问:
Proving that grassmannians are smooth manifolds
I'm trying to show that real grassmannians $G(k, n)$ are smooth manifolds of dimension $k(n-k)$.
The problem is set in this way:

Identify the set of all real matrices with $n$ rows and $k$ columns with $\mathbb{R}^{kn}$. Then consider the following equivalence relation on $\mathbb{R}^{kn}$:
$M_1 \sim M_2$ if and only if there exists $L \in GL(k,\mathbb{R})$ such that $M_1=M_2L$. Then define $G(k, n)=\mathbb{R}^{kn}/ \sim$. $G(k, n)$ becomes a topological space whose topology is the quotient topology.
Now, for each subset of $J$ of $\{1, \ldots, n\}$ consisting of $k$ indices, consider the matrix $M_J$ obtained from $M \in \mathbb{R}^{kn}$ taking only the the $k$ rows whose indices are in $J$ and consider the set
$$U_J=\{[M]\in G(k, n): detM_J \neq 0\}. $$
Finally, for each $J$, define the map from $U_J$ to $\mathbb{R}^{k(n-k)}$ given by
$$\phi_J([M])= (MM^{-1}_J)_{J^c}$$
where $J^c$ denotes the complement of $J$ in $\{1, \ldots, n\}$.

I think I've managed to show that this map is a homeomorphism, but I'm in trouble when I hace to prove that transitions map between the charts are smooth, i.e. for each $J_1,J_2$ the composition $\phi_{J_1} \circ \phi_{J_2}^{-1}$ is smooth.

Any help would be greatly appreciated.

答:
Let $X\in\mathbb{R}^{(n-k)k}$. I will first denote $[M]=\phi_{J_2}^{-1}(X)$ and rewrite $\phi_{J_1}([M])$ as follow:
\begin{align}
\label{comp:1}
\phi_{J_1}([M])&=(MM_{J_1}^{-1})_{J_1^c} \\
&=M_{J_1^c}M_{J_1}^{-1} \\
&=M_{J_1^c}\cdot\frac{1}{\det M_{J_1}}\text{adj}(M_{j_1})
\end{align}
In case it is not clear, the second step is justified by noting that $M_J$ can be obtained from $M$ by a pre-multiplication of the $k\times n$ matrix $E_J$, whose $j$-th column is equal to the standard basis column vector $e_r\in\mathbb{R}^k$ if $j=i_r\in J$ and is equal to $0$ otherwise. For example, if $n=7$, $k=3$ and $J=(1,4,6)$, then
\begin{align}
E_j=\begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0
\end{pmatrix}
\end{align}
Then $M_J=E_JM$ as you can check easily. Thus
\begin{align}
(MM_{J_1}^{-1})_{J_1^c}&=E_{J_1^c}(MM_{J_1}^{-1}) \\
&=(E_{J_1^c}M)M_{J_1}^{-1} \\
&=M_{J_1^c}M_{J_1}^{-1}
\end{align}
which justified the second step.

By rewriting $\phi_{J_1}([M])$ in this way, it is clear that each entry of $\phi_{J_1}([M])$ is a rational function of entries of $M$. In particular, the denominator of this rational function is equal to $\det M_{J_1}$, which is nonzero since $[M]\in U_{J_1}$. Therefore, each entry of $\phi_{J_1}([M])$ depends smoothly on the entries of $M$.

Note that the whole computation does not depend on the choice of the representative of the equivalence class $[M]$. So we may choose the representative $M$ to be a convenient one. When you were proving the surjectivity of $\phi_{J_2}$, I believe you have already seen how to obtain such $M$ from $X$: Simply inserting back the row vectors $(0 \cdots 1 \cdots 0)$ to the position indexed by $J_2$. For example, when $n=4$, $k=2$ and $J_2=\{1,3\}$, if
\begin{align}
X=\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
\end{align}
then
\begin{align}
M=\begin{pmatrix}
1 & 0 \\
a & b \\
0 & 1 \\
c & d
\end{pmatrix}
\end{align}
Thus entries of $M$ are either $0$, $1$ or entries of $X$. Hence by choosing such $M$, it follows that
\begin{align}
\phi_{J_1}([M])=\phi_{J_1}\circ\phi_{J_2}^{-1}(X)
\end{align}
is a matrix whose entries depend smoothly on the entries of $X$. This proves the smoothness of $\phi_{J_1}\circ\phi_{J_2}^{-1}$.


问:
Express in symbols: "There is a real number that is not the cube of any real number besides itself."

I have the following statement, and I want to use symbols only to express its meaning.

There is a real number that is not the cube of any real number besides itself.

Here is my answer (assuming that $x$ and $y$ are in the universal set $R$ just to make the notation shorter):

$$(\exists x)(\forall y)(x \neq y^3)$$

Is this correct?

Thank you ^^

答:
Since this seems like a nonsense statement, I would be clear to translate verbatim

$$
\exists x, \forall y \neq x:  x \neq y^3
$$

There is a real number:  $\exists x$

For any real number besides itself:  $\forall y \neq x$

Is not the cube of:  $x \neq y^3$

ContentLicense: CC BY-SA 4.0


问:
$\{f_n\}$ integrable functions and $\{g_n\}$ uniformly bounded sequence of continuous functions, then is the following statement true?
Let $\{f_n\}_{n=1}^\infty$ and $f$ be integrable functions on [0,1] such that

$\displaystyle\lim_{n \rightarrow \infty} \int_{0}^1 |f_n(x) - f(x)| dx = 0$.

Then, is the following statement true?

If $\{g_n\}$ is uniformly bounded sequence of continuous functions converging pointwise to a function $g$ , then,

$\displaystyle\int_0^1 |f_n(x) g_n(x) - f(x) g(x) | dx \rightarrow 0 $ as $ n \rightarrow \infty$.

Could you please help me solve this problem?

答:
Notice that
\begin{align*}
\int | f_ng_n -fg| &\leq \int | f_n g_n - fg_n| + \int |f g_n -fg|\\
& \leq M \int |f_n-f| + \int |f| |g_n-g|
\end{align*}
where $M = \sup ||g_n||_\infty$. Furthermore, $| g_n -g| |f| \to 0$ almost everywhere, and it is dominated by an integrable function. Can you finish the proof? (apply dominated convergence thm)


问:
exercise on vectors of discrete random variables
How would you do the following exercise?

You toss a fair coin three times. Let X be the r.v. which counts the total number of heads and Y the r.v. which counts the number of heads of the last toss. Calculate the product density function of $(X,Y$ and the marginal ones.

答:
Assume $\mathbb{P}(\text{head}) =p$.

If $Y=0$, then $X$ can be at most $2$ therefore for $i=0,1,2$

$$
\mathbb{P}(X=i,Y=0) = \mathbb{P}(X=i|Y=0)\mathbb{P}(Y=0) = {2\choose i}p^{i}(1-p)^{2-i} (1-p),
$$
If $Y=1$, then $X$ has to be at least $1$ therefore for $i=1,2,3$

$$
\mathbb{P}(X=i,Y=1) = \mathbb{P}(X=i|Y=1)\mathbb{P}(Y=1) = {2\choose i-1}p^{i-1}(1-p)^{2-i+1} p.
$$


问:
Evaluating $\int_{-1}^{1}\frac{\arctan{x}}{1+x}\ln{\left(\frac{1+x^2}{2}\right)}dx$
This is a nice problem. I am trying to use nice methods to solve this integral, But I failed.

$$\int_{-1}^{1}\dfrac{\arctan{x}}{1+x}\ln{\left(\dfrac{1+x^2}{2}\right)}dx, $$

where  $\arctan{x}=\tan^{-1}{x}$

mark: this integral is my favorite one. Thanks to whoever has nice methods.

I have proved the following:

$$\int_{-1}^{1}\dfrac{\arctan{x}}{1+x}\ln{\left(\dfrac{1+x^2}{2}\right)}dx=\sum_{n=1}^{\infty}\dfrac{2^{n-1}H^2_{n-1}}{nC_{2n}^{n}}=\dfrac{\pi^3}{96}$$

where $$C_{m}^{n}=\dfrac{m}{(m-n)!n!},H_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}$$

I also have got a few by-products
$$\int_{-1}^{1}\dfrac{\arctan{x}}{1+x}\ln{\left(\dfrac{1+x^2}{2}\right)}dx=-I_{1}-2I_{2}$$

where $$I_{1}=\int_{0}^{1}\dfrac{\ln{(1-x^2)}}{1+x^2}\ln{\left(\dfrac{1+x^2}{2}\right)}dx=\dfrac{\pi}{4}\ln^2{2}+\dfrac{\pi^3}{32}-2K\times\ln{2}$$

and
$$I_{2}=\int_{0}^{1}\dfrac{x\arctan{x}}{1+x^2}\ln{(1-x^2)}dx=-\dfrac{\pi^3}{48}-\dfrac{\pi}{8}\ln^2{2}+K\times\ln{2}$$

and same methods,I have follow integral
$$\int_{0}^{1}\dfrac{\ln{(1-x^4)}\ln{x}}{1+x^2}dx=\dfrac{\pi^3}{16}-3K\times\ln{2}$$
where $ K $ denotes Catalan's Constant.

答:
FWIW, here's Maple:

```<code>> f:= arctan(x)/(1+x)*ln((1+x^2)/2);
> int(f, x=-1..1);
</code>```

$$ {\frac {7}{64}}\,{\pi }^{3}-{\frac {5}{16}}\,\pi \, \left( \ln
 \left( 2 \right)  \right) ^{2}-\ln  \left( 2 \right) {\it Catalan}+1/
2\, \left( \ln  \left( 1-i \right)  \right) ^{2}\pi -1/2\,i \left(
\ln  \left( 1+i \right)  \right) ^{2}\ln  \left( 2 \right) +\ln
 \left( 1+i \right) {\it Catalan}+1/2\, \left( \ln  \left( 1+i
 \right)  \right) ^{2}\pi -i{\it polylog} \left( 3,-i \right) +i{\it
polylog} \left( 3,i \right) +1/2\,i \left( \ln  \left( 1-i \right)
 \right) ^{2}\ln  \left( 2 \right) -1/2\,i \left( \ln  \left( 1-i
 \right)  \right) ^{3}-1/48\,i\ln  \left( 1+i \right) {\pi }^{2}+1/2\,
i \left( \ln  \left( 1+i \right)  \right) ^{3}+\ln  \left( 1-i
 \right) {\it Catalan}+1/48\,i\ln  \left( 1-i \right) {\pi }^{2}
$$

```<code>> simplify(%);
</code>```

$$ \frac{\pi^3}{96}
$$

I don't know if that qualifies as "nice", but it's certainly easy.

ContentLicense: CC BY-SA 3.0

Body:
FWIW, here's Maple:

> f:= arctan(x)/(1+x)*ln((1+x^2)/2);
> int(f, x=-1..1);

$$ {\frac {7}{64}}\,{\pi }^{3}-{\frac {5}{16}}\,\pi \, \left( \ln
 \left( 2 \right)  \right) ^{2}-\ln  \left( 2 \right) {\it Catalan}+1/
2\, \left( \ln  \left( 1-i \right)  \right) ^{2}\pi -1/2\,i \left(
\ln  \left( 1+i \right)  \right) ^{2}\ln  \left( 2 \right) +\ln
 \left( 1+i \right) {\it Catalan}+1/2\, \left( \ln  \left( 1+i
 \right)  \right) ^{2}\pi -i{\it polylog} \left( 3,-i \right) +i{\it
polylog} \left( 3,i \right) +1/2\,i \left( \ln  \left( 1-i \right)
 \right) ^{2}\ln  \left( 2 \right) -1/2\,i \left( \ln  \left( 1-i
 \right)  \right) ^{3}-1/48\,i\ln  \left( 1+i \right) {\pi }^{2}+1/2\,
i \left( \ln  \left( 1+i \right)  \right) ^{3}+\ln  \left( 1-i
 \right) {\it Catalan}+1/48\,i\ln  \left( 1-i \right) {\pi }^{2}
$$

> simplify(%);

$$ \frac{\pi^3}{96}
$$

I don't know if that qualifies as "nice", but it's certainly easy.


问:
Poisson's, Bernoulli's and Laplace's formulas
I'm not sure if i got correct answers as i often choose q and p wrong

It is known that a team is equally likely to win three games out of five and
two out of four. Find the probability of winning in one game.

My attempt: $$P_5(3)=P_4(2)$$ and using Bernoulli's formula I've got $$10p^3=6p^2$$ and answer p=0,6

The probability that the part will not pass the quality test is equal to 0.2. Which
the probability that out of 400 randomly selected parts will be defective:

a) 80 parts;

b) from 30 to 80 parts

My attempt: p=0,2 q=0,8 n=400

A)k=80 $$P_{400}(80)=0,05$$ (Laplace's formulas)

B) $$P_{400}(30<=k<=80)=x(0)+x(6,25)=0,5$$

The probability of failure for each call is equal to 0.00008. Identify
probability that at 1500 calls there will be 6 failures.

My try:

$$P_{1500}(6)=(0,12^6/6!)*exp(-0,12)=3,68*10^{-9}$$

答:
ok

I do not know what do you mean with Laplace's formula.

Given that calculating the exact probability (with the calculator) it is $\approx 52.987\%$, using a gaussian approx I get

$$\mathbb{P}(30\leq k\leq 80)=\Phi\left( \frac{80.5-80}{8} \right)-\Phi\left( \frac{29.5-80}{8} \right)\approx 52.492\%$$

which is a better approximation of the actual value w.r.t. your 50%

3.Ok, the poisson is good for your purpose...


问:
book recommendation numerical double integration
I am looking for references (books, articles, or forums) that treat the approximation of a double numerical integration by a double sum and how to implement it on Matlab. Thank you.</p>

答:
No book recommendation (really, any text-book on numerical integration should cover it). In general, it is possible to iterate the rule for a single integral. For example, trapezoid:

$$\int_a^{a+h}f(x)dx\approx \frac{h}{2}(f(a)+f(a+h))$$

$$\int_b^{b+h}\int_a^{a+h} f(x,y)dxdy\approx \int_b^{b+h} \frac{h}{2}(f(a,y)+f(a+h,y))dy\approx \frac{h^2}{2^2}(f(a,b)+f(a+h,b)+f(a,b+h)+f(a+h,b+h))$$

It's also possible to integrate the whole area using Monte-Carlo. In this case for example, envelope the area into a multi-dimensional cube, and count the hits from an multidimensional uniform variable covering the cube. The area is proportional to the hits. Then take the average of the function values on the hits and scale by the area.


问:
Is there a way to further simplify this using of the Chinese Remainder Theorem?
Both $p, q$ are prime

$x \equiv y \pmod p$

$x \equiv y \pmod q$

Using Chinese remainder theorem,

$M = p*q$

$x = a1*b1*M/p + a2*b2*M/q$

$x = a1*b1*q + a2*b2*p$

$a1 = y$

$a2 = y$

$b1$ is solution of $(M/p)*x \equiv 1 \pmod p$

i.e. soln of $qx \equiv 1 \pmod p$

Likewise
$b2$ is solution of $px \equiv 1 \pmod q$

$q$ & $q$ are prime - so they are co-prime to each other.

So
$b1 \equiv (1/q) \pmod p$  ----> (1)

$b2 \equiv (1/p) \pmod q$  ----> (2)

So
$x \equiv y*b1*q + a2*b2*q$  ---> (3)

From (1), (2) & (3) is it possible to arrive at

$x \equiv y \pmod{p*q}$

I know it's possible to prove the above in a much more easy way, but I was wondering if it's possible to prove this using CRT?

答:
Yes, rotely applying the common CRT formula yields the unique solution

$$ x\,\equiv\, y\,\overbrace{(p(p^{-1}\bmod q) + q(q^{-1}\bmod p))}^{\textstyle u} \pmod{pq}$$

Note $\,u\equiv 1\,$ both mod $\,p$ & $q.\,$ This has obvious solution $\,\color{#c00}{u\equiv 1}\pmod{pq}$ which is unique $\!\bmod pq\,$  by CRT. $ $ Thus $\,x\equiv y\color{#c00}u\equiv y\cdot \color{#c00}1\equiv y\pmod{\!pq},\,$ as claimed.

So rotely applying the CRT formula does indeed solve this constant case of CRT, but it's wasteful. The formula yields the existence of a solution, but that is trivial here since there is an obvious solution $\,x\equiv y\pmod{\!pq}.\,$ So here we need only uniqueness, i.e. proof that this obvious solution is the only solution $\!\bmod pq,\,$  for which see the various proofs  of CCRT = Constant-case CRT.


问:
Intuition behind Group Action
I am going over the group action and unfortunately I am not sure if I understand it very well. So for this reason, I have a few questions that possibly could help me to understand this concept better. A group action is defined as: Group $G$ acts on a set $X$ if there is a homomorphism $\sigma:G \rightarrow S_X$. I am aware group action has other definitions but for me this one is easier to understand.

Now $S_X$ is the group of permutations of $X$. It is not crystal clear what is meant by this to me. Is $S_X$ the same thing as $S_{|X|}$? When we talk about $S_n$, I have a clear mental picture; $n$ is an integer and I think of all the permutations with $n$ elements. But in $S_X$, $X$ is not an integer but a set. So suppose $X=\{1,2,3\}$. Is $S_X$ the same thing as $S_3$ in this case?

The other part of my question is about what the definition implies: it says "... if there is a homomorphism". Is it possible that homomorphism does not exist? Is the homomorphism unique? Or is it possible to have more than one homomorphism? Once we specify a group and a set, how do we find such homomorphism(s)? Is there a requirement for the size of the group and the set to make this definition work, for instance $|G| \geq |X|$?

Please let me know if you would like me to clarify my question. Any help with making a better intuition is appreciated!

答:
As for the first part of your question. If $X$ is finite of size $n\in\Bbb N$, then there is a bijection $f:\{1,\dots,n\}\to X$; build the map $\varphi\colon S_X\to S_n$ by $\varphi(\sigma):=f^{-1}\sigma f$; note that:

$f^{-1}\sigma f\in S_n$ (good definition);

$\varphi(\sigma)=\varphi(\tau)\Rightarrow f^{-1}\sigma f=f^{-1}\tau f\Rightarrow \sigma=\tau$ (injectivity);

for every $\alpha\in S_n$, $\alpha=\varphi({f\alpha f^{-1}})$ (surjectivity);

$\varphi(\sigma\tau)=f^{-1}\sigma\tau f=f^{-1}\sigma(ff^{-1})\tau f=(f^{-1}\sigma f)(f^{-1}\tau f)=\varphi(\sigma)\varphi(\tau)$ (group homomorphism)

Therefore, $S_X\cong S_n$ (noncanonically, being the isomorphism dependent on the bijection $f$).

Title: Intuition behind Group Action


问:
Matrix sums along rows and columns
I have two vectors $x$ and $y$ of length $m$ and $n$ respectively.

I'm trying to determine whether a matrix $N$ of dimension $(m \times n)$ exists, such that the sum along its i'th row is equal to $x_i$ and the sum along every column $j$ (possibly with elements weighted by $\omega_{ij}$) is equal to or greater than $y_j$

Can anyone give me some pointers?

答:
The general case will be as difficult as solving a linear system of $m+n$ equations.

For a special case with all entries of $y$ are positive, and the sum of entries of $x$ is at least the sum of entries of $y$, let
$$m_{ij}=x_i \cdot \frac{y_j}{\sum\limits_{k=1}^n y_k}, \ \ 1\leq i\leq m,  \ \ 1\leq j\leq n.$$

Title: Matrix sums along rows and columns


问:
Existence of a $k$-coloring of a complete graph with no monochromatic subgraph
I'm currently working on the exercises of Noga Alon's book "The Probabilistic method". One exercise said that we are given a graph $H$ with $p$ vertices and that there exists a graph $G$ with $n > p$ vertices and $m$ edges containing no copy of $H$.

We have to prove that if we have $k > \frac{n^2ln(n)}m$ and we can find a $k$-coloring of a complete graph $K_n$ with $k$ colors then there exists a graph with no monochromatic $H$. I know this should be proved with probabilistic method, I just don't know how. Could any one give me a hint?

答:
Suppose we have $k$ copies of $G$, $i.e. $ $G_{1}$,$G_{2}$, $...$ $G_{k}$, where all edges of  $G_{i}$ are colored by color $i$. Now successively and randomly lay $G_{i}$ on top of $K_{n}$.  $i.e. $ color a random copy of $G_{i}$ in $K_{n}$ with color $i$. We ignore the previous coloring of the edge that has been colored more than once. The very last coloring of the edge matters. If there is a monochromatic subgraph of $K_{n}$, then the subgraph  must be a subgraph of $G_{i}$ which contains no subgraph $H$. That is to say, coloring the $K_{n}$ this way, there can be no monochromatic $H$.  Thus, it is sufficient to show it is possible to color the whole $K_{n}$ in this way.

Let $X_{e}$ be the random variable which indicates whether  $e\in{E}$ is colored by some $G_{i}$.

$X_e=\cases {1,&if $e\in{E}\wedge{e\notin{\cup{G_{i}}}}$\cr
0,&\text{else}\cr}$

It's easy to see $E[X_e]=\left(1-\frac{m}{\binom{n}{2}}\right)^{k}.$  Let $X$ be the random variable which indicates the number of such edges. By the linearity of expectation, we have $E[X]=\sum_{e\in{E}}[X_e]=\binom{n}{2}\left(1-\frac{m}{\binom{n}{2}}\right)^{k}$<=$\frac{n^2}{2}\left(1-\frac{2m}{n^2}\right)^k$<=$\frac{n^2}{2}e^\frac{-2km}{n^2}$, given that $k>\frac{n^2\ln{n}}{m}$, the above inequality$\frac{n^2}{2}e^\frac{-2m}{n^2}$<$\frac{n^2}{2}e^{-2\ln{n}}=\frac{1}{2}$.
 So we have a $X$ such that $X<\frac{1}{2}$, $i.e. X=0$, which means it's possible to color all of the edges of $K_{n}$ in this way.


问:
Measurable cover
For any bounded set $E$ there is a $G_{\delta}$ set $G$ s.t.
$$E \subseteq G \text{ and } m^*(G)=m^*(E)$$
This is a problem from Royden's Real Analysis. I have finished the proof for this but I was wondering:

Does $E$ need to be bounded?

Does $E$ need to be measurable?

Since in my proof, I didn't use the condition that $E$ is bounded or measurable. So can I conclude that for $any$ set $E$, the result still holds?

Thanks!

答:
I think here is another proof when $E$ is a bounded set of real numbers.

As $E$ is bounded, so $E \subset [\alpha, \beta]$, where $\alpha := \inf E$ and $\beta := \sup E$. Let us put $G := [\alpha, \beta]$. Then we have $E \subset G$, which implies that
$$
m^*(E) \leq m^*(G); \tag{1}
$$
moreover, for any $\epsilon > 0$, $E \not\subset [ \alpha + \epsilon/2, \beta - \epsilon / 2 ]$, and so
$$ m^*(E) \not\leq m^*\big( [ \alpha + \epsilon/2, \beta - \epsilon / 2 ] \big). $$
But since both $m^*(E)$ and $m^*\big( [ \alpha + \epsilon/2, \beta - \epsilon / 2 ] \big)$ are real numbers, we must have
$$ m^*(E) >  m^*\big( [ \alpha + \epsilon/2, \beta - \epsilon / 2 ] \big) = (\beta - \epsilon/2) - (\alpha + \epsilon / 2) = \beta - \alpha - \epsilon = m^*(G) - \epsilon, $$
which in turn implies
$$
m^*(G) < m^*(E) + \epsilon.
$$
Since the last inequality holds for any $\epsilon > 0$, we must have
$$
m^*(G) \leq m^*(E). \tag{2}
$$
From (1) and (2) we have
$$
m^*(E) = m^*(G).
$$

Moreover, since we cna write this set $G$ as
$$
G = [\alpha, \beta] = \bigcap_{n=1}^\infty \left( \alpha - \frac{1}{n}, \beta + \frac{1}{n} \right),
$$
therefore $G$ is indeed a $G_\delta$-set.

Title: Measurable cover


问:
closed subscheme of $\mathbb{P}_Z^1$ which is affine over $Z$
Given an algebraically closed field $k$ and a scheme $Z$ over $k$. Consider the projective scheme $X=\mathbb{P}^1_Z=\mathbb{P}_k^1\times_kZ$ and $Y\subset X$ a closed subscheme. Assume that $Y$ does not contain any closed fibre of the projection $p:X\rightarrow Z$, try to show that $\pi:=p|_Y:Y\rightarrow Z$ is an affine morphism, i.e. one can find an affine covering $\{U_i\}$ of $Z$ such that each $\pi^{-1}(U_i)$ is affine.

My attempt:

In the extremely special case $Z=\text{Spec}(k)$, one has $X=\mathbb{P}_k^1$ and $X_z\cong\mathbb{P}_k^1$. Here any closed subscheme $Y$ of $X$ satisfies the condition that $X_z\not\subseteq Y$. So we want the canonical map $Y\hookrightarrow\mathbb{P}_k^1$ to be affine, which is clearly true.

Now let $Z=\text{Spec}(A)$ over $k$, we also have $X_z\cong\mathbb{P}_k^1$ and now $X=\mathbb{P}_A^1$. We know that any closed subscheme of $\mathbb{P}_A^1=\text{Proj}(A[x_0,x_1])$ is determined by a homogeneous ideal $I\subset A[x_0,x_1]$, i.e.
$$Y=V_+(I)=\text{Proj}(A[x_0,x_1]/(I))$$
Then I want to define a scheme $Y_k$ such that $Y=Y_k\times_kZ$. If I can do it, them I am done. However,  I have no ideal how to apply the setting $X_z\not\subseteq Y$ here.

答:
In order to make it complete, I combine Pol van Hoften's comment and hm2020's answer here.

The map $\pi:Y\rightarrow X\rightarrow Z$ is proper because the projection $p:X=\mathbb{P}^1_Z\rightarrow Z$ is proper and $Y$ is closed in $X$ (the closed inclusion $i:Y\subset X$ is proper).

Pick any closed point $z\in Z$, one can see that the fibre $X_z\cong\mathbb{P}_k^1$. By our setting, $Y$ does not contain $X_z$, so the fiber $Y_z$ of $\pi$ over $z$ is a real closed subset of $X_z\cong\mathbb{P}_k^1$. It follows that $\dim(Y_z)=0$. It also follows that $Y_z$ is noetherian (closed subscheme of a noetherian scheme is noetherian). Therefore, $Y_z$ must be finite and thus the morphism $\pi:Y\rightarrow Z$ is quasi-finite.

A proper and quasi-finite morphism is finite, and finite morphisms are affine.


问:
Every natural number $n$ can be written as $n=s-t$ with $\omega(s)=\omega(t)$
Can we prove the following statement ?

Let $\omega(m)$ denote the number of all distinct prime factors of a positive integer $m$.

Prove that every  natural  number $n$ can  be  written  as $n=s-t$, where $s$ and $t$ are positive integers with $\omega(s)=\omega(t)$, in other words, as the difference of two positive integers with the same number of distinct prime factors.

If $n=1$ , we can choose $\ s=3\ $ and $\ t=2$.

If $n$ is even , we can choose $\ s=2n\ $ and $\ t=n$.

答:
If $\ n\ $ is even, see user3733558's comments or Servaes's answer for a much simply argument than the one I originally gave.

If $\ n\ $ is odd, let $  q\ $ be the smallest odd prime not dividing $\ n\ $.  Then all the odd prime factors of $\ q-1\ $ must be factors of $\ n\ $, and both $\ qn\ $ and $\ (q-1)n\ $ must therefore have exactly one more prime factor than $\ n\ $.


问:
Gram-Schmidt algorithm to find orthonormal basis
I must use the Gram-Schmidt algorithm to find an orthonormal basis for Sp$\{1,x,x^2 \}$ in $L^2[-1,1]$.

Suppose the elements of the orthonormal basis are $e_1, e_2$ and $e_3$. Then they must be pairwise orthogonal and $\|e_i \|=1$ for each $i$.

I thought of trying with the standard inner product $<f,g>=\int_X \mid f \bar{g} \mid d\mu$ on $L^2$, where $ X=[-1,1]$. However I have no idea of how to determine the orthogonal basis.

答:
See Gram-Schmidt's ortogonalization process for a description.

You get (note that everything is real, so you don't have to worry about complex conjugates)
\begin{align}
u_2 &= v_2 - \operatorname{proj}_{e_1} v_2 \\
    &= v_2 - \langle e_1,v_2 \rangle e_1 \\
    &= x - \left( \int_{-1}^1 x \cdot \frac{1}{\sqrt2} \,dx\right) = x
\end{align}
and
$$
e_2 = \frac1{\|v_2\|}v_2 = \left( \int_{-1}^1 x^2\,dx \right)^{-1/2} x = \frac{x\sqrt3}{\sqrt2}.$$

I'll leave the computation of $e_3$ to you.


问:
Percentage calculation: What's the difference between using 1.08875 and 0.08875?
Use case is for tax calculation.

Tax is at 8.875%. Using $100 tax-inclusive amount, I get different tax amounts using these formulas:

First:
$$\mathrm{taxAmount} = \$100\times 0.08875 = \$8.875$$
Second:
$$\mathrm{taxAmount} = \$100 - (\frac{\$100}{1.08875}) = \$8.15$$

I find the first formula is the correct one and straightforward but a colleague suggests to use the second.

What's the difference and which one is correct?

答:
The first formula is what you should use if $\$100$ is the price before tax, and you are calculating how much tax you need to add on. Thus you have prices of $\$100$ (net) and $\$108.875$ (gross, i.e. net plus tax).

The second formula is what you should use if $\$100$ is the price after tax, and you want to work out how much of this was tax. So here you have prices of $\$100$ (gross) and $\$91.85$ (net, i.e. gross minus tax).

As a check, if you apply the first formula to a net price of $\$91.85$ you should get back to $\$100$ gross (there is a small error because the $\$8.15$ in your post is rounded to the nearest cent).


问:
Change in the vertex connectivity of a given graph
This is a homework problem I am trying to solve:
What impact does the following operations (prove some bounds on the amount of change and provide examples showing that the bounds are tight) has on the vertex connectivity of a graph:
(a) adding a single edge to a graph (without adding vertices),
(b) removing a single edge,
(c) removing a single vertex,
(d) contracting an edge.

so what I have tried is that I have taken a random graph and have tried to see what happens to the vertex connectivity. In 1st case, I observed that when a single edge (without adding vertices) is added depending on where it gets added the vertex connectivity can be increased but it does not decrease. In the 2nd case, removing a single edge might decrease the vertex connectivity i.e make it 0 (like in the case of a star-like structure) same can be seen in the case of removal of a single vertex. I do not much understand contracting an edge. Even though through various examples I can understand what impacts is having but I am unable to prove or find the bounds are tight or not. I do understand a few specific bounds like $k(G)\leq\lambda(G)\leq\delta (G)$. But I still find myself stuck here.

Any help is appreciated.

答:
To think about this problem, instead of starting with an arbitrary graph and trying these operations, I suggest imagining a graph and a cut in the graph and doing these operations. I'll walk you through the reasoning for one of them: removing a vertex.

There are three cases:

If we remove a vertex that's part of the cut, then the number of vertices in the cut goes down by $1$.

If we remove a vertex that's not part of the cut, then the number of vertices in the cut is not changed.

An exception: if we remove the last vertex on one side of the cut, then the cut stops being a cut.

Each graph has a lot of different cuts, some of which are the minimum cuts. If the above things happen to every single one of the cuts, then the vertex connectivity could decrease by $1$ (if all minimum cuts ended up in the first case), stay the same (if some minimum cuts are in the second case), or increase arbitrarily (if all minimum cuts end up in the third case, and the next smallest cut is very large).

You should then be able to draw examples of each of these things happening. (For the arbitrary increase, you'll need an infinite family of similar examples to show that the increase can be arbitrary.)

Contracting a vertex is the weirdest operation and it's hard to get intuition for. The cases are going to be if you

contract an edge between two vertices in the cut;

contract an edge between two vertices not in the cut;

contract an edge between a vertex in the cut and a vertex not in the cut.

These are actually going to act a lot like deleting a vertex; your examples might look slightly different.


答:
If $n$ is even then $n$ and $2n$ have the same number of prime factors, and so
$$n=2n−n,$$
shows that $n$ is the difference of two natural numbers having the same number of prime factors.

If $n$ is odd then let $q$ be the smallest odd prime number not dividing $n$. Then $q+1$ is even, and its odd factors all divide $n$. It follows that $qn$ and $(q+1)n$ have the same number of prime factors, and so
$$n=(q+1)n−qn,$$
shows that $n$ is the difference of two natural numbers having the same number of prime factors.

Title: Every natural number $n$ can be written as $n=s-t$ with $\omega(s)=\omega(t)$

Body:
Can we prove the following statement ?

Let $\omega(m)$ denote the number of all distinct prime factors of a positive integer $m$.

Prove that every  natural  number $n$ can  be  written  as $n=s-t$, where $s$ and $t$ are positive integers with $\omega(s)=\omega(t)$, in other words, as the difference of two positive integers with the same number of distinct prime factors.

If $n=1$ , we can choose $\ s=3\ $ and $\ t=2$.

If $n$ is even , we can choose $\ s=2n\ $ and $\ t=n$.


问:
Borel set in an uncountably product space with product topology
Let $I=[0,1]$ be with its usual topology and $I^I$ be with product topology.

Let $1_A(x)$ be the indicator function of set $A$, i.e., $1_A(x)=1$ for $x\in A$ and $1_A(x)=0$ for $x\notin A$.

Let $f(x):=1_{\{x\}}$, so that $f$ defines a function from $I$ into $I^I$. Show that the range of $f$ is a Borel set in $I^I$.

My efforts:

For any $x\in I$, $U_x:=\{g\in I^I:g(x)\neq0\}$ is open. Thus $U_x^C:=\{g\in I^I:g(x)=0\}$ is closed and Borel. Similarly, $V_x^C:=\{g\in I^I:g(x)=1\}$ is closed and Borel.

Since arbitrary intersection of closed sets is closed, $V_y^C\cap\bigcap_{x\in I,x\neq y}U_x^C=\{1_{\{y\}}\}$ is closed and Borel for any $y\in I$.

Then I don't know how to continue.

答:
Let the image of $f$ be $A$. We will show that $A$ is closed and hence Borel.

To show $A$ is closed, we will show $A$ equals its closure, i.e., $A=\bar{A}$.

To show $A=\bar{A}$, we will show: if $y\in\bar{A}$, then $y\in A$. It is easier to show the contrapositive: if $y\notin A$, then $y\notin\bar{A}$.

For $y\notin A$, there are two cases.

(1) $y(x)=a$ with $a\notin\{0,1\}$ for some $x\in I$. There exists a neighborhood (nbhd) $U_a$ of $a$ such that $U_a\cap\{0,1\}=\emptyset$. Let $U=\prod_{i\in I}X_i$ with $X_i=U_a$ for $i=x$, $X_i$ is open for finitely many $i$, and $X_i=I$ for the remaining $i$. Then $U$ is a nbhd of $y$ and $U\cap A=\emptyset$. Therefore, $y\notin\bar{A}$.

(2) $y(x)\in\{0,1\}$ for all $x\in I$ but $y(x)=1$ for more than two $x\in I$, say, $x_1$ and $x_2$, i.e., $y(x_1)=y(x_2)=1$. There exists a neighborhood (nbhd) $U_1$ of $1$ such that $0\notin U_1$. Let $U=\prod_{i\in I}X_i$ with $X_i=U_1$ for $i=x_1,x_2$, $X_i$ is open for finitely many $i$, and $X_i=I$ for the remaining $i$. Then $U$ is a nbhd of $y$ and $U\cap A=\emptyset$. Therefore, $y\notin\bar{A}$.


问:
Clarifying Polygonal Paths in Complex Analysis
Introduction to Complex Analysis by H. A. Priestley (Revised ed.) shows a diagram (page 40) which is a polygonal path in a set $G$ (link below). Their proof claims we can polygonally pave our way from $z$ to $a$ by using the fact that open discs centred at $z$ enclose a point $w$ such that the line segment $[z,w]\subset D(z;r),r>0$. I am confident in that notion, but maybe it's their diagram which confuses me.

It displays large line segments paving through the set $G$ which if you were to draw open discs centred from the original point, the open disc is then not contained by $G$. I assume they could have used smaller open discs, therefore, more $w$s to get to $a$.

Any clarification on this topic would help me out a lot. Thank you very much.

答:
I assume the statement you're after really is that $G$ open and connected implies $G$ path-connected which, in turn, implies points in $G$ can, in fact, be connected by polygonal paths.

The first implication is simply the fact that $G$ is locally path-connected (connected and locally path-connected spaces are, necessarily, path-connected).

To get the second implication, pick some path $\gamma$ from $z$ to $a$. Since $\gamma$ is compact and $\partial G$ is closed, there exists some $\varepsilon>0$ such that $d(\gamma,\partial G)>\varepsilon$. Accordingly, $(D(x,\varepsilon))_{x\in \gamma}$ is an open cover of $\gamma$ in $G$ and we can apply compactness to get that $z,x_1,...,x_n,a$ such that $(D(x_j,\varepsilon))_{1\leq j \leq n}$ forms a cover of $\gamma$ in $G$. We may, without loss of generality, assume that the ordering of the $x_j$ is such that $D(x_j,\varepsilon)\cap D(x_{j+1},\varepsilon)\neq \emptyset$ for every $j$, $z\in D(x_1,\varepsilon)$ and such that $a\in D(x_n,\varepsilon)$. Note that if our initial path was poorly chosen, this might imply that the $x_j$'s are non-distinct but that doesn't matter for the argument.

At this point, however, it's clear that there is a straight line from $z$ to $x_1$ in $G$, a straight line of $a$ to $x_n$ in $G$ and that there is a polygonal path in $G$ consisting of two line segments between $x_j$ and $x_{j+1}$ for every $j$. Concatenating these paths yields a polygonal path between $z$ and $a$.


问:
Not sure what this notation means
I have the following equation attached. I know that the double lines means that I need to find the Euclidean distance between $P$ and $P'$ however I am not sure what the upper and lower value of $2$ mean. I think that the upper value of $2$ means that I should perform a power of $2$ after finding the Euclidean distance, but what does the lower value of $2$ indicate that I should do?

答:
The value $\|x\|_2$ is defined as the Euclidean norm of the vector $x$, i.e.

$$\|x\|_2 = \sqrt{\langle x,x\rangle}=\sqrt{x^Tx} = \sqrt{x\cdot x}.$$

Note that every norm also induces a metric, and the metric, induced by the Euclidean norm, i.e. the mapping $$(x, y) \mapsto \|x-y\|_2$$

is called the Euclidean metric.

Title: Not sure what this notation means


答:
The Euclidean norm is a specific example of a family of norms indexed by non negative integers. For $\mathbb{R}^n$ define the norm $||\cdot ||_p$ by

$||x||_p = \left(\sum_{i=1}^n |x_i|^p\right)^{\frac{1}{p}}$.

Then we get the Euclidean norm for $p=2$.

Title: Not sure what this notation means

Body: I have the following equation attached. I know that the double lines means that I need to find the Euclidean distance between $P$ and $P'$ however I am not sure what the upper and lower value of $2$ mean. I think that the upper value of $2$ means that I should perform a power of $2$ after finding the Euclidean distance, but what does the lower value of $2$ indicate that I should do?


问:
Does Eisenstein's criterion apply when some coefficients are $0$?
I've just been reading the Wikipedia page on Eisenstein's criterion; in summary it says that the polynomial with integer coefficients
$$a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$$
is irreducible over the rationals if there exists a prime number $p$ such that

$p$ divides each $a_i$ for $0\leqslant i<n$

$p$ does not divide $a_n$ and

$p^2$ does not divide $a_0$

I've just been wondering:  does Eisenstein's criterion also apply even when some of the coefficients are $0$? Do we consider $p$ to always divide $0$?

Thank you for your help.

答:
Both in general, and in the context of Eisenstein in particular, we always have $n$ divides $0$ for any integer $n$.

The definition of $a\mid b$ is

There is an integer $x$ such that $b=xa$.

For $b=0$, it always works to set $x=0$, no matter what $a$ is.


答:
A number $a$ divides a number $b$ if there exists some value $x$ such that $a\cdot x = b$.

By definition, all numbers divide the number $0$, since the value $x=0$ satisfies the condition for all of them.

Title: Does Eisenstein's criterion apply when some coefficients are $0$?

Body:
I've just been reading the Wikipedia page on Eisenstein's criterion; in summary it says that the polynomial with integer coefficients
$$a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$$
is irreducible over the rationals if there exists a prime number $p$ such that

$p$ divides each $a_i$ for $0\leqslant i<n$

$p$ does not divide $a_n$ and

$p^2$ does not divide $a_0$

I've just been wondering:  does Eisenstein's criterion also apply even when some of the coefficients are $0$? Do we consider $p$ to always divide $0$?

Thank you for your help.


问:
sine function is not closed
I have searched on MSE but did not find any clear solution for this problem. I have found in a blog that

sin map from $\mathbb R$ into $\mathbb R$ (or $[-1,1]$) is not closed. He/She claimed that

The set $F=\displaystyle \bigcup_{n \in \mathbb{N}} [2n\pi + \frac{1}{n+1},2n\pi + \pi – \frac{1}{n+1}]$ is closed, but $f(F)=(0,1]$

Can someone explain this to me as I think $F$ is not closed in this representation? Is this true?

Special thanks go to Jean-Pierre Merx and here is the question

答:
Based on your definition of $F$, we have
$$\Bbb R\setminus F=(-\infty,1)\cup\bigcup_{n\in\Bbb N}((2n+1)\pi-\frac{1}{n+1},(2n+2)\pi+\frac{1}{n+1})$$
Since the complement of $F$ is a union of open intervals (open in $\Bbb R$), $F$ is closed, but on those intervals, $\sin(F)$ runs from $0$ (exclusive) to $1$ (inclusive), i.e., $(0,1]$. This is clearly NOT closed in $\Bbb R$ equipped with the Euclidean topology. This proves the claim.

I'm assuming that $\Bbb N$ contains $0$.

Title: sine function is not closed


问:
How to prove that the Gelfand transform is determined by absolutely convergent Laurent series?
A self-contained context can be found in the picture below (from the book An Introduction to Local Spectral Theory by Kjeld Bagger Laursen, Michael Neumann). I'm stuck at the authors' claim that "a simple argument shows that there are no others" (bold text). The term "Beurling algebra" is irrelevant here but I'll add the definition at the bottom.

Here's my attempt before being stuck: Suppose there are others, namely $\varphi:\ell^1(\omega) \to \mathbb{C}$ is another nontrivial multiplicative linear functional (complex homomorphism) that cannot be represented in the form $\varphi_\lambda$, then there exists $y \in \ell^1(\omega)$ such that for all $z \in R_\omega$, we have $\varphi(y) \neq \sum\limits_{n = -\infty}^{\infty} y_n z^n$... Which should gives a contradiction but I don't know how.

Is my attempt sensible at all? I'm afraid my assumption is not perfect for a contradiction and maybe there is something wrong with it. If my attempt is OK so far, how could I go further? Is it possible to "solve" the Laurent series for given the complex number $\varphi(y)$ and a sequence $\{y_n\}$? I have never seen that in any complex analysis theory.

Anyways, any hint or solution would be sincerely appreciated!

We now turn to the case of Beurling algebras on the additive group of integers. In this case, the weight function appears as a two-sided sequence $\omega=(\omega_n)_{n\in\mathbb{Z}}$ of positive real numbers for which $\omega_{m+n}\leq\omega_m\omega_n$, for all $m,n\in\mathbb{Z}$, and $L^1(\omega)$ becomes the space $\ell^1(\omega)$ of all two-sided complex sequences $x=(x_n)_{n\in\mathbb{Z}}$ for which $\sum_{n\in\mathbb{Z}}|x_n|\omega_n$ is finite.

For $x,y\in\ell^1(\omega)$, the convolution product $x*y$ is the sequence given by
$$(x*y)_n=\sum_{j=-\infty}^\infty x_{n-j}y_j\quad\text{for all}\ n\in\mathbb{Z}.$$
Obviously, $\ell^1(\omega)$ is a commutative Banach algebra with identity, and therefore coincides with its multiplier algebra. We note that the spectrum of the two-sided Beurling sequence algebra $\ell^1(\omega)$ may be identified with the annulus
$$R_\omega:=\{\lambda\in\mathbb{C}:\ \rho_\omega^-\leq|\lambda|\leq\rho_\omega^+\},$$
where, due to the submultiplicativity of the weight sequence $\omega$, the two limits
$$\rho_\omega^+:=\lim_{n\to\infty}\omega_n^{1/n}=\inf_{n>0}\omega_n^{1/n}\quad\text{and}\quad\rho_\omega^-:=\lim_{n\to-\infty}\omega_n^{1/n}=\sup_{n<0}\omega_n^{1/n}$$
exist and satisfy $0<\rho_\omega^-\leq\rho_\omega^+<\infty$.

In fact, given an arbitrary $\lambda\in R_\omega$, it is immediate that the definition
$$\varphi_\lambda(x):=\sum_{n=-\infty}^\infty x_n\lambda^n\quad\text{for all}\ x\in\ell^1(\omega)$$
yields a non-trivial multiplicative linear functional $\varphi_\lambda$ on $\ell^1(\omega)$, and a simple argument shows that there are no others. Thus elementary Gelfand theory allows us to think of the elements of $\ell^1(\omega)$ as absolutely convergent Laurent series on the annulus $R_\omega$.

Let $G$ be a locally compact abelian group endowed with a weight function $\omega$, where $\omega$ is positive and for all $s,t \in G$, $\omega(s+t) \leq \omega(s)\omega(t)$ (i.e., $\omega$ is submultiplicative). Let $L^1(\omega)$ denote the space of all (equivalence classes of) Borel measurable functions $f:G \to \mathbb{C}$ for which $f\omega \in L^1(G)$. Endowed with convolution as multiplication and the weighted $L^1$-norm $\lVert \cdot \rVert_\omega$ given by

$$
    \lVert f \rVert_\omega = \int_G |f(t)|\omega(t)dm(t),
$$

the space $L^1(\omega)$ turns into a commutative Banach algebra, the Beurling algebra for the weight $\omega$. In this question, $G$ is actually $\mathbb{Z}$.

答:
I found the solution and hope my solution will help some random people.

The key is the generator. Note we by this product rule, we have the unit:

$$
e:e_n = \begin{cases} 1 & n=0, \\ 0 &\text{otherwise}. \end{cases}
$$

Also we have the generator

$$
g:g_n = \begin{cases} 1 & n = 1, \\ 0 &\text{otherwise}. \end{cases}
$$

and its inverse is given by

$$
g^{-1}:g_n^{-1} = \begin{cases} 1 & n = -1, \\ 0 &\text{otherwise}. \end{cases}
$$

Note for $k>0$ and $g^{\ast k} = g \ast \cdots \ast g$ ($k$ times), the $k$th component is $1$ and other components are $0$. Similarly, the $-k$ th component of $g^{\ast -k} = g^{-1} \ast \cdots \ast g^{-1}$ (also $k$ times) is $1$ and other components are $0$. Now for any $x \in \ell^1(\omega)$, we have a unique expansion (in the form of Laurent series) given by

$$
x = \sum_{k=-\infty}^{\infty}x_k \cdot g^{\ast k}
$$

where $\cdot$ is merely the scalar multiplication.

For any complex homomorphism $\varphi$, suppose $\varphi(g)=\lambda$, then the correspondence of addition and multiplication gives us

$$
\varphi(x)= \sum_{k=-\infty}^{\infty}x_k\lambda^k.
$$

This is exactly what we want. $\lambda$ must be in $R_\omega$, since if not the absolute convergence of $\ell^1(\omega)$ would be violated.


问:
Relative scheme structure on a specific scheme related to the symmetric product of curves.
Given a smooth algebraic curve $C$ with a closed point $p$. The symmetric product $\text{Sym}^i(C)$ has a closed subvariety $\text{Sym}^{i-1}(C)$ which its embedding is given by adding the point $p$. Let $\text{Spec}(A)$ be an affine in $C$ containing $p$. Is the $n$-th nilpotent thickening of $\text{Sym}^{i-1}(C)$ in $\text{Sym}^i(C)$ a scheme over $\text{Spec}(A/m^{n+1})$ where $m$ is the maximal/prime ideal corresponding to $p$?

Note that zeroth nilpotent thickening is the subscheme itself, the first nilpotent is given by second powers and so on.

This problem seems trivial intuitively, since symmetric product is the quotient of product, if we look at the product and the inverse image of $\text{Sym}^{i-1}(C)$, its $n$-th nilpotent thickening is the union of bunch of $\text{Spec}(A/m^{n+1})$-schemes (each irreducible component). It seems that gluing these schemes (The ideal corresponding to the gluing is going to be the product of ideals) will have a natural $\text{Spec}(A/m^{n+1})$-scheme structure. There also seems to be symmetry to this $\text{Spec}(A/m^{n+1})$-scheme structure that descends to the symmetric product.

答:
Question: "Is the n-th nilpotent thickening of $Sym^{i−1}(C)$ in $Sym^i(C)$ a scheme over $Spec(A/m^{n+1})$ where $m$ is the maximal/prime ideal corresponding to p?"

Answer:  Note that the symmetric product is not functorial for prime ideals as you claim if you use the symmetric group to define the symmetric product:

If $k$ is the base field and $C$ is a curve over $k$, let $U:=Spec(A) \subseteq C$ be an affine open subscheme. Let us define the symmetric product locally using the following construction: Define

$$A_i:=A^{\otimes_k i}:=A\otimes_k A\otimes \cdots \otimes_k A.$$

There is a canonical action of the symmetric group $S_i$ on $i$ elements on $A_i$, and by definition

$$ Sym^i(U):= Spec(A_i^{S_i})$$

is the spectrum of the invariant ring $A_i^{S_i}$. Given an arbitrary closed point $\mathfrak{m} \subseteq A$ you get a canonical map

$$ U^{\times_k (i-1)} \times_k Spec(\kappa(\mathfrak{m})) \rightarrow U^{\times_k i}$$

and if $\mathfrak{m}$ is a $k$-rational point you get a canonical map

$$i_{\mathfrak{m}}: U^{\times_k (i-1)} \rightarrow U^{\times_k i}$$

by "adding the point $\mathfrak{m}$", but this map does not induce a map of symmetric products for general points as you claim:

Example: Let $A:=k[x]$ with $A_3:=A^{\otimes_k 3}\cong k[x_1,x_2,x_3]$ and let $\mathfrak{m}:=(x_3-a_3)\subseteq k[x_3]$ with $a_3\in k$ be a $k$-rational point.
It follows

$$A_3^{S_3}\cong k[s_1,s_2,s_3]$$

where $s_1:=x_1+x_2+x_3, s_2:=x_1x_2+x_1x_3+x_2x_3, s_3:=x_1x_2x_3$

and $A_2^{S_2}\cong k[t_1,t_2]$ with

$$t_1:=x_1+x_2, t_2:=x_1x_2$$

and it follows the map

$\phi: A_3 \rightarrow A_2$ defined by $\phi(x_3):=a_3$ maps as follows:

$$\phi(s_1):=x_1+x_2+a_3:= t_1+a_3$$

$$\phi(s_2):= a_3t_1+t_2$$

$$\phi(s_3):= a_3t_2,$$

hence you do get an induced map of symmetric products

$$\phi^*: Sym^2(U) \rightarrow Sym^3(U)$$

in the case when your point $\mathfrak{m}$ is $k$-rational. You do not get an induced map $\phi^*$ for a general point:

Example: Let $k$ be the real numbers and let $K$ be the complex numbers. Let $(p(x)) \subseteq k[x_3]$ be a maximal ideal with residue field $K$ (ie we map $x_3$ to $z_3\in K$). In the above construction you get a map

$$k[x_1,x_2,x_3] \rightarrow k[x_1,x_2]\otimes_k K \cong K[x_1,x_2]$$

and an induced map

$$k[x_1,x_2,x_3]\rightarrow K[x_1,x_2]$$

inducing a map

$$k[s_1,s_2,s_3] \rightarrow K[t_1,t_2]$$

and

$$Sym^2(K\times_k U) \rightarrow Sym^3(U).$$

Hence for an arbitrary point $p\in C$, you may not view $Sym^{i-1}(C)$ as a closed subscheme of $Sym^i(C)$ by "adding the point $p$": There is a canonical map

$$ Sym^2(K\times_k U) \rightarrow Sym^2(U)$$

but no closed immersion

$$Sym^2(U) \rightarrow Sym^3(U)$$

in general.


问:
Cardinality of the set of all finite subsets of an infinite set.
Let $I$ be an infinite set. And let $\mathcal{I}$ be a set of finite subsets of $I$, this is $\mathcal{I}=\{J\subset I: |J|<\infty\}$. It is a known fact that  $|I|=|\mathcal{I}|$. This is equivalent to saying that there exists injective functions $f:I\longrightarrow \mathcal{I}$ and $g:\mathcal{I}\longrightarrow I$. One is very easy to find:

\begin{align*}
f:I&\longrightarrow \mathcal{I}\\
i&\longmapsto\{i\}
\end{align*}

But how can I define an injective function:
\begin{align*}
g:\mathcal{I}&\longrightarrow I\\
J&\longmapsto g(J)
\end{align*}

Any hint or idea would be appreciated.

答:
This fact, in the generality presented here, is equivalent to the axiom of choice. So a simple injection from $\cal I$ into $I$ cannot be written out explicitly when you're taking the whole generality.

Let's limit ourselves to sets that can be well-ordered, or rather to ordinals, or rather to initial ordinals (since we care about cardinality). Assuming the axiom of choice every set is equipotent with an initial ordinal, so that's fine.

If $\omega_\alpha$ is an initial ordinal, then every finite set can be thought of as a finite sequence. The knee-jerk reaction is to look at the ordinal $(\omega_\alpha)^\omega$, here the exponent is an ordinal exponentiation. That's absolutely fine, and we can prove that if $\alpha$ is an infinite ordinal then $\alpha$ and $\alpha^\omega$ have the same cardinality (again, ordinal exponentiation!)

But I want to go to a different route. Let's look at $\omega^{\omega_\alpha}$. We can prove, quite easily depending on the definitions and theorems at hand, that this ordinal is $\omega_\alpha$ itself. So it is enough to find an injection from the finite subsets of $\omega_\alpha$ into $\omega^{\omega_\alpha}$.

A direct definition of $\alpha^\beta$ is the order type of finite partial functions from $\beta$ to $\alpha$ which are decreasing. But this gives us a dead giveaway for an injection: given a finite set $J\subseteq\omega_\alpha$, write it in increasing order as $\{\xi_1,\dots,\xi_n\}$, and then simply map $\xi_i$ to $i$.

Therefore we found an injection from $\cal I$ to $\omega^{\omega_\alpha}$ as wanted.


问:
Projection expressed as a Matrix * a Vector
So, let's say we want the projection of vector v [2, 1] on to L (line L) L = C[2, 1]), where vector x is a vector that lies along the Line.

I know that one way to figure out the projection would be  C*vectorX (C representing some scaled version of vector x).

Then, a second way to figure out the projection would be to normalize vector x. then C = (vector v * vector x) * vector x (or vector x as vector u (normalized).

But the final way to figure it out would be to see the projection as a linear transformation. Projection is Matrix A * vector v. For some reason when expressed as a Matrix / Vector product, it is less clear/intuitive for me.

The key point of confusion: I'm not sure for example how we get from a normalized version of vector x to expressing it as a Matrix (let's say Matrix A). Where then we would simply multiply it by vector v.

Hopefully this is clear.

答:
A fact about projection matrices i hope you would find interesting is that they are always positive semi-definite. All eigenvalues are all ones (for a vector that is in column space) or zeros (for a vector that is not in column space). I am very new here. Sorry for my answer being improper.

I got your question better after reading it again. For $$P = A(A^T A)^{-1}A$$. $$ (A^TA)$$ does the weighting job here. If A were an orthogonal matrix it would be the identity. You know that for a vector v,$$v^Tv$$ is the square of the 2-norm of the vector. This could be the relation you are looking for.


问:
why do we need to prove that the limit of a contraction is a fixed point in the contraction mapping theorem
In this chapter they first prove that iteratively applying a contraction is a cauchy sequence. Since the metric space is complete we know that sequence converges, and it does so to a unique point. So why do they then additionally prove that the point of convergence is a fixed point of the contraction, isn't this already implied by the proving the sequence is cauchy. I feel like it's more of a corollary since it applies to any arbitrary function that is continuous in the domain.</p>

答:
So why do they then additionally prove that the point of convergence
is a fixed point of the contraction

Because that's what the theorem states, and therefore, that's what the proof should conclude with.

isn't this already implied by the original proof.

What do you mean by "original proof"? There is only one proof and it can be divided into the following parts:

Prove that the sequence, defined by $x_{n+1} = T(x_n)$ converges.

Prove that if a sequence, defined by $x_{n+1}=T(x_n)$, converges, then the limit of the sequence is a fixed point of $T$.

Conclude from 1 and 2 that $T$ has a fixed point.

Prove that $T$ can have at most one fixed point.

Conclude from 3 and 4 that $T$ has exactly one fixed point.

Now, sure, alternatively, the authors could cite point number 2 above as a separate lemma, and prove it before proving the main theorem, but I'd say it's cleaner this way, for two reasons:

The proof is short, so splitting it into subtheorems would not help readability

The lemma itself is not all that interesting.

Edit:

It is, of course, also possible to prove point 3 directly from 1 and from the fact that $T$ is continuous. However, this would most probably lead to a proof that is harder to follow, and possibly even longer than original.

But the mere fact that you can prove something without a particular sub-lemma does not mean that that particular lemma is "useless" for the proof. Technically, every proof in mathematics can be trimmed down to base principles, but at a great cost of readability.


问:
Obtain $\lim _{n \rightarrow \infty} \mathbb{E}\left\{\left|1-X_{n}\right|\right\},$ if it exists for $X_{1},X_{2},X_{3}.. RVs$ and under conditions
$X_{1}, X_{2}, \ldots .$ are nonnegative $\mathrm{RVs}$, such that $ \boldsymbol{X}_{n} \stackrel{\text { a.s. }}{\rightarrow} \mathbf{0}$ (a.s stands for almost sure convergence) and $\lim\limits _{n \rightarrow \infty} \mathbb{E}\left\{\boldsymbol{X}_{\boldsymbol{n}}\right\}=\mathbf{2}$, where $\mathbb{E}$ stands for expectation of a random variable RV.
I want to obtain
$$
\lim _{n \rightarrow \infty} \mathbb{E}\left\{\left|1-X_{n}\right|\right\}, \text { if it exists }
$$
There is no independence supposed here, I tried to go from the a.s convergence definition or simply by performing upper-bounds but I still fail and can't seem to see the use of the expectation convergence to $2$ which is obviously important here.

Any suggestions?

答:
Note that for $M\geq 1$,
$$
\mathbb{E}[|1-X_n|]=\mathbb{E}[(X_n -1)1_{X_n\geq M}]+\mathbb{E}[|1-X_n| 1_{|X_n|\leq M}]
$$
Now, for uniformly bounded variables, a.s. convergence implies $L^1$-convergence by the dominated convergence theorem. Thus,

$$
\lim_{n\to\infty} \mathbb{E}[|1-X_n| 1_{|X_n|\leq M}]=1
$$
and
$$
\mathbb{E}[(X_n-1)1_{X_n\geq M}]=\mathbb{E}[X_n 1_{X_n\geq M}]-\mathbb{P}[X_n\geq M]
$$
since $X_n\to 0$ a.s., we also have $X_n\to 0$ in probability and hence,
$$
\lim_{n\to\infty}\left(\mathbb{E}[X_n 1_{X_n \geq M}]-\mathbb{P}[X_n\geq M]\right)=\lim_{n\to\infty}\mathbb{E}[X_n 1_{X_n\geq M}]=\lim_{n\to\infty}\mathbb{E}[X_n]-\lim_{n\to\infty}\mathbb{E}[X_n1_{X_n<M}]=2.
$$

Hence,
$$
\lim_{n\to\infty}\mathbb{E}[|1-X_n|]=3.
$$


问:
Mathematics for Pleasure of a Beginner
I've just read "The Music of the Primes" by Marcus du Sautoy, it is worth a read. I'm not from a maths background, but I'd like to develop a deeper understanding of the concepts. The poetry of math is something I'd like  to experience and I feel like I didn't find as much joy in math as there is in the world of numbers.

Elliptical curve crypto-systems sounds like a direction I might want to head towards. I have no concept of where to start, or how in depth I would need to go to understand elliptical curves. Geometry was the most interesting math to me, but I am willing to move in different directions based on advice and materials.

In sum, I want to sit down with math again to give myself the chance at something I haven't experienced, the music of mathematics. There is no timeline or test on the horizon. What captivates you about math? What are the most interesting questions? Where would you recommend I start?

Thank you for any thoughts ahead of time.

答: That book is really inspiring. I have a background in physical chemistry but this book made me studying mathematics just because I found it pleasurable. I've started from basic undergraduate mathematics books and currently I'm studying G.C.Smith's "Introductory mathematics algebra and analysis" book.</p>

Title: Mathematics for Pleasure of a Beginner


问:
Proof of Boole's inequality in two event case
If we have the probability space $(\Omega, \mathscr{F}, \mathbb{P})$, how can we show that, for any $A, B \in \mathscr{F}$ we have: $$\mathbb{P}(A \cup B) \leq \mathbb{P}(A) + \mathbb{P}(B)$$

Forgive me if I have missed another question asking this, but I haven't been able to find a resource that demonstrates this proof using measure/ probability theory axioms. It is a basic question but I struggle to understand how to approach it.

答:
Here is one way to do it:
$$ \begin{align*}\mathbb{P}(A \cup B) &\leq \mathbb{P}(A \cup B) + \mathbb{P}(A \cap B) \\ &=
\mathbb{P}(A) + \mathbb{P}(B \setminus A) + \mathbb{P}(A \cap B) \\
&=\mathbb{P}(A) + \mathbb{P}(B)\end{align*}$$
using that
$$A\cup B = A \mathbin{\dot{\cup}} (B\setminus A), \quad B = (B\setminus A)\mathbin{\dot{\cup}} (A\cap B)$$


问:
Pointwise convergence implies $L^{2}$ convergence
If we have  a sequence of bounded functions $f_{n}$ converging almost everywhere to another bounded function $f$ in a finite measure space such that

$$ |f_{n}(t)| \leq c$$ for some constant $c$. Then

$$ \int |f_{n}(t) - f(t)|^{2} d\mu$$ goes to zero.

Since we are in a finite measure space, the constants are integrable and hence DCT applies. However, it only gives the result that

$$ \int f_{n}(t)d\mu \rightarrow \int f(t)d\mu$$ and
$$ \int |f_{n}(t) - f(t)| d\mu \rightarrow 0$$.
How do we prove the convergence in $L^{2}$ norm?

Here is what I tried:

$$ \int |f_{n}(t) - f(t)|^{2} d\mu = \int (f_{n}-f)(\overline{f_{n}}-\overline{f})$$
$$ = \int |f_{n}|^{2} - \int f\overline{f_{n}} - \int f_{n}\overline{f} + \int |f|^{2} $$
which goes to zero by DCT (as $f_{n}\overline{f} \rightarrow |f|^{2}$ and so on).
Am I right in all this or am I missing something? I am apprehensive because I intuitively feel that pointwise convergence should not imply $L^{2}$ convergence.

答:
You are right, but you can also argue more directly. Let $g_n(t) = |f_n(t) - f(t)|^2$. Then $g_n \to 0$ almost everywhere and $|g_n| \le 4c^2$ almost everywhere. As we are in a finite measure space,
$$ \|f_n - f\|_{L^2}^2 =  \int g_n  \, d\mu \to 0. $$</p>


答:
By Menger's Theorem, a graph $G$ is $k$-connected iff for each pair of vertices, there are $k$ internally-disjoint paths between every pair of distinct vertices $u, v \in V(G)$.  By internally-disjoint, I mean the paths only intersect at their endpoints $u$ and $v$.

Suppose $e$ is an edge between $x$ and $y$ and it's contracted, giving the graph $G'$.  Assume the connectivity of $G$ is $k$.  If $w$ is an vertex or edge of $G$ (other than $e$) denote its image under contraction by $w'$.  Note then that $x' = y'$.

Case 1: First assume neither $u', v'$ is $x'$.  If $P_1, \dots, P_n$ are a maximal family of internally-disjoint paths from $u$ to $v$, then at most two of these paths contain either $x$ or $y$.  Thus the connectivity between $u'$ and $v'$ is reduced by at most one.

The connectivity between $u$ and $v$ is not increased by contraction: Assume BWOC that it is, so there are internally-disjoint paths $P_1', \dots, P_{n+1}'$ from $u'$ to $v'$, with at most one containing $x' = y'$.  If none contained $x'$ then the graph spanned by the paths pulls back homeomorphically, a contradiction, so assume $x' \in P_1'$.  Then the other paths pull back homeomorphically, so it just remains to show that $P_1'$ pulls back to a path that's internally-disjoint from the rest to finish the case.

To do this, let $w'x'z'$ be a subpath of $P_1'$, where the preimages of $w'$ and $z'$ under contraction are $w, z \in G$.  If $wxz$ (or $wyz$) is a path in $G$, then $P_1'$ pulls back homeomorphically to a path $P_1$ from $u$ to $v$ by mapping $x'$ to $x$ (or $y'$ to $y$ respectively), and since it's internally-disjoint from the other $P_j'$ paths so is $P_1$ from the other $P_j$ paths.

If not, then without loss of generality assume $wx, yz \in E(G)$.  Since $xy = e \in E(G)$ the path $wxyz$ is contained in $G$.  Since none of the other $P_j'$ paths contain $x'$ their pullbacks don't contain $x$ or $y$, and since they're internally disjoint from the rest of $P_1'$ they are internally disjoint to the path $P_1 = Q_1wxyzQ_2$, where $Q_1$ is the pullback of $P_1'$ from $u'$ to $w'$ and $Q_2$ is the pullback of $P_1'$ from $z'$ to $v'$.

Thus edge contraction can decrease connectivity by at most one (two triangles connected along an edge gives an example where it does decrease) and does not increase it between vertices not on $e$.

Case 2: Let $u'=x'$ and $v' \neq x'$.  Let $P_j$ be a maximal collection of internally-disjoint paths from $v$ to $x$, and let $Q_j$ be a maximal collection of internally-disjoint paths from $v$ to $y$.  Then at most one of the $P_j$ paths contains $y$ and vice-versa, so the connectivity between $v'$ and $x'$ can be reduced by at most one (within the previous bound).

However - as pointed out in the comments below - it is possible to increase the connectivity here.  For if $v$ has many more paths to $x$ than $y$ (e.g. all paths to $x$ go through $y$ and thus can't be internally-disjoint), then the connectivity can be increased by $|vc(y) - vc(x)|$ where $vc(w)$ is the minimal connectivity between $w$ and any other vertex.  That this (crappy) bound is actually sharp is given by the same example below, namely a complete graph plus one pendant vertex.

Case 3: Suppose $u' = v' = x' = y'$.  In other words, $u = x$ and $v = y$.  Then assume that the connectivity between $x$ and $y$ is $k$, but for every other pair of vertices it is at least $k+1$.  But that's impossible because $e = xy \in E(G)$ so the connectivity between $x$ and $y$ is $|G| - 1$, which is maximal.

Thus if $vc(G)$ denotes vertex connectivity, $vc(G) -1 \leq vc(G') \leq vc(G) + |vc(y) - vc(x)|$.

After having to edit this one like 500 times, please double-check everything in case 2 before handing it in.

Title: Change in the vertex connectivity of a given graph

Body:
This is a homework problem I am trying to solve:
What impact does the following operations (prove some bounds on the amount of change and provide examples showing that the bounds are tight) has on the vertex connectivity of a graph:
(a) adding a single edge to a graph (without adding vertices),
(b) removing a single edge,
(c) removing a single vertex,
(d) contracting an edge.

so what I have tried is that I have taken a random graph and have tried to see what happens to the vertex connectivity. In 1st case, I observed that when a single edge (without adding vertices) is added depending on where it gets added the vertex connectivity can be increased but it does not decrease. In the 2nd case, removing a single edge might decrease the vertex connectivity i.e make it 0 (like in the case of a star-like structure) same can be seen in the case of removal of a single vertex. I do not much understand contracting an edge. Even though through various examples I can understand what impacts is having but I am unable to prove or find the bounds are tight or not. I do understand a few specific bounds like $k(G)\leq\lambda(G)\leq\delta (G)$. But I still find myself stuck here.

Any help is appreciated.


答: $$A\cup B=(A-B)\cup(A\cap B)\cup(B-A)$$The sets on RHS are disjoint so that:$$P(A\cup B)=P(A-B)+P(A\cap B)+P(B-A)\leq P(A-B)+2P(A\cap B)+P(B-A)=$$$$P(A)+P(B)$$

Title: Proof of Boole's inequality in two event case

Body:
If we have the probability space $(\Omega, \mathscr{F}, \mathbb{P})$, how can we show that, for any $A, B \in \mathscr{F}$ we have: $$\mathbb{P}(A \cup B) \leq \mathbb{P}(A) + \mathbb{P}(B)$$

Forgive me if I have missed another question asking this, but I haven't been able to find a resource that demonstrates this proof using measure/ probability theory axioms. It is a basic question but I struggle to understand how to approach it.


答: For any event $S$ let $$\chi_S=\begin{cases}1,&x\in S\\0,&x\notin S\end{cases}$$  so that $\Pr(S)=\int \chi_S$.  Note that $$\chi_{A\cup B}=\chi_A+\chi_B-\chi_{A\cap B}$$


问:
Question about a proof of Second Derivatve Test

Partial Statement: Let $f$ be defined in a neighborhood of $c$ with $f'(c) = 0$. Then $f''(c) > 0 \implies f(c)$ is a local minimum of $f$

Proof: Suppose  $f$ is defined in a neighborhood of $c$ with $f'(c) = 0, \ f''(c) > 0 $. $\color{red}{\text{It can be shown}}$ there's punctured neighborhood of $N = (c - r, c) \cup (c, c + s)$ of $c$ s.t. $\displaystyle{\frac{f'(x) - f'(c)}{x - c} > 0}$ for $x \in N$ meaning $f'(x) < 0$ for $x \in (c - r, c)$ and $f'(x) > 0$ for $x \in (c, c + s)$. It follows $c$ is a local minimum as $f$ dips towards $c$ from the left and grows back up on the right of $c$.

My question is about the $\color{red}{\text{part in red of the quote above}}$. How does that follow? Here below is my attempt. If incorrect, what would be a better way to show that? Thanks.

By definition, $f$ is differentiable at $c$ if $\displaystyle{\forall\varepsilon > 0, \exists \delta>0 \ \left|\frac{f(x) - f(c)}{x - c} - f'(c)\right| < \varepsilon}$ for any $x$ with $0<|x - c| < \delta.$ Then $\displaystyle{\frac{f(x) - f(c)}{x - c} = f'(c)} > 0$. Applying the definition above one more time we get $\displaystyle{\frac{f'(x) - f'(c)}{x - c} = f''(c)} > 0$.

答:
A prerequisite is missing: $ f$ should be a $C^2-$ function.

From $f''(c)>0$ and the continuity of $f''$ we get a punctured neighborhood $N = (c - r, c) \cup (c, c + s)$ such that $f''(t) >0$ for all $t \in N.$

If $x \in N$, then there is, by the mean value theorem $t \in $ such that

$$\frac{f'(x) - f'(c)}{x - c}=f''(t).$$

Hence

$$\frac{f'(x) - f'(c)}{x - c} >0.$$


问:
ergodic measures of rational rotation
Let $R: [0,1]\to[0,1]$, with $ R(x)=x+\alpha \mod 1 $ and $ \alpha = \frac{p}{q} $ a rational rotation map on the unit interval. I want to proof the following assertion

the ergodic measures for $R$ are of the form $ \mu =\sum\limits_{i=1}^q\frac{\delta_{\{x_i\}}}{q} $ for some $x_1,\dots,x_n\in[0,1]$ (and where $\delta_{\{x_i\}}$ denotes the dirac measure).

My understanding is, that to be $R$- ergodic, the measure $\mu$ has to be atomic and concentrated on a finite number of atoms, which - I'm guessing - are going to be the the measures that are concentrated on the finite orbits of points. My proof so far:

Let $ A $ be a $ R $-invariant set, i.e. $ R^{-1}A=A $. I can assume, that $ A $ is not empty (the empty set is a trivial case). So there exists a point $ x\in A $. Since $ \alpha $ is rational, it's clear, that the orbit of $ x $ is a finite set $ orb(x)=\{x_1,\dots,x_q\} $. Since $ R^{-1}A=A $ and clearly the points of the orbit map onto each other, it follows that $ \{x_1,\dots,x_n\}\subset A $. Since $ R $ is ergodic, $ \mu(A)\in\{0,1\} $. First, we assume, that $ \mu(A) = 1 $. It's easy to see, that the $ R $-Orbits of points in $ [0,1] $ are $ R $-invariant sets, this means, that $ \mu(\{x_1,\dots,x_q\})\in \{0,1\} $.

If $ \mu(\{x_1,\dots,x_q\}) = 1 $, then $ \mu = \sum\limits_{i=1}^q\frac{\delta_{\{x_i\}}}{q} $.

If $ \mu(\{x_1,\dots,x_q\})= 0 $, then $ \mu(A\setminus\{x_1,\dots,x_q\}) =1 $. This means there is another $ \hat{x}\in A\setminus\{x_1,\dots,x_q\} $. And since $ A\setminus\{x_1,\dots,x_q\} $ is still $ R $-invariant, the whole finite orbit of $ \hat{x} $ is a subset of $ A\setminus\{x_1,\dots,x_q\} $:  $ orb(\hat{x})=\{\hat{x}_1,\dots,\hat{x}_q\}\subset A\setminus\{x_1,\dots,x_q\}. $

Now we can proceed the same way as before:

$ \{\hat{x}_1,\dots,\hat{x}_q\} $ is a $ R $-invariant set, this means $ \mu(\{\hat{x}_1,\dots,\hat{x}_q\})\in\{0,1\} $. If $ \mu(\{\hat{x}_1,\dots,\hat{x}_q\})=1 $, then $ \mu= \sum\limits_{i=1}^q\frac{\delta_{\{\hat{x_i}\}}}{q} $. If $ \mu(\{\hat{x}_1,\dots,\hat{x}_q\})=0 $, we look again at $ A\setminus(\{x_1,\dots,x_n\}\cup\{\hat{x}_1,\dots,\hat{x}_q\}) $ and repeat the same procedure on this set.

So my questions are

is the proof correct so far? In particular: is it valid, that $ \mu(\{x_1,\dots,x_q\}) = 1 $ implies $ \mu = \sum\limits_{i=1}^q\frac{\delta_{\{x_i\}}}{q} $ ?

how can I handle the case $ \mu(A) = 0 $? (Then $ \mu(\{x_1,\dots,x_q\})= 0 $ as well, so I guess $ \mu $ has to be concentrated on some other orbit, but I don't know how to formalize that part).

答:
$\mu(\{x_1,\dots,x_q\}) = 1$ implies $\mu = \sum\limits_{i=1}^q\frac{\delta_{\{x_i\}}}{q}$ is correct. But I do not understant how this close the case $\mu (A)=1$ since you can repeat infinitely times your procedure without reaching a atom of the measure.

If $A$ is invariant and $\mu(A)=0$ then $A^c$ the complement, is also invariant and $\mu(A^c)=1$. That might help.

I think a working strategy is to do a dichotomy. Take a invariant ergodic measure $\mu$. If $A\subset[0,1/q[$ we will call $\hat{A}=\cup_{i=0}^{q-1}R^i(A)$. $\hat{A}$ is always invariant for $R$.

Let's called $A^0=[0,1/q[$. We have $\mu(A^0)=1/q$ (because of invariance and $\hat{A^0}$ is $[0,1]$ .

Now you can cut $A^0$ in two parts $A^1_1=[0,1/2q[$ and $A^1_2=[1/2q,1/q[$, $\hat{A^1_1}$ and  $\hat{A^1_2}$ are a partition of $[0,1]$ and are disjoint, by ergodicity there is $i_1 \in \{0,1 \}$ such that  $\mu(\hat{A^1_{i_1}})=1$.

Doing this again and again you will find a sequence of set $A^k \subset [0,1/q[$ with the following properties :

$A^{k+1} \subset A^K$

$diam(A^k)=\frac{1}{q*2^k}$

$\mu(\hat{A^k})=1$

Taking the intersection of all this set you find just a point $x$, and then $\mu=\sum\limits_{i=1}^q\frac{\delta_{\{x_i\}}}{q}$.

Title: ergodic measures of rational rotation


问:
Lagrange multiplier calculation issue
I have 8 Lagrange multipliers $\lambda_1, \lambda_2,  \lambda_3,\cdots, \lambda_8$.

And the Lagrange formulation is

$L(\lambda_1, \lambda_2,  \lambda_3,\cdots, \lambda_8) = \sum\limits_{i=1}^{8} \lambda_i - \sum\limits_{i=1}^{8} \sum\limits_{j=1}^{8} \lambda_i \lambda_j  a_{i,j}$ with constraint

$\lambda_i \geq 0$ for all $1 \le i \le 8$ and

And auxiliary information is

$\lambda_1 - \lambda_2 -\lambda_3-\lambda_4+ \lambda_5+ \lambda_6 -\lambda_7+\lambda_8 = 0$

The matrix that supplies $a_{i,j}$ is shown below

\begin{pmatrix}
  0.36 &-0.472 & -0.542& -0.699 & 0.340 & 0.320 & -0.73 & 0.126 \\
 -0.472 & 0.612 & 0.700 & 0.905 & -0.441 & -0.414 & 0.94 & -0.163 \\
 -0.542 & 0.700 & 1.014 & 1.045 & -0.403 & -0.520 & 1.187 & -0.234 \\
 -0.699 & 0.905 & 1.045 & 1.3397 &-0.649 & -0.614 & 1.409 & -0.244  \\
 0.340 &  -0.441 & -0.403 & -0.649 & 0.368 & 0.276 & -0.637 & 0.095 \\
 0.320 & -0.414 & -0.520 & -0.614 & 0.276 & 0.290 & -0.664 & 0.121  \\
 -0.734 & 0.949 & 1.187 & 1.409 & -0.637 & -0.664 & 1.521 & -0.276 \\
 0.126 & -0.163 & -0.234 & -0.244 & 0.095 & 0.121 & -0.2763 & 0.054 \\
        \end{pmatrix}

I need to find $\lambda_i$ for $1 \le i \le 8$

My approach is as follows:

Calculate $\frac{\partial L}{\partial \lambda_i} $ for $1 \le i \le 8$,

Finally, I get eight linear equations in eight unknowns after calculating derivatives and one extra equation (given auxiliary information). Solving them gives all the desired values.

But, my approach is not working and I am getting wrong answers. the correct answer is $\lambda_1 = \lambda_2  \approx 65.52$ and $\lambda_k = 0$ for all other $k$'s.

Where am I going wrong?

Note: Please ignore auxiliary information in case of any issue.

Title: Lagrange multiplier calculation issue

Body:
Considering the lagrangian

$$
L(\lambda,\mu,s) = f(\lambda)+\mu_0(\lambda_1 - \lambda_2 -\lambda_3-\lambda_4+ \lambda_5+ \lambda_6 -\lambda_7+\lambda_8)+\sum_{k=1}^8\mu_k(\lambda_k-s_k^2)
$$

with $f = \sum\limits_{i=1}^{8} \lambda_i - \sum\limits_{i=1}^{8} \sum\limits_{j=1}^{8} \lambda_i \lambda_j  a_{i,j}$

the stationary points are the solutions for

$$
\nabla L = 0
$$

Assuming $A=\{a_{i,j}\}$  as given, we obtain the stationary (feasible) points for $L$ as

$$
\left[
\begin{array}{ccccccccccccccccc}
f & \lambda_1&\lambda_2&\lambda_3&\lambda_4&\lambda_5&\lambda_6&\lambda_7&\lambda_8&s_1^2&s_2^2&s_3^2&s_4^2&s_5^2&s_6^2&s_7^2&s_8^2\\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0.977804 & 0 & 0 & 0 & 0 & 0 & 0 & 0.977804 & 0.977804 & 0 & 0 & 0 & 0 & 0 & 0 & 0.977804 & 0.977804 \\
 1.10412 & 0 & 0 & 0 & 1.10412 & 0 & 0 & 0 & 1.10412 & 0 & 0 & 0 & 1.10412 & 0 & 0 & 0 & 1.10412 \\
 1.62602 & 0 & 0 & 0 & 0 & 1.62602 & 0 & 1.62602 & 0 & 0 & 0 & 0 & 0 & 1.62602 & 0 & 1.62602 & 0 \\
 1.66667 & 0 & 0 & 1.66667 & 0 & 0 & 0 & 0 & 1.66667 & 0 & 0 & 1.66667 & 0 & 0 & 0 & 0 & 1.66667 \\
 1.73611 & 0 & 0 & 1.73611 & 0 & 1.73611 & 0 & 0 & 0 & 0 & 0 & 1.73611 & 0 & 1.73611 & 0 & 0 & 0 \\
 1.80868 & 0 & 0 & 1.12341 & 0 & 1.80868 & 0 & 0.685278 & 0 & 0 & 0 & 1.12341 & 0 & 1.80868 & 0 & 0.685278 & 0 \\
 1.889 & 0 & 0 & 1.889 & 0 & 1.04221 & 0 & 0 & 0.846795 & 0 & 0 & 1.889 & 0 & 1.04221 & 0 & 0 & 0.846795 \\
 2.07039 & 0 & 0 & 0 & 0 & 0 & 2.07039 & 2.07039 & 0 & 0 & 0 & 0 & 0 & 0 & 2.07039 & 2.07039 & 0 \\
 2.39808 & 2.39808 & 0 & 0 & 0 & 0 & 0 & 2.39808 & 0 & 2.39808 & 0 & 0 & 0 & 0 & 0 & 2.39808 & 0 \\
 2.44081 & 0 & 0 & 0 & 2.44081 & 2.44081 & 0 & 0 & 0 & 0 & 0 & 0 & 2.44081 & 2.44081 & 0 & 0 & 0 \\
 2.48942 & 0 & 0 & 0 & 2.48942 & 0 & 2.48942 & 0 & 0 & 0 & 0 & 0 & 2.48942 & 0 & 2.48942 & 0 & 0 \\
 2.49559 & 0 & 0 & 0.460885 & 2.03471 & 2.49559 & 0 & 0 & 0 & 0 & 0 & 0.460885 & 2.03471 & 2.49559 & 0 & 0 & 0 \\
 2.63818 & 0 & 0 & 0 & 2.63818 & 1.21954 & 1.41864 & 0 & 0 & 0 & 0 & 0 & 2.63818 & 1.21954 & 1.41864 & 0 & 0 \\
 2.94118 & 0 & 2.94118 & 0 & 0 & 0 & 0 & 0 & 2.94118 & 0 & 2.94118 & 0 & 0 & 0 & 0 & 0 & 2.94118 \\
 3.31455 & 3.31455 & 0 & 0 & 3.31455 & 0 & 0 & 0 & 0 & 3.31455 & 0 & 0 & 3.31455 & 0 & 0 & 0 & 0 \\
 3.44828 & 3.44828 & 0 & 3.44828 & 0 & 0 & 0 & 0 & 0 & 3.44828 & 0 & 3.44828 & 0 & 0 & 0 & 0 & 0 \\
 3.46975 & 3.46975 & 0 & 3.10338 & 0 & 0 & 0 & 0.366371 & 0 & 3.46975 & 0 & 3.10338 & 0 & 0 & 0 & 0.366371 & 0 \\
 3.78788 & 0 & 0 & 3.78788 & 0 & 0 & 3.78788 & 0 & 0 & 0 & 0 & 3.78788 & 0 & 0 & 3.78788 & 0 & 0 \\
 4.01689 & 0 & 0 & 3.05722 & 0.959667 & 0 & 4.01689 & 0 & 0 & 0 & 0 & 3.05722 & 0.959667 & 0 & 4.01689 & 0 & 0 \\
 4.3517 & 4.3517 & 0 & 2.27239 & 2.07931 & 0 & 0 & 0 & 0 & 4.3517 & 0 & 2.27239 & 2.07931 & 0 & 0 & 0 & 0 \\
 10.2041 & 0 & 10.2041 & 0 & 0 & 10.2041 & 0 & 0 & 0 & 0 & 10.2041 & 0 & 0 & 10.2041 & 0 & 0 & 0 \\
 13.5135 & 0 & 13.5135 & 0 & 0 & 0 & 13.5135 & 0 & 0 & 0 & 13.5135 & 0 & 0 & 0 & 13.5135 & 0 & 0 \\
 13.7805 & 0 & 12.6829 & 1.09756 & 0 & 0 & 13.7805 & 0 & 0 & 0 & 12.6829 & 1.09756 & 0 & 0 & 13.7805 & 0 & 0 \\
 17.1994 & 0 & 17.1994 & 0 & 0 & 6.6526 & 10.5468 & 0 & 0 & 0 & 17.1994 & 0 & 0 & 6.6526 & 10.5468 & 0 & 0 \\
 35.7143 & 35.7143 & 35.7143 & 0 & 0 & 0 & 0 & 0 & 0 & 35.7143 & 35.7143 & 0 & 0 & 0 & 0 & 0 & 0 \\
\end{array}
\right]
$$

here $s_k = 0$ indicates that the respective restriction is actuating.


问:
If $(a,b,c)=1$, is there $n\in \mathbb Z$ such that $(a,b+nc)=1$?
In the book Lectures on modular forms, one finds the statement at page 8 that

If $(a,b,c)=1$ then there is $n\in \mathbb Z$ such that $(a,b+nc)=1$.

I know that, if $(a,b)=1$, then we can take $n=0$. But, if $(a,b)\not=1$, then what could we do? Further, I tried to look at the linear combinations of $a, b, c$ which are $=1$, but to no avail have I discovered anything. The main difficulty I encountered is that the coefficient of $b$ might not be $1$.

Any hint is well-appreciated.

Thanks in advance.

答:
Let $\rm n$ be the product of all primes that divide $\rm a$ but not $\rm b$. Assume $\rm p\mid a,b+nc$ with $\rm p$ prime.

Suppose $\rm p\mid b$. Then $\rm p$ cannot divide $\rm c$ (since $\rm p\mid a,b,c\implies p\mid(a,b,c)$) nor does it divide $\rm n$, by definition, but $\rm p\mid b\implies\rm p\mid (b+nc)-b=nc\implies p\mid n$ or $\rm p\mid c$, impossible.

Suppose $\rm p\nmid b$. But then $\rm p\mid a,p\nmid b\implies p\mid n\implies\rm p\mid (b+nc)-nc=b$, impossible.

Therefore the gcd $\rm(a,b+nc)$ is $1$ as it is not divisible by any prime $\rm p$.


问:
How to uncurl $\mathbf{B}=\mathbf{\nabla\times A}$ if we only know $\mathbf{B}$ at $z=z_0$?
Let
$$\mathbf{B} = \mathbf{\nabla\times A},$$
where
$$\mathbf{B}(x,y,z) = B_x(x,y,z)\mathbf{\hat{x}} + B_y(x,y,z)\mathbf{\hat{y}} + B_z(x,y,z)\mathbf{\hat{z}},$$
$$\mathbf{A}(x,y,z) = A_x(x,y,z)\mathbf{\hat{x}} + A_y(x,y,z)\mathbf{\hat{y}} + A_z(x,y,z)\mathbf{\hat{z}},$$
Assume that
$$\mathbf{\nabla \cdot A} = 0.$$
This ensures that $\mathbf{A}$ is uniquely defined by the top equation.

Is it possible to calculate $A_x(x,y,z_0)$, $A_y(x,y,z_0)$ if we only know $\mathbf{B}(x,y,z)$ at $z=z_0$?

Here is my naive attempt.
Assume that
$$A_x(x,y,z_0) = \frac{\partial \phi}{\partial y},$$
$$A_y(x,y,z_0) = -\frac{\partial \phi}{\partial x},$$
where
$$\phi=\phi(x,y).$$
Substituing this into the $z$-component of the top equation gives the following 2D Poisson equation
$$\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right)\phi(x,y) = -B_z(x,y,z_0).$$
We can solve this for $\phi(x,y)$ which we can use to calculate $A_x(x,y,z_0)$ and $A_y(x,y,z_0)$. Is this a valid solution? There seems to be problems. For example, since
$$\mathbf{\nabla\cdot A} = 0,$$
this implies that
$$\left.\frac{\partial A_z}{\partial z}\right|_{z=z_0}=0,$$
but $z_0$ is arbitrary so $\partial A_z / \partial z = 0$ $\forall z$?

答:
Given $\mathbf{B}$ on $z=z_0$ we can extend the field to all on $\mathbb{R}^3$ by
$$\begin{cases}
B_x(x,y,z) := B_x(x,y,z_0) \\
B_y(x,y,z) := B_y(x,y,z_0) \\
B_z(x,y,z) := B_z(x,y,z_0) - z \left( \partial_x B_x(x,y,z_0) + \partial_y B_y(x,y,z_0)\right)
\end{cases}$$
This extensin satisfies $\nabla\cdot \mathbf{B} = 0$ but is not the only possible extension.

Then, given $\mathbf{B}$ in all of $\mathbb{R}$ we can construct $\mathbf{A}$ such that $\nabla\times \mathbf{A} = \mathbf{B}$ and $\nabla\cdot \mathbf{A}=0$ by the convolution
$$\mathbf{A} = -G * (\mathbf{\nabla\times B}),$$
where $\nabla^2 G = \delta,$ i.e. $G(x,y,z) = \frac{1}{4\pi\sqrt{x^2+y^2+z^2}}.$


问:
Combinatorics problem: # of arithmetic problems possible
"How many arithmetic problems of the following form are possible? You must use each of the digits 1 through 9, they must appear in numerical order from left to right, and you can use any combination of the + and * symbols you like, as long as the resulting expression makes mathematical sense. For example, 1234+5678+9 and 123456+789 and 123456789 are three possibilities, but 1**23456789 is not."

We know the numbers are ordered as: 1_2_3_4_5_6_7_8_9 where the _'s can be filled by one of the three operations '+', '*', and ''.

There are 2*3{n-1} ways to arrange these operations. I just don't know how to use the information correctly for the problem. Where 'n' is 8 I believe.

Advice/Hints?

edit: 2*3^{n-1} is totally wrong. Brain was stuck on previous problem. Sorry.

答: Each '_' may be either empty, $*$, or $+$ -- so a total of three possibilities. We have 8 such '_'. So by the fundamental principle of counting we are left with a total of $3^8$ possibilities.</p>


问:
Is the cokernel $\mathbb{Z}$?
I have  matrix (in SageMath notation):

```<code>[-1  0  0  0 -1  0  1  0  0]
[ 0 -1  0  0  0 -1  0  1  0]
[ 0  0 -1 -1  0  0  0  0  1]
</code>```

I have calculated its Smith normal form as follows:

```<code>[1 0 0 0 0 0 0 0 0]
[0 1 0 0 0 0 0 0 0]
[0 0 1 0 0 0 0 0 0]
</code>```

So is the cokernel $\mathbb{Z}$ or $\mathbb{Z}^3$?

答:
Neither. If I interpret your question correctly you are looking at a group homomorphism $\varphi\colon\mathbb Z^9\to\mathbb Z^3$ given by the matrix

$$M=\begin{pmatrix}-1 & 0 & 0 & 0 & -1 & 0 & 1 & 0 & 0\\ 0 & -1 & 0 & 0 & 0 & -1 & 0 & 1 & 0\\0 & 0 & -1 & -1 & 0 & 0 & 0 & 0 & 1\end{pmatrix}$$

with Smith normal form

$$\begin{pmatrix}1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\end{pmatrix}$$

Recall that $\operatorname{coker}\varphi=\mathbb Z^3/\operatorname{im}\varphi$ and the normal form tells us that $\varphi$ is surjective. Hence $\operatorname{coker}\varphi$ is trivial.

Title: Is the cokernel $\mathbb{Z}$?


问:
Why is the difference of two vectors parallel to the vector connecting the two vectors?
Say we have any two vectors $u$ and $v$. We want to find a vector $w$ that is parallel to the vector that goes from $u$ to $v$ (is there a standard notation for this vector?). Then I already know that such a vector can be found simply by subtracting one of the two vectors from the other. Can you give me some intuition why this works? Probably what I am not understanding is how you get the vector connecting $u$ and $v$ in the first place.

答: It's a consequence of how vector addition works. Say we're traveling along $v$ then going to $u$ and to get there I travel along some other vector $x$. Geometrically we would model this as $v+x=u$ and then solving for $x$ gives us $x=u-v$. Now it becomes $v + (u-v)=u$ and we see this works both algebraically and geometrically.


问:
Can you write this operation in matrix notation?
Consider the matrix $A$ and column vector $b$ where,

$A = \begin{bmatrix}1 & 2\\3 & 4\end{bmatrix}$ and  $b = \begin{bmatrix}1\\3\end{bmatrix}$

In the R statistics software, the the code A*b performs element-wise multiplication of the columns in $A$ with the column vector $b$ returning a matrix with the same dimensions as $A$, i.e.

$A*b = \begin{bmatrix}1 & 2\\3 & 4\end{bmatrix} * \begin{bmatrix}1\\3\end{bmatrix} = \begin{bmatrix}1\times1 & 2\times1\\3\times3 & 4\times3\end{bmatrix}=\begin{bmatrix}1 & 2\\9 & 12\end{bmatrix}$

Is there a name for this operation? Can this operation be expressed using common matrix notation?

答:
It can be written as

$A\circ(bu)$ where $u$ is row vector of ones and $\circ$ is the hadamard product


答:
The matrix in question (call it $A$) fits in the exact sequence:

$$
\mathbb{Z}^9 \overset{A}{\longrightarrow} \mathbb{Z}^3 \to \text{coker}(A).
$$

Here the cokernel is $\mathbb{Z}^3 / \text{Im}(A)$, but we know that $A$ is surjective (because it has full rank). So the cokernel is actually $0$.

In general, if your smith normal form has entries $a_1, \ldots, a_n$ on the leftmost diagonal (where the $a_k$ are allowed to be $0$), then the cokernel is exactly

$$\bigoplus_k \mathbb{Z} / a_k \mathbb{Z}$$

where $\mathbb{Z} / 1 \mathbb{Z} = 0$ and $\mathbb{Z} / 0\mathbb{Z} = \mathbb{Z}$. This is part of what makes the SNF useful! It allows us to quickly read off the decomposition guaranteed by the fundamental theorem of PID modules.

I hope this helps ^_^

Title: Is the cokernel $\mathbb{Z}$?

Body:
I have  matrix (in SageMath notation):

```<code>[-1  0  0  0 -1  0  1  0  0]
[ 0 -1  0  0  0 -1  0  1  0]
[ 0  0 -1 -1  0  0  0  0  1]
</code>```

I have calculated its Smith normal form as follows:

```<code>[1 0 0 0 0 0 0 0 0]
[0 1 0 0 0 0 0 0 0]
[0 0 1 0 0 0 0 0 0]
</code>```

So is the cokernel $\mathbb{Z}$ or $\mathbb{Z}^3$?


问:
Solving $\sin(\alpha)-\cos(\alpha)\tan(\beta) = -a\frac{\tan(\beta)}{b}$ for $\alpha$
I haven't practiced trigonometry for a while, and I'm wondering if there's a way to solve this analytically:

$$\sin(\alpha)-\cos(\alpha)\tan(\beta) = -a\frac{\tan(\beta)}{b}$$

I'm solving for $\alpha$, so $\beta$, $a$ and $b$ are known.

If there is, can someone point out the right way?

Thanks in advance!

答:
Move the cosine term to the other side:$$\sin(\alpha)=\cos(\alpha)\tan(\beta)  -a\frac{\tan(\beta)}{b}$$
Now square the equation (you will need to check if this introduce additional solutions)
$$\sin^2(\alpha)=1-\cos^2\alpha=\left(\cos(\alpha)\tan(\beta)  -a\frac{\tan(\beta)}{b}\right)^2$$
You can see now that you have a quadratic equation in $\cos\alpha$. Can you take it from here?


问:
Problem with example of $G_\delta$ set
It was known that every closed set is a $G_\delta$ set. Now, I am trying to make an example about that.
This is my example:
$$[1,2]=\bigcap_{n=1}^\infty\bigg(1-\frac{1}{n},2+\frac{1}{n}\bigg)$$
I just can show that by visualization on a real line. But, when I try to proof that mathematically, I get stuck. This is my attempt:

$[1,2]\subseteq \bigcap_{n=1}^\infty\bigg(1-\frac{1}{n},2+\frac{1}{n}\bigg)$. I think for this part is clear, because $[1,2]\subseteq \bigg(1-\frac{1}{n},2+\frac{1}{n}\bigg)$ for all $n\in\mathbb{N}$.

Conversely, i.e. $\bigcap_{n=1}^\infty\bigg(1-\frac{1}{n},2+\frac{1}{n}\bigg)\subseteq [1,2]$. I try to use contraposition method, take any $x\notin [1,2]$, so $x<1$ or $x>2$. Then I will proof that $x\notin \bigcap_{n=1}^\infty\bigg(1-\frac{1}{n},2+\frac{1}{n}\bigg)$, that means $x\in \bigg(\bigcap_{n=1}^\infty\bigg(1-\frac{1}{n},2+\frac{1}{n}\bigg)\bigg)^c=\bigcup_{n=1}^{\infty}\bigg(1-\frac{1}{n},2+\frac{1}{n}\bigg)^c$. Because $x \in  \bigcup_{n=1}^{\infty}\bigg(1-\frac{1}{n},2+\frac{1}{n}\bigg)^c$, there exists $n\in \mathbb{N}$ sucht that $x\in\bigg(1-\frac{1}{n},2+\frac{1}{n}\bigg)^c$, so $x\in\bigg(-\infty,1-\frac{1}{n}\bigg]$ or $x\in\bigg[2+\frac{1}{n},\infty\bigg)$. Take any $\varepsilon>0$, by Archimedian Property, we have $\frac{1}{n}<\varepsilon$, then
$$x\leq 1-\frac{1}{n}<1-\varepsilon<1$$
or $$x\geq2+\frac{1}{n}$$
I get stuck on inequality of $x\geq2+\frac{1}{n}$, because it's impossible that $x\geq2+\frac{1}{n}<2+\varepsilon$.
So, I am not sure about my attempt. Is there another way to prove that? Thanks for any help.

答: Okay so there is a slight mistake in the last steps you made. By the Archimedean property you get that $1/n<\epsilon$ but this means when you switch signs that you get $-\epsilon<-1/n$ and so the sequence of inequalities should really be $x\leq 1-\epsilon<1-\frac{1}{n}<1$. Also it's not clear what purpose $\epsilon$ is supposed to have. I am taking the liberty to assume you meant it to be $\epsilon = min\{\vert x-1 \vert, \vert x-2\vert\}$. Intuitively speaking this is the distance that $x$ has compared to the the set $[1,2]$. Then you need to break the proof down into two cases. The first case is the one in which $x<1$ and the second one when $x>2$. The first case besides the small error works since you showed if $x<1$ then by the definition of $\epsilon$ $x\leq 1-\epsilon<1-\frac{1}{n}$ and so it's for some $n\in\mathbb{N}$ in the the set $(-\infty,1-\frac{1}{n}]$. For the second case you consider the case $x>2$ so by the definition of $\epsilon$ you have $x\geq 2+\epsilon>2+\frac{1}{n}$ and so $x\in [2+\frac{1}{n},+\infty)$. You need to be careful not to mix the inequalities that appear in one of the cases with the inequalities in the other case.

Title: Problem with example of $G_\delta$ set


问:
Improper Riemann Integral equal to Lebesgue Integral
Let $f: [a,b] \to [0,\infty)$ and $f$ is Riemann Integrable on every subinterval $[a + \epsilon,b]$ for $\epsilon > 0$. Suppose that the improper Riemann integral exists. That is
$$I = \lim_{\epsilon \to 0} \int_{a + \epsilon}^{b} f(x) dx < \infty$$
exists. Prove that $f$ is Lebesgue integrable on $[a,b]$ and that $\int_{[a,b]} f(x) dx = I$.

I found another posting of this problem written in the same way, but it used dominated convergence theorem, which I have not covered. The others I saw were written slightly differently and/or weren't making too much sense.

Also, maybe I'm not seeing something clearly, but here's something I thought:

The improper integral I is finite (it exists). If the improper integral exists, isn't it equal to the "proper" integral? That would mean the "normal" proof of showing that every Riemann Integrable function is Lebesgue integrable would apply. However, I have a feeling this is not the way to prove it or else this question wouldn't be asked.

Thanks for the help!

答:
Consider the Riemann integral
$$  \int_0^1 x^{-2} \,\mathrm{d}x  $$
This is an improper Riemann integral due to the unbounded behaviour at the left endpoint of the interval of integration.  It's value is defined to be (if this limit exists)
$$  \lim_{\varepsilon \rightarrow 0^+} \int_{\varepsilon}^1 x^{-2} \,\mathrm{d}x  \text{.}  $$

Now think of a decreasing sequence $(\varepsilon_i)_{i \in \Bbb{Z}_{>0}}$ where each $\varepsilon_i \in (0, 1]$.  This sequence provides a way to get at that limit:
$$  I_i = \int_{\varepsilon_i}^1 x^{-2} \,\mathrm{d}x  \text{.}  $$

Every $I_i$ is a proper integral.  But just because every proper integral in the sequence $(I_i)_i$ converges does not mean the limit $\lim_{i \rightarrow \infty} I_i$ exists.  Contrast with $\displaystyle J_i = \int_{\varepsilon_i}^1 x^{-1} \,\mathrm{d}x$, for which each $J_i$ is some finite number, but the limit of $J_i$ as $i \rightarrow \infty$ does not exist.

So to make progress, you will use some idea like $|I - I_i| < \delta$, which idea is based on the fact that $I$ exists and is finite and that each $I_i$ exists, is finite, and is approaching the integral over the whole interval.  You might use this to show $f$ is Lebesgue integrable on $(0,1]$, then observe that $[0,1] \smallsetminus (0,1]$ is a set of Lebesgue measure zero, obtaining your goal.  (I'm having to guess at what tools are available to you -- you assert that DCT is unavailable, but you don't say what is available.)


答:
You want to solve $\sin(x)+u\cos(x)+v=0$. Writing $\cos(x)=y+y^{-1}$, $\sin(x)=\frac{y-y^{-1}}{i}$ with $y=e^{ix}$, giving you a degree two polynomial equation.$$
\frac{y-y^{-1}}{i} + u(y+y^{-1})+v=0
$$
So
$$
y^2(u-i)+yv+(u+i)=0
$$
I think you can finish yourself.

Title: Solving $\sin(\alpha)-\cos(\alpha)\tan(\beta) = -a\frac{\tan(\beta)}{b}$ for $\alpha$

Body:
I haven't practiced trigonometry for a while, and I'm wondering if there's a way to solve this analytically:

$$\sin(\alpha)-\cos(\alpha)\tan(\beta) = -a\frac{\tan(\beta)}{b}$$

I'm solving for $\alpha$, so $\beta$, $a$ and $b$ are known.

If there is, can someone point out the right way?

Thanks in advance!


问:
Numerically compute intersection of infinite line and rectangle
I'm trying to numerically compute where an infinite line defined by two points intersects a (finite) rectangle also defined by two points. Here's a illustration of the various ways that the line can intersect the rectangle:

How can I determine the number and location of the places where the infinite line intersects the rectangle? I know how to do this when the line is finite, but I can't figure out how to generalize that to an infinite context.

答:
Idea: find out on which side of the line the four corners lie

no corners on the line and

all four corners on one side: $0$ intersections

one corner on one side, three corners on the other side: $2$ intersections

two corners on one side, two on the other side: $2$ intersections

exactly one corner on the line and

three on one side: $1$ intersection

one on one side, two on the other sides: $2$ intersections

exactly two corners on the line and

they are diagonally opposite ones: $2$ intersections

they are adjacent ones: infinitely many intersections


问:
In how many ways can you divide $12$ people in $5$ distinct groups?
Problem:
$12$ students are asked to go into the groups $A$, $B$, $C$, $D$, and $E$. In how many ways can this division be done if there must at least one person per group?

I tried to approach this problem first by going through each possible group size. So for example, there could be group sizes $8, 1, 1, 1, 1$. There are $5$ ways to have this type of division. Then there are $12\cdot11\cdot10\cdot9=\frac{12!}{8!}$ ways of dividing the students into this type of arrangement. So in total, we have $5\cdot\frac{12!}{8!}$ ways to do this.

However, this approach is problematic as it is quite cumbersome and bruteforcey as there are quite a lot of ways to do the group division (how many persons are in each group).

So, is there a more general way of doing/approaching this type of problem?

答:
Apply inclusion - exclusion.

There are $5$ choices for each person, so $5^{12}$ total ways.

Applying inclusion-exclusion,

Total ways - at least one group empty + at least two groups empty -...

$5^{12} - \binom51 4^{12} + \binom52 3^{12} - \binom532^{12} + \binom541^{12}$


答:
See the Twelvefold Way and Stirling Numbers of the Second Kind.

First, count the number of ways to partition your $12$ distinct persons into $5$ unlabeled parts which is precisely what the Stirling Numbers of the Second Kind are used to count, and then choose a labeling of the parts in the partition in $5!$ ways for a total of:
$$5!\cdot {12\brace 5}$$

The articles above give information on how you might choose to calculate the Stirling Numbers of the Second Kind through various means, be it inclusion-exclusion or recurrence relations, or what else.

Title: In how many ways can you divide $12$ people in $5$ distinct groups?

Body:
Problem:
$12$ students are asked to go into the groups $A$, $B$, $C$, $D$, and $E$. In how many ways can this division be done if there must at least one person per group?

I tried to approach this problem first by going through each possible group size. So for example, there could be group sizes $8, 1, 1, 1, 1$. There are $5$ ways to have this type of division. Then there are $12\cdot11\cdot10\cdot9=\frac{12!}{8!}$ ways of dividing the students into this type of arrangement. So in total, we have $5\cdot\frac{12!}{8!}$ ways to do this.

However, this approach is problematic as it is quite cumbersome and bruteforcey as there are quite a lot of ways to do the group division (how many persons are in each group).

So, is there a more general way of doing/approaching this type of problem?


问:
Volume of parabolic cylinder bound by plane
Make a sketch of the solid in the first octant bounded by the plane

$x + y = 1$

and the parabolic cylinder

$x^{2} + z = 1$

Calculate the volume of the solid..

I have no idea where to even start :(

答: As can be seen from the plot below, we have a flat triangle with $(0,0),(0,1),(1,0)$ on $xy-$plane. So the required limits here is as $$x|_0^1,y|_0^{1-x},z|_0^{1-x^2}$$


问:
Show that $f''(\xi)≥8$

Let $f:[0,1]\rightarrow\mathbb{R}$ be a $\mathcal{C}^{2}([0,1])$ function such that $f(0)=f(1)=1$ and such that the minimum of $f$ is negative. Prove that there must exist a point $\xi\in [0,1]$ such that $f''(\xi)≥8$

Maybe I can use the fact that there exists maximum and minimum in $[0,1]$ so there are two critical points. But I don't know how to proceed. Can someone give me any advice? Thanks before!

答:
I can get to $4$, by mean value theorem.

Proof:
Let $x_{\min}$ denote the minimum point. By mean value theorem there is $c_1,c_2$ such that

$$f'(c_1)=\frac{f(x_{\min})-f(0)}{x_{\min}},f'(c_2)=\frac{f(1)-f(x_{\min})}{1-x_{\min}}.$$

And then, by mean value theorem there is $c_3$:

$$f''(c_3)=\frac{f'(c_2)-f'(c_1)}{c_2-c_1}$$

$$=\frac{\frac{f(1)-f(x_{\min})}{1-x_{\min}}-\frac{f(x_{\min})-f(0)}{x_{\min}}}{c_2-c_1}$$
$$>{\frac{1}{1-x_{\min}}+\frac{1}{x_{\min}}}$$
$$=\frac{1}{x_{\min}(1-x_{\min})}\geq 4$$

You can likely get a tighter bound following the comments.

EDIT: The correct way...

By Taylor's theorem, there is $c$ such that

$$f(x)=f(x_{min})+\frac{f''(c)(x-x_{min})^2}{2}.$$

Then

$$f''(c)=2\frac{f(x)-f(x_{min})}{(x-x_{min})^2}$$

Now develop this at $x=0,x=1$ and sum those, then there is $c_1$, $c_2$

$$\frac{f''(c_1)+f''(c_2)}{2}=\frac{1-f(x_{min})}{x_{min}^2}+\frac{1-f(x_{min})}{(1-x_{min})^2}>\frac{1}{x_{min}^2}+\frac{1}{(1-x_{min})^2}$$
$$=\frac{1-2x_{min}+2x_{min}^2}{(1-x_{min})^2x_{min}^2}\geq8$$

which can only hold if $f''(c_1)$ or $f''(c_2)$ is greater than $8$.

Title: Show that $f''(\xi)≥8$


问:
Using quotient to show that a map $\mathbb{RP}^2\rightarrow\mathbb{R}^3$ is an immersion at all but six points
We want to show that $\phi:\mathbb{RP}^2\rightarrow \mathbb{R}^3$ given by $\phi([x,y,z])=\frac{(yz,xz,xy)}{x^2+y^2+z^2}$ is an immersion at all but 6 points. Recall that a map $f$ is an immersion at $x$ if the matrix $Df|_x$ is injective.

We know that $\mathbb{RP}^2$ can be given by the quotient map $q:\mathbb{R}^3-\{0\}\rightarrow\mathbb{RP}^2$ by identifying one dimensional subspaces of $\mathbb{R}^3$. I believe that we have a natural map $(\phi\circ q)(x,y,z)=\frac{(yz,xz,xy)}{x^2+y^2+z^2}$. I think that the points where $\phi\circ q$ fails to be an immersion correspond to the points where $\phi$ fails to be an immersion. Then I wrote down the Jacobian of $D(\phi\circ q)|_{(x,y,z)}$ in hopes of obtaining $6$ one-dimensional subspaces where the matrix was non-injective, that is, has zero determinant. However, setting the determinant to be zero, I obtained the equation $xyz(-x^2+y^2+z^2)(x^2-y^2+z^2)(x^2+y^2-z^2)=0$, which in fact gives us three two-dimensional subspaces and $3$ cone-shapes in $\mathbb{R}^3$. This does not seem to correspond to $6$ points in $\mathbb{RP}^2$.

What am I misunderstanding here?

This was a "simple" question in a qualifying topology/geometry exam, so I do not believe the solution is too involved.

答:
Let us write $\xi = (x,y,z)$.

$\phi$ is an immersion at $[\xi]$ if the rank of $D\phi \mid_{[\xi]}$ is  $2$. Solving $\det D(\phi \circ q) \mid_{\xi} = 0$ gives you points at which $D(\phi \circ q) \mid_{\xi}$ has rank $< 3$. This does of course not exclude that the rank of $D\phi \mid_{[\xi]}$ is  $2$. In fact, we have
$$D(\phi \circ q) \mid_\xi = D\phi \mid_{[\xi]} \circ Dq \mid_\xi \tag{1}$$
which shows that $D(\phi \circ q) \mid_{\xi}$ will always have rank $\le 2$ (note that $D\phi \mid_{[\xi]}$ has rank $\le 2$) so that the determinant of the Jacobian of $D(\phi \circ q) \mid_{\xi}$ is always $0$.

Let $n(\xi) = x^2+y^2+z^2 = \lVert \xi \rVert^2$. Then the Jacobian of $D(\phi \circ q) \mid_{\xi}$ is
$$J(\xi) = \begin{pmatrix}
\frac{-2xyz}{n(\xi)^2} & \frac{z(n(\xi)-2y^2)}{n(\xi)^2} & \frac{y(n(\xi)-2z^2)}{n(\xi)^2}\\
\frac{z(n(\xi)-2x^2)}{n(\xi)^2} & \frac{-2xyz}{n(\xi)^2} & \frac{x(n(\xi)-2z^2)}{n(\xi)^2} \\
\frac{y(n(\xi)-2x^2)}{n(\xi)^2} & \frac{x(n(\xi)-2y^2)}{n(\xi)^2} & \frac{-2xyz}{n(\xi)^2}
\end{pmatrix}$$
It is easy to see that $\sqrt{n(\xi)} J(\xi) = J(\xi/\sqrt{n(\xi)}) = J(\xi/\lVert \xi \rVert)$. Thus $J(\xi)$ and $J(\xi/\lVert \xi \rVert)$ have the same rank. It therefore suffices to determine the rank for $\xi $ with $\lVert \xi \rVert = 1$. In that case
$$J(\xi) = \begin{pmatrix}
-2xyz & z(1-2y^2) & y(1-2z^2)\\
z(1-2x^2) & -2xyz & x(1-2z^2) \\
y(1-2x^2) & x(1-2y^2) & -2xyz
\end{pmatrix}$$

Case 1. $x,y, z \ne 0$.

Assume that $J(\xi)$ has rank $< 2$. Then all determinants of $2 \times 2$-submatrices must vanish. In particular
$$0 = (-2xyz)^2 - z(1-2x^2)z(1-2y^2) = -(1 -2(x^2 + y^2))z^2 = -(1 -2(1-z^2))z^2 = -(2z^2-1)z^2$$
which implies $2z^2-1 = 0$, i.e. $z^2 = 1/2$. Similarly $x^2 = 1/2$ and $y^2 = 1/2$. But then $1 = x^2 +y^2 + z^2 = 3/2$ which is a contradiction. Therefore $J(\xi)$ has rank $2$.

Case 2. Exactly one of $x,y,z$ is $0$.

Let us first consider $z = 0$. Then
$$J(\xi) = \begin{pmatrix}
0 & 0 & y\\
0  & 0 & x \\
y(1-2x^2) & x(1-2y^2) & 0
\end{pmatrix}$$
This matrix has rank $2$ unless $y(1-2x^2) = x(1-2y^2) = 0$. This means $x^2 = y^2 = 1/2$. Thus we get four points $\xi = (\pm \sqrt{1/2},\pm \sqrt{1/2},0)$ in which $J(\xi)$ has rank $1$.

The same is true for the four points $\xi = (\pm \sqrt{1/2},0,\pm \sqrt{1/2})$ and the four points $\xi = (0,\pm \sqrt{1/2},\pm \sqrt{1/2})$.

Case 3. Exactly two of $x,y,z$ are $0$.

Consider $y = z = 0$. Then
$$J(\xi) = \begin{pmatrix}
0 & 0 & 0\\
0  & 0 & x \\
0 & x & 0
\end{pmatrix}$$
which has rank $2$. The same is true if $x = z = 0$ and $x = y = 0$.

We conclude from $(1)$ that $D\phi \mid_{[\xi]}$ must have rank $2$ with the possible exception of six points $[\xi]$, the equivalence classes of the $12$ points $\xi$ detected in case 2. In fact the rank of $D\phi \mid_{[\xi]}$ is $1$ at these points. This follows from the fact $Dq \mid_{\xi}$ has rank $2$ in all points. This is fairly obvious, but let us prove it. Again it suffices to consider $\lVert \xi \rVert = 1$. With the possible exception of the $12$ points of case 2 this follows from $(1)$ because we know that $D(\phi \circ q) \mid_\xi$ has rank $2$ for all other points. We can now argue by symmetry that also in the above $12$ points $\xi_i$ the rank must be $2$. Formally, we can chose a rotation $R$ on $\mathbb R^3$ such that $R(\xi_i)$ is a non-exceptional point. Since $q(R(-\xi)) = q(-R(\xi)) = q(R(\xi))$, we see that $q \circ R$ induces a map $R' : \mathbb{RP}^2 \to \mathbb{RP}^2$ such that $q \circ R = R' \circ q$. Now $R$ and $R'$ are diffeomorphisms and it follows that the rank of $D(\phi \circ q) \mid_{\xi_i}$ agrees with the rank of $D(\phi \circ q) \mid_{R(\xi_i)}$.