问:
How to calculate the inverse of Hessian matrix?
In the question Calculating the determinant of the Hessian of a function, we know that the Hessian of function
$$f(x_1,\dots,x_n) = g\bigg(x_1,\bigg(\sum_{i=2}^n x_i^2\bigg)^{1/2}\bigg), $$

and denote
\begin{equation}
t:=x_1 \quad \text{and}\quad r:= \bigg(\sum_{i=2}^n x_i^2\bigg)^{1/2},
\end{equation}
is given by

\begin{align}
D^2 f &= J_1^T\big(H_1 - \frac{g_r}{r}I_2\big)J_1 + \frac{g_r}{r}I_n ,
\end{align}
where $I_k$ is the $k\times k$ identity matrix,
$$J_1 = \left(
        \begin{array}{ccccc}
          1 & 0 & 0 & \cdots & 0 \\
          0 & \frac{x_2}{r} & \frac{x_3}{r} & \cdots & \frac{x_n}{r} \\
        \end{array}
      \right),
$$
and
$$H_1 = \left(
          \begin{array}{cc}
            g_{tt} & g_{tr} \\
            g_{rt} & g_{rr} \\
          \end{array}
        \right).
$$

Then the determinant of the Hessian of $f$ is given by
\begin{equation}
\det D^2f = (g_{tt}g_{rr}-g_{tr}^2)\bigg(\frac{g_r}{r}\bigg)^{n-2}.
\end{equation}

My question is how to calculate the inverse of Hessian $ D^2 f$, or how to get the following formula
$$\sum_{i,j=1}^nf^{ij}w_{ij}=(n-2)\frac{r}{g_r}\frac{w_r}{r}+\frac{g_{tt}w_{rr}+g_{rr}w_{tt}-2g_{rt}w_{rt}}{g_{rr}g_{tt}-g_{rt}^2},$$
where $\{f^{ij}\}$ is the inverse matrix of $ D^2 f$ and $w=(\det D^2 f)^{-\frac{n+1}{n+2}}.$

This question come from the paper The Bernstein problem for affine maximal hypersurfaces by Neil S. Trudinger and Xu-Jia Wang, I have made efforts to calculate it, but failed. I hope I can find some good ideas here, thank you.

答:
Finally, I figured it out. I will use the well-known formula
$$\left.\frac{d}{ds}\det(A+sB)\right|_{s=0}=\det A \cdot\text{tr}(A^{-1}B)\,.$$

We denote $D^2f=(f_{ij})$ and $D^2 w=(w_{ij})$, and note that
$$\det D^2 f=\left(\frac{g_r}{r}\right)^{n-2}\left(g_{rr}g_{tt}-g_{rt}^2\right),$$
then we have
$$\begin{aligned}\det\big((f_{ij})&+s(w_{ij})\big)\\&=\left(\frac{g_r+sw_r}{r}\right)^{n-2}\left((g_{rr}+sw_{rr})(g_{tt}+sw_{tt})-(g_{rt}+sw_{rt})^2\right),\end{aligned}$$
differentiating it at $s=0$ directly yields
$$\begin{aligned}(n-2)\left(\frac{g_r}{r}\right)^{n-3}\frac{w_r}{r} \left(g_{rr}g_{tt}-g_{rt}^2\right) &+\left(\frac{g_r}{r}\right)^{n-2}(g_{rr}w_{tt}+g_{tt}w_{rr}-2g_{rt}w_{rt})\\&=\det D^2f\cdot\text{tr}\big((D^2f)^{-1}D^2w\big)\,.\end{aligned}$$
Hence, we obtain what we want.


问:
How do I show the improper integral exist?
Let $a_0=0$, and define for $k \geq 1$ $a_k=\sqrt{1+\frac{1}{2}+\cdots+\frac{1}{k}}$

Assume further that $b_k$ is sequence of positive real numbers such that $\sum_{n=1}^\infty b_k^2 < \infty$, and that $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ is a continuous and positive functions so that $f(x) \leq b_k$ when $a_{k-1} \leq |x| \leq a_k$ for $k=1,2,3,\ldots$. Show that the improper integral $\int_{\mathbb{R}^2} f(x) \, dx $ exists.

Can anyone give some direction to solve this question?

答:
We have
$$
\iint_{a_{k-1}\leq |x|\leq a_k} f \leq b_k (\pi a_k^2-\pi a_{k-1}^2)=\frac{\pi b_k}{k},
$$
so apply the Cauchy-Shwartz inequality,
$$
\iint_{{\mathbb R}^2} f \leq \sum_{k=1}^\infty \iint_{a_{k-1}\leq |x|\leq a_k} f
\leq \sum_{k=1}^\infty\frac{\pi b_k}{k}\leq \pi \Big(\sum_{k=1}^\infty b_k^2\Big)^{1/2}
\Big(\sum_{k=1}^\infty \frac{1}{k^2}\Big)^{1/2}<\infty.
$$

ContentLicense: CC BY-SA 4.0


问:
interesting set of integrals
When integrating functions $\dfrac {\sin x}{\sin x}$, as well as $\dfrac {\sin3x}{\sin x}$ , $\dfrac {\sin 5x}{\sin x}$, $\dfrac {\sin 7x}{\sin x}$ etc. on interval from $0$ to $\dfrac {\pi}{2}$, the answer turns out to be $\dfrac {\pi}{2}$. How can this be proven?</p>

ContentLicense: CC BY-SA 3.0

Title: interesting set of integrals

Body:
Hint:

$$1+2 \cos{2 x} + 2 \cos{4 x} + \ldots+2 \cos{2 n x} = \frac{\sin{(2 n+1) x}}{\sin{x}} $$


问:
Probability of geometric random variable
Problem: A bag contains  $K$  blue balls and $N−K$ red balls. Find the expected number of blue balls observed when $n$ balls are randomly drawn.

Trying to figure out how to solve and there are 4 options:

$n\frac{K}{N}$

$(n-1)\frac{K}{N}$

$(n-1)\frac{K-1}{N-1}$

$(n)\frac{K-1}{N-1}$

When you pick ball 1 it would have $p(K)=\frac{K}{N}$.  How do you find the $p(K)$ when the $p(K)$ keeps decreasing?  Do you need to assume that $n$ decreases also? My first guess would be that the answer is: $(n-1)\frac{K-1}{N-1}$.  But that is wrong.  Please help.

答:
your distribution is not a geometric but it can by an Hypergeometric or a Binomial, it depends if you draw the balls without or with replacement, respectively.

In both cases, the Expectation of the number of Blue balls drawn is $n\frac{K}{N}$


问:
finite field with characteristic not equal to 2
How many elements are squares for a finite field with characteristic $p \ne 2$?

Not sure how to begin this problem.

答: Let $K$ be that finite field of prime characteristic $p > 2,$ and let $K^*$ be its multiplicative group. Then $K$ has $p^n$ elements for some positive integer $n,$ and $K^*$ is a cyclic group with $p^n - 1$ elements. Consider the set $H := \{t^2 \mid t \in K^*\}.$ It consists precisely of the non-zero square numbers in $K.$ Show that $H$ is a subgroup of $K^*$ of index $2.$


问:
How much can we tell about $\det(X)$ if we know $\det(I + X)$?

What can we tell about $\det(X)$ if we know $\det(I + X)$? Will it
give some kind of bound for $\det(X)$?

In general, if we know the determinant of matrix $A + X$, where $A$ is a constant matrix, how much can we say about $\det(X)$?

Thank you for the attention.

答:
Nothing To document that "nothing if $d>1$" where $d$ is the dimension is the answer (as said above) :

Claim: For any $\lambda\in \mathbb R$ there exists a matrix $X$ such that $\det X  = \lambda$, while $\det(I+X)=0$.

As a corollary (take $\lambda = \pm n\in\mathbb N$ in the first claim),

Corollary: there exists a sequence $X_n$ such that $\det X_n \to \pm \infty$ while $\det(I+X_n)=0$.

This tells you that nothing general can be said.

Proof. If $d$ is odd take $X$ to be be diagonal matrix with $X_{11}=\lambda$ and $X_{kk}=1$ for $2\leq k\leq d-1$ and $X_{dd}=-1$.  If $d$ is even and greater or equal to $4$, $X_{kk}=1$ for $2\leq k\leq d-2$ and $X_{d-1,d-1}= X_{dd}=-1$. if$d=2$, take $X=\begin{pmatrix} -1 & \lambda \\ -1 & -1\end{pmatrix}$.

Something

On the other hand, it could be that you meant something else than what you asked. For example, what is true is Hadamard's inequality:
$$
\| \det(A+X) -\det(A) \| \leq C(d) \|X\| \max(\|A+X\|^{d-1},\|A\|^{d-1})
$$


答: $\sin((k+2)x)-\sin kx=2\cos (k+1)x \sin x$ Use this to reduce the problem.</p>

Title: interesting set of integrals

Body: When integrating functions $\dfrac {\sin x}{\sin x}$, as well as $\dfrac {\sin3x}{\sin x}$ , $\dfrac {\sin 5x}{\sin x}$, $\dfrac {\sin 7x}{\sin x}$ etc. on interval from $0$ to $\dfrac {\pi}{2}$, the answer turns out to be $\dfrac {\pi}{2}$. How can this be proven?</p>


问:
Geometric significance of 3 linear equations of three unknowns.
I am trying to learn the basic concepts of linear algebra and matrices.
What is the geometric significance of the equations?

$$\left\{\begin{array}{l}
    x + y + z = 2 \\
    x + 2y + z = 3 \\
    2x + 3y + 2z = 5 \\
\end{array}\right.$$

The first two equations add up to give the third one. But I do not know how they are related in terms of geometry.

答: Each of the equations describes a plane. Generally, three such equation describes the only point that belongs to those three planes. But since, in your case, the third equation adds no new information, what you have there describes that intersection of two distinct non-parallel planes, which is a line.</p>


问:
Arc length Integral from 2 point

For $x  = \frac{1}{3}y^{3} + \frac{1}{4}y$, the arc length from $(\frac{7}{12},1)$ to $(a,b)$ is $\frac{53}{6}$. What is the value of $a + b$?

I've tried to solve this but I can't find the solution to because I didn't really understand what it means

答:
You know that
$$a=\frac 13 b^3+\frac 1 4b$$ The formula for this specific arclength is

$$L=\int_1^b \sqrt{1+\left(y^2+\frac{1}{4}\right)^2}\,dy$$ which is really
messy (it involves elliptic functions of the first and second kinds).

Let us assume (this looks reasonable taking into account the target value) that $b$ is not too large. So, using series
$$\sqrt{1+\left(y^2+\frac{1}{4}\right)^2}=\frac{\sqrt{41}}{4}+\frac{10 (y-1)}{\sqrt{41}}+\frac{333 (y-1)^2}{41
   \sqrt{41}}+O\left((y-1)^3\right)$$ and integrating termwise
$$L\sim -\frac{1305}{164 \sqrt{41}}+\frac{1373 b}{164 \sqrt{41}}-\frac{128 b^2}{41
   \sqrt{41}}+\frac{111 b^3}{41 \sqrt{41}}+O\left(b^4\right)$$

Now, use $L=\frac{53}6$. Solving the cubic gives, as an estimate, $b=2.90794$.

Repeat the process around $y=3$
$$\sqrt{1+\left(y^2+\frac{1}{4}\right)^2}=\frac{\sqrt{1385}}{4}+\frac{222 (y-3)}{\sqrt{1385}}+\frac{52397 (y-3)^2}{1385
   \sqrt{1385}}+O\left((y-3)^3\right)$$

$$L\sim -\frac{512747}{16620 \sqrt{1385}}+\frac{114877 b}{5540 \sqrt{1385}}-\frac{3456
   b^2}{1385 \sqrt{1385}}+\frac{52397 b^3}{4155 \sqrt{1385}}+O\left(b^4\right)$$ Solving this new cubic gives $b=2.93919$.

At this point, I stop ! The exact value of $b$ was given by
@DavidP.

Title: Arc length Integral from 2 point


问:
If $\beta$ is a basis of $V$ then does that mean that $V=span{\beta}$?
I am confused with the definition of 'basis'.

A basis $\beta$ for a vector space $V$ is a linearly independent subset of $V$ that generates $V$. And span($\beta$) is the set consisting of all linear combinations of the vectors in $\beta$.
So from my understanding, just because all vectors in $V$ can be generated by $\beta$ doesn't necessarily mean that V=span($\beta$), since there might exist $b\in span(\beta)$  s.t. $b\notin V$
But I have learned that if $W\leq V$ and $\beta$ is a basis for both $V, W$, then $V=W$ since $W=V=span(\beta)$. ($V, W$ are finite dimensional vector space)
 This seems to imply that a vector space is equal to the span of its basis, which contradicts my understanding of its definition.
 I'd like to know which part of my understanding above is flawed.

答: Since $V$ is the whole space, the possibility “there might exist $b\in\operatorname{span}(\beta)$  s.t. $b\notin V$” isn't real. So, $V$ being generated by $\beta$ is the same thing as $V=\operatorname{span}(\beta)$.


答:
Yes, it is true that $$V = \text{span}\ \beta$$
To address your concern, suppose $\beta = \{v_1,v_2,\ldots,v_n\}$. If $b\in\text{span}\ \beta$, then $b = \alpha_1v_1 + \ldots + \alpha_nv_n$ for scalars $\alpha_1,\alpha_2,\ldots,\alpha_n$. Since $V$ is a vector space, and $v_1,v_2,\ldots,v_n \in V$, $\alpha_1v_1 + \ldots + \alpha_nv_n\in V$. This is from vector space axioms!

Not only is a vector space a span of its basis, but the basis of a vector space is also oftentimes defined as a minimal spanning set! In other words, the basis of a vector space (finite-dimensional) is a set of minimal/least possible size, such that the span of vectors in this set is exactly the entire space $V$.

Title: If $\beta$ is a basis of $V$ then does that mean that $V=span{\beta}$?

Body:
I am confused with the definition of 'basis'.

A basis $\beta$ for a vector space $V$ is a linearly independent subset of $V$ that generates $V$. And span($\beta$) is the set consisting of all linear combinations of the vectors in $\beta$.
So from my understanding, just because all vectors in $V$ can be generated by $\beta$ doesn't necessarily mean that V=span($\beta$), since there might exist $b\in span(\beta)$  s.t. $b\notin V$
But I have learned that if $W\leq V$ and $\beta$ is a basis for both $V, W$, then $V=W$ since $W=V=span(\beta)$. ($V, W$ are finite dimensional vector space)
 This seems to imply that a vector space is equal to the span of its basis, which contradicts my understanding of its definition.
 I'd like to know which part of my understanding above is flawed.


问:
$\overline{A \cup B}$ is read as "neither $A$ nor $B$", how about $\overline{A \cap B}$?
$\overline{A \cup B}$ is read as "neither $A$ nor $B$", how about $\overline{A \cap B}$?

I know that $\overline{A \cap B}$ equals to $\overline{A}\cup \overline{B}$ which can be read as "not A or not B". But it is not what I am looking for.

答:
Neither or only one of A and B

At most one of A and B

Not both A and B

‘NAND’ A and B  (borrowing the terminology of logic gates)


问:
Computing $\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$
I'm trying to compute
$$I=\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$$

The following is my effort,
$$I(a)=\int_0^\infty\frac{\ln x}{x^2+a^2}dx$$
Let $x=a^2/y$ so that $dx=-(a^2/y^2)dy$ which leads to
$$I(a)=\int_0^\infty \frac{\ln(a/y)}{a^2+y^2}dy=\int_0^\infty\frac{\ln a}{y^2+a^2}dy-I(a)$$
$$I(a)=\frac{1}{2}\int_0^\infty\frac{\ln a}{y^2+a^2}dy=\frac{1}{2}\frac{\ln a}{a}\arctan\left( \frac{y}{a}\right)_0^\infty=\frac{\ln a}{a}\frac{\pi }{4}$$
Differentiating with respect to $a$ then
$$\frac{dI(a)}{a}=-2aI'(a)=\frac{\pi}{4}\left( \frac{1}{a^2}-\frac{\ln a}{a^2}\right)$$
where $$I'(a)=\int_0^\infty \frac{\ln y}{(y^2+a^2)^2}dx$$
$$I'(a)=\frac{\pi}{-8a}\left( \frac{1}{a^2}-\frac{\ln a}{a^2}\right)$$
$$I'(a=1)=-\frac{\pi}{8}$$
But the correct answer is $-\pi/4$.

Can you help me figure where I mistake? Please give some method if there is which is much better than what I have done?

答:
Contour Integration Approach

Consider the keyhole contour, $\gamma$, that goes from $0$ to $\infty$ just above the positive real axis, circles the complex plane counterclockwise, then returns to $0$ just below the positive real axis.

Now compute the integral
$$\newcommand{\Res}{\operatorname*{Res}}
\begin{align}
\int_\gamma\frac{\log(z)^2}{\left(1+z^2\right)^2}\,\mathrm{d}z
&=\int_0^\infty\frac{\log(x)^2}{\left(1+x^2\right)^2}\,\mathrm{d}x
-\int_0^\infty\frac{(\log(x)+2\pi i)^2}{\left(1+x^2\right)^2}\,\mathrm{d}x\tag1\\
&=-4\pi i\int_0^\infty\frac{\log(x)}{\left(1+x^2\right)^2}\,\mathrm{d}x
+4\pi^2\int_0^\infty\frac1{\left(1+x^2\right)^2}\,\mathrm{d}x\tag2\\
&=2\pi i\left(\color{#C00}{\Res_{z=i}\left(\frac{\log(z)^2}{\left(1+z^2\right)^2}\right)}+\color{#090}{\Res_{z=-i}\left(\frac{\log(z)^2}{\left(1+z^2\right)^2}\right)}\right)\tag3\\
&=2\pi i\left(\color{#C00}{-\frac\pi4+i\frac{\pi^2}{16}}+\color{#090}{\frac{3\pi}4-i\frac{9\pi^2}{16}}\right)\tag4\\[9pt]
&=i\pi^2+\pi^3\tag5
\end{align}
$$
Explanation:

$(1)$: integrate along the contour above and below the positive real axis

$\phantom{\text{(1):}}$ the integral along the huge circle vanishes

$\phantom{\text{(1):}}$ the integral near zero also vanishes

$(2)$: expand $(\log(x)+2\pi i)^2$ and combine

$(3)$: the original integral is $2\pi i$ times the sum of the residues of the integrand

$(4)$: compute the residues

$(5)$: simplify

Thus, we get not only that
$$
\int_0^\infty\frac{\log(x)}{\left(1+x^2\right)^2}\,\mathrm{d}x=-\frac\pi4\tag6
$$
but also that
$$
\int_0^\infty\frac1{\left(1+x^2\right)^2}\,\mathrm{d}x=\frac\pi4\tag7
$$

A Real Approach

This is similar to what is attempted in the question.
$$
\begin{align}
\int_0^\infty\frac{\log(x)}{x^2+a^2}\,\mathrm{d}x
&=\frac1a\int_0^\infty\frac{\log(x)+\log(a)}{x^2+1}\,\mathrm{d}x\tag8\\
&=\frac{\log(a)}a\int_0^\infty\frac1{x^2+1}\,\mathrm{d}x\tag9\\
&=\frac{\pi\log(a)}{2a}\tag{10}
\end{align}
$$
Explanation:

$\phantom{1}\text{(8)}$: substitute $x\mapsto ax$

$\phantom{1}\text{(9)}$: substituting $x\mapsto1/x$ shows that $\int_0^\infty\frac{\log(x)}{x^2+1}\,\mathrm{d}x=0$

$(10)$: evaluate the integral

Taking the derivative of $(10)$ in $a$ and dividing by $-2a$ gives
$$
\int_0^\infty\frac{\log(x)}{\left(x^2+a^2\right)^2}\,\mathrm{d}x=\frac\pi4\frac{\log(a)-1}{a^3}\tag{11}
$$
Plug in $a=1$


问:
Prove that totally bounded metric spaces are separable

Prove that a totally bounded metric space $M$ is separable.

Please help me complete my proof, I think I'm almost there.

Since $M$ is totally bounded, for every $1/n$, there exists a finite set $D_n$ for which $$M \subset \bigcup_{x\in D_n} B\left(x,\frac{1}{n}\right)$$
Let $$D = \bigcup_{n=1}^\infty D_n$$
Since all $D_n$ are finite, $D$ is countable. Showing that $D$ is also dense will complete the proof. We know that $D\subset M$ is dense iff $\overline D = M$. So we must show $\overline D \supset M$. In other words, $x\in M\implies x\in \overline D$. We also know that $x\in\overline D$ iff $\forall\epsilon >0$, $B(x,\epsilon)\cap D\ne \varnothing$.

Suppose $x\in M$. For a contradiction, assume that there exists some $\epsilon > 0$ such that $B(x,\epsilon) \cap D = \varnothing$. I'm unable to find a contradiction from here, and would appreciate any help! Thank you.

答: Choose $n$ such that $\frac  1n <\epsilon$. Then $x \in B(d,\frac  1n)$ for some $d \in D_n \subseteq D$. This implies that $d \in B(x,\epsilon) \cap D$ so $B(x,\epsilon) \cap D$ is not empty.


问:
Compute integral without using derivative
I want to calculate the following integral without using derivatives

$$\frac{1}{b-a}\int_a^b e^{ty}\,{\rm d} y$$

where $t, a, b \in \Bbb R$. I know that the result is

$$\frac{1}{b-a} \left( \frac{e^{tb}-e^{ta}}{t} \right)$$

My first idea was to transform it into a sum:
$$ \lim_{n\to \infty} \sum_{i=1}^n \frac{b-a}{n}e^{t(a+\frac{b-a}{n}i)}*\frac{1}{b-a} $$
But it doesn't seems to work or the transformation is wrong.Any suggestions?

答:
Let us start where you stopped:
$$ \begin {align}
&\lim_{n\to \infty} \sum_{i=0}^{n-1} \frac{b-a}{n}e^{t(a+\frac{b-a}{n}i)}\frac{1}{b-a}\\
&=e^{ta}\lim_{n\to \infty}\frac1n \sum_{i=0}^{n-1} e^{t\frac{b-a}{n}i}\\
&=e^{ta}\lim_{n\to \infty}\frac1n  \frac{e^{t(b-a)}-1}  {e^{t\frac{b-a}n}-1}
\end {align} $$

Can you take it from here?


答:
Let $\Bbb F$ be a finite field with characteristic $> 2$. Then, $\Bbb F^\times = \Bbb F \setminus \{0\}$ is an abelian group. Thus, the map $$\varphi : \Bbb F^\times \to \Bbb F^\times$$
defined by $\varphi(x) = x^2$ is a homomorphism.

Now, note that $\varphi(x) = 1$ has two distinct solutions $1$ and $-1$. (Why are they distinct?)

Since $\Bbb F$ is a field, $x^2 = 1$ cannot have more solutions.

Thus, $\ker \varphi$ has cardinality exactly $2$ and thus, $\varphi(\Bbb F^\times)$ has cardinality $\frac{1}{2}\lvert \Bbb F^\times \rvert.$

Note that $\varphi(\Bbb F^\times)$ is precisely the set of non-zero squares. Since $0$ is indeed a square, the total number of squares are $$\frac{1}{2}\lvert \Bbb F^\times \rvert + 1 = \frac{1}{2}(\lvert \Bbb F \rvert - 1) + 1 = \boxed{\frac{1}{2}(\lvert \Bbb F \rvert + 1)}.$$

Title: finite field with characteristic not equal to 2

Body:
How many elements are squares for a finite field with characteristic $p \ne 2$?

Not sure how to begin this problem.


问:
Show that if $A_{\alpha}$ is closed in $X_{α}$, then $\prod A_\alpha$ is closed in $\prod X_\alpha$

Show that if $A_{\alpha}$ is closed in $X_{α}$, then $\prod A_\alpha$ is closed in $\prod X_\alpha$.

Please could you help me?
I have no idea how to do it.
I appreciate any help, hint or solution.
Thanks a lot.

答:
The projections $p_\alpha: \prod_{\alpha \in A} X_\alpha \to X_\alpha$ are all continuous and so

$$\prod_{\alpha \in A} A_\alpha = \bigcap_{\alpha \in A} p_\alpha^{-1}[A_\alpha]$$

is an intersection of closed sets in $\prod_{\alpha \in A} X_\alpha$ and hence closed too.


问:
References for Infinite Galois Theory
What are some "standard texts" on infinite Galois theory? I feel that Dummit&Foote only has a really brief introduction to infinite Galois theory. At the very least, I need a reference to clarify the Krull topology on the groups, preferably with an introduction to topological groups. Introducing that topology in a somewhat categorical way could also be helpful.

If knowledge about topological groups are a "prerequisite," I would also like to see some references specifically about that. I have studied a bit about pointset topology but nothing in particular about topological groups.

答: Thanks to everyone who suggested a reference. I eventually used J.S. Milne's Fields and Galois Theory, which is freely available online and very readable.</p>

Title: References for Infinite Galois Theory


问:
A basic intuition on a probability problem
Two players take turns shooting at a target, with each shot by player $i$ hitting the target with probability $p_i$, $i=1,2$. Shooting ends when two consecutive shots hit the target. Let $\mu_i$ denote the mean number of shots taken when player $i$ shoots first, $i=1,2$.

Now, I have calculated that $$\mu_1 - \mu_2 = \frac{p_2 - p_1}{p_1+p_2-p_1p_2-2}=\frac{q_2-q_1}{1+q_1q_2}$$

where $q_i=1-p_i, i=1,2 $

i.e. if the player with higher winning probability starts the game then the mean number of shots taken is higher than the case when the the player with lower winning probability starts the game. I don't understand the intuition behind this.

答:
If the better player starts, then the better player's first shot does not have opportunity to end the game. So the better player serves to end the game sooner if they come second. That is, if the better player plays first, the game is expected to last longer.

I'm not sure that your formula is right though. Consider the special case where $p_1=1$, so that it all rests on player 2, with probability of success $p=p_2$ and failure $q=1-p$. We compute $$\begin{align}\mu_1&=2p+4qp+6q^2p+\cdots\\&=2p\sum_{n=1}^\infty nq^{n-1}\\&=\frac{2p}{(1-q)^2}=\frac{2}{p_2}\end{align}$$
And
$$\begin{align}\mu_2&=2p+3qp+5q^2p+\cdots\\&=p+p\sum_{n=1}^\infty (2n-1)q^{n-1}\\&=p+p\sum_{n=1}^\infty n\sqrt{q}^{n-1}-p(2\sqrt{q})\sum_{n=1}^\infty nq^{n-1}\\&=p+p\frac{1}{(1-\sqrt{q})^2}-p(2\sqrt{q})\frac{1}{(1-q)^2}\\
&=p+p\frac{(1+\sqrt{q})^2}{(1-q)^2}-\frac{2\sqrt{q}}{p}\\
&=p+\frac{(1+\sqrt{q})^2}{p}-\frac{2\sqrt{q}}{p}\\
&=p+\frac{1+q}{p}=p_2+\frac{2}{p_2}-1
\end{align}$$

So $\mu_1-\mu_2=1-p_2$. This is not consistent with your formula when $p_1=1$.


问:
Are there non-continuous kernels?
In Mercer's theorem context, the kernel has to be a continuous function. But if you look at the RKHS theory, the kernel does not have to be continuous.

Is it possible to find a symmetric, positive semidefinite kernel which is non-continuous?

Thanks

答:
Yes there are non-continuous positive semi-definite kernels. One actually useful example is the kernel
\begin{equation} K(x,y)=
\begin{cases} 1, & \text{ if } x=y \\
0, & \text{ if }x\neq y.
\end{cases}
\end{equation}
This is positive definite as it is the covariance function of the Gaussian white noise process, i.e. independent unit normal. Adding this kernel to a continuous one is the standard way to include noise/uncertainty in Gaussian process regression.

Title: Are there non-continuous kernels?


问:
If a finite field has characteristic 2 why is every element a square?

If a finite field has characteristic 2 why is every element a square?

I've attempted this problem by calculating the square root as follows: $\mathbb{F}$ has $q = 2^m$ elements, so if $a \in \mathbb{F}$ then $a = a^q = (a^{2^{(n-1)}})^2$. Since, $a \in \mathbb{F}$ is a square elements if there exist $b \in \mathbb{F}$ s.t $b^2 = a$, I believe the next step would be to show that the square root $b = a^{2^{(n-1)}} \in \mathbb{F}$. Is this reasonable or should I take a different approach to the problem (and if so what approach?)?

答:
Consider the function $f : \Bbb F \to \Bbb F$ given by $f(x) = x^2$.

Then, $$f(x) = f(y) \iff x^2 = y^2 \iff x^2 - y^2 = 0 \color{red}{\iff} (x - y)^2 = 0 \iff x - y = 0.$$
The $\color{red}{\iff}$ is true since the field has characteristic $2$. Thus, $f$ is actually injective. An injective map from a finite set to itself is always surjective and thus, every element of $\Bbb F$ is a square.

In general, the above argument shows that if $\Bbb F$ is a finite field with characteristic $p$, then $x \mapsto x^p$ is a bijection. In fact, this is actually an isomorphism since $(x + y)^p = x^p + y^p$.

Note that finiteness is crucial. For example, consider the field $\Bbb F = \Bbb F_{2}(T)$ of rational functions over $\Bbb F_{2}$ in the indeterminate $T$. In this case, the map above is an injection but the image is strictly smaller than $\Bbb F$. For example, $T$ is not in the image.


问:
prove that the map $d:\Bbb R \times\Bbb R \to\Bbb R $ defined by.... Is a metric
Prove that the map $d:\Bbb R \times\Bbb R \to\Bbb R $ defined by

$$d(x,y)=\begin{cases}
0,&\text{if }x=y\\
|x|+|y|,&\text{if }x\ne y\,.
\end{cases}$$

is a metric.

If $d(x,y)≥0 $  for all  $x, y \in\Bbb R$.

If $d(x,y) = 0 $ indicates that if $x ≠ y : |x|+|y| = 0$

therefore $0 = x = y $ then $x = y$, if we assume that $x ≠ y$ then

$$\begin{align*}
d(x,y) &= |x|+|y|\\
& \le |x|+|y|+2|z|\\
&= (|x|+|z|) + (|z|+|y|)\\
& = d(x,z) + d(z,y)
\end{align*}$$

Is correct?

答:
$d(x,y) = 0 = d(y,x)$ if $x=y$ and if $x \neq y$ then $d(x,y)= |x|+|y| = |y|+|x|=d(y,x)$ too, so symmetry is fulfilled.

If $x \neq y$, then $x \neq 0$ or $y \neq 0$ (they cannot both be $0$ of course) and then $d(x,y) = |x| + |y| >0$. This shows (by contrapositive) that $d(x,y)=0 \to x=y$, and the reverse was already true by definition.

In checking the triangle inequality we can always assume WLOG that all 3 points $x,y,z$ are distinct (as the other cases then follow from the remaining axioms), so

indeed $$d(x,z)=|x|+|z| \le (|x|+|y|) + (|y| + |z|) (\text{ as } |y| \ge 0) = d(x,y)+d(y,z)$$ shows the triangle inequality.

Note that balls around $0$ are the same as in the usual metric, as $d(0,x)=|x|$ for all $x$ and all points $p \neq 0$ are isolated as $B(p, |p|)= \{p\}$ for all such $p$. The metric is often called the post-office metric wrt $0$.


问:
Solving equation that involves the sum of linear terms and the $L_2$ norm
$\require{begingroup}\begingroup
\DeclareMathOperator{\sign}{sign}$

I'm trying to solve the following optimization problem:

$$
    \min_{x \in \mathbb{R}^n} \frac{1}{2} x^{\mathrm{T}} A x + b^{\mathrm{T}}x + \lVert Cx + d \rVert_2
$$

where $b \in \mathbb{R}^n$, $d \in \mathbb{R}^m$, $A \in \mathbb{R}^{n \times n}$, $C \in \mathbb{R}^{m \times n}$.

By applying the first order optimality conditions, I obtain that if $\lVert Cx + d \rVert > 0$, then
$$
Ax + b + C^\mathrm{T} \frac{Cx + d}{\lVert Cx + d \rVert_2} = 0
$$
Otherwise, for the case of $\lVert Cx + d \rVert_2 = 0$, by taking the subgradient I obtain
$$
  Ax + b + C^\mathrm{T} g = 0
$$
where $g$ is any vector such that $\lVert g \rVert_2 \leq 1$.

I'm not sure how I should approach solving this problem. In the case where $n=m=1$, assuming that $\lvert Cx + d \rvert > 0$, then
$$
\frac{Cx + d}{ \lVert Cx + d \rVert_2 } = \frac{Cx + d}{ \lvert Cx + d \rvert } = \sign(Cx + d)
$$
Hence,
$$
x = \begin{cases}
    -\frac{b + C}{A} & Cx + d > 0 \\
    -\frac{b - C}{A} & Cx + d < 0 \\
\end{cases}
$$
and this solution is applicable for $C(-\frac{b+C}{A}) + d > 0$ or $C(-\frac{b-C}{A}) + d < 0$. Otherwise, we apply $Cx + d = 0$ to obtain that $x = -\frac{d}{C}$.
However, I'm not sure how to solve this for the case of $m, n > 1$.
$\endgroup$

答:
Your minimization problem is best handled in another set of coordinates.  Done right, one can avoid the square roots from the norm term.

Firstly, your work appears to tacitly assume that $A^{\mathsf{T}}=A$.  I will assume $A$ is symmetric going forward, just for simplicity of notation.  If not, then note that since the transpose is the identity on scalars, when we take the transpose of $x^{\mathsf{T}}Ax$, we obtain $$x^{\mathsf{T}}Ax=x^{\mathsf{T}}A^{\mathsf{T}}x$$  Adding $x^{\mathsf{T}}Ax$ to both sides and dividing by $2$ shows that $$x^{\mathsf{T}}Ax=x^{\mathsf{T}}\left(\frac{A+A^{\mathsf{T}}}{2}\right)x$$  Thus we can replace $A$ by its symmetric part.

Suppose $C$ is invertible.  Let $y=Cx+d$, so that $x=C^{-1}y-C^{-1}d$ and let $P=(C^{-1})^{\mathsf{T}}AC^{-1}$.  Note that $P$ is symmetric.  You seek to minimize the expression $$f(y)=\frac{1}{2}y^{\mathsf{T}}Py-\frac{1}{2}y^{\mathsf{T}}Pd-\frac{1}{2}d^{\mathsf{T}}Py+b^{\mathsf{T}}y+\|y\|_2$$ in $y$, where I have dropped the constant terms $\frac{1}{2}d^{\mathsf{T}}Pd$ and $-b^{\mathsf{T}}C^{-1}d$.  Rearranging a little, $$f(y)=\frac{1}{2}y^{\mathsf{T}}Py+\left(b^{\mathsf{T}}-d^{\mathsf{T}}P\right)y+\|y\|_2$$  To avoid writing that monstrosity inside the parentheses, I call it $\phi^{\mathsf{T}}=b^{\mathsf{T}}-d^{\mathsf{T}}P$.

Write $y=au$, where $u$ is a unit vector.  We can now describe our problem as constrained minimization: we want to minimize $$f(au)=\frac{1}{2}a^2u^{\mathsf{T}}Pu+a\phi^{\mathsf{T}}u+a$$  subject to the constraint that $1=u^{\mathsf{T}}u$.

Applying Lagrange multipliers, an extremum occurs when:
\begin{gather*}
au^{\mathsf{T}}Pu+\phi^{\mathsf{T}}u+1=0\quad\quad(\partial_a) \\
a^2(Pu)^{\mathsf{T}}+a\phi^{\mathsf{T}}=2\lambda u^{\mathsf{T}}\quad\quad(\nabla_u) \\
1=u^{\mathsf{T}}u\quad\quad(\text{constraint})
\end{gather*}  Suppose we examine ($\nabla_u$) in the $u$ direction: then $$2\lambda=2\lambda u^{\mathsf{T}}u=a^2u^{\mathsf{T}}Pu+a\phi^{\mathsf{T}}u=-a$$ where the first equality comes from substituting (constraint) and the last from substituting ($\partial_a$).

Eliminating $\lambda$ and canceling a common factor of $a$, we now find $$aPu+\phi=-u$$  This is an eigenvalue problem, which I leave to you.

If $C$ is not invertible, then we can use the same tricks as above, but the algebra is more complicated.  I will only sketch these cases.  We can't say that $x=C^{-1}y-C^{-1}d$, but we can say that there is some linear transformation $T$ and (nonlinear!) function $v$ such that $x=Ty-Td+v(y)$  and $v(y)\in\ker{(C)}$ for all $y\in d+\text{im}{(C)}$.  (This follows from the first isomorphism theorem of vector spaces, or, more prosaically, the definitions of image and kernel plus some algebra.)

To describe how to handle this, I would recommend changing basis to write $C$ as a block matrix: either $C=\begin{bmatrix}I_n&0\end{bmatrix}$ or $C=\left[\begin{smallmatrix}I_n\\0\end{smallmatrix}\right]$.  (Note that this will change your values of $A$ and $b$.)

In the former case, where $C$ is surjective, we can choose a pseudoinverse to make $v$ linear.  Consider $x$ as two separate vectors $(au-d,z)$ adjoined.  Then $Cx+d=au$, and so your problem is to minimize $a+(\text{a polynomial in $a$, $u$, and $z$})$ subject to $1=u^{\mathsf{T}}u$.

In the latter case, $C$ is only injective.  This is the hardest scenario.  We can write $Cx+d$ as two vectors adjoined, but this does not help us make progress on eliminating the square roots.  Instead, I would interpret your optimization problem as a highly-constrained one.  To wit, let $\pi_1=\begin{bmatrix}I_n&0\end{bmatrix}$ be the projection $\mathbb{R}^m\to\mathbb{R}^n$ and $\pi_2$ the complementary projection, so that $\pi_1C=I_n$ and $\pi_2C=0$.  Then you want to minimize $$a^2u^{\mathsf{T}}\pi_1^{\mathsf{T}}A\pi_1u+ab^{\mathsf{T}}\pi_1u+a$$ subject to the constraints $a\pi_2u=\pi_2d$ and $1=u^{\mathsf{T}}u$.  (Note that this requires $m-n+1$ many Lagrange multipliers!)


问:
A function in usual topology
How can the function in usual topology $f:\mathbb{R} \to \mathbb{R}, f(x) = 1/(1+x^2)$ be neither open nor closed. I found that the interval (0,1) is mapped to [1/2,1] therefore it is not open. But I could not think of an example for the closed part. Can somebody help me

答: $f[\Bbb Z] = \{\frac{1}{n^2+1}\mid n=0,1,2,3,\ldots\}$ which is not closed as it does not contain $0$, which is a limit point of the set.

Title: A function in usual topology


答:
Consider the function $\displaystyle f(x) = \frac{1-x^2}{(1+x^2)^2} $. The Maclaurin series of $f$ is given by:

$$\displaystyle \sum_{k \ge 0}  [k(-1)^k x^{2k} - k(-1)^k x^{2k-2}]$$

The integral is equal to $\displaystyle \int_0^1 f(x)\log{x}\,\mathrm{d}x $. Using the series expansion you should get

$$\displaystyle I = \sum_{k \ge 0} \bigg(\frac{k(-1)^k}{(2k-1)^2} + \frac{k(-1)^{k+1}}{(2k+1)^2}\bigg) = \sum_{k \ge 0} \frac{(-1)^{k+1}}{(2k+1)} = -\frac{\pi}{4}.$$

Title: Computing $\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$

Body:
I'm trying to compute
$$I=\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$$

The following is my effort,
$$I(a)=\int_0^\infty\frac{\ln x}{x^2+a^2}dx$$
Let $x=a^2/y$ so that $dx=-(a^2/y^2)dy$ which leads to
$$I(a)=\int_0^\infty \frac{\ln(a/y)}{a^2+y^2}dy=\int_0^\infty\frac{\ln a}{y^2+a^2}dy-I(a)$$
$$I(a)=\frac{1}{2}\int_0^\infty\frac{\ln a}{y^2+a^2}dy=\frac{1}{2}\frac{\ln a}{a}\arctan\left( \frac{y}{a}\right)_0^\infty=\frac{\ln a}{a}\frac{\pi }{4}$$
Differentiating with respect to $a$ then
$$\frac{dI(a)}{a}=-2aI'(a)=\frac{\pi}{4}\left( \frac{1}{a^2}-\frac{\ln a}{a^2}\right)$$
where $$I'(a)=\int_0^\infty \frac{\ln y}{(y^2+a^2)^2}dx$$
$$I'(a)=\frac{\pi}{-8a}\left( \frac{1}{a^2}-\frac{\ln a}{a^2}\right)$$
$$I'(a=1)=-\frac{\pi}{8}$$
But the correct answer is $-\pi/4$.

Can you help me figure where I mistake? Please give some method if there is which is much better than what I have done?


答:
Here is an easy way to compute it :

Let's substitute $ \left\lbrace\begin{matrix}y=\frac{1}{x}\ \ \\ \mathrm{d}x=-\frac{\mathrm{d}y}{y^{2}}\end{matrix}\right. $, we get : \begin{aligned}\int_{0}^{+\infty}{\frac{\ln{x}}{\left(x^{2}+1\right)^{2}}\,\mathrm{d}x}&=-\int_{0}^{+\infty}{\frac{y^{2}\ln{y}}{\left(1+y^{2}\right)^{2}}\,\mathrm{d}y}\\ &=\left[\frac{y\ln{y}}{2\left(1+y^{2}\right)}\right]_{0}^{+\infty}-\frac{1}{2}\int_{0}^{+\infty}{\frac{1+\ln{y}}{1+y^{2}}\,\mathrm{d}y}\\ &=-\frac{1}{2}\int_{0}^{+\infty}{\frac{\ln{y}}{1+y^{2}}\,\mathrm{d}y}-\frac{1}{2}\int_{0}^{+\infty}{\frac{\mathrm{d}y}{1+y^{2}}}\\ &=-\frac{1}{2}\left[\arctan{y}\right]_{0}^{+\infty}\\ \int_{0}^{+\infty}{\frac{\ln{x}}{\left(x^{2}+1\right)^{2}}\,\mathrm{d}x}&=-\frac{\pi}{4}\end{aligned}

$\textbf{Note :}$ We first integrated by parts, setting $ u':y\mapsto -\frac{y}{\left(1+y^{2}\right)^{2}} $, and $ v:y\mapsto y\ln{y} $. From the third to the fourth line, we used the fact that $ \int_{0}^{+\infty}{\frac{\ln{y}}{1+y^{2}}\,\mathrm{d}y}=0 $, which can be proven substituting $ y=\frac{1}{x} $.

$\textbf{Bonus :}$ Using the same method, prove that $ \left(\forall x>0 \right) $, we have : $$ \int_{\frac{1}{x}}^{x}{\frac{\ln{y}}{\left(1+y^{2}\right)^{2}}\,\mathrm{d}y}=\frac{1}{2}\left(\frac{\pi}{2}-2\arctan{x}\right)+\frac{x\ln{x}}{1+x^{2}} $$


问:
Maximize $\langle {\bf A} , {\bf X} \rangle$ subject to $\| {\bf X} \|_* \leq 1$

Given ${\bf A} \in \mathbb R^{m \times n}$, $$\begin{array}{ll} \text{maximize} & \langle {\bf A} , {\bf X} \rangle\\ \text{subject to} & \| {\bf X} \|_* \leq 1\end{array}$$ where $\| \cdot \|_*$ denotes the nuclear norm.

Though I know something about the spectral norm, I know almost nothing about the nuclear norm,  dual norms, convex analysis, etc.  Since I am utterly unqualified to answer this on my own, I post this question.

Related:

Show that the dual norm of the spectral norm is the nuclear norm

Maximize $\langle \mathrm A , \mathrm X \rangle$ subject to $\| \mathrm X \|_2 \leq 1$

答:
I assume that $\| X \|_*$ is the nuclear norm of $X$. Then the function $f$ defined by
$$
f(A) = \sup_{\| X \|_* \leq 1} \langle A, X \rangle
$$
is by definition the dual of the nuclear norm. But a standard result is that the dual of the nuclear norm is the spectral norm. Thus
$$
f(A) = \sigma_\max(A),
$$ the largest singular value of .


问:
Expected value of hypergeometric distribution
There's this derivation of the formula for the expected value of hypergeometric distribution which I'm trying to understand:

$$\begin{align}E(X) & =\sum\limits _{x=0}^{n}xP(x)\\
(1) & =\sum\limits _{x=0}^{n}\dfrac{x\left(_{s}C_{x}\right)\left(_{N-s}C_{n-x}\right)}{_{N}C_{n}}\\
(2) & =\sum\limits _{x=0}^{n}\left(_{N-s}C_{n-x}\right)\dfrac{xs!}{x!(s-x)!}\dfrac{n!(N-n)!}{N!}\\
(3) & =\sum\limits _{x=1}^{n}\left(_{N-s}C_{n-x}\right)\dfrac{s(s-1)!}{(x-1)!(s-x)!}\dfrac{n(n-1)!(N-n)!}{N(N-1)!}\\
(4) & =\dfrac{ns}{N}\sum\limits _{x=1}^{n}\dfrac{\left(_{N-s}C_{n-x}\right)\left(_{s-1}C_{x-1}\right)}{_{N-1}C_{n-1}}\\
(5) & =\dfrac{ns}{N}\dfrac{1}{_{N-1}C_{n-1}}\sum\limits _{x=1}^{n}\left(_{N-s}C_{n-x}\right)\left(_{s-1}C_{x-1}\right)\\
(6) & =\dfrac{ns}{N}\dfrac{1}{_{N-1}C_{n-1}}\sum\limits _{x=0}^{n-1}\left(_{N-s}C_{n-1-x}\right)\left(_{s-1}C_{x}\right)\\
(7) & =\dfrac{ns}{N}\dfrac{_{N-1}C_{n-1}}{_{N-1}C_{n-1}}\\
(8) & =\dfrac{ns}{N}
\end{align}
$$

Could you please explain the transition from (5) to (6) and from (6) to (7)?

I don't understand three things:

Why in (6) the summation limit is from $0$ to $n-1$, why is it possible to do so?

Why in (6) we have $_{s-1}C_{x}$, not $_{s-1}C_{x-1}$, why we added $1$ to $x$?

How can we go from (6) to (7), why is $\sum\limits _{x=0}^{n-1}\left(_{N-s}C_{n-1-x}\right)\left(_{s-1}C_{x}\right) = _{N-1}C_{n-1}$?

I appreciate any help.

答:
To do it more clearly and avoid confusion, they could have done the step (5) to (6) this way: introduce the new summation variable $x'$ with $x' = x - 1$. Then,

\begin{align}
(5) & =\dfrac{ns}{N}\dfrac{1}{_{N-1}C_{n-1}}\sum\limits _{x=1}^{n}\left(_{N-s}C_{n-x}\right)\left(_{s-1}C_{x-1}\right)\\
(6) & =\dfrac{ns}{N}\dfrac{1}{_{N-1}C_{n-1}}\sum\limits _{x'=0}^{n-1}\left(_{N-s}C_{n-1-x'}\right)\left(_{s-1}C_{x'}\right)\\
\end{align}

But you see, the way you name the summation label is really irrelevant, it is what we call a "dummy variable", so you can actually relabel it as $x$. They just skipped explaining that step.

For the step from (6) to (7), the following identity

$$(1+z)^{N-1} = (1+z)^{N-s}(1+z)^{s-1}$$

is developed with the binomium of Newton,

$$\sum\limits _{n=0}^{N-1} {_{N-1}C_{n}}z^{n} = \left(\sum\limits _{k=0}^{N-s} {_{N-s}C_{k}}z^{k}\right)\left(\sum\limits _{l=0}^{s-1} {_{s-1}C_{l}}z^{l}\right)$$

You have to work out the product on the right hand and arrange terms such that you regroup terms with the same power, or what is equivalent such that $n=k+l$, so that you can compare with the left hand side. Doing that and appropriate relabeling will give you the formula you're looking after.


问:
Calculating the volume of an oblique ellipse cone
I am trying to calculate the volume of an oblique cone that is an ellipse (rather than a circular cone). I have the following measurements

Perimiter of the Ellipse (in cm)

Slant Height of longer distance from ellipse circumference to height point (in cm)

Slant Height of shorter distance from ellipse circumference to height point (in cm)

Any ideas on how to calculate the volume?

答:
The perimeter of the ellipse isn't necessary. your answer is $\frac{1}{3}\pi abh$ ($\frac{1}{3}$ base's area * height); where $a$ and $b$ are the ellipse's semi-major and semi-minor axes respectively; and $h$ is the cone's height.

The perimeter is necessary just for calculating the cone's lateral area.


问:
In the history of mathematics, has there ever been a mistake?
I was just wondering whether or not there have been mistakes in mathematics. Not a conjecture that ended up being false, but a theorem which had a proof that was accepted for a nontrivial amount of time before someone found a hole in the argument. Does this happen anymore now that we have computers? I imagine not. But it seems totally possible that this could have happened back in the Enlightenment.

Feel free to interpret this how you wish!

答:
Poincaré's discovery of homoclinic points grew out of a extremely serious mistake he made in his original submission for a prize essay contest sponsored by Acta Mathematica in 1888. His original 200 page manuscript, on the restricted three-body problem, was evaluated by Weierstrass, Mittag-Leffler, and Phragmén, who had great difficulty following his arguments. Poincaré responded with a dozen further explanations, totaling 100 pages. After many further exchanges, the editors finally decided to accept the manuscript (this was, after all, Poincaré, and he must know what he's doing) and awarded him the prize.

But around the time of publication, Phragmén was still puzzled by some points and Mittag-Leffler wrote to Poincaré. They received back a telegram from Poincaré asking that publication be stopped immediately! Poincaré realized that his belief that the stable and unstable manifolds could not intersect transversally was wrong, and that such intersection points, which he later called homoclinic points, immediately forced very complicated dynamically behavior, invalidating much of his work. He wrote to Mittag-Leffler:

"I have written this morning to Mr. Phragmén to tell him about an
error which I have committed and he has undoubtedly informed you of my
letter. But the consequences of this error are more serious than I
first thought. It is not true that the asymptotic surfaces are closed,
at least not in the sense that I meant before. What is true, is that
if one considers the two parts of that surface (which I yesterday
still believed coincided with each other) they intersect along
infinitely many asymptotic trajectories and furthermore their distance
is an infinitesimal of higher order than μp however big p is.

I don't conceal from you the trouble this discovery gives me."

Mittag-Leffler immediately halted the presses and recalled all copies of this issue he could get, destroying them all (except for a few, one of which remains in the library of the Mittag-Leffler Institute). They asked Poincare to pay for the suppression of this issue, which he did.

Poincare then wrote a new essay, incorporating many of the added notes from the original, and this was the version that Acta Mathematica published (with no mention of the earlier one). Eventually Poincaré used this as the basis of his three volume classic Les méthodes nouvelles de la mécanique céleste.

A riveting account of this story is contained in Poincaré's discovery of homoclinic points by K. G. Anderson, Archive History of Exact Sciences, 48(2) (1994), 133–147.


问:
$f:\mathbb{R}\to\mathbb{R}$ is continuous and $\int_{0}^{\infty} f(x)dx$ exists
$f:\mathbb{R}\to\mathbb{R}$ is continuous and $\int_{0}^{\infty} f(x)dx$ exists  could any one tell me which of the following statements are correct?

$1. \text{if } \lim_{x\to\infty} f(x) \text{ exists, then it is 0}$

$2.  \lim_{x\to\infty} f(x) \text{ must exists, and it is 0}$

$3. \text{ in case if f is non negative } \lim_{x\to\infty} f(x) \text{ exists, and it is 0}$

$4. \text{ in case if f is differentiable } \lim_{x\to\infty} f'(x) \text{ exists, and it is 0}$

I solved one problem in past which says: if $f$ is uniformly continuos and $\int_{0}^{\infty} f(x)dx$ exists then $\lim_{x\to\infty} f(x)=0$, so the condition in one says $f$ is uniformly continuous? and hence $1$ is true? well I have no idea about the other statements. will be pleased for your help.

答:
True and this a proof by contradiction if $\lim_{x\to\infty}f(x)=\ell\neq 0$  and suppose $\ell>0$ (the case $\ell<0$ is similar) then there's $A>0$ s.t
$$\forall x\geq A\quad f(x)\geq \frac{\ell}{2}$$
so we have
$$\int_A^\infty f(x)dx\geq \frac{\ell}{2}\int_A^\infty dx=+\infty \quad \Rightarrow\Leftarrow$$

and 3. aren't correct and we can construct a counterexample as suggested by J.J


问:
show whether or not the following statement is true: "The set of rational numbers Q is connected"
if we consider the set $G =\Bbb Q$ where $\Bbb Q$ is the set of rational numbers, it can be shown that $G$ is not connected

Let  $A_1 =\{x \in{R} : x< \sqrt{2} \}$ and
$B_1 = \{x \in{R} : x> \sqrt{2}\}$ it is then that $A_1 $ and $B_1$ open disjoint and their union is $G$ therefore $G$ is not connected. Will it be like this?

I do not pretend that they solve my tasks, I have been investigating. I need to explain to me step by step how I can show that a discrete metric space of more than one point is totally disconnected and locally connected.
Let $A$ and $B$ be connected and $ A \subset B$ if $C$ is a component of $A - B$, show that $B - C$ is connected.
Please, I know you can help me.

答:
Yes, the solution will be "like that", but you have to be more precise:

Use $A_1 = \{x \in \Bbb Q: x < \sqrt{2})=(-\infty, \sqrt{2}) \cap \Bbb Q$, where the open segment is taken in $\Bbb R$ and is open there so that $A_1$ is open in the subspace topology on $\Bbb Q$. And similarly $A_2 = \{x \in \Bbb Q: x > \sqrt{2})=(\sqrt{2}, +\infty) \cap \Bbb Q$ is open in $\Bbb Q$ too. You could also use $\{x \in \Bbb Q\mid x^2 < 2\}$ and $\{x \in \Bbb Q\mid x^2 > 2\}$ if don't want to refer to a non-rational like $\sqrt{2}$. But you can assume we have constructed and know about $\Bbb R$ here and use that to our advantage..


问:
Represent this ordered pair using a function

Suppose we have two boxes with Box 1 containing $p$ number of
chocolates and Box 2 containing $q$ number of chocolates, $p$ and $q$
being positive integers.

A game consists of taking one chocolate from Box 1 at every odd turn
and leaving Box 2 intact in that turn, and taking 1 chocolate from Box
2 at every even turn and leaving Box 1 intact in that turn.

Represent this setup using an ordered pair of suitable functions.

I tried taking the ordered pair as $(f(n), g(n))$, where it goes like,

$$ (f(n), g(n)) \xrightarrow{\text{odd turn}} (f(n)-1, g(n)) \xrightarrow{\text{even turn}} (f(n)-1, g(n)-1) \xrightarrow{\text{odd turn}} \ldots$$

And then tried setting $f(n)$ as some combination of $(-1)^n$, but it didn't work. I don't know what else to do now.

Any help will be appreciated.

答:
Does this work:

$$\big(f(n), g(n)\big) \xrightarrow{\text{$(n+1)$-th turn}} \left(f(n)-\frac {1-(-1)^{n+1}}2, g(n)-\frac {1+(-1)^{n+1}}2\right) $$
?

This assumes $\big(f(n), g(n)\big)$ represents the number of chocolates in Boxes $(1,2)$ after the $n$-th turn.


问:
Convexity proof of $h$ function
Let $h: \mathbb{R} \rightarrow \mathbb{R}$ be a convex function. Assume that $h'(0)$ exists and is bigger than $1$. Moreover, $h(0) \geq 0$. Show that $h(x) > x$ for $x > 0$.

I tried using monotonicity of slopes of convex function but I am not sure if this is correct and not sure about this info about derivative.

答: Let $0<y<x$. Then $y=tx+(1-t)0$ where $t=\frac  y  x$. So $h(y) \leq th(x)+(1-t)h(0)$. I will let you show using this and some simple algebra that $\frac {h(x)-h(0)} {x-0} \geq  \frac {h(y)-h(0)} {y-0}$. Letting $y \to 0+$ we get $\frac {h(x)-h(0)} {x-0}\geq h'(0)>1$. Hence, $h(x)> h(0)+x\ge x$

Title: Convexity proof of $h$ function


问:
Find the probability density function of a dart.
Someone plays darts that has a join density function based on the surface z = 3 - r where z is the likelihood of the dart landing at $(r,\theta)$. How can I find the value of C so that f is a joint density function and then find the probability that the thrown dart lands in S?

$$g(r,\theta)=
\begin{cases}
C_{1}(3-r), & \text{if} (r,\theta)∈ R\\
0, & \text{otherwise}
\end{cases}$$

R is the region inside of r = 1 + tan $(\theta /4)$ (which includes the little loop) and S is just the region inside the inner loop.

So far this is how I am solving for C but how can I actually get the probability? Also is this correct so far?

答:
Limits of your integral do not seem correct.

The outer loop of the region $R$ forms for
$\theta \in  \displaystyle \small  \bigg(\big (0, \frac{4\pi}{3} \big), \big(\frac{7\pi}{3}, \frac{8\pi}{3} \big) , \big(\frac{11\pi}{3}, 4\pi\big)\bigg)$

And the inner loop $S$ forms for $ \ \theta \in \displaystyle \small \big(\frac{8\pi}{3}, \frac{11\pi}{3}\big)$.

So the integral to find $\small C$ should be,

$\displaystyle \small \int_{0}^{4 \pi/3} \int_{0}^{|1 + \tan (\theta/4)|} C(3 - r) \ r \ dr \ d\theta + \int_{7 \pi /3}^{8 \pi/3} \int_{0}^{|1 + \tan (\theta/4)|} C(3 - r) \ r \ dr \ d\theta$

$+ \displaystyle \small \int_{11 \pi /3}^{4 \pi} \int_{0}^{|1 + \tan (\theta/4)|} C(3 - r) \ r \ dr \ d\theta$

$ \small \approx 14.043 C \implies \small C \approx 0.0712 $

Then the desired probability is,

$\displaystyle \small \int_{8 \pi /3}^{11 \pi / 3} \int_{0}^{|1 + \tan (\theta/4)|} C(3 - r) \ r \ dr \ d\theta \approx 0.0551 $


问:
Algebraic Solution of Inequalities
Ok I'm doing some Inequalities revision and I know what the signs mean and stuff but this question I'm a bit stuck on. You need to find the x such that $\frac{1}{4} (2x + 1) \leq \frac{3}{5} (3x - 1)$. </p>

答:
"$<$" it is less than symbol means the value $a<b$ is called as a is less than b.To solve this inequality, simply go as it is "$=$" sign but remember when you reciprocal both side, inequality sign changes and also when you multiply with $(-1)$.
$$\dfrac{1}{4} (2x + 1) \leq \dfrac{3}{5} (3x - 1)$$
$$\dfrac{(2x + 1)}{4}  \leq \dfrac{(9x - 3)}{5} $$
cross multiply:
$$5\times(2x + 1)  \leq 4\times(9x - 3) $$
$$10x + 5  \leq 36x - 12 $$

$$10x-36x\le-12-5$$
$$-26x\le-17$$
$$-x\le\dfrac {-17}{26}$$
$$x\ge\dfrac {17}{26}$$ (inequality changes due to multiplication of $-1$)

Title: Algebraic Solution of Inequalities


问:
finding big theta notation with recursion and nlg(n)
given the recursion formula: $T(n) = T(n-1) + n\cdot \lg{n}$ I need to find $g(n)$ so that $T(n)=\theta(g(n))$. I know that $T(n) = n\cdot \lg n + (n-1)\cdot \lg (n-1) + (n-2)\cdot \lg (n-2 + ... = \sum_{i=2}^{n}i\cdot \lg{n} + \theta(1) \le n^2\cdot\lg{n} + \theta(1)\Longrightarrow T(n)=O(n^2\cdot\lg{n})$ but I can't find a way to prove $T(n)=\Omega(n^2\cdot\lg{n})$ and I don't see another function that can be $g(n)$ other than $n^2\cdot\lg{n}$

答:
Assuming that $T(n) = \sum\limits_{i=2}^n i \lg i + \Theta(1)$, we have
$$
T(n) > O(n)+ \sum\limits_{i=n/2}^n i \lg i > O(n) + \sum\limits_{i=n/2}^{n} \frac{n}{2} \lg \frac{n}{2} > O(n)+ \sum\limits_{i=n/2}^n \Omega( n \lg n) = O(n) + \Omega(n^2 \lg n).
$$
Basically, the sum included at least roughly $n/2$ summands that exceed $n/2 \lg (n/2) > (n \lg n)/2 + O(n)$


问:
Obtain the $N$-point IDFT of $X(k)=\frac{3}{5-4 \cos \left(2 \pi \frac{k}{N}\right)}$
I want to obtain the $N$-point IDFT of  $X(k)=\frac{3}{5-4 \cos \left(2 \pi \frac{k}{N}\right)}$. My idea was to go with the discrete Poisson summation formula.
$$
x_{s}(n)=\sum_{m=-\infty}^{\infty} x(n-m N)=\frac{1}{N} \sum_{k=0}^{N-1} X(k) e^{i 2 \pi k n / N}
$$
but I didn't manage to solve it.

Any suggestions??

答:
Following this answer of mine, introduce $\omega=\exp(2\pi i/N)$ and consider $$S=\sum_{k=0}^{N-1}\frac{\exp(2nk\pi i/N)}{1-2z\cos(2k\pi/N)+z^2}=\sum_{k=0}^{N-1}\frac{\omega^{kn}}{(1-z\omega^k)(1-z\omega^{-k})}$$ where $|z|\neq 1$ for simplicity (you're computing $3S/N$ at $z=2$). Now use $$\sum_{p=0}^{N-1}(z\omega^{\pm k})^p=\frac{1-(z\omega^{\pm k})^N}{1-z\omega^{\pm k}}=\frac{1-z^N}{1-z\omega^{\pm k}}$$ to get $$(1-z^N)^2 S=\sum_{k=0}^{N-1}\omega^{kn}\sum_{p=0}^{N-1}(z\omega^k)^p\sum_{q=0}^{N-1}(z\omega^{-k})^q=\sum_{p,q=0}^{N-1}z^{p+q}\sum_{k=0}^{N-1}\omega^{k(n+p-q)}.$$

The inner sum is $N$ if $n+p-q$ is a multiple of $N$ (assuming $\color{blue}{0\leqslant n<N}$, this happens iff $0\leqslant p<N-n$ and $q=p+n$, or $N-n\leqslant p<N-n$ and $q=p+n-N$).

Otherwise, this sum is $0$. And finally
$$S=\frac{N}{(1-z^N)^2}\left(\sum_{p=0}^{N-n-1}z^{2p+n}+\sum_{p=N-n}^{N-1}z^{2p+n-N}\right)=\frac{N(z^n+z^{N-n})}{(1-z^2)(1-z^N)}$$ (note that $n=N$ is also acceptable here, so that the result holds for $0\leqslant n\leqslant N$).


问:
Why adding to x in quadratic equation subtracts the value from x axis.
Let suppose we have a simple U shaped vertical parabola whose vertex lies on $(0,0)$.

To move it $h$ units up, we subtract $2$ from $x$ and vice versa. Same for $k$ and $y$ axis.

This seems pretty natural now.

When I learned it for the first time, I thought - to go up we should add to $x$ (not subtract). The logic was never explained (or maybe I didn't recognize it). The teacher gave no reason (and even I didn't ask).

Today this question again popped up in my brain. So,

Why adding $h$ in $x$ in the equation, subtracts $h$ from the value of $x$ axis?

What is the logic behind this?

答:
By going up or down what we are doing is translating the axis, let say you want to go $h$ units up and $k$ units right, let our new coordinate system be $(X,Y)$, by our translation $(x,y)$ become $(X,Y)$, i.e $$(x+k,y+h) =(X,Y)$$

equating the respective coordinates we get $$X-k=x$$ $$Y-h=y$$

notice our new coordinates system is in ($X,Y$), but your equation was in ($x,y$) replace $x$ and $y$ in that equation with the above-obtained expression to get the equation in our new coordinate system, now you can rename $X$ and $Y$ as $x$ and $y$ respectively.

A quick and easy way to see this is to observe where the origin has shifted.


问:
What is the cardinality of $(\aleph_\omega)^{\aleph_0}$?
I want to show that $\aleph_0 \cdot \aleph_1 \cdot \aleph_2 \cdots = (\aleph_\omega)^{\aleph_0}$. I am not sure why this is true. My thinking is that $\aleph_n \cdot \aleph_{n+1} = \max\{\aleph_n,\aleph_{n+1}\} = \aleph_{n+1}$ and so $\aleph_0 \cdot \aleph_1 \cdot \aleph_2 \cdots = \sup\{\aleph_n \mid n \in \mathbb{N}\} = \aleph_\omega$. Now we want to show that $\aleph_\omega = (\aleph_\omega)^{\aleph_0}$. I read in Jech that for cardinals $\kappa$ and $\lambda$ with $\lambda \geq cf(\kappa)$ we have $\kappa^\lambda > \kappa$. Since the  cofinality of $\aleph_\omega$ is $\aleph_0$ then we conclude that $(\aleph_\omega)^{\aleph_0} > \aleph_\omega$ and so the result I am trying to prove does not hold. Clearly I have made a mistake somewhere but cannot figure out what it is. Any help is much appreciated.

答:
Your mistake is assuming that infinite products are continuous, i.e. that the product is the supremum of its initial segments.

By that logic, $\aleph_0^{\aleph_0}$ is $\sup\{\aleph_0^n\mid n<\omega\}=\aleph_0$.

As to your question, what is the cardinality of $(\aleph_\omega)^{\aleph_0}$? Well, we only know that it is at most $2^{\aleph_0}\cdot\aleph_{\omega_4}$. But it could be that $2^{\aleph_0}<\aleph_\omega$ and $(\aleph_\omega)^{\aleph_0}$ is, assuming the consistency of large cardinal axioms, any $\aleph_{\alpha+1}$ where $\alpha$ is a countably infinite ordinal.


问:
Suppose that $V$ is a complex vector space with $\text{dim} V= n$ and $T \in L(V)$ is such that $\text{Ker}(T^{n−2}) \neq \text{Ker}(T^{n−1})$

Suppose that $V$ is a complex vector space with $\dim V= n$ and $T \in L(V)$ is such that $\ker(T^{n−2}) \neq \ker(T^{n−1})$
Prove that $T$ has at most two distinct eigenvalues.

The solution said that $0$ is an eigenvalue but I am not sure why that has to be the case. What if $0$ is not an eigenvalue? What would be the contradiction here?

答: If $0$ is not  an eigenvalue then $T$ is invertible and so is any power of $T$. Hence, $ker (T^{n-2})=ker(T^{n-1})=\{0\}$.


问:
Compound random variables -- expected value
Let $N$ be a discrete random variable such that for $n=0,1,2,...$ $$ \mathbb{P}(N=n)=(n+1)(3/4)^2(1/4)^n $$ Let $N, X_{1},X_{2},...$ be indepedent random variables; $\mathbb{P}(X_{i}=1)=1-\mathbb{P}(X_{i}=0)=p$

Now define $N_{1}=\sum\limits_{i=1}^{N}X_{i}$, if $N>0$ and $N_{1}=0$ for $N=0$. Let $N_{0}=N-N_{1}$. Find
$$\mathbb{E}[\frac{N_{1}}{N_{0}+1}] $$

I suppose the first step is to use the law of total probability, but what next?

答:
For every $n\geqslant0$, let $U_n=\mathbb E\left[\frac{N_1}{N_0+1}\mid N=n\right]$, then $U_n=n\mathbb E\left[\frac{X_1}{N_0+1}\mid N=n\right]$ by symmetry, hence $U_n=np\mathbb E\left[\frac1{N_0+1}\mid X_1=1,N=n\right]=npV_n$ with $V_n=\mathbb E\left[\frac1{N_0+1}\mid N=n-1\right]$.

Conditionally on $N=n-1$, $N_0$ is binomial $(n-1,1-p)$ hence, by definition, $V_n=\sum\limits_{k=0}^{n-1}{n-1\choose k}(1-p)^kp^{n-1-k}\frac1{k+1}$. Using ${n-1\choose k}\frac1{k+1}=\frac1n{n\choose k+1}$, one gets
$$
V_n=\frac1n\sum\limits_{k=0}^{n-1}{n\choose k+1}(1-p)^{k+1}p^{n-(k+1)}\frac1{1-p}=\frac1n\frac{1-p^n}{1-p}.
$$
Finally, $U_n=p\frac{1-p^n}{1-p}$, hence
$$
\mathbb E\left[\frac{N_1}{N_0+1}\right]=\frac{p}{1-p}(1-\mathbb E[p^N]),
$$
and surely you can compute $\mathbb E[p^N]$.


问:
Representation $cos(\pi z )$
I already know that
$$
\frac{\cos(\pi z)}{\sin(\pi z)}:= \frac{1}{z}+\sum_{n\neq 0} ( \frac{1}{z-n}+\frac{1}{n})\ .
$$

However, I am wondering if it also true that

$$
\frac{\cos(\pi z)}{\sin(\pi z)}:= \sum_{n} ( \frac{1}{z-n}+\frac{1}{n+i})\ ?
$$

答:
You can re-write your second sum in the form
$$
\frac{1}{z} + \frac{1}{i} + \sum\limits_{n \ne 0} {\left( {\frac{1}{{z - n}} + \frac{1}{{n + i}}} \right)} .
$$
If your guess was true, subtracting the first formula from the second would give
$$
0 = 1 + 2\sum\limits_{n = 1}^\infty  {\frac{1}{{n^2  + 1}}}  > 0,
$$
which is a contradiction.

Title: Representation $cos(\pi z )$


问:
Non-trivial solutions implies row of zeros?
If there exist non trivial solutions, the row echelon matrix of homogenous augmented matrix A has a row of zeros.

True or False?

I'm not sure where to begin as to see why this would be true or false. I know that if there are a row of zeros it means that there are infinitely many solutions, but not sure how I can tell if that means there are non-trivial solutions though. Any help would be appreciated. Thanks in advance.

答:
Recall that a system can have either $0$, $1$, or infinitely many solutions. Thus, the fact that there is at least one nontrivial solution (other than the trivial solution consisting of the zero vector) implies that there are infinitely many solutions. Thus, your statement is false; as a counterexample, consider the folloring homogeneous augmented matrix (conveniently in reduced row echelon form):
$$ A=
\left[ \begin{array}{ccc|c}
1 & 0 & 2 & 0 \\
0 & 1 & 3 & 0
  \end{array}\right]
$$
Notice that $A$ has infinitely many solutions (the third column has no pivot, so the system has one free variable), yet there is no row of zeroes.

Note: The converse is not necessarily true either. That is, it is NOT the case that: if the row echelon matrix of a homogenous augmented matrix A has a row of zeroes, then there exists a nontrivial solution. As a counterexample, consider:
$$ A=
\left[ \begin{array}{cc|c}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0
  \end{array}\right]
$$

Notice that $A$ has only the trivial solution (every column has a pivot, so the system has no free variables), yet $A$ has a row of zeroes.


问:
Markov chain - probability - stochastic process
I got stuck here:

The probability that it will rain today is that it did not rain in the previous two days is $0.3$, but if it rained in one of the last two days then the probability of rain today is $0.6$.

$W(n)$ is a random variable that receives the value $1$ if in the $n\ge1$ day it rained, and the value $0$, if it's not.

I need to explain why the series $(W(n))_{n\ge 0}$ is not a Markov chain, but when they define $X_n=(W_n,W_{n-1})$, so $(W(n))_{n\ge 0}$ it is a Markov chain!

答: Markov chains only depend on the previous state.  But W_n depends on W_(n-1) and W_(n-2).  What does X_n depend on?</p>


问:
A fair die and some ties in a wardrobe - formal argument needed

A wardrobe $W_1$ contains $m$ black and $m$ blue ties. A fair die with $m$ sides is rolled. If the $r$th face appears, then $r$ ties are removed from the wardrobe $W_1$, and placed in another wardrobe $W_2$ (empty to begin with). Now, pick a tie from $W_2$ at random. What is the probability that it is a tie of black color?

Quick guess: $\displaystyle \frac{1}{2}$

Suppose $r$ ties are removed from $W_1$, and the tuple $(x,y)$ denotes how many ties of each color are removed. Here, $x+y = r$, $x$ represents black and $y$ represents blue ties. The possibilities for $(x,y)$ are as follows:
$$(0,r)\\ (1,r-1)\\ (2,r-2)\\ \vdots \\(r-2,2)\\(r-1,1)\\(r,0)\\\text{where } 1\le r\le m$$
There is an obvious inherent symmetry involved, i.e. if $(x_1,y_1)$ is a possibility, then so is $(y_1,x_1)$. This is what led me to think that the probability of picking a tie of black color from the second wardrobe is exactly $\frac{1}{2}$, since I found no reason to bias myself towards either of the colors. Is that some good intuition to count upon?

How would we formally evaluate $$\mathbf P(\text{Black tie is picked from }W_2) = ?$$
The following comes to mind:
$$\mathbf P(\text{Black tie is picked from }W_2) = \sum_{r=1}^m \mathbf P(\text{Black tie is picked from }W_2|r^{\text{th}} \text{ face appears on the die})$$
Edit: Still looking for a formal argument.

答: The probability is ${1\over2}$, by symmetry: Whatever calculation you do to answer the question for "black" is also valid for your friend, who was asked to compute the probability for "blue".


问:
Convergence almost surely implies convergence in probability sup
I'm trying to prove that convergence almost surely implies convergence in probability. I know this proof can be found everywhere online, but I don't find one using the fact that:

$$ X_n \rightarrow_{n+\infty}^{p.s.} X \Rightarrow \forall \epsilon > 0, \lim_{n+\infty} \mathbb{P}(\sup_{k\geq n} \lvert X_k - X \rvert \geq \epsilon) = 0 $$

My attempt:

Let's fix $\epsilon > 0$. Let $n \in \mathbb{N}$.

$$\lvert X_n - X \rvert \leq \sup_{n} \lvert X_n - X \rvert$$

Hence,

$$ (\sup_n \lvert X_n - X \rvert < \epsilon) \subset (\lvert X_n - X \rvert < \epsilon)$$

Using the fact that the probability mapping is monotonically increasing,

$$\mathbb{P}(\sup_n \lvert X_n - X \rvert < \epsilon) < \mathbb{P}(\lvert X_n - X \rvert < \epsilon)$$

yet,

$$\lim_{n+\infty} \mathbb{P}(\sup_{k\geq n} \lvert X_k - X \rvert < \epsilon) = 1$$

So,

$$\lim_{n+\infty} \mathbb{P}(\lvert X_n - X \rvert < \epsilon) = 1$$

Eventually,

$$\lim_{n+\infty} \mathbb{P}(\lvert X_n - X \rvert \geq \epsilon) = 0$$

So, $(X_n)$ converges in probability towards $X$.

I don't know if it is correct, I feel like I don't manage well the supremum.

答:
Your proof is (almost) correct. Below is the slightly revised proof for mathematical rigor:

Let's fix $\epsilon > 0$. Let $n \in \mathbb{N}$.

$$\lvert X_n - X \rvert \leq \sup_{\color{blue}{k \geq n}} \lvert X_{\color{blue}{k}} - X \rvert$$

Hence,

$$ (\sup_{\color{blue}{k \geq n}} \lvert X_{\color{blue}{k}} - X \rvert < \epsilon) \color{blue}{\subseteq} (\lvert X_n - X \rvert < \epsilon)$$

Using the fact that the probability mapping is monotonically increasing,

$$\mathbb{P}(\sup_{\color{blue}{k \geq n}} \lvert X_{\color{blue}{k}} - X \rvert < \epsilon) \color{blue}{\leq} \mathbb{P}(\lvert X_n - X \rvert < \epsilon)$$

yet,

$$\lim_{n+\infty} \mathbb{P}(\sup_{k\geq n} \lvert X_k - X \rvert < \epsilon) = 1$$

So,

$$\lim_{n+\infty} \mathbb{P}(\lvert X_n - X \rvert < \epsilon) = 1$$

Eventually,

$$\lim_{n+\infty} \mathbb{P}(\lvert X_n - X \rvert \geq \epsilon) = 0$$

So, $(X_n)$ converges in probability towards $X$.


问:
If $\varnothing \ne A \subseteq X$ with $X$ path connected then $H_0(X,A) = 0$
Since $H_0(\mathbb{D}^{n},\mathbb{S}^{n-1}) = 0$ I was wondering wheter there was a simple generalization of the fact that $H_0(X,A) = 0$. It should be true if $X$ is path connected with $ \varnothing \ne A \subseteq X$ ? $H_0(X,A) = C_0(X)/i_{\sharp}C_0(A)$ so if $H_0(X)$ is generated by points $H_0(X,A)$ is generated by the "costant" which doesn't belong to $A$.

If I take $\gamma$ with $\gamma(0) = x_0 \in X \backslash A, \gamma(1) = y \in A$ if I think to $\gamma$ as $1-$simplex with $\partial_1 \gamma = [C_{x_0}] - [C_y] = [C_{x_{0}}]$, then $\partial_1$ is surjective so every point is homologous to $[0]$.

Is this Correct ? In any case, any type of comment about this proof of or giving a real proof of this fact would be appreciated.

答:
Your proof is correct. It uses the construction of singular homology groups and has thus a "geometric" flavor.

If you know the long exact homology sequence for reduced homology, then it is very easy to see. It ends with

$$ \dots \to \tilde H_0(X) \to H_0(X,A) \to 0$$

For path-connected $X$ we have $\tilde H_0(X) = 0$ and therefore $H_0(X,A) = 0$.


问:
Expected Value of Binomial Distribution
What is the expected value of the absolute value of the difference between the number of incoming tails and the number of incoming heads  when a coin is tossed 5 times?

Here is what I think :  Let  $X$  is random variable of number of incoming heads then
$p=\frac{1}{2}$ and $n=5$ so $X\sim Binom(5,\frac{1}{2})$

$X$ and $Y=5-X$ ($Y$ is random variable of number of incoming tails) then $E(K)=E(\left|X-(5-X)\right|)=E(\left|2X-5\right|)=E(2X-5)=2E(X)-5=2.\frac{5}{2}-5=0$ where $K=\left|X-Y\right|=\left|2X-5\right|$ and $E(X)=n.p=5.\frac{1}{2}=\frac{5}{2}$

But the right answer is $\frac{15}{8}$. Where am I making a mistake? Any help will be appreciated.

答: Since $X\sim Binom(5,\frac{1}{2})$, the range of $X$ is $\{0,1,2,3,4,5\}$, which implies that $2X - 5$ is not necessarily always positive. Therefore in your proof, it is not true that $E(\left|2X - 5\right|) = E(2X - 5)$.

Title: Expected Value of Binomial Distribution


问:
Find the time span of snow plow operation, given that its speed is inversely proportional to the height of the snow

One day snow began to fall before dawn and continued to fall at a constant rate. At midday a snowplot set out to clear a road. At 2pm it turned back, arriving to the starting point at 3pm.
  If we suppose that the snowplow speed is inversely proportional to the height of the snow, at what time did it start snowing? At what time should the snowplot turn back in order to arrive to the starting point at 2pm?

Let:

$x(t)$ be the position of the snowplow at time $t$.

$h(t)$ the height of the snow at time $t$

We denote $\Delta V $ the volume of snow removed by the snowplow in a time $\Delta t$ small enough to suppose that $h(t)$ constant in $\Delta x$. If L is the widht of the shovel of the snowplow we have:

$\frac{\Delta V}{\Delta t} = h L \frac{\Delta x}{\Delta t}$

If we take  $\Delta t \rightarrow 0$:

$\frac{dV}{dt} = h L \frac{dx}{dt}$

We know that $\frac{dV}{dt} = \alpha$ is constant, so we have

$\frac{dx}{dt} = \frac{\alpha}{L h}$

As snow falls at a constant rate, we have $h(t) = c t$ then:

$\frac{dx}{dt} = \frac{\alpha}{L c t}$

Calling $ A = \frac{\alpha}{L c}$ we have:

$\frac{dx}{dt} = \frac{A}{t}$

and then:

$x(t) = A Log(t) + C$

We have $x(T) = 0 \Longrightarrow x(t) = A Log(\frac{t}{T})$

This is clearly the position of the snowplow between 12pm and 2pm. I don't know how to continue from here because in the way back h(t) isn't as simple and depends on when the snowplow went over there the first time.

答:
The only sense I can make of the problem is to assume that the plow covers exactly the same route, so the snow depth between $2$ and $3$ is only the snow that fell since the plow passed over the point the time before.  The logic for this comes from the fact that the plow made better time between $2$ and $3$ than it did between $12$ and $2$, so there must have been less snow on the ground.  A rough approach is that there was an average of $1\frac 12$ hours of snow on the second pass, so there must have been an average of $3$ hours on the first pass, so it started snowing at $10$ AM.  The alternate, that the plow turns around and comes back over the same route, fails because the speed would be infinite at the turnaround and the singularity is not integrable.

It would be good to define your variables explicitly.  You seem to define $t=0$ as when it starts snowing, then $t=T$ as noon when the plow sets out.  You are fine with $x(t)=A \log \frac tT$ between $12$ and $2$.  Then between $2$ and $3$ you have the height is $c$ times the time between the first pass and the second pass.  This should allow you to integrate it.


答:
To help you understand just how complicated the integral setup is, here is the polar plot for the curve $r(\theta) = 1 + \tan \frac{\theta}{4}$ and the Cartesian plot of the same curve, with $\theta$ on the horizontal axis and $r$ on the vertical axis:

I have limited the domain of $\theta$ to the area of interest, namely $\theta \in [-5\pi/3, 4\pi/3]$.  I leave it as an exercise to show that that this corresponds to the point of intersection in Quadrant III.  The other intersection point, in Quadrant IV, corresponds to $\theta = -4\pi/3$ and $\theta = -\pi/3$.  I have color coded the segments.  Note the red and green segments correspond to $r < 0$; the blue and orange correspond to $r > 0$.  The red and orange segments correspond to the larger (outer) lobe, and the green and blue to the smaller (inner) lobe.  In this way, we can see that the region enclosed by the outer lobe is the sum of an integral on $\theta \in \color{red}{[-5\pi/3, -4\pi/3]}$ and on $\theta \in \color{orange}{[-\pi/3, 4\pi/3]}$, but because the red interval has $r < 0$, we need to integrate $(3 - |r|)r = (3 + r)r$, so $$\frac{1}{C} = \int_{\theta = -5\pi/3}^{-4\pi/3} \int_{r=0}^{1 + \tan \theta/4} (3+r)r \, dr \, d\theta + \int_{\theta = -\pi/3}^{4\pi/3} \int_{r = 0}^{1 + \tan \theta/4} (3-r)r \, dr \, d\theta.$$  Then the inner lobe is handled similarly, with $\theta \in \color{green}{[-4\pi/3, -\pi]}$ and $\theta \in \color{blue}{[-\pi, -\pi/3]}$:
$$\int_{\theta = -4\pi/3}^{-\pi} \int_{r = 0}^{1 + \tan \theta/4} (3+r)r \, dr \, d\theta + \int_{\theta = -\pi}^{-\pi/3} \int_{r = 0}^{1 + \tan \theta/4} (3-r)r \, dr \, d\theta.$$  This expression is divided by $C$ to get the desired probability.

Title: Find the probability density function of a dart.

Body:
Someone plays darts that has a join density function based on the surface z = 3 - r where z is the likelihood of the dart landing at $(r,\theta)$. How can I find the value of C so that f is a joint density function and then find the probability that the thrown dart lands in S?

$$g(r,\theta)=
\begin{cases}
C_{1}(3-r), & \text{if} (r,\theta)∈ R\\
0, & \text{otherwise}
\end{cases}$$

R is the region inside of r = 1 + tan $(\theta /4)$ (which includes the little loop) and S is just the region inside the inner loop.

So far this is how I am solving for C but how can I actually get the probability? Also is this correct so far?


问:
How to calculate the intersection of two hyperplanes in MATLAB?
Assume you have two affine spaces defined as follows:

$$S_1:  v^{*} + \sum \alpha_i \hat{v}_{i}$$

where $v^{*}$ is a vector in $R^n$ and $\alpha$'s are coefficients and $\hat{v}$'s are basis vectors for spanning.

and the second space is defined similarly

$S_1:  u^{*} + \sum \beta_j \hat{u}_{i}$

So the question is, how to calculate the intersection of the two spaces in MATLAB. I know how to calculate it by hand: just set $S_1=S_2$ and derive some relationships for $\alpha$ and $\beta$.

The solution (or let's say the output of the matlab script) should be the intersection basis. That is,

$S_1\bigcap S_2 : z^{*} + \sum \gamma_k\hat{z}_k$.

So the output is vectors $z^{*}$ and $\hat{z}_k$. Of course it is not unique.

Let me make it absolutely clear. Here's an example:

$S_1 :  \left(
  \begin{array}{c}
     1\\
     1\\
    1 \\
  \end{array}
\right)+ \alpha_1\left(
  \begin{array}{c}
     1\\
     2\\
     1\\
  \end{array}
\right)+\alpha_2 \left(
  \begin{array}{c}
     1\\
     1\\
     2\\
  \end{array}
\right)$.

And

$S_2 :  \left(
  \begin{array}{c}
     0\\
     0\\
    1 \\
  \end{array}
\right)+ \beta_1\left(
  \begin{array}{c}
     0\\
     1\\
     1\\
  \end{array}
\right)+\beta_2 \left(
  \begin{array}{c}
     1\\
     0\\
     1\\
  \end{array}
\right)$.

Basically two planes, so the intersection would be a line.
And when I work out (on paper!) I come up with the intersection (after row-elimination) :

$S_1\bigcap S_2 : \left(
  \begin{array}{c}
     1\\
     0\\
     2\\
  \end{array}
\right)+ \gamma_1 \left(
  \begin{array}{c}
     1\\
     1\\
     2\\
  \end{array}
\right)$.

How can you do the whole procedure automatically in Matlab?

答:
you can do something similar to bellow

```<code>    s1 = [1 1; 2 1; 1 2;];
    s1 = s1';
    n1 = null(s1);
    x1 = [1; 1; 1];

    s2 = [ 0  1; 1 0; 1 1;];
    s2 = s2';
    n2 = null(s2);
    x2 = [0; 0; 1];

    drawVector(n1, {'n1'});
    hold on
    drawSpan(s1', 'r');

    drawVector(n2, {'n2'});
    drawSpan(s2', 'r');

    n3 = null([n1 n2]')
    drawVector(n3, {'n3'});
</code>```

For visualization, I used: drawLA

Body:
you can do something similar to bellow

    s1 = [1 1; 2 1; 1 2;];
    s1 = s1';
    n1 = null(s1);
    x1 = [1; 1; 1];

    s2 = [ 0  1; 1 0; 1 1;];
    s2 = s2';
    n2 = null(s2);
    x2 = [0; 0; 1];

    drawVector(n1, {'n1'});
    hold on
    drawSpan(s1', 'r');

    drawVector(n2, {'n2'});
    drawSpan(s2', 'r');

    n3 = null([n1 n2]')
    drawVector(n3, {'n3'});

For visualization, I used: drawLA


问:
What is the overall probability of success of infinite draws with exponentially decreasing odds of success per draw?
If I flip a coin until I get "heads", sooner or later, I will get one occurrence and stop. Probability of success is 1.
But what if after each "tails" (failed) draw, odds for the next draw are divided by 2 ? Instead of having successive odds like 0.5, 0,5, 0.5, etc... I have 0.5, 0.25, 0.125, etc...
Is it possible to never get "heads"?

Trying to figure out this, I ran the following rust code:

```<code>use rand;

fn draw() -> bool {
    const MIN_MAX: f64 = 1e-20;
    let mut max = 1.0;
    while max > MIN_MAX {
        max /= 2.0;
        if rand::random::<f64>() < max {
            return true
        }
    }
    false
}

fn main() {
    const TOTAL: i32 = 10_000_000;
    let mut success = 0;
    for _ in 0..TOTAL {
        if draw() {
            success += 1;
        }
    }
    println!("{} / {}", success, TOTAL);
}
</code>```

Some comments:

I found an overall probability of 0.711

The MIN_MAXis the minimum value of individual draw odds at which the program gives up and return false. Changing this value between 1e-20 and 1e-40 does not change the result much.

I would be interested to know if the overall probablity is below 1, and if so, what are the fancy maths to explain the result.

EDIT 1 2021-03-20:

Probability to have success between 1 and n draws would be:
$\frac{1}{2}+\frac{1}{2\times4}+\frac{1\times3}{2\times4\times8}+\frac{1\times3\times7}{2\times4\times8\times16}...$
(too bad at jaxmath for formula with n)
which converges to 0.7112119049133813

EDIT 3 2021-03-20:

Googling 7112119049I found this:
https://forumserver.twoplustwo.com/47/science-math-philosophy/538-riddler-question-about-coin-flipping-game-1719467/

答:
For each positive integer $n$, let $x_n$ be the probability that experiment terminates in at most $n$ flips.

Then we have the recursion
$$
x_n=x_{n-1}+(1-x_{n-1})\Bigl({\small{\frac{1}{2}}}\Bigr)^n
$$
for $n > 1$, together with the initial value $x_1={\large{\frac{1}{2}}}$.

Claim:$\;$For all $n$ we have
$
{\large{\frac{1}{2}}}
\le
x_n <
{\large{\frac{3}{4}}}
$.

Proof:

From the definition of $x_n$ (or from the recursion), it's immediate that $x_n\ge x_{n-1}\;$for all $n > 1$.

It follows that $x_n\ge {\large{\frac{1}{2}}}\;$for all $n$.

For the upper bound, we prove the more precise claim:
$
x_n
\le
{\large{\frac{3}{4}}}-{\large{\frac{1}{2^{n+1}}}}
$.

Proceed by induction on $n$ . . .

The base case $n=1$ is easily verified.

Next let $n > 1$ and assume the claim holds for the previous value of $n$.$\;$Then
\begin{align*}
x_n&=x_{n-1}+(1-x_{n-1})\Bigl({\small{\frac{1}{2}}}\Bigr)^n
\\[4pt]
&\le
\left(\frac{3}{4}-\frac{1}{2^n}\right)
+
\Bigl(
\frac{1}{2}
\Bigr)
\Bigl(\frac{1}{2}\Bigr)^n
\\[4pt]
&=
\frac{3}{4}-\frac{1}{2^{n+1}}
\\[4pt]
\end{align*}
which completes the induction.


问:
Young's inequality for three variables
Let $x, y, z \geqslant 0$ and let $p, q, r > 1$ be such that
$$
\frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 1.
$$
How can one show that under these hypotheses we have
$$
xyz \leqslant \frac{x^p}{p} + \frac{y^q}{q} + \frac{z^r}{r}
$$
with equality if and only if $x^p = y^q = z^r$, using twice the standard two-parameters Young's inequality which says that for all $x, y \geq 0$ and for all $p, q > 1$ for which $\frac{1}{p} + \frac{1}{q} = 1$ we have
$$
xy \leqslant \frac{x^p}{p} + \frac{y^q}{q}
$$
with equality if and only if $x^p = y^q$ ?

I've tried to apply it twice directly, to multiply two inequalities and to add two inequalities, but in each case it gets quite messy and I can't get the desired result, even though I'm sure it should be quite simple.

答:
Here's a solution by applying Young's inequality twice.

First apply the inequality to $x$ and $yz$ with $p$ and $\frac{p}{p-1}$ to get
$$xyz \le \frac{x^p}{p} + \frac{(yz)^{\frac{p}{p-1}}}{\frac{p}{p-1}}.$$
Then apply it to $y^{\frac{p}{p-1}}$ and $z^{\frac{p}{p-1}}$ with $\frac{p-1}{p} q$ and $\frac{\frac{p-1}{p}q}{\frac{p-1}{p}q - 1} = \frac{(p-1)q}{pq-p-q}$ to get
$$xyz \le \frac{x^p}{p} + \frac{p-1}{p} \left(\frac{y^q}{\frac{p-1}{p}q} + \frac{z^{\frac{pq}{pq-p-q}}}{\frac{(p-1)q}{pq-p-q}}\right)$$
Notice that $\frac{pq}{pq-p-q} = r$, so you get
$$xyz \le \frac{x^p}{p} + \frac{y^q}{q} + \frac{z^r}{r}$$
as wanted.

Title: Young's inequality for three variables


问:
How to prove that interval $(x-1,x]$ contains exactly one integer?
The exersise says that "Show that for any real number x, the interval $(x-1,x]$ contains exactly one integer", though there's a hint which says that consider three separately cases : $x\in \mathbb{Z}$,$(x \in \mathbb{R\setminus Z})\wedge(x>0)$, $(x \in \mathbb{R\setminus Z})\wedge(x<0)$, I don't know how to write them explicitly.

My attempt are belows:

Step1: try to prove the uniqueness.

If there are two integers $t$,$s$ and $t$ is not equal to $s$. So, |s-t|>=1, but distance between every two real numbers in (x-1,x] should strictly less than 1, it's contradict, so at most one integer in (x-1,x].

Step2: try to prove the existance.

Think about set S={$s \in \mathbb{Z}$|s<=x}, due to archimedian's property, we can conclude that S is not empty.And S has an upper bound x, so due to "maximum exists in subset of $\mathbb{Z}$", S has a maximum $l \in \mathbb{Z}$. Now I try to prove that l is contained in (x-1,x]. From definition of S, we know that $l$<=$x$.Now we try to prove that $l$>$x-1$. If not, that is $l$<=$x-1$, so $l+1$>=$l$,and $l+1$<=$x$, so $l+1 \in S$, it contradicts that $l$ is the maximum.

Can anyone tell me if my proof is true? Or if there some place where need to improve?

答: For every $y,z \in (x-1,x]$ we have $|y-z|<1$ because size of this interval is $1$ and you can't take elements on it's edge. For every two integers $n,m$ we have $|n-m|\geq 1$ so it would be contradiction if they were both in this interval.


问:
How to prove that the limit of $f(x,y)=\frac{xy}{2(x+y)}$ at $(0,0)$ is 0 using the definition of limits?
I got to the point

$|f(x,y)|=\frac{|x||y|}{2\,|x+y|} \leq \frac{x^2+y^2}{2|x|+2|y|}$

but then I'm stuck as I can't see how I could make the denominator smaller such that I create a $\sqrt{x^2+y^2}$ type expression. Any help would be appreciated.

答: But the limit is not $0$. Note that$$\lim_{x\to0}f(x,-x+x^2)=\lim_{x\to0}\frac{x-1}2=-\frac12.$$


答:
I can not explicitly write the solution in Logic notation but I have the answer.

Existence of the integer: Suppose $x$ is not an integer. Take $[x]$ as the greatest integer function. Clearly $x-1<[x]<x$ so the existence is justified.

Uniqueness of the integer: Clearly there is no integer between $x$ and $[x]$ as then it would contradict the definition of greatest integer function and also there can not be any integer between $x-1$ and $[x]$ (why? Because if there is an integer $y$ such that $x-1<y<[x]<x$ then we have $0<[x]-y<x-y<1$, now the difference between two integers can never be non-integer! hence it is a contradiction).

This completes the proof.

Title: How to prove that interval $(x-1,x]$ contains exactly one integer?

Body:
The exersise says that "Show that for any real number x, the interval $(x-1,x]$ contains exactly one integer", though there's a hint which says that consider three separately cases : $x\in \mathbb{Z}$,$(x \in \mathbb{R\setminus Z})\wedge(x>0)$, $(x \in \mathbb{R\setminus Z})\wedge(x<0)$, I don't know how to write them explicitly.

My attempt are belows:

Step1: try to prove the uniqueness.

If there are two integers $t$,$s$ and $t$ is not equal to $s$. So, |s-t|>=1, but distance between every two real numbers in (x-1,x] should strictly less than 1, it's contradict, so at most one integer in (x-1,x].

Step2: try to prove the existance.

Think about set S={$s \in \mathbb{Z}$|s<=x}, due to archimedian's property, we can conclude that S is not empty.And S has an upper bound x, so due to "maximum exists in subset of $\mathbb{Z}$", S has a maximum $l \in \mathbb{Z}$. Now I try to prove that l is contained in (x-1,x]. From definition of S, we know that $l$<=$x$.Now we try to prove that $l$>$x-1$. If not, that is $l$<=$x-1$, so $l+1$>=$l$,and $l+1$<=$x$, so $l+1 \in S$, it contradicts that $l$ is the maximum.

Can anyone tell me if my proof is true? Or if there some place where need to improve?


问:
Prove that $S^n$ is $(n-1)$-connected.
I heard someone mention that since $S^n$ is $(n-1)$-connected (i.e. $\pi_k (S^n) = 0$ for $k<n$), $\pi_2 (S^3)=0$. However, I can't seem to imagine how this is the case. The person who said it made it seem trivial, so I've come to the conclusion that this is probably a stupid question, but are there any easy/intuitive ways to understand this?

答:
It suffices to show that any map $S^n\to S^{n+k}$ is nullhomotopic for any $k\geq 1$. Give $S^n$ and $S^{n-k}$ the a CW structure consisting of only one $0$-cell and one $n$-cell (or $n-k$ cell respectfully). Use cellular approximation to obtain a cellular map $S^n\to S^{n+k}$. By definition, for any CW complex $X$, we have $X^{k-1}\subset X^k$, so by definition of cellular maps the image $S^n$ must be contained in the $n$-skeleton of $S^{n+k}$. But this is just a point (namely the $0$-skeleton of $S^{n+k}$).

The proof can also be cast in terms of Whitney's approximation and Sard's theorem (as the other answers indicated).

Title: Prove that $S^n$ is $(n-1)$-connected.


问:
Conditional probability problem for 5 red balls and 5 blue balls in a bag
A bag contains 5 red balls and 5 blue balls. Three balls are drawn randomly without replacement.

Find: the conditional probability of drawing at least 1 blue ball if we had already drew at least one red ball.

答:
⚠ I wrote this assuming that you are not satisfied with this answer and want a different answer

 This answer is with a different approach and may be wrong.

Probability = 2/3

According to the condition,

it is fixed that we had already drawn a red ball.

So we can say that - now we are left only with the possible combination from 2 balls.

These 2 balls can be in combination of

red-blue (or blue-red both are same)

red-red

blue-blue

Only in 2 out of 3 combinations, we get at least 1 blue ball.

(Just don't forget that we are ignoring the 3rd ball as it is already been told that it is red.)


问:
Divisibility and factors
1) Can factors be negative? Please prove your opinion.
2)If prime factorization is given to you, how will you find out how many composite factors are there? Not the factors, just how many.

For 2), my teacher gave me an easy method:

let x = $a^{a_1} . b^{b_1}. c^{c_1}..$

Then your answer is : $(a_1 +1)(b_1 + 1)(c_1 +1)...$

But again, he didnt prove it, and I didnt have any time left in the class to ask him, so please help me prove it...

答:
1) Prime factors are certainly positive only, since prime numbers are positive by definition. But a divisor of a number can certainly be negative. For example, all numbers in the set $\{-10,-5,-2,-1,1,2,5,10\}$ divide $10$.

2) Say $x = p_1^{e_1} * p_2^{e_2} *p_3^{e_3} * \dots * p_n^{e_n}$, where $p_i$ is the $i$th prime factor of $x$ and $e_i$ the exponent of that factor. If $a$ divides $n$, then all prime factors of $a$ have to be in the prime factorisation of $x$, and the exponent of such a factor in the factorisation in $a$ can be no more than the exponent of that factor in the factorisation of $x$. Therefore, the exponent of $p_1$ in the prime factorisation of $a$ is in the set $\{0,1,2,\dots , e_1\}$. More specifically, it is one of $e_1 + 1$ different values. Similarly, the exponent of $p_2$ in the prime factorisation of $a$ can take up $e_2 + 1$ values. Multiplying all these numbers together shows that the number of (positive) divisors of $a$ is indeed $(e_1+1)*(e_2+1)*(e_3+1)*\dots *(e_n+1)$.

For example, if $x = 12 = 2^2 * 3^1$, a divisor $a$ of $x$ can have a factor $2$ in its prime factorisation, two of them, or zero of them. Three cases for the factor $2$. Also, it can either have or have not a factor $3$ in its prime factorisation. Two cases for the factor $3$. Therefore, the possible total number of values $a$ can have is $3 * 2 = 6$ and indeed this is true: the positive factors of $12$ are $1,2,3,4,6,12$.

As I said above, a divisor can also be negative. If you want to count not only the number of positive divisors but the number of positive and negative divisors combined, you just have to multiply the outcome by $2$, because the number of positive and negative divisors are the same.

Title: Divisibility and factors


问:
Math Kangaroo Problem 3D Geometry
"Question 24
How many planes pass through exactly three vertices of a given cube?
(A) 1 (B) 2 (C) 4 (D) 8 (E) 12"

This was one only problems that did not make sense to me on the test although it seems so easy. The reason I do not understand it is that a plane can go through 1 vertex (tangent), it can go through no vertices, it can go through 2 vertices (tangent to a edge), and it can go through 4 vertices by being on or diagonal to one of the sides. How can a plane go through 3 vertices? By plane does it have to be an infinitely endless rectangular plane? Any help through diagrams or answers would be helpful!

答:
The only way a plane can go through exactly $3$ vertices in a cube is through $3$ face diagonals that form a triangle (see the diagram).

The moment a plane includes an edge, it either goes through $2$ vertices or through $4$ vertices. Same is true for a plane going through a space diagonal.

Now the counting:

i) Start with one of the vertices, say vertex $B$. There are $3$ face diagonals through $B$. There are $3$ ways to choose two face diagonals that will uniquely identify a plane.

ii) Then vertex $A$. As $AB$ is an edge, we had no plane counted in (i) that included vertex $A$. So we again have $3$ desired planes through vertex $A$.

iii) Out of $3$ planes through vertex $C$, two of them include diagonal $AC$ so those two are duplicates as they were counted in $(ii)$.

iv) Same with planes through vertex $D$, two of them include diagonal $BD$ so two duplicates, already counted in $(i)$.

Please note that we do not have to count planes through vertices $E, F, G, H$ meeting the criteria as any plane through these vertices meeting the criteria would involve a vertex from the bottom face $ABCD$ which we already counted.

So total number of planes going through exactly $3$ vertices,

$ = 3 + 3 + (3-2) + (3-2) = 8$

Title: Math Kangaroo Problem 3D Geometry


问:
Expressing $\frac{x-y}{w-z}$ as a function of $\frac{x}{w}$ and $\frac{y}{z}$
I have the expression $\frac{x-y}{w-z}$ and I would like to rewrite it as a function of two fractions, $\frac{x}{w}$ and $\frac{y}{z}$. Ideally a linear combination of $\frac{x}{w}$ and $\frac{y}{z}$ and their powers. If that is not possible, a non-linear expression in $\frac{x}{w}$ and $\frac{y}{z}$ will also do.

(Apologies in advance for possible tagging this improperly ..)

答: @Babado has written it as a function of $x/w,\,y/z,\,w,\,z$. It can't be written as a function of $x/w,\,y/z$ alone, because e.g. $x/w=2,\,y/z=1$ is compatible with $\frac{x-y}{w-z}=\frac{2w-z}{w-z}$ taking any value $k\notin\{1,\,2\}$ via $w/z=(1-k)/(2-k)$.


问:
How to prove that $\vdash \neg \neg (p \vee \neg p)$? (Natural Deduction)
I do know that double negation and LEM are equivalent, but can we prove
$$\vdash \neg \neg (p \vee \neg p)$$
without using either of them, in a Fitch-style proof?

答:
What you may use is the fact that in intuitionistic logic one can derive from $\lnot(p\lor q)$ that $\lnot p\land\lnot q$ (and vice versa: the two ways to interpret $p$ NOR $q$ are equivalent)*. And substituting $q:=\lnot p$ into those expressions, the second form easily leads to a contradiction. With this, you should be able to prove what you want to without much difficulty.

*The same is not true for NAND: while from $\lnot p\lor\lnot q$ one easily deduces $\lnot(p\land q)$, the opposite deduction is not possible in intuitionistic logic.


问:
Showing Unitary operator in $L^2(\mathbb C)$
I'm studying Lie group and functional analysis.

In $L^2(\mathbb{C})$, with respect to the measure ${(1+|z|^2)^w} dxdy$ which $x,y$ are real, imaginary part.

Let, $A : L^2(\mathbb{C}) \to L^2(\mathbb{C})$ be an operator such that $ad-bc= 1$, ($a,b,c,d\in \mathbb C$) and

$$Af(z) = |-bz + d|^{-2 - w} f\left(\frac{az-c}{-bz + d}\right)$$

Then, $A$ is unitary operator under $0<w<2$,

$$\langle f, g \rangle = \int_\mathbb{C}\int_\mathbb{C}\frac{f(z)\overline{g(\zeta)}dzd\zeta}{|z-\zeta|^{2-w}}$$

Firstly, I cannot prove the surjectiveness of $A$, but I think this prove might use some kinds of linear fractional transformation.

Second, I tried below formula, but I got some different results.

\begin{align*}
\langle f, g \rangle &= \int_\mathbb{C}\int_\mathbb{C}\frac{f(z)\overline{g(\zeta)}dzd\zeta}{|z-\zeta|^{2-w}}
\\
&=\int_\mathbb{C}\int_\mathbb{C}\frac{f(\frac{az-c}{-bz + d})\overline{g(\frac{a\zeta-c}{-b\zeta + d})}d\left(\frac{az-c}{-bz + d}\right)d\left(\frac{a\zeta-c}{-b\zeta + d}\right)}{|\frac{az-c}{-bz + d}-\frac{a\zeta-c}{-b\zeta + d}|^{2-w}}
\\
&= \int_\mathbb{C}\int_\mathbb{C}\frac{f(\frac{az-c}{-bz + d})\overline{g(\frac{a\zeta-c}{-b\zeta + d})}\frac{1}{|-bz + d|^2}dz\frac{1}{|-b\zeta + d|^2}d\zeta}{|\frac{z-\zeta}{(-bz + d)(-b\zeta + d)}|^{2-w}}
\\
&= \int_\mathbb{C}\int_\mathbb{C}\frac{f(\frac{az-c}{-bz + d})\overline{g(\frac{a\zeta-c}{-b\zeta + d})}|-bz + d|^{-w}|-b\zeta + d|^{-w}dzd\zeta}{|z-\zeta|^{2-w}}
\\&\neq \langle Af, Ag\rangle
\end{align*}

What's wrong with my calculation. Thank you.

In fact, this is concerend with Knapp, Representation theory of semisimple groups Page 34 (complementary series).

答:
The error is when you are doing the coordinate transform, the Möbius transform changes the volume element $dxdy$ to
$$\dfrac1{|-bz+d|^4} dxdy$$
so the exponent is a "$4$" not a "$2$".

This means that your $|-bz+d|^{-w}$ type expressions should be $|-bz+d|^{-2-w}$ in the final expression. As a consequence you find that $A$ does preserve the inner product.

To see bijectivity let $B$ be map corresponding to $\begin{pmatrix}d & -b\\ -c &a \end{pmatrix}$, ie the inverse Möbius transform corresponding to $A$. You may check that $AB = BA = \Bbb1$ by evaluating on an arbitrary function and looking at the equation pointwise. Remember that the $|-bz+d|^{-2-w}$ term must change to (for $BA(f)$):

$$\left|-b \left(\frac{dz+c}{bz+a}\right)+d\right|^{-2-w}\cdot|bz+a|^{-2-w}= |z\cdot (bd-db)+-bc+da|^{-2-w}=1$$

Whats left is to check that the map is well defined, which I'll leave to you.


问:
Suppose that $V$ is a complex vector space with $\text{dim} V= n$ and $T \in L(V)$ is such that $\text{Ker}(T^{n−2}) \neq \text{Ker}(T^{n−1})$

Suppose that $V$ is a complex vector space with $\dim V= n$ and $T \in L(V)$ is such that $\ker(T^{n−2}) \neq \ker(T^{n−1})$
Prove that $T$ has at most two distinct eigenvalues.

The solution said that $0$ is an eigenvalue but I am not sure why that has to be the case. What if $0$ is not an eigenvalue? What would be the contradiction here?

答:
First, notice that $\text{Ker}(T^i) \subseteq \text{Ker}(T^{i+1})$, for any $i\ge 0$.

Remark that if $\text{Ker}(T^k) = \text{Ker}(T^{k+1})$ for some $k\ge 0$ then one has $\text{Ker}(T^i) = \text{Ker}(T^{i+1})$ for every $i\ge k$. This fact can be proved easily by an induction.

By the hypothesis, $\text{Ker}(T^{n-2}) \subsetneq \text{Ker}(T^{n-1})$, hence there is no $0\le k \le n-2$ such that $\text{Ker}(T^k) = \text{Ker}(T^{k+1})$. Therefore, we get the following strictly ascending chain:
$$ 0 = \text{Ker}(T^0) \subsetneq \text{Ker}(T^1) \subsetneq \ldots \subsetneq \text{Ker}(T^{n-1})$$
Hence, $\text{dim} \ \text{Ker}(T^{n-1})$ can be either $n-1$ or $n$.

If $\text{dim} \ \text{Ker}(T^{n-1}) = n$, then actually $T^{n-1} = 0$, so all the eigenvalues of $T$ are $0$.

We remain with the case $\text{dim} \ \text{Ker}(T^{n-1}) = n-1$. By the Rank-Nullity theorem, $\text{dim} \ \text{Im}(T^{n-1}) = 1$. Since $0 = \text{Ker}(T^0) \subsetneq \text{Ker}(T^1)$, we get that $0$ is an eigenvalue of $T$. Assume that $T$ has two nonzero eigenvalues $\lambda_1 \neq \lambda_2$. Consider $v_1$ and $v_2$ the corresponding eigenvectors. Since $\lambda_1 \neq \lambda_2$, the vectors $v_1$ and $v_2$ are linearly independent. (It is a pretty well-known fact. If you don't know it, try to prove it as an exercise.) Now, just note that $T(v_i) = \lambda_i v_i$, implies that $T^{n-1}(v_i) = \lambda_i^{n-1} v_i$, with $i=1,2$. Thus, $\text{Im}(T^{n-1})$ contains two linearly independent vectors, which is a contradiction to the fact that $\text{dim} \ \text{Im}(T^{n-1}) = 1$.


问:
The 'sine and cosine theorem' - formulas for the sum and difference
I've read somewhere that the sine and cosine functions can be fully described by this theorem:

$\sin(0) = 0, \cos(0) = 1$

$\sin(a-b) = \sin(a)\cos(b) - \sin(b)\cos(a)$

$\cos(a-b) = \cos(a)\cos(b) + \sin(a)\sin(b)$

There is na $r>0$ such that:
$$0<\sin(x)<x<\tan(x), x \in(0, r), \tan(x) = \frac{\sin(x)}{\cos(x)} $$

With this theorem, we can prove things like:

$\sin^2(x) + \cos^2(x) = 1$ by doing $\cos(a-a) = \cos(a)\cos(a) + \sin(a)\sin(a)$

$\sin(-x) = -\sin(x)$ by doing $\sin(0-x) = \sin(0)\cos(x) - \sin(x)\cos(0)$

$\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b)$
by doing
 $\cos(a-(-b)) = \cos(a)\cos(-b) + \sin(a)\sin(-b)$

$\sin(a+b) = \sin(a)\cos(b)+\sin(b)\cos(a)$
 by doing
$\sin(a-(-b)) = \sin(a)\cos(-b) - \sin(-b)\cos(a)$

And other trigonometric identities that also follows from what I've already done.

The problem is that there are many definitions for the sine and cosine function. Let's begin by the classical definition:

Classical Definition

The sine function is defined as the ratio between the opposite side of the angle, and the hipotenuse of this right triangle.

The cosine function is defined as the ratio between the adjacent side of the angle, and the hipotenuse of this right triangle.

The tangent function is defined as the ratio between the sine function and the cosine function (with $\cos (x) \neq 0)$

The other trigonometric identities can be proven geometrically for an angle less than or equal $\frac{\pi}{2}rad$ because it's a right triangle. So, we can't prove $\sin(a-b)$ geometrically and then prove $\sin(a+b)$ analytically like I did, because we assumed a negative $b$, something that's not defined geometrically in the right triangle.

The identity $\sin^2(x) + \cos^2(x) = 1$ can be proven with a simple pythagorean theorem in a triangle with hypotenuse 1.

The formulas for the sum and difference of sines and cosines can be proven geometrically like in this images I found in this answer:

Unit circle definition

Imagine a circle centered at the origin of the cartesian plane, then:

The sine function for an $x \in R, x>0$ can be defined as the $y$ position of point of the circle where the angle stops if we travel anti-clockwise inside the circle's line.

The cosine for an $x \in R, x>0$ function can be defined as the $x$ position of point of the circle where the angle stops if we travel anti-clockwise inside the circle's line.

We can make the same definition for negative angles, so for an $x \in R, x<0$ the same holds, but we're know travelling clockwise.

The tangent function is defined as the ratio between the sine function and the cosine function (with $\cos (x) \neq 0)$

Then, we can define these functions for all real numbers, since when we travel $2\pi$ we get back to the initial point. So we defined sine and cosine as periodic functions.

Open question: How do I prove, with the unit circle periodic definition and without being circular, the $\sin(a-b)$ and $\cos(a-b)$ formulas? (same for $\sin(a+b)$ and $\cos(a+b)$.

Taylor Series definition

$$\sin x = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n+1}$$
$$\cos x = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} x^{2n}$$
$$\tan x = \sum^{\infty}_{n=1} \frac{B_{2n} (-4)^n (1-4^n)}{(2n)!} x^{2n-1}$$

Here, in the same question, there's an analytic proof of the trigonometric identites, for these sums.

What's the best definition for calculus?

Well, in calculus we use trigonometric functions a LOT: in integral substitutions, in series, taylor series (like the ones I showed now), derivatives, convergence tests (like the Euler onde in the basel problem) and other things...

All the definitions I see, are kinda circular or not rigorous enough to make me feel good taking some derivatives or integral substitution, because I always care about the domain of these things. So I want to define it very nicely and be able to use all the trigonometric identities.

I've seen many geometrical proofs of $\sin(a-b)$, $\sin(a+b)$, $\cos(a-b)$, $\cos(a+b)$ using a right triangle and then suddenly the person starts using this formula for all real numbers. I need a complete definition of the trigonometric functions that works periodically and for all reals. The taylor series definition seems good but they're generated using trigonometric identities that are not yet proven (assuming this definition).

ps: I know that I used some primitive words in some definitions, like 'travel' so I let them in emphasis, but I hope you guys understand. And sorry by the long post, but I needed to do it, because I've never seen a complete definition in any book. Thanks.

Body:
I created this shortcut proof for the cos addition formula, it uses the cosine and sine formula
The proof I know usually involves complex triangle shapes, so I'll be working with a basic triangle here

$a+b+c = \pi$, and $c = \pi -(a+b)$
$$ C^2 = A^2+B^2-2AB\cos{c}$$
$$ C^2 = A^2+B^2-2AB\cos{(\pi-(a+b))}$$
$$\frac{C^2}{A^2} = \frac{A^2}{A^2}+\frac{B^2}{A^2}-2\frac{AB}{A^2}\cos{(\pi-(a+b))}$$
$${\frac{C}{A}}^2 = 1+{\frac{B}{A}}^2-2\frac{B}{A}\cos{(\pi-(a+b))}$$
$$\frac{\sin{a}}{A} = \frac{\sin{b}}{B} = \frac{\sin{c}}{C} $$
$$\frac{\sin^2{c}}{\sin^2{a}} = 1+\frac{\sin^2{b}}{\sin^2{a}}-2\frac{\sin{b}}{\sin{a}}\cos{(\pi-(a+b))}$$
$$\sin^2{c} = \sin^2{a} \cdot ( 1+\frac{\sin^2{b}}{\sin^2{a}}-2\frac{\sin{b}}{\sin{a}}\cos{(\pi-(a+b))} )$$
$$1-\sin^2{c} = 1-\sin^2{a} \cdot ( 1+\frac{\sin^2{b}}{\sin^2{a}}-2\frac{\sin{b}}{\sin{a}}\cos{(\pi-(a+b))} )$$
$$\cos^2{c} = 1-\sin^2{a} \cdot ( 1+\frac{\sin^2{b}}{\sin^2{a}}-2\frac{\sin{b}}{\sin{a}}\cos{(\pi-(a+b))} )$$
$$\cos^2{c} = 1-\sin^2{a}-\sin^2{b}+2\sin{a}\sin{b}\cos{c}$$
$$\cos^2{c} - 2\sin{a}\sin{b}\cos{c}+\sin^2{a}+\sin^2{b}-1 = 0$$
$$\cos{c} = \frac{ 2\sin{a}\sin{b} \pm \sqrt{ (2\sin{a}\sin{b})^2-4(\sin^2{a}+\sin^2{b}-1) } }{2}$$
$$\cos{c} = \frac{ 2\sin{a}\sin{b} \pm \sqrt{ 4\sin^2{a}\sin^2{b}-4\sin^2{a}-4\sin^2{b}+4 } }{2}$$
$$\cos{c} = \frac{ 2\sin{a}\sin{b} \pm 2\sqrt{ \sin^2{a}\sin^2{b}-\sin^2{a}-\sin^2{b}+1 } }{2}$$
$$\cos{c} = \frac{ 2\sin{a}\sin{b} \pm 2\sqrt{ (1-\sin^2{a})(1-\sin^2{b}) } }{2}$$
$$\cos{c} = \frac{ 2\sin{a}\sin{b} \pm 2\sqrt{ \cos^2{a}\cos^2{b} } }{2}$$
$$\cos{c} = \sin{a}\sin{b} \pm \cos{a}\cos{b}$$
$$\cos{(\pi-(a+b))} = \sin{a}\sin{b} - \cos{a}\cos{b}$$
$$-\cos{(a+b)} = \sin{a}\sin{b} - \cos{a}\cos{b}$$
$$\cos{(a+b)} = \cos{a}\cos{b}- \sin{a}\sin{b}$$


问:
Evaluate : $\int\limits_0^{\pi/2} \cot x \ln(\sec x) dx$
I want to solve this integral :
$$\int_0^{\pi/2} \cot x \ln(\sec x) dx$$
I tried the following substitution : $\ln(\sec x)=t$ which means $dt=\tan x dx$
$$I=\int_0^\infty \frac{\cot x}{\tan x}tdt=\int_0^\infty \frac{t}{\tan^2 x}dt$$
I'm really disturbed by the $\tan^2 x$, I tried also to substitute $\sec x =t$ but it's not helpful either. Any helpful approach to solve this problem ?

答:
\begin{align}
\int_0^{\pi/2} &\cot x\ln(\sec x) \, dx
=\frac14\int_0^{\pi/2}\frac{-2\cos x\sin x \ln(\cos^2 x)}{1-\cos^2 x} \, dx
\\
& \overset{\color{blue}{\overset{\scriptstyle t=\cos^2 x}{dt = -2\cos x\sin x\,dx}}}{=}\quad
\frac14\int_0^1\frac{\ln t}{t-1}\, dt
=\frac14\cdot\frac{\pi^2}6=\frac{\pi^2}{24}
.
\end{align}