问:
A few problems on sup and nested intervals
I've been doing these 3 problems for a `proof´ oriented class, one i have found a solution (in fact has been asked here before but the threads are all closed), and checked a correct solution in the Analysis book from Apostol. The thing is, I think im making a mistake and i cant find it. The other 2 problems are using nested closed intervals and i dont really know where to start to solve them. Thanks for your help in advance, and here they go:

The first one is

If $A$, $B\subseteq\mathbb{R}$ are non empty and bounded above (so sup exists) we consider $A+B =\{a+b:a\in A,b\in B\}$.
Show the relation between $sup(A)$ $sup(B)$ and $sup(A+B)$.

My solution:
First i show $A+B$ is bounded. Consider an arbitrary element of $A+B$ is of the form
$a+b$.

Now  $a \leq sup(A)$  and  $b \leq sup(B)$  from this we get

$a+b \leq sup(A) + sup(B)$. $  $ $ $ So $sup(A+B)$ exists, as $A+B$ is non-empty and bounded (completeness axiom). Also $sup(A+B) \leq sup(A)+sup(B)$ because $sup(A+B)$ is the smallest bound of the set.

Now i want to study the relation between supremums.

I know that for any $a+b$,

$a+b \leq sup(A+B)$ and hence $a \leq sup(A+B) - b$
So $sup(A+B)-b$ is an bound for $a$, and as $sup(A)$ is the supremum for any $a$
$sup(A) \leq sup(A+B) - b$

Doing the same for b, $b \leq sup(A+B) - sup(A)$ and as this is a above bound for $b$ and $sup(B)$ is the smallest one, $sup(B) \leq sup(A+B) - sup(A)$ then $sup(A) + sup(B) \leq sup(A+B)$ and $sup(A+B) \leq sup(A)+sup(B)$

And because of this $sup(A+B) = sup(A)+sup(B)$

And here are the other two excercises that i really dont have idea where to start, specially the second one

Second one:

Be $(I_n)_{n\in\mathbb{N}}$ a sequence of closed intervals and for every
$n\in\mathbb{N}$, be $\lambda_n$ the lenght of $I_n$; lets supose that $I_1\supseteq
I_2\supseteq\dots\supseteq I_n\supseteq I_{n+1}\supseteq\dots$. Show that
$\lim \lambda_n$ exists and that if $\lim\lambda_n>0$ then
$\bigcap_{n\geq1}I_n$ is a closed interval of lenght $\lim\lambda_n$.

My attemp to solution For this one i know that they're not empty and bounded above and below (they're closed intervals) and i know that if two sets $A \subseteq B$ are bounded an non-empty, then $sup A \leq sup(B)$ and as they're closed $max(A) \leq max(B)$ and the same with their minimums so the lenght of $[P_n, Q_n]$ $|Pn-Qn|$ is decreasing.

And the last one is

Be $x_1$, $a\in\mathbb{R}$ Such that $x_1>0$ y $a>x_1^2+x_1$,
and for every $n\in\mathbb{N}$ be $x_{n+1}=a/(1+x_n)$.
Then every interval $[x_1,x_2]$, $[x_3,x_4]$, $\dots$ and so on, form  an intersection of intervals and the only point in common $\xi$ common to every one of them is A root of the equation $x^2+x=a$.

sugestion (this comes with the excercise too)
Show that for every $n\in\mathbb{N}$$x_{n+1}-x_n$ and $x_n-x_{n-1}$ have opposite signs and that exists $\theta\in(0,1)$ such that $|{x_{n+1}-x_n}|\le\theta|{x_n-x_{n-1}}|$ for every $n\in\mathbb{N}$.

答:
Note: This is my first time putting a bit of effort into answering a post on MSE so I apologize in advanced if anything I write is false or unclear

1) Your proof seems fine to me other than justifying that $\sup(A+B) \le \sup(A) + \sup(B)$ (i.e. its an upper bound...)

2) First off, to show that $\lambda_n$ is decreasing I would observe that since $\{I_n\}_{n \in \mathbb{N}}$ is a sequence of nested intervals $\exists \{a_n\}_{n \in \mathbb{N}} , \{b_n\}_{n \in \mathbb{N}} : I_n = [a_n , b_n]$ and it must be true that $a_n \le a_{n+1} \le b_{n+1} \le b_n$ so it's clear that $b_{n+1} - a_n \le b_n - a_n$ and $-a_n \ge -a_{n+1}$ so $b_{n+1} - a_{n+1} \le b_n - a_n$. We can see that $\lambda_n = b_n - a_n$ so $\{\lambda_n\}$ must be decreasing. Now if
$$
\lim_{n \to \infty} \lambda_n = \lambda > 0
$$
then it's clear that $\lim_{n \to \infty} b_n - a_n = \lambda$. So it must be true that $\lim_{n \to \infty} a_n = a , \lim_{n \to \infty} b_n = b$ where $a,b$ are finite and $\lambda = b - a$. now we observe that $\bigcap_{n \in \mathbb{N}} I_n = [a , b]$ (it's as if you take do $\lim_{n \to \infty} [a_n , b_n]$ - I've never seen this actually done, considering this isn't really a "limit" anymore) and then the length of $[a , b]$ is clearly $\lambda$.

3) Now I don't see exactly how to do this but it appears that this sequence oscillates between increasing and decreasing. I would try to notice a pattern with the oscillations between increasing and decreasing and then form subsequences so that you form this inequality $a_n \ge a_{n+1} \ge b_{n+1} \ge b_n$ and then form a sequence of nested intervals then the limit exists. From this you can pass the limit through your recursive definition so you get
$$
L = \frac{a}{L + 1} \iff L^2 + L - a = 0
$$
which shows that this sequence converges to the solution to the equation $x^2 + x = a$
I will keep thinking of how to find the proper subsequences here and update this answer accordingly

Hope this helps!


问:
A question on probability theory with 3 urns with return
Each of $3$ urns contains twenty balls. First urn contains ten white balls, second urn contains six white balls and third urn contains two white balls. All other balls are black.
One ball is drawn from the random urn with return in the same urn. The ball's color is white.
What is the probability that the second ball drawn from the same urn is white?

I think, this is $\frac{1}{9} \cdot \frac{2}{20} + \frac{3}{9} \cdot \frac{6}{20} + \frac{5}{9} \cdot \frac{10}{20}$ by Bayes'theorem and Law of total probability, but can't be sure.

Thanks for any help.

答:
Welcome.

Your answer is correct, it does need Bayes' Theorem and total probability, but you haven't shown how you arrived at it and I don't know what form of Bayes' Theorem you have learnt.

To keep it very rudimentary, the probability that the first white ball was found in a particular box chosen at random will equal the ratio of white balls it contains to total white balls.

Then, applying the law of total probability, we get the formula

$\Sigma$ [P(1st ball white from box $k$)$\times$ P(2nd ball white from box $k$)] = $\frac{10}{18}\cdot\frac{10}{20} + \frac6{18}\cdot\frac6{20} + \frac2{18}\cdot\frac2{20}$

ContentLicense: CC BY-SA 4.0


问:
How to prove that f is convex?
This is from Bela Bollobas's book on functional analysis.

Given:

$f: (a,b) \to (c,b) \text{ and } \phi: (c,b) \to \mathbb{R} \quad \phi^{-1}f, \phi \text{ are both convex }$

To show:

$f$ is convex

What I've tried:

Looked at the definitions a lot of times, and the fact that $\frac{f(x_2)-f(x_1)}{x_2 - x_1}$ is non-decreasing in one variable when the other is held fixed. I also tried looking at the convex set. But I'm stuck. I also know that if $\phi$ is non-decreasing then we're done. I'm just looking for a hint.

答: This is not true. On $(0,1)$ let $\phi (x)=\frac 1 x$ and $f(x)=\sqrt x$. Then $\phi^{-1}\circ f$ and $\phi$ are convex but $f$ is not. If $\phi$ is also increasing then the result is true.

Title: How to prove that f is convex?


问:
Prove the diagonalizing matrix $P$ of a hermitian matrix $A$ is unitary
Let $A\in\mathbb{C}^{n\times n}$ be a hermitian matrix ($A=A^{\dagger}$) with $n$ eigenvectors $\{\vec{v}_k\}_{k=1}^{n}\in\mathbb{C}^{n\times1}$ and corresponding eigenvalues $\{\lambda_k\}_{k=1}^n$. It is given that $\lambda_j\neq\lambda_k$ for all $j,k\in[1,n]_{\mathbb{N}},j\neq k$, and that:

$$||\vec{v}_k||=1\quad \forall k\in[1,n]_{\mathbb{N}}$$

Let $P\in\mathbb{C}^{n\times n}$ be a square matrix, such that its columns are the eigenvectors of $A$:

$$P=\left(
    \begin{matrix}
    | & | & \cdots & | \\
    \vec{v}_1 & \vec{v}_2 & \cdots & \vec{v}_n \\
    | & | & \cdots & | \\
    \end{matrix}
\right)$$

Prove that $P$ is unitary, meaning $P^{-1}=P^\dagger$.

I'm sure that there is more than one proof somewhere in the forum. I didn't bother to search because I don't want to know how to prove the statement, I just need help completing my proof (or altering what has be to changed).

Proof: First, define a diagonal matrix $D\in\mathbb{C}^{n\times n}$ such that:

$$D=\text{diag}\{\lambda_k\}_{k=1}^n$$

It is clear that $P^{-1}$ exists. Let $\{\vec{u}_k\}_{k=1}^n\in\mathbb{C}^{1\times n}$ be the row vectors that build $P^{-1}$, such that:

$$P^{-1}=\left(
    \begin{matrix}
    - & \vec{u}_1 & - \\
    - & \vec{u}_2 & - \\
    \vdots & \vdots & \vdots \\
    - & \vec{u}_n & - \\
    \end{matrix}
\right)$$

Then, we can see that:

$$A_{jk}=\sum_{m=1}^{n}\sum_{l=1}^{n}P_{jm}D_{ml}P^{-1}_{lk}=\sum_{m=1}^{n}\sum_{l=1}^{n}(\vec{v}_m)_j\delta_{ml}\lambda_{m}(\vec{u}_l)_k=\sum_{m=1}^{n}(\vec{v}_m)_j\lambda_{m}(\vec{u}_m)_k$$

Since $A$ is hermitian, its eigenvalues are real, and $A_{jk}=A^\star_{kj}$ for all $j,k\in[1,n]_\mathbb{N}$. We can see that:

$$A^\star_{kj}=\sum_{m=1}^{n}(\vec{u}^\star_m)_j\lambda_{m}(\vec{v}^\star_m)_k$$

Therefore:

$$\sum_{m=1}^{n}(\vec{u}^\star_m)_j\lambda_{m}(\vec{v}^\star_m)_k=\sum_{m=1}^{n}(\vec{v}_m)_j\lambda_{m}(\vec{u}_m)_k$$

In order to prove $P$ is unitary, we need to show that $\vec{v}_m=(\vec{u}^T_m)^\star$ for all $m\in[1,n]_\mathbb{N}$. We can see clearly that if this equality is true, then $A_{jk}=A^\star_{kj}$ as we expect. But I need the opposite direction - I need to prove that $\vec{v}_m=(\vec{u}^T_m)^\star$ just using the equation above. Not sure how to do that. Also, I haven't used the fact that the eigenvectors are normalized. I'm not sure if that's necessary.

Thanks!

答:
If $ \lambda_j \ne \lambda_k$, then $\vec{v}_j$ and $\vec{v}_k$ are othonormal, thus

$\vec{v}_1,...,\vec{v}_n$ is an orthonormal basis of $ \mathbb C^n.$

Can you take it from here ?


问:
drawing balls from a bin but the balls are changing over time
I am doing this problem:

An urn initially has w white balls and k black balls. The balls are randomly drawn and replaced in the bin. Whenever a white ball is selected, it is painted black before it is replaced. What is the probability that the ball selected on the (n+1)st draw is white?

What I am thinking is to conditional on there are j times that a white ball is selected in the first n draws. So

$$P(the\ (n+1)st\ draw\ is\ white) = \sum_{j=0}^{n}P(the\ (n+1)st\ draw\ is\ white\ |\ j\ whites\ in\ the\ first\ n\ draws)P(j\ whites\ in\ the\ first\ n\ draws) $$
And I know

$$P(the\ (n+1)st\ draw\ is\ white\ |\ j\ whites\ in\ the\ first\ n\ draws) = \frac{w-j}{k+w}$$

But I am wondering how to calculate the probability that there are j whites in the first n draws. I think it should be $\frac{w(w-1)...(w-j+1)}{(w+k)^j}$, but I am not sure if it is correct.

Can anyone give me a hint? Thank you.

答:
$\boldsymbol{n^\text{th}}$ Pick is a White Ball

The probability that the $n^\text{th}$ pick will be a white ball equals the probability of choosing a white ball, then choosing $n-1$ other balls. That is,
$$
\bbox[5px,border:2px solid #C0A000]{\frac{w}{k+w}\left(\frac{k+w-1}{k+w}\right)^{n-1}}\tag1
$$

Drawing the $\boldsymbol{j^\text{th}}$ White Ball

Here is a typical sequence for drawing $j$ white balls:
$$
\overbrace{\left(\frac{k}{k+w}\right)^{n_0}}^{\substack{\text{draw $n_0$}\\\text{black balls}}}\overbrace{\ \frac{w\vphantom{k^n}}{k+w}\ }^{\substack{\text{first}\\\text{white ball}}}\cdot\overbrace{\left(\frac{k+1}{k+w}\right)^{n_1}}^{\substack{\text{draw $n_1$}\\\text{black balls}}}\overbrace{\ \frac{w-1\vphantom{k^n}}{k+w}\ }^{\substack{\text{second}\\\text{white ball}}}\cdots\overbrace{\left(\frac{k+j-1}{k+w}\right)^{n_{j-1}}}^{\substack{\text{draw $n_{j-1}$}\\\text{black balls}}}\overbrace{\ \frac{w-j+1\vphantom{k^n}}{k+w}\ }^{\substack{j^\text{th}\\\text{white ball}}}\tag2
$$
we need to consider sums where $n_0+n_1+\dots+n_{j-1}=n-j$. One way to keep track of things is with generating functions.
$$
\left[x^n\right]\frac1{1-\frac{kx}{k+w}}\frac{wx}{k+w}\cdot\frac1{1-\frac{(k+1)x}{k+w}}\frac{(w-1)x}{k+w}\cdots\frac1{1-\frac{(k+j-1)x}{k+w}}\frac{(w-j+1)x}{k+w}\tag3
$$
Thus, the generating function for the probability of drawing the $j^\text{th}$ white ball on a given pick is
$$
\bbox[5px,border:2px solid #C0A000]{\prod_{i=0}^{j-1}\frac{(w-i)x}{k+w-(k+i)x}}\tag4
$$

Verifying $\boldsymbol{(1)}$ Using $\boldsymbol{(4)}$

If we sum $(4)$ over all the possible white balls, we should get the generating function for $(1)$.
$$
\begin{align}
\sum_{j=1}^w\prod_{i=0}^{j-1}\frac{(w-i)x}{k+w-(k+i)x}
&=\sum_{j=1}^w\prod_{i=0}^{j-1}\frac{w-i}{\frac{k+w}x-(k+i)}\tag5\\
&=\sum_{j=1}^w\frac{\binom{w}{j}}{\left(\frac{k(1-x)+w}x\atop{j}\right)}\tag6\\
&=\frac1{\binom{v}{w}}\sum_{j=1}^w\binom{v-j}{v-w}\tag7\\[3pt]
&=\frac1{\binom{v}{w}}\binom{v}{w-1}\tag8\\[6pt]
&=\frac{w}{v-w+1}\tag9\\[9pt]
&=\frac{wx}{k+w}\frac1{1-\frac{k+w-1}{k+w}x}\tag{10}\\
&=\sum_{n=1}^\infty\color{#C00}{\frac{w}{k+w}\left(\frac{k+w-1}{k+w}\right)^{n-1}}x^n\tag{11}
\end{align}
$$
Explanation:

$\phantom{1}\text{(5)}$: divide numerator and denominator by $x$

$\phantom{1}\text{(6)}$: write products as binomial coefficients

$\phantom{1}\text{(7)}$: set $v=\frac{k(1-x)+w}x$

$\phantom{1}\text{(8)}$: Hockey-stick identity

$\phantom{1}\text{(9)}$: simplify

$(10)$: undo $(7)$

$(11)$: write as a series

Indeed, $(11)$ is the generating function for $(1)$.

How Long Will the White Balls Last?

Suppose we start with $10$ white and $10$ black balls. After a possibly large number of picks, we will have drawn all the white balls and be left with a bin of black balls.  We can use $(4)$ to compute the probability of the various durations until the white balls disappear.

The average duration after the $j-1^\text{st}$ white ball is drawn until the $j^\text{th}$ white ball is drawn is $\frac{k+w}{w-j+1}$. Thus, the average duration until all the white balls have been drawn is
$$
(k+w)H_w\tag{12}
$$
where $H_w$ is the $w^\text{th}$ Harmonic Number.

Plotting the coefficients of the generating function
$$
g(x)=\prod_{i=0}^9\frac{(10-i)x}{20-(10+i)x}\tag{13}
$$
we get

where the maximum occurs at $n=46$. The average can be computed by evaluating
$$
g'(1)=\frac{7381}{126}\tag{14}
$$
which matches $20H_{10}$ from $(12)$.


问:
How can I evaluate $\sum_{n=0}^\infty \frac{(2n)!}{(n!)^2(2n+1)4^n}$?

Evaluate:
$$\sum_{n=0}^\infty \frac{(2n)!}{(n!)^2(2n+1)4^n}$$

I'm trying to solve this infinite series of problems, but I'm not making any progress. I'd appreciate your help.

I'm sorry if my writing is rude because I'm not good at English.

Thank you for reading.

答:
[I see you have brought a little change to your question].

Let $$f(x):=\sum_{n=0}^{\infty}\frac{x^n}{2n+1}\binom{2n}{n}.$$

(it will be time to change $x$ into $\frac14$ at the very last step).

Let:

$$g(x):=xf(x^2)=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}\binom{2n}{n}$$

is a primitive function of

$$h(x):=\sum_{n=0}^{\infty}{(x^2)}^n\binom{2n}{n}=\dfrac{1}{\sqrt{1-4x^2}} \tag{1}$$

the last equality coming from the classical generating function for central binomial coefficients (Generating functions and central binomial coefficient).

It remains now to work backwards, the remaining difficulty being the integration of expression (1).


问:
Proof of the identity $2^n = \sum\limits_{k=0}^n 2^{-k} \binom{n+k}{k}$
I just found this identity but without any proof, could you just give me an hint how I could prove it?
$$2^n = \sum\limits_{k=0}^n 2^{-k} \cdot \binom{n+k}{k}$$

I know that $$2^n = \sum\limits_{k=0}^n \binom{n}{k}$$ but that didn't help me

ContentLicense: CC BY-SA 3.0

Body:
You can prove it by induction on $n$. Here’s a start. Let

$$f(n)=\sum_{k=0}^n2^{-k}\binom{n+k}k\;;$$

clearly $f(0)=1=2^0$. Assume that $f(n)=2^n$ for some $n\ge 0$. Then

$$\begin{align*}
f(n+1)&=\sum_{k=0}^{n+1}2^{-k}\binom{n+1+k}k\\
&=\sum_{k=0}^{n+1}2^{-k}\left(\binom{n+k}k+\binom{n+k}{k-1}\right)\\
&=\sum_{k=0}^{n+1}2^{-k}\binom{n+k}k+\sum_{k=1}^{n+1}2^{-k}\binom{n+k}{k-1}\\
&=2^{-(n+1)}\binom{2n+1}{n+1}+\sum_{k=0}^n2^{-k}\binom{n+k}k+\sum_{k=0}^n2^{-(k+1)}\binom{n+1+k}k\\
&=2^{-(n+1)}\binom{2n+1}{n+1}+2^n+\frac12\sum_{k=0}^n2^{-k}\binom{n+1+k}k\;,
\end{align*}$$

and

$$f(n+1)=2^{-(n+1)}\binom{2n+2}{n+1}+\sum_{k=0}^n2^{-k}\binom{n+1+k}k\;.$$

Now combine these results and use the fact that

$$\binom{2n+2}{n+1}=\binom{2n+1}{n+1}+\binom{2n+1}n=2\binom{2n+1}n$$

to complete the induction step.


Title: How can I evaluate $\sum_{n=0}^\infty \frac{(2n)!}{(n!)^2(2n+1)4^n}$?

Body:
Evaluate:
$$\sum_{n=0}^\infty \frac{(2n)!}{(n!)^2(2n+1)4^n}$$

I'm trying to solve this infinite series of problems, but I'm not making any progress. I'd appreciate your help.

I'm sorry if my writing is rude because I'm not good at English.

Thank you for reading.

Body:
Your series can be written as $$S=\sum_{k=0}^{\infty} {2k\choose k} \frac{2^{-2k}}{2k+1}$$
Note that $$\sum_{k=0}^{\infty} {2k \choose k} x^{2k}=\frac{1}{\sqrt{1-4x^2}}$$
Integrate both sides w.r.t. $x$ from $x=0$ to $x=1/2$, then you get
$$\frac{1}{2}\sum_{k=0}^{\infty} {2k \choose k} \frac{2^{-2k}}{2k+1}=\int_{0}^{1/2} \frac{dx}{\sqrt{1-4x^2}}=\frac{1}{2}\sin^{-1} 2x|_{0}^{1/2}=\frac{\pi}{4}.$$
Finally,$$S=\pi/2$$


问:
How to solve $\ddot{f}=f^{-2}$
I am facing a question where I need to find the solution for this differential equation:

$$\frac{d^2}{dt^2}f(t)=\frac{1}{f(t)^2}$$

What type of differential equation is this? What is the standard method for solving it?

答:
Rewrite the equation
$$y''(t)=\frac 1{[y(t)]^2}$$ as
$$-\frac {t''(y)}{[t'(y)]^3}=\frac 1 {y^2}$$ Reduction of order $p=t'(y)$
$$\frac {p'}{p^3}=-\frac 1 {y^2}\implies -\frac 1{2p^2}=\frac 1 y+C\implies p=\pm\frac{\sqrt{y}}{\sqrt{2} \sqrt{-C y-1}}$$ which is quite nasty but ... doable.

Title: How to solve $\ddot{f}=f^{-2}$


问:
Finding the area between $f(x)=e^{x/2}-2$ and $g(x)=x^2+2x-1$, from $x=0$ to $x=2$

I need to find the area between the graphs of
$$f(x)=e^{x/2}-2 \quad\text{and}\quad g(x)=x^2+2x-1$$
and the lines $x = 0$ and $x = 2$.

I think I need to do $g(x)-f(x)$ with $0$ and $2$ as the lower and upper limits on the integrals.

But how do I go forward with it?

答:
Yes your idea is fine, firstly we need to show that $g(x)\ge f(x)$ within the domain

then the area between the curves is given by

$$\int_0^2 (g(x)-f(x))dx$$

which is the sum of trivial integrals.

Note that if we were interested in the signed area we need to divide the integral according to the different sign of the funtion in the domain of integration.


问:
Vector direction
I have a vector in the 2nd Coordinate of the Cartesian plane. I want to know that how can I find out the direction of the vector that whether it is towards the origin or away from it whether its in increasing direction of Y-Coordinate or in the decreasing direction of Y-Coordinate.

Thanks

答:
1) If you only know the second coordinate and not the first, you cannot know if the vector points to or away the origin.

2) If you have a function that describes the y coordinate (but you didnt tell which on it is), you can know if it grows or not in the y-coordinate

Title: Vector direction


问:
A quick question about limit of $[u]$
Suppose $u$ is a function. when $u\to \infty$ ,can we say : $\lim[u]=u=\infty$?

My justification is:

$$u-1<[u]\le u$$

Therefor if $u\to \infty$ then $u-1\to \infty$ too. and because of Sandwich Theorem we conclude $\lim[u]=\infty$ . Am I right?

答:
Yes you are absolutely right and since $u \to \infty$ in this case the left inequality suffices that is

$$[u]>u-1 \to \infty \implies [u] \to \infty$$

Title: A quick question about limit of $[u]$


问:
Epistemic sentence "Agent A knows that he doesn't know the answer, but Agent A knows that Agent B also doesn't know the answer"
For that sentence, I have translated it to an epistemic sentence. Can someone see if I did it right? If not, where did I go wrong?

Thanks

$K_a\land\lnot K_a\land K_a\land K_b$

答:
Knowledge operators index an agent and operate on a subject (a proposition, or fact).

$\mathcal K_c\,\varphi$ means "$c$ knows $\varphi$", and here $c$ is the agent, and the proposition $\varphi$ is the subject.

You have agents $a$ and $b$ in your indices.  You are missing subjects of their knowledge, such as "the answer is whatever it is."

Let's use $\varphi$ for that.  Then $\mathcal K_a\mathcal K_b\varphi$ reads "$a$ knows $b$ knows the answer is whatever it is", and so forth.

Thus, to modify your attempt:

$$\mathcal K_a\neg\mathcal K_a\varphi~\land~\mathcal K_a\neg\mathcal K_b\varphi$$


答:
Also note that

$$[x]=x-\{ x \}=\infty-t \;\;\;,\;\;\;\;t\in[0,1)$$

$$\to\infty$$

Where $\{x\}$ denotes the fractional part of x

Title: A quick question about limit of $[u]$

Body:
Suppose $u$ is a function. when $u\to \infty$ ,can we say : $\lim[u]=u=\infty$?

My justification is:

$$u-1<[u]\le u$$

Therefor if $u\to \infty$ then $u-1\to \infty$ too. and because of Sandwich Theorem we conclude $\lim[u]=\infty$ . Am I right?


问:
Prove that $\sqrt[3]{2}+\sqrt[3]{4}$ is irrational
As the question says, how can I prove that $\sqrt[3]{2}+\sqrt[3]{4}$ is irrational?

I have tried setting it to be equal to $a$, and $\sqrt[3]{4}$ equal to $a^2$, but I haven't gotten anywhere. The solution does not have to use the above. Any help is appreciated. (I know this is a duplicate, but I haven't seen any answers that I have gotten a full solution from)

答:
Since, $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$$ and
$$\sum_{cyc}(a^2-ab)=\frac{1}{2}\sum_{cyc}(a-b)^2=0$$ for $a=b=c$ only and $$\sqrt[3]2\neq\sqrt[3]4,$$ we obtain:
$$\sqrt[3]2+\sqrt[3]4-x=0$$ is equivalent to $$2+4-x^3+3\sqrt[3]2\cdot\sqrt[3]4\cdot x=0$$ or
$$x^3-6x-6=0.$$
But  the polynomial $x^3-6x-6$ is irreducible by Eisenstein (just take $p=2$, for example): https://en.wikipedia.org/wiki/Eisenstein%27s_criterion

Thus, the polynomial $x^3-6x-6$ has no  rational roots.


问:
Span of a finite set of linearly dependent vectors.
Show that a linear dependent set of $n$ vectors does not span $V$, where $\dim(V)=n$.

For example, show that $\{(1,0,0),(0,1,0),(2,0,0)\}$ does not span $\mathbb{R}^3$.

答: If there are some $a,b,c\in \Bbb R$ such that $(x_1,x_2,x_3)=a(1,0,0)+b(0,1,0)+c(2,0,0)$, then $x_3=0$. Therefore $(0,0,1)$ is not in that span.


Title: Span of a finite set of linearly dependent vectors.

Body:
Show that a linear dependent set of $n$ vectors does not span $V$, where $\dim(V)=n$.

For example, show that $\{(1,0,0),(0,1,0),(2,0,0)\}$ does not span $\mathbb{R}^3$.

Body: $(0,0,1)$ is not in the span of $\{(1,0,0),(0,1,0),(2,0,0)\} $


问:
Testing of significance for a single proportion
Suppose a large number $n$ of independent Bernoullian trials is performed and $x$ successes are obtained. Suppose we wish to test the hypothesis that probability of success in each trial is $p$, assuming the hypothesis to be correct, the mean and variance of the sampling distribution of the number of success are $np$ and $npq$ respectively. Then for large $n$ $z=({x-np})/{\sqrt {npq}}$ is distributed as a standard normal variate. Hence we get $P(\vert z \vert >3)=0.0027$
Where did the $3$ come from?

答:
It is arbitrary, illustrating the probability with a normal distribution of being more than $3$ standard deviations from the mean in the two tails.

It could have said

"Hence we get $P(\vert z \vert >2)=0.0455$"

"Hence we get $P(\vert z \vert >1)=0.3173$"

or various similar things.  This is similar to the so-called 68–95–99.7 rule but with the probabilities of being in the tails rather than the middle of the distribution


问:
The perimeter of a segment of a circle is 22 inches. The arc is 2 radians. What is the area of the segment?
The perimeter of a segment of a circle is 22 inches. The arc is 2 radians. What is the area of the segment?

I am having a hard time answering this question since the given are the perimeter of the segment as well as the arc measure. On the other hand, I am positive that the answer can be found using the given but I don't know-how. Can you please help me? It will be a big help. Thank you.

答:
Let $r$ be a radius of the circle.

Thus, $$2r\sin1+2r=22,$$ which gives $$r=\frac{11}{1+\sin1}.$$
Id est, the are of the segment is equal to
$$\frac{1}{2}\cdot2\left(\frac{11}{1+\sin1}\right)^2-\frac{1}{2}\left(\frac{11}{1+\sin1}\right)^2\sin2=\left(\frac{11}{1+\sin1}\right)^2(1-\sin1\cos1).$$


答:
This statement uses the normal approximation to a binomial probability, which works well for sufficiently large $n$ and $p$ not too far from 1/2.

The idea is that you will reject the null hypothesis that $p = p_0$ at the 0.3% level if $|Z| > 3.$

For example, if $n = 36$ and $p_0 = 1/2,$ then you would
reject $H_0: p = 1/2,$ if you got only $X=1$ Success.

The probability that an observation from a standard normal distribution is in the interval $(-3,3)$ is about 0.997. In R:

```<code>diff(pnorm(c(-3,3)))
[1] 0.9973002
</code>```

Alternatively, perhaps you can find a printed table of the standard normal PDF that goes that far out from $0.$

Also, perhaps google the Empirical Rule.

Title: Testing of significance for a single proportion

Body:
Suppose a large number $n$ of independent Bernoullian trials is performed and $x$ successes are obtained. Suppose we wish to test the hypothesis that probability of success in each trial is $p$, assuming the hypothesis to be correct, the mean and variance of the sampling distribution of the number of success are $np$ and $npq$ respectively. Then for large $n$ $z=({x-np})/{\sqrt {npq}}$ is distributed as a standard normal variate. Hence we get $P(\vert z \vert >3)=0.0027$
Where did the $3$ come from?

Body:
This statement uses the normal approximation to a binomial probability, which works well for sufficiently large $n$ and $p$ not too far from 1/2.

The idea is that you will reject the null hypothesis that $p = p_0$ at the 0.3% level if $|Z| > 3.$

For example, if $n = 36$ and $p_0 = 1/2,$ then you would
reject $H_0: p = 1/2,$ if you got only $X=1$ Success.

The probability that an observation from a standard normal distribution is in the interval $(-3,3)$ is about 0.997. In R:

diff(pnorm(c(-3,3)))
[1] 0.9973002

Alternatively, perhaps you can find a printed table of the standard normal PDF that goes that far out from $0.$

Also, perhaps google the Empirical Rule.


问:
Compute the limit $\lim\limits_{t \to + \infty} \int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin tx} $
I have been working on this limit for days, but I am not getting it. The question is

Compute the limit $$\lim_{t \to + \infty} \int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin (tx)}$$

Note that the integral is well defined and convergent for every $t >0$. Indeed the integrand function is a positive function for every $t >0$ since
$$e^x + \sin tx > e^x-1 > x>0$$
And as $x \to + \infty$ the integrand function behaves like $e^{-x}$.

WHAT I TRIED:

I consider $t=2n \pi$ a multiple of $2 \pi$, and see what happens:
$$\int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin (2n \pi x)} = \sum_{k=0}^\infty \int_{k /n}^{(k+1) /n} \frac{ \mathrm d x}{e^x+ \sin (2n \pi x)}$$
Making the change of variables $u = 2n \pi x$ I get
\begin{align}\sum_{k=0}^\infty \frac{1}{2n \pi} \int_{2k \pi}^{(2k+2) \pi} \frac{ \mathrm d u}{e^{u/2n \pi}+ \sin (u)} &\ge
\sum_{k=0}^\infty \frac{1}{2n \pi} \int_{2k \pi}^{(2k+2) \pi} \frac{ \mathrm d u}{e^{(2k+2) \pi/2n \pi}+ \sin (u)} \\&=
\sum_{k=0}^\infty \frac{1}{2n \pi} \int_{2k \pi}^{(2k+2) \pi} \frac{ \mathrm d u}{e^{(k+1)/n}+ \sin (u)}\end{align} where I write the lower bound with the minimum of the function at $u=(2k+2) \pi$. Now I use the fact that the integrand function does is integrated over a period of $2 \pi$, and using the result for $C>1$
$$\int_0^{2 \pi} \frac{ \mathrm d u}{C+ \sin (u)} = \frac{2 \pi}{\sqrt{C^2-1}}$$
I get the estimate
\begin{align}\sum_{k=0}^\infty \frac{1}{2n \pi} \int_{2k \pi}^{(2k+2) \pi} \frac{ \mathrm d u}{e^{(k+1)/n}+ \sin (u)} &= \sum_{k=0}^\infty \frac{1}{2n \pi} \frac{2 \pi}{\sqrt{e^{2(k+1)/n} -1 }} \\&= \frac{1}{n} \sum_{k=0}^\infty \frac{1}{\sqrt{e^{2(k+1)/n} -1 }}\end{align}
Summing all up, I got that
$$\int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin (2n \pi x)} \ge \frac{1}{n} \sum_{k=0}^\infty \frac{1}{\sqrt{e^{2(k+1)/n} -1 }}$$
As $n \to \infty$ the series converges to the Riemann integral
$$\int_0^{+ \infty} \frac{\mathrm d y}{\sqrt{e^{2y}-1}} = \frac{\pi}{2}$$

Hence the limit should be a number larger than $\pi/2$, or $+ \infty$.

Using WA I got for large values of $t$ that the integral is between $1$ and $2$, thus $\pi/2$ could be the actual limit.

答:
A slight modification of OP's attempt will lead to a solution. Indeed, write $I(t)$ for the integral and note that

$$ I(t) = \int_{0}^{\infty} \frac{\mathrm{d}x}{e^x + \sin(tx)}
\stackrel{(y=tx)}= \frac{1}{t} \int_{0}^{\infty} \frac{\mathrm{d}y}{e^{y/t} + \sin y}. $$

Also, define $J(t)$ by

$$ J(t)
= \frac{1}{t} \sum_{k=1}^{\infty} \int_{0}^{2\pi} \frac{\mathrm{d}y}{e^{2\pi k/t} + \sin y}
= \sum_{k=1}^{\infty} \frac{2\pi/t}{\sqrt{e^{4\pi k/t} - 1}}, $$

where the second step follows from the integration formula

$$ \int_{0}^{2\pi} \frac{\mathrm{d}y}{c + \sin y} = \frac{2\pi}{\sqrt{c^2 - 1}}, \qquad c > 1. \tag{1} $$

Then similarly as in OP's attempt, we obtain

$$ J(t) \leq I(t) \leq J(t) + \frac{1}{t} \int_{0}^{2\pi} \frac{\mathrm{d}y}{e^{y/t} + \sin y}. \tag{2} $$

Now we observe:

Since the map $ u \mapsto \frac{1}{\sqrt{e^{2u} - 1}} $ is monotone decreasing, we have
$$ \int_{2\pi/t}^{\infty} \frac{\mathrm{d}u}{\sqrt{e^{2u} - 1}} \leq J(t) \leq \int_{0}^{\infty} \frac{\mathrm{d}u}{\sqrt{e^{2u} - 1}}. $$
So by the squeezing theorem, we get
$$ \lim_{t \to \infty} J(t) = \int_{0}^{\infty} \frac{\mathrm{d}u}{\sqrt{e^{2u} - 1}} = \frac{\pi}{2}. $$

Observe that
$$ \int_{\pi}^{2\pi} \frac{\mathrm{d}y}{c + \sin y}
\stackrel{\text{(1)}}\leq \frac{2\pi}{\sqrt{c^2 - 1}}. $$
From this, we have
\begin{align*}
\frac{1}{t} \int_{0}^{2\pi} \frac{\mathrm{d}y}{e^{y/t} + \sin y}
&\leq \frac{1}{t} \left( \int_{0}^{\pi} \mathrm{d}y + \int_{\pi}^{2\pi} \frac{\mathrm{d}y}{e^{\pi/t} + \sin y} \right) \\
&\leq \frac{1}{t} \left( \pi + \frac{2\pi}{\sqrt{e^{2\pi/t} - 1}} \right).
\end{align*}
It is not hard to check that this bound converges to $0$ as $t \to \infty$.

Combining altogether and applying the squeezing theorem to $\text{(2)}$, we get

$$ \lim_{t\to\infty} I(t) = \frac{\pi}{2}. $$

Further Discussion:

This proof can actually show that $I(t) = \frac{\pi}{2} + \mathcal{O}(t^{-1/2})$ as $t \to \infty$. Can we do better? It is reasonable to suspect that the asymptotic formula takes the form
$$I(t) = \frac{\pi}{2} + \frac{c}{\sqrt{t}} + \cdots \tag{3} $$
Is this true? If so, then what will be the value of $c$? I do not have enough time and energy to pursue in this direction right now (I need to crawl into the bed right now), but it seems an interesting question.

Addendum. Regarding the extra question, the following heuristic approach gives a guess on the value of the constant $c$ in the asymptotic expansion $\text{(3)}$:

Note that, for $x > 0$ and $\theta \in \mathbb{R}$,

\begin{align*}
\frac{1}{e^x + \sin\theta}
&= \frac{1}{\sqrt{e^{2x}-1}} \biggl( 1 + 2 \sum_{k=1}^{\infty} \frac{(-1)^k}{(e^x + \sqrt{e^{2x}-1})^{2k-1}} \sin((2k-1)\theta) \\
&\hspace{7em}  + 2 \sum_{k=1}^{\infty} \frac{(-1)^k}{(e^x + \sqrt{e^{2x}-1})^{2k}} \cos (2k\theta) \biggr).
\end{align*}

Using this and substituting $\epsilon = 1/t$, we have

\begin{align*}
I(t)
&= \int_{0}^{\infty} \frac{\mathrm{d}x}{\sqrt{e^{2x}-1}} \\
&\quad + 2 \sum_{k=1}^{\infty} (-1)^k \int_{0}^{\infty} \frac{\sin((2k-1)x/\epsilon)}{\sqrt{e^{2x}-1}(e^x + \sqrt{e^{2x}-1})^{2k-1}} \, \mathrm{d}x \\
&\quad +2 \sum_{k=1}^{\infty} (-1)^k \int_{0}^{\infty} \frac{\cos(2kx/\epsilon)}{\sqrt{e^{2x}-1}(e^x + \sqrt{e^{2x}-1})^{2k}} \, \mathrm{d}x \\
&= \frac{\pi}{2} + 2 \sqrt{\epsilon} \sum_{k=1}^{\infty} (-1)^k \int_{0}^{\infty} \sqrt{\frac{\epsilon}{e^{2\epsilon u} - 1}} \frac{\sin((2k-1)u)}{(e^{\epsilon u} + \sqrt{e^{2\epsilon u}-1})^{2k-1}} \, \mathrm{d}u \\
&\hspace{3em} + 2 \sqrt{\epsilon} \sum_{k=1}^{\infty} (-1)^k \int_{0}^{\infty} \sqrt{\frac{\epsilon}{e^{2\epsilon u} - 1}} \frac{\cos(2ku)}{(e^{\epsilon u} + \sqrt{e^{2\epsilon u}-1})^{2k-1}} \, \mathrm{d}u,
\end{align*}

where we utilized the substitution $x = \epsilon u$ in the last step. So it is reasonable to expect that $c$ in $\text{(3)}$ is given by:

\begin{align*}
c
&= 2 \sum_{k=1}^{\infty} (-1)^k \int_{0}^{\infty} \lim_{\epsilon \to 0^+} \sqrt{\frac{\epsilon}{e^{2\epsilon u} - 1}} \frac{\sin((2k-1)u)}{(e^{\epsilon u} + \sqrt{e^{2\epsilon u}-1})^{2k-1}} \, \mathrm{d}u \\
&\quad + 2 \sum_{k=1}^{\infty} (-1)^k \int_{0}^{\infty} \lim_{\epsilon \to 0^+}  \sqrt{\frac{\epsilon}{e^{2\epsilon u} - 1}} \frac{\cos(2ku)}{(e^{\epsilon u} + \sqrt{e^{2\epsilon u}-1})^{2k-1}} \, \mathrm{d}u \\
&= 2 \sum_{k=1}^{\infty} (-1)^k \biggl( \int_{0}^{\infty} \frac{\sin((2k-1)u)}{\sqrt{2u}} \, \mathrm{d}u + \int_{0}^{\infty} \frac{\cos(2ku)}{\sqrt{2u}} \, \mathrm{d}u \biggr) \\
&= \sum_{k=1}^{\infty} (-1)^k \biggl( \sqrt{\frac{\pi}{2k-1}} + \sqrt{\frac{\pi}{2k}} \biggr),
\end{align*}

where we utilized the identity

$$ \int_{0}^{\infty} \frac{\sin(a u)}{\sqrt{u}} \, \mathrm{d}u = \int_{0}^{\infty} \frac{\cos(a u)}{\sqrt{u}} \, \mathrm{d}u = \sqrt{\frac{\pi}{2a}}, \qquad a > 0. $$

Of course, interchanging the order of limit operators requires a great deal of care, especially in the situation like this where the absolute convergence fails. So this is not a proof yet, but rather a hand-waving heuristics. (Even the possibility that this guess is not true at all is still open to question!)


问:
4 quarters, 7 nickels, and 2 pennies in a wallet. One by one a coin is removed.
This question came up in my textbook, and I simply couldn't get my head around it. The question is as follows:
What is the probability that the fourth coin that is taken out is a nickel and the seventh is a quarter?

I tried listing cases, but I soon found out that the process was far too tedious. Does anyone have any ideas as to how to solve this problem?

Another question that came from this is the following:
What is the probability that the last coin taken out is a penny?

I knew that there was some form of symmetry going on, and I could see why the solution would be $\frac{2}{13}$. However, I couldn't fully find a proof or some suitable reasoning behind this solution. Any help would be appreciated.

答:
the probability that a coin is drawn at the first, second, third....last draw is always the same, it cannot change.

Same reasoning for the rest of the questions.

You can verify my statement in an easy way:

Penny drawn at first draw:

$$\frac{2}{13}$$

Penny drawn at the second draw:

$$\frac{11}{13}\times \frac{2}{12}+\frac{2}{13}\times\frac{1}{12}=\frac{24}{13\times 12}=\frac{2}{13}$$

and so on...


问:
The set of differences for a set of positive Lebesgue measure
Quite a while ago, I heard about a statement in measure theory, that goes as follows:

Let $A \subset \mathbb R^n$ be a Lebesgue-measurable set of positive measure. Then we follow that $A-A = \{ x-y \mid x,y\in A\}$ is a neighborhood of zero, i.e. contains an open ball around zero.

I now got reminded of that statement as I have the homework problem (Kolmogorov, Introductory Real Analysis, p. 268, Problem 5):

Prove that every set of positive measure in the interval $[0,1]$ contains a pair of points whose distance apart is a rational number.

The above statement would obviously prove the homework problem and I would like to prove the more general statement. I think that assuming the opposite and taking a sequence $\{x_n\}$ converging to zero such that none of the elements are contained in $A$, we might be able to define an ascending/descending chain $A_n$ such that the union/intersection is $A$ but the limit of its measures zero. I am in lack of ideas for the definition on those $A_n$.

I am asking specifically not for an answer but a hint on the problem. Especially if my idea turns out to be fruitful for somebody, a notice would be great. Or if another well-known theorem is needed, I surely would want to know. Thank you for your help.

答:
Here's an attempt at a hint for the first result you ask about: Assume without loss of generality that $A$ has finite measure. Let $f$ be the characteristic function of $A$ and let $\tilde{f}$ be the one of $-A$. The convolution $g = f \ast \tilde{f}$ is continuous and $0$ is in the support of $g$.

Added later: One nice standard application is that every measurable homomorphism $\phi: \mathbb{R} \to \mathbb{R}$ is continuous. For more on that and related matters have a look at these two MO-threads:

On measurable homomorphisms $\mathbb{C} \to \mathbb{C}$.

On measurable automorphisms of locally compact groups

They might elucidate what is mentioned in another answer.

Update:

What I wrote above is the way I prefer to prove this.

Another approach is to appeal to regularity of Lebesgue measure $\lambda$ (used in $1$ and $2$ below).

Since $A$ contains a compact set of positive measure, we can assume $A$ to be compact right away (as $B-B \subset A - A$ if $B \subset A$).

There is an open set $U \supset A$ such that $\lambda(U) \lt 2 \lambda(A)$.

Since $A$ is compact there is $I = (-\varepsilon, \varepsilon)^{n}$ such that $A + x \subset U$ for all $x \in I$.

Since $\lambda (U) \lt 2\lambda(A)$ we must have $\lambda((A + x) \cap A) \gt 0$.

This is of course very closely related to the argument given by Chandru1 below.


问:
Prove that eigenvalue of matrix A is the same as the eigenvalue of $BAB^{-1}$
So let $\lambda$ be an eigenvalue of $A$ and let $B$  be invertible. How do I prove that $\lambda$ is also an eigenvalue for $BAB^{-1}$

答:
Hint:

$$ \det( \lambda I -BAB^{-1})= \det(B( \lambda I -A)B^{-1}).$$


问:
Solution theory for parabolic problems with irregular boundary data
Let $\Omega \subset \mathbb R^n$, $n \in \mathbb N$, be a smooth bounded domain, $\mathcal A$ an elliptic operator (for instance $\mathcal A = - \Delta$), $u_0 \in C^0(\overline \Omega)$ and $g \in C^\alpha(\partial \Omega)$ for some $\alpha \in (0, 1)$. What is known about the equation

$$\begin{cases}
 u_t + \mathcal A u = 0 & \text{in $(0, \infty) \times \Omega$}, \\
 u = g & \text{in $(0, \infty) \times \partial \Omega$}, \\
 u(\cdot, t) = u_0 & \text{in $\Omega$}
\end{cases}$$

(or also about the corresponding elliptic equation)?

If $g \in C^{2+\beta}(\partial \Omega)$ for some $\beta \in (0, 1)$, then one can extend $g$ to a function $\tilde g \in C^{2+\beta}(\overline \Omega)$ and consider $u-\tilde g$ instead of $u$. Of course, this is no longer possible for less regular $g$.

However, I would still expect that the problem above (or at least the elliptic version) has a (unique?) classical solution – which may even be $C^\alpha$ up to the boundary (but of course not more).

I guess this has been treated somewhere and hence I am thankful for any references.

答:
I think Theorem 9, Chapter 3.4, in Avner Friedman's book, Partial Differential Equations of Parabolic Type, might help you. Basically it says that with suitable assumptions on $\mathcal{A}$ and the boundary of $\Omega$, you get a unique solution that is in $C^{2+\alpha}$ for a $0<\alpha<1$, but only in the interior of $\Omega\times(0,T)$.

For the proof you don't even need that $g\in C^\alpha$. It is enough that $u=\psi$ on the parabolic boundary $\Omega\times\{t=0\} \cup \partial\Omega\times(0,T)$, where $\psi$ is a continuous function on that boundary. For a bounded domanin $\Omega$ we can for example apply Tietze's extension theorem to get a $\psi\in C(\bar{\Omega}\times[0,T]$). Let $N$ be the rectangle in $\mathbb{R}^{d+1}$ that contains $\bar{\Omega}\times[0,T]$. By the Weierstrass approximation theorem there exists a sequence of polynomilas $\psi_k$ that approximate $\psi$ uniformly in $N$. As you already suggested, you can now apply interior parabolic schauder estimates to $u-\psi_k$. You get a sequence of solutions $u_k$, by interior parabolic Schauder estimates you can show that $u_k$ converges uniformly to a $u$. $u$ is continuous in $\bar{\Omega}\times[0,T]$ and $u\in C^{2+\alpha}$ locally in $\Omega\times(0,T)$.

Tl;dr, you got the right idea. You consider $u-g$, but since $g$ is not differentiable, you approximate $g$ by a sequence of polynomials $g_k$. Interior Schauder estimates show that the corresponding sequence $u_k$ converges uniformly to your solution $u$.

If you want $u\in C^{2+\alpha}(\bar{\Omega}\times[0,T])$, i.e. differentiability up to the boundary, then you need that the boundary condition $\psi$ is in $C^{2+\alpha}$, see for instance Ladyzenskaja-Solonnikov-Ural'ceva's book Linear and Quasi-linear Equations of Parabolic Type, Theorem 5.2,Chapter IV Section 5.


问:
How to integrate a function with an unknown constant
How do I integrate $(e^x + e^{-x} -5)^p$ with respect to $x$ where $p>0$?

I have tried integrating by substitution with $u=e^x$ and $u= e^x + e^{-x} -5$ but I can’t make progress because of the unknown $p$

Is this the right approach?

答:
Trying to make the problem more general, consider
$$I_p=\int(e^x + e^{-x} +2k)^p\,dx=2^p\int(\cosh(x)+k)^p\,dx$$ So, using the binomial expansion, the integrand is a polynomial in $\cosh(x)$
$$(\cosh(x)+k)^p=\sum_{m=0}^p \binom{m}{p}\cosh^m(x)\, k^{p-m}$$

and the integrals
$$J_{2m}=\int \cosh^{2m}(x)\,dx \qquad \text{and} \qquad J_{2m+1}=\int \cosh^{2m+1}(x)\,dx$$ are known (have a look at equations $2.413.3$ and $2.413.4$ in the  "Table of Integrals, Series, and Products"
(seventh edition) by I.S. Gradshteyn and I.M. Ryzhik.

As a general result, you should have something like
$$\frac{I_p}{2^p}=a_p x+ \sum_{n=1}^p b_{p,n} \sinh(n x)$$

I do not see any pattern in the results but they could be expressed in terms of the gaussian hypergeometric function. Using a CAS
$$J_{p}=\int \cosh^{p}(x)\,dx=$$ $$\frac{\sqrt{-\sinh ^2(x)} \,\text{csch}(x)\, \cosh ^{p+1}(x)}{p+1}\,\, _2F_1\left(\frac{1}{2},\frac{p+1}{2};\frac{p+3}{2};\cosh ^2(x)\right)$$


问:
A problem about independent, i.i.d, random variables and exponential distribution.

Let $X_{1}, \ldots, X_{n}$ be independent and identically distributed random variables, with exponential distribution of the parameter $\lambda$. Consider $Y_{(1)}, \ldots, Y_{(n)}$ the associated order statistics and define $$Z_{1} = Y_{(1)}  \quad \text{and} \quad Z_{j} = Y_{(j)} - Y_{(j - 1)}, j = 2, \ldots , n$$ Show that $Z_{1},\ldots, Z_{n}$ are independent and $Z_{j} \sim \textbf{Exp} (\lambda (n -j +1)), j= 1,\ldots, n$.

My approach:  Let, $X_{i}\overset{i.i.d}{\sim}$ with $i=1,2,\ldots,n$ and $X_{i}\sim \textbf{Exp}(\lambda)$. So, since that $X_{i}'s$ are independent and indentically distributed, then $f$ is the common density function (this mean that $X_{i}'s$ has the same density).

Since $Y_{(1)},\ldots, Y_{(n)}$ are the associated order statistic to $X_{1},\ldots,X_{n}$, so the joint density is $$f_{Y_{(1)},\ldots,Y_{(n)}}(y_{1},\ldots,y_{n})=n!f(y_{1})\cdots f(y_{n}), \quad y_{1}<y_{2}<\cdots<y_{n}$$

How can I continue or solve this problem?

Really, I don't know how to continue.

I think I can use some transformation theorem, but I don't sure how can use that.

I was studying a book and I found similar problem where using that ${n!}$ =$\prod_{1}^{n}(n-i+1)$ and $\sum_{1}^{n}Y_{i}$ = $\sum_{1}^{n}(n-i+1)(Y_{i}-Y_{i-1})$ ,so the joint p.d.f of the $Y_{i}$

${n!}\lambda^nexp(-{\lambda}\sum_{1}^{n}y_{i})$  may be expressed as $\prod_{1}^{n}{\lambda}(n-i+1)exp(-{\lambda}(n-i+1)z_{i})$ ($z_{0}=0$ )

But I don't know how to adapt that to the solution of my problem.

答:
I answered quite the same question here. The only differences are that in the original post the distribution was $Exp(1)$ and the requeste distribution was the law of $Z_j=Y_{(j+1)}-Y_{(j)}$

Unfortunately this forum is an Italian forum...but I think you can understand the solution because formulas are formulas....in any language.

On the contrary please advice and I will repost the solution in English

Solution details

Just to simplify the notation, let's set $Y$ and $X$ the two order statistics and $Z=Y-X$ the rv we are looking for distribution ($Y>X$).

Using the known result on the joint density of two Order Statistics we get

$$f_{XY}(x,y)=\frac{n!}{(j-2)!(n-j)!}(1-e^{-\lambda x})^{j-2}e^{-\lambda(n-j)y}\lambda e^{-\lambda x}\lambda e^{-\lambda y}\cdot\mathbb{1}_{(x;\infty)}(y)$$

In order to get the CDF of the rv $Z=Y-X$ we have to integrate the joint density over the desired support, say:

$$F_Z(z)=\frac{n!}{(j-2)!(n-j)!}\int_0^{\infty}(1-e^{-\lambda x})^{j-2}\lambda e^{-\lambda x} \Bigg[ \int_x^{x+z}\lambda e^{-\lambda(n-j+1)y} dy\Bigg] dx$$

The integral is not difficult...it is enough to sove the inner integral in $dy$ (very easy) and get out of the sign of integral all what does not depend on x.

You get

$$\frac{n!}{(j-2)!(n-j+1)!}[1-e^{-\lambda(n-j+1)z}]\int_0^{\infty}(1-e^{-\lambda x})^{j-2}\lambda (e^{-\lambda x})^{n-j+2}dx$$

Now substitute

$1-e^{-\lambda x}=u$

and you get

$$F_Z(z)=\frac{n!}{(j-2)!(n-j+1)!}[1-e^{-\lambda(n-j+1)z}]\underbrace{\int_0^1 u^{j-2}(1-u)^{n-j+1}du}_{=Beta(j-1;n-j+2)}=1-e^{-\lambda(n-j+1)z}$$

As

$$Beta(j-1;n-j+2)=\frac{\Gamma(j-1)\Gamma(n-j+2)}{\Gamma(n+1)}=\frac{(j-2)!(n-j+1)!}{n!}$$

That is exactly the CDF they ask you to show.


问:
Proof of the identity $2^n = \sum\limits_{k=0}^n 2^{-k} \binom{n+k}{k}$
I just found this identity but without any proof, could you just give me an hint how I could prove it?
$$2^n = \sum\limits_{k=0}^n 2^{-k} \cdot \binom{n+k}{k}$$

I know that $$2^n = \sum\limits_{k=0}^n \binom{n}{k}$$ but that didn't help me

答:
Let's do this in lazy, stream of consciousness fashion, adjusting little details one by one until it looks like a recognizable enumeration.

First combinatorialize it by multiplying the whole thing by $2^n$, to

$$2^{2n} = \sum\limits_{k=0}^n 2^{n-k} \cdot \binom{n+k}{k}$$

This looks better, because $(n-k)$ and $(n+k)$ look like a decomposition of $2n$.  The left side counts all subsets of a $2n$ element set, so maybe something about the larger and the smaller subset in the partition of $2n$ elements into a subset and its complement.  Some fuzzy feeling about needing to break the symmetry when the sets are of equal size tells to multiply by $2$ and consider subsets of a $2n+1$ element set, which as a partition of the set has a larger and a smaller part.  So,

$$2^{2n+1} = \sum\limits_{k=0}^n 2 \times 2^{n-k} \cdot \binom{n+k}{k}$$  or

$$2^{2n+1} = \sum\limits_{k=0}^n 2 \times 2^{n-k} \cdot \binom{n+k}{n}$$

Better!  A subset of a size $2n+1$ set is equivalent to a partition into (larger set with more than $n$ elements) $\cup$ (smaller set with less than $n$ elements), and a choice (the $\times 2$) of which one is the subset.  And we can pick the larger subset by first selecting its largest $n+1$ elements, then adding some random subset from the lower indexed elements (this is the $2^{n-k}$), and... ok, it is all clear now:

From the first $2n+1$ integers, select a subset by partitioning into a large and a small subset and picking ($\times 2$) one of them as the set.  Determine the large subset, which has at least $n+1$ elements, by selecting an index $j$ from $1$ to $n+1$ as the location of the $(n+1)$th-largest element of the set, selecting the $n$ elements above $j$ (there are ${2n+1 - j} \choose {n}$ options) and adding any subset of the elements below $j$ (there are $2^{j-1}$ of those).  This is the last sum with $k = n+1 - j$.

In the opposite direction, divide by $2^n$ to get

$$1 = \sum\limits_{k=0}^n 2^{-(n+k)} \cdot \binom{n+k}{n}$$

and look for a natural probability process that delivers this distribution. One answer:

Throw a coin to determine whether voters $1, 2, \ldots$ are supporting candidate $A$ or $B$ in an election, and stop the process when one candidate reaches $n+1$ votes.  Stopping happens at a time $t$ between $n+1$ and $2n+1$ steps, and if $k=t-1-n$, the set of times at which the victorious candidate received votes before time $t$ is one of ${{t-1} \choose n} = {{n+k} \choose n}$ possibilities.

I think this process is essentially the same as what is sometimes called Banach's Matchbox Problem.


答:
In absence of any new information, the probability would remain the same as it was at the beginning.

One way to look at it is as below.

To start with, you have $4$ quarters, $7$ nickels, $2$ pennies (ratio of $4:7:2$).

How many quarters, nickels and pennies after first coin is taken out? You do not know but here are the expected number of coins based on their probability of being taken out.

Expected Number of Quarters $= 4 - \frac{4}{13} = \frac{48}{13}$

Expected Number of Nickels $= 7 - \frac{7}{13} = \frac{84}{13}$

Expected Number of Pennies $= 2 - \frac{2}{13} = \frac{24}{13}$

Their sum $= 12$ (coins left)

Their ratio based on expected numbers after one coin is taken out $= 4:7:2$ which is same as how you started. You can take these expected numbers and now try the second coin out but this ratio would not change.

Title: 4 quarters, 7 nickels, and 2 pennies in a wallet. One by one a coin is removed.

Body:
This question came up in my textbook, and I simply couldn't get my head around it. The question is as follows:
What is the probability that the fourth coin that is taken out is a nickel and the seventh is a quarter?

I tried listing cases, but I soon found out that the process was far too tedious. Does anyone have any ideas as to how to solve this problem?

Another question that came from this is the following:
What is the probability that the last coin taken out is a penny?

I knew that there was some form of symmetry going on, and I could see why the solution would be $\frac{2}{13}$. However, I couldn't fully find a proof or some suitable reasoning behind this solution. Any help would be appreciated.


问:
show that unit $n$-sphere is closed in $\mathbb{R}^{n+1}$
Let $S^n=\{(x_1,\dots,x_n,x_{n+1})\in \mathbb{R}^{n+1}: x_1^2+\dots+x_{n+1}^2=1\}$.

I want to show that unit $n$-sphere is closed in $\mathbb{R}^{n+1}$ this way:

Let
$A=\{(x_1,\dots,x_n,x_{n+1})\in \mathbb{R}^{n+1}: x_1^2+\dots+x_{n+1}^2<1\}$ and
$B=\{(x_1,\dots,x_n,x_{n+1})\in \mathbb{R}^{n+1}: x_1^2+\dots+x_{n+1}^2>1\}$.

Then I want use that the union of two open sets is open and the complementary of the open set is closed, hence $S^n$ is closed, but I cant understand how rigorously show that $A$ and $B$ are open.

Is it a bad path? What is the easiest way to do it (I mean not the shortness but using just as elementary notions as it possible).
Thank you for your time!

答:
Method 1.

We have $A=\{(x_1,\dots,x_n,x_{n+1})\in \mathbb{R}^{n+1}: x_1^2+\dots+x_{n+1}^2<1\}= f^{-1}(]-\infty,1[)$, with $f(x)= x_1^2+\dots+x_{n+1}^2$ is a continuous function and $]-\infty,1[$ is open, so $A$ is an open set.

By the same way, we show that $B=f^{-1}(]1,+\infty[)$ is open.

So  $S^n=A^c \cap B^c$ is a closed set.

Method 2.

$S^n=f^{-1}(\{1\})$, with $\{1\}$ a closed set and $f$ the same function as in the Method 1 .


问:
Question in Ch-12 Apostol's Number theory (Vol1)
I am trying some exercises from Apostol's Introduction to Analytic number theory and I could not solve this particular problem (number 16) of textbook and need help.

I am sorry, I wouldn't be able to provide anything as attempt as I have no ideas on which result to use.

For background, I am taking number theory course this sem where bernaulli polynomials were taught and also read them from Apostol.

Kindly shed some light on this!!

答:
Provided that Cauchy product is defined as

$$
\sum_{n=0}^\infty a_nx^n\sum_{k=0}^\infty b_kx^k=\sum_{m=0}^\infty x^m\sum_{r=0}^ma_rb_{m-r}
$$

We may apply it to the left handside of what the problem states:

$$
\begin{aligned}
\sum_{r=0}^\infty z^{2r}\sum_{k=0}^r{2^{2k}B_{2k}\over(2k)![2(r-k)+1]!}
&=\sum_{r=0}^\infty{(2z)^{2r}B_{2r}\over(2r)!}\sum_{n=0}^\infty{z^{2n}\over(2n+1)!} \\
&=\left[{2z\over e^{2z}-1}+z\right]\cdot{e^z-e^{-z}\over2z} \\
&={e^z-e^{-z}\over e^z(e^z-e^{-z})}+{e^z-e^{-z}\over2}=\cosh z \\
&=\sum_{k=0}^\infty{z^{2r}\over(2r)!}
\end{aligned}
$$

Comparing the coefficients of the $z^{2r}$ terms yields

$$
\sum_{k=0}^r{2^{2k}B_{2k}\over(2k)!(2r+1-2k)!}={1\over(2r)!}
$$

which is exactly what the problem asks us to prove.


问:
Finding the change in position and angle after moving the rectangle
enter image description here

I am curious about how to approach the problem mathematically, so I write.
There are 4 dots on the square and I know the location. The rectangle moves and the positions and angles of the four points change.
I also know the location of the four points that have changed. I don't know the changed angle.
How can I find the angle and value to move to the initial position?
Please advise.

答:
Let us denote $$u_{12}=\binom{x_2-x_1}{y_2-y_1}, \ u'_{12}=\binom{x'_2-x'_1}{y'_2-y'_1},$$

Dot product $u_{12}.u'_{12}$ can be expressed in two ways:

$$u_{12}.u'_{12}=\|u_{12}\| \|u'_{12}\| \cos \theta=(x_2-x_1)(x'_2-x'_1)+(y_2-y_1)(y'_2-y'_1)$$

Therefore:

$$\cos \theta= \dfrac{(x_2-x_1)(x'_2-x'_1)+(y_2-y_1)(y'_2-y'_1)}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\sqrt{(x'_2-x'_1)^2+(y'_2-y'_1)^2}}\tag{1}$$

from which you can extract $\theta$.

This works well for $0 \le \theta \le \pi$.

But the main problem when you have only the cosine is that you cannot distiguish for exemple a $+30°$ move from a $-30°$ move !

If you want to eliminate this ambiguity, i.e., work, say, in interval $[-180°;+180°]$, you have to determine $\sin \theta$ as well, by using this formula:

$$\sin \theta= \dfrac{(x_2-x_1)(y_2-y_1)-(x'_2-x'_1)(y'_2-y'_1)}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\sqrt{(x'_2-x'_1)^2+(y'_2-y'_1)^2}}\tag{2}$$

[(2), with striking similarity with (1) can be justified in a similar manner by expressing the cross product $u \times u'$ in two ways.]

Once you have the value of $\theta$, the formula sending any point $\binom{x}{y}$ connected to the first rectangle onto its homologous points $\binom{x'}{y'}$ connected to the second rectangle.

$$\binom{x'}{y'}=\begin{pmatrix}\cos \theta & -\sin \theta\\\sin \theta & \ \ \ \cos \theta\\\end{pmatrix}\binom{x-x_1}{y-y_1}+\binom{x'_1}{y'_1}$$

Remark: The work done for indices $1$ and $2$ could have been done with any index "twinning", for example $1$ and $4$.

It is even a good idea to do it for all "twinings" and average the $\theta_{ij}$ values thus obtained ; the error will be less...


问:
Rotate a plane along a line in 3d space
I have a plane A in 3D which can be defined either by the scalar plane equation or by its normal n. I want to find a new plane B which is orthogonal to plane A but shares an edge with plane A. I.e. I want to rotate plane A 90 degrees along a line defined by one of its edges. </p>

<p>I may be asking how do I use Rodrigues formula to do this but at this point I need some clarification of the process of figuring out what the Rotation matrix is going to be.
Any initial suggestions on how to go about this? </p>

答:
TL;DR The easiest way to go is to ask yourself, what points stay fixed under rotation. The answer is that these are exactly the points of the axis of rotation. If the axis of rotation lies in your plane, then surely it's going to be the intersection of this plane and it's rotated copy, provided that the angle of rotation is not a multiple of $\pi$.

Now, I want to fill in the blanks. Let $A$ be a 3-by-3 rotation matrix, that is a matrix satisfying the equation $A^T A = 1$ (where by $1$ I denote the unit matrix) and $\det A = 1$. This matrix has a real eigenvalue $\lambda$: indeed, its characteristic polynomial has degree $3$, and such polynomials always have a real root.

$A^{-1} = A^T$, therefore $A^T$ has an eigenvalue $1/\lambda$. But $A^T$ has the same characteristic polynomial as $A$. If the other two eigenvalues of $A$ are complex, we can immediately conclude that $\lambda = 1$ (and the complex eigenvalues are complex conjugate of length $1$). If they are real, in general we can conclude that $A$ has eigenvalues $\lambda$, $1/\lambda$ and $\mu = \det A = 1$. Take an eigenvector $x$ corresponding to $\lambda$. $x \cdot x = Ax \cdot Ax = \lambda^2 x \cdot x$, therefore $\lambda = \pm 1$.

The case when eigenvalues are all $1$ corresponds to trivial rotation. The case when eigenvalues are $1$, $-1$, $-1$ corresponds to rotation by $\pi$ (because eigenvectors of a rotation matrix corresponding to different real eigenvalues are orthogonal: $x \cdot y = Ax \cdot Ay = - x \cdot y$ can only hold if $x \cdot y = 0$).

The eigenspace of eigenvectors $\mu$ and $1/\mu$ is orthogonal (proven above), and it is easy to see that birestriction of $A$ on it is still a rotation. It is now easy to see that any plane orthogonal to the axis of rotation is preserved by $A$, so a rotation in 3D is indeed rotation around the axis of rotation.

Title: Rotate a plane along a line in 3d space


问:
Computing integral of $f(z)$
Suppose $f(z)=1+z+z^2+z^3.$ My question is

(1) How can we calculate $\int\limits_{[-1, i]}f(z)dz$

(2) How do we see $|\int\limits_{[-1, i]}(f(z))^ndz|\leq k(\frac{4}{3\sqrt{3}})^n$ where $k$ is a positive constant.

My idea:

(1) since $z$ is a complex number we can't just integrate it as usual.

(2) For this problem maybe we can use part (1) to prove it?

I'm not sure how to get a grip on this problem.

答:
(1) In this case, you can integrate in the exact same way...

$$
\int_{[-1,i]} (1+z+z^2+z^3) dz = \left[z+\frac 12 z^2 + \frac 13 z^3 + \frac 14 z^4\right]_{-1}^i= \frac 13 + \frac{2i}{3}
$$

you get the same answer as going through the computation of the line integral over the path $\gamma(t) = -1+t + ti, \quad t \in [0,1]$

$$
\int_{\Gamma} f(z) dz = \int_0^1 \gamma'(t) f(\gamma(t)) dt = \int_0^1 (1+i) f(-1+t+ti)dt = \frac 13 +\frac{2i}{3}.
$$

[edit: comments on (2)]

The second part of the question seems trickier. What you can easily establish is that
\begin{align*}
\int_{[-1,i]} (1+z+z^2+z^3)^n dz= &\int_{[-1,i]}(1+z)^n(1+z^2)^n \\
= & \int_0^1 (1+i) (t(1+i)^n)(2(1-t)-2t(1-t)i)^ndt\\
=& (1+i)^{n+1} 2^n \int_0^1 t^n(1-t)^n(1-ti)^n dt
\end{align*}

and so,
$$
\left| \int_{[-1,i]} f(z)^n dz \right|\leq \sqrt{2} (2\sqrt{2})^n \int_0^1 t^n (1-t)^n \left(\sqrt{1+t^2}\right)^n dt
$$

Although it is true that this bound is smaller than $\left(\frac{4}{3 \sqrt{3}}\right)^n$, and the result follows, I feel that there must be some simple trick to get the suggested bound.

Title: Computing integral of $f(z)$


问:
How to solve this non linear and second order ODE
$$y'' + ay(y')² + by = 0$$

Here, $a$ and $b$ are constants. I can't solve this!

答:
$$\frac{d^2y}{dx^2} + ay\left(\frac{dy}{dx}\right)^2 + by = 0$$
This is an ODE of autonomous kind. The usual method is :

$$\frac{dy}{dx}=u(y)\quad;\quad \frac{d^2y}{dx^2}=\frac{du}{dy}\frac{dy}{dx}=u\frac{du}{dy}$$
$$u\frac{du}{dy} + ayu^2 + by = 0$$
Let $v(y)=(u(y))^2$
$$\frac12\frac{dv}{dy} + ayv + by = 0$$
This is a first order linear ODE. Solving it leads to :
$$v(y)=-\frac{b}{a}+c_1e^{ay^2}$$
$$u(y)=\pm\sqrt{-\frac{b}{a}+c_1e^{ay^2}}$$
$$\frac{dy}{dx}=\pm\sqrt{-\frac{b}{a}+c_1e^{ay^2}}$$
$$x(y)=\pm\int \frac{dy}{\sqrt{-\frac{b}{a}+c_1e^{ay^2}}}+c_2$$
There is no closed form for this integral with standard functions. Numerical calculus is requiered as well as for the inverse function $y(x)$ .


问:
Reccursive formula as a non-recursive
The recursive formula is give as follows:

$$ s_n = n s_{n-1} + (n-1)!$$ and $s_2 = 1$. How to construct a non-recursive formula?

答:
By working out the first few cases, you observe that the sequence satisfies
$$s_n=n!(s_0+H_n)\,,$$ where $H_n=\sum_{k=1}^n\frac{1}{k}$ is the $n$th harmonic number. One can then confirm that it is indeed the case by observing that $$s_{n+1}=(n+1)s_n+n!=(n+1)!(s_0+H_n)+n!=(n+1)!\left(s_0+H_n+\frac{1}{n+1}\right)=(n+1)!(s_0+H_{n+1})\,.$$
You can obtain your particular sequence by substituting the appropriate value for $s_0$.

Title: Reccursive formula as a non-recursive


问:
Explanation of why $\int \sin^2 x\cos^2 x\; dx = 1/3 \sin^3 x - 2/5 \sin^5 x + 1/7 \sin^7 x +c$

verify the solution
  $$\int \sin^2 x\cos^5 x\; dx = 1/3 \sin^3 x - 2/5 \sin^5 x + 1/7 \sin^7 x +c$$

I have hit this in my book and can't work it out. Does anyone have any ideas or a walk-through that might help?

Body:
Hint:

1) $$2\sin(x)\cos(x)=\sin(2x),$$

2) $$\sin^2(x)=\frac{1-\cos(2x)}{2} \implies \sin^2(2x)=\frac{1-\cos(4x)}{2}. $$


问:
Bijection between two sets of permutations
I'm studying discrete math and in my book the definition of permutation is as follows :

Let $X$ be a non-empty set, it is called permutation of $X$ a bijective function $f: X \to X$.

The set of all permutations is denoted by $S_X$.

which I get , then it is written

To study $S_X$ where $|X| = n$ , it's enough to study $S_n$ (where $X = I_n = \{1,2,\dots,n\}$)

This is good to me because they have the same number of elements so it is intuitively true, but next there is the following proposition which should clarify the relation between $S_X$ and $S_n$ :

Let $X$ be a finite set with $|X| = n$. Then there's a biejection $f : S_X \to S_n$ so that for all $\sigma , \pi \in S_X$ it is true that $f(\sigma \circ \pi ) = f(\sigma) \circ f(\pi) $. In particular  $|S_X |= n!$.

However I don't understand what does the part 'so that for all...$f(\sigma \circ \pi ) = f(\sigma) \circ f(\pi)$'. What's the intuitive and conceptual meaning for this formula?  can you help me to understand this?
Thanks in advance

答: Here, $\circ$ denotes the composition of permutations, i.e. $\sigma \circ \pi$ is the permutation obtained by applying $\pi$, then applying $\sigma$. The statement says that there exists a "natural" bijection between $S_X$ and $S_n$, in the sense that it preserves the law of composition. If you have already heard about group morphisms, this just says that there exists a group isomorphism between $(S_X, \circ)$ and $(S_n, \circ)$.


答:
By definition of $\lambda$, $\exists x\neq 0$ such that $Ax=\lambda x$

$BAB^{-1}(Bx)$

$=BAB^{-1}Bx$

$=BAIx$

$=BAx$

$=B \lambda x$

$=\lambda Bx$

$=\lambda (Bx)$

If $BAB^{-1}(Bx)=\lambda (Bx)$ then clearly $\lambda$ is an eigenvalue of $BAB^{-1}$ (with $Bx$ is an eigenvector; invertibility of $B$ ensures $Bx$ is not zero).

Title: Prove that eigenvalue of matrix A is the same as the eigenvalue of $BAB^{-1}$

Body: So let $\lambda$ be an eigenvalue of $A$ and let $B$  be invertible. How do I prove that $\lambda$ is also an eigenvalue for $BAB^{-1}$


答:
Suppose we have
$$ AX = \lambda X.$$
We show that  $Y= BX$  is an eigenvector of $ BAB^{-1}$ associated to the eigenvalue $\lambda.$
$$ BAB^{-1}(BX) = BAX = B(\lambda. X)= \lambda BX.$$


问:
How prove this matrix $rank (A_{11}A_{12})=rank(A_{12})$
let $A_{n\times n}$ is a positive semidefinite matrix,
and
$$A=\begin{bmatrix}
A_{11}&A_{12}\\
A_{21}&A_{22}
\end{bmatrix}$$
where $A_{11}$ is $m\times m,m<n $ matrix
prove or disprove
$$rank(A_{11}A_{12})=rank(A_{12})$$</p>

答:
Hint: the proof is comprised of two steps.

Show that $\operatorname{Im}(A_{12})\subseteq \operatorname{Im}(A_{11})$. Suppose the contrary. Then there is a nonzero $v$ such that $A_{12}v \notin \operatorname{Im}(A_{11})$. Since $A_{11}$ is positive semidefinite, its kernel (i.e. nullspace) is precisely the orthogonal complement of its image. It follows that the orthogonal projection of $A_{12}v$ on $\ker(A_{11})$ is nonzero. Let $x$ be this projection and let $k\in\mathbb{R}$. Then
$$(kx^T,v^T)\,A\pmatrix{kx\\ v}=2kx^TA_{12}v + v^TA_{22}v.\tag{1}$$
Argue that when $|k|$ is sufficiently large and the sign of $k$ is chosen appropriately, $(1)$ will contradict the assumption that $A$ is positive semidefinite. Therefore $\operatorname{Im}(A_{12})\subseteq \operatorname{Im}(A_{11})$.

Show that $A_{11}$ is invertible on $\operatorname{Im}(A_{11})$ (hint: $A_{11}$ is positive semidefinite). Hence the assertion follows from the result of step 1.

Alternatively, note that positive semidefinite matrices can be orthogonally diagonalised. So, WLOG you may assume that $A_{11}=D\oplus0$, where $D$ is an $r\times r$ positive diagonal matrix ($r=\operatorname{rank}(A_{11})$) and the zero block is $(m-r)\times(m-r)$. In this case, it is not hard to show that the last $m-r$ rows of $A_{12}$ must be zero and hence the assertion follows. This actually is the same proof as the above, but presented in a less linear algebraic but more matrix theoretic fashion.


问:
minimax inequality
Given any function $\varphi : X \times Y \rightarrow \mathbb{R}$ it holds that

$$\max_{y}\min_x \varphi(x,y) \le \min_y\max_y \varphi(x,y).$$

The proof I'm trying to read states the following thing: 1)the above inequality can be proven by considering that
$$For \,\, all \,\, x,y \qquad \min_x \varphi(x,y) \le \max_y \varphi(x,y)$$

(Why is this true?)

Then I can take $\max_y$ on left-hand side since the above ineq is true for all $y$ values, getting:

$$\max_y\min_x \varphi(x,y) \le \max_y \varphi(x,y)$$

3)Finally I can take the $\min_x$ on the 'inner' part of the equality, getting
$$\max_y\min_x \varphi(x,y) \le \max_y (\min_x)\varphi(x,y)$$

Are these steps correct? Also why is the first inequality (1) true? Thanks

答:
I think your statement isn't well stated. Maybe you want prove the following?: $\max_\limits {y}\min_\limits x \varphi(x,y) \le \min_\limits x\max_\limits y \varphi(x,y)$ (note that on the RHS you have taken the minimum on $y$ after taking the maximum on $y$ too, which doesn't make much sense).

For the first part, note that for all $x,y$ we have $\min_\limits x \varphi(x,y) \le\varphi(x,y)$ and $\varphi(x,y) \le \max_\limits y \varphi(x,y)$, so for all $x,y$ we have $\min_\limits x \varphi(x,y) \le \max_\limits y \varphi(x,y)$.

The second step seems ok, but it should be pointed out that the inequality holds for all $x$.

The third step is just taking minimum on $x$, not on the inner part or anything, it's just like you did with the maximum in the second step.

Title: minimax inequality


问:
Show $\lim_{x \to \infty}{\bigr( (x+1)^{2H} - x^{2H}}\bigr) = 0$ for $0 \lt H \lt 1/2$.
I want to show $\lim_{x \to \infty}{\bigr( (x+1)^{2H} - x^{2H}}\bigr) = 0$ for $0 \lt H \lt 1/2$.

I believe this has something to do with the fact $x \mapsto x^{2H}$ is concave for $0 \lt H \lt 1/2$ but I cant seem to use the concavity property to prove this.

答:
We can use Newton's generalized binomial theorem
$$(1+x)^{2H}=\sum\limits_{k=0}^{\infty}\binom {2H} k x^k$$
where $\binom m k = \frac{m(m-1)\cdots(m-k+1)}{k!}$. So Al remained members will have power  $\leqslant 2H-1$ and limit gives zero. One moment, that I want leave for you, with hope you can manage it, is interchanging limit and summation.


问:
A binomial expansion problem I am having problem with.
When $ (1+cx)^n$ is expanded a series in ascending powers of $x$, the first three terms are given by $1+20x+150x^2$. Calculate the value of the constants $c$ and $n$.

If possible please include working so I could perhaps understand how you got to the answer.

答:
You need to use the Binomial formula

Applying it you get:

$$(1+cx)^n = {n \choose 0} \cdot 1^n \cdot (cx)^{0} + {n \choose 1} \cdot 1^{n-1} \cdot
(cx)^{1} + {n \choose 2} \cdot 1^{n-2} \cdot (cx)^{2} +\ ... $$

$$(1+cx)^n = {n \choose 0} + {n \choose 1} \cdot (cx)^{1} + {n \choose 2} \cdot (cx)^{2} +\ ... $$

$$(1+cx)^n = {n \choose 0} + {n \choose 1} \cdot cx + {n \choose 2} \cdot c^2x^{2} +\ ... $$

Now you can construct equations for $c,n$ because you are given the coefficients before $x$ and $x^2$.

$${n \choose 1} \cdot c = 20 $$
$${n \choose 2} \cdot c^2 = 150 $$

Let us solve this system of two equations.

$$nc= 20$$
$$\frac{n(n-1)}{2}c^2 = 150$$

We get

$$nc = 20$$
$$(n-1)c = 15$$

This means

$$nc = 20$$
$$nc-c = 15$$

This gives us the answer which is $c=5, n=4$


答:
For concave differentiable $f$ we have: $$f(y)-f(x)\leq f'(x)(y-x).$$
In this case $f(x)=x^{2H}$ and $f'(x)=2Hx^{2H-1}$ decreases to $0$ at $\infty$ for $0<H<1/2$. Taking $y=x+1$ gives: $$0\leq(x+1)^{2H} - x^{2H}\leq2Hx^{2H-1}(x+1-x)\\=\frac{2H}{x^{1-2H}}\to0$$ when $x\to\infty$. By the squeeze theorem, the limit is $0$.

Title: Show $\lim_{x \to \infty}{\bigr( (x+1)^{2H} - x^{2H}}\bigr) = 0$ for $0 \lt H \lt 1/2$.

Body:
I want to show $\lim_{x \to \infty}{\bigr( (x+1)^{2H} - x^{2H}}\bigr) = 0$ for $0 \lt H \lt 1/2$.

I believe this has something to do with the fact $x \mapsto x^{2H}$ is concave for $0 \lt H \lt 1/2$ but I cant seem to use the concavity property to prove this.


问:
Proving $\int_0^1 \frac{(\ln(x))^5}{1+x} \mathrm{d}x = -\frac{31\pi^6}{252}$
I would like to show the following identity:

$$\boxed{
I :=
\int_0^1 \dfrac{(\ln(x))^5}{1+x} \mathrm{d}x =
-\dfrac{31\pi^6}{252}
}$$

Here is what I tried.
The change of variables $u=1/x$ yields
$$I=
\int_1^{\infty} \dfrac{(\ln(x))^5}{1+1/u} \dfrac{1}{u^2} \mathrm{d}u =
\int_1^{\infty} \dfrac{(\ln(x))^5}{u^2+u} \mathrm{d}u$$
Then $z=u-1$ gives
$$I = \int_{0}^{\infty} \dfrac{(\ln(z+1))^5}{z^2+3z+2} \mathrm{d}z $$
with $z^2+3z+2=(z+1)(z+2)$.
I wanted to use contour integration like here, but I was not sure how to proceed in this case. Anyway, the computations of the residues (of which "well-chosen" function? Maybe something like this?) seem to be difficult.

I believe that we can generalize to $\frac{(\ln(x))^n}{1+x}$, or maybe even more (e.g. $\frac{(\ln(x))^n}{ax^2+bx+c}$).
Related computations are: (1), (2), (3), (4).

Thank you for your detailed help.

Body:
\begin{align}J&=\int_0^1 \frac{(\ln(x))^5}{1+x} \mathrm{d}x \\
&=\int_0^1 \frac{(\ln(x))^5}{1-x} \mathrm{d}x-\int_0^1 \frac{2u(\ln(u))^5}{1-u^2} \mathrm{d}u\\
&\overset{x=u^2}=\int_0^1 \frac{(\ln(x))^5}{1-x} \mathrm{d}x-\frac{1}{2^5}\int_0^1 \frac{(\ln(x))^5}{1-x} \mathrm{d}x\\
&=\frac{31}{32}\int_0^1 \frac{(\ln(x))^5}{1-x} \mathrm{d}x\\
&=-\frac{31}{32}\times \frac{5!}{945}\times \zeta(6)\\
&=\boxed{-\frac{31\pi^6}{252}}
\end{align}
I assume: $\displaystyle \int_0^1 \frac{\ln^5 x}{1-x}dx=-5!\zeta(6)$ and $\displaystyle\zeta(6)=\frac{\pi^6}{945}$


问:
How to represent the floor function using mathematical notation?
I'm curious as to how the floor function can be defined using mathematical notation.

What I mean by this, is, instead of a word-based explanation (i.e. "The closest integer that is not greater than x"), I'm curious to see the mathematical equivalent of the definition, if that is even possible.

For example, a word-based explanation of the factorial function would be "multiply the number by all integers below it", whereas a mathematical equivalent of the definition would be $n! = \prod^n_{i=1}i$.

So, in summary: how can I show what $\lfloor x \rfloor$ means without words?

答: $\lfloor x \rfloor = x - \arctan(\tan(\pi x))/\pi$ ?... </p>


问:
Representations of finite p-groups G over a field K
Let $K$ be any field of characteristic $p > 0$ and let $G$ be a finite $p$-group. Let $V$ be any $K$-linear
representation of $G$. Then prove that there exists a non-zero $v \in V$ such that $gv = v$ for all $g \in G$.

If $G$ is a finite group and $V$ is a vector space. Suppose $GL(V)$ denote the group of invertible $K-$linear transformation from $V$ to $V$. Then
I know that a linear representation of $G$ in $V$ is a homomorphism $p$ from the group $G$ into the group $GL(V)$.

Can anyone give me any hint for this problem. I am supposed to determine an element $v \in V$ such that it is doesn't change for any map $p_g \in GL(V)$ for any $g \in G$. That is $p_g(v)=v$ for all $g \in G$. Here $p_g \in GL(V)$.

How am I supposed to use the fact that $G$ is a of order $p^k$?

答:
There is probably a much simpler way but I think this solution is fun.

Show that a finite $p$-group has a non-trivial center.

$$W=\bigcap_{z\in Z(G)} \ker(z-1)$$ $\quad$ (it is non-empty because they commute and $(z-1)^{p^k}=z^{p^k}-1=0\in K[G]$)

$W$ is a subrepresentation of $G$ (ie. $\forall g, gW\subset W$), since $Z(G)$ acts trivially it is also a representation of $G/Z(G)$.

Then replace $G,V$ by $G/Z(G),W$ and repeat until $G$ is trivial.

Any $v$ in the last $V$ will work.


问:
Minkowski dimension of the subset
Suppose $A\subset R^N$ is a manifold with topological dimension d.

(To be precise, the case I need is

A is the set of m by n matrices with rank r, N=mn.

It is known that its codimension is $(m-r)(n-r)$.)

Now can we say any (compact) subset of A has

Hausdorff dimension less than or equal to d?

Minkowski dimension(box covering dimension) less than or equal to d?

If not, is there any bounding inequality for the Minkowski dimension using d?

(For example, $dim_{M} (B) \leq 10d + 100 $ holds for all $B \subset A$.

If that upper bound is $O(d)$ then it would be fantastic)

答:
The short answer is that charts on smooth manifold can be taken to be bi-Lipschitz and then they preserve dimension (see, for example, Hausdorff Dimension of a manifold of dimension n?).

In more details one could argue as follows:

If $A^d$ is a smooth submanifold of $\mathbb{R}^n$ then for every point $a\in A$ there exists a neighborhood $U$ of $a$, an open neighborhood $V$ of the origin in $\mathbb{R}^n=\{x_1, \ldots, x_n\}$  and diffeomorphism $f_a(U, U\cap A, a) \to (V, V\cap H, 0)$ where $H$ is the coordinate space $\{x_1, \ldots, x_n| x_{d+1}=0,\ldots, x_n=0\}$. This is either a definition of a submanifold, or  basic theorem in the subject.  Moreover, we can take $f(U)$ to be an open ball and this diffeomorphism can be taken to have derivative bounded in operator norm $|Df_u|<C, |Df^{-1}_u|<C$ for all $u\in U$.  (Proof: Pick $V$ to be open ball inside closed ball inside $f(U)$ and take new $U$ to be $f^{-1}(V)$; then by compactness the bounds hold on the closed ball for some $C$, and hence on the open ball as well.)

Now pick a compact subset $K$ of $A$. The sets $W_a=U_a\cap K$ with $a\in K$ form a cover, so there is a finite subcover $W_1, \ldots, W_N$ with $W_j=U_{a_j}\cap K\subset  U_{a_j}\cap A$. If we show that $\dim  U_{a_j}\cap A \leq d$, then the same is true for $W_j$s, and hence for their union $K$, which is what we want.

We prove that upper box counting (Minkowski) dimension of $ A_j=U_{a_j}\cap A$ is at most $d$. The bound you want then follows for Hausdorff dimension as well since it is bounded by the box counting dimension.

But the fact that that upper box counting dimension of $A_j$ is at most $d$ follows from two observations: 1) the image of $A_j$ under $f$ being an open disc of $d$-dimensional plane has box counting dimension $d$ 2) if we have  any covering of $f(A_j)$ by $\varepsilon$ balls centered at $h_l$s, then the $C\varepsilon$ balls centered at $f^{-1}(h_l)$ cover $A_j$. Indeed, suppose $a\in A_j$ is arbitrary. Then $h=f(a)$ is connected to some $h_j$ by a segment of length at most $\varepsilon$ (the length is the integral of velocity, and by our bound on $|Df^{-1}|$ the velocity is rescaled by at most $C$, so integrates to at most $C$ times orgonal lenght; i.e. we ensure $f^{-1}$ is Lipschitz). Then the preimage of this segment connects $a$ to $f^{-1}(h_l)$, and has length at most $C\varepsilon$. So the balls around $f^{-1}(h_j)$ of size $C\varepsilon$ cover $A_j$.

These two observations together imply that the upper box counting dimension of $A_j$ is at most $d$.

In fact, by taking charts where each $U$ is a ball (end ensuring $f$ is Lipschitz) and running similar argument one can show that lower box counting dimension of $A_j$s is at least $d$.

Title: Minkowski dimension of the subset


Title: A binomial expansion problem I am having problem with.

Body:
When $ (1+cx)^n$ is expanded a series in ascending powers of $x$, the first three terms are given by $1+20x+150x^2$. Calculate the value of the constants $c$ and $n$.

If possible please include working so I could perhaps understand how you got to the answer.

Body:
$$(1+cx)^n=1+ncx+\frac{1}{2}n(n-1)(cx)^2+...$$
So we have $nc=20~~~(1)$ and $\frac{1}{2}n(n-1)c^2 =150~~~(2)$
$$\implies n(n-1)c^2=300 \implies 20 (n-1)c=300 \implies (n-1)C=15 \implies 20-c=15 $$
$$\implies c=5, n=4$$


答:
Dividing by $n!$,

$$\frac{s_n}{n!}=\frac{s_{n-1}}{(n-1)!}+\frac1n,$$

which we can write

$$h_n=h_{n-1}+\frac1n.$$

So obviously

$$h_n=H_n+h_0$$ and $$s_n=n!(H_n+h_0)=n!(H_n+s_0).$$

Unfortunately, there is no shortcut to computing the Harmonic numbers exactly, so this result is not much better than the initial recursion.

Title: Reccursive formula as a non-recursive

Body:
The recursive formula is give as follows:

$$ s_n = n s_{n-1} + (n-1)!$$ and $s_2 = 1$. How to construct a non-recursive formula?


问:
Evaluating $\lim_{n \to +\infty} ((n+1)^a-n^a)$
I am trying to determine the limit of the following expression which arose when looking at correlations for a stochastic process.

$\lim_{n \to +\infty} ((n+1)^a-n^a)$ for values of $a\in (0,2)$.

I'm not sure how to approach this. I am aware that for $a>1$ the limit equals $\infty$ and for $a<1$ the limit is $-1$ (and $a=0 \implies$ limit $=0$).

Any ideas of what to try here would be greatly appreciated.

答:
As I wrote in the comments, we can also use the mean-value theorem. By the MVT, there exists $x_n$ in $(n, n+1)$ such that
$$ (n+1)^\alpha - n^\alpha = \alpha x_n^{\alpha-1}. $$
Now as $x_n\in (n, n+1)$ we have $n^{\alpha-1}\leq x_{n}^{\alpha-1}\leq (n+1)^{\alpha-1}$ and hence,
$$ \alpha n^{\alpha-1}\leq (n+1)^\alpha - n^\alpha \leq \alpha (n+1)^{\alpha-1}. $$
For $\alpha \in (1,2)$ we have
$$\infty = \limsup_{n\rightarrow \infty} \alpha n^{\alpha-1} \leq \limsup_{n\rightarrow \infty} ((n+1)^\alpha - n^\alpha).$$
On the other hand, we have for $\alpha \in (0,1)$
$$ \lim_{n\rightarrow \infty} (n+1)^{\alpha-1}=0 $$
and thus by the squeeze theorem we have $\lim_{n\rightarrow \infty} ((n+1)^\alpha - n^\alpha) =0. $


问:
Number of balls inside a ball - metric space version
Suppose we have a metric space $(X,d)$ and let $a>1$. Also, let $x\in X$ and $r>0$ small enough.

Is there an upper bound $N(a)> 0$ (independent of $x,r$) so that there can fit at most $N(a)$ disjoint balls of radius $r/a$ inside the ball $B(x,r)$?

I can see that this happens if $X=\mathbb R^n$... but I use volume of balls to check this. What can we say in the general case? What about more geometric $X$? I see that this is also true, for small balls, if $X$ is a manifold.

Thanks!

答:
Here's an easy construction where you can have infinity:

Let $X$ be a metric space, and let $d(x, y)=1 \forall x \neq y \in X$. Of course, $d(x,x) = 0$. It's easy to check that $d$ is a metric.

Now, $B(x, 1) = X$. On the other hand, $B(x, a) = {x}$ if $a < 1$, so for any $a < 1$, $\cup_{y \in X} B(y,a) \subset B(x, 1)$ and the union is disjoint. So if $X$ is not finite, you can pack infinitely many balls.


问:
Has the congruence $x^3\equiv 84 \mod (139)$ more than one solution?
Has the congruence $x^3\equiv 84 \mod (139)$ more than one solution?

I could find $x=27$ by brute force. I could not find more solutions:

$x^{138}=(x^3)^{46}\equiv 1 \mod (139)$

Is this correct:

$x^3\equiv 1 \mod (139)$

If yes how should we continue to find more solutions?

答:
Since $139$ is prime, $(\mathbb Z/139\mathbb Z)^\times$ is cyclic of order $138$. This is the key point. The solution below uses a generator of this group, also known as a primitive root mod $139$.

Since $3$ divides $138$, the equation $x^3\equiv 1 \bmod 139$ has exactly $3$ solutions. They are $1,h,h^2$, where $h=g^{46}$ and $g$ is a primitive root mod $139$.

Therefore, the equation $x^3\equiv 84 \bmod 139$ has three solutions: $27, 27h, 27h^2$.

For instance, the primitive root $g=2$ gives $h=96$ and the three solutions: $27, 90, 22$.


答: Consider a Hilbert space with an orthonormal basis $(e_n)$. Let $\epsilon >0$. Take $a=2$ and $ r=\epsilon $. Consider  the balls $B(\frac {\epsilon e_i} {3}, \frac {\epsilon } {3\sqrt 2})$. These balls are disjoint and they are all contained in $B(0,r)$. So $N(a)$ need not be finite.

Title: Number of balls inside a ball - metric space version

Body:
Suppose we have a metric space $(X,d)$ and let $a>1$. Also, let $x\in X$ and $r>0$ small enough.

Is there an upper bound $N(a)> 0$ (independent of $x,r$) so that there can fit at most $N(a)$ disjoint balls of radius $r/a$ inside the ball $B(x,r)$?

I can see that this happens if $X=\mathbb R^n$... but I use volume of balls to check this. What can we say in the general case? What about more geometric $X$? I see that this is also true, for small balls, if $X$ is a manifold.

Thanks!


答: No. E.g., consider the space $X$ comprising the coordinate axes in the space $\Bbb{R}^* = \bigcup_{n=1}^\infty\Bbb{R}^n$. $X$ looks like countably many copies of the real line identified at the origin metrized such that each pair of lines has the same metric as the union of the coordinate axes in $\Bbb{R}^2$. The open unit ball centred at the origin contains infinitely many pairwise disjoint open balls of radius $\frac{1}{2}$ centred at the points at distance $\frac{1}{2}$ from the origin. Likewise the closed unit ball centred at the origin contains infinitely many pairwise disjoint closed balls of radius $\frac{1}{3}$.


问:
Vitali set of outer measure 1
How to construct a Vitali set of outer measure 1. I couldn't understand the argument given here. Isn't there any easier way? I would also like if someone explains that to me.

Thank you in advance!

答:
Since  $\mathbb{R}$ is a vector space over $\mathbb{R}$, there exists a
$\mathbb{Q}$ vector subspace $V$ of $\mathbb{R}$ such that $V\oplus \mathbb{Q} = \mathbb{R}$ ( it uses the existence of a basis of $\mathbb{R}$ extending $\{1\}$, so it uses the Zorn lemma).

Now, $V$ is a Vitali set, and moreover $\mathbb{Q}^{\times}\cdot  V = V$.  This is the only thing we will use.

Now, since $\cup_{q\in \mathbb{Q}} (V+ q) = \mathbb{R}$, the exterior measure of $V$ is $>0$. (in fact, since $V$ is invariant under multiplication by non-zero rationals, $\mu^{*}(V) = \infty$). Nevertheless, since $\mu^{*}(V)>0$, for every $\epsilon > 0$ there exists an interval of form $I=[\frac{m}{n}, \frac{m+1}{n}] $ such that $\mu^{*}(V\cap I) > (1-\epsilon) \mu(I)$. Now, since $n V= V$, we get
$$\mu^{*}(V\cap I) > (1-\epsilon) \mu(I)$$
for $I=[m, m+1]$.

Now consider the Vitali set $\tilde V$ consisting of all the fractional parts of the elements of $V$
$$\tilde V \colon = \{ \{x\} \ | \ x \in V\}$$

that is
$$\tilde V = \bigcup_{m\in \mathbb{Z}} (V\cap [m, m+1)) - m$$

Clearly $\tilde V \subset [0,1)$, and moreover, from the above, for every $\epsilon > 0$, we have $\mu^{*}(\tilde V) > 1-\epsilon$. We conclude that $\mu^*(\tilde V) = 1$.

Note that if we define $\tilde V_q \colon= \{ \{x\} \ | \ x \in V+ q\}$, then we have the partition of $[0,1)$ into countably many disjoint subsets of exterior measure $1$
$$[0,1) = \bigcup_{q \in \mathbb{Q}\cap [0,1)} \tilde V_q$$

Title: Vitali set of outer measure 1


问:
Proof that $\sum_{k=0}^m \binom{m}{k}\frac{1}{k+1} = \frac{2^{m+1}-1}{m+1}$
Recently I needed to compute $E[\frac{1}{X+1}]$ where $X\sim Bin(m, \frac 1 2)$.
While expanding, I came across the sum $\sum_{k=0}^m \binom{m}{k}\frac{1}{k+1}$, which I was unable to solve. Plugging into Mathematica gives $\frac{2^{m+1}-1}{m+1}$, but I don't see how to derive this result.

Can anyone give a proof of this identity or an alternate way to compute $E[\frac{1}{X+1}]$?

Body:
HINT:

$$\binom mk \frac1{k+1}=\frac{m!}{(k+1)\cdot k! (m-k)!}=\frac1{m+1}\cdot \frac{(m+1)\cdot m!}{(k+1)!\{(m+1)-(k+1)\}!}=\frac1{m+1}\cdot \binom {m+1}{k+1}$$

So, $$\sum_{0\le k\le m}\frac1{k+1}\cdot \binom mk=\frac1{m+1}\cdot \sum_{0\le k\le m}\binom {m+1}{k+1}=\frac1{m+1}\cdot\sum_{1\le k\le m+1}\binom {m+1}k$$

Now,  $$\sum_{1\le k\le n}\binom nk=(1+1)^n-1$$


问:
Can an arc be greater than 360 degrees or less than 0 degrees.
Can an arc be greater than 360 degrees or less than 0 degrees? Is the (angular) arc length a signed scalar?

I'm creating a Graphics and basic geometry software library. I believe an arc can be defined by 5 scalars, the x and y components of the start point, the x and y components of the centre point and the signed value of the rotation. However I'm concerned that this data definition could lead to unnecessary rounding errors on the end point of the arc, the precise value of which is generally of greater concern than the precise value of the arc's centre point. So I'm leaning towards defining arcs by the 6 scalar definition of the start point, the end point, and the mid point of the circumference of the arc. This data definition has the added advantage that you don't have to worry about reflections, but would exclude arcs of greater than 360 degrees. Is this mathematically sound?

A successful software library will get used in ways unimagined by the library authors. So this where I place a strong priority or weighting on following mathematical definitions, because those more rigorous,robust and tried and tested definitions can hopefully pre-empt problems further down the road.

答:
The first definition sounds more natural to me. You are speaking too roughly here so it's hard to tell what issues you're foreseeing with it. Of course the 2nd definition is also valid i.e. is minimal and is sufficient to define that arc.

In math in principle an arc can be greater that 360 degrees or negative but for practical purposes probably you don't need such arcs. It depends on the goals of that library you're developing.


答:
In infinite-dimensional spaces, there’s a limit to the radii at which balls of unit radius can fit infinitely many non-overlapping balls of those radii; precisely, let $(X,\|\cdot\|)$ be Banach space with unit ball centered at the origin $B_X=B(0,1)$ and define the packing constant
$$P(X):=\sup\{r>0:B_X\text{ contains infinitely many non-overlapping } B(x,r)\}$$ then we have a formula given by $$P(X)=\frac{K(X)}{2+K(X)}\,$$ where $K(X)$ is the Kottman separation constant defined as $$K(X)=\sup\left\lbrace r:\|x_n-x_m\|\ge r\,,\forall n\ne m\,, \{x_n\}_n\subset B_X\right\rbrace\,.$$
For some spaces, the Kottman constant values are well-known; for instance, given the sequence space $\ell^p$ one obtains $$K(X)=2^{1/p}\,,~\,~\,~\,~\text{if $1\le p<\infty$}\,.$$ Thus, as example, the space $\ell^1$ of absolutely summable sequences can contain infinitely many non-overlapping balls of radius $r<\frac{1}{2}$ in its unit ball.

(For further reading, see for instance, https://pdfs.semanticscholar.org/f857/7f978843efd0b7b96cd7cd7b94770907032a.pdf).


问:
Examples of functions with a natural domain $\Bbb R\setminus 2\Bbb N$

I would like to find a few examples of a function $f$ with a natural domain $\Bbb R\setminus 2\Bbb N$ which can be defined in a single expression.

Thoughts:

Since $\lfloor x\rfloor=\lceil x\rceil\iff x\in\Bbb Z$, I took
$$f(x)=\begin{cases}\frac1{\left|\left\lfloor\frac{x}2\right\rfloor\right|-\left\lceil\frac{x}2\right\rceil},&x\ne 0\\0,&x=0\end{cases}\space$$
Similarly, it can be defined as:
$$f(x)=\begin{cases}\ln\left(\left|\left|\left\lfloor\frac{x}2\right\rfloor\right|-\left\lceil\frac{x}2\right\rceil\right|\right),&x\ne 0\\0,&x=0\end{cases}$$
Another attempt is:

$$f(x)=\begin{cases}\tan\left(\left\lfloor\left(\frac32\right)^{\operatorname{sgn}(x)}\right\rfloor\frac{x+1}2\pi\right),&x\ne0\\0,&x=0\end{cases}$$

but $0\notin\Bbb N$, so I had to define $f(0)$ separately (however, $\tan\left(\operatorname{sgn}(x)\left\lfloor\left(\frac32\right)^{\operatorname{sgn}(x)}\right\rfloor\frac{x+1}2\pi\right)$ works if one takes $\operatorname{sgn}(0)=0$).
Furthermore, it is constant on $(-\infty,0]$, which isn't quite interesting.

How can I fix this problem if I want $f$ to be composed of $\tan(x)$? Are there any other examples of such functions defined by one expression?

Thank you in advance!

Body: What about $$f(x)=\sum_{k=0}^{+\infty} \frac{1}{(x-2k)^2} \quad ?$$


问:
Show that the matrix $A^2 + I$ is invertible for all matrices $A$, where $A$ is an $n \times n$ symmetric matrix.
I'm a little stuck on this problem. I know that since $A$ is symmetric, $A=A^{T}$. I'm also pretty sure that $AA^{T}$ is invertible. Therefore $A^2$ would be invertible. I'm not really sure how to account for the identity matrix (In).

Thanks in advance!

答: Hint: Since $A$ is a symmetric matrix, its eigen value are real ( it can be zero as well ) however when the identity matrix is added, the eigen value will become non-zero real values.</p>


答:
Use a geometric series.
$$\frac{1}{1+x}=\sum_{k=0}^\infty(-1)^kx^k$$
So
$$\mathcal{I}_n=\int_0^1\frac{\ln(x)^n}{1+x}\mathrm{d}x=\sum_{m=0}^\infty\left[(-1)^mI_{m,n}\right]$$
Where
$$I_{m,n}=\int_0^1 x^m\ln(x)^n\mathrm{d}x$$
Now use a substitution $x=e^{-t}$ which changes our region of integration to $(\infty,0]$.
$$I_{m,n}=\int_\infty^0(e^{-t})^m(-t)^n(-e^{-t})\mathrm{d}t$$
$$=(-1)^n\int_0^\infty t^ne^{-t(m+1)}\mathrm{d}t$$
Now use another substitution $z=t(m+1)$:
$$I_{m,n}=(-1)^n\int_0^\infty\left(\frac{z}{m+1}\right)^ne^{-z}\frac{\mathrm{d}z}{m+1}$$
$$=\frac{(-1)^n ~n!}{(m+1)^{n+1}}$$
So then
$$\mathcal{I}_n=(-1)^nn!\sum_{m=0}^\infty\frac{(-1)^m}{(m+1)^{n+1}}$$
$$=(-1)^nn!\sum_{m=1}^\infty\frac{(-1)^{m-1}}{m^{n+1}}$$
$$=(-1)^n~n!~\eta(n+1)$$
Using the Dirichlet eta function. In the case of $n=5$ we get
$$\mathcal{I}_5=-5!\cdot\eta(6)=\frac{-31\pi^6}{252}$$

Which is confirmed numerically by Mathematica:

Title: Proving $\int_0^1 \frac{(\ln(x))^5}{1+x} \mathrm{d}x = -\frac{31\pi^6}{252}$

Body:
I would like to show the following identity:

$$\boxed{
I :=
\int_0^1 \dfrac{(\ln(x))^5}{1+x} \mathrm{d}x =
-\dfrac{31\pi^6}{252}
}$$

Here is what I tried.
The change of variables $u=1/x$ yields
$$I=
\int_1^{\infty} \dfrac{(\ln(x))^5}{1+1/u} \dfrac{1}{u^2} \mathrm{d}u =
\int_1^{\infty} \dfrac{(\ln(x))^5}{u^2+u} \mathrm{d}u$$
Then $z=u-1$ gives
$$I = \int_{0}^{\infty} \dfrac{(\ln(z+1))^5}{z^2+3z+2} \mathrm{d}z $$
with $z^2+3z+2=(z+1)(z+2)$.
I wanted to use contour integration like here, but I was not sure how to proceed in this case. Anyway, the computations of the residues (of which "well-chosen" function? Maybe something like this?) seem to be difficult.

I believe that we can generalize to $\frac{(\ln(x))^n}{1+x}$, or maybe even more (e.g. $\frac{(\ln(x))^n}{ax^2+bx+c}$).
Related computations are: (1), (2), (3), (4).

Thank you for your detailed help.


问:
How algebra differs in complex plane and cartesian plane?
I noticed the difference first when I tried to convert the complex form of the equation of circle in cartesian form, this is how I approached-

$$\tag{1}|z-a|=r \qquad (\text{here},z= x+iy)  $$

I thought, $iy$ corresponds to $Y$-axis in the Cartesian plane,
and, my converted equation was this-
$$\tag{2}
                        |x+y-a|=r
$$
this was simple, then I went to proof

1)

$$
                                         |z-a|=r
$$
$$
                                        (z-a)^2= r^2
$$
$$
                                       z^2-2za+a^2=r^2
$$

Another form of it is- zz¯−z0¯z−z0z¯+(|z0|2−R2)=0, but I did not approach to that,
Now the prof of-

$$
                              |x+y|=r
$$
$$
                            x^2+2xy+y^2=r^2
$$

this looks very similar to 1. But, it does not give a circle, when I plot it. The 2xy seems to get in my way of circle.

another thing that I learned is-
iy.i= -y
it is a 90-degree tilt of y

but in the cartesian plane, you have to multiply -1.

```<code>                         y.(-1)=y
</code>```

I lack information in this field, I want to know how the cartesian plane is different from the complex plane? what distinguishes them?
how to make a connection with the complex geometry and cartesian geometry?

答:
One of the differences is $|z|^2 \ne z^2$ for most complex numbers $z$. Take $z = i$ as an example: $$z^2 = i^2 = -1 \ne 1 = 1^2 = |z|^2.$$

I think that you misunderstood the concept of an absolute value for complex numbers which led to several mistakes in your proof. One possible definition goes as follows: for any complex number $z = x + iy$ $$|z| = \sqrt{x^2 + y^2}.$$ The absolute value of a complex number is also sometimes called its norm to highlight the difference from real numbers.

Using this definition the proof becomes simple: first, shift $a$ to $0$ so that we don't have to worry about it, then (for $r \ge 0$ ofc) $$|z| = r \iff \sqrt{x^2 + y^2} = r \iff x^2 + y^2 = r^2,$$ which is exactly the equation of a circle in a real plane.


问:
Is It Hopeless for Me to Try to Understand the Derivations of All Formulas I Use?
Recently, I've been doing some relatively more advanced math (multivariable and vector calculus) and the derivations of formulas are becoming increasingly complicated. I don't like using a formula if I don't understand why it works and where it comes from, so I have understood the derivation of all formulas I've used so far. My question is: is it hopeless for me to keep learning derivations? Will I eventually reach a point where understanding why the formulas I use work becomes too time-consuming and unreasonable? To clarify, I prefer to understand proofs in order use them, I don't memorize them. Any help is appreciated. Thank you.

答: Personally I don't think this is reasonable at all. You can easily spend 1 hour - 2 hours (or more in some cases) to fully understand a proof, and in multivariable calculus and linear algebra textbooks there might be a proof every 3 pages depending on how advanced the text is. In a Real Analysis text there might be a proof 2 pages. My point is, you should learn the proof if the theorem is important.</p>


问:
Explanation of why $\int \sin^2 x\cos^2 x\; dx = 1/3 \sin^3 x - 2/5 \sin^5 x + 1/7 \sin^7 x +c$

verify the solution
  $$\int \sin^2 x\cos^5 x\; dx = 1/3 \sin^3 x - 2/5 \sin^5 x + 1/7 \sin^7 x +c$$

I have hit this in my book and can't work it out. Does anyone have any ideas or a walk-through that might help?

答:
It's easier to start from the right hand side. Derive to get:
$$ (\sin^2 x - 2\sin^4 x + \sin^6 x)\cos x$$
Note that this is just:
$$(\sin x - \sin^3 x)^2\cos x=\sin^2 x(1-\sin^2x)^2\cos x=\sin^2 x\cos^5x$$
So, it seems you are missing a couple of powers of $\cos x$ in the integral..</p>


问:
Is the space of all sequences on $[0,1]$ sequentially compact?
We are in the space of sequences on [0,1] under the metric $d(x,y) = \Sigma^{\infty}_{i=1} 2^{-k} \mid x_k - y_k \mid $

The hint I've been given is to use a diagonal argument.

I'm thinking to take a sequence of sequences in the space, and by Bolzano-Weierstrass each of these sequences must have a convergent subsequence. If I take the first term of the first sequence, the second of the second etc. then that seems to be a 'diagonal' argument as told to use in the hint - but then I only have one sequence. Am I not supposed to find a sequence of sequences which converges?

答:
Hint : Consider a sequence $(x^k)_{k \in \mathbb{N}}$ of elements of $[0,1]^{\mathbb{N}}$ (that is, each $x^k$ is a sequence $(x^k_n)_{n \in \mathbb{N}}$ of real numbers between $0$ and $1$).

Then $(x^0_n)_{n \in \mathbb{N}}$ is a bounded sequence of real numbers, so there exists $\varphi_0 : \mathbb{N} \rightarrow \mathbb{N}$ such that $(x^0_{\varphi_0(n)})_{n \in \mathbb{N}}$ converges.

Now $(x^1_{\varphi_0(n)})_{n \in \mathbb{N}}$ is a bounded sequence of real numbers, so there exists $\varphi_1 : \mathbb{N} \rightarrow \mathbb{N}$ such that $(x^0_{\varphi_0 \circ \varphi_1(n)})_{n \in \mathbb{N}}$ converges.

Now $(x^2_{\varphi_0 \circ \varphi_1(n)})_{n \in \mathbb{N}}$ is a bounded sequence of real numbers, so there exists $\varphi_2 : \mathbb{N} \rightarrow \mathbb{N}$ such that $(x^2_{\varphi_0 \circ \varphi_1 \circ \varphi_2(n)})_{n \in \mathbb{N}}$ converges.

And so on, you can construct $\varphi_0, \varphi_1, ..., \varphi_n, ...$ such that for all $k \in \mathbb{N}$, the sequence $(x^k_{\varphi_0 \circ ... \circ \varphi_k(n)})_{n \in \mathbb{N}}$ converges.

Finally, let $\varphi : \mathbb{N}  \rightarrow \mathbb{N}$ defined for all $n \in \mathbb{N}$ by $\varphi(n) = \varphi_0 \circ ... \circ \varphi_n(n)$ (here is the diagonal argument !). You can show that $\varphi$ is strictly increasing, and that the sequence $(x^{\varphi(n)})_{n \in \mathbb{N}}$ converges.


答: Let $(x_i^{n})$ be a sequence in this space. The diagonal argument gives you integers $n_1,n_2,..$ such that $\lim_{k \to \infty} x_i^{n_k}$ exists for each $i$. Call this limit $x_i$. We would like to show that $(x_i^{k})$ tends to $(x_i)$. Given $\epsilon >0$ choose $N$ such that $ \sum\limits_{k=N+1}^{\infty} \frac 1 {2^{k}} <\epsilon$. Split the sum in $d((x_i^{k}), (x_i))$ into the sum from $1$ to $N$ and the sum from $N+1$ to $\infty$. Can you finish?

Title: Is the space of all sequences on $[0,1]$ sequentially compact?

Body:
We are in the space of sequences on [0,1] under the metric $d(x,y) = \Sigma^{\infty}_{i=1} 2^{-k} \mid x_k - y_k \mid $

The hint I've been given is to use a diagonal argument.

I'm thinking to take a sequence of sequences in the space, and by Bolzano-Weierstrass each of these sequences must have a convergent subsequence. If I take the first term of the first sequence, the second of the second etc. then that seems to be a 'diagonal' argument as told to use in the hint - but then I only have one sequence. Am I not supposed to find a sequence of sequences which converges?


问:
Asymptotic of prime number pairs
I am not an expert in number theory, so if someone might know the answer or a lead for the following, it would be greatly appreciated. Let $\mathcal{P}$ be the primes, and let $x$ stand for a natural number. What is the large $x$ behavior of $$\# \{ (p, q) : p, q \in \mathcal{P}, \, p < q \, , \, p \, q \leq x \},$$ where $\#$ is the number of elements of a set, as usual? In other words, how many (unordered) pairs of primes with product at most $x$ are there as $x \rightarrow \infty$? For example, it is easy to see that, asymptotically, the number should be $$\sum_{p \in \mathcal{P}, \, p \leq \sqrt{x}}\left( \frac{x/p}{\log x/p} - \frac{p}{\log p} \right),$$but, of course, the question is whether more meaning can be given to this sum, or, written as an expression involving only $x$. Thanks.

答:
$$\sum_{p<q,pq\le x} 1 =\sum_{p\le \sqrt{x}} \pi(x/p)-\pi(p)$$

$$\sum_{p\le \sqrt{x}} \pi(x/p)=\sum_{p\le x/2} \frac{x/p}{\log x/p}+O(\frac{x/p}{\log^2 x/p})$$ (I used the PNT)
$$=\sum_{p\le \sqrt{x}}\frac{x/p}{\log x}(1+O(\frac{\log p}{\log x}))+O(\frac{x/p}{\log^2 x}(1+\frac{\log p}{\log x}))$$

$$=\sum_{p\le \sqrt{x}}\frac{x/p}{\log x}+O(\frac{x \log p}{p\log^2 x})= \frac{x \log\log x}{\log x}+O(\frac{x}{\log x}) $$
(Mertens theorem)

$$\sum_{p\le \sqrt{x}} \pi(p) = O(\sum_{n\le \sqrt{x}} \frac{n}{\log n}) = O(\frac{x}{\log x})$$

Title: Asymptotic of prime number pairs


问:
Prove that $\lim_{x\to\infty} (\ln x) = \infty$
Can someone help me prove that the function $\ln(x)$ diverges to infinity as $x$ approaches infinity. I tried using the definition to show that $\lvert \ln(x) -∞ \rvert < \epsilon $ where $\epsilon > 0$ and there exists a number $N$ which is in the set of natural numbers where $x > N$, proving that it does indeed diverge to infinity but I can't get any further.

Any help is appreaciated.

答:
One can use $\ln(x)=\int_1^x\frac{1}{t}dt\,$ (differentiable with $\ln'(x)=\frac{1}{x}$) and L'Hopital's (actually Bernoulli's) rule as follows:

Suppose that $\lim\limits_{x\rightarrow\infty}\ln(x)=L<\infty$ (as $\ln$ is a strictly increasing function, we only have two options: $\lim\limits_{x\rightarrow\infty}\ln(x)=\infty$ or $\lim\limits_{x\rightarrow\infty}\ln(x)=L<\infty$) and define $f(x)=L-\ln(x)$, which satisfies $\lim\limits_{x\rightarrow\infty}f(x)=0$. By L'Hopital's rule $$\lim\limits_{x\rightarrow\infty}xf(x)=\lim\limits_{x\rightarrow\infty}\frac{f(x)}{\frac{1}{x}}=\lim\limits_{x\rightarrow\infty}\frac{-\frac{1}{x}}{-\frac{1}{x^2}}=\lim\limits_{x\rightarrow\infty}x=\infty.$$

Hence, using L'Hopital's again $$0=\lim\limits_{x\rightarrow\infty}f(x)=\lim\limits_{x\rightarrow\infty}\frac{xf(x)}{x}=\lim\limits_{x\rightarrow\infty}\frac{f(x)-1}{1}=-1.$$ Therefore, $\lim\limits_{x\rightarrow\infty}\ln(x)=\infty.$