问:
Application of Sylow Theorems to groups of order $pq^2r$

Suppose G is a group of order $pq^2r$, where $p,q,r$ are primes and $p  < q < r.$ Suppose further that $q^2 \not\equiv 1\pmod{r}$, $pq^2 \not\equiv 1\pmod{r}$, and $\gcd(q,r-1) = 1$. Show that either $G$ contains a unique Sylow $r$-subgroup, or that $Z(G)$ contains an element of order $q$.

Here is my attempt at the problem. Striving for a contradiction, suppose that neither $G$ has a unique Sylow $r$-subgroup, nor does $Z(G)$ have an element of order $q$. By the third part of Sylow's Theorem, $n_r \equiv 1\pmod{r}$ and $n_r \in \{1,p,q,q^2,pq,pq^2\}$. By our assumption that $G$ does not have a unique Sylow $r$-subgroup, we cannot have $n_r = 1$. We cannot have $n_r = p$, since $p \not\equiv 1\pmod{r}$, and similarly $n_r \neq q.$ Furthermore, by our hypotheses that $q^2, pq^2 \not\equiv 1\pmod{r}$, we deduce by exhaustion that $n_r = pq.$

Next, we look at possible values for $|Z(G)|.$ By Lagrange's Theorem, we have $|Z(G)| \in \{1,p,q,q^2,r,pq,pq^2,pr,qr,q^2r,pqr,pq^2r\}.$ By Cauchy's Theorem, combined with our assumption that $Z(G)$ contains no element of order q, we immediately reduce this set of possibilities to $\{1,p,r,pr\}.$ Now suppose $R \in \text{Syl}_r(G)$. Then $[G:N_G(R)] = n_r = pq$ implies $|N_G(R)| = qr$. Since $Z(G) \leq N_G(R)$, we must have $|Z(G)|$ divides $qr$. Thus $|Z(G)| \in \{1,r\}$. But if $|Z(G)| = r$, then $Z(G)$ is a normal Sylow $r$-subgroup, which would imply $n_r = 1$, a contradiction. We deduce that $G$ has trivial center.

Finally, if we look at $n_q$, we see that $n_q \in \{1,p,r,pr\}$. Since $n_q \equiv 1\pmod{q}$, we cannot have $n_q = p$. If $n_q = r$, then we may write $r = 1 + kq$ for some $k \in \mathbb{Z}$. But then $r-1 = kq$, which implies that $q = \gcd(q,r-q)$, a contradiction. Thus either $n_q = 1$ or $n_q = pr.$

From here I'm kind of stuck.

答:
You are off to a good start. I will jump in at the point were you had concluded that
if $n_r>1$ then $n_r=pq$ and $|N_G(R)|=qr$ for any Sylow $r$-subgroup $R$.

Let us fix one $R$. We know that there exists an element $z\in N_G(R)$ such that $\operatorname{ord}_G(z)=q$. The automorphism group of $R$ is cyclic of order $r-1$, so it has no elements of order $q$. Therefore we can conclude that $z$ actually centralizes $R$.

Getting warmer.

Next, we also know that $z$ is an element of at least one Sylow $q$-subgroup. Let $Q$ be one such. Because $Q$ has order $q^2$ we know that it is abelian. Also, the $Q$-orbit of conjugates of $R$ must have exactly $q$ Sylow $r$-subgroups. This is because

$Q$ is not contained in $N_G(R)$, so the orbit has size $>1$, and

$\operatorname{Stab}_R(Q)$ contains $z$, so the orbit has size $<q^2$.

Let $x\in Q$ be arbitrary. We have $C_G(xRx^{-1})=xC_G(R)x^{-1}$. It follows that
$xzx^{-1}\in C_G(xRx^{-1})$. But $Q$ is abelian, so we have shown that
$$
z\in C_G(xRx^{-1})\ \text{for all $x\in Q$.}
$$

Let $Z=C_G(z)$:

Because $Q$ is abelian $q^2\mid |Z|$.

Because $z$ centralizes $R$ we have $r\mid |Z|$, so $q^2r\mid |Z|$.

But, we also know that $Z$ contains at least those $q$ conjugates of $R$. As $q^2\not\equiv1\pmod r$ the number of Sylow $r$-subgroups of $Z$ cannot be $1,q$ or $q^2$ either.

It follows that $Z$ must have more than $q^2r$ elements Hence $Z=G$, and $z\in Z(G)$.


问:
Counting number of distinct regions with intersecting circles
Given $n$ circles of possibly different radii, how many distinct regions can there be?
For small $n$, I can work it out with pictures. (I'm pretty sure $n=4$ can yield 13 distinct regions, but not positive.)

Just curious what an approach to the problem could be.

E.g. I have considered that is could be a quadratic sequence based on skimpy evidence, but don't know why that would be, or how to prove it.

I would also be interested to know if there was a way of doing this problem with regular $n$-gons. (I am assuming that there would be more regions, but am not sure.)

答: OEIS gives the number of regions for circles.  It gets 14 for 4 circles, including the exterior.  The general formula is $n^2-n+2$.  Allowing different size circles doesn't help.  The proof is by induction.  With one circle there are $2$ regions.  Assuming the formula is true for $n$, the new circle can cross old ones in at most $2n$ points.  Each segment can cut an existing region in two parts, adding $2n$ regions.  So the maximum number of regions for $n+1$ circles is $n^2-n+2+2n=(n+1)^2-(n+1)+2$</p>

ContentLicense: CC BY-SA 3.0


问:
Diagonalizable Operators: An Operational Extension
Let $T$ be a diagonalizable operator on a vector space $V$. Prove that the operator
$$a_nT^n + a_{n-1}T^{n-1}+\cdots+a_1T+a_0 Id_V$$
 on $V$ is also diagonalizable for any scalars $a_1, a_1,\dots,a_n$.

First off, what is $Id_V$? I've never seen this before.

答:
Hint:

$$\exists\,P\;\,\text{an invertible operator}\,\;s.t.\;\, P^{-1}TP=D\;\,\text{is diagonal}\implies \forall\,n\in\Bbb N:$$

$$P^{-1}T^nP=\left(P^{-1}TP\right)^n=D^n\;\;\text{is diagonal,  and from here the claim follows}$$


问:
Find an ordered basis $\beta$ such that $[T]_{\beta}^{\beta}$ is diagonal
Consider the basis $\epsilon$ for $V=M_{2 \times 2}(\mathbb{R})$ with the following basis vectors:

$$ e_1 = \left( \begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix} \right) $$

$$ e_2 = \left( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right) $$

$$ e_3 = \left( \begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix} \right) $$

$$ e_4 = \left( \begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix} \right) $$

And consider the linear transformation $T: V \rightarrow V$:

$T(\left( \begin{matrix} a & b \\ c & d \end{matrix} \right)) = \left( \begin{matrix} 2a-2b & -a+3b \\ 4c-2d & 3c-d \end{matrix} \right)  $

Can somebody help me to find an ordered basis $\beta$ such that $[T]_{\beta}^{\beta}$ is diagonal? Thank you!

答:
$[T]_\beta^\beta=M_{\epsilon\to\beta}[T]_\epsilon^\epsilon M_{\beta\to\epsilon}=M^{-1}[T]_\epsilon^\epsilon M$ where $M=M_{\beta\to\epsilon}$ is the basis change matrix from basis $\beta$ to $\epsilon$.

You want $M^{-1}[T]_\epsilon^\epsilon M=\delta$, a diagonal matrix. So $\delta$ must contain the eigenvalues of $[T]_\epsilon^\epsilon$ along its diagonals and $M$ is the matrix whose column vectors are the corresponding linearly independent eigenvectors.

Once we have obtained $M=M_{\beta\to\epsilon}$, we can determine the $\beta$ basis vectors like this: the first vector in the basis has coordinates $[1,0,0,0]^T$ in the same basis, so its coordinates in the $\epsilon$ basis are given by $M[1,0,0,0]^T$ which is the first column vector of $M$, i.e. the first eigenvector of $[T]_\epsilon^\epsilon$. Similarly the $i^\text{th}$ basis vector is the $i^\text{th}$ column of $M$, i.e. the $i^\text{th}$ eigenvector of $[T]_\epsilon^\epsilon$.

For $[T]_\epsilon^\epsilon=\begin{bmatrix}2&-2&0&0\\-1&3&0&0\\0&0&4&-2\\0&0&3&-1\end{bmatrix}$ the eigenvalues and corresponding eigenvectors are $(1,[2,1,0,0]^T),(1,[0,0,2,3]^T),(2,[0,0,1,1]^T),(4,[-1,1,0,0]^T)$. You may verify that the eigenvectors are linearly independent. Thus, one option for $\beta$ is$$\{[2,1,0,0]^T,[0,0,2,3]^T,[0,0,1,1]^T,[-1,1,0,0]^T\}$$

ContentLicense: CC BY-SA 4.0


问:
Bounded linear operator norm
Given T a bounded linear operator $T:X \to Y$ where $X$ and $Y$ are normed spaces,

I'm trying to show that $\|T\|_*=\sup\{ \frac{\|Tx\|_Y}{\|x\|_X},x\neq0\}$  equals to $\sup\{\|T(x)\|_Y\,:\, \|x\|_X< 1\}$.

I can show that  $\|T\|_*=\sup\{\|T(x)\|_Y\,:\, \|x\|_X\le 1\}$, but I can't go further.
Do you have any suggestions?

答: Any $x_0$ such that $\|x_0\|_X=1$ can be approached by a sequence of elements in $\{x \in X : \|x\|_X < 1\}$. Since $x \mapsto \|T(x)\|_Y$ is a continuous function, you can approximate $\|T(x_0)\|$ arbitrarily closely with some $\|T(x)\|$ for $\|x\|_X < 1$.

Title: Bounded linear operator norm


答:
Right so what you can do is find an $x \in\overline{B_1(0)}$ that is $\epsilon >0$ close to the supremum. i.e.
$$\| Tx\| \geq \|T\| - \epsilon.$$
Then we can find $y \in B_1(0)$ such that $y = (1-\epsilon)x$ and so by linearity it follows that,
$$\| Ty\| \geq (1-\epsilon)(\|T\| - \epsilon)$$
and as $\epsilon$ was arbitrary we can conclude that,
$$\sup \{\|Tx\| : \|x\| \leq 1\} \leq \sup \{\|Tx\| : \|x\| < 1\}.$$
The other inequality should be clear.

Title: Bounded linear operator norm

Body:
Given T a bounded linear operator $T:X \to Y$ where $X$ and $Y$ are normed spaces,

I'm trying to show that $\|T\|_*=\sup\{ \frac{\|Tx\|_Y}{\|x\|_X},x\neq0\}$  equals to $\sup\{\|T(x)\|_Y\,:\, \|x\|_X< 1\}$.

I can show that  $\|T\|_*=\sup\{\|T(x)\|_Y\,:\, \|x\|_X\le 1\}$, but I can't go further.
Do you have any suggestions?


问:
$X_n$ independent variables if equations hold
Let $X_1,X_2,\dots ,X_n$ be random variables.

Statement 1

If for any $B_i \in \mathcal{B}^n, i=1,2,\dots,n$,

$$P\left( \overset{n}{\underset{i=1}{\bigcap}}X_i^{-1}\left(B_i \right) \right) = \prod_{i=1}^{n} P\left(X_i^{-1} \left(B_i \right) \right)$$

then $X_i$ are independent.

Statement 2

If we set $X=\left( X_1, X_2, \dots , X_n \right)$

and since
$$\overset{n}{\underset{i=1}{\bigcap}}X_i^{-1}\left(B_i \right) = X^{-1}\left(B_1 \times B_2 \times \dots X_n \right)$$

then the equation in Statement 1 is equivalent to

$$P_X \left( B_1 \times B_2 \times \dots \times B_n \right) = P_{X_1}(B_1) \times P_{X_2}(B_2) \times \dots \times P_{X_n}(B_n)$$

Does anyone have guidance on how to proceed proving these two statements?

EDIT

I want to show that if either Statement 1 OR Statement 2 is true for $B_i \in \mathcal{P}^n, i=1,2,...,n$, then $X_i$ are mutually independent.

$\mathcal{P}^n = \{ (a_1,b_1]\times(a_2,b_2]\times \dots \times (a_n,b_n]\} \cup \{ \emptyset \}$

Attempt

My idea is that if Statement 1 is true for $B_i \in \mathcal{P}^n, i=1,2,...,n$ then, then by setting some $B_i = \mathbb{R}^n \in \mathcal{P}^n$ we get that for every subset of indices $I = \{i_1, \dots, i_k \} \subset \{1,2, \dots, n\}$  we have

$$P\left( \overset{k}{\underset{m=1}{\bigcap}}X_{i_m}^{-1}\left(B_{i_m} \right) \right) = \prod_{m=1}^{k} P\left(X_{i_m}^{-1} \left(B_{i_m} \right) \right)$$

for any $B_{i_m} \in \mathcal{P}^n$.

But the above does not imply that $X_i$ are mutually independent since we need it to be true for any $B_{i_m} \in \mathcal{B}^n$. How can I proceed from here?

If Statement 2 is true I don't know how to proceed to prove independence.

答:
Let $(\Omega,\Sigma,\mathbb{P})$ be a probability space and $X:(\Omega,\Sigma,\mathbb{P})\to(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}))$ be a random vector.

Based on the assumption of independence, we can proceed as follows:

\begin{align*}
\mathbb{P}_{X}(B_{1}\times B_{2}\times\ldots\times B_{n}) & = \mathbb{P}(X^{-1}(B_{1}\times B_{2}\times\ldots\times B_{n}))\\\\
& = \mathbb{P}(\left\{\omega\in\Omega\mid X(\omega)\in B_{1}\times B_{2}\times\ldots\times B_{n}\right\})\\\\
& = \mathbb{P}(\{\omega\in\Omega \mid (X_{1}(\omega)\in B_{1})\wedge(X_{2}(\omega)\in B_{2})\wedge\ldots\wedge(X_{n}(\omega)\in B_{n})\}\\\\
& = \mathbb{P}(\{\omega\in\Omega \mid X_{1}(\omega)\in B_{1}\}\cap\ldots\cap\{\omega\in\Omega \mid X_{n}(\omega)\in B_{n}\})\\\\
& = \mathbb{P}\left(X^{-1}_{1}(B_{1})\cap X^{-1}_{2}(B_{2})\cap\ldots\cap X^{-1}_{n}(B_{n})\right)\\\\
& = \mathbb{P}(X^{-1}_{1}(B_{1}))\times\mathbb{P}(X^{-1}_{2}(B_{2}))\times\ldots\times\mathbb{P}(X^{-1}_{n}(B_{n}))\\\\
& = \mathbb{P}_{X_{1}}(B_{1})\times\mathbb{P}_{X_{2}}(B_{2})\times\ldots\times\mathbb{P}_{X_{n}}(B_{n})
\end{align*}

Hopefully this helps!


问:
How does $\frac{\nabla_\theta p(\theta\mid x)}{p(\theta\mid x)}$ become $\nabla_\theta \log(p(\theta \mid x))$ in this proof?
I'm trying to understand this proof of the gradient of the expected value of a certain function. Here is the full text of the proof:

Let $f(x)$ be any function, and $p(\theta \mid x)$ be any parameterized distribution over $x$ which is differentiable with respect to $\theta$. The gradient of the expectation of $f(x)$ can be derived as:

$$ \nabla_\theta \mathbb{E}_{x\sim p(\theta \mid x)}[f(x)] = \nabla_\theta \int_\Omega f(x)p(\theta \mid x)dx \quad \text{(makes sense. Just the definition of E)} $$
$$ = \int_\Omega f(x)\frac{\nabla_\theta p(\theta\mid x)}{p(\theta\mid x)}p(\theta\mid x) dx \quad \text{(Push in the gradient, and mult by 1=p(.)/p(.))}$$
$$ =\int_\Omega f(x)\nabla_\theta (\log p(\theta\mid x))p(\theta \mid x)dx \quad \text{(What??)}$$

(Here $\Omega$ just represents the domain of x)

The main paper says this proof is somehow derived from policy gradients, but I think that's for the later lines of the proof. I'm totally stumped by where that $\log$ comes from. Is this one of those simple 10th-grade algebra identities I just forgot? My stats knowledge is admittedly rusty, but I just don't understand how they got from line 2 to line 3.

答: This is a generalization of the fact that, by the chain rule, $$\frac{d}{dx}\log(f(x)) = \frac{f'(x)}{f(x)}$$ where here $\log(x)$ is the natural logarithm.


问:
Solve, $[3x + 1] = 2x - \frac{1}{2}$ , and find the sum of all roots.

Solve, $[3x + 1] = 2x - \frac{1}{2}$ , and find the sum of all roots.

What I Tried: I have :-
$$\rightarrow [3x + 1] = \frac{4x - 1}{2}$$
Now, the LHS is an integer, so $2|(4x-1)$ .
I can see that if $x$ is an integer, then $(4x - 1)$ is odd, so there are no integer solutions. I am not sure about fractions, as $x = \frac{1}{2}$ still makes it odd, but $x = \frac{1}{4}$ makes it even, although $2$ does not divided $0$.

This is where I am stuck, how should I proceed next?

Can anyone help me?

答:
As $2x-\dfrac{1}{2}$ is an integer, $x$ is either $n+\dfrac{1}{4}$ or $n+\dfrac{3}{4}$, $n \in \mathbb{Z}$. (Recall any real $x$ can be written as sum of an integer and a positive fraction).

Substitute this in original equation to solve for $n$.

$$n=-1 \Rightarrow x=-1+\dfrac{1}{4}$$

$$n=-2 \Rightarrow x=-2+\dfrac{3}{4}$$


问:
Find a closed form for the recurrence relation: $a_{n+2}-a_n=\sin\frac{n \pi }{2} \;\;\;a_0=a_1=1$
Find a closed form for the following recurrence relation:

$$a_{n+2}-a_n=\sin\frac{n \pi }{2} \;\;\;a_0=a_1=1\tag{$n \ge 0$}$$

This is what I've done so far:

Define a function $f(x):=\sum_{n\ge0}^{ }a_{n}x^{n}$,then:

$$\sum_{n\ge0}^{ }a_{n+2}x^{n+2}-x^{2}\sum_{n\ge0}^{ }a_{n}x^{n}=x^{2}\sum_{n\ge0}^{ }\sin\left(\frac{n\pi}{2}\right)x^{n}$$

The RHS is $\sum_{n\ge0}^{ }\sin\left(\frac{n\pi}{2}\right)x^{n}=\sum_{n\ge0}^{ }\left(-1\right)^{n}x^{2n+1}$
$$f\left(x\right)-a_{0}-a_{1}x-x^{2}f\left(x\right)=x^{3}\sum_{n\ge0}^{ }\left(-x^{2}\right)^{n}$$

Considering the initial values the equality transforms to:

$$f\left(x\right)-1-x-x^{2}f\left(x\right)=x^{3}\sum_{n\ge0}^{ }\left(-x^{2}\right)^{n}$$

\begin{align}
a_n=[x^n]f\left(x\right)
&=[x^n]\frac{2x^{3}+x^{2}+x+1}{\left(1+x^{2}\right)\left(1-x^{2}\right)}\\
&=\left[x^{n}\right]\left(\frac{5}{4}\frac{1}{1-x}-\frac{1}{4}\frac{1}{1+x}-\frac{1}{2}\frac{x}{1+x^{2}}\right)\\
&=\left[x^{n}\right]\left(\frac{5}{4}\sum_{n\ge0}^{ }x^{n}-\frac{1}{4}\sum_{n\ge0}^{ }\left(-1\right)^{n}x^{n}-\frac{1}{2}x\sum_{n\ge0}^{ }\underbrace{\binom{-1}{n}}_\text{=$(-1)^n$}x^{2n}\right)\\
&=\frac{5}{4}-\frac{1}{4}\left(-1\right)^{n}\end{align}

But the problem is that the answer is $$a_n=\frac{5}{4}-\frac{1}{4}\left(-1\right)^{n}-\frac{1}{2}\sin\frac{n\pi}{2}$$

And I don't understand why.

答:
The main question is why the $\sin\left(\frac{n\pi}{2}\right)$ term is missing.  There are two reasons:

You are using $n$ as both a real index and a dummy index when you write things like
$$ a_n = [x^n] \left( \frac{5}{4} \sum_{n \ge 0} x^n + \dots \right) $$
In $a_n$ and $[x^n]$, this refers to a specific index $n$, but in the summation, it is a dummy variable.  Instead you should write
$$ a_n = [x^n] \left( \frac{5}{4} \sum_{k \ge 0} x^k + \dots \right) $$
This is an extremely common mistake that even experienced mathematicians occasionally make.

Because of the confusion introduced by (1), you effectively wrote
$$ [x^n] \sum_{n \ge 0} (-1)^n x^{2n} = 0 $$
by dropping the last term on the right-hand side in your second-to-last equation.  But actually the correct way to do this is
$$ [x^n] \sum_{k \ge 0} (-1)^k x^{2k} = \begin{cases} (-1)^{n/2} & n\ \text{even} \\ 0 & n \ \text{odd} \end{cases} $$
If you try again with this change then you should get the correct answer.  (I didn't check for algebra errors, but you can figure that part out.)


问:
Must there be an invertible matrix $Q$ such that $Q^{-1}MQ=N$?
Let $M$ be a complex square matrix such that it has the following:

Minimal polynomial: $\mu(\lambda)= (\lambda-3)^2(\lambda+2)^3(\lambda-5)$

Characteristic polynomial: $\chi(\lambda)=(\lambda-3)^4(\lambda+2)^5(\lambda-5)^2$

$\dim(E_3)=2$ and $\dim(E_{-2})=3$, where $E_\lambda$ is the eigenspace for $\lambda \in \mathbb{R}\,$.

Now, my question is this:

If $N$ is another complex square matrix with the same characteristic polynomial, minimal polynomial, and geometric multiplicities as stated above, must there be an invertible matrix $Q$ such that $Q^{-1}MQ=N$ ?

If this is true, I would like to understand why. And if not, a counter-example would be much appreciated.

答:
Warning: This answer is incorrect, as currently given.

Correction is underway.

If you are given both the characteristic and the minimal polynomial of a matrix, then you know its Jordan canonical form (which is usually not unique because permutations of Jordan blocks are still allowed).

A matrix $M$ and its Jordan normal form $J$ are similar, which means $Q^{-1}MQ=J$ for some invertible matrix $Q$.

Your question is answered in the positive, because $M$ and $N$, sharing characteristic and minimal polynom, then also share the Jordan normal form, and both $M,N$ are similar to it. Now recall that similarity is an equivalence relation.

If you strive for uniqueness, then look after the Frobenius normal form of a square matrix.


问:
Area between parabolas
I would like to calculate the area between these 2 curves. As you can see the first one is

$x^2 = αy$ and the second one is $y^2 = 2αx$

What I tried is to solve the equation when the two curves are equal $x^2/α = \sqrt{2αx}$. After, that I would calculate the integral of the two functions in all the solutions I found from the above equation.

Do you think this is a correct approach or I need to see it differently. My question is mostly asked because it has the parameter $α$ and I am not quite sure how that changes the things.

答:
We will first rewrite the equations as functions of $x$:

$$f(x) = \frac{x^{2}}{a}$$

$$g(x)=\sqrt{2ax}$$

We may take the square root for the second because we only care about the positive branch. Setting $f(x) = g(x)$:

$$\frac{x^{2}}{\alpha} = \sqrt{2\alpha x}$$

$$x^{4} = \alpha^{2}\cdot 2\alpha x$$

$$x^{4} - 2\alpha^{3}x = 0$$

$$x(x^{3} - 2\alpha^{3}) = 0$$

$$x = 0, \sqrt[3]{2}\alpha$$

Note that $g(x) > f(x)$ for $0 < x < \sqrt[3]{2}\alpha$. Then, we find the area between the curves:

$$A = \int_{0}^{\sqrt[3]{2}\alpha}g(x) - f(x)\ dx = \int_{0}^{\sqrt[3]{2}\alpha}\sqrt{2\alpha x} - \frac{x^{2}}{\alpha}\ dx = \bigg(\frac{2\sqrt{2\alpha}x^{\frac{3}{2}}}{3} - \frac{x^{3}}{3\alpha}\bigg)\bigg\vert_{0}^{\sqrt[3]{2}\alpha}$$

$$=\frac{2\sqrt{2\alpha}}{3}(\sqrt[3]{2}\alpha)^{\frac{3}{2}} - \frac{1}{3\alpha}(\sqrt[3]{2}\alpha)^{3} = \frac{4\alpha^{2}}{3} - \frac{2\alpha^{2}}{3} = \boxed{\frac{2\alpha^{2}}{3}}$$

When you think about it, it makes sense that the area scales with $\alpha^{2}$ because one of the curves scales linearly with $\alpha$ in the $x$-direction, and the other scales in the $y$-direction.

Title: Area between parabolas


问:
Solving this Differential equation.
I have been given this task as a assignment but I this far not able to come anywhere near a solution.

Let $p_1,p_2$ continous gunctions from $\mathbb{R}$ to $\mathbb{R}$. The following Differential equation
$$y''+p_1(x)y'+p_2(x)y=0 \ \ \ \ \ (*)$$
has as a solution $y_1(x)=\sqrt{1+x^2}$ and the Wronskian of every 2 solutions of (*) is constant.

Find the basis solutions for (*) and the functions $p_1$ and $p_2$.

I tried using that the derivative of the Wronskian would be 0 but it didn't leed to anything.

Can anyone give me any hints into finding the solution for both tasks. Thank you very much in advance.

答:
$W$ is constant so:
$$y'_2y_1-y_2y'_1=C$$
$$ \left( \dfrac {y_2}{y_1} \right)'=\dfrac C {y_1^2}$$
This is a first order Differential equation where $y_1$ is given.
$$y_1= \sqrt {1+x^2}$$
Then you will have two equations to solve in $p_1,p_2$:
$$y_2''+p_1(x)y_2'+p_2(x)y_2=0$$
$$y_1''+p_1(x)y_1'+p_2(x)y_1=0$$

Title: Solving this Differential equation.


问:
prove $(\hat{f}g)^\lor$=$f*g^{\lor}$
I want to prove that $(\hat{f}g)^\lor$=$f*g^{\lor}$ give that $f$,$g$:$\mathbb{R}\to \mathbb{C}$ are integratble functions.

I know that $\hat{f}$ is bounded an the function $g$ is integratble zo the product $\hat{f}.g$ is integratble. This means that the left side of the equation makes sense. Now the function $f$ is intgratble and $\check{g}$ is bounded. This means that $f*\check{g}$ is well defined.

So I tried to fill in the left side of the formula and then with the Fubini positive , I tried to rewrite it to the right side of the formula but it never works out. I think I don't get how you need to fill in with several variables. Is there someone who can prove it?

答:
\begin{align}
(\hat f g)^{\vee} (x)&=\int_{\mathbb{R}}\hat f(t) g(t) e^{2\pi ity}dy \\
     &= \int_{\mathbb{R}}\int_{\mathbb{R}} f(y)  e^{-2\pi ity} g(t) e^{2\pi ity} dtdy
\\    &=\int_{\mathbb{R}}f(y)\int_{\mathbb{R}}   g(t) e^{2\pi it(x-y)} dtdy
\\&=\int_{\mathbb{R}}f(y)\check g(x-y)dy
\\&=(\hat fg)^{\vee}(x)
\end{align}

Title: prove $(\hat{f}g)^\lor$=$f*g^{\lor}$


问:
Problem finding determinant using elementary row or column operations.
I'm trying to find this determinant using row and column operations, but I got $-9$ as an answer and the right answer is $9$ and I couldn't figure out my mistake.  \begin{vmatrix}
&{1}&&{3}&&{4}&\\
\\
&{3}&&{6}&&{9}&\\
\\
&{1}&&{6}&&{4}&\\
\end{vmatrix} So what I did was I removed $3$ times column $1$ from column $2$, and after that I removed $4$ times column $1$ from column $3$. and I got:  \begin{vmatrix}
&{1}&&{0}&&{0}&\\
\\
&{3}&&{-3}&&{-3}&\\
\\
&{1}&&{3}&&{0}&\\
\end{vmatrix} but now, I tried to calculate the determinant according to first row and got $(-1)^2 *1 *(-9)=-9$
 Am I missing something? I would be happy if someone can tell me where my mistake is. Thanks in advance.

Body:
$$\begin{vmatrix}
1&0&0\\
3&-3&-3\\
1&3&0
\end{vmatrix}=1\begin{vmatrix}
-3&-3\\
3&0
\end{vmatrix}=-3(0)-3(-3)=+9$$


问:
Coming up with explicit formula for $f_\ast$
Suppose $f:S^1\times S^1\rightarrow S^1\times S^1$ is given by $f(z,w)=(z^3, w^2)$. What can be the explicit formula for $f_\ast$ then? Since $\pi_1(S^1)=\mathbb{Z}$, can I say it is $f_\ast(z,w)=(z+3, w+2)$? How can I check if this formula is correct?

答:
$\pi_1(S^1 \times S^1) = \mathbb{Z} \times \mathbb{Z}$. A homomorphism $f_* : \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}$ is determined by where it sends $(1,0)$ and $(0,1)$. Thus, it's enough to check that $f_* (1,0) = (3,0)$ and $f_* (0,1) = (0,2)$ to confirm that $f_*(z,w) = (3z, 2w)$.

To check $f_*(1,0) = (3,0)$, it's enough to look at a representative. Let $\gamma : S^1 \to S^1 \times S^1$ be defined by $\gamma(t) = (e^{2\pi i t} , 0)$. Then $[\gamma] = (1,0)$. Now $f_*([\gamma]) = [f\circ \gamma]$. We have
$$
(f \circ \gamma)(t) = (e^{6\pi i t} , 0)
$$
At this point it depends on your course how you should finish the argument, but the point is that $(e^{6\pi i t} , 0)$ winds $3$ times around the first circle and $0$ times around the second circle. Thus $[f\circ \gamma] = (3, 0)$.


问:
Since dividing $x=x^6$ by $x$ gives $1=x^5$, how can I get to $x=0$ as a root?
this might sound like a stupid question,
bear with me it probably is.

I know the solutions for $x=x^6$ are 1 and 0.

Now, since $1 \cdot x = 1 \cdot x^6 $ and it follows $ 1 \cdot x = 1 \cdot x \cdot x^5$ and I divide both sides by $ 1 \cdot x$ to get $ 1=x^5 $, how can I find $x=0$?

Thank you so much in advance.

答:
As said, you don't divide, but you factorize. Factorization gets all the solutions. Division, especially by $x$, as shown here:

$$ x = x^6 $$

then
$$ \frac{x}{x} = \frac{x^6}{x} $$

doesn't follow. In the first case, $x = 0$ is a solution, but then, in this case, $x = 0$ gets "washed away", because in this new equation, both sides evaluate to $\frac{0}{0}$, which is indeterminate.


答:
By using the expansion of $\gamma(k)$ we have:
$$\gamma(k)=-\frac{\sqrt{\frac{\pi }{2}}}{\sqrt{k}}+\frac{1}{8} \sqrt{k} \left(-3 \sqrt{2} \pi ^{3/2}-2 \gamma  \sqrt{2 \pi }+8 \sqrt{2 \pi }-2 \sqrt{2 \pi } \psi ^{(0)}\left(\frac{1}{2}\right)\right)+O\left(k^{3/2}\right)$$ therefore $\gamma(k)\to -\infty$ as $k\to 0^+$

Title: what is $\lim_{k \to 0}\frac{(k+1)A-3kA+2A^3}{(k-A^2)^{\frac{3}{2}}}$ where $A=-\sqrt{2} \frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k}{2})}$

Body:
let $k>0$ and define
$$A=-\sqrt{2} \frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k}{2})}$$

and

$$\gamma(k)=\frac{(k+1)A-3kA+2A^3}{(k-A^2)^{\frac{3}{2}}}.$$
Where $\Gamma$ is the Gamma function.

What is

$$\lim_{k \to 0^{+}} \gamma(k)?$$

First try

So

$$\gamma(k)=\frac{A-2kA+2A^3}{(k-A^2)^{\frac{3}{2}}}=\frac{A^{3}}{A^{3}}
\frac{1/A^2-2k/A^2+2}{(k/A^2-1)^{\frac{3}{2}}}=\frac{1/A^2-2k/A^2+2}{(k/A^2-1)^{\frac{3}{2}}}$$

But I  stuck in $\lim_{k \to 0}k/A^2$?

Since $\Gamma(\frac{1}{2})=\sqrt{\pi}$ so $A\sim -\frac{\sqrt{2\pi}}{\Gamma(\frac{k}{2})}$ so

$$\gamma(k)=\frac{\frac{\Gamma^2(\frac{k}{2})}{2\pi}-2k\frac{\Gamma^2(\frac{k}{2})}{2\pi}+2}{(k\frac{\Gamma^2(\frac{k}{2})}{2\pi}-1)^{\frac{3}{2}}}=
\frac{
\frac{\Gamma^2(\frac{k}{2})(1-2k)+4\pi}{2\pi}
}{
\frac{
(k\Gamma^2(\frac{k}{2}) -2\pi)^{3/2}
}{(2\pi)^{3/2}}
}
$$

$$=\sqrt(2\pi)\frac{\Gamma^2(\frac{k}{2})(1-2k)+4\pi}{(k\Gamma^2(\frac{k}{2}) -2\pi)^{3/2}}$$

By using $\Gamma(z)=\frac1z - \gamma + o(1)$ (mentioned by @Brian Tung)

$$\Gamma^2(\frac{k}{2})=(\frac{2}{k} - \gamma + o(1))^2
=\frac{4}{k^2}+\gamma^2+o^2(1)-2\gamma \frac{2}{k} +2\frac{2}{k}o(1)-2\gamma o(1) $$

Second try

By knowing that  $\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}$  [(4.1 The complement formula)][2]

$$A\sim -\frac{\sqrt{2\pi}}{\Gamma(\frac{k}{2})} =-\sqrt{2\pi} \frac{\Gamma(1-\frac{k}{2})\sin(\pi \frac{k}{2})}{\pi} \sim -\sqrt{2\pi} \frac{\sin(\pi \frac{k}{2})}{\pi}=-\sqrt{\frac{2}{\pi}} \sin (\frac{k\pi}{2}) $$

so by Taylor expansion
$$A\sim -\sqrt{\frac{2}{\pi}} \left\{ (\frac{k\pi}{2}) -\frac{(\frac{k\pi}{2})^3}{3!} \right\} $$ and so

$$\gamma(k)=\frac{(1-2k)A+2A^3}{(k-A^2)^{\frac{3}{2}}}=
\frac{(1-2k)\left(
-\sqrt{\frac{2}{\pi}} \left\{ (\frac{k\pi}{2}) -\frac{(\frac{k\pi}{2})^3}{3!} \right\}
\right)+2
\left(
-\sqrt{\frac{2}{\pi}} \left\{ (\frac{k\pi}{2}) -\frac{(\frac{k\pi}{2})^3}{3!} \right\}
\right)^3}{\left(k-
\left(
-\sqrt{\frac{2}{\pi}} \left\{ (\frac{k\pi}{2}) -\frac{(\frac{k\pi}{2})^3}{3!} \right\}
\right)^2\right)^{\frac{3}{2}}}
$$


问:
How to show that $\int_p^x dT \int_{p}^T dt \ f(t) = \int_{p}^x dt\ (x - t) f(t)$
Suppose $f(t)$ is continuous on the interval $[p,q]$ and that $p \leq x \leq q$. Formula (4.631) of Gradshteyn and Ryzhik then states that
$$
\int_p^x dT \int_{p}^T dt \ f(t) = \int_{p}^x dt\ (x - t) f(t) \ .
$$
How does one prove this result?

答:
(Gradshteyn/Rhyzik 4.631)

$$ \int_p^x dt_{n-1} \int_p^{t_{n-1}} dt_{n-2}\   \dots \dots \int_p^{t_1} \ f(t)\ dt  = \frac{1}{(n-1)!} \ \int_p^x\ (t-x)^{n-1} \ f(t)\ dt
$$

Using the indicator function $\mathbf{1}_{t \leqslant T}(t,T) = \begin{cases}1, & t \leqslant T\\0, &t > T \end{cases}$ and Fubini's theorem,

$$\int_p^x\ \left(\int_p^Tf(t)\, dt \right)\  dT = \int_p^x \ \left(\int_p^x\ f(t) \ \mathbf{1}_{t \leqslant T}\  dt \right)\ dT=$$
$$\int_p^x \ \left(\int_p^xf(t)\ \mathbf{1}_{t \leqslant T}\ dT\  \right)\ dt = \int_p^x\left(\int_t^xf(t)\, dT
\right)\  dt  = \int_p^x \ (x-t) \ f(t) \ dt$$


问:
Limit using Riemann sums
I'm having some trouble solving the following limit:

$$\lim_{n \to +\infty} \frac{\sqrt[n] e + \sqrt[n] {e^2} + ... + \sqrt[n] {e^n}}{n}$$

This question is in the "Riemann Sum" section so I think that we're supposed to turn this into an integral, so:

$$\lim_{n \to +\infty} \frac{\sqrt[n] e + \sqrt[n] {e^2} + ... + \sqrt[n] {e^n}}{n} = \lim_{n \to +\infty} \sum_{k=1}^n \dfrac{1}{n} \sqrt [n] {e^k} = \int_a^b f(x) dx$$

I think that $n$ is the number of partitions and $1/n$ is the length of each one, so this means that $b - a = 1$ or $b = a+1$, meaning that we only need to find a value for $a$ and $b$ will be that $+1$. But now I can't seem to find the value of  $a$ nor $f(x)$. How can I solve this?

答: Note that $\sqrt[n]{e^k}=e^{k/n}$ and that therefore$$\lim_{n\to\infty}\frac1n\sum_{k=1}^ne^{k/n}=\int_0^1e^x\,\mathrm dx=e-1.$$

Title: Limit using Riemann sums


问:
How many 5 digit numbers are there that doesn't have "13" in them?
What is the general formula for N digits?
1 and 3 can exist in the number but they can't be next to each other.
For example, 13598 or 54135 isn't allowed because they have "13" in them but 51431 is allowed because there is no "13" in the number. Am assuming that this can be solved with permutations and combinations but I just can't seem to figure out the general formula for N digits.

答:
Let $a_N$ be the number of $N$-digit numbers which does not have $13$ in them.

Let $b_N$ be the number of $N$-digit numbers which ends with $1$ and does not have $13$ in them.

Then for $N \geq 1$ we have:

$a_{N + 1} = 10 a_N - b_N$

because one can append any digit from $0$ to $9$ to every $N$-digit number. The only case where a number containing $13$ is created is when the original number ends with $1$ and one appends $3$ to it.

$b_{N + 1} = a_N$

because removing the last $1$ gives an $N$-digit number and vice-versa.

The rest is simply solving this linear recursion, which is standard.


问:
How to solve $\int_0^\infty J_0(x)\ \text{sinc}(\pi\,x)\ e^{-x}\,\mathrm dx$?
I need some help with solving this integral involving Bessel function:

$\hspace{2in}\displaystyle\int_0^\infty$$J_0(x)\ $$\text{sinc}(\pi\,x)\ $$e^{-x}\,\mathrm dx.$

答:
O.L.'s answer is absolutely correct. I just want to generalize the formula a little:
$$\forall a>0,\hspace{5mm}\int_0^\infty J_0(x)\ \text{sinc}(a\,x)\ e^{-x}\,\mathrm dx=\frac1a\arctan\sqrt{\frac{2}{\sqrt{a^4+4}-a^2}-1}.$$</p>


问:
How to develop $(n+1)^{p+1}-1=(p+1)\sum_{k=1}^{n}k^p + \text{terms for } \sum_{k=1}^{n}k^r \text{, for r < p}$?
So, I am really sorry if this is a stupid question, but I am trying to solve a problem (Spivak $2-7$) for which there is a part where I have
$$
(k+1)^{p+1}-k^{p+1} = (p+1)k^p + \text{terms involving lower powers of k}.
$$
as in
$$
(k+1)^{p+1}-k^{p+1} = (p+1)k^p + Ak^{p-1} + Bk^{p-2}+...
$$
where the author develops this expression for k = $1,...,n$.
From which I get
$$
(n+1)^{p+1}-1=(p+1)\sum_{k=1}^{n}k^p + \text{terms for } \sum_{k=1}^{n}k^r \text{, for r < p}
\tag{1}
$$
the latter part of the equation from which I stole from the author, but I understand that part, the issue is that the author gets:
$$
\frac{(n+1)^{p+1}}{p+1} = \sum_{k=1}^{n}k^p + \text{terms for } \sum_{k=1}^{n}k^r \text{, for r < p}
\tag{2}
$$
My question is, how do we get from (1) to (2)? A
similar
question I have looked at derives this:
$$
\frac{(n+1)^{p+1}}{p+1} - \frac{1}{p+1}
\tag{3} =\sum_{k=1}^n k^p + \sum_{k=1}^n \mathcal{O}(k^r)
$$
(where $\mathcal{O}(k^r)$ represents the terms for $\sum_{k=1}^{n}k^r \text{, for r < p}$)

which seems more understandable than (2) as in (3) I see that the answerer just divided the identity by $(p+1)$, from which I also mean to ask, can you divide the terms in $\mathcal{O}(k^r)$ and still keep $\mathcal{O}(k^r)$? What I mean by this is that when the term $\mathcal{O}(k^r)$ represents $Ak^{p-1} + Bk^{p-2}+...$, can you still call $\frac{\mathcal{O}(k^r)$}{p+1}$, $\mathcal{O}(k^r)$? The reason I find (3) more understandable than (2) is that at least in (3), the "$-1$" term remains, where did the "$-1$" term go in (2)!?

I am very sorry if this question represents the utmost stupidity in me, and if the question is stupid in itself, thank you.

答:
I hope that this is the answer that you are expecting.

First, notice that
$$1=k^0.$$
Since we only focus on $k^p$ and any $k^r$ with $r<p$ will be put in the remainder term. Thus you can ignore $1=k^0$ to move from (1) to (2).

Second, for fixed natural number $r$ and fixed real constant $c$
$$O(k^r)=O(c k^r)=c O(k^r)$$
as $O(k^r)$ means any polynomial of $k$ of order less than or equal to $r$.


问:
Show that exist $p\in \mathbb{R} $ such that $p^3+p=1$
Show that there exists $p\in \mathbb{R} $ such that $p^3+p=1$.

How can I prove it? I was thinking in what can I do to prove it but I really stuck .

I just see supreme axiom, Archimedean property and sequences.

Any hint of what I can do?

答: From algebra we know that real coefficient polynomial equations if a complex $\alpha$ is a root, then  its conjugate $\overline{\alpha}$ is a root, too. Therefore odd degree polynomial equations have at least one real root.


问:
Computing explicitly three integrals involving radial functions
How can I compute the following three (similar)
integrals?
$$
\int_{\mathbb R^N} \left(\frac{1}{1+|x|^2} \right)^{\beta} dx
$$
$$
\int_{\mathbb R^N} \left(\frac{1}{1+|x|^2} \right)^{(N+2\alpha)/2} \left(\frac{1}{|x|} -\frac{1}{(1+|x|^2)^{1/2}} \right) dx
$$
$$
\int_{\mathbb R^N} \left(\frac{1}{1+|x|^2} \right)^{(N+2\alpha)/2} |x|^{4\alpha - N} dx
$$
where $\alpha, \beta >0$. Since the involved functions are radial, one can make a change of variables to reduce the problems to 1-d problems, but then I don't know how the integral can be computed explicitly or it if can be done in an easy way at all or not.

答:
All these integral can be expressed through Beta function. Let's take, for instance, the first one: $I(n,\beta)=\int_{\mathbb R^N} \left(\frac{1}{1+|x|^2} \right)^{\beta} dx_1..dx_n=\int_{\mathbb \Omega}d{\Omega}\int^\infty_0\left(\frac{1}{1+r^2} \right)^{\beta} r^{n-1}dr$, where we integrate over all angles ($d{\Omega}$) in a polar system of coordinate.

Because the integrand is a scalar (and does not depend on angles), integration over $d{\Omega}$ can be done straightforward. We can use, for instance, the following trick:
$\int_{\mathbb x_1}..\int_{\mathbb x_n}e^\left(-x_1^2-..-x_n^2\right)d{x_1}..d{x_n}=(\int_{\mathbb x}e^{-x^2}dx)^n=\left(\sqrt{\pi}\right)^n=\int_{\mathbb \Omega}d{\Omega}\int^\infty_0e^{-r^2} r^{n-1}dr=\frac{1}{2}\int_{\mathbb \Omega}d{\Omega}\int^\infty_0e^{-t} t^{\frac{n}{2}-1}dt  (t=r^2)=\frac{1}{2}\Gamma(\frac{n}{2})\int_{\mathbb \Omega}d{\Omega} \Rightarrow $$\int_{\mathbb \Omega}d{\Omega}=\frac{2(\sqrt{\pi})^n}{\Gamma(\frac{n}{2})}$ (where ${\Gamma}(x)$ is Gamma function).

$I(n,\beta)=\frac{2(\sqrt{\pi})^n}{\Gamma(\frac{n}{2})}\int^\infty_0\left(\frac{1}{1+r^2}\right)^{\beta} r^{n-1}dr= \frac{(\sqrt{\pi})^n}{\Gamma(\frac{n}{2})}\int^1_0(\frac{1-t}{t})^{\frac{n}{2}-1}t^{\beta-2} dt (t=\frac{1}{1+r^2})=\frac{(\sqrt{\pi})^n}{\Gamma(\frac{n}{2})}B(\frac{n}{2};\beta-\frac{n}{2})=(\sqrt{\pi})^n\frac{\Gamma(\beta-\frac{n}{2})}{\Gamma(\beta)}$

The formula gives itself the the answer when it is valid: ${\beta}>\frac{n}{2}$.
The second and third integral can be evaluated in the same fashion - just make designated substitutions and use formula for Beta function (and its relation with Gamma function) https://en.wikipedia.org/wiki/Beta_function.

PS  Third integral $=\frac{(\sqrt{\pi})^n}{\Gamma(\frac{n}{2})}B(2{\alpha};\frac{n}{2}-{\alpha})=(\sqrt{\pi})^n\frac{\Gamma(2\alpha)\Gamma(\frac{n}{2}-\alpha)}{\Gamma(\frac{n}{2})\Gamma(\frac{n}{2}+\alpha)}$ - valid for $0<\alpha<\frac{n}{2}$

Second integral $=\frac{(\sqrt{\pi})^n}{\Gamma(\frac{n}{2})}\left(B({\alpha}+\frac{1}{2};\frac{n}{2}-\frac{1}{2})-B({\alpha}+\frac{1}{2};\frac{n}{2})\right)=(\sqrt{\pi})^n\frac{\Gamma(\frac{1}{2}+\alpha)}{\Gamma(\frac{n}{2})}\left(\frac{\Gamma(\frac{n}{2}-\frac{1}{2})}{\Gamma(\frac{n}{2}+\alpha)}-\frac{\Gamma(\frac{n}{2})}{\Gamma(\frac{n}{2}+\alpha+\frac{1}{2})}\right)$. In this case, we have uncertainty at $\alpha =-\frac{1}{2}$, so it is easier to get from the original integral at $x\rightarrow \infty$ that integral is valid for $\alpha>-\frac{3}{2}$ and $n>1$.


问:
Suspicious corollary of Lusin’s Theorem
Suppose $f : [0,1] \to \mathbb R$ is integrable. According to Lusin’s Theorem (as I understand it), for every $\epsilon > 0$, there is a closed subset $C \subset [0,1]$ for which $f|_C : C \to \mathbb R$ is continuous and $\lambda([0,1] \setminus C) < \epsilon$, where $\lambda$ is the Lebesgue measure on $[0,1]$. Furthermore, if $B \subset [0,1]$ is open, then there is a closed $C’ \subset B$ for which $f_{C’} : C’ \to \mathbb R$ is continuous and $\lambda(B \setminus C’) < \epsilon$.

Playing with this, I arrived at the following corollary, which seems way too powerful to be true:

Corollary: There is a subset $C \subset [0,1]$ of full measure, for which $f|_C : C \to \mathbb R$ is continuous (in the induced subspace topology on $C$).

Proof. Let $C_1 \subset [0,1]$ be a closed set for which $f|_{C_1}$ is continuous, with $\lambda([0,1] \setminus C_1) < 2^{-1}$, so $\lambda(C_1) > 2^{-1}$. The existence of such a $C_1$ follows from Lusin’s Theorem. Now let $C_2 \subset [0,1] \setminus C_1$ be such that $f|_{C_2}$ is also continuous, and $\lambda\left(\left([0,1]\setminus C_1\right)\setminus C_2\right) < 2^{-2}$, or $\lambda(C_1 \sqcup C_2) > 2^{-2}$. The existence of $C_2$ also follows from Lusin’s theorem. Continuing in this way, we get a countable collection of disjoint closed sets $(C_n)_{n \geq 1}$ for which $f|_{C_n} : C_n \to \mathbb R$ is continuous for each $n$, and for which $\lambda\left(C_1 \sqcup C_2 \sqcup \cdots \sqcup C_n\right) > 1-2^{-n}$. By lower semicontinuity of $\lambda$, denoting $C = \bigsqcup_{n\geq 1} C_n$, we have
$$
1 \geq \lambda(C) = \lim_{n \to \infty} \lambda\left(\mathop{\bigsqcup}_{k=1}^n C_k\right) \geq \lim_{n \to \infty}(1-2^{-n}) = 1.
$$
So $\lambda(C) = 1$. and since $C$ is the countable disjoint union of closed sets, on each of which $f$ is continuous in the subspace topology, it follows that $f$ is continuous on $C$ in its subspace topology. QED.

I don’t see an obvious flaw with this proof, but I find this hard to believe. The only place I can see an issue possibly is with the existence of $C_n$ for $n \geq 2$ using Lusin’s Theorem, but Lusin’s Theorem should hold on any Radon measure space $(X, \mathcal A, \mu)$—in particular it should hold on any measurable subset of $[0,1]$. Where is the flaw in this proof? Or is this really a corollary of Lusin’s Theorem?

EDIT: See the answer below.

答: The flaw is in assuming $f|_C$ is continuous. As a friend pointed out to me, a function that is continuous on an exhaustive sequence of closed sets need not be continuous: you’d need to show that given $x \in C$ and $\epsilon>0$, we have $|f(x) - f(y)| < \epsilon$ for all $y$ sufficiently close to $x$ in the set $C$. But if we have $x \in C_n$, then we could have $|x-y|<\delta$ for small $\delta>0$ but $y \in C_m$, $m \neq n$. Since there are infinitely many $C_m$, this can happen infinitely often.

Title: Suspicious corollary of Lusin’s Theorem


问:
Solve, $[3x + 1] = 2x - \frac{1}{2}$ , and find the sum of all roots.

Solve, $[3x + 1] = 2x - \frac{1}{2}$ , and find the sum of all roots.

What I Tried: I have :-
$$\rightarrow [3x + 1] = \frac{4x - 1}{2}$$
Now, the LHS is an integer, so $2|(4x-1)$ .
I can see that if $x$ is an integer, then $(4x - 1)$ is odd, so there are no integer solutions. I am not sure about fractions, as $x = \frac{1}{2}$ still makes it odd, but $x = \frac{1}{4}$ makes it even, although $2$ does not divided $0$.

This is where I am stuck, how should I proceed next?

Can anyone help me?

答:
$y=3x+1 \implies [y] = (4y-7)/6$.

Let $[y]=n, \{y\}=r$ then $n=(4n+4r-7)/6, 2n=4r-7$.

Since $0\le r < 1$, $0 \le 4r < 4$, we must have $4r=1,3$, so either $r=\frac 14, n=-3$ or $r=\frac 34, n=-2$.

Finally, $3x+1 = -\frac{11}{4}$ or $-\frac 54$, $x=-\frac 54$ or $-\frac 34$.

(You don't need $y$ but it simplifies things a bit.)


问:
Diagonalizable Operators: An Operational Extension
Let $T$ be a diagonalizable operator on a vector space $V$. Prove that the operator
$$a_nT^n + a_{n-1}T^{n-1}+\cdots+a_1T+a_0 Id_V$$
 on $V$ is also diagonalizable for any scalars $a_1, a_1,\dots,a_n$.

First off, what is $Id_V$? I've never seen this before.

答:
I suppose $V$ is $d$-dimensional and $T$ is represented by a square matrix. Then $T$ diagonalizable if it is similar to a diagonal matrix $D=\mathop{\rm diag}(\lambda_1,\ldots,\lambda_d)$, i.e., there is an invertible matrix $S$ such that $D=S^{-1}TS$. Now
\begin{equation}
S^{-1}T^nS=S^{-1}TS\cdot S^{-1}TS\cdots S^{-1}TS=D^n,
\end{equation}
and since $D^n=\mathop{\rm diag}(\lambda_1^n,\ldots,\lambda_d^n)$ is again diagonal, it follows that $T^n$ is diagonal for any $n$. Hence, by linearity, your operator is diagonal.

If $T$ is considered as an endomorphism on an abstract $d$-dimensional vector space it is called diagonalizable if there is a basis $\{v_1,\ldots,v_d\}$ of eigenvectors (notice that then the matrix of $T$ wrt. $\{v_1,\ldots,v_d\}$ will be diagonal, and the matrix of $T$ wrt. any other basis is diagonalizable in the above sense). Now if $T(v_i)=\lambda_iv_i$ we get $T^n(v_i)=\lambda_i^nv_i$, $i=1,\ldots,d$, thus $\{v_1,\ldots,v_d\}$ is a basis of eigenvectors of $T^n$, and we conclude.


问:
Number of ideals with norm equals to a certain number in number field.
I can't follow the claim in the lecture notes below about the number of ideals with norm $n$. Any help or hint is appreciated.

答:
You meant $n^k$. If $N(I)=p^k$ then $I$ is a product of at most $k$ prime ideals of norm a power of $p$ and there are at most $n$ of those.

If $p$ splits completely then the exact number is ${n+k-1 \choose k}$ (a combination with repetition).


问:
Different versions of the entropy term in the entropy-regularized Wasserstein distance
\begin{equation}
\mathcal{W}_\epsilon(\alpha, \beta) = \min_{\pi\in \Pi(\alpha\beta)} \int c(x,y) \mathrm{d}\pi(x,y) + \epsilon H(\pi \| \alpha \otimes \beta)
\end{equation}
Cuturi (2013) introduced the entropy-regularized Wasserstein distance, or Sinkhorn distance, shown above, where $\epsilon $ is the regularization parameter and $H(\pi \| \alpha \otimes \beta)$ is the relative entropy, or KL-divergence, between the transport plan and the marginal probabilities.

But I have seen the $H(\cdot)$ term shown in two different ways, one with entropy and the other with relative entropy:

\begin{align}
H(\pi) &= \int \pi(x,y) \ln \pi(x,y) \\
H(\pi \| \alpha \otimes \beta)  &= \int \ln \left(\frac{\mathrm{d}\pi (x,y)}{\mathrm{d}\alpha(x)  \mathrm{d}\beta(y)  } \right) \mathrm{d}\pi (x,y)
\end{align}

How are the last two lines equal or connected to each other? Obviously they're not the same, so why are there two different versions running around?

答:
I would like to add a couple of points here which I think shouldn't be overlooked.

Neither of the choices are "wrong". In the 2013 Cuturi paper you reference he chooses to regularise with "entropy" (notice this is actually the negative Boltzmann entropy) :

$$ H(\pi)= \begin{cases} \int \pi \log \pi~~&\text{when}~\pi~\text{has a density} \\
\infty & o.w \end{cases}. $$

The reason this is a natural choice for regularisation is that it does the "smoothing" or "relaxing" job that regularisation is intended to do. Adding $H$ into the optimal transport problem gives the mass "freedom to spread out".  This can be seen in this example let $\mu$ be concentrated on the two points $x_1,x_2 \in \mathbb{R}$ such that $\mu(x_1)=\mu(x_2)=\frac{1}{2}$, and $\nu$ be concentrated on the two points $y_1,y_2 \in \mathbb{R}$ such that $\nu(y_1)=\mu(y_2)=\frac{1}{2}$, then the optimal coupling $\pi$ which maximises $H$ is

$$ \pi(x_i,y_j)=\frac{1}{4},\forall~i,j.$$

Since we have a minimisation problem its beneficial to be adding a uniformly convex term, again $H$ ticks that box!

The choice of adding $H(\pi~||~\alpha\otimes\beta)$, the entropy conditioned on the product measure, has its advantages as outlined in https://audeg.github.io/publications/these_aude.pdf. As far as I understand it lets you rephrase the dual problem in a neat way.

Now to comparing the two choices: I don't think it matters too much either way, both do the same job. As the other answer points out, the minimiser is the same, and they differ by a constant $C$. Lastly remember that, usually when "doing regularisation" you have a small parameter $\epsilon \ll 1$ multiplying the regularisation term, hence

$$ \epsilon \Big(H(\pi~||~\alpha\otimes\beta)-H(\pi)\Big)=\epsilon C \ll 1 .$$

edit I have changes to make about what choice to take.


问:
A less known definition of the definite integral of a continuous function
The definite integral of a continuous function can be defined using the bounded monotone sequence property: see Osgood's Functions of Real Variables, p.110.

(link to full book) (screenshots: page 110, page 111, page 112)

Please can you give some other reference of the same type (or, better, a link or a book where the matter is approached in the same manner) ?

答: I think you are trying to ask for the distinction between defining the Riemann integral as a limit of supremum and infimum partitions vs just using a fixed family of partitions, in this case something akin to the dyadic intervals as Osgood does. The point is that both methods yield precisely the same result when the function $f$ is continuous on the whole interval. In fact Osgood seems to prove this fact calling it the "Convergence Theorem" on the bottom of page 113 to page 114. In other words, the type of partition and which values to sample $f$ at within the partitions are irrelevant. If you try to extend this definition to badly discontinuous functions, you'll get nonsense, whereas with the usual definition, it'll imply the supremum and infimum sums do not agree thereby not being Riemann Integrable.


问:
Let $G=\langle \mathbb R \times \mathbb R, +\rangle$ and $H=\{(x,y): y=2x\}$. Show that $H$ is a subgroup of $G$.

Let $G=\langle\mathbb R \times \mathbb R, +\rangle$ and $H=\{(x,y): y=2x\}$. Show that $H$ is a subgroup of $G$.

I'm trying to show that $H$ is closed and contains inverses (individually). So the ordered pairs have the form $(x,2x)$. Let $(x,y)$ and $(z,w)$ be in $H$. Then

$$\begin{align}
(x,y)+(z,w)&=(x,2x)+(z,2z)\\
&=(x+z,2x+2z)\\
&=(x+z,2(x+z)).
\end{align}$$ Thus $H$ is closed. Now, if $(x,y)=(x,2x)\in H$, then $(-x,-y)=(-x,-2x)$ implies that their sum is $(0,0)$ which is the identity element. Thus $(-x,-y)\in H$ and $H$ is a subgroup of $G$.

Is this correct? Could I also have consider ordered pairs of the form $({1\over 2}y,2x)$ or just consider two of the same elements, $(x,y)$ and done the proof that way? How do I know that I'm setting the problem up right in the first place. I feel like I don't really understand it and I'm just getting lucky sometimes. Any other solutions, hints, advice is greatly appreciated.

答:
I will use the one-step subgroup test.

Since $0=2\times 0$, we have $(0,0)\in H$. Hence $H\neq \varnothing$.

Because $H=\{\color{red}{(x,y)}\mid y=2x\}$, we have $H\subseteq G$.

Let $a=(x,y), b=(z, t)\in H$. Then $y=2x$, $t=2z$, and

$$\begin{align}
a-b&=(x-z, y-t)\\
&=(x-z, 2x-2z)\\
&=(\color{green}{x-z}, \color{blue}{2(x-z)}),
\end{align}$$

which is in $H$ since $\color{blue}{2(x-z)}=2(\color{green}{x-z})$; so $a-b\in H$.

Hence $H\le G$.


问:
$f \geqslant 0$ and $f''\leqslant 0$. Show that $\lim_{x\to\infty} \frac{f(x)}{x}$ exist
Let $f:\mathbb{R} \rightarrow \mathbb{R}$ twice differentiable. Suppose $f \geqslant 0$ and $f''\leqslant 0$. Show that $\lim_{x\to\infty} \frac{f(x)}{x}$ exist.

My attempt was, as $f$ is concave, to pass by next inequalities:

Let $a<b<x$ so we have

$\frac{f(b)-f(a)}{b-a}>\frac{f(x)-f(a)}{x-a}>\frac{f(x)-f(b)}{x-b}$

and then pass to limit and observe that $\frac{f(x)}{x}$ is bounded as $x$ goes to $\infty$. But, then i recognised that it doesn't hold...

So i need some help please. Thanks in advance

答:
Hint Since $f'(x)$ is decreasing the limit
$$
L=\lim_{x \to \infty} f'(x)
$$
exists, and it is either a number or $- \infty$.

Now, by L'Hopital (note that you need the version $\frac{\mbox{something}}{\infty}$) , the limit
$$
\lim_{x \to \infty} \frac{f(x)}{x}
$$
exists and
$$
\lim_{x \to \infty} \frac{f(x)}{x}= \lim_{x \to \infty} f'(x)=L
$$
Finally, since $f \geq 0$ you have $L \geq 0$, which implies that it is finite.


问:
How do I solve de ODE $y' = \frac xy\ln\left(\frac xy\right)$?
I need some help finishing the ODE, please.
This is what I have so far:

$w=\frac xy$, or $y=\frac xw$. Then we have

$$y’=\frac1w-\frac{xw'}{w^2}.$$

Then the DE becomes

$$y’=\frac1w-\frac{xw'}{w^2}=w\log w$$

$$\frac1w-w\log w=\frac{xw'}{w^2}$$

and clear $w^2$ out of denominator to get

$$w-w^3\log w=x\frac{dw}{dx}.$$

The equation is now separable as

$$\frac{dw}{w-w^3\log w}=\frac{dx}x$$

so integrate both sides to yield

$$\int\frac{dw}{w-w^3\log w}=\log x+k_1$$

and exponentiate to get

$$\exp\left(\int\frac{dw}{w-w^3\log w}\right)=k_2x.$$

However, I can't go any further than this.

My professor said I might be ignoring other solutions.
He also said there is another way to solve the ODE that is way easier.

I appreciate your help, thank you!

答:
A particular solution can be obtained of the form:

$$y=kx$$

Substituting this into the ODE gives:

$$k = - \frac{1}{k} \ln{k}$$

Which is satisfied when $k\approx0.653$.

I plotted a few numerical simulations and it appears that they asymptotically approach this particular solution.


Title: How to develop $(n+1)^{p+1}-1=(p+1)\sum_{k=1}^{n}k^p + \text{terms for } \sum_{k=1}^{n}k^r \text{, for r < p}$?

Body:
So, I am really sorry if this is a stupid question, but I am trying to solve a problem (Spivak $2-7$) for which there is a part where I have
$$
(k+1)^{p+1}-k^{p+1} = (p+1)k^p + \text{terms involving lower powers of k}.
$$
as in
$$
(k+1)^{p+1}-k^{p+1} = (p+1)k^p + Ak^{p-1} + Bk^{p-2}+...
$$
where the author develops this expression for k = $1,...,n$.
From which I get
$$
(n+1)^{p+1}-1=(p+1)\sum_{k=1}^{n}k^p + \text{terms for } \sum_{k=1}^{n}k^r \text{, for r < p}
\tag{1}
$$
the latter part of the equation from which I stole from the author, but I understand that part, the issue is that the author gets:
$$
\frac{(n+1)^{p+1}}{p+1} = \sum_{k=1}^{n}k^p + \text{terms for } \sum_{k=1}^{n}k^r \text{, for r < p}
\tag{2}
$$
My question is, how do we get from (1) to (2)? A
similar
question I have looked at derives this:
$$
\frac{(n+1)^{p+1}}{p+1} - \frac{1}{p+1}
\tag{3} =\sum_{k=1}^n k^p + \sum_{k=1}^n \mathcal{O}(k^r)
$$
(where $\mathcal{O}(k^r)$ represents the terms for $\sum_{k=1}^{n}k^r \text{, for r < p}$)

which seems more understandable than (2) as in (3) I see that the answerer just divided the identity by $(p+1)$, from which I also mean to ask, can you divide the terms in $\mathcal{O}(k^r)$ and still keep $\mathcal{O}(k^r)$? What I mean by this is that when the term $\mathcal{O}(k^r)$ represents $Ak^{p-1} + Bk^{p-2}+...$, can you still call $\frac{\mathcal{O}(k^r)$}{p+1}$, $\mathcal{O}(k^r)$? The reason I find (3) more understandable than (2) is that at least in (3), the "$-1$" term remains, where did the "$-1$" term go in (2)!?

I am very sorry if this question represents the utmost stupidity in me, and if the question is stupid in itself, thank you.

Body:
Here's a way
to get the recurrence
for the power sums.

If
$s_m(n)
=\sum_{k=0}^n k^m
$
then

$\begin{array}\\
(n+1)^m
&=\sum_{k=0}^n ((k+1)^m-k^m)\\
&=\sum_{k=0}^n (\sum_{j=0}^{m-1}\binom{m}{j}k^j)\\
&= \sum_{j=0}^{m-1}\binom{m}{j}\sum_{k=0}^nk^j\\
&= \sum_{j=0}^{m-1}\binom{m}{j}s_j(n)\\
&= ms_{m-1}(n)+\sum_{j=0}^{m-2}\binom{m}{j}s_j(n)\\
\text{so}\\
s_{m-1}(n)
&=\dfrac1{m}\left((n+1)^m-\sum_{j=0}^{m-2}\binom{m}{j}s_j(n)\right)\\
&=\dfrac1{m}\left(\sum_{j=0}^m \binom{m}{j}n^j-\sum_{j=0}^{m-2}\binom{m}{j}s_j(n)\right)\\
&=\dfrac1{m}\left(n^m+mn^{m-1}+\sum_{j=0}^{m-2} \binom{m}{j}n^j-\sum_{j=0}^{m-2}\binom{m}{j}s_j(n)\right)\\
&=\dfrac1{m}\left(n^m+mn^{m-1}+\sum_{j=0}^{m-2} \binom{m}{j}(n^j-s_j(n))\right)\\
\end{array}
$

Immediately,
$s_m(n)
\lt \dfrac1{m+1}(n+1)^{m+1}
$.


答:
As you noticed, $f$ is concave, so that for $0 < x < y$
$$
 f(x) \ge \frac{y-x}{y-0} f(0) + \frac{x-0}{y-0} f(y) \ge \frac x y f(y) \\
\implies \frac{f(x)}{x} \ge \frac{f(y)}{y} \, .
$$
So $f(x)/x$ is decreasing and bounded below (by zero) and therefore convergent.

Note that the conclusion holds for all non-negative concave functions, even if they are not differentiable.

Body:
Let $f:\mathbb{R} \rightarrow \mathbb{R}$ twice differentiable. Suppose $f \geqslant 0$ and $f''\leqslant 0$. Show that $\lim_{x\to\infty} \frac{f(x)}{x}$ exist.

My attempt was, as $f$ is concave, to pass by next inequalities:

Let $a<b<x$ so we have

$\frac{f(b)-f(a)}{b-a}>\frac{f(x)-f(a)}{x-a}>\frac{f(x)-f(b)}{x-b}$

and then pass to limit and observe that $\frac{f(x)}{x}$ is bounded as $x$ goes to $\infty$. But, then i recognised that it doesn't hold...

So i need some help please. Thanks in advance


问:
How to tell $\lim\limits_{x \to \infty} f(x)$ exists
There is a test to check if $\lim\limits_{x \to a} f(x)$ exists. One needs to check to see if both right limit and left limit is same.

As there is no right limit when $x \to \infty$ the above test would be not useful. Is there any other test to confirm one is not calculating non existent limit to, e.g. say a function that keeps oscillating is there a test?

答:
There are multiple ways to show it.

Compute the limit itself by using some tricks, L'Hopital, etc. (but I don't think it will help you)

Cauchy's(epsilon-delta) definition:

$$
\lim_\limits{x \to \infty} f(x)=a \iff \forall \space \varepsilon >0 \space \exists \space \delta>0: \space \space x > \frac{1}{\delta} \implies |f(x)-a|< \varepsilon
$$

Heine's definition:
$$
\lim_\limits{x \to \infty} f(x)=a \iff \text{ for every sequence $x_n$ s.t. $x_n$ converges to $\infty$, the sequence $f(x_n)$ converges to $a$ }
$$

Actually the Cauchy's is useful to show the existence of limit. But the Heine's is more useful(in my experience) proving that limit doesn't exist(just take to sequences $x_n'$ and $x_n''$ that converge to infinity and show that sequences $f(x_n')$ and $f(x_n'')$ converge to different points).

But you should always understand that the actual way to solve any math problem depends on the problem itself. Maybe, for your case what I've described won't be useful. The whole math is such a thing that there always are lot's of shortcuts, you just need to recognize them.


问:
Prove $\int_{-\pi}^\pi F_n(y) \, dy=1$
Prove $\int_{-\pi}^\pi F_n(y)\,dy=1$, with
$$F_n(y)=\frac{1}{2\pi (n+1)}\frac{\sin^2 \left( \frac{(n+1)y}{2} \right)}{\sin^2(\frac{y}{2})}$$

I tried a similar question but there I had given a serie of the function. This time I don't know if there is a serie that can help me with this.
I tried it without a serie:

\begin{align}
\int_{-\pi}^{\pi}F_n(y) \, dy &= \int_{-\pi}^\pi \frac{1}{2\pi (n+1)} \frac{\sin^2\left(\frac{(n+1)y}{2}\right)}{\sin^2(\frac{y}{2})} \, dy\\
&=\int_{-\pi}^\pi \frac{1}{2\pi (n+1)}\frac{(e^{\frac{i(n+1)y}{2}}-e^{\frac{-i(n+1)y}{2}})^2}{(e^\frac{iy}{2}-e^\frac{-iy}{2})^2} \, dy\\
&=\int_{-\pi}^\pi \frac{1}{2\pi (n+1)}\frac{(e^{i(n+1)y}+e^{-i(n+1)y}-2)}{(e^{iy}+e^{-iy}-2)} \, dy
\end{align}

But now I'm stuck again. I think there needs to be an easier way to prove this. Can someone help me out?

答:
Hint:
Let $$D_n(x)= \sum_{k=-n}^n e^{ikx},$$ and let
$$F_N(x) = \sum_{n=0}^{N-1} D_n(x).$$
Prove that
$$ F_N(x) = \frac{1}{N}\frac{\sin^2 (Nx/2)}{\sin^2 (x/2)}.$$

$D_n$ is known as the Dirichlet kernel and $F_N$ is known as the Fejér Kernel.

Title: Prove $\int_{-\pi}^\pi F_n(y) \, dy=1$


问:
Checking if the complement of an undirected graph is bipartite
Is there any linear time algorithm that says if the complement of a given undirected graph is bipartite?

I've tried using König's theorem, with no success.

答:
To check whether a connected graph is bipartite:

Start at a vertex $v$ and colour it black. Colour all its neighbours white; mark $v$ as done and all its neighbours as pending. For each pending vertex $w$, colour its neighbours the opposite colour of $w$ and mark them as pending if they are not already done; mark $w$ as done. Repeat.

If ever there is a conflict, the graph is not bipartite. If there is no conflict, then you have a colouring.

If the graph is not connected, do the above for each component.


问:
How to calculate the implicit derivative for this equation
Basically, I am trying to find the implicit derivative for this problem, but I am stuck after a few steps.

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

I have calculated the derivatives of each term and used the product rule to combine everything, which has gotten me to this step.

$$(x^2)\left(2y\frac{dy}{dx}\right)+(y^2)(2x)=1+\frac{dy}{dx}$$

However, after reaching this step, I am stuck. I have tried isolating the terms but when I do so, something does not look right, and this confuses me as there are $2 dy/dx$'s. How can I solve this problem? Any help is appreciated. Thanks.

答: Bring all terms with $dy/dx$ to one side,$$2xy^2-1=\frac{dy}{dx}[1-2x^2y]$$Now divide both sides by $1-2x^2y$ to get$$\frac{dy}{dx}=\frac{2xy^2-1}{1-2x^2y}$$


问:
Show $Z_n= \frac{X_1 +\cdots + X_n}{\sqrt{n}}$ is uniformly sub-gaussian.
As described above I would like to establish a tail bound for the $Z_n$, using that $X$ is bounded with zero mean and unit variance and $X_n$ are iid copies of $X$. So namely I would like to show that $P(\vert Z_n \vert \geq \lambda) \leq C\exp(-c\lambda^{2})$ for all $\lambda >0$ and $C,c>0$ independent of $n$ (uniformly sub-gaussian). I think that should be possible using some kind of Chernoff bound. I would really appreciate some Help.

答:
Let the $X_i$ lie in $[a,b]$ with probability $1$. Then Hoeffding's Lemma implies $E[e^{tX_1 / \sqrt{n}}] \le \exp(\frac{t^2(b-a)^2}{n})$.

The Chernoff bound implies $P(Z_n \ge \lambda) \le \min_{t > 0} e^{-t\lambda} (E[e^{tX_1 / \sqrt{n}}])^n$. Plug in the above inequality and solve for $t$ to get a bound on the probability.

$P(Z_n \le -\lambda)$ can be handled similarly.


问:
Radioactive isotopes differential equationa
I am having a hard time finding the correct differential equation to my problem. The problem is :

There's 2 isotopes: A and B. A is is transforming into B to a rate proportional to its quantity and B is decreasing to a rate proportional to its quantity.

I need to find an equation that gives the quantity of the B isotope at time t.

I've found the the equation of the lost of A over time, but I don't know how to integrate it to the differential equation of B.

The quantity of A isotope that has turned into B over time is y(t) = yi*e^(kt)

edit
B is not a fixed quantity. It's the lost quantity of A so the lost rate of B is proportional to the quantity B at the time which is equals to the lost quantity of A - minus the quantity of B that has been lost over time.

Thanks for the help

答:
Let $A(t)$ be the amount of substance A at time $t$, and let $B(t)$ be the amount of substance B at time $t$.

For suitable decay constants $a$ and $b$, we have
$$A'(t)=-aA(t),$$
and
$$B'(t)=-A'(t)-bB(t)=aA(t)-bB(t).$$
The second equation comes from the fact that B atoms are "born" (through decay) at rate $aA(t)$, and die at rate $bB(t)$.

How we handle these is a matter of taste. If it is familiar, we can express the system in matrix notation. The more straightforward approach is to solve the first equation explicitly for $A(t)$, and substitute in the second equation. We obtain a fairly simple but non-homogeneous linear equation.


答:
Here is to integrate without resorting to series

\begin{align}
\int_{-\pi}^{\pi}F_{n}(y)dy
& = \frac1{\pi(n+1)}\int_{0}^{\pi} \frac{\sin^{2}\frac{(n+1)y}{2}}{\sin^{2}\frac{y}{2}}dy\\
& = \frac1{\pi(n+1)}\int_{0}^{\pi}  \frac{ \cos(n+1)y-1}{\cos y-1}  dy \\
 & = \frac1{\pi(n+1)}\cdot \lim_{a\to 0}\int_{0}^{\pi}  \frac{ \cos(n+1)y-\cos(n+1)a}{\cos y-\cos a}  dy\\
 & = \frac1{\pi(n+1)}\cdot \lim_{a\to 0}\frac{\pi\sin(n+1)a}{\sin a}\\
&=1
\end{align}
where the result
$$\int_{0}^{\pi}{\frac{\cos(nx)-\cos(na)}{\cos x-\cos a}}dx
= \frac{\pi \sin(na )}{\sin a}$$
derived in Parametric trigonometric integral $\int_{0}^{\pi}{\frac{\cos(nx)-\cos(na)}{\cos x-\cos a}}dx$ is used.

Title: Prove $\int_{-\pi}^\pi F_n(y) \, dy=1$

Body:
Prove $\int_{-\pi}^\pi F_n(y)\,dy=1$, with
$$F_n(y)=\frac{1}{2\pi (n+1)}\frac{\sin^2 \left( \frac{(n+1)y}{2} \right)}{\sin^2(\frac{y}{2})}$$

I tried a similar question but there I had given a serie of the function. This time I don't know if there is a serie that can help me with this.
I tried it without a serie:

\begin{align}
\int_{-\pi}^{\pi}F_n(y) \, dy &= \int_{-\pi}^\pi \frac{1}{2\pi (n+1)} \frac{\sin^2\left(\frac{(n+1)y}{2}\right)}{\sin^2(\frac{y}{2})} \, dy\\
&=\int_{-\pi}^\pi \frac{1}{2\pi (n+1)}\frac{(e^{\frac{i(n+1)y}{2}}-e^{\frac{-i(n+1)y}{2}})^2}{(e^\frac{iy}{2}-e^\frac{-iy}{2})^2} \, dy\\
&=\int_{-\pi}^\pi \frac{1}{2\pi (n+1)}\frac{(e^{i(n+1)y}+e^{-i(n+1)y}-2)}{(e^{iy}+e^{-iy}-2)} \, dy
\end{align}

But now I'm stuck again. I think there needs to be an easier way to prove this. Can someone help me out?


答:
You have

$$F_n(x) = \frac{2\pi}{n+1}D_n^2(x)=\frac{1}{2\pi (n+1)}\left(\sum_{k=-n}^n e^{ikx} \right)^2= \frac{1}{2\pi (n+1)}\frac{\sin^2 \left(\frac{(n+1)x}{2}\right)}{\sin^2 \left(\frac{x}{2}\right)}.$$

from which follows by switching $\int$ and $\sum$ the equalities

$$\begin{aligned}\int_{-\pi}^{\pi}F_n(x) \ dx&= \frac{1}{2\pi (n+1)}\int_{-\pi}^{\pi}\left(\sum_{k=-n}^n e^{ikx}\right)^2 \ dx\\
&= \frac{1}{2\pi (n+1)}\int_{-\pi}^{\pi}\sum_{k=-n}^n \sum_{l=-n}^ne^{i(k+l)x}\ dx\\
&= \frac{1}{2\pi (n+1)}\sum_{k=-n}^n \sum_{l=-n}^n \int_{-\pi}^{\pi}e^{i(k+l)x}\ dx\\
&=1
\end{aligned}$$

as in the double sum, the only non vanishing terms are for $k=-l$ and there are $n+1$ such terms.

Title: Prove $\int_{-\pi}^\pi F_n(y) \, dy=1$


问:
If f$(x) = x \arcsin\left(\frac{2x}{1+x^2}\right)$ how can I divide it into a composition of two function like $h(g(x))?$
The full question is: Let $f(x) = x \arcsin\left(\frac{2x}{1+x^2}\right),$ $x \in\mathbb R.$  Find the derived function $f'(x)$ by considering it as a composition of two functions, and provide the domain of the existence of $f'(x)$.

I tried doing it as $g(x)=\tan x$ and figuring out something for $h(x)$ but the boundary conditions for the conversion of the inverse trigonometric functions keep confusing me.

答:
It's simpler than looks. Firstly, by the product rule, if $f(x)=xg(x)$, then: $$f'(x)=xg'(x)+g(x)$$
Now, we have $g(x)=\arcsin(\frac{2x}{1+x^2})$. We set $\alpha(x)=\arcsin(x), \beta(x)=\frac{2x}{1+x^2}$, then $g(x)=\alpha(\beta(x))$, and by the chain rule: $$\frac{d[\alpha(\beta(x))]}{dx}=\frac{d[\beta(x)]}{dx}\cdot\frac{d[\alpha(\beta(x))]}{d[\beta(x)]}\implies(\alpha(\beta))'=\beta'\cdot\alpha'(\beta)$$

We can calculate  $\alpha', \beta'$ and force through the first equation to find $f'$:
$$f'=x\beta'\alpha'(\beta)+\alpha(\beta)$$


答:
One can define a function $f:[a,b]\to\Bbb R$ to be regulated if there exists a sequence $\{s_n\}$ of step  functions such that $s_n\to f$ uniformly. Recall that if $s$ is a step function with constants $c_i$ and intervals of partition $[x_{i-1},x_i]$ for say $i=1,\dots, r$ we define its integral as over $[a,b]$ as $$\int_a^b s=\sum_{i=1}^r c_i\Delta x_i$$

One can prove that if $s_n,t_n$ are two sequences of step functions that converge uniformly to $f$ then $$\lim \int_a^b s_n=\lim \int_a^b t_n$$
 by first proving that for each $\epsilon >0$ there exists $N$ such that $n\geq N$ gives $$|s_n(x)-t_n(x)|<\epsilon$$
We then are led to define the integral of a regulated function as $$\int_a^b f=\lim \int_a^b s_n$$ where $\{s_n\}$ is any sequence of step functions that converge to $f$ uniformly over $[a,b]$.

It is a nice task to show that continuous functions over closed and bounded intervals are indeed regulated in the sense of the definition above.

SKETCH OF PROOF

Let $f:[a,b]\to \Bbb R$ be continuous, $\epsilon >0$be given. Let $P_\epsilon(y)$ mean  $-$ for each $\epsilon >0$ and each $y\in[a,b]$ $-$ that there exists a step function $s$ such that $|s-f|<\epsilon$ for $x\in[a,y]$. Define $A=\{y\in[a,b]:P_\epsilon(y)\}$

Then $a\in A$ (trivially) and $A$ is bounded above by $b$, so $\sup A=\alpha$ exists.

$(i)$ First, prove $\alpha=b$ as follows. We know $\alpha\leq b$, since $b$ is an upper bound. Suppose that $\alpha 0$ such that $\alpha+\delta\in A$.

$(ii)$ To show $b\in A$, use again that $f$ is continuous, but now at $x=b$.

It is proven that for each $\epsilon >0$ there exists said step function over $[a,b]$.
Now take $\epsilon =1,\frac 1 2,\frac 13,\dots,\frac 1n,\dots $ to obtain said sequence. $\blacktriangle$.

ADD In the above, we needn't proceed by contradiction. What we really are doing is proving that for each $\lambda 0$ such that $\lambda+\delta\in A$, thus showing that $\sup A\geq b$, which gives that $\sup A=b$.

Title: A less known definition of the definite integral of a continuous function

Body:
The definite integral of a continuous function can be defined using the bounded monotone sequence property: see Osgood's Functions of Real Variables, p.110.

(link to full book) (screenshots: page 110, page 111, page 112)

Please can you give some other reference of the same type (or, better, a link or a book where the matter is approached in the same manner) ?

Body:
One can define a function $f:[a,b]\to\Bbb R$ to be regulated if there exists a sequence $\{s_n\}$ of step  functions such that $s_n\to f$ uniformly. Recall that if $s$ is a step function with constants $c_i$ and intervals of partition $[x_{i-1},x_i]$ for say $i=1,\dots, r$ we define its integral as over $[a,b]$ as $$\int_a^b s=\sum_{i=1}^r c_i\Delta x_i$$

One can prove that if $s_n,t_n$ are two sequences of step functions that converge uniformly to $f$ then $$\lim \int_a^b s_n=\lim \int_a^b t_n$$
 by first proving that for each $\epsilon >0$ there exists $N$ such that $n\geq N$ gives $$|s_n(x)-t_n(x)|<\epsilon$$
We then are led to define the integral of a regulated function as $$\int_a^b f=\lim \int_a^b s_n$$ where $\{s_n\}$ is any sequence of step functions that converge to $f$ uniformly over $[a,b]$.

It is a nice task to show that continuous functions over closed and bounded intervals are indeed regulated in the sense of the definition above.

SKETCH OF PROOF

Let $f:[a,b]\to \Bbb R$ be continuous, $\epsilon >0$be given. Let $P_\epsilon(y)$ mean  $-$ for each $\epsilon >0$ and each $y\in[a,b]$ $-$ that there exists a step function $s$ such that $|s-f|<\epsilon$ for $x\in[a,y]$. Define $A=\{y\in[a,b]:P_\epsilon(y)\}$

Then $a\in A$ (trivially) and $A$ is bounded above by $b$, so $\sup A=\alpha$ exists.

$(i)$ First, prove $\alpha=b$ as follows. We know $\alpha\leq b$, since $b$ is an upper bound. Suppose that $\alpha <b$. Then there exists a step $s$ such that approximates $f$ within $\epsilon$ over $[a,\alpha]$. Use that $f$ is continuous at $x=\alpha$ to obtain a $\delta>0$ such that $\alpha+\delta\in A$.

$(ii)$ To show $b\in A$, use again that $f$ is continuous, but now at $x=b$.

It is proven that for each $\epsilon >0$ there exists said step function over $[a,b]$.
Now take $\epsilon =1,\frac 1 2,\frac 13,\dots,\frac 1n,\dots $ to obtain said sequence. $\blacktriangle$.

ADD In the above, we needn't proceed by contradiction. What we really are doing is proving that for each $\lambda <b$, there exists a $\delta >0$ such that $\lambda+\delta\in A$, thus showing that $\sup A\geq b$, which gives that $\sup A=b$.


问:
Line equation from ray form
Given a ray r = u + (v)t where u is a bounded vector representing an origin point and v is a free vector representing direction, how can the equation for the line in the form y = mx + b be derived?

Given:

$$u = \langle 0, \frac{-3}{4} \rangle$$
$$v = \langle 1, \frac{8}{9} \rangle$$

How does one arrive at:

$$y = \frac{8}{9}x + \frac{3}{4}$$

答:
So you have a vector $v$ that starts from point $u$. It means that the line is going to be parallel to $v$. We know that any vector ($a$) parallel to $v$ will have cross product $a \times v= 0$. But $a$ needs to be shifted by $u$ meaning:

$$
a=\langle x - 0, y + \frac{3}{4}\rangle
$$

Now cross product (a.k.a. $2\times2$ matrix determinant) would be:

$$
\begin{vmatrix}
x && 1 \\
y + \frac{3}{4} && \frac{8}{9}
\end{vmatrix} = 0
$$
$$
\frac{8}{9}x - (y + \frac{3}{4}) = 0
$$
$$
y = \frac{8}{9}x - \frac{3}{4}
$$
The sign in your answer was wrong, you can check that if you set $x=0$ , then the expression should give $- \frac{3}{4}$ since $u=\langle 0, - \frac{3}{4}\rangle$

Title: Line equation from ray form


问:
prove that $\mathcal{F} (\partial_{x_j}f)(\xi) = i \xi_j \mathcal{F} f(\xi) $
Let $f \in L^{1}(\mathbb{R}^n) \cap C^{1}(\mathbb{R}^n)$, and $\partial_{x_j}f \in L^{1}(\mathbb{R}^n)$. Then, prove that $\mathcal{F} (\partial_{x_j}f)(\xi) = i \xi_j \mathcal{F} f(\xi) $ where $\mathcal{F}f$ is the Fourier transform of $f$.

My attempt: $\mathcal{F} (\partial_{x_j}f)(\xi) = \int_{\mathbb{R}^n} e^{-i x \cdot \xi} (\partial_{x_j}f) dx$. I don't know what to do with this partial derivative.

答:
I had rewritten completely the answer for a cleanest and shorter one. For fixed $\xi\in \mathbb{R}^n$ set
$$
G(y):=\int_{\mathbb{R}^n}e^{i\langle y-x,\xi \rangle}f(x)\mathop{}\!d x=\int_{\mathbb{R}^n}e^{i\langle x,\xi \rangle}f(y-x)\mathop{}\!d x
$$
and note that the second identity follow by the substitution rule. Now as the integrand of both integrals belongs to $L^1 \cap C^1(\mathbb{R}^n)$ then by the measure-theoretic version of the Leibniz integral rule we can differentiate $G$ and we get
$$
\partial _jG(y)=\int_{\mathbb{R}^n}\partial _je^{i\langle y-x,\xi \rangle}f(x)\mathop{}\!d x=\int_{\mathbb{R}^n}i\xi_je^{i\langle y-x,\xi \rangle}f(x)\mathop{}\!d x
$$
and
$$
\partial _jG(y)=\int_{\mathbb{R}^n}e^{i\langle x,\xi \rangle}\partial _jf(y-x)\mathop{}\!d x=\int_{\mathbb{R}^n}e^{i\langle y-x,\xi \rangle}\partial _jf(x)\mathop{}\!d x
$$
where $\partial_j f(v)$ is the partial derivative of $f$ at $v$ in direction $e_j$ and the last identity follows again from the substitution rule. Then setting $y=0$ we find the desired identity.∎


问:
What does the Comma Notation in the textbook "Serious Cryptography" describing DRBGs mean
I'm going through Serious Cryptography: A Practical Introduction to Modern Encryption by Jean-Philippe Aumasson. In the first chapter we have this:

One of the simplest constructions of a semantically secure cipher uses a deterministic  random  bit  generator ($DRBG$),  an  algorithm  that  returns random-looking bits given some secret value:

$E(K, R, P) = (DRBG(KR) ⊕P, R)$

Here, $R$ is a string randomly chosen for each new encryption and givento a DRBG along with the key ($K || R$ denotes the string consisting of $K$ followed by $R$)

Can anyone explain to me what the $(DRBG(KR) ⊕P, R)$ part means? (K = Key, R = random bits, P = Plaintext)?

In particular what is confusing me is the following:

Is the second paragraph trying to say that DRBG(KR) is actually DRBG(K || R)? (in which case I assume '||' means concatenate, not 'OR')

What does the '$, R$' mean in the left hand side of the formula? There's no function prefix at the start of it so I'm not sure what the R is being used for, is it trying to say that $E(K, R, P) = E(DRBG(KR) ⊕P, R)$? If that's the case then it doesn't match the arguments anymore for E

答:
A Deterministic Random Bit Generator (DRBG) is a random bit generator such that when the same seed is given the output will be always the same.

Can anyone explain to me what the  $E(K, R, P) = (\operatorname{DRBG}(K||R) \oplus P, R)$ part means?

It means

seed the $\operatorname{DRBG}$ with concatenating the key $K$ and random $R$

output $len(P)$ amount of random bits ( call it as Key Stream (KS))

$\oplus$ the key stream with the plaintext $C = P \oplus KS$

return $C$ and $R$.

send or store $(C,R)$.

Is the second paragraph trying to say that DRBG(KR) is actually DRBG(K || R)? (in which case I assume '||' means concatenate, not 'OR')

Yes, there is a typo there since $DRBG(K \oplus R)$ is distinguishable by $DRBG(K \oplus R)$ and $DRBG(R \oplus K)$ but $||$ is not with a fixed key.

What does the '$, R$' mean in the left hand side of the formula? There's no function prefix at the start of it so I'm not sure what the R is being used for, is it trying to say that $E(K, R, P) = E(DRBG(KR) ⊕P, R)$? If that's the case then it doesn't match the arguments anymore for E

The $R$ is similar to nonce/IV of the block/stream ciphers so that one can use the same key $K$ more than once. As in the nonce/IV case you need to send it so that the receiver will use it during the decryption where here as in OTP, it is the same process.

The crucial part is the $K||R_i$ and $K||R_j$ must not overlap for $i \neq j$. If overlapping occurs (two-time pad), the crib-dragging technique can be used to reveal the plaintexts and it can be automated;

A Natural Language Approach to Automated Cryptanalysis of Two-time Pads


问:
diophantine equation $(1+\ldots+n)^k=1^l+\ldots+n^l$
Here's my equation I wrote and can't solve for over 3 months: $$(1+\ldots+n)^k=1^l+\ldots+n^l$$ in positive integers of course.

For any $n$ there are two obvious solutions: $k=l=1$ and $k=2,\ l=3$. Are there any others?

--

I tried to use some congruences, e.g. LHS is divisible by $n$ for odd $n$, but without any sensible result.

答:
It is easy to see that $(1+2+3,\cdots+n)^1=(1^1+2^1+3^1,\cdots+n^1)$ and, for the sums of integers $I(n)$ and cubes $C(n)$

$$I(n)=(1+2+3,\cdots+n)=\frac{n(n+1)}{2}\implies (I(n))^2=\frac{n^2(n+1)^2}{2^2}$$
$$C(n)=(1^3+2^3+3^3,\cdots+n^3)=\frac{n^2(n+1)^2}{2^2}$$
$$\therefore (1+2+3,\cdots+n)^2=(1^3+2^3+3^3,\cdots+n^3)$$

$$\text{In all cases }\quad (1+2+3,\cdots+n)^k=\frac{n^k(n+1)^k}{2^k}$$

The sums of the first $10$ powers are shown here
and I have worked out formulas up to $\sum_{x=1}^n x^{31}$.

None of these are even divisible by any power of $I(n)$ except for $l-3$ so I do not believe that are any other solutions.


问:
Each person is working in some department
From: Vilnis Detlovs and Karlis Podnieks, Introduction to Mathematical Logic,
in the the usual predicate logic context, with $Person$, $Department$ and $working$ as predicates, they advise:

For “each person is working in some department”, the correct option is:
$${
\forall x (Person(x) \to \exists y (Department(y) \land working(x,y)))
}$$
While the following is incorrect:
$${
\forall x \exists y (Person(x) \land Department(y) \to working(x,y))
}$$

I'm looking for an explanation as to why the incorrect version is incorrect, e.g. what sentence it's equivalent to, and in which cases it's different from the correct one.

Reference http://podnieks.id.lv/

答: The problem with the second statement is that it is true when $y$ is not a department and $x$ is a person. This is because when $\text{Department}(y)$ is false, the hypothesis is false and the implication is true. Thus it does not ensure that each person works in some department. We may have some person not working in any department at all but the second statement will still be true.


问:
Let $A= \{x: 0<|x|<1$ and $\frac{1}{x}\notin \mathbb{Z}_{>0}\}$.
Consider the set $Y=[-1.1]$ a subspace of $\mathbb{R}$.  Let  $A= \{x: 0<|x|<1 $ and $\frac{1}{x}\notin \mathbb{Z}_{>0}\}$.

I know that $A$ is  open in the straight, can you help me prove that?

But because $A$ would be open in $Y$, I haven't been able to see that yet. I know that those opened in $Y$ would be the intersections of the common open of the straight with the $Y$. So how do I prove $A$ is open at $Y$?

答: As I understood $Y=[-1,1]$ is a subspace of $\Bbb R$ endowed with the usual topology. The set $A$ is open in $\Bbb R$, being a union of an open set $(-1,0)$ and a family $\{(1/(n+1),1/n):n\in\Bbb N\}$ of open sets. Since $A\subset Y$, $A$ is an open subspace of $Y$.


问:
Show that exist $p\in \mathbb{R} $ such that $p^3+p=1$
Show that there exists $p\in \mathbb{R} $ such that $p^3+p=1$.

How can I prove it? I was thinking in what can I do to prove it but I really stuck .

I just see supreme axiom, Archimedean property and sequences.

Any hint of what I can do?

答: Look at the function $$p^3+p-1.$$ Its derivative wrt $p$ is $$3p^2+1,$$ which is always positive. Hence the function is always increasing. Since for $p\to-\infty$ the function is negative, it follows that it vanishes at some value of $p.$


问:
How is the inner product of a matrix space defined?
I will use a link which will save a lot of writing and possibility of typo's on my part. Page 4 first paragraph the textbook defines the matrix space and the last sentence is :

"This inner product is identical to the dot product on Rmn if an m × n matrix
is viewed as an mn × 1 matrix by stacking its columns."

If I use matrix multiplication I come up with a vector not a string of numbers.
Does anyone have any ideas?

Here is the link by the way.

https://web.auburn.edu/holmerr/2660/Textbook/innerproduct-print.pdf.

答:
The inner product that is being described is

$\langle A, B \rangle=\mbox{tr}(A^{T}B)=\sum_{i=1}^{m} \sum_{j=1}^{n} A_{i,j}B_{i,j} $.


答: The inner product defined there $\langle A, B \rangle = \sum_{i=1}^m \sum_{j=1}^n a_{ij} b_{ij}$ is not matrix multiplication. It is as described in the quote (the usual inner product on $\mathbb{R}^{mn}$ if you convert $A$ and $B$ into tall column vectors)

Title: How is the inner product of a matrix space defined?

Body:
I will use a link which will save a lot of writing and possibility of typo's on my part. Page 4 first paragraph the textbook defines the matrix space and the last sentence is :

"This inner product is identical to the dot product on Rmn if an m × n matrix
is viewed as an mn × 1 matrix by stacking its columns."

If I use matrix multiplication I come up with a vector not a string of numbers.
Does anyone have any ideas?

Here is the link by the way.

https://web.auburn.edu/holmerr/2660/Textbook/innerproduct-print.pdf.


问:
Monotonic integral proof
Let $f$ be a continuous function on $[a,b]$ such that $f(x) \geq 0 $
for every $x\in [a,b]$. Suppose $\int_a^b f = 0$ and show that $f (x) = 0$ for every $x\in [a,b]$.
 obv this is monotonic ( non-decreasing definitionaly) i would liek to show monotonic non increaseing.

We know that $\int_a^b f = 0$
i have a theorem that says S(P) = s(P) = $\sigma$ over a Partition P on [a,b] or my integral does not exists.

So we have $(S(P) = \sigma) \leq 0$  is this enough to make a statement that my function is monotonic non increaseing?

if so how can we combine monotonic non-increase and monotonic non-decreasing to show that $f(x) = 0, \forall x \in [a,b]$?

EDIT:
my only other trick / idea i can come up with is that $\int_a^b |f| = 0$

答:
Suppose that $f(x_1)=A>0$ for some $x\in [a,b]$. Then, by continuity, for $\varepsilon:=A/2$ there is a $\delta>0$ such that $x\in (x_1-\delta,\,x_1+\delta) \,\implies\, |f(x)-f(x_1)|<\varepsilon$, which in this case implies $f(x)>A/2$.

As the interval $(x_1-\delta,\,x_1+\delta)$ intersects $[a,b]$ at least on a length of $\delta$, and as $f(x)\ge 0$ everywhere, we have
$$\int_a^bf\ge \frac{A\delta}2 \ >0\,.$$

Title: Monotonic integral proof


问:
Moment and maximum likelihood estimator of a parameter of a distribution
Question: Suppose $n$ samples are taken from the following distribution:

$f(x)=\frac{x}{\beta} e^{-x^{2} /(2 \beta)}$

where $x$ and $\beta$ are positive. $E(X_i^{2}) = 2\beta$ and $E(X_i) = \sqrt{\frac{\beta \pi}{2}}$, now we are asked to find moment estimator and maximum likelihood estimator for $\beta$.

Here are my approaches that the results of which for some reason are not consistent between the two approaches (My maximum likelihood estimator is the negative of the moment estimator!):

Moment estimator method:

$E[x]=\bar{x}
\rightarrow \sqrt{\frac{\beta \pi}{2}}=\bar{x} \rightarrow \frac{\beta \pi}{2}=\bar{x}^{2} \rightarrow \beta=2 \bar{x}^{2} / \pi$

Maximum likelihood method:

$L(\beta)=\prod_{i=1}^{n} f\left(x_{i} ; \beta\right)=\prod_{i=1}^{n} \frac{x_{i}}{\beta} e^{-x_{i}^{2}/(2 \beta)}$

$ \ \ \ \ \ \ \ \ =\beta^{-n}\left(\prod_{i=1}^{n} x_{i}\right) \exp \left(\frac{-\sum_{i=1}^{n} x_{i}^{2}}{2 \beta}\right)$

$\operatorname{lag} L(\beta)=-n \log \beta+\sum_{i=1}^{n} \log x_{i}-\frac{\sum_{i=1}^{n} x_{i}^{2}}{2 \beta}$

$\frac{d}{d \beta} \log L(\beta)=\frac{-n}{\beta}+\frac{\sum_{i=1}^{n} x_{i}^{2}}{2 \beta^{2}}=0$

$\frac{\sum_{i=1}^{n} x_{i}^{2}}{2 \beta}=n \rightarrow \beta=\frac{1}{2} \times \frac{\sum_{i=1}^{n} x_{i}^{2}}{n} = \frac{E(X_i^{2})}{2}$

$E(X_i^{2}) = Var(X_i) - E(X_i)^2 = 2\beta - \frac{\beta \pi}{2} - \bar{X}^2$

Hence $\beta=-2 \bar{x}^{2} / \pi$

Can't find the problem of my solutions, can anyone help?

答:
Your method of moments estimator is correct.  Clearly, your maximum likelihood calculation cannot be correct.

The likelihood function for a sample $\boldsymbol x = (x_1, \ldots, x_n)$ is proportional to $$\mathcal L(\beta \mid \boldsymbol x) \propto \beta^{-n} \exp \left(- \frac{1}{2\beta} \sum_{i=1}^n x_i^2 \right)$$ so that the log-likelihood is $$\ell(\beta \mid \boldsymbol x) = -n \log \beta - \frac{n \overline{x^2}}{2\beta} ,$$ where $\overline{x^2}$ is the sample second moment; i.e., the mean of the square of the sample.  Consequently we search for critical points satisfying $$0 = \frac{\partial \ell}{\partial \beta} = -\frac{1}{\beta} + \frac{\overline{x^2}}{2\beta^2},$$ hence $$\hat \beta = \frac{\overline{x^2}}{2}.$$  This of course is different than the method of moments estimator, but they are both asymptotically consistent.


问:
Normal Bundle Orientable and inclusion null homotopic then M is orientable
I am trying to do the following exercise :

Let $M$ be a submanifold of $N$, both without boundary. Show that if the normal bundle of $M$ in $N$ is orientable and $i: M\rightarrow N$ is null-homotopic then $M$ is orientable.

I am bit lost and have no clue where to start. Since the normal bundle is orientable I guess proving the orientability of $M$ would be equivalent to that of $N$ but I am not sure this helps. Also I have no idea where to use the fact that $i: M\rightarrow N$ is null-homotopic, thought about using degree theory but this is only defined when we are working with orientable manifolds.

So we have a map $H:M\times [0,1] \rightarrow N$ such that $H(0,m)=i(m)$ and $H(1,m)=p$ for some $p\in N$. I have no idea what to do with this affirmation , I have thought about tubular neighborhoods , or doing the pull-back by the homotopy but none of this seems useful.

Any hint to get me started is appreciated , just trying to understand how to tackle this.

Thanks in advance.

答:
Hint: given a vector bundle $E \to X$ show that there is a corresponding group homomorphism $w_E: \pi_1 X \to \Bbb Z/2$ so that $w_E(\gamma) = 0$ iff $\gamma^*E$ is the trivial bundle over the circle (equivalently, iff $\gamma^*E$ is orientable).

Then show that one has $w_{E \oplus F} = w_E + w_F$, and $E$ is orientable iff $w_E = 0$. From here, determine how to conclude.


问:
Is this statement true or false ?:$\forall \epsilon >0,(\left|a\right|\le\epsilon\implies a=0)$
Can someone help me demonstrate whether the following statement is true or false using pure logic:
$$\forall \epsilon >0,\forall a\in\mathbb{R}(\left|a\right|\le\epsilon\implies a=0)$$

My thought process: since we know that $\epsilon\neq0$ and $a=\epsilon$ that can't imply $a=0$, Therefore the statement is false and it should be $\left|a\right|<\epsilon\implies a=0$, but I'm not sure how to properly put this in logical math sense.

-Thanks for reading...

答:
This statement is false.

Take $ a=\frac{\epsilon}{2}$.

we have $ |a|\le \epsilon $ but $ a\ne 0 $.

to be true, you should change the position of the parenthesis :

$$\forall a\in \Bbb R \;\; (\forall \epsilon>0 \;\; |a|\le \epsilon \implies a=0)$$

to prove it is true, assume that $ a\ne 0 $.

then there exist $ \epsilon =\frac{|a|}{2} $ such that

$$|a|>\epsilon$$


答:
Your statement should contain a quanitifier for $a$,$$\forall\epsilon>0,\forall a\in\Bbb R(|a|\le\epsilon\implies a=0)$$

The negation of your statement is$$\exists\epsilon>0,\exists a\in\Bbb R(|a|\le\epsilon\wedge a\ne0)$$The negation is true for $\epsilon=a=1$ hence the original statement is not true. A single counter-example is enough to falsify a statement.

Title: Is this statement true or false ?:$\forall \epsilon >0,(\left|a\right|\le\epsilon\implies a=0)$

Body:
Can someone help me demonstrate whether the following statement is true or false using pure logic:
$$\forall \epsilon >0,\forall a\in\mathbb{R}(\left|a\right|\le\epsilon\implies a=0)$$

My thought process: since we know that $\epsilon\neq0$ and $a=\epsilon$ that can't imply $a=0$, Therefore the statement is false and it should be $\left|a\right|<\epsilon\implies a=0$, but I'm not sure how to properly put this in logical math sense.

-Thanks for reading...


问:
Triple integral with spherical coordinates.
I'm working in this triple integral:

$ \int_R ln(x^2 + y^2 + z^2 )dV $ at the domaine R $ \{(x,y,z) | z>0  \text{ and } x^2 + y^2 < z^2 \text{ and } x^2 + y^2 + z^2 < 1 \} $

So I've been suggested spherical coordinates:

$\int_0^{\pi/4}\int_0^{2\pi} \int_0^1 \ln(x^2 + y^2 + z^2)drd\phi d\theta$

I'm quite unsure with regards to the order of the integrals. Now I just been thinking setting them in the x,y,z order was the most logical. But I can explain why. Would be very grateful if someone could explain if its correct and why/why not.

答:
So you have $x=\rho\cos\theta\sin\varphi$, $y=\rho\sin\theta\sin\varphi$, and $z=\rho\cos\varphi$. Then$$x^2+y^2+z^2\leqslant1\iff\rho^2\leqslant1$$and\begin{align}x^2+y^2\leqslant z^2&\iff\rho^2\sin^2\varphi\leqslant\rho^2\cos^2\varphi\\&\iff\sin^2\varphi\leqslant\cos^2\varphi\\&\iff\varphi\in\left[0,\frac\pi4\right]\cup\left[\frac{3\pi}4,\pi\right].\end{align}Since, furthermore, we have $z>0$, we have in fact that $\varphi\in\left[0,\frac\pi4\right]$.

So, you do\begin{align}\int_0^{\pi/4}\int_0^{2\pi}\int_0^1\rho^2\sin(\varphi)\log(\rho^2)\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi&=2\int_0^{\pi/4}\int_0^{2\pi}\int_0^1\rho^2\log(\rho)\sin(\varphi)\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi\\&=4\pi\left(\int_0^{\pi/4}\sin(\varphi)\,\mathrm d\varphi\right)\left(\int_0^1\rho^2\log(\rho)\,\mathrm d\rho\right).\end{align}


问:
how to solve in $\Bbb{R}$ the inequality $(x-3)(x+1)<0$

Solve the following inequality in $\Bbb{R}$:
$$(x-3)(x+1)<0$$

My answer was:
$$x^2+x-3x-3<0 \\
x^2+x-3x<+3 \\
x^2-2x<3
$$
what should i do next,is it a remarkable identity?
i dont know how to continue.

答: Another way is to think of when the factors vanish. This helps you form an idea of the range where they are negative, and where positive. Then you're interested in finding the range where their product is negative.</p>


问:
How to show in a clean way that $z^4 + (x^2 + y^2 - 1)(2x^2 + 3y^2-1) = 0$ is a torus?
How to show in a clean way that the zero-locus of $$z^4 + (x^2 + y^2 - 1)(2x^2 + 3y^2-1) = 0$$ is a torus? </p>

答:
I don't know how "clean" this is, but here's my try.

Write $x=r \cos \alpha$, $y=r\sin\alpha$. The equation becomes
$$z^4+\frac{5- \cos2 \alpha }{2}r^4+ \frac{\cos2 \alpha -7}{2}r^2+1=0$$
The details don't matter too much as long as nothing changes sign, which is indeed the case. So we'll write the above as
$$z^4+f(\alpha)r^4-g(\alpha)r^2+1=0$$
where $\forall\alpha \ f>0,g>0$. By rescaling $r$ we can further simplify that to
$$z^4+\tilde{r}^4-\tilde g(\alpha)\tilde{r}^2+1=0$$
where $\tilde r=f^{1/4} r$ and $\tilde g=\frac{g}{\sqrt f}>0$.

This is an equation for an ellipse in the $z^2$-$r^2$ plane. To see this you can complete the square to get
$$
\left(z^2\right)^2+\left(\tilde{r}^2-\frac{\tilde g(\alpha)}{2}\right)^2=\left(\frac{\tilde g(\alpha)}{2}\right)^2-1$$
It is easy to check that $(\tilde g/2)^2-1>0$ for all $\alpha$.

We therefore get that for any fixed $\alpha$, $z$ and $r$ are related by some closed curve $z(\beta), r(\beta)$ which is homeomorphic to $S_1$. In other words, the surface is homeomorphic to $S_1\times S_1$. QED.

Generalization

This was a bit ugly. What about that was essential and what was details? The crux of the matter is the term $f(x,y)=\left(x^2+y^2-1\right)\left(3x^2+2y^2-1\right)$. The contour $f(x,y)=0$ looks like this:

and any other term with the same topology (two copies of $S_1$ where one is in the interior of the other) would do the trick (also, replacing $z^4$ with any other even power).


问:
To prove that a given transformation is linear
The question is as follows:

Let $V$ denote the vector space of all real valued functions on $\mathbb{R}$. Define $T_a \colon V\to V$ by
$(T_a(h))(x) = h(x + a)$, $h \in V$, $x \in \mathbb{R}$.

Prove that $T_a$ is a linear operator on $V$

Here we have no information whatsoever about the properties of the transformation, so how do we show that the transformation is linear? We must be given a known property about the transformation in order to proceed, right?

答:
You were given a function whose domain and codomain are vector spaces. You need to verify whether that function is linear. There is no more "information" that you need, assuming that the vector space structure of your domain and codomain are clear.

So, what are those structures? The elements of $V$ are functions $h\colon\mathbb{R}\to\mathbb{R}$, a vector space over $\mathbb{R}$. The addition of vectors is defined pointwise, so that if $h$ and $g$ are functions, then the function "$h+g$" is defined as follows: (to describe a function, I can just tell you how it is defined on every real number, which is what the equation below does):
$$(h+g)(x) = h(x)+g(x).$$
Note that (i) the $+$ on the left is the sum of vectors in $V$; the $+$  on the right is the sum of real numbers. Same symbol, but two different operations! And (ii) this makes sense, because $h(x)$ and $g(x)$ are both real numbers and we can add them; so this formula tells you how to evaluate "$h+g$" at every real number to get a real number. That is, this formula defines a function from $\mathbb{R}$ to $\mathbb{R}$.

Scalar multiplication in $V$ is defined as follows: given a vector $h\in V$ (that is, a function from $\mathbb{R}$ to $\mathbb{R}$), and a scalar $c$ (which is, presumably, a real number), the function "$c\cdot h$" is defined by the rule
$$(c\cdot h)(x) = ch(x),$$
where the right hand side is the real number product of the real number $c$ and the real number $h(x)$.

I leave it to you to verify this gives $V$ the structure of a vector space.

So now, we fix a real number $a$. We define an operator $T_a\colon V\to V$. This is an operator that takes functions as inputs, and returns functions as outputs. Given a function $h\in V$, $T_a(h)$ needs to be a function from $\mathbb{R}$ to $\mathbb{R}$. To describe this function, we tell you what its value is at any given real number $x$. And that definition is:
$$\Bigl(T_a(h)\Bigr)(x) = h(x+a).$$
Again, this makes sense: $x+a$ is a real number, $h$ is a function that takes real numbers as inputs and gives real numbers as outputs, so $h(x+a)$ is a real number. Thus, $T_a(h)$, evaluated at the real number $x$, yields a real number. That is, $T_a(h)$ is in fact a function from $\mathbb{R}$ to $\mathbb{R}$, and so is an element of $V$.

So $T_a$ defines an operator form $V$ to itself. Now you want to prove that it is a linear operator. That is, you need to verify that:

$T_a(h+g) = T_a(h)+T_a(g)$; and

$T_a(c\cdot h) = c\cdot T_a(h)$.

This requires you to verify equalities of functions. Two functions are equal when they have the same domain, the same codomain, and the same value at every point of the domain. These are all functions from $\mathbb{R}$ to $\mathbb{R}$, so you just need to verify they take the same value at every real number. So you need to check that, for every real number $x$, the result you get from evaluating
$$\Bigl(T_a(h+g)\Bigr)(x)$$
is the same as the result you get from evaluating
$$\Bigl(T_a(h)\Bigr)(x) + \Bigl(T_a(g)\Bigr)(x).$$
This will require you to unwind a few definitions.

Similarly, you need to verify that evaluating
$$\Bigl( T_a(c\cdot h)\Bigr)(x)$$
gives the same result as evaluating
$$\Bigl(c\cdot T_a(h)\Bigr)(x).$$
Again, this requires unwinding a few definitions.

If you get equalities, you have proven $T_a$ is a linear operator.

You have all the information required.


问:
How to generate uniformly distributed points within an ellipse?
I am give a sequence of tasks. The each next task is the continuation of the previous one. I managed to solve all of them, except the last. In the first two tasks I was asked to find marginal distributions, expected value of the random vector and variance-covariance matrix.
$$
\newcommand{\u}[1]{\underline{#1}}
p(x) =
\begin{cases}
1/\pi,\text{  if }x_1^2 + x_2^2 < 1 \\
0, \text{  otherwise}
\end{cases}
$$
$$ p_\underline{x_1}  = \int_{-\sqrt{1 - x_1^2}}^{\sqrt{1 - x_1^2}} \frac{1}{\pi} dx_2 = \frac{2}{\pi}\sqrt{1 - x_1^2} $$

$$ p_\underline{x_2}  = \int_{-\sqrt{1 - x_2^2}}^{\sqrt{1 - x_2^2}} \frac{1}{\pi} dx_1 = \frac{2}{\pi}\sqrt{1 - x_2^2} $$
$$
\mathbb{E}(\underline x) = \begin{bmatrix}
0 \\
0
\end{bmatrix}
$$
$$
Var(\underline{x}) = \begin{bmatrix}
\frac{1}{4} & 0 \\
0 & \frac{1}{4}
\end{bmatrix}
$$
For the next tasks I have the linear transformation of the random vector. Again I have to find the expected value, variance-covariance matrix and using Chebyshev inequality find the radius where at least 90% of probability is:
$$
\u{y} =
\begin{bmatrix}
1 & -1 \\
0 & 2
\end{bmatrix} \u{x} +
\begin{bmatrix}
2 \\
3
\end{bmatrix} \\
$$
$$
\mathbb{E}(b + A\u{x}) = \begin{bmatrix}
2 \\
3
\end{bmatrix}
$$
$$
Var(b + A\u{x}) = \begin{bmatrix}
\frac{1}{2} & -\frac{1}{2} \\
-\frac{1}{2} & 1
\end{bmatrix}
$$
$$
P(||\u{x} - \mathbb{E}(\u{x})||^2 < \frac{tr(var(\u{x}))}{1 - \alpha}) \ge \alpha => \text{ radius } = \sqrt{15}
$$
Lastly I am asked to plot with equal axis (it will look like a circle) the points of the linearly transformed distribution and the circle with the calculated radius. This is the task where I am stuck. I generated points uniformly within Chebyshev's radius. Plot on the left is the stretched circle and on the right is the same with equal axes. The plots contain 90% of the data. So I have to plot the rest 10%, which must be outside the circle/ellipse.

I imagined that this transformation stretches the unit circle into an ellipse. In the case of unit circle I can generate points from uniform distribution values between -1 and 1. How to determine the range in the case of the ellipse? What is this radius then? Is there any connection with eigen vectors? Sure, I can generate points within the unit circle and then transform them with the matrix given, but I think this is not the solution, which I have to provide. I provided all previous tasks and answers because there must be connection (I think).

答:
In a 2D case, generating a random point in a bounding rectangle and throwing away points outside the ellipse isn't that ineffective. However, it is very ineffective if you generalise it for high dimension. I will tell you a way which is intuitive, effective for high dimension and it is guaranteed to terminate in fixed time.

You imagine that the ellipse is a bottle. You calculate the volume and you spill random amount of water into the bottle. The random point you are looking for is somewhere on the level of water. In this case, it is a line. This way, the task reduced by one dimension. Next thing you have to do is choosing a random point of the line you got, which is trivial.

The same thing more formally: You choose the $x$-coordinate by the marginal distribution $p(x)$. After that, you choose the $y$-coordinate by the distribution given $x$, i.e. $p(y|x)$. And you are done.

You can use the same trick for other shapes also, for example simplexes.


问:
$D$ is not open at $Y$, as $D$ is not open at $\mathbb{R}$.
Let    $D=\{x\hspace{0.2cm}|\hspace{0.2cm} \frac{1}{2}\le|x|\le 1\}=\left[-1,-\frac{1}{2}\right]\cup \left[\frac{1}{2},1\right]$, $Y=[-1,1]$

$D$ is not open at $\mathbb{R}$, as there is no $\epsilon>0$ such as $B(-1,\epsilon)\subset D$, $B\left(-\frac{1}{2},\epsilon\right)\subset D$, $B\left(\frac{1}{2},\epsilon\right)\subset D$ and $B(1,\epsilon)\subset D$.

$D$ is not open at $Y$, as $D$ is not open at $\mathbb{R}$.

Is that correct? Obs: Usual topology of the straight and $Y$ induced topology

答: Hint: you can argue that $D = \overline D$, where $\overline D$ stands for the closure of $D$.


问:
$\int_{0}^{\pi} D_{n}(y)dy=\frac{1}{2}$ Dirichlet
I need to calculate that $\int_{0}^{\pi} D_{n}(y)dy=\frac{1}{2}$ with $D_{n}(y)= \frac{1}{2\pi}\frac{\sin((n+\frac{1}{2})y)}{\sin(\frac{y}{2})}$ from Dirichlet.

Now I tried to do this with the known that $D_{n}(y)=\frac{1}{2\pi}\sum_{k=-1}^{n}e^{iky}$.

I tried to things but I never become 1/2.

First I tried to solve the somation and then integrate: $\int_{0}^{\pi} D_{n}(y)dy=\int_{0}^{\pi} \frac{1}{2\pi}\sum_{k=-1}^{n}e^{iky}dy=\int_{0}^{\pi} \frac{1}{2\pi}(\frac{e^{-niy}-e^{iy(n+1)}}{1-e^{iy}})dy$. This integral doesn't converges so I did something wrong but I don't know what?

Second I tried to switch the integral and the sommetion and then solve it: $\int_{0}^{\pi} D_{n}(y)dy=\int_{0}^{\pi} \frac{1}{2\pi}\sum_{k=-1}^{n}e^{iky}dy=\int_{0}^{\pi} D_{n}(y)dy=\frac{1}{2\pi}\sum_{k=-1}^{n}\int_{0}^{\pi}e^{iky}dy=\frac{1}{2\pi}\sum_{k=-1}^{n}\frac{1}{ik}(1-e^{-\pi i k})$. But this sometion will give $i\infty -i \infty$ so this doesn't work eather.

Can someone eplain me what I do wrong and how I can fix it?

Body:
Note

$$2\sin\frac y2(\cos y + \cos2y+\cos3y+...+\cos ny)
= \sin\frac{(2n+1)y}2-\sin\frac y2 $$

Then

\begin{align}
 \int_{0}^{\pi} D_{n}(y)dy=&\frac{1}{2\pi}
\int_0^\pi \frac{\sin\frac{(2n+1)y}{2}}{\sin \frac y2}{\rm d}y\\
=&\frac{1}{2\pi} \int_0^\pi(1+2\cos x + 2\cos2x+...+2\cos nx){\rm d}x\\
=&\frac1{2\pi}[\pi + (0+0+...+0)]=\frac12 \end{align}