问:
Find the radius of curvature at any point $(r, \theta)$
Find the radius of curvature at any point $(r, \theta)$ for the curve $a^2 = r^2\cos (2\theta)$

My Attempt:
$$a^2 = r^2 \cos (2\theta)$$
$$r^2 = a^2 \sec (2\theta)$$
Differentiating both sides
$$2rr_{1} = 2a^2 sec(2\theta)\cdot \tan (2\theta)$$
$$rr_{1} = a^2 \sec (2\theta) \tan (2\theta)$$

答: In polar coordinates, the radius of curvature is given by (have a look here)$$\large R=\frac{\left(r^2+\left(\frac{dr}{d\theta}\right)^2\right)^{3/2}}{\left|r^2+2\left(\frac{dr}{d\theta}\right)^2-r\left(\frac{d^2r}{d\theta^2}\right)\right|}$$


问:
How to find the closed form of $\sum_{k=0}^n\frac{1}{k+1} {\binom n k}^2$
How to sum the following identity: $$\sum_{k=0}^n\frac{1}{k+1}{\binom n k}^2$$  I know the answer is $\frac{(2n+1)!}{{(n+1)!}^2}$ but I have tried a lot to find the closed form through any algebraic way or combinatorial but not able to find it out, any help?

ContentLicense: CC BY-SA 4.0

Body:
Binomial theorem:
$$(1+t)^n=\sum_{k=0}^{n}{n \choose k} t^k~~~~(1)$$
Integrate w. r. t. $t$ from $t=0$ to $t=x$,
$$\frac{(1+x)^{n+1}-1}{n+1}=\sum_{k=0}^{n} \frac{n \choose k}{1+k}x^{1+k}~~~~(2)$$
Let $t=1/x$ in (1), to get
$$x^{-n}(1+x)^n=\sum_{k=0}^{n}{ n \choose k} x^{-k}~~~~(3)$$
Multiply (2) and (3) and retain only the terms of $x$
$$x \sum_{k=0}^{n}\frac{{n\choose k}^2}{1+k}+...+...+...=\frac{x^{-n}(1+x)^{2n+1}-x^{-n}(1+x)^n}{n+1}.$$
$$\implies  \sum_{k=0}^{n}\frac{{n\choose k}^2}{1+k}=[x^{n+1}] \frac{(1+x)^{2n+1}-(1+x)^n}{n+1}=\frac{2n+1\choose n+1}{n+1}=\frac{{2n+1\choose n}}{n+1}$$

Note that $[x^j] f(x)$ means coefficient of $x^j$ in $f(x)$.


问:
What is the Cov(A,B)?
I am working on a problem: $Y = A+B+C$.
I know that $Cov(A,B+C)=0$,$Cov(B,C)=0$, $E(B)=0$ and $E(C)=0$.
Can I get the $Cov(A,B)=0$?
Now I can only get $Cov(A,B)=-Cov(A,C)$.
But it seems to be intuitive that both $Cov(A,B)$ and $Cov(A,C)$ should be zero?

答:
For a counterexample, suppose $var(B) = var(C) > 0$, and $A = B - C$.

Then
$$
\begin{aligned}
Cov(A, B+C) &= Cov(B-C, B+C) \\
&= E((B-C)(B+C)) \\
&= E(B^2 - C^2) \\
&= E(B^2) - E(C^2) \\
&= var(B) - var(C) \\
&= 0
\end{aligned}$$
but
$$\begin{aligned}
Cov(A,B) &= Cov(B-C, B) \\
&= E((B-C)B) \\
&= E(B^2 - BC) \\
&= E(B^2) - E(BC) \\
&= var(B) - Cov(B,C) \\
&= var(B) \\
&> 0
\end{aligned}$$

Title: What is the Cov(A,B)?


问:
Use Fubini's Theorem to verify the integration by parts formula, and justify that the hypotheses of Fubini's Theorem are satisfied.
Let $f$ and $g$ be Lebesgue integrable real-valued functions on $[0,1]$, and define

$$F(x) = \int_0^x f(t)dt, \quad \text{and} \quad G(x) = \int_0^x g(t) dt.$$

Use Fubini's Theorem to verify the integration by parts formula

$$\int_0^1 F(x)g(x)dx = F(1)G(1) - \int_0^1 f(x)G(x)dx.$$

Justify that the hypotheses of Fubini's Theorem are satisfied.

This one seems straightforward to us, but for some reason we can't get anything nice to appear.

Here are our attempts for the left-hand side and the right-hand side respectively after the hint:

$$\begin{aligned} \int_0^1 F(x)g(x)dx &= \int_0^1 \int_0^x f(t)dtg(x)dx\\
&= \int_0^1\int_0^1 1_{[0,x]}f(t)g(x)dtdx\\
&= \int_0^1 f(t) \int_0^1 1_{[0,x]}g(x)dxdt. \end{aligned}$$

$$\begin{aligned} F(1)G(1) - \int_0^1 f(x)G(x)dx &= F(1)G(1) - \int_0^1 f(x)\int_0^x g(t)dtdx\\
&= F(1)G(1) - \int_0^1\int_0^1 1_{[0,x]}f(x)g(t)dtdx\\
&= F(1)G(1) - \int_0^1g(t)\int_0^1 1_{[0,x]}f(x)dxdt.\end{aligned}$$

答:
The first few steps just involves rewriting the terms as they are defined and moving constants inside or outside of an integral as needed.
$$
\begin{align*}F(1)G(1)&=\Big(\int_0^1f(t)dt\Big)\Big(\int_0^1 g(x)dx\Big)= \int_0^1g(x)\Big(\int_0^1 f(t)dt\Big)dx\\
&= \int_0^1g(x)\Big(\int_0^x f(t)dt\Big)dx + \int_0^1g(x)\Big(\int_x^1 f(t)dt\Big)dx\\
&= \int_0^1 g(x)F(x)dx + \int_0^1g(x)\Big(\int_x^1 f(t)dt\Big)dx\end{align*}$$
Now we use Fubini's Theorem to show that the second integral takes the appropriate form. Well
$$\begin{align*}\int_0^1 g(x)\Big(\int_x^1 f(t)dt\Big)dx &= \int_0^1g(x)\int_0^1 1_{[0\leq x\leq t\leq 1]}(x,t)f(t)dt dx\\
&= \int_0^1\int_0^1 1_{[0\leq x\leq t\leq 1]}(x,t)g(x)f(t)dt dx\\
&= \int_0^1\int_0^1 1_{[0\leq x\leq t\leq 1]}(x,t)g(x)f(t)dxdt\\
&= \int_0^1f(t)\int_0^1 1_{[0\leq x\leq t\leq 1]}(x,t)g(x)dx dt \\
&= \int_0^1f(t)\Big(\int_0^1 1_{[0\leq x\leq t]}(x)g(x)dx\Big) dt \\
&= \int_0^1f(t)\Big(\int_0^t g(x)dx\Big) dt \\
&= \int_0^1 f(t)G(t)dt\end{align*}$$
Hence after a change of variables we have that
$$F(1)G(1) = \int_0^1 g(x)F(x)dx + \int_0^1 f(x)G(x)dx $$


问:
$I=[a,b]$ and $f:[a,b] \to \mathbb R$ is continuous and for all $x$ in $I$ such property does hold
Assume $I=[a,b]$ and $f:[a,b] \to \mathbb R$ is continuous on $I$ and for all $x$ in $I$ exists  $y \in I$ such that $\left|f\left(y\right)\right|\le\frac{1}{2}\left|f\left(x\right)\right|$ prove that there exists a number $c$ in $I$ such that $f(c)=0$.

For $x_1 \in I$ exists $x_2 \in I$ such that $\left|f\left(x_2\right)\right|\le\frac{1}{2}\left|f\left(x_1\right)\right|$,inductively it's possible to find a sequence $(x_n) \in I$ such that for all $n \in \mathbb N$:$$\left|f\left(x_{n+1}\right)\right|\le\frac{1}{2}\left|f\left(x_n\right)\right|$$
And so
$$\left|f\left(x_{n+1}\right)\right|\le\frac{1}{2}\left|f\left(x_{n}\right)\right|\le\left(\frac{1}{2}\right)^{2}\left|f\left(x_{n-1}\right)\right|\le...\le\left(\frac{1}{2}\right)^{n-1}\left|f\left(x_{2}\right)\right|\le\left(\frac{1}{2}\right)^{n}\left|f\left(x_{1}\right)\right|$$

From here we see that the sequence $\left\{f\left(x_{n}\right)\right\}_n$ is a bounded sequence in $\mathbb R$ and by Bolzano–Weierstrass theorem there exists a subsequence $\left\{f\left(x_{n_k}\right)\right\}_k$ of $\left\{f\left(x_{n}\right)\right\}_n$ which converges to $f(c)$,on the other hand $$f(c)=\lim f(x_{n_k})=0 $$

I have two problems here,first how do we really "inductively" can find a sequence $(x_n) \in I$ such that for all $n \in \mathbb N$:$\left|f\left(x_{n+1}\right)\right|\le\frac{1}{2}\left|f\left(x_n\right)\right|$

Besides I have not used the fact that $f$ is continuous ,if I could show that $(x_n)$ is bounded then I could use  Bolzano–Weierstrass theorem to conclude that there exists a subsequence $\left(x_{n_k}\right)$  which converges to $c$,and continuity of $f$ implies that $f(c)=\lim f(x_{n_k})=0 $,but I cannot show that $(x_n)$ is bounded.

答: Let $m=\inf \{|f(x)|: a \leq x \leq b\}$. There exists $c$ such that $|f(c)|=m$. There exist $y$ such that $|f(y)| \leq \frac 1 2 |f(c)| =\frac  m 2$. This is a contradicton unless $m=0$. Hence $f(c)=0$.


问:
What's the most elementary way to solve this trigonometric problem?

A bead is threaded onto a light, inextensible string of length $4m$. One end of the string is fixed to a point, $A$, on a (vertical) wall. The other end of the string is attached to a point $B$ on the wall exactly $2m$ directly below $A.$ The bead is held in place so that it is at a distance of $1m$ from the wall, such that the string is taut and the plane the string is in is perpendicular to the plane the wall is in. Find the two possible vertical components of the displacement from $B$ to the bead.

This is a question I came up with, and I thought there should be some relatively simple trigonometric methods to get to the answer. I tried lots of stuff but nothing seemed to work.

Of course, we could find the equation of the ellipse that corresponds to the locus of points the bead could be when the string is taut, and then find the two values of $y$ when $x = 1$. But I'm looking for more elementary methods involving only trigonometry and Pythagoras. This is because I want the answer to be aimed at secondary school students who know elementary trigonometry only (Pythagoras, addition angle formulae, R addition formulae etc).

Thanks in advance.

答:
This is a slight variation on the answer of @leoli1 which omits the square roots. So start with $$\begin{eqnarray}
y^2+1&=& \lvert BC \rvert^2 \\
(y+2)^2+1&=& \lvert AC \rvert^2
\end{eqnarray}$$ but now note instead that $$ \lvert AC \rvert^2 = (4 - \lvert BC \rvert)^2.$$ Subtract the first equality above from the second to find $$4y + 4 = 16 - 8 \lvert BC \rvert$$ so $\lvert BC \rvert = (3 - y)/2$. Substitute this back in the first equality to get a quadratic equation for $y$.


答:
I don't know if it's a bad manner to post the solution of own question but anyway thanks to Brian M. scott for his comment and using his suggestion: $\frac1{k+1}\binom{n}k=\frac1{n+1}\binom{n+1}{k+1}$ \begin{align}
\sum_{k=0}^n\frac{1}{k+1}{\binom n k}^2&=\sum_{k=0}^n\frac{1}{n+1}\binom{n+1}{k+1}\binom{n}{k}\\
&=\frac{1}{n+1}\sum_{k=0}^n\binom{n+1}{n-k}\binom{n}{k}\\
&=\frac{1}{n+1}\times\frac{(2n+1)!}{n!(n+1)!}\\
&=\frac{(2n+1)!}{{(n+1)!}^2}
\end{align}

Title: How to find the closed form of $\sum_{k=0}^n\frac{1}{k+1} {\binom n k}^2$

Body: How to sum the following identity: $$\sum_{k=0}^n\frac{1}{k+1}{\binom n k}^2$$  I know the answer is $\frac{(2n+1)!}{{(n+1)!}^2}$ but I have tried a lot to find the closed form through any algebraic way or combinatorial but not able to find it out, any help?


Body:
consider this polynomial (or OGF).
$$
\sum_{k=0}^n \binom{n}{k}^2 \frac{1}{k+1}=[x^{n+1}]\sum_{k=0}^n \binom{n}{n-k}x^{n-k}\times \binom{n}{k}\frac{x^{k+1}}{k+1}
$$

It's the convolution(product) of $(1+x)^n$ and $F(x)$ where

$$
\begin{aligned}
F(x)&=\sum_{k=0}^n \binom{n}{k}\frac{x^{k+1}}{k+1}\\
F'(x)&=\sum_{k=0}^n \binom{n}{k}x^k=(1+x)^n\\
F(x)&=\frac{(1+x)^{n+1}}{n+1}
\end{aligned}
$$

Thus the polynomial is $(1+x)^n\frac{(1+x)^{n+1}}{n+1}$ and the $(n+1)$ -th coefficient is $\frac{1}{n+1}\binom{2n+1}{n+1}=\frac{(2n+1)!}{(n+1)!(2n+1-n-1)!}\frac{1}{n+1}=\frac{(2n+1)!}{{\left((n+1)!\right)}^2}$


问:
Probability of all elements of a subset being coprime
Let $S=\{1,..,n\}$ and $R \subset S$ ($|R|=k$, $k<n$) -- $R$ is a random subset of $S$. Let $m=min(R)$, and $R'=\{x-m: x \in R, x \neq m\}$, so $|R'|=k-1$.

What's the probability that $gcd(R')=1$?

EDIT:

I know that probability of $gcd(R)=1$ is $1/\zeta(k)$ when $n = \infty$ and I was wondering about this variation. When $n=\infty$, $R'$ is just a random subset of natural numbes, just like $R$ is, so in both cases elements have a uniform distribution. When $n$ is finite, elements of $R'$ don't have a uniform distribution -- for example $1 \in R'$ is more probable than $n-1 \in R'$.

答:
Hints:

Inclusion-exclusion.

Let $A$ denote the event that the elements of $R$ are coprime. For every integer $i\gt n/k$, let $a(i)=0$. For every positive integer $i\leqslant n/k$,
$$
a(i)=\frac{\lfloor\frac{n}i\rfloor!}{n^k\left(\lfloor\frac{n}i\rfloor-k\right)!},
$$
then
$$
P[A]=1-\sum_pa(p)+\sum_{p,q}a(pq)-\sum_{p,q,r}a(pqr)+\cdots
$$
where each sum is on the distinct prime numbers $p$, $q$, $r$, and so on (the sum is finite since, in fact, at most the first $\log_2(n/k)$ terms of this series are nonzero).

ContentLicense: CC BY-SA 3.0


问:
Prove that $11,111,111.....$ is not a perfect square
I know there are many other same related questions, but as my profile says I don't have a mentor, I just wanted you guys to check my solution.

First I noticed that,
$11$ = $10 + 1$,

$111$ = $10^2+10+1$,

Similarly, $\underbrace{111\cdots111}_{n\,\text{times}}$ = $10^{n-1}+10^{n-2}+...+10+1$,

which  can easily be evaluated as it is a finite geometric series.

$$\sum_{i=0}^{n-1} 10^i=\frac{10^n-1}{9}$$ for $n \ge
2$.

After this, I checked perfect square modulo $4$,

Let $j$ be any integer, then if $j$ is even, then $j=2k \Rightarrow j^2 = 4k^2 \equiv 0 \mod 4$

and if $j$ is odd then, $j = 2k+1 \Rightarrow j^2 = 4k^2+4k+1 \equiv 1 (\mod4)$

Then, Since $\underbrace{111\cdots111}_{n\,\text{times}} = \frac{10^n-1}{9} \Rightarrow (9) (\underbrace{111\cdots111}_{n\,\text{times}}) = {10^n-1}$ $\equiv$ $2^n-1$ $(\mod 4)$

$\Rightarrow (\underbrace{111\cdots111}_{n\,\text{times}}) \equiv 2^n-1 \mod 4$

Since for $n \ge 2, 2^n$ is always a multiple of $4$. Hence,

$\Rightarrow (\underbrace{111\cdots111}_{n\,\text{times}}) \equiv -1 \mod 4$

$\Rightarrow (\underbrace{111\cdots111}_{n\,\text{times}}) \equiv 3\mod 4$

But a perfect square leaves remainder $0$ if it is even and $1$ if it is odd. So, clearly for

$n \ge 2,$  $(\underbrace{111\cdots111}_{n\,\text{times}})$ is not a perfect square.

I know there are other questions related to this but I just wanted to verify if my solution was correct. Thank you and suggestions would be appreciated!

答:
base 3:  102 * 102 = 11111
base 7:   26 *  26 = 1111

These disprove the assertion.

In decimal, any number 11 modulo 100 is 3 mod 4, and is thus not square.


问:
Problem calculating a complex integral
I'm trying to calculate the integral from $(0,0)$ to $(1,1)$ of the following function: $z\operatorname{Im}(z) + \operatorname{Re}(z) + 2z\overline z$ on the parabola $y=x^2$.

The problem is that when I switch to $y$, I end up with the integral $d(x+ix^2)$...and I don't know what to do with the $x^2$.

Any help would be appreciated, thanks!

答:
Put $\;z=x+iy\sim(x,y)\;$ , and then your function is

$$f(z)=(x+iy)y+x+2x^2+2y^2\;\;(\text{ since}\;\;z\overline z=|z|^2=x^2+y^2\;$ )$$

The parametrization of that chunk of that parabola is $\;(t, t^2)\,,\,\,\;$  so

$$\int_{y=x^2}f(z)\,dz=\int_\alpha^\beta(t^3+2t^2+2t^4, t^4)\cdot(1, 2t)dt$$

and you get a rather nice improper integral if $\;\alpha,\,\beta\;$ or both are infinite...

Title: Problem calculating a complex integral


问:
Representation of exponential by a finite sum
In the simplest case, I know that the expression $\exp({\frac{x_i + x_j}{2}})$, where $x_{i,j} \in \mathbb{R}$ can be represented by the following

$$
\sum_{k=1}^nf_{i,k}f_{j,k} = \exp({\frac{x_i + x_j}{2}})
$$

I know this is possible, I just can't see how to do it.

Something I have been considering, for n = 2, is the following

$$
\sum_{k=0}^{\infty}\frac{x^{2k}}{(2k)!} + \sum_{k=1}^{\infty}\frac{x^{2k-1}}{(2k-1)!} = \exp(x)
$$

but I can't see a way of generalising this, or even if this is the correct route.  Any thoughts on this would be appreciated.

答:
Note that $\sum_k e^{(x_i+x_k)\over2}e^{(x_j+x_k)\over2}=e^{(x_i+x_j)\over 2}\sum_k e^{x_k}$, so that choosing a normalization $Z=\sum_k e^{x_k}$ you can solve for $f_{ik}={1\over \sqrt Z}e^{(x_i+x_k)\over2}$ .

More generally, if you think of $e^{(x_i+x_j)\over 2}$ as a matrix $A_{ij} = u^T u$ with $u_i = e^{x_i\over2}$, you see that your equation is $F^2 = u^Tu$, and assuming $F={1\over \sqrt Z} u^Tu$ will get you $F^2={1\over Z}u^T(uu^T)u$, and $uu^T$ is a scalar, so setting $Z=uu^T$ solves.


问:
Algorithm based Combinatorics problem IMO style
The problem is described in the picture, the solution is suppose to be a constructive proof, so you need to find an algorithm that works regardless of stuff like running time. Any help would be grate I can’t solve this and I’ve trying for a while

EDIT:
My main direction was changing the sign on every raw with negative sum, now every raw is complete and the sum of all entries is nonnegative (since the sum of all raws is now nonnegative)

答:
If any row or column has a negative sum, choose one and reverse the signs.

Iterate the process.

After each transition, the sum of all the entries in the matrix is increased.

But the absolute value of the entries remains unchanged, hence there are only finitely many possibilities for the total sum.

It follows that the process will eventually terminate.


问:
Calculate the following integral: $\oint \frac{1}{z^{n}-1}dz$ where $\gamma ={|z-1|=\frac{1}{n} , n\in \mathbb{N}}$
Calculate the following integral: $\oint \frac{1}{z^{n}-1}dz$ where $\gamma ={|z-1|=\frac{1}{n} ,  n\in \mathbb{N}}$.

I used the following identity: $a^{n}-b^{n}=(a-b)(a^{n-1}b+a^{n-2}b^2...+a^{2}b^{n-2}+ab^{n-1})$ and the formula of the geometric sum to reach the following integral: $\oint \frac{\frac{z-1}{z^n-1}}{z-1}dz$, then I multiplied and divided by $2\pi i $.

My thinking was that this is a difficult Integral, so I decided to use Cauchy's integral formula, but the function $f(z)=\frac{z-1}{z^n-1}$ is undefined for z=1, my question is whether defining a new function that is defined for z=1 could solve this problem?

For example: $g(z)=\left\{\begin{matrix}
 & \\ \frac{z-1}{z^n-1}, if: z\neq 1& \\ 1,   if: z=1\end{matrix}\right.$

答: Yes, your function is nothing but $g(z)=\frac 1 {1+z+z^{2}+\cdots z^{n-1}}$. Then $g$ is analytic inside and on the contour  and the given integral is $\int_{\gamma} \frac {g(z)} {z-1}dz$. So you can apply Cauchy's Integral formula.


答:
Just use the following simple algorithm:
In each turn, reverse the signs of a random row or column with negative sum and we'll prove that this procedure ends in finite moves:
Note that by using this algorithm the sum of numbers in all of the board will always increase and because this sum is bounded from above and also there are finite possibilities for this sum, our algorithm will stop somewhere and there is the point that the sum of all rows and columns are non-negative.</p>

Title: Algorithm based Combinatorics problem IMO style

Body:
The problem is described in the picture, the solution is suppose to be a constructive proof, so you need to find an algorithm that works regardless of stuff like running time. Any help would be grate I can’t solve this and I’ve trying for a while

EDIT:
My main direction was changing the sign on every raw with negative sum, now every raw is complete and the sum of all entries is nonnegative (since the sum of all raws is now nonnegative)


问:
How do you find the fixed and period-$2$ points of $f(x)=x^2-3x+3$?
I am trying to do this question using the Fixed Point Factor Theorem. I keep getting an answer $>0$ at the end of my long division of $f(x)-x$ into $f^2(x)-x$ therefore I must using the wrong divider. Can somebody help me please?</p>

答:
Fixed points ( period 1 points): $f(x)=x$

so solve $x^2-4x+3=0$ (which leads to $x=3$ or $x=1$)

Period two points: $f(f(x))=x$

so

$f(f(x))= x^4-6x^3+12x^2-9x+3$, so to get only period 2 points we want to divide the fixed points out of $x^4-6x^3+12x^2-10x+3=(x-3)(x-1)(x^2-2x+1)$.

As it turns out, $x^2-2x+1$ has only $x=1$ as roots, so there are no proper period-2 points.


问:
$f\circ f$ attains its minimum at at least two points
Let $f$ be continuous on $[a,+\infty)$. Assume $\lim_{x\to+\infty}f(x)=+\infty$, and $f$ attain its minimum at $c>a$. Assume further $a\leq f(c)<c<f(a)$. How to prove that $f\circ f$ attains its minimum at at least two points? What extra function shall we construct then?

答: Note that $f\circ f$ is defined on $[a,\infty),$ $\lim_{x\to\infty} f\circ f(x) = \infty$ and it attains its minimum at points $b$ with $f(b)=c.$ So you need to show that $f$ obtains the value $c$ twice. Does this help?


问:
Why the community of two subspace is not necessarily a whole space of the set R?
If W1 and W2 are the subspace of W, prove that

(a): The Intersection of W1 and W2 is a subspace of W

(b): The community of W1 and W2 may not necessarily be subspace W

答:
For instance, consider two lines in the 2D or 3D real space; a line is a 1D-subspace.

If the lines are distinct, the intersection is given by the zero point $\{0\}$ which is a 0D-subspace.
The union of two distinct lines is not a subspace. The smallest subspace containing these lines is the plane (2D-space) generated by the two lines.


问:
Why is $\frac{987654321}{123456789} = 8.0000000729?!$
Many years ago,
I noticed that $987654321/123456789 = 8.0000000729\ldots$.

I sent it in to Martin Gardner at Scientific American
and he published it in his column!!!

My life has gone downhill since then:)

My questions are:

Why is this so?

What happens beyond the "$729$"?

What happens in bases other than $10$?

答:
$98765432 / 12345679 = 8$, exactly. You can see how the pattern works by multiplying out $12345679 * 8$ starting at either end.

This explains why your fraction is close to an integer. If you think the $729$ is interesting (I don't), it can be explained by some of the other answers here.

Edit:

What can we say about the fact that $12345679 * 8 = 98765432$? I have been aware of this 'factlet' for about 20 years, and remember it being used to 'demonstrate' calculators (which often had 8 digit displays back in the day).

I just recently realised that:

$$
\frac{1}{81} = \left(\frac{1}{9}\right)^2 = \left(\sum_{k=1}^{\infty}\frac{1}{10^k}\right)^2 = \sum_{k=1}^{\infty} \sum_{m=1}^{k-1} \frac{1}{10^m} \frac{1}{10^{k-m}} = \sum_{k=1}^{\infty} \frac{k-1}{10^k}
$$

In other words, while $\frac{1}{9} = 0.1111111\ldots$
$$
\frac{1}{81} = 0.01 + 0.002 + 0.0003 + 0.0004 + 0.00005 \ldots
$$

It is pretty easy to see that this infinite sum is going to converge to something starting $0.012345\ldots$. If you keep on adding, or work out $\frac{1}{81}$ by division, you get
$$
0.012345679012345679012345679\ldots
$$
When you get to the point where you add $\frac{10}{10^{11}}$, the first carry happens, which leads to the 9 where you might expect an 8. After that every addition carries and the decimal expansion repeats every 9 digits (not every ten - because the amount we carry keeps on getting bigger and bigger).

Now, $\frac{8}{81} = \frac{9}{81} - \frac{1}{81}$, or
$$
\frac{8}{81} = 0.11111111\ldots - 0.012345679012345\ldots
$$
Think of each '1' digit in $0.111\ldots$ as being a '10' in the next column. This means that we can work out $\frac{8}{81}$ as the "10's complement" of $\frac{1}{81}$, since we are subtracting a digit between $1$ and $9$ from $10$, to get another single digit which appears in the same place. So $\frac{8}{81}$ starts $0.098765\ldots$. The only break in the pattern is when you get to the digit '0' - subtracting 0 from 10 leaves you with 10, or a '1' in the next digit on the left, changing the 1 to a 2.

So
$$
\frac{8}{81} = 0.098765432098765432098765\ldots
$$

and therefore
$$
0.0123456790123456790\ldots * 8 = 0.0987654320987654320\ldots
$$
and clearly this gets you that
$$
12345679 * 8 = 98765432
$$


问:
$\lim_{n \rightarrow \infty} \frac{2^n+(-1)^n}{2^{n+1}+(-1)^{n+1}}$
I have to find this limit but I'm not sure how can I do it

Find this limit:

$\lim_{n \rightarrow \infty} \frac{2^n+(-1)^n}{2^{n+1}+(-1)^{n+1}}$

I analyzed it by this way:

\begin{align}
\underbrace{\lim_{n \rightarrow \infty}\left ( \frac{2^n}{2^{n+1}+(-1)^{n+1}} \right )}_{\text{This must go to zero}} + \underbrace{\lim_{n \rightarrow \infty} \left ( \frac{(-1)^n}{2^{n+1}+(-1)^{n+1}} \right )}_{\text{This, also must converge to zero}}
\end{align}

Am I correct? How can I proove it?

答: When $n \to \infty$, the dominant term up is $2^n$ and down it is $2^{n+1}$ so the required limit is $$L=\lim_{n \to \infty} \frac{2^n}{2^{n+1}}=\frac{1}{2}$$


答: As a consequence of the binomial theorem we have that $$\sum_{k=0}^{n}\dbinom{n}{k}\frac{x^{k+1}}{k+1}=\frac{\left(1+x\right)^{n+1}-1}{n+1}$$ hence, by the integral representation of the binomial coefficient, we have $$\sum_{k=0}^{n}\dbinom{n}{k}^{2}\frac{1}{k+1}=\frac{1}{2\pi i}\oint_{\left|z\right|=\varepsilon}\left(1+z\right)^{n}\sum_{k=0}^{n}\dbinom{n}{k}\frac{z^{-k-1}}{k+1}dz$$ $$=\frac{1}{n+1}\left[\frac{1}{2\pi i}\oint_{\left|z\right|=\varepsilon}\frac{\left(1+z\right)^{2n+1}}{z^{n+1}}dz-\frac{1}{2\pi i}\oint_{\left|z\right|=\varepsilon}\left(1+z\right)^{n}dz\right]=\color{red}{\frac{\dbinom{2n+1}{n}}{n+1}}.$$


问:
Prove $Area(T(\mathcal{R}))=|Det(A)|.Area(\mathcal{R})$

Let $T:\mathbb{R}^2\to\mathbb{R}^2$ be a linear transformation with matrix $A$. Let
$\mathcal{R}$ be a region in $\mathbb{R}^2$. Then:
$$
Area(T(\mathcal{R}))=|Det(A)|.Area(\mathcal{R})
$$
It is also statemed that this can be extended to higher dimensions too.

Is there a formal proof for this ? Or where does this actually come from ?

I just know that the above statements can be verified for specific examples and $|Det(A)|$ is the area of the parallelogram formed by the column vectors of $A$.

答:
As an illustration to my comment on your post I give you the calculation for any invertible linear transformation $T:\mathbb{R}^n\to\mathbb{R}^n$ with matrix $A$. In your case ($n = 2$) the n-dimensional "$Volume$" would be the "$Area$":

Startig with $y = Tx$ we have

\begin{eqnarray*} Volume(T(\mathcal{R}))
& = & \int_{T(\mathcal{R})}\;dy \\
& = & \int_{\mathcal{R}}\;d(Tx) \\
& = & \int_{\mathcal{R}}|D(Tx)|\;dx\\
& = & \int_{\mathcal{R}}|det(A)|\;dx\\
& = & |\det(A)|\int_{\mathcal{R}}\;dx\\
& = & |\det(A)|Volume(\mathcal{R})\\
\end{eqnarray*}


问:
Finding all normal subgroups of a group
On my homework today, we had to find all the normal subgroups of $D_{n}$, the dihedral group of order 2n. I solved the problem by looking at how the conjugacy classes change based on whether n is even or odd and then constructed the normal subgroups as unions of the conjugacy classes.

I have 2 questions:

(i) Is there a better way to approach the problem than looking at the conjugacy classes?

(ii) Could someone explain why we want to find all the normal subgroups of a particular group?   How does this provide us additional insight into the structure of the group we are studying? (Right now, this exercise feels more like a 'computation' to me than a way of understanding $D_{n}$)

Thanks :)

答:
I'm assuming you want to know more about group theory, so here is an answer to point you to some of the interesting things that you can learn if you are interested in computing with finite groups.

(i) If you have the conjugacy classes and there are not too many, then writing normal subgroups as unions of conjugacy classes is fast and easy.  This is often the case when you only have access to a group through its character table, such as when the Suzuki and the sporadic simple groups were being discovered, and we wanted to understand some of their (non-simple) subgroups.

For a typical finite group given concretely as a permutation group, you use a special type of induction working along what is called a chief series, where you find a maximal chain of normal subgroups.  It turns out there are some fairly easy ways to find these: for a solvable group, or any group G with an abelian quotient group, you can fairly easily and concretely find the derived subgroup, [G,G].  The quotient group is an abelian group, so every subgroup between the whole group and the derived subgroup is normal.  You then take the derived subgroup of the derived subgroup, but now you only take subgroups that are normalized by G/[G,G].  The action of G/[G,G] is relatively plain and easy to understand, so these "invariant" subgroups are pretty easy to find.  To finish the inductive step, you need to find the normal subgroups of G/[[G,G],[G,G]] that don't contain [G,G], and this can be done by a slightly easier version of the general subgroup lattice algorithm using what is called the first cohomology group, or "derivations" (basically derivatives for groups, and related in a way to the derived subgroup).  When you reach the point where the group is its own derived subgroup, a perfect group, then a second algorithm begins, that identifies the simple groups involved in the top of the group, and then constructs the solvable group below them.  Sometimes you even have to repeat those last two steps, but not for groups of order less than 6024 ≈ 4.7e42 or so.

(ii) Often in mathematics, you want to use "induction".  A normal subgroup is one of the two main ways to do induction in group theory.  Usually it is not necessary to find all normal subgroups, but rather a single (nice) chief series will do.  Some groups only have a single chief series (a fair number of dihedral groups are like this), and so finding the chief series and finding all normal subgroups is the same question.  You might try the symmetric group of degree 4 and order 24 as an example.

ContentLicense: CC BY-SA 2.5

Title: Finding all normal subgroups of a group


问:
Landmarks of subjects of mathematics
In order to have a good view of the whole mathematical landscape one might want to know a deep theorem from the main subjects (I think my view is too narrow so I want to extend it).

For example in natural number theory it is good to know quadratic reciprocity and in linear algebra it's good to know the Cayley-Hamilton theorem (to give two examples).

So, what is one (per post) deep and representative theorem of each subject that one can spend a couple of months or so to learn about? (In Combinatorics, Graph theory, Real Analysis, Logic, Differential Geometry, etc.)

答: In Probability theory I was told by my professor, that that 3 most important theorems are Central Limit Theorem, Law of Large Numbers and Law of the Iterated Logarithm.

Title: Landmarks of subjects of mathematics


问:
Convergence of a sequence defined by recursion
Let $x_1\in(0,\pi/2]$ and
$$x_{n+1}=x_n-\frac{1}{n+1}\sin x_n,\ \forall n\in\mathbb{Z}_{\geq1}$$
It could be seen that the sequence $\{x_n\}_{n=1}^{\infty}$ is positive and strictly decreasing, so it has a limit $\lim_{n\to\infty}x_n=c\geq0$. And then I could calculate that $c=0$ from this point of view:
$$
x_{n}-x_1=\sum_{k=2}^n(x_k-x_{k-1})=-\sum_{k=2}^n\frac{1}{k}\sin x_{k-1}<-\sin c\sum_{k=2}^nk^{-1}
$$
Because of the divergence of the harmonic series and the convergence of $\{x_n\}_{n=1}^{\infty}$, we must have $c=0$.

And my question raised during I did the derivation above: How quickly does the sequence $\{x_n\}_{n=1}^{\infty}$ converge? For example, will it satisfy $x_n\sim O(1/n)$? (i.e. Does the limit $\lim_{n\to\infty}nx_n$ exist?) Thanks for your help!

答:
On the region of interest, $\sin(x) \geq x - \frac{x^3}{6}$, and this approximation is extremely good for $x \ll 1$. Since it's easily seen that $0 \leq x_2 \leq 0.3006$, this approximation is basically exact if we start applying at at $x_2$.

Applying this approximation to the recurrence gives

$$
x_{n+1}
\leq
\frac{n}{n+1} x_n \left ( 1 + \frac{x_n^2}{6n} \right )
\leq
\frac{n}{n+1} x_n \left ( 1 + \frac{1}{6n} \right )
$$

The estimate $x_n^2 \leq 1$ is very bad. In fact, $x_n^2 \approx 0$. But this is good enough to show (by induction) that

$$
x_{n+1}
\leq
\frac{x_0}{n+1} \prod_{i = 1}^n \left ( 1 + \frac{1}{6i} \right )
$$

Applying our favorite estimate $1+x \leq e^x$, we see

$$x_{n+1} \leq \frac{x_0}{n+1} e^{\frac{1}{6} H_n}$$

where $H_n = \sum_{i = 1}^n \frac{1}{i}$ is the $n$th harmonic number.

Then since $H_n \leq \ln(n) + 1$, we see

$$x_{n+1} \leq \frac{x_0}{n+1} e^{\frac{1}{6} (\ln(n) + 1)}$$

And lastly

$$x_{n+1} \leq \frac{x_0}{n+1} n^{\frac{1}{6}}e^{\frac{1}{6}}$$

So $x_n = O(n^{-5/6})$. This might be good enough, but we can do better. After all, the approximation $x_n^2 \leq 1$ isn't very good.

Now that we know $x_n = O(n^{-5/6})$, we can write

$$x_{n+1} \leq \frac{n}{n+1} x_n \left ( 1 + O(n^{-8/3}) \right )$$

Then as before

$$x_{n+1} \leq \frac{x_0}{n+1} \exp \left ( \sum_{i=1}^n O(i^{-8/3}) \right )$$

We can upper bound this sum by the infinite version of it, which converges, so we have

$$x_{n+1} \leq \frac{x_0}{n+1} e^{O(1)}.$$

This gives $x_n = O \left ( \frac{1}{n} \right )$.

As a (fun?) exercise:

Do you see why iterating this idea again won't improve our big Oh past
$O \left ( \frac{1}{n} \right )$?

Lastly, with sage one can compute the constant $k$ (depending on $x_0$) so that $x_n \approx \frac{k x_0}{n}$.

For $x_0 = \frac{\pi}{2}$, for instance, we see $k \approx 0.387$. Then

$x_{1000}                     = 0.000608664796133123$

$\frac{k \frac{\pi}{2}}{1000} = 0.000608664342810232$

which is quite a good approximation.

I hope this helps ^_^


问:
Assume $f: \mathbb R \to \mathbb R$ is continuous on $\mathbb R$ and $\lim_{x \to +\infty} f(x)=0=\lim_{x \to -\infty} f(x)$
Assume $f: \mathbb R \to \mathbb R$ is continuous on $\mathbb R$ and $$\lim_{x \to +\infty} f(x)=0=\lim_{x \to -\infty} f(x)$$

Show that $f$ is bounded on $\mathbb R$ and attains either its maximum or its minimum on $\mathbb R$.

$$\lim_{x \to +\infty} f(x)=0$$
$$\iff \forall \epsilon >0(\exists a \in \mathbb R(\forall x(x> a \implies \left|f\left(x\right)\right|<\epsilon)))$$

And
$$\lim_{x \to -\infty} f(x)=0$$
$$\iff \forall \epsilon >0(\exists b \in \mathbb R(\forall x(x< b \implies \left|f\left(x\right)\right|<\epsilon)))$$

Now if for all $x \in \mathbb R:f(x)=0$ then we are done ,otherwise we are able to find a $c \in \mathbb R$ such that $f(c) \ne 0$ and setting $\epsilon \mapsto \left|f\left(c\right)\right|$ follows that exists $a<x<b$ such that $\left|f\left(x\right)\right|<\left|f\left(c\right)\right|$.

On the other hand continuity of $f$ over $\mathbb R$ implies continuity of $f$ over $[a,b]$ and by extreme value theorem $f$ attains its maximum and minimum .( I don't understand why the question say or)

But I only know that for $x \in [a,b]$ the function $f$ is bounded and  it's not enough,since the domain of $f$ is not $[a,b]$ and I need to show that for all real $x$ the function is bounded.

Besides I want to know the validity of my proof about the max-min part.

答:
Sometimes there is only one of max and min. Take for example  $f(x)=\frac{1}{x^2+1}$, which answers your question about "and" / "or".

Also your argumentation does not work: Every finite interval has both extrema, but they are mostly in $a$ and $b$. Now we've got the problem, because our interval is infinite and can't be approximated well enough by finite intervals.


问:
Are the reals genuinely a subset of the complex numbers?
In Michael Spivak's Calculus, he defines a complex number as an ordered pair of real numbers: $z=(a,b)$ with $a,b \in \mathbb{R}$. The imaginary unit $i$ is then just a shorthand for the ordered pair $(0,1)$. Spivak goes on to say

When complex numbers were first introduced, it was understood that real numbers were, in particular, complex numbers; if our definition is to be taken seriously then this is not true—a real number is not a pair of real numbers, after all.

Although the complex number $(a,0)$ behaves in pretty much the same way as the real number $a$, they are still not identical. There seem to be some non-trivial differences, as well: we can't write $(5,0)>(3,0)$ in the way we can write $5>3$. With this in mind, I ask the following questions:

Can the real numbers be said to be a subset of the complex numbers if complex numbers are defined as ordered pairs?

If the complex numbers are constructed in some other way, then is it meaningful to write $\mathbb R \subset \mathbb C$?

答:
After defining complex numbers you can realize that the subset $X=\{(r,0)\ |\ r\in\mathbb{R}\}$ with inherited addition and multiplication is isomorphic to $\mathbb{R}$. Inherited metric agrees as well. Hence we can replace our standard $\mathbb{R}$ with $X$ and now the inclusion property holds with other properties preserved as well.

In other words $X$ is a model of real numbers as well. And that's exactly what we do.

we can't write $(5,0)>(3,0)$ in the way we can write $5>3$.

Of course we can. What exactly stops you from doing that? This only depends on how (and on what) the "$>$" is defined. In fact for the $X$ I defined earlier this is a well defined total order, which agrees with the classical ordering of reals.

Note that reals can be defined by axioms. We often use this axiomatic approach for reals, not concrete model, because we usually care about properties, not how exactly things are constructed. After all reals can be constructed in many ways, e.g. Dedekind cuts or metric completion of $\mathbb{Q}$ or as the $X$ I defined earlier.


问:
Bounds on $S = \frac{1}{1001} + \frac{1}{1002}+ \frac{1}{1003}+\dots+\frac{1}{3001}$

$S =  \frac{1}{1001} +   \frac{1}{1002}+ \frac{1}{1003}+ \dots+\frac{1}{3001}$.

Prove that $\dfrac{29}{27}<S<\dfrac{7}{6}$.

My Attempt:

$S<\dfrac{500}{1000} +   \dfrac{500}{1500}+ \dfrac{500}{2000}+ \dfrac{500}{2500}+\dfrac{1}{3000} =\dfrac{3851}{3000}$

(Taking 250 terms together involves many fractions and is difficult to calculate by hand.)

Using AM-HM inequality gave me $S > 1$, but the bounds are weak.

Prove that $1<\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+\dots+\frac{1}{3001}<\frac43$

Inequality with sum of inverses of consecutive numbers

The answers to these questions are nice, but the bounds are weak.

Any help without calculus and without calculations involving calculators would be appreciated.

(I encountered this question when I was preparing for a contest which neither allows calculators nor calculus(Only high-school mathematics.))

答:
We can find an approximation of the result.

Consider
$$\sum_{i=1}^{2001}\frac1{1000+i}=\sum_{i=1}^{3001}\frac1{i}-\sum_{i=1}^{1000}\frac1{i}$$ and we shall use the fact that
$$\sum_{i=1}^{n}\frac1{i}=H_n$$

Now, the asymptotics of harmonic numbers
$$H_n=\gamma +\log \left({n}\right)+\frac{1}{2
   n}-\frac{1}{12 n^2}+O\left(\frac{1}{n^4}\right)$$ Using it twice and you will have sharper bounds.


问:
Prove for all $1<n\in \mathbb{N}$ there are infinity natural numbers $t \in \mathbb{N}$ such that $\dfrac{\varphi(n)}{n} = \dfrac{\varphi(t)}{t}$
Prove for all $1<n\in \mathbb{N}$ there are infinity natural numbers $t \in \mathbb{N}$ such that $\dfrac{\varphi(n)}{n} = \dfrac{\varphi(t)}{t}$.

we know that $$n=p_1^{a_1}\cdots p_k^{a_k}$$

therefore, $$\varphi(n) = n(1-\dfrac{1}{p_1})\cdots (1-\dfrac{1}{p_k})$$

so, $$\dfrac{\varphi(n)}{n} =\prod_{i=1}^{k}(1-\dfrac{1}{p_i})$$

that's means $$\varphi(t)=\dfrac{t\cdot\varphi(n)}{n}=t\cdot\prod_{i=1}^{k}(1-\dfrac{1}{p_i})$$

答:
You are almost there! If $p_1,..., p_k$ are the primes dividing $n$, you just said that

$$ \frac{\phi(n)}{n} = \prod_{i=1}^k \left( 1-\frac{1}{p_i} \right) $$

Hence, to get $\phi(n)/n = \phi(t)/t$ you need $t$ to be divisible by exactly the same primes as $n$. If $n = \prod_{i=1}^k p_i ^{\alpha_i}$ (where $\alpha_i > 0$) then you need $t = \prod_{i=1}^k p_i^{\beta_i}$ for $\beta_i > 0$.


答:
Consider the one-point compactified real line $\widehat{\mathbb{R}}=\mathbb{R}\cup\infty$, where ‘$\infty$’ here denotes the identification of $+\infty$ and $-\infty$. It is clear (and in any case well-known or easy to prove) that $\widehat{\mathbb{R}}$ is topologically equivalent to the unit circle, thus it is compact. Now consider the function
$$
\hat{f}\colon \widehat{\mathbb{R}}\to\mathbb{R}\,,~\,~\,~ x\mapsto\left\lbrace
\begin{array}{ll}
f(x)&\mbox{if $x\in\mathbb{R}$}\\
0&\mbox{if $x=\infty$}
\end{array}\,.
\right.
$$
It follows, from your hypothesis, that $\hat{f}$ is continuous on the compact space $\widehat{\mathbb{R}}$, and therefore it is bounded and attains both its minimum and maximum in $\widehat{\mathbb{R}}$. Now since both extrema cannot possibly obtained at $\infty$, it follows that either one of them must be obtained in $\mathbb{R}$, as desired.

Added Later

Because the OP has provided a comment below regarding the slight advanced nature of the proof above, I’ve decided to rattle it down to a simpler one as follows: Define a new function $$g(t):=f(\frac{t}{1-|t|})\,.$$
Now it’s easy to see that the domain of $g$ is the open interval $(-1,1)$ and in particular the map $t\mapsto \frac{t}{1-|t|}$ is a one-to-one correspondence between $(-1,1)$ and $\mathbb{R}$. However, by your hypothesis on $f$ as $x\to\pm\infty$, it follows that $g$ extends to a continuous function on the closed interval $[-1,1]$ with
$$g(1):=\lim_{t\to1} f(\frac{t}{1-|t|})=0\,,~\,~\,~g(-1):=\lim_{t\to-1} f(\frac{t}{1-|t|})=0\,.$$
Thus $g$ is a continuous function on the closed bounded interval $[-1,1]$ and is therefore bounded—hence $f$ is also bounded. By the Extreme Value Theorem, $g$ attains both it’s minimum and maximum in $[-1,1]$. On the one hand, since $g(-1)=g(1)=0$, then if $0$ is simultaneously the minimum and maximum of $g$, then clearly $g(t)=0$, or equivalently, $f(x)=0$, which solves your problem trivially; on the other hand, if both extrema values are different or one of them is not $0$, then one of them is attained in $(-1,1)$ which means one of the extrema of $f$ is attained in $\mathbb{R}$.

Title: Assume $f: \mathbb R \to \mathbb R$ is continuous on $\mathbb R$ and $\lim_{x \to +\infty} f(x)=0=\lim_{x \to -\infty} f(x)$

Body:
Assume $f: \mathbb R \to \mathbb R$ is continuous on $\mathbb R$ and $$\lim_{x \to +\infty} f(x)=0=\lim_{x \to -\infty} f(x)$$

Show that $f$ is bounded on $\mathbb R$ and attains either its maximum or its minimum on $\mathbb R$.

$$\lim_{x \to +\infty} f(x)=0$$
$$\iff \forall \epsilon >0(\exists a \in \mathbb R(\forall x(x> a \implies \left|f\left(x\right)\right|<\epsilon)))$$

And
$$\lim_{x \to -\infty} f(x)=0$$
$$\iff \forall \epsilon >0(\exists b \in \mathbb R(\forall x(x< b \implies \left|f\left(x\right)\right|<\epsilon)))$$

Now if for all $x \in \mathbb R:f(x)=0$ then we are done ,otherwise we are able to find a $c \in \mathbb R$ such that $f(c) \ne 0$ and setting $\epsilon \mapsto \left|f\left(c\right)\right|$ follows that exists $a<x<b$ such that $\left|f\left(x\right)\right|<\left|f\left(c\right)\right|$.

On the other hand continuity of $f$ over $\mathbb R$ implies continuity of $f$ over $[a,b]$ and by extreme value theorem $f$ attains its maximum and minimum .( I don't understand why the question say or)

But I only know that for $x \in [a,b]$ the function $f$ is bounded and  it's not enough,since the domain of $f$ is not $[a,b]$ and I need to show that for all real $x$ the function is bounded.

Besides I want to know the validity of my proof about the max-min part.


答:
When you believe the formula for squares, and ${\cal R}$ is a nice region you can argue as follows (assume $\det T\ne 0$ for simplicity):

Given an $\epsilon>0$ draw a lattice in the plane which is so fine that there are  two "buildings" $B_0$, $B_1$ of lattice squares such that
$$B_0\subset {\cal R}\subset B_1,\qquad{\rm area}(B_1\setminus B_0)<\epsilon\ .\tag{1}$$
We then have $T(B_0)\subset T({\cal R})\subset T(B_1)$ and therefore
$$|\det T|\ \>{\rm area}(B_0)\leq{\rm area}\bigl(T({\cal R})\bigr)\leq|\det T|\ {\rm area}(B_1)\ .$$
Because of $(1)$ this implies
$${\rm area}({\cal R})-\epsilon\leq{{\rm area}\bigl(T({\cal R})\bigr)\over |\det T|}\leq{\rm area}({\cal R})+\epsilon\ ,$$
and as $\epsilon>0$ was arbitrary we obtain the claim.

Title: Prove $Area(T(\mathcal{R}))=|Det(A)|.Area(\mathcal{R})$

Body:
Let $T:\mathbb{R}^2\to\mathbb{R}^2$ be a linear transformation with matrix $A$. Let
$\mathcal{R}$ be a region in $\mathbb{R}^2$. Then:
$$
Area(T(\mathcal{R}))=|Det(A)|.Area(\mathcal{R})
$$
It is also statemed that this can be extended to higher dimensions too.

Is there a formal proof for this ? Or where does this actually come from ?

I just know that the above statements can be verified for specific examples and $|Det(A)|$ is the area of the parallelogram formed by the column vectors of $A$.


问:
$I=[a,b]$ and $f:[a,b] \to \mathbb R$ is continuous and for all $x$ in $I$ such property does hold
Assume $I=[a,b]$ and $f:[a,b] \to \mathbb R$ is continuous on $I$ and for all $x$ in $I$ exists  $y \in I$ such that $\left|f\left(y\right)\right|\le\frac{1}{2}\left|f\left(x\right)\right|$ prove that there exists a number $c$ in $I$ such that $f(c)=0$.

For $x_1 \in I$ exists $x_2 \in I$ such that $\left|f\left(x_2\right)\right|\le\frac{1}{2}\left|f\left(x_1\right)\right|$,inductively it's possible to find a sequence $(x_n) \in I$ such that for all $n \in \mathbb N$:$$\left|f\left(x_{n+1}\right)\right|\le\frac{1}{2}\left|f\left(x_n\right)\right|$$
And so
$$\left|f\left(x_{n+1}\right)\right|\le\frac{1}{2}\left|f\left(x_{n}\right)\right|\le\left(\frac{1}{2}\right)^{2}\left|f\left(x_{n-1}\right)\right|\le...\le\left(\frac{1}{2}\right)^{n-1}\left|f\left(x_{2}\right)\right|\le\left(\frac{1}{2}\right)^{n}\left|f\left(x_{1}\right)\right|$$

From here we see that the sequence $\left\{f\left(x_{n}\right)\right\}_n$ is a bounded sequence in $\mathbb R$ and by Bolzano–Weierstrass theorem there exists a subsequence $\left\{f\left(x_{n_k}\right)\right\}_k$ of $\left\{f\left(x_{n}\right)\right\}_n$ which converges to $f(c)$,on the other hand $$f(c)=\lim f(x_{n_k})=0 $$

I have two problems here,first how do we really "inductively" can find a sequence $(x_n) \in I$ such that for all $n \in \mathbb N$:$\left|f\left(x_{n+1}\right)\right|\le\frac{1}{2}\left|f\left(x_n\right)\right|$

Besides I have not used the fact that $f$ is continuous ,if I could show that $(x_n)$ is bounded then I could use  Bolzano–Weierstrass theorem to conclude that there exists a subsequence $\left(x_{n_k}\right)$  which converges to $c$,and continuity of $f$ implies that $f(c)=\lim f(x_{n_k})=0 $,but I cannot show that $(x_n)$ is bounded.

答:
how do we really "inductively" can find a sequence

This is a procedure that's used quite frequently, but the full details are rarely spelled out. What you're doing is showing the following claim:

Claim: There exists a sequence $\{x_k\}$ in $[0,1]$ such that
$$ |f(x_{k+1})| \leq \frac12 |f(x_k)| $$
for all $k.$

Proof: We will prove by induction that for all $n,$ there exists $\{x_1,\dots,x_n\}$ satisfying the above inequality for all $1 \leq k \leq n-1.$

For $n=1$ we can take $\{x_1\}$ with any $x_1 \in [0,1].$ By inductive hypothesis there exists $\{x_1,\dots,x_{n}\}$ exists satisfying the above for all $1 \leq k \leq n-1.$ By assumption on $f$ we know there exists some $x_{n+1}$ such that $|f(x_{n+1})|\leq \frac12 |f(x_n)|.$ Then the collection $\{x_1,\dots,x_{n+1}\}$ verifies the $(n+1)$ case.


问:
integration bounds for triple integral
Im still working on finding the right bounds for my triple integrals, with good help from this forum I have come a bit further, I think.

$ \int_D \text{ at the domaine } \{(x,y,z) | 0 \leq z \leq 1 - x^2 - \frac{y^2}{4} \} $

So I have plugged in the shape in Geogebra: $x^2+\frac{y^2}{4}$ and then two planes: $z=0 \text{ and } z = 1$.  So I clearly see the shape I have to integrate and from which points. But Im not quite shure how to write the bounds, feel like I'm guessing. My initial line of thinking is that x,y,z all share a beginning point at (0,0,0) and then they move in their respective directions until z = 1:

$ \int_0^1 \int_{0}^{x^2} \int_{0}^{y^2/4} dzdxdy $

Is that the correct way to conceptualize is?

Would be very happy for feedback and help!

答:
Sorry if this is more advanced than what you're used to, but this is how I would go about it:

Normally, if the upper bound for $z$ was $1 - x^2 - y^2$, we would convert to cylindrical coordinates using the transformations $x = r\cos(\theta), y=r\sin(\theta), z=z$. This works very nicely because the $x^2 + y^2$ in the bound turns into $r^2$.

Here, our upper bound is an elliptical paraboloid, with $z = 1 - x^2 - \dfrac{y^2}{4}$, so if we want the same nice transformation, we need $x^2 + \dfrac{y^2}{4} = 1$. We can achieve this with the transformation $x = r\cos(\theta), y = 2r\sin(\theta), z = z$, sort of an elliptical version of our standard cylindrical coordinates.

After this transformation, our bounds of integration change to $0 \leq z \leq 1 - r^2$. Solving $1- r^2 \geq 0$ should give you $0 \leq r \leq 1$, and then our angle makes a full rotation with $0 \leq \theta \leq 2\pi$.

(Edit: The $r \geq 0$ comes as a matter of convention, you could alternatively use $-1 \leq r \leq 1$ and $0 \leq \theta \leq \pi$ but it confuses things a little.)

The only thing left to do is to calculate the determinant of the Jacobian. If you're not familiar with the Jacobian, it's basically where the $r dr d\theta$ comes from when you integrate with polar coordinates: it accounts for the transformation of the area or volume element itself.

The Jacobian of our transformation is:
\begin{bmatrix}
\cos(\theta) & 2\sin(\theta) & 0 \\
-r\sin(\theta) & 2r\cos(\theta) & 0 \\
0 & 0 & 1
\end{bmatrix}

and its determinant is $\cos(\theta)\cdot2r\cos(\theta) - 2\sin(\theta)\cdot-r\sin(\theta)+0 = 2r(\cos^2(\theta) + \sin^2(\theta)) = 2r$.

This result should make sense intuitively if you think about it as first making the transformation $y = 2y$ and then converting to cylindrical.

Now we should be able to put everything together to get our integral:
$\int_0^{2\pi}\int_0^1\int_0^{1-r^2}2rdzdrd\theta$.

Hope this helps!

Title: integration bounds for triple integral


问:
Show $\|f\|_p\leq \lim\inf\|f_n\|$
$\Omega$ is a bounded domain of $\mathbb R^n$. If $\{f_n\}\subset L^p(\Omega)$ and $f_n\rightarrow f\in L^p(\Omega)$ weakly, then

$$\|f\|_p\leq \lim_{n\rightarrow\infty}\inf\|f_n\|$$

答:
You are asking to prove why $$\left(\int  |f|^p\right)^{1/p} \leq \lim_{n\rightarrow \infty} \inf \left(\int {|f_n|^p}\right)^{1/p},$$

so it is enough to prove that $\left(\int  |f|^p\right)\leq \lim_{n\rightarrow \infty} \inf \left(\int {|f_n|^p}\right)$. Now if $f_n \to f$ pointwise a.e. then $|f_n|^p \to |f|^p$ pointwise a.e. Thus
$$\begin{eqnarray*} \int |f|^p &=& \int \lim_{n \rightarrow \infty} |f_n|^p \\
&=&\int \lim_{n \rightarrow \infty} \inf |f_n|^p \\
&\leq& \lim_{n \rightarrow \infty} \inf \int |f_n|^p \end{eqnarray*}$$

where the last step was made using Fatou's lemma. Q.E.D.

Title: Show $\|f\|_p\leq \lim\inf\|f_n\|$


问:
Writes in polar coordinates $x''-(1-x^2-(x')^2)x'+4x=0$.
I need to determine a periodic solution for :$x''-(1-x^2-(x')^2)x'+4x=0$.
We have the equivalent system:
$$\begin{cases}
       x'=y \\
       y'=(1-x^2-y^2)y-4x.
   \end{cases}
$$

We determined the stationary points for the equivalent system: $(0,0)$.

I saw that this solution is unstable.

But now I have to turn this system into polar coordinates. Unfortunately, I tried in all possibilities and I failed to bring the system to a beautiful shape depending on the polar coordinates.

I present what I tried:

Polar coordinates:
$\begin{cases}
       x(r,\theta)=r\cos(\theta) \\
       y(r,\theta)=r\sin(\theta)
   \end{cases}
$

We have that $x^2+y^2=r^2$ and $\tan(\theta)=\frac{y}{x}$. So $r'=\frac{xx'+yy'}{r}$ and $\theta'=\frac{xy'-x'y}{r^2}.$
So $$r'=\frac{xy+y[(1-x^2-y^2)y-4x]}{r}=\frac{xy+y^2-x^2y^2-y^4-4xy}{r}.$$
I tried to calculate, but I have no idea how to bring the system into polar coordinates to continue my work.
Thanks!

答:
The term in the middle acts as friction for $(x,x')$ outside the unit circle and as anti-friction inside the unit circle. This means that any particle moving under this dynamic will move away from the origin if close to it, and towards the unit circle if far away from it. The periodic solution will thus be close to the unit circle.

The numerical search for a periodic orbit can be formulated as boundary value problem, having the end point of the integration equal to the initial point. To account for the unknown period, map the whole problem to the constant interval $[0,1]$ using a parameter transformation.

In python/scipy this can be realized as follows

```<code>def odesys(t,u,p):
    x,v = u
    T, = p
    a = (1-x**2-v**2)*v-4*x
    return T*np.array([v,a])

def bc(u0, u1, p): return [u0[0], u1[0], u1[1]-u0[1]]

T = 2*np.pi
s = np.linspace(0,1,21);
u = [ np.sin(T*s), np.cos(T*s) ]

res = solve_bvp(odesys,bc,s,u,[T])
print(res.message, res.p)

s = np.linspace(0,1,301)
u = res.sol(s)
plt.plot(u[0],u[1],'b')
plt.plot(res.y[0],res.y[1],'xb')
</code>```

This gives the result

Body:
The term in the middle acts as friction for $(x,x')$ outside the unit circle and as anti-friction inside the unit circle. This means that any particle moving under this dynamic will move away from the origin if close to it, and towards the unit circle if far away from it. The periodic solution will thus be close to the unit circle.

The numerical search for a periodic orbit can be formulated as boundary value problem, having the end point of the integration equal to the initial point. To account for the unknown period, map the whole problem to the constant interval $[0,1]$ using a parameter transformation.

In python/scipy this can be realized as follows

def odesys(t,u,p):
    x,v = u
    T, = p
    a = (1-x**2-v**2)*v-4*x
    return T*np.array([v,a])

def bc(u0, u1, p): return [u0[0], u1[0], u1[1]-u0[1]]

T = 2*np.pi
s = np.linspace(0,1,21);
u = [ np.sin(T*s), np.cos(T*s) ]

res = solve_bvp(odesys,bc,s,u,[T])
print(res.message, res.p)

s = np.linspace(0,1,301)
u = res.sol(s)
plt.plot(u[0],u[1],'b')
plt.plot(res.y[0],res.y[1],'xb')

This gives the result


问:
Existence of the limit of an integral
This is a question related to the problem I asked in  Show that $|f'(z)|>\frac{n_k}{10k}$ for $1-\frac{1}{n_k}<|z|<1-\frac{1}{2n_k}$.. Suppose that
\begin{equation}
f(z)=\sum_{k=1}^n \frac{z^{n_k}}{k},
\end{equation} where $\{n_k\}$ is a sequence of positive integers with $n_k>e^{n_{k-1}}$ for every $k \ge 2$. Does the limit
\begin{equation}
\lim_{R \to 1} \int_0^R f'(re^{it}) dr
\end{equation} exist for almost all $t$ on $[-\pi,\pi]$? My idea is that
\begin{equation}
f(re^{it})=\sum_{k=1}^\infty \frac{n_k r^{n_k-1}}{k}e^{i(n_k-1)t}
\end{equation} implies
\begin{equation}
\int_0^R f'(re^{it}) dr=\sum_{k=1}^\infty \frac{n_k}{k} \Big[\int_0^R r^{n_k-1} dr\Big] e^{i(n_k-1)t}=e^{-it}f(Re^{it}),
\end{equation} so it seems that the limit exist $\textit{for all}$ $t$. However, I feel that the question is not so easy. What mistake(s) have I made?

答: I think I have got a solution. In fact, since $\sum \frac{1}{k^2}<\infty$, $f \in H^2$ and thus the nontangential limits $f^\ast(e^{it})$ exists a.e. on $T$. Hence we are done.

Title: Existence of the limit of an integral


问:
How to solve Logarithm in case of addition with same bases
How to solve $\log (15+9)=24$
and how to get an Exponential value of 24 with base 10 if
$\log 2$ value is given,
$\log 3$ value is given,
and $\log 5$ is given?

and suppose if I start with

$\log (5×3+3^2)=24$

$\log(5)+\log(3)+2\log3=24$

then how to solve further?

Is this correct?

$\log(5×3+3^2)=\log 5+\log 3+2\log3$????

答:
To get the value of $\log_{10}{24}$ by just knowing the valus of $\log_{10}(2)$ and $\log_{10}(3)$ you have to write $24$ as a product $24=2^3\cdot3$ now you can use what I wrote in the comments what holds for products inside the logarithm $$\log_{10}(24)=\log_{10}(2^3\cdot3)=\log_{10}(2^3)+\log_{10}(3)=3\cdot\log_{10}(2)+\log_{10}(3).$$ Thus knowing the value of $\log_{10}(5)$ is not relevant at this time, which can be seen by all numbers $2,3,5$ are prime and $5$ is not a primefactor of $24$.

Since I'm not sure if this was exactly your question maybe this helps you to solve your problem: for every base $b$ of the logarithm holds
$$\log_b(24)=\frac{\log_{10}(24)}{\log_{10}(b)}.$$ So if your first question was to solve $$\log_b(24)=24$$ for $b$ you have to calculate
$$\log_{10}(b)=\frac{\log_{10}(24)}{24}$$ as above and then taking its exponetial with respect to $10$ to get the value of $b$.


问:
Recurrence relations of the form $T(n) = T(n-1) + (n-1)! n^k$
In the context of recursive permutation-related algorithms, the following form of recurrence relations often comes up:
$$
T(n) = 1 \\
T(n) = T(n-1) + (n-1)! \cdot n^k
$$
where $k$ is usually 0, 1, or 2.

What can be said about the asymptotic behaviour of $T(n)$ (in terms of $\Theta$ or $O$ notation, and in general or for specific $k$)?

答:
Put $a_n = T(n)$, $b_n = n^k (n-1)!$, $\Delta a_n = a_n - a_{n-1} = b_n$, $\Delta b_n = b_n - b_{n-1}$.

Hence
$$\Delta b_n = n^k (n-1)! - (n-1)^k (n-2)! = n^k (n-1)! \cdot ( 1 - \frac{(\frac{n-1}{n})^k}{n-1})$$
$$= n^k (n-1)! \cdot ( 1 - \frac{1+\text{o}(1)}{n-1}) = b_n (1+\text{o}(1)).$$
By Stolz–Cesàro theorem

$$\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\Delta a_n}{\Delta b_n} = \lim_{n \to \infty} \frac{b_n}{ b_n(1+\text{o}(1))}  = 1.$$
It follows that $T_n \sim n^k (n-1)! = n^{k-1} n!$ for $n \to \infty$.


问:
$f$ is partially derivable if and only if $f$ is derivable in the direction of the standard unit vectors $e_{1},...,e_{n}$
$f$ is a function defined by, $f:U \to \mathbb{R}$ with $U \subset \mathbb{R^{n}}$ open. v $\in \mathbb{R^{n}}$ a unitvector then we say that $f$ is derivatible in a point $a \in U$ in the direction of v if the function $t\to f(a+t$v) is derivatble in $zero$

I want to proof that $f$ is partially derivable if and only if $f$ is derivable in the direction of the standard unit vectors $e_{1},...,e_{n}$ with $e_{i}=(0,...,0,1,0,....,0)$ with $1$ at place $i$.

My definition of partial derivatble is:

If $U\subset \mathbb{R^{n}}$ open and $f:U \to \mathbb{R}$ a function. We say that $f$ is partial derivatible in a point $(a_{1},...,a_{n}) \in U$ if for every $i:1,...,n$ the function $h_{i}: x\to f(a_{1},...,a_{i-1},x,a_{i+1},...,a_{n})$ is derivatible in $a_{i}$

I tried the following. Can someone check if it's correct and help me to understand my mistakes.

$=>$ $f$ is partial derivable. We need to prove that a function $t \to f(e_{i}+tv)$ is derivable in zero. $v$ is also a standard unit vector, with $v \in \mathbb{R^{n}}$. Let's call the previous function $g: \mathbb{R} \to \mathbb{R}$, then $g$'(t)=$f'(e_{i}+tv)v$. This exist beceause $f$ is partial derivative. Now $g'(0)=f'(e_{i})v$ which is also derivable beceause $f$ is partial derivative.

$<=$ if $f$ is derivative in all the directions of the standard unit vector, then we can write every other direction as a lineaire combination of the standard unit vectors. This implicate that $f$ will be partial derivative.

答:
This is false. For $n \ge 2$, the following three conditions on a function $f : \mathbb{R}^n \to \mathbb{R}$ and a point $p \in \mathbb{R}^n$ are in strictly increasing order by strength, and in particular are inequivalent:

The partial derivatives $\frac{\partial f}{\partial x_i}$ exist at $p$.

The directional derivatives $\frac{\partial f}{\partial v}$ exist for every $v \in \mathbb{R}^n$ at $p$.

$f$ is differentiable at $p$.

(I'm not sure whether by "partially derivable" you mean 2 or 3, but the statement is false either way.)

Your argument for $\Leftarrow$ does not work; you've just assumed that the partial derivative exists which is exactly what you were trying to prove. It is true that if $f$ is differentiable then every directional derivative is a linear combination of partial derivatives $\frac{\partial f}{\partial x_i}$, but you can't assume that here.

ancientmathematician's nice counterexample in the comments of the function $f$ which is equal to $1$ on the coordinate axes and $0$ otherwise shows that 1 does not imply 2 (which also shows that 1 does not imply 3); the partial derivatives $\frac{\partial f}{\partial x_i}$ exist at the origin and are equal to $0$, but the directional derivatives in all other directions fail to exist at the origin.

You can find a counterexample showing that 2 does not imply 3 at this MO question, namely the function

$$f(x, y) =
  \begin{cases}
        0 & \text{for } (x,y)=(0,0) \\
   \frac{x^3}{x^2+y^2}       & \text{for } (x,y) \neq (0,0)
  \end{cases}$$

which is also a counterexample showing that "the directional derivatives exist for every $v$ at every point" does not imply "$f$ is differentiable at every point."


问:
A clarification in a limit
While reading this answer I do not understand that how can we claim that there exist a polynomial with the properties, even if it exist then how is it same for the function and the inverse.</p>

答:
Start with $ f ( x ) = x + a _ 2 x ^ 2 + a _ 3 x ^ 3 + a _ 4 x ^ 4 \cdots $ and $ f ^ { − 1 } ( x ) = x + A _ 2 x ^ 2 + A _ 3 x ^ 3 + A _ 4 x ^ 4 \cdots $ and substitute into $ f ^ { - 1 } ( f ( x ) ) = x $ (or into $ f ( f ^ { - 1 } ( x ) ) = x $ for the same final result):
$$ \eqalign { x & = ( x + a _ 2 x ^ 2 + a _ 3 x ^ 3 + a _ 4 x ^ 4 + \cdots ) + A _ 2 ( x + a _ 2 x ^ 2 + a _ 3 x ^ 3 + a _ 4 x ^ 4 + \cdots ) ^ 2 \\ & \quad { } + A _ 3 ( x + a _ 2 x ^ 2 + a _ 3 x ^ 3 + a _ 4 x ^ 4 + \cdots ) ^ 3 + A _ 4 ( x + a _ 2 x ^ 2 + a _ 3 x ^ 3 + a _ 4 x ^ 4 + \cdots ) ^ 4 + \cdots \\
& = x + ( a _ 2 + A _ 2 ) x ^ 2 + ( a _ 3 + 2 a _ 2 A _ 2 + A _ 3 ) x ^ 3 + ( a _ 4 + 2 a _ 3 A _ 2 + a _ 2 ^ 2 A _ 2 + 3 a _ 2 A _ 3 + A _ 4 ) x ^ 4 + \cdots \text . } $$
This holds for all $ x $ (at least sufficiently close to $ 0 $), so
$$ \displaylines { a _ 2 + A _ 2 = 0 \text , \\ a _ 3 + 2 a _ 2 A _ 2 + A _ 3 = 0 \text , \\ a _ 4 + 2 a _ 3 A _ 2 + a _ 2 ^ 2 A _ 2 + 3 a _ 2 A _ 3 + A _ 4 = 0 \text , \\ \cdots \text . } $$
We can now start solving these:
$$ \displaylines { A _ 2 = - a _ 2 \text , \\ A _ 3 = - a _ 3 - 2 a _ 2 A _ 2 = - a _ 3 + 2 a _ 2 ^ 2 \text , \\ A _ 4 = - a _ 4 - 2 a _ 3 A _ 2 - a _ 2 ^ 2 A _ 2 - 3 a _ 2 A _ 3 = - a _ 4 + 5 a _ 2 a _ 3 - 5 a _ 2 ^ 3 \text , \\ \cdots \text . } $$
So we have $ A _ n = - a _ n + P _ n ( a _ 2 , \ldots , a _ { n - 1 } ) $, where $ P _ 2 = 0 $, $ P _ 3 ( a _ 2 ) = 2 a _ 2 ^ 2 $, $ P _ 4 ( a _ 2 , a _ 3 ) = 5 a _ 2 a _ 3 - 5 a _ 2 ^ 3 $, etc.

So these polynomials exist; and it doesn't matter whether you call the coefficient sequences $ a $ and $ A $, or $ b $ and $ B $; it's the same polynomial sequence $ P $ either way.  As long as a function $ f $ is analytic at $ 0 $ with $ f ( 0 ) = 0 $ and $ f ' ( 0 ) = 1 $, then the coefficients of the function and its inverse will be related by these same polynomials.

Title: A clarification in a limit


答:
We have
\begin{align*}
T(n) & = \sum\limits_{j = 1}^n {(j - 1)!j^k }  = \sum\limits_{j = 0}^{n - 1} {(n - j - 1)!(n - j)^k }
\\
& = (n - 1)!n^k \left( {1 + \frac{1}{{(n - 1)}}\left( {1 - \frac{1}{n}} \right)^k  + \sum\limits_{j = 2}^{n - 1} {\frac{1}{{(n - j) \cdots (n - 1)}}\left( {1 - \frac{j}{n}} \right)^k } } \right).
\end{align*}
From this it follows that
$$
T(n) = (n - 1)!n^k \left( {1 + \mathcal{O}\!\left( {\frac{1}{n}} \right)} \right),
$$
as $n\to +\infty$.

Title: Recurrence relations of the form $T(n) = T(n-1) + (n-1)! n^k$

Body:
In the context of recursive permutation-related algorithms, the following form of recurrence relations often comes up:
$$
T(n) = 1 \\
T(n) = T(n-1) + (n-1)! \cdot n^k
$$
where $k$ is usually 0, 1, or 2.

What can be said about the asymptotic behaviour of $T(n)$ (in terms of $\Theta$ or $O$ notation, and in general or for specific $k$)?


问:
marginal density problem
Let $m(t)$ be a moment generating function of some random variable. Which of the following are Moment generating functions of some other random variables?

a. $m(t)m(6t)$

b. $-3m(t)$

c. $\mathrm{e}^{-t}m(t)$

d. All of these

答:
Consider $\xi$ such that $m(t) = Ee^{t \xi}$. Suppose that $\xi$ and $\eta$ are independent and have the same distribution. Then m.g.f. of $\xi + 6 \eta$ is equal to
$$Ee^{t (\xi + 6 \eta)} = Ee^{t \xi}Ee^{6 t  \eta} = m(t) m(6t).$$
Also m.g.f. of $\xi - 1$ is equal to
$$Ee^{t (\xi -1)} = Ee^{t \xi} e^{-t}.$$
Any m.g.f. is equal to $1$ when it's parameter is equal to $0$. Thus a) and c) are m.g.f. and b) is not.

Title: marginal density problem


问:
Prove that the system has a single limit cycle
We have the following system:
\begin{cases}
      x'=x(x^2+y^2-2x-3)-y\\
      y'=y(x^2+y^2-2x-3)+x
   \end{cases}
We want to determine that it has a single limit cycle.
To prove this we had to use the Poincare Bendixson theorem.

We have the linearized system:
\begin{cases}
   x'=-y\\
   y'=x
\end{cases}
$A=\begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}$. We can see that we have a center point.

After that I found the stationary points. My problem arose when I had to determine the set A, which does not contain stationary points and is invariant to flow. I tried to solve this problem, but without any success.

答:
Caution: In its present state, this "hint" provides only a part of the answer. I am going to revise it in a short while.

Hint: consider dot product

$$(x,y).(x',y')=xx'+yy'=(x^2+y^2)(x^2+y^2-2x-3)$$

It is negative inside the circle with equation $x^2+y^2-2x-3=0$, i.e. with center $(1;0)$ and radius $2$,

This geometric property, means that a particle submitted to the corresponding field is "directed" in the half plane (defined by the tangent) where the circle's center is situated. Only a partial conclusion can be derived here.

Matlab program:

```<code>function edh
clear all;close all;hold on;axis equal off;
axis([-1.5,4.5,-3,3]);
t=0:pi/100:2*pi;
plot(1+2*exp(i*t),'c','linewidth',2);% circle
plot([-1.5,4.5],[0,0]);plot([0,0],[-3,3]);
tspan = 0:0.01:10;  % temporal span
ode = @(t,y) sc(y);
[x,y] = meshgrid(-1.5:0.25:4.5,-3:0.25:3);
w=x.^2+y.^2-2*x-3;% a way to describe the diff. system
u=x.*w-y;
v=y.*w+x;
quiver(x,y,u,v,2);% arrows
for r=1.8:0.1:2.2
    for k=-pi:pi/20:pi
        y0 = [1+r*cos(k);r*sin(k)]; % initial points
        [t,y] = ode45(ode, tspan, y0); % Runge Kutta
        plot(y(:,1),y(:,2),'color','r');
    end
end
%%%
function yp = sc(y)
w=y(1).^2+y(2).^2-2*y(1)-3; % another way to
u=y(1).*w-y(2); % describe the diff. system
v=y(2).*w+y(1);
yp=[u;v];
</code>```

Body:
Caution: In its present state, this "hint" provides only a part of the answer. I am going to revise it in a short while.

Hint: consider dot product

$$(x,y).(x',y')=xx'+yy'=(x^2+y^2)(x^2+y^2-2x-3)$$

It is negative inside the circle with equation $x^2+y^2-2x-3=0$, i.e. with center $(1;0)$ and radius $2$,

This geometric property, means that a particle submitted to the corresponding field is "directed" in the half plane (defined by the tangent) where the circle's center is situated. Only a partial conclusion can be derived here.

Matlab program:

function edh
clear all;close all;hold on;axis equal off;
axis([-1.5,4.5,-3,3]);
t=0:pi/100:2*pi;
plot(1+2*exp(i*t),'c','linewidth',2);% circle
plot([-1.5,4.5],[0,0]);plot([0,0],[-3,3]);
tspan = 0:0.01:10;  % temporal span
ode = @(t,y) sc(y);
[x,y] = meshgrid(-1.5:0.25:4.5,-3:0.25:3);
w=x.^2+y.^2-2*x-3;% a way to describe the diff. system
u=x.*w-y;
v=y.*w+x;
quiver(x,y,u,v,2);% arrows
for r=1.8:0.1:2.2
    for k=-pi:pi/20:pi
        y0 = [1+r*cos(k);r*sin(k)]; % initial points
        [t,y] = ode45(ode, tspan, y0); % Runge Kutta
        plot(y(:,1),y(:,2),'color','r');
    end
end
%%%
function yp = sc(y)
w=y(1).^2+y(2).^2-2*y(1)-3; % another way to
u=y(1).*w-y(2); % describe the diff. system
v=y(2).*w+y(1);
yp=[u;v];


问:
How can I find $k$ for an inequality involving this probability density function?
The diagram is as shown. The area under $f(x)$ for a given probability density function is $1$. I am thinking maybe I can use the equation of the second "half" of the triangle (the diagonal sloping downwards) to find the value of k, but I keep getting the wrong answer. The equation I got for the second "half" of the triangle is $y=-x+2$. This is the definite integral I tried:

$$\int_{1}^{k} (-x+2)dx=0.02$$

Is there an easier way to get the answer? The book's solution says $k=1.8$, but I don't understand how to get there. Any hints / tips would be appreciated!

答:
If $k \le 2$ we have $$P(X > k) = \int_{k}^{\infty} f(x)dx = \int_{k}^2 f(x)dx$$ For $k \in [1,2]$ we have
$$P(X > k) = \int_{k}^{\infty} f(x)dx = \int_{k}^2 (-x + 2) dx $$
You may find $k \ge 1$ such that  $P(X > k) =  \int_{k}^2 (-x + 2) dx = 0.02$ and it will give you answer.


问:
About a property of Reynolds operator (Invariant theory)
I am learning about classical invariant theory, and I have a question about Reynolds operator. The book I am reading is $\ulcorner$Classical Invariant Theory$\lrcorner$ written by Hansepter Kraft, Claudio Procesi. If you want, you can download it here.

Consider an infinite field $K$ and a group $G$ and a finite dimensional G-module W and its coordinate ring $K[W]$, and assume that the representation of $G$ on $K[W]$ is completely reducible.
In the textbook, they define a canonical G-equivariant linear projection $R:K[W]\rightarrow K[W]^G$ and call it a Reynolds operator. ($K[W]^G$ is a ring of invariants)

They asserts that $R(hf)=hR(f)$ holds for $h\in K[W]^G$, $f\in K[W]$, and this property is crucial in proof of Hilbert's finiteness result for invariants. (p.13 Theorem 1 in the textbook) But I cannot figure out why this relation holds. Actually, it is one of its exercise to prove the equation in a little more general setting. (p.13 Exercise 30)

Let $A$ be a (commutative) algebra and let $G$ be a group of algebra automorphisms
of $A$. Assume that the representation of $G$ on $A$ is completely
reducible. Then the subalgebra $A^G$ of invariants has a canonical $G$-stable
complement and the corresponding $G$-equivariant projection $p:A\rightarrow A^G$ satisfies the relation $p(hf)=hp(f)$ for $h\in A^G$, $f\in A$.

I have two questions. First, what does a canonical $G$-stable complement mean? Indeed complete reducibility ensures the existence of a $G$-stable complement, but how can we 'construct' such complement of $A^G$?

Second, how can we show that $p(hf)=hp(f)$ for $h\in A^G$, $f\in A$ ? I tried to prove that the complement is stable under multiplication by $h\in A^G$, but couldn't make any progress.

Any helps, comments, suggestions will be appreciated. Thank you for reading.

答:
First, 'constructing' this complement in such generality certainly cannot be done. The point is just that A can be broken up into a direct sum $A^G \oplus \bigoplus_{i} V_i$, where each $V_i$ is an isotypic $G$-submodule of $A$, i.e., a submodule such that any two simple submodules of $V_i$ are $G$-isomorphic. Moreover, this decomposition is unique up to relabeling, so in particular there is a unique $G$-module complement to $A^G$ in $A$, and it makes sense to define this projection without reference to an auxiliary decomposition. So in this case 'canonical' means something more like 'uniquely determined by the data' than 'directly describable from the data'.

For your second question, the main point is that it suffices to show that if $W$ is a simple $G$-submodule of $A$ not lying in $A^G$ and $a \in A^G$, then $aW$ is also a $G$-submodule intersecting $A^G$ trivially. So let $w \in W$, $a \in A^G$, and $g \in G$. We have
\begin{align}
g(aw) = g(a)g(w) = ag(w),
\end{align}
so $aW$ is a $G$-submodule and the map $W \to aW$, $w \mapsto aw$, is a surjective $G$-module homomorphism. Thus either $aW$ is trivial or it is isomorphic to $W$, and in either case it intersects $A^G$ trivially. So multiplication by elements of $A^G$ preserves the $G$-stable complement to $A^G$ in $A$ and this gives the desired property for the projection.


问:
Question about integrability
Let f be a continious function on [a,b] and exist a partition P of [a,b] such that $\bar{S}(f,p)=\int_a^b f(x)dx$. Prove that f is a constant function. I thought stratting assuming the claim is not correct and then must exist $x_0,x_1 \in [a,b]$ such that without the loss of genraility $f(x_0)<f(x_1)$. Then looking at the sum of darboux, somehow the $\bar S(f,p)$ does not converge to the integral. Why is this correct?</p>

答:
Start out from the given partition $P$. The indirect assumption that $f$ is not constant implies that it is not constant over at least one interval $I_n=[a_n,a_{n+1}]$ of the partition. Then choose $x_1\in I_n$ such that $f(x_1)$ is maximal. Then there exists $x_0\in I_n$ such that
 $f(x_0)<f(x_1)$.

Then, by continuity, for $\varepsilon:=\displaystyle\frac{f(x_1)-f(x_0)}2$ there is a $\delta>0$ such that $f(x)<f(x_1)-\varepsilon$ if $\ |x-x_0|<\delta$. This interval $(x_0-\delta,x_0+\delta)$ intersects $I$ in an interval with positive length, say $\varrho>0$. Over this interval we estimate $f$ from above by $f(x_1)-\varepsilon$, and on the rest of $I$ we estimate $f$ by $f(x_1)$. Then we have
$$ \begin{align}
\int_{a_n}^{a_{n+1}}f\  &\le\ \varrho\,(f(x_1)-\varepsilon)+(a_{n+1}-a_{n}-\varrho)\,f(x_1)\ < \\
&<\ (a_{n+1}-a_n)\,f(x_1)
\end{align} $$
Since over each interval $I_k$ we always have $\displaystyle\int_{I_k}f\le (a_{k+1}-a_k)\,\max_{I_k}f$, it means that
$$\int_a^bf < \bar S(f,P)\,.$$

Title: Question about integrability


问:
Prove that the multitude of polinomials $f \in F^4[x]$, for which $f(91)=0$ form a linear subspace of $F^4[x]$
$\mathbb{V}$ is the multitude of all polinomials of degree less than or equal to 3. In this problem, I have to:

prove that $\mathbb{V}$ is a linear space over $\mathbb{R}$.

find the dimension of $\mathbb{V}$

prove that the polinomials $1, x-91, \frac{(x-91)^2}{2!}, \frac{(x-91)^3}{3!}$ form a basis of $\mathbb{V}$

prove that the multitude of polinomials $f \in \mathbb{V}$, for which $f(91)=0$ form a linear subspace of $\mathbb{V}$

$\mathbb{V}$ is the multitude of all polinomials of degree less than or equal to 3. Therefore, $\mathbb{V}=F^4[x] \subset F[x]$. We can present it as$$F^4[x]=\{f(x)=a_0x^0+a_1x^1+a_2x^2+a_3x^3\mid a_0, a_1, a_2, a_3\in F\}.$$The standart basis for this linear space is $e_1=1, e_2=x, e_3=x^2, e_4=x^3$.
So far, I have proven the first and second points. For the third point, I expressed the four polinomials using the standart vectors $e_1, e_2, e_3, e_4$ and formed the matrix$$\begin{pmatrix}
1&0&0&0\\\
-91 & 1 & 0 & 0\\\\
\frac{{91}^2}{2}&-91&\frac{1}{2}&0\\\\
-\frac{{91}^3}{6}&\frac{{91}^2}{2}&\frac{91}{2}&\frac{1}{6}
\end{pmatrix}$$
The determinant of this matrix is non-zero, therefore the four polinomials are linearly independent, therefore, they form a basis of $\mathbb{V}$.
The fourth point however is problematic for me. I can't think of any ideas how to solve it.
Thank you in advance!

Body: If $f(91)=g(91)=0$, then$$(f+g)(91)=f(91)+g(91)=0.$$And, if $\alpha\in F$ and if $f(91)=0$,$$(\alpha f)(91)=\alpha f(91)=0.$$So, that set is indeed a subspace.


问:
Find the limit $\lim_{x \to 1}(\frac{3x}{2+x})^\frac{x}{1-x}$
Find the limit
$$
\lim_{x \to 1}\left(\frac{3x}{2+x}\right)^\frac{x}{1-x}
$$

I've transformed the function by changing limit of 1 to zero and become the following:
$$
\lim_{x \to 0}\left(1 + \frac{3x-3}{x-3}\right)^\frac{1 - x}{x}
$$

I spent many time and tried to transform the power part $\frac{1 - x}{x}$ to $\frac{x-3}{3x-3}$ but still can't remove the $x$'s on final. How can I find this limit? Thanks a lot!

Body:
$$
L=\lim_{x \to a}(f(x))^{g(x)}
$$
So, if $\lim_{x \to a}(f(x))=1,\lim_{x \to a}(g(x))= \infty $
$$
\ln L=\lim_{x \to a}(f(x)-1)({g(x)})
$$
In your case,
$$
L=\lim_{x \to 1}(\frac{3x}{2+x})^{\frac{x}{1-x}}
$$
Therefore,
$$
\ln L=\lim_{x \to 1}(\frac{3x}{2+x}-1)({\frac{x}{1-x}})
$$
$$
\ln L=\lim_{x \to 1}(\frac{2x-2}{2+x})({\frac{x}{1-x}})
$$
$$
\ln L=\lim_{x \to 1}(\frac{-2x}{2+x})
$$
$$
L= e^{\frac{-2}{3}}
$$

Proof of the first claim is here: 1 to the power of infinity formula


问:
Show that $\sum_{n=1}^{\infty} {(-1)^n \sin(\frac{x}{n})}$ converges for $a > 0$ in $[-a,a]$
Given:
$$\sum_{n=1}^{\infty} {(-1)^n \sin(\frac{x}{n})}$$

We need to show that it converges in $[-a,a]$ $\forall a >0$.

Now what I figured out:

This is a function series summation. we need first to know what $C_n$ actually is and then do the Radius test.

But how do I exactly get it in the general form ?

答: for $n \ge |a|$,  $\sin \left( \frac{x}{n} \right)$ has same sign as $x$, so we can use Jordan's inequality and then use Leibniz criterion to argue that the series converges.</p>


问:
Are the reals genuinely a subset of the complex numbers?
In Michael Spivak's Calculus, he defines a complex number as an ordered pair of real numbers: $z=(a,b)$ with $a,b \in \mathbb{R}$. The imaginary unit $i$ is then just a shorthand for the ordered pair $(0,1)$. Spivak goes on to say

When complex numbers were first introduced, it was understood that real numbers were, in particular, complex numbers; if our definition is to be taken seriously then this is not true—a real number is not a pair of real numbers, after all.

Although the complex number $(a,0)$ behaves in pretty much the same way as the real number $a$, they are still not identical. There seem to be some non-trivial differences, as well: we can't write $(5,0)>(3,0)$ in the way we can write $5>3$. With this in mind, I ask the following questions:

Can the real numbers be said to be a subset of the complex numbers if complex numbers are defined as ordered pairs?

If the complex numbers are constructed in some other way, then is it meaningful to write $\mathbb R \subset \mathbb C$?

答:
It depends on how you define real numbers.

$\mathbb{R}$ can be defined by a set of axioms (a totally ordered field with the section separation element postulate).

In this setting, the construction you referred to is one of the many possible instances (technically called models) of "the real numbers", because it satisfies those axioms. All proposition you can construct about $\mathbb{R}$ that are proven true or false from those axioms will be the same for all models of $\mathbb{R}$, so distinguishing between those is not needed for such statements.

The subset of $\mathbb{C}$ you mentioned also satisfies the axioms, so it can also be called $\mathbb{R}$, making the inclusion statement true.

If, on the other hand, you define $\mathbb{R}$ to be one particular construction, all other isomorphic instances will be "just isomorphic". This approach has little use though, so it is usually not preferred.


问:
Pattern string with repetition
How many license plate with the following pattern?

$$LL DDD LL$$

where L is a letter chosen from $\{A,B,\dots,Z\}$ (26 letters) and D is a digit chosen from $\{0,1,2,\dots,9\}$ (10 digits, zero is allowed).

The answer is $26^2 \times 10^3 \times 26^2$ because repetition of letters and digits are permitted. But what about double counting, why not divide for all cases like $AA$ $111$ $AA$ where we count more that one permutation of the same letter and digit  (in this case A and one )?

My solution is $\frac{26^2 \times 10^3 \times 26^2}{2! \times 3! \times 2!}$.

Exemples of valid string:

$$AB123CD$$
$$AA000AA$$
$$AA111BC$$

答: The given answer is correct. Think about generating the two letters in front like this: first you can pick any of the available letters to go into the first slot, a total of $26$ options. Then, pick any letter again to go in the second slot, again $26$ choices. If you think about choosing one letter before the other, it should be clear that we're only counting each combination once.

Title: Pattern string with repetition


问:
Showing that if $\lim_{n \to \infty} x_n=a$, then for any $k\in \mathbb{N}$, $\lim_{n \to \infty} x_n^{2k}=a^{2k}$
$x_n$ is a sequence. The only thing I can do here is just write the definition of $\lim_{n \to \infty} x_n=a$, but that doesn' t seem helpful to me. I tried looking for some inequalities that would help, and the only thing  that I found that I thought could be a little bit helpful, was the reverse triangle inequality, but I don' t know how to use that here.Could you please give some advice or a hint?

Thanks in advance

答: This is an immediate consequence of the continuity of the map $f_n(x)=x^{n}$.


答:
SeanC has already addressed your second question so I'll just address your first one. I think Sean is being a bit too pessimistic. In general if $V$ is any $G$-representation then we have a canonical inclusion $i : V^G \to V$ of invariants and also a canonical quotient $q : V \to V_G$ to coinvariants (the quotient of $V$ by the equivalence relation $gv \sim v$). These compose to give a canonical map

$$q \circ i : V^G \to V_G.$$

If $V$ is completely reducible then this map is an isomorphism (exercise), so we can consider its inverse $(q \circ i)^{-1} : V_G \to V^G$. This lets us define a map

$$(q \circ i)^{-1} \circ q : V \to V^G$$

which, by construction, splits the inclusion $i : V^G \to V$. This means we have a canonical idempotent

$$R = i \circ (q \circ i)^{-1} \circ q : V \to V$$

whose image is $V^G$; this is the Reynolds operator. The kernel of $R$ is then the desired canonical $G$-equivariant complement of $V^G$. ("Canonical" here can be strengthened to "unique.") To connect this up with Sean's answer, $\text{ker}(R)$ can equivalently be defined as the sum of all nontrivial simple subrepresentations of $V$, or as the maximal subrepresentation of $V$ containing no nonzero invariant vector, and this (together with the fact that $R$ is an idempotent with image $V^G$) uniquely determines $R$.

This is quite abstract, so it's worth knowing for concreteness that if $G$ is a finite group and $K$ is a field of characteristic not dividing the order of $G$ (which implies that every $G$-representation is completely reducible by Maschke's theorem) then $R$ is given explicitly by the averaging operator

$$R(v) = \frac{1}{|G|} \sum_{g \in G} gv.$$

You can also gain some intuition for the relationship between invariants and coinvariants, and for the map $(q \circ i)^{-1}$, by considering the following special case: suppose $G$ is a finite group and $K$ a field of characteristic not dividing $|G|$ as above, and let $V$ be a permutation representation given by taking the free $K$-vector space on a set $X$ on which $G$ acts. Then $V^G$ has a basis given by sums over the orbits of $G$ acting on $X$, while $V_G$ has a basis given by the orbits themselves, each member of which has been identified. The map $(q \circ i)^{-1} : V_G \to V^G$ then sends an orbit $O$ to $\frac{1}{|O|} \sum_{o \in O} o$, which is equal to $\frac{1}{|G|} \sum_{g \in G} go$ for any $o \in O$. (This special case is also a nice example of what happens if we don't assume complete reducibility: if $G$ and $X$ are infinite in this construction then $V^G$ has a basis given by sums over finite orbits while $V_G$ still has a basis given by all orbits, so $q \circ i : V^G \to V_G$ is not an isomorphism if there are infinite orbits.)

When $K = \mathbb{R}$ or $\mathbb{C}$ and $G$ is a compact Lie group acting on a finite-dimensional $V$ then there is a generalization of the above formula involving averaging over $G$ with respect to Haar measure.

Title: About a property of Reynolds operator (Invariant theory)

Body:
I am learning about classical invariant theory, and I have a question about Reynolds operator. The book I am reading is $\ulcorner$Classical Invariant Theory$\lrcorner$ written by Hansepter Kraft, Claudio Procesi. If you want, you can download it here.

Consider an infinite field $K$ and a group $G$ and a finite dimensional G-module W and its coordinate ring $K[W]$, and assume that the representation of $G$ on $K[W]$ is completely reducible.
In the textbook, they define a canonical G-equivariant linear projection $R:K[W]\rightarrow K[W]^G$ and call it a Reynolds operator. ($K[W]^G$ is a ring of invariants)

They asserts that $R(hf)=hR(f)$ holds for $h\in K[W]^G$, $f\in K[W]$, and this property is crucial in proof of Hilbert's finiteness result for invariants. (p.13 Theorem 1 in the textbook) But I cannot figure out why this relation holds. Actually, it is one of its exercise to prove the equation in a little more general setting. (p.13 Exercise 30)

Let $A$ be a (commutative) algebra and let $G$ be a group of algebra automorphisms
of $A$. Assume that the representation of $G$ on $A$ is completely
reducible. Then the subalgebra $A^G$ of invariants has a canonical $G$-stable
complement and the corresponding $G$-equivariant projection $p:A\rightarrow A^G$ satisfies the relation $p(hf)=hp(f)$ for $h\in A^G$, $f\in A$.

I have two questions. First, what does a canonical $G$-stable complement mean? Indeed complete reducibility ensures the existence of a $G$-stable complement, but how can we 'construct' such complement of $A^G$?

Second, how can we show that $p(hf)=hp(f)$ for $h\in A^G$, $f\in A$ ? I tried to prove that the complement is stable under multiplication by $h\in A^G$, but couldn't make any progress.

Any helps, comments, suggestions will be appreciated. Thank you for reading.


Title: Find the limit $\lim_{x \to 1}(\frac{3x}{2+x})^\frac{x}{1-x}$

Body:
Find the limit
$$
\lim_{x \to 1}\left(\frac{3x}{2+x}\right)^\frac{x}{1-x}
$$

I've transformed the function by changing limit of 1 to zero and become the following:
$$
\lim_{x \to 0}\left(1 + \frac{3x-3}{x-3}\right)^\frac{1 - x}{x}
$$

I spent many time and tried to transform the power part $\frac{1 - x}{x}$ to $\frac{x-3}{3x-3}$ but still can't remove the $x$'s on final. How can I find this limit? Thanks a lot!

Body:
$$A=\Bigg[\frac{3x}{2+x}\Bigg]^\frac{x}{1-x}$$ Make $x=1+y$ to work around $y=0$. So
$$A=\Bigg[\frac{3y+3}{y+3}\Bigg]^{-\frac{y+1}{y}}=\Bigg[3-\frac{6}{y+3}\Bigg]^{-\frac{y+1}{y}}$$
$$\log(A)=-\frac{y+1}{y}\log\Bigg[3-\frac{6}{y+3}\Bigg]$$ Now, by Taylor
$$\log\Bigg[3-\frac{6}{y+3}\Bigg]=\frac{2 y}{3}-\frac{4 y^2}{9}+O\left(y^3\right)$$
$$\log(A)=-\frac{2}{3}-\frac{2 y}{9}+O\left(y^2\right)$$
$$A=e^{\log(A)}=\frac{1}{e^{2/3}}-\frac{2 y}{9 e^{2/3}}+O\left(y^2\right)$$


问:
Show that $v\mapsto \|v\|^p$ for $1\le p<\infty$ is strictly convex
I want to show that $f(v)=\|v\|^p$ for $1\leq p<\infty$ is strictly convex.

In the simplest case when $p=2$, we have:
$$\left\|\frac{v_1+v_2}{2}\right\|^2=\left(\frac{v_1+v_2}{2},\frac{v_1+v_2}{2}\right)=$$
$$\frac{\|v_1^2\|}{4}+\frac{\|v_2^2\|}{4}+\frac{1}{2}(v_1,v_2)$$
$$\leq \frac{\|v_1^2\|}{4}+\frac{\|v_2^2\|}{4}+ \frac{\|v_1^2\|}{4}+\frac{\|v_2^2\|}{4}$$
(by Cauchy inequality). That is convex, but not strictly convex - where did I make a mistake?

Please help me. Thanks.

答:
The $p=2$ case:

If $f(\frac12(u+v))=\frac12(f(u)+f(v))$ then $\frac14\|u\|^2+\frac14\|v\|^2+\frac12\langle u,v\rangle=\frac12\|u\|^2+\frac12\|v\|^2$ by the computation in your post, hence $\|u\|^2+\|v\|^2-2\langle u,v\rangle=0$, that is, $\|u-v\|^2=0$. Thus, for every $u\ne v$, $f(\frac12(u+v))\lt\frac12(f(u)+f(v))$, QED.


问:
How to reduce to affine case to determine whether a given functor is representable
[Definition] These are the contents of Gortz and Wedhorn , Algebraic Geometry.

$\widehat{(Sch)}$ is the category of functors $(Sch)^{opp} \rightarrow (Sets)$

For scheme $X$, define the functor $h_X := Hom(\_\_ , X)$ and identify $X$ with $h_X$ in $\widehat{(Sch)}$

A morphism $f : F \rightarrow G$ of functors in $\widehat{(Sch)}$ is called representable if for all schemes $X$ and all morphisms $g : X \rightarrow G$ in $\widehat{(Sch)}$  the functor $F \; {\times_G} X$ is representable

An open subfunctor $F'$ of $F$ is a representable morphism $f:F' \rightarrow F$ that is an open immersion.

A family $(f_i:F_i \rightarrow F)_{i \in I}$ of open subfunctors is called a Zariski open covering of $F$ if for every $S$-scheme $X$ and every $S$-morphism $g:X \rightarrow F$ the images of the $(f_i)_{(X)}$ form a covering of $X$, where $(f_i)_{(X)} : F_i \; \times_F X \rightarrow X $ is the second projection map of fiber product.

[Theorem 8.9] Let $S$ be a scheme and $F:(Sch/S)^{opp} \rightarrow (Sets)$ be a functor such that

 (i) $F$ is a sheaf for the Zariski topology
(ii) $F$ has a Zariski open covering $(f_i:F_i \rightarrow F)_{i \in I}$ by representable functors $F_i$.
Then $F$ is representable.

[Theorem 11.1] Let $X$ be a scheme and let $\mathscr{R}$ be a quasi-coherent $\mathscr{O}_{X}$-algebra. Then there exsist an $X$-scheme $\text{Spec}(\mathscr{R})$ which is represents the functor
$$ F: (Sch/X)^{opp} \rightarrow (Sets), \hspace{1cm} (f:T \rightarrow X) \mapsto \text{Hom}_{(\mathscr{O}_X-Alg)}(\mathscr{R},f_{*}\mathscr{O}_T)$$

[Question]

In the proof of the theorem 11.1 , they say that " To show that $F$ is representable, by theorem 8.9 we may assume that $X=\text{Spec}{A}$ is affine."

I wonder how I can exactly take a family of open subfunctors $(f_i:F_i \rightarrow F)_{i \in I}$ so that we can assume $X$ is affine.

Let $X = \bigcup U_i$ and $U_i = \text{Spec}A_i$.
$$ F_i: (Sch/X)^{opp} \rightarrow (Sets), \hspace{1cm} (f:T \rightarrow X) \mapsto \text{Hom}_{(\mathscr{O}_{U_i}-Alg)}(\mathscr{R}\vert_{U_i},f_{*}\mathscr{O}_T)$$

Then, are these $F_i$ open subfunctors of $F$ ?

I can't prove this $(f_i:F_i \rightarrow F)_{i \in I}$ is a Zariski open covering of $F$

答:
Here's a rewritten answer for better comprehension. I also apologise for the delay in responding, I was very busy for various reasons.

The answer below is long, but fills in most of the details.

Objective- We want to show that given a scheme $X$ and $\mathcal{B}$ a quasi-coherent $\mathcal{O}_X$-algebra, we want to show that the contravariant functor on $\operatorname{Sch}_{X}$ which sends any $X$-scheme $f\colon T\to X$ to the set $$F(T)\colon= \operatorname{Hom}_{\mathcal{O}_X-\text{Alg}}(\mathcal{B},f_*\mathcal{O}_T)$$ is in fact representable.

I will leave the verification that this functor is a Zariski sheaf as an exercise.

Observation 1- Let's note that this functor is affinely representable.
This means that if $X=\operatorname{Spec}(A)$ for some ring $A$ and $B$ an algebra over $A$, then the functor on $\operatorname{Sch}_X\to \operatorname{Sets}$ which sends
$$\big(f\colon T\to X \big)\mapsto \operatorname{Hom}_{A-\text{Alg}}(B,\Gamma(T,\mathcal{O}_T))$$ is representable. In this case it is representable by $\operatorname{Spec}(B)$.

This is Exercise $2.4$ in Hartshorne in the aboslute case (i.e. when $A=\mathbf{Z}$) and Prop $3.4$ in Gortz-Wedhorn in the relative case. You can also probably find it somewhere in the Stacks Project.

Observation 2- Now let $X$ be a general scheme and let $j\colon U \hookrightarrow X$ be an affine open subscheme of $X$.

Let us try to understand the category $\operatorname{Sch}_U$ in terms of $\operatorname{Sch}_X$. Note that there is a functor $j_!\colon \operatorname{Sch}_U\to \operatorname{Sch}_X$ which sends every $U$-scheme $g\colon S\to U$ to an $X$-scheme via the composition $$j\circ g\colon S\to U\hookrightarrow X.$$

Conversely, every $X$ scheme $f\colon T\to X$ for which the structure morphism factors as $T\to U\hookrightarrow X$ is canonically a $U$-scheme. Thus $$j_!\colon \operatorname{Sch}_U\to \operatorname{Sch}_X$$ embeds $\operatorname{Sch}_U$ as a full subcategory of $\operatorname{Sch}_X$.

Observation 3- Now note that since $\mathcal{B}$ is a quasi-coherent $\mathcal{O}_X$-algebra, it's restriction $\mathcal{B}|_U$ to $U$ is a quasi-coherent $\mathcal{O}_U$-algebra. Since $U$ is affine $\mathcal{B}|_{U}=\Gamma(U,\mathcal{B})$ which is an algebra over $\Gamma(U,\mathcal{O}_U)$.

So consider the functor $F_U(-)\colon \operatorname{Sch}^{op}_U\to \operatorname{Sets}$ which sends $$\big(g\colon S\to U \big)\to \operatorname{Hom}_{\mathcal{O}_U-\text{Alg}}(\mathcal{B}|_{U},g_*\mathcal{O}_S).$$

By Observation $1$, this functor is representable by $\operatorname{Spec}(\Gamma(U,\mathcal{B}))$ since $U$ is affine.

Composing $F_U$ with $j_!$ of Observation $2$, we see that $$F_U\circ j_!\colon \operatorname{Sch}^{op}_U\to \operatorname{Sch}^{op}_X\to \operatorname{Sets}$$ is a functor on the full subcategory $\operatorname{Sch}_U$ of $\operatorname{Sch}_X$ and we extend it to a functor on all of $\operatorname{Sch}_X$ by sending everything not in $\operatorname{Sch}_U$ to $\emptyset$.

This defines a functor on $\operatorname{Sch}_X$ which we again denote by abuse of notation as $$F_U\colon \operatorname{Sch}^{op}_X\to \operatorname{Set}.$$

Observation 4-

There is an adujunction morphism $\eta\colon \mathcal{B}\to j_*j^{*}\mathcal{B}=j_*\mathcal{B}|_U.$ On an open set $V\subset X$, the morphism is the restriction morphism from $\eta_V\colon \Gamma(V,\mathcal{B})\to \Gamma(U\cap V,\mathcal{B})$. This is a morphism of algebras since $\mathcal{B}$ is a sheaf of algebras.
In particular, for any $\mathcal{O}_U$ algebra $\mathcal{A}$, and a morphism of $\varphi \colon \mathcal{B}|_U\to \mathcal{A}$ of $\mathcal{O}_U$ algebras, the composition
$$\mathcal{B}\xrightarrow{\eta} j_*\mathcal{B}|_U\xrightarrow{j_*\varphi}j_*\mathcal{A}$$ is a morphism of $\mathcal{O}_X$-algebras, and so you get an induced morphism

$$\eta^*\colon\operatorname{Hom}_{\mathcal{O}_X-\text{Alg}}(j_*\mathcal{B}|_U,j_*\mathcal{A} )\to \operatorname{Hom}_{\mathcal{O}_X-\text{Alg}}(\mathcal{B},j_*\mathcal{A})$$ by pre-composition. Note that this is functorial in $\mathcal{A}$ i.e. $\eta^*$ is a natural transformation between two functors in the variable $\mathcal{A}$.

But note that any $\mathcal{O}_X$-algebra morphism $j_*\mathcal{B}|_U\to j_*\mathcal{A}$ is just an $\mathcal{O}_U$-algebra morphism since $j\colon U\hookrightarrow X$ is an open embedding. So in fact the morphism we get is
$$\eta^*\colon \operatorname{Hom}_{\mathcal{O}_U-\text{Alg}}(\mathcal{B}|_U,\mathcal{A} )\to \operatorname{Hom}_{\mathcal{O}_X-\text{Alg}}(\mathcal{B},j_*\mathcal{A}).$$

Observation 5- Note that the morphism $\eta^*$ is a monomorphism of functors.

Indeed suppose $\alpha,\beta\colon \mathcal{B}|_U\to \mathcal{A}$ are morphisms of $\mathcal{O}_U$-algebras which satisfy $\alpha\circ \eta=\beta\circ \eta$, then in particular at stalks for all $x\in U$ they must satisfy $\alpha_x=\beta_x$ and so $\alpha=\beta$.

Thus we see that $\eta^*$ as defined in Observation $4$ is a monomorphism.

Observation 6- Now using the definition of $F_U$ in Observation $3$ and $F$ as in Objective, we get a monomorphism of functors $$\eta^*\colon F_U\to F.$$

Indeed for any scheme $f\colon T\to X$, we have two conditions. Either $f$ factors through $U$ or it doesn't. In case it does, then what we discussed above shows that $\eta^*_T\colon F_U(T)\to F(T)$ is an injection. In case it doesn't, then $F_U(T)$ is $\emptyset$ and so tautologically a subfunctor of $F(T)$.

Observation 7- Let $g\colon S\to X$ be any $X$-scheme. An $S$-valued point of $F$ is an object of $F(S)=\operatorname{Hom}_{\mathcal{O}_X-\text{Alg}}(\mathcal{B},g_*\mathcal{O}_S)$ i.e. a morphism $\varphi\colon \mathcal{B}\to g_*\mathcal{O}_S$. By Yoneda this is the data of a morphism $\varphi^*\colon h_S\to F$.

So we get a morphism of functors $\eta^*\colon F_U\to F$ and $\varphi^*\colon h_S\to F$ whose fiber product is $F_U\times_F h_S$ in the category of presheaves on $\operatorname{Sch}_X$.

Observation 8- Since limits can be computed pointwise, let $f\colon T\to X$ be an object of $\operatorname{Sch}_X$ and so $$F_U\times_F h_S(T)=F_U(T)\times_{F(T)}h_S(T).$$

Assuming $F_U(T)\neq \emptyset$, the object on the right consists of pairs $(\alpha, h)$ where $\alpha\colon \mathcal{B}|_U\to f_*\mathcal{O}_T$ and $h\colon T\to S$ an $X$ morphism (i.e. $g\circ h=f$). These pairs have to satisfy the condition that they agree over $F(T)$ i.e. they define the same $T$ point of $F$ via the natural maps $\varphi^*\colon h_S\to F$ and $\eta^*\colon F_U\to F$.

More explicitly, this is the condition that the composition $\mathcal{B}\xrightarrow{\varphi} g_*\mathcal{O}_S\xrightarrow{F(h)}f_*\mathcal{O}_T$ agrees with the composition $\mathcal{B}\xrightarrow{\eta}\mathcal{B}|_U\xrightarrow{\alpha}f_*\mathcal{O}_T.$

Observation 9- We need to show now that the functor $F_U\times_F h_S$ is representable by an open subscheme of $S$.

It is enough to consider the open subscheme $V\hookrightarrow S$ so that $h\colon T\to S$ factors as $T\to V\hookrightarrow S$ if and only if $f\colon T\to X$ factors as $T\to U\hookrightarrow X$. Explicitly this subscheme is the preimage $V\colon=g^{-1}U$ where $g\colon S\to X$ was the structure morphism.

Indeed any $T$ point of $h_V$ i.e. a morphism $f  \colon T\to V$ would define a $T$ point of $h_S$ by composition with the open immersion $V\hookrightarrow S$. Moreover, it defines a $T$ point of $F_U$ by definition of $V$. I leave it as an exercise in unwinding definitions to check that these $T$-points agree on $F(T)$.

Thus the pullback is representable by $V$ as defined above.

Observation 10- If $\bigcup_i U_i$ is a Zariski cover of $X$ by affine opens, then for each $i$ the functor $$F_i(-) \colon=\operatorname{Hom}_{\mathcal{O}_{U_i}-\text{Alg}}(\mathcal{B}|_{U_i}, )$$ satisfies the conditions above.

The $V_i\colon= g^{-1}U_i$ constructed as above cover $S$ for each $X$-scheme $g\colon S\to X$. Thus these functors form a Zariski open covering of the functor $F$ defined as in Objective.


问:
Assume $f: \mathbb R \to \mathbb R$ is continuous on $\mathbb R$ and $\lim_{x \to +\infty} f(x)=0=\lim_{x \to -\infty} f(x)$
Assume $f: \mathbb R \to \mathbb R$ is continuous on $\mathbb R$ and $$\lim_{x \to +\infty} f(x)=0=\lim_{x \to -\infty} f(x)$$

Show that $f$ is bounded on $\mathbb R$ and attains either its maximum or its minimum on $\mathbb R$.

$$\lim_{x \to +\infty} f(x)=0$$
$$\iff \forall \epsilon >0(\exists a \in \mathbb R(\forall x(x> a \implies \left|f\left(x\right)\right|<\epsilon)))$$

And
$$\lim_{x \to -\infty} f(x)=0$$
$$\iff \forall \epsilon >0(\exists b \in \mathbb R(\forall x(x< b \implies \left|f\left(x\right)\right|<\epsilon)))$$

Now if for all $x \in \mathbb R:f(x)=0$ then we are done ,otherwise we are able to find a $c \in \mathbb R$ such that $f(c) \ne 0$ and setting $\epsilon \mapsto \left|f\left(c\right)\right|$ follows that exists $a<x<b$ such that $\left|f\left(x\right)\right|<\left|f\left(c\right)\right|$.

On the other hand continuity of $f$ over $\mathbb R$ implies continuity of $f$ over $[a,b]$ and by extreme value theorem $f$ attains its maximum and minimum .( I don't understand why the question say or)

But I only know that for $x \in [a,b]$ the function $f$ is bounded and  it's not enough,since the domain of $f$ is not $[a,b]$ and I need to show that for all real $x$ the function is bounded.

Besides I want to know the validity of my proof about the max-min part.

答:
Please allow me copy and modify your argument. The incorrect point in your argument is to claim that $a<b$, it should be $b<a$!

We know
\begin{align}
\lim_{x \to +\infty} f(x)=0
\iff \forall \epsilon >0(\exists a >0 (\forall x\in \mathbb R
(x> a \implies \left|f\left(x\right)\right|<\epsilon)))
\end{align}
and \begin{align}
\lim_{x \to -\infty} f(x)=0
\iff \forall \epsilon >0(\exists b<0 (\forall x\in \mathbb R
(x<b \implies \left|f\left(x\right)\right|<\epsilon))).
\end{align}

If $f(x)=0, \forall x\in \mathbb R$ then we are done.
Otherwise we are able to find a $c\in \mathbb R$ such that $f(c)\neq 0$ and setting $\epsilon:=|f(c)|$.

Thus, we conclude that for $x\in (-\infty, b)\cup(a,+\infty)$, we have $|f(x)|<\epsilon$.

On the other hand continuity of $f$ over $\mathbb R$ implies continuity of $f$ over $[a,b]$ and by extreme value theorem $f$ attains its maximum $M_{\text{max}}$ and minimum $M_{\text{min}}$ on $[a,b]$, thus bounded on $[a,b]$.

Hence $f$  is bounded on $\mathbb R$.

Regarding the maximum and minimum of $f$ on $\mathbb R$, we consider the following cases

If $M_{\text{max}}>\epsilon$, then $M_{\text{max}}$ is a maximum value of $f$ on $\mathbb R$.

If $M_{\text{min}}<-\epsilon$, then $M_{\text{min}}$ is a minimum value of $f$ on $\mathbb R$.

Otherwise, $-\epsilon\leq M_{\text{min}} \leq M_{\text{max}}\leq \epsilon$, then $f(c)$ is a maximum if $f(c)>0$ and $f(c)$ is a minimum if $f(c)<0$.

Now, we are able to conclude that $f$  attains either its maximum or its minimum on $\mathbb R$.


问:
How to prove that $\log |\sin(x/2)| = -\log 2 -\sum_{n=1}^{\infty} \frac{\cos nx} {n} $ ( if $ x\neq 2 k\pi)$, $k$ is an integer
This question number 11.15(a) Tom M Apostol's MAthematical Analysis From Chapter Fourier analysis.

Prove that  $ \log |\sin(x/2)| = -\log 2 -\sum_{n=1}^{\infty} \frac{\cos nx} {n} $ ( if $ x\neq  2 k\pi)$.

So, I tried by using $a_0 = 2/\pi \int_{0}^{\pi} \log|\sin(x/2)| dx$ and I used $\sin(2x) =2\sin x\cos x$.

So Using it I get $a_0 /2 =\log 2 + 2/\pi [\int_{0}^{\pi}\log|\sin(x/4)|dx +\int_{0}^{\pi}\log|\cos (x/4) |dx]$.

But the problem is how to compute $\int_{0}^{\pi}\log|\cos (x/4) |dx$. The limit of integral can be taken from $0$ to $2 \pi $

Similar problem will also arise when I will compute $a_n = \int_{0}^{\pi} \log|\sin(x/2)| \times\cos(nx) dx$.

Can you please tell how to compute this integral OR If there is alternate way of computing the fourier series.

Thank you.

答: Note that$$\ln|2\sin(x/2)|=\ln|1-e^{ix}|=\tfrac12(\ln(1-e^{ix})+\text{c.c.})=-\tfrac12\sum_{n\ge1}\tfrac{e^{inx}+\text{c.c.}}{n}=-\sum_n\tfrac{\cos nx}{n},$$where $z+\text{c.c.}$ is short for $z+\bar{z}$.