问:
if $a_{n+1}=xa_{n}+ya_{n-1}$ then have this results :$a_{m+n}=Xa_{m}a_{n+1}+Ya_{m-1}a_{n}$？
it is well  known:
if $F_{1}=1,F_{2}=1$,and $F_{n+1}=F_{n}+F_{n-1}$.then we have
$$F_{m+n}=F_{m}F_{n+1}+F_{m-1}F_{n}$$
prove 1

and I have know this :if $a_{0}=0,a_{1}=1$ and $x,y$ be give postive integers,and $$a_{n+1}=xa_{n}+ya_{n-1}$$
then we have
$$a_{m+n}=a_{m}a_{n+1}+ya_{m-1}a_{n}$$
prove 2

My question is
if the sequence such $a,b,x,y$ be give postive integers,and such that  $a_{1}=a,a_{2}=b,a,b\in N^{+}$,and $$a_{n+1}=xa_{n}+ya_{n-1}$$,then is also have this  Similar results
$$a_{m+n}=Xa_{m}a_{n+1}+Ya_{m-1}a_{n}?$$
if there exist,Find the $X,Y$?Thanks

答:
So, the answer is indeed, yes, such a relation exists (with some caveats), and we can prove it via induction, just as in the 1st answer you have linked in the question.

MAJOR EDITS: My original answer was incorrect, as pointed out by @mathlove. Here is a revised answer.

First, let us prove the induction step. For it, assuming the relation is true upto $(m,n)$, we have to prove it for $(m+1,n)$ and $(m,n+1)$, just as mentioned in the $1^{st}$ answer.

For $(m+1,n)$, we have
\begin{align}
a_{m+1+n} &= xa_{m+n}+ya_{(m-1)+n} \\
&= x(Xa_ma_{n+1}+Ya_{m-1}a_n) + y(Xa_{m-1}a_{n+1}+Ya_{m-2}a_n) \\
&= Xa_{n+1}(xa_m+ya_{m-1}) + Ya_n(xa_{m-1}+ya_{m-2}) \\
&= Xa_{m+1}a_{n+1} + Ya_ma_n
\end{align}

Similarly we can do so for $(m,n+1)$. Thus, the inductive step is proved.

For the base case, as the induction relation involves $m-1$ and $n$, the lowest values that these can have is $1$ (As $a_0$ is not defined). So, the base case for the induction must be for $(m,n) = (2,1)$.

For the base case to hold, the following relation must hold -
\begin{align}
a_{2+1} &= Xa_2a_{1+1} + Ya_{2-1}a_1 \\
\implies a_3 &= Xb^2 + Ya^2
\end{align}

From the given recursion relation, we get
\begin{align}
a_3 &= xa_2 + ya_1 \\
&= xb + ya
\end{align}

Hence, the given relation holds iff $X,Y$ satisfy
$$Xb^2 + Ya^2 = xb + ya$$

Addendum: I added this as this was part of my original answer, and I wanted to show where I originally went wrong.

Now, let us determine what restrictions $X$ and $Y$ must have, if such a result was to exist.

Let $m=2$. Then, the induction gives
\begin{align}
a_{n+2} &= Xa_2a_{n+1}+Ya_1a_n \\
&= (Xb) a_{n+1} + (Ya) a_n
\end{align}

Now, we have already been given that $a_{n+2}=xa_{n+1}+ya_{n}$. Thus, we get
\begin{align}
(Xb) a_{n+1} + (Ya) a_n &= xa_{n+1}+ya_{n}\\
\implies (Xb-x)a_{n+1} &= (y-Ya)a_n
\end{align}

Now, if either $X = \frac{x}{b}$ or $Y = \frac{y}{a}$, then this relation shows that the other must be zero as well. Hence, we have either $(X,Y) = \left(\frac{x}{b}, \frac{y}{a}\right)$, or
\begin{align}
a_{n+1} &= \frac{y-Ya}{Xb-x}a_n \\
\implies a_n &= \left(\frac{y-Ya}{Xb-x}\right)^{n-1} a_1
\end{align}

Note: We can easily see that in both the cases, $Xb^2 + Ya^2 = xb + ya$ holds (as it should).


问:
Question about periodic functions.

Assume $f:\Bbb R\to \Bbb R$ is periodic and differential.

I need to show $f'(0)=f'(p)$, $p$ being the periodic componenet i.e
$f(x) = f(x+p)$. and that there exist two different $x,y$ in $[0,p]$
such that $f'(x)=f'(y)=0$

How do I approach this question? I'm having trouble even getting started...

答: For the second part of the question (I already gave a hint for part one in the comments), we can throw out the case where $f'(0)=f'(p) = 0$, as the lemma will then be trivially true. So WLOG assume that $f'(0) = a > 0$. Because $a$ is positive, there exists some $\delta_1,\delta_2>0$ for which $f(\delta_1)-f(0)/\delta_1 > 0$ and $f(p)-f(p-\delta)/\delta_2 > 0$. Consequently, $f(\delta_1) > f(0)$ and $f(p-\delta_2)<f(p)$. Then by the Intermediate Value Theorem, there exists some $x'\in [\delta_1,p-\delta_2]$ such that $f(x')=f(0) = f(p)$. Apply Rolle's theorem to the intervals $[0,x']$ and $[x',p]$ to finish off the proof. The case for $f'(0)<0$ follows a very similar argument.

ContentLicense: CC BY-SA 4.0

Title: Question about periodic functions.


问:
Chomsky–Schützenberger representation theorem in terms of groups
There is the Chomsky–Schützenberger representation theorem on a decomposition of context-free languages:
 a language $L$ over the alphabet $\Sigma$ is context-free if and only if there exists

a matched alphabet $T\cup \overline T$

a regular language $R$ over $T\cup \overline T$

a homomorphism $h:(T\cup \overline T)^{*}\to \Sigma ^{*}$

such that $L=h(D_{T}\cap R)$, where $D_{T}$ is a Dyck language. $D_{T}=\{\,w\in (T\cup \overline T)^{*}\mid w{\text{ is a correctly nested sequence of parentheses}}\,\} $. $T$ contains the opening parenthesis, $\overline T$ contains the closing parenthesis, $|T| = |\overline T|$.

On the other hand, there is a set of works which describes groups with a context-free word or co-word problem:

Holt D. F. et al. Groups with context-free co-word problem

Muller D. E., Schupp P. E. Groups, the theory of ends, and context-free languages

Finite groups are groups with regular word and co-word problem(Anisimov, "The group languages").

Is there exist variation of the Chomsky–Schützenberger representation theorem in terms of groups composition (or in terms of word or co-word problem)?

答:
I think the Wikipedia article needs to be updated (but I'm not going to do so):

The version stated concerns the one-sided Dyck language, whose transductions correspond directly to push-down automata. If you look at the Chomsky–Schützenberger article (itself referring to source articles by Chomsky and Schützenberger that I haven't chased down), you'll see Prop. 2 stated for the 2-sided Dyck language, i.e., the word problem for the free group. I think this version is more properly called the Chomsky–Schützenberger theorem, as the construction required to simulate pushdown-automata using free group automata needs some cleverness. See Kambites for a nice proof.

Chomsky–Schützenberger, The algebraic theory of context-free languages, 118–161, In: Computer Programming & Formal Systems (Eds: P. And Hirschberg, D. Braffort),
Studies in Logic and the Foundations of Mathematics 35,
North-Holland, 1963

Kambites, Formal languages and groups as memory, Communications in Algebra, 37:1, 193–208, 2009, DOI:10.1080/00927870802243580, arXiv:math/0601061

ContentLicense: CC BY-SA 3.0


问:
Prime Degree Field Extension Separable
Let $p$ be a prime and $L$ is a prime degree field extension of $K$, therefore $[K: L]=p$. Why is then $L$ separable over $K$?

答:
From Serge Lang, Algebra, 2002, p.254, exercise 14:

Let $\text{char}\; K = p.$ Let $L$ be a finite extension of $K$, and suppose $[L : K]$ prime to
$p$. Show that $L$ is separable over $K$.

Maybe you meant this? In which case, it follows from the fact that the minimal polynomial of any purely inseparable element $\alpha\in L$ over $K$ has degree $p^{\mu}$ for some integer $\mu$, and hence $p^{\mu}\mid [L:F]$, which can only be the case when $\mu=0$ i.e. $L$ is separable over $K$.

Title: Prime Degree Field Extension Separable


问:
Schur's Lemma in Group Theory
The analogue of well celebrated Schur's Lemma in group theory will be

"If $G$ is a finite simple group, and $\phi$ is a non-identity homomorphism from $G$ to $G$, then $\phi$ is an isomorphism".

The proof follows exactly same lines as it is in representation theory.

I would like to ask the following question, which is a converse of Schur's lemma in some sense.

Question 1: If $G$ is a finite group, $|G|>2$, such that every non-identity homomorphism $\phi\colon G\rightarrow G$ is an automorphism, should $G$ necessarily simple?

The progress towards the solution I made is the following: if $G$ satisfies the hypothesis in the question and $G$ is non-abelian, then $G$ is not solvable: if $G$ is non-abelian, solvable, then it has non-identity abelian quotient $G^{ab}=G/[G,G]$. Let $p$ be a prime divisor of $|G^{ab}|$ (hence $p$ also divides $|G|$). Then there is a surjective homomorphism from $G^{ab}$ to $C_p$ (cyclic group of order $p$). We can embedd $C_p$ in $G$. Then, consider the composition of the homomorphisms:

$G\rightarrow G^{ab} \rightarrow C_p\rightarrow G$. It is a non-identity homomorphism from $G$ to $G$ which is clearly not an automorphism.

Note that if $G$ is abelian, which is not cyclic of prime order, then we can easily find non-identity homomorphisms $G\rightarrow G$ which are not automorphisms.

I would like to ask a general question also (remove the condition of finiteness in above question).

Question 2: If $G$ is an infinite group, such that every non-identity homomorphism $\phi\colon G\rightarrow G$ is an automorphism, should $G$ necessarily simple?

答:
By request, posting a counterexample for question 2 as an answer.

The group of rational numbers $\mathbb{Q}$ under addition is clearly not simple. But the endomorphism ring $\mathrm{End}(\mathbb{Q})$ is isomorphic to $\mathbb{Q}$ itself, and is a division ring, which means that all non-zero endomorphisms are automorphisms.

Title: Schur's Lemma in Group Theory


问:
M/M/3/4 queue in R, translating code from mathematica.
I have the following problem:

Consider an M/M/3/4 queuing system with $\lambda=\mu=1$ that is the
arrival time is exponentially distributed with parameter $\lambda = 1$
and the service times are exponentially distributed with parameter
$\mu=1$. A busy time $B$ for the system is the random time it takes
from a customer comes into the system when it is empty until the
system gets empty again. Write a computer program that by means of
stochastic simulation finds an approximate value of the probability
$\Pr(B>4)$.

So the basic idea of the code below is to count the occurrences of $B>4$ and divide by the number of simulations. X denotes the possible states. My problem is that the last if-statement does not seem to work as it should. If runif(1)<1/2 then X=0, so the while loop should exit. Now it's very unlikely that time is greater than 4 so count should not be updated. But it's always updated no matter what. Note that time is the same thing as $B$. I can't find my error. The answer should be

$$\Pr(B>4)\approx 0.117121$$

```<code>nrsim = 100000
count = 0

for (k in 1:nrsim) {
  time = 0
  X = 1
  while (X > 0 && time <= 4) {

    if (X == 1){
      time = time + rexp(1,2)
      if (runif(1) < 1/2) {
        X = 0
      }
      else {X = 2}
    }

    if (X == 2){
      time = time + rexp(1,3)
      if (runif(1) < 2/3) {
        X = 1
      }
      else {X = 3}
    }

    if (X == 3){
      time = time + rexp(1,4)
      if (runif(1) < 3/4) {
        X = 2
      }
      else {X = 4}
    }

    if (X == 4){
      time = time + rexp(1,3)
    }
      else {X = 3}
  }

  if (time > 4){
    count = count + 1
  }
}

print(count/nrsim)
</code>```

My lecturer only solved this in Mathematica with the code below but I want to do it in Matlab or R, however I can't seem to translate the code properly.

```<code>In[1]:={repetitions, count}={1000000, 0};

For[i=1, i<=repetitions, i++, X=1; time=0;

While[X>0 && time<=4,

If[X==1, time=time+Random[ExponentialDistribution[2]];

If[Random[UniformDistribution[{0,1}]]<1/2, X=0, X=2],If[X==2, time=time+Random[ExponentialDistribution[3]];

If[Random[UniformDistribution[{0,1}]]<2/3, X=1, X=3],If[X==3, time=time+Random[ExponentialDistribution[4]];

If[Random[UniformDistribution[{0,1}]]<3/4, X=2, X=4],If[X==4, time=time+Random[ExponentialDistribution[3]]; X=3]]]]];

If[time>4, count=count+1]];

N[count/repetitions]

    Out[1]= 0.117121
</code>```

答:
Figured it out. The last if-statement is wrong. If x = 4, I should update the time and set X = 3. So it should be

```<code>if (X == 4){
  time = time + rexp(1,3)
  X = 3
}
</code>```

instead of what I've written in my question.

Body:
Figured it out. The last if-statement is wrong. If x = 4, I should update the time and set X = 3. So it should be

if (X == 4){
  time = time + rexp(1,3)
  X = 3
}

instead of what I've written in my question.


问:
Intuition on non-unique ideal factorization if not integrally closed
I'm aware of the result that integral domains admit unique factorization of ideals iff they are Dedekind domains.

It's clear that $\mathbb{Z}[\sqrt{-3}]$ is not a Dedekind domain, as it is not integrally closed.  I'm having difficulty demonstrating directly that $\mathbb{Z}[\sqrt{-3}]$ does not admit unique factorization of ideals.

Obviously we only need one example of an ideal that doesn't have a unique factorization into prime ideals, but a good answer would provide either a process or some intuition in finding such an example.

答:
The arithmetic of nonmaximal quadratic orders comes up naturally when studying binary quadratic forms, so in particular Cox's book contains some discussion of this.

As it happens, when I taught a course based (sometimes closely, sometimes not) on Cox's book, I included a "case study" of factorization in the ring $\mathbb{Z}[\sqrt{-3}]$: see Section 3 of these notes.


问:
Generalized nontransitive dice
Let $X_1, \ldots, X_n$ be a collection of random variables. Consider the directed graph with vertex set $\{ 1, 2, \ldots, n \}$ where there is a directed edge $i \to j$ if $\mathbb{P}(X_i > X_j) > \frac{1}{2}$.

Question 1: What directed graphs can arise in this way? Certainly they must be simple and have no loops. Is that the only restriction? Alternatively, $\mathbb{P}(X_i > X_j) > \frac{1}{2}$ defines an irreflexive antisymmetric relation on $\{ 1, 2, ... n \}$. Which such relations arise in this way?

Question 2: Does the answer change if we require the $X_i$ to all be defined on a finite sample space?

Question 3: What if we require the $X_i$ to be independent?

It is known (see nontransitive dice) that this graph can have directed cycles even if the $X_i$ are independent; in particular, the corresponding relation need not be transitive.

答:
Note: I wrote this answer before question 3 became a separate part of the question; this answer only answers questions 1 and 2. Question 3 seems harder.

Every simple directed graph arises in this way. Consider probability distributions over all $n!$ possible total orders of the values of the $X_i$, and associate a potential directed edge $i\to j$ with the distribution that has probability zero for orders with $X_i\lt X_j$ and $2/n!$ for orders with $X_i\gt X_j$. Given a directed graph, choose any network flow with positive flow along all the directed edges (you can always find one by taking any flow from the sources to the sinks and adding circular flow to the cycles if necessary), and form the normalized linear combination of the distributions associated with the edges, with coefficients given by the flow.

Now consider an edge $i\to j$. The contributions from any edges not incident at $i$ or $j$ don't affect whether $\mathbb P(X_i\gt X_j)\gt\frac12$, since they assign weight to equally many orders with $X_i\gt X_j$ as with $X_i\lt X_j$. Thus we only need to consider edges incident at $i$ or $j$. For an edge $j\to k$, consider the six possible orders of $X_i$, $X_j$ and $X_k$. Three of these are assigned weight by the edge $j\to k$, namely $X_i\gt X_j\gt X_k$, $X_j\gt X_i\gt X_k$ and $X_j\gt X_k\gt X_i$, and of these, one has $X_i\gt X_j$ and two have $X_i\lt X_j$. For an edge $k\to j$, it's the other way around.

Thus, considering the effect of all edges on $\mathbb P(X_i\gt X_j)-\frac12$, the edge $i\to j$ itself contributes $1-\frac12=\frac12$, edges not incident at $i$ or $j$ contribute $\frac12-\frac12=0$, edges "aligned with" $i\to j$, of the forms $k\to j$ and $i\to k$, contribute $\frac23-\frac12=\frac16$, and edges "opposed to" $i\to j$, of the forms $j\to k$ and $k\to i$, contribute $\frac13-\frac12=-\frac16$. Equal but opposite contributions from "aligned" and "opposed" edges cancel, so for a non-terminal edge the net result is equivalent by flow conservation to that of one incoming "opposed" edge incident at $i$ and one outgoing "opposed" edge incident at $j$ with net flows equal to that of $i\to j$. Thus the net result is that flow times $\frac12-\frac16-\frac16=\frac16\gt0$. The situation is only improved by sources and sinks.

If neither $i\to j$ nor $j\to i$ is an edge in the graph, then if $i$ and $j$ are not sources or sinks all contributions cancel and $\mathbb P(X_i\gt X_j)=\frac12\not\gt\frac12$ as desired; and again the situation is only improved by sources and sinks.

Qiaochu, I should add that I probably wouldn't have come up with this solution if I hadn't learned from you to try to reduce everything to linear algebra. :-)

Title: Generalized nontransitive dice


问:
What is a Herbrand disjunction?
I am supposed to find a Herbrand Disjunction for the following formula:

$$(\exists x)(P(f(f(x)))\supset P(x))$$

I'm still confused; what exactly is a Herbrand disjunction? Is it the same as a Herbrandization? I Googled Herbrand disjunction, but I found not a single page which defines it.

答: There are differences between satisfiable (true some interpretation) and valid (true under all interpretations). A Herbrand disjunction is valid, not invalid (not false under some interpretation). More accurately a Herbrand disjunction is a tautology. The important thing is that Herbrand's fundamental theorem is that every     first-order quantified valid formula can be reduced to a finite tautological propositional quantifier-free logic formula. The creation of these disjuncts is a Herbrand expansion. When it is a tautology it is a Herbrand disjunction.</p>

Title: What is a Herbrand disjunction?


问:
Noetherian Hausdorff Topological Spaces
What are the Noetherian Hausdorff topological spaces? (We know that Zariski topology is not Hausdorff, so we should look for another topological spaces).

I would be thankful for any comments.

答: You might know that in Noetherian space $X$ every subset is compact. So if $a \in X$ then $X-\{a\}$ is compact. For every $y \in X-\{a\}$ choose $y\in U_y$ and $a \in V_y$ open subsets of $X$ such that $U_y \cap V_y = \emptyset$ by Hausdorffness. By compactness of $X-\{a\}$ finite $U_{y_i}$s cover $X-\{a\}$. Consider $\cap V_{y_i} $- this is a open set such that $a \in \cap V_{y_i}$ and $\cap V_{y_i} \bigcap (X-\{a\})= \emptyset$. So $\{a\}$ is a open set. The result then follows from compactness of $X$.

Title: Noetherian Hausdorff Topological Spaces


问:
What is the "average slope"?

Derivative

This article says the following:

To find the slope at the desired point, the choice of the second point needed to calculate the ratio represents a difficulty because, in general, the ratio will represent only an average slope between the points, rather than the actual slope at either point (see figure).

I have simplified this as follows:

To find the slope at the desired point we need a second point to calculate the ratio. The choice of the second point represents difficulty. Because, in general, the ratio will not represent the actual slope at either point. Rather, it will represent an average slope between the points.

What is the "average slope"? What is the "actual slope"? What is the difference between these two?

答:
If we have two points $(x_1,y_1)$ and $(x_2,y_2)$, then the average slope is referring to $$\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}$$
and the actual slope at the point $(x_1,y_1)$ is simply $$\left.\frac{dy}{dx}\right|_{x=x_1}$$
Take this graph as an example:

the slope of the red line segment represents the average slope between points $B$ and $C$; the slopes of the orange and the green lines represent the actual slope at the points $C$ and $B$ respectively.

Title: What is the "average slope"?


答: Every subspace of a Noetherian space is Noetherian and hence compact. In a Hausdorff space, all compact subspaces are closed. Thus every subspace is closed and hence the topology is discrete. By compactness, the space is also finite.</p>

Title: Noetherian Hausdorff Topological Spaces

Body:
What are the Noetherian Hausdorff topological spaces? (We know that Zariski topology is not Hausdorff, so we should look for another topological spaces).

I would be thankful for any comments.


问:
Convergence of sequence in p-adic integers
Let $a\in Z_p^{*}$. I need to prove the sequence $\{a^{p^n}\}_{n\geq 0}$ converges in $\mathbb{Z}_p$ to $w$, where $$a=wb$$ for $w$ is the $p-1$ root of unity in $\mathbb{Z}_p^*$ and $b\in 1+p\mathbb{Z}_p$.

I feel like I need to use Hensel's Lemma for this question but I do not know how to start it. Is there any particular hint for this question?

答:
We already know from Hensel's lemma that there's an element $w$ that satisfies $w^p=w$. Further we know that $a$ shares at least one digit in common with it, so we can write for some $k\ge 1$ and $|z_k|_p \le 1$,

$$a = w + p^k z_k$$

Raising this to the $p$th power and expanding the binomial,

$$a^p = w^p +\sum_{k=1}^p \binom{p}{n} (pz_k)^n w^{p-n}$$

Simplifying $w^p=w$ and noticing we have at least a factor of $p^{n+1}$ we can factor out, for some other p-adic integer with $|z_{k+1}|_p \le 1$

$$a^p = w + p^{k+1} z_{k+1}$$

In essence raising it to the $p$th power has "uncovered" a digit of $w$. So we can repeat this process make the sequence of terms now,

$$a^{p^n} = w + p^{k+n} z_{k+n}$$

$$\lim_{n \to \infty} a^{p^n} = \lim_{n \to \infty} w + \lim_{n \to \infty} p^{k+n} z_{k+n} = w$$

Because the $z_k$ are bounded and $p^k\to 0$ and the actual terms of the $z_k$ sequence are irrelevant.


Title: function upper than $-2$

Body:
How can I demonstrate this inequation in IR:
$$x-1+\sqrt{{x^2}+2x+3}≥-2\\$$

I did this:
$$x-1+\sqrt{{x^2}+2x+3}+2≥0\\$$
$$x+1+\sqrt{{x^2}+2x+3}≥0\\$$
$$x+1+\sqrt{{x^2}+2x+1+2}≥0\\$$

$$x+1+\sqrt{{(x+1)^2}+2}≥0\\$$

$$\text{if}: x≥0 \implies x+1+\sqrt{{(x+1)^2}+2}≥0\\$$

$$\text{if}: x<0 \implies??\\$$ I don't know what I can do.

Body:
$$x-1+\sqrt{(x+1)^2+2}>x-1+\sqrt{(x+1)^2}=x-1+|x+1|$$

If $x\ge-1$, we have $x-1+|x+1|=x-1+x+1=2x\ge2(-1)$.

If $x<-1$, we have $x-1+|x+1|=x-1-1-x=-2$.


问:
Closedness in subspace topology
Let $(E,\tau)$ be a topological space, $\Omega\in\tau$ and $$\left.\tau\right|_\Omega=\left\{O\cap\Omega:O\in\tau\right\}$$ denote the subspace topology on $\Omega$ inherited from $\tau$. Since $\Omega\in\tau$, it's easy to see that $$\left.\tau\right|_\Omega=\left\{O\subseteq\Omega:O\in\tau\right\}\tag1.$$

By $(1)$, a subset of $\Omega$ is $\left.\tau\right|_\Omega$-open if and only if it is $\tau$-open. But does the analogue result hold for "open" replaced by "closed"?

I thought this should clearly be true, but it seems like it is not. Let $C\subseteq\Omega$ be $\left.\tau\right|_\Omega$-closed. Then, by $(1)$, $O:=\Omega\setminus C$ is $\tau$-open. However, $$E\setminus C=\underbrace{\Omega\setminus C}_{\:=O}\cup\underbrace{(E\setminus\Omega)\setminus C)}_{=\:E\setminus\Omega}=O\cup E\setminus\Omega\tag2$$ is not necessarily $\tau$-open, since $E\setminus\Omega$ is $\tau$-closed. Am I missing something?

答:
The statement does not hold if you replace by "closed". Take as counter-example $(E, \tau) := (\mathbb{R}, \mathcal{B}(\mathbb{R}))$ and $\Omega = (0,1)$. Then $(0,1/2]$ is $\tau \vert_{\Omega}$-closed, but it is not $\tau$-closed.

However, the statement is true if $\Omega$ is $\tau$-closed. Can you show this?

Title: Closedness in subspace topology


问:
If $L/K$ normal and $H = \operatorname{Aut}(L/K)$, then $L/L^H$ is separable and $L^H/K$ is purely inseparable.
I need to prove the following:

Let $L/K$ be a normal field extension. Denote by $H=\operatorname{Aut}(L/K)$ the Galois group of the extension, and by $L^H$ the fixed field of $H$ in $L$. Prove that $L/L^H$ is separable, and that $L^H/K$ is purely inseparable.

For now, we tried to use the fact that an extension $M/F$ is separable (purely inseparable) if the number of homomorphisms $\phi:M\to \overline{M}$ that preserve $k$ is $[M:F]$ ($1$). We did not manage to conclude any results, and moreover this works only for finite extensions. How should we proceed?

答:
This is in Lang's Algebra, 2002, p. 251, Proposition 6.11. It is not assumed that $L/K$ is Galois. The first part of the proof is identical to Stefen4024's proof.

For the second part, let $\beta\in L^H$ be separable over $K$. Then the identity map on $K$ may be extended to an embedding $\tau$ of $K(\beta)$ in $L$. Since $L/K$ is normal, this embedding is an automorphism of $K$, but $\tau\beta=\beta$ by definition and hence $\tau$ is the identity on $K(\beta)$. Therefore, $L^H/K$ is purely inseparable.


问:
Are $A^T B - I$ and $I - A B^T$ equal in Frobenius norm?
Is $\text{frob}(A^T B - I) = \text{frob}(I - A B^T)$, where $\text{frob}$ is the Frobenius norm?

Can you show why?

I ask because, I am curious what is the intuitive meaning of Eq. 4, on page 3 here:https://arxiv.org/pdf/1908.07640.pdf

It seems to be just finding the $S$ matrix s.t. $S^{-1}$ is "closest" to $R$, because the Frobenius norm of the difference between two rotations is a rotation metric known as the deviation from the identity matrix: Eq 21, https://www.cs.cmu.edu/~cga/dynopt/readings/Rmetric.pdf

答:
For the purposes of this question, I'll assume that 1) we're only interested in real-valued matrices, and 2) $A,B$ have the same dimensions (since otherwise $A^TB$ isn't square). As such, the Frobenius norm is defined as $\|A \|_F=\sqrt{\operatorname{tr}(A^T A)}$ where $\text{tr}$ is the trace operator. So two matrices $A,B$ are equal in Frobenius norm if $A^T A,B^T B$ have the same trace.

In particular, note that the cyclic property of trace ($\operatorname{tr}(AB)=\operatorname{tr}(BA)$) means that

$$\operatorname{tr}(A^T A)=\operatorname{tr}(AA^T)=\operatorname{tr}((A^T)^T (A^T)).$$
Hence $A^T$ has the same Frobenius norm as $A$ itself. In addition, we have $(-A)^T(-A)=A^T A$ and thus $-A$ also has the same Frobenius norm as $A$. Applying this to $A^TB-I$, we have

$$\|A^T B-I\|_F=\|-(B^T A-I)^T\|_F=\|I-B^T A^T\|_F$$ as asserted.


答:
You may write:

$$(x+1)^2+2\geq(x+1)^2$$

$$\sqrt {x^2+2x+3}\geq -(x+1)$$

$$x+1 +\sqrt {x^2+2x+3}\geq 0$$

This is what you found by reducing.

Title: function upper than $-2$


问:
Understanding formula to find all solutions of a linear congruence in $\mathbb Z_n$
Assume we are given a linear congruence $ax \equiv b \mod n$.

Now let $d = \gcd(a, n) | b$ and let $d > 1$. Then I learned that there are in total $d$ solutions to the linear congruence in $\mathbb Z_n$, all of which can be found in the following way:

Create a simpler congruence from the originally given one like so: $(a/d)x \equiv (b/d) \mod (n/d)$.

Let $x=c$ be a solution to this simpler congruence. Then all solutions to the originally given congruence are: $c, c + n/d, c + (2n)/d, ..., c + [(d-1)n] / d$.

We can demonstrate that if we find $c$ is a solution to the simpler congruence, then it has to be also a solution to the original congruence.

Assuming $x=c$ is a solution, translate $(a/d)c \equiv (b/d) \mod (n/d)$ to the linear Diophantine equation $a/d \times c = b/d + k \times n/d$ for some integer $k$.

Multiply through with $d$ gives $ac = b + kn$ or $ac -b = kn$ or $ac \equiv b \mod n$.

Two questions here:

If $c$ is a solution to the simpler congruence, it's not obvious to me that $c + n/d, ...$ are also automatically solutions to the original congruence. Is there a way to show this?

Is there a way to show that there are only $d$ solutions to the original linear congruence in $\mathbb Z_n$? I have found the formula for finding all solutions for a linear Diophantine equation, but I wasn't able to understand it in context of $\mathbb Z_n$.

答:
1.

$a(c+\frac nd)\equiv b\ \ \text{  mod } n \iff ac+n\frac ad\equiv b \ \ \text{ mod } n $

This is valid, since $d|a$. You have that $ac$ is congruent to $b$ modulo $n$, so:

$\iff b+n\frac ad\equiv b \ \ \text{ mod } n  \iff n\frac ad\equiv 0 \ \ \text{ mod } n $

And this is trivially true.

2.

You are already in the context of $\mathbb{Z}_n$ if you think about it. In fact if we want to reason in $\mathbb{Z}$, the solution that you wrote are not the only ones. You wrote these solutions:
$c+\frac{jn}{d} \ \ \ j=0,...,d-1$

But working in $\mathbb{Z}$ the general solution is:
$c+\frac{jn}{d}+kn \ \ \ j=0,...,d-1 \ \ k\in \mathbb{Z}$
You can easily verify that these are all solutions to the initial equation. The fact is that the solutions that differ of a multiple of $n$ are not interesting because they are congruent modulo $n$. So you can synthetize all the solutions in equivalence classes of the congruence modulo $n$:

$[c+\frac{jn}{d}]_n\ \ \ j=0,...,d-1$

And this are the only solutions if you look at the equation not as a congruence in $\mathbb{Z}$

($ax \equiv b \ \ \text{mod } n$), but as an equality in $\mathbb{Z}_n$ ($[a]_n[x]_n = [b]_n$).

If you want a formal proof that those classes are the only classes that satisfy the congruential equation then let's do it. We have already a solution $c$. Let $c'$ be a solution to our congruential equation. Then:

$a(c-c') \equiv ac-ac' \ \ \text{mod } n$

Since $c,c'$ are both solutions:

$ac-ac'\equiv b-b \equiv 0 \ \ \text{mod } n$

So:

$a(c-c') \equiv 0  \ \ \text{mod } n \iff n|a(c-c')$

Since $d=\text{gcd}(a,n)$, then $n=dk$ and $a=dh$ with $k$ and $h$ coprime:

$\iff dk|dh(c-c') \iff k|h(c-c')$

Since h and k are coprimes:

$\iff k|c-c'$

But $n=dk$ so $k=\frac nd$:

$\iff \frac nd |c-c'$

So:

$\iff c-c'=j \frac nd \iff c'=c+\frac {jn}{d}$

So a general solution is of the form:

$c'=c+\frac {jn}{d}$

But this representation is not unique in $\mathbb{Z}_n$, indeed if $j=0$ or $j=d$, we obtain the same equivalence class(because $c$ and $c'=c+n$ will differ of a multiple of $n$).
Now you can easily verify that the solutions with $j=0,1,...,d-1$ are all different(in the sense that they are not congruent modulo $n$) indeed if $k,j \in \{0,1,...,d-1\}$:

$(c+\frac {jn}{d})-(c+\frac {kn}{d})=\frac {(k-j)}{d}n$

And this is not divisible by $n$ because if it was :

$\frac {(k-j)}{d} \in \mathbb{Z} \Rightarrow |k-j| \geq d$

But it's not because $k,j \in \{0,1,...,d-1\}$.
So the solutions for $j \in \{0,1,...,d-1\}$ are all different, now we have to prove that they are the only ones(any other solution is congruent to one of them modulo $n$). So let's take a solution:

$c'=c+\frac{qn}{d} \ \  q\neq 0,1,...,d-1$

Clearly $q$ is integer so it must be in one(and only one) congruence class modulo $d$, let's say $q \in [Q]_d$(where we can consider $Q\in \{0,1,...,d-1\}$, because $\mathbb{Z}_d=\{[0]_d,...,[d-1]_d\}$). I want to show that $c+\frac{qn}{d}$ and $c+\frac{Qn}{d}$ are congruent modulo $n$:

$(c+\frac{qn}{d})-(c+\frac{Qn}{d})=\frac{(q-Q)n}{d}$

But since $q \in [Q]_d$ then $q \equiv Q \ \ \ \text{mod  }d$, and by definition $q-Q=zd$ , where $z$ is integer:

$\frac{zdn}{d}=zn$

And this proves they are congruent modulo $n$. So a generic solution of the equation is always congruent modulo $n$ to one(and only one) solution of the kind:
$c+\frac{jn}{d} \ \ \ j=0,...,d-1$
And so they completely(and withouth repetitions) represent the set of solutions of the equation modulo $n$.


问:
Why we get different answers for this integral by parts?
The integral is :$$I=\int e^{\alpha x}\cos\beta xdx$$

To evaluate the integral I used integral by parts method twice. for first integral I used substitution $u=e^{\alpha x}$ and $dv=\cos\beta x dx$ and for second one $dv=\sin\beta x$ My final answer is $$I=\frac{\beta}{\alpha^2+\beta^2}e^{\alpha x}(\sin\beta x+\frac{\alpha}{\beta}\cos\beta x)+C$$

But the answer in the book I am studying is :

$$I=\frac{\alpha}{\alpha^2+\beta^2}e^{\alpha x}(\cos\beta x+\frac{\beta}{\alpha}\sin\beta x)+C$$

First I thought my answer is wrong but I took derivative of that and obtained $e^{\alpha x}\cos\beta x$. so I realized my answer is also correct.

I find out the reason for different answers is different substitution the book used $u=\cos\beta x$ and $dv=e^{\alpha x}dx$. My question is why we got different answers here for different $u$ and $dv$? Is there any mathematically logic that can explain this?

答:
If $F'(x)=G'(x)$ then we have $F(x)-G(x)=constant$ so the difference related to constant.
For example since we have $$\sin^2x+\cos^2x=1$$ then  one can let $\sin^2x=1-\cos^2x$
or $\cos^2x=1-\sin^2x$ that is no basic differnce between $\sin^2x $ and $\cos^2x$ as answer of integrals.


问:
representations of a non-separable $C^*$-algebra
If $A$ is a non-separable $C^*$-algebra, can we deduce that all the representations of $A$ are infinie dimensional?

答:
Let $I$ be an uncountable set and fix $j \in I$. Define

$$\pi: \mathbb{C}^{(I)} \to B(\mathbb{C}) = \mathbb{C}:  (\lambda_i)_{i \in I} \mapsto \lambda_j$$

Then $\pi$ is a one-dimensional representation of $A$. Moreover, $\mathbb{C}^{(I)}$ is non-separable because the set $\{\delta_i : i \in I\}$ of Dirac-functions satisfies $\| \delta_i - \delta_j\| = 1$ when $i \neq j$, and thus we can build an uncountable set of disjoint open balls, which shows that any dense set must be uncountable. (Credit goes to @Ruy for the above argument).

If you want something non-abelian simply take the direct sum with a non-abelian $C^*$-algebra.


问:
Jordan matrices and isomorphisms
We just proved in class that if $A$ is a linear transformation of a finite dimensional $K$-vector space $V$ (where $K$ is algebraically closed), then there exists a basis of $V$ such that the matrix of $A$ relative to that basis is a Jordan matrix.

I was now wondering how I could prove the following theorem under the same circumstances where $V$ and $V'$ are the vector spaces of the same dimension and $A$ and $A'$ are linear transformations on these vector spaces:

$A$ and $A'$ have the same Jordan matrix iff there exists and isomorphism $f: V \rightarrow V'$ so that the following diagram is commutative:

答:
Let $B=\{v_1,\dots,v_n\}$ be a basis of $V$. When $f\colon V\to V'$ is an isomorphism satisfying $A'\circ f=f\circ A$, it follows that $B':=\{f(v_1),\dots,f(v_n)\}$ is a basis of $V'$ and the transformation matrices $[A]_{B,B}=[A']_{B',B'}$ are identical. In particular they have the same Jordan canonical form.

When $A$ and $A'$ have the same Jordan canonical forms, take Jordan bases $B=\{v_1,\dots,v_n\}$ of $V$ with respect to $A$ and $B'=\{v'_1,\dots,v'_n\}$ of $V'$ with respect to $A'$. Define $f\colon V\to V'$ by $f(v_i)=v'_i$ and note that this yields $A'\circ f=f\circ A$.

Note that this is just a corollary of matrices being similar iff they have the same JCF.

Title: Jordan matrices and isomorphisms


问:
Is vector dot product and vector cross product allowed in Vector space?
As per the definition of Vector space, for a set of vectors and scalars, the only operations allowed are the 2 closure properties(for addition and scalar multiplication) and 10 axioms(for associativity commutativity, inverse and identity) for the vectors, scalars involved.
If so, where does dot product and cross product(which is vector time vector multiplication) fit in?
These 2 vector times vector multiplication operations are not defined and hence not allowed as per  the definition of Vector space. Is this understanding correct or am I missing something here?</p>

答:
You are correct, an abstract vector space (assume all spaces I mention have $\mathbb R$ as scaars) does not have such operations (dot product, cross product) defined.

An "inner product" on a vector space $V$ is an operation with the properties of a dot product.  Any abstract vector space admits many different inner products.  A vector space, together with a specified inner product, is called an inner product space .  This name exists because mathematicians have found this type of structure useful.

An unexpected result says that an operation like "cross product" only exists for $3$-dimensional spaces.  There is an "analog" of the cross product in $\mathbb R^n$, but in that case you multiply $n-1$ vectors to get a vector answer.  So only the case $\mathbb R^3$ is where you multiply two vectors to get a vector answer.  See HERE for various generalizations of the cross product to other vector spaces.


问:
How to integrate $\int\frac{dx}{\sqrt{ax^2+bx+c}}$
Let
$$I=\int\frac{dx}{\sqrt{ax^2+bx+c}}$$
I know this can be either
$$\displaystyle I=\frac{1}{\sqrt{a}}\ln\left({2\sqrt{a}\sqrt{ax^2+bx+c}+2ax+b}\right)+C$$
$$\displaystyle I=-\frac{1}{\sqrt{-a}}\arcsin{\left(\frac{2ax+b}{\sqrt{b^2-4ac}}\right)}+C$$
Or
$$I=\frac1{\sqrt{a}}\mathrm{arcsinh}\left(\frac{2ax+b}{\sqrt{4ac-b^2}}\right)+C$$

Can you tell me the difference between these three solutions?

Thanks a lot

答:
$$
\displaystyle I=\frac{1}{\sqrt{a}}\ln\left({2\sqrt{a}\sqrt{ax^2+bx+c}+2ax+b}\right)+C
\label{eq1} \tag{eq1}\\
$$

$$
\displaystyle I=-\frac{1}{\sqrt{-a}}\arcsin{\left(\frac{2ax+b}{\sqrt{b^2-4ac}}\right)}+C
\label{eq2} \tag{eq2}\\
$$

$$
I=\frac1{\sqrt{a}}\mathrm{arcsinh}\left(\frac{2ax+b}{\sqrt{4ac-b^2}}\right)+C
\label{eq3} \tag{eq3}\\
$$

First, \eqref{eq1} and \eqref{eq3} are the same.
$$\text{arcsinh} (x/a) = \ln(x + \sqrt{x^2 + a^2})-\ln a.$$

Second, \eqref{eq2} and \eqref{eq3} are different.

\eqref{eq2} holds, when $a<0$ and $b^2-4ac> 0$.

\eqref{eq2} holds, when $a>0$ and $b^2-4ac< 0$.


答:
When $f:\>[a,b]\to{\mathbb R}$ is convex then $$f'(a)<{f(b)-f(a)\over b-a}<f'(b)\ .$$
In this way the difference quotient ${f(b)-f(a)\over b-a}$ can be seen as a sort of "mean" or "average" between $f'(a)$ and $f'(b)$. But more is true: There is actually a point $\xi\in\>]a,b[\>$ such that
$${f(b)-f(a)\over b-a}=f'(\xi)\ .$$
The word average slope is also justified by the following equation:
$${f(b)-f(a)\over b-a}={1\over b-a}\int_a^b f'(t)\>dt\ .$$
Here on the RHS the actual integral mean, or average,  of the pointwise slopes between $a$ and $b$ is computed.

Title: What is the "average slope"?

Body:
Derivative

This article says the following:

To find the slope at the desired point, the choice of the second point needed to calculate the ratio represents a difficulty because, in general, the ratio will represent only an average slope between the points, rather than the actual slope at either point (see figure).

I have simplified this as follows:

To find the slope at the desired point we need a second point to calculate the ratio. The choice of the second point represents difficulty. Because, in general, the ratio will not represent the actual slope at either point. Rather, it will represent an average slope between the points.

What is the "average slope"? What is the "actual slope"? What is the difference between these two?


答: No.  Take a non-metrizable compact space $X$ so that $C(X)$ is non-separable, but it has lots of one-dimensional representations.

Title: representations of a non-separable $C^*$-algebra

Body: If $A$ is a non-separable $C^*$-algebra, can we deduce that all the representations of $A$ are infinie dimensional?


问:
Intuitive way to understand the square wave spectrum?
What is a intuitive way to understand that a transform of a square wave can result into something like this?

答:
Is it intuitive to switch the time and frequency axes? We can do it because the Fourier transform is essentially its own inverse. So, the spectrum of a square wave looks the same way as the time profile of a signal that contains all frequencies up to $\omega$, with equal amplitude (because rectangular wave has constant modulus) and phase (because rectangular wave is a positive function).

Let's simplify further and replace continuous superposition of frequencies by discrete. So, at time $0$ we strike $n$ strings that vibrate with angular frequencies $$ \frac{1}{N}\omega,\frac{2}{N}\omega,\frac{3}{N}\omega,\dots, \frac{N-1}{N}\omega, \omega$$
The first moment is the loudest, when all strings are in phase. (This is the peak on your picture). But that quickly changes. Half of all strings have frequencies between $\omega/2$ and $\omega$. They form a  packet that stays together for a while; certainly for more than half of one period $T=2\pi/\omega$. Indeed, at time $t\approx 2T/3$ most of high-frequency contributions are negative. On the graph below this happens around $0.4$.

This is why we see a negative minimum when adding these frequencies together:

After that the pattern becomes less pronounced. But the basic idea remains: the behavior of the sum is driven  by its high-frequency components, simply because there are more of them (which becomes apparent if one groups the frequencies on logarithmic scale).


问:
A problem about a rank's inequality of a complex matrix
Let $X,Y \in M_{n\times n}(\mathbb{C})$ such that $X^{2}=Y^{2}=I_{n}$. Prove that

$$\operatorname{rank}((X+I_n)(Y-I_n))+\operatorname{rank}((X-I_n)(Y+I_n))=\operatorname{rank}(X-Y)$$

My approach: Using the well-known inequality: $\operatorname{rank}(X)+\operatorname{rank}(Y)\geq \operatorname{rank}(X+Y)$, we can see that $$\operatorname{rank}(X-Y)=\operatorname{rank}(X-XY+XY-Y)\leq \operatorname{rank}(X-XY)+\operatorname{rank}(XY-Y)$$

but, how can I continue from here?

答:
As $X$ and $Y$ are involutions we have that
$$
P_x=\frac12(X+I)\qquad\text{and}\qquad P_y=\frac12(Y+I)
$$
are idempotent (projections). It is straightforward to see that the rank identity is equivalent to
$$
\operatorname{rank} P_x(I-P_y)+\operatorname{rank}(I-P_x)P_y=\operatorname{rank}(P_x-P_y)
$$
which follows easily from the fact that
$$
P_x-P_y=P_x(I-P_y)-(I-P_x)P_y
$$
together with the standard result that the images of $P_x$ and $I-P_x$ are complementary.


答:
Suppose $a>0$.  Complete the square.  You get one of these cases:
$$
\frac{1}{\sqrt{a}}\int\frac{dx}{\sqrt{(x-\beta)^2}},\qquad \beta\in \mathbb R,\\
\frac{1}{\sqrt{a}}\int\frac{dx}{\sqrt{(x-\beta)^2+\gamma^2}},\qquad \beta\in \mathbb R, \gamma > 0,\\
\frac{1}{\sqrt{a}}\int\frac{dx}{\sqrt{(x-\beta)^2-\gamma^2}},\qquad \beta\in \mathbb R, \gamma > 0 .
$$
The first one has a "$\log$" solution, the second one has an "$\arcsin$" solution, the third one has an "$\text{asinh}$" solution.

There are three similar cases when $a<0$.

Title: How to integrate $\int\frac{dx}{\sqrt{ax^2+bx+c}}$

Body:
Let
$$I=\int\frac{dx}{\sqrt{ax^2+bx+c}}$$
I know this can be either
$$\displaystyle I=\frac{1}{\sqrt{a}}\ln\left({2\sqrt{a}\sqrt{ax^2+bx+c}+2ax+b}\right)+C$$
$$\displaystyle I=-\frac{1}{\sqrt{-a}}\arcsin{\left(\frac{2ax+b}{\sqrt{b^2-4ac}}\right)}+C$$
Or
$$I=\frac1{\sqrt{a}}\mathrm{arcsinh}\left(\frac{2ax+b}{\sqrt{4ac-b^2}}\right)+C$$

Can you tell me the difference between these three solutions?

Thanks a lot


问:
Approximating $\int \log\big[1+\sin^2(t)\big]\,dt$
On yesterday, a now closed question asked for a while how to compute the antiderivative
$$I=\int \log\big[1+\sin^2(t)\big]\,dt$$ which, as given by a CAS, is just awful (see @Forester's comment and @Turing's answer). During the time the question was open, I tried on my side and gave up since being unable to find anything simpler.

Then, I tried to approximate the integrand since we just need a good representation of it for $0 \leq t \leq \frac \pi 2$ because of the symmetries and periodicity. What I am proposing is a polynomial function matching the function and its first and second derivative values at a minimum number of data points. For the time being, I chose $t=0$, $t=\frac \pi 4$ and $t=\frac \pi 2$. The final result is
$$f(t)=\log\big[1+\sin^2(t)\big]\sim t^2+\frac 19 \sum_{n=3}^8 \frac{a_n}{\pi^n} t^n=g(t)$$ the $a_n$'s being
$$\left(
\begin{array}{ccc}
n & a_n & \frac{a_n}{9\pi^n}\\
 3 & -768 \pi -217 \pi ^2-22608 \log (2)+18432 \log (3) &         +0.087806356 \\
 4 & 7680 \pi +1940 \pi ^2+286416 \log (2)-221184 \log (3) &      -1.360303642 \\
 5 & -27648 \pi -8500 \pi ^2-1501632 \log (2)+1105920 \log (3) &  +1.225383826 \\
 6 & 43008 \pi +19568 \pi ^2+3960576 \log (2)-2801664 \log (3) &  -0.512958585 \\
 7 & -24576 \pi -22720 \pi ^2-5170176 \log (2)+3538944 \log (3) & +0.102611702 \\
 8 & 10496 \pi ^2+2654208 \log (2)-1769472 \log (3) &             -0.007207813
\end{array}
\right)$$

About the norm
$$\Phi=\int_0^{\frac \pi 2} \big[f(t)-g(t)\big]^2 \, dt=1.79 \times 10^{-8}$$ while the beautiful approximation
$$\sin(t) \simeq \frac{16 (\pi -t) t}{5 \pi ^2-4 (\pi -t) t}\qquad \text{for} \qquad 0\leq t\leq\pi$$ would lead to  $\Phi=8.74 \times 10^{-7}$ (almost $50$ times larger).

From this
$$J(x)=\int_0^x \log\big[1+\sin^2(t)\big]\,dt\sim \frac 13 x^3+\frac 19 \sum_{n=3}^8 \frac{a_n}{(n+1)\,\pi^n} x^{n+1}$$

As a test
$$J\left(\frac{\pi }{2}\right)=\frac{\pi }{181440}\left(432 (128 \log (3)-87 \log (2))-55 \pi ^2 \right)\approx 0.591383$$ to be compared to the exact value
$$ \pi \log\left(\frac{1+\sqrt{2}}{2}\right)\approx 0.591331$$ recalled by @projectilemotion in comments.

My questions

Without adding more terms, could it be possible to improve the approximation ? Would changes of base points could have a significant impact (if yes, how to optimize them ?)

Would, by chance, be known (in a simple form) the exact values of the integrals
$$K_n=\int_{-\frac \pi n}^{\frac \pi n} \log\big[1+\sin^2(t)\big]\,dt$$ beside the case of $n=2$ ? If there were, I suppose that much better approximations could be done.

Body:
After @Martin Gales's, I think that I had an idea.

Rewrite
$$\log\big[1+\sin^2(t)\big]=\log\Bigg[\frac {1+ \frac{2 \sin ^2(t)}{5-\cos (2 t)}} {1- \frac{2 \sin ^2(t)}{5-\cos (2 t)}}\Bigg]$$

and use
$$\log \left(\frac{1+a}{1-a}\right)=2 \sum_{n=0}^\infty \frac{a^{2n+1}}{2n+1}$$ to make
$$\log\big[1+\sin^2(t)\big]=\sum_{n=0}^\infty \frac{4^{n+1}}{2n+1} \Bigg[\frac{ \sin ^2(t)}{5-\cos (2 t)}\Bigg]^{2n+1}$$
$$\int\log\big[1+\sin^2(t)\big]\,dt=\sum_{n=0}^\infty \frac{4^{n+1}}{2n+1} \int\Bigg[\frac{ \sin ^2(t)}{5-\cos (2 t)}\Bigg]^{2n+1}dt$$
All these integrals can be analytically computed (an example here) in terms of elementary trigonometric functions.

Computing
$$S_p=\sum_{n=0}^p \frac{4^{n+1}}{2n+1} \int_0^{\frac \pi 2}\Bigg[\frac{ \sin ^2(t)}{5-\cos (2 t)}\Bigg]^{2n+1}dt$$ it looks better to write it as
$$S_p=-\sqrt{\pi }\,\,\sum_{n=0}^p \frac{\Gamma \left(2 n+\frac{3}{2}\right)}{\Gamma (2 n+2)}\,B_{-\frac{1}{2}}\left(2 n+1,-2 n-\frac{1}{2}\right)$$
$$\left(
\begin{array}{cc}
 p & S_p \\
 0 & 0.5764929933 \\
 1 & 0.5905254035 \\
 2 & 0.5912752808 \\
 3 & 0.5913264052 \\
 4 & 0.5913303391 \\
 5 & 0.5913306647 \\
 6 & 0.5913306929 \\
 7 & 0.5913306955 \\
 8 & 0.5913306957
\end{array}
\right)$$
The last value in the table is
$$S_8=\left(\frac{1593269}{765765}-\frac{2025528882598026950447}{436968810224560373760
   \sqrt{6}}\right) \pi$$ (error= $2.45\times 10^{-11}$).

The next one
$$S_9=\left(\frac{31037876}{14549535}-\frac{8543966267760579986943883}{179331999716159577
   3911040 \sqrt{6}}\right) \pi$$ (error= $2.35\times 10^{-12}$).

This fast convergence is explained by the fact that, for large $n$
$$\frac{a_{n+1}}{a_n}=\frac{(4 n+3) (4 n+5)}{8 (n+1) (2 n+3)}\frac{B_{-\frac{1}{2}}\left(2 n+3,-2 n-\frac{5}{2}\right)}{B_{-\frac{1}{2}}\left(2
   n+1,-2 n-\frac{1}{2}\right)}\to \sim \frac 19$$

Edit

If this could help

$$I_n=\frac{4^{n+1}}{2n+1} \int_0^{x}\Bigg[\frac{ \sin ^2(t)}{5-\cos (2 t)}\Bigg]^{2n+1}dt$$ can write
$$I_n=\frac {z ^{4 n+3}}{4^n \,(2n+1)(4n+3)}F_1\left(2 n+\frac{3}{2};2 n+1,\frac{1}{2};2
   n+\frac{5}{2};-\frac{z^2}{2},z^2\right)$$ where $z=\sin(x)$ and $F_1(.)$ is the Appell hypergeometric function of two variables.


问:
Proof equivalence of $X\succcurlyeq Y \iff Y^{-1}\succcurlyeq X^{-1}$
For to symmetric matrices $X,Y$ I would like to prove this via Schur. However, I'm not fully understand the solution provided.

$$ X\succcurlyeq Y \iff X-Y\succcurlyeq 0 \iff \begin{pmatrix}X & 1\\1 & Y^{-1}
\end{pmatrix}\succcurlyeq 0\iff Y^{-1}-1X^{-1}1\succcurlyeq 0$$

The very last two $\iff$ are the ones which trouble me. How can we say
$$X-Y\succcurlyeq 0 \iff \begin{pmatrix}X & 1\\1 & Y^{-1}
\end{pmatrix}\succcurlyeq 0$$

and

$$\begin{pmatrix}X & 1\\1 & Y^{-1}
\end{pmatrix}\succcurlyeq 0\iff Y^{-1}-1X^{-1}1\succcurlyeq 0$$

答:
According to Schur Conditions for positive definiteness and semi-definiteness for a matrix

$$X=\begin{pmatrix}
A & B\\
B^T & C
\end{pmatrix}$$

where $A$ is invertible, we have
$$X \succcurlyeq 0 \iff A \succcurlyeq 0, X /A = C - B^T A^{-1} B \succcurlyeq 0.$$

Applied to the case of your question, this provides the equivalence you're looking for.

Note: wikipedia provides a link about the equivalence I mention above.


问:
The man with two boys
I have recently seen a probability question which says

"i am asking randomly the persons I met if they are having two chidren and one of them is a boy who was born on tuesday. At last I met one whose answer is yes. What is the probability that the other child is also a boy. Assume equal probability to either gender and equal probability to be born on each day of the week"

I could actually solve it to 2/21. Did I do it right or can some one help me solve it?

答:
This is going to get a bit long for a comment.

Suppose you are asking three questions in sequence:

1 Do you have exactly two children?

If the answer is yes, you include the person on your sample, and ask

2 Do you have a child who is a boy born on a Tuesday?

If the answer is yes, the person continues in the sample - call this $BT$ and you ask

3 Are both of your children boys?

Say the yes answers to this are $BB$, you are looking for $\frac {BB}{BT}$

Note that (given reasonable assumptions) one fourteenth of all children are tuesday-born boys. However, at stage 2, there will be $0$ or $1$ or $2$ Tuesday-born boys in the family. Case $0$ is discarded, and the other two are not equally likely.

Taking the day and the sex into account, there are 14 possibilities for each child - 196 altogether. In one of these, both are Tuesday-boys. There are 26 more possibilities containing a Tuesday-boy, making 27 in all (the $2$ case takes two of the expected 28) - twelve of which have two boys (two of the fourteen having been taken into account already), and fourteen (as expected) with a boy and a girl.

Counting, one sees the proportion with two boys is $\frac {13}{27}$.

I noted this, because it was not explained in the answers to the closely related question Zev Chonoles linked.

Title: The man with two boys


问:
Properties of topological vector space.

Let $X$ be a topological vector space. Let $\mathcal B$ be a local base at $0.$ Then the following statements hold $:$

$(1)$ $\mathcal U, \mathcal V \in \mathcal B \implies \exists\ \mathcal W \in \mathcal B$ such that $\mathcal W \subset \mathcal U \cap \mathcal V.$

$(2)$ $x \in \mathcal U \in \mathcal B \implies \exists\ V \in \mathcal B$ such that $x + \mathcal V \in \mathcal U.$

$(3)$ $\mathcal V \in \mathcal B, x \in X \implies \exists\ 0 \neq \lambda \in \Bbb C\ (\text {or}\ \Bbb R)$ such that $\lambda x \in \mathcal V.$

$(4)$ $\mathcal U \in \mathcal B \implies \exists\ \mathcal V \in \mathcal B$ such that $\mathcal V + \mathcal V \subseteq \mathcal U.$

$(5)$ $\mathcal U \in \mathcal B \implies \lambda\ \mathcal U \subseteq \mathcal U,\ \forall\ \lambda$ with $|\lambda| \lt 1.$

$(1)$ $:$ It follows from the fact that $\mathcal U \cap \mathcal V$ is an open neighbourhood of $0.$ Now since $\mathcal B$ is a local base at $0$ it follows that there exists $\mathcal W \in \mathcal B$ such that $\mathcal W \subseteq \mathcal U \cap \mathcal V.$

$(2)$ $:$ Since $X$ is a topological vector space the map $f_x : X \longrightarrow X$ defined by $y \mapsto (y - x)$ is a homeomorphism and hence open. Thus $\mathcal U - x$ is an open neighbourhood of $0.$ So there exists $\mathcal V \in \mathcal B$ such that $\mathcal V \subseteq \mathcal U - x \iff x + \mathcal V \subseteq \mathcal U.$

$(3)$ $:$ Since $X$ is a topological vector space the map $g_x : \Bbb C \longrightarrow X$ defined by $\lambda \mapsto \lambda x$ is continuous. In particular $g_x$ is continuous at $0.$ Hence there exists $\varepsilon \gt 0$ such that for all $\lambda \in \Bbb C$ with $|\lambda| \lt \varepsilon$ we have $\lambda x \in \mathcal V.$ Now take any $\lambda \in \Bbb C$ with $0 \lt |\lambda| \lt \varepsilon$ and then we are through.

$(4)$ $:$ Since $X$ is a topological vector space the binary operation $(x,y) \mapsto (x+y)$ is continuous. Using the continuity of the binary operation at $(0,0)$ we get open neighbourhoods $V_1$ and $V_2$ of $0$ such that $V_1 + V_2 \subseteq \mathcal U.$ Find $\mathcal W_1, \mathcal W_2 \in \mathcal B$ such that $\mathcal W_1 \subseteq V_1$ and $\mathcal W_2 \subseteq V_2.$ Then $\mathcal W_1 \cap \mathcal W_2$ is an open neighbourhood of $0.$ So there exists $\mathcal V \in \mathcal B$ such that $\mathcal V \subseteq \mathcal W_1 \cap \mathcal W_2.$ Then by the construction $\mathcal V \subseteq V_1 \cap V_2$ and hence $\mathcal V + \mathcal V \subseteq V_1 + V_2 \subseteq \mathcal U.$

Can anybody please help me proving the last statement i.e. $(5)\ $? I am having hard time in proving this. Also please check my arguments from $(1)$-$(4).$

Thanks in advance.

答:
It is not true under your assumption "Let $\mathcal B$ be a local base at $0$." As Sumanta Das comments, take $X = \mathbb R$ and $\mathcal B = \{(−1/n,2/n)\mid n \in \mathbb N\}$. Then no $U \in \mathcal B$ has the property $(−3/4)U \subset U$ although $|−3/4|<1$.

You can show that there exists a local base $\mathcal B'$
at $0$ with properties $(1)−(5)$. Given any $\mathcal B$, for each $U \in \mathcal B$ define $U' = \bigcup_{|λ| \le 1} λU$ and let $\mathcal B'$ be the collection of these $U'$.

Title: Properties of topological vector space.


问:
No of possible binary sequences with at most 2 consecutive zeroes
The problem:

How many such sequences of ten $1's$ and ten $0's$ are possible such that they don't contain three or more consecutive zeroes? (Eg. $00101001001101110011$)

I know there are $20!/10!10!$ total sequences. How do I filter out the ones which have more than two consecutive zeroes somewhere?

答:
The ten $1's$ can be looked upon as $10$ bars forming $11$ "compartments" in which a maximum of two $1's$ can be put.

This can be solved using stars and bars  with inclusion-exclusion

$N = \binom{20}{10} - \binom{11}1\binom{17}{10} +\binom{11}2\binom{14}{10} - \binom{11}3\binom{11}{10} = 24068$


问:
compute$\iint_D \sqrt{\frac{x^2}{y}+\frac{y^2}{x}} \ dxdy$
The question is:
$$\iint_D \sqrt{\frac{x^2}{y}+\frac{y^2}{x}} \ dxdy \quad D=y<x^2<2y,\ x<y^2<2x$$

The only thing that i can think of is that maybe change of variable would work but i'm not able to solve it i think that i'm on the wrong path;
I've tried this change of variable:
$$u=\frac{x}{\sqrt{y}}, \  v=\frac{y}{\sqrt{x}},\quad \iint(\sqrt{u^2+v^2})(\sqrt{xy}) \ dudv$$

Any suggestion would be great, thanks

答:
With your change of variables, you can see that $uv = \sqrt{xy}$. Also, the bounds for the integral are $0<u,v<\sqrt{2}$. As correctly pointed out by MathLover, the Jacobian for this transformation equals $4/3$. Hence the integral becomes

$$ \frac{4}{3}\int_1 ^\sqrt{2} \int_1 ^\sqrt{2} (\sqrt{u^2+v^2})uv \mbox{ d}u\mbox{d}v = \int_1 ^\sqrt{2} \frac{4}{9}v\big( (2 + v^2)^{3/2} - (1+v^2)^{3/2} \big)\mbox{ d}v = \frac{4}{45}((4^{5/2} - 3^{5/2})-(3^{5/2}-2^{5/2}) = \frac{4}{45}(32+4\sqrt{2}-18\sqrt{3}). $$


问:
What exactly is the complex plane, and how is it useful?
A lot of functions are defined on the complex plane, like the Gamma function:

the Lambert W function,

etc.

But I have no idea about what the complex plane means and how it's useful, or just how they graph it. If someone has an idea, please tell me about it. Thanks to all.

答:
The complex plane is simply an x-y grid, where the x axis is the reals, and the y axis is the complex number $\sqrt{-1}$.  What makes this sort of representation is that rotation around the centre is a multiplication, and thus the re-scaling and rotation required to bring $(1,0)$ to the point in question, is the multiplication table by this.

One use of this rotation-as-multiplication, is the cyclic functions are represented as a finite curve in the polar notation.  It is used extensively in electrics because inductances and capacitances behave as complex resistances, leading and laging the voltage.

The graps shown above are coloured, because they are trying to show information at each point of the plane.  That is, you can't just have a plot of the result against an input on a plane: the input is the plane, and we leave it to other devises to indicate what the output is doing.


问:
$\lambda((a,b)/E)\geq\frac{a-b}{2}$ show that $\lambda(E) = 0$
Given that $E$ is a Lebesgue measurable set of $\Re$ and $\lambda((a,b)\setminus E)\geq\frac{a-b}{2}$(1) (for every $(a,b)\subset \Re$)

show that $\lambda(E)=0$

What I have tried so far is try to contradict the property

-$E$ is union of a $F_$ and a set of Lebesgue measure zero

and specifically show that there can't be $F_$ with the property (1) thus leaving us with a zero measure

答:
For every measurable set $E$ and $\varepsilon>0$ there is an open set $O\supset E$ such that $\lambda(O\setminus E)\leq\varepsilon $ (this comes directly from de definition of lebesgue measurable), now $O$ is the disjoint union of countable open intervals so $\lambda(O\setminus E)\geq\lambda (O)/2$ so this implies $$\varepsilon\geq\lambda(O\setminus E)\geq\lambda(O)/2\geq\lambda (E)/2$$
Since $\lambda(E)\leq 2\varepsilon$ for every $\varepsilon>0$, $\lambda(E)=0$.


问:
Evaluating a summation involving binomial coefficients and gaussian CDF
I am trying to evaluate the following summation:

$\hspace{2in}f=\sum_{z=0}^{N-1}Q(z)\phi(z)$

Here, $\hspace{1in}Q(z)=\left(\begin{array}{c}N-1 \\ z\end{array}\right)\left(\frac{ q}{N}\right)^{z}\left(1-\frac{ q}{N}\right)^{N-1-z}$

$\hspace{1.5in}\phi({z})$ is CDF of gaussian RV ~$N(0,a+z)$

Is there a way of evaluating $f$ analytically?

Any help is appreciated.

If it is impossible, is there a good approximation that can be helpful. I can assume $N$ is a large positive integer.

答:
Here's a heuristic approach that can easily be done strict.

Claim: If $N \to \infty$ then $$f \approx \sum_{z=\max([q]-10, 0)}^{[q]+10} \phi(z) e^{-q}\frac{q^z}{z!} \in (0,1)$$
for $q \le 10$
and error may be estimated easily.
In case when $q \ge 10$  we have $f \approx \sum_{q - \sqrt{10q} < z < q+\sqrt{10q}} \phi(z) e^{-q}\frac{q^z}{z!} \in (0,1)$.

Corollary 1. If $q - \sqrt{10q} > 2 + 2\sqrt{a+1}$ and $ q > 10$ then  $f \approx 1$.

As $q$ is fixed then number of terms in sums is fixed and not bigger than $2\max(10, \sqrt{10q}) + 1$.

Proof of the claim.
We suppose that $N >> a$ and $N >>q$. Consider the case $q \le 10$.

As we have $z \ge 0$ and $N(0,a+z)$ exists we will suppose $a > 0$ (case $a=0$ is close to case $a=0.001$).

Put $\Phi(u) = \int_{-\infty}^u \frac{1}{\sqrt{2\pi}}e^{-\frac{(x-a)^2}{2}} dx$, then $\phi(z) = \Phi(\frac{z}{\sqrt{z+a}})$. For $z\ge 0$ function $\frac{z}{\sqrt{z+a}}$ is non-decreasing, positive and tends to 1 as $z \to +\infty$. Hence $\phi(k)$, $k = 0, 1, 2, \ldots$ is non-decreasing, positive and tends to 1 as $z \to +\infty$.

We have
$$f = \sum_{z=\max([q]-10, 0)}^{[q]+10} \phi(z) Q(z) + \sum_{z \in A} \phi(z) Q(z) $$
where $A = \{ z \ge 0: (z < \max([q]-10, 0))  \vee  (z > [q]+10)\}$
But
$$\sum_{z \in A} \phi(z) Q(z) \le  \sum_{z \ge 0, |z-q| \ge 9} \phi(z) Q(z) \le \sum_{z \ge 0, |z-q| \ge 9} Q(z). $$

Put $\xi \sim Bin(\frac{q}{N}, N-1)$. Hence
$$\sum_{z \in A} \phi(z) Q(z) \le P(|\xi - q| \ge 9) = P(|\xi - E\xi - \frac{q}N| \ge 9) \le $$
$$P(|\xi - E\xi| \ge 9 - \frac{q}N) \approx P(Pois(q) \ge 9) \le \frac{D Pois(q)}{81} \approx 0 $$
as $q \le 10$.

Hence $$f \approx \sum_{z=\max([q]-10, 0)}^{[q]+10} \phi(z) Q(z)$$.

If $N \to \infty$ then $Q(k) \to P(Posi(q)=k) = e^{-q}\frac{q^k}{k!}$ by Poisson limit theorem. When we write  $Q(k) \approx e^{-q}\frac{q^k}{k!}$ and error may be estimated easily from Stirling's formula.

Thus $$f \approx \sum_{z=\max([q]-10, 0)}^{[q]+10} \phi(z) e^{-q}\frac{q^z}{z!}$$.

Case $q \ge 10$ is similar. Instead  of $A = \{ z \ge 0: (z < \max([q]-10, 0))  \vee  (z > [q]+10)\}$ and
$$\sum_{z \in A} \phi(z) Q(z) \le P(|\xi - q| \ge 9) \le P(|\xi - E\xi| \ge 9 - \frac{q}N) \approx P(Pois(q) \ge 9) \le \frac{D Pois(q)}{81} \approx 0 $$
we will have
$A = \{ z \ge 0: (z > q + \sqrt{10q}) \vee z< q - \sqrt{10q}\}$ and
$$\sum_{z \in A} \phi(z) Q(z) \le P(|\xi - q| \ge \sqrt{10q}-1) \approx P(|\xi - E\xi| \ge \sqrt{10q} ) \approx P(Pois(q) \ge \sqrt{10q}) \le \frac{D Pois(q)}{10q} \approx 0 $$.

Corollary2. In case when $q \ge 0$ is arbitrary we have $f \approx \sum_{q - \max(\sqrt{10q},10) <z < q+\max(\sqrt{10q},10)} \phi(z) e^{-q}\frac{q^z}{z!} \in (0,1)$.

Proof of the corollary 1: $\Phi(z) \approx 1$ for $z \ge 2$, thus $\phi(z) \approx 1$ for $\frac{z}{\sqrt{z+a}} \ge 2$. As we have $\frac{z}{\sqrt{z+a}} \ge 2$ iff $z \ge 2 + 2 \sqrt{a+1}$ then $\phi(z) \approx 1$ for all $z \ge 2 + 2 \sqrt{a+1}$ .

If $q - \sqrt{10q} > 2 + 2\sqrt{a+1}$ and $ q > 10$ then  $$f \approx\sum_{q - \sqrt{10q} < z < q+\sqrt{10q}}\phi(z) e^{-q}\frac{q^z}{z!} \approx \sum_{q - \sqrt{10q} < z < q+\sqrt{10q}} e^{-q}\frac{q^z}{z!}  = P(q -
 \sqrt{10q} < Pois(q) < q +\sqrt{10q})$$.

But $$P(q -
 \sqrt{10q} < Pois(q) < q +\sqrt{10q}) \approx 1$$ by Chebyshev's inequality.


问:
Find Fourier half range sine Series for $f(x)=x(\pi-x)$ on $(0,\pi)$.
I tried finding the Fourier coefficient  $ b_n$  but it came zero.

My attempt:

$ b_n=\frac{4}{\pi}\int_0^\pi x(\pi-x)\sin(2nx)dx\\
=\frac{4}{\pi}\left\{\pi\int_0^\pi x\sin(2nx)dx-\int_0^\pi x^2\sin(2nx)dx\right\}\\
=0$

Correct answer is:

$\frac{8}{\pi}\left(\frac{\sin x}{1^3}+\frac{\sin3x}{3^3}+\frac{\sin5x}{5^3}+...\right)$

答:
You should integrate
$$\frac{2}{\pi} \int_0^{\pi } x (\pi -x) \sin (n x) \, dx=\frac{4-2 \pi  n \sin (\pi  n)-4 \cos (\pi  n)}{\pi  n^3}$$
to get coefficients
$$\frac{8}{\pi },0,\frac{8}{27 \pi },0,\frac{8}{125 \pi },\ldots$$
and finally
$$\frac{8}{\pi}\left(\sin x +\frac{\sin 3x}{3^3}+\frac{\sin 5x}{5^3}\ldots\right)$$


答: The correct formula for $b_n$ is $$\frac{2}{\pi}\int_0^\pi x(\pi-x)\sin(nx)\,dx$$

Title: Find Fourier half range sine Series for $f(x)=x(\pi-x)$ on $(0,\pi)$.

Body:
I tried finding the Fourier coefficient  $ b_n$  but it came zero.

My attempt:

$ b_n=\frac{4}{\pi}\int_0^\pi x(\pi-x)\sin(2nx)dx\\
=\frac{4}{\pi}\left\{\pi\int_0^\pi x\sin(2nx)dx-\int_0^\pi x^2\sin(2nx)dx\right\}\\
=0$

Correct answer is:

$\frac{8}{\pi}\left(\frac{\sin x}{1^3}+\frac{\sin3x}{3^3}+\frac{\sin5x}{5^3}+...\right)$


答:
$\displaystyle \iint_D \sqrt{\frac{x^2}{y}+\frac{y^2}{x}} \ dx \ dy \ , \quad D=y<x^2<2y,\ x<y^2<2x$

I would recommend the change of variable as $u = \frac{x^2}{y}, v = \frac{y^2}{x}$.

So, $x = u^{2/3} \ v^{1/3}, \ y = u^{1/3} \ v^{2/3}$

Jacobian is simply $|J| = \frac{1}{3}$

So the integral becomes,

$\displaystyle \frac{1}{3} \int_1^2 \int_1^2 \sqrt{u+v} \, du \, dv$

Title: compute$\iint_D \sqrt{\frac{x^2}{y}+\frac{y^2}{x}} \ dxdy$

Body:
The question is:
$$\iint_D \sqrt{\frac{x^2}{y}+\frac{y^2}{x}} \ dxdy \quad D=y<x^2<2y,\ x<y^2<2x$$

The only thing that i can think of is that maybe change of variable would work but i'm not able to solve it i think that i'm on the wrong path;
I've tried this change of variable:
$$u=\frac{x}{\sqrt{y}}, \  v=\frac{y}{\sqrt{x}},\quad \iint(\sqrt{u^2+v^2})(\sqrt{xy}) \ dudv$$

Any suggestion would be great, thanks


问:
Show that F is continuous with the following property
Let $f:[0,1]$x$[0,1]$ $\rightarrow$ $\mathbb{R}$ be continuous and let F:[0,1] $\rightarrow$ $\mathbb{R}$ with $F(x)=\int_{0}^{1}f(x,y)dy$. Show that F ist continuous. I don't really have an idea, so help is appreciated.

答:
Let $\varepsilon >0$. You have that $f$ is uniformly continuous, and thus, there is $\delta >0$ s.t. for all $(x,y),(a,b)\in [0,1]\times [0,1]$, $\|(x,y)-(a,b)\|<\delta $ implies that $|f(x,y)-f(a,b)|<\varepsilon $. Let $x\in [0,1]$ and $|h|<\delta $. Then,

$$|F(x+h)-F(x)|\leq \int_0^1|f(x+h,y)-f(x,y)|\,\mathrm d y<\varepsilon .$$
Therefore, $F$ is continuous.


问:
Proving coefficient $a_n = O(1/n) $ and $b_n =O(1/n)$ in fourier series
This question was asked in my real analysis quiz and I was unable to solve it.

So, I am asking for help here.

Question : If $f(x) \sim a_0 /2 + \sum_{n=1}^{\infty} \left(a_n \cos(nx) + b_n \sin (nx)\right) $ and if f is of bounded variation on $[0,2\pi]$ show that $a_n =O(1/|n |) $  and $b_n =O(1/|n|)$.

Using the condition of bounded variation i wrote $f =g-h$ where $g$ and $h$ are increasing on $[0,2 \pi]$. But  I am not able to use any result of fourier series to proceed.

Do you mind giving me some hints on how to prove it?

Thanks for your time.

答:
Since $f$ is of bounded variation on a compact interval $[0,2\pi]$ it is differentiable almost everywhere and so there exists a function, denoted $f'\in L^1[0,2\pi]$ such that

$$f(x)=f(a)+\int_a^x f'(y) \lambda(d y),\quad x,a\in  [0,2\pi]$$

Let $\hat{f}_n$ denote the Fourier coefficient of a function $f$. Then using integration by parts
\begin{align}
    i n \hat{f}_n &= -  \int_0^{2\pi} f(x) (- in) e^{-i n x} \frac{1}{2\pi} d y \\
    &= - \frac{1}{2\pi} [f(2\pi) e^{-2\pi i n}-f(0)] + \int_0^{2\pi} f'(x) e^{-i n x} \frac{1}{2\pi} d y \\
    &=c+ \int_0^{2\pi} f'(x) e^{-i n x}\frac{1}{2\pi} d y \\
    &= c+ \hat{(f')}_n
\end{align}
divide both sides by $i n$ for $n\neq 0$ in $\mathbb Z$.
Ignoring the constant $c$ we have
\begin{align*}
   |\hat f_n| =\frac{1}{|in|} \left| \int_{[0,2\pi]} f'(x) e^{-i n x}  dx\right| \leq \frac{1}{|n|} \int_{[0,2\pi]} |f'| |e^{-i n x }| dx\leq \frac{1}{|n|} \int_{[0,2\pi]} |f'| dx =\frac{1}{|n|}\|f'\|_{L^1}
   \end{align*}
since $i$ and $e^{-i n x}$ have norm one and $f'$ is in $L^1$ by assumption we are done.


问:
How to prove that $S^{2n+1}/S^1$ is homeomorphic to $\mathbb CP^n$ under a given identification
We represent an element $(x_1,y_1,...,x_n,y_n,x_{n+1},y_{n+1})\in S^{2n+1}$ as an element $(z_1,...,z_{n+1})\in \mathbb C^{n+1}$ where $z_k = x_k+iy_k$. Now I'm considering the action of $S^1$ on $S^{2n+1}$ as $e^{iθ}·(z_1,...,z_{n+1}) = (e^{iθ}z_1,...,e^{iθ}z_{n+1})$. Under this action we identify any two elements as $(z_1,...,z_{n+1}) ∼ e^{iθ} ·(z_1,...,z_{n+1})\forall (z_1,...,z_{n+1})\in S^{2n+1}\subseteq \mathbb C^{n+1}$ and $e^{iθ}\in S^1$. For the resulting quotient space $S^{2n+1}/S^1$, I'm trying to prove that $S^{2n+1}/S^1$ is homeomorphic to $\mathbb CP^n$. How do I prove this? Any solution will be highly appreciated.</p>

答:
What is your definition of $\mathbb{C}P^n$? Is it

the quotient space of $\mathbb{C}^{n+1}− {0} $ under the equivalence
relation $ v \sim \lambda v$ for $ \lambda \neq 0 $

? I assume it is defined as above. I used $\pi$ to denote the quotient space. So  $\pi: \mathbb{R}^{n+1} \setminus \{0\} \to \mathbb{R}P^n$.

Here we identify $\mathbb{C}^{n+1} = \mathbb{R}^{2n+2}$

You essentially just need to apply the universal property of quotients.

Consider the inclusion map $S^{
2n+1} \hookrightarrow \mathbb{R}^{2n+2}$. This is of course continuous, and thus the composition
$$\begin{array}{ccccccccc} S^{2n+1} & \xrightarrow{i} & \mathbb{R}^{2n+2} \backslash \{0\} &\\
 & \searrow{f} & \downarrow{\pi} \\
 & & \mathbb{C}P^n \end{array}$$
is continuous. Note that $f$ is defined on $S^{(2n+1)}$,  $f(x)=f(\lambda x)$, where $|\lambda| = 1$. So $f$ makes the same identifications as your map: $e^{i\theta}$.

Thus, based on passing to the quotient, $f$ induces a map $F$ in the quotient by
$$\begin{array}{ccccccccc} S^{2n+1} & \xrightarrow{i} & \mathbb{R}^{2n+2}\backslash \{0\} &\\
\downarrow{e^{i\theta}} & \searrow{f} & \downarrow{\pi} \\
S^{2n+1}/\sim & \xrightarrow{F} & \mathbb{C}P^n .\end{array}$$
We have that $F$ is continuous. Note that it is also a bijection. Since $S^{2n+1}/\sim$ is compact (it is the image by $e^{i\theta}$ of $S^{2n+1}$, which is compact) and $\mathbb{C}P^n$ is Hausdorff, it is a homeomorphism.

Actually, I just copied the proof from this answer, making some modifications.


问:
Length of parametric curve in polar coordinates
I have to find the length of this parametric curve: $R(\theta)= \theta^4 $  with $0 < \theta < 1$

So, We have the formula: $ ds = \sqrt{r^2 + (\frac{dr}{d\theta})^2} d\theta $.

My question: As I have understood. I have to convert the $\theta$  into trig functions. How do I do that and proceed with my calculations?

答:
We have
\begin{align}
ds &= \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} d\theta \\
&= \sqrt{\theta^8 + 16\theta^6} \ d\theta \\
&= \theta^3 \sqrt{\theta^2 + 16} \ d\theta
\end{align}

Now, take $\theta = 4\tan\alpha$, and integrate.


问:
How to prove $\lim_{n\to \infty} \int_{0}^{+\infty} \frac{dx}{e^x+x^n}=1-\frac{1}{e}$?
How to prove $$\lim_{n\to \infty} \int_{0}^{+\infty} \frac{1}{e^x+x^n}\ \mathrm{d}x=1-\frac{1}{e}$$?

答: Hint: explain the integral-limit exchanges in$$\lim_{n\to\infty}\int_0^1\frac{dx}{e^x+x^n}+\lim_{n\to\infty}\int_1^\infty\frac{dx}{e^x+x^n}=\int_0^1e^{-x}dx+\int_1^\infty0dx.$$


问:
Law of constant multiple in case limit doesn't exist
I needed to show that $\lim_\limits{x\to 0} asin(1/x)$ doesn't exist when $a \neq 0$. This can be clearly done using
$\epsilon-\delta$ method.

But in one of the proposed solutions following argument was used.

$\lim_\limits{x\to 0} asin(1/x)=a\lim_\limits{x\to 0} sin(1/x) \cdots\cdot (1)$

And since $\lim_\limits{x\to 0} sin(1/x)$ doesn't exist hence above limit also doesn't exist.

I have some reservations in the above step (1) which seems very obvious.  Is it valid to separate constant as such when the limit doesn't exist?

Also, if it is valid then how to reason on above lines when $a=0$.

答:
I believe the Theorem which you must use is this (and note that it has hypotheses as well as conclusions):

Theorem If $\lim_{x\to x_0}f(x)$ exists, and $\lambda\in\mathbb{R}$, then $\lim_{x\to x_0}\lambda f(x)$ exists and equals $\lambda\lim_{x\to x_0} f(x)$.

Now argue as follows when $a\not=0$.

Suppose that $\lim_{x\to 0} a \sin\frac{1}{x}$ exists. Then by the Theorem (with $\lambda=\frac{1}{a}$, $f(x)=a \sin\frac{1}{x}$) we get that $\lim_{x\to 0}\sin\frac{1}{x}$ exists. But we already know that this limit does not exist, so neither does $\lim_{x\to 0} a \sin\frac{1}{x}$.

When $a=0$ then $a\sin\frac{1}{x}=0$, and clearly $\lim_{x\to 0}0=0$.


问:
Why do $\exists x\in S: p(x)$ and $\forall x\in S: p(x)$ have different logical structures?
Consider the following two statements:

Statement 1.  There exists a $x\in S$ such that $p(x)$ holds, which can be written by notation as $\exists x\in S: p(x).$

Statement 2.  For any $x\in S$, $p(x)$ holds, which can be written by notation as $\forall x\in S: p(x).$

But Zorich , in his book, interpretes the first one as
$$\exists x\in S: p(x)=\exists x( x\in S \color{red}{\wedge} p(x)),$$ and the second one as $$\forall x\in S: p(x)=\forall x( x\in S \color{red}{\to} p(x)).$$

Why one is "AND" and another is "IMPLY"?

The cited webpage has not answer my question，because it just tell that all things are arranged like this， but do not give the reason why to make such arrangment.

答:
I think it is better to view the question in the universe of discourse, so in the first statement $\exists x\in U:p(x)$, in the case of this element it has two properties, one it belongs to the universe of discourse AND it satisfies the conditions of $p(x)$.

For an analogy, say the universe of discourse would be planets in the Milky Way and $p(x)$ is that the planet has radius bigger than (insert some number). Here x just stands for a planet, but x could be any planet but putting it a slightly modified syntax of your first statement.

$$\exists x\in U:p(x) = \exists(x\in U \land p(x))$$

Now we see clearer that we need the two properties to hold. For the case of the latter statement, if we stick with the analogy of the former example, $$\forall x\in U: p(x) = \forall x(x\in U \rightarrow p(x))$$
This may make it clearer that the right hand of the equal sign start by saying out of ALL planets that would be all planets in the entirety of space, if that planet belongs to the Milky Way then it also has a radius bigger than (insert some number)

So to summarise, in the confusing can stem when you don't think in terms of your universe of discourse. Since all planets of the Milky Way have a radius bigger than (insert some number) then simply if a planet resides there it is satisfied but not all planets with a radius bigger than (insert some number) reside in the milky way.

I have recently self learned some logic from Daniel J. Velleman's book "How to prove it", I hope this crude analogy helps your understanding as I am not capable of giving you a rigorous answer.


问:
Showing that the Morley rank is the same.
Let $A$ be an L-structure with some formula $\phi(x)$ and some $t(x)$ as an L-term, where $x=x_1,...,x_m$. I wanna show that the Morley rank of $\phi(x) \wedge y$ = $t(x)$ and $\phi(x)$ are the same. I'm wondering what this property in general would be called?(some kind of Morley rank-invariance). And, how would you go about the proof?

答:
The more general property is that Morley rank is preserved under definable bijections. That is, if we have definable sets $X$ and $Y$ and a definable bijection $f: X \to Y$ (i.e. the graph of $f$ is a definable set) then the Morley rank of $X$ is the same as the Morley rank of $Y$. Definable here means definable with parameters.

In your case the following defines a bijection between $\phi(x)$ and $\phi(x') \wedge y = t(x')$, where $x'$ is a copy of the variables in $x$:
$$
\phi(x) \wedge x' = x \wedge y = t(x).
$$


问:
Why is $2\pi i \neq 0?$
We know that $e^{\pi i} = -1$ because of de Moivre's formula. ($e^{\pi i} = \cos \pi + i\sin \pi = -1).$

Suppose we square both sides and get $e^{2\pi i} = 1$(which you also get from de Moivre's formula), then shouldn't $2\pi i=0$? What am I missing here?

答:
Actually, the following is true:

$ e^{2n\pi i} = 1  \forall  n \in \mathbb{Z} $

As a special case of $n = 0$ this gives what you know: $e^0 = 1$, but as mentioned, this is only a special case of the general formula given above.

You could as well assume that $2\pi = 0$ because of

$\sin 2\pi = \sin 0 \land \cos 2\pi = \cos 0$.

As others already mentioned, this stuff just isn't injective, so conclusions relying on the injectivity can be wrong.

Title: Why is $2\pi i \neq 0?$


问:
True or false: There is only one $n \times n$ matrix such that $(I_n-A)^{-1}=I_n-\frac{1}{n-1}A$
There is only one $n \times n$ matrix that suits:

$(I_n-A)^{-1}=I_n-\frac{1}{n-1}A$

We know that:$(I_n-A)^{-1}\neq I_n-A^{-1} $ but I dont know what to do next:

Thanks for any tip.

答: There are many matrices with that property. For example, $$A=\begin{pmatrix}nI_k&O\\O&O\end{pmatrix}$$ Then $$I-A=\begin{pmatrix}-(n-1)I_k&O\\O&I\end{pmatrix}$$ $$(I-A)^{-1}=\begin{pmatrix}-\frac{1}{n-1}I_k&O\\O&I\end{pmatrix}=I-\frac{1}{n-1}\begin{pmatrix}nI_k&O\\O&O\end{pmatrix}$$ This is just one example. The size of $k$ can be varied, the diagonal terms can be permuted, and in general one can take any matrix of the form $P^{-1}AP$.


问:
Show that the support of the derivative is contained in the support of the function
Preliminary remark: If you don't know an answer in this general setting, I'm also interested in an answer for the case $E_1=\mathbb R^d$, $d\in\mathbb N$, and $E_2=\mathbb R$.

If $E_1$ is a topological space and $E_2$ is a $\mathbb R$-vector space, remember that $$\operatorname{supp}f:=\overline{\left\{f\ne0\right\}}\;\;\;\text{for }f:E_1\to E_2.$$

Assume $E_1$ is a $\mathbb R$-Banach space, $\Omega\subseteq E_1$ is open and $E_2$ is a normed $\mathbb R$-vector space. Please consider the following trivial result:

Lemma 1: If $f:\Omega\to E_2$ is differentiable at $x\in\Omega$ in direction $h\in E_1$ and $\{f=0\}$ is a neighborhood of $x$, then $${\rm D}_hf(x)=\lim_{t\to0}\frac{f(x+th)-f(x)}t=0\tag1.$$

Question

Are we able to conclude the following corollary?

Corollary 2: If $f:\Omega\to E_2$ is differentiable in direction $h\in E_2$, then $$\operatorname{supp}{\rm D}_hf\subseteq\operatorname{supp}f\tag2.$$

From Lemma 1, we immediately obtain $$\{f=0\}^\circ\subseteq\{{\rm D}_hf=0\}\tag3.$$ It's tempting to conclude $$\{{\rm D}_hf\ne0\}\subseteq\Omega\setminus\{f=0\}^\circ=\overline{\{f\ne0\}}=\operatorname{supp}f\tag4$$ and argue that the closure of the leftmost side must be contained in the rightmost side as well, since the latter is closed.

However, as thinking about this I realized that I'm not sure whether I'm considering the correct definition of the support. To be precise, in the definition of the support of $f$, do we need to take the closure with respect to the topology on $E_1$ or with respect to the subspace topology on $\Omega$?

Note that this question is crucial, since these closures do not coincide.

Regarding to my definition of the support, we would need to take the closure with respect to the subspace topology on $\Omega$ (and every closure which is taken in this question has to be understood with respect to this topology), but I wonder whether this is what's usually done.

Remark: Feel free to add any kind of continuity assumption, if necessary to obtain the desired result.

答:
Yes, that's the definition of support. You have to consider a topogical space, a function defined in there with values in a vector space, and you define the support of the function as the complement of the nullity set of the function. And the nullity set is the biggest open set where the function is zero, that is, the interior of the set where the function is zero.

But anyhow, to talk about nullity and support, you have to fix a function, and this function has a domain. This domain has a topology. Depending on the topology you put on that domain you have different answers. But the correct definition is the one that first consider the nullity set and then defines the support as the complement of it.


问:
Trouble evaluating $\lim_{x \to 0} \frac{1-\cos^4(x)}{x^2}$ without Hôpital's rule
I'm having trouble evaluating this limit as $x$ approaches $0$ without using Hôpital's rule.

$$\lim_{x \to 0} \frac{1-\cos^4(x)}{x^2}$$

I'd appreciate any hints on how to proceed.

EDIT:
I totally forgot that

$\lim_{x \to 0} \frac{\sin(x)}{x} = 1$

I shall write my solution as punishment:

$$\lim_{x \to 0} \frac{1-\cos^4(x)}{x^2}=\lim_{x \to 0} \frac{(1-\cos^2(x))\cdot(1+\cos^2(x))}{x^2}=$$
$$=\lim_{x \to 0} \frac{1-\cos^2(x)}{x^2} \cdot \lim_{x \to 0} 1+\cos^2(x)=$$
$$=\lim_{x \to 0}\Bigl(\frac{\sin(x)}{x}\Bigr)^2\cdot2=1\cdot2=$$
$$=2$$

答:
Hint: We have

$$1-\cos^{4}x = \left(1+\cos^2 x\right)\left(1-\cos^{2}x\right) .$$


问:
Kernel and Range of a Linear transformation Inclusions
For a linear algebra problem, I want to proof three things. The first one is that, for a linear transformation $f: \mathbb{R}^n \to \mathbb{R}^n$, $Im(f^2) \subset Im(f)$. The second one is that, for that same linear transformation $f$, $Ker(f) \subset Ker(f^2)$. The last one is a bit more tricky: $Ker(f) = Ker(f^2) \iff \mathbb{R}^n = Im(f) \oplus Ker(f)$, for $f: \mathbb{R}^n \to \mathbb{R}^n$.

For the first one, I am stuck. I think it might go like this: let $v\in Im(f^2)$, then we define $w = f(v)$ and therefore $f(f(v)) = f(w)$, thus $v \in Im(f)$. I don't know if it is correct or eve "legal". Besides, if it is, I don't see WHY this should be true.

For the second one, here is what I did: let $v \in Ker(f)$, then $f(v) = 0$ and hence $f(f(v)) = f(0) = 0$, since $f$ is linear. Therefore $v \in Ker(f^2)$. Again, I think this is correct but I don't see WHY it should be like that.

For the third one I have no clue. Can somebody explain all of this? Why and how to prove it? And by why I mean some intuition, I think I lack that.

答:
Your idea is good. Just need to write in a proper way.
Hints

Let $u\in Im(f)$, then $f(w)=u$ for some $w\in \Bbb{R}^n$.if $v\in Im(f^2)$, then $v=f^2(w)$ for some $w \in \Bbb{R}^n$, $v=f(f(w))$, $v=f(u)$, thus $v\in Im(f)$. Thus $Im(f^2)\subseteq Im(f)$.
Similarly, $Ker(f)\subseteq Ker(f^2)$.

Your second proof is absolutely correct.

If $x\in Ker (f)$, then $f(x)=0\Rightarrow f(f(x))=f(0)=0$. Therefore, $x\in Ker(f^2).$ Thus $Ker(f)\subseteq Ker(f^2)$

Go through this proposition", You can prove the third.

Proposition: Let $f:\Bbb{R}^n\to \Bbb{R}^n$ satisfy $fof=f$. Show that $\Bbb{R}^n= Ker(f)\oplus Im(f)$.

Write $v\in\Bbb{R}^n$, as $v=fv+(v-fv)$ and $f(v)\in Im(f)$, $v-f(v)\in Ker(f)$.
Now proceed...

Note Here, $f^2=fof$.


问:
Probability Game Optimal Strategy
You're playing a game where you have a .45 probability of winning and .55 probability of losing. You start out with 2000 chips and the winning condition is that if you bet a cumulative total of 10000 chips, then you win. Otherwise if you run out of chips, you lose.

For example, if you bet 20 chips, no matter the outcome, that will contribute 20 to the total. And you'll be left with either 1980 or 2020 current chips.

The min bet size is 1$ and max is as much as you currently have. What is the best strategy to maximize your probability of winning?

What I said was, you'd want to use a strategy akin to betting as much as possible at each round. Since otherwise, from LLN, you are losing in general and just prolonging with small bets will make you lose more in the long run. I said you'd be able to maybe calculate the exact strategy using dynamic programming, with f(s,t) representing what you should bet at each combination of (current money, total cumulative so far). However I do not know if this is correct nor did I have time to fully solve it.

Thank you!

答:
You're quite correct that the optimal strategy can be calculated by using dynamic programming.  As in your question, let $\ s\ $ be your current bank, $\ t\ $ the total amount you've so far bet, and $\ f(s,t)\ $ the optimal bet in this situation.  Let $\ p^*(s,t)\ $ be the probability that you will eventually succeed if you play optimally from then on. Let $\ S_0\ $ be your initial bank $(\ S_0=\$2,000\ $ in the question as posed$)$, and $\ T\ $ be the target value for $\ t\ $ ($ T=\$10,000\ $ in the question as posed).

First observe that if $\ s+t\ge T\ $ (and $\ t<T\ $) then you can win immediately on the next round by betting $\ T-t\ $, so $\ p^*(s,t)=1\ $ for all such values of $\ s\ $ and $\ t\ $.  Also $\ p^*(0,t)=0\ $ for all $\ t<T\ $.

Next, if $\ s+t<T\ $ and you bet $\ b\ $ then if you win, the new value of $\ s\ $ will be $\ s+b\ $ and the new value of $\ t\ $ will be $\ t+b\ $ and since these sum to $\ s+t+2b\ $  you will be able to win with certainty on your next go if $\ s+t+2b\ge T\ $.  It is therefore never advantageous to bet more than  $\ \frac{T+1-s-t}{2}\ $, because this is sufficient to guarantee a win on the next round if you win this one, and if you bet any more, you will have fewer funds available on the next round if you happen to lose this one.

If you bet $\ b\ $ and lose, then the new values of $\ s\ $ and $\ t\ $ will be $\ s-b\ $ and $\ t+b\ $ respectively.  Thus, if $\ p\ $ is the probability of winning a single round ($ p=0.45\ $ in the question as posed), and $\ q=1-p\ $ the probability of losing it, then
\begin{align}
p^*(s,t) &= \max_{1\le b\le \min\left(s,\frac{T+1-s-t}{2}\right)}\big(p\,p^*(s+b,t+b)+q\,p^*(s-b,t+b)\big)\\
f(s,t) &= \underset{1\le b\le \min\left(s,\frac{T+1-s-t}{2}\right)}{\arg\max}\big(p\,p^*(s+b,t+b)+q\,p^*(s-b,t+b)\big)\ .
\end{align}
By working backwards from $\ t=T-2\ $ you can use the above equations to succesively calculate $\ p^*(s,t)\ $ and $\ f(s,t)\ $ for $\ s\in \big\{1, 2,\dots,$$ T-t-1\big\}\ $ and $\ t=T-2,T-3,\dots,1,0\ $.

I've written the following Magma script to carry out these calculations for $\ T=1000\ $.

```<code>SetDefaultRealField(RealField(48));
R:=RealField(48);
p:=0.45;
q:=1-p;
T:= 1000;
optstrat:=ZeroMatrix(Integers(),T+2,T+2);
bestprob:=ZeroMatrix(R,T+1,T+2);
for s in [1..T-1] do
  bestprob[s+1,T-s+1] := 1.0;
  bestprob[s+1,T-s+2] := 1.0;
end for;;
for r in [2..T] do
  t:=T-r;
  for s in [1..T-t-1] do
    bp:= 0.0;
    os:=0;
    for b in [1..Minimum(s,(T+1-s-t) div 2)] do
       cp:=p*bestprob[s+b+1,t+b+1]+q*bestprob[s-b+1,t+b+1];
      if cp gt bp then
        bp:=cp;
        os:=b;
      end if;
    end for;
    bestprob[s+1,t+1]:= bp;
    optstrat[s+1,t+1]:= os;
  end for;
end for;
for s in [1..200] do
  print s, bestprob[s+1,1], optstrat[s+1,1];
end for;
</code>```

The script can be run by copying and pasting it into the online Magma calculator here.  Unfortunately, for $\ T=10000\ $, the script will fail by exceeding the time and memory limits of the online Magma calculator.  For $\ T=1000\ $, however, the output tells us that for a starting bank of $\$64$ or less, the optimal first bet is the whole of one's bank, and for starting banks of between $\$117$ and $\$200$ inclusive, the optimal first bet is $\$1$.  With optimal play, the chance of ultimate success is better than $\ 99\%\ $  for any starting bank of $\$175$ or more.  With optimal play, the probability of success for $\ S_0=200\ $turns out to be
$$
0.99913348949325492439801621030571175189427531\dots\ .
$$
The results of these calculations suggest that bets of $\$1$ are optimal unless one's bank is fairly small relative to the current target.

The above script can easily be modified to calculate the probability of success for the strategy B described in user2661923's answer (as amended per the correction suggested in my third comment to the answer).  If $\ p_1(s,t)\ $ is the probability of success playing that strategy from a position where the  current bank is $\ s\ $, and the total so far bet is $\ t\ $, then $\ p_1\ $ satisfies the recursion
$$
p_1(s,t) =p\,p_1(s+1,t+1)+q\,p_1(s-1,t+1)\ .
$$
The above script can be modified to tabulate the values of $\ p_1\ $ for $\ 1\le s\le 1000\ $ and $\ 0\le t\le 1000\ $ by replacing the code

```<code>     bp:= 0.0;
     os:=0;
     for b in [1..Minimum(s,(T+1-s-t) div 2)] do
       cp:=p*bestprob[s+b+1,t+b+1]+q*bestprob[s-b+1,t+b+1];
       if cp gt bp then
         bp:=cp;
         os:=b;
       end if;
     end for;
     bestprob[s+1,t+1]:= bp;
     optstrat[s+1,t+1]:= os;
</code>```

with the single instruction

```<code>bestprob[s+1,t+1]:= p*bestprob[s+2,t+2]+q*bestprob[s,t+2];
</code>```

For $\ T=1000\ $ and $\ S_0=200\ $, the modified script gives a probability of
$$
0.999051356533203111486952131756939063716486183\dots
$$
of ultimate success when  using this strategy.  While this probability is  smaller than that guaranteed by the optimal strategy, thus showing that this strategy is not quite optimal, the difference is so small ($\ \approx0.000082\ $) as to be effectively negligible.

For smaller values of $\ S_0\ $, the strategy of using $\$1$ bets can be very far from optimal.  For $\ S_0=50\ $ and $\ T=1000\ $, for instance, the optimal strategy gives a probability of
$$
0.30041494580652103310049100479730984926032991\dots
$$
of ultimate success, whereas the strategy of betting just $\$1$ all the time has a probability of success of only
$$
0.03353384615182354449189130202943683024299148\dots\ .
$$

The following script calculates the probability of ultimate success with a starting bank of $\ S_0=\$2000\ $, and target $\ T=\$10000\ $ when you  always bet just $\$1$ unless either $\ T\ $ rounds have been played, or $\ s=T\ $, in which case you bet $\ T-t\ $.  If you survive for $\ T\ $ rounds, even if your bank drops to zero on the last round, you will win, because the total amount you have then bet will be exactly $\ T\ $. On the other hand, if $\ s=T\ $ before you've played $\ T\ $ rounds, then the total amount $\ t\ $ that you've bet so far is just the number of rounds you've played, so $ 0\le T-t\le s\ $ and you can therefore win on the very next round by betting $\ T-t\ $.

This scenario can be modelled as a time-homogeneous Markov chain with absorbing states at $\ s=0\ $ and $\ s=T\ $, and initial state $\ s=S_0\ $ with probability $1$.  If $\ p_a(s,t)\ $ is the probability that the chain is in state $\ s\ $ at time $\ t\ $, $\ w(t)=p_a(T,t)\ $, and $\ l(t) = p_a(0,t)\ $, then $\ p_a, w\ $ and $\ l\ $ satisfy the following equations:

\begin{align}
p_a(s,0)&=\cases{1&if $\ s=S_0\ $, or\\
0&otherwise}\\
l(t+1)&=l(t)+ q\,p_a(1,t)\\
w(t+1)&=w(t)+p\,p_a(T-1,t)\\
p_a(1,t+1)&=q\,p_a(2,t)\\
p_a(T-1,t+1)&=p\ p_a(T-2,t)\\
p_a(s,t+1)&=q\,p_a(s+1,t)+p\,p_a(s-1,t)\ \ \text{ for }\ 2\le s\le T-2\ .
\end{align}
The probability of ultimate success is then
$$
\sum_{s=0}^Tp_a(s,T) - p_a(0,T-1)=1-l(T-1)\ ,
$$
which the following script calculates for $\ T=10000\ $ and $\ S_0=2000\ $.

```<code>SetDefaultRealField(RealField(48));
R:=RealField(48);
pa:=[* Zero(R) *];
qa:=[* Zero(R) *];
T:=10000;
S0:=2000;
p:=0.45;
q:=1-p;
for s in [1..T-1] do
  pa:=pa cat [* Zero(R) *];
  qa:=qa cat [* Zero(R) *];
end for;
w:=Zero(R);
l:=Zero(R);
pa[S0]:=One(R);
for t in [1..T-1] do
  w:=w+p*pa[T-1];
  lp:=l;
  l:=l+q*pa[1];
  w0:=q*pa[1];
  qa[T-1]:=p*pa[T-2];
  qa[1]:=q*pa[2];
  for s in [Maximum(2,S0-t)..Minimum(T-2,S0+t)] do
    qa[s]:=q*pa[s+1]+p*pa[s-1];
  end for;
  pa:=qa;
end for;
print 1-lp;
</code>```

For $\ T=10000\ $ and $\ S_0=2000\ $ and the strategy of repeatedly betting $\$1$ the above script gives
$$
0.9999999999999999999999960876990789091728946\dots
$$
as the probability of ultimate success.   While this strategy is unlikely to be "optimal" in the strictest sense of the word, its probability of success is so close to $1$ that the discrepancy is completely irrelevant.

While this strategy is slightly different from the strategies A and B described in user2661923's answer, all three answers are completely equivalent, and have exactly the same probability of success.  This is because $\ s+t\ $ never decreases, so once $\ s+t\ge T\ $ (thus allowing strategy B to terminate earlier than the other two would), even if you continue betting $\$1$, your bank $\ s\ $ cannot decrease to zero before $\ t\ $ reaches $\ T\ $ , so you are still certain to succeed.

For $\ T=1000\ $ and $\ S_0=200\ $ this script gives a probability of ultimate success of
$$
0.999051356533203111486952131756939063716486183\dots,
$$
identical to the answer which the preceding modified script gives for strategy B, thus providing some confirmation that the algorithms have been correctly implemented.

Title: Probability Game Optimal Strategy

Body:
You're quite correct that the optimal strategy can be calculated by using dynamic programming.  As in your question, let $\ s\ $ be your current bank, $\ t\ $ the total amount you've so far bet, and $\ f(s,t)\ $ the optimal bet in this situation.  Let $\ p^*(s,t)\ $ be the probability that you will eventually succeed if you play optimally from then on. Let $\ S_0\ $ be your initial bank $(\ S_0=\$2,000\ $ in the question as posed$)$, and $\ T\ $ be the target value for $\ t\ $ ($ T=\$10,000\ $ in the question as posed).

First observe that if $\ s+t\ge T\ $ (and $\ t<T\ $) then you can win immediately on the next round by betting $\ T-t\ $, so $\ p^*(s,t)=1\ $ for all such values of $\ s\ $ and $\ t\ $.  Also $\ p^*(0,t)=0\ $ for all $\ t<T\ $.

Next, if $\ s+t<T\ $ and you bet $\ b\ $ then if you win, the new value of $\ s\ $ will be $\ s+b\ $ and the new value of $\ t\ $ will be $\ t+b\ $ and since these sum to $\ s+t+2b\ $  you will be able to win with certainty on your next go if $\ s+t+2b\ge T\ $.  It is therefore never advantageous to bet more than  $\ \frac{T+1-s-t}{2}\ $, because this is sufficient to guarantee a win on the next round if you win this one, and if you bet any more, you will have fewer funds available on the next round if you happen to lose this one.

If you bet $\ b\ $ and lose, then the new values of $\ s\ $ and $\ t\ $ will be $\ s-b\ $ and $\ t+b\ $ respectively.  Thus, if $\ p\ $ is the probability of winning a single round ($ p=0.45\ $ in the question as posed), and $\ q=1-p\ $ the probability of losing it, then
\begin{align}
p^*(s,t) &= \max_{1\le b\le \min\left(s,\frac{T+1-s-t}{2}\right)}\big(p\,p^*(s+b,t+b)+q\,p^*(s-b,t+b)\big)\\
f(s,t) &= \underset{1\le b\le \min\left(s,\frac{T+1-s-t}{2}\right)}{\arg\max}\big(p\,p^*(s+b,t+b)+q\,p^*(s-b,t+b)\big)\ .
\end{align}
By working backwards from $\ t=T-2\ $ you can use the above equations to succesively calculate $\ p^*(s,t)\ $ and $\ f(s,t)\ $ for $\ s\in \big\{1, 2,\dots,$$ T-t-1\big\}\ $ and $\ t=T-2,T-3,\dots,1,0\ $.

I've written the following Magma script to carry out these calculations for $\ T=1000\ $.

SetDefaultRealField(RealField(48));
R:=RealField(48);
p:=0.45;
q:=1-p;
T:= 1000;
optstrat:=ZeroMatrix(Integers(),T+2,T+2);
bestprob:=ZeroMatrix(R,T+1,T+2);
for s in [1..T-1] do
  bestprob[s+1,T-s+1] := 1.0;
  bestprob[s+1,T-s+2] := 1.0;
end for;;
for r in [2..T] do
  t:=T-r;
  for s in [1..T-t-1] do
    bp:= 0.0;
    os:=0;
    for b in [1..Minimum(s,(T+1-s-t) div 2)] do
       cp:=p*bestprob[s+b+1,t+b+1]+q*bestprob[s-b+1,t+b+1];
      if cp gt bp then
        bp:=cp;
        os:=b;
      end if;
    end for;
    bestprob[s+1,t+1]:= bp;
    optstrat[s+1,t+1]:= os;
  end for;
end for;
for s in [1..200] do
  print s, bestprob[s+1,1], optstrat[s+1,1];
end for;

The script can be run by copying and pasting it into the online Magma calculator here.  Unfortunately, for $\ T=10000\ $, the script will fail by exceeding the time and memory limits of the online Magma calculator.  For $\ T=1000\ $, however, the output tells us that for a starting bank of $\$64$ or less, the optimal first bet is the whole of one's bank, and for starting banks of between $\$117$ and $\$200$ inclusive, the optimal first bet is $\$1$.  With optimal play, the chance of ultimate success is better than $\ 99\%\ $  for any starting bank of $\$175$ or more.  With optimal play, the probability of success for $\ S_0=200\ $turns out to be
$$
0.99913348949325492439801621030571175189427531\dots\ .
$$
The results of these calculations suggest that bets of $\$1$ are optimal unless one's bank is fairly small relative to the current target.

The above script can easily be modified to calculate the probability of success for the strategy B described in user2661923's answer (as amended per the correction suggested in my third comment to the answer).  If $\ p_1(s,t)\ $ is the probability of success playing that strategy from a position where the  current bank is $\ s\ $, and the total so far bet is $\ t\ $, then $\ p_1\ $ satisfies the recursion
$$
p_1(s,t) =p\,p_1(s+1,t+1)+q\,p_1(s-1,t+1)\ .
$$
The above script can be modified to tabulate the values of $\ p_1\ $ for $\ 1\le s\le 1000\ $ and $\ 0\le t\le 1000\ $ by replacing the code

     bp:= 0.0;
     os:=0;
     for b in [1..Minimum(s,(T+1-s-t) div 2)] do
       cp:=p*bestprob[s+b+1,t+b+1]+q*bestprob[s-b+1,t+b+1];
       if cp gt bp then
         bp:=cp;
         os:=b;
       end if;
     end for;
     bestprob[s+1,t+1]:= bp;
     optstrat[s+1,t+1]:= os;

with the single instruction

bestprob[s+1,t+1]:= p*bestprob[s+2,t+2]+q*bestprob[s,t+2];

For $\ T=1000\ $ and $\ S_0=200\ $, the modified script gives a probability of
$$
0.999051356533203111486952131756939063716486183\dots
$$
of ultimate success when  using this strategy.  While this probability is  smaller than that guaranteed by the optimal strategy, thus showing that this strategy is not quite optimal, the difference is so small ($\ \approx0.000082\ $) as to be effectively negligible.

For smaller values of $\ S_0\ $, the strategy of using $\$1$ bets can be very far from optimal.  For $\ S_0=50\ $ and $\ T=1000\ $, for instance, the optimal strategy gives a probability of
$$
0.30041494580652103310049100479730984926032991\dots
$$
of ultimate success, whereas the strategy of betting just $\$1$ all the time has a probability of success of only
$$
0.03353384615182354449189130202943683024299148\dots\ .
$$

The following script calculates the probability of ultimate success with a starting bank of $\ S_0=\$2000\ $, and target $\ T=\$10000\ $ when you  always bet just $\$1$ unless either $\ T\ $ rounds have been played, or $\ s=T\ $, in which case you bet $\ T-t\ $.  If you survive for $\ T\ $ rounds, even if your bank drops to zero on the last round, you will win, because the total amount you have then bet will be exactly $\ T\ $. On the other hand, if $\ s=T\ $ before you've played $\ T\ $ rounds, then the total amount $\ t\ $ that you've bet so far is just the number of rounds you've played, so $ 0\le T-t\le s\ $ and you can therefore win on the very next round by betting $\ T-t\ $.

This scenario can be modelled as a time-homogeneous Markov chain with absorbing states at $\ s=0\ $ and $\ s=T\ $, and initial state $\ s=S_0\ $ with probability $1$.  If $\ p_a(s,t)\ $ is the probability that the chain is in state $\ s\ $ at time $\ t\ $, $\ w(t)=p_a(T,t)\ $, and $\ l(t) = p_a(0,t)\ $, then $\ p_a, w\ $ and $\ l\ $ satisfy the following equations:

\begin{align}
p_a(s,0)&=\cases{1&if $\ s=S_0\ $, or\\
0&otherwise}\\
l(t+1)&=l(t)+ q\,p_a(1,t)\\
w(t+1)&=w(t)+p\,p_a(T-1,t)\\
p_a(1,t+1)&=q\,p_a(2,t)\\
p_a(T-1,t+1)&=p\ p_a(T-2,t)\\
p_a(s,t+1)&=q\,p_a(s+1,t)+p\,p_a(s-1,t)\ \ \text{ for }\ 2\le s\le T-2\ .
\end{align}
The probability of ultimate success is then
$$
\sum_{s=0}^Tp_a(s,T) - p_a(0,T-1)=1-l(T-1)\ ,
$$
which the following script calculates for $\ T=10000\ $ and $\ S_0=2000\ $.

SetDefaultRealField(RealField(48));
R:=RealField(48);
pa:=[* Zero(R) *];
qa:=[* Zero(R) *];
T:=10000;
S0:=2000;
p:=0.45;
q:=1-p;
for s in [1..T-1] do
  pa:=pa cat [* Zero(R) *];
  qa:=qa cat [* Zero(R) *];
end for;
w:=Zero(R);
l:=Zero(R);
pa[S0]:=One(R);
for t in [1..T-1] do
  w:=w+p*pa[T-1];
  lp:=l;
  l:=l+q*pa[1];
  w0:=q*pa[1];
  qa[T-1]:=p*pa[T-2];
  qa[1]:=q*pa[2];
  for s in [Maximum(2,S0-t)..Minimum(T-2,S0+t)] do
    qa[s]:=q*pa[s+1]+p*pa[s-1];
  end for;
  pa:=qa;
end for;
print 1-lp;

For $\ T=10000\ $ and $\ S_0=2000\ $ and the strategy of repeatedly betting $\$1$ the above script gives
$$
0.9999999999999999999999960876990789091728946\dots
$$
as the probability of ultimate success.   While this strategy is unlikely to be "optimal" in the strictest sense of the word, its probability of success is so close to $1$ that the discrepancy is completely irrelevant.

While this strategy is slightly different from the strategies A and B described in user2661923's answer, all three answers are completely equivalent, and have exactly the same probability of success.  This is because $\ s+t\ $ never decreases, so once $\ s+t\ge T\ $ (thus allowing strategy B to terminate earlier than the other two would), even if you continue betting $\$1$, your bank $\ s\ $ cannot decrease to zero before $\ t\ $ reaches $\ T\ $ , so you are still certain to succeed.

For $\ T=1000\ $ and $\ S_0=200\ $ this script gives a probability of ultimate success of
$$
0.999051356533203111486952131756939063716486183\dots,
$$
identical to the answer which the preceding modified script gives for strategy B, thus providing some confirmation that the algorithms have been correctly implemented.


问:
Every subsequence of $x_n$ has a further subsequence which converges to $x$. Then the sequence $x_n$ converges to $x$.
Is the following true?

Let $x_n$ be a sequence with the following property: Every subsequence of $x_n$ has a
further subsequence which converges to $x$. Then the sequence $x_n$ converges to $x$.

I guess that it is true but I am not sure how to prove this.

答: True. If not, there exists an $\epsilon > 0$, such that for all $k$, there exists an $n_k > k$ satisfying $|x_{n_k}−x| \ge \epsilon$ since if there is some $k$ which doesn't have such $n_k$, then we can take it as $N$, so $x_n$ converges to $x$. The subsequence $x_{n_{k}}$ does not have any subsequence converging to $x$.</p>