问:
Prove that a sequence of random variables is almost surely a Cauchy sequence
Let $ \left(X_{n}\right)_{n} $ be a sequence of random variables such that for any $ \varepsilon >0 $  we have:

$ \lim_{n,m\to\infty}\mathbb{P}\left(|X_{n}-X_{m}|>\varepsilon\right)>0 $

I want to prove that $ X_n $ has a subsequence wich is Cauchy.

So, here's what I've tried:

For any $ k $ we'll choose $ n_k $ such that for any $ n,m \geq n_k $ we have:

$ \mathbb{P}\left(|X_{n}-X_{m}|>2^{-k}\right)<2^{-k} $.

We could also choose the $ n_k$'s such that $ n_{k+1} > n_k $. So denote: $ Y_k=X_{n_k}$.

I want to show that $ (Y_k)_k $ is a cauchy sequence.

Define $ F=\limsup_{k}\left\{ |Y_{k}-Y_{k+1}|>2^{-k}\right\} =\bigcap_{n}\bigcup_{k\geq n}\left\{ |Y_{k}-Y_{k+1}|>2^{-k}\right\}  $

(This is the event which $ \left\{ |Y_{k}-Y_{k+1}|>2^{-k}\right\}  $ occurs infinitly many times).

I proved, using Borel-Cantelli Lemma, that $ \mathbb{P}\left(F\right)=0 $.

All I have left to do is to conclude that $ Y_k $ is cauchy, so it sufficies to show that the probability that $ Y_k $ is not cauchy is $ 0 $. Denote the event that $(Y_k) $ is not a Cauchy sequence as $ B $.

If I'll show that $ B\subset F $ that will finish the proof. But actually I cannot see how's this even true; Given a sequence, such that for any $ n\in \mathbb{N} $ we can find $ k\geq n $ such that $ |Y_{k+1}-Y_{k}|>2^{-k} $ dosent seem to imply that $ Y_k $ is not Cauchy.

So what can I do to conclude the desirable conclusion?

Thanks in advance.

答:
Indeed $B \subset F$ but it seems you are trying to prove that $F \subset B$...
You might find it easier to check that $F^c \subset B^c$. Namely, assuming $F^c$ means that only finitely many bad events $|Y_k-Y_{k+1}|>2^{-k}$ hold. Now given $\epsilon$ choose $k_0$ larger than the indices of all  these bad events such that also $2^{-k_0}<\epsilon$ and now you can verify $|Y_n-Y_m| <\epsilon$ for $n,m>k_0$.


问:
Solve for $x$, $3\sqrt{x+13} = x+9$
Solve equation: $3\sqrt{x+13} = x+9$

I squared both sides and got $9 + x + 13 = x^2 + 18x + 81$

I then combined like terms $x^2 + 17x + 59 = 0$

I then used the quadratic equation $x= -\frac{17}2 \pm \sqrt{\left(-\frac{17}2\right)^2-59}$

However, the answer is 3

答: When you square the LHS, it should be $9(x+13)$ not $9+x+13$.</p>

ContentLicense: CC BY-SA 3.0

Title: Solve for $x$, $3\sqrt{x+13} = x+9$


问:
Condition for probability of even number heads to be $1/2$
Suppose we flip $n$ coins (each flip is independent, but not necessarily identically distributed). Under what conditions will the probability that we flip an even number of heads be exactly $1/2$?

I know that when all the coins are fair, then the probability is half (see e.g. this thread). But I'm not sure how to handle the case if the coins aren't fair.

答:
Pick a coin, suppose it comes up heads with probability $p$. Suppose the other $n-1$ coins give an even number of heads with probability $q$. Then we need
$$p(1-q)+(1-p)q=\frac{1}{2}\implies p+q-2pq=\frac{1}{2}\implies(2p-1)(2q-1)=0.$$
So either $p=\frac{1}{2}$ or $q=\frac{1}{2}$. Inducting downwards implies that the probability of an even number of heads is $\frac{1}{2}$ iff at least one of the coins is fair.

ContentLicense: CC BY-SA 4.0


问:
If $Z = X + Y$ where $X$ and $Y$ are two independent, uniform random variables, then what is the conditional distribution of $X | Z$?
So I have been struggling with this question for a while. Suppose $X$ is uniformly distributed over an interval $(a, b)$ and $Y$ is uniformly distributed over $(-\sigma, \sigma)$, $X$ and $Y$ are independent. Consider a random variable $Z = X + Y$. What would be the conditional distribution of $X | Z$, i.e. $F_{X|Z}(x)$?

I know that $F_X(x) = \frac{x - a}{b-a}$ and $F_y(y) = \frac{y + \sigma}{2\sigma}$ but I am having trouble figuring out $F_{X|Z}(x)$. Could anyone help me? My gut says that $X|Z$ should be uniformly distributed over $(Z - \sigma, Z + \sigma)$ but I can't figure out how to prove this.

答:
You can show that the conditional density of $X$ given $X+Y$ is uniform by applying the change of variables formula. Define the transformation $(W,Z):=(X,X+Y)$. The joint density of $(W,Z)$ is
$$
f_{W,Z}(w,z)=f_{X,Y}(x(w,z),y(w,z)) = f_{X,Y}(w, z-w).\tag1$$ (Check that the Jacobian of the transformation is $1$.) Since the joint density of $(X,Y)$ is
$$f_{X,Y}(x,y)=I_{(a,b)}(x)I_{(-\sigma,\sigma)}(y),\tag2$$ write (1) as
$$
f_{W,Z}(w,z)=I_{(a,b)}(w)I_{(-\sigma,\sigma)}(z-w).\tag3
$$
Plotting in the $(w,z)$-plane, we find the density (3) is uniform over a skewed diamond shape. In fact, for fixed $z$ the value of (3) is zero unless $a<w<b$ and $z-\sigma<w<z+\sigma$, in which case the value of (3) is $1$. The key observation is that for fixed $z$, the expression (3), when nonzero, is constant as a function of $w$. This means that when we divide (3) by the marginal density $f_Z(z)$ (note we don't even have to compute the value!), the resulting conditional density $f_{W\mid Z}(w\mid z)$ remains constant as a function of $w$.


问:
Find the equation of hyperbola centered at (0,0), transverse axis horizontal, passing through points (3,5) and (2,-3).
With this problem, I substuted the given points to the standard form of a horizontal hyperbola, $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$

and got
$$\frac{3^2}{a^2}-\frac{5^2}{b^2}=1→\frac{9}{a^2}-\frac{25}{b^2}=1 $$
$$\frac{2^2}{a^2}-\frac{-3^2}{b^2}=1→ \frac{4}{a^2}-\frac{9}{b^2}=1$$

What should I do next?

Thank you very much!

答: Hint: You have $2$ equations in the $2$ unknowns of $a^2$ and $b^2$. The usual thing to try to do is to eliminate one of the variables so you can then solve for the other variable. You can do that here by, for example, subtracting $4$ times the first equation from $9$ times the second equation to eliminate the term with $a^2$.


问:
How do we solve the equation $3\sec^{2}(x) - 4 = 0$?
I am confused about the answer to this question:

$$3\sec^{2}(x) - 4 = 0$$

I got the answer

$$x = \frac{\pi}{6} + 2n\pi\quad\text{or}\quad x = \frac{5\pi}{6} + 2n\pi$$

However, the book's answer is $\pi/6 + n\pi, 5\pi/6 + n\pi$

Isn't the period of a secant function $2\pi$? If so, why is the answer saying $\pi$?

答:
WolframAlpha offers $2$ solutions here
where

$$x = \pi n - \frac{5 \pi}{6}, n \in\mathbb{Z}$$

$$x = \pi n - \frac{\pi}{6}, n \in\mathbb{Z}$$

Your addition of $2n\pi$ is always a multiple of $360^\circ$ so the results would be the same but $\quad n\pi\quad$ would take you, for example from quadrant $(1)$ to quadrant $(3)$ where the $\sin$ and $\cos$ are both negative. In any case the $\sec^2$ would still be  positive so a rotation of only $180^ \circ$ will not affect the solution.


问:
Prove there is a sequence in $x_n$ in $A$ with $x_n \rightarrow x$.
Let $(X,d)$ be a metric space and let $A \subset X$. Let $x \in \overline{A}-A$.

Prove there is a sequence in $x_n$ in $A$ with $x_n \rightarrow x$.

I am not sure how to even start this. I am not good at building sequences. Any hints would be helpful as so I can start to write something up. Even the first line would help or maybe what the goal is.

$\textbf{Convergent sequence:}$ $x_n \rightarrow x$ means $\forall \, \text{open} \, U \subset X$ with $x \in U \,\, \exists N$ such that $x_n \in U \,\, \forall n \geq N$.

答: By definition, for every $n \in \mathbb{N}$, $B(x, 1/n) \cap A \not = \emptyset$. So there is a $x_n \in B(x, 1/n) \cap A$ for every  $n \in \mathbb{N}$. I will leave it to you to check that the sequence $\{x_n\}_{n \in \mathbb{N}}$ works.


问:
Unsigned Integer Binary Subtraction
So I am having a bit of an issue. First, what is the difference between doing an unsigned binary integer subtraction and doing a signed integer subtraction? I think that is what is confusing me. For an unsigned integer, I was told that you have to add a 0 next to the most significant bit (I may be wrong), but I don't feel comfortable with that. After watching a couple of videos on youtube, I tried to the subtraction.

0011 0001 – 1010 1011

My thought process:
I decided to take 1010 1011 and do one's complement and then two's complement and then add it to 0011 0001.

Click here to see my work

答:
This is a reasonable answer from a mathematical point of view, but not a great answer from computer science point of view.  To understand this you want to look at the most common two's complement binary math  https://notesformsc.org/2s-complement-subtraction/.  This also overlaps an interesting subject in math, modular arithmetic (since register sizes are fixed) used to prevent rollover issues when doing timing math see http://www.gammon.com.au/forum/?id=12127.

The C++ program below shows that if you interpret the number as time subtracting T2-T1 works even if the counter has rolled over (gone from large to small) once. It also shows that if you allow promotion to large types you get, what may be initially, unexpected values.

```<code>#include <iostream>
#include <cmath>
#include <bitset>

void take_short_diff(unsigned short T1, unsigned short T2)
{
    //Force all math to happen without promotion
    unsigned char diff = (unsigned char)(T2 - T1);
    //Ints below are just so it doesn't print as characters, they don't impact the problem
    std::cout << "\n\nT1  =" << (int)T1 << "\nT2  =" << (int)T2 << "\ndiff=" << (int)diff;
    //Show the binary pattern
    std::cout << "\nT1  =" << std::bitset<8>(T1) << "\nT2  =" << std::bitset<8>(T2) << "\ndiff=" << std::bitset<8>(diff);
    //Allow promotion to short
    short diff_short = T2 - T1;
    std::cout << "\ndiff short=" << diff_short;
    std::cout << "\n" << std::bitset<16>(diff_short);
    //Allow promotion to unsigned short
    unsigned short diff_short_u = T2 - T1;
    std::cout << "\ndiff short unsigned=" << diff_short_u;
    std::cout << "\n" << std::bitset<16>(diff_short_u);
}

void main()
{
    take_short_diff(0, 2);
    take_short_diff(253, 255);
    take_short_diff(254, 1);
}
</code>```

Title: Unsigned Integer Binary Subtraction

Body:
This is a reasonable answer from a mathematical point of view, but not a great answer from computer science point of view.  To understand this you want to look at the most common two's complement binary math  https://notesformsc.org/2s-complement-subtraction/.  This also overlaps an interesting subject in math, modular arithmetic (since register sizes are fixed) used to prevent rollover issues when doing timing math see http://www.gammon.com.au/forum/?id=12127.

The C++ program below shows that if you interpret the number as time subtracting T2-T1 works even if the counter has rolled over (gone from large to small) once. It also shows that if you allow promotion to large types you get, what may be initially, unexpected values.

#include <iostream>
#include <cmath>
#include <bitset>

void take_short_diff(unsigned short T1, unsigned short T2)
{
    //Force all math to happen without promotion
    unsigned char diff = (unsigned char)(T2 - T1);
    //Ints below are just so it doesn't print as characters, they don't impact the problem
    std::cout << "\n\nT1  =" << (int)T1 << "\nT2  =" << (int)T2 << "\ndiff=" << (int)diff;
    //Show the binary pattern
    std::cout << "\nT1  =" << std::bitset<8>(T1) << "\nT2  =" << std::bitset<8>(T2) << "\ndiff=" << std::bitset<8>(diff);
    //Allow promotion to short
    short diff_short = T2 - T1;
    std::cout << "\ndiff short=" << diff_short;
    std::cout << "\n" << std::bitset<16>(diff_short);
    //Allow promotion to unsigned short
    unsigned short diff_short_u = T2 - T1;
    std::cout << "\ndiff short unsigned=" << diff_short_u;
    std::cout << "\n" << std::bitset<16>(diff_short_u);
}

void main()
{
    take_short_diff(0, 2);
    take_short_diff(253, 255);
    take_short_diff(254, 1);
}


问:
derivative of square root of $x_1$ and $x_2$
I am confusing with calculation of derivative of $$\sqrt{x_1x_2}$$ I am very thankful if anyone help me out of this problem.
my question is that,

(1) $x_1$ and $x_2$ are two different variable i just want to calculate the derivative with respect to $x$.

(2) if $x_1=x_2$ what would be the value of derivative.

Thanks in advance

Body:
Ok then: so $\,x_i=x_i(x)\;,\;i=1,2\,$ , and then:

$$\frac d{dx}\left(\sqrt{x_1x_2}\right)=\frac{x_1' x_2}{2\sqrt{x_1x_2}}+\frac{x_1x_2'}{2\sqrt{x_1x_2}}=\frac{x_1'x_2+x_1x_2'}{2\sqrt{x_1x_2}}$$


答: Observe that $\bar A-A\subseteq\partial A$, where $\partial A$ is the boundary of $A$. From there, you can use the definition of the boundary to prove the existence of the sequence.

Body:
Let $(X,d)$ be a metric space and let $A \subset X$. Let $x \in \overline{A}-A$.

Prove there is a sequence in $x_n$ in $A$ with $x_n \rightarrow x$.

I am not sure how to even start this. I am not good at building sequences. Any hints would be helpful as so I can start to write something up. Even the first line would help or maybe what the goal is.

$\textbf{Convergent sequence:}$ $x_n \rightarrow x$ means $\forall \, \text{open} \, U \subset X$ with $x \in U \,\, \exists N$ such that $x_n \in U \,\, \forall n \geq N$.


问:
Solutions of $f(k,s)=0$ as $k\to\infty$
Let $f(k,s) = D(s) + kN(s)$ where $D(s)$ and $N(s)$ are polynomials of $s \in \mathbb{C}$ such that $\text{Deg}(D) = n, \ \text{Deg}(N) = m$ and $n\ge m$. Also $D(s)$ and $N(s)$ doesn't have common factor. Find the values of $s$ such that $$\lim_{k\to \infty}f(k,s) = 0  \tag{1}$$ My try: Assume that $N(s) \not = 0$. Obviously, in this case $\lim_{k\to \infty}f(k,s)$ is divergent for any $s$. Now suppose $N(s) = 0$ and we have $\lim_{k\to \infty}f(k,s) = D(s)$. So the answer is $D(s) = N(s) = 0$ which is impossible since $D(s)$ and $N(s)$ doesn't have common factor. What's wrong in my reasoning?

Also is $(1)$ equivalent to find $s$ such that $f(k,s) = 0$ and then $\lim_{k \to \infty} s(k)$? I mean that we first find the roots of $f(k,s)$ with respect to $k$ and see what happens to the roots when $k \to \infty$.

Edit: The correct answer is as follows. Divide the equation $D(s) + kN(s) = 0$ by $k$ and let $k \to \infty$. So we have $$\lim_{k \to \infty}(\frac{D(s)}{k} + N(s)) = N(s) = 0$$ I don't know why my argument is flawed and whether the mentioned equivalence holds.

Edit2: Thanks to @Alex Ravsky, it turns out that the true question is

What are the solutions of $f(k,s)=0$ as $k\to\infty$?

答:
If $N$ and $D$ are relatively prime, then $q(s)N(s)+r(s)D(s)=1$ for some $q(s),r(s)\in \mathbb{C}[s]$.

Using this, $r(s)f(k,s)$ can be re–written as $r(s)f(k,s)=1+(kr(s)-q(s))N(s)$. Now if $N(s)\not =0$, then $\lim_{k\rightarrow \infty} q(s)f(k,s)\not =0$ because $kr(s)-q(s)\not =0$ by taking $k$ large enough for a fixed $s$. Furthermore, as $k\rightarrow \infty$ for any fixed value the RHS blows up.

After writing this, I am wondering why you are interested in this question since it numerical experimentation should indicate that the polynomial blows up.

Title: Solutions of $f(k,s)=0$ as $k\to\infty$


问:
Can I get a light hint for this binomial proof?

Consider the geometric series

$$S=1+(1+x)+(1+x)^2+\dots+(1+x)^n$$

(a) Show that
$$S=\frac{(1+x)^{n+1}-1}{x}$$

(b) Hence show that

$$S={n+1\choose1}+{n+1\choose2}x+\dots+{n+1\choose r+1}x^r+\dots+{n+1\choose n+1}x^n$$

(c) Hence prove that

$${n\choose r}+{n-1\choose r}+\dots+{r\choose r}={n+1\choose r+1}$$

I have successfully solved parts a) and b). However, I am struggling with c). I can tell it's the hockey stick identity but the thing that's getting me is the hence. How can I use the previous parts to prove it? I just need a little push. Perhaps it has to be a specific binomials relationship. I tried using the definition of symmetry but that didn't work.

答:
(A) Ok so from $S=1+(1+x)+(1+x)^2+\dots+(1+x)^n$, consider the coefficient of $x^r$, which gives the following ways:
${ 0 \choose r} + { 1 \choose r } + { 2 \choose r} + \ldots { n \choose r} = 0 + 0 + ... + 0 + { r \choose r } + { r+1 \choose r } + { r+2 \choose r} + \ldots  + { n \choose r }$ ie. the LHS of what we want to prove in part (c).

(B) Also, the coefficient of $x^r$ from $
S={n+1\choose1}+{n+1\choose2}x+\dots+{n+1\choose r+1}x^r+\dots+{n+1\choose n+1}x^n
$
gives ${n+1\choose r+1}$. Equating coeffiecnts with (A) gives $
{n\choose r}+{n-1\choose r}+\dots+{r\choose r}={n+1\choose r+1}
$

Calvin's edit: This is exactly it. However, you have skipped several steps (of which were listed out in the problem), so I still am unable to fully determine if you truly understand what is happening, or have skipped through those parts.

(C) OR the part which I am getting wrong: from $
S=\frac{(1+x)^{n+1}-1}{x}
$ to get the coefficient of $x^r$ consider $(1+x)^{n+1}-1$. It would be ${n+1\choose {r}}x^r -1$. But since there is and $x$ in the denominator, it needs to be $ \frac{{n+1\choose {r+1}}x^{r+1} -1}{x} = {n+1\choose {r+1}}x^{r} - x^{-1}$. Then equating coefficients of $x^r$ from (A) also gives $
{n\choose r}+{n-1\choose r}+\dots+{r\choose r}={n+1\choose r+1}
$

Calvin's edit: This part has several errors in logic / reasoning, or at least errors in not writing up exactly what you meant. E.g. to get the coefficient of $x^r$, we would want the coefficient of $ x^{r+1}$ in the numerator.

(C)$_2$: Consider $(1+x)^{n+1}-1$. The general term is ${{n+1}\choose{k}}x^k$ and we want it to be $x^r$. But we can't sub that in straight away since the full statement is $S=\frac{(1+x)^{n+1}-1}{x}$. Therefore from ${{n+1}\choose{k}}x^k$

${{n+1}\choose{k}}x^k$ --> ${{n+1}\choose{k}}x^k -1$ --> $\frac{{{n+1}\choose{k}}x^k -1}{x}$ --> ${{n+1}\choose{k}}x^{k-1} - x^{-1}$

and then $k - 1 = r$ so ${n+1\choose r+1}x^r - x^{-1}$

Calvin's edit:

No, the general term is not ${n+1 \choose k } x^k$. Check $k= 0 $.

I do not know what you mean by "we want (the general term) to be $x^r$". AFAIK "We want to find the coefficient of the $x^r$ term". This is a distinction that you don't seem to be making.

You seem to be claiming that $(1+x)^{n+1} - 1  = {n+1 \choose k } x^k -1$. This is not true. Please explain what you are thinking.

The goal of this part is to show that the coefficient of $ \frac{ ( 1 + x) ^ {r+1} - 1 }{ x} $ is ${n+1 \choose r+1}$. You skipped showing in part (B), and claimed that you know why it's true.


问:
Subset of countable set is countable proof
I'm self studying Understanding Analysis and there's an exercise that asks to prove that if $ A\subseteq B $ and $B$ is countable, then $A$ is either countable or finite. I want to be sure that my proof is not wrong or badly written.

Being finite is trivial, so let's suppose it's not.
In the exercise, we assume there exists $f:N \to B$ which is 1-1 and onto and $n_1=\min\{n\in N:f(n)\in A\}$. As a start to a definition of $g:N \to A$, we set $g(1)=f(n_1)$. These are given by the exercise and we're asked to continue the process inductively.

I construct sets $a_n=\{n_c \text{ for } c \le n\}$ for example $a_1=\{n_1\}$, $a_2=\{n_1, n_2\}$ etc.

Then I construct: $$n_c=\min\{n\in N:f(n)\in A\ and\ n\notin a_{n_{c-1}}\}$$
and thus $g(c)=f(n_c)$ for every $c\in N$ gives us a $1-1$ function from $N$ onto $A$.

I would love to hear any sort of feedback, thank you!

答:
It's clear that you have an effective solution in mind, but your proof doesn't accurately convey it. Informally, the crucial idea you have is to define $n_c$ to be the smallest $n$ so that $f(n)$ is in $A$ and was not yet enumerated - but the structure of your proof doesn't effectively communicate this.

A good way to start proofs is just to succinctly state your goal, introducing all variables necessary to do so. In this case, I'd begin with:

Let $B$ be a countable set and $A\subseteq B$ be some subset thereof. We claim that $A$ is either countable or finite.

You've started by writing, a bit confusingly, that "being finite is trivial" - but when you say something is trivial (which you usually shouldn't!), it means you have some statement in mind that you're leaving out the proof for. That's not what you're doing here. I think you mean to say is something along these lines:

We will prove this by showing that, if $A$ is not finite, then it is countable.

Where we note that proving "X or Y" is the same as proving "not X implies Y" - and we're not claiming that it's easy to prove that $A$ is finite, but rather that we're going to assume it's not - and that the form of the statement we're trying to show justifies a proof using this assumption. Also, note that the assumption we extract is that $A$ is not finite (i.e. $A$ is infinite) whereas you've written "suppose it's uncountable" (which means "neither finite nor countable") - although I suspect this is just a slip of language more than a conceptual inaccuracy.

Your next step is fine:

We assume there exists $f:N \to B$ which is 1-1...

Though, as a minor nitpick, I'd note that you should write "let" instead of "assume" or even "Since $B$ is countable, there exists..." since there is no assuming going on - you're just writing what it means to be countable.

Then, you sort of define a bunch of things a bit aimlessly - you should always tell your reader what you're doing and try to build things in steps. My inclination would be to write something like:

We will now define a sequence of values $n_i$ with the intention that $f(n_1),\,f(n_2),\ldots$ will enumerate $A$. We first define
$$n_1=\min\{n\in \mathbb N: f(n)\in A \}.$$
To describe further values,  we define the set $A_k=\{n_i : i \leq  k\}$ to be the set of prior values of the sequence and let
$$n_k = \min\{n\in\mathbb N : f(n)\in A\text{ and }n\not\in A_{k-1}\}.$$

This is more or less what you've written, but expanded out with set builder notation and with some minor changes to conventions (e.g. using capital letters for sets and using variables $i,j,k$ for indices). Your proof is missing something here, however: you need to show that the definition makes sense - in particular, that a minimum exists. This is where you use that $A$ was assumed infinite:

Note that the set $\{n\in\mathbb N : f(n)\in A\text{ and }n\not\in A_{k-1}\}$ must be non-empty since every element of $A$ can be written as $f(n)$ for some $n$ since $f$ is onto and, since we assumed $A$ is infinite, it cannot be the case that $n_1,n_2,\ldots,n_{k-1}$ are the only indices so that $f(n)\in A$.

This is a rather crucial step since otherwise you fail to use the assumption that $A$ is infinite - which is an immediate red flag. In more nitpicking terms, you might notice that you could avoid ever assuming $A$ is infinite by splitting into cases here; if that set were ever empty, we would have directly proved that $A$ is finite - and mathematicians often feel that a direct proof is preferable to proofs by contradiction*.

Only after defining this sequence would I define $g$ - and I would be sure to tell the reader the purpose of $g$ when introducing it. Note that saying somewhere that $g$ is 1-1 is really important since it's literally the whole point of constructing it!

We now define a 1-1 function $g:\mathbb N\rightarrow A$ by the following rule:
$$g(k)=f(n_k)$$

Your proof has a problem where you write $g(n_c)=f(n_c)$ - the input to $g$ needs to be the index in the sequence, not the value, but I assume this is a typo.

The proof you wrote ends there. This might be okay as long as everyone involved understands that there is a claim left unproven and agrees that it is trivial - but let me be pedantic: We never showed that $g$ is 1-1! We have a good definition, but a definition is not a proof. I won't spoil this for you, but there are basically three claims you should finish, which I'll list in order of how difficult they are to prove:

$g(k)\in A$ for every $k\in\mathbb N$.

$g$ is one-to-one (as in "injective"). A proof of this ought to start with:

Let $i,j\in\mathbb N$ be such that $g(i)=g(j)$. We will show that $i=j$.

and then proceed to some convincing argument that $i$ really does equal $j$.

$g$ is onto. This proof is more difficult since you have to find a preimage for an arbitrary $a\in A$. The proof should begin:

Let $a\in A$. We will construct some $k\in\mathbb N$ such that $g(k)=a$.

If you haven't already thought this through, this is a somewhat interesting challenge. You'll need to use that $f$ is onto in proving this, and likely also that the sequence $n_1,n_2,\ldots$ is strictly increasing.

Fixing up the writing of the first half of the proof and filling in the proof that $g$ is 1-1 should give you a nice solution to the exercise.

(*Okay, strictly speaking, the form of the proof you have wouldn't usually be called a proof by contradiction - but it's also not a direct proof. Loosely speaking, some mathematicians find it objectionable or inelegant to assume that either "X or not X" and when you can avoid doing so, it's often wise to)


问:
Do I use combinations or permutations when calculating the probability of something happening?
Alexander runs a doggy daycare, where 11 dogs are currently enrolled. The collection of dogs includes 4 of Alexander's favourite type of dog, golden retrievers.

If Alexander randomly picks 7 dogs to play in the park during the first time slot, what is the probability that exactly 3 of the chosen dogs are golden retrievers?

Write your answer as a decimal rounded to four decimal places.

答:
Do I use combinations or permutations when calculating the probability of something happening?

Typically, such a problem is best solved by expressing the probability as

$$\frac{N\text{(umerator)}}{D\text{(enominator)}}.$$

Here, the most critical point is that whichever approach that you use, you must be consistent re computing $N$ and computing $D$.  That is, if $N$ is computed as the total # of favorable combinations (rather than permutations), then $D$ must be computed as the total number of possible combinations.

Factors that I consider when choosing whether to use permutations or combinations:

Whichever is easiest.  If for example, computing $D$ as permutations is immediate, without much thought, that will influence me.  However, I sometimes will choose a moderately harder choice for $D$ (e.g. combinations instead of permutations), if it significantly eases the computation of $N$ (as combinations instead of permutations).

Whichever facilitates mutually exclusive cases, or helps avoid over-counting.  This issue is tricky, varies from problem to problem, often requires creativity, and is too broad a subject to detail here.  I will simply say that this issue can easily affect my choice of whether to use permutations or combinations.

Whichever approach promotes elegance.  Sometimes, I will be able to visualize an approach (either combinations or permutations) that simplifies computing $N$ and $D$ consistently, and avoids delving into sub-cases.  In such a situation, I will opt for elegance because it significantly reduces the analysis/computations required.


问:
Interesting cyclic infinite nested square roots of 2 and cosine values
It is interesting to note that any angle between 45° to 90° satisfying $1\over4$ < $p \over q$ <$1\over2$ where $ p \over q$ is of form $p = 2^n $ and $q$ is an odd number satisfying $2^{n+1} <q <2^{n+2}$ can be represented as cyclic infinite nested square roots of 2 (Hereafter referred as $cin\sqrt2$) refer here for an example of interesting cyclic infinite nested radical

Let us consider certain cosine angles with odd numbers starting from 5 and its exponents and see the number of $'+'$ and $'-'$ signs in single cycle

$2\cos(\frac{2\pi}{5})=\sqrt{2+2\cos(\frac{4\pi}{5}})=\sqrt{2-2\cos(\frac{\pi}{5}})=\sqrt{2-\sqrt{2+2\cos(\frac{2\pi}{5}}})$. We can observe expansion goes infinite and results in cyclic infinite nested square roots of 2 for $2\cos(\frac{2\pi}{5})$ or $2\cos72^\circ$

$2\cos\frac{2\pi}{5} = cin\sqrt2[1-1+]$ single cycle contains $1 -$ sign and $1 +$ sign and is simple representation of $\sqrt{2-\sqrt{2+...}}$

$2\cos(\frac{8\pi}{25}) = \sqrt{2+2\cos(\frac{16\pi}{25}}) = \sqrt{2-2\cos(\frac{9\pi}{25}}) = \sqrt{2-\sqrt{2+2\cos(\frac{18\pi}{25}}}) = \sqrt{2-\sqrt{2-2\cos(\frac{7\pi}{25}}}) = \sqrt{2-\sqrt{2-\sqrt{2+2\cos(\frac{14\pi}{25}}}}) = \sqrt{2-\sqrt{2-\sqrt{2-2\cos(\frac{11\pi}{25}}}}) = \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+2\cos(\frac{22\pi}{25}}}}}) = \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-2\cos(\frac{3\pi}{25}}}}}) = \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+2\cos(\frac{6\pi}{25}}}}}}) = \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+2\cos(\frac{12\pi}{25}}}}}}}) = \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+2\cos(\frac{24\pi}{25}}}}}}}}) = \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-2\cos(\frac{\pi}{25}}}}}}}}) = \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+2\cos(\frac{2\pi}{25}}}}}}}}}) = \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2+2\cos(\frac{4\pi}{25}}}}}}}}}}) = \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+2\cos(\frac{8\pi}{25}}}}}}}}}}})$

Above cycle repeats infinitely

$2\cos\frac{8\pi}{25} = cin\sqrt2[4-2+1-3+]$ single cycle contains $5 +$ signs and $5 -$ signs and is simple representation of $\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+...}}}}}}}}}}$

Steps can be simplified as follows.

When the angle exceeds $\frac{\pi}{2}$ we are applying the basic cosine angle identity $2\cos(\pi-\theta) = -2\cos(\theta)$

Doubling happens infinitely with $\frac{\pi}{4} < \theta < \pi$

As the denominator in the angle is odd number, when the $\frac{numerator}{denominator} > \frac{1}{2}$ the signs changes while doubling the cosine angle expansion as nested radical.

Above steps are programmable.

I have created a program to calculate the number of $'+'$ and $'-'$ signs in cyclic infinite nested square roots of 2 for cosine values in Python as follows and anyone can verify which will provide the result for any odd number other than 3 (I am not expert in this)

```<code>import time
n = int(input("Enter an odd number to get single cycle cinsqrt2 ")) #odd number is denominator
# steps to calculate numerator as <span class="math-container">$2^n$</span> so that fraction lies between 0.25 and 0.5
i = 0
for i in range(n):
    if 2 ** i > n and n < 2 ** (i + 1):
        break
numerator = 2 ** (i - 2)
print("Numerator is", numerator)
#print("The Angle generated is", (numerator*180)/n, (numerator*180)//n,'+', ((numberator*180)%n)/n)
halfway_of_n = (n - 1) // 2 #to decide the sign, we need halfway number
# print("Half way to n is", halfway_of_n)
lst = []
r = numerator * 2
begin = time.time()
while r != numerator:
    if r > halfway_of_n:
        r = n - r
        r = r * 2
        lst = lst + ['-']

    else:
        r = r * 2
        lst = lst + ['+']
lst = lst + ['+']
print(len(lst))
count = lst.count('-')
print('No of minus signs is/are ', count)
count = lst.count('+')
print('No of plus signs is/are ', count)
print(lst) #list containing '+' and '-' signs

end = time.time()

print('Program execution time is', end - begin)
</code>```

$2\cos\frac{32\pi}{125} = cin\sqrt2[2-4+1-1+1-2+2-1+2-1+1-1+2-2+3-1+3-3+1-2+1-1+5-1+1-5+]$ single cycle contains $25 +$ signs and $25 -$ signs and (representing this as a infinite nested radical will be big and occupy big space. Therefore I'm representing in the simplified form as shown above)

Exciting pattern emerges for cosine values having exponent of 5 in the denominator and number of + and - signs in the $cin\sqrt2$ which also exponentially growing as follows

for $5^1$ in denominator $5^0 + 5^0$ signs in total (in single cycle)

for $5^2$ in denominator  $5^1 + 5^1$ signs in total (in single cycle)

for $5^3$ in denominator  $5^2 + 5^2$ signs in total (in single cycle)

for $5^4$ in denominator  $5^3 + 5^3$ signs in total (in single cycle)

and so on

Question:
Any other way to get these signs inside the nested radicals in a better way?

答:
Main Question(s)

I'd like to generalize the setup of the OP a bit, and strip out some of the context to focus on the main idea.

Setup

Let $q>2$ be an odd integer and let $p$ be an integer with $0<p<\dfrac{q}2$.

Then note that $0<\cos\left(\dfrac{p\pi}{q}\right)$. We can use a version of the cosine half angle formula to find that $2\cos\left(\dfrac{p\pi}{q}\right)=\sqrt{2+2\cos\left(\dfrac{2p\pi}{q}\right)}$. If $2p>\dfrac{q}2$, the inner cosine is no longer positive; in that case, we can first use $\cos(\pi-x)=-\cos(x)$ to write $2\cos\left(\dfrac{2p\pi}{q}\right)=-2\cos\left(\dfrac{(q-2p)\pi}{q}\right)$, and have $\cos\left(\dfrac{(q-2p)\pi}{q}\right)>0$.

This means that, for $0<p_1<\dfrac{q}2$, we can choose $0<p_2<\dfrac{q}2$ so that $2\cos\left(\dfrac{p_1\pi}{q}\right)=\sqrt{2\pm2\cos\left(\dfrac{p_2\pi}{q}\right)}$ where we have a $+$ sign and $p_2=2p_1$ if $2p_1<\dfrac{q}2$ and we have a $-$ sign and $p_2=q-2p_1$ otherwise. Repeating this step gives rise to a "$p$ sequence": $p_1,p_2,\ldots$.

But by doing calculations, it seems empirically that all the sequences for a given $q$ and starting $p$ are actually periodic, which would mean that we can always write $2\cos\left(\dfrac{p\pi}{q}\right)$ as a nested radical of the form described above, with $2\cos\left(\dfrac{p\pi}{q}\right)$ on the inside (suggesting a sort of infinite nesting).

For example, with $q=17$ and an initial $p=8$, we obtain $$2\cos\left(\dfrac{8\pi}{17}\right)=\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+2\cos\left(\dfrac{8\pi}{17}\right)}}}}$$

Questions

Is the sequence of $p$ values guaranteed to be periodic (so that we can always write these cosines as these special nested radicals)?

When it is periodic, is there another way to find/understand the pattern of $+$ and $-$ signs in the radical expression?

Main Answers

Answer 1

Yes, under our assumption that $q$ is odd, the sequence of $p$ values is guaranteed to be periodic. Note that if $q$ is not odd, then this would not be guaranteed as in the calculation of $2\cos(1^\circ)=2\cos\left(\dfrac{1\pi}{180}\right)$ from the previous question.

Answer 2

Starting with $p$, the sequence of $+$/$-$ signs can be written with the modulo operation, borrowing % from Python, as follows:
$$n^\text{th}\text{ sign}=\begin{cases}
+ & \text{ if }\left(2^{n}p\right)\%\left(2q\right)<\dfrac{q}{2}\\
+ & \text{ if }\left(2^{n}p\right)\%\left(2q\right)>\dfrac{3q}{2}\\
- & \text{ otherwise}
\end{cases}$$

We can also phrase this in an equivalent and suggestive way using quadrants of the plane and the standard position of angles greater than $2\pi$.
$$n^\text{th}\text{ sign}=\begin{cases}
+ & \text{ if }\dfrac{2^{n}p\pi}{q}\text{ is in Quadrant I or IV}\\
- & \text{ otherwise}
\end{cases}$$
Using $+1$ and $-1$ to represent the signs, and the sign function, this is equivalent to
$$n^\text{th}\text{ sign}=\mathop{\mathrm{sgn}}\left(\cos\left(\dfrac{2^{n}p\pi}{q}\right)\right)$$

Proof of 1

Even without our assumption that $q$ is odd, since there are only finitely many values that the $p$s can take on, the sequence must be eventually periodic by the pigeonhole principle.

But if $q$ is odd, the sequence is guaranteed to be periodic. It may be easiest to show this using some facts about groups.

Let $G_q$ be the multiplicative group of integers modulo $q$. Then we can quotient out by the $2$-element subgroup represented by $\{1,-1\}$ to get a new group $G_q^*$, with elements of the form $[n]$ where $0<n<q/2$ and $n$ is relatively prime to $q$. Since $q$ is odd, $2$ is a representative of an element of $G_q$, and hence $G_q^*$. And since $p$ and $q$ only appear in a fraction, we can assume WLOG that $p$ and $q$ are relatively prime, so that $[p]\in G_q^*$.

Since $[2]$ generates a subgroup of the finite group $G_q^*$, the sequence $[p],[2][p],[2]^2[p],\ldots$ must be periodic. But note that if $[n]\in G_q^*$, then $[2][n]=[2n]=[-2n]=[q-2n]$, so that the smallest positive representatives of $[p],[2][p],[2]^2[p],\ldots$ are exactly the terms of the original $p$ sequence.

Proof of 2

Note that the first sign for $p$ corresponds to $\mathop{\mathrm{sgn}}\left(\cos\left(\dfrac{2p\pi}{q}\right)\right)$, and then for the next sign we start from a new value of $p$ based on that sign. By the $2\pi$ periodicity of $\cos$ and the denominator of $q$ in $\cos\left(\dfrac{m\pi}{q}\right)$, the signs can only depend on $(2^np)\%(2q)$.

Certainly, if $(2^np)\%(2q)<\dfrac{q}{2}$, then the residue $r=(2^np)\%(2q)$ of $2^np$ modulo $2q$ is the same as the residue modulo $q$; and this residue is small enough that there is no $q-r$ step needed, so the sign is $+1$.

Similarly, if $\dfrac{q}{2}<(2^np)\%(2q)<q$, then the residue $r$ of $2^np$ modulo $2q$ is the same as the residue modulo $q$; but this residue is large enough we have a $q-r$ step and the sign is $-1$.

And if $\dfrac{3q}{2}<(2^np)\%(2q)$, then if $R,r$ are the residues modulo $2q,q$, respectively, we have $R-q=r$. And since $R>\dfrac{3q}{2}$, $r>\dfrac{q}2$, so that we are dealing with $q-r=q-(R-q)=2q-R<\dfrac{q}{2}$. Thus, the sign is $+1$.

Similarly, if $q<(2^np)\%(2q)<\dfrac{3q}{2}$, then the sign is $-1$.

Context

I'd like to share how I found references covering these results, and some other related facts about this construction. For $q=25$, I noted the $p$ sequence (starting from $p=1$ instead of $p=8$) $1,2,4,8,9_{=25-16},7,11,3,6,12,1,\ldots$. If you search Sloane's for 1,2,4,8,9,7, you find just one sequence: A334430. This sequence basically lists the all the $p$ sequences ("cycles") for all the odd $q$, except that they start cycles at twice the lowest number of the cycle, by convention. The $1$, in "$1,2,4,8,9,7$", that was found was actually the end of the sole cycle for $q=23$, not the beginning of a cycle for $q=25$.

References

The page for A334430 mentions and links many relevant texts, including (in chronological order of first publication):

"Polygons and Chaos" by Jay Kappraff and Gary W. Adamson, published in Journal of Dynamical Systems and Geometric Theories and earlier in the freely-available proceedings of the Bridges conference

Trigonometrie und unterhaltsame Zahlentheorie by Carl Schick (Table of Contents scan provided by Deutsche Nationalbibliothek.

A Mathematical Tapestry: Demonstrating the Beautiful Unity of Mathematics by Peter Hilton and Jean Pedersen

"Modified Congruence Modulo $n$ with Half The Amount of Residues" by Tim Beyne and Gerold Brändli (arXiv)

"On the Equivalence of Three Complete Cyclic Systems of Integers" by Wolfdieter Lang (arXiv)

The above approach to proving periodicity came from sections 3.1 and 3.2 of Beyne and Brändli. And the proof of the sign pattern was a direct adaptation of the proof of Lemma 19 in Lang.

mod* Congruence

Building on the work of Schick, Beyne and Brändli investigate properties of the group $G_n^*$, where we consider integers $a,b$ to be congruent $(\mathrm{mod}^*n)$ if $a-b$ or $a+b$ is a multiple of $n$. Many theorems and conjectures of elementary number theory carry over from $\mathrm{mod}$ to $\mathrm{mod}^*$

Period length

One missing ingredient to finding the exact sign sequence for a single period of the $p$ sequence is the length of the period. Otherwise, it's not clear when to cut off the infinite sign sequence from answer 2 above. This period is mentioned in all 5 references above, and is A003558. It could be written "the order of $[2]$ in $G_q^*$" or "the least number $m>0$ such that $2^m\equiv\pm1\pmod q$".

Cosine Polynomials

Rather than turning a nested radical representation into an infinite one, we can use it to find a polynomial that a given cosine value is a root of. For example, $$2\cos\left(\dfrac{8\pi}{17}\right)=\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+2\cos\left(\dfrac{8\pi}{17}\right)}}}}$$ means that $2\cos\left(\dfrac{8\pi}{17}\right)$ is a root of $x=\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+x}}}}$, and hence of $\left(\left(\left(2-x^{2}\right)^{2}-2\right)^{2}-2\right)^{2}-2=x$. Rewriting, it is a zero of $x^{16}- 16x^{14}+ 104x^{12}- 352x^{10}+ 660x^8 - 672x^6+ 336x^4- 64x^2 +x+2$.

Unfortunately, this is not the minimal polynomial for $2\cos\left(\dfrac{8\pi}{17}\right)$. That would be $x^8-x^7-7x^6+6x^5+15x^4-10x^3-10x^2+4x+1$, whose coefficients are listed in A136571. Minimal polynomials for the cosines whose $p$ sequence contains $1$ are discussed in Lang, as well as its reference "The field $\bf \mathbb Q(2\, cos\left(\frac{\boldsymbol{\pi}}{n}\right))$, its Galois group, and length ratios in the regular $\bf n$-gon" (arXiv), also by Lang.


问:
Prove continuity of a function constructed from a Borel measure
If $\mu$ is a finite Borel measure on $\mathbb{R}$, then the function $F^{\mu}: \mathbb{R} \rightarrow \mathbb{R}$ ,  $F^{\mu}(\lambda) = \mu((-\infty, \lambda])$ is right-continuous.

Attempt at solution

I need to show that $\lim \limits_{\lambda \to c^{+}} F^{\mu}(\lambda) = F^{\mu}(c)$. Using the $(\epsilon, \delta)$ formulation of limits, I consider the inequality $$F^{\mu}(\lambda) - F^{\mu}(c) < \epsilon$$ $$\mu((-\infty, \lambda]) - \mu((-\infty, c]) < \epsilon$$ $$\mu((c, \lambda]) < \epsilon$$

where I obtained the last step using $\mu(A) = \mu(B) + \mu(A\setminus B)$, which in turn derives from the union-sum propriety of Borel measures for disjoint sets.

But now I seem to be stuck due to the generality of $\mu$: if it were the usual measure $\mu((a,b]) = b-a$, then the result would immediately follow with $\delta = \epsilon$, but since here I don't have an explicit expression for $\mu$ I don't know how to get to an inequality involving the argument of $\mu$, which is what one usually does in similar situations. What general propriety of Borel measures am I missing which allows one to arrive at the result?

答:
It follows from continuity of measure, that is if $A_n \downarrow A$ (with $\mu A_1 $ finite, which is the case here) we have $\mu A_n \to \mu A$.

Let $\lambda_n \downarrow \lambda$, and $A_n = (-\infty, \lambda_n]$, then $A=\cap_n A_n = (-\infty, \lambda]$ and so $F^\mu(\lambda_n) \downarrow F^\mu(\lambda)$.


答:
When you square both sides you should get $9(x+13) = (x+9)^2$ which rearranges to

\[x^2+9x-36 = 0 \ ,\]
which has the solutions
\[x_{1,2} = -\frac 92 \pm \sqrt{\frac{81}{4}+36} = \frac{-9\pm 15}{2} \: ,\]
i.e. $\begin{cases}x_1 = 3 \\ x_2 = -12\end{cases}$. By putting these into the original equation $3\sqrt{x+13} = x+9$ you realize that $x_2 = -12$ is not a solution, but $x_1 = 3$ is.

Title: Solve for $x$, $3\sqrt{x+13} = x+9$

Body:
Solve equation: $3\sqrt{x+13} = x+9$

I squared both sides and got $9 + x + 13 = x^2 + 18x + 81$

I then combined like terms $x^2 + 17x + 59 = 0$

I then used the quadratic equation $x= -\frac{17}2 \pm \sqrt{\left(-\frac{17}2\right)^2-59}$

However, the answer is 3


问:
Let $K \subset L$ be a finite extension with $p = \text{char}(K) > 0.$ Prove $[L : K]_s \cdot p^k= [L : K]$ for $k \in \mathbb{Z}_{\geq 0}$.
This was part of a two part question which first asked to show that $[L : K]_s= [K_s: K],$ where $K_s$ is the separable closure of $K$ in $L$. I use this result in my answer. I'm looking for help showing
\begin{equation*}
[L:K]=p^k [L:K]_s
\end{equation*}
where $k$ is a non-negative integer. I've attached my attempt at a proof below. I would appreciate if you can point out where it goes wrong, or advise how it could be repaired. Thanks in advance.

We assume $K$ has order $p^n.$ Let $|L|=p^j$ and $|K_s|=p^i.$ Since $K$ is a subfield of $K_s$ which is a subfield of $L,$ we know that $n \mid i$ and $i \mid j$ i.e. $i=nx$ and $j=iy$ for $x,y \in \mathbb{N}.$ By the tower law for finite extensions,
\begin{equation*}
    [K_s:K]=\frac{[L:K]}{[L:K_s]}.
\end{equation*}
Combining this with $[L:K]_s=[K_s:K]$ yields
\begin{equation*}
    [L:K]=[L:K_s]\cdot [L:K]_s.
\end{equation*}
To continue, it seems necessary to compute $[L:K_s]$. I note that again by the tower law,
\begin{equation*}
[L:K_s]=\frac{[L:K]}{[K_s:K]}.
\end{equation*}
I get stuck now on how to compute this term. My approach was that if $\mathbb{F}_p$ was the base field, then
\begin{equation*}
[K_s:K]=\frac{[K_s:\mathbb{F}_p]}{[K:\mathbb{F}_p]}=\frac{i}{n}=\frac{nx}{n}=x.
\end{equation*}
Trying the same for $[L:K]$ gives
\begin{equation*}
[L:K]=\frac{[L:\mathbb{F}_p]}{[K:\mathbb{F}_p]}=\frac{j}{n}=\frac{nxy}{n}=xy.
\end{equation*}
Therefore, $[L:K_s]=\frac{xy}{x}=y$ however as I can't seem to prove that this $y$ is a power of $p$ I assume I've gone wrong somewhere.

答:
Your argument leading up to showing the equivalence between the claim in the title and $ [L : K_s] $ being a power of $ p $ is correct. From there, you know that $ L/K_s $ is a purely inseparable extension, so it becomes a matter of knowing what are the possible degrees for such extensions.

You can show by a simple induction argument that a finite purely inseparable extension $ N(\alpha)/N $ with a primitive element $ \alpha $ must have degree a power of $ p $, where $ p = \operatorname{char} N $. It's obviously true if the extension is trivial, which gives the base case for the induction. In that case, since $ \alpha $ is inseparable over $ N $, its minimal polynomial $ f \in N[X] $ must be of the form $ g(X^p) $ for some $ g \in N[X] $. $ \alpha $ is a root of the polynomial $ X^p - \alpha^p $ in $ N(\alpha^p)[X] $, which is irreducible since otherwise we would have $ \alpha \in N(\alpha^p) $, so $ [N(\alpha) : N(\alpha^p)] = p $. We know by the induction hypothesis that $ [N(\alpha^p) : N] $ has degree a power of $ p $, so this shows by the tower law that $ [N(\alpha) : N] $ is a power of $ N $ as well.

For general finite purely inseparable extensions $ M/N $, you just pick any element $ \alpha \in M $ with degree $ > 1 $ over $ N $. $ [N(\alpha) : N] $ is a power of $ p $ by the above argument, and $ M/N(\alpha) $ is still purely inseparable of lower degree, so another induction argument along with the tower law finishes the proof.


问:
Evaluate $y'''+3y''+3y'+y = e^{it}$
$y'''+3y''+3y'+y = e^{it}$

I've got this method of undermined coefficients but i am unable to find the particular solution. I got the complimentary solution as $$Y_c = {C_1}{e^{it}}+ {C_2}{t}{e^{it}}+{C_3}{t^{2}}{e^{it}}$$ but keep getting the wrong answer when I try to find the particular.

答:
HINT

\begin{align*}
y''' + 3y'' + 3y' + y = e^{it} & \Longleftrightarrow (y'''+ y'') + 2(y'' + y') + (y' + y) = e^{it}\\\\
& \Longleftrightarrow (y' + y)'' + 2(y' + y)' + (y' + y) = e^{it}\\\\
& \Longleftrightarrow u'' + 2u' + u = e^{it}\\\\
& \Longleftrightarrow (u' + u)' + (u' + u) = e^{it}\\\\
& \Longleftrightarrow v' + v = e^{it}\\\\
& \Longleftrightarrow (ve^{t})' = e^{t + it}\\\\
& \Longleftrightarrow (ue^{t})'' = e^{t + it}\\\\
& \Longleftrightarrow (ye^{t})''' = e^{t + it}
\end{align*}

Title: Evaluate $y'''+3y''+3y'+y = e^{it}$


问:
Necessary and sufficient condition for having a density in $L^2$
Let $X$ be a real random variable, $\phi(x)=E[e^{ixX}],x \in \mathbb{R},$ its characteristic function.

Prove that $\phi \in L^2$ if and only if $X$ is absolutely continuous and having a density $f \in L^2.$ In this case $$\int_{\mathbb{R}}(f(x))^2dx=\frac{1}{2\pi}\int_{\mathbb{R}}|\phi(x)|^2dx.$$
Is there an inversion formula for $f?$

The above equality seems to be Plancherel identity, any suggestions are welcomed.

答:
Suppose $\mu$ is a positive Borel measure on $\mathbb R$ with $\mu(\mathbb R)=1.$ Define the "Fourier transform" of $\mu$ to be

$$\phi(x)=\hat {\mu}(x) = \int e^{ixt}\,d\mu(t).$$

(All integrals are over $\mathbb R .$) We want to show $\phi\in L^2$ iff $\mu$ is absolutely continuous with $d\mu = \psi\,dx$ and $\psi\in L^2.$ Let's concentrate on the $\implies$ part of the proof, since $\impliedby$ is pretty obvious.

Suppose $\phi\in L^2.$ Then by Plancherel, $\phi=\hat{\psi}$ for some $\psi\in L^2.$ Let $g$ be in the Schwartz space $\mathcal S.$ Using standard identities, we have

$$ \int g\phi   = \int g \hat {\mu} = \int \hat{g}\,d\mu.$$

Similarly,

$$ \int g\phi = \int g\hat {\psi} = \int \hat {g}\psi.$$

So we have $\int \hat{g}\,d\mu = \int \hat {g}\psi$ for all $g\in\mathcal S.$ That implies $\int g\,d\mu = \int g\psi$ for all $g\in\mathcal S.$ Since $\mathcal S$ is dense in $C_0,$ the Riesz Representation theorem shows that $d\mu= \psi(x)\,dx,$ which is the desired result.


答:
I like jlammy's solution. Here's an alternative proof without induction.

Denote $p_i, q_i$ the probability of the $i^{th}$ coin getting a head or tail, respectively. Then $p_i+q_i=1$. Consider the following product:

$$T=(q_1-p_1) (q_2 -p_2) \cdots (q_n - p_n) = P-M \tag 1$$

where $P$ is the sum of all positive terms after expansion, and $M$ is the absolute value of the sum of all negative terms. Then $P$ is the probability of getting an even number of heads, $M$ is the probability of getting an odd number of heads, and $P+M=1$. Therefore,

$$P=M = \frac 12 \iff T=0  \iff p_i=q_i=\frac 12, \text{ for some }i. \blacksquare$$

Title: Condition for probability of even number heads to be $1/2$

Body:
Suppose we flip $n$ coins (each flip is independent, but not necessarily identically distributed). Under what conditions will the probability that we flip an even number of heads be exactly $1/2$?

I know that when all the coins are fair, then the probability is half (see e.g. this thread). But I'm not sure how to handle the case if the coins aren't fair.


问:
Solving for x in $\frac{1+y}{1-y}=(1+x)^{1+x}(1-x)^{1-x}$

How can I solve for x?
$$\frac{1+y}{1-y}=(1+x)^{1+x}(1-x)^{1-x}$$

I have tried a few methods, but to be honest, I do not think I have what it takes to find the solutions. I'm hoping someone has has experience with these sorts of equations and can help me out!

Body:
Let ${y+1\over y-1}=M$

Then $M-1={2\over y-1}\implies y={2\over M-1}+1$ or $y={M+1\over M-1}={(1+x)^{1+x}(1-x)^{1-x}+1 \over (1+x)^{1+x}(1-x)^{1-x}-1}$


问:
If $(X,d)$ is a metric space and $x_0 \in X $ ,show that the function $f :X \to \mathbb R ; f(x)=d(x_0,x)$ is Lipschitz continuous
I was writting
$d(x_0,x_1) \le d(x_0,x_2)+d(x_2,x_1)$
Since $\displaystyle d(x_2,x_1) > 0;d(x_0,x_1) - d(x_2,x_1) \le d(x_0,x_2);
|d(x_0,x_1) - d(x_2,x_1)| \le |d(x_0,x_2)|$

But I don't know how to keep going.

Body: From $d(x_0,x_1) \leq d(x_0,x_2)+d(x_1,x_2)$ we get $d(x_0,x_1)-d(x_0,x_2) \leq d(x_1,x_2)$, and from $d(x_0,x_2) \leq d(x_0,x_1)+d(x_1,x_2)$ we get $-d(x_1,x_2) \leq d(x_0,x_1)-d(x_0,x_2)$. Thus $$-d(x_1,x_2) \leq d(x_0,x_1)-d(x_0,x_2) \leq d(x_1,x_2)$$ and this means $|d(x_0,x_1)-d(x_0,x_2)| \leq d(x_1,x_2)$, that is, $|f(x_1)-f(x_2)| \leq d(x_1,x_2)$.


问:
A question regarding the Power set
In the proofs that I have seen so far for showing that the power set $2^X$ of a set $X$ cannot be in bijection to $X$, the common idea is to assume that there exists a surjection $f \colon X \to 2^X$ and then consider the set
$$
B = \{ x \in X \mid x \notin f(x)\}
$$

Then the argument proceeds by saying that this set is contradictory, because its pre-image (say $y \in X$, i.e. $f(y) = B$) satisfies both $y \in B$ and $y \notin B$.

On Wikipedia, this method is said to be analogous to Cantor's diagonal argument that is used to show that the interval $(0,1)$ of real numbers between $0$ and $1$ is uncountable. Here we assume that there exists an enumeration. Representing a number $x \in (0,1)$ in its unique decimal expansion, we obtain a list
\begin{align}
x_1 &= 0.a_{11}a_{12}a_{13}... \\
x_2 &= 0.a_{21}a_{22}a_{23}... \\
x_2 &= 0.a_{31}a_{32}a_{33}... \\
x_2 &=0.a_{11}a_{12}a_{13}... \\
\vdots
\end{align}
where $a_{ij} \in \{0,1,\dots,9\}$ for each $i \in \mathbb{N}$ and $j \in \mathbb{N}$. Then one constructs an element from $(0,1)$ that is not in this list. For example, one can take the number $x = 0.b_{1}b_{2}b_{3}\dots$ where
\begin{equation}
b_i = \begin{cases} 1 & \text{if } a_{ii} = 5 \\ 5 & \text{if } a_{ii} \ne 5
\end{cases}
\end{equation}

Here, the number $x$ that is generated is well defined as an element of $(0,1)$, and it does not appear in the list above so we've reached a contradiction.

On the other hand, for the argument above regarding the power set, assuming $f$ exists implies the object $B$ is not well defined as an element of $2^X$ since it is not the image of a function from $X$ to the set $\{0,1\}$ (it is multi-valued, namely the preimage of $B$ evaluates both to $0$ and $1$). In other words, I cannot use this object $B$ to derive a contradiction. What I can derive is that, if $f$ exists then $B$ is not a set, and if $B$ is a set for each function $f \colon 2^X \to X$ then no such $f$ can be surjective.

The latter is what the proofs claim to be the only option, in other words this object $B$ must be a well defined set by some other reason - what am I missing?

答:
Cantor's theorem is indeed very close to the diagonal argument.

The idea is a generalization of the following concept. We write a table: $$\begin{array}{|c|c|c|c|c}
\hline
\quad & f(x_1) & f(x_2) & f(x_3) & \ldots\\\hline
x_1 & 0 & 1 & 0 & \ldots\\\hline
x_2 & 1 & 1 & 0 & \ldots\\\hline
x_3 & 0 & 0 & 1 & \ldots\\\hline
\vdots
\end{array}$$
Where $(x,f(x))$ is $0$ if $x\notin f(x)$ and $1$ otherwise. The diagonal argument is to traverse across the diagonal, and consider those which have $0$ there. Collect this set, and you can in fact show that it is not $f(x)$ for any $x$.

This $B$, as you denote it, or rather its indicator function (if you prefer considering $2^X$ rather than $\mathcal P(X)$ for one reason or another) exists because we define it. We explicitly gave a description of its members. Ordered pairs of the form $\langle x,i\rangle$ where $i=0$ if $x\in f(x)$ and $1$ otherwise.

One of the earliest principles of set theory is comprehension. It's an important principle, mathematically and philosophically. If we can describe a collection then we want it to exist. And while the unrestricted comprehension (all the describable collections are sets) is inconsistent, when axiomatic set theory was formulated this was limited in the following sense:

If $A$ is a set, and $\varphi(x,u_1,\ldots,u_n)$ is a formula, then for every choice of parameters, $p_1,\ldots,p_n$ the set $\{a\in A\mid\varphi(a,p_1,\ldots,p_n)\}$ exists.

This is known as the axiom schema of specification as mentioned in the comments, and also as "restricted comprehension" and "separation" sometimes. Why does that help us? Well, if we assume that $X$ is a set, then we have defined a subset (from the parameter $f$) and therefore it exists. If you prefer the functional version, then consider the set $X\times\{0,1\}$ and apply the same argument.

Therefore we proved that $B$ exists, and there is no surjection from a set onto its power set.

Title: A question regarding the Power set


问:
How to get the distance between two anchoring points in a tv mast?
The problem is as follows:

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{4 m}\\
2.&\textrm{5 m}\\
3.&\textrm{6 m}\\
4.&\textrm{7 m}\\
\end{array}$

How exactly should this be solved relying only in euclidean geometry?.

I'm confused exactly if this should use congruence or what?. The part which I'm stuck is what to do with that the double of the angle on $\angle ACD$?.

I'm stuck can someone help me here?. Please include a drawing in your answer because I'm not very savvy with understanding where exactly congruence could be

答:
A trigonometric solution is straightforward. We have:

$$\frac{2}{AB} = \tan \alpha = \frac {\sin \alpha}{\cos \alpha}$$

$$\frac {AB}{10} = \sin 2\alpha = 2\sin \alpha \cos\alpha$$

Multiplying the two equations, we have:

$$\frac2{10} = 2\sin^2\alpha$$

which gives $\sin \alpha = \dfrac {1}{\sqrt{10}}$. This gives $AD = 2\sqrt{10}$, and by applications of the Pythagorean Theorem we can find the lengths of $AB$, $BC$ and finally $DC$.


问:
Prove that a set is an event
Let $ (X_k)_k $ be a sequence of events and define:

$ Z\left(\omega\right)=\limsup_{k\to\infty}X_{k}\left(\omega\right),\thinspace\thinspace\thinspace\thinspace\thinspace W\left(\omega\right)=\liminf_{k\to\infty}X_{k}\left(\omega\right) $

and:

$ Y\left(\omega\right)=\begin{cases}
\lim_{k\to\infty}X_{k}\left(\omega\right) & if\thinspace\thinspace the\thinspace\thinspace\thinspace limit\thinspace\thinspace\thinspace exist\\
0 & else
\end{cases} $

Now I want to prove that $ Y $ is a random varaible.

What Im allowed to use:

1.$ Z $ is a random variable.

2.$ W $ is a random variable.

3.An indicator of an event is a random variable.

4.A multipication of random variables is a random variable.

Now define
$ A=\left\{ Z=W\right\}  $ that is, all the $ \omega $ such that the limit exists.

If I'll manage to show thta $ A $ is an event, then I could write

$ Y=Z1_{A} $ where $ 1_A $ is the indicator of $ A $ and we're done.

I have tried something and I would like to know if its true. (Im aware that there is much simpler solution):

$ A=\left\{ \omega:Z\left(\omega\right)=W\left(\omega\right)\right\} =\left\{ \omega:\lim_{k}X_{k}\left(\omega\right)\thinspace\thinspace\thinspace exist\right\} =\left\{ \omega:X_{k}\left(\omega\right)\thinspace\thinspace\thinspace is\thinspace\thinspace\thinspace a\thinspace\thinspace\thinspace Cauchy\thinspace\thinspace\thinspace\thinspace sequence\right\}  $

$ =\left\{ \omega|\forall m\in\mathbb{N},\exists N\in\mathbb{N:\forall}k_{1},k_{2}\geq N:|X_{k_{1}}\left(\omega\right)-X_{k_{2}}\left(\omega\right)<\frac{1}{m}\right\}  $

$ =\bigcap_{m\in\mathbb{N}}\bigcup_{N\in\mathbb{N}}\bigcap_{k_{1},k_{2}\geq N}\left\{ |X_{k_{1}}-X_{k_{2}}|<\frac{1}{m}\right\}  $

Since this is just a countable intersection of a countable union of a countable intersection of events, all in all $ A $ is also event.

Do you agree?

答: Yes, that works.  You could also use that $$A=\{\omega : Z(\omega)=W(\omega)\} = \bigcap_{n \in \mathbb{N}} \left\{\omega : |Z(\omega)-W(\omega)| < \frac 1n\right\}$$ for a slightly faster proof that $A$ is a countable intersection of events.

Title: Prove that a set is an event


答: If you are just trying to solve for $y$ (and not have a particular value, which I think there are multiple valid values), then it is pretty straightforward.  Multiply both sides by $y - 1$.  Then, collect all of the $y$ terms to one side.  Then, pull out the $y$ term.  Finally, divide by the rest to isolate $y$ by itself.

Title: Solving for x in $\frac{1+y}{1-y}=(1+x)^{1+x}(1-x)^{1-x}$

Body:
How can I solve for x?
$$\frac{1+y}{1-y}=(1+x)^{1+x}(1-x)^{1-x}$$

I have tried a few methods, but to be honest, I do not think I have what it takes to find the solutions. I'm hoping someone has has experience with these sorts of equations and can help me out!


问:
How do even and odd functions relate to even and odd numbers?
How do the notions of oddness and evenness apply to both functions and numbers?

If Even and Odd functions share nothing in common with Even and Odd numbers, then why were Even and Odd adopted for functions? Why not use other adjectives?

James Stewart, Calculus 7th ed. 2011. This isn't the Early Transcendentals version.

答:
They share lots in common!

The product of two even functions is even, the product of an even function and an odd function is odd, and the product of an odd and an even function is odd. This is exactly how even and odd numbers add, and is a nice motivating example of an isomorphism.

The Taylor series (at $x = 0$) of an even function has only even terms $x^{2n}$, and the Taylor series of an odd function has only odd terms $x^{2n+1}$. If you don't know what a Taylor series is, we can restrict our attention to polynomials: a polynomial is even iff it has only even terms, and a polynomial is odd iff it has only odd terms.


答:
For an even function $f(-x)=f(x).$  Now you know that $(-x)^n=x^n$ if $n$ is even.

For an odd function $g(-x)=-g(x)$. Now you also know that $(-x)^n=-x^n$  if $n$ is odd.

Title: How do even and odd functions relate to even and odd numbers?

Body:
How do the notions of oddness and evenness apply to both functions and numbers?

If Even and Odd functions share nothing in common with Even and Odd numbers, then why were Even and Odd adopted for functions? Why not use other adjectives?

James Stewart, Calculus 7th ed. 2011. This isn't the Early Transcendentals version.


问:
How to derive this exponential generating function?
Anyone can help to solve this problem?

Let $\mathcal{F}$ be the combinatorial class of all functions $f : [1,n] \rightarrow [1,n]$. Derive the exponential generating function and use
it to compute (or verify) the number of such functions.

答:
Going the other way might be easier: the number $|\mathcal{F}(n)|$ is not hard to compute. Consider the function as a vector $(f(1),\ldots,f(n))$, where the coordinates can take the values $[1;n]$. How many such vectors are there?

Once you have this number, you can plug it in to the formula for an exponential generating function.


问:
Are these variations of complex multiplication studied topics?
Complex multiplication is very well understood geometrically and algebraically, but I wonder what about the following operators -angles assumed to be randians $[0,2\pi)$:

Complex multiplication(muladd):  $$ x_1 \cdot x_2 = |x_1||x_2|e^{(\arg(x_1) + \arg(x_2))i}$$

Complex mulmul: $$ x_1 \bigodot x_2 = |x_1||x_2|e^{\arg(x_1) \cdot \arg(x_2)i} $$

Complex addadd: $$ x_1 \bigoplus x_2 = (|x_1|+|x_2|)e^{(\arg(x_1) + \arg(x_2))i} $$

Complex addmul: $$ x_1 \bigotimes x_2 = (|x_1|+|x_2|)e^{(\arg(x_1) \cdot \arg(x_2))i} $$

Have these operators been studied? Are there any books or papers on their properties? Do they have names?

答: I haven't heard of any of those (besides ordinary complex multiplication) being studied. The problem with the ones that multiply arguments is that usually when taking the arg of a complex number, we get an angle out. Angles are cyclical, so that $2\pi$ is treated the same as 0. But multiplying angles doesn't preserve this cyclical nature. $0 \times \frac{\pi}{2}$ is not the same as $2\pi\times\frac{\pi}{2}$. So those operations will be discontinuous. And finally, addadd is pretty much vector addition.


问:
When can you define addition on a set of equivalence classes?
Suppose I am given an equivalence relation on set A which is endowed with addition & closure under addition.

Then, to have addition on equivalence classes of A to be well-defined, it is to show:

$[a]+[b] = [a+b]$

Suppose I have shown that if $\forall a,b \in A, a \sim a', b\sim b'$ then $a+b \sim a'+b'$, then to my understanding, I proved that there exist an equivalence class $[a+b]$ so that $[a]+[b] \subset [a+b]$.

But then, now I need to show the other direction. I found it is easiest to define $c \in A$ so $b+c = 0$. So if $x \in [a+b]$ then $x \sim a+b$, and using what I proved, $x+c \sim a+(b+c) = a$. Letting $r = x-b$ and $s = b$ shows the other direction.

My question now is, to do this, (A,+) seems to need to have the property that all elements must have an identity, an inverse, and is associative, so what exactly must (A,+) be for all this to work, a group? abelian group? and why?

Note: Also I think I'm failing to consider a lot of things when thinking about this, so it will be very helpful if anyone could point many of those things out.

答:
So you have a semigroup (a single binary operation that is associative on your set $A$). The magic words are “congruence relation”, which are exactly what drhab gives in his answer. You can see a lot of this at play in this previous answer about groups.

Definition. Let $S$ be a semigroup. A congruence on $S$ is an equivalence relation $\sim$ on $S$ such that for all $x,y,x’,y’\in S$, if $x\sim x’$ and $y\sim y’$, then $xy\sim x’y’$.

Proposition. An equivalence relation on $S$ is a congruence if and only if, when considered a subset of $S\times S$, it is a subsemigroup of the latter under coordinatewise operation.

Proof. If $\sim$ is a congruence, then by definition the subset is closed under the coordinatewise operation, hence is a subsemigroup. Conversely, if an equivalence relation is a subsemigroup, then $x\sim x’$ and $y\sim y’$ mean $(x,x’),(y,y’)\in\sim$, and by closure we get $(xy,x’y’)=(x,x’)(y,y’)\in \sim$, hence $xy\sim x’y’$, hence $\sim$ is a congruence. $\Box$

Given an equivalence relation $\sim$ on $S$, let $S/\sim$ be the set of equivalence classes, and let $[s]$ denote the equivalence class of $s\in S$ under $\sim$.

Proposition. If $\sim$ is a congruence on $S$, then $S/\sim$ is a semigroup under the operation $[s][t]=[st]$.

Proof. The operation is well defined: if $[s]=[s’]$ and $[t]=[t’]$, then $s\sim s’$ and $t\sim t’$, hence $st\sim s’t’$, so $[st]=[s’t’]$. The operation is associative, as
$$([s][t])([v]) = [st][v] = [(st)v] = [s(tv)] = [s][tv] = [s]([t][v]).$$
Thus, $S/\sim$ is a semigroup. $\Box$

And conversely:

Theorem. Let $S$ be a semigroup, and let $\sim$ be an equivalence relation on $S$. Then the operation on $S/\sim$ given by $[s][t]=[st]$ is well-defined if and only if $\sim$ is a congruence on $S$.

Proof. That the operation is well-defined when $\sim$ is a congruence is given above. Conversely, assume the operation is well defined: that means that if $s\sim s’$ and $t\sim t’$, then $[st]=[s][t] = [s’][t]’] = [s’t’]$, hence $st\sim s’t’$, proving that $\sim$ is a congruence on $S$. $\Box$

Congruences play the role of normal subgroups for groups and two-sided ideals for rings in the context of general algebras, the study of Universal Algebra. The isomorphism theorems have counterparts for general algebras using congruences.

Thus, you don’t need this to be a monoid, or a group, for this to work; you just need a subsemigroup.


问:
Integer/Binary Programming - Convert nonlinear constraint to linear constraint
I have a constraint based on matrix multiplication.  It involves summation of a product.  The product multiplies two entries of a variable matrix named order, and an entry of a constant matrix named guess.

All of the variables and constants are binary.  All of the indexes have the same arbitrary range $N$.

($\forall n,m$) result$_{n,m}$ = $\sum_r \sum_s$ (order$_{n,r}$ * guess$_{r,s}$ * order$_{m,s}$)

Is it possible to turn this into ~$N^2$ linear constraints?

答:
You can linearize each product of binary variables by introducing a new binary variable and three linear constraints:
https://or.stackexchange.com/questions/37/how-to-linearize-the-product-of-two-binary-variables

That is $O(N^4)$ linear constraints.  See the references in the linked answer for formulations that use fewer constraints.


问:
if $G$ has average degree $d$, then there exists a subgraph with minimum degree at least $d/2$
Say we have a graph $G$ of average degree at least $d$. I want to show that we can find an induced subgraph with minimum degree at least $d/2$. The idea is to repeatedly delete vertices of degree smaller than $d/2$ from $G$, as long as there are such vertices. This should increase the average degree, but I’m not sure how to show that. First we have
$$
\frac{\sum_{i=1}^nd_i}{n}=d,
$$
where $d_i$ is the degree of the $i$th vertex. Say there exists a vertex of degree $k<d/2$. If we delete this vertex, our average degree becomes
$$
\frac{\sum_{i=1}^nd_i-2k}{n-1}.
$$
I need to show that this number is greater, using the fact that $2k<d$, but I'm not sure how. Any ideas?</p>

答:
In fact, there is an algorithm that returns a subgraph $G'$ of $G$ where every vertex in $G'$ has degree strictly larger than $d/2$.

Let us assume that there is a vertex in $G$ of degree no larger than $d/2$ or we are done.  Let us see what happens to the average degree if we were to remove a vertex $v$ of degree $d/2$ or less from $G$. In fact, we claim that the average degree of $G' \doteq G \setminus \{v\}$ is at least $d$ if the average degree of $G$ is at least $d$. [Indeed let us write the number of vertices in $G$ as $n$. Then as $G$ has average degree at least $d$ it follows that $G$ has at least $nd/2$ edges. Then $G'$ has at least $nd/2 - d_G(v) \le$ $nd-d/2 = (n-1)d/2$ edges and $n-1$ vertices, so the average degree of $G'$ is at least $2 \times \frac{(n-1)d}{n-1} \ge d$ by the Handshaking Lemma.]

This gives an algorithm to find a subgraph $G'$ of $G$ with degree greater than $d/2$:

$G' \leftarrow G$.

WHILE there is a remove a vertex $v$ of degree $d/2$ or less in $G$:  $G' \leftarrow G' \setminus \{v\}$

RETURN $G'$.

As each time a vertex of degree $d/2$ or less is removed from $G'$ the number of vertices goes down by 1 but the average degree stays above $d$ it follows that this will conclude with a graph where every vertex has degree strictly larger than $d/2$.


问:
What is a proof?
I am just a high school student, and I haven't seen much in mathematics (calculus and abstract algebra).

Mathematics is a system of axioms which you choose yourself for a set of undefined entities, such that those entities satisfy certain basic rules you laid down in the first place on your own.

Now using these laid-down rules and a set of other rules for a subject called logic which was established similarly, you define certain quantities and name them using the undefined entities and then go on to prove certain statements called theorems.

Now what is a proof exactly? Suppose in an exam, I am asked to prove Pythagoras' theorem. Then I prove it using only one certain system of axioms and logic. It isn't proved in all the axiom-systems in which it could possibly hold true, and what stops me from making another set of axioms that have Pythagoras' theorem as an axiom, and then just state in my system/exam "this is an axiom, hence can't be proven".

EDIT : How is the term "wrong" defined in mathematics then? You can say that proving Fermat's Last Theorem using the number theory axioms was a difficult task but then it can be taken as an axiom in another set of axioms.

Is mathematics as rigorous and as thought-through as it is believed and expected to be? It seems to me that there many loopholes in problems as well as the subject in-itself, but there is a false backbone of rigour that seems true until you start questioning the very fundamentals.

答:
Since you're a high-school student, here's an answer that's less sophisticated and much less rigorous:

I suppose you could make up any set of axioms you want, and start using them to prove theorems. So, as you say, you could make Pythagoras' theorem an axiom in your world, and then you wouldn't need to "prove" it.

But, if you're going to start making up your own system of axioms, and doing mathematics in this private world, there are a few things you need to worry about:

(1) If no-one else uses the same axioms as you, then no-one will be very interested in your "theorems", since they are only true in your private world. Your private world might be a bit lonely. So, better to use the same axioms as everyone else.

(2) It's useful (though not absolutely necessary) to have a system of axioms that bears some relationship to reality. That way, the theorems you prove will sometimes give you information that has value in the "real" world -- in fields like economics and engineering, for example. Your private world might be quite different from physical reality, if you don't choose the axioms carefully. So, your results could be misleading or even dangerous, even though they are provably "true" in your world.

(3) If you're not careful, the system of axioms you invent might lead to contradictions, or it might have other fundamental logical flaws. The axioms can't be completely arbitrary (as far as I know).

There are some areas of mathematics where part of the game is making up modified systems of axioms and seeing what happens. But most of us play by a fairly well established set of rules, for the reasons outlined above (and for other reasons, too, I expect).

Additions

Regarding your added comment that "there is a false backbone of rigour that seems true until you start questioning the very fundamentals". It seems to me that the rigour is in the reasoning that's used to derive theorems from the chosen set of axioms. I don't think this rigour is "false".

What's bothering you, I suppose, is that there is some freedom when choosing the set of axioms, and, depending on what choices you make, you get a different set of theorems -- a different version of the truth, and different statements of what is "right" and "wrong". I understand your concern -- I can see how it might be disturbing to find out that the axioms of mathematics are not universally agreed. One example of a debatable axiom is the "Axiom of Choice" (read more here). Most mathematicians assume that this axiom is true, but some don't, and, of course, the two groups get a different set of theorems. Not entirely different, but different.

But, on the other hand, the choice of axioms is not completely arbitrary, and there is a very large overlap in the sets of axioms that are in common use. So, in practice, things typically work just fine, despite the fact that the foundations are not entirely cast in stone.

Questioning the fundamentals, as you are doing, is a valid thing to do, and mathematicians have been doing it for a long time. If you want to know more about this,  from sources that are at least somewhat "credible and reliable", then this Wikipedia page might be a good place to start.

Title: What is a proof?


问:
The global sections of the affine plane with the origin removed
Consider the affine plane with the origin removed $U:=\mathbb{A}_k^2-\{(x,y)\}$.

First, This post asserts that the map $j^\sharp:\mathcal{O}(\mathbb{A}^2_k)\to \mathcal{O}(U)$ induced by the embedding $j:U\hookrightarrow \mathbb{A}^2_k$ is an isomorphism. I do not quite see why this follows from the fact that $\mathcal{O}(\mathbb{A}^2_k) \cong k[x,y]$. (This fact is computed in this lecture note, page 140.) I mean a a priori there can be many maps between $k[x,y] \to k[x,y]$ which might not be an isomorphism.

Second, the post also claims that if $U$ is affine, then $j^\sharp$ being an isomorphism would imply that $U \cong \mathbb A_{k}^{2}$, why is that? My thinking is that will imply the ring underlying both "affine" schemes will be isomorphic and so their spectrum is isomorphic(bijective as sets).

答:
For any integral scheme $X$ with generic point $\xi$,  open subset $U\subset X$, and point $x\in U\subset X$, there are canonical injections $\mathcal{O}_X(X)\to \mathcal{O}_X(U) \to \mathcal{O}_{X,x} \to \mathcal{O}_{X,\xi} = K(X)$, where $K(X)$ denotes the field of fractions of $X$. This means that $\mathcal{O}_X(U)\subset \bigcap_{x\in U} \mathcal{O}_{X,x}$ as subsets of $K(X)$.

In our particular case, we have $\bigcap_{x\in U} \mathcal{O}_{X,x} \subset \bigcap_{x\in U, \operatorname{ht} x=1} \mathcal{O}_{X,x} = \bigcap_{\mathfrak{p}\subset k[x,y], \operatorname{ht} \mathfrak{p}=1} k[x,y]_\mathfrak{p}$, and this last term is exactly $k[x,y]$ by algebraic Hartog's. So we have the following series of inclusions (after identifying everything with it's image inside $K(X)$): $$k[x,y]\subset \mathcal{O}_X(U) \subset  \bigcap_{x\in U} \mathcal{O}_{X,x} \subset \bigcap_{x\in U, \operatorname{codim} x=1} \mathcal{O}_{X,x} = \bigcap_{\mathfrak{p}\subset k[x,y], \operatorname{ht} \mathfrak{p}=1} k[x,y]_\mathfrak{p} = k[x,y]$$ and so $\mathcal{O}_X(U)=k[x,y]$.

The second question is a simple consequence of the fact that affine schemes and commutative rings are contravariantly equivalent categories: if $X$ and $Y$ are affine schemes, then $X\cong \operatorname{Spec} \mathcal{O}_X(X)$ and $Y\cong \operatorname{Spec} \mathcal{O}_Y(Y)$, so if $f:\mathcal{O}_X(X)\to\mathcal{O}_Y(Y)$ is an isomorphism, then $\operatorname{Spec} f$ induces an isomorphism $Y\to X$. (It is important to notice that "isomorphic as schemes" is much stronger than simply bijective as sets. All quasicompact schemes of dimension one over a field are bijective as sets, and all irreducible schemes of dimension one over a field are homeomorphic as topological spaces, but the classification of such objects is much more interesting than that.)


问:
A map $f : sp(1) \to sp(1)$ which commutes with the adjoint representation is a multiple of the identity
Let $f : \operatorname{sp}(1) \to \operatorname{sp}(1)$ be a linear map which commutes with the adjoint representation of $\operatorname{Sp}(1)$, i.e. $f(\operatorname{Ad}_g X) = \operatorname{Ad}_g f(X)$ for all $g \in \operatorname{Sp}(1)$, $X \in \operatorname{sp}(1)$. I have managed to prove that necessarily $f = \lambda I$ by direct inspection (using the adjoint representation of $\operatorname{sp}(1)$ on itself, and the isomorphism $\operatorname{sp}(1) \cong (\mathbb{R}^3, \times)$).

However, the adjoint rep. of $\operatorname{Sp}(1)$ is irreducible, and that reminds me of Schur's lemma. It is not directly applicable, however, because $\operatorname{sp}(1)$ is a real vector space and not a complex one. Is it possible to argue in this way regardless? Perhaps by complexifying $\operatorname{sp}(1)$? I'm interested if there is some representation theory machinery that could squash this bug.

答:
Schur's lemma over a commutative ring $k$ says that the endomorphism $k$-algebra of an irreducible representation (more precisely, a simple object in any $k$-linear abelian category - the argument is very general) is a division algebra over $k$, which is finite-dimensional if the representation is finite-dimensional. When $k = \mathbb{R}$ this implies that the endomorphism algebra of a finite-dimensional irreducible representation is $\mathbb{R}$ or $\mathbb{C}$ or $\mathbb{H}$ (by the Frobenius theorem). But $\dim_{\mathbb{R}} \mathfrak{sp}(1) = 3$ is odd so it can't be a module over $\mathbb{C}$ or $\mathbb{H}$.

Using complexification will also work. It's a nice exercise to show that if $V$ is a representation over $\mathbb{R}$ then

$$\text{End}_{\mathbb{C}}(V \otimes_{\mathbb{R}} \mathbb{C}) \cong \text{End}_{\mathbb{R}}(V) \otimes_{\mathbb{R}} \mathbb{C}$$

and a computation of the complexifications of $\mathbb{C}$ and $\mathbb{H}$ will then show that a real irreducible representation has endomorphism algebra $\mathbb{R}$ iff its complexification is irreducible (provided that the complexification is semisimple - that’s true in this case and I think it’s true in general but it might require an additional argument), so you can try to show that. It amounts to showing that the adjoint representation of the complexification

$$\mathfrak{sp}(1) \otimes_{\mathbb{R}} \mathbb{C} \cong \mathfrak{sp}_2(\mathbb{C}) \cong \mathfrak{sl}_2(\mathbb{C})$$

is irreducible. You can check out Frobenius-Schur indicators for more on this sort of thing.


问:
A matrix with invertible pricinple minor can be left multiplier by a lower trianglular matrix to be an upper triangular matrix
Let $A$ be a $n\times n$ matrix, with the principle minor being nonzro, that is, $det\begin{pmatrix}
a_{11}&\cdots&a_{1n}\\
\vdots&\ddots&\vdots\\
a_{k1}&\cdots&a_{kk}
\end{pmatrix}\neq 0$ for $1\leq k\leq n$. Show that there exists a lower triangular matrix $B$ such that $BA$ is upper triangular.

How to do this? What I know and what I can do is as follows: by Gram-Schmidt orthogonal process, there exists an orthogonal matrix $P$ such that $PA$ is upper triangular. How this to be modified?

答:
A weaker hypothesis will suffice . . .

Claim:

If $K$ is a field and $A\in GL_n(K)$, there exists a nonzero lower triangular matrix $B\in M_n(K)$ such that $BA$ is upper triangular.

Proof:

Let $L\subset M_n(K)$ be the set of lower triangular matrices, and let $U\subset M_n(K)$ be the set of upper triangular matrices.

Regarding $M_n(K)$ as a vector space over $K$, and $L,U$ as vector subspaces, we have $$\dim(M_n(K))=n^2$$
and
$$\dim(L)=\dim(U)=\frac{n^2+n}{2}$$

Then the set $S=\{BA{\,\mid\,}B\in L\}\;$is a vector subspace of $M_n(K)$, and
since $A$ is invertible, we get
$$\dim(S)=\dim(L)=\frac{n^2+n}{2}$$

Then we have
$$
\dim(S)+\dim(U)=n^2+n > n^2 = \dim(M_n(K))
$$
hence $S\cap U\ne \{0\}$.

The conclusion follows.


问:
Any sober space is $T_0$
Recall that a sober space is a topological space $X$ with the property that each closed irreducible set is the form $cl \{x\} $  for some $x \in X$.

How to prove that each sober space is necessarilly $T_0$?

答:
Your characterization of sober spaces is slightly different than the usual characterization, which is that every closed irreducible set is of the form $\overline{\{x\}}$ for some unique $x\in X$, where we use an overline to denote the topological closure. (See eg here or here, where this is the definition given.)

With this definition the result is clear. Indeed, suppose $X$ is not $T_0$; then there are $x,y\in X$ such that every open set of $X$ either contains both $x$ and $y$ or contains neither of them. Taking complements, this means that every closed set of $X$ contains either both or neither of $x$ and $y$, and so $\overline{\{x\}}=\overline{\{y\}}$ is a closed irreducible set expressable as the closure of two distinct points of $X$, whence $X$ is not sober.

If we get rid of the uniqueness demand in our definition, then there are sober spaces that are not $T_0$. For instance, consider the space $\{\star,\bullet\}$ with the indiscrete topology. This is in some sense a minimal example, as it embeds into every non-$T_0$ space.

Title: Any sober space is $T_0$


问:
Modular Numbers: Not Accounting for Decimal Portion in Decimal Expansion
Let n = 10, s = 4.

[s] = {s' $\in \mathbb{z}$ | s' $\sim$ s}
= {s' $\in \mathbb{z}$ | 10 | s' - 4}
= {10k+4 | k $\in \mathbb{z}$ }
= {s' $\in \mathbb{z}$ | decimal expansion of s' ends in a 4 if s' is pos, ends in a 6 if s' is
neg}

How does this last line make sense? We know that 4 $\sim$ 14. But, from my understanding of decimal expansions, isn't the decimal expansion of 4 4.0000 (repeating) and 3.9999 (repeating). So how can we say that decimal expansion of 4 ends in a 4? Similarly, we know that -6 $\sim$ -16. But shouldn't the decimal expansion of -6 be -6.0000 (repeating) or -5.99999 (repeating)? So how can we say that decimal expansion of -6 ends in a 6?

It seems like my professor is stopping at the ones digit and not accounting for the "decimal" portion. I think the problem is that I lack a clear understanding of what a "decimal expansion" is.

Edit: Is my teacher just considering one decimal representation out of two?

答:
The term "decimal expansion" here is just the way you would write an integer as $267456$ or $127$ or any other string of digits - where there are no decimal points at all to worry about. It's the ordinary way of writing an integer.

Formally, the idea is that the notation $83521$ is really just shorthand for:
$$8\cdot 10^4 +3\cdot 10^3 + 5\cdot 10^2 + 2\cdot 10 + 1$$
where we have a sum over powers of $10$ times digits from $\{0,1,2,3,4,5,6,7,8,9\}$. Every non-negative integer can be uniquely expressed in this form (and every negative integer is just the negation of some positive integer) - and, when you're talking only about integers, this is what is meant by "decimal expansion."

There's a generalization of this that applies to all real numbers, where something like
$$13.1415\ldots = 1\cdot 10^1 + 3 + 1\cdot 10^{-1} + 4\cdot 10^{-2}+1\cdot 10^{-3}+5\cdot 10^{-4}+\ldots$$
where we allow negative exponents of $10$ as well (indicated by the decimal point) and also allow infinitely many terms with these negative exponents if we desire. Some numbers can be represented in multiple ways in this notation since $0.999\ldots = 1.000\ldots$. If people are talking about real numbers, this is what "decimal expansion" would mean - but it's not what your professor is referring to.

(Note: The word "decimal" here refers to the fact that we write a number as a sum of powers of ten - which is probably what you were going to do anyways. Other bases can be indicated in the same terminology - for instance, the term "ternary expansion" refers to writing a number as a sum of powers of $3$ times values in $\{0,1,2\}$ and generalizes in the same ways)


问:
Doubt in finding critical point by partial differentiation.
For a given $G=X_{a}G_{a}+X_{b}G_{b}+\Omega X_{a}X_{b}+RT(X_{a} lnX_{a}+X_{b}ln X_{b} ) $;$X_{a}=1-X_{b}$,I am having trouble with finding the second order derivative$\partial ^2G/\partial x_{b}^2$.Somebody told me,"the first two term disappear because the are linear. The second term is $x-x^2$ and the 2nd derivative is $-2$ and the last term is basically $x \log x + (1-x) \log(x-1)$ and the 2nd derivative is ${1 \over x} + { 1\over x-1}$".However the final value that I get is given by the picture(this is after taking the first derivative).$T_{Cr}=2\Omega x(1-x)/R.$Now the value of x is set to 0.5 for getting the critical value of the function that I found in terms of T. Firstly,how do I calculate x=0.5 which gets me the critical value of temperature.Also how do I judge whether the critical value is maximum or minimum?Last but not the least these are all partial derivatives,how is the differentiation technique going to be different than from a normal derivative?

答:
My understanding (since I am the 'somebody') is that we are looking at the expression
$G(x) = G_a (1-x)+ G_b x + \Omega x(1-x) + RT(x \ln x + (1-x) \ln (1-x))$.

If you differentiate $x \mapsto G_a (1-x)$ twice you get zero.

If you differentiate $x \mapsto G_b x$ twice you get zero.

If you differentiate $x \mapsto x(1-x)$ twice you get $-2$.

If you differentiate $x \mapsto x \ln x + (1-x) \ln (1-x) $ twice you get
${1 \over x}+{1 \over 1-x}$.

Hence
$G''(x) = -2 \Omega + RT ({1 \over x}+{1 \over 1-x} ) = -2 \Omega + RT {1 \over x(1-x)} $.

I have no idea what the $x=0.5$ part of the question means, but it is straightforward to check (assuming $x \in (0,1)$) that $x={1 \over2}$ minimises the function $x \mapsto -2 \Omega + RT {1 \over x(1-x)} $.

I am not exactly sure what the last part of the questions means, but the calculus of differentiation is the same for differentiation & partial differentiation.

If You differentiate one more time and set to zero you get
$G'''(x) = RT {(2x-1) \over x^2 (1-x)^2} = 0$ and we see that $x={1 \over 2}$ is the only solution.


问:
Beginner's question about homotopy type in Milnor's Morse Theory
In the introductory section of Milnor's Morse Theory, he gives an example of the torus.

The set of points of the torus which have height less than $a$ for some $p<a<q$ is just a solid disk, i.e., a 2-cell. When we move from $q$ to $r$, we start getting a cylinder. Milnor notes that moving from a 2-cell to a cylinder is the same (homotopically) as attaching a 1-cell:

I can sort of understand why: A cylinder can be squashed into a circle, and the bottom half of the left side is homotopically a point, so both sides are homotopically the same as a circle.

But at the same time, moving from $q$ to $r$ means attaching something which looks kind of like the pair of pants:

We can squash the legs together to get a circle, so doesn't this mean that we are attaching a circle to our disk, instead of just a line?

答:
A pair of pants doesn't have the same homotopy type as a circle. You can flatten it and then contract it to the shape of the digit 8, i.e. two circles glued at a point.

In any case, who says that the cancellation law holds when gluing topological spaces together? You know that
$$\text{disk} + \text{pair of pants} \approx \text{disk} + \text{line.}$$
Why should you expect that
$$\text{pair of pants} \approx \text{line?}$$


问:
Solving a differential equation with constants
In Gilbert Strang's Differential Equations and Linear algebra notebook, he mentions that, $$y(t) = e^{at}y(0) + q/a(e^{at}-1)$$ can be used to solve $$y'- ay = q(t)$$ if q(t) is constant.
Well I am doing a practice problem (not from Gilbert's book) and my answer doesn't match the correct answer. I am required to solve $$x'=Ix-q$$ where I is a constant interest rate per year and q is constant rate of withdrawing money from the bank. And x is the amount of money in the bank.
Now using the formula from Gilbert's book I came up with $$x=Ce^{It}-q/I(e^{It})+q/I$$ but but the right solution is apparently $$Ce^{It}+q/I$$ My solution has that extra middle term, can anyone tell me what I am doing wrong?

答:
Since $q$ is a constant, couldn't you just use

$$\int\frac{dx}{Ix-q} = \int dt$$

$$\implies \frac{\ln(Ix-q)}{I} = t+A$$

$$\implies Ix-q = Ce^{It}$$

$$\implies x = C'e^{It} + q/I$$

Now if you apply the formula, what you get is

$$x = (C-q/I)e^{It} + q/I$$

What the book has done is called the constant $(C-q/I)$ as another constant $C$, if you get what I mean


问:
How to create successive layers of Hexagonal Closed Packing?

How do I create a HCP using the bottom A layer? In other words, if I take the first layer and make a second layer, how much do I shift it vertically (z direction) and how much do I shift it backwards and forwards (along y axis) each time I make a new layer? In terms of r, the radius of a sphere, would be best.

答:
If we pick any sphere in the bottom layer and an adjacent sphere in the layer above, the displacement vector between their centers tells us how to translate the bottom layer in 3-space to the layer above.

For simplicity, let's pick three adjacent spheres whose centers are nicely situated in the $xy$-plane. Specifically, let one be at the origin $(0,0,0)$, another at $(2,0,0)$, and a third at $(1,\sqrt{3},0)$. (It is well-known that if you cut an equilateral triangle in half, you get a $1$-$\sqrt{3}$-$2$ triangle.)

This triangle sits at the base of a pyramid (regular tetrahedron), the top vertex of which is the center of a sphere in the above layer that is adjacent to the original three.

Exercise. Repeatedly use the Pythagorean theorem to show the centroid of the equilateral triangle must be $(1,\frac{1}{\sqrt{3}},0)$ and the top vertex of the tetrahedron is $(1,\frac{1}{\sqrt{3}},\sqrt{\frac{8}{3}})$.

You can also figure out the angles between various features of this pyramid.


问:
Mapping property of free abelian group, from Artin's Algebra
The question (exercise 7.10.4) is as follows:
The group $G = \langle x, y; xyx^{-1}y^{-1}\rangle$ is called a free abelian group. Prove that if $u, v$ are elements of an abelian group $A$, then there is a unique homomorphism $\phi:G\to A$ such that $\phi(x)=u, \phi(y)=v$.

So at first I thought that this was pretty straightforward. If $u,v$ are elements of an abelian group $A$ then they clearly satisfy the relation $uvu^{-1}v^{-1}=1$ and so if we let $S=\{u,v\}, R=\{uvu^{-1}v^{-1}\}$, $\mathcal{F}$ the free group on $S$, and $\mathcal{R}$ to be the normal subgroup of $\mathcal{F}$ generated by $R$, and finally $\mathcal{A} = \mathcal{F}/\mathcal{R}$, then we have a homomorphism $\phi:\mathcal{A}\to A$ that sends $u,v$ to $u,v$ (I believe this is just a direct application of corollary 7.10.4 from Artin). And $\mathcal{A}$ and $G$ are the same group, so we're done.

But, the solution here seems to do a lot of extra work that I don't understand the need for.

Thanks!

答: Consider the mapping $\varphi : \{x,y\} \to A$ such that $x \mapsto u$ and $y \mapsto v$. Then by the universal property of the free group there is a group homomorphism $\widetilde\varphi : F(\{x,y\}) \to A$ that extends $\varphi$. Now, as $A$ is abelian, $$\widetilde\varphi(xyx^{-1}y^{-1}) = uvu^{-1}v^{-1} = 1$$ meaning that $xyx^{-1}y^{-1} \in \ker \widetilde\varphi$, and then $\ker \widetilde\varphi$ contains the normal subgroup $N$ generated by $xyx^{-1}y^{-1}$. In fact, $N = \ker \widetilde\varphi$ (this is the lot of extra work), and then $\phi : G \to A$ is exactly ($G=F(\{x,y\})/N$) the induced map $F(\{x,y\})/N \to A$ of $\widetilde\varphi$, that is, the one that satisfies that $\phi \circ \pi = \widetilde\varphi$, where $\pi : F(\{x,y\}) \to F(\{x,y\})/N$ is the canonical projection.


问:
Show that $\int_0^\pi \log( 1 - 2r\cos(t) + r^2)\, dt=0$
Show that for $r \in (-1,1)$

$$ \int_0^\pi \log( 1 - 2r\cos(t) + r^2)\, dt = 0$$

Here's what I did so far:

$$f(r,t) = \log(1 - 2r\cos(t) + r^2) = \log( (1-re^{it})(1-re^{-it}))$$
The Leibniz rule states that $$\dfrac{d}{dr} \int_0^\pi f(r,t)\ dt = \int_0^\pi \dfrac{\partial}{\partial r} f(r,t) \ dt$$

After calculating the right part I found $2\pi r$ which means that $\displaystyle\int_0^\pi f(r,t)\ dt = \pi r^2$ when it should be $0$.

Thanks in advance.

答:
Let $ n\in\mathbb{N} $, first of all, we wanna factor the polynomial $ X^{2n}-1 $, it has $ 2n-1 $ zeros which are $ \mathrm{e}^{\mathrm{i}\frac{k\pi}{n}} ,\ k\in\left[\!\left[0,2n-1\right]\!\right] $. Thus : \begin{aligned}X^{2n}-1=\prod_{k=0}^{2n-1}{\left(X-\mathrm{e}^{\mathrm{i}\frac{k\pi}{n}}\right)}&=\left(X-1\right)\prod_{k=1}^{n-1}{\left(X-\mathrm{e}^{\mathrm{i}\frac{k\pi}{n}}\right)}\left(X+1\right)\prod_{k=n+1}^{2n-1}{\left(X-\mathrm{e}^{\mathrm{i}\frac{k\pi}{n}}\right)}\\ &=\left(X^{2}-1\right)\prod_{k=1}^{n}{\left(X-\mathrm{e}^{\mathrm{i}\frac{k\pi}{n}}\right)}\prod_{k=1}^{n-1}{\left(X-\mathrm{e}^{\mathrm{i}\frac{\left(2n-k\right)\pi}{n}}\right)}\\ &=\left(X^{2}-1\right)\prod_{k=1}^{n}{\left(X-\mathrm{e}^{\mathrm{i}\frac{k\pi}{n}}\right)\left(X-\mathrm{e}^{-\mathrm{i}\frac{k\pi}{n}}\right)}\\ X^{2n}-1&=\left(X^{2}-1\right)\prod_{k=1}^{n-1}{\left(X^{2}-2X\cos{\left(\frac{k\pi}{n}\right)}+1\right)}\end{aligned}

Hence, if $ r\in\left(-1,1\right) $, we have : $$ \prod_{k=1}^{n-1}{\left(r^{2}-2r\cos{\left(\frac{k\pi}{n}\right)}+1\right)}=\frac{r^{2n}-1}{r^{2}-1} $$

Using Riemann sum theorem, we have the following : \begin{aligned}\int_{0}^{\pi}{\ln{\left(r^{2}-2r\cos{x}+1\right)}\,\mathrm{d}x}&=\lim_{n\to +\infty}{\frac{\pi}{n}\sum_{k=0}^{n-1}{\ln{\left(r^{2}-2r\cos{\left(\frac{k\pi}{n}\right)}+1\right)}}}\\ &=\lim_{n\to +\infty}{\left(\frac{2\pi}{n}\ln{\left(1-r\right)}+\frac{\pi}{n}\ln{\left(\prod_{k=1}^{n-1}{\left(r^{2}-2r\cos{\left(\frac{k\pi}{n}\right)}+1\right)}\right)}\right)}\\ &=\lim_{n\to +\infty}{\left(\frac{2\pi}{n}\ln{\left(1-r\right)}+\frac{\pi}{n}\ln{\left(\frac{1-r^{2n}}{1-r^{2}}\right)}\right)}\\ \int_{0}^{\pi}{\ln{\left(r^{2}-2r\cos{x}+1\right)}\,\mathrm{d}x}&=0\end{aligned}


问:
Identifying a prime number: Why only check for factors up to $\sqrt{N}$?
My math professor told the class that

For a positive integer $N$, if $N$ is not divisible by prime less than $|\sqrt{N}|$, then $N$ is prime.

For example, for $139$,  $12>|\sqrt{139}|>11$. But $139$ is not divisible by $2, 3, 5, 7, 11$ so $139$ is prime.

Why is it necessary that it has to be $\sqrt{N}$? Is the same property applicable for some other $f(N)$ (here $f(N)=\sqrt{N}$. I am asking if there is some other $f(N)$ besides that)?

答:
If $N=pq$, then $p$ or $q$ is at least $\sqrt{N}$, since otherwise $pq<N$.  That is why $f(N)=\sqrt{N}$ works.

It's also true that, if $N$ is not divisible by a prime less than $f(N)=N$ or $f(N)=\sqrt{N}+1, $

then $N$ is prime, but those are not as helpful.


答:
Note that $1 - 2r\cos t + r^2 = (1-re^{it})(1-re^{-it})$

\begin{align}
&\int_0^\pi \log( 1 - 2r\cos t + r^2) dt\\
 = &\ 2Re \int_0^\pi \log(1-re^{it}) dt
= - 2Re \int_0^\pi \sum_{k=1}^\infty \frac{(re^{it} )^k }kdt\\
 = &\>-2\sum_{k=1}^\infty \frac{r^k}k  \int_0^\pi \cos kt\ dt =0
\end{align}

Title: Show that $\int_0^\pi \log( 1 - 2r\cos(t) + r^2)\, dt=0$

Body:
Show that for $r \in (-1,1)$

$$ \int_0^\pi \log( 1 - 2r\cos(t) + r^2)\, dt = 0$$

Here's what I did so far:

$$f(r,t) = \log(1 - 2r\cos(t) + r^2) = \log( (1-re^{it})(1-re^{-it}))$$
The Leibniz rule states that $$\dfrac{d}{dr} \int_0^\pi f(r,t)\ dt = \int_0^\pi \dfrac{\partial}{\partial r} f(r,t) \ dt$$

After calculating the right part I found $2\pi r$ which means that $\displaystyle\int_0^\pi f(r,t)\ dt = \pi r^2$ when it should be $0$.

Thanks in advance.


问:
How can I define $A$ in according to these demands?
I'm given $M\subseteq P(\mathbb{N}\times\mathbb{N})$, such that $M$ is countable set.

So: $M=\{A_0 , A_1 ,A_2 , \ldots\}$  .

How can I define $A \in P(\mathbb{N}\times\mathbb{N})$ so that, $\forall i\in\mathbb{N}$, $A\neq A_i$ and $(0,0)\notin A$?

I tried to define $A=\{v_i \mid i\in\mathbb{N}\}$ while:

if $A_i=\emptyset$ so $v_i=(1,1)$.

if from all the elements in $A_i$, the minimal coordinate in the $x$ element is equal to $0$, and the minimal coordinate in the $y$ element is bigger than $0$ so: $v_i=(1,\mathrm{minCoordinateY}-1)$.

if from all the elements in $A_i$, the minimal coordinate in the $x$ element is bigger than $0$, and the minimal coordinate in the $y$ element is equal to $0$ so: $v_i=\mathrm{minCoordinateX}-1,1)$.

if from all the elements in $A_i$, the minimal coordinate in the $x$ element is bigger than $0$, and the minimal coordinate in the $y$ element is bigger than $0$ so: $v_i=(\mathrm{minCoordinateX}-1,1)$.

But what can I do in the case that from all the elements in $A_i$, the minimal coordinate in the $x$ element is equal to $0$ and the minimal coordinate in the $y$ element is equal to $0$ too?

答:
An alternative approach per Brian M. Scott's answer here would be to instead simply define $A$ by

$$A := \{(1,n) \mid (1,n) \not \in A_n\}$$

This ensures that $A \ne A_i$ for every $i \in \Bbb N$ by a sort of diagonalization argument. If $(1,i) \in A_i$, then $(1,i) \not \in A$, so $A \ne A_i$, and similar occurs the other way around.

Mostly just posting this to get this out of the unanswered queue. Posting as Community Wiki in particular since I have nothing further to add.


问:
Why is this etale cohomology computation going wrong?
Suppose $f: X \to Y$ is a finite, galois map (with galois group $\Gamma$) of curves over a finite field $\mathbb F_q$. If I pick a constant sheaf $\Lambda (\cong \mathbb Z/\ell^n$, say), then $f_*(\Lambda) \cong \Lambda^\Gamma$ (by which I simply mean $\deg(f) = |\Gamma|$ many copies of $\Lambda$) since $X\times_Y X \cong \bigsqcup_{\Gamma}X$. Moreover, the Leray spectral sequence converges immediately since all the higher pushforwards are zero (since $f$ is finite) and so
$$H^p(X,\Lambda) \cong H^p(Y,f_*\Lambda) \cong H^p(Y,\Lambda)^\Gamma.$$

However, I don't think this will be true. I don't think this is true even for $p=0$. Where am I going wrong?

答:
The problem is with the statement that $f_*(\Lambda) = \Lambda^\Gamma$. While $f_*(\Lambda)(Y) = \Lambda^\Gamma$ as claimed, it is not true that the pushforward of a constant sheaf under a finite map has to be constant.

It is easy to compute the stalks of the pushforward of a finite map and we have:
$$(f_*\Lambda)_y = \bigoplus_{x \to y}\Lambda_x$$
for $x,y$ geometric points on $X,Y$ and we see that over a ramified point, we have fewer points.


答:
Let $N=pq$. If $p=k\sqrt{N}$ where $k\in\mathbb{Q}^+$, then $q=\dfrac{\sqrt{N}}{k}$.

If $k=1$, clearly $\sqrt{N}$ is a factor of $N$.

If $k<1$, then $p<\sqrt{N}$, so there must be at least one factor of $N$ less than $\sqrt{N}$.

If $k>1$, then $q<\sqrt{N}$, so there must be at least one factor of $N$ less than $\sqrt{N}$.

Therefore, there always exists at leastone number less than or equal to $\sqrt{N}$ that divides $N$. So, if there is no such number, $N$ is definitely prime.

Hope this is clear. Ask anything if not clear :)

Title: Identifying a prime number: Why only check for factors up to $\sqrt{N}$?

Body:
My math professor told the class that

For a positive integer $N$, if $N$ is not divisible by prime less than $|\sqrt{N}|$, then $N$ is prime.

For example, for $139$,  $12>|\sqrt{139}|>11$. But $139$ is not divisible by $2, 3, 5, 7, 11$ so $139$ is prime.

Why is it necessary that it has to be $\sqrt{N}$? Is the same property applicable for some other $f(N)$ (here $f(N)=\sqrt{N}$. I am asking if there is some other $f(N)$ besides that)?


答:
Doubling the integral, we may integrate over $[-\pi,\pi)$. The function
$u(z)=\log(1-z)$ is analytic in the unit disk, so averaging it on the circle $\{|z|=r\}$ yields $u(0)=0$.