问:
partial derivative with respect to a vector?
I encountered what I can only understand as a partial derivative with respect to a vector, used in a taylor series expansion, and would like to understand it better.

In the book I'm reading, $x(t)$ is an $n$-dimensional real vector, that is $x: \mathbb{R} \to \mathbb{R}^n$, and $u(t)$ is a real valued function, $t\in \mathbb{R}$.

We have the equation
$$
\dot{x} = f(x, u)
$$

The author writes that $f$ can be linearized around a stationary point, that is a point $(x_0, u_0)$ such that $f(x_0, u_0) = 0$:

By looking at small deviations from $x_0$ and $u_0$
$$
x = x_0 + \Delta x\\
u = u_0 + \Delta u
 $$
we get with taylor expansion

$$
\dot{x} = \Delta \dot{x} = f(x_0 + \Delta x, u_0 + \Delta u) \approx \\
f(x_0, u_0) + f_x(x_0, u_0)\Delta x + f_u(x_0, u_0)\Delta u
$$

where higher order terms in $\Delta x$ and $\Delta u$ are discarded. $f_x, f_u, h_x$ and $h_u$ denotes partial derivatives with respect to $x$ and $u$. $f_x$ is a $n \times n$ matrix who's $i, j$ element is
$$
\dfrac{\partial}{\partial x_j}f_i(x, u)
$$
where $f_i$ is the i:th row in $f$, and the corresponding is true for $f_u$.

I have never seen this type of partial derivative before, $f_x$ is a partial derivative with respect to $x$ and $x$ is a vector. I've only seen partial derivatives with respect to a real variable before, and the wikipedia article on partial derivatives does not seem to mention it. Not surprisingly I've also never seen this kind of taylor expansion, the wikipedia article does not seem to use partial derivatives with respect to vectors, but I've never even worked with taylor series of several variables so I might be misunderstanding something.

I can't find much information about this, am I right that this is a special kind of partial derivatives and taylor series, and in that case does anyone know of some relatively easy to understand source of information?

答:
First of all, here is a pdf answering your question in a more detailed fashion: https://www.cs.huji.ac.il/~csip/tirgul3_derivatives.pdf

Now, how I would make sense of this, is well suppose that we have a vector $x$ with coordinates $(x_1,...,x_n)$. Then if one were to take partial derivatives $\frac{\partial}{\partial x} f$, we would have to take derivatives of all possible coordinates, because recall that if you wish to evaluate the gradient of a function $f$, one would have to take partial derivatives depending on the number of variables $f$ takes. And so one could suppose the same thing for derivatives with respect to vectors. (I am sorry if this answer isn't clear. I am trying to explain things intuitively).

Maybe that was the authors intention. Namely, instead of righting down "blah-blah-blah partial derivative with respect to $x_i$" and so on and so forth, he just wrote partial derivative with respect to a vector $x$.


问:
Finding the coefficient of $x^2$ in an expansion
I was given this question as a practise assignment and I am unsure of my answer.

The coefficient of $x^2$ in the expansion of $(x+\frac{1}{ax})^8$ is 7. Find the possible value of $a$.

I did $(x+\frac{1}{ax})^8$ = $(x (1+\frac{1}{ax^2}))^8$

Given my answer, does that mean that $a=7$?

Edit: Thank you for everyone's help! I think I understand it a little bit now. I will have to review this question again and practise more.

答:
Two options:

With the binomial theorem: the expansion of $(a+b)^n$ is
$$(a+b)^n = \binom{n}{0}a^nb^0 + \binom{n}{1}a^{n-1}b^1 + \binom{n}{2}a^{n-2}b^2+\cdots + \binom{n}{n-1}a^1b^{n-1} + \binom{n}{n}a^0b^n.$$
Set $a=x$, $b=\frac{1}{ax}$, $n=8$, and figure out which of the terms is the $x^2$ term. This will give you an expression involving $a$, which you can then solve for $a$.

With derivatives. Rewrite the binomial as
$$\left(x + \frac{1}{ax}\right) = \frac{1}{x}\left(x^2 + \frac{1}{a}\right).$$
Therefore,
$$\left(x + \frac{1}{ax}\right)^8  = \frac{1}{x^8}\left(x^2+\frac{1}{a}\right)^8.$$
The coefficient of $x^2$ will be the coefficient of $x^{10}$ in $(x^2+\frac{1}{a})^{8}$.

This can be found using derivatives: if $p(x)$ is a polynomial, then the constant term of $p^{(k)}(x)$ is $k!$ times the coefficient of $x^k$ in $p(x)$. So $p^{(k)}(0)$ will give you $k!$ times the coefficient of $x^k$ in $p(x)$. Replace $x^2$ with $y$, and look for the coefficient of $y^5$ in the expansion of $(y+\frac{1}{a})^8$ to find the coefficient of $x^{10}$ in the original, and from there you can get the one for $x^2$ in the original expression. This will give you an expression involving $a$, which you can then solve for $a$.

ContentLicense: CC BY-SA 4.0


问:
8. Suppose that two balanced dice are rolled. Define the three events A, B and C as follows:
A= the first die shows an even number.
B= the sum in the two dice is 4.
C= the two outcomes in the two dice differ at most by 2. That is, C = {(x, y) : |x − y| ≤ 2}.
(a) Show that P(A ∩ B ∩ C) = P(A)P(B)P(A).
(b) Show that P(A ∩ B) does not equal P(A)P(B)

I think P(A) is 1/2
P(B) is 1/12 and P(C) is 1/2. Unsure how to find (a) and (b)

答:
You have found that $A\cap B$ is the event where: the first die shows an even number, and the sum of the two dice is $4$. So $A\cap B=\{(2,2)\}$, and hence $\mathsf P(A\cap B)= 1/36\neq (1/2)(1/12)= \mathsf P(A)\,\mathsf P(B)$.

$A\cap B\cap C$ is the event where: the first die shows an even number, the sum of the two dice is $4$, and the two outcomes in the two dice differ at most by $2$.

 $$A\cap B\cap C=\{(2,2)\}$$

PS:
$$\mathsf P(C)~{= {\small\mathsf P(\{{(1,1),(1,2),(1,3)},{(2,1),(2,2),(2,3),(2,4)},{(3,1),(3,2),(3,3),(3,4),(3,5)},{(4,2),(4,3),(4,4),(4,5),(4,6)}, {(5,3),(5,4),(5,5),(5,6)}, {(6,4),(6,5),(6,6)}\})}\\=24/36}$$


问:
Sequence queue - converges or diverges?
I have the next sequence queue: $\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$. Does the queue converges or diverge?

My attempt:
I have tried to show that $\frac{1}{n^2}>|\frac{n^2-n-1}{n^4+n^2+1}|$ for any $n>0$, and then by the comparation test we get that since $\sum_{n=1}^{\infty}{\frac{1}{n^2}}$ converges, we have that $\sum_{n=1}^{\infty}{|\frac{n^2-n-1}{n^4+n^2+1}|}$ converges too, and by a theorem we have that since $\sum_{n=1}^{\infty}{|\frac{n^2-n-1}{n^4+n^2+1}|}$ $\implies$ $\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$ converges. Is that right?

答:
I continue from where you stopped,

Easy to show that :

$$\frac{n^2-n-1}{n^4+n^2+1} < \frac{n^2-1}{n^4+1} < \frac{n^2}{n^4} < \frac{1}{n^2} $$

When we know that $\sum_{n=1}^{\infty}{\frac{1}{n^2}}$ converges.

Therefore, by comparison test :

$$\underset{converges}{\underbrace{\frac{n^2-n-1}{n^4+n^2+1}}} < \underset{converges}{\underbrace{\frac{1}{n^2}}}$$

Title: Sequence queue - converges or diverges?


问:
Show that the sequence $5, 12, 19, 26, ...$ contains no terms of the form $2^n$ or $2^n-1$.
Show that the sequence $5, 12, 19, 26, ...$ contains no terms of the form $2^n$ or $2^n-1$.

I know that this is an arithmetic progression, $a_n=7n-2$. In order to show what is asked I tried setting $7n-2=2^n$ and $7n-2=2^n-1$ and tried to show that either no $n$ satisfies either equation or a contradiction arises, but I'm having a hard time. I have a feeling this problem is really simple too. For $7n-2=2^n$ I found that $n$ needs to be even and for $7n-2=2^n-1$ I found that $n$ needs to be odd. I also think that this approach might not be valid. Any hints or solutions are greatly appreciated.

答:
You are confusing showing that the $n$th term isn't equal to the $n$th power of $2$ (or $1$ less).

That's not what's being asked. (And it's easy to show $7n -2 \ne 2^n$ or $2^n - 1$.)  You need to show that no term is equal to any power.

So you have to show that we can't ever have $7n -2 = 2^m$ nor $7n -2 = 2^m-1$ for any pair of $n, m$ which are not necessarily equal.

....

I'd note that $2^3 \equiv 1 \pmod 7$ if $m = 3k + i$ where $i = 0,1,2$ then $2^m=2^{3k+i} (2^3)^k2^i=8^k2^i\equiv 1\cdot 2^i\mod 7$.

So $2^m\equiv 1,2$ or $4\pmod 7$ and $2^m\not \equiv 0,3,5,6\pmod 7$.

And so $2^m = 7n-2\equiv 5\pmod 7$ is not possible.

Nor is $2^m - 1=7n -2\implies 2^m = 7n -1\equiv 6\pmod 7$.

(We can't have $7n-2\equiv 2^m-2$ or $7n-2\equiv 2^m-5$ either BTW).


问:
circumcenter of the $n$-simplex
Given $m$ points $v_i\in\mathbb{R}^n$, $m<n$. How to find the circumcenter of the simplex formed by the points?

答:
Let us first focus on one face of the cell and assume that we know the circumcenter $c$ and the normal $u$ pointing towards the opposing vertex. Then the circumcenter of the cell $C$ is
$$
C = c + \lambda u
$$
for some $\lambda$. Of course $C$ must have the same distance $R$ to all vertices $v_i$ of the cell,
$$
\begin{split}
R^2
&= \|v_i - (c + \lambda u)\|^2 \quad\forall i,\\
&= \|v_i - c\|^2 + \lambda^2 - 2 \langle v_i - c, u\rangle \lambda \quad\forall i.
\end{split}
$$
Since $u$ is orthonormal on the face, the last term disappears for all vertices on the face. The first is the circumradius $r$ of the face, so
$$
\lambda^2 = R^2 - r^2.
$$
For the opposing vertex $v_o$, the term $\langle v_i - c, u\rangle$ really is the altitude $h$ of $v_o$ over the face, so
$$
\lambda^2 - 2 h \lambda + \|v_o - c\|^2 = R^2.
$$
From these two equations, $\lambda$ can be determined to be
$$
\lambda = \frac{\|v_o -c\|^2 - r^2}{2h}.
$$
This is the elevation of the circumcenter over the face. Note that it is 0 if the distance of the opposing vertex from the circumcenter of the face is exactly the circumradius of the face -- makes sense, says Thales.

This can be done for every face, and we end up with the $n$-linear coordinates $\lambda_k$ of the circumcenter. These can be converted to barycentric coordinates by multiplication with the face area $s_k$,
$$
b_k = \frac{\|v_k - c_k\|^2 - r_k^2}{2h_k} s_k
$$
or equivalently (because of $V=\frac{1}{n}h_k s_k$)
$$
b_k = \frac{\|v_k - c_k\|^2 - r_k^2}{2h_k^2}.
$$
(The constant terms $n$ and $V$ can be discarded.)

Hence, the circumcenter of an $n$-simplex is given by the circumcenters and circumradii of all faces, their altitudes, and the distances of the circumcenters to the opposing vertices. This way, the circumcenter can be computed iteratively "from the ground up", i.e., first compute the circumcenters of the edges, then of the triangles etc. Setting the circumradius of a single point to 0 and the circumcenter to that point, the barycentric coordinates of an edge circumcenter are indeed
$$
b_k = \frac{\|e\|^2 - 0}{2 \|e\|^2} = \frac{1}{2}.
$$

Title: circumcenter of the $n$-simplex


问:
Find an example of a sequence of functions $\{f_n\}$ and $\{g_n\}$
Q. Find an example of a sequence of functions $\{f_n\}$ and $\{g_n\}$ that converge uniformly to some $f$ and $g$ on some set $A$, but such that $\{f_ng_n\}$ (the multiple) does not converge uniformly to $fg$ on $A$. Hint: Let $A=R$, let $f(x)=g(x)=x.$ You can even pick $f_n=g_n.$

As it is stated in Hint, let $A=R,$ and $f(x)=g(x)=x,$ and $f_n=g_n $

Since $f_n$ converges uniformly to $x$. $\forall\varepsilon>0, \exists M\in  s.t. $ for $n\ge M, |f_n-x|<\varepsilon \hspace{0.5cm}\forall x \in R.$

But, $|f_n^2-x^2| = |f_n+x||f_n-x|=|f_n+x|\varepsilon\not < \varepsilon$ for some $x$.

I don't know how to pick $f_n$ satisfying these conditions. Could you give some hint?

答: Take $A=[0,1]$, and define$$\begin{array}{rccc}f\colon&[0,1]&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}0&\text{ if }x=0\\\frac1x&\text{ otherwise.}\end{cases}\end{array}$$For each $n\in\Bbb N$, define $f_n=f$ and let $g_n$ be the constant function $\frac1n$. Then $(f_n)_{n\in\Bbb N}$ converges uniformly to $f$ and $(g_n)_{n\in\Bbb N}$ converges uniformly to the null function, but $(f_ng_n)_{n\in\Bbb N}$ does not converge uniformly.


问:
Proving a property of convex quadrilateral and the Thales-circle via the Pigeonhole Principle
I'm asked to prove the following:

In a convex quadrilateral, the Thales circle of every side, (meaning the half circle, where the side is the diameter) is drawn. Prove that the four half-circles cover the entire quadrilateral.

I am given the hint that I should use the Pigeonhole Principle to prove this. However I have never applied the Pigeonhole Principle to Geometry and don't know where to start. Any help would be appreciated!

答:
Suppose the convex quadrilateral $ABCD$. Let's $O$ any point inside $ABCD$.

We have $\widehat{AOB}+\widehat{BOC}+\widehat{COD}+\widehat{DOA} = 360°$ then there exists an angle (WOLG, suppose $\widehat{AOB}$) less than or equal to $90°$.

Because $\widehat{AOB}\le 90$ then $O$ must inside the circle where its diameter is the side $AB$.


问:
$I_1,I_2,\dots$ disjoint sequence of open intervals $\Rightarrow |\bigcup_{k=1}^{\infty} I_k|=\sum_{k=1}^{\infty}\ell(I_k)$
I have proved the following statement and I would like to know if I have made any mistakes:

$I_1,I_2,\dots$ disjoint sequence of open intervals $\Rightarrow |\bigcup_{k=1}^{\infty} I_k|=\sum_{k=1}^{\infty}\ell(I_k)$

($|\cdot|$ denotes outer measure)

My proof:

By definition of outer measure we know that $\sum_{n=1}^{\infty}\ell(I_k)\geq |\bigcup_{k=1}^{\infty}I_k|$. Now, since $\bigcup_{k=1}^{N}I_k\subset \bigcup_{k=1}^{\infty}I_k$ for $N\geq 1$, from the fact that outer measure preserves order we have that $|\bigcup_{k=1}^{N}I_k|\leq |\bigcup_{k=1}^{\infty}I_k|$ for $N\geq 1$ which implies that the following chain of inequalities holds: $$\sum_{n=1}^{\infty}\ell(I_k)\geq |\bigcup_{k=1}^{\infty}I_k|\geq |\bigcup_{k=1}^{N}I_k|$$
So, if we prove that $|\bigcup_{k=1}^{N} I_k|=\sum_{k=1}^{N} \ell(I_k)$, by taking the limit we would have the desired claim.

We proceed by induction on $N$.

The base case $N=1$ follows from the fact that $|I|=\ell(I)$ if $I=(a,b)$,  $a,b\in\mathbb{R}, a<b$.

Suppose now that the claim is valid for $N\geq 1$: we prove it for $N+1$.

First, $|\bigcup_{k=1}^{N+1}I_k|=|\bigcup_{k=1}^{N}I_k \cup I_{N+1}|\leq |\bigcup_{k=1}^{N}I_k|+|I_{N+1}|\overset{\text{ind.hyp.+def.length open interval}}{=}\sum_{k=1}^{N}\ell(I_k)+\ell(I_{N+1})=\sum_{k=1}^{N+1}\ell(I_k)$.

Now, it remains to prove that $\sum_{k=1}^{N+1}\ell(I_k)\leq |\bigcup_{k=1}^{N+1}I_k|$. So, let $U_1,U_2,\dots$ be a sequence of open intervals whose union contains $\bigcup_{k=1}^{N+1} I_k$ and
for $n\geq 1$ let $J_n:=U_n\cap (-\infty,a_{N+1}),\ K_n:=U_n\cap (a_{N+1},b_{N+1}),\ L_n:=U_n\cap (b_{N+1},\infty)$. $K_1, K_2,\dots$ is a sequence of open intervals whose union contains $I_{N+1}=(a_{N+1},b_{N+1})$ and $J_1,L_1,J_2,L_2,\dots$ is a sequence of open intervals whose union contains $\bigcup_{k=1}^{N} I_k$. Thus $$\sum_{n=1}^{\infty}\ell(U_n)=\sum_{n=1}^{\infty}(\ell(J_n)+\ell(L_n))+\sum_{n=1}^{\infty}\ell(K_n)\geq |\bigcup_{k=1}^{N}I_k|+|I_{N+1}|\overset{\text{ind.hyp.}}{=}\sum_{k=1}^{N}\ell(I_k)+\ell(I_{N+1})=\sum_{k=1}^{N+1}\ell(I_k)$$ and taking the $\inf$ on both sides we have $|\bigcup_{k=1}^{N+1} I_k|\geq\sum_{k=1}^{N+1}\ell(I_k)$, as desired.

Thank you.

答: One way to proceed is to note that the outer measure of a compact set $K$ is equal to the Darboux upper integral of $\chi_K$, $U(\chi_K)$, aka the Jordan upper content of $K$, due to the finite subcover property for compact sets. Let $\varepsilon > 0$. Choose $N$ large enough and closed intervals $J_k \subset I_k$, $1 \leq k \leq N$, such that $\sum_{k = 1}^{N}l(J_k) \geq \sum_{k = 1}^{\infty}l(I_k) - \varepsilon$. Then appeal to the result that the Jordan content of a disjoint union of a finite number closed intervals is the sum of their lengths, which is simple to prove using theory of the Darboux integral. This proof has the advantage that it does not assume beforehand that $|I| = l(I)$.


问:
Integral of the inverse of a "simple" sum
$$\Large \int_{0}^{\infty} \left (\frac{1}{ \sum_{k=0}^{\infty} x^k} \right)~ dx$$

It seems like this converges to $\approx \frac{1}{2}$ but when I try to calculate this I get really stuck, and for no reason!

$$ \sum_{k=0}^{\infty} x^k = \frac{1}{1-x} $$

And so the integral becomes:

$$\large \int_{0}^{\infty} (1-x)~ dx = \text{Does not converge}$$

Even if I use the "regular" geometric progression sum $$ \frac{ 1 \cdot (x^n - 1)}{x-1} $$

Where $n \to \infty$ (number of "elements" in the progression) it seems to actually converge to $0$.

Why these $2$ tries failed? I don't get it... what is the correct way to find the value?

答:
The limit of the partial sums evaluates to different things depending on where in the domain of integration we are

$$\int_0^1 \lim_{n\to\infty}\frac{1-x}{1-x^n}\:dx + \int_1^\infty \lim_{n\to\infty}\frac{1-x}{1-x^n}\:dx = \int_0^1 1-x\:dx + \int_1^\infty 0\:dx = \frac{1}{2}$$


问:
Definite integration includes Exponential and Logarithmic functions
Does anyone have an idea of having solution for the following integration?

$$I=\int_0^\infty x^a e^{-b x} \log(1+cx)dx$$

where $a$ is a positive integer and $b,c>0$.

I found similar integral in eq. (2.6.23.4) in book " Integrals and Series: elementary functions. Vol.1", which is only valid for non-integer $a$.

Further, if I give same integral for known $a$ values such as $a=1,2,3,..$, then MATHEMATICA gives answers one by one. But it did not give for general $a$.

答:
I was able to derive a recursive formula for the integral based purely on b in a simpler case, then expanded to the current problem.

Through IBP,

Where a is any positive integer and b is a positive number. Then, you can solve for the first two cases individually through IBP.

Where  is the Exponential Integral function

Your problem can be derived from this through substitution as the relation:

Where the b/c term replaces the b in the I integral.

I couldn't find an explicit formula for the integral, but the recursion works from what I could tell and try. Let me know if someone finds one or I did something wrong.


问:
Need help with the Laplace Equation.
Hello I'm John and I'm new here. I need help with the fundamental solutions with the Laplace equations.

First Question: Can anyone explain how the chain rule gets Equation #3?

Second Question: Can anyone explain where Equation #4 comes from and compute the second order partial derivatives $f_{xx}$, $f_{xy}$, and $f_{yy}$?

Third Question: Can anyone also explain how Equation #4 equals zero?

答:
Those are differential operators in Equation #3 (meaning gadgets that take a function and differentiate it, returning a function), so to make sense of them, you need to use a test function. With $\eta = x + iy$ and $\xi = x - iy$, for any function $f$,
$$
\def\dd#1#2{\frac{d#1}{d#2}}
\dd{f}{x} = \dd{f}{\eta} \dd{\eta}{x} + \dd{f}{\xi} \dd{\xi}{x}
= \dd{f}{\eta} \dd{}{x}(x + iy) + \dd{f}{\xi} \dd{}{x}(x - iy)
= \dd{f}{\eta} + \dd{f}{\xi}.
$$
Thus, at the level of operators,
$$
\dd{}{x} = \dd{}{\eta} + \dd{f}{\xi}.
$$
The calculation
$$
\dd{}{y} = i \, \biggl( \dd{}{\eta} - \dd{f}{\xi} \biggr)
$$
is quite similar.

For Equation #4, substitute formulas from Equation #3:
$$
\begin{align*}
\dd{}{x} \dd{}{x} + \dd{}{y}\dd{}{y}
&= \biggl( \dd{}{\eta} + \dd{}{\xi} \biggr) \biggl( \dd{}{\eta} + \dd{}{\xi} \biggr)
+ i^2 \, \biggl( \dd{}{\eta} - \dd{}{\xi} \biggr)
\biggl( \dd{}{\eta} - \dd{}{\xi} \biggr) \\
&= \biggl( \dd{}{\eta}\dd{}{\eta} + 2\dd{}{\eta}\dd{}{\xi} + \dd{}{\xi}\dd{}{\xi} \biggr)
- \biggl( \dd{}{\eta}\dd{}{\eta} - 2\dd{}{\eta}\dd{}{\xi} + \dd{}{\xi}\dd{}{\xi} \biggr) \\
&= 4\dd{}{\eta}\dd{}{\xi}.
\end{align*}
$$

Title: Need help with the Laplace Equation.


答:
You have to break it up into two cases: for $x \in (0,1)$, and $x \in (1,\infty)$. Then you have that

$$\int_0^\infty \frac{1}{\sum_{n=0}^\infty x^n} \, dx = \int_0^1 \frac{1}{\sum_{n=0}^\infty x^n} \, dx + \int_1^\infty \frac{1}{\sum_{n=0}^\infty x^n} \, dx$$

For the first one, we can use your approach to get that

$$\int_0^1 \frac{1}{\sum_{n=0}^\infty x^n} dx = \int_0^1 \frac{1}{1/(1-x)} dx = \int_0^1 (1-x) dx = x - \frac{x^2}{2} \bigg|_{x=0}^1 = \frac 1 2$$

This holds because (whenever $|x|<1$: note that it doesn't hold if $|x| \ge 1$!), then

$$\sum_{n=0}^\infty x^n = \frac{1}{1-x}$$

If $x > 1$, however, then $\sum_{n=0}^N x^n \to \infty \implies 1 / \sum_{n=0}^N x^n \to 0$, and so

$$\int_1^\infty \frac{1}{\sum_{n=0}^Nx^n} \, dx \stackrel{N \to \infty}{\longrightarrow} 0$$

though I think being able to do this limiting process requires additional justification (e.g. dominated convergence, perhaps?). But either way with that in mind, then we should be able to see that (if just intuitively), that the second integral should be $0$, and, overall, you get

$$\int_0^\infty \frac{1}{\sum_{n=0}^\infty x^n} \, dx = \frac 1 2$$

Title: Integral of the inverse of a "simple" sum

Body:
$$\Large \int_{0}^{\infty} \left (\frac{1}{ \sum_{k=0}^{\infty} x^k} \right)~ dx$$

It seems like this converges to $\approx \frac{1}{2}$ but when I try to calculate this I get really stuck, and for no reason!

$$ \sum_{k=0}^{\infty} x^k = \frac{1}{1-x} $$

And so the integral becomes:

$$\large \int_{0}^{\infty} (1-x)~ dx = \text{Does not converge}$$

Even if I use the "regular" geometric progression sum $$ \frac{ 1 \cdot (x^n - 1)}{x-1} $$

Where $n \to \infty$ (number of "elements" in the progression) it seems to actually converge to $0$.

Why these $2$ tries failed? I don't get it... what is the correct way to find the value?


答:
You want to solve

$${8 \choose k}x^{8-k}\left(\frac{1}{ax^2}\right)^k=7x^2$$

This can be expressed in the form

$${8 \choose k}\cdot\frac{1}{a^k}x^{8-3k}=7x^2$$

Solving $8-3k=2$ gives $k=2$. Solving $\displaystyle{8\choose2}\cdot\dfrac{1}{a^2}=7x^2$ gives $a^2=4$.

However, $-2$ turns out to be extraneous, so $2$ is the only solution.

Title: Finding the coefficient of $x^2$ in an expansion

Body:
I was given this question as a practise assignment and I am unsure of my answer.

The coefficient of $x^2$ in the expansion of $(x+\frac{1}{ax})^8$ is 7. Find the possible value of $a$.

I did $(x+\frac{1}{ax})^8$ = $(x (1+\frac{1}{ax^2}))^8$

Given my answer, does that mean that $a=7$?

Edit: Thank you for everyone's help! I think I understand it a little bit now. I will have to review this question again and practise more.


问:
$(V,\omega)$ with $W,W'$ isotropic, then there exists $\phi \in Aut(V)$ such that $\phi(W)=W'$ iff $\dim(W)=\dim(W')$
Let $(V,\omega)$ be a symplectic vector space. Let $W,W'$ be two isotropic subspaces. I want to see that there is symplectomorphism $\phi:V\rightarrow V$ such that $\phi(W)=W'$ if and only if $\dim(W)=\dim(W')$.

I know that every isotropic subspace is contained in a lagrangian subspace. I also know that if $L\subset V$ is a lagranian subspace, there is a symplectomorphism $W\times W^*\rightarrow V$

We have a nice result about lagrangian subspaces:

Let $L\subset V$ be a Lagrangian subspace. Consider $(L\times L^*,\Omega)$. Then there is a symplectomorphism $\Psi:L\times L^*\rightarrow V$ such that $\Psi(L\times\{0\})=L$.

It seems the idea is to show that for $W\subset L$ we have $\Psi(W\times\{0\})=W$. For $W'\subset L'$ we have $\Psi'(W'\times\{0\})=W'$.

Then by passing through $\mathbb R^{2n}$ we can get a symplectomorphism $L\times L^* \cong \mathbb R^{2n} \cong L'\times L'^*$.

I don't see why we have $\Psi(W\times\{0\})=W$ or why the symplectomorphism $L\times L^* \cong \mathbb R^{2n} \cong L'\times L'^*$ will send $W$ to $W'$

答:
Let $k=\dim W=\dim W'\leq n$ where $\dim V=2n$.

You may use this fact

Symplectic basis $(A_i,B_i)$ such that $S= $ span$(A_1,...,A_k)$ for some $k$ when $S$ is isotropic

to give bases $A_1,...,A_k$ and $A_1',...,A_k'$ of $W$ and $W'$ respectively, where $A_1,...,A_n,B_1,...,B_n$ and $A'_1,...,A'_n,B'_1,...,B'_n$ are two symplectic bases of $V$, which therefore differ by a symplectic transformation.


答:
We are trying to prove that the series, $$\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$$
converges.

Let us use the Limit Comparison Test. That is, if $a_n>0$ and $b_n>0$, then if $$\lim_{n\to \infty}\frac{a_n}{b_n}=L$$ for some real number $L$. Then, if $b_n$ converges, by the limit comparison test, $a_n$ must also converge.

Let $a_n=\frac{n^2-n-1}{n^4+n^2+1}$ and we say $b_n=\frac{1}{n^2}$. Then, $$\lim_{n\to \infty} \frac{\frac{n^2-n-1}{n^4+n^2+1}}{\frac{1}{n^2}}\\
\lim_{n\to \infty}\left(\frac{(n^2-n-1)(n^2)}{n^4+n^2+1}\right)\\
\lim_{n\to \infty}\frac{n^4-n^3-n^2}{n^4+n^2+1}\cdot \frac{\frac{1}{n^4}}{\frac{1}{n^4}}\\
\lim_{n\to \infty} \left(\frac{{1-\frac{1}{n}}-\frac{1}{n^2}}{1+\frac{1}{n^2}+\frac{1}{n^4}}\right)\\
\lim_{n\to \infty}\frac{1-0-0}{1+0+0}=\frac{1}{1}=1$$

Hence, we have shown that the $\lim_{n\to \infty}\frac{a_n}{b_n}=L$ converges because it equals some constant; in this case, $1$.

Now, because $\sum_{n=1}^{\infty}\frac{1}{n^2}$ converges by the p-test, (since $p=2>1$), we can conclude that the series, $$\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$$ must also converge by the limit comparison test.

Title: Sequence queue - converges or diverges?

Body:
I have the next sequence queue: $\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$. Does the queue converges or diverge?

My attempt:
I have tried to show that $\frac{1}{n^2}>|\frac{n^2-n-1}{n^4+n^2+1}|$ for any $n>0$, and then by the comparation test we get that since $\sum_{n=1}^{\infty}{\frac{1}{n^2}}$ converges, we have that $\sum_{n=1}^{\infty}{|\frac{n^2-n-1}{n^4+n^2+1}|}$ converges too, and by a theorem we have that since $\sum_{n=1}^{\infty}{|\frac{n^2-n-1}{n^4+n^2+1}|}$ $\implies$ $\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$ converges. Is that right?


问:
Specific branch of complex logarithm
Find a branch of $\log(3z-2)$ which is analytic at $z=0$ and takes the value $\ln 2+\pi i$ there. Specify the domain on which this branch is analytic.

These are my thoughts. I know that $\log z$ is multivalued function and $\log(3z-2)=\log|3z-2|+i(\text{Arg} z+2\pi n)$, where $n\in \mathbb{Z}$.

Can anyone show the solution, please?

答: For each $z\in\Bbb C\setminus[0,\infty)$, let $\operatorname{Log}(z)=\log|z|+\operatorname{arg}(z)i$, where $\operatorname{arg}(z)$ is the only argument of $z$ such that $\operatorname{arg}(z)\in(0,2\pi)$. Then $\operatorname{Log}$ is analytic. Now, let $l(z)=\operatorname{Log}(3z-2)$. Then $l$ is a branch of $\log(3z-2)$ and $l(0)=\operatorname{Log}(-2)=\log2+\pi i$.

Title: Specific branch of complex logarithm


问:
Why there's a sin and cos here? Shouldn't the obvious guess be $Ae^{-x}$?
image shows the solution for the differential equation y double prime minus 4 y prime plus 5 y equal to e powered to the minus x

I solved this $y'' - 4y' + 5y = e^{-x}$ equation with the guess of:

$Ae^{-x}$, such that A ended up being $A = 1/2$;

The solution given by math calculator websites was the one of the image —

$$y(x)=(C_1 \sin(x)+C_2 \cos(x))e^{2x}+\frac{e^{-x}}{10},$$

but I can't understand how could I ever get to that solution on my own.
Can someone explain the logic to me?

答:
The auxiliary equation of the given ODE is,

$m^2-4m+5 = 0 \Rightarrow (m-2)^2 +1 = 0 \Rightarrow \boxed{m = 2 \pm i}$

So the solution (complementary function) will be,

$$y_{CF}(x) = k_1 e^{(2 + i)x} + k_2 e^{(2 - i)x} =e^{2x}(k_1 e^{ix} + k_2 e^{-ix}) \equiv \boxed{e^{2x}(C_1 \sin x + C_2\cos x)}$$

Now you have to add this with the particular solution to get the complete solution.


问:
Multiplicity of 0 eigenvalue of directed graph Laplacian matrix
I am looking for a result (if it exists) for directed graphs relating the multiplicity of 0 eigenvalues of the directed Laplacian matrix.

Consider a directed graph $\mathcal{G}=(\mathcal{V},\mathcal{E})$ and define the in-degree Laplacian as $L_{in}(\mathcal{G}) = \Delta_{in}(\mathcal{G}) - A_{in}(\mathcal{G})$, where $\Delta_{in}(\mathcal{G})$ is the diagonal in-degree matrix, and $A_{in}(\mathcal{G})$ is the adjacency matrix.

A well-known result states that $L_{in}(\mathcal{G})$ has rank $|\mathcal{V}|-1$ if and only if $\mathcal{G}$ contains a rooted out-tree.

For undirected graphs, is is also known that the rank of $L(\mathcal{G})$ is $|\mathcal{V}|-c$ where $c$ is the number of connected components.

Is there a similar result for directed graphs?  It is straight-forward to show, for example, that $L_{in}(\mathcal{G})$ loses rank for every node with in-degree 0 that is, if there are $p$ nodes with in-degree 0, then $\mbox{rk}[L_{in}(\mathcal{G})] \leq n-p$.  However, I am not sure if this is in fact an equality.

I hope this is clear.  Thanks!

答:
I'm a few years late but I think tst conjecture is correct, according to https://arxiv.org/pdf/2002.02605.pdf (which also cites other references).

They use the notion of Reach :

Definition 2.3 :

i)  Let $i\in V$.  The reachable  set $R(i)$ consists  of  all $j\in V$ such that there exists a path from $i$ to $j$

ii) A reach $R$ is a maximal reachable set

so basically the number of reach is the number of (oriented) trees needed to cover the graph (notice that the number of reaches might not be the number of strongly connected components).

The paper then claim

Theorem 4.6:

Given a digraph $G$.  The algebraic and geometric multiplicity of  the  eigenvalue  $0$  of $L$ equals  the  number  of  reaches.

The paper does contain a proof (which is actually quite nice) and detailled examples

ContentLicense: CC BY-SA 3.0


问:
Number of strongly connected components of a digraph and the Laplacian
The multiplicity of the zero eigenvalue of the Laplacian matrix of an undirected graph is equal to the number of its connected components. I am wondering if a similar result holds for directed graphs? In other words, is the number of strongly connected components also equal to the multiplicity of the zero eigenvalue of the digraph Laplacian? I only found a couple of papers that discuss the spectral properties of digraphs based on their Laplacians (e.g., Chung 2005), but I see no mention of such result.

Chung, F. (2005). Laplacians and the Cheeger inequality for directed graphs. Annals of Combinatorics, 9(1), 1-19.

答:
Your question is answered Multiplicity of 0 eigenvalue of directed graph Laplacian matrix , I'm just rewriting it there.

A nice reference seems to be https://arxiv.org/pdf/2002.02605.pdf

They use the notion of Reach :

Definition 2.3 :

i)  Let $i\in V$.  The reachable  set $R(i)$ consists  of  all $j\in V$ such that there exists a path from $i$ to $j$

ii) A reach $R$ is a maximal reachable set

so basically the number of reach is the number of (oriented) trees needed to cover the graph (notice that the number of reaches might not be the number of strongly connected components).

The paper then claim

Theorem 4.6:

Given a digraph $G$.  The algebraic and geometric multiplicity of  the  eigenvalue  $0$  of $L$ equals  the  number  of  reaches.

The paper does contain a proof (which is actually quite nice) and detailed examples


问:
Counter-example to the Krull intersection theorem
In this overflow post some counter-examples to the Krull intersection theorem are given. I'm interested in the counter-example given by Dustin Cartwright. He takes $$R = \mathbb{Q}[x,z,y_1, y_2, \dots]/ (x - y_1 z, x - y_2 z^2, x - y_3z^3, ...)$$ and $I = (\overline{z})$. Then he claims that $\cap_{j \in \mathbb{N}} I^j = (\overline{x})$, but $(\overline{z})(\overline{x}) \neq (\overline{x})$, where $\overline{x}$ and $\overline{z}$ are the images of $x$ and $z$ in the quotient.

I would like to know how to prove both of these last statements. The inclusion $\cap_{j \in \mathbb{N}} I^j \supseteq (\overline{x})$ is clear, but I don't see how to get the reverse inclusion. I tried writing letting $f \in \mathbb{Q}[x,z, y_1, y_2\dots]$ be a polynomial such that $\overline{f} \in \cap_{j \in \mathbb{N}} I^j$ and using the fact that $\overline{f y_j} \in (\overline{x})$ for every $j \in \mathbb{N}$, but I didn't manage to get anything from this. Similarly, I couldn't sort out why $(\overline{z})(\overline{x}) \neq (\overline{x})$ is true.

答:
Let $A=\mathbb{Q}[x,z,y_1,y_2,\dots]$ and consider $J:=(x-y_1z,x-y_2z^2,x-y_3z^3,\dots)$ as an ideal of $A$. Pulling back along the projection map $A\to R$ turns the desired fact about the ideal $I<R$ into a statement about the ideal $(J+Az)<A$; namely, what you wish to show for your first question is that $$\bigcap_{k\in\mathbb{N}}\left[J+(J+Az)^k\right]\subseteq J+Ax.$$ (Personally, I think things become clearer in this reframing, since $A$ is a polynomial ring over a field and hence must more intuitive and familiar than the ring $R$. I will thus prove the statements in the context of $A$ rather than $R$.) Now, first note that $$J+Ax=(x,y_1z,y_2z^2,y_3z^3,\dots)$$ and that $J+Az=(x,z)$, and thus that $$(J+Az)^k=(x,z)^k=(x^k,x^{k-1}z,\dots,xz^{k-1},z^k)$$ for each $k\in\mathbb{N}$. In particular, $J+(J+Az)^k=(x,y_1z,\dots,y_{k-1}z^{k-1},z^k)$. Now suppose that $f$ lies in the intersection on the left hand side. To show $f\in J+Ax$, it suffices to show that every monomial term of $f$ is either divisible by $x$ or divisible by $y_nz^n$ for some $n\in\mathbb{N}$. Suppose therefore for contradiction that this is not the case; then there is a monomial term in $f$ of form $y_{m_1}\dots y_{m_l}z^n$, where $m_i>n$ for each $i$, the $m_i$ are not necessarily distinct, and one of $n$ and $l$ may be zero. But this means that $f\notin\left[J+(J+Az)^{n+1}\right]$, since every generator of that ideal is a monomial and either divisible by $x$, divisible by $y_m$ for some $m\leqslant n$, or divisible by $z^{n+1}$. Thus $f$ cannot lie in the intersection on the left, and this gives the desired contradiction.

Now let's pull back the second statement about the ideals $(\overline{x}),(\overline{z})<R$ into a statement about the ideals $J+Ax,J+Az<A$. What you wish to show is that $$J+(J+Ax)(J+Az)\neq J+Ax.$$ It suffices to show that $x\notin J+(J+Ax)(J+Az)$, so recall that $J+Az=(x,z)$ and that $J+Ax=(x,y_1z,y_2z^2,y_3z^3,\dots)$. In particular, we have
\begin{align}
(J+Ax)(J+Az)=(x^2,xy_1z,xy_2z^2,\dots xz,y_1z^2,y_2z^3,\dots).
\end{align} In particular, every element of $(J+Ax)(J+Az)$ has degree at least $2$. Also, every monomial term appearing in each generator of $J$ has degree at least $1$. Thus, if some $A$-linear combination of the generators of $J$ and the generators of $(J+Ax)(J+Az)$ were equal to $x$, the coefficient of at least one of the generators of $J$ would have to have a non-zero constant term from $\mathbb{Q}$. Suppose for contradiction that there is such an $A$-linear combination, and let $x-y_{n}z^{n}$ be one of the generators of $J$ whose coefficient from $A$ has a non-zero constant term; call this constant term $\lambda\in\mathbb{Q}^\times$.

Note that $\lambda y_nz^n$ cannot be a monomial term in any element of $(J+Ax)(J+Az)$, since every generator of that ideal is a monomial and is either divisible by $x$, divisible by some $y_m$ where $m\neq n$, or divisible by $z^{n+1}$. On the other hand, if $J'$ is the ideal generated by all of the generators of $J$ except $x-y_nz^n$, then no element of $J'$ can have a term $\lambda y_nz^n$ either, since every monomial term of every generator of $J'$ is either divisible by $x$ or divisible by some $y_m$ where $m\neq n$. Thus $\lambda y_nz^n$ must appear as a monomial term in any $A$-linear combination of the generators of $J$ and $(J+Ax)(J+Az)$ for which the $A$-coefficient of $x-y_nz^n$ has constant term $\lambda$. In particular, no such $A$-linear combination can be equal to $x$ if $\lambda\neq 0$, thus giving the desired contradiction.


问:
Maximum of $\sum_{n=1}^\infty \sin(n) \cdot (x_n - x_{n+1})$
I want to find supremum of |L| where $$L = \sum_{n=1}^\infty sin(n)\cdot (x_n - x_{n+1})$$
where $x_n$ is such a sequence that $\sum_{n=1}^\infty |x_n| \le 1$.

My work so far

Let's expand expression of $|L|$ to get more knowledge about possible candidates for maximum.

$$|L| = |\sum_{n=1}^\infty sin(n)\cdot (x_n - x_{n+1})|\le\sum_{n=1}^\infty |sin(n)\cdot (x_n - x_{n+1})|\le$$
$$\le\sum_{n=1}^\infty |(x_n - x_{n+1})|$$

And here I'm not sure what to do next. I'm sure I have to somehow use that $\sum_{n=1}^\infty|x_n| \le 1$ but I don't see how it can be forced in expression $\sum_{n=1}^\infty|x_n-x_{n+1}|$. Could you give me a hint how to bound this expression to obtain maximum ?

EDIT

Due to the hints mentioned in comment section I got that:

$$|L| \le \sum_{n=1}^\infty|x_n - x_{n+1}| \le \sum_{n=1}^\infty |x_n| + |x_{n+1}| \le 2$$

So the candidate for my supremum is $2$. But to prove it I have to point exactly the sequence $x_n$ such that $\sum_{n=1}^\infty |x_n| \le 1$ and $|L| = 2$. Am I correct ? If so, could you please also give a hint how to choose this sequence ? I tried to choose $\frac{1}{2^n}$ but value of L is far away from $2$

答:
First, we solve the following problem

The supremum of $$P = \left|\sum_{n=1}^{+\infty}a_n(x_n-x_{n+1})\right|$$ given
$$\sum_{n=1}^{+\infty}|x_n| \le 1$$ and $|a_n| \le 1 $ for all $n\in
 \Bbb N^*$

is  $$\sup_{k\in \Bbb N^*}|a_k-a_{k-1}|$$

Proof: We have

\begin{align}
P &= \left|\sum_{n=1}^{+\infty}a_n(x_n-x_{n+1})\right| \\
&= \left|\sum_{n=1}^{+\infty}a_nx_n-\sum_{n=2}^{+\infty}a_{n-1}x_n\right| \\
&= \left|a_1x_1+\sum_{n=2}^{+\infty}(a_n-a_{n-1})x_n\right| \le |a_1| |x_1|+\sum_{n=2}^{+\infty}(|a_n-a_{n-1}| |x_n|)
\end{align}
For the sake of simplicity, denote $a_0 = 0$, then

$$P \le \sum_{n=1}^{+\infty}(|a_n-a_{n-1}| |x_n|)$$

By definition of supremum, $|a_n-a_{n-1}| \le \sup_{k\in \Bbb N^*}|a_k-a_{k-1}|$ for all $n\in \Bbb N^*$.
Then

\begin{align}
P \le \sum_{n=1}^{+\infty}(|a_n-a_{n-1}| |x_n|) \le \sum_{n=1}^{+\infty}(\sup_{k\in \Bbb N^*}|a_k-a_{k-1}| |x_n|) &= \sup_{k\in \Bbb N^*}|a_k-a_{k-1}|\sum_{n=1}^{+\infty}|x_n| \\
&\le \sup_{k\in \Bbb N^*}|a_k-a_{k-1}|
\end{align}

$$$$

Second, it suffice to find the supremum $|\sin(n) - \sin(n-1)|$ for $n\in \Bbb N^*$

We have
$$|\sin(n) - \sin(n-1)| = 2\sin \left( \frac{1}{2} \right) \left|\cos \left( n-\frac{1}{2} \right) \right|$$

We notice that $\sup_{n\in \Bbb N^*} \left|\cos \left( n-\frac{1}{2} \right) \right| = 1$ because there always exists a sequence $(u_n,v_n) \in \Bbb N^* \times \mathbb{N}^*$ such that $$\frac{u_n-\frac{1}{2}}{v_n} \to \pi$$

or
$$ \left|\cos \left( u_n-\frac{1}{2} \right) \right| \to 1$$

Then,
$$ \sup_{n\in \Bbb N^*}|\sin(n) - \sin(n-1)| \le  2\sin \left( \frac{1}{2} \right)$$

Third, we prove that the supremum of $|L|$ is $2\sin \left( \frac{1}{2} \right)$

For each value $u_k$ of the sequence $\{ u_k\}_{k\in \Bbb N^*}$ (constructed in the second step), let's construct the sequence $\{x_n\}_{n\in \Bbb N^*}$ as follows: $x_{u_n} = 1$ and $x_k = 0$ for $k\ne u_n$. Hence for each construction of the sequence $\{x_n\}_{n\in \Bbb N^*}$, the value of $L$ (we denote $L(u_k)$) is

$$L(u_k) = |\sin(u_k) - \sin (u_k-1) | = 2\sin \left( \frac{1}{2} \right) \left|\cos \left( u_k-\frac{1}{2} \right) \right|  \to 2\sin \left( \frac{1}{2} \right) $$

Hence, this supremum can be atteined.

Conclusion:

$$ \sup |L| = 2\sin \left( \frac{1}{2} \right)$$


问:
Modified sum of three squares problem
My teacher proposed this problem some time ago:

We have four different integers $x,y,z,k$ such that $x^2+y^2+z^2=3k^2$. Prove that the difference between the largest one of $x,y,z$ and the smallest one of $x,y,z$ is greater than $\sqrt{k\,}$.

The time for submissions is over so I thought I might ask here. I'm not sure it's ok with the rules here, especially that I haven't made any progress. I will still share what I've tried:

$k$ is a quadratic mean, however I could not find any mean bigger than it;

sub $y = x + t$ and $z = x+v$ doesn't seem to help show $t > \sqrt{k\,}$ or $v > \sqrt{k\,}$

geometric thinking of a sphere equation and looking for integer points is even more complicated

答:
First of all let (WLOG) $x>y>z$.

Let's say that we may assume $gcd(x,y,z)=1$. Otherwise if $x=dX , y=dY , z=dZ , k=dK$ then we'll have $X^2+Y^2+Z^2=3K^2$. Now it is sufficient to prove the claim for the latter equation since :

$$X-Z\geqslant\sqrt{K} \iff dX-dZ\geqslant d\sqrt{K}\geqslant\sqrt{dK} \iff x-z\geqslant\sqrt{k}$$
Now since $gcd(x,y,z)=1$ and $3|x^2+y^2+z^2$ , one can deduce that $3\not|xyz$.

Now rewrite the equation as follows:

$$(x+y+z)^2+(x-y)^2+(y-z)^2+(x-z)^2=9k^2$$
Let $x-z=a$ , $y-z=b$ , $x+y+z=t$ $\implies$ $x-y=a-b$ and $t=3z+(a+b)$

And the equation will now be:
$$a^2+b^2+(a-b)^2=(3k)^2-t^2=(3k-t)(3k+t)$$
Let $3k+t=\alpha$ and $3k-t=\beta$

Now note that since $x>y>z$, we'll have $a>b$.

Then one can see that $\alpha\beta=(a-b)^2+a^2+b^2<2a^2$. On the other hand clearly since $\alpha,\beta$ are positive integers, $\alpha\beta\geqslant\alpha+\beta-1$. Therefore, $2a^2\geqslant\alpha\beta\geqslant\alpha+\beta-1$. Finally:
$$k=\frac{\alpha+\beta}{6}<\frac{2a^2+1}{6} \iff \sqrt{3k-\frac12}<a=x-z=\max\{x,y,z\}-\min\{x,y,z\}$$

Which is a better bound.

Title: Modified sum of three squares problem


问:
show that $f_n$ converges to $f \in L_1$ in the weak topology
Problem: Let $f_n$ be a sequence of functions in $L_1[0,1]$ such that there exists a positive $g \in L_1$ such that $\lambda(|f_n| > R) \leq \lambda (|g| > R)$ for all $R$ (where $\lambda$ is the Lebesgue measure) and
$$\phi(h) = \lim_n \int f_n(t)h(t) \mathrm dt$$ exists for all bounded measurable functions.

Show that there exists an $f \in L_1$ such that $f_n$ converges to $f$ in the weak topology.

Attempt: We know that $\lambda$ is a probability measure on $[0,1]$. The goal (I think) is to show that
$$\int f_n h \to \int f h$$
for some $f \in L_1$ and all $h \in L_\infty$ as $n \to \infty$. I am not sure how to proceed.

答:
In the title weak* should be changed to weak.

We have $\int |f_n|d\lambda=\int_0^{\infty} \lambda (|f_n| >t)dt \leq \int_0^{\infty} \lambda (|g| >t)dt=\int |g|d\lambda$ so $(f_n)$ is bounded in $L^{1}$.

Let $\mu (A) =\lim \int_A f_n d\lambda$. Then $\mu$ is real measure which is absolutely continuous w.r.t. $\lambda$. Let $f$ be its Radon Nikodym derivative. Then $\int_A f_n \to \int_A f$ for all $A$. Can you now complete the proof using the fact that any $h \in L^{\infty}[0,1]$ is a limit of simple functions in $L^{\infty}[0,1]$ norm?


问:
Counting strings of $\{ 0,1 \}$'s of length $n$ s.t. last $k$ bits are equal
I'm trying to derive a closed formula for the number of strings of $0$'s and $1$'s of length $n$ such that the last $k \leq n$ bits are all zeros or all ones, and such that there is no other place in the string with $k$ consecutive zeros or ones. </p>

<p>I have tried constructing a recursive formula by "reasoning backward", but the formula is still a mystery.  </p>

答:
The question is equivalent to asking how many strings of length $n-k$ do not contain any string of $k$ consecutive zeros or ones. Half of these will end in $0$ and require appending $1^k$ to make the desired string of length $n$, and the other half will end in $1$ and require appending $0^k$. Thus, it suffices to find $f(m,k)$, the number of strings of length $m$ that do not contain any string of $k$ consecutive zeros or ones.

Such a string can be thought of as a composition of $m$ into parts of size less than $k$, the parts being the blocks of consecutive zeros or consecutive ones. Each composition is represented twice, however, once with a string starting with a block of zeros and once with a string starting with a block of ones. E.g., the composition $3+1+2$ of $6$ is represented by $000100$ and by $111011$. Thus, $f(m,k)$ is the twice the number of compositions of $m$ into parts of size less than $k$.

If define the $\ell$-nacci numbers $a_i^{(\ell)}$ by $a_0^{(\ell)}=a_1^{(\ell)}=\ldots=a_{\ell-2}^{(\ell)}0$, $a_{\ell-1}^{(\ell)}=1$, and

$$a_i^{(\ell)}=a_{i-1}^{(\ell)}+\ldots+a_{i-\ell}^{(\ell)}\tag{1}$$

for $i\ge\ell$, it turns out that $f(m,k)=2a_{m+k-2}^{(k-1)}$, which is easily computed from the recurrence $(1)$.

When $k=2$, for instance, $a_i^{(1)}=1$ for all $i$, and $f(m,2)=2$: the only acceptable sequences are the two alternating sequences of length $m$. When $k=3$, $a_i^{(2)}=F_i$, the $i$-th Fibonacci number, and $f(m,3)=2F_{m+1}$; in this case Binet’s formula gives a closed form, and OEIS A000073 has a really ugly closed form for the $k=4$ case, but for $k>4$ there’s nothing much better than generating functions.

For more information and references see OEIS A048887 and its references; the specific case $k=5$ is OEIS A000078.


问:
Does every every non-empty family of non-empty sets have a choice function whose fibers are clutters (sperner families)?
We call any map $f$ a choice function of the family of sets $\mathcal{K}$ iff $f$ is a map from $\mathcal{K}$ to $(\cup_{S\in\mathcal{K}}S)$ such that $\forall X\in\mathcal{K}(f(X)\in X)$ likewise we call $f$ an anti-choice function of $\mathcal{K}$ iff $f$ is a choice function of $\mathcal{K}$ such that $\small\forall X,Y\in\mathcal{K}(X\subset Y\rightarrow f(X)\neq f(Y))\iff \forall X\in\mathcal{K}(A,B\in f^{-1}[\{f(X)\}]\rightarrow A\not\subset B)$.

Now if $\forall X,Y\in\mathcal{K}(X\not\subset Y)$ then all choice functions of $\mathcal{K}$ are 'anti-choice' functions. But what about other families? Does AC imply all families of non-empty sets have anti-choice functions?

答:
The concept is entirely inconsistent.

Take $\{\{0\},\{1\},\{0,1\}\}$ to be a family of non-empty sets. Then any choice function must map $\{0\}$ to $0$ and $\{1\}$ to $1$. But now $\{0,1\}$ cannot be mapped to either of its elements.

For an example with infinite sets, consider all the initial segments of $\Bbb Q$ except $\varnothing$. There are uncountably many of them (each one defining a unique $r\in\Bbb R$, and $\Bbb Q$ itself), so by cardinality reasons alone uncountably many must have chosen the same element. And since the collection as a whole is a chain, "anti-choice" is necessarily injective, which is impossible.


问:
Equation describing an elliptical orbit using time and angle
I'm working on a project where I'm trying to describe the orbits of the planets in the solar system using the polar equation of an ellipse.

Below is an equation I got from here.
$$
r = \frac{b^{2}}{a - c \cos \theta}
$$

Where
$a$ is the semi-major axis,
$b$ is the semi-minor axis,
$c$ is the distance between the center of the orbit and a focus point of the orbit $(c^2 = a^2 - b^2)$. The sun is at one of these foci, so $c$ is the distance between the center of the orbit/ellipse and the sun, the origin of the graph. This form of the equation also assumes no displacement of the ellipse in the $y$ direction from the origin.

This equation does relate the radius $r$ of a point on the ellipse as a function of the angle $\theta$ it makes with the origin.

However, I am trying to look for an equation that models the angle $\theta$ as a function of time $t$. For example, if $T$ was the period of one full orbit, then after $T$ seconds, the change in angle should be $2\pi$ radians. In other words, where if $f(t) = \theta$, $f(T) = 2\pi$ or $0$.

Considering that I have the values for $a$, $b$, and $c$, the orbital period $T$, and the linear speed $v$ of each planet, and since I cannot use the radius directly (because it is not constant across the ellipse), how could I derive the polar angle $\theta$ as a function of time $t$ without $r$?

答:
You need to use some physics here, in particular, the conservation of angular momentum. $$L = \overrightarrow r \times \overrightarrow p$$
where $\overrightarrow r$ is the position vector and $\overrightarrow p$ is the momentum, or mass times velocity.  Essentially, you can't bring time into the problem without describing how the particle moves.

To shortcut some of that, you could use Kepler's observation that the planet sweeps out equal areas in equal times.  That could give you a relationship between $d\theta$ and $dt$.


答:
okay lets solve it, first:
$$y''-4y'+5y=0$$
lets make an educated guess that the solutions will be of the form:
$$y=Ae^{\lambda x}$$
now sub in:
$$Ae^{\lambda x}(\lambda^2-4\lambda+5)=0$$
solving for $\lambda$ the non-trivial solutions are going to be in the quadratic:
$$\lambda^2-4\lambda+5=0$$
$$\lambda=2\pm i$$
now we add these two solutions together:
$$y=Ae^{(2+i)x}+Be^{(2-i)x}=e^{2x}(Ae^{ix}+Be^{-ix})$$
now notice that this will just form sines and cosines so lets redefine our constants and we arrive at:
$$y=e^{2x}\left(C_1\sin x+C_2 \cos x\right)$$
This part forms the "complementary function" so now you just need to find the particular solution. Hope this helps

Title: Why there's a sin and cos here? Shouldn't the obvious guess be $Ae^{-x}$?

Body:
image shows the solution for the differential equation y double prime minus 4 y prime plus 5 y equal to e powered to the minus x

I solved this $y'' - 4y' + 5y = e^{-x}$ equation with the guess of:

$Ae^{-x}$, such that A ended up being $A = 1/2$;

The solution given by math calculator websites was the one of the image —

$$y(x)=(C_1 \sin(x)+C_2 \cos(x))e^{2x}+\frac{e^{-x}}{10},$$

but I can't understand how could I ever get to that solution on my own.
Can someone explain the logic to me?


问:
Diophantine equation : $N= \frac{x^2+y}{x+y^2}$
I am looking for information about the following diophantine equation :

$N = \displaystyle\frac{x^2+y}{x+y^2}$

Has it been studied ?
Is there any efficient algorithm to solve it?

Any links?

I have tried to solve it by myself this week-end, but haven't made any progress ...

Thanks in advance

Philippe

P.S:

My first post. Sorry for being unclear.

Does this equation have solutions in integers x,y for all integer N > 0 ?

答:
Follow user9325's suggestion about completing the squares, and then (look up and) apply the theory of Pell equations.

Edit: OK, I guess you didn't get anything out of user9325's suggestion, so I'll take it up for you.

$N=(x^2+y)/(x+y^2)$, $Ny^2-y=x^2-Nx$, $U^2-NV^2=1-N^3$ where $U=2Ny-1$, $V=2x-N$. This has the solution $U=-1$, $V=\pm N$. The solution $U=-1$, $V=N$ corresponds to $x=N$, $y=0$, which already shows that there's a solution for each $N$, but maybe that's too trivial. Then take any solution to $a^2-Nb^2=1$ and you get another solution, $U=-a\pm bN^2$, $V=-b\pm aN$. Now $a^2-Nb^2=1$ has lots of solutions - that's the Pellian I alluded to. For $y$ to be an integer, you need $a\equiv 1\pmod N$, so you have to study enough of the theory to see if that can be made to happen.

If $N$ is a square, say, $N=m^2$, then the Pellian doesn't apply, but you have something simpler; $(U+mV)(U-mV)=1-m^6$. Now you'll get at most finitely many solutions, since there are only finitely many ways to factor $1-m^6$. Here's one example; take $N=4=2^2$ so $m=2$ and $1-m^6=-63$; take $U+2V=63$, $U-2V=-1$ to get $U=31$, $V=16$; then $x=10$, $y=4$.


问:
Proof of Zero density of the Riemann Zeta function
I am currently trying to understand the proof of the simplest zero-density theorem for $\zeta$ there is, namely
$$N(T+1)-N(T-1) \ll \log(T),$$
where $T>2$ and  $N(T) = \#\{\rho \in \mathcal N : |\Im(\rho)| < T \}$ with $\mathcal N$ the set of non-trivial zeroes of $\zeta$. I consulted three books on analytic number theory so far (by Brüdern, Vaugan and Tenenbaum) and in all three of those the proof is (more or less) the same:
We use Jensen's Formula to obtain
$$\int_0^1 \log | \zeta(2+iT+re(\theta))| d\theta = \log|\zeta(2+iT)| + \sum_{\rho \in \mathcal N \text{ and } |\rho - (2+iT)| < r} \log \left(\frac r {|\rho-(2+iT)|} \right)$$
for some $r \in [3,4]$ and $e(\theta) = \text{e}^{2\pi i \theta}$. In this formula, it is relatively easy to see that the sum on the right hand side is $\approx N(T+1)-N(T-1)$. So far so good.  It is now stated that a bound of the form $$\zeta(\sigma + it) \ll t^k$$ for $t>1$, $\sigma \in \mathbb R$ and some $k$ of fitting size implies the bound $$\log|\zeta(s)| \ll \log |t|.$$ Tenenbaum even explicitly mentions the bound
$$\log |\xi(s)| \ll |s|\log|s| \text{ (as |s| }\rightarrow \infty \text{ for }\Re s \geq \tfrac 12) $$
for the complete Riemann-Zeta function in his book Introduction to analytic and probabilistic number theory as Formula (41) on page 152. It is clear that these bounds imply the desired result.

I could agree with the authors if they stated (and used) these bounds as $\log |\zeta(s)| \leq O(\log |s|)$, but in general these bounds are clearly wrong (as long as I've not completely lost it)! $\zeta$ (and $\xi$) have zeroes with arbitrarily large imaginary part, and $\log |\zeta|$ ($\log |\xi|$ resp.) has to be large near those.

Am I missing something? It might well be that I am misinterpreting some notations here, it's hard to believe that all three (or even one) of these authors would have published this without thinking about my issue.

答:
Copying Conrad's comment so that more people see the method:

For $T$ large and $r=3$

Jensen's formula gives $$\int_0^1 \log | \zeta(2+iT+re^{2i\pi \theta})| d\theta = \log|\zeta(2+iT)| + \sum_{\zeta(\rho)=0, |\rho - (2+iT)| < r} \log \left(\frac r {|\rho-(2+iT)|} \right)$$
The RHS is $\ge  a(N(T+1)-N(T))$ and it is $$\le \int_0^1\max(0, \log | \zeta(2+iT+re^{2i\pi \theta})|) d\theta=\int_0^1 O(\log(T^3))d\theta=O(\log T)$$
where $\zeta(2+iT+re^{2i\pi \theta})=O(T^3)$ follows from $\zeta(s)=\frac{s}{s-1}+s\int_1^\infty(\lfloor x\rfloor-x)x^{-s-1}dx$ and say the functional equation.


问:
How to show $\int_{\mathbb{C}}|f(z)|^2(1+|z|)^{-\alpha}dV(z) <\infty$ and $f$ entire then $f$ is a polynomial
I read in a textbook and see the following problem:

Let $f:\mathbb{C}\rightarrow \mathbb{C}$ be an entire function such that for some  $\alpha >0$, we have
$$\int_{\mathbb{C}}|f(z)|^2(1+|z|)^{-\alpha}dxdy <\infty.$$
Then $f$ is a polynomial.

I don't know how to verify this. In order to prove that $f$ is a polynomial we need to show that $f^{(n)}\equiv 0$ for some $n\in \mathbb{N}$. I think Liouville theorem might help but I have no idea.

Could anyone give me a hint how to relate am integral inequality with a certain estimate on the derivative?

答:
Let $f(z)=\sum a_nz^n$ and note that for $0 \le |z| \le R$ we have $(1+|z|)^{-\alpha} \ge (1+|R|)^{-\alpha}$ so:

$I=\int_{\mathbb{C}}|f(z)|^2(1+|z|)^{-\alpha}dV(z) \ge (1+|R|)^{-\alpha}\int_{|z| \le R}|f(z)|^2dV(z) =$

$=(1+|R|)^{-\alpha}\int_0^R\int_0^{2\pi}(\sum a_nr^ne^{in\theta})(\sum \bar a_n r^ne^{-in\theta})rd\theta dr=$

$=2\pi (1+|R|)^{-\alpha} \sum_{n \ge 0}\frac{|a_n|^2R^{2n+2}}{2n+2}$

(noting that we can switch series and integral by uniform convergence and that non-diagonal terms integrate to zero $d\theta$)

The hypothesis implies:

$2\pi (1+|R|)^{-\alpha} \sum_{n \ge 0}\frac{|a_n|^2R^{2n+2}}{2n+2} \le I < \infty$

Letting $R \to \infty$ we immediately get that $a_n=0$ if $2n+2- \alpha >0$ so $f$ must be a polynomial with degree at most $\frac{\alpha-2}{2}$ if that is nonnegative (or $f$ identically zero otherwise)


问:
The $p$-completion of a knot group
In Serre's LNM 5, in the section on profinite groups, he give the exercise:

Let $k$ be a knot in $ \mathbb{R}^3$, and let $G= \pi _1( \mathbb{R}^3-k)$ be the knot group of $k$. Show that the $p$-completion of $G$ is isomorphic to $\mathbb{Z}_p$.

Here $ \pi_1 $ is the fundamental group, and by $p$-adic completion I mean the inverse limit of quotients of $G$ by finite $p$-index normal subgroups.

I'm sort of unsure where to start with this, since I think the problem of actually calculating knot groups can be quite a hard one...

答:
The essential property of knot groups is the following.

The group $G$ has an element $x$ so that every element or $G$ can be written as a product of conjugates of $x$ and $x^{-1}$ (that is, it is normally generated by a single element).

This can be proved by analysis of the Wirtinger presentation.

The essential lemma is the following.

A finite $p$-group which is normally generated by a single element is cyclic.

Proof: Any element $x$ of order a power of p in a finite group is contained in a p-Sylow subgroup, which is a proper normal subgroup, and hence the smallest normal subgroup of $G$ is still a proper subgroup.

It follows that a finite p-group quotient of a knot group is isomorphic to $\Bbb Z/p^k$. Because this is Abelian it is a quotient of the abelianization $H_1(\Bbb R^3 - k) \cong \Bbb Z$, and so it is the p-completion of $\Bbb Z$. Conclude.

Title: The $p$-completion of a knot group


问:
Area of surface over $xy$-plane
In my calculus book, it is claimed that the area $dS$ below, can be expressed as $$\frac{1}{\cos\gamma}dxdy.$$

Here $\vec{n}$ is perpendicular to the surface element, and $\vec{k}$ is given as $(0,0,1)$.

I understand how this can be concluded if either the area element's $x$- or $y$-component is parallell to the coordinate system. I do however have a hard time seeing how this can be concluded if the element is tilted in both the $x$ and $y$-direction.

Any tips on how this can be shown?

答:
WLOG we can translate things around so that the base of our rectangular prism has vertex $(0, 0, 0), (a, 0, 0), (0, b, 0), (a, b, 0)$ and $(0, 0, 0)$ is the vertex with the smallest $z$-coordinate where the plane cut the rectangular prism.

Also, WLOG we can assume our normal vector $n = \begin{pmatrix} p \\ q \\ r \end{pmatrix}$ has length $|n| = 1$.

With this setup, all we need to do is to find the other 2 vertices $(a, 0, z_1), (0, b, z_2)$ where the plane cut the rectangular prism. Since the cross section is a parallelogram, the area is then simply the length of the cross product of the position vectors of the 2 vertices.

This is the hint part. You can stop reading now and try to solve it yourself.

Since $\gamma$ is the angle between the $k = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ and $n$, by the property of dot product, $\cos \gamma = |k||n| \cos \gamma = k \cdot n = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} p \\ q \\ r \end{pmatrix} = r$

From now on, we can write $n = \begin{pmatrix} p \\ q \\ \cos \gamma \end{pmatrix}$

Observe $1 = |n| = \sqrt{p^2 + q^2 + \cos^2 \gamma}$

Therefore $p^2 + q^2 + \cos^2 \gamma = 1$ and hence $$p^2 + q^2 = 1 - \cos^2 \gamma = \sin^2 \gamma$$

Since the plane cut the rectangular prism at $(0, 0, 0)$, it passes throught the origin. Hence we can write its equation as $$\begin{pmatrix} p \\ q \\ \cos \gamma \end{pmatrix} \cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix} = 0$$ which is the same as $px + qy + (\cos \gamma)z = 0$.

As the plane cut through the rectangular prism at vertice $(a, 0, z_1)$, the vertice satisfies the equation of the plane. Therefore, $p a + q 0 + (\cos \gamma)z_1 = 0$ and hence $$z_1 = -\frac{pa}{\cos \gamma}$$

Similarly, at vertice $(0, b, z_2)$, we have $p 0 + q b + (\cos \gamma)z_2 = 0$ and hence $$z_2 = -\frac{qb}{\cos \gamma}$$

With these two $z$-coordinates, we now know the position vectors of the 2 vertices are $\begin{pmatrix} a \\ 0 \\ -\frac{pa}{\cos \gamma} \end{pmatrix}$ and $\begin{pmatrix} 0 \\ b \\ -\frac{qb}{\cos \gamma} \end{pmatrix}$.

Since $(0, 0, 0)$ is the vertex with the smallest $z$-coordinate where the plane cut the rectangular prism, these 2 position vectors are the sides of the parallelogram.

Now area $$\begin{align*} A
&= \left| \begin{pmatrix} a \\ 0 \\ -\frac{pa}{\cos \gamma} \end{pmatrix} \times \begin{pmatrix} 0 \\ b \\ -\frac{qb}{\cos \gamma} \end{pmatrix} \right| \\
&= \left| \det \begin{pmatrix}
i & j & k \\
a & 0 & -\frac{pa}{\cos \gamma} \\
0 & b & -\frac{qb}{\cos \gamma}
\end{pmatrix} \right| \\
&= \left| i \det \begin{pmatrix}
0 & -\frac{pa}{\cos \gamma} \\
b & -\frac{qb}{\cos \gamma}
\end{pmatrix} - a \det \begin{pmatrix}
j & k \\
b & -\frac{qb}{\cos \gamma}
\end{pmatrix}\right| \\
&= \left| \frac{pab}{\cos \gamma} i - a \left( -\frac{qb}{\cos \gamma} j - bk \right) \right| \\
&= \left| \frac{pab}{\cos \gamma} i + \frac{qab}{\cos \gamma} j + ab k \right| \\
&=  \sqrt{\frac{p^2 a^2 b^2}{\cos^2 \gamma} + \frac{q^2 a^2 b^2}{\cos^2 \gamma} + a^2 b^2} \\
&= a b \sqrt{\frac{p^2}{\cos^2 \gamma} + \frac{q^2}{\cos^2 \gamma} + 1} \\
&= ab \sqrt{\frac{p^2 + q^2}{\cos^2 \gamma} + 1} \\
&= ab \sqrt{\frac{\sin^2 \gamma}{\cos^2 \gamma} + 1} \\
&= ab \sqrt{\tan^2 \gamma + 1} \\
&= ab \sqrt{\sec^2 \gamma} \\
&= ab \sec \gamma \\
&= \frac{ab}{\cos \gamma}
\end{align*}$$
which is exactly what we want to show.

Title: Area of surface over $xy$-plane


问:
States on the reduced group $C^\ast$-algebra of $\mathbb{F}_2$
Consider the reduced group $C^\ast$-algebra $C_{\text{r}}^\ast (\mathbb{F}_2)$ of the free group in two generators $a$ and $b$. Further, let $\phi$ be a state on $C_{\text{r}}^\ast (\mathbb{F}_2)$. Let $(g_i)_i \subseteq \mathbb{F}_2$ be a sequence of group elements for which the word length (with respect to the generators $a$ and $b$) of $g_i$ is equal to $i$. Is it then true that $\phi(\lambda_{g_i}) \rightarrow 0$ for $i\rightarrow \infty $? Here $\lambda$ denotes the left-regular representation of $\mathbb{F}_2$. The statement looks kind of obvious, but I don't see how to prove it.

答:
The answer is no.

Let $A$ be the closed *-subalgebra of $C_{\text{red}}^*(\mathbb F_2)$ generated by $\lambda _a$.  It is easy to see that $A$
is isomorphic to $C(S^1)$, the continuous functions on the unit circle, via an isomorphism that maps $\lambda _a$ to the identity
function "$f(z)=z$".

Consider the state $\varphi $ (actually a character) of $A$ defined by
$$
  \varphi :g\in A\mapsto g(1)\in {\mathbb C}.
  $$

By a well known result, $\varphi $ extends to a state, say $\tilde\varphi $,  of $C_{\text{red}}^*(\mathbb F_2)$.

Setting $g_n = a^n$, one has that the word length of $g_n$ is $n$, but
$$
  \tilde \varphi (\lambda _{g_n}) = \varphi (\lambda _a^n) = z^n|_1 = 1,
  $$
which doesn't converge to zero.


问:
Solving differential equation
I have the following differential equation
\begin{align}
  \left[-\partial^{2}_{x}+ x^{2} -1-2\alpha_{+}\right] \left[-\partial^{2}_{x}+ x^{2} -1-2\alpha_{-}\right] \phi(x)=0, \quad \alpha_{\pm}=\dfrac{-1}{2}+\dfrac{E^{2}}{2}\pm \dfrac{\sqrt{a^{4}+(\beta E)^{2}}}{2a^{2}}
  \end{align}
My question is : what is the method used to find the eigenvalues of this equation.
Thank you in advance.

答:
This looks too much  like a homework problem to warrant an answer, so I will remind you a couple of facts about the quantum harmonic oscillator and its solutions, the Hermite functions used in physics problems; your posting this here assumes you are intrigued by physics applications. Your second specification appears meaningless without further information.

You are interested for all all eigenfunctions $\phi(x)$ in the kernel of this operator, for allowable constants $\alpha_{\pm}$,
$$
\hat O=  \left[-\partial^{2}_{x}+ x^{2} -1-2\alpha_{+}\right] \left[-\partial^{2}_{x}+ x^{2} -1-2\alpha_{-}\right].
$$
As a physicist, you surely recognize in each square bracket Hermite's operator
$$
D= (-\partial^{2}_{x}+ x^{2} -1)/2~,
$$
trivially related to the oscillator hamiltonian. You know its eigenvalues are the integers, 0,1,2,...,n,... for eigenfunctions $\psi_n(x)$, the Hermite functions, linked, $D\psi_n = n\psi_n$, but I will not insist on integer ns in the absence of further (boundary) information.

Your operator then presents as
$$
\hat O/4=(D-\alpha_+) (D-\alpha_-),
$$
and, acting on these Hermite functions, $\psi_n(x)   = (-1)^n \left (2^n n! \sqrt{\pi} \right)^{- 1/2} e^{ {x^2}/{2}}   \partial_x^n e^{-x^2}$, it yields
$$
(n-\alpha_+)(n-\alpha_-),
$$
which must vanish for your kernel, so $n=\alpha_{\pm}$ will do.

You have to fuss the details and choices yourself, subject to the physical constraints of your problem, possibly continuing off the integers.

OwnerDisplayName: Elmanara

Title: Solving differential equation

OwnerDisplayName: Cosmas Zachos


问:
Why is $(12345) = (54)(53)(52)(51)$?
I'm trying to understand the concept of permutations being decomposed into transpositions in group theory. In Pinter's Book of Abstract Algebra pg. 83, he says

$(12345) = (54)(53)(52)(51)$

I just can't make any sense of this. The permutation $(12345)$ is the identity, right? So if I apply those transpositions to $(12345)$ starting from the right, I should end up with $(12345)$ again, correct?

When I write out each permutation in full, I get

$(52341)\\
(15342)\\
(12543)\\
(12354)$

But when I trace what happens to each of the numbers 1, 2, 3, 4 and 5, I end up with $(23451)$. For instance, 1 goes to 5, then 2, then 2, then 2. And 2 goes to 2, then 5, then 3, then 3. And so on.

He provides a number of other decompositions, and I get the same end result from all of them.

What am I doing wrong?

答:
No, the permutation (1 2 3 4 5) is not the identity.  It is the cyclic permutation that sends: 1 to 2, 2 to 3, 3 to 4, 4 to 5, 5 to 1.

To simplify the explanation of decomposing into transpositions, let's instead use a simpler cyclic permutation: (1 2 3).  It sends 1 to 2, 2 to 3, and 3 to 1.

Now consider the product of transpositions: (1 3) (2 3).  It sends:

*) 1 to 3 to 2; i.e., ultimately, 1 to 2;

*) 3 to 1;

*) 2 to 3.

I.e., it has the same ultimate effect as permutation (1  2  3).

You will get a clear exposition to groups and, in particular, to permutations, from Part 1 of Abel's theorem in problems and solutions.

Title: Why is $(12345) = (54)(53)(52)(51)$?


问:
Derivative of limited convolution
If I take infinite convolution, I know that its derivative is the convolution between one term and other term's derivative.

However, I am interested on derivative of

$$\dfrac {d}{dt}\int_0^t f (t) g (t-a) da$$

i.e. in terms of limited convolution.

Many thanks in advance!

======

EDIT: Please review comments on answer

答:
I think @tossimmar 's answer gives some preliminary remarks, but I think this answer will provide the asker with more of what he/she is looking for. First of all, Leibniz's integral rule, in most general form, is
$$\frac{\mathrm{d}}{\mathrm{d}t}\left(\int_{a(t)}^{b(t)}f(t,s)\mathrm{d}s\right)=f\left(t,b(t)\right)b'(t)-f\left(t,a(t)\right)a'(t)+\int_{a(t)}^{b(t)}\frac{\partial}{\partial t}f(t,s)\mathrm{d}s$$
In our case, this gives us
$$\frac{\mathrm{d}}{\mathrm{d}t}\left(\int_{0}^{t}f(t)g(t-s)\mathrm{d}s\right)=f(t)g(0)+\int_0^tf'(t)g(t-s)\mathrm{d}s+\int_0^tf(t)g'(t-s)\mathrm{d}s$$
Of course, we could write this in terms of this partial convolution which I'll denote with an tilded asterisk:
$$\mathrm{D}(f~\tilde{*}~g)=g(0)\cdot f+f'~\tilde{*}~g+f~\tilde{*}~g'$$

Title: Derivative of limited convolution


问:
Probability question, confusion about with vs without replacement
Say this is the question:
An bowl contains 6 green and 8 yellow balls.  Five balls are drawn without replacement.  What is the probability that 3 are green and 2 are yellow? Order does not matter.

The way I interpretted this was: out of 6 green balls, I can choose 3, and then out of 8 yellow, choose 2. so I can calculate this as :

$$\frac {\binom 63\times \binom 82}{\binom {14}{5}}$$

First, is this correct? and second, how would this change if it were with replacement?

答:
Indeed the probability for obtaining 3 from 6 green balls and 2 from 8 yellow balls when selecting any 5 from all 14 balls without replacement is as you had.
The count for green balls has a hypergeometric distribution$$\dfrac{\dbinom 63\dbinom 82}{\dbinom {14}5}$$

When that selection is made without replacement, the probability is for any arrangement of 3 successes and 2 failures when the success rate is $6/14$. The count for successes has a binomial distribution.$$\dbinom53\dfrac{6^3~8^2}{14^5}$$


问:
Is complement of an algebraic set also algebraic in $\mathbb{A}^n(k)$?
Is complement of an algebraic set also algebraic in $\mathbb{A}^n(k)$, where $k$ is arbitrary field?

My Solution: Let $V$ be an algebraic set i.e. $V = V(F)$ for some $F =\{f_{\alpha}: \alpha \in \Lambda \}$. Assume that $U := \mathbb{A}^n(k) -V$  is also an algebraic set. Then, $U$ and $V$ are both Zariski open and closed set in $\mathbb{A}^n(k)$. Thus, $\mathbb{A}^n(k)$ will be disconnected.

Doubt: I am not sure how to conclude after this. Is it true that $\mathbb{A}^n(k)$ is connected in Zariski topology? If yes, why?

答:
This is false. For example, consider the case $n=1$. If $X\subsetneq\mathbb{A^1}(k)$ is algebraic then $X=V(I)$ for some ideal $0\ne I\subseteq k[x]$. Since $k[x]$ is a Noetherian ring we know $I$ is finitely generated, and so $X=V(f_1,...,f_r)$ for some nonzero polynomials $f_1,...,f_r\in k[x]$. Every polynomial in one variable has only finitely many roots, and so $X$ is finite.

Conversely, if $X\subseteq\mathbb{A^1}(k)$ is a one point set $X=\{x_0\}$ then obviously $X=V(x-x_0)$, and so $X$ is algebraic. It follows that every finite set is algebraic, because a finite union of closed sets is closed.

So if $n=1$ then the Zariski topology is simply the cofinite topology on $k$. If $k$ is an infinite field then the complement of a finite set is not finite, and so not algebraic.


问:
Why is the knot group of the trefoil isomorphic to the group of 3-braids?
I apologise in advance for the vagueness of this question but I have not been able to find very much info on the topic and have made very little progress on my own.

I am trying to understand why the knot group $\pi_1 (S^3 - K)$ of the trefoil is isomorphic to Artin's 3-strand braid group $B_3$. I know that the Wirtinger presentation for $\pi_1 (S^3 - K)$ gives Artin's presentation for $B_3$ directly but I was hoping someone could paint a more topological picture which takes homotopy classes directly to braids (or vice-versa) without using group presentations as the middle man. Thanks in advance.

Edit: Thanks for the replies guys. I should have stated that $S^3-K$ is diffeomorphic to the space $SL(2,\mathbb{R}) / SL(2,\mathbb{Z})$ (John Baez says so in his blog). It is possible that the disk with 3 holes is a deformation retract of $SL(2,\mathbb{R}) / SL(2,\mathbb{Z})$ but I don't know much about this space. I'll update if I find the answer myself.

答:
The 3-strand braid group is the fundamental group of $UC_3(\mathbb{C}) = \{\{z_1,z_2,z_3\} \subset \mathbb{C} \mid z_i \not= z_j\}$, the unordered configuration space of distinct triples in the complex plane. By translating, this space deformation retracts to the subspace of triples where $z_1+z_2+z_3=0$. By positive real scaling, this deformation retracts further so the space where $|z_1|^2+|z_2|^2+|z_3|^2=1$. (Note that these conditions are invariant up to permutation of the $z_i$.) Call this new space $X$.

Now we see that $X$ is a 3-manifold. That it is a manifold should be clear. As for the dimension, you can heuristically think of there being $2$ degrees of freedom for where to pick a first point in the plane. The normalization tells you that you can pick one more point (very nearly) freely on a circle, so $1$ degree of freedom, and then the last point is uniquely determined.

Not only is $X$ a 3-manifold, it's also a Seifert fiber space: there is a circle action $S^1 \curvearrowright X$ given by rotation, or multiplication by $e^{i\theta}$ if you prefer. This action is almost free. There are just two exceptional orbits, given by the triples of points $\{\pm 1/\sqrt{2},0\}$ (collinear) and $\{1/\sqrt{3},e^{\pm 2\pi i/3}/\sqrt{3}\}$ (equilateral triangle). From here, you see that $X$ has exceptional fibers of order 2 and order 3. There is also one "missing" generic fiber, corresponding to the collapse of two points. (Our normalization prevents the collapse of three points, so don't need to worry about that). Maybe this is enough to make you start to suspect that you're looking at the trefoil knot complement.

There's another explanation along the same lines where you view the points in $\mathbb{C}$ as roots of complex polynomials. Steve Trettel has a nice exposition of this at his website.


答:
This is cycle notation, which is ideal for many types of calculations with permutations. For example, the notation $(1\,2\,3\,4\,5)$ stands for the transformation
that sends
$$
1 \mapsto 2 \mapsto 3 \mapsto 4 \mapsto 5 \mapsto 1.
$$
It's not unique. In fact, for a $5$-cycle such as this, there are $5$ ways to represent the permutation, depending on which one you put first. Also, any symbol that doesn't appear in a cycle is understood to be fixed by that cycle.

Now, permutations are bijective functions, so you compose them to form the permutation group. This is denoted by concatenation. For example, just composing the first two (reading from right to left, as you do with functions), let's calculate:
$$
(5\,2)(5\,1).
$$
Starting on the right, the first permutation maps $5 \mapsto 1$, then the second permutation leaves $1$ fixed. So in the composition, $5 \mapsto 1$. Repeating this for other elements, you can see that
\begin{align*}
&1 \mapsto 5 \mapsto 2 \\
&2 \mapsto 2 \mapsto 5 \\
&5 \mapsto 1 \mapsto 1
\end{align*}
Any other symbol is fixed (e.g. $3 \mapsto 3 \mapsto 3$ under the composition).
Thus,
$$
(5\,2)(5\,1) = (5\,1\,2)
$$
as cycles. With this understanding you should be able to reproduce the other calculations.

Title: Why is $(12345) = (54)(53)(52)(51)$?

Body:
I'm trying to understand the concept of permutations being decomposed into transpositions in group theory. In Pinter's Book of Abstract Algebra pg. 83, he says

$(12345) = (54)(53)(52)(51)$

I just can't make any sense of this. The permutation $(12345)$ is the identity, right? So if I apply those transpositions to $(12345)$ starting from the right, I should end up with $(12345)$ again, correct?

When I write out each permutation in full, I get

$(52341)\\
(15342)\\
(12543)\\
(12354)$

But when I trace what happens to each of the numbers 1, 2, 3, 4 and 5, I end up with $(23451)$. For instance, 1 goes to 5, then 2, then 2, then 2. And 2 goes to 2, then 5, then 3, then 3. And so on.

He provides a number of other decompositions, and I get the same end result from all of them.

What am I doing wrong?


问:
Showing that losing positions in Wythoff's game are generated by $(\lfloor n\phi\rfloor, \lfloor n\phi^2\rfloor)$, where $\phi$ is the golden ratio
A very neat problem in combinatorial game theory:

In Wythoff's Game, how do I show that all the losing positions are generated by the formula

$$\left(\lfloor n\phi\rfloor, \lfloor n\phi^2\rfloor\right)$$
where $n \in \mathbb{N}$ and $\phi$ is the golden ratio?

More precisely, the $k^{\text{th}}$ losing position is $\left(\lfloor k\phi\rfloor, \lfloor k\phi^2\rfloor\right)$ where $k \in \mathbb{N}$ and $\phi$ is the golden ratio.

I have tried this problem for a while and got a way to show that the sequences $\lfloor n\phi\rfloor$ and $\lfloor n\phi^2\rfloor$ are disjoint and their union is $\mathbb{N}$, this can be easily showed using the Beatty's theorem, but I'm unsure how to prove that all the losing positions are generated by the formula $\left(\lfloor n\phi\rfloor, \lfloor n\phi^2\rfloor\right)$.

I have gone through wikipedia and its references for the proof, but it doesn't seem like the proof for the formula is present anywhere in the sources or references.

Any help or hint would be highly appreciated. Thanks.

答:
Wythoff’s argument is actually quite straightforward. He first identifies the $P$-positions (which he calls safe positions), i.e., the positions in which the previous player wins; equivalently, they are the positions in which the player whose turn it is to move must lose, assuming that the other player plays correctly. He lists the first few:

$$\begin{array}{rcc}
n:&0&1&2&3&4&5&6&7&8&9&10\\\hline
u_n:&0&1&3&4&6&8&9&11&12&14&16\\
\ell_n:&0&2&5&7&10&13&15&18&20&23&26
\end{array}$$

(I’ve shortened his table a bit.) Clearly $\langle 0,0\rangle$ is a $P$-position: if you face this, the previous player has already won. The others in the table can be described as follows. For $n\in\Bbb Z^+$ let $U_n=\{u_k:0\le k<n\}$ and $L_n=\{\ell_k:0\le k<n\}$; then

$$u_n=\min\big(\Bbb Z^+\setminus(U_n\cup L_n)\big)\,,$$

and $\ell_n=u_n+n$.

Let $\mathscr{P}$ be the set of positions generated in this way. To show that $\mathscr{P}$ really is the set of $P$-positions, we need only show that every legal move from a position in $\mathscr{P}$ yields a position not in $\mathscr{P}$, while from every position not in $\mathscr{P}$ there is a legal move to a position in $\mathscr{P}$.

Suppose first that $\langle u,\ell\rangle$ is a position not in $\mathscr{P}$ such that $u\le\ell$, and let $n=\ell-u$. If $u>u_n$, let $d=u-u_n$; taking $d$ from each pile leaves the position $\langle u_n,\ell_n\rangle\in\mathscr{P}$. If $u<u_n$, there is a $k<n$ such that $u\in\{u_k,\ell_k\}$. If $u=u_k$, removing $\ell-\ell_k$ counters from the larger pile leaves the position $\langle u_k,\ell_k\rangle\in\mathscr{P}$. If $u=\ell_k$, removing $\ell-u_k$ counters from the larger pile also leaves the position $\langle u_k,\ell_k\rangle\in\mathscr{P}$.

Now let $n\in\Bbb Z^+$, and consider moves from the position $\langle u_n,\ell_n\rangle$. If the result of such a move is in $\mathscr{P}$, it clearly must be $\langle u_k,\ell_k\rangle$ for some $k<n$. But then $$\begin{align*}\ell_n-\ell_k&=(u_n+n)-(u_k+k)\\&=(u_n-u_k)+(n-k)\\&>u_n-u_k\\&>0\,,\end{align*}$$ which is not possible with a legal move. Thus, there is no move from $\langle u_n,\ell_n\rangle$ to a position in $\mathscr{P}$.

This shows that $\mathscr{P}$ really is the set of $P$-positions. (Wythoff did not include the arguments given at the bullet points, merely noting that are easy.)

He then claims essentially that $u_n=\lfloor n\varphi\rfloor$ and $\ell_n=\left\lfloor n\varphi^2\right\rfloor$. Certainly

$$\begin{align*}
\left\lfloor n\varphi^2\right\rfloor-\lfloor n\varphi\rfloor&=\lfloor n(\varphi+1)\rfloor-\lfloor n\varphi\rfloor\\
&=\lfloor n\varphi+n\rfloor-\lfloor n\varphi\rfloor\\
&=\lfloor n\varphi\rfloor+n-\lfloor n\varphi\rfloor\\
&=n\\
&=\ell_n-u_n\,,
\end{align*}$$

and $\lfloor n\varphi\rfloor=\left\lfloor n\varphi^2\right\rfloor=0$ when $n=0$, so it only remains to show that when $n\ge 1$, $\lfloor n\varphi\rfloor$ is the smallest positive integer not equal to $\lfloor k\varphi\rfloor$ or $\left\lfloor k\varphi^2\right\rfloor$ for any $k\in\{0,\ldots,n-1\}$. He does this by showing that $\langle n\varphi:n\in\Bbb N\rangle$ and $\left\langle n\varphi^2:n\in\Bbb N\right\rangle$ are complementary Beatty sequences, something that you’ve already done.


问:
Truth of a sentence in predicate logic
I'm working through the exercises in this Logic problem set and I'm struggling to understand ex. 7.5:

Determine whether the following sentence is logically true in predicate logic:

There is someone such that, if he or she is asleep, everyone is asleep.

The solutions section suggests this is True and gives a formal 'Natural Deduction' proof that I don't really understand (it probably doesn't help that I don't have a copy of the book the solutions keep referring to).

Is there another way to approach this?

答:
Assume we have a non-empty universe of people, i.e. there exists at least one person. Using a simple form of natural deduction, we have the following proof by contradiction (lines 2 -13) where

P(x) means x is a person

S(x) means x is sleeping (or drinking?)

(Screen print from my proof checker)

Title: Truth of a sentence in predicate logic


问:
Proving a function is Lipschitz is symmetrical
When we want to prove that a function is Lipschitz, we want to prove $|f(x)-f(y)|\le L|x-y|$. But we could just prove $f(x)-f(y)\le L|x-y|$ right? The problem is kind of symmetrical.

答: If you show $f(x)-f(y) \leq L|x-y|$ is true for all $x$ and $y$, then $|f(x)-f(y)| \leq L|x-y|$ for all $x$ and $y$. $f(x)-f(y) \leq L|x-y|$ implies $f(y)-f(x) \leq L|y-x| = L|x-y|$. One of $f(x)-f(y)$ and $f(y)-f(x)$ is equal to $|f(x)-f(y)|$, so $|f(x)-f(y)| \leq L|x-y|$.


答: I think it helps to understand what we're trying to say here. What we really mean when we say a function is lipschitz is that the distance between two points of the function is a constant times the distance between the points. In a sense, the distance between points of the function depends linearly on the distance of the points themselves. For instance, a linear function is like this. Let $f(x) = ax$ for a positive real number $a$. Try to show that $f$ is Lipschitz. What is $L$ here?

Title: Proving a function is Lipschitz is symmetrical

Body: When we want to prove that a function is Lipschitz, we want to prove $|f(x)-f(y)|\le L|x-y|$. But we could just prove $f(x)-f(y)\le L|x-y|$ right? The problem is kind of symmetrical.


问:
subgroups of finite cyclic group
Let $G=(g)$ be a finite cyclic group generated by $g$ with $|G|=n$, and let $d \in \mathbb{N}$ with $d|n$, then an unique subgroup $H$ of $G$ with $|H|=d$ exists.

Proof of existence: $\exists m \in \mathbb{N}: md=N$. Since $G=(g)$ we have $g^k\neq g^l\ \forall 0< k\neq l\leq N$ and $g^k=g^{k+nl}\ \forall k,l \in \mathbb{N}$.

Hence $g^{mk} \neq g^{ml}\ \forall 0< k\neq l \leq d$ and $g^{mk}=g^{m(k+nl)}\ \forall k,l \in \mathbb{N}$. Therefore $H=(g^m)=\{g^{m1},...,g^{md}=e\}$ is a (cyclic) subgroup of $G$ with $|H|=d$.

Is this ok?

Proof of uniqueness: Let $H_1,H_2$ be two subgroups with order $d$. As subgroups of a cyclic group $H_1,H_2$ are also cyclic. Set $k_1,k_2$ to be the smallest positive exponents so that $g^{k_1} \in H_1, g^{k_2} \in H_2$ respectively.

I do not know how I can continue from there to proof uniqueness..any help? Or is the approach wrong?

答:
Here is another solution to this old problem:

As every finite cyclic group of order $n$ is isomorphic to $\mathbb{Z}_n:=\mathbb{Z}/n\mathbb{Z}$ with the sum of integers modulo $n$, it is suffices to show this for $G=\mathbb{Z}_n$

Suppose $n=kp$, for $p,k\in\mathbb{N}$. If $p=1$ or $p=n$ then the corresponding subgroups of order $p$ are $(0)$ and $\mathbb{Z}_n$. for $1<p<n$ consider
$H_p:=(0,p,2p,\ldots,(k-1)p)$. This is clearly a subgroup for order $k$ for if
$$ap\equiv bp\mod(kp)$$
then $a\equiv b\mod k$ which means that the elements in $H_p$ do not repeat.

As for uniqueness, suppose $H'<\mathbb{Z}_n$ with $|H'|=k$. Let $p'$ be the smallest positive integer in $H'\cap\{1,1,\ldots,n-1\}$. Then $H'=\{0,p',2p',\ldots,(k-1)p'\}$ and $p'k=pk=n$. That is, $p=p'$.

Title: subgroups of finite cyclic group


问:
Given a field $F\subset R$ contained in a ring which is not an integral domain. How can R fail to be a vector space over F?
My algebra notes mention being an integral domain as a requirement, but I can't figure out which vector space axioms are dependent on it.</p>

答: It isn't necessary. $R$ is indeed a vector space over $F$ where the scalar multiplication is obtained by restricting the multiplication of $R$ to $F\times R$.


问:
Possible Playable Chords on a Guitar

Fingerstyle Guitar Chord Diversity Check

Considering a $20$-fret $6$-string acoustic guitar and supposing that the fretting range (inclusive of the fingered notes) for an average hand is $4$ frets in the first $8$ frets, $5$ frets in the $9-14$ fret region and $6$ frets on the remaining of the fretboard, how many combinations of at least $3$ different notes can be strummed if we ignore barring and use at most $4$ fingers to fret?

We would have the worst fretting range if our fingered frets involve any of the $3$ prescribed regions down the fretboard (i.e. $8$ fret region, $9−14$ fret region and the remaining of the fretboard). Suppose you decide to fret $12$ on the first string, then, the lowest other fret you can finger is $9$ since fretting $8$ would require a fret range of $5$ while playing $8$ reduces your fret range to $4$ since we consider the worst fret range on overlapping fret regions.

In our case, a chord is defined as a collection of $3$ to $6$ distinct notes strummed together. Each chord is unique. For example, fretting $3$ on sixth string, $2$ on the fifth string, $3$ on the first string and strumming all 6 strings is the same as fretting $3$ on sixth string, $2$ on the fifth string, $3$ on the second string and $3$ on the first string and strumming all 6 strings as both would produce a $\text{G}$ chord.

答:
The following is not really specific to guitar music, but a general approach that could be specialized to guitar music with further restrictions.

Dmitri Tymoczko has tried to develop a kind of geometrical theory of chord spaces. I found it interesting insofar as it managed to get some structure, I didn't find it at all useful when it comes to making music though. But I'm sure this is something you want to look into. His is not the only approach though.

Anyway, while trying to understand his papers, I started to do my own computations to determine how many different chords of 3, 4, 5,... notes there exist. The key here is Burnside's lemma from group theory. Let me illustrate for the case of 3 note chords.

First, we have to establish the space we are working on. We'll start from the conventional Western 12 tones in equal temperament, but if you want to make quarter tone or microtonal music or use different temperaments, you can easily adapt my story. Each tone will get a number from 0 to 11, we'll work modulo 12 since the twelfth tone is just the 0th an octave higher. So that's a first symmetry that is already implicit in the definition of our space. In modern harmony, this is called a pitch class. Now, since we consider 3-tone chords, we are going to present them as follows

$$\text{CMaj}=\{\text{C,E,G}\}=\{0,4,7\} \\
\text{Cmin}=\{\text{C,Eb,G}\}=\{0,3,7\} \\
\text{FMaj 2nd inversion}=\{\text{C,F,A}\}=\{0,5,9\} \\$$

Then we are going to introduce additional symmetries. We consider all chords that can be transposed into each other to be the same chord. For instance

$$\text{FMaj}=\{\text{F,A,C}\}=\{5,9,0\}\overset{T}{=}\{5-5,9-5,0-5\}=\{0,4,7\}=\text{CMaj} \; .$$

Next, we are also going to identify inversions, which mathematically amounts to permutations of the elements

$$\text{FMaj 2nd inversion}=\{\text{C,F,A}\}=\{0,5,9\}\overset{P}{=}\{5,9,0\}\overset{T}{=}\{0,4,7\}=\text{CMaj} \; .$$

One could go further and add additional symmetries, for instance

$$\text{Cmin}=\{\text{C,Eb,G}\}=\{0,3,7\}\overset{I}{=}\{7-0,7-3,7-7\}\overset{P}{=}\{0,4,7\}=\text{CMaj} \; .$$

But since this turns major chords into minor chords and visa versa and these are usually considered distinct in conventional harmony, we're not going that way. One might argue that even inversions of chords should be considered separately because of their different functions, but for the sake of the present argument, I'm gonna stick to two symmetries $P$ and $T$.

Now, let's see how we can apply Burnside's lemma. Our space of 3-tone chords is $X$ and is acted upon by a group $G$ generated by transpositions $T$ on one hand and permutations $P$ on the other. Burnside's lemma tells us that

$$|X/G|=\frac{1}{|G|}\sum_{g\in G}|X^g|$$

or in human words: the number of 3-tone chords up to transpositions and permutations is found by adding for each symmetry element the number of chords fixed by that certain symmetry and dividing by the total number of symmetries acting upon the space. $X/G$ is called in jargon the set of orbits. Whereas $X^g$ is the set of fixed elements under the action of $g$.

Applying this to our 3-tone chords set, we really only have to look at what happens under permutations since transpositions can never fix elements. So, we have 6 possible permutations:

$$ \{\text{Id},(12),(13),(23),(123),(132)\} \; .$$

The identity fixes all $12^3$ elements of our space. The three following permutations fix elements of the form $\{a,a,b\},\{a,b,a\}$ and $\{b,a,a\}$ resp. thus accounting each for $12^2$ fixed elements. The last two are more tricky. Obviously elements of the form $\{a,a,a\}$ are fixed. But also elements of the form $\{a,a+4,a+8\}$ and $\{a,a+8,a+16\}$ are. This accounts for a total of $6 \times 12$ elements. Applying Burnside's lemma, the amount of distinct 3-tone chords is

$$\frac{12^3+3\times 12^2+6\times 12}{12\times 6}=31 \; .$$

Among these 31 chords are however a bunch of chords containing repeated tones, 12 to be precise. If we leave those 12 out, we are left with 19 chords. Leaving out 9 more that contain a minor second interval leaves us with the following 10 chords

$$\{\text{C,E,G#}\},\{\text{C,E,G}\},\{\text{C,Eb,G}\},\{\text{C,D,F#}\},\{\text{C,E,Gb}\},\{\text{C,D,G}\},\{\text{C,D,E}\},\{\text{C,Eb,F}\},\{\text{C,Eb,Gb}\},\{\text{C,D,F}\}$$

The first three chords are resp. the augmented triad, the major triad and the minor triad. The second to last triad is the diminished triad. Those are very common in classical harmony. Chords like $\{\text{C,D,G}\}$ are used in quartal and quintal harmony. It is however also understood as a suspended fourth in classical and jazz harmony.

Note that it is possible in my analysis to exclude repeated notes from the onset. The sequence of integers for the number of chords consisting of several distinct notes is listed on OEIS with number A035495 and continues as follows:

$$1, 1, 6, 19, 43, 66, 80, 66, 43, 19, 6, 1, 1$$

Finally, a small "dirty" picture I made to illustrate the 10 chords and how you can move by stepwise motions between them, i.e. by altering just one tone a half step up or down.

Title: Possible Playable Chords on a Guitar


答:
There is no reason $R$ has to be an integral domain at all. I'm assuming you're saying that $F$ is a subring of $R$ when you write $F \subset R$; it's irrelevant whether $F$ is a subset of $R$ when determining whether $R$ has an $F$-vector space structure.

The most general result is that whenever we have a ring homomorphism $i : F \to R$, $R$ can be given an $F$-vector space structure, where the additive structure is $R$'s additive structure and addition is given by

$f \cdot r = i(f) r$

We must check the axioms:

$f \cdot (r + s) = i(f) (r + s) = i(f) r + i(f) s = f \cdot r + f \cdot s$

$(f + q) \cdot r = i(f + q) r = (i(f) + i(q)) r = i(f) r + i(q) r = f \cdot r + q \cdot r$

$1 \cdot r = i(1) r = 1r = r$

$(fq) \cdot r = i(fq) r = i(f) i(q) r = i(f) (q \cdot r) = f \cdot (q \cdot r)$

So $R$ is an $F$-vector space.

In the special case that $F$ is a subring of $R$, we take $i : F \to R$ to be the inclusion ring homomorphism. We see that there is no reason at all to demand $R$ be an integral domain.

Title: Given a field $F\subset R$ contained in a ring which is not an integral domain. How can R fail to be a vector space over F?

Body: My algebra notes mention being an integral domain as a requirement, but I can't figure out which vector space axioms are dependent on it.</p>


问:
Let $G=(\Bbb{Q}\setminus\{1\},*)$, where $*: x+y-xy$, and $H=\{0,\pm 2,\pm 4,...\}$. Show that $H\le G$.
I am having trouble checking algebraically if one group $H$ is a subgroup of the other group $G$. In an abstract way I can do it (I show that it says closing, neutral and inverse element), but when the example is algebraic I get confused, can you help me?

My question: $G = (\mathbb{Q}\setminus\{1\},*)$, where $*: x+y-xy$, and $H=\{0,\pm 2,\pm 4,...\}$. Show that $H$ is a subgroup of $G$.

答: This is false.  The group identity under this operation is $0$, but $4$ has no inverse in $H$:  $4+y-4y=0 \Rightarrow 4=3y \Rightarrow y = \frac 43 \notin H$.  I suspect that $H$ is supposed to be $ \{ r \in \Bbb Q \setminus \{ 1 \} \mid r = \frac{2n}{k}, k, n \in \Bbb Z, k \text{ odd} \}.$


问:
$\sum_{n=1}^{\infty}\frac{\sin(n^p)+5}{n^2}$ and find $p$
Does $\sum_{n=1}^{\infty}\frac{\sin(n^p)+5}{n^2}$ converges for which $p \in \mathbb{R}$. I'm starting to think it's every value of p converges because it's sin function. I can compare to $\sum \frac{1}{n^2}$ and it converges everytime.

Does $\sum_{n=1}^{\infty}\frac{\sqrt{n^p +1}}{n^4}$ converge for what values of $p \in \mathbb{R}$? i think it's squareroot function, so $n^p$ is slow to increase, and we can compare it to $\sum \frac{1}{n^4}$, but it's might grow so fast and it diverges. So finally I think for values $p < 6$ it converges.

答:
Yes. It converges absolutely because $|\sin(n^p) + 5| \leq 6$ and $\sum\limits_{n = 1}^\infty {6 \over n^2}$ converges (and in fact equals $\pi^2$).

We derive that $\sum\limits_{n = 1}^\infty {6 \over n^2}$ converges by the fact that $\sum\limits_{n = 1}^\infty {1 \over n^k}$ converges if and only if $k > 1$.

If we consider your second problem of $\sum\limits_{n = 1}^\infty {\sqrt{n^p + 1} \over n^4}$, we see that ${\sqrt{n^p + 1} \over n^4}$ ~ ${1 \over n^{4 - {p \over 2}}}$. And therefore, this sum converges iff $4 - {p \over 2} > 1$; that is, iff $p < 6$.


问:
Proof of Zero density of the Riemann Zeta function
I am currently trying to understand the proof of the simplest zero-density theorem for $\zeta$ there is, namely
$$N(T+1)-N(T-1) \ll \log(T),$$
where $T>2$ and  $N(T) = \#\{\rho \in \mathcal N : |\Im(\rho)| < T \}$ with $\mathcal N$ the set of non-trivial zeroes of $\zeta$. I consulted three books on analytic number theory so far (by Brüdern, Vaugan and Tenenbaum) and in all three of those the proof is (more or less) the same:
We use Jensen's Formula to obtain
$$\int_0^1 \log | \zeta(2+iT+re(\theta))| d\theta = \log|\zeta(2+iT)| + \sum_{\rho \in \mathcal N \text{ and } |\rho - (2+iT)| < r} \log \left(\frac r {|\rho-(2+iT)|} \right)$$
for some $r \in [3,4]$ and $e(\theta) = \text{e}^{2\pi i \theta}$. In this formula, it is relatively easy to see that the sum on the right hand side is $\approx N(T+1)-N(T-1)$. So far so good.  It is now stated that a bound of the form $$\zeta(\sigma + it) \ll t^k$$ for $t>1$, $\sigma \in \mathbb R$ and some $k$ of fitting size implies the bound $$\log|\zeta(s)| \ll \log |t|.$$ Tenenbaum even explicitly mentions the bound
$$\log |\xi(s)| \ll |s|\log|s| \text{ (as |s| }\rightarrow \infty \text{ for }\Re s \geq \tfrac 12) $$
for the complete Riemann-Zeta function in his book Introduction to analytic and probabilistic number theory as Formula (41) on page 152. It is clear that these bounds imply the desired result.

I could agree with the authors if they stated (and used) these bounds as $\log |\zeta(s)| \leq O(\log |s|)$, but in general these bounds are clearly wrong (as long as I've not completely lost it)! $\zeta$ (and $\xi$) have zeroes with arbitrarily large imaginary part, and $\log |\zeta|$ ($\log |\xi|$ resp.) has to be large near those.

Am I missing something? It might well be that I am misinterpreting some notations here, it's hard to believe that all three (or even one) of these authors would have published this without thinking about my issue.

答:
It follows from

$$
|\zeta^2(2)\zeta^4(2+iT)\zeta(2+2iT)|\ge1
$$

and $|\zeta(2+it)|\le\zeta(2)$ for all $t\in\mathbb R$ that

$$
1/\zeta(2+iT)=\mathcal O(1)
$$

This implies that $-\log\zeta(2+iT)$ is both upper bounded and lower bounded.

Consequently, by Jensen's inequality the number of roots within the region $|s-2-iT|\le r<R$ is bounded above by

$$
n(r)\le{\max_{|z|=R}|\log\zeta(2+iT+z)|-\log\zeta(2+iT)\over R/r}
$$

Now, it follows from Titchmarsh's book on Riemann zeta, we have $\log\zeta(2+iT+z)\ll\log T$. Choosing proper $R$ and $r$, we deduce

$$
N(T+1)-N(T-1)\ll\log T
$$


问:
How can I cover $\mathbb{Z}_5$ as a vector space over itself with hyperplanes
Consider $V=\mathbb{Z}_5$ a vector space over itself, How can I write $\mathbb{Z}_5$ as a union of finite hyperplanes? I am just looking for an example of I can do it because people told me that the case of my field is finite is trivial, but still don't know the process to conclude that. Thanks.

答:
The hyperplanes defined by
$$
x = 0\\
x = 1\\
x = 2\\
x = 3\\
x = 4
$$
cover $\Bbb Z / 5 \Bbb Z$, assuming that's what you mean by $\Bbb Z_5$.


答: Adding on @KReiser comment, it is sufficient to prove that if $\mathbb{A}^n(k)$ is irreducible, then $\mathbb{A}^n(k)$ is connected. Use the result that $V$ is irreducible $\iff I(V)$ is prime to get the result.

Title: Is complement of an algebraic set also algebraic in $\mathbb{A}^n(k)$?

Body:
Is complement of an algebraic set also algebraic in $\mathbb{A}^n(k)$, where $k$ is arbitrary field?

My Solution: Let $V$ be an algebraic set i.e. $V = V(F)$ for some $F =\{f_{\alpha}: \alpha \in \Lambda \}$. Assume that $U := \mathbb{A}^n(k) -V$  is also an algebraic set. Then, $U$ and $V$ are both Zariski open and closed set in $\mathbb{A}^n(k)$. Thus, $\mathbb{A}^n(k)$ will be disconnected.

Doubt: I am not sure how to conclude after this. Is it true that $\mathbb{A}^n(k)$ is connected in Zariski topology? If yes, why?