问:
Please show that $f(x)=\sum \limits_{n=1}^\infty \frac{x^2}{x^2+n^2}$ is continuous on $\Bbb R$

Let $f(x)=\sum \limits_{n=1}^\infty \frac{x^2}{x^2+n^2}$ on $\Bbb R$. Please show that $f$ is continuous on $\Bbb R$.

I thought if I can prove that $f$ converges uniformly on $R$, then we are done. But I found that Show that $\sum_{n=1}^{\infty}\frac{x^{2}}{x^{2}+n^{2}}$ does not converge uniformly on $(-\infty,\infty)$., we have to find some other ways to prove it.

From Is $f(x)=\sum_{n=1}^\infty\frac{nx^2}{n^3+x^3}$ uniformly continuous on $[0,\infty)$? and Show $ f(x) = \sum_{n=1}^{\infty} \frac{nx}{n^3 + x^3}$ ,$\ g(x) = \sum_{n=1}^{\infty} \frac{x^4n}{(n^3 + x^3)^2}$ are bounded on $[0, \infty)$., I think I can use the same method to prove this question.

I think $f$ converges uniformly on any bounded intervals, hence $$f'(x)=2x\sum \limits_{n=1}^\infty \frac{1}{n^2+x^2}+x^2\sum \limits_{n=1}^\infty \frac{-2x}{(n^2+x^2)^2}$$

Let $g(x)=\sum \limits_{n=1}^\infty \frac{x}{n^2+x^2}$ and $h(x)=\sum \limits_{n=1}^\infty \frac{x^3}{(n^2+x^2)^2}$ on $\Bbb R$. We need to show that $g$ and $h$ are bounded on $\Bbb R$.

First, we discuss $g$.
For $x\in [-1,1]$, $$\left|\frac{x}{n^2+x^2} \right| \le \frac{1}{n^2}$$, so $\left|g(x) \right|\le \sum \limits_{n=1}^\infty \left|\frac{x}{n^2+x^2} \right|\le \sum \limits_{n=1}^\infty \frac{1}{n^2} < \infty$.

For $|x|>1$ be fixed and $m\in \Bbb N$, for any $n\in \Bbb N$ such that $$x(m-1)\le n < mx$$, we have $$\left|\frac{x}{n^2+x^2} \right| \le \left|\frac{x}{x^2(m-1)^2+x^2}\right|=\left|\frac{1}{x}\frac{1}{(m-1)^2+1}\right|$$ and there are at most $\lfloor x \rfloor +1$ of such $n$. $$\left|g(x) \right|\le \sum \limits_{n=1}^\infty \left|\frac{x}{n^2+x^2} \right|\le \sum \limits_{n=1}^\infty \left|\frac{\lfloor x \rfloor +1}{x}\right|\left|\frac{1}{(m-1)^2+1}\right|\le \sum \limits_{n=1}^\infty 2\frac{1}{(m-1)^2+1} < \infty  $$

Hence, $g$ is bounded.

Now, consider $h$, if $x\in [-1,1]$, then $$\left|\frac{x^3}{(n^2+x^2)^2} \right|\le \frac{1}{n^4}$$

For $|x|>1$ and $m\in \Bbb N$ be fixed, for any $n\in \Bbb N$ such that  $$x(m-1)\le n < mx$$, we have $$\left|\frac{x^3}{(n^2+x^2)^2} \right| \le \left|\frac{x^3}{(x^2+(m-1)^2x^2)^2} \right|=\left|\frac{1}{x}\frac{1}{(1+(m-1)^2)^2} \right|$$ and there are at most $\lfloor x \rfloor +1$ of such $n$.

$$\left|h(x) \right|\le \sum \limits_{n=1}^\infty \left|\frac{x^3}{(n^2+x^2)^2} \right|\le \sum \limits_{n=1}^\infty \left|\frac{\lfloor x \rfloor +1}{x}\right|\left|\frac{1}{((m-1)^2+1)^2}\right|\le \sum \limits_{n=1}^\infty 2\frac{1}{((m-1)^2+1)^2} < \infty$$ Hence, $h$ is bounded.

Finally, by mean value theorem, we can prove that $f$ is uniformly continuous on $\Bbb R$.

Appreciate any suggestion.

答: A much shorter proof: continuity is local property. To prove that $f$ is continuous it is enough to prove that it is  continuous at each point. Take any $x \in \mathbb R$. In $[x-1,x+1]$ the series converges uniformly by M-test since it is dominated by a constant times $\sum \frac  1 {n^{2}}$. Hence $f$ is continuous on $[x-1,x+1]$ and we are done.


问:
Basis for $L(D)$ over the hyperelliptic curve $y^2=x^6-1$
Let $X$ be the Riemann Surface associated to the curve $y^2=x^6-1$ and let $\pi$ be the x-projection. Let $p+q=\pi^{-1}(\infty)$ and $r_{k}=(e^{ki\pi/3},0)$. I need to find a basis for the Riemann-Roch space $L(D)$ with $D=p+q+r_0+r_3$.I have found out that $dimL(D)=3$ and I think that the function $x$ is in $L(D)$. How could I complete the basis? I mean, how can I find another function in $L(D)$?

答:
You've already found that $x \in L(D)$. Since $1 \in L(D)$, too, it suffices to find one more linearly independent function.

Let $\DeclareMathOperator{\div}{div} \infty_+ = [1:1:0]$ and $\infty_- = [1:-1:0]$ be the points of $C$ at infinity, and let $\newcommand{\iinfty}{\underline{\infty}} \iinfty = \infty_+ + \infty_-$. Let $\zeta = e^{2\pi i/6}$ be a primitive $6^\text{th}$ root of $1$. Then
\begin{align*}
\div(y) &= r_0 + \cdots + r_5 - 3 \iinfty \\\
\div(x - \zeta^k) &= 2 r_k - \iinfty \, .
\end{align*}
So to construct a function with simple poles at $r_0$ and $r_3$, we should take a ratio of $y$, $x - 1$, and $x - \zeta^3 = x + 1$. So we compute
\begin{align*}
\div\left(\frac{y}{x^2-1}\right) &= \div\left(\frac{y}{(x-1)(x+1)}\right) = r_0 + \cdots + r_5 - 3 \iinfty - \left(2r_0 - \iinfty + 2r_3 - \iinfty\right)\\
&= r_1 + r_2 + r_4 + r_5 - r_0 - r_3 - \iinfty \, .
\end{align*}
Thus $\frac{y}{x^2 - 1} \in L(D)$, and by examining the orders of poles at $\infty_+$ and $r_0$, we find that $\left\{1, x, \frac{y}{x^2 - 1}\right\}$ is linearly independent.

ContentLicense: CC BY-SA 4.0


问:
Difference of two standard Brownian motions
How can one prove that with probability 1 $X(t)=Y(t)$  for infinitely many values of $t\in [0, \infty)$ , where $\{X (t); t \geq 0\}$ and $\{Y (t); t \geq 0\}$ are independent, standard Brownian motions? In a previous step we are asked to show that $Z(t) = X(t) - Y(t)$ is a Brownian motions with zero drift and volatility parameter 2t.

This is just a practice problem as I prepare for my exam.

答:
You know that $Z(t) = X(t) - Y(t)$ and hence $\hat{W}(t) = \frac{Z(t)}{\sqrt{2}}$ is a standard Brownian motion. It follows from the law of iterated logarithm that

$$\overline{lim}_{t \to \infty} \frac{ \hat{W}_t}{
\sqrt{2t \ln \ln t} } = 1$$
and
$$\underline{lim}_{t \to \infty} \frac{\hat{W}_t}{
\sqrt{2t \ln \ln t} } = -1.$$
Hence $\hat{W}(t) = 0$ for infinitely many values of $t$, q.e.d.


问:
Checking if a set is algebraic
Given a subset of $\mathbb C^n,$ is there some way to check if it is the zero set of a collection of polynomials of degree $\le d?$ Certainly when $d=1$ this is easy, and for larger $d$ there are many necessary conditions - but is any of them sufficient?

答: So here is one way to check: Let $I(X)\subset \mathbb C[x_1,\ldots,x_n]$ be the ideal of polynomials vanishing on $X$ and let $V(I(X))\subset \mathbb C^n$ be the collection of common zeros of the polynomials in $I(X).$ Then $X$ is an algebraic set if and only if $V(I(X)) = X.$ Then to find the degree we have to find generators for $I(X)$ of minimal degrees. What bothers me about this criterion is that we have to know what the global polynomial functions are and check if they vanish on $X.$ On the other hand, to test if $X$ is a subspace we just have to check that it's closed under scalars and addition which are conditions on the points of $X$ not involving the global polynomials. I'm wondering if there is a similar condition on the points of $X$ that we can check for higher degrees?

Title: Checking if a set is algebraic


问:
System of equations, containing irrational one.
I have been recently given such a system containing irrational equation. Unfortunately for me, i was unable to solve this.
$$\begin{cases}
\sqrt{x^2+y^2}+\sqrt{x-y}=6 \\
x^2-y^2-x+y=12
\end{cases}
$$
I could only find the (4;0) solution.

I would really like to know if there is some pattern you should follow in order to solve this sort of problems.

Thanking in advance.

答:
That's absolutely a lot of algebra work.
$$
x^2+y^2=\left( 6-\sqrt{x-y} \right) ^2
\\
x^2+y^2=36-12\sqrt{x-y}+x-y
\\
\left\{ \begin{array}{c}
 x^2+y^2+y-x=36-12\sqrt{x-y}  \quad(1)\\
 x^2-y^2+y-x=12 \quad(2)\\
\end{array} \right.
$$
$(1)-(2)$, we have
$$
2y^2=24-12\sqrt{x-y}
\\
y^2=12-6\sqrt{x-y}
\\
y^2-12=6\sqrt{x-y}
\\
y^4-24y^2+144=36\left( x-y \right)
\\
x=\frac{1}{36}\left( y^4-24y^2+36y+144 \right)
\\
\frac{y^8}{1296}-\frac{y^6}{27}+\frac{y^5}{18}+\frac{23 y^4}{36}-\frac{4 y^3}{3}-\frac{14 y^2}{3}+8 y=0
$$
We can also get a polynomial for $x$ in a similar way. Generally speaking, for high degree polynomials we can hardly find a nice way to write the roots down. According to Mathematica's computation, there's another real root, numerically $(4.12...,1.54...)$ and cannot be represented using radicals. So I think you have reached the limit: $(4,0)$ is the only root that can be computed by hand.


问:
Roots of the equation $2^x+2^{-x}=2 - {(x - 2)}^2$
The number of real-valued solutions of the equation

$2^x + 2^{-x} = 2 - {(x - 2)}^2$

I am not sure how to solve this question and I am stuck.

My first guess was $2$ as it seems to me like a quadratic equation but this was not the answer. Can someone please give me an approach to this question?

答: Hints:-Note that $2^x+2^{-x}\geq 2$ which occurs at $x=0$ and also the maximum value of $2-(x-2)^2$ occurs at $x=2$ . Make a graph for the two plots. I hope you know the basic curves. From that you'll see that there is no x for which LHS=RHS.


Title: Roots of the equation $2^x+2^{-x}=2 - {(x - 2)}^2$

Body:
The number of real-valued solutions of the equation

$2^x + 2^{-x} = 2 - {(x - 2)}^2$

I am not sure how to solve this question and I am stuck.

My first guess was $2$ as it seems to me like a quadratic equation but this was not the answer. Can someone please give me an approach to this question?

Body:
Notice that $a + \frac1{a} \ge 2$ for positive $a$ and hence $2^x + 2^{-x} \ge 2$.

Notice that $2 - (x-2)^2 \le 2$. Hence $$2 \le 2^x + 2^{-x} = 2 - {(x - 2)}^2 \le 2.$$

It follows that  $$2^x + 2^{-x}  =2 \text{${}$ ${}$ ${}$ ${}$  (1)}$$ and $$2 = 2 - {(x - 2)}^2 \text{${}$ ${}$ ${}$ ${}$  (2)}$$.

It follows from (2) that $x= 2$.

It follows form (1) that $x \ne 2$.

Hence there's no solutions.


答:
Ahh well as you can see in LHS there is a equation which has a minimum value 2

You can prove this using A.M-G.M inequality

That A.M > G.M for any two numbers

For this case the A.M of $2^x and 1/2^x$ is

$(2^x +(1/2^x))/2$

And the G.M will be 1

So it follows LHS $>= 2 $

And it can be easily found that the range of the function on RHS is $<=$ 2
So here exists only one solution that is $2^x$ = 1 for LHS which is x = 0

And for RHS at x = 2

We got contradictory values of x so this equation has no real solution.

See the graph {LHS = Blue } and { RHS = Red}


问:
Absolute continuity of induced measures
Let $g : [a,b] \rightarrow \mathbb{C}$ be of bounded variation and let $1_A$ denote the indicator function of $A$ for measurable subsets of $[a,b]$.

We put
$$m_g(A) := \int 1_A dg,$$
i. e. $m_g(A)$ is the Lebesgue-Stieltjes integral of $1_A$ with respect to $g$. Then $m_g$ is a Borel measure on $[a,b]$.

I know that if $g$ is differentiable, then $m_g$ is absolutely continuous (=dominated by Lebesgue measure).

Now I wonder if this is always true? If not, what are conditions on $g$ that ensure that $m_g$ is absolutelly continuous?

Note: This is not a homework but I'm just interested and I am a little bit rusty on measure theory. Any reference or hint is welcome.

答:
Lat $a<c<b$. If $g(x)=0$ for $x <c$ and $1$ for $x \geq c$ then $g$ is of bounded variation and $m_g$ is the delta measure at $c$. [This means $m_g(A)=1$ if $c \in A$ and $0$ otherwise].  This measure is singular w.r.t. Lebesgue measure. A sufficient condition  for absolute continuity is $g$ continuously differentiable, as you have observed.

A necessary and sufficient condition is the so-called absolute continuity of the function $g$: For every $\epsilon >0$ there exists $\delta >0$ such that for any finite disjoint collection $(a_i,b_i):  1\leq i \leq n$ of intervals with $\sum (b_i-a_i) <\delta $ we have $\sum |g(b_i)-g(a_i)| <\epsilon$.

Title: Absolute continuity of induced measures


问:
Finding the number of non-negative integral solution of a general equation.
Let there be an equation,
$$a_1x_1+a_2x_2+\dots+a_nx_n=S$$
Here, $a_i$ where $i=1,2,\dots,n$ are positive integers, also S is a positive integer too.
How many non-negative integral solutions does the above equation have?

For example $3x+y+z=24$ has 117 integral solutions.  Can you also verify it with this result?

I am not looking for a Java program to solve it and I am in 10 standard so a bit of explanation will help a lot.

答:
We can solve it with the help of "finding a probability using a multinomial".
If in the equation,
$$a_1x_1+a_2x_2+...+a_nx_n=S$$
a)Zeroes are not considered blanks, the number of solution of the equation
=coefficient of $x^n$ in $$(1+x^{a_1}+x^{2a_1}+..)(1+x^{a_2}+x^{2a_2}+..)...(1+x^{a_n}+x^{2a_n}+..)$$
$$=(1-x^{a_1})^{-1}(1-x^{a_2})^{-1}...(1-x^{a_n})^{-1}$$
b)Zeroes are considered as blanks, the number of solution of the equation
=coefficient of $x^n$ in $$(x^{a_1}+x^{2a_1}+..)(x^{a_2}+x^{2a_2}+..)...(x^{a_n}+x^{2a_n}+..)$$
$$=(x^{a_1+a_2+a_3+...+a_n})(1-x^{a_1})^{-1}(1-x^{a_2})^{-1}...(1-x^{a_n})^{-1}$$

Now there are many ways to find the coefficient i.e.,synthetic multiplication of polynomials, calculating the $n^{th}$ derivative of the multinomial using cauchy integral formula and many other ways.


问:
Double integral polar coordinates, limits of angle
We would like to calculate using polar coordinates the double integral of f(x,y)=x^2*y in R, where R is the surface between x+y-2=0,the circle with (0,0) as its center and radius equals to 2 and x>0,y>0.
I used polar coordinates r,t, but I face difficulty in finding the limits of angle t.

I would appreciate for your quidance.

Thank you very much in advance.

答:
$\textbf{Hint:}$ you have to evaluate the integral of the function $f(x,y)=x^2y$ in the domain $E=\{(x,y)\in \mathbb R^2:x,y\geq 0,y>2-x\text{ and }x^2+y^2<4\}$.

If you look at the area bounded by these two curves you can observe that it can be obtained by a subtraction of the area defined by the inequality $y>2-x$, which is a triangle, from the area defined by the inequality $x^2+y^2<4$, which is a circular sector.

Let $T$ be the area of the triangle with vertices in $(0,0),(2,0)$ and $(0,2)$, and $C$ the area of the circular sector. Since we have the restriction $x,y\geq0$, we have to consider $C\cap \mathbb R^+\times\mathbb R^+$ and it's easy to see that, applying the polar coordinates' substitution, we get
$$\begin{cases}\rho\cos\theta\geq0\\\rho\sin\theta\geq0\\\rho^2<4\end{cases}\implies\rho \in(0,2)\text{ and }\theta\in(0,\dfrac{\pi}{2})\implies$$$$\iint_Ef(x,y)dxdy=\iint_Cf(x,y)dxdy-\iint_Tf(x,y)dxdy=$$$$=\int_{(0,\frac{\pi}{2})}\int_{(0,2)}(\rho\cos\theta)^2\rho\sin\theta\rho d\rho d\theta-\iint_T x^2ydxdy=$$$$=\int_{(0,\frac{\pi}{2})}\int_{(0,2)}\rho^4\cos^2\theta\sin\theta-\int_0^2\int_0^{2-x} x^2ydydx.$$


问:
$\min(X,Y)$ and $X-Y$ with $X,Y$ iid exponential distributions
Let :

$X$ and $Y$ be two independent exponential distributions of parameter $\lambda$

$T=X-Y$

$Z =\min(X,Y)$

We want to show that $T$ and $Z$ are independant, without any memoryless property.

My attempt :

$
\begin{align*}
f_{T}(s) & = \int_{ \mathbb{ R} }^ {} f_X(s-u) f_{-Y}( u) du \\
                 &= \int_{ \mathbb{ R} }^ {} e^{ - \lambda (s-u) } e^{ - \lambda u} \mathbb{1}_{ s-u \geq 0}   \mathbb{1}_{ u \leq 0}   \\
               & = e^{ - \lambda s }   \int_{ \mathbb{ R} }^ {}  e^{ 2 \lambda u}  \mathbb{1}_{ u \leq  (0 \wedge s)} \\
              & = \frac{1} {2}  e^{ - \lambda  (2 (0 \wedge s)     -s  )} \\
             &= \frac{1} {2}  e^{  -  \lambda |s|} \\
\end{align*}
$

$f_Z(z)= 2 e^{- 2 \lambda z}$

$
\begin{cases}
X-Y <t & \\
\min(X,Y) <z\\
\end{cases}
\iff
\begin{cases}
X<t+Y & \\
\{ \min(X,Y) <z, Y \leq Z  \} \cup  \{ \min(X,Y) <z, Y \geq Z  \} \\
\end{cases}
\iff
\begin{cases}
\{ X< Y+t ,Y \leq Z  \} \cup \\
 \{ X<z, Y \geq z  \} \\
\end{cases}
$

For $t>0$ and $z>0$
$
\begin{align*}
F_W(t,z)&= P(T<t,Z \leq z) \\ &= \int \int \mathbb{1}_{(X-Y) <t , \min(X,Y) <z } dP \\
 &= \int_{0}^{z} \int_{0}^{y+t} f_X(x) dx  f_Y(y) dy + \int_{z}^{ + \infty} \int_{0}^{z} f_X(x) dx ) f_Y(y) dy \\
\end{align*}
$

For $t>0$ and $z>0$, $F_{Z,T}(z,t) =1 - \frac{1}{2} e^{-t}(1- e^{-2z}) - e^{ -2 z} $

For $z >0$ and $t<0$, $ Y \geq X-t$ and $X \leq z$, therefore,

$
\begin{align*}
F_W(t,z) &= \int_{0}^{z} \lambda e^{ - \lambda x}  ( \int_{x-t}^{ + \infty} \lambda e^{ - \lambda y } dy ) dx \\
&=\int_{0}^{z} \lambda e^{ - \lambda x}  e^{ - \lambda (x-t)} \\
&=e^{ \lambda t} \frac{1}{2} (1- e^{ - 2 \lambda z })\\
\end{align*}
$

答:
You only need to prove the PDF is separable.

If $x\ge y$, $z=x$ and $z_x=1,\,z_y=0$. In terms of Iverson brackets$$z_x=[x\ge y],\,z_y=1-[x\ge y],$$ so$$\frac{dtdz}{dxdy}=\left|\begin{array}{cc}
1 & -1\\
\left[x\ge y\right] & 1-\left[x\ge y\right]
\end{array}\right|=1.$$Since $x+y=|t|+2z$, the infinitesimal probability is$$\lambda^2e^{-\lambda(x+y)}dxdy=\tfrac12\lambda e^{-\lambda|t|}dt\cdot2\lambda e^{-2\lambda z}dz.$$This also obtains the distributions of $T\sim\operatorname{Laplace}(0,\,1/\lambda),\,Z\sim\operatorname{Exp}(2\lambda)$.


问:
Infinite paths that connect two vertices?
This is a follow-up to another question concerning infinite paths which was admittedly ill-posed. I hope this question is posed better.

The graph $N$ with vertex set $V(N) = \mathbb{N}$ and $(x,y) \in E(N)$ iff $x < y$ contains paths of arbitrary length connecting 0 and an appropriate $n$ .

The graph $Q$ with vertex set $V(Q) = \mathbb{Q}$ and $(x,y) \in E(Q)$ iff $x < y$ contains paths of arbitrary length connecting 0 and 1.

Of course, both graphs contain infinite paths, starting from 0, but ending nowhere.

It's more or less obvious, that $N$ doesn't contain a path of infinite length connecting 0 and an $n\in \mathbb{N}$ (because all $n$ are finite).

But it's hard (for me) to "see" and get a feeling why $Q$ doesn't contain a path of infinite length connecting 0 and 1: each finite path between 0 and 1 is a finite subset of $E(Q)$: $\lbrace (0,q_1),(q_1,q_2),...,(q_n,1)\rbrace$. Why cannot there be an infinite subset of that kind?

Is it impossible to define "that kind", i.e. the
  property "being a path
  connecting 0 and 1", or can we define
  it (in second order logic maybe) but
  prove, that no infinite subset of $E(Q)$ with this property exists?

答:
I, for one, think I understand what you mean.  To see why there are no infinite sets, try to formulate "that kind of path" (the ones connecting 0 to 1) as a set: We could try letting $P = \{\text{paths }\alpha \text{ in } Q| \alpha \text{ starts at 0 and ends at 1}\}$ but an infinite set never ends, so our vocabulary is lacking.  Let's try something more precise: Let $P=\{(v_0,v_1),(v_1,v_2),\ldots,(v_{n-1},v_n)|v_0=0,v_n=1\}$ but this is inherently assuming that the length of the path is finite, namely, $n$.  I know this isn't a proof, but it seems to me that "connectedness" (at least in graphs) is a finite property.

In order to distinguish between the two paths brought up by @tomcuchta and @Theo, you would need to assign weights to the edges, and then the definition of minimal path changes drastically. The reason we see the related sequences of real numbers as different is because the distance between the values is shrinking (or staying constant).  But the distance between two vertices in $Q$ is still always 1.  In $Q$, the vertices are just labelled with the elements of $\mathbb{Q}$.  So, since the vertices are not getting closer to one another, you never get any closer to 1, following the path given by tomcuchta.

There may be other ways to describe the possibility of infinite paths, and I'd be interested to hear them, but given the way you've formulated your question above, I don't think you can speak of an infinite path with two endpoints.  The path given by tomcuchta is infinite, but it does not end at 1 (or anywhere else).

ContentLicense: CC BY-SA 3.0


问:
Finding $E[X\mid Y]$ and $\operatorname{Var}(X\mid Y)$ given mean and variance of $X$ and $Y$
Suppose we have 2 normal distributions $X$ and $Y$ with mean $u_1$ and $u_2$ and variance $\sigma_1^2$ and $\sigma_2^2$; find $E[X\mid Y]$ and $\operatorname{Var}(X\mid Y)$.

I know
$$E[X\mid Y] = \mu_1 + \rho\sigma_1 \frac{Y - u_2}{\sigma_2} $$
and
$$\operatorname{Var}[X\mid Y] = \sigma_1 (1 - \rho^2)$$
but I can't prove it.

For $E[X\mid Y]$ I start with
$$E[X\mid Y] = \int_{-\infty}^{+\infty} x f_{X|Y}(x\mid y)\ dx$$
but that doesn't work because for calculating $f_{X\mid Y}(x\mid y)$ I need $f_{X,Y}(x,y)$ I don't have that.
Can any one help me?

答:
The density approach will work. In the simplest case, assume that $X$ and $Y$ are each standard normal, with correlation $\rho$, so that the joint density of $(X,Y)$ is
$$
f(x,y)=\frac1{2\pi\sqrt{1-\rho^2}}\exp \left[-\frac1{2(1-\rho^2)}(x^2-2\rho xy+y^2)\right]
$$
while the marginal density of $Y$ is
$$f(y)=\frac1{\sqrt{2\pi}}\exp\left[-\left(\frac{y^2}2\right)\right].
$$
The conditional density $\displaystyle f(x\mid y)=\frac{f(x,y)}{f(y)}$ is the ratio of these. So conditional on $Y=y$, the density of $X$  is
$$\begin{align}
f(x\mid y)&=\frac1{\sqrt{2\pi(1-\rho^2)}}\exp\left[-\frac1{2(1-\rho^2)}(x^2-2\rho xy+y^2-(1-\rho^2)y^2)\right]\\
&=
\frac1{\sqrt{2\pi(1-\rho^2)}}\exp\left[-\frac1{2(1-\rho^2)}(x-\rho y)^2\right]\end{align}
$$ which we recognize as the density of a normal random variable with mean $\rho y$ and variance $1-\rho^2$. It follows that
$$
E(X\mid Y=y) = \rho y\qquad{\rm and}\qquad \operatorname{Var}(X\mid Y=y)=1-\rho^2.$$

For the general case, write $\displaystyle X':=\frac{X-\mu_1}{\sigma_1}$ and $\displaystyle Y':=\frac{Y-\mu_2}{\sigma_2}$. Apply the previous case to $X'$ and $Y'$, and conclude
$$\begin{aligned}
E\left (X\mid Y=y\right)&=E\left(\mu_1+\sigma_1 X'\biggm| Y'=\frac{y-\mu_2}{\sigma_2}\right)=\mu_1+\sigma_1 E\left(X'\biggm| Y'=\frac{y-\mu_2}{\sigma_2}\right)\\
&=\mu_1+\sigma_1\rho\left(\frac{y-\mu_2}{\sigma_2}\right)
=\mu_1+\rho\frac{\sigma_1}{\sigma_2}(y-\mu_2)
\end{aligned}
$$
and
$$\begin{aligned}
\operatorname{Var}(X\mid Y=y)&=\operatorname{Var}\left(\mu_1+\sigma_1 X'\biggm| Y'=\frac{y-\mu_2}{\sigma_2}\right)
=\sigma_1^2\operatorname{Var}\left( X'\biggm| Y'=\frac{y-\mu_2}{\sigma_2}\right)\\
&=\sigma_1^2(1-\rho^2).\end{aligned}
$$


问:
Finding probability of all outcomes occurring in n repetitions
Suppose you have numbers 1, 2, 3, 4. You can pick each of these numbers with probability 1/4. After n iid selections, what is the probability that you have seen all 4 numbers? Express this as a function of n.

For example, if n=5, and you draw the numbers 1,1,3,2,1 then this is unsuccessful, as 4 was never picked. If you draw 3,4,2,3,4, this is also unsuccessful as 1 was never picked. However, 3,2,2,1,4 is successful.

I am aware you can use a discrete Markov chain and find the n-step transition probability matrix on Wolfram, but is there a way to find this without a calculator and with reasoning alone?

答:
Let $S(i)$ be the set of outcomes where $i$ does not appear in $n$ picks.

Let $N(j)$ be the number of outcomes in the intersection of $j$ of the $S(i)$. Then
$$
N(j)=\overbrace{\quad\binom{4}{j}\quad}^{\substack{\text{number of}\\\text{ways to}\\\text{choose the}\\\text{$j$ missing}}}\overbrace{\ (4-j)^n\ \vphantom{\binom41}}^{\substack{\text{number of}\\\text{ways to pick}\\\text{$n$ from the}\\\text{remainder}}}
$$
According to the Generalized Inclusion-Exclusion Principle, the number of outcomes in none of the $S(i)$ is
$$
\begin{align}
&N(0)-N(1)+N(2)-N(3)+N(4)\\
&=\binom{4}{0}(4-0)^n-\binom{4}{1}(4-1)^n+\binom{4}{2}(4-2)^n-\binom{4}{3}(4-3)^n+\binom{4}{4}(4-4)^n\\
&=4^n-4\cdot3^n+6\cdot2^n-4\cdot1^n+0^n
\end{align}
$$
The total number of outcomes is $4^n$, so the probability that we see all of the possible numbers is
$$
1-4\ \left(\frac34\right)^n+6\ \left(\frac12\right)^n-4\ \left(\frac14\right)^n+[n=0]
$$


问:
Stein's Complex Analysis Page 46 Theorem 4.1
In the proof of the Theorem 4.1 in the Stein's Complex Analysis book, he first constructed a keyhole. Then by previous knowledge, the integral of $F$ along the boundary of keyhole is zero. Then he said: make the corridor narrower by letting the $\delta$ tend to $0$, and use the continuity of $F$ to see that in the limit, the integrals over the two sides of the corridor cancel out...

I don't understand why while letting $\delta$ tend to $0$, the Integral can still keep $0$. It seems that additional conditions are needed. For example, something like 'uniformly convergence' to allow the interchange of integration and limiting.

Let me explain what I am confused about: for example, I integral an entire function $f(z)$ from $-1+iy$ to $1+iy$ horizontally (called the segment $T_y$). So $\int_{T_y}f(z)dz=\int_{-1}^{1}f(x+iy)dx$. But however, I cannot deduce that $\int_{T_{y}}f(z)dz\rightarrow\int_{-1}^{1}f(x)dx$ as $y\rightarrow 0$.

So similarly, I don't think we can get $\int_{\varGamma_{0,\epsilon}}F(\zeta)d\zeta=0$...

答:
You are right. You need uniform continuity. Do you know the following theorem?

The Heine–Cantor theorem: Every continuous function on a compact set is uniformly continuous.

Now to solve your problem:

The function $f$ is continuous on the solid rectangle containing both lines and  the rectangle is compact, thus $f$ is uniformly continous on the rectangle. Now using uniform continuity, for $y$ small you can write $ f(x+iy) = f(x) + \epsilon(x,y) $ and $|\epsilon(x,y)|\to 0$ as $y \to 0$.

Now $$\left|\int (f(x+iy)-f(x))dx\right| \le \int \left|f(x+iy) -f(x)\right|dx = \int |\epsilon(x,y)|dx,$$

and by choosing $y$ small you can make the last integral as small as you want.

Heads up: Because of continuity, the values of the two integrals are within $\epsilon$ of one another, and so in the limit they are equal, but their values are not necessarily zero.


问:
Evalutate $\int\frac{dx}{x\sqrt{1-\frac{4}{3}x^4}}$
How to integrate
$$\int\frac{dx}{x\sqrt{1-\frac{4}{3}x^4}}$$
Mathematica found
$$\ln x-\frac{1}{2}\ln\left(1+\sqrt{1-\frac{4}{3}x^4}\right)$$
but I can't find a method to arrive at this solution.</p>

Body: Try $u=\sqrt{1-(4/3)x^4}$, $u^2=1-(4/3)x^4$, $2u\,du=-(16/3)x^3\,dx$, together with $dx/x=x^3\,dx/x^4$.


问:
Prove that $f(x)= \int_{0}^\infty\sin t \cdot x(t)dt$, is linear and continuous and find $\Vert f \Vert$
Given $a>0$, we denote $X$ is a space of real function $x=x(t)$ continuous on $[0,\infty)$ and satisfy $\sup e^{at} \vert x(t) \vert <\infty$.

On $X$, we define $\Vert x \Vert =\displaystyle \sup_{0 \le t < +\infty} e^{at}\vert x(t) \vert $.

Prove that $f:X \rightarrow \mathbb{R}$, $f(x)=\displaystyle \int\limits_{0}^\infty\sin t \cdot x(t)dt$, $x \in X$ is linear and continuous and find $\Vert f \Vert$.

I proved $f$ is linear already and when it comes to "continuous" I have tried to prove $\vert f(x) \vert \le M \Vert x \Vert$ but have no idea.

Any hint or solution is greatly appreciated. Thank you.

答: Hint: $|f(x)|=|\int_0^{\infty} e^{-at} \sin t \, e^{at}x(t)dt| \leq \|x\| \int _0^{\infty} e^{-at} |\sin t|dt$. If $x(t)=e^{-at} \text{sgn} (\sin t)$ then $\|x\|=1$ and  $f(x)= \int _0^{\infty} e^{-at} |\sin t|dt$. This would have proved that $\|f\| =\int _0^{\infty} e^{-at} |\sin t|dt$ but for the fact that $x$ is not continuous. You know exactly where $\sin t$ changes sign so modify $x$ suitably near those points to show that $\|f\| =\int _0^{\infty} e^{-at} |\sin t|dt$


问:
$f:X \to Y$ is a continuous map. Suppose it induces an isomorphism $f_*:H_*(X) \to H_*(Y)$. Does that imply $f:X \to Y$ is a homotopy equivalence?
Suppose that $f:X \to Y$ is a continuous map. Suppose it induces an isomorphism $f_*:H_*(X) \to H_*(Y)$. Does that imply $f:X \to Y$ is a homotopy equivalence?

I think that this is not the case. This is because the following is not True.

Let $f_*:C_*\to D_*$ be a chain map such that $f_*$ induces an isomorphism in homology. Then $f_*$ is a chain homotopy equivalence,

Can anyone give me a simple example.

答:
No; for example, there are non-contractible spaces that are acyclic (have trivial reduced homology), such as the Poincare homology sphere minus a point. You can take $X$ to be any such space and $Y$ to be a point.

However, this is true if $X$ and $Y$ are simply connected CW complexes, by a combination of the relative Hurewicz theorem and Whitehead's theorem. This is "Whitehead's theorem for homology."

Edit: As for chain complexes, there's a Whitehead's theorem in this setting also: a quasi-isomorphism between two bounded-below complexes of projective modules is a chain homotopy equivalence (see, for example, Proposition 4.1 in these notes). These complexes are roughly the analogues of CW complexes.


问:
Conditional probability hard problem
Suppose that we have three coins. One has two heads, while the others are standard. Two coins are flipped. In each case, the result is heads.

If you pick one of these two coins randomly, what is the probability that you will get the two headed coin?

If you pick the unflipped coin, what is the probability that you will get the two headed coin?

My attempt at 1:

Let A be the event: "We obtain two heads" and let B be the event: "We got the two-headed coin." We are looking for $$P(B|A)=\frac{P(B\bigcap A)}{P(A)}$$
Now, from Baye's Theorem: $$P(B\bigcap A)=P(A|B)P(B)=(\frac{1}{2})(\frac{1}{3})=\frac{1}{6}$$
I still need to find $P(A)$, could anyone help?

答:
For the first question, ask yourself what's the probability that one of the coins picked are two-headed (event $A$) given that we obtained $2$ heads (event $B$)? There are two competing statements here:

$1-$Both of the coins are fair coins.

$2-$ One of the coins is a fair coin.

According to Bayes theorem, we have:

$P(A| B) = \frac{P(B| A) P(A)}{ P(B| A) P(A) + P(B| \neg A) P(\neg A)}$

Here we have $P(B|A)=\frac{1}{2}$ since flipping the coins are independent and the probability of getting head with the two-headed coin is $1$ while the probability of getting head with the other coin is $\frac{1}{2}$, we multiply these two.

We have $P(A)=\frac{\binom{1}{3}}{\binom{2}{3}}$ since we are picking two coins out of three in total and we need one of them to be the two-headed coin. This gives us that $P(\neg A)=1- \frac{\binom{3}{1}}{\binom{3}{2}}$

$P(B| \neg A)=\frac{1}{4}$ Since event $A$ not happening means that both of the coins picked are fair coins where the probability of getting head in each one of them is $\frac{1}{2}$ so we multiply this probability by itself then get $\frac{1}{4}$

I will leave the arithmetic to you, however, this gives us the probability that one of the coins picked is the two headed coin, then we need to multiply this probability  by $\frac{1}{2}$ since we are picking on these $2$ coins at random and we need the probability we the two headed one.

Title: Conditional probability hard problem


问:
What is the maximum number of consecutive sides whose lengths you can choose without uniquely determining the polygon?

For every $n$-sided equiangular polygon, define $f(n)$ to be the maximum number of consecutive sides whose lengths you can choose without uniquely determining the polygon. Find the last two digits of:$$\sum_{n=3}^{2019} f(n)$$

Can someone verify this ? I think $(n-3)$ works, and I posted it in AOPS, but the guy is saying that it is wrong.

Here's the solution:

Answer : $36$, $f(n)=n-3 $

clearly for any $n , n-2$ doesn't satisfy because then equiangular property fixes the polygon.

For $n=3, 4$ , clearly $f(n)=n-3$ works .
For $n\ge 5$ , consider the following construction , which i took for $n=7$ ,( but one can understand what's the construction) . It's just extending $2$ sides and making a parallel side . So $n-3$ consecutives doesn't fix.

So $n-3$ works .

And answer is  $\frac{2016\cdot 2017}{2}=2033136 \equiv 36 \pmod {100}$

Here's the other guy's solution, which many people are saying it to be true :
Equilateral triangle which is regular, square but we can have a rectangle, pentagon where all it's sides are equal and so on. Then I see a pattern. This was the crucial observation.

In Euclidean Geometry, if we have an equiangular polygon with odd sides then it's all the sides will be equal, i.e it will be a regular polygon. So, we have the regular polygons with side lengths $3,5,\cdots$ and so on. Here, in this case we can choose only one of it's side length and all the side lengths will be automatically chosen since it is an equiangular and hence regular polygon. Therefore, $f(n)=\boxed{1}$, in the case when we have an equiangular polygon with odd sides.
Now, we see what happens when we choose an equiangular polygon with odd sides, so we can choose half of the sides of the polygon(these needs to be consecutive sides). Therefore $f(n)=\frac{n}{2}$ in the case where $n=\text{even}$.
$\text{Sum}=(1+2)+(1+3)+(1+4)+\cdots+(1+1009)+1=510553 $.

Therefore Last two digits of the sum$\boxed{53}$.

Is there a flaw in the solution , I wrote ?

答:
Your solution is correct.

You can have an equiangular pentagon that is not regular. Take the regular pentagon and extend any two of its non-adjacent alternate sides. You can now shift the intermediate side parallel to itself such that it remains between the extensions.

You now have an equiangular non-regular pentagon.


答:
In case when $Y=\{*\}$ is a singleton your question can be stated as:

Suppose $X$ has all homology groups trivial. Is it contractible?

Unfortunately this is false. The double comb space is famously known to have all homotopy, homology and cohomology groups trivial, even though it is not contractible.

Body:
Suppose that $f:X \to Y$ is a continuous map. Suppose it induces an isomorphism $f_*:H_*(X) \to H_*(Y)$. Does that imply $f:X \to Y$ is a homotopy equivalence?

I think that this is not the case. This is because the following is not True.

Let $f_*:C_*\to D_*$ be a chain map such that $f_*$ induces an isomorphism in homology. Then $f_*$ is a chain homotopy equivalence,

Can anyone give me a simple example.


问:
Combinatorics on numbers.

Find the integer $n$ which has the following property: If the numbers from $1$ to $n$ are all written down in decimal notation the total number of digit written down is $1998$. What is the next higher number (instead of $1998$) for which this problem has an answer?

My try:

This problem was given under exercise Inclusion-Exclusion Principle, I think we should first find that $n$ for which above holds, then we'll find the number of digits (same manner as in problem) in $n+1$.

Like digits from: $\begin{cases}1-9 &=9\\ 10-99 &=2\times 90\\ 100-999 &=3\times 900\\ & \vdots\\ \end{cases}$

We can easily see the pattern above. But I don't see whether we seriously need IEP.

How to take it from here?

答:
The number of digits when we write down all numbers till 100 is 192.

Now adding 100 more will cause total number of digits to increase by 300, as every number added was a three digit number.

Then for n=700 gives 1992 total digits
Because the first 100 gives 192 and the rest of the 100s gives 300 digits.
Then n=701 will have 3 more and n = 702 will have 6 more i.e will have 1998

Then n = 702 is required number
and next possible will be 1998+3 = 2001.

Title: Combinatorics on numbers.


问:
Finding the MLE of a multinomial distribution (uneven probabilities)
I am trying to simulate loaded die where the face probabilities are:

$$
p_1=p_2=p_3=p_4=1/6+\theta\text{ and }p_5=p_6=1/6-2\theta
$$

And so using the multinomial distribution I have:

$$
\binom{n}{x_i}\prod_{i=1}^6 p_i^{\displaystyle x_i}=\binom{n}{x_i}\left ( \frac{1}{6} + \theta  \right)^{\displaystyle\sum_{i=1}^4 x_i} \left(\frac{1}{6}-2\theta \right)^{ \displaystyle\sum_{i=5}^6 x_i}
$$

How do I find the MLE w.r.t $\theta$?

答:
Your notation is rather confused.  You have $\displaystyle\prod_{i=1}^6 p_i^{x_i}$, and that means
$$
p_1^{x_1} p_2^{x_2} p_3^{x_3} p_4^{x_4} p_5^{x_5} p_6^{x_6}.
$$
In the first factor, $p_1^{x_1}$ you have $i=1$, in the next you have $i=2$, and so on.  But what is $i$ equal to in the factor $\dbinom{n}{x_i}$ that you've written in front of the product??

What needs to be there is not $\dbinom{n}{x_i}$, but $\dbinom{n}{x_1,\ldots,x_n}=\dfrac{n!}{x_1!\cdots x_n!}$, i.e. a multinomial coefficient.

You need the value of $\theta\in[0,1/12]$ that maximizes the product. First check that $\theta$ must be between $0$ and $1/12$. The multinomial coefficient does not depend on $\theta$.  So you can just maximize
$$
L(\theta)=\left(\frac16+\theta\right)^{x_1+x_2+x_3+x_4}\left(\frac16-2\theta\right)^{x_5+x_6}.
$$
Since $\log$ is an increasing function this is the same as maximizing
$$
\ell(\theta)=\log L(\theta) = (x_1+x_2+x_3+x_4)\log\left(\frac16+\theta\right) + (x_5+x_6)\log\left(\frac16-2\theta\right).
$$
We have
\begin{align}
\ell\,'(\theta) & = \frac{x_1+x_2+x_3+x_4}{\frac16+\theta} - \frac{2(x_5+x_6)}{\frac16-2\theta} \\[8pt]
& = 6\left( \frac{x_1+x_2+x_3+x_4}{1+6\theta} - \frac{2(x_5+x_6)}{1-12\theta} \right) \\[8pt]
& = (\text{a positive number}) \cdot \left( (1-12\theta)(x_1+x_2+x_3+x_4)-(1+6\theta)(x_5+x_6) \right) \\[8pt]
& = (\text{pos.}) \cdot (x_1+x_2+x_3+x_4-(x_5+x_6) - 6\theta(2(x_1+x_2+x_3+x_4)+x_5+x_6)).
\end{align}

Find the value of $\theta\in[0,1/12]$ for which $\ell\,'(\theta)=0$, and show that for smaller values of $\theta\in[0,1/12]$ this derivative is positive and for larger ones it is negative.


问:
Can I assume linear independence for range $T = \{Tv_1, ..., Tv_n\}$ where $\{v_1, ..., v_n\}$ is linearly independent and doesn't belong to null $T$.
Assume $T \in \mathcal{L}(V, W)$. If $\{u_1, ..., u_m, v_1, ..., v_n\}$ is a basis of $V$, where $\{u_1, ..., u_n\}$ spans the null space, can I automatically assume that $\{Tv_1, ..., Tv_n\}$ is linearly independent?

I have seen this several times now but found no convincing proof  for it. Perhaps I am overlooking the obvious.

Body:
We have:

$c_1Tv_1+c_2Tv_2+...+c_nTv_n = 0 \\
\implies T(c_1v_1+c_2v_2+...+c_nv_n) = 0\\
\implies c_1v_1+c_2v_2+...+c_nv_n \in Null(T)$

So we can write:

$c_1v_1+c_2v_2+...+c_nv_n = d_1u_1+d_2u_2+...+d_nu_n$

Since the $\{u_1, ..., u_m, v_1, ..., v_n\}$ is independent, all $c$'s and $d$'s are 0.

So $\{Tv_1, ..., Tv_n\}$ is independent.


答:
Suppose $Tv_i$ are linearly dependent. Then $\exists c_i$ not all $0$ such that $\sum_{i=1}^n c_iTv_i=0$.

This gives $T(\sum_{i=1}^n c_iv_i)=0$, i.e. $\sum_{i=1}^n c_iv_i$ belongs to the null-space of $T$.

Since $u_i$ span the null-space, this would mean $\sum_{i=1}^nc_iv_i=\sum_{j=1}^mC_ju_j$ which violates the linear independence of basis vectors.

Title: Can I assume linear independence for range $T = \{Tv_1, ..., Tv_n\}$ where $\{v_1, ..., v_n\}$ is linearly independent and doesn't belong to null $T$.

Body:
Assume $T \in \mathcal{L}(V, W)$. If $\{u_1, ..., u_m, v_1, ..., v_n\}$ is a basis of $V$, where $\{u_1, ..., u_n\}$ spans the null space, can I automatically assume that $\{Tv_1, ..., Tv_n\}$ is linearly independent?

I have seen this several times now but found no convincing proof  for it. Perhaps I am overlooking the obvious.


答:
<s> When calculating $P(B\bigcap A)$, I think you made a mistake: $P(B) = 2/3$, not $1/3$.</s>

Sorry I think you are right: $P(B) = 1/3$. I misunderstood what $B$ is. (I thought it is the event "the two headed coin is among the two flipped", but it is the event "you picked up the two headed coin", which has probability 1/3.)

To find P(A), use the total probability theorem:
$P(A) = P(A|B)P(B) + P(A|B^c)P(B^c)$

Title: Conditional probability hard problem

Body:
Suppose that we have three coins. One has two heads, while the others are standard. Two coins are flipped. In each case, the result is heads.

If you pick one of these two coins randomly, what is the probability that you will get the two headed coin?

If you pick the unflipped coin, what is the probability that you will get the two headed coin?

My attempt at 1:

Let A be the event: "We obtain two heads" and let B be the event: "We got the two-headed coin." We are looking for $$P(B|A)=\frac{P(B\bigcap A)}{P(A)}$$
Now, from Baye's Theorem: $$P(B\bigcap A)=P(A|B)P(B)=(\frac{1}{2})(\frac{1}{3})=\frac{1}{6}$$
I still need to find $P(A)$, could anyone help?


问:
Is the symmetric closure of a relation which is already transitive, itself transitive?
Let $R$ be a transitive relation. Is the symmetric closure of $R$ also transitive?

答: Counterexample, adapted from @amWhy's attempt based on an observation by @aefrrs: if $a\ne b\ne c\ne a$, the transitive relation $\{(a,\,b),\,(b,\,c),\,(a,\,c)\}$ has symmetric closure $\{(a,\,b),\,(b,\,c),\,(a,\,c),\,(b,\,a),\,(c,\,b),\,(c,\,a)\}$, which isn't transitive as it doesn't own $\{a,\,a\}$. As @NoahSchweber notes, a simpler example, whose transitivity is vacuous, is $\{(a,\,b)\}$ with $a\ne b$.


问:
Finding mclaurin expansion for $f(x,y) = \log(1-x-y+xy)$.
As the title mentions, I am trying to find the McLaurin Expansion of $f(x,y)$. My approach is finding the expansion for the function $f(x) = \log(1+x)$, the replacing $x$ with $(-x-y+xy)$. Is this the correct approach?

答:
The standard logarithm expansion is given by

\begin{equation}
\log(1-x) = -\sum_{n=1}^{\infty}\frac{x^n}{n}, \qquad |x|<1
\end{equation}

Write $$\log (1-x-y+xy) = \log (1-x) + \log (1-y)$$


问:
Question on showing quotient space is not Hausdorff
Let $A = \{\frac{1}{n}: n \in \Bbb{N}\}$ and $X = \Bbb{R}$ with topology
$$\tau = \{U\smallsetminus B: \text{$U$ is open in $\Bbb{R}$ with usual topology and }B\subset A\}.$$

I need to show that $X/A$ is not Hausdorff.

For this, I am trying to show $\{0\}$ and $A$ can't be seperated by open sets in $X/A$.

So I let $U$ and $V$ to be open sets containing $\{0\}$ and $A$.

If $p$ is the projection map, then by definition of quotient topology, $p^{-1}$$(U)$ and $p^{-1}(V)$ are open in $\Bbb{R}$. They will contain $0$ and all points of $A$.

How can I show that these two sets will not be disjoint?

Help, please!

答:
$X$ is a Hausdorff space that is not regular as $A$ cannot be separated from $0$ by open sets.

Sketch of proof: suppose that $0$ is contained in some $U-A$, with $U$ standard-open, and $V \supseteq A$  is also standard open. Then $U$ contains some $\frac1n \in A$ and for some standard open $W$ we will have $\frac1n \in W \subseteq U \cap V$. This shows that $U-A$ and $V$ intersect.

Now we identify $A$ to a point to make $X{/}A$, and then $A$ becomes a point that still cannot be separated from $0$, hence the quotient is not Hausdorff. We just use the fact from the first paragraph about $X$ to show it.


问:
A little bit confused about group rings
According to Wikipedia:

Let G be a group, written multiplicatively, and let $R$ be a ring. The group ring of $G$ over $R$, which we will denote by $R[G]$ (or simply $RG$), is the set of mappings $f : G \to R$ of finite support, where the module scalar product $\alpha f$ of a scalar $\alpha$ in $R$ and a vector (or mapping) $f$ is defined as the vector $x\mapsto \alpha \cdot f(x)$.

My current understanding is that the set $R[G]$ contains all the possible mappings $f : G \to R$ with the following property: $f(x)= 0$ for all but finitely many $x\in G$. I'm not sure how to interpret "the module scalar product $\alpha f$ of a scalar $\alpha$ in $R$ and a vector (or mapping) $f$ ".
Moreover, what is the domain of these mappings? The group $G$ itself?

The product of two mappings $f$ and $g$ is defined as:
$x\mapsto \sum\limits_{u\in G}{f\left( u \right)g\left( {{u}^{-1}}x \right)}=\sum\limits_{uv=x}{f\left( u \right)g\left( v \right)}$.
I understand that, for instance, $f\left( u \right)\ne 0$ only for a finite number of $u$'s, therefore this definition is legitimate. What I don't understand is why those two definitions are equal.

答:
I'm not sure how to interpret "the module scalar product αf of a scalar α in R and a vector (or mapping) f ".

This is just an intricate way to say that $\alpha\cdot f$ is the function $\alpha f$ (defined in the last part of your sentence).

Moreover, what is the domain of these mappings? The group G itself?

Yes.

What I don't understand is why those two definitions are equal.

Well, just set $v=u^{-1}x$ (if if $uv=x$, then $v=u^{-1}x$ ande vice versa).

Title: A little bit confused about group rings


问:
Relationship between Partial Harmonic Sum and Logarithm.
Let $H_n$ denote the partial harmonic sum

$$H_n = \sum_{j=1}^n \frac{1}{j} \ \ .$$

I saw in some lecture notes that $H_n \approx \ln(n)$ without any explanation given. I tried to understand this approximation using Taylor series. Recall that the Taylor expansion of $\ln(x)$ for $x>0$ is

$$\ln(x) = 2 \sum_{j=1}^{\infty} \frac{ \bigg[ \displaystyle\frac{(x-1)}{(x+1)} \bigg]^{(2j-1)} }{2j-1} \ \ .$$

I tried to use this to obtain the connection, however I was unable to see the relationship between $H_n$ and $\ln(n)$.

I performed some crude numerical calculations and I'm not sure whether the approximation is correct because $H_n$ does not appear to be close to $\ln(n)$ for the values of $n$ that I have tried.

Can someone please help me understand this approximation or confirm my suspicions that there is an error in the notes.

If there is an error in the notes, does a relationship exist between the partial harmonic sum and the logarithm however?

答:
Claim:
$$H_n = \ln n + C + O\left(\frac{1}{n}\right)$$
where $C$ is the Euler–Mascheroni constant ($=0.5772156649...$)

Proof: For any natural $k$ and $x$ such that $k-1 \le x < k$, we have $[x] = k-1$. Hence, we also have $$\frac{1}{k} = \frac{1}{[x] + 1}$$
Therefore, $$\frac 1k = \int_{k-1}^k \frac 1k dx = \int_{k-1}^k \frac{dx}{[x]+1}$$
Now, summing the above equality from $1$ to $n$, we get
$$\sum_{k=1}^n\frac 1k = \sum_{k=1}^n\int_{k-1}^k \frac 1k dx = \int_{0}^n \frac{dx}{[x]+1}$$
We subtract the following equality from the above one:
$$\ln(n+1) = \int_0^n \frac{dx}{x+1}$$
to get
$$\sum_{k=1}^n \frac 1k - \ln(n+1) = \int_0^n \left( \frac{1}{[x]+1} - \frac{1}{x+1}\right)dx = \int_0^n \frac{x - [x]}{([x]+1)(x+1)}dx =: C_n$$ We denote $C = \lim C_n$ since $0 \le x-[x] < 1$ implies $$0 \le \frac{x - [x]}{([x]+1)(x+1)} < \frac{1}{x^2}, \ \ x\ge 1$$
which implies the improper integral (note: $\lim C_n$ was an improper integral) is convergent. Now, if we integrate the above inequality from $n$ to $\infty$, we get $$0 \le \int_n^\infty \frac{x - [x]}{([x]+1)(x+1)}dx \le \int_n^\infty \frac{dx}{x^2} = \frac 1n$$
Hence, we have
$$\int_0^n \frac{x - [x]}{([x]+1)(x+1)}dx = C + O\left(\frac 1n\right), \ \ n\in \mathbb N$$
Therefore, we conclude
$$\sum_{k=1}^n \frac{1}{k} = \ln(n+1) + C + O\left(\frac 1n\right)$$ Finally, noting that $\ln(n+1) = \ln n + O(1/n)$ completes the proof. $\square$


问:
Matrix $P$ with conjugate columns and real vector $x \implies P^{-1} x$ has conjugate entries
[Background]

I came accross this problem while studying real matrices with complex Eigenvalues. If $A$ is a real matrix, the nonreal Eigenvalues occur in conjugate pairs, and so do the corresponding Eigenvectors. We can construct a transformation matrix $P$ with the Eigenvectors as columns such that $A=PDP^{-1}$ with $D$ diagonal. Now, the book used by our university (Toegepaste Algebra, oefeningen en extra's - V. Rijmen, H. Van Hamme) introduces a complex vector $w$ such that $Pw$ is real and claims that the entries in $w$ corresponding to conjugate columns are conjugate. As we can construct an $A$ to fit any $P$, I deduced the following must hold:

Suppose that $P$ is a complex, non-singular $n\times n$-matrix of which the
nonreal columns occur in conjugate pairs and that $x$ is a real
vector.
 Then if $P$ has conjugate columns i and j, the i'th and
j'th entries of $P^{-1}x$ are conjugate.

How could one prove this?

e.g.

$$
P =
\begin{bmatrix}
1 & 1 & 1 & 2+i & 2-i \\
0 & i & -i & 0 & 0 \\
0 & 0 & 0 & -1+2i & -1-2i \\
2 & 1+i & 1-i & 0 & 0 \\
3 & -1 & -1 & 1 & 1
\end{bmatrix}
$$

and

$$
x =
\begin{bmatrix}
1 \\
5 \\
0 \\
0 \\
2
\end{bmatrix}
$$

Then
$$
P^{-1}x =
\begin{bmatrix}
-1 \\
-3/2-5/2i \\
-3/2+5/2i \\
1-1/2i \\
1+1/2i
\end{bmatrix}
$$

As you can see, the 2nd and 3rd and the 4th and 5th columns of $P$ are conjugate, and so are the 2nd and 3rd and the 4th and 5th entries of $P^{-1}x$.

Thanks in advance

答:
I solved my own question overnight, and I will leave the answer here for future reference.

As $P$ has conjugate columns, there is a permutation matrix $Q$ that switches the conjugate pairs. I.e. $$\overline{P} = P\cdot Q$$
In my example with
$$
P =
\begin{bmatrix}
1 & 1 & 1 & 2+i & 2-i \\
0 & i & -i & 0 & 0 \\
0 & 0 & 0 & -1+2i & -1-2i \\
2 & 1+i & 1-i & 0 & 0 \\
3 & -1 & -1 & 1 & 1
\end{bmatrix}
$$
$Q$ would be
$$
Q =
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 1 & 0
\end{bmatrix}
$$
Switching the 2nd and 3rd column and the 4th and 5th column of $P$. Note, in particular, that $Q^T=Q$.

Now suppose $x$ is a real vector and call $w=P^{-1}x$. Then $Pw=x$ is real, so
$$Pw = \overline{Pw} = \overline{P}\overline{w} = PQ\overline{w}$$
P is invertible (question statement), so this implies
$$w = Q\overline{w}$$

I.e. if the $i$'th and $j$'th column of $P$ are conjugate, then so are the $i$'th and $j$'th entry of $w$.


问:
Using Cauchy's theorem on $1/z$ in the unit disk
The problem is as follows:

For $|z| < 1,$ evaluate the integral $\int_{|\zeta|=1} \frac{1}{\zeta (\zeta - z)} d\zeta.$

I've been trying to utilize the Cauchy's theorem on annulus on the choice of $f(\zeta) = \frac{1}{\zeta},$ but so far this has led me to nowhere.

The problem additionally asks,
for $|z_0| = 1,$ does $\lim_{z \to z_0, |z| < 1} \frac{1}{2\pi i}\int_{|\zeta|=1} \frac{1}{\zeta(\zeta -z)} d\zeta = \frac{1}{z_0}?$

So I guess I would have to use $\epsilon$ radius smaller annulus around $0$ and the unit disk as the larger annulus to apply Cauchy's theorem to compare $f(z_0)$ with, and if the integral on $\epsilon$ radius disk vanishes to 0, we get the equality on the fourth line. But this didn't work out as I thought.

答: Note that, unless $z=0$,$$\frac1{\zeta(\zeta-z)}=\frac1z\left(\frac1{\zeta-z}-\frac1\zeta\right).$$Therefore\begin{align}\int_{|\zeta|=1}\frac{\mathrm dz}{\zeta(\zeta-z)}&=\frac1z\left(\int_{|\zeta|=1}\frac{\mathrm dz}{\zeta-z}-\int_{|\zeta|=1}\frac{\mathrm dz}{\zeta}\right)\\&=\frac{2\pi i}z(1-1)\text{ (since $|z|<1$)}\\&=0.\end{align}And, if $z=0$, the function that you are integrating is $\frac1{\zeta^2}$, which has a primitive ($-\frac1\zeta$), and therefore the integral will be $0$ in that case as well.


问:
All affine sets are affine subspaces (and vice-versa)

Show that a set $∅ \neq A ⊂ \mathbb{R}^n$ is affine if and only if it is an affine subspace.

Definitions I know:

Affine set: If a set $A \subset\mathbb{R}^n$ is affine, then for all $x,y\in A$, we have $tx + (1-t)y \in A$ where $t\in\mathbb{R}$.

Affine subspace: A set $A$ is an affine subspace of $\mathbb{R}^n$ if every element $a\in A$ can be written as $p + V$, where $p\in\mathbb{R}^n$ is a fixed vector, and $V$ is a subspace of $\mathbb{R}^n$.

My work:
$[\Leftarrow]$ seems easier. Consider two elements $p+v_1$ and $p+v_2$ in $A = p+V$ as defined above. Then $t(p+v_1) + (1-t)(p + v_2) = p + tv_1 + (1-t)v_2 \in p+V$ because $V$ is a subspace. So all affine subspaces are affine sets.

How do I show the other direction, namely $[\Rightarrow]$?

I started with a non-empty affine set $A\subset\mathbb{R}^n$, and I'm trying to construct $p$ and $V$ so that $p+V = A$, but I don't know how to! I'd appreciate any help!

答: Hint: Fix $x \in A$ and put $V:= A-x$. Check that $V$ is a vector space.


问:
Finding probability of all outcomes occurring in n repetitions
Suppose you have numbers 1, 2, 3, 4. You can pick each of these numbers with probability 1/4. After n iid selections, what is the probability that you have seen all 4 numbers? Express this as a function of n.

For example, if n=5, and you draw the numbers 1,1,3,2,1 then this is unsuccessful, as 4 was never picked. If you draw 3,4,2,3,4, this is also unsuccessful as 1 was never picked. However, 3,2,2,1,4 is successful.

I am aware you can use a discrete Markov chain and find the n-step transition probability matrix on Wolfram, but is there a way to find this without a calculator and with reasoning alone?

答:
You may just count systematically as follows:

Number of sequences of length $n$ containing only $1$ digit: $$\color{blue}{\Rightarrow 4}$$

Number of sequences of length $n$ containing exactly $2$ digits:

Choose two of the digits: $\color{blue}{\binom 42}$

Number of sequences of length $n$ containing at least one of these digits: $\color{blue}{2^n}$

Number of sequences of length $n$ containing only one of the $2$ selected digits: $\color{blue}{2}$
$$\color{blue}{\Rightarrow \binom 42\left(2^n-2\right)}$$

Number of sequences of length $n$ containing exactly $3$ digits:

Choose three of the digits: $\color{blue}{\binom 43}$

Number of sequences of length $n$ containing at least one of these digits: $\color{blue}{3^n}$

Number of sequences of length $n$ containing at least one of only two of the $3$ selected digits: $\color{blue}{\binom{3}{2}2^n}$

Number of sequences of length $n$ containing only one of the $3$ selected digits: $\color{blue}{3}$

$$\color{blue}{\Rightarrow \binom 43\left(3^n- \binom{3}{2}2^n+3\right)}$$

All together, number of sequences where at least one digit is missing:
$$\color{blue}{4 + \binom 42\left(2^n-2\right) + \binom 43\left(3^n- \binom{3}{2}2^n+3\right)=4-6\cdot 2^n+4\cdot 3^n}$$

Now, the probability you are looking for is

$$\frac{4^n - \left(4-6\cdot 2^n+4\cdot 3^n\right)}{4^n}=\boxed{1-\frac 1{4^{n-1}}+6\frac 1{2^n}-4\left(\frac 34\right)^n}$$


问:
Calculating $\sqrt{28\cdot 29 \cdot 30\cdot 31+1}$
Is it possible to calculate $\sqrt{28 \cdot 29 \cdot 30 \cdot 31 +1}$ without any kind of electronic aid?

I tried to factor it using equations like $(x+y)^2=x^2+2xy+y^2$ but it didn't work.

答: If you are willing to rely on the problem setter to make sure it is a natural, it has to  be close to $29.5^2=841+29+.25=870.25$  The one's digit of the stuff under the square root sign is $1$, so it is either $869$ or $871$.  You can either calculate and check, or note that two of the factors are below $30$ and only one above, which should convince you it is $869$.


答:
Let $p$ be any point in $A$. Let $V=A-p$ (i.e. $\{a-p: a \in A\}$). Let $x \in V$ so $x=a-p$ with $a \in A$ and take $c \in \mathbb R$. Let us show that $cx \in V$. $cx=ca-cp=(ca-cp+p) -p$ and $ca-cp+p=ca+(1-c)p \in A$. So $cx \in V$.

Now consider $(a_1-p)+(a_2 -p)$ where $a_1.a_2 \in A$. Write this as $ 2(\frac {a_1+a_2} 2 -p)$. can you finish? .

Title: All affine sets are affine subspaces (and vice-versa)

Body:
Show that a set $∅ \neq A ⊂ \mathbb{R}^n$ is affine if and only if it is an affine subspace.

Definitions I know:

Affine set: If a set $A \subset\mathbb{R}^n$ is affine, then for all $x,y\in A$, we have $tx + (1-t)y \in A$ where $t\in\mathbb{R}$.

Affine subspace: A set $A$ is an affine subspace of $\mathbb{R}^n$ if every element $a\in A$ can be written as $p + V$, where $p\in\mathbb{R}^n$ is a fixed vector, and $V$ is a subspace of $\mathbb{R}^n$.

My work:
$[\Leftarrow]$ seems easier. Consider two elements $p+v_1$ and $p+v_2$ in $A = p+V$ as defined above. Then $t(p+v_1) + (1-t)(p + v_2) = p + tv_1 + (1-t)v_2 \in p+V$ because $V$ is a subspace. So all affine subspaces are affine sets.

How do I show the other direction, namely $[\Rightarrow]$?

I started with a non-empty affine set $A\subset\mathbb{R}^n$, and I'm trying to construct $p$ and $V$ so that $p+V = A$, but I don't know how to! I'd appreciate any help!


问:
Calculate $\lim\limits_{x \to 0}(\frac{\sin x}{x})^{\cot ^2x}$

Calculate
$$\lim\limits_{x \to 0}\left(\frac{\sin x}{x}\right)^{\cot ^2x}$$

I tried to use the natural logarithm and L'hopital rule but it didn't help me.
$\lim\limits_{x \to 0}\cot ^2x\ln(\frac{\sin x}{x})=\ln L$

$\lim\limits_{x \to 0}\cot ^2x\ln(\frac{\sin x}{x})=\lim\limits_{x \to 0}\frac{\ln\frac{\sin x}{x}}{\tan ^2x}=\lim\limits_{x \to 0}\frac{x}{2\sin x\tan x \sec^2x}$

答:
Partial solution (fill in details): Note that $$\left(\frac{\sin x}{x}\right)^{\cot ^2x}=e^{\ln \left(\frac{\sin x}{x} \right)^{\cot^{2}(x)}}=e^{\cot^{2}(x)\ln\left( \frac{\sin x}{x} \right)}$$
and since that $$\cot(x)=\frac{\cos(x)}{\sin(x)} \implies \cot^{2}(x)=\frac{\cos^{2}(x)}{\sin^{2}(x)}$$
By, L'Hópital theorem we know that $$\lim_{x\to 0} \frac{\ln\left(\frac{\sin(x)}{x} \right)}{\sin^{2}(x)}=\lim_{x\to 0} \frac{x\cos(x)-\sin(x)}{2x\cos(x)\sin^{2}(x)}$$
and we know that $$\lim_{x\to 0} \frac{\sin(x)}{x}=1$$
So, we can see that $$\lim_{x\to 0}\left(\frac{\sin x}{x}\right)^{\cot ^2x}=e^{\lim_{x\to 0}\cot^{2}(x)\ln\left( \frac{\sin x}{x} \right)}=e^{-\frac{1}{6}}=\boxed{\frac{1}{\sqrt[6]{e}}}$$


Title: Calculate $\lim\limits_{x \to 0}(\frac{\sin x}{x})^{\cot ^2x}$

Body:
Calculate
$$\lim\limits_{x \to 0}\left(\frac{\sin x}{x}\right)^{\cot ^2x}$$

I tried to use the natural logarithm and L'hopital rule but it didn't help me.
$\lim\limits_{x \to 0}\cot ^2x\ln(\frac{\sin x}{x})=\ln L$

$\lim\limits_{x \to 0}\cot ^2x\ln(\frac{\sin x}{x})=\lim\limits_{x \to 0}\frac{\ln\frac{\sin x}{x}}{\tan ^2x}=\lim\limits_{x \to 0}\frac{x}{2\sin x\tan x \sec^2x}$

Body:
If $$L=\lim_{x\to a} f(x)^{g(x)} \to 1^{\infty}$$,
then $$L=\exp[\lim_{x\to a} g(x)(f(x)-1)]$$
So here
$$L=\exp[\lim_{x \to 0} \cot^2 x(\frac{\sin x}{x}-1)]=\exp[\frac{(\sin x-x)\cos^2x}{\sin^2 x}].$$
$$L=\lim_{x \to 0} \exp\frac{\sin x -x}{x}\frac{1+\cos 2x}{1-\cos 2x}=\lim_{x \to 0} \exp\frac{-x^3/6+...}{x}\frac{2-2x^2+...}{2x^2+..}=e^{-\frac{1}{6}}.$$
We have used $\sin z \to z-z^3/6+...$ and $\cos z=1-z^2/2+..$ When $|z|$ is very small.


答:
Here's a direct way with Taylor Series:
$$
\begin{aligned}\lim _{x\rightarrow 0}\frac{\ln \frac{\sin x}{x}}{\tan ^2x}
=\lim _{x\rightarrow 0}\frac{-\frac{x^2}{6}+o\left( x^4 \right)}{x^2+o\left( x^4 \right)}=-\frac{1}{6}\end{aligned}
$$
Thus
$$
\lim\limits_{x \to 0}\left(\frac{\sin x}{x}\right)^{\cot ^2x} = e^{-\frac16}
$$


问:
Probablity of all different socks pulled out.
There were 5 pairs of socks in a pile of clothes. On pulling out one by one and after pulling out 5 socks, I found they were all different. What is the probability of such an event?</p>

答:
A simple way is to use the double factorial  which will give

$Pr=\dfrac{10\cdot8\cdot6\cdot4\cdot 2}{10\cdot9\cdot8\cdot7\cdot6} = \dfrac8{63}$

The numerator skips pairs, while the denominator doesn't.


问:
Solve $(2y^{2}+xy+y)dx-(3x^{2}+4yx+x)dy=0$

Solve $(2y^{2}+xy+y)dx-(3x^{2}+4yx+x)dy=0$

My approach: Since that $$P(x,y)=2y^{2}+xy+y \implies P_{y}(x,y)=4y+x+1$$
and $$Q(x,y)=-3x^{2}-4yx-x \implies Q_{x}(x,y)=-6x-4y-1$$
So, we can see that $P_{y}\not=Q_{x}$. So, the ODE is not exact equation.

Also, we can see that we cannot to rewrite the ODE as $f(x)dx=g(y)dy$, so the ODE is not separable equation. I think it's clear that the ODE is not Bernoulli's equation. Since that we don't have the form $y'=f(y/x)$ or $y'=f((a_{1}x+a_{2}y+a_{3})/(b_{0}x+b_{1}y+b_{3}))$ so we can't use the homogeneous method of solutions for the ODE.

How can I find a integrate factor for to solve this ODE-nonlinear or maybe a substitution?

答:
$$(2y^2+xy+y)dx-(3x^2+4yx+x)dy=0$$
Since one cannot find a simple form of integrating factor $\mu(x,y)$ , we try some more complicated forms. For example

$$\mu(x,y)=x^py^q(y+cx)^r$$
$$\begin{cases}
N(x,y)=(2y^2+xy+y)x^py^q(y+cx)^r\\
M(x,y)=-(3x^2+4yx+x)x^py^q(y+cx)^r
\end{cases}$$
$$N(x,y)dx+M(x,y)dy=0$$
The condition to be an exact DE is :
$$\frac{\partial N(x,y)}{\partial y}=\frac{\partial M(x,y)}{\partial x}$$
So we have to compute $\frac{\partial N(x,y)}{\partial y}$ and $\frac{\partial M(x,y)}{\partial x}$. After simplification of $\frac{\partial N(x,y)}{\partial y}=\frac{\partial M(x,y)}{\partial x}$ we have a system of rather simple equations which solving leads to :
$$p=-2 \quad;\quad q=\frac12 \quad;\quad r=-\frac12 \quad;\quad c=1.$$
So, an integrating factor is
$$\boxed{\mu(x,y)=\frac{y^{1/2}}{x^2(x+y)^{1/2}}}$$
The exact DE is :
$$(2y+x+1)\frac{y^{3/2}}{x^2(x+y)^{1/2}}dx-(3x+4y+1)\frac{y^{1/2}}{x(x+y)^{1/2}}dy=0$$
Then one can integrate.


问:
Proof of the boundedness theorem
Proof of the boundedness theorem

I am new to this subject so can anyone please explain $f(x_{n_k})>n_k \geq k$ ? I can't understand  $n_k \geq k$

https://en.wikipedia.org/wiki/Extreme_value_theorem

Thank you!

答:
Since you are choosing a subsequence, you may "jump" some terms, thus $n_k\geq k$.

For example, the orignal sequence (the first line represents $k$)$$
\left[ \begin{matrix}
 1&  2&  3&  4&  5&  6&  7&  \cdots\\
 a_1&  a_2&  a_3&  a_4&  a_5&  a_6&  a_7&  \cdots\\
\end{matrix} \right]
$$
One subsequece of it (also the first line represents $k$ and second line is $a_{n_k}$):
$$
\\
\left[ \begin{matrix}
 1&  2&  3&  4&  5&  6&  7&  \cdots\\
 a_1&  a_2&  a_4&  a_5&  a_6&  a_7&  a_9&  \cdots\\
\end{matrix} \right]
\\$$
Then it's clear that $1=1,2=2,4>3,5>4,6>5,7>6,9>7...$

Title: Proof of the boundedness theorem


问:
The special function $P(s)=\int^\infty_0 \frac{\ln(x)dx}{1+x^s}$

Evaluate the integral
$$P(s)=\int^\infty_0 \frac{\ln(x) \,dx}{1+x^s}$$

I tried integration by parts, using $u=\ln(x),du=\frac{x}{dx},v=\frac{1}{1+x^s}$, but I did not manage to get anything.

I thought about using a power series, but I did not find one that converges on the entire domain.

Using the substitution $u=\ln(x),du=\frac{dx}{x}$ we get $$\int^\infty_{-\infty} \frac{ue^{-u}}{1+e^{su}} ,$$
which doesn't seem useful either.

The previous similar question Integral of $\int^{\infty}_0 \frac{x^n}{x^s+1}dx$ gives
$$\int^\infty_0 \frac{x^n \,dx}{1+x^s}=\frac{\Gamma(\frac{1}{s})\Gamma\left(1-\frac{1}{s}\right)}{s(n+1)} .$$

Another possibly relevant question is Integral of $\int^{\infty}_0 \frac{e^{-x}}{x^s+1}\,dx$.

答: Differentiating$$\int_0^\infty\frac{x^{t-1}dx}{1+x^s}=\frac1s\int_0^\infty\frac{y^{t/s-1}dy}{1+y}=\frac{\pi}{s}\csc\frac{\pi t}{s}$$with respect to $t$ gives$$\int_0^\infty\frac{x^{t-1}\ln x\,dx}{1+x^s}=-\frac{\pi^2}{s^2}\csc\frac{\pi t}{s}\cot\frac{\pi t}{s}.$$Set $t=1$ to get$$\int_0^\infty\frac{\ln x\,dx}{1+x^s}=-\frac{\pi^2}{s^2}\csc\frac{\pi}{s}\cot\frac{\pi}{s}.$$The case $s=2$ is a famous sanity check, for which the integral is $0$.


答:
An algorithm to get a solution.

Step 1: As it was mentioned in math.stackexchange.com/questions/3709298  by Calvin Khor for natural $n$, it's easy to make a substituion $y = x^{n+1}$ in the integral $\int_{0}^{\infty }\frac{x^n}{x^s + 1}$ and get the intergal of such type: $\int_{0}^{\infty }\frac{1}{y^s + 1}dy$, which is known (case $n=0$). But this idea works not only for natural $n$. Hence, we may find

$$I(a) = \int_{0}^{\infty }\frac{x^a}{x^s + 1}dx$$
for real $a$.

Step 2. We have
$$I'(a) = \int_{0}^{\infty }\frac{x^a \ln x}{x^s + 1}dx.$$

So it's sufficient to put $a=1$.

Title: The special function $P(s)=\int^\infty_0 \frac{\ln(x)dx}{1+x^s}$

Body:
Evaluate the integral
$$P(s)=\int^\infty_0 \frac{\ln(x) \,dx}{1+x^s}$$

I tried integration by parts, using $u=\ln(x),du=\frac{x}{dx},v=\frac{1}{1+x^s}$, but I did not manage to get anything.

I thought about using a power series, but I did not find one that converges on the entire domain.

Using the substitution $u=\ln(x),du=\frac{dx}{x}$ we get $$\int^\infty_{-\infty} \frac{ue^{-u}}{1+e^{su}} ,$$
which doesn't seem useful either.

The previous similar question Integral of $\int^{\infty}_0 \frac{x^n}{x^s+1}dx$ gives
$$\int^\infty_0 \frac{x^n \,dx}{1+x^s}=\frac{\Gamma(\frac{1}{s})\Gamma\left(1-\frac{1}{s}\right)}{s(n+1)} .$$

Another possibly relevant question is Integral of $\int^{\infty}_0 \frac{e^{-x}}{x^s+1}\,dx$.


问:
Huffman Tree and Three Ambiguous facts
I see three facts on Huffman Algorithms:

Two of them is True:

Fact A: The average length of an optimal code is always at most $\lceil \log n \rceil$, where $n$ is the number of codewords.

Fact B: The average codeword length in Huffman's algorithm is $\Omega(\log n)$.

and one Fact that say this is False:

Fact C: The average codeword length in Huffman's algorithm is $O(\log n)$.

Really I couldn't get the point, what is the difference among these
three facts. Is there any example or simple intuitive idea about these
three facts? (I means what is the role of "optimal" words in first
fact, or what is difference among these three facts in very short sentence)

答:
First fact speaks about average length of optimal code, while second and third speaks about average codeword length of some specific (Huffman's) code.

Given probability distribution on $n$ symbols and some code $c$, average codeword length is just average length of words in code - $\frac{1}{n} \sum |c(i)|$ (expectation wrt uniform distribution), while average length is expectation wrt given probability distribution - $\sum |c(i)| \cdot p_i$.

The first fact, informally, says that if we need to transmit many symbols sampled from some distribution, we well need at most $\lceil \log n\rceil$ bits in average for each symbol if we use good code.

It's also known that Huffman's code is optimal. However, average codeword length can be significantly greater than average length - length of codes of rare symbols don't significantly affect average length, but they do affect average length as much as lengths of codes of frequent symbols.

Fact B says that Huffman's code definitely have "many" "long" codewords (so average length is quite large).

Fact C claims that all words in Huffman's code are "short", but it's not necessary true. For example, if we have symbols with probabilities $\frac{1}{2}$, $\frac{1}{4}$, $\frac{1}{8}$, ..., $\frac{1}{2^n}$, $\frac{1}{2^n}$ - for a total of $n + 1$ symbols - Huffman's code will assign them codewords with lengths $1$, $2$, $3$, ..., $n$, $n$ respectively - so average codeword length will be $\frac{\frac{n(n + 1)}{2} + n}{n + 1} \approx \frac{n}{2} \gg \log n$.

Average length of code will be $\sum\limits_{i=1}^n \frac{i}{2^n} + \frac{n}{2^n} = 2 - 2^{1 - n} = O(\log n)$.

Title: Huffman Tree and Three Ambiguous facts


答:
To see a possible pattern, I think that we need to explore for larger values of $s$.

For example
$$P(7)=-\frac{4 \pi ^2 \left(1-3 \sin \left(\frac{\pi }{14}\right)+3 \sin \left(\frac{3 \pi
   }{14}\right)\right)}{49 \left(3+6 \sin \left(\frac{\pi }{14}\right)-4 \sin
   \left(\frac{3 \pi }{14}\right)\right)}$$ which can be simplified nicely.

In fact, a CAS gives the beautiful

$$\color{blue}{P(s)=\int^\infty_0 \frac{\log(x)}{1+x^s}dx=-\pi ^2\frac{ \cot \left(\frac{\pi }{s}\right) \csc \left(\frac{\pi
   }{s}\right)}{s^2}}$$


问:
calculate $\mathbb{E}[X]$ . what is wrong in my attempt?
friend A choose 3 cards from 10 cards packet, then he return the cards to the pack. friend B choose 3 cards from the same pack in an independent way from friend A. Let $X$ be the number of cards that didnt choose by any of the friends, calculate $\mathbb{E}[X]$

First I can see that $4\leq X \leq 7$
$${p(X=4)}
~=~ \frac{\binom{10}{3}\binom{7}{3}}{\binom{10}{3}\binom{10}{3}}.$$
$${p(X=5)}
~=~ \frac{\binom{10}{3}\binom{7}{2}\binom{3}{1}}{\binom{10}{3}\binom{10}{3}}.$$
$${p(X=6)}
~=~ \frac{\binom{10}{3}\binom{7}{1}\binom{3}{2}}{\binom{10}{3}\binom{10}{3}}.$$
$${p(X=7)}
~=~ \frac{\binom{10}{3}\binom{3}{3}}{\binom{10}{3}\binom{10}{3}}.$$

but I dont get the right answer when I calculate it according to $$
\mathbb{E}(X)=\sum_{t \in R_{X}} t \cdot \mathbb{P}(X=t)
$$

答:
Put $A_i= \{$ the card number $i$ was not chose by any friend $\}$. Then
$X = \sum_{i=1}^{10} I_{A_i}$, where $I$ is indicator function.
Hence $EX = \sum_i P(A_i) = 10 P(A_1)$.

Put $B = \{$ the card number $1$ was not chose by the first friend $\}$ and $C = \{$ the card number $1$ was not chose by the second friend $\}$. Then
$A_1=  B\cap C$, where $B$ and $C$ are independent and $P(B) = P(C) =0.7$.
Hence $P(A_1) = 0.49$ and the answer is $4.9$.


问:
Solving coupled nonlinear ODEs
I am trying to find $z$ such that
$$\dot{z} = -1 + e^{-iz^*},$$
where $*$ denotes complex conjugate and the dots represent derivatives with respect to time. The time dependence of the dependent variables is suppressed for clarity of presentation. Letting $z=x+iy$,
$$\dot{x} = -1 -  e^{-y}\cos x, \quad \dot{y}=  e^{-y}\sin x,$$
together with the constraint that
$$A = -y+ e^{-y}\cos x$$
is a constant, i.e. $\dot{A}=0$.

The normal way I would proceed would be to eliminate $y$ in favor of $x$, which yields product log (ie Lambert W) functions, but the resulting ODE is complicated, and I cannot solve it in closed form. Note, if we take $z = i \log \zeta$, we can equate this to a problem involving a point vortex in a uniform stream, but I'm not sure this identification helps.

答:
$$\dot{x} = -1 -  e^{-y}\cos x, \quad \dot{y}=  e^{-y}\sin x,$$
$$\sin x \dfrac {dx}{dy}=-e^y- \cos x$$
$$(\cos x)'-\cos x=e^y$$
$$(e^{-y}\cos x)'=1$$
That you can integrate. But the system is not linear so it isn't going to be easy to integrate.

Title: Solving coupled nonlinear ODEs


问:
Deriving kinetic energy from Newton's Second Law (step I don't understand)
Hello so there is a step in deriving newton's second law into the Kinetic Energy I don't quite understand it goes like this:
$$\overrightarrow{F}\overrightarrow{v}=m\frac{d\overrightarrow{v}}{dt}\overrightarrow{v})$$
goes to this:
$$\overrightarrow{F}\overrightarrow{v}=\frac{d}{dt}(\frac{1}{2}m\overrightarrow{v}\overrightarrow{v})$$
First I thought they just factored out $\frac{d}{dt}$, but then I couldn't explain why there is a 1/2 between the brackets.

If someone knows how to get from the first equation to the second it would be very much appreciated to tell me your thought process or whrite it out clearly

Thanks in advance

答:
By Newton's second Law, we know that $$\sum_{i=1}^{n}\vec{F}=\vec{F}_{neta}=m\cdot \vec{a}$$
also we know that $$\frac{d\vec{x}}{dt}=\vec{v}(t)$$ $$\frac{d\vec{v}}{dt}=\vec{a}(t)$$
So, we have $$\vec{F}_{neta}\vec{v}=(m \cdot \vec{a})\vec{v}=m\frac{d\vec{v}}{dt}\vec{v}$$

By other hand, we know that $$K=\frac{1}{2}m(|\vec{v}|)^{2}$$
where $K$ is the kinetic energy and $K$ is a scalar magnitude.

Note that $$\vec{v}\cdot \vec{v}=|\vec{v}|^{2}$$where $|\vec{v}|$ denote the magnitude of the vector $\vec{v}$.

Also, note that  $$\frac{d}{dt}\left( \frac{1}{2}m  \vec{v}\cdot \vec{v}\right)=\frac{1}{2}m \frac{d}{dt}(\vec{v}\cdot \vec{v})=m\frac{d|\vec{v}|}{dt}$$


问:
Let $X=\mathbb{R}$ and $\tau$ be a finite complement topology. Is the funtion $f: X\to X$ defined by $f(x)=x^2$ continuous?

Let $X=\mathbb{R}$ and $\tau$ be a finite complement topology. Is the funtion $f: X\to X$ defined by  $f(x)=x^2$ continuous?

I think it is continuous. Because the open sets of $\mathbb{R}$ with respect to finite complement topology are infinite. I am not exactly sure, can anyone help me?

答:
I assume you consider the finite complement (a.k.a. cofinite) topology on both the domain and the codomain.

A subset $A\subseteq \mathbb{R}$ in the cofinite topology is closed if and only if $A$ is either finite or $A=\mathbb{R}$.

Now clearly $f^{-1}(\mathbb{R})=\mathbb{R}$. And so the question boils down to: if $A$ is a finite set then is $f^{-1}(A)$ finite as well? And this, on the other hand, translates to the algebraic question: if $p\in\mathbb{R}[X]$ is a polynomial, then how many solutions $p(X)=0$ has (also note that the preimage of union is the union of preimages)? The answer is always finite. In particular $f$ is continuous.

Note that not every classical function will be continuous in the cofinite topology. For example $f(x)=\sin(x)$ is not continuous, because $f^{-1}(\{0\})$ is infinite.


答:
In addition do @BotnakovN. 's answer, the following Python session shows that the OP's original computation is correct</p>
```<code>>>> from scipy.special import binom as b
>>> x = b(10, 3)
>>> 4 * b(7, 3)/x + 5 * b(7, 2) * 3/x + 6 * 7 *3/x + 7 * 1/x
4.9
</code>```

Title: calculate $\mathbb{E}[X]$ . what is wrong in my attempt?

Body:
friend A choose 3 cards from 10 cards packet, then he return the cards to the pack. friend B choose 3 cards from the same pack in an independent way from friend A. Let $X$ be the number of cards that didnt choose by any of the friends, calculate $\mathbb{E}[X]$

First I can see that $4\leq X \leq 7$
$${p(X=4)}
~=~ \frac{\binom{10}{3}\binom{7}{3}}{\binom{10}{3}\binom{10}{3}}.$$
$${p(X=5)}
~=~ \frac{\binom{10}{3}\binom{7}{2}\binom{3}{1}}{\binom{10}{3}\binom{10}{3}}.$$
$${p(X=6)}
~=~ \frac{\binom{10}{3}\binom{7}{1}\binom{3}{2}}{\binom{10}{3}\binom{10}{3}}.$$
$${p(X=7)}
~=~ \frac{\binom{10}{3}\binom{3}{3}}{\binom{10}{3}\binom{10}{3}}.$$

but I dont get the right answer when I calculate it according to $$
\mathbb{E}(X)=\sum_{t \in R_{X}} t \cdot \mathbb{P}(X=t)
$$

Body:
In addition do @BotnakovN. 's answer, the following Python session shows that the OP's original computation is correct

>>> from scipy.special import binom as b
>>> x = b(10, 3)
>>> 4 * b(7, 3)/x + 5 * b(7, 2) * 3/x + 6 * 7 *3/x + 7 * 1/x
4.9


Body:
Composing Taylor series one piece at the time
$$y=\Bigg[\frac{\sin (x)}{x}\Bigg]^{\cot ^2(x)} \implies \log(y)={\cot ^2(x)}\log\Bigg[\frac{\sin (x)}{x}\Bigg]$$
$$\frac{\sin (x)}{x}=1-\frac{x^2}{6}+\frac{x^4}{120}+O\left(x^6\right)$$
$$\log\Bigg[\frac{\sin (x)}{x}\Bigg]=-\frac{x^2}{6}-\frac{x^4}{180}+O\left(x^6\right)$$
$$\cot(x)=\frac{1}{x}-\frac{x}{3}-\frac{x^3}{45}-\frac{2 x^5}{945}+O\left(x^6\right)$$
$$\cot^2(x)=\frac{1}{x^2}-\frac{2}{3}+\frac{x^2}{15}+\frac{2 x^4}{189}+O\left(x^5\right)$$
$$\log(y)=-\frac{1}{6}+\frac{19 x^2}{180}-\frac{22 x^4}{2835}+O\left(x^6\right)$$
$$y=e^{\log(y)}=\frac{1}{\sqrt[6]{e}}+\frac{19 x^2}{180 \sqrt[6]{e}}+O\left(x^4\right)$$


问:
Absolutely convergence of the Fourier coefficients for cosine terms
Suppose $f$ is a merely continuous function on $[0,\pi]$. Let $c_n$ be the cosine Fourier coefficients of $f$. Then does $\sum |c_n|$ converge or not?

I can prove the situation when $f$ is derivable.
$$\left. \begin{array}{l}{ c _ n = \int _ { 0 } ^ { \pi } f ( x ) \cos n x d x }\\{ = - \frac { 1 } { n } \cdot \int _ { 0 } ^ { \pi } f ^ { \prime } ( x ) \sin n x d x }\\{ = - \frac { A _ n } { n } }\end{array} \right.$$
Then $\sum | c _ { n } | = \sum | \frac { A _ { n } } { n } | \leq \frac { 1 } { 2 } ( \sum\frac { 1 } { n ^ { 2 } } + \sum A _ { n } ^ { 2 } )<+\infty$.

$\sum_{n>1}\frac{\sin nx}{n\ln n}$is a
counterexample for the sine term and I want to know how about the cosine term.

答: Fejer's example of a continuous function with a divergent FS (see 10.3.1, p. 154 of 'Fourier Series' by Edwards) is a cosine series. It is divergent at $0$. But if $f$ is any function whose Fourier coefficients $(c_n)$ satisfy $\sum |c_n| <\infty$ then the function is continuous and the Fourier series converges to $f$ at every point.


问:
What is a good Geometry computer library? (Language Agnostic)
I'm trying to perform the following actions with the library:

determine if two polygons intersect?- scale/re-size a polygon

minimum distance between the edges of two polygons

calculate the vertices of a polygon that contains a set of smaller non-rectangular 4-corner polygons, given the vertices of the smaller polygons. (ie. is contained)

The library may not have all the functionality I'm looking for, but the closer, the better.

答:
CGAL, full name is The Computational Geometry Algorithms Library,
https://www.cgal.org/</p>


问:
An infinite sequence of inscribed squares alternate black and white. Find the black area as a fraction of the first square's area.

There is an infinite sequence of squares such that vertices of every next square lie in the center of the sides of the previous square. Odd-numbered squares are filled with white color, and even-numbered are filled with black. Find the black area as a fraction of the first square.

Suggestions : 1/3 or 1/2 or 2/3 or 3/4

Please help.

答:
If the side length of a square is $\ell$, then the side length of the inner square whose vertices are the centers of the original square is $\dfrac{\ell}{\sqrt{2}}$, and thus its area is $\dfrac{\ell^2}{2}$, which represents half of the area of the original square. Since the first square is colored in white, then the number we are looking for is :
$1-\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\cdots\right)=1-\displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^n}{2^n}$

It is well known that $\displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^n}{2^n}=\dfrac{2}{3}$ and therefore the answer is $\dfrac{1}{3}$


答:
The $p$-test implies that this integral diverges for $s \leq 1$, so we assume that $s > 1$.

Hint This integral is a standard application of the Residue Theorem. In this case, we can take the contours $\Gamma_R$ to be the boundaries of the sectors, centered at the origin, of radius $R$ and central angle $\frac{2 \pi}{s}$. (A convenient choice is to take one boundary line segment along the positive real axis and the other along the ray through $e^{2 \pi i / s}$.) Then, the contour contains a single pole, at $e^{\pi i / s}$. Proceeding as usual by rewriting the contour integral as a sum of three integrals, taking the limit as $R \to \infty$ (which eliminates one of the integrals), rearranging, and taking real and imaginary parts gives values both of the given integral,
$$\int_0^\infty \frac{\log x \,dx}{1 + x^s} ,$$
and, as a welcome bonus, the related integral,
$$\int_0^\infty \frac{\,dx}{1 + x^s} .$$

Carrying out the above procedure gives that the relevant residue is $$\operatorname{Res}\left(\frac{\log z}{1 + z^s}, z = e^{\pi i / s}\right) = -\frac{\pi}{s^2} \exp \left(\frac{s + 2}{2 s} \pi i\right)$$ and then that the integral has value $$\int_0^\infty \frac{\log x \,dx}{1 + x^s} = -\frac{\pi^2}{s^2} \cot \frac{\pi}{s} \csc \frac{\pi}{s} .$$

The above technique is essentially robjohn's approach in his answer to this question, which treats the special case $s = 3$. Ron Gordon's approach there, that is, using instead a keyhole contour, applies at least in the special case that $s$ is an integer (necessarily $\geq 2$). Marko Riedel's approach there is similar in spirit to J.G.'s answer to this question.

Remark This integral takes on special values where $\frac{\pi}{s}$ does, including at various rational numbers with small numerator and denominator. In particular for $s = 2$ the integral vanishes, which can be shown using a slick but easier argument.


问:
question on summation?
Please, I need to know the proof that

$$\left(\sum_{k=0}^{\infty }\frac{n^{k+1}}{k+1}\frac{x^k}{k!}\right)\left(\sum_{\ell=0}^{\infty }B_\ell\frac{x^\ell}{\ell!}\right)=\sum_{k=0}^{\infty }\left(\sum_{i=0}^{k}\frac{1}{k+1-i}\binom{k}{i}B_in^{k+1-i}\right)\frac{x^k}{k!}$$

where $B_\ell$, $B_i$ are Bernoulli  numbers.

Maybe we should replace $k$ with $j$?

Anyway, I need to prove how to move from the left to right.

Thanks for all help.

答:
The right-hand side is just the Cauchy product of the two series on the left-hand side. Generally, the Cauchy product of $\sum_{k=0}^\infty a_k$ and $\sum_{l}^\infty b_l$ is $$\sum_{j=0}^\infty\sum_{i=0}^j b_i a_{j-i}.$$
 Now you should have no problem finishing the problem. (Change $j$ to $k$ at the end to match your right hand side.)</p>

Title: question on summation?


问:
If $a^2 + b^2 +c^2 =4$ and $a^3 + b^3 + c^3 = 8$ then find $a^4+b^4+c^4$
This is a popular question, but I can find $a^4+b^4+c^4$ if I know $a+b+c, a^2+b^2+c^2, a^3 +b^3 +c^3$ or in special case $a+b+c=0$ we only need a more condition like $a+b+c$. I have tried in many times and can't find out the way to archive it.

答:
Well, it sounds logical that we need three equations, because three variables and stuff. This is wrong, though. Sometimes one equation can give us two variables, like $x^2+y^2=0\Rightarrow x=0,\;y=0$ (if we stay in $\mathbb R$, that is; but if we don't, then the original question can't be answered either). Same thing here.

Obviously†, $a^3+b^3+c^3\leqslant(a^2+b^2+c^2)^{3/2}$, and the equality is only possible when two of the variables are $0$. So $(0,0,2)$ and its permutations are indeed the only real solutions, from where you can easily find the answer.

So it goes.

†
$$(a^2+b^2+c^2)^3\geqslant \overbrace{a^6+b^6+c^6+\underbrace{a^4b^2+a^2b^4}_\text{apply AM-GM}+\underbrace{b^4c^2+b^2c^4}_\text{here too}+\underbrace{a^4c^2+a^2c^4}_\text{and here}}^\text{skipped quite a lot of terms, all of them quadratic}\geqslant \\\geqslant a^6+b^6+c^6+2a^3b^3+2b^3c^3+2a^3c^3=(a^3+b^3+c^3)^2,\text{ hence}\\
a^3+b^3+c^3\leqslant (a^2+b^2+c^2)^{3/2}=4^{3/2}=8$$


问:
Why is the Frobenius norm (energy) of a rectangular matrix $A$ equal to the trace of $A^T A$ or $AA^T$
Reading through a book, it is mentioned that $||{A}||^2_F=\text{Energy}(A)=\text{tr}(AA^T)=\text{tr}(A^TA)$. I understand that for square matrices the square Frobenius norm would be the the squared sum of all elements within the matrix, but I cannot intuitivley get why for rectangular matrices it would be the trace of the matrix multiplied by its transpose (or the other way around). For instance it would be that $\text{tr}(CD^T) = \text{tr}(DC^T) = \displaystyle\sum_{i=1}^n\sum_{j=1}^dc_{ij}d_{ij}$ for some matrices $D, C$. of size $n \times d$. Maybe some sort of proof would help?

SOURCE: Linear Algebra and Optimization for Machine Learning: A Textbook (page 20)

答:
Note that:
$$
\|A\|_F^2 = \sum_{j=1}^n\sum_{k=1}^d a_{jk} a_{jk} = \mathrm{tr}(A A^T),
$$
using your definition of the trace (take $D = A, C = A$). But then since you note the trace is cyclic, we also have equality with $\mathrm{tr}(A^TA)$.


问:
Presentation of the symmetric group of a cartesian product of finite sets
Let $X$ and $Y$ be finite sets. For any set $A$, I will write $\mathfrak{S}(A)$ for its symmetric group, ie the group of bijections $A\to A$. Then $\mathfrak{S}(X\times Y)$ is generated by the subgroups $\mathfrak{S}(X)^Y$ (the permutations which perserve each $X\times \{y\}$) and $\mathfrak{S}(Y)^X$ (the permutations which perserve each $\{x\}\times Y$).

Is there a reference for finding the relations between the elements of those subgroups so that we get a presentation of $\mathfrak{S}(X\times Y)$? So, in other terms, a nice set of generators of the kernel of $\mathfrak{S}(X)^Y\ast \mathfrak{S}(Y)^X\to \mathfrak{S}(X\times Y)$ (where $G\ast H$ is the free product of $G$ and $H$)?

I would already be completely happy with the case $|Y|=2$, so that the big group is $\mathfrak{S}(X\coprod X)$, and the subgroups are the Young subgroup $\mathfrak{S}(X)\times \mathfrak{S}(X)$ and the group $(\mathbb{Z}/2\mathbb{Z})^X$ acting by exchanging corresponding elements in the two copies of $X$. The presentation could for instance be in terms of the standard transpositions $(i,\, i+1)$ in each $\mathfrak{S}(X)$ (if we order $X$) and the standard basis elements of $(\mathbb{Z}/2\mathbb{Z})^X$ (so the transpositions exchanging two copies of one element).

It seems to me that this should be a somewhat classical consideration in combinatorics or representation theory, but it is not that easy, and I have not been able to find any reference so far.

Edit: It occurred to me that perhaps a nice way to formulate the issue is the following: given a connected simple graph with set of vertices $A$, the transpositions corresponding to each edge generate the symmetric group $\mathfrak{S}(A)$. How to describe the relations between those generators in terms of the geometry of the graph? This is very wide, but the question above corresponds to the graph being an $n\times m$ grid, with an edge between any adjacent vertices in the grid, vertically and horizontally.

答:
Obviously I should have thought about this a little more before posting the question, it is actually not that hard. It turns out that in the graph formulation of the question, the global geometry of the graph does not matter at all: if $\tau_e$ is the transposition attached to the edge $e$, then we get a presentation of the full symmetric group of the vertices using the relations: $\tau_e^2=1$, $\tau_e\tau_f=\tau_f\tau_e$ if $e$ and $f$ do not have a common vertex, and $(\tau_e\tau_f)^3=1$ if $e$ and $f$ have one common vertex.

This is the classical presentation when the graph is just a linear chain, but I initially thought that more complicated graphs would add extra relations. It is actually not the case. To see this, let $A$ be a finite set, and let $G_A$ be the set of all simple graphs with set of vertices $A$. To each such graph $\mathcal{G}$, one can associate the group presentation
$$ S_\mathcal{G} = \langle \tau_e, e \text{ edge in $\mathcal{G}$} \,|\, \tau_e^2=1,
(\tau_e\tau_f)^3=1 \text{ if $e$ and $f$ are adjacent}, \tau_e\tau_f = \tau_f\tau_e \text{ otherwise} \rangle $$
and there is a canonical injective morphism $S_\mathcal{G}\to \mathfrak{S}(A)$ where $\tau_e$ is sent to the transposition exchanging the vertices of $e$. Now let $H_A\subset G_A$ be the subset of graphs such that this morphism is an isomorphism. Then we have the following facts:

if $\mathcal{G}$ is a subgraph of $\mathcal{G}'$ and $\mathcal{G}\in H_A$, then $\mathcal{G}'\in H_A$;

given subsets $A_i\subset A$ such that any pair of elements of $A$ is in at least one $A_i$, and $\mathcal{G}_i\in H_{A_i}$, then $\mathcal{G}=\bigcup_i \mathcal{G}_i\in H_A$;

If $\mathcal{G}\in G_A$ is a linear chain, then $\mathcal{G}\in H_A$.

Only the last claim is nontrivial, but it is a reformulation of the classical presentation of the symmetric group (as a Coxeter group). Then we can prove that $\mathcal{G}\in H_A$ if and only if it is connected: the fact that it is necessary to be connected is obvious, and if it is the case, first choose a covering subtree $\mathcal{T}\subset \mathcal{G}$, and then for each pair of endpoints of $\mathcal{T}$ consider the unique path $\mathcal{T}_i\subset \mathcal{T}$ between those endpoints. By basic properties of trees, any pair of points is in such a path, and those paths are linear (there is no loop involved). Then the third claim shows that $\mathcal{T_i}$ is in $H_{A_i}$ (where $A_i$ is its set of vertices), the second claim implies that $\mathcal{T}\in H_A$, and the first allows us to conclude that $\mathcal{G}\in H_A$.

For my original question, this means that we get a presentation of $\mathfrak{S}(X\times Y)$ from the transpositions $\tau_{x_1,x_2;y}$ exchanging $(x_1,y)$ and $(x_2,y)$ (which generate $\mathfrak{S}(X)^Y$), and $\tau'_{x;y_1,y_2}$ exchanging $(x,y_1)$ and $(x,y_2)$ (which generate $\mathfrak{S}(Y)^X$), by simply adding the relations that $\tau_{x_1,x_2;y}$ and $\tau'_{x;y_1,y_2}$ commute when
$x\neq x_1,x_2$ or $y\neq y_1,y_2$, and $(\tau_{x_1,x_2;y}\tau'_{x;y_1,y_2})^3=1$
when $x=x_i$ and $y=y_j$.


问:
Prove that $\sigma_2 = U \sigma_2 U^\top$ for all $U\in\operatorname{SU}(2)$
Let $\sigma_1,\sigma_2,\sigma_3$ be the Pauli matrices. Prove that $U\sigma_2 U^\top=\sigma_2$ for all $U\in\operatorname{SU}(2)$.

I can prove that $U\sigma_2 U^\top= \pm \sigma_2$ using the fact that $U\sigma_2 U^\top$ is skew-symmetric and that the determinant is $1$, but I can't differentiate between $+1,-1$.

答: Let $u,v\in\Bbb C$ such that $|u|^2+|v|^2=1$. Then $$U\sigma_2U^\top=\begin{pmatrix}u&v\\-\bar v&\bar u\end{pmatrix}\begin{pmatrix}0&-i\\i&0\end{pmatrix}\begin{pmatrix}u&-\bar v\\v&\bar u\end{pmatrix}=\begin{pmatrix}i(-uv+uv)&-i(u\bar u+v\bar v)\\i(u\bar u+v\bar v)&i(\bar u\bar v-\bar u\bar v)\end{pmatrix}=\sigma_2.$$


问:
Integration by parts disconnect
I'm trying to integrate $\displaystyle E(Y^2) = \int^\infty_0 y^2\lambda e^{-\lambda y} dy$
doing it by parts this is my logic.

$\displaystyle E(Y^2) = \int^\infty_0 y^2\lambda e^{-\lambda y} dy$ where $u=y^2$, $du=2y\,dy$, $dv=\lambda e^{-\lambda y}\, dy$ and $v = -e^{-\lambda y}$ solving by parts we get $\displaystyle (y^2)(-e^{-\lambda y})-\int^\infty_0 -e^{-\lambda y} 2y \,dy$ which then simplifies to $\displaystyle 0 + 2\int^\infty_0y e^{-\lambda y} dy$ which is the same as $0 + 2 E(Y) = \dfrac{2}{\lambda}$ where $E(Y)=\dfrac{1}{\lambda}$

however the solutions shows a different answer of $\dfrac{2}{\lambda^2}$ can someone explain why this is?
http://oi40.tinypic.com/b5fwxs.jpg

答:
We have:

$$E(Y^2) = \int^\infty_0 y^2\lambda e^{-\lambda y} dy$$

Lets use Integration by Parts to do the indefinite integral and then we'll the limits of integration.

We have:

$$\int y^2\lambda e^{-\lambda y} dy = \lambda \int y^2 e^{-\lambda y} dy$$

using IBP, we let:

$u = y^2 \rightarrow du = 2 y dy$

$v = e^{-\lambda y} dy \rightarrow = -\dfrac{1}{\lambda e^{\lambda y}}$

So, we have:

$$2 \int \dfrac{y}{e^{\lambda y}} dy-\dfrac{y^2}{e^{\lambda y}}$$

Repeating this process, we have:

$u = y \rightarrow  du = dy$

$dv = e^{-(\lambda y)}  dy \rightarrow v = -\dfrac{1}{\lambda e^{\lambda y}}$

So, we have:

$$-\dfrac{y^2}{e^{\lambda y}}-\dfrac{2 y}{\lambda e^{\lambda y}}+\dfrac{2 \int e^{-\lambda y}  dy}{\lambda}$$

This is equal to:

$$ -\dfrac{e^{-\lambda y} (\lambda^2 y^2+2 \lambda y+2)}{\lambda^2}+ \text{constant}$$

Over the limits of integration $(0 , \infty)$, we see that this yields:

$$\dfrac{2}{\lambda^2}$$

Title: Integration by parts disconnect


问:
Minimum value of $\sin^{-1}x \cos^{-1}x$
Let us consider the function $~~f(x)= \sin^{-1}x \cos^{-1}x.$ I want to find the minimum value of $f$.

So, first I find the derivative $df$ of $f$ and put $df=0$ to get $\cos^{-1}x = \sin^{-1}x$ and from this critical point I got the maximum value of $f$ which is $\frac{\pi^2}{16}.$ But I want to find the minimum value of $f$.

Please help me to find this.

答:
As $-\dfrac\pi2\le\sin^{-1}x\le\dfrac\pi2$

and $0\le\cos^{-1}x\le\pi$

Clearly $x=-1$ yields the minimum value

Alternatively,

$$f(x)=\sin^{-1}x\cos^{-1}x=\left(\dfrac\pi2-\cos^{-1}x\right)\cos^{-1}x=\left(\dfrac\pi4\right)^2-\left(\cos^{-1}x-\dfrac\pi4\right)^2$$

Now $f(x)$ will be minimum if $\left(\cos^{-1}x-\dfrac\pi4\right)^2$ is maximum

$0\le\cos^{-1}x\le\pi\implies0-\dfrac\pi4\le\cos^{-1}x-\dfrac\pi4\le\pi-\dfrac\pi4$

$\implies\left(\cos^{-1}x-\dfrac\pi4\right)^2\le\left(\pi-\dfrac\pi4\right)^2$


问:
Difference of Positive Semidefinite Matrices
Suppose I have two matrices:
$$
A\succeq 0\\
B\succ 0
$$
and I know that
$$
\langle v_i,Bv_i\rangle - \lambda_i \geq 0
$$
for every normalized eigenpair $(v_i,\lambda_i)$ of $A$.  Is this enough to conclude
$$
B-A\succeq 0
$$
i.e., the difference is positive semi-definite? I can't seem to prove this or find a counterexample, but I suspect it is true.  Any help would be appreciated.</p>

答: As a counterexample, take $$B=\pmatrix{1&0\\0&1} \qquad \textrm{and} \qquad A=\pmatrix{1&1\\1&1} ;$$ then $B-A$ is not positive semi-definite.


问:
Divisible totally ordered additive abelian groups
Let $(G,+,\leq)$ be a divisible totally ordered additive abelian group and $g_{1},g_{2}\in G$. If for every integer $n>1$,  $g_{1}\geq (1-\frac{1}{n})g_{2}$, Can we have $g_{1}\geq g_{2}$? Thanks for any answer.

答:
Hint: Consider $G=\Bbb Q\times\Bbb Q$ with the lexicographic order $(a,b)\le (c,d)\iff a<c$ or $(a=c)\land (b\le d)$.

Here viewing the first coordinate as the ordinary rational numbers, $\varepsilon:=(0,1)$ will behave like an infinitesimal: it's positive but less than $\frac1n(1,0)$ for every $n\ge 1$.


答: As you can see by the fact that at $\frac{\pi^2}{16}$ it is the only maximum (or stationary point of any kind), that means it is strictly decreasing in both directions. Taking the range of $\arcsin(x)$ and $\arccos(x)$, you see that the function has a range of [-1,1]. Checking both -1 and 1 (because you are strictly decreasing in both directions from $\frac{\pi^2}{16}$ but at different rates), you find that f(-1) is smaller and thus the minimum.

Title: Minimum value of $\sin^{-1}x \cos^{-1}x$

Body:
Let us consider the function $~~f(x)= \sin^{-1}x \cos^{-1}x.$ I want to find the minimum value of $f$.

So, first I find the derivative $df$ of $f$ and put $df=0$ to get $\cos^{-1}x = \sin^{-1}x$ and from this critical point I got the maximum value of $f$ which is $\frac{\pi^2}{16}.$ But I want to find the minimum value of $f$.

Please help me to find this.


问:
Approximate a positive Schwartz function
Context: let $f \in \mathcal{S}(\mathbb{R}^+) $ be a function of the Schwartz space (all functions whose derivatives are rapidly decreasing) on $\mathbb{R}^+$.

We already know that a generic such $f$ can be pointwise approximated by a linear combination of Laguerre functions because they form a basis of the Schwartz space. Note that $\text{n}^{\text{th}}$ Laguerre function $l_n$ is defined as:

$$ l_n(x) = e^{\frac{-x}2} L_n(x)$$

where $L_n$ is the $\text{n}^{\text{th}}$ Laguerre polynomials.

Question: now suppose that $f \in \mathcal{S}(\mathbb{R}^+)$ is positive on $\mathbb{R}^+$. Is it true that I can always pointwise converge to $f$ with a sequence of positive polynomials multiplied by a decreasing exponential?

Any reference will be appreciated.

答:
If $f\ge 0$ then $f^{1/2}= \lim_{n\to \infty} P_n(x)e^{-x/2}$ and $f =\lim_{n\to \infty} P_n(x)^2 e^{-x}$

If you want it to be strictly positive then add $e^{-x}/n$.

If you meant that $f\ge 0 \implies \sum_{n=1}^N L_n\frac{\langle f,L_n\rangle}{\langle L_n,L_n\rangle} \ge 0$  then no, $L_2-L_1$ is bounded so $(c+L_2-L_1)e^{-x/2}\ge 0$ but $(c-L_1)e^{-x/2}$ is not $\ge 0$.


问:
Complex polynomials and lines of constant argument
Here are three domain plots of a complex polynomial of degree $5$. The left picture is very zoomed out, and the right picture is more zoomed into the zeroes. (Pictures are taken from Elias Wegert's book "Visual Complex Functions".) The color indicates the argument of the function; the modulus is not featured.

Say we focus our attention on one color, like yellow. Then we can see that the yellow lines coming in from infinity seem to "end" at the zeroes of the polynomial. My confusion is: I cannot justify why this should be the case in general. Why is it true that every yellow line (that is, a line of constant argument) coming in from infinity should terminate at a zero of the polynomial?

It makes sense to me that around a zero of order $n$, the polynomial should look like $z^n$ -- that is, there should be $n$ yellow lines emanating from that point. What is not clear to me, though, is why every yellow line coming in from infinity should terminate at a zero in particular. Why can't it terminate at any other point? Any suggestions / hints would be greatly appreciated. Thanks!

答:
Not quite, there're continous lines (not only yellow, but also cyan and magenta) cutting through the right-most zeroes in the pictures.

Also, the cyan or magenta line comes to one side of other zeros but leaves out with other colours.

Yellow looks brighter than other colours.  See more about luminous flux.

Note that Hue[1/6]=yellow, Hue[1/2]=cyan and Hue[5/6]=magenta, it depends on the values of arguments for field lines lying on (usually differ by $180^\circ$ before and after cutting).

See also how the electrostatic field lines passing through the poles here.

Considering $f(z)$ near zero $c$ with multiplicity $m$,

\begin{align}
  z &= c+re^{i\theta} \\
  f(z) & \approx f(c)+\frac{f^{(m)}(c)}{m!}(re^{i\theta})^m \\
  & = \frac{r^m}{m!} f^{(m)}(c) e^{im\theta} \\
  \arg f(z) & \approx m\theta+\arg f^{(m)}(c)
\end{align}

The last term governs how the colour wheel is rotated about $c$.