问:
Particular Integral for $x\sin2x$
$$ \frac{d^2y}{dx^2} + 4 y = x\sin2x$$
tried solving by $$ {e}^{ix} = \cos x + i\sin x $$
$$ P.I. = \frac{1}{D^2+4}xe^{2ix} $$

答:
Try the method of undetermined coefficients.  The solution to the homogeneous equation is straightforward; call it $h(x)$.  The method of undetermined coefficients then recommends trying the particular solution, where I have already found a large enough power of $x$ to use to produce a particular part that is linearly independent of the homogeneous solution,
$$  h(x) + (a+bx+cx^2)\cos(2x) + (d+ex+fx^2)\sin(2x)  \text{.}  $$

(Generically, how did I find the power of $x$?  I tried $h(x) + (a+bx)\cos(2x) + (d+ex)\sin(2x)$, the base version of the solution by the method of undetermined coefficients, in the differential equation.  The resulting coefficients, $a$, $b$, $c$, and $d$, either (1) did not have a consistent solution or (2) produced a particular solution in the span of the homogeneous solution.  Then I tried increasing the degrees of the polynomials by $1$ and found that that the coefficient equations could be solved and resulted in a particular solution not in the span of the homogeneous solution.  So this is a usable particular solution.)

Substituting this back into the differential equation, it will turn out that three or more of $a$, $b$, $c$, $d$, $e$, and $f$ are zero.

Title: Particular Integral for $x\sin2x$


问:
Python coding with numpy sympy
New to python, and I am working on a problem with the ultimate goal of being able to evaluate

$$\frac{\sqrt{x}\bigg[\prod_{k=0}^{n-1}(\sqrt{x}+2^k)- \prod_{k=0}^{n-1}(-\sqrt{x}+2^k)\bigg]}{\prod_{k=0}^{n-1}(\sqrt{x}+2^k)+ \prod_{k=0}^{n-1}(-\sqrt{x}+2^k)}$$

at whichever value of $x, n$ I choose.

First, I probably need to expand the function: how should I go about doing so? Perhaps sympy?
Then, how would I evaluate the sympy function?

ContentLicense: CC BY-SA 4.0

Title: Python coding with numpy sympy

Body:
As the infinite product $\prod_{k=0}^\infty(1 + 2^{-k} a)$ converges for all $a$, it seems better numerically to compute the expression as
$$\frac{\sqrt{x}\bigg[\prod_{k=0}^{n-1}(1 + 2^{-k}\sqrt{x})- \prod_{k=0}^{n-1}(1 - 2^{-k}\sqrt{x})\bigg]}{\prod_{k=0}^{n-1}(1 + 2^{-k}\sqrt{x})+ \prod_{k=0}^{n-1}(1 - 2^{-k}\sqrt{x})}$$
The following code does that

import numpy as np

def prod(a, n):
    res = 1.0
    y = a
    for k in range(n):
        res *= (1.0 + y)
        y /= 2.0
    return res

def expr(x, n):
    r = np.sqrt(x)
    u, v = prod(r, n), prod(-r, n)
    return r * (u - v)/(u + v)

if __name__ == '__main__':
    for x in range(10):
        print(expr(x, 30))

""" output ->
0.0
1.0
1.433872292244485
1.7430889640419365
2.0
2.2274334143759584
2.4351764585875912
2.6283079353312218
2.809840294960271
2.9817458083387267
"""


问:
Trigonometric Equation - can this be solved using complex numbers
How do I solve the following:

$\cos (12x) = 5 \sin (3x) + 9 \tan^2( x )+ \cot ^2 (x)$ for $x \in (0,360)$

I tried converting cos and sin term into single angle i.e. into x but the equation becomes messy. Any elegant solution at hand?

答:
The minimum value of $9 \tan^2 x + \cot^2 x$ is $2 \sqrt{9 \tan^2 x \cot^2 x} = 6$ by AM-GM for all real $x$, as $u^2 ≥ 0, u \in \mathbb R$. Thus $\cos(12x) - \left(5 \sin(3x) + 9 \tan^2 x + \cot^2 x \right)$ is bounded above by $g(x)=\cos(12x) - \left(5 \sin(3x) + 6 \right)$.

Now $g(x)$ is again bounded above by $1 -5 (-1) - 6 = 0$. Hence you just need to solve $g(x) = 0$, but as $\cos(12x) ≤ 1$, $5 \sin(3x) + 6 ≥ 1$ for all $x \in \mathbb R$, this implies:

$$\cos(12x) = 1, 5 \sin(3x) + 6 = 1$$


问:
Delta function $\delta_{n,m}$ with two variables $n, m$
I am trying to understand the equations below. What seems to be confusing me is the use of two variables $n, m$ in the delta function: $\delta_{n,m}$. I understand that d_n would just be the spoke at x=n, but I have no idea how to use m.

I hope the question makes sense. I'm rather new to this so excuse the potentially basic question.

Body: $\delta_{n,m}=\begin{cases}1,&n=m\\0,&n\ne m\end{cases}$


问:
All roots of the polynomial $x^n +a_{n-1} x^{n-1} + a_{n-2} x^{n-2}+...+a_1 x+a_0$ lie in the interval $(0,1)$. Prove $\sum_{k=0}^{n-2} k a_{k}>0$
The polynomial $x^n +a_{n-1} x^{n-1} + a_{n-2} x^{n-2}+...+a_1 x+a_0$ has $n$ distinct real roots in the interval $(0,1)$ where $n\geq 3$.

Prove that $$\sum_{k=0}^{n-2} ka_{k} \gt 0$$

My Attempt:
I tried using Cauchy Schwarz inequality but it isn't working out

答:
Inequality $\bf1$

For $a_k\gt0$, if $|S|\ge2$, then
$$
\prod_{k\in S}(1-a_k)\gt1-\sum_{k\in S}a_k\tag1
$$
$(1)$ follows by induction using $(1-a)(1-b)=1-a-b+ab\gt1-(a+b)$.

Answer

Let
$$
\begin{align}
p(x)
&=x^n+\sum_{k=0}^{n-1}a_kx^k\tag{2a}\\
&=\prod_{k=1}^n(x-r_k)\tag{2b}
\end{align}
$$
Then
$$
\begin{align}
p'(x)
&=nx^{n-1}+\sum_{k=0}^{n-1}ka_kx^{k-1}\tag{3a}\\
&=\sum_{j=1}^n\prod_{k\ne j}(x-r_k)\tag{3b}
\end{align}
$$
The question is equivalent to showing
$$
\begin{align}
p'(1)
&\gt n+(n-1)a_{n-1}\tag{4a}\\
&=n-(n-1)\sum_{k=1}^nr_k\tag{4b}
\end{align}
$$
Since
$$
p'(1)=\sum_{j=1}^n\prod_{k\ne j}(1-r_k)\tag5
$$
we need to show
$$
\sum_{j=1}^n\prod_{k\ne j}(1-r_k)\gt n-(n-1)\sum_{k=1}^nr_k\tag6
$$
However, if $n\ge3$, Inequality $1$ implies
$$
\begin{align}
\sum_{j=1}^n\prod_{k\ne j}(1-r_k)
&\gt\sum_{j=1}^n\left(1-\sum_{k\ne j}r_k\right)\tag{7a}\\
&=n-(n-1)\sum_{j=1}^nr_j\tag{7b}
\end{align}
$$


问:
Let $f$, $g$ be multiplicative functions, not identically $0$ such that $f(p^k)=g(p^k)$ for each prime $p$ and $k \ge 1$. Prove that $f = g$.
Let $f$ and $g$ be multiplicative functions that are not identically $0$ and such that $f(p^k)=g(p^k)$ for each prime $p$ and $k\ge1$. Prove that $f=g$.

Source: Elementary Number Theory by David M. Burton

I just know that any function is multiplicative if $f(mn)=f(m)f(n)$ where $gcd(m,n) = 1$ ,I just need some ideas or hints to solve problems like these.

答:
As the other answers have shown, $f(n) = g(n)$ for all $n > 1$ follows from prime factorisation. (You only need existence, not even uniqueness.) There was no use of $f$ and $g$ not being identically zero so far.

However, for $n = 1$, the above argument does not work. Now, we know that $$f(n) = f(n)f(1)$$ for all $n \in \Bbb N$. Since $f$ is not identically $0$, there exists $n \in \Bbb N$ such that $f(n) \neq 0$. From that, we can conclude that $f(1) = 1.$ Similarly, $g(1) = 1$. This lets you conclude that $f = g$.

What if you weren't given $f$ and $g$ are not identically zero?

Well, consider $f \equiv 0$ but $g$ to be defined as $$g(n) = \begin{cases}1 & n = 1\\ 0 & n > 1\end{cases}.$$

Then, $f(p^k) = g(p^k)$ for all primes $p$ and $k \ge 1$ (and both are (completely!) multiplicative) but $f \neq g$.


问:
Looking for old YouTube math video - and curious about what field of statistics this is
I have been looking for a YouTube video I watched quite a while ago that I found pretty fascinating.

Suppose you had a game where you had to try and pick the largest number in a list. You only get to see the value of the next number in the list and you decide if that is the number your are going to pick, or decide to see the next number. If there are 100 numbers, how many numbers should you observe before you commit to picking the next largest number you find.  I remember that the youtube video explained that you should observe some fraction of the list and the fraction was related to e somehow.

Does anyone know what type of statistics this is or possibly a youtube video on this subject?

答:
I know Numberphile did a video on it a while back. I do not know whether that's the one you're looking for.

The solution, if your goal is to maximize the probability of finding the highest number, is that you should look through the first $1/e$ fraction of the numbers, and only after that take the first number that's higher than any other you've previously seen.

They call it the "toilet problem" in that video, but I think the more conventional name is the secretary problem. That ought to help you in your search for resources.


答: This is actually the Kronecker delta, as @ShubhamJohri defined. You can prove these equations by writing each integrand as half a sum or difference. (Note these equations implicitly assume $m,\,n$ are integers.) For example, the second equation admits the proof$$\begin{align}\int_0^T\tfrac12\left[\cos\tfrac{\pi(n-m)t}{T}-\cos\tfrac{\pi(n+m)t}{T}\right]dt&=\tfrac{1}{2\pi}\left[\tfrac{\pi t}{T}\operatorname{sinc}\tfrac{\pi(n-m)t}{T}-\tfrac{T}{n+m}\sin\tfrac{\pi(n+m)t}{T}\right]_0^T\\&=\tfrac12T\delta_{nm}-0,\end{align}$$where $\operatorname{sinc}$ is the unnormalized sinc function. In particular, it helps to evaluate the leftmost contribution separately in the cases $m=n,\,m\ne n$.

Title: Delta function $\delta_{n,m}$ with two variables $n, m$

Body:
I am trying to understand the equations below. What seems to be confusing me is the use of two variables $n, m$ in the delta function: $\delta_{n,m}$. I understand that d_n would just be the spoke at x=n, but I have no idea how to use m.

I hope the question makes sense. I'm rather new to this so excuse the potentially basic question.


答:
$$a+b\geq 2\sqrt{ab}\\9tan^2x+1cot^2x\geq2\sqrt{9tan^2x\times 1cot^2x}=6$$so
$$cos(12x)-5sin(3x)\geq 6 $$Implicit $$cos(12x)-5sin(3x)\leq \max\{cos(12x)\}+\max \{ -5sin(3x)\}\leq 1+5$$ and only possibilities : $$\begin{cases}cos(12x)=1\\sin(3x)=-1\end{cases}$$

Title: Trigonometric Equation - can this be solved using complex numbers

Body:
How do I solve the following:

$\cos (12x) = 5 \sin (3x) + 9 \tan^2( x )+ \cot ^2 (x)$ for $x \in (0,360)$

I tried converting cos and sin term into single angle i.e. into x but the equation becomes messy. Any elegant solution at hand?


答:
Not much different form Toby Mak's answer but  We could infact rewrite the given equaltion in the following form  $$1-\cos 12x+5(\sin 3x+1)+{\left( 3\tan x-\cot x\right)}^2=0$$ as each terms is $\ge 0$ we must have that each term must be zero ,

can you take it from here?


问:
Separating hyperplanes of a convex cone
Let $W$ be a fixed matrix.
Define

$posW \triangleq \{t |Wy =t , y≥ 0\}$,

It is called the positive hull of $W$. It represents
the set of right-hand sides that can be obtained by a non-negative combination
of the columns of $W$ . The positive hull is easily seen to be a convex cone.

Let $p$ be a point not in the set $posW$. Then, there exist a hyperplane

$H \triangleq \{x |\sigma^Tx =0\}$ that separates $p$ and $posW$.

How can we prove that the number of possible separating hyperplanes (separating $p$  and $posW$)
is finite based on the fact that $posW$ is finitely generated?

答: The result is wrong in general. I.e. for $W=0$, $\text{pos }W= \{0\}$. And if $p \neq 0$, there is an infinite number of hyperplanes separating $p$ and $\text{pos } W$.

Title: Separating hyperplanes of a convex cone


问:
Is my interpretation of integration correct?
I thought of this when I was thinking about the $dx$ under the integral sign. So we have a function $y=f(x)$. Therefore, $dy/dx=f'(x)$, so $dy=f'(x)dx$. Now the graph of $f$ is split into small $constant$ widths called $dx$. Now we $define$ an operation $"∫"$ such that if I input a small change in $y$ over one of these constant interval of length $dx$, the operation will spit out the y-coordinate at which this change occurred, so $∫dy=y=f(x)$. Therefore, $f(x)=∫f'(x)dx$. Is this interpretation correct? Thanks!</p>

答:
It sounds like you're trying to start out by assuming the Fundamental Theorem, which while not wrong, is not a particularly good place to start. The classic intuitive interpretation of integration approaches this idea a little differently:

You want to know how much a function F(x) changes from a to b, so you sample the function f(x) representing how much F(x) is changing at x at one $x_i^*$ in each interval $[x_i, x_{i+1}]$. Over the interval $[x_i, x_{i+1}]$, F(x) changed by about  $f(x_i^*)(x_{i+1}-x_i)$.

You add up all these little changes from $x_0=a$ to $x_n=b$, which it can be shown will give you F(b)-F(a).

ContentLicense: CC BY-SA 3.0


问:
Variant of the Collatz Conjecture - Why is the following happening:
I am a web developer that programs in PHP which is limited to large math calculations.

I am also a math enthusiastic that likes learning math through problems that seem simple (but are anything like simple), and the best example is the Collatz Conjecture.

I am not dilusional and I am well aware that there is no chances for someone at my level to solve the Collatz Conjecture, but I like to try and solve other elements that may be related.

Let's begin with the description of the conjecture: According to the rules of the Collatz Conjecture if $$ is odd then execute $3+1$ and when $$ is even execute $/2$. Repeat until (supposedly) reaching $=1$.

So I was trying to prove the most basic variant: If $$ is odd then execute $+1$ and when $$ is even execute $/2$. Repeat until (supposedly) reaching =1.

I was able to proof it in my own head,

So I progressed into trying to prove: If $$ is odd then execute $+$ and when $$ is even execute $/2$. Repeat until (supposedly) reaching $=1$. So obviously sometimes it reached 1 and some times it reached .

I was able to proof it in my own head.

So I realized that I had to find a different approach for when  is odd that is more than just adding a constant.

So I decided to add $1$ for the first odd occurrence, $3$ to the second odd occurrence, $5$ to the third odd occurrence , $7$ to the fourth odd occurrence  .... (with increments of $+ 2$)

I have then ran a PHP script with a list of the first 2500 odd numbers, with a limit accepted of 10000 steps.

I immediately noticed that most numbers ended in 1, example:

$13 → 14 → 7 → 10 → 5 → 10 → 5 → 12 → 6 → 3 → 12 → 6 → 3 → 14 → 7 → 20 → 10 → e5 → o20 → 10 → 5 → 22 → 11 → 30 → 15 → 36 → 18 → 9 → 32 → 16 → 8 → 4 → 2 → 1$

but then there were numbers that had so many steps and were cut by the accepted $10000$ steps of my script.

I have then checked a specific number that seemed to have endless steps, which is the beginning number $21$.

To my surprise there were no loops but I have noticed that there seemed to be a pattern that shows growing into infinity.

Fo example the beginning number of $21$:

$21 → 22 → 11 → 14 → 7 → 12 → 6 → 3 → 10 → 5 → 14 → 7 → 18 → 9 → 22 → 11 → 26 → 13 → 30 → 15 → o34 → 17 → 38 → 19 → 42 → 21 → 46 → 23 → 50 → 25 → 54 → 27...$

I have then noticed that all the numbers that seem to have endless steps reach $6  → 3  → 10  → 5  → 14  → 7  → 18....$

Which seems to be a pattern that grows into infinity,

So my question is how can I prove that:

Starting with $6$ and then first odd occurrence ($3$) add $7$, second odd occurrence add $9$, third odd occurrence add $11$... will always grow infinitely (*Note the sequence/pattern of odd → even → odd → even...)?

Do all these numbers that seem to have endless steps, begin the endless path through $6  → 3  → 10  → 5  → 14  → 7  → 18....$ ? In other words is that the only path to infinite steps in my presented variant?

答:
This is the sequence https://oeis.org/A066070, and yes it continues indefinitely. The pattern is

$$2x \to x \to 2x+4 \to x+2 \to 2x+8 \to x + 4 \to \dots$$

Notice that whenever an odd number $n$ is in the sequence, the number $n+4$ is added to it. This can be formally proven by induction.

EDIT: incomplete answer to follow-up question: this is the only loop of this "kind". If we have:

$$n \to 2n+2a \to n+a \to 2n+3a+2 \to n+\frac32a+1\to\dots$$

we have $n+2a = n + \dfrac32 a + 1$, giving $a=2$. If we have:

$$n \to 4n + 4a \to\to n+a \to 4n+4a + 2\to2n + 2a + 1$$

we already start breaking the pattern. Therefore there are no more "short, double-arithmetic-sequence loops". Whether a larger or irregular loop exists requires some more thought. For all values that fell into a loop that I have checked, it eventually falls into this loop.


问:
Behavior of the solution to a linear system of differential equations when $t\rightarrow \infty$
Consider the system of linear DEs given below:
$$ x' =2x+y \\
y' = 3x +4y\\
x(0) =-y(0) = 1.
$$
What would be $L = \lim_{t\rightarrow \infty} ||x(t)||$?

Attempt: The given system matrix
$\begin{pmatrix}
2 & 1\\
3 & 4
\end{pmatrix}$
has eigenvalues $\lambda = 1,5$ and a corresponding pair of eigenvectors $v_1 = \begin{pmatrix}
1\\-1
\end{pmatrix}$ and
$v_5 = \begin{pmatrix}
1\\3
\end{pmatrix}$. As a result, the general solution may be written as
$$ \begin{pmatrix}
x(t)\\y(t)
\end{pmatrix} = c_1e^t \begin{pmatrix}
1\\-1
\end{pmatrix} + c_2 e^{5t} \begin{pmatrix}
1\\3
\end{pmatrix}.$$
Applying the initial conditions leads to
$$ \begin{pmatrix}
x(t)\\y(t)
\end{pmatrix} = \begin{pmatrix}
e^t \\ -e^t
\end{pmatrix}.$$

So, naturally, $L = \infty$. However, the solution manual says $L=0$ but I don't understand it's mumbo-jumbo reasoning. Can you help me figure out what's wrong with my solution?

Thank you in advance for your help!

答:
Your computations of the eigenvectors and eigenvalues seem correct.

However, the solutions you computed are the good ones in the basis of the eigenvectors. But you forgot to switch back to the canonical basis. For example, the initial conditions are given in the canonical basis.

Note: you can by the way easily verify that your « solution » of the IVP is in fact not one by computing the derivatives.


问:
If a complex number has infinitely many solutions why is it a line?
Let $s,t,r$ be non-zero complex numbers and $L$ be the set of solutions $z = x+iy \text{ },\:x,y \in \mathbb{R}$ of the equation $sz+t\bar{z}+r=0$ where $\bar{z} = x-iy$.

Prove the number of elements in $L \cap \{z: |z-1+i|=5\}$ is at most $2$.

What they did in the solution is say $z(|s|^2-|t|^2) = \bar{r}t-r\bar{s}$. They did this by taking conjugate of the original equation and eliminating $\bar{z}.$

Now how is $L \cap \{z: |z-1+i|=5\}$ is at most $2$?

If $|s| \neq |t|$, $z = \dfrac{\bar{r}t-r\bar{s}}{(|s|^2-|t|^2)}$ which is one solution and it can intersect the circle once. If $|s| = |t|$ and  $\bar{r}t-r\bar{s}\neq0$ then $L$ is empty. But if $|s| = |t|$ and  $\bar{r}t-r\bar{s}=0$, then $z\cdot0=0 \implies$ $z$ has infinitely many solutions. Why is $L$ a line in this case? Isn't it the entire plane? Why does it intersect the circle at most twice?

Here Z Ahmed told its a line but I can't seem to wrap my head around why this is the case.

答:
Your equation is never true on the entire plane (as you've stated the hypothesis that the coefficients $s,t,r$ are nonzero), otherwise, you've defined the function $f(z) = z^{*}$ (* denotes conjugate) as a linear function $z$, which is absurd, not least because $z^{*}$ isn't holomorphic.

You've incorrectly stated -- given a simultaneous system, the equation yielded by subtracting one equation from another provides a complete description of the solution set, but clearly it just gives a new constraint. (well, to be exact not new, just incomplete) Therefore, in the case $|s|=|t|$, you found this new constraint tells you nothing new at all, which just means you have to go back to the original system and solve it some other way.

Perhaps the easiest way to see the solution is a point or a line is to expand the equation into its real and imaginary parts, i.e. you get two linear equations in $x$,$y$. Their solutions are lines. It turns out when $|s|=|t|$, those two lines coincide, and when $|s|\neq |t|$, they do not, therefore those two lines are parallel or intersect at a point. In fact, you've shown they must intersect at a point. To see why $|s|=|t|$ is the right condition for those two lines to coincide, consider a geometric description of a line in the Argand diagram.


问:
Solving a Second Order Linear Homogeneous ODE Without an Ansatz/Characteristic Polynomial.
I wanted to try solving the following differential equation:
$$y''+ay'+by=0$$
without using the ansatz of $y=e^{kt}$. This means using separation of variables, substitutions, or an integrating factor to possibly find the solution. If we let $a=0$, it becomes easier to find a solution with that constraint:
$$y''=-by$$
$$\int y''\cdot y'dt=-b\int y\cdot y'dt$$
$$\left(y'\right)^{2}=-by^{2}+c$$
$$\frac{dy}{dt}=\sqrt{c-by^{2}}$$
$$\frac{1}{\sqrt{c}}\int \frac{dy}{\sqrt{1-\frac{b}{c}y^{2}}}=\int_{ }^{ }dt$$
let $\omega=\sqrt{b}$, and $R=\sqrt{\frac{c}{b}}$:
$$\frac{1}{\omega}\sin^{-1}\left(\frac{1}{R}y\right)=t+D$$
let $D\cdot\omega=\delta$
$$y(t)=R\sin\left(\omega t+\delta\right)$$
In fact any equation of the form
$$y''=-by^{n}$$
can be simplified into a solution containing an indefinite integral at least (since u sub can be done on the right).
If I tried the same technique, but when a is nonzero. I end up with:
$$\int_{ }^{ }y''\cdot y'dt+\int_{ }^{ }a\left(y'\right)^{2}dt+\int_{ }^{ }by\cdot y'dt=0$$
I don't know how to integrate the $\left(y'\right)^{2}$ term though. I then tried using an integration factor. I wrote the equation as:
$$-\frac{1}{a}y''=y'+\frac{b}{a}y$$
let $\mu\left(t\right)=e^{\frac{b}{a}t}$. By doing some rearranging:
$$-\frac{1}{a}\int_{ }^{ }e^{\frac{b}{a}t}y''dt=e^{\frac{b}{a}t}y$$
I don't know how to the integral on the left either :S

I know that the actual solution ends up being an exponential function times a trig function (through the ansatz method, which is admittedly much faster), but I would like to derive the solution without using that method. Thanks

EDIT: I think I specifically want to preserve the part where you integrate to get arcsin. Using an ansatz, and then eulers method sort of gets around that step. I guess I wanted to see specifically where that trig function comes in, and that was easy to see when a=0.

答:
Let set $\begin{cases}\omega_1+\omega_2=-a\\\omega_1\omega_2=b\end{cases}\quad$ solutions of sum and product $x^2+ax+b=0$

The ODE can be rewritten $$y''-(\omega_1+\omega_2)y'+\omega_1\omega_2y=0\iff (\underbrace{y''-\omega_1y'}_{z'})=\omega_2(\underbrace{y'-\omega_1y}_z)$$

You can solve $z'=\omega_2 z$ then go on solving $y'-\omega_1y=z$ with RHS $z$.


问:
Find the Pattern/Sequence
(1+2)

(1+2)(4+2)

(1+2)(4+2)(7+2)

(1+2)(4+2)(7+2)(10+2)

I can see that each term in the sequence is multiplied by an additional term that is 3 greater than the previous 2 terms.

How can I represent this as a sequence? (e.g. a_n= f(n) )

Title: Find the Pattern/Sequence

Body: $$a_n=\Pi_{k=0}^{n} (1+2)\cdot (4+2) \cdot (7+2) \cdot \cdot\cdot\cdot\cdot ((3k+1)+2) $$


问:
If two matrices A and B such that that AB is invertible, but A and B are not, then A and B can't be square matrix?
"Find two matrices A and B that AB is invertible, but A and B are not. Hint:square matrices A and B would not work."

I was able to complete this problem easily finding simple rectangle matrix, but I'm curious as to why A and B cannot be square matrix?

答: If $A$ and $B$ are square matrices they must be of the same order (so that $AB$ is defined). But then $0 \neq det(AB)=det(A) det (B)$ which implies $det (A) \neq 0$ and $det(B) \neq 0$. So $A$ and $B$ are both invertible.


问:
What makes $5$ and $6$ so special that taking powers doesn't change the last digit?
Why are $5$ and $6$ (and numbers ending with these respective last digits) the only (nonzero, non-one) numbers such that no matter what (integer) power you raise them to, the last digit stays the same? (by the way please avoid modular arithmetic) Thanks!</p>

答:
You want $x^2 - x$ to be a multiple of 10. Since $x^2-x = (x-1)x$, and $x-1$ and $x$ are coprime, we have the following cases:

Case 1: $x$ is a multiple of 2. $x$ is a multiple of 5.

In this case, $x$ is a multiple of 10, and ends with 0.

Case 2: $x$ is a multiple of 2. $x-1$ is a multiple of 5.

You can show that this is satisfied when $x$ leaves a remainder of 6 when divided by 10.

Case 3: $x-1$ is a multiple of 2. $x$ is a multiple of 5.

you can show that this is satisfied when $x$ leaves a remainder of 5 when divided by 10.

Case 4: $x-1$ is a multiple of 2. $x-1$ is a multiple of 5.

You can show that this is satisfied when $x$ leaves a remainder of 1 when divided by 10.


答: Take $W=I$, then the positive hull is $Q=\{x | x \ge 0 \}$. Take $p= (-1,...,-1)$ and note that the hyperplanes generated by $H_q = \{ x | q^Tx = 0 \}$, for any non zero $q \in Q$ (which is also the dual cone of $Q$) all separate $p$ from $Q$. There are infinitely many of these.

Title: Separating hyperplanes of a convex cone

Body:
Let $W$ be a fixed matrix.
Define

$posW \triangleq \{t |Wy =t , y≥ 0\}$,

It is called the positive hull of $W$. It represents
the set of right-hand sides that can be obtained by a non-negative combination
of the columns of $W$ . The positive hull is easily seen to be a convex cone.

Let $p$ be a point not in the set $posW$. Then, there exist a hyperplane

$H \triangleq \{x |\sigma^Tx =0\}$ that separates $p$ and $posW$.

How can we prove that the number of possible separating hyperplanes (separating $p$  and $posW$)
is finite based on the fact that $posW$ is finitely generated?


问:
The mean of the integral of an adapted process
I am a newbie in the stochastic theory. I read in a lecture note that the mean of an integral of an adapted process is zero. For instance,
Let
$$
X_{t}=\int_{0}^{t} s \mathrm{~d} B_{s}
$$
So, we have$\mathbb{E}\left(X_{t}\right)=0.$

I cannot make an argument on this.

答: If $f$ is step function it is easy to see that $\int f(s)dB_s$ has mean $0$. [The integral is a finite sum of terms of the type $c(B_v-B_u)$ where $c$ is a constant and $0 \leq u <v$]. By definition of the stochastic integral you can write the given intergal as  a limit in mean square (hence also in mean) of integrals of this type. Hence $E\int_0^{t}sdB_s=0$.


答:
If possible let, $A,B$ are square matrices of order $n$.( They will have the same order, otherwise $AB$ is not possible when $A,B$ are square matrices).

As, $rank(AB) \le min\{rank(A),rank(B)\} $

If one of $A,B$ be not invertible then $rank(AB)\le (n-1)$

So, then $AB$ can't remain invertible.

Hence , atleast one of $A,B$ should be not square matrix.

Title: If two matrices A and B such that that AB is invertible, but A and B are not, then A and B can't be square matrix?

Body:
"Find two matrices A and B that AB is invertible, but A and B are not. Hint:square matrices A and B would not work."

I was able to complete this problem easily finding simple rectangle matrix, but I'm curious as to why A and B cannot be square matrix?


问:
Sum of squares of the dimension of irreducible characters and conjugacy classes
While working through a prior problem on classifying a group of order $625$, I stumbled upon (by virtue of a mistaken answer by a user), the following problem. Since the dimension of irreducible representations divide the order of a group and have their sum of squares equal to the order of the group, how can we determine the number of conjugacy classes of the $625$ order group with center of order $25$?

Specifically, the center of the group would correspond to $25$ trivial representations of dimension $1$. Now, how to show that there are $120$ other irreducible representations using only representation/ character properties, as we must have $625=1^2A+5^2B+25^2C+125^2D+625^2E$, ($A,B,C,D,E$ are number of irreducible representations of degrees $1, 5, 25, 125, 625$ respectively) which does not give us anything special? Any hints? Thanks beforehand.

答:
It is not the center, but the index $|G:G'|$ that corresponds to the number of linear characters.

If $|G|=625$ and $Z(G)=25$, then for every $g \in G-Z(G)$ we must have $|Cl_G(g)|=5$. How can you see this? Well, if $g$ is non-central, then $Z(G) \subset C_G(g) \subset G$, where the inclusions are all strict. So the only way this can happen is when $|Cl_G(g)|=|G:C_G(g)|=|C_G(g):Z(G)|=5$. Hence the class equation amounts to $625=25+5a$, whence $a=120$ and $k(G)=25+120=145$ is the number of conjugacy classes.

Note that $|G/Z(G)|=25$, so $G/Z(G)$ is abelian, hence $G' \subseteq Z(G)$. $G'$ cannot be trivial ($G$ is not abelian), so $|G:G'|=25$ or $=125$.

Now, if $|G|=625=5^4$, $\chi \in Irr(G)$, non-linear, then $\chi(1) \mid 625$ and $1 \lt \chi(1)^2 \lt 625$, hence $1 \lt \chi(1) \lt 25$. So $\chi(1)=5$.

If $G'=Z(G)$, then $|G:G'|=25$ and we would have $625=25+25a$, so $a=24$ and thus $k(G)=24+25=49$. This rules out $|G:G'|=25$.

So $|G:G'|=125$ and a similar calculation shows that the number of non-linear irreducible characters is $20$ and indeed $k(G)=20+125=145$ as calculated above.


问:
Showing the Fibonacci inequality $f_1^{f_1}f_2^{f_2}f_3^{f_3}\cdots f_n^{f_n}\leq f_1!f_2!f_3!\cdots f_n!\;e^{({f_{n+2}-n-1)}}$ without induction.
Today I saw the following 2 beautiful inequalities on a facebook page.

$$1^12^23^34^4\cdots n^n\leq 1!2!3!4!\cdots n!e^{\frac{n(n-1)}{2}}$$

$$f_{1}^{f_{1}}f_{2}^{f_{2}}f_{3}^{f_{3}}\cdots f_{n}^{f_{n}}\leq f_{1}!f_{2}!f_{3}!\cdots f_{n}!e^{({f_{n+2}-n-1)}}$$

Here $f_{n}$ denotes the fibonacci numbers $f_{1}=f_{2}=1$ and $f_{n+2}=f_{n+1}+f_{n}$ for $n\in N$

Here is How I proved the first one.

Method 1:-

$$1^12^23^34^4...n^n\leq 1!2!3!4!...n!e^{\frac{n(n-1)}{2}}$$

$$\implies\bigg(\frac{1^1}{1!}\bigg)\bigg(\frac{2^2}{2!}\bigg)\bigg(\frac{3^3}{3!}\bigg)...\bigg(\frac{n^n}{n!}\bigg)\leq e^{\frac{n(n-1)}{2}}$$

$\implies \displaystyle\prod_{x=1}^{n} \frac{x^x}{x!}\leq
e^{\frac{n(n-1)}{2}}$

Now taking logarithm on both sides of inequality.

$\implies \displaystyle\sum_{x=1}^{n} \ln\bigg(\frac{x^x}{x!}\bigg)\leq
{\frac{n(n-1)}{2}}$

$\implies \displaystyle\sum_{x=1}^{n} \ln\bigg(\frac{x^x}{x!}\bigg)\leq
{\frac{n(n+1)}{2}}-n$

$\implies \displaystyle\sum_{x=1}^{n} \ln\bigg(\frac{x^x}{x!}\bigg)\leq
\bigg(\sum_{x=1}^{n} x \bigg)-n$

$\implies \displaystyle \ \sum_{x=1}^{n} x -\displaystyle\sum_{x=1}^{n} \ln\bigg(\frac{x^x}{x!}\bigg)\geq n$

$\implies \displaystyle \ \sum_{x=1}^{n} x + \ln\bigg(\frac{x!}{x^x}\bigg)\geq n$

$$\implies 1+1.30+1.49+1.63+1.74+1.82+1.90+1.97+2.02+...\geq n$$

$\implies 1+(1+0.30)+(1+0.49)+(1+0.63)+(1+0.74)+...\geq  {\smash[b]{1+\! \underbrace{1+1+\cdots1\,}_\text{$n$ times}}}$

This proves our first inequality.

Method 2:-

Using $\frac{n^n}{n!}\leq e^{n-1}$ for $n\in N$.

$\frac{1^1}{1!}\leq e^{0}$

$\frac{2^2}{2!}\leq e^{1}$

$\frac{3^3}{3!}\leq e^{2}$

$\vdots\\$

$\frac{n^n}{n!}\leq e^{n-1}$

Now multiply all above inequalities we get,

$\implies \displaystyle\prod_{x=1}^{n} \frac{x^x}{x!}\leq
e^{\frac{n(n-1)}{2}}$

Using the same approach for $2^{nd}$ inequality we get,

$$\sum_{x=1}^{n} \ln\bigg(\frac{{f_{x}}^{f_{x}}}{f_{x}!}\bigg)\leq f_{n+2}-n-1$$

Now I got stuck on this step.

How can we prove the $2^{nd}$ inequality. Moreover can we show without numerically calculating(as I do above) that $\displaystyle \ \sum_{x=1}^{n} x + \ln\bigg(\frac{x!}{x^x}\bigg)\geq n$

答: The case $n=0$ reduces to $1\le e^{f_2-1}$, as its left-hand side is an empty product. Increasing $n$ from $k$ to $k+1$ multiplies the left- (right)-hand side by $f_{k+1}^{f_{k+1}}$ ($f_{k+1}!e^{f_{k+1}-1}$), so just prove $f_{k+1}^{f_{k+1}}\le f_{k+1}!e^{f_{k+1}-1}$. Indeed, you can prove by induction $m^m\le m!e^{m-1}$ for $m\ge1$. In particular,$$u_m:=\frac{m!e^{m-1}}{m^m}\implies u_1=1,\,\frac{u_{m+1}}{u_m}=\frac{e}{(1+1/m)^m}>1.$$


问:
Length of submodule
Does there exists a module $M$ of a finite length(greater than $3$) over a ring $R$ such that each submodule of $M$ has different length?(not a simple module)

Let $M$ be a module and $0=M_0\subset M_1 \subset M_2 \subset ......\subset M_n =M$ is chain of submodules of $M$ we say that $n$ is the length of the chain. The length of module $M$ is the largest length of any of its chains.

In order to get such module, what I think is there must be only one such chain exists of largest length in module and all submodules appear in that chain.

I am looking for example of module of length greater than $3$ such that each submodule has different length.

答:
Let $k$ be a field and consider the ring $k[x]/(x^3)$. The regular module $k[x]/(x^3)$ has length 3 and contains a unique submodule of length $r$ for $r=0,1,2,3$.

This can be generalized to any $n$, not just $n=3$. A module having a unique composition series is called uniserial. A ring is uniserial if each (finite length) indecomposable module is uniserial. Examples include all (not necessarily commutative) dvrs.

If there is a unique module of each length, then there is a unique simple submodule, each nonzero submodule contains this simple, and by induction on length we see that the module is uniserial.

Title: Length of submodule


问:
Does a semicircle have 2 right angles?
If you have a line passing through the middle of a circle, does it create a right angle at the intersection of the line and curve?

More generally, is it valid to define an angle created between a line and a curve? Is the tangent to the curve at the point of intersection a valid interpretation (I.e a semi circle has 2 right angles)

I saw it in a debate thread and it got me curious now.

答:
More generally, is it valid to define an angle created between a line and a curve?

Yes. It can be done with a little calculus. Calculus gives us a workable definition of a unique tangent vector at each point on a curve. So, we can use calculus to translate questions about curves into equivalent questions about lines.

The line $\overline{AH}$ crosses the circle at $A$ and $H$. It is intuitively plausible that the line $\overline{GH}$ from the point $G$ on the circle to the point $H$ becomes perpendicular to $\overline{AH}$ as $G$ approaches $H$. If there is a line through the circle perpendicular to $\overline{AH}$ then it must be the line that $\overline{GH}$ approaches as $G$ goes to $H$. It is possible to rigorously define such a line using the mathematically precise definitions of limit and derivative from calculus.

The equation for the curve of the circle in polar coordinates is

$(x(\theta), y(\theta))=(r\cos(\theta), r\sin(\theta))$ with $r$ is constant.

The equation for the line through the circle in polar coordinates is

$(x(r), y(r))=(r\cos(\theta), r\sin(\theta))$ with $\theta$ is constant.

The vector tangent to the circle at $\theta$ is  $\vec{a}=lim_{h\rightarrow 0} \frac{(x(\theta+h), y(\theta+h))-(x(\theta), y(\theta))}{h}$.

This is a mathematically rigorous definition of the notion of the tangent to the curve at a point.

$\vec{a}=\frac{d(x(r,\theta),y(r\theta))}{d\theta}=\frac{d(r\cos(\theta),(r\sin(\theta))}{d\theta}=(-r\sin\theta , r\cos\theta)$

Similarly, the vector tangent to the line at $r$ is

$\vec{b}=\frac{d(x(r,\theta),y(r\theta))}{dr}=\frac{d(r\cos(\theta),(r\sin(\theta))}{dr}=(\cos \theta , \sin\theta)$

The Dot Product.

$\vec{a} \cdot \vec{b}= |\vec{a}| |\vec{b}| \cos(\phi)$ where $\phi$ is the angle between $\vec{a}$ and $\vec{b}$.

$\implies$

$\vec{a} \cdot \vec{b}=0 \rightarrow \vec{a} \perp \vec{b}$.

$\vec{a} \cdot \vec{b} = -r \sin \theta \cos\theta + r \cos \theta \sin \theta=0$

Therefore, the vectors are orthogonal.

Assume the line is crossing the unit circle where  $\theta=0$ then

$a=(0,1)|_{\text{where the line crosses}}$ and $b=(1,0)|_{\text{where the line crosses}}$.

And you can always define any circle to be a unit circle by changing your units of measurement. So this particular choice of $a$ and $b$ really holds for any circle.

Title: Does a semicircle have 2 right angles?


问:
Finding an exponential generating function in it's 'closed' form
I was going through an article on counting principles and came across exponential generating functions.

 is defined recursively as -

,

This is the exp. generating function -

When differentiated, I was confused as to how this -

Changes to this -

Here are the complete steps before the differential equation part (for your reference) -

Could anyone help me here? I know how to differentiate, but this specifically wasn't clear to me.

Note: My markdown editor is acting up at the moment, so I'll update the post using MathJax.

答:
Line (2) is application of the recurrence for $I$.

In line (3), write down the first few terms of the first series to help you see what is going on:
$$  \frac{I(0)t^0}{0!} + \frac{I(1)t^1}{1!} + \frac{I(2)t^2}{2!} + \cdots  \text{,}  $$
which is $\theta(t)$

To get to line (4), pull out one of the factors to $t$ so the argument to $I$, the power of $t$, and the argument to the factorial are all the same, to try to do the same thing we did on the previous line.  ... which works.

Another way to see what is going on is to change the index.  For
$$  \sum_{k \geq 1} \frac{I(k-1)t^{k-1}}{(k-1)!}  $$
set $j = k-1$.  Since $k$ ranges over $[1,\infty)$, $j$ ranges over $[0, \infty)$.  We also have $k = j+1$, so \begin{align*}
\sum_{k \geq 1} \frac{I(k-1)t^{k-1}}{(k-1)!}
    &= \sum_{j \geq 0} \frac{I((j+1)-1)t^{(j+1)-1}}{((j+1)-1)!}  \\
    &= \sum_{j \geq 0} \frac{I(j)t^{j}}{j!}  \\
    &= \theta(t)  \text{,}
\end{align*}
by definition.  Pulling out a power of $t$ from the second sum and making the substitution $j = k-2$ does the same thing with the second sum.


问:
Set of cylinder sets is closed under finite intersections
Notation: In what follows, $T$ is an arbitrary parameter set and $(X, \mathbb{X})$ is a measurable space. Moreover, the $n$-fold cartesian products of $X$ (resp. $T$) with itself is denoted by $X^{n}$ (resp. $T^{n}$). When $X^{n}$ is considered, the associate $\sigma$-algebra $\mathbb{X}^{n}$ is just the usual product $\sigma$-algebra. Finally, the set of all functions $f: X \to T$ is denoted by $X^{T}$.

Definition: Let $t:=(t_{1},...,t_{n}) \in T^{n}$ and $B \in \mathbb{X}^{n}$. We define a cylinder set with coordinates $t_{1},...,t_{n}$ and basis $B$ by:
$$C_{n}(t, B) := \{f \in X^{T}: f(t) := (f(t_{1}),...,f(t_{n})) \in B\}$$
Moreover, the set of all such cylinder sets is denoted by $\mathcal{A}$. In other words:
$$\mathcal{A} := \{C_{n}(t, B): \hspace{0.1cm} t \in T^{n}, B\in \mathbb{X}^{n}, n \in \mathbb{N}\}$$

I would like to prove that $\mathcal{A}$ is an algebra of subsets of $E^{T}$. It is easy to see that $\emptyset, E^{T}\in \mathcal{A}$.

[$\mathcal{A}$ is closed under complementations] Given $B \in \mathbb{X}^{n}$ and $\tau = (t_{1},...,t_{n})\in T^{n}$, we can see that $C_{n}(\tau, B)^{c} = C_{n}(\tau, B^{c})$. In fact
\begin{align}
    C_{n}(\tau,B)^{c} &= \{f\in X^{T}: \hspace{0.1cm} f(\tau) := (f(t_{1}),...,f(t_{n}))\not\in B\} \\
    &= \{f\in X^{T}: \hspace{0.1cm} f(\tau) := (f(t_{1}),...,f(t_{n}))\in B^{c}\} \\
    &= C_{n}(\tau, B^{c}). \nonumber
\end{align}
Since $B^{c} \in \mathbb{X}^{n}$, it follows that $C_{n}(\tau, B)^{c} = C_{n}(\tau, B^{c}) \in \mathcal{A}$.

My problem is proving $\mathcal{A}$ is closed under finite intersections. Any help is appreciated! Thanks.

答:
Observe that if  $\tau=\{t\}$ and $\tau'=\{t'\}$ and $B,B' \in \mathcal{B}(\mathbb{R})$ with $t<t'$,then we have :

$C(\tau,B)\cap C(\tau',B') = C(\{t,t'\},B\times B')\in \mathcal{A}$

and if $t=t'$ then :

$C(\tau,B)\cap C(\tau',B') = C(\tau,B\cap B')\in \mathcal{A}$

From those two observations you can build any intersection of set in $\mathcal{A}$ and observe that they still in $\mathcal{A}$.


问:
Coupling defined upon marginals with different mass
Let us say we define a coupling of two measures $\mu,\nu$ (reference : http://websites.math.leidenuniv.nl/probability/lecturenotes/CouplingLectures.pdf). It seems that a coupling can't be defined if $\int d \mu \neq \int d \nu$. For example, it is said here in the definition of a Frechet class  which seems to me to be the class of all couplings : https://warwick.ac.uk/fac/sci/statistics/staff/academic-research/kendall/personal/talks/CouplingReview-handout-2x2.pdf

my question is why so?

答: The answer comes down to the definition of a coupling. One wondering the answer should look up the definition of product space, marginals and projection. Also, intuitively, how would one couple two measures with different masses?</p>


问:
Understanding the symmetries of f(x)=7x (mod 12)
I found something interesting while studying the circle of fifths.

Define a map $f(x) = 7x$ (mod 12), this models the circle of fifths as a table. The input is the location on the circle of fifths, the output is the note at that location on the circle of fifths.

You could also read the input as the number of sharps or flats in the scale. So $f(2) = 14 \equiv 2$ (mod 12) reads "the scale with two sharps is D" or "two o'clock on the circle of fifths is D". And $f(-1) = -7 \equiv 5$ (mod 12) reads "the scale with one flat is F".

The function f is an involution, because $f(f(x)) = 7(7x) = 49x= (48 + 1)x \equiv x$ (mod 12).

While pairing up the elements, a beautiful relation appeared:

A line between two numbers means those numbers map to each other, no line means the number maps to itself. It's easy to verify this picture, Proof:

Evens are fixed because:
$f(2n) = 7(2n) = 14n = (12 + 2)n \equiv 2n$ (mod 12)

Odds map to the opposite side of the circle because:
$f(2n + 1) = f(2n) + f(1) = 2n + f(1) = 2n + 7 = (2n + 1) + 6$

My goal is to understand the above picture better, I'm open to general suggestions about how to do this, but I got stuck trying something very specific.

I'm trying to articulate the question "Prove the symmetries of this function are $D_{12}$" I've begun by defining:
$i(n) = n$ (mod 12)
$r(n) = n + 2$ (mod 12)
$s(n) = -n$ (mod 12)
These satisfy: $r^6 = s^2 = (rs)^2 = i$ which make them generators of $D_{12}$.

I'm struggling to continue my argument, here is my current attempt:

Now let $D_{12}$ act on $\mathbf{H}(\mathbf{Z}/12\mathbf{Z}, \mathbf{Z}/12\mathbf{Z})$ by conjugation. So for $g \in \mathbf{H}(\mathbf{Z}/12\mathbf{Z}, \mathbf{Z}/12\mathbf{Z})$ and $p \in D_{12}$ we have
$p \cdot g = p g p^{-1}$.

Now we can show $D_{12}$ fixes $f$, it follows from $r \cdot f (x) = r f r^{-1} (x) = - f(- x) = - 7(- x) = 7x = f(x)$ and $s \cdot f(x) = s f s^{-1} (x) = f(x - 2) + 2 = f(x) - 2 + 2 = f(x)$

I have a handful of misgivings with my thoughts so far. I don't know what larger group should be thought of as containing $D_{12}$. And $f$ being a fixed point of a group doesn't prove it isn't a fixed point of some larger group.

Right now, my current approach for a fix of this is to prove that the stabilizer subgroup of the permutations with respect to $f$ equals the dihedral group: ${S_{12}}_f = D_{12}$.

Is there a more specific ambient space than just permutations though? Aren't I cheating by applying an arbitrary permutation to a ring isomorphism. And that's what my map is right? Or is it actually a $Z/12Z$ - module isomorphism? I feel very stuck on what I think are some pretty basic concepts.

I have a strong intuition for the symmetries of this circle now, For example from the above it's clear that f having one flat is equivalent to b having five sharps. But I'm struggling translating these intuitions into algebraic proofs.

I would appreciate any help incorporating $D_{12}$ into wherever it belongs.

Best,

Edit: I tried to clarify my definition.

I'm getting suggestions that I either meant, or should be using the function $m(k) = k + 7$ (mod 12). I labeled it "$m$" for modulate. It has the advantage that the inputs and the outputs are both semitones. It's related to my function by:

$f(n) = m^n(0)$

Or "the nth location on the circle of fifths is labeled by the note you get by finding n fifths from C"

答:
A partial answer:

Your clock picture is saying that for $x$ odd then $7x \equiv x+6\bmod 12$.

This is clearly true $\bmod 3$ (for all $x$) so it remains to check $\bmod 4$ where this is clearly true again (for $x$ odd, and it fails for $x$ even).

From the music perspective it is just saying that $2n$ perfect fifts = $2n$ semitones.

I don't see any reason to restrict to $x$ odd and to seek $D_{12}$ symmetries in there. $D_{12}$ is a subgroup of $Aff(\Bbb{Z/12Z})$ and the natural embedding of $D_6$ acts on $\Bbb{2Z/12Z}$ thus on your $6$ points.


答:
Yes. The semi-circle can even be referred to sometimes as a ( Curvilinear) Diangle, sum of the two shown right angles is $\pi$.

If Q and P are opposite points of a diameter in a circle then the tangent at Q makes a right angle to the line PQ. Similarly tangent at P makes a right angle to the line QP.

The question is about two possibilities of the "diangle" and these two cases are possible, included in a rough sketch. The second region however cannot be be called a semicircle, but a lens shape etc..

Title: Does a semicircle have 2 right angles?

Body:
If you have a line passing through the middle of a circle, does it create a right angle at the intersection of the line and curve?

More generally, is it valid to define an angle created between a line and a curve? Is the tangent to the curve at the point of intersection a valid interpretation (I.e a semi circle has 2 right angles)

I saw it in a debate thread and it got me curious now.


问:
Motivation and some examples of Quantifier Rules
I am reading A Friendly Introduction to Mathematical Logic by  Christopher Leary and Lars Kristiansen. The authors define rules of inference of type (QR):

Definition. Suppose that the variable $x$ is not free in the formula $\psi$. Then both of the following are rules of inference of type (QR):

$$\left( \{ \psi \to \phi \}, \psi \to \left( \forall x\phi \right) \right)$$
$$\left( \{ \phi \to \psi \}, \left( \exists x\phi\right)  \to  \psi \right)$$

The authors write the following as a motivation which is intuitive but I cannot seem to relate the motivation and the definition:

The motivation behind our quantifier rules is very simple. Suppose, without
making any particular assumptions about x, that you were able to prove
“x is an ambitious aardvark.” Then it seems reasonable to claim that you
have proved “(∀x)x is an ambitious aardvark.” Dually, if you were able
to prove the Riemann Hypothesis from the assumption that “x is a bossy
bullfrog,” then from the assumption “(∃x)x is a bossy bullfrog,” you should
still be able to prove the Riemann Hypothesis.

This is intuitive but how does all this motivate the definition? Also, can someone explain the requirement of $x$ not being free in $\psi$ in the definition?

答:
See Universal generalization.

The rule formalize the intuitive sound argument that, if we can assert that a "generic" object $x$, i.e. an object whatever, has property $P$, then everything has that property.

The "generic" restriction must be expressed with the restriction that nothing must be asserted in the "context" [the premises] of the argument about the object $x$:

Assume that $\Gamma$  is a set of formulas, the set of premises, and that $\Gamma \vdash \varphi(x)$ has been derived. The generalization rule states that $\Gamma \vdash \forall y \varphi (y)$ can be derived, provided that $x$ is not used [is not free] in $\Gamma$.

Here are examples of invalid use of Generalization: "Socrates is a Philosopher, therefore everyone is a Philosopher".

The above fallacy shows that $P(x) \vdash \forall y P(y)$ is invalid.

Why so? Because we have not satisfied the restriction in the formulation of the Generalization rule: the variable $x$ occurs free in the premise of the argument ($\Gamma = \{ P(x) \}$).

The second rule is the contraposed of the first one; thus the "motivation" is the same.


问:
Normal open subgroup of $\operatorname{Gal}(L / K)$
Let $L / K$ be Galois extension and $H$ be a subgroup of $\operatorname{Gal}(L / K)$.
Suppose that $K$ is a subfield of $L$ that is invariant by $H$.
Show that $HN = \operatorname{Gal}(L / K)$ for any normal open subgroup $N$ of $\operatorname{Gal}(L / K)$.

答:
As it is your question doesn't make sense. My guess to repair it:

Let $L / K$ be an infinite Galois extension and $H$ a subgroup of $\operatorname{Gal}(L / K)$.
$L^H$ is the subfield of $L$ fixed by $H$.
Show that $HN = \operatorname{Gal}(L / L^H)$ for any normal open subgroup $N$ of $\operatorname{Gal}(L / L^H)$.

Proof: $N=\operatorname{Gal}(L/L^N)$ and $L^N/L^H$ is a finite extension and $L^H$ is the subfield fixed by $HN/N$, whence $\operatorname{Gal}(L/L^H)/\operatorname{Gal}(L/L^N)=\operatorname{Gal}(L^N/L^H)=HN/N$ so that $HN\to\operatorname{Gal}(L/L^H)$ is surjective ie. $HN=\operatorname{Gal}(L/L^H)$.


问:
How to turn this algorithm into something without a loop
I wasn't sure if this was a better place than Stack Overflow to ask this question.  On a basic level, I have an algorithm I've written in Java to compute a number, but I feel like there's a better way to write it without a loop that just involves use of floor and things like that.

So here's my algorithm:

```<code>float startAngle = getRotation();
double dAngle = (startAngle < 0 ? -Math.PI : Math.PI);
double endAngle = (float)((destination % 2 == 0) ? 0 : dAngle/2);
while (Math.abs(endAngle - startAngle) > Math.PI/2)
  endAngle += dAngle;
</code>```

In simpler terms it does the following:

Get the current angle of the object

Get a delta angle based on whether the object has been rotated left or right (further left or further right)

Define an angle to rotate the object to, based on whether the destination should be rotated parallel or perpendicular to 0

As long as the difference between the end angle and start angle is greater than 90 degrees, change the end angle by the delta from step 2.

The reason behind all this is I've got playing cards on the screen at random angles, which any of four players can take (where destination 0 is the bottom of the screen, 1 is left, 2 is top, and 3 is right).  I want to rotate the playing cards the shortest distance to face that player as they take them.  This algorithm works correctly, but like I said before, it seems like there should be a way to do this without a loop.  Is there?

答:
Once you have an angle you want to rotate, you are trying to reduce it to the range $(-\pi,\pi)$.  I don't know if Java will do modulo on reals, nor what it returns if you do a%$\pi$ for $a \lt 0$.  It looks like dAngle is in the range $(-2\pi,2\pi)$ in which case you can do
it with if statements:

if dAngle < -PI: dAngle +=2PI

elseif dAngle > PI:  dAngle-=2PI

endif


问:
Suppose $AB=0$ for some non-zero matrix $B$. Can $A$ be invertible?
"Suppose $AB=0$ for some non-zero matrix $B$.  Can $A$ be invertible?"

I think $A$ is not invertible by proof by contradiction. Assume $A$ is invertible, then $A$ has a left inverse. Multiplying both sides we get $B= (A^{-1})(0)$. Now if the right side is compatible then it should be $0$ and we get a contradiction because $B$ is suppose to be a nonzero matrix. However, because the dimensions of the matrix $A$ and $B$ are not given, I'm not sure if this works. The zero matrix may not be matrix multiplication compatible with $A$ inverse.

However, this does tell us that in the case where $A$ and $B$ are square matrix, $A$ is not invertible if $AB=0$ where $B$ is nonzero.

Now what do we do about the cases where $A$ and $B$ are not square matrices?

答: Since $A$ is invertible it is a square matrix. Now each column $y$ of $B$ satisfies $Ay=0$ which impllies $y=0$. So $B$ has to be the zero matrix.


问:
Subgroups of finite simple groups.
Let $G$ be a nontrivial finite group and $X_G$ the set of all the proper subgroups of $G$, $X_G:=\{H\subseteq G\mid H\le G \wedge H\ne G\}$. Lagrange's theorem put a limitation on the order of the elements of $X_G$, which must be a proper divisor of $|G|$. If $G$ is simple, then an additional limitation on the order of the elements of $X_G$ gets in, namely:

$$G\space\text{simple}\Longrightarrow[G:H]!\ge|G|, \space\forall H\in X_G \tag 1$$

In fact, for $H\in X_G$, the group $G$ acts by left multiplication on the left quotient set $G/H$, and this action has trivial kernel$^\dagger$. So, $G$ embeds into $S_{[G:H]}$, whence $(1)$.

The constraint $(1)$ means that finite simple groups can't have "relatively big" subgroups. For example, from $(1)$ follows that $A_5$ can't have subgroups of order $15$, $20$ and $30$ (this latter is ruled out by the very simplicity of $A_5$, though), because $[60:k]!<60$ for $k=15, 20, 30$.

Does the inverse implication in $(1)$ hold, too? Some thoughts. A counterexample would be a nonsimple finite group with all its proper subgroups fulfilling $(1)$. $|G|=2,3$: it is $G\cong C_2,C_3$, both simple, so we have to move upwards. $|G|=4$: both classes ($C_2\times C_2$ and $C_4$) are nonsimple, but both have subgroups of index $2$ (and $2!<4$). $|G|=5$: $C_5$ is simple. $|G|=6$: likewise $|G|=4$ case, both classes ($S_3$ and $C_6$) are nonsimple, but both have subgroups of index $2$ (and $2!<6$). $|G|=7$: $C_7$ is simple. Therefore, counterexamples (if any) must have $|G|\ge 8$.

$^\dagger$For $H\in X$, we get: $K:=\bigcap_{g\in G}gHg^{-1}\lneq G$, and thence $K=\{1\}$ for the simplicity of $G$.

答:
Let $p$ be a prime and $k > 1$ such that $p^k \leq p!$. Then any finite group $G$ of order $p^k$ satisfies your condition, since any subgroup of $G$ has index $\geq p$. (For example consider $G$ elementary abelian of order $25 = 5^2$.)

You could also look any group of order $n = p_1^{k_1} \cdots p_t^{k_t}$, where $p_i$ are distinct primes such that $n \leq p_i!$ for all $1 \leq i \leq t$.

Title: Subgroups of finite simple groups.


问:
Need a good method for computing $\int _{\infty }^a\:\frac{1}{\left(x^2-1\right)^p}dx$
I need a good method for computing $$\int _{\infty }^a\:\frac{1}{\left(x^2-1\right)^p}dx$$
Where $p>0$ and $a>1$ are real. Maybe you have a good numerical algorithm for such a computation? That would really help me, thank you.

Body:
For integer values of $p$, there are very nica patterns for
$$I_p=\int\limits_{a}^{\infty}\frac{dx}{(x^2-1)^p}$$ We have
$$\color{blue}{I_p=(-1)^{p}\,\frac {a\, J_p}{2^{p-1}\, (p-1)!\,(a^2-1)^{p-1}}  +}$$ $$\color{blue}{(-1)^{p+1}\, 4^{1-p}\, \binom{2 p-2}{p-1}\,\coth ^{-1}(a)}$$ where $J_p$ represents a polynomial of degree $(p-2)$ in $a^2$.

The first of these are
$$\left(
\begin{array}{cc}
p & J_p \\
 2 & 1 \\
 3 & 3 a^2-5 \\
 4 & 15 a^4-40 a^2+33 \\
 5 & 105 a^6-385 a^4+511 a^2-279 \\
 6 & 945 a^8-4410 a^6+8064 a^4-7110 a^2+2895 \\
 7 & 10395 a^{10}-58905 a^8+137214 a^6-166914 a^4+110055 a^2-35685
\end{array}
\right)$$ in which more patterns could be found.

For example, the coefficient of the largest power is $(2n-3)!!$ and the constant term is $(-1)^n \big[(2 n-2)!!-(2 n-3)!!\big]$


答:
Your argument is good: you proved that $A$ cannot have a left inverse. In particular it cannot have a two-sided inverse, which is the condition for being invertible.

Knowing that an invertibile matrix is necessarily square is not required.

Just use a different name for the left inverse, say $L$.

Sizes are not of a concern: if $A$ is $m\times n$ and $B$ is $n\times p$, the the zero matrix is $m\times p$ and $L$ is $n\times m$.

Body:
"Suppose $AB=0$ for some non-zero matrix $B$.  Can $A$ be invertible?"

I think $A$ is not invertible by proof by contradiction. Assume $A$ is invertible, then $A$ has a left inverse. Multiplying both sides we get $B= (A^{-1})(0)$. Now if the right side is compatible then it should be $0$ and we get a contradiction because $B$ is suppose to be a nonzero matrix. However, because the dimensions of the matrix $A$ and $B$ are not given, I'm not sure if this works. The zero matrix may not be matrix multiplication compatible with $A$ inverse.

However, this does tell us that in the case where $A$ and $B$ are square matrix, $A$ is not invertible if $AB=0$ where $B$ is nonzero.

Now what do we do about the cases where $A$ and $B$ are not square matrices?


问:
How do I simplify $34\csc{\frac{2\pi}{17}}$?
I have $$ 34\csc\left(\dfrac{2\pi}{17}\right)$$ is equal to $$\dfrac{136}{\sqrt{8-\sqrt{15+\sqrt{17} +  \sqrt{34 + 6\sqrt{17} - \sqrt{34-2\sqrt{17} } + 2\sqrt{ 578-34\sqrt{17}} - 16\sqrt{34-2\sqrt{17}}    }  }}}.$$

I want to rationalize it, but I am not sure where to start. Could anyone provide an explanation on how to rationalize this, and what should the answer be?


问:
solved(Measure theory)
We say that a subset $A$ of $(0, 1)$ is open if $A = B \cap (0, 1)$, where $B$ is some open set in $\mathbb{R}$. Let $C$ denote the collection of open sets in $(0, 1)$. Prove that the Borel $\sigma$-field on $(0, 1)$, i.e. $\sigma(C)$, is the same as $\mathcal{B}_\mathbb{R} \cap (0, 1) := \{B \cap (0, 1) \mid B \in \mathcal{B}_\mathbb{R}\}$.

答:
Note that $\sigma(C)$ is contained in $\mathscr{B}_{\mathbb{R}}\cap(0,1)$. To prove the other inclusion, note that $\mathscr{B}_{\mathbb{R}}\cap(0,1)$ is a restriction of the Borel sigma algebra on $\mathbb{R}$. You can use the $\pi-\lambda$ theorem in the following way:

Let $S=\{B\in \mathscr{B}_{\mathbb{R}}: B\cap(0,1)\in\sigma(C)\}$. Then $S$ is a sigma algebra (and therefore  $\pi$ system) which contains the open sets of $\mathbb{R}$. The $\pi-\lambda$ theorem states that $S=\sigma(S)=\mathscr{B}_{\mathbb{R}}$. This proves that the other inclusion is true.

$\pi-\lambda$ theorem:

https://en.m.wikipedia.org/wiki/Dynkin_system#:~:text=the%20π-λ%20theorem%20is,determine%20the%20joint%20law%20of&text=have%20the%20same%20distribution%20if,same%20joint%20cumulative%20distribution%20function.

This is  my first post here so I apologize in advance if the text is unreadable.

Title: solved(Measure theory)


问:
Why cannot $\varepsilon$ be negative?
I understand why $\varepsilon$ can not be equal to zero, but why cannot it be negative? One possible explanation might be $|f(x)-L|<\varepsilon$. But why can't we remove the modulus (even though distance cannot be negative)?

答: Because since the modulus is always nonnegative, if we allowed negative epsilon values then it would never be possible to define the limit.</p>

Title: Why cannot $\varepsilon$ be negative?


问:
Process' mean, covariance and stationarity
Let $X_t=Yt+Zt^2$ be random process, where $Y$,$Z$ are uncorrelated random variables, with characteristics: $EY=3$, $EZ=0.5$, $DY=1$, $DZ=0.05$. Find $X_t$ mean and covariance and prove whether process is wide-sense stationary.

Covariance:

$\Gamma(Y,Z)=E(YZ)-EYEZ=EYEZ-EYEZ=0$

How about its mean? Any hint?

答:
$\newcommand{\cov}{\operatorname{cov}}$

You've seen the answer to the first question in comments.  As for covariance, you're making it unreasonably complicated.
$$
\begin{align}
\cov(X_t,X_s) & = \cov(Yt+Zt^2,Ys+Zs^2) = \cov(Yt,Ys+Zs^2)+\cov(Zt^2,Ys+Zs^2) \\[8pt]
& =\cov(Yt,Ys)+\cov(Yt,Zs^2)+\cov(Zt^2,Ys)+\cov(Zt^2,Zs^2) \\[8pt]
& = st\cov(Y,Y) + ts^2\cov(Y,Z) + t^2 s\cov(Y,Z) + t^2 s^2 \cov (Z,Z) \\[8pt]
& = st\operatorname{var}(Y) + 0 + 0 + t^2s^2 \operatorname{var}(Z).
\end{align}
$$


答: It can not be negative because we are taking convergence informally distance between the term and its limit. The distance is informally taken as an absolute value</p>

Title: Why cannot $\varepsilon$ be negative?

Body: I understand why $\varepsilon$ can not be equal to zero, but why cannot it be negative? One possible explanation might be $|f(x)-L|<\varepsilon$. But why can't we remove the modulus (even though distance cannot be negative)?


问:
Why can't you solve $0 = -5\sqrt{x+2}-2$
So i did this,
\begin{aligned}
2&=-5\sqrt{x+2},\\
\tfrac{2}{-5}&=\sqrt{x+2},\\
\tfrac{4}{25}&=x+2,\\
x&=-\tfrac{46}{25}.
\end{aligned}
Is there something wrong with my algebra?

答:
When we have an equation, let's say something simple like $x+1=0$, then a number $a$ is a solution of the equation if the following logic holds:

If $x=a$, then $x+1=0$.

There is only one number for which this statement is true, namely $-1$, so that's the only solution of the equation. Essentially, if we plug in a solution, then the equation should be a true statement ($-1+1=0$ is such a true statement). Now the issue here is that this logical statement "if this, then that" starts with the solution: "If $x=a$, then [...]". But we don't usually know the solution right off the bat, we have to find it first. We usually do this be using equivalence transformations. These are ways to manipulate equations in such a way that the original equation is a true statement if and only if the transformed equation is true. For instance, two numbers are equal if and only if those same numbers decreased by 1 are also equal. If $a=b$, then also $a-1=b-1$, and if $a-1=b-1$, then $a=b$, no matter what $a$ and $b$ are. So subtracting $1$ is an equivalence transformation, which allows us to say the following:

$x+1=0$ if and only if $x+1-1=0-1$, which just says that $x=-1$.

What's the most interesting to us is the "if" part, which says that $x+1=0$ if $x=-1$. Exactly what it means for $-1$ to be a solution of the equation. The "only if" part then tells us that there is no other number $x$ could be to yield a true statement, so $-1$ is the only solution. When solving more complicated equations, we write each transformation step below the next and suppress the "if and only if", because we're lazy. But in your mind, between any two lines in your attempt to solve an equation, there should be an "if and only if" hidden.

Now back to your original equation. You didn't use equivalence transformations. Your second step already isn't one. You transform $\frac{2}{-5}=\sqrt{x+2}$ by squaring to $\frac{4}{25}=x+2$. But squaring is not an equivalence transformation. If $a=b$, then $a^2=b^2$, true. But if $a^2=b^2$, then it doesn't follow that $a=b$. So if we insert logical statements into your algebra, we could say that if $2=-5\sqrt{x+2}$, then $x=-\frac{46}{25}$. But for $-\frac{46}{25}$ to be a solution, the reverse should hold: If $x= -\frac{46}{25}$, then $2=-5\sqrt{x+2}$.

So what you showed is that if there is a solution, then it is $-\frac{46}{25}$. But you didn't show that this number is in fact a solution (and you can't, because it isn't).


答:
You want "the distance" between $f(x)$ and $L$ to be small.  You do not care whether $L$ is larger or $f(x)$ is larger as long as the distance between them is small.  Distance is never negative, so $0$ is automatically a hard lower bound for the distance between $f(x)$ and $L$.

Absolute value of a difference exactly captures "the distance" between the subtrahend and minuend (by ignoring the sign of the difference).  So the symbolic translation of "the distance between $f(x)$ and $L$ is small" is
$$  |f(x) - L| < \varepsilon  \text{,}  $$
which can never be satisfied if $\varepsilon < 0$.

Title: Why cannot $\varepsilon$ be negative?


问:
Prove using Fermat's Little Theorem
Prove using Fermat's Little Theorem that if $p\in \mathbb{P}$ then:
$$p \mid\left(\underbrace{1\ldots1}_{p}\underbrace{2\ldots2}_{p}\ldots\underbrace{9\ldots9}_{p} - 123456789  \right)$$

答:
Let us assume $p \neq 3$. (I leave $p = 3$ up to you. Use the divisibility test.)

By Fermat's Little Theorem, we have $10^p \equiv 10 \bmod p$.

Let $k = \underbrace{1\ldots1}_{p}$. Then,
$$k = 1 + 10 + \cdots + 10^{p-1} = \frac{10^p - 1}{9} \equiv \dfrac{10 - 1}{9} \equiv 1 \bmod p.$$
(Note that $p \neq 3$ and thus, $9^{-1}$ makes sense.)

Now, let
$$N = \underbrace{1\ldots1}_{p}\underbrace{2\ldots2}_{p}\ldots\underbrace{9\ldots9}_{p} = k\cdot10^{8p} + 2k\cdot10^{7p} + \cdots + 9k.$$

Going modulo $p$, we get
$$N \equiv k\cdot10^8 + 2k\cdot10^7 + \cdots + 9k \bmod p.$$

Using $k \equiv 1 \bmod p$ gives

$$N \equiv 10^8 + 2\cdot10^7 + \cdots + 9 \equiv 123456789 \bmod p.$$

Title: Prove using Fermat's Little Theorem


问:
If we restrict the Heaviside step function to $\mathbb{R}\setminus\{0\},$ does it suddenly become continuous?
The Heaviside step function is discontinuous, despite that its continuous at every point except $0$.

Supposing we restricted it to $\mathbb{R}\setminus\{0\},$ does it suddenly become continuous?

I think 'yes,' but it seems a bit counterintuitive. I just want someone to confirm that I'm not fooling myself.

答: Yep! Continuous at every point in the domain means continuous! Think of $\mathbb{R} \setminus 0$ as being two completely separate chunks of space. You probably wouldn't have a problem with the function that's zero on $[0,1]$ and one on $[3,4]$ being continuous. Something similar happens here.</p>


问:
Analytic formula for $\sup_{|x| \le 1} ax + |x-b|$ as a function of $a,b \in \mathbb R$.
Disclaimer: This should be "easy" but, I'm startled by the explosion of subcases.

Question. What is an analytic formula for $\sup_{|x| \le 1} ax + |x-b|$ as a function of $a,b \in \mathbb R$.

答:
By simplifying the apparent explosion of subcases, it turns out that the problem has a nice analytic solution.

Lemma. For any $a,b \in \mathbb R$, it holds that
$$
\sup_{|\theta| \le 1}a\theta +|\theta-b| = |a-b/\max(|b|,1)|+\max(|b|,1).
$$

Proof.
Let $f(a,b):=\sup_{|\theta| \le 1}a\theta+|\theta-b|$.
First, suppose $b \le -1$. Then, $|\theta-b| = \theta-b$ for all $\theta \in [-1,1]$, and so $f(a,b) = \sup_{|\theta| \le 1}(a+1)\theta-b = |a+1|-b$. Next, suppose $b \ge 1$. Then $|\theta-b|=b-\theta$, and so $f(a,b) = \sup_{|\theta| \le 1}(a-1)\theta+b = |a-1|+b$. Finally, suppose $|b| < 1$. One computes
\begin{eqnarray*}
\begin{split}
f(a,b) &= \max((1+a)b-b,1+a-b,b-(a-1),b+b(a-1))\\
&=\max(ab,1+a-b,b-a+1,ab)\\
&=\max(ab,1+a-b,1-a+b)\\
&= \max(ab,|a-b|+1)\\
&= |a-b|+1,
\end{split}
\end{eqnarray*}
where the last step is because $ab \le |ab| \le |a| \le |a-b|+1$, since $|b| \le 1$. Putting things together then gives the result. $\quad\quad\Box$


问:
If $X_n$ is an $L_1$-bounded martingale, show that $\sum_n(X_n-X_{n-1})^2< \infty$ a.s.
I had asked this question previously in the following post but there were no replies.

Recently, I found a two page article(free download) with a possible alternate line of proof to the one suggested in the first link

Note: $X_n$ is an $L_1$-bounded martingale means that $\sup_nE[|X_n|]<\infty$

However in the link, the author shows that $\sum_n(X_n-X_{n-1})^2< \infty$ a.s. on a set $E$ which is defined as follows:$$E_n:=\bigcap_{k\leq n}\{\omega||X_n(\omega)|\leq M\}\text{ and }E=\bigcap_n E_n.$$ He then states that for large $M$ the measure of $E$ is near 1, since by the martingale convergence theorem, $\sup_n|X_n|<\infty$ which completes the proof.

However, I have two issues.

Why can the measure of $E$ be made close to 1?

Shouldn't we require a set $E$ with measure exactly 1 to show a.s convergence. If so how do we find a set E of measure 1?

Can someone please explain how to rectify these issues?

答:
1.  In Austin's note, the following was proved:
\begin{align*}
 &\{\omega: \sup_n|X_n(\omega)|<M\}\\
 &\subset \Big\{\omega: \sum_n(X_n(\omega)-X_{n-1}(\omega))^2<\infty\Big\}, \quad \forall M>0 \tag{1}
\end{align*}
(or exactly
\begin{equation*}
 \mathsf{P}\Big(\{\sup_n|X_n|<M\}\setminus
 \Big\{ \sum_n(X_n-X_{n-1})^2<\infty\Big\}\Big)=0,\quad  \forall M>0.)
\end{equation*}
Using (1),
\begin{align*}
\{\omega: \sup_n|X_n(\omega)|<\infty\}&= \bigcup_{M=1}^\infty\{\omega: \sup_n|X_n(\omega)|<M\}\\
 &\subset \Big\{\omega: \sum_n(X_n(\omega)-X_{n-1}(\omega))^2<\infty\Big\},
\end{align*}
This is the reason of last sentence in Austin's note(the measure of $ E $ is near 1(if the $ M $ is large enough)).

2. In following two texts, the conclusion you are interesting in is proved also:

Y. S. Chow & H. Teicher, Probability Theory, 3Ed, Springer Verlag, 1997, Theorem 11.1.2, p.408.

P. Malliavin, Integration and Probability, Springer Verlag, 1995. Prop. IV.5.6.2, p.214.


问:
Why does the definition of topological space require that the empty set be open?
I'm taking my first course in general topology, and so obviously I came across the definition of topological space $(T, \tau)$:

$\tau \in \mathcal P \mathcal P (T)$

$\emptyset, T \in \tau$

$(A_i)_{i \in I} \in \tau \implies \bigcup_{i \in I} A_i \in \tau$

$A_1, \dots, A_n \in \tau \implies \bigcap_{i=1}^n A_i \in \tau$

This is very abstract, so I tried to find a few examples, and these examples I found caught my eye:

$T = \mathbb R^n$, $\tau = \{ \text{every set whose complement is compact} \}$

$T = [0,1]$ $\tau = \{ \text{every set whose complement is a finite set}  \}$

Now these satisfy every part of the definition, except that one has to add the empty set to $\tau$ explicitly as well. This gives the natural question: why is this needed?

答: One of the most useful properties of the closed sets is that they contain all of their limit points since space itself also satisfies this condition it is natural to want the entire space to be closed as well as open, the only this can be true is if the empty set is closed, moreover as pointed out earlier without considering the empty set open it is impossible to have finite intersection property for open sets.</p>


问:
Best spline method for closed curves
What is the best spline method I can use to obtain a closed curve for the following data?

Note that I will need to obtain the derivative as in every point there will be a vector that is perpendicular to the contour curve (the shape does not have to be a "circle like" as the example, but can be any closed shape).

答:
For an interpolating closed curve you can use a parametric spline with periodic boundary condition:

Create a time-coordinate, e.g. proportional to the distance of two adjacent grid points: $t_{i+1}=t_i+\sqrt{(x_{i+1}-x_i)^2+(y_{i+1}-y_i)^2})$

Create two splines for each $x$ and $y$ coordinate as a function of time. Set boundary conditions to be periodic, i.e. $f'(t_1)=f'(t_n)$.

In R you can do something like

```<code># generate some input data
n=10
phi=sort(runif(n,0,2*pi))
r=rnorm(n,10,0.5)
X=r*cos(phi)
Y=r*sin(phi)
plot(X,Y)
# add first point to the end to create a closed loop
X=append(X,X[1])
Y=append(Y,Y[1])

# create time vector (traverse at constant speed)
T=vector(length=length(T))
T[1]=0
for(i in 2:length(Y)) T[i]=T[i-1]+norm(c(X[i]-X[i-1], Y[i]-Y[i-1]), type="2")

# setup periodic splines for both x and y coordinate
sx=splinefun(T,X,method="periodic")
sy=splinefun(T,Y,method="periodic")

# plot splines
Tplot=seq(T[1],tail(T,n=1),by=0.02)
Xplot=sx(Tplot)
Yplot=sy(Tplot)
lines(Xplot,Yplot)
</code>```

The spline also contains information regarding derivatives: $\text{derivative} = (sx'(t), sy'(t))$.

```<code># plot some derivatives and normals
Tplot=seq(T[1],tail(T,n=1),by=2.0)
for(t in Tplot)
{
    x = sx(t)
    y = sy(t)
    arrows(x,y,x+sx(t,deriv=1),y+sy(t,deriv=1),length=0.05) # derivative
    arrows(x,y,x+sy(t,deriv=1),y-sx(t,deriv=1),length=0.05) # normal
}
</code>```

This gives

closed spline
closed spline with derivative and normal arrows

Title: Best spline method for closed curves

Body:
For an interpolating closed curve you can use a parametric spline with periodic boundary condition:

Create a time-coordinate, e.g. proportional to the distance of two adjacent grid points: $t_{i+1}=t_i+\sqrt{(x_{i+1}-x_i)^2+(y_{i+1}-y_i)^2})$

Create two splines for each $x$ and $y$ coordinate as a function of time. Set boundary conditions to be periodic, i.e. $f'(t_1)=f'(t_n)$.

In R you can do something like

# generate some input data
n=10
phi=sort(runif(n,0,2*pi))
r=rnorm(n,10,0.5)
X=r*cos(phi)
Y=r*sin(phi)
plot(X,Y)
# add first point to the end to create a closed loop
X=append(X,X[1])
Y=append(Y,Y[1])

# create time vector (traverse at constant speed)
T=vector(length=length(T))
T[1]=0
for(i in 2:length(Y)) T[i]=T[i-1]+norm(c(X[i]-X[i-1], Y[i]-Y[i-1]), type="2")

# setup periodic splines for both x and y coordinate
sx=splinefun(T,X,method="periodic")
sy=splinefun(T,Y,method="periodic")

# plot splines
Tplot=seq(T[1],tail(T,n=1),by=0.02)
Xplot=sx(Tplot)
Yplot=sy(Tplot)
lines(Xplot,Yplot)

The spline also contains information regarding derivatives: $\text{derivative} = (sx'(t), sy'(t))$.

# plot some derivatives and normals
Tplot=seq(T[1],tail(T,n=1),by=2.0)
for(t in Tplot)
{
    x = sx(t)
    y = sy(t)
    arrows(x,y,x+sx(t,deriv=1),y+sy(t,deriv=1),length=0.05) # derivative
    arrows(x,y,x+sy(t,deriv=1),y-sx(t,deriv=1),length=0.05) # normal
}

This gives

closed spline
closed spline with derivative and normal arrows


问:
Are the angles in a halved circle right?
Imagine a circle cut in half. Now, are the two angles on it right? I can imagine when going limitedly close that they would be.

Got curious about this question after seeing this picture on the internet.

答: That depends upon how you define angle. But if you define the angle between two intersecting curves at a point $P$ at which they intersect as the angle between their tangent lines at $P$ (which is a reasonable definition) then, yes, there are two right angles in that picture.


问:
How to convert a geometric series so that exponent matches index of sum?
I need to convert the following series into a form that works for the equation $$\frac{a}{1-r}$$ so that I can calculate its sum. But the relevant laws of exponents are eluding me right now.

$$\sum_{n=1}^{\infty}\left(\frac{4}{10}\right)^{3n-1}$$

How do I get the 3 out of the exponent?

Body: Hint: $$\left(\frac{4}{10}\right)^{3n-1} = \left(\frac{4}{10}\right)^{3n}\left(\frac{4}{10}\right)^{-1} =  \frac{10}{4}\cdot\left(\left(\frac{4}{10}\right)^{3}\right)^{n}$$


问:
$1+px^2$ is a square iff $x\in\mathbb Z_p$
I'm fairly new to number theory and have just started to study p-adic numbers. In a book I'm reading about the fundamentals of number theory, there is the following statement that is supposed to be easy. However I did not find a proof.

Let $p \neq 2$ be a prime number, then $1+px^2$ is a square iff $x\in\mathbb Z_p.$

One of my first thoughts was to use the unique representation of $x\in\mathbb Z_p,$ i.e. $x = u p^n,$ where $u\in\mathbb Z_p^\times,\ n\in \mathbb N_0.$

I'd be really glad if someone might hint me into the right direction here.

答:
For $p=2$ it is not true, $3$ isn't a square $\bmod 4$.

For $p$ odd and $v_p(x)<0$ then $v_p(1+px^2)=v_p(px^2)$ is odd so $1+px^2$ can't be a square (in $\Bbb{Q}_p$). For $v_p(x)\ge 0$ then $1+px^2$ is a square due to Hensel lemma (or the binomial series).


问:
Let $u_n=\sum_{r=0}^n (-1)^r \frac{\binom nr}{x+r}$ at $x=2$. Then find the sum to infinity of $u_1+u_2..$
Can I get a hint on how to begin the solution? I tried all possible derivative/integral methods of forming the general sum using the expansion of $(1+x)^n$, but I am just not able to the get $2+r$ in the denominator

Body:
We know that,

\begin{gather*}
( 1-x)^{n} =\sum ^{n}_{r=0}( -1)^{r} \cdot ^{n} C_{r} x^{r}\\
x( 1-x)^{n} =\sum ^{n}_{r=0}( -1)^{r} \cdot ^{n} C_{r} x^{r+1}\\
\int ^{1}_{0} x( 1-x)^{n} dx=\int ^{1}_{0}\sum ^{n}_{r=0}( -1)^{r} \cdot ^{n} C_{r} x^{r+1} dx\\
=\sum ^{n}_{r=0}\int ^{1}_{0}( -1)^{r} \cdot ^{n} C_{r} x^{r+1} dx\\
For\ LHS,\\
\int ^{1}_{0} x( 1-x)^{n} dx=\int ^{1}_{0}( 1-x) \cdot ( 1-( 1-x))^{n} dx\\
=\int ^{1}_{0} x^{n}( 1-x) dx=\frac{1}{n+1} -\frac{1}{n+2}\\
For\ RHS,\\
\sum ^{n}_{r=0}\int ^{1}_{0}( -1)^{r} \cdot ^{n} C_{r} x^{r+1} dx=\sum ^{n}_{r=0}( -1)^{r} \cdot \frac{^{n} C_{r}}{r+2}\\
So,\ as\ per\ your\ question,\ \\
u_{n} =\frac{1}{n+1} -\frac{1}{n+2}\\
for\ n=1\ to\ n=\infty .
\end{gather*}
Can you now complete the telescoping series and show that the result is 1/2?


问:
If two matrices A and B such that that AB is invertible, but A and B are not, then A and B can't be square matrix?
"Find two matrices A and B that AB is invertible, but A and B are not. Hint:square matrices A and B would not work."

I was able to complete this problem easily finding simple rectangle matrix, but I'm curious as to why A and B cannot be square matrix?

答:
You don't need determinants or ranks to see why this cannot the case.

A matrix $A$ is invertible $\iff$ it is a square matrix and there is a matrix $P$ such that $AP = I$.

If $AB$ is invertible, then there is a matrix $P$ such that $ABP = I$.  Can you see why $A$ cannot be a square matrix now?


问:
What would be the smallest/lowest Big-O set possible of these $2$ functions
Here i have $2$ functions and i was just wanting to know what the smallest big-oh set possible is for these functions, my guess would be the lowest possible of 1 would be the big-ph set $O(n^3)$ but was unsure if it may be $O(1)$, and for the 2nd function i was unsure if it was $O(n \log n)$ or $O(n)$

$\frac {n^3} 4 + 6n^2 – 10000$

$5n\log 2n + 6n$

Any help and explanation is appreciated.

答:
I am not sure that the smallest $O$ does exist.

For example,

$$\frac{n^3} 3 + 6 n^2 - 10000= O(n^3) $$

but it is also true that

$$\frac{n^3} 3 + 6 n^2 - 10000= O((n-1)^3) $$

and
$$\frac{n^3} 3 + 6 n^2 - 10000= O(T_n)$$
while $T_n<(n-1)^3<n^3$. [$T_n$ denotes the $n^{th}$ tetrahedral number.]

For the second,

$$5n\log 2n + 6n=O(\log_{2021}(n!))$$ is the same as
$$O(n\log(n)).$$


问:
Is there a mistake to the answer given to this problem: Find the matrix of the reflection through the line y=−2x/3.
This is the answer given:

I think this answer is not correct because the line given by y=-2x/3 makes an angle that is below the x axis, so the order is incorrect right? We need to start  with a counterclockwise rotation first and end with the clockwise rotation, but here, we started with the clockwise rotation first.

答:
Here the slope of the line = $m = tan \gamma = -\frac{2}{3}$.

$\therefore sin \gamma = \frac{m}{\sqrt{1+m^2}},\; cos \gamma = \frac{1}{\sqrt{1+m^2}}$.

The transformation matrix for reflection around the line is given by the product of 3 matrices (rotate by angle -$\gamma$ to make it horizontal, reflect w.r.t the horizontal line and rotate back by angle $\gamma$)

\begin{align}T&=T_{-\gamma}.R_0.T_{\gamma}\\
&=\begin{pmatrix}cos \gamma & -sin\gamma\\ sin\gamma  & cos \gamma\end{pmatrix}
\begin{pmatrix}1&0\\ 0  & -1\end{pmatrix}
\begin{pmatrix}cos \gamma & sin\gamma\\ -sin\gamma  & cos \gamma\end{pmatrix} \\
&=\begin{pmatrix}\frac{1}{\sqrt{1+m^2}} & -\frac{m}{\sqrt{1+m^2}}\\ \frac{m}{\sqrt{1+m^2}}  & \frac{1}{\sqrt{1+m^2}}\end{pmatrix}
\begin{pmatrix}1&0\\ 0  & -1\end{pmatrix}
\begin{pmatrix}\frac{1}{\sqrt{1+m^2}} & \frac{m}{\sqrt{1+m^2}}\\ -\frac{m}{\sqrt{1+m^2}}  & \frac{1}{\sqrt{1+m^2}}\end{pmatrix} \\
&=\begin{pmatrix}\frac{1}{\sqrt{1+m^2}} & -\frac{m}{\sqrt{1+m^2}}\\ \frac{m}{\sqrt{1+m^2}}  & \frac{1}{\sqrt{1+m^2}}\end{pmatrix}
\begin{pmatrix}\frac{1}{\sqrt{1+m^2}} & \frac{m}{\sqrt{1+m^2}}\\ \frac{m}{\sqrt{1+m^2}}  & -\frac{m}{\sqrt{1+m^2}}\end{pmatrix} \\
&=\frac{1}{1 + m^2}\begin{pmatrix}1-m^2&2m\\2m&m^2-1\end{pmatrix} \\
&=\begin{pmatrix}\frac{5}{13}&-\frac{12}{13}\\ -\frac{12}{13}&-\frac{5}{13}\end{pmatrix}
\end{align}

The following example in R shows how the black point (1,1) is reflected w.r.t. the life to new location (red point)

```<code>x <- seq(-10,10,0.01)
y <- -2*x/3
m <- -2/3
T <- matrix(c((1-m^2)/(1+m^2),2*m/(1+m^2),2*m/(1+m^2),(m^2-1)/(1+m^2)), nrow=2, byrow=T)
p <- c(1,2)
p1 <- T %*% p
plot(x,y, type='l')
points(p[1],p[2],pch=19)
points(p1[1],p1[2],pch=19, col='red')
</code>```

Body:
Here the slope of the line = $m = tan \gamma = -\frac{2}{3}$.

$\therefore sin \gamma = \frac{m}{\sqrt{1+m^2}},\; cos \gamma = \frac{1}{\sqrt{1+m^2}}$.

The transformation matrix for reflection around the line is given by the product of 3 matrices (rotate by angle -$\gamma$ to make it horizontal, reflect w.r.t the horizontal line and rotate back by angle $\gamma$)

\begin{align}T&=T_{-\gamma}.R_0.T_{\gamma}\\
&=\begin{pmatrix}cos \gamma & -sin\gamma\\ sin\gamma  & cos \gamma\end{pmatrix}
\begin{pmatrix}1&0\\ 0  & -1\end{pmatrix}
\begin{pmatrix}cos \gamma & sin\gamma\\ -sin\gamma  & cos \gamma\end{pmatrix} \\
&=\begin{pmatrix}\frac{1}{\sqrt{1+m^2}} & -\frac{m}{\sqrt{1+m^2}}\\ \frac{m}{\sqrt{1+m^2}}  & \frac{1}{\sqrt{1+m^2}}\end{pmatrix}
\begin{pmatrix}1&0\\ 0  & -1\end{pmatrix}
\begin{pmatrix}\frac{1}{\sqrt{1+m^2}} & \frac{m}{\sqrt{1+m^2}}\\ -\frac{m}{\sqrt{1+m^2}}  & \frac{1}{\sqrt{1+m^2}}\end{pmatrix} \\
&=\begin{pmatrix}\frac{1}{\sqrt{1+m^2}} & -\frac{m}{\sqrt{1+m^2}}\\ \frac{m}{\sqrt{1+m^2}}  & \frac{1}{\sqrt{1+m^2}}\end{pmatrix}
\begin{pmatrix}\frac{1}{\sqrt{1+m^2}} & \frac{m}{\sqrt{1+m^2}}\\ \frac{m}{\sqrt{1+m^2}}  & -\frac{m}{\sqrt{1+m^2}}\end{pmatrix} \\
&=\frac{1}{1 + m^2}\begin{pmatrix}1-m^2&2m\\2m&m^2-1\end{pmatrix} \\
&=\begin{pmatrix}\frac{5}{13}&-\frac{12}{13}\\ -\frac{12}{13}&-\frac{5}{13}\end{pmatrix}
\end{align}

The following example in R shows how the black point (1,1) is reflected w.r.t. the life to new location (red point)

x <- seq(-10,10,0.01)
y <- -2*x/3
m <- -2/3
T <- matrix(c((1-m^2)/(1+m^2),2*m/(1+m^2),2*m/(1+m^2),(m^2-1)/(1+m^2)), nrow=2, byrow=T)
p <- c(1,2)
p1 <- T %*% p
plot(x,y, type='l')
points(p[1],p[2],pch=19)
points(p1[1],p1[2],pch=19, col='red')


问:
Why this integral is $0$? $ \int_{0}^{2 \pi} \frac{2 \cos(2t) (\sqrt{2}ie^{it})}{4e^{i4t}-2\sqrt{2}e^{i2t}+1}dt$
I was unable to calculate this integral directly $$ \int_{0}^{2 \pi} \frac{2 \cos(2t) (\sqrt{2}ie^{it})}{4e^{i4t}-2\sqrt{2}e^{i2t}+1}dt$$

so I put it into wolfram and it says that it is equal to zero. Does anyone know why?

EDIT:

The original integral was $$\int_{|z| = \sqrt{2}} \frac{\operatorname{Re}(z^2)}{z^4-\sqrt{2}z^2+1} dz$$ and I did the substitution $z = \sqrt{2}e^{it}$.

答:
Due to the $e^{it}$ factor, the integrand multiplies by $-1$ when you apply $t\mapsto t+\pi$ (i.e. $z\mapsto-z$), so the $[\pi,\,2\pi]$ part of the integral cancels the $[0,\,\pi]$ part (as long as you can show it's finite).

Alternative argument:

Note that $\left|z\right|=\sqrt{2}\implies\Re\left(z^{2}\right)=\frac{z^{2}+4z^{-2}}{2}$. Since $f\left(z\right):=\frac{z^{4}+4}{2z^{2}\left(z^{4}-\sqrt{2}z^{2}+1\right)}$ has a second-order pole $0$ and a set $S$ of first-order poles closed under $z\mapsto-z$,$$\oint_{\left|z\right|=2}f\left(z\right)dz=2\pi i\left(\lim_{z\to0}\frac{d}{dz}\left(z^{2}f\left(z\right)\right)+\sum_{w\in S}\lim_{z\to w}\left(z-w\right)f\left(z\right)\right).$$Since $\lim_{z\to-w}\left(z+w\right)f\left(z\right)=-\lim_{z\to w}\left(z-w\right)f\left(z\right)$ because $f$ is even, poles in $S$ don't contribute. But $\frac{d}{dz}\left(z^{2}f\left(z\right)\right)$ is odd, with $z\to0$ limit $0$, so the integral is $0$. Or if you want to be explicit,$$\frac{d}{dz}\frac{z^{4}+4}{2\left(z^{4}-\sqrt{2}z^{2}+1\right)}=\frac{2z^{3}}{z^{4}-\sqrt{2}z^{2}+1}-\frac{z\left(z^{2}-\sqrt{2}\right)\left(z^{4}+4\right)}{\left(z^{4}-\sqrt{2}z^{2}+1\right)^{2}}.$$


问:
A family of subsets with finite intersection property has finite character
I'm trying to prove that given a set $X$ the set $\mathfrak{C}$ whose elements are families of subsets of $X$ with the finite intersection property has finite character.

To add more context, we say that a family of sets $\{F_i\}_{i \in I}$ has the finite intersection property if for every finite $J \subseteq I$ we have that $\bigcap_{j \in J} F_j \neq \emptyset$

And we say that a family of sets $\mathfrak{C}$ has finite character if it holds that: $Y \in \mathfrak{C} \Leftrightarrow \forall A \subseteq Y \text{ finite, } A \in \mathfrak{C}$

I'm having trouble proving that given a set $Y$ such that $\forall A \subseteq Y \text{ finite, } A \in \mathfrak{C}$, then $Y$ must be in $\mathfrak{C}$. I've tried that in this conditions every singleton of $Y$ must be in $\mathfrak{C}$ but I don't know how to conclude that $Y$ is in $\mathfrak{C}$. I would appreciate any help.

答:
If  $\mathcal F$ has FIP then it is immediate that every finite subcollection of $\mathcal F$ has FIP:$$\mathcal F\in\mathfrak C\implies\text{ all finite subsets of }\mathcal F\text{ are elements of }\mathfrak C$$

For the converse:

If $\mathcal F$ does not have FIP then this implies the existence of a finite subcollection $\mathcal F_0\subseteq\mathcal F$ such that $\bigcap\mathcal F_0=\varnothing$. But then $\mathcal F_0\notin\mathfrak C$ so we can state that:$$\mathcal F\notin\mathfrak C\implies\text{ finite subsets of }\mathcal F\text{ exists that are not in }\mathfrak C$$


问:
Solving first order non-linear ODE with logarithm
I want to solve for $t \in \mathbb{R}, u'(t)=-u(t)\ln \lvert u(t) \rvert$.

I defined two cases: $\mathbb{R^*_+}$ and $\mathbb{R^*_-}$.

For $\mathbb{R^*_+}$:

$$\frac{du}{u}=-\ln(u(t))dt$$

And by integrating, it follows that:

$$\ln\lvert u \rvert = -u\ln(u) + u$$

But I'm stuck here, I don't know what to do next.

答:
$$\frac{du}{u}=-\ln(u(t))dt$$
You cannot integrate $\int-\ln(u(t))dt$ directly.

The mistake is : You integrate $\quad \int-\ln(u(t))du= -u\ln(u)+u\quad$ but this is not $\quad \int -\ln(u(t))dt$.

Back to the original ODE :
$$\frac{du}{dt}=u\ln(u)$$
This is a separable ODE thus
$$\frac{du}{u\ln(u)}=dt$$
$$\int\frac{du}{u\ln(u)}=\int dt$$
$$\ln(\ln(u))=t+c$$
$$u=\exp(\exp(t+c))$$


问:
Is infinite a infinite or finite
Long back I have watched a documentary based on a mathematician named Cantor. According to that documentary, Cantor claimed that infinite does not exists, it is only finite. Is that True?</p>

答:
you cannot have an infinity as a set of parts in the real world like you do in math.

You cannot have infinity apples. You can have a process where apples are added indefinitely every second heading towards infinity but you'll never reach the amount of infinity apples. Hence, to think of an actual infinity existing in the here and now as being comprised of an infinite set of finites is an idea based on a fallacy. There will never be a point in time where this infinity will exist because such an infinity is not a number but rather, an indefinite process.

Title: Is infinite a infinite or finite


问:
Determining the points of intersection of a parametric line with the coordinate planes

Find the points where the line given by $x = 7t + 9$, $y = 8t - 7$, and $z = -7t - 2$ intersect the $xy$, $xz$, and $yz$ planes.

Attempted solution:

This seems as though a very simple problem to me. Take, for example, finding the intersection with the $xz$ plane. It must be that $y = 0$. I substitute $y = 0$ to obtain $t = \frac{7}{8}$, then $x = \frac{121}{8}$ and $z = -\frac{65}{8}$. The point of intersection is $(\frac{121}{8}, 0, -\frac{65}{8})$.

It seems, however, that my solution is wrong, which completely baffles me.

What is the correct approach?

答:
Find the points where the line given by $x = 7t + 9$, $y = 8t - 7$, and $z = -7t - 2$ intersect the $xy$, $xz$, and $yz$ planes.
Your approach is correct, You can also write the equation of line in symmetric form:

$$\frac{x-9}7=\frac{y+7}8=\frac{z+2}{-7}$$

Now intersection of this line with xy, yz and zx is respectively :

xy): $z=0\Rightarrow x-9=-2\Rightarrow x=7, y+7=-\frac{16}7\Rightarrow y=-\frac{65}7$

YZ): $x=0$

zx): $y=0$


问:
Express the Laplacian in $u-v$ coordinates. How?
I am working on becoming more agile using different coordinates to solve problems.

Depending on which coordinates one uses the "Laplacian" takes different forms:

Coordinate expressions for the Laplacian

I have (1) watched online tutorials (2) read the wikipedia section, and tried to use the information I know to express the "Laplacian" in $u-v$ coordinates s.t. $u=e^x$ and $v=e^y.$

How do you write the "Laplacian" in $u-v$ coordinates?

I understand pretty well the conversion of the "Laplacian" from cartesian to cylindrical coordinates. I don't quite understand the conversion for a less used coordinate system such as the $u-v$ one.

Here's my attempt:

答:
Here is a hint:
$$ \frac{\partial}{\partial x}( \frac{\partial u}{\partial x} \frac{\partial f(u,v)}{\partial u}) = \frac{\partial^2 u}{\partial x^2} \frac{\partial f}{\partial u} +( \frac{\partial u}{\partial x} )  \frac{\partial}{\partial x}\left[ \frac{\partial f}{\partial u} \right]$$

Now,

$$ \frac{\partial }{\partial x} (\frac{\partial f}{\partial u}) = [\frac{\partial u}{\partial x} \frac{\partial^2 f}{\partial u^2} + \frac{\partial v}{\partial x} \frac{\partial^2 f } {\partial v \partial u}] \tag{1}$$

To understand this, consider $ \frac{\partial f}{\partial u} = g(u,v)$ then,

$$ \frac{\partial }{\partial x} g(u,v) = \left[ \frac{\partial u}{\partial x} \frac{\partial g}{\partial u} + \frac{\partial v}{\partial x} \frac{\partial g}{\partial v} \right]$$

And then plug back in $ g(u,v) = \frac{\partial f}{\partial u}$

And plug that back in. It's much more easier to track the dependencies when you have an actual function attached than to work with abstract operators. A comment on (1) is that it  reminiscent of the parameteric derivative:
$$ \frac{dy}{dx} = \frac{dt}{dx} \frac{dy}{dt}$$

Or, in operator form:

$$ \frac{d}{dx} = \frac{dt}{dx} \frac{d}{dt}$$

The 'trouble' comes when we try to compose the operator that is:

$$ \frac{d^2}{dx^2} = \frac{dt}{dx} \frac{d}{dt} ( \frac{dt}{dx} \frac{d}{dt})$$

The right hand simplfies too:

$$ \frac{dt}{dx} \left[ \frac{d}{dt} (\frac{dt}{dx}) + \frac{d^2 }{dt^2}\right]$$

The confusing term at hand is $ \frac{d}{dt} \frac{dt}{dx}$ similar to $ \frac{\partial}{\partial x}\left[ \frac{\partial f}{\partial u} \right]$ [see here]


问:
How prove this integral inequality $\int_{0}^{1}f^2(x)dx\ge 24\left(\int_{0}^{1}f(x)dx\right)^2$?
let $f:[0,1]\longrightarrow R $ be a continuous function, if
$$\int_{0}^{1}x^2f(x)dx=-2\int_{\frac{1}{2}}^{1}F(x)dx$$
where $F(x)=\displaystyle\int_{0}^{x}f(t)dt,x\in [0,1]$,then prove that
$$\int_{0}^{1}f^2(x)dx\ge 24\left(\int_{0}^{1}f(x)dx\right)^2$$

I think this inequality is very interesting, becasue not long ago,I have solve this a little same problem:Funny integral inequality

I believe this two problem have same methods.Thank you everyone

答:
Thought：by Cauchy-Schwarz inequality,
$$\bigg(\int_0^1g(x)f(x)\,\mathrm{d} x\bigg)^2\leqslant\int_0^1g^2(x)\,\mathrm{d} x\int_0^1f^2(x)\,\mathrm{d} x$$
We just need to prove that there is a constant K.
$$\int_0^1g(x)f(x)\,\mathrm{d} x=K\int_0^1f(x)\,\mathrm{d} x$$
Proof. by integration by parts,
\begin{align*}
-2\int_{\frac{1}{2}}^{1}F(x)\,\mathrm{d} x&=\int_{\frac{1}{2}}^{1}F(x)\,\mathrm{d}(C-2x)
=(C-2x)F(x)\bigg|_{\frac{1}{2}}^1+\int_{\frac{1}{2}}^{1}(2x-C)f(x)\,\mathrm{d} x\\
&=\int_0^1(C-2)f(x)\,\mathrm{d} x+\int_0^{\frac{1}{2}}(1-C)f(x)\,\mathrm{d} x+\int_{\frac{1}{2}}^{1}(2x-C)f(x)\,\mathrm{d} x
\end{align*}
Since $\textstyle\int_0^1x^2f(x)\,\mathrm{d} x=-2\int_{\frac{1}{2}}^{1}F(x)\,\mathrm{d} x$,
$$\int_0^1(C-2)f(x)\,\mathrm{d} x=\int_0^{\frac{1}{2}}(x^2+C-1)f(x)\,\mathrm{d} x+\int_{\frac{1}{2}}^{1}(C-2x+x^2)f(x)\,\mathrm{d} x$$
Consequently, let
$$g(x)=\begin{cases}
x^2+C-1,& x\in[0,\tfrac{1}{2})\\
C-2x+x^2,& x\in[\tfrac{1}{2},1)
\end{cases}$$
by Cauchy-Schwarz inequality,
\begin{align*}
\left(\int_0^1(C-2)f(x)\,\mathrm{d} x\right)^2&=(C-2)^2\left(\int_0^1f(x)\,\mathrm{d} x\right)^2=\left(\int_0^1g(x)f(x)\,\mathrm{d} x\right)\\
&\leqslant\int_0^1g^2(x)\,\mathrm{d} x\int_0^1f^2(x)\,\mathrm{d} x
\end{align*}
After calculation, we get
$$\int_0^1g^2(x)\,\mathrm{d} x=\frac{240c^2-440c+203}{240}$$
Hence, we have
$$\left(\int_0^1f(x)\,\mathrm{d} x\right)^2\leqslant\frac{240c^2-440c+203}{240(C-2)^2}\int_0^1f^2(x)\,\mathrm{d} x$$
Let $h(C)=\frac{240C^2-440C+203}{240(C-2)^2}\Rightarrow \lim_{C\to2}h(C)=+\infty$
$$h'(C)=\frac{237-260C}{120(C-2)^3} \Leftrightarrow h'(C)=0\Rightarrow C=\frac{237}{260}\Rightarrow h(\tfrac{237}{260})=\frac{4}{849}$$
$$\begin{array}{cccccc}
\hline
 & (-\infty,\tfrac{237}{260}) & \tfrac{237}{260} & (\tfrac{237}{260},2) & (2,+\infty) & +\infty\\\hline
h'(C) & <0 & 0 & >0  & <0 & <0 \\
h(C)  & \downarrow & \text{local minimum} & \uparrow & \downarrow & 1\\\hline
\end{array}$$
So, we have
$$\min_{C\neq2}h(C)=h(\tfrac{237}{260})=\frac{4}{849}$$
Therefore,
$$\int_0^1f^2(x)\,\mathrm{d} x\geqslant\frac{849}{4}\bigg(\int_0^1f(x)\,\mathrm{d} x\bigg)^2$$
Obviously, $\tfrac{849}{4}$ is the best constant, the equality holds if and only if $C=\frac{237}{260}$ and $f(x)=g(x)$


问:
Does f'(x)g'(x)=0 for every x imply f=const. or g=const.?
f,g are smooth functions of x in R.

f=const. or g=const. is obviously solution to this equation. I wonder if there exists a function which somewhere f'=0 and g'≠0 and otherwhere g'=0. f'≠0. It seems to be against to the smoothness of f and g intuitively.

Is this statement true? If it is, how to prove?

答:
Not necessarily. Take any smooth functions $\phi,\psi$ such that
$$
\phi(x) = \begin{cases}
0 &\text{ if } x \leq 0\\
1 &\text{ if } x \geq 1\\
\end{cases}
$$
and
$$
\psi(x) = \begin{cases}
0 &\text{ if } x \geq 0\\
1 &\text{ if } x \leq -1\\
\end{cases}
$$
They exist: for instance, build $\phi$ through a mollifier such as a "bump function" and set $\psi(x)=\phi(-x)$. Note that $\phi(x)\psi(x)=0$ for all $x$.

Then take $f,g$ to be primitives of $\phi,\psi$.


Title: How to convert a geometric series so that exponent matches index of sum?

Body:
I need to convert the following series into a form that works for the equation $$\frac{a}{1-r}$$ so that I can calculate its sum. But the relevant laws of exponents are eluding me right now.

$$\sum_{n=1}^{\infty}\left(\frac{4}{10}\right)^{3n-1}$$

How do I get the 3 out of the exponent?

Body: $$\sum_{n=1}^{\infty}\left(\frac{4}{10}\right)^{3n-1}=\frac{10}{4}\sum_{n=1}^{\infty}\left(\frac{64}{1000}\right)^{n}=\frac{10}{4}\times\frac{\frac{64}{1000}}{1-\frac{64}{1000}}$$


答: If$$f(x)=\begin{cases}e^{-1/x^2}&\text{ if }x\geqslant 0\\0&\text{ if }x<0\end{cases}$$and $g(x)=f(-x)$, then $f'(x)g'(x)$ is always equal to $0$, but neither $f$ nor $g$ is constant.

Title: Does f'(x)g'(x)=0 for every x imply f=const. or g=const.?

Body:
f,g are smooth functions of x in R.

f=const. or g=const. is obviously solution to this equation. I wonder if there exists a function which somewhere f'=0 and g'≠0 and otherwhere g'=0. f'≠0. It seems to be against to the smoothness of f and g intuitively.

Is this statement true? If it is, how to prove?


答:
If you define the angle between two vectors $\bf{x}$ and $\bf{y}$ as
$$
\arccos\left(\frac{\bf{x \cdot y}}{\bf{|x||y|}}\right) \, ,
$$
and the angle between two intersecting curves as the angle between their tangent lines, then the angle pictured in the diagram is a right angle:

To prove this, take any two vectors that act along the tangent lines, for instance $u=-\mathbf{i}$ and $v=\mathbf{j}$. Then,
$$
\mathbf{u} \cdot \mathbf{v} =
\begin{bmatrix}
-1 \\
0
\end{bmatrix}
\cdot
\begin{bmatrix}
0 \\
1
\end{bmatrix}
= 0
$$
and $\arccos(0)=\pi/2$. This approach to defining what an angle is requires you to define $\cos$ and $\arccos$ analytically, as otherwise the definition would be circular.

Title: Are the angles in a halved circle right?

Body:
Imagine a circle cut in half. Now, are the two angles on it right? I can imagine when going limitedly close that they would be.

Got curious about this question after seeing this picture on the internet.


问:
Show that $\forall n\ge3,(2 n-1) \times(2 n-3) \times(2 n-5) \times \cdots \times 5 \times 3 \times 1\ge2 \times 7^{n-2}$
Here's my question: Let $f(n)=(2 n-1) \times(2 n-3) \times(2 n-5) \times \cdots \times 5 \times 3 \times 1$, and I need to prove that for all $n\ge3,f(n)\ge2\times7^{n-2}$.

I have figured out some interesting points, but still I have no clue how to make use of them.

$f(n+1)=2n\times(2n-2)\times\dots\times4\times2=2^n\times n!$

$f(n+1)\times f(n)=(2n)(2n-1)\dots(3)(2)(1)=(2n)!$

Then everything stops here. I'm trying to make sense with the $7^{n-2}$ with the statement, but failed. So is there any other hints for this question? Thanks a lot for your help!

答: Just bound the terms this way: $$f(n)=\underbrace{(2n-1)}_{\geq 7}\underbrace{(2n-3)}_{\geq 7}\ldots\underbrace{(7)}_{\geq 7}(5)\underbrace{(3)}_{\geq 2}$$ and take into account that there are $n$ terms.


问:
Math Induction Proof: $(1+\frac1n)^n < n$
So I have to prove:

For each natural number greater than or equal to 3,
  $$(1+\frac1n)^n<n$$

My work:
Basis step: $n=3$
$$\left(1+\frac13\right)^3<3$$
$$\left(\frac43\right)^3<3$$
$$\left(\frac{64}{27}\right)<3$$
which is true.

Now the inductive step, assume $P(k)=\left(1+\frac1k\right)^k<k$ to be true and prove $P(k+1)=\left(1+\frac1{k+1}\right)^{k+1}<k+1$.

This is where I am stuck because usually you add or multiply by $k+1$ or some similar term.

答:
Two approaches

Bernoulli-like Inductive Step
$$
\begin{align}
\frac{\left(1+\frac1{n+1}\right)^{n+1}}{\left(1+\frac1n\right)^n}
&=\overbrace{\left(\frac{n(n+2)}{(n+1)^2}\right)^{n+1}}^{\lt1}\frac{n+1}n\\
&\lt\frac{n+1}n
\end{align}
$$
This shows that if $\left(1+\frac1n\right)^n\lt n$, then $\left(1+\frac1{n+1}\right)^{n+1}\lt n+1$

Non-Inductive Bernoulli Approach

Bernoulli's Inequality says that for $n\ge2$,
$$
\begin{align}
\left(1-\frac1{n+1}\right)^n
&=\left(\left(1-\frac1{n+1}\right)^{n/2}\right)^2\\
&\ge\left(1-\frac{n/2}{n+1}\right)^2\\[3pt]
&=\left(\frac{n/2+1}{n+1}\right)^2
\end{align}
$$
which is the reciprocal of
$$
\begin{align}
\left(1+\frac1n\right)^n
&\le\left(\frac{n+1}{n/2+1}\right)^2\\[6pt]
&\lt2^2
\end{align}
$$
Thus, $\left(1+\frac1n\right)^n\lt n$ if $n\ge4$.

We just need to verify the inequality for $n=3$:
$$
\left(1+\frac13\right)^3=\frac{64}{27}\lt3
$$


问:
Prove that there is a unique $A\in\mathscr{P}(U)$ such that for every $B\in\mathscr{P}(U), A\cup B = B$
$U$ can be any set.

For the existence element of this proof, I have $A = \varnothing$

But it's for the uniqueness element of this proof where I am having trouble.  So far I have:

$\forall(C\in\mathscr{P}(U))(\forall(B\in\mathscr{P}(U))(C\cup B=B)\rightarrow C\in\mathscr{P}(U) = A\in\mathscr{P}(U))$

Assume $C\in\mathscr{P}(U)$ is arbitrary and assume $\forall(B\in\mathscr{P}(U))(C\cup B=B)$ then:

What I have left to prove is:
$C\in\mathscr{P}(U) = A\in\mathscr{P}(U) $

which is where I am having trouble.  I can say that $C = \varnothing$ which would then equal $A$ but it is not certain that $C = \varnothing$.

答: For uniqueness: The claim has to hold in particular for $B = \varnothing$. So if $A \neq \varnothing$, what would happen?</p>


问:
Prove that $n^{31} \equiv n$ mod $77$ for all integers $n$
Question:

Prove that $n^{31} \equiv n$ mod $77$ for all integers $n$

I've read this question: How to prove $n^5 - n$ is divisible by $30$ without reduction But I don't know how to use the same method to prove my example.

Induction seems very long and unnecessary.

I tried to come up with something using:

$n^7 \equiv n \mod{7}$ and

$n^{11} \equiv n \mod{11}$

but I am not sure that's useful.

I am now assuming I have to show that $n^{31} - n$ is a multiple of 7 and 11 and thus a multiple of 77. How do I arrive in this conclusion?

答:
Let's work $\pmod {7}$: in this case you have $a^6 \equiv 1$ for all $a \neq 0$. Then
$$n^{31} \equiv (n^5)^{6} \cdot n \equiv 1 \cdot n \equiv n$$

If we work $\pmod {11}$, we have $a^{10} \equiv 1$ for all $a \neq 0$. Then
$$n^{31} \equiv (n^3)^{10} \cdot n \equiv 1 \cdot n \equiv n$$