问:
Solving for a matrix from its quadratic form
I have a set of vectors that I am trying to predict from another set of vectors using a matrix $W$.  To find this matrix, I decide I want to minimize the $\ell^2$ norm of the error, e.g.:

$$
\text{find} \min_W \|y - Wx\|_2 \\
x,y \in \mathbb{C}^N \quad W\in \mathbb{C}^{N \times N}
$$

Where $x$ and $y$ are respective vectors from their sets, and each pair is (I hypothesize) related to eachother by $W$. I start by expanding this out:

$$
\min_W \left[ (y - Wx)^H (y - Wx) \right] \\
= \min_W \left[ (y - Wx)^H y - (y - Wx)^H Wx \right] \\
= \min_W \left[ (y^H y - y^H Wx)^H - (x^H W^H y - x^H W^H Wx)^H \right] \\
= \min_W \left[ y^H y - x^H W^H y - y^H W^H x + x^H W^H W x \right]
$$

Where $(\cdot)^H$ denotes Hermitian transpose.  I want to take the derivative with respect to $W$ and set it equal to zero, but I'm having a hard time with the derivative as I'm not sure exactly how that works with Hermitian transposes.  Taking a look at this page, it looks like I might be in trouble (e.g. I'm going about this the wrong way), but I wanted to pick your brains to see if you all had any ideas on how to move forward from here.

Thank you all!

EDIT
Michael C. Grant has pointed out that this question is underdetermined, and he is right, if I only have a single $x$ and $y$.  However I have a set of $x$ and $y$ vectors that I assume are related to eachother by $W$ (And since right now I am working with simulations, I can generate as many pairs of $x$'s and $y$'s as I want). This is a kind of "system identification" problem, where I have input-output correspondences and I'm trying to understand how the math is derived.

答:
Yes, I'd suggest that you are in trouble, but not for the reason you may think. :-) You can certainly take the derivative of that expression with respect to $W$.

But more fundamentally, your problem is underdetermined: you have $2N$ complex quantities $x$ and $y$ that you are using to construct a matrix $W$ with $N^2$ complex quantities.

If $x\neq 0$, the minimum value of the norm is zero, and any matrix that satisfies $Wx=y$ is going to attain this minimum value. For instance, $W=(x^Hx)^{-1}yx^H$ will do the trick. But you can add any other matrix $V$ satisfying $Vx=0$ to $W$ and achieve the same result.

If $x=0$, then the minimum value of the norm is $\|y\|$, and any $W$ will achieve it.

EDIT: OK, now we see the problem. What you really want to do is something like this:

$$\begin{array}{ll}\text{minimize}_W & \| Y - W X \|_F\end{array}$$
where $X,Y\in\mathbb{C}^{N\times M}$, with $M\geq N$. This is actually a standard least squares problem, though it may not seem so since it's expressed in matrix form. Using Kronecker products, we can write this as
$$\begin{array}{ll}\text{minimize}_W & \| \mathop{\textrm{vec}}(Y) - (X^H\otimes I) \mathop{\textrm{vec}}(W) \|_2\end{array}$$
where $\otimes$ denotes the Kronecker product, and $\mathop{\textrm{vec}}$ stacks the columns of its matrix argument on top of one another. The normal equations for this least squares problem are
$$
\begin{aligned}
&(X^H\otimes I)^H (X^H\otimes I) \mathop{\textrm{vec}}(W) = (X^H\otimes I)^H \mathop{\textrm{vec}}(Y) \\
&\qquad \Longrightarrow\quad (X\otimes I) (X^H\otimes I) \mathop{\textrm{vec}}(W) = (X\otimes I) \mathop{\textrm{vec}}(Y) \\
&\qquad \Longrightarrow\quad (XX^H\otimes I) \mathop{\textrm{vec}}(W) = (X\otimes I) \mathop{\textrm{vec}}(Y) \\
&\qquad \Longrightarrow\quad WXX^H = YX^H
\end{aligned}
$$
Therefore, if $XX^H\in\mathbb{C}^{N\times N}$ is invertible---and note that's an important condition that is not guaranteed to be satisfied even if $M\geq N$---then the solution is
$$W = YX^H(XX^H)^{-1}$$
The similarity to the single-vector case is not too surprising. If $XX^H$ is singular, then  $$W=YX^H(XX^H)^\dagger$$ is a minimizer, where $(XX^H)^\dagger$ is the pseudoinverse of $XX^H$.


问:
Using Steinitz Exchange Lemma to prove that a set of four vectors in $\mathbb{R}^2$ is linearly dependent
Cheers, so I am asked to prove that in the vector space of $V = \mathbb{R}^2$, every set of 4 vectors is linearly dependent. I tried solving it using Steinintz Exchange Lemma, so:

Let $V = \mathbb{R}^2$ be the vector space of interest, and a set $\{ v_1, v_2 \}$ be a basis for our vector space (e.g. $\{ (1,0), (0,1) \}$). Now let $A \subset V, A = \{ a_1 , a_2 , a_3 , a_4 \} $. As $A \subset V$ we can say that: $$
a_i=\sum_{j=1}^2 \gamma_{ij}v_j \qquad(i=1,\dots,4)
$$ I then got a bit stuck, and saw a solution that proceeded by saying that:

If $\alpha_1a_1+\dots+\alpha_4a_4=0$, then
$$
0=\sum_{i=1}^4\alpha_ia_i=
\sum_{i=1}^4\alpha_i\biggl(\,\sum_{j=1}^2\gamma_{ij}v_j\biggr)=
\sum_{j=1}^2\biggl(\,\sum_{i=1}^4\alpha_i\gamma_{ij}\biggr)v_j
$$
so
$$
\sum_{i=1}^4\alpha_i\gamma_{ij}=0 \qquad(j=1,2)
$$ and since $4 > 2$ there are infinetely many solutions so A is not linearly independent.

Although, I understand the whole logic here, and why the result solves our question, why did we suppose that $\alpha_1a_1+\dots+\alpha_4a_4=0$, and especially why did we use the elements of A in tuples? Could it be done with any other way? Thanks for the help!

答:
You need to find scalars $\alpha_1, \alpha_2,\alpha_3,\alpha_4$ not all zero such that $\alpha_1a_1+\dots+\alpha_4a_4=0$. You found infinitely many solutions for $\alpha_1, \alpha_2,\alpha_3,\alpha_4$ so $\alpha_1=\alpha_2=\alpha_3=\alpha_4=0$ is not the only solution.

Here is another way, which doesn't use any theory regarding linear systems of equations. Let $$\begin{bmatrix} x_1 \\ y_1\end{bmatrix}, \begin{bmatrix} x_2 \\ y_2\end{bmatrix}, \begin{bmatrix} x_3 \\ y_3\end{bmatrix}, \begin{bmatrix} x_4 \\ y_4\end{bmatrix} \in \Bbb{R}^2$$
be arbitrary.

If $x_1=x_2=y_1=y_2 = 0$, they are clearly linearly dependent.

If any of $x_1,x_2,y_1,y_2$ is nonzero and $x_2y_1 = x_1y_2$ then from
$$x_2 \begin{bmatrix} x_1 \\ y_1\end{bmatrix} - x_1 \begin{bmatrix} x_2 \\ y_2\end{bmatrix} = y_2 \begin{bmatrix} x_1 \\ y_1\end{bmatrix} - y_1 \begin{bmatrix} x_2 \\ y_2\end{bmatrix} = 0$$
it follows that they are linearly dependent.

If $x_2y_1 \ne x_1y_2$ then
$$(x_2y_4-x_4y_2)\begin{bmatrix} x_1 \\ y_1\end{bmatrix} + (x_4y_1-x_1y_4)\begin{bmatrix} x_2 \\ y_2\end{bmatrix} + \underbrace{(x_1y_2-x_2y_1)}_{\ne 0}\begin{bmatrix} x_4 \\ y_4\end{bmatrix} = 0$$
so they are linearly dependent.

ContentLicense: CC BY-SA 4.0


问:
picking a color game probability
The game has 5 colors red yellow green blue and black
A player randomly picks 4 colors and sorts them in a line the second player has to guess how the first player sorted them.

Mike picked and sorted four random colors (the color can be used more than once) what is the probability of guessing it right from the first try?

since Mike has to guess it right from the first time it is going to be $\frac {1} {n(Ω)}$ and our $n(Ω)$ is $5*5*5*5$ since we have 5 colors and each color can be used more than once so my answer was $\frac {1} {5^4}$

Julia knows that the color yellow appears atleast once what is the probability she guesses it right from the next guess try? (the answer should be $\frac {1} {369}$ according to the book

what I tried was $n(Ω)$=${4 \choose 1}$ $*$ $5*5*5$ but this is wrong and it felt more like guessing to me and I also could not figure out what they mean by "next try" does it mean she tried 2 times or just once?

Mike told Julia that in his order no color appears more than once what is the probability that Julia gets it right from the first try?

What I tried was $\frac {1} {5*4*3*2}$ and it also seems to be right according to the book.

Much appreciation for any help and tips on the 2nd question and if anyone can confirm that the 1st and 3rd questions are right. thank you

答:
The first and third answers look correct on the assumption all possible patterns in order are equally likely.

A hint for the second question:

Consider the number of patterns which do not include any yellows, i.e. the number selecting from red, green, blue and black.  Then subtract this number from the overall number of possible patterns (your answer to the first question) to give the number of patterns in which yellow appears at least once.

Title: picking a color game probability


问:
Understanding $\left(\sum_{n=1}^N x_n^2\right) - \bar{x}^2 = \sum_{n=1}^N (x_n - \bar{x})^2$
Equation 3.19 from this document claims (at the bottom of page 5) that
$$
\left(\sum_{n=1}^N x_n^2\right) - \bar{x}^2 = \sum_{n=1}^N (x_n - \bar{x})^2.
$$

I struggle to understand this. What rules were used to group $\bar{x}^2$ into the sum?

答:
The statement you posted is wrong.

Expand the square in the right member and find the correct one:

$$\Sigma_n[X_n-\overline{X}]^2=\Sigma_n[X_n^2-2\overline{X}X_n+\overline{X}^2]=\Sigma_nX_n^2-2\overline{X}\cdot n\overline{X}+n\overline{X}^2=\Sigma_nX_n^2-n\overline{X}^2$$

And this is exactly what the document you linked claims.


问:
Differentiating a boundary condition at infinity
A typical boundary condition for an initial boundary value problem is
$$ \lim_{x\rightarrow\infty} T(x,t) = T_\infty.$$

For example, this might be the temperature at the end of a very long rod.  Under what conditions is it equivalent to instead enforce the boundary condition
$$ \lim_{x\rightarrow\infty} \frac{\partial T}{\partial x} = 0$$

or, perhaps worded differently, is it ever the case, given $T\rightarrow T_\infty$ as $x\rightarrow\infty$, that $\partial T/\partial x\ne0$ as $x\rightarrow\infty$?  The alternative boundary condition makes sense physically, but does there exist a formal mathematical treatment of this?

答:
Take $T(x) = \sin(x^2)/x$. Then, $T(x) \to 0$ as $x \to +\infty$, but $T'(x) = (2\,x^2\,\cos(x^2)+ \sin(x^2))/x^2 \not\to 0$ as $x \to +\infty$.

Also, the reverse implication is not true: take $T(x) = \ln(x)$. Then, $T'(x) \to 0$, but $T(x) \not\to 0$.

Hence, those two conditions are not related.

ContentLicense: CC BY-SA 3.0


问:
How to solve $(1+y^2)+(x-e^{-\arctan y})y' = 0 $?
$(1+y^2)+(x-e^{-\arctan y})y' = 0    $

I have to solve this differential equation. Normally  I would just put all the terms with y variable on one side and x on the other and integrate. However when I do that there is y' in both numerator and denominator. Or I will make it in a form of $y + Px =  Q$ and find integrating factor.  There is other method of using auxiliary equation but that does not work. None of these methods work.

答:
Hint:

$$(1+y^2)+(x-e^{-\arctan y})y' = 0$$
Rewrite the DE as:
$$(1+y^2)x'+x=e^{-\arctan y} $$
Now it's a first order linear differential equation.


问:
If $2^{2k}-x^2\bigm|2^{2k}-1$ then $x=1$
This is the $y=2^k$ case of this question.

Suppose that $k\geq1$ and $0<x<2^k$ and $2^{2k}-x^2\bigm|2^{2k}-1$. Is it necessarily the case that $x=1$?

Equivalently: Suppose that there are two positive divisors of $2^{2k}-1$ which average to $2^k$. Is it necessarily the case that these two divisors are $2^k-1$ and $2^k+1$?

答:
Following from the OP's self-answer, the quadratic Diophantine equation $$nx^2-4(n-1)y^2=1\tag1$$ means that $nx^2\equiv1\pmod4$ which forces $n=4N+1$ as $x^2\equiv0,1\pmod4$. Thus $$(4N+1)x^2-NY^2=1$$ where $Y=4y$ so $Y^2\equiv4\pmod{4N+1}$ and $x^2\equiv1\pmod N$.

Suppose that $Y\equiv\pm2\pmod{4N+1}$. Then $y=(rn\pm1)/2$ and substituting into $(1)$ gives $$nx^2-(n-1)(rn\pm1)^2=1\implies x^2=r^2n^2-(r^2\mp2r)n\mp2r+1.$$ Let $x=rn-a$ so $$n=\frac{a^2\pm2r-1}{2ra-r^2\pm2r}=\frac1{4r^2}\left(2ra+r^2\mp2r+\frac{r^2(r\pm4)}{2a-r\pm2}\right)$$ which reduces to $$2^{k+3}=m+2r+\frac{r(r\pm4)}m$$ since $rn\pm1=2y$ and $y=2^k$.

Notice that this formulation is quite similar to your equivalency statement. The latter is derived from the system $sx=ty=4^k-1$ and $x+y=2^{k+1}$, which in turn is equivalent to solving $st=c(4^k-1)$ and $s+t=c\cdot2^{k+1}$. In the formulation above, we are looking for integers $s,t$ such that $st=r(r\pm4)$ and $s+t=2^K-2r$.


问:
Curve $\psi(t) = (t \cos\frac{1}{t}, t\sin\frac{1}{t})$ is simple, bounded and not rectifiable
Let $\psi(t)$ be a curve defined on $t\in[-1,0)$ by

$\psi(t) = (t \cos\frac{1}{t}, t\sin\frac{1}{t})$

Prove that $\psi(t)$ is simple, bounded and not rectifiable.

I know this curve is simple (if you take $t_1\ne t_2 \in [-1,0)$ then $\psi(t_1)\ne \psi(t_2)$) but I don't know how to prove it properly.

Instead, I really don't know where to start to show that is bounded and not rectifiable. To show that isn't rectifiable I need to prove that the length of the curve is infinitive, but I can't find any inscribed chord-polygons to use. I just need some hint.

答:
Building off of Raffaele's answer, use the fact that
$$
L(\psi|_{[-1, -\epsilon]}) = \int_{-1}^{-\epsilon} \| \psi'(t) \| \, dt
$$
for all $\epsilon > 0$. Then, suppose for contradiction that $L(\psi)$ is finite, i.e. $\psi$ is rectifiable. Observe
$$
L(\psi) \geq L(\psi|_{[-1, -\epsilon]} )
$$
for all $\epsilon > 0$. Since $L(\psi|_{[-1, -\epsilon]})$ gets arbitrarily large as $\epsilon \to 0^+$, we obtain a contradiction. Thus, $\psi$ is not rectifiable.


问:
Why there are only 3 automorphisms of $\mathbb F_3 [X] / (x^3 + 2x -1)$?
The following question was part of my abstract algebra assignment and I was clueless on how to solve the problem.

So, I am posting it here.

Let $F= \mathbb F_3 [X] / (x^3 + 2x -1)$, where $\mathbb F_3$ is the field with 3 elements. Then find the automorphism group of $F$.

I had no idea on how this question should be approached

So, I posted it here and it was linked with another question:
This question has a answer here: Automorphism group of $F=\mathbb F_3[x]/(x^3+2x-1)$.
But I have a question I answer of Shivani Goel and she has been unactive for quite few days and I urgently need to figure out my question in her answer.

My question is I know why $x \to x $, $x \to x^3 $ and $x\to x^9$ are automorphisms but why $x \to x^{27} $ and $x\to x^{81}$ and so higher powers of 3 are not automorphisms ?

Kindly give reason for that.

答:
The general theory goes as follows.
Let $p$ be a prime number. Then for all powers $q=p^r$ there is only one field $\mathbb{F}_q$ with $q$ elements up to isomorphism. The automorphism group
$$
{\rm Aut}(\mathbb{F}_q)={\rm Gal}(\mathbb{F}_q/\mathbb{F}_p)
$$
is cyclic of order $r$, generated by the Frobenius automorphism
$$
{\rm Frob}_p:\mathbb{F}_q\longrightarrow\mathbb{F}_q,\qquad
{\rm Frob}_p(x)=x^p.
$$
The fact that ${\rm Frob}_p^r={\rm id}$ is very consistent with the fact that $\mathbb{F}_q^\times$ is a group, in fact cyclic, of order $q-1$.

In the question's case $p=3$ and $q=3^3=27$ since the given polynomial is irreducible. Thus $x\to x^{27}$ is simply the identity map.


问:
Prove $f(1)-f(-1)>f'\left(-\frac{1}{\sqrt{3}}\right)+f'\left(\frac{1}{\sqrt{3}}\right)$.
Suppose the 5th order derivative function of $f(x)$ ,say, $f^{(5)}(x)>0$ for $x\in [-1,1]$.Prove $f(1)-f(-1)>f'\left(-\dfrac{1}{\sqrt{3}}\right)+f'\left(\dfrac{1}{\sqrt{3}}\right)$.

I know this can be done directly by Gauss-Legendre formula, but does there exist a more elementary proof?

Body:
Its quite a rough idea but:
$$f^{(5)}(x)>0$$
$$f^{(4)}(x)>C_1$$
$$f^{(3)}(x)>C_1x+C_2$$
$$f''(x)>\frac {C_1}2x^2+C_2x+C_3$$
$$f'(x)>\frac{C_1}6x^3+\frac{C_2}2x^2+C_3x+C_4$$
$$f(x)>\frac{C_1}{24}x^4+\frac{C_2}6x^3+\frac{C_3}2x^2+C_4x+C_5$$

so:
$$f(1)-f(-1)>\frac{C_2}3+2C_4$$
$$f'(-3^{-1/2})+f'(3^{-1/2})>\frac{C_2}6+2C_4$$
and so this suggests it can be true, however I have no upper limit so this is not really a proof :(


问:
Does such a family of sets exist?
Here is a question on set theory:

Does there exist a family $S$ of closed discs in $\mathbb{R^2}$ having positive radii such that each of these discs has at most one point in common and $\mathbb{R^2}- {\cup D}$ is countable.

I have no idea how to approach this question.

Please tell me a hint how to start this one

答:
It's a well known theorem (of Sierpiński I believe) that the real line does not admit a nontrivial partition into countably many closed sets; see my answer to this question.

Now assume for a contradiction that the plane is the union of a (necessarily countable) collection of closed discs (of positive radius) with disjoint interiors, plus countably many single points.

Let $L$ be a line in the plane which does not pass through any of the countably many points where two of those closed discs touch. The intersection of each of the given discs with $L$ is either a closed interval or a single point. Thus $L$ is the union of a countable disjoint collection of closed intervals and singletons, contradicting the theorem mentioned above.

Remark. If $S$ is a family of closed discs of positive radius with disjoint interiors in $\mathbb R^2$, then $\mathbb R^2\setminus\bigcup S$ is an uncountable $G_\delta$ set, so it has cardinality $2^{\aleph_0}$.

Title: Does such a family of sets exist?


问:
Inaccessible cardinal and von Neuman Hierarchy
I read a couple of days ago that if $k$ is inaccessible, then $V_k=H(k)$, where $V$ is the von Neumann hierarchy and $H(k)$ the class of sets that are heredititarily of cardinality $< k$.

My question would be: is there any singular $k$ for which $V_k=H(k)$? I managed to show that if $K$ is regular then $V_k=H(k)$ iff $k$ inaccessible or $k=\omega$, but I'm clueless about an example?

答:
You do not need the regular part. If $\kappa > \omega$, $V_\kappa = H_\kappa$ if and only if $\kappa = \beth_\kappa$. (Kunen \textit{Set Theory} 78.)

Define $a_0 = \aleph_0$. Define recursively, $a_{i + 1} = \beth(a_i)$. Let $\alpha = \lim_{i< \omega} a_i$. $\beth(\alpha) = \alpha$. $\alpha$ has cofinality $\omega$. So $\alpha$ is a singular cardinal with $\beth(\alpha) = \alpha$.

I hope it is correct now.


答: Hint: As mentioned in the other answer, the statement is wrong, but you can easily show that the determinant is positive with the Cauchy-Schwartz inequality $\sum_{k = 1}^{N} x_n y_n \le \sqrt{\left(\sum_{k = 1}^{N} x_k^2 \right)\left(\sum_{k = 1}^{N} y_k^2 \right)}$ by choosing $y_k = 1$.

Title: Understanding $\left(\sum_{n=1}^N x_n^2\right) - \bar{x}^2 = \sum_{n=1}^N (x_n - \bar{x})^2$

Body:
Equation 3.19 from this document claims (at the bottom of page 5) that
$$
\left(\sum_{n=1}^N x_n^2\right) - \bar{x}^2 = \sum_{n=1}^N (x_n - \bar{x})^2.
$$

I struggle to understand this. What rules were used to group $\bar{x}^2$ into the sum?


问:
Evaluate $\int_{0}^{\pi/2}\frac{x\cdot \sin(2x)}{(1+\cos(4x^2))}\,\mathrm{d}x$
Do you have a tip to accomplish this integral?

$$\int_{0}^{\pi/2}\frac{x\cdot \sin(2x)}{(1+\cos(4x^2))}\,\mathrm{d}x$$
I already tried by substitution and by parts, neither gave me a light how to solve it.

This is a multiple choice question from an Olympiad. The options are A) $\pi/2$, B) $\pi^2/4$, C) $\pi/8$ and D) $\pi^2/16$.

答:
The integration over the odd integrand after the variable change vanishes
\begin{align}
\int_0^{\pi /2} \frac{x \sin 2x}{1+\cos^2 2x}dx
&\overset{x=t+\frac\pi4}=\int_{-\frac\pi4}^{\frac\pi4} \frac{(t+\frac\pi4)\cos 2t}{1+\sin^2 2t}dt\\
&= \frac\pi4\int_{-\frac\pi4}^{\frac\pi4} \frac{\cos 2t}{1+\sin^2 2t}dt=\frac\pi8\int_{-\frac\pi4}^{\frac\pi4} \frac{d(\sin 2t)}{1+\sin^2 2t}dt\\
&= \frac\pi8 \tan^{-1}\sin2t\bigg|_{-\frac\pi4}^{\frac\pi4}
=\frac\pi8\cdot \frac\pi2=\frac{\pi^2}{16}
\end{align}


问:
Show that $\mathbb{C}(t)/\mathbb{C}(t^{n} + t^{−n})$ is a Galois extension .

Show that $\mathbb{C}(t)$ is a Galois extension of $\mathbb{C}(t^{n} + t^{−n})$, where $t$ denotes an indeterminate. Find the Galois group of this extension.

Let $F = \mathbb{C}(t^{n} + t^{−n})$ and $E = \mathbb{C}(t)$. Since $t \in E$, we have $t^n \, , 1/t^n \, \in E$ and thus $F \subset E$. Now we need to show that $E/F$ is an algebraic extension. For $t^{n} + t^{−n} \in F$, $(t^{n} + t^{−n})^2 = t^{2n}+t^{-2n}+2$, therefore $t^{2n}+t^{-2n} \in F$, using strong induction one can show that $t^{kn}+t^{-kn} \in F$ for all $k \in \mathbb{N}$. Using this we can see that $Irr(t^m,F) = (t^{mn}+t^{-mn})^{-1}(x^{2n}+1)-x^n$ for $1\le m \le n-1$. I am not sure how to find Irr polynomials for a general $p(t) \in E$ ($t$ is algebraic).

This field extension is also seperable since $Char(F) = 0$. To show that this is a normal extension, it is enough to show that every $F$-embedding $\sigma:  E \to \bar{F}$ is an automorphism of $E$. But i am unable to show this.

Edit 1: For $m=1$, let $f(x)= (t^{n}/(t^{2n}+1))(x^{2n}+1)-x^n$, roots of $f(x)$ are $x = t\omega^{i}$ and $t^{-1}\omega^{i}$, where $\omega$ is an nth root of unity. Therefore $E$ is a splitting field of a seperable polynomial in $F$, thus it is Galois.

Edit 2: Define maps $\sigma(t)= \omega t$ and $\tau(t) = \omega t^{-1}$, then $\sigma^{n} =Id$, $\tau^{2} = Id$ and $\sigma \circ \tau = \tau \circ \sigma^{n-1}$. The galois group must be $D_{2n}$  :)

答:
$$(X^n-t^n)(X^n-\sigma(t^{n}))\in \mathbb{C}(t^{n} + t^{−n})[X]$$ where $\sigma(t^n)$ is the $\mathbb{C}(t^{n} + t^{−n})$-conjugate of $t^n$.

So $$\sigma(t^n) =\ ?$$


问:
Second-largest singular value of a doubly stochastic matrix
When I read a paper about distributed optimization, I saw a lemma says :
suppose $X^{k+1} = AX^k$ where $X^k$ is a vector from $\mathbb{R}^n$ whose elements is $x_i^k$ and $A$ is a doubly stochastic matrix (its elements are non-negative and its rows and colums sum to 1), then we have \begin{equation}
||X^k - \frac{\sum_{i=1}^n x_i^0}{n}\textbf{1}||_2 \leq \sigma_2(A)^k ||X^0 - \frac{\sum_{i=1}^n x_i^0}{n}\textbf{1}||_2 ,
\end{equation}
where $\textbf{1}$ is a vector with all 1 element and $\sigma_2(A)$ is the second-largest singular value of $A$.

My thought is that
\begin{equation}
||X^k - \frac{\sum_{i=1}^n x_i^0}{n}\textbf{1}||_2 = ||A^kX^0 - \frac{\sum_{i=1}^n x_i^0}{n}\textbf{1}||_2 = ||A^kX^0 - \frac{\sum_{i=1}^n x_i^0}{n}A^k\textbf{1}||_2 = ||A^k(X^0 - \frac{\sum_{i=1}^n x_i^0}{n}\textbf{1})||_2
\end{equation}

The second equality comes from $A\textbf{1} = \textbf{1}$.

But I don't know how to proceed and where does the $\sigma_2(A)$ come from. Why isn't it the biggest singular value(which is 1)? Any hits? Thanks.

答:
Since $A$ is doubly stochastic, it has a singular value decomposition $USV^T$ where $\mathbf v_1$, the first column of $V$, is parallel to $\mathbf 1$ and orthogonal to the other $n-1$ columns $\mathbf v_2,\mathbf v_3,\ldots,\mathbf v_n$.

Now let $\mathbf x=X^0-\frac{\sum_ix_i^0}{n}\mathbf1$ and $\mathbf y_k=A^k\mathbf x$ for every integer $k\ge0$. Then $\mathbf 1^T\mathbf y_k=(\mathbf 1^TA^k)\mathbf x=\mathbf1^T\mathbf x=0$, i.e. $\mathbf y_k\in\mathbf1^\perp=\operatorname{span}\{\mathbf v_2,\ldots,\mathbf v_n\}$. It follows that $\|A\mathbf y_k\|_2\le\sigma_2(A)\|\mathbf y_k\|_2$. So, inductively, we have $\|A^k\mathbf x\|_2\le\sigma_2(A)^k\|\mathbf x\|_2$.


Body:
From the definition of the mean $\bar{x}=\frac{1}{N}\sum_{n=1}^{N}x_{n}$, we have $N\bar{x}=\sum_{n=1}^{N}x_n$. At the bottom of page $5$ :
$$\det M=\sum_{n=1}^{N}x_n^2\cdot\sum_{n=1}^{N}1-\sum_{n=1}^{N}x_n\cdot\sum_{n=1}^{N}x_n$$
$$=N\sum_{n=1}^{N}x_{n}^2-\big(\sum_{n=1}^{N}x_n\big)^2=N\sum_{n=1}^{N}x_{n}^2-N^2\bar{x}^2$$ and $$\det M=N\sum_{n=1}^{N}x_{n}^2-(N\bar{x}^2)$$
$$=N^2\big(\frac{1}{N}\sum_{n=1}^{N}x_n^2-\bar{x}^2\big)=N^2\cdot\frac{1}{N}\sum_{n=1}^{N}(x_n-\bar{x})^2$$
$$=N\sum_{n=1}^{N}(x_n-\bar{x})^2$$
So you want to show that
$$\sum_{n=1}^{N}x_{n}^2-\color{red}{N}\bar{x}^2=\sum_{n=1}^{N}(x_n-\bar{x})^2$$

Indeed we have

$$\sum_{n=1}^N (x_n - \bar{x})^2
=\sum_{n=1}^{N}(x_{n}^2-2x_{n}\bar{x}+\bar{x}^2)$$
$$=\sum_{n=1}^{N}x_{n}^2-2\bar{x}\sum_{n=1}^{N}x_{n}+\bar{x}^2\sum_{n=1}^{N}1$$
$$=\sum_{n=1}^{N}x_{n}^2-2N\bar{x}^2+N\bar{x}^2=\sum_{n=1}^{N}x_n^2-N\bar{x}^2$$


问:
How to show that$(X,d)$ is totally bounded
How to show that a metric space $(X,d)$ is totally bounded $\iff$ every infinite subsets of $X$ contains distinct points which distinct points that arbitrarily close to each other.

I don't know how to prove $\Leftarrow.$ Please help.

BTW I'm looking for a little hint !

答:
Suppose that $\langle X,d\rangle$ is not totally bounded. Then there is an $\epsilon>0$ such that for all finite $F\subseteq X$, $\{B(x,\epsilon):x\in F\}$ does not cover $X$, i.e., $X\setminus\bigcup_{x\in F}B(x,\epsilon)\ne\varnothing$. This implies that we can recursively construct a set $A=\{x_n:n\in\Bbb N\}\subseteq X$ such that

$$x_{n+1}\in X\setminus\bigcup_{k\le n}B(x_k,\epsilon)$$

for each $n\in\Bbb N$. $A$ is an infinite subset of $X$; does it contain distinct points that are arbitrarily close to each other?


问:
What is the closed-form expression for inverse of geometric distribution CDF?
The pmf of geometric distribution is given by $p(1-p)^{x-1}$.

I want to generate the samples for a random variable which has geometric distribution.

So, the formula I found is $X=F^{-1}(u)$ where $F$ is the cdf of geometric distribution.

Is it possible to find the mathematical expression for $F^{-1}(u)$ so that by I can easily calculate the sample.

答:
Yes, very simple.

Start from the fact that

$$F_X(x)=1-(1-p)^x$$


问:
Convergence of a monotone increasing sequence.
So here is the question. There is a real number $\beta$. Show that there exists a monotone increasing sequence of rational numbers which converges to $\beta$. I don't understand how to start the proof. I need to show that $\beta$ is the supremum of the sequence such that it converges to $\beta$. So can someone explain how should I go about proving this? Also, can this proof be modified to suit a monotone increasing sequence of irrational nos. as well?
This is my approach- Suppose I consider a sequence {an} (monotone increasing) which has the supremum= $\beta$. Thus, an<=$\beta$ for all n in N. Thus, for some N, $\epsilon$+aN> $\beta$. Also, by denseness, we always have rational nos. in (aN,$\beta$). Thus, for this N>N(natural numbers), (beta-an)>$\epsilon$, thus satisfying convergence at $\beta$. Can someone say if this is a rigorous enough proof?

答: Just construct the sequence.  There's a rational number $\beta_1$ in the interval $(\beta-1, \beta)$.  There's a rational number $\beta_2$ in the interval $(\beta_1,\beta)$ and so on.  Construct a sequence of irrationals in a similar way.


问:
Abelian categories with direct sums
Does any abelian category admits direct sums?

If not, categories admiting direct sums have a special name?

I'm asking this since I am writing a proof that requires direct sums but I only know that the category is abelian.

Thank you very much for your answer!

Edit. In the tag description is written that indeed they posses finite direct sums.

答: You should clarify what you mean by direct sum. If you mean 'finite direct sum' then this is included in the definition of an abelian category. Arbitrary direct sums don't necessarily exist, like the category of finite abelian groups. An abelian category is called 'cocomplete' if it possesses all direct sums.</p>

Title: Abelian categories with direct sums


答:
Define:

$A_n = \{k/2^n \mid k \in \mathbb Z\}$

$a_n = \sup A_n \cap \{x \in \mathbb R \mid x \lt \beta\}$

$b_n = \sup\{a_k \mid 1\le k \le n\}$

$\{b_n\}$ is a possible sequence you're looking for.

And if you're looking for a sequence of irrational numbers, replace $A_n$ with $A_n^\prime = \{k \sqrt 2/2^n \mid k \in \mathbb Z\}$.

Title: Convergence of a monotone increasing sequence.

Body:
So here is the question. There is a real number $\beta$. Show that there exists a monotone increasing sequence of rational numbers which converges to $\beta$. I don't understand how to start the proof. I need to show that $\beta$ is the supremum of the sequence such that it converges to $\beta$. So can someone explain how should I go about proving this? Also, can this proof be modified to suit a monotone increasing sequence of irrational nos. as well?
This is my approach- Suppose I consider a sequence {an} (monotone increasing) which has the supremum= $\beta$. Thus, an<=$\beta$ for all n in N. Thus, for some N, $\epsilon$+aN> $\beta$. Also, by denseness, we always have rational nos. in (aN,$\beta$). Thus, for this N>N(natural numbers), (beta-an)>$\epsilon$, thus satisfying convergence at $\beta$. Can someone say if this is a rigorous enough proof?


问:
The definition of simplicial hyperplane arrangements
I'm struggling to understand a statement in Sec 2.2 of "A simplicial complex of Nichols algebras" by Cuntz and Lentner. Simplicial arrangements are sets of hyperplanes in real vector spaces satisfying various properties, and arise in the study of Weyl groupoids. They are defined as:

A simplicial arrangement is a finite set $A=\{H_1,\dots,H_n\}$ of hyperplanes in $V=\mathbb{R}^r$ such that every chamber $K$ (i.e. connected component in $V\backslash \cup_{H\in A}H$) is an open simplicial cone, meaning $\exists \alpha_1^\vee,\dots,\alpha_r^\vee\in V$ s.t.
$$K=\{\sum_{i=1}^r\lambda_i \alpha_i^\vee|\ \lambda_i>0\ \forall i\}=:\langle \alpha_1^\vee,\dots,\alpha_r^\vee \rangle_{>0}$$

Now for each hyperplane $H_i$ we can choose $x_i\in V^*$ s.t. $H=\ker(x_i)$. Let $R:=\{\pm x_1,\dots,\pm x_n\}$. It is then stated that for each chamber $K$ there is a unique choice of basis $\{\alpha_1^\vee,\dots,\alpha_r^\vee\}$ of $V$ such that $K=\langle \alpha_1^\vee,\dots,\alpha_r^\vee \rangle_{>0}$ and the dual basis $\{\alpha_1,\dots,\alpha_r\}$ for $V^*$ is a subset of $R$.

I'm trying to verify this statement with a simple example, but running into difficulties. Let $V=\mathbb{R}^2$, and take hyperplanes $H_1,H_2$ (i.e. just lines) given by the span of vectors $v_1=(1,0),v_2=(1,1)$ respectively. Then all the chambers can be given as $$K_1=\langle v_1,v_2 \rangle_{>0},\ K_2=\langle -v_1,v_2 \rangle_{>0},\ K_3=\langle v_1,-v_2 \rangle_{>0},\ K_4=\langle -v_1,-v_2 \rangle_{>0}$$
So by choosing $\alpha_i^\vee=\pm v_i$ with signs depending on chamber chosen, I believe it then follows that this is a simplicial arrangement? Any pair from $\{\pm v_1,\pm v_2\}$ is basis as well, but considering the corresponding dual bases, these have kernels on the hyperplanes/lines orthogonal to $H_1$ or $H_2$, which do not coincide with $H_1,H_2$, and therefore these dual basis vectors cannot be in $R$, so something isn't working here.

Perhaps my example is not a simplicial arrangement for some reason, I expect I've overlooked something quite trivial.

Edit based on answer: So now if $\triangle^K$ denotes the dual basis $\{\alpha_1,\dots,\alpha_r\}$ of $V^*$ corresponding to chamber $K$, then we have ``crystallographic arrangement" if for all chambers $K$, $R\subset \sum_{\alpha\in \triangle^K} \mathbb{Z}\alpha$. If I now understand things better my example should be crystallographic too, because, for instance $\triangle^{K_2}=\{-f_2,f_1\}$, and indeed the elements of $R=\{\pm f_1,\pm f_2\}$ are $\mathbb{Z}$-linear combinations of $f_1,-f_2$. And the other chambers follow similarly.

答:
Your example is a simplicial arrangement, but you are mixing up $V$ and $V^*$ or have a wrong notion of dual bases in mind. In your example in $\mathbb R^2$ you have for $i=1,2$
$$
H_i = \langle v_i \rangle = \operatorname{ker}(f_i)
$$
where $v_1=(1,0)$, $v_2=(1,1)$, $f_1(x,y)=y$ and $f_2(x,y)=x-y$. So $R = \{\pm f_1,\pm f_2\}$.

Now consider some chamber, say $K_2 = \langle -v_1, v_2\rangle_{>0}$. The basis $\{-v_1, v_2\}$ of $\mathbb R^2$ has dual basis $\{-f_2, f_1\}$ of $(\mathbb R^2)^*$ and indeed both $-f_2$ and $f_1$ are elements of $R$.


问:
Question on Combinatorics
How many non-negative integer solutions of $x_1+x_2+x_3+x_4+x_5= 20$ are there with $x_i\leq 8, i=1,2,3,4,5$
Is there any way to solve this question without considering multiple cases?

答: HINT:we basically seek the coefficient of $x^{20}$ in the expansion $${(1+x+x^2..+x^8)}^5={(1-x^9)}^5{(1-x)}^{-5}=?$$

Title: Question on Combinatorics


问:
Error in Spivak's proof that an odd degree polynomial has a root?
I'm reading Spivak's Calculus 3rd edition. On page 123 (chapter 7, Three Hard Theorems) he gives a proof of the theorem that states if $n$ is odd, then any equation $x^n+a_{n-1}x^{n-1}+...+a_0=0$ has a root.

Essentially he's showing that there's an interval $[a, b]$ such that $f(a)<0<f(b)$ where  $f(x)=x^n+a_{n-1}x^{n-1}+...+a_0$, concluding that there's $x \in [a,b]$ such that $f(x)=0$.

The proof begins as follows:

$$f(x)=x^n\left(1+\frac{a_{n-1}}{x}+...+\frac{a_0}{x^n}\right)$$

$$\left|\frac{a_{n-1}}{x}+...+\frac{a_0}{x^n}\right|\leq\frac{|a_{n-1}|}{|x|}+...+\frac{|a_0|}{|x^n|}$$

If we choose $x$ satisfying

$$|x|>1, 2n|a_{n-1}|,...,2n|a_0|$$
then
$$\mbox{ (1) } |x^k|>|x|$$
and
$$\mbox{ (2) }\frac{|a_{n-k}|}{|x^k|}<\frac{|a_{n-k}|}{|x|}<\frac{|a_{n-k}|}{2n|a_{n-k}|}=\frac1{2n}$$

Now he doesn't say anything about $k$. Since he stated that $|x^k|>|x|$, I assumed that $k\in[2,n]$ because if $k$ were in $[1,n]$ then it wouldn't have been true that $|x^k|>|x|$ - we'd have had $|x^1|>|x|$ which is false. But if $k\in[2,n]$, how does it follow that $\frac{|a_{n-1}|}{|x|}<\frac1{2n}$? If this doesn't, does (3) still holds?

In the next step he says that

$$\mbox{ (3) } \left|\frac{a_{n-1}}{x}+...+\frac{a_0}{x^n}\right|\leq\sum_{1}^n\frac1{2n}=\frac12$$

where "less than or equals" doesn't really make sense - it should have been "strictly less" since we have "strictly less" in (2).

But everything would've made sense if he did the following:

$$\mbox{ (1*) } |x^k|\geq|x|, k\in [1,n]$$
$$\mbox{ (2*) }\frac{|a_{n-k}|}{|x^k|}\leq\frac{|a_{n-k}|}{|x|}\leq\frac{|a_{n-k}|}{2n|a_{n-k}|}=\frac1{2n}$$
$$\mbox{ (3*) } \left|\frac{a_{n-1}}{x}+...+\frac{a_0}{x^n}\right|\leq\sum_{1}^n\frac1{2n}=\frac12$$

Am I missing something, or it's indeed an error in the book?

答:
You are correct that
$$
\frac{\lvert a_{n-k}\rvert}{\lvert x^k\rvert}<\frac{\lvert a_{n-k}\rvert}{\lvert x\rvert}<\frac{\lvert a_{n-k}\rvert}{2n\lvert a_{n-k}\rvert}=\frac{1}{2n}
$$
only holds for $k\ge2$. However, for $k=1$ we have
$$
\frac{\lvert a_{n-k}\rvert}{\lvert x^k\rvert}\le\frac{\lvert a_{n-k}\rvert}{\lvert x\rvert}<\frac{\lvert a_{n-k}\rvert}{2n\lvert a_{n-k}\rvert}=\frac{1}{2n},
$$
where the strict part of the inequality holds by the assumption $\lvert x\rvert>2n\lvert a_{n-1}\rvert$. So for all $k\ge1$ the inequality
$$
\frac{\lvert a_{n-k}\rvert}{\lvert x^k\rvert}<\frac{1}{2n}
$$
holds. From this you can conclude that
$$
\left\lvert\frac{a_{n-1}}{x}+\ldots+\frac{a_0}{x^n}\right\lvert<\sum_1^n\frac{1}{2n}=\frac{1}{2}.
$$
That he writes "$\le$" here instead of "$<$" does not make his inequality false, only weaker than it could be (and potentially confusing to readers). In the remainder of the proof he doesn't need the strict form of this inequality; the non-strict form he writes suffices.


问:
Proof Verification: $aRb$ iff $\exists c \in A: a,b \in [c]$
In an exercise I'm asked to prove the following:

Let $R$ be an equivalence relation in a set $A$, and let $a,b \in A$, then:

$$aRb \  \text{ iff } \ \exists c \in A: a,b \in [c]$$
Where $[c]$ is the equivalence class of $c$.

I proved that $\exists c \in A: a,b \in [c] \Rightarrow aRb$, using the fact that $R$ is transitive, but I'm not sure whether my proof for $aRb \Rightarrow \exists c \in A: a,b \in [c]$ is valid. This is what I did:

My proof:

Let's assume that $aRb$. I'll do this part by contradiction. Let's assume that $\nexists c \in A: a,b \in [c]$. This means that , one of the following is true for all $c \in A$:

$c R a  \wedge c\not R b$ $\ \ \ $ (1)

$c \not R a  \wedge c R b$  $\ \ \ $(2)

$c \not R a  \wedge c\not R b$ $\ \ \ $(3)

If (1):

$c R a \wedge a R b$, so by transitivity $c R b \to$ contradiction

If (2):

$c R b \wedge b R a$, so by transitivity $ c R a \to $ contraditcion

If (3):

$\forall c \in A \setminus \{a,b\}, c \notin [a] \wedge c \notin [b]$. This means that $[a] = [b] = \{a,b\}$, so if we set $c = a$, we have $c R a \wedge c R b \to$ contradiction.

So assuming that $\nexists c \in A: a,b \in [c]$ leads to a contradiction.

I think that (1) and (2) are correct but I'm a little sceptical about (3). Did I made some mistake or is this correct?

答: Your proof is incorrect. In point (3) you stated that "$c \in A \setminus \{a,b\}$" but then you arrived at the conclusion that $c = a$.


问:
Why does the Denominator of the Denominator go to the Numerator?
Quite blindly I've learnt a basic rule about fractions:

The Denominator of the Denominator goes to the numerator.

I'm confused about it and I'll give an example as to why.
Imagine the following:

1/2/2

Now, if the denominator of the denominator went to the numerator this would be 2/2 which is 1

and that's the what everyone agrees upon.

But what I feel is that since 1/2 is 0.5 ,

1/2/2 = 0.5/2 = 0.25

So, I hope you can see why I feel confused.

I apologize if this question seems stupid

答:
There are two confusions here. One is that the expression 1/2/3, when there are no parentheses, is defined to be (1/2)/3, which is different from the expression 1/(2/3), and that is why you got the wrong answer in your post. The second is why "the denominator of the denominator goes to the numerator", and it is not something you need to memorize, instead you can do the following

$$\dfrac1{\left(\frac23\right)} = \dfrac1{\left(\frac23\right)}\cdot 1 = \dfrac1{\left(\frac23\right)}\cdot \dfrac33 = \dfrac{1\cdot 3}{\left(\frac23\right)\cdot 3} = \dfrac{3}{\frac{2}3\cdot 3} = \dfrac32.$$

That is, if the denominator in the denominator is 3, just multiply the entire fraction by $\frac33$, and this will cancel the denominator in the denominator.


问:
Automorphism of direct product of rings

Let $$R=A_1\times A_2\times\dots\times A_n.$$ Let's assume there are no non trivial automorphisms of these $A_i$s. I claim that the only possible automorphisms of this structure are the permutations of the $A_i$s. So, there are a total of $n!$ automorphism possible over this structure.

Question is how to prove this. I am proceeding in this way: let $R\cong A\times (0)$, $(0)$ being the zero ring. Now, $(a,0)$, $a\in A$, can only get mapped to $(a,0)$ or $(0,a)$. This can be established via some simple isomorphism argument. But I am not been able to generalize this to any ring product. Any help are welcome.

答:
You meant that there are at most $n!$ possible automorphisms all given by permuting the isomorphic $A_j$.

RghtHndSd showed a counter-example, eg. $A_1=A_2=\Bbb{F}_p\times \Bbb{Q}$ then $((a,b),(c,d))\to ((a,d),(c,b))$ is an automorphism of $A_1\times A_2$.

If the $A_j$ are integral domains then your claim is true, because the $e_j = (1,\ldots,\underbrace{0}_{j \ th},1,\ldots)$ are the only elements such that $e_j R$ is a prime ideal and $e_j^2=e_j$. An automorphism  $\sigma\in Aut(R)$ must permute them. So it permutes the $1-e_j$, thus it induces an isomorphism $$A_j\cong (1-e_j) R \quad to\quad A_{\sigma_j} \cong (1-e_{\sigma_j}) R$$ And two distinct isomorphisms $A_j\to A_{\sigma_j}$ give a non-trivial automorphism of $A_j$, that you assume to not exist. Thus $\sigma$ is determined by the $\sigma_j$.

Title: Automorphism of direct product of rings


问:
$AG,AF$ are perpendiculars to the internal bisectors $CG,BF$ and $GF$ extended intersects $AB,AC$ at $H,I$
In the figure $AG,AF$ are perpendiculars to the internal bisectors $CG,BF$ and GF extended  intersects $AB,AC$ at $H,I$ respectively.Find $5GF-HI$ if $c=AB=7,a=BC=8,b=CA=5$

I know we can calculate the internal bisectors length by stewarts formula as $$BD=\frac{2\sqrt{s(s-b)}}{a+c}$$ And similarly for the other  bisector.From here we can also calculate $AF,AG$.I am stuck otherwise..it  looks like $GF$ is parallel to $BC$ but  i cant prove it..

答:
Extend $AF$ and $AG$ until it touches $BC$.

Since $CG, BF$ bisects $\angle ACB$ and $\angle ABC$, and $AF \perp BF$, $AG \perp GC$:

$\triangle AGC \cong \triangle G'GC$ and $\triangle AFB \cong \triangle F'FB$ by considering common sides and AAS.

Hence $BF'=7, CG'= 5$ and $F,G$ are midpoints of $AF', AG'$ respectively.

By midpoint/midline theorem, $FG // F'G'$ and $FG = \frac12 F'G' = \frac12(8-(8-7)-(8-5))=2$.

By considering the parallel lines, similar triangles and using midpoint/midline theorem again, we see that $HI$ is also half of $BC$, giving $HI = 4$. Therefore $5GF - HI =6$.


问:
Optimizing $\rm \frac{x^TAx}{c^T x}$ subject to $\rm1^T x = 1$
I need to optimize the following quadratic-over-linear objective:

$$
\frac{x^TAx}{c^T x}
$$

subject to

$$\mathbf{1}^Tx = 1$$

Where $A$ is a diagonal (with all positive entries ) matrix and $c$ (also with strictly positive entries ) and $x$ are vectors. $\mathbf{1}$ is a vector of all ones.

There is a problem in Boyd and Vandenberghe book that solves the unconstraint problem but the solution I have found is not constructive and I am not able to generalize to the constraint case.

I have tried to follow the standard approach, computing the gradient of the Lagrangian and making it equal to zero but I am not able to get the solution. I get stuck with some cubic matrix equation.

Do anybody know if this is a typical problem in any field??

By the way, I am trying to find a closed form solution because I need it to use somewhere else.

答:
$\def\p#1#2{\frac{\partial #1}{\partial #2}} \def\o{{\tt1}} \def\a{\lambda} \def\b{\ell}$

Use an unconstrained vector $w$ to construct $x$ such that it automatically satisfies the constraint. Treat its the length and direction as independent variables.
$$\eqalign{
\b &= \o:w\quad&\implies\quad&d\b&=\o:dw \\
x &= \b^{-1}w\quad&\implies\quad&dx&=\b^{-1}dw - w\b^{-2}d\b \\
 &&&&= \b^{-1}(I-x\o^T)\,dw \\
}$$
Consider the numerator/denominator of the objective function.
$$\eqalign{
\alpha &= A:xx^T \quad&\implies d\alpha = 2Ax:dx \\
\beta &= c:x
 &\implies d\beta = c:dx \\
\a &= \beta^{-1}\alpha
 &\implies \beta\a = \alpha \\
}$$
In the above, a colon has been used to denote the inner product
$$A:B\;=\;{\rm Tr}(A^TB)\;=\;\sum_{i=1}^m\sum_{j=1}^n A_{ij}B_{ij}$$
(which is also valid for vectors if you treat them as rectangular matrices)

Calculate the gradient of the objective function with respect to
the unconstrained variable by substituting the above fragments.
$$\eqalign{
d\a &= \beta^{-2}(\beta\,d\alpha-\alpha\,d\beta) \\
 &= \beta^{-1}(2Ax-\a c):dx \\
 &= \beta^{-1}(2Ax-\a c):\b^{-1}(I-x\o^T)\,dw \\
 &= (\beta\b)^{-1}(I-\o x^T)\,(2Ax-\a c):dw \\
\p{\a}{w} &= (\beta\b)^{-1}(I-\o x^T)\,(2Ax-\a c) \\
}$$
Set the gradient to zero and solve.
$$\eqalign{
2Ax + \a\beta\o &= 2\alpha\o + \a c \\
2Ax - \a\beta\o &= \a c \\
\left(\tfrac 2\a A - \o c^T\right)x  &= c \quad
\implies x = \left(\tfrac 2\a A-\o c^T\right)^{-1}c \\
}$$
The Sherman-Morrison formula for the inverse of a rank-1 perturbation is
$$\eqalign{
\Big(B+uv^T\Big)^{-1} = B^{-1} - \frac{B^{-1}uv^TB^{-1}}{1+v^TB^{-1}u}
}$$
Substitute $B=\left(\tfrac 2\a A\right),\,u=\o,\,v=-c\;$ to obtain
$$\eqalign{
\Big(\tfrac 2\a A - \o c^T\Big)^{-1}
 &= \tfrac\a 2
\left(A^{-1} + \frac{A^{-1}\o c^TA^{-1}}{\tfrac 2\a-c^TA^{-1}\o}\right)
}$$
and to reduce clutter for the next calculation,
define the scalars
$$\theta=c^TA^{-1}c,\quad\rho=c^TA^{-1}\o,\quad\mu=\o^TA^{-1}\o$$
Now use the constraint to evaluate $\a\;$
(via a scalar quadratic equation).
$$\eqalign{
\o^Tx &= \tfrac\a 2 \left(\o^TA^{-1}c + \frac{\o^TA^{-1}\o\;c^TA^{-1}c}{\tfrac 2\a-\rho}\right) \\
2 &= \a\rho+\left(\frac{\a\theta\mu}{\tfrac 2\a-\rho}\right)\left(\frac \a\a\right) \\
0 &= (\theta\mu-\rho^2)\a^2 + 4\rho\a - 4 \\
\a_\pm &= \left(\frac{2}{\rho\pm\sqrt{\theta\mu}}\right) \\
}$$
Substituting yields two possible $x$ vectors
$$\eqalign{
x_\pm  &= \left(\frac{1}{\rho\pm\sqrt{\theta\mu}}\right)
A^{-1}\left(c\pm\sqrt{\frac{\theta}{\mu}}\o\right) \\
x_\pm  &= \frac{A^{-1}c\pm\Big(\frac{c^TA^{-1}c}{\o^TA^{-1}\o}\Big)^{\frac 12}A^{-1}\o}{c^TA^{-1}\o\pm\Big((c^TA^{-1}c)\;(\o^TA^{-1}\o)\Big)^{\frac 12}} \\
}$$


问:
How to compute the infinite product?
Given the following infinite product:
$$\lim_{n\to \infty} \frac{1}{2}\frac{3}{4}\cdots \frac{2^n-1}{2^n}.$$

It is easy to see that above infinite product is convergent by the convergence of the following series:
$$\sum_{n=1}^{\infty}\ln \left(1-\frac{1}{2^n}\right)$$.

But I do not know how to calculate the infinite product. I would appreciate it if someone can give any suggestions and comments.

答: You maybe could simplify the denominator of the infinite product since $\prod_{n=1}^{m}2^n=2^{\sum_{n=1}^{m}n}=2^{\frac{m(m+1)}{2}}$ which would mean that $\lim_{m \rightarrow \infty} \prod_{n=1}^{m}\frac{2^n-1}{2^n}=\lim_{m \rightarrow \infty}\frac{1}{2^{\frac{m(m+1)}{2}}} \prod_{n=1}^{m}(2^n-1)$

Title: How to compute the infinite product?


问:
Prove non existence of semi-conjugacy between $E_{k}$ and $E_{-k}$.
I've been trying to prove the non existence of a semi-conjugacy (intuitively assuming this is true) between the circle maps $E_k$ and $E_{-k}$ for $k = 2,3,\dots$ defined as
$$
E_k: S^1 \to S^1,\qquad
E_k(x \mod 1) = kx \mod 1.
$$
where $S^1 \simeq [0,1)$. Conjugacy is clearly not an option as the number of fixed points varies, i.e. $P_n(E_{k}) = |k^n -1|$, where $P_n(f)$ is the number of fixed points of $f^n$. What about semi-conjugacy however?

Any help is more than welcome!

答:
Assume $$ h \circ E_{-2}= E_2 \circ h,$$ where $h$ is the semi cojnugacy. Applying the degree both sides you get $deg(h)=0$. Let $H$ denote the  lifting of h such that $$H(0) \in [0,1), $$ then,
$$H(x+1)=H(x) $$
i.e H is periodic. Moreover, lifting both sides of the above equation, we easily obtain that
$2H(x)  $ and $H(2-2x)$ are both lifing of the same map. A well known result implies that
$$2H(x)= H(2-2x) + c$$
where c is an integer.
Periodicity implies
$$2H(0)=H(2)+c= H(0)+c$$
and the only possibile solution to this is $c=0$ and $H(0)=0$.
Applying weiestrass theorem and periodicity we get that H attains a global maximum  and minimum and the first  is nonnegative, the second is nonpositive. Since the global extrema of $2H(x) , H(2-2x) $ must coincide, we get their extrema  are zero. In details:
let $M,m$ be resp. the  maximum and minimum of H. Assume $M>0$. Then, since $c=0$, for proper $y$:
$$2M=2H(y)=H(2-2y)>M, $$
contradicting the fact that M is a maximum. This implies $M=0.$
Now assume $m<0$,say $m=-q$, with $q$ strictly positive. Again, for proper $y$:
$$-2q=2H(y)=H(2-2y)<-q.$$
This implies $c=0$.
Then,  H is constant, contradiction.
This is  generalizable for different $k$, but  not so  trivially , though.


问:
Simple Triangle-like Inequality for a general norm
I came up with this inequality as I was thinking about norms. I don't know if it's true though.

Given three points $x,y,z$ in a normed space $(X,\|\cdot\|)$, is it true that if

$$
\|x-y\|\le\|y\|
$$

and

$$
\|x-z\|\le\|z\|
$$

then

$$
\|x-(y+z)\|\le\|y+z\| ?
$$

In other words, for a given vector $x$ in a normed space $X$, is the set $K:=\{y:\|x-y\|\le\|y\|\}$ additive?

First, if it were true, it almost looks like we could change the $y$ and $z$ around so that, e.g. the 1st and 3rd inequalities imply the 2nd. As it is, given any arrangement $x+y=z$ (none of which is $0$), we can tell which is which by simply adding.

I thought proving $X=\mathbb{R}$ would easily turn into $X=l_p$, but that seems naive.

Inner Product Space

The first two inequalities reduce to

$$
\langle x,x\rangle\le 2\langle y,x\rangle
$$

and

$$
\langle x,x\rangle\le 2\langle z,x\rangle
$$

Adding these together gets

$$
\|x\|^2\le \langle y+z,x\rangle\le 2\langle y+z,x\rangle
$$

Working the same algebra backwards then gets the third inequality.

Other norms?

Note that if $y\in K$, then $cy\in K$ for all $c\ge 1$. So there is a curve/surface in $X$, symmetric about $0$ (AKA odd), that divides $X$ into two parts, one part being $K$. It looks like if the unit ball $B$ of $X$ is oblong and $x$ is not perpendicular to an axis of symmetry of B, the curve/surface will not be linear and hence not convex (symmetric about $0$ and convex implies linear).

So, the factor of $2$ cannot be dismissed for general $X$ as it was for inner product spaces (?). I feel like that argument says the 2 cannot be dismissed for some inner product spaces as well. Mistake?

Thanks!

答:
We wish to investigate the statement
$$
\|y-x\|\le\|y\|\land\|z-x\|\le\|z\|\implies\|y+z-x\|\le\|y+z\|\tag0
$$

Inner Product Spaces

Suppose that we are in an inner product space: $|u|^2=\langle u,u\rangle$.

Proposition 1:
$$
|y-x|\le|y-o|\land|z-x|\le|z-o|\implies\left|\frac{y+z}2-x\,\right|\le\left|\frac{y+z}2-o\,\right|\tag1
$$
Proof:
$$
\begin{align}
&\left|\frac{y+z}2-o\,\right|^2-\left|\frac{y+z}2-x\,\right|^2\\[6pt]
&=\langle(y+z)-(x+o),x-o\rangle\tag{2a}\\[6pt]
&=\left\langle y-\frac{o+x}2,x-o\right\rangle+\left\langle z-\frac{x+o}2,x-o\right\rangle\tag{2b}\\
&=\frac12\left(|y-o|^2-|y-x|^2\right)+\frac12\left(|z-o|^2-|z-x|^2\right)\tag{2c}\\[6pt]
&\ge0\tag{2d}
\end{align}
$$
Explanation:

$\text{(2a)}$: $\langle u+v,u-v\rangle=|u|^2-|v|^2$

$\text{(2b)}$: linearity of the inner product

$\text{(2c)}$: $\langle u+v,u-v\rangle=|u|^2-|v|^2$

$\text{(2d)}$: assumptions

$\large\square$

Proposition 2:
$$
|p-x|\le|p-o|\implies\left|p-\frac{x+o}2\,\right|\le|p-o|\tag3
$$
Proof:
$$
\begin{align}
\left|p-\frac{x+o}2\,\right|^2
&=\left\langle p-\frac{x+o}2,p-\frac{x+o}2\right\rangle\tag{4a}\\
&=\left\langle\frac{p-x}2+\frac{p-o}2,\frac{p-x}2+\frac{p-o}2\right\rangle\tag{4b}\\
&=\frac14\langle p-x,p-x\rangle+\frac12\langle p-x,p-o\rangle+\frac14\langle p-o,p-o\rangle\tag{4c}\\
&\le\frac14|p-x|^2+\frac12|p-x||p-o|+\frac14|p-o|^2\tag{4d}\\[4pt]
&\le|p-o|^2\tag{4e}
\end{align}
$$
Explanation:

$\text{(4a)}$: write norm in terms of inner product

$\text{(4b)}$: rewrite the terms in the inner product

$\text{(4c)}$: use the linearity of the inner product

$\text{(4d)}$: $\langle u,v\rangle\le|u||v|$

$\text{(4e)}$: $|x-p|\le|o-p|$

$\large\square$

Statement $\textbf{(0)}$ Holds in an Inner Product Space

In general, a counterexample is hard to visualize because Statement $(0)$ is true in an inner product space:
$$
\begin{align}
\overbrace{\color{#090}{|y-x|\le|y-o|}}^{|y-x|\le|y|}\land\overbrace{\color{#00F}{|z-x|\le|z-o|}}^{|z-x|\le|z|}
&\implies\left|\frac{y+z}2-x\,\right|\le\left|\frac{y+z}2-o\,\right|\tag{5a}\\
&\implies\underbrace{\color{#C00}{\left|\frac{y+z}2-\frac{x+o}2\,\right|\le\left|\frac{y+z}2-o\right|}}_{|y+z-x|\le|y+z|}\tag{5b}
\end{align}
$$
Explanation:

$\text{(5a)}$: Proposition 1

$\text{(5b)}$: Proposition 2 with $p=\frac{y+z}2$

Importance of Convexity

Proposition 1 says that the region which is closer to $x$ than to $o$ (the origin) is convex. Since $y$ and $z$ are in that region (to the right of the vertical black line), so is $\frac{y+z}2$.

Proposition 2 says that any point that is closer to $x$ than to $o$ is closer to $\frac{x+o}2$ than to $o$. That is, the region to the right of the vertical black line is also to the right of the vertical violet line, and anything to the right of the vertical violet line is closer to $\frac{x+o}2$ than to $o$.

Other Norms

For inner product spaces, convexity plays an important part in proving Statement $(0)$. Using the $\|\cdot\|_\infty$ norm,
$$
\|(x,y)\|_\infty=\max(|x|,|y|)\tag6
$$
the region where $\|p-x\|\le\|p\|$ need not be convex, and the statement fails.

The violet region is where $\|p-x\|\le\|p\|$ and the green region is where $\|p-x/2\|\gt\|p\|$. Since $y$ and $z$ are in the violet region, we have
$$
\|y-x\|\le\|y\|\land\|z-x\|\le\|z\|\tag7
$$
However, since $\frac{y+z}2=(0,0)$ is in the green region, we have
$$
\|y+z-x\|\gt\|y+z\|\tag8
$$
Inequalities $(7)$ and $(8)$ show that $(0)$ fails.


答:
The number-theoretic partitions are well-studied.

A $k$-partition of $n$ is a word $x_1x_2 . . . x_k$ of length $k$ over the alphabet $\Bbb N$ where $x_1 + x_2 + . . . + x_k = n$.

Let $p(n,k)$ be the number of such words.

Then $p(n,1)=1$ (its the word $n$) and $p(n,n) = 1$ (its the word $11...1$).

Moreover, if $1<k<n$, then $p(n,k) = {n-1\choose k-1}$.

For instance, if $n=5$, then for $k=2$ we have the words $14,23,32,41$; for $k=3$, $113,131,311, 122, 212, 221$, and for $k=4$, $1112, 1121, 1211, 2111$.

Now consider correspondingly words over $\Bbb N_0$ by including $0$'s.

Title: Question on Combinatorics

Body:
How many non-negative integer solutions of $x_1+x_2+x_3+x_4+x_5= 20$ are there with $x_i\leq 8, i=1,2,3,4,5$
Is there any way to solve this question without considering multiple cases?


问:
Interpretation of eigenvalues and associated eigenvalues
I am given a 4x4 matrix such and a basis for $R^4$, $\{u_1,u_2,u_3,u_4\}$. Suppose the following conditions:
$$Bu_1 = 2u_1 \\Bu_2 = 0\\Bu_3 = u_4\\Bu_4 = u_3$$

First, I am asked to find the eigenvalues. It's fairly obvious that $1$ and $2$ are eigenvalues. Then I manipulate the lower 2 equations as such:

$$Bu_3 = u_4\\B(Bu_3) = Bu_4\\B^2u_3 = u_3$$

Hence an eigenvalue of $B^2$ is supposed to be 1. Square rooting this, I obtain eigenvalues of $1$ and $-1$ for $B$.

Thereafter, I need to find the associated eigenvectors. Again, for $\lambda = 1$ is $u_1$ and $\lambda = 0$ is $u_2$. But for the other two, I cannot figure it out. The answer key says that that of $\lambda = 1$ is $u_3+u_4$, and that of  $\lambda = -1$ is $u_3-u_4$. I cannot comprehend why. Am I missing out on come manipulation?

Any advice is appreciated! Thank you!

答:
$Bu_3=u_4$ and $Bu_4=u_3$ $\Rightarrow$ $Bu_3+Bu_4=B(u_3+u_4)=u_4+u_3$ hence 1 is an eigenvalue for $u_3+u_4$

$Bu_3=u_4$  $\Rightarrow$ $-Bu_3=B(-u_3)=-u_4$

$B(-u_3)=-u_4$ and $Bu_4=u_3$ $\Rightarrow$ $B(u_3-u_4)=-(u_3-u_4)$ hence -1 is an eigenvalue


问:
I can't find the analytical expression of the sum from this probability problem
It originates from this problem:

We have a box with $n$ balls that are evenly mixed, all are
$\color{blue}{blue}$ except for one that is $\color{red}{red}$.

If we pick blindly one ball per try and after each try we don't put
it back in the box, what is the average number of tries it will take
us to take get the $\color{red}{red}$ ball $?$.

If I'm correct the probability of picking the $\color{red}{red}$ ball at the try number $x$ is: $$
P[X=x]=\frac{(n-x)!}{n!},\quad X = \mbox{number of tries.}
$$

Thus the expected value of $X$ is:
$$
 E(X)=\sum_{x=1}^{n}x\frac{(n-x)!}{n!}
$$
whose 'clean' analytical expression I can't find.

Could you help me advance in the problem please $?$. Unfortunately the book where I found the problem doesn't have the solution.

答:
The red ball is equally likely to come in any position.  There are $(n+1)!$ orders is which the balls may be drawn, and in $n!$ of them the red ball is in position $k$, for any $1\leq k\leq n+1$ so the probability, that the red ball is in position $k$ is $\frac1{n+1}$

Now I'm sure you'll find it easy to finish.


问:
Does the set $A$, of all permutations of $\{1,2,3,4\}$ that fixes $2$ and $4$ form a subgroup of $S_4$?
Does the set $A$, of all permutations of $\{1,2,3,4\}$ that fixes $2$ and $4$ form a subgroup of $S_4$?

I’ve got as far,
The set $A$ will be the identity where all $4$ numbers map to themselves (as all fixed) and $(1, 3)$ as $2$ and $4$ are fixed. So the set $A =\{\operatorname{id}, (1 3)\}$

Now I need to show that this is a sub group of $S_4$, by identity closure and inverses.

Identity is in both so ok but I’m stuck on the other $2$.

Any help would be great

Kara x

答:
Hint: Each such permutation has the form $\pi = \begin{pmatrix} 1 & 2 & 3 & 4 \\ x_1 & x_2 & x_3 &x_4 \end{pmatrix}$ with $x_2=2$ and $x_4=4$.

Show that the composition of two such permutations has the same shape.

Moreover show that the inverse permutation of such a permutation has the same format.

You're right about the structure of $A$, consisting of $(1)(2)(3)(4)$ and $(13)(2)(4)$.


问:
How to determine $f$ if $f_n(x) = \frac{nx}{1+nx^3} , n=1,2...$ and $f(x)=\lim_{n \to \infty}f_n(x)$?
If I have that  $f_n(x) = \frac{nx}{1+nx^3} , n=1,2...$ and $f(x)=\lim_{n \to \infty}f_n(x)$ and the question is: "determine $f$", does that mean that I have to compute the limit?, i.e. to compute $f(x)=\lim_{n \to \infty}f_n(x)$ ?

And also how should I think about the convergence uniform on an interval?

答:
I'm assuming you're working on $[0,+\infty\rangle$ to avoid zeroes in the denominator.

Pointwise for $x \in [0,+\infty\rangle$ we have
$$\lim_{n\to\infty} f_n(x)=\lim_{n\to\infty} \frac{nx}{1+nx^3} = \begin{cases} \frac1{x^2}, &\text{ if } x > 0\\
0, &\text{ if }x=0\end{cases}$$

so $f(x) = \frac1{x^2} \chi_{\langle 0,+\infty\rangle}(x)$.

On an interval $[a,b] \subseteq \langle 0, +\infty\rangle$ the convergence is uniform. This follows from the fact that $(f_n)_n$ is monotonically increasing to $f$ (i.e. $f_n(x) \le f_{n+1}(x) \le f(x)$ for all $n \in \Bbb{N}$ and $x \in [a,b]$) and $f$ is continuous on $[a,b]$. Uniform convergence follows from Dini's theorem.

On an interval of the form $[0,b]$ we cannot make such a conclusion since $f$ is not continuous. Indeed, the convergence on $[0,b]$ is not uniform since for large enough $n\in\Bbb{N}$ we have $\frac1n \in [0,b]$ and
$$\left|f\left(\frac1n\right)-f_n\left(\frac1n\right)\right| = n^2 - \frac1{1+\frac1{n^2}} \xrightarrow{n\to\infty} +\infty.$$


答:
$\DeclareMathOperator\id{id}$Note that
$$
\id\circ \id = \id,\ \id \circ (13) = (13),\ (13)\circ\id = (13),\ \text{and}\ (13)\circ (13) = \id.
$$
So any product of elements of $A$ is an element of $A$.
Furthermore
$$
\id^{-1} = \id,\ \text{and}\ (13)^{-1} = (13).
$$
So also any inverse of an element of $A$ is an element of $A$.

You can also see that without even considering the elements of $A$ one by one. If $\sigma$ and $\tau$ are permutations fixing $2$ and $4$, note that
$$
(\sigma\circ \tau)(2) = \sigma(\tau(2)) = \sigma(2) = 2,
$$
so $\sigma\circ \tau$ fixes $2$. By the same reasoning $\sigma\circ\tau$ fixes $4$ as well and $\tau\circ\sigma$ also fixes $2$ and $4$.

Furthermore, since $\sigma(2)=2$ you also have $\sigma^{-1}(2)=2$ and since $\sigma(4)=4$ you also have $\sigma^{-1}(4)=4$, so the inverse of any element of $A$ also fixes $2$ and $4$ and hence is an element of $A$ as well.

Body:
Does the set $A$, of all permutations of $\{1,2,3,4\}$ that fixes $2$ and $4$ form a subgroup of $S_4$?

I’ve got as far,
The set $A$ will be the identity where all $4$ numbers map to themselves (as all fixed) and $(1, 3)$ as $2$ and $4$ are fixed. So the set $A =\{\operatorname{id}, (1 3)\}$

Now I need to show that this is a sub group of $S_4$, by identity closure and inverses.

Identity is in both so ok but I’m stuck on the other $2$.

Any help would be great

Kara x


答:
You're correct: $A=\{{\rm id}, (13)\}$.

I will use the two-step subgroup test.

Observe that $\varnothing \neq A\subseteq S_4$.

Consider the multiplication table for $A$:

$$\begin{array}{c|cc}
\circ & {\rm id} & (13) \\
\hline
{\rm id} & {\rm id} & (13) \\
(13) & (13) & {\rm id}.
\end{array}$$

From the table, one can see that for any $a,b\in A$, we have $a\circ b\in A.$

Since $(13)^{-1}=(13)\in A$ and ${\rm id}^{-1}={\rm id}\in A$, $A$ is closed under inverses.

Hence $A\le S_4$.

Title: Does the set $A$, of all permutations of $\{1,2,3,4\}$ that fixes $2$ and $4$ form a subgroup of $S_4$?


问:
A simple yet tricky binomial question

What are the number of dissimilar terms in the expansion of $(x+\frac{1}{x}+x^2+\frac{1}{x^2})^{15}$?

I know how to solve this kind of problem.

First I would arrange the term into a binomial expression. The expansion will have $(n+1)$ dissimilar terms.

But how can I arrange it into a binomial expression?

答:
Hint :

$$x+\dfrac{1}{x} = t \Rightarrow x^2+\dfrac{1}{x^2} = t^2 - 2$$

$$\therefore \Big(x+\dfrac{1}{x}+x^2+\dfrac{1}{x^2}\Big)^{15} = (t^2+t-2)^{15} $$

Now consider @lulu's precious advice : "What's the highest degree term? What's the lowest? Do all the intermediate terms have non-zero coefficients?"

Title: A simple yet tricky binomial question


答:
This is how I would proceed with the question:

$(x+\frac{1}{x}+x^2+\frac{1}{x^2})^{15}$ = $(\frac{1+x+x^3+x^4}{x^2})^{15}$
= $(\frac{1}{x^30})(1+x+x^3+x^4)^{15}$
= $(\frac{1}{x^30})$(1+.......+x^60)

That expression has 61 different powers. So the answer should be 61. Hope it helps!

Title: A simple yet tricky binomial question

Body:
What are the number of dissimilar terms in the expansion of $(x+\frac{1}{x}+x^2+\frac{1}{x^2})^{15}$?

I know how to solve this kind of problem.

First I would arrange the term into a binomial expression. The expansion will have $(n+1)$ dissimilar terms.

But how can I arrange it into a binomial expression?


问:
Proof analytic functions $f:\mathbb{C}_\infty \to \mathbb{C}_\infty$ are constant
How can I prove that analytic functions defined $f: \mathbb{C}_\infty \to \mathbb{C}_\infty$ are constant? I know about Picard Little Theorem that states that entire functions in the complex are either constant or their image is the whole complex plane or the complex plane minus one point, but I don't understand how the addition of $\infty$ in the domain let me discard all these options. I have also thought about using the maximum modulus principle, so I can state that either $\lim_{z \to \infty} |f(z)| = +\infty$ or $\lim_{z \to \infty} |f(z)|$ doesn't exist. Would this be enought to state that I would have problems at the infinite?

答:
Let $U = \{ z\in \Bbb{C},|z|<2\}$ and $\phi_j: U\to \Bbb{C}_\infty, \phi_1(z)=z,\phi_2(z)=1/z$.

Let $f(z) = 1/z$. Would you mind telling us how do we check if it is analytic $\Bbb{C}_\infty\to \Bbb{C}_\infty$ ?


问:
Subsequence of a bounded sequence converging to a specific limit
In my Real Analysis textbook, it proved the Bolzano-Weierstrass theorem which states that every bounded sequence in $\mathbb{R}$ has a convergent subsequence. The author proved this theorem by splitting the interval in which the bounded sequence sits into two, and taking the half with infinitely many terms. He continued to split the interval with infinitely many terms into two. He then used the nested intervals theorem to conclude that the intersection of all the intervals is a singleton, and constructed a sequence that converges to that point by taking an element from every "half interval" (half in the sense that it was obtained by cutting a larger interval in half).

As an exercise, it asks me to prove the same theorem in the following way:

Let $A = \{x \in \mathbb{R} \, | \, \exists \, \text{infinitely many} \, a_n\,\text{'s such that} \, x < a_n\}$. Show that $\{a_n\}$ has a subsequence converging to $\sup A$.

Since this also proves the Bolzano-Weierstrass theorem, I thought I could apply the same procedure, but either it is a dead-end, or I got confused because I failed.

答:
Since the sequence $(a_n)$ is bounded it is clear that $A$ is bounded. Hence $\sup A$ exists. Now for each $k \in \mathbb{N}$, by the definition of supremum, there is $x_k \in A$ such that $x_k >\sup A - \frac{1}{k}$. Hence there is an infinite family $J_k \subset \mathbb{N}$ such that $a_n >x$, for all $n \in J_k$. If $a_n> \sup A$ for infinitely many  $n \in J_k$ (there is infinite set $I_k \subset J_k$ such that this happens) you have that $(a_n)_{n \in I_k}$ converges to $\sup A$. If not there is $\epsilon>0$ such that $a_n>\sup A +\epsilon$ for all $n \in I_k$ contradicting the definition of supremum because in this way $\sup A +\epsilon \in A$. So if some $k \in \mathbb{N}$ we have that $a_n> \sup A$ for infinitely many $n \in J_k$ we know that there is a subsequence converging to $\sup A$. Suppose this is not the case for all $k \in \mathbb{N}$. Hence, for each $k \in \mathbb{N}$, there is an infinite family (again denoted by $J_k$) such that $\sup A \geq a_n >x_k > \sup A -\frac{1}{k}$ for all $n \in J_k$. Now, choose $n_1 \in J_1$. Choose $n_2 \in J_2 \setminus \{n_1\}$. And for the $k-th$ case choose $n_k \in J_k \setminus \{n_1, \cdots, n_{k-1}\}$ (This is always possible since the sets are infinite). Now notice the subsequence $a_{n_k}$ converges to $\sup A$ because for all $k\in \mathbb{N}$
$$
\sup A \geq a_{n_k} > \sup A -\frac{1}{k}.
$$


问:
How to show $T$ is diagonalizable?
Let $T\colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be linear with distinct eigenvalues $\lambda_1, \lambda_2, \lambda_3$.  Show that $T$ is diagonalizable.

It seems as if this is a very simple question but I'm a bit confused about how to start!

答:
You have $3$ distinct  eigenvalues  (as $\lambda_1 , \lambda_2 , \lambda_3$) and eigenvector corresponding to each of them must have $\dim$ $1$ (if you have doubt, see Hint below) then we have $3$ eigenvector spaces s that they are subspace of $\mathbb R^3$ and sum of them can construct $\mathbb R^3$.

So if we choice basis from these sub-spaces (choice one base from each sub spaces as $v_1$ from eigenvector space correspond $\lambda_1$ eigen value and  $v_2$ from eigen vector space correspond $\lambda_2$ eigen value and similar about $v_3$ ) since each of thses basis ARE eigen vector then our matrix (represented T) will be diagonalizable because $$T(v_1)=\lambda_1 v_1,T(v_2)=\lambda_2 v_2,T(v_3)=\lambda_3 v_3,$$ And so:

our matrix with basis$\{v_1,v_2,v_3\}$ will be:

$$\begin{pmatrix}\lambda_1&0&0\\0& \lambda_2&0\\0&0&\lambda_3\end{pmatrix}$$

Hint:
any eigenvalue as $\lambda$ have at least one $\dim$ because :
$$\det(A-\lambda I)=0$$

$(A-\lambda I)x=0$ has at least one non trivial answer. And at this problem it can not has more than one $\dim$ because sum of $\dim$ of $3$ sub-spaces must be $3$.

Title: How to show $T$ is diagonalizable?


问:
How to solve this difference equations in the non-homogenous case
Let $0<\alpha<1$ and $N\in\mathbb{N}$. I would like to solve the following difference equation

$$n=0: s_0 = 1+\alpha s_1 $$
$$n\in\{1,\dots,N-2\}: s_n = 1+\alpha s_{n+1}+(1-\alpha)s_n$$
$$n=N-1:s_{N-1}= 1+\alpha s_0 + (1-\alpha)s_{N-1}$$

and the boundary condition $s_N=0$. I was able to calculate the homogeneous solution, i.e. the solution to the equations:

$$n=0: s_0 = 1+\alpha s_1 $$
$$n\in\{1,\dots,N-2\}: s_n = \alpha s_{n+1}+(1-\alpha)s_n$$
$$n=N-1:s_{N-1}= 1+\alpha s_0 + (1-\alpha)s_{N-1}$$

to be $s_n = \frac{1}{\alpha},0<n<N$, $s_0=0$. Is this correct and how can I now solve this for a non-homogneous casse? To get the homogeneous solution I used that for $0<n<N$ we see that $s_n$ are all equal. Then I sued the equation for $n-1$ and $0$ to find the particular values for $s_0$ and all other values

答:
$$n\in\{1,\dots,N-2\}: s_n = 1+\alpha s_{n+1}+(1-\alpha)s_n \\
\implies -1 = \alpha (s_{n+1}-s_n).
$$

Therefore $s_1, s_2, \ldots, s_{N-1}$ is an arithmetic sequence.

Can you end it now?


问:
Dimension of rational subspace ($\mathbb{Q}[\sqrt{2}]^4$)
For the rational space say $\mathbb{Q}[\sqrt{2}]$ I know that this has dimension $2$ as $(1,\sqrt{2})$ forms a basis of dimension $2$. But what if you take $\mathbb{Q}[\sqrt{2}]^4$? Would the dimension then be $4\times 2=8$?

答:
The dimension of a finite cartesian product of finite dimensional linear spaces is the sum of their dimension.

So if by $\mathbb{Q}[\sqrt{2}]^4$ you mean $\mathbb{Q}[\sqrt{2}] \times \mathbb{Q}[\sqrt{2}] \times \mathbb{Q}[\sqrt{2}]\times \mathbb{Q}[\sqrt{2}]$, then yes its dimension as a linear space over $\mathbb Q$ is equal to $8$.


问:
I need help with extracting x and y from this formula
this is my first time posting, I couldn't do this by myself (its way out of my league), I need to extract x and y from the formula here:

$$ vpa = 0.6112 \times \mathit e ^ \left(\frac{18.678-\mathit x}{234.5} \times x \over {x + 257.14}\right)\times y $$

that superscript is:

$$ {{18.678 - x \over 234.5}  \times x} \over x+ 257.14 $$

Is it even possible? Thank you in advance, i appreciate this much!

答:
Divide both sides by $0.6112y$ and take the $\log$.  Define $z=\log \frac{vpa}{0.6112y}$.  Then you have
$$z={{{18.678 - x \over 234.5}  \times x} \over x+ 257.14}\\
(x+257.14)z\cdot234.5=(18.678-x)x=18.678x-x^2$$
Which is a quadratic you can feed to the formula to get $x$ in terms of $z$.  You can substitute back in to get it in terms of $y$


问:
unique solution to system of two equations
$$\begin{cases}
ax+by=0&\\
cx+dy=0&
\end{cases}$$
How do the coefficients a, b, c and d have to be chosen for this system of equations to have one unique solution $$ \begin{cases}
x=0&\\
y=0&
\end{cases}$$?
Or what kind of conditions does one have to give to those coefficients for this system to have the unique solution?

答: You need the determinant of $\begin {pmatrix} a&b\\c&d \end {pmatrix}$ to be nonzero to get a unique solution.  That determinant is $ad-bc$


问:
Is this projection matrix?
What we know about projection matrices:

$A^T=A $

$A=A^2 $

I want to prove that $(I-A)^T $is also projection matrix :

My solution:

$(I-A)^T =I-A^T=I-A=I-A^2$

we can see that $ I-A=I-A^2 => 0=A-A^2$,
$A^2=A$ then $A-A=0$

Does it look correct?

Thanks for help!

答:
Hint:

You need to prove that
$$\left((I-A)^T\right)^T = (I-A)^T, \quad \left((I-A)^T\right)^2 = (I-A)^T.$$

Title: Is this projection matrix?


答:
The solution is unique unless the two equations say the same (and there is in fact a single equation), which occurs if the coefficients are proportional.

Hence

$$\frac ac\ne\frac bd,$$ also written $$ad\ne bc.$$

(The second form is preferred, as it works even with zero coefficients.)

Title: unique solution to system of two equations

Body:
$$\begin{cases}
ax+by=0&\\
cx+dy=0&
\end{cases}$$
How do the coefficients a, b, c and d have to be chosen for this system of equations to have one unique solution $$ \begin{cases}
x=0&\\
y=0&
\end{cases}$$?
Or what kind of conditions does one have to give to those coefficients for this system to have the unique solution?


答:
I don't understand what you mean by "we can see that $ I-A=I-A^2 => 0=A-A^2$
$A^2=A$ then $A-A=0$".

However

$$\begin{aligned}
\left(I - A^T\right)^2 & = (I-A)^T(I-A)^T\\
&=\left((I-A)(I-A)\right)^T\\
&= (I-2A + A^2)^T\\
&=(I-A)^T
\end{aligned}$$

So $(I-A)^T$ which satisfies the two required identities to be a projection is indeed a projection.

Title: Is this projection matrix?

Body:
What we know about projection matrices:

$A^T=A $

$A=A^2 $

I want to prove that $(I-A)^T $is also projection matrix :

My solution:

$(I-A)^T =I-A^T=I-A=I-A^2$

we can see that $ I-A=I-A^2 => 0=A-A^2$,
$A^2=A$ then $A-A=0$

Does it look correct?

Thanks for help!


问:
Parametric Transfinite Recursion Theorem implies Double Recursion Theorem?
The Parametric Transfinite Recursion Theorem (PTRT) states:

For each operation $\mathbf{G}$, there is an operation $\mathbf{F}$ such that $\mathbf{F}(z,\alpha)=\mathbf{G}(z,\mathbf{F}_z\restriction \alpha)$.

$\mathbf{F}_z$ is just the operation given by $\mathbf{F}_z(x)=\mathbf{F}(z,x)$.

The Transfinite Double Recursion Theorem (TDRT) states:

For each operation $\mathbf{G}$, there is an operation $\mathbf{F}$ such that $\mathbf{F}(\alpha,\beta)=\mathbf{G}(\mathbf{F} \restriction (\alpha \times \beta))$ for all ordinals $\alpha, \beta$.

Is there any hope to use the PTRT to prove the TDRT? I've been trying for quite some time now, but have not come up with any useful idea yet. I'm starting to wonder if the PTRT can even be used to prove the TDRT.

答:
Yes, you can do this.  The idea is to construct the operation $\alpha\mapsto\mathbf{F}\restriction(\alpha\times\alpha)$, which intuitively you can do by ordinary transfinite recursion since the rule $\mathbf{F}(\gamma,\delta)=\mathbf{G}(\mathbf{F} \restriction (\delta \times \gamma))$ tells you how to define $\mathbf{F}$ on $\alpha\times\alpha$ if you already know it on $\beta\times\beta$ for all $\beta<\alpha$ (since $\delta\times\gamma$ is contained in some such $\beta\times\beta$).

To make this precise, let $\mathbf{G}'$ be the operation defined as follows.  Suppose $f$ is a function whose domain is an ordinal $\alpha$, and moreover assume that $f(\beta)$ is a function with domain $\beta\times\beta$ for each $\beta<\alpha$.  We then define $\mathbf{G}'(f)$ be the function $g$ with domain $\alpha\times\alpha$ defined by $g(\gamma,\delta)=\mathbf{G}(f(\beta)\restriction(\delta\times\gamma))$ where $\beta=\max(\gamma,\delta)$.  (If $f$ is not such a function, we define $\mathbf{G}'(f)=\emptyset$.)

By the PTRT (ignoring the parameter), there is an operation $\mathbf{F}'$ such that $\mathbf{F}'(\alpha)=\mathbf{G}'(\mathbf{F}'\restriction\alpha)$ for all ordinals $\alpha$.  It is straightforward to prove by induction that $\mathbf{F}'(\alpha)$ is a function with domain $\alpha\times\alpha$ for each $\alpha$, and if $\beta<\alpha$ then $\mathbf{F}'(\beta)$ is the restriction of $\mathbf{F}'(\alpha)$ to $\beta\times\beta$.  Now define $\mathbf{F}(\alpha,\beta)=\mathbf{F}'(\gamma)(\alpha,\beta)$ where $\gamma=\max(\alpha,\beta)+1$.  It is now easy to verify that $\mathbf{F}$ is the desired operation.


问:
How to find the limit of this function
We have the function $$\dfrac{\sqrt{n^4 + 100}}{4n}$$

I think the best method is by dividing by $n$, but I have no idea what that yields, mainly because of the square root.

答:
Dividing by $n$ is a great idea: Note that $$\sqrt{n^4 + 100}\cdot\frac 1n = \sqrt{(n^4 + 100)\cdot\frac 1{n^2}} = \sqrt{n^2 + \frac {100}{n^2}}$$

So your limit is, after dividing numerator and denominator by $n$:

$$\lim_{n\to \infty} \frac{\sqrt{n^2 + \frac {100}{n^2}}}{4} \to \infty$$

Title: How to find the limit of this function


问:
Why is $\operatorname{Spec}\mathbb{C}[x^{\pm 1},y^{\pm 1},z^{\pm 1},w^{\pm 1}] \cong (\mathbb{C}^{*})^4$

Why is $\operatorname{Spec}\mathbb{C}[x^{\pm 1},y^{\pm 1},z^{\pm 1},w^{\pm 1}] \cong (\mathbb{C}^{*})^4$?

I'm trying to get a basic understanding of this $\operatorname{Spec}$ after reading a bit of algebraic geometry from Dummit and Foote. So, $\operatorname{Spec}$ is going to give us all the prime ideals of $\mathbb{C}[x^{\pm 1},y^{\pm 1},z^{\pm 1},w^{\pm 1}]$... Is this because a prime ideal is of the form $\langle cx^{\pm 1} \rangle$ for $c \in \mathbb{C}$? Thanks

答:
Consider the simpler case of $\operatorname{Spec}(\mathbb C[x^{\pm 1}])$. Note that we have the $\mathbb C$-algebra isomorphism
$$
\mathbb C[x^{\pm 1}]=\mathbb C[x,x^{-1}] \cong \mathbb C[x,y]/(xy-1).
$$
Now $\mathbb C[x,y]/(xy-1)$ is the coordinate ring of the variety in $\mathbb C^2$ given by $xy=1$. However, this variety is isomorphic to $\mathbb C^*$ by the isomorphism $V(xy-1)\to\mathbb C^*$ given by $(x,y)\mapsto x$. In this way $\mathbb C[x^{\pm 1}]$ becomes the coordinate ring of $\mathbb C^*$, in other words $\operatorname{Spec}(\mathbb C[x^{\pm 1}]) = \mathbb C^*$.


问:
What is the maximum value of $\frac{x-x^3}{x^4+2x^2+1}$?
What is the maximum value of $\frac{x-x^3}{x^4+2x^2+1}$  ?

I've got the hint that this problem can be solved by replacing $x$ with tangent $K$.

So it became $\frac{\tan K(1-\tan^2K)}{(\tan^2K+1)^2}=\frac{\tan K}{\tan^2K+1}\cdot\frac{1-\tan^2 K}{\tan^2K+1}=\frac{\sin(2K)}{2(1-\sin^2K)}$
then I have no idea what to do next.

Any help is appreciated, thank you.^^

答:
Your method yields the answer. The only thing is that you have assumed $(\frac{1-tan^{2}k}{1+tan^{2}k})$ to be equal to $(sec^{2} x)$ while it is equal to cos 2x

Hope it helps!


答: Both $ax+by = 0$ and $cx+dy=0$ are straight lines passing through origin, $(0,0)$. For unique solution, these two shouldn't be coincident; their slopes should be distinct. $$-\dfrac{a}{b} \neq  -\dfrac{c}{d} $$


问:
Is it possible for an increasing function to tend to zero?
I have to plot a graph of a real function $f(x)$ such that
$$\lim_{x\to\infty}f(x)=0$$
and for $x\geq3$, $f$ increases ($x=3$ is supposed to be the minimum point of $f$).

So, is it even possible? For me, since $f$ increases, then
$$\lim_{x\to\infty}f(x)=\infty\neq0$$

答:
Hint: $f(3)$ can be negative.

About your final "So ...": an increasing function can be bounded. Consider $1-e^{-x}$.


答: $$f(x)=-\frac1{x^2-6x+10}$$ has a single minimum at $x=3$ and tends to $0$.

Title: Is it possible for an increasing function to tend to zero?

Body:
I have to plot a graph of a real function $f(x)$ such that
$$\lim_{x\to\infty}f(x)=0$$
and for $x\geq3$, $f$ increases ($x=3$ is supposed to be the minimum point of $f$).

So, is it even possible? For me, since $f$ increases, then
$$\lim_{x\to\infty}f(x)=\infty\neq0$$


问:
Frobenius map as a ring homomorphism
I have to prove that the frobenius map given by $Fr: GF27 \to GF27$ where $Fr([y]) = [y]^3$ for $[y] \in GF27$. Here $GF27$ denotes a field made as $\mathbb{Z}_3/f(x)$ where $f$ is an irreducible polynomial of degree 3.
I have to show $Fr([y_1] + [y_2]) = Fr([y_1])+Fr([y_2])$ and $Fr([y_1]\cdot[y_2]) = Fr([y_1])\cdot Fr([y_2])$ for $[y_1], [y_2] \in GF27$. I do not have access to Freshmans dream, is there an easy way to prove this without that access, or do i have to show it the hard way doing all the calculations and showing it that way?

答:
Any function $x\mapsto x^k$ with arbitrary $k$ preserves multiplication in a commutative ring.

And, use the binomial theorem to prove $x\mapsto x^p$ preserves addition in a ring with characteristic $p$, where $p$ is a prime (specifically $3$ in your example).

 $$(x+y)^p=\sum_{i=0}^p{p\choose i}x^iy^{p-i}$$
 and $p$ divides ${p\choose i}$ whenever $1\le i< p$, so the intermediate terms all vanish.

Title: Frobenius map as a ring homomorphism


问:
Finding the limit of $a_n = \frac{n+1}{2^{n+1}}\left(\frac{2}{1}+\frac{2^2}{2}+...+\frac{2^{n}}{n}\right)$
We have to prove the convergence and find the limit of $a_n = \frac{n+1}{2^{n+1}}\left(\frac{2}{1}+\frac{2^2}{2}+...+\frac{2^{n}}{n}\right)$. I tried to find $a_{n+1}$ in terms of $a_{n}$ and got $a_{n+1}=\frac{n+2}{2(n+1)}\left(a_n+1\right)$ and could not proceed further.Any help is appreciated.

Edit : Following K.defaoite's answer limit appears to be 1 . But we still need to prove convergence.

答:
Assuming $a_n$ has a limit, which it appears to have, and your recursive expression is correct, which it appears to be, we can write
$$\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\frac{n+2}{2(n+1)}(a_n+1)$$
$$x=(x+1)\lim_{n\to\infty}\left(\frac{n+1}{2(n+1)}+\frac{1}{2(n+1)}\right)$$
$$x=\frac{x+1}{2}$$
$$x=1$$


问:
Find radius of convergence and center of this complex series.
$$\sum_{n=0}^{\infty} \frac{n(n-1)}{3^n} (z-i)^{2n}$$

My book says the answer is that the center = i, and the radius = $\sqrt{3}$

But I keep getting radius 3.

$$\sum_{n=0}^{\infty} \frac{n(n-1)}{3^n} (z-i)^{2n}$$
so $a_{n} = \frac{n(n-1)}{3^n}$
so $a_{n+1} = \frac{(n+1)(n)}{3^{n+1}}$

So the $\lim_{n \to \infty} \left | \frac{n(n-1)}{3^n} \cdot \frac{3^{n+1}}{(n+1)(n)}\right | = \lim_{n \to \infty} \frac{3(n-1)}{(n+1} $

Isn't this just 3 So the radius is 3?

答:
Let $$f(z) = \sum_{n=0}^{\infty} \frac{n(n-1)}{3^n} (z-i)^{2n}$$

Then denoting $Z=(z-i)^2$, $$f(Z) = \sum_{n=0}^{\infty} \frac{n(n-1)}{3^n} Z^n$$ is a power series centered on the origin.

Using the argument you developed in the question, $f(Z)$ has a radius of convergence equal to $3$.

And $$\vert Z \vert \lt 3 \iff \left\vert (z-i)^2 \right\vert \lt 3 \iff \vert z- i \vert \lt \sqrt 3$$

which means that the power series is centered on $i$ and has a radius of convergence equal to $\sqrt 3$.


Title: How to find the limit of this function

Body:
We have the function $$\dfrac{\sqrt{n^4 + 100}}{4n}$$

I think the best method is by dividing by $n$, but I have no idea what that yields, mainly because of the square root.

Body:
You have that
$$\frac{\sqrt{n^4 + 100}}{4n} = \frac{n^2\sqrt{1 + \frac{100}{n^4}}}{4n}=\frac{n\sqrt{1 + \frac{100}{n^4}}}{4}$$
and $\frac{\sqrt{1 + \frac{100}{n^4}}}{4}\xrightarrow[n\to\infty]{}\frac{1}{4}$.


问:
Solve the ODE with Laplace Transforms: $y'(t) + y(t-1) = t^2$
The exercise is the following:

Using Laplace Transforms, find the solution to $y'(t) + y(t-1) = t^2$, with $y(t) = 0$ for $t \leq 0$.

Being the first exercise I saw that showed something like $y(t-1)$ and knowing that $\mathcal{L}\{u_1(t)y(t-1)\}(s)=e^{-s}\mathcal{L}\{y(t)\}(s)$ where $u_1(t)$ is the Heaviside step function, I tried multiplying the whole expression by $u_1(t)$, but then didn't know how to proceed with $\mathcal{L}\{u_1(t)y(t)\}(s)$.

答:
Performing the Laplace transform of the equation you obtain
$$sY(s)-\underbrace{y(0^{-})}_{=0}+e^{-s}Y(s)=\frac{2}{s^{3}}$$
It follows
$$Y(s)=\frac{2}{s^{3}}\frac{1}{s+e^{-s}}=\frac{2}{s^{4}}\frac{1}{1+\frac{1}{s}e^{-s}}=2\sum_{n=0}^{\infty}\frac{(-1)^{n}}{s^{n+4}}e^{-ns}$$
It follows
$$y(t)=\int_{\eta-i\infty}^{\eta+i\infty}\frac{ds}{2\pi i}e^{st}Y(s)=2\sum_{n=0}^{\infty}(-1)^{n}\int_{\eta-i\infty}^{\eta+i\infty}\frac{ds}{2\pi i}e^{st}\frac{1}{s^{n+4}}e^{-ns}$$
$$=\sum_{n=0}^{\infty}\frac{2(-1)^{n}}{(n+3)!}(t-n)^{n+3}\Theta(t-n).$$
Where we've used Cauchy theorem in the last line.


问:
generalised likelihood ratio test
Given the p.m.f of a negative binomial random variable X is given by $$P(X=x|\theta)={r+x-1 \choose x}(1-\theta)^x\theta^r, x=0,1,2... $$ with parameters $\theta\in[0,1]$ and $r>0$ consider the maximum likelihood estimator $$\hat{\theta}=\frac{r}{r+x}$$

State the form of a generalised likelihood ratio test based on $\hat{\theta}$ for the hypothesis testing problem:

$$\begin{cases}
\mathcal{H}_0,  & \theta=0.5 \\
\mathcal{H}_1,  & \theta\ne 0.5
\end{cases}$$

What I know:

I know I need to compute the likelihood ratio but because there are no values I'm confused as to how I do this. The likelihood function is $$L(\theta)= {r+x-1 \choose x}(1-\theta)^x(\theta)^r$$

Then does the ratio become : $\frac{{r+x-1 \choose x}(1-0.5)^x(0.5)^r}{{r+x-1 \choose x}(1-\theta\neq0.5)^x(\theta\neq0.5)^r}$

答:
No, it's wrong.

The definition of generalized likelihood ratio is the following

$$\lambda(\mathbf{x})=\frac{sup_{\theta=\theta_0}L(\theta|\mathbf{x})}{sup_{\theta \in\Theta}L(\theta|\mathbf{x})}$$

Thus in the denominator of your ratio you have $\theta=\hat{\theta}_{ML}$

Substituting you get

$$\lambda(\mathbf{x})=\frac{sup_{\theta=\theta_0}L(\theta|\mathbf{x})}{sup_{\theta \in\Theta}L(\theta|\mathbf{x})}=\frac{\Big(\frac{x+r}{2}\Big)^{x+r}}{r^r x^x}$$

The exercise ends here because no other questions are asked. Going on,

$$0\leq \lambda(\mathbf{x})\leq 1$$

And you will reject $H_0$ when $ \lambda(\mathbf{x})\leq c$ for a certain $c$

In Bayesian Statistics (just FYK; it is not your case because you are dealing with Classical Statistics), under certain assumptions, the ratio you posted is not completely wrong....it becomes

$$\frac{0.5^{x+r}}{\int_0^1 \theta^{r}(1-\theta)^x d\theta}$$

Title: generalised likelihood ratio test


问:
Show that any two circles in the plane with the same center are Bertrand curves.
$\alpha(s)$ and $\beta(s)$ are called Bertrand curves if for each $s_0$, the normal line to $\alpha$ at $s = s_0$ is the same as the normal line to $\beta(s)$ at $s = s_0$. ($s$ need not be arc length on both $\alpha$ and $\beta$.) We say that $\beta$ is a Bertrand mate for $\alpha$ if $\alpha$ and $\beta$ are Bertrand curves.

答: A line is normal to a circle if and only if it passes through its center. As the two circles are supposed to have the same center, you're done taking for parameter the angle of the points on the circles.</p>


问:
Choose the correct one(s). Differentiability of complex function.
Choose the correct one(s).

$f(z)=|z|$ is differentiable exactly at origin.

$f(z)=|z|^2$ is differentiable exactly at origin.

$f(z)=z|z|$ is differentiable exactly at origin.

$f(z)=z Re(z)$ is differentiable exactly at origin.

My attempt

I could prove that $f(z) $ is not differentiable at $z=0.$ I am not able to prove the differentiability of $f(z)$ for $z\neq 0$
$\lim _{h\to 0\ }\frac{f\left(z+h\right)-f\left(z\right)}{h}=\lim _{h\to 0\ }\frac{\sqrt{(z+h)\overline{(z+h)}}-\sqrt{(z)\overline{z}}}{h}$

I am stucked here.

I could prove that $f(z) $ is  differentiable at $z=0.$ To to prove the differentiability of $f(z)$ for $z\neq 0. $ Consider
$\lim _{h\to 0\ }\frac{f\left(z+h\right)-f\left(z\right)}{h}=\lim _{h\to 0\ }\frac{(z+h)\overline{(z+h)-(z)\overline{z}}}{h}=\lim _{h\to 0\ }\frac{z\overline z+h\overline{z}+z\overline{h}+h\overline{h}-z\overline{z}}{h}=\lim _{h\to 0\ }\frac{h\overline{z}+z\overline{h}+h\overline{h}}{h}.$ Here limit doesnot exists. Since $\lim _{h\to 0\ }\frac{z\overline{h}}{h}$ does not exists.

I could prove that $f(z) $ is  differentiable at $z=0.$ To to prove the differentiability of $f(z)$ for $z\neq 0. $ Consider
$\lim _{h\to 0\ }\frac{f\left(z+h\right)-f\left(z\right)}{h}=\lim _{h\to 0\ }\frac{(z+h) \sqrt{(z+h)\overline{(z+h)}}-z\sqrt{(z)\overline{z}}}{h}$

I am stucked here.

I could prove that $f(z) $ is  differentiable at $z=0.$ To to prove the differentiability of $f(z)$ for $z\neq 0. $ Consider
$\lim _{h\to 0\ }\frac{f\left(z+h\right)-f\left(z\right)}{h}=\lim _{h\to 0\ }\frac{(z+h)Re(z+h)-(z)Re(z)}{h}=\lim _{h\to 0\ }\frac{(z)Re(z)+zRe(h)+hRe(z)+hRe(h)-(z)Re(z)}{h}=\lim _{h\to 0\ }\frac{zRe(h)+hRe(z)+hRe(h)}{h}.$

I am stucked here. I am not able to solve further. May I get help in 1.,3.,4..

答:
If $z_0\ne0$, then $z\mapsto|z|$ is not differentiable at $z_0$. To see why, see that$$\lim_{z\to z_0,\ z\in\Bbb R^+z_0}\frac{|z|-|z_0|}{z-z_0}=\lim_{\lambda\to1, \lambda\in\Bbb R^+}\frac{|\lambda z_0|-|z_0|}{\lambda z_0-z_0}=\frac{|z_0|}{z_0}\ne0$$whereas$$\lim_{z\to z_0,\ |z|=|z_0|}\frac{|z|-|z_0|}{z-z_0}=0.$$

What you did is fine.

Let $f$ be your function. If $f$ was differentiable at a point $z_0$ other than $0$, then the same would be true for $\frac{f(z)}z$. But, we have already seen that that does not happen.

The function $\operatorname{Re}$ is differentiable nowhere, because$$\lim_{z\to z_0,\ \operatorname{Re}(z)=\operatorname{Re}(z_0)}\frac{\operatorname{Re}(z)-\operatorname{Re}(z_0)}{z-z_0}=0$$whereas$$\lim_{z\to z_0,\ \operatorname{Im}(z)=\operatorname{Im}(z_0)}\frac{\operatorname{Re}(z)-\operatorname{Re}(z_0)}{z-z_0}=1.$$It's now easy do deduce from this that $z\mapsto z\operatorname{Re}(z)$ is diffferentiable at the origin and only there.


问:
Equivalent of completely multiplicative function for dot product
Does there exist an equivalent to the set of completely multiplicative functions ie functions for which $f(ab) = f(a)f(b)$ for the dot product?

Ie I'm looking for equations satisfying $$f(X\cdot Y) = f(X)\cdot f(Y)$$ for some vectors $X,Y$ (ideally with $X,Y \in \mathbb{R}^d$) or potentially any other operation such that $$f(X\cdot Y) = f(X)\circ f(Y)$$ Apart from obvious solutions like constant functions or identity, I can't come up with any as the other solutions to completely multiplicative functions, most importantly power functions (and also obviously Cauchy functional equations) do not work.

One issue is the domain of the function $f(X)$, so either we define that to be $\mathbb{R} \cup \mathbb{R}^d$ or we need two functions $f(X): \mathbb{R} \rightarrow \mathbb{R}$ and $g(x): \mathbb{R}^d \rightarrow \mathbb{R}$ s.t. $f((X \cdot Y)) = g(X) \cdot g(Y)$ - or, as mentioned above, we use another operation for $f(X) \circ f(Y)$ s.t. $f(X) \circ f(Y) \in \mathbb{R}$

答:
So, the problem is to find functions $f, g, h$ s.t. $f((X, Y)) = h(g(X), g(Y))$. The first variant is $h(a, b) = a \cdot b$.

Assume $d > 1$ and $g \not \equiv 0$. Choose $x$ s.t. $g(x) \neq 0$, $y \perp x$ and let $X = \lambda x$, $Y = \mu y$. We get $f(0) = f((X, Y)) = g(\lambda x) \cdot g(\mu y)$. As left part doesn't depend on $\mu$ and $\lambda$, right part doesn't depend on them either - implying $g(\lambda x)$ and $g(\mu y)$ are constants. Taking $\lambda = \mu = 0$ we get $g(\lambda x) = g(0)$ and $g(\lambda y) = g(0)$.

If there exists $x \neq 0$ s.t. $g(x) \neq 0$, taking $X = \lambda x$, $Y = \mu x$ we get $f(\lambda \mu \|x\|^2) = g(\lambda x) g(\mu x) = g(0)^2$. As $\lambda \mu \|x\|^2$ is arbitrary number, $f \equiv g(0)^2$. And then $g$ is constant too.

So, if $g(x) \neq 0$ for some $x \neq 0$, then $g$ is constant. If $\forall x \neq 0: g(x) = 0$ then $f(0) = f((X, 0)) = g(X) \cdot g(0)$. If $X \neq 0$ then right part is equal to $0$, thus $f(0) = 0$ and so $g(0) = 0$ too.

So, if $h(a, b) = a \cdot b$, then $g$ is constant.

If $h$ is arbitrary, it's simple - choose any $f$, let $g$ be your favorite injection $\mathbb R^d \to \mathbb R$ and say $h(a, b) = f((g^{-1}(a), g^{-1}(b)))$.


Title: What is the maximum value of $\frac{x-x^3}{x^4+2x^2+1}$?

Body:
What is the maximum value of $\frac{x-x^3}{x^4+2x^2+1}$  ?

I've got the hint that this problem can be solved by replacing $x$ with tangent $K$.

So it became $\frac{\tan K(1-\tan^2K)}{(\tan^2K+1)^2}=\frac{\tan K}{\tan^2K+1}\cdot\frac{1-\tan^2 K}{\tan^2K+1}=\frac{\sin(2K)}{2(1-\sin^2K)}$
then I have no idea what to do next.

Any help is appreciated, thank you.^^

Body:
With $t=\frac1x - x$
$$\frac{x-x^3}{x^4+2x^2+1}=\frac{\frac1x - x}{x^2+\frac1{x^2}+2}
= \frac{t}{t^2+4}= \frac{\text{sgn}(t)}{(\sqrt{|t|}-\frac2{\sqrt{|t|}})^2+4}\in (-\frac14, \frac14)
$$
Thus, the maximum  is $\frac14$.


答:
I assume you mean the limit as $n\to \infty$.  In that case, the limit is easily seen to be infinity as follows:

The numerator scales as $\sim n^2$ for large $n$, but the denominator scales as $n$ only.

The ratio thus scales as $\sim n$.

So the limit as $n\to \infty$ will be $\infty$.

Title: How to find the limit of this function


问:
Partial derivative of a multivariable piecewise function
I have this piecewise function:

$$f(x,y)= \left\{ \begin{array}{lcc}
             \frac{(x+y)^2}{x^2+y^2};  &  (x,y)\neq(0,0) \\
             \\ 1; & (x,y)=(0,0) \\
             \\
             \end{array}
   \right.$$

We know that it is not continuous at $(0,0)$ and we are asked about whether this statement about partial derivative is true:
$$f_x(0,0)=0$$
I checked that $f$ function and got that it is not differentiable at $(0,0)$. The derivative using the definition based on limits, also shows that this statement is wrong. But my professor wrote that it is true and quals to zero as said. Here is my problem. Do I miss something, or that partial derivative by x really equals to zero?

Body:
For $x \neq 0$ $$\frac{f(x,0) - f(0,0)}{x}= \frac{x^2/x^2 -1}{x} = 0.$$

So indeed $f_x(0,0) = 0$.


问:
Is $S\cup\{v\}$ is linearly dependent or linearly independent?
Is the following statement is true/false ?

Let $S$ be a linearly independent subset of a vector space $V.$
Suppose $v$ is a vector in $V$ which is not in the subspace spanned by $S$. Then
the set obtained by adjoining $v$ to $S$ is linearly dependent

My attempt : no

It given that $S$ is linearly independent . we will denote set obtained by adjoining $v$ to $S$  by  $S\cup\{v\}$ so  there is a  distinct vectors $u_1,u_2,...,u_n\in S\cup\{v\}$ and a collection of non-zero scalars $c_1,c_2,...,c_n$ such that $$c_1u_1+c_2u_2+\cdots+c_nu_n=0.\tag{1}$$ Since the $c_i$s are non-zero and $S$ is linearly independent, then we can't have all the $u_i$s in $S,$ so one of them is $v$. Without loss of generality , say $u_1=v$, and so the rest are in $S$. Since $c_1\neq 0,$ we can solve $(1)$ to get $$v=u_1=-\frac1{c_1}\left(c_2u_2+\cdots+c_nu_n\right)\in\text{span}\{u_2,...,u_n\}\subseteq\text{span }S.$$

This implies that $S\cup\{v\}$ is linearly independent

Edit : Proof verification

答:
Well, I'd consider a set of vectors $u_1,\ldots,u_n$ in $S$ and the vector $v$.

Suppose $c_0 v + c_1u_1+\ldots+c_nu_n=0$ with $c_0\ne 0$; the case $c_0$ is uninteresting.

Then $v = -\frac{1}{c_0}(c_1u_1+\ldots+c_nu_n)$ lies in the subspace spanned by $S$, a contradiction.

Thus $c_0=0$ and since $u_1,\ldots,u_n\in S$ are linearly independent, it follows that all $c_i$'s are $0$. Hence, $S\cup\{v\}$ is a linearly independent set.


问:
$θ_2$ is better than $θ_1$ to estimate $μ$?
We offer two estimators for the average concentration $μ$ of lead in the atmosphere of a region of Quebec where factories manufacturing dyes are located. The first estimator $θ_1$ has a bias equal to $0.2$ and a variance of $0.02$. The second estimator $θ_2$ is unbiased and has a variance equal to $0.06$.

Which one is the best estimator?

I think $θ_2$ is better than $θ_1$ to estimate $μ$, but I am not sure.

EDIT

A PhD student in statistics explained to me that if $MSE(\theta_1) = MSE(\theta_2)$, then we cannot conclude. In other words, $\theta_2$ is not preferred over $\theta_1$ or inversely. I am not sure about that.

答:
In my experience, a situation like this can end up depending on the context, i.e. what you are trying to estimate. Since you are discussing the average concentration of lead in the atmosphere of a particular region, the results of this could be severe, in which case I have been taught taking the unbiased estimator is a better idea here (since your variance is still quite small).

Hope this helps!


问:
$f$ is strictly increasing, $f (a)  b$. Number of solutions to $f(c) = c$.
Let $f$ be a strictly monotonic continuous real valued function defined on $[a, b]$ such that $f (a) < a$ and $f(b) > b$.

My claim is that, $f(c)=c$ for $c \in (a,b)$, their is either exactly one such $c$ or infinitely many such $c's$.

How should I prove this?

答:
Hint

If your claim was true, some polynomials of degree three would have an infinite number of roots.

Title: $f$ is strictly increasing, $f (a) < a$, $ f(b) > b$. Number of solutions to $f(c) = c$.


答:
$\text{Note that $\sqrt{n^4+100} > \sqrt{n^4}$. Hence, we have}$
$$\dfrac{\sqrt{n^4+100}}{4n} > \dfrac{\sqrt{n^4}}{4n} = \dfrac{n^2}{4n} = \dfrac{n}4$$
$\text{I trust you can finish it from here.}$</p>

Title: How to find the limit of this function


问:
Nilpotent matrix?
I have the following question regarding the matrix
$$
        \mathbf{A} =
        \left[\begin{array}{rrr}
             x-3&-x+3&-x+5\\
             x-2&-x+2&-x+4\\
             -1&1&-1
        \end{array}\right]
$$
I have been told that such a matrix $\mathbf{A}$ is called nilpotent. Looking at https://en.wikipedia.org/wiki/Nilpotent_matrix it should for some power $k$ give a zero matrix. I am asked to compute $\mathbf{A}^2$ and $\mathbf{A}^3$. I tried computing them in maple but none of them give a zero matrix? Why would you then call $\mathbf{A}$ a nilpotent matrix?

(If requested I could write the results in latex code :) )

Here is the exercise definition:

Is the author of the exercise wrong?

答:
The characteristic polynomial of the matrix is $\lambda^3+2\lambda^2+2\lambda$, so the matrix is not nilpotent. This can also be checked with the trace not being zero.

There is a typo in the exercise: if the $(3,3)$ coefficient is changed into $1$, then the statement becomes true, because the characteristic polynomial is $\lambda^3$.

I fed the matrix to Pari-GP, so to get the characteristic polynomial. Changing $-1$ into $1$ in the bottom right corner makes the trace zero, so the necessary condition is fulfilled. And the characteristic polynomial confirms the typo.

By the way, the square of the (modified) matrix is
$$
\begin{bmatrix}
-2 & 2 & 2 \\
-2 & 2 & 2 \\
 0 & 0 & 0
\end{bmatrix}
$$
which is independent on $x$.

The square of the original matrix is instead
$$
\begin{bmatrix}
-2 &  2 & 2x - 8 \\
-2 &  2 & 2x - 6 \\
 2 & -2 & 0
\end{bmatrix}
$$
and the cube is
$$
\begin{bmatrix}
-2x + 10 & 2x - 10 & -2x + 6 \\
-2x + 8  & 2x - 8  & -2x + 4 \\
-2       & 2       &  2
\end{bmatrix}
$$

Title: Nilpotent matrix?


问:
How to find the angle formed by two joined triangles when they are multiples?
The problem is as follows:

The figure from below shows two triangles. Assume $PR=QS$. Using this
information find $x$.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&20^{\circ}\\
2.&18^{\circ}\\
3.&22^{\circ}\\
4.&24^{\circ}\\
4.&30^{\circ}\\
\end{array}$

What I attempted to do to solve this problem was to use the identity which relates the exterior sum of the two angles in a triangle. In other words:

$\angle PQS= 6x$

But that's it, I came stuck there. Since I found this problem in a section belonging to triangle congruency, I think it should be solved using such approach. Can someone help me here?. If so, what sort of the cases related with congruency of triangles is it?. Could it be that is side-angle-side?.

Please include a drawing in the solution as I dont know how to spot those in triangles.

答:
Construct point $A$ (as shown in the picture) such that $\overline{PS}=\overline{RA}$. Note that $$\overline{QS}=\overline{PR}=\overline{PS}+\overline{SR}=\overline{RA}+\overline{SR}=\overline{SA}$$ Therefore $$\angle SQA=\angle SAQ\implies 2x+\angle RQA=\angle RAQ\quad\quad(1)$$ Also note that $$\angle RQA+\angle RAQ=4x\quad\quad(2)$$

Solve system of equations and we have $$\angle RQA=x,\quad\angle RAQ=3x$$

Now since $\angle SPQ=\angle RAQ=3x$, we have $\overline{QP}=\overline{QA}$. Hence we can say that $$\triangle PQS\cong\triangle AQR\quad(S.A.S.)$$
so $$\overline{QS}=\overline{QR}\implies \angle QSR=\angle QRS=4x$$

Finally $$4x+4x+2x=180^{\circ}\implies \color{red}{x=18^{\circ}}$$


问:
Closed form for $I= \int_{-\infty}^{\infty} e^{-e^x-e^{-x}+x}~ dx$
Is there a closed form for this integral?

$$I= \int_{-\infty}^{\infty} e^{-e^x-e^{-x}+x}~ dx$$

I tried $u=e^x$ but it didn't simplify things much.

I think the closed form is a Bessel function, but I can't verify this directly.

I'd like to see the steps involved to reach the closed form.

答:
You can use the integral representation
$$
K_\nu  (z) = \frac{1}{2}\int_{ - \infty }^{ + \infty } {\mathrm{e}^{ - z\cosh x + \nu x} \mathrm{d}x} \quad (\Re z>0)
$$
of the modified Bessel function to conclude that $I = 2K_1 (2)$.


问:
Field of definition of representations of symmetric groups
Can one show in an elementary way, without recourse to Young tableaux etc., that the complex representations of symmetric groups are realisable over $\mathbb{R}$? It is easy to show that they are all self-dual, since the conjugacy classes of symmetric groups are self-dual, so one just has to exclude the possibility of quaternionic representations. Surely, there must be a similarly elementary argument? If there is, it is escaping me at the moment.</p>

答:
Some thoughts. Let $r_2(g)$ denote the number of square roots of $g$. Then $\langle r_2, \chi \rangle$ is the Frobenius-Schur indicator of $\chi$, and we want to show that this is equal to $1$ for all irreducibles $\chi$. This would follow if we could directly construct a representation of $S_n$ with character $r_2$, since then we would immediately have $\langle r_2, \chi \rangle \ge 0$.

The representation associated to this character would have dimension $r_2(e)$, or the number of involutions of $S_n$. And, indeed, there is a natural permutation representation of $S_n$ on the set of involutions (by conjugation), but it has the wrong character. In fact $r_2$ cannot in general be the character of a permutation representation: since it contains the trivial representation only once, it must be transitive, but its degree does not divide $n!$ in general.

But it still might be possible to construct this representation in a reasonably elementary way without going through the full construction of the irreducible representations of $S_n$.