答: HINT: If $p\nmid m$, then $\frac{n}m\in R$.</p>

Title: Subrings of $\mathbb{Q}$

Body:
Let $p$ be prime. Suppose $R$ is the set of all rational numbers of the form $\frac{m}{n}$ where $m,n$ are integers and $p$ does not divide $n$.

Clearly then $R$ is a subring of $\mathbb{Q}$.

I now want to show that if $\frac{m}{n}$ belongs to any proper ideal of $R$ then $p|m$.

Can someone point me in the right direction.

Thanks


问:
The letter A appears an even number of times. How many different sequences could Dr. Lizardo have written down?

Dr. Lizardo writes down a sequence of $n$ letters, where each letter is A, B, or C, and the letter A appears an even number of times. How many different sequences could Dr. Lizardo have written down?

What I have tried so far

The last letter is predefined in terms of "A or not" based on the first $n-1$ letters. If there are an odd number of As so far, the last one has to be A, otherwise, B or C. Therefore, there are $3^{n-1}$ ways to write the first $n-1$ letters. However, the last letter can either have 1 choice or 2 choices.

I am thinking maybe we need to do $S_n = 3^{n-1} + S_{n-1}$. However, I need a closed form, not a recurrance. How can I do this?

Thank you!

Note: $S_n = 3^{n-1} + S_{n-1}$ seems to be working for the first few values of $n$ (I tested through $n=4$), so finding a closed form of that recurrance would work as well.

答:
Here we show the equality of two already given answers. The number $S_n$ is given by

\begin{align*}
\color{blue}{S_n=\frac{1}{2}\left(3^n+1\right)=\sum_{k=0}^n\binom{n}{2k}2^{n-2k}\qquad\qquad  n\geq 0}\tag{1}
\end{align*}

The following is valid
\begin{align*}
\sum_{k=0}^{n}\binom{n}{2k}x^{2k}+\sum_{k=0}^n\binom{n}{2k+1}x^{2k+1}&=(1+x)^n\\
\sum_{k=0}^{n}\binom{n}{2k}x^{2k}-\sum_{k=0}^n\binom{n}{2k+1}x^{2k+1}&=(1-x)^n\\
\sum_{k=0}^n\binom{n}{2k}x^{2k}&=\frac{1}{2}\left((1+x)^n+(1-x)^n\right)\tag{2}
\end{align*}

We obtain from (2)
\begin{align*}
\color{blue}{\sum_{k=0}^n\binom{n}{2k}2^{n-2k}}
&=2^n\sum_{k=0}^n\binom{n}{2k}\left(\frac{1}{2}\right)^{2k}\\
&=2^n\,\frac{1}{2}\left(\left(1+\frac{1}{2}\right)^n+\left(1-\frac{1}{2}\right)^n\right)\\
&=2^n\,\frac{1}{2}\left(\left(\frac{3}{2}\right)^n+\left(\frac{1}{2}\right)^n\right)\\
&\,\,\color{blue}{=\frac{1}{2}\left(3^n+1\right)}
\end{align*}
and the claim (1) follows.

ContentLicense: CC BY-SA 4.0


问:
Let a,b and c be the three sides of a triangle. Show that $\dfrac 1a+\dfrac 1b+\dfrac 1c\le \dfrac 1{a+b-c}+\dfrac 1{a-b+c}+\dfrac 1{-a+b+c}.$
Let a,b and c be the three sides of a triangle. Show that $\dfrac 1a+\dfrac 1b+\dfrac 1c\le \dfrac 1{a+b-c}+\dfrac 1{a-b+c}+\dfrac 1{-a+b+c}.$ When does equality occur?

I have used the Ravi-transformation and obtained

$\frac{1}{x+y} + \frac{1}{x+z} + \frac{1}{y+z} \leq \frac{1}{(x+y)+(y+z)-(x+z)} + \frac{1}{(x+y)-(y+z)+(x+z)} \\ +  \frac{1}{-(x+y)+(y+z)+(x+z)} \Longleftrightarrow  \frac{1}{x+y} + \frac{1}{x+z} + \frac{1}{y+z} \leq \frac{1}{2x} + \frac{1}{2y} + \frac{1}{2z}.$

But I get stuck here. Does anyone have an idea to proceed?

答:
You are almost there. For $x, y >0$, observe that
$$(x-y)^2 \geqslant 0\implies (x+y)^2\geqslant 4xy\implies x+y\geqslant \frac{4xy}{x+y}.$$
Reciprocating yields
$$\frac{1}{x+y}\leqslant \frac{x+y}{4xy}\implies \frac{1}{x} + \frac{1}{y} \geqslant \frac{4}{x+y}.$$
Analogous inequalities for the other variables also hold. Summing them up solves the problem.


问:
Homotopy classes of maps out of a wedge of circles
Let $W_n$ be the wedge on $n$ circle, i.e.,
$$
W_n=\bigvee_{k=1}^nS^1
$$
Let $X$ be a path connected space, then how do I calculate the homotopy classes of mas from $W_n$ to $X$? The based maps are easy as the wedge is the coproduct in the category of pointed spaces, thus
$$
\mathrm{hTop}_*\left[\bigvee_{k=1}^nS^1,X\right]=\prod_{k=1}^n\mathrm{hTop}_*\left[S^1,X\right]=\prod_{k=1}^n\pi_1\left(X\right)
$$
Is something similat true for non-pointed maps?

答:
When $X, Y$ are path-connected, the set $[X, Y]$ of homotopy classes of maps relates to the set $\langle X, Y \rangle$ of pointed homotopy classes of pointed maps by $$[X, Y] = \langle X, Y\rangle /\pi_1(Y),$$ where the action of $\pi_1(Y)$ is given by finding a map $\gamma \cdot f: X \to Y$ with an unbased homotopy $F_t: f \to \gamma \cdot f$ (that is, $F_0 = f$ and $F_1 = \gamma \cdot f$) so that $F_t(x_0) = \gamma(t)$. See Hatcher 4A.1.

In your case you already know that $\langle W_n, Y\rangle = \pi_1(Y)^n$. The action of $\pi_1(Y)$ on this set is the diagonal action where we have the conjugation action in each component; so $[W_n, Y]$ is the set of ordered $n$-tuples $(\gamma_1, \cdots, \gamma_n)$ considered up to conjugacy --- so $$(\gamma_1, \cdots, \gamma_n) \sim (\eta \gamma_1 \eta^{-1}, \cdots, \eta \gamma_n \eta^{-1}).$$


问:
Relation between averages in the sphere and the ball
Let $\Omega$ be a domain in $\mathbb R^N$ and $u \in C(\Omega)$. If $\omega_N$ denotes the volume of $B_1(0) \subset \mathbb R^N$ and $0 < r < d(x, \partial \Omega)$, the averages are
$$
A_b(x) = \frac{1}{\omega_N r^N} \int_{B_r(x)} u(y) \ dy
$$
in the ball, and
$$
A_s(x) = \frac{1}{N \omega_N r^{N - 1}} \int_{\partial B_r(x)} u(y) \ d\sigma(y)
$$
in the sphere.

We know that if $u$ is harmonic, then $A_b(x) = A_s(x) = u(x)$.

In general, can one say something like $A_s \leq (\geq) A_b$?

I kindly ask for a hint, in case the answer is positive, or for examples, in case the answer is negative.

EDIT:
Ted's hint made me think of the following: take $u$ to be negative inside the ball and $0$ on the boundary. Then of course $A_b < 0 = A_s$. On the other hand, if $u$ is positive, it is clear that $A_b > 0 = A_s$. We conclude that the answer to the question is negative in general.

This seems to be very simple, am I missing something?

答:
The answer seems to be no in the case $N=3$. Fix $\epsilon > 0$ small, and
take $u$ to be a
continuous function with $u(x,y,z) = 1$ when $x > \epsilon$
and $u(x,y,z) = 0$ when
$x < 0$. Consider averages over $B = B_2(1,0,0)$ and $B' = B_2(-1,0,0)$ and their boundaries.

The sphere $\partial B$ has, modulo $\epsilon$,  $3/4$ of a diameter
in the region where $u= 1$, and by a result of Archimedes also has
$3/4$ of its area there, so that average is $3/4$. But $B$ itself has more
than $3/4$ of its volume in the region where $u=1$, so
$$ A_b > A_s $$
for this case.

The sphere $\partial B'$ has, modulo $\epsilon$, $1/4$ of its area in the
region where $u = 1$, so that average is $1/4$. But $B'$ has less than
$1/4$ of its volume in that region, so
$$
 A_b < A_s
$$
for this case.

I don't know what happens in other dimensions.


问:
Confused on what a series means.
I have been reading the book Relativity: The special & General Theory where in chapter XV, the author develops the expression of kinetic energy
$$
\frac{1}{2} mv^2
$$
or
$$
\text{ }m\frac{v^2}{2}
$$
in the form of a series,
$$
mc^2+\text{ }m\frac{v^2}{2}+\frac{3}{8}m\frac{v^4}{c^2}+...
$$
Can someone explain to me what a series is? Or at the very least show me where I can find more information about this? Thank you.

答:
I'll explain what a series is based on my studies.

A series of general term $ a_{n} $ denoted $ \sum\limits_{n\geq 0}{a_{n}} $ is a mathematical object, precisely the couple formed by the sequence $ \left(a_{n}\right)_{n\geq 0} $ and the sequence of partial sums $ \left(\sum\limits_{k=0}^{n}{a_{k}}\right)_{n\geq 0} $ : $$ \left(\left(a_{n}\right)_{n\geq 0},\left(\sum\limits_{k=0}^{n}{a_{k}}\right)_{n\geq 0}\right) $$

Whenever the sequence $ \left(\sum\limits_{k=0}^{n}{a_{k}}\right)_{n\geq 0} $ converges to some $ \ell $, we say that the series $ \sum\limits_{n\geq 0}{a_{n}} $ is convergent, we call $ \ell $ the sum of the series, and we write : $$ \sum_{n=0}^{+\infty}{a_{n}}=\ell $$

If the sequence $ \left(\sum\limits_{k=0}^{n}{a_{k}}\right)_{n\geq 0} $ diverges, we say that the series $ \sum\limits_{n\geq 0}{a_{n}} $ is divergent.

This is basically the formal definition of a series.

OwnerDisplayName: anon

LastEditorDisplayName: anon

Title: Confused on what a series means.


问:
How to solve this integral equation? (if possible using functional analysis)
I have the following integral equation:
$$u(x)=1+2 \int_{0}^{\pi} u(t) \cos (x+t) d t,$$
and I need to solve it for $u$, this is an exercise from my course of functional analysis, but I have never seen before an integral equation, and I have no idea how to approach this. Can you help me here?

答:
You have
$$u''(x) = 2 \int_{0}^{\pi} u(t) \frac{\partial^2}{\partial x^2} \cos (x+t) dt =-2 \int_{0}^{\pi} u(t)  \cos (x+t) dt =-u(x)+1$$

$$\frac{d^2}{dx^2}(u(x)-1) = -u(x)+1 $$
The solution will be
$$u(x) = 1 + a \sin(x)+b\cos(x)$$
with $a,b\in \mathbb{R}$


问:
Related to Discontinuity Property of Floor Function
For each $x \in \mathbb R$ let $\lfloor x \rfloor := \max\{n ∈ \mathbb Z : n ≤ x\}$ and consider
$f: \mathbb R → \mathbb R$ such that
$$f(x) = \begin{cases}
0 & \textrm{when } x=0, \\
x^2\left\lfloor \tfrac1x \right\rfloor & \textrm{otherwise}.
\end{cases}$$

Prove that $f$ is discontinuous at all points of the form $x_0 = \tfrac1n$, with $n \in \mathbb Z \setminus \{0\}$.

If possible, can you please help solve it in epsilon-delta criterion? Thank you very much.

答:
hint

Assume $ n>1$.

$$\begin{align*}
x\in\left(\frac1n,\frac1{n-1}\right]&\implies n-1\le\frac1x<n\\
&\implies f(x)=x^2(n-1)\\
&\implies\lim_{x\to\frac1n^+}f(x)=\frac{n-1}{n^2}
\end{align*}$$

$$x\in\left(\frac{1}{n+1},\frac 1n\right]\implies $$
$$n\le\frac 1x<n+1 \implies$$
$$f(x)=x^2n\implies$$
$$\lim_{x\to\frac 1n^-}f(x)=\frac 1n$$

...


问:
Sequence of continuous functions with a bounded derivative and specific conditions
Let $\{f_n\}$ be a sequence of continuously differentiable functions on $[0,1]$ with $f_n(0) = f'_n(0)$ and $|f'_n(x)| \leq 1 \ \forall x \in [0,1]$ and $n \in N$. Show that if $\lim_{n \to \infty} f_n(x) = f(x) \ \forall x \in [0,1]$, then $f$ is continuous on $[0,1]$. Does the sequence have to converge? Must there be a convergent subsequence?

I've seen similar questions to this, but they all seem to have slightly different conditions.  Normally I like looking at similar problems, but the thing I'm having trouble with is the conditions. I'm just saying this because I know someone will mark this as a duplicate, but I do not believe it is if you read carefully. What I'm not sure about is how these conditions can be applied to show the things being asked. I'm just not sure how to start. If anyone could get me started it would be great!

Edit: Have continued to try this and still not having any luck. If anyone could help I'd really appreciate it! This is one of the last questions I'd like to figure out for my class!

答:
By the Mean Value Theorem, $|f_n(x)-f_n(y)|\leq|x-y|$ for all $x,y,n$, so taking the limit $n\to\infty$ yields $|f(x)-f(y)|\leq|x-y|$ as well for all $x,y$, and $f$ is continuous (actually, Lipschitz).

For an example of a sequence of such functions which does not converge, simply take $f_n(x)=(-1)^nx^2$. But yes - there will always be a subsequence of the $f_n$ which converges uniformly to some $f$ by Arzela-Ascoli: The argument above shows that all functions $f_n$ are Lipschitz, so the family $\left\{f_n\right\}_n$ is uniformly equicontinuous, and you can use the hypothesis that $|f_n(0)|=|f_n'(0)|\leq 1$ and the Mean Value Theorem again to show that these functions are uniformly bounded, so you can apply Arzela-Ascoli.

If you have not seen the Arzela-Ascoli Theorem, you can use the argument above with the Mean Value Theorem to show that $\left\{f_n(x)\right\}_n$ is bounded for all $x$ (in fact, $|f_n(x)|\leq 2$ for all $n$ and $x\in[0,1]$). Then use this fact for rational $x$, and take a diagonal subsequence which makes $f_n(x)$ converge to some $f(x)$ for all rational $x$. This function $f$ -- defined over the rationals in $[0,1]$ -- will be Lipschitz, so it extends uniquely to a Lipschitz function on all of $[0,1]$. Using the Lipschitz property again for both $f_n$ and $f$ you can actually show that $f_n\to f$ pointwise on all of $[0,1]$.


问:
Question about identifying pairs of edges of disjoint $2$ simplices
This exercise $2.1.10$ in page $131$ of Hatcher's book Algebraic topology.

(a) Show the quotient space of a finite collection of disjoint $2$-simplices obtained by identifying pairs of edges is always a surface, locally homeomorphic to $\mathbb{R^2}$.

(b) Show the edges can always be oriented so as to define a $\Delta$-complex structure on the quotient surface. [author: This is more dificult.]

I've done the 1st one, but I got stuck while solving the 2nd part. Currently I've no clue for this.

I am not able to find any kind of algorithm which fits for it for any arbitrary $n$. For example, if we take this operation using two 2-simplex as a result of various quotien,t we can get different spaces like torus, Klein bottle, projective plane etc...and for all this spaces the ordering will be different.

So as a result I cannot guess the algorithm and I need some serious help, a way of thinking. Thank you.

答: The resulting $\Delta$-complex does not need to be orientable as shown by your examples. If you start from the ordered list of vertices, edges created from identified pairs of edges would each have two faces (vertices), possibly identified as one vertex in the final space. You can orient each non-loop edge by pointing from the vertex with the smaller index to the other. For any loop edge, you can make its orientation consistent with the orientation of the original 2-simpler with a smaller index (assuming that you also order the original 2-simplices).


Title: Confused on what a series means.

Body:
I have been reading the book Relativity: The special & General Theory where in chapter XV, the author develops the expression of kinetic energy
$$
\frac{1}{2} mv^2
$$
or
$$
\text{ }m\frac{v^2}{2}
$$
in the form of a series,
$$
mc^2+\text{ }m\frac{v^2}{2}+\frac{3}{8}m\frac{v^4}{c^2}+...
$$
Can someone explain to me what a series is? Or at the very least show me where I can find more information about this? Thank you.

Body:
With respect to the original formula,and

usually $v<<c$
$$E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}=\\mc^2(1-\frac{v^2}{c^2})^{-\frac12}\\
\sim mc^2 (1-(-\frac12)\frac{v^2}{c^2}+...)\\=mc^2+\frac 12mv^2+...$$


问:
Subrings of $\mathbb{Q}$
Let $p$ be prime. Suppose $R$ is the set of all rational numbers of the form $\frac{m}{n}$ where $m,n$ are integers and $p$ does not divide $n$.

Clearly then $R$ is a subring of $\mathbb{Q}$.

I now want to show that if $\frac{m}{n}$ belongs to any proper ideal of $R$ then $p|m$.

Can someone point me in the right direction.

Thanks

答: Hint:  What happens if you assume that $m$ is divisible by $p$?</p>

ContentLicense: CC BY-SA 3.0

Title: Subrings of $\mathbb{Q}$


问:
Kernel and scalar product
I need to prove that $k(x,y)=f(x)f(y),\ f: \mathbb R^d \rightarrow \mathbb R$ continuous function and $x,y \in \mathbb R^d$, is a Mercer kernel, i.e. for any $x_1,...x_n \in \mathbb R^d$ and any $c_1,...c_n\in \mathbb R,\, $ the inequality holds $$\sum_{i=1}^n \sum_{j=1}^n c_ic_jk(x_i, x_j) \geq 0,$$
whereby $k:\mathbb R^d$x $ \mathbb R^d \rightarrow \mathbb R$ represents the kernel.

In the case $f$ is linear, we could define the scalar product $<x_i,x_j>:=f(x_i)f(x_j).$ Given that the scalar product is a positive definite symmetric bilinear form, the inequality will hold. I am having difficulty in considering non linear continuous functions $f$.

Can somebody provide an idea or a solution proposal ?

答: $\sum_i\sum_j c_ic_jk(x_i,x_j)=\sum_i\sum_j c_ic_j f(x_i)f(x_j)=(\sum_i c_i f(x_i))(\sum_j c_j f(x_j))=(\sum_k c_k f(x_k))^2$.

Title: Kernel and scalar product


问:
Strong minimality of theory of a model that is minimal and $\omega$-saturated
I want to show that if a model $M$ is minimal and $\omega$-saturated, then the Theory($M$) is strongly minimal.

I know that the reverse statement would be trivial. However, I'm not sure about the forward statement. More specifically, how is it that just the minimality and $\omega$-saturatedeness of one model result in the whole theory being strongly minimal.

答: Hint: If $T = \text{Th}(M)$ is not strongly minimal, then there is some model $N\models T$ and some formula $\varphi(x;\bar{b})$ with $\bar{b}$ from $N$ such that $\varphi(N;\bar{b})$ is infinite and coinfinite. Now use the fact that $M$ is $\omega$-saturated to find such a witness in $M$ (it is enough to realize $\text{tp}(\bar{b}/\emptyset)$ in $M$!), contradicting minimality of $M$.


问:
Is $\operatorname{Aut}(\mathbb{Z}_3\times \mathbb{Z}_3)$ isomorphic with $\operatorname{Aut}(\mathbb{Z}_3)\times \operatorname{Aut}(\mathbb{Z}_3)$?
Is  $\operatorname{Aut}(\mathbb{Z}_3\times \mathbb{Z}_3)$ isomorphic with $\operatorname{Aut}(\mathbb{Z}_3)\times \operatorname{Aut}\left(\mathbb{Z}_3\right)$?

From the lecture I know that $\operatorname{Aut}\left(G\times H\right)\simeq \operatorname{Aut}\left(G\right)\times \operatorname{Aut}\left(H\right)$ if gcd$(G,H)=1$ but I don't know what I can use when $G=H$ so $\gcd(G,H)\neq1$.

Can I have any tips?

答:
No. Note that we have $\text{Aut}(\mathbb{Z_3}) \cong \mathbb{Z}_2$ so we can safely write
$$\text{Aut}(\mathbb{Z_3}) \times \text{Aut}(\mathbb{Z_3}) \cong \mathbb{Z}_2 \times \mathbb{Z}_2$$
However, if we consider an automorphism $\sigma$ in $\text{Aut}(\mathbb{Z_3} \times \mathbb{Z_3})$ with $\sigma(1,0) = (0,2)$ and $\sigma(0,1) = (1,0)$, we have $o(\sigma) = 4$ but there is no element of order $4$ in $\mathbb{Z}_2 \times \mathbb{Z}_2$.


问:
$ A: \{\frac{m^2n^3 + 3m^2n^2+2mn^3+6mn^2}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}, n, m \in \mathbb{N} \}$ Got limitation need superemum.
I need to find supremum of:

$$ A: \{\frac{m^2n^3 + 3m^2n^2+2mn^3+6mn^2}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}, n, m \in \mathbb{N} \}$$

I found out that:

$$ \frac{m^2n^3 + 3m^2n^2+2mn^3+6mn^2}{(m^2 + 2m)^2+(n^3 + 3n^2)^2} = \frac{m^2(n^3 + 3n^2) + 2m(n^3+3n^2)}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}= \frac{(m^2+2m)(n^3 + 3n^2)}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}$$

$$\frac{(m^2+2m)(n^3 + 3n^2)}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}= \frac{\sqrt{(m^2+2m)^2(n^3 + 3n^2)^2}}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}$$
And from inequality of means (AM - GM) I know that:

$$\frac{\sqrt{(m^2+2m)^2(n^3 + 3n^2)^2}}{(m^2 + 2m)^2+(n^3 + 3n^2)^2} \leq \frac{1}{2}$$

Therefore I found a limitation of my set A. Now I need a sequence of elements belonging in $A$ that would be convergent to $\frac{1}{2}$. That is the part I can't do myself.

From graph I know that if I take $m = n$ then for $n \in [1 ; \infty)$ I get decreasing function and for that function $\frac{1}{2}$ is reached by $n < 1$. A maximal value of that function for $n \in  [1 ; \infty)$ is equal to $\frac{12}{25}$ and is reached by $n = 1$. But that is not a solution.

There is no ANY fact that would legitimise value $\frac{12}{25}$ as the supremum of that set since there always might by something bigger if $n \neq m$.

What I need is a prove that in the supremum of set $A$: $m = n$ (I don't know why that would be the case) or that there is a bigger value reached for $n$ and $m$ that are not equal.

It might be important to add that I can not use integrals nor derivatives.

答:
We pick a sub-sequence of $A$ in the following fashion: $n=1, 2, \cdots, $, and $m$ depends on $n$ via $m=\lfloor \sqrt{a_n+1} \rfloor$.

Define $a_n = n^3+3n^2, b_m=m^2+2m=(m+1)^2-1=\lfloor\sqrt{a_n+1}+1 \rfloor^2-1$. Then $\frac{a_n b_m}{a_n^2+b_m^2} \in A$ and we will prove that $\lim_{n\to \infty} \frac{a_n b_m}{a_n^2+b_m^2} \to \frac 12.$

Note that $x-1 < \lfloor x \rfloor \le x, \forall x$, then

$$a_n=(\sqrt{a_n+1})^2-1 < \lfloor \sqrt{a_n+1}+1\rfloor^2-1 \le (\sqrt{a_n+1}+1)^2-1 \\
\implies 1 < \frac{b_m}{a_n} \le \frac{(\sqrt{a_n+1}+1)^2-1}{a_n} = 1 + \frac{1}{a_n}+\frac{2\sqrt{a_n+1}}{a_n}$$

Therefore $\lim_{n\to\infty} \frac{b_n}{a_n} = 1 \implies \lim_{n\to\infty} \frac{a_n b_n}{a_n^2+b_n^2} = \frac 12. \blacksquare$


问:
Non-Noetherian ring with a single prime ideal

What are the most simple examples of a commutative ring $R$ satisfying both of the following two properties:

  1. $R$ is not Noetherian.

  2. $R$ has exactly one prime ideal.

答:
$$R:= k[x_1, x_2, x_3, \cdots ]/\langle x_1^2,\ x_2^2=x_1,\ x_3^2=x_2,\ x_4^2=x_3,\ \cdots \rangle$$

Remember that prime ideals are precisely the kernels of maps to fields. If $K$ is any field, and we have a map $\phi: R \to K$, then $\phi(x_1)=0$ and hence $\phi(x_2) =0$ and hence $\phi(x_3)=0$ and so forth, so the only prime ideal is $\langle x_1, x_2, x_3, \ldots \rangle$.

To see that this is not noetherian, note that $(0) \subsetneq (x_1) \subsetneq (x_2) \subsetneq (x_3) \subsetneq \cdots$ is an ascending chain of ideals that doesn't terminate.

Of course, the specific exponent $2$ isn't important. The key is to make a sequence of variables $x_1$, $x_2$, ..., all of which are nilpotent, but so that you can kill $x_1$, $x_2$, ..., $x_k$ while not killing the later $x$'s.


问:
solve $y(x)=\cos \left(y'(x)\right) + y'(x)\sin (y'(x)), y(0)=1$
solve $$y(x)=\cos (y'(x)) + y'(x)\sin (y'(x)), y(0)=1$$

with wolfram alpha I got that a solution is $y(x)=x\arcsin x+\cos (\arcsin x)$

but I have no idea how to find it.

I tried transforming into an exact equation by letting $u=y'$ then I get $y=\frac{u^2}{2}$

I just realised this is wrong (because $\frac{d}{dx}\frac{u^2}{2}=u'u\neq y'$) so there's no point posting the rest of my work.

any hint?

答: Hint:  Apply the chain rule of calculus.</p>

Body: Hint:  Apply the chain rule of calculus.


问:
$u_t+u^2u_x=0$ with somewhat arbitrary initial conditions
I have the following PDE with initial values as stated.
$$u_t+u^2u_x=0\hspace{1 cm} t>0, -\infty<x<\infty$$
$$u(x,t=0)=\begin{cases}
      -a & x< 0 \\
      0.5 &  x\geq 0
   \end{cases}$$ where $a>0$.

I know that we can solve this using method of characteristics so we have
$$\frac{dX}{dT}=u^2=\begin{cases}
      a^2 & x< 0 \\
      0.5^2 &  x\geq 0
   \end{cases}$$
and from this I know that when $a>0.5$ the characteristic in the middle will be a shock wave with a certain shock speed defining the slope of that shock characteristic. For this, I found $x=\frac{A(u^+)-A(u^-)}{u^{+}-u^-}t=\frac{0.5^2-0.5a+a^2}{3}t$ to be the characteristic line for the shock. I am not sure if this is correct for the characteristic lines in the fan region?

Also, when $a<0.5$ the characteristics between the two sides will be like a fan and we can find the solution for $u(x,t)$ in this fan portion, but I have no idea how I can find a $u(x,t)$ that satisfies this PDE inside of this fan portion of the characteristics. Can someone please help me find this?

答:
The problem at hand is a Riemann problem for the conservation law $u_t+f(u)_x =0$ where the flux function $f:u\mapsto \frac13 u^3$ is non-convex. Here, we have the value $-a<0$ on the left side of the initial discontinuity, and the value $0.5>0$ on the right side. Following this post, the entropy-satisfying solution might be of the following type:

• if $a>1$, then the solution is a shock wave with speed $\frac1{3}(a -\frac12)^2$.

• if $0< a\leq 1$, then the solution is a semi-shock, with intermediate state $u^*=a/2$ satisfying $0< u^*\leq \frac12$.


答:
As the other answer is rather formal, let me provide my own short one, dealing with the problem at hand.

The exact expression for the relativistic kinetic energy is:

$$K=\frac{mc^2}{\sqrt{1-v^2/c^2}}-mc^2$$

That is, total energy minus the rest energy.

In the non-relativistic limit we have the following assumption:

$$v^2 \ll c^2$$

Or:

$$\beta=\frac{v^2}{c^2} \ll 1$$

Turns out that there's a useful sequence of approximations that converge to the square root, provided that $\beta$ is small in absolute value as indicated above.

$$\sqrt{1+\beta}=1+\frac{\beta}{2}-\frac{\beta^2}{8}+\cdots$$

Where $\cdots$ should be replaced by as many terms as it takes to get the desired accuracy.

This is a case of so called Taylor series, and I encourage the OP to look it up.

In this particular case, it's enough to take only the first term, which gives us:

$$K \approx mc^2 \left( \frac{1}{1-v^2/(2c^2)}-1 \right)$$

But this is still too complicated.

However, we know another Taylor series, or rather a simple case of geometric progression sum:

$$\frac{1}{1-\beta} = 1+\beta+\beta^2+\cdots$$

Again, it's enough to use the first term, which gives us:

$$K \approx mc^2 \cdot \frac{v^2}{2c^2}=\frac{mv^2}{2}$$

For what it's worth, if you don't want to get too formal, a convergent series is a sequence of sums with increasing number of terms that gives a good approximation to some finite value. Usually, we can even estimate the error of the approximation.

Example:

$$\frac{\pi}{4}=1-\frac13+\frac15-\frac17+\cdots$$

the famous Leibniz series for Pi.

A good estimation for the error is half the last term.

Title: Confused on what a series means.


问:
Don't know how to solve a question regarding sets
The next definition is giving in the book the exercise was proposed on.

The definition reads:

Let $a=n(A)$ and $b=n(B)$ where $A$ and $B$ are two disjoint finite sets. Then $a+b=n(A∪B).$

I understand the fact of A and B being disjoint is a must. Yet, what should I do in case A and B have at least an element in common?

I was proposed an exercise in which I am to compute the cardinal of two non-disjoint sets given. But the definition on my mathbook does not cover non-disjoint sets. What am I supposed to do?

答:
The reason the formula $|A \cup B| = |A| + |B|$ is only applicable to the case when $A$ and $B$ are disjoint is that it would double-count the elements in the intersection $A \cap B$. Thus, it can be generalized to the case of not necessarily disjoint sets as

$$
|A \cup B| = |A| + |B| - |A \cap B|\tag1.
$$

This can be seen by considering the three possible cases for $x \in A \cup B$:

$x \in A$ and $x \notin B$

$x \notin A$ and $x \in B$

$x \in A$ and $x \in B$

and verifying that $x$  is counted exactly once in $(1)$.

This reasoning is called the inclusion-exclusion principle and can be generalized to more than two sets.


问:
Having trouble of reaching a specific expression for Maxwell's rheological model with 2 modes from a given formula.
I have an expression for a Maxwell viscoelastic rheological model with 2 modes.

I have tried derivation for both sides but I always seem to reach an impasse.
I tried to get second derivation as well but im missing some terms that should have stayed behind.
I know have to get the second derivation of the original expression to get to the first (I CAN SEE THAT) but i cant really get it to match.

Any opinions, inputs on the matter?

Tau is a stress tensor thus irrelevant with time and λ1, λ2 are relevant as they are relaxation times.

original expression
$\boldsymbol{\tau}=\int_{-\infty}^{t}\left(\mathrm{G}_{1} \mathrm{e}^{-\left(t-t^{\prime}\right) / \lambda_{1}}+\mathrm{G}_{2} \mathrm{e}^{-\left(t-t^{\prime}\right) / \lambda_{2}}\right) \dot{\gamma}\left(\mathrm{t}^{\prime}\right) \mathrm{dt}^{\prime}$

final expression $\boldsymbol{\tau}+\left(\lambda_{1}+\lambda_{2}\right) \frac{\partial \boldsymbol{\tau}}{\partial \mathrm{t}}+\lambda_{1} \lambda_{2} \frac{\partial^{2} \boldsymbol{\tau}}{\partial \mathrm{t}^{2}}=\left(\eta_{1}+\eta_{2}\right)\left[\dot{\gamma}+\left(\frac{\lambda_{2} \eta_{1}+\lambda_{1} \eta_{2}}{\eta_{1}+\eta_{2}}\right) \frac{\partial \dot{\boldsymbol{\gamma}}}{\partial t}\right]$

For

$\eta_{1}=\lambda_{1} *{G}_{1}$

$\eta_{2}=\lambda_{2} *{G}_{2}$

答:
Let $\phi_1(t) = \int_{-\infty}^{t}\mathrm{G}_{1} \mathrm{e}^{-\left(t-t^{\prime}\right) / \lambda_{1}} \dot{\gamma}\left(\mathrm{t}^{\prime}\right) \mathrm{dt}^{\prime}$ and  $ \phi_2(t) = \int_{-\infty}^{t}\mathrm{G}_{2} \mathrm{e}^{-\left(t-t^{\prime}\right) / \lambda_{2}} \dot{\gamma}\left(\mathrm{t}^{\prime}\right) \mathrm{dt}^{\prime}$.

Using Leibniz's rule $\frac{\partial }{\partial t} \int_{-\infty}^t f(t,t') \, dt' =f(t,t) + \int_{-\infty}^t \frac{\partial }{\partial t}f(t,t') \, dt'$, we get for $j = 1,2$,

$$\frac{\partial \phi_j}{\partial t} = G_j\dot{\gamma}\left(\mathrm{t}\right)- \frac{1}{\lambda_j}\phi_j(t),\quad  \frac{\partial^2 \phi_j}{\partial t^2} = G_j \frac{\partial \dot{\gamma}}{\partial t} + \frac{1}{\lambda_j^2}\phi_j(t)$$

Thus,

$$\tau = \phi_1 + \phi_2 ,\\\frac{\partial \tau}{\partial t} = \frac{\partial \phi_1}{\partial t}+ \frac{\partial \phi_2}{\partial t} = G_1\dot{\gamma}+ G_2 \dot{\gamma} - \frac{\phi_1}{\lambda_1} - \frac{\phi_2}{\lambda_2},$$

and
$$\frac{\partial^2 \tau}{\partial t^2} = \frac{\partial^2 \phi_1}{\partial t^2}+ \frac{\partial \phi_2^2}{\partial t^2} = G_1\frac{\partial \dot{\gamma}}{\partial t}+ G_2\frac{\partial \dot{\gamma}}{\partial t } - \frac{1}{\lambda_1}\frac{\partial \phi_1}{\partial t} - \frac{1}{\lambda_2}\frac{\partial \phi_2}{\partial t} = \\G_1\frac{\partial \dot{\gamma}}{\partial t}+ G_2\frac{\partial \dot{\gamma}}{\partial t } - \frac{G_1 \dot{\gamma}}{\lambda_1}+  \frac{\phi_1}{\lambda_1^2}- \frac{G_2 \dot{\gamma}}{\lambda_2}+  \frac{\phi_2}{\lambda_2^2}$$

We then have

$$(\lambda_1+ \lambda_2)\frac{\partial \tau}{\partial t}  = -(\phi_1 + \phi_2) - \frac{\lambda_2}{\lambda_1} \phi_1 - \frac{\lambda_1}{\lambda_2} \phi_2 + (\lambda_1+\lambda_2)(G_1+G_2)\dot{\gamma} ,\\\lambda_1 \lambda_2 \frac{\partial^2 \tau}{\partial t^2} = \lambda_1\lambda_2(G_1 + G_2)\frac{\partial \dot{\gamma}}{\partial t } +  \frac{\lambda_2}{\lambda_1} \phi_1 + \frac{\lambda_1}{\lambda_2} \phi_2-  \lambda_2G_1 \dot{\gamma} - \lambda_1G_2 \dot{\gamma}$$

Adding terms, we get
$$ \tau + (\lambda_1+ \lambda_2)\frac{\partial \tau}{\partial t}+ \lambda_1 \lambda_2 \frac{\partial^2 \tau}{\partial t^2} \\=  (\lambda_1 G_1+ \lambda_1G_2 + \lambda_2G_1 + \lambda_2 G_2) \dot{\gamma} - (\lambda_2G_1 + \lambda_1 G_2) \dot{\gamma} + (\lambda_2 \lambda_1 G_1 + \lambda_1 \lambda_2 G_2) \frac{\partial \dot{\gamma}}{\partial t} \\ = (\lambda_1G_1+ \lambda_2 G_2) \dot{\gamma} +  (\lambda_2 \lambda_1 G_1 + \lambda_1 \lambda_2 G_2) \frac{\partial \dot{\gamma}}{\partial t}\\ = (\eta_1 + \eta_2) \dot{\gamma} + (\lambda_2 \eta_1 + \lambda_1 \eta_2)\frac{\partial \dot{\gamma}}{\partial t} = (\eta_1 + \eta_2)  \left[\dot{\gamma} +  \frac{\lambda_2 \eta_1 + \lambda_1 \eta_2}{\eta_1+ \eta_2}\frac{\partial \dot{\gamma}}{\partial t}\right]$$


问:
How many edge-disjoint Hamilton cycles are there in the complete graph $K_{n}$ for $n$ odd
I was originally working on a problem that asked to find all edge-disjoint Hamilton cycles of $K_{21}$.

My first idea was to make an arbitrary cycle:

$v_{21}-v_{20}-v_{19}-v_{18}-v_{17}-v_{16}-v_{15}-v_{14}-v_{13}-v_{12}-v_{11}-v_{10}-v_{9}-v_{8}-v_{7}-v_{6}-v_{5}-v_{4}-v_{3}-v_{2}-v_{1}$

There are $21$ edges, since we are looking at edge-disjoint, we can't use those edges anymore, thus there are $210-21=189$ edges left to make more Hamilton cycles. For every Hamilton cycle, I use up $21$ edges each time, thus at most $10=\frac{(21-1)}{2}$ cycles?

Is there a better method for the general case $n$ odd?

I tried with some simple cases like $|V|=5,7$

For $n$ even, this is covered here.

答:
That is correct.  The result is mentioned in the question you link to.  Each vertex of $K_n$ has $n-1$ edges incident on it.  Each cycle uses up two of those, so if you can find $\frac {n-1}2$ disjoint Hamiltonian cycles you have used up all the edges.  That is clearly an upper bound.

To prove that the upper bound can be achieved you need to find that many cycles that do not reuse an edge.  If $n$ is prime you can just have one cycle where each vertex connect to the one before and the one after in your list, one cycle where they connect two before and two after $\left(\bmod n\right)$, one where they connect three before and three after, and so on.  For composite $n$ you need to be a little more clever because when you try to add a number not coprime to $n$ you will come back to where you started before you have visited all the vertices.  You can do this by starting that way and getting a number of loops, then connecting the loops.


问:
Understanding a coin tossing problem
The question: If we toss a coin 4 times, what are the chances of seeing 2 heads?

The approach: Introducing a random variable $X, S_{x}=\{0,1,2,3,4\}$ would easily show us that $P(X=2)=\frac{\binom{4}{2}}{2^4}=0.375$ (basically the PMF of a binomial distribution).

Now the problem is that my intuition, considering the 50% chance of seeing a head in a single toss, suggests a 50% chance of seeing head/tails being evenly split regardless of the number of tosses.

Obviously my intuition is wrong as the chance of seeing head/tails being evenly split decreases as we increase the number of tosses. So can someone explain the situation so that I understand what I'm "doing" wrong?

答: There are many more possible results when you throw more coins.  If you throw $100$ coins, the most likely result is $50$ heads, but $49$ heads and $51$ heads are not far behind.  Clearly $50$ cannot be half the time, or the sum of the three would be greater than $1$.

Title: Understanding a coin tossing problem


问:
A simple proof for Glasser: $\int_{-\infty}^{\infty} f(x-a/x) dx=\int_{-\infty}^{\infty} f(x) dx, a>0$
$$\int_{-\infty}^{\infty} f(x-a/x) dx=\int_{-\infty}^{\infty} f(x) dx~~~~(1)$$
Several difficult integral can be handled using (1) easily. For instance we know that $$\int_{-\infty}^{\infty} \text{sech}^2 x~dx=2~~~~~~~~~~~~(2)$$ but what is interesting is that one may verify numerically that $$\int_{-\infty}^{\infty} \text{sech}^2(x-a/x)~dx=2, a>0.~~~~~~~~~(3)$$ This interesting property (1) follows from Glasser's Master Theorem. See one more interesting integral and its solutions in MSE post:s
https://en.wikipedia.org/wiki/Glasser%27s_master_theorem.

In (1) what is also interesting is that LHS is independent of $a(>0)$. I have tried proving (1) or even (3) by the ordinary rules of integration ( Real Analysis) without a success. Can someone help me?

EDIT: One more friendly (doable by hand) twin is
$$\int_{-\infty}^{\infty} \frac{dx}{1+x^2}=\pi=\int_{-\infty}^{\infty} \frac{dx}{1+(x-a/x)^2} dx , a >0$$

答:
Note
\begin{align}
\int_{-\infty}^{\infty} f\left(x-\frac ax \right) dx
&= \int_{-\infty}^{0}
\overset{x=-\sqrt a e^{-t}} {f\left(x-\frac ax \right) dx }+ \int_{0}^{\infty} \overset{x=\sqrt a e^t}{f\left(x-\frac ax \right) dx}\\
&= \int_{-\infty}^{\infty} \underset{ x=\sqrt a (e^t-e^{-t})}{f[\sqrt a(e^t-e^{-t})]\sqrt a(e^t +e^{-t})} dt =\int_{-\infty}^{\infty} f(x) dx
\end{align}


问:
Proof that the set of Pochhammer numbers satisfies Benford's law
Consider the set $S_x$ of the following Pochhammer numbers:

$$(x)_n := \frac{\Gamma(x+n)}{\Gamma(x)}\,, \tag{1}$$

with the gamma function:

$$\Gamma(n) := (n-1)!\,. \tag{2}$$

From "experiment" I noticed that the distribution of digits in the sets defined above satisfies Benford's law (see in particular the generalization to digits beyond the first). Consider e.g.:

$$S_1 = \left\lbrace 1,2,6,24,120,720,5040, \ldots \right\rbrace\,. \tag{3}$$

Is it possible to prove non-experimentally that such a set satisfies Benford's law? I am okay with using scaling invariance if it helps.

答:
By definition, Benford's law is honoured when the fractional part of a positive random variable's logarithm is uniformly distributed. This is one of those "often approximately true" results, like a large sample's mean being almost Normal.

You mentioned scaling invariance. If we measured lengths over orders of magnitude, our Bayesian prior for the fractional part will be the same whether we work in feet or yards, and switching between these scales the raw value and shifts its logarithm. The fractional part "wraps around", implying uniformity (see also here).

When $n=1$, your experiment looks at factorials; when $n=2$, it looks at them again, but with the sequence shifted; with any larger fixed value of $n$, the sequence is also rescaled. So we expect the distribution not to depend on $n$, which forces it to (approximately) follow Benford's law.


问:
Example of non-separable subspace of a separable Hausdorff space
The following question was asked in my topology assignment and I was unable to solve it, so I am looking for help here.

Example of non separable subspace of a separable hausdorff space.

I couldn't construct such example and I looked for an answer by searching mathstackexchange and found this answer :Density character of a subspace of a topological space.

But this is beyond my understanding as I  have just started 1st course in topology (2 weeks back) and I am unable to understand that terminology.

So, kindly give a answer which is for a person that has just started learning topology. I am following Lecture notes and Assignment sheets of the course and not any textbook.

Thank you!

答:
bof has given you the useful example of the Michael line; here are two more examples that are also useful examples of a variety of things.

The Sorgenfrey plane $\Bbb S^2$ is the product of two copies of the Sorgenfrey line $\Bbb S$, which is $\Bbb R$ with the topology generated by the base $\big\{[a,b):a,b\in\Bbb R\text{ and }a<b\big\}$. (The Sorgenfrey line is also known as the reals with the lower-limit topology.) The topology on $\Bbb S^2$ is finer than the usual topology on $\Bbb R^2$, so it is Hausdorff. $\Bbb Q\times\Bbb Q$ is a countable dense subset of $\Bbb S^2$, which is therefore separable. But $\{\langle x,-x\rangle:x\in\Bbb R\}$ is an uncountable discrete subspace of $\Bbb S^2$, so it is not separable.

The Niemytzki plane is another useful example. Let $X=\{\langle x,y\rangle\in\Bbb R^2:y\ge 0\}$, the closed upper half-plane. Let $p=\langle x,y\rangle\in X$.

If $y>0$, $\{B(p,r):r\le y\}$ is a local base at $p$, where $B(p,r)$ is the usual open ball of radius $r$ centred at $p$. (In other words, we take as basic open nbhds of $p$ the usual open balls that lie entirely above the $x$-axis.)

If $y=0$, the basic open nbhds of $p$ are the sets of the form $\{p\}\cup B(\langle x,r\rangle,r)$ for $r>0$, consisting of $p$ and an open disk tangent to the $x$-axis at $p$. (This is why the space is also known as the tangent disk space.)

It’s straightforward to check that $X$ is Hausdorff, and clearly $X\cap(\Bbb Q\times\Bbb Q)$ is a countable dense subset of $X$, so $X$ is separable. However, the $x$-axis is a discrete subspace of $X$ that is uncountable and therefore not separable.

Two other important examples are noted in this answer: $\beta\Bbb N$, the Čech-Stone compactification of the natural numbers, and the Mrówka space $\Psi$.


问:
Showing that $\sum_{n=3}^\infty\frac1{n(\log n)(\log\log n)}$ diverges
I know that the series diverge, I'm just having hard time showing it.

$$\sum_{n=3}^\infty\frac1{n(\log n)(\log\log n)}$$

Thanks in advance

答: HINT: Use the integral test. There’s a nice $u$-substitution available to handle the integral.</p>


问:
Closed form expression for the given map?
Say we have a map that sends an intger to a finite tuple as follows:

$3 \rightarrow (3,2,4,1,5,2,4,3)$

$4 \rightarrow (4,3,5,2,6,1,7,2,6,3,5,4)$

and in general:

$n \rightarrow (n,n-1,n+1,n-2,n+2,\cdots,1,2n-1,2,2n-2,\cdots,n-1,n+1,n)$

Is there a more compact way I could write this? Thank you.

答:
Let $a_k$, $k=0,\ldots,4n-5$ be your tuple. This expression should do the job:

$$
a_k =
\begin{cases}
-\frac{\lvert k-2n+2 \rvert}{2} + 2n-1,  & \text{if $k$ is even} \\
\frac{\lvert k-2n+3 \rvert}{2} + 1,  & \text{if $k$ is odd} \\
\end{cases}
$$

Or, if you like a more convoluted form:

$$
a_k = (-1)^{k+1}\frac{\lvert k-2n+2+\frac{1-(-1)^k}{2}\rvert}{2}+2n-1-(1-(-1)^k)(n-1) $$

I checked the cases $n=3$ here and $n=4$ here.


答:
For every integer $n \geq 2$, the homomorphisms of $\mathbf Z/(n) \times \mathbf Z/(n)$ to itself are the linear maps given by $2 \times 2$ mod $n$ matrices $A = (\begin{smallmatrix}a&b\\c&d\end{smallmatrix})$, which act on $\mathbf Z/(n) \times \mathbf Z/(n)$ by $A\binom{x}{y} = \binom{ax+cy}{bx+dy}$.  This homomorphism is invertible if and only if the matrix is invertible, which means $\det A$ is invertible mod $n$. Such $A$ form the group ${\rm GL}_2(\mathbf Z/(n))$ of invertible $2 \times 2$ mod $n$ matrices. So ${\rm Aut}(\mathbf Z/(n) \times \mathbf Z/(n)) = {\rm GL}_2(\mathbf Z/(n))$.

By similar reasoning, for all $d \geq 1$ we have ${\rm Aut}(({\mathbf Z}/(n))^d) = {\rm GL}_d(\mathbf Z/(n))$, where $d \times d$ mod $n$ matrices act on vectors in $(\mathbf Z/(n))^d$ in the natural way.

A very simple reason ${\rm Aut}(G \times G)$ is not pairs of automorphisms of $G$, when $G$ is a nontrivial group, is that there is always the swap automorphism $f(g,g') = (g',g)$.

Title: Is $\operatorname{Aut}(\mathbb{Z}_3\times \mathbb{Z}_3)$ isomorphic with $\operatorname{Aut}(\mathbb{Z}_3)\times \operatorname{Aut}(\mathbb{Z}_3)$?

Body:
Is  $\operatorname{Aut}(\mathbb{Z}_3\times \mathbb{Z}_3)$ isomorphic with $\operatorname{Aut}(\mathbb{Z}_3)\times \operatorname{Aut}\left(\mathbb{Z}_3\right)$?

From the lecture I know that $\operatorname{Aut}\left(G\times H\right)\simeq \operatorname{Aut}\left(G\right)\times \operatorname{Aut}\left(H\right)$ if gcd$(G,H)=1$ but I don't know what I can use when $G=H$ so $\gcd(G,H)\neq1$.

Can I have any tips?


问:
Applying Green's Theorem to evaluate line integral.
I have to evaluate the line integral of the vector field
$$F(x,y)=\left(\frac{-2y}{x^2+y^{2}/4}+2x,\frac{2x}{x^{2}+y^{2}/4}+xy+1\right)$$
through $\gamma$ where $\gamma$ is the ounterclockwise unit circle.

I just want to know if I can use Green's Theorem here; I don't remember if I could apply it directly due to the fact that, if $x=y=0$ then the denominator above is equal to zero.

答: Green's theorem requires that the function $F$ has partial derivatives which exist and are continuous over the entire interior of the interior of $\gamma$.  Since the partial derivatives of $F$ do not exist at $(0,0)$, this means that the conditions required for Green's theorem do not hold.


问:
Number array divided into several parts, genelize $a>b>c>d>0$ so $ab+cd>ac+bd$ to more numbers
Now, we have an original number array: $$a_1 > a_2 > a_3 > ... > a_{mn} > 0$$, I wonder whether the following inequality is the truth, if so, could you give me the proof or some instruction?

inequality:

$$
a_1 \times a_2 \times \cdots \times a_n + a_{n+1} \times a_{n+2} \times \cdots \times a_{2n} + \cdots + a_{i \times n+1} \times a_{i\times n+2} \times \cdots \times a_{i\times n+n} + \cdots + a_{(m-1)  n+1} \times a_{(m-1)  n+2} \times \cdots \times a_{(m-1)  n+n} > a_1 \times a_{1+m} \times \cdots \times a_{1+(n-1)m} + a_2 \times a_{2+m} \times \cdots \times a_{2+(n-1)m} + \cdots + a_i \times a_{i+m} \times \cdots \times a_{i+(n-1)m} + \cdots + a_m \times a_{m+m} \times \cdots \times a_{m+(n-1)m}
$$
That is,
$$
\sum_{i=1}^m\prod_{j=1}^n a_{j+n(i-1)} > \sum_{i=1}^m\prod_{j=1}^n a_{i+m(j-1)}
$$

Namely, the left partitions the number array into $m$ parts each of which has $n$ numbers and the numbers in each part are next to each other in the original array. While, the right partitions the number array into $m$ parts each of which has $n$ numbers and the adjacent two numbers in each part have a distance of $m$ in original array.

For example, we have six numbers: $$a>b>c>d>e>f>0$$, we want to partition it into 3 blocks, each block has 2 numbers, namely $m=3, n=2$, I want to know whether $$ab+cd+ef>ad+be+cf $$and prove it. Certainly, this is easy. The question is I want to generalize it to $m\times n$ numbers divided into $m$ blocks.

Thank you!

答:
In general, given $a_1>a_2>a_3>...>a_{mn}>0$ and from all the ways to partition the number array into m parts each of which has n numbers, the maximum value of the sum of product of each part will be $\sum\limits_{i=1}^m\prod\limits_{j=1}^na_{n(i-1)+j}$.

To prove it, we need to find a series of inequality which shows that the sum of product of any other partition is smaller than this one.

First, given $a_k>a_l$ and sum=$a_k\prod\limits_{p}a_p+a_l\prod\limits_{q}a_q$. If we exchange $\prod\limits_{p}a_p$ and $\prod\limits_{q}a_q$, will the sum become larger? It depends on the products, if $\prod\limits_{p}a_p<\prod\limits_{q}a_q$, exchanging them will increase the sum.

Now we need to find a way to put $a_1,a_2...a_n$ together. First we find the part which contains $a_1$, simply call it part X. Then we find the part which contains $a_2$. If it's also part X, we look for $a_3$ and so on. If it's a different part, let's call it part Y.

We need to put $a_2$ into part X and increase the sum. Now we know that $a_1$ is the largest number, but we don't know about the others, so we cannot simply exchange $a_2$ and one number from X. To solve this problem, we have to exchange more numbers.

So we list all the 2n numbers from X and Y on a line,

　　　$a_1　...$　(numbers from X)

Large ---------- Small

　　　　$a_2　...$(numbers from Y)

And then we put a dot to the left of $a_1$. Then we move it to a position between $a_1$ and $a_2$, and then move it to the right of $a_2$, then move it rightward to skip another number and so on. Define $c(i)$ to be the number of numbers from X which are to the right of the dot after i moves, and $d(i)$ to be the number of numbers from Y which are to the left of the dot. At first we have $c(1)=(n-1)$ and $d(1)=0$, when i increase by 1, $c(i)$ decrease by 1 or $d(i)$ increase by 1. At last $c(2n)=0$ and $d(2n)=n$. We can see that $c(1)>d(1)$ and $c(2n)<d(2n)$, so that there must be a moment when $c(j)=d(j)$, say this number is $t$.

Now we can exchange these $t$ numbers. Define $X_l$ to be the set and $x_l$ to be the product of the numbers from X which are to the left of the dot, and $X_r$ to be the set and $x_r$ to be the product of the right ones, $Y_l$ to be the set and $y_l$ to be the product of the numbers from Y which are to the left of the dot, and $Y_r$ to be the set and $y_r$ to be the product of the right ones.

$X_l$ and $Y_r$ each contains $n-t$ numbers, $X_r$ and $Y_l$ each contains $t$ numbers, and since each number from $X_l$ is larger than any number from $Y_r$ and each number from $Y_l$ is larger than any number from $X_r$, we have $x_l>y_r$ and $y_l>x_r$. It's easy to prove that $x_lx_r+y_ly_r<x_ly_l+x_ry_r$. Now we can exchange $Y_l$ and $X_r$ and the sum will increase.

Since $a_2$ is in $Y_l$, now we have moved it into part X. Then we do the same thing to the part which contains $a_3$ and so on, at last we will have $a_1, a_2, a_3...a_n$ in X. After this one is done, we start with the part which contains $a_{n+1}$ and do the same thing. At last the sum will be $\sum\limits_{i=1}^m\prod\limits_{j=1}^na_{n(i-1)+j}$ and that will be the maximum.


问:
If $u$ is an unbounded sequence and $\lim_{n\to\infty} \frac{u_{n+1}}{u_n} = 1$, is $\lim_{n\to\infty} u_n = +\infty$ true?
Let $(u_n)_n$ be an unbounded (i.e. $(|u_n|)_n$ is not bounded from above) sequence of real numbers such that
$
\frac{u_{n+1}}{u_n} \to_{n\to\infty} 1
$. Does this necessarily imply that $(u_n)_n$ diverges to $+\infty$? I can't find any counter-examples, so I think this might be true. Using Cesaro's mean sequence, we get that $\sqrt[n]{u_n} \to_{n\to\infty}1$, which leads to the fact that $n=o(\ln(u_n))$. So, if we write $u_n=f(n)$, $f$ would be a polynomial or weaker than a polynomial (in the sense of limits in infinity) like $\ln$ and $\ln \circ \ln$, and more generally $\ln \circ \dots \circ \ln$. I feel like my counter-example function should be a very "weak" function that is unbounded but still doesn't tend to infinity, it seems to me intuitively that this function lies in the limit between bounded functions and weak unbounded functions (such as $\ln$), so the problem would be to show that there is no such limit function, or that there is. (Although, I think my reasoning is a little bit flawed as sequences are discrete and by translating the problem to functions, I ignore this whole property.)

答:
A simple counterexample would be $u_n = -n$.

But it is wrong even if we require that all $u_n$ are positive. A counterexample is
$$
 u_n = 1 + n^{1/4} \sin^2(n^{1/2})
$$
which is unbounded, but also approaches the value $1$ infinitely often.

With $f(x) = x^{1/4} \sin^2(x^{1/2})$ we have
$$
 \left| \frac{u_{n+1}}{u_n} - 1\right | \le |f(n+1) - f(n)| = |f'(c_n)|
$$
for some $c_n \in (n, n+1)$, and that converges to zero because
$$
 f'(x) = \frac 14 x^{-3/4} \sin^2(x^{1/2}) + x^{-1/4} \sin(x^{1/2}) \cos (x^{1/2}) \to 0
$$
for $x \to \infty$.

Roughly speaking, the idea is to construct a sequence which oscillates with increasing amplitude, but decreasing frequency.

Another way (inspired by this) is to start with the sequence
$$
\begin{align}
 x_n =\; & 0, 1, \\
 & 0, \frac 12, \frac 22, \frac 32, 2,  \frac 32, \frac 22, \frac 12, \\
& 0, \frac 13, \frac 23, \ldots, \frac 83, 3, \frac 83, \ldots, \frac 13, \\ & 0, \frac 14, \ldots \, .
\end{align}
$$
For $k=1, 2, 3, \ldots$ we have “segments” where $x_n$ increases from $0$ to $k$ in increments of $1/k$, and back to $0$ in decrements of $-1/k$. $(x_n)$ is unbounded  with $\lim_{n\to \infty} (x_{n+1} - x_n) = 0$. Every non-negative real number is a limit point of this sequence.

Then define $u_n = e^{x_n}$.


问:
Question about elementary limits
Suppose $\;f(x)\;$ is a continuous function, and that
$$\lim_{x\rightarrow 2017}\big(x+f(x)\big)=2018$$

I want to know if $\;y=2017\;$ intersects the graph of function $\;y=\left(2018x^2-x^3\right)f(x)\;$ multiple times.

Is it correct to say that $\;f(x)=2018-x\;$, since if we separate the limit into two limits, we get
$$\lim_{x\rightarrow 2017}\big(x+f(x)\big)=\lim_{x\rightarrow2017}x+\lim_{x\rightarrow2017}f(x)=2018\Rightarrow \lim_{x\rightarrow 2017}f(x)=1,$$

and because of this the limit

$$\lim_{x\rightarrow n}x=n$$

Is this correct?

答:
As you said, $\lim_{x\to2017}f(x)=1$ and since $f$ is continuous, $f(2017)=1$. This is all we can say about $f$ which is not enough information to determine the number of intersection points or what $f$ actually is. $f(x)=2018-x$ is only one possibility and there's infinitely many.

However, we can guarantee at least $2$ intersection points. If you let $g(x)=(2018x^2-x^3)f(x)$, then $g(0)=0, g(2017)=2017^2$, and $g(2018)=0$. $g$ is continuous (since $f$ is continuous) and by the Intermediate Value Theorem there are numbers $a\in(0,2017)$ and $b\in(2017,2018)$ such that $g(a)=2017, g(b)=2017$. So $2$ intersection points are $(a,2017)$ and $(b,2017)$.

An example with exactly $2$ intersection points is $f(x)=x/2017$. $f(x)=1$ would give you exactly $3$ intersection points. And there's an example with infinitely many intersection points.

Title: Question about elementary limits


问:
Does $A^{-1} = A_0^{-1} + O_p(n^{-1})$ where $A = A_0 + O_p(n^{-1})$?
Let $A$ be a random matrix that converges in probability to a constant matrix $A_0$ at the rate $O_p(n^{-1})$:
$$
A = A_0 + O_p(n^{-1}).
$$

I want to know if $A^{-1} = A_0^{-1} + O_p(n^{-1})$?

Let $I$ be the identity matrix. Here is my attempt at proving it, which I'm not sure is valid mainly because I don't know if we are allowed to define a series expansion in the form of $O_p$ variables.

$$
\begin{align}
A^{-1} &= (A_0 + O_p(n^{-1}))^{-1} \\
&= (A_0(I + A_0^{-1}O_p(n^{-1}))^{-1} \\
&= (A_0(I + O_p(n^{-1}))^{-1} \\
&= (I + O_p(n^{-1}))^{-1}A_0^{-1}.
\end{align}
$$
Now we perform a Neumann series expansion which gives
$$
\begin{align}
(I + O_p(n^{-1}))^{-1}
&= \sum_{k=0}^\infty (-1)^{k} (O_p(n^{-1}))^{k} \\
&= I - O_p(n^{-1}) \\
&= I + O_p(n^{-1}).
\end{align}
$$
I have assumed here that $(O_p(n^{-1}))^0$, is this true?

Then we substitute this into $(I + O_p(n^{-1}))^{-1}A_0^{-1}$ to get

$$
\begin{align}
A^{-1} &= (I + O_p(n^{-1}))A_0^{-1} \\
&= A_0^{-1} + O_p(n^{-1})A_0^{-1} \\
&= A_0^{-1} + O_p(n^{-1}).
\end{align}
$$

In the deterministic case, if we wanted to invert the matrix $(I+K)$ we need to have $||K|| < 1$ to ensure convergence. So does this mean that in the probabilistic case, to invert the matrix $(I + O_p(n^{-1}))$ we need to have $||O_p(n^{-1})|| < 1$? So the norm of an $O_p(n^{-1})$ term is required to be less than one..does this phrase even make sense?

It seems an $O_p(n^{-1})$ term will be increasingly likely to have a value less than $1$ as $n \to $\infty$, since the probability is becoming more concentrated around zero, but I don't think it can ever be fully guaranteed to be below one?

So does $A^{-1} = A_0^{-1} + O_p(n^{-1})$ really hold and if it does is my proof valid? If not, what is a valid proof?

答:
Given a stochastic $A$ and fixed $A_0$ of size $d\times d$ such that $A=A_0+O_p(n^{-1})$, we can show $A^{-1}=A_0^{-1}+O_p(n^{-1})$ by noticing that matrix inverse is differentiable, so by local linearity rates are preserved.$\DeclareMathOperator\vect{vec}$

Consider the vector-valued vector-input function $f$ over flattened invertible matrices where $ f(\vect X)=\vect X^{-1}$. This vector-valued function is differentiable on the open set of invertible matrices, such that, refering to the indices $ij, kl$ from the unflattened matrices,
$$
\partial_{ij}f_{kl}(\vect X)=-(X^{-1})_{jk}(X^{-1})_{li}\,.
$$
For any convex combination $A_t$ of $A_0,A$, we have on the event $A_t$ is invertible,
$$\|Df(\vect A_t)\|\le \|Df (A_t)\|_F\le \|\vect Df (A_t)\|_1=\|\vect A_t^{-1}\|_1^2\le d^2\|A_t^{-1}\|_F^2\,.$$

Then since $\|A-A_t\|_F=O_p(n^{-1})$ by the assumption, as the other answer shows with Neumann series expansion, we have $A_t^{-1}$ exists whp and converges entrywise in probability to $A^{-1}$, so $\|A_t^{-1}\|_F=O_p(\|A^{-1}\|_F)=O_p(1)$.

Thus whp $\|Df(\vect A_t)\|$ is bounded by some constant $M$ and by the mean value inequality on $f$, $\|A_0^{-1}-A^{-1}\|_F\le M\|A_0-A\|_F$, which implies the conclusion.


问:
given a square matrix, what's the expected number of cells that is simultaneously the biggest value in its row and column?
the NxN square matrix represents a graph actually. the number in a non-zero cell represent the weight of the edge. If a cell is zero then there is no edge between the corresponding nodes.

If the matrix is generated randomly. We count the number X of cells who are both the biggest value in its row and the biggest value in its column.

what's $E[x]$ ?

I am clueless. Can anyone share some ideas?

答:
Assuming the weights are all i.i.d. and from a distribution that makes probability 0 for any entries being equal.

Let $X_{ij}$ be the random variable that is $1$ if entry $(i,j)$ is larger than any other entries in its row and its column, and $0$ otherwise. We have $E[X_{ij}]=\frac1{2n-1}$.

Now note that your $X$ is the sum of all the $X_{ij}$, and expectation is linear. So you can just add all the expectations for each of them.


问:
Why open sets in $\mathbf R^k$ is Polish?
I am reading van der Vaart and Wellner . The footnote $\dagger$ on p.17 says that $(0,1)$ is Polish. But seems to me that $(0,1)$ is not complete. Do I miss something here?</p>

答: $(0,1)$ is topologically complete (also referred to as completely metrizable): it’s homeomorphic to $\Bbb R$ and therefore has a complete metric that generates its topology.


问:
Prime vs Maximal Ideals of Modules over Polynomial Rings
Let $M$ be an $R[X]$ algebra and finitely generated as an $R[X]$-module, where $R$ is a field (so $R[X]$ is a PID). Is it generically true that nonzero prime ideals in $M$ are maximal? If not, what additional conditions are necessary (on either $M$ or $R$)?

答:
Let $M=R[X]^2$, and $I=R[X]\times\{0\}$. Then $M$ is a finitely generated $R[X]$-module, and an $R[X]$-algebra. Since $M/I\simeq R[X]$ is a domain (since $R$ is a field), $I$ is a non zero prime ideal, which is not maximal, since $R[X]$ is not a field.

Note that the argument works if we replace $R$ by any domain, so there is no real hope to get a positive result by strengthening assumptions on $R$.

Anyway, what you really want is $M$ to have Krull dimension $\leq 1$. Since $R[X]$ has dimension $1$ already, what you wish for is probably equivalent to ask $M$ to be integral over $R[X]$ (meaning any element of $M$ is the zero of some monic polynomial with coefficients in $R[X]$).


问:
Notation for maps: is "$A$" in "$\alpha : A \to B$" always the domain?
According to the Wikipedia page on functions / maps (emphasis added):

A function is a process or a relation that associates each element x of a set X, the domain of the function, to a single element y of another set Y (possibly the same set), the codomain of the function.

It then goes on to say, in the subsection Specifying a function By a formula, from the same page (emphasis added):

For example, ${\displaystyle f(x)={\sqrt {1+x^{2}}}}$ defines a function $f\colon \mathbb{R} \to \mathbb{R}$ whose domain is ${\displaystyle \mathbb {R} ,}$ because ${\displaystyle 1+x^{2}}$ is always positive if $x$ is a real number. On the other hand, $f(x)=\sqrt{1-x^2}$ defines a function from the reals to the reals whose domain is reduced to the interval $[–1, 1]$.

which would seem to suggest that, for the latter function, one should write:
"let $f \colon \mathbb{R} \to \mathbb{R}$ be the function defined by the equation $f(x)  = \sqrt{1-x^2}$, valid for all $x \in [-1,1]$", i.e. $\mathbb{R}$ is not the domain, but some set containing the domain; however, in another subsection, Notation Arrow notation it says:

For explicitly expressing [the] domain $X$ and the codomain $Y$ of a function $f$, the arrow notation is often used (read: "the function $f$ from $X$ to $Y$" or "the function $f$ mapping elements of $X$ to elements of $Y$"):
$$f\colon X\to Y$$

which seems to imply that it is not correct to write $f: \mathbb{R} \to \mathbb{R}$ for the function $x \mapsto \sqrt{1-x^2}$, as the domain is not $\mathbb{R}$, nor is the codomain $\mathbb{R}$.

Should one write

$$ \begin{array}{ll}f\colon \mathbb{R} &\hspace{-0.6em} \to \mathbb{R} \\[5pt] & \hspace{-0.6em} x \mapsto \sqrt{1-x^2}, \end{array}$$

where $\mathbb{R} \to \mathbb{R}$ implies that the domain and image are subsets of $\mathbb{R}$, respectively? or should one write:

$$
\begin{array}{ll}f:[-1,1] & \hspace{-0.6em} \to [0,1] \\[5pt] & \hspace{-0.6em} x \mapsto \sqrt{1-x^2}, \end{array}
$$

N.B. My interest is not just pedantic: I have never studied maps, formally, and am trying to clarify whether a map from the set $A$ onto the set $B$ requires only that every element in $B$ be an image, or whether it also requires that every element of $A$ have an image, as well. I've consulted 2 different textbooks, as well as the Wikipedia page; unfortunately, although great care is given to distinguish the codomain and the image, all the explanations and examples provided do not address whether the domain is $A$ or if it may be a subset of $A$.

Moreover, several textbooks I've read have made much ado about the fact that one ought to distinguish between a function and a function of and imply that the distinction is to be found in these definitions (unfortunately, none of them consider it important enough to actually clarify within their own text).

答:
If you write $f : A \to B$ that is usually understood to mean that:

for every $a \in A$, the value $f(a)$ is defined

this value is an element of $B$.

So the domain has to be $A$. However, $B$ is not necessarily the image but just the codomain (i.e. it is acceptable for $f$ to not “hit” every value in $B$; however, if $f$ does "hit" every value in $B$, then the image is equal to the codomain).

For your definition $f(x) = \sqrt{1 - x^2}$, writing $f : \mathbb{R} \to \mathbb{R}$ would be considered wrong in most circumstances (people get sloppy sometimes, or figuring out the exact domain might be something they want you to do; writing $f : \mathbb{R} \to \mathbb{R}$ still sloppy in the latter case, though). So you have to be more precise in your domain, i.e. $f: [-1, 1] \to \mathbb{R}$ would be fine (note: you do not have to specify the image, just a superset of it), as would $f : [0,1] \to [0, 1]$ would (i.e. you can consider the function restricted to a subset of its maximal domain).


问:
Compelx plane : Draw each step of $z$
Let $z_1=2+i$, $z_2=2$, $z_3=1-\frac{i}{2}$ and $z_4=\frac{i}{3}$.

Draw these numbers in the complex plane.

Calculate $z=\overline{(z_2\cdot z_1+z_3)\cdot z_4}$ and do each step also graphically.

$$$$

As for the second part, to draw each step graphically do we just calculate for example $z_2\cdot z_1$ and then we draw that in the complex plane? Or is it meant to do the calculations on the graph?

答:
Yes, I would draw $z_2\cdot z_1$  and label it on the graph as such, and then do the same for each step of the rest of the calculation.

It's a rather tedious procedure, but it is good for understanding the graphical impact of complex transformations.

Title: Compelx plane : Draw each step of $z$


问:
Could this conjecture be proved ? (sum of even powers of cotangents in arithmetic progression )
Having tried (in vain) to answer this question, I worked the explicit formulae of
$$\color{blue}{S_k=\sum _{n=1}^m \Big[\cot \left(\frac{n \,\pi  }{2 m+1}\right)\Big]^k}$$ where $k$ is an even integer.

To my surprise, the CAS I used were able to produce explicit results only for $k=2$ and $k=4$. All the remaining was done by hand using pen and paper.

What I noticed is that apparently
$$\color{red}{S_{2k}=\frac{m (2 m-1)}{a_k} \left( \sum_{n=0}^{2(k-1)} b_n\, m^n \right)}$$ where the  $a_k$'s
$$\color{blue}{\{3,45,945,14175,467775\}}$$ are the first terms of sequence $A171080$ in $OEIS$:
$$a_k = \prod_{q\;\text{odd prime}\\ \;\;\leq 2k+1}q^{\lfloor 2k/(q-1) \rfloor}$$

Concerning the coefficients $b_n$, they are given in the following table
$$\color{blue}{\left(
\begin{array}{cc}
 k & \text{list of the } b_n \\
 2 & \{1\} \\
 4 & \{-9,10,4\} \\
 6 & \{135,-252,8,112,32\} \\
 8 & \{-1575,3834,-1388,-2248,496,864,192\} \\
 10 & \{42525,-122760,77040,70400,-57920,-38720,14720,14080,2560\}
\end{array}
\right)}$$

My questions are :

Why cannot we compute explicit forms for odd values of $k$ ?

Could this conjecture be proved ?

What are these polynomials ?

Could we find their generic formula ?

答:
In an article by Ejsmont and Lehner a method using operator trace is applied to evaluate finite sums involving the cotangent. The paper gives numerous references concerning finite trigonometric sums which seem to have a rich recent history. We use here a result obtained by Berndt and Yeap by a contour integration technique. It reads
\begin{equation}
\sum_{n=1}^{q-1}\cot^{2k}\frac{n\pi}{q}=(-1)^kq-(-1)^k2^{2k}\sum_{\substack{j_0,j_1,\ldots,j_{2k}\ge0 \\j_0+j_1+\ldots+j_{2k}=k}}q^{2j_0}\prod_{p=0}^{2k}\frac{B_{2j_p}}{(2j_p)!}\tag{BY}\label{BY}
\end{equation}
where $B_s$ are the Bernoulli numbers. It seems that no similar results with odd powers of the function were found.

To evaluate
\begin{equation}
 S_{2k}=\sum _{n=1}^m \left[\cot \left(\frac{n \,\pi  }{2 m+1}\right)\right]^{2k}
\end{equation}
we remark that, by symmetry,
\begin{equation}
 S_{2k}=\frac{1}{2}\sum _{n=1}^{2m} \left[\cot \left(\frac{n \,\pi  }{2 m+1}\right)\right]^{2k}
\end{equation}
then, by (\ref{BY}),
\begin{equation}
 (-1)^kS_{2k}=m+\frac{1}{2}-2^{2k-1}\sum_{\substack{j_0,j_1,\ldots,j_{2k}\ge0 \\j_0+j_1+\ldots+j_{2k}=k}}(2m+1)^{2j_0}\prod_{p=0}^{2k}\frac{B_{2j_p}}{(2j_p)!}
\end{equation}
Under this form, it appears that the sum is polynomial in $m$.

The summation is however difficult to perform. It can be transformed by the use of the series expansion of $\coth Z$
\begin{equation}
 Z\coth Z=1+\sum_{s=1}^\infty\frac{B_{2s}}{(2s)!}(2Z)^{2s}
\end{equation}
One has then
\begin{align}
 S_{2k}&=(-1)^k\left( m+\frac{1}{2}\right)\left[ 1-\left[Z^{-1}\right]\left( \coth\left( (2m+1)Z \right)\coth^{2k}(Z) \right) \right]\\
 &=(-1)^k\left( m+\frac{1}{2}\right)\left[ 1-\frac{1}{2i\pi}\int_{\left|z\right|=\varepsilon} \coth\left( (2m+1)z \right)\coth^{2k}(z)\,dz \right]\label{eq:Cauchy}
\end{align}
where $\left[z^r\right]\left( F(z) \right)$ represents the coefficient of $z^r$ in the Laurent series of $F(z)$ at $z=0$.

The latter expression for $S_{2k}$ vanishes when $m=0$, as $\operatorname{Res}\left( \coth^{2k+1}z,z=0 \right)=1$. Moreover, from the identity $\coth 2z=(\coth z+(\coth z)^{-1})/2$, it also vanishes for $m=1/2$ as proposed. Then,  polynomials $Q_k(m)$ exist such that
\begin{equation}
 S_{2k}=m(2m-1)Q_k(m)
\end{equation}
where
\begin{align}
 Q_k(m)&=\frac{(-1)^k(2m+1)}{2m(2m-1)}\left[ 1-\frac{1}{2i\pi}\int_{\left|z\right|=\varepsilon} \coth\left( (2m+1)z \right)\coth^{2k}(z)\,dz \right]\\
 &=\frac{(-1)^k(2m+1)}{2m(2m-1)}\left[ 1-\left[Z^{-1}\right]\left( \coth\left( (2m+1)Z \right)\coth^{2k}(Z) \right) \right]
\end{align}

Evaluated with a CAS, the latter formula gives the results which are given in the OP up to $k=5$. Higher values of $k$ were numerically tested up to $S_{20}$ for several values of $m$.

Some values of $m$ are remarkable. For instance, it is not difficult to show that $Q_k(-1)=(-1)^{k+1}/3$ and $Q_k\left( -\frac{3}{2} \right)=\frac{(-1)^{k+1}}{3}$. Moreover $Q_k\left( \frac{3}{2} \right)=\frac{1}{3}$. Indeed, for $m=3/2$,
\begin{align}
 \coth\left( (2m+1)z \right)&=\coth(4z)\\
 &=\frac{1}{2}\left( \coth 2z+\frac{1}{\coth 2z} \right)\\
 &=\frac{1}{4}\left( \coth z+\frac{1}{\coth z} \right)+\frac{\coth z}{1+\coth^2z}
\end{align}

the integral reads
\begin{align}
 \frac{1}{2i\pi}\int_{\left|z\right|=\varepsilon} \coth\left( 4z \right)\coth^{2k}(z)\,dz &=\frac{1}{2i\pi}\int_{\left|z\right|=\varepsilon} \frac14\left( \coth^{2k+1}(z)+\coth^{2k-1}(z) \right)\,dz+I_k\\
 &=\frac{1}{2}+I_k
\end{align}
where
\begin{equation}
 I_k=\frac{1}{2i\pi}\int_{\left|z\right|=\varepsilon} \frac{\coth^{2k+1}(z)}{1+\coth^2z}\,dz
\end{equation}
We have the recurrence relation
\begin{equation}
 I_k=1-I_{k-1}
\end{equation}
and
\begin{align}
 I_0&=\frac{1}{2i\pi}\int_{\left|z\right|=\varepsilon} \frac{\coth(z)}{1+\coth^2z}\,dz\\
 &=\frac{1}{2}\frac{1}{2i\pi}\int_{\left|z\right|=\varepsilon}\tanh(2z)\,dz\\
 &=0
\end{align}
Thus $I_{2s}=0$ and $I_{2s+1}=1$. Finally,
\begin{equation}
Q_k\left( \frac32 \right)=
\begin{cases}
\frac23\left( 1-\frac{1}{2}-0\right)=\frac{1}{3}&\text{ when } k=2s\\
-\frac23\left( 1-\frac{1}{2}-1 \right)=\frac{1}{3}&\text{ when } k=2s+1
                                \end{cases}
\end{equation}
Thus $Q_k\left( \frac{3}{2} \right)=\frac{1}{3}$.
This also shows, by adapting the signs, that, for $m=-5/2$, the values of the polynomials oscillate between $1/3$ and $-1/5$. Adapting to the decomposition of $\coth 3z$, should also make it possible to show (as suggested by @Blue in a comment) that $Q_k(1)=3^{-k}$, which is numerically observed.


问:
How to prove a surface is orientable
Given a hyperboloid $M=(x^2+y^2-z^2=1)$

how do I prove it is orientable, I know that a surface is orientable is there exists a non zero 2 form on M or if there is a unit normal vector field on M, or even if M if compact or simply connected, but I don't know which one of these to choose, or how to go about actually proving this. Any help would be great!

答:
The surface is defined by an equation $g(x,y,z)=0$. The gradient of $g$ gives a field of normal vectors. They can be normalized to unit length.

More specifically, at the point $(x,y,z)$ on the hyperboloid, the vector
$$
\frac{(2x,2y,-2z)}{\sqrt{4x^2+4y^2+4z^2}}=\frac{(x,y,-z)}{\sqrt{x^2+y^2+z^2}}
$$
is normal to the surface and has unit length. (Note that neither the numerator nor the denominator is ever zero on the surface.)

Title: How to prove a surface is orientable


答:
Consider this definition of function taken from Mathematical Proofs: A Transition to Advanced Mathematics by Chartrand et. al.:

Let $A$ and $B$ be nonempty sets. By a function from $A$ to $B$, written $f:A\rightarrow B$, we mean a relation from $A$ to $B$ with the property that every element $A$ is the first coordinate of exactly one ordered pair in $f$. Since $f$ is a relation, the set $A$ in this case is the domain of $f$... . The set $B$ is called the codomain of $f$. (216)

This particular definition assumes you know some things about relations. In short, a relation from a set $A$ to a set $B$ is a subset of the cartesian product of $A$ and $B$. Since a cartesian product is a set of ordered pairs, the authors choose to use the ordered-pair-formalism in their definition of a function, rather than referring to a function as a mapping or a process. I like this, and you might find it helpful.

Case in point, if $f:A\rightarrow B$ is a function, the set $A$ is always the domain. Furthermore, if not every element in $A$ is mapped to some element in $B$, $f$ is not a function. So I would disagree with the Wikipedia article you cited which claims the validity of the statement "$f(x)=\sqrt{1-x^2}$ defines a function from the reals to the reals whose domain is reduced to the interval $[-1,1]$." Since there exist elements in $\mathbb{R}$ $-$ namely every $x\in\mathbb{R}$ with $|x|>1$ $-$ which are not mapped to any element of $\mathbb{R}$ (the codomain), $f:\mathbb{R}\rightarrow\mathbb{R}$ defined by $f(x)=\sqrt{1-x^2}$ is not a function. Since the domain of $f$ should be $[-1,1]$ (as indicated in the Wiki article), the proper notation is $f:[-1,1]\rightarrow\mathbb{R}$; in this case, if $f$ is defined by $f(x)=\sqrt{1-x^2}$, $f$ is a function.

Another thing. The codomain is not necessarily equal to the image/range. As indicated in the definition I provided, the codomain is simply the set that all the second coordinates of the function $f$ belong to. The image/range, on the other hand, is the set of all second coordinates of the ordered pairs in $f$. The image is always a subset of the codomain, but it is not necessarily the same as the codomain. In other words, there could be elements in the codomain that are not second coordinates of any ordered pair in $f$. Considering the same function as in the previous paragraph, no $x\in\mathbb{R}$ is mapped to 2 (or any $x<0$ or $x>1$ for that matter), but it is still proper to say that the codomain of $f$ is $\mathbb{R}$.

The special case in which the image equals the codomain happens when $f$ is a surjection. The function $f:A\rightarrow B$ is a surjection when every element of $B$ is the second coordinate of an ordered pair in $f$. More formally, $f$ is a surjection when for all $b\in B$ there exists an $a\in A$ such that $f(a)=b$. Surjections are only special cases, though.

I hope this was helpful.

Title: Notation for maps: is "$A$" in "$\alpha : A \to B$" always the domain?

Body:
According to the Wikipedia page on functions / maps (emphasis added):

A function is a process or a relation that associates each element x of a set X, the domain of the function, to a single element y of another set Y (possibly the same set), the codomain of the function.

It then goes on to say, in the subsection Specifying a function By a formula, from the same page (emphasis added):

For example, ${\displaystyle f(x)={\sqrt {1+x^{2}}}}$ defines a function $f\colon \mathbb{R} \to \mathbb{R}$ whose domain is ${\displaystyle \mathbb {R} ,}$ because ${\displaystyle 1+x^{2}}$ is always positive if $x$ is a real number. On the other hand, $f(x)=\sqrt{1-x^2}$ defines a function from the reals to the reals whose domain is reduced to the interval $[–1, 1]$.

which would seem to suggest that, for the latter function, one should write:
"let $f \colon \mathbb{R} \to \mathbb{R}$ be the function defined by the equation $f(x)  = \sqrt{1-x^2}$, valid for all $x \in [-1,1]$", i.e. $\mathbb{R}$ is not the domain, but some set containing the domain; however, in another subsection, Notation Arrow notation it says:

For explicitly expressing [the] domain $X$ and the codomain $Y$ of a function $f$, the arrow notation is often used (read: "the function $f$ from $X$ to $Y$" or "the function $f$ mapping elements of $X$ to elements of $Y$"):
$$f\colon X\to Y$$

which seems to imply that it is not correct to write $f: \mathbb{R} \to \mathbb{R}$ for the function $x \mapsto \sqrt{1-x^2}$, as the domain is not $\mathbb{R}$, nor is the codomain $\mathbb{R}$.

Should one write

$$ \begin{array}{ll}f\colon \mathbb{R} &\hspace{-0.6em} \to \mathbb{R} \\[5pt] & \hspace{-0.6em} x \mapsto \sqrt{1-x^2}, \end{array}$$

where $\mathbb{R} \to \mathbb{R}$ implies that the domain and image are subsets of $\mathbb{R}$, respectively? or should one write:

$$
\begin{array}{ll}f:[-1,1] & \hspace{-0.6em} \to [0,1] \\[5pt] & \hspace{-0.6em} x \mapsto \sqrt{1-x^2}, \end{array}
$$

N.B. My interest is not just pedantic: I have never studied maps, formally, and am trying to clarify whether a map from the set $A$ onto the set $B$ requires only that every element in $B$ be an image, or whether it also requires that every element of $A$ have an image, as well. I've consulted 2 different textbooks, as well as the Wikipedia page; unfortunately, although great care is given to distinguish the codomain and the image, all the explanations and examples provided do not address whether the domain is $A$ or if it may be a subset of $A$.

Moreover, several textbooks I've read have made much ado about the fact that one ought to distinguish between a function and a function of and imply that the distinction is to be found in these definitions (unfortunately, none of them consider it important enough to actually clarify within their own text).


问:
Endomorphisms of a direct sum of Lie groups
Let $G = \mathbb{T}^n \oplus A = \mathbb{T}^n \times A$, where $\mathbb{T}^n = \left( \mathbb{R} / \mathbb{Z} \right)^n$ and $A$ is a finite discrete abelian group. I'm trying to characterize $\operatorname{End}(G)$.

I know that $\operatorname{End} \left( \mathbb{T}^n \right) = \operatorname{Mat}(n, \mathbb{Z})$. I also know that if $T \in \operatorname{End}(G)$, then $T \left( \mathbb{T}^n \right) \subseteq \mathbb{T}^n$, since $\mathbb{T}^n$ is the identity component of $G$. However, we can't necessarily say that $T(A) \subseteq A$. Is there any kind of nice structure theorem for this?

Thanks!

答:
If $A$ and $B$ are objects in any category with finite biproducts, which includes the category of abelian Lie groups, as well as the ordinary category of abelian groups, then the endomorphism ring of $A \oplus B$ can be written

$$\text{End}(A \oplus B) \cong \left[ \begin{array}{cc} \text{End}(A) & \text{Hom}(B, A) \\ \text{Hom}(A, B) & \text{End}(B) \end{array} \right].$$

This notation refers to matrices acting on "vectors" in $\left[ \begin{array}{c} A \\ B \end{array} \right]$ in such a way that every multiplication operation refers to an evaluation homomorphism, and similarly matrices multiply in such a way that every multiplication operation refers to a composition. Addition refers to the canonical commutative monoid structure on the homsets of any category with finite biproducts (which for abelian Lie groups and abelian groups is pointwise addition). See the "matrix multiplication" section of A meditation on semiadditive categories for more details on how finite biproducts let you do this.

A direct sum of the form $\mathbb{T}^n \oplus A$ where $A$ is finite has the additional property that $\text{Hom}(\mathbb{T}^n, A) = 0$ because $\mathbb{T}^n$ is divisible. So the above simplifies to the upper triangular form

$$\text{End}(\mathbb{T}^n \oplus A) \cong \left[ \begin{array}{cc} M_n(\mathbb{Z}) & \text{Hom}(A, \mathbb{T}^n) \cong (A^{\ast})^n \\ 0 & \text{End}(A) \end{array} \right].$$

Here $A^{\ast} \cong \text{Hom}(A, \mathbb{T})$ is the Pontryagin dual of $A$. Concretely what this says is that an endomorphism of $\mathbb{T}^n \oplus A$ takes the form

$$\mathbb{T}^n \oplus A \ni (t, a) \mapsto (f(t) + g(a), h(a)) \in \mathbb{T}^n \oplus A$$

where $f : \mathbb{T}^n \to \mathbb{T}^n$ is an endomorphism, $g : A \to \mathbb{T}^n$ is a homomorphism, and $h : A \to A$ is an endomorphism. One of the nice things about writing this in matrix form is it's immediately clear what the automorphisms are: an upper-triangular matrix is invertible iff its diagonal entries are and this property continues to hold for block matrices like the above.


问:
Powers of complex numbers : Which is the general formula?
Let $z=\frac{1}{2}\left (\sqrt{3}+i\right )\in \mathbb{C}$.

I want to calcualte the powers $z^n$ for $n\in \mathbb{Z}$.

I have done the following: \begin{align*}&z^1=\frac{1}{2}\left (\sqrt{3}+i\right ) \\ &z^2=\left (\frac{1}{2}\left (\sqrt{3}+i\right )\right )^2=\frac{1}{4}\left (3+2\sqrt{3}i-1\right )=\frac{1}{4}\left (2+2\sqrt{3}i\right )=\frac{1}{2}\left (1+\sqrt{3}i\right ) \\ &z^3=z^2\cdot z=\frac{1}{2}\left (1+\sqrt{3}i\right )\cdot \frac{1}{2}\left (\sqrt{3}+i\right )=\frac{1}{4}\left (\sqrt{3}+i+3i-\sqrt{3}\right )=\frac{1}{4}\cdot 4i=i \\ & z^4=z^3\cdot z=i\cdot \frac{1}{2}\left (1+\sqrt{3}i\right )=\frac{1}{2}\left (-\sqrt{3}+i\right ) \end{align*} Is therea specific formula that we get for an arbitrary $n$ ?

答:
It is easiest to express the general formula using the polar representation of $z = r e^{i\theta}$

$$
z^n = r^n e^{in\theta} = r^n\left(\cos(n\theta) + i \sin(n\theta)\right).
$$

See also Euler formula and De Moivre's formula.


问:
Solving $\frac{\cos^23t}{\tan t}+\frac{\cos^2t}{\tan3t}=0$
I am having trouble solving the trigonometric equation below:
$$\frac{\cos^2\left(3t\right)}{\tan\left(t\right)}+\frac{\cos^2\left(t\right)}{\tan\left(3t\right)}=0$$

I would appreciate any help you can give me!
Thank you!

答:
I will do the first one, hopefully, it will give you some ideas how to tackle the rest.

$\frac {\cos^2 3t}{\tan t} + \frac {\cos^2 t}{\tan 3t} = 0$

Find the common denominator and make this into one fraction.

$\frac {\cos^2 3t\tan 3t + \cos^2t\tan t}{\tan 3t\tan t} = 0$

Lets focus on the numerator.  If this expression equals $0$ then the numerator equals $0$ and the denominator does not equal $0.$

$\cos^2 3t\tan 3t + \cos^2t\tan t = 0$

use: $\tan t = \frac {\sin t}{\cos t}$ and cancel.

$\cos 3t\sin 3t + \cos t\tan t = 0$

I want to use $2\cos t\sin t = \sin 2t$

$2\cos 3t\sin 3t + 2\cos t\tan t = 0\\
\sin 6t + \cos t\tan t = 0$

The next step is a little bit obscure... $\sin 3t = (4\sin t - 3\sin^3 t)$

$3\sin 2t - 4\sin^3 2t + \sin 2t = 0$

Everything is in terms of $\sin 2t$!

$u=\sin 2t$ will make what we have look more like a polynomial.

$4u - 4u^3 = 0\\
u(1-u)(1+u)= 0\\
u = 0, \pm 1$

Reverse the substitution.

$\sin 2t = 0 \text { or } \pm 1$

$t = 0, \pm\frac {\pi}{2}, \pm \frac {\pi}{4}, \pm \frac{3\pi}{4}+ n\pi$

But if $t=0$ the denominator in the original expression equals 0, and we are going to throw that answer away.  And $\tan t$ is not defined at $t = \pm\frac {\pi}{2}$ so we should disregard that solution, too.

$t = \frac {(2n+1)\pi}{4}$


问:
how to find the optimal solution of two circles
Let us have the following problem with the parameter $a\in [-3,1]$: $$\min x $$ $$(x-1)^{2}+(y-1)^{2}\le 1$$ $$(x-1)^{2}+(y+a)^{2}\le 1.$$ How to find the optimal solution? It should be the left intersection of the circles....

答:
Here is a picture.

Can you see now how to use your equations to solve for x?


问:
Is joint PDF always less than or equals to marginal PDF?
The question here is whether the statement below is true for PDFs:

ℙ(,) ≤ ℙ()

The answer is no, due to the fact that PDFs are not probabilities. There is an intuitive answer with an example here, but I came across a more theoretical answer which I can't understand. The answer given is that by using the product rule:

ℙ(,) = ℙ(|) ℙ()

It continues the proof by stating that the conditional probability is not bounded, and if x is forced to be a function of y then the conditional probability is infinite at x=f(y). The source of the answer is at part 3(ii) here.

How does the conditional probability becomes infinite if x is a function of y?

答:
That link does not say it is the conditional probability which is infinite, but the conditional density.  Even then it is stretching a point.

A conditional probability is a probability so is between $0$ and $1$.  If $X=f(y)$ when $Y=y$ then $x$ is fixed by $y$ and you can say $\mathbb P(X=f(y) \mid Y=y)=1$ as a conditional probability.  This is clearly finite.

But your quote is trying to say this in the language of densities, and so it says the conditional density is $p(x \mid y) = \delta(x-f(y))$ using a Dirac delta function, which despite the name is not a proper function $\mathbb R \to \mathbb R$. This  $\delta(w)$ might be seen as zero when $w \not =0$ but with the property $\int\limits_{-\infty}^{\infty} \delta(w)\, dw = 1$, so some people might describe $\delta(0)$ as infinite in a special way. This same special way is what leads to mention of the infinite conditional density in your quote.


问:
Fair way to split fees
So I have a very simple problem, but I am unable to find a proper solution top it.

Four friends rented and apartment for 1 year. They paid \$590 of deposit each (\$2360 total).

By the end of the year, they received \$1470 back from their landlord because there were some unexpected bills. They calculated that each one was responsible for 44%, 34%, 15% and 7% of that extra charge (imagine the difference is in extra energy consumption or damages for example).

If they were to divide the deposit in 4 equal parts, they each would get \$367,5 (\$1470/4), but that is not fair as one of them only overspent 7% .

What would be the fair value each one should receive?

(I'm sorry for the probably wrong tags, I didn't know which to pick)

Thanks

答:
The overage is $\$890$ ($2360-1470=890$).

Suppose I am responsible for $44\%$ of the overage.  That makes $0.44\times\$890=\$391.60.$  Then, I should get back $\$198.40$ (my $\$590$ deposit less the overage for which I am responsible).

Use this method to calculate the rest.

Title: Fair way to split fees


答:
Here it is a slightly different approach from @Doug M. One has that
\begin{align*}
\frac{\cos^{2}(3t)}{\tan(t)} + \frac{\cos^{2}(t)}{\tan(3t)} = 0 & \Rightarrow \cos^{2}(3t)\tan(3t) + \cos^{2}(t)\tan(t) = 0\\\\
& \Rightarrow \cos(3t)\sin(3t) + \cos(t)\sin(t) = 0\\\\
& \Rightarrow \sin(6t) + \sin(2t) = 0\\\\
& \Rightarrow \sin(6t) = \sin(-2t)
\end{align*}

After solving such equation, verify whether the solutions satisfy $\tan(t)\tan(3t)\neq 0$.

Can you take it from here?

Title: Solving $\frac{\cos^23t}{\tan t}+\frac{\cos^2t}{\tan3t}=0$

Body:
I am having trouble solving the trigonometric equation below:
$$\frac{\cos^2\left(3t\right)}{\tan\left(t\right)}+\frac{\cos^2\left(t\right)}{\tan\left(3t\right)}=0$$

I would appreciate any help you can give me!
Thank you!


问:
$\int_a^b\sum_{n=1}^\infty f_n(x) dx=\sum_{n=1}^\infty \int_a^b f_n(x) dx$
$\{f_n\}$ is a sequence of continuous (so also integrable) real functions on $[a,b]$ that is uniformly convergent and uniformly equicontinuous. I want to show $$\int_a^b\sum_{n=1}^\infty \frac{f_n(x)}{{\color{red}{n^2}}} dx=\sum_{n=1}^\infty \frac{1}{{\color{red}{n^2}}}\int_a^b f_n(x) dx,$$
and I cannot figure out how to do this. I think to take the limit as $n\to\infty$ and use uniform convergence to the limit $f$ which will somehow allow the sum and integrals to switch, but I cannot figure out how to do this. Help would be greatly appreciated.

Edit: In red.

答:
First note that $$\int_a^b\sum_{n=1}^{N} \frac{f_n(x)}{{{n^2}}} dx=\sum_{n=1}^{N} \frac{1}{{{n^2}}}\int_a^b f_n(x) dx$$ for each $N$ and that RHS of this equation tends to RHS of the equation we want to prove. Hence, it remains only to prove that LHS of this equation tends to LHS of the given equation. For this we use the following:

If $g_n \to g$ uniformly (with each $g_n$ continuous) then $\int_a^{b}g_n(x)dx \to \int_a^{b}g(x)dx$.

So it reamins only to prove that $\sum_{n=1}^{N} \frac{f_n(x)}{{{n^2}}}$ converges uniformly to $\sum_{n=1}^{\infty} \frac{f_n(x)}{{{n^2}}}$ as $N \to \infty$.

By M-test it is enough to show that $(f_n(x))$ is uniformly bounded. There exists $n_0$ such that $|f_n(x)-f_{n_0}(x)| <1$ for all $n \geq n_0$. Let $M$ be an upper bound for $|f_{n_0}|$ Then $|f_n(x)| <1+M$ for all $x$ for all $n \geq n_0$. Can you finish the proof now using the fact that each of the functions $f_1,f_2,...,f_{n_0-1}$ are bounded?


问:
Compute the probability that $\Pr(\tau<\infty)$.
Let $B$ and $W$ be independent symmetric random walks starting at $0$ with filtration $\mathcal F_0=\{\emptyset,\Omega\}$ and $\mathcal F_n=\sigma(B_1,...,B_n,W_1,...,W_n)$.

Define $\mathcal E(B)_n=e^{B_n}(\cosh 1)^{-n}$ and $\mathcal E(W)_n=e^{W_n}(\cosh 1)^{-n}$ for all $n\ge 0$.

Also let $\tau=\inf\{n:\mathcal E(B)_n\le\frac 12\},\sigma=\inf\{n:\mathcal E(W)_n\ge 2\}$.

(a) Calculate $\Pr(\tau<\infty)$ and show that $\Pr(\sigma<\infty)\in (0,1)$.

(b) Which of the following three processes are uniformly integrable martingales:
$$\mathcal E (B)^{\sigma\wedge\tau},\mathcal E(W)^{\sigma\wedge\tau},\mathcal E (B)^{\sigma\wedge\tau}\mathcal E(W)^{\sigma\wedge\tau}.$$

We can write
\begin{align}
\Pr(\tau<\infty)&=1-\Pr(\mathcal E(B)_n>\frac 12\text{ for all }n).
\end{align}
But then how to proceed. I have no idea why we are trying to compute $\mathcal E(B)_n$ and neither do I know how to compute this probability. Can someone give me a hint? Thank you!

Edit:

I have a rough idea but it still does not work. We need to involve $\Pr(\tau<\infty)$.

Firstly, it is easy to show that $\mathcal E(B)_n,\mathcal E(W)_n$ are martingales. Secondly, it is easy to see that $\tau,\sigma$ are stopping times with respect to the filtration $\mathcal \{F_n\}_{n\in\mathbb Z_{>0}} $ . So I want to use the Optimal Stopping Theorem. Note that
\begin{align}
\mathbb E[\mathcal E(B)_\tau]&=\mathbb E[\lim_{n\to\infty}\mathcal E(B)_{n\wedge\tau}].
\end{align}
By Fatou's lemma, $$\mathbb E[\lim_{n\to\infty}\mathcal E(B)_{n\wedge\tau}]\le\liminf_{n\to\infty}\mathbb E[\mathcal E(B)_{n\wedge\tau}].$$
Since $\mathcal E(B)_n$ is a martingale, we have
$$ \mathbb E[\mathcal E(B)_{n\wedge\tau}]=\mathbb E[\mathcal E(B)_{0\wedge\tau}]=1. $$

So $1\ge \mathbb E[\mathcal E(B)_\tau].$

Write $\mathbb E[\mathcal E(B)_\tau]=\mathbb E[\mathcal E(B)_\tau 1_{\tau<\infty}+\mathcal E(B)_\tau 1_{\tau=\infty}]=\mathbb E[\mathcal E(B)_\tau;\tau<\infty]+\mathbb E[\mathcal E(B)_\tau;\tau=\infty].$

Remark: Note that we have applied Fatou's lemma, but we cannot apply the dominated convergence to get equality, since the random walk can visit any big number before it returns to some point such that $\mathcal E(B)_n\le\frac 12$.

答:
I will write $M_n$ for $\mathcal{E}(B)_n$ to save typing.  I claim $M_n \to 0$ almost surely.

Note that $M_n$ is a positive martingale so it converges almost surely to some limiting nonnegative random variable $M$.  On the event $\{M > 0\}$, we have $\frac{M_{n+1}}{M_n} \to 1$ a.s.  But $\frac{M_{n+1}}{M_n}$ can only have the values $\frac{1}{\cosh 1} e^{\pm 1}$ so it cannot converge to 1.  Therefore $P(M > 0) = 0$ and so $M_n \to 0$ almost surely.

In particular, almost surely we have $M_n \le 1/2$ for all sufficiently large $n$, and thus $\tau < \infty$.

As for $\sigma$ (which I will think of as a stopping time for $B$ instead of $W$), by optional stopping we have $E[M_{\sigma \wedge n}] = 1$, and so by dominated convergence $E[\lim_{n \to \infty} M_{\sigma \wedge n}] = 1$ as well.  If $\sigma < \infty$ a.s., we have $M_{\sigma \wedge n} \to M_\sigma \ge 2$ and this is absurd, as you noted.  If $\sigma = \infty$ a.s., we have $\lim_{n \to \infty} M_{\sigma \wedge n} = \lim_{n \to \infty} M_n =0$ which is also absurd.

Some hints on (b):

Suppose $\mathcal{E}(B)^{\sigma \wedge \tau}$ were uniformly integrable. Since $\sigma$ is independent of $B$ and $\tau$, for each $0 \le k \le \infty$ we have $E[\mathcal{E}(B)^{\sigma \wedge \tau}_\infty \mid \sigma=k] = E[\mathcal{E}(B)_\infty^{k \wedge \tau}]$.  Compute this for each $k$ and write $$E[\mathcal{E}(B)_\infty^{\sigma \wedge \tau}] = \sum_{0 \le k \le \infty} E[\mathcal{E}(B)^{\sigma \wedge \tau}_\infty \mid \sigma=k] P(\sigma = k).$$

$\mathcal{E}(W)^\sigma$ is uniformly integrable, and $\sigma \wedge \tau \le \sigma$.

No ideas about the third one.


问:
matrix interesting problem about determinant
Hi guys i am a new member and i need help about this problem .
Let  $A=(a_{ij}) \in M_{n,n}(\mathbb{Z})$ ,  $p$  a prime, and  $\tilde{A}\in M_{n,n}(\mathbb{F}_p)$  be the matrix obtained by reducing every $a_{ij}$  modulo  $p$ . If  $rank(\tilde{A})=r$ , how can I show that  $\det(A)$  is divisible by  $p^{n−r}$ ?

答:
$\tilde{A}$ has rank $r$ if and only if there exists invertible matrices $g,h\in M_{n,n}(\mathbb{F}_p)$ such that $$
g\tilde{A}h=\begin{bmatrix}
I_r & 0\\
0 & 0
\end{bmatrix}
$$ where $I_r$ denotes the $r\times r$-identity matrix and $0$'s denote $0$ matrices of correct sizes. This matrix is called the Smith Normal Form of $\tilde{A}$.

Pick $G,H\in M_{n,n}(\mathbb{Z})$ such that $\tilde{G}=g$ and $\tilde{H}=h$. Then, $\tilde{GAH}=g\tilde{A}h$, so each entry of $GAH$ is divisible by $p$, except the first $r$ diagonal entries. In particular, $n-r$ columns of $GAH$ is divisible by $p$ and the multilinearity of $\det$ implies that $\det(GAH)$ is divisible by $p^{n-r}$. On the other hand, $\det(GAH)=\det(G)\det(A)\det(H)$ and $\det(G),\det(H)$ are not divisible by $p$ since $g,h$ are invertible. Thus, $p^{n-r}$ divides $\det(A)$.


问:
(Ir)reducibility criteria for homogeneous polynomials
Suppose I have a homogeneous polynomial in at least 3 variables over some algebraically closed field (of characteristic 0, if need be). Question: How may I test — by hand — whether it is irreducible?

I understand that there aren't even yet algorithms which can decide whether a univariate polynomial is irreducible, but nonetheless there are tools like the Eisenstein criterion, reduction modulo $p$, and Gauss's lemma which come in useful.

Example: How can I show that $X^n + Y^n + Z^n$ is irreducible over a field of characteristic 0? The following argument works for $n$ a power of 2 and base field $\mathbb{Q}$: we know that $Y^n + Z^n$ is irreducible, since otherwise we could produce a $n$-th root of $-1$ in $\mathbb{Q}$. Thus $\mathfrak{p} = (Y^n + Z^n)$ is a prime ideal, and we can apply the generalised Eisenstein criterion. But this argument breaks down when $n$ is divisible by an odd number (because then we could factor out ($X^m + Y^m$, where $m$ is the highest power of 2 dividing $n$) or when there are $n$-th roots of $-1$. Alternatively, for $n = 2$, I can use an ad hoc approach: if $G(X, Y, Z) H(X, Y, Z) = X^2 + Y^2 + Z^2$, treating $Y$ and $Z$ as constants, we see that we must have a difference-of-squares factorisation, so we have $- (Y^2 + Z^2)$ a square, and iterating the argument, we see this cannot possibly be the case, even if the field is algebraically closed.

That kind of trick works well when there aren't any cross terms, but what about, say, the polynomial $X^2 Y^3 + Y^2 Z^3 + Z^2 X^3$? Here I'm quite lost. If the base field is algebraically closed, then substituting constants for $Z$ don't yield anything useful, since the resulting polynomial is always reducible. While it remains true that the polynomial is irreducible if if I can find an ideal $\mathfrak{I}$ such that the polynomial modulo $\mathfrak{I}$ is irreducible, it's not obvious to me whether any such ideals exist or not.

Context: Some of my exercises / exam problems ask me to show that a given hypersurface in projective space is irreducible. I'm not sure how much weight is placed on it, but it would be reassuring if I could do it properly.

答:
In exactly $3$ variables, suppose a given homogeneous polynomial $F$ is reducible and suppose that the base field is algebraically closed. Then its factors describe projective curves which intersect by Bezout's theorem, and you can determine these intersection points by looking at the points in $\mathbb{P}^2$ where the partial derivatives of $F$ simultaneously vanish. If there are no such points, $F$ must be irreducible. </p>

<p>For example, $X^n + Y^n + Z^n$ must be irreducible by this criterion because the partial derivatives only simultaneously vanish at $(0 : 0 : 0)$. </p>

Body:
In exactly $3$ variables, suppose a given homogeneous polynomial $F$ is reducible and suppose that the base field is algebraically closed. Then its factors describe projective curves which intersect by Bezout's theorem, and you can determine these intersection points by looking at the points in $\mathbb{P}^2$ where the partial derivatives of $F$ simultaneously vanish. If there are no such points, $F$ must be irreducible.

For example, $X^n + Y^n + Z^n$ must be irreducible by this criterion because the partial derivatives only simultaneously vanish at $(0 : 0 : 0)$.


问:
Conjugacy classes of the rotations of the cube
Could anyone help calculate the conjugacy classes in the rotation group of the cube (in
terms of the type of axis of rotation and angle of rotation)? I know that they are related to $S_4$, and the axis of rotation are the main diagonals of the cube, but how do I describe it?

答:
The group of rotational symmetries (also sometimes referred to as the orientation preserving symmetries) is indeed isomorphic to $S_4$, the trouble is figuring what kind of isomorphism there is from the cube to the symmetric group. The cube has eight faces and six vertices so its not immediately clear where a set of four objects will come from. However, if one consider the set of opposing corner of the cube, you can convince yourself that there are in fact, four of them. Thus the rotations of the cube are precisely the permutations of there four sets of opposite corners (or diagonals if you like). If you're having a hard time seeing this, there's really nothing better than picking up a cube (I like to use tissue boxes since they're a decent size and can be written on) and start turning it around.

Once you've convinced yourself that this really is an isomorphism, you've reduced the problem to finding the conjugacy classes of $S_4$ which is trivially easy (think about cycle type).

If you're struggling with realizing the isomorphism from the cube to $S_4$, you can also think about the group of rotations of an octahedron, which is the same as that of a cube, since they are dual, although this need justification if you want to use it.


问:
Number of failures until first r^th success - Expectation, Variance, Probability
I was given the following problem:

Given:

For the r.v $\displaystyle F=number\ of\ failures\ before\ first\ success$.

$\displaystyle E[ F] \ =\ \frac{1-p}{p}$ and $\displaystyle Var[ F] \ =\ \frac{1-p}{p^{2}}$.

Let's now define $\displaystyle X=number\ of\ failure\ before\ r^{th} \ success$, calculate the $\displaystyle E[ X] ,\ Var[ X]$.

My solution was:

Let's notice that if $\displaystyle X=the\ number\ of\ failures\ till\ r^{th} \ success$, then we can define it using the previous question. We have $\displaystyle r$ successes, each success has an $\displaystyle i$ number of falilures before it, for $\displaystyle 0\leqslant i\leqslant k$.

So we can define:

$\displaystyle X\ =F_{1} +F_{2} +F_{3} +...+F_{r}$.

Where each $\displaystyle F_{i} =the\ number\ of\ failures\ after\ the\ ( i-1)^{th} \ success\ and\ before\ the\ i^{th} \ success$.

For example:

$\displaystyle F_{1}$ will be the number of failures before the $\displaystyle 1^{st}$ success.

$\displaystyle F_{2}$ will be the number of failure between the $\displaystyle 1^{st}$ and $\displaystyle 2^{nd}$ success.

$\displaystyle F_{3}$ will be the number of failure between the $\displaystyle 2^{nd}$ and $\displaystyle 3^{rd}$ success.

etc.

Note that each $\displaystyle F_{i}$ is simply the number of failures before some first success, meaning, it is exactly the random variable $\displaystyle F$ we were given.

Expectation:

$\displaystyle E[ X] =E[ F_{1} +F_{2} +F_{3} +...+F_{r}] =E[ F_{1}] +[ F_{2}] +[ F_{3}] +...+[ F_{r}] =r\cdotp \frac{1-p}{p} .$

Variance:

As per the $\displaystyle Var[ X]$, I'm having some trouble calculating it.

I can't use linearity with Variance, and I fail to understand wether I can simply say that:

$\displaystyle Var[ X] \ =\ Var[ rF] =r^{2} Var[ F]$.

Or, I should prove that each $\displaystyle F_{1} +F_{2} +F_{3} +...+F_{r}$ is indepentent, and so deduct that:

$\displaystyle Var[ X] \ =rVar[ F]$.

Which one of the two is correct? And how would I prove that the $\displaystyle F_{i}$ are independet?

答: You are correct. This is the negative binomial distribution and if you think a bit, you'll realise that any failure will be independent of the previous succeses that have occurred so the variance would be the $r\frac{1-p}{p^2}$


问:
How can I start trying to solve currently unsolved math problems?
I'm sorry if this is a stupid question, but I really do want some guidance on this.

I really wanted to start helping solve unsolved math problems. (Like integer factoring, discrete log, etc. but I do want to expand to other problems too) How could I get a grasp of how far a particular problem is to being solved? How can I get a grasp of what math skills I would need to start solving a particular problem? Are there any good resources for problems like factorization?

Any help or guidance is appreciated.

答:
Most if not all open questions in mathematics have at least some previously published work that can be found through a standard literature review.  In fact, the first step in doing any kind of mathematical research is literature review, which addresses all of your questions:

Does this problem have a name?  What do people call this problem?

What is the current state of knowledge regarding this problem or topic?

What are the mathematical prerequisites for research in this area?

Who has worked on the problem should I want to collaborate or ask questions?

If you do not know how to perform this type of scholarly review, then you're probably not in a position to make progress on most open questions, especially the more well-known ones.  It is exceedingly rare that someone, either due to age, lack of prior experience, or lack of access to published research, can make meaningful headway on such problems in complete isolation.  Even when young and inexperienced but extremely talented mathematicians make significant breakthroughs (one of which recently happened in the field of Ramsey theory), such discoveries do not occur in an intellectual vacuum.