问:
Find the sum of all possible values of $\cos 2x + \cos 2y + \cos 2z.$
Let $x$, $y$, and $z$ be real numbers such that $ \cos x + \cos y + \cos z = \sin x + \sin y + \sin z = 0 $. Find the sum of all possible values of $ \cos 2x + \cos 2y + \cos 2z $.

Here is what I have done so far

$$ \cos x + \cos y = -\cos z $$
$$ (\cos x + \cos y)^{2} = (-\cos z)^{2} $$
$$ \cos^{2} x + \cos^{2} y + 2\cos x \cos y = \cos^{2}z $$

likewise,
$$ \sin x + \sin y = -\sin z $$
$$ (\sin x + \sin y)^{2} = (-\sin z)^{2} $$
$$ \sin^{2} x + \sin^{2} y + 2 \sin x \sin y = \sin^{2}z $$

from this, you get

$$ \cos^{2} x + \cos^{2} y + 2\cos x \cos y + \sin^{2} x + \sin^{2} y + 2\sin x \sin y = \sin^{2}z + cos^{2}z $$
$$ 1 + 1 + 2(\cos x \cos y + \sin x \sin y) = 1 $$
$$ \cos x \cos y + \sin x \sin y = -\frac{1}{2} $$
$$ \cos(x-y) = -\frac{1}{2} $$
$$ x-y = \frac{2 \pi}{3}, \frac{4 \pi}{3} $$

likewise,
$$ \cos (x-z) = -\frac{1}{2} $$
$$ x-z = \frac{2 \pi}{3}, \frac{4 \pi}{3} $$

where do I go from here?

答:
Notice that $\begin{cases}\cos(x)+\cos(y)+\cos(z)=0\\\sin(x)+\sin(y)+\sin(z)=0\end{cases}\iff \underbrace{e^{ix}}_a+\underbrace{e^{iy}}_b+\underbrace{e^{iz}}_c=0$

We have $a+b+c=0\implies \bar a+\bar b+\bar c=0$

But since $|a|=|b|=|c|=1$ they verify $\bar a=\frac 1a$ and so on.

Therefore we get $$\frac 1a+\frac 1b+\frac 1c=0\implies \dfrac{ab+bc+ca}{abc}=0$$

Note that $|abc|=1$ so the numerator should be zero.

But then $$\overbrace{(a+b+c)^2}^{=0}=\overbrace{(a^2+b^2+c^2)}^{e^{i2x}+e^{i2y}+e^{i2z}}+2\overbrace{(ab+bc+ca)}^{=0}$$

And your sum is simply $0$ too.


问:
Famous or common mathematical identities that yield $1$
To me the most common identity that comes to mind that results in $1$ is the trigonometric sum of squared cosine and sine of an angle:

$$
\cos^2{\theta} + \sin^2{}\theta = 1 \tag{1}
$$
and maybe

$$
-e^{i\pi} = 1 \tag{2}
$$

Are there other famous (as in commonly used) identities that yield $1$ in particular?

答:
Pick your favorite probability density function $f$, then
$$
\int_{-\infty}^\infty f(x)dx=1.
$$

ContentLicense: CC BY-SA 4.0


问:
A problem on matrices : Sum of elements of skew-matrix

If $A=[a_{ij}]$ is a skew-symmetric matrix, then write the value of $$ \sum_i \sum_j a_{ij}$$

My doubt is that what is the meaning of $ \sum_i \sum_j ?$ Is it the same as $\sum_{ij}?$

Please offer your assistance.

Thank you

答:
As for the notaion. It is easier to understand on concrete example. Say $n=3$, then
$$
\begin{align}
\sum\limits_{i=1}^3\sum\limits_{j=1}^3a_{ij}
&=\sum\limits_{i=1}^3(a_{i1}+a_{i2}+a_{i3})\\
&=(a_{11}+a_{12}+a_{13})+(a_{21}+a_{22}+a_{23})+(a_{31}+a_{32}+a_{33})\\
&=(a_{11}+a_{12}+a_{13}+a_{21}+a_{22}+a_{23}+a_{31}+a_{32}+a_{33})\\
&=\sum\limits_{i,j=1}^3 a_{ij}
\end{align}
$$

As for the original problem. Elements of the diagonal are zeros (why?). So you need to perform calculations only for nondiagonal entries. Divide them into pairs $a_{ij}+a_{ji}$, recall definition of skew symmetric matrix and conclude...

ContentLicense: CC BY-SA 3.0


Title: Famous or common mathematical identities that yield $1$

Body:
To me the most common identity that comes to mind that results in $1$ is the trigonometric sum of squared cosine and sine of an angle:

$$
\cos^2{\theta} + \sin^2{}\theta = 1 \tag{1}
$$
and maybe

$$
-e^{i\pi} = 1 \tag{2}
$$

Are there other famous (as in commonly used) identities that yield $1$ in particular?

Body: $$(\forall x\in \Bbb R)\;\;\cosh^2(x)-\sinh^2(x)=1$$


问:
Show there must be two numbers in the list whose difference is $12$.
This question came up while tutoring: "$55$ numbers are randomly selected from the first $100$ positive integers. Show there must be two numbers in the list whose difference is $12$."

It definitely requires the Pigeonhole Principle but we were unsure how to set it up. My idea was to list all of the possible pairs whose difference is $12$:

$$(1,13), (2,14), \ldots (87, 99), (88, 100)$$ of which there are $88$ pairs. I'm not sure what to do from here.

I know similar questions have been asked here, like this one. Is there a way to apply those techniques to this problem?

答:
HINT: Note that $55 = 4\times 12 + 7$. So either

there is an integer $k$ so that there are at least 6 integers
that are $k \mod 12$, or

there has to be at least 7 integers $k \in \{0,1,\ldots, 11\}$ such that there are 5 numbers picked that are $k \mod 12$.

So fix such a $k$. If 1. is satisfied then there are 6 integers
$j \in \{0,1,\ldots, 8\}$ such that $12j+k$ was picked. If 2. was satisfied then fix such a $k$, and we may also assume $k \in \{7,8,\ldots, 11\}$. So here there would be 5 integers $j \in \{0,1,2,\ldots,7\}$ so that $12j+k$ was picked, so there must be an $i$ s
t. both $12i+k$ and $12(i+1)+k$ were picked.


答:
The set of all integers between $1$ and $100$ can be written as
$$\lbrace 1,..., 24 \rbrace \cup \lbrace 25, ..., 48 \rbrace \cup \lbrace 49,..., 72\rbrace \cup \lbrace 73, ..., 96 \rbrace \cup \lbrace 97,..., 100 \rbrace $$

If you choose $55$ numbers, then at least $51$ of them are taken in the first four sets. So at least $13$ of them are taken in one of the four sets. Now you can conclude.

Body:
This question came up while tutoring: "$55$ numbers are randomly selected from the first $100$ positive integers. Show there must be two numbers in the list whose difference is $12$."

It definitely requires the Pigeonhole Principle but we were unsure how to set it up. My idea was to list all of the possible pairs whose difference is $12$:

$$(1,13), (2,14), \ldots (87, 99), (88, 100)$$ of which there are $88$ pairs. I'm not sure what to do from here.

I know similar questions have been asked here, like this one. Is there a way to apply those techniques to this problem?


问:
$g'(t)=\frac{\partial \rho}{\partial t}(x(t),t)+ (u(x,t).\nabla)\rho(x,t)$.
We are Studying the behavior of the fluid using Navier Stokes' Equation, and I am new to this lesson, but I am stuck on my below question:

If I have $g(t)=\rho(x(t),t)$, then using Chain rule, we get:
$$g'(t)=\frac{\partial \rho}{\partial t}(x(t),t)+\frac{\partial \rho}{\partial x}(x(t),t)x'(t)$$

So how to continue in order to get the latter is equal to : $$\frac{\partial \rho}{\partial t}(x(t),t)+ (u(x,t).\nabla)\rho(x,t).$$

Remark that $u(x,t)=x'(t)$ is the velocity field of the flow of the fluid (here we are dealing with the Eulerian description), and $\rho:\Omega×[0,T] \rightarrow \mathbb{R}$ is the mass density, $\nabla$ is the nabla function

I am really stuck and I need some help please!!!

答:
Observe that $x(t)=(x_1(t),x_2(t),x_3(t))$ and $u(t,x)=(x'_1(t),x'_2(t),x'_3(t))$
by the chain rules results

$$g'(t)=\frac{\partial \rho}{\partial t}(x(t),t)+\frac{\partial \rho}{\partial x_1}(x(t),t)x_1'(t)+\frac{\partial \rho}{\partial x_2}(x(t),t)x_2'(t)+\frac{\partial \rho}{\partial x_3}(x(t),t)x_3'(t)$$

$$g'(t)=\frac{\partial \rho}{\partial t}(x(t),t)+(x'_1(t),x'_2(t),x'_3(t)) \cdot
\left(\frac{\partial \rho}{\partial x_1}(x(t),t),\frac{\partial \rho}{\partial x_2}(x(t),t),\frac{\partial \rho}{\partial x_3}(x(t),t)\right )  $$

$$g'(t)=\frac{\partial \rho}{\partial t}(x(t),t)+ u(x,t) \cdot
\left(\frac{\partial \rho}{\partial x_1}(x(t),t),\frac{\partial \rho}{\partial x_2}(x(t),t),\frac{\partial \rho}{\partial x_3}(x(t),t)\right )$$

$$g'(t)=\frac{\partial \rho}{\partial t}(x(t),t)+ u(x,t) \cdot \nabla \rho(x(t),t).
$$


问:
How to evaluate $\int_{0}^{1}\int_{0}^{1} \sqrt{1 + 4(x^2 + y^2)}\,dx\, dy$?
I need to solve the integral

$$\int_{0}^{1}\int_{0}^{1} \sqrt{1 + 4(x^2 + y^2)}\,dx\,dy$$

I am using polar coordinates here to get :

$$ \int_{0} ^{\pi/4}\int_{0}^{\sec \theta} \sqrt{(1 + 4r^2)} r \,dr \,d\theta +  \int_{\pi/4} ^{\pi/2}\int_{0}^{\operatorname{cosec}\theta} \sqrt{(1 + 4r^2)} r \,dr \,d\theta$$

After this integral becomes too complex to solve further . For eg : the first integral gives :

$$\int_{0}^{\pi/4}\frac1{12}{((1 + 4\sec^2\theta)^{3/2} - 1)}\, d\theta$$

After this I am stuck how to proceed further, Please help.

Thank You.

Body:
It boils down to integrate

$$I= \int_0^{\pi/4}(1+4\sec^2\theta)^{3/2}\,d\theta$$
Let $\sinh t= {\frac{2}{\sqrt{5}}}\tan\theta$ to proceed

\begin{align}
I &=\int_0^{\sinh^{-1}\frac{2}{\sqrt{5}}}
\frac{50\cosh^4 t}{4+{5}\sinh^2t }dt
=\int_0^{\sinh^{-1}\frac{2}{\sqrt{5}}}
\frac{25(1+\cosh 2t)^2}{3+5\cosh2t}dt\\
 &=\int_0^{\sinh^{-1}\frac{2}{\sqrt{5}}}
\left(5 \cosh2t + 7 + \frac{4}{3+5\cosh2t}\right)dt\\
 &= 6 + \frac72 \ln{5}+\cot^{-1}3
\end{align}


答:
Continuing from your work, we have $y-z\equiv z-x\equiv x-y\equiv\frac23 r\pi\;\,(\!\!\!\!\mod2\pi)$, with $r\in\{1,2\}$ ($r$ is constant, since the three differences sum to zero, modulo $2\pi$). So $$\cos2x+\cos2y+\cos2z=\cos2x+\cos(2x-\tfrac43r\pi)+\cos(2x+\tfrac43r\pi)\qquad\qquad$$
$$=\cos2x+2\cos2x\cos\tfrac43r\pi$$
$$=0.\qquad\qquad\qquad\qquad\quad$$

Title: Find the sum of all possible values of $\cos 2x + \cos 2y + \cos 2z.$

Body:
Let $x$, $y$, and $z$ be real numbers such that $ \cos x + \cos y + \cos z = \sin x + \sin y + \sin z = 0 $. Find the sum of all possible values of $ \cos 2x + \cos 2y + \cos 2z $.

Here is what I have done so far

$$ \cos x + \cos y = -\cos z $$
$$ (\cos x + \cos y)^{2} = (-\cos z)^{2} $$
$$ \cos^{2} x + \cos^{2} y + 2\cos x \cos y = \cos^{2}z $$

likewise,
$$ \sin x + \sin y = -\sin z $$
$$ (\sin x + \sin y)^{2} = (-\sin z)^{2} $$
$$ \sin^{2} x + \sin^{2} y + 2 \sin x \sin y = \sin^{2}z $$

from this, you get

$$ \cos^{2} x + \cos^{2} y + 2\cos x \cos y + \sin^{2} x + \sin^{2} y + 2\sin x \sin y = \sin^{2}z + cos^{2}z $$
$$ 1 + 1 + 2(\cos x \cos y + \sin x \sin y) = 1 $$
$$ \cos x \cos y + \sin x \sin y = -\frac{1}{2} $$
$$ \cos(x-y) = -\frac{1}{2} $$
$$ x-y = \frac{2 \pi}{3}, \frac{4 \pi}{3} $$

likewise,
$$ \cos (x-z) = -\frac{1}{2} $$
$$ x-z = \frac{2 \pi}{3}, \frac{4 \pi}{3} $$

where do I go from here?


问:
Solve $f''(x) - f'(2x) - f(x) = 0$
I was trying to solve a problem and fall into that differential equation. I'm curious because I couldn't find a way to solve it.
What can work to solve it?</p>

答:
Here is my attempt:

The first thing we can see is that this equation describes the kernel of the linear operator $L : C^{\infty} ( \Bbb R ) \to C^{\infty} ( \Bbb R )$
$$Lf(x)=f''(x) - f'(2x)-f(x)$$
In other words the space of solutions is a vector space. In particular $f(x)=0$ is a trivial solution.

Now, suppose $f(x)$ is analytic at $x=0$, and write
$$f(x)= \sum_{n=0}^{\infty} a_n x^n$$
Then the equation becomes
$$\sum_{n=0}^{\infty} ((n+2)(n+1)a_{n+2} - (n+1)a_{n+1}2^n - a_n )x^n=0$$
Hence for all $n$ we need
$$(n+2)(n+1)a_{n+2} - (n+1)a_{n+1}2^n - a_n =0$$
which can be written as a recursive sequence for all $n \ge 0$
$$a_{n+2}= \frac{(n+1)a_{n+1}2^n + a_n}{(n+2)(n+1)}$$
Clearly here $a_0$ and $a_1$ are free (they can be anything). Hence the space of solutions has dimension 2.

HOWEVER there is a problem:

Assuming $a_0, a_1 \ge 0$, one can compute the radius of convergence; indeed:
$$a_{n+2} = \frac{(n+1)a_{n+1}2^n + a_n}{(n+2)(n+1)} \ge \frac{a_{n+1}2^n }{(n+2)}$$
And so
$$\frac{a_{n+2}}{a_{n+1}} \ge \frac{2^n }{(n+2)} \to \infty$$
this means that the radius of convergence is zero: we have no function at all.

Here my attempt concludes. My suspect is that there is no solution other than the trivial zero function.

Title: Solve $f''(x) - f'(2x) - f(x) = 0$


Body:
$$\sec^2x-\tan^2x = 1$$
$$\csc^2x-\cot^2x = 1$$
$$\sin x \csc x = 1$$
$$\tan x\cot x = 1$$
$$\sec x\cos x = 1$$
$$ \frac{\sin x + \cos x}{\cos ^3x} - \tan^3x - \tan^2x-\tan x = 1$$


问:
Proof of tensor identity
I am asked to prove
$$\nabla \cdot (T \cdot v) = T : \nabla v + v\cdot (\nabla \cdot T)$$
Where $T$ is a order-2 tensor and $v$ is vector, in an orthogonal basis. Let $\delta _{ij}$ denote the Kronecker delta.

My proof:
$$T = T_{ij} e_i e_j, v = v_k e_k$$
where $e_i$ are unit vectors in an orthogonal basis.
So,
$$T \cdot v = T_{ij}v_k e_i e_j \cdot e_k = T_{ij}v_k e_i \delta _{jk} = T_{im}v_me_i \\
\implies \nabla \cdot (T \cdot v) = \frac{\partial (T_{im}v_m e_i)}{\partial x_n}\cdot e_n = \frac{\partial (T_{im}v_m)}{\partial x_n} \delta_{in}=\frac{\partial (T_{pm}v_m)}{\partial x_p} = \frac{\partial (T_{pq}v_q)}{\partial x_p} $$

Moving on to the RHS:
$$T:\nabla v = T_{ij}e_i e_j : \frac{\partial (v_m e_m)}{\partial x_n} e_n = T_{ij}\frac{\partial v_m}{x_n}e_ie_j:e_me_n = T_{ij}V_{mn}\delta_{jm}\delta_{in}=T_{pq}V_{qp}=T_{pq}\frac{\partial v_q}{\partial x_p}$$
The second term of the RHS:
$$v\cdot(\nabla \cdot T)=(v_ie_i)\cdot\left(\frac{\partial}{\partial x_j}(T_{kl} e_k e_l)\cdot e_j\right)=(v_ie_i)\cdot\left(\frac{\partial T_{kl}}{\partial x_j} e_k\delta _{lj}\right) = (v_ie_i)\cdot\left(\frac{\partial T_{km}}{\partial x_m} e_k\right)=v_i\frac{\partial T_{km}}{\partial x_m} \delta _{ik}=v_q\frac{\partial T_{qp}}{\partial x_p}$$
Where I replaced $m$ with $p$, and contracted $i$,$k$ with the index $p$ in the last statement.

As you can see, $$v_q\frac{\partial T_{qp}}{\partial x_{p}} + T_{pq}\frac{\partial v_{q}}{\partial x_{p}} \neq \frac{\partial T_{pq}v_q}{x_p}$$

What am I doing wrong here? Any advice you have would be appreciated!

答:
Okay, I figured it out...

I was making a mistake when taking the divergence of the tensor.

$$\nabla \cdot T = e_j \cdot \frac{\partial (T_{kl} e_k e_l)}{\partial x_j} \neq \frac{\partial (T_{kl} e_k e_l)}{\partial x_j} \cdot e_j$$

This resolves the issue

Title: Proof of tensor identity


问:
Logical expression problem
How can I solve this?

$ \varphi = (\neg x_1 \lor \neg x_2 \lor \neg x_3)\land(\neg x_2 \lor \neg x_3 \lor \neg x_4)\land(\neg x_3 \lor \neg x_4 \lor \neg x_1)\land(\neg x_4 \lor \neg x_1 \lor \neg x_2)\land(x_1 \lor x_2)\land(x_2 \lor x_3)\land(x_3 \lor x_4)\land(x_4 \lor x_1).$

Also write $\varphi$ as $\varphi(x_1,x_2,x_3,x_4)$.
Give an interpretation of $\varphi$ in English. What does $\varphi$ compute? Explain and justify your answer.

答:
If we put the $4$ different variables $x_i$ variables in a circle, then the last four terms express that for any two 'neighboring' variables, at least one of them is 'true' or 'on' (or a $1$, if that's what you prefer), but the first four terms specify that out of ever three successive ones, at least one must be false.

So note: if you have two neighboring ones being both 'true' ('on'), then on either side of them there must be a 'false' ('off') one (e.g. if $x_1$ and $x_2$ are both true, then $x_3$ and $x_4$ must both be false). However, those other two are neighbors in the circle themselves, and so we have a problem, since out of any two neighboring ones there must at least be one true one.

So, this tells us that you cannot have two true ones being next to each other ... and hence for any two neighboring ones, one must also be true and one must be false. Hence, the values must alternate going around the circle: either $x_1=True$, $x_2=False$, $x_3=True$, and $x_4=False$, or $x_1=False$, $x_2=True$, $x_3=False$, and $x_4=True$.  I assume that reflects the two $1$'s that you found in your truth-table as well ... so that's probably not news to you.

However, with the above analysis, I think you do reach a more conceptual 'picture' of what the function is computing: it is effectively true if and only if the values are alternating (again, assuming an order of $x_1, x_2, x_3, x_4$ and then possibly circling back to $x_1$) ... and that makes for a nice and concise English description/interpretation... which is what the question was asking for.

Title: Logical expression problem


问:
A problem on matrices : Powers of a matrix

If $ A=
        \begin{bmatrix}
        i & 0 \\
        0 & i \\
        \end{bmatrix}
, n \in \mathbb N$, then $A^{4n}$ equals?

I guessed the answer as $ A^{4n}=
        \begin{bmatrix}
        i^{4n} & 0 \\
        0 & i^{4n} \\
        \end{bmatrix}
=\begin{bmatrix}
        1 & 0 \\
        0 & 1 \\
        \end{bmatrix}$ which actually was the answer.

Can you provide me with the correct method of getting the answer?

 Thank you.

答:
you can see $A^2=\begin {pmatrix}i^2&0\\0&i^2 \end{pmatrix}$

you assume that it satisfy fo n-1 so :

$A^{n-1}=\begin {pmatrix}i^{n-1}&0\\0&i^{n-1} \end{pmatrix}$

then :

$A^n=A*A^{n-1}=\begin {pmatrix}i^n&0\\0&i^n \end{pmatrix}$


问:
How to access variable value without the header?
If anyone can tell me how to access the individual numbers from this list without the label, that would be wonderful. Thank you. This is Maple 2020 software. I have included the screenshot here. For example, in this case, I would like to access -2.913439763

答:
You could use</p>
```<code>subs(values[1], x2);
</code>```

Body:
You could use

subs(values[1], x2);


问:
How to use an anonymous function to solve Stirling series n! using MATLAB
I want to be able to use Stirling Series to approximate n! inputing a vector. So far, I can do this entry wise. And I'm not sure why I cannot input a vector.

```<code>F = @(n) sqrt(2.*pi.*n).*(n/exp(1)).^n .* (12.*n+1)/(12.*n)

>> F(1), F(2), F(3), F(4)
ans =
   0.998981759637105
ans =
   1.998962866134357
ans =
   5.998326524438801
ans =
  23.995887114828559
</code>```

When I try

```<code>n = [1:6]
F(n)

n =
     1     2     3     4     5     6
ans =
     3.226859511810607e+02
</code>```

I would like to get the solution of

```<code>F(n)
ans =
0.9990 1.9990 5.9983 23.9959 119.9862 719.9404
</code>```

Can someone offer some advice?

答:
You are not using right array division. Using "/" instead of "./" yields to a single scalar.

Change your function handle to

F = @(n) sqrt(2.pi.n).(n./exp(1)).^n . (12.*n+1)./(12.*n)

For the input

n=[1:6]

we then get

ans = 0.9990     1.9990     5.9983    23.9959   119.9862   719.9404

as desired.


问:
Relations between Cayley-Hamilton theorem and the classification theorem for finitely generated modules over a PID
There is a sentence in the book "Algebra Chapter 0" that the Cayley-Hamilton theorem would be evident if we connect it with the classification theorem for finitely generated modules over a PID. I don't think I fully understand this sentence. Given an integral domain $R$ and its free module $F$, let $\alpha\in End_{R}(F)$, I guess $\{f\in R[t]|f(\alpha)=0\}$ would somehow be related to annihilator ideal of a finite generated module, and $P_{\alpha}(t)$ is related to the characteristic ideal, and since the characteristic ideal is contained in the annihilator, we have $P_{\alpha}(t)\in \{f\in R[t]|f(\alpha)=0\}$. Unfortunately, I don't have a good idea about that. Can someone help me with that?

Thank you!

答: I find that actually section VI.7 of this book "Algebra Chapter 0" has somehow answered this question.</p>


答:
Well you're in luck because what you did actually work with diagonal matrices. Let's look at an example :

Let $A = \begin{bmatrix}
        a & 0 \\
        0 & b \\
        \end{bmatrix}$ be a diagonal matrix then $ A^{2}=
        \begin{bmatrix}
        a & 0 \\
        0 & b \\
        \end{bmatrix}
\begin{bmatrix}
        a & 0 \\
        0 & b \\
        \end{bmatrix} = \begin{bmatrix}
        a^2& 0 \\
        0 & b^2 \\
        \end{bmatrix}$ and more generally :

$$A^n = \begin{bmatrix}
        a^n & 0 \\
        0 & b^n \\
        \end{bmatrix}$$

So you guess wasn't wrong at all, it's just that when you have a diagonal matrix, raising it to a power $n$ equals to raising each of it's coefficients to the power $n$.

But keep in mind that this only works for diagonal matrices. If you have a matrix $B$ that is not diagonal, the best way to find $B^n$ is to diagonalize it (if possible), raise it's coefficients to the power $n$ and go back to your first basis.

Title: A problem on matrices : Powers of a matrix

Body:
If $ A=
        \begin{bmatrix}
        i & 0 \\
        0 & i \\
        \end{bmatrix}
, n \in \mathbb N$, then $A^{4n}$ equals?

I guessed the answer as $ A^{4n}=
        \begin{bmatrix}
        i^{4n} & 0 \\
        0 & i^{4n} \\
        \end{bmatrix}
=\begin{bmatrix}
        1 & 0 \\
        0 & 1 \\
        \end{bmatrix}$ which actually was the answer.

Can you provide me with the correct method of getting the answer?

 Thank you.


问:
Why does a concentration inequality imply equivalence of moments and exponential integrability of the square?
I have the following basic question about a concentration inequality:

Let $X$ be a random variable, denote by $m$ a median and assume that for every $t>0$ one has

$$
P \left( \left| X - m \right| > t \right)
\leq
C e^{-t^2/\lambda}
$$

for some positive constants $C$ and $\lambda$.

Why does this imply that $\operatorname{E} \left( e^{a X^2} \right) < \infty$ for some $a>0$ and that all moments are equivalent, i.e. for all $0<p,q<\infty$ one has $\left\lVert X \right\rVert_p \leq K_{p,q} \left\lVert X \right\rVert_q$, where $\left\lVert X \right\rVert_p^p = \operatorname{E} \left( \left| X \right|^p \right)$ for positive constants $K_{p,q}$?

Background:
I have condensed the above statement from Theorem 4.7 on page 100 of the book Probability on Banach spaces by Ledoux and Talagrand. However, in case there are necessary assumptions I have stripped, here is the full statement for reference:

Let $X$ be a Rademacher series in a Banach space $B$ such that for some countable subset $D$ in the unit ball of $B^{\ast}$ it holds that $\left\lVert x \right\rVert = \sup_{f \in D} \left| f(x) \right|$. Denote by $M$ a median of $\left\lVert X \right\rVert$. Then

$$
P \left( \left| \left\lVert X \right\rVert - M \right| > t \right)
\leq
4 \exp \left( -t^2/8\sigma^2 \right),
$$

where $\sigma = \sup_{f \in D} \sqrt{\operatorname{E} \left( f^2(X) \right)}$.
In particular, there exists $\alpha>0$ such that $\operatorname{E} \left( \exp \left( \alpha \left\lVert X \right\rVert^{2} \right) \right) < \infty$ and all moments of $X$ are equivalent: that is, for any $0<p,q<\infty$, there is a constant $K_{p,q}$ depending on $p,q$ only such that $\left\lVert X \right\rVert_p \leq K_{p,q} \left\lVert X \right\rVert_q$.

答:
The exponential tails follow by applying Markov inequality and the finiteness of $\mathbb E e^{aX^2}$. Such random variables are called subgaussian random variables. For more details, see [1, Proposition 2.5.2].

The equivalence of moments is a rather subtle issue. The subgaussian assumption would imply that $\|X\|_p = O(\sqrt{p})$ for all $p\geq 1$. However, the equivalence of moments does not hold in general without additional assumptions, and that is where it is crucial that $X$ is a Rademacher series. This equivalence of moments for Rademacher Series is also related to the famous Khinchine-Kahane inequality. See, for example, Wikipidea, A short proof by  Rafał Latała; Krzysztof Oleszkiewicz, a recent talk by Tomasz Tkocz.

[1] High-Dimensional Probability, Roman Vershynin. https://www.math.uci.edu/~rvershyn/papers/HDP-book/HDP-book.html


问:
Determining Independence in Bayesian Network

Consider the Bayesian Network Structure Below, decide whether the statements are true or false.

b) $G \perp \!\!\! \perp A$ (G is independent of A)

c) $E \perp \!\!\! \perp H | \{D,G\}$ (E and H are conditionally independent given D and G).

d) $E\perp \!\!\! \perp H | \{C,D,F\}$ (E and H are conditionally independent given C,D and F).

b) For these problems I applied the d-separation algorithm as explained here http://web.mit.edu/jmn/www/6.034/d-separation.pdf
Using this approach I found that G and A are indeed independent.

c) True (using d-separation).

d) Also true (d-separation).

Could anyone please verify/correct my answers? I feel like at least one of the problems $b,c$ or $d$ should be false.

答: d) is false. H and E are connected through G in the d-seperation method.</p>


问:
Curl in spherical coordinates without Del operator
Trying to get practice with differential forms and related topics I had tried to calculate curl in spherical coordinates without mystic "del" operator.

Let $rot\: F = (\star d (F^\sharp))^\flat$;

$x = r \cos \phi \sin \theta$

$y = r \sin \phi \sin \theta$

$z = r \cos \theta $

Non normalized basis of tangent space $T$:

$\bar e_r = (\frac{dx}{dr}, \frac{dy}{dr}, \frac{dz}{dr}) = (\cos \phi \sin \theta, \sin \phi \sin \theta, \cos \theta)$

$\bar e_\phi = (\frac{dx}{d\phi},\frac{dx}{d\phi},\frac{dz}{d\phi}) = (-r\sin \phi \sin \theta, r\cos \phi \sin \theta, 0)$

$\bar e_\theta = (\frac{dx}{d\theta},\frac{dx}{d\theta},\frac{dz}{d\theta}) = (r \cos \phi \cos \theta, r \sin \phi \cos \theta, -r\sin \theta)$

Norms:

$\|\bar e_r\| = 1$

$\|\bar e_\phi\| = r \sin \theta$

$\|\bar e_\theta\| = r$

Orthonormal basis of tangent space $T$:

$ e_r = \frac{\bar e_r}{\| \bar e_r\|} = (\cos \phi \sin \theta, \sin \phi \sin \theta, \cos \theta)$

$ e_\phi = \frac{\bar e_r}{\| \bar e_r\|} = (-r\sin \phi \sin \theta, r\cos \phi \sin \theta, 0)$

$ e_\theta = \frac{\bar e_r}{\| \bar e_r\|} = (r \cos \phi \cos \theta, r \sin \phi \cos \theta, -r\sin \theta)$

Let $w_r, w_\phi , w_\theta$ be corresponding basis in $T^*$. Then

$dr = w_r$

$d\phi = \frac{1}{r\sin \theta} w_\phi$

$d\theta = \frac{1}{r} w_\theta$

So, going back to the $rot$:

$F = F^1 e_r + F^2 e_\phi + F^3 e_\theta$

$F^\sharp = F^1 w_r + F^2 w_\phi + F^3 w_\theta$

$d (F^\sharp) = $
$$ = F^1_\theta d\theta \wedge w_r + F^1_\phi d\phi \wedge w_r $$
$$ + F^2_\theta d\theta \wedge w_\phi + F^2_r dr \wedge w_\phi$$
$$ + F^3_\phi d\phi \wedge w_\theta + F^3_r dr \wedge w_\theta$$

$$ = F^1_\theta \frac{1}{r} w_\theta \wedge w_r + F^1_\phi \frac{1}{r\sin \theta} w_\phi \wedge w_r $$
$$ + F^2_\theta \frac{1}{r} w_\theta \wedge w_\phi + F^2_r w_r \wedge w_\phi$$
$$ + F^3_\phi \frac{1}{r\sin \theta} w_\phi \wedge w_\theta + F^3_r w_r \wedge w_\theta$$

$$ = (F^2_r - F^1_\phi \frac{1}{r\sin \theta}) w_r \wedge w_\phi$$
$$ + (F^3_r - F^1_\theta \frac{1}{r}) w_r \wedge w_\theta$$
$$ + (F^3_\phi \frac{1}{r\sin \theta} - F^2_\theta \frac{1}{r}) w_\phi \wedge w_\theta$$

$\star d (F^\sharp) = $
$$ = (F^2_r - F^1_\phi \frac{1}{r\sin \theta}) w_\theta$$
$$ - (F^3_r - F^1_\theta \frac{1}{r}) w_\phi$$
$$ + (F^3_\phi \frac{1}{r\sin \theta} - F^2_\theta \frac{1}{r}) w_r$$

$rot F  = (\star d (F^\sharp))^\flat = $
$$= (F^3_\phi \frac{1}{r\sin \theta} - F^2_\theta \frac{1}{r}) e_r$$
$$- (F^3_r - F^1_\theta \frac{1}{r}) e_\phi$$
$$+ (F^2_r - F^1_\phi \frac{1}{r\sin \theta}) e_\theta$$

But in wikipedia the answer is completely different. Please, help me to find which operations i had understood wrongly.

答:
First of all, your notation for $\flat$ and $\sharp$ is reversed: if $F$ is a vector field, $F^{\flat}$ is a covector field (1-form); and if $\omega$ is a 1-form, then $\omega^{\sharp}$ is a vector field, so the correct way of writing the curl definition is $\text{curl}(F)= (\star d(F^{\flat}))^{\sharp}$.

Your mistake is in calculating the exterior derivatives. Because you expressed your 1-forms in terms of $w_r, w_{\theta},w_{\phi}$, you need to apply the product rule when calculating $d$. You can't just take the $d$ of the coefficient and wedge it. For example, if $f$ is a smooth function and you have a 1-form $\alpha=fw_{\theta} = f\cdot rd\theta$, then the way you wrote the calculation is $d\alpha = df\wedge w_{\theta} = \frac{\partial f}{\partial r}\,dr \wedge w_{\theta} + \frac{\partial f}{\partial \phi}\,d\phi \wedge w_{\theta}$; but this is wrong. The correct way to do it is
\begin{align}
d\alpha &= df\wedge w_{\theta} +f (dw_{\theta}) = df\wedge w_{\theta} + f\,dr\wedge d\theta
\end{align}
Or, another way of saying it is that $\alpha = (rf)\,d\theta$, so
\begin{align}
d\alpha &= \frac{\partial (rf)}{\partial r}\,dr\wedge d\theta + \frac{\partial (rf)}{\partial \phi}\,d\phi\wedge d\theta
\end{align}
You made this mistake for each term.

Calculating the curl is annoying for the following reason: exterior derivatives are easy to calculate when you express everything in terms of the coordinate 1-forms (i.e in terms of $dr,d\theta,d\phi$). On the other hand, when you do the musical isomorphisms, various factors will pop up, and then again when you want to use the hodge star, that is simplest when using an orthonormal coframe (i.e the $w_r,w_{\theta},w_{\phi}$ you mention). Because of this, there's a lot of factors to keep track of.

The following works for any orthogonal system of coordinates $(x^1,x^2,x^3)$ (i.e such that the metric tensor has components $g_{ij}:=g\left(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right)=0$ if $i\neq j$). Start by writing $F=F^1e_1 + F^2e_2 + F^3e_3$, then you'll find
\begin{align}
F^{\flat}&= F^1 \omega^1 + F^2\omega^2 + F^3\omega^3\\
&= (F^1\sqrt{g_{11}})\,dx^1+
(F^2\sqrt{g_{22}})\,dx^2 +(F^3\sqrt{g_{33}})\,dx^3
\end{align}
Or, if you define $f_i:= F^i\cdot \sqrt{g_{ii}}=F^ih_i$ (no summation convention) then $F^{\flat}=\sum_{i=1}^3f_i\,dx^i$. Now, it should be easy to calculate $d(F^{\flat})$.

You should of course get terms involving $dx^1\wedge dx^2$ etc. Now, write
\begin{align}
dx^1\wedge dx^2 = \left(\frac{\omega^1}{h_1}\right)\wedge \left(\frac{\omega^2}{h_2}\right) = \frac{1}{h_1h_2}\omega^1\wedge\omega^2
\end{align}
And now, since $\{\omega^1,\omega^2,\omega^3\}$ are an orthonormal coframe, the Hodge star is easy to calculate (be careful with signs of course): $\star(\omega^1\wedge \omega^2)=\omega^3$ and so on. Now, you have computed $\star d(F^{\flat})$, and your answer should be in terms of $\omega^1,\omega^2,\omega^3$, so taking the $\sharp$ of this just changes $\omega^i$ to $e_i$. Then you should be done (recall that $f_i= F^ih_i$; no summation). You should notice that the formula can be expressed as a nice determinant as in @K.defaoite's answer.

Then, of course, you can just take the special case of spherical coordinates where you already know the various scale factors $h_r=1$, $h_{\theta}=r, h_{\phi}=r\sin\theta$.


问:
What are all the integral solutions of $n!=m(m^2-1)$?
Observe that:

$3!=2(2^2-1)$

$4!=3(3^2-1)$

$5!=5(5^2-1)$

$6!=9(9^2-1)$

Question: What are all the integral solutions of $n!=m(m^2-1)$?

I guess it is just $(n,m) = (3,2),(4,3),(5,5),(6,9)$, but how to prove that there is no other one?

I checked that there is no other one for $n<20$.

Let reformulate the problem using Cardano's formula:    consider the cubic equation $$x^3+px+q=0,$$
its discriminant is $\Delta = -(4p^3+27p^2)$. Our case corresponds to $(p,q) = (-1,-n!)$, so $\Delta = 4-27(n!)^2 <0$. Thus the cubic has one real root and two non-real complex conjugate roots, and by Cardano's formula, the real root is $$ \left(-\frac q2+\sqrt{\frac{q^2}4+\frac{p^3}{27}} \right)^{1/3} +\left(-\frac q2-\sqrt{\frac{q^2}4+\frac{p^3}{27}}\right)^{1/3} $$
So the problem "reduces" to ask whether there exists an integer $n>6$ such that

$$ \left(\frac{n!}{2} +\sqrt{\frac{(n!)^2}{4} - \frac{1}{27}} \right)^{1/3} +\left(\frac{n!}{2} - \sqrt{\frac{(n!)^2}{4} - \frac{1}{27}} \right)^{1/3}  $$
is also an integer. It is not clear that it helps...

答:
Interestingly enough, there might be something even deeper going on. Let us define a family of elliptic curves given by the Weierstrass equations

$$E_n: y^2=x^3-x-n!$$

Your question is asking whether $E_n$ has any integer solutions $(x,0)$, which by the Nagell-Lutz theorem is equivalent to asking if $E_n$ has any points of 2-torsion. A natural generalization of this conjecture is to ask whether or not the torsion group $E_n(\mathbb{Q})$ is trivial for any $n\neq3,4,5,6$.

Using Sage one can compute the torsion groups of elliptic curves, and computationally it checks out that $E_n(\mathbb{Q})$ is trivial for any $n\leq10,000$, except for $n=3,4,5,6$ where we have that the only torsion points were the points of 2-torsion.

It turns out, in fact, that this generalized problem is $\mathbf{equivalent}$ to your conjecture. This is true because $x^3-x$ is always a multiple of $3$ (check by hand of by Fermat's Little Theorem) so any non $2$-torsion solution $y^2=x^3-x-n!$ will also be a multiple of $3$. By Nagell Lutz this would imply that the discriminant $\Delta_{E_n/\mathbb{Q}}=-4(27n!^2-4)$ would be a multiple of $3$ which as we can clearly see it is not.


问:
Sum of independent Binomial and Geometric distributions
This is a question from a mock test in my Intro to Probability coure:

Let $X$~$Bin(26,\frac{1}{23})$ and $Y$~$Geo(\frac{1}{2})$ be two independent random variables. Let $Z=X+Y$. Calculate $\Bbb{P}[Z=46]$.

My initial solution:

$\Bbb{P}[Z=46] = \Bbb{P}[X+Y=46] = \sum_{k=0}^{26}{\Bbb{P}[X+Y=46, X=k]} = \sum_{k=0}^{26}{\Bbb{P}[Y=46-k]\cdot\Bbb{P}[X=k]} = \sum_{k=0}^{26}{(\frac{1}{2})^{46-k}\binom{26}{k}(\frac{1}{23})^{k}(\frac{22}{23})^{26-k}} $

I got stuck calculating this sum. If anyone could help me, it would be much appreciated.

Thank you

答:
You are very close to get the desired answer. Note the following: $$\sum_{0 \leq k \leq 26} (\frac{1}{2})^{46 - k} \binom{26}{k} (\frac{1}{23})^k (\frac{22}{23})^{26 - k} = (\frac{1}{2})^{20} \sum_{0 \leq k \leq 26} (\frac{1}{2})^{26 - k} \binom{26}{k} (\frac{1}{23})^k (\frac{22}{23})^{26 - k} = \frac{1}{2^{20}} \sum_{0 \leq k \leq 26} \binom{26}{k} (\frac{1}{23})^k (\frac{11}{23})^{26 - k} = \frac{1}{2^{20}} (\frac{1}{23} + \frac{11}{23})^{26} = \frac{1}{2^{20}} \frac{12^{26}}{23^{26}},$$ which can be further simplified to give $\dfrac{2^{32} 3^{26}}{23^{26}}.$

Digressing a little from your very particular example, the question of determining the distribution of the sum of two continuous random variables is a very interesting one and it is one of the places where the convolution of two functions proves itself once again to be quite an useful tool. For references, you might want to start by reading this wikipedia page.


问:
Why is the intersections between $f(x)=x^{2}+x^{-2}-3$ and the x-axis resembles $\varphi$?
The story behind this is quite silly, I was messing around with functions in geogebra, inputting several functions to see their properties, until I was amazed by something i don't know about.

Apparently, the intersections (unfortunately I don't know how the're named in english) between $f(x)=x^{2}+x^{-2}-3$ and the x-axis resembles $\varphi$, in the screenshot i've attached you can see it reveals the first 8 digits of $\varphi$ correctly. Can someone explain to me why this happens? I don't believe it's just a coincidence and I'm definitely missing something.

答:
The intersection points are the solutions to the equation
$$x^2+x^{-2}-3=0.$$
The solutions are precisely the (nonzero) solutions to the equation
$$x^2(x^2+x^{-2}-3)=0,$$
or equivalently
$$x^4-3x^2+1=0.$$
It turns out that this polynomial factors as the product of two quadratic polynomials:
$$x^4-3x^2+1=(x^2-x-1)(x^2+x-1).$$
And of course the golden ratio $\varphi$ and its conjugate $\varphi^{-1}=\varphi-1$ are precisely the roots of
$$x^2+x-1=0.$$
And similarly their negatives $-\varphi$ and $-\varphi^{-1}$ are precisely the roots of
$$(-x)^2+(-x)-1=0,$$
or equivalently
$$x^2-x-1=0.$$
So the four solutions to the original equation are precisely $\pm\varphi$ and $\pm\varphi^{-1}$.


问:
Number of lattice paths passing through a point
If I have a $7 \times 5$ grid, and want to go from (0,0) to (7,5) we of course have $\binom{13}{7}$ possible lattice paths. lets say we want to find all lattice paths that pass through (1,4).

I thought this would simply by $\binom{5}{1}+\binom{7}{6}$ but this does not appear to be the correct solution

Furthermore if i wanted to find all paths that go through points (a,5-a) where a is an int from 1 to 5 (this should give me all possible paths) why is the following equation incorrect?

$$\sum_{i=0}^{5} {\binom{5}{i}+\binom{7}{7-i}}$$

答:
The choices of paths from $(0,0)$ to $(1,4)$ and from $(1,4)$ to $(7,5)$ are independent. So multiplication principle applies.

For a given path from origin to $(1,4)$ you can take every possible path from $(1,4)$ to $(7,5)$. The options multiply. So correct are

$$\binom{5}{1} \cdot \binom{7}{6}$$
$$\sum_{i=0}^{5} {\binom{5}{i} \cdot \binom{7}{7-i}}$$


问:
Need help with $\int_0^\pi\arctan^2\left(\frac{\sin x}{2+\cos x}\right)dx$
Please help me to evaluate this integral:
$$\int_0^\pi\arctan^2\left(\frac{\sin x}{2+\cos x}\right)dx$$
Using substitution $x=2\arctan t$ it can be transformed to:
$$\int_0^\infty\frac{2}{1+t^2}\arctan^2\left(\frac{2t}{3+t^2}\right)dt$$
Then I tried integration by parts, but without any success...</p>

答:
Generalizing Jack's answer. Take $a\in\Bbb R\setminus\{0\}$, and write
$$J(a)=\int_0^\pi\arctan^2\left(\frac{\sin x}{a+\cos x}\right)dx.$$
From symmetry, we can write
$$J(a)=\frac12\int_{-\pi}^\pi\arctan^2\left(\frac{\sin x}{a+\cos x}\right)dx$$
Then
$$\arctan\left(\frac{\sin x}{a+\cos x}\right)=\Im\log(a+e^{ix})=\sum_{n\ge1}\frac{(-1)^{n+1}}{na^n}\sin(nx).$$
Then from Parseval's theorem,
$$\frac1\pi\int_{-\pi}^\pi\arctan^2\left(\frac{\sin x}{a+\cos x}\right)dx=\sum_{n\ge1}\left(\frac{(-1)^{n+1}}{na^n}\right)^2,$$
so that
$$J(a)=\frac\pi2\mathrm{Li}_2\left(\tfrac1{a^2}\right).$$
Unfortunately, your integral, given by $J(2)=\frac\pi2\mathrm{Li}_2(1/4)$, does not appear to have a closed form.


答: Now you've found your mistake, let me suggest a slight change in proof technique that saves these difficulties. Rather than working with explicit $e_i$, write$$\nabla\cdot(T\cdot v)=\partial_i(T_{ij}v_j)=T_{ij}\partial_iv_j+v_j\partial_iT_{ij}=T:\nabla v+v\cdot(\nabla\cdot T).$$

Title: Proof of tensor identity

Body:
I am asked to prove
$$\nabla \cdot (T \cdot v) = T : \nabla v + v\cdot (\nabla \cdot T)$$
Where $T$ is a order-2 tensor and $v$ is vector, in an orthogonal basis. Let $\delta _{ij}$ denote the Kronecker delta.

My proof:
$$T = T_{ij} e_i e_j, v = v_k e_k$$
where $e_i$ are unit vectors in an orthogonal basis.
So,
$$T \cdot v = T_{ij}v_k e_i e_j \cdot e_k = T_{ij}v_k e_i \delta _{jk} = T_{im}v_me_i \\
\implies \nabla \cdot (T \cdot v) = \frac{\partial (T_{im}v_m e_i)}{\partial x_n}\cdot e_n = \frac{\partial (T_{im}v_m)}{\partial x_n} \delta_{in}=\frac{\partial (T_{pm}v_m)}{\partial x_p} = \frac{\partial (T_{pq}v_q)}{\partial x_p} $$

Moving on to the RHS:
$$T:\nabla v = T_{ij}e_i e_j : \frac{\partial (v_m e_m)}{\partial x_n} e_n = T_{ij}\frac{\partial v_m}{x_n}e_ie_j:e_me_n = T_{ij}V_{mn}\delta_{jm}\delta_{in}=T_{pq}V_{qp}=T_{pq}\frac{\partial v_q}{\partial x_p}$$
The second term of the RHS:
$$v\cdot(\nabla \cdot T)=(v_ie_i)\cdot\left(\frac{\partial}{\partial x_j}(T_{kl} e_k e_l)\cdot e_j\right)=(v_ie_i)\cdot\left(\frac{\partial T_{kl}}{\partial x_j} e_k\delta _{lj}\right) = (v_ie_i)\cdot\left(\frac{\partial T_{km}}{\partial x_m} e_k\right)=v_i\frac{\partial T_{km}}{\partial x_m} \delta _{ik}=v_q\frac{\partial T_{qp}}{\partial x_p}$$
Where I replaced $m$ with $p$, and contracted $i$,$k$ with the index $p$ in the last statement.

As you can see, $$v_q\frac{\partial T_{qp}}{\partial x_{p}} + T_{pq}\frac{\partial v_{q}}{\partial x_{p}} \neq \frac{\partial T_{pq}v_q}{x_p}$$

What am I doing wrong here? Any advice you have would be appreciated!


问:
The complement of a neighborhood of a divisor.
Given a projective variety $X$ and a hyperplane section $H$ (intersection of some hyperplane in the ambient projective space with $X$). Is it true that complement of any neighborhood of $H$ is a finite number of closed points? My reasoning is that since $H$ is ample by Nakai–Moishezon it has non-zero intersection number with closed sub-varieties in $X$ that are not points. Since the complement has zero intersection with $H$, it implies it has to be points.

答: Yes, this is correct.  More simply, if $X$ is a closed subvariety of $\mathbb{P}^n$, then so is any closed subvariety of $X$, so you just need to use the fact that a hyperplane in $\mathbb{P}^n$ has nontrivial intersection with any closed subvariety of $\mathbb{P}^n$ of positive dimension.


问:
How to check continuity of a function?
I know that a function is continuous at a point if the limit from left and right side exists and are equal and for a function to be continuous, the function should be continuous at all points. My question is that if I want to check continuity of a function, I cannot practically check continuity at each and every point in $\mathbb{R}$. So, how to do it?

答:
One way is to take an arbitrary point $x_0$ with no prior assumptions and show that for every $\varepsilon>0$ there exists $\delta>0$ (which may depend on $x_0$) s.t for every $x\in (x_0-\delta,x_0+\delta)$, $|f(x)-f(x_0)|<\varepsilon$. This is the case in $\Bbb{R}$, but this is the basic definition.

For example if we wanted to show that the function $f(x)=x$ is continuous in $\Bbb{R}$, we may say the following: let $x_0\in\Bbb{R}$, $\varepsilon>0$. So we take $\delta=\varepsilon$, so for every $x$ s.t $|x-x_0|<\delta$ we get $|f(x)-f(x_0)|=|x-x_0|<\delta=\varepsilon$, so $x$ is continuous in $x_0$ and therefore because we made no prior assumptions, it's continuous in $\Bbb{R}$.

Title: How to check continuity of a function?


问:
How to prove differentiability of the following unknown function
We have $f:\Bbb{R}^n\rightarrow \Bbb{R}^m$ s.t $f(x)=o(\|x-x_0\|)$, and I have to prove $f$ is differentiable at $x_0$.

So by definition, I know that $f$ is differentiable if there exists a linear transformation $L_{x_0}$ s.t $\lim_{x\rightarrow x_0}\frac{f(x)-f(x_0)-L_{x_0}(x-x_0) }{\|x-x_0\|}=0$. Also I know that $L_{x_0}(x-x_0)$ satisfies $L_{x_0}(x-x_0)\in O(\|x-x_0\|)$. Any help would be appreciated.

答:
You can prove that $f$ is differentiable at $x_0$ if and only if there is a linear transformation $L_{x_0}$ such that

$$f(x)-f(x_0)-L_{x_0}(x-x_0)\in o(\Vert x-x_0\Vert).$$

It's really just a slight reformulation of your definition. Now if $f\in o(\Vert x-x_0\Vert)$, then you'll quickly see that this condition is fulfilled.


问:
An example for a complex-valued test function
Does anyone have an easy example for a complex-valued test function $f \in C_c^\infty(\mathbb{R},\mathbb{C})$? By test function I mean a $C^\infty$-function with compact support. </p>

答: Why don't you take usual test function and multiply it by complex number? If you want something more reasonable take two real valued test functions $f_1$ and $f_2$ and consider $f=f_1+i f_2$. This is complex valued test function because its support is contained in $\operatorname{supp}(f_1)\cup\operatorname{supp}(f_2)$. In fact every complex valued test function can be obtained by this procedure.</p>


答:
Well, that is not the rigurous definition of continuity but it works for most UNDERGRADUATE functions.

Some tricks to check continuity:

You may know that elementary fucntions ($\sin$, $\cos$, $\exp$, polynomials,...) are continuous everywhere.

$\log(\cdot)$ is continuous on its domain (whenever what is inside is strictly positive).

$\sqrt{\cdot}$ is continuous on its domain (wheneever what is inside is positive (not stricly).

Moreover, any linear combination of continuous functions is also continuous (say, the addition of subtraction, and multiplication by a number). Also multiplication of continuous functions is also continuous.

For quotient of contuinuous functions, everything works okay EXECEPT for those points that cancel the denominator.

These are just a few tricks; they won’t prove continuity in every case, but for undegraduate students they may be enough.

Title: How to check continuity of a function?

Body: I know that a function is continuous at a point if the limit from left and right side exists and are equal and for a function to be continuous, the function should be continuous at all points. My question is that if I want to check continuity of a function, I cannot practically check continuity at each and every point in $\mathbb{R}$. So, how to do it?


问:
Common false/unproven "popular mathematics" claims
Motivated by the common unproven claim in memes that every finite sequence of digits appears in the digits of $\pi$ (see Does $\pi$ contain all possible number combinations?), I started wondering what other similar common claims are out there?

Criteria

A claim about math that's "commonly" stated as true by math enthusiasts who are non-mathematicians. For instance, something that one might find posted somewhere as a meme.

The claim must be either false or unproven.

答:
I add this as an answer, instead of a comment, for it gains more visibility by people. So please, treat this as a comment.

https://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics


问:
The collection of unions of finite disjoint collections of sets in a semiring is closed under finite unions and finite intersections.
Define a semiring to be a collection $S$ of subsets of $X$ such that if $A$, $B \in S$, then $A \cap B \in S$, and there is a finite disjoint collection $\{C_k\}$ of sets in $S$ such that $A - B = \bigcup_{k=1}^n C_k$.

Define $S'$ to be the collection of unions of finite disjoint collections of sets in $S$, how to prove that $S'$ is closed with respect to the finite unions and finite intersections.

(please do not assume $S'$ is closed under relative complements, cause later I need to use the above property to prove $S'$ is closed under relative complements.)

Can anyone help me on this please?

答:
Let $A$ and $B$ be sets in $S'$. Then there are finite disjoint collections $\{A_i\}_{i=1}^n$ and $\{B_j\}_{j=1}^m$of sets in $S$, such that $A= \bigcup_{i=1}^n A_i$ and
$B= \bigcup_{j=1}^m B_j$.

It is easy to see that $\{A_i\cap B_j\}_{i,j=1}^n$ is a finite disjoint collection of sets in $S$. So,
$$ A\cap B= \bigcup_{i,j=1}^n (A_i\cap B_j) \in S' \tag{1}$$

Now note that
$$A \cup B_1 = \left (\bigcup_{i=1}^n A_i \right )\cup B_1 = \left (\bigcup_{i=1}^n (A_i \setminus B_1) \right )\cup B_1 $$

For each $i$, $1\leqslant i \leqslant n$ there is a finite disjoint collection $\{D_{i,k}\}_{k=1}^{n_i}$ of set in $S$ such that $A_i \setminus B_1= \bigcup_{k=1}^{n_i}D_{i,k}$. Since $\{A_i\}_{i=1}^n$ is a finite disjoint collection of sets in $S$, we can conclude that the collection $\{ D_{i,k} : 1 \leqslant i \leqslant n, 1 \leqslant k \leqslant n_i \} \cup \{B_1\} $  is a finite disjoint collection of sets in $S$. So, we have
$$ A \cup B_1 = \left (\bigcup_{i=1}^n (A_i \setminus B_1) \right )\cup B_1 =
 \left (\bigcup_{i=1}^n \bigcup_{k=1}^{n_i} D_{i,k} \right )\cup B_1 \in S'
$$
So, we have that $ A \cup B_1  \in S'$. By finite induction, we have that
$$ A\cup B = A \cup \left (\bigcup_{j=1}^m B_j \right ) =( ( ( (A\cup B_1)\cup B_2) \cdots )\cup B_m) \in S' \tag{2} $$

From $(1)$ and $(2)$, by finite induction, we have that $S'$ is closed with respect to the finite unions and finite intersections.

Remark: Since the question is about semirings, we must be careful about the term "finite intersection". If we take an empty collection of sets in $S'$, the intersection of such empty family is, by definition, $X$ which may not be in $S$ nor in $S'$. So, we understand here "finite intersection" as being the intersection of a non-empty family of sets in $S'$.


问:
Proof by mathematical induction in sets
How would I go about writing a proof for this question? Is there too little information given about set A itself?

Prove by mathematical induction that |2^A| = 2^|A| for every finite set A.

I'm new to the idea of sets but I do know that 2^A is the powerset of A and |2^A| is the number of elements in the powerset of A.

E.g. A = {1,2,3}
2^A = { {},{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3} }
|2^A| = 8 = 2^|A|

答:
A proof by induction involves proving a statement is true for all natural numbers. How do we turn your statement into a statement about natural numbers? For a finite set $A$, we must have $|A|=n$ for some $n\in\mathbb{N}$. This way, we can say "for all sets $A$ such that $|A|=n$ for some $n\in\mathbb{N}$, we have that $|2^A|=2^{|A|}=2^n$.

Now, for any proof by induction, you need two distinct subproofs: i) you have to prove the base case. ii) you have to prove the inductive step.

i) The base case is simply proving that your statement is true for $n=0$. Since $|A|=0$ implies $A=\emptyset$, this boils down to showing that $|2^\emptyset|=2^0$.

ii) This inductive step is the so called "lining up the dominoes". In this subproof, you have to show that when ever your statement is true for a natural number $k$, the statement has to be true for the next natural number, $k+1$. This requires proving an implication statement for all natural numbers. The implication statement you want to prove is that if $|2^A|=2^{|A|}$ is true for any set such that $|A|=k$, then it is true for any set $|A|=k+1$. How do you do this? You must assume $|2^A|=2^{|A|}$ is true for $k=|A|$ and then deduce $|2^A|=2^|A|$ for any set such that $k+1=|A|$.

Title: Proof by mathematical induction in sets


答:
There are several „methods“ to check continuity of a function $f:\Bbb R \longrightarrow \mathbb{R}$:

show that given an arbitrary point $x$ and any sequence $x_n \rightarrow x$ converging to $x$ you have that $f(x_n) \rightarrow f(x)$. This is feasible, if your function itself is given by a formula closely related to limits, like $\exp, \sin, \cos, x\mapsto x^2$ etc. Here you may use facts like (under certain hypotheses) limits commute or are compatible with multiplication, addition etc.

show that it is given by a composition of continuous functions. For example $x \mapsto e^{x^2}$ is continuous due to $x \mapsto e^x$ and $x\mapsto x^2$ being continuous.

show that it is induced by some sort of universal mapping property, by which I mean some sort of general theorem telling you that for example having continuous functions $f:\Bbb R \rightarrow \Bbb R, g: \Bbb R \rightarrow \Bbb R$ induces a unique continuous function $(f,g):\Bbb R \rightarrow \Bbb R^2, x \mapsto (f(x),g(x))$.

Title: How to check continuity of a function?


问:
How many solutions does this equation have $x^2 \equiv 1017 (\mod 2^k)$
How many solutions does this equation have $x^2 \equiv 1017 \  \mod 2^k$?

I know that $1017 \equiv 1 \mod 8$?

I think that for $k=1$ we have $x^2 \equiv 1017 \mod 2$ and the solution is $x=1 \mod 2$

and for $k=2$ we have $x^2 \equiv 1017 \mod 4$, so  $x^2 \equiv 1 \mod 4$.

(Previously, I wrote that $x^2 \equiv 3 \mod 4$)

What can I do for $k \ge 3$?

I've read somewhere that if $x_k ^2 \equiv a \mod 2^k, \ \ k \ge 3$, this means that $2^k | x^2-a$.

If $\frac{x^2 - a}{2^k}$ is odd, let $i=1$, if it os even, let $i=0$.

Then $x_{k+1}:= x_k + i\cdot 2^{k-1}$ is a solution to $x^2 _{k+1} \equiv a \mod 2^{k+1}$.

Do you think this could be useful?

Could you help me?

Thank you.

答:
Modulo 8 you have four pairwise noncongruent solutions $x=1,3,5,7$. You can "lift" all of those to solutions modulo a higher power of two by a process called Hensel lifting. That process is  unique, so you end up with four pairwise noncongruent solutions modulo $2^k$, $k\ge3$.

Another way of seeing is that the solutions must form four residue classes modulo $2^k$ is to observe that the ratio of any two solutions must be a square root of $1$ modulo $2^k$, and it is well known that there are exactly four such residue classes, namely $\pm1$ and $\pm1+2^{k-1}$.

The lifting is very closely related to the process you describe. Assume that you have a solutions $x_k$ modulo $2^k, k\ge3$. In other words,
$$
x_k^2\equiv1017\pmod{2^k}.
$$
Obviously here $x_k$ is then an odd integer.
Let's try to find a solution $x_{k+1}=x_k+a2^{k-1}$ that would work modulo $2^{k+1}$. We calculate
$$
x_{k+1}^2=x_k^2+2^kax_k+2^{2k-2}a_k^2,
$$
so
$$
x_{k+1}^2-1017=(x_k^2-1017)+2^kax_k+2^{2k-2}a_k^2.
$$
Here $1017-x_k^2$ is divisible $2^k$, and the last term is divisible by $2^{2k-2}$ and hence by $2^{k+2}$ (here we need the assumption $k\ge3$).
So choosing $a=0$ or $a=1$ will make the right hand side divisible by $2^{k+1}$, which is exactly what we wanted.

Once we get this going, we can rinse and repeat as many times as needed.

For example starting with the solution $x_3=1$ modulo $2^3$ ($k=3$). We check that
$$x_3^2-1017\equiv1-9\not\equiv0\pmod{16},$$
so we fix the solution by adding a multiple of $2^{k-1}$, so $x_4=x_3+4=5$, which works modulo $16$ as $5^2=25\equiv9\equiv1017\pmod{16}$. As $1017\equiv25\pmod{32}$, it also works modulo $32$, so we choose $x_5=x_4=5$.
Modulo $64$ we have $1017\equiv57\not\equiv5^2$, so we "improve" our solution
$x_6=x_5+2^{5-1}=5+16=21$ again by adding $2^{k-1}$ to it. As
$21^2\equiv57\pmod{128}$, but $1017\equiv121\pmod{128}$, we need to again improve the solution to $x_7=x_6+2^5=53$ to get one that works modulo 128.
We continue the sequence as $x_8=x_7=53$, $x_9=x_8+2^7=181$, which is also
a solution modulo $2^10=1024$.

The other square roots of $1017$ modulo $1024$ are, as promised,
$$-181\equiv843,\qquad (1+512)(181)\equiv693\qquad\text{and}\qquad-693\equiv331.$$

It is hopefully clear that you can keep doing this as long as you wish. We actually get an infinite sequence of square roots of $1017$ modulo an ever increasing sequence of powers of two. As a limit, we get a $2$-adic square root of $1017$, but if you have never heard of $2$-adic numbers, that may not mean anything to you.


答:
Most people (probably outside mathematics) think that continuous functions are those that can be drawn without lifting the pencil from the paper.

But this is actually false. For example it is well known that the set of continuous nowhere differentiable functions is prevalent (its complement has, in some sense, of infinite-dimensional-measure equals zero), residual (its complement is of first Baire category), it is spaceable (it contains, except for the null-function, a closed infinite dimensional vector space) and many more properties that suggest to us it is a VERY BIG set.

And what have this to do with being drawable? Well, any of these functions cannot be drawn with a pencil: If you draw some (enough) small line, you do it upwards, downwards or horizontally, so it must have positive, negative or null derivative (on its whole domain, say a small interval).... BUT the functions is NOT derivable.

I used to call this, "The great lie of continuity".

Title: Common false/unproven "popular mathematics" claims

Body:
Motivated by the common unproven claim in memes that every finite sequence of digits appears in the digits of $\pi$ (see Does $\pi$ contain all possible number combinations?), I started wondering what other similar common claims are out there?

Criteria

A claim about math that's "commonly" stated as true by math enthusiasts who are non-mathematicians. For instance, something that one might find posted somewhere as a meme.

The claim must be either false or unproven.


问:
Does the average of i.i.d. Poisson random variables have the same moments as the said variables
This question might have been asked here before, but I have not managed to found an answer to my question. So for $k = 1,2,\dots$, let $X_i \sim \mathrm{Poi}(\lambda)$ be i.i.d. r.v.s and $Y = \frac{1}{K}\sum_{k=1}^{K}X_k$.

Is the theoretical expected value of $Y$ equal to $\lambda$?

Supposing $\mathrm{E}(Y) = \lambda$, does it follow that $\mathrm{Var}(Y) = \lambda$?

If all $X_k$ would still be independent, but now $X_k \sim \mathrm{Poi}(\lambda_k)$ it is true that $\sum_{k=1}^{K}X_k \sim \mathrm{Poi}(\sum_{k=1}^{K}\lambda_k)$. But is now $Y \sim \mathrm{Poi}(\frac{1}{K}\sum_{k=1}^{K}\lambda_k)$? How would the change in the intensity of some $X_k$ affect the theoretical expected value and variance of $Y$?

答: All you need is the first part of 3. (which is a correct conclusion), and the fact that the mean and variance of Poisson-type RV are both equal to its only parameter. But you have to be more careful when dividing your sum by $K$; this divides the mean of the sum by $K$, but its variance by $K^2$. Also, the resulting Y has then non-integer values and cannot thus be Poisson.


问:
Weak* convergence equivalent to boundness, Banach-Steinhaus theorem
Let $(a_n)$ be a sequence of real numbers.
How to prove that weak* convergence to $0$ of $(a_ne_n)\in l_\infty(\mathbb{R})$ is equivalent to $(a_n)$ being bounded?

I'm asking for help, how to relate this to Banach-Steinhaus theorem of uniform boundness?

答:
With regard to QuantumSpace's solution, we prove the $\Leftarrow$ direction:

Let $(a_{n})$ be a sequence of real numbers. For each $n$, define $\theta_{n}\in l^{\infty}$
by $\theta_{n}=(0,0,\ldots,0,a_{n},0,0,\ldots).$ Then, $\{\theta_{n}\mid n\in\mathbb{N}\}$
is a family of bounded linear functionals on $l^{1}$. We go to show
that $\sup_{n}||\theta_{n}||_{\infty}<\infty$ by Uniform Boundedness
Principle. Let $g\in l^{1}.$ Denote $g=(g_{1},g_{2},\ldots)$. By
assumption, it is given that $\lim_{n}a_{n}g_{n}=0$. Note that $\langle\theta_{n},g\rangle=a_{n}g_{n}$.
It follows that $\{\langle\theta_{n},g\rangle\mid n\in\mathbb{N}\}$
is bounded for each $g\in l^{1}.$ By Uniform Boundedness Principle,
$\sup_{n}||\theta_{n}||_{\infty}<\infty$. However, $||\theta_{n}||_{\infty}=|a_{n}|$.
Therefore, $\sup_{n}|a_{n}|<\infty$.


问:
From DNF to CNF
What is the most efficient way to switch from DNF to CNF?.</p>

答:
A method with complete enumeration of 2^n terms is described here.

A suitable tool would be bc2cnf (http://users.ics.aalto.fi/tjunttil/bcsat/).
To use bc2cnf, you'd have to express the DNF terms as "AND gates" in bc2cnf syntax.

Example for a EXOR function with four inputs and eight DNF minterms:

```<code>BC1.1
_m1 := !A & !B & !C &  D;
_m2 := !A & !B &  C & !D;
_m4 := !A &  B & !C & !D;
_m7 := !A &  B &  C &  D;
_m8 :=  A & !B & !C & !D;
_m11 := A & !B &  C &  D;
_m13 := A &  B & !C &  D;
_m14 := A &  B &  C & !D;
_F := _m1 | _m2 | _m4 | _m7 | _m8 | _m11 | _m13 | _m14;
ASSIGN _F;
</code>```

Title: From DNF to CNF

Body:
A method with complete enumeration of 2^n terms is described here.

A suitable tool would be bc2cnf (http://users.ics.aalto.fi/tjunttil/bcsat/).
To use bc2cnf, you'd have to express the DNF terms as "AND gates" in bc2cnf syntax.

Example for a EXOR function with four inputs and eight DNF minterms:

BC1.1
_m1 := !A & !B & !C &  D;
_m2 := !A & !B &  C & !D;
_m4 := !A &  B & !C & !D;
_m7 := !A &  B &  C &  D;
_m8 :=  A & !B & !C & !D;
_m11 := A & !B &  C &  D;
_m13 := A &  B & !C &  D;
_m14 := A &  B &  C & !D;
_F := _m1 | _m2 | _m4 | _m7 | _m8 | _m11 | _m13 | _m14;
ASSIGN _F;


问:
Simplifying matrix equation $A^{-1}B^T(BA^{-1}B^T)^{-1}BA^{-1}$
I want to simplify the inverse of a $2\times 2$ block matrix:
$$
\begin{bmatrix}
A_{n\times n} & B^T_{n\times m}\\
B_{m\times n} & 0_{m\times m}
\end{bmatrix}
$$
$A_{n\times n}$ is square and invertible (and possibly positive definite), and $B_{m\times n} (m < n) $ is not square but full row rank. Using this eq. (2.2), the inverse can be written as:
$$
\begin{bmatrix}
A & B^T\\
B & 0
\end{bmatrix} ^{-1}=
\begin{bmatrix}
A^{-1}-A^{-1}B^T(BA^{-1}B^T)^{-1}BA^{-1}  &  A^{-1}B^T(BA^{-1}B^T)^{-1} \\
(BA^{-1}B^T)^{-1}BA^{-1} & -(BA^{-1}B^T)^{-1}
\end{bmatrix}
$$
I want to see is there a way to simplify this matrix? For example, can we simplify this term $A^{-1}B^T(BA^{-1}B^T)^{-1}BA^{-1}$, or other terms?

答:
This is really just an extended comment. One can derive the desired expression for the inverse by considering the matrix equation $$\begin{bmatrix} y_1 \\ y_2 \end{bmatrix}=\begin{bmatrix} A & B^T \\ B & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$$
which is equivalent to the system of equations $y_1=Ax_1+B^T x_2,\, y_2=Bx_1$. To solve this for $x_1,x_2$, observe that

$$BA^{-1}y_1 = BA^{-1}(Ax_1+B^T x_2)=Bx_1+(BA^{-1}B^T)x_2=y_2+(BA^{-1}B^T)x_2.$$

Since $A^{-1}$ exists and $B$ has full row rank, the matrix $BA^{-1}B^T$ is full rank. Hence we may solve for $x_2$ by inverting $BA^{-1}B^T$:

\begin{align}
x_2&=(BA^{-1}B^T)^{-1}(BA^{-1}y_1-y_2)\\
&=(BA^{-1}B^T)^{-1}BA^{-1}y_1-(BA^{-1}B^T)^{-1}y_2.
\end{align}
Furthermore, we have $A^{-1}y_1=x_1+A^{-1}B^T x_2$ and therefore
\begin{align}
x_1
&=A^{-1}y_1-A^{-1}B^T x_2\\
&=[A^{-1}-A^{-1}B^T (BA^{-1}B^T)^{-1}BA^{-1}]y_1+A^{-1}B^T (BA^{-1}B^T)^{-1}BA^{-1}y_2
\end{align}

Converting this system of equations back to a matrix equation yields precisely the stated matrix elements of $\begin{bmatrix} A & B^T \\ B & 0\end{bmatrix}^{-1}$.


问:
Translating into predicate logical
Every odd number is a sum of two even numbers.

The above translates to $∀_{x∈ℤ}$ $∃_{a,b∈ℤ}$ such that $2x+1=2a+2b$.

The negation of the above is $∃_{x∈ℤ}$ $∀_{a,b∈ℤ}$ such that $2x+1≠2a+2b$

The negation is true, but the original statement is false.

Is my translation valid?

答:
You have a working translation but there are a few notes I might add.

First, your sentences are not entirely in predicate logic, they are still using the English convention "such that" and they are using restricted quantifiers. These are phrases like $\forall x\in\mathbb{Z}$ as opposed to $\forall x$. They are not a part of the language of predicate logic, it is a linguistic convention in mathematics.

Your class might accept those conventions. In case they don't, I'll describe how to work around them.

Removing "such that" is easy. If you have a phrase "$\forall x$ such that $\phi(x)$, it is the case that $\psi(x)$", what this is doing is putting a restriction on $x$ before asserting that $\psi(x)$. In other words, if an $x$ is $\phi$ then we have to have that $x$ is $\psi$. But, we can't deduce that $x$ is $\psi$ unless we have $x$ is $\phi$. This leads the translation $\forall x(\phi(x)\rightarrow \psi(x))$.

Restricted quanitfers are doing the same thing. $\forall x\in\mathbb{Z}\psi(x)$ asserts that if $x\in\mathbb{Z}$, then $\psi(x)$. This is the previous case but $\phi\equiv x\in\mathbb{Z}$ and, well,  $\psi\equiv\psi$.

Title: Translating into predicate logical


问:
Can there be a holomorphic on a disc for which $f(z)=f(\alpha z)$ whereas $\alpha$ is transcendental?
Can there be a holomorphic on a disc for which $f(z)=f(\alpha z)$ whereas $\alpha$ is transcendental and $\alpha z \in D$? where D is the disc.
I believe it to be true as I find no reason for contradiction.
But I find no example.

Edit: it seems that the constant function does hold for this question with any transcendental $|\alpha|=1$. Is there a non constant function?

答:
Choosing an holomorphic branch of $\log(z)$ then $f(z) =\exp(2i\pi \log(z)/\log(\alpha))$ works on any disk not containing $0$.

If the disk contains $0$ then write the power series of $f(z)$ and $f(\alpha z)$ and equate the coefficients.


问:
Famous or common mathematical identities that yield $1$
To me the most common identity that comes to mind that results in $1$ is the trigonometric sum of squared cosine and sine of an angle:

$$
\cos^2{\theta} + \sin^2{}\theta = 1 \tag{1}
$$
and maybe

$$
-e^{i\pi} = 1 \tag{2}
$$

Are there other famous (as in commonly used) identities that yield $1$ in particular?

答:
Limits
$$\lim_{x\to0}\frac{\sin x}{x}=1$$
This is actually a special case of the more general identity: If $f(0)=0$ and $f'(0)=1$, then
$$\lim_{x\to0}\frac{f(x)}{x}=1$$
which includes things like
$$\lim_{x\to0}\frac{\arctan x}{x}=1$$

Series

Any geometric series (which converges) will do as long as the first term, $a$, is equal to $1-r$, where $r$ is the common ratio. For example:
$$\begin{align}1&=\frac{1}{3}+\frac{1}{3}\left(\frac{2}{3}\right)+\frac{1}{3}\left(\frac{2}{3}\right)^2+\frac{1}{3}\left(\frac{2}{3}\right)^3+\cdots\\
&=\frac{1}{4}+\frac{1}{4}\left(\frac{3}{4}\right)+\frac{1}{4}\left(\frac{3}{4}\right)^2+\frac{1}{4}\left(\frac{3}{4}\right)^3+\cdots\\
&=\sum_{k=1}^\infty (1-r)r^{k-1},~~~~~~~\lvert r\rvert <1\end{align}$$
Another quite interesting series:
$$\sum_{k=1}^\infty \frac{k}{(k+1)!}=1$$

I can also list many integrals, such as
$$\begin{align}
\int_0^\frac{\pi}{2}\cos x~dx=1
\end{align}$$
And just for fun:
$$0!=1$$
I will try to add some more interesting identities if I can remember/come across them.


答: Start with the empty set, so |A| = 0. Your formula can be shown to be true for this set. Then move to a set with cardinality $n$, and assume your hypothesis to be true. If you add one element to the set, then the power set doubles, because you start with the power set of the set with $n$ elements, then add same set, but to every set in the power set, add the $(n + 1)$st element. This doubles the number of sets in the original power set, hence showing that by adding one element to a set, you double the cardinality of its power set.

Title: Proof by mathematical induction in sets

Body:
How would I go about writing a proof for this question? Is there too little information given about set A itself?

Prove by mathematical induction that |2^A| = 2^|A| for every finite set A.

I'm new to the idea of sets but I do know that 2^A is the powerset of A and |2^A| is the number of elements in the powerset of A.

E.g. A = {1,2,3}
2^A = { {},{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3} }
|2^A| = 8 = 2^|A|


问:
Alternatives to Cantor's pairing that produces numbers between 0 and the number of combinations?
I have many pairs of numbers (A,B), with both numbers in the pair falling between 0 and N. As an example, let's say N=100.

I need a way to uniquely map these pairs to another number. We assume pairs of (A,A) or (B,B), etc are not valid pairs.

Since I don't care about the order, these are combinations. So the total number of such pairs should be N(N-1)/2. For this example, that means there are 4950 combinations.

However, if I were to use Cantor's algorithm, I would get numbers between 0 and 20,000. That's no good! It's even worse than permutations, which you could imagine as a table of N by N (100^2).

Because this is needed for programming, I need the produced integer number from the pair to be as small as possible while still remaining unique. I also ideally need the math to be fast and simple as this will be done many times.

Is there another pairing function I can use, or did I misunderstand something along the way? Forgive me, my math is rusty.

答:
There are tons of different pairing functions you can use. Cantor's function is particularly useful since you don't need to know what $N$ is, but if you know $N$ beforehand there are more options.

If you do know $N$, possibly the simplest thing to consider is something like
$$(a,b)\mapsto aN+b.$$
This is essentially a lexicographic ordering: $(0,0)$ comes first, then $(0,1)$, etc. up to $(0,N-1)$, and then $(1,0)$, $(1,1)$, et cetera. This does depend on order, so one thing you could do is
$$(a,b)\mapsto \min(a,b)N+\max(a,b);$$
this misses some numbers, but the calculations are quite simple.

Another thing you can do is just sort the pairs: order
$$(0,1),(0,2),\dots,(0,N-1),(1,2),(1,3),\dots,(1,N-1),\dots,(N-2,N-1),(N-2,N),(N-1,N).$$
This bijects your pairs to the numbers in the interval $\left[0,\frac{N(N-1)}{2}\right),$ so it doesn't use any "extra" numbers. It's not immediately obvious, but this function can be represented as
$$\frac{a(2N-1-a)}{2}+b-a-1;$$
there might be a few ways to speed up this computation. In general, though, I'd imagine it's better to use something like $\min(a,b)N+\max(a,b)$; using numbers that are twice as big is probably not that much of an issue, and it's also easily invertible.


问:
Is every linear transformation also technically a change of basis matrix?
I don't really see the difference between a linear transformation and a change of basis matrix.

Let's take this formulation:

$$ Ax = v$$

I can interpret it in two different ways.

1.) The matrix $A$ encodes a linear transformation that converts vector $x$ to vector $v$.

2.) The matrix $A$ is a change of basis matrix whose span includes the vector v. Scaling the new basis $A$ by the values stored in $x$, followed by linearly combining the basis, will give us $v$.

Could all linear transformations also be considered as a change of basis?

Edit: A comment pointed out that a basis by definition must be a set of linearly independent vectors. So not all linear transformations are a change of basis. In that case, what's the point of a change of basis matrix? It just seems like a label for a specific type of linear transformation.

答:
In the setting of abstract (finite-dimensional, real) vector space $V$ and a linear transformation $T:V\to V$ on $V$, there is no notion of matrices involved.

If you choose an ordered basis $\beta$, then $T$ can be represented as an $n\times n$ matrix $[T]_{\beta}$.

On the other hand, if you have two bases $\alpha$ and $\beta$, given any vector $x\in V$, the two different coordinates of $x$ (wrt $\alpha$ and $\beta$),  $[x]_\alpha$ and $[x]_\beta$ are related by a matrix $A_{\alpha}^\beta$:
$$
[x]_\alpha=A_{\alpha}^\beta[x]_{\beta}\tag{1}
$$
This matrix $A_{\alpha}^\beta$ is a linear transformation on $\mathbb{R}^n$, where $n$ is the dimension of $V$. One can show that  $A_{\alpha}^\beta$ must be invertible. So the answer to your question in the title is no, since no every linear transformation (and thus its matrix under any basis) is invertible.

The change-of-basis matrix gives you a formula that expresses the coordinates relative to one basis in terms of coordinates relative to the other basis.


问:
Details of the proof of supermartingale theorem
Theorem 5.17. If $X_{n} \geq 0$ is a supermartingale then $X_{\infty}=\lim _{n \rightarrow \infty} X_{n}$ exists and $E X_{\infty} \leq E X_{0}$
The bad martingale in Example 5.9 shows that we can have $X_{0}=1$ and $X_{\infty}=0$. The key to the proof of this is the following maximal inequality.

Lemma 5.18. Let $X_{n} \geq 0$ be a supermartingale and $\lambda>0 .$
$$
P\left(\max _{n \geq 0} X_{n}>\lambda\right) \leq E X_{0} / \lambda
$$
Proof. Let $T=\min \left\{n \geq 0: X_{n}>\lambda\right\} .$ Theorem 5.13 implies that
$$
E X_{0} \geq E\left(X_{T \wedge n}\right) \geq \lambda P(T \leq n)
$$
i.e., $P(T \leq n) \leq E X_{0} / \lambda .$ Since this holds for all $n$ the desired result follows.
Proof of Theorem 5.17 Let $a<b,$ let $S_{0}=0$ and define stopping times for $k \geq$ 1 by
$$
\begin{array}{l}
R_{k}=\min \left\{n \geq S_{k-1}: X_{n} \leq a\right\} \\
S_{k}=\min \left\{n \geq R_{k}: X_{n} \geq b\right\}
\end{array}
$$
Using the reasoning that led to Lemma 5.18
$$
P\left(S_{k}<\infty \mid R_{k}<\infty\right) \leq a / b
$$
Iterating we see that $P\left(S_{k}<\infty\right) \leq(a / b)^{k} .$ Since this tends to 0 as $k \rightarrow \infty X_{n}$ crosses from below $a$ to above $b$ only finitely many times. To conclude from this that $\lim _{n \rightarrow \infty} X_{n}$ exists, let
$$
Y=\liminf _{n \rightarrow \infty} X_{n} \quad \text { and } \quad Z=\limsup _{n \rightarrow \infty} X_{n}
$$
If $P(Y<Z)>0$ then there are numbers $a<b$ so that $P(Y<a<b<Z)>0$ but in this case $X_{n}$ crosses from below $a$ to above $b$ infinitely many times with positive probability, a contradiction. To prove $E X_{\infty} \leq E X_{0}$ note that for any time $n$ and positive real number $M$
$$
E X_{0}=E X_{n} \geq E\left(X_{n} \wedge M\right) \rightarrow E\left(X_{\infty} \wedge M\right)
$$
where the last conclusion follows from the reasoning in the proof of Theorem $5.14 .$ The last conclusion implies $E X_{0} \geq E\left(X_{\infty} \wedge M\right) \uparrow E X_{\infty}$ as $M \uparrow \infty$

$\\$
$\\$
$\\$

This theorem 5.17 is from Richard Durrett's 《Essentials of Stochastic the Processes Second Edition》(Page 200-201).Now there are two problems with this theorem and I still can't prove it after trying.

1)How to use Lemma5.18 to derive$P\left(S_{k}<\infty \mid R_{k}<\infty\right) \leq a / b$.

2)How to derive $P\left(S_{k}<\infty\right) \leq(a / b)^{k}$

Could anyone show me detailed proof procedure of these two problem?Thank you very much!

答:
Let $Y_n={X_{n+R_k}}$ for $n\ge 0$, so this is the process after $R_k$. $$P(S_k<\infty|R_k<\infty)=P(\max_n Y_n\ge b)\le E[Y_0]/b= E[X_{R_k}]/b \le a/b,$$ since $X_{R_k}\le a$ almost surely.

You can then prove $P(S_k<\infty)<(a/b)^k$ by induction on $k$. Indeed,

$$
P(S_k<\infty)=P(S_{k-1}<\infty)\cdot P(R_k<\infty |S_{k-1}<\infty)\cdot P(S_k<\infty |R_k<\infty, S_{k-1}<\infty)
\\\le \underbrace{(a/b)^{k-1}}_{\text{induction hypothesis}}\cdot 1\cdot (a/b)
$$


答:
Many people don't believe that $$0.\overline{9}=1$$

They will argue $0.999999999...$ cannot equal $1$.


问:
When a reflection over the y axis is applied first, does it apply to a horizontal translation?
For example, if I reflect something over the y axis and it becomes $f(x) = -x$, does horizontally shifting it $2$ units right cause it to look like this:
$$f(x) = -(x-2)?$$
Furthermore, if the horizontal shift is applied first, does the equation become $-(x-2)$ or $-x-2?$ My main question is, does the order of transformations matter when it is just horizontal translation, reflection in the y axis, or horizontal stretches.

答:
A translation of a function horizontally two units to the right and then reflected about the y-axis:

Original Function: f(x)

Translated 2 to the right: f(x-2) = g(x)

Reflected about the y-axis: g(-x) = f(-x-2)

To understand more clearly, we can take a point on a function as an example. Let's use the function f(x) = x and take the point (3,3) on that function.

Beginning Point: (3,3)

Translated 2 units to the right: (5,3)

Reflected about y-axis: (-5,3)

Since our original function was f(x) = x, our new function should be f(-x-2) = -x-2.
Since h(-5) = -(-5)-2 = 3, our function demonstrates these translations.

A function reflected about the y-axis and then shifted horizontally:

Original function: f(x)

Reflected about the y-axis: f(-x) = g(x)

Translated 2 units to the right: g(x-2) = f(-(x-2)) = f(-x+2)

If we take the point (3,3) again:

Reflected about y-axis: (-3,3)

Translated 2 units to the right: (-1,3)

Since our original function was f(x) = x, our new function should be f(-x+2) = -x+2.

Since h(-1) = -(-1) + 2 = 3, our function demonstrates these translations.

Although most problems arise when transforming functions horizontally, order does matter when transforming vertically as well. For example, an upward shift of a function 3, then a reflection about the x-axis: -(f(x) + 3) is much different than a reflection about the x-axis, then vertical shift 3: -f(x) + 3


问:
Calculate the Work in a cylinder tank, through integral
"A vertical cylindrical tank has a height of 9 m and a diameter of 6 m. The tank is full of oil that weighs 8042 N per cubic meter. The pumping is done from the top of the tank. How much work is required to pump all the oil to the rim of the tank?"

We are asked to calculate the work though integral.
I tried to do so, but we need the limits of variable x so that to calculate the integral of the force.
I would appreciate for your guidance.
Thank you very much in advance.

答: Your integral should be with respect to $y$, with $0 \leq y \leq 9$. Then take a horizontal slab of the tank, and calculate the weight of the slab of oil. The thickness of the slab will be $\Delta y$, and because you have a cylindrical tank, the radius does not change. This simplifies the calculation of the weight of the slab. Then multiply this weight by $(9 - y)$, which is the distance the slab must be moved to the top of the tank (remember work = force times distance). Then integrate this quantity, with $\Delta y$ replaced by $dy$.


问:
Compactness about fiber bundle
I am working on a problem (problem 10.19 (d)) in John M. Lee's Introduction to Smooth Manifold.

Assume that $\pi$: $E$ $\rightarrow$ $M$ is a fiber bundle with model fiber $F$, I need to prove if $E$ is compact, then so are $M$ and $F$.
Clearly, $M$ is compact and if we assume that $M$ is Hausdorff, then it follows easily from part (c) of this problem.

($\pi$: $E$ $\rightarrow$ $M$ is proper iff $F$ is compact.)

(Edited: I think it suffices to assume that $M$ is $T_1$, i.e., singletons are closed, then it follows from the continuity that the fibers are compact.)

However, I have no idea how to proceed in the general case.
Any hints are appreciated.

答:
Since the author chimed in in the comments, let me record the full answer here.

This is still true without separation assumptions on the base. Indeed, by compactness of $M$ there exists a finite cover $\{U_i \subseteq M\}_{i \in I}$ over which $E$ is trivialised. Without loss of generality we may assume that no strict subset of $\{U_i \subseteq M\}$ covers $M$. Then
$$Z = M \setminus \bigcup_{j \neq i} U_j \subseteq U_i$$
is a nonempty closed subset over which $E$ is trivial, i.e. $E|_Z \cong F \times Z$. Since $E|_Z \subseteq E$ is closed, it is compact, hence so is $F$ since it is the image of the first projection $F \times Z \to F$. $\square$

Title: Compactness about fiber bundle


问:
Number of integer points inside an equilateral triangle with side length $n\in\mathbb{N}$
Does there exist any $n\in\mathbb{N}$ such that there exist at least one point inside of(not on the border) an equilateral triangle with side length $n$ ,which its distances to the vertices be integers?

If yes; can anyone give a formula for the number of such points for each $n\in\mathbb{N}$ ?

Notice: We already know by this post and for further results here ,that there are infinite number of points with rational distances inside an equilateral triangle with side length one, but what about integers?

答: Yes, infinitely many such triangles exist. Possible n are 112, 147, 185, 224, 273, 283, 294, 331...... See https://oeis.org/A061281</p>


问:
Partial cycles in projective resolutions of square-free algebra

Short version: Over a square-free algebra must every projective resolution of a simple module eventually terminate or contain a shift of itself as a direct summand?

I suspect not, but have not found a counterexample and have checked a few thousand non-isomorphic algebras.

Let $R$ be a finite-dimensional, associative, unital, connected, basic algebra over an algebraically closed field such no projective indecomposable module has two isomorphic composition factors. In other words, $R$ is a square-free algebra.

For an (single-sided) finitely generated $R$-module $M$, define $\Omega^0(M) = M$ and $\Omega^{n+1}(M)$ to be the kernel of the projective cover of $\Omega^n(M)$.

Since $\Omega^n(M)$ has finite-length, it has a unique Krull-Schmidt decomposition involving some isomorphism classes of indecomposable modules $M_{n,i}$.

There are two mutually exclusive possibilities: $\Omega^n(M)=0$ for some $n$, or $M \cong M_{n,i}$ for some $n \geq 1$ and some $i$. In the former case, $M$ has finite projective dimension, and in the latter the projective resolution “repeats” insofar as $\Omega^{n+k}(M)$ contains $\Omega^k(M)$ as a direct summand.

Are those the only two possibilities?

If $R$ is a square-free algebra, and $M$ is a simple module then must there be $n>0$ and $i$ such that either $\Omega^n(M) = 0$ or $M_{n,i} \cong M$?

Even more strongly,

If $R$ is a square-free algebra, and $M$ is a simple module, then is $\{ [ M_{n,i} ] \}$ a finite set of isomorphism classes of indecomposable modules?

If so, then $\Omega$ can be represented as a non-negative integer matrix in the Burnside ring, and we are asking if a certain diagonal entry in every proper power is 0, then must the matrix be nilpotent. I haven't found a quick way to verify this, but in the cases I've checked it works, which strikes me as crazy.

答:
If I understood your question correctly then the following provides a counterexample:
$$1\to 2\to 3\to 4\to 2$$
where the two $2$'s should be identified. Take $A$ to be the path algebra of this quiver modulo $\operatorname{rad}^2 A$. If you take the projective dimension of $S_1$, then $P_1$ just appears in the beginning as $P_0$ and then $P_2, P_3$ and $P_4$ repeat.</p>


问:
When the digits in the number $2005$ are reversed we obtain the number $5002$, and $5002 = a \cdot b \cdot c$, such that $a$, $b$ and $c$

When the digits in the number $2005$ are reversed we obtain the number $5002$, and $5002 = a \cdot b \cdot c$, such that $a$, $b$ and $c$ are three distinct primes. How many other positive integers are the products of exactly three distinct primes $p_1$, $p_2$ and $p_3$ such that $p_1 + p_2 + p_3 = a+b+c$?

I'm not sure how to approach this. I found that $a, b, c$ are $2, 41, 61$. Since there is only $1$ even prime, the other two primes must be odd. But short of going through every combination of primes that sum to $41+61=102$, how can I cleverly find the answer?

Thanks!

The answer to this is $\boxed{7}.$

答:
So $5002 = 2\cdot 41\cdot 61$ and $2+41+61=104$. We need to partition $104$ into three distinct primes, which means we need two odd primes and one (the only) even prime.

So partitioning $102 = 2\cdot3\cdot 17$  into two primes; probably the easiest way is descending through the sparser higher primes, if you remember those; otherwise work up, it's not hugely different.

$101+1 \quad\!\color{red}{\Large\times}$ (maybe too obvious $\ddot\smile$)

$97+5 \;\quad\color{green}\checkmark$ (and $97\cdot 5 \cdot 2 = 970$, etc.)

$89+13 \quad\color{green}\checkmark$

$83+19 \quad\color{green}\checkmark$

$79+23 \quad\color{green}\checkmark$

$73+29 \quad\color{green}\checkmark$

$71+31 \quad\color{green}\checkmark$

$67+35 \quad\!\color{red}{\Large\times}$

$61+41 \quad\checkmark$ (but already used)

$59+43 \quad\color{green}\checkmark$

$53+49 \quad\!\color{red}{\Large\times}$

giving $7$ alternatives, as you found.


问:
Prove $\lim_{x \to b} f(x) = c \iff \lim_{k \to \infty} f(x_k) = c $
Let $A \subseteq \mathbb{R^n}$ and $f\colon A \rightarrow \mathbb{R}$. Let $b \in \mathbb{R}^n$ s.t. there exists $\dot{B}_r(b) \subseteq A$. Prove $$\lim_{x \to b} f(x) = c \iff \lim_{k \to \infty} f(x_k) = c \text{ for all } \{x_k\} \subseteq A \text{ with } x_k \neq b \ \forall k \text{ and } \lim_{k \to \infty} x_k = b.$$

I have shown "$\Rightarrow$" but the problem is "$\Leftarrow$". My idea is the following:

Let $\varepsilon > 0 $. Let $g:A \cup\{b\} \rightarrow \mathbb{R}$ with
$$g(x) = \begin{cases} f(x) & x \in A \\ c & x =b \end{cases}$$

If one could show that $g$ is sequentially continuous at $b$ then $g$ is continuous at $b$ and therefore we can pick a $\delta > 0 $ s.t. $0 < \Vert x-b\Vert < \delta \implies \vert g(x) - g(b) \vert = \vert f(x) - c \vert < \varepsilon $.

The trouble is showing sequentially continuous at $b$. Is it possible to get this since we only have the desired convergence for sequences with $x_k \neq b $. And if so, how? Any help is appreciated!

答:
Here it is the proof in the context of metric spaces.

Proposition

Let $(X,d_{X})$ and $(Y,d_{Y})$ be metric spaces, $f:X\to Y$ a function and $E\subseteq X$ s.t. $x_{0}\in X$ is an adherent point of $E$ and $L\in Y$. The following statements are logically equivalent:

(a) The limit of $f$ as $x$ approaches $x_{0}$ along $E$ is $L$.

(b) For every sequence $x_{n}\in E$ which converges to $x_{0}$ according to $d_{X}$, the sequence $f(x_{n})$ converges to $f(x_{0})$ according to $d_{Y}$.

Proof

Let us prove $(a)\Rightarrow(b)$ first.

Let $\varepsilon > 0$. Then there exists $\delta_{\varepsilon} > 0$ s.t. for every $x\in E$ one has that
\begin{align*}
d_{X}(x,x_{0}) < \delta_{\varepsilon} \Rightarrow d_{Y}(f(x),f(x_{0})) < \varepsilon
\end{align*}

On the other hand, for every $\delta_{\varepsilon}$, there corresponds a $n_{\varepsilon}\in\mathbb{N}$ s.t.
\begin{align*}
n > n_{\varepsilon} \Rightarrow d_{X}(x_{n},x_{0}) < \delta_{\varepsilon} \Rightarrow d_{Y}(f(x_{n}),f(x_{0})) < \varepsilon
\end{align*}

hence $f(x_{n})$ converges to $f(x_{0})$, and we are done.

Let us prove now the implication $(b)\Rightarrow(a)$.

We shall do so by contradiction.

Let us assume that $f(x)$ does not converge to $f(x_{0})$.

Hence we conclude there is a $\varepsilon > 0$ s.t. for every $\delta > 0$ there corresponds a $x_{\delta}\in E$ satisfying
\begin{align*}
(d_{X}(x_{\delta},x_{0}) < \delta)\wedge(d_{Y}(f(x_{\delta}),f(x_{0})) \geq \varepsilon)
\end{align*}

Thus if we take $\delta = 1/n$, we conclude that $x_{n}\to x_{0}$ according to the squeeze theorem.

This means there exists a sequence $x_{n}$ which converges to $x_{0}$, but $f(x_{n})$ does not converge to $f(x_{0})$.

This contradicts our assumptions.

Hence the proposed claim is true.

Hopefully this helps!


问:
How to find the coefficient of $x^2y^3z^6$ in the expression $(x+y+xz)^{4}(y+z)^3$?
I know this is the binomial theorm but not too sure on how to progress. The $(y+z)^3$ and $xz$ is throwing me off as I've not done such a complicated one, but rather just with one expression bracket.

Do I multiply $(x+y+xz)^{4}(y+z)^3$ to form one big expression?

答:
Summing the degrees we get $2+3+6 = 11$

All the terms of  $(y+z)^3$ are of combined degree $3.$

That means we need a degree $8$ term from $(x+y+xz)^4$ and the only term of degree $8$ is $(xz)^4$

But then there would be a $x^4$ factor on the other side, which there isn't.

There is no $x^2y^3z^6$ term.


答:
Here are two well-known infinite series whose sum is $1$:
$$\sum_{n=1}^{\infty} \frac{1}{2^n} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \cdots = 1$$
$$\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \frac{1}{4 \cdot 5} + \frac{1}{5 \cdot 6} + \cdots = 1$$
The following series is less known, interesting although:
$$\sum_{n=1}^{\infty} \frac{1}{s_n} = \frac{1}{2} + \frac{1}{3} + \frac{1}{7} + \frac{1}{43} +  \frac{1}{1807} + \cdots = 1$$
where denominators form Sylvester sequence: every term is equal to the product of all previous terms, plus one. For example
$$\begin{array}{rcl}
2&  & \\
3 & = & 2+1\\
7 & = & 2 \cdot 3 +1 \\
43 & = & 2 \cdot 3 \cdot 7 +1 \\
1807 & = & 2 \cdot 3 \cdot 7 \cdot 43 +1 \\
\end{array}$$
and so on.


问:
Divisors of n! including 1 which sum to n!
I believe that given any positive integer $n$, we can find a set of divisors of $n!$, including $1$, which sum to $n!$ I checked for the first 6 positive integers, and tried to prove it by induction on $n$. However I was unable to formalise the proof, can anybody present a method of proving this? Or maybe without induction?

答:
Write $n!=2 \times \prod_{k=2}^{n-1}(1+k)$, then expand the product to get a subset of the divisors of $\frac{1}{2}n!$. Now you just need to multiply each term by $2$. Since you want one summand to be one, you should split at the end the least term greater than two in $1+(x-1)$. By the restriction 'greater than two' the other summand will be unique in the sum.

Comment: Take a closer look at the case that the expansion gives some term multiple times, but I think that should not be hard to solve.


问:
Show there must be two numbers in the list whose difference is $12$.
This question came up while tutoring: "$55$ numbers are randomly selected from the first $100$ positive integers. Show there must be two numbers in the list whose difference is $12$."

It definitely requires the Pigeonhole Principle but we were unsure how to set it up. My idea was to list all of the possible pairs whose difference is $12$:

$$(1,13), (2,14), \ldots (87, 99), (88, 100)$$ of which there are $88$ pairs. I'm not sure what to do from here.

I know similar questions have been asked here, like this one. Is there a way to apply those techniques to this problem?

答:
actually the puzzle works with 53 numbers chosen.  Look at arithmetic progressions with common difference 12, up to max 100. In
$$  1, 13, 25, 37, 49, 61, 73, 85, 97 $$
we can choose five members that are not adjacent, namely
$$ 1, 25, 49, 73, 97.$$
The same happens for the sequences beginning $1,2,3,4,$  so we can accommodate $4 \cdot 5 = 20$ numbers without forcing any difference of exactly 12.

For the sequences that begin $5,6,7,8,9,10,11,12$ we get only four spots each;
$$5,17,29, 41, 53, 65, 77, 89 $$
stops early because the next number 101 is above 100.
This sequence can accommodate just four socially distant element, for instance
$$  5, 29, 53, 77. $$
This adds  $8 \cdot 4 = 32$ more numbers: we can accommodate $20+32= 52$ socially distant numbers, but not 53


问:
Show an example of a finite abelian $p$-group that has $p^2 + p + 1$ subgroups of order $p$
Show an example of a finite abelian $p$-group that has $p^2 + p + 1$ subgroups of order $p$

My Path: I tried with the abelian group of $\mathbb{Z}_{63}=\mathbb{Z}_{9} \times \mathbb{Z}_{7} \times \mathbb{Z}_{1}$ but I'm not sure how to say if it's an abelian group

答:
Take the elementary abelian group $\;V:=\left(\Bbb Z/3\Bbb Z\right)^3 =\Bbb Z/3\Bbb Z\times\Bbb Z/3\Bbb Z\times\Bbb Z/3\Bbb Z\;$ .

This is also a vector space of dimension $\;3\;$ over the field $\;\Bbb F_p\cong\Bbb Z/3\Bbb Z\;$, and its subspaces are the same as its subgroups (when we flip from the group to the linear space structure or back).

Let us see how many subspaces of dimension $\;1\;$ this linear space has: for any non-zero vector $\;v\in V\;$, and there are $\;p^3-1\;$ of these, you get that Span$\,v\;$ is a linear subspace of dimension 1.

To count all the different subspaces of dimension $\;1\;$
(subgroups 0f order $\;p\;$), we must divide by all the non-zero multiple scalars of $\;v\;$, when $\;0\neq v\in V\;$ ...well, there you go.

The idea, of course, is for you to complete details in the above.


问:
Finding a real function using boundary conditions
If I have a region $R=\{z\in \mathbb{C}:1<|z|<2\}$. I want to find a real valued function $f$ which takes values $x$, $y$ where $f$ has boundary conditions: $1$ when $|z| = 1$ and $2$ when $|z| = 2$. We also have that $f$ is harmonic.

To start this I know $f$ has to be real, the real part of a complex logarithm is constant if $|z|$ is constant, so using the $\operatorname{Log}(z)=\ln|z|+i\operatorname{Arg}(z)$ the only way this can be real is if we consider $\ln|z|$. We also know that $f$ is harmonic so $f_{xx}+f_{yy}=0$. I think I need to find a branch cut for $\ln|z|$ but I am unsure actually how this helps? Could someone help me out I don't really know how branch cuts come into this?

答:
$\ln |z|$ is harmonic in $\Bbb C \setminus \{ 0 \}$ because it is locally the real part of any complex logarithm. Therefore
$$
 f(z) = 1 + \frac{\ln |z|}{\ln 2}
$$
is harmonic in $R$ and satisfies the given boundary conditions.


问:
Let $X$ be a metric space and $A \subset X$ a bounded subset. If $d(A)$ equals $r$, show that $A$ is a subset of some ball $\bar{B}(a,r)$.

Let $X$ be a metric space and $A \subset X$ a bounded subset. If the diameter of $A$ denoted $d(A)$  equals $r$, show that $A$ is a subset of some ball $\bar{B}(a,r)$. Give an example of a scenario where $X=\mathbb{R}^2$ and $A$ doesn't belong to any $\bar{B}(a,r/2)$.

Let $a \in A$, then I have that $\bar{B}(a,r) = \{x \in X \mid d(x,a) \leqslant r = d(A) \}$. So essentially I would need to show that for any $x \in X$ I have that the distance between $x$ and $a$ is less than or equal to the diameter of $A$? I'm slightly confused about this I want to show that $A$ belong to an closed ball that has it's radius equal to the diameter of $A$?

答:
Let $(X,\rho)$ be a metric space.
I'll denote by $B(x,r) = \{y \in X\mid \rho(x,y) < r\}$ the open ball of radius $r>0$ around $x \in X$ and by $\overline{B}(x,r)= \{y \in X\mid \rho(x,y) \le r\}$ the corresponding closed ball.

For a non-empty subset $A \subset X$ which is bounded we can define  the diameter $d(A) = \sup \{\rho(a,a')\mid a,a' \in A\}$, which is a well-defined real number precisely because $A$ is assumed to be bounded.

If $A$ is bounded non-empty with $d(A)=r$, pick some fixed $a_0 \in A$.
Then for any $a \in A$: $\rho(a_0,a) \le d(A)  =r$ by the definition of $d(A)$ as a supremum (it's an upperbound for the set of all distances between points of $A$).
So by definition, $a \in \overline{B}(a_0r)$ and so $A \subseteq \overline{B}(a_0,r)$, as required.

In some cases we can say that there is in fact some $a_0 \in A$ such that $A \subseteq \overline{B}(a_0,\frac{r}{2})$, e.g. in the plane with $A=\{(x,y)\mid x^2 + y^2 \le 1$, where $d(A)=2$ and $A= \overline{B}((0,0), 1)$ etc.

But e.g. if we take $A=\{(x,y)\mid x^2+y^2= 1\}$ we cannot find such a point in $A$ (!) such that all points lie within the closed ball of radius $1$ around it. Outside of $A$ we can in this case, but the problem as stated suggests to me that $a_0 \in A$ is meant..


问:
Minimizing trigonometric expression with the hint of substitution
I have asked this question earlier for minimization of this expression.

$$
\left[\cos ^{2}\left(\dfrac{\theta }{2}\right) +
\,\sqrt{\,{1 - \gamma}\,}\,\sin^{2}\left(\dfrac{\theta }{2}\right)\right] ^{2} + \dfrac{\gamma }{4}\,\sin^{2}\left(\theta\right)\quad
\mbox{where}\quad 0 \le \gamma \le 1.
$$

But then I am not sure if we can reach any relevant minimization (closed form) from the strategy mentioned here. What is the way to minize this with the following hint?

Hint is given in the first question itself.

答: Try starting with the half angle formulas for sine and cosine, so the entire expression will  be with respect to $\theta$. Then multiply everything out and convert the $\sin^2 \theta$ to $1 - \cos^2 \theta$. After simplifying, you can differentiate with respect to $\theta$ and find the critical values.


问:
Homeomorphic images of endpoints of the Cantor set.
A Cantor set is a homeomorphic image of the standard ternary Cantor set $T$. Suppose that we have a Cantor set $C$ on the plane. It is well know that $C$ is in fact ambiently homeomorphic to $T$, meaning that the homeomorphism $h:C\to T$ extends to a homeomorphism of the entire plane $\mathbb{R}^2$.

My question is: Suppose that $E$ is the set of endpoints of the removed intervals in the construction of $T$. Is there a property that points in $C$ must satisfy so that they get mapped to points in $E$ under $h$? In other words given a point in $C$ can someone tell, before knowing where it gets mapped under $h$, whether it will get mapped to an endpoint or not?

If the answer is no in general is there any property we could impose on any point in $C$ so that it would be impossible for it to get mapped to a point in $E$ under $h$?

Added later:
For example if the Cantor set $T$ lies on the real line then a point of $T$ that can be approximated by both sides from points in $T$ cannot get mapped by $h$ to a point in $E$ since those points can be approximated only by one side by points in $C$.

答: The Cantor set is countable dense homogeneous, so we can map the endpoints of the ternary Cantor set to any countable dense subset of a Cantor set.</p>


问:
An Euler-Lagrange Equation
I have an action with a Lagrangian which I would like to apply the Euler-Lagrange equations to https://en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation but have spent hours really struggling with it. That is I define

$$
L\Big(\begin{pmatrix}x \\ y\end{pmatrix},\begin{pmatrix} \dot{x} \\ \dot{y}\end{pmatrix} \Big):= \Big\| A \Big( \begin{pmatrix}x \\ y\end{pmatrix} +  \begin{pmatrix} -\dot{y} \\ \beta\dot{y}\end{pmatrix}  \Big) \Big\|^2
$$

where

$$
A=\begin{pmatrix} c_1 & c_2 \\ c_2 & c_3 \end{pmatrix}, \text{for some }~c_i~\text{for which $A$ is invertible}.
$$

Which $x,y:[0,T]\to \mathbb{R}$ solve the Euler -Lagrange equation

$$
\nabla_{\begin{pmatrix}x \\ y\end{pmatrix}} L-\partial_t\nabla_{\begin{pmatrix} \dot{x} \\ \dot{y}\end{pmatrix} }L=0
$$
with initial and terminal conditions $x(0)=q,x(T)=q',y(0)=p,y(T)=p'$. ? Please help !

$\Big($would more information on how the constants $c_1,c_2,c_3$ relate to eachother be useful? because infact $c_1=\sigma^2+\gamma^2, c_2=-\gamma(\sigma+\alpha),c_3=\alpha^2+\gamma^2$, where these constants $\alpha,\sigma,\gamma$ are positive and $\alpha\sigma>\gamma^2$ $\Big)$.

EDIT : I got the E.L as with the above choices for $c_1,c_2,c_3$

\begin{equation}
    \nabla_{(x,y)} L = (\frac{1}{\alpha\sigma-\gamma^2})^2\begin{pmatrix}
    0 \\
    \nabla_{y} B_1+\nabla_{y} B_2
    \end{pmatrix}
\end{equation}

with

\begin{equation}
    \nabla_{y} B_1=2(\sigma+\gamma \beta) y + \frac{1}{\sigma+\gamma\beta}(\gamma \dot{y}-\sigma \dot{x})
\end{equation}

and

\begin{equation}
    \nabla_{y} B_2=2(\gamma + \alpha \beta) y + \frac{1}{\gamma+\alpha \beta}(\alpha \dot{x}-\gamma \dot{y}).
\end{equation}

Also

\begin{equation}
    \partial_t \nabla_{(\dot{x},\dot{y})} L=2 (A^{-1})^2(\begin{pmatrix}\ddot{x}\\ \ddot{y} \end{pmatrix}-\begin{pmatrix}
    \dot{y}
    \\
    -\beta \dot{y}
    \end{pmatrix})
\end{equation}

Which is a coupled ODE

答:
Here is a sketched derivation.

First of all, notice that the Lagrangian $L$ does not depend $\dot{x}$. Then we don't need the 2 Dirichlet boundary conditions (BCs) for $x$ to deduce the EL equation for $x$. The EL equation for $x$ reduces to
$$0~=~\chi(x,y,\dot{y})~:=~\frac{1}{2}\frac{\partial L(x,y,\dot{y})}{\partial x}~=~c_1(x-\dot{y}) +c_2(y+\beta\dot{y}) \tag{1}$$
from which we can determine $x$.

If we eliminate$^1$ $x$ in the Lagrangian $L$, the Lagrangian becomes (up to an overall non-zero multiplicative normalization)
$$ L_0~=~\frac{1}{2}(y+\beta\dot{y})^2. \tag{2}$$
The momentum is
$$ p ~=~ \frac{\partial L_0}{\partial \dot{y}}
~=~\beta(y+\beta\dot{y}).\tag{3}$$
The energy function
$$ E~=~p\dot{y}-L_0~=~\frac{1}{2}(\beta\dot{y}-y)(\beta\dot{y}+y) ~=~\frac{1}{2}(\beta^2\dot{y}^2-y^2) \tag{4}$$
is a constant, since there is no explicit $t$-dependence.

In other words, we get a 1st-order ODE
$$ \beta^2\dot{y}^2~=~2E +y^2 \tag{5}$$
Eq. (5) can easily be solved via separation of variables. This produces 1 integration constant. Together with $E$ we then have 2 integration constants, which can be matched with the 2 Dirichlet BCs for $y$.

Generically, it is impossible to match 4 boundary conditions (BCc), as already noted in Cesareo's answer. It would be natural to discard the 2 Dirichlet BCs for $x$.

--

$^1$ Normally, one is not allowed to use a EL equation (1) inside the Lagrangian, but one may show that it is okay for a quadratic $x$-dependence. We can complete the square
$$ L~=~L_0+L_2 \tag{6} $$
where
$$ L_n ~\propto~ \chi^n .\tag{7}$$
It is not difficult to see that $L$ and $L_0$ lead to the same EL equation for $y$ modulo the constraint (1).

Title: An Euler-Lagrange Equation


问:
Interpreting a homotopy as a function from $[0,1]$ to a function space?
Let $X,Y$ be topological spaces. A homotopy is a continuous $H:X\times[0,1]\to Y$. Currying, we get $H:[0,1]\to X\to Y$. Is there a way to interpret $H$ as a continuous function of type $H:[0,1]\to (X\to Y)$, where $(X\to Y)$ is the space of continuous function from $X$ to $Y$?

There are a few questions.

Given topological spaces $X,Y$, is there natural space of continuous functions from $X$ to $Y$?

Given a function out of a product space $f:X\times Y\to Z$, is there a natural way to curry?

If the above doesn't work for general spaces, is it possible for $Y=[0,1]$?

答:
Q: Given two topological spaces, $X$ and $Y$, is there a natural space of continuous functions from one space to the other?

A: Yes.  It is the collection of all continuous functions from $X$ to $Y$.  (From another vantage point, one sometimes encounters an arbitrary set $X$, a topological space $Y$, and a set ${\cal F}$ of functions from $X$ to $Y$.  One can then ask, What minimal topology will make all the functions in ${\cal F}$ continuous?)

Since you call this collection of functions a ``space'', I suspect you might be looking to find out whether there is some natural (topological?) structure on this collection.  If so, yes, there is: the compact-open topology.  I think it leads to a general answer to your question about regarding a homotopy as a continuous (in some sense) mapping from $[0, 1]$ to the set of continuous mappings from $X$ to $Y$.  An efficient source to grasp this from is the book "Homotopic topology" by Fomenko, Fuchs, and Gutenmaher.