问:
Convergence in $L^1$ of ratio of random variables with denominator positive but not bounded away from zero?
Let $\{X_n\}_{n=1}^{\infty}, \{Y_n\}_{n=1}^{\infty}$ be such that $X_n$ converges to a constant $a$ in L1 (and hence in probability); $Y_n\geq c > 0$ for all $n$, and $Y_n$ converges in probability to a constant $b\geq 1$.  In this post it was proved that $E[\frac{X_n}{Y_n}]\rightarrow a/b$. Just to note, I am interested in continuous random variables, if that makes any difference.

I am wondering can the condition that $Y_n \ge c > 0$ for all $n$ be weakened to $Y_n > 0$ for all $n$? In particular because in practise $Y_n$ could be the sample variance which is a positive random variable but not generally bounded away from $0$.

Intuitively it seems like if the density of $Y_n$ decays fast enough at the origin for all $n$ we might still be able to say $E[\frac{X_n}{Y_n}] \to a/b$ even if there is no $c >0$ such that $Y_n \ge c$ for all $n$?

答:
You cannot, in general, remove the intermediate $c$ such that $Y_n \geq c > 0$. Imagine taking, independent of $X_n$, $Y_n = 1$ with probability $1-1/n$ and $Y_n = 1/n^2$ with probability $1/n$. We have $Y_n \to 1$ in probability since $P(|Y_n - 1| \neq 0) = 1/n$.
However, $$E[X_n/Y_n] = E[X_n] E[1/Y_n] = E[X_n](n^2\cdot \frac1n + 1\cdot(1-\frac1n)) \to a \cdot \infty = \infty$$ if $a \neq 0$.

Recall Vitali's convergence theorem that tells us that $L^1$ convergence is equivalent to uniform integrability + convergence in probability. The $Y_n \geq c > 0$ condition guarantees both UI and convergence in probability, but $Y_n > 0$ does not.


问:
How to prove $A=\mathcal{O}_n$, use trace of matrix
Let $A=B+C$, and $B=B^T$, $C=-C^T$, $BC=\mathcal{O}_n$. Prove that if $A^2=\mathcal{O}_n$, then $A=\mathcal{O}_n$.
I want to prove $B=C=
\mathcal{O}_n$, and I got that
$$A^2=B^2+CB+BC+C^2\implies0=\mathrm{tr}(A^2)=\mathrm{tr}(B^2)+\mathrm{tr}(C^2)$$
So$$\mathrm{tr}(BB^T)=\mathrm{tr}(CC^T)\geqslant0.$$
But I don't know how to go on.

答:
if you extend what you have in the original post

$\mathbf 0 = A^4 = B^4+C^4 \implies \text{trace}\big(B^4 + C^4\big)=0\implies \text{trace}\big(B^4\big) =  -1\cdot\text{trace}\big(C^4\big)$

But $C^4 =  (-C^TC)^2= (C^TC)^2$ is necessarily symmetric positive semi-definite, as is $B^4$

$\implies\text{trace}\big(B^4\big)=0=\text{trace}\big(C^4\big)\implies B^4=\mathbf 0 = C^4 \implies B=\mathbf 0=C$

since the only normal matrix that is nilpotent is the zero matrix

ContentLicense: CC BY-SA 4.0


问:
How to solve $x\log(x) = 10^6$
I am trying to solve

$$x\log(x) = 10^6$$

but can't find an elegant solution. Any ideas ?

答:
You won't find a "nice" answer, since this is a transcendental equation (no "algebraic" solution).  There is a special function related to this called the Lambert W-function, defined by  $ \ z = W(z) \cdot e^{W(z)} \ $ .  The "exact" answer to your equation is
$ \ x = e^{W( [\ln 10] \cdot 10^6)} \ . $  (I'm assuming you're using the base-10 logarithm here; otherwise you can drop the $ \ln 10 $ factor.)

ContentLicense: CC BY-SA 3.0

Title: How to solve $x\log(x) = 10^6$


问:
How to derive LPP problem from the auxiliary problem using simplex method?
The original LPP: $f = 2x_1+3x_2 \rightarrow \max$. The constrains:
\begin{align}
-x_1-x_2 &\le -1 \\
 x_1+x_2&\le 6\\
x_1+2x_2&\le 8\\
x_2&\le 3\\
x_1\ge 0, x_2&\ge 0\\
\end{align}

In order to solve the original LPP, the auxiliary problem must be solved, because there is one constraint with negative RHS. It is done using simplex algorithm. Here is the final tableau:

\begin{array}{rr:rrrrrrr:r}
T^1_p& h & x_1 & x_2 & x_3 & u & x_4 & x_5 & x_6 & b \\ \hline
 x_1 & 0 & 1 & 1 & -1& 1 & 0 & 0 & 0&1\\
 x_4 & 0 & 0 & 0 & 1& -1 & 1 & 0 & 0&5\\
 x_5 & 0 & 0 & 1 & 1& -1 & 0 & 1 & 0&7\\
 x_6 & 0 & 0 & 1 & 0& 0 & 0 & 0 & 1&3\\ \hdashline
 h & 1 & 0 & 0 & 0& 1 & 0 & 0 & 0&0\\
\end{array}

From there, it is stated, that we can solve the original LPP by rewriting the objective function $f$ as a linear combination of variables outside of the base ( in this case $x_2$ and $x_3$) of the auxiliary problems' last simplex tableau. The function is rewritten as : $f = 2x_3+x_2+2 \rightarrow max$. All the constraints for this problem can be derived from the last simplex tableau ($T^1_p)$. I don't understand how is the objective function obtained in this instance? The original objective function for this problem is $f:2x_1+3x_2\rightarrow \max$. Maybe someone knows how to derive the objective function from the tableau above?

答:
I cannot relate the variables $x_3,\ldots,x6$, $h$ and $u$ to your original problem, so it would have helped to provide the starting tableau.

There are two ways of starting the second phase of two phase simplex. The first one is to keep the original objective as a separate row and resume from there.

The second one (and this is what you seem to be doing) is to take the original objective $f=2x_1+3x_2$ and express it using nonbasic variables. To eliminate $x_1$, we use $x_1+x_2-x_3+u=1$ from the final tableau ( $x_2$ is already nonbasic), so the objective is $f = 2(1-x_2+x_3-u)+3x_2 = 2 + x_2 + 2x_3 - 2u$. Just put this as the final row and continue. This method is demonstrated here.


问:
Expected size of largest connected component
I have a sample space of all the undirected graphs having n vertices and m edges. Then what is the expected value of the size of the largest connected component?

My original problem is that using the G(n, p) model (family of graphs having n vertices and any two vertices would be joined by an edge with probability p. See: Erdős–Rényi model), I need to calculate the expected size of the largest connected component in G(n, p). If the above problem can be solved then this original problem can be easily solved.

PS: I need to exactly calculate the value of the expectation for any fixed n and p given in the input and not just upper or lower bounds.

答:
I am answering my own question as I have solved the original problem.

The solution is extremely computationally intensive.

Let S be our required expected size of the largest connected component in G(n, p). and let $P_i^n$ be the probability that a graph of n vertices formed from G(n, p) have the largest connected component(s) of size $i$. then

$$S = \sum_{i = 1}^{i = n}iP_i^n$$

Now, for the calculation for $P_i^n$, first choose a set of $i$ vertices and define the below terms

$f_i$ = the probability that a specific set of already chosen $i$ vertices is connected

$g_i^n$ = the probability that no vertex from the set of $i$ chosen vertices make an edge to other remaining $(n-i)$ vertices.

Then $P_i^n$ would be
$$P_i^n = \binom{n}{i} f_i g_i^n \sum_{j = 1}^{j = i} P_j^{n - i} $$

This gives a recursive relation for $P_i^n$ (Note that in the summation term would go for $j = i$ and not for $j = i - 1$ as there can be more than one connected component of size i can be possible.

Calculation for $g_i^n$

$g_i$ can be easily calculated as the probability of not forming an edge between two vertices is $1 - p$ and there are total $i * (n - i)$ possible edges which shouldn't be formed, thus
$$g_i^n = (1 - p)^{i(n - i)}$$

Calculation for $f_i$

The calculation of $f_i$ is described here: Exact probability of random graph being connected

$$f_i = 1 - \sum_{k = 1}^{i - 1}f_k \binom{i - 1}{k - 1} (1 - p)^{k(i - k)}$$

The calculation of $f_i$ is also recursive. Overall the exact calculation for the required expectation $S$ is much computationally intensive.

Practically, generating a random set of graphs from the G(n, p) model, and calculating the average size of the largest connected component using DFS/BFS would be much easier than the exact method.


问:
Convergence of $\sum_{n=1}^{\infty}o\left(\frac{1}{n}\right)$
We already know that  $\forall \alpha>1$, we have  $$\sum_{n=1}^{+\infty}\frac{1}{n^\alpha}<+\infty$$ I’m considering adding some little amounts, say$$\sum_{n=1}^{+\infty}o\left(\frac{1}{n}\right) $$ I’m not sure if series $\sum_{n=1}^{+\infty}o\left(\frac{1}{n}\right)$ is convergent. Intuitively speaking, $o(1/n)$ is considered “smaller” than $1/n$, so it should converge. However I’m quite sure this statement is not correct(lol). Can anybody give me a counter example?

答:
I encountered a similar exercise problem in Zorich's textbook:

If $a_{n}\ge a_{n+1}>0$ for all $n\in\mathbb{N}$ and the series $\displaystyle\sum_{n=1}^{\infty}a_{n}$ converges then $a_{n}=o\left(\frac{1}{n}\right)$ as $n\to\infty$.

So I believe $o\left(\frac{1}{n}\right)$ is only a necessary condition to converge.


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

$$y^2−x^2−2y+2+i(2x(1−y))$$

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

Any suggestions would be much appreciated

答: If you know in advance that the function is analytic (which you do if you've checked that it satisfies the Cauchy–Riemann equations), you can obtain its expression in terms of $z$ simply by setting $x=z$ and $y=0$. This works because of the uniqueness theorem: two analytic functions which agree on the real axis must agree everywhere.</p>

Title: Rearranging Cauchy Riemann equations


问:
Integrating $\frac{z}{1-\cos(z)}$ over the unit circle
I want to evaluate  $ \int_C f(z)\,dz $  where $f(z)$ is :

$$f(z)=\frac{z}{1-\cos(z)}$$

and $C$ is the unit circle, counterclockwise.

I kept having problems with it. If someone can help, it would really be appreciated .

答:
The function $f$ is holomorphic except at points where $\cos z = 1$, i.e. at points where $z=2\pi k$, $k \in \mathbb{Z}$. You can check that $1-\cos z$ has a double zero at $z=0$, so $f$ has a simple pole there. By the residue theorem
$$
\int_{C} \frac{z}{1-\cos z}\,dz = 2\pi i \operatorname{Res}\limits_{z=0} \frac{z}{1-\cos z}
= 2\pi i \lim_{z\to 0} \frac{z^2}{1-\cos z} = 4\pi i
$$
(since $1-\cos z = \dfrac{z^2}{2} + O(z^3)$).


问:
polynomial of degree $n$ with real coefficient takes integral values on a certain set $n+1$ consecutive integers , then $f(x)$ is integer-valued
Show that if $f(x)$ is a polynomial of $deg n$ with real coefficient ,which takes integral values on a certain set $n+1$ consecutive integers , then $f(x)$ is integer-valued.

How to prove this I thought of this like $P(x) =\sum_{i=0}^{\infty}a_{i}x^{i}$ where $a_{n}=\frac{P^{n}(0)}{n!}$
Your answers are really nice. But , I don't know or I want answer from strong principle of induction.

答:
►Let $P_{n+1}(x)=A_0+A_1x+A_2x^2+\cdots+A_{n+1}x^{n+1}$ where $A_i\in\mathbb R$ for $i=0,1,...,n+1$ and $a_1,a_2,...,a_{n+2}$ a set of $n+2$ consecutive integers.We have the equation
$$\begin{pmatrix}1&a_1&a_1^2&\cdots&a_1^{n+1}\\1&a_2&a_2^2&\cdots&a_2^{n+1}\\\vdots\\1&a_{n+2}&a_{n+2}^2&\cdots&a_{n+2}^{n+1}\end{pmatrix}*\begin{pmatrix} A_0\\A_1\\\vdots\\A_{n+1}\end{pmatrix}=\begin{pmatrix} m_1\\m_2\\\vdots\\m_{n+1}\end{pmatrix}$$ where each $m_i$ is  integer.This leads to a Vandermonde's determinant (VD) of order $n+2$ which like as for a proof of it (there are several ones) can be reduced to a VD of order $n+1$. We do this to go to a polynomial $P_n$ of degree $n$ so we can apply induction.

We have $$\det\begin{pmatrix}1&a_1&a_1^2&\cdots&a_1^{n+1}\\1&a_2&a_2^2&\cdots&a_2^{n+1}\\\vdots\\1&a_{n+2}&a_{n+2}^2&\cdots&a_{n+2}^{n+1}\end{pmatrix}=\det\begin{pmatrix}1&0&0&\cdots&0\\1&a_2-a_1&a_2(a_2-a_1)&\cdots&a_2^n(a_2-a_1)\\\vdots\\1&a_{n+2}-a_1&a_{n+2}(a_{n+2}-a_1)&\cdots&a_{n+2}^n(a_{n+2}-a_1)\end{pmatrix}=$$
$$(n+1)!\det\begin{pmatrix}a_2&a_2^2&\cdots&a_2^{n+1}\\\vdots\\a_{n+2}&a_{n+2}^2&\cdots&a_{n+2}^{n+1}\end{pmatrix} $$
►Now for $n=1$ we have $P_1(x)=A_0+A_1x$ so if $$A_0+A_1a=m_1\\A_0+A_1(a_1+1)=m_2$$ we have $$A_1=m_2-m_1\in\mathbb Z\\A_0=m_1-A_1a\in\mathbb Z$$
►Suposse it is true for $P_n(x)$ and prove that this is so for $P_{n+1}(x)$.

Note that $$\det\begin{pmatrix}a_2&a_2^2&\cdots&a_2^{n+1}\\\vdots\\a_{n+2}&a_{n+2}^2&\cdots&a_{n+2}^{n+1}\end{pmatrix}=a_2a_3\cdots a_{n+2}\det\begin{pmatrix}1&a_2&\cdots&a_2^n\\\vdots\\1&a_{n+2}&\cdots&a_{n+2}^n\end{pmatrix} $$ and you are done.


问:
How many combinations are there to factorize a given integer into two numbers.
Hey guys I'm trying to solve a number theory Problem on codeforces. Part of the solution is to figure out how many ways are there to factorize a given number $N$ into $N=AB$. I can illustrate with an example what I mean. Let's look at the number 30. We have that $30=2*3*5$. Thus we find that $30=6*5$, $30=10*3$, $30=15*2$, $30=30*1$. So we have four different ways Since $*$ is commutative we can switch these two numbers and we find 8. If in the Prime Factorization are no prime numbers appearing twice the anwser for the Problem is ways to factorize N=$2^{number\ of \ prime factors \ of \ N}$. But what is the anwser if there are more than one prime factors appearing like $20=2*2*5$?

thank you for the anwsers!

答:
This is equivalent to dividing the number of divisors by $2$, since if $N =AB$, $A$ determines $B$. However, you have to be careful about squares.

If $n = \prod p_i ^{\alpha_i}$ not a square for the $p_i$ distinct primes then the answer is
$$\frac{\tau(n)}{2} = \frac{\prod (\alpha_i + i) }{2}$$

If $n$ is a square, then you have the case where the pair $(\sqrt{n}, \sqrt{n})$ is not counted twice so the answer is

$$\frac{\tau (n)+1}{2}$$

Hence, the answer in general is

$$\lceil \frac{\tau(n)}{2} \rceil$$


问:
Zeroth degree part of localization of a flat graded $A$-module is flat.
This question arises from the proof of Thm 3.9.9 in Hartshorne. Let $T$ be an integral noetherian scheme and $X \subseteq \mathbb{P}^n_T$ be a closed subscheme. The proof shows that the following are equivalent

(i) $\mathscr{F}$ is flat over $T$;

(ii)$H^0(X,\mathscr{F}(m))$ is a free $A$-module of finite rank, for all $m \gg 0$.

In showing that (ii) implies (i), he let $M = \oplus_{m\ge m_0}H^0(X,\mathscr{F}(m))$ where $m_0$ is a large integer s.t.  $H^0(X,\mathscr{F}(m))$ are all free. Then $\mathscr{F} = \tilde{M}$. But then he says that since $M$ is a free hence flat $A$-module, $\mathscr{F}$ is also flat over $A$. However, the stalk of $\mathscr{F}$ is not obtained simply by localization. We must localize $M$ and take the $0$th degree part. So how can we show that zeroth degree part of localization of a flat graded $A$-module is flat.

答:
Lemma (Stacks 00JR): Let $S$ be a graded ring. Let $M$ be a graded $S$-module. Let $\mathfrak{p}$ be an element of $\operatorname{Proj} S$. Let $f\in S$ a homogeneous element of positive degree so that $f\notin\mathfrak{p}$, i.e., $\mathfrak{p}\in D_+(f)$. Let $\mathfrak{p}'\in S_{(f)}$ be the element of $\operatorname{Spec}(S_{(f)})$ corresponding to $\mathfrak{p}$. Then $S_{(\mathfrak{p})} = (S_{(f)})_{\mathfrak{p}'}$ and compatibly $M_{(\mathfrak{p})} = (M_{(f)})_{\mathfrak{p}'}$.

Hartshorne's reduced to the case $T=\operatorname{Spec} A$ for $A$ a local noetherian domain. Let $S=A[x_0,\cdots,x_n]$, $\mathfrak{p}\subset S$ a graded prime, and $M$ a graded flat $S$-module. Since $D_+(x_i)$ cover $\Bbb P^n_A$, we may choose $x_i$ so that $\mathfrak{p}\in D_+(x_i)$. Applying the lemma, we see that $M_{(\mathfrak{p})} = (M_{(x_i)})_{\mathfrak{p}'}$.

Next, we note that $S_{x_i}=S_{(x_i)}[x_i,x_i^{-1}]$, so $S_{(x_i)}\to S_{x_i}$ is a flat extension of rings. Therefore as $M_{x_i}$ is a flat $S_{x_i}$-module, it is also a flat $S_{(x_i)}$-module. Since a direct sum of modules is flat iff each summand is and$M_{x_i}$ is the direct sum of its graded pieces, we see that $M_{(x_i)}$ is a flat $S_{(x_i)}$-module. Localizing at $\mathfrak{p'}$, we see that $(M_{(x_i)})_{\mathfrak{p}'}$ is a flat $(S_{(x_i)})_{\mathfrak{p}'}$-module, and applying the lemma again we see that this means that $M_{(\mathfrak{p})}$ is a flat $S_{(\mathfrak{p})}$-module.


问:
Proving that the union of compact sets is compact (in $\mathbb{R}$)
Proof

Let $\{C_i\}_{i=1}^n$ be a collection of nonempty compact sets in $\mathbb{R}$

Let $\mathcal{U}$ be an open cover of $\displaystyle\bigcup_{i=1}^n C_i$

Since each $C_i$ is compact, there are finitely many open sets $O_1,O_2,...,O_m \in \mathcal{U}$ such that $\displaystyle\bigcup_{j=1}^m O_j \supseteq C_i$

Since the finite union of finite sets is finite, there exists an finite subcover of $\mathcal{U}$, $\mathcal{U}_o$ such that $\mathcal{U}_o \supseteq\displaystyle\bigcup_{i=1}^n C_i$.

Thus, $\displaystyle\bigcup_{i=1}^n C_i$ is compact.  $\hspace{1cm} \square$

Did I leave anything out or did I make any jumps in my logic? Let me know.

答:
You are on the good tract, you just have to consider $\cup O_{i_n}^i$

where $C_i$ is a subset of the union of open subsets $O^i_{i_1},...,O^i_{i_{n_i}}$.


问:
Find maximize value of this expression
Let $a;b;c \in [0;2]$

Find maximum value of: $P=ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2)$

After some factoring, I found:

$P=(a-b)(b-c)(a-c)(a+b+c)$

Then square it: $P^2=(a-b)^2(b-c)^2(c-a)^2(a+b+c)^2$

Now $P^2$ be a symmetric expression. But it' still hard. Can someone help, please? Thank you.

答:
Since $K=[0,2]^3$ is a compact set, and $P=P(a,b,c)$ is continuous on $K$ with real values, there is a maximal value. It is taken at some point $(a^*,b^*,c^*)$, which is

(i) either in the interior $(0,2)^3$ of $K$, which is then a local extremal value, so that in this point
$$
\tag{$\dagger$}
P'_a=P'_b=P'_c=0\ ,$$

(ii) or at the boundary, and then by symmetry, we may and do assume $c^*=0$. Else if $c^*=2$ we replace $a^*$, $b^*$, $c^*$ by $2-a^*$, $2-b^*$, $2-c^*$ and possibly also exchange $a^*\leftrightarrow b^*$ to have a positive value.

We consider both cases individually.

(i) The system $(\dagger)$ becomes:
$$
\left\{
\begin{aligned}
0 &=3(a^2 - 6b^2 - 6bc - 6c^2)(b - c)
\ ,\\
0 &= 3(b^2 - 6c^2 - 6ca - 6a^2)(c - a)
\ ,\\
0 &= 3(c^2 - 6a^2 - 6ab - 6b^2)(a - b)
\ .
\end{aligned}
\right.
$$
If $a\ne b\ne c\ne a$, then all factors of the shape $(a^2 - 6b^2 - 6bc - 6c^2)$ with the cyclicly permuted cousins must vanish. In sum we obtain $11a^2 + 11b^2 + 11c^2 + 6ab +6ac + 6ca=0$. So $3(a-b)^2+3(b-c)^2+3(c-a)^2+5(a^2+b^2+c^2)=0$. This gives $a=b=c=0$, which is not an interior point. So at least two values among $a,b,c$ coincide, we may and do assume by symmetry that $a=b$. Then from the other two equations, since $a^2-6c^2-6ca-6a^2\le 0$ - with equality at the boundary of $K$ - we get $a=b=c$, so in such a point $P$ is zero, which does not correspond to a maximal value. (For instance, the point $(2,1,0)$ leads to a bigger value.)  So there is no interior maximal point.

(ii) Me may and do assume $c=0$, so the function to be maximized on $L=[0,2]^2$ is
$$
Q(a,b)=ab(a-b)(a+b)=ab(a^2-b^2)=a^3b-b^3a\ .
$$
Again, this function is defined on a compact set $L$, is real valued, so there is a maximal value taken at some point $(a^*,b^*)\in L$. And we distinguish the two cases, (ii,1) the point is an interior point, and (ii,2) the point is a boundary point. For an interior point, we must have $Q'_a=Q'_b=0$, but the corresponding system $3a^2b-b^3=0=-(3b^2a-a^3)$ has no solution with $ab\ne 0$. So the case (ii,1) is excluded. The maximum is thus taken in some point $(a^*,b^*)$ with either $a^*=2$, case (ii,2,A), or $b^*=0$, case (ii,2,B).

The case (ii,2,B) is immediately excluded, since $Q(a,0)=0$.

The case (ii,2,A) leads to the function $R(b)=2b(2-b)(2+b)=8b-2b^3$, which should be maximized on the interval $J=[0,2]$. At the boundary the values are zero, and in the interior the values are $>0$, so we have an interior point $b^*\in J$ with maximal value for $R$, and its is a solution of the equation $R'=0$, so $8-6b^2=0$. So
$$
b^*=\sqrt{\frac 86}=\frac 2{\sqrt 3}\ .
$$
So the maximal value of $P$ is reached in the point
$$
(a^*,b^*,c^*)=\left(2,\frac 2{\sqrt 3},0\right)
$$
and
$$
\max P=
P(a^*,b^*,c^*)
=
Q(a^*,b^*)
=R(b^*)
=2b^*(4-{b^*}^2)
=2\cdot\frac 2{\sqrt 3}\left(4-\frac 43\right)
=\frac{32}{3\sqrt 3}
\approx 6.158402871356\dots\ .
$$

Computer checks:

Sage code:

```<code>sage: def minusP(point): a, b, c = point; return -(a*b*(a^2 - b^2) + b*c*(b^2 - c^2) + c*a*(c^2 - a^2))
sage: point = minimize_constrained(minusP, [[0,2], [0,2], [0,2]], (2, 1, 0))
sage: point
(2.0, 1.1547005363453975, 0.0)
sage: 2/sqrt(3.)
1.15470053837925
sage: -minusP( point )
6.1584028713560075
sage: 32. / 3 / sqrt(3.)
6.15840287135601
</code>```

The twist via minimize_constrained was needed since there is no maximize_constrained.

Mathematica (TM): Link to wolframalpha code

Title: Find maximize value of this expression

Body:
Since $K=[0,2]^3$ is a compact set, and $P=P(a,b,c)$ is continuous on $K$ with real values, there is a maximal value. It is taken at some point $(a^*,b^*,c^*)$, which is

(i) either in the interior $(0,2)^3$ of $K$, which is then a local extremal value, so that in this point
$$
\tag{$\dagger$}
P'_a=P'_b=P'_c=0\ ,$$

(ii) or at the boundary, and then by symmetry, we may and do assume $c^*=0$. Else if $c^*=2$ we replace $a^*$, $b^*$, $c^*$ by $2-a^*$, $2-b^*$, $2-c^*$ and possibly also exchange $a^*\leftrightarrow b^*$ to have a positive value.

We consider both cases individually.

(i) The system $(\dagger)$ becomes:
$$
\left\{
\begin{aligned}
0 &=3(a^2 - 6b^2 - 6bc - 6c^2)(b - c)
\ ,\\
0 &= 3(b^2 - 6c^2 - 6ca - 6a^2)(c - a)
\ ,\\
0 &= 3(c^2 - 6a^2 - 6ab - 6b^2)(a - b)
\ .
\end{aligned}
\right.
$$
If $a\ne b\ne c\ne a$, then all factors of the shape $(a^2 - 6b^2 - 6bc - 6c^2)$ with the cyclicly permuted cousins must vanish. In sum we obtain $11a^2 + 11b^2 + 11c^2 + 6ab +6ac + 6ca=0$. So $3(a-b)^2+3(b-c)^2+3(c-a)^2+5(a^2+b^2+c^2)=0$. This gives $a=b=c=0$, which is not an interior point. So at least two values among $a,b,c$ coincide, we may and do assume by symmetry that $a=b$. Then from the other two equations, since $a^2-6c^2-6ca-6a^2\le 0$ - with equality at the boundary of $K$ - we get $a=b=c$, so in such a point $P$ is zero, which does not correspond to a maximal value. (For instance, the point $(2,1,0)$ leads to a bigger value.)  So there is no interior maximal point.

(ii) Me may and do assume $c=0$, so the function to be maximized on $L=[0,2]^2$ is
$$
Q(a,b)=ab(a-b)(a+b)=ab(a^2-b^2)=a^3b-b^3a\ .
$$
Again, this function is defined on a compact set $L$, is real valued, so there is a maximal value taken at some point $(a^*,b^*)\in L$. And we distinguish the two cases, (ii,1) the point is an interior point, and (ii,2) the point is a boundary point. For an interior point, we must have $Q'_a=Q'_b=0$, but the corresponding system $3a^2b-b^3=0=-(3b^2a-a^3)$ has no solution with $ab\ne 0$. So the case (ii,1) is excluded. The maximum is thus taken in some point $(a^*,b^*)$ with either $a^*=2$, case (ii,2,A), or $b^*=0$, case (ii,2,B).

The case (ii,2,B) is immediately excluded, since $Q(a,0)=0$.

The case (ii,2,A) leads to the function $R(b)=2b(2-b)(2+b)=8b-2b^3$, which should be maximized on the interval $J=[0,2]$. At the boundary the values are zero, and in the interior the values are $>0$, so we have an interior point $b^*\in J$ with maximal value for $R$, and its is a solution of the equation $R'=0$, so $8-6b^2=0$. So
$$
b^*=\sqrt{\frac 86}=\frac 2{\sqrt 3}\ .
$$
So the maximal value of $P$ is reached in the point
$$
(a^*,b^*,c^*)=\left(2,\frac 2{\sqrt 3},0\right)
$$
and
$$
\max P=
P(a^*,b^*,c^*)
=
Q(a^*,b^*)
=R(b^*)
=2b^*(4-{b^*}^2)
=2\cdot\frac 2{\sqrt 3}\left(4-\frac 43\right)
=\frac{32}{3\sqrt 3}
\approx 6.158402871356\dots\ .
$$

Computer checks:

Sage code:

sage: def minusP(point): a, b, c = point; return -(a*b*(a^2 - b^2) + b*c*(b^2 - c^2) + c*a*(c^2 - a^2))
sage: point = minimize_constrained(minusP, [[0,2], [0,2], [0,2]], (2, 1, 0))
sage: point
(2.0, 1.1547005363453975, 0.0)
sage: 2/sqrt(3.)
1.15470053837925
sage: -minusP( point )
6.1584028713560075
sage: 32. / 3 / sqrt(3.)
6.15840287135601

The twist via minimize_constrained was needed since there is no maximize_constrained.

Mathematica (TM): Link to wolframalpha code


问:
It is given that $x.y.z=12^3$ where $x,y,z$ are positive integers.Then , how many different solutions are there for $x+y+z$?

It is given that $x.y.z=12^3$ where $x,y,z$ are positive integers.Then , how many different solutions are there for $x+y+z$?

Its answer is easy if it were $x.y=12^3$ , because there are $28$ possible $(a,b)$ pairs by $x.y=2^6 \times 3^3$ and $C(8,2) \times C(4,1) =28$.

Moreover, half of $28$ will give us the number of $x+y$ thanks to symmetry property.

Let's come to original question:

Firtly , i found the number of how many possible $x,y,x$ there exist by combination with repetition such that $C(6+3-1,6) \times C(3+3-1,3)=280$.

However , when i want to find the number of different values of $x+y+z$ , i stuck in.Because , i thought that if i divide $280$ by $3$ , i can obtain the result.However , it did not work and i dont know what i should do.

Therefore, i hope to find tricks or solutions for my problem. Moreover , what can i do for the expanding versions of this question such as $x+y+z+t$

答:
Let's define an order that $a < b$ if the power of $3$ that divides $a$ is less than the power of $3$ that divide $a$ or those powers are equal then if power of $2$ that divides $a$ is larger then the power than divides $b$

Then wolog $x \le y \le z$ and we need to consider the cases

$x=y=z$

$x=y < z$

$x < y = z$

$x < y < z$.

Case 1: $x = y = z$ so $3^1|x,y,z$ and $2^2|x,y,z$ and $x=y=z=12$ and $x+y+z = 36$.

Case 2: $x=y < z$ then either $3|x,y,z$ and $2^3|x,y$ or $3^3|z$ and $1,2,4|x,y$

That gives us $4$ more options $x=y=24;z=3$ or $x=y=1,2,4; z = 27\cdot(2^6,2^3,1)$

So that's $5$ options so far.

Case 3: $x< y = z$ then $3|x,y,z$ and either $2^6|x$ or $2^4|x, 2|y,z$

So that's $2$ more options. $7$ so far.

Case 4: $x < y < z$.

either $3|y; 3^2|z$ or $3|x,y,z$ and $2^4|x; 2^2|y$ or $2^3|x; 2^2|y; 2|z$

The latter two give us $2$ more for $9$.

And finally The first case we can have the powers of $2$ distributed any way.

That $3$ times $1,1,2^6$.

$1$ times $1,2^2,2^4$ and $2$ times $1,2^3,2^3$.

And $2$ times $2,2,2^4$.  And $6$ times $2,2^2,2^3$.

And $1$ times $2^2,2^2,2^2$ so that is $3+1+2+2+6+1=15$.

So a total of $24$.

I'm taking it on faith, perhaps incorrectly,  that we can show these solutions are pairwise relatively prime so distinct.

And as Robert Isreal got $49$ sums I'm obviously making an error. Probably missing some options.

Still I'll leave this up as a strategy that I think is valid.


答:
Your idea looks correct, though you may want to be more careful about your indexing. Another way to state what you've done is the following.

Let $\{C_i\}_{i=1}^n$ be a family of compact sets, and let $\mathcal{U}$ be an open cover of $\bigcup_{i=1}^nC_i$. Thus $\mathcal{U}$ is an open cover for each $C_i$. So by compactness, each $C_i$ is covered by some $U^i_1,\ldots,U^i_{j_i}\in\mathcal{U}$. Now the collection
$$
\bigcup_{i=1}^n\{U_1^i,\ldots,U_{j_i}^i\}\subset\mathcal{U}
$$
forms a finite subcover for $\bigcup_{i=1}^nC_i$.

Title: Proving that the union of compact sets is compact (in $\mathbb{R}$)

Body:
Proof

Let $\{C_i\}_{i=1}^n$ be a collection of nonempty compact sets in $\mathbb{R}$

Let $\mathcal{U}$ be an open cover of $\displaystyle\bigcup_{i=1}^n C_i$

Since each $C_i$ is compact, there are finitely many open sets $O_1,O_2,...,O_m \in \mathcal{U}$ such that $\displaystyle\bigcup_{j=1}^m O_j \supseteq C_i$

Since the finite union of finite sets is finite, there exists an finite subcover of $\mathcal{U}$, $\mathcal{U}_o$ such that $\mathcal{U}_o \supseteq\displaystyle\bigcup_{i=1}^n C_i$.

Thus, $\displaystyle\bigcup_{i=1}^n C_i$ is compact.  $\hspace{1cm} \square$

Did I leave anything out or did I make any jumps in my logic? Let me know.


问:
Show that in a group of four people it is more likely than not that two of them were born on the same day of the week.
I've started the working :

Possible combinations : AB AC AD BC BD CD = 6

P(2 people born same day of week) = 1 - P(No people born on same day of week)

=1−((7/7)⋅(6/7)⋅(5/7)⋅(4/7))=1−(120/343)=223/343≈0.65

I'm not sure if this is correct and would really appreciate any working or answers.

答:
As N. F. Taussig commented, you have too many $7$s and should have

$$= 1 - \left({\frac{7}{7} \cdot \frac{6}{7} \cdot \frac{5}{7} \cdot \frac{4}{7}}\right) = 1 - \frac{120}{343} = \frac{223}{343} \approx 0.65%%$$


问:
Is it true that a smooth, harmonic functions with compactly supported gradient is trivial?
Let $u \in C^{\infty}(\mathbb{R}^n)$ be such that $\nabla u \in C^{\infty}_c(\mathbb{R}^3; \mathbb{R}^3)$ and $\Delta u=0$. I wish to prove or disprove that $u = c$ for some $c \in \mathbb{R}^n$; note the statement would be true with $c=0$ (by Liouville classical theorem) if $u$ itself is compactly supported.

I had the idea of exploiting a representation formula like:

$$ u(x) = \frac{1}{n \omega_n} \int_{\mathbb{R}^3} \nabla u(y) \cdot \frac{x-y}{|x-y|^n} dy $$

together with an integration by parts in order to get the Laplacian. Unfortunately, the latter is only valid if $u \in W^{1,1}(\mathbb{R}^n)$ and here we don't have the hypothesis $u \in L^1(\mathbb{R}^n)$. I am stuck here, but at same time I cannot produce a counterexample.

答:
I think that you can prove that the function is constant, but not necessarily zero: since $\nabla u$ has compact support, there exists $R>0$ such that $\nabla u \equiv 0$ in $\mathbb{R}^{n} \setminus \overline{B(0;R)}$, and then $u \equiv c$ ($c \in \mathbb{R}$ is a constant) in $\mathbb{R}^{n} \setminus \overline{B(0;R)}$. From this,
$$ |u(x)| \leq \max \left \lbrace c, \Bigg| \max_{x \in \overline{B(0,R)}}u \Bigg|\right \rbrace, $$
that is, $u$ is bounded. By Liouville theorem, you obtain that $u$ is constant.


答: Note that the gradient has compact support, thus $u$ is constant outside a compact set. Hence, there exists a constant $c$ such that $v:=u-c$ has compact support. But $v$ is harmonic and thus vanishes. This implies that $u$ is constant.

Title: Is it true that a smooth, harmonic functions with compactly supported gradient is trivial?

Body:
Let $u \in C^{\infty}(\mathbb{R}^n)$ be such that $\nabla u \in C^{\infty}_c(\mathbb{R}^3; \mathbb{R}^3)$ and $\Delta u=0$. I wish to prove or disprove that $u = c$ for some $c \in \mathbb{R}^n$; note the statement would be true with $c=0$ (by Liouville classical theorem) if $u$ itself is compactly supported.

I had the idea of exploiting a representation formula like:

$$ u(x) = \frac{1}{n \omega_n} \int_{\mathbb{R}^3} \nabla u(y) \cdot \frac{x-y}{|x-y|^n} dy $$

together with an integration by parts in order to get the Laplacian. Unfortunately, the latter is only valid if $u \in W^{1,1}(\mathbb{R}^n)$ and here we don't have the hypothesis $u \in L^1(\mathbb{R}^n)$. I am stuck here, but at same time I cannot produce a counterexample.


问:
Find the n-th base 2 palindrome
So I want to compute the n-th base 2 palindrome, where n is a number less than or equal to 5000, and we can pick it.

So we know that in base 2 only odd numbers can be palindromes. Okay so that saves us some computation power, but still not enough.

My first thought was, to compute all the possible odd numbers in base 2 and check for every one of them if it is a palindrome. Then I would count in order how many palindromes there are, and when we reach the target $n$ we can print out the n-th palindrome. I think this is not enough, or my method of finding such numbers is bad. My question is, is there any other property of base 2 palindromes, besides that they are odd, that could help my search ? What kind of technique would you pick for search instead of a simple iteration (that's whagt I have done).
(I am coding in python.)

答:
For $n\ge 0$ there are $2^n$ palindromes of length $2(n+1)$, one for each possible choice of the second through $(n+1)$-st bits. For $n\ge 1$ there are also $2^n$ palindromes of length $2n+1$, again one for each possible choice of the second through $(n+1)$-st bits. Finally, there are $2$ palindromes of length $1$ (instead of just $1$). Thus, the total number of palindromes of length at most $2n$ is

$$\sum_{k=0}^{n-1}2^k+\sum_{k=0}^{n-1}2^k+1=2(2^n-1)+1=2^{n+1}-1\,.$$

The least $n$ such that the $k$-th palindrome has at most $2n$ bits is then the least $n$ such that $k\le 2^{n+1}-1$, i.e., $n=\lfloor\lg k\rfloor$. (In case you’ve not seen it before, $\lg$ is $\log_2$.) We can now determine the $k$-th palindrome as follows.

Let $n=\lfloor\lg k\rfloor$, so that $2^n\le k<2^n-1$; the $k$-th palindrome must have $2n-1$ or $2n$ bits.

The $2^n$-th palindrome is the first palindrome with $2n-1$ bits, and there are $2^{n-1}$ palindromes of length $2n-1$, so the $(2^n+2^{n-1})$-st palindrome is the first one with $2n$ bits.

Let $\ell=k-2^n$.

If $\ell<2^{n-1}$, let $b_1b_2\ldots b_{n-1}$ be the $(n-1)$-bit binary representation of $\ell$; then the $k$-th palindrome is $1b_1b_2\ldots b_{n-2}b_{n-1}b_{n-2}\ldots b_2b_11$, of length $2n-1$.

If $\ell\ge 2^{n-1}$, let $b_1b_2\ldots b_{n-1}$ be the $(n-1)$-bit binary representation of $\ell-2^{n-1}$; then the $k$-th palindrome is $1b_1\ldots b_{n-1}b_{n-1}\ldots b_11$, of length $2n$.

Examples: Suppose that we want the $11$-th palindrome; $\lg 11\approx 3.46$, so $n=3$, and $\ell=11-2^3=3<2^2$. The $2$-bit representation of $3$ is $11$, so the $11$-th palindrome is $11111$.

If instead we want the $13$-th palindrome, $n$ is still $3$, but $\ell=5\ge 2^2$. The $2$-bit representation of $\ell-2^2$ is $01$, so the $13$-th palindrome is $101101$.

Title: Find the n-th base 2 palindrome


答:
By residue theorem
$$\int_C f(z)dz=2i\pi Res(f,0)=$$

and $Res(f,0)$ is the coefficient $a_{-1}$ in the Laurent series of $\frac{z}{1-\cos z}=\sum_{-\infty}^\infty a_n z^n$ so we find $Res(f,0)=2$
and then
$$\int_C f(z)dz=4i\pi $$

Title: Integrating $\frac{z}{1-\cos(z)}$ over the unit circle

Body:
I want to evaluate  $ \int_C f(z)\,dz $  where $f(z)$ is :

$$f(z)=\frac{z}{1-\cos(z)}$$

and $C$ is the unit circle, counterclockwise.

I kept having problems with it. If someone can help, it would really be appreciated .


问:
Bioche's rules in the integration
Looking this question Indefinite integral $\int \frac{1}{1+\sin^4\left(x\right)}dx$ and the comment of the nice user @Bernard for the use of the substitution

$$t=\tan x$$

On Wikipedia there are some examples when it is necessary to use the

If
$$\omega (-t)=\omega (t)$$
a good change of variables is $u=\cos t$ (even function).

If
$$\omega (\pi -t)=\omega (t)$$
a good change of variables is $u=\sin t$ (odd function).
If
$$\omega (\pi +t)=\omega (t)$$
a good change of variables is $u=\tan t$.

If two of the preceding relations both hold, a good change of variables is
$$u=\cos 2t$$

In all other cases, use
$$u=\tan(t/2)$$

Why @Bernard have used $t=\tan x$ if have an even integrand function? Shouldn't you use cosine $u=\cos t$?

PS: Never I have studied the Bioche's rules in the integration during the period at the university.

答: As Wikipedia notes, $\omega$ includes the $dt$ factor, say $\omega=fdt$, so $\omega(c-t)=\omega(t)$ if $f(c-t)=-f(t)$.

Title: Bioche's rules in the integration


答:
Your proof is good but:

I'd explicitly point out that $\mathscr U$ being a cover of $\cup C_i$ then $\mathscr U$ is therefore a cover of each $C_i$.  (You implicitly assume that is clear but I think it should be deliberately noted.)

And in manipulating the finite subcover I'd be a little more exact and formal.

I'd point that for each $C_k$ we have a finite number of $_kO_{1},....,_kO_{m_k}\in \mathscr U$ that form an open subcover, $\mathscr U_k=\cup_{i=1}^{m_k} $$_kO_i$, of $C_k$.

The $\overline {\mathscr U}=\cup \mathscr U_k$ (or whatever notation you like) is a finite open cover.

(We could further state, but don't need to, that $\overline {\mathscr U}=\cup_{k=1}^n (\cup_{i=1}^{m_k} $$_kO_i)$.)

(As you currently have it written it is implicit but not actually stated how you get the $O_m$ for a particular $C_i$ to getting a collection of ... sets indexed how? ... for the union of all the $C_i$. It's better to be able to spell it out explicitly.)

....

Also, this is me nitpicking.... really nitpicking.  You should specify as finite union in the title of the post.  That it is a finite union and not an infinite union is essential.


问:
Number of index $p^k$ subgroups
I'm trying to find the number of subgroups of $\mathbb{Z}^n$ such that the quotient is a $\bf{cyclic}$ group of order $p^k$ ($p$ a prime).

I guess I don't know enough about the finitely generated abelian groups, though I consulted some results about that. So I hope for a hint, thank you very much.

答:
This is a much-studied question. For the general theory see the paper:

Grunewald, F. J.; Segal, D.; Smith, G. C., Subgroups of finite index in nilpotent groups, Invent. Math. 93, No. 1, 185-223 (1988). ZBL0651.20040.

Their proposition 1.1 shows that for subgroups of prime power index we have the following zeta function:
$$\prod_{i=1}^{d-1}(1-p^j p^{-s})^{-1},$$
where the coefficient of $p^{-ks}$ is the number you seek.

Title: Number of index $p^k$ subgroups


问:
Evaluate this limit $ \lim_{x\to\infty}\left (\frac{1}{x}\frac{a^x - 1}{a - 1} \right)^\frac{1}{x}$
Please help to evaluate this limit

$$ \lim_{x\to\infty} \left(\frac{1}{x}\frac{a^x - 1}{a - 1} \right)^\frac{1}{x},$$

where $0 \leq a$ and $a \not= 1$.

I tried to logarithm from both sides, and apply taylor series but so far without success.

答:
Take the (natural) logarithm. For $a>1$, we have
$$
\lim_{x\to\infty}\frac{-\log x+\log(a^x-1)-\log(a-1)}{x}=
\lim_{x\to\infty}\frac{\log(a^x-1)}{x}=\lim_{x\to\infty}\frac{a^x\log a}{a^x-1}=\log a
$$
with a simple application of l'Hôpital.

For $0<a<1$, you get
$$
\lim_{x\to\infty}\frac{-\log x+\log(1-a^x)-\log(1-a)}{x}=0
$$
which needs nothing special.


问:
On state-transition matrix and fundamental theorem of local theory of curves
Lately I was faced by a problem and I can't think some way to solve it:

Let $A(t)$ be an antisymmetryc matrix, for all $t \in I$. Prove that $\phi(t,t_{0})$ (the state-transition matrix) is an orthogonal matrix $\forall t \in I$ in the ODE
$$
u'(t)=A(t)u(t).
$$

Then, one must prove, using this result, the fundamental theorem of local theory of curves, the same described by Manfredo.

I've tried to write $\phi(t,t_{0}) = \phi(t)(\phi(t_{0}))^{-1} = \phi(t_{0},t)$, but I can't related this to the transpose. Any help will be very appreciated! Thanks in advance!

答:
You know that $ϕ(t_0,t_0)=I_d$, which is, among other things, orthogonal. Now consider the evolution of the orthogonality condition
$$
\frac{d}{dt}[ϕ(t,t_0)^Tϕ(t,t_0)-I_d]=[A(t)ϕ(t,t_0)]^Tϕ(t,t_0)+ϕ(t,t_0)^T[A(t)ϕ(t,t_0)]
$$
and use the anti-symmetry of $A$.

Another approach to the same calculation is to take a fixed vector $C$ and compute the derivative of the norm square $\|ϕ(t,t_0)C\|_2^2$.


问:
Set of all subset is a sigma algebra?
In the book Measure, Integration and Real Analysis (Axler, 2020), the author claims that:

Suppose $X$ is a set. Then the set of all subsets of $X$ is a $\sigma-$ algebra on $X$

Then he defines the measurable set:

A measurable space is an ordered pair $(X,S)$ where $X$ is a set and $S$ is a $\sigma -$ algebra on $X$.

An element of $S$ is called an $S$ measurable set (or just a measurable set if $S$ is clear from the context.

I have a question concerning the measurablity of every subset of $\mathbb{R}$.

As we know, not every subset of $\mathbb{R}$ is measurable ($i.e$ it does not belong to any $\sigma -$ algebra. So the set of all subset of $\mathbb{R}$ can't be a $\sigma -$ algebra on $\mathbb{R}$ which contradicts the claim of the author.

Could you please explain me what does the author mean ? Or is there any misunderstanding in my mind when reading the book ?

Thank you very much for your help!

答: The assertion “not every subset of $\Bbb R$ is measurable” is valid for the Lebesgue measure. Anyway, here Axler is talking about $\sigma$-algebras, not about measures. And, yes, $\mathcal P(\Bbb R)$ is a $\sigma$-algebra. And, on this $\sigma$-algebra you can define the measure $m(X)=\#X$; with respect to this measure, $\Bbb R$ is measurable.

Title: Set of all subset is a sigma algebra?


问:
If $\sum^\infty_n a_n$ is convergent, then is $\sum^\infty_n\sqrt{a_n}$ also convergent?

Let $\displaystyle\sum^\infty_n a_n$ be a convergent series, where $a_n>0$ for every $n\in\mathbb{N}$. Is it true that $\displaystyle\sum^\infty_n\sqrt{a_n}$ is also convergent?

My first thought was to use a comparison test: We know that for $a_n\geq 1$. But this seems wrong to me since $\displaystyle \lim_{n\to\infty}a_n=0$ implies that for big enough $n$, $a_n<1$ (which causes the test to fail).

I have a feeling that this is not a true statement. I just can't come up with any counter examples.

答:
No. Take

$$a_n =\frac{1}{n^2}$$ for which $$\sum\limits_{n=1}^{\infty} a_n = \sum\limits_{n=1}^{\infty} \frac{1}{n^2}$$

is convergent*.
Of course
$$\sum\limits_{n=1}^{\infty} \sqrt{a_n}=\sum\limits_{n=1}^{\infty} \frac{1}{n}$$ diverges as it is a harmonic series.

*Convergence can be shown by computing partial sums $$\sum\limits_{n=2}^{k} \frac{1}{n(n-1)}=\frac{k-1}{k}$$ as $$\frac{1}{n^2} \leq \frac{1}{n(n-1)}$$ for $n>1$ and then taking limit: $$\lim\limits_{k \to \infty} \sum\limits_{n=2}^{k} \frac{1}{n(n-1)}=\lim\limits_{k \to \infty} \frac{k-1}{k}=1$$


答:
The set   $\mathcal{P}(X)$ of all subsets of $X$ is a sigma algebra because it satisfies the axioms for that kind of structure. That is independent of any discussion of measures.

The definition of a measure space is a pair $(X,S)$ where $S$ is some subset of $\mathcal{P}(X)$ that happens to be a sigma algebra. It may or may not be all of $\mathcal{P}(X)$.

When you set out to define Lebesgue measure on $\mathbb{R}$ you soon discover that you can't get all the properties you want if you try to assign a "length" to every subset of $\mathbb{R}$. So you restrict the definition of "length" to a subalgebra $S$ of $\mathcal{P}(\mathbb{R})$ and call those the measurable sets.

Title: Set of all subset is a sigma algebra?

Body:
In the book Measure, Integration and Real Analysis (Axler, 2020), the author claims that:

Suppose $X$ is a set. Then the set of all subsets of $X$ is a $\sigma-$ algebra on $X$

Then he defines the measurable set:

A measurable space is an ordered pair $(X,S)$ where $X$ is a set and $S$ is a $\sigma -$ algebra on $X$.

An element of $S$ is called an $S$ measurable set (or just a measurable set if $S$ is clear from the context.

I have a question concerning the measurablity of every subset of $\mathbb{R}$.

As we know, not every subset of $\mathbb{R}$ is measurable ($i.e$ it does not belong to any $\sigma -$ algebra. So the set of all subset of $\mathbb{R}$ can't be a $\sigma -$ algebra on $\mathbb{R}$ which contradicts the claim of the author.

Could you please explain me what does the author mean ? Or is there any misunderstanding in my mind when reading the book ?

Thank you very much for your help!


问:
Trace-class operators between different Hilbert spaces
Let $H$ be a (separable) Hilbert space and $B(H)$ be the space of bounded linear operators defined on $H$. Suppose $\{ e_i \}$ be an orthonormal basis of $H$, we say $T \in B(H)$ is a trace-class operator if

$$ \|T\|_1 := \sum_i \langle |T| e_i, e_i \rangle < \infty, $$

where $|T|$ is the positive operator such that $|T|^2 = T^*T$ and $T^*$ is the adjoint operator of $T$.
If $T \in B(H)$ is a trace-class operator, we can define its trace as

$$ \newcommand{\Tr}{\mathrm{Tr}}  \Tr(T) := \sum_i \langle Te_i, e_i \rangle. $$

From Wikipedia, or Conway's book A course in functional analysis, we know that trace-class operators have some nice properties:

If $T \in B(H)$ is a trace-class operator, then for any $A \in B(H)$, we know both $AT$, $TA$ are trace-class, and $\Tr(AT) = \Tr(TA)$, and $\|AT\|_1 \leq \|A\| \|T\|_1$.

However, I'm curious about the definition of trace-class operators between two different Hilbert spaces $H_1$ and $H_2$. Let $B(H_1, H_2)$ be the space of bounded linear operators from $H_1$ to $H_2$.

It seems the above definition can be generalized to this case as follows: We say an operator $T \in B(H_1, H_2)$ is trace-class if $\sqrt{T^*T} : H_1 \to H_1$ is trace-class and define its trace-norm as $\|T\|_1 := \| \sqrt{T^*T} \|_1$. This generalization looks reasonable to me, but I am not sure whether there exist some subtleties.

I am trying to prove the aforementioned properties of trace-class operators under such definition. Suppose $H_3$ is another Hilbert space, and $A \in B(H_2, H_3)$. To show that $AT$ is trace-class, by definition it suffices to show that $\sqrt{T^* A^* AT}$ is trace-class. But I don't know how to proceed.

I guess that the definition of trace-class operators can be generalized and the aforementioned properties also hold. I would appreciate it if you could provide some references about these.

(BTW: It seems that the Hilbert-Schmidt operator can be easily generalized to different Hilbert spaces.)

Update: Thanks for the answer of Ruy. There is another question about these trace-class operators: When $A \in B(H_2, H_1)$, can we conclude that $\Tr(AT) = \Tr(TA)$?

答:
Notice that
$$
  |AT|^2 = T^*A^*AT \leq \|A\|^2T^*T = \|A\|^2|T|^2,
  $$
so, by Proposition 1.3.8 in [1] (square-root is an operator monotone function), one has that
$$
  |AT| \leq   \|A\||T|,
  $$
and it follows that $AT$ is trace class according to the OP's definition.

[1] Pedersen, Gert K., C*-algebras and their automorphism groups, London Mathematical Society Monographs. 14. London - New York -San Francisco: Academic Press. X, 416 p. $ 60.00 (1979). ZBL0416.46043.


问:
Show that $\sigma(1+p)<1+p+p^2$.
Show that $\sigma(1+p)<1+p+p^2$, where $p$ is prime.

I know that $\sigma(p^2)=1+p+p^2={p^3-1 \over p-1}$. So this is the same as proving that $\sigma(1+p)<\sigma(p^2)$. I know that $\sigma(p)>p$ for all primes $p$, so $\sigma(1+p)>1+p$ and $\sigma(p^2)>p^2$.

答:
It has nothing to do with $p$ being prime.

$\sigma(1+p)$ is the sum of divisors of $1+p$, so it is certainly at most the sum of all natural numbers $\le 1+p$, which is $(p+1)(p+2)/2$, and it is easy to see that this $<1+p+p^2$ as long as $p > 1$.

Title: Show that $\sigma(1+p)<1+p+p^2$.


问:
polynomial of degree $n$ with real coefficient takes integral values on a certain set $n+1$ consecutive integers , then $f(x)$ is integer-valued
Show that if $f(x)$ is a polynomial of $deg n$ with real coefficient ,which takes integral values on a certain set $n+1$ consecutive integers , then $f(x)$ is integer-valued.

How to prove this I thought of this like $P(x) =\sum_{i=0}^{\infty}a_{i}x^{i}$ where $a_{n}=\frac{P^{n}(0)}{n!}$
Your answers are really nice. But , I don't know or I want answer from strong principle of induction.

答: Along the same lines as GEdgar’s answer: use the fact that if $f(x)$ has degree $n>1$ then $g(x) := f(x+1) - f(x)$ has degree $n-1$ (this is easy to see by looking at the leading term, but ultimately we’re still appealing to binomial expansion).  Now note that if $f$ has integer values at $n+1$ consecutive integers, then $g$ has integer values at $n$ consecutive integers.  Induct from the fact that the claim is trivial for the degree $0$ case (constant polynomials).


问:
Explicit expression for a periodic function with repeated lines.
I know there HAS to be a better way to describe the function I'm looking for, but I'm looking for an explicit form for the following function with domain $(0,4)$:

$$f(x)=  \left\{
\begin{array}{ll}
      x & 0\leq x\leq 1 \\
      2-x & 1< x\leq 3 \\
x-4 & 3<x\leq 4
\end{array}
\right. $$
Tile this function with period $4$ over $(0,4),(4,8),...$

Essentially, it's just a spiky wave function. I would like an explicit form of this including floor functions, maybe something of the form $\max \{f(x),g(x)\},$ I have no idea. How would I do this?

答:
On [0,4] :
$$
f : x\to \min(\max(0, x), 1) + \min(0, 1 - x) + 2\max(0, x - 3)
$$
By defining
$$
g:x\to f\left(x-4\text{floor}\left(\frac x4\right)\right)
$$
you get your periodic function


问:
Intersection of two submanifolds with dimension condition
Consider a manifold of dimension $n$, we have two submanifolds $A$ and $B$, with dimension $n_1$ and $n_2$ that satisfies $n_1+n_2=n$. Can we conclude they only intersect at finite points (or isolated points)? How does one prove this?

Motivation: The reason that I asked this question is I was looking at the definition of the intersection number. In wiki it says the following

So I wonder why we have they intersect generically at isolated points when the dimension adds up to the total dimension.

答:
This isn't true even if you restrict yourself to $\mathbb{R}^n$.
For example, consider these two 2-dimensional planes in $\mathbb{R^4}$:
$$A = \{(x,y,0,0) \mid x,y \in \mathbb{R}\}$$
$$B = \{(x,0,z,0) \mid x,z \in \mathbb{R}\}$$
Then, $A\cap B$ is the line $\{(x,0,0,0)\mid x \in \mathbb{R}\}$.


问:
How many combinations are there to factorize a given integer into two numbers.
Hey guys I'm trying to solve a number theory Problem on codeforces. Part of the solution is to figure out how many ways are there to factorize a given number $N$ into $N=AB$. I can illustrate with an example what I mean. Let's look at the number 30. We have that $30=2*3*5$. Thus we find that $30=6*5$, $30=10*3$, $30=15*2$, $30=30*1$. So we have four different ways Since $*$ is commutative we can switch these two numbers and we find 8. If in the Prime Factorization are no prime numbers appearing twice the anwser for the Problem is ways to factorize N=$2^{number\ of \ prime factors \ of \ N}$. But what is the anwser if there are more than one prime factors appearing like $20=2*2*5$?

thank you for the anwsers!

答:
The number of divisors of a number
$$n=\prod_i p_i^{a_i}$$
where $p_i$ are the prime numbers and $a_i$ are their multiplicities is determined by the expression:$$\sigma_0(n)=\prod_{i}(a_i+1).$$

Particularly if the number $n$ is square-free ($\forall p_i:\ a_i=0,1$) one obtains the formula $\sigma_0(n)=2^k$ where $k$ is the number of prime factors of $n$ as you have already found.


问:
Find the Image of $\tanh{x}$
Problem

Find the Image of $\tanh{x}$

Attempt

I know that $\tanh{x} \rightarrow 1$ for $x \rightarrow \infty$ and $\tanh{x} \rightarrow -1$ for $x \rightarrow -\infty$. I know that $\tanh{x}$ is continuous over the real numbers and I know that I have to argue from the intermediate value thereom. However, I do not know how to apply the IVT to show that the image is indeed $(-1,1)$. I now that I have to pick an element $y\in (-1,1)$ and....

This is where I am stuck. I do not see how the argue for the double inclusion for

$$\tanh{((-\infty,\infty))}=((-1,1))$$

答:
Here's the intended argument:

Clearly, the range of $\tanh$ is a subset of $(-1, 1)$, since $|e^x - e^{-x}| < |e^x + e^{-x}|$.

Now, let us show that $(-1, 1)$ is a subset of the range of $\tanh$.

You have already stated that $\lim\limits_{x \to \infty} \tanh(x) = 1$ and $\lim\limits_{x \to -\infty} \tanh(x) = -1$. You also know that $\tanh$ is continuous.

Consider an arbitrary $y \in (-1, 1)$. Then $-1 < y$. Then $y + 1 > 0$. Define $\epsilon = y + 1$. Then consider some $k$ such that for all $x < k$, we have $|\tanh(x) - (-1)| < \epsilon$. Such a $k$ exists by the definition of $\lim\limits_{x \to -\infty}$.

Then in particular, for $x < k$, we have $\tanh(x) - (-1) < \epsilon = y + 1$. Then $\tanh(x) < y$. Take the particular value $x = k - 1$; then $\tanh(x) < y$.

Similarly, $y < 1$. A completely analogous argument allows us to find some $z$ such that $\tanh(z) > y$.

So we have $\tanh(x) < y < \tanh(z)$. Thus, by the intermediate value theorem, we must have some $j$ such that $\tanh(j) = y$.

Therefore, we see that the range of $\tanh$ is indeed $(-1, 1)$.

Title: Find the Image of $\tanh{x}$


问:
How to solve $\int_0^x f(t) dt = g(f(x)) $?
How to solve an equation of this type :

$$\int_0^x f(t) dt = g(f(x)) $$

for a given function $g$.

Im not so good with differential equations and integral equations.

I know how to solve $h'(x) = j(h(x)) $ in general and some approximations of specific cases of $k'(x) = k(l(x))$. But even my knowledge of those is limited.

However for this $$\int_0^x f(t) dt = g(f(x)) $$ I do not even know how to begin.

It seems very fundamental and simple though so my intuition assumes there must be a way.

I considered differentiating both sides but that got me to

$$ f(x) = g'(f(x)) f'(x) $$

Many thanks !

答: Let $F(x):=\int_0^xf(t)dt=g(dF/dx)$ so, if $g$ has a functional inverse $g^{-1}$,$$dF/dx=g^{-1}(F)\implies x+C=\int dx=\int\frac{dF}{g^{-1}(F)}$$with $C$ an integration constant. In theory, this obtains $x+C$ as a function of $F$, say $h(F)$. If $h$ is invertible, $F(x)=h^{-1}(x+C)$.


问:
Number of possible pairs from $\{1,\dots, n\}$ with $i < j$
I am not sure how to find the above number of possible pairs. For example, for $\Sigma = \{1,2,3,4\}$, the desired number would be $6$. I see that this is half of ${4 \choose 2}$ and I am wondering if this is in fact the way to go about this: take all possible two element selections from a set of $n$ elements, and then equate choosing $(i, j)$ with choosing $(j, i)$. </p>

答: ${4\choose 2}$ is 6! ${n\choose 2}$ is the number of ways of choosing 2 elements from a set of n in any order. So it will just give you the number of pairs $(i, j)$ - but that is exactly what you want, as in any pair $(i,j)$ one of the elements must be greater than the other.


问:
Formula for number of surjective functions in case where codomain is greater than domain
We were deriving the formula for number of surjective functions from f: $X \to Y$, where $|X| = n$ and $|Y| = m$. The formula that we have is:

$$\sum_{k=0}^m(-1)^k (m-k)^n {m \choose k}$$

Now, if $m>n$ we have no surjective functions, as codomain is greater than domain and the answer should be zero. The formula gives the right answer, but I cannot figure out why does it work algebraically.

Any hints on why does this sum equals to zero when $m > n$?

答:
In this answer, $3$ proofs of
$$
\sum_{j=0}^m(-1)^{m-j}\binom{m}{j}\binom{j}{k}=[m=k]\tag1
$$
are given, where $[\cdots]$ are Iverson brackets.

Any polynomial, $P(x)$, of degree $n$ can be written as
$$
P(x)=\sum_{k=0}^nc_k\binom{x}{k}\tag2
$$
for some constants $c_k$.

Then the $m^\text{th}$ forward difference of $P(x)$ is
$$
\begin{align}
\sum_{j=0}^m(-1)^{m-j}\binom{m}{j}P(x+j)
&=\sum_{j=0}^m(-1)^{m-j}\binom{m}{j}\sum_{k=0}^nc_k\sum_{i=0}^k\binom{x}{i}\binom{j}{k-i}\tag{3a}\\
&=\sum_{k=0}^nc_k\sum_{i=0}^k\binom{x}{i}\sum_{j=0}^m(-1)^{m-j}\binom{m}{j}\binom{j}{k-i}\tag{3b}\\
&=\sum_{k=0}^nc_k\sum_{i=0}^k\binom{x}{i}[m=k-i]\tag{3c}\\
&=\sum_{k=0}^nc_k\binom{x}{k-m}\tag{3d}
\end{align}
$$
Explanation:

$\text{(3a)}$: apply $(2)$ and Vandermonde's Identity

$\text{(3b)}$: change the order of summation

$\text{(3c)}$: apply $(1)$

$\text{(3d)}$: apply the Iverson brackets

If $n\ge m$, $(3)$ shows that the $m^\text{th}$ forward difference of a polynomial of degree $n$, is a polynomial of degree $n-m$.

If $n\lt m$, then each term of the sum in $\text{(3d)}$ will be $0$. Therefore,
$$
\begin{align}
\sum_{k=0}^m(-1)^k(m-k)^n\binom{m}{k}
&=\sum_{k=0}^m(-1)^{m-k}\binom{m}{k}k^n\tag{4a}\\
&=\sum_{k=0}^m(-1)^{m-k}\binom{m}{k}P(0+k)\tag{4b}
\end{align}
$$
Explanation:

$\text{(4a)}$: substitute $k\mapsto m-k$

$\text{(4b)}$: $P(x)=x^n$

$\text{(4b)}$ is the $m^\text{th}$ forward difference of $P(x)=x^n$ at $x=0$, which, by $(3)$ is $0$ if $n\lt m$.


问:
Show that a number belongs or does not belong to a Cantor set
I have trouble showing that $\frac{10}{31}$ and $\frac{45}{69}$ belong/do not belong to the Cantor set. I tried to do the following
$$\sum_{n = x}^\infty \frac{2}{3^n} = \frac{2}{3^x} + \frac{2}{3^{(x + 1)}} + \frac{2}{3^{(x + 2)}} + \dots = \frac{2}{3^x} \left[ 1 + \frac{1}{3^1} + \frac{1}{3^2} + \dots \right] = \frac{2}{3^x} \sum_{k = 0}^\infty \frac{1}{x^k} =  \frac{2}{3^x} \times \frac{1}{1- \frac{1}{3}} = \frac{1}{3^{x- 1}}$$
and then plug different values of $x = 1, 2, \dots$, but I could not seem to get these numbers. Does it mean they do not belong to the Cantor set? Please help if you can. Thank you so much.

答:
Here is the start of a long division base $3$ for the first problem:

```<code>         0.0222010
      ------------
  1011)101.0000000
        20 22
        -----
        10 010
         2 022
         -----
           2110
           2022
           ----
             1100
             1011
             ----
               120
</code>```

At this point it is clear that $\frac{10}{31}$ is not in the Cantor set, since no later digits can change that $1$ into a $2$. One step earlier it wasn’t clear  from the quotient digits alone, since $(0.022201)_3$ could conceivably have continued $$(0.02220122222\ldots)_3\,,$$ which is equal to $(0.022202)_3$, which is in the Cantor set. However, if you realize that the only numbers that have two ternary expansions are rational numbers that can be written with a denominator that is a power of $3$, you can stop as soon as you get that $1$.

What we have at this point says that $(0.022201)_3<\frac{10}{31}<(0.022202)_3$, so it is in one of the deleted open intervals.

The second one is even easier.

Body:
Here is the start of a long division base $3$ for the first problem:

         0.0222010
      ------------
  1011)101.0000000
        20 22
        -----
        10 010
         2 022
         -----
           2110
           2022
           ----
             1100
             1011
             ----
               120

At this point it is clear that $\frac{10}{31}$ is not in the Cantor set, since no later digits can change that $1$ into a $2$. One step earlier it wasn’t clear  from the quotient digits alone, since $(0.022201)_3$ could conceivably have continued $$(0.02220122222\ldots)_3\,,$$ which is equal to $(0.022202)_3$, which is in the Cantor set. However, if you realize that the only numbers that have two ternary expansions are rational numbers that can be written with a denominator that is a power of $3$, you can stop as soon as you get that $1$.

What we have at this point says that $(0.022201)_3<\frac{10}{31}<(0.022202)_3$, so it is in one of the deleted open intervals.

The second one is even easier.


问:
Idempotent endomorphism: show $M=\text{Range}(f) \oplus \text{Range}(\text{id} - f)$
let $f$ be in idempotent endomorphism. Show that $M=\text{Range}(f) \oplus \text{Range}(\text{id} - f)$.

I want to show the intersection of both range spaces is $\{0\}$.

I currently have let $v \in \text{Range}(f) \cap {\text{Range}(\text{id} - f)}$

$f(v) = \text{id}(v)  - f(v)$

$f(v) = v - f(v)$

and I want to show that $v=0$. But I do not know how to progress. any advice is much appreciated!

答:
Suppose $z = f(y) = x - f(x)$. Then we have $f(x + y) = x$. Then we have $f(f(x + y)) = f(x)$. Since $f$ is idempotent, $f(x + y) = f(x)$. Then $f(y) = 0$. Then $z = 0$. $\DeclareMathOperator{id}{id}$ $\DeclareMathOperator{range}{range}$

To get that the union of the ranges is $M$, just note that $x = x - f(x) + f(x)$. Thus, $M = \range(f) \oplus \range(\id - f)$.


问:
Find maximize value of this expression
Let $a;b;c \in [0;2]$

Find maximum value of: $P=ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2)$

After some factoring, I found:

$P=(a-b)(b-c)(a-c)(a+b+c)$

Then square it: $P^2=(a-b)^2(b-c)^2(c-a)^2(a+b+c)^2$

Now $P^2$ be a symmetric expression. But it' still hard. Can someone help, please? Thank you.

答:
Let $P=f(a,b,c)$ and note that for $k\le \min\{2-a,2-b,2-c\}$

$$f(a+k,b+k,c+k)\ge f(a,b,c)$$

Thus, we can choose $a=2$ (WLOG). Now, we need to maximize

\begin{align}f(2,b,c)&=2b(4-b^2)+bc(b^2-c^2)+2c(c^2-4)
\\
&=2b(4-b^2)+c(b^3-8)+c^3(2-b)
\\
&=2b(4-b^2)+c(b-2)(b^2+2b+4-c^2)
\end{align}

Observe that $$b-2\le 0$$

$$b^2+2b+(4-c^2)\ge 0$$

Thus,

$$f(2,b,0)\ge f(2,b,c)$$

Now, we need to maximize

$$f(2,b,0)=2b(4-b^2)$$

Candidates are boundary values and values that make the first order equal to $0$. Checking all values gives us

$$\frac{32}{3\sqrt{3}}=f(2,\frac{2}{\sqrt{3}},0)\ge f(a,b,c)$$

Remember $f$ is symmetric.

Done!

Title: Find maximize value of this expression


Title: Show that a number belongs or does not belong to a Cantor set

Body:
I have trouble showing that $\frac{10}{31}$ and $\frac{45}{69}$ belong/do not belong to the Cantor set. I tried to do the following
$$\sum_{n = x}^\infty \frac{2}{3^n} = \frac{2}{3^x} + \frac{2}{3^{(x + 1)}} + \frac{2}{3^{(x + 2)}} + \dots = \frac{2}{3^x} \left[ 1 + \frac{1}{3^1} + \frac{1}{3^2} + \dots \right] = \frac{2}{3^x} \sum_{k = 0}^\infty \frac{1}{x^k} =  \frac{2}{3^x} \times \frac{1}{1- \frac{1}{3}} = \frac{1}{3^{x- 1}}$$
and then plug different values of $x = 1, 2, \dots$, but I could not seem to get these numbers. Does it mean they do not belong to the Cantor set? Please help if you can. Thank you so much.

Body: Neither one is in the Cantor set. $\dfrac{10}{31} = 0.02220101110012..._{3}$, and $\dfrac{45}{69} = \dfrac{15}{23} = 0.12212110220122..._{3}$. Both has $1$ in their ternary presentation.


答:
As you wrote,
$$ f(x) = g'(f(x)) f'(x)$$
That is, $y = f(x)$ satisfies the differential equation
$$ \dfrac{dy}{dx} = \frac{y}{g'(y)}$$
at least when $g'(y) \ne 0$.
For $x=0$ we have $g(y(0)) = 0$.  Depending on $g$, this may have no solution, or one, or several.    If it happens that $g(0) = 0$, one solution of the integral equation is $f(x) = 0$.

The differential equation is separable, with solutions in implicit form

$$\int \frac{g'(y)}{y}\; dy = x + c$$

Title: How to solve $\int_0^x f(t) dt = g(f(x)) $?

Body:
How to solve an equation of this type :

$$\int_0^x f(t) dt = g(f(x)) $$

for a given function $g$.

Im not so good with differential equations and integral equations.

I know how to solve $h'(x) = j(h(x)) $ in general and some approximations of specific cases of $k'(x) = k(l(x))$. But even my knowledge of those is limited.

However for this $$\int_0^x f(t) dt = g(f(x)) $$ I do not even know how to begin.

It seems very fundamental and simple though so my intuition assumes there must be a way.

I considered differentiating both sides but that got me to

$$ f(x) = g'(f(x)) f'(x) $$

Many thanks !


问:
Average number of order flips in an array of random numbers
I'm looking for something I can cite.  Published work.
Computationally, it seems that the average number of times the order changes in an array of random numbers of size $n$ is $$2\cdot\frac{n-2}{3}.$$  At least this works for $n=2,3,4,5,\dots$.
Thoughts?

答:
This follows immediately from Linearity of Expectation.

Let $\{a_1, \cdots, a_n\}$ denote a permutation, selected uniformly at random.

Each of the positions $i\in \{3, \cdots, n\}$ could represent an order change,  Let $X_i$ denote the indicator variable for that event.  The $E[X_i]=\frac 23$ since exactly $2$ out of the six possible orders of $\{a_{i-2},a_{i-1},a_i\}$ fail to change order.  But then $$E=E\left[\sum X_i\right] = \sum E[X_i]=\frac 23\times (n-2)$$ and we are done.


问:
Prove $\alpha(x,y) = (x(x^2+y^2), y(x^2 + y^2), x^2 + y^2)$ has a continuous inverse
Let $\alpha : \mathbb{R}^2 \to \mathbb{R}^3$ defined by $\alpha(x,y) = (x(x^2+y^2), y(x^2 + y^2), x^2 + y^2)$.  How do I show that $\alpha$ as a map from $\mathbb{R}^2$ to its image, it has a continuous inverse - ie. $\alpha^{-1}: \alpha(\mathbb{R}^2) \to \mathbb{R}^2$ is continuous?

This was given as an example in Munkres' Analysis on Manifolds, regarding a function that was almost a coordinate patch (it fails to be due to the rank of the matrix $D\alpha (x)$ failing to be of rank 2 at 0), but I can't seem to prove rigourously why the inverse should in fact be continuous.

答:
Yes, your solution is fine. Just to finish this off slightly differently, note that $M$ is a (topological) $2$-dimensional manifold parametrized by $(a,b)$. I.e., $M= \{(a,b,c): c=(a^2+b^2)^{1/3}\}$. Now, for $(a,b,c)\in M$ with $c\ne 0$,
$$\alpha^{-1}(a,b,c) = \left(\frac ac,\frac bc\right) = \frac1{(a^2+b^2)^{1/3}}(a,b).$$
Note that
$$\left\|\frac{(a,b)}{(a^2+b^2)^{1/3}}\right\| = \frac{(a^2+b^2)^{1/2}}{(a^2+b^2)^{1/3}} = (a^2+b^2)^{1/6},$$
whose limit as $(a,b)\to (0,0)$ is clearly $0$. (Yes, if $\|(a,b)\|<\delta$, then $\|\alpha^{-1}(a,b,c)\|<\delta^{1/3}$, as you said.)


问:
Ways to number a row of trees
On the median strip of an infinite road there are n trees in a row. We want to mark all of them with the three numbers 1, 2 and 3, in such a way that all trees have one number and no two adjacent trees can have the same number. In how many ways is this possible, if one of the numbers (say, 3) must be used at least twice?

I tried to find a recursive relation by starting from the small numbers:
For $n=4$ I found that only 4 numberings are possible.
For $n=5$, I got 34 and for $n=6$, 88. I think the general formula for all possible numberings without the exclusion of the ones which have only 1 or 2 3's, is $3*2 ^{n-1}$ but I don't know how to calculate the invalid numberings.

Any help?

答: Hint: disregarding the requirement for $3$ to be used at least twice, there would be $3 \cdot 2^{n-1}$ ways.  Now subtract the number of ways with $0$ or $1$ $3$'s.  Each configuration with no $3$'s consists of alternating $1$'s and $2$'s.  Each configuration with one $3$ consists either of a $3$ followed by alternating $1$'s and $2$'s, or alternating $1$'s and $2$'s followed by a $3$ and then (if there are some trees left) alternating $1$'s and $2$'s.

Title: Ways to number a row of trees


问:
Sum=Integral? Find $m$ so that $\sum _{n=1}^{\infty } \frac{n^m}{e^{2 \pi n}-1}=\int_0^{\infty } \frac{x^m}{e^{2 \pi x}-1} \, dx$
This question asked for a derivation of the interesting relation

$$\sum_{n=1}^{\infty}  \frac{n^{13}}{e^{2 \pi n}-1} = \frac{1}{24}$$

In an attempt to answer this problem I played around a bit and was surprised that the related integral

$$\int_{0}^{\infty}  \frac{n^{13}}{e^{2 \pi n}-1} \,dn= \frac{1}{24}$$

has the same nice result. Notice that the sum starts at $n=1$ and the integral starts at its "natural" boundary $0$.

Defining more generally

$$S(m) = \sum_{n=1}^{\infty}  \frac{n^m}{e^{2 \pi n}-1}\tag{1a}$$
$$i(m) = \int_{0}^{\infty}  \frac{n^m}{e^{2 \pi n}-1} \,dn\tag{1b}$$

We can ask for which $m$ sum and integral lead to equal values, i.e.

$$S(m) = i(m)\tag{2}$$

It turns out that equality seems to happen if and only if $m=4k+1, k=1,2,3...$.

This picture shows a discrete plot of the quotient $\frac{S(m)}{i(m)}$ of sum and integral.

Questions:

(1) Find all natural solutions $m$ of equation $(2)$

(2) Find an analytic continuation of $S(m)$ to real $m$ and study if the quotient $S(m)/i(m)$ is unity for other values

Hints (for derivations see):

a) the integral has a closed expression which is not restricted to natural values of $m$

$$i(m)= \frac{1}{(2 \pi )^{m+1} } \zeta (m+1) \Gamma (m+1)\tag{3}$$

Here $\zeta(x)$ is the Riemann zeta function.

b) the sum has a closed expression for $m=1,2,3,...$ as well

$$S(m)=\frac{(-1)^{m+1} \psi _{e^{-2 \pi }}^{(m)}(1)}{(2 \pi )^{m+1}}\tag{4}$$

Here $\psi _{q}^{(m)}(x)$ is the $m$-th derivative of the $q$-digamma function with respect to x.

EDIT: after a comment of @Jean Marie I saw that Marko Riedel has derived as a special case of $(4)$ for $m=4k+1$ the simplified formula

$$S(4 k+1)=\frac{B_{4 k+2}}{2 (4 k+2)}, k=1,2,3,...\tag{4a}$$

Here $B_{n}$ is the Bernoulli number.

答:
The standard method is to use that $$\frac{\pi^2}{\sin^2(\pi z)}-\sum_m \frac1{(z+m)^2}=0$$ (it is a $1$-periodic entire function which is bounded on $\Re(z)\in [0,1]$ and which vanishes at $i\infty$)

to get that for $k\ge 4$ even and $\Im(z)>0$ $$G_k(z)=\sum_{(n,m)\ne (0,0)} \frac1{(zn+m)^k}= 2\zeta(k)+\frac{4i\pi}{(k-1)!}\sum_{n\ge 1}\sum_{d\ge 1}(2i\pi d)^{k-1}e^{2i\pi dnz}$$

When $k\equiv 2\bmod 4$ we get that $G_k(i)=0$.

On the other hand $$\int_0^\infty \frac{x^{k-1}}{e^{2\pi x}-1}dx=(k-1)!(2\pi)^{-k} \zeta(k),\qquad\sum_{d\ge 1}\frac{d^{k-1}}{e^{2\pi d}-1}=\sum_{d\ge 1}\sum_{n\ge 1}d^{k-1}e^{-2\pi dn}$$

The case $k\equiv 0\bmod 4$ has a solution too. From the theory of modular forms (mainly that $E_4(z)^3-E_6(z)^2$ is a non-vanishing cusp form),  with $E_k(z)=\frac{G_k(z)}{2\zeta(k)}$: $$E_k(z)=\sum_{4a+6b=k} c_{k,a,b}E_4(z)^a E_6(z)^b$$ for some rational coefficients $c_{k,a,b}$. As $E_6(i)=0$ we get $$E_k(i) = c_{k,k/4,0}E_4(i)^{k/4}$$ The known closed-form of $E_4(i)$ in term of $\Gamma(1/4)$ implies that $E_k(i)$ is irrational, so it is $\ne 2$ and hence $\int_0^\infty \frac{x^{k-1}}{e^{2\pi x}-1}dx\ne \sum_{d\ge 1}\frac{d^{k-1}}{e^{2\pi d}-1}$.


答:
Yes your intuition is correct. Think of having the permutation $p = (2, 1, 3, 5, 4)$ of the elements $(1, 2, 3, 4, 5)$. There are two changes of order. If you plot your permutation (like in the image) and the indices of the permutation you will see that the changes of order are at "peaks" or "bottoms" of the plot.

The answer lies to finding a way to compute the expected number of these "bottoms" and "peaks" (excluding the first and the last element of the permutation).

There are $n-2$ spots where a "peak" or "bottom" can occur.

A "peak" occurs at a given index $i$ of the permutation $p$ with probability $1/3$. This is because the probability of $max(p(i-1), p(i), p(i+1)) = p(i)$ is $1/3$.

Using the same argument, one can see that a "bottom" occurs at index $i$ with probability $1/3$.

A "peak" can't be a "bottom" at the same time (and the opposite), so the expected number of "peaks" and "bottoms" is:

$(n-2)*(\frac{1}{3}+\frac{1}{3}) = 2\frac{n-2}{3}$

Title: Average number of order flips in an array of random numbers

Body:
I'm looking for something I can cite.  Published work.
Computationally, it seems that the average number of times the order changes in an array of random numbers of size $n$ is $$2\cdot\frac{n-2}{3}.$$  At least this works for $n=2,3,4,5,\dots$.
Thoughts?


答:
I found two ways of making an explicit continuous periodic function.

$$f\left(x\right)=\frac{2\arcsin{\left(\sin{\left(\frac{x\pi}{2}\right)}\right)}}{\pi}$$

Which works fine for programs like Desmos, despite the $\arcsin(x)$ and $\sin(x)$ cancelling out in general math. This still works because $\arcsin{x}$ is the principle branch. If you need something that doesn't require using a principle branch, I suggest something like this:

$$f\left(x\right)=-\mathrm{sign}\left((x-1)\bmod 4\right)\left(2-\left((x-1)\bmod 4\right)\right)-1$$

This one works works without the need of a principle branch.

Title: Explicit expression for a periodic function with repeated lines.

Body:
I know there HAS to be a better way to describe the function I'm looking for, but I'm looking for an explicit form for the following function with domain $(0,4)$:

$$f(x)=  \left\{
\begin{array}{ll}
      x & 0\leq x\leq 1 \\
      2-x & 1< x\leq 3 \\
x-4 & 3<x\leq 4
\end{array}
\right. $$
Tile this function with period $4$ over $(0,4),(4,8),...$

Essentially, it's just a spiky wave function. I would like an explicit form of this including floor functions, maybe something of the form $\max \{f(x),g(x)\},$ I have no idea. How would I do this?


问:
Closed form of $\sum_{m=1}^{\infty} \frac{(-1)^mH_{\frac{3m}{4}}}{3m}$
I've been working on an integral, namely: $${\displaystyle \int_0^1 \frac{x^2}{1 + x^3}\ln(1 - x^4)dx}$$

Which I managed to narrow down to the following expression: $$\sum_{m=1}^{\infty} \frac{(-1)^mH_{\frac{3m}{4}}}{3m}$$

Where $H_n$ is the n-th harmonic number. I managed to get here after converting the integrand into a double summation, recognizing the digamma function hidden inside, and summing back to a harmonic number, but couldn't get past that.

Is there another way to solve the integral, or to solve the summation? I've tried breaking the sum down to its parts, but the best I could do was to find a few terms and was left with
$$\sum_{m=1}^{\infty} \frac{(-1)^m\psi(\frac{3m}{4})}{3m}$$
Where $\psi(x)$ is the digamma function, and still couldn't solve that.

答:
Let $\mathcal{I}$ denote the value of the following logarithmic integral:

$$\mathcal{I}:=\int_{0}^{1}\mathrm{d}x\,\frac{3x^{2}}{1+x^{3}}\ln{\left(1-x^{4}\right)}\approx-0.47159.$$

It will be shown that

$$\mathcal{I}=-\frac12\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{5\pi^{2}}{144}+\frac34\ln^{2}{\left(2\right)}.$$

Making use of the factorizations $\left(1+x^{3}\right)=\left(1+x\right)\left(1-x+x^{2}\right)$ and $\left(1-x^{4}\right)=\left(1-x\right)\left(1+x\right)\left(1+x^{2}\right)$, we can split up the integral into simpler components as

$$\begin{align}
\mathcal{I}
&=\int_{0}^{1}\mathrm{d}x\,\frac{3x^{2}}{1+x^{3}}\ln{\left(1-x^{4}\right)}\\
&=\int_{0}^{1}\mathrm{d}x\,\left[\frac{1}{1+x}+\frac{2x-1}{1-x+x^{2}}\right]\left[\ln{\left(1-x\right)}+\ln{\left(1+x\right)}+\ln{\left(1+x^{2}\right)}\right]\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{1}{1+x}\left[\ln{\left(1-x\right)}+\ln{\left(1+x\right)}+\ln{\left(1+x^{2}\right)}\right]\\
&~~~~~+\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\left[\ln{\left(1-x\right)}+\ln{\left(1+x\right)}+\ln{\left(1+x^{2}\right)}\right]\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-x\right)}}{1+x}+\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+x\right)}}{1+x}+\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+x^{2}\right)}}{1+x}\\
&~~~~~+\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1-x\right)}+\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1+x\right)}\\
&~~~~~+\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1+x^{2}\right)}\\
&=:\mathcal{I}_{1}+\mathcal{I}_{2}+\mathcal{I}_{3}+\mathcal{I}_{4}+\mathcal{I}_{5}+\mathcal{I}_{6}.\\
\end{align}$$

Let's focus on the last integral first.

Using the derivatives

$$\frac{d}{dx}\ln{\left(1-x+x^{2}\right)}=\frac{2x-1}{1-x+x^{2}},$$

and

$$\frac{d}{dx}\frac{\ln^{2}{\left(1-x+x^{2}\right)}}{2}=\frac{2x-1}{1-x+x^{2}}\ln{\left(1-x+x^{2}\right)},$$

we show that the following integral vanishes identically:

$$\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1-x+x^{2}\right)}=\frac{\ln^{2}{\left(1-x+x^{2}\right)}}{2}\bigg{|}_{0}^{1}=0.$$

Then, $\mathcal{I}_{6}$ can be rewritten in the following way:

$$\begin{align}
\mathcal{I}_{6}
&=\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1+x^{2}\right)}\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1+x^{2}\right)}-\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1-x+x^{2}\right)}\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\left[\ln{\left(1+x^{2}\right)}-\ln{\left(1-x+x^{2}\right)}\right]\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(\frac{1+x^{2}}{1-x+x^{2}}\right)}\\
&=\int_{1}^{0}\mathrm{d}y\,\frac{\left(-2\right)}{\left(1+y\right)^{2}}\cdot\frac{\left(1+y\right)\left(1-3y\right)}{\left(1+3y^{2}\right)}\ln{\left(\frac{2+2y^{2}}{1+3y^{2}}\right)};~~~\small{\left[x=\frac{1-y}{1+y}\right]}\\
&=\int_{0}^{1}\mathrm{d}y\,\frac{2}{\left(1+y\right)}\cdot\frac{\left(1-3y\right)}{\left(1+3y^{2}\right)}\ln{\left(\frac{2+2y^{2}}{1+3y^{2}}\right)}\\
&=\int_{0}^{1}\mathrm{d}y\,\left[\frac{2}{\left(1+y\right)}-\frac{6y}{\left(1+3y^{2}\right)}\right]\ln{\left(\frac{2+2y^{2}}{1+3y^{2}}\right)}\\
&=\int_{0}^{1}\mathrm{d}y\,\frac{2}{\left(1+y\right)}\ln{\left(\frac{2+2y^{2}}{1+3y^{2}}\right)}-\int_{0}^{1}\mathrm{d}y\,\frac{6y}{\left(1+3y^{2}\right)}\ln{\left(\frac{2+2y^{2}}{1+3y^{2}}\right)}\\
&=\int_{0}^{1}\mathrm{d}y\,\frac{2}{\left(1+y\right)}\left[\ln{\left(2\right)}+\ln{\left(1+y^{2}\right)}-\ln{\left(1+3y^{2}\right)}\right]\\
&~~~~~-\int_{0}^{1}\mathrm{d}t\,\frac{3}{\left(1+3t\right)}\ln{\left(\frac{2+2t}{1+3t}\right)};~~~\small{\left[y^{2}=t\right]}\\
&=2\ln{\left(2\right)}\int_{0}^{1}\mathrm{d}y\,\frac{1}{\left(1+y\right)}+2\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y^{2}\right)}}{\left(1+y\right)}-2\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+3y^{2}\right)}}{\left(1+y\right)}\\
&~~~~~-\int_{0}^{1}\mathrm{d}u\,\frac{3}{\left(1+3u\right)}\ln{\left(1+u\right)};~~~\small{\left[t=\frac{1-u}{1+3u}\right]}\\
&=2\ln^{2}{\left(2\right)}+2\mathcal{I}_{3}-2\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+3y^{2}\right)}}{\left(1+y\right)}\\
&~~~~~+\operatorname{Li}_{2}{\left(\frac34\right)}-\operatorname{Li}_{2}{\left(\frac12\right)}+\operatorname{Li}_{2}{\left(-1\right)}\\
&=\operatorname{Li}_{2}{\left(\frac34\right)}-\operatorname{Li}_{2}{\left(\frac12\right)}+\operatorname{Li}_{2}{\left(-1\right)}+2\ln^{2}{\left(2\right)}+2\mathcal{I}_{3}\\
&~~~~~-2\int_{1}^{2}\mathrm{d}t\,\frac{\ln{\left(1+3(t-1)^{2}\right)}}{t};~~~\small{\left[1+y=t\right]}\\
&=\operatorname{Li}_{2}{\left(\frac34\right)}-\operatorname{Li}_{2}{\left(\frac12\right)}+\operatorname{Li}_{2}{\left(-1\right)}+2\ln^{2}{\left(2\right)}+2\mathcal{I}_{3}\\
&~~~~~-2\int_{1}^{2}\mathrm{d}t\,\frac{\ln{\left(3t^{2}-6t+4\right)}}{t}\\
&=\operatorname{Li}_{2}{\left(\frac34\right)}-\operatorname{Li}_{2}{\left(\frac12\right)}+\operatorname{Li}_{2}{\left(-1\right)}+2\ln^{2}{\left(2\right)}+2\mathcal{I}_{3}\\
&~~~~~-2\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}u\,\frac{\ln{\left(4u^{2}-4\sqrt{3}\,u+4\right)}}{u};~~~\small{\left[t=\frac{2u}{\sqrt{3}}\right]}\\
&=\operatorname{Li}_{2}{\left(\frac34\right)}-\operatorname{Li}_{2}{\left(\frac12\right)}+\operatorname{Li}_{2}{\left(-1\right)}+2\ln^{2}{\left(2\right)}+2\mathcal{I}_{3}\\
&~~~~~-2\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}u\,\frac{\ln{\left(4\right)}}{u}-2\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}u\,\frac{\ln{\left(u^{2}-\sqrt{3}\,u+1\right)}}{u}\\
&=2\mathcal{I}_{3}-\frac{\pi^{2}}{6}-\frac32\ln^{2}{\left(2\right)}+\operatorname{Li}_{2}{\left(\frac34\right)}-2\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}.\\
\end{align}$$

The integral $\mathcal{I}_{3}$ can in turn be reduced to

$$\begin{align}
\mathcal{I}_{3}
&=\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+x^{2}\right)}}{1+x}\\
&=\int_{1}^{2}\mathrm{d}y\,\frac{\ln{\left(1+(y-1)^{2}\right)}}{y};~~~\small{\left[1+x=y\right]}\\
&=\int_{1}^{2}\mathrm{d}y\,\frac{\ln{\left(2-2y+y^{2}\right)}}{y}\\
&=\int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}\mathrm{d}t\,\frac{\ln{\left(2-2t\sqrt{2}+2t^{2}\right)}}{t};~~~\small{\left[y=t\sqrt{2}\right]}\\
&=\int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}\mathrm{d}t\,\frac{\ln{\left(2\right)}}{t}+\int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}\mathrm{d}t\,\frac{\ln{\left(1-\sqrt{2}\,t+t^{2}\right)}}{t}\\
&=\ln^{2}{\left(2\right)}+\int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{2}\,x+x^{2}\right)}}{x}.\\
\end{align}$$

Next up, $\mathcal{I}_{5}$ can be reduced to a similar expression involving the same undetermined integral in the expression for $\mathcal{I}_{6}$:

$$\begin{align}
\mathcal{I}_{5}
&=\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1+x\right)}\\
&=-\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-x+x^{2}\right)}}{1+x};~~~\small{I.B.P.s}\\
&=-\int_{1}^{2}\mathrm{d}y\,\frac{\ln{\left(3-3y+y^{2}\right)}}{y};~~~\small{\left[x=y-1\right]}\\
&=-\int_{\frac{1}{\sqrt{3}}}^{\frac{2}{\sqrt{3}}}\mathrm{d}t\,\frac{\ln{\left(3-3t\sqrt{3}+3t^{2}\right)}}{t};~~~\small{\left[y=t\sqrt{3}\right]}\\
&=-\int_{\frac{1}{\sqrt{3}}}^{\frac{2}{\sqrt{3}}}\mathrm{d}t\,\frac{\ln{\left(3\right)}}{t}-\int_{\frac{1}{\sqrt{3}}}^{\frac{2}{\sqrt{3}}}\mathrm{d}t\,\frac{\ln{\left(1-\sqrt{3}\,t+t^{2}\right)}}{t}\\
&=-\ln{\left(2\right)}\ln{\left(3\right)}-\int_{\frac{1}{\sqrt{3}}}^{\frac{2}{\sqrt{3}}}\mathrm{d}t\,\frac{\ln{\left(1-\sqrt{3}\,t+t^{2}\right)}}{t}\\
&=-\ln{\left(2\right)}\ln{\left(3\right)}-\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}u\,\frac{\ln{\left(\frac{1-\sqrt{3}\,u+u^{2}}{u^{2}}\right)}}{u};~~~\small{\left[t=u^{-1}\right]}\\
&=-\ln{\left(2\right)}\ln{\left(3\right)}+\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}u\,\frac{2\ln{\left(u\right)}-\ln{\left(1-\sqrt{3}\,u+u^{2}\right)}}{u}\\
&=-\ln{\left(2\right)}\ln{\left(3\right)}+\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}u\,\frac{2\ln{\left(u\right)}}{u}-\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}u\,\frac{\ln{\left(1-\sqrt{3}\,u+u^{2}\right)}}{u}\\
&=-\ln^{2}{\left(2\right)}-\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}.\\
\end{align}$$

The remaining three integrals $\mathcal{I}_{1,2,4}$ each have nice exact values in terms of well-known constants: we have

$$\mathcal{I}_{1}=\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-x\right)}}{1+x}=-\operatorname{Li}_{2}{\left(\frac12\right)}=\frac12\ln^{2}{\left(2\right)}-\frac{\pi^{2}}{12},$$

$$\mathcal{I}_{2}=\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+x\right)}}{1+x}=\frac12\ln^{2}{\left(2\right)},$$

and

$$\begin{align}
\mathcal{I}_{4}
&=\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1-x\right)}\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-x+x^{2}\right)}}{1-x};~~~\small{I.B.P.s}\\
&=\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1-y+y^{2}\right)}}{y};~~~\small{\left[x=1-y\right]}\\
&=\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y^{3}\right)}}{y}-\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}}{y}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1+t\right)}}{3t};~~~\small{\left[y^{3}=t\right]}\\
&~~~~~-\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}}{y}\\
&=-\frac23\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1+t\right)}}{t}\\
&=\frac23\operatorname{Li}_{2}{\left(-1\right)}\\
&=-\frac{\pi^{2}}{18}.\\
\end{align}$$

Then,

$$\begin{align}
\mathcal{I}
&=\mathcal{I}_{1}+\mathcal{I}_{2}+\mathcal{I}_{3}+\mathcal{I}_{4}+\mathcal{I}_{5}+\mathcal{I}_{6}\\
&=\left[\frac12\ln^{2}{\left(2\right)}-\frac{\pi^{2}}{12}\right]+\frac12\ln^{2}{\left(2\right)}+\mathcal{I}_{3}-\frac{\pi^{2}}{18}\\
&~~~~~-\ln^{2}{\left(2\right)}-\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}\\
&~~~~~+2\mathcal{I}_{3}-\frac{\pi^{2}}{6}-\frac32\ln^{2}{\left(2\right)}+\operatorname{Li}_{2}{\left(\frac34\right)}-2\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}\\
&=\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{11\pi^{2}}{36}-\frac32\ln^{2}{\left(2\right)}+3\mathcal{I}_{3}\\
&~~~~~-3\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}\\
&=\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{11\pi^{2}}{36}-\frac32\ln^{2}{\left(2\right)}\\
&~~~~~+3\left[\ln^{2}{\left(2\right)}+\int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{2}\,x+x^{2}\right)}}{x}\right]\\
&~~~~~-3\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}\\
&=\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{11\pi^{2}}{36}+\frac32\ln^{2}{\left(2\right)}\\
&~~~~~+3\int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{2}\,x+x^{2}\right)}}{x}-3\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}.\\
\end{align}$$

At this point, it will be helpful to introduce a two-variable variant of the dilogarithm function:

$$\operatorname{Li}_{2}{\left(r,\theta\right)}:=-\frac12\int_{0}^{r}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\theta\right)}+x^{2}\right)}}{x};~~~\small{\left(r,\theta\right)\in\mathbb{R}^{2}}.$$

This function can be shown to satisfy the following pair of formulas useful to the problem at hand:

$$\operatorname{Li}_{2}{\left(\cos{\left(\theta\right)},\theta\right)}=\frac12\left(\frac{\pi}{2}-\theta\right)^{2}+\frac14\operatorname{Li}_{2}{\left(\cos^{2}{\left(\theta\right)}\right)};~~~\small{\theta\in\left[0,\pi\right]},$$

$$\operatorname{Li}_{2}{\left(2\cos{\left(\theta\right)},\theta\right)}=\left(\frac{\pi}{2}-\theta\right)^{2};~~~\small{\theta\in\left[0,\pi\right]}.$$

Then, for $\theta\in\left(0,\frac{\pi}{2}\right)$, we have

$$\begin{align}
\int_{\cos{\left(\theta\right)}}^{2\cos{\left(\theta\right)}}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\theta\right)}+x^{2}\right)}}{x}
&=\int_{0}^{2\cos{\left(\theta\right)}}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\theta\right)}+x^{2}\right)}}{x}\\
&~~~~~-\int_{0}^{\cos{\left(\theta\right)}}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\theta\right)}+x^{2}\right)}}{x}\\
&=-2\operatorname{Li}_{2}{\left(2\cos{\left(\theta\right)},\theta\right)}+2\operatorname{Li}_{2}{\left(\cos{\left(\theta\right)},\theta\right)}\\
&=-2\left(\frac{\pi}{2}-\theta\right)^{2}+2\left[\frac12\left(\frac{\pi}{2}-\theta\right)^{2}+\frac14\operatorname{Li}_{2}{\left(\cos^{2}{\left(\theta\right)}\right)}\right]\\
&=\frac12\operatorname{Li}_{2}{\left(\cos^{2}{\left(\theta\right)}\right)}-\left(\frac{\pi}{2}-\theta\right)^{2}.\\
\end{align}$$

We can now complete our calculation for $\mathcal{I}$:

$$\begin{align}
\mathcal{I}
&=\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{11\pi^{2}}{36}+\frac32\ln^{2}{\left(2\right)}\\
&~~~~~+3\int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{2}\,x+x^{2}\right)}}{x}-3\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}\\
&=\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{11\pi^{2}}{36}+\frac32\ln^{2}{\left(2\right)}\\
&~~~~~+3\int_{\cos{\left(\frac{\pi}{4}\right)}}^{2\cos{\left(\frac{\pi}{4}\right)}}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\frac{\pi}{4}\right)}+x^{2}\right)}}{x}\\
&~~~~~-3\int_{\cos{\left(\frac{\pi}{6}\right)}}^{2\cos{\left(\frac{\pi}{6}\right)}}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\frac{\pi}{6}\right)}+x^{2}\right)}}{x}\\
&=\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{11\pi^{2}}{36}+\frac32\ln^{2}{\left(2\right)}\\
&~~~~~+3\left[\frac12\operatorname{Li}_{2}{\left(\cos^{2}{\left(\frac{\pi}{4}\right)}\right)}-\left(\frac{\pi}{2}-\frac{\pi}{4}\right)^{2}\right]\\
&~~~~~-3\left[\frac12\operatorname{Li}_{2}{\left(\cos^{2}{\left(\frac{\pi}{6}\right)}\right)}-\left(\frac{\pi}{2}-\frac{\pi}{6}\right)^{2}\right]\\
&=\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{11\pi^{2}}{36}+\frac32\ln^{2}{\left(2\right)}\\
&~~~~~+\frac32\operatorname{Li}_{2}{\left(\frac12\right)}-\frac{3\pi^{2}}{16}\\
&~~~~~-\frac32\operatorname{Li}_{2}{\left(\frac34\right)}+\frac{\pi^{2}}{3}\\
&=-\frac12\operatorname{Li}_{2}{\left(\frac34\right)}+\frac{\pi^{2}}{36}+\frac32\ln^{2}{\left(2\right)}\\
&~~~~~+\frac32\left[\frac{\pi^{2}}{12}-\frac12\ln^{2}{\left(2\right)}\right]-\frac{3\pi^{2}}{16}\\
&=-\frac12\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{5\pi^{2}}{144}+\frac34\ln^{2}{\left(2\right)}.\blacksquare\\
\end{align}$$


问:
Let $G=\Bbb Z\times\Bbb Z$ be additive and $H=\{0\}\times\Bbb Z$. Show that: a) $H\le G$. b) $H\cong \Bbb Z$.
I did more or less the item

a)
$H$ is a subset of $G$ so we need to show that $H$ is a group.

$H$ is associative because

$$\begin{align}
[(a, b) + (c, d)] + (e, f) &= (a + c, b + d) + (e, f) \\
&= (a + c + e, b + d + f) \\
&= (a, b) + [(c + e, d + f)] \\
&= (a, b) + [(c, d) + (e, f)].
\end{align}$$

$H$ has a neutral element because
$$(0,0) + (a, b) = (a, b)$$

Does $H$ have an inverse? (Not because $H$ is of the form $(0, x)$ where $x\in\Bbb Z$
$$(-a, -b) + (a, b) = (0.0)$$

Thank you in advance for any help.

答:
For a)

I will use the one-step subgroup test.

Since $(0,0)\in H$, we have $H\neq \varnothing$.

By definition of $H$, we have $H\subseteq G$.

Let $x,y\in H$. Then $x=(0,a), y=(0,b)$ for some $a,b\in\Bbb Z$. Now

$$\begin{align}
x-y&=(0,a)-(0,b)\\
&=(0,a)+(0,-b)\\
&=(0,a-b).
\end{align}$$

But $a-b\in\Bbb Z$, so $x-y\in H$.

Hence $H\le G$.

For b), consider

$$\begin{align}
\varphi: H&\to \Bbb Z\\
(0,h)&\mapsto h.
\end{align}$$


问:
Show that two $R$-algebras are not isomorphic.

Consider the $\mathbb{R}$-algebras $A := \mathbb{R}\left[a,b,c,d\right]\left[\frac{1}{a^2+b^2+c^2+d^2}\right]$ and $B := \mathbb{R}\left[a,b,c,d\right]\left[\frac{1}{ad-bc}\right]$ (so we are adjoining the inverses of $a^2 + b^2 + c^2 + d^2$ and $ad-bc$, respectively).
I claim that these are not isomorphic as $\mathbb{R}$-algebras, but I'm not sure how to formalize this.

My intuition says that this is because $$a^2 + b^2 + c^2 + d^2 = 0 \iff a=b=c=d=0,$$ while $ad-bc=0$ has nonzero solutions. I tried to formalize this using evaluation homomorphisms, but to no avail. I also tried assuming they were isomorphic and obtaining a contradiction, but haven't been able to.

Does anyone have any pointers on how to approach this?

答:
I don't know a purely algebraic proof. The following proof involves a little algebraic geometry and algebraic topology, so it may not be the intended answer to your question. In that case, please specify your context (i.e. where did you get the problem from).

Suppose $A$ and $B$ are isomorphic as $\Bbb R$-algebras. Then their corresponding $\Bbb R$-points, viewed as real manifolds, are isomorphic, as polynomial maps are continuous (and even $C^\infty$).

However, these two manifolds are: $A(\Bbb R) = \Bbb R^4 \backslash \{0\}$ and $B(\Bbb R) = \operatorname{GL}(2, \Bbb R)$. It is obvious that they are not isomorphic as topological spaces, by e.g. comparing their simplicial (co)homology groups.


问:
Circle of infinite radius
I've read a lot of content saying a circle of infinite radius is a line
And likewise that a plane has no boundary points

Now if I put the problem as such
Let $D(C,R)$ be the disk (open or closed) of center C and radius R

It's boundary is the circle of same center and radius

Now what is the boundary of this same disk when $R \rightarrow \infty$ ??

A disk of infinite radius is the whole plane and I think it's bringing me back to the original problem

Any explanation would be appreciated
Thanks.

答: To get a line, you need both $R \to \infty$ and $C$ goes to $\infty$.  For example, in $\mathbb R^2$, take $R > 0$ and $C = (R,0)$.  Then as $R \to \infty$, the boundary of $D(C,R)$ approaches the $y$ axis.

Title: Circle of infinite radius


答: $H $ is a subset and a group  (easy).  $\varphi:\Bbb Z\to\{0\}\times\Bbb Z $ by $x\mapsto (0,x) $ is an isomorphism.

Title: Let $G=\Bbb Z\times\Bbb Z$ be additive and $H=\{0\}\times\Bbb Z$. Show that: a) $H\le G$. b) $H\cong \Bbb Z$.

Body:
I did more or less the item

a)
$H$ is a subset of $G$ so we need to show that $H$ is a group.

$H$ is associative because

$$\begin{align}
[(a, b) + (c, d)] + (e, f) &= (a + c, b + d) + (e, f) \\
&= (a + c + e, b + d + f) \\
&= (a, b) + [(c + e, d + f)] \\
&= (a, b) + [(c, d) + (e, f)].
\end{align}$$

$H$ has a neutral element because
$$(0,0) + (a, b) = (a, b)$$

Does $H$ have an inverse? (Not because $H$ is of the form $(0, x)$ where $x\in\Bbb Z$
$$(-a, -b) + (a, b) = (0.0)$$

Thank you in advance for any help.


问:
Does every manifold with boundary admit a flat metric?
There are several obstructions for a closed manifold $M$ admit a flat metric. for example Cartan-Hadamard Theorem implies the universal cover of $M$ must be $\mathbb R^n$. So if the manifold $M$ is allowed to have nonempty boundary and we do not put any restrictions on the boundary(say we do not ask it to be convex). Then can we say something?

答:
You should explain what you mean by a flat metric is there is nonempty boundary. Whatever definition you have in mind, it should imply the existence of a flat metric on the interior of your manifold. You should also assume that the manifold is connected. For connected oriented manifolds of dimension $n\le 3$ there are no obstructions since every such manifold admits an immersion in $R^n$.

However, starting with dimension $n\ge 4$ there are obstructions. The simplest one is the following. Suppose that your manifold $M$ is also simply-connected. Then the existence of a flat metric on the interior of $M$ implies that $M$ admits an (isometric) immersion in $R^n$. This, in turn, implies that $M$ is parallelizable. (Actually, an open $n$-manifold is parallelizable if and only if it admits an immersion in $R^n$.) Now, take your favorite simply-connected non-parallelizable $n$-manifold with nonempty boundary, for instance, let $M$ be the complement to an open ball in $CP^2$. (It is not parallelizable because $CP^2$ is not a spin-manifold.) The interior of this manifold will not have a flat metric.


问:
$L^p(\mu)$ space relations
Let $(X,M, \mu)$ be a measure space. Suppose $1 \leq p < r \leq \infty$.

Suppose $f \in L^p (\mu) \cap L^r(\mu)$. Show that $f\in L^s(\mu)$ for $p \leq s \leq r$.

Attempt:

Suppose $f \in L^p (\mu) \cap L^r(\mu)$. Then $\int |f| ^p d\mu < \infty$ and $\int |f|^r d\mu < \infty$.

Isn't it obvious that $\int |f|^s d\mu < \infty$ already?

I'm bit confused on what to show.

Any help will be appreciated!

Thank you for your help!

答:
It seems obvious and it is to some degree but it requires a small amount of work to get there. Kavi's comment almost completely gives it away. If $|f(x)| \le 1$, then $|f(x)|^s \le |f(x)|^p$ since $p < s$. Similarly if $|f(x)| > 1$, then $|f(x)|^s \le |f(x)|^r$ since $s < r$. Can you put these together to show that $\int |f(x)|^s \,d\mu(x) < \infty$?

This is called ($L^p$) interpolation for what it's worth and is a very useful technique in functional analysis.

Title: $L^p(\mu)$ space relations


问:
Deriving the centripetal acceleration in circular motion
I tried to make sure I understand uniform circular motion by deriving the formula for centripetal acceleration, $a_c = v^2/r$, rigorously from first principles. I think I did it correctly, but it seems quite complicated - can you show me better ways?

Given a particle moving uniformly in a circule of radius $r$ around $O$ with period $T$, I take it as given that it has tangential velocity at each point with the same constant magnitude. So let's say its velocity at the topmost point, broken down into Cartesian components, is $v_0=(v,0)$. Now some small amount of time $\Delta t$ passes, and the particle rotates by the angle $\alpha = \frac{2\pi}{T} \Delta t$. The magnitude of velocity remains $v$, and it's now pointed in the new tangential direction, so it's easy to see its components as $v_1=(v\cos\alpha, v\sin\alpha)$.

Now acceleration is the limit of $\frac{v_1-v_0}{\Delta t}$ when $\Delta t \to 0$. I just want the magnitude of acceleration which will be

$$\lim_{\Delta t \to 0}\frac{\sqrt{(v\cos\alpha - v)^2 + (v\sin\alpha)^2}}{\Delta t}$$

I can simplify this, by passing to $\alpha$ as the dependent variable ($\Delta t=\frac{T}{2\pi}\alpha$), to

$$v \frac{2\pi}{T} \lim_{\alpha \to 0}\frac{\sqrt{2-2\cos\alpha}}{\alpha}        $$

To evaluate the limit, I bring the denominator under the square sign and break down $\cos\alpha$ by its Taylor series:

$$\lim_{\alpha \to 0}\sqrt{\frac{2-2\cos\alpha}{\alpha^2}} = \lim_{\alpha \to 0}\sqrt{\frac{2-2(1-\frac{\alpha^2}{2} + \frac{\alpha^4}{24} - \cdots}{\alpha^2}} =  \lim_{\alpha \to 0}\sqrt{1 - \frac{\alpha^2}{12} + \cdots} = 1$$

So in the end the magnitude of acceleration is $v\frac{2\pi}{T}$. But since the particle traces the circumference of the circle, of length $2\pi r$, under constant speed $v$ in time $T$, we have $vT = 2\pi r$, so $\frac{2\pi}{T} = \frac{v}{r}$, and finally the acceleration is the familiar formula $$a_c=\frac{v^2}{r}$$

Did I do this right? And even if I did, was it necessary to do the Taylor series stuff to obtain the trigonometric limit? What are some ways to make this derivation easier, but still rigorous under the formal definition of acceleration in Cartesian coordinates?

答:
You are making this much more difficult than it has to be. Hints:

$1).\ $Write $\vec r(t)=r\cos \theta(t)\vec i+r\sin\theta(t)\vec j$, note that $r$ is constant, differentiate once to obtain $\vec v(t)$, again to get $\vec a(t).$ You will use the chain rule to do this.

$2).\ $Now, with $\frac{d\theta}{dt}:=\omega, $ the angular velocity, you will get $\vec a(t)=-\omega^2\vec r(t).$

$3).\ $To finish, take the magnitude of both sides, and the formula you want will drop out as soon as you observe that since the speed is constant (why?), one has $\omega=\|\vec v(t)\|/r=v/r.$


问:
Evaluating $\sum_{k=1}^{a}\frac{-1-H_k}{k(1-e)^k}$
Question :

My attempt:

Let $a=17399172$
$$\begin{align}
&\sum_{k=1}^{a}\frac{-1-H_k}{\log_2\left(\sum_{j=0}^{k}\left(\ln\left(e^{C_j^k}\right)\right)\right)(1-e)^k} \\
&= \sum_{k=1}^{a}\frac{-1-H_k}{\log_2\left(\sum_{j=0}^{k}\left(C_j^k\right)\right)(1-e)^k} \\
&= \sum_{k=1}^{a}\frac{-1-H_k}{\log_2\left(2^k\right)(1-e)^k} \\
&= \sum_{k=1}^{a}\frac{-1-H_k}{k(1-e)^k} \\
\end{align}$$

Can you give me some hints what to do afterwards? I'll try by myself.

答:
This is not a full answer but I will elaborate on @Dr.WolfgangHintze comments. We have
$$
\begin{align}
S
&= \sum_{k=1}^{a}\frac{-1-H_k}{k(1-e)^k}\\
&= -\sum_{k=0}^{a-1}(1+H_{k+1})\frac{(\frac{1}{1-e})^{k+1}}{k+1} \\
&= -\int_0^{\frac{1}{1-e}}\sum_{k=0}^{a-1}(1+H_{k+1})t^k\,\mathrm dt\\
&=: -\int_0^{\frac{1}{1-e}}S^\ast(t)\,\mathrm dt.
\end{align}
$$
Now write
$$
\begin{align}
S^\ast(t)
&= \sum_{k=0}^{a-1}(1+H_{k+1})t^k\\
&= \frac{1-t^a}{1-t}+\sum_{k=0}^{a-1}H_{k+1}t^k\\
&= \frac{1-t^a}{1-t}+\sum_{k=0}^\infty H_{k+1}t^k-\sum_{k=a}^\infty H_{k+1}t^k\\
&= \frac{1-t^a}{1-t}+\sum_{k=0}^\infty H_{k+1}t^k-\sum_{k=0}^\infty H_{a+k+1}t^{a+k}.
\end{align}
$$
Working with the recurrence relations for the harmonic numbers as well as the Cauchy product we find
$$
\begin{align}
\sum_{k=0}^\infty H_{a+k+1}t^{a+k}
&=t^a\sum_{k=0}^\infty (H_a+\sum_{\ell=1}^{k+1}\frac{1}{a+\ell})t^k\\
&=t^a\left(\frac{H_a}{1-t}+\sum_{k=0}^\infty\sum_{\ell=0}^k\frac{1}{a+1+\ell}t^k\right)\\
&=\frac{t^a}{1-t}\left(H_a+\sum_{k=0}^\infty\frac{1}{a+1+k}t^k\right)\\
&=\frac{t^a}{1-t}\left(H_a+\sum_{k=0}^\infty\frac{(1)_k\Gamma(a+1+k)}{\Gamma(a+2+k)}\frac{t^k}{k!}\right)\\
&=\frac{t^a}{1-t}\left(H_a+\frac{1}{1+a}F\left({1,1+a\atop 2+a};t\right)\right).
\end{align}
$$
Setting $a=0$ gives the closed form for the other series term and so
$$
\begin{align}
S^\ast(t)
&= \frac{1-t^a}{1-t}-\frac{\log(1-t)}{1-t}-\frac{t^a}{1-t}\left(H_a+\frac{1}{1+a}F\left({1,1+a\atop 2+a};t\right)\right).
\end{align}
$$
Therefore,
$$
\begin{align}
S
&= -\int_0^{\frac{1}{1-e}} \frac{1-t^a}{1-t}-\frac{\log(1-t)}{1-t}-\frac{t^a}{1-t}\left(H_a+\frac{1}{1+a}F\left({1,1+a\atop 2+a};t\right)\right)\,\mathrm dt.
\end{align}
$$
Most of the integrals are easy to evaluate with the exception of the integral involving the hypergeometric function which will take a little extra work.


问:
Why does $\lim_{n \to \infty} P_{X}((y-a_{n},y])=F_{X}(y)-\lim_{n \to \infty}F_{X}(y-a_{n})$?
I am currently working on the following problem for my measure theory class and have a question about the hint given:

Show that $F_{X}$ (the cumulative distribution function of a random variable $X : \Omega \to \Bbb{R}$) is continuos if and only if $P_{X}(\{y\})=0 \ \text{(were $P_X$ is the probability distribution of the random variable $X$.)} \ \forall y.$

Where the hint states: note that $\lim_{n \to \infty} P_{X}((y-a_{n},y])=F_{X}(y)-\lim_{n \to \infty}F_{X}(y-a_{n})$ for a sequence $a_n$ such that $a_n \nearrow 0$.

My questions is why can we do the step in $\color{blue}{blue}$? $$\lim_{n \to \infty} P_{X}((y-a_{n},y])=\color{blue}{F_{X}(y)-\lim_{n \to \infty}F_{X}(y-a_{n})}.$$I understand how this hint can be used to prove that $P_{X}(\{y\})=0$, but I am unsure as to how it is formulated.

Note my definition of the CDF of a random variable is $F_{X}(y) :=P(\{\omega: X(\omega) \leq y\}) =P_{X}((-\infty,y]).$

答:
$$P(X \in (y-a_n,y])$$ $$=P(X \in (-\infty, y])-P(X \in (-\infty, y-a_n)=F_X(y)-F_X(y-a_n).$$

[$X^{-1}(y-a_n,y]=X^{-1}(-\infty, y] \setminus X^{-1}(-\infty, y-a_n)]$ and $B \subset A$ implies $P(A\setminus B )=P(A)-P(B)$].