问:
Calculate cosh 1 correct to 6 decimal places.
Can anyone help me with this problem?

Calculate $\cosh 1$ correct to 6 decimal places.

Body:
Notice $$\cosh x = \sum_{n=0}^\infty \frac{x^{2n}}{(2n)!}$$
We have:
$$\begin{align} &\left|\;\cosh 1 - \left(1 + \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \frac{1}{8!}\right)\right|\\
= & \frac{1}{10!}\left( 1 + \frac{1}{11\cdot 12} + \frac{1}{11\cdot 12 \cdot 13 \cdot 14} + \cdots\right)\\
\le &\frac{1}{10!}\left(1 + 10^{-2} + 10^{-4} + \cdots\right)\\
= &\frac{100}{99\cdot 10!} = \frac{1}{3592512} \sim 2.7835675983824134\times10^{-7}
\end{align}
$$
So up to 6 decimal places,

$$\cosh 1 \sim 1 + \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \frac{1}{8!} = \frac{6913}{4480} \sim \color{red}{1.543080}357142857$$


问:
Cantor set and evaluating an integral
I'm trying to prove that the following function is Riemann integrable on $[0,1]$: $$g(x) = \begin{cases} 5, &x\in C \\ x, &x\not \in C\end{cases}$$
where $C$ denotes the Cantor set. It's well known that characteristic function of the Cantor set is Riemann integrable. I think we can use the same proof for showing integrability of $g(x)$ but I'm not sure about that. The other idea that I have is as follows: Since characteristic function is integrable we can conclude reverse characteristic function is integrable. $f(x) = x$ is integrable and product of integrable functions is integrable. So $h(x)$ which is defined as$$h(x) = \begin{cases} 0, &x\in C \\ x, &x\not \in C\end{cases}$$ is integrable. This shows $g(x)$ is integrable since $g(x)$ and $h(x)$ differ by a set of measure $0$. I don't know whether this solution is acceptable or not. It would be nice to see other methods to solve this problem.

答: Sums and products of RI functions are RI. Let $g$ be he charactetrstic function of $C$ and $f(x)=x$. Then $f(x)g(x)$ is RI. Hence $f(x)-f(x)g(x)$ is RI. Also $5 g(x)$ is RI. Now $5g(x)+f(x)-f(x)g(x)$ is RI. But $5g(x)+f(x)-f(x)g(x)$  is exactly your function.

ContentLicense: CC BY-SA 4.0

Title: Cantor set and evaluating an integral


问:
How to solve the following recurrence relation involving a floor function
Consider the following recurrence:
$$
f(x) = \lfloor f(x-1)\rfloor \cdot(f(x-1) - \lfloor f(x-1)\rfloor) + 1 \space(x\geq 2, \space x\in \mathbb{N}), \space f(1) = c
$$

How can one solve such a recurrence, i.e. finding a closed formula? I know some basic methods how to solve linear homogenous recurrence relations, e.g. with a spectral decomposition or by finding the characteristic equation, but I don't really know whether this can be applied to the above.

Edit: $c\in \mathbb{R}$.

答:
For convenience I’ll write $a_n$ for $f(n)$ and $\{x\}$ for the fractional part of $x$. For real $x$ let $g(x)=\lfloor x\rfloor\{x\}+1$. Then the recurrence can be written

$$a_n=\lfloor a_{n-1}\rfloor\{a_{n-1}\}+1=g(a_{n-1})$$

for $n\ge 2$, where $a_1=c$. Clearly $g(x)=1$ if $x\in\Bbb Z\cup[0,1)$, and it’s not hard to prove that $g$ is continuous from the right (because the floor function is).

Suppose that $n\le x<n+1$, where $n\in\Bbb Z$, so that $\lfloor x\rfloor=n$, and let $\alpha=\{x\}=x-n$; then

$$x-g(x)=n+\alpha-n\alpha-1=(n-1)(1-\alpha)\,.\tag{1}$$

This is $0$ iff $n=1$, so $g(x)=x$ iff $x\in[1,2)$. In particular, if $a_n\in[1,2)$ for some $n\in\Bbb Z^+$, then $a_k=a_n$ for all $k\ge n$.

We have now shown that if $a_n\in\Bbb Z\cup[0,2)$ for some $n\in\Bbb Z^+$, then $\langle a_k:k\in\Bbb Z^+\rangle$ is eventually constant: at $1$ if $a_n\in\Bbb Z\cup[0,1)$, and at $a_n$ if $a_n\in[1,2)$. In particular, this is the case if $c=a_1\in\Bbb Z\cup[0,2)$.

Now suppose that $c>2$; it follows from $(1)$ that $\langle a_n:n\in\Bbb Z^+\rangle$ is strictly decreasing as long as $a_n\ge 2$. Clearly $a_n\ge 1$ for all $n\ge 2$, so either some $a_n<2$, or $\langle a_n:n\in\Bbb Z^+\rangle$ converges monotonically down to some $a\ge 2$. But then

$$g(a)=\lim\limits_{n\to\infty}g(a_n)=\lim\limits_{n\to\infty}a_{n+1}=a\ge 2\,,$$

since $g$ is continuous from the right, so $a\in[1,2)$. This contradiction shows that some $a_n<2$, and it follows that the sequence is eventually constant at some value in $[1,2)$.

Now suppose that some $a_n<0$, and let $m=\lfloor a_n\rfloor$. Then $m<m\{a_n\}$, so $m+1<a_{n+1}$, and an easy induction shows that $0=m+|m|<a_{n+|m|}$. In particular, if $c<0$, there is an $n\in\Bbb Z^+$ such that $a_n>0$, and it follows again that the sequence is eventually constant at some value in $[1,2)$.


问:
How do I compute the distribution of $\min\{X,Y,1-Y,1-X$ $,|X-Y|,1-|X-Y|\}$, where $X,Y \overset{\text{iid}}{\sim} \operatorname{Unif}(0,1)$?
Let $X,Y \overset{\text{iid}}{\sim} \operatorname{Unif}(0,1)$. Let $$M = \min\{X,Y,1-Y,1-X,|X-Y|,1-|X-Y|\}.$$ Supposedly $\operatorname{image}(M) \subseteq (0,\frac13)$ and distribution of $M$ is $$F_M(m)= (3m(2-3m))1_{(0,\frac13)}(m) + 1_{(\frac13,\infty)}(m).$$ See here.

Question: How do I compute the distribution of $M$? I wonder if there's a clever way of doing this like with $W$ here. Below is what I've tried so far.

Well $|X-Y| \ge m$ and $1-|X-Y| \ge m$ tell me $m \le |X-Y| \le 1-m$. So it looks like I have 4 cases

$x+(m-1) \le y \le x-m$

$x+m \le y$

$y \le x-m$

$x-(m-1) \le y \le x+m$ (but I think I rule out this last case because here I have $m \ge \frac12$.

Then $X,Y,1-Y,1-X \ge m$ tell me $m \le X \le 1-m$ and $m \le Y \le 1-m$. So I guess the 3 cases become

$x+m-1 = \max\{m,x+m-1\} \le y \le \min\{x-m,1-m\} = x-m$

$x+m = \max\{m,x+m\} \le y \le 1-m$

$m \le y \le \min\{x-m,1-m\} = x-m$

But it seems if I do $\int_0^1 \int_{\cdot}^{\cdot} 1\, dy \,dx$ for each case and then add them up I get only $2-6m$.

答:
Ok here's what I did on wolfram:

By trying out values of $m=\frac15, \frac16, \frac17$ (Btw: $\frac14$ was kind of an outlier. See below re 'only issue'. Also, as expected, $\frac13$ gave no solutions), I got 4 cases

$x \in (m,2m)$, $y \in (x+m,1-m)$

$x \in (2m,1-2m)$, $y \in (m, x-m)$

$x \in (2m,1-2m)$, $y \in (x+m, 1-m)$

$x \in (1-2m,1-m)$, $y \in (m,x-m)$

When sketching this, I ended up getting 4 trapezoid figures.

Anyhoo, I got

$\int_m^{2m} \int_{x+m}^{1-m} 1 dy dx$

$\int_{2m}^{1-2m} \int_{m}^{x-m} 1 dy dx$

$\int_{2m}^{1-2m} \int_{x+m}^{1-m} 1 dy dx$

$\int_{1-2m}^{1-m} \int_{m}^{x-m} 1 dy dx$

Combining (1) and (3), we get $\int_m^{1-2m} \int_{x+m}^{1-m} 1 dy dx$

Combining (2) and (4), we get $\int_{2m}^{1-m} \int_{x+m}^{1-m} 1 dy dx$

Combining (5) and (6) we get our desired distribution.

My one issue however is the very fact that we say $(2m,1-2m)$ in the first place suggests $m \in (0,\frac14)$ only! (See above re 'outlier'.) I mean, I don't think $2m$ and $1-2m$ should be compared, but I think $m$ and $1-m$ should be compared.

I think this can be fixed if we just do 2 cases

$x \in (m,1-2m)$, $y \in (x+m,1-m)$, combining (1) and (3)

$x \in (2m,1-m)$, $y \in (m, x-m)$, combining (2) and (4)


问:
Factorising a principal ideal where its norm is $p^k$
I was considering the following example:

Let $K=\mathbb{Q}(\sqrt{-29})$ and so $O_K=\mathbb{Z}[\sqrt{-29}].$ So by Dedekind's theorem, I wanted to decompose the principal ideal $(5)$ and I obtained $(5)=(5,\sqrt{-29}-1)(5,\sqrt{-29}+1):=P_1P_2 $ and that both prime ideals have norm $5$.

Then my book said

"Consider the ideal $(3+2\sqrt{-29})$, this has norm $125$ and so must be either $P_1^3, P_1^2P_2, P_1P_2^2$ or $P_2^3.$"

Why does this follow? I know that we can prime decompose this ideal and that $P_1$ and $P_2$ are the only ideals with norm $5$. But why did we rule out the possibility that $(3+2\sqrt{-29})$ is already prime and the possibility that $(3+2\sqrt{-29})=QP_1$ where $Q$ is a prime ideal with norm $25$, say?

I am thinking I must have missed some small details, many thanks in advance for lending a hand!

答:
In a quadratic field there can be at most two prime ideals above any rational prime see: https://en.wikipedia.org/wiki/Splitting_of_prime_ideals_in_Galois_extensions#The_Galois_situation. Using the notation there $\text{degree} = e \cdot f \cdot g$ we have the degree is $2 = 1\cdot1\cdot2$ already, having any more primes would contradict this, so the fact you found two distinct  primes above 5 means these are the only primes above 5.

Even if your field wasn't Galois some similar arguments can be made and we always have that the degree of the field is always greater or equal to the number of primes above any prime in the base field.

I don't know if this has already been introduced in your book though :) which book is it?


问:
Finding the number of spanning tree of a particular graph
I have the following graph: start with a graph $C_n$(a n-cycle), where $n$ is at least 4, and we label the vertices from $1$ to $n$. Now, let an extra vertex $x$ be connected to the vertex with label 1 in $C_n$, and also connected to the vertex with label $m$ in $C_n$, and denote this graph as $G_{n,m}$, and find a closed formula of spanning trees in $G_{n,m}$, in terms of $n,m$.
Here are an example of what this graph looks like:

For $G_{4,3}:$

For $G_{6,4}:$

Immediately I think about edge contraction recurrence, $\tau(G)=\tau(G-e)+\tau(G*e)$. Take $G_{6,4}$ as example: suppose the edge between $x$ and vertex labelled 1 is $e_1$ and the edge between $x$ and vertex labelled 4 is $e_2$. Apply edge contraction recurrence to $e_2$, we get $\tau(G_{6,4})=\tau(G_{6,4}-e)+\tau(G_{6,4}*e)$. Apply the edge contraction recurrence to $e_1$on $\tau(G_{6,4}-e)$, it becomes $\tau(C_n)$-- so thats is $n$. But the tricky part is $\tau(G_{6,4}*e)$. A picture of $G_{6,4}*e$ is provided:

Now contract the edge between vertex $1$ and $4,x$ and use edge contraction recurrence, we get a graph that is $C_n$ so that gives $n$ spanning subtrees, summing it with a graph that looks like this:

and I'm really not sure how many spanning subtree this graph has. After doing some small examples, I think for the specific case here there is $2*3$ spanning subtrees, but how can I generalize this to a general $m,n$? Thank you in advance!

答:
Another useful property of $\tau(G)$, the number of spanning trees of $G$, is the following.

Suppose $G$ consists of two subgraphs $G_1, G_2$ which have only one vertex, $v$, in common. All edges of $G$ are either edges of $G_1$ or edges of $G_2$. Then to pick a spanning tree of $G$, we must pick a spanning tree of $G_1$ and a spanning tree of $G_2$ - and whichever way you combine those works. This tells us that $$\tau(G) = \tau(G_1) \cdot \tau(G_2).$$

In your case, when you contract both edges out of $x$, you get exactly this structure: the graph you get has two cycles that meet at one vertex. The rule above should help you deal with this graph. When you delete one edge out of $x$ and contract the other, whichever way you do it, you get $C_n$, and you can find $\tau(C_n)$. Finally, you get no spanning trees if you delete both edges out of $x$, because then the graph is disconnected.


答:
HINT: $G_{n,m}$ has three cycles: the original $C_n$, a $C_{n-m+3}$, and a $C_{m+1}$. In order to get a spanning tree, you must remove enough edges to break each of these cycles without disconnecting the graph. You have $n+1$ vertices and $n+2$ edges, while a tree with $n+1$ vertices must have $n$ edges, so you have to remove exactly $2$ edges.

If you remove one of the two new edges incident at $x$, you can remove any of the $n$ original edges to get a spanning tree. How many spanning trees can you get in this way?

The only other possibility is to remove one old edge from each of the shorter cycles. How many old edges are there in the $C_{n-m+3}$ cycle, and how many are there in the $C_{m+1}$ cycle? How many spanning trees altogether can you get in this way?

Title: Finding the number of spanning tree of a particular graph

Body:
I have the following graph: start with a graph $C_n$(a n-cycle), where $n$ is at least 4, and we label the vertices from $1$ to $n$. Now, let an extra vertex $x$ be connected to the vertex with label 1 in $C_n$, and also connected to the vertex with label $m$ in $C_n$, and denote this graph as $G_{n,m}$, and find a closed formula of spanning trees in $G_{n,m}$, in terms of $n,m$.
Here are an example of what this graph looks like:

For $G_{4,3}:$

For $G_{6,4}:$

Immediately I think about edge contraction recurrence, $\tau(G)=\tau(G-e)+\tau(G*e)$. Take $G_{6,4}$ as example: suppose the edge between $x$ and vertex labelled 1 is $e_1$ and the edge between $x$ and vertex labelled 4 is $e_2$. Apply edge contraction recurrence to $e_2$, we get $\tau(G_{6,4})=\tau(G_{6,4}-e)+\tau(G_{6,4}*e)$. Apply the edge contraction recurrence to $e_1$on $\tau(G_{6,4}-e)$, it becomes $\tau(C_n)$-- so thats is $n$. But the tricky part is $\tau(G_{6,4}*e)$. A picture of $G_{6,4}*e$ is provided:

Now contract the edge between vertex $1$ and $4,x$ and use edge contraction recurrence, we get a graph that is $C_n$ so that gives $n$ spanning subtrees, summing it with a graph that looks like this:

and I'm really not sure how many spanning subtree this graph has. After doing some small examples, I think for the specific case here there is $2*3$ spanning subtrees, but how can I generalize this to a general $m,n$? Thank you in advance!


问:
Find the sum $\sum_{n=1}^\infty\frac{\sqrt{n}}{2^{n/2}}$
Find the sum $\sum_{n=1}^\infty\frac{\sqrt{n}}{2^{n/2}}$. Is it possible to find a closed form for this sum? I was trying to upper bound this sum by another series, but could not find one.
Can anyone help?

答:
The polylogarithm is defined $$\mathrm{Li}_s(z)=\sum_{n\ge1}\frac{z^n}{n^s}$$
for $s\in\Bbb C$ and $|z|<1$.This being the case, we can write
$$S=\sum_{n\ge1}\frac{\sqrt n}{2^{n/2}}=\mathrm{Li}_{-1/2}(\tfrac{1}{\sqrt2}).$$
From here, we have
$$\mathrm{Li}_s(z)=\frac{z}{2}+\frac{\Gamma(1-s,-\ln z)}{(-\ln z)^{1-s}}+2z\int_0^\infty \frac{\sin(s\arctan t-t\ln z)}{(1+t^2)^{s/2}(e^{2\pi t}-1)}dt$$
for complex $z,s$, where $\Gamma(a,z)$ is the upper incomplete gamma function. So,
$$S=\frac{1}{2\sqrt2}+\left(\frac{2}{\ln2}\right)^{3/2}\Gamma(\tfrac32,\tfrac12\ln2)+\sqrt2\int_0^\infty\frac{\sin(\tfrac{t}2\ln2-\tfrac12\arctan t)}{e^{2\pi t}-1}(1+t^2)^{1/4}dt.$$
The $\Gamma$ integral is
$$\Gamma(\tfrac32,\tfrac12\ln2)=\int_{\tfrac12\ln2}^\infty t^{1/2}e^{-t}dt=\left(\int_0^\infty-\int_0^{\tfrac12\ln2}\right)t^{1/2}e^{-t}dt=\frac{\sqrt\pi}{2}-\int_0^{\tfrac12\ln2}t^{1/2}e^{-t}dt=\frac{\sqrt\pi}2-I_1.$$
The other integral is, with $t\mapsto \tan x$,
$$I_2=\int_0^\infty\frac{\sin(\tfrac{t}2\ln2-\tfrac12\arctan t)}{e^{2\pi t}-1}(1+t^2)^{1/4}dt=\int_0^{\pi/2}\frac{\sin(\tfrac{\ln2}{2}\tan x-x/2)}{e^{2\pi\tan x}-1}\sec(x)^{5/2}dx.$$
So,
$$S=\frac1{2\sqrt2}+\sqrt{\frac{2\pi}{\ln^32}}-\left(\frac{2}{\ln2}\right)^{3/2}I_1+\sqrt2 I_2.$$
Then from here, we have
$$-\left(\frac{2}{\ln2}\right)^{3/2}I_1+\sqrt2 I_2\approx −0.552128172429,$$
so
$$S<\frac1{2\sqrt2}+\sqrt{\frac{2\pi}{\ln^32}}-\frac12.$$
I would be very surprised if there is a closed form for $S$ which does not rely on $\mathrm{Li}$ special values.


问:
find density function of Z = (X - 1)^2 when X is a exponential distribution
let X be an exponential distribution with lambda as parameter and want to find the density function of $Z = (X - 1)^2$

I tried that: $F(Z) = P(Z < z) = P((X - 1)^2 <z)= P(1 - \sqrt{z} < x < 1 + \sqrt{z})$

for the next part, do I have to assume that $\sqrt{z}$ is greater than one or we have to determine for both cases?

of $\sqrt{z} > 1$: $F(Z) = P(X < \sqrt{z}+1)$ since exponential distribution have x always greater than 0.

if $\sqrt{z} < 1$: $F(Z) = P(1 - \sqrt{z}< x < 1 + \sqrt{z})$

and then find the pdf?

thank you!

答:
As you have: $F_{\small Z}(z)=\mathsf P(1-\surd z\lt X\leqslant 1+\surd z)$

Which you have identified will have a piecewise partition at $z=1$

And thus: $\quad F_{\small Z}(z)~=(\mathrm e^{-\lambda(1-\surd z)}-\mathrm e^{-\lambda(1+\surd z)})\mathbf 1_{0\leqslant z\lt 1}+(1-\mathrm e^{-\lambda(1+\surd z)})\mathbf 1_{1\leqslant z}$

Now just find the derivative.

So:  $\qquad\qquad f_{\small Z}(z)~ = \dfrac{\lambda(\mathrm e^{-\lambda(1+\surd z)}+\mathrm e^{-\lambda(1-\surd z)})}{2\surd z}\mathbf 1_{0\leqslant z\lt 1}+\dfrac{\lambda\mathrm e^{-\lambda(1+\surd z)}}{2\surd z}\mathbf 1_{1\leqslant z}$


问:
Every nonzero right ideal in an associative F-algebra A contains a minimal (nonzero) right ideal?
I'm trying to prove the title statement. As usual in these cases, I'm using Zorn's lemma. This is my advance:

Consider a nonzero right ideal $I$ of $A$.

Let $G = \{ C \in P(A) : \text{C is a right ideal of A} \land \{0\} \neq C \subset I\}$. Suppose $B$ is a chain of G. Then, if $ B = \phi$, $I$ is trivially a lower bound; by the other side, if $B \neq \phi$, Consider $\bigcap B$. It is a right ideal, but I can't see that it is not the zero ideal. With that detail at hand, I can apply Zorn's lemma, and it will cast the desired minimal right ideal.

Edit: $\mathbb{F}$ is a field so that A is an algebra over a field $\mathbb{F}$. On the other hand, This is a detail that my professor let me for solving, which appear in his proof of the following result:

$\textbf{Proposition:}$Let $A$ be a finite-dimensional simple and associative algebra over a field
$\mathbb{F}$, then every nonzero right ideal contains a nonzero
idempotent element.

答:
Consider the $\Bbb{R}$-algebra $A = \Bbb{R}[x]$ comprising polynomials in the variable $x$ with coefficients in the real numbers $\Bbb{R}$. (As $A$ is commutative, we don't have to worry about the distinction between right ideals and left ideals.) If $I$ is a non-zero ideal in $A$, then $xI$, the ideal comprising all polynomials $xf$, where $f \in I$, is properly contained in $I$ (because, if $f$ is a polynomial in $I$ of minimal degree, then $f \not\in xI$). So in $A$, there are no minimal non-zero ideals - any non-zero ideal contains a smaller one. Hence the statement you are trying to prove does not hold in $A$, without some additional restrictions.

Edit: the above relates to the question before the edit telling us that $A$ was expected to be finite-dimensional over $F$. If you know that $A$ is finite-dimensional, any ideal is a finite-dimensional vector space over $F$. So, if $I = I_1 \supset I_2 \supset \ldots$ is a maximal strictly descending sequence of right ideals in $A$, then it is also a strictly decreasing sequence of finite-dimensional vector spaces over $F$. Putting $d_i = \dim_F(I_i)$, the $d_i$ are a strictly decreasing sequence of natural numbers. So the sequence $d_i$ must be finite and must have a least non-zero element (given that $I \neq \{0\}$ so $d_1 > 0$). If $d_k$ is that least non-zero element, then $I_k$ is a minimal non-zero right ideal contained in $I$.


问:
Prove by induction that for all $n \geq 3$: $n^{n+1} > (n+1)^n$
I am currently helping a friend of mine with his preperations for his next exam. A big topic of the exam will be induction, thus I told him he should practice this a lot. As at the beginning he had no idea how induction worked, I showed him some typical examples.

Now he showed me an exercise he was having trouble with, which states that one should prove
$$n^{n+1} > (n+1)^n$$
for all $n \geq 3$. I have to admit that I too have trouble showing this inequality, as in all of my attempts, my lower bound is too low. Also, I have not yet figured out, how one gets the $(n+2)^{n+1}$, primarely the number $2$ is a problem. I think this might be solved using the binomial theorem, however, I don't think they have already seen the binomial theorem in school.

Is there an easy method to show this inequality by induction not using the binomial theorem? If no: How can one show it using the binomial theorem?

Thanks for answers in advance.

ContentLicense: CC BY-SA 3.0

Body:
Hint: show
$$\frac{(n+1)^{n+2}}{(n+2)^{n+1}}> \frac{n^{n+1}}{(n+1)^n}$$
for $n>3$, or
$$(n+2)^{n+1}n^{n+1}<(n+1)^{2n+2}.$$
(This is simply $(n+2)n<(n+1)^2$.)


问:
About the transference of perfect sets under Polish group actions
Let $G$ be a Polish group, $X$ a Polsih space on which $G$ acts continously and consider the orbit equivalence relation on $X$ with respect to $G$.

Suppose $A\subseteq X$ is a perfect set of pairwise non orbit equivalent elements, $B\subseteq X$ is closed and for all $a\in A$, the $G$ orbit of $a$ contains exactly one $G$ orbit of $B$.

My question: Does it follow that $B$ contains a perfect set of pairwise non orbit equivalent elements?

I assume the answer is positive but I cannot prove it. One idea I have been entertaining is to show that for some $g\in G$, there are uncountably many $a\in A$ such that $g\circ a\in B$, where $\circ$ denotes the action of $G$ on $X$. Clearly, if this can be done, then by the perfect set theorem the question is answered.

Any help is appreciated.

答:
Sorry about my earlier false attempt. I think this one now is slightly better.

Consider the Polish space $A\times G$ with a subset $S$, such that $(a,g)\in S$ iff $g\circ a\in B$. Clearly $S$ is closed, and $\operatorname{proj}_A(S)=A$ by assumption. By Jankov-von-Neumann uniformization, there is a $\sigma(\mathbf{\Sigma^1_1})$ measurable function $f:A\to G$ such that $f\subseteq S$. In particular, $f$ is Baire mesurable. Recall that every Baire measurable function is continuous on a comeager set, i.e. we may fix some comeager $G_\delta$ set $C\subseteq A$ such that $f\upharpoonright C$ is continuous. Now because that for any $a\ne a'\in A$ we have $f(a)\circ a\nsim_G f(a')\circ a'$, the set $T=\{f(a)\circ a\mid a\in C\}\subseteq B$ is $G$-independent. As $C$ is comeager $G_\delta$, the set $T$ is also Borel uncountable. By the perfect set property of Borel sets, $T$ has a perfect $G$-independent subset as desired.


问:
poles and geometric series
I am having trouble understanding how to find the poles of a function. I feel like I understand what a pole is but I am not sure about the order of the pole.

I think the reason I struggle to find poles is because I do not have a good understanding of geometric series. Are there different geometric series? If so, how do I know when to use them?

For example, $$\int_{C}{\frac{e^{z}}{(z^{2}+\pi^{2})^{2}}}dz,$$
$C:|z|=4$

How would I go about finding the poles and their order?

Even if I do not need geometric series to solve this could someone please still explain them to me?

答:
In the case you're citing there, $\int_C \frac{e^z}{(z^2+\pi^2)^2}dz$, you have to factor the denominator to find the roots of the polynomial. So since $(z^2+\pi^2)^2 = (z-i\pi)^2(z+i\pi)^2$ we see the poles of $f(z)$ are $\pm i\pi$, and these poles have multiplicity 2. Since your circle has radius $4$ and $|i\pi|=\pi<4$ your poles are inside the contour. In general, you just have to find where the poles are (e.g. by factoring like above) and then, if your contour is a circle, seeing if the distance between the pole and the center of your circle is less than the radius.

I should warn that in this case you still need to find the residues at $\pm i\pi$ to evaluate the integral. Factoring is just the first step.

Title: poles and geometric series


问:
What's the difference between $(\mathbb Z_n,+)$ and $(\mathbb Z_n,*)$?
I noticed that $(\mathbb Z_n,+)$ and $(\mathbb Z_n,*)$ are not the same thing. For example 2 is not invertible in $(\mathbb Z_6,*)$. However, since $\bar2+\bar4=\bar0$, thus it is invertible in  $(\mathbb Z_6,+)$.

I found the following problems:

Am I right? If so, why does any book not mention this difference which is very important?

When we study groups, what's the standard operation of $\mathbb Z_n$ when the operation is omitted?

When we study rings, do we generally use $(\mathbb Z_n,+)$ to be the abelian group of the ring $(\mathbb Z_n,+,*)$?

If I work with $(\mathbb Z_n,+)$ I would lose these theorems which I proved for $(\mathbb Z_n,*)$?

An element $a$ in $(\mathbb Z_n,*)$ is an unit iff $a$ and $n$ are  coprimes

An element $a$ in $(\mathbb Z_n,*)$ is an zero divisor iff $a$ and
  $n$ aren't coprimes

I'm sure there are a lot of student with this same doubt.

I really need help.

Thanks a lot.

答:
$(\mathbb Z_n,+)$ is cyclic. It is a regular polygon in the euclidian plane.

$(\mathbb Z_n,*)$ is a much complicated thing that makes the wonders of great mathematicians (Pythagoras, Gauss, Euler,...).


问:
Can eigenvectors of a real matrix with complex eigenvalues be linearly independent?
I am a bit confused about whether a real matrix with complex eigenvalues can have linearly independent eigenvectors? I do not understand how we can invert this real matrix if it has complex conjugates eigenvalues.</p>

答: A matrix some of whose eigenvalues are complex will not generally have a basis of real eigenvectors (example: the rotation matrix $\begin{pmatrix} \cos \theta & \sin \theta\\ -\sin \theta & \cos \theta\end{pmatrix}.$ As for inverses, these have nothing to do with eigenvalues, so use Cramer's rule, and be happy.


问:
Prove $f: \mathbb{R} \rightarrow \mathbb{Z}$ is an increasing function
Let $f$ be the following function from $\mathbb{R} \rightarrow \mathbb{Z}: f(x) = m$ where m is the unique integer such that $m \le x < m+1$. In other words, $f(x)$ is equal to the largest integer that is less than or equals to x. For instance, $f(2.3) = 2$, $f(1) = 1$, $f(-1.37) = -2$. Prove f is an increasing function.

I know that the definition of increasing function is "if for any $x,y \in A$ such that $x \le y$, we have $f(x) \le f(y)$." I have this unorganized ideas how to prove this problem, and I'm not sure how to proceed.

Let $x \le y$. Set $f(x) = m$ and $f(y) = n$ where $m \le x < m+1$ and $n \le y < n+1$. Then, we have $f(y)-f(x) = n - m$. Since $y \ge x$, we have $n - m \le y - x < n + 1 - m -1 \Rightarrow n - m \le y - x < n - m$. Hence, $f(y)-f(x) = n - m = y - x$.

I don't know how to connect this idea to show that $f(x) \le f(y)$. How do I go to next part? Do I even have the right idea?

EDIT

the definition of increasing function is "if for any $x,y \in A$ such that $x < y$, we have $f(x) \le f(y)$."

Here's my proof with the right definition.

Let $x < y$. Set $f(x) = m$ and $f(y) = n$ where $m \le x < m+1$ and $n \le y < n+1$. Then, we have $f(y)-f(x) = n - m$. Since $y > x$, we have $n - m \le y - x < n + 1 - m -1 \Rightarrow n - m \le y - x < n - m$. Hence, $f(y)-f(x) = n - m = y - x$. Since $y > x$, we have $f(y)-f(x) > 0$. Therefore, $f$ is an increasing function.

I don't know if this does make sense..

答: $m \leq x \leq y <n+1$. But $n$ and $m$ are integers so $m <n+1$ implies $m \leq n$.


问:
maximum cliques of a subgraph of an interval graph
I have a question: It is known that it is easy to find the maximum cliques of an interval graph. I would like to ask whether the same task is also easy when considering the following subgraph of an interval graph:

Let $ G=(\mathcal{I},E) $ be an interval graph and let $ k:\mathcal{I}\rightarrow (0,1) $ be a vertex weight function. Now we delete every edge $ \lbrace i,j \rbrace \in E $ with $ k_i+k_j \le 1 $ from $ E $. The remaining edge set is called $ \tilde{E} $. Moreover, we remove vertices that are not part of an edge anymore, so that we end up with a subgraph $ \tilde{G}=(\tilde{\mathcal{I}},\tilde{E}) $ of $ G $.

It is also easy to find the maximum cliques of this graph? (I would call it a 'mixture' between an interval graph and a threshold graph.)

Thanks for your help!

答:
Each clique in $\tilde G$ contains at most one vertex with weight $\le \frac12$. So we can do casework on which of these vertices, if any, we include:

Find a maximum clique in the subgraph of $G$ containing all vertices with weight more than $\frac12$. (This clique survives in $\tilde{G}$.)

For each vertex $i$ with $k_i \le \frac12$, find a maximum clique in the subgraph of $G$ containing all vertices adjacent to $i$ and with weight more than $1 - k_i$. (Such a clique, together with vertex $i$ itself, is still a clique in $\tilde{G}$.)

Then, just take the largest clique of any found in this process. Note that an induced subgraph of an interval graph is also an interval graph, so all of these steps should be easy.


问:
Two harmonic functions with positive ratio are multiple of each-other?
Suppose $f,g,h:\mathbb{R}^n \to \mathbb{R}$ are functions so that $f$ and $g$ are harmonic and not identically zero, $f=g\cdot h$ and $h\geq 0$. Is $h$ a constant function?

EDIT: Someone voted to close presumably because of a lack of context. The context is that I bounced on this problem while trying to flesh out a derivation of the Ginibre formula of the GUE ensemble. As for arguments why the statement is plausible, let me prove that $h$ cannot reach a strict local maximum or strict local minimum at a (isolated) point $x_0\in O:=\{x\in \mathbb{R}^n:\,f(x)> 0\}$. Suppose, anticipating a contradiction, that $h$ did reach such an (isolated) local minimum at $x_0$, then by the mean value property (fixing the radius $R>0$ sufficiently small so that $B(x_0,R)\subseteq O$)
$$g(x_0)h(x_0)=f(x_0)=\frac{\int_{B(x_0,R)}dx\,f(x)}{\int_{B(x_0,R)}dx\,} = \frac{\int_{B(x_0,R)}dx\,g(x)h(x)}{\int_{B(x_0,R)}dx\,}>\frac{\int_{B(x_0,R)}dx\,g(x)h(x_0)}{\int_{B(x_0,R)}dx\,}=\frac{h(x_0)\int_{B(x_0,R)}dx\,g(x)}{\int_{B(x_0,R)}dx\,}=g(x_0)h(x_0),$$
a contradiction.

答:
For $n>2$, the family of entire harmonic functions $$h_{a,b}:\mathbb{R}^n \to \mathbb{R}:(x_1,...,x_n)\mapsto \sin(x_1)\exp(ax_2)\exp(bx_3),\qquad a^2+b^2=1$$
provides a counterexample. This example came from this paper, which alleges on p.255-256 that such a counterexample does not exist for $n=2$.


问:
Some apparently simple algebraic manipulation I just cant figure out ....
I found the following equation in a paper im reading with the caption, that it is a simple algebraic manipulation without any furhter explanation. I just cant figure out how they did it.

In the following it is $c\in(0,1)$ and $a,b \geq 0$ with $a\neq b$
\begin{align*}
   &
   \frac{2(1-c)}{a-b}
   \left(
      \frac{a}{
         a+1-c+\sqrt{2a+(1-c)^2}
      }
      -
      \frac{b}{
         b+1-c+\sqrt{2b+(1-c)^2}
      }
   \right)
   \\
   =\,\,\,\,\,&
   \frac{1}{a-b}
   \left(
      \frac{4(1-c)}{
         \sqrt{2b+(1-c)^2} + 1 + c
      }
      -
      \frac{4(1-c)}{
         \sqrt{2a+(1-c)^2} + 1+c
      }
   \right)
   \\
   =\,\,\,\,\,&
   \frac{8(1-c)}{
       \sqrt{2a+(1-c)^2}+\sqrt{2b+(1-c)^2}
   }
   \,\,
   \cdot
   \,\,
   \frac{1}{
      (\sqrt{2a+(1-c)^2} + 1+ c)
      (\sqrt{2b+(1-c)^2} + 1+ c)
   }
\end{align*}

Body:
Let me fill in some of the gaps.

$$\frac{2(1-c)}{a-b}\Bigg(\frac{a}{a+1-c+\sqrt{2a+(1-c)^2}} - \frac{b}{b+1-c+\sqrt{2b+(1-c)^2}}\Bigg)$$

$$ = \frac{2(1-c)}{a-b}\Bigg(\Bigg(1-\frac{1-c+\sqrt{2a+(1-c)^2}}{a+1-c+\sqrt{2a+(1-c)^2}}\Bigg) - \Bigg(1-\frac{1-c+\sqrt{2b+(1-c)^2}}{b+1-c+\sqrt{2b+(1-c)^2}}\Bigg)\Bigg)$$

$$ = \frac{2(1-c)}{a-b}\Bigg(\frac{1-c+\sqrt{2b+(1-c)^2}}{b+1-c+\sqrt{2b+(1-c)^2}} - \frac{1-c+\sqrt{2a+(1-c)^2}}{a+1-c+\sqrt{2a+(1-c)^2}}\Bigg)$$

Consider:

$$ \frac{1-c+\sqrt{2b+(1-c)^2}}{b+1-c+\sqrt{2b+(1-c)^2}}$$

$$ =\frac{1-c+\sqrt{2b+(1-c)^2}}{b+1-c+\sqrt{2b+(1-c)^2}}\cdot\Bigg(\frac{1+c+\sqrt{2b+(1-c)^2}}{1+c+\sqrt{2b+(1-c)^2}}\Bigg)$$

I won't write out the expansion (of the numerator), but this simplifies to:

$$\frac{2\big(b+1-c+\sqrt{2b+(1-c)^2}\big)}{\big(b+1-c+\sqrt{2b+(1-c)^2}\big)\big(1+c+\sqrt{2b+(1-c)^2}\big)}$$

$$ = \frac{2}{\big(1+c+\sqrt{2b+(1-c)^2}\big)}.$$

By symmetry, we can do the same to the fraction involving $a$. Substituting into where we left off, we get:

$$ = \frac{2(1-c)}{a-b}\Bigg(\frac{2}{\big(1+c+\sqrt{2b+(1-c)^2}\big)} - \frac{2}{\big(1+c+\sqrt{2a+(1-c)^2}\big)}\Bigg)$$

$$ = \frac{1}{a-b}\Bigg(\frac{4(1-c)}{\big(1+c+\sqrt{2b+(1-c)^2}\big)} - \frac{4(1-c)}{\big(1+c+\sqrt{2a+(1-c)^2}\big)}\Bigg)$$

In the home stretch now:

$$ = \frac{4(1-c)}{a-b}\Bigg(\frac{1}{\big(1+c+\sqrt{2b+(1-c)^2}\big)} - \frac{1}{\big(1+c+\sqrt{2a+(1-c)^2}\big)}\Bigg)$$

$$ = \frac{4(1-c)}{a-b}\Bigg(\frac{\sqrt{2a+(1-c)^2} - \sqrt{2b+(1-c)^2}}{\big(1+c+\sqrt{2b+(1-c)^2}\big)\big(1+c+\sqrt{2a+(1-c)^2}\big)}\Bigg)$$

$$ = \frac{4(1-c)}{a-b}\Bigg(\sqrt{2a+(1-c)^2} - \sqrt{2b+(1-c)^2}\Bigg)\Bigg(\frac{1}{\big(1+c+\sqrt{2b+(1-c)^2}\big)\big(1+c+\sqrt{2a+(1-c)^2}\big)}\Bigg)$$

Consider

$$\sqrt{2a+(1-c)^2} - \sqrt{2b+(1-c)^2}$$

By the difference of two squares again,

$$ = \frac{\big(2a+(1-c)^2\big)-\big(2b+(1-c)^2\big)}{\sqrt{2b+(1-c)^2}\big) + \sqrt{2a+(1-c)^2}}$$

$$ = \frac{2(a-b)}{\sqrt{2b+(1-c)^2} + \sqrt{2a+(1-c)^2}}$$

Substituting in to where we left off again, we get:

$$ = \frac{4(1-c)}{a-b}\Bigg(\frac{2(a-b)}{\sqrt{2b+(1-c)^2} + \sqrt{2a+(1-c)^2}}\Bigg)\Bigg(\frac{1}{\big(1+c+\sqrt{2b+(1-c)^2}\big)\big(1+c+\sqrt{2a+(1-c)^2}\big)}\Bigg)$$

$$ = \Bigg(\frac{8(1-c)}{\sqrt{2b+(1-c)^2} + \sqrt{2a+(1-c)^2}}\Bigg)\Bigg(\frac{1}{\big(1+c+\sqrt{2b+(1-c)^2}\big)\big(1+c+\sqrt{2a+(1-c)^2}\big)}\Bigg)$$

as required.

Not completely trivial (and even worse to write out in LaTeX), so don't worry that you didn't spot it. When doing algebra, always just hold your nerve. Often, it doesn't look anything like the result -- until it suddenly does. Corrections are welcome.


答:
Suppose that the claim holds for $n$, so we have $$n^{n+1} > (n+1)^n$$

For a proof by contradiction, suppose that the claim fails for $n+1$, so we have: $$(n+2)^{n+1} \geq (n+1)^{n+2}$$

Mulpitlying these two inequalities gives:
$$ (n^2 + 2n)^{n+1} = (n(n+2))^{n+1} > (n+1)^{2n+2} = (n^2+2n+1)^{n+1}$$
Clearly, this is impossible. So, the claim for $n+1$ has to hold as well.

Title: Prove by induction that for all $n \geq 3$: $n^{n+1} > (n+1)^n$

Body:
I am currently helping a friend of mine with his preperations for his next exam. A big topic of the exam will be induction, thus I told him he should practice this a lot. As at the beginning he had no idea how induction worked, I showed him some typical examples.

Now he showed me an exercise he was having trouble with, which states that one should prove
$$n^{n+1} > (n+1)^n$$
for all $n \geq 3$. I have to admit that I too have trouble showing this inequality, as in all of my attempts, my lower bound is too low. Also, I have not yet figured out, how one gets the $(n+2)^{n+1}$, primarely the number $2$ is a problem. I think this might be solved using the binomial theorem, however, I don't think they have already seen the binomial theorem in school.

Is there an easy method to show this inequality by induction not using the binomial theorem? If no: How can one show it using the binomial theorem?

Thanks for answers in advance.


问:
How many equivalence classes are there in $J$.?
Let $G$ be the symmetric group $S_5$ of permutations of five symbols. Consider the set $J$ of subgroups of $G$
that are isomorphic to the non-cyclic group of order $4$. Let us call two subgroups $H$ and $K$ belonging to $J$
as equivalent if they are conjugate (that is,  there exists $g $ such that $gHg^{-1}=K$). How many equivalence
classes are there in $J$.?

I have tried a lot but i found only one. But correct answer is two. But how ?

答: Here's a hint: in one class, all the elements of order $2$ are double transpositions, in the other class, there are single transpositions.


答:
First of all, it's clear that the function is analytic when the denominator doesn't vanish, so the only places there could possibly be poles are $z=\pm\pi i$.  We know that $e^z$ never vanishes, so these are indeed poles.

Take $p=\pi i$.  Since the numerator doesn't vanish, we need only be concerned with the denominator.  For the denominator, we have $$\begin{align}
z^2+\pi^2&=(z-\pi i)^2+2\pi i(z-\pi i)\\
&=(z-\pi i)(z-\pi i +2\pi i)\\
&=(z-\pi i)2\pi i\left(1+\frac{z-\pi i}{2\pi i}\right)
\end{align}$$ so that $$\left(z^2+\pi^2\right)^{-2}=\frac{-1}{4\pi^2}(z-\pi i)^{-2}\left(1+\frac{z-\pi i}{2\pi i}\right)^{-2}\tag1$$

Now, by the the binomial formula, $$\left(1+\frac{z-\pi i}{2\pi i}\right)^{-2}= 1 - \frac{z-\pi i}{\pi i}+\text{higher order terms}$$ and we see that the leading term in the Laurent expansion has degree $-2$ so that $\pi i$ is a pole of order $2$.

For a pole of order $2$ we have the formula $$\operatorname{Res}_{z=\pi i} f(z)=\lim_{z\to\pi i}\frac d{dz}\left(f(z)(z-\pi i)^2\right),$$ where $f(z)$ is the integrand.  From $(1)$,
$$\operatorname{Res}_{z=\pi i} f(z)=\lim_{z\to\pi i} \frac d{dz}\frac{-e^z}{4\pi ^2}\left(1+\frac{z-\pi i}{2\pi i}\right)^{-2}$$

Take it from here.

Title: poles and geometric series

Body:
I am having trouble understanding how to find the poles of a function. I feel like I understand what a pole is but I am not sure about the order of the pole.

I think the reason I struggle to find poles is because I do not have a good understanding of geometric series. Are there different geometric series? If so, how do I know when to use them?

For example, $$\int_{C}{\frac{e^{z}}{(z^{2}+\pi^{2})^{2}}}dz,$$
$C:|z|=4$

How would I go about finding the poles and their order?

Even if I do not need geometric series to solve this could someone please still explain them to me?


问:
Vector component paradox
This is taken from the famous 3 polarisation filter experiment on light and got my thinking about coordinate transformations. and vector component consistancy and would really like an answer.

Suppose in the regular coordinate system a vector which is 0i+yj
obviously has a magnitude y

Now, lets say I switch coordinates  such that my new axis are 45 degrees rotated anticlockwise.  with basis vectors i' and j'

my vector is simply ycos(45)i'+ysin(45)j'

now this is where the physics comes in. lets say something gets rid of this j' component of the vector( acting as a polarisation filter)

my new vector is now simply ycos(45)i'

now... when we switch back to my original coordinates. this vector now has an i component and a j component...

which if im not mistaken will be ycos^2(45)i + ycos(45)*sin(45)j

So my question is. how can it be that by switching coordinates. and REMOVING a component in a different frame, not adding anything to it at all. and then switching back to our original coordinates we have magically produced an i component when it never had any before?

this works with light and im totally unsure about how this makes any sense

答:
One metaphorical way to grasp it is that the filter changes the vector.  You could imagine it as rotating and shrinking the vector, instead of just cutting it and removing one component.

Another way to think about it is that the filter removes the $j'$ component, which is like subtracting that much of the vector.  But if you start with zero and subtract you don't get zero.  That's why you can end up with an $x$ component, because the thing you subtracted had an $x$ component.

It comes down to precisely describing "get rid of the $j'$ component."  I have given two ways that might be done.  There are surely others as well.

Title: Vector component paradox


问:
Rearranging Cauchy Riemann equations
I've used the Cauchy-Riemann equations to find the analytic function:

$$y^2−x^2−2y+2+i(2x(1−y))$$

But I'm having a slight rearranging problem and need to write it in terms of $z$, where $z=x+iy$.

Any suggestions would be much appreciated

答: In addition to the formula for $z^2$, note that $iz = ix - y$.</p>

Title: Rearranging Cauchy Riemann equations


问:
Understanding a proof that $\delta(x-x') = \mathrm{d}/\mathrm{d}x\, \theta(x-x')$
A solution to an exercise in Principles of Quantum Mechanics:

My question concerns the leap

This is apparently an application of Integration by Parts; however, it seems to me it is conflating
$$
\int g(x)\,\mathrm{d}\theta(x-x')
$$

with

$$
\int g(x) \theta'(x-x')\,\mathrm{d}x
$$

which is the form required for us to able to apply Integration by Parts.
Is this not the case?

答:
Okay well in terms of understanding why this is the case, visualising it graphically, $\theta(x-x')$ is going to be a step from $0$ to $1$ at the point $x=x'$. Now clearly this is a discontinuity but we can visualise this as a vertical line i.e. infinite gradient at the point $x=x'$. Now if we are to plot the gradient of $\theta$ we would see that it is clearly $0$ everywhere other than at this point as it is a constant function over the two domains. As for the integral you mentioned, it is completely fine to use it in that form. the dirac delta is commonly defined as:
$$\int_{-\infty}^\infty g(x)\delta(x-x')dx=g(x')$$
so bearing this in mind:
$$\int_{-\infty}^\infty g(x)d\theta(x-x')=\int_{-\infty}^\infty g(x)\frac{d\theta(x-x')}{dx}dx$$
$$=\int_{-\infty}^\infty g(x)\delta(x-x')dx=g(x')$$
which is effectively what this is proving


问:
Probability proof of indicator random variable$I_AI_B = I_{A \cap B}$
If I have events $A$ and $B$ over some probability space $(\Omega, Pr)$

And $I$ is the indicator random variable such that $I_A$ is $1$ if $A$ occurs and $0$ if $A$ does not occur

How can I prove that

$I_AI_B = I_{A \cap B}$

I know that $I_AI_B(\omega) = I_A(\omega) \cdot I_B(\omega)$. What do i do from here?

答:
Let $\Gamma = \{ w \in \Omega : w \in A\}$ and $\Xi = \{ w \in \Omega : w \in B\}$ and notice that $\Gamma \cap \Xi = \{ w \in \Omega : w \in A \cap B\}$

\begin{equation*}
\begin{split}
I_A I_B (w) & =
\begin{cases}
1, & \text{ if } I_A(w) = I_B(w) =1\\
0, & \text{ if } I_A(w) = 0 \text{ or } I_B(w) =0
\end{cases}
=
\begin{cases}
1, & \text{ if } w \in \Gamma \text{ and } w \in \Xi\\
0, & \text{ otherwise }
\end{cases}
\\
& =
\begin{cases}
1, & \text{ if } w \in \Gamma \cap \Xi\\
0, & \text{ otherwise }
\end{cases}
=
\begin{cases}
1, & \text{ if } I_{A \cap B}(w) = 1\\
0, & \text{ if } I_{A \cap B}(w) = 0
\end{cases}.
\end{split}
\end{equation*}

In conclusion,
$$I_AI_B = I_{A \cap B}.$$


问:
Amalgamation of matching family for empty cover
Let $\mathcal{J}$ be a Grothendieck topology on a small category $\mathbb{C}$. If $C$ is an object of $\mathbb{C}$ for which $\mathcal{J}(C)$ contains the empty cover/sieve, then if $F : \mathbb{C}^{\mathsf{op}} \to \mathsf{Set}$ is any $\mathcal{J}$-sheaf, it is my understanding that $F(C)$ must be a singleton, since any element of $F(C)$ will be an amalgamation for the (empty) matching family for the empty cover of $C$, and such amalgamations are unique. Is it a 'common' situation for an object to have an empty cover in a Grothendieck topology? Is it possible to assume without loss of generality that no object is covered by the empty sieve, in the sense that there is another site $(\mathbb{D}, \mathcal{K})$ with this property for which the topos of $\mathcal{J}$-sheaves is equivalent to the topos of $\mathcal{K}$-sheaves?

答:
It's extremely common for there to be objects $C$ for which $\mathcal{J}(C)$ contains the empty sieve. Take the Grothendieck topology arising from any topological space in the standard way: then the empty set has an empty cover.

You have the same situation with the join Grothendieck topology on a distributive lattice with all joins and meets - a partial order with all joins will have a least element, and this element will have an "empty cover" since it is the join of the empty set. Not surprising, given that this is a generalisation of the above example.

I'm not sure whether it is always possible to assume WLOG that there is no empty covering of any element.


问:
Find a set of vectors that spans $A$.
Let $A$ be a set of points that satisfies $\begin{bmatrix}a-b\\b-c\\c-a\\b \end{bmatrix}$, with $a,b,c \in \mathbb{R}$. Decide if $A$ is a vector space and if so, choose a set of vectors that span $A$. It seems a bit tedious to check for all the properties of a vector space, so I thought that it might be easier to check if $A$ is a subspace of another space, like $\mathbb{R}^4$.

Id: $A$ contains the zero vector (e.g let $a=b=c=0$)

Closure under addition: For $x,y,z\in\mathbb{R}$ we have $\begin{bmatrix}a-b\\b-c\\c-a\\b \end{bmatrix}$ + $\begin{bmatrix}z-y\\y-z\\z-x\\y \end{bmatrix}$=$\begin{bmatrix}(a+x)-(b+y)\\(b+y)-(c+z)\\(c+z)-(a+x)\\b+y \end{bmatrix}$.

For scalar (real) multiplication I suppose it is quite straight forward.

I am having difficulties when it comes to finding a set of vectors of the given form that spans $A$. A hint or solution would be appreciated.

答: You can write $A = \{ae_1 + be_2 + ce_3 \mid a,b, c \in \mathbb R\},$ where $e_1 = (1,0,-1,0), e_2 = (-1,1,0,1),$ and $e_3 = (0,-1,1,0).$ Then it should be very easy to see that $A$ is a vector space, namely the space spanned by those three vectors. They are even linearly independent, so they are a basis for $A,$ and $A$ is a 3-dimensional subspace of $\mathbb R^4.$

Title: Find a set of vectors that spans $A$.


问:
Dual basis example and the isomorphism $V\cong V^{\ast }$
I am trying to understand the dual spaces and working on some problems.

Let $V=\mathbb{R}^{3} $ with the base $b=\left \{ (1,0,1),(1,2,1),(0,0,1) \right \}$ find the dual basis $ b^{\ast }$ of the dual space $ V^{\ast }$

let $ b^{\ast }=\left \{ f_1,f_2,f_3 \right \}$

i did some standar calculations and found that

$ f_1(x,y,z)=x-\frac{y}{2}$

$f_2(x,y,z)=\frac{y}{2}$

$f_3(x,y,z)=-x+z$

Could someone tell what this exactly means? who is the dual space $V^{\ast }$ in the end? Which function takes us from $V$ to $V^{\ast }$

I hope my questions have a meaning because i am a little confused at the moment.

答:
The dual space $V^*$ is the set of all linear maps $V\to\Bbb R$, endowed with the usual (=pointwise) addition and scalar multiplication.

A specific isomorphism $V\to V^*$ through the given example can be given, specifically map  the basis elements $u_1,u_2,u_3$ in $b$ to the basis elements $f_1,f_2,f_3$ in $b^*$. More specifically,
$$(\alpha u_1+\beta u_2+\gamma u_3)\ \mapsto \ (\alpha f_1+\beta f_2+\gamma f_3)\,.$$

Nevertheless, it's easier to establish an isomorphism through the standard basis and its dual, and actually the standard inner product may also come handy, as using that, the standard isomorphism can simply be written as:
$$v\mapsto\ (w\mapsto v^Tw)\,.$$


问:
$x_n: |\sum_{n=1}^\infty \sin(n)(x_n - x_{n-1})| = 2\sin(\frac12)$
Let's consider functional $$L(x_n) = \sum_{n=1}^\infty \sin(n)(x_n - x_{n-1})$$

I'm looking for such a sequence $x_n$ such that $\sum_{n=1}^\infty |x_n| \le 1$ and $|L(x_n)| = 2\sin(\frac12)$

I was mostly trying to point a sequnece which has only one value at certain point and zeroes elsewhere. (easiest thing to calculate and easiest way to match condition of bounded sum). Apart from that I also tried sequneces with several points but those also bring me to nothing (I was using R software to check a lot of possibilites). Could you please give me a hand with figuring out the accurate example ?

EDIT
I got your answers but I don't understand one thing within it:

You are defining sequence $x_n$ in such way that this is $x_n = 1$ for $n = u_k$ and $0$ otherwise, and then you are describing $L(x_n)$.

But why $\sum_{n=1}^\infty |x_n| \le 1$ i.e. $\exists!n_0: x_{n_0} = u_k$ ?

Is this obvious ? I was trying to find argument for that but I wasn't able to

答:
Rearranging your series, it becomes
$$\sum_{n\ge 1}x_n(\sin(n) - \sin(n+1)).$$
Applying the trigonometric identity,
$$\sin(n) - \sin(n+1) = -2\sin(1/2)\cos(n+1/2),$$
we find that you seek for a family $x_n$ for which
$$\sum_{n\ge 1}x_n\cos(n+1/2) = -1.$$
Now we see why there isn't any solution. We may bound
$$|\sum_{n\ge 1}x_n\cos(n+1/2)| \le \sum_{n\ge 1}|x_n\cos(n+1/2)|,$$
since $|\cos(n+1/2)| < 1$ for all $n\in \mathbb{N}$, (why?), we have $|x_n\cos(n+1/2)| < |x_n|$ for all $n\in \mathbb{N}$, such that
$$|\sum_{n\ge 1}x_n\cos(n+1/2)| < \sum_{n\ge 1}|x_n| \le 1.$$
Therefore there aren't any solutions $x_n$ to your equation.

Is this a homework question? If so, please add the tag homework.


问:
How to find the Taylor series of $\frac{1}{1+x^3}$ at $0$
I can't find a way to calculate the n-th derivative of $$f(x)=\frac{1}{1+x^3}$$ in order to find its Taylor series at $0$. How could one find its Taylor series?

答:
The Taylor series of $\frac {1}{1+x} = 1 - x + x^2 - x^3 + \cdots$

To find the Taylor series of $\frac {1}{1+x^3}$ substitute $x^3$ for $x$ in the series above.


问:
How many letters to be generated to get a specific string?
I was reading a post about how the digits of pi contain every known sequence of numbers, was curious if I could narrow this to solve this puzzle?

I have a computer program (hypothetical) that generates a string of letters, chosen randomly, independently, and uniformly from the alphabet.

How many letters have to be printed so that the expected number of occurrences of the string "ABCDEFGH" are printed is 1?

答: This has nothing to do with the (presumed) normality of $\pi$.  The probability that a given $8$-character string is ABCDEFGH is $26^{-8}$, so for the expected number of such strings to be $1$, by linearity of expectation, we need $26^8$ $8$-character strings, so we must have $26^8+7$ characters.


问:
Intuition behind conjugation in group theory
I am learning group theory, and while learning automorphisms, I came across conjugation as an example in many textbooks. Though the definition itself, (and when considering the case of abelian groups), it seems pretty innocent, I have to admit that I have no intuition about whats happening, or why even such a map is important. Did anyone feel this way when you learnt it? Any examples/references that you have found useful when you learnt it?</p>

答:
Let $ G $ be a group. With no further specifications, can we form any interesting isomorphisms from $ G $ to $ G $ ?

For each $ g \in G $, we can try to form an isomorphism $ \varphi_g $ from $ G $ to $ G $ using $ g $.

Let's try maps of the form $ t \mapsto g^k t^l g^m $ $(k,l,m \in \mathbb{Z})$. We want $ g^k (t_1 t_2)^l g^m $ $ = g^k t_1^{l} g^m g^k t_2^{l} g^m $ to hold for all $ t_1, t_2 \in G $, and this can be made to happen by setting $ \{ k = (-m), l = 1 \} $. Maps $ t \mapsto g^{-m}tg^{m} $ are bijections too, giving us a family $ \{ t \mapsto g^{-m}tg^{m} : g \in G, m \in \mathbb{Z} \} $ $ ( = \{ t \mapsto g^{-1}tg : g \in G \} ) $ of isomorphisms from $ G $ to $ G $. These isomorphisms are called conjugations.

[ We could've tried, instead of $ g^k t^l g^m $, some longer more general string of $ g $s and $ t $s, but I can't seem to get any more isomorphisms this way ]

ContentLicense: CC BY-SA 2.5


答:
Few different ways to consider:

Brute force differentiation of $\frac{1}{1+x^3}$ - not impossible (esp. if Q asks for only low terms).

Find the expansion of $\frac{1}{1+x}$ and substitute x^3 in (easier)

Notice that $\frac{1}{1+x^3}$ looks suspiciously like the form $\frac{a}{1-r}$ which is the closed form of a geometric series when r < 1. Perhaps this would give you a series representation with suitable substitution.

Title: How to find the Taylor series of $\frac{1}{1+x^3}$ at $0$

Body: I can't find a way to calculate the n-th derivative of $$f(x)=\frac{1}{1+x^3}$$ in order to find its Taylor series at $0$. How could one find its Taylor series?


问:
How prove this and for any non-negative integer $k$, there exists a non-negative integer $k'$ such that $v_p(f(k))<v_p(f(k'))\leq v_p(f(k))+m.$
For a given positive integer $n$ and prime number $p$, find the minimum value of positive integer $m$ that satisfies the following property: for any polynomial $$f(x)=(x+a_1)(x+a_2)\ldots(x+a_n)$$ ($a_1,a_2,\ldots,a_n$ are positive integers), and for any non-negative integer $k$, there exists a non-negative integer $k'$ such that $$v_p(f(k))<v_p(f(k'))\leq v_p(f(k))+m.$$ Note: for non-zero integer $N$,$v_p(N)$ is the largest non-zero integer $t$ that satisfies $p^t\mid N$.

The answer is $\boxed{n+\nu_p(n!)}$. To show that $m\ge n+\nu_p(n!)$ is necessary (which is the important motivation), take
$$f(x)=(x+p)(x+2p)...(x+np).$$
and $k=1$. It's easy to see that whenever $p$ divides $f(k')$ then $p\mid k'$. Letting $k' = pr$, we notices that
$$f(k') = p^n(r+1)(r+2)...(r+n)\implies p^nn!\mid f(k')$$
so $m\ge n+\nu_p(n!)$ as desired.

Now we show that $m=n+\nu_p(n!)$ works. then I can't it

答:
Note: In this answer, I will ignore the requirement that $k'$ be positive. This is not an issue, since we may always replace $k'$ with $k' + p^s$ for a sufficiently large $s$.

Lemma: For any positive integer $n$ and any monic polynomial $f \in \Bbb Z[x]$ of degree $n$, there exists an integer $k$ such that $v_p(f(k)) \leq v_p(n!)$.

Proof: It is an easy exercise to show that $$\sum_{k = 0}^n (-1)^{n - k}\binom n k f(k) = n!$$ and hence at least one of $k = 0, 1, \dots n$ satisfies $v_p(f(k)) \leq v_p(n!)$.

Proposition: Let $a_1, \dots, a_n$ be integers, and let $k$ be an integer such that $k + a_i\neq 0$ for every $i$. Then there exists an integer $k'$ such that $$v_p\left(\prod_{i = 1}^n(k + a_i)\right) < v_p\left(\prod_{i = 1}^n(k' + a_i)\right) \leq v_p\left(\prod_{i = 1}^n(k + a_i)\right) + n + v_p(n!).$$

Proof: Without loss of generality, we assume that $v = v_p(k + a_1)$ is the maximum among all $v_p(k + a_i)$. We also assume that $v_p(a_i - a_1) > v$ for $i = 1, \dots, r$ and $v_p(a_i - a_1) \leq v$ for $i = r + 1, \dots, n$.

I claim that there exists $k'$ of the form $k' = p^{v + 1}x - a_1$ which satisfies the required property.

Indeed, if $k' = p^{v + 1}x - a_1$ for some integer $x$, then it is easy to show that $v_p(k' + a_i) = v_p(k + a_i)$ for $i = r + 1, \dots, n$. Moreover, we have $$\prod_{i = 1}^r(k' + a_i) = \prod_{i = 1}^r (p^{v + 1}x + a_i - a_1) = p^{r(v + 1)}\prod_{i = 1}^r(x + \frac{a_i - a_1}{p^{v + 1}}).$$
By the Lemma, we may choose $x$ such that the $v_p$ of the right hand side is $\leq r(v + 1) + v_p(r!)$.

Since $v_p(x + a_i)$ is equal to $v$ for $i = 1, \dots, r$, we get $$r + v_p\left(\prod_{i = 1}^n(k + a_i)\right) \leq v_p\left(\prod_{i = 1}^n(k' + a_i)\right) \leq r + v_p(r!) + v_p\left(\prod_{i = 1}^n(k + a_i)\right).$$ This finishes the proof, as $1 \leq r \leq n$ holds.


问:
Obscurity Calculating Partial Derivative at a Point Using Limit Definition
In the beginning of a chapter that speaks about second order directional derivatives, my textbook presents
the following example that shows that generally, the order by which we derive a certain function may matter greatly.
Unfortunately I got stumped by some step along the way. The example goes as follows:

$$
f:\mathbb{R}^2 \rightarrow \mathbb{R}
$$
$$
  (x,y) \longrightarrow \begin{cases}
      \frac{x^2 - y^2}{x^2 + y^2}xy & (x,y)\neq(0,0)\\
       0 & (x,y)=(0,0)
      \end{cases}
$$
As the function is broken up at the point $(0,0)$ we must use the limit definition
to calculate the partial derivatives at $(0,0)$. So, the way I thought it should be done is by calulating as follows:
$$
    \frac{f_{|(0,0),(1,0)}(t) - f_{|(0,0),(1,0)}(0)}{t}=\frac{\frac{t^2 - 0}{t^2}t\cdot0 - 0}{t} = 0\xrightarrow[t \rightarrow 0]{}0
$$
I thought that at this point we can safely say $\frac{\partial f}{\partial x}(0,0) = 0$.
However, my textbook, though it starts as I did, goes further to say:
$$
    \text{when } y \neq 0 \text{ we have } \frac{f(x,y) - f(0,y)}{x - 0} =
    \frac{x^2 - y^2}{x^2 + y^2}y \xrightarrow[x \rightarrow 0]{}-y
$$

And thus we have $\frac{\partial f}{\partial x}(0,y) = -y$.
Next the second order is calculated: $\frac{\partial^2 f}{\partial y \partial x}(0,0) = -1$.
The same process is repeated in the reverse order and we get $\frac{\partial^2 f}{\partial x \partial y}(0,0) = 1$.

I went over the definitions of directional derivative and partial derivative
yet I still don't fully understand the process. Why do we calculate $\frac{\partial f}{\partial x}(0,y)$, that is for all $y$?
Isn't $\frac{\partial f}{\partial x}(0,0)$ suppose to be enough to find the second order in the $y$ direction?

Thanks!

答:
Your function $f$ is a classic example. You should contrast it with
$$
  G(x,y) = \begin{cases}
       xy &\,\,\,\, |y|\leq|x|\\
       -xy &\,\,\,\, |y|>|x|.
      \end{cases}
$$

The point of each example is to show that, in your words, "the order may matter greatly"; this despite the fact that each of $f$ and $G$ is continuous at the origin. One of several impressive things about $f$ is that both (first-order) partial derivatives are continuous at the origin, as functions of two variables. (You could investigate if this is also true for $G$.)

As for your question, I think you answered it yourself. The derivative of $-y$ with respect to $y$ is $-1$.

And thus we have $\frac{\partial f}{\partial x}(0,y) = -y$. Next the
second order is calculated: $\frac{\partial^2 f}{\partial y \partial
 x}(0,0) = -1$.

The object $\frac{\partial f}{\partial x}(0,0)$ is just a number, a constant! You need a function of an independent quantity (for us, $y$) to find a derivative with respect to $y$.

$\frac{\partial f}{\partial x}(0,0)$ alone is not useful for finding a second-order partial.


答:
Let us define the event
$$ E_i = \textrm{ "the string $s$ occurs from letter $i$ to letter $i+7$" }. $$

The event $E_i$ are not independent of each other. However, we can still use the linearity of the expected value, since the independence is not needed.

Let us compute
$$ \mathbb{E}[\mathbb{1}_{E_i}] = \mathbb{P}(E_i) = \frac{1}{26^8} . $$

Then notice that the expected number of occurrence after $N$ letters is (by linearity)
$$ \mathbb{E}\left[\sum_{i=1}^{N-7} \mathbb{1}_{E_i} \right] = \sum_{i=1}^{N-7} \mathbb{E}[\mathbb{1}_{E_i}] = (N-7) * \frac{1}{26^8} .$$

Thus, in order to have
$$ 1 = (N-7) * \frac{1}{26^8} $$
we should choose
$$ N = 26^8 + 7 . $$

Title: How many letters to be generated to get a specific string?

Body:
I was reading a post about how the digits of pi contain every known sequence of numbers, was curious if I could narrow this to solve this puzzle?

I have a computer program (hypothetical) that generates a string of letters, chosen randomly, independently, and uniformly from the alphabet.

How many letters have to be printed so that the expected number of occurrences of the string "ABCDEFGH" are printed is 1?


问:
Prove by induction that for all $n \geq 3$: $n^{n+1} > (n+1)^n$
I am currently helping a friend of mine with his preperations for his next exam. A big topic of the exam will be induction, thus I told him he should practice this a lot. As at the beginning he had no idea how induction worked, I showed him some typical examples.

Now he showed me an exercise he was having trouble with, which states that one should prove
$$n^{n+1} > (n+1)^n$$
for all $n \geq 3$. I have to admit that I too have trouble showing this inequality, as in all of my attempts, my lower bound is too low. Also, I have not yet figured out, how one gets the $(n+2)^{n+1}$, primarely the number $2$ is a problem. I think this might be solved using the binomial theorem, however, I don't think they have already seen the binomial theorem in school.

Is there an easy method to show this inequality by induction not using the binomial theorem? If no: How can one show it using the binomial theorem?

Thanks for answers in advance.

Body:
For the induction step, we want to prove that
 $$\frac{(k+1)^{k+2}}{(k+2)^{k+1}}\gt 1,$$
given that
$$\frac{k^{k+1}}{(k+1)^{k}}\gt 1.$$
Note that
$$ \frac{(k+1)^{k+2}}{(k+2)^{k+1}}=\left(\frac{(k+1)^{k+2}}{(k+2)^{k+1}}\cdot \frac{(k+1)^k}{k^{k+1}}\right)\left(\frac{k^{k+1}}{(k+1)^{k}}\right).$$
So it is enough to prove that
$$\frac{(k+1)^{k+2}}{(k+2)^{k+1}}\cdot \frac{(k+1)^k}{k^{k+1}}\gt 1.\tag{$1$}$$
Rewrite the left-hand side of $(1)$ as
$$\frac{(k+1)^{2k+2}}{(k+2)^{k+1}k^{k+1}}.\tag{$2$}$$
But the expression $(2)$ is the $(k+1)$-th power of $\dfrac{(k+1)^2}{k(k+2)}$, and $(k+1)^2\gt k(k+2)$.


问:
Simplifying an 8-bit boolean expression
I am asked to find a minimum SOP implementation of a Hex to Braille code converter. Assuming my truth table is accurate, and D stands for "Don't Care", an 8-bit K-map seems impractical. Are there any other tools for hand analysis other than basic boolean algebra simplification?

答:
Tools like Logic Friday 1 or Espresso are able to minimize a truth-table with multiple outputs and don't cares.

With Logic Friday 1, your truth-table looks as follows:

It is possible to minimize the number of product terms in different ways:

The result is one minimized sum-of-products expression per output:

You can turn this into a multi-level circuit of Boolean gates:

In case you want to experiment with Karnaugh maps including don't cares, the following web application is helpful:

Karnaugh-Veitch Map, Marburg University

Note that a four-bit map is sufficient. You need one map per output.

Title: Simplifying an 8-bit boolean expression


问:
Find length BC of triangle with incircle and circumcircle

Some thoughts... some chord theorem to get the angle between PQ and BC... AB and AC are tangent to the circle, there has to be another theorem about that, perhaps ADE is isosceles and that helps looking at the angled PDE and AEQ, and with all this one should be able to figure the angle at A, and once I have that probably I can use yet another theorem on chords to determine the length BC from the length PQ...idk

答:
Let $r$ and $R$ be the radii of the incircle and circumcircle, respectively. Then

$$\frac r{4R} =\sin \frac A2 \sin \frac B2 \sin \frac C2\tag1$$
Note that $ab= PD\cdot DQ =6$ and $ac= PE\cdot EQ =4$
$$r\cos \frac A2 =a \sin\frac A2 =\frac{DE}2=1$$
$$2R\sin A = BC = b+ c=\frac{6+4}{a}=10\sin\frac A2\tag2
$$
which leads to the ratio $\frac rR = \frac25$. Also

$$\tan \frac B2 = \frac r{b} = \frac{ra}6 = \frac1{3\sin A}, \>\>\>\>\>\>\>\tan \frac C2 = \frac1{2\sin A}
$$
Substitute above into (1)

$$\frac1{10} =\frac{\sin\frac A2}{\sqrt{(9\sin^2A+1)(4\sin^2A+1)}}
=\frac{\sqrt{\frac{1-\cos A}2}}{\sqrt{(10-9\cos^2 A)(5-4\cos^2A)}}\tag3
$$

which has the solution $\cos A=0$, or $\sin\frac A2 = \frac1{\sqrt2}$. Thus, per (2)
$$BC = 10\sin \frac A2 = 10\cdot \frac1{\sqrt2} = 5\sqrt2$$

Edit: as pointed out by @Intelligenti pauca, $\cos A= \frac56$ is also a valid solution to (3), which leads to
$BC = \frac5{\sqrt3}$.


问:
Evaluating the expected value of a stochastic process
Suppose we have a stochastic process

$$dX_t = X_t\mu dt+X_t\sigma dW_t$$
where $W_t$ is a Wiener process.
How does one evaluate the following expected value:

$$E\left[{\frac{1}{\sqrt{X_t}}}\right]$$

or in more detalized fashion,

$$E\left[X_0^\frac{-1}{2}e^{-\frac{1}{2}\left((\mu-\frac{1}{2}\sigma^{2})t+\sigma W_t\right)}
\right]$$

答:
To evaluate the expectation, we can rearrange the expression you have given us as
$$ E\left[X_0^{-\frac{1}{2}} e^{-\frac{1}{2}\left(\left(\mu - \frac{1}{2}\sigma^2\right)t + \sigma W_t\right)}\right] = E\left[X_0^{-\frac{1}{2}} e^{-\sigma\frac{1}{2}W_t - \frac{1}{4}\sigma^2t}e^{\frac{1}{2}\left(\sigma^2 - \mu\right)t}\right] = e^{\frac{1}{2}\left(\sigma^2 - \mu \right)t}E\left[X_0^{-\frac{1}{2}}\right]E\left[e^{-\frac{1}{2}\sigma W_t - \frac{1}{4}\sigma^2 t}\right]$$
The right most expectation is 1 as $\exp\left(-\frac{1}{2}\sigma W_t - \frac{1}{4}\sigma^2 t\right)$ is a martingale. This is an example of an exponential martingale. It is easy to check that for any $\theta \in \mathbb{R}$, the process $\left(\exp{\left(\theta W_t - \frac{1}{2}\theta^2 t\right)}\right)_{t\geq 0}$ is a martingale. Thus
$$E\left[\frac{1}{\sqrt{X_t}}\right] = e^{\frac{1}{2}\left(\sigma^2 - \mu\right)t}E\left[X_0^{-\frac{1}{2}}\right]$$


答:
For any event $E$, the indicator $\mathbf I_E$ has value $1$ when the event occurs, else $0$ when it does not.

Thus $\mathbf I_{A}\,\mathbf I_{B}$ will have value $1$ when both events $A$ and $B$ occur, else $0$ when one does not.

Also $\mathbf I_{A\cap B}$ will have value $1$ when both events $A$ and $B$ occur, else $0$ when one does not.

$\therefore\quad\mathbf I_{A\cap B}=\mathbf I_A\,\mathbf I_B$

That is all.

Title: Probability proof of indicator random variable$I_AI_B = I_{A \cap B}$

Body:
If I have events $A$ and $B$ over some probability space $(\Omega, Pr)$

And $I$ is the indicator random variable such that $I_A$ is $1$ if $A$ occurs and $0$ if $A$ does not occur

How can I prove that

$I_AI_B = I_{A \cap B}$

I know that $I_AI_B(\omega) = I_A(\omega) \cdot I_B(\omega)$. What do i do from here?


答:
Assume inequality holds for $k$
$$
k^{k+1} > (k+1)^k \\
k^k k > (k+1)^k \\
\left( \frac {k+1}k\right )^k < k \\
\left( 1 + \frac 1k \right )^k < k
$$
Now, one needs to check which side is larger for
$$
(k+1)^{k+2}\ ?\ (k+2)^{k+1} \\
(k+1)^{k+1} (k+1)\ ?\ (k+2)^{k+1} \\
k+1 \ ? \ \left( \frac {k+2}{k+1}\right )^{k+1} \\
k+1 \ ? \ \left( 1 + \frac 1{k+1}\right )^{k+1} \\
$$
Consider this chain of inequalities
$$
\left(1+\frac 1{k+1} \right)^{k+1} > \left(1+\frac 1k \right)^{k+1} = \left(1+\frac 1k \right)^k \left( 1+ \frac 1k\right)
$$
Now, use inequality that assumed to be true
$$
\left(1+\frac 1k \right)^k \left( 1+ \frac 1k\right) < k \left( 1+ \frac 1k\right) = k + 1
$$
So, the final inequality is
$$
k+1 > \left( 1 + \frac 1{k+1}\right )^{k+1}
$$
or alternatively
$$
(k+1)^{k+2} > (k+2)^{k+1}
$$</p>


问:
$|(f, Tf)| \leq (f, |T|f)$?
Let $H$ be a separable and complex Hilbelt space.
If $T$ is a bounded linear operator on $H$, then is it true that $|(f, Tf)| \leq (f, |T|f)$ for all $f \in H$ ?

Observation.
If $H = L^2(\mathbb{R}^d)$ and $T_g f:=gf$ with $g \in L^\infty$, the above statement is true because
$$
|(f, T_g f)| \leq \int_{\mathbb{R}^d} |f(x)|^2 |g(x)| dx = (f, |T_g|f).
$$
Note that $|T_g| = \sqrt{(T_g)^* T_g} = \sqrt{T_{\overline{g}} T_g} = \sqrt{T_{|g|^2}} = T_{|g|}$.

答: I believe this corresponds to  a (known) characterization of singular value decomposition as a trace maximization problem (this was certainly known to von Neumann). In finite dimension this is definitely true, in infinite dimension obvious for unitary and self-adjoint case, not 100% certain in general.</p>

Title: $|(f, Tf)| \leq (f, |T|f)$?


答:
You have three parameters $a,b,c$ in the general vector. So you want to split up the general vector into a sum of three vectors, each of which involving only one parameter. What I mean is, write
$$\begin{bmatrix}a-b\\b-c\\c-a\\b\end{bmatrix}=a\begin{bmatrix}1\\0\\-1\\0\end{bmatrix}+b\begin{bmatrix}-1\\1\\0\\1\end{bmatrix}+c\begin{bmatrix}0\\-1\\1\\0\end{bmatrix}$$
so the first vector only involves $a$, the second only involves $b$, and the third only involves $c$. Then you can see that these three vectors (on their own, without the $a,b,c$) span the subspace.

Title: Find a set of vectors that spans $A$.

Body:
Let $A$ be a set of points that satisfies $\begin{bmatrix}a-b\\b-c\\c-a\\b \end{bmatrix}$, with $a,b,c \in \mathbb{R}$. Decide if $A$ is a vector space and if so, choose a set of vectors that span $A$. It seems a bit tedious to check for all the properties of a vector space, so I thought that it might be easier to check if $A$ is a subspace of another space, like $\mathbb{R}^4$.

Id: $A$ contains the zero vector (e.g let $a=b=c=0$)

Closure under addition: For $x,y,z\in\mathbb{R}$ we have $\begin{bmatrix}a-b\\b-c\\c-a\\b \end{bmatrix}$ + $\begin{bmatrix}z-y\\y-z\\z-x\\y \end{bmatrix}$=$\begin{bmatrix}(a+x)-(b+y)\\(b+y)-(c+z)\\(c+z)-(a+x)\\b+y \end{bmatrix}$.

For scalar (real) multiplication I suppose it is quite straight forward.

I am having difficulties when it comes to finding a set of vectors of the given form that spans $A$. A hint or solution would be appreciated.


问:
How Do Mathematicians ‘make up’ or perhaps more correctly ‘derive’ equations
I know this is definitely a thing otherwise we wouldn’t have equations.

What are the methods for making up an equation from scratch and from making up an equation from an existing equation?

The only way I have been able to do this is using dimensional analysis and I’d like to learn other ways, if you want to see an example of me using dimensional analysis to ‘make up an equation’ see my answer to this question.

How for example can you make up the following without experimentation:

Speed equation:
$$\text v=\frac{\text d}{\text t}$$

Density equation:
$$\text p=\frac{\text m}{\text v}$$

After seeing some answers, how do you make up an equation for a definition? @littleO tells me the velocity equation is a definition and I’d like to know how the guy who made the equation went about it.

Or how do you look at a square and come up with a quadratic equation?

Are there any equations you can make up without experimentation?

If this is too general, specific examples are welcome.

答: Do an experiment. Now, do it again, but change just one of the parameters (e.g., the temperature, or the amount of substance, or the volume) while keeping all the other parameters the same. Now do it fifty more times, recording the results of all these experiments. Now look at your table of all the different results you got, each one with its corresponding input value. Draw a graph showing the dependence of the output on the input. Compare it to all the graphs you know about. If the graph fits one of the ones in your arsenal, congratulations! You have discovered a law of Nature.</p>


问:
How to prove $\sum_{n=0}^\infty \frac{(2n)!}{(2^nn!(2n+1))^2} = \frac{\pi}{2}\ln(2)$?
I was messing around with some integrals and series when I arrived at the result (and WolframAlpha agrees):

$$\sum_{n=0}^\infty \frac{(2n)!}{(2^nn!(2n+1))^2} = \sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{(2n+1)^2} = \frac{\pi}{2}\ln(2)$$

I therefore have a proof, but I would like to see how a proof would be done the other way around (i.e. starting with this series). Notice that the series could also be expressed with double factorials:

$$\sum_{n=0}^\infty \frac{(2n)!}{(2^nn!(2n+1))^2} = \sum_{n=0}^\infty \frac{(2n-1)!!}{(2n)!!}\frac{1}{(2n+1)^2}$$

My Derivation

My derivation is quite simple. I was considering (and trying to solve) the integral

$$\int_0^1 \frac{\arcsin(x)}{x} \, dx$$

I couldn't work it out, however I found the exact same question here on math exchange and the answers show that the integral equals $\frac{\pi}{2}\ln(2)$. What I did found however, was that I could plug the Taylor series for $\arcsin(x)$ (the proof of the taylor series can be found here) into the original series to only be left with a series. Here's the Taylor series for $\arcsin(x)$:

$$
  \arcsin(x) = \sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{ x^{2n+1}}{2n+1}
$$

So plugging in the series above into the integral gives us:

$$\int_0^1 \frac{1}{x} \sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{ x^{2n+1}}{2n+1} \, dx$$

Now here we have to be careful, but since the series works on the integral's interval $[0,1]$ and the summand is only positive, we can interchange the the integral and the sum, leaving us with:

$$\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{2n+1} \, \int_0^1 \frac{x^{2n+1}}{x} \, dx$$

And then solving the integral in the sum gives us the final result:

$$\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{(2n+1)^2} = \frac{\pi}{2}\ln(2)$$

And then expanding the binomial gives the original series in the title.

答:
$$\frac{(2n)!}{(2^nn!(2n+1))^2}=\frac{(2n)!}{2^{2n}(n!)^2(2n+1)^2}={2n\choose n}\frac{1}{2^{2n}(2n+1)^2}={2n\choose n}\frac{1}{(2n+1)^2}\left(\frac12\right)^{2n}$$
Now try and find a function whose taylor series fits this summation


问:
Finding the determinant of an $n\times n$ matrix where $det(A)=det(M)\cdot det(N)$.

Consider an $n\times n$ matrix of the form
$$
A=\begin{pmatrix} M & P \\ 0 & N \\\end{pmatrix}
$$
where $M\in K^{r\times r},P\in K^{r\times s},N\in K^{s\times s}$, and $0\in K^{s\times r}$ is the zero $s\times r$ matrix, $n=r+s$. Show that for such matrices, we have
$$
det(A)=det(M)\cdot det(N).
$$

To answer this question, I think that I want to use row operations on $M$ and $N$ that bring $M, N$ and $A$ into upper triangular form, but I'm not sure on how to do this with "unknown" matrices.

答:
@RobertTheTutor is right here. Remember, for upper triangular matrices, the product of the values on the diagonal equals the determinant. You can reduce any matrix into an upper triangular form. This is because as shown here in method 2, that there always exist a $P$ such that:

$$A_{tri} = P^{-1} * A * P $$

By determinant properties,

$$det(A_{tri}) = det(P^{-1}) * det(A) * det(P) = \frac{1}{det(P)} * det(A) * det(P) = det(N)$$

$$det(A_{tri}) == det(A)$$

You just need ot show that the value of $P$ doesn't affect anything. This works since we don't care what happens to $P$, so just reduce $\begin{bmatrix} M & P\end{bmatrix}$ as if you are reducing $M$ alone, and you will get $M_{tri}$ on the left. Do the same with $\begin{bmatrix}0 & N\end{bmatrix}$, and since there are only zeros on the left, you will still just have zeros on the left. Now you have:

$\begin{bmatrix} M_{tri} & P_{junk} \\
 0 & N_{tri}\end{bmatrix}$

Which is an upper triangular matrix where the product along the diagonal equals the product along $M_{tri}$ times the product along $N_{tri}$, which is $det(A) = det(M)*det(N)$


问:
Mean value theorem for essentially bounded functions
I have the problem with the following:

Let $f \in L^{\infty}(\mathbb{R}_+)$ is the mean value theorem in the following form:

Let $0 \leq a < b < \infty$, then

$\int_{a}^{b} |f| \ d\mu = m(b-a)$, for some $m \leq \mathrm{ess} \sup |f|$.

valid for $f$?

We can see that
$$\int_{a}^b |f| \ d\mu \leq \mathrm{ess} \sup |f|(b-a)$$

thus $f \in L^1([a,b])$. Then $\mathrm{ess} \inf |f| = - \mathrm{ess} \sup |f|$, so

$$\mathrm{ess} \inf |f|(b-a) \leq \int_{a}^b |f| \ d\mu.$$ Thus the intermidiate value theorem applied to $|f|$ shows that there exists $x \in [a,b]$ such that $|f(x)| \in [\mathrm{ess} \inf |f|,\mathrm{ess} \sup|f| ]$ and
$$ \int_{a}^b |f| \ d\mu = |f(x)| (b-a) .$$
Is it correct?

How about the following:

Let $0 \leq a < b < \infty$,  does it exist a complex number $m$ such that

$\int_{a}^{b} f \ d\mu = m(b-a)$, for some $|m| \leq \mathrm{ess} \sup |f|$.

I think for that we could use that $f= \mathrm{Re}f + i \mathrm{Im} f$ and then apply thesis from the first question (if it is true).
Thank you for any help.

答:
Some parts are correct, others are not. You observed correctly that $\def\esssup{\mathop{\mathrm{ess\,sup}}}\def\essinf{\mathop{\mathrm{ess\,inf}}}\def\abs#1{\left|#1\right|}$
$$
 \essinf \abs f \cdot (b-a) \le \int_a^b \abs f \, dx \le \esssup \abs f \cdot (b-a) \tag 1
$$
But in in gerenal doesn't that $\essinf \abs f = -\esssup \abs f$, note for example, that $f \ge 0$. This inequality follows from
$$ \essinf \abs f \le \abs{f(x)} \le \esssup\abs f, \quad \text{a. e. $x \in [a,b]$} $$
Moreover note, that $\abs f$ will in general not be continuous, so you cannot just apply the MVT to it.

To prove you theorem, divide (1) by $b-a$ and obtain
$$ \essinf \abs f \le \frac 1{(b-a)}\int_a^b \abs f \, dx \le \esssup \abs f $$
Now let $m := (b-a)^{-1}\int_a^b \abs f \, dx$, then $m \in [\essinf \abs f, \esssup \abs f]$ and
$$ \int_a^b \abs f\, dx = m(b-a).$$


问:
Surjectivity of Exponential Map of a Lie group
I'm trying to learn some Lie algebra theory with bare bones knowledge of differentiable manifolds, and little knowledge of Lie groups. I see why the exponential map $\exp: \mathfrak{g} \to G$ is surjective if $G$ is a Lie subgroup of $GL_n(\mathbf{C})$. However, my intuition is a little loose when it comes to isomorphisms of Lie groups. If $H$ is a Lie group isomorphic to some matrix Lie group, does this imply that the exponential map $\exp: \mathfrak{h} \to H$ is surjective. Furthermore, is there a nicer condition to guarantee that a Lie group is isomorphic to a matrix Lie group (it seems like almost all interesting examples of Lie groups are, except for the covering spaces of certain matrix Lie groups)?</p>

答:
In this blog post Terry Tao mentions a nice criterion:

For a (connected) Lie group $G$, if the exponential map $\mathfrak{g} \rightarrow G$ is surjective, then every element of $G$ is a square in $G$ (in fact, is an $n^{\text{th}}$ power in $G$ for every $n\in\mathbb{Z}_{>0}$).

Proof: Indeed, for $g \in G$, if there exists $x \in \mathfrak{g}$ with $\mathrm{exp}(x) = g$,
then for each $n\in\mathbb{Z}_{>0}$ we have $\exp(x/n)^n = \exp(x)= g$, hence $g$ is the $n^{\text{th}}$ power of $\mathrm{exp}(x/n) \in G$.

As is mentioned (in a specific example) in the above blog post, one can show if $a\in\mathbb{R} , \,\&\,a>0 ,\,\&\,a\neq1$, then $\bigl(\begin{smallmatrix}
-a & 0 \\
0 & -1/a
\end{smallmatrix}\bigr) \in \mathrm{SL}_2(\mathbb{R})$ is not a square in (the Lie group) $\mathrm{SL}_2(\mathbb{R})$, or even in $\mathrm{GL}_{2}(\mathbb{R})$ as proved below.

One proof:
Let $A := \bigl(\begin{smallmatrix}
-a & 0 \\
0 & -1/a
\end{smallmatrix}\bigr)$.
Aiming for a contradiction, suppose there exists $B \in \mathrm{Mat}_{2\times 2}(\mathbb{R})$
with $B^2 = A$.
Note $A$ has minimal polynomial $(X+a)(X+1/a)$ over $\mathbb{R}$ (indeed this polynomial does vanish on $X=A$; and $A$'s minimal polynomial cannot have degree 1 since $A$ is not a scalar multiple of the identity matrix, since $a\neq\pm1$).
Therefore the monic minimal polynomial $P_B \in \mathbb{R}[X]$ of $B$ (over $\mathbb{R}$) divides $(X^2 + a)(X^2+1/a)$. Note $X^2+a,\,X^2+1/a$ are irreducible in $\mathbb{R}[X]$ (since $a>0$), and $P_B$ has degree $\leq2$ (since $B$ is 2-by-2). Therefore $P_B \in \{X^2+a,X^2+1/a\}$, and $P_B$ also equals the (monic) characteristic polynomial of $B$ (since e.g. the char poly has degree 2 since $B$ is 2-by-2, and the min poly $P_B$ already has degree 2 and must divide the char poly).
The complex roots of $P_B$ (which are the complex eigvals of $B$) are then either $\{\pm i\sqrt{a}\}$ or $\{\pm i\sqrt{1/a}\}$, both with multiplicity 1;
but then $\mathrm{det}(B)$, which equals the product w/ multiplicity of its complex eigenvalues, is either $a$ or $1/a$;
so then $\mathrm{det}(B)^2 \in \{a^2,1/a^2\}$.
However $\mathrm{det}(B)^2 = \mathrm{det}(B^2) = \mathrm{det}(A) = 1$,
and $1 \notin \{a^2,1/a^2\}$ (since $a\neq\pm1$);
this gives the contradiction.

For your question about a condition to guarantee that a Lie group is isomorphic to a matrix Lie group: this is true for compact Lie groups; see the following Stackexchange post: Are all Lie groups Matrix Lie groups?


问:
Do I got this change of basis problem right?

I posted parts of this problem earlier, but I have now managed to solve it. Since I don't get any respons in my earlier post if my solution is right or not, I post it again here. My friends and I don't have the same answer for this task, more spesificly problem (ii) and (iii). I wondered if someone could check if I got it right?

So I have solved (i) and think I got it right:

$$P_{\beta\rightarrow\epsilon}=\pmatrix{cos\theta & -sin\theta\\sin\theta&cos\theta}$$

$$P_{\epsilon\rightarrow\beta}=\pmatrix{cos\theta & sin\theta\\-sin\theta&cos\theta}$$

For (ii) I have done this:

$\pmatrix{x'\\y'}=P_{\epsilon\rightarrow\beta}\pmatrix{x\\y}=\pmatrix{cos\theta & sin\theta\\-sin\theta&cos\theta}\pmatrix{x\\y}$ that gives us that

$x'=xcos\theta+ysin\theta$

$y'=-xsin\theta+ycos\theta$

For (iii) I have done this:

$\pmatrix{x\\y}=P_{\beta\rightarrow\epsilon}\pmatrix{x'\\y'}=\pmatrix{cos\theta & -sin\theta\\sin\theta&cos\theta}\pmatrix{x'\\y'}$

So we must have that:

$x=x'cos\theta-y'sin\theta$

$y=x'sin\theta+y'cos\theta$

And we use in these expressions for y and x in the equation $xy=1$ and get that:

$(x'cos\theta-y'sin\theta)(x'sin\theta+y'cos\theta)=1$ Rembering that $\theta=\frac{\pi}{4}$ the equation is equal to

$(x'\frac{\sqrt{2}}{2}-y'\frac{\sqrt{2}}{2})(x'\frac{\sqrt{2}}{2}+y'\frac{\sqrt{2}}{2})=1$

The hyperbola is then given by the equation:

$\frac{1}{2}(x')^2-\frac{1}{2}(y')^2=1$

答:
Your methods and answers look correct.

We can check these answers:

For part (ii), saying that $\vec u = \left(\begin{array}{c} x' \\ y' \end{array}\right)$ in $\mathcal{B}$-coordinates means

$$ \vec u = x' \vec u_1 + y' \vec u_2 $$

So in $\mathcal{E}$-coordinates,

$$ \vec u = (x \cos \theta + y \sin \theta)
 \left(\begin{array}{c} \cos \theta \\ \sin \theta \end{array}\right)
 + (-x \sin \theta + y \cos \theta)
 \left(\begin{array}{c} -\sin \theta \\ \cos \theta \end{array}\right) \\
\vec u = \left(\begin{array}{c}
x \cos^2 \theta + y \sin \theta \cos \theta + x \sin^2 \theta - y \sin \theta \cos \theta \\
x \sin \theta \cos \theta + y \sin^2 \theta - x \sin \theta \cos \theta + y \cos^2 \theta
 \end{array}\right) = \left(\begin{array}{c} x \\ y \end{array}\right) $$

Good, it's still the same vector.

For part (iii), we could transform the whole equation back to $x$-$y$ coordinates, but instead let's take a sample point and see if it works. $(2, \frac{1}{2})$ in $\mathcal{E}$-coordinates is on the curve. In $\mathcal{B}$-coordinates, we have

$$ \left(\begin{array}{c} x' \\ y' \end{array}\right) =
P_{\mathcal{E} \to \mathcal{B}} \left(\begin{array}{c} x \\ y \end{array}\right)
= \left(\begin{array}{cc} \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\
\frac{-\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{array}\right)
\! \left(\begin{array}{c} 2 \\ \frac{1}{2} \end{array}\right)
= \sqrt{2} \left(\begin{array}{c} 1 + \frac{1}{4} \\
 -1 + \frac{1}{4} \end{array} \right)
= \frac{\sqrt{2}}{4} \left(\begin{array}{c} 5 \\ -3 \end{array}\right) $$

Plugging into the left side of your hyperbola equation,

$$ \frac 12 (x')^2 - \frac 12 (y')^2
 = \frac 12 \left(\frac{5 \sqrt 2}{4}\right)^2 -
      \frac 12 \left(\frac{-3 \sqrt 2}{4}\right)^2
 = \frac{5^2 - 3^2}{4^2} = 1 $$

Checks out!


问:
Definite integral $\int_0^{\pi} \sin^{n} x \ln(\sin x) dx$?
Is the integral
$$I(n) = \int_0^{\pi} \sin^{n}(x) \ln(\sin(x)) dx$$
analytically tractable for $n \in \mathbb{N}$? If not, are there good upper and lower bounds?

To clarify, although this has a fairly standard-looking form, I haven't been able to find an answer anywhere and I've tried a few of the usual tricks (although I'm a tad rusty).

I have tried integration by parts to reduce to a simpler or recursive form but this seems to make the expression more complex, and likewise with the substitution $u=\sin(x)$.

Equally, it is relatively easy to see that the expression under the integral is non-positive. Taking a derivative
$$ \frac{d}{dx} \sin^{n}(x) \ln(\sin(x)) = \sin^{n-1}(x) \cos(x) (1 + n \ln(\sin(x))) = 0$$
gives a has minimum at $-\frac{1}{en}$.

Together these give the reasonably simple bounds: $$-\frac{\pi}{en} < I(n) < 0.$$

But numerically these don't appear to be particularly tight, so I'm still hopeful of an exact solution or an improvement.

答:
We have
$$
I_n=2\int_0^{\pi/2}\sin^n(x)\log(\sin(x)) dx=
\int_0^{\pi/2}\sin^n(x)\log(1-\cos^2(x)) dx
$$
$$
=-\sum_{k\geq 1}\frac 1k\int_0^{\pi/2}\sin^n(x)\cos^{2k}(x) dx.
$$

With the help of Euler's Beta function
$$
\frac{\Gamma(u)\Gamma(v)}{\Gamma(u+v)}=B(u,v)=2\int_0^{\pi/2}\sin^{2u-1}(x)\cos^{2v-1}(x) dx
$$
we obtain
$$
I_n=-\Gamma(\frac {n+1}2)\sum_{k\geq 1}\frac 1{2k}\frac{\Gamma(k+\frac 12)}{\Gamma(\frac n2+k+1)}.
$$
This can be further evaluated in closed terms by means of Harmonic Numbers of integer / half integer arguments, as in Igor Rivin's answer, but it may be already good for estimates.


问:
How to interpret the $\wedge$ in $\left[(-2 i \pi x)^{k} f(x)\right]^{\wedge}(\xi)$
I have seen this notation multiple times and I am not aware if the "$^{\wedge}$" is an operation or just a symbol in this case.

答:
That denotes the Fourier transform.

$$\hat{f} (y) = \int_{\mathbb{R}} e^{-2\pi i xy} f(x) \, dx$$


问:
Is there one way to write the permutation that takes $(1,2,...,n) \rightarrow (n,n-1,...,1)$ that is the same whether $n$ is even or odd?
Is there one way to write the permutation that takes $(1,2,...,n) \rightarrow (n,n-1,...,1)$ that is the same whether $n$ is even or odd?

Are these formulas correct?

If $n$ is even then this permutation is given by $(1,n)(2,n-1)...(\frac{n}{2},\frac{n}{2}+1)=\prod_{k=1}^{\frac{n}{2}}(k,n-(k+1))$

If $n$ is odd then this permutation is given by $(1,n)(2,n-1)...(\frac{n-1}{2},\frac{n+2}{2})=\prod_{k=1}^{\frac{n-1}{2}}(k,n+1-k)$

Is there another way of expressing this permutation that is independent of whether $n$ is even or odd?? Perhaps without breaking it down to transpositions, using cycle notation or something? Thank you.

答: You could write it as $\sigma(k)=n-k+1.$


问:
Where did I go wrong with my differentiation here?
This question involves the product rule and power of a function rule for differentiation. I often make silly algebra errors, but here I cannot find them. The answer is supposed to be y'=22. Thanks!

When $x=-1$, $y=(2x+1)^5(3x+2)^4$.
Hence, the derivative,
$$y'=5(2x+1)^4 \cdot (2)(3x+2)^4 + (2x+1)^5 \cdot 4(3x+2)^3 \cdot 3$$
Plugging in $x=-1$,
$$y'= 5(2(-1)+1)^4 \cdot (2)(3(-1)+2)^4 + (2(-1)+1)^5 \cdot 4(3x+2)^3 \cdot 3\\
y' = 10-12\\
y' = -2$$

Body:
We are computing the derivative at $x=-1$ when $y=(2x+1)^5 (3x+2)^4$. To find the derivative, we will be using chain rule and the product rule in correlation.

The product rule: $f'g+fg'$

In our case, Let $f=(2x+1)^5$ and $g=(3x+2)^4$

Hence, $f'=5(2x+1)^{5-1}\cdot(2x+1)^{'}\\ \;\;\;=5(2x+1)^4\cdot 2\\ \;\;\;=10(2x+1)^4$ and $g'=4(3x+2)^{4-1}\cdot (3x+2)^{'}\\ \;\;\;=4(3x+2)^3\cdot3\\ \;\;\;=12(3x+2)^3$

Completing the product rule ($f'g+fg'$): $$y'=10(2x+1)^4*(3x+2)^4+(2x+1)^5*12(3x+2)^3$$

Evaluating at $x=-1$, $$y'(-1)=10(2(-1)+1)^4\cdot (3(-1)+2)^4 +(2(-1)+1)^5\cdot 12(3(-1)+2)^3\\y'(-1) =10+12
\\y'(-1)=22$$


答:
Define sequence $f(n)$=$\frac{n^{n+1}}{(n+1)^n}$. The inequality $n^{n+1}>(n+1)^n$ is true for a given $n$ if and only if $f(n)>1$.

The inequality holds for n=3, so we now prove our inductive step. Assume that for a given $k$, $f(k)>1$. We now have to show that $f(k+1)>1$.

We see that $\frac{f(k+1)}{f(k)}$=$\frac{\frac{(k+1)^{k+2}}{(k+2)^{k+1}}}{\frac{k^{k+1}}{(k+1)^k}}$=$\frac{(k+1)^{(k+2)+k}}{k^{k+1}(k+2)^{k+1}}$=$\frac{(k+1)^{2k+2}}{(k^2+2k)^{k+1}}$=$\frac{(k^2+2k+1)^{k+1}}{(k^2+2k)^{k+1}}$>1. Thus, $f(n)$ is an increasing sequence, and so $f(k+1)>1$, thus completing our proof by induction.


问:
Modelling with Higher-order DEs
I would like to know if there are any models that make use of higher-order (5th, 6th and so on...) linear differential equations. The highest that I know is the Euler-Bernoulli beam equation. Do you know any? Or how to model with higher-order ones?

Thanks

答: How about the bi-Laplace equation? I have not studied them in details and I am only aware of the existence of such equations.</p>

Title: Modelling with Higher-order DEs


问:
Write the equations as a single equation in $x$ and $y$
Given the parameterization,

\begin{cases}
x=at\cos 2t\\
y=a\sin t
\end{cases}

I need to write $y$ (or $x$) as a function of $x$ (or $y$).

What I tried, there was always the variable $t$, so I isolated $t$,

$$t=\frac{ax}{a^2-2y^2}$$

But when I substitute I always come in $x=x$ or $y=y$. Could someone give me a hint on how to proceed.

答:
The answer is pretty messy.

$$t = \arcsin {\frac{y}{a}}$$
$$ x = at \cos(2t)= at(1-2\sin^2t) = a \arcsin \frac{y}{a} \left(1-2\left(\frac{y}{a}\right)^2\right)$$


问:
Can a notation for ordered pairs in ZFC set theory be introduced as a conservative extension of ZFC without junk theorems?
Currently, a standard way to introduce ordered pairs in ZFC set theory is to define
an ordered pair $(a,b)$ as $\{\{a\},\{a,b\}\}$. From this definition one can deduce the main property of ordered pairs:
$(a,b) = (x,y) \rightarrow a=x \,\&\, b=y$.

But some junk theorems can also be proved, for example, $(a,a) = \{\{a\}\}, \{a\} \in (a,b)$, etc.

see also https://mathoverflow.net/questions/90820/set-theories-without-junk-theorems

https://mathoverflow.net/questions/386756/a-better-way-to-introduce-ordered-pairs-in-zf

For a different approach see the paper "Adding an Abstraction Barrier to ZF Set Theory"

https://link.springer.com/content/pdf/10.1007%2F978-3-030-53518-6_6.pdf

An excerpt from it:

" Such ‘accidental theorems’ do not match how people think about the mathematics and also cause practical difficulties when using set theory in machine-assisted theorem proving."

So a natural question arises: is it possible to introduce ordered pairs in a more safe way, without possibility to get
such strange "theorems".

A rather natural approach is to extend ZFC just by adding a new function symbol (say $\pi$, it is usually omitted)
and a defining axiom such as  $\pi(a,b) = \pi(x,y) \rightarrow a=x \,\&\, b=y$
and prove that such theory extension is conservative (without such proof we can get an undesirable result, in a worse case - an inconsistent theory).

So my question:

Is this extension of ZFC (by adding the function symbol $\pi$ and its defining axiom)  conservative?

答:
Yes, in the strongest possible sense: if $M\models\mathsf{ZFC}$ then $M$ has an expansion to a model $M'$ of the theory you describe (that is, there is an $M'$ satisfying your theory whose reduct to $\{\in\}$ is exactly $M$ itself). Specifically, any of the usual set-theoretic implementations of ordered pairs can be taken to be $\pi^{M'}$.

EDIT: in response to a comment below, let me explain in detail why this fact about extensions implies conservativity. Suppose your theory were to prove some sentence $\varphi$ purely in the language of set theory (i.e. no reference to $\pi$). I claim $\mathsf{ZFC}$ must already prove that sentence. For otherwise take some $M\models\mathsf{ZFC}+\neg\varphi$. By the previous paragraph it has some expansion $M'$ satisfying your theory. But since your theory entails $\varphi$ this means $M'\models \varphi$. But $M$ is the reduct of $M'$ to the language of pure set theory, so we get $M\models\varphi$ as well, a contradiction.

This may feel silly but it's not: we avoid junk theorems about $\pi$ even if we consider such expansions because of the different ways ordered pairs can be implemented in set theory. E.g. considering the difference between the Kuratowski ordered pair $\langle a,b\rangle_{K}=\{\{a\},\{a,b\}\}$ and the Hausdorff ordered pair $\langle a,b\rangle_H=\{\{a,1\},\{b,2\}\}$ we have that neither $\forall x(\{x\}\in\pi(x,x))$ nor $\exists x(\{x\}\not\in\pi(x,x))$ is a theorem of the new system.


答:
It is indeed conservative. We can prove that it is conservative quite simply. In fact, it's easy to show that $ZFC + \pi + Q$ is conservative, where $Q$ is the axiom $\forall a \forall b (a, b) = \pi(a, b)$.

This follows from a general result in logic which states the following: consider a theory $T$ and a statement $P$, where $FV(P) \subseteq \{x_1, x_2, ..., x_n, y\}$. Suppose that $T \vdash \forall x_1 \forall x_2 ... \forall x_n \exists ! y P$. Then consider the new theory $T$, which is $T$ with added $n$-ary function symbol $f$ and axiom $\forall x_1 \forall x_2 ... \forall x_n P[y \mapsto f(x_1, x_2, ..., x_n)]$, where $[y \mapsto f(x_1, x_2, ..., x_n)]$ is capture-avoiding substitution. Then $T'$ is a conservative extension of $T$.

Therefore, $ZFC + \pi$ is conservative.

Title: Can a notation for ordered pairs in ZFC set theory be introduced as a conservative extension of ZFC without junk theorems?

Body:
Currently, a standard way to introduce ordered pairs in ZFC set theory is to define
an ordered pair $(a,b)$ as $\{\{a\},\{a,b\}\}$. From this definition one can deduce the main property of ordered pairs:
$(a,b) = (x,y) \rightarrow a=x \,\&\, b=y$.

But some junk theorems can also be proved, for example, $(a,a) = \{\{a\}\}, \{a\} \in (a,b)$, etc.

see also https://mathoverflow.net/questions/90820/set-theories-without-junk-theorems

https://mathoverflow.net/questions/386756/a-better-way-to-introduce-ordered-pairs-in-zf

For a different approach see the paper "Adding an Abstraction Barrier to ZF Set Theory"

https://link.springer.com/content/pdf/10.1007%2F978-3-030-53518-6_6.pdf

An excerpt from it:

" Such ‘accidental theorems’ do not match how people think about the mathematics and also cause practical difficulties when using set theory in machine-assisted theorem proving."

So a natural question arises: is it possible to introduce ordered pairs in a more safe way, without possibility to get
such strange "theorems".

A rather natural approach is to extend ZFC just by adding a new function symbol (say $\pi$, it is usually omitted)
and a defining axiom such as  $\pi(a,b) = \pi(x,y) \rightarrow a=x \,\&\, b=y$
and prove that such theory extension is conservative (without such proof we can get an undesirable result, in a worse case - an inconsistent theory).

So my question:

Is this extension of ZFC (by adding the function symbol $\pi$ and its defining axiom)  conservative?


问:
if $\log_xa+\log_ya=4\log_{xy}a$ prove that $x=y$
Let $x,y$ be numbers in the interval $(0,1)$ with the property that there exists a positive number $a\ne1$ such that
$$\log_xa+\log_ya=4\log_{xy}a$$
I have used the property
$$\log _ba=\frac{\log a}{\log b}$$

and I have
$$\log_xa+\log_ya=\frac{\log a}{\log x}+\frac{\log a}{\log y}$$

$$\log a\left(\frac{\log x+\log y}{\log x \log y}\right)=4\frac{\log a}{\log(xy)} \implies (\log(xy))^2=4 \log x \log y$$

I don't see how this implies $x=y$

Body:
$$\log_xa+\log_ya=4\log_{xy}a$$
$$\dfrac{1}{\log_ax}+\dfrac{1}{\log_ay}=4\dfrac{1}{\log_a{xy}}$$
$$\dfrac{{\log_ay}+{\log_ax}}{{\log_ax}\cdot{\log_ay}}=4\dfrac{1}{{\log_ay}+{\log_ax}}$$
$$\dfrac{({\log_ay}+{\log_ax})^2}{{\log_ax}\cdot{\log_ay}}=4$$
$$({\log_ay})^2+({\log_ax})^2+2\cdot\log_ax\cdot\log_ay=4{{\log_ax}\cdot{\log_ay}}$$
$$({\log_ay})^2+({\log_ax})^2-2\cdot\log_ax\cdot\log_ay=0$$
$$({\log_ay}-{\log_ax})^2=0$$
$$({\log_ay}-{\log_ax})=0$$
$${\log_ay}={\log_ax}$$
$$x=y$$


问:
How to find the Taylor series of $\frac{1}{1+x^3}$ at $0$
I can't find a way to calculate the n-th derivative of $$f(x)=\frac{1}{1+x^3}$$ in order to find its Taylor series at $0$. How could one find its Taylor series?

答:
The most straightforward method is to look at a database of usual Taylor expansions and combine them to develop the function at hand, here that would be $\frac 1{1+u}$ applied for $u=x^3$.

Nevertheless, a common method to find an a priori unknown Taylor expansion is to find an ODE verified by $f(x)$ and solve it via power series.

This is a bit overkill here, since $\frac 1{1+x}$ has a very basic and known development (listed in the compendium above), which can also be determined as a sum of a geometric series, but let assume we start from scratch.

Here is the general method, and it can be applied to many functions:

$f(x)=\dfrac 1{1+x^3}\implies f'(x)=\dfrac{-3x^2}{(1+x^3)^2}$

We could have $f'(x)=-3x^2f(x)^2$ but this is not interesting, as power series multiplication (i.e. $f(x)^2$) is not something that can be easily calculated.

Instead we prefer the form $$(1+x^3)f'(x)+3x^2f(x)=0$$

So let search for the coefficients of $f(x)=\sum\limits_{n=0}^{\infty} a_nx^n$

$\begin{align}\text{ODE}
&=(1+x^3)\sum\limits_{n=1}^{\infty} na_nx^{n-1}+3x^2\sum\limits_{n=0}^{\infty} a_nx^n\\\\
&=\sum\limits_{n=1}^{\infty} na_nx^{n-1}+\sum\limits_{n=1}^{\infty} na_nx^{n+2}+\sum\limits_{n=0}^{\infty} 3a_nx^{n+2}\\\\
&=\sum\limits_{n=0}^{\infty} (n+1)a_{n+1}x^n+\sum\limits_{n=3}^{\infty} (n-2)a_{n-2}x^n+\sum\limits_{n=2}^{\infty} 3a_{n-2}x^n\\\\
&=a_1+(2a_2)x+(3a_3+3a_0)x^2+\sum\limits_{n=3}^{\infty} \Big((n+1)a_{n+1}+(n-2)a_{n-2}+3a_{n-2}\big)x^n
\end{align}$

Identifying the coefficients to the zero series gives:

$\begin{cases}a_1=0\\2a_2=0\\3a_3+3a_0=0\\(n+1)a_{n+1}+(n+1)a_{n-2}=0\end{cases}\iff\begin{cases}a_1=0\\a_2=0\\a_3=-a_0\\a_{n+1}=-a_{n-2}\end{cases}$

Since $a_0=f(0)=1\, $ we get $\, \begin{cases}a_{3n}=(-1)^n\\a_{3n+1}=0\\a_{3n+2}=0\end{cases}$

This gives the Taylor series $\dfrac{1}{1+x^3}=1-x^3+x^6-x^9+x^{12}-x^{15}+\cdots$