D$,
giving the equation

$$(D^2+1)(D-1)Dy=0$$

or

$$y''''-y'''+y''-y'=0$$ like said in the comments.

Title: Number of arbitrary constants in an ODE

Body:
Consider the general solution of a differential equation-
$$y=(C_1+C_2)\cos(x+C_3)+C_4\exp(x)+C_5$$
Without differentiating this equation and finding the differential equation for it, how can we say what the order of that DE is. Ofcourse, we can always count the number of arbitrary constants, which in this case I think are $4$ (since $C_1+ C_2$ is just one). But my textbook says that the DE has order $3$. Am I missing something? I think my mistake is in not absorbing one more constant, which I tried doing in the following way-
$$y=K (\cos C_3 \cos x-\sin C_3 \sin x)+C_4\exp(x)+C_5$$
and then we can also absorb the cosine and sine of $C_3$. But even here, it appears to me that we still have $4$ constants.
Any help is appreciated.


答:
Yes you are integrating over a square bound by $4$ given line segments. Your integral should be

$\displaystyle \int_{-1}^{0}\int_{-1-y}^{1+y}  e^{x+y} \, dx \, dy + \displaystyle \int_{0}^{1}\int_{y-1}^{1-y}  e^{x+y} \, dx \, dy$

You do not need substitution.

Title: Finding $\iint_D e^{x+y} \,\mathrm{d}x\,\mathrm{d}y$ without substitution

Body:
Is it possible to set this integral up without using substitution?

$$\iint_D e^{x+y} \,\mathrm{d}x\,\mathrm{d}y\,,$$ where

$$D = \left\{-1\le x+y \le 1, -1 \le -x + y \le 1\right\}$$

The answer is: $e-\frac{1}{e}$


答:
Suppose $i\ne j$ and $x_i>x_j.$ For $y\in [0,x_i-x_j]$ let $g(y)=(x_i-y)\ln (x_i-y)+(x_j+y)\ln (x_j+y).$ Then $g'(0)<0.$ So for some (sufficiently small) $y_1\in (0,x_i-x_j)$ we have $g(y_1)<g(0).$

So, in the expression for $f,$ if we replace $x_i,x_j$ with $x_i-y_1,x_j+y_1,$ we get a lower value for $f$. So $f$ is not minimized if some $x_i>x_j.$

So (i)$\,\min f$ exists,and (ii) $\,\min f$ does not occur if any two $x_i,x_j$ are unequal. Therefore $\min f$ occurs when and only when every $x_i=1/m.$


问:
General version of Hoeffding's inequality
I know that if we have $X_{1},\dots,X_{n}$ independant random variables such that $\left|X_{i}-\mathbb{E}(X_{i})\right| \leq M \;$ for all $1 \leq i \leq n$ then if we define $X=\sum_{i=1}^{n}X_{i}\;$ we have for all $a>0$:

$$\mathbb{P}(X-\mathbb{E}(X)\geq a)\leq \exp\left(\frac{-a^{2}}{2nM^{2}}\right).$$

How can I derive from that the general version that says that:

$$\mathbb{P}(\left|X-\mathbb{E}(X)\right|\geq a)\leq2\cdot \exp\left(\frac{-a^{2}}{2nM^{2}}\right)?$$

答:
Apply the fact you know to $X'_i:=-X_i$, which satisfies the same assumptions: this gives
$$
\mathbb{P}(-X+\mathbb{E}(X)\geq a)\leqslant \exp\left(\frac{-a^{2}}{2nM^{2}}\right)
$$
or in other words,
$$
\mathbb{P}( X- \mathbb{E}(X)\leqslant  -a)\leqslant \exp\left(\frac{-a^{2}}{2nM^{2}}\right).
$$
Now, the event $\left\{\left|X-\mathbb{E}(X)\right|\geq a\right\}$ is the union of the events $A:=\left\{ X-\mathbb{E}(X) \geq a\right\}$ and $B:=\left\{ X-\mathbb{E}(X)\leqslant -a\right\}$; bound the probability using of $\left\{\left|X-\mathbb{E}(X)\right|\geq a\right\}$ by the sum of probabilities of the   events $A$ and $B$.


问:
Addition vs multiplication as binary operators
It is known that arithmetic operations of addition and multiplication are binary operations that take in two inputs and give out a single output. However, consider the following scenario:

Suppose I have a jar containing some number of balls, and I give you some balls to add to the jar. If I asked you to add three balls into the jar, then that's easy, you'd just toss em in there. However, let's say I asked you to double the number of balls in the jar, you wouldn't be able to do that unless I showed you many balls were in the jar.

This is sort of strange: We require to know both inputs for multiplication but we don't require to know both to do addition. Why is this so when they are both known to be binary operators?

答: You still won't know how many balls are inside the jar until you look inside the jar. You can't add two numbers when you only know one of the numbers...</p>


问:
Relation between $n$ and $m$ with $\mathfrak{g}^{(n)} = \mathfrak{g}^{(n+1)}$ and $\mathrm{rad}(\mathfrak{g})^{(m)} = 0$
Let $\mathfrak{g}$ be a finite-dimensional Lie algebra, let $\mathfrak{g}^{(i)}$ with $i \geq 0$ denote the derived series of $\mathfrak{g}$ (with $\mathfrak{g}^{(0)} = \mathfrak{g}$) and let $\operatorname{rad}(\mathfrak{g})$ be the solvable radical of $\mathfrak{g}$, i.e. the unique maxmimal solvable ideal of $\mathfrak{g}$.
There exist minimal natural numbers $n$ and $m$ with
$$
  \mathfrak{g}^{(n)} = \mathfrak{g}^{(n+1)} \,
  \qquad\text{and}\qquad
  \operatorname{rad}(\mathfrak{g})^{(m)} = 0 \,.
$$

Question.
Is there some relation between the numbers $n$ and $m$?

Some thoughts:

If $\mathfrak{g}$ is solvable, then $\operatorname{rad}(\mathfrak{g}) = \mathfrak{g}$ and $n = m$.

If $\mathfrak{g}$ is semisimple, then $n = 0 = m$.

There exist perfect, non-semisimple Lie algebras. It may therefore happen that $n = 0 < m$.
But the only examples I know have $m = 1$.

答:
Benayadi has constructed a perfect non-semisimple Lie algebra $L$
in dimension $25$, where $\operatorname{rad}(L)$ is $2$-step solvable, so with $m=2$. It is given by
\begin{align*}
L & =\mathfrak{sl}_2(\Bbb C)\ltimes (V(7)\oplus V(5)\oplus V(7)\oplus V(3)) \\
   & =\mathfrak{sl}_2(\Bbb C)\ltimes (L_1\oplus L_2\oplus L_3\oplus L_4)
\end{align*}
where $V(n)$ is the $n$-dimensional irreducible representation
of $\mathfrak{sl}_2(\Bbb C)$ and
$$
[L_1,L_1]=L_3,\; [L_1,L_2]=L_4,\; [L_2,L_2]=L_4,\; [L_1,L_3]=L_4.
$$
So the radical is not abelian. This can be generalized.

Benayadi had another motivation of course, he wanted to find a non-semisimple perfect Lie algebra of lowest dimension satisfying

$$
H^0(L,L)=H^1(L,L)=0.
$$

Reference: S. Benayadi: Structure of perfect Lie algebras without center and outer derivations. Annales de la faculte des sciences de Toulouse $6^e$ serie, tiome 5, no.2, (1996), p. 203-231.


问:
Computing limits in matrix exponential
If $A \in B(V)$ where $V$ is a Banach space, and $B(V)$ is set of all bounded linear operators on $V$, then show if $lim_{n→\infty} A_n=A$, then $lim_{n→\infty}(I+\frac{A_n}{n})^n = exp(A)$

I am very new in learning matrix exponentials and norm linear spaces, and but I can't approach.

Obviously $(I+\frac{A_n}{n})^n$ is bounded .

A small help is appreciated .

答:
Here's one way to prove it:

Step 1): For each fixed complex number $z$ have have $(1 + \frac{z}{n})^n \to \exp(z)$; this is simply Euler's formula.

Step 2): If $|z| \le R$ for a real number $R \ge 0$, then
$$
  \big|(1 + \frac{z}{n})^n\big| \le \big(1 + \frac{R}{n}\big)^n \to \exp(R).
$$
This shows that the sequence $\big((1 + \frac{z}{n})^n\big)_{n \in \mathbb{N}}$ is uniformly bounded for $z$ in any fixed compact subset of the complex plane.

Hence, the convergence $(1 + \frac{z}{n})^n \to \exp(z)$ is uniform on compact subsets of $\mathbb{C}$ by Step 1) and Montel's theorem.

Step 3): General observation:

If $(f_n)$ is a sequence of entire functions which converges to $0$ uniformly on compact subsets of $\mathbb{C}$, then for each number $R \ge 0$,
$$
  \sup\big\{ \|f_n(C)\|: \, C \in B(V), \; \|C\| \le R \big\} \to 0
  \quad \text{as } n \to \infty.
$$
Indeed, this follows from the Cauchy formula representation of the functional calculus, i.e.,
$$
  f_n(C) = \frac{1}{2\pi i} \oint_{|z| = {R+1}} f_n(z) (z-C)^{-1} \, dz.
$$
(Use that boundedness of the resolvent $(z-C)^{-1}$ on the circle $|z| = R+1$ follows from $\|C\| \le R$ and from the Neumann series representation of the resolvent.

Step 4):
Use a $2\varepsilon$-argument: For each index $n$, we have
$$
  \left\|(1 + \frac{A_n}{n})^n - \exp(A)\right\|
  \le
  \left\|(1 + \frac{A_n}{n})^n - \exp(A_n)\right\|
  +
  \left\| \exp(A_n) - \exp(A) \right\|.
$$
The first summand converges to $0$ according to Steps 2) and 3) above, and the second summand converges to $0$ due to the continuity of the operator exponential function.

Title: Computing limits in matrix exponential


问:
$\int_0^\infty(\log x)^2(\mathrm{sech}\,x)^2\mathrm dx$
Is there any closed-form representation for the following integral?
$$\int_0^\infty(\log x)^2(\mathrm{sech}\,x)^2\mathrm dx,$$
where $\mathrm{sech}\,x$ is the hyperbolic secant, $\mathrm{sech}\,x=\dfrac{2}{e^x+e^{-x}}$.</p>

答:
Hint: Use the integral:
$$\int_0^\infty x^n\,\text{sech}^2x\,\mathrm dx=(2^{1-n}-4^{1-n})\,\Gamma(n+1)\,\zeta(n)$$
and take the 2nd derivative with respect to $n$.


问:
$T_f:L^2[0,1] \to L^2[0,1]$ defined as $T_f=fg$ is bounded and compact only if $f=0$.
I am trying to solve the following problem:

Let $f:[0,1] \to \mathbb{C}$ be a continuous function and $T_f:L^2[0,1] \to L^2[0,1]$ be the operator given by $T_f(g)=fg, \; g \in L^2[0,1].$ Prove that $T_f$ is a bounded linear operator on $L^2[0,1]$ and that $T_f$ is compact iff $f=0$.

Now, the linearity of $T_f$ follows by linearity of the usual operations. Indeed
$$
T_f(\alpha g_1 + g_2)= f(\alpha g_1 + g_2)= \alpha fg_1 + fg_2= \alpha T_f g_1 + T_f g_2.
$$
For boundedness, we have that
\begin{align*}
\|T_f\|^2_{L^2[0,1]} = \int_{0}^{1} |f|^2|g|^2 &\leq  \int_{0}^{1} |f|^2 \quad (\|g\|\leq 1) \\
&\leq \|f\|^2_{L^2[0,1]}.
\end{align*}

Then if $f=0$ clearly $T_f$ is compact (being the range just $\{0\}$), but I am having trubles showing that if $f\neq 0$ then $T_f$ is not compact. Any help is appreciated.

答:
(As noted, your proof of boundedness is somewhat confused...)

Say $f\ne0$. Since $f$ is continuous there exist an open interval $I\ne\emptyset$ and $\delta>0$ such that $|f(x)|\ge\delta$ for all $x\in I$. Define an "extension" operator $E:L^2(I)\to L^2([0,1])$ by  $$Eg(x)=\begin{cases}g(x),&(x\in I),\\0,&(x\notin I).\end{cases}$$

It follows that for every $h\in L^2(I)$ there exists $g\in L^2([0,1])$ with $$Eh=T_fg$$and $||g||_2\ge\frac1\delta||h||_2$, and the non-compactness of $T_f(B)$ folllows.


问:
The ring of integers in a number field is a maximal order.
Let $F$ be a number field and $O_F$ be the ring of integers i.e. the integral closure of $\Bbb Z$ in $F$. I know that it is an order in $F$, i.e. a subring which is finitely generated as $\Bbb Z$-module and contains a $\Bbb Q$-basis of $F$.

I want to show that $O_F$ is a maximal order i.e. $R \subset F$ order $\implies R \subset O_F$.

Since $R$ is an order, it is a $\Bbb Z$-module generated by some $b_1, ..., b_n$. Then how do I show that the characteristic polynomials of the $b_i$ lie in $\Bbb Z[T]$ ?

答: The multiplication by $b_k$ is represented in the $b_j$ generating set by an integer matrix $B_k\in M_n(\Bbb{Z})$, it is obvious that $f_k(x)=\det(xI-B_k)\in \Bbb{Z}[x]_{monic}$. Caley Hamilton gives that $f_k(B_k)=0$ in $M_n(\Bbb{Z})$ thus $f_k(b_k)=0$ in the order.


问:
Does there exist $u \in V$ so that $\left\langle u, e_{n}\right\rangle=1 / \sqrt{n(n+2)} ?$
Let $V$ be an inner product space, and let $\left\{e_{n}\right\}_{n=1}^{\infty}$ be an orthonormal system.

Assume that $V$ is a complete inner product space (a Hilbert space). Does there exist $u \in V$ so that $\left\langle u, e_{n}\right\rangle=1 / \sqrt{n(n+2)} ?$

I am a little bit confused because of course I can say that $$u=\large{\sum_{n=1}^{n=\infty}}\frac{ e_n}{\sqrt{n(n+2)}}$$ But Is it fine that I am assuming that my vector is infity linear combination?

答: Yes, that is correct. Note that the series$$\sum_{n=1}^\infty\frac{e_n}{\sqrt{n(n+2)}}$$is a auchy series, and therefore it converges, since your space is a Hilbert space. And it is a Cauchy series since, if $m,n\in\Bbb N$ and $m\geqslant n$,$$\left\|\sum_{k=n}^m\frac{e_n}{\sqrt{k(k+2)}}\right\|^2=\sum_{k=n}^m\frac1{k(k+2)}$$and the series $\sum_{n=1}^\infty\frac1{n(n+2)}$ converges.


问:
Finding matrices for which $x^2+1$ is the minimal polynomial

Show that there is no $A\in M_3(\mathbb{R})$ whose minimal polynomial is $x^2+1$, but there is $B\in M_2(\mathbb{R})$ and $C\in M_3(\mathbb{C})$ whose minimal polynomials are both $x^2+1$. $M_n(\mathbb{F})$ denotes the vector space of $n\times n$ matrices over the field $\mathbb{F}$.

My work:

So if there exists $A\in M_3(\mathbb{R})$ whose minimal polynomial is $x^2+1$, its characteristic polynomial must be of the form $(x-a)(x^2+1)$ for some $a$, since the minimal polynomial divides the characteristic polynomial and the characteristic polynomial has degree $3$. I'm not able to find a contradiction from here.

For $B\in M_2(\mathbb{R})$, by a similar argument, $x^2+1$ must be $B$'s characteristic polynomial. Another thing I noticed is that $B$ cannot be diagonalizable, since if it were, $x^2+1$ should have split into distinct linear factors over $\mathbb{R}$, but that is clearly not the case. Doing a brute force calculation, considering $$B = \left[\begin{matrix} a & b \\ c & d  \end{matrix}\right]$$ we get $x^2 - (a+d)x + ad - bc = x^2 + 1$, so $a+d = 0$ and $ad-bc=1$. This gives $a^2+1 = -bc$. Choosing $(a,b,c,d) = (1,-2,1,-1)$ yields a desirable $B$.

I am planning on doing something similar for $C\in M_3(\mathbb{C})$, but seems like it would be overkill. Is there an easier way around this? It'd be great if someone could help me out with disproving the existence of $A$, proving the existence of $C$, and finding a somewhat better solution for the existence of $B$.

Thanks!

答:
Hint: Consider these:

$A\in M_3(\mathbb{R}) \implies \det A \in\mathbb{R}$

$A^2=-I \implies (\det A)^2 = \det (A^2) = \det(-I) = -1$

The simplest $B$ is
$
\begin{pmatrix}0&-1\\1&\hphantom-0\end{pmatrix}
$
(see also Matrix representation of complex numbers).

The simplest $C$ is
$
\begin{pmatrix}i&0&0\\0&i&0\\0&0&-i\end{pmatrix}
$


问:
Average of a sequence of r.v.s converge to 0 a.s.
We have $\left\{Y_{n}\right\}$ to be a sequence of independent random variables. $X_{n}$ is defined by $\mathbb{P}\left(Y_{n}=1\right)=\mathbb{P}\left(Y_{n}=-1\right)=\frac{1}{2}\left(1-2^{-n}\right), \mathbb{P}\left(Y_{n}=2^{n}\right)=\mathbb{P}\left(Y_{n}=-2^{n}\right)=2^{-(n+1)}$
$n=1,2, \ldots$

How can we prove:
$$
\frac{1}{n}\left(Y_{1}+\cdots+Y_{n}\right) \stackrel{a . s .}{\rightarrow} 0
$$

I can show $Y_n/n$ converges to 0 a.s., but I am not sure does it imply the sum is also convergent to 0 a.s.

答:
Let $Z_k:=Y_k\mathbb{1}_{\{\lvert Y_k\rvert\leqslant 1\}}$; then $\{Y_k\neq Z_k\}$ is an event having probability $2^{-k}$ and by the Borel-Cantelli lemma, we can show that $n^{-1}\sum_{k=1}^n(Y_k-Z_k)$ converges to $0$ almost surely. It thus suffices to prove that $n^{-1}\sum_{k=1}^nZ_k$ to $0$ almost surely, which is simpler than the original task, as $Z_k$ takes the values $-1$ and $1$ with probability $1/2-2^{-k-1}$ and $0$ with probability $2^{-k}$.

Nevertheless, we cannot use directly the strong law of large numbers because $(Z_k)$ is not identically distributed. Nevertheless, the fact that the $Z_k$ have bounded variance can be used in order to bound
$$
\mathbb P\left(2^{-N}\max_{1\leqslant n\leqslant 2^N}\left\lvert \sum_{k=1}^nZ_k\right\rvert>\varepsilon\right)
$$
thanks to Kolmogorov's maximal inequality.


问:
Simplify with fractional exponents and negative exponents
I am trying to simplify

$$ \left(\frac{3x ^{3/2}y^3}{x^2 y^{-1/2}}\right)^{-2} $$

It seems pretty simple at first. I know that a negative exponent means you flip a fraction. So I flip it.

$$  \left(\frac{x^2 y^{-1/2}}{3x^{3/2}y^3}\right)^2$$

Now I need to square it, which is tricky because there are a lot of weird rules with squaring. This is probably where I went wrong.

$$  \left(\frac{x^2 y^{-1/2}}{9x^{3/2}y^3}\right)$$

Now I need to try and simplify things. I know that I can get rid of the $x$ on top since there is a larger one on the bottom.

$4/2 - 3/2 = 1/2$

$$  \left(\frac{x^{1/2} y^{-1/2}}{9y^3}\right)$$

Now I need to get rid of the $y$ exponent.

I am not sure how that is possible.

答:
The main problem is that you didn’t actually perform the squaring:

$$\left(\frac{x^2y^{-1/2}}{3x^{3/2}y^3}\right)^2=\frac{(x^2y^{-1/2})^2}{(3x^{3/2}y^3)^2}=\frac{x^4y^{-1}}{9x^3y^6}\;,$$

since squaring entails multiplying each exponent by $2$. Can you finish it from there?