问:
Precision on the Cantor space
Let $C=\{0,1\}^\mathbb{N}$ be the Cantor space. Is $C$ homeomorphic to $\{0,1\}^m$ when $m>\aleph_0$?

答:
If $\kappa>\aleph_0$, then $X=\{0,1\}^\kappa$ is not second countable. Let $\mathscr{U}=\{U_n:n\in\omega\}$ be any countable family of open sets in $X$. For each $n\in\omega$ there is a finite $F_n\subseteq\kappa$ and a function $s_n:F_n\to\{0,1\}$ such that $B_n=\{x\in X:x\upharpoonright F_n=s_n\}\subseteq U_n\}$.

Let $C=\bigcup_{n\in\omega}F_n$; $C$ is countable, so there is an $\alpha\in\kappa\setminus C$. Let $$V=\{x\in X:x_\alpha=0\}\,.$$ For each $n\in\omega$ there is a $y^{(n)}\in B_n\subseteq U_n$ such that $y_\alpha^{(n)}=1$, so $y^{(n)}\in U_n\setminus V$. Thus, $V$ is an open set that contains no member of $\mathscr{U}$, and $\mathscr{U}$ is therefore not a base for $X$.

Title: Precision on the Cantor space


问:
If $A$ is open, $\operatorname{Int}(\operatorname{Bdry}(A))=\emptyset$
I'm trying to prove that if $A\subset X$ is open then $\operatorname{int}(\operatorname{Bdry}(A))=\emptyset$. I have that $\operatorname{Bdry}(A) = \operatorname{Cl}(A)\cap (X-\operatorname{int}(A))=\operatorname{Cl}(A)\cap(X-A)$ because $A$ is open, but I'm stuck in that.

答: Suppose $x \in O:=\rm{Int}(\rm{Bd}(A))$. Then $O$ would be a neighbourhood of $x$ that intersects $A$ (as $x \in \rm{Bd}(A)$) and then $O \cap A$ would be a non-empty open set that sits inside $A$ and also intersects $A^\complement$ (as $x \in O \subseteq \rm{Bd}A)$, and this is absurd.

ContentLicense: CC BY-SA 4.0


问:
Prove if $n^2$ is even, then $n$ is even.
I am just learning maths, and would like someone to verify my proof.

Suppose $n$ is an integer, and that $n^2$ is even.  If we add $n$ to $n^2$, we have $n^2 + n = n(n+1)$, and it follows that $n(n+1)$ is even.  Since $n^2$ is even, $n$ is even.

Is this valid?

答: It is valid for sure. You can simplify it just by noting that for every odd $n$ $n^2$ is odd. So the square can be even just with even $n$. But this is the proof just for the implication.</p>

ContentLicense: CC BY-SA 3.0


问:
Can anyone show me the process to get the result Undefined (1/0) from Tangent inverse one (-2/0) [ tan−1(-2/0) ]?
I know that $\tan^{-1}(1/0)$ is undefined. But I'm getting a little trouble figuring out this $\tan^{−1}(-2/0)$. The answer of $\tan^{−1}(-2/0)$ will also be Undefined.
By the way, I got the problem while solving some complex numbers and finding their argument. The complex number was $0-2i$.
Thanks in advance.
Asif Touhid.

答:
If we think about a generic complex number as $z=a+ib$ then it is possible to observe that:

$a+ib=r(\cos\theta+i\sin\theta)$ (Argand-Gauss representation), where $r=\sqrt{a^2+b^2}$.

In our case we have $\displaystyle a=0$ (and $b=-2$),  and so from the equality above $r\cos\theta=0\implies \theta=\pm\frac{\pi}{2}\implies \sin{\theta}=\pm1$.

In our case we have to look at the negative case since $b<0$ and $r>0$, so: $\displaystyle \sin\theta=-1\iff  \theta=-\frac\pi2$


问:
Derivation of Modified Newton Method
I have been stuck on this question for quite some time, and can't seem to figure the right piece of information given other sources on the internet primarily due to notational differences.

This is for the case of multiple roots, specifically
$$m = 2$$

I am currently at the following step:
$$e_{n+1} = x_{n+1} - r = e_n - 2 \cdot \frac{f(r)+f'(r)e_n+\frac{1}{2}f''(r)e_n^2+\frac{1}{6}f'''(\xi_n)e^3_n}{f'(r)+f''(r)e_n+\frac{1}{2}f'''(\zeta_n)e^2_n}$$

I'm supposed to end up with something on the form of:
$$e_{n+1} = \bigg(\frac{[]}{[]}\bigg)e_n^2$$

Any help or pointers would be greatly appreciated.

答:
A hint to get you moving: once you substitute $f(r)=0,f'(r)=0,f''(r) \neq 0$ (which is what it means that $r$ is a root of $f$ with multiplicity $m=2$) you have

$$e_{n+1}=e_n - \frac{f''(r) e_n^2 + \frac{1}{3} f'''(\xi_n) e_n^3}{f''(r) e_n + \frac{1}{2} f'''(\zeta_n) e_n^2}.$$

Now get a common denominator between these two terms and you should see several terms cancel. Note here $\xi_n$ and $\zeta_n$ are just unknown numbers; they depend on $x_n$, but not in a way that you can or need to resolve. Instead you will just replace the associated terms with a bound on $|f'''|$.

Title: Derivation of Modified Newton Method


问:
Convergence of $\sum \frac{ \cos\sqrt n} {\sqrt n}$
As you undestand from the topic title, I am wondering how to determine whether the series
$$
\sum_{n=1}^{\infty} \frac{ \cos\sqrt n} {\sqrt n}
$$
converges or not(it diverges actually).

You know, the problem is, if it would be something like

$$
\sum_{n=1}^{\infty} \frac{ \cos n} {\sqrt n}
$$

Then the task can be easily solved using Dirichlet's test: we just need to show that $|\sum_{n=1}^{N} \cos n| \le K$ for all $N$ having fixed $K$. This was described here

Also, if it was like

$$
\sum_{n=1}^{\infty} \frac{ \cos\sqrt n} {n}
$$

then we are in much more complicated situation. Every reasonable solution I found on the internet involves approximating the series with an integral. This type of task implies very cute mathematical background(which I don't have:))

In my school we didn't study improper integrals(even just integrals). And actually our problem distincts from two others described above:
$$
\sum_{n=1}^{\infty} \frac{ \cos\sqrt n} {\sqrt n}
$$

This is why I believe there must be a solution which is simpler than using improper integrals.

We studied lots of different convergence tests at school, but they seems useless here. Well, actually I think we should apply them, but after some mathematical magic.
Also I tried to expand terms using Taylor's formula, but it didn't help.

Any ideas?

答:
Let $M\in\mathbb{N}$. Let us count how many $n\in\mathbb{N}$ satisfy $2\pi M - \frac{\pi}{4} \le \sqrt{n} \le 2\pi M + \frac{\pi}{4}$.

This condition is satisfied by $n \in [N_1,N_2]$ where
$$ N_1 = \lceil (2\pi M - \frac{\pi}{4})^2\rceil = \lceil 4\pi^2 M^2 - \pi^2M - \frac{\pi^2}{16}\rceil $$
$$ N_2 = \lfloor (2\pi M + \frac{\pi}{4})^2\rfloor = \lfloor 4\pi^2 M^2 + \pi^2M - \frac{\pi^2}{16} \rfloor $$
We have
$$ N_1 \le 4\pi^2 M^2 - \pi^2M - \frac{\pi^2}{16} + 1 $$
$$ N_2 \ge 4\pi^2 M^2 + \pi^2M - \frac{\pi^2}{16} -1 $$
so
$$ N_2-N_1 \ge 2\pi^2M -2 $$
For $n\in[N_1,N_2]$ we have
$$ 2\pi M - \frac{\pi}{4} \le \sqrt{n} \le 2\pi M + \frac{\pi}{4}$$
$$ \cos \sqrt{n} \ge \frac{1}{\sqrt{2}}$$
$$ \frac{\cos \sqrt{n}}{\sqrt{n}} \ge \frac{1}{\sqrt{2}(2\pi M + \frac{\pi}{4})}$$
so
$$ \sum_{n=N_1}^{N_2} \frac{\cos \sqrt{n}}{\sqrt{n}} \ge \frac{N_2-N_1+1}{\sqrt{2}(2\pi M + \frac{\pi}{4})} \ge \frac{2\pi^2M -1}{\sqrt{2}(2\pi M + \frac{\pi}{4})} $$
For large $M$ this bound tends to $\frac{\pi}{\sqrt{2}} > 0$ and from this we can conclude that the sequence of partial sums of the series isn't a Cauchy sequence, and therefore it's not convergent.

This proof relies on the fact that for large $n$ the square rootsts of natural numbers get close together, and therefore $\cos\sqrt{n}$ changes slowly - there are long stretches of natural numbers for which $\cos\sqrt{n}$ is positive (followed by long stretches for which it is negative, but it is enough to consider the 'positive ones'). During these long stretches the  partial sums of the series significantly increase, and a lower bound can be put on this increase, thus proving that the partial sums do not create a Cauchy sequence.


问:
Let $f_n(x)=\frac{1}{nx+1}$ and $g_n(x)=\frac{x}{nx+1}$ for $x\in (0,1)$. Show that $f_n$ doesn't converge uniformly but that $g_n$ does
Let $f_n(x)=\frac{1}{nx+1}$ and $g_n(x)=\frac{x}{nx+1}$ for $x\in (0,1)$. Show that $f_n$ doesn't converge uniformly on $(0,1)$ but that $g_n$ does. I know that it is not something difficult, but I still would like to know if my proofs are correct, please.

First of all, we remark that $\lim_{n\to\infty}f_n(x)=1$ and $\lim_{n\to\infty}g_n(x)=x$. We prove now that $f_n$ doesn't converge uniformly to $1$:

$\bullet$ First approach: By definition, to show that $f_n$ doesn't converge uniformly to $1$, we have to satisfy the following: $ \exists \epsilon>0 \ \forall N \ \exists n\ge N \ \exists x\in (0,1)$: $|\frac{1}{nx+1}-1|>\epsilon$. If we take $x=1/n$ and $\epsilon=1/3$, we satisfy the definition. Therefore, $f_n(x)$ doesn't converge uniformly to $1$ on $(0,1)$

$\bullet$ Second approach: Suppose by absurd that $f_n$ converges uniformly to $1$. Then, we have the following: $\forall \epsilon>0 \ \exists N \ \forall n\ge N \ \forall x\in(0,1)$:

$|\frac{1}{nx+1}-1|<\epsilon$

But, $|\frac{1}{nx+1}-1|=|\frac{-nx}{nx+1}|\le|\frac{nx}{nx}|=1$. Taking $\epsilon=2$ we got the contradiction. So, $f_n$ doesn't converge uniformly to $1$.

Now, we show that $g_n$ converges uniofrmly to $x$ on $(0,1)$. By definition we have to satisfy the following: $\forall \epsilon>0 \ \exists N \ \forall n\ge N \ \forall x\in(0,1)$:

$|\frac{x}{nx+1}-x|<\epsilon$.

But,

$|\frac{x}{nx+1}-x|\le|\frac{x}{nx+1}|\le|\frac{x}{nx}|\le|\frac{1}{n}|$.

If we pick $N$ such that $\frac{1}{N}< \epsilon$ we obtain that $f_n$ converges uniforomly to $x$ on $(0,1)$

答:
hint

You made a mistake in your $\lim_{n\to+\infty}f_n(x)=1$.

In fact,
$$\lim_{n\to +\infty}f_n(x)=0$$

and
$$M_n=\sup_{0<x<1}|f_n(x)-0|=1$$

thus, the convergence is not uniform at $(0,1)$.
Or

$$M_n\ge f_n(\frac 1n)=\frac 12$$

By the same, let
$$G_n(x)=|g_n(x)-0|$$
then
$$G_n'(x)=\frac{1}{(nx+1)^2}$$
and
$$\sup_{0<x<1}G_n(x)\le g_n(1)=\frac{1}{n+1}\to 0$$


问:
Minimax problem inversion of sign
Consider the optimization problem with $x,y\in\mathbb{R}^n$, with $f:\mathbb{R}^{2n}\to \mathbb{R}$:

$$
\min_{x}\min_{y} - f(x,y).
$$

If I want to exchange the second minimum with a maximum is this the correct way to do it?

$$
\min_{x}\max_{y}f(x,y),$$

or I need to change also the second minimum into a maximum? Can you  explain the reasoning behind these change of min/max? Thanks!

答:
Since $\min_{y} - f(x,y) = -\max_{y} f(x,y)$:

$$
\min_{x}\min_{y} - f(x,y) = \min_{x} - \max_{y} f(x,y).
$$
Applying the same rule again, you get $-\max_{x} \max_{y} f(x,y)$. Alternatively, you can do it in one step:

$$
\min_x \min_y - f(x,y) = \min_{x,y} - f(x,y) = - \max_{x,y} f(x,y).
$$

Title: Minimax problem inversion of sign


问:
Tricky integral limit $\lim_{n \to \infty} \int_0^x \frac{(x-t)^{n}}{(1-t)^{n+1}} dt=0$
I have $x \in (-1,1)$ and I need to show that
$$\lim_{n \to \infty} \int_0^x \frac{(x-t)^{n}}{(1-t)^{n+1}} dt=0$$
How do I even start?

The context to this problem is the remainder of the Taylor series for $\ln{(1-x)}$. If there is a way to show that the Taylor series is identical to the function in $(-1,1)$ without taking this limit: I'm also interested.

答:
Let $y= \frac{x-t}{1-t}\in (0,x)$ and recognize $\lim_{n \to \infty}  y^{n}=0$ to obtain

$$\lim_{n \to \infty} \int_0^x \frac{(x-t)^{n}}{(1-t)^{n+1}} dt
= \lim_{n \to \infty} \int_0^x \frac{y^{n}}{1-y} dy
= \int_0^x \frac{\lim_{n \to \infty}  y^{n}}{1-y} dy=0
$$


问:
What to think about this limit trick $\lim_{x\to 1}f(x)=\lim_{x\to 1}f(\frac 1x)$?
While searching for duplicates to a question asked recently I stumbled onto this specific answer:

Show that $\lim\limits_{x\to 1}\left(\frac p{1-x^p}-\frac q{1-x^q}\right)=\frac {p-q}2$

I reproduce the sketch here:

$$L=\lim\limits_{x\to 1}\left(\frac p{1-x^p}-\frac q{1-x^q}\right)=\lim\limits_{x\to 1}\left(\frac p{1-(\frac 1x)^p}-\frac q{1-(\frac 1x)^q}\right)=\lim\limits_{x\to 1}\left(\frac {-px^p}{1-x^p}-\frac {-qx^q}{1-x^q}\right)$$

So by adding first and third term we get $2L=p-q$.

What annoys me is that it looks like an answer we could put in the "bad math that gets away with it" thread...

It is true that $\lim\limits_{x\to 1}f(x)=\lim\limits_{x\to 1}f(\frac 1x)$ when the limit exists, however here, I think it is part of the problem to show that this limit actually exists.

Looking at it more closely we have that:

$f_p(x)=\dfrac{p}{1-x^p}=\frac 1{1-x}+\frac{n-1}2+O(1-x)$

$f_p(\frac 1x)=\dfrac{p}{1-(\frac 1x)^p}=-\frac 1{1-x}+\frac{n+1}2+O(1-x)$

So if we were to apply the same process to $\ell_p=\lim\limits_{x\to 1}\dfrac{p}{1-x^p}$ we would get $2\ell_p=p\iff \ell_p=\frac p2$ which is wrong since $1$ is a pole here (limit does not exist).

But the solution proposed in the cited answer let us think that we got the result from $\ell_p-\ell _q=\frac p2-\frac q2=\frac{p-q}2$

While in reality we get it from cancellation of the divergent parts in:

$\require{cancel}f_p(x)-f_q(x)=\left(\cancel{\frac 1{1-x}}+\frac 12(p-1)+O(1-x)\right)-\left(\cancel{\frac 1{1-x}}+\frac 12(q-1)+O(1-x)\right)=\frac{(p-\cancel{1})-(q-\cancel{1})}2+O(1-x)\to \frac{p-q}2$

So what is your opinion on this answer, do you agree with my point of view ?

How would you convince a student that does this, that his method is flawed, or at least that he should prove first that the limit exists before carrying the $\lim$ operator from equality to equality like this (which BTW seems to be a very common habit among posters...).

答:
I wouldn't quite call this bad math, sometimes you just have to read from right to left.
Often people write $\lim_{x\to a} f(x) = \lim_{x \to a} g(x)$ to mean that the limits either both exist or both don't exists, and eventually continue the chain showing the right one exists, meaning they both exist.

Recall that if $\lim_{x \to a} g(y) = L$ and $\lim_{y \to L}f(y) = M$, then $\lim_{x \to a} f(g(x)) = M$. Apply to $g(x) = 1/x$ at $x=1$. We see that if $\lim_{y \to 1} f(y)$ exists and equals $M$, then $\lim_{x \to 1} f(1/x)$ exists and equals $M$. Now apply this again to  $h(x) = f(1/x)$ to see that if $\lim_{y \to 1}f(1/y)$ exists and equals $M$, then $\lim_{x \to 1} f(1/(1/x)) = \lim_{x \to 1}f(x)$ exists and equals $M$.

Thus we see that $\lim_{x \to 1} f(x)$ exists and equals $M$ if and only if $\lim_{x \to 1} f(1/x)$ exists and equals $M$. Thus it is not really "bad math" to write $\lim_{x \to 1} f(x) = \lim_{x \to 1} f(1/x)$ as long as you eventually show that one of the two sides exists.


问:
The nonexistence of any "good" norm on $\ell^\infty(\mathbb R)$
I'd like to prove by counterexample that there is no norm on the space of all real-valued bounded sequences such that $\lim_{n \to \infty} \|f_n - f\| = 0$ is equivalent to $\lim_{n \to \infty} f_n = f$. In other words, I want to find a sequence $(f_n)_{n\in \mathbb N} \subset \ell^\infty(\mathbb R)$ which converges to $f$ (pointwise) but for which
$$
\lim_{n \to \infty} \|f_n - f\| \neq 0.
$$

This question is inspired by the paper A Note on Pointwise Convergence, which proves that pointwise convergence is not "metrizable" on $\mathcal{C}[0,1].$ I haven't been able to find any proofs that this result holds in the space of bounded sequences, but since the space is not even complete under any norm by Baire, I'm quite sure a counterexample exists. I'd really appreciate any ideas y'all have.

答:
Let $\lVert \bullet \rVert$ be any norm on $\ell^\infty$. For each $i \in \mathbb{N}$, let $\delta_i$ be the Kronecker delta sequence (that is, $\delta_i(j) = 1$ if $i = j$, and $\delta_i(j) = 0$ if $i \neq j$).

Define $f_i = \frac{1}{\lVert \delta_i \rVert} \delta_i$, so that $\lVert f_i \rVert = 1$ for all $i$. Then $f_i \to 0$ pointwise, but

$$\lVert f_i - 0 \rVert = 1$$

does not converge to $0$.


问:
If $a^3b-ab^2$ is divisible by $3$ then either $a$ is divisible by $3$ or $b$ is divisible by $3$ or $(b-1)$ is divisible by $3$
Question:
Prove for all integers $a,b$ that if $a^3b-ab^2$ is divisible by $3$ then either $a$ is divisible by $3$ or $b$ is divisible by $3$ or $(b-1)$ is divisible by $3$

Attempt:

Suppose by contradiction that $3 \nmid a$ and $3 \nmid b$ and $3 \nmid (b-1)$,

Then $a^3b-ab^2=(a)(b)(a^2-b)$, and so since $3$ is prime by Euclid's lemma we have

$3|a$ or $3|b$ or $3|(a^2-b)$, but by assumption $3 \nmid a$ and $3 \nmid b$ so then
$3|(a^2-b)$

Now since $3 \nmid a$ and $3 \nmid b$ and $3 \nmid b-1$ then there exists some integers $k,j$ with

$a = 3k+1$ or $a=3k+2$ and $b=3j+1$

Here is where I am not sure how to continue, Consider
$a^2-b \mod 3 \equiv 9k^2+6k+1 - (3j+1) \equiv 0 \mod 3$ or
$a^2-b \mod 3 \equiv 9k^2+12k+4 - (3j+1) \equiv 0 \mod 3$

where now I can't seem to find a contradiction. I think I have made a small mistake somewhere along the way but I can't seem to find it.

Any insights appreciated.

答:
Your error is that if $3\not \mid b$ (So $b\not \equiv 0\pmod 3$) and $3\not \mid b-1$ (So $b-1\not \equiv 0 \pmod 3$ and $b\not \equiv 1 \pmod 3$ then $b \equiv 2 \pmod 3$ (and not $b \equiv 1 \pmod 3$). So $b = 3j + 2\ne 3j + 1$.

(If $b = 3k$ then $3\mid b$.  If $b = 3j+ 1$ then $3\mid b-1$.  So to have $3\not \mid b, b-1$ we must have $b= 3j + 2$.)

And then you contradiction is right in front of our eyes.

$a^2 - b = 9k^2 + \begin{cases}6k+1\\12k + 4\end{cases} - (3j+2)\equiv -1\pmod 3$.

(Minor error:  $(3k+2)^2 = 9k^2 + 12k + 4$ and not $9k^2 +6k + 4$.  But it's better to use $a\equiv -1\pmod 3; a =3k -1; (3k-1)^2 = 9k^2 -6k + 1$.)

=======

But I'd do it directly.  With Euclids lemma we have if $3|a^3b-ab^2 = ab(a^2-b)$ we must have either $3|a$ or $3|b$ or $3|a^2-b$.

If $3|a$ or $3|b$ we are done.  SO it suffices to show: Assume $3\not\mid a$ and $3\not \mid b$ then $3|a^2 -b\implies 3|b-1$.

And if $3\not \mid a$ then $a\equiv 1,2\pmod 3$ and $a^2 \equiv 1 \pmod 3$ so

So $b-1 \equiv b - a^2 \equiv -(a^2 -b)\equiv 0 \pmod 3$.

And we are done.


答:
For any infinite cardinal $\kappa$ we have that $w(\{0,1\}^\kappa) = \kappa$, where the weight $w(X)$ of a space is the minimal size of a base for $X$.

So in particular $\{0,1\}^\kappa \not\simeq \{0,1\}^\lambda$ when $\kappa \neq \lambda$ are infinite cardinals. So we have a lot ( a class of) of topologically distinct Cantor cubes. The only metrisable one is $\{0,1\}^{\aleph_0}$ as a compact metrisable space has countable weight.

I write more on these cubes and dense subsets of them here.

Title: Precision on the Cantor space

Body: Let $C=\{0,1\}^\mathbb{N}$ be the Cantor space. Is $C$ homeomorphic to $\{0,1\}^m$ when $m>\aleph_0$?


问:
Proving $\sum\limits_{n = 0}^{\infty} \frac{x^n}{n!}$ converges
I am trying to prove that the series $\sum\limits_{n = 0}^{\infty} \frac{x^n}{n!}$ converges. It's enough to show that the series of partial sums $S_N = \sum\limits_{n=0}^N a_n$ converges. We have
$$s_N = \sum\limits_{n=0}^N \frac{x^n}{n!}.$$
We propose that its limit is $\sum\limits_{n=0}^{\infty} \frac{x^n}{n!}$. We have to show that for all $\epsilon > 0$, there exists a $K$ such that $N > K$ implies $|s_N - \sum\limits_{n=0}^{\infty} \frac{x^n}{n!} | < \epsilon$. I'm stuck on showing this, however. The set-up is:
\begin{align*}
\left \lvert s_N - \sum\limits_{n=0}^{\infty} \frac{x^n}{n!} \right \rvert & = \left \lvert \sum\limits_{n=0}^N \frac{x^n}{n!} - \sum\limits_{n=0}^{\infty} \frac{x^n}{n!} \right \rvert \\
& = \left \lvert \sum\limits_{n=N+1}^{\infty} \frac{x^n}{n!} \right \rvert \\
& = \sum\limits_{n=N+1}^{\infty} \frac{|x|^n}{n!}
\end{align*}
As $n \to \infty$, each of these terms go to $0$, so I should be able to pick such a $K$ to make this less than $\epsilon$, but I can't figure out how to solve for it and to "isolate" $n$.

答:
One way to prove it is as follow : $(s_n)_{n\geqslant 1}$ is a non-decreasing sequence therefore it suffices to show that it is bounded above. There exists $C>0$ such that $\frac{|x|^n}{n!}\leqslant\frac{C}{2^n}$, this is because $\lim\limits_{n\rightarrow +\infty}\frac{(2x)^n}{n!}=0$, thus the sequence $\left(\frac{(2x)^n}{n!}\right)_{n\geqslant 1}$ is bounded. Finally, $$ s_n\leqslant\sum_{k=1}^n\frac{C}{2^k}\leqslant C $$
and this ends the proof.


答:
Both your pointwise limits for $f_n$ and $g_n$ are wrong.

Hint.

For every $x\in (0,1)$,
$$
\lim_{n\to\infty}\frac{1}{nx+1}=0,\quad \lim_{n\to\infty}\frac{x}{nx+1}=x\lim_{n\to\infty}\frac{1}{nx+1}=0
$$

So both $f_n$ and $g_n$ converges to $0$ pointwise on $(0,1)$.

To show that $f_n$ does not converge uniformly to zero, consider the sequence $x_n=\frac1n$ and estimate
$$
|f_n(x_n)-f(x_n)|
$$
It is easy to see that this quantity does not go to zero.

For $g_n$, consider the estimate
$$
|\frac{x}{nx+1}|=|\frac{1}{n+\frac{1}{x}}|\le\frac1n
$$

Title: Let $f_n(x)=\frac{1}{nx+1}$ and $g_n(x)=\frac{x}{nx+1}$ for $x\in (0,1)$. Show that $f_n$ doesn't converge uniformly but that $g_n$ does

Body:
Let $f_n(x)=\frac{1}{nx+1}$ and $g_n(x)=\frac{x}{nx+1}$ for $x\in (0,1)$. Show that $f_n$ doesn't converge uniformly on $(0,1)$ but that $g_n$ does. I know that it is not something difficult, but I still would like to know if my proofs are correct, please.

First of all, we remark that $\lim_{n\to\infty}f_n(x)=1$ and $\lim_{n\to\infty}g_n(x)=x$. We prove now that $f_n$ doesn't converge uniformly to $1$:

$\bullet$ First approach: By definition, to show that $f_n$ doesn't converge uniformly to $1$, we have to satisfy the following: $ \exists \epsilon>0 \ \forall N \ \exists n\ge N \ \exists x\in (0,1)$: $|\frac{1}{nx+1}-1|>\epsilon$. If we take $x=1/n$ and $\epsilon=1/3$, we satisfy the definition. Therefore, $f_n(x)$ doesn't converge uniformly to $1$ on $(0,1)$

$\bullet$ Second approach: Suppose by absurd that $f_n$ converges uniformly to $1$. Then, we have the following: $\forall \epsilon>0 \ \exists N \ \forall n\ge N \ \forall x\in(0,1)$:

$|\frac{1}{nx+1}-1|<\epsilon$

But, $|\frac{1}{nx+1}-1|=|\frac{-nx}{nx+1}|\le|\frac{nx}{nx}|=1$. Taking $\epsilon=2$ we got the contradiction. So, $f_n$ doesn't converge uniformly to $1$.

Now, we show that $g_n$ converges uniofrmly to $x$ on $(0,1)$. By definition we have to satisfy the following: $\forall \epsilon>0 \ \exists N \ \forall n\ge N \ \forall x\in(0,1)$:

$|\frac{x}{nx+1}-x|<\epsilon$.

But,

$|\frac{x}{nx+1}-x|\le|\frac{x}{nx+1}|\le|\frac{x}{nx}|\le|\frac{1}{n}|$.

If we pick $N$ such that $\frac{1}{N}< \epsilon$ we obtain that $f_n$ converges uniforomly to $x$ on $(0,1)$


问:
On the existence of abundant Carmichael numbers
A natural number $n$ is called Carmichael if it is squarefree and for each prime factor $p$ of $n$, $p-1$ divides $n-1$. Moreover, $n$ is abundant if the sum of its divisors is $>2n$. I have checked the first Carmichael numbers and I have seen that none of them is abundant. So my question is: does there exist any abundant Carmichael number? If so, which is the smallest one?

答: See the article by Daniel Șuteu here: https://trizenx.blogspot.com/2020/08/pseudoprimes-construction-methods-and.html. Abundant Carmichael numbers exist (2059832906607460252767290568443059994787898033540634712711845135488141590979778401392385 is one), but finding the smallest one is mentioned by Șuteu as an open problem.</p>


问:
finding a unit vector of maximum distance from a subspace in a banach space
Let $W$ be a strict subspace of a Banach space $X$ and let the distance between a vector $v$ and the subspace $W$ be defined as $d(v,W)=\inf_{w\in W}|w-v|$.

If $X$ is a Hilbert space, then we can always find a unit vector $v$ st. $d(W,v)=1$. Does this also hold for Banach spaces? If not, what is a counter example?

答:
This does not hold for  Banach spaces in general. For example, take a look at this (notice that $X$ is not reflexive.)

However, it is true for reflexive Banach spaces:

Let $X$ be a reflexive Banach space (i.e. the closed unit ball of $X$ is weakly compact) and let $W$ be a closed  subspace of $X$. Take any $x \in X \setminus W$. Pick a sequence  $(w_n) \subset W$  such that $ ||x-w_n|| \downarrow d(x,W)  \ $ (by the definition of the infimum). Since $||w_n|| \leq ||x|| + ||x-w_n|| $ we see that $(w_n)$ is bounded. By the EBerlein - Smulian theorem, we infer that $(w_n) \subset W $ has a weakly convergent subsequence. Hence, we may assume that $ w_n \xrightarrow{w} w $ for some $w \in W$ (notice that $W$ is weakly closed (by Mazur's theorem) and thus $w \in W$.) By the weak lower semi continuity of the norm $ || \cdot || \colon (X,w) \to \mathbb R$ we get that
$$ d(x,W) \leq ||x-w|| \leq \liminf ||x_n -w|| = d(x,W) $$
and thus $d(x,W)=||x-w||$. We essentially showed that every closed subspace of $X$ is a proximinal set (the same proof applies for closed convex subsets). Now, since $d(x,W) = d(x-w,W) $ we conclude that
$$d \left (\frac{x-w}{||x-w||}  ,W \right)=1 $$


问:
Tensor products and natural isomorphisms
Let $U$ and $V$ be two finite-dimensional vector spaces.  How to prove that $U \otimes V \cong V \otimes U$?

The equivalence symbol generally refers to natural isomorphisms – i.e. isomorphisms defined without any reference to the representation of the underlying vector spaces. This is the point that I try to understand.

A straightforward proof is derived from the universality property of the tensor product definition. Once the bilinear function $h_1 : U \times V \rightarrow U \otimes V$ (from the tensor definition) and $h_2 : U \times V \rightarrow V \otimes U$ (from the tensor definition) are defined, there is a unique linear function $f$ between $U \otimes V$ and  $V \otimes U$. BUT this function depends on $h_1$ and $h_2$. How can we conclude there is a natural isomorphism between both vector spaces?

A related problem can be found in the proof that two tensor products definitions are isomorphic. Again the proof relies on the universality property of the tensor product definition. Thus, the isomorphism between the two tensor products depends on $h_1$ and $h_2$.

Is there any explanation on the aspect of the natural isomorphism without relying on category theory (and functors and Yoneda lemma) or defining the tensor product  “as the quotient space of a vector space having U × V as a basis”? Many references mention the natural isomorphism, use the universality property, but never mention those more formal set-ups.

答:
Wasn’t category theory developed to make sense of the word natural isomorphism? Anyway this is more of a lengthy comment than an answer, but maybe it helps anyway.

As far as I am concerned there can only be one definition of the tensor product and the other characteristic properties are, well, properties. What is natural/canonical depends on the definition.

If you define the tensor product $U \otimes V$ as a quotient, you have an explicit generating system consisting of pure tensors $u\otimes v$. Without invoking the universal property you may check that the natural/canonical = obvious/immediate assignment
$$\begin{array}{ccc}
U \otimes V & \rightarrow & V \otimes U\\
u \otimes v & \mapsto & v \otimes u
\end{array}$$
extends to a welldefined isomorphism $U \otimes V \cong V
\otimes V$.

If you define the tensor product via the universal property (note that the bilinear map $U \times V \rightarrow U \otimes V$ is part of the datum of a tensor product!), you are doing category theory. In this case canonical/natural is commonly used to denote morphisms induced by universal properties. But that means in particular that you need to have some data, by which they are induced. In our case this happens by using the bilinear maps $h_1,h_2$ you mention. Note that they can be defined in a canonical way as well in that the universal property of the product gives a canonical isomorphism $U \times V \cong V \times U$ and $h_1,h_2$ are just bilinear maps obtained from postcomposing this isomorphism with the bilinear map defining the tensor product.

The point of the latter perspective is that we don’t need to specify a basis and use linear combinations, extending linear maps etc. In other words it is independent from any particular representation of our vector spaces as $K^I$. It is not about not having to specify anything and getting isomorphisms for free. It is rather about getting things with minimal (but nonzero!) effort...


问:
How do one solve a nonlinear combinatoric problem?
I am an undergraduate CS student and I am struggling with a problem.

$Qx = b$ where $Q$ is a constant $m \times n$ matrix (with $m>n$), $x$ is a $n \times 1$ vector and $b$ is a $m\times 1$ vector.

I want to maximize the number of zeros in vector $b$.

subject to: $x(i)>0$ for $i=1,2, \ldots,n$

How do one tackle a problem like this ?

答: Note that there is no guarantee that you will be able to create any zeros at all, for instance if all entries of $Q$ are strictly positive. So a first thing I would do is to forget about those coordinates $b_j$ that cannot ever be made zero, namely those for which row $j$ of $Q$ does not have both a positive and a negative entry (this includes the rows that are entirely zero, which are uninteresting for a different reason). Then for each remaining row you've got a linear equation in $x$ that gives the condition for that coordinate to become zero, and which is such that the hyperplane of its solutions meets the strictly positive cone given by $x_i>0$ for all $i$. Now the task is to find the largest collection of rows so that the intersection of their solution hyperplanes still meets that cone. You might search for sets of $n$ rows whose equations are linearly dependent, as that is a necessary condition for having a collection of at least $n$ coordinates that can be made zero simultaneously. It there are no sets of $n$ linearly dependent rows, then you just need to find the largest collection of at most $n-1$ rows whose intersection of hyperplanes meets the positive cone. If sets of $n$ linearly dependent rows do exist, you could check their intersection of hyperplanes, and see if it meets the positive cone; if it does you can get those $n$ coordinates to be zero, if it doesn't then that set of $n$ linearly dependent rows is not productive after all. You are probably left with a few productive $n$-tuples at best, which you might compare to see if even more than $n$ simultaneous zeros can be achieved.</p>


问:
Show that $f_1,f_2 \not\models f_3$
Let $f_1=(\forall x(r(x) \vee \neg p(x))$, $f_2=(\exists x(r(x)\rightarrow q(x)))$ and $f_3 = (\forall x(\neg p(x)\rightarrow q(x)))$. Show that $f_1,f_2 \not\models f_3$.

My attempt: lets assume that $\mathcal{A}=(\{a\},S)$ such that $r^{\mathcal{A}}=p^{\mathcal{A}}=q^{\mathcal{A}} = \{a\}$. Then obviously $\mathcal{A} \models f_1$ and $\mathcal{A} \models f_2$. Im not certain how i should continue. Any help is appreciated. Should i consider cases?

For example $I[a / x] \models r(x)$ and $I[a / x] \models p(x)$ then $I[a / x] \not\models \neg p(x)$ so $f_1,f_2 \not\models f_3$ in this case.

答:
Work backwards. You want a model which makes $f_3$ false, while making $f_1$ and $f_2$ true.

All it takes to make $f_3$ false is (i) for there to be  something in the domain which satisfies neither $p(x)$ nor $q(x)$. Hold on to that thought!

Suppose in fact (ii) nothing satisfies $p$, then $f_1$ will be true.

Suppose (iii) nothing satisfies $r$, then $f_2$ will be true (since for any $x$, the conditional $(r(x) \to q(x))$ has a false antecedent and will be true).

It is easy to make (i), (ii) and (iii) true together!

Title: Show that $f_1,f_2 \not\models f_3$


问:
Subsets using Binomial coefficients and the Binomial theorem
I've been trying this question but with no luck. Could someone please help with the following question? Thanks!

Prove that if $n \geq 2$, we can find a collection $S$ of at least $\frac{2^n}{n}$ subsets of $[n]$ such that no set in $S$ contains another one.

答: HINT: Show that if two subsets of $[n]$ have the same cardinality, neither of them is contained in the other. For $k\in[n]$ let $s_k$ be the number of subsets of $[n]$ of cardinality $k$. The mean value of the numbers $s_1,\ldots,s_n$ is $\frac{2^n-1}n$. Use the fact that $[n]$ has only one subset of cardinality $n$ to show that there is some $k\in[n]$ such that $s_k>\frac{2^n-1}n$. Then show that because $\binom{n}k$ is an integer, $s_k\ge\frac{2^n}n$.


问:
I need visual examples of topological concepts
I'm trying to understand basic concepts of topology, unfortunately I'm a very visual person, and as much documentation there is on how to come up with closed/open/clopen/etc. There are very few visual examples (using actual sets of integers, shapes, etc). So it's very hard for me to understand. Not everyone learns through text or mathematical definitions, so I figured this could help other people that have the same difficulties I have. Thanks for your help!

Can anyone get an example of a topology over a set.

I need one for a topology over a set, an open set, closed set, and clopen set. It needs to be something real. Like, say a set contained numbers { 1,2,3,4 }

It doesn't have to be that set, but it has to be a actual set of something (numbers, letters, geometric shapes), no labels (set X is the union of set Y and set Z), etc.

答:
If you're looking for visuals on topology, look no further than Wolfram MathWorld:

The blood types (O+, AB+, etc) form a topology under the power set topology: http://tumblr.com/xp12cy25v0.

That's a visual for the power set topology on {1,2,3}. And as Asaf said, you can topologise {1,2,3} with the trivial topology or something else (as long as your $\cup$s and $\cap$s commute), though the above visual wouldn't work (nor would the logic).

Also @Xaan, a tip: you might want to be more polite, since people are helping you here for free.


答:
So what you're asking is to show that $f_1$ and $f_2$ being valid does not imply that $f_3$ is valid. I'm assuming $r$,$p$, and $q$ are all unary predicates.

I'm not sure if this is exactly what you want, but all you have to show is that there exists some model with an element $a$ such that $r(a) \vee \neg p(a)$ and $r(a) \to q(a)$ are valid, but $\neg p(a) \to q(a)$ is not valid. Before you read on, I recommend seeing if you can find this counterexample for yourself, however I have included my reasoning under this spoiler for you.

 Breaking what we have down further we have that $r(a)$ and $q(a)$ are false and $p(a)$ is true. This gives you a counterexample model. The domain of our model is $D = \{a\}$, and our interpretation function is defined with $I(r) = \emptyset$, $I(q) =\emptyset$, and $I(p) = \{a\}$

Therefore, since this is not true in every model, we have that $f_3$ does not follow from those premises.

Title: Show that $f_1,f_2 \not\models f_3$

Body:
Let $f_1=(\forall x(r(x) \vee \neg p(x))$, $f_2=(\exists x(r(x)\rightarrow q(x)))$ and $f_3 = (\forall x(\neg p(x)\rightarrow q(x)))$. Show that $f_1,f_2 \not\models f_3$.

My attempt: lets assume that $\mathcal{A}=(\{a\},S)$ such that $r^{\mathcal{A}}=p^{\mathcal{A}}=q^{\mathcal{A}} = \{a\}$. Then obviously $\mathcal{A} \models f_1$ and $\mathcal{A} \models f_2$. Im not certain how i should continue. Any help is appreciated. Should i consider cases?

For example $I[a / x] \models r(x)$ and $I[a / x] \models p(x)$ then $I[a / x] \not\models \neg p(x)$ so $f_1,f_2 \not\models f_3$ in this case.


问:
Sampling from a distribution which can't be normalized
Suppose you sample a point $x$ from a standard random variable and then you sample a point $y$ from a normal distribution of variance $1/x^2$. With probability $1$, $x\neq 0$, so it is possible to sample $y$. However if we try to write the distribution of the pair $x,y$, it turns out that it is not normalizable. Indeed we have

$$\pi(dx,dy) = \tfrac{1}{Z|x|}e^{-\frac{x^2}{2}-\frac{y^2}{2x^2}}dxdy$$

which gives infinity if integrated. However the sampling is possible. But how does it actually means to sample pairs $(x,y)$ if I'm sampling from something which is not actually a distribution?

答: The formula you give does integrate to 1, not $\infty$, which can be seen by first integrating in $y$ and then integrating in $x$. If you are worried about the slice where $x= 0$, the integral does not see single slices of functions and the value of the integral along a single slice (even if it is undefined or infinite) does not affect the value of the integral. Of course we must be using either an improper Riemann integral or a Lebesgue integral to make sense of this.


问:
Why does the second equality hold?

A sequence of n independent experiments is performed. Each experiment
is a success with probability p and a failure with probability q = 1 -
p. Show that conditional on the number of successes, all possibilities
for the list of outcomes of the experiment are equally likely (of
course, we only consider lists of outcomes where the number of
successes is consistent with the information being conditioned on).

Why does the second inequality hold?

答:
The second equality holds because $k$ is chosen so that $\sum a_i = k$. E.g. saying that $X_1 = 1, X_2 = 0, X_3 = 1, X_1+X_2+X_3 = 2$ is the same as saying that $X_1 = 1, X_2 = 0, X_3 = 1$, the last part is redundant information because on the event that $X_1 = 1, X_2 = 0, X_3 = 1$ we necessarily have $X_1+X_2 + X_3 = 1+0+1 = 2$. Do you see how this applies in general?

As for your main question, the issue with your reasoning is that you do not know (or have not stated why you know) that the $\binom{n}{k}$ paths that are possible are equally likely given the condition that the total number of heads is $k$. This follows from the independence of the $X_i$, which your reasoning did not mention, but which is used in the third equality.

Title: Why does the second equality hold?


问:
Does $f(x)$ is continuous and $f = 0$ a.e. imply $f=0$ everywhere?
I wanna prove that

"if $f: \mathbb{R}^n \to \mathbb{R}$ is continuous and satisfies $f=0$ almost everywhere (in the sense of Lebesgue measure), then, $f=0$ everywhere."

I am confident that the statement is true, but stuck with the proof. Also, is the statement true if the domain $\mathbb{R}^n$ is restricted to $\Omega \subseteq \mathbb{R}^n$ that contains a neighborhood of the origin "$0$"?

答:
Here is a generalization of the result that you want:

Theorem: Let $f,g$ be two continuous functions such that $f = g$ a.e. Then $f = g$ everywhere.

Proof: Let $E$ be the set of all $x$ such that $f(x) \neq g(x)$. Suppose $E$ is not empty and so contains some $x$. Then $E$ being the complement of a closed set is open and so we can find $\epsilon > 0$ such that $B_\epsilon(x) \subseteq E$. But now this means
    $$0 < \mu(B_\epsilon(x)) \leq \mu(E)$$
    contradicting $\mu(E) = 0$. It follows that $E$ has to be empty so that $f = g$ everywhere.


问:
Is this Inclusion-Exclusion: $m$ balls colored $n$ colors such that each color is represented?
Suppose you have $m$ balls each of which can be colored by $n<m$ different colors, and your paint cans are unlimited.  I want to count the number of ways that you can color balls such that each color is represented at least once.  I reasoned that this can be counted by something that seems ... "philosophically" similar to inclusion-exclusion.  But when I try to actually align this counting method with the actual inclusion-exclusion method, I can't seem to perfectly describe how this counting method is an application of inclusion-exclusion.

Here's my solution:  First over-estimate the number of colorings by all possible assignments of colors to balls:  $n^m$.

Next we subtract out those colorings in which some color is absent.  There are $\binom n 1$ choices of one color, and for each, we can color each ball by $(n-1)^m$ colors.  So our new estimate is

$$ n^m - \binom n 1 (n-1)^m $$

However, there are some colorings that are double-counted by this subtraction.  In particular the colorings which are lacking two colors are double-counted (and the colorings missing three colors are triple-counted, and so on).  So to correct for this we add back in $\binom n 2 (n-2)^m$.

$$ n^m - \binom n 1 (n-1)^m + \binom n 2 (n-2)^m $$

We continue the process until, with just one color left, double-countings no longer occur.  So the number is

$$ \sum_{i=0}^{n-1} (-1)^i \binom{n}{n-i}(n-i)^m $$

Edit: I tried searching for this in the recommendations that StackExchange gives, and didn't see anything that looked very relevant.  But now on the sidebar I see this link:  Coloring m things with k colors, using each color at least once.

So this is a known problem, but is there a way to view it as an instance of inclusion-exclusion, or is the similarity just superficial?

答:
I would leave it as

$$\sum_{i=0}^{n-1}(-1)^i\binom{n}i(n-i)^m$$

or even

$$\sum_{i=0}^n(-1)^i\binom{n}i(n-i)^m\,,$$

but what you’ve done is fine and is exactly an inclusion-exclusion argument. The Stirling numbers mentioned in the answer to the earlier question that you found can be calculated by essentially the same inclusion-exclusion argument.


问:
Continuous implies surjective
Suppose $g(x)$ is a function from the set of non negative reals to the set of non negative reals such that $g(x)=x^3+cx$ where $c$ is a constant.Does there exist an $a$ such that $a^3+ca=1$?
I tried proving that using Intermediate Value theorem but can't actually prove.

答:
Continuity in general does not imply surjectivity. In this case, we are closer to the following situation: suppose that $f:\mathbb{R}_+ \to \mathbb{R}_+$ is continuous and is such that $f(0) = 0$ and $\lim_{x \to \infty} f(x) = \infty$. Then $f$ is surjective.

To prove this, let $a \geq 0$ be given. Since $f \to \infty$ we can choose $x$ with $f(x) > a$. Then since $f$ is continuous we have by the intermediate value theorem that $f$ attains all values in the range $[f(0), f(x)] = [0, f(x)]$. But $a \in [0, f(x)]$ since $0 \leq a < f(x)$. Thus $a$ is in the image of $f$. This shows $f$ is surjective.

For your specific function, it is easy to verify the hypotheses.

Title: Continuous implies surjective


问:
How to resolve the question below using polar or cylindrical coordinates?
$\oint_{c} xy ds$ where $C$ is the intersection of the surfaces $x^2+y^2=4$ and $ y+z=8 $

Hi. I tried to resolve the issue below using the definition of line integral, but I couldn’t solve it. Could someone appreciate my resolution and help me finish it? Below follows what I did:

$$\oint_{c} xy ds = {\int }_{C}f\left(x,y,z\right)ds= \int_{a}^{b}f\left(\text{r}\left(t\right)\right)\sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime }\left(t\right)\right)}^{2}+{\left({z}^{\prime }\left(t\right)\right)}^{2}}dt.$$

However, I could not continue.

Thanks!

答:
Hint:

You will need to find the parametrization $\textbf{r}(t)$ for the intersection.

First, in order to be on the surface $x^2+y^2=4$, it must look like $x=...$ and $y=...$ (review parametrization of circles).

Then, in order to be on the surface $y+z=8$, i.e. $z=8-y$, we get $z=...$ (simply substitute what you got for y=).

So then $\textbf{r}(t)=...$


问:
probability and simulation for baseball games
In order for a baseball team to win the world series,the team must win four of seven games. assume the two teams are equally likely to win. 1) Using a simulation with 25 trials,what is the probaility that the teams in the world series will play all seven games? 2) Often we do the same simulation to answer different question.Using the same simulation data,what is the average number of games that must be played to win the world series? 3) the following data represent the number of games played in each world series from 1923 to 2010. compute the average number of games played and compare that number to your result for question 2 from your simulation.

X(games played)| probability

4|0.1954

5| 0.2069

6 | 0.2184

7 | 0.3793

Note: include an explanation of the simulation process in context for each question and include the data for each trial.

答:
Try looking here, it is a nice lesson that will answer your questions and give you the answers.  Its a lesson by Jay Hill. </p>

<p>It will show you how to solve the problem by both simulation and also analytically. </p>

Body:
Try looking here, it is a nice lesson that will answer your questions and give you the answers.  Its a lesson by Jay Hill.

It will show you how to solve the problem by both simulation and also analytically.


问:
Derivative Matrix with general function and product
Given $F,f:\mathbb R\to\mathbb R$ such that $F'=f$ and $\pmb a,\pmb b\in\mathbb R^n$, compute
$$\frac{d}{d\pmb X}\left(\pmb a^T F\left(\pmb X\right)\pmb b\right)$$
where $\pmb X\in\mathbb R^{n\times n}$.

My guess is $\operatorname{diag}(\pmb b) f(\pmb X)^T\operatorname{diag}(\pmb a)$ but I would like to

(i) confirm it and

(ii) see if there's a better way than using indices.

答:
First, note that $a^TF(X)b=\operatorname{tr}(a^TF(X)b)=\operatorname{tr}(ba^TF(X))$. So using differentials we have

$$d\operatorname{tr}(ba^TF(X))=\operatorname{tr}(ba^TdF(X))=\operatorname{tr}(ba^Tf(X)\odot dX)=ab^T\cdot f(X)\odot dX=$$
$$=ab^T\odot f(X)\cdot dX$$
As $\operatorname{tr}(A^TB)=A\cdot B$,  where $A\cdot B=\sum_{i,j}A_{i,j}B_{i,j}$ for matrices of the same dimensions. $\odot$ - Hadamard product.

From this we see that
$$\frac{d}{dX}\left(a^T F\left(X\right)b\right)=ab^T\odot f(X)$$
Also note $ab^T\odot f(X)=\operatorname{diag}(a)f(X)\operatorname{diag}(b)$.

I have used notations and properties of operations from this paper.


问:
If $p\geq 5$ is Prime, Prove that $\binom{p^2}{p}-p$ is a multiple of $p^5$
If $p\geq 5$ is Prime, Prove that $$\binom{p^2}{p}-p$$ is a multiple of $p^5$

I started using the fact that:
$$\binom{n}{r}=\frac{n}{r}\binom{n-1}{r-1}$$
So we have:
$$\binom{p^2}{p}-p=\frac{p^2}{p}\binom{p^2-1}{p-1}-p=p\left(\binom{p^2-1}{p-1}-1\right)$$
Now it suffices to prove that:
$$\binom{p^2-1}{p-1}-1$$ is divisible by $p^4$
Any help from here?

答:
It is perhaps interesting, that this congruence follows from other well known congruences in the literature. The result is attributed to Schwarz, who seems to have proven it first in $1959$ (see the link above).

First,
Jacobsthal proved $1952$ in the paper On the divisibility of the difference between two binomial coefficients the following result, for all $c\le b\le a$, and any prime $p\ge 5$,
$$
\binom{p^a}{p^b}\equiv \binom{p^{a-c}}{p^{b-c}}\bmod p^{3+a+2b-3c}.
$$
For $(a, b,c)=(2,1,1)$ we obtain
$$
\binom{p^2}{p}\equiv p \bmod p^{4}.
$$
This can be found in the paper Wolstenholme's Theorem. There one can also find the result by Helou and Terjanian
in $2008$, which says that

$$
\binom{np}{mp}\equiv \binom{n}{m}\bmod p^k,
$$
where $k$ is the exact power of $p$ dividing $p^3mn(n-m)\binom{n}{m}$.
For $n=p$ and $m=1$ this yields
$$
\binom{p^2}{p}\equiv p \bmod p^{5}.
$$


问:
$ \int_c^\infty \frac{e^{-a x^2 + bx}}{x} \, dx $
Suppose $a > 0$, $b$ is an arbitrary real number, and $c$ is a positive real number. Is there any closed form solution for the following integral

$$
\int_c^\infty \frac{e^{-a x^2 + bx}}{x} \, dx \text{ ?}
$$

答:
By changing variables, we may set one constant to unity, so it is enough to consider

$$
I=\int\limits_c^\infty \frac{dx}{x} e^{-x^2+bx}
$$

Differentiating w.r.t $b$ looks promising, but leads to another tricky integral after the $x$ integration

$$
J=\int\limits_0^b dt \ e^{t^2} \operatorname{erf}(\alpha t + \beta)
$$

If $J$ has a 'nice' form, then so does $I$. If that was a $e^{-t^2}$ instead, then it could be done, but even so the result is not pretty and involves the Owen T special function. As an exact solution is not forthcoming, let us investigate asymptotic solutions.

For small $b$ we may expand $e^{bx}$ and integrate term by term. It's useful to note

$$
\frac{e^{bx}}{x}= \sum\limits_{n=0}^\infty \frac{b^n x^{n-1}}{n!}
$$

$$
\int\limits_c^\infty dx \ e^{-x^2} x^n =\frac{1}{2} \Gamma \left( \frac{n+1}{2} ,c^2 \right)
$$

Where $\Gamma$ is the incomplete gamma function. Thus all the coefficients in the series are known!

$$
I(b)=\frac{1}{2} \sum\limits_{n=0}^\infty \frac{b^n}{n!} \Gamma(n/2,c^2)
$$

This series is an exact representation of your integral. We may truncate the series to find an approximation as $b\to 0$. Here is a plot:

For large $b$, we note the integrand has a maxima at $x=b/2$. So long as $c<<b/2$, we may simply do the Gaussian integral (Laplace's method), setting $x=b/2$ in the denominator

$$
I(b) \sim \frac{2\sqrt{\pi}}{b}e^{b^2/4} \ \ , \ b >> 2c
$$

For large $c$, integrate by parts

$$
K(c) = \int\limits_c^\infty \frac{dx}{x} e^{-x^2+bx}=  \int\limits_c^\infty dx \ \frac{e^{-x^2}}{bx} \frac{d}{dx} \left( e^{bx} \right) \\
$$

$$
K(c)=\frac{e^{-x^2 +bx}}{bc} \bigg\vert_c^\infty +\frac{2}{b} \int\limits_c^\infty  \frac{dx}{x} e^{-x^2+bx} + \frac{1}{b} \int\limits_c^\infty \frac{dx}{x^3} e^{-x^2+bx}
$$

The second term on the right is $K(c)$. For fixed $b$ and large $c$, the third term differs from $K(c)$ by at least $1/c^2$, so we have

$$
K(c)>>\frac{1}{b} \int\limits_c^\infty \frac{dx}{x^3} e^{-x^2+bx} \ \ , \ \ c \to \infty
$$

Neglecting this term, solving for $K$, and simplifying, we find

$$
K(c) \sim \frac{e^{-c^2+bc}}{c(2-b)} \ \ , \ \ c \to \infty
$$

Here is a plot:

Finally for $c\to 0$, the best way I've come up with is to use the series representation, split off the $n=0$ term, then use the identities to write

$$
I(c) \sim -\frac{1}{2} \operatorname{Ei}(-c^2) + \sum\limits_{n=1}^N \left[ \frac{b^n \Gamma(n/2)}{2 \cdot n!} -\frac{b^n c^n}{n!n} \right] \ \ , \ \ c\to 0 \text{ and } N \to \infty
$$

Where $\operatorname{Ei}$ is the exponential integral, and we have the 'complete' gamma function $\Gamma$. At leading order we have

$$
I(c) \sim -\frac{1}{2} \operatorname{Ei}(-c^2) + b \left( \frac{\sqrt{\pi}}{2}-c \right) \ \ , \ \ c\to 0 \text{ and } b \to 0
$$

Here is a plot:


问:
Let $x \in \mathbb{N}$. Prove that if $x$ is odd, then $\sqrt{2x}$ is not an integer.
I need help verifying my proof the following theorem:

Let $x$ be a natural number. Prove that if $x$ is odd, then $\sqrt{2x}$ is not an integer.

The following are the definitions of even and odd numbers:

Definition 3.1. An integer $x$ is odd if there is an integer $n$ such that $x = 2n+1$

Definition 3.2. An integer $x$ is even if there is an integer $n$ such that $x = 2n$

$Proof.$

The contrapositive of the statement is: If $\sqrt{2x}$ is an integer, then $x$ is even.

Because an integer is odd or even and not both and $x$ is a natural number,
\begin{align*}
\sqrt{2x} &= 2n+1 \qquad &\text{for some positive integer n (Definition 3.1)};\\
\qquad &\text{or}\\
\sqrt{2x} &= 2m \qquad &\text{for some positive integer m (Definition 3.2)};\\
\qquad &\text{and not both.}
\end{align*}

In case $\sqrt{2x}$ is odd:
\begin{align*}
(\sqrt{2x})^{2} &= (2n+1)^{2}\\
2x &= 4n^2 + 4n +1\\
x &= 2n^2 + 2n + \frac{1}{2}\\
x &= 2(n^2 + n) + \frac{1}{2}
\end{align*}
Because $n^2 + n$ is some integer, $2(n^2 + n)$ is even (Definition 3.2).
Since no even number added by $\frac{1}{2}$ is an integer, $\sqrt{2x}$ is not odd.

In case $\sqrt{2x}$ is even:
\begin{align*}
(\sqrt{2x})^{2} &= (2m)^{2}\\
2x &= 4m^2\\
x &= 2m^2
\end{align*}

Because $m^2$ is some integer, $2m^2$ is even (Definition 3.2).

Because each implications and its contrapositive are equivalent, given that $x$ is a natural number, if $x$ is odd, then $\sqrt{2x}$ is not an integer.

The question is from the book:
Daepp, U., & Gorkin, P. (2011). Reading, writing, and proving: A closer look at mathematics. In Reading, writing, and proving: A closer look at mathematics (2nd ed., p. 31). New York: Springer.

答:
Your proof is correct but you can make a shorter proof.

If $x$ is odd and $\sqrt{2x}$ is an integer, then $2x$ is even, which implies $\sqrt{2x}$ is also even.  So, you have

$$\sqrt{2x}=2m, ~ m\in \mathbb Z^{+}$$

$$\implies 2x=4m^2$$

$$\implies x=2m^2$$ which gives a contradiction.


答:
So we have $x^3 +cx < 0$ for all $x < 0$.  So $cx < -x^3$ and $c > -x^2$ for all $x<0$ so $c\ge 0$.

So $0^3 + 0\cdot c = 0 < 1$ and $1^3 + 1\cdot c \ge 1$ so by Intermediate Value Theorem there is an $a: 0 < a \le 1$ where $a^3 + ac = 1$.

What does your title have to do with anything though.

Title: Continuous implies surjective

Body:
Suppose $g(x)$ is a function from the set of non negative reals to the set of non negative reals such that $g(x)=x^3+cx$ where $c$ is a constant.Does there exist an $a$ such that $a^3+ca=1$?
I tried proving that using Intermediate Value theorem but can't actually prove.


问:
Projective modules restricted to smooth curves
I want to prove a coherent sheaf $M$ on $X$ is locally free if and only if this is true for $M|_{X'}$ ,
for all smooth curves $X'$ mapping to $X$. I think the only if direction is obvious. For the if direction, a coherent module is flat if and only if it is projective, for Dedekind domains if and only if torsion free as well. So I am thinking of using Tor$_1$. There is local criterion for flatness, but I am not sure if this will help.

This is used at the beginning of the proof of Proposition 5.13 of Gaitsgory's lecture notes.

答:
This question has been asked and answered here on MO. This is a community-wiki recording of the accepted answer there by user abx so that this question can also be marked as answered.

Assume that $X$ is integral and smooth (as in Gaitsgory's notes). For any two points $x,y$ in $X$, there is a smooth connected curve $C$ passing through $x$ and $y$. Since $M_{|C}$ is locally free, this implies $\dim M_x/\mathfrak{m}_xM_x=\dim M_y/\mathfrak{m}_yM_y$, where  $\mathfrak{m}_x$ is the maximal ideal of $\mathscr{O}_{X,x}$. But this implies that $M$ is locally free.


问:
Random Variables: Show that conditional on the number of successes, all possibilities for the list of outcomes of the experiment are equally likely

A sequence of n independent experiments is performed. Each experiment
is a success with probability p and a failure with probability q = 1 -
p. Show that conditional on the number of successes, all possibilities
for the list of outcomes of the experiment are equally likely (of
course, we only consider lists of outcomes where the number of
successes is consistent with the information being conditioned on).

I just learned about random variables and my thinking is: if we know there are k successes, we know there are $\binom{n}{k}$ different sequences of 1's and 0's such that there are k 1's. And these sequences are all equally likely by symmetry. Thus, given the number of successes, k, all possibilities for the list of outcomes of the experiment are equally likely.

But I know that's wrong because I didn't use any random variables.

So my main question is: what's wrong in my thinking (which involves no random variables)?

答: The issue with your reasoning is that you do not know (or have not stated why you know) that the $\binom{n}{k}$ paths that are possible are equally likely given the condition that the total number of heads is $k$. This follows from the independence of the $X_i$, which your reasoning did not mention.


问:
Combinatorial proof of a sum with binomial coefficients
I would like to prove $\sum_{k=0}^{n}{k {n \choose k}}=n2^{n-1}$ with a combinatorial proof, once I already know to prove it algebrically. I thought about it in the following way:

Consider a set $S=\{x_1,...,x_n\}$ with $n$ elements. How many ways can I choose a subset of S with a leader element?

I) Consider a subset of $S$ with $k+1$ elements. I can start by choosing the $k$ non-leader elements ,${n \choose k}$, and then from the other $n-k$ elements I can choose the leader, ${n-k \choose 1}$. Therefore there exist ${n-k \choose 1}{n \choose k}=(n-k){n \choose k}=(n-k){n \choose n-k}$ ways of choosing a subset of $S$ with $k+1$ elements. Therefore there are $\sum_{k=0}^{n-1}{(n-k) {n \choose n-k}}$.

II)Every  element can be leader. Therefore for each leader I have to join a subset of $S$ that does not include the leader, $n2^{n-1}$.

Is it a combinatorial proof? I would be very happy if anyone could share a different one.

答:
Your reasoning is correct, but what it proves is that

$$\sum_{k=0}^{n-1}(n-k)\binom{n}{n-k}=n2^{n-1}\,;$$

in order to complete the proof, you have to show that

$$\sum_{k=0}^{n-1}(n-k)\binom{n}{n-k}=\sum_{k=0}^nk\binom{n}k\,.$$

This isn’t hard:

$$\sum_{k=0}^{n-1}(n-k)\binom{n}{n-k}=\sum_{k=1}^nk\binom{n}k=\sum_{k=0}^nk\binom{n}k\,.$$

But you can avoid this step by using the story suggested in the comments: $k\binom{n}k$ is the number of ways to choose $k$ elements and then choose one of those $k$ to be the leader.


问:
Solving a wave equation with boundary conditions on a circle?
How might one solve a wave equation which had boundary conditions on a circle?

i.e. given $$(\partial^2_x+\partial^2_y)\phi(x,y)=0$$

And known values, $f$ on a circle:

$$\phi(\cos(\theta),\sin(\theta)) = f(\theta)$$

How would one find a solution for $\phi$ in terms of $f$?

答:
There are a few techniques for this, but the most plug-and-chug one is separation of variables in polar coordinates.   Most books on PDEs or mathematical methods for scientists will devote a good chunk of a chapter to explaining this, but here's a summary that will fit in a StackExchange answer:

"Guess" a solution of the form
$$
\phi(r, \theta) = R(r) \Theta(\theta).
$$
If we rewrite the Laplacian in terms of $r$ and $\theta$, it becomes
$$
\frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial \phi}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 \phi}{\partial \theta^2} = 0,
$$
and after some algebra this can be rearranged to read
$$
\frac{r}{R} \frac{\partial}{\partial r} \left( r \frac{\partial R}{\partial r} \right) + \frac{1}{\Theta}\frac{\partial^2 \Theta}{\partial \theta^2} = 0.
$$

Realize that since $R$ doesn't depend on $\theta$, and $\Theta$ doesn't depend on $r$, the two terms in the above equations must be constants.  Moreover, since $\Theta(0) = \Theta(2\pi)$, the solution for $\Theta$ must be sinusoidal rather than exponential, with an integer number of periods in $2 \pi$ radians.  Taken together, this implies that
$$
\frac{r}{R} \frac{\partial}{\partial r} \left( r \frac{\partial R}{\partial r} \right) = n^2 \qquad \frac{1}{\Theta} \frac{\partial^2 \Theta}{\partial \theta^2} = -n^2.
$$
for some $n \in \mathbb{Z}$.

Solve the above two ODEs (not PDEs any more!) to obtain $R$ and $\Theta$ for a fixed value of $n$.  These solutions turn out to be
$$
R(r) = A_n r^n + B_n r^{-n} \qquad \Theta(\theta) = C_n \sin (n \theta) + D_n \cos (n \theta).
$$
If the solution is to be valid at $r = 0$, then we cannot have $|R| \to \infty$ there, and so we must have $B_n = 0$.  This then implies that the solution is of the form
$$
\phi(r, \theta) = r^n [C_n \sin (n \theta) + D_n \cos (n \theta)].
$$

Use the linearity of the PDE to write a more general solution as the sum of these individual solutions:
$$
\phi(r, \theta) = \sum_{n = 0}^\infty r^n \left[C_n \sin (n \theta) + D_n \cos (n \theta)\right] \tag{1}
$$

Note that on the boundary $r = 1$, this ansatz becomes
$$
\phi(1, \theta) = f(\theta) = \sum_{n = 0}^\infty \left[C_n \sin (n \theta) + D_n \cos (n \theta) \right]
$$
and thus the coefficients $C_n$ and $D_n$ can be found by expanding the function $f(\theta)$ in a Fourier series.  The full solution to Laplace's equation with the given boundary conditions is then just Eq. (1) above with the coefficients $C_n$ and $D_n$ plugged in appropriately.

Depending on the function $f(\theta)$, the solution so obtained may be a finite series or an infinite series.  In particular, if $\phi(x,y)$ on the boundary can be written in terms of a polynomial in $x$ and $y$, then $f(\theta)$ will be expressible as a set of powers of $\cos \theta$ and $\sin \theta$, and so the Fourier series will have a finite number of terms.  IIRC this then leads to $\phi(x,y)$ being a polynomial everywhere on the disc.

An infinite series solution, on the other hand, is often not as useful.  It can sometimes (but not always) be re-summed into an explicit form—but when it can, this is often a signal that a different technique (e.g., conformal mapping) could also have been used to find the solution.


答:
Another proof is that if $y=2x$ with $x$ odd, then $2|y$ but $4 \nmid y$.

Hence $\sqrt{y}=k \implies y=k^{2}={p_{1}}^{2}\cdot\cdot\cdot {p_{n}}^{2}$, a contradiction since $p_{i}=2$ for some $i$.

Title: Let $x \in \mathbb{N}$. Prove that if $x$ is odd, then $\sqrt{2x}$ is not an integer.

Body:
I need help verifying my proof the following theorem:

Let $x$ be a natural number. Prove that if $x$ is odd, then $\sqrt{2x}$ is not an integer.

The following are the definitions of even and odd numbers:

Definition 3.1. An integer $x$ is odd if there is an integer $n$ such that $x = 2n+1$

Definition 3.2. An integer $x$ is even if there is an integer $n$ such that $x = 2n$

$Proof.$

The contrapositive of the statement is: If $\sqrt{2x}$ is an integer, then $x$ is even.

Because an integer is odd or even and not both and $x$ is a natural number,
\begin{align*}
\sqrt{2x} &= 2n+1 \qquad &\text{for some positive integer n (Definition 3.1)};\\
\qquad &\text{or}\\
\sqrt{2x} &= 2m \qquad &\text{for some positive integer m (Definition 3.2)};\\
\qquad &\text{and not both.}
\end{align*}

In case $\sqrt{2x}$ is odd:
\begin{align*}
(\sqrt{2x})^{2} &= (2n+1)^{2}\\
2x &= 4n^2 + 4n +1\\
x &= 2n^2 + 2n + \frac{1}{2}\\
x &= 2(n^2 + n) + \frac{1}{2}
\end{align*}
Because $n^2 + n$ is some integer, $2(n^2 + n)$ is even (Definition 3.2).
Since no even number added by $\frac{1}{2}$ is an integer, $\sqrt{2x}$ is not odd.

In case $\sqrt{2x}$ is even:
\begin{align*}
(\sqrt{2x})^{2} &= (2m)^{2}\\
2x &= 4m^2\\
x &= 2m^2
\end{align*}

Because $m^2$ is some integer, $2m^2$ is even (Definition 3.2).

Because each implications and its contrapositive are equivalent, given that $x$ is a natural number, if $x$ is odd, then $\sqrt{2x}$ is not an integer.

The question is from the book:
Daepp, U., & Gorkin, P. (2011). Reading, writing, and proving: A closer look at mathematics. In Reading, writing, and proving: A closer look at mathematics (2nd ed., p. 31). New York: Springer.


问:
Map $z^2=\frac{\frac{1}{2}+it}{ \frac{1}{2}-it },t\in\mathbb{R}$ maps the real axis $(-\infty,\infty)$ onto the unit circle $\mid z\mid=1$.

Show that the mapping
$$
z^2 = \frac{\frac{1}{2}+it}{\frac{1}{2}-it}, \quad t\in\mathbb{R}
$$
maps the real axis $(-\infty,\infty)$ to the unit circle $|z|=1$.

My try-
$$z^2=\frac{\frac{1}{2}+it}{ \frac{1}{2}-it  }$$
$$|z^2|=\frac{|\frac{1}{2}+it|}{ |\frac{1}{2}-it|  }$$
$$|z^2|=\frac{\sqrt{1/4+t^2}}{  \sqrt{1/4+t^2} }$$
$$|z^2|=1$$
$$|z|^2=1$$
$$|z|=1$$...

答:
Let the unit circle be $z=e^{i\theta}$, $\theta\in[0,2\pi]$. Then,

$$t= \frac1{2i} \frac{z^2-1}{z^2+1} = \frac1{2i} \frac{e^{i 2\theta}-1}{e^{i 2\theta} +1} =\frac12\tan\theta
$$
which reveals that $\theta\in[0,\pi)$ and $\theta\in[\pi,2\pi)$ separately map to $t\in(-\infty,\infty)$, thus, not bijective.


问:
Difference between two integrals with PDFs
I am trying to evaluate $$\int_{E'}^{\bar{v}}(v-c)g(v)dv-\int_{E'}^{\bar{v}}(v-E')g(v)dv$$ where $v$ is a random variable with support $[\underline{v}, \bar{v}]$, and $c<E'$ (both of which are in $[\underline{v},\bar{v}]$). Is it acceptable to ignore the pdfs (as they are the same in each term) and simply treat the problem as one with two linear functions? Thank you.

答:
The integrals have the same boundaries so we can write them as one integral:
$$\int\limits_{E'}^{\bar{v}}(v-c)g(v)\mathrm{d}v-\int\limits_{E'}^{\bar{v}}(v-E')g(v)\mathrm{d}v
=\int\limits_{E'}^{\bar{v}}(v-c)g(v)-(v-E')g(v)\,\mathrm{d}v$$
Factoring $g$ out yields
$$\int\limits_{E'}^{\bar{v}}\Big[(v-c)-(v-E')\Big]g(v)\,\mathrm{d}v=\int\limits_{E'}^{\bar{v}}\Big[E'-c\Big]g(v)\,\mathrm{d}v.$$


问:
Martingale inequality - generalized Kolmogorov
The below inequality is a more general case of Kolmogorov maximal inequality.

$\mathbf{Theorem}:$ Let $I \subset \mathbb{R}$ be an interval and $U, U_1, \dots, U_n$ be real RV's such that $U \in \mathbb{L}_1, \mathbb{P}[U \in I] =1$ and $\forall j=1, \dots, n \text{ holds that} \space U_j \le \mathbb{E}[U | U_1,\dots,U_j]$ a.s. and $\mathbb{P}[U \in I] =1$. Denote $M_n = \max[U_1,\dots,U_n]$. Then for $\psi:I\to[0,\infty]$ a nondecreasing convex function and an arbitrary $t \in I$ the following is true:
$$\psi(t) \mathbb{P}[M_n > t] \le \int_{[M_n >t]} \psi(U) d \mathbb{P}$$

$\mathbf{Proof}:$ Let $t \in I$ and denote $\tau=\min\{i=1,2,\dots,n: U_i>t \}$. Clearly, $\tau$ is a stopping time w.r.t. the filtration $(\sigma(U_1,\dots,U_i),i \in \{1,\dots, n\}) $. and thus we obtain the estimate

\begin{align}
\int_{[M_n >t]} \psi(U) d \mathbb{P} &=\sum_{i=1}^n \int_{[\tau=i]} \psi(U) d \mathbb{P}\\
&\ge \sum_{i=1}^n \int_{[\tau=i]} \psi(\mathbb{E}[U | U_1,\dots,U_i]) d \mathbb{P}\\
& \ge \sum_{i=1}^n \int_{[\tau=i]} \psi(U_i) d \mathbb{P}\\
& \ge \sum_{i=1}^n \mathbb{P}[\tau=i] \psi(t) \\
&=\psi(t) \mathbb{P}[M_n>t] \space \space \square
\end{align}

$\mathbf{Q1}$ I don't understand how we come around to use the summation representation of the integral in the first line.

$\mathbf{A1}$ Since $U_i \le \mathbb{E}[U | U_1,\dots,U_i]$ and $[\tau=i]$ requires $U_i>t$ then we know that $t<\mathbb{E}[U | U_1,\dots,U_i]$. $M_n>t$ requires the $$\max_{i \in N, i \le n} U_i >t$$
So this way the stopping time is giving us the first (and the summation any subsequent) time that $M_n>t$. Smart.

$\mathbf{Q2}$ I don't get how we go from first to second. Jensen inequality?
$$\mathbb{E}\psi(U) \ge \psi(\mathbb{E}U).$$ But I can't see any expectation in the first line.

Going from second to the third line merely exploits the fact, that $U_i \le \mathbb{E}[U | U_1,\dots,U_i]$ and the rest is easy.

I'd need a confirmation of my self-supplied A1 and an answer to Q2.

答:
Q1: $M_n(\omega)>t$ if and only if there exists an $i\in \{1,\ldots,n\}$ such that $U_i(\omega)>t$. Hence
$$
\{M_n>t\}=\bigcup_{i=1}^n \{U_i>t\}.
$$
The problem here is that the union on the right-hand side is not a disjoint union, i.e. we can have that $U_i(\omega)>t$ and $U_j(\omega)>t$ for the same $\omega$. The way to go about this is to choose the minimum $i$ such that $U_i>t$, and hence
$$
\{M_n>t\}=\bigcup_{i=1}^n \{\tau =i\}.
$$
If you're still not convinced that the equality holds, you can just show the equality by showing both inclusions. Now that we have a disjoint union we can use that
$$
\int_{\{M_n>t\}}\cdots\,\mathrm dP=\sum_{i=1}^n\int_{\{\tau=i\}}\cdots\,\mathrm dP.
$$

Q2: Jensen's inequality for conditional expectations states that $\psi({\rm E}[U\mid \mathcal{F}_n])\leq {\rm E}[\psi(U)\mid \mathcal{F}_n]$ with $\mathcal{F}_n=\sigma(U_1,\ldots,U_n)$. Hence for $i=1,\ldots,n$ we have
$$
\begin{align*}
\int_{\{\tau=i\}}\psi({\rm E}[U\mid\mathcal{F}_n])\,\mathrm dP&\leq \int_{\{\tau=i\}}{\rm E}[\psi(U)\mid\mathcal{F}_n]\,\mathrm dP=\int_\Omega {\rm E}[\psi(U)\mid\mathcal{F}_n]1_{\{\tau=i\}}\,\mathrm dP\\
&=\int_\Omega {\rm E}[\psi(U)1_{\{\tau=i\}}\mid\mathcal{F}_n]\,\mathrm dP={\rm E}[{\rm E}[\psi(U)1_{\{\tau=i\}}\mid\mathcal{F}_n]]\\
&={\rm E}[\psi(U)1_{\{\tau=i\}}]=\int_{\{\tau=i\}}\psi(U)\,\mathrm dP.
\end{align*}
$$
Summing over $i=1,\ldots,n$ yields the desired result.


问:
Show that $n^2 + m^2 = 2nm$ is an equivalent relation
How can I show this is an equivalence relation:

$nRm ⟺ n^2 + m^2  = 2nm$

I had no problem proving the first two steps (reflexivity and symetric). However, for the transitivity part, I can't get to show it. Anyone can help me?

答:
This is a solution without the fact that this relation is equivalent to standard equality

let $n^2 + m^2 = 2mn$ and  $m^2 + p^2 = 2mp$. This is equivalent to mRn and pRm

Subtracting the equations, $$n^2 - p^2 = 2m(n-p)$$
$$(n-p)(n+p) = 2m(n-p)$$
So either n = p and nRp trivially or we can divide by $n - p$
$$n+p = 2m$$
$$\frac{n+p}{2} = m$$
$$n^2 + (\frac{n+p}{2})^2 = 2n\frac{n+p}{2}$$
$$n^2 + \frac{n^2+2np+p^2}{4} = n^2+np$$
$$\frac{n^2+2np+p^2}{4} = np$$
$$n^2+2np+p^2 = 4np$$
$$n^2+p^2 = 2np$$
so nRp

Therefore nRm and mRp $\Rightarrow$ nRp


问:
Proving $\sum\limits_{n = 0}^{\infty} \frac{x^n}{n!}$ converges
I am trying to prove that the series $\sum\limits_{n = 0}^{\infty} \frac{x^n}{n!}$ converges. It's enough to show that the series of partial sums $S_N = \sum\limits_{n=0}^N a_n$ converges. We have
$$s_N = \sum\limits_{n=0}^N \frac{x^n}{n!}.$$
We propose that its limit is $\sum\limits_{n=0}^{\infty} \frac{x^n}{n!}$. We have to show that for all $\epsilon > 0$, there exists a $K$ such that $N > K$ implies $|s_N - \sum\limits_{n=0}^{\infty} \frac{x^n}{n!} | < \epsilon$. I'm stuck on showing this, however. The set-up is:
\begin{align*}
\left \lvert s_N - \sum\limits_{n=0}^{\infty} \frac{x^n}{n!} \right \rvert & = \left \lvert \sum\limits_{n=0}^N \frac{x^n}{n!} - \sum\limits_{n=0}^{\infty} \frac{x^n}{n!} \right \rvert \\
& = \left \lvert \sum\limits_{n=N+1}^{\infty} \frac{x^n}{n!} \right \rvert \\
& = \sum\limits_{n=N+1}^{\infty} \frac{|x|^n}{n!}
\end{align*}
As $n \to \infty$, each of these terms go to $0$, so I should be able to pick such a $K$ to make this less than $\epsilon$, but I can't figure out how to solve for it and to "isolate" $n$.

答:
Fix $x\in \mathbb{R}$. Observe that if $\left| x \right| \le \left\lfloor {\frac{{N + 2}}{2}} \right\rfloor$,
\begin{align*}
\sum\limits_{n = N + 1}^\infty  {\frac{{\left| x \right|^n }}{{n!}}} & = \frac{{\left| x \right|^{N + 1} }}{{(N + 1)!}}\left( {1 + \frac{{\left| x \right|}}{{N + 2}} + \frac{{\left| x \right|^2 }}{{(N + 1)(N + 3)}} +  \cdots } \right)
\\ &
 \le \frac{{\left| x \right|^{N + 1} }}{{(N + 1)!}}\frac{1}{{1 - \left| x \right|/(N + 2)}} \le 2\frac{{\left| x \right|^{N + 1} }}{{(N + 1)!}}.
\end{align*}
Since
$$
\frac{{\left| x \right|^{N + 1} }}{{(N + 1)!}} \to 0,
$$
for every $\varepsilon>0$, there is a $K>0$ such that
$$
2\frac{{\left| x \right|^{N + 1} }}{{(N + 1)!}} < \varepsilon
$$
for all $N>K$. Hence,
$$
\sum\limits_{n = N + 1}^\infty  {\frac{{\left| x \right|^n }}{{n!}}}  < \varepsilon
$$
for all $N> \max (2\left| x \right| - 2,K)$.


问:
There exists a $\mu\in C_0(\Omega\times \mathbb{R}^n)^*$ such that $\mu_{j_k}(f) → \mu(f)$
Let $\Omega$ be an open subset of $\mathbb{R}^n$. I have a sequence $\mu_j\in C_0(\Omega\times \mathbb{R}^n)^*$ (topological dual). Now I am told that $(\mu_j)_j$ is weakly* sequentially precompact. I am not even sure what this means but I think it means $\mu_j$ has a subsequence $\mu_{j_k}$ that is weakly* Cauchy. I am not sure either on the meaning of this but I guess it means $\|\mu_{j_k}(f)-\mu_{j_l}(f)\|<\epsilon$ when $k,l$ big enough. Now from here I have to show that there exists some $\mu\in C_0(\Omega\times \mathbb{R}^n)^*$ such that
$\mu_{j_k}(f) → \mu(f)$. But I am not sure on why this is the case.

What I know is that $\mathbb{R}$ is complete so we $\mu_{j_k}(f)$ converges to some real number that I can call $\mu(f)$ but then it is not so clear that $\mu\in C_0(\Omega\times \mathbb{R}^n)^*$
for all $f ∈ C_0(Ω \times \mathbb{R}^n)$.

答:
Definition: We say that a sequence $\mu_j$ is weakly* sequentially pre-compact if the weak* closure of that sequence is weakly* sequentially compact.

Definition: We say that $\mu_k\rightarrow \mu$ if and only if $\mu_{jk}(f)\rightarrow \mu(f)$, $\forall f\in C_0(\Omega \times \mathbb{R}^n)$. This is a consequence of the Riesz Representation Theorem and hence what it means to be weakly* convergent.

The result follows once you write down the definition of everything. Let $A:=\{\mu_j\}$ just the set containing every element of the sequence. Since A is weakly* sequentially pre-compact, this implies that $\overline{A}$, the weak* closure of A, is weakly* sequentially compact. Therefore, there exists a weakly* convergent subsequence $\mu_{nk}$ such that $\mu_{nk}\rightarrow \mu\in \overline{A}$.

The only thing that remains to show is that $\mu \in A$ is actually in $C_0(\Omega \times \mathbb{R}^n)^*$. However, the space $C_0(\Omega \times \mathbb{R}^n)^*$ is weak* closed as a topological space and hence $\overline{A}\subset C_0(\Omega \times \mathbb{R}^n)^*$.


问:
How does canonical form relate to a particular $\xi$ and $\eta$?
Given an equation, $$u_{xx} + u_{xt} -20u_{tt}= 0$$ there exists a $\xi$ and a $\eta$ such that $$u_{\xi \eta} = 0$$ I know that because this equation is in the first canonical form of a hyperbolic, seeing that  $\Delta = b^2 - c > 0$. However I don't know how that relates to finding out what $\xi$ and $\eta$ are in terms of $x$ and $y$? I have seen $\xi = x + ct$ and $\eta = x - ct$ often, but I don't know if this is always the case, and/or if we can go further.

How do I solve for $\xi$ and $\eta$ in terms of $x$ and $y$, and in general?

答:
We compose the equation of characteristics
$$
(dt)^2 - dx dt -20 (dx)^2 = 0.
$$
We solve it with respect to $dt$. And we get $dt = -4 dx$ or $dt = 5 dx$, so we have $t + 4 x = \operatorname{const}_1$ and $t - 5 x = \operatorname{const}_2$. These lines are characteristics of your equation.

So, change of variables $\xi =  t + 4 x$ and $\eta =  t - 5 x$ leads to the second canonical form $\partial_\xi\partial_\eta u = 0$.

We can conclude that $u(x, t) = \phi(t + 4 x) + \psi(t - 5 x)$ is general solution of your equation, where $\phi$ and $\psi$ are some twice continuously differentiable functions.


答:
If we take constants $A,B,C,D$  with
$$  x = A \xi + B \eta $$
$$ t = C \xi + D \eta$$
we get
$$ \frac{\partial}{ \partial \xi} = A  \frac{\partial}{ \partial x} + C\frac{\partial}{ \partial t} $$

$$ \frac{\partial}{ \partial \eta} = B  \frac{\partial}{ \partial x} + D\frac{\partial}{ \partial t} $$

and
$$ u_{\xi \eta} = AB U_{xx}  + (AD+BC) u_{xt} + CDu_{tt} $$
Factoring $p^2 + pq - 20 q^2 = (p+5q)(p-4q)$ tells us that we want $A=1, B=1, C= -4, D=5$

$$\left(
\begin{array}{rr}
4 & 1 \\
-5  & 1 \\
\end{array}
\right)
\left(
\begin{array}{rr}
2 & 1 \\
1 &  - 40 \\
\end{array}
\right)
\left(
\begin{array}{rr}
4 &  -  5  \\
1 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rc}
0 & -81 \\
-81 &  0  \\
\end{array}
\right) = -81
\left(
\begin{array}{rc}
0 & 1 \\
1 &  0  \\
\end{array}
\right)
$$

This matrix relation is called congruence. The first symmetric matrix is the Hessian matrix of second partials (by the variables $p,q$) of the quadratic form $p^2 + pq - 20 q^2.$

Title: How does canonical form relate to a particular $\xi$ and $\eta$?

Body:
Given an equation, $$u_{xx} + u_{xt} -20u_{tt}= 0$$ there exists a $\xi$ and a $\eta$ such that $$u_{\xi \eta} = 0$$ I know that because this equation is in the first canonical form of a hyperbolic, seeing that  $\Delta = b^2 - c > 0$. However I don't know how that relates to finding out what $\xi$ and $\eta$ are in terms of $x$ and $y$? I have seen $\xi = x + ct$ and $\eta = x - ct$ often, but I don't know if this is always the case, and/or if we can go further.

How do I solve for $\xi$ and $\eta$ in terms of $x$ and $y$, and in general?


问:
How can I prove that we have $∠(r(u), r(v)) = ∠(u, v)$
How can I prove in geometry that for any $r \in O^+(\mathbb{E})$ we have $∠(r(u), r(v)) = ∠(u, v)$? And what can I say about the case where $r$ is a reflection? Where we write  $O^+(\mathbb{E})$ for the orientation preserving elements of $O(\mathbb{E})$ and $\mathbb{E}$ is a two-dimensional vector space with inner product with a fixed orientation. The group of rotations $O^+(\mathbb{E})$ is generated by pairs of
reflections in lines through the origin.

Any help or tip would be grateful. Thanks in advance.

答:
Rephrasing Berci's comment/NOT an answer.

It depends on your definition. Normally orthogonal transformation is defined as the inner product (and hence norm) preserving linear map. In that case, there is nothing to prove.

If you define orthogonal transformations as the linear transformations $T$ on $E$ (finite-dimensional) such that $TT^{*}=I,$ then it is also clear from the definition. It would be great if you could write your definition of $SO(\mathbb{E}).$ In that case, people would be able to help you.

NB: If $r$ is a reflection, then $r \in O(\mathbb{E})$ but $r \notin SO(\mathbb{E}).$


问:
Quick way to see that $\sqrt{1+x^2+y^2}^{-4}$ is $L^2$-integrable
Is there a quick trick to determine that a function like $$f(x,y):=\sqrt{1+x^2+y^2}^{-4}$$ (which is essentially a negative power of the norm on $\mathbb R^d$ with the singularity at $0$ removed) is in $L^2(\mathbb R^2)$? $f$ is clearly bounded (it takes valeus in $[0,1]$), but that doesn't help here.

答: Hint $$\int_{\Bbb{R}^2} f^2=\sigma(S^1)\int_0^{\infty}\frac{r}{(\sqrt{1+r^2})^8}dr=\sigma(S^1)\int_0^{\infty}\frac{r}{(1+r^2)^4}dr$$ by polar coordinates.


问:
Change of variables (proof of Morrey's inequality)
I have one question of one of the steps in the proof of Morrey's inequality in Evans second edition.
That is in the changing of variable, with $y = x+tw$, $t = |x-y|$. Evans do the following:

$$ \int_0^s \int_{ \partial B(0,1) } |Du(x+tw)|dS(w) dt = \int_0^s \int_{\partial B(x,t)} \frac{|Du(y)|}{t^{n-1}}dS(y)dt.$$

Here $x,y,w \in \mathbb{R}^n$.I'm wondering how the $t^{n-1}$ appears in the denominator? There should be a jacobian, due to the change of variable, however I am unsure how this is treated in the change from $dS(w)$ to $dS(y)$.

答:
Hausdorff Measure and The Coarea formula are what you need to make it really rigorous. The  Coarea formula though is in the appendix and working through Evans PDE book I reckon you can take it as a sort of 'black box' result. As a reference though have a look at sections 2 and 3 of Gariepy and Evans book: Measure Theory and Fine Properties of Functions.

In Evans's notation in the surface integral over an n-dim ball in $\mathbb{R}^n$ the $dS$ corresponds to the Hausdorff $n-1$ dim measure of the surface of the ball. In other words, you get the surface area of the $n$ dim ball. In particular,
\begin{equation}
\mathcal{H}^{n-1}(\partial B(x, r))=\int_{\partial B(x, r)}\mathrm{d}S=n\alpha(n)r^{n-1}.
\end{equation}

So I guess one way to 'see' the change of variables which is just a scaling map in your work above is as follows. We have
\begin{equation}
\int_{0}^{s}\int_{\partial B(0, 1)}\vert Du(x+tw)\vert\mathrm{d}S\mathrm{d}t= \int_{0}^{s}\int_{\partial B(0, 1)}\vert Du(x+tw)\vert\frac{t^{n-1}}{t^{n-1}}\mathrm{d}S\mathrm{d}t.
\end{equation}
Let $y=x+tw$, then
\begin{equation}
t=\vert x-y\vert\Rightarrow y\in\partial B(x, t)
\end{equation}
and (symbolically)
\begin{equation}
dS_y=n\alpha(n)t^{n-1}=t^{n-1}dS
\end{equation}
because
\begin{equation}
\mathcal{H}^{n-1}(\partial B(0, 1))=\int_{\partial B(0, 1)}\mathrm{d}S=n\alpha(n).
\end{equation}
So now putting everything together we have
\begin{equation}
\int_{0}^{s}\int_{\partial B(0, 1)}\vert Du(x+tw)\frac{t^{n-1}}{t^{n-1}}\mathrm{d}S\mathrm{d}t=\int_{0}^{s}\int_{\partial B(x, t)}\vert\frac{\vert Du(y)\vert}{t^{n-1}}\mathrm{d}S_y\mathrm{d}t
\end{equation}


问:
probability distribution of $X=U^{-\alpha}$
Let $U \sim \operatorname{Unif}[0,1],$ and $\frac{1}{2}<\alpha<1 .$ Set $X=U^{-\alpha}$. What is the probability distribution of X? Is X integrable?

My attempt:
$\mathbb{P}(X \leq a) = \mathbb{P}(U^{-\alpha} \leq a)\\
 = \mathbb{P}(e^{-\alpha \ln{U}} \leq a) = \mathbb{P}(- \alpha \ln{U} \leq \ln{a}) \\
= \mathbb{P}(\ln{U} \geq - \frac{\ln{a}}{\alpha}) = \mathbb{P}(U \geq e^{- \frac{\ln{a}}{\alpha}}) = 1 - \mathbb{P}(U \leq e^{- \frac{\ln{a}}{\alpha}})\\
 = 1 - e^{- \frac{\ln{a}}{\alpha}}$

答: You are almost done. Note that $U^\alpha\in(0,1)$ and so $U^{-\alpha}\in(1,\infty)$. Thus$$P(U^{-\alpha}\le a)=\begin{cases}0,&a\le1\\1-a^{-1/\alpha},&a>1\end{cases}$$differentiating which we obtain $f(a)=\frac1\alpha a^{-1/\alpha-1},a>1$. Fairly easy to observe $\Bbb E[U^{-\alpha}]=1/(1-\alpha)$.


问:
Gradient Descent on differential equation
I have a differential equation of the form
$$\frac{d}{dx}f=f^2$$
I want to find a root of the second derivative of $f$, in order to maximize the derivative $df/dx$.

I could of course simply solve the differential equation and apply gradient descent on the second derivative to find the root, but I would like to find this root without having to solve the differential. Any thoughts?

答:
$\frac{df}{dx}=f^2 \Rightarrow \frac{d^2f}{dx^2}=2f\frac{df}{dx}=2f^3$

$f(x)=0 \Rightarrow \frac{d^2f}{dx^2}=0$


答:
You have

$$f^\prime(x)= f^2(x)$$ hence taking the derivative on both sides

$$f^{\prime \prime}(x)=2f(x)f^\prime(x)=2f^3(x)$$which proves that the second derivative vanishes if and only $f$ vanishes at the same point.

But if $f$ vanishes at $x_0$, you also have $f^\prime(x_0)=0$. And according to the Picard-Lindelöf theorem, $f$ is the always vanishing map.

Conclusion: either $f$ is the always vanishing map or its second derivative never vanishes as $f$ itself.

Title: Gradient Descent on differential equation

Body:
I have a differential equation of the form
$$\frac{d}{dx}f=f^2$$
I want to find a root of the second derivative of $f$, in order to maximize the derivative $df/dx$.

I could of course simply solve the differential equation and apply gradient descent on the second derivative to find the root, but I would like to find this root without having to solve the differential. Any thoughts?


问:
Is there a notation for the inverse image function itself?
Let $f$ be a function.

If $B \subset im(f)$ then the inverse image of $B$ under $f$ in the current standard notation is

$f^{-1}(B) = \{x \in dom(f) : f(x) \in B\}$.

But $f^{-1}$ does not denote the inverse image function !

Using a kind of the typed lambda notation the inverse image function
-- the name $preim$ (from preimage) may be used - can be defined as follows:

$preim := F[B \in P(im(f)), \{x \in dom(f): f(x) \in B \}]$.

But in the common case (if $f$ is not necessary bijective):

$f^{-1} = F[y \in im(f)), \{x \in dom(f): f(x) = y \}]$.

答:
Some common notations for the converse of a relation $R$:
$R^\text{T}$, $\breve{R}$, $\check{R}$.

Now these aren't quite what you're asking for. As you note, if $f:X\rightarrow Y$, then the function mapping $P(X)\rightarrow P(Y)$ (powersets) is different. However, another standard notation for $f(A)$, with $A\subseteq X$, is $f``A$. So you might write
$$\breve{f}``$$
for your function. I admit I've never seen this particular combination in print.


问:
A Calculus Question on onto functions with a specified range.
The following question was from a mock test of a competitive exam.

Suppose $f:\mathbb{R} \to [-8,8]$ is an onto function and $f(x) = \dfrac{bx}{(a-3)x^3 + x^2 + 4}$ where $a,b \in \mathbb{R}^+$. If the set of all values of $m$ for which the equation $f(x) = mx$ has three distinct real solutions in the open interval $(p,q)$, then find the value of $a+b+p+q$.

I tried the problem and obtained an answer. According to me, it is $43$. I will post the answer later. My solution was quite involved and long. I want to know if there is a quick solution to this problem.

Thank you.

答:
First off, the function $f(x)=\dfrac{bx}{(a-3)x^3 + x^2 + 4}$ has a cubic polynomial in the denominator (assuming $a\ne3$), and since we assume $a\in\Bbb{R}$, this equation must have a real root, which cannot be canceled by the $x$ in the numerator, because $x=0$ is not a root of the cubic (which instead evaluates to $4$ at $x=0$). In the vicinity of this root, $f(x)$ is unbounded above and below, so clearly it doesn't satisfy the criterion that $\operatorname{ran}f=[-8,8]$. Thus $a=3$ (in order to make the denominator have no real roots).

The new function $f(x)=\dfrac{bx}{x^2 + 4}$ has no poles, and goes to $0$ at $x\to\pm\infty$, and it's self-evidently continuous, so it takes on a maximum and minimum somewhere, which can be found using the first-derivative test. $f'(x)=-\dfrac{b(x^2-4)}{(x^2+4)^2}=0$ when $x=\pm2$, and at these points $f(x)=\pm\frac b4$. Thus $\operatorname{ran}f=\big[\!-\frac b4,\frac b4\!\big]=[-8,8]$ implies $b=32$.

The function is now $f(x)=\dfrac{32x}{x^2 + 4}$. Note that $f$ is odd. Thus $f(0)=0=m\cdot0$ is always a solution to $f(x)=mx$, and nontrivial solutions come in pairs $\pm x$, since $mx$ is also an odd function. Having discarded $x=0$, we can divide by $x$ and rearrange the equation to get $x^2 + 4=32/m$. This equation has a solution when $32/m > 4$ (noting that $32/m=4$ merely yields a triple root at $0$ which is stated to be inadmissible), which is equivalent to $0<m<8$. The case $m=0$ must be analyzed separately, but $\dfrac{32x}{x^2 + 4}=0$ only when $x=0$, so it is not in the case of interest. Thus $m\in(0,8)=(p,q)$ implies $p=0$ and $q=8$.

Putting it all together, we have $a+b+p+q=3+32+0+8=43$, so it looks like you got the answer right.


问:
Determine if $x^3+y^3+z^3+t^3 = 10^{2021}$ has a solution
I want to know if the equation $x^3+y^3+z^3+t^3=10^{2021}$ has distinct positive integer solutions
PowersRepresentations[10^2021, 4, 3]

return

PowersRepresentations::ovfl: Overflow occurred in computation.

```<code>FindInstance[{x^3 + y^3 + z^3 + t^3 == 10^2021, 0 < x < y < z < t}, {x,y,z,t}, Integers]
</code>```

My computer runs too long. How can I reduce timing to solve this equation?

OwnerDisplayName: expression

Body:
Two solutions may be found starting from:
$$10^{2021}=10^5\times10^{2016}=12500\times2^3\times10^{3\times672}$$

Since $12500=19^3+17^3+8^3+6^3=18^3+17^3+12^3+3^3$ we have:

$$10^{2021}=(38\times10^{672})^3+(34\times10^{672})^3+(16\times10^{672})^3+(12\times10^{672})^3$$
$$10^{2021}=(36\times10^{672})^3+(34\times10^{672})^3+(24\times10^{672})^3+(6\times10^{672})^3$$


问:
Multiplying Argument

I'm solving the following problem: Sketch the image of the set $$\{ Arg(z) \geq \frac{\pi}{2} \text{ and } 1 \leq |z| \leq 2\}$$ by the function $f(z) = z^3$.

What I tried so far: $z^3 =  r^3e^{3\theta i}$. Then,

\begin{equation}
 \begin{split}
        1 &\leq r^3 \leq 8 \\
        3\pi/2 &\leq 3Arg(\theta)
    \end{split}
\end{equation}

This is where I am stuck. How do you know where to stop starting from $3\pi/2$? 3rd quadrant can be  $\geq 3/2\pi$.

I know the correct solution is the image above by plugging in three random but easy points from the domain. Is there any way to know where the graph should stop (radian-wise) without having to plug in values?

答: The first condition can be written as: $$\{\pi/2\le \arg(z)\le \pi/2 \text{ and }1\le|z|\le2\}.$$ Hence, the image of the domain would be $$\{3\pi/2\le \arg(z)\le 3\pi/2 \text{ and }1\le|z|\le8\} .$$

Title: Multiplying Argument


Title: Quick way to see that $\sqrt{1+x^2+y^2}^{-4}$ is $L^2$-integrable

Body: Is there a quick trick to determine that a function like $$f(x,y):=\sqrt{1+x^2+y^2}^{-4}$$ (which is essentially a negative power of the norm on $\mathbb R^d$ with the singularity at $0$ removed) is in $L^2(\mathbb R^2)$? $f$ is clearly bounded (it takes valeus in $[0,1]$), but that doesn't help here.

Body:
Hint

$$
\sqrt{1+x^2+y^2}^{-4}=\frac{1}{(1+x^2+y^2)^2}=\frac{1}{1+x^2+y^2} \cdot \frac{1}{1+x^2+y^2}\leq \frac{1}{1+x^2} \cdot \frac{1}{1+y^2}
$$


问:
Continuous coloring of a Mandelbrot fractal
I've recently started making a small fractal app in Javascript using the famous Mandelbrot bulb $(z = z^2 + c)$. I've been trying to find the best method of coloring the points on the complex plane, and I've come across some very interesting ideas:

http://linas.org/art-gallery/escape/escape.html

http://en.wikibooks.org/wiki/Fractals/Iterations_in_the_complex_plane/Mandelbrot_set#Real_Escape_Time

So I've been focusing on this "normalized iteration count algorithm" and I've run into a small mental hurdle sorting out what needs to be calculated. Basically, I'm confused about the first equation that shows up in that second link. Assuming we have a complex number $Z$ which is the end result of $n$ iterations. I can't quite figure out what "Zx2" or "zy2" is and how how one might add them together or take the square root of their sum. Is that just shorthand for some basic complex operators, or is there something else going on? Basically, my problem is that I'm not particularly good at reading mathematical notation in this area.

Anyway...any directional guidance you can offer would be extremely helpful. Thanks in advance!

答:
edit: "Zx2" and "Zy2" are defined in the C code in the second code block at this point in your wikibooks link.  They are the squares of the real and imaginary parts, respectively, of $z$.

I'm not sure if it's exactly the same, but what I've used is:
$$n-\log\left(\frac{\log(|z|)}{\log(r)}\right)$$
where $z$ is the result of $n$ iterations and has just escaped as noted by $|z|>r$ (where $r$ is the escape radius).  This generates real numbers in $[0,n)$, so you could divide by $n$ to end up with a number in $[0,1)$.

ContentLicense: CC BY-SA 2.5


问:
Struggling to determine the bounds for this volume by integration problem
I have
$$y=|x|+1, 2x+3y=6\text{ about }x=-20.$$

I need to set up an integral (but not evaluate it) that will give the volume of the solid when the region enclosed by those curves is revolved about the given line. I have to use either the shell or washer's method.

I have been trying to use the shell method (I think) but I am having trouble setting up the integral. How do I figure out the bounds for it? I'm not sure how I get the numbers for "top" and "bottom" either. I solved for y in the $2x+3y=6$ equation and got $y=-(2/3)x+2$ but how do I know where that belongs in the integral?

Thank you for any help

答:
Whatever the suggested methods, the simplest way to find this volume $V$ is to use Guldin's rule: $V$ is equal to the area $A$ of the given triangle $\Delta$  times the circumference of the circle described by the centre of gravity $S$ of $\Delta$. The vertices of $\Delta$ being $(0,1)$, $({3\over5}, {8\over5})$ and $(-3,4)$ we get $A={9\over5}$ and $S=(-{4\over5},{11\over5})$. This implies
$$V= {9\over5} \cdot 2\pi (-{4\over5}+20) ={1728\over25}\ \pi\ .$$</p>


问:
Units of the quotient of an order
Let $n$ be a positive integer and $R$ be an order in a imaginary quadratic number field such that $disc(R)$ is prime to $n$. Further suppose that for every prime $p$ dividing $n$, $p$ is inert in $R$. Now let $A = R/nR$. I want to show that $\#A^\times = n^2\prod\limits_{p|n}(1-\frac 1 {p^2})$.

If i'm not mistaken by the chinese remainder theorem we have $A \cong \bigoplus\limits_{p|n} A/p^{v_p(n)}A$. So what we want to show is that $\# (A/p^{v_p(n)}A)^\times = p^{2v_p(n)}(1-\frac 1 {p^2})$.

When $v_p(n) = 1$ this is true because $p$ is inert in $R$ and so $A/pA \cong \mathbb{F}_{p^2}$.

I guess we have to use the fact that $disc(R)$ is prime to $n$ but I can't see how. In fact I would be quite interested to know what this condition implies in general.

答:
Let $K$ be the imaginary quadratic field and $S$ be its ring of integers.
Let $\tau$ and $d$ be such that $S = \Bbb Z[\tau]$ and $R = \Bbb Z[d\tau]$.

$R/nR = \Bbb Z[d\tau]/n\Bbb Z[d\tau] \approx (\Bbb Z/n\Bbb Z)[d\tau]$.

Since $n$ and $d$ are coprime, $d$ is invertible in $\Bbb Z/n\Bbb Z$, and we have
$(\Bbb Z/n\Bbb Z)[d\tau] \approx (\Bbb Z/n\Bbb Z)[\tau] \approx S/nS$.

So $R/nR \approx S/nS \approx \prod S/p^{v_p(n)}S$.

Since $p$ is inert, it is a prime in $S$, and the non-invertible elements of $(S/p^{v_p(n)}S)$ are those that are a multiple of $p$. Their density is $1/p^2$, which proves that $\# A^* = \prod (1 - p^{-2})$

For a more explicit calculation,
if $a+bd\tau \in (\Bbb Z/n\Bbb Z)[d\tau]$ with $\gcd(a,b,n) = 1$, then
$(a+bd\tau)(a+bd\bar\tau) = a^2+abd(\tau+\bar\tau)+b^2d^2\tau\bar\tau = N$ is invertible in $\Bbb Z/n\Bbb Z$, and so $((a/N)+(b/N)d\bar\tau) \in (\Bbb Z/n\Bbb Z)[d\tau]$ is the inverse of $a+bd\tau$

Title: Units of the quotient of an order


问:
Prove that it doesn’t exist $n\geq3$ Such that $n$ is divisible by all $m\le n$
A Claim:

Let $A$ be a set contains all $n\geq3$ Such that $n$ is divisible by all the numbers less or equal to $n$.
$$A=\{n\geq3 \mid m\mid n, \forall m\le n \}$$

Show that $A=\emptyset$.

This is just another way to explain the question in the title, and why $n\geq3$? It’s because $1,2$ are divisible by all numbers less than themselves.

I don’t have any idea how to prove this claim, i’ve just tried a lot of values and it seems valid, maybe the best way to prove it it’s by contradiction.

答:
Suppose that $A\ne \emptyset$, then exists $n\in A$.

Since every $m\leq n$ divides $n$ then also $n-1\mid n$. But then $$n-1\mid n-(n-1) =1$$ so $n-1 =1$ a contradiction, since $n\geq 3$. Thus $A= \emptyset$.


答: $n-1$ does not divide $n$ for any $n>2$, because $\frac{n}{n-1}$ gives a value between 1 and 2.

Title: Prove that it doesn’t exist $n\geq3$ Such that $n$ is divisible by all $m\le n$

Body:
A Claim:

Let $A$ be a set contains all $n\geq3$ Such that $n$ is divisible by all the numbers less or equal to $n$.
$$A=\{n\geq3 \mid m\mid n, \forall m\le n \}$$

Show that $A=\emptyset$.

This is just another way to explain the question in the title, and why $n\geq3$? It’s because $1,2$ are divisible by all numbers less than themselves.

I don’t have any idea how to prove this claim, i’ve just tried a lot of values and it seems valid, maybe the best way to prove it it’s by contradiction.


问:
Why is $T \Rightarrow T$ false, but $T \stackrel{*}{\Rightarrow} T$ true? I'm not sure how to derived this.
```<code>R -> XRX | S
S -> aTb | bTa
T -> XTX | X | ε (empty string)
X -> a | b
</code>```

$T \Rightarrow T$

$T \stackrel{*}{\Rightarrow} T$

I know the first one is false since you cannot derived it to T but for the second one my friend said it is true while I thought it is false. I was wondering how do you eventually derived from T to T.

答: The star in $\overset{*}\Rightarrow$ works like the Kleene star in regular expressions: it includes $\overset{n}\Rightarrow$ for all $n\ge 0$, so it includes $0$-step derivations $u\Rightarrow v$ in which $u=v$. You’re correct in thinking that there is no derivation of $T$ from $T$ that actually uses at least one production.


问:
Convergence of Riemann maps
For $\epsilon > 0$, let $K_\epsilon = \{e^{i\theta} : \theta \in [0,\pi-\epsilon]\}$ be an almost-semicircle. Let $D_\epsilon = H \setminus K_\epsilon$, where $H$ is the upper half plane. Let $\phi_\epsilon : D_\epsilon \rightarrow H$ be the Riemann map normalized so that $\lim_{z\rightarrow\infty} \phi_\epsilon(z) -z = 0$. Let $\psi_\epsilon = \phi_\epsilon/\phi_\epsilon'(0)$ on the intersection of $A$ of the unit disk with the upper half plane. I am trying to show that $\psi_\epsilon$ converges uniformly on compact subsets of $A$ to a conformal map from $A$ to $H$ as $\epsilon \rightarrow 0$.

Intuitively, as $\epsilon \rightarrow 0$, $D_\epsilon$ approaches a domain with two connected components, and an appropriately normalize map should approach a biholomorphism on each. But, I do not have much of an idea as to how to turn this intuition into a proof. Does anyone have any suggestions? Even better, are there any general results about the limits of Riemann maps on domains which are "almost disconnected" which can be applied here?

答:
A standard tool is the Carathéodory kernel theorem. It operates with a notion of the limit of domains with a distinguished interior point; the same point that is also used to normalize the Riemann map.

So, if you normalize $\phi_\epsilon$ so that $\phi_\epsilon(i/2)=i$, they will converge to a conformal map of $H\cap \{z:|z|<1\}$ onto $H$. And if you normalize them so that $\phi_\epsilon(2i)=i$, they will converge to a conformal map of $H\cap \{z:|z|>1\}$ onto $H$.

Normalization by a boundary value of a derivative is not as reliable: it gives less control on the map. If you have  a reason to use this particular normalization, comment and I'll try to come up with something.

Additional details:

In Wikipedia, the Carathéodory kernel theorem is stated for maps from the unit disk $D$ to a sequence of domains. It applies just as well to maps from halfplane $H$ to a sequence of domains, because they can composed with a fixed map $D\to H$.

So, the theorem applies to the inverses of $\phi_\epsilon$, normalized, for example, by $\phi_\epsilon(i)=z_0$. The kernel of the sequence $\phi_\epsilon^{-1}(H)$ depends on the choice of this normalization: if $|z_0|<1$, it is the upper half-disk, if $|z_0|>1$, it is the upper half-plane minus the half-disk.

One can infer the convergence of $\phi_\epsilon$ from the convergence of $\phi_\epsilon^{-1}$. Rouche's theorem and Hurwitz's theorems come to mind as  possible tools here. But it may be more practical to read a more detailed discussion of kernel convergence, either in Geometric theory of functions of a complex variable by Goluzin or in Univalent functions by Duren.

Title: Convergence of Riemann maps


问:
Why does $2$ appear in $95\%$ confidence intervals?
I was told that, in linear regression, the $95\%$ CI for $\beta$ is $$\beta \in \left(\hat\beta-2 \sigma,\hat\beta+2\sigma\right), \text{ where } \sigma = \text{standard error}(\hat\beta).$$

My question: Where does the $2$ come from?

Also, would that $2$ change if, say, we were looking for a $90\%$ CI, or $40\%$ CI?

I recall that the level of confidence affects the $t$-value cutoff for hypothesis testing, but I can't recall how it changes in the above context.

答:
It's because $$2\approx 1.96 \approx \Phi(0.975),$$

where $\Phi$ is the cumulative distribution function of the standard normal distribution $Z\sim N(0,1)$, that is,

$$\mathbb{P}(Z<1.96)\approx 0.975$$

The number even has its own wikipedia article. Note: always draw the picture.

If you want a $100(1-\alpha)\%$ confidence interval, you should use

$$\mathbb{P}\left(-\Phi\left(1-\frac{\alpha}{2}\right)<Z<\Phi\left(1-\frac{\alpha}{2}\right)\right)=1-\alpha.$$


问:
"$\Sigma_1^1$-Peano arithmetic" - does it pin down $\mathbb{N}$?
Let $\mathsf{PA}_{\Sigma^1_1}$ be the theory in second-order logic gotten by extending the usual first-order Peano axioms to include arbitrary $\Sigma^1_1$ formulas in the induction scheme. My question is:

Does $\mathsf{PA}_{\Sigma^1_1}$ have any nonstandard models?

Note that a model of $\mathsf{PA}_{\Sigma^1_1}$ is exactly a model of $\mathsf{PA}$ with no (nontrivial proper) $\Sigma^1_1$-definable cuts.

If we replace $\Sigma^1_1$ with $\Pi^1_1$ the answer is immediately negative, since the set of standard elements of a model of $\mathsf{PA}$ is $\Pi^1_1$. However, nothing similar seems to work for $\Sigma^1_1$ (although I could easily be missing something obvious).

One quick observation is that $\mathsf{PA}_{\Sigma^1_1}$ does entail true first-order arithmetic. Given a first-order formula $\varphi(x)$, let $\hat{\varphi}(x)$ be the $\Sigma^1_1$ formula "There is a cut containing $x$ such that every element of the cut satisfies $\varphi$." If $M\models\mathsf{PA}_{\Sigma^1_1}$ we trivially have $\hat{\varphi}^M\in\{\emptyset,M\}$; by induction on the complexity of $\varphi$ we can show that if every standard natural number satisfies $\varphi$ then $0\in\hat{\varphi}^M$ and consequently $M\models\forall x\varphi(x)$ (which then gives $M\equiv\mathbb{N}$). However, I don't see how to use this to get categoricity. In fact, as far as I know it's possible that e.g. every nontrivial ultrapower of $\mathbb{N}$ satisfies $\mathsf{PA}_{\Sigma^1_1}$. (Note that $\Sigma^1_1$ sentences are preserved under taking ultrapowers; however, an instance of induction for a $\Sigma^1_1$ formula is $\Sigma^1_1\vee\Pi^1_1$ and $\Pi^1_1$ sentences are not preserved under taking ultrapowers, so this doesn't seem to help.)

答: If you allow your $\Sigma^1_1$ formulas to have parameters, then PA$_{\Sigma^1_1}$ has only the standard model. To prove it, use the $\Pi^1_1$ definition of standardness to produce a $\Sigma^1_1$ formula $\sigma(x,y)$ saying that $x<y$ and $y-x$ is not standard, i.e., $x$ is infinitely far below $y$. It's easy to show that $\sigma(x,y)$ implies $\sigma(x+1,y)$. So, by $\Sigma^1_1$ induction, if $\sigma(0,y)$ then $\forall x\,\sigma(x,y)$ and, in particular, $\sigma(y,y)$, which is absurd. So $\neg\sigma(0,y)$. But that means $y$ is standard.