问:
Finding laplace inverse of $\frac{1}{(p+2)^2(p^2+\omega^2)}$
I am finding the laplace inverse of the below :-
$$\frac{1}{(p+2)^2(p^2+\omega^2)}$$

I can apply convolution theorem by taking $$f(p)=\frac{1}{(p+2)^2}, g(p)=\frac{1}
{(p^2+\omega^2)}$$
and calculating the function which yield the above functions on laplace transform  i.e. essentially ,
$$F(t)=e^{-2t}t,G(t)=\frac{\sin{\omega}t}{\omega}$$
then solve by routine integration .

But this seems a little cumbersome . Is there a better way to solve the above ?

Thanks for reading !

答:
The inverse Laplace transformation is define as $f(t)=\frac{1}{2\pi{i}}\int_{a-i\infty}^{a+i\infty}F(p)e^{pt}dp$, where integral is taken in the complex plane along some line $(a-i\infty,a+i\infty)$, where $a$ is some real number (greater than the real part of all singularities of $F(p)$)
https://en.wikipedia.org/wiki/Inverse_Laplace_transform

We can close the contour, integrating along a part of the circle of big radius $R$.
Integral along this circle $\to0$ at $R\to\infty$  if $|F(p)|\to 0$ at $|p|\to\infty$ ; this is the case for $F(p)=\frac{1}{(p+2)^2(p^2+\omega^2)}$

$$f(t)=\oint_C=\frac{2\pi{i}}{2\pi{i}}\sum_{\text{all poles}}Res\frac{e^{pt}}{(p+2)^2(p^2+\omega^2)}=\frac{e^{i\omega t}}{2i\omega(2+i\omega)^2}+\frac{e^{-i\omega t}}{-2i\omega(2-i\omega)^2}+\frac{d}{dp}\bigl(\frac{e^{pt}}{p^2+\omega^2}\bigr)|_{p=-2}$$
$$=\frac{1}{\omega}\Im\frac{e^{i\omega t}}{(2+i\omega)^2}+\frac{e^{-2t}}{4+\omega^2}\bigl(t+\frac{4}{4+\omega^2}\bigr)$$
$$=\frac{(4-\omega^2)\sin \omega t-4\omega\cos\omega t}{\omega(4+\omega^2)^2}+\frac{e^{-2t}}{4+\omega^2}\bigl(t+\frac{4}{4+\omega^2}\bigr)$$

Please, check.


问:
A basic question on the derivative of a continuous function
Is the following condition necessary for the existence of derivative of a continuous function at point $x$: </p>

<p>$$\lim_{h \to 0^+}\frac{f(x+h)-f(x)}{h} = \lim_{h \to 0^-}\frac{f(x)-f(x+h)}{h}$$</p>

答: If the derivative of $f$ exists, then the left-hand side there will be $f'(x)$ and the right-hand side will be $-f'(x)$.  If you fix this by putting a minus sign on the right, then your statement is a necessary and sufficient condition for the derivative to exist, since it's just the definition of the derivative combined with the fact that a limit exists iff it exists from each side and the one-sided limits are equal.

ContentLicense: CC BY-SA 3.0


问:
Prove: $(\forall m, n\in\Bbb N_{>0})(\exists x\in\Bbb R)$ s. t. $2\sin n x \cos m x \ge 1$

Problem 1:
Prove that for any $m,n\in\Bbb N_{>0}$, there exists $x \in\Bbb R$ such that
$2\sin n x \cos m x \ge 1$.

Four months ago, someone asked the above question. However, when I wanted to post my answer,
the question was deleted. I searched by Approach0 without any result.
I think that it is a nice question. I don't know why it was deleted. I post it here. I don't remember who posted it before.

Edit 2021/02/20: We may restrict $x$ to be the form $x = r\pi$ where $r$ is a rational number. We give the following problem:

Problem 2: Prove that for any $m, n \in \mathbb{N}_{>0}$, there exists rational $r$ such that
$2\sin (n r \pi) \cos (m r\pi) \ge 1$.

Any comments and solutions are welcome and appreciated.

Partial results are as follows.

If $n = m$, let $x = \frac{\pi}{4n}$ and we have $2\sin n x \cos m x = \sin 2n x = 1$.

If $n > m$, let $x = \frac{\pi}{2(2n-1)}$. Since $0 < n x < \pi$ and $0 < m x \le (n-1)x < \pi$, we have
\begin{align}
2\sin n x \cos m x &\ge 2\sin n x \cos (n-1)x \\
&= \sin (2n-1)x + \sin x \\
&= 1 + \sin \frac{\pi}{2(2n-1)}\\
& \ge 1.
\end{align}

答:
This answer shows that, for any given two positive integers $m,n$ with $m\gt n$,  there exists a rational number $r$ such that
$$2\sin(nr\pi)\cos(mr\pi)\geq 1$$

Proof :

We see that $$2\sin(nx)\cos(mx)\ge 1$$ is equivalent to
$$\sin((m+n)x)-\sin((m-n)x)\ge 1\tag1$$

To get $x$ satisfying $(1)$, let us consider $x$ such that
$$\sin((m+n)x)=1\qquad\text{and}\qquad \sin((m-n)x)\le 0\tag2$$

We have
$$\sin((m+n)x)=1\iff (m+n)x=\frac{\pi}{2}+2s\pi\iff x=\frac{4s+1}{2(m+n)}\pi$$
where $s$ is an integer. So, we get
$$\begin{align}(2)&\iff \pi+2t\pi\le (m-n)\frac{4s+1}{2(m+n)}\pi\le 2\pi+2t\pi
\\\\&\iff 4t+2\le (4s+1)\frac{m-n}{m+n}\le 4t+4\end{align}$$
where $t$ is an integer.

For any given $m,n$ with $m\gt n$, we want to find a pair of integers $(s,t)$ such that
$$4t+2\le f(s)\le 4t+4$$
where
$$f(s):=(4s+1)\frac{m-n}{m+n}$$

We know that $s=0$ doesn't work since $0\lt f(0)=\dfrac{m-n}{m+n}\lt 1$.

If either $s=1$ or $s=2$ works, then we are happy.

In the following, let us consider the case where $s=1,2$ don't work.

Suppose that $8\lt f(1)$. Then, we have $3m+13n\lt 0$ which is impossible. So, we get $f(1)\le 8$.

Suppose that $12\lt f(2)$. Then, we have $m+7n\lt 0$ which is impossible. So, we get $f(2)\le 12$.

Suppose that $0\lt f(1)\lt 2$ and $4\lt f(2)$. Then, we have $\frac 37\lt \frac nm\lt \frac{5}{13}$ which is impossible. So, if $0\lt f(1)\lt 2$, then $0\lt f(2)\lt 2$.

Suppose that $4\lt f(1)\lt f(2)\lt 6$. Then, we have $\frac 15\lt \frac nm\lt \frac 19$ which is impossible. So, if $4\lt f(1)\lt 6$, then $8\lt f(2)\lt 10$.

So, we have two cases to consider :

Case 1 : $0\lt f(1)\lt f(2)\lt 2$

We have $f(2)\lt 2\implies \dfrac{7}{11}\lt \dfrac nm$ which implies
$$0\lt f(s+1)-f(s)=\frac{4(m-n)}{m+n}=\frac{4-4\frac nm}{1+\frac nm}\lt \frac{4-4\cdot \frac{7}{11}}{1+\frac{7}{11}}=\frac 89\lt 1$$
So, it is sufficient to find the smallest integer $s$ such that $f(s)\ge 2$, i.e. $\dfrac{m+3n}{4(m-n)}\le s$. So, we see that $s=\left\lceil \dfrac{m+3n}{4(m-n)}\right\rceil$ works where we can take $t=0$ since $2\le f(s)\le 4$.

Case 2 : $4\lt f(1)\lt 6$ and $8\lt f(2)\lt 10$

We have $\dfrac{m}{n}\gt 17$ which implies
$$3=4-1\lt f(1)-f(0)= f(s+1)-f(s)=\frac{4+4\frac mn}{\frac mn-1}\lt \frac{4+4\times 17}{17-1}=\frac 92$$ So, it is sufficient to find the smallest $s$ such that $f(s)\le 4s$, i.e. $\dfrac{m-n}{8n}\le s$. So, we see that $s=\left\lceil\dfrac{m-n}{8n}\right\rceil$ works where we can take $t=s-1$ since $4(s-1)+2\le f(s)\le 4(s-1)+4$.

It follows from these that we can take
$$x=\begin{cases}\dfrac{5}{2(m+n)}\pi&\text{if $s=1$ works}
\\\\\dfrac{9}{2(m+n)}\pi&\text{if $s=1$ doesn't work, and $s=2$ works}
\\\\\dfrac{4\left\lceil \dfrac{m+3n}{4(m-n)}\right\rceil+1}{2(m+n)}\pi&\text{if $s=1,2$ don't work, and $f(2)\lt 2$}
\\\\\dfrac{4\left\lceil\dfrac{m-n}{8n}\right\rceil+1}{2(m+n)}\pi&\text{otherwise}\end{cases}$$
which can be simplified as
$$x=\begin{cases}\dfrac{5}{2(m+n)}\pi&\text{if $\ \dfrac nm\in \bigg[\dfrac 19,\dfrac 37\bigg]$}
\\\\\dfrac{9}{2(m+n)}\pi&\text{if $\ \dfrac nm\in \bigg[\dfrac{1}{17},\dfrac 19\bigg)\cup\bigg(\dfrac 37,\dfrac{7}{11}\bigg]$}
\\\\\dfrac{4\left\lceil \dfrac{m+3n}{4(m-n)}\right\rceil+1}{2(m+n)}\pi&\text{if $\ \dfrac nm\in \bigg(\dfrac{7}{11},1\bigg)$}
\\\\\dfrac{4\left\lceil\dfrac{m-n}{8n}\right\rceil+1}{2(m+n)}\pi&\text{if $\ \dfrac nm\in\bigg(0,\dfrac{1}{17}\bigg)$}\end{cases}$$

ContentLicense: CC BY-SA 4.0


问:
Average Length of Longest Segment: What am I Doing Wrong?
I'm following this problem Average length of the longest segment, but I have a mistake in how I compute part of it. The problem is that a rope of 1m is divided into three pieces by two random points. Find the average length of the largest segment.

They start by noting (their text copied)

Here we have two independent random variables $X,Y$, both uniform on $[0,1]$. Let
$A=\min (X,Y), B=\max (X,Y)$ and $C=\max (A, 1-B, B-A)$. First we want to find the probability
density function $f_C(a)$ of $C$. Let $F_C(a)$ be the cumulative distribution function. Then
$$ F_C(a) = P(C\le a)=P(A\le a, 1-B\le a, B-A\le a).$$  By rewriting this probability as area in the unit square, I get
$$F_C(a)=\left\{\begin{array}{ll} (3a-1)^2 & \frac{1}{3}\le a\le \frac{1}{2}\\ 1-3(1-a)^2 & \frac{1}{2}\le a\le 1\end{array}\right.$$

(my text) However, I can't follow how they get those probabilities. I noted (thanks saulspatz) that the inequalities can be written as
\begin{align}
A&\leq a\\
B&\geq 1-a\\
B&\leq A+a\\
B&\geq A
\end{align}
and we also have that $A\in [0,1],B\in [0,1]$. For $\frac{1}{2}\leq a\leq 1$, I 'think' that the first two inequalities give us a square of area $a^2$, and the third forces us to remove a triangle with two sides of length $1-a$ and thus area $\frac{(1-a)^2}{2}$, while the fourth forces us to remove a triangle with two sides of length $a-(1-a)=2a-1$, which has area $\frac{(2a-1)^2}{2}$. This is shown below for $a=2/3$ in the shaded area with the x-axis being $A$ and the y-axis being $B$ using wolfram alpha.

However, this doesn't match what they did or anyone else's solution: what am I doing wrong?

答:
One way to describe your error is that by moving from the $X,Y$-space to the $A,B$-space, you are no longer dealing with the unit square $[0,1] \times [0,1]$ as your possible space for the valid $(X,Y)$ coordinates. You are now dealing with the "upper left diagonal half" of that square as the possible space for your $(A,B)$ coordinates, the triangle formed by the points $(0,0), (1,1)$ and $(0,1)$.

So while your construction of the geometric polygon where the maximum length of the parts is at most $a$ is correct, the incorrect part is that you use that as the probability of it occuring. You are building the quotient of the area of the constructed polygon with the area of all possible values, which you assume is the whole $[0,1] \times [0,1]$ square, with area $1$. But corect would be to use the area of the triangle I described, which has area $\frac12$.

So the geometric area of your polygon is $a^2 - \frac{(1-a)^2}2 - \frac {(2a-1)^2}2$, as you correctly calculated, but the probability is actually double that, as you need to divide by $\frac12$:

$$
\begin{eqnarray}
2a^2 \color{red}{-(1-a)^2} \color{blue}{- (2a-1)^2} &=& 2a^2 \color{red}{- 1 + 2a - a^2} \color{blue}{- 4a^2 + 4a - 1}\\ &=&-3a^2+6a-2 \\ &=& 1 - (3a^2-6a+3)\\ &=& 1-3(a-1)^2
\end{eqnarray}
$$

which is the result in the other answer!

But there is still a hidden assumption made here that turns out to be correct, but is worth pointing out because it may not always be true.

By divding the area of the found polygon by the area of the triangle, we implictely assume that the way we arrive at the points in that triangle makes them uniformly distributed over that triangle.

It is true, but it isn't clear or trivial at all, because we are now no longer dealing with $(X,Y)$ coordinates that come from independent, uniformly distributed random variables $X,Y$. We are know dealing with dependent, non-uniformly distributed random variables $A,B$.

That they are dependent should be clear, if you know that that $B=0.7$, $A$ can't be $0.8$ any more. That is "captured" by the $B \ge A$ condition.

That they are non-uniform can be seen by finding the probability that $B \le \frac12$. Since $B$ is the maximum of $A,B$, this is equivalent of finding the probability that $X \le \frac12, Y \le \frac12$. Since $X$ and $Y$ are independent, that probability is just the product of the separate probabilities, so we get

$$
\begin{eqnarray}
P\left(B \le \frac12\right) &=& P\left(X \le \frac12, Y \le \frac12\right) \\
&=&P\left(X \le \frac12\right) P\left( Y \le \frac12\right)\\
&=& \frac12\frac12 = \frac14
\end{eqnarray}
$$

If $B$ would be uniformly distributed over $[0,1]$, that probability would have to be $\frac12$.

There is a way to transform your solution to work in the 'easy to understand/argue about' X,Y space. Simply divide the $[0,1] \times [0,1]$ square into the "upper left diagonal" triangle from above and the "lower right diagonal" part formed by $(0,0),(1,1)$ and $(1,0)$. Then realize that the upper left diagonal part is where $A=X, B=Y$ and the lower right diagonal part is where $A=Y, B=X$.

Your construction in the upper left diagonal part from above works exactly the same as before, only now you get to apply it again to the lower right diagonal part, where it produces a polygonal area of the same shape (I'm sorry, I'm not that good with making a picture of that).

So now your two "good" areas combined have double the area as before, so this again leads to the same probability as above.


问:
Einstein summation convention of 2-forms: $T_{ik}\omega_{kj}$
I am familiar with Einstein summation convention that means

if the same index name appears exactly twice, once as an
upper index and once as a lower index, then that term is understood to be summed
over all possible values of that index.

But what is the meaning of the following? Let $\omega=\omega_{ij}E^i\wedge E^j$ and $T$ a (symmetric) $2$-tensor
$$T_{ik}\omega_{kj}.$$
Which index should I raise then contract it? index of $T$ or of $\omega$? in the global notation, it is $\omega(T)$ or $T(\omega)$ with appropriate converted $(1,1)$-tensors?

Also I have another problem with Einstein summation convention for 2-forms $\omega=\omega_{ij}E^i\wedge E^j$, that is the sum cannot range over all possible values of its index because $\omega=\sum_{i<j}\omega_{ij}E^i\wedge E^j$ so $i$ cannot be $n$ for example, while in $T$ it can. Unless we see $\omega$ as 2-tensors as well then the notation $\omega=\omega_{ij}E^i\wedge E^j$ is meaningless and it should be $\omega=\omega_{ij}E^i\otimes E^j$ with $\omega_{ij}=-\omega_{ji}$.

答:
I take it that you are working in a Riemannian setting so you have a metric $g_{ij}$ and its inverse $g^{ij}$ around. Then the question "which index should I raise and the contract" does not really occur, since what you want to form is $g^{k\ell}T_{ik}\omega_{\ell j}$ and by symmetry of $g^{k\ell}$ there is no choice involved. The notations $\omega(T)$ and $T(\omega)$ tend to be ambiguous anyway, in the end, the index notation is more precise here.

The two interpretations of index notation for two-forms (i.e. via $\sum_{i<j}$ or via $\sum_{i,j}$ and $\omega_{ji}=-\omega_{ij}$) are indeed there and they both are meaningful. In an interpretation involving contractions you have to take the interpretation as a tensor field with skew symmetric coefficients. Observe that also the result of your operation is a tensor field and not a two-form. I think it is also good to keep in mind that there actually is a linear algebra meaning hidden behind the index expression. You have bilinear maps $T(x),\omega(x):T_xM\times T_xM\to\mathbb R$ for which in addition you know that $\omega(x)$ is skew symmetric. Your contraction then expresses the bilinear map $(X,Y)\mapsto \sum_{i=1}^nT(X,e_i)\omega(e_i,Y)$ where $\{e_i\}$ is an orthonormal basis of $T_xM$ (and the result is independen from which orthonormal basis you uses).


问:
bisection method - quantiles of the normal distribution
I got stuck on this question, I can't understand it.

I need to solve
$$
\Phi(a)=\int_{-\infty}^a\frac1{\sqrt{2\pi}}\exp\left(-\frac12x^2\right)dx=\alpha
$$
by the bisection method, the purpose of the question is to calculate the 30th, 70th, and 95th percentages of a standard normal distribution.

In other words, in order to calculate the appropriate fracture the equation must be solved:
$$
f(a)=\Phi(a)-\alpha=0
$$

I know the algorithm of the method, but I don't succeed to do it.
Any explanation, or any help, will be great.

答:
Your teacher will not appreciate this answer, which is more appropriate than bisection.

See the problem as a differential equation

$$\frac{d\Phi}{dx}=\frac{e^{-x^2/2}}{\sqrt{2\pi}}$$ or

$$\frac{dx}{d\Phi}=\sqrt{2\pi}\,e^{x^2/2}=f(x,\Phi)$$ and integrate by Runge-Kutta (or similar) from $\Phi=\frac12$ to $\alpha$. (Starting from $\frac12$ is numerically safer, and we know that $\Phi(0)=\frac12$.)


问:
Prove closed of dimension one of $X\times I$.
Suppose $F(x,t): X\times I \rightarrow R$ is a homotopy of Morse functions. That is, $f_t: X \rightarrow R$ is Morse for every $t$. Show that the set $C = \{(x,t)\in X\times I : d(f_t)_x = 0\}$ forms a closed, smooth submanifold of dimension one of $X\times I$. Assume the homotopy is constant near the ends of I and use an open interval.

What I have done so far is that showed it is a smooth manifold by inverse function theorem. But I don't know how to show it is closed and has dimension 1.

Thank you very much for your guidance!

答:
Thank you for @DanielMcLaury's comment, I finally figured it out after his very wise guidance.

Submanifold:
When $dF=0, d^2 F \neq 0$, we know the Jacobian of $dF$ is not singular. Hence by inverse function theorem, it is a submanifold.

Closed:
The given condition {$(x,t) \in X \times I:d(f_t)_x=0$} is closed, since all the limit points satisfy the condition and therefore is included in the set.

1-dimension: By Morse function condition, because $d^2 F \neq 0$, that means given $df_x = 0$, that means $df_{x^\prime} \neq 0$ for all $x^\prime$ in any neighborhood of $x$. Therefore, the set can be only varied by $t$, which suggests it is 1-dimensional.


问:
When can we make $H$ separable in $A\subset B(H)$?
I came across the fundamental result in operator algebras that any $C^*$-algebra can be embedded in some $B(H)$.

Is there some characterization of a $C^*$-algebra (or von Neumann algebra) $A$ such that $H$ can be chosen to be

separable?

finite dimensional?

How rare/common/special are such situations?

(Apologies if this is a trivial question as I'm not advanced in my studies for this.)

答:
I am only adding details to what has been already stated in the comment section.

If $A$ is finite dimensional, then $A=\bigoplus_{i=1}^nM_{k_i}$, and since $M_k\cong\mathbb{B}(\mathbb{C}^k)$ we have that $A$ can be embedded as the diagonal operators in $\mathbb{B}(\bigoplus_{i=1}^n\mathbb{C}^{k_i})$. On the other hand, if $A\subset\mathbb{B}(\mathcal{H})$ where $\mathcal{H}$ is finite dimensional, then $\mathbb{B}(\mathcal{H})$ is finite dimensional so $A$ is finite dimensional. Thus a $C^*$-algebra admits a faithful representation on a finite dimensional Hilbert space if and only if it is of finite dimension.

The interesting part is question (1). If $A$ is separable for example, then the state space $S(A)$ is separable in the weak-star topology, so, if $\{\sigma_n\}$ is a collection of states that is weak-star dense in $S(A)$, then $\sigma=\sum_{n=1}^\infty\frac{\sigma_n}{2^n}$ defines a faithful state on $A$. It is not hard to verify that the GNS representation associated to $\sigma$, namely $(\mathcal{H}_\sigma,\pi_\sigma)$, is faithful; Also, recall that the GNS space $\mathcal{H}_\sigma$ is defined as the completion of $A$ modulo $\{a\in A:\sigma(a^*a)=0\}\equiv0$ (since $\sigma$ is a faithful state), so $\mathcal{H}_\sigma$ is separable. Thus question (1) is always true for separable $C^*$-algebras.

The converse is not true: not every $C^*$-algebra that is faithfully represented on a separable Hilbert space is separable. This is the easiest thing to see: for example, $\ell^\infty\subset\mathbb{B}(\ell^2)$ as multiplication operators, $$\ell^\infty\ni x:=(x_n)\mapsto M_x:\ell^2\to\ell^2,\;\;\;M_x(y_n)=(x_ny_n)\text{ for all }(y_n)\in\ell^2,$$
but $\ell^\infty$ is well-known to be non-separable. Or, if you prefer, take any $\delta>0$ and an uncountable subset $S$ of $\mathbb{B}(\ell^2)$ so that $\|x-y\|\geq\delta$ for all $x,y\in S$ and simply consider $C^*(S)\subset\mathbb{B}(\ell^2)$; this $C^*$-algebra cannot be separable.


问:
How to write this fraction as sum of $2^{-n}$?
I am trying to write the fraction $$\frac{29}{160}$$ as a sum of $2^{-n}$ such that $0\leq n\leq22$ but I can use each n only one time, wich means I cannot write the following  $$\frac{1}{2^3}+\frac{1}{2^3}+...$$

Is it even possible to do that? what Is the actuall mathematic equation I should constract?

P.S

I dont have to use all the values for $n$

答: It is not be possible, because the number equals $\frac{29}{ 2^4 \cdot 10}$, and the binary expansion of $\frac{1}{ 10}$ doesn't terminate. (And $10$ is coprime with $29$)


问:
tangent space of the manifold $\operatorname{Imm}(S^1,\mathbb{R}^d)$
The space of smooth, periodic, immersed curves
$$\operatorname{Imm}(S^1,\mathbb{R}^d)=\{c \in C^\infty(S^1,\mathbb{R}^d) : c'(\theta) \neq 0, \forall \theta \in S^1\}$$
forms an infinite-dimensional manifold. Supposedly the tangent space is the modelling space itself, i.e. $T_c\operatorname{Imm}(S^1,\mathbb{R}^d)=C^\infty(S^1,\mathbb{R}^d).$ I understand why $T_c\operatorname{Imm}(S^1,\mathbb{R}^d) \subset C^\infty(S^1,\mathbb{R}^d)$, but I don't see why we get the reverse inclusion because surely the trivial curve $\gamma(x)=0$ is in $C^\infty(S^1,\mathbb{R}^d),$ but it cant be in $T_c\operatorname{Imm}(S^1,\mathbb{R}^d)$ by definition of $\operatorname{Imm}(S^1,\mathbb{R}^d)$. What am I missing?

答: I think the non-technical answer to your question is that the subset $Imm(S^1,\mathbb R^d)\subset C^{\infty}(S^1,\mathbb R^d)$ is open. Similarly to the finite dimensional situation the tangent spaces to an open subset is the modelling vector space. Otherwise put, for $c\in Imm(S^1,\mathbb R^d)$, $f\in C^{\infty}(S^1,\mathbb R^d)$ and $t\in\mathbb R$, you get $c+tf\in Imm(S^1,\mathbb R^d)$ is $t$ is suffieciently small. Thus you get a curve emanating from $c$ in direction $f$, and hence $f$ should be in the tangent space at $c$.


问:
Winding number in complex plane homework
If $\gamma_1,\gamma_2$ are closed paths s.t. $|\gamma_1(t)|>|\gamma_2(t)|$ for all $t$, and $\Gamma(t) := \gamma_1(t) + \gamma_2(t)$, show that $n(\Gamma,0) = n(\gamma_1,0)$, where $n$ is the winding number defined by $n(\Gamma,0) :=\frac{1}{2\pi i} \int_\Gamma \frac{dz}{z}$.

I have tried looking at $n(\Gamma,0) - n(\gamma_1,0)$ but am still stuck. Please can somebody give me a hint??

答:
To develop some intuition about this problem, imagine $\gamma_1$ as the orbit of the Earth around the Sun, and $\gamma_2$ as the orbit of the Moon around the Earth.  Then, $\Gamma$ describes the motion of the Moon around the Sun.  (Suppose, for simplicity that a year is exactly 12 months so that these are closed curves.)

Notice that the Moon winds around the Sun once for each time that the Earth does.  The motion of the Moon relative to the Earth doesn't affect these winding numbers.


问:
Finding maxima and minima of $f(x,y) = x^{4} + y^{4} - x^{3}$
Finding maxima and minima of $$f(x,y) = x^{4} + y^{4} - x^{3}$$
I tried solving this question but need help.
In this question, I calculated the partial derivatives of first and second order.
I found
$$f_{x} = 4x^{3} - 3x^{2}\\   f_{y}= 4y^{3} \\ f_{xx} = 12x^{2}-6x \\ f_{yy}= 0$$
On equating $$f_{x} = 0 \hspace{0.5cm}  and\hspace{0.5cm} f_{y} = 0 $$ I get the critical points as (0,0) and (0.75,0) but I am not able to understand whether these points constitute a maxima or minima because $$f_{xx}f_{yy} - f_{xy}f_{yx}=0$$

Can somebody please guide?

答:
$$f(x,y)=x^4+y^4-x^3\\\qquad\quad ~~ ~=x^3(x-1)+y^4$$

Apply $x\to\  +\infty$ and $y=0$, you get $f(x,y)\to +\infty.$

So, global maxima doesn't exist.

Now, we will prove that global minimum exists and is equal to

$$\text{min}[x^4-x^3+y^2]=-\frac{27}{256}.$$

In this case , of course, $y=0$ must be. Then observe that,

$$\begin{align}x^4-x^3+\frac{27}{256}=\frac{1}{256}(256x^4-256x^3+27)=\frac{1}{256}(4 x - 3)^2 (16 x^2 + 8 x + 3)≥0. \end{align}$$

Because,

$\Delta_{\text{half}}=16-16\times 3=-32<0$.

This means, $16x^2+8x+3>0$  for all $x\in\mathbb R.$

Finally, we deduce that

$$\text{min}\left[x^4-x^3+\frac{27}{256}\right]=0$$

at $x=\dfrac 34.$

which follows

$$\text{min}\left[x^4-x^3+y^4\right]=-\frac{27}{256}.$$


问:
Given two natural numbers $m, n$ so that $m> n.$ Prove that there exist any real numbers $x$ so that $2\sin nx\cos mx\geq 1$

Given two natural numbers $m, n$ so that $m> n.$ Prove that there exist any real numbers $x$ so that (unsolved in a case..)
$$2\sin nx\cos mx\geq 1$$
Source: StackMath/@RiverLi_ Prove: $(\forall m, n\in\Bbb N_{>0})(\exists x\in\Bbb R)$ s. t. $2\sin n x \cos m x \ge 1$ .

I couldn't find a concrete value of $x,$ but I succeeded at proving the problem by using Intermediate value theorem (My friend has taught me a solving plan..). Let $x$ on the interval $I= \left [ -\frac{5}{6}\pi, \frac{\pi}{6} \right ],$ we call $f\left ( x \right ):=\frac{1}{2\sin x}, g\left ( x \right ):=\cos\alpha x, \alpha:=\frac{m}{n}> 1,$ I devide my problem into 2 cases as followings

When $1< \alpha< \frac{6}{5},$ we have $f, g$ continuous both and $f\left ( -\frac{3}{4}\pi \right )\geq g\left ( -\frac{3}{4}\pi \right ),$ by Intermediate value theorem $\exists x\in I$ so that $f\left ( x \right )= g\left ( x \right )$

When $\alpha\geq\frac{6}{5},$ we have $f, g$ continuous and $\exists x\in I$ so $g\left ( x \right )= -1\Rightarrow\exists x\in I$ so that $f\left ( x \right )= g\left ( x \right )$

Conclusion.
$$\forall\alpha= \frac{m}{n}> 1, \exists x\in I: f\left ( x \right )= g\left ( x \right )\Leftrightarrow\forall m, n, \exists x\in I: 2\sin x\cos\alpha x= 1$$
$$\Rightarrow\forall m, n, \exists x\in \mathbb{R}: 2\sin nx\cos mx\geq 1$$
I'm waiting for @RiverLi's complete solution and want to see such a value of $x$ for my proof, thanks

答:
The following approach comes from looking at the graphs of the involved functions.

Given $m>n>0$ put  $$m+n=:p\geq3,\qquad m-n=:q\geq1\ .\tag{1}$$ We have to study the function
$$f(t):=2   \sin(nt)\cos(mt)=\sin(pt)-\sin (qt)$$
on the  circle ${\mathbb T}:={\mathbb R}/(2\pi{\mathbb Z})$. After scaling the $t$-axis appropriately we may assume ${\rm gcd}(p,q)=1$. The function $t\mapsto\sin(pt)$ has $p$ equidistant local maxima (value $=1$) on  ${\mathbb T}$, color them red, and the  function $t\mapsto-\sin(qt)$ has $q$ equidistant local maxima (value $=1$) on  ${\mathbb T}$, color them blue. Denote by $\delta\geq0$ the minimal distance between a red and a blue maximal point. I claim that $$\delta\leq{\pi\over p q}\ .\tag{2}$$
Proof. When a red and a blue point coincide we have $\delta=0$. In this case it follows from Bézout's identity that there are two more red/blue pairs,  with opposite order along ${\mathbb T}$, having distance ${2\pi\over pq}$. When the blue pattern of points is then rigidly rotated vs.\ the red pattern the two coinciding points are disassociating, while one pair of distance ${2\pi\over pq}$ points is going to combine. The worst case is in the middle, when the rotation angle is ${\pi\over pq}$. At that moment  the $\delta$ of the situation is just covered by $(2).\quad{\square}$

We therefore have a red maximum at $\xi\in{\mathbb T}$ and a blue maximum at $\eta\in{\mathbb T}$ with $$\delta:=|\xi-\eta|\leq{\pi\over pq}\ .$$ Let $\tau$ be the point between $\xi$ and $\eta$ having distance ${q^2\over p^2+q^2}\delta$ from $\xi$ and  ${p^2\over p^2+q^2}\delta$ from $\eta$. We are now going to estimate $f(\tau)$.

In the neighborhood of $t=\xi$ the graph of $t\mapsto\sin(pt)$ is congruent to the graph of
$$t\mapsto\cos(pt)\geq1-{p^2\over2}t^2$$
in the neighborhood of $t=0$. It follows that
$$\sin(p\tau)\geq 1-{p^2\over2}(\tau-\xi)^2\geq1-{p^2\over2}\left({q^2\over p^2+q^2}\delta\right)^2\ .$$
Similarly we have
$$-\sin(q\tau)\geq 1-{q^2\over2}(\tau-\eta)^2\geq1-{q^2\over2}\left({p^2\over p^2+q^2}\delta\right)^2\ .$$
Using $(1)$ and $(2)$ we now obtain
$$f(\tau)\geq2-{p^2q^2\delta^2\over 2(p^2+q^2)}\geq2-{\pi^2\over 2(p^2+q^2)}>1\ .$$

Example: When $m=3$ and $n=2$ then $p=5$, $q=1$. Our estimate then gives $f(\tau)\geq1.81$. This value is indeed realized by $f$:


答:
The function does not have a global maximum because $y^4$ grows without bound as $y\to\pm\infty$; the same holds for the $x$ variable. This can be shown by just taking the limits separately, because $f$ splits nicely between $x$ and $y$. However, because of this behavior at infinity and due to its continuity, it must have a global minimum thanks to a variant of Weierstrass' theorem.

As for $(0,0)$ and $(3/4,0)$, which annihilate the gradient, the Hessian-determinant condition you mention implies that they are degenerate points (which means analyzing the Hessian is not sufficient to conclude that they are local maxima, local minima, or saddle points). Indeed, the Hessian is
$$\mathbf D^2f(x,y) = \left(\begin{array}{cc}
12 x^{2}-6 x & 0 \\
0 & 12 y^{2}
\end{array}\right), $$
so at those points
$$\mathbf D^2f(0,0) = \left(\begin{array}{cc}
0 & 0 \\
0 & 0
\end{array}\right), \qquad \mathbf D^2f(3/4,0) = \left(\begin{array}{cc}
9/4 & 0 \\
0 & 0
\end{array}\right). $$
To decide their nature as critical points, we must take restrictions of the function near those points, for example along lines. Take $(0,0)$: we know $f(0,y) \geqslant 0$ in a neighborhood of $y=0$, whereas $f(x,0)\geqslant 0$ in a left-neighborhood of $x=0$ and $f(x,0) \leqslant 0$ in a right-neighborhood of $x=0$, so $f$ must have a (generalized?) saddle point at the origin.

As for $(3/4,0)$, it is sufficient to observe that $f(x,y) = g(x) + h(y)$ with $g(x)=x^4-x^3$ and $h(y)= y^4$, and the degeneracy of the Hessian at that point is only due to the fact that $h''(0) = 0$. Since we know $h \geqslant 0$ for all $y$, and $g''(3/4) > 0$ (it is the positive eigenvalue of $\mathbf D^2 f(3/4,0)$ that was written before), we obtain that $f$ has a local minimum at $(3/4,0)$. This is also a global minimum because there are no other options.

Title: Finding maxima and minima of $f(x,y) = x^{4} + y^{4} - x^{3}$

Body:
Finding maxima and minima of $$f(x,y) = x^{4} + y^{4} - x^{3}$$
I tried solving this question but need help.
In this question, I calculated the partial derivatives of first and second order.
I found
$$f_{x} = 4x^{3} - 3x^{2}\\   f_{y}= 4y^{3} \\ f_{xx} = 12x^{2}-6x \\ f_{yy}= 0$$
On equating $$f_{x} = 0 \hspace{0.5cm}  and\hspace{0.5cm} f_{y} = 0 $$ I get the critical points as (0,0) and (0.75,0) but I am not able to understand whether these points constitute a maxima or minima because $$f_{xx}f_{yy} - f_{xy}f_{yx}=0$$

Can somebody please guide?


答:
No. In the current form, the condition says the derivative from the left and from the right have opposing signs. For a simple counter-example, consider the function $f(x)=x$ at any point; the left-hand limit will be equal to 1, while the right-hand one would be equal to (-1).

If reverse the order of $f(x+h)$ and $f(x)$ in the limit on the right-hand side, the condition would be both necessary and sufficient since it'd basically be the definition of a derivative (in order for a limit in $0$ to exist, it must be the same regardless of the way you approach $0$).

Title: A basic question on the derivative of a continuous function

Body:
Is the following condition necessary for the existence of derivative of a continuous function at point $x$: </p>

<p>$$\lim_{h \to 0^+}\frac{f(x+h)-f(x)}{h} = \lim_{h \to 0^-}\frac{f(x)-f(x+h)}{h}$$</p>


答: In XY plane you'll observe that in some direction (neighborhood points)from critical point  function is greater while in some other direction value of function is Less than that of value of function at critical point (0,0)and hence no local maxima or minima at (0,0)</p>


问:
Proving the following identity
Prove that

$${ \left( \frac { \cos  (\alpha +\beta ) }{ \cos  (\alpha -\beta ) } -\frac { \cos  (\alpha +\gamma ) }{ \cos  (\alpha -\gamma ) }  \right)  }^{ 2 }+{ \left( \frac { \sin  (\alpha +\beta ) }{ \cos  (\alpha -\beta ) } -\frac { \sin  (\alpha +\gamma ) }{ \cos  (\alpha -\gamma ) }  \right)  }^{ 2 }={ \left( \sec ^{ 2 }{ (\alpha -\beta ) } \sin ^{ 2 }{ (\beta -\gamma ) } \sec ^{ 2 }{ (\gamma -\alpha ) }  \right)  }$$

I brought everything to the common denominator noticing bot the fractions had the same denominator but after that I find I am stick with a mess

答:
We easily get LHS $ = \sec^2 (\alpha - \beta)+\sec^2 (\alpha-\gamma) - \dfrac{2\cos (\beta -\gamma)}{\cos (\alpha - \beta) \cos (\alpha - \gamma)}$

$= \sec^2 (\alpha - \beta)+\sec^2 (\alpha-\gamma) - \dfrac{2\cos (\beta -\alpha+\alpha-\gamma)}{\cos (\alpha - \beta) \cos (\alpha - \gamma)}$

$=\sec^2 (\alpha - \beta)+\sec^2 (\alpha-\gamma) -2 + 2\tan (\beta-\alpha) \tan (\alpha-\gamma)$

$=\tan^2 (\alpha - \beta)+\tan^2 (\alpha-\gamma)  + 2\tan (\beta-\alpha) \tan (\alpha-\gamma)$

$=\left(\tan (\beta-\alpha)+\tan  (\alpha-\gamma)\right)^2$

Now using $\tan A + \tan B = \dfrac{\sin (A+B)}{\cos A \cos B}$ we get further

$\left(\dfrac{\sin (\beta - \gamma)}{\cos (\beta -\alpha) \cos (\alpha-\gamma)}\right)^2$

$=\sec^2 (\beta -\alpha) \sin^2 (\beta - \gamma) \sec^2 (\alpha-\gamma)$ = RHS

Title: Proving the following identity


问:
Evaluate $\int_{0}^{\pi} \frac{x\coth x-1}{x^2}dx$
I've been trying to evaluate certain series recently, and I found that
$$\sum_{r=1}^{\infty}\frac{1}{r}\arctan\frac{1}{r}=\frac{\pi}{2}\int_{0}^{\pi} \frac{x\coth x-1}{x^2} \, dx$$
Therefore, I would like to know how to evaluate $$\int_{0}^{\pi} \frac{x\coth x-1}{x^2} \, dx$$
The integral is approximately equal to $0.895004...$ but even Wolfram Alpha gives no closed form solution. Any ideas? Part of the problem is that the indefinite integral has no elementary antiderivative. I've tried using 'Feynman's trick' on this integral too but without success.

Thank you for your help.

Body:
Approximations.

Since
$$\frac{x \coth (x)-1}{x^2}=\sum_{n=0}^\infty \frac{2^{2 n+2} B_{2 n+2} }{(2 n+2)!}x^{2 n}$$ it is simple to build a $[2k,4]$ Padé approximant of it. For example
$$P_1= \frac{7 \left(2 x^2+45\right)}{x^4+105 x^2+945}$$
$$P_2=\frac{1}{21}+\frac{43 x^2+990}{7 \left(x^4+60 x^2+495\right)}$$
$$P_3=\frac{7613}{117978}-\frac{x^2}{6678}+\frac{121 \left(166969 x^2+3935295\right)}{39326 \left(106
   x^4+5610 x^2+45045\right)}$$ and so on.
So, we face the problem of
$$I(a,b,c,d)=\int_0^\pi \frac {a x^2+b}{(x^2+c)(x^2+d)}\,dx$$
$$I(a,b,c,d)=\frac 1{c-d} \Bigg[ \frac{ (b-a d)}{\sqrt{d}}\tan ^{-1}\left(\frac{\pi }{\sqrt{d}}\right)-\frac{ (b-a c)}{\sqrt{c}}\tan ^{-1}\left(\frac{\pi }{\sqrt{c}}\right)\Bigg]$$ Now, some numerical results
$$\left(
\begin{array}{cc}
 k & \text{approximation} \\
 1 & 0.8949240197 \\
 2 & 0.8950080586 \\
 3 & 0.8950039337 \\
 4 & 0.8950042166 \\
 5 & 0.8950041937 \\
 6 & 0.8950041958 \\
 7 & 0.8950041956
\end{array}
\right)$$

Just for the fun of it, an inverse symbolic calculator proposed as an approximation
$$I= \Big[e^\alpha \, \pi^\beta \cot ^4(e \pi ) \csc ^{10}(e \pi )\Big]^{\frac 1 {25}}$$
$$\alpha =38+e-\frac{5}{e}-\frac{47}{\pi }-16 \pi\qquad \text{and} \qquad \beta=14 e-19$$ which is in error of $2.29\times 10^{-20}$.


问:
How to minimize a summation function
My question is: How do we minimize a function like this with find a constant number for x?

in another word, we have a set and "i" is a member of that set and we should find a constant number as x to minimize that function
My problem function

答:
As @Yves says, take the median. Proof:
$$f(x)=\sum_i |i-x|=\sum_{i \geqslant x} (i-x) + \sum_{i < x} (x-i)  $$
$$f'(x)=\sum_{i \geqslant x} (-1) + \sum_{i < x} (+1)  $$
which is the number of data points less than $x$ LESS the number of data points greater than $x$.
To minimize $f(x)$ we want its derivative to be zero, which gives us the median.

Title: How to minimize a summation function


问:
Elegant way to solve $n\log_2(n) \le 10^6$
I'm studying Tomas Cormen Algorithms book and solve tasks listed after each chapter.

I'm curious about task 1-1. that is right after Chapter #1.

The question is: what is the best way to solve: $n\lg(n) \le 10^6$, $n \in \mathbb Z$, $\lg(n) = \log_2(n)$; ?

The simplest but longest one is substitution. Are there some elegant ways to solve this? Thank you!

Some explanations: $n$ - is what I should calculate - total quantity of input elements, $10^6$ - time in microseconds - total algorithm running time. I should figure out nmax.

答:
I think Newton's Method and Fixed-Point Iteration are too complicated when you're looking for an integer. A simple Binary Search would be much more effective,
and probably faster if you don't need as many floating-point operations.

I give more details on my answer on this question's duplicate on CS.SE.


问:
closed subscheme of $\mathbb{P}_Z^1$ which is affine over $Z$
Given an algebraically closed field $k$ and a scheme $Z$ over $k$. Consider the projective scheme $X=\mathbb{P}^1_Z=\mathbb{P}_k^1\times_kZ$ and $Y\subset X$ a closed subscheme. Assume that $Y$ does not contain any closed fibre of the projection $p:X\rightarrow Z$, try to show that $\pi:=p|_Y:Y\rightarrow Z$ is an affine morphism, i.e. one can find an affine covering $\{U_i\}$ of $Z$ such that each $\pi^{-1}(U_i)$ is affine.

My attempt:

In the extremely special case $Z=\text{Spec}(k)$, one has $X=\mathbb{P}_k^1$ and $X_z\cong\mathbb{P}_k^1$. Here any closed subscheme $Y$ of $X$ satisfies the condition that $X_z\not\subseteq Y$. So we want the canonical map $Y\hookrightarrow\mathbb{P}_k^1$ to be affine, which is clearly true.

Now let $Z=\text{Spec}(A)$ over $k$, we also have $X_z\cong\mathbb{P}_k^1$ and now $X=\mathbb{P}_A^1$. We know that any closed subscheme of $\mathbb{P}_A^1=\text{Proj}(A[x_0,x_1])$ is determined by a homogeneous ideal $I\subset A[x_0,x_1]$, i.e.
$$Y=V_+(I)=\text{Proj}(A[x_0,x_1]/(I))$$
Then I want to define a scheme $Y_k$ such that $Y=Y_k\times_kZ$. If I can do it, them I am done. However,  I have no ideal how to apply the setting $X_z\not\subseteq Y$ here.

答: Answer: For any closed point $t\in Z$ it follows the fiber $Y_t$ by definition is a strict closed subscheme $Y_t \subsetneq \mathbb{P}^1_{\kappa(t)} \subseteq \mathbb{P}^1_Z$, and since $dim(\mathbb{P}^1_{\kappa(t)})=1$ it follows $dim(Y_t)=0$ and hence $Y_t$ is finite.


问:
Relation of probability of a random variable with its standard deviation and mean?
I just came across this question:

What is the probability that the value of a normal random variable X exceeds its mean by more than 1.9 times its standard deviation?

I'm not really sure how to approach this to be honest. Is there any equation I need to know that related all of them?

答:
you have to calculate the following probability

$$\mathbb{P}[X>\mu+1.9\sigma]$$

now standardize and find

$$\mathbb{P}\left[Z>\frac{\mu+1.9\sigma-\mu}{\sigma}\right]=\mathbb{P}[Z>1.9]=1-\Phi(1.9)\approx0.029$$


问:
$\tau$ is a pure state and $\tau(a)=0$. Then $a=b+c$ for some $b,c$ with $\tau (b^*b)=\tau(cc^*)=0$.
A pure state $\tau $ is a state such that every positive linear functional  $\rho\leq \tau$ satysfies $\rho=t\tau$ for some $t\in\mathbb R$.

$a$ is an element in a $C^*$-algebra and $\tau$ is a pure state. Prove that if $\tau(a)=0$ then there are $b,c$ such that $\tau(b^*b)=\tau(cc^*)=0$ and $a=b+c$.

I have no idea how I use pureness here. None of the propoties of pure state I know seem to help. Could someone please give me a hint?

答:
This is a very nice question. Believe it or not, the result follows from Kadison's transitivity theorem (I'm not sure if someone can avoid it).

Proposition: Let $A$ be a C$^*$-algebra and $\varphi\in S(A)$ a state. Set $N_\varphi:=\{a\in A:\varphi(a^*a)=0\}$ for the closed left ideal appearing in the GNS construction. Then $\varphi$ is a pure state if and only if $$\ker(\varphi)=N_\varphi+N_\varphi^*.$$

Proof:

The converse is easy and no big theorem is used, so I am leaving it as an exercise. Also, observe that the inclusion $N_\varphi+N_\varphi^*\subset\ker(\varphi)$ is true for any state $\varphi$. So the direct implication (which is what you are interested in) is about showing the other inclusion and it goes as follows.

Let $\varphi$ be a pure state and take the associated GNS representation $(H_\varphi,\pi_\varphi,\xi_\varphi)$, where $\xi_\varphi$ is the unique cyclic vector. Let $a\in\ker(\varphi)$. By the properties of the GNS construction we have that $$0=\varphi(a)=\langle\pi_\varphi(a)\xi_\varphi,\xi_\varphi\rangle $$
so $\pi_\varphi(a)\xi_\varphi$ and $\xi_\varphi$ are two orthogonal vectors of $\mathcal{H}_\varphi$. Since $\varphi$ is pure, $\pi_\varphi(A)$ acts irreducibly on $\mathcal{H}_\varphi$, so we can apply Kadison's transitivity theorem: the vecotrs $\pi_\varphi(a)\xi_\varphi$ and $\xi_\varphi$ are linearly independent and the orthogonal projection $p\in\mathbb{B}(H_\varphi)$ onto the one-dimensional subspace $\text{span}\{\pi_\varphi(a)\xi_\varphi\}$ is a self-adjoint operator that maps $\pi_\varphi(a)\xi_\varphi$ to itself and it maps $\xi_\varphi$ to $0$. By Kadison's theorem, we can find a self-adjoint element $b\in A$ so that
$$\pi_\varphi(b)\big(\pi_\varphi(a)\xi_\varphi\big)=\pi_\varphi(a)\xi_\varphi\text{ and }\pi_\varphi(b)\big(\xi_\varphi\big)=0.$$

Now $b\in N_\varphi^*$: indeed, $\varphi(bb^*)=\varphi(b^2)=\langle\pi_\varphi(b^2)\xi_\varphi,\xi_\varphi\rangle=\|\pi_\varphi(b)\xi_\varphi\|^2=0$. Since $N_\varphi$ is a left ideal, $N_\varphi^*$ is a right ideal, thus $ba\in N_\varphi^*$. Now
$$a=(a-ba)+ba.$$
To conclude the proof, one observes that $a-ba\in N_\varphi$:
$$\varphi((a-ba)^*(a-ba))=\varphi((a^*-a^*b)(a-ba))=\varphi(a^*a-2a^*ba+a^*b^2a)= $$
$$=\langle\pi_\varphi(a^*a)\xi_\varphi-2\pi_\varphi(a^*)\pi_\varphi(b)\pi_\varphi(a)\xi_\varphi+\pi_\varphi(a^*)\pi_\varphi(b)^2\pi_\varphi(a)\xi_\varphi,\xi_\varphi\rangle=$$
$$\langle\pi_\varphi(a)^*\pi_\varphi(a)\xi_\varphi-2\pi_\varphi(a^*)\pi_\varphi(a)\xi_\varphi+\pi_\varphi(a^*)\pi_\varphi(a)\xi_\varphi,\xi_\varphi\rangle=\langle0,\xi_\varphi\rangle=0.$$

References: I have seen this proposition in Ilijas Farah's book Combinatorial Set Theory of C$^*$-algebras. The idea of the proof is illustrated there.


问:
Eigenvalues of block matrix comprising diagonal matrices
This seems simple, but I can't seem to solve it.
Let
$$A = \begin{pmatrix} D_{11} & \dots & D_{1n} \\ D_{21} & \dots & D_{2n} \\ \vdots & \vdots & \vdots\\ D_{n1} & \dots & D_{nn} \end{pmatrix}$$
be a block matrix such that
(1) $A$ is symmetric
(2) each $D_{ij}$ is diagonal.
The goal is to find the eigenspectrum of $A$.

Special case) If $D_{ij} = c_{ij}I$, would replacing each block $D_{ij}$ by the scalar $c_{ij}$ give the same eigenspectrum?

General) Without this additional assumption, can we compute the eigenspectrum of $A$ using a surrogate matrix $\tilde{A}$ with block entries $A_{ij}$ replaced by a characterization of $D_{ij}$? If so, what would be this characterization?

I have seen this post where we can reorder the rows/columns to preserve eigenspectrum and reduce computation, but I think this question is a bit more specific and hopefully there exists a more analytical answer.

答:
In the special case where $D_{ij} = c_{ij}I_d$, there is an answer using Kronecker product, because you can write your matrix:

$$A=c \otimes I_d$$

As a consequence, the eigenvalues of $A$ are all the possible products of eigenvalues of $c$ with eigenvalues of $I_d$; otherwise said, the spectrum of $D$ is the spectrum of $c$, each eigenvalue having multiplicity $d$.


问:
Formulas for lcm of three integers
I know that both of the following formulas equal to $\text{lcm}(a,b,c)$, but how can we show that

$\dfrac{abc \cdot \gcd(a,b,c)}{\gcd(a,b) \cdot \gcd(a,c) \cdot  \gcd(b,c)} = \dfrac{abc }{\gcd(ab,ac,bc)}$ ?

答:
As here, cancelling $\,abc\,$ then clearing denominators shows it is equivalent to

$$(a,b,c)(ab,bc,ca) = (a,b)(b,c)(c,a)\qquad$$

This equality is easy by gcd "polynomial" arithmetic: expanding the products on both sides (apply gcd associative, commutative, distributive laws), shows both $= (aab,aac,abb,abc,acc,bbc,bcc)$.

Note: as shown here, it is equivalent to gcd distributes over lcm (or vice-versa).

Title: Formulas for lcm of three integers


问:
How to show that $f(x) = 5x^{5}-8x^{4}+6x^{3}-4x^{2}+7x+9$ is irreducible over $\mathbb{Q}$?
Rational Roots Theorem cannot be used since $\pm \frac{9,3,1}{5,1}$ are not zeroes of $f(x)$.

Eisentein's criterion cannot be used since the non-leading coefficients have no common factors.

For mod $p$ test, I tried $p=2$, but $\overline{f}(x) = x^{5} + x +1 = (x^{2}+x+1)(x^{3}+x^{2}+1)$.

For $p=3$, $\overline{f}(x)$ has a factor of $x$. $p=5$ is not possible since $deg \overline{f}=4 \neq 5 = deg f$.

I tried $p=7$ but got a linear factor.

Should I keep on looking for primes that could satisfy the mod p test or are there more efficient ways to show the desired conclusion?

答:
The polynomial is indeed irreducible over $\Bbb F_{13}$, and
$5$ is a unit in this field. We have
$$
f=5(x^5 + x^4 + 9x^3 + 7x^2 + 4x + 7).
$$


问:
Von Neumann Algebra built from the bilateral shift
Let $T: \ell^2(\mathbb{Z}) \rightarrow \ell^2(\mathbb{Z})$ be the bilateral shift, and let $S= \{ I, T, T^* \}$.

Why do we have $S'= S'' = \{ W\in B(\ell^2(\mathbb{Z})) : \langle e_n, We_k \rangle = \langle e_{n-k}, We_0 \rangle\} $?

I know that if we think of it in terms of matrices then this is $w_{{n+j},{k+j}} = w_{n,k}$. But I still don't really see what's going on here. I wonder if there is way to see this with out the use of matrices.

Thank you in advance!

答:
Let $\mathcal{W}:=\{W\in B(H):\langle e_n,We_k\rangle=\langle e_{n-k},We_0\rangle\}$.

Then $\mathcal{S}'=\mathcal{W}$.

Proof: If $W\in\mathcal{S}'$, that is, $WT=TW$, then $$\langle e_n, We_k\rangle=\langle e_n,WT^ke_0\rangle=\langle e_n,T^kWe_0\rangle=\langle T^{-k}e_n,We_0\rangle=\langle e_{n-k},We_0\rangle$$
Conversely, if $W\in\mathcal{W}$, then $$\langle e_n,T^{-1}WTe_k\rangle=\langle e_{n+1},We_{k+1}\rangle=\langle e_{n-k},We_0\rangle=\langle e_n,We_k\rangle$$ Since this is true of all $n,k$, then $T^{-1}WT=W$, so $W\in\mathcal{S}'$.

Since $T$ and $T^*=T^{-1}$ commute, it follows that $\mathcal{S}\subset\mathcal{S}'$ and thus $\mathcal{S}''\subseteq\mathcal{S}'$.

To show $\mathcal{S}''=\mathcal{W}$, let $W,V\in\mathcal{S}'$, and let $We_0=\sum_i\alpha_ie_i$. Then $W^*e_0=\sum_i\alpha_i^*e_{-i}$ since $$\langle W^*e_0,e_n\rangle=\langle e_0,We_n\rangle=\langle e_{-n},We_0\rangle=\alpha_{-n}^*$$

Then since $\mathcal{W}$ is an algebra $$\langle e_n,VWe_k\rangle=\langle e_{n-k},VWe_0\rangle=\sum_i\alpha_i^*\langle e_{-i},Ve_{k-n}\rangle=\langle W^*e_0,Ve_{k-n}\rangle=\langle e_n,WVe_k\rangle$$ so $VW=WV$ and thus $\mathcal{W}\subseteq \mathcal{S}''$.


问:
If $a_{ij}=|A_i\cap A_j|$ and let $A=(a_{ij})\in M_n(\mathbb{R})$, show that $\det(A)\geq0$

Let $n\in\mathbb{N}$, $n\neq0$ and $A_1,A_2,...,A_n$ finite sets. For any $i,j\in\{1,2,...,n\}$, we define $a_{ij}=|A_i\cap A_j|$ and let $A=(a_{ij})\in M_n(\mathbb{R})$. Show that $\det(A)\geq0$

I took a simple example of two sets, $A_1={1,2}$, $A_2={2,3}$, $A_1\cap A_1=\{1,2\}$, $A_1\cap A_2=\{2\}=A_2\cap A_1$ and $A_2\cap A_2=\{2,3\}$. Now, I tried to calculate the determinant but I am stuck and as you may notice I don't see how this problem unfolds. Any help, please?

答:
A similar idea was used in my previous answer here.

We enumerate elements of $\bigcup_{i=1}^n A_i = \{b_1,\ldots, b_k\}$. Let $v_j\in\mathbb{R}^k$ with  $v_j=\mathbf{1}_{A_j}(b_i)$ which is $1$ if $b_i\in A_j$, and $0$ otherwise. Then each $v_j$ encodes the elements of $A_j$. Moreover, the $k\times n$ matrix $M$ whose $j$ th column is $v_j$, satisfies
$$
M^T M=(m_{ij})=A, \ \ m_{ij} = |A_i\cap A_j|.
$$
As darij grinberg commented, such matrix is positive semidefinite. Therefore, $\det A \geq 0$.


问:
If M is a linear subspace of X such that $M^⊥$ = {0}, then M is dense - counterexample
I am trying to show that the following statement is false: if $M$ is a linear subspace of a hilbert space $X$ such that $M^⊥$ = {0}, then $M$ is dense $M^⊥=\{0\}$.

My counterexample: the infinite sequence space $l^2$ over the reals is a closed linear subspace of $l^2$ over complex numbers. The former has orthogonal complement $\{0\}$ but is not dense in the latter.

Does this work?

答:
Your counter-example does not work because you are using complex scalars and real sequences do not form a linear subspace.

Actually there is no such example. If $M$ is not dense then Hahn Banach theorem gives a non-zero continuous linear functional which is $0$ on $M$. By Riesz Theorem this means ther exist $y \in H$ such that $ \langle x, y \rangle=0$ for all $x \in M$ but $y \neq 0$. But the $y \in M^{\perp} $ and $y \neq 0$.


问:
How to for loop in Math like C?
In C, we can for loop easily. But I don't know how we can loop in Math like C.

For example, I want to loop in Math like this C code:

```<code>int i;
int a = 10;
for (i = 1; i < 100; i++)
{
    a -= 2;
}
</code>```

答:
In general, we don't need to loop in mathematics, because we have alternative constructs.

Two examples, the first one is probably the best translation of yours:

$u_0 = 10, u_{n+1} = u_n - 2$, an arithmetic sequence, of which you're asking the 99-th value ($u_{99}$).

But often, you'll want to have an operation like the functional programming "fold/reduce"; which acts like a for-loop over an array, for a given operator.

For this, we have the "indexed $n$-ary operators", also called "generalized operators".

For addition:
$$\sum_{i=1}^{i=n} a^i = a^1 + a^2 + \dots + a^n$$

For multiplication:

$$\prod_{i=0}^{i=n} i+4 = 4 \times 5 \times \dots \times (n + 4)$$

You can also define an $n$-ary version for basically every binary operator you can find: $\bigcup$, $\bigcap$, etc.

The rest of the syntax consists in picking your indices intelligently. Say you wanted to sum over all elements of a matrix (generally done with a nested for-loop with $i$ and $j$ iterators), you could define it as something like this:

$$\sum_{i=1, j=1}^{i=n, j=m} a_{ij}$$

Title: How to for loop in Math like C?


答:
Let $a_i$ be the value of the variable a after the $i$-iteration, then
$$a_0 = 10, \qquad a_{i+1} = a_i - 2$$

One could show by induction that
$$a_i = 10-2i$$
for all $i\in\mathbb{N}$.

Title: How to for loop in Math like C?

Body:
In C, we can for loop easily. But I don't know how we can loop in Math like C.

For example, I want to loop in Math like this C code:

```<code>int i;
int a = 10;
for (i = 1; i < 100; i++)
{
    a -= 2;
}
</code>```


问:
Is the integral over a convex mapping of the solution to the heat equation monotonic?
I have the following problem:

Suppose $u = u(t,x)$ solves the Cauchy-Problem for the (one-dimensional) heat equation: $$\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}$$ $$u(0,x) = g(x) $$ and satisifes $\int_{\mathbb{R}} u(t,x) \,\text{d}x = 0$ for all $t \geq 0$. Furthermore, define for a convex function $\phi$ satisfying $\phi(0) = 0$ the following:  $$F(t) := \int_{\mathbb{R}} \phi (u(t,x)) \,\text{d}x $$
The goal is to show that $F$ is monotonic (I suppose here that it is decreasing).

What I have tried so far:

The key to this problem seems to be Jensen's inequality, since it relates integrals and convex functions. Therefore, define $T>t$ and apply the inequality:

$$
F(T) - F(t) \geq \phi\left(\int_{\mathbb{R}} u(T,x) \,\text{d}x\right) - F(t) = - F(t)
$$

The minuend being zero by assumtions on $u$ and $\phi$. I don't know how to follow from here. I have also tried representing $u$ with the heat kernel, i.e.:
$$ u(t,x) = \int_{\mathbb{R}} \frac{1}{\sqrt{4\pi t}} \exp{\left\{{\frac{-(x-y)^2}{4t}}\right\}} g(y) \,\text{d}y$$

but I am not able to follow from there either. Any help is appreciated!

答:
So I have come to the following, which I think is a proof:

Firstly, I used the generalized solution formula for any starting time $t_0$ (so not only $0$) and value at time $t_0$: $u(t_0,x) = u_{t_0}(x)$. For any time $t>t_0$ we get:

$$
u(x,t) = \int_{\mathbb{R}} u_{t_0} \Phi(t-t_0, x-y) \,\text{d}y
$$
where $\Phi(x,t)= \frac{1}{\sqrt{4\pi t}} \exp{\left\{{\frac{-x^2}{4t}}\right\}}$. This formula is the same as above for $t_0 = 0$. Crucially, however, we may use it to show the monotonicity, by treating the smaller value as starting value. Let $t>t_0$ arbitrary, then we get:
$$
\begin{align}
F(t) = &\int_{\mathbb{R}} \phi(u(t,x)) \,\text{d}x \\= &\int_{\mathbb{R}} \phi\left(\int_{\mathbb{R}} u_{t_0}(x-y) \Phi(t-t_0,y) \,\text{d}y\right) \,\text{d}x \quad \text{(Convolution)}\\ \leq  &\int_{\mathbb{R}} \int_{\mathbb{R}} \phi\left(u_0(x-y)\right) \Phi(t-t_0,y)   \,\text{d}y\,\text{d}x \quad \text{(Jensen)}\\
=  &\int_{\mathbb{R}} \int_{\mathbb{R}} \phi\left(u_0(y)\right) \Phi(t-t_0,x-y)   \,\text{d}y\,\text{d}x \quad \text{(Convolution)}\\
=  &\int_{\mathbb{R}} \phi\left(u_0(y)\right) \int_{\mathbb{R}}  \Phi(t-t_0,x-y)   \,\text{d}x\,\text{d}y \quad \text{(Fubini)}\\
=  &\int_{\mathbb{R}} \phi\left(u_0(y)\right) \,\text{d}y = F(t_0)
\end{align}
$$

We may use Jensen in the second step, since $\Phi(t,x)$ integrates over $x$ to $1$, therefore formally, $\Phi(t,x) \,\text{d}y$ is a probability mass.


问:
How to describe the stalk on fiber $X_y$?
Let $f:X\rightarrow Y$ be a morphism of schemes. (May assume more if necessary) Let $x\in X$ and $y\in Y$ such that $f(x) = y$ and $X_y$ be fiber over $y$.

My question: Is there any relation in general between the stalk $\mathcal O_{X_y,x}$ and $\mathcal O_{X,x}$? If the answer is no, what could we say or how to understand the stalk $\mathcal O_{X_y,x}$?

My question is from the following a proof in Qing Liu's book:

Lemma 4.3.20. Let $f:X\rightarrow Y$ be a morphism of finite type between locally Noetherian schemes. Then $f$ is unramified if and only if for every $y\in Y$, the fiber $X_y$ is finite, reduced, and if $k(x)$ is separable over $k(y)$ for every $x\in X_y$.

The author says: 'Indeed, the quotient $\mathcal O_{X,x}/m_y\mathcal O_{X,x}$ remains unaltered when we pass from $X$ to the fiber $X_y$'. I'm confused about the meaning of this statement.

答:
Question: "My question: Q1. Is there any relation in general between the stalk $O_{X_y,x}$ and $O_{X,x}$? Q2. If the answer is no, what could we say or how to understand the stalk $O_{X_y,x}$?"

Answer: The following seems to hold: If $\phi: X:=Spec(B) \rightarrow Y:=Spec(A)$ and  $\mathfrak{p}\subseteq A$ is a prime ideal corresponding to the point $y\in Y$ we get canonical maps

$$ A \rightarrow B \rightarrow B/\mathfrak{p}B\cong A/\mathfrak{p}\otimes_A B.$$ If $Z:=Spec(A/\mathfrak{p})$, we get $X_Z:=\phi^{-1}(Z)\cong Spec(B/\mathfrak{p})$. If $\mathfrak{p}$ is a maximal ideal corresponding to the closed point $y$, it follows
$Z\cong Spec(\kappa(\mathfrak{p}))$ hence $X_y\cong Spec(B/\mathfrak{p}B)$.

Let $\mathfrak{m}\subseteq B$ be a maximal ideal in $B$ corresponding to a point $x \in X$ with $\mathfrak{p} \subseteq \mathfrak{m}\cap A$, hence $x\in X_y$ - the fiber at $y$.  It follows $\mathfrak{p}B \subseteq \mathfrak{m}$ . Define $\overline{\mathfrak{m}}:=f(\mathfrak{m}) \subseteq B/\mathfrak{p}B$ where $f$ is the canonical map. We get by standard properties of localization a canonical exact sequence

$$ 0 \rightarrow (\mathfrak{p}B)_{\mathfrak{m}} \rightarrow B_{\mathfrak{m}} \rightarrow^i (B/\mathfrak{p}B)_{\overline{\mathfrak{m}}} \rightarrow 0$$

but the map $i$ is not an isomorphism in general - its kernel is the ideal

$$(\mathfrak{p}B)_{\mathfrak{m}}.$$

When we pass to the residue fields
we get an isomorphism

$$ \kappa(\mathfrak{m})\cong \kappa(\overline{\mathfrak{m}}).$$

Question Q1: There is when $y$ is a closed point a canonical surjective map

$$M1.\text{  }i^{\#}_x: \mathcal{O}_{X,x} \rightarrow \mathcal{O}_{X_y,x},$$

with kernel described in the above comment: If we choose open affine subschemes $x\in Spec(B), y\in Spec(A)$ it follows the kernel equals

$$ ker(i^{\#}_x)=(\mathfrak{p}B)_{\mathfrak{m}} \subseteq B_{\mathfrak{m}}.$$

Hence the map $i^{\#}_x$ is not an isomorphism in general.

Question Q2: The map $i^{\#}_x$ induce an isomorphism at residue fields, hence the two local rings in M1 have the same residue field.

To the inclusion map $i: X_y \rightarrow X$ there is a canonical map of sheaves

$$i^{\#}: \mathcal{O}_X \rightarrow i_*\mathcal{O}_{X_y}$$ and passing to the stalk at $x$ you recover the map

$$ i^{\#}_x: \mathcal{O}_{X,x}\rightarrow \mathcal{O}_{X_y,x}.$$

It seems the kernel of the map $i^{\#}_x$ is the ideal $(\mathfrak{p}B)_{\mathfrak{m}}$. Note that $\mathcal{O}_{X,x}$ is a local ring with maximal ideal $\mathfrak{m}_x$ hence $\mathfrak{m}_y\mathcal{O}_{X,x} \subseteq \mathfrak{m}_x$.

Note 1: There is a canonical map

$$f^{\#}_y: \mathcal{O}_{Y,y} \rightarrow \mathcal{O}_{X,x}$$

and you may consider the ideal

$$\mathfrak{m}_y\mathcal{O}_{X,x}\subseteq \mathcal{O}_{X,x}.$$

The ideal $\mathfrak{m}_y\mathcal{O}_{X,x}$ is not a prime-ideal in general.
If $X_y$ is an integral scheme it follows $\mathcal{O}_{X_y,x}$ is an integral domain, hence there cannot be an isomorphism

$$ \mathcal{O}_{X_y,x} \cong \mathcal{O}_{X,x}/\mathfrak{m}_y\mathcal{O}_{X,x}$$

in general as is being claimed above: This is beacuse the quotient $\mathcal{O}_{X,x}/\mathfrak{m}_y\mathcal{O}_{X,x}$ is not an integral domain. An integral domain cannot be isomorphic to a non-integral domain.


问:
Find the matrix of the linear transformation $T$
Let $A=\begin{pmatrix}0\\0\end{pmatrix},B=\begin{pmatrix}1\\0\end{pmatrix},C=\begin{pmatrix}0\\1\end{pmatrix}$ and $D=\begin{pmatrix}1\\1\end{pmatrix}$. The images of these vectors under the linear transformation is $A^*=\begin{pmatrix}0\\0\end{pmatrix},B^*=\begin{pmatrix}2\\2\end{pmatrix},C^*=\begin{pmatrix}0\\1\end{pmatrix}$ and $D^*=\begin{pmatrix}2\\3\end{pmatrix}$ respectively. Find the matrix of the linear transformation $T$.

Let $T=\begin{pmatrix}a & b\\c & d\end{pmatrix}$

$$
T(A)=A^*\implies\begin{pmatrix}a & b\\c & d\end{pmatrix}
\begin{pmatrix}0\\0\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}
$$

$$
T(B)=B^*\implies\begin{pmatrix}a & b\\c & d\end{pmatrix}
\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}a\\c\end{pmatrix}=
\begin{pmatrix}2\\2\end{pmatrix}
$$

$$
T(C)=C^*\implies\begin{pmatrix}a & b\\c & d\end{pmatrix}
\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}b\\d\end{pmatrix}=
\begin{pmatrix}0\\1\end{pmatrix}
$$

$$
T(D)=D^*\implies\begin{pmatrix}a & b\\c & d\end{pmatrix}
\begin{pmatrix}1\\1\end{pmatrix}=\begin{pmatrix}a+b\\c+d\end{pmatrix}=
\begin{pmatrix}1\\1\end{pmatrix}
$$

$(2)$ and $(3)$ implies $T=\begin{pmatrix}2&0\\2&1\end{pmatrix}$ but it's not satisfy $(4)$. Am I doing it in wrong way?

答: It's correct! Note $D^*=\begin{pmatrix}2\\3\end{pmatrix}$


问:
Prove that there are no maximum and minimum of $\frac{\sin(x_1)-\sin(x_2)}{x_1-x_2}$
The problem I'm in stuck is

For $\forall x_1, x_2\in\mathbb R$, prove that $$\frac{\sin(x_1)-\sin(x_2)}{x_1-x_2}$$ has neither maximum nor minimum.

I tried to fix $x_2$ to make a single variable function $$f(x):=\frac{\sin(x)-\sin(x_2)}{x-x_2}$$ and find the point $x_0$ such that $f'(x_0)=0$. I wanted to find some properties of $f(x_0)$, so that if $x_2$ varies, I wanted to show that $g(x_2):=f(x_0(x_2))$ has no maximum and minimum. One that I found is
$$\frac{\sin(x_0)-\sin(x_2)}{x_0-x_2}=\cos(x_0)$$
so it somehow implies that $f(x_0)=\cos(x_0)$, which is nothing helpful to me. How can I make the direction from here? Thanks in advance.

答: $\frac {\sin x_1-\sin x_2} {x_1-x_2} \leq 1$ whenveer $x _1 \neq x_2$ by MVT. Also, taking $x_2=0$ and letting $x_1 \to 0$ we  see that the expresion takes value as close to $1$ as we want. Hence, the supremum is $1$. To show that this supremum  is never attained note that if $x_1>x_2$ and  $\frac {\sin x_1-\sin x_2} {x_1-x_2} = 1$ implies that $1=\frac 1{ x_1-x_2}\int_{x_2}^{x_1} \cos t \, dt\leq \frac 1{ x_1-x_2}\int_{x_2}^{x_1} dt=1$. This forces equality to hold through oot so $\cos t=1$ for all $t \in (x_2,x_1)$. This is obviously a contradiction. So the maximum value is not attained . A similar argument works for the minimum.


答:
The last equation should be $$\begin{pmatrix}a+b\\c+d\end{pmatrix}=
\begin{pmatrix}2\\3\end{pmatrix}.$$
The right hand side of the equation is $D^*$, not $D$.

Body:
Let $A=\begin{pmatrix}0\\0\end{pmatrix},B=\begin{pmatrix}1\\0\end{pmatrix},C=\begin{pmatrix}0\\1\end{pmatrix}$ and $D=\begin{pmatrix}1\\1\end{pmatrix}$. The images of these vectors under the linear transformation is $A^*=\begin{pmatrix}0\\0\end{pmatrix},B^*=\begin{pmatrix}2\\2\end{pmatrix},C^*=\begin{pmatrix}0\\1\end{pmatrix}$ and $D^*=\begin{pmatrix}2\\3\end{pmatrix}$ respectively. Find the matrix of the linear transformation $T$.

Let $T=\begin{pmatrix}a & b\\c & d\end{pmatrix}$

$$
T(A)=A^*\implies\begin{pmatrix}a & b\\c & d\end{pmatrix}
\begin{pmatrix}0\\0\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}
$$

$$
T(B)=B^*\implies\begin{pmatrix}a & b\\c & d\end{pmatrix}
\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}a\\c\end{pmatrix}=
\begin{pmatrix}2\\2\end{pmatrix}
$$

$$
T(C)=C^*\implies\begin{pmatrix}a & b\\c & d\end{pmatrix}
\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}b\\d\end{pmatrix}=
\begin{pmatrix}0\\1\end{pmatrix}
$$

$$
T(D)=D^*\implies\begin{pmatrix}a & b\\c & d\end{pmatrix}
\begin{pmatrix}1\\1\end{pmatrix}=\begin{pmatrix}a+b\\c+d\end{pmatrix}=
\begin{pmatrix}1\\1\end{pmatrix}
$$

$(2)$ and $(3)$ implies $T=\begin{pmatrix}2&0\\2&1\end{pmatrix}$ but it's not satisfy $(4)$. Am I doing it in wrong way?


问:
How to find integral bounds in double integral?

$\iint_{D} 4 x y-y^{3} \; dA$, $ D$ is the region bounded by
$y=\sqrt{x}$ and $y=x^{3} .$

How do we find the bounds for the outer integral? In the solution they sketch a graph and the outer bounds are where $y=\sqrt{x}$ and $y=x^{3}$ intersect. I have two questions:

Algebraically this can be solved by solving $\sqrt{x} =x^{3}$ which is essentially the problem of finding the roots of $x^3-\sqrt{x}$, right? How can this be done?

Why is it obvious that the bound of the outer integral is the intersection of those two terms?

答:
Initially, keep in mind that the solution you saw is not the only one possible because you can first integrate in relation to $x$ and then in $y$ or the opposite.

As you can see for a sketch of the graphics of the functions involved, a graphic is above another. However, these graphics meet in two distinct points. Such points are essential to know the limits of the region in question and, consequently, of the integral. To find out these points, just solve the equation $\sqrt{x}=x^{3}$. You can raise both sides of this equation to the square and get the values 0 and 1 (for we are disregarding the possible complex roots).

To solve a double integral in a given region, you must know the limits of this region. The limits of the region are determined by the two curves (the graphs of the two functions). In the internal integral you consider the two functions involved, taking into account which curve is above and which is below (to determine which will be the lower limit and what will be the upper limit of the integral). The external integral will only be the numbers that correspond to the (numerical) limits of the region. If you do the opposite, leaving the expressions of the curves (as functions) in the external integral, the final result of your double integral will depend on the $x$ and $y$ variables and will not be a number, which would be an incongruity.


问:
Closed form of inverse of $(0,1) \ni x\mapsto \int_{-\infty }^{- \sqrt{-\log(x)}} e^{-\frac{z^2}{2}} dz + \sqrt{-x\log(x)}$
The function  $x\mapsto \frac 1 {\sqrt{2\pi}} \int_{-\infty }^{- \sqrt{-\log(x)}} e^{-\frac{z^2}{2}} dz + \frac 1 {\sqrt{2\pi}} \sqrt{-x\log(x)}$ appeared as cumulative distribution function of a random variable. For an algorithm I need the quantile transformation of this cdf, i.e. the generalized inverse. Is there a way to obtain a closed form of the inverse function, which may contain the inverse of $\Phi (x) = \frac 1 {\sqrt{2\pi}} \int_{-\infty}^x e^{-\frac{z^2} 2 } dz$ or any other special function?

What I tried:

The function $x\mapsto \frac 1 {\sqrt{2\pi}} \int_{-\infty }^{- \sqrt{-\log(x)}} e^{-\frac{z^2}{2}} dz$ has inverse function $f(y) = \exp (- (\Phi ^{-1}(1 - y))^2)$. Thus the problem reduces to inverte $y \mapsto y + \sqrt{-f(y) \log (f(y))} = y + \Phi^{-1}(1-y) e^{- \frac{- (\Phi^{-1} (1-y))^2}{2}}$, but here I am stuck.

Body:
Too long for a comment:

With $f(y)$ the inverse function, the first few coefficients of the series for $f(y)$ about $f(y)=e^{-2}$ are

$$a_n=\left\{\frac{\sqrt{2}}{e},\frac{3}{2},\sqrt{2} e,2 e^2,\frac{25 e^3}{4 \sqrt{2}},\frac{203 e^4}{16},\frac{129 e^5}{2 \sqrt{2}},\frac{1559 e^6}{8},\frac{2743
   e^7}{2 \sqrt{2}},\frac{87833 e^8}{16},\frac{1579043 e^9}{32 \sqrt{2}},\frac{31453069 e^{10}}{128},\frac{172014693 e^{11}}{64 \sqrt{2}},\frac{4097307607
   e^{12}}{256},\frac{13198808237 e^{13}}{64 \sqrt{2}},\frac{365808959801 e^{14}}{256},\frac{5425581922717 e^{15}}{256 \sqrt{2}},\frac{171492791499467
   e^{16}}{1024},\frac{1438592935184979 e^{17}}{512 \sqrt{2}},\frac{51068804745322799 e^{18}}{2048},\frac{956121875317413283 e^{19}}{2048
   \sqrt{2}},\frac{37660423557711966481 e^{20}}{8192},\frac{194578848405494636675 e^{21}}{2048 \sqrt{2}},\frac{8420850948835904133145
   e^{22}}{8192},\frac{11900089625422782750375 e^{23}}{512 \sqrt{2}},\frac{140317375915480012838165 e^{24}}{512},\frac{27563321019141948054192515 e^{25}}{4096
   \sqrt{2}},\frac{1407173174419688575420970525 e^{26}}{16384},\frac{18643581423362891004782922125 e^{27}}{8192 \sqrt{2}},\frac{1024262203390061020445525577235
   e^{28}}{32768},\frac{14565616213719545439974663575335 e^{29}}{16384 \sqrt{2}},\ldots\right\}$$
where my convention is $f(y)=f^{-1}(e^{-2})+\sum\frac{a_n(x-f^{-1}(e^{-2}))^n}{n!}$ and $a_1=\sqrt2/e$.  A call to FindSequenceFunction (in Mathematica) times out.  A nice closed form for these coefficients probably exists, and perhaps that closed form is informative.


问:
Area under polar curve inside cardioid and circle
I'm asked to find the area of the polar region inside $r(\theta)=5(1-\sin(\theta))$ and $r=5$. Below is a plot of the situation. THe black area is the area I want.

I then found the intersection points. So $5=5(1-\sin(\theta))$ or $1=1-\sin(\theta)$ which omplies that $\sin(\theta)=0$ or $\theta=0,\pi,2\pi$.

So the area is $\int_{\pi}^{2\pi} \frac{1}{2} [(5(1-\sin(\theta)))^2-5^2] d\theta=\frac{25\pi}{2}$.

However, the answer should be $\frac{25\pi}{2}+\frac{75\pi}{4}-50$.

Can someone point out my mistake and how I can fix this? Thanks!

答:
I don't quite understand why you though that your integral was the correct answer (as an aside, the integral you were looking at actually is equal to $6.25(8+\pi)$). I feel like you may have a difficulty with the formula we use to find areas bounded by polar curves; if you are having difficulties with it then I will be happy to explain it to you.

The way to find this area is to think of the area split into $2$ regions: the first above the $x$ axis-let's call it $A_1$- and the second below the $x$ axis-let's call it $A_2$.

To find the first area, $A_1$:
$$A_1=\frac{1}{2}\int_0^{\pi}25(1-\sin\theta)^2d\theta$$
or note that by symmetry,
$$A_1=2\left(\frac{1}{2}\int_0^{\pi/2}25(1-\sin\theta)^2d\theta\right)=\int_0^{\pi/2}25(1-\sin\theta)^2d\theta$$

And the value of the second area, $A_2$ is equal to the area of half a semicircle of radius $5$, which is just $25\pi/2$. If you really wanted, you could also calculate $A_2$ via an integral:
$$A_2=\frac{1}{2}\int_{\pi}^{2\pi}5^2d\theta$$
Add $A_1$ and $A_2$ together and you have your answer.

I hope that helps. If you have any questions please don't hesitate to ask :)


问:
A problem on smallest $\sigma-$ algebra in Axler's Measure theory book
Could you please give me some hints on this problem please (this problem is from Measure, Integration and Real Analysis, section 2.28, Axler, 2020)

$X$ is a set. $A$ is the set of subset of $X$ such that $A$ = {{$x$}: $x \in X$}.

Prove that the smallest $\sigma-$ algebra on $X$ that contains $A$ is the set of all subset $E$ of $X$ such that $E$ is countable or $X$\ $E$ is countable (which I will call the set $B$ from now on).

My idea is to prove that $B$ is the intersection of all $\sigma-$algebra on $X$ that contains $A$. To do it, I must prove that every element of $B$ $\in$ an arbitrary $\sigma-$ algebra on X that contains $A$.

But from here, I'm stuck at how to prove a set is in a $\sigma-$ algebra.

Thank you very much for any comments or any hints.

答:
$X$ is a set. $A$ is the set of subset of $X$ such that $A$ = {{$x$}: $x \in X$}.

Prove that the smallest $\sigma-$ algebra on $X$ that contains $A$ is the set of all subset $E$ of $X$ such that $E$ is countable or $X$\ $E$ is countable (which I will call the set $B$ from now on).

Proof: Let us write $\sigma(A)$  to indicate the smallest $\sigma-$ algebra on $X$ that contains $A$. Let $B$ be the set of all subset $E$ of $X$ such that $E$ is countable or $X \setminus E$ is countable

Item a. In this exercise, it is need to prove that $B$ is, in fact, a $\sigma$-algebra. It can be done by just checking the definition.

$1$. $\emptyset$ is countable, so $\emptyset \in B$.

$2$. If $E \in B$ we have two possibilities: $E$ is countable or $X$\ $E$ is countable.

$2$.a.If $E$ is countable, then $X \setminus E^c= X \setminus (X \setminus E)=E$ is countable, so $E^c\in B$.

$2$.b. If $X \setminus E$ is countable, then $ E^c = X \setminus E$ is countable, so $E^c\in B$.

So, in both cases, we have that $E^c\in B$.

$3$. Given a countable collection $\{ E_n\}_n$ such that, for all $n$, $E_n \in B$, we have that:

$3$.a If for all $n$, $E_n$ is countable, then $\bigcup_n E_n$ is countable and so, $\bigcup_n E_n \in B$.

$3$.b If there is one $n_0$ such that $X \setminus E_{n_0}$ is countable, then
$X \setminus \bigcup_n E_n= \bigcap (X\setminus E_n) \subseteq X \setminus E_{n_0}$ is countable and so, $\bigcup_n E_n \in B$.

So, in both cases, we have that $\bigcup_n E_n \in B$.

So $B$ is a $\sigma$-algebra.

Item b. Let us now prove that $\sigma(A) \subseteq B$.
Since for each $x \in X$ , $\{x\}$ is countable, we have that $ A \subseteq B$. Since $B$ is a $\sigma$-algebra, it follows that $$ \sigma(A) \subseteq B \tag{1}$$

Item c. Now let us prove that $ B \subseteq \sigma(A) $.

Let $E \in B$.

If $E$ is countable, then $E$ can be written as a countable union of single-point subsets (that is $E=\bigcup_{x \in E} \{x\}$). Since all single-point subsets are in $A$ and $\sigma(A)$ is a $\sigma$-algebra, we have that $E \in \sigma(A) $.

If $X \setminus E$ is countable, then, by the previous paragraph, $X \setminus E \in \sigma(A) $. Since $\sigma(A)$ is a $\sigma$-algebra, we have that $E \in \sigma(A) $.

So, in both cases, $E \in \sigma(A) $. So, $$ B \subseteq \sigma(A)  \tag{2}$$

From $(1)$ and $(2)$, we have $\sigma(A) = B$.


问:
Two path in a graph form a cycle when they have the same starting and ending point
I have a problem with two path $P$ and $Q$ in a graph with same starting and ending point. How I can show that $P \cup Q$ contain a cycle?
I have also an hint to do this exercise but I don't know how to manage with it.

Hint: consider symmetrical difference on $P$ and $Q$.

答:
Alternative hint: do it by induction on the total length.

If $P$ and $Q$ don't intersect (except at their start and end vertices) then they form a cycle. If they do intersect, can you find two shorter paths between some pair of vertices?


问:
Hamlet to be or not to be in Primes?
It is conjectured that, if you read $\pi$ long enough you'll find Hamlet. Since other numbers, like the Copeland–Erdős constant are known to be normal in base $10$, it should be true at least there. I wondered, if it also possible to find Hamlet or something else in the sequence of halved differences of subsequent primes:
$$
L_k = \frac{p_{k}-p_{k-1}}{2} \bmod 26
$$
($k>2$ for the nit-pickers ;-). This gives a sequence of number from $0$ to $25$, that one maps to the letters of the alphabet.
I checked the first million, but I couldn't even find my username. So before wasting more time, can one show Hamlet to be or not to be in Primes?

This might boil down to the question, if we get enough 'A's, i.e. are there infinitely many twin primes, but maybe Hamlet or at least my username comes before the very last if there is one...

答:
The answer is NO.  Almost any contiguous fragment of the play will do, so let's use “The slings and arrows of outrageous fortune”, or numerically 20 8 5 19 12 9 14 7 19 1 14 4 1 18 18 15 23 19 15 6 15 21 20 18 1 7 5 15 21 19 6 15 18 20 21 14 5.  Doubling these determines the spacings modulo 52, and hence also modulo 13, which are:

$1,3,10,12,11,5,2,1,12,2,2,8,2,10,10,4,7,12,4,12,4,3,1,10,2,1,10,4,3,12,12,4,10,1,3,2,10.$

The partial sums of this table modulo 13 are

$ \color{red}1, \color{red}4, 1, \color{red}0, \color{red}{11}, \color{red}3, \color{red}5, \color{red}6, 5, \color{red}7, \color{red}9, 4, 6, 3, 0, 4, 11, \color{red}{10}, 1, 0, 4, 7, \color{red}8, 5, 7, 8, 5, 9, \color{red}{12}, 11, 10, 1, 11, 12, \color{red}2, 4, 1,$

which contains a complete system of residues mod $13$.  Aye, there's the rub.  For, in that sequence encoding this phrase, what remainders may occur, when we have cast off multiples of $13$, must give us zero.  This will not be prime unless it is very small (in the first few letters of the sequence).

As you can see, the sequence of prime gaps is not normal.  For similar reasons you won't find AAAAAAAAAAAA anywhere.  On the other hand, Shiu proved that there are arbitrarily long strings of consecutive primes all lying in the same congruence class mod $n$ (one can even prescribe the congruence class).  Therefore we can find strings of ZZZZZZZZZZZZ, as long as devoutly to be wished.  To sleep, perchance to dream...

Title: Hamlet to be or not to be in Primes?


问:
find limit using polar coordinates
I know that when we are approaching the origin, we can always take the limit as r approaches 0. But what if we are not approaching zero? For example
$\lim \limits_{(x,y)\rightarrow (4,3)}\frac{x ^2-1}{3x+y}=1$ by direct ssubstitution.

If I were to use polar coordinates instead, $x=r\cos \theta$, $y=r\sin \theta$, $r=x^2+y^2$.

leads me to $\frac{r^2\cos^2 \theta-1}{3r\cos \theta +r\sin \theta}=\frac{25\cos^2\theta -1}{15\cos \theta +5\sin \theta}$. This and 1?, how can this possible be 1?

答:
If you insist on polar coordinates, then $(4,3)$ in polar coordinates gives $r=5$, $\sin \theta =\frac{3}{5}$ and $\cos \theta =\frac{4}{5}$. As denominator is not zero, then you can directly insert values and obtain
$$\frac{5^2 \frac{4^2}{5^2}-1}{3\cdot 5 \cdot \frac{4}{5} +5 \frac{3}{5} }=1$$
As I wrote in comment same can be obtained directly in $x,y$ coordinates.

Title: find limit using polar coordinates


问:
Every natural number $n$ can be written as $n=s-t$ with $\omega(s)=\omega(t)$
Can we prove the following statement ?

Let $\omega(m)$ denote the number of all distinct prime factors of a positive integer $m$.

Prove that every  natural  number $n$ can  be  written  as $n=s-t$, where $s$ and $t$ are positive integers with $\omega(s)=\omega(t)$, in other words, as the difference of two positive integers with the same number of distinct prime factors.

If $n=1$ , we can choose $\ s=3\ $ and $\ t=2$.

If $n$ is even , we can choose $\ s=2n\ $ and $\ t=n$.

答:
If $d$ is any even number, Schinzel's hypothesis implies the existence of a prime $p$, such that $p+d$ is prime as well. In this case , the pair $(p,p+d)$ does the job.

Let $d$ be an odd positive integer.

$$[dx(x+y),2d\frac{x^2+xy+1}{2}]$$ is a suitable pair with difference $d$ , if all the numbers $$x,x+y,\frac{x^2+xy+1}{2}$$ are odd distinct prime numbers , all larger than the largest prime factor of $d$.

With $$p=x,q=\frac{x^2+1}{2},y=2k$$ where $p$ is an odd prime , we get the numbers $$p,p+2k,q+pk$$ Note that $q$ is an integer, if $p$ is an odd prime. Schinzel's hypothesis predicts infinite many $k$ such that all the $3$ numbers are prime because $p$ and $q$ must be coprime. Since $p$ is arbitary , we can establish large enough primes and hence get the desired solution.

So, if Schinzel's hypothesis holds, we can find a pair with every given difference.


问:
Finding maxima and minima of $f(x,y) = x^{4} + y^{4} - x^{3}$
Finding maxima and minima of $$f(x,y) = x^{4} + y^{4} - x^{3}$$
I tried solving this question but need help.
In this question, I calculated the partial derivatives of first and second order.
I found
$$f_{x} = 4x^{3} - 3x^{2}\\   f_{y}= 4y^{3} \\ f_{xx} = 12x^{2}-6x \\ f_{yy}= 0$$
On equating $$f_{x} = 0 \hspace{0.5cm}  and\hspace{0.5cm} f_{y} = 0 $$ I get the critical points as (0,0) and (0.75,0) but I am not able to understand whether these points constitute a maxima or minima because $$f_{xx}f_{yy} - f_{xy}f_{yx}=0$$

Can somebody please guide?

答:
For the minimum, if $x\ge 1$ or $x\le 0$ we have $x^4-x^3+y^4=x^3(x-1)+y^4\ge 0$.  If $0<x<1$ we have by AM-GM
$$x^3(1-x)=27(x/3)^3(1-x)\le 27\left(\frac{3(x/3)+(1-x)}{4}\right)^4=27/256$$
so $x^4-x^3\ge -27/256$ and so $x^4-x^3+y^4\ge x^4-x^3\ge -27/256$.  The minimum value is reached at $x=3/4$, $y=0$.

For the maximum, clearly it doesn't exist.


问:
Limits and digit sums
Let $x\in  \mathbb{N}, f(x)$ - digit sum of $x$ , $f^{[n]}(x)$ - digit sum of digit sum of digit sum of $x$.... ($n$ times)

$radix = 10$

Does this limits for different $n$ exists and if it's true when what they are?

$$\lim_{x\to \infty} f^{[n]}(x) $$

$$\lim_{x\to \infty} \frac{f^{[n-1]}(x)}{f^{[n]}(x)} $$

$$\lim_{x\to \infty} \frac{f^{}(x)}{f^{[n]}(x)} $$

$$n= 1, 2, radix, \infty$$

What if a $radix$ is a prime number? Is it somehow affects to the existence/value of the limit?

答:
Note that $f(x)\equiv x\pmod{9}$ (or in generally modulo radix minus one).
For fixed $n$, $f^{[n]}$ becomes arbitrarily large, so the first limit does not exist (not even $\to\infty$, as we have $f(x)=1$ every now and then again).
The second limit does not exist either and even less the last.</p>

Title: Limits and digit sums


问:
Let $f: X \to Y$ be a homeomorphism and $A \subset X$. Show that $\overline{f(A)} = f(\overline{A})$.

Let $f: X \to Y$ be a homeomorphism and $A \subset X$. Show that $\overline{f(A)} = f(\overline{A})$.

In essence I want to show that two sets are the same so I should show that $\overline{f(A)} \subset f(\overline{A})$ and $f(\overline{A}) \subset \overline{f(A)}$?

I'm stuck on how I should go about this. If I let $x \in \overline{f(A)}$, then what properties of the homeomorphism can I use to show that $x \in f(\overline{A})$? Initially I assumed I could use continuity to say that if $f(x) \in f(\overline{A})$, then there exist $x \in \overline{A}$ and we have open neighborhoods $U_x$ and $V_{f(x)}$, but this doesn't seem to be helpful. Is there something else I should consider? $f$ is certainly bijective also since it's homeomorphic.

答:
Note $A \subseteq \overline{A}$ and thus $f(A) \subseteq f(\overline{A})$. Now take closure, $\overline{f(A)}\subseteq \overline{f(\overline{A})}$. But, $f(\overline{A})$ is closed by $f$ a homeomorphism, thus $\overline{f(\overline{A})}=f(\overline{A})$. Therefore, $\overline{f(A)}\subseteq f(\overline{A})$.

The harder direction is showing the other direction; however, it follows similarly.


答:
To which it seems to me, it is enough that the function $f$ is continuous. For this case I see that it is more convenient to work with limits than with neighborhoods. Note that, as the $f$ function is continuous, we have the property: $$ \lim f(a_n) = f (\lim a_n) $$

$f(\overline{A})\subset\overline{f(A)}$

If $x\in f(\overline{A})$, then $x=f(a)$ with $a\in \overline{A}$. That is, $a=\lim a_n$ and $$x=f\left(\lim a_n\right)=\lim f(a_n) \Rightarrow x\in\overline{f(A)}$$

$\overline{f(A)}\subset f(\overline{A})$

Similarly, on the other hand, if $x\in \overline{f(A)}$, then $$ x=\lim f(a_n)=f(\lim a_n)=f(a_0) $$

where $a_0=\lim a_n$. Note that if $f(a_ n)$ diverges, then $a_ n$ also diverges, because $f$ is a continuous function. So, from 1 and 2 we have $\overline{f(A)} = f(\overline{A})$.

Body:
Let $f: X \to Y$ be a homeomorphism and $A \subset X$. Show that $\overline{f(A)} = f(\overline{A})$.

In essence I want to show that two sets are the same so I should show that $\overline{f(A)} \subset f(\overline{A})$ and $f(\overline{A}) \subset \overline{f(A)}$?

I'm stuck on how I should go about this. If I let $x \in \overline{f(A)}$, then what properties of the homeomorphism can I use to show that $x \in f(\overline{A})$? Initially I assumed I could use continuity to say that if $f(x) \in f(\overline{A})$, then there exist $x \in \overline{A}$ and we have open neighborhoods $U_x$ and $V_{f(x)}$, but this doesn't seem to be helpful. Is there something else I should consider? $f$ is certainly bijective also since it's homeomorphic.


问:
Question about inhomogeneous linear differential equations
I have a feeling there is a trick to this question, but I do not know what it is.. Please help :)

This is the question

答:
Let $\gamma:= \alpha +i \beta= \pi e^{i \frac{5}{7} \pi}.$

Observe that the constant function $1$ is a special solution of the inhomogenous equation. Thus

$h$ is a solution of the inhomogenous equation $ \iff h-1$ is a solution of the homogenous equation $ \iff \gamma $ is a root of the charactristic polynomial of the homogenous equation $\iff \sum_{k=0}^6 (-1)^k \gamma^k=0.$


问:
Wedge product of permutations in definition of pairing?
In May's definition of a pairing of operads, he states that a pairing of operads $\tau: (A,O)\rightarrow C$ consists of a collection of maps $\tau: A(i)\times O(j)\rightarrow C(ij)$ that satisfies the condition that for $a\in A(i)$, $o\in O(j)$, and for $\sigma\in\Sigma_i$ and $\mu\in\Sigma_j$, we have $\tau(a\sigma,o\mu)=\tau(a,o)(\sigma\wedge\mu)$, where $\sigma\wedge\mu\in \Sigma_{ij}$.

I know of the block permutations used in the definition of operads, but I've never come across a wedge product of permutations before; so my question is what exactly is the permutation $\sigma\wedge\mu$?

答:
In the Appendix: Pairings of operads (starting on page 57) of Equivariant iterated loop space theory and permutative G-categories
by B. Guillou and J. P. May it is written explicitly:

Write $\boldsymbol{j}=\{1,\ldots,j\}$ and let $$\otimes: \Sigma_j \times \Sigma_k \to \Sigma_{jk}$$ be the homomorphism obtained by identifying $\boldsymbol{j}\times \boldsymbol{k}$ with $\boldsymbol{jk}$ by ordering the set of $jk$ elements $(q,r)$, $1 \leq q \leq j$ and $1 \leq r \leq k$ lexicographically. More precisely, let $\lambda_{j,k}: \boldsymbol{jk} \to \boldsymbol{j}\times\boldsymbol{k}$ be the lexicographic ordering. Then, given $\rho \in \Sigma_j$ and $\sigma \in \Sigma_k$, $\rho \otimes \sigma$ is defined by $$\boldsymbol{jk} \xrightarrow{\lambda_{j,k}} \boldsymbol{j}\times\boldsymbol{k} \xrightarrow{\rho\times\sigma} \boldsymbol{j}\times\boldsymbol{k} \xrightarrow{\lambda_{j,k}^{-1}} \boldsymbol{jk}.$$

Here, the notation $\rho \otimes \sigma$ is used instead of your $\rho \wedge \sigma$.


问:
Nonnegative vs non-negative
When you write about integers that are not negative, do you write "non-negative integers" or "nonnegative integers" in papers in US English?

EDIT:

In https://ieeecs-media.computer.org/media/conferences/cpstemplates/8.5x11x2/instruct8.5x11x2.pdf we see

The prefix “non” is not a word; it should be joined to the word it modifies, usually without a hyphen.

The really diffuse word above is "usually". :-)

答:
The American Mathematical Society Style Guide states on page 92, section 12.7.5.3

Do not hyphenate words formed with the prefixes ... “non-” ...

Similarly, the Society for Industrial and Applied Mathematics Style Manual states on page 40, section 3.6.6

Don’t hyphenate  words formed with the prefixes ... "non-" ...

Both guides refer to The Chicago Manual of Style for exceptions.

Title: Nonnegative vs non-negative


问:
The set of continuous linear operators is a vector subspace of the space of all linear operators.
I am trying to proof that the set $L(X,Y)$ of continuous linear operators between to topological vector spaces is a vector subspace of the space of all linear operators $Hom(X,Y)$.
I proved that linear operator $$T:X \longrightarrow Y$$ is continuous iff it's continuous at 0. Using this i am trying to prove that $f + h$ is continuous, where $f, h$ are continuous linear operators, but i can't find for an neighborhood of zero $W(0) \subset Y$ any open set $U(0) \subset X$ such that $$(f+h)(U(0)) \subset W(0)$$
I know that sum and scalar multiplications are continuous, but i don't understand how to prove that the topology is shift invariant and using this proof the statement.

答:
If $V$ is an open set in $Y$ containing $0$ then there exist open sets $V_1$ and $V_2$ containing $0$  such that $V_1+V_2 \subseteq V$. [This is by continuity of addition].
Choose open sets $U_1$ and $U_2$ in $X$  containing $0$ such that  $f(U_1) \subseteq V_1$ and $h(U_2) \subseteq V_2$. If $U=U_1\cap U_2$ then $(f+h)(U)\subseteq V$.


答:
You just need to use the properties of topological vector spaces. The sum operator

$+: X^2 \rightarrow X$

is continuous, and since $f$ and $g$ are continous, their cartisian product

$(f,g): X^2 \rightarrow X^2$

is continuous, where $(f,g)(x,y) = (f(x),g(y))$. We also have that the function

$i: X \rightarrow X^2$

defined by $i(x)=(x,x)$ is continuous, so since composition of functions is continuous we get

$f+g = + \circ (f,g) \circ i$

is continous.

Title: The set of continuous linear operators is a vector subspace of the space of all linear operators.

Body:
I am trying to proof that the set $L(X,Y)$ of continuous linear operators between to topological vector spaces is a vector subspace of the space of all linear operators $Hom(X,Y)$.
I proved that linear operator $$T:X \longrightarrow Y$$ is continuous iff it's continuous at 0. Using this i am trying to prove that $f + h$ is continuous, where $f, h$ are continuous linear operators, but i can't find for an neighborhood of zero $W(0) \subset Y$ any open set $U(0) \subset X$ such that $$(f+h)(U(0)) \subset W(0)$$
I know that sum and scalar multiplications are continuous, but i don't understand how to prove that the topology is shift invariant and using this proof the statement.


问:
Limiting value of iteration $x(k+1) = A x(k) + B u(k)$ for summable $u(k)$
A matrix $A$ is known to converge such that $\lim_{k\rightarrow \infty} A^k  = \bar{A} \neq 0$. We have an iteration defined as
$$x(k+1) = A x(k) + B u(k), \ \ k\in \mathbb{Z}_+.$$

$\{u(k), k=0,1,..\}$ is an infinite sequence. Under what kind of sequence $\{u(k)\}$, will $x(k)$ converge as $k \rightarrow \infty$?

I suspect the sequence converges if $\{u(k)\}$ is absolutely summable and, in that case, converges to $\bar{A}x_0 + \bar{A} B \sum_{k=0}^\infty u(k)$ where $x_0 = x(0)$. However, I am not sure how to prove it. Would really appreciate any help on this.

答:
A sufficient condition is summability of $u$.

The solution to the difference equation is $x_k = A^k x_0 + \sum_{i=0}^k A^{k-i} B u_i$.

Since $A^k \to \overline{A}$, we see that $\|A^k\| \le M$ for some $M$ and for all $k \ge 0$. (It follows that $\|\overline{A}\| \le M$.)

We have$\sum_{i=0}^k \|A^{k-i}\| \|B\| \|u_i\| \le M \|B\| \sum_{i=0}^k \|u_i\| \le M \|B\| \sum_{i=0}^\infty \|u_i\|$, hence $x_1 = \lim_{k \to \infty} \sum_{i=0}^k A^{k-i} B u_i$
exists.
It follows that $x_k \to \overline{A} x_0 + x_1$.

If $u$ is summable, then $\|u_k \| \le M'$ for some $M'$ and for all $k \ge 0$.
Let $ \epsilon >0$ and choose $N$ such that  $2 M \|B\| \sum_{i>N} \|u_i\| < \frac{1}{2} \epsilon$. Now choose $N' \ge N$ such that for $k \ge N'$, $\|B\| M' \max(\|A^k-\overline{A}\|, \|A^{k-1}-\overline{A}\|,..., \|A^{k-N}-\overline{A}\|) < \frac{1}{2(N+1)} \epsilon$.

If $k \ge N'$, then
\begin{eqnarray}
\| \sum_{i=0}^k (A^{k-i} - \overline{A}) B u_i \| &\le& \| \sum_{i=0}^N (A^{k-i} - \overline{A}) B u_i \| + \| \sum_{i>N} (A^{k-i} - \overline{A}) B u_i \| \\
&\le& \|B\|M' \sum_{i=0}^N \| A^{k-i} - \overline{A} \| + 2 M \|B\| \sum_{i>N} \|u_i\| \\
&<& (N+1)\frac{1}{2(N+1)} \epsilon + \frac{1}{2} \epsilon \\
&=& \epsilon
\end{eqnarray}
It follows that $x_1 = \lim_{k \to \infty} \sum_{i=0}^k A^{k-i} B u_i
= \lim_{k \to \infty} \sum_{i=0}^k \overline{A} B u_i  = \overline{A} B \sum_{i=0}^\infty  u_i$, and so
$\lim_{k \to \infty} x_k = \overline{A} x_0 + \overline{A} B \sum_{i=0}^\infty  u_i$.


问:
Books for Hyperbolic Geometry.
I want to read hyperbolic geometry.
Can any one suggest some good books on the topic.</p>

答: The book by martelli "An introduction to geometric topology" is a nice one on Hyperbolic geometry.</p>

Title: Books for Hyperbolic Geometry.


问:
Integration application to volume
I'm trying to find out the volume of:
Area enclosed by $y=\arctan x, \ y=0, \ x=1$ rotated about y-axis.

I tried to use both the disc and cylindrical method, and I can get it with the cylindrical shell method, but not with the disc method. Here's my cylindrical shell solution:

$$
2\pi rh\delta t=2\pi x\arctan x\delta x \\
\int^1_0 2\pi x\arctan xdx=\pi(\frac{\pi}{2}-1)
$$
And here's my attempt at a disc solution:

Inner shell$$
\pi r^2\delta h=\pi x^2\delta y=\pi(1-\tan y)^2\delta y \\
\int^{\frac{\pi}{4}}_0{\pi(1-tany)^2)}
$$Outer shell$$
\pi r^2 \delta h=\pi1^2\delta h \\
\pi\int^{\frac{\pi}{4}}_0{1}dy
$$However,$$
\pi\int^{\frac{\pi}{4}}_0{1}dy-\int^{\frac{\pi}{4}}_0{\pi(1-tany)^2)}\neq\pi(\frac{\pi}{2}-1)
$$Could someone help me pick out where I have gone wrong? Thank you very much for your help!

答:
Yes your working for the cylindrical shell method is correct. In disk method, you have a mistake in the volume integral for the inner shell.

It should be $\displaystyle \int_0^{\pi/4} \pi x^2 \ dy = \int_0^{\pi/4} \pi  \ \color {blue} {(\tan y)}^2  \ dy = \frac{\pi}{4}(4-\pi) \ $.

So the volume using disk method $ = \displaystyle \frac{\pi^2}{4} -  \frac{\pi}{4}(4-\pi) = \pi (\frac{\pi}{2}-1)$

Title: Integration application to volume


问:
Express in symbols: "There is a real number that is not the cube of any real number besides itself."

I have the following statement, and I want to use symbols only to express its meaning.

There is a real number that is not the cube of any real number besides itself.

Here is my answer (assuming that $x$ and $y$ are in the universal set $R$ just to make the notation shorter):

$$(\exists x)(\forall y)(x \neq y^3)$$

Is this correct?

Thank you ^^

答:
You are close...but not quit. Assuming we all know from where $\;x,\,y\;$ come, we can simply write

$$\exists\,x\,\forall\,y\left((y\neq x\implies x\neq y^3)\;\wedge x^3=x\right)$$


答: I´rather use $$\exists x\in\mathbb{R}\quad \text{ s.t. }\quad x\not=y^3 \wedge x^3=x  \quad\forall y\not=x \in \mathbb{R}.$$ If you add $x$ and $y$ belongs in $\mathbb{R}$, your answer will look fine to me.

Title: Express in symbols: "There is a real number that is not the cube of any real number besides itself."

Body:
I have the following statement, and I want to use symbols only to express its meaning.

There is a real number that is not the cube of any real number besides itself.

Here is my answer (assuming that $x$ and $y$ are in the universal set $R$ just to make the notation shorter):

$$(\exists x)(\forall y)(x \neq y^3)$$

Is this correct?

Thank you ^^


答: A map is continuous if and only if $f\left (\overline A\right ) \subseteq \overline {f(A)}$ for all subsets $A$ of the domain. If $f$ is a homeomorphism then $f^{-1}$ is defined and continuous, so $f^{-1}\left (\overline B\right ) \subseteq \overline {f^{-1}(B)}$ for all subsets $B$ of the codomain. Applying $f$ to the inclusion gives $\overline B = f\left (f^{-1}\left (\overline B\right ) \right) \subseteq f\left (\overline {f^{-1}(B)}\right )$, if we now take $B = f(A)$ we get the opposite inclusion $\overline {f(A)}\subseteq f\left (\overline A\right )$.

Title: Let $f: X \to Y$ be a homeomorphism and $A \subset X$. Show that $\overline{f(A)} = f(\overline{A})$.


问:
How can I find the surface area of a normal chicken egg?
This morning, I had eggs for breakfast, and I was looking at the pieces of broken shells and thought "What is the surface area of this egg?" The problem is that I have no real idea about how to find the surface area.

I have learned formulas for circles, and I know the equation for an ellipse; however, I don't know how to apply that.

The only idea I can think of is to put an egg on a sheet of paper and trace it, and then measure the outline drawn, and then try to find an equation for that ellipse and rotate that about the $x$-axis.  Now, my problem is how I can find the equation of the ellipse from the graph, and will my tracing method really be the edge of the egg?  Also, can I use the standard surface area integral? Will I have to use some techniques to solve the integral that are not covered in the AP BC Calculus?

There has to be a better method for finding the surface area. Please, help me understand how to find the surface area of an egg; i.e., how to use my mathematical knowledge for something other than passing exams.

答:
There is a nice equation describing the equation of an egg curve credit to Nobuo Yamamoto :
$$
(x^2+y^2)^2 = ax^3 + \frac{3a}{10}xy^2, \tag{1}
$$
where $0\leq x\leq a$, $a$ is the length of the major axis of symmetry for an egg.

In other words, we could get it by cutting a boiled egg in half and measure the distance from tip to the bottom. I just drew it in MATLAB, and the curve looks like the following for $a=1$:

I must say this curve fits pretty well with an egg. Now we have a curve, then the method of computing surface area by revolution, which is taught in Calculus II I believe, can be used to computing the surface. We just revolve the curve above around the egg's major axis of symmetry and get a surface, here is what looks like when we revolve it around the $x$-axis by degree $\pi$, we can get the lower half by revolve another $\pi$:

First we solve for $y$ in (1) when $y>0$:
$$
y = \sqrt{\frac{3ax}{20} - x^2 + x\sqrt{\frac{7ax}{10} + \frac{9a^2}{400}} },
$$
The formula of computing surface area by revolution is:
$$
S = 2\pi\int_0^a y\sqrt{1+\left(\frac{dy}{dx}\right)^2} \,dx,\tag{2}
$$
The derivative is:
\begin{align}
\frac{dy}{dx} = \frac{1}{2\sqrt{\frac{3ax}{20} - x^2 + x\sqrt{\frac{7ax}{10} + \frac{9a^2}{400}} }} \left( \frac{3a}{20}- 2x+ \sqrt{\frac{7ax}{10} + \frac{9a^2}{400}} + \frac{7ax}{20\sqrt{\frac{7ax}{10} + \frac{9a^2}{400}}}\right),
\end{align}
Plugging $dy/dx$ and $y$ into (2), then you could use your favorite tool of numerical integration to perform the computing for you(Octave, MATLAB, Mathematica, etc).

A more tweakable/numerical/experimental approach:

As Ｊ. M. suggests in the comments, the shape looks like an egg, but is a real egg being approximated nicely by this curve? I guess the answer is that "it really depends on that specific egg"!

Let's say we still want to use surface of revolution to compute the surface area.

But this time, we handle it more numerically from the very start, instead of looking for a curve to fit one thing for all.

Two assumptions:

All eggs are axial symmetric with respect to its major axis, i.e., if $x$-axis is its major axis, its surface can be obtained by revolving a curve $y= f(x)$, for $0\leq x\leq a$.

That curve $y = f(x)$ has certain smoothness: $f$ and $f'$ are continuous for $x\in (0,a)$ .

Now we want to compute the integral (2) using Simpson's rule or Trapezoidal rule, which is also taught in Calculus II in most colleges I believe. A remark is that $|f'|\to \infty$ when $x\to 0$ and $x\to a$, it would be much better if we use adaptive qudrature by putting more sample points near $0$ and $a$.

The steps are:

Boil an egg, cut it by half, hold it against a paper, use a pencil to outline its boundary (upper half is enough).

Draw the major axis, set it to be the $x$-axis, and measure its length $a$.

Choose $(n+1)$-sample points (including the end points) so that the points are equidistant to their neighbor on the curve. $n$ is chosen to be even, measure the distance to the major axis ($y$-coordinates) like the above figure.

Denote the sample point as $(x_i,y_i)$, $x_0=0$, $x_n = a$.

Approximate
$$\frac{dy}{dx}\Big|_{x_i} \approx s_i = \frac{1}{2}\left( \frac{y_{i+1} - y_i}{x_{i+1} - x_i} +  \frac{y_{i} - y_{i-1}}{x_{i} - x_{i-1}} \right).$$
For two end points:
$$
s_0 = \frac{y_1 - y_0}{x_1 - x_0},\quad \text{ and }\quad s_n = \frac{y_n- y_{n-1}}{x_n - x_{n-1}}.
$$
This step may be problematic, we can use other methods to approximate $dy/dx$: for example, approximating the curve by cubic spline using sample points $(x_i,y_i)$, but it would be beyond the content of college calculus.

Let $h = x_{i+1} - x_i = a/n$, approximate (2) by computing:
$$
\frac{2\pi h}{3}\bigg[g(x_0)+2\sum_{j=1}^{n/2-1}g(x_{2j})+
4\sum_{j=1}^{n/2}g(x_{2j-1})+g(x_n)
\bigg], \quad \text{ where } g(x_i) = y_i \sqrt{1+s_i^2}.
$$

Some results comparison:

Amzoti gave a link in his comments above that has two semi-empirical formulas:
$$
S_1 = (3.155 − 0.0136a + 0.0115b)ab, \;\text{ and }\;S_2 = \left(0.9658\frac{b}{a}+2.1378\right)ab
$$
where $a$ and $b$ are the length for major and minor axis of the real eggs. If there exists an egg's shape like (1), $a = 1$, and $b\approx 0.7242629$, the surface area computed by above formula is:
$$
S_1 \approx 2.278215  \;\text{ and }\;S_2 \approx 2.054946.
$$
Using the surface by revolution formula (2), we have:
$$
S \approx 2.042087.
$$