问:
Can there be involutions in the unit group of $\mathbb{Z}_n$?
Let $x\neq 1$ be an element of the multiplicative group $\mathbb{Z}_n^*$.  Then my question is, is it possible for $x^2$ to equal $1$?

For what $n$ is it possible or impossible?

答:
Firstly, and most importantly, your title is incorrect.  Idempotents are elements $x\in G$ such that $x^2=x$, and it is an exercise in elementary group theory that the only idempotent in a group is the identity.  As has been mentioned in the comments, it looks like you mean to say involutions instead of idempotents, so I answer on this assumption.

In general this is very possible, consider, for example, $4$ modulo $15$ (or indeed $(-1)^2=1$ always as is noted above).  The important thing is that involutions are elements of order $2$.  The general structure of $\mathbb{Z}/n\mathbb{Z}^\times$ is well known, (see e.g. the wikipedia page or most good introduction to groups/number theory books).  I give it below:

Firstly, for an odd prime $p$ and positive integer $k$ we have
$$\mathbb{Z}/p^k\mathbb{Z}^\times\cong \mathbb{Z}/(p^k-p^{k-1})\mathbb{Z}.$$

Secondly, at powers of $2$ we have
$$\mathbb{Z}/2^k\mathbb{Z}^\times\cong
\begin{cases}
\{1\}&k=1\\
\mathbb{Z}/2&k=2\\
\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2^{k-2}\mathbb{Z} &k\geq 3
\end{cases}$$

The general case then follows from the Chinese remainder theorem, which says that if $n=\prod_{i=1}^rp_i^{k_i}$ is the unique prime decomposition of $n$ then
$$\mathbb{Z}/n\mathbb{Z}^\times=\prod_{i=1}^r\mathbb{Z}/p_i^{k_i}\mathbb{Z}^\times$$

From the structure of the unit group as a product of cyclic groups in this way, it is easy to see how often (and even how many) involutions appear.


问:
Variation problem
How many strings (no longer than 8 letters) can be created from letters A, B and C so that the letter B appears as many times as A and C appear together?

I realised that if letter A appears i times, and B appears k times, then C has to appear k-i times. But since a problem takes in consideration strings of length 0-8 I am not sure how to proceed.

答:
Nice, so your string will have $2k$ of length.

Hint:

What if you go $k=0,1,2,3,4.$ Notice that fixing $k$ you have to choose $k$ spots for $B$ and then you are free to place $A'$s and $B'$s in $2^{2k-k}$ ways.

ContentLicense: CC BY-SA 4.0

Title: Variation problem


问:
Problem of minimum time for a falling mass.
A mass falls from a point A (height h) to a point P (height $0$) and then it continues to move with the speed acquired in P.
I have to find the position of the point P on the ground in order to have the minimum time.
I take x the lenght on the ground of A from P.

$E_A=mgh$ (the mass falls down and it has initial speed $0$).

$E_P= \frac{1}{2}m (v_P)^2$ (P is on the ground amd I've taken the ground as my reference plane).

$E_A=E_P \Rightarrow (v_P)^2=2gh \Rightarrow v_P= \sqrt{2gh}$.

This velocity remains constant until the mass reaches the final point B.

$v_P=v_A+gt_1 \Rightarrow t_1= \frac{v_P-v_A}{g}= \frac{\sqrt{2gh}-0}{g}=\sqrt{\frac{2h}{g}}$

$t_2= \frac{L-x}{\sqrt{2gh}}$

$ t_{tot}=t_1+t_2= \sqrt{\frac{2h}{g}} + \frac{L-x}{\sqrt{2gh}} = f(x)$

But  $\frac{df}{dx}=0 $ is never verified
and I don't know how to procede.

In the solution on the book I don't understand why it says that $x$ is the minimum of $L$ and $\frac{h}{\sqrt  3}$

and it analyze the cases of $h \le \sqrt3 l $ and of  $h > \sqrt3 l $.

答:
The issue is in your equation $$v_P=v_A+gt_1$$
The acceleration is not $g$, but the component of $g$ along the $AP$ line. You need to decompose $g$ into components parallel and perpendicular to $AP$. If $\alpha=\angle APC$, then $$g_{||}=g\sin\alpha$$
From $\triangle APC$, $$\sin\alpha=\frac h{\sqrt{h^2+x^2}}$$
Can you continue from here?


问:
Series of probabilities
Consider $Y_k=X_1+...+X_k $, where $X_k \in \mathbb{N}_0$ are i.i.d random variables and $E[X_1]<1$

$$\sum_{j=1}^{\infty} P(Y_j=j) \overset{!}{=}1 $$

How can I verify that this equation is true or does this not hold?

答:
Take $X_i$ iid Bernoulli s.t. $$\mathbb P\{X_i=1\}=p\in (0,1)\setminus \left\{\frac{1}{2}\right\}.$$

Then, $\mathbb E[X_1]=p<1$ and $Y_k$ are Binomial $(k,p)$. Therefore, $$\mathbb P\{Y_j=j\}=p^j.$$

Furthermore $$\sum_{j=1}^\infty p^j=\frac{p}{1-p}\neq 1.$$

Title: Series of probabilities


问:
Royden's proof of a fixed-point theorem
As in the title, Royden proves a fixed-point theorem as a prelude to the Kakutani's fixed point theorem:

$\pi:G\to GL(E)$ is a representation of the compact group $G$ on $E$, a Banach space.  Suppose that $g\mapsto \pi(g)x$ is continuous for each $x\in E$, where we give $E$ the norm topology. Further, suppose that there is a non-empty convex, weak$^*$- compact $K^*\subseteq E^*$ such that $\pi^*(g)(K^*)\subseteq K^*.$  Then the conclusion is that there is a functional $\psi\in K^*$ such that $\pi^*(g)(\psi)=\psi.$ (Note that $\pi^*(g)(\psi)=\psi\circ \pi(g^{-1})).$

The proof proceeds along the lines of Krein-Milman, and is fairly straightforward, but there is an estimate that I can't verify. Having chosen a particular $x_0\in E$ (with a property that, for the purposes of my question, is not necessary to state), we have that

$1).\ K^*$ is bounded and $p$ is weak$^*$- continuous, where

$2).\ $ the functional $p(\psi)=\sup\{\psi(\pi(g)x_0):g\in G\}$ for $\psi\in K^*.\ p$ is positive homogeneous and subadditive. Consequently, the balls

$3).\ B_0(\eta,r)=\{\psi\in K^*: p(\eta -\psi)<r\}$ are open in the weak$^*$- topology on $K^*.$ So,

$4).\ d=\sup \{p(\psi-\varphi):\psi,\varphi\in K^*\}$ is properly defined. Now, compactness gives us $\{\psi_1,\cdots,\psi_n\}$ such that $K^*=\bigcup B(\psi_k,d/2).$ Put

$5).\ \psi^*=\frac{\psi_1+\cdots + \psi_n}{n}\in K^*$ Then, if $\psi\in K^*,$ we must have $\psi\in B(\psi_{k_0},d/2)$ for some $1\le k_0\le n$ so

$6).\ p(\psi-\psi^*)\le p(\psi-\psi_{k_0})+p(\psi^*-\psi_{k_0})\le \frac{d}{2}+\frac{n}{n-1}d\color{red}{<d}$, where the red indicates what Royden has. It's obviously a typo. I guess he means $\frac{d}{2}+\frac{n}{n-1}\frac{d}{2}<d$ but when I do the calculation I get $\frac{d}{2}+p(\frac{\psi_1+\cdots + \psi_n}{n}-\psi_{k_0})=\frac{d}{2}+\frac{1}{n}p(\psi_1+\cdots + \psi_n-n\psi_{k_0})<\frac{d}{2}+\frac{n-1}{n}d$ which is what Royden has. How can we patch this up?

答:
The step sketched is indeed incorrect. A correct version could read:

$$p(\psi - \psi^*) = p\left(\frac1n\sum_k (\psi-\psi_k)\right)≤ \frac1np(\psi - \psi_{k_0})+ \frac1n \sum_{k\neq k_0}p(\psi -\psi_k)≤\frac1n \frac d2+\frac{n-1}nd <d$$

Title: Royden's proof of a fixed-point theorem


问:
Obtain $x$ from exponential equation
I'm looking for $x$ from the following equation:

$$\frac1{(1+x)^{\frac1{p}}}+\frac1{(1+x)^{\frac2{p}}} + \cdots \frac1{(1+x)^{\frac{n}{p}}} = 0$$

Or, what's the same:

$$\sum_{i=1}^n\frac1{(1+x)^{\frac{i}{p}}} = 0 $$

where $n$ and $p$ are given (say, $n=16$ and $p=52$, for example)

Is there a way to get $x$ from this?

ContentLicense: CC BY-SA 3.0

Title: Obtain $x$ from exponential equation

Body: $$\sum_{k=1}^{n}\frac{1}{(1+x)^{k/p}}=\sum_{k=1}^{n}((1+x)^{-1/p})^k=(1+x)^{-1/p}\frac{1-(1+x)^{-(n+1)/p}}{1-(1+x)^{-1/p}}$$


问:
Eigenvalues and eigenvectors of the operator $J$.
I am reading a linear algebra book. Assume $V$ is a real vector space. If $J : V \rightarrow V$ is linear such that $J^2 = -\mathit{id}$, it follows that the eigenvalues are positive or negative $i$.

What is $i$ in this case? How did he calculate the eigenvectors? What about the eigenvalues? The matrix of $J$ isn't known. This essentially is what is confusing me.

答: That $i$ is the complex number $i$. If $\lambda$ is an eigen value of $J$ and if $v$ is an eigenvector corresponding to the eigenvalue $\lambda$, then$$J\bigl(J(v)\bigr)=J(\lambda v)=\lambda J(v)=\lambda^2 v.$$But $J\bigl(J(v)\bigr)=-v$ too. So, $\lambda^2=-1$, and therefore $\lambda=\pm i$.


问:
Proving the uniform convergence for a series
Let $a_n$ be a sequence which tend to $0$. I want to prove that $$\sum_{n=1}^{\infty}a_nx\left(\frac{\sin(nx)}{nx}\right)^2$$
converges uniformly for all $x \ne 0$. I thought that using summation by parts could help here, but I'm not sure how to proceed any further. What do you suggest?

There's a similar question here, where $a_n$ is replaced by $1$. It's proven that in this case the series does not converge uniformly.

答:
Your question comes from the following series inequality when $0 <x$,
$$ 0\le \sum_{n=1}^{\infty} x \left( \dfrac{ \sin (nx)}{nx} \right)^2  \le 3$$
From that, you can bound your uniform converge by $2\sup_{k \ge n} |a_k|$

As this ineq is pretty loose, it's now up to your choice to decide what to do with it. Either you find the explicit formula for it, or either you can evaluate it as follows:

Proof for the above ineq

 If $x \ge 1$, the inequality is straight forward.

 Then for any $0<x<1$, we can find an positive integer $N_x \ge 2$, such that $$N_x\ge \dfrac{1}{x}  \ge N_x-1$$
By then, you'll observe that:
$$ \sum = \sum_{n=1}^{N_x} x \left( \dfrac{ \sin (nx)}{nx} \right)^2 +\sum_{n=N_x+1}^{\infty} x \left( \dfrac{ \sin (nx)}{nx} \right)^2  \le (N_x x ) + \dfrac{1}{x}\left( \sum_{n \ge N_x+1} \frac{1}{n^2} \right) \le \frac{N_x}{N_x-1} +N_x\dfrac{1}{N_x}\le 3$$
Done.


问:
Show that if $u_n + u_{n+1} \sim \dfrac{1}{n}$ then $u_n \to 0$
Saw this question in my textbook : Let $(u_n)$ be a real decreasing sequence such that :

$$u_n + u_{n+1} \sim \dfrac{1}{n}$$
Show that $(u_n)$ converges to $0^{+}$ and give a simple sequence equivalent to $(u_n)$

My attempt : for every $\epsilon > 0$, we have for every $n\ge N $ :$$2u_n < u_n + u_{n+1} \le |u_n +u_{n+1}|\le \epsilon$$
It seems like I need to show that the terms of this sequence are all positive for this to work, I just don't know how to procede.

答:
Your intuition on non-negativeness of the terms of $u_n$ is correct, since if $u_{n_0}<0$ for some $n_0$ then
$$
u_n+u_{n+1}<2u_{n_0}\implies u_n+u_{n+1} \not\sim{1\over n}\quad,\quad n>n_0
$$


问:
Disc with two points identified
Is a disc $D^2$ with two points on the boundary identified, same as $D^2 \vee D^2$ ? They both have boundary $S^1 \vee S^1$. I am confused because an exercise in Hatcher seems to ask the same question about the two spaces (i.e. to prove that there doesn't exist a retraction from each of this space to $S^1 \vee S^1$).</p>

答: If you delete one point from the first space it is still connected. However if you delete the wedge point from $D^2 \vee D^2$, the resulting space is disconnected.</p>

Title: Disc with two points identified


问:
Minimal value of $\mathbb{E}[(X-g(Y))^2]$.
In a test I have two true/false question and I'm not sure if I got it properly:

Given X and Y as two random variable (we don't know if indipendent), with $g:\mathbb{R}\to\mathbb{R}$:

$g(Y)=\mathbb{E}[X|Y]$ makes $\mathbb{E}[(X-g(Y))^2]$ minimal

$c=var[X]$ makes $\mathbb{E}[(X-c)^2]$ minimal

Now, in both cases the second part resemble the formula of the variance, but I don't think I need it.

Due to the fact that both $\mathbb{E}[(X-g(Y))^2]$ and $\mathbb{E}[(X-c)^2]$ are the expected value of a square, the minimal value that they can get is zero, and they get it when $g(Y)=X$ and $c=X$.

So both the the answers should be false, but I'm not really sure if I'm missing something, can you confirm that my logic is right?

答: The key point here is that when trying to define an estimator function such as $g(\cdot )$, we should note it to be deterministic rather than stochastic. The substitution $g(X)=X$ may sound reasonably good, but impractical due to it uselessness because we don't know the exact amount of $X$. Instead, we can exploit additional information, say $Y$, to make a best possible (from the aspect of minimum error variance) estimation of $X$. The more $Y$ is dependent to $X$, the better the estimator is .


问:
How can we find a good upper bound for $\sum_i \frac{a_i}{b_i}$ in terms of $\frac{\sum_i a_i}{\sum_i b_i}$?
I've been thinking about inequalities of the kind:

$$\frac{\sum_i a_i}{\sum_i b_i}\leq \sum_i \frac{a_i}{b_i} \qquad a_i,b_i>0$$

For some cases, it is easy to see why it works, for example:

$$\frac{a_1+a_2}{b_1+b_2}\leq \frac{a_1}{b_1}+\frac{a_2}{b_2}\qquad a_i,b_i>0$$

We write

$$0\leq a_1b_2^2+a_2b_1^2$$

And then:

$$a_1b_1b_2+a_2b_1b_2 \leq a_1b_1b_2+a_2b_1b_2+ a_1b_2^2+a_2b_1^2$$

Finally:

$$b_1b_2(a_1+a_2)\leq (a_1b_2+a_2b_1)(b_1+b_2)$$

Which yields the desired inequality:

$$\frac{a_1+a_2}{b_1+b_2}\leq \frac{a_1}{b_1}+\frac{a_2}{b_2}$$

I am curious about the following:

How can we find a good upper bound for $\sum_i \frac{a_i}{b_i}$ in terms of $\frac{\sum_i a_i}{\sum_i b_i}$? In my hypothetical application for this, obtaining the number $\frac{\sum_i a_i}{\sum_i b_i}$  is very easy but obtaining $\sum_i \frac{a_i}{b_i}$ would be very messy. Having a good upper bound in terms of $\frac{\sum_i a_i}{\sum_i b_i}$ would be desirable. I know next to nothing about inequalities. I guessed about squares, cubes of $\frac{\sum_i a_i}{\sum_i b_i}$ but this would perhaps be too large. I guess I don't know how to find better upper bounds in terms of $\frac{\sum_i a_i}{\sum_i b_i}$.

答:
If you want an upper bound that only depends on $\frac{\sum a_i}{\sum b_i}$ then no.

The quantity $\frac{\sum a_i}{\sum b_i}$ is called the mediant, or freshman sum, of the fractions $\frac{a_i}{b_i}$. It is a weighted average of the fractions so it's always between $\min \frac{a_i}{b_i}$ and $\max \frac{a_i}{b_i}$ and can be arbitrarily close to either. And this is exactly how Simpson's Paradox works.

https://en.wikipedia.org/wiki/Mediant_(mathematics)

Now if $n=2$, WLOG assume $\frac{a_1}{b_1} < \frac{a_2}{b_2}$. The mediant can be arbitrarily close to $\frac{a_1}{b_1}$ while the ratio of $\frac{a_2}{b_2}$ to $\frac{a_1}{b_1}$ can also be arbitrarily large. Therefore $\frac{\sum \frac{a_i}{b_i}}{\frac{\sum a_i}{\sum b_i}}$ can be arbitrarily large.

For example, if $N$ is very large, then
$$
\frac{\frac 1N + \frac{N}{1}}{\frac{1+N}{N+1}} = N+\frac 1N > N.
$$


问:
Hitting set expected value
Let's consider variable that satisfies $P(X_i=1)=P(X_i=-1)=\frac12$  and define $\tau_k=\inf\{n : |S_n|=k\}$. I want to calculate $E[\tau_k]$.

I thought it would be good to hit this with wolds decomposition : $$E[S_\tau]=E[\tau_k] E[X_1]$$

But $S_\tau$ can be either $k$ or $-k$ with equal probability so $E[S_\tau]=0$ as well as $E[X_1]$. It seems that I have equality $0=0$ which is meaningless. Is there any simple way how can I calculate $E[\tau_k] $ ?

答:
Why don't you consider the martingale $( n-S_n^2 ; n \ge 1 )$

Update
Firstly, by any mean possible, you can prove that :
$$\mathbb{P}( \tau < +\infty) =1$$
(Mine is CLT, but if you can find a better approach for that, I'd be happy to get informed)

So now, let's consider the stopped martingale $(M_n) := ( n \wedge \tau -S_{n\wedge \tau }^2$  )
We see that:

$\lim_{n \rightarrow +\infty} \tau \wedge n = \tau \text{ a.e}$ (as $\mathbb{P}( \tau <+\infty) =1$ ).

So $\lim_{n \rightarrow+ \infty} \mathbb{E}( \tau\wedge n ) =\mathbb{E}( \tau )$ ( monotone convergence)

$\lim_{n \rightarrow +\infty} S^2_{n \wedge \tau } =  S^2_{ \tau }=k^2 $ ( also because $\tau$ is finite a.s).

Thus $\lim_{n \rightarrow+ \infty} \mathbb{E}( S^2_{\tau\wedge n} ) =k^2$( Dominated Convergence)

Besides, $M_n$ is a martingale, then. $\mathbb{E} (M_n) = M_0=0$.
So,
$$ 0 = \lim_{n \rightarrow \infty} \mathbb{E}(M_n) =\lim_{n \rightarrow \infty} \left( \mathbb{E}(n\wedge\tau)-  \mathbb{E}(S^2_{n\wedge\tau}) \right) =\mathbb{E}(\tau) -k^2$$

Conclusion,
$$ \mathbb{E}(\tau) =k^2$$

Title: Hitting set expected value


问:
Using the precise definition of limits to prove a limit
Question

Let $a$ be a positive real number. Use the $\epsilon-\delta$ definition of limits to prove that $$\lim\limits_{x\to a}\ \frac {3x^2 + 2ax + a^2} {2x + a} = 2a\ .$$

My working

Let $\epsilon > 0$ be given.

For $|x - a| < \delta$, choose $\delta = \frac 1 3 \epsilon - 2a$.

\begin{align}
|x + a| & =
|x - a + 2a|
\\[5 mm] & \leq
|x - a| + 2a
\\[5 mm] & <
\frac 1 3 \epsilon
\end{align}

\begin{align}
\left|\frac {3x^2 + 2ax + a^2} {2x + a} - 2a\right| & =
\left|\frac {3x^2 - 2ax - a^2} {2x + a}\right|
\\[5 mm] & =
\left|\frac {(3x + a)(x - a)} {2x + a}\right|
\\[5 mm] & <
\left|\frac {3(x + a)(x - a)} {2(x - a)}\right|
\\[5 mm] & =
\frac 3 2 |x + a|
\\[5 mm] & <
\frac 1 2 \epsilon
\\[5 mm] & <
\epsilon
\end{align}

May I know if I have chosen a suitable $\delta$ and if my proof is logical? Any comments/suggestions to improve my working will be greatly appreciated :)

Edit

Thank you everyone for your inputs! I have learnt a lot :)

答:
Your $\delta$ doesn't work and it would be strange if it did, because in almost every case $\delta$ must be very close to $0$ when $\varepsilon$ is very close to $0$. But your $\delta$ is always greater than $2a$.

Note that$$\frac{3x^2+2ax+a^2}{2x+a}-2a=\frac{(3x+a)(x-a)}{2x+a}$$and that therefore$$\left|\frac{3x^2+2ax+a^2}{2x+a}-2a\right|=\left|\frac{3x+a}{2x+a}\right||x-a|.\tag1$$If $|x-a|<\frac a4$, then $|x|<\frac54a$ and so $|3x+a|<\frac{19}4a$. On the other hand,\begin{align}|2x+a|&=|2(x-a)+3a|\\&\geqslant3a-2|x-a|\\&>\frac52a.\end{align}So,$$\left|\frac{3x+a}{2x+a}\right|<\frac{19}{10}.$$Therefore, it follows from $(1)$ that $\delta=\min\left\{\frac a4,\frac{10}{19}\varepsilon\right\}$ will work.


问:
Posterior of an iid sample withh distribution $Pois(\theta)$ and a prior of $Exp(\lambda)$
I have a sample $X_1,…, X_n$ that are iid with $Pois(\theta)$ as their distribution where $Θ = ℝ^+$. Now, suppose that the prior distribution of $$ is $Exp(\lambda)$. I am given the solution as follows:

The likelihood function is:

$L(\theta) = \prod_{i=1}^{n} \frac{e^{-\theta}\theta^{x_i}}{x_i!} = \frac{e^{-n\theta} \theta^{\sum_{i=1}^{n}}}{\prod_{i=1}^{n} x_i!}$

As such, the posterior distribution is:
$f(\theta|\mathbf{x}) = \frac{\frac{e^{-n\theta} \theta^{\sum_{i=1}^{n}}}{\prod_{i=1}^{n} x_i!} \times \frac{1}{\lambda}e^\frac{-\theta}{\lambda}}{\int_0^\infty  \frac{e^{-n\theta} \theta^{\sum_{i=1}^{n}}}{\prod_{i=1}^{n} x_i!} \times \frac{1}{\lambda}e^\frac{-\theta}{\lambda} d\theta}$

My question is, why is the prior distribution written as $\frac{1}{\lambda}e^\frac{-\theta}{\lambda}$ and not $\frac{1}{\lambda}e^\frac{-x}{\lambda}$? Isn't the probability density function of $Exp(\lambda)$ distribution $\frac{1}{\lambda}e^\frac{-x}{\lambda}$? So where did $\theta$ come from?

答:
You are dealing with Bayesian Statistic. Your parameter (unknown parameter $\theta$) has its own distribution. In particular your prior is an $Exp(\lambda)$ and the variable is exactly $\theta$

Moreover, the solution can be derive in a faster way

$$\pi(\theta|\mathbf{x})\propto \pi(\theta)p(\mathbf{x}|\theta)$$

Thus

$$\pi(\theta|\mathbf{x})\propto e^{-n \theta}\theta^{\Sigma x} e^{-\frac{\theta}{\lambda}}=\theta^{\Sigma x}e^{-\theta(n+\frac{1}{\lambda})}$$

We immediately recognize that

$$ \bbox[5px,border:2px solid black]
{
\pi(\theta|\mathbf{x})\sim \mathrm{Gamma}\Big(\Sigma x+1;n+\frac{1}{\lambda}\Big)
\qquad (1)
}
$$

All the calculations you showed are not necessary and tedious

In fact all the quantities not depending on $\theta$ can be removed because they are included in the normalization constant...also you denominator does not depend on $\theta$ because it is integrated over all $\Theta$

Having derived that the posterior (1) is a Gamma with the specified parameters, the normalization constant is immediately given, thus the complete expression of your posterior is

$$ \bbox[5px,border:2px solid black]
{
\pi(\theta|\mathbf{x})=\frac{(n+\frac{1}{\lambda})^{\Sigma x+1}}{\Gamma(\Sigma x+1)}\theta^{\Sigma x}e^{-\theta(n+\frac{1}{\lambda})}
\qquad (2)
}
$$

In the density (2) the variable is $\theta \in (0;+\infty)$; $\lambda$ is a parameter, while $n$ and $\Sigma x$ are given observed data.


问:
Prove that Gauss map on M is surjective
Let $M$ be a closed, orientable, and bounded surface in $\mathbb{R}^3$.

(a) Prove that the Gauss map on $M$ is surjective.

(b) Let $K_+(p) = \max \{0, K(p)\}$.  Show that
$$
\int K_+dA \ge 4\pi.
$$
in the area of $M$.  Do not use the Gauss-Bonnet Theorem to prove this.

答:
While the answers given are wonderful I thought it would be interesting to look at the first question from the point of view of the Euler characteristic of the surface.

the tangent bundle of the surface is the pull back bundle of the tangent bundle to the sphere via the Gauss mapping. This is easy to see. This means that the pull back of the volume form of the sphere integrates to 2pi times the Euler characteristic of the surface. If the Gauss map were not surjective then the pull back would be an exact form and would integrate to zero. So the Gauss map must be surjective for any surface of non zero Euler charateristic.

This leaves the question of the torus. Not sure what to do.

Title: Prove that Gauss map on M is surjective


问:
Distance between two circles with known size and intersection area
Caution - biologist at work

I am trying to plot some circles and want to work out how far apart my circles should be with a target in mind. The ditance between the centre points of the two circles is $d$. The first circle has a radius of $r_1$ = 5 cm and the second has a radius of $r_2$ = 4 cm. The area of these circles is 78.54 cm$^2$ and 50.27 cm$^2$ respectively, making the joint area $a_{1,2}$ = 128.81 cm$^2$. I want 30% of that area, $0.3 \times 128.81$ cm$^2$ to be intersection area (area shared by the two circles). Therefore I have a known pair of radii and a known intersection area, and an unknown distance. By trial and error with various values of $d$ in the equation

$$ a_{1,2} = r_1^2 cos^{-1} \frac{d^2 + r_1^2 - r_2^2}{2dr_1} + r_2^2 cos^{-1} \frac{d^2 + r_2^2 - r_1^2}{2dr_2} - \frac{1}{2} \sqrt{(-d+r_1+r_2)(d+r_1-r_2)(d-r_1+r_2)(d+r_1+r_2)} $$

I find that, to get 30% of the total area into the intersection area, when the radii are 5 and 4, $d$ needs to be approximately 2.71. However, is there a way to solve $d$ from known radii and intersection area? I've been searching for hours and come up short - the internet is dominated by solutions to find the intersection are from known radii and distance.

FYI, I'm programming this in R - bonus kudos for answers providing accompanying script!

```<code>r1 <- 5
r2 <- 4
a1 <- pi*r1^2
a2 <- pi*r2^2
aJ <- 0.3 * (a1+a2)

intersectionArea <- function(r1, r2, d){
  (r1^2 * acos((d^2 + r1^2 - r2^2)/(2*d*r1))) +
  (r2^2 * acos((d^2 + r2^2 - r1^2)/(2*d*r2))) -
  (0.5 * sqrt((-d+r1+r2)*(d+r1-r2)*(d-r1+r2)*(d+r1+r2)))
}

aJ/intersectionArea(r1,r2, 2.709)
</code>```

Body:
Given $r_1,r_2$ and

\begin{align}
k&=\frac{\mathrm{intersection\ area}}{\pi(r_1^2+r_2^2)}
\end{align}

you can try
Halley's method
for approximation, starting with some reasonable initial guess for $d$, for example, $d_0=\tfrac12(r_1+r_2)$:

\begin{align}
d_{{n+1}}
&=d_{n}-{\frac  {2f(d_{n},r_1,r_2,k)f'(d_{n},r_1,r_2)}{2{[f'(d_{n},r_1,r_2)]}^{2}-f(d_{n},r_1,r_2,k)f''(d_{n},r_1,r_2)}}
,
\end{align}

where

\begin{align}
f(d,r_1,r_2,k)&=
-\pi(r_1^2+r_2^2)\,k
+r_1^2\arccos\left(\frac{d^2+r_1^2-r_2^2}{2d r_1}\right)
\\
&+r_2^2\arccos\left(\frac{d^2+r_2^2-r_1^2}{2d r_2}\right)
\\
&-\tfrac12\sqrt{(d+r_1+r_2)(-d+r_1+r_2)(d-r_1+r_2)(d+r_1-r_2)}
,\\
f'(d,r_1,r_2)&=
-\frac{\sqrt{2d^2 r_1^2-d^4+2d^2r_2^2-r_1^4+2r_1^2 r_2^2-r2^4}}d
,\\
f''(d,r_1,r_2)&=
\frac{(d^2+r_2^2-r_1^2)(d^2+r_1^2-r_2^2)}{d^2\sqrt{(d+r_1+r_2)(-d+r_1+r_2)(d-r_1+r_2)(d+r_1-r_2)}}
.
\end{align}

For example, this R-code

##
r1 <- 7
r2 <- 9
k <- 0.0702529004662957  ## fraction (intersection area)/(pi*(r1^2+r2^2))

f <- function(d,r1,r2,k){
  -pi*(r1^2+r2^2)*k+r1^2*acos(1/2*(d^2+r1^2-r2^2)/d/r1)+r2^2*acos(1/2*(d^2+r2^2-r1^2)/d/r2)-1/2*sqrt((d+r1+r2)*(-d+r1+r2)*(d-r1+r2)*(d+r1-r2))
}

df <- function(d,r1,r2){
  -(2*d^2*r1^2-d^4+2*d^2*r2^2-r1^4+2*r1^2*r2^2-r2^4)^(1/2)/d
}

ddf <- function(d,r1,r2){
  (d^2+r2^2-r1^2)*(d^2+r1^2-r2^2)/(d+r1+r2)^(1/2)/(-(d-r1+r2)*(-r1+d-r2)*(d+r1-r2))^(1/2)/d^2
}

F <- function(x,r1,r2,k){
  x-2*f(x,r1,r2,k)*df(x,r1,r2)/(2*df(x,r1,r2)^2-f(x,r1,r2,k)*ddf(x,r1,r2))
}

d <- (r1+r2)/2  ## initial approximation
d <- F(d,r1,r2,k)
d <- F(d,r1,r2,k)
d <- F(d,r1,r2,k)
d <- F(d,r1,r2,k)

results in consequent approimations:

8
11.85015
11.99998
12
12

Edit

You are correct that the total area is not the same
as the sum of the two circles, but this can be easily corrected.
Let $S_x$ and $S_t$ be the intersection area and the total area, respectively,
then
\begin{align}
S_t&=\pi(r_1^2+r_2^2)-S_x
.
\end{align}

Consider two fractions, $k$ and $q$,

\begin{align}
k&=\frac{S_x}{\pi(r_1^2+r_2^2)}
,\\
q&=\frac{S_x}{S_t}
,
\end{align}

so, given $q$ tou can find corresponding $k$ as

\begin{align}
k&=1-\frac1{q+1}
\end{align}

and use the procedure above as before.

Just add a function

Fq <- function(x,r1,r2,q){
  vf <- f(x,r1,r2,1-1/(q+1))
  x-2*vf*df(x,r1,r2)/(2*df(x,r1,r2)^2-vf*ddf(x,r1,r2))
}

and use it instead of $F$.


问:
Contour integral in presence of a branch cut.
I'm trying to evaluate

$$\int_0^{\infty} \frac{\log x}{1+x^3}dx$$

and I have to use Residue Theorem. I chose the classical pacman contour centered in the origin, and by using little/big circle theorems I can state that:

$$(2+2\pi i)\int_0^{\infty}\frac{\log x}{1+x^3}dx=2\pi i\sum_{z_i} Res(f,z_i)$$
where $z_i$ are the poles. Computing the residues we have:

$$Res(f,-1)=\frac{i\pi}{3}$$
$$Res(f,e^{i\frac{\pi}{3}})=\frac{i\pi}{9}e^{-i\frac{2\pi}{3}}$$
$$Res(f,e^{i\frac{\pi}{3}})=-\frac{i\pi}{9}e^{i\frac{2\pi}{3}}$$

And so
$$2\pi i\sum_{z_i}Res(f,z_i)=2\pi i(\frac{i\pi}{3}+\frac{i\pi}{9}e^{-i\frac{2\pi}{3}}-\frac{i\pi}{9}e^{i\frac{2\pi}{3}})=-\frac{2\pi^2}{3}+\frac{2\pi^2}{9}(2i\sin(\frac{2pi}{3}))$$

But clearly I'm making mistakes as the result doesn't match. Can you please help me spot it?

答:
If you use
$$ f(z)=\frac{\log}{z^3+1} $$
the integral becomes
$$ \int_0^\infty f(x)dx+\int_{\infty}^0f(xe^{2\pi i})dx=2\pi i\text{Res}(f,z_1,z_2,z_3). \tag1$$
Since
$$ \int_0^\infty f(x)dx=\int_0^\infty \frac{\log x}{x^3+1}dx $$
and
$$ \int_{\infty}^0f(xe^{2\pi i})dx=-\int_0^{\infty}\frac{\log x+2\pi i}{x^3+1}dx $$
putting them in (1), the integral you wanted will be cancelled. For this type of integrals, you have
$$ f(z)=\frac{\log^2z}{z^3+1} $$
instead of
$$ f(z)=\frac{\log z}{z^3+1}. $$


问:
What does the $\cdot$ mean in $X \to T(X, \cdot)\in V^∗$
Let $T$ be a $(0, 2)$-tensor, and let $X, Y$ be vectors.

$T$ can act as $(X, Y) \mapsto T(X, Y )\in\mathbb R $. I see

$$X \to T(X, \cdot)\in V^∗$$

$$X \to  T(\cdot, X) \in V^∗$$

What does the  $\cdot$ mean?

答: A dual vector in $V^*$ takes vectors to real numbers. The dot is where the vector would go, i.e. $T(X,\cdot) (Y) = T(X,Y).$ It's not hard to see that this defines a linear map from $V$ into $\mathbb{R},$ i.e. a dual vector.


问:
Applying reduction formula to integrals
Question

For $n \geq 0$, let $$I_n = \int_{0}^{8} x^n(8 - x)^\frac 1 3\ \mathrm{d}x\ .$$

Show that, for $n \geq 1$, $$I_n = \frac {24n} {3n + 4} I_{n - 1}\ .$$

My working

Let $u = x^n$ and $v' = (8 - x)^{\frac 1 3}$

$\implies u' = nx^{n - 1}$ and $v = -\frac 3 4 (8 - x)^{\frac 4 3}$

$\implies I_n = [-\frac 3 4 x^n(8 - x)^{\frac 4 3}]^{x = 8}_{x = 0} + \frac 3 4 n \int_{0}^{8} x^{n - 1}(8 - x)^\frac 4 3\ \mathrm{d}x$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac 3 4 n \int_{0}^{8} x^{n - 1}(8 - x)^\frac 1 3(8 - x)\ \mathrm{d}x$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac 3 4 n \int_{0}^{8} [8x^{n - 1}(8 - x)^\frac 1 3 - x^n(8 - x)^{\frac 1 3}]\ \mathrm{d}x$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ = 6nI_{n - 1} - \frac 3 4 n I_n$

$\implies (1 + \frac 3 4)I_n = 6nI_{n - 1}$

$\therefore I_n = \frac {24n} {3n + 4} I_{n - 1}$

I try to use integration by parts, but with my choices of $u$ and $v'$, I do not seem to be getting anywhere. I am currently taking an introductory module to calculus in college. In class, we have only gone through the reduction formula for $\sin$ and $\cos$ and I have been stuck at this question for quite some time. Any hints/intuitions as to how to proceed will be greatly appreciated :)

Edit

Following some help from Phicar, I have finally got it! :) (I have edited my post to include my updated working)

Title: Applying reduction formula to integrals

Body:
Take $v'=(8-x)^{1/3},$ integrating you get, as implicit in your comment,$$v=-\frac{3}{4}(8-x)^{1/3+1}=-\frac{3}{4}(8-x)^{1/3}(8-x)=\frac{-24}{4}(8-x)^{1/3}+\frac{3x}{4}(8-x)^{1/3}.$$
Split the integral in the addition sign.


问:
I Don't get the solution. I need an explanation or Is there any videos on you tube to understand
A man buys 20 pens and 12 books for tk 400.He sells pens at a profit of 40% and books at a gain of 25%.If his overall profit was to 112.The cost price of the book is?detail pls
Ans

40%400 - 112 = 48
(40-25) i.e. 15% of 32 B = 48
So, B = 48100/(15*32) = 10

答:
Let $b$ be the cost of a book, and let $p$ be the cost of a pen. From the first statement in the problem, we must have:

$$20p + 12b = 400$$

Now, since the man sells the pens for $1.4$ times their price, and sells books for $1.25$ times their price, making a profit of $112$, we find:

$$1.4\cdot 20p + 1.25\cdot 12b = 400 + 112$$

We find the system of equations:

$$20p + 12b = 400$$

$$28p + 15b = 512$$

Can you take it from here?

If you're still stuck:

Multiplying the first equation by $5$ and the second by $4$:

$$100p + 60b = 2000$$

$$112p + 60b = 2048$$

Subtracting the first equation from the second:

$$12p = 48$$

$$p = 4$$

Substituting into the original cost equation:

$$80 + 12b = 400$$

$$12b = 320$$

$$b = \frac{80}{3}$$

Thus:

$$\boxed{\text{A book costs approximately }26.67}$$


问:
Probability defective process
Certain mechanical pieces undergo two independent processes by two distinct machines A and B. Stain A introduces defects in 5% of the processed pieces. a) Knowing that 7% of the final pieces are defective (ie that they have undergone both processes), calculate the value of the probability P (DB) that machine B introduces defects. b) If a randomly selected final piece is defective, calculate the probability that it has suffered defects from both machines. To carry out the procedure, use the following events: DA = {Machine A has introduced a defect} DB = {Machine B has introduced a defect} D = {The final piece is defective} {Hint: first express D using DA and DB}

I found that $ P(D_B) = 0,09 $ that’s because $ P(D) = P(D|D_A ) P(D_A) + P(D|D_B) P(D_B) $ but I think this is not correct because I also should have $ + P(D|AB) P(AB) $ but I don’t understand how to find this , to find after $P_B$

答:
$P(D_A) = 0.05, P(D) = 0.07$ and we need to find $P(D_B)$.

There are $3$ ways for a product to be defective - defect by machine $A$, defect by machine $B$ or by both. There is only one way for the product to be non-defective - no defect introduced by either of the machines. So we can go about our calculation in two ways. I will use the latter.

$(1-P(D_A)) \times (1-P(D_B)) = 1 - P(D)$

i.e $0.95 \times (1-P(D_B)) = 0.93 \implies P(D_B) \approx 0.021$

Now to find the conditional probability please note that

$P(D_A \cap D_B) = P(D_A) \times P(D_B)$ and you already know $P(D)$.

Title: Probability defective process


问:
Computing the curvature associated to "quaternionic" connection
I am attempting to verify that for a matrix $A=\begin{pmatrix}\alpha &\beta\\\gamma & \delta\end{pmatrix}\in\text{Sp}(2)$ where $\text{Sp}(2)$ is the compact sympletic group and the matrix entries are quaternions, the connection defined by $\omega=\text{Im}(\bar{\alpha}d\alpha+\bar{\gamma}d\gamma)$, which takes values in $S^3\cong \text{SU}(2)$ satisfies the Yang-Mills equation (I think these solutions are known as BPST instantons in physics). Given a section of the principal bundle $S^7\xrightarrow{\pi}\mathbb{HP}^1$ for a local trivialisation, i.e. $\sigma:U_1\to \pi^{-1}(U_1)$, it is given that for $q\in\mathbb{H}$ we have $A_q=(\sigma^*\omega)_q=\text{Im}(\frac{\bar{q}}{1+|q|^2}dq)$. Let $q=x+iy+jv+kw$, and $X=X^1+iX^2+jX^3+kX^4$, likewise for $Y$. I computed that
$$dA_q=\frac{1}{1+|q|^2}d\bar{q}\wedge dq \\ [A_q,A_q](X,Y)=\frac{2}{(1+|q|^2)^2}\text{Im}(\text{Im}(\bar{q}X)\text{Im}(\bar{q}Y))
$$
I feel like I am quite close to being able to express the "field strength" $F^A$ of the pulled back connection this way, since $F^A_q=A_q+\frac{1}{2}[A_q,A_q]$. After I do so, I think I can show that it is self-dual, and therefore satisfies the Yang-Mills equation. The issue is that I have not been able to write the expression I computed for the Lie bracket independent of the arguments $X$ and $Y$. Could anyone help me by giving a hint, or explaining how I can convert the given expression into an actual differential form? I considered $\text{Im}(\text{Im}(\bar{q}dq)\wedge\text{Im}(\bar{q}dr))$ for some $r$ but I did not quite see how to make it work. Thanks for your time.

答:
I obtained a solution. I first showed that $\text{Im}(\bar{q}dq)\wedge\text{Im}(\bar{q}dq)=\text{Im}(\bar{q}dq\wedge\bar{q}dq)$, and using this fact along with $\bar{q}\wedge q$ being anti-self-dual,

$$F^A_q=dA_q+A_q\wedge A_q=\text{Im}(\frac{d\bar{q}\wedge dq}{1+|q|^2}+\bar{q}d(\frac{1}{1+\bar{q}q})\wedge dq+\frac{\bar{q}dq\wedge\bar{q}dq}{(1+|q|^2)^2})=\\
    \text{Im}((1+|q|^2)\frac{d\bar{q}\wedge dq}{(1+|q|^2)^2}-\bar{q}\frac{qd\bar{q}\wedge dq}{(1+|q|^2)^2}-\bar{q}\frac{dq\wedge \bar{q} dq}{(1+|q|^2)^2}+\frac{\bar{q}dq\wedge\bar{q}dq}{(1+|q|^2)^2})=\\
    \text{Im}((1+|q|^2)\frac{d\bar{q}\wedge dq}{(1+|q|^2)^2)}-|q|^2\frac{d\bar{q}\wedge dq}{(1+|q|^2)^2})=\\
    \text{Im}(\frac{d\bar{q}\wedge dq}{(1+|q|^2)^2})$$
Which is anti-self-dual by the above remark.


问:
Solving Inhomogeneous Recurrence Relations
Equation is: ${(1/2)(a_n + a_{(n-1)})} = 2n + 5$ when $a_0 = 3$.

So, I got ${a_n = A(-1)^n}$ for the homogenous solution

I got 2n + 1 for a particular solution to ${a_n = -a_{(n-1)}  + 4n}$

I got 5 for  a particular solution to ${a_n = -a_{(n-1)}  + 10}$.

Combined, I used $a_0 = 3$ to get A = -8.
So, I ended up with ${a_n = -8(-1)^n + 2n + 11}$

Is that correct? Or, if not, where did I go wrong?

Update: In my original work, I mistakenly added $10$ instead of $5$. So now, I instead got A = -3 with ${a_n = -3(-1)^n + 2n + 6}$. Is this correct?

答:
Here is a more direct solution. Rewrite your initial relationship under the form:

$$\frac12\left(\underbrace{(a_n-(2n-6))}_{b_n}+\underbrace{(a_{n-1}-(2(n-1)-6))}_{b_{n-1}}\right)=0$$

In this way, you eliminate at once the "non-linearity", giving, as you have done:

$$b_n=A(-1)^n=a_n-(2n-6) \ \iff \ a_n=(2n-6)+A(-1)^n$$

It remains to adjust $A$ in order to comply with the initial value of $a_0$.

We find back your final solution!


问:
How to move/rotate a line segment intersecting another, just enough so that they become orthogonal?
I'm trying to find a way to move or rotate a line segment intersecting another until they become orthogonal (the angle formed in the intersection is tuned into 90 degrees) while keeping the length of the line and the intersection point intact, a sample problem is displayed in the image below:

Sample problem

```<code>### Sample lines:
lineA = ([2, 4], [1, 1])
lineB = ([1, 1], [4, 1])
# Intersection point: (1,1)
</code>```

In the sample our target would be the point "a1". Keep in mind while the angle in the sample above is less than 90 degrees there may be cases in which the angle may be more than 90. To make the problem simpler, suppose we know from before hand which line we need to modify (for the sample above, line "a").

I know how to calculate the angle between the lines but don't know where to go from here.

Appreciate your help.

P.S: Sorry for not making the image inline! I'm new here and one needs 15 points to be able to paste inline images.

答:
You could calculate a transformation of your coordinate system that rotates lineA around it second endpoint until it is orthogonal to lineB.

A simpler approach is: the transformation will preserve the length of lineA and the position of one endpoint, but will wipe out other properties, such as the original direction of the line segment.
But the transformation must also add a property, that the transformed segment is orthogonal to lineB.
So instead of computing the transformation, we simply build the result with the known properties it must have:

Identify the common endpoint of lineA and lineB, call it $b_1$.

Identify the line through $b_1$ orthogonal to lineB.

Construct a line segment lying on the line in step 2, with one endpoint at $b_1$, the other endpoint at a distance from $b_1$ equal to the length of lineA.

In coordinates, the direction of lineB is represented by the vector
$[4,1] - [1,1] = [3,0],$ and an orthogonal line has the direction given by a vector orthogonal to $[3,0].$ To find a vector orthogonal to $[x,y]$ we can take $[-y,x]$ or any non-zero multiple of $[-y,x]$, including $[y,-x]$.
So let's take $[0,3]$ as the vector orthogonal to $[3,0].$

We need the vector from $b_1$ to the other endpoint of the new segment to have the same length as lineA, which is $\sqrt{(2-1)^2 + (4-1)^2} = \sqrt{10}.$
So we multiply $[0,3]$ (which has length $3$) by $\frac{\sqrt{10}}{3}$ to obtain a vector of length $\sqrt{10}$ in the same direction: $[0, \sqrt{10}].$
(Actually in this example we could have seen directly that the answer would be
$[0, \sqrt{10}]$ without working out the multiplication factor,
but the method used here will also work in the case where lineB is not parallel to a coordinate axis, when it is not so obvious how to substitute coordinates to obtain the desired length in the same direction.)

Finally we find the other endpoint of the new segment by adding the vector we just found to the coordinates of the known endpoint, $b_1$:

$$ [1,1] + [0, \sqrt{10}] = [1, 1 + \sqrt{10}] . $$

The rotated copy of lineA is therefore $([1,1], [1, 1 + \sqrt{10}]).$

There is one more detail you might like to enforce: to choose which of the two possible answers to produce. There are, after all, two places you could rotate lineA to so that it is orthogonal to lineB, one on each side of lineB.
If you want to minimize the amount of rotation you want to choose the answer on the same side of lineB as the original lineA.

To get the other possible result in this example, instead of simply adding
$[1, 1 + \sqrt{10}]$ to $b_1$ we multiply by $-1$ first and then add;
or equivalently, we simply subtract $[1, 1 + \sqrt{10}]$, with the result

$$ [1,1] - [0, \sqrt{10}] = [1, 1 - \sqrt{10}] . $$

If you are doing this graphically, you can eyeball the directions of lineB, lineA, and the vector you find orthogonal to lineB, and choose to add or subtract that vector according to whichever way produces a segment on the same side of lineB as lineA.

If you are doing this numerically without the aid of graphics and a human eye
(for example writing a simple computer program to solve the problem),
you first find a vector of the correct length orthogonal to lineB,
which vector may be in the correct direction or the opposite direction.
Then you can use the "dot" product of this vector with the vector from $a_2$ to $a_1$ to decide whether to add or subtract.
If the "dot" product is positive then the angle between vectors is less than $90$ degrees and you should add the new vector to $b_1,$
but if the "dot" product is negative than the angle is greater than $90$ degrees and you really want the "other" orthogonal vector, so you should subtract.


答:
Your book's answer is wrong. I get the cost price of a book to be $\frac{80}3$ and that of a pen to be $4$ in whatever currency you're using. These numbers check out when substituted back into the conditions of the problem statement.

Let the cost price of a single book be $b$. Then the cost price of a single pen is $\frac{400 - 12b}{20}$. The total cost (outlay) is $400$. The total selling price (revenue) is $1.4 \times 20 \times \frac{400 - 12b}{20} + 1.25 \times 12 \times b = 560 - 1.8b$.

The difference between selling price and cost is the profit, so we form the equation $560-1.8b - 400 = 112$, the solution of which is $b = \frac{80}3$.

Title: I Don't get the solution. I need an explanation or Is there any videos on you tube to understand

Body:
A man buys 20 pens and 12 books for tk 400.He sells pens at a profit of 40% and books at a gain of 25%.If his overall profit was to 112.The cost price of the book is?detail pls
Ans

40%400 - 112 = 48
(40-25) i.e. 15% of 32 B = 48
So, B = 48100/(15*32) = 10


问:
Evaluate $\lim_{x\to 0^+}\frac{\ln x}{(1+x)^2}-\ln\left(\frac{x}{1+x}\right)$
I tried :$$\lim_{x\to 0^+}\frac{\ln x}{(1+x)^2}-\ln\left(\frac{x}{1+x}\right)=\lim_{x\to 0^+}\frac{\ln x}{(1+x)^2}-\ln x+\ln (x+1)=\lim_{x\to 0^+}\frac{-x^2\ln x}{(1+x)^2}+\ln(x+1)$$

Here I can use $\ln(x+1)\sim x-\frac{x^2}{2}+\frac{x^3}{3}$. but it seems It doesn't work because I don't know how to simplify $\cfrac{-x^2\ln x}{(1+x)^2}$.

答: Actually, $(1+x)^2-1$ isn't $x^2$. I should be$$\lim_{x\to0^+}\left(\frac{-x(x+2)\ln x}{(1+x)^2}+\ln(x+1)\right)=-2\lim_{x\to0^+}x\ln x=0.$$(See here for proofs that $\lim_{x\to0^+}x\ln x=0$.)


问:
Can we know the equation of line " $L$ " from one point $P(a,b)$ in this line?
Suppose we have a point $ p(a,b)$ a and b are given and let $L$ be a line s.t $p(a,b) $ belong to $L$
By this information can we know the equation of L in terms of $P(a.b)$ ?can it gives as more information about $ L$?

答: If we just have one point $p(a,b)$ known then there are $\text{infinitely (uncountably) many }$ straight lines passing through it. We have to know at least two points to identify the equation of a line. See Euclid's first postulate here and here


问:
find the value $AB$, for which $\log_{A} B =\log_{B} A$
Suppose $A$ and $B$ are positive real numbers for which $\log_{A} B = \log_{B} A$ . If neither $A$ nor $B$ is $1$ and $A$ $\not\equiv B$, find the value of $AB$.

We can rewrite as

$A^{\log_{A} B} = B$.

$B^{\log_{B} A} = A$.

$(B^{\log_{B} A})^{A^{\log_{A} B}} =B$.

$B^{2 (\log_{B} A)} =B$.

$B^{2 (\frac{1}{2})} =B$.

I'm unable to deduce a correct solution with $\frac{1}{2}$.
I wanted to try a different approach by converting the powers to multiplication but can't figure out the base. i.e.

${\log_{A} B (\log_{?}A)} $.

Body:
\begin{eqnarray*}
\log_A(B) = \log_B(A) \\
\frac{\ln(B)}{\ln(A)}=\frac{\ln(A)}{\ln(B)} \\
(\ln(B))^2=(\ln(A))^2 \\
\ln(B)=-\ln(A) \text{     or ...} \\
\ln(AB)=0.
\end{eqnarray*}
S0 $AB=?$


答:
To change from the base $b$ to the base $c$ in the logarithm below
$$\log_{b}a$$
we can use the following formula
$$\log_{b}a=\frac{\log_{c}a}{\log_{c}b}$$

So, to change from the base $B$ to base $A$:
$$\log_{B}A=\frac{\log_{A}A}{\log_{B}A}$$
and, since $\log_{A}A=1$, we get
$$\log_{B}A=\frac{1}{\log_{A}B}$$
and
$$\log_{B}A=\log_{A}B$$
then
$$(\log_{A}B)^{2}=1\Rightarrow\log_{A}B=\pm1$$
Therefore,
$$A=B$$
or
$$A=B^{-1}$$

Title: find the value $AB$, for which $\log_{A} B =\log_{B} A$

Body:
Suppose $A$ and $B$ are positive real numbers for which $\log_{A} B = \log_{B} A$ . If neither $A$ nor $B$ is $1$ and $A$ $\not\equiv B$, find the value of $AB$.

We can rewrite as

$A^{\log_{A} B} = B$.

$B^{\log_{B} A} = A$.

$(B^{\log_{B} A})^{A^{\log_{A} B}} =B$.

$B^{2 (\log_{B} A)} =B$.

$B^{2 (\frac{1}{2})} =B$.

I'm unable to deduce a correct solution with $\frac{1}{2}$.
I wanted to try a different approach by converting the powers to multiplication but can't figure out the base. i.e.

${\log_{A} B (\log_{?}A)} $.


问:
An extension of Baire's category theorem
In a topological space, a set is said to be rare if its closure has empty interior, and a set is said to be meager if it is a countable union of rare sets. If meager sets all have empty interior, then the topological space is said to be a Baire space. The following fact is known as Baire's category theorem.

Theorem. Complete metric spaces and locally compact Hausdorff spaces are Baire spaces.

[Zălinescu 2002] provides the following extension of this theorem for complete metric spaces, attributing it to C. Ursescu.

Theorem. ([Zălinescu 2002, Theorem 1.4.5]) Let $X$ be a complete metric space, and $\{S_n\}$ be a sequence of open sets in $X$. Then $\mathrm{cl}(\cap_{n=1}^\infty S_n)$ and $\cap_{n=1}^\infty \mathrm{cl}(S_n)$ have the same interior.

It is easy to check that a topological space is a Baire space if and only if any countable intersection of dense open sets is still dense in this space. Consequently, the theorem above implies that complete metric spaces are Baire spaces. Thus it is considered as an extension.

Question. Does the extension hold for locally compact Hausdorff spaces?

Update: As pointed out by @Alex Kruckman, the property in Ursescu's theorem is indeed equivalent to the fact that $X$ is a Baire space.

Theorem. A topological space is a Baire space if and only if $\mathrm{cl}(\cap_{n=1}^\infty S_n)$ and $\cap_{n=1}^\infty \mathrm{cl}(S_n)$ have the same interior for any sequence of open sets $\{S_n\}$.

Recall that a space is a Baire space if and only if any countable intersection of dense open sets is still dense. The theorem is essentially another way to state that a space is a Baire space if and only if all its open subspaces are Baire spaces.

Combined with Baire's category theorem, this observation by Alex proves the desired theorem. My answer below proves it from scratch, the proof being essentially the same as that of Baire's category theorem for locally compact Hausdorff spaces.

I will accept Alex's answer a few days later if nobody else has anything to add.

Thanks.

答:
Here is my attempt to prove it. It is inspired by the proof of [Zălinescu 2002, Theorem 1.4.5] and the classical proof of Baire's theorem for locally compact Hausdorff spaces. Any comments or criticism will be appreciated. Thanks.

Theorem. Let $X$ be a locally compact Hausdorff space, and $\{S_n\}$ be a sequence of open sets in X. Then $\mathrm{cl}(\cap_{n=1}^\infty S_n)$ and $\cap_{n=1}^\infty \mathrm{cl}(S_n)$ have the same interior.

proof. It suffices to show that $\mathrm{int}(\cap_{n=1}^\infty \mathrm{cl}(S_n))\subset \mathrm{cl}(\cap_{n=1}^\infty S_n)$. To this end, we only need to prove for any given nonempty open set $U \subset \cap_{n=1}^\infty \mathrm{cl}(S_n)$ that
\begin{equation}
    \label{eq:usnonempty}
    U\cap(\cap_{n=1}^\infty S_n) \neq \emptyset.
  \end{equation}
In the sequel, we will define a sequence of compact sets $\{C_n\}_{n=0}^\infty$ such that $C_n$ has nonempty interior and
\begin{equation}
  C_{n+1} \subset C_{n} \subset  U\cap(\cap_{k=1}^n S_k)
  \quad \text{ for each } \quad  n \ge 0,
  \end{equation}
where $\cap_{k=1}^0 S_k=X$. Once this is done,
Cantor's Theorem will yield
\begin{equation}
  \label{eq:nestnonempty}
  \emptyset \;\neq \;\cap_{n=1}^\infty C_n \;\subset\; U\cap (\cap_{n=1}^\infty S_n),
\end{equation}
which gives us what we want.

We define $\{C_n\}$ inductively.
As $U$ is a nonempty open set and $X$ is locally compact,
we can take a compact set $C_0\subset U$
such that $\mathrm{int}(C_0)\neq \emptyset$.
Assume that $C_n$ is already defined for an $n\ge 0$ so that
$C_n\subset U\cap(\cap_{k=1}^n S_k)$ and $\mathrm{int}(C_n)\neq \emptyset$.
Recalling that $U\subset\mathrm{cl}(S_{n+1})$,
we have $\mathrm{int}(C_n)\subset \mathrm{cl}(S_{n+1})$, which implies
that $\mathrm{int}(C_n) \cap S_{n+1} \neq\emptyset$.
Since $\mathrm{int}(C_n)\cap S_{n+1}$ is open,
invoking again the local compactness of $X$, we
can take a compact set $C_{n+1}\subset \mathrm{int}(C_n) \cap S_{n+1}$
such that $\mathrm{int}(C_{n+1})\neq \emptyset$.
Clearly, $C_{n+1}\subset C_n$ and
$C_{n+1}\subset C_{n}\cap S_{n+1} \subset U\cap(\cap_{k=1}^{n+1} S_k)$.
This finishes the induction and completes the proof.


答:
Answer is $0$

And the last step should give you

$\ln(x)-\ln(\frac{x}{1+x})(1+x^2)$ and as $x=0$ it gives you $0$!

Title: Evaluate $\lim_{x\to 0^+}\frac{\ln x}{(1+x)^2}-\ln\left(\frac{x}{1+x}\right)$

Body:
I tried :$$\lim_{x\to 0^+}\frac{\ln x}{(1+x)^2}-\ln\left(\frac{x}{1+x}\right)=\lim_{x\to 0^+}\frac{\ln x}{(1+x)^2}-\ln x+\ln (x+1)=\lim_{x\to 0^+}\frac{-x^2\ln x}{(1+x)^2}+\ln(x+1)$$

Here I can use $\ln(x+1)\sim x-\frac{x^2}{2}+\frac{x^3}{3}$. but it seems It doesn't work because I don't know how to simplify $\cfrac{-x^2\ln x}{(1+x)^2}$.


答:
From $A^{\log_A B} = B$ and $B^{\log_B A} = A$ we can write:

$$(A^{\log_A B})^{\log_B A} = A = A^1$$

Hence $(\log_A B) (\log_B A) = 1$.

But we are given $\log_A B = \log_B A$. Hence $(\log_A B)^2 = 1$.

If $\log_A B = 1$ then $A = B$, which contradicts an assumption.

Hence $\log_A B = -1$. What does that imply for $AB$?


答:
"Suppose we have a point p(a,b) a and b are given and let L be a line s.t p(a,b) belong to L By this information can we know the equation of L in terms of P(a.b) ?can it gives as more information about L?"
I'm not sure I understand this.  You say "p(a, b)" is a point but what are a and b?  Do you mean that p is a "point valued function of a and b"?  If a and b are independent variables, then p(a, b) will, in general, be two dimensional, a plane, not a line.</p>

Title: Can we know the equation of line " $L$ " from one point $P(a,b)$ in this line?

Body:
Suppose we have a point $ p(a,b)$ a and b are given and let $L$ be a line s.t $p(a,b) $ belong to $L$
By this information can we know the equation of L in terms of $P(a.b)$ ?can it gives as more information about $ L$?


问:
Any ideas how to solve this log inequality?
$\log_{\frac 12 }x > \log_{\frac 13 }x $

I don't know how to move with it :/

答:
We wish to prove:

$$\log_{\frac{1}{2}}(x) > \log_{\frac{1}{3}}(x)$$

By the change-of-base formula:

$$\frac{\ln(x)}{\ln(\frac{1}{2})} > \frac{\ln(x)}{\ln(\frac{1}{3})}$$

$$-\frac{\ln(x)}{\ln(2)} > -\frac{\ln(x)}{\ln(3)}$$

Multiplying both sides by $-\ln(2)\ln(3)$ (and flipping the inequality sign):

$$\ln(3)\ln(x) > \ln(2)\ln(x)$$

For $x\neq 1$:

$$\ln(3)>\ln(2)$$

Which is true. Thus, the original inequality is true for $x\neq 1$. If $x = 1$, then $\log_{\frac{1}{2}}(x) = \log_{\frac{1}{3}}(x) = 0$. Thus:

$$\log_{\frac{1}{2}}(x) > \log_{\frac{1}{3}}(x)\text{ for }x\in(0, 1)\cup(1,\infty)$$

($x$ must also be positive for its logarithms to exist.)


问:
What is a geometric relation?

We will call a relation geometric if $∀x, y, z(xRy ∧ xRz → yRz)$

Prove that if a relation is geometric and reflexive, then it is also
symmetric.

I'm trying to solve this problem but I don't understand the question. I've heard of Reflexive, Symmetric, Transitive, and equivalence relation, but not geometric?

答:
You want to prove that $xRy$ implies $yRx$.

So suppose that $xRy$;
since $R$ is reflexive, you have that $xRx$.

So, just take $z=x$ in the definition and obtain $yRx$.

Title: What is a geometric relation?


问:
Box topology first countability
I want to show that $\mathbb{R}^2$ with the topology $\tau_\mathcal{B}$ generated by the base $\mathcal{B} = \{U \times V | U \in \tau_1, V \in \tau_2\}$ where $\tau_1$ and $\tau_2$ are topologies for $\mathbb{R}$ with $\tau_1$ not first countable, is not first countable without using sequences.

答:
For convenience let $X$ denote $\Bbb R$ with the topology $\tau_1$, and let $Y$ denote $\Bbb R$ with the topology $\tau_2$. Let $x\in X$ have no countable local base, and let $y\in Y$ be arbitrary. Let $p=\langle x,y\rangle\in X\times Y$, and suppose that $\{B_n:n\in\Bbb N\}$ is a countable local base at $p$. For each $n\in\Bbb N$ there are $V_n\in\tau_1$ and $W_n\in\tau_2$ such that $p\in V_n\times W_n\subseteq B_n$, so $\mathscr{B}_p=\{V_n\times W_n:n\in\Bbb N\}$ is also a countable local base at $p$.

$\{V_n:n\in\Bbb N\}$ is a countable family of open nbhds of $x$ in $X$, so it is not a local base at $x$, and there is an open nbhd $V$ of $x$ such that $V_n\setminus V\ne\varnothing$ for each $n\in\Bbb N$. But then $V\times Y$ is an open nbhd of $p$ that contains no member of $\mathscr{B}_p$, which therefore cannot be a local base at $p$. This contradiction shows that $X\times Y$ cannot be first countable at $p$.

Title: Box topology first countability


答:
This is not really a strengthening of the Baire category theorem - it is equivalent to it. That is, any Baire space (and in particular any compact Hausdorff space) satisfies Ursescu's theorem.

Let's assume $X$ is a Baire space. Suppose $(S_n)_{n\in \mathbb{N}}$ is a sequence of open sets in $X$. We always have $\text{int}(\text{cl}(\bigcap_{n=1}^\infty S_n))\subseteq \text{int}(\bigcap_{n=1}^\infty \text{cl}(S_n))$, so it suffices to prove the reverse inclusion. And to do this, it suffices to prove that for any open $U\subseteq \bigcap_{n=1}^\infty \text{cl}(S_n)$, we have $U\subseteq \text{cl}(\bigcap_{n=1}^\infty S_n)$.

So fix some $U$. Note that an open subspace of a Baire space is Baire. [Why? If $(V_n)_{n\in\mathbb{N}}$ is a sequence of dense open subsets of $U$, then $(V_n\cup (X\setminus \text{cl}(U)))_{n\in\mathbb{N}}$ is a sequence of dense open subsets of $X$. The intersection of these sets is $(\bigcap_{n=1}^\infty V_n)\cup (X\setminus \text{cl}(U))$, and the fact that this is dense in $X$ implies that $(\bigcap_{n=1}^\infty V_n)$ is dense in $U$.]

So defining $V_n = U\cap S_n$, we have that each $V_n$ is open in $U$, and since $U\subseteq \text{cl}(S_n)$ for all $n$, each $V_n$ is dense in $U$. Since $U$ is Baire, $\bigcap_{n=1}^\infty V_n$ is dense in $U$, which implies $U\subseteq \text{cl}(\bigcap_{n=0}^\infty S_n)$, as was to be shown.

Title: An extension of Baire's category theorem

Body:
In a topological space, a set is said to be rare if its closure has empty interior, and a set is said to be meager if it is a countable union of rare sets. If meager sets all have empty interior, then the topological space is said to be a Baire space. The following fact is known as Baire's category theorem.

Theorem. Complete metric spaces and locally compact Hausdorff spaces are Baire spaces.

[Zălinescu 2002] provides the following extension of this theorem for complete metric spaces, attributing it to C. Ursescu.

Theorem. ([Zălinescu 2002, Theorem 1.4.5]) Let $X$ be a complete metric space, and $\{S_n\}$ be a sequence of open sets in $X$. Then $\mathrm{cl}(\cap_{n=1}^\infty S_n)$ and $\cap_{n=1}^\infty \mathrm{cl}(S_n)$ have the same interior.

It is easy to check that a topological space is a Baire space if and only if any countable intersection of dense open sets is still dense in this space. Consequently, the theorem above implies that complete metric spaces are Baire spaces. Thus it is considered as an extension.

Question. Does the extension hold for locally compact Hausdorff spaces?

Update: As pointed out by @Alex Kruckman, the property in Ursescu's theorem is indeed equivalent to the fact that $X$ is a Baire space.

Theorem. A topological space is a Baire space if and only if $\mathrm{cl}(\cap_{n=1}^\infty S_n)$ and $\cap_{n=1}^\infty \mathrm{cl}(S_n)$ have the same interior for any sequence of open sets $\{S_n\}$.

Recall that a space is a Baire space if and only if any countable intersection of dense open sets is still dense. The theorem is essentially another way to state that a space is a Baire space if and only if all its open subspaces are Baire spaces.

Combined with Baire's category theorem, this observation by Alex proves the desired theorem. My answer below proves it from scratch, the proof being essentially the same as that of Baire's category theorem for locally compact Hausdorff spaces.

I will accept Alex's answer a few days later if nobody else has anything to add.

Thanks.


问:
27 lines on a cubic surface - which proof is better?
There is a famous fact that any smooth cubic surface has exactly 27 lines.

I am a undergraduate students who want to view some detailed proofs of this result. I know there are several different approaches to it. Can anyone recommend me a proof which is both beautiful and useful for my future math study?

答:
There is a question on MathOverflow for interesting results in algebraic geometry that's both enticing and accessible to undergrads, and among the answers this outline of a proof of the 27 line theorem is given.
In the comments, the author points to a more detailed presentation (page 222), which unfortunately is in Italian.
(The link provided in the MO answer is dead, but I have linked to a more recent version of the same text here.)
Another commenter points to a few other sources for some of the more technical results used in the proof.</p>


问:
Proving $f(z)=\frac{1}{z}- \frac{1}{z-1}$ is not holomorphic on $0<|z|<1/2$ (question is wrong f is indeed holomorphic)
I want to prove that $$f(z)=\frac{1}{z}- \frac{1}{z-1}$$ is not holomorphic on $0<|z|< 1/2$.

My original thought was to use Morera's theorem, to prove that $\oint_\gamma f \neq 0$ for some path $\gamma$. I tried $\gamma$ as the circle of radius $\frac{1}{2}$ and I got

\begin{align}
\oint_\gamma f = \oint_\gamma \frac{1}{z} - \oint_\gamma \frac{1}{z-1} =2\pi i -2\pi i=0
\end{align}

where each of the integrals is respectively $2\pi i$ as a result of cauchies integral theorem. Does anybody have a suggestion as to what path to try? Moreover, what is the intuition between the choice of this path?

答: But $f$ is holomorphic, since both $\frac1z$ and $\frac1{z-1}$ are holomorphic.


问:
Monotone nature of sequences used to define 'lim sup' and 'lim inf'

Hi guys, this is from the definition of lim sup and lim inf.

I don't quite understand the second paragraph which goes "As n increases...".

How exactly could (yₙ) be monotone decreasing? Doesn't it depend on the original sequence?

For example when the original sequence diverges to +infinity...since we are taking the sup the set of values for yₙ will always be +infinity and hence not monotone decreasing right?

Please help me get around this,is my intution correct or have I got the definitions wrong...really appreciate your help.

Thanks for your time.

答:
Let $E_n=\{x_k: k\geq n\}$. Then $E_{n+1}\subset E_n$ for all $n$.

Claim: If $A \subset B$, then $\sup A \leq \sup B$.

Proof. For all $b\in B$, we know $b \leq \sup B$. Since $A\subset B$, we know that for all $a\in A,$ we have $a\leq \sup B$. Hence, $\sup B$ is an upper bound for $A$. By definition of supremum as the least upper bound, it follows that $\sup A \leq \sup B$.

From here, it should be obvious that $y_n$ is monotone decreasing. Your remark about blow up at infinity is valid, as in this case, we just have $y_n=+\infty$ for all $n$, which for the convenience of notation, we say approaches $+\infty$.

A similar argument holds for the lim inf.

One of the really nice things about lim inf and lim sup in analysis is that they always exist in some intuitive way, and this is helpful when one wants to work with subsequences (which gets more at the heart of compactness, a big topic in research).

The nice thing about monotone sequences is that even when the limit "doesn't exist," it always doesn't exist because of blow up, not because it just oscillates.


答:
What you have to realize is that if $I\subseteq J\subseteq\Bbb N$, then

$$\sup\{x_n:n\in I\}\le\sup\{x_n:n\in J\}\,.\tag{1}$$

This is because $\{x_n:n\in J\}$ includes every $x_n$ with $n\in I$, so its supremum is at least as big as the supremum of $\{x_n:n\in I\}$, and it also contains $x_n$ for each $n\in J\setminus I$. These extra members cannot decrease the supremum and might increase it.

In the present instance we have $n_1,n_2\in\Bbb N$ with $n_1\le n_2$, and we’re looking at $I=\{k\in\Bbb N:k\ge n_2\}$ and $J=\{k\in\Bbb N:k\ge n_1\}$. Since $n_1\le n_2$, $I\subseteq J$, and $(1)$ holds:

$$y_{n_2}=\sup\{x_k:k\ge n_2\}\le\sup\{x_k:k\ge n_1\}=y_{n_1}\,,$$

and $\langle y_n:n\in\Bbb N\rangle$ is indeed non-increasing. The specific values $y_n$ certainly depend on the original sequence, but the fact that they are non-increasing does not.

If the original sequence increases without bound, then $y_n=\infty$ for each $n\in\Bbb N$, so $\langle y_n:n\in\Bbb N\rangle$ is a constant sequence (in the extended reals), and all constant sequences are non-increasing. (They are also non-decreasing.) Note that the claim is not that $\langle y_n:n\in\Bbb N\rangle$ is strictly decreasing: the term decreasing in this context generally means non-strictly decreasing, i.e., non-increasing.

Title: Monotone nature of sequences used to define 'lim sup' and 'lim inf'

Body:
Hi guys, this is from the definition of lim sup and lim inf.

I don't quite understand the second paragraph which goes "As n increases...".

How exactly could (yₙ) be monotone decreasing? Doesn't it depend on the original sequence?

For example when the original sequence diverges to +infinity...since we are taking the sup the set of values for yₙ will always be +infinity and hence not monotone decreasing right?

Please help me get around this,is my intution correct or have I got the definitions wrong...really appreciate your help.

Thanks for your time.


答:
Think about listing out the elements in $(x_n)$:
$$ x_1, x_2, x_3, x_4, x_5, \dots $$
Then
\begin{align*}
y_1 & = \sup\{ x_1, x_2, \dots \},\\
y_2 &= \sup\{ x_2, x_3, x_4, \dots \},\\
\end{align*}
and so on.

So $y_1$ is the "maximum" over all $x_1, x_2, \dots$ and $y_2$ is the "maximum" over all $x_2, x_3, x_4, \dots$, and so on. This means that $y_1$ can only be greater or equal to $y_2$: if $x_1 \leq y_2$, then $y_1 = y_2$, and if $x_1 > y_2$, then $y_1 = x_1 > y_2$.

You can draw this relationship like this:
$$ \underbrace{x_1, \underbrace{x_2, x_3, x_4, \dots}_{\sup \text{of these elements} = y_2}}_{\sup \text{of these elements}= y_1} $$
This nesting effect is why we can say $(y_n)$ is monotone decreasing, without knowing anything about $(x_n)$.

Similar logic applies to general $y_k$. You can compare $y_{k}$ and $y_{k+1}$ by noting that $y_k = \max \{ y_{k+1}, x_k \} \geq y_{k+1}$. Thus, $(y_k)$ is monotone decreasing.


问:
Eigenvalues and eigenvectors of the operator $J$.
I am reading a linear algebra book. Assume $V$ is a real vector space. If $J : V \rightarrow V$ is linear such that $J^2 = -\mathit{id}$, it follows that the eigenvalues are positive or negative $i$.

What is $i$ in this case? How did he calculate the eigenvectors? What about the eigenvalues? The matrix of $J$ isn't known. This essentially is what is confusing me.

答:
We are given that J^2= -1.  That is, given any vector u, applying operator J to it twice gives -u.

v is an eigenvector of J, with eigenvalue $\lambda$, if and only if $Jv= \lambda v$.  Given that, applying J again, $J^2v= J(\lambda v)= \lambda JV= \lambda(\lambda v)= \lambda^2 v$ and, in this case, $J^2v=-1$ for all vectors so we have $\lambda^2= -1$.  Since i is defined as a root of -1, the two eigenvalues are i and -i as those are the only two complex numbers whose square is -1.


问:
Is a $\Omega$-Spectrum a connective one?
I can't find this result anywhere, but it seems pretty straightforward. I want to avoid silly mistakes, but I can't see any fault. I'd love to receive some feedback

Let $X$ be a $\Omega$-spectrum (of compactly generated topological spaces). By definition for $q<0$, $$\pi_q(X)=\lim_k \pi_{k+q <k}X_k$$

But since $\pi_i(X_k)\cong \pi_{i}(\Omega X_{k+1})\cong \pi_{i+1}(X_{k+1})$ the spaces are more and connected, in particular, $X_k$ is $k-1$-connected and therefore $\pi_q(X)=0$.

Is that correct?

答: Direct answers would be helpful. I believe you are saying for connective $\Omega$-spectrum $X$, $\pi_i(X)=0$ for all $i\leq -1$. For non connective $\Omega$-spectrum, there are nontrivial negative homotopy groups.


问:
If ${a_n}\rightarrow\infty\Rightarrow\frac{a_n}{e^{a_n}}\rightarrow 0$
This is driving me crazy. How to show that? Even though it looks intuitive, I cannot find a proper way to prove it. I tried to use the definition, to proof by absurdity, but still cannot. I do not unterstand what I am missing. Any hint?</p>

答:
For other two methods given below, you have to assume that $f$ is differentiable. But if you use
$$ e^x>x^2 $$
for large $x$, you don't need this assumption. In fact
$$ 0\le \frac{a_n}{e^{a_n}}\le \frac{1}{a_n} $$
for large $n$. This gives
$$ \lim_{n\to\infty}\frac{a_n}{e^{a_n}}=0. $$


问:
A confusion in the $\epsilon-\delta$ proof of $\lim_{x\to 2}\frac{x - 2}{1 + x ^ 2} = 0$
I want to prove
$$
\lim_{x\to 2}\frac{x - 2}{1 + x ^ 2} = 0
$$

The proof is: Let $\epsilon > 0$ be given. Then
$$\left|\frac{x - 2}{1 + x ^ 2} - 0\right| = \frac{|x - 2|}{1 + x ^ 2} \leqslant |x - 2| < \epsilon$$
provided $|x - 2| < \delta = \epsilon$.

My question is we have
$$\left|\frac{x - 2}{1 + x ^ 2} - 0\right| < \epsilon \text{ and }\frac{|x - 2|}{1 + x ^ 2} \leqslant |x - 2|.$$
How do we conclude
$$\left|\frac{x - 2}{1 + x ^ 2} - 0\right| = \frac{|x - 2|}{1 + x ^ 2} \leqslant |x - 2| < \epsilon$$
from these two inequalities? What if $\epsilon < |x - 2|$?

答:
You have made an incorrect statement which I think needs to be corrected:
it is not true that you "have"

$$ \left\lvert \frac{x - 2}{1 + x ^ 2} - 0\right\rvert < \epsilon  \tag1 $$

at the point in the proof you are questioning.
Instead, this inequality is the thing you have to prove at that point.

The real point of difficulty, however, seems to be how to recognize the implications that are used in the proof, and to understand which one follows from which.

Note that when we write, "Blah blah blah provided $|x - 2| < \delta = \epsilon$,"
it means

If $\delta = \epsilon$ and $|x - 2| < \delta,$ then blah blah blah.

So the steps of the proof are:

For whatever $\epsilon$ you are given, choose $\delta = \epsilon.$

Now you know that if $|x - 2| < \delta,$ then $|x - 2| < \epsilon.$

As separate facts, we see that $\left\lvert \frac{x - 2}{1 + x ^ 2} - 0\right\rvert = \frac{\left\lvert x - 2\right\rvert}{1 + x ^ 2}$ and that $\frac{\left\lvert x - 2\right\rvert}{1 + x ^ 2} \leq \left\lvert x - 2\right\rvert.$

Putting these facts together, if $|x - 2| < \delta,$ then $\left\lvert \frac{x - 2}{1 + x ^ 2} - 0\right\rvert = \frac{\left\lvert x - 2\right\rvert}{1 + x ^ 2} \leq \left\lvert x - 2\right\rvert < \epsilon.$

Notice that if $\left\lvert \frac{x - 2}{1 + x ^ 2} - 0\right\rvert = \frac{\left\lvert x - 2\right\rvert}{1 + x ^ 2} \leq \left\lvert x - 2\right\rvert < \epsilon$ then $\left\lvert \frac{x - 2}{1 + x ^ 2} - 0\right\rvert < \epsilon.$

From the two implications in the previous two steps, you can conclude that if $|x - 2| < \delta,$ then $\left\lvert \frac{x - 2}{1 + x ^ 2} - 0\right\rvert < \epsilon,$ which is what you need to show for the critical step of your delta-epsilon proof.

In the proof as you saw it, many of the steps above are rephrased or condensed, which might cause confusion.

You may well ask, "What if $\left\lvert x - 2\right\rvert > \epsilon$?"
In that case, since you have already said that $\delta = \epsilon,$
if $\left\lvert x - 2\right\rvert > \epsilon$ then
$\left\lvert x - 2\right\rvert > \delta$
and the "if $\left\lvert x - 2\right\rvert < \delta$" condition is false.
For a delta-epsilon proof where $x\to 2$ you only need to show what happens if
$\left\lvert x - 2\right\rvert < \delta$ is true;
it does not matter what happens when $\left\lvert x - 2\right\rvert < \delta$ is false.
So if $\left\lvert x - 2\right\rvert > \epsilon$ then we are exploring cases that are completely irrelevant to whether the limit exists and what it is.


问:
Is the absolute value on matrices well-defined with respect to PSD cone?
Consider for two matrices $A, B$ we say $A \geq B $ if $ A - B$ is positive semidefinite.
Also, there exists a definition of absolute value for a matrix $A$ that is $|A| = \sqrt{A^\dagger A}$.

I was studying Hermitian matrices(matrices like $A$ with complex entries such that $A^\dagger = A$) and faced with a problem, is the following statement true for two arbitrary Hermitian matrices?
$$ A \geq B \geq -A \implies |A| \geq |B|$$

Note that if $A$ is a Hermitian matrix then $|A|$ is PSD and the eigenvalues of $|A|$ are equal to the absolute value of the eigenvalues of $A$.

答:
The statement is false. Consider
$$A=\begin{bmatrix} 6 & 0 \\ 0 & 2\end{bmatrix}, \quad B=\begin{bmatrix} 1 & 2 \\ 2 & 1\end{bmatrix}.$$

We have $$A+B = \begin{bmatrix} 7 & 2 \\ 2 & 3\end{bmatrix} \ge 0, \quad A-B = \begin{bmatrix} 5 & -2 \\ -2 & 1\end{bmatrix} \ge 0$$
so $A \ge B \ge -A$. On the other hand, we have
$$|A| = A, \quad |B| = \sqrt{B^*B} = \sqrt{\begin{bmatrix} 5 & 4 \\ 4 & 5\end{bmatrix}} = \begin{bmatrix} 2 & 1 \\ 1 & 2\end{bmatrix}$$
and therefore $$|A|-|B| = A-|B| = \begin{bmatrix} 4 & -1 \\ -1 & 0\end{bmatrix} \not\ge 0$$
so $|A| \ge |B|$ is false.


问:
Introduction to modular arithmetic (residues)
I have been recently using the book introduction to Number Theory from Art of Problem Solving as I am preparing for the upcoming AMC 10 (I'm an 8th grader).

In section 12.3 problem 12.3.3 it says

Determine the residue of each of the following within the given modulus.

It says $71\pmod{3}$ and that the answer is $71 = 23 \cdot 3 +2$, hence $71 \equiv 2 \pmod{3}$.

Can someone explain why it has to be $71 = 23 \cdot 3 +2$? I'm re-reading the passage but I don't get the fundamental concept of a residue.

答:
The residue of $a$ in the modulus $n$ is the remainder of the division of $a$ by $b$, hence it must be a number $r$ such that $0\le r<b$.

The “division with remainder” can be written as $a=bq+r$, where $q$ is the quotient and $r$ is the remainder.

In your case, the quotient of the division of $71$ by $3$ is $23$, and
$$
71=3\cdot23+2
$$
Since $m\equiv n\pmod{b}$ means that $m-n$ is divisible by $b$, you surely have
$$
71-2=3\cdot23
$$
and therefore $71\equiv2\pmod{3}$.

It's nothing else than primary school division, written in a slightly different way.


问:
Proving a statement in the form $A \implies (B \implies C )$
Say I have a statement in the form

$$A \implies (B \implies C )$$

how can I prove it? Should I assume $A$ and $B$ and then try to prove $C$ from them ?

It seems that it could be the right thing because if $A = \bot $ then the statement is always true, so I assume $A$, now if $B = \bot$ then again I have an always true statement because $A$ is true and $B \implies C$ is true so I assume also $B$.

Is my reasoning right? Is there an intuitive explanation of this 'iterative' process?

答:
Should I assume $A$ and $B$ and then try to prove $C$ from them?

Using a form of natural deduction, the key is to make two separate assumptions...

Assume $A$

Assume $B$

Prove $C$

Discharge assumption (2) to obtain $B\implies C$

Discharge assumption (1) to obtain $A\implies(B\implies C)$


问:
Unique Minimal Basis?
Suppose there exists a collection of finite sets which is finite.

We pick up the minimal sub-collection such that any set in the collection can be expressed as a union of sets in the sub-collection.

Is the minimal sub-collection unique?

答: Yes it is unique. Let us consider two minimal sub-collections $\mathcal A, \mathcal B$. Neither contains the other by their minimality. Since everything is finite, let $A\in \mathcal A\setminus\mathcal B$ be an element of minimal cardinality. Now $A$ can be expressed as a union of element of $\mathcal B$, which all need to be of smaller cardinality than $A$ (or same but $A\not\in\mathcal B$), but $\mathcal A$ then contains all of them, letting $A$ be expressed by a union of elements of $\mathcal A$ contradicting the minimality of $\mathcal A$.</p>

Title: Unique Minimal Basis?


问:
Obtain $x$ from exponential equation
I'm looking for $x$ from the following equation:

$$\frac1{(1+x)^{\frac1{p}}}+\frac1{(1+x)^{\frac2{p}}} + \cdots \frac1{(1+x)^{\frac{n}{p}}} = 0$$

Or, what's the same:

$$\sum_{i=1}^n\frac1{(1+x)^{\frac{i}{p}}} = 0 $$

where $n$ and $p$ are given (say, $n=16$ and $p=52$, for example)

Is there a way to get $x$ from this?

答:
Let $z=\frac{1}{(1+x)^{1/p}}$. Let $$S=\sum_{i=1}^{n} z^i = z \frac{z^n-1}{z-1}$$

Solving for $z$ in terms of $S$ is not easy. You are trying to solve the polynomial equation:

$$z^{n+1} -(S+1)z+S = 0, 0<z<1$$

That is going to be difficult to come up with an exact solution, only approximation is going to be possible, I suspect.

Once you've solved for $z$, you get $x=z^{-p}-1$.

Title: Obtain $x$ from exponential equation


问:
Why is isomorphism defined as it is in category theory?
In terms of an intuitive understanding of categories, I've been told to think of a category as "objects with some structure" and a  morphism as "a transformation preserving the structure". (This understanding obviously falls apart when considering abstract categories, but seems to me to be unrelated to my problem, which is with its interpretation on concrete categories)

An isomorphism is then meant to indicate that the two objects in question have exactly identical structure. It seems to me that we have no guarantee that this notion is captured in general by our general definition of isomorphism.

If you asked someone who only knew of algebraic structures what an isomorphism on concrete categories ought to be, the answer would obviously be "a bijective morphism", but this fails for topological spaces.

By the same token, while it's certainly true that "having a morphism with inverse" ensures that two objects have identical structure for all the concrete categories I can think of, how do we know there isn't some kind of structure that violates our current notion of isomorphism in the same way that topological spaces violate the "naive" algebraic notion?

答:
This answer addresses (or I hope it does) your concern in the second paragraph, at least for groups. Maybe within category framework this approach becomes "trivial", but I can't say more, as category theory is far beyond my knowledge.

So, a group $G$ shows its structure as soon as we allow the internal operation to fully deploy its effects. Accordingly, we can reasonably state the following:

Definition 1. The structure of a group $G$ is the set $\theta_G:=\{\theta_a, a\in G\}\subseteq \operatorname{Sym}(G)$, where $\theta_a$ is the bijection on $G$ defined by: $g\mapsto \theta_a(g):=ag$.

Here a problem arises, if we want to determine whether two groups, $G$ and $\tilde G$, "have the same structure", since in general $\operatorname{Sym}(G)\cap\operatorname{Sym}(\tilde G)=\emptyset$, and then any attempt to "compare by overlapping" the structures $\theta_G$ and $\tilde\theta_\tilde G$ is doomed to fail. We can overcome this issue "by transporting" the structure of $G$ in $\operatorname{Sym}(\tilde G)$, and see whether we can made the "transported $\theta_G$" to overlap with $\tilde\theta_{\tilde G}$. If we succeed, then we can rightly say that $G$ and $\tilde G$ are isomorphic, since we have been able to bring the structure of one onto precisely that of the other. So, with reference to the following diagram:

$$
\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\las}[1]{\kern-1.5ex\xleftarrow{\ \ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
\begin{array}{c}
 &  & \\
G & \ras{\space\space\space f \space\space\space}   &   \tilde G \\
\da{\theta} &  &  \da{\tilde\theta}  \\
\operatorname{Sym}(G) & \ras{\varphi^{(f)}} & \operatorname{Sym}(\tilde G) \\
\end{array}
$$

we set forth this other:

Definition 2. Two groups, $G$ and $\tilde G$, are isomorphic if there is a bijection $f\colon G\to \tilde G$ such that the above diagram commutes, namely:

$$\tilde\theta=\varphi^{(f)}\theta f^{-1}\tag 1$$

where $\varphi^{(f)}\colon \operatorname{Sym}(G)\to \operatorname{Sym}(\tilde G)$ is the "structure transporting" bijection defined by $\sigma\mapsto f\sigma f^{-1}$.

Now, as a caracterization of such an "enabling" bijection $f$, the following holds:

Claim. Two groups $G$ and $\tilde G$ are isomorphic (as per Definition 2) if and only if there is a bijection $\psi\colon \tilde G \to G$ such that:

$$\psi(\tilde a\tilde g)=\psi(\tilde a)\psi(\tilde g), \space\space\forall \tilde a,\tilde g\in \tilde G \tag 2$$

Proof.

\begin{alignat}{1}
&\tilde\theta=\varphi^{(f)}\theta f^{-1} &\iff \\
&\tilde\theta_\tilde a(\tilde g)=(\varphi^{(f)}\theta f^{-1})(\tilde a)(\tilde g), \space\space\forall \tilde a,\tilde g\in \tilde G &\iff \\
&\tilde a\tilde g=(\varphi^{(f)}\theta f^{-1})(\tilde a)(\tilde g), \space\space\forall \tilde a,\tilde g\in \tilde G &\iff \\
&\tilde a\tilde g=(\varphi^{(f)}(\theta_{f^{-1}(\tilde a)}))(\tilde g), \space\space\forall \tilde a,\tilde g\in \tilde G &\iff \\
&\tilde a\tilde g=(f\theta_{f^{-1}(\tilde a)}f^{-1})(\tilde g), \space\space\forall \tilde a,\tilde g\in \tilde G &\iff \\
&\tilde a\tilde g=f(\theta_{f^{-1}(\tilde a)}(f^{-1}(\tilde g)), \space\space\forall \tilde a,\tilde g\in \tilde G &\iff \\
&\tilde a\tilde g=f(f^{-1}(\tilde a)f^{-1}(\tilde g)), \space\space\forall \tilde a,\tilde g\in \tilde G &\iff \\
&f^{-1}(\tilde a\tilde g)=f^{-1}(\tilde a)f^{-1}(\tilde g), \space\space\forall \tilde a,\tilde g\in \tilde G\\
\tag 3
\end{alignat}

So, $(1)\Rightarrow (2)$, by setting $\psi:=f^{-1}$, and $(2)\Rightarrow (1)$, by setting $f:=\psi^{-1}$. $\space\space\Box$

Therefore, a bijection between two groups with the property $(2)$ is rightly called isomorphism.


问:
Factor $t^3-6t^2+9t-4$

Factor $t^3-6t^2+9t-4.$

I have studied how to factor only quadratic polynomials. Can you give me a hint?

答: To my knowledge, there is no 'easy algorithm' to factor cubics or higher order polynomials into factors. The general method is to try to find a root by trial and error, as @Lion Hear already wrote in the comment.</p>

Title: Factor $t^3-6t^2+9t-4$


答:
You've encountered a very common problem with proofs of this sort, namely that the logical order of the steps is different from the order in which one does the work of finding the proof.

First, here's the proof for your question, in logical order. According to the definition of "limit", we're supposed to prove that, for every $\epsilon>0$, there exists some $\delta>0$ such that whenever $|x-2|<\delta$ then $|\frac{x-2}{1+x^2}-0|<\epsilon$. So consider an arbitrary $\epsilon>0$. We need to produce an appropriate $\delta$, and we choose $\delta=\epsilon$. [This choice looks completely arbitrary here, and in other problems the formula for $\delta$ would be more complicated.] Now we need to prove that this choice of $\delta$ works, i.e., that whenever $|x-2|<\delta$ then $|\frac{x-2}{1+x^2}|<\epsilon$.So consider any $x$ that satisfies $|x-2|<\delta$ and compute (as in your question) that $|\frac{x-2}{1+x^2}|<\delta=\epsilon$.  So our arbitrary-looking choice of $\delta$ has succeeded and the proof is complete.

It's reasonable to ask, though, where that arbitrary choice of $\delta$ came from. (It's even more reasonable to ask that in other problems, where the formula for $\delta$ can be a complicated mess rather than a nice simple $\delta=\epsilon$.) It came from the computation at the end of the logical proof. To figure out the right choice of $\delta$, one does the computation first, finding that when $|x-2|<\delta$ then $|\frac{x-2}{1+x^2}|<\delta$, and then saying "That would be what I want, namely $|\frac{x-2}{1+x^2}|<\epsilon$, if $\delta$ were equal to $\epsilon$." So then one can go back and fill in the part of the proof that says "choose $\delta=\epsilon$."

After going through this process in several examples, people often just write the computation that leads to $|\frac{x-2}{1+x^2}|<\delta$, and then say we're done if $\delta=\epsilon$", without bothering to say that, in logical order, this formula for $\delta$ should have been way earlier, near the beginning of the proof. That's what apparently happened in the proof that you quoted.

Title: A confusion in the $\epsilon-\delta$ proof of $\lim_{x\to 2}\frac{x - 2}{1 + x ^ 2} = 0$

Body:
I want to prove
$$
\lim_{x\to 2}\frac{x - 2}{1 + x ^ 2} = 0
$$

The proof is: Let $\epsilon > 0$ be given. Then
$$\left|\frac{x - 2}{1 + x ^ 2} - 0\right| = \frac{|x - 2|}{1 + x ^ 2} \leqslant |x - 2| < \epsilon$$
provided $|x - 2| < \delta = \epsilon$.

My question is we have
$$\left|\frac{x - 2}{1 + x ^ 2} - 0\right| < \epsilon \text{ and }\frac{|x - 2|}{1 + x ^ 2} \leqslant |x - 2|.$$
How do we conclude
$$\left|\frac{x - 2}{1 + x ^ 2} - 0\right| = \frac{|x - 2|}{1 + x ^ 2} \leqslant |x - 2| < \epsilon$$
from these two inequalities? What if $\epsilon < |x - 2|$?


答:
Although xpaul gave an excellent answer,

I wanted to post a reason why given that:

$$\lim_{n \to \infty} a_n = \infty$$

and

$$\lim_{x \to \infty} f(x) = 0$$ (where f is a function from the reals)

we can deduce that

$$\lim_{n \to \infty} f(a_n) = 0$$

The proof is from Theorem 2.4 here: https://www.math.ucdavis.edu/~hunter/m125a/intro_analysis_ch2.pdf

'sequential characterization of the limit'

So applying it to this problem $f(x)$ would be $\frac{x}{e^x}$

I will just prove the forward direction for infinity here.

So we're given that:

$$\lim_{x \to \infty} f(x) = 0$$

So given an $\epsilon > 0$, there exists a real number $R>0$ such that $x>R$ implies $|f(x)-0| < \epsilon$

We're also given that:

$$\lim_{n \to \infty} a_n = \infty$$

which means that for every real number $M>0$, there is a natural number $N$ such that $n>N$ implies $a_n > M$. So we pick $M=R$. So we have an $n>N$ such that $a_n>R$

But that implies that if $n>N$ then:

$|f(a_n)-0| < \epsilon$

So given an $\epsilon > 0$, there exists an N such that $n>N$ implies $|f(a_n)-0|<\epsilon$

That proves that:

$$\lim_{n \to \infty} f(a_n) = 0$$

So we can use l'Hopital's rule here:

$$\lim_{x \to \infty} \frac{x}{e^x} = \lim_{x \to \infty} \frac{1}{e^x} = 0$$

and deduce that given $$\lim_{n \to \infty} a_n=\infty$$,

$$\lim_{n \to \infty}\frac{a_n}{e^{a_n}} = 0$$

Title: If ${a_n}\rightarrow\infty\Rightarrow\frac{a_n}{e^{a_n}}\rightarrow 0$

Body: This is driving me crazy. How to show that? Even though it looks intuitive, I cannot find a proper way to prove it. I tried to use the definition, to proof by absurdity, but still cannot. I do not unterstand what I am missing. Any hint?</p>


答:
Whenever you see a polynomial of more than degree $2$ that you have to factor, first use the Rational Root Theorem. It states that, for an integer-coefficient polynomial $a_{n}x^{n} + a_{n-1}x^{n-1} +...+a_{1}x + a_{0}$, all rational rational roots are of the form $\frac{d_{0}}{d_{n}}$, where $d_{0}$ is a factor of $a_{0}$ and $d_{n}$ is a factor of $a_{n}$. (Note: factors can be positive or negative). Then, all possible rational roots of your polynomial are $-4, -2, -1, 1, 2, 4$. We find that $1$ and $4$ are roots, and thus:

$$t^{3}-6t^{2} + 9t - 4 = (t - 1)(t - 4)(t - r)$$

Where $r$ is the third root. Note that $(-1)(-4)(-r)$ must equal $-4$, so $-r = -1$ and thus $r = 1$. Then, the only roots are $\boxed{1\text{ and }4.}$

Title: Factor $t^3-6t^2+9t-4$

Body:
Factor $t^3-6t^2+9t-4.$

I have studied how to factor only quadratic polynomials. Can you give me a hint?


问:
Neat way to prove that $x^2\gt (1+x)(\ln(1+x))^{2}, ~\forall x\gt 0$
I am supposed to prove the following inequality. One standard method that can be put to use is considering a function $f(x)=x^2-(1+x)(\ln(1+x))^2$ and take its derivative to comment on the behaviour of its increase and arrive at concluding the inequality using the fact that $f(0)=0$.

$$x^2\gt (1+x)(\ln(1+x))^2,~~ \forall x\gt 0$$

I was wondering if there was a neater way to come up with the proof, especially as this is a contest-practice problem. Any hints are appreciated. Thanks.

Body:
Denote $y=\ln(1+x)$, then

$$x^2 - (1+x)(\ln(1+x))^2 = (e^y -1)^2 - e^y y^2\\
= e^{2y}-2e^{-y}+1 - e^y y^2 = e^y (e^y + e^{-y}-2 - y^2)\\
= e^y\left(\sum_{n=0}^\infty \frac{2 y^{2n}}{(2n)!}  - 2 - y^2 \right) = e^y\sum_{n=2}^\infty \frac{2 y^{2n}}{(2n)!} >0.\blacksquare
$$


问:
Classification of the surface of revolution with -1 constant Gaussian curvature
I know that under the arc length parametrization for the curve $t\mapsto (x(t),0,z(t))$, the Gaussian curvature of the surface of revolution is $K = -\frac{x''}{x}$. So, $K = -1\iff x'' = x$. Now, I know that the pseudosphere $\gamma(t) = (sin(t),0,cos(t)+ln(tan(t/2)))$ for $t\in (\pi/2,\pi)$ has -1 Gaussian curvature. As the arc length of $\gamma(t)$ is $Arclength(\gamma) =cot(s)$, $\gamma(cot^{-1}(s))$ is an arc length parameterization of it. But, $\frac{d^2}{ds^2}sin(cot^{-1}(s))\neq sin(cot^{-1}(s))$. What's wrong in here?

答: First, the arclength $s$ satisfies $ds/dt =-\cot t$. You lost a sign and an integral. So, in fact, $s=-\log\sin t$. Now finish correctly.


答:
Let me put $$r=\frac1{(1+x)^{1/p}},$$ so that your sum is $$\sum_{i=1}^n r^i.\tag{1}$$ Now, in the case that $x=-1,$ $r$ isn't even defined, so we must have $x\ne-1$. Moreover, if $p$ is even, then we must have $x>-1$. Now, multiplying $(1)$ by the non-zero value $r-1$ gives us $$\begin{align}(r-1)\sum_{i=1}^nr^i &= \sum_{i=1}^n(r-1)r^i\\ &= \sum_{i=1}^nr^{i+1}-r^i\\ &= \left(\sum_{i=1}^nr^{i+1}\right)- \left(\sum_{i=1}^nr^i\right)\\ &= \left(\sum_{i=2}^{n+1}r^i\right)- \left(\sum_{i=1}^nr^i\right)\\ &= r^{n+1}-r,\end{align}$$ so $$\sum_{i=1}^nr^i = \frac{r^{n+1}-r}{r-1},$$ and resubstitution yields $$\sum_{i=1}^n\frac1{(1+x)^{i/p}}=\cfrac{\frac1{(1+x)^{(n+1)/p}}-\frac1{(1+x)^{1/p}}}{\frac1{(1+x)^{1/p}}-1}.\tag{2}$$

Since you want your expression equal to $0$, then we need the numerator in $(2)$ to be $0$ (and the denominator to be nonzero), so we need $$ \frac1{(1+x)^{(n+1)/p}}-\frac1{(1+x)^{1/p}}=0.$$ Note that we can't have $$ \frac1{(1+x)^{1/p}}=0,$$ no matter what we pick for $x,n,p.$ (Why?) Factoring gets us $$\frac1{(1+x)^{1/p}}\left(\frac1{(1+x)^{n/p}}-1\right)=0,$$ whence $$\frac1{(1+x)^{n/p}}=1,$$ so $$(1+x)^{n/p}=1.$$ Can you get the rest of the way? Remember that our solution must not make the denominator of $(2)$ become $0$, so depending on $n$ and $p$, we may have no solution.

Title: Obtain $x$ from exponential equation

Body:
I'm looking for $x$ from the following equation:

$$\frac1{(1+x)^{\frac1{p}}}+\frac1{(1+x)^{\frac2{p}}} + \cdots \frac1{(1+x)^{\frac{n}{p}}} = 0$$

Or, what's the same:

$$\sum_{i=1}^n\frac1{(1+x)^{\frac{i}{p}}} = 0 $$

where $n$ and $p$ are given (say, $n=16$ and $p=52$, for example)

Is there a way to get $x$ from this?


问:
problem of maximum in physics
Aa elastic ball with a mass $m_1$ is moving at the speed of $v$ and strikes a first elastic ball at rest.

The second ball, with a mass $m_2$ acquires a speed $v_2$ and strikes a third elastic ball at rest, with a mass $m_3$, giving it a speed $v_3$.

If the masses of the first and the third balls are fixed, find what mass should be the second ball in order to have the highest velocity for the third ball.

I've tried with the conservation of the momentum and kinetic energy.

$m_1* (v_{m1})^i=m_1* (v_{m1})^f+m_2* (v_{m2})^f+m_3* (v_{m3})^f$

$(v_{m1})^i=(v_{m1})^f+(v_{m2})^f+(v_{m3})^f$

but I'm afraid I'm lost

答: Suppose the first ball has speed $v^\prime$ after the first collision. We combine momentum conservation $m_1v=m_1v^\prime+m_2v_2$ with energy conservation (since the collision is elastic) $\tfrac12m_1v^2=\tfrac12m_1v^{\prime2}+\tfrac12m_2v_2^2$ to prove $v_2=\frac{2m_1v}{m_1+m_2}$ (I'll leave that as an exercise). By the same logic,$$v_3=\frac{2m_2v_2}{m_2+m_3}=\frac{4m_1m_2v}{(m_1+m_2)(m_2+m_3)}.$$So we choose $m_2$ to minimize$$\frac{(m_1+m_2)(m_2+m_3)}{m_2}=m_1+m_3+m_2+\frac{m_1m_3}{m_2}$$in terms of fixed $m_1,\,m_3$. By the AM-GM inequality, we take $m_2=\sqrt{m_1m_3}$.

Title: problem of maximum in physics


答:
There are a few points to note. The operator $J$ cannot have any eigenvector in $V$, because such an eigenvector $v\ne0$, with respect to an eigenvalue $\lambda$, has to satisfy $J(v)=\lambda v$ and therefore
$$
-v=J(J(v))=\lambda^2v
$$
and so $\lambda^2+1=0$ (as $v\ne0$ by assumption). There is no real number $\lambda$ such that $\lambda^2+1=0$, and therefore no eigenvector we can speak of, given that $V$ is a real vector space and it makes no sense to do $iv$ (where $i$ is the imaginary unit, of course).

So it seems that your book is making some confusion.

However, attached to any real vector space there is a complex vector space $\hat{V}$. One can define it as $\mathbb{C}\otimes_{\mathbb{R}}V$ (tensor product). Basically, the set of $\hat{V}$ is $V\times V$, with addition defined in the obvious way and, for the complex number $a+ib$, with $a,b\in\mathbb{R}$,
$$
(a+ib)(u,v)=(au-bv,bu+av)
$$
In the case of $V=\mathbb{R}^n$, this is just seeing $\mathbb{R}^n$ as a subset of $\mathbb{C}^n$: if you take $w\in\mathbb{C}^n$, you can write
$$
w=u+iv
$$
where $u,v\in\mathbb{R}^n$ and we can identify the pair $(u,v)$ with the vector $u+iv$.

The operator $J$ induces an operator $\hat{J}\colon\hat{V}\to\hat{V}$ in the obvious way, namely $\hat{J}(u,v)=(J(u),J(v))$ and you can check linearity (as complex vector space).

This new operator $\hat{J}$ also satisfies $\hat{J}^2=-\mathit{id}$ and here the eigenvalues $i$ and $-i$ appear, with their eigenvectors.

Title: Eigenvalues and eigenvectors of the operator $J$.

Body:
I am reading a linear algebra book. Assume $V$ is a real vector space. If $J : V \rightarrow V$ is linear such that $J^2 = -\mathit{id}$, it follows that the eigenvalues are positive or negative $i$.

What is $i$ in this case? How did he calculate the eigenvectors? What about the eigenvalues? The matrix of $J$ isn't known. This essentially is what is confusing me.


问:
When there exists an element of $S_4$ in $H$ such that it is fixed by each element of $H$. ($H$ is a subgroup of $S_4$)
The following question was part of my Group Theory assignment and I am not able to get an idea on which result I should use to solve this question.

So, I am posting it here.

Let $S_4$ be the group of 4 element permutations and $H$ be any subgroup of $S_4$. Then which of following options are true. Give proper reasons.

If order of $H$ is 4, then there exists $i\in X$ such that $i$ is fixed by each element of $H$.

If order of $H$ is 6, then there exists $i\in X$ such that $i$ is fixed by each element of $H$.

If order of $H$ is 8, then there exists $i\in X$ such that $i$ is fixed by each element of $H$.

If order of $H$ is 12, then there exists $i\in X$ such that $i$ is fixed by each element of $H$.

I can't think of which result to use despite thinking a lot and would really like someone helping me. I have studied abstract algebra from Thomas Hungerford and did a lot of problems but couldn't did it.

答:
Hints:

If $H$ fixes $i\in X$, that means that $H$ must be a subgroup of the stabilizer of $i$ which is just $S_{X\setminus\{i\}}\cong S_3$.

This can be directly applied to get answer for problems 1., 3. and 4.

For problem 2., you can use Cauchy's theorem to obtain a 3-cycle $\sigma\in H$ and also a permutation $\tau\in H$ of order 2. Without loss of generality, we can assume $\sigma=(1\,2\,3)$.

Then show that either if $\tau$ is a transposition involving $4$ or if it is a product of two disjoint transpositions, then the subgroup generated by $\sigma$ and $\tau$ has more than 6 elements.


答:
In the first collision, the initial velocity of the first and second ball are $v_1$ and $0$ respectively, and the final ones are $v_1'$ and $v_2$. The conservation of momentum and kinetic energy gives:
$$\left.\begin{array}{c}
m_1v_1+m_2\cdot 0 = m_1v_1' + m_2v_2\\
\frac12 m_1v_1^2+\frac12 m_2\cdot 0^2 = \frac12m_1v_1'^2 + \frac12m_2v_2^2
  \end{array}\right\} \implies v_1' = \frac{m_1-m_2}{m_1+m_2}v_1,\quad v_2 = \frac{2m_1}{m_1+m_2}v_1.
$$

In the second collision, the initial velocity of the second and third ball are $v_2$ and $0$ respectively, and the final ones are $v_2'$ and $v_3$. The conservation of momentum and kinetic energy gives:
$$\left.\begin{array}{c}
m_2v_2+m_3\cdot 0 = m_2v_2' + m_3v_3\\
\frac12 m_2v_2^2+\frac12 m_3\cdot 0^2 = \frac12 m_2v_2'^2 + \frac12m_3v_3^2
  \end{array}\right\} \implies v_2' = \frac{m_2-m_3}{m_2+m_3}v_2,\quad v_3 = \frac{2m_2}{m_2+m_3}v_2.
$$
$v_3$ in terms of constants $m_1, m_3, v_1$ expressed as a function of $m_2$ is
$$v_3(m_2) = \frac{4m_1m_2}{(m_1+m_2)(m_2+m_3)}v_1.$$
Calculating $v_3'(m_2) = 0$ yields that $m_2 = \sqrt{m_1m_3}$.

Title: problem of maximum in physics

Body:
Aa elastic ball with a mass $m_1$ is moving at the speed of $v$ and strikes a first elastic ball at rest.

The second ball, with a mass $m_2$ acquires a speed $v_2$ and strikes a third elastic ball at rest, with a mass $m_3$, giving it a speed $v_3$.

If the masses of the first and the third balls are fixed, find what mass should be the second ball in order to have the highest velocity for the third ball.

I've tried with the conservation of the momentum and kinetic energy.

$m_1* (v_{m1})^i=m_1* (v_{m1})^f+m_2* (v_{m2})^f+m_3* (v_{m3})^f$

$(v_{m1})^i=(v_{m1})^f+(v_{m2})^f+(v_{m3})^f$

but I'm afraid I'm lost


问:
prove X and Y are uncorrelated
Prove that if E(X|Y) = E(X) then X and Y are uncorrelated and give an example where A and B are uncorrelated but E(A|B) isn't equal to E(A)

I know that for X and Y to be uncorrelated, correlation has to be 0. Which means Covariance has to be 0. And,

Cov(X,Y) = E(XY) - E(X) E(Y)

But this equation doesn't help get anywhere.

答:
But this equation doesn't help get anywhere.

That equation doesn't help but if you focus on the first addend you can observe that

$$\mathbb{E}[XY]=\mathbb{E}[\mathbb{E}[XY|Y]]=\mathbb{E}[Y\underbrace{\mathbb{E}[X|Y]}_{=\mathbb{E}[X]}]=\mathbb{E}[Y]\mathbb{E}[X]$$

Thus you are all set...

Title: prove X and Y are uncorrelated


问:
Quotient Ring of $M_n(\mathbb Z)$
I have an question.

How can I show that the quotient rings of $M_n(\mathbb Z)$ other than itself have finitely many elements?

If it was division ring instead of $\mathbb{Z}$,  I could say something. But know I don't know how to start.

答: For $I$ a non-zero two-sided ideal of $M_n(\Bbb{Z})$ then $\Bbb{Q} I$ is a two sided ideal of $M_n(\Bbb{Q})$ thus the whole it, a $n^2$-dimensional $\Bbb{Q}$-vector space, generated by $n^2$ elements, once multiplied by the right integer they are in $I$ and they generate a finite index sub-$\Bbb{Z}$-module of $M_n(\Bbb{Z})$.

Title: Quotient Ring of $M_n(\mathbb Z)$


问:
Show that $\omega'(x_k)\neq 0$
For $x_0, \ldots , x_n\in \mathbb{R}, \ x_j\neq x_k$, we define the polynomial $$\omega (x)=(x-x_0)\cdots (x-x_n)\in \Pi_{n+1}$$ Show that $\omega'(x_k)\neq 0$ and that the polynomials $$\ell_j(x)=\frac{\omega(x)}{(x-x_j)\omega'(x_j)}, \ j=0, \ldots , n$$ have the property $\ell_j(x_k)=\delta_{jk}, \ j,k=0, \ldots , n$.

We have that $$\omega'(x)=\sum_{i=1}^n\prod_{j=1, j\neq i}^n(x-x_j)$$ right? Each $x_i$ makes one term of the product to zero, but not the whole expression, right?

Can we just say that to prove that $\omega'(x_k)\neq 0$ or is there an other way to prove that formally?

答:
Fix $j\leq n,$ then
$$\ell_j(x)=\frac{\omega(x)}{(x-x_j)\omega'(x_j)}=\frac{\prod _{i=0}^n(x-x_i)}{(x-x_j)\omega'(x_j)}=\frac{\prod _{i\neq j}(x-x_i)}{\prod _{i\neq j}(x_j-x_i)},$$
so if $x=x_k$ there are two options i.e., either $k=j$ and everything cancels out or $k\neq j$ and the numerator vanishes.

Title: Show that $\omega'(x_k)\neq 0$


问:
Irreducible polynomial in $\mathbb{F}_p[x]$ divides $x^{p^n}-x$
Let $p(x)=x^{p^n}-x \in \mathbb{F}_p[x]$, $n \in \mathbb{N}$. I want to prove that any irreducible polynomial in $\mathbb{F}_p[x]$ with degree $n$ divides $p(x)$.

Let $q(x)$ be this irreducible polynomial:
$$q(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0 \quad , \quad a_i \in \mathbb{F}_p$$
I know that $q(x) \vert p(x)$ means that $q(x) \vert (x^{p^n-2}+x^{p^n-3}+\ldots +1)$ but I don't think this leads me anywhere. Can someone help me?

答:
Let $q(x)$ be an irreducible polynomial of degree $d$ dividing $n$. Then $F=\mathbb{F}_p[x]/(q(x))=\mathbb{F}_p[\alpha]$ is a field of $p^d$ elements, where $\alpha$ is a primitive element of $F$  and $q(x)$ is the minimal polynomial of $\alpha$. The set of non-zero elements $\left( \mathbb{F}_p[x]/(q(x)) \right)^{*}$ is a multiplicative group of order $ p^d-1$ and $\alpha$ is the generator of this multiplicative group. By Lagrange's theorem, the order of any element of this multiplicative group divide $p^d-1$ and hence, we have $\alpha^{p^d-1}=1$ or $\alpha^{p^d}=\alpha$. (One can use the Fermat's Little Theorem also)

Since $d~|~n$, we have  $n=dm$, for some integer $m$,  we have  $$\alpha^{p^n}=\alpha^{p^{dm}}=\alpha,$$ and we see $\alpha$ is a rootof $x^{p^n}-x.$

Now $q(x)$ is irreducible in $\mathbb{F}_p[x]$, so either $q(x)$ divides $x^{p^n}-x$ or by Bezout's identity, we have $$f(x)q(x)+g(x)(x^{p^n}-x)=1, ....(*)$$   for some polynomials $f(x),~ g(x) \in \mathbb{F}_p[x]$. In the second case, setting $x=\alpha$ in equation $(*)$, we get $0=1$, which is not possible. Hence the $d$-degree irreducible polynomial $q(x)$ must divide $x^{p^n}-x$. As $n ~|~n$, your question is answered.