问:
Confusion on analytic continuation.
The definition of analytic continuation of holomorphic function is stated as follows:

Let $f_{1}$ and $f_{2}$ be two analytic functions on two domains (open and connected) $\Omega_{1}$ and $\Omega_{2}$ such that $\Omega_{1}\cap\Omega_{2}\neq\varnothing$. If $f_{1}$ and $f_{2}$ agree on $\Omega_{1}\cap \Omega_{2}$, we say $f_{2}$ is the analytic continuation of $f_{1}$ on $\Omega_{2}$, and vice versa.

A smaller version of it is that:

If $f$ is analytic on a domain $D\subset\mathbb{C}$ and $F$ is analytic on a bigger domain $E\subset\mathbb{C}$ such that $f=F$ on $D\subset E,$ then $F$ is the analytic continuation of $f$ on $E$.

From what I read, this kind of technique allows us to define a function in a smaller domain and extend it analytically to a larger domain. But I don't understand why this definition allows us to do so.

What confuses me is that the definition only guarantees $f=F$ on the intersection $\Omega_{1}\cap\Omega_{2}$, so perhaps $f\neq F$ on $\Omega_{2}$, then how do I know $f$ is analytic on $\Omega_{2}\setminus\Omega_{1}$?

I tried to use the identity theorem as follows:

Let $f$ and $g$ be two holomorphic functions on a domain $D$ such that $f=g$ on a subset $S\subset D$ that contains a limit point, then $f=g$ on the whole $D$.

But this seems backward. By the hypothesis of analytic continuation, we only have $f=g$ on $S$, and $g$ is analytic on $D$, we do not really know if $f$ is analytic on the whole $D$ (this is the purpose of analytic continuation, right? to extend $f$ analytically to the whole $D$.)

Am I overthinking this and confusing myself?? I guess we should have, say $f_{1}=f_{2}$ on the whole $\Omega_{1}\cup\Omega_{2}$, but I don't know how to prove it.

Edit 1: (Some Clarification, Possible Answer and Reference)

I am sorry if I am asking a confusion (bad) question. My confusion is that, even though the analytic continuation exists, I don't think that means anything helpful. It only gives us an analytic function $F$ on a bigger domain $\Omega_{2}$ such that $F|_{\Omega_{1}}=f$ for $\Omega_{1}\subset\Omega_{2}$. But it does not say anything about $f$, $f$ is still in $\Omega_{1}$. So I do not understand why analytic continuation can extend the domain on which $f$ is analytic.

The book "Complex Analysis and Applications" by Hemant Kumar Pathak, has a chapter about analytic continuation.

As Jose suggested, it does not make sense to say $f=F$ on $\Omega_{2}$, because $f$ is on $\Omega_{1}$.

The book explains that if we have an analytic continuation of $f_{1}$ from $\Omega_{1}$ into $\Omega_{2}$ via $\Omega_{1}\cap\Omega_{2}$, then the aggregate value of $f_{1}$ in $\Omega_{1}$ and $f_{2}$ in $\Omega_{2}$ can be regarded as a single function $f(z)$ analytic in $D_{1}\cup D_{2}$ such that $$f(z)=\left\{
\begin{array}{ll}
f_{1}(z), z\in D_{1}\\
f_{2}(z), z\in D_{2}
\end{array}
\right.$$

This actually clarifies things out. This is like what we did when we want to remove the singularity: if $f_{1}$ has a removable singularity at $z_{0}$, then we actually extend $f_{1}$ to $f$ by defining $$f(z)=f_{1}(z), z\neq z_{0}\ \ \text{and}\ \ f(z_{0})=\lim_{z\rightarrow z_{0}}f_{1}(z).$$

Thus, we are actually extending $f_{1}(z)$ to $f(z)$, not to $f_{2}(z)$. We sort of complete $f_{1}(z)$ into $\Omega_{2}$ by defining $f(z)$.

I hope my explanation can help other people who study complex analysis and find analytic continuation confusing.

Feel free to add anythings more!

答:
Those theorems are not about extending analytic functions, in the sense that they are not about the possibility of extending such a function. What they say is that you can extend an analytic function in, at most a single way. So, they are about uniqueness of extensions, not about their existence.

To be more precise, they say that if $\Omega_1$ and $\Omega_2$ are domains, with $\Omega_1\subset\Omega_2$, and if $f\colon\Omega_1\longrightarrow\Bbb C$ is an analytic function, then there is at most an analytic function $F\colon\Omega_2\longrightarrow\Bbb C$ whose restriction to $\Omega_1$ is $f$. But it is perfectly possible that there is none! That's the case if, for instance, $\Omega_1=D(0,1)$, $\Omega_2=\Bbb C$ and $f\colon\Omega_1\longrightarrow\Bbb C$ is defined by $f(z)=\frac1{z-2}$.

Title: Confusion on analytic continuation.


问:
Synthetic solution needed: What is $\angle MBC$?
Consider $\Delta ABC$ and a point $M$ in its interior so that $\angle MAB = 10^{\circ}, \angle MBA = 20^{\circ}, \angle MCA =30^{\circ}$ and $\angle MAC = 40^{\circ}$. What is $\angle MBC$?

I actually got a solution, with trig with answer $60$. However can someone give a pure synthetic solution?

答:
I hope this helps. I happen to write in my blog. Anyways!

Walkthrough: a. Take $D$ as a point on $CM$ such that $\angle DAC=30^{\circ}$, and define $BD\cap AC=E$. So $\Delta DAC$ is isosceles.

b. Show M is the incentre of $\Delta ABD$

c. Show $\angle EDC=60^{\circ}$.

d. Show $\Delta BAC$ is isosceles .

e. So $\boxed{\angle MBC=60^{\circ}}. $

ContentLicense: CC BY-SA 4.0


问:
Epimorphism from $3\mathbb{Z}$ to $U_8$
I'm trying to disprove/prove that an epimorphism exists from $3\mathbb{Z} $ to $U_8$

$U_8$ is acyclic, and is composed out of $\{1,3,5,7\}$. Therefore I can't map every $3k \in 3\mathbb{Z}$ to a cycle, so I'm thinking of creating a composition $f(x) = h(g(x))$ such that:
$$g(x) = x \mod 12 \\ h(0) = 1, \: h(3) = 3,\: h(6) = 5,\: h(9) = 7 $$

However I'm stuck proving this is even an homomorphism.

答:
Consider these:

$3 \mathbb Z$ is a cyclic additive group.

A homomorphic image of a cyclic group is cyclic.

$U_8 \cong C_2 \times C_2$ is not cyclic.

Title: Epimorphism from $3\mathbb{Z}$ to $U_8$


问:
Understanding eigenvectors and eigenvalues
I am trying to follow the discussion on eigenvectors and eigenvectors in the QM book by Cohen-Tannoudji, et. al.

If $\mathscr{A}$ is a $N \times N$ matrix with $A_{ij}$ as its entry at $i$th row and $j$th column, $c_j$ the $j$th component of the eigenvector corresponding to the eigenvalue $\lambda$, then $$\sum_j [A_{ij} - \lambda \delta_{ij}]c_j = 0\tag{1}\label{1}.$$ The above equation can be considered to be a system of $N$ linear and homonegeneous equations. I understand till here. What I don't understand is the following.

Consider $\lambda_0$ to be a simple eigenvalue of the characteristic equation $\operatorname{Det}[\mathscr A - \lambda I] = 0$. Now the authors say that \ref{1} is comprised of $N-1$ linear equations, the $N$th one following from the preceeding ones and hence redudant. How did we get $N-1$ linear equations?

答:
$\lambda_0$ is a simple eigenvalue, which means that its algebraic multiplicity is equal to its geometric multiplicity which is equal to one.

Let $\mathcal M$ be the representative matrix of the linear map $f$, with $\mathcal M \in \operatorname{M}(N\times N,\mathbb K)$.

You know that an eigenvector $\underline v$ associated to an eigenvalue $\bar \lambda$ belongs to $Ker(\mathcal f-\bar \lambda id)$, and the geometric multiplicity gives us also its dimension $\implies \underline v\in Sol\Big(\big(\mathcal M-\bar \lambda I_N)\underline x|$ $\underline 0\big)\Big)$.

So, in your case, $1=dim(Ker(f-\bar \lambda id))=N-\text rk(\mathcal M-\lambda_0 I_N)$ which means that $\text rk(\mathcal M-\lambda_0 I_N)$ is equal to $N-1$ that is also the number of linearly independent equations of the system $\big(\big(\mathcal M-\lambda_0I_N\big)\underline x|\underline 0\big)$.


问:
Computing the limit of an integral of a function series
I am trying to figure out how to compute the limit

$$\lim_{t \rightarrow \infty} \int_{-1}^1 \frac{\cos^2(t^3x^{10})}{tx^2 + 1} \, dx.$$

If I exchange the limit and integral (when is this allowed?) then I get the integral of 0.  This seems too easy.  What should I be looking for?

答:
Exchanging integral and limit is allowed here:
Your integrand is bounded by 1 so you can use dominated convergence theorem.</p>


问:
How to solve first order linear recurrence relation using characteristic equation method
Consider the recurrence relation
$$ a(n) = 3\cdot a(n-1) + 2 ,\quad a(0) = 1.$$

Using the method of iteration, I found the solution to be

$$a(n) = 2\cdot 3^n - 1.$$

I would like to know how to form the characteristic equation for the given recurrence relation and solve it using that.

答: The homogeneous part is $a_n-3a_{n-1}=0$, yielding characteristic polynomial $x-3$ with a single root $x=3$.  So the homogeneous solution is $a_n=A \cdot 3^n$, and the general solution is $a_n=A \cdot 3^n + B$.  Now $1=a_0=A+B$ and $5=a_1=3A+B$, so $A=2$ and $B=-1$.


问:
Puzzle: Maximum amount of rice
Farmer Felix has just had a bountiful harvest with 1 ton of rice, and now he wants to sell all of it. Before getting his rice to the market, his rice needs to go through inspection. Unfortunately, the inspector is a greedy tyrant, and he wants a part of Felix's harvest. The inspector's price is as follow:

The first inspection costs all the rice.

If Felix gives the inspector x ton of rice, then the next inspection costs x less.

For example, if the inspector is given 0.2, 0.3, 0.5 ton portions in that order, then finally the total amount of rice gets to the market is:
$$0.2\times0 + 0.3\times0.2 + 0.5\times(0.2 + 0.3) = 0.31$$
What is the maximum amount of rice Felix can get to the market (given that the inspector doesn't mind even when he's requested to inspect infinite times)?

答:
If Felix takes rice to the inspector $n$ times with $a_i$ tons of rice at $i$-th step, then the rice that goes to the market is given by the following formula:$$
x = \sum_{i=1}^n a_i\cdot \big(\sum_{j=1}^{i-1} a_j\big)=\sum_{i=1}^n\sum_{j=1}^{i-1}a_i\cdot a_j = \frac{(a_1+a_2+\dots+a_n)^2-a_1^2-a_2^2-\dots-a_n^2}{2}=\frac{1-\sum a_i^2}{2}.
$$ The last equality holds since $\sum a_i=1$. By picking $n$ to be arbitrarily large and $a_i$ to be arbitrarily small, you can see that the amount of rice taken to the market can be made arbitrarily close to $0.5$ tons. Also by the formula you can see that there is no way to take more than $0.5$ tons to the market.

In your example, you made a mistake since $0.2\times 0+0.3\times 0.2+0.5\times (0.2+0.3)=0.31$, not $0.51$.

Edit: The $0.5$ bound also works if you allow Felix to go to the inspector infinitely many times. Say $\sum_{i=1}^\infty a_i=1$ (with, of course, $a_i>0$). Then $\sum_{i=1}^\infty a_i^2$ is also convergent, thus the above formula still holds.

Title: Puzzle: Maximum amount of rice


问:
What is the derivative of a delayed function in discrete time?
How do you calculate such derivative?

$$\frac{d\ \ x_{1}\left(k+1\right)\ }{d\ \ x_{1}(k)}$$

Sorry for the probably silly question but I have been having some serious trouble with this

答:
I've never really seen this notation, but it probably refers to recurrences of the form $x (k+1) = g(x(k))$. In this framework, I'd say that
$$
\frac{d x(k+1)}{d x(k)} := g'(x(k)).
$$


问:
Proof of Schur's test via Young's inequality
I am able to prove the following generalization of Schur's test using the Riesz-Thorin interpolation theorem, however I have been stuck for days now trying to prove it using Young's inequality:

Let the integral operator $T$ from functions $f: X \to \Bbb C$ to functions $Tf: Y \to \Bbb C$ be defined via the kernel $K: X \times Y \to \Bbb C$ , which is some measurable function.  Moreover, let

$$ \| K(x,\cdot) \| _{L^{q_0}(Y) } \leq 1 $$

$$\| K(\cdot,y)\|_{L^{p_1'}(X) } \leq 1$$

for all $x\in X$ and all $y\in Y$.

Then for every $0<\theta<1$ and all $f\in L^{p_\theta}$

$$\| Tf \| _{L^{q_\theta}(Y) } \leq \| f\|_{L^{p_\theta}(X) } $$

where

$${1\over p_\theta} = {1-\theta\over p_0} + {\theta\over p_1}   $$

$${1\over q_\theta} = {1-\theta\over q_0} + {\theta\over q_1}   $$

$${1} = {1\over p_1} + {1\over p_1'}   $$

and $p_0=1$ and $q_1=+\infty$.

I am able to prove the special case $p_1'=q_0=1$  via Hölder's as well as Young inequalities.  However, I am making zero progress trying to prove the general case.
I am struggling with this now for almost a week and I would greatly appreciate help! From online sources I know that a proof based on Young's inequality is possible. Young's inequality for non-negative reals $x,y$ is
$$ xy \leq x^r/r + y^s/s$$ for dual exponents satisfying $1/r+1/s=1$ for $1<r<\infty$.  Thanks in advance.

答:
[EDIT: This answer, in fact, is wrong. See comments.]

I was finally able to use a three-way Young's inequality to prove the theorem. However, I was able to see a very slightly more direct proof using Hölder's inequality. Both theorems exploit convexity, whereas Riesz-Thorin relies on complex analyticity.

Here is the sketch of the proof using Hölder's inequality:

Below, $\mu$ and $\nu$ are the measures on $X$ and $Y$ respectively.

The idea is to prove the theorem for simple functions $f$ of finite measure support and then use monotone convergence.

$$
\|~T f~\|_{L^{q_\theta} } \leq \sup_{\|{h}\|_{L^{q_\theta'}(Y) }\leq 1}
| \int_Y \int_X |K(x,y)| |f(x)| ~d\mu(x) ~h(y) ~d\nu(y) ~|
$$

so that it suffices to show that

$$
| \int_Y \int_X |K(x,y)f(x)  h(y)| ~d\mu(X) ~d\nu(y) ~|
\leq \|{f}\|_{L^{p_\theta}(X) }    \|{h}\| _{L^{q_\theta'}(Y) }
~.
$$

We can exploit homogenization symmetry in the claim.
Let us normalize
$\|{f}\|_{L^{p_\theta}(X) }=\|{h}\| _{L^{q_\theta'}(Y) }=1$.

Denoting $L^p(X\times Y)$ as $L^p$ for convenience, we can use
Hölder's inequality for multiple exponents as follows:

\begin{align}
\|{ K f h }\|_{L^1}
&\leq
\|{ K^{p_1'/r_1}  f^{p_\theta/r_1} } \|_{L^{r_1}}
\|{ K^{q_0/r_2}  h^{q_\theta'/r_2} }\|_{L^{r_2}}
\|{ f^{p_\theta/r_3}  h^{q_\theta'/r_3} } \|_{L^{r_3}}
\\
1&=
{1\over r_1}
+{1\over r_2}
+{1\over r_3}
\\
1&=
{p_1'\over r_1} + {q_0\over r_2}\\
1&=
{p_\theta\over r_1} + {p_\theta\over r_3}
\end{align}

which has solution the solution

\begin{align}
{1\over r_1} &={1\over q_0}\\
{1\over r_2} &= {q_\theta-q_0 \over q_\theta p_1' }\\
{1\over r_3} &= {1\over q_0 } - {1\over p_\theta} ~.
\end{align}

Since $q_\theta >q_0$ exponents $r_1,~r_2$ are finite. If $1/r_3$ becomes zero then the three-way Hölder's inequality becomes the standard Hölder's inequality.

Each of the $L^r$ norms can be evaluated by the Fubini-Tonelli
theorem.  Since each of 3 the norms on the RHS equals 1,
we finally get, after putting everything together,
$$ \| Tf \|_{q_\theta}  \leq 1$$
and the claim follows.

ContentLicense: CC BY-SA 3.0


问:
Prove that there exists $A \in F$ (σ-algebra) such that $P(A) = a$
The question

Let $\left( \Omega,F,P \right)$ be a probability space.

A set $E\in F $ is called  K-set iff

$P(E)>0$ and $B \subseteq E$ with $B \in F$ $\Rightarrow P(B) = P(E)$ or $P(B) = 0$

Note: I don't know how the above term is translated in English. I looked it up on Google but found nothing so I will name it for the sake of this post as K-set. If anyone knows the name of this term please tell me.

Suppose that the probability measure $P$ has no $K$-sets. Prove that for every $a \in [0,1]$ there exists $A \in F$ such that $P(A) = a$

Attempt

So $P$ has not $K$-sets. From this I understand that

$\forall A \in F$ with $P(A)>0$ and $\forall B \subseteq A$ with $B \in F$ we have $P(B) \neq P(A)$ and $P(B)>0$

For $a=0$, $P(\emptyset ) = 0$ and $\emptyset \in F$ since $F$ is $\sigma$-algebra.

For $a=1$, $P(\Omega) = 1$ and $\Omega \in F$ since $F$ is $\sigma$-algebra.

For $a \in (0,1)$ I thought of the idea that we can find a sequence $\{B_{n} \}$ of sets in $A$ such that $B_{n+1} \subset B_{n} \forall n $

I also tried using the Borel-Cantelli Theorem and other Propositions involving $\lim \inf B_n $ and $\lim \sup B_n $ but couldn't get to the desired result. I know that $P(B_n)$ will get smaller and smaller as $n$ gets bigger. But how do I know that $\forall a \in (0,1)$ I can find such $B_n$ such that $P(B_n) =a$ ? If this is the approach anyway.

答:
For any $E\in F$ with $P(E)>0$ we can construct $E\supset E_1\supset E_2\supset\dots$ such that $P(E_n)\to 0$.

Taking their relative complements, we get $E'_n:=E\setminus E_n$ such that $P(E'_n)\to P(E)$.

Apply this construction first for $E=X$ to find $A_1:=E_n$ where $n$ is the smallest index with $P(E_n)\le a$.

Then apply it for $E=X\setminus A_1$ and define $B_1:=A_1\cup E'_m$ where $m$ is the smallest index with $P(E'_m)\ge a-P(A_1)$,

so that $A_1\subset B_1$ and $P(A_1)\le a\le P(B_1)$.

Repeat this procedure for $E=E'_m$ to construct the required sequence $A_i$.


问:
Prove $|(a+b) \log (a+b) - a \log a| \leq b \log (\frac{1}{b})$
Can you prove or point me where I can reference the following assertion?

$|(a+b) \log (a+b) - a \log a| \leq b \log (\frac{1}{b}); a \in [0,1), b \in (0,1], a+b \in [0,1]$

答:
I agree with Vlad in the conclusion that the inequality does not always hold true. Here I try to visualize the problem.

All the terms in the inequality are of the form $ x.ln x $ (even on the right side, $ b.ln(1/b) = - b.ln b $). Note that since $ a , b , a+b $ are less than or equal to 1, their logarithms are negative. Also note that logarithms are undefined at zero, so the domains of $ a $ and $ a+b $ should exclude zero. Let's take a look at the function $ f(x) = x.lnx $ in the interval $ (0,1] $ :

<img src="https://i.stack.imgur.com/yKd5n.png" alt="$ f(x) = x.lnx $" />

The function has a minimum somewhere in the interval. We can calculate where it is, and we will see that it helps us find a counterexample, a set of numbers for which the inequality does not hold. The minimum is where the first derivative is zero.
$$ f'(x) = lnx + 1 = 0 $$
This happens at $ x = 1/e $  (because $ ln(e^{-1})=-1 $). In the above graph I have marked this vertical line in red.

We know this point is a minimum because (in addition to the visualization) $ f''(x) = 1/x $ is positive in the interval $ (0,1] $ , which means that $ f'(x) $ is increasing all through the interval. This, in turn, means that the slope is negative before it reaches zero, and positive afterwards. Hence, minimum!

At this $ x $ the function has the greatest absolute value in the interval.

Now let's look at another point: $ f(1) = 1.ln(1) = 0 $. At $ x = 1 $ the function has the least absolute value in the interval. Therefore,  $ f(1) - f(1/e) $ has the largest absolute value for the difference between function values of two points in the interval. This leads us to a conclusion about the validity of the inequality. Any other value for the function over this interval is smaller, in absolute value, than $ |f(1) - f(1/e)| $ . This is also true for $ b = 1 - a = 1 - 1/e $ . The line $ x = 1 - 1/e $ is shown in blue in the graph, and it visually confirms that in this case, $ |f(a+b) - f(a)| > |f(b| $ , or, for $a = 1/e$ and $b=1 - 1/e$ :
$$ |(a+b).ln(a+b) - a.ln a| > |b.ln b| = b.ln(1/b) $$
Voila our counterexample.


问:
Importance of Neatness / Organization / Speed in Math?
Pretty simple question here but it does relate to math. I ask this as my writing is quite messy, possibly a cause of silly mistakes.

How important is neatness in math?

Does having messy writing put you at a disadvantage?

What is the tradeoff of speed vs. neatness?

Should I look to increase the neatness of my writing?

What are some simple tips to prevent mistakes that can be solved by simple organization / tidyness/structure?

In timed settings (exams), neat vs. speed?

答:
I find this interesting myself as I also have messy writing.

For me I always try to keep equal signs aligned and leave equal spacing. I also prefer using paper landscape as opposed to portrait but this is all just personal preference.

As for speed vs. neatness, I think it really is just about finding a fine line between them, neatness is important but you don't want to jeopardize leaving things incomplete.


问:
Polynomial function of simple symmetric random walk and martingales
Problem: Let $S_n$ denote the position of a simple random walk after $n$ steps, that is, $S_n=X_1+\cdots+X_n$ with $X_i$ i.i.d. and $P(X_i=1)=P(X_i=-1)=1/2.$

For which polynomials $g(x)$ will $g(S_n)$ be a martingale? (Here the coefficients of $g$ cannot depend on $n$.)

My Thoughts: I have the following. Let $g(x)=a_mx^m+a_{m-1}x^{m-1}+\cdots+a_0$ be a generic polynomial. Then $g(S_n)$ is a martingale if and only if
$$E[g(S_{n+1})\mid\mathcal F_n]=g(S_n)\quad\text{for all }n\geq1.$$
Hence if and only if
$$E\left[a_mS_{n+1}^m+a_{m-1}S_{n+1}^{m-1}+\cdots+a_0\mid\mathcal F_n\right]=a_mS_n^m+a_{m-1}S_n^{m-1}+\cdots+a_0.$$
So we need to compute the conditional expectation on the left. By linearity, we have
$$E\left[g(S_{n+1})\mid\mathcal F_n\right]=a_mE\left[S_{n+1}^m\mid\mathcal F_n\right]+a_{m-1}E\left[S_{n+1}^{m-1}\mid\mathcal F_n\right]+\cdots+a_0.$$
Now we look at each term separately. Using the binomial theorem and the fact that $X_{n+1}$ is independent of $\mathcal F_n$ we get
\begin{align*}
a_mE\left[S_{n+1}^m\mid\mathcal F_n\right]
&=a_mE\left[(S_n+X_{n+1})^m\mid\mathcal F_n\right]\\
&=a_mE\left[\sum_{k=0}^m\binom{m}{k}S_n^kX_{n+1}^{m-k}\mid\mathcal F_n\right]\\
&=a_m\sum_{k=0}^m\binom{m}{k}S_n^{k}E\left[X_{n+1}^{m-k}\mid\mathcal F_n\right]\\
&=a_m\sum_{k=0}^m\binom{m}{k}S_n^kE\left[X_{n+1}^{m-k}\right].
\end{align*}
From the calculation above we see that in order for the martingale property to be satisfied, the coefficients will have to depend on $n$, since so do the higher moments of $X_{n+1}$. Therefore, the only polynomials that satisfy the martingale property are $g(x)=x+a$ for constant $a$.

Do you agree with my reasoning above?

Thank you for your feedback.

答:
Actually, $X_{n+1}^{m-k}$ has the same distribution as $X_{1}^{m-k}$ hence the expectation of this term does not depend on $n$. We get that $E\left[X_{n+1}^{m-k}\right]=0$ if $m-k$ is odd and $E\left[X_{n+1}^{m-k}\right]=1$ if $m-k$ is even.

One finds that $2E\left[g(S_{n+1})\mid\mathcal F_n\right]=g(S_n-1)+g(S_n+1)$ hence $g(S_n)$ is a martingale if and only if
$$
2g(S_n)=g(S_n-1)+g(S_n+1)\mbox{ a.s.}.
$$
As $S_n$ takes the values $2k-n$, $0\leqslant k\leqslant n$, with positive probability, we get in particular that
$2g(n)=g(n-1)+g(n+1)$ for each $n$. Letting $P\colon x\mapsto 2g(x)-g(x-1)-g(x+1)$, we can see that $P$ is a polynomial having infinitely many roots hence
$2g(x)=g(x-1)+g(x+1)$ for each real number $x$. Now, defining $Q(x)=P(x)-P(x+1)$, we can see that $Q(x+1)=Q(x)$ hence $Q$ is constant and this implies that $P$ is a polynomial of degree one.


问:
Solve differential equasion $(y'')^2-5y'+6=0$
Solve differential equation  $(y'')^2-5y'+6=0$

For there is no "x" in equation, I made substitution $y'=p(y)$ and $y''=p'p$

$(p'p)^2-5p+6=0$

What can I do with the squared member?

答:
$$(p'p)^2-5p+6=0$$
Continue....
$$\dfrac {dp}{dy}p=\pm\sqrt {5p-6}$$
$$\int \dfrac {p}{\sqrt {5p-6}}dp=\pm \int dy$$

But you don't need any substitution:
$$(y'')^2-5y'+6=0$$
$$\dfrac {dy'}{dx}=\pm \sqrt {5y'-6}$$
it's separable:
$$\int \dfrac {dy'}{ \sqrt {5y'-6}}=\pm \int dx$$
$$\dfrac 1 5\int \dfrac {d(5y'-6)}{ \sqrt {5y'-6}}=\pm (x+C)$$
$$\dfrac 2 5{ \sqrt {5y'-6}}=\pm (x+C)$$
$$\dfrac 4 {25}(  {5y'-6})= (x+C)^2$$
That you can easily solve.


问:
Finding the covariance when given joint pdf
If the joint probability density of X and Y given by

$$
  f(x,y) =
\begin{cases}
1/4(2x+y),  & 0<x<1, 0<y<1\\
0, & \text{elsewhere}
\end{cases}
$$

Can I get some hints on how to find the Cov(x,y)?

答:
Maybe we could just consider the question through the relationship between marginal distribution and joint distribution.

The marginal probability density of y is $f_Y(y)=\int_0^1f(x,y)dx$=1/4(1+y),

so $E(Y)=\int_0^1yf_Y(y)dy=5/24$

then just calculate $E(X)$ and $E(XY)=\int_0^1\int_0^1xyf(x,y)dydx$.


问:
A problem of (Convex?) Optimization about compressive sensing
In fact, I'm not sure if the following question is convex.

I am processing an optimization problem about compressive sensing:

$$\arg\min_{A,x} \quad \lVert Ax-y \rVert _{2}^{2}+\lambda _1\lVert A^{\top}Ax-x \rVert _{2}^{2}+\lambda _2\lVert x \rVert _1$$

where $A$ should be an $m\times n$ $(m\ll n)$ matrix, $x$ should be an $n$-d vector, and $y$, $\lambda_1$, $\lambda_2$ are given. Especially, $A$ is variable intead of constant matrix.

Does we can solve this problem by using an iterative method like ISTA?

答:
You can definitely use a proximal gradient method such as ISTA. Recall that the general framework is $$\min_\mathbf{x} f(\mathbf{x}) + g(\mathbf{x})$$
where $f(\mathbf{x})$ is differentiable and $g(\mathbf{x})$ is convex. Then actually it's possible to solve this in two ways:

Define:

$f(\mathbf{x})=\frac{1}{2}\|A\mathbf{x}-\mathbf{y}\|^2_2+\frac{\lambda_1}{2}\|A^TA\mathbf{x}-\mathbf{x}\|^2_2$

$g(\mathbf{x})= \lambda_2\|\mathbf{x}\|_1$

Define:

$f(\mathbf{x})=\frac{1}{2}\|A\mathbf{x}-\mathbf{y}\|^2_2$

$g(\mathbf{x})= \frac{\lambda_1}{2}\|A^TA\mathbf{x}-\mathbf{x}\|^2_2+\lambda_2\|\mathbf{x}\|_1$

Notice that $\|A^TA\mathbf{x}-\mathbf{x}\|^2_2=\|(A^TA-I)\mathbf{x}\|^2_2$. This will make it easier to get the derivative. Also, if you take the second approach, then deriving a proximal operator for $g(\cdot)$ is not difficult, since under certain conditions, which apply in this case, $\text{prox}_{g_1+g_2}=\text{prox}_{g_1}\circ\text{prox}_{g_2}$. The proximal operator of the $l_1$ norm is the well known soft thresholding operator, and getting an explicit prox formula for the second term in $g(\cdot)$ is also not complicated. Then all you have to do is compose them.

Lastly, I would recommend to use an accelerated prox-grad method such as FISTA to get much better performance than ISTA.


问:
Basic Homotopy Question
I am starting to read the book "Rational Homotopy Theory" by Yves Felix, Stephen Halperin, J.-C. Thomas and I have a quick question about the very beginning (which only concerns basic homotopy theory in spaces and not even rational homotopy theory). The book proves a result referred to as "Whitehead's Lifting Lemma" as Lemma 1.5 (p. 12):

Suppose given a (not necessarily commutative) diagram:
\begin{array}{ccc}
A &\xrightarrow{\varphi} &Y \\
\ \downarrow i & &\ \downarrow f\\
X &\xrightarrow{\psi} &Z,
\end{array}
together with a with a homotopy $H: A \times I \rightarrow Z$ from $\psi i$ to $f\varphi$.

Assume $(X,A)$ is a relative CW-complex and $f$ is a weak homotopy equivalence. Then $\varphi$ and $H$ can be extended respectively to a map $\Phi: X \rightarrow Y$ and a homotopy $K: X \times I: \rightarrow Z$ from $\psi$ to $f \Phi$.

Then book continues with some corollaries, and my question is: How is the following statement a corollary of Whitehead's Lifting Lemma ?

If $(X, A)$ is a relative CW-complex and $A$ has the homotopy type of a CW-complex, then $X$ has the homotopy type of a CW-complex.

I think I could manage to prove this result by constructing a CW-complex $\tilde{X}$ from $\tilde{A}$ (a complex equivalent to $A$) by gluing cells using the attaching maps from $(X, A)$, and using a result of preservation of equivalences in pushouts (like this one Homotopy equivalences in pushout square with cofibration.) at each skeleton, but I do not see how that uses the Lemma above, and the result I would need about pushouts and equivalences appears later in the book I think.

Any insight is welcome, cheers!

答:
Let $A$ be a CW complex and $X$ obtained from $A$ by inductively attaching cells. Write $i:A\hookrightarrow X$ for the inclusion.

To begin let $p:\widetilde X\rightarrow X$ be a CW approximation (aka cellular model, see Th.1.4). Since $A$ is a CW complex the weak equivalence $p$ induces a bijection $p_*:[A,\widetilde X]\xrightarrow\cong[A,X]$ (see Co.1.6). Thus there is a map $\widetilde i:A\rightarrow\widetilde X$ together with a homotopy $H:p\widetilde i\simeq i$. Now consider the diagram
\begin{array}{ccc}
A &\xrightarrow{\widetilde i} &\widetilde X \\
\ i\downarrow & &\ \downarrow p\\
X &\xrightarrow{=} &X.
\end{array}
The assumptions of Lemma 1.5 are satisfied, so there is a map $\varphi:X\rightarrow\widetilde X$ such that $\varphi i=\widetilde i$ and $p\varphi\simeq id_X$. Thus $X$ is a (homotopy) retract of the CW complex $\widetilde X$, and it follows immediately from this that $X$ has CW homotopy type.

Now the last fact is true in the generality stated, but we will establish a more precise statement for the current situation: we will show that $X$ is homotopy equivalent to $\widetilde X$ as expected.

For this notice that $p_*:[\widetilde X,\widetilde X]\rightarrow [\widetilde X,X]$ takes $\varphi p$ to $p(\varphi p)=(p\varphi)p\simeq p$. But because $p$ is a weak equivalence the induced map is bijective, so the equation $p_*(\varphi p)=p_*(id_{\widetilde X})$ implies that $\varphi p\simeq id_{\widetilde X}$. Thus we have the claim.

Title: Basic Homotopy Question


问:
Does that series converge or diverge?
Does the series $$\sum \limits _{n=3}^\infty \frac{(-1)^{[\log n]}}{\sqrt{n}}$$ converge or diverge? As usually, $[x]$ denotes the integer part of $x.$ </p>

答: Calculate the sum, $n=\lceil e^k\rceil$ to $\lfloor e^{k+1}\rfloor$. There are about $(e-1)e^k$ terms. Each has the same sign, and  each has absolute value $\ge \frac{1}{\sqrt{e^{k+1}}}$. So the Cauchy Criterion for the partial sums fails, and the series does not converge. </p>

Title: Does that series converge or diverge?


问:
Complex number such that affixes of itself, its squared and its cube form a right triangle. $z,z^2,z^3$ form a right triangle in argand plane
$\DeclareMathOperator{\aff}{Aff}$
Here is a funny problem  I stumbled upon recently

I call the affix of a complex number the corresponding point in the argand plane, that is

$$\aff(\mathrm{i})=(0;1)$$

Find all $z \in \mathbf{C}$ such that the affixes of $z,z^2\ \& \ z^3$ form a right triangle.

I decided to Set $A=\aff(z),B=\aff(z^2),C=\aff(z^3)$

I have proven that :

If $z$ is a solution, then $z\neq 0,\  z\neq 1, \ \& \ z\neq-1$

If the triangle is right angled at A, $z$ must be a non zero imaginary number.

If the triangle is rigth angled at B, $\aff(z)$ has to lie on the vertical line defined by $x=(-1)$, except for $z=(-1)$.

(I'll post my proof when I get some time to write it in $\mathrm{\LaTeX}$ soon.)

I'm down to the third case : I have proven that if the triangle is right angled at $C$, then :
$$\Re \left(\frac{1+z}{z}\right)=0$$

Thanks to some experimentation, I know that $\aff(z)$ should lie on the circle centered at $(-\frac12;0)$ with radius $\frac12$, with $z$ again respecting the first condition above.

Now,how do I prove this ?

Thanks for the help.

Here is an animation.

答:
Oh dear,

It turns out I'm too tired to think, I found the following :
let's set $z=a+ib$ as in the question

then :
$$\frac{z+1}{z}=\frac{(a+1)+ib}{a+ib}=\frac{(a+1)+ib)(a-ib)}{a^2+b^2}$$
From which one gets
$$\frac{z+1}{z}=\frac{a+a^2+b^2-ib}{a^2+b^2}=\frac{a+a^2+b^2}{a^2+b^2}-\frac{ib}{a^2+b^2}$$

now, because this is an imaginary number we get
$$C(a,b)=a+a^2+b^2=0$$
which is the equation of an ellipse because it is a polynomial of two variables, of degree two. We see that there is no division, therefore, this is the equation of a circle

Let's rewrite it to see the circle's centre and radius :
$$C(a,b)=a+a^2+(b-0)^2$$
That is with two canonical forms
$$C(a,b)=\left(a+\frac{1}{2}\right)^2+(b-0)^2-\frac14=0$$
But then we get
$$\left(a-\left(-\frac{1}{2}\right)\right)^2+(b-0)^2=\frac14$$
We just need a square after the equal sign to get the radius
$$\left(a-\left(-\frac{1}{2}\right)\right)^2+(b-0)^2=\left(\frac12\right)^2$$

which is the equation of the circle mentioned in the question / OP.

Please note however that $(-1)$ and $0$ both belong to this circle.


问:
Expected number of heads, stopping at n heads
I flip a coin once a second, for $t$ seconds. If I get $n$ heads before time is up, I stop.

What is the expected number of heads, assuming the coin is fair (probability of a head is 0.5).

I believe the answer is $$E[\text{number of heads}] = \min\left(n,\frac{t}{2}\right),$$

but I'm not sure of a formal proof.

答: In effect you have a truncated binomial distribution so the expected number of heads is $$\sum\limits_{i=0}^t {t \choose i}\frac{\min(i,n)}{2^t}= n - \sum\limits_{i=n+1}^t {t \choose i}\frac{i-n}{2^t} = \frac t2 - \sum\limits_{i=0}^{n-1} {t \choose i}\frac{n-i}{2^t}$$  which is smaller than your expression


问:
Solving a matrix of the form $XA = cX$
I am trying to solve a matrix of the form $XA = cX$ where $A$ is a square matrix of size $N$ with known values and $c$ is a scalar representing the largest eigenvalue of matrix $A$ (though I would think a solution would hold for any scalar). $X$ is a $1 x N$ row vector of variables. I want to solve for the values of $X$

$A$ is the infinitesimal generator for a continuous time markov chain and what I am trying to accomplish here is solving for the system of linear equations (of the form described above) that characterize the steady-state solution.

My thoughts so far:

We can write $B$ = $A/c$ to move the scalar to the left hand side

$XB = X$ can be written as $B'X = X$ where $B'$ is the transpose of $B$

$B'X = X$ can be written as $(B'-I)X = 0$, where $I$ is the identity matrix of size $N$

答:
You can simply write:

$XA=cI_NX$

Where $I_N$ is the identity matrix:

$XA=X(cI_N) \Rightarrow X(A-cI_N)=0 \Rightarrow (A-cI_N)^t X^t=0 $

Since $c$ is an eigenvalue of $A$($\det(A-cI_N)=\det(A-cI_N)^t=0$) and the system is homogeneous, it admits non-trivial solutions.Notice that:
$(A-cI_N)^t=(A^t-cI_N)$. So the system becomes:

$(A^t-cI_N) X^t=0 $

So $X^t$ has to be in the eigenspace of the eigenvalue $c$ of the matrix $A^t$. I think nothing more can be said(if $c$ is not an eigenvalue of $A$ there aren't non-trivial solutions)

Title: Solving a matrix of the form $XA = cX$


问:
inner product and conjuction
I have a question regarding a general inner product space.

V is a inner product space.

() is the inner product.

can I say that :

$\overline{(a,b)}$ $\overline{(c,d)}$ = $\overline{(a,b)(c,d)}$

I am asking that because I know that over the complex It is true.

but the field not always will be the complex so...

can I say it ?

Is it true ?

if not how can I proof that ?

答:
How do you define the conjugate of a number if you are not in a complex or real field? Or the inner product?

But over the reals or complex numbers, the following equality is of course true, as (a,b) are just numbers.

Title: inner product and conjuction


问:
Can a certain monoid exist?
Is it possible to have an uncountable commutative monoid, where for every $a$ in the monoid, $a+a=a$?

I have a set which I am trying to define a group structure on (I am settling for a monoid structure though). the set is a bunch of equivalence classes of the real numbers, and I'm trying to see if it would be possible to define addition of classes by addition of representatives. The problem is adding an element of any class to itself, will always get you in the same class, so i'm not sure if this would mean I cant define the structure this way.

Also I should add, when I say i'm defining addition by addition of representatives, each class has to have a canonical representative which is used in the addition operation, as addition definitely is dependant on the choice of representative.

答:
Sure. Take $M = [0..1)$ within $ℝ$ and as operation taking maxima
$$∨~ \colon M × M → M,~(x, y) ↦ \max (x,y).$$
This yields an uncountable, commutative monoid with zero as identity and $a ∨ a = a$ for all $a ∈ M$.

Title: Can a certain monoid exist?


问:
Is [0.1, 1[ countable? (Or, why can a decimal number have an infinite of digits, but a whole number cannot?)
Note: I'm not asking about [0,1], I'm explicitly asking about [0.1, 1].

I'm being told [0.1, 1] is not countable. It's clear for me that [0,1] isn't, but it feels less clear for [0.1, 1], and I'll try to explain my understanding:

Essentially the idea is that if I can map every number in [0.1, 1] on a unique whole number it's countable, otherwise it's not. So the idea would be to just remove the 0. and what remains is the unique whole number.

0.123 maps to 123, 0.123456789101112 maps to 123456789101112, 0.1 maps to 1. Since we're already mapping 0.1 to 1 we can't map 1 to it too, so we map 1 to 0. For any example you can write down in it's decimal form, this would work for. However...

An obvious example where this might/does break down is for instance the number pi/4, which has an infinite number of decimals. So we can't map a whole number onto it. Or can we? I don't know.

So why is it allowed for a decimal number to have infinite digits, but a whole number cannot? Or is that just the definition of whole numbers and the "countability" of a set, like the difference between rational and irrational numbers?

答:
It is not countable. The quickest way to see this, if you already accept that $[0,1]$ is uncountable, is with a map $f(x) = \frac{x-0.1}{0.9}$, which is a bijection from $[0.1,1]$ to $[0,1]$.

An obvious example where this might/does break down is for instance the number $\pi/4$, which has an infinite number of decimals. So we can't map a whole number onto it. Or can we? I don't know.

Correct. Or take a number like $\frac19 = 0.\bar1$: what should this map to?

So why is it allowed for a decimal number to have infinite digits, but a whole number cannot? Or is that just the definition of whole numbers and the "countability" of a set, like the difference between rational and irrational numbers?

Informally, a number can have infinitely digits after the decimal point because each new digit makes it more and more precise: $3, 3.1, 3.14, 3.141, \ldots$ is getting closer and closer to some number (perhaps $\pi$). Adding digits to the left of the decimal point, on the other hand, makes the number explode: $1, 21, 321, 9321, \ldots$ isn't approaching any given number, it's shooting off to infinity.


答:
A simpler example than $\frac\pi4$ is $\frac13=0.3333333\ldots$

Anyway, our way of representing natural numbers is defined in such a way that any such number can be represented using only finitely many digits. And therefore that function that you mentioned is undefined for infinitely many elements of $[0.1,1]$.

Title: Is [0.1, 1[ countable? (Or, why can a decimal number have an infinite of digits, but a whole number cannot?)

Body:
Note: I'm not asking about [0,1], I'm explicitly asking about [0.1, 1].

I'm being told [0.1, 1] is not countable. It's clear for me that [0,1] isn't, but it feels less clear for [0.1, 1], and I'll try to explain my understanding:

Essentially the idea is that if I can map every number in [0.1, 1] on a unique whole number it's countable, otherwise it's not. So the idea would be to just remove the 0. and what remains is the unique whole number.

0.123 maps to 123, 0.123456789101112 maps to 123456789101112, 0.1 maps to 1. Since we're already mapping 0.1 to 1 we can't map 1 to it too, so we map 1 to 0. For any example you can write down in it's decimal form, this would work for. However...

An obvious example where this might/does break down is for instance the number pi/4, which has an infinite number of decimals. So we can't map a whole number onto it. Or can we? I don't know.

So why is it allowed for a decimal number to have infinite digits, but a whole number cannot? Or is that just the definition of whole numbers and the "countability" of a set, like the difference between rational and irrational numbers?


问:
Convergence of a series by comparsion test
My book says, that
$$\sum_{n=1}^{\infty}\frac{3 + (-1)^k}{k^2}$$
is convergent. I used comparsion test to determine it. $\frac{3 + (-1)^k}{k^2} \leq 3 + (-1)^k$, but $\sum_{n=1}^{\infty}3 + (-1)^k$ is divergent.  What am I doing wrong?

答: That approximation misses the point: we need to keep the $\frac{1}{k^2}$ bit - we know that $\sum \frac{1}{k^2}$ converges, so it's helping us. What we need to do is sort out that mess on the top. In particular, we bound $$\dfrac{3 + (-1)^k}{k^2}\leq \dfrac{4}{k^2},$$ and hence the result.


问:
Suppose A is an m-by-n matrix with $A\ne0$. Prove that the rank of A is 1 if and only if ....
Suppose $A$ is an $m \times n$ matrix with $A\ne 0$. Prove that the rank of A is $1$ if and only if there exist $(c_1,...,c_m) \in {F^m}$ and $(d_1,...,d_n) \in {F^n}$ Such that ${A_{j,k}}$ = ${c_j}{d_k}$ for every $j = 1,...,m$ and every $k = 1,...,n$.
It is problem from linear algebra done right(3.F.11)

I think for firs part if we take all entries $1$ and all i,j ${c_{j}=1}$ , ${d_{k}=1}$  it will work.
but for other direction??

答:
If $\mathbf{v}_1,\dots,\mathbf{v}_r$ is a basis for the image of $A$ (aka the column space), then for any $\mathbf{x} \in \operatorname{Domain}A$,

$$A\mathbf{x} \in \operatorname{Span}\{\mathbf{v}_1,\dots,\mathbf{v}_r\} \iff A\mathbf{x} = f_1(\mathbf{x})\mathbf{v}_1+\dots+f_r(\mathbf{x})\mathbf{v}_r$$

for some coefficients $f_i(\mathbf{x})$ that depend on $\mathbf{x}$.

I claim, and you should check, these coefficients $f_i(\mathbf{x})$ are linear functionals in $\mathbf{x}$. (i.e. linear maps from $\mathbf{F}^n \to \mathbf{F}$).

If you have a linear functional $f : \mathbf{F}^n \to \mathbf{F}$ then it's associated matrix is a row vector. Because, by definition, the matrix is $[f(\mathbf{e}_1),\dots,f(\mathbf{e}_n)]$. If I call this matrix $\mathbf{w}^\top$ then $f(\mathbf{x}) = \mathbf{w}^\top\mathbf{x}$.

So if $A$ has rank $r$, then I can write $A = \sum_{i = 1}^r \mathbf{v}_i\mathbf{w}_i^\top$. So we can check:

$$A\mathbf{x} = \left(\sum_{i = 1}^r \mathbf{v}_i\mathbf{w}_i^\top \right) \mathbf{x} = \sum_{i = 1}^r \mathbf{v}_i(\mathbf{w}_i^\top\mathbf{x}) = \sum_{i=1}^r f_i(\mathbf{x})\mathbf{v}_i. $$

$r = 1$ is the exercise in question.


问:
Minimization problem associated to a PDE $-\Delta y + f(y) = 0$?
I want to write the PDE
$$-\Delta y + f(y) = 0$$
$$y|_{\partial\Omega} = 0$$
where $f$ is some smooth function from $\mathbb{R}$ to $\mathbb{R}$ as the solution of
$$\min_{y \in H^1_0(\Omega)} \frac{1}{2}\int_\Omega |\nabla y|^2 + \int_\Omega F(y).$$
How can I relate $F$ to $f$?

答:
Define $J(y)=\int_\Omega \frac12 |\nabla y(x)|^2 + F(y(x)) ~dx$. Let $y$ be a minimizer of $J$ on $H_0^1$, i.e., $y \in H_0^1$ and $J(y)=\min_z J(z)$. Then $y+\lambda \varphi \in H_0^1$ for all $\varphi \in C_0^\infty$ and all $\lambda \in \mathbb{R}$. Then

$$\tilde J(\lambda):=J(y+\lambda \varphi)=\int_\Omega \frac12 |\nabla(y+\lambda \varphi)|^2 + F(y+\lambda\varphi) ~dx.$$

Since $0$ is a minimzer of $\tilde J$ it yields $\tilde{J}'(0)=0$. In other words,

$$\tilde J'(0)=\int_\Omega \nabla y \cdot \nabla \varphi + F'(y)\varphi ~dx=\int_\Omega (-\Delta y+ F'(y))\varphi ~dx.$$

By the fundamental lemma of calculus of variations $-\Delta y+ F'(y)=0$. Now, compare it to your question. It gives $F'(y)=f(y)$.


问:
The function $f:(0,1)\to(0,1)$ is defined by the rule $f(x)= \frac{2x}{1+x^2}$. Show that f is a one-to-one and onto function.
The function $f:(0,1)\to(0,1)$ is defined by the rule $f(x)= \frac{2x}{1+x^2}$. Show that f is a one-to-one and onto function. Find the rule $f^{-1}(x)$ of the inverse function $f^{-1}$

I tried to prove this is an injective function by using $f(m) = f(n)$, but the result is false. Help guys!!!

答: $\lim_{x\to0}f(x)=0$ and $\lim_{x\to1}f(x)=1$, use the Intermediate Value Property of continuous functions to see that  $f$ takes all values in $(0,1)$. For one-one see that you have two conditions $$(x-y)(xy-1)=0$$ now the case $xy=1$ is never satisfied by two numbers in $(0,1)$ can you observe this?. After you show one-one and onto, it remains trivial to show the inverse.


问:
Expectation of uniform random variable conditioned on an event
Let $D \in \mathbb R^n$ and let  $X \sim \mathrm{Unif}(D)$ be a uniform random variable on $D$. Moreover, let $K \subset D$. I am struggling to write a closed form expression for the quantity
$$
\mathbb E[g(X)\mid X \in K]
$$
where $g\colon \mathbb R^n \to \mathbb R$ is a sufficiently smooth function. I guess I have to go through the formula
$$
f_{X\mid Y}(x\mid y) = \frac{f_{X,Y}(x,y)}{f_Y(y)},
$$
with $Y = \chi_K(X)$, but I cannot conclude from here.

答:
Pose $Z= \mathbb{E}[g(X)|\chi_{X \in K}]$. If we have a guess for $Z$, we just need to show that $\mathbb{E}[Z] = \mathbb{E}[g(X)]$ and that $\mathbb{E}[Z\chi_{X \in K}] = \mathbb{E}[g(X)\chi_{X \in K}]$ (of course we also need the measurability of $Z$ wrt the right $\sigma$-algebra).

I claim that $Z = \frac{\mathbb{E}[g(X)\chi_{X \in K}]}{\mathbb{P}(X \in K))} \chi_{X \in K} + \frac{\mathbb{E}[g(X)\chi_{X \in K^c}]}{\mathbb{P}(X \in K^c))} \chi_{X \in K^c} $ works (easy to verify).

So $$\mathbb{E}[g(X)|X \in K] = \frac{\mathbb{E}[g(X)\chi_{X \in K}]}{\mathbb{P}(X \in K))}.$$ Id est $X$ conditioned on $X \in K$ has law $U(K)$.


问:
Generating functions with power series
How to find the series expansion for generating function $\frac {1} {1-2x-x^2}$?

I have got so far $$\frac {1} {1-2x-x^2}=-\frac {1} {2\sqrt2} (\frac {1} {1-\sqrt2+x} -\frac {1} {1+\sqrt2+x})$$

$\frac {1} {1-\sqrt2+x}$ is expanded power series of Sum of $(\sqrt2-x)^k$ for $k=0$ to infinity, but how would I express it in terms of $x^k$?

答:
If you call

$$
P(x) = \frac{1}{1-2x-x^2} = \sum_{i=0}^{\infty} a_i x^i
$$

then:

$$
(1-2x-x^2) P(x) = 1
$$
$$
P(x) = 1 + 2x P(x) + x^2 P(x)
$$

Then we can see that $a_0$ = 1, $a_1 = 2a_0 = 2$, and $a_{n+2} =  2a_{n+1} + a_n$

You can then find the closed form for $a_n$ by finding the roots $r_1, r_2$ of the characteristic polynomial $x^2 - 2x -1 = 0$, namely $r_1 = 1 + \sqrt{2}, r_2 = 1-\sqrt{2}$, and calculating $A,B$ in $a_n = Ar_1^n + Br_2^n$ for the initial conditions
$$
a_0 = A+B = 1
$$
$$a_1 = A(1+\sqrt{2} + B(1-\sqrt{2} = 2
$$
Solving for $A,B$ gives $A=\frac{2+\sqrt{2}}{4}, B = \frac{2-\sqrt{2}}{4}$, and
$$
a_n = \frac{2+\sqrt{2}}{4} (1+\sqrt{2})^n + \frac{2-\sqrt{2}}{4}(1-\sqrt{2})^n
$$

Title: Generating functions with power series


问:
Find the sum of the series $ \sum_{n=1}^{\infty} \frac{n^2}{n!} $
Find the sum of the series $ \sum_{n=1}^{\infty} \frac{n^2}{n!} $

$My \ attempt :$ $$
 \sum_{n=1}^{\infty} \frac{n^2}{n!} \\
 = \sum_{n=1}^{\infty} \frac{n}{(n-1)!} \\
 = \sum_{n=0}^{\infty} \frac{n+1}{n!}
 = \sum_{n=0}^{\infty} \frac{1}{(n-1)!} + \frac{1}{n!}
$$
But I don't know if this will take me anywhere.

答:
Since

$$\sum_{n=0}^{\infty}\frac{x^n}{n!}=e^x$$

We can differentiate to get

$$\sum_{n=1}^{\infty}\frac{nx^{n-1}}{n!}=e^x\implies \sum_{n=1}^{\infty}\frac{nx^{n}}{n!}=xe^x$$

differentiating again, we get that

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

plugging in $x=1$, we get that

$$\sum_{n=1}^{\infty}\frac{n^2}{n!}=2e$$


问:
When will two cars pass each other
There was a question in my math text book the other day that stated:

$2$ cars each travelling at a constant velocity around a ring ,
complete exactly $4$ and $7$ rounds in one hour. If they start at exactly the same time from the same place but travel in opposite directions around the track, how many times will they
pass each other in the one hour?

I tried various things using $\displaystyle v = \frac{d}{t}$ but couldnt get very far. Any help is appreciated, thanks.

答:
Consider alternative case when cars complete exactly $4$ and $8$ rounds.  It's easily seen that the number of times they pass is

$$2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 = 12$$

So for $4$ and $7$ it would be one less than that which is $11$.

Title: When will two cars pass each other


问:
Let $a+b\mid ab$; extrapolates the $\gcd(a,b)$
Let $a$, $b$ be a positive integers such that $(a+b)\mid ab$. Prove that $\gcd(a,b)\geq\sqrt{{a}+{b}}$. I have no idea how to approach this problem. Any hints or solutions are appreciated.

答:
Let $a=d\cdot x$ and $b=d\cdot y$ with $x$ and $y$ coprime (i.e. $\gcd(x,y)=1$). This means that $\gcd(a,b)=d$. So your question translates to $d\geq \sqrt{d(x+y})$ knowing that $d(x+y)|d^2xy$.

In other words, if $x+y|dxy$ then $d\geq x+y$.

Suppose $k|xy$ and $k|x+y$. Then $k|(x+y)\cdot x=xy+x^2$, so $k$ also divides $x^2$. Analogously, $k(x+y)\cdot y$ so $k|y^2$. But $x$ and $y$ are coprime, so $x^2$ and $y^2$ are coprime, so $K=1$, meaning that $x+y$ and $xy$ are coprime.

So if $x+y|dxy$, then $x+y|d$, so $d\geq x+y$


问:
Prove $|a − x| < |a − b|$
Prove that if $a, b, c$ are distinct points of $\mathbb R^d$
such that $∠abc < π/2$, then there is a $x$ on the open segment $(b, c)$ with $|a − x| < |a − b|$.

I tried to take these points as three vertices of an acute-angled triangle such that $∠abc < \pi/2$, and tried to take the property of sum of two sides is greater than the third side i.e. $ab +bc > ac$ then I draw a line from the point $a$ which intersect $bc$ at $x$. Then again I took the sum of two sides property such that $ab + ax > bx$ and after trying all the possibilities I am not unable to prove the above.

Can someone please help me?
Thanks in advance.

答:
Consider the triangle $abx$.  You are asking that $a$ is closer to $x$ than it is to $b$, or in other words that the line segment $ax$ is shorter than the line segment $ab$.  This would be true if and only if the measure of $\angle x$ is greater than the measure of $\angle b$.

There is certainly a point $x$ on the line segment $bc$ that satisfies that.  As noted in comments, the projection would work.  Alternatively, you could just choose an $x$ so close to $b$ that $m\angle a$ is insignificant, which would make $\angle x$ the largest angle in the triangle (since we are given that $\angle b$ is acute).

Title: Prove $|a − x| < |a − b|$


答:
I think you can explain everithing using this observation

if a prime $p$ divides $a+b$ then $p$ divides $a$ or $b$ (WLOG $p|a$). At this point you
$p|(a+b)$ and $p|a$ but then $p|b$ wich means $p|gcd(a,b)$.

That said if you put $d=gcd(a,b)$ and $a=da'$, $b=db'$ you have $a+b|d^2a'b'$ but from the observation $a+b|d^2$ and from this the thesis follow.

Title: Let $a+b\mid ab$; extrapolates the $\gcd(a,b)$

Body: Let $a$, $b$ be a positive integers such that $(a+b)\mid ab$. Prove that $\gcd(a,b)\geq\sqrt{{a}+{b}}$. I have no idea how to approach this problem. Any hints or solutions are appreciated.


Body:
Find the sum of the series $ \sum_{n=1}^{\infty} \frac{n^2}{n!} $

$My \ attempt :$ $$
 \sum_{n=1}^{\infty} \frac{n^2}{n!} \\
 = \sum_{n=1}^{\infty} \frac{n}{(n-1)!} \\
 = \sum_{n=0}^{\infty} \frac{n+1}{n!}
 = \sum_{n=0}^{\infty} \frac{1}{(n-1)!} + \frac{1}{n!}
$$
But I don't know if this will take me anywhere.

Body:
$$S=\sum_{n=1}^{\infty} \frac{n^2}{n!}=\sum_{n=1}^{\infty} \frac{n(n-1)+n}{n!}=\sum_{n=1}^{\infty} \left(\frac{1}{(n-2)!}+\frac{1}{(n=1)!} \right)$$ $$=\sum_{p=-1}^{\infty} \frac{1}{p!}+\sum_{m=0}^{\infty} \frac{1}{m!}=e+e=2e$$
Note that $(-1)!=\infty$.


问:
A quadrilateral $ABCD$ with $AB = 10$ , $BC = 6$ , $CD = 8$ and $DA = 2$ has diagonals $AC$ and $BD$ meeting at $O$.

A quadrilateral $ABCD$ with $AB = 10$ , $BC = 6$ , $CD = 8$ and $DA = 2$ has diagonals $AC$ and $BD$ meeting at $O$, such that $\angle COB = 45^\circ$. Find the area of $ABCD$.

What I Tried: Here is a picture :-

The only thing missing to find the area is one of the diagonals, and I can find the area easily by Heron's Formula. Also I thought that since some angles are there, I might have to use trigonometry, and I tried but without success.

We have that $\angle DOC = \angle AOB = 135^\circ$ . I know that the area of a triangle can be found from the formula :- $$\rightarrow \frac{1}{2}ab\sin \gamma$$

Where $\gamma$ is the angle between $a$ and $b$, but I cannot use it sadly as I don't know any of the lengths of $AO$ , $BO$ , $CO$ and $DO$. Next I thought of the formula :-
$$\rightarrow \frac{a}{\sin a} = \frac{b}{\sin b} = \frac{c}{\sin c}$$

But this dosen't work here either due to the same reason.

I think that Law of Cosines might work, but I am actually new to Trigonometry and not sure if it will really work or not, and then I have to consider $4$ equations for $4$ sides, which will be complicated.

Can anyone help me? Thank You.

Note:- Solutions without Trigonometry and mostly welcome. I hate using Trigonometry.

答:
Here is a solution without trigonometry. Drop altitudes DE and BF to the base AC, with E and F as the feet. The area is

\begin{align}
Area&= \frac12 AC \cdot ( DE + BF)\\
& =\frac12(AO \cdot ED +CO\cdot ED+AO\cdot BF + CO\cdot BF)\tag1
\end{align}

Per ED = EO, alone with the Pythagorean’s theorem
\begin{align}
2AO\cdot ED &= -(AO -EO)^2 +AO^2 +ED^2\\
&=-AE^2 +AO^2 +ED^2\\
&= -AD^2 +AO^2+ 2ED^2\\
\\
2 CO\cdot ED&= CD^2 -CO^2-2ED^2\\
2AO\cdot BF &= AB^2 -AO^2- 2BF^2 \\
2 CO\cdot BF&= CO^2-BC^2+2BF^2\\
\end{align}

Substitute above into (1) to obtain
$$Area= \frac{AB^2-AD^2-BC^2+CD^2}4$$


问:
$n$-linear form: An Interpretation
What is a good example of an $n$-linear form that is more familiar to a student learning at an elementary level?

EDIT:

I'm just trying to show that every $n$-linear alternating form on a vector space of dimension
less than $n$ is the zero form.

答:
Well, let's make it simple. Since you just want an example, consider the vector space $\mathbb{R}^n$ with usual operations. Then $\det : \mathbb{R}^n \times \cdots \times \mathbb{R}^n \to \mathbb{R}$ is one $n$-linear form. It's even one special case (one alternating $n$-linear form), that is it has more properties. But it doesn't matter, what it matters is that if you look at the usual definition of $\det$ you'll see that it's linear on each of it's arguments with the others held fixed.

You might ask: "what the hell you're talking about?! $\det$ is a function of a single matrix!" but if you think a little it's the same as being a function of $n$ vectors: you just have to think of the matrix as the result of joining $n$ vectors as the columns. So a function of a matrix is exact the same as a function of the $n$ columns when we think of each of them at a time.

Now if $v_1, \cdots, v_n$ are the columns of a matrix, you know that multiplying the $i$-th row $v_i$ by $k$ results the following:

$$\det(v_1,\dots ,kv_i,\dots,v_n)=k\det(v_1,\dots,v_i,\dots,v_n)$$

And if you sum some column $w$ to the column $v_i$ you have:

$$\det(v_1, \dots, v_i + w,\dots,v_n)=\det(v_1,\dots,v_i,\dots,v_n)+\det(v_1,\dots,w,\dots,v_n)$$

So, the determinant is really one $n$-form. And also it has a cool interpretation: volume. Indeed we usually define the determinant with axioms that grants him the property of being able to tell the the volume of the solid whose sides are precisely the $n$ vectors given to it. So, that it's properties reflect the fact that if $\{v_1, \dots, v_n\}$ is a linearly dependent set, it'll have $n$-volume zero, so that we get $\det(v_1, \dots, v_n) = 0$ as expected.

Also, you can think of the association of some $k$-linear alternating form of each point of the euclidean $n$-space. This is a function which we call differential form and it has very interesting properties (mainly when talking about integrals). So if you want a little more on this topic (which is too extensive to be treated here) you can see Spivak's Calculus on Manifolds or Munkres' Analysis on Manifolds. I hope this helps you. Good luck!

Title: $n$-linear form: An Interpretation


问:
Let $f$ be a continuous and bijective function such as $f(\frac{a}{b}) = \frac{f(a)}{f(b}$, what are those possible $f$?
Let $f$ be a continuous and bijective function such as $f(\frac{a}{b}) = \frac{f(a)}{f(b}$.

What are the possible functions for $f$?

My guess is that it can only be a polynomial of the shape $f : x \mapsto x^n$.

I have already found a few properties:

$\forall n,m \in \mathbb{Z}, f(x^n) \cdot f(x^m) = f(x^{n+m})$

$\forall n \in \mathbb{Z}, f^n(\frac{a}{b}) = \frac{f^n(a)}{f^n(b)}$

$f(1) = 1$

$f(0) = 0 \text{ if } f \neq x \mapsto 1$

Any ideas on how to prove that it is only polynomials $X^n$ that works ?

答:
If $f$ is defined on $(0,\infty)$, take $g(x) := \ln(f(e^x))$. Then since $f$ is continuous, $g$ is continuous
$$
g(a-b) = \ln(f(e^a/e^b)) = \ln(f(e^a)/f(e^b)) = g(a) - g(b)
$$
Since $f(1) = 1$, then $g(0) = 0$, so $g(-b) = -g(b)$ and
$$
g(a+b) = g(a)+g(b)
$$
The problem then reduces to Cauchy's functional equation. , and so $g$ is linear, i.e.
$$
f(x) = e^{g(\ln(x))} = e^{a\ln(x)} = x^a
$$
for all $x>0$.

If $f$ is defined on $\mathbb R$, then for $x<0$, $f(x) = f(|x|/(-1)) = f(x)/f(-1) = |x|^a / f(-1)$. Since $f$ is bijective, $f(-1)<0$.

Since $f(-1) = f(-1/1) = f(1/-1) = f(-1) = 1/f(-1)$ we obtain $f(-1) = -1$. Finally
$$
\boxed{f(x) = |x|^{a-1} x}
$$
for some $a>0$.


问:
how can I solve this type of problem?
Given the following function $$\psi: \begin{cases}  \mathbb{R} \rightarrow \mathbb{R}^2 \\  t \mapsto (2\cos(t-\frac{\pi}{2}),\sin(2(t-\frac{\pi}{2}))) \end{cases}$$
show that $$M = \psi(\mathbb{R}) = \bigl\{ (x,y)\in \mathbb{R}^2: x^4 -4x^2+
4y^2 = 0 \bigr\} $$

答:
Step 1: Take $x(t)=2\cos(t-\pi/2)=2\sin t$ and $y(t)=\sin(2t-\pi)=-\sin 2t$.

Now verify that $x(t)^4-4x(t)^2+4y(t)^2=0$.

$$4(x(t)^2-y(t)^2)=4(4\sin^2t-\sin^22t)=4(4\sin^2t-4\sin^2t\cos^2t)=16\sin^4t$$

This gives $\psi(\mathbb R)\subseteq S=\{(x,y)\in\Bbb R^2:x^4-4x^2+4y^2=0\}$.

Step 2: Now we need to show $\psi$ is onto over $S$.

Let $(X,Y)\in S$, then $Y=\pm X\sqrt{1-X^2/4}$ and $X\in[-2,2]$. So $X/2\in[-1,1]$.

Take $t=\arcsin(X/2)\in\Bbb R$ and verify that $x(t)=X,y(t)=-X\sqrt{1-X^2/4}$.

Take $t=\pi-\arcsin(X/2)\in\Bbb R$ and verify that $x(t)=X,y(t)=X\sqrt{1-X^2/4}$.

 Hint: Use the fact that $\cos(\arcsin(z))=\sqrt{1-z^2}$.

Title: how can I solve this type of problem?


问:
Is the partial trace identity $\mathrm{Tr}_2 [(A\otimes B)C]=A\cdot\mathrm{Tr}_2[(\mathbb{I}_1\otimes B)C]$ true?
Suppose we have two (finite dimensional) Hilbert spaces $H_1$ and $H_2$. Let $A$ be an operator on $H_1$, and $B$ an operator on $H_2$, and $C$ an operator on $H_1 \otimes H_2$. Is the quantity (where $\mathrm{Tr}_2[\cdot]$ is the partial trace over the Hilbert space $H_2$)
$$
\mathrm{LHS} := \mathrm{Tr}_2 \left[ ( A \otimes B ) C \right]
$$
the same as
$$
\mathrm{RHS} := A \cdot \mathrm{Tr}_2 \left[ ( \mathbb{I}_1 \otimes B ) C \right] \ ?
$$
That is to say, is it true that $\mathrm{LHS} =\mathrm{RHS}$? It seems correct, but I am having difficulty proving it.

答:
Yes, this identity is correct. The key is to note that $H_1 \otimes H_2$ is spanned by simple tensors, which is to say that $C$ can be expressed in the form
$$
C = \sum_{j=1}^k P_j \otimes Q_j.
$$
With that, we can write
$$
\begin{align}
\mathrm{Tr}_2 \left[ ( A \otimes B ) C \right]
&=
\mathrm{Tr}_2 \left[ ( A \otimes B ) \sum_{j=1}^k P_j \otimes Q_j \right]
\\&=
\mathrm{Tr}_2 \left[ \sum_{j=1}^k (AP_j) \otimes (BQ_j) \right]
\\ &= \sum_{j=1}^k \mathrm{Tr}(BQ_j)(AP_j)
\\ &= A \sum_{j=1}^k \mathrm{Tr}(BQ_j)(P_j)
\\ &= A \cdot \mathrm{Tr}_2\left[ \sum_{j=1}^k P_j \otimes BQ_j\right]
\\ &= A \cdot \mathrm{Tr}_2\left[ (I \otimes B)\sum_{j=1}^k P_j \otimes Q_j\right]
 = A \cdot \mathrm{Tr}_2\left[ (I \otimes B)C\right].
\end{align}
$$


问:
Linear combination of vectors in $\mathbb{R}^3$
Show that any linear combination of $\pmatrix{1\\\frac{3}{2}\\0}$ and $\pmatrix{0\\3\\6}$ is also a linear combination of $\pmatrix{2\\3\\0}$ and $\pmatrix{0\\1\\2}$

I'm not sure how to do this. I have a proof sketch for showing that any 2-dimensional vector is a linear combination of any other two nonparallel 2d vectors, but as far as I understand, this is not the case for 3d vectors. So how should I go about showing this?

Thanks.

答:
every combination of $\pmatrix{1\\\frac{3}{2}\\0}$ and $\pmatrix{0\\3\\6}$ will be as $c_1\pmatrix{1\\\frac{3}{2}\\0}+c_2\pmatrix{0\\3\\6}=\frac{c_1}{2}\pmatrix{2\\3\\0}+3c_3\pmatrix{0\\1\\2} $

so :

any linear combination of$ \pmatrix{1\\\frac{3}{2}\\0}$ and $\pmatrix{0\\3\\6}$ is also a linear combination of $ \pmatrix{2\\3\\0}$and$\pmatrix{0\\1\\2}$


问:
In a queue of $n$ people, what is the average number of spaces in between two people?

There are $n$ people in a queue. You randomly select two of them. What is the expected number of people standing in between them?

I noticed that the question can be reformulated in the following way. Let $X$ and $Y$ be discrete random variables that can take values between $1$ and $n$. What is $\mathbb{E}(|X-Y|-1 \text{ given that $X\neq Y$})$?

Unfortunately, this didn't seem to get me anywhere. I also considered what the probability was for small values of $n$. Let $Z$ be the number of people standing in between the two people. For $n=2$, $Z=0$ trivially. For $n=3$, there is a two-thirds chance that the people are standing next to each other, and a one-third chance that they are on either side of the queue. So $E(Z)=\frac{2}{3}\cdot0+\frac{1}{3}\cdot1=\frac{1}{3}$. I also calculated it for $n=4$, but stopped after that point since the calculations were getting unwieldy.

Could anyone hint at what approach I could use instead?

答:
Let $p_i$ be the probability of a pair of randomly selected people having distance $i$.

You can easily observe that $$p_i=\frac{n-i-1}{\binom{n}{2}}$$ because the good cases are pairs $(1;i),(2;i+1),...,(n-i-1;n)$ and all the cases are $\binom{n}{2}$ (because that is the number of pairs.

Your expected value is $$\sum_{i=0}^{n-1}i\cdot p_i=\sum_{i=0}^{n-1}i\cdot\frac{n-i-1}{\binom{n}{2}}=\frac{1}{\binom{n}{2}}\cdot\sum_{i=1}^{n-1}i(n-i-1)=\frac{1}{\binom{n}{2}}\cdot\bigg(\frac{n(n-1)^2}{2}-\frac{(n-1)n(2n-1)}{6}\bigg)=$$$$=\frac{2}{n(n-1)}\cdot\frac{n(n-1)}{2}\cdot\bigg(n-1-\frac{2n-1}{3}\bigg)=\frac{n-2}{3}$$


问:
How to find the PMF of a random variable given it's PGF
How do I go about finding the pmf of a discrete positive random variable X if its pgf is =1/3 (z + z$^2$ + z$^3$)?

答:
Consider that your PGF is, by definition,

$$\mathbb{E}[z^X]$$

Thus $X$ takes the values $\{1;2;3\}$, each value with probability $\frac{1}{3}$


问:
Convexity of root
I need some help proving that
$$
\sqrt{x^T \Sigma x},
$$
with $\Sigma$ being a positive definite matrix, is convex. I already found some questions about this topic, indicating that one should use a conic representation but I did not understand these approaches, I was hoping maybe someone has another idea or could explain it better (maybe with a link to a resp. paper or book).

答:
When $\Sigma\succeq 0$ then we can define $\sqrt{\mathbf{x}\Sigma\mathbf{x}}=\|\mathbf{x}\|_{\Sigma}$ to get a normed vector space, which has all the properties of the Euclidean norm. Specifically, it is absolutely homogeneous and the triangle inequality holds. Then the solution becomes very easy:
\begin{aligned}
f(\lambda\mathbf{x}+(1-\lambda)\mathbf{y})&=\|\lambda\mathbf{x}+(1-\lambda)\mathbf{y}\|_{\Sigma}\\&\leq \|\lambda\mathbf{x}\|_{\Sigma}+\|(1-\lambda)\mathbf{y}\|_{\Sigma}\\
&=|\lambda|\|\mathbf{x}\|_{\Sigma}+|(1-\lambda)|\|\mathbf{y}\|_{\Sigma}\\
&=\lambda f(\mathbf{x})+(1-\lambda)f(\mathbf{y}) &\square
\end{aligned}

Title: Convexity of root


问:
When do units plus 0 form a subring?
For which (commutative, unital) rings $R$ is it the case that $R^*\cup \{0\}$ is a subring of $R$?
Here are the only examples I could think of:

a field

$R[x_1,\ldots,x_n]$ for $R$ with the above property

$(\mathbb F_2)^\alpha$ for any $\alpha$; here the only unit is $1=(1,1,\ldots)$. This doesn't work for any other field, because if $a\in k^*$, $a\neq 1$, then $(a,1,1,\ldots)-(1,1,1,\ldots)$ is not a unit.

Any other ring with characteristic 2 and no units besides 1 (is there any way to categorize these?)

Are there any others?

答:
In your notation $k=R^*\cup\{0\}$ is a subfield of the ring $R$. So, you are asking for $k$-algebras $R$ with the property that  $R^*=k^*$. There are many rings with this property.

For any smooth projective absolutely irreducible algebraic curve $C$ over $k$
and any point $P$ of $C$, consider the subring $R$ of the function field $C(k)$
that consists of functions without poles outside $P$.
The ring $R$ has the property that $R^*=k^*$.

Title: When do units plus 0 form a subring?


问:
Spiral function with constantly increasing separation between its loops
Spirals following the polar equation $r=a+bθ$ are known to have constant separation between each of its loops, I show this by transforming the polar equation into Cartesian equation by using the transformations $r=(x^2+y^2)^{0.5}$ and $ θ=tan^{-1}\left(\frac{y}{x}\right)$. For computing the separation between loops $n$ and $n+1$, I get the following formula

$x_{n+1}-x_{n}=b π [(n+1)-(n)]= b π$

As shown, the separation between the loops is a constant. When I use the equations $r=a+bθ^p$ or $r=a+bp^θ$, and then compute the separation between loops, for instance, $n$ and $n+1$, I get the following results;

$x_{n+1}-x_{n}=b 2π^p [(n+1)^p-(n)^p]$ for $r=a+bθ^p$

$x_{n+1}-x_{n}=b(p^{2π}-1)p^{n 2π }$ for $r=a+bp^θ$

As one would realize, the separation between loops of those two spirals increase, but not with a constant rate. I tried many more formulae, but I didn’t get any spiral with constant increase rate between its loops, i.e., stratifying the formula;

$x_{n+1}-x_{n}=An$

Wherein $A$ is a constant. Does anyone here know which equation would give me the spiral shape I’m looking for? Thank you so much guys.

答:
Placing $ \theta $ in position of exponent makes it a log spiral, making geometric progression of radial ray lengths.
Taking
$$ r= r_0 + \dfrac{ p \theta }{2 \pi}$$
we have  $p$ constant as radial increase or pitch for each revolution. It is in arithmetic progression.

Since $$ r =\sqrt {x^2+y^2}$$
$$ \sqrt{x^2+y^2} = \sqrt{x_0^2+y_0^2} + \dfrac{ p \theta }{2 \pi}$$
Setting $y=0$ we must have for x-axis start of Archimedean spiral
$$ x= x_0 + \dfrac{ p \theta }{2 \pi}$$
and setting $x=0$ we must have for y-axis start of the spiral
$$ y= y_0 + \dfrac{ p \theta }{2 \pi}$$
and the same holds good for any radial ray $ \theta = \alpha$.

So $p$ is radial pitch at any arbitrarily chosen $\alpha.$


问:
I need some clarification regarding functions.
If $f(x) = 3 + 5$ and $() = 4^2 − 2$, find:

$(g(3))$

$g(3)$

Would you get the same answer for one and two? I think I read or heard somewhere that it doesn't matter whether you put a function within brackets or not.

I want to be absolutely sure because I'm not exactly a math genius.

答: Yes, those are exactly the same.</p>

Body: Yes, those are exactly the same.


问:
Martingale problem in probability theory
I am stuck on the exercise below from an exam I took. I am going through it again and trying to figure it out because I couldn't during the exam.

In a), from the definition of a martingale I already proved that the $\mathbb{E}[M_n] < \infty$ but couldn't show that $\mathbb{E}[M_{n+1} \mid X_1, \ldots, X_n] = M_n$

For b) and c) however I couldn't get much done.

Any help would be greatly appreciated!

Thanks in advance.

PS: I am interested in understanding the exercise rather than simply getting the solution.

答:
To understand the loss of the casino $M_n,$ first observe that the moment $X_n=-1,$ everyone leaves the Casino empty-handed--that is, the casino doesn't lose anything.  In other words, it is a fresh start for the casino every time we see $X_n=-1$ (except for the fact that the Casino has gained $n$ unit of money by this time). So to calculate the loss of the Casino at step $n,$ the only thing that matters is the last consecutive sequence of $+1,$ that is, $K_n.$

With this information, now let us see how much the casino loses given $K_n$ many consecutive $+1.$ In the first step (since the beginning of the misfortune of the casino), there is one gambler and the gambler pays $1$ unit of money and gets $1/p$ amount of money back. The casino has lost $1/p-1$ unit. In the next step, there is another gambler, who wins. So now the loss of the casino is $$\frac{1}{p}\left(\frac{1}{p}-1\right)+(1/p-1)=\frac{1}{p^2}-1.$$
If this misfortune continues for $K_n$ time, it is clear that the Casino has lost $$M_n=\sum_{i=1}^{K_n}\left(\frac{1}{p}\right)^i-n.$$

For part $b)$, I usually prefer the 'conditioning trick'. The idea here is that every time you see $X_n=-1,$ the past does not matter and you have to start all over again. So it makes sense to condition on the 'first time you see $-1$'. In detail, let us think about what can we say about $ET$ when $X_1=-1.$ This means, we already have wasted one step and we have to wait for $T$ more time, that is,
$$E(T|X_1=-1)=1+E(T).$$

Continuing this idea, we can write
$$E(T|X_1=1=X_2=\ldots=X_{k-1}, X_k=-1)=k+E(T), \quad k\le L.$$
And finally on the event $C=(X_1=\ldots=X_L=1),$ we have $E(T|C)=L.$
Combining all this we get
$$E(T)=(1-p)\sum_{j=0}^{L-1}p^{j}(j+1+E(T))+Lp^L=\frac{p^{-L}-1}{1-p}.$$

Also, once I have got the above answer it immediately suggests that $T$ has come connected with $M_n.$ And, notice that $T$ is a stopping time and $M_n$ is a martingale, therefore (given that $E(T)<\infty$) we obtain $$E(M_{T})=E(M_1)=0.$$ On the other hand, we also have that
$$E(M_T)=\frac{p^{-L}-1}{1-p}-E(T).$$
This gives the expression for $E(T).$

For $(c)$, repeat the part $a)$ and $(b)$ when each new gambler has initial amount $s^{n-1}.$ it would be a good idea to see what is the loss of the casino. In fact, writing the correct loss would immediately tell you what to do next!


问:
Proving $\prod_{n=1}^{\infty} [1+ (-1)^n a_n] $ converges iff $\sum_{n=1}^{\infty}{a_n}^2$ uder some conditions
This question is from Ponnusamy and silvermann complex variables with applicatio Subsection ( Infinite Products ) and I am struck in 1 part of it.

Suppose that ${a_n}$ is a decreasing sequence of real numbers with lim $ n\to \infty$ $ a_n =0$ . Show that
$$\prod_{n=1}^{\infty}[1+ (-1)^n a_n] $$
converges iff
$$\sum_{n=1}^{\infty} {a_n}^2 $$
converges.

I have done the part which assumes convergence of
$$\sum_{n=1}^{\infty} {a_n}^2 .$$

For converse , convergence of
$$\prod_{n=1}^{\infty}[1+ (-1)^n a_n]$$
imples convergence of series
$$\sum_{n=1}^{\infty} (-1)^n a_n$$
converges (let to a). Now , I am unable to use convergence of $$\sum_{n=1}^{\infty} (-1)^n a_n$$ to a to prove convergence of $$\sum_{n=1}^{\infty} {a_n}^2 .$$

Kindly help me with that

答:
We have $a_n \searrow 0$ and, hence,
$$\lim_{n \to \infty}\frac{(-1)^na_n - \log(1+(-1)^n a_n)}{a_n^2} = \frac{1}{2}$$

This implies that for all sufficiently large $n \geqslant  N$,

$$\frac{(-1)^n a_n - \log(1+(-1)^na_n)}{a_n^2} > \frac{1}{4},$$

and

$$\sum_{n=N}^M(-1)^n a_n - \sum_{n=N}^M\log(1+(-1)^na_n) > \frac{1}{4}\sum_{n-N}^Ma_n^2$$

Note that $\sum (-1)^n a_n$ converges by the alternating series test.  If $\sum a_n^2$ diverges, then it diverges to $+\infty$ since each term is nonnegative. This would imply

$$\log \prod_{n=N}^\infty (1+ (-1)^na_n) =\sum_{n=N}^\infty\log(1+(-1)^na_n) = -\infty,$$

and the infinite product diverges to zero, a contradiction.

The convention for infinite products is that $\prod_{n=1}^\infty (1+a_n) = 0$ is considered divergence.


问:
Production Function
Yes.. I know this is a math forums but there is no economics.stackexchange.com :(.

Since this is also a math problem, I thought I'd post it here. Please help.

An undeveloped economy produces goods and services which rely heavily on natural resources. It is described by the following production function:

$$Y = T * K^{1/4} * Z^{1/2} * L^{1/4}$$

where Y is real GDP, T is technology, K is physical capital, Z is natural resources, and L is aggregate hours of work.

If there is no growth in labor productivity, and both the capital stock (K) and natural resources (Z) are constant, with population and labor hours growing at 2 percent per year, what is the growth rate of technological progress in this economy?

So, what I got it that there is no growth in labor productivity meaning Y/L = a constant. K and Z are constant meaning K/L and Z/L is 0. However, I dont know how to use the fact that population and labor hours are growing at 2%/yr.

Thanks!

答:
"K and Z are constant" does not mean "K/L and Z/L is 0".  It might be true that $\frac{dK}{dt}$ and $\frac{dZ}{dt}$ are 0, but you don't really need that restatement here.

You seem to have decided that you want to look at $Y/L$ so you get

$$Y/L = T \times K^{1/4} \times Z^{1/2} \times L^{-3/4}.$$

But some of these are constant so you have

$$\text{constant} = T \times \text{constant} \times \text{constant} \times L^{-3/4}$$

and that gives you a relationship between $T$ and $L$ so you can work out how $T$ changes if $L$ changes by a specified fraction each year.

Title: Production Function


答:
It's a little ambiguous to write $fg(3)$, because it could either mean $$(f\circ g)(3) = f(g(3))$$ or it could be a sloppy way of writing $$(fg)(3)= f(3)\cdot g(3),$$ and these two things are not generally the same:
$$
(fg)(3) = f(g(3)) = f(30) = 95$$
whereas
$$f(3)\cdot g(3) = 14\cdot 30 = 420$$

If this is given in an assignment it's probably best to ask your instructor what they mean.  If you are looking for a shorter way of writing $f(g(3))$ you should go with $(f\circ g)(3)$

Title: I need some clarification regarding functions.

Body:
If $f(x) = 3 + 5$ and $() = 4^2 − 2$, find:

$(g(3))$

$g(3)$

Would you get the same answer for one and two? I think I read or heard somewhere that it doesn't matter whether you put a function within brackets or not.

I want to be absolutely sure because I'm not exactly a math genius.


问:
Show that $\sum_{k=0}^n \binom{2n}{2k} = 2^{2n-1}$
I have played around with this question for a bit, but my answer doesn't feel right. I'm not happy with how I converted the binomial. it feels a bit hand-wavy. Is there any way I can make this proof better?

Let's say that we have two teams that are both size $n$. Let's call one team the Blue team and the other team Red. We are going to send a mixture of Blue and Red players to an event (they can also send no one). There are $\binom{2n}{a}$ ways we can select players, but if we want to only send an even sum of players then $a$ must be even. Lets rewrite it as $\binom{2n}{2k}$. Considering all possible $k$-values tells us that the total ways to send an even sum of randomly composed Blue and Red players is
$$\sum_{k=0}^n \binom{2n}{2k}$$
Since both teams have $n$ players, we know that there are $2^{2n}$ subgroups that could be sent. But again we only want to consider even groups so we divide by 2 to get rid of all odd groups. Thus we have

$$2^{2n}/2 = 2^{2n-1}$$

Both methods give the same result.

答:
Using Pascal's identity:

$$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$$

you can write this as

$$ \binom{2n-1}{0} + \binom{2n-1}{1} + \binom{2n-1}{2} + \dots + \binom{2n-1}{2n-1} = 2^{2n-1}. $$

This proof can be written purely combinatorialy if you wish, just use the combinatorial proofs of Pascal's identity and $2^n = \sum_k \binom{n}{k}$.

You can also get a bijection between even/odd subsets of $\{1,\dots,2n-1\}$ by mapping a subset to its complement (the complement of an even subset is odd and vice versa). This is enough to prove the identity as you've seen and is explained more succinctly in Vlad's answer.


问:
The function $f:(0,1)\to(0,1)$ is defined by the rule $f(x)= \frac{2x}{1+x^2}$. Show that f is a one-to-one and onto function.
The function $f:(0,1)\to(0,1)$ is defined by the rule $f(x)= \frac{2x}{1+x^2}$. Show that f is a one-to-one and onto function. Find the rule $f^{-1}(x)$ of the inverse function $f^{-1}$

I tried to prove this is an injective function by using $f(m) = f(n)$, but the result is false. Help guys!!!

答:
It depends on which analytic methods you are able to use. The most basic, and calculation heavy, would be to give the inverse function.

It is rather standard to compute the inverse.

We have to solve $x=\dfrac{2y}{1+y^2}$ for $y$.

After a few steps we get

$xy^2-2y+x=0$

For $x\neq 0$, where we note that $x\in (0,1)$ in the first place, we can divide by $x$ and get

$y^2-\dfrac2xy+1=0$.

Now we can use the quadratic formula to solve for $y$.

$y_{1,2}=\dfrac1x\pm\sqrt{\dfrac1{x^2}-1}$

As $\dfrac1{x^2}-1\geq 0$ holds for $x\in (0,1)$.

We have to find the correct choice for $y$.
This is $y=\dfrac{1}{x}-\sqrt{\dfrac{1}{x^2}-1}$ as this expression takes values in $(0,1)$.

So our inverse function is given by

$f^{-1}: (0,1)\to (0,1), x\mapsto \frac{1}{x}-\sqrt{\frac1{x^2}-1}$

We would have to calculate now $f(f^{-1}(x))$ and $f^{-1}(f(x))$ and show that is is $x$ respectivly.

I leave that to you.

The calculation might seem complicated, but it is straight forward.
Nevertheless time consuming and there are somethings to look out for. So it is not the nicest solution, but everything involved is taught in school, so it is rather elementary.

Also keep in mind that for the proof of surjectivity, when you go by definition, you have to give an inverse function anyways, or better said, when you try to find a preimage, you will calculate the inverse function anyways. So when you go straight by definition, you kinda have to do this calculation anyways, and can avoid to show that the function is injective, as you get that 'for free' when you find the left and right inverse.


问:
one optimization problem in matrix multiplication and related recurrence
We want to calculate $A_1 \times A_2 \times \cdots \times A_n$, where $A_i$ has dimensions $d_{i-1} \times d_i$.

In the classical matrix chain multiplication problem, we wish to minimize the total number of scalar multiplications. In this problem, we consider the all intermediate matrices arising in the computation (including the final result but excluding the original matrices), and the cost of a specific order is the maximal number of entries of such an intermediate matrix. As usual, we want to minimize the cost.

As an example, if $n = 2$ the answer is simply $d_0d_2$, and if $n = 3$, then there are two orders:

$(A_1A_2)A_3$, in which the intermediate matrices are $A_1A_2,A_1A_2A_3$. The cost is therefore $\max(d_0d_2,d_0d_3)$.

$A_1(A_2A_3)$, in which the intermediate matrices are $A_2A_3,A_1A_2A_3$. The cost is therefore $\max(d_1d_3,d_0d_3)$.

Denote by $C_{ij}$ the optimal cost of multiplying $A_i,\ldots,A_j$. We can write the following recurrence for $C_{ij}$:

$$C_{ij}=\min_{i \leq k <j} \max \{C_{i,k},C_{k+1,j},d_{i-1}d_j\},$$

with base case $C_{ii} = 0$.

How do we get this recurrence?

答:
The recurrence arises by considering the product $A_i \dots A_k$ that includes $A_i$.  If you compute $A_i \dots A_j$ as $(A_i \dots A_k)(A_{k+1} \dots A_j)$, then the first product has cost $C_{i,k}$, the second product has cost $C_{k+1,j}$, and the cost of multiplying the resulting $d_{i-1} \times d_k$ and $d_k \times d_j$ matrices is $d_{i-1} d_j$.  The maximum number of nonzero entries from this choice of $k$ is $\max(C_{i,k},C_{k+1,j},d_{i-1} d_j)$.

Because you are minimizing, choose the $k\in\{i,\dots,j-1\}$ that minimizes this cost:
$$C_{i,j}=\min_{i \le k < j}\max(C_{i,k},C_{k+1,j},d_{i-1} d_j)$$

Note that the third argument of the $\max$ does not depend on $k$, so you could rewrite the recurrence as
$$C_{i,j}=\max\left(\min_{i \le k < j}\max(C_{i,k},C_{k+1,j}),d_{i-1} d_j\right)$$

The $j=i$ case requires no multiplication, so $C_{i,i}=0$.  You want to compute $C_{1,n}$.

For example, when $n=2$, you have $$C_{1,n} = C_{1,2} = \min_{1 \le k < 2} \max(C_{1,k},C_{k+1,2},d_0 d_2) = \max(C_{1,1},C_{2,2},d_0 d_2)=d_0 d_2.$$

For $n=3$,
\begin{align}
C_{1,3} &= \min_{1 \le k < 3} \max(C_{1,k},C_{k+1,3},d_0 d_3) \\
&= \min(\max(C_{1,1},C_{2,3},d_0 d_3),\max(C_{1,2},C_{3,3},d_0 d_3)) \\
&= \min(\max(0,C_{2,3},d_0 d_3),\max(d_0 d_2,0,d_0 d_3)) \\
&= \min(\max(C_{2,3},d_0 d_3),\max(d_0 d_2,d_0 d_3)) \\
\end{align}
You would need to compute $C_{2,3}$ to finish the computation.


答:
No, your approach is incorrect. When you say "we divide by $2$ and get rid of all odd groups" you assert that the number of even groups is equal to the number of odd groups, in other words $$\sum_{k=0}^{n}\binom{2n}{2k}=\sum_{k=0}^{n-1}\binom{2n}{2k+1}$$ which is equivalent to $$2\cdot\sum_{k=0}^{n}\binom{2n}{2k}=\sum_{k=0}^{n}\binom{2n}{2k}+\sum_{k=0}^{n-1}\binom{2n}{2k+1}=2^{2n}$$

You essentially proved that $$\sum_{k=0}^{n}\binom{2n}{2k}=2^{2n-1}\text{ assuming }\sum_{k=0}^{n}\binom{2n}{2k}=2^{2n-1}$$

As @Trevor Gunn pointed out, Pascal's indetity can be used to get the result.

Title: Show that $\sum_{k=0}^n \binom{2n}{2k} = 2^{2n-1}$

Body:
I have played around with this question for a bit, but my answer doesn't feel right. I'm not happy with how I converted the binomial. it feels a bit hand-wavy. Is there any way I can make this proof better?

Let's say that we have two teams that are both size $n$. Let's call one team the Blue team and the other team Red. We are going to send a mixture of Blue and Red players to an event (they can also send no one). There are $\binom{2n}{a}$ ways we can select players, but if we want to only send an even sum of players then $a$ must be even. Lets rewrite it as $\binom{2n}{2k}$. Considering all possible $k$-values tells us that the total ways to send an even sum of randomly composed Blue and Red players is
$$\sum_{k=0}^n \binom{2n}{2k}$$
Since both teams have $n$ players, we know that there are $2^{2n}$ subgroups that could be sent. But again we only want to consider even groups so we divide by 2 to get rid of all odd groups. Thus we have

$$2^{2n}/2 = 2^{2n-1}$$

Both methods give the same result.


问:
Contour Integral: $\int^{\infty}_{0}(1+z^n)^{-1}dz$
I'm working through Priestley's Complex Analysis (really good book by the way) and this Ex 20.2:

Evaluate $\int^{\infty}_{0}(1+z^n)^{-1}dz$ round a suitable sector of angle $\frac{2\pi}{n}$ for $n=1,2,3,...$

Can someone advise what the contour may be? If we use a sector that includes zero, surely we'll have to indent it to avoid the singularity (since it isn't a simple pole on account of the negative power).

Thanks.

答:
The suitable contour is already given: a circle sector, with center at the origin, with one "edge" along the positive real axis and an opening angle of $2\pi/n$.

The function has simple poles at points where $z^n = -1$, i.e. at $z = \exp(i\pi/n + 2\pi i k/n)$, and only one of these lie inside the suggested contour. (None on the boundary.)


问:
Where did I go wrong differentiating arccos(4/x)?
I've been trying to differentiate $y=\arccos(4/x)$ and ended up with the answer $\frac{dy}{dx} = \frac{4}{x\sqrt{x^2-16}}$ however the solution to this question is marked as $\frac{dy}{dx}= \frac{4}{|x|\sqrt{x^2-16}}$. I opted to use implicit differentiation and honestly have no idea where the $|x|$ could of come from. Any help would be greatly appreciated.

答: Clearly,\begin{align}\frac{\mathrm dy}{\mathrm dx}&=\left(-\frac4{x^2}\right)\left(-\frac1{\sqrt{1-\frac{16}{x^2}}}\right)\\&=\frac4{x^2}\cdot\frac1{\sqrt{1-\frac{16}{x^2}}}\end{align}and\begin{align}\frac4{x^2}\cdot\frac1{\sqrt{1-\frac{16}{x^2}}}&=\frac4{|x|}\frac1{|x|\sqrt{1-\frac{16}{x^2}}}\\&=\frac4{|x|\sqrt{x^2-16}}.\end{align}Note that the equality $a\sqrt b=\sqrt{a^2b}$ only holds if $a\geqslant0$.


问:
Showing that $G^n$ is a normal subgroup of $G$
This is from exercise 2.55 "An Introduction to the Theory of Groups" by Joseph J. Rotman.

Let $G$ be a finite group, such that for some fixed integer $n>1$,for all $x,y\in G$, $(xy)^n = x^ny^n$. I need to prove that $G^n=\{x^n:x\in G\}$ is a normal subgroup of $G$.

I could prove that $G^n$ is a subgroup but i can't figure out the normal part.

答: Let $x \in G^n$. For $h \in G$, consider $h^{-1}xh.$ Write $x=g^n$ for some $g$. Then $h^{-1} x h=h^{-1}g^nh =(h^{-1} g h)^n$ because all of the $h$ terms in the middle cancel. Hence, $h^{-1} x h\in G^n$, and so $G^n$ is normal.


问:
Use of exponentiation in topos logic
I'm learning about topos theory and its many uses in (intuitionistic) logic. The definition of an elementary topos $\mathbf{C}$ requires the category to have exponentiation, that is, given $\mathbf{C}$-objects $A$ and $B$, there exists another $\mathbf{C}$-object $B^A$ which is "the object of all $A\to B$ morphisms".

Now, when developing the propositional logic, the valuations, the meet/join/negation/implication operations in the algebra of subobjects I haven't seen the exponential construction at all.

I'm wondering why is it crucial that a categroy has exponential objects (or power objects) in order to "do" strong enough internal logic, or similar enough to the $\mathbf{Set}$ topos internal logic.

What construction, rule of inference, quantification, formula or whatever, needs exponentiation/power objects?

答:
Your observation that Cartesian closedness of $\mathbf C$ is not relevant to getting a Heyting algebra structure on the subobject lattices, and even the adjoints $\forall$ and $\exists$ to the substitution morphisms between subobject lattices, is entirely correct. Indeed, many non-Cartesian closed $\mathbf C$ have a perfectly good internal first-order logic (namely, Heyting categories.)

Instead, the core reason for the cartesian closedness of a topos is to allow the internalization of higher-order logic. In a mere Heyting category, you can't define, for instance, a topological space. In a topos, you can formalize things like "a family of subsets of $X$ closed under intersection and finite union" as a nice subobject of $\mathcal P(X)$, where the "powerobject" of $X$ is defined as $\Omega^X$, $\Omega$ being the subobject classifier required in the definition of a topos.

Title: Use of exponentiation in topos logic


答:
Equation
$$f(x)=\frac{2x}{1+x^2}=\frac{2y}{1+y^2}=f(y)$$
is same as $xy^2-(1+x^2)y+x=0$  and have solutions only $(1,1)$ and $(0,0)$.

For any $y \in $ $0<\frac{2x}{1+x^2}=y<1$ we have $yx^2-2x+y=0$, $x_{1,2}=\frac{1}{y}(1\pm\sqrt{1-y^2})$ on $(0,1)$ and we have $x_1\cdot x_2=1$, which means, that only one from them is on $(0,1)$.

Body:
The function $f:(0,1)\to(0,1)$ is defined by the rule $f(x)= \frac{2x}{1+x^2}$. Show that f is a one-to-one and onto function. Find the rule $f^{-1}(x)$ of the inverse function $f^{-1}$

I tried to prove this is an injective function by using $f(m) = f(n)$, but the result is false. Help guys!!!


问:
Morphism of vector bundles in algebraic geometric terms
Hartshorne doesn't give a definition of morphism of vector bundles but only that of isomorphism between vector bundles. After looking at a few online references, I think I understand the following:
Let $V\xrightarrow{f}X$ and $V'\xrightarrow{f'}X$ be two vector bundles of rank $n$ and $m$ respectively on a scheme $X$. Then a morphism of vector bundles is a morphism of $X$-schemes $\phi:V \longrightarrow V'$ such that the induced map $V|_{U_i}\cong\mathbb{A}_{U_i}^n \longrightarrow \mathbb{A}_{U_i}^m \cong V'|_{U_i}$ is linear for all $U_i$  where $\{U_i\}_{i \in I}$ is a cover of $X$ which provides local trivializations for both $V$ and $V'$.

I am not sure if my understanding is correct. Please correct me if I am wrong. I do know the differential geometry version of the definition but I want a definition of morphism of vector bundles corresponding to the way Hartshorne has defined vector bundles (II,Ex:5.18).

Thanks in advance!

答:
Let $X$ be a scheme and be $E,F$ finite rank locally trivial $\mathcal{O}_X$-modules.
Assume for simplicity that $X:=Spec(A)$ and $E:=A^n,F=A^m$ are free $A$-modules of rank $n,m$. An $A$-linear map $\phi: E \rightarrow F$ induce an $A$-linear map $\phi^*: F^* \rightarrow E^*$ by $\phi^*(f):= f\circ \phi$. We get an induced map of $A$-algebras

$\phi^*: Sym_A(F^*) \rightarrow Sym_A(E^*)$.

Passing to schemes we get an induced map of "geometric vector bundles" (a morphism over $X$)

$\mathbb{V}(\phi): \mathbb{V}(E):=Spec(Sym_A(E^*)) \rightarrow \mathbb{V}(F):=Spec(Sym_A(F^*))$

and I believe the map $\mathbb{V}(\phi)$ is a "map of geometric vector bundles". It is a "linear map" in the sense that if you choose an element $f\in F^*$ it follows $\phi^*(f) \in E^*$. This construction is "functorial" in the sense that if $E,F$ are $A$-modules and $\phi$ is an arbitrary $A$-linear map, you may always define $\mathbb{V}(\phi)$. In Hartshorne Ex.II.5.18 some results on this construction are proved. It may be that with this definition it follows the category of finite rank locally trivial sheaves and morphism of $\mathcal{O}_X$-modules is equivalent to the category of geometric vector bundles and maps of geometric vector bundles, but this has to be checked. I guess you may find it in the litterature but at the moment I do not have a precise reference.

Example: If $k$ is an algebraically closed field of characteristic zero and if $G:=SL(V)$
is the special linear group on a finite dimensional $k$-vector space $V$ and $H \subseteq G$ a closed subgroup, you may always realize the "quotient" $G/H$ as a smooth quasi projective scheme of finite type over $k$. There is a canonical projection morphism $\pi: G \rightarrow G/H$ and $\pi$ is locally trivial in the etale topology in general. If $H$ is a parabolic subgroup of $G$, it follows $\pi$ is locally trivial in the Zariski topology with fiber $H$. In this case it follows the quotient $G/H$ is a smooth projective scheme of finite type over $k$. The map $\pi$ is a principal fiber bundle with fiber $H$, hence $\pi$ is not a "geometric vector bundle" in the above sense since $H$ is not an affine space.

Example: In the case when $H$ is a parabolic group it follows $H$ is classified. There is a flag of $k$-vector spaces $V_{\bullet} \subseteq V$ with the property that $H$ is the subgroup of $SL(V)$ fixing the flag. In this case the quotient $SL(V)/H$ is the flag variety $F(V_{\bullet})$ of $V$ of type $V_{\bullet}$. The $k$-rational points of $F(V_{\bullet})$ parametrize the set of flags in $V$ of type $V_{\bullet}$. In particular if $V_{\bullet}:=W \subseteq V$ where $W$ is a $d$-dimensional sub space, it follows $F(V_{\bullet})\cong \mathbb{G}(d,V)$ is the grassmannian of $d$-planes in $V$. Hence for the grasmannian here is a canonical projection map

$\pi: SL(V) \rightarrow \mathbb{G}(d,V)$

which is locally trivial in the Zariski topology with fiber $H$. Using $\pi$ me may associate to any linear representation $\rho: H \rightarrow GL_k(W)$
a vector bundle $\pi(\rho): E(\rho)\rightarrow \mathbb{G}(d,V)$ with a left $SL(V)$-action such that the map $\pi(\rho)$ is $SL(V)$-invariant. This sets up an equivalence of categories between the category of finite dimensional $k$-linear $H$-representations and the category of $SL(V)$-linearized vector bundles on the grassmannian. Because $\pi$ is locally trivial in the Zariski topology, it follows $\pi(\rho)$ is locally trivial in the Zariski topology. And from Hartshorne exercise II.5.18 there is a finite rank locally trivial sheaf $\mathcal{E}$ on $\mathbb{G}(d,V)$ with $E(\rho) \cong \mathbb{V}(\mathcal{E}^*)$.

Example. This construction relates the representation theory of $SL(V)$ to the geometry of the flag variety $F(V_{\bullet})$. Given any finite dimensional irreducible $SL(V)$-module $(W,\rho)$, there is a parabolic subgroup (not unique) $P \subseteq SL(V)$ and an invertible sheaf $\mathcal{L}(\rho)$ in $Pic(SL(V)/P)$ with an $SL(V)$-linearization and an isomorphism

$W \cong \operatorname{H}^0(SL(V)/P, \mathcal{L}(\rho))$

of $SL(V)$-modules - this is the Borel-Weil-Bott theorem.

Example. Let $k$ be a field of characteristic zero and let $V:=k\{e_0,e_1$} and $V^*:=k\{x_0,x_1\}.$ Let $l \subseteq V$ be a line and let $P \subseteq SL(V)$ be the parabolic subgroup fixing $l$. It follows the quotient $SL(V)/P$ is isomorphic to the projective line $\mathbb{P}^1_k$. For any integer  $d \geq 1$ it follows the symmetric product $Sym^d(V^*)$ is an irreducible finite dimensional $SL(V)$-module, and these are all such modules. Let $\mathcal{O}(d)$ be the invertible sheaf corresponding to $d$. It follows $\mathcal{O}(d)$ has a canonical $SL(V)$-linearization and there is a canonical isomorphism ($d\geq 1$)

$Sym^d(V^*) \cong \operatorname{H}^0(SL(V)/P,\mathcal{O}(d))$

of $SL(V)$-modules. Hence all irreducible $SL(V)$-modules may be realized as global sections of invertible sheaves on the projective line.

In general all finite dimensional irreducible $SL(V)$-modules for any $V$ may be realized geometrically as global sections of certain invertible sheaves on the flag variety $F(V_{\bullet})$ for some flag $V_{\bullet}$. This is a much studied subject involving combinatorics, geometry and representations.

Question: "So what would be a correct definition of morphism of vector bundles in algebraic geometry terms in Zariski topology? – Sam"

Answer: The map

$\phi: f^{-1}(U) \cong \mathbb{A}^n_{U} \cong f'^{-1}(U) $

should be a map

$f:A[x_1,..,x_n] \rightarrow A[x_1,..,x_n]$

with $f(x_k)=\sum_i a_{ik}x_i$ with $a_{ik} \in A$ and $U:=Spec(A)$ a local trivialization of $V,V'$. To specify a map $f$ of $A$-algebras is equivalent to giving a set of $n$ polynomials $p_1(x_i),..,p_n(x_i)$. You define

$f(x_l):=p_l(x_i)$.

The ring homomorphism $f$ is "linear" iff $p_l(x_i)\in A[x_1,..,x_n]$ is a "linear polynomial" for $i=1,..,n$.


问:
How do I integrate this function? I've tried every method I can think of, help!
I want to calculate the value of the integral $\int_{40}^{64} y^2(100y-4000)^{1/2} dy$

答:
It is pretty simple. Substitute $u = 100y - 4000$.
$$du = 100 dy$$
$$dy = du/100$$
(I didn't want to change the bounds because big number look bad).
Also, we can say $y^{2} = ((u+4000)/100)^{2}$.
Now, your new integral:
$$1/100\int ((u+4000)/100)^{2}\sqrt(u) du$$
Now, can you integrate this and sub u back and then definitely integrate from 40 to 64?


答:
We have, for any $g\in G$, $h=k^n\in G^n$ for $k\in G$, that

$$\begin{align}
ghg^{-1}&=gk^ng^{-1}\\
&=g\underbrace{k\dots k}_{n\text{ times}}g^{-1}\\
&=\underbrace{(gkg^{-1})\dots (gkg^{-1})}_{n\text{ times}}\tag{1}\\
&=(gkg^{-1})^n\\
&\in G^n,
\end{align}$$

where $(1)$ holds by inserting $e=g^{-1}g$ between the $k$s. Hence $ghg^{-1}\in G^n$.

Hence $G^n\unlhd G$.

Title: Showing that $G^n$ is a normal subgroup of $G$

Body:
This is from exercise 2.55 "An Introduction to the Theory of Groups" by Joseph J. Rotman.

Let $G$ be a finite group, such that for some fixed integer $n>1$,for all $x,y\in G$, $(xy)^n = x^ny^n$. I need to prove that $G^n=\{x^n:x\in G\}$ is a normal subgroup of $G$.

I could prove that $G^n$ is a subgroup but i can't figure out the normal part.


问:
$O$ is not dense in $\mathbb R^n$
$A$ is a matrix of order $n$, and $x$ is an $n$-dimensional column vector. The set
$$
O=\{x,Ax,A^2x,\cdots,A^mx,\cdots\}
$$
Prove that, $O$ is not dense in $\mathbb R^n$.

This claim seems to be trivial: it just looks like $\mathbb N$ in $\mathbb R$. I recalled how we verify the denseness of $\mathbb Q$ in $\mathbb R$, but it seeems that the same method doesn't work. Can anyone help?

答:
Suppose $O=\{A^mx:m\in\mathbb N_0\}$ is dense in $\mathbb R$. There are two possibilities:

$A^T$ has a real eigenvector $y$ corresponding to a real eigenvalue $\lambda$. Then
\begin{aligned}
&\ O=\{A^mx:m\in\mathbb N_0\}\text{ is dense in $\mathbb R^n$}\\
\Rightarrow&\ y^TO=\{\lambda^m y^Tx:m\in\mathbb N_0\}\text{ is dense in $\mathbb R$}\\
\Rightarrow&\ Y=\{\lambda^m:m\in\mathbb N_0\}\text{ is dense in $\mathbb R$}.
\end{aligned}

$A^T=A^\ast$ has an eigevector $z$ corresponding to some non-real eigenvalue $\overline{\lambda}$. Then the $\mathbb R$-linear span of the entries of $z$ must be two-dimensional and hence it is equal to $\mathbb C$. Therefore,
\begin{aligned}
&\ O=\{A^mx:m\in\mathbb N_0\}\text{ is dense in $\mathbb R^n$}\\
\Rightarrow&\ z^\ast O=\{\lambda^mz^\ast x:m\in\mathbb N_0\}\text{ is dense in $\mathbb C$}\\
\Rightarrow&\ Z=\{\lambda^m:m\in\mathbb N_0\}\text{ is dense in $\mathbb C$}.
\end{aligned}

It remains to show that $Y$ and $Z$ aren't really dense in $\mathbb R$ and $\mathbb C$ respectively. You may continue from here.

Title: $O$ is not dense in $\mathbb R^n$


问:
Prove that $\cos2\theta−\sqrt{3}\sin2\theta \equiv2 \cos (2\theta+\pi/3 )\equiv−2\sin(2\theta−\pi/6)$
Going from $2\cos(2\theta+\pi/3)$ to $\cos2\theta−\sqrt{3}\sin2\theta$ is simple enough, however I'm stuck on going from $2\cos(2\theta+\pi/3)$ to $−2\sin(2\theta−\pi /6)$. How do i do this?</p>

答:
First, let's recap:

$\sin{\pi\over3}=\frac{\sqrt{3}}{2}$

$\cos{\pi\over3}=\frac12$

$\sin{\pi\over6}=\frac12$

$\cos{\pi\over6}=\frac{\sqrt{3}}{2}$

Let's rock 'n roll!

Solution 1:
Going from $\cos2\theta−\sqrt{3}\sin2\theta$ to $2\cos\left(2\theta+\frac\pi3\right)$.
$$
\require{cancel}
\begin{align}
\cos2\theta−\sqrt{3}\sin2\theta&=\cos^2\theta-\sin^2\theta-\sqrt{3}\cdot2\sin\theta\cos\theta\\
&=2\cdot\frac12\left(\cos^2\theta-\sin^2\theta-\sqrt{3}\cdot2\sin\theta\cos\theta\right)\\
&=2\left(\frac12\left(\cos^2\theta-\sin^2\theta\right)-\frac12\sqrt{3}\cdot2\sin\theta\cos\theta\right)\\
&=2\left(\left(\cos^2\theta-\sin^2\theta\right)\cos\frac{\pi}{3}-2\sin\theta\cos\theta\sin\frac\pi3\right)\\
&=2\left(\cos2\theta\cos\frac\pi3-\sin2\theta\sin\frac\pi3\right)\\
&=2\cos\left(2\theta+\frac\pi3\right)
\end{align}
$$
Solution 2:
Going from $\cos2\theta−\sqrt{3}\sin2\theta$ to $\sin(2\theta−\frac\pi6)$.
$$
\begin{align}
\cos2\theta−\sqrt{3}\sin2\theta&=\cos^2\theta-\sin^2\theta-\sqrt{3}\cdot2\sin\theta\cos\theta\\
&=-2\cdot\left(-\frac12\right)\left(\cos^2\theta-\sin^2\theta-\sqrt{3}\cdot2\sin\theta\cos\theta\right)\\
&=-2\left(\left(-\frac12\right)\left(\cos^2\theta-\sin^2\theta\right)-\left(-\frac12\right)\left(\sqrt{3}\cdot2\sin\theta\cos\theta\right)\right)\\
&=-2\left(\left(-\frac12\right)\left(\cos^2\theta-\sin^2\theta\right)-\left(-\frac12\sqrt{3}\right)\left(2\sin\theta\cos\theta\right)\right)\\
&=-2\left(-\sin\frac\pi6\left(\cos^2\theta-\sin^2\theta\right)-\left(-\cos\frac\pi6\right)\left(\sin\theta\cos\theta\right)\right)\\
&=-2\left(-\sin\frac\pi6\left(\cos^2\theta-\sin^2\theta\right)+\cos\frac\pi6\left(2\sin\theta\cos\theta\right)\right)\\
&=-2\left(\cos\frac\pi6\left(2\sin\theta\cos\theta\right)-\sin\frac\pi6\left(\cos^2\theta-\sin^2\theta\right)\right)\\
&=-2\left(\cos\frac\pi6\sin2\theta-\sin\frac\pi6\cos2\theta\right)\\
&=-2\left(\sin2\theta\cos\frac\pi6-\cos2\theta\sin\frac\pi6\right)\\
&=-2\sin\left(2\theta-\frac\pi6\right)
\end{align}
$$
I hope this helps.


问:
2020 Australian Maths Competition Senior Problem 30
When I drive to school every day, I pass eight traffic lights, each either green, yellow,
or red. I find that, because of synchronization, a green light is always followed
immediately by a yellow, and a red light is never immediately followed by a red.
Thus a sequence of lights may start with GYRY, but not RRGG.
How many possible sequences of the eight lights are there?</p>

答:
I’m not good at math or English. I hope the following notations are not confusing, and my answer helps.

Let $f(n)$ represents the number of all possible sequences of $n$ lights.

If $n=0$, then the sequences are $\{ \emptyset \}$. Thus

$$
f(0) = 1
$$

If $n=1$, then the sequences are $\{ R , G, Y \}$. Thus

$$
f(1) = 3
$$

If $n=2$, then the sequences are $\{ RG, RB, GY, YR, RG, YY \}$. Thus,

$$
f(2) = 6
$$

If $n \ge 3$, then

$$\begin{align}
f(n) &= f_{R}(n-1) + f_{G}(n-1) + f_{Y}(n-1)\\
&= f_{RG}(n-2) + f_{RY}(n-2) + f_{GY}(n-2) + f_{Y}(n-1)\\
&= f_{RGY}(n-3) + f_{RY}(n-2) + f_{GY}(n-2) + f_{Y}(n-1)\\
&= f(n-3) + 2f(n-2) + f(n-1)\\
\end{align}
$$

I wrote a function in Python to compute the result. It told me $f(8)=595$.