问:
Prove that there are no prime numbers in $[n! + 2; n! + n]$
I am struggling trying to prove that there are no prime numbers in $[n! + 2; n! + n]$ gap. Could someone please help?

答:
Odd seeming things happen for $n=1,2$ so I will explicitly restrict my answer to $n>2$

To flesh out the answer given by fleablood: For $2< k \le n$ we have $k>1$, and every number in the interval is of the form $n!+k$ for $n,k$ as defined. Also, since $n!$ has every $k$ as an explicit factor, but not its only factor, every number $\frac{n!}{k}=a_k>1$ is an integer.

Thus for every $k$ as defined, $\frac{n!+k}{k}=(a_k+1) \Rightarrow n!+k=k\cdot (a_k+1)$. Since $n!+k$ has two factors, each greater than $1$, it is not prime, and this is true for every $k$. So no number in the interval is prime.


问:
Probability Question with a Dartboard and Darts with diameter 18 in and bullseye 2 in from center
Suppose a circular dart board is $18$ inches in diameter, and the
bull’s eye in the center is $2$ inches in diameter. What is the probability that out of
four darts thrown at the dart board, exactly two hit the bull’s eye? Assume that
each dart lands at a uniformly random point on the board, independently of the
other darts.

I am just wondering if my method was correct, I found that the total area was $81\pi$ and the area of the center was $4\pi$, then I divide $\frac{4\pi}{81\pi}$ and then squared it to get $.0024$. Is that correct? and if it is not, Can you show where I went wrong.

Thanks.

答:
Your process started out correct: The area of the board is

$$A_b=\pi r^2=81\pi$$

but the area of the bulls-eye is

$$A_e=\pi r^2=\pi$$

So the probability that any dart hits the bulls-eye is

$$P=\frac{A_e}{A_b}=\frac{\pi}{81\pi}=\frac{1}{81}$$

Now, to find the probability that exactly two darts hits the bulls-eye, first note there are exactly $\binom{4}{2}$ (read $4$ choose $2$) ways to choose exactly two darts from the four. This is defined as

$$\binom{4}{2}=\frac{4!}{2!(4-2)!}=\frac{24}{2\cdot 2}=6$$

Second, the probability that exactly two darts (in any order) hit the center is given by

$$P\cdot P\cdot (1-P)\cdot (1-P)=\frac{6400}{43046721}$$

Since there six ways to find arrange the darts and have this happen, the final probability of exactly two darts hitting the bulls-eye is given by

$$6\frac{6400}{43046721}=\frac{12800}{14348907}=0.0892054\%$$

ContentLicense: CC BY-SA 4.0


问:
Contradictory definitions of Open And Closed set
According to Erwin Kreyszig, The Open and closed set are defined as

A subset $M$ of a metric space $X$ is said to be open if it contains a ball about each of its points. A  subset $K$ of $X$ is said to be closed if its complement (in $X$) is open, that is, $K^C=X-K$ is open.

According to Ravi P. Agrawal in his An Introduction to Complex Analysis

A set $S$ is said to be closed if it contains all of its boundary points; i.e., $\partial S \subseteq S. $. It follows that $S$ is open if and only if its complement $\mathbf{C}-S$ is closed. The sets $\mathbf{C}$ and $\emptyset$ are both open and closed.

I found a contradictory example. Suppose $X=(1,5)$ ($X$ is usual metric space)  and $K=(1,2]\cup [4,5)$.

From the second definition, it's clear that $K$ is not a closed set.

From the first,

$$K^C=X-K=(2,4)$$
which is open, so $K$ must be closed follows from the first definition.

But the two definitions seem to be contradicting each other. Please, Point out flaws in my thinking.

As an additional doubt. In the second definition, Will the  statement

$S$ is open if and only if its complement $\mathbf{C}-S$ is closed.

is follow independent of

The sets $\mathbf{C}$ and $\emptyset$ are both open and closed.

?

答:
$K$ does contain all its boundary points because $\partial K = \{2, 4\}$.

You seem to erroneously believe that $1$ and $5$ are boundary points. But the boundary of $K$ must be a subset of $X$. Since neither $1$ nor $5$ is in $X$, $K$ is a closed subset of $X$.

To be more precise, the boundary of $K$ is defined as $\{x \in X | x$ is a limit point of $K$ and $x$ is a limit point of $X \setminus K\}$. The very definition of the boundary of $K$ precludes us from saying that $1 \in \partial K$.


问:
Lebesgue measure on set difference
While studying a book on measure theoretic probability (by KR Parthasarthy), I came across a proof which employs the following equality:

$
L((A + x) \Delta (A + y)) = L(A \Delta [A + (y - x)])
$

where $L$ is the Lebesgue measure on real line.

I am not able to see how this equality holds. Any ideas? Thanks in advance.

答: $A+x$ is the set $A$ translated by $x$, and $A+y$ is the set $A+(y-x)$ translated by $x$, so $(A+x)\mathop{\triangle}(A+y)$ is just the set $A\mathop{\triangle}\big(A+(y-x)\big)$ translated by $x$. Translation preserves Lebesgue measure.

Title: Lebesgue measure on set difference


问:
Why does $\sin(90 + x) = \cos(x)$ and $\sin(90 - x) = \cos(x)$?
Why are both statements below true in regards of a trigonometric circumference?

$\sin(90 + x) = \cos(x)$

$\sin(90 - x) = \cos(x)$

答:
Here is an animation that might help explain this.

The horizontal line on the $x$-axis is the adjacent side of the right triangle whose base is the $x$-axis. The length of that side is the length of the hypotenuse times the cosine of the counter-clockwise angle between the positive $x$-axis and the hypotenuse.

The moving horizontal line is the opposite side of the right triangle whose base is the $y$-axis. The length of that side is the length of the hypotenuse times the sine of the counter-clockwise angle between the negative $y$-axis and the hypotenuse. The length of that side is also the length of the hypotenuse times the sine of the clockwise angle between the positive $y$-axis and the hypotenuse.


问:
Help finding partial derivative with chain rule
I'm having trouble finding the partial derivative of $z$ with respect to $x$ of this function:

$$z(x,y) = f(x + y) + f(x - y)$$

I saw this exercise online, but I can't figure it out. How can I solve it? Because the functions $f(x + y)$ and $f(x - y)$ only depend on one variable, but I don't understand why their input is an expression involving $x$ and $y$.
Thank you!

答:
To address your second concern first, let's consider the function $f(x) = x^2$.  Then, there is nothing strange about evaluating, say, $f(2 - 3) = f(-1) = 1$ or $f(15 - 10) = f(5) = 25$.  In the same way, we could define a new function $h(x,y) = f(x-y)$, in which case we could write the above calculations as $h(2,3) = 1$ and $h(15,10) = 25$.  So, we are using a single-variable function $f$ to carry out the final computation, but if we are always considering a calculation 'inside' the function it is easier to think of the function of two variable, $h$.

Now, to your original question.  Given $$z(x,y) = f(x + y) + f(x - y),$$ if we want to compute the partial derivative of $z$ with respect to $x$, we will need to use the chain rule:
$$\frac{\partial z}{\partial x}(x,y) = f'(x+y)\frac{\partial(x+y)}{\partial x} + f'(x-y)\frac{\partial(x-y)}{\partial x} = \boxed{f'(x+y) + f'(x-y)}.$$

Now, if you wanted to compute the partial derivative of $z$ with respect to $y$ then the chain rule would actually do something more obvious:
$$\frac{\partial z}{\partial y}(x,y) = f'(x+y)\frac{\partial(x+y)}{\partial y} + f'(x-y)\frac{\partial(x-y)}{\partial y} = \boxed{f'(x+y) - f'(x-y)}.$$


问:
Showing Uniform Continuity of An Infinite Series and Termwise Differentiation
This is a problem from Kenneth A Ross 2nd Edition Elementary Analysis:

Show that the infinite series,$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n+x^2}$$ converges uniformly for all $x$, and by termwise differentiation, compute $f '(x)$.

My work so far involves the Weierstrass M Test - essentially $\lvert\frac{(-1)^n}{n+x^2}\rvert \leq \lvert\frac{1}{k^2}\rvert \forall x \in \mathbb{R}$ and since $\sum_{n=1}^{\infty} \frac{1}{k^2} \lt \infty$, the original series must be uniformly convergent for all x.

I'm not exactly sure if I did that right, or if the alternating negative one changes the requirements for the M-test.

Edit - based on comments, the alternating series test seems useful here. Because of the $(-1)^n$ component, and for each term $a_n$ in the series $a_{n+1} \leq a_n$, that means the series converges by the alternating series test. But does that show absolute convergence, and is that enough, or does it need more?

答: HINT: Let $f_k(x)=\displaystyle{\sum_{n=1}^k \frac{(-1)^n}{x^2+n}}$. You know that there is $f(x)$ so that $f_k(x)\to f(x)$ for each $x$. What is an estimate on $|f(x)-f_k(x)|$? Can you get that estimate independent of $x$?


答:
Uniform convergence of $f(x) = \sum_{n=1}^{\infty} \frac{(-1)^n}{n+x^2}$ on $\mathbb{R}$ follows from the Dirichlet test. Another way to prove this is to observe that for all $x \in \mathbb{R}$,

$$\left|\sum_{k=n+1}^{m} \frac{(-1)^n}{k+x^2}\right| = \frac{1}{n+1+x^2}- \left(\frac{1}{n+2+x^2}- \frac{1}{n+3+x^2} \right) - \ldots \\\leqslant \frac{1}{n+1+x^2} \leqslant \frac{1}{n+1}$$

and uniform convergence follows by the uniform Cauchy criterion.

The termwise differentiated series is

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

We have

$$\left|\frac{(-1)^{n+1}2x}{(n+x^2)^2}\right| = \frac{1}{\sqrt{n}}\frac{1}{n+|x|^2}\frac{2\sqrt{n}|x|}{n+|x|^2} \leqslant \frac{1}{n^{3/2}},$$

and this series converges uniformly by the Weierstrass M-test.

By a well-known theorem it follows that $f$ is differentiable and $f'(x)$ equals the sum of the termwise differentiated series.

Title: Showing Uniform Continuity of An Infinite Series and Termwise Differentiation

Body:
This is a problem from Kenneth A Ross 2nd Edition Elementary Analysis:

Show that the infinite series,$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n+x^2}$$ converges uniformly for all $x$, and by termwise differentiation, compute $f '(x)$.

My work so far involves the Weierstrass M Test - essentially $\lvert\frac{(-1)^n}{n+x^2}\rvert \leq \lvert\frac{1}{k^2}\rvert \forall x \in \mathbb{R}$ and since $\sum_{n=1}^{\infty} \frac{1}{k^2} \lt \infty$, the original series must be uniformly convergent for all x.

I'm not exactly sure if I did that right, or if the alternating negative one changes the requirements for the M-test.

Edit - based on comments, the alternating series test seems useful here. Because of the $(-1)^n$ component, and for each term $a_n$ in the series $a_{n+1} \leq a_n$, that means the series converges by the alternating series test. But does that show absolute convergence, and is that enough, or does it need more?


问:
Boolean space enveloping an infinite family of Boolean spaces
I am puzzling over the following problem, which I have been couldn't solve yet.

Suppose we are given an infinite family $\{S_i\}_{i\in I}$ of disjoint Boolean spaces - compact, Hausdorff topological spaces with base of clopen (simultaneously closed and open) subsets (alternatively called Stone spaces).

Is it always possible to find a Boolean space $S$ containing $\bigcup S_i$; and such that  for each $i\in I$ subsest $S_i\subseteq S$ be a closed subset of $S$?

I have tried to consider a topology on $\bigcup S_i$ generated by $(\bigcup_{k\neq i}S_k)\cup U_j$, where $U_j\in \mathcal O(S_i)$ an open subset of $S_i$, but it is easy to see that in that case we have no $T_2$ separation. Apart that one I have no other ideas yet. Any suggestions, references and any help is appreciated very well.

答:
Sure.  For instance, you can just take $S$ to be the 1-point compactification of the disjoint union $\coprod S_i$.  Or, you could take the coproduct of the $S_i$ in the category of Stone spaces, which is dual to the product $\prod B_i$ of the clopen algebras $B_i$ of each $S_i$.  (This coproduct is in fact the Stone-Cech compactification of the disjoint union $\coprod S_i$, though that takes some work to prove.)

You may find it easier to think about in terms of Boolean algebras.  If $B_i$ is the clopen algebra of $S_i$ and we assume for convenience that we want $\bigcup S_i$ to be dense in $S$, then we're just looking for an $S$ that is the Stone space of a subalgebra $B\subseteq \prod B_i$ with the following properties.  First, the projection of $B$ onto each $B_i$ is surjective (this gives an embedding of each $S_i$ into $S$).  Second, and for each distinct $i,j$, there exists $b\in B$ whose $i$th coordinate is $1$ and $j$th coordinate is $0$ (this means the images of $S_i$ and $S_j$ in $S$ are disjoint).  Alternatively, you can combine these two conditions as the following condition: for each finite $F\subseteq I$, the projection $B\to\prod_F B_i$ is surjective.

The second construction of the first paragraph is given by taking $B$ to just be the full product algebra $\prod B_i$, and the first (the one-point compactification) is given by taking $B$ to be the subalgebra of $\prod B_i$ consisting of elements that are either $0$ on all but finitely many coordinates or $1$ on all but finitely many coordinates.


问:
How to write sym sum $i^\text{th}$ in $\text{MATLAB}$?
How to write the function below in MATLAB?

Having hard time to express $i^\text{th}$ in sym sum. Trying to minimize $x$ for the optimization, so $f(x)$ is the objective function.
$$
 f_0(x)=\sum_{i=1}^nx_i\log(x_i).
$$

答:
Vectorised operations in MATLAB are a really good practise if you want to evaluate that sum fast.

Let $x$ denote the vector $(x_1,x_2,\dots,x_n)$. You can do it as follows

```<code>f0 = sum(x.*log(x))
</code>```

Body:
Vectorised operations in MATLAB are a really good practise if you want to evaluate that sum fast.

Let $x$ denote the vector $(x_1,x_2,\dots,x_n)$. You can do it as follows

f0 = sum(x.*log(x))


答:
To answer one part of your question, how $f(x-y)$ isn't simply a function of 1 variable, consider the most simple case of $f(w)=w$. If I substitute $x-y$ for $x$ then I get $f(x-y)=x-y$. So I had started with a function indeterminate in 1 variable, and obtained a function that is indeterminate in 2 variables.

So let's look at $\frac{\partial z}{\partial x}$. It will probably be easier if I just substitute $x+y$ with $u$ and $x-y$ with $v$. Let's note that $\frac{\partial u}{\partial x}=1$, and similarly for $v$. Then
$$\frac{\partial z}{\partial x} = \frac{\partial f(u)}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f(v)}{\partial v}\frac{\partial v}{\partial x}\\=\frac{\partial f(u)}{\partial u}+\frac{\partial f(v)}{\partial v}.$$ It's not possible to get much more definitive than this for your example. Just note that our result depends on what $u,v$ are, and if we took, say, the derivative with respect to $y$ then $\frac{\partial v}{\partial y}=-1$ which would change our answer.

Title: Help finding partial derivative with chain rule

Body:
I'm having trouble finding the partial derivative of $z$ with respect to $x$ of this function:

$$z(x,y) = f(x + y) + f(x - y)$$

I saw this exercise online, but I can't figure it out. How can I solve it? Because the functions $f(x + y)$ and $f(x - y)$ only depend on one variable, but I don't understand why their input is an expression involving $x$ and $y$.
Thank you!


问:
Show that the Gamma Function can be expressed as $\Gamma(z) =\int_1^\infty e^{-t}t^{z-1}dt+\sum_{n=0}^\infty\frac{(-1)^n}{n!(z+n)}$
This question is from an assignment sheet which I am solving and I need help in solving following problem.

Show that Gamma Function can be expressed
$$\Gamma(z) =\int_{1}^{\infty} e^{-t} t^{z-1} dt + \sum_{n=0}^{\infty} \frac{(-1)^n } { n! (z+n)}$$

Using Definition of Gamma Function , $\Gamma(z) = \int_{1}^{\infty} e^{-t} t^{z-1} dt + \int_{0}^{1} e^{-t} t^{z-1} dt $. but I am not able to co-orelate $\int_{0}^{1} e^{-t} t^{z-1} dt $ with $\sum_{n=0}^{\infty} \frac{(-1)^n } { n! (z+n)}$ .

So, Can you please shed some light on this.

I shall be really thankful.

答:
From the integral definition of the gamma function we have
$$
\Gamma(z)
=\int_0^\infty e^{-t} t^{z-1}\, \mathrm dt
=\int_1^\infty e^{-t} t^{z-1}\, \mathrm dt+\int_0^1 e^{-t} t^{z-1}\, \mathrm dt.
$$
All that is left is to find the series expansion for the second integral. Expanding the exponential term gives
$$
\int_0^1 e^{-t} t^{z-1}\, \mathrm dt=\int_0^1 \lim_{N\to\infty}\underbrace{\sum_{n=0}^N\frac{(-t)^n}{n!} t^{z-1}}_{f_n(z)}\, \mathrm dt.
$$
Now note that on $t\in[0,1]$:
$$
|f_n(z)|<\sum_{n=0}^N\frac{t^n}{n!} |t^{z-1}|<e^t |t^{z-1}|\leq e t^{\Re z-1},
$$
which is integrable in $t\in[0,1]$ for all $\Re z>0$. Hence, by argument of dominated convergence
$$
\int_0^1 e^{-t} t^{z-1}\, \mathrm dt=\sum_{n=0}^\infty\frac{(-1)^n}{n!}\int_0^1 t^{z+n-1}\, \mathrm dt=\sum_{n=0}^\infty\frac{(-1)^n}{n! (z+n)},
$$
which is the desired result.


问:
Existence of a zero of a function $f:(0;1) \to \mathbb{R}$
Let $f:(0;1) \to \mathbb{R}$ be $f(x):=-\ln(1-x)-cx$ for some $c>1$.

How does one show the existence of a zero $0 < a < 1$, $f(a)=0$? I can't work out a development for the intermediate value theorem or the fixpoint theorem (Banach)...

答:
Hint :

Study the variation of $f$ on $\left]0;1\right[$ and show there exists $x_0$ such as $f\left(x_0\right)<0$.
Then note that
$$
\lim\limits_{x \rightarrow 1}f\left(x\right)=+\infty
$$


问:
Assuming that $P(x)$ is a polynomial and it is true that $P(3x)=26 \cdot P(x+1)$, then of what degree is the polynomial?
Assuming that $P(x)$ is a polynomial and it is true that $P(3x)=26 \cdot P(x+1)$, then of what degree is the polynomial?

This is the first time that I've seen such a question and hence I don't know how to attack it. What I attempted to do was the following:

If it is of degree 1, then $P(x)=ax+b$ hence $P(3x)=3ax+b$, but $P(3x)=26*P(x+1)$, hence
\begin{align*}
3ax+b &= 26 \cdot 3a(x+1)+b \\
3ax &= 26 \cdot 3a(x+1) \\
x & = 26 \cdot (x+1) \\
x &= 26 \cdot x+26,
\end{align*}

Hence $x$ needs to be a certain value, something which can't be true.

I then tried the same thing for degree $2, 3, 4$ in the hope of finding some sort of pattern, but I couldn't find one. Could you please explain to me how to solve this question as well as how to solve other similar style questions?

答:
When $P(x)\not\equiv0$ then there is an $n\geq0$ with  $$P(x)=a_n x^n+\ldots+a_1 x+a_0,\qquad a_n\ne0\ .$$ It follows that $$Q(x):=P(x+1)=a_n x^n+b_{n-1}x^{n-1}+\ldots+ b_1 x+b_0$$ for certain $b_k$ determined by the $a_j$. When $x\ne0$ we can divide
$$P(3x)\equiv26 P(x+1)=26 Q(x)$$
by $x^n$ and obtain
$$3^n a_n +{3^{n-1}a_{n-1}\over x}+\ldots+{3a_1\over x^{n-1}}+{a_0\over x^n}=26\left(a_n+{b_{n-1}\over x}+\ldots+{b_0\over x^n}\right)\ .$$
This is valid for all $x\ne0$, in particular for $x\to\infty$. It follows that necessarily $(3^n-26)a_n=0$, hence $a_n=0$, contrary to assumption. This implies that in fact $P(x)\equiv0$. Therefore the degree of $P$ is $-\infty$.


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

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

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

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

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

答:
First, I found a power of 2, $n=2^p$, such that $n\log_2 n=2^p\cdot p\approx 10^6$. Since $2^{20}\approx 10^6$, it's probably going to be a click or two below that. Sure enough, $16\cdot 2^{16}=2^{20}=1048576$.

Now I figured that $\log_2 n \approx 16$, so $n \approx 10^6/16=62500$. Sure enough, for that $n$, $n\log_2 n \approx 995723$.

Now some simple searching or bisection will do the trick: $63000$ and $62750$ are too large; $62625$, $62700$, $62725$ are too small; etc. The number of bisection steps, if you use strict halving of the interval, is bounded by $\log_2 n$. So you know you'll get there quickly. Eventually I found that if $n=62746$, $n\cdot\log_2 n \approx 999998$.

That is the largest possible integer $n$; so $n\leq 62746$ is your answer.

EDIT: earlier I suggested that Newton's method was too beefy for the job. Well, I think I'm wrong. After all, the derivative of $f(n)=n\log_2 n$ is just $f'(n)=\log_2 n+\log_2 e$, and you have to compute $\log_2(n)$ anyway every time you do a bisection step. So the derivative information is there, free to use. If I were coding up an algorithm, yes I think I would do Newton's method. A similar case could be made for using Newton's even for the initial $p\cdot 2^p$ search.

ContentLicense: CC BY-SA 3.0


问:
Necessary and Sufficient Conditions for which an affine mapping with a convex function is convex?
For a convex function $h(x)$, what conditions must hold for $$g_2(x) = ah(x) - b, \\ a, b \in R$$ is also convex?

My intuition is that for a convex $h(x)$ then $a > 0$ because if $a < 0$ then we 'flip' the function and it becomes concave. I also think that $b \in R$ because this will only shift the function vertically. How can I formalise this in a mathematical manner?

答:
Here's a hint:

All convex functions $f : A \rightarrow B$ satisfy, for $t \in [0,1]$ and $x\leqslant y \in A$:

$$f((1-t)x+ty) \leqslant (1-t)f(x) + tf(y)$$

Can you verify that $g_2$ is convex given that $h$ satisfies this condition?

Your intuition that $a \geqslant 0$ is something you should listen to.


问:
Question regarding galois extension and galois groups
From what I have read, for a field extension $E/F$, the automorphism group $Aut(E/F)$ is a Galois group when $E/F$ is Galois, that is, the extension is normal and separable, or equivalently, when $E$ is a splitting field of a separable polynomial with coefficients in $F$.

My question is, is there any intuitive way to explain why this specific definition of a group results in the bijection in the fundamental theorem of Galois theory?

Moreover, I believe the tower of fields $F \subset F_{1} \subset F_{2} \cdots \subset F_{m} = E$, where each intermediate fields are galois extensions of previous ones, correspond to a tower of galois groups, where each subgroup extension are normal. I am confused by the use of word "normal", where in the tower of groups, they essentially say that the left coset is equal to the right coset, while in the tower of fields, the "normal" means every polynomial that is irreducible in $F$ either has no roots in $E$ or splits in $E$. Is there a one-to-one correspondence between the concept of "normal extension" or are they simply different concepts?

Any clarification will be much appreciated.

答:
If $E/F$ is finite normal and separable then so is $E/K$ for any $E/K/F$. Then $K=E^{Aut(E/K)}$ (the subfield of $E$ fixed by $Aut(E/K)$): clearly  $K\subset E^{Aut(E/K)}$. If $a\in E^{Aut(E/K)},\not \in K$, because $E/K$ is separable then $a$'s $K$-minimal polynomial has another root $b$. The natural isomorphism $K(a)\to K(b)$ extends to some $K$-homomorphism $\sigma: K(a,a_2)\to \overline{E},K(a,a_2,a_3)\to \overline{E}$ and so on until we get $\sigma:E\to \overline{E}$. Then $\sigma(\overline{E})\subset E$ because $E/K$ is normal. Whence $\sigma\in Aut(E/K),\sigma(a)\ne a$  so in fact $a$ doesn't exist.

$K/F$ is normal iff $g(K)=K$ for all $g\in Aut(E/F)$ iff $E^{Aut(E/K)}=E^{g^{-1}Aut(E/K)g}$ iff $Aut(E/K)$ is normal in $Aut(E/F)$.


问:
Is the existence of the supremum and the existence of the infimum equal?
Let $X$ is ordered set.

for every $A \subset X$, if $A$ has upper bound, $A$ has supremum

for every $A\subset X$, if $A$ has lower bound, $A$ has infimum

Are these two propositions equal?

答:
You may know how to prove the infimum of a subset of $\mathbb{R}$ that has a lower bound exists has an infimum, from the completeness axiom (which states every subset which has an upper bound has a supremum.) We can see that the same argument works for any partially ordered sets.

I will sketch the proof of 2 from 1. The remaining direction is analogous. Let $(P,\le)$ be a partially ordered set, and $X\subseteq P$ be a subset that has a lower bound. Let $L$ be the set of all lower bounds of $X$. We can see that $L$ has an upper bound: in fact, any element of $X$ is an upper bound of $L$.

By the assumption, $L$ has the supremum, say it $l$. You can see that if $m$ is a lower bound of $X$, then $m\le l$. We claim that $l$ is a lower bound of $X$, so $l$ is actually the infimum of $X$.

Assume that $l$ is not the lower bound of $X$. Then there is $a\in X$ such that $l\nleq a$. For any $m\in L$, $m\le a$. However, $l=\sup L$ shows $l\le a$, a contradiction.


问:
Does $n-4\nmid2n$ for all $n\in\mathbb{N}, n>12$?
From the table of values of $\frac{2n}{n-4}$ it seems to me that the expression never evaluates to an integer for all natural numbers $n$ greater than $12$. How should I go about proving it?

答: Division algorithm: write $$\frac{2n}{n-4} = 2 + \frac{8}{n-4}.$$ The fraction is between $0$ and $1$ strictly for large $n > 12$, as needed.


问:
Evaluating integral using residue
Show that $\int_{0}^{2 \pi} \frac{d \theta}{(5-3\sin(\theta))^2}= \frac{5 \pi}{32}$.

I don't quite know what zi sm doing wrong.

I simplified the expression as follows
$$\int_{0}^{2 \pi} \frac{d \theta}{(5-3\sin(\theta))^2}= \frac{1}{i} \int_{0}^{2 \pi} \frac{4z}{(3iz^{2}+10z-3i)^2}$$

From here I found the zeros to be $\frac{-1}{3i}, \frac{-3}{i}$.

I then went on to say that $\frac{-1}{3i}$ is our only pole.

Then when I plug it into the residue formula I get $$\lim_{z \rightarrow \frac{-1}{3i}}\frac{d}{dz}\left[\left(z-\left(\frac{-1}{3i}\right)\right)^{2} \frac{4z}{(3iz^{2}+10z-3i)^2}\right]$$.

And from here I am stuck. I am not sure if I messed up in my calculations or used the wrong pole or what? Can anyone point me in the right direction?

答:
You are almost done.

Now factor the denominator of the last expression using:
$$
(3i z^2+10z -3 i)=3i\left(z-\frac i3\right)(z-3 i)
$$

Can you take it from here?

$$\begin{align}&\lim_{z \rightarrow \frac{i}{3}}\frac{d}{dz}\left[\left(z-\frac{i}{3}\right)^{2} \frac{4z}{(3iz^{2}+10z-3i)^2}\right]\\&=-\frac49\lim_{z \rightarrow \frac{i}{3}}\frac{d}{dz}\left[\frac{z}{(z-3i)^2}\right]\\&=-\frac49\lim_{z \rightarrow \frac{i}{3}}\left[\frac{-z-3i}{(z-3i)^3}\right]=\frac{5}{64}.\end{align}$$

A side note: the integral over $z$ is not from $0$ to $2\pi$. It is $\oint_{|z|=1}f(z)dz$.

Title: Evaluating integral using residue


答:
$\frac {2n}{n-4} = \frac {2n-8}{n-4} + \frac 8{n-4} = 2 + \frac 8{n-4}$.  But if $n > 12$ then $n -4 > 8$ and $\frac 8{n-4}$ is not an integer.

.... or....

For sufficiently large $n$ (i.e. if $n > 4$) we have $\frac {2n}{n-4} > \frac {2n}n = 2$.

But if $n > 12$ then $4 < \frac 13n$ and $n-4 > (n - \frac 13 n)=\frac 23n$ and $\frac {2n}{n-4} < \frac {2n}{\frac 23 n}=3$ so

So if $n > 12$ then $2<\frac {2n}{n-4}< 3$.

......  OR....

If $n = 12$ we have $\frac {2n}{n-4} = \frac {24}8 = 3$.  $\lim_{n\to \infty} \frac {2n}{n-4} = 2$ and .... well, I was going to say as $\frac {2n}{n-4}$ is clearly decreasing ... but showing that $\frac {2(n+1)}{(n+1)-4} < \frac  {2n}{n-4}$ is not as obvious as my blase statement warrants.

... but....

Note: for $n > 4$ then $2 < \frac {2n}{n-4} < 3 \iff 2(n-4) < 2n < 3(n-4)$ and if you expand that out you get $2n -8 < 2n < 3n -12\implies -8< 0 < n-12 \implies 4< 12 < n$.

Title: Does $n-4\nmid2n$ for all $n\in\mathbb{N}, n>12$?

Body: From the table of values of $\frac{2n}{n-4}$ it seems to me that the expression never evaluates to an integer for all natural numbers $n$ greater than $12$. How should I go about proving it?


答: Say $k(n-4)=2n$ for positive $k$. Then $n=\frac{4k}{k-2}$. This is only positive when $k>2$. At $k=3$ you have $n=12$. For larger $k$, $n$ just decreases, tending to $0$. So $n\leq12$.


答:
$f $ is differentiable at $ (0,1)$ and

$$f'(x)=\frac{1}{1-x}-c=\frac{cx+1-c}{1-x}$$

So,

$$\forall x\in (0,1-\frac 1c)\; f'(x)<0$$

$$\implies f(1-\frac 1c)-f(0)=(1-\frac 1c)f'(y)<0$$
By MVT.

On the other side,
$$(\forall n\ge 0)\;\;f(1-\frac{1}{2^n})\le 0$$
$$\implies \lim_{x\to1^-}f(x)\le 0$$
but
$$\lim_{x\to1^-}f(x)=+\infty$$

thus, apply IVT at $ [1-\frac 1c,1-\frac{1}{2^n}]$.

Title: Existence of a zero of a function $f:(0;1) \to \mathbb{R}$

Body:
Let $f:(0;1) \to \mathbb{R}$ be $f(x):=-\ln(1-x)-cx$ for some $c>1$.

How does one show the existence of a zero $0 < a < 1$, $f(a)=0$? I can't work out a development for the intermediate value theorem or the fixpoint theorem (Banach)...


问:
On calculating the residue of some series containing log of Zeta function:
Consider the following function:

$$F(s)= \sum_m \mu(m) \sum_n \frac{e^{-n/2}\zeta^\prime (mns)}{n \zeta(mns)}$$

Now, we can see, that function has simple poles ${\left[\frac{1}{n}\right]}_{n=1}^\infty$ due to each log derivative of Zeta factor and singularities ${\left[\frac{\alpha_t+i\beta_t}{n}\right]}_{n=1}^\infty$ belonging to each Zeta at denominator where $\alpha_t+i\beta_t$ is zero of $\zeta(s)$.i.e. imaginary axis is 'natural boundary'.

Also I can see some other elementary properties of this.

I want to find following residue :

$$Res_{s=\frac{1}{2}} \frac{\int F(s)ds+(\frac{1}{4e}-\frac{1}{2√e})\ln(s-\frac{1}{2})}{(s-\frac{1}{2})²}$$

I think the value of this residue is zero but it's under Riemann hypothesis. Need more insight into this

As we can see $\int F(s)ds$ has logarithmic singularity at s=1/2 so, the log term in the numerator is to cancel that singularity.

答:
Is $G(s)=\int F(s)ds+(\frac{1}{4e}-\frac{1}{2√e})\log(s-\frac{1}{2})$ analytic at $s=1/2$ ? If so $\frac{G(s)-G(1/2)}{s-1/2}$ is analytic at $1/2$ so
$$Res(\frac{G(s)-G(1/2)}{s-1/2},1/2)=0, \qquad Res(\frac{G(s)}{s-1/2},1/2)=Res(\frac{G(1/2)}{s-1/2},1/2)=G(1/2)$$
Note that $G(1/2)$ depends on the constant of integration in $\int F(s)ds$ (as well as the branch of $\log(s-1/2)$)


问:
A question about the function field of the cuspidal cubic
I'm trying to understand function fields of affine varieties. In the book I'm reading, a function field of a variety $V = V(I)$ was defined to be the set

$$
K_V = \left\{\frac{f}{g}, f,g\in \mathcal{O}(V) : g \notin I\right\}
$$

where $\frac{f_1}{f_2} \sim \frac{g_1}{g_2} \iff f_1 g_2 - f_2 g_1 \in I$.

Now to understand this better I'm trying to the following exercise:

First of all I'm struggling to understand of the domain of the function $\phi^*$, why is it the function field of $V$, $K_V$? The map $\phi$ has codomain $V$ right, but I don't quite what is $F$ here. Also to show that $\phi^*$ is a homomorphism I need to show that $\phi^*(FG) = \phi^*(F)\phi^*(G)$, but not really sure how to proceed here.

答:
An element $F$ of the function field $K_V$ is equivalent to a rational map $\hat{F}: V \dashrightarrow \mathbb{P}^1$. We can pre-compose this with $\phi$ to get a map
$$
\mathbb{A}^1 \overset{\phi}{\to} V \overset{\hat{F}}{\dashrightarrow} \mathbb{P}^1 \, .
$$
So we get a rational map (and in fact, a morphism) $\mathbb{A}^1 \dashrightarrow \mathbb{P}^1$, which as above, is the same thing as an element of the function field of $\mathbb{A}^1$, which is $k(t)$, the field of rational functions in one variable. Thus the morphism $\phi: \mathbb{A}^1 \to V$ induces a $k$-algebra homomorphism by pre-composition:
\begin{align*}
\phi^*: K_V &\to k(t)\\
F &\mapsto F \circ \phi
\end{align*}

In this way, we get a contravariant functor from the category of varieties over $k$ to the category of function fields over $k$.

For instance, in your example where $V: y^2 = x^3$ is the cuspidal cubic and $\phi: \mathbb{A}^1 \to V, t \mapsto (t^2, t^3)$, then
$$
\phi^*(f) = f(\phi(t)) = f(t^2,t^3)
$$
for all $f \in K_V$.


问:
Prove the equation for the point of intersection of two vector lines
This is my first question here and I'm not sure if it's the right place but I'm kinda desperate. I was learning about vector lines and parametric equations. My teacher gave me the assignment below and it is nothing like I saw in my lessons and when I asked him for help he didn't really help (it's an online school so communication is difficult). Here is the question word for word:

*Prove that (if exists) the point of intersection between two lines L1: r→=r01→+tu1→ , t∈R and L2: r→=r02→+su2→ , s∈R is given by the vector formula:

r→=r01→+[( r02→-r01→ )u2→] (u1→×u2→)u1→×u2→2 u1→*

Like I said, nothing similar to that was talked about in the lessons and I really have no idea of what to do or how to prove it so I was hoping someone could help me here. The arrows in the equations above mean that they are vectors.

答:
First, you write your lines into vector form or in homogeneous coordinate. According to the definition of the cross product, if the two lines are non-parallel then the resulting vector $\mathbf{p}=\mathbf{a}\times\mathbf{b}$ is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$. In addition from the definition of dot product, the dot product between the vertical vectors is $0$, so you can get:
$$\mathbf{a}^\top\cdot(\mathbf{a}\times\mathbf{b})=\mathbf{a}^\top\cdot\mathbf{p}=0$$
$$\mathbf{b}^\top\cdot(\mathbf{a}\times\mathbf{b})=\mathbf{b}^\top\cdot\mathbf{p}=0$$
Therefore, the point $\mathbf{p}$ lies on the plane of $\mathbf{a}$ and $\mathbf{b}$, and on the line $\mathbf{a}$ and $\mathbf{b}$, and $\mathbf{p}$ is the intersection of the two lines. Note $\mathbf{p}$ is the homogeneous coordinates.


问:
A question on a periodic function
Let $f(x)$ be a bounded real function on $\mathbb R$ and for any $x \in \mathbb R$
$$
f(x+\frac{13} {42})+f(x)=f(x+\frac16)+f(x+\frac17) \tag1.
$$

What is the fastest way to compute the period of the function?

Thanks for your help.

答:
Without any extra assumptions on $f$, you can take this example: Let $$f(x) = \begin{cases} 1 & x \in \mathbb{Q} \\ 0 & x \notin \mathbb{Q}\end{cases}$$

This function satisfies your equation and is periodic with period $r$ for any rational number $r$. (And has no smallest period.)

Title: A question on a periodic function


问:
How to prove Big O, Omega and Theta asymptotic notations?
I know the definitions of this notations

Big $\mathcal{O} \; : \enspace T(n) \in \mathcal{O}(f(n))$ if and only if $∃ \, c, n_0$, such that $T(n) \leq c \cdot f(n) \enspace \forall \,n \geq n_0$.

Big $\Omega \; : \enspace T(n) \in \Omega(f(n))$ if and only if $∃ \, c, n_0$, such that $T(n) \geq c \cdot f(n) \enspace \forall \,n \geq n_0$.

Big $\Theta \; : \enspace T(n) \in \Theta(f(n))$ if and only if $∃ \, c_1, c_2, n_0$ such that $c_1 f(n) \leq T(n) \leq c_2 f(n) \enspace$ $\forall \,n \geq n_0$.

I'm having trouble manipulating this definitions to prove some notation for example:

Suppose that $f = \Theta(g)$ and $g = \Theta(h)$.
Prove that $f = \Theta(h)$

Suppose that $f$ and $g$ are two non-negative functions such that  $g = \mathcal{O}(f)$.
Prove that $f + g = \Theta(f)$.

Is there anything that could help me? I've read, watch videos but nothing helps me clarifying to prove these notations. Anything helps thanks. Also if you're a tutor and there's a way to set up a meeting it would help.

答:
Considering the first problem:

1.) Because $f \in \Theta(g)$ there must exist $c_1, c_2$ and $n_0$ such that

$$ c_1 g(n) \leq f(n) \leq c_2 g(n) \tag{1}$$

for all $n \geq n_0$.

2.) Because $g \in \Theta(h)$ there must exist $\tilde{c}_1, \tilde{c}_2$ and $\tilde{n}_0$ such that

$$ \tilde{c}_1 h(n) \leq g(n) \leq \tilde{c}_2 h(n) \tag{2}$$

for all $n \geq \tilde{n}_0$.

3.) Combining 1.) and 2.) yields

$$c_1 \tilde{c}_1 h(n) \leq f(n) \leq c_2 \tilde{c_2} h(n) \tag{3}$$

Therefore there exist $\hat{c}_1, \hat{c}_2$, namely $\hat{c}_1 = c_1 \tilde{c}_1$ and $\hat{c}_2 = c_2 \tilde{c}_2$ and a $\hat{n}_0 = \max \{ n_0, \tilde{n_0} \}$ such that $f$ satisfies the conditions for being an element of $\Theta(h)$, i.e. $f \in \Theta(h)$

P.S.: Why $\hat{n}_0 = \max \{ n_0, \tilde{n_0} \}$? Because you want both inequalites to hold, in order to insert them into each other. The first one holds for all $n \geq n_0$, the second one holds for all $n \geq \tilde{n}_0$ and therefore a combination of those two can only hold for all $n \geq \hat{n}_0 = \max \{ n_0, \tilde{n}_0 \}$

Edit 1 I numbered the inequalities for better reference.

Row $(1)$ consists of two inequalities, namely $c_1 g(n) \leq f(n)$ and $f(n) \leq c_2 g(n)$. These inequalities are true for all $n \geq n_0$.

Row $(2)$ consists again of two inequalities, namely $\tilde{c}_1 h(n) \leq g(n)$ and $g(n) \leq \tilde{c}_2 h(n)$. These inequalities are true for all $n \geq \tilde{n}_0$.

Note, that $n_0$ and $\tilde{n}_0$ do not necessarily have to be the same.

We insert the inequalities of $(1)$ and $(2)$ into each other to obtain $(3)$. This only makes sense if the inequalities we are inserting are indeed true.
But if $n_0 < \tilde{n}_0$, then row $(1)$ is still true for all $n \geq \tilde{n}_0$. Otherwise, if $n_0 > \tilde{n}_0$ then row $(2)$ is still true for all $n \geq n_0$. So you choose the greater of these two, i.e. $\max \{ n_0, \tilde{n}_0 \}$ in order for both rows $(1)$ and $(2)$ to hold. Then you have no problems combining them.


问:
F-Homomorphism between two field extensions
If $K$ and $L$ are two extensions of a field $F$ then they are said to  be $F$-isomorphic if there exists an isomorphism $\phi : K\rightarrow L$ which when restricted to the subfield $F$ forms an identity map.

Similarly I assume that $F$-homomorphism is a similar type of homomorphism between two field extensions which when restricted to the subfield $F$ forms an identity map.

I have two statements about field homomorphisms both of them false. But I don't know how they are false

$a)$ Let $K$ and $L$ be two field extensions of the field of rational numbers $Q$ then there exits a field homomorphism $\phi :K \rightarrow L$ which is not a $Q$-homomorphism

$b)$ Let $K$ and $L$ be two field extensions of the field $F=Q(\sqrt{2})$ then any field homomorphism $\phi :K \rightarrow L$ is a $F$-homomorphism

Can anyone please show me how these statements are false ?

答:
A somewhat expansion of @reuns’s comment:

Your a) is wrong because if $K$ is a quadratic extension and $L$ is a cubic extension of $\Bbb Q$, then there are no field homomorphisms from $K$ to $L$.

Your b) is wrong because, for a minimal example, you can take $K=L=F$ and the field homomorphism $\varphi(a+b\sqrt2\,)=a-b\sqrt2$, for $a,b\in\Bbb Q$.


问:
convergent or divergent $\sum_{n=1}^\infty \frac{1}{n^2 + 9}$
I got to the point where I get $1/3(\arctan(∞/3)-\arctan(1/3))$ , so since its gonna be a finite number right? Meaning converges

Body:
Hint

$$\sum_{n=1}^\infty \frac{1}{n^2 + 9}\le\sum_{n=1}^\infty \frac{1}{n^2}$$


答:
Notice that

$$\frac{1}{n^2+9}<\frac{1}{n^2}$$

Can you continue?

Title: convergent or divergent $\sum_{n=1}^\infty \frac{1}{n^2 + 9}$

Body: I got to the point where I get $1/3(\arctan(∞/3)-\arctan(1/3))$ , so since its gonna be a finite number right? Meaning converges


问:
Magnitude of projection proof
I don't get how the magnitude of projection formula follows from the projection vector.

$proj_v(u) = \frac{u \cdot v}{|v|} \frac{v}{|v|}$

$|proj_v(u)| = \frac{|u \cdot v|}{|v|}$

答:
Your magnitude is incorrect since the magnitude of a vector is the length of the vector, denoted as $∥a∥$ which is non-negative.
It should be $|proj_v(u)| = \frac{|u \cdot v|}{|v|}=|u\cos(u,v)|.$

Title: Magnitude of projection proof


问:
Prove that if $X, Y \sim N(0, 1)$ and $\alpha \sim U[0, 1])$ are independent, then $X\cos{\alpha} + Y\sin{\alpha} \sim N(0, 1)$
Let $X, Y \sim N(0, 1)$ and $\alpha \sim U[0, 1])$, and suppose $X, Y, \alpha$ are independent. Then how to prove that $X\cos{\alpha} + Y\sin{\alpha} \sim N(0, 1)$?

For a constant $\alpha$ the claim is obvious but in the case of $\alpha \sim U[0, 1])$, the problem appears to be trickier.

I think, using the formulas for convolution of probability distributions, the following plan could work:

We find the distribution of $X\cos{\alpha}$ and $Y\sin{\alpha}$ separately.

We find the distribution of their sum.

But the execution of this plan would probably be overly complicated, and I feel like there must be a better way.

I would appreciate any help with this problem.

答:
Using the characteristic function of $Z=X\cos \alpha+Y\sin \alpha$, we have

\begin{align}
\phi_Z(t) &= E(e^{it(X\cos \alpha+Y\sin \alpha)}) \\
          &= E\left(E(e^{it(X\cos \alpha+Y\sin \alpha)}|\alpha) \right) \\
          &= E\left(E(e^{it(X\cos \alpha)}|\alpha)E(e^{it(Y\sin \alpha)}|\alpha) \right) \\
          &= E\left(e^{-\frac{1}{2}t^2\cos^2 \alpha}e^{-\frac{1}{2}t^2\sin^2 \alpha} \right) \\
    &= E\left(e^{-\frac{1}{2}t^2} \right) \\
&= e^{-\frac{1}{2}t^2}  \\
\end{align}

Because  $\phi_Z(t) = e^{-\frac{1}{2}t^2}$ then $Z$ follows the standard normal distribution $\mathcal{N}(0,1)$.

Remark: whatever distribution $\alpha$ follows, $Z$ still follows the standard normal distribution $\mathcal{N}(0,1)$


答:
Compute the characteristic function, since we know the answer for constant $\alpha$
$$
\begin{align}
\phi(t) &= E[\exp(it(X\cos\alpha + Y \sin \alpha))]\\
&= E[E[\exp(it(X\cos\alpha + Y \sin \alpha))| \alpha] ]\\
&= E[\exp(-\frac{1}{2}t^2)]\\
&=\exp(-\frac{1}{2}t^2)
\end{align}
$$
and we are done.

Title: Prove that if $X, Y \sim N(0, 1)$ and $\alpha \sim U[0, 1])$ are independent, then $X\cos{\alpha} + Y\sin{\alpha} \sim N(0, 1)$

Body:
Let $X, Y \sim N(0, 1)$ and $\alpha \sim U[0, 1])$, and suppose $X, Y, \alpha$ are independent. Then how to prove that $X\cos{\alpha} + Y\sin{\alpha} \sim N(0, 1)$?

For a constant $\alpha$ the claim is obvious but in the case of $\alpha \sim U[0, 1])$, the problem appears to be trickier.

I think, using the formulas for convolution of probability distributions, the following plan could work:

We find the distribution of $X\cos{\alpha}$ and $Y\sin{\alpha}$ separately.

We find the distribution of their sum.

But the execution of this plan would probably be overly complicated, and I feel like there must be a better way.

I would appreciate any help with this problem.


问:
6-letter permutations in MISSISSIPPI
How many 6-letter permutations can be formed using only the letters of the word, MISSISSIPPI? I understand the trivial case where there are no repeating letters in the word (for arranging smaller words) but I stumble when this isn't satisfied. I'm wondering if there's a simpler way to computing all the possible permutations, as an alternative to manually calculating all the possible 6-letter combinations and summing the corresponding permutations of each. In addition, is there a method to generalize the result based on any P number of letters in a set with repeating letters, to calculate the number of n-letter permutations?

Sorry if this question (or similar) has been asked before, but if it has, please link me to the relevant page. I also apologize if the explanation is unclear, or if I've incorrectly used any terminology (I'm only a high-school student.) If so, please comment. :-)

答:
I'll add a "basic" approach to complement the excellent generating function solution given by @svenkatr above.

The difficulty attaches to the repeated letters; getting the number of 6-letter permutations from $ABCDEFGHIJK$ is simply the falling factorial $(11)_6 = 332640$. However taking $6$ letters from the multiset $\{M, I^4, S^4,P^2\}$ means the combinations are far fewer, since repeats are inevitable (there are only $4$ different letters) and may be quite high multiplicity.

For a given unordered repetition pattern in the chosen 6 letters, say $aaaabc$, we can fill this from the choice of letters based on which letters occur at a suitable multiplicity. For the example $aaaabc$, $a$ can be either $I$ or $S$ while then $b$ and $c$ are a free choice from the remaining letters, $\binom 32 = 3$ giving $6$ options to fill this pattern. Then the arrangements of this pattern are the multinomial $\binom {6}{4,1,1} = 30$.

So once we identify all patterns we can assess each in turn to get a total answer:
$$\begin{array}{|c|c|} \hline
\text{For this pattern:} & \text{options to fill} & \text{arrangements} &  \text{total options} \\[1ex] \hline
 aaaabb & \binom 21 \binom 21 = 4  & \binom{6}{4,2} = 15  &  60 \\[1ex]
 aaaabc & \binom 21 \binom 32  = 6  & \binom{6}{4,1,1} = 30  &  180 \\[1ex]
 aaabbb & \binom 22 = 1  & \binom{6}{3,3} = 20  &  20 \\[1ex]
 aaabbc & \binom 21 \binom 21 \binom 21 = 8  & \binom{6}{3,2,1} = 60  &  480 \\[1ex]
 aaabcd & \binom 21 \cdot \binom 33 = 2  & \binom{6}{3,1,1,1} = 120  &  240 \\[1ex]
 aabbcc & \binom 33 = 1  & \binom{6}{2,2,2} = 90  &  90 \\[1ex]
 aabbcd & \binom 32\cdot \binom 22 = 3  & \binom{6}{2,2,1,1} = 180  &  540 \\[1ex]
\hline
\end{array}$$

summing to $\fbox{1610}$ overall options.

ContentLicense: CC BY-SA 2.5

Title: 6-letter permutations in MISSISSIPPI


答:
Hint.

Assuming your lines are in $\mathbb{R}^2$ we have

$$
\cases{
L_1\to r = r_{01}+t u_1\\
L_2\to r = r_{02}+s u_2
}
$$

the possible intersection happens when $r_{01}+t u_1 = r_{02}+s u_2$ and now multiplying conveniently we have

$$
u_1\cdot r_{01}+t\|u_1\|^2 = u_1\cdot r_{02}+su_1\cdot u_2\\
u_2\cdot r_{01}+t u_2\cdot u_1 =   u_2\cdot r_{02}+s\|u_2\|^2\\
$$

and now we can try to solve for $t,s$

$$
\left(
\begin{array}{cc}
\|u_1\|^2 & u_1\cdot u_2\\
u_1\cdot u_2 & \|u_2\|^2
\end{array}
\right)
\left(
\begin{array}{c}
t\\
s
\end{array}
\right)=\left(
\begin{array}{c}
u_1\cdot(r_{02}-r_{01})\\
u_2\cdot(r_{02}-r_{01})
\end{array}
\right)
$$

NOTE

If $u_1 = \lambda u_2$ then $\det\left(
\begin{array}{cc}
\|u_1\|^2 & u_1\cdot u_2\\
u_1\cdot u_2 & \|u_2\|^2
\end{array}
\right) = 0$

Title: Prove the equation for the point of intersection of two vector lines

Body:
This is my first question here and I'm not sure if it's the right place but I'm kinda desperate. I was learning about vector lines and parametric equations. My teacher gave me the assignment below and it is nothing like I saw in my lessons and when I asked him for help he didn't really help (it's an online school so communication is difficult). Here is the question word for word:

*Prove that (if exists) the point of intersection between two lines L1: r→=r01→+tu1→ , t∈R and L2: r→=r02→+su2→ , s∈R is given by the vector formula:

r→=r01→+[( r02→-r01→ )u2→] (u1→×u2→)u1→×u2→2 u1→*

Like I said, nothing similar to that was talked about in the lessons and I really have no idea of what to do or how to prove it so I was hoping someone could help me here. The arrows in the equations above mean that they are vectors.


问:
What is the name of algorithm that generates a subgraph given a subset of nodes, where direct paths between nodes are retained?

Problem

I have a problem that requires a Graph algorithm, which given

$G = (V, E)$ and $V' \subset V$

returns

$G' =(V', E')$ where $E' = \{ (start, end) \in (V' \times V') : directpath(G, V', start,end) \}$

where directpath(Graph,V', node1, node2) means that there is a path between node1 and node2 in Graph where every node in the path is in $V \setminus V' \cup {start, end}$

More informally:
Given a subset of interesting nodes, the new graph should have an edge between two interesting nodes iff there was a path between those nodes in the original graph that only traversed non interesting nodes.

A simple example would be the graph $G = (V, \{ (R, 1), (1,C), (C,2), (2, E) \})$

Given the subset $V'=\{ R, C, E \}$ the resulting graph would be $G'$ =  $(V', { (R, C), (C,E)}$ )

Context

I already have the actual algorithm itself working, but I am fairly sure that I just reinvented something that I can't find the name of despite searching and asking around.

Also I am wondering what a reasonably efficient way to do this is, depending on the algorithmic complexity I might have to design the code that this is part of differently. My intuition would be that this should be at least doable in linear time and I could probably get my algorithm down to that, but maybe I missing something and am wrong in either direction.

The context of this is program analysis, specifically dataflow analysis.
My use case is only concerned with directed Graphs, but this is a problem that would probably also exist for undirected graphs.

Code

```<code>def filter_graph(
    graph: DiGraph[T],
    subset: Iterable[T],
    self_loops=False,
) -> DiGraph[T]:
    result: DiGraph[T] = DiGraph()

    subset = set(subset)

    for node in graph.nodes():
        if node in subset:
            t = filter_search(graph, node, subset)
            for bound in t:
                if node != bound or self_loops:
                    result.add_edge(node, bound)

    return result

def filter_search(
    graph: DiGraph[T], source: T, subset: Set[T], visited=None
) -> Set[T]:
    visited = visited or set()
    visited.add(source)

    bound: Set[T] = set()
    for succ in graph.successors(source):
        if succ in subset:
            bound.add(succ)
        else:
            if not succ in visited:
                bound |= filter_search(graph, succ, subset, visited)
    return bound
</code>```

答:
The best answer I have gotten to this is that "kind of a special case of Tarjan’s path expression algorithm"

Relevant Citation:

Robert Endre Tarjan. 1981. Fast Algorithms for Solving Path Problems. J. ACM 28, 3 (July 1981), 594–614. DOI:https://doi.org/10.1145/322261.322273

Credit to the person who told me about this: https://twitter.com/BanjoTragedy/status/1250796978994323456


问:
Is $\nabla \cdot (\nabla\times \textbf{F})$ = div $\textbf{F} $?
Is $\nabla \cdot (\nabla\times \textbf{F})$ = div $\textbf{F} $?

Where $(\nabla\times \textbf{F})$ = rot $\textbf{F}$. If yes, why?

答: In general, no. In fact, $\nabla\cdot(\nabla\times F)=\sum_{ijk}\epsilon_{ijk}\partial_i\partial_j F_k=0$, because $\epsilon_{ijk}$ ($\partial_i\partial_j$) is antisymmetric (symmetric) under $i\leftrightarrow j$.


问:
Let $X, Y, Z \sim N(0, 1)$ and are independent. How to show that $\frac{X + ZY}{\sqrt{1 + Z^2}} \sim N(0, 1)$?
Let $X, Y, Z$ be i.i.d. random variables distributed as $N(0, 1)$. How to show that $\dfrac{X + ZY}{\sqrt{1 + Z^2}} \sim N(0, 1)$?

The only solution I see is to progressively find the distributions of

First, $ZY$

Then $U := X + ZY$

Then $V := \sqrt{1 + Z^2}$

Then $\dfrac{U}{V}$

But finding these distributions appears to be a nightmare in terms of the number of calculations to be done.

And this solution doesn't really explain the intuition behind this problem: how could this expression randomly end up being $N(0, 1)$-distributed? That makes me wonder if there is a more nice solution. I also think, this is probably a well-known problem, but I couldn't find a solution anywhere.

Any help would be greatly appreciated!

答:
As you did in this question, it suffices to define the random variable $\alpha$ such that $$\cos \alpha =\frac{1}{\sqrt{1+Z^2}}$$
(we don't need to know its distribution) and calculate the characteristic distribution.

Or a more direct approach is to calculate the characteristic function with the conditional expectation $E( ...|Z)$ on the random variable $Z$. You'll have the same result.


问:
Diffeomorphism in a group action
Considering the group action: $\Omega = \{ \mathbb{Z},+  \}$ on $E = \mathbb{R}^2$ and:

$$ \omega \cdot (x,y)=(x+ \omega, (-1)^{\omega} y)$$

knowing that it is a free and proper action, how can we show that $ \mathbb{R}^2\ / \Omega$ is diffeomorphic to the quocient of $S^1 \times \mathbb{R}$ by the action of the group with two elements given by the antipode transformation:

$$ (x,y) \longrightarrow (-x,-y)$$

and what is the manifold we obtain?

EDIT:

Writting the following diagram:

how can I justify that $\alpha$ is a diffeomorfism? If it is, then by a covering I can get to $S^1/\{+1,-1\}$ correct?

答: First, we note that the quotient of $\mathbb{R}^2$ by the subgroup of even numbers is $S^1 \times \mathbb{R}.$ The we are left with the $\mathbb{Z}/2 \mathbb{Z}$ action, which is easily seen to be the antipode action. You should be able to figure out which manifold is obtaied.

Title: Diffeomorphism in a group action


问:
On the property of Gauss sum
In the book of a classical introduction to modern number theory by Kenneth Ireland and Michael Rosen page 92
in the proof of proposition 8.2.2 of gauss sum, I don't understand how he gets this

$ \sum_{x} \sum_{y} \chi(x) \bar{\chi(y)} \delta(x,y)p =p(p-1)$

Where $\delta(x, y)$ is one if $x\equiv y \pmod p$ and zero otherwise , x, y $\in$ $\mathbb{F}_{p}$ , and $\chi $ is dirichlet character mod $p $.

答:
Divide both sides of the equation
$$ \sum_x \sum_y \chi(x) \overline{\chi(y)} \delta(x,y)p = p(p-1) \tag{1} $$
by $\,p\,$ to get
$$ \sum_x \sum_y \chi(x) \overline{\chi(y)} \delta(x,y)  = (p-1). \tag{2} $$
The double sum in this new equation is essentially the trace of a matrix.

More precisely, if $\,A:=\{A_{i,j}\}\,$ is any matrix, then by definition,
its trace
$$ \text{tr}(A) := \sum_i A_{i,i} = \sum_i \sum_j A_{i,j}\delta_{ij} \tag{3} $$
can be written either as a single sum or else, using the
Kronecker delta,
as an equivalent double sum. The
same holds for equation $(2)$.

An alternative method is the following
$$ \sum_y \chi(x) \overline{\chi(y)} \delta(x,y)\,p
 = \chi(x) \overline{\chi(x)} \,p
 = \sum_y |\chi(x)|^2\ \tag{4} $$
where the first sum is equivalent to a single value since only
terms with $\,y=x\,$ are non-zero.

Equation $(1)$ can now be rewritten as
$$ S:=\sum_x \sum_y \chi(x) \overline{\chi(y)} \delta(x,y)p =
  \sum_x \sum_y |\chi(x)|^2\ = \sum_y\left(\sum_x |\chi(x)|^2\right). \tag{5}$$
Now, by definition of $\,\chi\,$ we have $\,|\chi(x)|^2=1\,$ iff $\,x\ne 0,\,$
or written another way, $\,|\chi(x)|^2 = 1 - \delta(x,0).\,$ Thus we have
$$ S = \sum_y \left(\sum_x 1-\delta(x,0)\right) = p(p-1) \tag{6} $$
where the right side is $\,p(p-1)\,$ because $\,1-\delta(x,0)\,$ is $1$
iff $\,x\ne 1\,$ and there are exactly $\,p(p-1)\,$ such pairs $\,(y,x).$

Title: On the property of Gauss sum


问:
Show that $L^1_\mathcal{F}$ is separable
Let $(\Omega, P)$ be a separable, perfect probability space. Let $\mathcal{F}$ be a closed $\sigma$-algebra on $\Omega$. Let $L^1_\mathcal{F}$ be the subspace of $L^1(\Omega)$ consisting of all the $\mathcal{F}$-measurable random variables. Show that $L^1_\mathcal{F}$ is a separable space.

I know that $L^1(\Omega)$ is separable since $(\Omega, P)$ is strongly isomorphic to $(S, \mu)$ for some $S \subset \mathbb{R}^1$ and some $K$-regular probability measure $\mu$. We can take
$$
\sum_{i = 0}^n a_i 1_{[\alpha_i, \beta_i)}
$$
for $a_i, \alpha_i, \beta_i \in \mathbb{Q}$ on $L^1(S)$. I wonder how we can apply this idea for $L^1_\mathcal{F}$, or if there is any other ideas.

答:
Here, $L_\mathcal{F}^1$ is a subspace of $L^1$. Therefore, we can use the following fact.

Subspace of a separable space is separable


答:
Continuing on the hint given above, we know the following is true: $$\sum_{n=1}^\infty \frac{1}{n^2 + 9}\le\sum_{n=1}^\infty \frac{1}{n^2}$$

The series, $\sum_{n=1}^\infty \frac{1}{n^2}$, is convergent by the p-test (since $p=2\geq1$).

Now because the series $$\sum_{n=1}^\infty \frac{1}{n^2} \geq \sum_{n=1}^\infty\frac{1}{n^2 + 9}$$ converges, by the direct comparison test, the series $\sum_{n=1}^\infty\frac{1}{n^2 + 9}$ converges as well.


问:
Simple non-closed geodesic.
In torus there exists simple non-closed geodesic. One example is take the irrational slope curves in $\mathbb{R}^2$ and project it down to torus. Is this thing can happen in closed hyperbolic surface i.e.

Does there exists a simple non-closed geodesic in a closed (i.e. compact without boundary) hyperbolic surface?

答:
For closed hyperbolic surfaces (with constant negative curvature) the answer is yes. This is shown by exhibiting a transitive geodesic flow on the surface. In this paper by Hedlund, the sufficient property of the flow is called regional transitivity and implies the existence of a simple dense geodesic on the surface $M$. As the geodesic is dense in $M$, it can not be closed.

In the above paper, it is stated in Theorem 3.1 that there are a countable number of periodic geodesics and so in some sense 'most' geodesics on $M$ are non-closed. To actually construct such a geodesic (although not necessarily simple), the standard technique is to use 'symbolic trajectories' which Hedlund hints at in the very last section of the paper. Essentially, you can lift any geodesic on $M$ to its universal cover by the hyperbolic plane and create a bi-infinite sequence associated to the geodesic which tells you which edges of the fundamental regions the geodesic passes through as you 'walk along' the geodesic. Such a sequence is sometimes called a cutting sequence. Admissible cutting sequences can be totally classified by local configuration rules and so it is then just a matter of finding a non-periodic admissible cutting sequences, which will correspond to non-periodic (and hence non-closed) geodesics on the surface $M$.

For an introduction to cutting sequences, and their beautiful links with continued fractions (as well as a direct construction of a non-periodic geodesic on the modular surface) I would strongly recommend reading this wonderful and easy to read article by Caroline Series.

Title: Simple non-closed geodesic.


问:
Fourier Transform of a harmonic polynomial
Let $x \in S^n$ and $P_d$ be a homogenous harmonic polynomial of degree $d$ in $n+1$ variables. I would like to find Fourier transform of the following function:
$$
\frac{1}{|x|^{n+d}}P_d(x).
$$

答:
Of course we treat this as a tempered distribution, since it's not in $L^1$.

As your notation already maybe suggests, writing the function as ${1\over |x|^s}P\Big({x\over |x|}\Big)$ usefully separates the radial and spherical coordinates.

Fourier transform behaves well with respect to homogeneity, sending a tempered distribution of degree $-s$ (thinking of $1/|x|^s$ to normalize) to a tempered distribution of degree $-(\hbox{dim}-s)$. Fourier transform commutes with rotations. We do somehow know that the space of harmonic degree $d$ polynomials (with or without dividing by $|x|^d$) is an irreducible representation of the rotation group, so (Schur's lemma) the only automorphisms of that space, commuting with rotations, are scalars.

Thus, up to some constant $C=C(s,n,d)$, the Fourier transform of the (tempered distribution, possibly requiring meromorphic continuation for some values of $s\in\mathbb C$), is
$$
C\cdot {1\over |x|^{(n+1)-s}}\cdot P\bigg({x\over |x|}\bigg)
\;=\;
C \cdot {1\over |x|^{(n+1)+d-s}} \cdot P(x)
$$
At $s=n$, this is
$$
C\cdot {1\over |x|^{1+d}}\cdot P(x)
$$
The constant(s) can be determined by evaluation at aptly chosen Schwartz functions, such as $P(x)e^{-\pi |x|^2}$.

EDIT: in fact, since Fourier transform is an isomorphism on spaces of harmonic functions, and these are irreducible, it suffices to take $P(x_1,\ldots,x_n)=(x_1+ix_2)^k$, which reduces the computation of the constant to a quite tractable elementary issue.


问:
Is $\int_{-\pi/2}^{\pi/2}\sec(x)$ bounded?
Is $\int_{-\pi/2}^{\pi/2}\sec(x)$ bounded?

It seems like it shouldn't be, since:

$$\int \sec(x) = \ln |\sec(x) + \tan(x)|+C$$

and $\sec(x)\to\infty$ as $x \to (\pi/2)^-$. But, I know integrals can sometimes be counter-intuitive and it may be possible for an integral to be bounded, even if the function itself goes to infinity.

Can anyone confirm?

答: Correct, the integral diverges to $\infty.$


答: The integrand is even, so your integral is $2\lim_{x\to(\pi/2)^-}[\ln|\sec x+\tan x|]_0^x=\infty$.

Body:
Is $\int_{-\pi/2}^{\pi/2}\sec(x)$ bounded?

It seems like it shouldn't be, since:

$$\int \sec(x) = \ln |\sec(x) + \tan(x)|+C$$

and $\sec(x)\to\infty$ as $x \to (\pi/2)^-$. But, I know integrals can sometimes be counter-intuitive and it may be possible for an integral to be bounded, even if the function itself goes to infinity.

Can anyone confirm?


答: Note that$$\lim_{x\to\pi/2}\frac{\sec(x)}{\frac1{\pi/2-x}}=1.$$So, since the integral $\int_0^{\pi/2}\frac1{\pi/2-x}\,\mathrm dx$ diverges, then so does $\int_0^{\pi/2}\sec(x)\,\mathrm dx$.

Title: Is $\int_{-\pi/2}^{\pi/2}\sec(x)$ bounded?


问:
Determine whether $\sum_{n=1}^{+\infty}c_n$ for a given $c_n$ is divergent or convergent
I want to determine whether the series below is absolutely convergent, conditionally convergent or divergent.
  $$\sum_{n=1}^{+\infty}c_n\text{, where }c_n =\begin{cases}
           -\frac{1}{n} \text{, if $\frac{1}{4}n$ is an integer} \\
           \frac{1}{n^2} \text{, if $\frac{1}{4}n$ is not an integer}\end{cases}$$

The terms $c_n$ of this series, for some initial values of $n$, are as follows:

$$\begin{array}{c|c|c|}
n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline
c_n  & \frac{1}{1^2} & \frac{1}{2^2} & \frac{1}{3^2} & -\frac{1}{4} & \frac{1}{5^2} & \frac{1}{6^2} & \frac{1}{7^2} & -\frac{1}{8} \\ \hline
\end{array}$$

This series appears to be divergent, and I will make an attempt to show it.

First (and probably wrong) attempt

(Edit) NOTE: I think this attempt is wrong, because, when the series is divergent, I can't necessarily group the terms arbitrarily to form a new infinite series.

See second attempt below.

I will use the following reasoning: I can group the terms into groups of four to form a new infinite series which is equivalent to the previous one, as follows:

$$\left(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4}\right) + \left(\frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} - \frac{1}{8}\right)+\cdots$$

The $N$th partial sum of the series formed as such can be represented as:

$$\sum_{i=1}^{N}\left(\frac{1}{(4n-3)^2}+\frac{1}{(4n-2)^2}+\frac{1}{(4n-1)^2}-\frac{1}{4n}\right)\ \ \ \ (*)$$

The partial sum above can be rewritten as a sum of two partial sums:

$$\sum_{i=1}^{N}\left(\frac{1}{(4n-3)^2}+\frac{1}{(4n-2)^2}+\frac{1}{(4n-1)^2}\right)+\sum_{i=1}^{N}\left(-\frac{1}{4n}\right) \ \ (**)$$

If we take the limit of the above expression as $N\to+\infty$, we see that the sum in the left converges, but the sum in the right diverges. Therefore, for $N\to+\infty$, the sum is divergent.

Also, since the expression marked above with one asterisk can be seen as the term-by-term sum of the two series in the expression above marked with two asterisks, I can also use the fact that if we sum a convergent infinite series and a divergent infinite series term-by-term, the resulting series is divergent.

Update: Second attempt

I will try to make use of the following theorem:

Theorem: Let $\{s_n\}$ be the sequence of partial sums of a given convergent series $\sum_{n=1}^{+\infty}u_n$. Then, for any $\epsilon>0$, there is a number $N$ such that:

$$|s_R - s_T| < \epsilon \text{ whenever } R>N \text{ and } T>N$$

Below, I will assume that the series is convergent and try to apply the above theorem, to show by contradiction that the series is divergent. Thus, I will try to find a positive lower bound for $|s_{8n}-s_{4n}|$, which would show that the series diverges.

The expression for $s_{8n}-s_{4n}$ is:

$$s_{8n}-s_{4n} = \left(\frac{1}{(4n+1)^2} + \frac{1}{(4n+2)^2} + \frac{1}{(4n+3)^2} - \frac{1}{4n+4}\right)+\left(\frac{1}{(4n+5)^2} + \frac{1}{(4n+6)^2} + \frac{1}{(4n+7)^2} - \frac{1}{4n+8}\right)+\cdots+\left(\frac{1}{(8n-3)^2} + \frac{1}{(8n-2)^2} + \frac{1}{(8n-1)^2} - \frac{1}{8n}\right)$$

The sum above has $4n$ terms. Separating the positive terms from the negative terms, and forming two groups (one with all the positive terms, and other one with all the negative terms), we get:

$$s_{8n}-s_{4n} = \left[\left(\frac{1}{(4n+1)^2} + \frac{1}{(4n+2)^2} + \frac{1}{(4n+3)^2}\right)+\left(\frac{1}{(4n+5)^2} + \frac{1}{(4n+6)^2} + \frac{1}{(4n+7)^2}\right)+\cdots+\left(\frac{1}{(8n-3)^2} + \frac{1}{(8n-2)^2} + \frac{1}{(8n-1)^2}\right)\right]-\left(\frac{1}{4n+4}+\frac{1}{4n+8}+\cdots+\frac{1}{8n}\right)$$

For simplicity, I will rewrite the equation above as follows:

$$s_{8n}-s_{4n} = f(n) - g(n)$$

where

$$f(n) = \left(\frac{1}{(4n+1)^2} + \frac{1}{(4n+2)^2} + \frac{1}{(4n+3)^2}\right)+\left(\frac{1}{(4n+5)^2} + \frac{1}{(4n+6)^2} + \frac{1}{(4n+7)^2}\right)+\cdots+\left(\frac{1}{(8n-3)^2} + \frac{1}{(8n-2)^2} + \frac{1}{(8n-1)^2}\right)$$

and

$$g(n) = \frac{1}{4n+4}+\frac{1}{4n+8}+\cdots+\frac{1}{8n}$$

Taking the absolute value of both sides:

$$|s_{8n}-s_{4n}| = |f(n) - g(n)| = |g(n) - f(n)|$$

Looking at the terms of $g(n)$, we can see that it has $n$ terms and the smallest term is $\frac{1}{8n}$; so, we can say that $g(n) \geq \frac{n}{8n} = \frac{1}{8}$. So, $\frac{1}{8}$ is a lower bound for $g(n)$. Looking at $f(n)$, we can see that it has $4n - n = 3n$ terms and that its greatest term is $\frac{1}{(4n+1)^2}$. So, we can say that $f(n) \leq \frac{3n}{(4n+1)^2}$. And, since $\frac{3n}{(4n+1)^2}$ is decreasing for $n\geq 1$, it has a maximum at $n = 1$ in this interval; so, substituting $n=1$ into the expression, we get that $f(n) \leq \frac{3}{(4+1)^2} = \frac{3}{25}$. Thus, $\frac{3}{25}$ is an upper bound for $f(n)$. Now, since $\frac{1}{8} > \frac{3}{25}$, the lower bound of $g(n)$ is always greater than the upper bound of $f(n)$ if $n\geq 1$. Therefore, $g(n) > f(n)$ for all $n\geq 1$, and:

$$|s_{8n}-s_{4n}| = g(n) - f(n) \geq \frac{1}{8} - \frac{3}{25} = \frac{1}{200}$$

The conclusion above is because $g(n)\geq\frac{1}{8}$ and $f(n)\leq \frac{3}{25}$ (so, $-f(n)\geq -\frac{3}{25}$).

So, we can conclude that:

$$|s_{8n}-s_{4n}| \geq \frac{1}{200} \text{ whenever } n\geq 1$$

But the theorem above requires that, for any $\epsilon>0$, there is a number $N$ such that:

$$|s_{8n} - s_{4n}| < \epsilon \text{ whenever } 4n>N$$

In particular, if we choose $\epsilon=\frac{1}{200}$, we get a contradiction, because $|s_{8n} - s_{4n}|$ can't get arbitrarily smaller than $\frac{1}{200}$. Thus, the series is divergent.

Is this correct? Is there a different argument?

Body:
Starting at a natural divisible by $4$, the terms are

$$c_{4n}+c_{4n+1}+c_{4n+2}+c_{4n+3}=-\frac{1}{4n}+\frac{1}{(4n+1)^2}+\frac{1}{(4n+2)^2}+\frac{1}{(4n+3)^2}$$

$$=-\frac{1024 n^6 +2304 n^5 + 2176 n^4 + 1152 n^3+  388 n^2 + 83 n + 9}{4 n (2 n + 1)^2 (4 n + 1)^2 (4 n + 3)^2}$$

Then the sum becomes

$$\sum_{n=1}^\infty c_n=c_1+c_2+c_3 +\sum_{n=1}^\infty(c_{4n}+c_{4n+1}+c_{4n+2}+c_{4n+3})$$

$$=1+\frac{1}{4}+\frac{1}{9}-\sum_{n=1}^\infty \frac{1024 n^6 +2304 n^5 + 2176 n^4 + 1152 n^3+  388 n^2 + 83 n + 9}{4 n (2 n + 1)^2 (4 n + 1)^2 (4 n + 3)^2}$$

Now, let us take a look at this sum: we have that

$$\frac{1024 n^6 +2304 n^5 + 2176 n^4 + 1152 n^3+  388 n^2 + 83 n + 9}{4 n (2 n + 1)^2 (4 n + 1)^2 (4 n + 3)^2}$$

$$\geq \frac{1024 n^6 +0 + 0 + 0+ 0 + 0 + 0}{4 n (2 n + 2n)^2 (4 n + 4n)^2 (4 n + 4n)^2}$$

$$=\frac{1024n^6}{4n\cdot 16n^2\cdot 64 n^2\cdot 64 n^2}=\frac{1}{256n}$$

Thus, the total sum is

$$1+\frac{1}{4}+\frac{1}{9}-\sum_{n=1}^\infty \frac{1024 n^6 +2304 n^5 + 2176 n^4 + 1152 n^3+  388 n^2 + 83 n + 9}{4 n (2 n + 1)^2 (4 n + 1)^2 (4 n + 3)^2}$$

$$\leq 1+\frac{1}{4}+\frac{1}{9}-\sum_{n=1}^\infty \frac{1}{256n}=-\infty$$

We conclude the overall sum is divergent to negative infinity.


问:
Applying CLT and Law of Large Numbers to a non-standard sequence involving normal distribution
Suppose that $\{X_n\}_{n=1}^{+\infty}$ is an i.i.d sequence distributed as $N(0, \sigma^2)$.

Let
$$U_n := \dfrac{1}{n}\sum\limits_{j=1}^{n} |X_j|$$
$$V_n := \dfrac{1}{n}\sum\limits_{j=1}^{n} X_j^2$$

The question is where does $P_n := \sqrt{n}\left(\sqrt{\frac{2}{\pi}}\frac{V_n}{U_n} - \sigma\right)$ converge by distribution?

What I did:

By the law of large numbers, $U_n \to E|X_1| = \sqrt{\frac{2}{\pi}}\sigma$ and $V_n \to EX_1^2 = \sigma^2$ (convergence in probability and, hence, weak convergence as well).

I managed to represent $P_n$ as $\alpha A_n + \beta B_n$ where $\alpha$ and $\beta$ are constants, and $A_n$, $B_n$ both converge to $A, B \sim N(0, 1)$ by CLT. However, that doesn't mean that the sequence converges to $\alpha A + \beta B$, as far as I understand (in general, we can't add it like that).

Maybe the delta method could help? But not sure, how to apply it here.

I would greatly appreciate any help.

答:
For simplicity's sake, let's denote the symbol $\sim$ as "be asymptotically equal to ... in distribution" for $n \to +\infty$. Example: $X_n \sim Y_n$ means $X_n$ follows asymptotically the distribution of $Y_n$ when $n \to +\infty$.

Let's recall the distribution of $U_n$ and $V_n$ (by using the CLT):

$$U_n \sim E(|X|)+\frac{1}{\sqrt{n}} \mathcal{N}(0,V(|X|))$$
$$V_n \sim E(X^2)+\frac{1}{\sqrt{n}} \mathcal{N}(0,V(X^2))$$

Applying the Taylor developpement on $\frac{V_n}{U_n}$
\begin{align}
\frac{V_n}{U_n} &\sim \left(E(|X|)+\frac{1}{\sqrt{n}} \mathcal{N}(0,V(|X|))\right) \left(E(X^2)+\frac{1}{\sqrt{n}} \mathcal{N}(0,V(X^2))\right)^{-1} \\
&\sim \frac{1}{E(X^2)}\left(E(|X|)+\frac{1}{\sqrt{n}} \mathcal{N}(0,V(|X|))\right) \left(1+\frac{1}{\sqrt{n}} \frac{\mathcal{N}(0,V(X^2))}{E(X^2)}\right)^{-1} \\
&\sim \frac{1}{E(X^2)}\left(E(|X|)+\frac{1}{\sqrt{n}} \mathcal{N}(0,V(|X|))\right) \left(1-\frac{1}{\sqrt{n}} \frac{\mathcal{N}(0,V(X^2))}{E(X^2)}\right) \\
&\sim \frac{1}{E(X^2)}\left(E(|X|)+\frac{1}{\sqrt{n}} \left( \mathcal{N}(0,V(|X|)) +\frac{\mathcal{N}(0,V(X^2))}{E(X^2)}\right) +\mathcal{O}(\frac{1}{n})
 \right)  \\
&\sim \frac{E(|X|)}{E(X^2)}+\frac{1}{\sqrt{n}} \left(\frac{ \mathcal{N}(0,V(|X|)) }{E(X^2)}+\frac{\mathcal{N}(0,V(X^2))}{E^2(X^2)}\right) +\mathcal{O}(\frac{1}{n})
 \\
\end{align}

Hence,
$$P_n = \sqrt{n}\left(\sqrt{\frac{2}{\pi}}\frac{V_n}{U_n}-\sigma \right) \sim \sqrt{\frac{2}{\pi}}\mathcal{N}\left(0,\frac{V(|X|)}{E^2(X^2)} +\frac{V(X^2)}{E^4(X^2)} \right) $$
or

$$P_n \xrightarrow{n\to +\infty} \mathcal{N}\left(0,\frac{2}{\pi}\left(\frac{V(|X|)}{E^2(X^2)} +\frac{V(X^2)}{E^4(X^2)} \right)\right) $$


问:
how to prove that $(n^k) \in \mathcal{O}(n^l)$
Can someone help me to prove that

$(n^k) \in  \mathcal{O}(n^l)$  for

$1 \lt  k  \le l$

I tried to prove it with L'Hopital Rule and I got  $ \lim_{n\to 0} \frac{\infty}{\infty} = 0$, but it is still not totally clear to me how it works.

答:
The definition of $f(n) \in \mathcal{O}(g(n))$ is that there are constants $N \geq 0$ and $c \in [0, \infty)$ such that $f(n) \leq c g(n)$ for all $n \geq N$.
Let's just apply the definition.

If $1 \leq k \leq l$ then $n^k \leq n^l$ for all $n \geq 1$, so $N = 1$ and $c = 1$ suffice.


问:
Laurent Series and annular regions
I'm a bit stuck on this question:

Identify the annular regions  centered on $z = 0$ in which the function :

$$f(z) = \frac{1}{z(z+1)}$$

is analytic.  Find the Laurent series about $z=0$ which is valid for each of the regions.

I've found the Taylor series for $\displaystyle \frac{1}{z+1}$ which gives me a principle part of $\displaystyle \frac{1}{z}$, so I assume the region will then be $0 < |z| < 1$.  I'm not sure how to progress from there. Do I then find the Taylor series for $\displaystyle \frac{1}{z}$?

Thanks in advance.

答:
the function has poles at $0$ and $-1$ so the regions you're looking for are $0<|z|<1$ and $1<|z|$.  in the smaller annulus around $0$ we have
$$
\frac{1}{z(z+1)}=\frac{1}{z}-\frac{1}{z+1}=\frac{1}{z}-\sum_{n=0}^{\infty}(-z)^n.
$$
which converges on $0<|z|<1$.  on the bigger annulus use
$$
\frac{1}{z+1}=\frac{1}{1+\frac{1}{z}}\frac{1}{z}=\frac{1}{z}\sum\frac{(-1)^n}{z^n}
$$
(a geometric series in $1/z$) which converges on $1<|z|$.</p>

Title: Laurent Series and annular regions


问:
Structure Descriptions (GAP) in semigroups
As can be easily seen through GAP manual:

37.6 Structure Descriptions

```<code>StructureDescription( G ) A
</code>```

The method for StructureDescription exhibits the structure of the given group to some extend using the strategy outlined below. The idea is to return a possibly short string which gives some insight in the structure of the considered group and can be computed reasonably quickly.

One can search the site to see this code has been used in some Computational group theory problems, for example here.

A.H.Clifford gives a prime constructive theorem in his book, The Algebraic Theory of Semigroups as follows:

Theorem 1.9. Let $a$ be an element of a semigroup $S$, and let $\langle a\rangle$ be the cyclic subsemigroup of $S$ generated by $a$. If $\langle a\rangle$ is infinite, all the powers of $a$ are distict. If $\langle a\rangle$ is finite, there exist two positive integers, the index $r$ and the period $m$ of $a$, such that $a^{m+r}=a^r$ and the order of $\langle a\rangle$ being $m+r-1$. The set $$K_a=\{a^r,a^{r+1},...,a^{m+r-1}\}$$ is a cyclic subgroup of $S$ of order $m$.

My question: We see that for any semigroup $S$ and $a\in S$ of finite order, we have an structure like $K_a$ which is a group. How can be investigated which finite group(s) has(have) the same structure(s) with a given $K_a$. It seems that cause of existing words in $K_a$ it is so difficult maybe impossible to use above code. Is there any code I don't missing? Any hint is welcomed and is appreciated.

答:
I don't understand the question. Are you asking:

If $S$ is a semigroup and $a$ is an element of $S$ with finite order, then what groups can occur as $K_a$?

The answer to this question is given in the theorem you quote: the cyclic groups.

Perhaps you meant to ask: which groups occur as subgroups of semigroups? Then the answer is all groups, since every group is also a semigroup.

Or maybe you meant: in GAP how can I determine the `structure' of a maximal subgroup of a finite semigroups? Here is how:

```<code>gap> LoadPackage("citrus", false);;
gap> s:=RandomTransformationSemigroup(4,4);
<semigroup with 4 generators>
gap> d:=Filtered(DClasses(s), IsRegularDClass);
[ {Transformation( [ 3, 1, 1, 3 ] )}, {Transformation( [ 3, 3, 4, 1 ] )},
  {Transformation( [ 3, 3, 3, 3 ] )} ]
gap> List(d, x-> StructureDescription(GroupHClass(x)));
[ "C2", "C3", "1" ]
</code>```

Body:
I don't understand the question. Are you asking:

If $S$ is a semigroup and $a$ is an element of $S$ with finite order, then what groups can occur as $K_a$?

The answer to this question is given in the theorem you quote: the cyclic groups.

Perhaps you meant to ask: which groups occur as subgroups of semigroups? Then the answer is all groups, since every group is also a semigroup.

Or maybe you meant: in GAP how can I determine the `structure' of a maximal subgroup of a finite semigroups? Here is how:

gap> LoadPackage("citrus", false);;
gap> s:=RandomTransformationSemigroup(4,4);
<semigroup with 4 generators>
gap> d:=Filtered(DClasses(s), IsRegularDClass);
[ {Transformation( [ 3, 1, 1, 3 ] )}, {Transformation( [ 3, 3, 4, 1 ] )},
  {Transformation( [ 3, 3, 3, 3 ] )} ]
gap> List(d, x-> StructureDescription(GroupHClass(x)));
[ "C2", "C3", "1" ]


问:
How to evaluate $ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$?
I was given the series:
$$ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$$
Making some observations I realized that the $ a_{n} $ term would be the following:
$$ a_{n} = \frac{1}{ 3^{n} } \left(\frac{3}{2}+ \frac{(-1)^{n}}{2} \right)$$
What I wanted to do is to find the result of the series, so the answer would be:
$$\sum_{n=1}^{ \infty } \frac{1}{ 3^{n} } \left(\frac{3}{2}+ \frac{(-1)^{n}}{2} \right) = \frac{3}{2} \sum_{n=1}^{ \infty } \left[ \frac{1}{3^{n} } \right]+ \sum_{n=1}^{ \infty } \left[\frac{1}{3^{n} } \frac{(-1)^{n}}{2}\right]$$
I can tell that the first term is convergent because it is a geometric series, in fact, the result is $\frac{3}{4}$. However, I have no clue in how to solve the second term series. I should say that the series given in the beginning is convert and its result is 5/8. How to arrive to it is a mystery to me.

Body:
HINT:

$\frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$

Separate odd and even terms.

$\frac{1}{3} +\frac{1}{3^{3}}  + \frac{1}{3^{5}} + ...$

$ \frac{2}{3^{2} } + \frac{2}{3^{4}} +\frac{2}{3^{6}}  ...$


问:
Residue field of splitting field is splitting field of residue field.
Let $f$ be an irreducible monic polynomial over $\Bbb Z$ and $L$ its splitting field. Let $p$ be a prime number and $\frak p$ be a prime ideal of ${\cal O} _L$ lying over $p$ (the question also makes sense in a more general setting, but lets consider this case for simplicity).

My question is more or less:

Is ${\cal O} _L/\frak p$ the splitting field of $\bar{f}$, the reduction of $f$ modulo $p$?

After thinking about this a bit I figured out that this is not true in general. Consider e.g. $f=X^2-5, p=2$.

But I think the statement is true if we impose the additional condition that $f$ is separable modulo $p$, i.e. $p$ doesn't divide the discriminant $D(f)$ of $f$.

Let $R=\Bbb Z[\alpha_1,\dots,\alpha_n]$ where the $\alpha_i$ are the roots of $f$. Then clearly $R/R\cap\frak p$ is the splitting field of $\bar{f}$ over $\Bbb F_p$. Our claim would follow if we could prove that $p$ does not divide the index $[{\cal O}_L : R]$, because in this case the map $R/R\cap{\frak p} \to {\cal O}_L / \frak p$ would be an isomorphism.

I tried out some examples and every time the discriminant of $f$ had the same prime factors as the discriminant of $R$, so the above seems to be true.

Maybe there is some formula for the discriminant of the product of orders? In this case we could use the fact that the discriminant of $\Bbb Z[\alpha_i]$ is (up to sign) the discriminant of $f$.

Note: I also saw this question but it didn't really answer my question, I think the answer there refers to something linked in the comments which is no longer accessible.

答:
There is a $p$-adic way, possibly more enlightening than the relative discriminant and the index $[O_L:R]$:

The completion $L_\mathfrak{p}$ is the splitting field of $f\in \Bbb{Q}_p[x]$ and $(O_L)_\mathfrak{p}/\mathfrak{p}\cong O_L/\mathfrak{p}$.

When $f\in \Bbb{Z}_p[x]_{monic}$ is separable $\bmod p$ then Hensel lemma gives that its splitting field is $\Bbb{Q}_p(\zeta_n)$ where $\Bbb{F}_p(\zeta_n)$ is the splitting field of $f\bmod p$.

Whence $$O_L/\mathfrak{p}\cong O_{\Bbb{Q}_p(\zeta_n)}/(p)\cong \Bbb{F}_p(\zeta_n)$$


Title: How to evaluate $ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$?

Body:
I was given the series:
$$ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$$
Making some observations I realized that the $ a_{n} $ term would be the following:
$$ a_{n} = \frac{1}{ 3^{n} } \left(\frac{3}{2}+ \frac{(-1)^{n}}{2} \right)$$
What I wanted to do is to find the result of the series, so the answer would be:
$$\sum_{n=1}^{ \infty } \frac{1}{ 3^{n} } \left(\frac{3}{2}+ \frac{(-1)^{n}}{2} \right) = \frac{3}{2} \sum_{n=1}^{ \infty } \left[ \frac{1}{3^{n} } \right]+ \sum_{n=1}^{ \infty } \left[\frac{1}{3^{n} } \frac{(-1)^{n}}{2}\right]$$
I can tell that the first term is convergent because it is a geometric series, in fact, the result is $\frac{3}{4}$. However, I have no clue in how to solve the second term series. I should say that the series given in the beginning is convert and its result is 5/8. How to arrive to it is a mystery to me.

Body:
It is acceptable to split one convergent series in two as long as both are convergent.

$$ 0 < \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ... < \frac{2}{3} + \frac{2}{3^{2} } + \frac{2}{3^{3}} + \frac{2}{3^{4}} + \frac{2}{3^{5}} + ...$$

and since

$$\frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$$ is monotonically increasing as you are adding terms, and it is bounded by another convergent series (above), it is convergent itself.

Now:

$$ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...=\sum_{n=1}^{\infty} \frac{1}{3^{2n-1}}+\sum_{n=1}^{ \infty }\frac{2}{3^{2n} }=3\sum_{n=1}^{\infty} \frac{1}{3^{2n}}+2\sum_{n=1}^{ \infty }\frac{1}{3^{2n} }=5\sum_{n=1}^{ \infty }\frac{1}{9^{n} }=\frac{5}{8}$$


问:
Tangent to Parametric Polar Curve
If we have some
$$\gamma(t)=r(t)e^{i\theta(t)}$$

Where $\gamma(t)$ is some complex parametric curve; how would one express the tangent vector to that curve, without just converting straight to rectangular cordinates?

答:
Suppose you are looking for the tangent in $t_0$. There is two cases :

If $r(t_0) = 0$ (it means that we pass by the pole) then the tangente is the line with equation $t = t_0$ (line with angle $t_0$ to the abscisse axe).

If $r(t_0) \neq 0$ (we're not in the pole) the the tangent is directed by the vector $r'(t_0) \vec{u} + r(t_0) \vec{v}$ where :
$$u = \vec{i} \cos t_0 + \vec{j} \sin t_0$$
and :
$$v = - \vec{i} \sin t_0 + \vec{j} \cos t_0$$
$\vec{u}$ is the unit vector passing by the pole and the point $\gamma(t_0)$ where we are looking for the tangent and $\vec{v}$ orthogonal to $\vec{u}$.

Title: Tangent to Parametric Polar Curve


答:
If $X,Y\sim N(0,1)$ then the rv $a(X+bY)$ is $ N(0,a^2(1+b^2))$. In particular, if $a=1/\sqrt{1+b^2}$, then it's  $N(0,1)$.

Then, letting $U=\frac{X + ZY}{\sqrt{1 + Z^2}}$, we know $U | Z  \sim N(0,1)$ : it does not depend on $Z$).

Hence $f_U=f_{U|Z}$ ($U$ and $Z$ are independent) and  $U \sim N(0,1)$

Title: Let $X, Y, Z \sim N(0, 1)$ and are independent. How to show that $\frac{X + ZY}{\sqrt{1 + Z^2}} \sim N(0, 1)$?

Body:
Let $X, Y, Z$ be i.i.d. random variables distributed as $N(0, 1)$. How to show that $\dfrac{X + ZY}{\sqrt{1 + Z^2}} \sim N(0, 1)$?

The only solution I see is to progressively find the distributions of

First, $ZY$

Then $U := X + ZY$

Then $V := \sqrt{1 + Z^2}$

Then $\dfrac{U}{V}$

But finding these distributions appears to be a nightmare in terms of the number of calculations to be done.

And this solution doesn't really explain the intuition behind this problem: how could this expression randomly end up being $N(0, 1)$-distributed? That makes me wonder if there is a more nice solution. I also think, this is probably a well-known problem, but I couldn't find a solution anywhere.

Any help would be greatly appreciated!


问:
Is $K$ is compact ? Yes/No
Taken from Rudin books

Let $K ⊂ \mathbb{R}$ consists of $0$ and the numbers $1 /n$ for $n = 1 , 2 ,... $.Prove that $K$ is compact directly from the definition (without Heine-Borel theorem)

My attempt : I found the  answer here.But im not satisfied with this answer

My thinking : I can make $$K=\{ 1, \frac{1}{2},......, \frac{1}{N-1}\} \cup \{0, \frac{1}{N} , \frac{1}{N+1},....\}$$

$$K=\underbrace{\{ 1, \frac{1}{2},......, \frac{1}{N-1}\}}_{\text{finite}} \cup \underbrace{\{0, \frac{1}{N} , \frac{1}{N+1},....\}}_{\text{infinite}}$$

we know that  $$\text{finite} \cup \text{infinite} = \text{infinite}$$

this  implies  that  $K$ doesn't contain finite subcover $\implies$  $K$  is not compact

答: Yes, $K$ is compact.  When we talk about whether a subcover is finite or infinite, we're talking about the number of open sets in the subcover, not whether any of those sets are themselves finite or infinite.  Your proposed cover has two open sets, so it is itself a finite subcover (because two is a finite number).

Title: Is $K$ is compact ? Yes/No


问:
If I already have eigenvectors and eigenvalues, how do I calculate T (x, y)?
I know all the process to how to get eigenvectors and eigenvalues. But how do I know A(x, y) already having these values?

"Let T: R² → R² be a linear transformation that has eigenvectors (3, 1) and (−2, 1) associated with eigenvalues -2 and 3, respectively. Calculate T(x, y)."

答:
Try writing an arbitrary $(x,y) \in \mathbb{R}^2$ as a combination of the eigenvectors.
You want to write
$$
\pmatrix{x\\y} = c_1 \pmatrix{3\\1} + c_2 \pmatrix{-2\\1}.
$$
If you could find such $c_1, c_2$, then $A(x,y) = c_1 A(3,1) + c_2 A(-2,1)$.
We can either try to solve it by hand, or recognize that the above equation can be rewritten in matrix form
$$
\pmatrix{3 & -2\\1 & 1}\pmatrix{c_1 \\ c_2} = \pmatrix{x \\ y}
$$
so in general the answer is
$$
\pmatrix{c_1 \\ c_2} = \pmatrix{3 & -2\\1 & 1}^{-1}\pmatrix{x \\ y}.
$$
I assume that you are able to calculate a matrix inverse, so we find the answer is
$$
\pmatrix{c_1 \\ c_2} = \frac{1}{5}\pmatrix{1 & 2\\-1 & 3}\pmatrix{x \\ y}.
$$
Hence $c_1 = \frac{1}{5}(x+2y)$ and $c_2 = \frac{1}{5}(-x + 3y)$.
You can now plug in to $A(x,y) = c_1 A(3,1) + c_2 A(-2,1)$ to get your answer.


问:
Any subset of the rationals is mappable to all the rationals
I want to prove that there is a bijection between any interval in the rationals and all the rationals. I'll try to prove a special case and I'm hoping to know if this proof is correct. This is not a homework problem, it is merely a curiosity. I'm not so experienced in math proofs.

I need to prove a bijection exists between any enumeration of all rationals in $(0,1]$ and any enumeration of the rationals in $(0,M]$.

If $s_n$ is an enumeration of the set of all rational numbers in the interval $(0,M]$ where $M$ is a rational number, then:

I am able to define $u_n = s_n/M$ with the property that $0 < u_n \le $. $u_n$ is an enumeration of all the rationals between $0$ and $M$ because if there were any empty space $w$, where $0 < w \le 1$ and $\forall n, u_n \ne w$ it would imply that $Mw$ is not in the set $s_n$ which is a contradiction.

Moreover the interval $(0,1]$ is wholly mappable onto any interval $(0,M]$ since any enumeration of $(0,1]$ multiplied by $M$ is a subset of $(0,M]$. Assume there is an empty space $u$ in the interval $(0,M]$. But $u/M \in (0,1]$. So there are no empty spaces.

Since a map exists in both directions a bijection exists between all the rationals and the subset $(0,1]$ from which I am able to prove that if a bijection exists between $A$ and $(0,1]$ and so too for $B$ then a bijection exists between $A$ and $B$.

If this is true I would be very excited.

答:
The problem with your proof is that if $M$ is itself not a rational number, then $s_n/M$ will not be a rational number.

What you can do instead is note that the rationals in $(0,1]$ and the rationals in $(0,M]$ are both countably infinite, so there are enumerations $(q_n)_n$ and $(r_n)_n$ enumerating them respectively. Then the map sending $q_n$ to $r_n$ for each $n$ is a bijection between the two sets of rationals in question.


问:
Transposition Distance between two permutations.
I'm working on the following problem.

Suppose that $A \subseteq S_n$ is a subset of at least $n!/2$ permutations, and let $A(t)$ be the set of permutations that can be obtained from starting at some element of $A$, and then applying at most $t$ transpositions. Prove that there is some absolute constant $c > 0$ so that $$A(t) \geq (1-e^{-ct^2/n})n!$$

My first thoughts are that I want to use the Talagrand concentration theorem, but for that I need to write $S_n$ as a product space. Since I can write any permutation $\sigma$ as a product of transpositions, I thought maybe I could think of $S_n$ as the product of $m = \binom{n}{2}$ independent identically distributed random variables taking values either $0$ or $1$, and a vector of $0$'s and $1$'s would correspond to whether or not the indicated transposition was in a minimal representation of $\sigma$ as a product of transpositions.

This has some problems however. First of all, I doubt this is well defined since there are several ways we can write $\sigma$ as a product of transpositions. The second problem is that order matters, since $(12)(23)$ and $(23)(12)$ give different permutations.

Is there any way I can use Talagrand concentration to approach this problem, probably with a different setup? Does anyone have any thoughts on a different approach?

答:
See Section 5.8 in my PhD thesis, in particular the last display on p. 65:
https://www.cs.bgu.ac.il/~karyeh/thesis.pdf

There is a 1-to-1 correspondence between $\Omega_1\times\ldots\times\Omega_n$ and $S_n$, so the normalized counting measure on the two sets is identical. The aforementioned result gives subgaussian concentration on this set under the Hamming metric, which immediately implies the result you're seeking (see Sec. 2.3.1 ibid. on Levy families and the connection between subgaussian concentration and the type of result you've stated).


问:
Does this disprove the Riemann Hypothesis?
I found interesting results when I put in certain values for z into zeta(z). I believe my findings do not disprove the Riemann hypothesis, but I am not so sure.

When I input z = 1/3 + i into the function, I got a result with a positive real component and a negative imaginary component, which can be written as a - bi.

When I input z = 1/3 - i, I got a result with a positive real component and a positive imaginary component, which can be written as c + di.

For z = 2/5 - i/2, I got -f + gi.

For z = 2/5 + i/2, I got -h - li.

Because the Riemann Zeta Function is continuous, these results imply that there is a number in this range z for which zeta(z) has a real component of zero. In fact, there would be infinitely many of these numbers in an unbroken chain. It also implies that there is a number in this range z for which zeta(z) has a complex component of zero and there there are an infinitely many of these in a broken chain. Also, these two chains should meet with each other at some point. This seems like it would disprove the Riemann Hypothesis.

答:
Yes, you can draw a curve separating $\{\frac13 + i, \frac13 - i\}$ from $\{\frac25 + \frac i2, \frac25 - \frac i2\}$ where the real component of the zeta function is $0$.

Yes, you can draw a curve separating $\{\frac13 + i, \frac25 + \frac i2\}$ from $\{\frac13 - i, \frac25 - \frac i2\}$ where the imaginanary component of the zeta function is $0$.

But you have no control over where these curves intersect; they do not have to intersect inside the trapezoid whose corners are at $\{\frac13  + i, \frac13 - i, \frac25 + \frac i2, \frac25 - \frac i2\}$, because the curves we drew may leave that trapezoid. So the zero you find may be a trivial zero, or it may be a nontrivial zero with real part $\frac12$, neither of which would disprove the Riemann hypothesis.

In fact, here is the picture we get if we actually draw these curves: in blue is the curve with real component $0$, and in orange is the curve with imaginary component $0$. (There's more going on outside the range $[-3,3] \times [-3,3]$, of course.) We see them intersecting once at $z=-2$, and we see them pretending to intersect at $z=1$. (But they actually don't, because the zeta function has a singularity at $1$, corresponding to the harmonic series $1 + \frac12 + \frac13 + \dots$.)


问:
The Method of Frobenius for $x^2y'' + x^2y' - \frac{3}{4}y = 0.$
This is from Introduction to Ordinary Differential Equations, by Shepley Ross:

Use the method of Frobenius to find solutions near $x = 0$ of each of the differential equations in Exercises 5-32.

$x^2y'' + x^2y' - \frac{3}{4}y = 0.$

I've gone through a good portion of the steps but I'm not sure how to proceed. I found that when solving for $y$ corresponding to both roots $r_1 = \frac{3}{2}$ and $r_2 = -\frac{1}{2}$, you end up with essentially the same solution.

Letting $r = r_1 =  \frac{3}{2}$, we obtain

$$y = \sum_{n = 0}^{\infty} c_nx^{n + \frac{3}{2}} = c_0x^{\frac{3}{2}}\big(1 - \frac{1}{2}x + \frac{5}{32}x^2 - \frac{7}{192}x^3 + \dots \big).$$

Letting $r = r_2 = -\frac{1}{2}$, we obtain

$$y = \sum_{n = 0}^{\infty} c_nx^{n - \frac{1}{2}} = c_2x^{\frac{3}{2}}\big(1 - \frac{1}{2}x + \frac{5}{32}x^2 - \frac{7}{192}x^3 + \dots \big).$$

What I'm trying to find is a solution of the form

$$ \sum_{n = 0}^{\infty} c_n^{*}x^{n - \frac{1}{2}} + Cy_1(x) \ln x,$$

where $c_0^{*} \neq 0$ and $C \neq 0$. Using the method of reduction (letting $y = f(x)v$) is extremely tedious and just doesn't seem practical given this solution (Please note: I haven't gone over Bessel's Equation and Bessel Functions yet).

Any advice would be appreciated.

答:
First we check whether Frobenius's method can succeed.  The given equation reduces to
$$  y'' + y' − \frac{3}{4x^2} y = 0  \text{.}  $$
The coefficient of $y'$, $1$ has a pole of order less than or equal to $1$ at $x = 0$ (in fact it has no pole at $x = 0$, so has a pole of order $0$ there).  The coefficient of $y$ has a pole of order less than or equal to $2$ at $x = 0$.  So $x = 0$ is a regular singular point.

So we solve the indicial equation:
$$  \left. \left(r(r-1) + x^2 r - \frac{3}{4} \right)\right|_{x = 0} = 0  \text{,}  $$
giving $r_1 = 3/2$ and $r_2 = -1/2$.

Since the difference of these two roots is an integer, inserting a power series and solving the recurrence for the coefficients does not yield two linearly independent solutions, we only get one.  Use the larger root to get a solution $y_1$.

Since the two roots are separated by an integer (including zero, so if a root is repeated), a second solution is of the form
$$  y_2 = C y_1 \ln x + \sum_{k=0}^\infty b_k x^{k+r_2}  \text{,}  $$
where $y_1$ is the solution you already found and the coefficients $C$ and the $b_k$ are to be determined.  (Once $b_0$ is chosen, typically by setting it to $1$, then $C$ and the $b_k$ up to but not including $b_{r_1 - r_2}$ are determined.  Then declare $b_{r_1 - r_2}$ arbitrarily, which forces values for all the remaining coefficients.)  (It is possible for $C = 0$ to happen.  I haven't checked whether this happens here.)


问:
Density of BM conditioned on start and end (with drift and diffusion)
I am trying to calculate the conditional density $X_t|X_{t_1},X_{t_2}$ for $0<t_1<t<t_2$ of a BM $X_t$ with drift and scaling given by $\mathrm{d}X_t=a\mathrm{d}t+b\mathrm{d}W_t$ (so Brownian bridge with fixed ends).

My approach is to follow the usual derivation for a normal BM. I know that since $f_{W_u|W_v}(x|y)=\frac{f_{W_u,W_v}(x,y)}{f_{W_v}(y)}=\frac{f_{W_u,W_{v-u}}(x,y-x)}{f_{W_v}(y)}=\frac{f_{W_u}(x)f_{W_{v-u}}(y-x)}{f_{W_v}(y)}$ where the second equality follows from $W_v=W_u+W_{v-u}$ and the third equality follows from independence of increments. I therefore tried to apply this to $X_t$ above as follows.

\begin{align*}
f_{X_t|X_{t_1},X_{t_2}}&=\frac{f_{X_t,X_{t_2}|X_{t_1}}(x,x_{t_2}|x_{t_1})}{f_{X_{t_2}|X_{t_1}}(x_{t_2}|x_{t_1})}\\
&=\frac
{\frac1{\sqrt{2\pi b^2(t-t_1)}}\frac1{\sqrt{2\pi b^2(t_2-t)}}\mathrm{exp}\left(-\frac{[(x-x_{t_1})-a(t-t_1)]^2}{2b^2(t-t_1)}\right)\mathrm{exp}\left(-\frac{[(x_{t_2}-x-x_{t_1})-a(t_2-t-t_1)]^2}{2b^2(t_2-t)}\right)}{\frac1{\sqrt{2\pi b^2(t_2-t_1)}}\mathrm{exp}\left(-\frac{[(x_{t_2}-x_{t_1})-a(t_2-t_1)]^2}{2b^2(t_2-t_1)}\right)}.
\end{align*}

where I have used $f_{X_t|X_{t_1}}(x|x_{t_1})=\frac1{\sqrt{2\pi b^2(t-t_1)}}\mathrm{exp}\left(-\frac{[(x-x_{t_1})-a(t-t_1)]^2}{2b^2(t-t_1)}\right)$. However, this doesn't seem right. I have seen the link here, but the density I have derived doesn't seem to match. Furthermore, by substituting $t_1=0$ and $t_2=T$, I don't get the correct answer of $\hat{\mu}=(tX_T+( T-t )X_0)/T$ and $\hat{\sigma}^2=b^2t(T-t)/T$ in the usual distribution of a BM. I also tried to use this conditional probability to find the probability of crossing a barrier (which I have asked in another question), but got nowhere near the correct answer. Where have I gone wrong in my derivation? Thanks!

Edit: in attempting to convert this to a normal BM problem as described in the current answer I have $f(\cdot)=\frac1bg(\cdot)$ where $g$ is the distribution of the standard BM $W_t$, although I haven’t progressed much (still getting complicated fractions).

答:
First let's establish the interpolation property of Brownian Motion.

Theorem. Let $W$ be a standard Brownian Motion, and $0<t_1 < t < t_2$. Then the conditional distribution of $W_t$ given $W_{t_1}=x_1$ and $W_{t_2}=x_2$ is $N(\mu, \sigma^2)$ with $\mu = x_1 + \frac{t-t_1}{t_2-t_1}(x_2-x_1)$ and $\sigma^2 = \frac{(t_2-t)(t-t_1)}{t_2-t_1}$.

Note for intuition: $\mu$ is the linear interpolation of the values $(t_1, x_1), (t_2, x_2)$ at time $t$.

Proof. Let $p(t; x,y) := \frac{1}{\sqrt{2\pi t}}e^{-(x-y)^2/2t}$ be the Gaussian kernel.
We have by the independence of increments
$$
P(B_{t_1} \in dx, B_{t} \in dy, B_{t_2} \in dz)
= p(t_1; 0, x)p(t-t_1; x, y)p(t_2-t; y, z)\,dx\,dy\,dz.
$$
Dividing by the density above by the density of
$$
P(B_{t_1} \in dx, B_{t_2} \in dz) = p(t_1; 0, x) p(t_2-t_1; x, z)\,dx\,dz
$$
we obtain the conditional density
$$
P(B_{t} \in dy | B_{t_1} = x, B_{t_2}=z) = \frac{p(t-t_1; x, y)p(t_2-t; y, z)}{ p(t_2-t_1; x, z)}\,dy
$$
so all that remains is to the simplify the expression above. I will try to go through the calculation without too much magic so you can see where the $\mu$ and $\sigma^2$ formulas come from. First let's handle the normalization constant:
$$
\frac{\frac{1}{\sqrt{2\pi(t-t_1)}} \frac{1}{\sqrt{2\pi(t_2-t)}}}{\frac{1}{\sqrt{2\pi(t_2-t_1)}}} = \frac{1}{\sqrt{2\pi \frac{(t-t_1)(t_2-t)}{t_2-t_1}}} = \frac{1}{\sqrt{2\pi\sigma^2}}.
$$
Then lets look at the exponentials. This would be a huge mess except we will introduce a new variable to simplify things considerably. Since $t_1 < t < t_2$, we can write $t$ as a convex combination of $t_1, t_2$, say
$t = \alpha t_1 + (1-\alpha)t_2$ with $\alpha \in (0,1)$. Then
$t_2-t = (t_2-t_1)\alpha$ and
$t - t_1 = (t_2-t_1)(1-\alpha)$ so we begin the calculation with the exponentials
$$\begin{align}
e^{-\frac12(\frac{(x-y)^2}{t-t_1} + \frac{(y-z)^2}{t_2-t} - \frac{(x-z)^2}{t_2-t_1})}
&= e^{-\frac{1}{2(t_2-t_1)\alpha(1-\alpha)}((x-y)^2\alpha + (y-z)^2(1-\alpha) - \alpha(1-\alpha)(x-z)^2)} \\
&= e^{-\frac{1}{2(t_2-t_1)\alpha(1-\alpha)}(y^2 - 2y(\alpha x + (1-\alpha)z) + (\alpha x + (1-\alpha)z)^2} \\
&= e^{-\frac{1}{2(t_2-t_1)\alpha(1-\alpha)}(y-(\alpha x + (1-\alpha)z))^2} \\
&= e^{-\frac{1}{2\sigma^2}(y-\mu)^2}
\end{align}
$$
completing the proof.

Now we move on to your question.

Note that because $a$ and $b\neq 0$ are constants, knowing $X$ is the same as knowing $W$.
We have $X = a t + b W$, and $W = \frac{1}{b}(X-at)$. Thus conditioning on $X_{t_1}=x_1, X_{t_2}=x_2$ is the same as conditioning on $W_{t_1} = \frac{1}{b}(x_1-at_1), W_{t_2} = \frac{1}{b}(x_2-at_2)$. Since you know the conditional distribution of
$W_t$ given $W_{t_1} = \frac{1}{b}(x_1-at_1), W_{t_2} = \frac{1}{b}(x_2-at_2)$, then you know the conditional distribution of $X_t$ given $W_{t_1} = \frac{1}{b}(x_1-at_1), W_{t_2} = \frac{1}{b}(x_2-at_2)$ is whatever your answer for $W_t$ was, rescaled. Note that for fixed $t$,  $g_t(w) := at + bw$ with $g_t^{-1}(x) = \frac{1}{b}(x-at)$ has $g_t(W) = X_t$ and $g_t^{-1}(X_t) = W_t$, so

$$
\begin{align}
&P(X_t \in dy | X_{t_1}=x_1, X_{t_2}=x_2) \\
&= P(g_t(W_t) \in dy | W_{t_1} = \frac{1}{b}(x_1-at_1), W_{t_2} = \frac{1}{b}(x_2-at_2)) \\
&=|g_t^{-1}(y)'|P(W_t \in d(g_t^{-1}(y)) | W_{t_1} = \frac{1}{b}(x_1-at_1), W_{t_2} = \frac{1}{b}(x_2-at_2)) \\
&=\frac{1}{b}P(W_t \in d(g_t^{-1}(y)) | W_{t_1} = \frac{1}{b}(x_1-at_1), W_{t_2} = \frac{1}{b}(x_2-at_2)) \\
&= \frac{1}{b}\frac{1}{\sqrt{2\pi \sigma^2}} e^{-\frac{1}{2}(g_t^{-1}(y)-\mu)^2/2\sigma^2}
\end{align}
$$
where $\mu = \frac{1}{b}(x_1-at_1) + \frac{t-t_1}{t_2-t_1}( \frac{1}{b}(x_2-at_2) - \frac{1}{b}(x_1-at_1))$ and $\sigma^2 =  \frac{(t_2-t)(t-t_1)}{t_2-t_1}$. You can simplify it further if you like, I'm not sure whether it gets rid of much complexity.