问:
Let $f(x)=x^6-2x^3-8$ and $g(x)=x^2+2x+4$. Let $a_1$ through $a_6$ be its roots. Find the value of $\prod_{n=1}^{6} (g(a_n))$
Question:

Let $f(x)=x^6-2x^3-8$ and $g(x)=x^2+2x+4$. Let $a_1$ through $a_6$ be its roots. Find the value of $\displaystyle \prod_{n=1}^{6} (g(a_n))$

My process:

I first thought of setting $z=a_i^2+2a_i+4$ and then putting the value back into $f(z)$ and then from there finding the product using Vieta's. But the process turned incredibly long and tedious and I found it to be impossible to do by hand.

So I just went to to the brute force method, expanding the whole expression out. Writing all the terms would be really tedious so I would like to abbreviate it. Let $P= a_1 \cdot a_2 \cdots \cdot a_6$ and let $e_n$ denote the sum of roots taken $n$ at a time.

Here is the expression I got:

$$\prod_{i=1}^{6} (a_i^2+2i+4)=P^2+2e_5P+2^2e_4P+2^3e_3P+2^4e_2P+2^5e_1P+2^6P+2^5*4*e_5+2^4*4^2*e^4+2^3*4^3*e_3+2^2*4^4*e_2+2*4^5*e_1+4_6+4(\sum_{cyc}(a_1a_2a_3a_4a_5)^2))^+4^2(\sum_{cyc}(a_1a_2a_3a_4)^2)+4^3\sum_{cyc}(a_1a_2a_3)^2+4^4\sum_{cyc}(a_1a_2)^2+4^5\sum_{cyc}(a_1)^2$$

Which is a monster in its own right. However, I was extremely interested in the patterns that kept popping up. Like the $2e_5P+2^2e_4P+2^3e_3P+2^4e_2P+2^5e_1P+2^6P$ where the power of 2 and the index of $e$ keeps increasing and decreasing with each other. Therefore I wondered if there could be any general formula for any $f(x)$ and $g(x)$.

And that is my question, is there any way to find the answer efficiently without all this unnecessary calculation; and also can this process be generalised to any polynomials $f(x)$ and $g(x)$. And if it cannot be generalised, then is there any sort of algorithm or methodical way that one can take while solving this type of problem?

答:
$g(x)=(x-a)(x-b)$ where $a=-1+\sqrt 3 i, b=-1-\sqrt 3 i$ are cubic roots of 8.

For $n=1, 2, \ldots, 6, a_n-a$ are the roots of $h_1 (x) = f(x+a)$, $a_n-b$ are the roots of $h_2(x)=f(x+b)$.

Recall that $a^3=b^3=8$, then the desired product is the product of $h_1(x)$ and $h_2(x)$'s constant terms

$$
h_1(0) \cdot h_2(0) = (a^6-2a^3-8)\cdot (b^6-2b^3-8) = (8^2-2 \cdot 8 - 8)^2 = 40^2=1600.
$$


答:
Note that $(x^2+2x+4)(x-2)=x^3-8$. So the product of the $g(a_i)$ is $P/Q$, where

$$P=\prod(a_i^3-8)$$
$$Q=\prod(a_i-2)$$

As @mathcounterexamples.net notes, the roots of $f$ are the cube roots of $4$ and $-2$; so $a_i^3-8$ takes the values $-4,-4,-4,-10,-10,-10$. Hence $P=64000$.

And the product of $a_i-2$ taken over all roots of $f(x)$ is the product of $a_i$ taken over all roots of
$$f(x+2)=(x+2)^6-2(x+2)^3-8$$
which is simply the constant term $2^6-2.2^3-8=40$.
So we get $P/Q=1600$.

ContentLicense: CC BY-SA 4.0

Title: Let $f(x)=x^6-2x^3-8$ and $g(x)=x^2+2x+4$. Let $a_1$ through $a_6$ be its roots. Find the value of $\prod_{n=1}^{6} (g(a_n))$

Body:
Question:

Let $f(x)=x^6-2x^3-8$ and $g(x)=x^2+2x+4$. Let $a_1$ through $a_6$ be its roots. Find the value of $\displaystyle \prod_{n=1}^{6} (g(a_n))$

My process:

I first thought of setting $z=a_i^2+2a_i+4$ and then putting the value back into $f(z)$ and then from there finding the product using Vieta's. But the process turned incredibly long and tedious and I found it to be impossible to do by hand.

So I just went to to the brute force method, expanding the whole expression out. Writing all the terms would be really tedious so I would like to abbreviate it. Let $P= a_1 \cdot a_2 \cdots \cdot a_6$ and let $e_n$ denote the sum of roots taken $n$ at a time.

Here is the expression I got:

$$\prod_{i=1}^{6} (a_i^2+2i+4)=P^2+2e_5P+2^2e_4P+2^3e_3P+2^4e_2P+2^5e_1P+2^6P+2^5*4*e_5+2^4*4^2*e^4+2^3*4^3*e_3+2^2*4^4*e_2+2*4^5*e_1+4_6+4(\sum_{cyc}(a_1a_2a_3a_4a_5)^2))^+4^2(\sum_{cyc}(a_1a_2a_3a_4)^2)+4^3\sum_{cyc}(a_1a_2a_3)^2+4^4\sum_{cyc}(a_1a_2)^2+4^5\sum_{cyc}(a_1)^2$$

Which is a monster in its own right. However, I was extremely interested in the patterns that kept popping up. Like the $2e_5P+2^2e_4P+2^3e_3P+2^4e_2P+2^5e_1P+2^6P$ where the power of 2 and the index of $e$ keeps increasing and decreasing with each other. Therefore I wondered if there could be any general formula for any $f(x)$ and $g(x)$.

And that is my question, is there any way to find the answer efficiently without all this unnecessary calculation; and also can this process be generalised to any polynomials $f(x)$ and $g(x)$. And if it cannot be generalised, then is there any sort of algorithm or methodical way that one can take while solving this type of problem?


问:
Maximum value of $\sin(A/2)+\sin(B/2)+\sin(C/2)$?
So I came across a question in my textbook:

In triangle ABC, if $A$,$B$,$C$ represent angles, then find the maximum value of $\sin(A/2)+\sin(B/2)+\sin(C/2)$?

So I have already tried and best and put my blood,sweat and tears into this question..But I'm not able to go solve further!

So here's my approach:
By Using $\sin(C)+\sin(D)$ and $A+B+C= \pi$;

$2\sin(\frac{A+B}{4})\cos(\frac{A-B}{4})+\cos(\frac{A+B}{2})$
Now,
Using $\cos(2A)$ formula i.e, $1-2\sin^2(A)
$

$2\sin(\frac{A+B}{4})\cos(\frac{A-B}{4})+1-2\sin^2(\frac{A+B}{4})$

So I got quadratic in variable $\sin(\frac{A+B}{4})$

$-2\sin^2(\frac{A+B}{4})+2\sin(\frac{A+B}{4})\cos(\frac{A-B}{4})+1$

But I dunno what to do After that

Can I solve this question using this method or I have to use a different approach!

BTW, the answer is 3/2

Edit:I have just finished my high school and preparing for entrance exam IIT-JEE,So please don't use hard terms to solve this question.

This solution is sent by my teacher, atleast make me understand this one
[https://i.stack.imgur.com/51pCB.png]

答:
Where you have stopped,
let $$z=-2\sin^2x+1+2\sin x\cos y$$

$$\iff2\sin^2x-2\sin x\cos y+z-1=0$$

As $\sin x$ is real, the discriminant must be $\ge0$

$\implies8(z-1)\le(-2\cos y)^2\le2^2$

$\implies8z\le4+8$

The equality occurs if $\cos^2y=1\iff\sin y=0$

and consequently $\sin x=\mp\dfrac{\cos y}2=\mp\dfrac12$


问:
Series Summation: $\sum\limits_{k=1}^{N-1}\frac{1}{z-w_k}$ where $w_k=e^{\frac{2\pi i k}{N}}$
I have the series

$$\sum_{k=1}^{N-1}\frac{1}{1-w_k} $$ where $w_k=e^{\frac{2\pi i k}{N}}$, how can I find the summation of this , another question related to this sum

$$\sum_{k=1}^{N-1}\frac{1}{z-w_k}   $$

Body:
Let $P(z)=z^{N}-1$. Then $$P(z)=\prod_{k=0}^{N-1}(z-w_{k})$$ where $w_{0}=1$.
From here, it is easy to check that
$$
\sum_{k=0}^{N-1}\frac{1}{z-w_{k}}=\frac{P^{\prime}(z)}{P(z)}%
$$
hence
$$
\sum_{k=1}^{N-1}\frac{1}{z-w_{k}}=\frac{P^{\prime}(z)}{P(z)}-\frac{1}{z-1}=\frac{Nz^{N-1}}{z^{N}-1}-\frac{1}{z-1}=\frac{(N-1)z^{N}-Nz^{N-1}+1}{(z^{N}-1)(z-1)}\text{.}
$$
In particular, for $z=1$:
$$
S:=\sum_{k=1}^{N-1}\frac{1}{1-w_{k}}=\lim_{z\rightarrow1}\frac{(N-1)z^{N}-Nz^{N-1}+1}{(z^{N}-1)(z-1)}.
$$
In order to compute the limit, we write is as a limit at $0$, i.e., let
$z=w+1$, $w\rightarrow0$, hence
$$
S=\lim_{w\rightarrow0}\frac{(N-1)(w+1)^{N}-N(w+1)^{N-1}+1}{w((w+1)^{N}-1)}.
$$
Next, recall that
$$
\lim_{w\rightarrow0}\frac{(w+1)^{N}-1}{w}=N
$$
hence
$$
S=\frac{1}{N}\cdot\lim_{w\rightarrow0}\frac{(N-1)(w+1)^{N}-N(w+1)^{N-1}+1}{w^{2}}.
$$
The last step is to apply l'Hopital's rule, hence
\begin{align*}
S  & =\frac{1}{N}\cdot\lim_{w\rightarrow0}\frac{\left(  (N-1)(w+1)^{N}-N(w+1)^{N-1}+1\right)  ^{\prime}}{\left(  w^{2}\right)  ^{\prime}}=\frac{1}{N}\cdot\lim
_{w\rightarrow0}\frac{N(N-1)(w+1)^{N-2}w}{2w}\\
& =\frac{1}{N}\cdot\lim_{w\rightarrow0}\frac{N(N-1)}{2}(w+1)^{N-2}=\frac{N-1}{2}.
\end{align*}

ContentLicense: CC BY-SA 3.0


问:
Continuous function on a compact interval is Riemann-integrable
We're talking about the $\Bbb{R^n}$ case here. This is the part of the proof, were we are choosing an appropriate partition so that we can arrive at the Riemann-condition for integrability.

Let $I=[a_1,b_1]\times \dots [a_n,b_n]$. Choose an even partition for $I$, such that $P_\epsilon=\{a_1,a_1+\frac{b_1-a_1}{N},\cdots,b_1\}\times \cdots\{a_n,a_n+\frac{b_n-a_n}{N},\cdots ,b_n\}$, where $N$ is chosen such that $\max_{1\leq j \leq n}\frac{b_j - a_j}{N}<\frac{\delta}{n}$ $\forall x,y\in I_\hat{j}$, and $\hat{j} = (j_1,...,j_n)$. The part I don't understand is this $$\vert\vert \hat{x} -\hat{y} \vert\vert \leq\sum_{j=1}^n\vert x_j-y_j\vert\leq\sum_{j=1}^n\frac{b_j-a_j}{N}<n\frac{\delta}{n}=\delta$$

How do we arrive at the second inequality?

答:
This is only true if the points $\hat{x} = (x_1,\ldots,x_n)$ and $\hat{y} = (y_1,\ldots,y_n)$ are in the same subrectangle, that is there exist $k_1, k_2, \ldots, k_n\in \{1,2,\ldots,n\}$ such that

$$\hat{x}, \hat{y} \in \left[a_1+ \frac{(b_1-a_1)(k_1-1)}{N}, a_1+ \frac{(b_1-a_1)k_1}{N}\right] \times \ldots \times \left[a_n+ \frac{(b_n-a_n)(k_n-1)}{N}, a_n+ \frac{(b_n-a_n)k_n}{N}\right] $$

In that case, both of the components $x_j,y_j$ lie in the same interval of length $\frac{b_j-a_j}{N}$, and we have

$$|x_j - y_j| \leqslant \frac{b_j-a_j}{N}$$


问:
Existence minimal projections compact operators
Consider the following fragment from Davidson's book on $C^*$-algebras:

Can someone explain why the sentence "Clearly, $P$ dominates a projection $E$ in $\mathfrak{A}$ with minimal positive rank." holds?

答:
If that were not the case, you would have an infinite strictly decreasing sequence of subprojections of $P$, and then $P$ would have infinite rank.

Or, if the above is not obvious enough: start with $P$ of rank $n$. Is it minimal? If it is, take $E=P$. If it isn't, let $P_1$ be a proper subprojection of $P$. Then the rank of $P_1$ is at most $n-1$. Is it minimal? If it is, take $E=P_1$. If it isn't, take a proper subprojection $P_2$ of $P_1$. Then the rank of $P_2$ is at most $n-2$. If you repeat this for at most $n$ steps, either you found a minimal subprojection along the way, or you got to $P_n$ of rank 1. Either case, you have a minimal projection in $\mathfrak A$.


问:
Solutions to equation of $6$ variables
Preliminaries: Let $M$ be any $4\times 4$ unitary matrix. Now, define the matrices $A$ and $B$ by

$$A=M^\dagger\left[ \begin{pmatrix}
\cos(\theta) & -e^{i\lambda}\sin(\theta)\\
e^{i\phi}\sin(\theta) & e^{i(\lambda+\phi)}\cos(\theta)
\end{pmatrix}\otimes \begin{pmatrix}
1 & 0\\
0 &1
\end{pmatrix}\right]M$$

$$B=M^\dagger\left[\begin{pmatrix}
1 & 0\\
0 &1
\end{pmatrix}\otimes \begin{pmatrix}
\cos(\theta) & -e^{i\lambda}\sin(\theta)\\
e^{i\phi}\sin(\theta) & e^{i(\lambda+\phi)}\cos(\theta)
\end{pmatrix}\right]M$$

We may then define $f:\mathbb{R}^6\to \mathbb{R}$ by

$$f(\theta,\lambda,\phi,\delta,\alpha,\beta)=\prod_{C\in\{A,B\}}\left(\left|\begin{pmatrix}
0\\
0\\
1\\
0
\end{pmatrix}^TC\begin{pmatrix}
\cos(\delta)e^{i\alpha}\\
\sin(\delta)e^{i\beta}\\\
0\\
0
\end{pmatrix}\right|+\left|\begin{pmatrix}
0\\
0\\
0\\
1
\end{pmatrix}^TC\begin{pmatrix}
\cos(\delta)e^{i\alpha}\\
\sin(\delta)e^{i\beta}\\\
0\\
0
\end{pmatrix}\right|\right)$$

Now, what are the zeros of this function? It is easy to show that any element of the set

$$S=\{(k_1\pi ,\lambda, 2\pi k_2-\lambda,\delta,\alpha,\beta):k_1,k_2\in\mathbb{Z}\text{ and }\lambda,\delta,\alpha,\beta\in\mathbb{R}\}$$

is a solution to $f(\theta,\lambda,\phi,\delta,\alpha,\beta)=0$. This is because

$$\begin{pmatrix}
\cos(\theta) & -e^{i\lambda}\sin(\theta)\\
e^{i\phi}\sin(\theta) & e^{i(\lambda+\phi)}\cos(\theta)
\end{pmatrix}\Bigg|_{\theta=k_1 \pi,\phi=2\pi k_2-\lambda}=(-1)^{k_1}\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}$$

and the matrix $A$ collapses down to

$$A=M^\dagger[(-1)^{k_1}I_4]M=(-1)^{k_1}M^\dagger M=(-1)^{k_1}I_4$$

(where $I_4$ is the $4\times 4$ identity matrix), which in turn means all the inner products are zero.

My question: Is there some $4\times 4$ unitary matrix $M$ such that $S$ is all the solutions to the equation $f(\theta,\lambda,\phi,\delta,\alpha,\beta)=0$?

Motivation: If I could prove this question in the negative (there are no such matrices $M$), then I could show that at least $3$ qubits are required for an error detection code I am developing for a quantum computer. I won't bore you with all the details (unless someone wants to know more), but my problem has simplified to the linear algebra question above.

Work: So far, I have tried a lot of different ways of proving the above without much success. For any set matrix $M$, I can always find solutions that are not in $S$. However, it does seem like almost all the degrees of freedom are required because there are some $M$ where the solution seems to lie on a line in $\mathbb{R}^6$ space (basically it depends on the difference between $\alpha$ and $\beta$). That is, it does not form a family of solutions on a plane/hyper-plane in the space. My current work is to try to use different generators of $U(4)$ and see if I can say something if I decompose $M$ into constituent parts (Here are two different papers I have been using for said generators). Unfortunately, this has not simplified my problem to the point where I can get a satisfactory solution.

答:
Not a full solution, an idea how to simplify the equation.

Vectors $\lvert e_i\rangle=\{(1,0,0,0)^{T}, (0,1,0,0)^{T}, (0,0,1,0)^{T}, (0,0,0,1)^{T}\}$ form a basis. $f$ is zero if for some vector $\lvert a\rangle=e^{i\alpha}\cos{\delta}\lvert e_1\rangle+e^{i\beta}\sin{\delta}\lvert e_2\rangle$:
$$\tag{1}
\langle e_3\rvert A\lvert a\rangle=\langle e_4\rvert A\lvert a\rangle=0 \; or \; \langle e_3\rvert B\lvert a\rangle=\langle e_4\rvert B\lvert a\rangle=0
$$
If we introduce matrix components $A_{ik}=\langle e_i \rvert A\lvert e_k\rangle$, equation (1) can be rewritten as:
$$
A_{31}e^{i\alpha}\cos{\delta}+A_{32}e^{i\beta}\sin{\delta}=A_{41}e^{i\alpha}\cos{\delta}+A_{42}e^{i\beta}\sin{\delta}=0
$$
(or similar equation for $B$, which is omitted); which is satisfied if:
$$
F(\theta,\lambda,\phi)\equiv A_{42}A_{31}-A_{41}A_{32}=0
$$
This is a single complex equation for 3 real parameters: $\theta,\lambda,\phi$. This way we can analyze an equation with just 3 parameters, which should be easier.

Title: Solutions to equation of $6$ variables


问:
Show that the following goes to $f'(0)$
Let $f: (-1,1) \to R$ and $f'(0)$. Suppose that $-1 < a_n < 0 < b_n < 1$ and $a_n \to 0, b_n \to 0$ Show that
$$D_n = \frac{f(b_n) - f(a_n)}{b_n - a_n}$$ goes to $f'(0)$.

This is a specific case where we need to show that $f'(0)$ is the result, not a general point where the derivative is evaluated at.

I'm not quite sure where to start. I'm not sure what $f'(0)$ is, all I'm told is that it exists. So I'm trying to show this quantity goes to that, when I don't even know what $f'(0)$ is equal to. That is throwing me off. I'm also not sure the relation to $f'(0)$ that $D_n$ has. Can anyone clarify the question for me and maybe get me started? Thank you!

答:
Hint

\begin{align*}
\frac{f(b_n)-f(a_n)}{b_n-a_n}&=\underbrace{\frac{b_n}{b_n-a_n}}_{=1+\frac{a_n}{b_n-a_n}}\cdot \frac{f(b_n)-f(0)}{b_n}-\frac{a_n}{b_n-a_n}\cdot \frac{f(a_n)-f(0)}{a_n}\\
&=\frac{f(b_n)-f(0)}{b_n}+\frac{a_n}{b_n-a_n}\left(\frac{f(b_n)-f(0)}{b_n}-\frac{f(a_n)-f(0)}{a_n}\right)
\end{align*}

Since $\left|\frac{a_n}{b_n-a_n}\right|\leq 1$, taking $n\to \infty $ gives the wished result.

Title: Show that the following goes to $f'(0)$


问:
Is it true that the $\lim_{n\to\infty}\frac{a_1b_n+a_2b_{n-1}+\ldots+a_nb_1}{n}=ab$?
I came across a question which went like this:

If $a_n\to a$ and $b_n\to b$, then prove or disprove the following: $$\lim_{n\to\infty}\frac{a_1b_n+a_2b_{n-1}+\ldots+a_nb_1}{n}=ab$$

Here's how I approached the problem (I tried to prove it cause I thought that it is true):

We already have $a_n\rightarrow a$ and $b_n\rightarrow b$. Thus,

$$\begin{align}\left|\frac{\sum_{i=1}^na_ib_{n-i+1}}{n}-ab\right|&=\left|\frac{\sum_{i=1}^na_ib_{n-i+1}-nab}{n}\right|\\
&=\left|\frac{\sum_{i=0}^n\left(a_ib_{n-i+1}-ab\right)}{n}\right|\\
&=\frac{1}{n}\left|\sum_{i=1}^n\left(a_ib_{n-i+1}-ab\right)\right|\\
&\leq\frac{1}{n}\sum_{i=1}^n\left|a_ib_{n-i+1}-ab\right|\\
&\leq\frac{1}{n}\sum_{i=1}^n\left|a_i\right|\left|b_{n-i+1}-b\right|+\frac{b}{n}\sum_{i=1}^n\left|a_i-a\right|\end{align}$$

Now as $a_n$ is convergent$\implies\left|a_n\right|\leq M$ for some $M\in\mathbb{R}$. Thus,
$$\left|\frac{\sum_{i=1}^na_ib_{n-i+1}}{n}-ab\right|\leq\frac{M}{n}\sum_{i=1}^n\left|b_i-b\right|+\frac{b}{n}\sum_{i=1}^n\left|a_i-a\right|$$.

Now I am stuck as to how can I show that this whole expression is $<\epsilon$. Any hints as well as alternative methods are totally welcome.

NOTE: This might be a similar question only that it is a special case with $a=0$.

答:
I'm going to assume $a,b$ are finite.

Let $N>0$ such that $|b_n-b|<\epsilon$ and $|a_n-a|<\epsilon$ for any $n>N$. Then for $n>N$ we have $\sum_{i=1}^n|b_i-b|\leq B_N+\epsilon(n-N)$ and $\sum_{i=1}^n|a_i-a|\leq a_N +\epsilon(n-N)$ where here $A_N=\sum_{i=1}^N|a_i-a|$ while $B_N=\sum_{i=1}^N|b_i-b|$. This implies for $n>N$ that $$\Bigg|\frac{\sum_{i=1}^na_ib_{n-i+1}}{n}-ab\Bigg|\leq \frac{M}{n}\Bigg[B_N+\epsilon(n-N)\Bigg]+\frac{b}{n}\Bigg[A_N+\epsilon(n-N)\Bigg]$$ Take $n \longrightarrow \infty$ to see how $$\lim_{n\rightarrow \infty}\Bigg|\frac{\sum_{i=1}^na_ib_{n-i+1}}{n}-ab\Bigg|\leq\epsilon(M+b)$$


问:
equality between the index between field with $p^{n}$ elements and $ \mathbb{F}_{p}$ and n?
can someone explain this?

$ \left[\mathbb{F}_{p^{n}}:\mathbb{F}_{p}\right]=n $

答:
$$|\Bbb F_p|=p\;,\;\;|\Bbb F_n^n|=p^n$$

amd since any element in the latter is a unique linear combination of some elements of it and scalars from the former, it must be that those some elements are exactly $\,n\,$ in number.


答:
Let $M> 0$ be such that $|a_i|, |b_i| \leq M$ for every $i$.

Given $\varepsilon > 0$, let $N\in\mathbb{N}$ be such that
$$
|a_i - a| < \varepsilon,\quad
|b_i - b| < \varepsilon, \qquad \forall i \geq N.
$$
Let $n > 2N$. We have that
$$
\sum_{i=1}^{n} a_i b_{n+1-i} =
\left( \sum_{i=1}^N + \sum_{i=n-N+1}^n\right) a_i b_{n+1-i}
+ \sum_{i=N+1}^{n-N} a_i b_{n+1-i}\,.
$$
The first term at r.h.s. is bounded by $2 N M^2$.
Moreover, for $i = N+1,\ldots, n-N$ we have that
$$
|a_i b_{n+1-i} - ab| \leq M|a_i - a| + M |b_i - b| \leq 2 M \varepsilon,
$$
hence the second term is bounded by $2 n M \epsilon$.
All in all, we get that
$$
\left|\frac{1}{n} \sum_{i=1}^{n} (a_i b_{n+1-i} - ab)\right|
\leq \frac{4 N M^2}{n} + 2 M \varepsilon,
$$
so that
$$
\limsup_{n\to +\infty}
\left|\frac{1}{n} \sum_{i=1}^{n} (a_i b_{n+1-i} - ab)\right| \leq 2M \varepsilon
$$
and the required convergence follows.

Title: Is it true that the $\lim_{n\to\infty}\frac{a_1b_n+a_2b_{n-1}+\ldots+a_nb_1}{n}=ab$?

Body:
I came across a question which went like this:

If $a_n\to a$ and $b_n\to b$, then prove or disprove the following: $$\lim_{n\to\infty}\frac{a_1b_n+a_2b_{n-1}+\ldots+a_nb_1}{n}=ab$$

Here's how I approached the problem (I tried to prove it cause I thought that it is true):

We already have $a_n\rightarrow a$ and $b_n\rightarrow b$. Thus,

$$\begin{align}\left|\frac{\sum_{i=1}^na_ib_{n-i+1}}{n}-ab\right|&=\left|\frac{\sum_{i=1}^na_ib_{n-i+1}-nab}{n}\right|\\
&=\left|\frac{\sum_{i=0}^n\left(a_ib_{n-i+1}-ab\right)}{n}\right|\\
&=\frac{1}{n}\left|\sum_{i=1}^n\left(a_ib_{n-i+1}-ab\right)\right|\\
&\leq\frac{1}{n}\sum_{i=1}^n\left|a_ib_{n-i+1}-ab\right|\\
&\leq\frac{1}{n}\sum_{i=1}^n\left|a_i\right|\left|b_{n-i+1}-b\right|+\frac{b}{n}\sum_{i=1}^n\left|a_i-a\right|\end{align}$$

Now as $a_n$ is convergent$\implies\left|a_n\right|\leq M$ for some $M\in\mathbb{R}$. Thus,
$$\left|\frac{\sum_{i=1}^na_ib_{n-i+1}}{n}-ab\right|\leq\frac{M}{n}\sum_{i=1}^n\left|b_i-b\right|+\frac{b}{n}\sum_{i=1}^n\left|a_i-a\right|$$.

Now I am stuck as to how can I show that this whole expression is $<\epsilon$. Any hints as well as alternative methods are totally welcome.

NOTE: This might be a similar question only that it is a special case with $a=0$.


Title: Show that the following goes to $f'(0)$

Body:
Let $f: (-1,1) \to R$ and $f'(0)$. Suppose that $-1 < a_n < 0 < b_n < 1$ and $a_n \to 0, b_n \to 0$ Show that
$$D_n = \frac{f(b_n) - f(a_n)}{b_n - a_n}$$ goes to $f'(0)$.

This is a specific case where we need to show that $f'(0)$ is the result, not a general point where the derivative is evaluated at.

I'm not quite sure where to start. I'm not sure what $f'(0)$ is, all I'm told is that it exists. So I'm trying to show this quantity goes to that, when I don't even know what $f'(0)$ is equal to. That is throwing me off. I'm also not sure the relation to $f'(0)$ that $D_n$ has. Can anyone clarify the question for me and maybe get me started? Thank you!

Body:
First write
$$
D_n  - f'(0) = \frac{{b_n }}{{b_n  - a_n }}\left( {\frac{{f(b_n ) - f(0)}}{{b_n }} - f'(0)} \right) \\ + \left( {1 - \frac{{b_n }}{{b_n  - a_n }}} \right)\left( {\frac{{f(a_n ) - f(0)}}{{a_n }} - f'(0)} \right).
$$
Using the fact that
$$
0 < \frac{{b_n }}{{b_n  - a_n }} < 1.
$$
we obtain
$$
\left| {D_n  - f'(0)} \right| \le \left| {\frac{{f(b_n ) - f(0)}}{{b_n }} - f'(0)} \right| + \left| {\frac{{f(a_n ) - f(0)}}{{a_n }} - f'(0)} \right|.
$$
Thus,
\begin{align*} &
\mathop {\lim }\limits_{n \to  + \infty } \left| {D_n  - f'(0)} \right| \le \mathop {\lim }\limits_{n \to  + \infty } \left| {\frac{{f(b_n ) - f(0)}}{{b_n }} - f'(0)} \right| + \mathop {\lim }\limits_{n \to  + \infty } \left| {\frac{{f(a_n ) - f(0)}}{{a_n }} - f'(0)} \right|
\\ &
 \le \left| {\mathop {\lim }\limits_{n \to  + \infty } \frac{{f(b_n ) - f(0)}}{{b_n }} - f'(0)} \right| + \left| {\mathop {\lim }\limits_{n \to  + \infty } \frac{{f(a_n ) - f(0)}}{{a_n }} - f'(0)} \right|
\\ &
 = \left| {f'(0) - f'(0)} \right| + \left| {f'(0) - f'(0)} \right| = 0.
\end{align*}


问:
Maximizing revenue for ticket price
A rail company, QNRail, is selling tickets for a special trip from Toronto to Winnipeg.
The train’s maximum capacity is 200 passengers. All tickets are sold at the same price p in Canadian
dollars. Based on market research, they know that the quantity q of tickets demanded depends on
the price p charged per ticket, according to the formula below.

q = (300 − 2p if p < 100

250 −
3/2
p if p ≥ 100)

In order to maximize their revenue for this trip, how much should QNRail charge per ticket? Please
show all of your reasoning! Round your final answer to the nearest cent.

答:
I speak about the first equation, the principle is the same for the second one.

$300 − 2p$ represents the number of tickets sold at a certain price.

So $(300 − 2p) * p$ represents the revenue.

From the derivative $300 - 4p$ you can see the maximum at $p = 75$

Putting this value in the function $(300 − 2p) * p$ you find the maximum revenue for a price $p < 100$

Do the same for the other case and compare the results.

Title: Maximizing revenue for ticket price


问:
subprojections in $C^*$-algebras
If $p,q$ are two projections in a $C^*$-algebra such that $p\leq q$, we have $p$ is subordinate to $q$. But the converse may not be true.

Question1.Can we add some additional conditions such that the following conclusion holds:
$p$ is subordinate to $q$, then $p\leq q$.

答:
Yes. Your condition is equivalent to the algebra being commutative, as long as the algebra is generated by projections (which every von Neumann algebra is).

If the algebra is generated by projections, given any unitary $u$ and any projection $p$, the projection $upu^*$ is equivalent to $p$, so $upu^*\leq p$. As you can do the same with $u^*$, you get that $upu^*=p$ for all unitaries and all projections, so the algebra is commutative.

For general C$^*$-algebras (i.e., non-von Neumann) the question is kind of meaningless. For instance take $B$ to be any projectionless C$^*$-algebra, like $C_0(\mathbb R)$ if you want commutative, or $C_0(\mathbb R)\otimes K(\ell^2(\mathbb N))$ if you want  non-commutative. Then your property holds (as there are no projections where it could fail) but you don't get anything out of it. Same is true with projectionless unital C$^*$-algebras, like $C_r^*(\mathbb F_2)$.

Title: subprojections in $C^*$-algebras


问:
Find corresponding digit string lengths between radices?
Each digit of base the base-16 representation of a number corresponds exactly to 4 digits in the base-2 representation, which allows for trivial conversion between these bases.

Other combinations of bases are less straightforward, but still permit short-cutting the generic remainder-and-divide algorithm for converting bases. E.g., 3 digits of base-16 correspond to 4 digits of base 8, and 3 digits of base-256 correspond to 4 digits of base-64 (hence the motivation for the base-64 encoding standard for binary data).

These sorts of relations are all fairly easily discoverable by comparing them to base-2 as an intermediary. But is there a more generic method of discovering whether or not a digit ratio exists between two radices, and if so, what it is?

答:
Yes!

Digits can be aligned between representations in different bases if there is a solution in integers for $x$ and $y$ in the equation $r_1^x = r_2^y$, where $r_1$ and $r_2$ are the radices of each base. This is equivalent to saying that an $x$-digit number in base $r_1$ can express the same range of values as a $y$-digit number is base $r_2$.

This in turn has a solution when each radix can be written as a power of a common base; i.e., $r_1 = c^w$ and $r_2 = c^z$, so $r_1^x = c^{wx} = c^{zy} = r_2^y$

The values of $x$ and $y$ then tell you how many digits in each base are needed to align with a block of digits in the other base.

The common base for 8, 16, 64, and 32 is base 2, which is why these can be converted between by lining up their bit patterns. However, a similar relationship holds between, e.g., base-27 and base-9 (where 2 base-27 digits correspond to 3 base-9 digits), which can be seen by aligning trit (base-3 digit) patterns.


问:
Laurent series centered at $z_0=1$ with $1<|z-1|<\infty$
There is a function $f(z)=\dfrac{1}{z(z-1)}$, and it can be expanded as $\dfrac{1}{z-1}=\dfrac{1}{z}\cdot \dfrac{1}{1-\frac{1}{z}}$. Furthermore, $\dfrac{1}{1-\frac{1}{z}}=1+\dfrac{1}{z}+\dfrac{1}{z^2}...$ I substituted $w=z-1$ and expanded to get my Laurent series: $\dfrac{1}{(z-1)^2}+\dfrac{1}{(z-1)^3}+\cdots$. Did I approach this correctly?

答:
When the function is rational, the standard procedure is to start with finding its expansion into partial fractions. In the present case, you get $$f(z)=\dfrac{1}{z(z-1)}=\dfrac{1}{z-1}-\dfrac{1}{z}.$$ Hence, making use of your substitution $w=z-1$ (or $z=1+w$), we obtain

\begin{align}
\dfrac{1}{z-1}-\dfrac{1}{z}&=\frac{1}{w}-\frac{1}{w+1}=\frac{1}{w}-\frac{1}{w}\frac{1}{1-(-\dfrac{1}{w} ) }\\&=\frac{1}{z-1}-
\frac{1}{z-1}\frac{1}{1-(-\dfrac{1}{z-1})}.\tag 1
\end{align}

And so, you just need to expand $\dfrac{1}{1-( -\dfrac{1}{w}
) }=\dfrac{1}{1-( -\dfrac{1}{z-1}) }$
for $|w|=|z-1|>1$:
$$\frac{1}{1-(-\dfrac{1}{z-1})}=\sum_{n\geq 0}\frac{(-1)^{n}}{(z-1)^{n}}=1+\sum_{n\geq 1}\frac{(-1)^{n}}{(z-1)^{n}}. \tag 2 $$
Please notice that this is an alternating series.

Therefore, combining $(1)$ and $(2)$, for $|z-1|>1$ the original function may be expanded into

\begin{align}
f(z)&=\frac{1}{z-1}-\frac{1}{z-1}-\frac{1}{z-1}\sum_{n\geq 1}\frac{(-1)^{n}}{\left( z-1\right) ^{n}} \tag 3\\&=\sum_{n\geq 1}\frac{(-1)^{n+1}}{\left( z-1\right) ^{n+1}}=\sum_{n\geq 2}\frac{(-1)^{n}}{\left( z-1\right) ^{n}}\\&=\frac{1}{(z-1)^2}-\frac{1}{(z-1)^3}\pm\cdots\tag{4}
\end{align}


问:
Solving limits involving factorials
Generally speaking, how does one handle a limit involving a factorial? Take, for example, the limit:

$$\lim_{n\to\infty} \frac{x^{n+1}}{(n+1)!}$$

According to WolframAlpha, this limit evaluates to $0$, which seems reasonable. How can this be determined?

答:
$x^n$ is a $o( n! )$ (little o) which means that exponentials grow much slower than factorials.

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

Title: Solving limits involving factorials


问:
are two random variables independent if their joint distribution can be written as the product of their univariate functions?
I know that, for random variables $X$ and $Y$, if $F_{X,Y}(x,y)$ is their joint distribution function, then if $F_{X,Y}(x,y) = F_X(x)F_Y(y)$, then $X$ and $Y$ are independent.

But what if I am just given that $F_{X,Y}(x,y) = G(x)H(y)$, where $G$ and $H$ are arbitrary univariate functions of $x$ and $y$? In this case, is $X$ and $Y$ still independent from each other? If yes, why is this? and if no, what is the counter example?

Thank you,

答:
The marginal of $X$ is $F_X=\int_{-\infty}^\infty Fdy=G\int_{-\infty}^\infty H~dy$.

The marginal of $Y$ is $F_Y=\int_{-\infty}^\infty Fdx=H\int_{-\infty}^\infty G~dx$.

Now$$\begin{align*}F_XF_Y&=GH\left(\int_{-\infty}^\infty G~dx\right)\left(\int_{-\infty}^\infty H~dy\right)\\&=GH\left(\int_{-\infty}^\infty\int_{-\infty}^\infty GH~dx~dy\right)\\&=GH\int\int F~dx~dy\\&=GH\\&=F\end{align*}$$since $F$ is a PDF and integrates to $1$.

Note that $F_X(x)=kG(x)$ and $F_Y=\frac1kH(y)$ where $k=\int_{-\infty}^\infty H~dy$ and $\frac1k=\int_{-\infty}^\infty G~dx$. You can extend this to discrete distributions.


问:
Regularity of balanced binary strings

How can one tell which number of propositional variables is necessary
  to express a Boolean function given as a sequence of 0s and 1s (a
  binary string) of length $2^n$ as a Boolean formula?

The extremal cases are clear: if no 0s or no 1s are present, we need no variables at all, if there is exactly one 0 or one 1, we need all $n$ variables. I am especially interested in the case that there are equally many 0s and 1s (balanced binary strings). This is the only case, where one variable may suffice – but also all may be necessary.

I start from the assumption that it's a matter of regularity or complexity of the binary string.

Assign to each binary string $\sigma$ of length $2^n$ recursively a sequence of symbols:

a + when $\sigma = xx$

a - when $\sigma = x\overline{x}$  with $\overline{x}$ the "negation" of $x$.

an S otherwise (S for stop)

Proceed with $x$ in the first two cases.

One needs exactly one variable if the sequence has length $n$ and contains only +'s with one single -. This corresponds to maximal regularity.

Conjecture: One needs $n - n_+$ variables, $n_+$ the number of +'s in the sequence.

Is this approach of a "recursively defined regularity" sensible, and if so: under which name is it already known, eventually?

答:
The first rule corresponds to checking if the first variable is needed in the formula or not... but not whether any of the others might be redundant.

As a simple counter-example, consider the function $f(A,B,C,D)=AB \vee \bar{A}C$. The corresponding binary string should be 0011001100001111, which is neither of the $xx$, nor of the $x\bar{x}$ form (so $n_{+}=n_{-}=0$). Yet, only three variables are needed to represent it.

Of course, the rules can be amended to check for additional types of redundancies (e.g. checking if every digit is doubled corresponds to checking if the last variable is relevant), but doing so pretty much boils the problem down to a tautology -- if the function value does not depend on a variable, we don't need to use that variable :-)

Title: Regularity of balanced binary strings


问:
Chebyshev's Estimates for residue classes: Is my solution correct?
I want to show that:

$$\psi(x;q,a) \asymp x$$

i.e. I want to show that $\psi(x; q, a) \ll x$ and $\psi(x; q, a) \gg x$

$\psi(x; q, a)$ is the Chebyschev's Second function in residue class a mod q.

In other words, $$\psi(x, q, a) = \sum_{n \leq x}\Lambda(n).\mathbf{1}_{q,a}(n)$$

where, $$\mathbf{1}_{q,a}(n) = \begin{cases}
       1 &\quad\ n = a \text{ mod }q \\
       0 &\quad\ n \neq a \text{ mod }q \\
     \end{cases}$$

and

$\Lambda$ is the von mangoldt function.

On pages 83-84 of the notes: Notes

$\psi(x) \asymp x$ has been proven. Therefore we know $\psi(x) \ll x$ and $\psi(x) \gg x$

Now,

it is simple to see that:

$$\psi(x) - \psi(x; q, a) \geq 0 \implies \psi(x; q, a) \leq \psi(x) \ll x \implies \psi(x; q, a) \ll x$$

Now, I do not know how to show $\psi(x; q, a) \gg x$

I have tried hard to solve this but I am unable to.

Any pointer towards a possible direction would be very helpful.

Thank you.

答:
I believe that this lower bound is quite a bit more difficult than the corresponding lower bound for $\psi(x)$ itself; by way of analogy, both lower bounds imply that there are infinitely many primes being counted, but it is far harder to prove that there are infinitely many primes congruent to $a$ (mod $q$) than just to prove that there are infinitely many primes.

Unless I'm missing something, I don't think there's a substantially easier way to prove $\psi(x;q,a) \gg x$ then to prove Dirichlet's theorem (on the Dirichlet density of such primes)—for example, equation (6.8) on page 181 of the (excellent) notes you linked—and then use partial summation starting from $\psi(x;q,a) = \int_1^x t \,dS_{a,q}(t)$.


问:
Finding the maximum of $G(x) = \int_{x}^{x+a} f(t) \,dt$ using FTOC.
Let's discuss with a simple example: $f(t) = -t^2.$

Given $a>0,$ define $G(x) = \int_{x}^{x+a} -t^2 \,dt$.

I'm interested in how to find the value of $x$ that maximises $G(x)$.

The direct method is:

$G(x) = \left[ -\frac13 t^3 \right]_x^{x+a} = -\frac13(x+a)^3 - (-\frac13x^3) = \ ...\ = -a\left[\left(x+\frac{a}{2}\right)^2 + \frac{a^2}{12} \right],$

which is maximised when $x = -\frac{a}{2},$ as expected.

But this method was only easy because I chose $f(t) = -t^2$. In general, it would appear we would have to use numerical methods to optimise $G(x),$ in particular if $G(x)$ is not easily integrable.

However, is there some trick we can use with the FTOC, that tells us the value of $x$ that maximises $G(x)$?

For example, $G(x) = \int_{x}^{x+a} -t^2 \,dt$ achieves a maximum when

$\frac{d}{dx}\left(\int_{x}^{x+a} -t^2 \,dt\right) = 0 \implies \frac{d}{dx}\left(\int_{q}^{x+a} -t^2 \,dt - \int_{q}^{x} -t^2 \,dt\right) = 0 \implies \frac{d}{dx}\left(\int_{q}^{x+a} -t^2 \,dt\right) - \frac{d}{dx}\left(\int_{q}^{x} -t^2 \,dt\right) = 0,$

where q is a sufficiently large negative constant. Now, the second term simplifies to $-x^2$ by FTOC. But I'm not certain if/how the first term reduces, or if there is some other trick I am missing here, which can also be applied for more difficult functions $f(t)$?

答:
If your $f$ is continuous then your $G$ is differentiable. Then applying the Leibnitz rule of differentiating under integral you calculate $G'(x)$ equate it to zero and then use the double derivative test to find out the point of extrema. Given your the function, we get
$$ G'(x)=0+(-(x+a)^2)-(-x^2)= -(x+a)^2+x^2= -a^2-2ax=0 \implies x=\dfrac{-a}{2}$$


问:
How many ways to pick 3 balls from 50 of 5 colors? (nPr nCr)
Confused with the following counting problem:

How many ways can I pick $3$ balls from a bag of $50$ balls of $5$ colors ($10$ of each color) using $nCr$ or $nPr$?

So far, my thinking is that I can pick...

All the same color $\binom51$

All different colors $\binom 53$

$2$ same $1$ different $\binom52\binom31$

To account for all these possibilities, I will add the probabilities together to make

$$\binom51+\binom53+\binom52\binom31$$

My question here is, well, first of all am I approaching this the right way? Further, do I need to multiply that sum by $50P3$ because of the number of balls that are available? Alternatively, would it be sufficient to simply solve by $4\cdot\binom{10}1^4$? I doubt this, but then again I have a way of overcomplicating things.

Much appreciated. PS answer must be achieved with nCr or nPr.

答:
Find number of selections of $3$ balls from a bag of $50$ balls of $5$ colors ($10$ of each color). Considering balls of the same color identical,

We need all possible combinations of $3$ balls from balls of $5$ different colors. Using stars and bars method, your answer is simply

$\displaystyle {3 + 5 - 1 \choose 5 - 1} = {7 \choose 4} = 35$

The way you did, it would be $\displaystyle \binom51+ 2 \, \binom52 + \binom53 = 35$


问:
tetration limit tending to 0
Let $x \uparrow \uparrow  n = x^{x^{x^{x}}}$ ex: $x \uparrow \uparrow 3 = x^{x^x}$

I'm trying to evaluate the limit

$$ \lim_{n \rightarrow \infty} \frac{3 \uparrow \uparrow \frac{1}{n} - 1 }{2 \uparrow \uparrow  \frac{1}{n} - 1 }$$

Obviously L'hopitals rule has some value here, but once applied i am stuck attempting to differentiate the tetration function and i'm not sure how to go about doing this.

This problem originates from attempting to find a homomorphism for exponentation. That is a complex function $g(x)$ such that $g(x^y) = g(x)^{g(y)}$

答:
If $a\uparrow\uparrow(1/n)$ means the $n$th superroot of $a$, usually rendered $\sqrt[n]a_s$, then for $a\ge 1$ there are two possible behaviors as $n\to\infty$:

If $1\le a\le e$ there is a unique base $s$, with $1\le s\le\exp(1/e)$, for which $s\uparrow\uparrow n$ converges to the selected value of $a$, namely $s=a^{1/a}$.

If $a>e$, then the superroot is always greater than $e^{1/e}$ but the tetrations of any fixed number greater than $e^{1/e}$ will blow up past the target value of $a$.  Therefore the $n$th superroot of $a$ is forced to converge to $e^{1/e}$ as $n\to\infty$.

So in the above example with $3>e$ but $2<e$, the limit is simply

$\dfrac{e^{1/e}-1}{\sqrt2-1}=(e^{1/e}-1)(\sqrt2+1)$.

Note that neither term of the fraction goes to zero because superroots do not in general converge to $1$.  Therefore the form is not indeterminate.

Title: tetration limit tending to 0


问:
Infinity norm of identity function
I would like to use Holder's extremal equality

$$
\|x||_{\infty} = \sup_{f \in L^{1}, \|f\|= 1}\int_{\mathbb{R}} |xf(x)|
$$
to compute the right side of this equation.

Cleary $\|x||_{\infty} = b$ when defined on a bounded interval $[a,b]$ but on all of $\mathbb{R}$ this is not defined.

答: The answer is $+\infty$. The function $f(x)$ equal to $1$ on $[n, n+1]$ and $0$ elsewhere is $L_1$, and the integral is at least $n$ for that function.

Title: Infinity norm of identity function


问:
Schatten norm when the product of the support projections is 0
I'm reading a paper that uses the following result: let $A \geq 0$ and $B \geq 0$ be elements of $B(H)$ ($H$ is a Hilbert space, and $B(H)$ is the space of bounded operators on $H$). If $s(A)$ and $s(B)$ are the support projections of $A$ and $B$, respectively, and $s(A)s(B) = 0$, then for every $1 < p < \infty$ we have
$$\|A + B\|_p^p = \|A\|_p^p + \|B\|_p^p$$
where $\|\cdot\|_p$ is the Schatten p-norm.

Can someone give a proof of this result?

答:
If either $A$ or $B$ is not compact, both sides of your equality are infinity and so the equality holds.

Assume $A,B$ are compact. By the Spectral Theorem
$$
A=\sum_n\alpha_n\,P_n,\qquad B=\sum_m\beta_m\,Q_n,
$$
where $P_1,P_2,\ldots$ are pairwise orthogonal and same for the $Q_m$. By removing terms if necessary, we may assume that $\alpha_n>0$ for all $n$, $\beta_m>0$ for all $m$. Since
$$
(A^*A)^{1/2}=\sum_n \alpha_n \,P_n,\qquad (B^*B)^{1/2}=\sum_m \beta_m \,Q_n
$$
we have that
$$
\sigma_n(A)= \alpha_n ,\qquad\sigma_m(B)= \beta_m .
$$
Also,
$$
s(A)=\sum_nP_n,\qquad s(B)=\sum_mQ_m.
$$
Thus the condition $s(A)s(B)=0$ translates to $P_nQ_m=0$ for all $n,m$. This implies $AB=BA=0$.
Then
$$
|A+B|=A+B=\sum_n\alpha_nP_n+\sum_m\beta_mQ_m
$$
and so the singular values of $A+B$ are $\{\alpha_n\}\cup\{\beta_m\}$. Finally,
$$
\|A+B\|_p^p=\sum_n\alpha_n^p+\sum_m\beta_m^p=\|A\|_p^p+\|B\|_p^p.
$$


答:
Hint :
If $$ \{a_n\} $$ is a sequence of positive real numbers and $$\lim_{n \to \infty} \frac {a_{n+1}}{a_n} = l. $$ where $$ 0 \lt l \lt 1$$ then $$ \lim a_n = 0$$.

Here $$a_n= \frac{|x|^{(n+1)}}{(n+1)!}.$$

Title: Solving limits involving factorials

Body:
Generally speaking, how does one handle a limit involving a factorial? Take, for example, the limit:

$$\lim_{n\to\infty} \frac{x^{n+1}}{(n+1)!}$$

According to WolframAlpha, this limit evaluates to $0$, which seems reasonable. How can this be determined?


问:
Converting equations to polar form
Convert the following rectangular equations to polar form:

\begin{align}\text{(i)  } y&=−x,\; y≥0 \\\text{(ii)  } 0&=x^2+6x+y^2−4y+12\end{align}

答:
I presume that you know that $x= r \cos(\theta)$ and $y= r \sin(\theta)$.

$y= -x$ so $r \sin(\theta)= r \cos(\theta)$ which reduces to $\frac{\sin(\theta)}{\cos(\theta)}= \tan(\theta)= 1$  That is true for $\theta= \frac{\pi}{4}$ or $\theta= \frac{3\pi}{4}$.  The condition that y> 0 means that $r \sin(\theta)> 0$.  Since r is never negative, $\sin(\theta)> 0$ which means $0< \theta< \pi$ so $\theta= \frac{\pi}{4}$.

$x^2+ 6x+ y^2- 4y= 12= 0$ becomes $r^2 \cos^2(\theta)+ 6r \cos(\theta)+ r^2 \sin^2(\theta)- 4r \sin(\theta)+ 12= r^2+ 6r \cos(\theta)- 4r \sin(\theta)+ 12= 0$.

Title: Converting equations to polar form


问:
Projection of linear combination of real numbers
Suppose a real number $\alpha$ is a rational linear combination of two linearly independent real numbers, such as $\pi$ and $\sqrt 3$. For instance, say $$\alpha = \tfrac12\pi + 5\sqrt 3.$$

By a simple linear algebra argument ($\{\pi,\sqrt3\}$ is a $\mathbb Q$-basis), it's clear that the coefficients of $\alpha$ are uniquely determined.

Question: Given $\alpha\in\operatorname{span}_\mathbb Q(\{\pi,\sqrt 3\})$, how do we determine the coefficients of $\pi$ and $\sqrt 3$?

It feels like there is enough information, but at the same time, I imagine that we can find different coefficients which give us numbers arbitrarily close to $\alpha$, so it is important that we have an exact expression for $\alpha$.

(In other words, if we only know that $\alpha\approx 10.23105$, say, I imagine it would be hopeless to expect that we can determine the coefficients, even if we restrict our view to the module $\operatorname{span}_{\mathbb Z}(\{\pi,\sqrt 3\})$.)

The reason I ask this question is because for each $n$, the integral $$I(n)=\int_0^{\pi/3}\sin^{2n}x\,dx$$ is a number of the form $a\pi+b\sqrt 3$. It's easy to obtain the recurrence relation
$$I(n) = \begin{cases} \hfil\frac\pi3\hfil & \text{if $n=0$}\\[3pt]
(1-\tfrac1{2n})\,I(n-1)-\frac1{4n}\big(\tfrac{\sqrt3}2\big)^{2n-1} & \text{otherwise}
\end{cases}$$
for $I(n)$, but I want to be able to compute the coefficients of $\pi$ and $\sqrt 3$ separately (in exact form) using a computer program.

I appreciate any help with solving either of the two problems, I'm assuming a solution to one necessarily sheds light on the other, that's why I haven't posted these as two separate questions.

答:
Using Mathematica

```<code>RSolve[{i[n] == (1 - 1/(2 n)) i[n - 1] -
    1/(4 n) (Sqrt[3]/2)^(2 n - 1), i[0] == Pi/3}, i[n], n]
</code>```

$$I_n=\frac{3^{n+\frac{1}{2}} \, _2F_1\left(1,n+1;n+\frac{3}{2};\frac{3}{4}\right)}{2^{2 n+2}(2 n+1) }$$
Where $_2F_1$ is the  hypergeometric function and has the series expansion

$$_2F_1(a,b;c;z)=\sum _{k=0}^{\infty } \frac{a_k b_k z^k}{k! c_k}$$

$a_k,b_k,c_k$ are the Pochhammer symbol

In the table below the first exact values of the integrals

$$
\begin{array}{c|r}
 n & I_n\\
\hline
 0 & \frac{\pi }{3} \\
 1 & \frac{4 \sqrt{3} \pi -9}{24 \sqrt{3}} \\
 2 & \frac{8 \sqrt{3} \pi -27}{64 \sqrt{3}} \\
 3 & \frac{20 \sqrt{3} \pi -81}{192 \sqrt{3}} \\
 4 & \frac{560 \sqrt{3} \pi -2511}{6144 \sqrt{3}} \\
 5 & \frac{\sqrt{3} \left(560 \sqrt{3} \pi -2673\right)}{20480} \\
 6 & \frac{7 \left(440 \sqrt{3} \pi -2187\right)}{40960 \sqrt{3}} \\
\ldots & \ldots\\
\end{array}
$$

Body:
Using Mathematica

RSolve[{i[n] == (1 - 1/(2 n)) i[n - 1] -
    1/(4 n) (Sqrt[3]/2)^(2 n - 1), i[0] == Pi/3}, i[n], n]

$$I_n=\frac{3^{n+\frac{1}{2}} \, _2F_1\left(1,n+1;n+\frac{3}{2};\frac{3}{4}\right)}{2^{2 n+2}(2 n+1) }$$
Where $_2F_1$ is the  hypergeometric function and has the series expansion

$$_2F_1(a,b;c;z)=\sum _{k=0}^{\infty } \frac{a_k b_k z^k}{k! c_k}$$

$a_k,b_k,c_k$ are the Pochhammer symbol

In the table below the first exact values of the integrals

$$
\begin{array}{c|r}
 n & I_n\\
\hline
 0 & \frac{\pi }{3} \\
 1 & \frac{4 \sqrt{3} \pi -9}{24 \sqrt{3}} \\
 2 & \frac{8 \sqrt{3} \pi -27}{64 \sqrt{3}} \\
 3 & \frac{20 \sqrt{3} \pi -81}{192 \sqrt{3}} \\
 4 & \frac{560 \sqrt{3} \pi -2511}{6144 \sqrt{3}} \\
 5 & \frac{\sqrt{3} \left(560 \sqrt{3} \pi -2673\right)}{20480} \\
 6 & \frac{7 \left(440 \sqrt{3} \pi -2187\right)}{40960 \sqrt{3}} \\
\ldots & \ldots\\
\end{array}
$$


问:
regarding the gradient of a smooth vectorial function
Proof that if $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ is a smooth function,  then d$f_p = D f(p)$, under the natural identification $T_a\mathbb{R}^n \backsimeq
 \mathbb{R}^n$.

I was considering that taking d$f_p : T_p (\mathbb{R}^n) \rightarrow T_p(\mathbb{R}^m$, $L_n(h): \mathbb{R}^n \rightarrow T_p(\mathbb{R}^m)$, where it is the isomorphism between vector spaces such that for $h \in \mathbb{R}^n$, $L_n(h) := D(\cdot)(p)\cdot h$, I mean, $L_n(h)$ is the derivation that to each smooth function $v:U \rightarrow \mathbb{R}$ (remember that U is a neighborhood of $p$), $L_n (h)v = Dv (p) \cdot h$, (where $Dv(p)\cdot h$ is the gradient of $v$ in $p$).
So, I'm proving that

$$L_m \circ Df(p) = \text{d}f_p \circ L_n$$

But I got stucked with the way that I need to define this compositions, I mean, tp proof the equality of the equation above and get what we need, thatd$f_p = D f(p)$.
If someone can help me to describe more or how can I achieve it...
Thanks

答:
Elements of $T_{f(p)}\mathbb{R}^n$ can be viewed as derivations on $C^\infty(\mathbb{R}^m, \mathbb{R})$. To prove $L_m \circ Df_p=df_p\circ L_n$, let $h\in \mathbb{R}^n$ and $w\in C^\infty(\mathbb{R}^m, \mathbb{R})$.

$\begin{equation}
df_p(L_n(h))(w) =
L_n(h)(w\circ f) =
\frac{d}{dt}\bigg\rvert_0 w(f(p+th))=
Dw_{f(p)}(Df_p(h))
\end{equation}$

$\begin{equation}
L_m(Df_p(h))(w)=
\frac{d}{dt}\bigg\rvert_0 w(f(p)+tDf_p(h)))=
Dw_{f(p)}(Df_p(h))
\end{equation}$


问:
Rotating vector versus dot product
Suppose that I have a vector

$$
v = [v_0, v_1, v_2]^T
$$

which I want to project to

$$
n = [n_0, n_1, n_2]^T
$$

I did this with dot product of $v$ and $n$. As an alternative way, I wanted to rotate coordinate system and express $v$ in new coordinate system. I rotated coordinate system such that x axis is parallel to $n$ (or $x^{\prime}$ in figure). In new coordinate system, I expected component of $v$ in $n$ or $x^{\prime}$ axis to be equal to $v \cdot n$ in old coordinate system. I did $Rv$ with rotation matrix

$$
R =
\begin{bmatrix}
\cos \theta \cos \phi && \cos \theta \sin \phi && \sin \theta\\
-\sin \phi && \cos \phi && 0\\
-\sin \theta \cos \phi && -\sin \theta \sin \phi && \cos \theta
\end{bmatrix}
$$

where, $\theta$ and $\phi$ are obtained with

$$
\theta = \text{atan2} \left(\frac{n_1}{n_0}\right)\\
\phi = \text{atan2} \left(\frac{n_2}{\lvert n_0 \rvert}\right)
$$

Then I wanted to try an example. $v=[111.6, 14.701, 0]^T$, $n=[0.349026, 0.919137, -0.182667]^T$. The results I get from two methods are different (app. 52 vs 47). Clearly, I am misunderstanding something. May you spot the mistake please.

答:
Problem is the definition of $\phi$ which should be

$$
\phi = \sin^{-1} \left(n_2\right)
$$

Title: Rotating vector versus dot product


问:
Splitting of prime ideals in algebraic extensions
I'm reading Algebraic Number Theory by Jurgen Neukirch. I have some problems with some of the exercises in Section 9 of Chapter 1.

They are:

1) If $L / K$ is a Galois extension of algebraic number fields with non-cyclic
Galois group, then there are at most finitely many non-split prime ideals of $K$.

2) Let $L / K$ be a finite (not necessarily Galois) extension of algebraic number
fields and $N / K$ the normal closure of $L / K$. Show that a prime ideal $p$ of $K$ is totally
split in $L$ if and only if it is totally split in $N$.

I have worked on them for a long time but couldn't get any idea of it.

Can you please help? Thank you!

答: 1)  Let $G=$ Gal$(L/K)$, $p \in K$. Suppose $p$ is unramified and nonsplit. (Since only finitely many primes are ramified, it suffices to show that this cannot occur.) Since $p$ is unramified and nonsplit and $efg=|G|$, we see that $f=|G|$ and the decomposition group $D_p$ is isomorphic to $G$. But we also have that $D_p$ is isomorphic to the Galois group of the residue field of $L/K$ at $p$, which is cyclic of order $f$. This contradicts our hypothesis on $G$.</p>


问:
properties of recursively enumerable sets
$A \times B$ is an r.e.(recursively enumerable) set,  I want to show that $A$ (or $B$) is  r.e. ($A$ and $B$ are nonempty)

I need to find a formula. I've got an idea that I should use the symbolic definition of an r.e. set. That is, writing a formula for the function that specifies $A$ or $B$, assuming a formula exists for $A \times B$. I guess I must use Gödel number somewhere.

I should've mentioned that I asked this question over there (Computer Science) but since I am more interested in mathematical arguments and looking for formulas, I'd ask it here as well.

答:
Here is a formalization or detailed exposition of Asaf Karagila's answer.

Because $A \times B$ is r.e., there is an $e$ such that for all $n$ and $m$,
$$
(n,m) \in A \times B \Leftrightarrow (\exists s)T(\underline{e},\underline{\langle n,m\rangle},s).
$$
Here $T$ is Kleene's $T$ predicate, $\underline{x}$ is the numeral for a number $x$, and $\langle \cdot,\cdot\rangle$ is a fixed effective pairing operation, e.g. $\langle n,m \rangle = 2^n3^m$.

Given $e$, we can define a unary computable function $f(n)$ which does the following: search for an $m$ and $s$ such that $T(\underline{e},\underline{\langle n,m\rangle},\underline{s})$ holds. If such an $m$ and $s$ are found, immediately halt and return $1$, otherwise keep searching forever.

Because $f$ is computable, it has an index $e'$. Then we have, for all $n$,
$$
n \in A \leftrightarrow (\exists s) T(\underline{e'},\underline{n},s)
$$
and so the formula on the right side of that equivalence is the desired formula. In particular, we can effectively produce $e'$ from $e$.

As a historical note, this argument is reminiscent of the way that arguments in computability were written in the earliest stages of the field, in the 1940s and 1950s. As the field progressed, we have stopped using this sort of argument - which needlessly brings up technical devices such as the $T$ predicate - and have moved to more intuitive arguments like the ones William and A.Schulz provided, which do not explicitly mention formulas. It is occasionally necessary to work with the $T$ predicate, especially when studying formal theories of arithmetic, but for pure computability theory it is not typical to write arguments in this style.


问:
Intersection of sets span a vectorspace
Let $V$ be a $\mathbb{K}$-Vectorspace, and let $B_1,B_2, ...$ be subsets of $V$ such that $B_1\supseteq B_2 \supseteq ...$ and for all $i\in\mathbb{N} \setminus \{ 0\} $ is $\langle B_i \rangle = V$. Prove or disprove that $\langle \cap_{i\in\mathbb{N}} B_i \rangle =V$.
The hint from our professor was that this is not true.
Let's choose $B_k$ such that it is a subset of all other $B_i$ and also the smallest set. So every vector of $B_k$ is also in every other $B_i$. Therefore, the intersection of all $B_i$ would be $B_k$ and we know that $\langle B_k \rangle = V$ which would make the statement true. Where am I wrong? Any solution or hint would be highly appreciated.

答: Nobody said that we have only finitely many $B_i$'s. Suppose that $V=\Bbb K$ and that$$B_n=\left\{x\in\Bbb K\mid|x|<\frac1n\right\}.$$Then $(\forall n\in\Bbb N):\langle B_n\rangle=\Bbb K$. However, $\bigcap_{n\in\Bbb N}B_n=\{0\}$, and so$$\left\langle\bigcap_{n\in\Bbb N}B_n\right\rangle=\{0\}.$$

Title: Intersection of sets span a vectorspace


问:
Why do chaotic systems need dense periodic orbits?
I know that chaotic systems have three properties (by the most cited definition):

Sensitive to initial conditions

Topological Mixing

Dense Periodic Orbits

I know that dense periodic orbits means that arbitrarily near any point is a periodic orbit.  What I'd like to understand is why mathematicians found it desirable to exclude systems which lack dense periodic orbits from chaos theory.  What about systems that fail (3) make them sufficiently different from chaotic systems that it's valuable to exclude them?

答:
There are a lot of defintions of chaos, some are weaker than other one. See https://hal.archives-ouvertes.fr/hal-00276838/document for a good survey.

The one you should be thinking of is Devaneys' definition of chaos.

He defined it in his book An Introduction to Chaotic Dynamical Systems, 1986.
Unfortunatly this book is not free so I cannot really give the motivation behind it. But we can found a book review by Floris TAKENS which quote the book :

DEFINITION 8.5. Let V be a set. F: V in V is said to be chaotic on V if

f has sensitive dependence on initial conditions.

f is topologically transitive.

periodic points are dense in V

To summarize, a chaotic map possesses three ingredients; unpredictability, indecomposability, and an element of regularity. A chaotic system is unpredictable because of the sensitive dependence on initial conditions. It cannot be broken down or decomposed into two subsystems (two invariant open subsets) which do not interact under f because of topological transitivity. And, in the midst of this random behavior, we nevertheless have an element of regularity, namely the periodic points which are dense.

So for him it is the melange of regularity and unpredicctability.

I have two opinion about why periodicity is important. The  first one is that it is in some meaning the opposite of minimal system.

A system is said to be minimal if it has no closed invariant subset which implied that

Every orbit is dense

It has no periodic point in meaningful situation(if it has one the dynamic is trivially finite)

So in a minimal system you can follow the orbit of any point to "travel" around all the space.

On the opposite on Devaneys chaotic system if you pick any point, there is "high probability" (in a topological sense, that is dense) that you pick a periodic point that won't teach you a lot about the rest of the dynamic.

The second idea is that it was inspired by work of Li and York on 1975 in the famous paper Period Three Implies Chaos which state that if a map $f$ from $\mathbb{R} \to \mathbb{R}$ have a period three point, then it has periodic point for every period. If you restricte this dynamic to a well choosen interval, it has dense periodic point.

So maybe Devaneys was thinking of this kind of phenomena when he wrote his book.

One last think, it has be proven on the paper, On Devaney's Definition of Chaos by Banks, Cairns and Stacey, 1992, (https://www.researchgate.net/publication/235605725_On_Devaney's_Definition_of_Chaos)  that (3)+(2) implies (1).


问:
Morse Homology and The Space of Broken Trajectories
I am reading Morse Homology from the book by Audin and Damian. I have read the proof about $\overline{\mathcal{L}(a,b)}$ (space of broken trajectories connecting two critical points) being a manifold with boundary when $ind(a)-ind(b)=2$ , but I don't get why $\mathcal{L}(a,b)$ is dense in $\overline{\mathcal{L}(a,b)}$. Actually, I don't even get why $\mathcal{L}(a,c) \times \mathcal{L}(c,b) \neq \emptyset$ implies $\mathcal{L}(a,b) \neq \emptyset$ and I was wondering how we could extend this argument in general and prove the fact that $\overline{\mathcal{L}(a,b)}$ is a manifold with corner with $dim=ind(a)-ind(b)-1$ .

Do you know other references on this subject?

答:
Proposition 3.2.8 in Audin-Damian is what you're looking for. For any element $\lambda = (\lambda_1,\lambda_2) \in \mathcal{L}(a,c) \times \mathcal{L}(c,b)$ there is a continuous map $\psi:[0,\delta) \to \overline{\mathcal{L}}(a,b)$ (for some $\delta>0$) such that $\psi(0) = \lambda$ and $\psi(s) \in \mathcal{L}(a,b)$. So you get density since $\mathcal{L}(a,b) \owns \psi(1/n) \to_{n \to \infty} \psi(0) = \lambda \in \mathcal{L}(a,c) \times \mathcal{L}(c,b)$. On the other hand, if there is any element $ \lambda$ as above, then $\psi(\delta/2) \in \mathcal{L}(a,b)$, proving $\mathcal{L}(a,b)$ is nonempty.

Also, if you accept the claim that $\overline{\mathcal{L}}(a,b)$ is a 1-dimensional manifold with boundary such that its interior is $\mathcal{L}(a,b)$, then both of your statements come from standard facts about manifolds: if $M$ is an $n$-dimensional ($n \geq 1$) manifold with boundary $\partial M$, then $\partial M$ is contained in the closure of $\mathrm{Int}\,M$ inside $M$, and also $\partial M \neq \varnothing \implies M \neq \varnothing$.
(Hint for proving that: firstly, $\mathbb{R}^{n-1}\times \{0\}$ is in the closure of $\mathbb{R}^{n-1} \times (0,\infty)$ inside $\mathbb{R}^n$; secondly, any open set containing an element of $\mathbb{R}^{n-1} \times \{0\}$ contains an element of $\mathbb{R}^{n-1} \times (0,\infty)$).

For higher difference of indices, the compactified moduli space is also a manifold with corners, in some sense, but the proof is much more involved. It is proven by Wehrheim: https://arxiv.org/abs/1205.0713 . But it is quite subtle: only after requiring some technical conditions there is a canonical structure of a manifold with corners on that space.


问:
Finding the value of $S$ given a matrix with $81$ terms
Suppose we have $81$ positive integers that form a $9 \times 9$ matrix so that the first row forms an arithmetic sequence and every column forms a geometric sequence with the same common ratio. If $a_{2,4} = 1, a_{4,2} = \frac18, a_{4,3} = \frac{3}{16},$ find the value of $S = a_{1,1} + a_{1,2} + \ldots + a_{9,9}.$

答:
Let $a_{1,5} = x$, the common difference of the arithmetic sequence be $d$ and the common ratio of the geometric sequence be $r$

Given: $a_{2,4} = 1, a_{4,2} = \frac{1}{8}, a_{4,3} = \frac{3}{16}$

Or, $$(x-d)r = 1$$ $$(x-3d)r^3 = \frac{1}{8}$$ $$(x-2d)r^3 = \frac{3}{16}$$

Solving the above system, we have $$x = \frac{5}{2}, d = \frac{1}{2}, r = \frac{1}{2}$$ (Since $x > 0$, we disregard the negative values...)

$$S = \sum_{i=1}^9 \sum_{j=1}^9 a_{i, j} = \left[\sum_{k = 0}^8 \frac{1}{2^k} \right] \left[\sum_{j=1}^9 a_{1, j} \right] = \frac{511}{256} \cdot\frac{45}{2} = \frac{22995}{512} \approx 44.912$$


问:
Algorithm to find all fields of orders $p^n$ for fixed $n$ and all primes $p$ simultaneously?
Is there an efficient way (an algorithm) to find, given a fixed $n\in\mathbb{N}_+$, all fields of order $p^n$ for all primes $p$ simultaneously?

There's no background behind this question, just an idea.

Given a prime $p$ and $n\in\mathbb{N}_+$ we can find a field of order $p^n$ using an irreducible polynomial $f$ of degree $n$ over $\mathbb{Z}_p[x]$ and considering the quotient $\mathbb{Z}_p[x]/(f)$. Can we achieve something like this but universally for all primes?

By that I mean, if we are able to find certain "thing" in finite time so that that thing will give us a description of all finite fields of order $p^n$ where $n$ is fixed.

答: I interpret your question as asking whether there is a polynomial $f\in\mathbb{Z}[x]$, of degree $n>1$, such that $\overline{f}$ is irreducible of degree $n$ in $(\mathbb{Z}/p)[x]$, for every prime $p$; unfortunately the answer to this is no. To see this, let $f\in\mathbb{Z}[x]$ be arbitrary. Find some $n\in\mathbb{Z}$ such that $\pm 1\neq f(n)=:m$, and let $p$ be any prime factor of $m$. Then $\overline n$ is a root of $\overline f$ in $(\mathbb{Z}/p)[x]$, since $\overline f(\overline n)=\overline m=\overline 0$, and thus $x-\overline n$ is a factor of $\overline f$, which is hence reducible (or of degree $\leqslant 1$) mod $p$.


问:
Are these the only real fixed points of the polynomials $(f_n)_{n\ge 1}$ given by $f_1=X^2-1$ and $f_{n+1}=f_n^2-1$?
Define a sequence of polynomials with integer coefficients $(f_n)_{n\ge 1}$ by $f_1 = X^2-1$ and
$$f_{n+1}=f_n^2-1.$$
Now, for any $n$, what are the $x\in\mathbb{R}$ with $f_n(x)=x$?

There are some obvious ones. Both roots of $X^2-X-1$ are fixed points of all $f_n$, and when $n$ is even, $0$ and $-1$ are as well. Computations with some (very) small values of $n$ suggest these are the only ones; is this correct? What if in the recursion we take a negative real number other than $-1$?

Edit: I believe that if we have a 'cycle' of $k$ distinct values $\alpha_1,\ldots,\alpha_k$ with $\alpha_{i+1}=\alpha_i^2-1$ and $\alpha_k^2-1=\alpha_1$, then the polynomial $g:=\prod_{i=1}^k(X-\alpha_i)$ satisfies $g(X^2-1)=(-1)^kg(X)g(-X)$. In particular, $g(X)\mid g(X^2-1)$, which might be useful.

答:
I'm going to define $f(x)=x^2-1$ so that your $f_n(x)$ are $f^{\circ n}(x) = f\left (f^{\circ(n-1)}(x)\right)$, the $n$-fold composition of $f$.

Now, for $0\leq x\lt \phi$ we have $f(x)\lt x$ and for $x\gt\phi$ we have $f(x)\gt x$, so by induction (also using the fact that $-1\lt f(x)\lt 0$ for $-1\lt x\lt 0$) the only positive fixed point of $f^{\circ n}$ is $\phi$. Since $f(x)\geq -1$ for all $x$, any remaining fixed points must be in $[-1, 0]$.

By the composition rule for derivatives, we have $\dfrac{d}{dx}f^{\circ n}(x)$ $=\dfrac{d}{dx}f\left(f^{\circ (n-1)}(x)\right)$ $=f'\left(f^{\circ(n-1)}(x)\right)\dfrac{d}{dx}f^{\circ(n-1)}(x)$; induction then gives that this is $\prod_{i=0}^{n-1}f'\left(f^{\circ i}(x)\right)$. Since $f'(x)=2x$, this is just $2^n\prod_{i=0}^{n-1}f^{\circ i}(x)$ . Now, if $n$ is odd then this product must be less than zero (each term in it is negative), so $\dfrac{d}{dx}\left(f^{\circ n}(x)-x\right)\lt 0$; this, combined with the values of $f^{\circ n}(-1)=0$ and $f^{\circ n}(0)=-1$ establishes that there's exactly one fixed point in the interval.

Finally, the case of $n$ even can be settled similarly to the case $x\gt 0$; for $-1\lt x\lt -\phi^{-1}$ we have $f^{\circ 2}(x)\lt x$ and for $-\phi^{-1}\lt x\lt 0$, $f^{\circ 2}(x)\gt x$, so induction yields $f^{\circ 2i}(x)=x$ only for $x=-\phi^{-1}$ in this range.

It's also pretty straightforward, using the above, to characterize the fixed points; both $-1$ and $0$ are attracting fixed points of $f^{\circ 2}()$ whereas $\phi$ and $-\phi^{-1}$ are repulsing fixed points of $f()$. In other words, for $-1\leq x\lt\phi$, iterating $f()$ converges to the cycle $\langle -1, 0\rangle$; for $x\gt\phi$, iterating $f()$ diverges. (The basin of attraction is somewhat larger than this, of course; in fact, the iterations converge to the cycle for exactly $x\in (-\phi, \phi)$. )

The case for general $c$ in $f_c(x)=x^2-c$ gets substantially more complicated, specifically as $c$ goes from $1$ to $2$. In fact, we can see that $c=1$ is a somewhat unusual point by studying convergence around the fixed cycle $\langle 0,1\rangle$. Consider a point $x=\epsilon$ for some small $\epsilon$; then after one iteration this goes to $\epsilon^2-1$ and after two this goes to $(\epsilon^2-1)^2-1=\epsilon^4-2\epsilon^2$. Since $\epsilon$ is small we can throw out the $\epsilon^4$ term and say that $f_1^{\circ 2}()$ maps $\epsilon$ to $-2\epsilon^2+o(\epsilon^2)$. But this is what's known as superlinear convergence; 'generically' near a fixed point, we would expect to have some constant $M$ such that $f_c^{\circ 2}()$ affects points around a fixed point $x_f$ by sending $x_f+\epsilon$ to $x_f+M\epsilon+o(\epsilon)$; the fixed point is then attracting if $|M|\lt 1$ and repelling if $|M|\gt 1$. The fact that our $M$ here is zero is a sign that we're not at a generic value of $c$. In fact, this $c$ is the exact center of the interval $(3/4, 5/4)$ where the dynamics of $f_c$ feature a period-2 cycle. For $c$ slightly larger than $5/4$, the cycle is instead period 4; e.g., if we look at $c=1.3$ then the iteration converges to a length-4 cycle roughly at $\langle -1.149, 0.019, -1.3, 0.389\rangle$.

These dynamics can be mapped to various features of the Mandelbrot set; the region $c\in(3/4, 5/4)$ where the iteration converges to a $2$-cycle corresponds exactly to the first circular bulb to the left of the main cardioid. They've also been studied even more extensively in another guise: by taking $x_n=ay_n+b$ we can write $ay_{n+1}+b$ $= x_{n+1}=x_n^2+c$ $=a^2y_n^2+2aby_n+b^2+c$, or on subtracting $b$ from both sides and dividing by $a$, $y_{n+1} = ay_n^2+2by_n+(b^2-b+c)/a$. Now, taking $b$ such that $b^2=b-c$ lets us eliminate the constant term and write $y_{n+1}=ay_n^2+2by_n$; further taking $a=-2b$ this can be written as $y_{n+1}=ay_n^2-ay_n$ $=ay_n(1-y_n)$. This reparametrization lets us look at the interval $y\in [0,1]$ specifically and iterations within this interval (as long as $a\leq 4$; for larger $a$ the map no longer carries the interval $[0,1]$ to a subset of itself and all bets are off). In this form, it's known as the Logistic Map and it's been very widely studied, including the famous 'bifurcation diagram' you may have seen that shows how the attracting fixed points/fixed cycles change as $a$ increases towards $4$ (or in your formulation of iterating $f_c=x^2-c$, as $c$ increases towards 2):


问:
Proving that $Y$ is independent of $X$ if $Y=XZ$ for some random variables $X,Y,Z$
Say $X$ is a random variable such that $P(X=1)=P(X=-1)=\frac{1}{2}$, and $Z \sim N(0,1)$.
If $Y=XZ$, then are $X$ and $Y$ independent from each other? If yes, how can I show this?

Intuitively I would say that $X$ and $Y$ are not independent, since $Y$ depends on $X$ by $Y=XZ$, but I know that my intuition may be wrong.

答:
$X$ and $Y$ are independent iff $\mathbb{P}(X=1,Y\leq y) = \mathbb{P}(X=1)\mathbb{P}(Y\leq y)$ for every real $y$.

Now, denoting by $\Phi(y) := \mathbb{P}(Z\leq y)$, $$\mathbb{P}(X=1) = \frac{1}{2}$$ $$\mathbb{P}(Y \leq y) = \mathbb{P}(Z\leq y,X=1) + \mathbb{P}(Z \geq -y,X=-1) = \frac{1}{2}(\Phi(y) + 1-\Phi(-y)) = \Phi(y)$$ $$\mathbb{P}(X=1,Y\leq y) = \mathbb{P}(X=1,Z\leq y) = \frac{1}{2} \Phi(y)$$


问:
An identity for $\binom{-1/2}{n}$
I would like to show the formula
$$\forall n \in \mathbb{N},\quad\sum_{p=0}^n \binom{n}{p} \frac{(-1)^p}{p!} \prod_{k=0}^{p-1} \left(n+\frac{1}{2} + k\right) =  \binom{-1/2}{n} \, , $$
which is true according to Mathematica. I tried to use the identity
$$\binom{-1/2}{n} = \frac{1}{n!} \sum_{k=0}^n s_{n,k}\left(-\frac{1}{2}\right)^k \, ,$$
where $s_{n,k}$ denotes the Stirling numbers of first kind, but without any success. We end up with the following identity to prove
$$ \frac{1}{n!} \sum_{p=k}^n \binom{n}{p}\frac{s_{p,k}}{p!} \left(n+\frac{1}{2}\right)^k = s_{n,k} \, , $$
which is not a piece of cake for me neither. Among all the formulas I could find in Concrete Mathematics book from Ronald Graham, Donald Knuth, and Oren Patashnik, none comes in a handy. I confess I don't have a large culture of this field.

I will be thankfull for any hint.

答:
Let $ \alpha $, $ \beta \in\mathbb{R} $. we know that, for any $ x\in\mathcal{B}\left(0,1\right)=\left(-1,1\right) $, we have : $$ \left(1+x\right)^{\alpha}=\sum_{n=0}^{+\infty}{\binom{\alpha}{n}x^{n}} $$ $$ \left(1+x\right)^{\beta}=\sum_{n=0}^{+\infty}{\binom{\beta}{n}x^{n}} $$

Which means : $$ \small\sum_{n=0}^{+\infty}{\binom{\alpha +\beta}{n}x^{n}}=\left(1+x\right)^{\alpha + \beta}=\left(\sum_{n=0}^{+\infty}{\binom{\alpha}{n}x^{n}}\right)\left(\sum_{n=0}^{+\infty}{\binom{\beta}{n}x^{n}}\right)=\sum_{n=0}^{+\infty}{\left(\sum_{k=0}^{n}{\binom{\alpha}{k}\binom{\beta}{n-k}}\right)x^{n}} $$

Thus, for any $ n\in\mathbb{N} $, we have : $$ \sum_{k=0}^{n}{\binom{\alpha}{k}\binom{\beta}{n-k}}=\binom{\alpha +\beta}{n} $$

Setting $ \beta = n $, $ \alpha = -n-\frac{1}{2} $, we get : $$ \sum_{k=0}^{n}{\binom{-n-\frac{1}{2}}{k}\binom{n}{n-k}}=\binom{-\frac{1}{2}}{n} $$

By definition $ \binom{-n-\frac{1}{2}}{k}=\frac{1}{k!}\prod\limits_{j=0}^{k-1}{\left(-n-\frac{1}{2}-j\right)} $, we also have that : $ \binom{n}{k}=\binom{n}{n-k} $. Thus : $$ \sum_{k=0}^{n}{\binom{n}{k}\frac{1}{k!}\prod_{j=0}^{k-1}{\left(-n-\frac{1}{2}-j\right)}}=\binom{-\frac{1}{2}}{n} =\frac{\left(-1\right)^{n}}{4^{n}}\binom{2n}{n}$$

Title: An identity for $\binom{-1/2}{n}$


问:
How prove this $ab+bc+cd\le\dfrac{5}{4}$
let $a,b,c,d\in \Bbb R$ and $a,b,c,d>-1,a+b+c+d=0$
prove that
$$ab+bc+cd\le\dfrac{5}{4}$$

I have this solution

if $b\le c$, then
$$ab+bc+cd=a(b-c)-c^2\le -(b-c)-c^2=-(c-\dfrac{1}{2})^2+\dfrac{1}{4}-b\le\dfrac{1}{4}-b\le \dfrac{5}{4}$$
and then
$b>c$,as the same methods.I think this equality have other nice methods.Thank you

答:
First we rewrite $ab+bc+cd= (a+c)(b+d) - ad = -(a+c)^2-ad$.

This expression can only be positive if either $a$ or $d$ is negative. Thus wlog we choose $d<0$ and $a>0$. Now we have $-(a+c)^2-ad 1$. In this case $a-(a+c)^2 < a-(a-1)^2 = -a^2+3a-1 = -\left(a-\frac{3}{2}\right)^2+\frac{5}{4}\leq \frac{5}{4}$.

Body:
First we rewrite $ab+bc+cd= (a+c)(b+d) - ad = -(a+c)^2-ad$.

This expression can only be positive if either $a$ or $d$ is negative. Thus wlog we choose $d<0$ and $a>0$. Now we have $-(a+c)^2-ad <a-(a+c)^2$. If $a\leq1$, then this expression is $\leq1$ as well, thus wlog we assume $a>1$. In this case $a-(a+c)^2 < a-(a-1)^2 = -a^2+3a-1 = -\left(a-\frac{3}{2}\right)^2+\frac{5}{4}\leq \frac{5}{4}$.


问:
Does the group of rationals under addition have a basis?
I'm trying to determine whether $\mathbb Q$ under addition has a basis. My naive idea, based on what I have covered so far, would be to try and show that if we assume $\mathbb Q$ has a basis then it would be finitely generated which I believe would then lead to a contradiction. But I don't think this is a valid approach and I'm not sure what other approaches I could take.

I'd appreciate some ideas/approaches about determining whether $\mathbb Q$ has a basis or not :)

答:
You haven't defined what a "basis" is.

Here's, however, a statement showing that $\mathbf{Q}$ is far to having a basis as abelian group for any reasonable notion of basis as group:

For every generating subset $S$ of $\mathbf{Q}$, for every finite subset $F$ of $S$, the subset $S-F$ generates $\mathbf{Q}$.

(I assume that a basis would at least have the property that it's a generating subset in which no element is redundant; the above shows it can't exist.)

The claim is true: indeed, the quotient $H=\langle S\rangle/\langle S-F\rangle$ is a finitely generated quotient of $\mathbf{Q}$. Hence, if nonzero, it admits for some prime $p$ a cyclic group $C_p$ of order $p$ as quotient. But since $p\mathbf{Q}=\mathbf{Q}$, it cannot admit $C_p$ as quotient, so $H=0$, that is, $\langle S-F\rangle=\langle S\rangle$.


问:
Invariant subspaces in Lie groups and algebras representations
I am trying to prove the following statement:

Let $G$ be a connected Lie group with Lie algebra $\mathfrak g$ and with a representation $\Phi: G \to \text{GL}(V)$, with $V$ finite-dimensional. Then, $\Phi$ induces a unique representation $\varphi: \mathfrak g \to \text{End}(V)$ of the Lie algebra and a subspace of $V$ is $G$-invariant if and only if it is $\mathfrak g$-invariant."

Could you give some references or any proof? I know there are several entries in this forum about it, but I don't see one which helped me.

答: The representation $\Phi$ of $G$ induces a representation of $\mathfrak g$ defined as follows: if $X\in\mathfrak g$ and $v\in V$, then$$\varphi(X)(v)=\left.\frac{\mathrm d}{\mathrm dt}\right|_{t=0}\Phi\bigl(\exp(tX)\bigr)(v).$$It is clear from this definition that if $W$ is a subspace of $V$ which is $G$-invariant, then it is also $\mathfrak g$-invariant. And, if $W$ is $\mathfrak g$-invariant, and if $v\in W$, then\begin{align}\Phi\bigl(\exp(X)\bigr)(v)&=\exp\bigl(\varphi(X)\bigr)(v)\\&=v+\varphi(X)(v)+\frac1{2!}\varphi(X)\bigl(\varphi(X)(v)\bigr)+\cdots\\&\in W.\end{align}Since $G$ is connected, $G$ is spanned by its elements of the form $\exp(X)$ ($X\in\mathfrak g$), from which it follows that $W$ is $G$-invariant.


问:
Solving limits involving factorials
Generally speaking, how does one handle a limit involving a factorial? Take, for example, the limit:

$$\lim_{n\to\infty} \frac{x^{n+1}}{(n+1)!}$$

According to WolframAlpha, this limit evaluates to $0$, which seems reasonable. How can this be determined?

答:
Let $\;a_n=\dfrac{x^{n+1}}{(n+1)!}\quad$ for all $\;n\in\mathbb{N}\;.$

It results that

$a_{n+1}=\dfrac{x^{n+2}}{(n+2)!}=\dfrac{x}{n+2}\cdot\dfrac{x^{n+1}}{(n+1)!}=\dfrac{x}{n+2}\cdot a_n$

for all $\;n\in\mathbb{N}\;,$

hence,

$|a_{n+1}|=\dfrac{|x|}{n+2}\cdot |a_n|\le|a_n|$

for all $\;n\in\mathbb{N}\land n\ge |x|-2\;,$

consequently the sequence $\;\big\{|a_n|\big\}_{n\in\mathbb{N}}\;$ is eventually decreasing and

$\exists\;\lim\limits_{n\to\infty}|a_n|=\inf\limits_{n\in\mathbb{N}\\n\ge|x|-2}\big\{|a_n|\big\}=l\ge0\;.$

Since $\;|a_{n+1}|=\dfrac{|x|}{n+2}\cdot |a_n|\;\;$ for all $\;n\in\mathbb{N}\;,\;$ then

$\lim\limits_{n\to\infty}|a_{n+1}|=\lim\limits_{n\to\infty}\dfrac{|x|}{n+2}\cdot\lim\limits_{n\to\infty}|a_n|\;,$

$l=0\cdot l\;,$

$l=0\;,\;$ therefore

$\exists\;\lim\limits_{n\to\infty}|a_n|=0\;,\;$ hence

$\exists\;\lim\limits_{n\to\infty}a_n=0\;.$

Title: Solving limits involving factorials


问:
Vector valued Mean value theorem: Norm for the gradient
The wikipedia article on the vector valued Mean value theorem, says

For $f:\mathbb R^n \to \mathbb R^n$, if the gradient is bounded,
  $$
\| \nabla f \| \le M,
$$
  then
  $$
\|f(x)-f(y) \| \le M \|x-y\|.
$$

What is the norm used for the gradient $\| \nabla f \|$?

I tried to look in some other references.

There the matrix-norm is mentionned.

They gave one example, where I don't understand how to get the bound of $\frac14$ on $\|\nabla f \|$.

Define $g:\mathbb R_+^2 \to \mathbb R_+^2$ with
$$
g(x,y)=\left( \frac{1}{4+x+y}, \frac{1}{4+x+y} \right),
$$
then, entry-wise,
$$
\nabla f \le \begin{pmatrix} \frac{1}{16} & \frac{1}{16} \\ \frac{1}{16} & \frac{1}{16} \end{pmatrix}=A,
$$
whence
$$
\|\nabla f \| \le \|A\|=\frac14.
$$

what norm reduces $\|A\|$ to $\frac14$?

答: For finite-dimensional spaces, matrix norms are equivalent, so the choice does not matter.</p>


问:
Is the expected value of determinant times random matrix proportional to expected value of random matrix.
Let $\mathbf{X}$ be a random matrix. Is it true that
$$
E[\det(\mathbf{X})^h \mathbf{X} ] \propto E[\mathbf{X}],
$$
where $h$ is a scalar and $\det(.)$ denotes the determinant?

答:
This is unfortunately not true. Consider the following counterexample for $h=1$:

\begin{align*}
 X(\omega_{1}) &= \begin{pmatrix} 2 & \ 1 \\ 0 & \ 2 \end{pmatrix} \\
    X(\omega_{2}) &= \begin{pmatrix} 1 & -1 \\ 0 & \ 1\end{pmatrix}
\end{align*}
with $P(\omega_{1}) = P(\omega_{2}) = \frac{1}{2}$.
Then we have
$$ \mathbb{E}[X] = \begin{pmatrix} \frac{3}{2} & 0 \\ 0 & \frac{3}{2} \end{pmatrix}.$$

On the other hand, you can see that $\det(X)(\omega_{1}) = 4$ and $\det(X)(\omega_{2}) = 1$, so
\begin{align*}
 [\det(X) \cdot X](\omega_{1}) &= \begin{pmatrix} 8 & \ 4 \\ 0 & \ 8 \end{pmatrix} \\
    [\det(X) \cdot X](\omega_{2}) &= \begin{pmatrix} 1 & -1 \\ 0 & \ 1\end{pmatrix}
\end{align*}

As such, we have
$$ \mathbb{E}[\det(X) \cdot X] = \begin{pmatrix} \frac{9}{2} & \frac{3}{2} \\ 0 & \frac{9}{2}\end{pmatrix}. $$


问:
Domain of $\sqrt{x}$ function when solving $x^2 - 2=0$
I've been going through this solution and got stuck on explaining myself why  $x^2 = 2 \Rightarrow \sqrt{x^2} =  \sqrt{2}\Rightarrow x = \pm \sqrt2$.  Domain of $f(x)=\sqrt{x}$ is $[0,\infty)$. How taking square root both sides give $\pm \sqrt{2}$ ?

答:
There are two distinct concepts at play here. $\sqrt{x}$ is a function denoting the nonnegative square root of $x$. However, the general solution to $a^2 = b^2$ is $a = \pm b$. So $\sqrt{25}=5$, but if $x^2=25$ then $x=\pm5$. As Gae. S. has already mentioned, the easiest way to see that $a^2=b^2 \iff a= \pm b$ is through factorisation:
\begin{align}
&a^2=b^2 \\
\iff&a^2-b^2=0 \\
\iff&(a+b)(a-b)=0 \\
\iff&a=-b \text{ or } a=b \, .
\end{align}
So, in your example, we have $x^2=(\sqrt{2})^2 \iff x=\pm\sqrt{2}$.


问:
$\lim_{x \to 0} \frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$
I've tried to solve this limit:
$$\lim_{x \to 0} \frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$$

Here,
$$\frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$$ $$\sim \frac{(x \arctan x-x^2)(\sqrt{1+2x^4}+1)}{2x^4}$$ $$\sim \frac{(x \arctan x-x^2)}{x^4}= \frac{ \arctan x-x}{x^3} \sim \frac{x-x}{x^3} \sim 0$$

But the final result should be $-\frac{4}{3}$.

Any help would be appreciated.

答:
The handling of denominator is easy. Multiplying the expression under limit with $$\frac{\sqrt{1+2x^4}+1}{\sqrt{1+2x^4}+1}$$ we can see that the limit in question is equal to the limit of the expression $$\frac{\log(1+x\arctan x) - e^{x^2}+1}{x^4}$$ Split this expression into three parts as $$\frac{\log(1+x\arctan x) - x\arctan x} {(x\arctan x) ^2}\cdot\frac{(\arctan x) ^2}{x^2}+\frac{\arctan x - x}{x^3}-\frac{e^{x^2}-1-x^2}{x^4}$$ For the first term substitute $u=x\arctan x$ and conclude that it tends to $-1/2$. You can also find the limit of second term easily as $-1/3$. For third term just put $t=x^2$ and observe that it tends to $1/2$. The desired answer is thus $$-\frac{1}{2}-\frac{1}{3}-\frac{1}{2}=-\frac{4}{3}$$

The issue with your approach is the wrong use of equivalents ($\sim$ notation). Unless you are familiar with the rules of this notation avoid using it and instead stick to the usual limit laws/theorems/rules.


问:
How to work with an expression involving matrix products and Hadamard products in least squares?
With the system $\boldsymbol{A} \boldsymbol{x} = \boldsymbol{b}$, where $\boldsymbol{A}$ is a known matrix, $\boldsymbol{b}$ is a measured vector and $\boldsymbol{x}$ the vector we want to estimate, the ordinary least squares solves our problem:
$$\hat{\boldsymbol{x}} = (\boldsymbol{A}^\mathrm{H} \boldsymbol{A})^{-1} \boldsymbol{A}^\mathrm{H} \boldsymbol{b} = (\boldsymbol{A}^\mathrm{H} \boldsymbol{A})^{-1} \boldsymbol{A}^\mathrm{H} \boldsymbol{A} \boldsymbol{x} = \boldsymbol{x}.$$

Now, I have a system slightly more complicated: $(\boldsymbol{A} \odot \boldsymbol{C}) \boldsymbol{x} = \boldsymbol{b}$, where $\odot$ denotes the Hadamard product, and $\boldsymbol{C}$ contains random values that I don't know.
In this case, if I try to solve it like the ordinary least squares, I have:

$$(\boldsymbol{A}^\mathrm{H} \boldsymbol{A})^{-1} \boldsymbol{A}^\mathrm{H} \boldsymbol{b} = (\boldsymbol{A}^\mathrm{H} \boldsymbol{A})^{-1} \boldsymbol{A}^\mathrm{H} (\boldsymbol{A} \odot \boldsymbol{C}) \boldsymbol{x} = \:?$$

Then, I am stuck here, because from what I know and what I read, we cannot simplify an expression where matrix products and Hadamard products are mixed (there are some things possible when vector or diagonal matrices are involved, but this is not the case here).

Can I go any further with this equation, and try to have an expression for the error introduced by the presence of $\boldsymbol{C}$ ?
Or can I do something else than the ordinary least squares ?

I have another similar problem where I am also stuck. I have $((\boldsymbol{A} \boldsymbol{C}) \odot \boldsymbol{D}) \boldsymbol{x} = \boldsymbol{b}$, where I know $\boldsymbol{D}$ but not $\boldsymbol{C}$,
and if I try to solve it like the ordinary least squares, I have:
$$((\boldsymbol{A} \odot \boldsymbol{D})^\mathrm{H} (\boldsymbol{A} \odot \boldsymbol{D}))^{-1} (\boldsymbol{A} \odot \boldsymbol{D})^\mathrm{H} \boldsymbol{b} = ((\boldsymbol{A} \odot \boldsymbol{D})^\mathrm{H} (\boldsymbol{A} \odot \boldsymbol{D}))^{-1} (\boldsymbol{A} \odot \boldsymbol{D})^\mathrm{H} ((\boldsymbol{A} \boldsymbol{C}) \odot \boldsymbol{D}) \boldsymbol{x} = \:?$$

答:
The best way to write the least squares solution of $Ax=b$ is
$$\hat x = A^+b$$
where $A^+$ denotes the Moore-Penrose pseudoinverse, because when you introduce a Hadamard factor and replace $A\to(A\odot C)$ it can destroy the full-rank property such that $(A^HA)$ becomes singular.

With this in mind, the solution to your first problem is
$$\hat x = (A\odot C)^+b$$
and the solution to the second one is
$$\hat x = (AC\odot D)^+b$$


问:
Question Regarding Riemann Mapping Theorem
I am asked this question and I think the Riemann mapping theorem can be used.

Let $D_1=\mathbb{C}\setminus\{a\},$ where $a\in \mathbb{C}$. Is there a map $f$ that maps $D_1$ analytically onto the unit disk? If yes, find such a map.

Also, the same question for $D_2=\mathbb{C}\setminus[0,\infty)$.

My thought:

The answer would be 'yes' if both $D_1$ and $D_2$ are simply connected, which is the condition in Riemann mapping theorem. However, I am not sure if it is the case.

If $D_1$ and/or $D_2$ is not simply connected, then we cannot use the Riemann mapping theorem. In this case, I am not sure how to proceed.

答:
Note that for $D_1$ any bounded holomorphic function $f$ extends to the full plane since the isolated singularity at $a$ is removable; by Liouville, $f$ is constant so cannot be onto any open set.

$D_2$ is simply connected and to find a Riemann map one can use that the (rotated) Koebe function $K(z)=\frac{z}{(1+z)^2}$ is a Riemann map from the unit disc to the plane from which we remove the ray $[1/4,\infty)$ so $K-1/4$ is a Riemann map from the unit disc to $D_2$ and its inverse, of course, sends $D_2$ onto the unit disc conformally


问:
Integral of a rational function: Proof of $\sqrt{C}\,\int_{0}^{+\infty }{{{y^2}\over{y^2\,C+y^4-2\,y^2+1}}\;\mathrm dy}= {{\pi}\over{2}}$?
I suspect that

$$\sqrt{C}\,\int_{0}^{+\infty }{{{y^2}\over{y^2\,C+y^4-2\,y^2+1}}\;\mathrm dy}=
 {{\pi}\over{2}}$$

for $C>0$.

I tried $C=1$, $C=2$, $C=42$, and $C=\frac{1}{1000}$ with Wolfram Alpha. But how to prove it?

答:
Note that
$$\frac{y^2}{y^4-(2-C)y^2+1}=\frac{1}{\alpha_+-\alpha_-}\left(\frac{\alpha_+}{y^2+\alpha_+}-\frac{\alpha_-}{y^2+\alpha_-}\right),$$
where $-\alpha_{\pm}$ denote two roots of the equation $z^2-(2-C)z+1=0$. For the initial integral to converge, $\alpha_{\pm}$ should be either both positive, either conjugate to each other. Consider for simplicity the former case. Then
\begin{align}
\int_{0}^{\infty}\frac{y^2dy}{y^4-(2-C)y^2+1}&=\frac{\alpha_+}{\alpha_+-\alpha_-}\int_0^{\infty}\frac{dy}{y^2+\alpha_+}-\frac{\alpha_-}{\alpha_+-\alpha_-}\int_0^{\infty}\frac{dy}{y^2+\alpha_-}=\\
&=\left(\frac{\sqrt{\alpha_+}}{\alpha_+-\alpha_-}-\frac{\sqrt{\alpha_-}}{\alpha_+-\alpha_-}\right)\int_0^{\infty}\frac{dt}{t^2+1}=\\ &=\frac{1}{\sqrt{\alpha_+}+\sqrt{\alpha_-}}\frac{\pi}{2}.
\end{align}
But we have $\left(\sqrt{\alpha_+}+\sqrt{\alpha_-}\right)^2=(\alpha_++\alpha_-)+2\sqrt{\alpha_+\alpha_-}=(C-2)+2\sqrt{1}=C$, so the statement follows.</p>


问:
Show that no matter how we colour all edges of K5,5 with red and blue you can find a copy of K2,2 which either has all edges red or all edges blue.

Show that no matter how we colour all edges of $K_{5,5}$ with red and blue you
can find a copy of $K_{2,2}$ which either has all edges red or all edges blue.

I have done this: $R_2(G_1,G_2)\le R_2(10,4)$($10$ and $4$ being the number of vertices of the graphs)

but it doesn't seem right as work after this step (calculating $R_2(10,4)$ is not a reasonable task to do). Any hint on this question?

答:
Each of the top vertices has at least three red edges or at least three blue edges. For one of the colours, there are at least three top vertices $a,b,c$ that have at least three edges each of that colour. Remove all edges of the other colour. So $a,b,c$ are still of degree at least $3$.
If $a$ and $b$ have two common neighbours, we are done. So assume otherwise, i.e., the neighbours of $a,b$ cover the bottom row completely. Then two of the at least three neighbours of $c$ are also neighbours of $a$ or two are neighbours of $b$.


问:
Finding the value of $S$ given a matrix with $81$ terms
Suppose we have $81$ positive integers that form a $9 \times 9$ matrix so that the first row forms an arithmetic sequence and every column forms a geometric sequence with the same common ratio. If $a_{2,4} = 1, a_{4,2} = \frac18, a_{4,3} = \frac{3}{16},$ find the value of $S = a_{1,1} + a_{1,2} + \ldots + a_{9,9}.$

答:
Let, the first entry $a_{11}=a$. Then the first row is $a, a+d, a+2d,...a+8d$.

Now, let $r$ be the common ratio for each column.

Then, the $i^{th}$ column looks like $$(a+ (i-1)d),(a+ (i-1)d)r,...,(a+ (i-1)d)r^8$$.

Therefore for the given values three equations arrive :
$$ (a+d)r^3 = \frac{1}{8} \\
   (a+2d)r^3 = \frac{3}{16} \\
   (a+3d)r = 1
$$

Now subtracting the first equation from the second equation gives $$ dr^3 = \frac{1}{16}
$$

As the entries are integers therefore $d= \frac{1}{2} = r$.

Which gives $ a= \frac{1}{2}$.

Now see that $$ S = \sum_{i=0}^8 \sum_{j=0}^8 (a+id) r^j \\
=  (9a+36d) \frac{1-r^9}{1-r}
$$.


问:
Showing subspaces are complete/ not complete
(i) Show that (2, 5] considered as a metric subspace of $(\Bbb R,d)$ is not complete.

(ii) Show that [6,+∞) considered as a metric subspace of $(\Bbb R,d)$ is complete

So for (i) I think its not complete because its neither open nor closed and only closed sets can be complete

Then for (ii) I think its complete because I believe $[6, +\infty)$ is technically closed so therefore its complete.

Does that make sense, will that work?  Just feel like its too little writing...

答: Yes, that works. I don't know why is it that you wrote that $(2,5]$ is not open. That is correct, of course, but irrelevant. What matters is that it is not closed. And $[6,+\infty)$ is simply closed, not “technically closed”, whatever that means.


问:
Why $\log_b(x)$ = $-\log_{1/b}(x)$?
IN a textbook I am asked to consider the graphs of $f(x) = \log_{1/2}(x)$ and $-\log_2(x)$. They appear identical and then the textbooks tells the rule that $\log_b(x)$ = $-\log_{1/b}(x)$.

I tried to prove this to myself on paper.

I can write both equations in exponential form. Pretend $x=4$:

$$\log_{1/2}(4)=\frac{1}{2}^y=4$$

Then write the same for the other version:

$$-\log_2(4)=-2^y=4$$

If I know these are equivalent then the two exponent versions should be the same no?

$$4=\frac{1}{2}^y=-2^y$$

It's not clicking and I don't 'get it' from here. Is my last equation true? Why are they equivalent?

答:
Let $y=-\log_{1/b}(x)$. We see that
\begin{align}
&y=\log_{1/b}(x^{-1}) \\[4pt]
\implies&\left(\frac{1}{b}\right)^y=\frac{1}{x} \\[4pt]
\implies& \frac{1}{b^y}=\frac{1}{x} \\[4pt]
\implies& x=b^y \\[4pt]
\implies& y=\log_b(x) \, .
\end{align}

Title: Why $\log_b(x)$ = $-\log_{1/b}(x)$?


问:
"Integral Milking"
I begin this post with a plea: please don't be too harsh with this post for being off topic or vague. It's a question about something I find myself doing as a mathematician, and wonder whether others do it as well. It is a soft question about recreational mathematics - in reality, I'm shooting for more of a conversation.

I know that a lot of users on this site (e.g. Cleo, Jack D'Aurizio, and so on) are really good at figuring out crafty ways of solving recreational definite integrals, like
$$\int_{\pi/2}^{\pi} \frac{x\sin(x)}{5-4\cos(x)} \, dx$$
or
$$\int_0^\infty \bigg(\frac{x-1}{\ln^2(x)}-\frac{1}{\ln(x)}\bigg)\frac{dx}{x^2+1}$$
When questions like this pop up on MSE, the OP provides an integral to evaluate, and the answerers can evaluate it using awesome tricks including (but certainly not limited to):

Clever substitution

Exploitation of symmetry in the integrand

Integration by parts

Expanding the integrand as a series

Differentiating a well-know integral-defined function, like the Gamma or Beta functions

Taking Laplace and Inverse Laplace transforms

But when I play around with integrals on my own, I don't always have a particular problem to work on. Instead, I start with a known integral, like
$$\int_0^\pi \cos(mx)\cos(nx) \, dx=\frac{\pi}{2}\delta_{mn},\space\space \forall m,n\in
\mathbb Z^+$$
and "milk" it, for lack of a better word, to see how many other obscure, rare, or aesthetically pleasing integrals I can derive from it using some of the above techniques. For example, using the above integral, one might divide both sides by $m$, getting
$$\int_0^\pi \frac{\cos(mx)}{m}\cos(nx) \, dx=\frac{\pi}{2m}\delta_{mn},\space\space \forall m,n,k\in
\mathbb Z^+$$
Then, summing both sides from $m=1$ to $\infty$, and exploiting a well-known Fourier Series, obtain
$$\int_0^\pi \cos(nx)\ln(2-2\cos(x)) \, dx=-\frac{\pi}{n},\space\space \forall n\in
\mathbb Z^+$$
or, after a bit of algebra, the aesthetically pleasing result
$$\int_0^{\pi/2} \cos(2nx)\ln(\sin(x)) \, dx=-\frac{\pi}{4n},\space\space \forall n\in
\mathbb Z^+$$
After pulling a trick like this, I look through all of my notebooks and integral tables for other known integrals on which I can get away with the same trick, just to see what integrals I can "milk" out of them in the same way. This is just an example - even using the same starting integral, countless others can be obtained by using other Fourier Series, Power Series, integral identities, etc. For example, some integrals derived from the very same starting integral include
$$\int_0^\pi \frac{\cos(nx)}{q-\cos(x)} \, dx=\frac{\pi(q-\sqrt{q^2-1})^{n+1}}{1-q^2+q\sqrt{q^2-1}}$$
$$\int_0^\pi \frac{dx}{(1+a^2-2a\cos(x))(1+b^2-2b\cos(mx))}=\frac{\pi(1+a^m b)}{(1-a^2)(1-b^2)(1-a^m b)}$$
and the astounding identity
$$\int_0^{\pi/2}\ln{\lvert\sin(mx)\rvert}\cdot \ln{\lvert\sin(nx)\rvert}\, dx=\frac{\pi^3}{24}\frac{\gcd^2(m,n)}{mn}+\frac{\pi\ln^2(2)}{2}$$
Everyone seems to be curious about the proof of this last identity. A proof can be found in my answer here.

I just pick a starting integral, and using every technique I know as many times as possible, try to come up with the most exotic integrals as I can, rather than picking a specific integral and trying to solve it.

Of course, integrals generated this way would be poor (or at least extremely difficult) candidates for contest problems or puzzles to evaluate given the integral, since they are derived "backwards," and determining the derivation given the integral is likely much harder than pursuing the vague goal of a "nice-looking integral" with no objective objective (ha ha).

QUESTION: Do you (residents of MSE who regularly answer/pose recreational definite integral questions) do this same activity, in which you try to generate, rather than solve, cool integrals? If so, what are some  integrals you have come up with in this way? What strategies do you use? Does anyone care to opine on the value (or perhaps lack of value) of seeking integrals in this way?

Cheers!

答:
A correctly re-done version of my deleted answer.

For $a_k\ne 0$, $k=1,2,...,n$, we have
$$\Pi_n:=u\left(\small{\prod_{k=1}^{n}}a_k\right)\prod_{k=1}^{n}u(a_k)=u(1)+\sum_{k=1}^{n}\sum_{\sigma\subset S_n\\ |\sigma|=k}u\left(\small{\prod_{j\in\sigma}}a_j^2\right),\tag{1}$$ where $S_n=\{1,2,...,n\}$
(from here), with $u(x)=x+1/x$. Letting $a_k=\exp(i\theta_k)$, we have $u(a_k)=2\cos\theta_k$ and
$$u\left(\small{\prod_{j\in\sigma}}a_j^2\right)=u\left(\small{\prod_{j\in\sigma}}e^{2i\theta_j}\right)=u\left(e^{2i\small{\sum_{j\in\sigma}}\theta_j}\right)=2\cos\left(2\small{\sum_{j\in\sigma}}\theta_j\right).$$
Thus
$$\Pi_n=2^{n+1}\cos\left(\small{\sum_{k=1}^n\theta_k}\right)\prod_{k=1}^{n}\cos\theta_k=2+2\sum_{k=1}^{n}\sum_{\sigma\subset S_n\\ |\sigma|=k}\cos\left(2\small{\sum_{j\in\sigma}}\theta_j\right).$$
Setting $\theta_k=x$,
$$\Pi_n=2^n\cos(nx)\cos(x)^n=1+\sum_{k=1}^{n}\sum_{\sigma\subset S_n\\ |\sigma|=k}\cos(2|\sigma|x)=1+\sum_{k=1}^{n}{n\choose k}\cos(2kx),$$
because $${n\choose k}=\left|\left\{\sigma\subset S_n:|\sigma|=k\right\}\right|.$$
Thus
$$\int_0^{a}\cos(nx)\cos(x)^ndx=\frac1{2^n}\left(a+\sum_{k=1}^{n}{n\choose k}\frac{\sin(2ka)}{2k}\right),$$
which, I don't doubt, can be used to develop more interesting identities.

An identity developed from the above:

$$\int_0^1t\sin\left(n(n+1)t\right)\cos(nt)^{n-1}dt=\frac1{2^n n^2}\sum_{k=1}^{n}{n\choose k}\left(\frac{\sin(2nk)}{2nk}-\cos(2nk)\right),$$
where $n\in\Bbb Z_{\ge1}$.

Another consequential identity:
$$\int_0^\infty \left\{x-\log(2\sinh x)\right\}\cos(x)\cos(x/n)^ndx\\=\frac{1}{2^{n+2}}\left[\frac{\pi^2}{3n^2}+\sum_{k=1}^{n}\frac{1}{k^2}{n\choose k}\left(\frac{\pi k}{n}\coth\frac{\pi k}{n}-1\right)\right]$$

Title: "Integral Milking"


问:
Confusion about definition of category using directed graph
A directed graph (diagram scheme, quiver) is a quadruple $(O, A, s, t)$, where $O$ is a set of objects, $A$ is a set of arrows and $s$ and $t$ are two mappings $s, t: A \to O$ ("source" and "target" of arrows respectively).

Then a category is defined as a directed graph with two additional mappings $id: O \to A$ (called identity) and $\circ : A \times_{O} A \to A$ (called composition), following usual rules of associativity, composition with identity etc.

The confusion with that definition for me is that it looks like any other definition in style of "a set with some extra stuff". I mean, for example, I can take any set and define group operations on it. But I can't take any directed graph and define a category structure on it. E.g. let $D = (\{1,2,3\}, \{a,b\}, s, t)$, where $s(a) = 1$, $s(b) = 2$ and $t(a) = 2$ and $t(b) = 3$ (i.e. there's no possible choices for identities and compositions). Identities and compositions restrict the choices of directed graphs.

What's wrong with such reasoning? And what is correct way of thinking about such definition of category?

NB. That's not a question about how to construct a free category on a given directed graph, it's about construction of a category according given definition.

UPDATE. Thanks to the answers, now I see where's the error in my reasoning. But confusion is still there.
Citing Categories for the Working Mathematician:

Every category $C$ determines a graph $UC$ with the same objects and
arrows, forgetting which arrows are composites and which are identities.
Every functor $F: C \to C'$ is also a morphism $UF: UC \to UC'$ between
the corresponding graphs. These observations define the forgetful functor
$U : \mathbf{Cat} \to \mathbf{Grph}$ from small categories to small graphs.

How this underlying graph of category relates with the one mentioned in definition of a category? They can't be the same just because underlying graph doesn't contain identities and compositions.

答:
If instead of groups you said e.g. Boolean algebras, then your argument would fail for one-sorted structures as well: every finite Boolean algebra has cardinality $2^n$, so not every set carries a Boolean algebra structure.

Similarly, not every graph is an underlying graph of a category: as you observed, one of the necessary conditions is that there's at least one loop $x\to x$ on each vertex.


问:
Prove that A is denumerable iff A~B for every infinite B$\subseteq A$ and understanding
Let A be an infinite set.
Prove that A is denumerable iff A~B for every infinite B$\subseteq A$

References

6.1 Definition By the set of the natural numbers we mean the intersection of all the successor sets. The set of the natural numbers is designated by the symbol ω; every element of ω is called a natural number.

7.14 Theorem A is an infinite set if and only if A has a denumerable subset.

Here are my thoughts on the matter (rewrite from notes)

A being countably infinite an bijection  exists
h:B$\mapsto  \omega$

B$\subset $A is infinite.Thus there is an injection f:$\omega\mapsto $B

We ahave the canonical injection:i :B$\mapsto $A:
b$\mapsto $b:

Thus h$\circ i \circ $f $:B\mapsto \omega$ is injective.
Thus B is equipotent it’s subset and B is infinite.

QED

Any help would be appreciated

答:
If $A$ is countably infinite, there is a bijection $h:A\to\omega$. Let $B$ be an infinite subset of $A$; then there is an injection $f:\omega\to B$, and of course we have the canonical injection $\iota:B\to A:b\mapsto b$, so $h\circ\iota:B\to\omega$ is an injection. Thus, $|B|=|\omega|$ by the Cantor-Schröder-Bernstein theorem. (Note that the composition $h\circ\iota\circ f$ is injective but not in general bijective.)

Now suppose that $|A|=|B|$ for each infinite $B\subseteq A$. $A$ is infinite, so there is an injection $f:\omega\to A$. Let $B=f[\omega]$; clearly $B$ is an infinite subset of $A$, so $|A|=|B|=|\omega|$, and $A$ is therefore countably infinite.


问:
Let $f:\Bbb R^2 \to \Bbb R$ be ${\cal C}^1$, such that $|f(x) − f(y)| ≤ ||x − y||^2$ for all $x,y\in \Bbb R^2$. Show that $f$ is constant.
Let $f:\Bbb R^2 \to \Bbb R$ be ${\cal C}^1$, such that  $|f(x) − f(y)| ≤ ||x − y||^2$ for all $x,y\in  \Bbb R^2$. Show that $f$ is constant.

I'm not sure about :

$=> \lim_{x\to y} \frac{|f(x) − f(y)|}
{\|x − y\|}  \le  \lim_{x\to y}\|x − y\|$

$=> |f '(x)| \le 0$

Is it true or not ?

答:
Select $y=x+\lambda \hat{v}\in \mathbb{R}^m$, where $\hat v$ is a unit norm vector. Now note that with $f\in\mathbb{R^n}$

$$\lim_{\lambda\to 0}\frac{||f(x+\lambda \hat{v})-f(x)||}{|\lambda|}=||\left(\hat{v}\cdot \nabla\right)f||\leq\lim_{\lambda\to0}||\lambda||=0$$
and therefore we conclude that for any unit vector $ \hat v$
$$||\left(\hat{v}\cdot \nabla\right)f||=0\iff\left(\hat{v}\cdot \nabla\right)f=0$$
Choosing $\hat v=\hat x_1, \hat x_2,..., \hat x_m$ we find
$$\frac{\partial f}{\partial x_1}=\frac{\partial f}{\partial x_2}=...=\frac{\partial f}{\partial x_m}=0$$
from which it is evident that $f$ can only be a constant vector in order to satisfy this inequality.

For the problem statement simply restrict $m=2, n=1$, but the nature of the arguments remains the same.


问:
In the definition of a Stopping Time, how important are the conditions on the Filtration being complete and right-continuous?
I'm reading a book and, in its section on the definition of a stopping time(continuous), the author declares at the start that for the whole section every filtration will be complete and right-continuous.

So, in the definition of a Stopping Time, how important are these conditions? Why would they matter?

答:
If the filtration is complete and right-continuous, the condition "$\{\tau \le t \} \in \mathcal F_t$ for all $t$" is equivalent to the condition "$\{\tau < t \} \in \mathcal F_t$ for all $t$."  However  I think the main importance is that if $(\tau_n)$ is a sequence of stopping times and $\mathcal F$ is right continuous and complete, then we can also say that $\sup_n (\tau_n)$, $\inf_n(\tau_n)$, $\limsup_n(\tau_n)$, and $\liminf_n(\tau_n)$ are stopping times.  If $\mathcal F$ does not satisfy the usual conditions, we can only say that $\sup_n (\tau_n)$ is a stopping time.  I would imagine that the book you're studying from included the proofs of those statements, but let me know if you would like me to add them (and point out where right-continuity of the filtration is used).

Proof of claims:

First, the condition "$\{\tau \le t \} \in \mathcal F_t$ for all $t$" always implies $\{\tau < t \} \in \mathcal F_t$ for all $t$ regardless of whether the filtration is right continuous because $$\{\tau < t \} = \bigcup_{n=1}^\infty \left\{\tau \le t - \frac 1n \right\}$$ and $\{\tau \le t - \frac 1n\} \in \mathcal F_{t-\frac 1n} \subset \mathcal F_t$.  If the filtration satisfies the usual conditions and $\{\tau < t \} \in \mathcal F_t$ for all $t$ then we have $$\{\tau \le t \} = \bigcap_{n=1}^\infty \left\{\tau < t + \frac 1n \right\}$$ and since this is the intersection of a decreasing sequence of sets in $\mathcal F_{t+\frac 1n}$ we have that $\{\tau \le t \} \in \mathcal F_{t^+} = \mathcal F_t$

To show $\inf_n(\tau_n)$ is a stopping time when $\mathcal F$ satisfies the usual conditions, we have $$\{\inf_n(\tau_n) \le t \} = \bigcap_{m=1}^\infty \bigcup_{n=1}^\infty \left\{\tau_n \le t + \frac 1m \right\} $$ (i.e. $\inf_n \tau_n \le t$ iff "for all $m \in \mathbb N$ there exists $n \in \mathbb N$ such that $\tau_n \le t + \frac 1m$").  The set $\bigcup_{n=1}^\infty \left\{\tau_n \le t + \frac 1m \right\}$ is in $\mathcal F_{t + \frac 1m}$, and so by the same argument as above we have $\{\inf_n \tau_n \le t \} \in \mathcal F_{t^+} = \mathcal F_t$.  The $\liminf$ and $\limsup$ cases are similar.  The reason $\sup$ is different is because $$\{\sup_n \tau_n \le t \} = \bigcap_{n = 1}^\infty \{\tau_n \le t\} \in \mathcal F_t$$ so we don't need to worry about right-continuity.