问:
Equality with Euler–Mascheroni constant
While trying to prove $\int_0^{\infty } \frac{\log (x)}{e^x+1} \, dx = -\frac{1}{2} \log ^2(2)$ How to show? in an alternative way, I came to this solution:

$$\sum_{k=0}^{+\infty}(-1)^{k+1}\frac{\log (k+1)+\gamma }{(k+1)}.$$

As both solutions have to be the same, the following equality should be valid:

$$\sum_{k=0}^{+\infty}(-1)^{k+1}\frac{\log (k+1)+\gamma }{(k+1)}=- \frac{1}{2}{{\log }^2(2)}. $$

Can anyone give me some advice on how to prove this equality.

p.s. You can be sure that the equality is correct, as I checked it numerically.

答:
The problem is equivalent to showing

$$
\sum_{n\ge1}{(-1)^n\log n\over n}=\gamma\log2+\frac12\log^22
$$

Now, let's first consider the finite case:

$$
\begin{aligned}
\sum_{n\le N}{(-1)^n\log n\over n}
&=2\sum_{n\le N/2}{\log(2n)\over2n}-\sum_{n\le N}{\log n\over n}+\mathcal O\left(\log N\over N\right) \\
&=\sum_{n\le N/2}{\log2+\log n\over n}-\sum_{n\le N}{\log n\over n}+\mathcal O\left(\log N\over N\right) \\
&=\log2\sum_{n\le N/2}\frac1n-\sum_{N/2<n\le N}{\log n\over n}+\mathcal O\left(\log N\over N\right) \\
\end{aligned}
$$

In fact, using Riemann-Stieltjes integral, we can show

$$
\begin{aligned}
\sum_{N/2<n\le N}{\log n\over n}
&=\int_{N/2}^N{\log x\over x}\mathrm d\lfloor x\rfloor \\
&={N\log(N)-N\log(N/2)\over N}-\int_{N/2}^N[x-\{x\}]\mathrm d\left(\log x\over x\right)+\mathcal O\left(\frac1n\right) \\
&=\log2-\int_{N/2}^N\left({1-\log x\over x}\right)\mathrm dx+\mathcal O\left(\log N\over N\right) \\
&=\int_{N/2}^N{\log x\over x}\mathrm dx+\mathcal O\left(\log N\over N\right) \\
&=\frac12[\log^2N-\log^2(N/2)]+\mathcal O\left(\log N\over N\right) \\
&=\frac12[\log N+\log(N/2)][\log N-\log(N/2)]+\mathcal O\left(\log N\over N\right) \\
&=\frac12\log2[2\log N-\log2]+\mathcal O\left(\log N\over N\right) \\
&=\log2\log N-\frac12\log^22+\mathcal O\left(\log N\over N\right)
\end{aligned}
$$

Now, employing this obtained identity and the asymptotic formula for harmonic series yields:

$$
\begin{aligned}
\sum_{n\le N}{(-1)^n\log n\over n}
&=\log2(\log N+\gamma)-\log2\log N+\frac12\log^22+\mathcal O\left(\log N\over N\right) \\
&=\gamma\log2+\frac12\log^22+\mathcal O\left(\log N\over N\right)
\end{aligned}
$$

Now, take the limit $N\to\infty$ on both side gives

$$
\sum_{n\ge1}{(-1)^n\log n\over n}=\gamma\log2+\frac12\log^22
$$

Title: Equality with Euler–Mascheroni constant


问:
Integral of functions of several variables
The function is defined from $[-1,1]\times[-1,1]$ to $\Bbb R$, given by $f(x,y)=\frac{x^2-y^2}{(x^2+y^2)^2}$ when $(x,y)≠0$ and $f(0,0)=0$.
I could find that the function is not continuous at origin, so not differentiable  and also partial derivatives does not exist at origin.
But how do we evaluate the integral of the function over the given domain. Being function of several variable how do we deal with the point of discontinuity? Can we convert it to polar coordinates and apply residue theorem? Do we have any other methods?
Any help would be appreciated. Thanks in advance.

答: The improper integral $$\int_{-1}^1 \int_{-1}^1 \frac{x^2-y^2}{(x^2+y^2)^2} \, dy \, dx$$ diverges. To see why, consider the region $\mathcal{R}$ contained in $[-1,1]\times [-1,1]$ defined in polar coordinates by$$\mathcal{R}=\Big\{(r,\theta)\Big| 0<r<1, 0<\theta<\pi/8\Big\}$$ Then $$\int \int _{\mathcal{R}}\frac{x^2-y^2}{(x^2+y^2)^2} \, dy \, dx=\int_{0}^{\pi/8} \int_0^1\frac{\cos(2\theta)}{r} \, dr \, d\theta=+\infty$$

ContentLicense: CC BY-SA 4.0


问:
$\sum\limits_{\text{prime }p} 2^{-p}$ is an irrational number
I need help to prove the following result.

$\displaystyle\sum_{\text{prime }p} 2^{-p}$ is an irrational number.

答:
One way to show that the sum is irrational
is the well-known fact that there
are arbitrarily long intervals
where there are no primes.
If the sum were rational,
chose the interval longer that the period,
and hello contradiction!

In particular,
$n!+k$ for $k = 2$ to $n$
has no primes, since each element is divisible
by a prime from $2$ to $n$.

This also shows that
$\sum_{\text{prime }p} m^{-p}$
is irrational for any integer $m \ge 2$.

ContentLicense: CC BY-SA 3.0


问:
If $U^T U = I_k$, $V^T V = I_{d-k}$, and $U^T V \approx 0$, is $U U^T + V V^T \approx I_d$?
Let $k$ and $d$ be integers with $1 \le k \le d-1$. Let $U \in \mathbb{R}^{d \times k}$ satisfy $U^T U = I_k$. Let $V, \in \mathbb{R}^{d \times (d-k)}$ satisfy $V^T V = I_{d-k}$.

A basic fact from linear algebra is that, if $U^T V = 0$, then $U U^T + V V^T = I_d$. In other words, $U$ represents a set of $k$ orthonormal vectors and $V$ represents a set of $d-k$ orthonormal vectors; if $U$ is orthogonal to $V$, then together they give $d$ orthonormal vectors, which form an orthonormal basis.

I'm looking for a quantitative version of this: Suppose $\| U^T V \| \le \varepsilon$ and $\| V^T U \| \le \varepsilon$. What can we say about $\| U U^T + V V^T - I_d \|$ in terms of $\varepsilon$?

For concreteness, let $\|\cdot\|$ be the operator norm: For $M \in \mathbb{R}^{a \times b}$, $\|M\| := \sup \{ \|Mx\|_2 : x \in \mathbb{R}^b, \|x\|_2 \le 1 \}$. However, I am interested in other norms too.

答:
We note two facts:

For any $A \in \mathbb{R}^{m \times n}$, the matrices $AA^T-I_m$ and $A^TA-I_n$ are both symmetric and have the same non-zero eigenvalues. Thus, $\|AA^T-I_m\| - \|A^TA-I_n\|$.

For any $B \in \mathbb{R}^{m \times n}$, the non-zero eigenvalues of the symmetric matrix $\begin{bmatrix}0 & B \\ B^T & 0\end{bmatrix}$ are $\pm\sigma_1,\ldots,\pm\sigma_{\min(m,n)}$ where $\sigma_1,\ldots,\sigma_{\min(m,n)}$ are the singular values of $B$. Therefore, $\left\|\begin{bmatrix}0 & B \\ B^T & 0\end{bmatrix}\right\| = \|B\|$.

Using these facts, we have
\begin{align*}
\|UU^T+VV^T-I_d\| &= \left\|\begin{bmatrix}U & V\end{bmatrix}\begin{bmatrix}U^T \\ V^T\end{bmatrix}-I_d\right\|
\\
&= \left\|\begin{bmatrix}U^T \\ V^T\end{bmatrix}\begin{bmatrix}U & V\end{bmatrix}-I_d\right\|
\\
&= \left\|\begin{bmatrix}U^TU & U^TV \\ V^TU & V^TV\end{bmatrix}-\begin{bmatrix}I_k & 0_{k,d-k} \\ 0_{d-k,k} & I_{d-k}\end{bmatrix}\right\|
\\
&= \left\|\begin{bmatrix}0 & U^TV \\ V^TU & 0\end{bmatrix}\right\|
\\
&= \|U^TV\|
\\
&\le \epsilon.
\end{align*}


问:
What is the meaning of "$\pm$"? Can we write $1=\pm1$? or $x^2=4\iff \pm x=\pm 2$?
Does "$\pm$" mean "positive or negative"?

Is it allowed to write: $1 = \pm1\;$?

Is it allowed to write: $X^2 = 4 \iff \pm X = \pm2\;$?

答:
$±$ means both the positive and the negative value. For example, $±1 = -1, 1$.

$X^2 = 4 \iff ±X = ±2$ is technically correct, but it is redundant, as $+X = +2$ is the same as $-X = -2$, and similarly $+X = -2 \iff -X = +2$. It is usually written as just $X = ±2$.

To remove ambiguity, it is perfectly fine to write $x \in \{-2, 2\}$.


问:
Does $\mathrm{correlation}(X, Y)=1$ imply $Y = aX + b$?
I feel confused for the past solution in Proving $Y = aX + b$ given correlation coefficient $|\rho(X, Y)| = 1$, which tells $\rm{correlation}(X,Y)=1$ implies $Y = aX + b$ in contrast with the solution below:

Let $X$ and $Y$ be two random variables, $\sigma_x = \sqrt{\rm{Var}(X)}$, then

$$\rm{Var}\left(\frac{X}{\sigma_x} - \frac{Y}{\sigma_y}\right) = \rm{Var}\left(\frac{X}{\sigma_x}\right) + \rm{Var}\left(\frac{Y}{\sigma_y}\right)-2\rm{Cov}\left(\frac{X}{\sigma_x}, \frac{Y}{\sigma_y}\right) = 2-2\rm{Corr(X,Y)}$$

and let $\rm{Corr}(X, Y)=1$, then

$$\rm{Var}\left(\frac{X}{\sigma_x} - \frac{Y}{\sigma_y}\right) = 0 \implies \frac{X}{\sigma_x} - \frac{Y}{\sigma_y} = c$$

It seems $\rm{correlation}(X,Y)=1$ can NOT imply $Y = aX + b$. Could someone please tell which version is correct?

答: Yes, we must have $Y=aX+b$ for some constants $a,b$ (with probability $1$). In your working, when you deduce that $$\dfrac{X}{\sigma_X} - \dfrac{Y}{\sigma_Y}=c,$$ this implies that $Y = aX+b$ for some constants $a,b$ (just rearrange the equation to express $Y$ in terms of $X$ to see this).


答:
In logic language, $1 = \pm 1$ is understood as "1 is equal to 1 or 1 is equal to -1". So it is true.

To answer your question on $x^{2} = 4$, first we get
\begin{equation}
\forall x \in \mathbb{C},\left(x = 2 \longrightarrow x^{2} = 4\right) \wedge \left(x = -2 \longrightarrow x^{2} = 4\right)
\end{equation}
Using logic rules, it leads to
\begin{equation}
\forall x \in \mathbb{C}, \left(x = 2 \vee x = -2\right) \longrightarrow x^{2} = 4
\end{equation}
At the same time, we have
\begin{equation}
\forall x \in \mathbb{C}, \textrm{there exist at most two distinct solutions for $ax^{2} + bx + c = 0$}.
\end{equation}
As a result,
\begin{equation}
\forall x \in \mathbb{C}, x^{2} = 4 \longleftrightarrow \left(x = 2 \vee x = -2\right),
\end{equation}
or, in brevity, we get $x = \pm 2$.

PS. who can help me write "there exist at most two distinct solutions for $ax^{2} + bx + c = 0$" in logic language?

Title: What is the meaning of "$\pm$"? Can we write $1=\pm1$? or $x^2=4\iff \pm x=\pm 2$?

Body:
Does "$\pm$" mean "positive or negative"?

Is it allowed to write: $1 = \pm1\;$?

Is it allowed to write: $X^2 = 4 \iff \pm X = \pm2\;$?


答: As suggested in comments (and the other answer), the expression $\pm$ generally refers to a set; in this way, it's very similar to another piece of notation, the so-called 'big O' notation, for representing a set of functions with a certain rate of growth. We often write $a=\pm b$ when talking about the set of possible solutions of an equation; but this is an equality among sets, not numbers. So, $1=\pm 1$ would generally not be considered accurate, whereas both $x^2=4\iff \pm x=\pm 2$ and $x^2=4\iff x=\pm 2$ would be correct (though the latter is less confusing). Also, it may be worth noting that multiple instances of $\pm$ 'multiply' combinatorially; $\pm a\pm b$ usually refers to the set $\{a+b, a-b, -a+b, -a-b\}$.


问:
With $X_1,...X_4 \stackrel{i.i.d}{\sim} Ber(p)$ find $\mathbb{P}(X_1=x_1,...,X_4=x_4|X_1+...+X_4=2)$
My attempt:
$\mathbb{P}(X_1,...,X_4|X_1+...+X_4=2)=\frac{\mathbb{P}(X_1,...,X_4,X_1+...+X_4=2)}{\mathbb{P}(X_1+...+X_4=2)}$

$\mathbb{P}(X_1+...+X_4)\sim Binom(4,p)\Leftrightarrow \mathbb{P}(X_1+...+X_4=2)= {4 \choose 2}p^2(1-p)^2$

And I dont know how to find the following.
$\mathbb{P}(X_1=x_1,...,X_4=x_4,X_1+...+X_4=2)$

edit:

By drawing an event table of $X_1,...,X_4$ against $\sum X_i$ we see that for
all the ${4 \choose 2}$ cases of two $\ 0s$ and two $1s$, each case have a probability of $p^2(1-p)^2$. Which is $\frac{1}{{4 \choose 2}}\mathbb{P}(\sum X_i=2)$.

Is this the answer for $\mathbb{P}(X_1=x_1,...,X_4=x_4|X_1+...+X_4=2)$?

答:
Your work is essentially correct, but there is a missing case (specifically the case $x_1+x_2+x_3+x_4 \ne 2$).

$$P(X_1=x_1, \ldots, X_4 = x_4 \mid X_1 + \cdots + X_4 = 2)
= \begin{cases}
1/\binom{4}{2} & x_1 + \cdots + x_4 = 2
\\
0 & \text{otherwise}.
\end{cases}$$


问:
Is there a perfect square number n, whose Euler Totient value is also a perfect square?
Start with a perfect square, denoted as a positive integer n.
The root of this square is k, another positive integer.
Thus n = k^2
Let t = totient(n)

Is there a way to prove there is no such number? I know that Fermat Primes have totient values that are squares, but what about any perfect squares who have a perfect square as their totient value?

And more generally, expanding this concept with an additional restriction, can we say that a totient value (derived from a square number) will not be part of a Pythagorean Triple?

n = t + a

Where a is another integer, and is the difference between n and its totient value.

a = n - t

In other words, can n, t, and a all be perfect squares?

Furthermore, if this is disprovable, then can we also disprove the case of Pythagorean Quadruples?
Where perfect squares must sum together to equal the totient value of a perfect square? Does removing the restriction of it being a Pythagorean Triple/Quadruple, and allowing a to be non-square make this possible?

I am not a mathematician. I'm an Electrical Engineer, and have been exploring the world of Number Theory. I find it fascinating and have posed this question for myself, as I have not seen an answer. I believe it's possible I've missed something simple here and there is an "easy" way to show that such things cannot be. I have my computer checking by brute force if any totients of n are also square, and got up to the first 10,000 perfect squares as having no totient number that is a perfect square.

Any insight is appreciated, and I am sorry for the lack of formatting and formal means of formulating my question. I hope this question is clear. I ultimately want to know if there is any relationship between the Pythagorean Triples/Quadruples(or higher sets) to Euler's Totient Function. Wondering if by "coincidence" there is such a number, or if it is impossible (and why). Thank you for taking the time to think about this with me. This is a related question, but is not my question: Pythagorean triples that "survive" Euler's totient function

答:
Besides $1$ there is no such example. If $p$ is any prime dividing $k^2$ with exponent $2m$ then $\varphi(k^2)$ is divided by $p$ with exponent $2m-1$.

It's easy to see from a direct calculation.


问:
Truncation of an injective resolution
This piece comes from the proof of corollary 10.5.11 in Weibel's book on homological algebra.

There we start with a cochain complex $X^{\bullet}$ and we are trying to construct a quasi-isomorphism $X^{\bullet} \to X'^{\bullet}$ with $X'^{\bullet}$ being complex of $F$-acyclic objects, where $F$ has cohomological dimension $n$.

We choose Cartan-Eilenberg resolution $X^{\bullet} \to I_{CE}^{\bullet, \bullet}$ and then Weibel suggests we take a (good) truncation $\tau_{\leq n}(I_{CE}^{p \bullet})$ to get a finite resolution by $F$-acyclic objects.

So I cannon see why the last object we are truncating at is actually $F$ acyclic.  I initially wanted to splice the exact sequence $X^p \to I_{CE}^{p \bullet}$ and apply that with  $A'$ and $A$ $F$-acyclic in $$0 \to A' \to A \to A'' \to 0,$$ $A''$ is also $F$-acyclic.

But than this doesn't work because $I^{p,0} \to I^{p,1}$ isn't injective. Can't quite see how cohomological dimension $n$ comes into play.  Any thoughts would be appreciated.

答:
I do not have the book at my hand, and am happy to accept any possible corrections.

Let $X$ be a general object of the category and $X\to I^{\bullet}$ be an injective resolution of $X$. If $J$ is the image of the map $I^{n-1}\to I^n$, then $J\to I^n\to I^{n+1}\to\dots$ is an injective resolution of $J$. Also, because $F$ has cohomological dimension $n$, $F_*I^n\to F_*I^{n+1}\to\dots$ is exact. These two facts show that $J$ is $F$-acyclic. Therefore, $\tau_{\leq n}(I^{\bullet})$, ending with $I^{n-1}\to J\to 0\to \dots$, is an $F$-acyclic resolution of $X$.

Title: Truncation of an injective resolution


问:
what is $U$ here and how it look like?
Quotient topology : let $X$ be a topological space and $\sim$ an equivalence relation on $X$. For every $x \in X$ , denote by $[x]$ its equivalence class .The quotient space  of $X$ modulo $ \sim$ is given by the set

$X/\sim~ =\{[x] : x \in X \}$

we have the projection map $p: X \to /\sim ,x \to [x]$
and we equip $X/\sim$ with the topology

$U\subseteq$ $X/ \sim~$ is open if and only if $p^{-1}(U)$ is an open subset  of $X$

From my previous post I got example

But I'm confused  that how can we colloborate the example with  definition of  quotient topology

My attempt : Here $X=[0,1]$  then projection  map  $p : [0,1] \to [0,1]/\sim~$

$p(x) =[x]$ where $[0,1]/\sim~ \cong S^1$

$U\subseteq [0,1]/\sim$ mean $U\subseteq S^1$ and $p^{-1}(U)$ is open

according to the definition I'm confused  about $U$ . I mean  what is  $U$ here and how it look like ?

答:
I imagine you're considering the euclidean topology on $[0,1]$. In that case, since open intervals generate the euclidean topology on $\Bbb R$, every open subset of $[0,1]$ is generated by intersections of open intervals with $[0,1]$ (by their union).

There are five types of intersections: $(a,b)\cap[0,1]=\begin{cases}(a,b)&\text{if }0\le a,b\le 1\\ [0,b)&\text{if }a<0,b\le1\\(a,1]&\text{if }0\le a,1<b\\ [0,1]&\text{if }a<0,1<b\\\emptyset&\text{otherwise}\end{cases}$.

If $0\le a,b\le1$ then $p^{-1}((a,b))=(a,b)$, so every $(a,b)$ with $0\le a,b\le1$ is open in $[0,1]/\sim$

Let $m$ denote $m=[0]=[1]=\{0,1\}$. If we have $U$ a nonempty open subset of $[0,1]/\sim$, then we have two options: $U$ contains $m$ or $U$ doesn't contain $m$. In the second case we have that $p^{-1}(U)=U$, so $U$ must be the union of intervals $(a,b)$ with $0\le a,b\le1$ (since $p^{-1}(U)$ is open in $[0,1]$ and it contains neither $0$ nor $1$) $\Rightarrow U=\bigcup_i(a_i,b_i)$ with $0\le a_i,b_i\le1\,\forall i$, so (by the previous paragraph) we have $U$ expressed as a union of open sets in $[0,1]/\sim$

If $U$ contains $m$ then $p^{-1}(U)$ contains $0$ and $1$, so $p^{-1}(U)$ must contain some intervals $(c,1]$ and $[0,d)$ (which are open in $[0,1]$) with $0\le c<1,0<d\le1$, so  in consequence $[0,d)\cup(c,1]\subset p^{-1}(U)$. This implies that $(0,d)\cup(c,1)\cup\{m\}\subset U$, since that's the only way we can obtain $p^{-1}((0,d)\cup(c,1)\cup\{m\})=[0,d)\cup(c,1]$. Note that this means $(0,d)\cup(c,1)\cup\{m\}$ is open in $[0,1]/\sim$

Note also $\{m\}$ is not open, since $p^{-1}(\{m\})=\{0,1\}$.

Of course, now we can say unions of intervals $(a,b)$ with $0\le a,b\le1$ and/or subsets of the form $(0,d)\cup(c,1)\cup\{m\}$ with $0\le c<1,0<d\le1$ are open in $[0,1]/\sim$

And actually that's how every nonempty open subset $U$ of $[0,1]/\sim$ is constructed:

If $U$ doesn't contain $m$ we've shown $U=\bigcup_i(a_i,b_i)$ for some intervals $(a_i,b_i)$.

If it does contain $m$ then $U=\{m\}\cup(U\setminus\{m\})=\{m\}\cup\bigcup_i(a_i,b_i)$, since $U\setminus\{m\}=U\cap\{m\}^\mathsf c=U\cap(0,1)$, which is an open subset not containing $m$; and we also know $U$ contains a set $(0,d)\cup(c,1)\cup\{m\}$, so $U=\big((0,d)\cup(c,1)\cup\{m\}\big)\cup\bigcup_i(a_i,b_i)$. In fact $(0,d)$ and $(c,1)$ must be in $\bigcup_i(a_i,b_i)$, so we are kind of writing them down twice in that expression, but I think it shows that, effectively, $U$ is union of the kind of open subsets we talked about.

Title: what is $U$ here and how it look like?


答:
Beside the other answer(s), care must be taken when used more than once in the same expression/equation or used along with “$\mp$”; usually that means that the same sign is understood throughout/respectively or that the signs are supposed to be arbitrary irrespective of multiplicity of appearance. The context will make it clear, or a diligent author will make this explicit prior.

So, on the one hand, for instance
$$(a^2-b^2)(c^2-d^2)=(ac\pm bd)^2-(ad\pm bc)^2$$
means either\both $$(a^2-b^2)(c^2-d^2)=(ac+bd)^2-(ad+bc)^2$$ or\and $$(a^2-b^2)(c^2-d^2)=(ac-bd)^2-(ad- bc)^2$$ is\are correct; whereas
$$(a^2+b^2)(c^2+d^2)=(ac\pm bd)^2+(ad\mp bc)^2$$ means  either\both $$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$$ or\and $$(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2$$ is\are correct.

However, on the other hand, the inequality
$$0\ne\pm 1\pm1\pm1$$
will usually mean any combination of the signs regardless of its double appearance is meant——that is,
$0\ne \pm 3$ and $0\ne\pm 1$.


问:
Prove that $x_0+\frac{1}{x_0-x_1}+\frac{1}{x_1-x_2}+\dots+\frac{1}{x_{n-1}-x_n}\ge x_n+2n$

St.Petersburg 1999 :Let $x_0>x_1>\dots >x_n$ be real numbers. Prove that $$ x_0+\frac{1}{x_0-x_1}+\frac{1}{x_1-x_2}+\dots+\frac{1}{x_{n-1}-x_n}\ge x_n+2n $$

I tried Titu's inequality.

So we get $$ x_0+\frac{1}{x_0-x_1}+\frac{1}{x_1-x_2}+\dots+\frac{1}{x_{n-1}-x_n}= x_0+\frac{1^2}{x_0-x_1}+\frac{1^2}{x_1-x_2}+\dots+\frac{1^2}{x_{n-1}-x_n} \ge x_0 +\frac{n^2}{x_0-x_n}.$$

So we need to show that $$x_0 +\frac{n^2}{x_0-x_n}\ge x_n+2n .$$

Now note that $$\frac{n^2}{x_0-x_n}=\frac{n^2}{x_0}+\frac{x_n \cdot n^2}{x_0\cdot(x_0- x_n)}$$

So we need to show that $$x_0 +\frac{n^2}{x_0}+\frac{x_n \cdot n^2}{x_0\cdot(x_0- x_n)}\ge x_n+2n .$$

Note that by AM-GM $$x_0 +\frac{n^2}{x_0}\ge 2n$$.

So it's enough to show that $\frac{x_n \cdot n^2}{x_0\cdot(x_0- x_n)}\ge x_n$ .

Enough to show that $$\frac{n^2}{x_0\cdot(x_0- x_n)}\ge 1.$$

Which I am not able to. Any hints? Thanks in advance!!

Body:
Hint: let $x_0=x_1+a_1,x_1=x_2+a_2....x_{n-1}=x_n+a_n$ where $a_i>0$

we have to prove $$x_{0}-x_{n}+\frac{1}{a_1}+\frac{1}{a_2}...+\frac{1}{a_n}\ge 2n$$ $$a_1+a_2+a_3+.....a_n+\frac{1}{a_1}+\frac{1}{a_2}..+\frac{1}{a_n}\ge 2n $$  $$\iff \left(a_1+\frac{1}{a_1} \right)+\left(a_2+\frac{1}{a_2} \right)......+\left(a_n+\frac{1}{a_n} \right)\ge 2n$$ which is true by AM-GM


问:
A little help with elementary set intuition, specifically on intersection
Simple example, $A\cup 0=A$ is true because $x\in A$ or $x\in 0$ and that $0\subseteq A$. This I can understand (Actually, I hope I understood it)

But what I am having difficulty with is something like $A\cup(B\cap C)$. I know this implies $x \in A$ or $x\in (B\cap C)$. If $x\in A$, then $x\in(A\cup B)$ and $x\in (A\cup C)$ or alternitavely, $(A\cup B) \cap (A\cup C)$. What I am confused by is how can it be both? Isn't it that the OR in $x \in A$ or $x\in (B\cap C)$ implies that  $x$ is either in $A$ OR $x$ is in $(B\cap C)$ and not both? Obviously not, but I am having a bit of difficulty wrapping my head on why that is exactly.

答:
Given a family of sets $A_i$, where $i$ ranges over some index set $I$, an element $x$ lies in the set $A=\bigcup_{i\in I}A_i$ if and only if there is some $i\in I$ such that $x\in A_i$. There is no condition that $x$ has to lie in only one $A_i$; it can lie in arbitrarily many.

On the other hand, given two sets $A$ and $B$, there is a construction $A\triangle B$, called the symmetric difference of the two, which is defined as $(A\setminus B)\cup (B\setminus A)$. So this consists of elements that lie in $A$ or in $B$, but not in both. It seems you are confusing the union construction with the symmetric difference construction; hopefully it helps.


问:
proving $\displaystyle\sum_{s,r=0,s\ne r}^{n}C_rC_s=\frac{1}{2}\left(2^{2n}-\frac{(2n)!}{(n!)^2}\right)$
Using $\displaystyle(1+x)^n=\sum_{r=0}^n C_rx^r$, find $\displaystyle\sum_{r=0}^n C_r$ and $\displaystyle\sum_{r=0}^n C_r^2$. Hence prove sum of product of two coefficients taken from $C_0,C_1,C_2\cdots,C_n$ is $\displaystyle\frac{1}{2}\left(2^{2n}-\frac{(2n)!}{(n!)^2}\right)$. i.e. $\displaystyle\sum_{s,r=0,s\ne r}^{n}C_rC_s$

Here's what I tried so far,

$$\sum_{r=0}^n C_r=2^n$$

using the fact that $(1+x)^{2n}$, I was able to derive,

$$\sum_{r=0}^n C_r^2=\frac{(2n)!}{(n!)^2}$$

But How can I proceed? Any Hint would be highly appreciated. Thank you.

Body: $\sum_{s,r=0.s\ne r}^nC_rC_s=(\sum_{r=0}^nC_r)^2-\sum_{r=0}^nC_r^2=2^{2n}-\frac{(2n)!}{(n!)^2}$  To get the $\frac{1}{2}$ you need to change $s\ne r$ to $s\lt r$.


答:
This is a (somewhat?) standard example of how rearranging an iterated integral makes it converge conditionally to either of two different numbers.

\begin{align}
& \int_0^1 \left( \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \, dy \right) \, dx = +\frac \pi 4. \\[10pt]
& \int_0^1 \left( \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \, dx \right) \, dy = -\frac \pi 4. \\[10pt]
& \iint\limits_{0\,<\,y\,<\,x\,<\,1} \frac{x^2-y^2}{(x^2+y^2)^2} \, d(x,y) = +\infty. \\[10pt]
& \iint\limits_{0\,<\,x\,<\,y\,<\,1} \frac{x^2-y^2}{(x^2+y^2)^2} \, d(x,y) = -\infty. \\[10pt]
& \iint\limits_{\varepsilon\,<\,x\,<\,1 \\ \varepsilon\,<\,y\,<\,1} \frac{x^2-y^2}{(x^2+y^2)^2} \, d(x,y) = 0.
\end{align}
Only when the integrals of the positive and negative parts are both infinite can the values of the two iterated integrals differ.

Let $y = x\tan \theta$, so that $dy = x\sec^2\theta\,d\theta$ and
$x^2 + y^2= x^2\sec^2\theta$, and as $y$ goes from $0$ to $1$
then $\theta$ goes from $0$ to $\arctan(1/x)$. Then
\begin{align*}
\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \, dy
= {} & \int_0^{\arctan(1/x)}
\frac{x^2 - x^2 \tan^2\theta}{(x^2 + x^2\tan^2\theta)^2} \big( x\sec^2\theta\,d\theta\big) \\[10pt]
= {} & \frac 1 x \int_0^{\arctan(1/x)} \frac{1-\tan^2 \theta}{\sec^2 \theta} \, d\theta \\[10pt]
= {} & \frac 1 x \int_0^{\arctan(1/x)} (\cos^2\theta-\sin^2\theta) \, d\theta \\[10pt]
= {} & \frac 1 x \int_0^{\arctan(1/x)} \cos(2\theta) \, d\theta \\[10pt]
= {} & \frac 1 {2x} \sin\left(2\arctan \frac 1 x\right) \\[10pt]
= {} & \frac 1 x \sin\left(\arctan \frac 1 x \right) \cos\left( \arctan \frac 1 x \right) \\[10pt]
= {} & \frac 1 x \cdot \frac 1 {\sqrt{1+x^2}} \cdot \frac x {\sqrt{1+x^2}} = \frac 1 {1+x^2}. \\[10pt]
\text{And then}
& \int_0^1 \frac{dx}{1+x^2} = \frac \pi 4.
\end{align*}

Title: Integral of functions of several variables

Body:
The function is defined from $[-1,1]\times[-1,1]$ to $\Bbb R$, given by $f(x,y)=\frac{x^2-y^2}{(x^2+y^2)^2}$ when $(x,y)≠0$ and $f(0,0)=0$.
I could find that the function is not continuous at origin, so not differentiable  and also partial derivatives does not exist at origin.
But how do we evaluate the integral of the function over the given domain. Being function of several variable how do we deal with the point of discontinuity? Can we convert it to polar coordinates and apply residue theorem? Do we have any other methods?
Any help would be appreciated. Thanks in advance.


答:
There are a number of ways to interpret this integral.

(1) The function is not Riemann integrable on $[-1,1]\times [-1,1]$ because it is unbounded.

(2)  It is not Lebesgue integrable since

$$\int_{[-1,1]^2}|f| \geqslant \int_0^{2\pi}\int_\delta ^1 \frac{|r^2 \cos^2\theta - r^2\sin^2\theta|}{r^4}\, r\, dr\, d\theta \underset{\delta \to 0+}\longrightarrow + \infty$$

(3) The iterated integrals exist but are unequal (the conditions of Fubini's theorem are violated).

Note that

$$\int_{-1}^1 \left(\int_{-1}^1\frac{x^2- y^2}{(x^2+y^2)^2}\, dy \right)\, dx = \int_{-1}^1 \left(\int_{-1}^1\frac{\partial}{\partial y}\frac{y}{x^2+y^2}\, dy \right)\, dx = \int_{-1}^1 \frac{2}{1+x^2}\, dx = \pi,$$

and by antisymmetry

$$\int_{-1}^1 \left(\int_{-1}^1\frac{x^2- y^2}{(x^2+y^2)^2}\, dx \right)\, dy = -\int_{-1}^1 \left(\int_{-1}^1\frac{x^2- y^2}{(x^2+y^2)^2}\, dy \right)\, dx = -\pi $$

(4) An improper (or extended) Riemann integral over a set $A$ in $\mathbb{R}^2$ is defined as

$$\tag{*} \lim_{n \to \infty} \int_{A_n}f(x,y) \,d(x,y),$$

where $(A_n)$ is a sequence of compact rectifiable sets such that $A_n \subset A_{n+1}$ and $\bigcup_{n=1}^\infty A_n = A$.

This will converge to a unique value independent of the choice for $(A_n)$ only if the function is absolutely integrable, and in this case it was shown in (2) that this does not hold.  It is possible to construct sequences $(A_n)$ where the limit (*) is $0$ or $+\infty$, for example.


问:
Upper bound on volume growth on Riemannian manifold
Let $M$ be a Riemmanian manifold such that for all $x\in M$ and large $R>0$, $$Vol(B(x, R))\leq \exp(R).$$ Let $p\in M$ and $d$ be the geodesic distance. Suppose that for large enough $r$ and all $x, y\in \partial B_{r}(p)$, $$d(x, y)\leq c\log r.$$ Is it then true that there exist constants $N>0$ and $C>0$ such that for all large enough $r$ we have $$Vol(B(p, r))\leq Cr^{N}?$$

Thanks in advance for your help!

答:
Yes.

In more details, we can pick $D>1$ be so that $\text{Vol}(B_R(x))<e^R$ for ALL $x$ whenever $R\geq D$.
Take the first integer $L>1$ so that $L>e^{D/c}$, then $c\log L>D$.
For any $s\geq L$, note $2c\log s-c\log s=c\log s\geq c\log L>D>1$.

Take $x\in \partial B_s(p)$, observe
$$
\{z\,|\, s\leq d(z, p)\leq s+1\}\subset B_{2c\log s}(x);
$$

in fact, for any $z$ with $s\leq d(z, p)\leq s+1$, take a minimal geodesic $\gamma$ from $p$ to $z$, then $z=\gamma(t)$ with $t=d(z, p)\in [s, s+1]$, so there is a point $y=\gamma(s)\in \partial B_s(p)$ so that $d(y, z)=d(\gamma(t), \gamma(s))\leq 1$. Now $d(x, y)\leq c\log s$, so
$$
d(z, x)\leq d(z, y)+d(y,x)\leq 1+c\log s\leq 2c\log s.
$$

In particular, let $N=2c+1$, we see
$$
\text{Vol}(\{z\,|\, s\leq d(z, p)\leq s+1\})
\leq \text{Vol}(B_{2c\log s}(x))\leq e^{2c\log s}=s^{2c}=s^{N-1}.
$$

Let $C_1$ be so that $\text{Vol}(B_L(p))=C_1L^N$.
Now for any $r>L$, let $k$ be $L+k\leq r<L+k+1$, so
$$
\begin{aligned}
\text{Vol}(B_r(p))\leq &\text{Vol}(B_L(p))+\text{Vol}(\{z\,|\, L\leq d(z, p)\leq L+1\})\\
&+\text{Vol}(\{z\,|\, L+1\leq d(z, p)\leq L+2\})+...\\
&+\text{Vol}(\{z\,|\, L+k\leq d(z, p)\leq L+k+1\})\\
\leq & C_1L^N+L^{N-1}+(L+1)^{n-1}+...+(L+k)^{N-1}\\
\leq & C_1L^N+(k+1)(L+k)^{N-1}\\
\leq & C_1(L+k)^N+(L+k)^{N}=(C_1+1)(L+k)^N\leq (C_1+1)r^N.
\end{aligned}
$$


问:
In ZF, does the ring of continuous functions $C([0,1], \mathbb{R})$ have prime ideals which is not maximal?
In ZFC, it is known that the ring of continuous functions $C([0,1], \mathbb{R})$ have prime ideals which is not maximal. But all proofs of this which I saw uses the axiom of choice.

Then, in ZF, does the same statement hold?

If ZF cannot prove this, how strong is $\mathrm{ZF}+(\text{$C([0,1], \mathbb{R})$ has prime ideals which is not maximal})$ as an intermediate between ZF and ZFC?

答:
No, the existence of such a nonmaximal prime ideal cannot be proved in ZF.  In fact, the existence of such a nonmaximal prime ideal is equivalent to the existence of a nonprincipal ultrafilter on $\mathbb{N}$.

First, suppose $F$ is a nonprincipal ultrafilter on $\mathbb{N}$.  Let $I$ be the set of continuous $f:[0,1]\to \mathbb{R}$ such that $\{n\in\mathbb{N}:f(1/n)=0\}\in F$.  It is then easy to see that $I$ is an ideal in $C([0,1],\mathbb{R})$, and it is prime since $F$ is an ultrafilter.  However, it is not maximal since it is strictly contained in the ideal of functions that vanish at $0$.

Conversely, suppose no nonprincipal ultrafilter on $\mathbb{N}$ exists and $I\subset C([0,1],\mathbb{R})$ is a prime ideal; we will prove $I$ is maximal.  For each $a\in[0,1]$ and each $n\in\mathbb{N}$, let $f_{n,a}:[0,1]\to\mathbb{R}$ be the function that is $0$ on $[a-1/n,a+1/n]$, $1$ off of $[a-2/n,a+2/n]$, and interpolates linearly in between.  I claim that there exists some $a\in[0,1]$ such that $f_{n,a}\in I$ for all $n$.  To prove this, suppose no such $a$ exists; for each $a$, let $n_a$ be minimal such that $f_{n_a,a}\not\in I$.  By compactness, finitely many of the intervals $(a-1/n_a,a+1/n_a)$ cover $I$.  But then the product of the corresponding $f_{n_a,a}$s is $0$, contradicting primeness of $I$ since each $f_{n_a,a}$ is not in $I$.

So, we have a point $a\in[0,1]$ such that $f_{n,a}\in I$ for all $n$.  It follows that every function that vanishes in a neighborhood of $a$ is in $I$, since every such function is divisible by some $f_{n,a}$.  I now claim that in fact $I$ contains every function that vanishes at $a$, and thus is maximal.  To prove this, suppose that $f\in C([0,1],\mathbb{R})$ is a function which vanishes at $a$.  To show that $f\in I$, it suffices to show that $f^2\in I$ since $I$ is prime.
Replacing $f$ by $f^2$, we may assume $f\geq 0$ and we wish to show $f\in I$.

Fix an increasing sequence $(a_n)$ converging to $a$ with $a_0=0$.  Let $F$ be the set of all $A\subseteq\mathbb{N}$ such that there exists $g\in I$ such that $g\geq 0$ everywhere and $g\geq f$ on $[a_{2n},a_{2n+1}]$ for all $n\not\in A$.  It is easy to see that $F$ is a filter on $\mathbb{N}$, and it contains all cofinite sets since $I$ contains all functions that vanish in a neighborhood of $a$.  Note also that if $A\subseteq\mathbb{N}$, then we can construct a function $g$ which is $f$ on $[a_{2n},a_{2n+1}]$ for all $n\not\in A$ and $0$ outside of small neighborhoods of these intervals, and similarly we can construct a function $h$ with the same property with respect to $\mathbb{N}\setminus A$.  Then $gh=0$ so either $g\in I$ or $h\in I$, so either $A\in F$ or $\mathbb{N}\setminus A\in F$.

Since by hypothesis, $F$ cannot be a nonprincipal ultrafilter on $\mathbb{N}$, the only remaining possibility is that $F$ is the improper filter, i.e. $\emptyset\in F$.  So there is a nonnegative function $g_1\in I$ such that $g_1\geq f$ on $[a_{2n},a_{2n+1}]$ for all $n$.  Similarly, there is a nonnegative function $g_2\in I$ such that $g_2\geq f$ on $[a_{2n-1},a_{2n}]$ for each $n$.  Adding up $g_1$ and $g_2$, we get a nonnegative element of $I$ which is bounded below by $f$ on all of $[0,a]$.  We can similarly get a nonnegative element of $I$ that is bounded below by $f$ on all of $[a,1]$.  Adding these functions together, we get a function $g\in I$ such that $g\geq f$ on all of $[0,1]$.  Now note that $f^2/g$ extends continuously to all the points where $g$ vanishes, since $f$ also vanishes at those points and $f^2/g$ is bounded above by $f$ outside them.  Thus $f^2$ is a multiple of $g$ and hence is in $I$.  Since $I$ is prime, this means $f\in I$, as desired.


问:
Approximating a Poisson distribution to a Normal distribution
I have the following problem I'm trying to solve:

I know that the quantity of complains in a call center is a Poisson variable with $\lambda=18 $ costumers/hour, and that the probability of being able to solve a complain is $0.35$. They ask me about the probability of having more than $50$ complaints without solution in $8$ hours.

I tried to solve it by obtaining a new Poisson process with $\lambda=0.35\times18$, which describes the quantity of complains unsolved in certain time.

So, to solve the problem I would have to find $F_x (x)$ and do this:

\begin{align}
P(\text{more  than 50 complaints in 8 hours}) &= 1 - P (\text{less  than 50 complaints in 8 hours}) \\&= 1 - F_x(x)
\end{align}

My problem is that I need to sum $50$ terms of this Poisson distribution in order to solve the problem (to find the $F_x(x)$, which is exhausting, and I quite doubt this is the objective of the problem).

I think the problem has more to do with using a Normal distribution to get some sort of approximation, but I don't know how many calls there will be (I can't find how many terms I should solve, if I think that the variables I'm adding are each one of the calls, that I can think as being Bernoulli Variables, each one with a probability of $0.35$ of being solved).

I'd appreciate any help. Thank you very much.

答:
The number of unresolved complaints in $1$ hour does indeed (under our assumptions) have Poisson distribution with parameter $(0.35)(18)$. This assumes essentially that a complaint is either resolved instantly or remains unresolved, which is not quite reasonable.

Let $X$ be the number of unresolved complaints in $8$ hours. This then also has Poisson distribution, with parameter $\lambda=(8)(0.35)(18)=50.4$.

The normal distribution with mean $\lambda$ and variance $\lambda$ is a reasonably good approximation to the Poisson with the large parameter $\lambda=50.4$. So now we have a standard normal calculation to do.


问:
Finding variance using convolution formula, gamma function

A company has two electric generators. The time until failure for each generator follows an exponential distribution with mean $10$. The company will begin using the second generator immediately after the first one fails. What is the variance of the total time that the generators produce electricity?

My attempt:

Let $S$ be the time until the first generator fails and $T$ be the time until the second generator fails. Let $X=S+T$. Then $Var(X) = E[X^2]- (E[X])^2$.

Finding the pdf of $X$ using the convolution formula:

$$f_X(x) = \int_0^x \frac{1}{10}e^{-s/10} \cdot \frac{1}{10}e^{(s-x)/10} \; ds = \frac{x\cdot e^{-x/10}}{100}$$

Now, utilizing the Gamma function:
$$E[X^2] = \int_0^\infty \frac{x^3e^{-x/10}}{100} = \frac{10^4 \times \Gamma(4)}{100} = 600$$

$$E[X] = \int_0^\infty \frac{x^2e^{-x/10}}{100} = \frac{10^2 \times \Gamma(2)}{100} = 1$$

So, $Var(X) = 599$, which is not even close to the correct answer. Where have I gone wrong?

Edit: The mean is supposed to be $20$, as pointed out in the comments, but the issue is that my current calculations are not leading to that.

答:
Method 1.  Linearity.

If $X = S+T$ where $S$, $T$ are the random lifetimes of each generator, respectively, then $$\operatorname{Var}[X] \overset{\text{ind}}{=} \operatorname{Var}[S] + \operatorname{Var}[T],$$ where 'ind' over the equality sign indicates that the equality holds under the condition that $S$ and $T$ are independent random variables.  Since the variance of an exponentially distributed random variable with mean $\operatorname{E}[S] = \operatorname{E}[T] = \mu$ is $\operatorname{Var}[S] = \operatorname{Var}[T] = \mu^2$, it follows that $\operatorname{Var}[X] = 2\mu^2 = 200.$

Method 2. Gamma distribution as a sum of iid exponential distributions.

Since $S, T$ are iid exponential with mean $\mu = 10$, then $X = S + T$ is Gamma with shape $n = 2$ and scale $\theta = \mu = 10$.  Specifically, $$f_X(x) = \frac{x^{n-1} e^{-x/\theta}}{\theta^n \Gamma(n)} = \frac{x e^{-x/10}}{100}, \quad x  > 0.$$  Then the $k^{\rm th}$ raw moment of $X$ is $$\begin{align}
\operatorname{E}[X^k] &= \int_{x=0}^\infty x^k f_X(x) \, dx \\ &= \int_{x=0}^\infty \frac{x^{k+1} e^{-x/10}}{10^2} \, dx \\ &= 10^k \Gamma(k+2) \int_{x=0}^\infty \frac{x^{(k+2) - 1} e^{-x/10}}{10^{k+2} \Gamma(k+2)} \, dx \\ &= 10^k \Gamma(k+2), \end{align}$$ where in the last equality, we observe that the integrand corresponds to the density of a gamma distribution with shape $k+2$ and scale $10$, thus integrates to $1$.  Consequently, $$\operatorname{E}[X^2] = 6 (10^2), \quad \operatorname{E}[X]^2 = (2(10))^2,$$ and $$\operatorname{Var}[X] = 600 - 400 = 200.$$


问:
Showing an ideal is prime in polynomial ring

Let $k=\mathbb{C}$ and let $J$ the ideal $(xw-yz,y^{3}-x^{2}z,z^{3}-yw^{2},y^{2}w-xz^{2})$. I want to see why $J$ is a prime ideal in $k[x,y,z,w]$.

I know that $Z(J)$ (the zero set of $J$) is exactly the image of the $4$-fold Veronese embedding from $\mathbb{P}^{1}$ to $\mathbb{P}^{3}$, i.e., the map given by $[s : t] \mapsto [s^{4}: s^{3}t : st^{3}: t^{4}]$. What I tried: if we consider the following ring homomomorphism: $k[x,y,z,w] \rightarrow k[s,t]$ given by $x \mapsto s^{4}$, $y \mapsto s^{3}t$, $z \mapsto st^{3}$ and $w \mapsto t^{4}$, then one can check that $J$ is contained in the kernel of this ring homomorphism. Now if we can show the the other inclusion we are done because then $k[x,y,z,w]/J$ embeds a subring of $k[s,t]$ and hence $J$ is prime. However I don't see the other inclusion, can you please help? Perhaps there's an easier way.

答:
One low-tech way to go from here is to write any polynomial $f$ in $k[x,y,z,w]$ as a sum of $j\in J$ and some other particularly nice $u$ and then apply the homomorphism $\varphi$ you've written down - if $f\in \ker\varphi$, then $\varphi(j)+\varphi(u)=0$, and you'll see that the only way for $\varphi(u)=0$ is if $u=0$ to begin, so $J$ is exactly the kernel of $\varphi$.

There's probably a more succinct way to write this down, but this is the way I would think through it:

$xw-yz$ is in $J$, so we can assume that we've chosen $j$ and $u$ so that $u$ contains no monomials with a $yz$ in them: if $p$ is a monomial and $pyz$ is in $u$, then $pyz=-p(xw-yz)+pxw$, so we can adjust $j$ and $u$ so there are no terms with a $yz$ by repeated applications of this formula.

$y^3-xz$ and $z^3-yw^2$ are in $J$, so via the same trick, we can assume there are no monomials in $u$ with a $y^3$ or a $z^3$ in them.

$y^2w-xz^2$ is in $J$, so there's no monomials in $u$ with $y^2w$ in them.

So we may take $u$ to consist of a $k$-linear combination of monomials of the following type:

$x^aw^b$,

$x^aw^by$,

$x^aw^bz$, and

$x^aw^bz^2$

where $a,b\geq 0$.

Now let's look at what happens when we apply $\varphi$:

$\varphi(x^aw^b) = s^{4a}t^{4b}$,

$\varphi(x^aw^by) = s^{4a+3}t^{4b+1}$,

$\varphi(x^aw^bz) = s^{4a+1}t^{4b+3}$, and

$\varphi(x^aw^bz^2) = s^{4a+2}t^{4b+6}$.

By examining the residue classes of the exponent of $s$ modulo $4$, we see that there can be no cancellation between any of these monomials, so the only way for $\varphi(f)=\varphi(j+u)=\varphi(j)+\varphi(u)=0+\varphi(u)$ to be equal to zero is if $u=0$.


问:
Rearranging a string such that specified characters appear in order
The letters in the word GUMTREE and KOALA are rearranged to form a 12-letter word where KOALA appears precisely in order but not necessarily together. How many ways can this happen?

So I attmepted it via this method:

Firstly arrange like this since KOALA must be in order but not necessarily together: (let the fullstops (.) be spots for the letters in GUMTREE. There are $7$ letters in GUMTREE and $6$ full stops so $6^7$.

.K.O.A.L.A.

Butthere are two E's so $\frac{6^7}{2!}$. The letters in KOALA are fixed so they have $1$ way each, except for A (there are two so the first A has two choices and the second A has one choice)

Therefore, $$2\cdot\frac{6^7}{2!}=279936$$

But the answer is 1995840 arrangements

I belive my method is very close but I am forgetting to multiply by something. Can someone point out my logical flaw? Otherwise the worked solutions propose $\frac{12!}{5!2!}$, but I don't get why you divide by 5! for the KOALA, since they are not identical letters... regardless it would be great to understand both ways!

Thanks

答:
You are assuming that the letters for the two words are interleaved, when they do not need to be.  For example, "KOALAGUMTREE" is a permissible arrangement, as is "GUMTREEKOALA" or "GUMKOALATREE."  The problem only stipulates that the letters in "KOALA" appear in sequence.

So, count the number of ways to order the letters in "GUMTREE."  There are $7$ letters, two of which are identical; therefore, there are $7!/2!$ such arrangements of the letters.  For "KOALA," there is only one permissible arrangement.  So now all that is left is to figure out how many combined arrangements exist; to do this, note there are a total of $7 + 5 = 12$ letters in total, hence there are $\binom{12}{5}$ ways to select the positions of the letters in "KOALA" among the $12$ total letters.

Hence the correct number of arrangements is $$\frac{7!}{2} \binom{12}{5} = 1995840,$$ as claimed.


问:
Maximum number of spanning trees of a planar graph with a fixed number of edges
Let $\mathcal{G}_m$ be the set of planar graphs with exactly $m$ edges. In this question, graphs are allowed to have multiple edges and/or loops.

I want to know what the maximum number of spanning trees of any graph in $\mathcal{G}_m$ is, as a function of $m.$ I suspect this question is too hard to have an exact formula, and so I am also interested in upper bounds for the maximum number of spanning trees of any graph in $\mathcal{G}_m$.

For small values of $m$, my (possibly incorrect) computations give me the following.

When $m=1$, the maximum number of spanning trees is $1$, realized by any connected graph with $1$ edge.

When $m=2$, the maximum number of spanning trees is $2$, realized by the graph with two vertices connected by two edges.

When $m=3$, the maximum number of spanning trees is $3$, realized by a $3$-cycle.

When $m=4$, the maximum number of spanning trees is $5$, realized by a $3$-cycle with one edge doubled.

When $m=5$, the maximum number of spanning trees is $8$, realized by a $3$-cycle with two edges doubled or by the complete graph $K_4$ with one edge deleted.

When $m=6$, the maximum number of spanning trees is $16$, realized by the complete graph $K_4$.

I could not find anything relevant in OEIS with respect to my computations so far. This is what leads me to believe that an exact formula may not be known.

答:
This is a very, very tough question. It is obvious that for any graph $G$ with $n$ edges the number of spanning trees $t(G)$ does not exceed $2^n$ (each edge is either included in or excluded from a subtree; this is also the upper bound of the number of connected subgraphs of $G$). We can somewhat improve this - if $G$ has $m$ vertices ($m \leqslant n+1$) then we have
$$
t(G) \leqslant \binom{n}{m-1} < 2^n ,
$$
since we need to choose $m-1$ edges out of $n$ (not arbitrarily, of course).

It is a bit easier for the planar graphs but you are correct that the exact formula is not known. I will say more, there is no exact formula even for such a "simple" planar graph as $Z_{n,n}$, which represents $n \times n$ rectangular lattice.

However, there are a few upper and lower bounds, most of them pretty complicated. I suspect there should be an elementary proof for the following conjecture.

Conjecture. For any finite connected planar graph $G$ with $n$ edges we have
$$
t(G) < \tau^n,
$$
where $\tau$ is... say, $1.8$.

Perhaps I am a bit optimistic about 1.8 but there has to be some value of $\tau < 2$ for which an elementary proof exists. I am actually doing some low-tech research into that area right now...

An example: for large values of $n$ it is known that $t(Z_{n,n})$ is pretty close to $\tau^n$, where $\tau = e^{2C/\pi} \approx 1.7916...$, where $C$ is the so-called Catalan constant
$$
C = 1 - 1/3^2 + 1/5^2 - 1/7^2 \pm ...
$$


答: Probably simplest to frame the problem as arranging 12 letters, in which 2 are identical of one kind (EE in GUMTREE), and 5 are identical of another kind (XXXXX substituting for KOALA, which is treated as a black box since its letters have already been pre-arranged). So, $$\frac{12!}{2!5!},$$ as suggested.

Title: Rearranging a string such that specified characters appear in order

Body:
The letters in the word GUMTREE and KOALA are rearranged to form a 12-letter word where KOALA appears precisely in order but not necessarily together. How many ways can this happen?

So I attmepted it via this method:

Firstly arrange like this since KOALA must be in order but not necessarily together: (let the fullstops (.) be spots for the letters in GUMTREE. There are $7$ letters in GUMTREE and $6$ full stops so $6^7$.

.K.O.A.L.A.

Butthere are two E's so $\frac{6^7}{2!}$. The letters in KOALA are fixed so they have $1$ way each, except for A (there are two so the first A has two choices and the second A has one choice)

Therefore, $$2\cdot\frac{6^7}{2!}=279936$$

But the answer is 1995840 arrangements

I belive my method is very close but I am forgetting to multiply by something. Can someone point out my logical flaw? Otherwise the worked solutions propose $\frac{12!}{5!2!}$, but I don't get why you divide by 5! for the KOALA, since they are not identical letters... regardless it would be great to understand both ways!

Thanks


问:
Combining indiscernible sequences
Fix some structure $M$ and $C\subseteq M$. Suppose $(a_i)_{i<\omega}\subset M$ and $(b_i)_{i<\omega}\subset M$ are $C$-indiscernible sequences. Can we conclude that $(a_ib_i)_{i\leq \omega}$ is $C$-indiscernible sequence as well?

答:
Here I'm going to answer my question based on Alex Kruckman's hint in the comments.

The answer is NO. Here is a counterexample!

In this picture $(a_i)_{i<\omega}$ and $(b_i)_{i<\omega}$ are indiscernible sequences but  $\text{tp}(a_0b_0)=\text{tp}(a_0)\neq \text{tp}(a_1b_1)$.

Title: Combining indiscernible sequences


问:
If $a^3=e$, then $a$ has a square root.
If $a^3=e$, then $a$ has a square root.

I have that $a^3=aaa=e$ and applying $a^{-1}$ twice on both sides of the equation gives us $a=a^{-1}a^{-1}$ and so $a=(a^{-1})^2$. Thus $a$ has a square root.

Is this correct? Is there another way to do this?

答:
Your attempt is correct, and as mentioned there is the other proof, that $a^3 = e$ implies $a^4 =(a^2)^2 =a$.

A more general statement is true, and is easy to prove.

If $G$ is a finite group of odd order then every element of $G$ has a unique square root.

Proof : Let $|G| = 2n-1$, then if $a = g^2 = h^2$ for two elements $g,h$, then $(g^{2})^{n} = (h^2)^n$ implies $g^{2n} = h^{2n}$ implies $g=h$ by Lagrange's theorem. Thus, $a$ has a unique square root.

Finally, the map $S : G \to G$ given by $S(x) = x^2$ is an injective map between finite sets of the same size, hence a surjective map. The result follows.

To apply this to your question, if $a^3 = e$ then the cyclic group generated by $a$ has three elements , and you apply the above statement to conclude.

(TASK : Prove that if $G$ is a group such that every element of $G$ has odd order, then every element of $G$ has a square root, using the method above).

Title: If $a^3=e$, then $a$ has a square root.


问:
Is a Noetherian graded ring also a finitely generated $R_0$-module?
Let $R=R_0\oplus R_1\oplus R_2\oplus\cdots$ be a graded ring. If $R$ is Noetherian, then I can see that $R_0$ is Noetherian and $R_1\oplus R_2\oplus\cdots$ is a finitely generated ideal of $R.$

But, is it also true that, $R$ is also a finitely generated $R_0$-module?

Edit: As mentioned by Alekos, it's not true. Then, can we say that $R$ is a finitely generated $R_0$-algebra?

答:
What if we take the graded ring $\Bbb{C}[x]=R$? $R$ is certainly Noetherian, and here $\Bbb{C}=R_0$ is indeed Noetherian. But, as we know $\Bbb{C}[x]$ is not a finitely generated $\Bbb{C}-$vector space. You can soup this example up to $k[x_1,\ldots, x_n]$ or you can use more exotic ground rings etc.

As for your second question, the answer is yes and is explained in detail in the first two lemmas of https://stacks.math.columbia.edu/tag/00JV.


问:
Why is the sum of all external angles in a convex polygon $360^\circ$ and not $720^\circ$?

Why is the sum of all external angles in a convex polygon $360^\circ$?

From my understanding, for each vertex in a convex polygon, there exist exactly $2$ exterior angles corresponding to it, which are both equal, vertically opposite, and add up to $180^\circ$ with the interior angle. If we take as true that sum of interior angles in a triangle is $(n-2)180^\circ$ degrees, then $$\sum_i 2\cdot (180^\circ-\alpha_i) = n\cdot 360^\circ - (n-2)\cdot 360^\circ = 720^\circ.$$ Am I missing something here?

答:
I recommend you look at Geometry by Jurgensen under the angles of a polygon section.

Draw a picture, it will help. Start out with pentagon to get some intuition. Now separate the inside of the polygon into non-overlapping triangles, and observe how you found the total sum of the interior angles of the polygon, and then of the sum of the exterior angles of it.

Basically, given a convex polygon, we can form a linear pair -- a pair of supplementary angles , at each vertex.

Taking all $n$ vertices of the polygon into account, we have $n \cdot 180 $ degrees.

The polygon can be partitioned/separated into exactly $n-2$ triangles.

So, the sum of all the interior angles of the polygon is $$(n-2) \cdot 180 \text{ degrees} $$

Hence, the sum of all the external angles is the difference

$$ n \cdot 180 - (n-2) \cdot 180 = 2\cdot 180  = 360 \text{ degrees} $$


问:
General integral for partial differential equation
I want to find a general integral for $xz_x + yz_y = z^2 + 1$, where $z = z(x,y)$ is implicitly a function of two variables. I'm not sure how to go about this.

答:
$$\dfrac {dx}{x}=\dfrac {dy}{y}=\dfrac {dz}{z^2+1}$$
First DE gives us:
$$d \ln x = d\ln y \implies \dfrac x y=C_1$$
Second DE gives us:
$$ d\ln y=d \arctan z$$
$$\ln y = \arctan z +C_2$$
The solution is therefore:
$$C_2= f(C_1)$$
$$\ln y - \arctan z = f\left (\dfrac x y \right)$$
$$ z (x,y)=\tan \left(\ln y+ f\left (\dfrac x y \right) \right)$$


答:
As I understand it, you are doing $7$ trials, one for each letter of GUMTREE. In the first trial, you decide which of the six slots, indicated by full stops, to use for G, in the second you decide which to use for U, and so on.

Here's the problem with that approach: let's say all seven letters of GUMTREE get put in the first slot. You still haven't said how the letters in that slot are to be ordered. There are $7!/2!$ orders, so you might think to multiply by that factor. But let's say that G got put in the first slot, U in the second, and so on up to R in the fifth slot, with both Es put in the sixth slot. Now there is no rearrangement possible, so the appropriate multiplicative factor is $1$, not $7!/2$.

Because of this issue I don't see any simple fix for your approach.


问:
Landmarks of subjects of mathematics
In order to have a good view of the whole mathematical landscape one might want to know a deep theorem from the main subjects (I think my view is too narrow so I want to extend it).

For example in natural number theory it is good to know quadratic reciprocity and in linear algebra it's good to know the Cayley-Hamilton theorem (to give two examples).

So, what is one (per post) deep and representative theorem of each subject that one can spend a couple of months or so to learn about? (In Combinatorics, Graph theory, Real Analysis, Logic, Differential Geometry, etc.)

答:
In Analytic Number Theory, the Prime Number Theorem is a great result.  </p>

<p>Riemann's 1859 paper, which outlined a possible approach to proving the PNT, is credited with motivating a large amount of the research done in Complex Analysis in the 19th century.   </p>

Title: Landmarks of subjects of mathematics

Body:
In Analytic Number Theory, the Prime Number Theorem is a great result.

Riemann's 1859 paper, which outlined a possible approach to proving the PNT, is credited with motivating a large amount of the research done in Complex Analysis in the 19th century.


问:
Summation involving subfactorial function
Inspired by this post:

Does the following series converge; if so, to what value does it converge? $$ \sum_{n = 2}^\infty \left|\frac{!n \cdot e}{n!} - 1\right|$$ I am looking for a closed form for the second question.

Note: !n denotes the subfactorial, also known as the number of derangments within $ \mathrm{S}_n $.

答:
(This is a corrected version)

I get that the sum converges,
but do not know the value.

Since $!n = n!\sum_{k=0}^n\frac{(-1)^k}{k!}$,
$\frac{!n \cdot e}{n!}
= e \sum_{k=0}^n\frac{(-1)^k}{k!}
= e \big(\frac1{e}-\sum_{k=n+1}^{\infty}\frac{(-1)^k}{k!}\big)
= 1-e\sum_{k=n+1}^{\infty}\frac{(-1)^k}{k!}
$.

Starting as Carl Najafi did,

$\begin{align*}
\sum_{n=2}^{\infty}\left|\frac{!n\cdot e}{n!}-1\right|
&=\sum_{n=2}^{\infty}\left|\big(1-e\sum_{k=n+1}^{\infty}\frac{(-1)^k}{k!}\big)-1\big)\right|\\
&=e\sum_{n=2}^{\infty}\left|\sum_{k=n+1}^{\infty}\frac{(-1)^{k+1}}{k!}\right|\\
&=e\sum_{n=2}^{\infty}\left|(-1)^{n}\sum_{k=n+1}^{\infty}\frac{(-1)^{k-n+1}}{k!}\right|\\
&=e\sum_{n=2}^{\infty}\left|(-1)^{n}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{(n+k)!}\right|\\
&=e\sum_{n=2}^{\infty}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{(n+k)!}\\
\end{align*}
$

The inner sum
$S(n)
= \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{(n+k)!}
$
is an alternating series
with terms decreasing in absolute value to $0$,
so its sum is bounded
by any two consecutive sums.
In particular,
taking the first two sums,
$\frac1{(n+1)!}
> S(n) > \frac1{(n+1)!} - \frac1{(n+2)!}
> \frac1{2(n+1)!}
$.

Therefore
$\sum_{n=2}^{\infty}\left|\frac{!n\cdot e}{n!}-1\right|
< \sum_{n=2}^{\infty}\frac1{(n+1)!}
$
which easily converges.

My original answer
actually looked at
$\sum_{n=2}^{\infty}\left|!n\cdot e-n!\right|
=\sum_{n=2}^{\infty}\left|n! S(n)\right|
$.
Since
$\frac1{n+1}
> |n!S(n)| > \frac1{n+1} - \frac1{(n+1)(n+2)}
\ge \frac1{2(n+1)}
$,
this sum diverges, but can be shown to converge
if the absolute value signs are removed,
since the resulting series is alternating
and decreasing.


答:
Here is the problem with your method of counting -

You are giving each letter of GUMTREE a choice of being in any of the $6$ places. So far so good.

But $6^7$ does not include permutations of "GUMTREE". For example, in the arrangements of all letters of GUMTREE being in the last place as KOALA {GUMTREE}, where are you considering permutations of GUMTREE?

So you may say we can multiply by $\frac{7!}{2!}$ but that brings in a different problem then. You overcount and to use P.I.E or to find multiplication factor for each type of arrangement separately is going to be a lot more work. Why does it overcount? Take an example,

K (G) O (UM) A (TR) L (EE) A

Now if you multiply as we mentioned above, it will also count cases,

K (U) O (GM) A (TR) L (EE) A but those have already been counted in $6^7$.

So it is easier to consider KOALA as letters that are same in total $12$ letters so we have $2$ $E$ and $5$ $X$ giving an answer of $\frac{12!}{5! \, 2!}$.


问:
$ H_n $ cardinal.
Let $ n \in \mathbb N $ and define $ H_n = \{(a, b) \in \mathbb Z ^ 2 \mid | a | + | b | = n + 1 \} $.  Show that $ | H_n | = 4n + 4 $.

My attempt was to give a bijective function between $ H_n$ and $ J_ {4n + 4} $ where $J_n=\{1,2,3\ldots,n\}$ is the prototype of the finite sets, that is, if a set $ A $ is equipotent to $ J_n $, it is said that $ A $ is finite and $ | A | = n $.

The function is as follows: Let $\psi: H_n \to J_ {4n + 4}$ given by:

$$ \psi (a, b) = \begin {cases} b & \text {if } a \geq 0 \ \wedge \ b> 0 \\ | a | + n + 2 & \text {if } a <0 \ \wedge \ b> 0 \\ | a | + 2n + 4 & \text {if } a \leq 0 \ \wedge \ b <0 \\ b + 4n + 5 & \text {if } a> 0 \ \wedge \ b  <0 \\ n + 2 & \text {if } b = 0 \ \wedge \ a> 0 \\ 2n + 3 & \text {if } b = 0 \ \wedge \ a <0 \end {cases}$$

I would like to know if this function is bijective, and how to show that it is bijective.  To show that $ | H_n | = 4n + 4 $.

答:
I think you are over-complicating this.  Try instead to reduce the problem to a simpler one.  If $(a,b)$ is a solution then so are $(\pm a, \pm b)$.  There is a complication in that the ambiguous sign makes no difference when $a$ or $b$ is $0$, so first we deal with that.

There are $3$ solutions when $a=0$ or $b= 0$, namely
$$(0,n+1),(0,-n-1),(n+1,0),(-n-1,0)$$

So now we know that the number of solutions is $4s+4$, where $s$ is the number of solutions in positive integers $a,b$.  These are of the form $(a, n+1-a)$ and in order that both entries are positive, $a$ can only take one the values $a=1,2,\dots, n$.

Title: $ H_n $ cardinal.


答: As you said, we have to fill $6$ gaps in $\text{_K_O_A_L_A_}$ with $7$ letters, so if $a$ letters go in the first gap, $b$ letters go in the second gap and so on and $f$ letters go in the sixth gap, we have the equation $a+b+c+d+e+f=7$ whose solution by stars and bars is given by $\displaystyle \binom{7+6-1}{6-1}=\binom{12}{5}$. The number of ways of arranging those $7$ letters as shown by other answers is $\dfrac{7!}{2!}$. Hence, the number of total combinations is $\displaystyle \dfrac{7!}{2!}\binom{12}{5}$.


问:
Analysing function at $x=0$
I am solving real analysis notes of my friend where i am having doubt in the below problem.

Show that if $x>0$, $\log(1+x)>\frac x{1+x}$ and hence prove that $x^{-1}\log(1+x)$ decreases monotonically as $x$ increases from $0$ to $\infty$.

Solution: Let $f(x)=\log(1+x)-\frac x{1+x}$, then
$$f'(x)=\frac1{1+x}-\frac{(1+x)\cdot1-x}{(1+x)^2}=\frac x{(1+x)^2}>0\ (\because x>0)$$
$$\therefore x>0\implies f'(x)>0$$
i.e $f$ is an increasing when $x>0$ and $f(x)>f(0)$.

In the text, the author has directly stated that $f(x) > f(0)$ even though  $f'(x) > 0 $ only when $x>0$. My understanding of the problem is as follows.

Technically speaking , given that $f'(x) > 0$ when $x > 0$ , $\implies$ $f'(x) > f(0^+)$ . Since the function is continuous at $x = 0$ , hence,  $\;f(0^+) = f(0) $ which leads to $f(x) > f(0)$.

My questions to the community are as follows :-

Does my reasoning holds good ?

If someone can explain / reason what is happening to the function $f(x)$ at $x=0$.

答: Your reasoning is not correct. Argue as follows:  Suppose $x<y$. The only case when we have difficulty in showing that $f(x) <f(y)$ is when $x=0$. In this case prove by contradiction. Suppose $f(0)(=f(x))\geq f(y)$. We have $f(\frac  1 n) < f(\frac  y 2) $ for all $n$ such that $\frac 1 n <\frac y 2$.   Letting $n \to \infty$ and using continuity at $0$ we get $f(0) \leq f(\frac y 2) <f(y)$. [Note that $<$ becomes $\leq $ in the limit. This is important]. We have now arrived at a contradiction.

Title: Analysing function at $x=0$


问:
If $\lfloor a-b\rfloor =2$, $\lfloor b-c\rfloor =3$, $\lfloor c-d\rfloor =4$, what is the sum of all possible values of $\lfloor a - d\rfloor?$

Let $a,b,c,d$ be real numbers with $\lfloor a - b\rfloor = 2 , \lfloor b - c\rfloor  = 3 , \lfloor c - d\rfloor = 4$. What is the sum of all possible values of $\lfloor a - d\rfloor ?$ ($\lfloor \cdot \rfloor$ denote the greatest integer function)

What I Tried:

I got that:-
$$\rightarrow 3 > a - b \geq 2$$
$$\rightarrow 4 > b - c \geq 3$$
$$\rightarrow 5 > c - d \geq 4$$
But I seem to find no way to use these. One think I thought is to first find the maximum $(a - d)$ , then find the minimum $(a - d)$, and then consider all the integer values between the maximum and the minimum. The question is, how do you find that?

Can anyone help me?

Body:
$a-d = (a-b) + (b-c) + (c-d)$

Now $[a-b]\le (a-b) < [a-b]+1$ etc.  so

$[a-b] +[b-c]+[c-d] \le (a-b) +(b-c)+(c-d) = (a-d) < [a-b] + [b-c]+[c-d] + 3$ so

$2 + 3 + 4 \le (a-d) < 2+3+4 + 3$

$9 \le a-d < 12$ so $a-d$ may be $9,10$ or $11$.

So the sum of all possible values (why do they ask these questions???) is $9+10+11 = 30$.


问:
Prove properties regarding the interior and frontier/boundary of a set
I know there properties:

where A $\subseteq R^m$ is a set, A' is the set of all limit points of A, called the derived set of A.

But now I'm wondering how can they be proven?

答:
$\newcommand{\fr}{\operatorname{fr}}$
$\newcommand{\int}{\operatorname{int}}$

By the definition, $\fr A=\overline{A}\setminus\int A$. Therefore $A\cap \fr A=\overline{A}\supset A’$. Since $\int A\subset A$, $\int (\Bbb R^m\setminus A)\subset  \Bbb R^m\setminus A$, and $A\cap (\Bbb R^m\setminus A)=\varnothing$, $\int A\cap \int (\Bbb R^m\setminus A)=\varnothing$ as well. It is easy to check that $\fr A=\overline{A}\setminus\int A=\overline{A}\cap \overline{\Bbb R^n\setminus A}=\fr(\Bbb R^n\setminus A)$. Finally, $A\setminus\fr A= A\setminus  (\overline{A}\setminus\int A)=\int A$.


答:
This might appeal to you as a simple way

First arrange $-K-O-A-L-A-$ with gaps between letters.

All we need to do is to  count how many ways GUMTREE can be interleaved.

$G$ can be inserted in $6$ ways, but then $U$ can now be inserted in $7$ ways,$M$ can be inserted in $8$ ways, and so on, and finally divide by $2!$ for the repeated $E$

Thus, $\dfrac{6\cdot 7\cdot 8\cdot 9\cdot {10}\cdot{11}\cdot{12}}{2!} = 1995840$


问:
Prove that there exist infinitely many primitive Pythagorean triples $x, y, z$ whose even member $x$ is a perfect square.
Prove that there exist infinitely many primitive Pythagorean triples $x, y, z$ whose even member $x$ is a perfect square. [Hint: consider the triple $4n^2, n^4-4, n^4+4$, where $n$ is an arbitraty odd integer.]

What I got:

Using the hint. $4n^2, n^4-4, n^4+4$ is a Phytagorean triple if $x^2+y^2=z^2$. Replacing and solving the equation it is clear that, $(4n^2)^2+(n^4-4)^2=(n^4+4)^2$ where $n$ is odd is indeed a Pythagorean triple with $x=4n^2=(2n^2)^2$ a perfect square.

Now I have to prove that $gcd(4n^2, n^4-4, n^4+4)=1$.

But I'm stuck here. I tried $gcd(n^4-4, n^4+4)=1$ without success. Any ideas?

答: If $n$ is odd, $\gcd (n^4-4, n^4+4)=\gcd (n^4-4,8)=1$ by the Euclidean algorithm.


问:
proving $\displaystyle\sum_{s,r=0,s\ne r}^{n}C_rC_s=\frac{1}{2}\left(2^{2n}-\frac{(2n)!}{(n!)^2}\right)$
Using $\displaystyle(1+x)^n=\sum_{r=0}^n C_rx^r$, find $\displaystyle\sum_{r=0}^n C_r$ and $\displaystyle\sum_{r=0}^n C_r^2$. Hence prove sum of product of two coefficients taken from $C_0,C_1,C_2\cdots,C_n$ is $\displaystyle\frac{1}{2}\left(2^{2n}-\frac{(2n)!}{(n!)^2}\right)$. i.e. $\displaystyle\sum_{s,r=0,s\ne r}^{n}C_rC_s$

Here's what I tried so far,

$$\sum_{r=0}^n C_r=2^n$$

using the fact that $(1+x)^{2n}$, I was able to derive,

$$\sum_{r=0}^n C_r^2=\frac{(2n)!}{(n!)^2}$$

But How can I proceed? Any Hint would be highly appreciated. Thank you.

答: In general for any sequence of real numbers $a_1,a_2,\cdots,a_n$, we have that $$2\sum_{1 \le i <j \le n} a_i a_j=\left(\sum_{1 \le i \le n} a_i\right)^2-\left(\sum_{1 \le i \le n} a_i^2\right)$$ and this property is used here.


问:
Does one have to be good at competition math to become a great mathematician in academia?
So I'm seventeen and currently a junior in high school. I love mathematics very much, it's developed into a strong passion over the past few years, after witnessing the president of the math club at my school give Euclid's proof on the infinitude of primes I've obsessed over finding beautiful arguments and proofs. The dilemma I've been having recently is regarding my relatively lackluster performance at mathematics competitions. I've worked my way through quite a few math books. (currently reading spivak's calculus on manifolds, have read his calculus, Shilov's linear algebra, apostol's calculus volumes and apostol's mathematical analysis.) I usually don't have insurmountable trouble doing the exercises here, but show me a problem on the AMC12 and I'll take an ungodly amount of time trying to make sense of it. My interests don't particularly lie in the discrete sort of mathematics that those sorts of competitions test though, but shouldn't the bit of mathematical maturity I seem to have be able to help me reason through these tests? Do these sorts of problem solving skills help me become a better mathematician? One last thing to note is I'm not a very competitive person so I feel that probably strongly affects the way I approach these competitions.</p>

答: To more directly answer your question, I used to think that math competitions were useless beyond something to write on a college application. It turns out that a ton of finance companies give interview tests with these sorts of problems, so yes it is important if you ever want to be a quant.</p>


问:
Solving the value of $x$ for the equation $\frac {(\sec x + \tan x)^2 - (\sec 2x + \tan x)^2}{\sin 2x - \sin x} = 2$
How to solve for the value of $x$ for the following equation $$\frac {(\sec x + \tan x)^2 - (\sec2x + \tan x)^2}{\sin2x - \sin x} = 2$$

My Attempt: $$ \frac {\sec^{2}x + 2\sec x\tan x + \tan^{2}x - (\frac {1 + \tan^{2}x}{1 - \tan^{2}x} + \tan x)^2}{\frac {2\tan x}{1 + \tan^{2}x} - \sin x} = 2$$

$$ \implies \frac {\sec^{2}x + 2\sec x\tan x + \tan^{2}x - (\frac {1 + \tan^{2}x}{1 - \tan^{2}x})^2 - 2\tan x(\frac {1 + \tan^{2}x}{1 - \tan^{2}x}) - \tan^{2}x}{\frac {2\tan x - \sin x(1 + \tan^{2}x)}{1 + \tan^{2}x}} = 2 $$

But here I stopped. I couldn't find any easier way for simplifying it more as it was growing more complicated while I was trying to solve. My other friends also tried to solve but their final equation turned into complex one.

Clarification: I thought I would lead to a conclusion with bringing an individual trig function. My intention was to solve it via using '$\tan x$' but couldn't transform '$\sec x$' & '$\sin x$' into $'\tan x'$ using any trig identities and formula.

Assumption: There is perhaps a basic in starting point I am missing for holding. Multiplying or dividing with something could have been lot more easier to end with a good conclusion.

Or if that not so, can we move straightforward from the last part of my approach to get the result?

答:
Not a pleasant solution but a nasty problem.

Consider that you look for the zero's of function
$$f(x)=(\sec (x) + \tan (x))^2 - (\sec (2x) + \tan (x))^2-2(\sin (2x) - \sin(x)) $$ If you try that tangent half-angle substitution, you will end with a polynomial of degree $14$
$$5 t^{14}-3 t^{13}-55 t^{12}-2 t^{11}+193 t^{10}+19 t^9-307 t^8+36
   t^7+367 t^6+19 t^5-85 t^4-2 t^3+11 t^2-3 t-1=0$$
which does not factor; so, forget such an approach.

A problem is the discontinuities because of the secants and the tangents. To remove all of them, multiply everything by $\big[\cos^2(x)\, \cos^2(2x)\big]$. Using the multiple angle formula, we end with
$$g(x)=9 \sin (x)-2 \sin (2 x)-3 \sin (3 x)-6 \sin (4 x)+5 \sin (5 x)-2 \sin (6x)+$$ $$\sin (7 x)-\sin (8 x)-4 \cos (2 x)+4 \cos (4 x)=0$$ which looks mor pleasant.

In the interval  $-\pi \leq x \leq \pi$, there are three obvious solutions $x=-\pi$, $x=0$ and $x=\pi$. Plotting $g(x)$ we can now notice "close" to $x=-0.42$, to $x=0.91$ and to $x=2.15$.

At this point, to polish the roots, I don not see anything elese than Newton method (or any other root-finding procedure). For each root, Newton iterates will be
$$\{-0.41,-0.417499,-0.417459\}$$
$$\{0.91,0.907893,0.907886\}$$
$$\{2.15,2.135306,2.135701\}$$


问:
Slightly weaker than twin prime conjecture, using $\Omega: \Bbb{Q}^{\times} \to \Bbb{Z}$
Necessary background.

Let $\Bbb{Q}^{\times} \xrightarrow{\Omega} \Bbb{Z}$ be the group hom given by $\Omega$ from the study of arithmetic functions (you can extend it uniquely (and homomorphically) to all of $\Bbb{Q}^{\times}$.)

It's kernel is generated by $\{\dfrac{p}{q} : p, q $ are prime numbers $\} = \ker \Omega$.  Does the weaker-than twin prime conjecture:

$$
\ker \Omega \cap A = \ker \Omega \cap \{\dfrac{n}{n+2} : n \in \Bbb{Z}\}
$$
is an infinite set hold?

Assume the intersection is finite, then $\ker \Omega \cap A' = \ker \Omega \cap \{\dfrac{n}{n-2} : n \in \Bbb{Z}\}$ is also finite.  The set $B = \{ \dfrac{n^2}{n^2 - 4} : n \in \Bbb{Z}\} \subset AA'$ elemenwise.  So that $\ker \Omega \cap B \subset \ker \Omega \cap AA'$.  That's as far as I've gotten.

What happens elementwise with respect to $\cap$ is:

$$
A(B\cap C) \subset AB \cap AC
$$

but not the other way around in general.  So that:

$$
(\ker \Omega \cap A')(\ker \Omega \cap A) \subset (\ker \Omega)^2 \cap AA' \cap (\ker \Omega)A \cap (\ker \Omega)A' = \\ \ker \Omega \cap A\ker \Omega \cap A'\ker \Omega \cap AA'
$$

So that in particular, the left side is a subset of $\ker \Omega \cap AA'$.  So maybe this route is ruled out?

Note that $A^{-1} = A'$ by construction.

Other facts: $\Bbb{Q}^{\times} = \biguplus_{k \in \Bbb{Z}} p^k G$, where $G = \ker \Omega$, for any fixed prime $p \in \Bbb{P}$.

Take $A = \{\dfrac{n -1}{n+1} : n \in \Bbb{Z}, n \neq -1\}$, then similarly $G \cap A$ would be finite.

$$\dfrac{r_1\cdots r_i}{q_1\cdots q_i}  = x \in G \cap A \iff \\
(n-1)q_1 \cdots q_i = (n+1) r_1 \cdots r_i, \\
q_j, r_j \in \Bbb{P} \\
\\ \iff \\
\Omega(\dfrac{\text{lcm}(n-1, n+1)}{n-1}) = \Omega(\dfrac{\text{lcm}(n-1, n+1)}{n+1}) \\
\iff \\
d := \gcd(n-1, n+1), \\
\Omega(\dfrac{n+1}{d}) = \Omega(\dfrac{n-1}{d})
$$

答:
A paper by Goldston, Graham, Pintz, and Yıldırım gives, among other interesting results, that (Theorem 6 in this paper)

For any $B\geq 5$ there are infinitely many integers $x$ with $\Omega(x) = \Omega(x + 2) = B$.

This gives a much stronger statement than you ask for; not only are there infinitely many rationals $\frac{x}{x+2}$ for which $$\Omega\left(\frac{x}{x+2}\right)= \Omega(x) - \Omega(x+2) = 0,$$
but there are infinitely many if you restrict the pair $(\Omega(x),\Omega(x+2))$ to be any $(B,B)$ with $B$ sufficiently large. This isn't quite enough to guarantee "twin semiprimes," but it's rather close.

Their results seem to be more general; in particular, it seems that they have also proven this where $2$ is replaced with any other positive integer, or where $\Omega$ is replaced with any sufficiently well-behaved number-theoretic function.


问:
Application of Urysohn's lemma
I am working on the following hw problem: If we have that $X$ is a compact Hausdorff space, with $\{U_\alpha\}_{\alpha\in A}$, then we can find a finite number of continous functions $f_1,...,f_k$, with $f_i:X\mapsto [0,1]$ such that $f_1(x)+...+f_k(x)=1$ for all $x$, and for each $i$ there exists a $\alpha_i$ such that $\overline{f_i^{-1}((0,1])}\subset U_{\alpha_i}$.

So this are my thoughts at the moment. Since $X$ is Hausdorff and compact it is normal, so we know that Urysohn's lemma applies. Meaning, if I have any two disjoint sets I can find a continous function that is $0$ is one of the sets and $1$ in the other one.

My intuition is telling me to use compactness and find $U_{\alpha_1},..., U_{\alpha_k}$ that cover $X$, and then by the last requirement of the problem we would need $f_i(x)=0$ if $x\notin U_{\alpha_i}$.

The problem is very clear if I can write my space $X$ as the disjoint union of finite closed sets cause then I could just apply Urysohns and get my functions very easily, but I do not see how to apply this lemma using the open cover given. Any hints would be greatly appreciated.

答:
Such a family of functions is called a partition of unity (where unity = 1, because of the summing to 1 property), and these exists for all paracompact Hausdorff spaces.

For a complete proof, see this, e.g. A compact (Hausdorff) space is paracompact Hausdorff (we can work with finite covers everywhere, but that makes the proof only slightly easier..).

Chapter 2 of the linked document handles the finite case, without going to paracompactness.

Title: Application of Urysohn's lemma


问:
Calculating Odds/Probability with Overlapping Type
Calculating Odds/Probability with Overlapping Type

I'm making a game where cards can have 1-2 types. Think pokemon types, where a 1-type pokemon is "normal" type and a two type pokemon is "normal/flying".

I'm trying to calculate odds of achieving certain hands given specific types. Something like: what are the odds of pulling 1 normal type in a draw of 5 cards, with a deck size of 50 and 4 normal types? Is relatively easy.

Where I'm running into trouble is where the types overlap. I'm trying to calculate something like:

What are the odds of pulling 2 normal types and 2 flying types in a draw of 5 cards with a deck size of 50, given 4 normal types, 3 flying types, and 4 normal/flying types (where normal/flying can count as a draw toward EITHER, but not BOTH, in other words, 1 normal/flying = 1 normal OR 1 flying, not 1 normal AND 1 flying).

Does anyone know where I could look for formulas or calculators that take into account the above scenario? Thank you.

答:
I know you are looking for a more general approach but you could do a case analysis for your stated example, and then perhaps use it to solve similar ones.

There are $\binom{50}{5}$ possible hands you can have. For the specific hand you're trying to create (two normal cards, two flying cards, and a fifth card that can be anything), we can enumerate the type selections. Let "P" stand for pure, "M" for mixed, "N" for normal, and "F" for flying. We then have for the four specific cards, the following ways they can be drawn from the normal or mixed sets along with the number of ways of occurrence.

$\{$PN, PN, PF, PF $\}$: $\binom{4}{2} \times \binom{3}{2} = 18$

$\{$PN, MN, PF, PF$\}$: $\binom{4}{1} \times \binom{4}{1} \times \binom{3}{2} = 48$

$\{$PN, MN, PF, MF$\}$: $\binom{4}{1} \times \binom{4}{2} \times \binom{3}{1} = 72$

$\{$PN, PN, PF, MF$\}$: $\binom{4}{2} \times \binom{3}{1} \times \binom{4}{1} = 72$

$\{$PN, MN, MF, MF$\}$: $\binom{4}{1} \times \binom{4}{3} = 16$

$\{$PN, PN, MF, MF$\}$: $\binom{4}{2} \times \binom{4}{2} = 36$

$\{$MN, MN, PF, PF$\}$: $\binom{4}{2} \times \binom{3}{2} = 18$

$\{$MN, MN, MF, PF$\}$: $\binom{4}{3} \times \binom{3}{1} = 12$

$\{$MN, MN, MF, MF$\}$: $\binom{4}{4} = 1$

After choosing four cards we have 46 cards to choose from for our final card. Thus the probability of getting two normal cards and two flying cards in a hand with five cards (given the type breakdown you gave) is
\begin{align}
 \text{Prob} & = \frac{1}{\binom{50}{5}}\left[\cdots \text{sum of terms}\cdots\right]\times 46\\
& = \frac{293}{46 060} \simeq 0.0064.
\end{align}


问:
Finding an MGF of discrete variable given some moments
Problem statement:

Let $X$ be a discrete random variable. Given that $E(X) = 0$, $E(X^2) = 2$ and $E(X^4) = 4$, find the moment-generating function (MGF) for $X$.

Now I know that the general formula for the MGF is $M_X(t) = \sum_{i = 0}^{\infty} E(X^i)\frac{t^i}{i!}$. However, I am provided with only three moment values, not a general formula. Perhaps, the fact that $X$ is discrete is an important clue.

How do I proceed with this problem?

答: Hint : $Var(X^2) = E[(X^2)^2] - (E[X^2])^2 = 0$. Thus, $X^2$ is a constant, but $E[X^2] = 2$ so $X^2 = 2$ almost surely. Then $X$ can take only two values, $\pm \sqrt 2$, each with some probability. Find these using the given constraints, and then find $E[e^{tX}]$, as desired.


答: $\newcommand{\E}{\mathbb{E}}$Hint: Let $Z=X^2$. Then we are given that $\E\left[Z^2\right] = \left(\E\left[Z\right]\right)^2$. What can we conclude?

Title: Finding an MGF of discrete variable given some moments

Body:
Problem statement:

Let $X$ be a discrete random variable. Given that $E(X) = 0$, $E(X^2) = 2$ and $E(X^4) = 4$, find the moment-generating function (MGF) for $X$.

Now I know that the general formula for the MGF is $M_X(t) = \sum_{i = 0}^{\infty} E(X^i)\frac{t^i}{i!}$. However, I am provided with only three moment values, not a general formula. Perhaps, the fact that $X$ is discrete is an important clue.

How do I proceed with this problem?


问:
Prove that $|\sum\limits_{k=1}^{n} a_{k}| \ge |a_{1}| - \sum\limits_{k=2}^{n} |a_{k}|$.
I am abjectly disappointed that I could not prove this statement on my own.  I have tried it directly and by contradiction but hit a wall.  Here is the statement (again) and my proof (thus far):

Prove that $|\sum\limits_{k=1}^{n} a_{k}| \ge |a_{1}| - \sum\limits_{k=2}^{n} |a_{k}|$.

Proof:
Let $a_{1}, ..., a_{n}$ be real numbers. Consider $|a_{1}| - \sum\limits_{k=2}^{n} |a_{k}|$. Then $|a_{1}| - \sum\limits_{k=2}^{n} |a_{k}| \leq ||a_{1}| - \sum\limits_{k=2}^{n} |a_{k}||$ holds by a property of the Absolute Value function $\forall a_{1},...,a_{n} \in \mathbb{R}$. Now, by the Reverse Triangle Inequality, we have $||a_{1}| - \sum\limits_{k=2}^{n} |a_{k}|| \leq |a_{1} - \sum\limits_{k=2}^{n} a_{k}|$.

I then proceeded to show that $|a_{1} - \sum\limits_{k=2}^{n} a_{k}| \leq \sum\limits_{k=1}^{n} a_{k} \leq |\sum\limits_{k=1}^{n} a_{k}|$ by considering different cases for $a_{1}$ and $\sum\limits_{k=1}^{n} a_{k}$ (e.g. $a_{1}, \sum\limits_{k=1}^{n} a_{k} > 0$ as one case out of four), and was able to show one case to hold but am struggling with the others.  Is there a more simplified way to write this proof?

Any help would be greatly appreciated! This has hurt my brain for weeks now :(

答:
We have
$$\left|\sum_{k = 1}^n a_k\right| = \left|a_1 + \sum_{k = 2}^n a_k\right| \ge |a_1| - \left|\sum_{k = 2}^n a_k\right|  \ge |a_1| - \sum_{k= 2}^n|a_k|,$$
where the first inequality comes from $|a+b| \ge |a| - |b|$ and the second one comes from $|a + b| \le |a| + |b|$.

EDIT: To prove that $|a+b| \ge |a| - |b|$ just remark that
$$|a| = |a + b - b| \le |a+b| + |b|$$
by triangular inequality.


问:
Regular $T_2$ space which is not completely regular.
Theorem 10. of
Pontryagin's Topological Groups says that:

Every Hausdorff topological group is completely regular.

But is there exists a Regular $T_2$ space which is not completely regular?

答: This answer has a complete description of such a space, due to John Thomas and published in A Regular Space, Not Completely Regular, The American Mathematical Monthly, Vol. 76, No. 2 (Feb., 1969), pp. 181-182.


问:
The existence of real SVD
I have known that if matrix A $\in C^{m*n}$,then there exists a SVD $A=U\Sigma V$.My question is if A is real,does there exist SVD which U and V real?</p>

答: As the comments and the other answer have pointed out, this follows from eigendecomposition. By transposing $A$ if necessary, we may assume that $A$ is tall, i.e. $A$ is $m\times n$ for some $m\ge n$. Let $VSV^T$ be an orthogonal diagonalisation of $A^TA$. Since $A^TA$ is real and positive semidefinite, $V$ can be taken to be real orthogonal and $S$ can be taken as a nonnegative diagonal matrix whose diagonal entries are arranged in decreasing order. Then $(AV)^T(AV)=S$, which implies that the columns of $AV$ are mutually orthogonal. Hence $AV=QS^{1/2}$ for some real $m\times n$ matrix $Q$ with orthonormal columns. Now complete $Q$ to an $m\times m$ real orthogonal matrix $U=\pmatrix{Q&\ast}$ and let $\Sigma=\pmatrix{S^{1/2}\\ 0}$. Then $AV=QS^{1/2}=U\Sigma$. Hence $A=U\Sigma V^T$ and we are done.

Title: The existence of real SVD


答:
First observe $$\begin{align} -\left|\sum_{k = 2}^{n}a_k \right| &\le 0\\
\left|\sum_{k = 1}^{n}a_k\right| - |a_1| &\ge0 \end{align}$$
And the second inequality follows from triangle inequality

Body:
I am abjectly disappointed that I could not prove this statement on my own.  I have tried it directly and by contradiction but hit a wall.  Here is the statement (again) and my proof (thus far):

Prove that $|\sum\limits_{k=1}^{n} a_{k}| \ge |a_{1}| - \sum\limits_{k=2}^{n} |a_{k}|$.

Proof:
Let $a_{1}, ..., a_{n}$ be real numbers. Consider $|a_{1}| - \sum\limits_{k=2}^{n} |a_{k}|$. Then $|a_{1}| - \sum\limits_{k=2}^{n} |a_{k}| \leq ||a_{1}| - \sum\limits_{k=2}^{n} |a_{k}||$ holds by a property of the Absolute Value function $\forall a_{1},...,a_{n} \in \mathbb{R}$. Now, by the Reverse Triangle Inequality, we have $||a_{1}| - \sum\limits_{k=2}^{n} |a_{k}|| \leq |a_{1} - \sum\limits_{k=2}^{n} a_{k}|$.

I then proceeded to show that $|a_{1} - \sum\limits_{k=2}^{n} a_{k}| \leq \sum\limits_{k=1}^{n} a_{k} \leq |\sum\limits_{k=1}^{n} a_{k}|$ by considering different cases for $a_{1}$ and $\sum\limits_{k=1}^{n} a_{k}$ (e.g. $a_{1}, \sum\limits_{k=1}^{n} a_{k} > 0$ as one case out of four), and was able to show one case to hold but am struggling with the others.  Is there a more simplified way to write this proof?

Any help would be greatly appreciated! This has hurt my brain for weeks now :(


问:
Proposition 8.23, Kechris (Classical Descriptive Set Theory) meager $F_\sigma$ set contains $A\Delta U$.
Let $X$ be a topological space and $A\subseteq X$. Then the following statements are equivalent:

I) $A$ has the Baire property (BP);

II) $A=G\cup M$, where $G$ is $G_\delta$ and $M$ is meager;

III) $A=F\setminus M$ where $F$ is $F_\sigma$ and $M$ is meager.

In the proof of $I)\Rightarrow II)$, it is noted that for $U$ open we have a meager $F_\sigma$ set $F$ such that $A\Delta U\subseteq F$.

I do not see why such a set exists in general, or how to proof the existence.

Can you help me out? Thanks in advance.

答:
If $A$ has the Baire property, there is an open set $U$ such that $M=A\mathrel{\triangle}U$ is meagre. Thus, there is a countable family $\mathscr{N}$ of nowhere dense sets such that $M=\bigcup\mathscr{N}$. For each $N\in\mathscr{N}$ the set $\operatorname{cl}N$ is closed and nowhere dense, so

$$F=\bigcup\{\operatorname{cl}N:N\in\mathscr{N}\}$$

is a meagre $F_\sigma$ set, and clearly $M\subseteq F$.


问:
Finding a PDF from a function
I have a function $y = f(x),\ x\in\mathbb{R}$ (assume $f(x)= \sin(x)/x$ if you need an example). How can I find the probability distribution function (PDF) of $y$, assuming $x\sim U(\mathbb{R})$ (uniformly distributed over the reals)? I'm a little rusty on whether the end points are problematic in defining it over the reals, so if that's an issue, let's make it simpler and say that $x\in\mathcal{D}\subset\mathbb{R}$ and $x\sim U(\mathcal{D})$ where $\mathcal{D}$ is an interval.

Do I explicitly need to know the inverse $x = f^{-1}(y)$ in order to be able to compute it?

答:
If $x$ is uniformly distributed over interval, its distribution can be written as $c\,|dx|$ where $c$ is  constant ($c=1/\ell$ where $\ell$ is the length of the interval). This suggests the  formula for distribution of $y$, namely $c\left|\frac{dx}{dy}\right|\,|dy|$. However, for  $\frac{dx}{dy}$ to make sense, we need the inverse function $x(y)$. If the given function $y(x)$ is not one-to-one, then its domain should be divided into subintervals on which it is one-to-one (that is, either increasing or decreasing).

Example: $x$ is uniformly distributed over $[-\pi/2,\pi]$, $y=\sin x$. The density of $x$ is $c=1/(3\pi/2)=2/(3\pi)$.

On the interval  $[-\pi/2,\pi/2]$ the values of $y$ run from $-1$ to $1$. The inverse function is $\sin^{-1}y$, with derivative $1/\sqrt{1-y^2}$. Therefore, this interval contributes $c\,(1-y^2)^{-1/2}\chi_{[-1,1]}$ to the pdf of $y$.

On the interval  $[\pi/2,\pi ]$ the values of $y$ run from $ 1$ to $ 0$. The inverse function is $\pi-\sin^{-1}y$, with derivative $-1/\sqrt{1-y^2}$. Therefore, this interval contributes $c\,(1-y^2)^{-1/2}\chi_{[0,1]}$ to the pdf of $y$.

Adding the above, we get the final answer: the pdf of $y$ is
$$\begin{cases}
\frac{2}{3\pi} (1-y^2)^{-1/2} \quad & -1\le y\le 0 \\ \\
\frac{4}{3\pi} (1-y^2)^{-1/2} \quad & 0\le y\le 1 \\ \\
0 & \text{otherwise}
\end{cases}$$

The example $\dfrac{\sin x}{x}$ is much less tractable, because there's no nice formula for the inverse.

Title: Finding a PDF from a function


问:
Local minimum free derivative algorithms?
I am looking for local minimum search algorithms free of derivative. More specifically, given a continuous and multi-modal $h:[0,1]\longrightarrow \mathbb{R}$ (note that we do not assume $h$ is smooth or Lipschitz), What free derivative algorithm can you recommend me to find a local minimum of $h$?

I know some stochastic adaptive algorithms, such as "HitAndRun", "Improved Hit and Run" and others, but these algorithms need, in general, a lot of functions evaluations to reach a local minimum.

Many thanks in advance for your comments.

答:
There are quite a few possibilities in terms of algorithms that you can use, and it’s difficult to give a recommendation without knowing anything about the specifics of you problem. How many variables does your objective function have? Do you have bounds on those parameters? Linear/nonlinear constraints? Is your function expensive to evaluate or very fast?

That said, depending on which programming language you are targeting, I would try (not necessarily in this order):

NLOpt: widely used optimization framework, has binding for multiple languages. In particular, I found the Subplex algorithm in NLOpt quite effective in many instances - but nothing stops you from trying others such as COBYLA, BOBYQA and Nelder-Mead Simplex (https://nlopt.readthedocs.io/en/latest/NLopt_Algorithms/)

Mystic: Python-based, I found the Powell algorithm to give good results in specific problems I had (https://mystic.readthedocs.io/en/latest/mystic.html)

A few more algorithms, some of those with less stellar performances- language and problem dependent, so your mileage may vary: https://www.gerad.ca/Sebastien.Le.Digabel/MTH8418/


问:
Question about the dual norm
Given a norm $\|\cdot\|$ on $\mathbb R^n$, the dual norm $\|\cdot\|_*$ is defined as
$$\|z\|_* = \sup \{z^{\mathrm T} x : \|x\| \le 1\}\text.$$

My question is: what is the $\arg\max$ of this quantity?

This is important to me because I wanted to show that if $K$ is a norm cone then its dual is the norm cone of the dual norm.
That is, if $K_{\|\cdot\|} = \{(x,r) : r \ge \|x\|\}$, then its dual cone is $K_* = K_{\|\cdot\|_*} = \{(y,t) : t \ge \|y\|_*\}$.

答:
The set of maximizers in the definition of the dual norm equals the sub-differential of the dual norm at the given point.

Given a norm $\varphi : \mathbb{R}^{n} \to [0,\infty)$ with dual norm $\varphi^{*}$, recall that $\varphi^{*}$ is a convex function and, therefore, to each $z \in \mathbb{R}^{n}$, we can associate the sub-differential
\begin{equation*}
\partial \varphi^{*}(z) = \left\{ q \in \mathbb{R}^{n} \, \mid \, \forall z' \in \mathbb{R}^{n} \, \, \varphi^{*}(z') \geq \varphi^{*}(z) + q^{T} (z' - z) \right\}.
\end{equation*}
It is worth noting that $\partial \varphi^{*}(z)$ consists of vectors determining supporting hyperplanes at $z$ to the set $\{z' \in \mathbb{R}^{d} \, \mid \, \varphi^{*}(z') \leq \varphi^{*}(z)\}$ (when $z \neq 0$).  Furthermore, since $\varphi^{*}$ is one-homogeneous, $\partial \varphi^{*}$ is zero-homogeneous (that is, $\partial \varphi^{*}(\alpha z) = \partial \varphi^{*}(z)$ for all $\alpha \in (0,\infty)$ and $z \in \mathbb{R}^{n}$).  Therefore, at least geometrically (for $z \neq 0$), we may as well think of $\partial \varphi^{*}(z)$ as encoding the supporting hyperplanes at $z$ to the set $\{z' \in \mathbb{R}^{n} \, \mid \, \varphi^{*}(z') \leq 1\}$ by replacing $z$ by $\frac{z}{\varphi^{*}(z)}$.

Back to your question, it is a nice exercise in convex analysis to prove the following identity:
\begin{equation*}
\partial \varphi^{*}(z) = \{q \in \mathbb{R}^{n} \, \mid \, \varphi(q) = 1, \, \, q^{T} z = \varphi^{*}(z) \}.
\end{equation*}
In particular, if $q \in \partial \varphi^{*}(z)$, then
\begin{equation*}
q^{T} z = \varphi^{*}(z) = \max\left\{ p^{T} z \, \mid \, \varphi(p) \leq 1 \right\}
\end{equation*}
and these are all the maximizers.

Conversely, if $q \in \mathbb{R}^{n}$ satisfies $\varphi(q) = 1$, then there is a $z \in \mathbb{R}^{n}$ such that $\varphi^{*}(z) = q^{T} z$.  This follows from the fact that $(\varphi^{*})^{*} = \varphi$; hence this holds if and only if $z \in \varphi^{*}(z) \partial \varphi(q)$.

To sum up, we see that the maximizers in the definition of the dual norm comprise the subdifferential of the dual norm at the corresponding point, which encodes the supporting hyperplanes to that point.  This can be a useful way of thinking about the unit ball $\{p \in \mathbb{R}^{n} \, \mid \, \varphi(p) \leq 1\}$ and the dual unit ball $\{z \in \mathbb{R}^{n} \, \mid \, \varphi^{*}(z) \leq 1\}$, including questions like the strict convexity of the boundary (e.g. does the boundary contain a line segment?) as well as its differentiability (e.g. does the boundary have a corner a la the "taxi-cab norm"?).

Finally, I would be remiss not to mention that the previous considerations show that the following inequality holds:
\begin{equation*}
\forall q \in \mathbb{R}^{n}, \, \, \forall z \in \mathbb{R}^{n} \quad q^{T} z \leq \varphi(q) \varphi^{*}(z)
\end{equation*}
with equality if and only if $q \in \varphi(q) \partial \varphi^{*}(z)$ and $z \in \varphi^{*}(z) \partial \varphi(q)$.  (Also, the discussion above applies to Finsler norms just as well as norms, modulo a few changes.)

Title: Question about the dual norm


问:
With $X_1,...X_4 \stackrel{i.i.d}{\sim} Ber(p)$ find $\mathbb{P}(X_1=x_1,...,X_4=x_4|X_1+...+X_4=2)$
My attempt:
$\mathbb{P}(X_1,...,X_4|X_1+...+X_4=2)=\frac{\mathbb{P}(X_1,...,X_4,X_1+...+X_4=2)}{\mathbb{P}(X_1+...+X_4=2)}$

$\mathbb{P}(X_1+...+X_4)\sim Binom(4,p)\Leftrightarrow \mathbb{P}(X_1+...+X_4=2)= {4 \choose 2}p^2(1-p)^2$

And I dont know how to find the following.
$\mathbb{P}(X_1=x_1,...,X_4=x_4,X_1+...+X_4=2)$

edit:

By drawing an event table of $X_1,...,X_4$ against $\sum X_i$ we see that for
all the ${4 \choose 2}$ cases of two $\ 0s$ and two $1s$, each case have a probability of $p^2(1-p)^2$. Which is $\frac{1}{{4 \choose 2}}\mathbb{P}(\sum X_i=2)$.

Is this the answer for $\mathbb{P}(X_1=x_1,...,X_4=x_4|X_1+...+X_4=2)$?

答:
This is the classical example to introduce the definition of Sufficient statistics.

They asked you to prove that, Given $S=\Sigma_i X_i=s$, the conditional distribution given $S=s$ does not depend anymore on the parameter $p$.

As shown, the conditional distribution is uniform over the 6 possible cases

$1 1 0 0 $

$1 0 1 0$

$1 0 0 1$

$0 1 1 0$

$0 1 0 1$

$0 0 1 1 $

... and zero elsewhere.

In other words, all the information needed to do inference about $p$ are included in the statistic $S=\Sigma_i X_i$


问:
Index of $M/N$, $M$,$N$ being modules
Let $A$ be a ring and $M$ be free $A$-module of rank $n$. Let $m_1,...,m_n$ be basis of $M$. We create elements $n_j=\sum_{i\in n}a_{ij}m_i$. Let $N=An_1+...+ An_n$. Then the index $[M:N]=det(a_{ij})$. Can anyone help me to proceed with the proof of this here?

Milne's notes has described one method. Let $A=\mathbb{Z}$, then we can find $a_i∈\mathbb{Z}$, s.t. $f′_i=a_ie′_i$, $e′_i,f′_i$ being bases of $M,N$ respectively, (which I am not sure how). Now let, $(e′_i)=E(e_i)$ and $f′_i=F(f_i)⇒(f_i)=F^{−1}AE(e_i)⇒det(F^{−1}AE)=∏_ia_i=[M:N]$. Here $A$ is the $diag(a_i)$ matrix..The last step is also doubtful to me as why $∏_ia_i=[M:N]$, Also why $det(E)=det(F)=1$. Can anyone help? $E,F$ is the transformation matrix of $e_i$ to $e_i'$ and $f_i$ to $f_i'$ respectively.

答:
I assume you want $A=\mathbb Z$ here since $[M:N]\in\mathbb Z\cup\{\infty\}$ while $\det(a_{ij})\in A$. Furthermore, we will need to assume that the $n_j$'s are linearly independent (otherwise $[M:N]=\infty$ while $\det(a_{ij})=0$), i.e. we are assuming that $N$ is free (submodules of free $\mathbb Z$-modules are always free) of rank $n$ over $\mathbb Z$. With these assumptions, my answer is basically an expanded version of what you read in Milne's notes.

Since $N\subset M$ are both free of rank $n$ over $\mathbb Z$, the structure theorem for modules over a PID tells us that there exists a basis $m_1',\dots,m_n'$ of $M$ along with scalers $c_1,\dots,c_n\in\mathbb Z$ such that $c_1m_1',\dots,c_nm_n'\in M$ form a basis for $N\subset M$. Using these bases, the inclusion $N\hookrightarrow M$ is given by the diagonal matrix $\mathrm{diag}(c_1,\dots,c_n)$. Since $(a_{ij})$ represents this inclusion in another basis, we have
$$\prod_{i=1}^nc_i=\det\mathrm{diag}(c_1,\dots,c_n)=\det(a_{ij}).$$
At the same time, with this nice choice of bases for $M,N$, one sees that
$$\frac MN\cong\prod_{i=1}^n\frac{\mathbb Z}{c_i\mathbb Z}$$
so also
$$\left|\frac MN\right|=\prod_{i=1}^nc_i=\det(a_{ij}).$$

All that remains is to explain where we get this choice of bases from. We have $N\subset M$ free $\mathbb Z$-modules of the same finite rank $n$. We want to construct a basis $e_1,\dots,e_n$ of $M$ along with elements $a_1,\dots,a_n\in\mathbb Z$ such that $a_1e_1,\dots,a_ne_n$ is a basis for $N$.

One argues by induction on $n=\mathrm{rank}(M)$. For any homomorphism $\lambda:M\to\mathbb Z$, let $I_\lambda=\lambda(N)\subset\mathbb Z$ be the ideal capaturing the image of $\lambda\vert_N$. Choose $\lambda_1$ such that $I_{\lambda_1}\subset\mathbb Z$ is maximal among $\{I_\lambda\}_{\lambda:M\to\mathbb Z}$, write $I_{\lambda_1}=a_1\mathbb Z$, and choose $x_1\in N$ such that $\lambda_1(x_1)=a_1$. Now, briefly consider any basis $f_1,\dots,f_n$ of $M$ with dual basis $\newcommand\dual[1]{#1^\vee}\dual f_1,\dots,\dual f_n$ of $\dual M=\mathrm{Hom}(M,\mathbb Z)$ (so $\dual f_i(f_j)=\delta_{ij}$); if we write $x_1=\sum_{i=1}^nb_if_i$, then
$$b_i=\dual f_i(x_1)\in(a_1)$$
by maximality of $I_{\lambda_1}$, so $a_1\mid b_i$ for all $i$. Hence, $x_1=a_1e_1$ for some $e_1\in M$ (in particular, $\lambda_1(e_1)=1$). We claim now that
$$M=\mathbb Ze_1\oplus\ker\lambda_1.$$
Indeed, for $m\in M$, one has $m=\lambda_1(m)e_1+(m-\lambda_1(m)e_1)\in\mathbb Ze_1+\ker\lambda_1.$ At the same time, if $m\in\mathbb Ze_1\cap\ker\lambda_1$, then $m=\lambda_1(m)e_1=0$ which shows $M=\mathbb Ze_1\oplus\ker\lambda_1$ as claimed. The inductive hypothesis then gives a basis $e_2,\dots,e_n$ of $\ker\lambda_1$ along with $a_2,\dots,a_n\in\mathbb Z$ such that $a_2e_2,\dots,a_ne_n$ is a basis of $N\cap\ker\lambda_1$, and so we win. Our desired bases are $e_1,e_2,\dots,e_n$ and $a_1e_1,a_2,e_2,\dots,a_ne_n.$

Title: Index of $M/N$, $M$,$N$ being modules


问:
Existence of a certain homeomorphism
Let $A$ be a compact topological space, and $X$ and $Y$ be homeomorphic Hausdorff spaces. Let $i_1$ and $i_2$ be continuous injections of $A$ into $X$ and $Y$ respectively. Does there exists a homeomorphism $g: X \to Y$ such that $i_2 = g \circ i_1$? If this is not true, what additional properties would be required to make this true?

My attempt at showing there exists such a $g$ didn't really lead anywhere, I think. I first tried showing that given two homeomoprhic compact subspaces of $X$ and $Y$, there exists a homeomorphisms $f$ from $X$ to $Y$ sending one subspace to the other. This I don't know if it is true. Assuming this, I took $i_1 (A)$ and $i_2 (A)$ to be these two subspaces, and I then tried to show there is an automorphism $h$ of $Y$ fixing $i_2 (A)$ such that $h \circ f \circ i_1 = i_2$, and finally taking $g=h \circ f$. This I don't know either if it's true.

Is this the correct approach so solving this problem?

答:
For a trivial counterexample, let $A=\{a\}$ be the one-point space and $X=Y=[0,2]$. Let $i_1(a)=0$ and $i_2(a)=1$. There is no autohomeomorphism of $[0,2]$ sending $0$ to $1$, since $1$ is a cut point, and $0$ is not.

A little less trivially, $A$ could be the subspace $\{0\}\cup\{2^{-n}:n\in\Bbb N\}$ of $\Bbb R$, with $X$ and $Y$ as before. Take $i_1(x)=x$ for $a\in A$ and $i_2(x)=x+1$, and you have the same problem as in the first example. As long as $X$ has points $p$ and $q$ such that no autohomeomorphism of $X$ sends $p$ to $q$, you can construct such a counterexample.

Finally, let

$$X=\big([-2,2]\times\{0\}\big)\cup\big(\{0\}\times[-2,2]\big)\,;$$

this is a nice compact, connected metric space. Let $A=[0,1]$. Let

$$i_1:A\to X:x\mapsto\langle x,0\rangle$$

and

$$i_2:A\to X:x\mapsto\langle x+1,0\rangle\,.$$

Both maps are embeddings of $A$ into $X$, but there is no autohomeomorphism of $X$ mapping the origin to $\langle 1,0\rangle$: deleting the origin from $X$ leaves $4$ components, while deleting $\langle 1,0\rangle$ leaves only $2$.

Title: Existence of a certain homeomorphism


问:
Topology on $[0,1]$ that makes it pseudocompact but not compact?
The space $[0,1]$ is compact with respect to its usual (metric) topology. I was wondering whether there is a known topology on it which makes it pseudocompact but not compact? Now sequentially compact and countably compact both imply pseudocompact so any such topology would work as well. Also, a metric space which is pseudocompact is automatically compact, so the topology cannot be induced by a metric. Any idea or reference would be appreciated.

I know we can always find such a topology. For instance, if we consider a pseudocompact space which is not compact which has the same cardinality as $[0,1]$ and transfer the topology to $[0,1]$ via bijection. I wanted to know if there is one which is frequently discussed and/or easy to understand.

答: Consider the topology on $[0,1]$ consisting of the empty set and of the sets containing $0$. Then every continuous function $f: [0,1] \to \Bbb R$ is constant, and thus $[0,1]$ is pseudocompact but not compact. Indeed, if $f(x) \not= f(0)$, there exists an open interval $I$ containing $f(x)$ but not $f(0)$. But since $f$ is continuous, $f^{-1}(I)$ is open, and hence contains $0$, a contradiction.


问:
Given $a > b > c > d > 0,$ and U $= \sqrt{ab} + \sqrt{cd}$ , V $= \sqrt{ac} + \sqrt{bd}$ , W $= \sqrt{ad} + \sqrt{bc}$.

Given $a > b > c > d > 0,$ and U $= \sqrt{ab} + \sqrt{cd}$, $V = \sqrt{ac} + \sqrt{bd}$, $ W = \sqrt{ad} + \sqrt{bc}$, arrange $U$,$V$,$W$ in ascending order.

What I Tried: I squared $U$,$V$,$W$ to get :-

$\rightarrow U^2 = ab + cd + 2\sqrt{abcd}$.

$\rightarrow V^2 = ac + bd + 2\sqrt{abcd}$.

$\rightarrow  W^2 = ad + bc + 2\sqrt{abcd}$.

We can cancel out the $2\sqrt{abcd}$ from each and we are only left to compare $(ab + cd) , (ac + bd) , (ad + bc)$ in order to compare $U$$,V$,$W$.

This is where I get stuck. I could say that $ab > ac$ , but I couldn't necessarily show that $cd > bd$ , as $b > c$. That way does not work for me, and I am not finding any other way to compare these.

Can anyone help?

答:
Hint: With your second & third expressions you're comparing, we get

$$\begin{equation}\begin{aligned}
(ac + bd) - (ad + bc) & = ac - ad + bd - bc \\
& = a(c - d) + b(d - c) \\
& = a(c - d) - b(c - d) \\
& = (a - b)(c - d) \\
& \gt 0
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$