问:
$\sin(4x)y''-2(1+\cos(4x))y'=\tan(x)$

looking for more friendly way for the following second order ode, thanks in advance

$$
\sin(4x)y''-2(1+\cos(4x))y'=\tan(x)\\
$$
integration factor

$$
\mu(x) = e^{-2\int\frac{1+\cos(4x)}{\sin(4x)}dx}
$$
make long short:

$$
\mu(x) = \sin(2x)\sin(4x)\cos(2x)
$$
now for the order reduction
$$
u=y'\\
\sin(4x)u'-2(1+\cos(4x))u=\tan(x)\\
$$
$$
\left[u \cdot (\sin(2x)\sin(4x)\cos(2x)\right]'= \frac{\tan(x)}{\sin(4x)}\sin(2x)\sin(4x)\cos(2x)
$$
integrating right side
$$
u \cdot (\sin(2x)\sin(4x)\cos(2x)) = \frac{\sin(2x)}{2}-\frac{x}{2}-\frac{1}{8}\sin(4x)
$$
now all left is to integrate u to get y, however it's seems it isn't the right way...

thanks.

答:
My approach was to solve the homogeneous equation:
$$\sin(4x)y_h^{\prime\prime}-2(1+\cos(4x))y_h^{\prime}=0$$
So
$$\frac{y_h^{\prime\prime}}{y_h^{\prime}}=\frac{2(1+\cos(4x))}{\sin(4x)}=2\cot(2x)$$
With integral $\ln\left(y_h^{\prime}\right)=\ln(\sin(2x))+C_1$ so $y_h^{\prime}=C_2\sin(2x)$. Then
$$y_h=-\frac12C_2\cos(2x)+C_3=C_4\cos(2x)+C_3$$
With this integral in hand we let $y(x)=u(x)\cos(2x)+v(x)$. Then
$$y^{\prime}=u^{\prime}\cos(2x)+v^{\prime}-2u\sin(2x)=-2u\sin(2x)$$
Where we have constrained $v^{\prime}=-u^{\prime}\cos(2x)$. Then $y^{\prime\prime}=-2u^{\prime}\sin(2x)-4u\cos(2x)$ and we have
$$\begin{align}\sin(4x)y^{\prime\prime}-2(1+\cos(4x))y^{\prime}&=-2u^{\prime}\sin(2x)\sin(4x)-8u\sin(2x)\cos^2(2x)\\
&\quad+8u\cos^2(2x)\sin(2x)\\
&=-4u^{\prime}\sin^2(2x)\cos(2x)=\tan x\end{align}$$
So we have
$$\begin{align}u^{\prime}&=-\frac14\frac{\sin x}{4\sin^2x\cos^3x(2\cos^2x-1)}\\
&=-\frac1{16}\frac{\sin x}{(1-\cos^2x)\cos^3x(2\cos^2x-1)}\\
&=-\frac{\sin x}{16}\left(-\frac3{\cos x}-\frac{\frac12}{1+\cos x}+\frac{\frac12}{1-\cos x}-\frac1{\cos^3x}+\frac{8\cos x}{2\cos^2x-1}\right)\end{align}$$
And
$$u(x)=\frac1{16}\left(-3\ln\cos x-\ln\sin x+\frac12\sec^2x+2\ln\cos(2x)\right)+C_5$$
Then
$$\begin{align}v^{\prime}&=-u^{\prime}\cos(2x)=\frac1{16}\frac{\sin x}{(1-\cos^2x)\cos^3x}\\
&=\frac{\sin x}{16}\left(\frac1{\cos^3x}+\frac1{\cos x}-\frac{\frac12}{1+\cos x}+\frac{\frac12}{1-\cos x}\right)\end{align}$$
And
$$v(x)=\frac1{16}\left(\frac12\sec^2x+\ln\tan x\right)+C_6$$
Which implies that
$$\begin{align}y(x)&=\frac1{16}\left(-3\ln\cos x-\ln\sin x+\frac12\sec^2x+2\ln\cos(2x)\right)\cos(2x)\\
&\quad+C_5\cos(2x)+\frac1{16}\left(\frac12\sec^2x+\ln\tan x\right)+C_6\\
&=\frac1{16}\left(-3\ln\cos x-\ln\sin x+2\ln\cos(2x)\right)\cos(2x)\\
&\quad+C_5\cos(2x)+\frac1{16}\ln\tan x+C_7\end{align}$$
I haven't been able to figure out how to check this with Wolfram Alpha; maybe you will have better luck than I did.

Title: $\sin(4x)y''-2(1+\cos(4x))y'=\tan(x)$


问:
Sequence of Continuous Functions converging pointwise
Basically, I'm wondering if the following function (defined over all the reals) is continuous:

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

The graph seems to be continuous, but I can't think of any way to prove that the sum itself is continuous.

Edit: Is there a way to do so with the M-test because I haven't covered Hyperbolic Identities

答:
Put $$f_N(x)=\sum_{n=1}^{N} \frac{1}{n^2+x^2}$$.

By Weierstrass criterion for uniform convergence we have that $f_N(x)$ converges uniformly on $\mathbb{R}$ because $\frac{1}{n^2+x^2} \le \frac{1}{n^2}$ and $\sum_n \frac{1}{n^2} < \infty$.

We know that $f_N(x)$, $N \ge 1$, are continuous. It follows from the uniform convergence that their limit $\lim_N f_N(x)$ is also continuous, q.e.d.

ContentLicense: CC BY-SA 4.0


问:
Compute the mean of a random variable
Imagine I have for a single individual some variable $X$ with mean $\lambda$ (for example the number of cars he has). Now I take a whole population of individuals. The parameter $\lambda$ for each of them is indeed different. Assume $\lambda \sim exp (\alpha)$. Let now $X$ be the variable $Y$ for a random selected individual in such population (i.e. $X$ is the number of cars for a randomly selected individual). What is $E X$ in this case?

Thanks a lot for your help!

答: Note that $E[X\mid\Lambda]=\Lambda$ hence $E[X]=E[E[X\mid\Lambda]]=E[\Lambda]$. If $\alpha$ denotes the mean of the exponential distribution, then $E[\Lambda]=\alpha$ hence $E[X]=\alpha$. If $\alpha$ denotes the rate of the exponential distribution, then $E[\Lambda]=1/\alpha$ hence $E[X]=1/\alpha$.

ContentLicense: CC BY-SA 3.0

Title: Compute the mean of a random variable


问:
If $\deg^+(u) = \deg^+(v)$ for all vertices in a simple digraph, then $\deg^-(u) = \deg^-(v)$ for all vertices. Is this a counterexample?
I was asked to prove the following statement:

If $\deg^+(u) = \deg^+(v)$ for all vertices in a simple digraph, then
$\deg^-(u) = \deg^-(v)$ for all vertices.

Where $\deg^+(u)$ is the number of vertices $v$ such that there is an edge from $u$ to $v$.

I have been drawing some graphs to figure out how I can prove that and I stumbled upon the following graph:

As you can see there is two outgoing edges from all nodes so $\deg^+(u) = \deg^+(v)$ for all vertices but $\deg^-(a) = 0$ while $\deg^-(d) = 2$. Is the statement true and my graph wrong or is the statement wrong?

答:
The statement is not true, as shown your by counterexample.

There is a simpler counterexample:

$ a \rightarrow b $

$ b \rightarrow c $

$ c \rightarrow b$

Writing it this way, you can see that conditioning what happens on the LHS (the number of times each vertex appears) doesn't give us much restriction on the RHS. As such, it's unlikely that the statement is true (unless something much deeper happens).


问:
about smooth manifold
Let $f:\mathbb{R^3} \to \mathbb{R}$
$$f(x, y, z)=\left(x^{2}+y^{2}+z^{2}+R^{2}-r^{2}\right)^2-4 R^{2}\left(x^{2}+y^{2}\right)
$$
and $R,r \in \mathbb{R} , 0<r<R$

now define :

$Z=\{ (x,y,z) \in \mathbb{R^3}  | f(x,y,z)=0\}$

(i)- Prove $Z$ is a smooth manifold.

(ii)-draw set $Z$ in $\mathbb{R^3} $.

(iii)- find coordinate charts on $Z$.

(iv) prove  $Z$ is diffeomorphic to $S^1 \times S^1$.

for (i) i think we can use following theorem :

theorem : if $M$ and $N$ are smooth manifolds, $f: M \to N$ is a smooth map and $n \in f(M)$ is a regular value for $f$, then $f^{-1} (\{n\})$ is a smooth submanifold of $M$ of dimension $\dim M - \dim N$.

for (ii) we must compute $Z$ . i think  $Z=\{ (x,y,z) \in \mathbb{R^3}  | x^2+y^2=R^2,z^2=r^2\}$ so $Z$ is two circle with radios $R$ and $z=-r,+r$

答:
Some hints:

$i).\ $ Fix an arbitrary $p\in Z$ and show there are no critical points for $f$ on $Z.$

$ii).\ $ Consider that  $\left(x^{2}+y^{2}+z^{2}+R^{2}-r^{2}\right)^2-4 R^{2}\left(x^{2}+y^{2}\right)=0\Rightarrow $

$z^2=\pm2R\sqrt{x^2+y^2}-(\sqrt{x^2+y^2})^2-(R^{2}-r^{2})$ and compare with the equations and arguments of this post.

$iii).\ $ From $i).,\ $ note that one of $\partial_x(p),\ \partial_y(p),\ \partial_z(p)$ is non-zero, say $\partial_x(p)$. Now, consider $\varphi:(x,y,z)\mapsto (f,y,z)$ defined in some $U(\subseteq \mathbb R^3)$-neighborhood of $p$ and show that $\varphi|_{U\cap Z}$ is a slice chart about $p\in Z.$

$iv).\ $ Apply $ii).$

Title: about smooth manifold


问:
Finding the nth integral of $\ln(1+z/k)$
I can find the nth integral of $\ln(z)$ as follows:
\begin{aligned}
\left(\frac d{dz}\right)^{-n}\ln(z)&=\frac1{\Gamma(n)}\int\limits_0^z(z-t)^{n-1}\ln(t)dt\\
&=\frac1{n!}\left[\int\limits_0^z\frac1t(z-t)^ndt-z^n\ln(0)\right]\\
&=\frac1{n!}\left[\int\limits_0^1\frac{z^nt^n-z^n+z^n}{1-t}dt-z^n\ln(0)\right]\\
&=\frac{\ln(z)-H_n}{n!}z^n,
\end{aligned}
but I can't get very far with $\ln(1+z/k)$. I was able to come up with this but it is only a conjecture:
\begin{aligned}
\frac1{\Gamma(n)}\int\limits_0^z(z-t)^{n-1}\ln\left(1+\frac tk\right)dt&=\frac{\ln\left(1+\frac{z}{k}\right)-H_n}{n!}(k+z)^n+\sum ^{n}_{i=0}\frac{H_{i}}{i!(n-i)!}z^{n-i}k^i\\
&=\frac1{n!}\ln\left( 1+\frac{z}{k}\right)( k+z)^{n} -\sum ^{n}_{i=0}\frac{H_{n} -H_{i}}{i!( n-i) !} z^{n-i} k^{i}.
\end{aligned}
I'm only using $k,n\in\mathbb{N}$ and $z\in\mathbb{R}$ for now. Any help proving this would be appreciated.

答:
This is not a derivation of the integral, but rather a proof of your conjecture.

Your expression is $$f_n(z)=\frac1{n!}\ln\left( 1+\frac{z}{k}\right)( k+z)^{n} -\sum ^{n}_{i=1}\frac{H_{n} -H_{n-i}}{i!( n-i) !} k^{n-i} z^{i}$$

I reversed the order of summation from yours and then got rid of the first term since $H_n - H_n = 0$. Then there are two things that need to be shown for each $n \ge 1$: $f_{n-1}(z) = \frac{d}{dz}f_n(z)$ and $f_n(z) = 0$. The second is true since each term is $0$ when $z = 0$. For the first, the derivative of $f_n(z)$ is $$\frac{1}{n!}\left( \ln\left( 1+\frac{z}{k}\right)\frac{d}{dz}(k+z)^n + (k+z)^n\frac{d}{dz}\ln\left( 1+\frac{z}{k}\right) \right) - \sum_{i=1}^n \frac{H_{n} -H_{n-i}}{i!( n-i) !} k^{n-i}\frac{d}{dz}z^{i}$$

This simplifies to $$\frac{1}{(n-1)!}\ln\left( 1+\frac{z}{k}\right)(k+z)^{n-1} + \frac{1}{n!}(k+z)^{n-1} - \sum_{i=1}^n \frac{H_{n} -H_{n-i}}{(i-1)!( n-i) !} k^{n-i}z^{i-1}$$

Then, what remains to be shown is $$\frac{1}{n!}(k+z)^{n-1} - \sum_{i=1}^n \frac{H_{n} -H_{n-i}}{(i-1)!( n-i) !} k^{n-i}z^{i-1} = -\sum ^{n-1}_{i=1}\frac{H_{n-1} -H_{n-1-i}}{i!( n-1-i) !} k^{n-1-i} z^{i}$$

Expanding $(k+z)^{n-1}$ by the binomial expansion and setting the coefficients of $z^i$ equal to each other yields $$\frac{1}{n!}\binom{n-1}{i}k^{n-1-i} - \frac{H_{n} -H_{n-i-1}}{i!( n-i-1) !} k^{n-i-1}=-\frac{H_{n-1} -H_{n-1-i}}{i!( n-1-i) !} k^{n-1-i}$$

Subtracting $\frac{H_{n-1-i}}{i!(n-1-i)!}k^{n-1-i}$, dividing by $k^{n-1-i}$, then multiplying by $i!(n-1-i)!$ on both sides makes it $$\frac{(n-1)!}{n!} - H_n = -H_{n-1}$$

This simplifies to $$\frac{1}{n}-H_n = -H_{n-1} \to H_n = H_{n-1} + \frac{1}{n}$$

which is exactly the recursive definition of the harmonic numbers. Therefore, your conjecture is true.


问:
If $\sum_{n=1}^\infty | x'(x_n) | < \infty$ for all $x' \in X'$ then $\sup_{\|x'\| \leq 1} \sum_{n=1}^\infty | x'(x_n) | < \infty$.
I have this question to solve, but I have no idea how to solve:

a) Let $X$ a normed space and $\{ x_n \}_{n \in \mathbb{N}} \subset X$ a sequence with $\sum_{n=1}^\infty | x'(x_n) | < \infty$ for all $x' \in X'$. Show that $\sup_{\|x'\| \leq 1} \sum_{n=1}^\infty | x'(x_n) | < \infty$.

b) Let $K \subset C([a,b])$ be $K = \{ f \in C([a,b]) : \int_{a}^{(a+b)/2} f(x) \mathrm{d}x = \int_{(a+b)/2}^{a} f(x) \mathrm{d}x \}$. Show that $K$ is a closed vector subspace and find $x' \in C([a,b])'$ so that $\ker (x')=K$.

Thanks for any help :)

答:
(a) Without loss of generality, we may assume that $X$ is complete.
For, if $X$ is not complete, let $\bar{X}$ be its completion. Note
that every bounded linear functional defined on $X$ has a unique
continuous extension on $\bar{X}$. Hence $x'\in X^{\ast}$ can be
regarded as a bounded linear functional on $\bar{X}$. Moreover, the
norm of $x'$ remains the same after extension. We also recall that the canonical
embedding $X\rightarrow X^{\ast\ast}$, $x\mapsto\hat{x}$, where
$\hat{x}(f)=f(x)$, $f\in\hat{x}$, is a linear isometry, i.e., $||\hat{x}||=||x||.$

Let $\mathcal{C}=\{\alpha = (\alpha_n)\in\mathbb{C}^{\mathbb{N}}\mid|\alpha_{n}|=1\}$
be the collection of all $\mathbb{C}$-valued (or $\mathbb{R}$-valued,
depending on the scalar field of $X$) sequences $\alpha=(\alpha_{n})$
that satisfies $|\alpha_{n}|=1$ for all $n\in\mathbb{N}.$ For each
$\alpha\in\mathcal{C}$ and $n\in\mathbb{N}$, define $\theta_{\alpha,n}\in X^{\ast\ast}$ by $\theta_{\alpha,n}=\sum_{k=1}^{n}\alpha_{k}\hat{x}_{k}.$
For each $f\in X^{\ast}$, we have
\begin{eqnarray*}
|\theta_{\alpha,n}(f)| & = & \left|\sum_{k=1}^{n}\alpha_{k}f(x_{k})\right|\\
 & \leq & \sum_{k=1}^{\infty}\left|f(x_{k})\right|\\
 & := & M_{f}\\
 & < & \infty.
\end{eqnarray*}
Hence, $\{\theta_{\alpha,n}(f)\mid n\in\mathbb{N},\,\,\alpha\in\mathcal{C}\}$
is bounded. By Uniform Boundedness Principle, $\eta:=\sup_{\alpha,n}||\theta_{\alpha,n}||<\infty.$
We go to show that $\sup_{||f||\leq1}M_{f}<\infty$ by contradiction.
Suppose the contrary that $\sup_{||f||\leq1}M_{f}=\infty$, then there
exists $f\in X^{\ast}$ with $||f||\leq1$ such that $M_{f}>\eta+1$.
Hence, there exists $n$ such that $\sum_{k=1}^{n}\left|f(x_{k})\right|>\eta+1.$
For each $k$, choose $\alpha_{k}\in\mathbb{C}$ with $|\alpha_{k}|=1$
such that $|f(x_{k})|=\alpha_{k}f(x_{k})$. Define $\alpha=(\alpha_{k})\in\mathcal{C}$.
Now we have that
\begin{eqnarray*}
\eta+1 & < & \sum_{k=1}^{n}\left|f(x_{k})\right|\\
 & = & \sum_{k=1}^{n}\alpha_{k}f(x_{k})\\
 & = & \theta_{\alpha,n}(f)\\
 & \leq & ||\theta_{\alpha,n}||\cdot||f||\\
 & \leq & \eta,
\end{eqnarray*}
which is a contradiction!


问:
Solving a non-linear integral equation
Is there any way to solve an integral equation of this type,

$s(1-b)f(s)=\int^s_{sb}f(x)f(x-sb)dx$

Any help would be highly appreciated.

答: You can solve this type of nonlinear equation by the Kantrivich method, it is a numerical method, so you start by linearizing the equation and approximate the linear equation obtained by the Nythrom method.</p>

Title: Solving a non-linear integral equation


问:
Maximal Abelian von Neumann algebra
Let $H$ be the Hilbert space of complex-valued functions on $\mathbb{R}$ that are square integrable w.r.t. the Lebesgue measure.

What is a concrete maximal von Neumann algebra of $\mathcal{B}(\mathcal{H})$?

答:
There are lots of masas in $B(H)$. They can be classified in two kinds, discrete and continuous (and and direct sums thereof).

The canonical example of a continuous masa in your setting would be $L^\infty(\mathbb R)$, seeing as multiplication operators.

The canonical example of a discrete masa is the diagonal masa: you fix an orthonormal basis $\{e_n\}$, and consider the corresponding orthogonal projections $\{E_n\}$. Then
$$
A=\{\sum_ka_kE_k:\ a\in\ell^\infty(\mathbb N)\}
$$
would be the diagonal masa corresponding to the orthonormal basis $\{e_n\}$. Not that you gain anything, but if you want to  make this concrete, you can take $\{e_n\}$ to be the Hermite Polynomials. Or you can use a double index and define
$$
e_{n,m}=e^{2\pi in(x-m)}\,1_{[m,m+1)},\qquad n,m\in\mathbb Z.
$$
This would make
$$
(E_{n,m}f)(x)=\langle f,e_{n,m}\rangle\,e_{n,m}=\bigg(\int_m^{m+1}f(t)\,e^{-2\pi i (t-n)}\,dt\bigg)\,e^{2\pi in(x-m)}\,1_{[m,m+1)}.
$$
In this case $A$ would consist of the operators
$$
(T_af)(x)=\sum_{n\in\mathbb Z}a_{n,m}\,\bigg(\int_m^{m+1}f(t)\,e^{-2\pi i (t-n)}\,dt\bigg)\,e^{2\pi in(x-m)},\qquad x\in[m,m+1),
$$
where $a\in\ell^\infty(\mathbb Z^2)$.

Title: Maximal Abelian von Neumann algebra


问:
Basis and transcendental basis
I need to have really good picture in my mind about transcendental basis and What are the similarity between basis and transcendental basis. Let $\mathbb R$  be the set of all really number. Let $\mathbb B$ be a basis of $\mathbb R$ over $\mathbb Q$ so it is a maximal linear independent set and  it spans all element of $\mathbb R.$ Also, each element of $\mathbb R$ can be written as linear combination of finitly many element of $\mathbb B.$

Let $T$ be a transcendental basis of $\mathbb R$ over $\mathbb Q$ so it is a maximal algebraically independent set. Also, $\mathbb Q(T)=\mathbb R.$

Is it true that each element of $\mathbb  R$ can  be spanned by finitely number algebraically independent elements of $T$? if so, why.

Can we choose for each element of $\mathbb R$ different element of $T$ that spans it?  if so , why ?

what we can also say about $T$?

I hope from those that have a deep understanding for this subject to say more about transcendental basis and algebraically independent set?

Any help will be appreciated greatly

答:
Is it true that each element of $\mathbb R$ can be spanned by finitely number algebraically independent elements of
$T$ ?

"Spanned" in the sense of linear combination, no.  But any element of $r \in \mathbb R$ is a rational function of finitely many elements of $T$.  That is what $\mathbb R = \mathbb Q(T)$ means.  So linear combinations are not enough.  Starting with $T$, you have to allow products and quotients as well as sums.

Can we choose for each element of $\mathbb R$ different element of
$T$ that spans it?

This seems nonsensical to me.  I do not know what "$r$ is spanned by $t$" means, where $r$ is a real number and $t$ is one element of $T$.

What can we say about $T$?

From $\mathbb R = \mathbb Q(T)$ it follows that $T$ is uncountable.

Title: Basis and transcendental basis


问:
Prove $\bigcup_{n=1}^{\infty} [1/n^{2}, \infty) = (0, \infty)$
Hi I am very confused about index sets.

My effort:

Let $x\in (0,\infty)$ since $(0,\infty)$ is open, $\exists y>0 \ni x\in (x-y,x+y)\subset (0,\infty)$. Since $x-y>0, \exists n_1 \in \mathbb{N}$ such that

$\frac{1}{n_1^2}<\frac{1}{n_1}<x-y$, $x\in \big[\frac{1}{n_1^2},\infty\big)$, $x\in \bigcup_{n=1\rightarrow \infty} \big[\frac{1}{n_1^2} , \infty\big)$

$\big[\frac{1}{n_1^2} , \infty\big)\subset (0,\infty), \forall n\in \mathbb{N}$, so $\bigcup_{n=1\rightarrow \infty} \big[\frac{1}{n_1^2} , \infty\big) \subseteq (0,\infty)$

Putting the two pieces together, we have $\bigcup_{n=1\rightarrow \infty} \big[\frac{1}{n_1^2} , \infty\big)=(0,\infty)$

Am I right? Please help, thank you!

答:
Your effort is quite difficult to read/understand, but here's perhaps a clearer explanation: we know that as $n\to\infty$, $\frac 1{n^2} \to 0$ (i.e. $\lim_{n\to\infty} \frac 1{n^2} = 0$). That is to say, for every positive number $\epsilon$, there will be some $N$ (depending on $\epsilon$) for which all $n\geq N \implies \frac 1{n^2} < \epsilon$. Applying this to any $x \in (0,\infty)$, we see that there must be some $N$ (depending on $x$) such that $\frac 1{N^2} < x$, i.e. $x \in [\frac{1}{N^2},\infty)$. This means that $(0,\infty)$ is contained within the infinite union $\bigcup_{n=1}^\infty [\frac{1}{n^2},\infty)$.

For the other direction of containment, consider any $x \leq 0$ -- we show that such $x$ can not be in the infinite union. Note that for any $n\in \mathbb N$, $x$ can not possibly be in $[\frac{1}{n^2},\infty)$ because $[\frac{1}{n^2},\infty)$ contains strictly positive numbers, and we said that $x$ was not a positive number. Thus, $x$ is not in the infinite union, because $x$ is not in any of the $[\frac{1}{n^2},\infty)$, and we are done.


问:
Weak convergence in probability implies uniform convergence in distribution functions

Exercise 1: Let $\mu_n$, $\mu$ be probability measures on $\left(\mathbb{R}, \mathcal{B}\left(\mathbb{R}\right)\right)$ with distribution functions $F_n$, $F$. Show: If $\left(\mu_n\right)$ converges weakly to $\mu$ and $F$ is continuous, then $\left(F_n\right)$ converges uniformly on $\mathbb{R}$ to $F$.

This is a problem that I am totally stuck at. I know the fact that $F_n$ converges pointwise to $F$ in this question. Also, I looked through Google and found out that I have to show first that $F_{n}(t_{n})$ converges to $F(t)$ if $t_{n}$ converges to $t$. But, no matter how I tried, I keep failing to prove the fact.
Also, I have no idea how to use the fact to get the uniform convergence. So I'm just stymied.

Could anyone please help me with this?

Add : I managed to show that $F_n$ converges uniformly to $F$ on any compact intervals. However, the generalization to the whole real line is still not solved...Could anyone at least help me with the generalization to the real line?

答:
Here's another proof for the compact case which I find a bit more intuitive. We are trying to show that $F_n \rightarrow F$ uniformly on an arbitrary closed interval $[a,b]$.

Fix $\epsilon > 0$. Let $d = F(b)-F(a)$, and take $k$ large enough so that $\frac{d}{k} \leq \frac{\epsilon}{5}$. By the continuity of $F$, we can apply the intermediate value theorem to show there exist real numbers $a := x_0 < x_1 < ... < x_k := b$ such that $F(x_i) = F(a) + i\frac{d}{k}$ for each $i \in \{0,1,...,k\}$.

Since $F_n \rightarrow F$ pointwise, for each $i$ in $\{0,1,...,k\}$ there exists $N_i$ such that $|F_n(x_i)-F(x_i)| \leq \frac{\epsilon}{5}$ for all $n \geq N_i$. Taking $N = \max(N_0,N_1,...,N_k)$, we conclude that $|F_n(x_i)-F(x_i)| \leq \frac{\epsilon}{5}$ for all $n \geq N$, $i \in \{0,1,...,k\}$. This gives
$F_n(x_{i+1})-F_n(x_i) \leq |F_n(x_{i+1})-F(x_{i+1})| + |F(x_{i+1})-F(x_i)| + |F(x_i)-F_n(x_i)| \leq \frac{\epsilon}{5} + \frac{d}{k} + \frac{\epsilon}{5} \leq \frac{3\epsilon}{5}$

For any $x \in [a,b)$, we have $x_i \leq x < x_{i+1}$ for some $i \in \{0,1,...,k\}$. Then for all $n \geq N$, we have
$$
\begin{align}
|F_n(x)-F(x)| & \leq |F_n(x)-F_n(x_i)| + |F_n(x_i) - F(x_i)| + |F(x_i)-F(x)| \\\\
& \leq |F_n(x_{i+1})-F_n(x_i)| + |F_n(x_i)-F(x_i)| + |F(x_{i+1})-F(x)| \\\\
& \leq \frac{3\epsilon}{5} + \frac{\epsilon}{5} + \frac{d}{k} \\\\
& \leq \epsilon \\\\
\end{align}
$$
where the second inequality uses monotonicity of $F$ and $F_n$.

Of course, $|F_n(b)-F(b)| \leq \frac{\epsilon}{5} \leq \epsilon$ since $b = x_k$.

Thus, we've shown $|F_n(x)-F(x)| \leq \frac{5\epsilon}{3}$ for all $x \in [a,b]$, concluding the proof.


问:
Calculating distribution of time until first event in a poisson process

Let crashes of a stock market be modelled as a Poisson process with rate 0.1 per year and let $N_t$ denote the number of crashes by time t. Then $N_t$ has a Poisson distribution with mean $0.1t$.

For $0<s<t$, calculate $P(N_s=0 \vert N_t=1)$. What is the distribution of the time until the first crash given $N_t=1$.

My solution for this is as follows:

\begin{align}
P(N_s=0 \vert N_t=1) &= \frac{P(\{N_s=0\} \cap \{N_t=1\} )}{P(\{N_t=1\})} \\
&=\frac{P(\{N_{t-s}=0\})}{P(\{N_t=1\})} \\
&= \frac{e^{-0.1(t-s)}}{0.1t\cdot e^{-0.1t}}\\
&= \frac{e^{0.1s}}{0.1t}
\end{align}

I do not know what to do past this point, consequently I cannot find the distribution required.

答:
Hints:

$P(\{N_s=0\} \cap \{N_t=1\} )$ is the probability is the probability of zero events from time $0$ to $s$ and one event from time $0$ to $t$; these are not independent

Instead, try saying $P(\{N_s=0\} \cap \{N_t=1\} )$ is the probability is the probability of zero events from time $0$ to $s$ and one event from time $s$ to $t$ (a time period of $t-s$; these are independent so you can find the probability of each and multiply them

$P(\{N_t=1\} )$ is the probability is the probability of zero events from time $0$ to $t$

You can calculate these from knowing over a time period $t$ saying "$N_t$ has a Poisson distribution with mean $0.1t$" and so the result from $P(N_s=0 \vert N_t=1) = \dfrac{P(\{N_s=0\} \cap \{N_t=1\} )}{P(\{N_t=1\})}$.

With tidying and cancellation, the final answer is remarkably simple

Your calculation is incorrect. $\dfrac{P(\{N_s=0\} \cap \{N_t=1\} )}{P(\{N_t=1\})} \not
=\dfrac{P(\{N_{t-s}=0\})}{P(\{N_t=1\})}$.  Instead $\dfrac{P(\{N_s=0\} \cap \{N_t=1\} )}{P(\{N_t=1\})}
=\dfrac{P(\{N_s=0\}\,P(\{N_{t-s}=1\})}{P(\{N_t=1\})}$


答:
Good effort, and thanks for using latex!

In regards to your attempt:

You use the fact that $(0,\infty)$ is an open interval to justify the fact that $\exists y \big(x-y > 0\big)$. This result is indeed true, but you haven't proved it. It's probably fine to assume that result (assuming you've seen it/proved it before). But you go on to say that because $x-y>0$, then $\exists n_1 \in \mathbb{N} \big(\frac{1}{n_1^2} < x-y\big)$. Again, this statement is true, but you have not proved it.

Instead, to prove it we should use an argument such as: Suppose $x\in (0, \infty)$. Suppose $n_1 > \frac{1}{\sqrt{x}}$. This implies that $n_1^2 > \frac{1}{x}$, which implies $\frac{1}{n_1^2} < x$. Hence for any $n_1 > \frac{1}{\sqrt{x}}$ (for example, take $\lceil \frac{1}{x^2}\rceil$), $x \in \big[\frac{1}{n_1^2}, \infty\big] \implies x \in \bigcup_{i=1}^\infty \big[\frac{1}{n_1^2}, \infty\big]$. Note that we don't need to use any result about the set being open in fact.

For the second part of your argument, I am not sure if that's a typo, or if you are trying to take a limit. Even if you did not mean to take a limit, I think your argument that since $\forall n \in \mathbb{N}\big(\big[\frac{1}{n^2},\infty\big]\subset (0,\infty)\big)$ then $\bigcup_{i=1}^\infty \big[\frac{1}{n_1^2}, \infty\big] \subset (0, \infty)$, is unjustified, and perhaps doesn't have details that you would be expected to include when writing these types of proofs.

Remember that we are ultimately trying to show that $\bigcup_{i=1}^\infty \big[\frac{1}{n_1^2}, \infty\big] \subset (0, \infty)$, so we should take an arbitrary element of $\bigcup_{i=1}^\infty \big[\frac{1}{n_1^2}, \infty\big]$ and show this is also an element of $(0,\infty)$.

Here is how we can do it: Suppose $x\in \bigcup_{i=1}^\infty \big[\frac{1}{n_1^2}, \infty\big]$. Then $\exists i \in \mathbb{N}\big(x \in \big[\frac{1}{i^2}, \infty \big]\big)$. But $\frac{1}{i^2} > 0$, so $x \in (0,\infty)$. Note here we use the definition of what the union is to justify the existence of some $i \in \mathbb{N}$

Title: Prove $\bigcup_{n=1}^{\infty} [1/n^{2}, \infty) = (0, \infty)$

Body:
Hi I am very confused about index sets.

My effort:

Let $x\in (0,\infty)$ since $(0,\infty)$ is open, $\exists y>0 \ni x\in (x-y,x+y)\subset (0,\infty)$. Since $x-y>0, \exists n_1 \in \mathbb{N}$ such that

$\frac{1}{n_1^2}<\frac{1}{n_1}<x-y$, $x\in \big[\frac{1}{n_1^2},\infty\big)$, $x\in \bigcup_{n=1\rightarrow \infty} \big[\frac{1}{n_1^2} , \infty\big)$

$\big[\frac{1}{n_1^2} , \infty\big)\subset (0,\infty), \forall n\in \mathbb{N}$, so $\bigcup_{n=1\rightarrow \infty} \big[\frac{1}{n_1^2} , \infty\big) \subseteq (0,\infty)$

Putting the two pieces together, we have $\bigcup_{n=1\rightarrow \infty} \big[\frac{1}{n_1^2} , \infty\big)=(0,\infty)$

Am I right? Please help, thank you!


答: Hint to help you start: $$\left(b+\frac1b\right)^2 = b^2+2+\frac1{b^2}$$

Title: Find $\sum_{k=1}^{1010}(a_k^k+2+\frac{1}{a_k^k})^{2020}$

Body:
Given a polynomial

$P(x)=x^{2020}+x^{2019}+x^{2018}+...+x^2+x+1$

with roots $a_1,a_2,a_3,...,a_{2020}$ Find the value of

$\sum_{k=1}^{1010}(a_k^k+2+\frac{1}{a_k^k})^{2020}$

we can write to be

$\sum_{k=1}^{1010}(a_k^{\frac{k}{2}}+\frac{1}{a_k^{\frac{k}{2}}})^{2020.2}$ = $\sum_{k=1}^{1010}(a_k^{\frac{k}{2}}+\frac{1}{a_k^{\frac{k}{2}}})^{4040}$ = $\sum_{k=1}^{1010}(a_k^k+1)^{4040}$ $(a_k^{-2020k})$

Please give me idea, i don't know where to start.


问:
Show that $a_1a_2 \cdots a_k$ has order $1$ or $2$.

Let $G = \{a_1, \cdots , a_k \}$ be a finite abelian group.  Show that $a_1a_2 \cdots a_k$ has order $1$ or $2$.

So far I have done this:

I think that the product $a_1a_2 \cdots a_k$ will have order $1$ when none of the elements of $G$ is inverse of itself. since by reorganizing the elements in an adequate way we arrive at the identity. Otherwise, if for some $a_i$ in $G$ $a_i$ is inverse of itself, then the product will have order $2$. If so, how could you formalize this?

答: Let $I$ be the subset of $G$ such that $a_i\in I$ implies that $a_i\neq a_i^{-1}$ and $J$ the subsets of $G$ such that $b_i\in J$ implies that $b_i=b_i^{-1}$. The product of elements of $G$ is the product $b_1...b_l$ of elements of $J$, we have $(b_1..b_l)^2=(b_1)^2..(b_l)^2=1$.


问:
Inverse Rotation of Axes
I have rotated point coordinates of original coordinates.
I am trying to find original coordinates from rotated coordinates..

For example;

Original Points: x=0, y=60000

Degree: -45

If i calculate with here:
https://keisan.casio.com/exec/system/1223522781

I found the x=-42,426.406871193, y=42,426.406871193

When I look here, the result is correct.

Now i want to calculate,

If i have,

```<code>degree = -45
rotated_x = -42,426.406871193
rotated_y = 42,426.406871193
</code>```

How can i find original coordinates with above values?

Result must be = (0, 60000)

答: In general, the 2 by 2 Rotation Matrix to rotate a point $\theta$ degrees counter clockwise, is given by $ \begin{bmatrix}cos\theta & -sin\theta\\sin\theta & cos\theta\end{bmatrix}$. The derivation of this matrix can be found elsewhere on the internet. So if I want to rotate $ \begin{bmatrix}-42426 \\42426 \end{bmatrix}$ "back" I can plug in $\theta=-45$ into the matrix and multiply $ \begin{bmatrix}cos(-45) & -sin(-45)\\sin(-45) & cos(-45)\end{bmatrix}$$ \begin{bmatrix}-42426 \\42426 \end{bmatrix}$ and this will give $ \begin{bmatrix}0 \\60000 \end{bmatrix}$ (Forgive my rounding)

Title: Inverse Rotation of Axes


问:
Is there a specific symbol for $\mathbb{N}\cup\lbrace 0\rbrace$?
It is well known that natural numbers start in 1.

However, sometimes people work with a "widened set" of natural numeres plus zero, $\mathbb{N}\cup\lbrace 0\rbrace$. That is, all non-negative integers.

Is there a specific symbol for this set? such as $\mathbb{N}^{*}$, $\mathbb{Z}^+$ or anything similar? Thank you

答: $\mathbb{N}_0$ is the most common choice.


答: I've seen $\mathbb{N}_0$.  But saying $n\ge 0$ is not heavy lifting.

Title: Is there a specific symbol for $\mathbb{N}\cup\lbrace 0\rbrace$?

Body:
It is well known that natural numbers start in 1.

However, sometimes people work with a "widened set" of natural numeres plus zero, $\mathbb{N}\cup\lbrace 0\rbrace$. That is, all non-negative integers.

Is there a specific symbol for this set? such as $\mathbb{N}^{*}$, $\mathbb{Z}^+$ or anything similar? Thank you


问:
Proof of Neumann Lemma
Prove that if $\|A\| < 1$, then $I-A$ is invertible. Here, $\|\cdot\|$ is a matrix norm induced by a vector norm.

This lemma is referred to as Neumann Lemma.

Any ideas on how to go ahead with this?

Thanks.

答:
Neumann Series:

For $A \in  \mathbb{C}^{n \times n},\rho(A)<1$ Since  $\rho(A)< 1.$

Then we know that:$$\lim_{k \to \infty}A^{k}=0$$

the below series converges:

$$(I-A)(I+A+A^{2}+\dots+A_{k}) = I - A^{k} =\sum _{k=0}^{\infty }A^{k},\ \text{(1)}$$

Proof :

\begin{align*}
(I-A)&(I+A+A^{2}+A^{3}+\dots +A^{k-1}) \\
&=I-A+A-A^{2}+A^{2}+\dots+A^{k-1}\\
&=I-A^{k}
\end{align*}

Now taking the limit as $k \to \infty$ and because of $(1)$ we have:

$$ I-\lim_{k \to \infty} A^{k} = I-0=I$$
Therefore coronary of the above,we have that:

\begin{align*}
(I-A)\sum_{k=0}^{\infty}A^{k}&=I \\
\sum_{k=0}^{\infty}A^{k}&=\dfrac{I}{(I-A)} \\
\sum_{k=0}^{\infty}A^{k} &= (I-A)^{-1}
\end{align*}

Since $||I|| =1$ holds we have that :
$$1 = ||I|| \leq ||I-A|| \cdot||(I-A)^{-1}|| \leq  (1+||A||)\cdot ||(I-A)^{-1}  || $$

Noting that $$I = I-A+A$$ and multiplying both sides on the right by $(I-A)^{-1}$ one gets $$(I-A)^{-1} =I+A(I-A)^{-1}.$$

Passing to the norms, we obtain
$$||(I-A)^{-1} || \leq 1+||A|| \cdot ||(I-A)^{-1} || $$ and thus $$||(I-A)^{-1} || \leq \frac{1}{1-||A||}$$

Title: Proof of Neumann Lemma


问:
Relating the Entries of a Block Operator Matrix to Compositions of Projections
I am having some difficulties proving the following statement:

If $V$ and $W$ are vector spaces over a field $F$ such that $V=V_1\oplus V_2$ and $W=W_1\oplus W_2$ for some subspaces $V_i\subseteq V$ and $W_i\subseteq W$ for $i=1,2$ and $A=\begin{bmatrix}A_{11}&A_{12}\\A_{21}&A_{22}\end{bmatrix}$ then $A_{ij}=\Gamma_{W_i}A\Gamma_{V_j}:V_j\to W_i$ for $i,j=1,2$.

Here $\Gamma_{V_j}$ is the projection of $v\in V$ onto $V_j$ along $V_i$ and similarly for $\Gamma_{W_i}$.

I have examined the behavior on an element of $V$ using the relation $Av=w$ but have been unable to establish the necessary equality. Any advice would be greatly appreciated.

答:
Your statement as written is incorrect. We have
$$
A_{ij} = \Gamma_{W_i} A \iota_{V_j},
$$
where $\iota_{V_j}$ is the inclusion map from $V_j$ to $V$.

I assume that $\oplus$ denotes an external direct sum, which is to say that we can write each $v \in V$ as a block vector $(v_1,v_2)$ with $v_i \in V_i$.

For any $v \in V_2$, we have
$$
\Gamma_{W_1} A \Gamma_{V_2} v = \pmatrix{A_{11} & A_{12}\\ A_{21} & A_{22}} \pmatrix{0\\v}
= \Gamma_{W_1}\pmatrix{A_{12}v\\ A_{22} v} = A_{12}v.
$$
Since $\Gamma_{W_1} A \Gamma_{V_2} v = A_{12}v$ for all $v \in V_2$, we conclude that $\Gamma_{W_1} A \Gamma_{V_2} = A_{12}$.

The general proof for $W_i,V_j$ is similar.


问:
$|f|$ has a global minimum where $f:\mathbb{C} \to \mathbb{C}$ is any polynomial function
The following Lemma is used in a proof for the Fundamental Theorem of Algebra, but I cannot understand some steps.

Lemma: Let  $f:\mathbb{C} \to \mathbb{C}$  be any polynomial function.
Then there exists a point  $z_0 \in \mathbb{C}$  where the function
$|f|$  attains its minimum value in  $\mathbb{R}$.

The proof starts like this:

If  $f$  is a constant polynomial function, then the statement of the
Lemma is trivially true since  $|f|$  attains its minimum value at
every point in  $\mathbb{C}$ . So choose, e.g.,  $z_0=0$.

If  $f$  is not constant, then the degree of the polynomial defining
$f$  is at least one. In this case, we can denote  $f$  explicitly by

$f(z)=a_n z^n + ... + a_1 z + a_0$

with $a_n \neq 0$. Now, assume  $z \neq 0$ , and set
$A=max\{|a_0|,…,|a_{n−1}|\}$ . We can obtain a lower bound for
$|f(z)|$  as follows:

$f(z)=|a_n| |z|^n |1+\frac{a_{n-1}}{a_n} \frac{1}{z}+...+\frac{a_0}{a_n} \frac{1}{z^n}| \leq |a_n| |z|^n (1 - \frac{A}{|a_n|} \sum \limits_{k=1}^{\infty} \frac{1}{|z|^k})=|a_n| |z|^n (1 - \frac{A}{|a_n|} \frac{1}{|z|-1})$.

I don't see how these two steps follow. We can distribute the absolute value over all the factors, but how does one turn the second factor from $1+...$ into a $1-...$ using an infinite sum? After that it seems like an application of a geometric series, but how do we know that $|z|>1$?

I've been stuck for quite some while now with no notable progress. Am I missing anything obvious here?

Edit:

The next step of the proof is a follows (so it justifies the use of the formula for the geometric series):

For all $z \in \mathbb{C}$ such that $|z| \geq 2$, we can further
simplify this expression and obtain

$ (*) |f(z)| \geq |a_n| |z|^n (1 - \frac{2A}{|a_n| |z|})$.

It follows from this inequality that there is an $R > 0$ such that $|f(z)| > |f(0)|$, for all $z \in \mathbb{C}$ satisfying $|z| > R$.

Now this confuses me once again. We have $|f(0)|=a_0$, so how does this follow from the inequality given above? I think it ought to have something to do with the definiton of $A$, but I can't quite figure it out.

Could anyone please provide a hint or an explanation?

答:
If $f$ is constant, then any complex number is a global minimum.

If $\deg f\geq 1$ and $a_0=0$, then $0$ is a global minimum.

If $\deg f\geq 1$ and $a_0\neq 0$, we need two lemmas:

$|f|\to \infty$ if $|t|\to \infty$ (Indeed, write $|f(t)|=|a_nt^n||1+...|$ with $|t|>0$).

The continuous map $|f|:\mathbb C\to \mathbb R$, $t\mapsto |f(t)|$ attains its minimum (and maximum) if $t$ is in the closed disc of radius $R$ for any $R>0$ (see Baby Rudin 4.16).

From 1., there is some $R_0>0$ such that $|f(t)|> |a_0|>0$ outside the closed disc of radius $R_0$. From 2., this closed disc contains $z_0$ such that $|f(z_0)|\leq |f(t)|$ for any $t$ in this disc. In particular, for $t=0$, $|f(z_0)|\leq |f(0)|=|a_0|$ so that, $|f(z_0)|\leq |a_0|<|f(t)|$ for any $t$ outside this disc. This shows that $z_0$ is a global minimum.


问:
For a polynomial $p(z)$, prove there exist $R>0$, such that if $|z|=R$, then $|p(z)|\geq |a_n|R^n/2$
For a polynomial $p(z)=a_0+\ldots+a_nz^n$, prove there exist $R>0$, such that if $|z|=R$, then $|p(z)|\geq |a_n|R^n/2$.

I get $|p(z)|\geq |a_n|R^n$, which makes that factor of $1/2$ useless. So I must have done something wrong.

$$|\frac{\partial^n{f}}{\partial z^n}(P)|\leq \frac{\sup_{z\in \overline{D}(P,r)} |f(z)|k!}{r^k}$$.
This is the Cauchy estimate.

$$|\frac{\partial^n{p}}{\partial z^n}(z)| = |a_n n!|$$
$$|\frac{\partial^n{p}}{\partial z^n}(0)|\leq \sup_{\zeta\in \overline{D}(0,|z|)} |p(\zeta)|n!/R^n$$
$$|a_n n!|\leq \sup_{\zeta\in \overline{D}(0,|z|)} |p(\zeta)|n!/R^n$$
$$R^n |a_n|\leq \sup_{\zeta\in \overline{D}(0,|z|)} |p(\zeta)|$$

There exist a $z_0$, such that $|z_0|>0$ and $$\sup_{\zeta\in \overline{D}(0,|z_0|)} |p(\zeta)| = |p(z_0)|$$.

This is saying there exist a disk centered at the origin, such that $|p|$ is maximum on some point on the boundary.

Let $R=|z_0|$, we have
$R^n |a_n|\leq |p(z_0)|$

Edit:
I know what I got wrong. $z_0$ is just one point on the circle $|z| = R$. One can chose another point $z_1$ on the circle, such that $p(z_1)<p(z_0)$.

答:
Let me prove a slightly stronger property than the one you ask about since this can be done in one go:

Let $p(z) = a_{n} z^n + a_{n-1} z^{n-1} + \cdots + a_{0}$ be a complex polynomial with $a_{n} \neq 0$. There exists $R \gt 0$ such that $\frac{1}{2}|a_{n}|\,|z|^n \leq |p(z)| \leq \frac{3}{2} |a_{n}|\,|z|^n$ for all $z$ with $|z| \geq R$.

This is clearly true if $n = 0$, so let's assume $n \geq 1$. Put $q(z) = |a_{n-1}|\,|z|^{n-1} + \cdots + |a_0|$. By the triangle inequality we have the estimates
$$|a_{n}|\,|z|^n - q(z) \leq |p(z)| \leq |a_{n}|\,|z|^n + q(z)
\qquad \text{for all } z \in \mathbb{C}.$$
For $k \leq n-1$ we have $|z|^{k} \leq |z|^{n-1}$ if $|z| \geq 1$, so
$$q(z) \leq (|a_{n-1}| + \cdots + |a_{0}|) |z|^{n-1}
\qquad \text{for all } |z| \geq 1.$$ Putting
$$R = \max{\left\{1, \frac{2}{|a_{n}|}(|a_{n-1}| + \cdots + |a_{0}|)\right\}}$$
we get the desired $R$ since for $|z| \geq R$ we then have $q(z) \leq q(z) \cdot \frac{|z|}{R} \leq \frac{|a_{n}|}{2} |z|^{n}$.


问:
Trigonometrical Question
the question is solve the following equation in the interval

$$0<\theta\leq 360$$
$$\tan(\theta) = \tan(\theta)(2+3\sin(\theta))$$

I got 199.5 and 340.5 as my answers like so:

$\tan(\theta) = \tan(\theta)(2+3\sin(\theta))$

$1=2+3\sin(\theta)$

$\sin(-1/3) = 199.5$ and $340.4$

However in the answer scheme it gives 180, 199.5, 340.5 and 360

How do they get the 2 extra values?

答: The $\theta = 180$ answer is the "trivial solution".  If you plug in $\theta=180$ or 360 you immediately get zeroes on both sides.  </p>

Title: Trigonometrical Question


问:
How to apply integration by parts to $\int_c^d dy \int_a^b dx\; f'(y-x) g(x+y)$
Consider the integral
$$
I(a,b,c,d) := \int_c^d dy \int_a^b dx\; f'(y-x) g(x+y)
$$
where I would like to apply integration by parts on the $x$ variable. My thinking is that I can straightforwardly apply integration by parts by defining $v'(x):=f'(y-x)$ and $u(x):= g(x+y)$ which implies that $v(x) = -f(y-x)$ and $u'(x):= g'(x+y)$, which would naively imply
$$
I(a,b,c,d) = \int_c^d dy \bigg[ -f(y-x) g(x+y) \bigg]_{x = a}^{x = b} - \int_c^d dy \int_a^b dx\; \big[ -f(y-x) \big] g'(x+y) \ . \ \ \ \ \  (**)
$$

However this link seems to imply that extra care needed to be taken with the boundary terms in the above. Is (**) correct here?

答:
The other link is dealing with a more general situation than you have here.

$$\int_c^d dy \int_a^b dx\; f'(y-x) g(x+y)$$

Literally means the integral $\int_c^d dy\;h(y)$, where the function $h$ is defined by $$h(y) = \int_a^bdx\; f'(y-x) g(x+y)$$
And in performing that integration, $y$ is a constant, so $f'(y-x)$ and $g(x + y)$ are, for its sake, just functions of $x$. You have applied integration by parts perfectly to show that $$h(y) = \bigg[ -f(y-x) g(x+y) \bigg]_{x = a}^{x = b} - \int_a^b dx\; \big[ -f(y-x) \big] g'(x+y)$$
for every $y$.
And therefore the rest of your expression follows naturally.


问:
Dirac delta function as a limit of sinc function
I'm looking for a rigorous proof of the statement:
$\delta(x) = \lim_{\epsilon->0} \frac{\sin(x/\epsilon)}{\pi x}$
(see (37)).

For any non-zero value of x, LHS of the above is by definition zero. But, for any non-zero value of x, the limit in RHS simply does not exist. So, how is this statement proved?

答:
In addition to the answer by Ian, one very important lemma you need to prove the second part, i.e
$$\lim_{\varepsilon \to 0^+} \int_{K \setminus (-\delta,\delta)} g_\varepsilon(x) dx = 0.$$
is the Riemann-Lesbegue lemma which shows how an increasingly oscillatory function under integration approaches zero measure. You can find it mentioned here :- https://onlinelibrary.wiley.com/doi/pdf/10.1002/9780470723876.app3(Page 498)


问:
Help me prove $K(k) = \frac{1}{1+k}K\left(\frac{2\sqrt{k}}{1+k}\right)$ where $K(k) = \int^{\pi/2}_0 \frac{du}{\sqrt{1-(k\sin{u})^2}}$
Hi this problem is posed in Sean M. Stewart's How to Integrate It book pg.143

$$K(k) = \int^{\pi/2}_0 \frac{du}{\sqrt{1-(k\sin{u})^2}}$$ where $ 0 \leq k < 1$

Based on the substitution:

$$ \tan{u} = \frac{\sin{2\theta}}{k+k\cos{2\theta}} =\frac{\tan{\theta}}{k}  $$

Show that:

$$K(k) = \frac{1}{1+k}K\left(\frac{2\sqrt{k}}{1+k}\right)$$

$K(k) $ is some function defined in term of definite-integral called complete elliptic integral which is not easily computable, so attacking $K(k)$ gave no fruitful result.

Well then proceeding with substitution ( It is easy to lose track as it  involved a lot of trigonometric manipulations )

I reduced the integral into this form :
$$K(k) = k\int^{\pi/2}_0 \frac{\sec^2{\theta}d\theta}{\sqrt{k^2 + \tan^2{\theta}}\sqrt{k^2 + \tan^2{\theta} - k^2\tan^2{\theta}}}$$

Using substitution : $\tan{\theta} = p$

$$K(k) = k\int^{\infty}_0 \frac{dp}{\sqrt{k^2 +p^2}\sqrt{k^2 +p^2 - (kp)^2}}$$

To my naive eyes this looked manageable but WolframAlpha says otherwise.

Any hint or ideas on how to attack this problem?

I thought  the trig manipulations to find a more suitable form but in the I end up transform $\sin {u}$ term into $\tan{u}$ to use the given substitution

EDIT:  It has been suggested in commens that there is typo in the substituion(given in book) , it should be :

$$ \tan{u} = \frac{\sin{2\theta}}{k+\cos{2\theta}} $$

答:
Hints=    make clever substitute
$$u=x+arctan(\frac{(1-k)tanx}{(1+k)})$$
$$then..u\rightarrow\pi;x\rightarrow\frac{\pi}{2}$$
$$u\rightarrow0;x\rightarrow0$$

Any others can you send s.m.Stewart's book pdf link


问:
Regularized sum of $1+(1+\frac1{2^2}) + (1+\frac1{2^2}+\frac1{3^2})+\cdots$, or $\sum_{k=1}^\infty \sum_{j=1}^k\frac1{j^2}$?
For my exercises on divergent summation I try to find a regularization of
$$   S^{^\star}_2 \underset{\mathcal R}= \sum_{k=1}^\infty \sum_{j=1}^k \frac1{j^2} \tag 1
$$
It is not difficult to derive the formula for the partial sums introducing the harmonic numbers of order $1$ and $2$
$$H_1(N)=\sum_{k=1}^N \frac1{k} \\
  H_2(N)=\sum_{k=1}^N \frac1{k^2} \tag {2.1}$$
and writing
$$ S_2(N) = (N+1) \cdot H_2(N) - H_1(N) \tag {2.2}$$
When going to the limit $N \to \infty$ and the serial summation according to eq 2.1 cannot be done, we can -using the $\psi$ (or "digamma", in Pari/GP  psi(1+N))-function-   approximate  $$H_1(N) \sim \gamma + \psi(1+N)  \tag {3.1}$$  and for $H_2(N)$ the derivative of the $\psi()$:
$$ H_2(N) \sim \zeta(2)-\psi^{(1)}(1+N) \tag {3.2}
$$

I don't see how I can proceed now. I'd assume for the regularization in the case for $N$ in the limit I can use $\gamma$ for the $H_1^{^\star}$.

Q:But how can I the do the $H_2$-part for the limit formula for eq (3.2) resp. (2.2)?

A bit background: In a treatize on the triangle of Eulerian numbers (pg.8) I hand-waving-generalize the triangle towards negative row numbers, and for the row-sum of the $-2$'nd row I get those formal sums.

update I just found my 2013-question about the same subject, and there was an answer but which I could not then believe it is a final answer, and still I'm unsure whether (and how) I can proceed to a "fixed statement" see earlierquestion The user @did gave the answer
$$
\lim_{n\to\infty}\left(s_2(n)-n\frac{\pi^2}6+\log n\right)=\frac{\pi^2}6-1-\gamma \qquad \text{introduced by me: } =\zeta(2)-1 - \gamma
$$

答:
Similair as above but more in line with the rest of the picure you posted

The eulerian numbers(s,n): $\Delta^{s+1}n^s$ cut-off at 1. (so you ignore 0 and bellow when delta'ing).

Upper limit when s is positive integer s btw.

$\sum_{n=1}^{\infty} \frac{ Eulerian.numbers(s,n) c^n}{(1-c)^{s+1}}=\sum_{n=1}^{\infty}c^n n^s$

Now at the negative side:

Eulerian numbers(-s,n): $.^{s-1}\sum_{m=1}^{n}m^{-s}$ (the s'th -1 'indefinite' sum, I aways struggled with notation for this cause it's just similair to a multiple integral).

In your case with s=-2.
Eulerian numbers(-2,n): $\sum_{n=1}^{k}n^{-2}$.

$\sum_{n=1}^{\infty}  \sum_{n=1}^{k}n^{-2} c^n(1-c)=\sum_{n=1}^{\infty}c^n n^{-2}$

You can "see it" by (1-c) which works like  a delta operator.
$$\sum_{n=1}^{\infty}  \sum_{n=1}^{k}n^{-2} c^n=\frac{\sum_{n=1}^{\infty}c^n n^{-2}}{1-c}$$

From here on figure the limit as c goes to 1, by expending the polynomal. What the answer previous stated before was that you could disregard the divergent parts, which is I think incorrect, e.g. zeta(1) which is also here the case I think.

An intuitive example:

$$\sum_{n=1}^{\infty}  \frac{c^n}{n^2 (1-c)}$$
Let  c goes to 1+h with h being small. Only looking at the $h^0$ order gives
$$\sum_{n=1}^{\infty} \frac{(1+h)^n}{n^2 (-h)}= \sum_{n=1}^{\infty} \frac{-1}{n}=-\zeta(1)$$

Now like you normally would (or see directly from it series it's going to contain a full zeta(1)).

$$\sum_{k=1}^{\infty}  \sum_{n=1}^{k}n^{-2}=\sum_{k=1}^{\infty}\pi^2/6-\frac{1}{k}+\frac{1}{2k^2}-\frac{1}{6k^3}+\frac{1}{30k^5}-...$$

$$\sum_{n=1}^kn^{-2}=\bigg(\pi^2/6-\sum_{j=-1}^{\infty}\zeta(-j)(k^{-2-j}) \frac{(-2)!}{(-2-j)!(j)!}\bigg)$$

$$\sum_{k=1}^{\infty}  \sum_{n=1}^{k}n^{-2}=-\pi^2/12-\sum_{j=-1}^{\infty}\zeta(-j)\zeta(2+j) \frac{(-2)!}{(-2-j)!(j)!}$$
And as you see $\zeta(1)$ is part of the sum at j=-1 twice and it will only gets reduced once by (-1)!. There for the regularized value got a pole, e.g. contains zeta(1).

Or better writen:

$$\sum_{k=1}^{S\infty}  \sum_{n=1}^{k}n^{-2}=-\zeta(1)-\sum_{j=1}^{\infty}\zeta(-j)\zeta(2+j) (-1)^{j+1} (j+1) $$

And even though the other sum also diverges it seems unlikely it contains zeta(1) but yeah it needs a more convincing argument/needs to be regularized first aswell. But it seems to be regulaziable.

Ps: doesn't it alternate in the picture? And thuse is trivial. It's used for c isn't 1 to very easily calculate the zeta function. I'd say it's the basics. Cause if it doesn't alternate you also get for sum 1/n just zeta(1) in the entry before.


问:
Estimating Work Done Line Integrals with Figure
I'm doing some problems in James Stewart's Calculus Concepts and Contexts (2nd ed.), and stumbled upon question 37 in section 13.2: Line Integrals, and it is giving me a bit of trouble. Here is a link to a screenshot of the problem: https://i.imgur.com/6Am6MPs.png.

My interpretation: We can estimate the work done, in joules, as the sum of the dot products between all the unit tangents and their respective force vectors provided by the diagram. This works because we only care about how much force these vectors act in the direction of the object that's moving.

So at quick glance, we have 2+2+2+2+1+1+1=11, but the answer book says that the answer is 22. I am a bit confused where this constant of 2 is coming from. Thanks for the help!

答:
$W = \Sigma F \cdot s$, or work = force $\times$ distance.

The distance moved by the object in each step is $2$, so you have $11 \times 2 = 22$.

Note to readers: the force in the direction of motion, not the total force.


问:
Do all LCG-based PRNGs suffer from predictable patterns?
I needed to produce trivial (low-quality) random integers and remembered how simple linear congruential generators were to implement from school:

Went to Wikipedia, found the first example which appeared to produce 32 bits (sourced from the ubiquitous Numerical Recipes):

Jammed this in as the following Perl one-liner, which takes the 10th iteration (n = 10) and the current Unix time as seed:

```<code>perl -le '$n = shift; for (1..10) { $n = int(1664525 * $n + 1013904223) % 2**32 } print $n' $(date +%s)
</code>```

And I was a little surprised to see the last digit in the output was always incrementing by 1. So I printed all of the iterates to find that this pattern occurs only in the 10th iteration:

```<code>~ perl -le '<span class="math-container">$n = shift; print "seed = $</span>n"; for (1..10) { <span class="math-container">$n = int(1664525 * $</span>n + 1013904223) % 2**32; printf "%2d = %d\n", <span class="math-container">$_, $</span>n }' $(date +%s)
 seed = 1603820918
  1 = 3975067229
  2 = 810304536
  3 = 3116890263
  4 = 2673982730
  5 = 3148974305
  6 = 3740634316
  7 = 619471291
  8 = 3601034206
  9 = 3361989029
 10 = 1048979136     # <-  6

~ perl -le '<span class="math-container">$n = shift; print "seed = $</span>n"; for (1..10) { <span class="math-container">$n = int(1664525 * $</span>n + 1013904223) % 2**32; printf "%2d = %d\n", <span class="math-container">$_, $</span>n }' $(date +%s)
 seed = 1603820919
  1 = 3976731754
  2 = 1199874241
  3 = 1762722604
  4 = 2832966811
  5 = 1716457790
  6 = 2657060741
  7 = 2470761248
  8 = 3240913919
  9 = 1251936594
 10 = 1295718537     # <-  7

~ perl -le '<span class="math-container">$n = shift; print "seed = $</span>n"; for (1..10) { <span class="math-container">$n = int(1664525 * $</span>n + 1013904223) % 2**32; printf "%2d = %d\n", <span class="math-container">$_, $</span>n }' $(date +%s)
 seed = 1603820920
  1 = 3978396279
  2 = 1589443946
  3 = 408554945
  4 = 2991950892
  5 = 283941275
  6 = 1573487166
  7 = 27083909
  8 = 2880793632
  9 = 3436851455
 10 = 1542457938     # <-  8

~ perl -le '<span class="math-container">$n = shift; print "seed = $</span>n"; for (1..10) { <span class="math-container">$n = int(1664525 * $</span>n + 1013904223) % 2**32; printf "%2d = %d\n", <span class="math-container">$_, $</span>n }' $(date +%s)
 seed = 1603820921
  1 = 3980060804
  2 = 1979013651
  3 = 3349354582
  4 = 3150934973
  5 = 3146392056
  6 = 489913591
  7 = 1878373866
  8 = 2520673345
  9 = 1326799020
 10 = 1789197339     # <-  9

~ perl -le '<span class="math-container">$n = shift; print "seed = $</span>n"; for (1..10) { <span class="math-container">$n = int(1664525 * $</span>n + 1013904223) % 2**32; printf "%2d = %d\n", <span class="math-container">$_, $</span>n }' $(date +%s)
 seed = 1603820922
  1 = 3981725329
  2 = 2368583356
  3 = 1995186923
  4 = 3309919054
  5 = 1713875541
  6 = 3701307312
  7 = 3729663823
  8 = 2160553058
  9 = 3511713881
 10 = 2035936740     # <-  0

~ perl -le '<span class="math-container">$n = shift; print "seed = $</span>n"; for (1..10) { <span class="math-container">$n = int(1664525 * $</span>n + 1013904223) % 2**32; printf "%2d = %d\n", <span class="math-container">$_, $</span>n }' $(date +%s)
 seed = 1603820923
  1 = 3983389854
  2 = 2758153061
  3 = 641019264
  4 = 3468903135
  5 = 281359026
  6 = 2617733737
  7 = 1285986484
  8 = 1800432771
  9 = 1401661446
 10 = 2282676141     # <-  1

~ perl -le '<span class="math-container">$n = shift; print "seed = $</span>n"; for (1..10) { <span class="math-container">$n = int(1664525 * $</span>n + 1013904223) % 2**32; printf "%2d = %d\n", <span class="math-container">$_, $</span>n }' $(date +%s)
 seed = 1603820924
  1 = 3985054379
  2 = 3147722766
  3 = 3581818901
  4 = 3627887216
  5 = 3143809807
  6 = 1534160162
  7 = 3137276441
  8 = 1440312484
  9 = 3586576307
 10 = 2529415542     # <-  2

~ perl -le '<span class="math-container">$n = shift; print "seed = $</span>n"; for (1..10) { <span class="math-container">$n = int(1664525 * $</span>n + 1013904223) % 2**32; printf "%2d = %d\n", <span class="math-container">$_, $</span>n }' $(date +%s)
 seed = 1603820925
  1 = 3986718904
  2 = 3537292471
  3 = 2227651242
  4 = 3786871297
  5 = 1711293292
  6 = 450586587
  7 = 693599102
  8 = 1080192197
  9 = 1476523872
 10 = 2776154943     # <-  3

~ perl -le '<span class="math-container">$n = shift; print "seed = $</span>n"; for (1..10) { <span class="math-container">$n = int(1664525 * $</span>n + 1013904223) % 2**32; printf "%2d = %d\n", <span class="math-container">$_, $</span>n }' $(date +%s)
 seed = 1603820926
  1 = 3988383429
  2 = 3926862176
  3 = 873483583
  4 = 3945855378
  5 = 278776777
  6 = 3661980308
  7 = 2544889059
  8 = 720071910
  9 = 3661438733
 10 = 3022894344     # <-  4

~ perl -le '<span class="math-container">$n = shift; print "seed = $</span>n"; for (1..10) { <span class="math-container">$n = int(1664525 * $</span>n + 1013904223) % 2**32; printf "%2d = %d\n", <span class="math-container">$_, $</span>n }' $(date +%s)
 seed = 1603820927
  1 = 3990047954
  2 = 21464585
  3 = 3814283220
  4 = 4104839459
  5 = 3141227558
  6 = 2578406733
  7 = 101211720
  8 = 359951623
  9 = 1551386298
 10 = 3269633745     # <-  5

~ perl -le '<span class="math-container">$n = shift; print "seed = $</span>n"; for (1..10) { <span class="math-container">$n = int(1664525 * $</span>n + 1013904223) % 2**32; printf "%2d = %d\n", <span class="math-container">$_, $</span>n }' $(date +%s)
 seed = 1603820928
  1 = 3991712479
  2 = 411034290
  3 = 2460115561
  4 = 4263823540
  5 = 1708711043
  6 = 1494833158
  7 = 1952501677
  8 = 4294798632
  9 = 3736301159
 10 = 3516373146     # <-  6

~ perl -le '<span class="math-container">$n = shift; print "seed = $</span>n"; for (1..10) { <span class="math-container">$n = int(1664525 * $</span>n + 1013904223) % 2**32; printf "%2d = %d\n", <span class="math-container">$_, $</span>n }' $(date +%s)
 seed = 1603820929
  1 = 3993377004
  2 = 800603995
  3 = 1105947902
  4 = 127840325
  5 = 276194528
  6 = 411259583
  7 = 3803791634
  8 = 3934678345
  9 = 1626248724
 10 = 3763112547     # <-  7

~ perl -le '<span class="math-container">$n = shift; print "seed = $</span>n"; for (1..10) { <span class="math-container">$n = int(1664525 * $</span>n + 1013904223) % 2**32; printf "%2d = %d\n", <span class="math-container">$_, $</span>n }' $(date +%s)
 seed = 1603820930
  1 = 3995041529
  2 = 1190173700
  3 = 4046747539
  4 = 286824406
  5 = 3138645309
  6 = 3622653304
  7 = 1360114295
  8 = 3574558058
  9 = 3811163585
 10 = 4009851948     # <-  8

~ perl -le '<span class="math-container">$n = shift; print "seed = $</span>n"; for (1..10) { <span class="math-container">$n = int(1664525 * $</span>n + 1013904223) % 2**32; printf "%2d = %d\n", <span class="math-container">$_, $</span>n }' $(date +%s)
 seed = 1603820931
  1 = 3996706054
  2 = 1579743405
  3 = 2692579880
  4 = 445808487
  5 = 1706128794
  6 = 2539079729
  7 = 3211404252
  8 = 3214437771
  9 = 1701111150
 10 = 4256591349     # <-  9
</code>```

So I'm curious if this is a quirk of the programming? Of the chosen a, c, and m parameters? Or do all LCGs suffer from these low-period patterns?

答:
You are comparing the same iteration number from different seeds, which is not the typical use of a random number generator.  Usually you fix a seed and take some number of iterations.

There are patterns in other iteration numbers too; for example, if you look at iteration 1 from all your seeds, the last digit alternates between 4 and 9.

Whether this pattern is a problem depends on your application, because it arises from a nonstandard use of the random number generator. If you are generating a statistical sample of a data set, for example, it is probably no big deal, because you will not be using multiple seeds anyway (or if you do, they will be separated by a random amount of time). If used in an adversarial context where someone might be trying to reverse-engineer your random numbers (e.g., cryptographic purposes), it might cause a problem.


问:
$T(n)=T(\frac{n}{3})+T(\frac{2n}{3})+cn$
I'm trying to understand the following proof from 'Introduction to Algorithms' that the recurrence $T(n)=T(n/3)+T(2n/3)+cn$ has solution $O(n lg n)$.

I'm stuck on the fourth step when he replaces $-d(2n/3)lg(3/2)$ with just $(2n/3)lg(3/2)$ i.e. he replaces $-d$ with $1$.

Should this be an inequality rather than an equality. It would make sense if it was less than or equal to.

$d \ge c/(lg(3)-2/3)$

答: They just factored out $-d$, and the coefficient is not replaced by $1$. Thus the sign $=$ is correct.


问:
Is there many statements of the universal coefficient theorem?
I have been using the following statement of the UCT:

$$H_{i}(C;G) \cong H_{i}(C) \otimes G \oplus \operatorname{Tor}(H_{i-1}(C),G).$$

But now I want to calculate the cohomology ring of a certain map and I know its homology groups, is there a statement of UCT that help me in calculating the cohomology groups knowing the homology groups? and how this form is related to the form of UCT I mentioned above?

答:
Given an abelian group $G$ and a chain complex $(C_\bullet, \partial_\bullet)$, the universal coefficient theorem for cohomology is given by the short exact sequence

$$0\to \operatorname{Ext}(H_{n-1}(C),G) \to H^n(C_n,G) \to \operatorname{Hom}(H_n(C),G)\to 0$$ where $H_*(C)$ is homology with integral coefficients, i.e. $$H_*(C) = H_*(C,\mathbb{Z})$$

The upshot of this is that it allows you to compute cohomology with coefficients in $G$ purely from homology with integral coefficients.

Note that obviously your homology and cohomology are dual exactly if $\operatorname{Ext}  (H_{n-1}(C),G)$ vanishes, which is precisely the case whenever $H_{n-1}(C)$ is free abelian.

Therefore, if your homology group $H_{n-1}(C)$ happens to be free abelian, you have

$$H^n(C_n,G) \cong \operatorname{Hom}(H_n(C),G).$$


问:
change of variable to get a quasi-cartesian laplace equation?
when writing the (vector) laplace equation in cylindrical coordinates in a $(r,\theta)$ plane, we get:
$$ \left( r^2\frac{\partial^2}{\partial r^2} + r\frac{\partial}{\partial r} + \frac{\partial^2}{\partial\theta^2} \right)A_z(r,\theta)=0 $$
the term $r^2$ is multiplied by the two sides of the equation to form the cauchy-Euler equation.

now through a change of variable of $v=\ln(r)$:
$$ \left( \frac{\partial^2}{\partial v^2} + \frac{\partial^2}{\partial\theta^2} \right)A_z(v,\theta)=0 $$
it is like a cartesian laplace operator, which has its advantages of simplification of computations.

however in $(r,z)$ plane the laplace equation becomes as following:
$$ \left( r^2\frac{\partial^2}{\partial r^2} + r\frac{\partial}{\partial r} -1 + \frac{\partial^2}{\partial z^2} \right)A_\theta(r,z)=0 $$
is it even possible through a change of $u=f(r)$ to get a result like the same?:
$$ \left( \frac{\partial^2}{\partial u^2} + \frac{\partial^2}{\partial z^2} \right)A_\theta(u,z)=0 $$

additional notes:it is worth noting that in the above equations, the 3D vector-laplace equation has been reduced to 2D plane due to symmetry of the problem.

答:
$$r^2\nabla^2 = \left( r^2\frac{\partial^2}{\partial r^2} + r\frac{\partial}{\partial r} + \frac{\partial^2}{\partial\theta^2} \right)$$
The Laplacien should be:
$$\nabla^2=\frac{\partial^2}{\partial r^2} + \dfrac 1r\frac{\partial}{\partial r} + \dfrac 1 {r^2}\frac{\partial^2}{\partial\theta^2} $$
And:
$$\nabla^2=\frac{\partial^2}{\partial r^2} + \dfrac 1r\frac{\partial}{\partial r} + \dfrac 1 {r^2}\frac{\partial^2}{\partial\theta^2} +\dfrac {\partial^2}{\partial z^2} $$


问:
Be $\tau_{1}$ and $\tau_{2}$ topologies over $\mathbb{N}$
Be $\tau_{1}$ and $\tau_{2}$ topologies over $\mathbb{N}$ defined by:

$\tau_{1}=\{\{m\in \mathbb{N}:m<n\}:n\in \mathbb{N}\}\cup \{\mathbb{N}\}\}$ and $\tau_{2}=\{A\subseteq \mathbb{N}: 0\in A\}\cup \{\emptyset\}$.

Check who are these topologies and prove which is the finest.

答: If $U\in\tau_1$, then since $0\in U$, or $U=\emptyset$,  $U\in\tau_2$.  That is, $\tau_2$ is finer.


答:
The topology consisting of all subsets containing a fixed point $x$ is called the particular/included point topology. In your case $x=0$. You can simply say that $\tau_1\subset \tau_2$ i.e. $\tau_2$ is strictly finer than $\tau_1$ since:

For all subsets $S\ne\emptyset\in\tau_1,0\in S\subseteq\Bbb N$ and thus $S\in\tau_2$, so $\tau_1\subseteq \tau_2$.

$\{0,2\}\in\tau_2$ but does not belong to $\tau_1$. Hence $\tau_1\ne\tau_2$.

Body:
Be $\tau_{1}$ and $\tau_{2}$ topologies over $\mathbb{N}$ defined by:

$\tau_{1}=\{\{m\in \mathbb{N}:m<n\}:n\in \mathbb{N}\}\cup \{\mathbb{N}\}\}$ and $\tau_{2}=\{A\subseteq \mathbb{N}: 0\in A\}\cup \{\emptyset\}$.

Check who are these topologies and prove which is the finest.


问:
Catalan numbers in polygons
I'm stuck on such problem: triangulation of the $n$-gon is division of said $n$-gon into $(n-2)$ triangles whose sides are either sides of the $n$-gon or certain non-intersecting diagonals. How many triangulations of the $n$-gon are there? How many of these are there such that every triangle has at least one side that is also a side of the $n$-gon?

The first part is easy - I've managed to deliver a Catalan number recursive expression and then found a formula for $n^{\text{th}}$ Catalan number which is $$ C_{n} =  \frac{1}{n+1}  {2n \choose n}.$$

What about the second part? Shall it be done by subtracting some set from $n$-th Catalan number ? What should it look like?

Thank you in advance :)

答:
I'll show how we can do this for an $n$-gon in $n2^{n-5}$ ways. First, pick two adjacent sides of the $n$-gon, and add a diagonal to complete the triangle. There are $n$ ways to pick these sides (just pick the vertex they share). The diagonal that we added shares a vertex with two other sides of the $n$-gon (excluding the ones we started with) and can form a triangle with either of them by adding a new diagonal. We can repeat this process on the newest diagonal added until we've triangulated the $n$-gon. The picture below shows an example for the hexagon.

There were $n$ ways to choose the starting diagonal, and we had two ways of adding each of the $n-4$ subsequent diagonals. This counts $n2^{n-4}$ triangulations. However, for any triangulation obtained this way, we could start with the last diagonal added and obtain the same triangulation, so we counted each triangulation twice. Hence, there are $n2^{n-5}$ such triangulations.

Title: Catalan numbers in polygons


答:
Adding color, and changing some of the brackets from round to square, the fourth step is
$$
\begin{aligned}
&\left[ \color{red}{d(n/3)\lg n} - d(n/3)\lg 3\right] +
\left[\color{red}{d(2n/3)\lg n}-d(2n/3)\lg(3/2)\right] + cn\\
&=\color{red}{dn\lg n}-d\left[(n/3)\lg 3+(2n/3)\lg(3/2)\right]+cn
\end{aligned}
$$
Looks like they factored out $-d$; it wasn't replaced by $1$.

Title: $T(n)=T(\frac{n}{3})+T(\frac{2n}{3})+cn$

Body:
I'm trying to understand the following proof from 'Introduction to Algorithms' that the recurrence $T(n)=T(n/3)+T(2n/3)+cn$ has solution $O(n lg n)$.

I'm stuck on the fourth step when he replaces $-d(2n/3)lg(3/2)$ with just $(2n/3)lg(3/2)$ i.e. he replaces $-d$ with $1$.

Should this be an inequality rather than an equality. It would make sense if it was less than or equal to.

$d \ge c/(lg(3)-2/3)$


答:
If $\tan\theta=0$, the equation obviously holds. That gives the solutions $180^\circ$ and $360^\circ$. We can only "cancel" $\tan\theta$ if $\tan\theta\ne 0$. When we cancel, we are potentially throwing away some roots.

When we cancel, we get $1=2+3\sin\theta$, or equivalently $\sin\theta=-\frac{1}{3}$. This has the solutions that you found.

Let's look at a related example, $x=x(x^2-3)$. This has the obvious solution $x=0$. For $x\ne 0$, the equation is equivalent to $1=x^2-3$, which has the solutions $x=\pm 2$.

Remark: Instead of immediately cancelling, we could rewrite the equation as $\tan\theta(+3\sin\theta)-\tan\theta=0$, or equivalently as $\tan\theta(1+3\sin\theta)=0$. A product is $0$ if and only if one of the terms is $0$. This gives us the possibilities $\tan\theta=0$ and $1+3\sin\theta=0$.

"Cancelling" is quicker, but carries a risk of throwing away some solutions.

Title: Trigonometrical Question

Body:
the question is solve the following equation in the interval

$$0<\theta\leq 360$$
$$\tan(\theta) = \tan(\theta)(2+3\sin(\theta))$$

I got 199.5 and 340.5 as my answers like so:

$\tan(\theta) = \tan(\theta)(2+3\sin(\theta))$

$1=2+3\sin(\theta)$

$\sin(-1/3) = 199.5$ and $340.4$

However in the answer scheme it gives 180, 199.5, 340.5 and 360

How do they get the 2 extra values?


问:
Why does this problem have an answer of 70 degrees instead of 110?
So I tried to solve a geometry problem. It says:

Triangle $PAB$ is formed by three tangents to circle $O$ and $\angle APB = 40\deg$, then $\angle AOB$ equals:

(A) 45 deg
(B) 50 deg
(C) 55 deg
(D) 60 deg
(E) 70 deg

I tried to solve the problem and came up with $110 \deg$. But the shown correct answer is $(E)$ $70 \deg$, which is exactly the supplement of my answer. And my interpretation of the problem is that circle with center $O$ is inscribed in triangle $PAB$. That leads to an answer of $110 \deg$ and an obviously obtuse angle of $\angle AOB$.

So which is the correct answer? Are both answers correct?
Geogebra picture of my interpretation:

答:
Assuming you have understood and communicated the question correctly, then your answer of $110^\circ$ is correct.  See the following diagram.

Since $\triangle PA'O \cong \triangle PB'O$ are right triangles, $\angle A'PO \cong \angle B'PO$ and both have measure $20^\circ$, thus $\angle A'OP \cong \angle B'OP = 90^\circ - 20^\circ = 70^\circ$.  Similarly, $\angle A'OB \cong \angle P'OB$ and $\angle B'OA \cong \angle P'OA$, so $$\angle AOB = \frac{360^\circ - 140^\circ}{2} = 110^\circ.$$


问:
$a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$

Q. If $a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$, prove that $a = b = c = 0$ ($a$, $b$, $c$ and $p$ are rational and $p$ is not a perfect cube.)

My approach:

Solving the quadratic, I get:

$p^\frac{1}{3} = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2c}$

Case 1: If the $b^2 - 4ac$ is a perfect square, I get the LHS as irrational and the RHS as rational, which is a contradiction.

Case 2: If $b^2 - 4ac$ is not a perfect square, $b =  \pm \sqrt{b^2 - 4ac} - 2cp^\frac{1}{3}$

Here, the LHS is rational and the RHS is irrational, contradiction again. (Edit: The answer of @GNUSupporter has the proper proof.)

So the equation is not quadratic and $c = 0$.

$a + bp^\frac{1}{3} = 0$

$-\dfrac{a}{b} = p^\frac{1}{3}$

This is a contradiction and hence $b = 0$ and $a = 0$

Is there any other way to solve this?

答:
Let $K= \mathbb{Q}(p^{\frac{1}{3}}) \cong \mathbb{Q}[x]/(x^3-p).$ We have the following:

$\text{Tr}_{K/\mathbb{Q}} (p^{\frac{1}{3}}) = 0 $, as the minimal polynomial of $p^{\frac{1}{3}}$ is  $x^3-p.$

$\text{Tr}_{K/\mathbb{Q}} (p^{\frac{2}{3}}) = 0 $, as the minimal polynomial of $p^{\frac{2}{3}}$ is  $x^3-p^2.$

$\text{Tr}_{K/\mathbb{Q}} (a) = 3a$, for all $a \in \mathbb{Q}.$

Now applying the trace to your equation we get
\begin{align*}
\text{Tr}_{K/\mathbb{Q}}(a + bp^\frac{1}{3} + cp^\frac{2}{3}) &= \text{Tr}_{K/\mathbb{Q}}(a)+ b \text{Tr}_{K/\mathbb{Q}}(p^\frac{1}{3} ) + c  \text{Tr}_{K/\mathbb{Q}}(p^\frac{2}{3})\\
 &= \text{Tr}_{K/\mathbb{Q}}(a)+0+0 = 3a = 0= \text{Tr}_{K/\mathbb{Q}}(0),\\
\end{align*}
thus $a=0.$ Next multiply your equation by $p^{\frac{1}{3}}$, apply the trace, and conclude that $c=0.$ Repeat.

Edit:
As Paramanand Singh correctly points out the problem reduces to showing that $f(x)= x^3-p$ is the minimal polynomial of $p^{\frac{1}{3}},$ which I assumed. However, this follows directly from Eisenstein's Criterion and the fact that $f(p^{\frac{1}{3}})=0.$ In light of this information, the polynomial given by the OP is of degree $2$, thus must be the zero polynomial.


问:
A question about pairwise mutual exclusive events
There is a common problem in the field of probability as follows:

"On a vessel containing 3 white and 5 black balls, 4 balls are transferred into an empty vessel. From this vessel a ball is drawn and it is found to be white. What is the probability that out of 4 balls transferred 3 are white and 1 is black?"

The solution is given on page 335 of the book "Engineering Mathematics by Sarveswara Rao Koneru (Universities Press, 2002)" as follows:

"The four balls transferred can be only one of the following combinations:

Event $A_1$ : 4B + 0W

Event $A_2$ : 3B + 1W

Event $A_3$ : 2B + 2W

Event $A_4$ : 1B + 3W

$A_1$, $A_2$, $A_3$ and $A_4$ are pairwise mutual exclusive since just one of them must happen so if E is the event of choosing a white ball from the second vessel,

$E = EA_1 \cup EA_2 \cup EA_3 \cup EA_4$

(and the answer continues to the end...)"

My question is how does the above equation hold and how does the author concluded that?

答:
$\{A_1,A_2,A_3,A_4\}$ are not just pairwise mutually exclusive events, they are also exhaustive.  This is a set of events which partitions the outcome set.

Let $\Omega$ be the set of all possible outcomes for drawing four balls from the vessel.   Since any outcome must be in one among the events from $\{A_1,A_2,A_3,A_4\}$, then $\Omega=A_1\cup A_2\cup A_3\cup A_4$.   Thus they are exhaustive over $\Omega$.

This exhaustive property is what is used to claim:

$$\begin{align}E&=E\cap\Omega\\&=E\cap(A_1\cup A_2\cup A_3\cup A_4)\\&=(E\cap A_1)\cup(E\cap A_2)\cup(E\cap A_3)\cup(E\cap A_4)\end{align}$$

Mutual exclusivity comes next.

Since any event in $E$, is an event in $\Omega$, it  must be in exactly one among the events from $\{A_1,A_2,A_3,A_4\}$, and therefore in exactly one among the events from $\{(E\cap A_1),(E\cap A_2),(E\cap A_3),(E\cap A_4)\}$.  Therefore this later set of events is also pairwise mutually exclusive.   The probability of a union of pairwise mutually exclusive events equals the sum of the probabilities of those events.

As we have already shown that this union equals $E$, thus:
$$\mathsf P(E)=\mathsf P(E\cap A_1)+\mathsf P(E\cap A_2)+\mathsf P(E\cap A_3)+\mathsf P(E\cap A_4)$$


问:
Derive half angle formula $ \cos\frac x 2= \sqrt{\frac {1 + \cos x} {2}}$

How  can the cosine of half angle formula
$$ \cos(\frac x 2) = \sqrt{\frac {1 + \cos x} {2}}$$
can be derived using the bisector angle theorem $ OA:OB = AM:MB$ for the $ \triangle AOM $?

Can anybody explain this to me?

答:
Beside  $\frac{AM}{BM} = \frac{OA}{OB}$ per the bisector angle theorem, $MN= \frac{BM}{OM}AM$ holds from similar triangles. Substitute them below

\begin{align}
\cos^2\frac x2= \frac{ON}{OA} \frac{OB}{OM}
 &=\left( \frac{OM}{OA} + \frac{MN}{OA} \right) \frac{OB}{OM}
= \frac{OB}{OA}  +  \frac{\frac{AM}{BM}}{\frac{OA}{OB}} \cdot \frac{BM^2}{OM^2}\\
&= {\cos x}+ 1\cdot\sin^2\frac x2 = {\cos x}+ 1-\cos^2\frac x2
\end{align}

which yields
$$ \cos^2\frac x2 = \frac{1+\cos x}2$$


问:
Completing the square with simple polys
I am suppose to rewrite $x^2 + x + 1$ by completing the square. I don't really know what that means but I know that if I add 3 at the end of this I get </p>

<p>$$(x + 2) (x - 1) - 3$$ this is the same as the original now but the answer isn't right. What is wrong with what I did? It seems a lot cleaner than the answer.</p>

Title: Completing the square with simple polys

Body:
$$
x^2 + x +1 = x^2 + x + \frac14 + \frac34 = (x+\frac12)^2 + \frac34
$$


问:
Inclusion of Sobolev spaces $H^2(\mathbb{R^2}) \subset W^{1,4}(\mathbb{R^2})$?
How do prove the inclusion of Sobolev spaces $H^2(\mathbb{R^2}) \subset W^{1,4}(\mathbb{R^2})$? Any ideas?

答:
This is an immediate consequence of the general Sobolev inequality. This implies the embedding theorem, which says that if you have $k,l,p,q,n$ satisfying $$\frac{1}{p} - \frac{k}{n} = \frac{1}{q} - \frac{l}{n}$$

then you have $W^{k,p}(\mathbb{R}^n) \subseteq W^{l,q}(\mathbb{R}^n)$.

Since $H^2(X) = W^{2,2}(X)$ by definition, you can do the arithmetic and see that this holds.


问:
Function with domain all real numbers and range $(0,1)$

Is there a one-to-one function whose domain is all real numbers and range is $(0,1)$?

I can't find any so I was thinking about trying to find a piece-wise function that meets the requirements, but I'm having a lot of trouble doing that too.

This is a part of the problem I'm trying to solve to show that two sets have the same cardinality.

Any help is appreciated!

答:
Consider $(0,1)\to (-1, 1)$ by $f(0,1) \to (-1,1):f(x) = 2x - 1$.

Now stretch $-1\mapsto -\infty$, $0\mapsto 0$, $1\mapsto \infty$ by .... how? Well as $x\to 0$ we have $\frac 1x \to \infty$ so $x\to 1^-$ we have $\frac 1{x-1}\to \infty$. But $\frac 1{x-1}$ will map $0\to 1$.  But we can do $g[0,1)\to [0,\infty):g(x) =\frac 1{x-1}-1$.

And we do the same for the negative values $h:(-1, 0]\mapsto (-\infty, 0]: h(x)=\frac 1{x+1}+1$.

So we have $G: (0,1)\to \mathbb R$ via $G(x)=\begin{cases}\frac 1{(2x-1)-1}-1&x\ge \frac 12\\\frac 1{(2x-1)+1}&x < \frac 12\end{cases}=\begin{cases}\frac 1{2x-2}-1&x\ge \frac 12\\\frac 1{2x}+1&x < \frac 12\end{cases}$

And so $G^{-1}\mathbb R\to (0,1)$ can be $\begin{cases} \frac 1{2(y+1)} + 1&x \ge 0\\\frac 1{2(y+1)}&x < 0\end{cases}$ will do it.

....

But there are better functions.

You shouldn't think to hard about these.  You should realize the must be possible so then just draw pictures and figure it out.

A better function is to note $\tan: (-\frac {\pi}2, \frac {\pi}2) \to \mathbb R$.

So $\arctan: \mathbb R \to (-\frac {\pi}2,\frac {\pi}2)$.

And $\frac 2{\pi}\arctan x: \mathbb R \to (-1, 1)$.

And $\frac {\frac 2{\pi}\arctan x + 2}2: \mathbb R \to (0,1)$.


问:
$\lVert A^TA\rVert_2$=$\lVert A\rVert_2^2$ and $\kappa_2(A^TA)=\kappa_2(A)^2$
I have a question from James Demmel's book. Chapter 2 Question 2.10

Let A be n $\times$ m matrix with n$\geq$m,  show that $\lVert A^TA\rVert_2$=$\lVert A\rVert_2^2$ and $\kappa_2(A^TA)=\kappa_2(A)^2$

and Let M be n $\times$ n and positive definite and L be its Cholesky factor so that $M = LL^T$. Show that $\lVert M \rVert_2$ = $\lVert L\rVert_2^2$ and $\kappa_2(M)=\kappa_2(L)^2$.

What I try is not correct and I cannot reach the proof.

答:
$\|Ax\|^2 = \langle Ax, Ax \rangle = \langle x, A^TAx \rangle \le \|x\| \|A^T A x\| \le \|A^TA\|\|x\|^2$. Taking the $\sup $ over $\|x\| \le 1$
shows that $\|A\|^2 \le \|A^TA\|$.

$\langle u, A^TA v\rangle = \langle A u, A v\rangle \le \|A\|^2 \|u\| \|v\|$. Taking the $\sup$ over $\|u\| \le 1, \|v\|\le 1$ gives $\|A^TA\| \le \|A\|^2$.

$\kappa_2(A^TA) = \|A^T A\| \|(A^T A)^{-1} \| = \|A^T A\| \| A^{-1} A^{-T} \| = \|A\|^2 \|A^{-1}\|^2 = \kappa_2^2(A)$.

The other statement follows immediately from the above.


问:
Recurrence of walks with growing range of step
Let $f:\Bbb{N}\to\Bbb{N}$ be a function where $f(n)\to\infty$ as $n\to \infty$.

Given $f$, we can define a random walk on $\Bbb{Z}$, $Y(f,1)=Y_0,Y_1,\dots$, where $Y_0=0$, and $Y_{n+1}-Y_n=X_n$, where $X_n$ is chosen uniformly at random from $[-f(n),f(n)]\cap\Bbb{Z}$ and independent of the outcomes of all other $X_i$.

Similarly, given $f$, we can define a random walk on $\Bbb{Z}^2$, $Y(f,2)=Y_0,Y_1,\dots$, where $Y_0=(0,0)$, and $Y_{n+1}-Y_n=X_n'$, with $X_n': = X_n\cdot v_n$, where $v_n$ uniformly samples the two vectors $(0,1),(1,0)$, and $X_n$ is distributed as before. (here, we should have the set of random variables $\{v_i: i\in \Bbb{N}\}\cup\{X_i: i\in \Bbb{N}\}$ be independent)

I believe that if $f$ "does not grow too fast", then the walks $Y(f,1)$ and $Y(f,2)$ will be recurrent. In particular, I believe it would be sufficient for $\sum_{n=1} 1/f(n)^2$ to diverge. I am interested in proofs or disproofs for either case, as well as any techniques/gadgets which are useful for analyzing these kinds of walks.

Here is a sketch of the first case which a grad student showed me, I don't fully understand it yet:

Let $f(n) = \sqrt{n}, Y_n = Y(f,1),$ with $X_n$ defined as above.

We have that $\textrm{Var}(X_n) = O(n),\textrm{Var}(Y_n) = \sigma_n^2 = O(n^2), \sigma_n = O(n)$.

Thus, as the standard deviation of $Y_n, \sigma_n$ is $O(n)$, almost surely, the long-run proportion of time where $|Y_n- 0|\le \sqrt{n+1}$ is $\Omega(1/\sqrt{n})$. (i.e. Let $\chi_n$ be the characteristic function that is one if $|Y_n-0|\le\sqrt{n+1}$ is true and zero otherwise. Let $V_n:= \sum_{i=0}^{n-1}\chi_i$. We have that there exists some absolute $C>0$ such that $\Bbb{P}\left(\lim_{n\to\infty} \frac{V_n}{n} > \frac{C}{\sqrt{n}}\right) =1$)

It follows that almost surely, the long-run proportion of the time where $Y_{n+1} = 0$ is $\Omega(1/n)$, as $\Bbb{P}(Y_{n+1} = 0|\chi_n = 1) = \frac{1}{2\lfloor\sqrt{n+1}\rfloor+1}$.

Through the same argument, replacing 0 with any $m\in \Bbb{Z}$, we have that almost surely, the long-run proportion of times where $Y_{n+1} = m$ is $\Omega(1/n)$. (alternatively, since $\Bbb{P}(Y_{n+1} = m| \chi_n=1, \textrm{sign}(Y_n)=\textrm{sign}(m),\sqrt{n+1}\ge m) = \frac{1}{2\lfloor\sqrt{n+1}\rfloor}$ so since the sign of $Y_n$ given $\chi_n$ should not be biased, and for all large $n$ we have $\sqrt{n+1}\ge m$ holds, the long-run proportion of times where $|Y_{n+1} = m|$ should also be $\Omega(1/n)$ almost surely)

As the are countably many integers, and the intersection of countably many events that occur with probability 1 still occurs with probability 1, we have that almost surely for each integer $m$, the long-run proportion of times where $Y_{n+1} =m$ is $\Omega(1/n)$, implying each integer will be visited infinitely often.

For the most part, this makes sense, but I would appreciate some clarification on Step 3. I would assume that step 3 is "morally" supposed to follow from Chebyshev's inequality. However, formally I don't know how to justify this for our walk. A sketch explaining what conditions (independence and perhaps other stuff) allow us to prove step 3 would be much appreciated. Any commentary about what the most sound/clear way to justify step 5 is also welcome!

Update: the grad student further clarified that Step 3 should follow by Hoeffding's Inequality.

答:
Let $G(n)=\sum_{k=1}^n f(k)^2$.

CLAIM: $Y(f,1)$ is recurrent iff
$\sum_n G(n)^{-1/2}=\infty$.

In particular, $f(n)=n^{1/2}\log(n)^\alpha$ yields recurrence of $Y(f,1) $ iff $\alpha \le 1$. (Note that for $\alpha>1/2$ the sum $\sum_n f(n)^{-2} $ converges.)

Proof sketch of CLAIM:
Let $\varphi_n(t)=E(e^{itXn})$ and $\psi_n(t)=E(e^{itY_n})=\prod_{k=1}^n \varphi_k(t)$.
Taylor expansion shows that $\phi_k(t)$ behaves like $e^{-cf(k)^2 t^2}$ near the origin, where $c$ is a constant. Since $\int_{-\infty}^\infty  e^{-a t^2}= (\pi/a)^{1/2}$, it follows that

$$P(Y_n=0)=\int_{-\infty}^\infty \psi_n(t)=\Theta(G(n)^{-1/2}) \,.$$
See [1] for a lovely account of such calculations. To conclude that $Y_n$ visits zero infinitely often iff the sum in the claim diverges, use the Kochen-Stone lemma [2].

For $Y(f,2)$ to be recurrent, $f$ has to grow much more slowly: $Y(f,2)$ is recurrent iff
$\sum_n G(n)^{-1}=\infty$. In particular, $f(n)=\log(n)^\beta$ yields recurrence of $Y(f,2)$  iff $\beta \le 1/2$. This can be verified by the same method.

For other methods to tackle related problems see [3].

[1] Kac, Mark. Statistical Independence in Probability, Analysis and Number Theory. Courier Dover Publications, 2018.

[2] https://projecteuclid.org/euclid.ijm/1256059668

[3] Benjamini, Itai, Robin Pemantle, and Yuval Peres. "Random walks in varying dimensions." Journal of Theoretical Probability 9, no. 1 (1996): 231-244.


问:
Is $q\to (p \vee \neg r)$ equivalent to $q \to p$?
I am hesitating if the following is correct:

$q\to (p \vee \neg r)$ Assume this is a premise

Therefore $q \to (\neg r \vee p)$

Therefore $q \to (r \to p)$

Therefore $q \to p$ - if this is correct, by which axiom/rule?

答:
It is not correct, because the inference from step 3 to step 4 does not follow.

If $q\to r$ and $r\to p$, then we have $q\to p$. But this is not the same as $q\to(r\to p)$.

As a concrete example, let $q$ be "It is raining", $r$ be "I do not bring my umbrella", and $p$ be "I get wet". If $q$ is true, then indeed $r\to p$ (if I leave my umbrella behind I will get wet), but it is not true that $q\to p$: it could rain without me getting wet (namely, if I remember my umbrella!).

This example suggests where the equivalence fails: if $q$ is true, $r$ is false, and $p$ is false. Indeed, you can check that the two expressions have different values in this case.


问:
cross product of a position vector and a vector field
The problem is to simplify the following expression: $$\nabla\times[\frac{(\vec{B}\cdot\vec{r})}{r^3}\vec{r}]$$ where $\vec{B}$ is a constant vector and $\vec{r}$ is a position vector.

So $(\vec{B}\cdot\vec{r})$ is scalar product and $\frac{(\vec{B}\cdot\vec{r})}{r^3}$ is a scalar field. Which means the identity $$\nabla \times (\Phi\vec{r}) = \Phi(\nabla\times\vec{r}) - \vec{r}\times(\nabla\Phi)$$ can be applied. And $\Phi = \frac{(\vec{B}\cdot\vec{r})}{r^3}\vec{r}$.

Then $(\nabla\times\vec{r})=0 $ and $\nabla\Phi$ is a vector field $\vec{A}$.

So the expression is left with $-\vec{r}\times\vec{A}$.

The question is: how do I find the cross product between a position vector and a vector field?

答:
If $\vec B$ is a constant vector, then we have

$$\begin{align}
 \nabla \left(\frac{\vec B\cdot \vec r}{r^3}\right)&=\nabla \left(\frac{ B_r}{r^2}\right)\\\\
&=B_r\nabla\left(\frac1{r^2}\right)+\frac1{r^2}\underbrace{\nabla (B_r)}_{=0}\\\\
&=B_r \left(-\frac{2\vec r}{r^4}\right)
\end{align}$$

Hence, we have

$$\vec r\times \nabla \left(\frac{\vec B\cdot \vec r}{r^3}\right)=0$$


问:
Proving an inequality: $|1-e^{i\theta}|\le|\theta|$
We have been using this result without proof in my class, but I don't know how to prove it. Could someone point me in the right direction?

$$|1-e^{i\theta}|\le|\theta|$$

I believe this is true for all $\theta\in\mathbb{R}$. It is easy to show that the left side is bounded by 2 (triangle inequality), but I'm stuck otherwise.

答:
Hint:
$$
\mathrm e^{\mathrm i\theta}-1=\int_0^\theta\mathrm i\mathrm e^{\mathrm ix}\mathrm dx,\qquad|\mathrm i\mathrm e^{\mathrm ix}|\leqslant1\ (x\in\mathbb R).
$$</p>


问:
Do the fiber bundles over an Abelian category form an Abelian category?
Assume I have an Abelian category $C$. Is the category of fiber bundles, where the fibers are objects of $C$ Abelian?

答:
It's not at all clear what this should mean in general. To define fiber bundles with fiber $F$ requires that you specify a topological group $G$ acting on $F$, and if we're just given that $F$ is an object in some abelian category then it just has some automorphism group with no obvious topology in play other than the discrete topology. Picking the discrete topology does not, for example, reproduce real vector bundles if we pick the abelian category to be $\text{Vect}(\mathbb{R})$; the construction of vector bundles is sensitive to the topology of $GL_n(\mathbb{R})$ but the bare abelian category structure of $\text{Vect}(\mathbb{R})$ is not.

In any case, the category of vector bundles already fails to be an abelian category, e.g. it does not have kernels; see, for example, this MO question.

In a positive direction, for a topological space $X$ and an abelian category $C$ we can define the category $\text{Sh}(X, C)$ of sheaves of objects of $C$ on $X$, and this is an abelian category; see, for example, this math.SE answer (Edit: or maybe not, the linked answer only cites a claim in Weibel without a proof so who knows).


问:
Help with truth statements "For all $x \in \varnothing, p(x)$" and "$\exists x \in \varnothing, p(x)$ "
Why is the statement "For all $x \in \varnothing, p(x)$" always true, and why is the statement "$\exists x \in \varnothing, p(x)$ always false for a statement $p(x)$ that depends on $x$?

答:
$\exists x\in\emptyset~.p(x)$ must be false, because there is no element in the emptyset that could satisfy predicate $p$.

$\exists x\in\emptyset~.\lnot p(x)$ must be false, because there is no element in the emptyset that could not satisfy predicate $p$.

So $\forall x\in\emptyset~.p(x)$ must be true, because there is no element in the emptyset that could not satisfy predicate $p$.


问:
Probability example A tosses coin B rolls a die
I have been given this example:

A those a coin and THEN B rolls a Die.
This is repeated independently until a head or one of the numbers 1,2,3,4 appears which the at this time the game is stopped. Note that winning is when the game is stopped and its a fair coin and dice

Compute the probability of of A winning the game.

I see this as follows:

A can win either by getting head on the first or by $$\sum _{n=1}^{\infty } \frac{1}{2}(P(5or6|tail)^n = \sum _{n=1}^{\infty } \frac{1}{2} \left(\frac{2}{12}\right)^n=1/10 $$
(using geometric series)

As these events are mutually exclusive I could sum them up and I get the probability of A given by P(A) is 1/2+1/10=60% chance of A winning.

Now this is the same answer as the book is giving, but in the example they have provided is this solution, which I don't follow:

Using independency, the probability of A wins is thus the conditional probability equals
$$(1/2)/((1/2)+(1/2)*(4/6))$$

Now could someone please show me how these numbers came about?

答: The point is that after a tail and a $5$ or $6$ we are in the same situation as the start of the game.  That essentially sums the geometric series for you.  If $a$ is the probability that $A$ wins, we have $a=\frac 12 + \frac 12 \cdot \frac 13 a$ because $A$ wins with a head, which is the $\frac 12$ or with a tail and a $5,6$ plus winning the game starting from there.


问:
Find $\int \frac{\ln(1-x)}{x}dx$
I would like to find $$\int \frac{\ln(1-x)}{x}dx$$

On the site where I found this the problem was actually slightly different; the question was to evaluate $$\int_0^1 \frac{\ln(1-x)}{x}dx$$ which can be solved using its Maclaurin series expansion and the solution for the Basel problem, but I was wondering if there's a way to directly find the antiderivative of the integrand. I have tried substitutions included hyperbolic and polynomial ones, such as $x=\sinh^2 x$, but I haven't managed to get anywhere with it.

Would you mind giving me a push in the right direction?

Thank you for your help.

答:
The Euler dilogarithm function $~Li_2(x)~$ defined for $0 ≤ x ≤ 1$ by
$$Li_{2}(x)=\sum_{n\ge1}\dfrac{x^n}{n^2}=-\int_{0}^{x}\frac{\ln{(1-t)}}{t}dt$$is one of the lesser transcendental function. Nonetheless, it has many intriguing properties
and has appeared in various branches of mathematics and physics.

So $$\int_0^1 \frac{\ln(1-x)}{x}dx=-Li_2(1)=-\dfrac{\pi^2}{6}$$

You may read the following links (and the references therein) for further knowledge about this type of integrals:

"Dilogarithm Identities" by Anatol N. Kirillov

Calculate a $\operatorname{Li}_{2}(-1)$ using Integral Representation

Dilogarithm by Wolfram MathWorld

The Dilogarithm Function by Don Zagier

Spence's function by Wikipedia

Title: Find $\int \frac{\ln(1-x)}{x}dx$


答:
$$ \left|1 - e^{i\theta}\right| = \left|\left(1 - \cos \theta\right) - i\sin\theta\right| = \sqrt{(1 - \cos \theta)^2 + \sin^2 \theta} = \sqrt{2 - 2 \cos \theta} = 2\left|\sin\frac{\theta}{2}\right| $$ Because $ \sin \theta $ is an odd function and $ \sin x < x\ \forall x \in \mathbb{R}^+ $, the result follows. </p>

<p>Here is a proof that $ \sin x < x $ using pre-calculus topics. It's a bit lengthy; the overall idea is using a geometrical interpretation. </p>

Title: Proving an inequality: $|1-e^{i\theta}|\le|\theta|$

Body:
We have been using this result without proof in my class, but I don't know how to prove it. Could someone point me in the right direction?

$$|1-e^{i\theta}|\le|\theta|$$

I believe this is true for all $\theta\in\mathbb{R}$. It is easy to show that the left side is bounded by 2 (triangle inequality), but I'm stuck otherwise.

Body:
$$ \left|1 - e^{i\theta}\right| = \left|\left(1 - \cos \theta\right) - i\sin\theta\right| = \sqrt{(1 - \cos \theta)^2 + \sin^2 \theta} = \sqrt{2 - 2 \cos \theta} = 2\left|\sin\frac{\theta}{2}\right| $$ Because $ \sin \theta $ is an odd function and $ \sin x < x\ \forall x \in \mathbb{R}^+ $, the result follows.

Here is a proof that $ \sin x < x $ using pre-calculus topics. It's a bit lengthy; the overall idea is using a geometrical interpretation.


答:
The proposition $\forall x \in \emptyset, p\left(x\right)$, in its formal form is:
$$\forall x, x \in \emptyset \rightarrow p\left(x\right).$$

As $x \in \emptyset$ is always false, the statement $x \in \emptyset \rightarrow p\left(x\right)$ is always true. In fact, You can think of $p \rightarrow q$ as equivalent to $\left(\lnot p\right) \vee q$.
On the other hand, the proposition $\exists x \in \emptyset, p\left(x\right)$ is formally written as
$$\exists x, x \in \emptyset \wedge p\left(x\right).$$
As $x \in \emptyset$ is always false, so is $x \in \emptyset \wedge p\left(x\right)$. As a result, the proposition $\exists x, x \in \emptyset \wedge p\left(x\right)$ can never be accepted.

Strictly speaking, what I have explained is not formal proof. It is based on intuitive semantics instead of rigorous syntax. The proof for these two propositions requires using deduction rules, which is another story in first order logic.

Title: Help with truth statements "For all $x \in \varnothing, p(x)$" and "$\exists x \in \varnothing, p(x)$ "

Body: Why is the statement "For all $x \in \varnothing, p(x)$" always true, and why is the statement "$\exists x \in \varnothing, p(x)$ always false for a statement $p(x)$ that depends on $x$?


问:
The indefinite integral of $\frac{e^{-\frac{(\alpha +\beta )^2}{4 \alpha }}}{2 \sqrt{\pi } \sqrt{\alpha }}$ w.r.t. $\alpha$
Mathematica tells me the indefinite integral is:

$$
\frac{1}{2} e^{-\frac{\sqrt{\beta ^2}}{2}-\frac{\beta }{2}} \left(e^{\sqrt{\beta ^2}} \left(\text{erf}\left(\frac{1}{2} \sqrt{\alpha } \left(\frac{\sqrt{\beta ^2}}{\alpha
   }+1\right)\right)-1\right)+\text{erf}\left(\frac{\sqrt{\alpha }}{2}-\frac{\sqrt{\beta ^2}}{2 \sqrt{\alpha }}\right)+1\right)
$$

I have tried integration by parts, a few substitutions, to no avail. What trick must I use to integrate this?

答:
I suppose that the first step is $\alpha=x^2$
$$I=\int\frac{e^{-\frac{(\alpha +\beta )^2}{4 \alpha }}}{2 \sqrt{\pi } \sqrt{\alpha }}\,d\alpha=\frac{1}{ \sqrt{\pi }}\int e^{-\frac{\left( x^2+\beta\right)^2}{4 x^2}}\,dx$$ and then
$$\frac{\left( x^2+\beta\right)^2}{ x^2}=x^2+2 \beta +\frac{\beta ^2}{x^2}$$

Then, I am stuck ! I did not find anythink looking like this integrand in tables of integrals.


问:
Question: Finding inverse when an integral is given
I am stuck in the following question:

$f(x)=\int_{0}^{\sin x}1+\sin(\sin(t))dt$

Find $(f^{-1})'(0)$

I've tried FTOC part I, and then using the formula of derivative of inverse functions find $(f^{-1})(0)$, but I am unable to solve it. How should I go about this? Any hints?

Thank you!

答:
\begin{gather*}
f\left( f^{-1}( x)\right) =x\\
f'\left( f^{-1}( x)\right) *\left( f^{-1}( x)\right) '=1\\
( from\ Chain\ Rule)\\
\left( f^{-1}( x)\right) '=\frac{1}{f'\left( f^{-1}( x)\right)}\\
( f^{-1}( 0) '=\frac{1}{f'\left( f^{-1}( 0)\right)}\\
Now,\ from\ the\ original\ eqn,\\
f( x) =\int ^{sinx}_{0}( 1+sin( sint)) dt\\
f'( x) =1+sin( sin( sin( x)) *cosx\\
( from\ Leibnitz\ formula)\\
Let\ f^{-1}( 0) =y\\
then\ f( y) =0,\ so\ y=0
\end{gather*}
Im sure you can take it from here. Just find f'(0) and then you are done.


答:
Completing the square takes a polynomial that looks like
$$x^2+ax+b$$
and says, what numbers $c$ and $d$ work to make
$$x^2+ax+b=(x-c)^2+d\quad ?$$
Note that
$$\begin{align*}
(x-c)^2+d&=\bigg[(x-c)(x-c)\bigg]+d\\\\\\
&=\bigg[x^2-2cx+c^2\bigg]+d\\\\\\
&=x^2-2cx+c^2+d\\\\\\
&=x^2+(-2c)x+(c^2+d).\\\\\\
\end{align*}$$
Since the polynomials $$x^2+ax+b\qquad\qquad x^2+(-2c)x+(c^2+d)$$ are supposed to be equal, we must have that
$$a=-2c\qquad b=c^2+d,$$
or after rewriting,
$$c=-\frac{a}{2}\qquad d=b-c^2.$$
When $a=1$ and $b=1$, what do you get?</p>

Title: Completing the square with simple polys

Body:
I am suppose to rewrite $x^2 + x + 1$ by completing the square. I don't really know what that means but I know that if I add 3 at the end of this I get </p>

<p>$$(x + 2) (x - 1) - 3$$ this is the same as the original now but the answer isn't right. What is wrong with what I did? It seems a lot cleaner than the answer.</p>


问:
Given acceleration relative to velocity, how to define both acceleration and velocity with respect to time
I am trying to come up with a formula for movement where the acceleration is very high at speeds near 0 and quickly falls off to a near constant as speed is gained. A maximum speed is set in the 400-500 range (units are pixels per second).

Following the logic, the equation I came up with gives an acceleration relative to speed:

$$a = 200000/s + 0.1s + 800$$

or, more exactly, to moderate the acceleration somewhat:

$$a = 200000/(s+100) + 0.1(s+100) + 800$$

The curve of acceleration vs. speed is as I want it. Different thresholds are as I want them: 2810 acceleration at 0 speed, 1820 acceleration at 100 speed, 1497 at 200 speed, etc. The problem is that I need to define speed and acceleration relative to time, and I do not know to reformulate/redefine the equation to get that.

答: Alpha will solve this for you.  I used $x$ for your $s$ and $x'$ for your $a$.  $c_1$ is chosen to match your initial condition.  The solution gives $t$ as a function of $s$, not the other way around.  You can plot it and find approximate solutions.


问:
Why do prime gaps tend to make these weird spirals?
I recently found out that Mathematica can plot polar coordinates, and I really liked polar plotting. Today, I found something surprising. The $n$ th prime gap is defined as
$$g(n)=p_{n+1}-p_n$$
I made a polar plot of $(g(n),n)$ and this is the result:

This surprised me, because prime gaps tend to be in some diagonals more than others, and there are some areas left untouched. Here is a plot with from $-10000$ to $10000$:

And this pattern continues to hold. My question is:

Why do prime gaps tend to be in some "lines" in the plot more than others? What is special in them?

Another thing:

Does this tell us anything about prime gaps?

答:
First, note that there are $22$ spokes since $\pi \approx 22/7$. If you were to zoom out (much) farther, you'd see another pattern emerge with $355$ spokes.

With that established, now observe that every increase by $2$ in the prime gap progresses $7$ spokes counterclockwise, since $$2 \cdot \frac{180}\pi \approx 2 \cdot \frac{180}{\frac{22}7} = 7 \cdot \frac{360}{22}.$$

If you were to lay out your data in a more standard histogram, you would see that there are many small gaps and fewer larger gaps. See this question, for example: Prime gaps distribution.


问:
Is $f(x)=-2/(x+c)$ the only solution to $f''=f'f$?
Given the equation $\displaystyle f''=f'f$, I came up with the solution $\displaystyle f(x)=\frac{-2}{x+c}\; c\in \mathbb{R}$ as follows.

$$f''=ff'\implies 2\int f''=\int 2ff'\iff2f'=f^2\implies \frac{f'}{f^2}=\frac{1}{2}\implies\int -\frac{f'}{f^2} = -\int \frac{1}{2}\implies \left(\frac{1}{f}\right)(x)=\frac{-x-c}{2}$$
$$\implies f(x)=-\frac{2}{x+c}\;\; c\in\mathbb{R}$$

Is there a way to know if this is the only solution? Is my process correct? I know nothing about differential equations, I sort of just thought of this one and decided to give it a go (I mention this so that the answers take that into account)

答:
Here you missed the red $C$: $$f''=ff'\implies 2\int f''=\int 2ff'\iff2f'=f^2 + \color{red}C$$ Then you have $$ \frac{d f}{d x} = \frac{1}{2}f^2 + \color{red}C \\ \frac{df}{\frac{f^2}{2}+\color{red}C} = dx$$
Now, integrate.

Added: As you said you are new to ODE, I will finish the work this time.

You have found the solution when $\color{red}C = 0$, and the other case is as follows:
$$ \int\frac{df}{\frac{f^2}{2}+\color{red}C} = \int dx \\ \sqrt{\frac{2}{\color{red}C}}\arctan\left(\frac{f}{2\color{red}C}\right)  = x + C$$ Now, you can find $f$ from the last equality if you wish or just leave it as it.


问:
Why is $\mathbb P( X(t) \in S\mid X(0) \in S)=1$ where $(X^{n})_{n \in \mathbb N}$ are a sequence of independent markov chains
Consider a sequence of $(X^{n})_{n \in \mathbb N}$ as a sequence of independent continuous-time Markov chains on state space $\{0,1\}$ for every $ n \in \mathbb N$ with $Q$ matrices:

$\begin{pmatrix}
-\beta_{n} & \beta_{n} \\
\delta_{n} & -\delta_{n}\end{pmatrix}$

such that $\sum\limits_{n}\frac{\beta_{n}}{\beta_{n}+\delta_{n}}<\infty.$

Now consider the process $\textbf{X}$ where for $ t \geq 0$ we have $\textbf{X}(t):=(X^{n}(t))_{n \in \mathbb N}$

Now consider the the set $S:=\{x\in \{0,1\}^{\mathbb N}: \sum\limits_{j} x_{j} < \infty\}$.

It is stated that $\mathbb P( \textbf{X}(t) \in S\mid \textbf{X}(0) \in S)=1$ where  $(X^n)_{n \in \mathbb N}$ and I am having trouble seeing why this is the case.

Clearly, if $X(0)\in S$, then we know that for $N$ large enough, we have that $X^n(0)=0$ for any $n \geq N$. I need to use the Markov property to show that there exists $M\in \mathbb N$ such that for all $n \geq M: \; X^n(t)=0$.

I assume I have to use the independence of the markov chains in some way. I'm stuck, any ideas?

答:
Given that for all  $n>N$ we have $X^n(0)=0$, we can infer that for all  $n>N$ and   $t>0$, we have $E(X^n(t)) =P(X^n(t)=1) \le \pi_n= \beta_n/(\beta_n+\delta_n)$.
Sum this over all $n>N$ and deduce that
$$E[\sum_{n=1}^\infty X^n(t) ]\le N+\sum_{n>N} \pi_n < \infty \,.$$