问:
find the maximal number of path graphs of size n fit in a grid graph
I am trying to find a way to calculate or approximate the number of path graphs that can fit in a grid graph. The path graph consists of different states of the nodes.

An example of this path graph is:
$1 \rightarrow 2 \rightarrow 3$
So state 1 wants to be connected to state 2 and so forth.

The goal is to find the optimal configuration of states on the nodes such that the path graph is satisfied the maximal times possible in a 10x10 grid graph with periodic boundaries.

In this example the path is satisfied 3 times.
\begin{matrix}
1 & 2 & 3\\
1 & 2 & 3\\
1 & 2 & 3
\end{matrix}

This is an example of a setting where this path graph is satisfied 8 times in a grid graph.
\begin{matrix}
1 & 2 & 1\\
2 & 3 & 2\\
1 & 2 & 1
\end{matrix}

Is there a way to approximate the maximal number of satisfactions of a specific path graphs in a grid graph? I think it has something to do with partitions/number theory, but I can not find a specific solution to this problem. I hope someone can give me a direction to find a solution.

答:
You can solve the problem via integer linear programming as follows. Let $P_n$ be the set of possible paths with $n$ nodes.  In particular, $|P_2|=400$, $|P_3|=1200$, $|P_4|=3600$, and $|P_5|=10000$. For path $p\in P_n$ and $k \in \{1,\dots,n\}$, let $(i_{p,k},j_{p,k})$ be the nodes in the path, in order.  For $(i,j)\in \{1,\dots,10\}^2$ and $k \in \{1,\dots,n\}$, let binary decision variable $x_{i,j,k}$ indicate whether node $(i,j)$ takes value $k$.  For $p\in P_n$, let binary decision variable $y_p$ indicate whether path $p$ is selected.  The problem is to maximize $\sum_p y_p$ subject to
\begin{align}
\sum_k x_{i,j,k} &= 1 &&\text{for all $i,j$} \tag1\\
y_p &\le x_{i_{p,k},j_{p,k},k} &&\text{for all $p,k,i_{p,k},j_{p,k}$} \tag2
\end{align}
Constraint $(1)$ assigns exactly one value to each node.
Constraint $(2)$ forces the node values to be consistent with every selected path.

Here are the best solutions I have found for small $n$.  Not sure whether the last three are optimal.

$200$ paths (optimal) with $2$ nodes:

```<code>1 2 1 2 1 2 1 2 1 2
2 1 2 1 2 1 2 1 2 1
1 2 1 2 1 2 1 2 1 2
2 1 2 1 2 1 2 1 2 1
1 2 1 2 1 2 1 2 1 2
2 1 2 1 2 1 2 1 2 1
1 2 1 2 1 2 1 2 1 2
2 1 2 1 2 1 2 1 2 1
1 2 1 2 1 2 1 2 1 2
2 1 2 1 2 1 2 1 2 1
</code>```

$200$ paths with $3$ nodes:

```<code>1 2 1 2 1 2 1 2 1 2
2 3 2 3 2 3 2 3 2 3
1 2 1 2 1 2 1 2 1 2
2 3 2 3 2 3 2 3 2 3
1 2 1 2 1 2 1 2 1 2
2 3 2 3 2 3 2 3 2 3
1 2 1 2 1 2 1 2 1 2
2 3 2 3 2 3 2 3 2 3
1 2 1 2 1 2 1 2 1 2
2 3 2 3 2 3 2 3 2 3
</code>```

$244$ paths with $4$ nodes:

```<code>1 2 3 4 3 4 3 2 1 4
2 1 4 3 2 1 2 1 2 3
3 4 3 2 1 2 3 4 3 4
4 3 2 1 2 3 4 3 2 1
3 2 1 2 3 4 3 2 1 2
4 1 2 3 4 3 2 1 2 3
3 2 3 4 3 2 1 2 3 4
2 1 4 3 2 1 2 3 4 3
1 2 3 2 1 2 3 4 1 2
4 3 4 1 2 3 4 3 2 3
</code>```

$316$ paths with $5$ nodes:

```<code>1 2 3 4 5 2 1 4 3 2
2 1 2 5 4 3 2 5 4 3
3 2 5 4 3 2 1 2 5 4
4 5 4 3 2 1 2 3 4 5
5 4 3 2 1 2 3 4 5 4
2 3 2 1 2 3 4 5 4 3
1 2 1 2 3 4 5 4 3 2
4 5 2 3 4 5 4 3 2 1
3 4 5 4 5 4 3 2 1 2
2 3 4 5 4 3 2 1 2 1
</code>```

Body:
You can solve the problem via integer linear programming as follows. Let $P_n$ be the set of possible paths with $n$ nodes.  In particular, $|P_2|=400$, $|P_3|=1200$, $|P_4|=3600$, and $|P_5|=10000$. For path $p\in P_n$ and $k \in \{1,\dots,n\}$, let $(i_{p,k},j_{p,k})$ be the nodes in the path, in order.  For $(i,j)\in \{1,\dots,10\}^2$ and $k \in \{1,\dots,n\}$, let binary decision variable $x_{i,j,k}$ indicate whether node $(i,j)$ takes value $k$.  For $p\in P_n$, let binary decision variable $y_p$ indicate whether path $p$ is selected.  The problem is to maximize $\sum_p y_p$ subject to
\begin{align}
\sum_k x_{i,j,k} &= 1 &&\text{for all $i,j$} \tag1\\
y_p &\le x_{i_{p,k},j_{p,k},k} &&\text{for all $p,k,i_{p,k},j_{p,k}$} \tag2
\end{align}
Constraint $(1)$ assigns exactly one value to each node.
Constraint $(2)$ forces the node values to be consistent with every selected path.

Here are the best solutions I have found for small $n$.  Not sure whether the last three are optimal.

$200$ paths (optimal) with $2$ nodes:

1 2 1 2 1 2 1 2 1 2
2 1 2 1 2 1 2 1 2 1
1 2 1 2 1 2 1 2 1 2
2 1 2 1 2 1 2 1 2 1
1 2 1 2 1 2 1 2 1 2
2 1 2 1 2 1 2 1 2 1
1 2 1 2 1 2 1 2 1 2
2 1 2 1 2 1 2 1 2 1
1 2 1 2 1 2 1 2 1 2
2 1 2 1 2 1 2 1 2 1

$200$ paths with $3$ nodes:

1 2 1 2 1 2 1 2 1 2
2 3 2 3 2 3 2 3 2 3
1 2 1 2 1 2 1 2 1 2
2 3 2 3 2 3 2 3 2 3
1 2 1 2 1 2 1 2 1 2
2 3 2 3 2 3 2 3 2 3
1 2 1 2 1 2 1 2 1 2
2 3 2 3 2 3 2 3 2 3
1 2 1 2 1 2 1 2 1 2
2 3 2 3 2 3 2 3 2 3

$244$ paths with $4$ nodes:

1 2 3 4 3 4 3 2 1 4
2 1 4 3 2 1 2 1 2 3
3 4 3 2 1 2 3 4 3 4
4 3 2 1 2 3 4 3 2 1
3 2 1 2 3 4 3 2 1 2
4 1 2 3 4 3 2 1 2 3
3 2 3 4 3 2 1 2 3 4
2 1 4 3 2 1 2 3 4 3
1 2 3 2 1 2 3 4 1 2
4 3 4 1 2 3 4 3 2 3

$316$ paths with $5$ nodes:

1 2 3 4 5 2 1 4 3 2
2 1 2 5 4 3 2 5 4 3
3 2 5 4 3 2 1 2 5 4
4 5 4 3 2 1 2 3 4 5
5 4 3 2 1 2 3 4 5 4
2 3 2 1 2 3 4 5 4 3
1 2 1 2 3 4 5 4 3 2
4 5 2 3 4 5 4 3 2 1
3 4 5 4 5 4 3 2 1 2
2 3 4 5 4 3 2 1 2 1


问:
Regular partition in the Riemann sum Openstax Calculus 2
I am currently studying some mathematics using openstax. I am having a difficult time understanding what the following means:

It is also worth noting here that we have retained the use of a
regular partition in the Riemann sums. This restriction is not
strictly necessary. Any partition can be used to form a Riemann sum.
However, if a nonregular partition is used to define the definite
integral, it is not sufficient to take the limit as the number of
subintervals goes to infinity. Instead, we must take the limit as the
width of the largest subinterval goes to zero.

I am guessing that the use of a regular partition means either (1) we have "blocks" of area or (2) we have a closed interval [a, b] that we are calculating on.

I am unsure what parition mean, I am again guessing that it is the closed interval [a, b].

My confusion is constant through the quote but I cannot extract even a single quess about what the largest subinterval goes to zero means. Can anybody explain in a different way what the text is trying to say?

答:
A partition of the interval $[a, b]$ is a set of points: $$\{ x_0 = a, x_1, x_2, \ldots x_n = b \}$$
Each pair of points defines a subinterval.

If the partition is regular, then the width of each subinterval is the same, so $\Delta x = \displaystyle\frac{b - a}{n}$. In this case, as the number of subintervals approaches infinity, the width of each subinterval approaches zero, and the sum approaches the value of the definite integral it represents.

If the partition is not regular, then the width of each subinterval does not have to be the same. In this case, we must force the width of the largest subinterval to approach zero in order for the Riemann sum to converge to the value of the definite integral it represents.

ContentLicense: CC BY-SA 4.0


问:
Necessary condition for local minima; non-negative Hessian matrix
The problem I have is the following. Any results on Taylor expansions etc. can be assumed:

Let F : R^n -> R  be a C2 function. Let x_0 be a local minimum of F. Prove that the Hessian matrix of F is non-negative.

Thanks for your help.

答:
Let $\def\R{\mathbb R}h \in \R^n$. Define $g \colon \R\to \R$ by $g(t) = F(x_0 + th)$. Then $g$ is a $C^2$-function, having a local minimum at $0$. Hence $g'(0) = 0$, $g''(0) \ge 0$. By the chain rule, we have for $t\in \R$:
\begin{align*}
  g'(t) &= F'(x_0 + th)h\\
  g''(t) &= F''(x_0 + th)[h,h]
\end{align*}
For $t = 0$, we get
$$ g'(0) = F'(x_0)h, g''(0) = F''(x_0)[h,h] $$
So we have $0 \le F''(x_0)[h,h]$. As $h$ was arbitrary, the bilinear mapping $F''(x_0)$ is non-negative, and hence it is representing matrix, the Hessian.</p>

ContentLicense: CC BY-SA 3.0


问:
Let $[0,\omega_1)$ be the set of all countable ordinals with the order topology. Prove that $[0,\omega_1)$ is star-Rothberger.
A space $X$ is said to be star-Rothberger if for each sequence $(\mathcal{U}_n)$ of open covers of $X$ there exists a sequence $(U_n)$ such that for each $n$ $U_n\in\mathcal{U}_n$ and $\{St(U_n,\mathcal{U}_n) : n\in\mathbb{N}\}$ is an open cover of $X$.

答:
HINT: Let $X=[0,\omega_1)$. Use the pressing-down lemma to show that if $\mathscr{U}$ is an open cover of $[0,\omega_1)$, then there is an $\alpha\in X$ such that

$$\bigcup\{U\in\mathscr{U}:\alpha\in U\}\supseteq[\alpha,\omega_1)\,,$$

and conclude that there is a $U\in\mathscr{U}$ such that $\operatorname{st}(U,\mathscr{U})\supseteq(\alpha,\omega_1]$ and hence $X\setminus\operatorname{st}(U,\mathscr{U})$ is compact.


问:
Probability measure is countably additive over almost disjoint sets.
Let $A_n, n\ge1,$ be events in a probability space $(\Omega,\mathcal{F}, P)$ that satisfy the property $$P(A_n\cap A_m)=\emptyset, m\not=n$$
Prove $P(\bigcup_{n\ge 1} A_n)=\sum_{n\ge 1}P(A_n).$

I can prove the result for finite unions by using the measure-theoretic version of the Inclusion-Exclusion Principal, but this is useless since $\lim P(\bigcup _{i=1} ^n A_i)$ isn't necessarily equal to $P(\bigcup _{i\ge 1}A_i).$
 Does the Inclusion-Exclusion Principle hold for infinite unions? I believe there's an easier way to solve than using that principal, but I am not sure how to do it.

I tried constructing a new sequence $B_n=A_n-\bigcup_{i=1}^{n-1}A_i$, but this doesn't allow me to use the assumption. The sequence $B_n=A_n-\bigcap_{i=1} ^nA_i$ doesn't suffice since it's not a disjoint sequence.

答:
Yes, Inclusion-Exclusion Principle holds for infinite unions.
Since you can show that $P(\bigcup_{i=1}^{n}A_i)=\sum_{i=1}^{n}P(A_i)$, then let's continue on here.

Since $\bigcup_{i=1}^{n}A_i \subseteq \bigcup_{i=1}^{\infty}A_i$, then we have
$$\sum_{i=1}^{n}P(A_i)=P(\bigcup_{i=1}^{n}A_i)\leq P(\bigcup_{i=1}^{\infty}A_i)$$ for all n.
Therefore, we have
$$\sum_{i=1}^{\infty}P(A_i)\leq P(\bigcup_{i=1}^{\infty}A_i)\leq \sum_{i=1}^{\infty}P(A_i)$$ where the second inequality is given by sub-additivity property. Thus, we got the equality.

Alternatively, you can apply continuity of probability, so
$$\lim_{n \rightarrow \infty} P(\bigcup_{i=1}^{n}A_i)=P\Big(\lim_{n \rightarrow \infty}(\bigcup_{i=1}^{n}A_i)\Big)=P(\bigcup_{i=1}^{\infty}A_i)$$
$$\lim_{n \rightarrow \infty}\sum_{i=1}^{n}P(A_i)=\sum_{i=1}^{\infty}P(A_i)$$


问:
Rewrite as conic optimization problem
Can I rewrite the convex optimization problem

\begin{align}
&\min \,\, f(x_1,x_2) \\
&\text{s.t.} \, \, x_1^2 + x_2^2 \leq r
\end{align}

with $r \in \mathbb{R}$ as a conic optimization problem? As far as I know, the set described by the constraint is not a cone, since for example it is not closed w.r.t. addition or a scalar multiplication. However, I found the post where a problem with a similar quadratic equality constraint is described as conic. What is the difference?

答:
For a problem to be conic, you do not require $x$ itself to be in a cone, but rather $Ax+b\in K$, where $K$ is a cone. The constraint in your case is

$$\begin{pmatrix}1 & 0  \\ 0 & 1 \\ 0 & 0 \end{pmatrix}x + \begin{pmatrix}0 \\ 0 \\ \sqrt{r}\end{pmatrix} \in K$$
where $K$ is the second order cone.

Whether the problem is conic depends on the objective function.

Title: Rewrite as conic optimization problem


问:
Statement about Weil representations being unramified over a certain extension
Let $K = \mathbb{Q}_p$ and $\rho: W_K \to \operatorname{GL}(V)$ be a Weil representation over a finite-dimensional complex vector space $V$ (where $W_K$ denotes the Weil group over $K$).

Let $L/K$ be a finite cyclic extension of degree $n$ and ramification index $e$ such that the restriction $\rho|_L: G_L \to \operatorname{GL}(V)$ is unramified (equivalently, $\rho(I_L)=1$ where $I_L$ is the inertia group over $L$).

Let $L'/K$ be a totally ramified of degree $e$ such that $LL'/L'$ is unramified of degree $n$.

Question: Is $\rho|_{L'}$ unramified, i.e. $\rho(I_{L'})=1$?

I am certain that if this is true, the only chance to show it is by using the fact that $LL'/L'$ is unramified. But I am not sure how this should work. Could you please help me with this problem?

Body:
I'd say use that
$$\overline{K}^{\displaystyle\ker(\rho|_{L'})}(\zeta_{p^\infty-1})=\overline{K}^{\displaystyle\ker(\rho|_{L'})\cap G_{L'(\zeta_{p^\infty-1})}}=\overline{K}^{\displaystyle\ker(\rho|_{L'(\zeta_{p^\infty-1})})}$$ $$=\overline{K}^{\displaystyle\ker(\rho|_{L(\zeta_{p^\infty-1})})}=L(\zeta_{p^\infty-1})=L'(\zeta_{p^\infty-1})$$


问:
The difference between log and ln
$$\dfrac{1}{2}\ln(x+7)-(2 \ln x+3 \ln y)$$

Our professor let's us solve this but i do not understand how $\ln$ works. He says it has same properties with $\log$ but i still don't get it. What's the difference of the two?

答:
The symbol $\log$, by itself and without other conventions, is meaningless, as is the word "logarithm."

The phrase "logarithm of $x$ to the base $b$" has meaning if $b$ and $x$ are positive.  The meaning is defined as follows: the logarithm of $x$ to the base $b$ (denoted by  $\log_{b} (x)$) is the exponent to which $b$ needs to be raised to obtain $x$.

That is, if $y = \log_{b} (x)$, then $x = b^y$.

By convention, the "natural logarithm" is the logarithm to the base $e$, where $e$ is Euler's constant:
$$
\ln(x) = \log_{e}(x).
$$

Title: The difference between log and ln


问:
Prove that $\sup(S^2)=\sup(S)^2$ where $S$ is a nonempty bounded set of nonnegative real numbers.
It seems like this statement is common sense so I don't know how to prove it using mathematical logic. S={x : x \in S }, and S^2={x^2 : x \in S }. Would it suffice to say that because the numbers in the set are nonnegative, that taking the supremum before or after squaring the set is irrelevant? How do I say this as a proof?</p>

答:
Let $s:=\sup S$. Then for all $x\in S$ we have $x\le s$. Squaring, we obtain $x^2\leq s^2$ and thus $\sup(S^2)\leq s^2=\sup(S)^2$.

Let $\epsilon>0$. Choose $x\in S$ with $s\leq \epsilon + x$. Then $$s^2\le (\epsilon+x)^2=\epsilon^2 +2\epsilon x + x^2\leq \epsilon^2+2\epsilon x +\sup(S^2) \leq \epsilon^2 + 2\epsilon s + \sup(S^2)$$ Letting $\epsilon\to 0$, we obtain $s^2\leq \sup(S^2)$.

Thanks to @Daniel Schepler for the useful comment.


问:
Question on two equivalent densities
I have two integrals $I_1$ and $I_2$ that are almost similar :
$$I_1=\int_K^{+\infty}(x-K)f_1(x)dx$$ $$I_2=\int_K^{+\infty}(x-K)f_2(x)dx$$ with$f_1$ and $f_2$ being two equivalent densities (so they are always positive or null, both integrate to 1 over their domain, and are "good" functions :their limit value when x goes to $-\infty$ or to $+\infty$ is null).

What conditions must respect $f_1$ and $f_2$ to have $I_1 < I_2$ whatever the value of K is ?

答: Unless I misunderstand your question, why not $f_2(x)=f_1(x-c)$ where $c\gt 0$?

Title: Question on two equivalent densities


问:
Divisors of n! including 1 which sum to n!
I believe that given any positive integer $n$, we can find a set of divisors of $n!$, including $1$, which sum to $n!$ I checked for the first 6 positive integers, and tried to prove it by induction on $n$. However I was unable to formalise the proof, can anybody present a method of proving this? Or maybe without induction?

答: You can do this by induction. For $n=3$, you already observed that $1+2+3$. Now take $n>3$, then $(n-1)!$ is sum of some of its divisors, with $1$ included among them. By multiplying this equality by $n$, you'll get sum of divisors of $n!$, summing to $n!$, and the smallest of them is $n$. Now we can just put $1+(n-1)$ instead of $n$ and we're done.


Title: Need help to prove

Body:
I got the result below during my research.

$$1=\frac{1}{1+a_1}+\frac{a_1}{(1+a_1)(1+a_2)}+\frac{a_1a_2}{(1+a_1)(1+a_2)(1+a_3)}+\frac{a_1a_2a_3}{(1+a_1)(1+a_2)(1+a_3)(1+a_4)}+... \tag 1$$

$$1=\frac{1}{1+a_1}+\sum\limits_{k=1}^ \infty\frac{\prod\limits_{j=1}^{k}a_j}{\prod\limits_{j=1}^{k+1} (1+a_j)}  $$

$$1+a_1=1+\frac{a_1}{1+a_2}+\frac{a_1a_2}{(1+a_2)(1+a_3)}+\frac{a_1a_2a_3}{(1+a_2)(1+a_3)(1+a_4)}+...$$

$$a_1=\frac{a_1}{1+a_2}+\frac{a_1a_2}{(1+a_2)(1+a_3)}+\frac{a_1a_2a_3}{(1+a_2)(1+a_3)(1+a_4)}+...$$

We can get the same relation but without $a_1$
$$1=\frac{1}{1+a_2}+\frac{a_2}{(1+a_2)(1+a_3)}+\frac{a_2a_3}{(1+a_2)(1+a_3)(1+a_4)}+...$$

Examples:

Example-1:
$a_n=c$

$$1=\frac{1}{1+c}+\frac{c}{(1+c)^2}+\frac{c^2}{(1+c)^3}+\frac{c^3}{(1+c)^4}+...$$

$$1+c=1+\frac{c}{1+c}+\frac{c^2}{(1+c)^2}+\frac{c^3}{(1+c)^3}+...$$

We know that if $\frac{c}{(1+c)}<1$

then $$1+\frac{c}{1+c}+\frac{c^2}{(1+c)^2}+\frac{c^3}{(1+c)^3}+...=\frac{1}{1-\frac{c}{(1+c)}}=1+c$$

Example-2:
$a_n=x^{2^{n-1}}$

$$1=\frac{1}{1+x}+\frac{x}{(1+x)(1+x^2)}+ \frac{x^3}{(1+x)(1+x^2)(1+x^4)}+ \frac{x^7}{(1+x)(1+x^2)(1+x^4)(1+x^8)}+...$$

$$1=\frac{1-x}{1-x^2}+\frac{x(1-x)}{1-x^4}+\frac{x^3(1-x)}{1-x^8} +\frac{x^7(1-x)}{1-x^{16}} +...$$

$$1=\frac{1-x}{1-x^2}+\frac{x(1-x)}{1-x^4}+\frac{x^3(1-x)}{1-x^8} +\frac{x^7(1-x)}{1-x^{16}} +...$$

$$\frac{x}{1-x}=\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8} +\frac{x^8}{1-x^{16}} +...$$

If we put $x---->x^2$

$$\frac{x^2}{1-x^2}=\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8} +\frac{x^8}{1-x^{16}} +....$$

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

Example-3:
$a_n=n$

$$1=\frac{1}{2}+\frac{1}{2.3}+\frac{1.2}{2.3.4}+\frac{1.2.3}{2.3.4.5}+... $$

$$1=\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+... $$

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

I wonder which general conditions are required for $a_n$ that Equation $ 1 $ is true.

Thanks for answers and comments.

Body:
I proved the relation and required condition for  $a_n$. I would like to share it.

$$(1+a_1)(1+a_2)(1+a_3)(1+a_4)(1+a_5)... =A\tag 1$$

$$A/A=1=\frac{(1+a_2)(1+a_3)(1+a_4)(1+a_5)...}{(1+a_1)(1+a_2)(1+a_3)(1+a_4)(1+a_5)... }+a_1\frac{(1+a_2)(1+a_3)(1+a_4)(1+a_5)...}{(1+a_1)(1+a_2)(1+a_3)(1+a_4)(1+a_5)... } \tag 2$$

$$1=\frac{1}{(1+a_1)}+a_1\frac{(1+a_2)(1+a_3)(1+a_4)(1+a_5)...}{(1+a_1)(1+a_2)(1+a_3)(1+a_4)(1+a_5)... } \tag 3$$

$$1=\frac{1}{(1+a_1)}+a_1\frac{(1+a_3)(1+a_4)(1+a_5)...+a_2(1+a_3)(1+a_4)(1+a_5)...}{(1+a_1)(1+a_2)(1+a_3)(1+a_4)(1+a_5)... } \tag 4$$

$$1=\frac{1}{(1+a_1)}+\frac{a_1}{(1+a_1)(1+a_2)}+a_1a_2\frac{(1+a_3)(1+a_4)(1+a_5)...}{(1+a_1)(1+a_2)(1+a_3)(1+a_4)(1+a_5)... } \tag 5$$

$$1=\frac{1}{(1+a_1)}+\frac{a_1}{(1+a_1)(1+a_2)}+a_1a_2\frac{(1+a_4)(1+a_5)...}{(1+a_1)(1+a_2)(1+a_3)(1+a_4)(1+a_5)... }+a_1a_2\frac{a_3(1+a_4)(1+a_5)...}{(1+a_1)(1+a_2)(1+a_3)(1+a_4)(1+a_5)... } \tag 6$$

$$1=\frac{1}{(1+a_1)}+\frac{a_1}{(1+a_1)(1+a_2)}+\frac{a_1a_2}{(1+a_1)(1+a_2)(1+a_3)}+a_1a_2a_3\frac{(1+a_4)(1+a_5)...}{(1+a_1)(1+a_2)(1+a_3)(1+a_4)(1+a_5)... } \tag 7$$

If we continue in that way to create series then the last term will be the same as  @polkjh wrote in comments.

$$\frac{a_1a_2a_3a_4a_5....}{(1+a_1)(1+a_2)(1+a_3)(1+a_4)(1+a_5)... }\tag 8$$

The last term must go to zero otherwise the relation I wrote above must be as shown below

$$1-\frac{\prod\limits_{j=1}^{\infty}a_j}{\prod\limits_{j=1}^{\infty} (1+a_j)}=\frac{1}{1+a_1}+\frac{a_1}{(1+a_1)(1+a_2)}+\frac{a_1a_2}{(1+a_1)(1+a_2)(1+a_3)}+\frac{a_1a_2a_3}{(1+a_1)(1+a_2)(1+a_3)(1+a_4)}+... \tag 9$$

$$1-\frac{\prod\limits_{j=1}^{\infty}a_j}{\prod\limits_{j=1}^{\infty} (1+a_j)}=\frac{1}{1+a_1}+\sum\limits_{k=1}^ \infty\frac{\prod\limits_{j=1}^{k}a_j}{\prod\limits_{j=1}^{k+1} (1+a_j)}  \tag {10}$$

If $\frac{\prod\limits_{j=1}^{\infty}a_j}{\prod\limits_{j=1}^{\infty} (1+a_j)}=0$ then the  relation will be

$$1=\frac{1}{1+a_1}+\frac{a_1}{(1+a_1)(1+a_2)}+\frac{a_1a_2}{(1+a_1)(1+a_2)(1+a_3)}+\frac{a_1a_2a_3}{(1+a_1)(1+a_2)(1+a_3)(1+a_4)}+... \tag {11}$$

Also The formula can be written for n terms.
$$1-\frac{\prod\limits_{j=1}^{n}a_j}{\prod\limits_{j=1}^{n} (1+a_j)}=\frac{1}{1+a_1}+\sum\limits_{k=1}^ {n-1} \frac{\prod\limits_{j=1}^{k}a_j}{\prod\limits_{j=1}^{k+1} (1+a_j)}  \tag {12}$$


问:
How do I rotate a polynomial algebraically?
How do I algebraically rotate a polynomial 90 degrees CW (clockwise) or CCW (counterclockwise) in the $xy$ plane?

For example, rotate $f(x) = x^2$ clockwise $90$ degrees. I understand that a CW rotation of 90 degrees takes a point $(x,y)$ to a point $(y,-x)$, but when I substitute and simplify this, I get $-x = y^2$, then $y = \pm \sqrt{-x}$. I know it should be $y = \pm \sqrt{x}$ and I don't understand how it rotated backwards. I teach algebra II to high schoolers and they do not yet know radians or parametric equations, so I would like a simple explanation if at all possible. Thank you!

答:
Short answer: There is a difference between rotating the plane, and rotating the coordinate grid that you've put on top of the plane. Moving one of them one way yields the same effect as moving the other the exact opposite way.

It's the same reason that $f(x-5)$ moves the graph of $f$ five units to the right compared to $f(x)$, when subtracting ought to move things to the left: it actually moves the coordinate grid to the left, keeping the graph in place. But you don't see that, as the coordinate grid is your only point of reference.


问:
Prove $\sum_{j=1}^{n}{j\cdot j!} = (n+ 1)!−1$
I got to the point in my proof by induction of (k+1+1)! -1 but I don't know where to go from there.</p>

Body:
Hint: the induction step is just:
$$
(n+1)\cdot(n+1)!+(n+1)!-1=[(n+1)+1]\cdot(n+1)!-1=(n+2)!-1.
$$


答:
you can use this
$$j.j! = (j+1)! - j! $$

Title: Prove $\sum_{j=1}^{n}{j\cdot j!} = (n+ 1)!−1$

Body: I got to the point in my proof by induction of (k+1+1)! -1 but I don't know where to go from there.</p>


问:
Exchangeability of a continuous linear operator with an integral
Let's consider a compact metric space X, $K:X\times X \to \mathbb{R}$ is a continuous kernel, $H_K$ is the induced RKHS and $\nu$ is a finite measure on X. We define the integral operator associated to K as:

\begin{align*}
  L_K \colon & L_2(\nu) \to H_K\\
  & f \mapsto \int_X K_x f(x) \nu(dx)
\end{align*}

I want to prove the following (which I am not sure is true): $$\| i(f) \|_2^2 = \langle f, L_K \circ i(f) \rangle_{H_K}$$ where $f$ is in $H_K$ and $i$ is the continuous embedding from $H_K$ to $L_2(\nu)$ . My proof is as follows, since $f$ is in $H_K$, using the reproducing property:

$$\| i(f) \|_2^2 = \int f(x)^2 \nu(dx) = \int_X f(x)\langle f, K(x,.) \rangle_{H_K} \nu(dx) = \langle f, \int_X f(x) K(x,.) \nu(dx) \rangle_{H_K} = \langle f, L_K \circ i(f) \rangle_{H_K}$$

I don't know how to justify the step with the exchange of the integral with the scalar product. Is it true that if $H$ is a Hilbert space, $X$ is a metric space, $\phi:X \to H$ and $A$ is a continuous linear operator from $H$ to $\mathbb{R}$ then $$A\left( \int_X \phi(x)\nu(dx) \right) =  \int_X A( \phi(x) ) \nu(dx)$$

And if true, what would be a proof?

答: Yes. An integral is a limit of linear combinations; those two are addressed precisely by the continuity and the linearity of $A$.


问:
What is the Mellin transform of $s^n \Gamma(s)$?
I am looking for the inverse Mellin Transform of $s^n \Gamma(s)$.  I found a table in which it is given as
$$\left(x \frac{d}{dx}\right)^n e^{-x},$$
but I do not know how to interpret the $(x \frac{d}{dx})^n$. Is it the product derivate of a function??
I know it is a silly question but it is bother me that in Mathematics using a imprecise symbols?

答: $(x\frac{d}{dx})^n$ is just a short notation for taking a derivative and multiplying by $x$ $n$ times in a row.


问:
Limiting Sum including Stirling numbers
I would like to simplify the following summation including $S(m,n)$, Stirling numbers of the Second kind.

$\begin{gather*}\\
&&
\frac{1}{n^{n+1}}\sum_{k=1}^{n-1}\binom{n}{n-k}S\left( n,n-k\right)
\left( n-k\right) !k\left( 1-\left( \frac{n-2}{n-1}\right) ^{k}\right)
&&
\end{gather*}$

I conjecture (in fact, computations show) that this sum approaches to $\frac{1}{e}(1-e^{-\frac{1}{e}})$ as $n$ approaches to $\infty$  . However, I do not have a proof.

It would be great if this can be simplified for any $n$ (I am not sure whether this is possible) or at least show the limiting result.

答:
A different approach, first do $k=n-k$ so make the summation go in the inverse way and distribute, we get
$$
\frac{1}{n^{n+1}}\sum_{k=1}^{n-1}\binom{n}{n-k}S\left( n,n-k\right)
\left( n-k\right) !k\left( 1-\left( \frac{n-2}{n-1}\right) ^{k}\right)=\frac{1}{n^{n+1}}\left ( \sum_{k=1}^{n-1}\binom{n}{k}S\left( n,k\right)
\left( k\right) !(n-k)\right )-\frac{1}{n^{n+1}}\left ( \sum_{k=1}^{n-1}\binom{n}{k}S\left( n,k\right)
\left( k\right) !(n-k) \left (1-\frac{1}{n-1}\right )^{n-k}\right ),
$$
Recall that $x^n=\sum _{k=1}^nS(n,k)x^{\underline{k}}=\sum _{k=1}^nS(n,k)\binom{x}{k}k!,$  so that the first part becomes
$$\frac{1}{n^{n+1}}(n^{n+1}-n(n^n-(n-1)^n)),$$
where the negative part comes from, for example, OEIS. the limit of this is $1/e.$

Now, for the second part, one has
$$\frac{1}{n^{n+1}}\left ( \sum_{k=1}^{n-1}\binom{n}{k}S\left( n,k\right)
\left( k\right) !(n-k) \left (1-\frac{1}{n-1}\right )^{n-k}\right )=\frac{1}{n^{n+1}}\left ( \sum_{k=1}^{n-1}\binom{n}{n-k}S\left( n,k\right)
\left( k\right) !(n-k) \sum _{l=0}^{\infty}\binom{n-k}{l}(-1/(n-1))^l\right ),$$
exchanging the two sums and using $\binom{a}{b}\binom{b}{c}=\binom{a}{c}\binom{a-c}{b-c},$ one gets
$$\frac{1}{n^{n+1}}\sum _{l=0}^{\infty}\binom{n}{l}(-1/(n-1))^l\sum _{k=1}^{n-1}\binom{n-l}{n-k-l}k!S(n,k)(n-k),$$
which looks a lot like the first problem. Except that now $x=n-l.$
So we can check that the inner sum is
$$n(n-l)^n-n((n-l)^n-(n-l-1)^n),$$
so if one pushes the limit inside (ALERT: I am used to combinatorics, probably one has to check something about that pushing the limit) one then can see using that $\binom{n}{l}\sim n^l/l!,$
that one gets $$-\sum _{l=0}^{\infty}\frac{(-1)^{l}}{l!}e^{-(l+1)}=\frac{-1}{e}e^{-e^{-1}}.$$
Which agrees with your computation.

Title: Limiting Sum including Stirling numbers


问:
Finding length of arc intercepted by two tangent lines given radius and distance from circle
I'm in the process of writing a script for work but realized that I need to calculate the length of a circular arc intersected by two tangent lines, and when I try finding it I get lost in a mess of trig that seems to go nowhere.

Here's what I have so far:

Tangent lines intersect each other at point $P$

Tangent lines intersect circle at points $A,B$

Circle radius is $R$

Point on circle nearest to $P$ is $Q$

Length of $\vec{PQ}$ is $T$

Using some trig, I found that $\vec{PA}=\vec{PB} = R\arccos(\frac{R}{R+T})$

But that's where I'm stuck. How can I find $\widehat{AQB}$ ?

答:
$OAP$ is a right triangle because $PA$ is tangent to the circle. With other given data we are able to find angle $\widehat O$ in this triangle:
$$\cos O = \frac{OA}{OP} = \frac{R}{R+T}$$
$$\widehat {AOQ} = arccos \frac{R}{R+T} $$
Now, it is easy to show that $\widehat{QOB}=\widehat{AOQ}$ and therefore
$$\widehat {AOB} = 2 arccos \frac{R}{R+T}$$
The above is the angle of the arc $AQB$. If you are looking for the other arc that connects $A$ to $B$, it is $2 \pi - 2 arccos \frac{R}{R+T}$. And if you are looking for the angle $\widehat{AQB}$ , it is half of the latter arc:
$$\widehat{AQB} = \pi - arccos \frac{R}{R+T} $$ .

May I ask, what kind of programming project are you involved in?


问:
Number of ways of traversing a graph through all of its nodes?

A lift has $N$ stops ($1,2,3,4,...,N$), hence have $N(N-1)$ distinct rides of travelling from floor $A$ to floor $B$ such that $A\neq B$. How many arrangements of these rides form a continuous trip that starts from floor 1 and ends at floor 1?

I was lost trying to solve it. Anyone has an idea?

答:
There are $N-2$ possible intermediate stopping points; at floors $2, 3, ..., N-1$.

Each one of these can either be chosen or not and so the total number of arrangements is $2^{N-2}.$


问:
Covariance of Linear Combinations
Can the covariance of a linear combination be written as:

$\sum_{i=1}^n\sum_{j=1}^n a_ia_jCov[X_i,Y_j]$

My problem is I'm trying to figure out an exact property for finding:

$Cov[\frac{1}{2}\epsilon_t + \frac{1}{2}\epsilon_{t-1}, \frac{1}{2}\epsilon_{t-1} + \frac{1}{2}\epsilon_{t-2}]$ where $\epsilon_t$ is i.i.d. $(0, \sigma^2)$ and independent white noise.

So I am thinking I can just do the following:

$(0.5 \cdot0.5)Cov[\epsilon_t, \epsilon_{t-1}] + (0.5 \cdot0.5)Cov[\epsilon_t, \epsilon_{t-2}] +
(0.5 \cdot0.5)Cov[\epsilon_{t-1}, \epsilon_{t-1}] + (0.5 \cdot0.5)Cov[\epsilon_{t-1}, \epsilon_{t-2}]$

Which then comes out to:

$(0.5 \cdot 0.5)Cov[\epsilon_{t-1}, \epsilon_{t-1}] = \frac{1}{4}\sigma^2$ because the covariance between lags should just be $0$ so three of the four terms fall out.

Is all of this correct or am I missing something?

答:
Yes, that's correct. Covariance has the property you mentioned.$Cov(aX+bY,cZ+dW)=abCov(X,Y)+acCov(X,Z)+bcCov(Y,Z)+bdCov(Y,W)$.

For your case, $$\begin{split}.25cov(e_t, e_{t-1})+.25cov(e_t, e_{t-2})+.25cov(e_{t-1}, e_{t-1})+ .25cov(e_{t-1}, e_{t-2})&=0+0+.25var(e_{t-1})+0\\
&=.25\sigma^2\end{split}$$

Title: Covariance of Linear Combinations


问:
Calculating the $L^1$-norm of a piecewise linear and continuous function
Let $0 = x_0 < x_1 < \dots x_n =1 $ (e.g. the $x_i$'s are equidistantly distributed) and $c_0, c_1, \dots, c_n \in \mathbb R$. Let $f : [0,1] \to \mathbb R$ be the unique continuous, piecewise linear function such that $f(x_i)=c_i$ for all $i$. I want to calculate the $L^1$ norm of $f$, i.e. the value
$$
I = \int_0^1|f(t)| \, dt.
$$
Is there an easy and direct formula for $I$ in terms of the $x_i$'s and $c_i$'s?

Thanks in advance

答:
You can split up the integral by using $\int_a^c f(x)dx = \int_a^b f(x)dx + \int_b^c f(x)dx$:

$$\int_0^1|f(t)| \, dt = \sum_{i=0}^{n-1} \int_{x_i}^{x_{i+1}} |f(t)| \, dt.$$

The part $\int_{x_i}^{x_{i+1}} |f(t)| \, dt$ is simple if both $c_i$ and $c_{i+1}$ have the same sign, because then this integral is just the average height multiplied with the width: $\int_{x_i}^{x_{i+1}} |f(t)| \, dt=0.5|c_i+c_{i+1}|(x_{i+1}-x_i)$.

If $c_i$ and $c_{i+1}$ have opposing signs, then $\int_{x_i}^{x_{i+1}} |f(t)| \, dt$ is the area of two triangles. The first triangle has width $(x_{i+1}-x_i)|c_i|/(|c_i|+|c_{i+1}|)$ and height $|c_i|$, the other triangle has width $(x_{i+1}-x_i)|c_{i+1}|/(|c_i|+|c_{i+1}|)$ and height $|c_{i+1}|$, resulting in:
$$\int_{x_i}^{x_{i+1}} |f(t)| \, dt = 0.5(x_{i+1}-x_i)\frac{|c_i|^2+|c_{i+1}|^2}{(|c_i|+|c_{i+1}|)}.$$
The factor $(|c_i|^2 + |c_{i+1}|^2) / (|c_i| + |c_{i+1}|)$ simplifies to:
\begin{align}
((c_i - c_{i+1})^2 + 2c_i c_{i+1}) / |c_i - c_{i+1}| &= |c_i - c_{i+1}|+2c_i c_{i+1} / |c_i - c_{i+1}| \\ &= |c_i| + |c_{i+1}|+2c_i c_{i+1} / |c_i - c_{i+1}|
\end{align}
The first two terms appear in the 'same sign' formula (as $|c_i + c_{i+1}|$) and the last term is negative. The formulas can therefore be combined into:
$$\int_0^1|f(t)| \, dt = \sum_{i=0}^{n-1} 0.5(x_{i+1}-x_i) \left(|c_i| + |c_{i+1}| + \min\left\{0, 2c_i c_{i+1} / |c_i - c_{i+1}|\right\}\right).$$


问:
How to solve $(z-1)^n=(z+1)^n$
I have problem with this equation because when I put
$\left(\dfrac{z+1}{z-1}\right)^n=1$ I find solutions different than when I put $\left(\dfrac{z-1}{z+1}\right)^n=1$
So which one should I use ?
For first one -icot(kπ/5) and icot(kπ/5) for the second one

答: The approaches agree. Whether you argue$$\frac{z+1}{z-1}=e^{2\pi ik/n},\,k\in\Bbb Z\implies z=\frac{e^{2\pi ik/n}+1}{e^{2\pi ik/n}-1}=-i\cot\frac{\pi k}{n}$$or$$\frac{z-1}{z+1}=e^{2\pi il/n},\,l\in\Bbb Z\implies\frac{z+1}{z-1}=e^{2\pi ik/n},\,k=-l,$$from which we can continue with the above logic, the solution sets are the same.

Title: How to solve $(z-1)^n=(z+1)^n$


问:
Proof of the expected value of $|X|$ for a stable distribution?
Nassim Nicholas Taleb's Statistical Consequences of Fat Tails has a proof (Section 7.7) of the expected value of the the mean deviation for a stable distribution. In the midst of the proof, Taleb says:

The Hilbert transform $H$

$(H f)(t)=\frac{2}{\pi i}\int_0^\infty \chi_S(u+t)-\chi_S(u-t) dt$

can be rewritten as

$(H f)(t)=-i \frac{\partial}{\partial u}(1 + \chi_S(u)+\frac{1}{\pi}\int_0^\infty \chi_S(u+t)-\chi_S(u-t)-\chi_S(t)+\chi_S(-t)dt)$

Should these be equations for $(H f)(u)$, rather than $(H f)(t)$? In the first equation, shouldn't we divide the integrand by $t$, as this is a Hilbert transform? And what is the justification for the expansion in the second equation?

More broadly, does anyone know an alternative proof of the expected value of $|X|$ (given that $X$ is from a stable distribution) that I could review?

答:
Taleb computes the expected value of the absolute value of a stable distribution with mean zero. The version below gives some of the details that are omitted in his book.

In what follows, let $\mathscr{F}(f)$ denote the Fourier transform of $f$, with the convention that $\mathscr{F}(f)(\omega)=\int_{-\infty}^{\infty}e^{i\omega t}f(t)dt$. Then,
$$\partial_u \mathscr{F}(f)(u)\rvert_{u=0}=i\int_{-\infty}^{\infty}x f(x)dx$$

Let $f(x)$ denote the PDF of the stable distribution. So,
$$E({\mid}X{\mid})=\int_{-\infty}^{\infty}{\mid}x{\mid}f(x)dx=\int_{-\infty}^{\infty}sgn(x) x f(x)dx=\frac{\partial_u \mathscr{F}(sgn(x) f(x))(u)\rvert_{u=0}}{i}$$

By the Convolution Theorem (with the Fourier transform convention given above),
$$\mathscr{F}(sgn(x) f(x))(u)=\frac{1}{2\pi}(\mathscr{F}(sgn)*\mathscr{F}(f))(u)$$

As the characteristic function of the distribution $\chi(t)$ is the Fourier transform of the PDF $f(x)$, and as $\mathscr{F}(sgn)(u)=\frac{2i}{u}$,

$$E({\mid}X{\mid})=\partial_u \frac{1}{2{\pi}i} \int_{-\infty}^{\infty}\frac{2i}{u} \chi(u-x) dx\rvert_{u=0}=\partial_u\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{\chi(u-x)}{u}dx\rvert_{u=0}$$
$$=\partial_u\frac{1}{\pi}\int_{0}^{\infty}\frac{\chi(u-x)-\chi(u+x)}{u}dx\rvert_{u=0}$$

The characteristic function of a stable distribution is of the form

$$\chi(t)=e^{i \mu t - {\mid}t c{\mid}^\alpha(1- i \beta tan(\frac{\pi \alpha}{2})sgn(t))}$$

As we are assuming $\mu=0$, this simplifies to:

$$\chi(t)=e^{- {\mid}t c{\mid}^\alpha(1- i \beta tan(\frac{\pi \alpha}{2})sgn(t))}$$

We reverse the order of differentiation and integration in the computation of $E({\mid}X{\mid})$. The derivative at $u=0$ of the quantity inside the integral is (according to Mathematica):

$$\frac{2 \alpha  e^{-(c t)^{\alpha }} (c t)^{\alpha } \left(\beta  \tan \left(\frac{\pi  \alpha }{2}\right) \sin \left(\beta  \tan \left(\frac{\pi  \alpha }{2}\right) (c t)^{\alpha }\right)+\cos \left(\beta  \tan \left(\frac{\pi  \alpha }{2}\right) (c t)^{\alpha }\right)\right)}{t^2}$$

Changing variables, letting $z=log(t)$ yields:

$$2 \alpha  e^{-\left(c e^z\right)^{\alpha }-z} \left(c e^z\right)^{\alpha } \left(\beta  \tan \left(\frac{\pi  \alpha }{2}\right) \sin \left(\beta  \tan \left(\frac{\pi  \alpha }{2}\right) \left(c e^z\right)^{\alpha }\right)+\cos \left(\beta  \tan \left(\frac{\pi  \alpha }{2}\right) \left(c e^z\right)^{\alpha }\right)\right)$$

Replacing $sin(x)$ with $\frac{e^{i x}-e^{-i x}}{2i}$ and $cos(x)$ with $\frac{e^{ix}+e^{-ix}}{2}$ and integrating (using the principal value) gives the solution:

$$E({\mid}X{\mid})=\frac{c \Gamma \left(\frac{\alpha -1}{\alpha }\right) \left(\left(1+i \beta  \tan \left(\frac{\pi  \alpha }{2}\right)\right)^{1/\alpha }+\left(1-i \beta  \tan \left(\frac{\pi  \alpha }{2}\right)\right)^{1/\alpha }\right)}{\pi }$$


问:
Power of adjacency matrix
What does ij element of the k-th power of adjacency matrix for a directed graph, i.e. A^k represent: number of paths with exact k length or number of paths with length k or less than k from i to j?</p>

答:
The issue here is the precise meaning of "path".

The standard definition (to the extent that there is a 'standard' one) is that a walk is any  alternating sequence of vertices and edges, consecutive elements of which are incident, that begins and ends with a vertex. With that definition:

The number of walks of length $k$ between nodes $i$ and $j$ in a graph is precisely given by the $i,j$ element of $A^k$.

The standard definition of path is that it is a walk without repeated vertices. That was the definition used in the post to which you refer. With this definition of path, having a non-zero $i,j$ element means that you can get from $i$ to $j$ in $k$ steps but you may have repeated a vertex. In that case you  know that there is a path from $i$ to $j$ (by shortening the walk of length $k$) but that it's length is less than $k$.

Title: Power of adjacency matrix


问:
$M = -I_2$ or $M$ is similar to $A=\begin{pmatrix} 1&0\\0&-1\end{pmatrix}$?
I read a property that I did not understand:

Let $M\in GL(2,\mathbb{F}_3)$ the matrix such that $M= I_2$ or $M^2=I_2$. As it has coefficients in a field of characteristic different from $2$, it is diagonalizable. So, $M = I_2$, $M = -I_2$ or $M$ is similar to $A=\begin{pmatrix} 1&0\\0&-1\end{pmatrix}$ .

Problem: Why  $M = -I_2$ or $M$ is similar to $A=\begin{pmatrix} 1&0\\0&-1\end{pmatrix}$ ?

答:
Suppose $M^2=I_2$. Then the eigenvalues of $M$ are among $1$ and $-1$.

If both are eigenvalues, then $M$ is diagonalizable, so it is similar to a diagonal matrix having the eigenvalues on the diagonal, in whatever order you prefer. Thus $M$ is similar to
\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}

It is also similar to
\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}
but there is no problem: these two matrices are similar to one another using row and column swapping.

The interesting part of the exercise is to show that if there is a single eigenvalue, then $M=I_2$ or $M=-I_2$. It's sufficient to prove that if the only eigenvalue is $1$, then $M=I_2$ (justify it).

Over the complex numbers one could use the Jordan canonical form, which could also be used here, by passing to the algebraic closure of $\mathbb{F}_3$, but it would be overkill. Suppose
$$
M=\begin{pmatrix} a & b \\ c & d \end{pmatrix}
$$
Then the characteristic polynomial is $X^2-(a+d)X+(ad-bc)$, so we have
$$
a+d=2,\quad ad-bc=1
$$
by the assumption on the eigenvalues.

Hence $a(2-a)-bc=1$. Now we have $a\in\{0,1,2\}$ and it's sufficient to check the various cases.

If $a=0$, then we need $bc=-1$, so $c=-1/b$, but this contradicts $M^2=I_2$ (check it).

If $a=1$, then $bc=0$; if $c=0$, we see that $M^2=I_2$ implies $2b=0$, so $b=0$. Similarly, $b=0$ implies $c=0$.

If $a=2$, then we're in the same situation as for $a=0$.


问:
How can we evaluate the characteristic polynomial with a matrix as the parameter?

For any polynomial p(x) = $a_0+a_1x+· · ·+a_kx^k$
and any square matrix A, p(A) is defined
as p(A) = $a_0I + a_1A + · · · + a_kA^k$
. Show that if v is any eigenvector of A and $χ_A(x)$ is
the characteristic polynomial of A, then $χ_A(A)v$ = 0, Deduce that if A is diagonalisable
then $χ_A(A)$ is the zero matrix

I don't get what it means here to apply the characteristic polynomial with the matrix as the parameter. Does it subtract from each term in p(A)?

答: The definition given for $p$ is just showing you how to treat $\chi_A(A)$. The characteristic polynomial typically takes scalar values and has roots at the eigenvalues of $A$, but now instead of a function from $\mathbb{R}\to\mathbb{R}$ we treat it as a function $M_{n\times n}(\mathbb{R})\to M_{n\times n}(\mathbb{R})$ where we plug in a matrix and obtain a matrix as a result. The function $p$ is just being used as an example.


问:
Inj, surj, equivalence relation. Is this exercise correctly resolved?
This is the problem that I've tried to solve. Let me know if I was right. It was written originally in italian, but I don't know how translate some mathematical words in english, so sorry in advice.

Given $S = \{1, 2, 3\}$, and the following function $$f:X\in\mathbb{P}(S)\longmapsto X\backslash\{2\} \in \mathbb{P}(S)$$

Talk about injectivity and surjectivity

Given the equivalence relation $\mathcal{R}_f$ induced by $f$ defined as following: $$X \mathcal{R}_f Y \Longleftrightarrow f(X)=f(Y)$$ describe every equivalence class in $\mathcal{R}_f$.

Define in $\mathbb{P}(S)\backslash\{\varnothing\}$ the following order relation $$ X \quad \Sigma \quad Y \Leftrightarrow X =Y \quad \text{or}\quad \max(X)<\max(Y).$$ Determine in
  $\{\mathbb{P}(S)\backslash\{\varnothing\}, \Sigma\}$ min, max, maximal
  and minimal element.

Is $\{\mathbb{P}(S)\backslash\{\varnothing\}, \Sigma\}$ a totally ordered set?

This is my solution:

Not injective. $$ \{1,3\}\neq\{1,2,3\}\quad \text{but} \quad f(\{1,3\})=f(\{1,2,3\}) $$ Not surjective. (e.g. $ \{1,2\}\in B$ have not corresponding element in $A$ )

Four equivalence classes:
$$[\{1\}]_{\mathcal{R}_f}=\{\{1\},\{1,2\}\}$$
$$[\{3\}]_{\mathcal{R}_f}=\{\{3\},\{2,3\}\}$$
$$[\{\varnothing\}]_{\mathcal{R}_f}=\{\{2\},\{\varnothing\}\}$$
$$[\{1,3\}]_{\mathcal{R}_f}=\{\{1,3\},\{1,2,3\}\}$$

Am I right? How can I start to resolve 3rd and 4th point?

Best regard

答:
This CW answer intends to remove this question from the Unanswered queue.

Your answers for 1. and 2. are correct.

Regarding 3., Neal's advice is good:

For 3, just write down all the elements of $\Bbb P(S)$ and draw a digraph with an arrow from $X$ to $Y$ exactly when $X\Sigma Y$. This should make the max, min, maximal, minimal elements obvious.

Regarding 4., André Nicolas's remark should do the trick:

On 4: If I understand the problem correctly, let $X = \{1,3\}$ and $Y=\{2,3\}$. then neither $X\;\Sigma\;Y$ nor $Y\;\Sigma\;X$ is true. So the order cannot be total.


问:
Is the function $f(z) = z$ continuous on $|z| = 1$?
Is the function $f(z) = z$ continuous on $|z| = 1$? Here $z$ is a complex number.

I'm scratching my head here as $z_{0}$ is a modulus.

How do I calculate this?

答: Since $f(z)=z$ is continuous everywhere, it is continuous on the circle $|z|=1.$


问:
Find general solution in radians of $\cos x + \cos 3x + \cos 5x = 0$
I study maths as a hobby. I have this problem:

Find in radians the general solution of:

$$\cos x + \cos 3x + \cos 5x = 0$$

I have said:

$\cos 3x + \cos 5x = 2\cos\left( \frac{1}{2}(3x + 5x)\right)\cos\left( \frac{1}{2} (3x - 5x)\right) = 2\cos 4x\cos x$

$\cos x + 2\cos 4x \cdot \cos (x) = 0$

$(1 + 2\cos 4x)\cos x = 0$

$\cos x = 0$ or  $\cos 4x = - \frac{1}{2}$

$x = \frac{\pi}{2}$ or $x = \frac{\pi}{3} $

My book gives the answers as:

$(2n + 1)\frac{\pi}{6}, n\pi \pm \frac{\pi}{3}$

So we agree on the second answer but is the book wrong when it says $(2n + 1)\frac{\pi}{6}$?

I get the answer as $x = n\pi + \frac{\pi}{2}$, which can be written as $(2n +1)\frac{\pi}{2}$

答:
You can rewrite the equation in this form: $$\cos(3x-2x)+\cos 3x+\cos(3x+2x)=0$$ This is equivalent to: $$2\cos 3x \cos 2x + \cos 3x = \cos 3x (2\cos 2x+1)=0$$ Eventually, we have the equations $\cos 3x=0$ and $\cos 2x=-\frac{1}{2}$. $$3x=\pi n +\frac{\pi}{2} \space \space \text{or} \space \space 2x=2\pi n \pm \frac{2\pi}{3}\space \rightarrow \space x=\frac{\pi}{6}(2n+1)\space \space \text{or} \space \space x=\pi n \pm \frac{\pi}{3}$$

This is just one way to look at it. You reached $\cos x=0$ and $\cos 4x =-\frac{1}{2}$ equations in your solution. In this case, your solution set is: $$x=\frac{\pi}{6}(6n+3) \space \space \text{or} \space \space x=\frac{\pi}{6}(3n\pm1) $$ The difference here is correlated with the horizontal compression of trigonometric functions.


问:
Prove a metric space generated by a $\sigma$-field is complete
The following question is from Linear Operators written by Nelson Dunford and Jacob T. Schwartz, Chapter III.7.1

Let $(S, \Sigma, \mu)$ be a positive measure space and $\Sigma$ (proper $\sigma$-algebra) will be defined as a field where the addition is the symmetric difference (denoted by $\Delta$) and multiplication the intersection (we just write $AB$ instead of $A\cap B$). Let $\mathcal{N}$ be the ideal of all $\mu$-null sets in $\Sigma$ and assign the quotient set $\Sigma/\mathcal{N}$ the following metric:

$$d(A, B) = \arctan\mu(A \Delta B)$$
Also, we have:
$$
\begin{aligned}
\begin{split}
&(AB)\Delta(A_1B_1)=[A(B\Delta B_1)]\Delta[B_1(A\Delta A_1]\\
&A^c\Delta A_1^c = A \Delta A_1\\
&(A \cup B)\Delta(A_1\cup B_1) = (A\Delta A_1)\Delta(B\Delta B_1)\Delta[A(B\Delta B_1)]\Delta[B_1(A\Delta A_1)]
\end{split}
\end{aligned}
$$

Then one can show operations $A \bigcup B, AB, A\Delta B, A^c$ are all (jointly) continuous.Now the question is to show that metric space $(\Sigma/\mathcal{N}, d)$ is complete.

Given $\{E_n\}$ where $\lim\sup d(E_n, E_m)=0$, my attempt is to use the fact that $L^1(S, \Sigma, \mu)$ is weakly complete, show $\chi_{E_n}$ converge to an $f \in L^1(S, \Sigma, \mu)$ weakly. Then show that $f$ is a characteristic function, say $\chi_E$. However, I do not know if $E \in \Sigma/\mathcal{N}$. Any hints will be appreciated.

答: Hints: If $\chi_{E_n} \to f$ in $L^{1}(\mu)$ then there is a subsequence $(\chi_{E_{n_k})}$ conveging to $f$ almost everywhere. Let $g(x)=\lim \chi_{E_{n_k}}(x)$ if the limit exists and $0$ otherwise. Then $g$ is measurbale and $f=g$ a.e.. Further, $g$ takes only the values $0$ and $1$ so it is $\chi_C$ where $C=\{x: g(x)=1\}$.


答:
In the basic topological definition, a function $f : X \to Y$ between two topological spaces $X$ and $Y$ is continuous if the preimage of any open set $U \subset Y$ is an open set $\textrm{preim}_f(U) \subset X$. Since $f$ is the identity map on $\mathbb{C}$, it is continuous on all of $\mathbb{C}$. This is easy to show since the preimage of any subset $U \in \mathbb{C}$ under $f$ is the set itself, $\textrm{preim}_f(U) = U$, so if $U$ was open, so is $\textrm{preim}_f(U) = U$. You can essentially prove that the identity is always continuous between two identical topological sets. But then this argument also holds for the circle $|z|=1$, since $f$ is also the identity on the circle.

If you're not familiar with topology, forget this answer and parametrize the circle as $z = e^{i\phi}$.

Body:
Is the function $f(z) = z$ continuous on $|z| = 1$? Here $z$ is a complex number.

I'm scratching my head here as $z_{0}$ is a modulus.

How do I calculate this?


问:
What is a natural bijection?
Here is the definition of adjoint functors taken from here:

and here is a property about universal arrows:

To me the concept of natural bijection is imprecise. Does it coincide with the notion of natural isomorphism (so a natural transformation that has an inverse)? Bijection because in the associated commutative diagram we get mappings between sets which should be perhaps bijections?

答:
It is, as you said, a bijective natural transformation and the naturality condition is one and the same as the defining condition for it to be a natural transformation.

In the following, I will use the conventions that for a category $$:

$||$ = the objects of $$,

for $X, X' ∈ ||$, $(X,X')$ = the set of arrows/morphisms $X → X'$,

along with the definitions

$^+$ is the dual category:

$|^+| = ||$,

for $X, X' ∈ ||$, $^+(X,X') = (X',X)$,

identities in $^+$ are the same as those in $$,

compositions in $^+$ are the reverse of those in $$.

$×$ is the direct product category:

$|×| = ||×||$,

for $(X,Y),(X',Y') ∈ |×|$, $(×)((X,Y),(X',Y')) = (X,X')×(Y,Y')$.

For there to be an adjunction relation between categories $$ and $$ with functors
$$:  → , :  → $$
means that you have polymorphic operators
$$⋀: (X, Y) → (X, Y),$$
$$⋁: (X, Y) → (X, Y),$$
such that for ($X ∈ ||$, $Y ∈ ||$)
$$f: X → Y ⇒ ⋁⋀f = f,$$
$$g: X → Y ⇒ ⋀⋁g = g,$$
thereby making them inverses, such that the following naturality conditions hold
$$⋀(l∘f∘k) = l∘⋀f∘k, ⋁(l∘g∘k) = l∘⋁g∘k,$$
where
$$k: X' → X, l:Y → Y'.$$

Make the parametrization of the polymorphic operators explicit $⋀_{X,Y}$ and $⋁_{X,Y}$ and there's your natural transformation.

The natural transformations are thus $⋀:  → $, $⋁:  → $, where the functors
$$,: ^+× → $$
are given by
$$(X,Y) ∈ |^+×| ↦ (X,Y) ≡ (X,Y),$$
$$(k,l) ∈ (^+×)((X,Y),(X',Y')) ↦ ((k,l): f ∈ (X,Y) ↦ (l∘f∘k) ∈ (X',Y')),$$
and
$$(X,Y) ∈ |^+×| ↦ (X,Y) ≡ (X,Y),$$
$$(k,l) ∈ (^+×)((X,Y),(X',Y')) ↦ ((k,l): g ∈ (X,Y) ↦ (l∘g∘k) ∈ (X',Y')).$$
The defining conditions for $⋀$ and $⋁$ to be natural transforms are:
$$⋀_{X',Y'}∘(k,l) = (k,l)∘⋀_{X,Y},$$
$$⋁_{X',Y'}∘(k,l) = (k,l)∘⋁_{X,Y},$$
where
$$(k,l) ∈ (^+×)((X,Y),(X',Y')) = (X',X) × (Y,Y'),$$
i.e. $k: X' → X$ and $l: Y → Y'$.

When applied to
$$f ∈ (X,Y), g ∈ (X, Y),$$
i.e. to $f: X → Y$ in $$ and $g: X → Y$ in $$, this yields the result:
$$⋀_{X',Y'} (k,l)(f) = (k,l)\left(⋀_{X,Y} f\right),$$
$$⋁_{X',Y'} (k,l)(g) = (k,l)\left(⋁_{X,Y} g\right),$$
or, upon application of the definitions:
$$⋀(l∘f∘k) = l∘⋀f∘k, ⋁(l∘g∘k) = l∘⋁g∘k.$$

Title: What is a natural bijection?


问:
Discrete Math - Calculating $\sum_{k}^{n}$
I have a question regarding on how to calculate expressions like (I need to write it without $\sum$ and without $k$):

$$\sum_{k=2}^{25}\binom{25}{k}\binom{k}{2}$$

And also like:

$$\sum_{k=1}^{10}k\binom{10}{k}\binom{20}{10-k}$$

Unfortunately I didn't learn how to do it, but this is a part of the material for my exam today.

Thank you so much for your help !!

Appreciate that!

答:
This will come too late to be of help before the exam, but the ideas may be useful to you at some later point.

Sometimes the easiest way to attack one of these questions is to find a useful combinatorial interpretation. For instance, imagine that you have $25$ white balls numbered $1$ through $25$. Then $\binom{25}k\binom{k}2$ can be thought of as the number of ways to choose $k$ of them to paint red and then to choose $2$ of the red balls and attach a gold star to each of them. In the end we have $25-k$ white balls and $k$ red balls, two of which have gold stars.

$$\sum_{k=2}^{25}\binom{25}k\binom{k}2$$

is then the total number of possible outcomes when we allow all possible values of $k$.

We can generate the same outcomes by first choosing $2$ balls to be painted red and given gold stars, and then choosing any subset of the remaining $23$ balls to be painted red: choosing $\ell$ balls in that last step gives the outcomes corresponding to $k=\ell+2$. And if we think of it this way, it’s easy to get the final result in a closed form: there are $\binom{25}2$ ways to choose the balls that get gold stars, and there are $2^{23}$ ways to choose a subset of the remaining $23$ balls to be painted red, so

$$\sum_{k=2}^{25}\binom{25}k\binom{k}2=2^{23}\binom{25}2=\frac{25\cdot24}2\cdot2^{23}=25\cdot24\cdot2^{22}\,.$$

For the second one I would first use a very basic identity to get rid of the annoying factor of $k$:

$$k\binom{10}k=10\binom9{k-1}\,.$$

(Combinatorial interpretation: Picking a team of $k$ people from a pool of $10$ and then designating one of them captain produces the same outcomes as first picking one of the $10$ to be captain and then picking $k-1$ more people from the $9$ remaining to fill out the team of $k$.)

Thus,

$$\begin{align*}
\sum_{k=1}^{10}k\binom{10}k\binom{20}{10-k}&=10\sum_{k=1}^{10}\binom9{k-1}\binom{20}{10-k}\\
&=\sum_{k=0}^9\binom9k\binom{20}{9-k}\,.\tag{1}
\end{align*}$$

If you know the Vandermonde identity, you can simply apply it. If not, imagine that you have a pool of $9$ men and $20$ women, from which you want to select a committee of $9$ people. There are $\binom9k\binom{20}{9-k}$ ways to choose a committee of $9$ people that contains exactly $k$ men, so $(1)$ is just the total number of possible committees of $9$ people, and that is clearly $\binom{29}9$.


问:
A Messy System of Linear Equations. (does anyone have a solver?)
Does anyone have a solver for the system of equations :

$$ v_1=A+B- (c_5 a / c_1(c_8+ c_3c_5/c_1 )), $$

$$  v_2=Ae^{r_1 h }+Be^{r_2 h}-  (c_5 a / c_1(c_8+ c_3c_5/c_1 )),$$

$$x_1=b-(1/c_1)(A(c_2+c_3/r_1)+B(c_2+c_3/r_2)),$$

$$ x_2=b+ah-(1/c_1)(Ae^{r_1 h}(c_2+c_3/r_1)+Be^{r_2 h}(c_2+c_3/r_2))$$

for unknowns $a, b, A, B$, and everything else known constants.

答:
I think what you have to do to see the point clearly is to put $(A, B, a, b) = (X, Y, Z, W)$ and not mess things up with fixed values of the constants. With this and representing the constant values, let's say with $t_i$, you have the following simple system:
$$X+Y+t_1Z=t_2\\t_3X+t_4Y+t_1Z=t_5\\t_6X+t_7Y+W=t_8\\t_9X+t_{10}Y+t_{11}Z+t_{12}W=t_{13}$$ or is you want (to see more clearly maybe) $$\begin{pmatrix}1&1&t_1&0\\t_3&t_4&t_1&0\\t_6&t_7&0&1\\t_9&t_{10}&t_{11}&t_{12}\end{pmatrix}*\begin{pmatrix}X\\Y\\Z\\W\end{pmatrix}=\begin{pmatrix}t_2\\t_5\\t_8\\t_{13}\end{pmatrix}$$
You solve this system and replace the corresponding values of the constants with which you end up.


问:
Proof of convex spaces are homotopy equivalent to a point
In order to prove that a convex subpace $X \subseteq \mathbb{R^n}$ is homotopy equivalent to a point I did the following:

Let $X \subseteq \mathbb{R^n}$ be a convex subspace in Euclidean space and $Y=\{0\}$ be a one point space. Let $x \in X$ and define $f(x) = 0$ and $g(0) = x_0$ for some $x_0 \in X$. Then we have $(f \circ g)(0) = 0$ and $(g \circ f)(x) = x_0$.

However this seems wrong to me since $(f \circ g) = 1_Y$ but $(g \circ f) \neq 1_X$ since $x_0 \neq x$. Where is my reasoning flawed and how can I fix this?

答:
You have to exhibit a homotopy between $h = g \circ f : X \to X$ and $1_X$, i.e., a continuous function $H : X \times I \to X$ such that $h(x) = H(x, 0)$ and $1_x(x) = H(x, 1)$ for every $x \in X$. To do this define $H(x, t) = tx + (1-t)x_0$, then $H : X \times I \to X$ because $X$ is convex and clearly $H(x, 0) = x_0 = h(x)$ and $H(x, 1) = x = 1_X(x)$ for every $x \in X$.

In more complex cases, you might also have to exhibit a homotopy between $f \circ g$ and $1_Y$, but, in this case, $f \circ g = 1_Y$ as you observed, so there is nothing to do.


答:
Your reasoning is actually correct. A homotopy equivalence is one in which $g \circ f$ is homotopic to $1_X$, not on the nose equal.

How can you use the convexity assumption to homotope $g \circ f$ to the identity?

Title: Proof of convex spaces are homotopy equivalent to a point

Body:
In order to prove that a convex subpace $X \subseteq \mathbb{R^n}$ is homotopy equivalent to a point I did the following:

Let $X \subseteq \mathbb{R^n}$ be a convex subspace in Euclidean space and $Y=\{0\}$ be a one point space. Let $x \in X$ and define $f(x) = 0$ and $g(0) = x_0$ for some $x_0 \in X$. Then we have $(f \circ g)(0) = 0$ and $(g \circ f)(x) = x_0$.

However this seems wrong to me since $(f \circ g) = 1_Y$ but $(g \circ f) \neq 1_X$ since $x_0 \neq x$. Where is my reasoning flawed and how can I fix this?


问:
Given $n=36$, $ p=20/36$, and $p_0=0.5$; calculate $p$-value
I know so far that:

$H_0:\ p= 0.5,$

$H_1:\ p> 0.5.$

And that $z$-score: $0.672$

I did $1-0.44 = 0.56$ for $p$-value but got it wrong, the correct answer is $0.2514$

答:
Your question suggests that you are using an approximate binomial test, and @tommik's answer (+1) follows that lead.

[Often, one cannot expect more than about two
place accuracy from normal approximations to binomial probabilities, especially after accounting for rounding errors from using printed tables. And here, a
slightly better z-score would use a continuity correction: $z = (19.5/36 - .5)12 = 0.5333$ instead of the answer book's $z = 2/3.$ With continuity correction the P-value is approximately $0.297.$ The 'approved' normal approximation of the P-value turns out
to be about $0.25$ as @tommik has shown. By contrast, the true binomial P-value is about $0.31,$ as shown below. In this problem the difference doesn't change the decision not to
reject $H_0$ at the 5% level. However, in practical applications, it is best to use exact binomial P-values.]

The exact P value from observing
$x = 20$ successes is $n = 36$ independent trials, under $H_0: p = 0.5.$ is
$$P(X \ge 20\,|\,n=36, p=0.5) = 1 - P(X \le 19\,|\,n=36,p=0.5) = 0.3089,$$
from R:

```<code>1 - pbinom(19, 36, 0.5)
[1] 0.3088597
</code>```

```<code>k = 0:36;  PDF = dbinom(k, 36, .5)
plot(k, PDF, type="h", lwd=3, col="blue")
 abline(h=0, col="green2");  abline(v=0, col="green2")
 abline(v = 19.5, col="red", lwd=2, lty="dotted")
 curve(dnorm(x, 18, 3), add=T, lwd=2)
</code>```

Body:
Your question suggests that you are using an approximate binomial test, and @tommik's answer (+1) follows that lead.

[Often, one cannot expect more than about two
place accuracy from normal approximations to binomial probabilities, especially after accounting for rounding errors from using printed tables. And here, a
slightly better z-score would use a continuity correction: $z = (19.5/36 - .5)12 = 0.5333$ instead of the answer book's $z = 2/3.$ With continuity correction the P-value is approximately $0.297.$ The 'approved' normal approximation of the P-value turns out
to be about $0.25$ as @tommik has shown. By contrast, the true binomial P-value is about $0.31,$ as shown below. In this problem the difference doesn't change the decision not to
reject $H_0$ at the 5% level. However, in practical applications, it is best to use exact binomial P-values.]

The exact P value from observing
$x = 20$ successes is $n = 36$ independent trials, under $H_0: p = 0.5.$ is
$$P(X \ge 20\,|\,n=36, p=0.5) = 1 - P(X \le 19\,|\,n=36,p=0.5) = 0.3089,$$
from R:

1 - pbinom(19, 36, 0.5)
[1] 0.3088597

k = 0:36;  PDF = dbinom(k, 36, .5)
plot(k, PDF, type="h", lwd=3, col="blue")
 abline(h=0, col="green2");  abline(v=0, col="green2")
 abline(v = 19.5, col="red", lwd=2, lty="dotted")
 curve(dnorm(x, 18, 3), add=T, lwd=2)


问:
A problem on matrices: Find the value of $k$

If $
        \begin{bmatrix}
        \cos \frac{2 \pi}{7} & -\sin \frac{2 \pi}{7} \\
        \sin \frac{2 \pi}{7} & \cos \frac{2 \pi}{7}  \\
        \end{bmatrix}^k
 =
        \begin{bmatrix}
        1 & 0 \\
        0 & 1 \\
        \end{bmatrix}
$, then the least positive integral value of $k$ is?

Actually, I got no idea how to solve this.

 I did trial and error method and got 7 as the answer.

how do i solve this?
 Can you please offer your assistance? Thank you

答:
Look if your matrix is diagonizable. If it is, write it's diagonal form. Let say that $A$ is the matrix you describe in your post, then we will write $B=S^{-1}AS$ its diagonal form where $S$ is the matrix that allows you to change from the canonical base $E$ to a base $G$ where $B$ is diagonal.

Now assume that $A$ is diagonizable you have a diagonal matrix so the problem is to find the smallest integer $k$ for which $\lambda^k_1=1$ and $\lambda^k_2=1$ where $\lambda_1$ and $\lambda_2$ are the eigenvalues of $A$. The answer you will find will be the one you are looking for. Indeed, let's say that $m$ is the smallest value for which $B^m=I_2$ then you have :

$$I_2=B^m=\underbrace{(S^{-1}AS)…(S^{-1}AS)}_{m \text{ times}}=S^{-1}A^mS \implies A^m=I_2$$


问:
Distribution of sum of product of normal random variables.
The distribution of a product $Z=XY$ of two normally distributed random variables is given by the product distribution https://mathworld.wolfram.com/NormalProductDistribution.html.

What is the distribution of $Q=\sum_{i=1}^n Z_i$, where $Z_i= X_i Y_i$, and $X_i,Y_i\sim \mathcal N(0,\sigma_i)$?

A nice answer was given in: Distribution of sum of product-normal distributions. by @wolfies, for the case where $\sigma_i=1$ (that is, all the $X_i,Y_i$ are identically distributed standard normal). The distribution can be expressed in terms of the modified Bessel function of the second kind. And as $n\to\infty$, the distribution approaches a normal distribution.

But I am interested in the case where the $\sigma_i$ are different. If no-closed form solution can be found, I am curious about when $p(Q)$ will approach a normal distribution. Is there an "effective" $n$ in terms of the $\sigma_i$? My first guess was the participation ratio $n_\text{eff} = (\sum_{i=1}^n \sigma_i^2)^2/\sum_{i=1}^n \sigma_i^4$.

Here is how I've approached it so far. The characteristic function of each $Z_i$ is given, I believe, by

$$\varphi_{Z_i}(t)=\frac{1}{\sqrt{t^2\sigma_i^2 + 1}}$$

So by independence the characteristic function of $Q$ is given by

$$ \varphi_Q(t) = \prod_{i=1}^n\varphi_{Z_i}(t) = \prod_{i=1}^n \frac{1}{\sqrt{t^2\sigma_i^2 + 1}}$$

And $p(Q)$ should be given by the inverse Fourier transform of $\varphi_Q(t)$. I am stuck trying to perform the inverse Fourier transform, and any help would be greatly appreciated!

edit:

@Henry gave a very nice answer regarding the asymptotic behavior of $p(Q)$ as $n\to\infty$. but I am still curious about the behavior of $p(Q)$ for $n$ small. Can $p(Q)$ be computed exactly? If not, how large must $n$ be before $p(Q)$ is approximately normal, as a function of the $\{\sigma_i\}$?

答:
If $X_i$ and $Y_i$ are independent and have zero mean then the variance of $X_iY_i$ will be $\sigma_i^2$ (or $\sigma_{X_i}\sigma_{Y_i}$ if these differ) so $\sum\limits_{i=1}^n X_iY_i$ will have zero mean and variance $s_n^2=\sum\limits_{i=1}^n  \sigma_i^2$

The Central Limit Theorem will apply in the sense that $\frac1{s_n} \sum\limits_{i=1}^n  X_iY_i$ will converge in distribution to $\mathcal N(0,1)$ as $n$ increases, providing that the $\sigma_i^2$ do not become too extreme, for example if they are bounded above and below by positive finite numbers, or if say the Lyapunov or Lindeberg conditions are met.


问:
The Kähler condition on a Riemann surface
A Hermitian metric $h$ on a complex manifold $X$ is Kähler if the associated $2$-form $\omega=\mathrm{Im} (h)$ is closed.

This condition is trivial on compact Riemann surface, implying that every Hermitian metric on a compact Riemann surface is Kähler.

An equivalent condition for a metric to be Kähler is that this metric "osculates to order 2" the standard metric, meaning that the components of the metric satisfy

$$
h_{ij}(z) = \delta_{ij} + O(|z|^2).
$$

The proof of the fact that $d\omega=0$ implies this condition that most books do (for example, Griffiths-Harris or Huybrechts) is clear when $\dim X\geq 2$. However, the proof given in those books does not work for a Riemann surface since the condition $d\omega=0$ is vacuous in that case.

My question is how does one prove that, if $h$ is a Hermitian metric on a Riemann surface, that locally can be written as

$$
h=h(z) dz\otimes d\bar{z},
$$

then $h(z)=1+O(|z|^2)$.

答:
You just follow the usual proof, but it's easier. $h$ is a positive (real analytic?) function and you write $h\,dz\otimes d\bar z = \phi\otimes\bar\phi$. Without loss of generality, we assume $h(0)=1$. Then the $(1,0)$-form $\phi$ is given by
$$\phi(z) = \left(1+az+O(|z|^2)\right)dz.$$
Set $w=z+\frac a2 z^2 + O(|z|^3)$, and note that
$$dw = \left(1+az+O(|z|^2)\right)dz,$$
so the $(1,0)$-form $\phi = \left(1+O(|w|^2)\right) dw$, as needed. The Kähler condition is totally redundant.


问:
Semantics for set theory
Every logic book I’ve ever come across, when talking about the semantics of first-order logic, defines concepts like structures and interpretations where the most important element is a set called the “domain of discourse” or the “universe”. Everything is then based on this set. This works very nicely for number theory, but what is the universe of set theory? The set of all sets does not exist. So how do you define all the concepts of semantics?</p>

答:
To be rigorous about this you need to understand the distinction between the metalanguage in which we are reasoning about the semantics and the object languages whose semantics we are reasoning about. The statement that the set of all sets does not exist is relative to the closure properties that we require of the set of all sets: we may assume stronger closure properties in our metalanguage (e.g., the existence of Grothendieck universes containing any set) than we do in our object language (e.g., closure under the axioms of ZF (Zermelo-Fraenkel set theory)). Note that the metalanguage and the object language may have exactly the same syntax: it is the way we are using them and the axioms that we assume in them that matters.

For many purposes in set theory, particularly independence results, it is enough to formulate results predicated on the existence of a model of set theory. E.g., to prove that the axiom of choice $AC$ is independent of the other axioms of $ZF$, we have to show that if $ZF$ has a model, then so does $ZFC = ZF+AC$. This statement does not assume the existence of a model of either $ZF$ or $ZFC$.

Likewise, the completeness theorem states that for any theory $T$ and any sentence $\phi$, if $T \models \phi$ (i.e., if $\phi$ holds in every model of $T$) then $T \vdash \phi$ (i.e., $\phi$ can be derived from $T$ using first-order logic). ZFC can prove this for all $T$, including $T = ZFC$, but that doesn't imply that a model for $ZFC$ can be constructed in $ZFC$.

Note also the assertion $T \models \phi$ in the statement of the completeness theorem is about all models of $T$, not just some chosen standard model with a given universe of discourse. Taking $T$ to be $PA$ (Peano arithmetic), we know from the incompleteness theorem that there are sentences $\phi$ (such as a sentence asserting that $PA$ is consistent) such that $PA \not\vdash \phi$, even though $\phi$ is true if we restrict our universe of discourse to the natural numbers. The completeness theorem then tells us that $PA$ has a model in which $\lnot\phi$ holds.

Title: Semantics for set theory


答:
As mentioned in the other post, if $p(x)$ is a polynomial then $p(A)$ is simply plugging in A. It works analogously (but remember to multiply the constant term by the identity matrix).

Now to help you with the second part, you need to know about the spectral mapping theorem. It says that $$spec(p(A))=p(spec(A))$$
That is, the eigenvalues of p(A) are the eigenvalues of A, evaluated in the polynomial, i.e. $p(\lambda_1),...,p(\lambda_n)$, where $\lambda$ are eigenvalues of p(A).

Here $p=\chi$, the characteristic polynomial. Then by the spectral mapping theorem you know that the eigenvalues of $\chi(A)=\chi(\lambda_1),...,\chi(\lambda_n)$. But by definition of the characteristic polynomial, these are all 0. So the only eigenvalue of $\chi(A)$ is 0. If A is diagonalizable, it can be written in the form $A=VDV^{-1}$ where $D=diag(\lambda)$ and V are the eigenvectors. Suppose the characteric polynomial of A has the form $a_0+a_1x+...+a_nx^n$. Then

$$\begin{split}\chi(A)&=a_0I+a_1A+...+a_nA^n\\
&=a_0I+a_1VDV^{-1}+...+a_nVD^nV^{-1}\text{ a nice property of diagonalization}\\
&=V(a_0I+a_1D+...+a_nD^n)V^{-1}\\
&=Vdiag(a_0+a_1\lambda_i+...+a_n\lambda_i^n)V^{-1}\text{ by nice property of diagonal matrices}\end{split}$$

But the characteristic polynomial of A evaluated at each of its eigenvectors is 0. Thus this becomes

$$\begin{split}\chi(A)&=V0V^{-1}\\&=0\end{split}$$ the nxn zero matrix.

This proves the second part.

I leave the first part to you.

Title: How can we evaluate the characteristic polynomial with a matrix as the parameter?

Body:
For any polynomial p(x) = $a_0+a_1x+· · ·+a_kx^k$
and any square matrix A, p(A) is defined
as p(A) = $a_0I + a_1A + · · · + a_kA^k$
. Show that if v is any eigenvector of A and $χ_A(x)$ is
the characteristic polynomial of A, then $χ_A(A)v$ = 0, Deduce that if A is diagonalisable
then $χ_A(A)$ is the zero matrix

I don't get what it means here to apply the characteristic polynomial with the matrix as the parameter. Does it subtract from each term in p(A)?


问:
$X$ be compact, $U$ open and $f: X → \mathbb{R}$ continuous such that $f (X) \subset U$. $n> n_0 ⇒ f_n (X) \subset U.$
Let $X$ be compact, $U$ open and $f: X → \mathbb{R}$ continuous such that $f (X) \subset U$. If a sequence of functions
$f_n: X → \mathbb{R}$ converge uniformly to $f: X → \mathbb{R}$, prove that there is $n_0 \in\mathbb{N}$ such that
$$n> n_0 ⇒ f_n (X) \subset U.$$

I am completely stuck in that statement. I tried to use the fact that

A sequence of functions $f_n: X → \mathbb{R}$ is uniformly
convergent if and only if it is a Cauchy sequence.

But I didn't find any better way to prove this statement. Does someone know how to prove it?

答: $f(X)$ is a  compact set disjoint from $U^{c}$. Hence, there is  a positive distance $r$ between these two. Choose $n_0$ such that $|f_n(x)-f(x)| <r$ for all $x$ if $n >n_0$. Can you now show that $f_n(x) \in U$ for all $x$ if $n >n_0$?


问:
Let $X$ and $Y$ be finite sets. Then the set $Y^X$ is finite and $\#(Y^X) = (\#Y)^{\#X}$.

Let $X$ and $Y$ be finite sets. Then the set $Y^X$ is finite and $\#(Y^X) = (\#Y)^{\#X}$.

I can see how to do it combinatorially. Let $y \in Y$. There are $\#(X)$ choices for such y to come from x, therefore we have $\#Y^{\#X}$ choices. I am not sure how to do it bijectively.

答:
Let $X=\{1,\ldots,N\}$ and $Y=\{1,\ldots,M\}$, then define the set of functions from $X$ to $Y$

$$Y^X=\{f:X\rightarrow Y\}.$$

Now, we try to construct a bijection with a new set $Z$ by $\phi:Y^X\rightarrow Z$ such that

$$f\mapsto \left(f(1), \ldots, f(N)\right).$$

There are $\underbrace{M\times \cdots \times M}_{N}$ choices in total to make. Each choice corresponds to a different function.


问:
Probability of getting absorbed in an absorbing Markov Chain
Given a Markov Chain with 5 states: 1, 2, 3, 4, and 5 where states 4 and 5 are absorbing. This is given by the transition diagram below. What is the probability of getting absorbed at state 5 given that you start at state 1.

I think I am supposed to use first-step analysis but I am confused. Any help would be great!

答:
Let $h_k$ be the probability of absorption in $5$, given that you start in state $k$. Doing the usual one step analysis, we get the equations
\begin{align*}
h_1 &= ph_2+(1-p)h_3 \\
h_2 &= (1-p)h_3 \\
h_3 &= ph_2 + (1-p).
\end{align*}
Solve the bottom two equations to get $h_3=\frac{1-p}{p^2-p+1}$, $h_2=\frac{(1-p)^2}{p^2-p+1}$ and so $\boxed{h_1=\tfrac{(1-p)^2(p+1)}{p^2-p+1}}$.


问:
Dini's theorem; relaxing the condition of compactness
Here's Dini's theorem:
If monotone sequence of continuous functions converges pointwise on a compact set $X$ and the limit function is also continuous, then the convergence is uniform.

I wonder if the compactness condition in Dini's theorem can be replaced by its dense set.
For example, suppose that there is a set $Y$ dense in a compact set $X$.
Can we replace $X$ by $Y$ in Dini's theorem?

答: $Y=[0,1)$ is dense in the compact set $X=[0,1]$, and the sequence of continuous functions $f_n(x) = x^n$ converges monotonically and pointwise on $Y$ to the continuous function 0, but not uniformly.


问:
Are absolute extrema only in continuous functions?
The Extreme Value Theorem says that if $f(x)$ is continuous on the interval $[a,b]$ then there are two numbers, $a≤c$ and $d≤b$, so that $f(c)$ is an absolute maximum for the function and $f(d)$ is an absolute minimum for the function.

So, if we have a continuous function on $[a,b]$ we're guaranteed to have both absolute maximum and absolute minimum, but functions that aren't continuous can still have either an absolute min or max?

For example, the function $f(x)=\frac{1}{x^2}$ on $[-1,1]$ isn't continuous at $x=0$ since the function is approaching infinity, so this function doesn't have an absolute maximum.
Another example: suppose a graph is on a closed interval and there is a jump discontinuity at a point $x=c$, and this point is the absolute minimum.

The extreme value theorem requires continuity in order for absolute extrema to exist, so why can there be extrema where the function isn't continuous?

答: Semi-continuous functions have this property.</p>


问:
probability of exactly 3 people choosing the same two numbers from n numbers

There are 30 people. Each of them will choose randomly two distinct numbers from the set $\{1,2,\cdots,20\}$.

What is the probability that exactly 3 of them will choose the same two numbers?

Let A denote the event that exactly 3 of them will choose the same two numbers.

Note that we have $\binom{20}{2}=190$ different ways to choose 2 distinct numbers from $\{1,2,\cdots,20\}$.

I see that there are basically two cases: all the remaining 27 choose distinct numbers or there might be groups of 2 people with the same numbers. I have difficulty with the second case!

Therefore, $P(A)=\frac{\binom{190}{1} \binom{30}{3}\binom{189}{27}27!+(???)}{190^{30}}$

答:
If you have $189$ selections that can be made by $27$ people with $p$ selections each chosen by two people  and $27-2p$ each chosen by one person, there are $\frac{189!\,27!}{p!\,(27-2p)!\,(162+p)!\,2^p}$ ways of doing that.

Sum that over $0\le p \le 13$ and you get about $2.7025\times 10^{61}$

Multiply that by your  $\binom{190}{1} \binom{30}{3}$ and you get about $2.0847\times 10^{67}$

Divide that by $190^{30}$ and you get a probability of about $0.09046$.  Simulation suggests this is sensible.

Doing something similar with $190$ selections and $30$ people with no triples and you would get a probability of a less extreme result of about $0.90257$, leaving about $0.00697$ for a more extreme result


问:
Terence Tao Volume 1 pigeonhole principle
I am self-studying Terence Tao Volume 1. I am having few issues. I don't understand the question about pigeonhole principle below.

答:
Here is the question:

Do you also need help with the proof?


问:
Prove a metric space generated by a $\sigma$-field is complete
The following question is from Linear Operators written by Nelson Dunford and Jacob T. Schwartz, Chapter III.7.1

Let $(S, \Sigma, \mu)$ be a positive measure space and $\Sigma$ (proper $\sigma$-algebra) will be defined as a field where the addition is the symmetric difference (denoted by $\Delta$) and multiplication the intersection (we just write $AB$ instead of $A\cap B$). Let $\mathcal{N}$ be the ideal of all $\mu$-null sets in $\Sigma$ and assign the quotient set $\Sigma/\mathcal{N}$ the following metric:

$$d(A, B) = \arctan\mu(A \Delta B)$$
Also, we have:
$$
\begin{aligned}
\begin{split}
&(AB)\Delta(A_1B_1)=[A(B\Delta B_1)]\Delta[B_1(A\Delta A_1]\\
&A^c\Delta A_1^c = A \Delta A_1\\
&(A \cup B)\Delta(A_1\cup B_1) = (A\Delta A_1)\Delta(B\Delta B_1)\Delta[A(B\Delta B_1)]\Delta[B_1(A\Delta A_1)]
\end{split}
\end{aligned}
$$

Then one can show operations $A \bigcup B, AB, A\Delta B, A^c$ are all (jointly) continuous.Now the question is to show that metric space $(\Sigma/\mathcal{N}, d)$ is complete.

Given $\{E_n\}$ where $\lim\sup d(E_n, E_m)=0$, my attempt is to use the fact that $L^1(S, \Sigma, \mu)$ is weakly complete, show $\chi_{E_n}$ converge to an $f \in L^1(S, \Sigma, \mu)$ weakly. Then show that $f$ is a characteristic function, say $\chi_E$. However, I do not know if $E \in \Sigma/\mathcal{N}$. Any hints will be appreciated.

答: Let $\{\epsilon_n\}$ be a sequence of positive real numbers and $\epsilon_n \overset{n \rightarrow \infty}{\longrightarrow} 0$. Supposed there is an $n \geq 1$ such that the set $S[(\chi_{E_k} - f), \epsilon_n] = \{s \in S\,\vert\,\vert\chi_{E_k}(s)-f(s)\,\vert > \epsilon_n\}$ has measure greater than $\epsilon_n$ for all $k \geq n$. For $k$ large enough such that $\int_S \vert\,\chi_{E_k}(s)-f(s)\,\vert\,\mu(ds) < \epsilon_n^2$ we obtain the contradiction. Hence we have for any $n \geq 1$ there is $k_n\geq n$ such that $\mu(S[(\chi_{E_{k_n}} - f), \epsilon_n]) \leq \epsilon_n$. Then $\chi_{E_{k_n}} \rightarrow f$ pointwise a.e.. The $\mu$-measure convergence further implies $f$ is $\mu$-measurable and hence a characteristic function of a set in $\Sigma$.


问:
Find the nth term of a geometric series
I realized I forgot most of my high school maths. Earlier today I asked a question and I thought I understood it until I looked at it again.

I'm trying to find the 100th term of the following sequence:

$$\bbox[5px,border:2px solid #C0A000]{\small a_0 = 1}$$
$$\bbox[5px,border:2px solid #C0A000]{\small a_n = a_{n-1} + 2^{-n}}$$

How can you quickly arrive to the fact that this is a geometric series? If I'm looking at the first few terms $1, \frac{3}{2}, \frac{7}{4}$, normally to find the common ratio you divide a term by its predecessor right? So $r = \frac{\frac{7}{4}}{\frac{3}{2}} = \frac{7}{6}$ which does not hold.

The answer for getting the 100th term is supposed to be $\frac{2^{101} - 1}{2^{100}}$ but I'm not quite sure how to arrive to this conclusion.

I would really appreciate some help, thanks in advance!

答:
$a_n=1+\frac12+\frac14+\cdots+\frac1{2^n}$,

so each term in the sequence $(a_n)$ is the sum of the first $n+1$ terms of a geometric series.

Using the formula for the sum of the first $n+1$ terms of a geometric series,

$a_n=\dfrac{1-\dfrac1{2^{n+1}}}{1-\dfrac12}=\dfrac{2^{n+1}-1}{2^n}$.

Title: Find the nth term of a geometric series


答:
I will try to paint a picture which will hopefully give you some intuition. If you don't know about metric spaces, just read $|x-y|$ anywhere I wrote $d(x,y)$.

In a metric space $X$, when $K \subseteq X$ is compact and $U \subseteq X$ is open, one can show that the "inflation of $K$ of radius $r$" is contained in $U$ for small enough $r$. This set is defined as
$$
\mathrm{infl}(K,r) = \{ y \in X \, | \, \exists x \in K \, \text{such that} \, d(x,y) < r \}.
$$
So basically the result is that if $K$ is compact and $U$ is open, then $\mathrm{infl}(K,r) \subseteq U$ for some small enough $r$. The proof by contradiction is fairly simple: since $U$ is open, $X \setminus U$ is closed, and $K \cap (X \setminus U) = \varnothing$. This implies that
$$
\inf_{x \in K, y \in X \setminus U} d(x,y) > 0.
$$
To see this, suppose this infimum is zero. Pick a sequence $(x_n,y_n) \in K \times (X \setminus U)$ satisfying $d(x_n,y_n) < \frac 1n$. Since $K$ is compact, the sequence $x_n$ admits a convergent subsequence; without loss of generality, we may assume $x_n$ is convergent since $\frac 1n$ is decreasing. So $x_n \to x \in K$, and since $d(x_n,y_n) < \frac 1n$, we also have $y_n \to x \in X \setminus U$ because $X \setminus U$ is closed. This contradicts that $K \cap (X \setminus U) = \varnothing$, so we must conclude that the infimum is positive.

Let $r$ be this infimum. By definition of the infimum, this implies $\mathrm{infl}(K,r) \cap (X \setminus U) = \varnothing$ because if $y \in X \setminus U$ would satisfy $d(x,y) < r$ for some $x \in X$, this would contradict the definition of our infimum above. It follows that $\mathrm{infl}(K,r) \subseteq U$ for some $r > 0$.

Now that we got this "metric space fact" out of the way, since $X$ is compact and $f$ is continuous, $f(X)$ is compact and contained in $U$, so we can choose some $r > 0$ small enough such that $\mathrm{infl}(f(X), r) \subseteq U$. The fact that $f_n$ converges uniformly to $f$ is equivalent to saying that under the supremum norm
$$
\|g\|_{\infty} = \sup_{x \in X} |g(x)|,
$$
we have $\|f_n - f\|_{\infty} \to 0$. Choose $n_0$ large enough so that $\|f_n - f\|_{\infty} < r$ for all $n \ge n_0$. I claim that $\|f_n - f\|_{\infty} < r$ implies $f_n(X) \subseteq \mathrm{infl}(f(X), r)$, which is itself a subset of $U$, completing the proof.

But this is clear: if $x \in X$, since $|f_n(x) - f(x)| \le \|f_n-f\|_{\infty} < r$, it follows that $f_n(x) \in \mathrm{infl}(f(X), r)$.

Hope that helps,

Title: $X$ be compact, $U$ open and $f: X → \mathbb{R}$ continuous such that $f (X) \subset U$. $n> n_0 ⇒ f_n (X) \subset U.$

Body:
Let $X$ be compact, $U$ open and $f: X → \mathbb{R}$ continuous such that $f (X) \subset U$. If a sequence of functions
$f_n: X → \mathbb{R}$ converge uniformly to $f: X → \mathbb{R}$, prove that there is $n_0 \in\mathbb{N}$ such that
$$n> n_0 ⇒ f_n (X) \subset U.$$

I am completely stuck in that statement. I tried to use the fact that

A sequence of functions $f_n: X → \mathbb{R}$ is uniformly
convergent if and only if it is a Cauchy sequence.

But I didn't find any better way to prove this statement. Does someone know how to prove it?


问:
Different answers for same integral.
Let, $x = \cos(\theta)$ and

$$s = \int^{1}_{-1}\frac{1 \ dx}{\sqrt{1-x^2}}.$$

If we let $1 = \cos(2\pi)$ and $-1 = \cos(\pi)$, we obtain:
$$s = -\int^{2\pi}_{\pi}1 dx  = -\pi.$$

However, if we let $1 = \cos(0)$ and $-1 = \cos(\pi)$, we obtain:
$$s = -\int^{0}_{\pi}1 dx = \pi.$$

答: Your substitution requires the use of the inverse cosine function in the transformed limits [ $ \theta \ = \ \arccos(x) \ $ ] .  So $ \ \arccos(1) \ = \ 0 \ $ and $ \ \arccos(-1) \ = \ \pi \ $ , making your second definite integral the correct one.

Title: Different answers for same integral.


问:
Questions on the symmetry of of the alexander polynomial and the rank of the Seifert matrix
For my bachelor thesis I am using the book "Lectures on the topology of 3-manifolds. an introduction to the Casson invariant"(1999) by Nikolai Saveliev.

Regarding the Alexander Polynomial as defined below I have two questions.

Why does the equation hold for rank $S$ and not only for $n$ with $n$ the size of the Seifert matrix $S$?

Why is the rank of the Seifert matrix $S$ $2g$ with $g$ the genus of the corresponding Seifert surface?

I know Seifert surfaces of genus $g$ are isotopic to a surface in disc-band form with $2g$ bands. Therefore the corresponding Seifert matrix is a $2g\times 2g$ matrix.
But this matrix doesn't need to have full rank.

Thanks in advance.

Alexander polynomial definition

答:
The author probably had in mind "dimension" instead of "rank", so the sentence simply means that $S$ is a $2g \times 2g$ matrix since $k$ is a knot (if it was a $n$-component link it would have dimension $2g+n-1$).

In particular, the determinant of a Seifert matrix can very well be zero, eg if $A$ is the Seifert matrix associated to a Seifert surface $F$ for a knot, then the Seifert surface obtained from $F$ by attaching a 1-handle has as associated matrix an enlargement of $A$ with a column/row of zeroes. So the usual "rank" of a matrix doesn't really work here.


问:
Questions about Definition of Vector Spaces
I know that a vector space is over a field and has the operations of addition (among elements, which are called vectors) and scalar multiplication (in which a field element acts on a vector to yield a vector). Now, I want to reconcile this understanding with my previous intuition by asking some questions.

Why can't a vector space have cross products/dot products defined inside it? What problems would be run into? In general, what kinds of "products" can we define in a vector space?

Over what fields is $\mathbf{R}^n$ considered a vector space?

Is it correct that vector space isomorphism only exists in the same field? Or can we claim that different vector spaces in different fields are isomorphic? What is the intuition used to extend vector spaces to infinite dimensions in functional analysis?

Why do we only have to check for closure under addition and scalar multiplication to prove that a set is a vector space when a vector space has six (or seven, depending on how you count) axioms?

I'm studying linear algebra for the second time in a proof-based context, so answers relevant to this level would be appreciated!

答:
You can define an inner product on every finite dimensional real vector space, in fact if $V$ is a n-dimensional vector space it is isomorphic to $R^n$ and you can use this isomorphism to define an inner product. This definition is not canonic(it depends on the basis you choose).
Cross product is defined only on $R^3$, you could define a generalization also in $R^n$ but it is not so simple(see Hodge star operator).

$\mathbb{R}^n$ is normally considered as a $\mathbb{R}$ vector space. Of course it can also be equipped with a structure of vector space over other fields(for example over $\mathbb{Q}$), but it is not the standard structure.

Because morphisms of vector spaces over different fields cannot be defined. Consider $V$ and $W$ vecotr spaces over $F$ and $E$. Let $f:V\rightarrow W$ be a linear map. Let $\lambda\in F$. The axiom $f(\lambda x)=\lambda f(x)$ doesn't make sense, in fact $\lambda \notin E$ and so it is not defined $\lambda f(x)$.

Because if $V$ is a vector space and $W$ is a subset of $V$, commutativity, associativity etc. are satisfied because elements of W are also elements of V


问:
Understanding Smash Product
Suppose I have pointed topological spaces $(X,x_0)$ and $(Y,y_0)$, then I would like to understand $(X,x_0) \land (Y,y_0)$. Since the quotient space can be defined by a partition, I will try to define it in that way.

From what I read, I think that the smash product is the quotient space of $X \times Y $ where the partition is given as follows. if $$ A=X \times \{y_0\} \cup \{x_0\} \times Y  $$ then the partition consists of the class $A$ and rest are singleton sets of the form $X \times \{y\}$  or  $\{x\} \times Y$ where $x \neq x_0$ and $y \neq y_0$. Am I right?

If I am right, can anyone explain why are we using such spaces? Are these really helpful?

答:
Regarding intuition:

I think its easier to think of the quotient as the wedge product, because at least then its easier to visualize what we are quoteinting out by. For example, with $S^1 \wedge S^1$ we are taking the product $S^1 \times S^1 $ which is the torus, and quotienting out by the wedge of two circles. You should draw a torus, with two loops on it, each representing the generators of H_1, and then work out what you get if yo collapse the two loops to a single point.

And since examples are super helpful for intuition, you should try to prove that smashing X by $S^1$ is the same as suspension of $X$.

Regarding why we care:

In most cases, I don't think we visualize the smash product, because its usually really hard to picture. The smash product is important though because it basically acts like tensor products in algebra. Namely, with the tensor product, we have the adjunction $Hom(X,Hom(Y,Z)) \cong  Hom(X \otimes Y, Z)$. We have the same thing with smash product in the category of pointed spaces: $Maps(X, Maps(Y,Z)) \cong Maps(X \wedge Y, Z)$.

From a more advanced and rather hand-wavy viewpoint, this product extends to spectra, which are basically sequences of spaces that stabilize after applying suspension enough. The example with it being suspension in the case that one of the spaces is $S^1$ indicates that it sort of "plays nicely" with spectra.

Title: Understanding Smash Product


问:
Solving a Gaussian-like integral in polar coordinates
I want to solve this Gaussian-like integral

$$ I =\int_0^1 e^{-x^2} \mathrm dx \tag{1} $$

This integral may be solved by considering x as a dummy variable and applying the Fubini's theorem to transform it into a double integral as follows:

$$ I =\int_0^1 e^{-y^2} \mathrm dy \tag{2} $$

$$ I^2 =\left(\int_0^1 e^{-x^2} \mathrm dx\right)\left(\int_0^1 e^{-y^2} \mathrm dy\right) = \int_0^1\int_0^1 e^{-\left(x^2+y^2\right)} \mathrm dx \mathrm dy \tag{3} $$

Here is when things get a bit interesting. The double integral may be solved by a variable change to polar coordinates:

$$ I^2 = \int_0^{\frac{\pi}{2}}\int_0^R e^{-r^2} \mathrm r \mathrm dr \mathrm d\theta \tag{4} $$

The above integral should yield the following result:

$$ I^2 = \frac{\pi}{4} \bigg[\text{erf}\left(1\right)\bigg]^2 \tag{5} $$

Or:

$$ I = \frac{\sqrt\pi}{2} \text{erf}\left(1\right) \tag{6} $$

Now comes my inquiry: I have not managed to figure out how to transform the integration region in (3) so I may find an appropriate value of R for which the integral result is (6). By doing some calculations, i.e., by making (4) = (5), finding the integral and solving for R, I found that:

$$ R = \sqrt{-\text{ln}\bigg[1-\left(\text{erf}\left(1\right)\right)^2\bigg]} \tag{7}  $$

Is there an analytical (inductive/non-trial-and-error) procedure to find R as in (7) from a transformation of the integration region in (3) to polar coordinates? I will be most grateful if anyone can help me figure this puzzle.

答:
This is perhaps more of a comment than an answer, because I'm not exactly sure what you're asking.  Just by drawing a picture, we have $$R(\theta)=\cases{
\frac1{\cos\theta},&$0\leq\theta\leq\frac\pi4$\\
\frac1{\sin\theta},&$\frac\pi4\leq\theta\leq\frac\pi2$\\
}$$
but you seem to be asking for an $R$ that doesn't depend on $\theta$.

EDIT


问:
Find $y$-Lipschitz constant
$$f(x,y)=x^3e^{-xy^2}, 0\leq x\leq a, y\in \mathbb R, a>0$$

I need to find $K>0$ such that $$|f(x,y_1)-f(x, y_2)|\leq K|y_1-y_2|$$ for all $0\leq x\leq a$ and $y_1,y_2\in \mathbb R$

I did this $$|x^3e^{-xy_1^2}-x^3e^{-xy_2^2}|=x^3|e^{-xy_1^2}-e^{-xy_2^2}|\leq a^3|e^{-xy_1^2}-e^{-xy_2^2}|$$

now I don't know what to do.

答: Lipschitz constant Lipschitz constant Lipschitz constant Lipschitz constant

Title: Find $y$-Lipschitz constant


答:
Question 1.

A general definition of a "product" is the following: suppose you have three vector spaces $X,Y,Z$ all over the same field. Then, a bilinear mapping $\beta:X\times Y\to Z$ is essentially what we mean by a "product". So you can certainly consider the notion of products/multiplication. But, what is important to note is that this is extra information that you have to provide; it is not part of the vector space axioms, hence there is no standard/canonical choice in general.

You certainly can look at vector spaces equipped with dot products (more commonly called inner products). More precisely, suppose $V$ is a vector space over $\Bbb{R}$. Then, an inner product on $V$ is a bilinear, symmetric positive-definite mapping $\langle\cdot,\cdot\rangle : V\times V \to \Bbb{R}$. By prescribing an inner product, we are in essence prescribing a geometry for the vector space $V$. If you want to generalize some more, we can weaken the conditions imposed. For instance, a bilinear, symmetric, non-degenerate mapping $g:V\times V\to \Bbb{R}$ is called a pseudo-inner product, and we refer to the pair $(V,g)$ as a pseudo-inner product space. (I intentionally left out the complex case since the definitions are slightly different). The notion of a cross product is more subtle so let me not go into the details; but really if you formulate an appropriate definition, you can define various types of "products".

Question 2.

If $\mathbb{F}$ is any subfield of $\Bbb{R}$, then we can consider $\Bbb{R}^n$ as a vector space over $\Bbb{F}$ (by restricting the "usual" operations to the field $\Bbb{F}$). The most obvious examples are $\Bbb{F}=\Bbb{R},\Bbb{Q}$.

Question 3.

Suppose $V,W$ are vector spaces over fields $F_1,F_2$ respectively. To define the notion of a linear map $T:V\to W$, we would like the following equation to hold for all $x,y\in V,\lambda\in F_1$: $T(\lambda x+y)=\lambda T(x)+T(y)$. Well, to talk about $\lambda T(x)$, we need the target space to be considered as a vector space over $F_1$, so we'd need $F_1\subset F_2$. If you want to talk about isomorphisms, then you would also need the inverse map to be linear, so we want $F_2\subset F_1$. Thus, we always consider them over the same field $F$.

Question 4.

If $V$ is a vector space over a field $F$, and $W\subset V$ is any non-empty subset, then of course, by definition, to say $W$ is a vector space over $F$ we have to verify a whole bunch of axioms. But, it's a nice fact that we actually don't have to do so much work: as long as we check $0\in W$ (or just that $W\neq \emptyset$), and that $W$ is closed under addition and scalar multiplication, then it follows that $W$ (equipped with the restricted addition and scalar multiplication as operations) is a vector space over $F$ in its own right (this is usually called the "subspace" criterion or something). The nice thing is all the axioms like associativity/commutativity or addition/ distributive laws etc all hold in $W$ because they hold in $V$ and since $W\subset V$.

Title: Questions about Definition of Vector Spaces

Body:
I know that a vector space is over a field and has the operations of addition (among elements, which are called vectors) and scalar multiplication (in which a field element acts on a vector to yield a vector). Now, I want to reconcile this understanding with my previous intuition by asking some questions.

Why can't a vector space have cross products/dot products defined inside it? What problems would be run into? In general, what kinds of "products" can we define in a vector space?

Over what fields is $\mathbf{R}^n$ considered a vector space?

Is it correct that vector space isomorphism only exists in the same field? Or can we claim that different vector spaces in different fields are isomorphic? What is the intuition used to extend vector spaces to infinite dimensions in functional analysis?

Why do we only have to check for closure under addition and scalar multiplication to prove that a set is a vector space when a vector space has six (or seven, depending on how you count) axioms?

I'm studying linear algebra for the second time in a proof-based context, so answers relevant to this level would be appreciated!


问:
Convergence of sequence in $\Bbb{R}^{n}$ in dot product
Let $\{\textbf{u}_{k}\}$ be a sequence in $\mathbb{R}^n$ that converges to the point $\textbf{u}$, Prove that $$\lim_{k \to \infty} \langle{\textbf{u}_k,\textbf{v}} \rangle = \langle \textbf{u,v} \rangle$$

for every point $\textbf{v}$ in $\mathbb{R}^n$.

attempt let $p_i(\textbf{u})=u_i$ be the $i$th coordinate of $\textbf{u}$. Since the sequence converges componentwise

$\lim_{k \to \infty} (p_1(\textbf{u}_k)v_1+\dots+p_n(\textbf{u}_k)v_n)=p_1(\textbf{u})v_1+\dots+p_n(\textbf{u})v_n=\langle \textbf{u},\textbf{v} \rangle$

Would this be the correct way to show this result?

答:
That's certainly one way. Another way would be to use Cauchy-Schwarz

$$ |\langle \mathbf u_k , \mathbf v\rangle - \langle \mathbf u , \mathbf v\rangle | =|\langle \mathbf u_k -\mathbf u,\mathbf  v\rangle|\le \| \mathbf u_k -\mathbf u \|\|\mathbf v\| \to 0.$$


问:
Is tetration and quintation etc. of infinite cardinals well defined?
I would like to know whether tetration and quintation functions are well defined for infinite cardinals, thus, for example, $$\aleph_0 \text{ [tet] } \aleph_0 =  \aleph_0 ^ {\aleph_0 ^ {\aleph_0^{\dots}}}$$ taken $\aleph_0$ times and compounded from top to bottom, to produce a higher cardinality power set upon every recursive iteration.

In an analogous manner, $$\aleph_0 \text{ [quint] }\aleph_0 = \aleph_0 \text{ [tet] } (\aleph_0  \text{ [tet] } (\aleph_0 \text{ [tet] } \dots))$$ taken recursively $\aleph_0$ times, again compounded from top to bottom.

Since as one moves to ever higher compouding operations like tetration, quintation, sextation etc. the series of new cardinals so producded would cleary increase MUCH faster than under an operation as "slow" as "mere" exponentially recursive power set taking, do we perhaps get the production of new huge cardinals that are of a different order and character than those generated by applying infinite power set operartions?  Would all of these much larger cardinals, including even $\aleph_0\, [\aleph_0\text{ compounding operation}] \,\aleph_0$, itself compounded $\aleph_0$ times etc. still be smaller than other better known large cardinals defined in some other manner?

I hope these questions are fairly easy to comprehend. They hardly go much beyond elementary set theory. Thanks for reading and let me know if you can answer any of them!

答:
Assuming the GCH, the tetration example is tantamount to $\aleph_\omega$. If you rewrite the GCH as a tetration formula, then the pentation gets you the sequence $\aleph_\omega$, $\aleph_{\omega_\omega}$... until you get the first fixed point of the aleph function. (This is because $\aleph_0 \uparrow\uparrow \aleph_0$ gets you $\aleph_{\aleph_0}$ by replacing the index with the given factor, so then $\aleph_0 \uparrow\uparrow \aleph_{\aleph_0}$ would be $\aleph_{\aleph_{\aleph_0}}$, etc.)* But even waiving the GCH, the other posters are right, you never get higher than various $\beth$ numbers and their like, based on higher operations, without presupposing an inaccessible number (as the index of your operator) in the first place.

Major edit: even a compounded operation like $\aleph_0 \uparrow^{\aleph_0 \uparrow^{\aleph_0}\aleph_0} \aleph_0$ or an infinite such tower would give a cardinal cofinal with $\aleph_0$ and hence not an inaccessible cardinal.

Second major edit: consider $2 \uparrow^n m$, for $m$ greater than 2. This tends towards $\omega$ as $n$ increases, but without presupposing $\omega$ in place of $n$, this sequence never actually reaches it. Likewise, the smallest uncountable inaccessible must be presupposed as the index of the operation that "accesses" it. Let $\kappa$ be the smallest uncountable inaccessible. Then write $\aleph_{\alpha<\kappa} \uparrow^{\beta<\kappa} \lambda < \kappa$, when $\lambda$ is < $\kappa$ also. This expresses how without starting from $\kappa$ somewhere in one's arithmetic, one will not reach it. So one might parse "inaccessible" as "inaccessible without presupposition," hence the need for an axiom of inaccessible cardinality.

*Start with the fact that $2^{\aleph_0} = \aleph_0^{\aleph_0} = \aleph_0 \uparrow\uparrow 2$, or $\aleph_0 \uparrow\uparrow 1+1$. Then the GCH (for the $\aleph_n$ anyway) in tetration format would be $\aleph_n \uparrow\uparrow 1 + m = \aleph_{n+m}.$ For example, $2^{\aleph_2} = \aleph_2^{\aleph_2} = \aleph_2 \uparrow\uparrow 2$, so the formula gives the successor cardinal in this case again, and so on. In the infinite case, $1 + \aleph_0 = \aleph_0$, so $\aleph_0 \uparrow\uparrow \aleph_0 = \aleph_{0+\aleph_0}$. Note that $n+\aleph_0 = \aleph_0$, for all natural numbers $n$. So here $\aleph_2 \uparrow\uparrow \aleph_0$ also gives $\aleph_{\omega}$, and so on for all the $\aleph_n$. But $\aleph_0 \uparrow\uparrow \aleph_2 = \aleph_{\omega_2}$, here, etc.

Last edit: it turns out, though, that these formulae break down at the first fixed point of the aleph function, $\aleph_{\omega_{\omega_{\omega...}}}$. I will call this $\aleph_f$ or just then $f$ since $\aleph_f = f$ (this is its fixed-point character). Even in ZFC, we can establish some hypothetical bounds on powersets, like Shelah's $2^{\aleph_n} < \aleph_{\omega} \rightarrow 2^{\aleph_{\omega}} < \aleph_{\omega_4}$. We can also express failures of the GCH at various points. See Moti Gitik's "No bound for the first fixed point" for a model where the GCH holds up to $f$ but there is no bound as such on the powerset of $f$. Analogously, on our tetration model, $f \uparrow\uparrow f = f \uparrow^5 2$; I will leave to you the exercise of showing that our tetration formula gives a directly impossible result thereby. Note that here $\aleph_0 \uparrow\uparrow \aleph_f = \aleph_{\aleph_f} = \aleph_f = f$. The GCH would have $f^f = f^+$, which I think would be $\aleph_{f+1}$, which I think isn't a fixed point. But all this gives us that $f \uparrow\uparrow 2 = \aleph_{f+1}$. Now what happens then if our formula adds $f$ to itself?