答:
Part II

Please look around for the first part, if this is incidentally first.

Finally, the most complicated integral, $J_{11}$. We have:
$$
\begin{aligned}
J_{11}
&=
\int_0^{\pi}x^2\;\cos x\; \Big(
  \operatorname{Li}_2(\cos x)
- \operatorname{Li}_2(-\cos x)
\Big)
\;dx
\\
&=
\int_0^{\pi/2}x^2\;\cos x\; \Big(
  \operatorname{Li}_2(\cos x)
- \operatorname{Li}_2(-\cos x)
\Big)
\;dx
\\
&\qquad\qquad
+
\int_0^{\pi/2}(\pi-x)^2\;\cos (\pi-x)\; \Big(
  \operatorname{Li}_2(\cos (\pi-x))
- \operatorname{Li}_2(-\cos (\pi-x))
\Big)
\;dx
\\
&=
\int_0^{\pi/2}(x^2+(\pi-x)^2)\;\cos x\; \Big(
  \operatorname{Li}_2(\cos x)
- \operatorname{Li}_2(-\cos x)
\Big)
\;dx
\\
&=
\int_0^{\pi/2}\Big( \ (4x-2\pi)\cos x + (2x^2-2\pi x +\pi^2-4) \sin x\ \Big)'\;
\\
&\qquad\qquad\cdot
\Big(
  \operatorname{Li}_2(\cos x)
- \operatorname{Li}_2(-\cos x)
\Big)
\;dx
\\
&=
\frac 12\pi^3
-
\int_0^{\pi/2}
\Big( \ (4x-2\pi)\cos x + (2x^2-2\pi x +\pi^2-4) \sin x\ \Big)
\\
&\qquad\qquad\cdot
\left(
\frac 1{\cos x}\operatorname{Li}_1(\cos x)\cdot (-\sin x)
-
\frac 1{-\cos x}\operatorname{Li}_1(-\cos x)\cdot \sin x
\right)\; dx
\\
&=
\frac 12\pi^3
-
\int_0^{\pi/2}
\Big( \ (4x-2\pi)\cos x + (2x^2-2\pi x +\pi^2-4) \sin x\ \Big)\;
\frac {\sin x}{\cos x}\log\frac{1-\cos x}{1+\cos x}
\; dx
\\
&=
\frac 12\pi^3
-
\int_0^{\pi/2}
(4x-2\pi)\;
\sin x\;\log\frac{1-\cos x}{1+\cos x}
\; dx
\\
&\qquad\qquad
-
\int_0^{\pi/2}
(2x^2-2\pi x +\pi^2-4) \;
\frac {\sin^2 x}{\cos x}\log\frac{1-\cos x}{1+\cos x}
\; dx
\\
&=
\frac 12\pi^3
-
\int_0^1
(8\arctan t-2\pi)\;
\frac{2t}{1+t^2}\;\log t^2\;\frac2{1+t^2}
\; dt
\\
&\qquad\qquad
-
\int_0^1
(8\arctan^2 t-4\pi \arctan t +\pi^2-4) \;
\frac {4t^2}{(1+t^2)(1-t^2)}\log t^2\;\frac2{1+t^2}
\; dt
\\
&=
\frac 12\pi^3
-
\underbrace{
\int_0^1
(8\arctan t-2\pi)\;
\left(
\frac{4t^2}{1+t^2}\log t - 2\log(1+t^2)
\right)'
\; dt}
_{=8\pi\log 2- 4\pi}
\\
&\qquad\qquad
+2
\underbrace{
\int_0^1
(8\arctan^2 t-4\pi \arctan t +\pi^2-4) \;
\left(
\frac{2t}{1+t^2} + \log\frac{1-t}{1+t}
\right)'
\;\log t
\; dt}_{=J_{111}}
\ .
\end{aligned}
$$
Some words on the value of the known integral above, the one with the value
$8\pi\log 2-4\pi$. We use partial integration. Then the piece in $(8\arctan t-2\pi)'=8/(1+t^2)$ is rational, we use partial fraction decomposition over $\Bbb C$, and thus the integral is broken into pieces, each piece
being of the form $\frac 1{t-b}\log(t-a)$ or $\frac 1{(t-b)^2}\log(t-a)$ with $a,b$ among $0,\pm i$. These integrals can be handled.
Dilogarithms appear, we use than the dilogarithm identities.

A numerical check:

```<code>? 8*Pi*log(2) - 4*Pi
%41 = 4.8543181080696440901549376527829000419
? intnum( t=0, 1, (8*atan(t) - 2*Pi) * 4*t/(1+t^2)^2 * 2*log(t) )
%42 = 4.8543181080696440901549376527829000419
? intnum( x=0, Pi/2, (4*x - 2*Pi) * sin(x) * log( (1-cos(x)) / (1+cos(x)) ) )
%43 = 4.8543181080696440901549376527829000419
</code>```

Now we look closer to the remained integral, $J_{111}$. Parts of it are rather easy. We have
$$
\begin{aligned}
J_{111s}
&:=
\int_0^1
\left(
\frac{2t}{1+t^2} + \log\frac{1-t}{1+t}
\right)'
\;\log t\; dt
\\
&=
\underbrace{\int_0^1
\left(
\frac{2t}{1+t^2}\right)'
\;\log t\; dt}_{=-2\arctan 1=-\pi/2}
-
\underbrace{
\int_0^1
\left(
\frac 1{1-t}+\frac 1{1+t}
\right)
\;\log t\; dt}_{=-\pi^2/4}
\ .
\end{aligned}
$$
The integral with value $-2\arctan 1$ is done by partial integration, the other one using the (real part of the) primitive $\operatorname{Li}_2(1\pm t)$.
The dilog vanishes in $0$, the two contributions in $t=0$ of $\operatorname{Li}_2(1\pm t)=\operatorname{Li}_2(1\pm t)$ cancel each other (different signs),
and there remains $-\Re\operatorname{Li}_2(1+1)=-\frac 14\pi^2$.

The part in $\arctan t\cdot \log t$ from $J_{111}$ can also be computed. The parts derived from
$\left(
\frac 1{1-t}+\frac 1{1+t}
\right)
$ are considered in $J_{1211a}$, $J_{1211b}$. The remained part in $\arctan t\cdot \log t$ is
$$
\begin{aligned}
J_{111t}
&:=
\int_0^1
\arctan t\;
\left(
\frac{2t}{1+t^2}
\right)'
\;\log t
\; dt
\\
&=
-
\int_0^1
\frac{2t}{1+t^2}
\left(
\frac 1{1+t^2}\cdot\log t
+
\arctan t\cdot\frac 1t
\right)
\; dt
\\
&=
\int_0^1
\left(
\frac 1{1+t^2}-1
\right)'
\cdot\log t
\; dt
-
\int_0^1
\frac{2}{1+t^2}
\arctan t
\; dt
\\
&=\frac 12\log 2-\frac 1{16}\pi^2\ .
\end{aligned}
$$
So the main issue is
$$
\tag{16}
$$
$$
\begin{aligned}
J_{111u}
&:= \int_0^1
\arctan^2 t\;
\left(
\frac{2t}{1+t^2}
+
\log\frac{1-t}{1+t}
\right)'
\;\log t
\; dt
\\
&=
\underbrace{
\int_0^1
\arctan^2 t\;
\left(
\frac{2t}{1+t^2}
\right)'
\;\log t
\; dt}_{J_{111u1}}
-
\underbrace{
\int_0^1
\arctan^2 t\;
\log t\;
\left(\frac 1{1-t} +\frac 1{1+t}\right)
\; dt}_{J_{111u2}}
\ .
\end{aligned}
$$
The term listed first is simpler.
$$
\begin{aligned}
J_{111u1}
&=
-
\int_0^1
\frac{2t}{1+t^2}
\left(
\arctan^2 t\cdot \frac 1t
+
2\arctan t\cdot\log t\cdot\frac 1{1+t^2}
\right)
\; dt
\\
&=
-
\left[\frac 23\arctan^3 t\right]_0^1
+
2\int_0^1
\left(\frac 1{1+t^2}\right)'\cdot\arctan t\cdot \log t\; dt
\\
&=
-\frac 23\cdot\frac 1{4^3}\pi^3
-
2\int_0^1
\frac 1{1+t^2}\;
\left(\frac 1{1+t^2}\cdot \log t + \arctan t\cdot\frac 1t\right)\; dt
\\
&=
-\frac 1{96}\pi^3
-2\underbrace{\int_0^1\frac{\log t}{(1+t^2)^2}\; dt}_{-(\pi+4G)/8}
-2\underbrace{\int_0^1 \frac 1t\;\arctan t\; dt}_{G}
+\underbrace{\int_0^1 \frac {2t}{1+t^2}\;\arctan t\; dt}_{-\frac 14\pi\log 2+G}
\\
&=
-\frac 1{96}\pi^3 +\frac 14\pi -\frac 14\pi\log 2
\ .
\end{aligned}
$$
For the integral of $\frac 1t\arctan t$ use maybe the Taylor expansion.
For the integral with the $\log t$ use the partial fraction decomposition over $\Bbb C$ for
$\frac 1{(t^2-a^2)^2}=\frac 1{4a^3}\left(\frac 1{t+a}-\frac 1{t-a}\right)+\frac 1{4a^2}\left(\frac 1{(t-a)^2}+\frac 1{(t+a)^2}\right)$,
where $a=i$.  For the remained integral, using partial integration we compute instead
$\int_0^1\frac{\log(1+t^2)}{1+t^2}\; dt$. We split again in $\log$, and in partial fractions, using $(t^2+1)=(t-i)(t+i)$.
Then $\int_0^1\frac{\log(t+i)}{t+i}\; dt=\int_0^1\log(t+i)\cdot(\log(t+i))'\; dt$, so we can integrate. And
$\int_0^1\frac{\log(t+i)}{t-i}\; dt$ leads to a dilog value, $\operatorname{Li}_2((1-i)/2)$, and we need finally from it
$\Im\operatorname{Li}_2((1-i)/2)=-G+\frac 18\pi\log 2$.

So up to mentioned white noise integrals, that can be handled, we are in position to start the solution.
We show, using $K$ for a shorter name:
$$
\tag{17}
$$
$$
\begin{aligned}
K:=J_{111u2}
&:=
\int_0^1
\arctan^2 t\cdot\log t\;
\left(
\frac 1{1-t}
+
\frac 1{1+t}
\right)
\; dt
\\
&=
\frac 3{128}\pi^4
+
\frac 1{32}\pi^2\log^2 2
-
\frac 12\pi G\log 2
-
\pi\Im\operatorname{Li}_3\left(\frac{1+i}2\right)\ .
\end{aligned}
$$
(This relation resisted to all standard attacks. I tried several ideas and tricks like
partial integration, and the substitutions $s=1/t$ and $t=(1-u)/(1+u)$, and the deformation
of the $\arctan t$ using the parameter $a$, thus replacing
$\arctan t=\int_0^1\frac {t\; da}{1+a^2t^2}$ and/or replacing
$\arctan^2 t=\iint_{[0,1]^2}\frac {t\; da}{1+a^2t^2}\cdot \frac {t\; db}{1+b^2t^2}$, and so on.
The complexity
of the resulted expressions was not reduced. Some of these lines are mentioned after the solution below,
which is something experimentally found and new, that i never saw before.
Well, when ideas and tricks do not work, only madness can help us...)

We will work using complex analysis and the first step is to write:
$$
\int_0^1=\int_0^i+\int_i^1\ .
$$
Since $K\in \Bbb R$, we compute only the real part of the integrals in the R.H.S. above. Using
$$
\arctan t=\frac 1{2i}\log\frac{1+it}{1-it}
$$
around zero we can compute by the parametrization $t=iu$, $u\in[0,1)$:
$$
\begin{aligned}
K_1&:=
\Re \int_0^i
\arctan^2 t\cdot\log t
\cdot\frac 2{1-t^2}
\\
&=
\Re\int_0^1
-\frac 14\cdot\log^2\frac{1+i\cdot iu}{1-i\cdot iu}\cdot\log(iu)\cdot\frac2{1+u^2}\; i\; du
\\
&=
\Re
\int_0^1
-\frac 14\cdot\log^2\frac{1-u}{1+u}\cdot\left(i\frac \pi 2+\log u\right)\cdot\frac2{1+u^2}\; i\; du
\\
&=
\frac \pi 4
\int_0^1
\log^2\frac{1-u}{1+u}\cdot\frac 1{1+u^2}\; du
\\
&=
\frac \pi 4
\int_0^1
\log^2 s\cdot\frac 1{1+\left(\frac{1-s}{1+s}\right)^2}\; \frac 2{(1+s)^2}\;ds
=
\frac \pi 4
\int_0^1
\log^2 s\cdot\frac 2{(1+s)^2 + (1-s)^2}\; ds
\\
&=
\frac \pi 4
\int_0^1
\log^2 s\cdot\frac 1{2i}\left(\frac 1{s-i}-\frac 1{s+i}\right)\; ds
\\
&=
\frac \pi 4
\cdot\frac 1{2i}
\left(-2\operatorname{Li}_3\left(\frac 1i\right)
+2\operatorname{Li}_3\left(\frac 1{-i}\right)
\right)
=\frac \pi 4\cdot 2\cdot\frac{\pi^3}{32}
\\
&=\frac{\pi^4}{64}\ .
\end{aligned}
$$
(Similarly to the $\log(1-t)$ factor from the original integral, that may show some integration problem in $u=1$,
but there isn't any, there also no issue for $\log(1+iu)$ in $i$.)
Numerical check, pari/gp can do it also under such unusual circumstances:

```<code>? real(intnum( t=0, i, atan(t)^2 * log(t) *2/(1-t^2) ))
%290 = 1.5220170474062880818193801982610173632769935261357097139291853029682946165
? pi^4 / 64
%291 = 1.5220170474062880818193801982610173632769935261357097139291853029682946165
</code>```

The other integral turns out to be also feasible by using a path from $i$ to $1$ on the unit circle,
explicitly we use the parametrization "$t=t(s)$" with
$$ t(s) = \sin 2x + i\cos 2s=i\cdot e^{-2is}\ ,\qquad s\in[0,\ \pi/4]\ .$$
Then
$$
\begin{aligned}
\frac
{1+i\; t(s)}
{1-i\; t(s)}
&=
\frac
{1-e^{-2is}}
{1+e^{-2is}}
=
\frac
{(1-e^{-2is})(1+e^{+2is})}
{(1+e^{-2is})(1+e^{+2is})}
=
\frac
{2i\; \sin 2s}
{2+2\cos 2s}
\\
&=i\cdot\frac{2\sin s\cos s}{2\cos ^2 s}
=i\;\tan s\ .
\end{aligned}
$$
The other factors in the integrand can be reshaped in a similar way. So...
$$
\begin{aligned}
K_2&:=
\Re
\int_i^1
\arctan^2 t\cdot\log t
\cdot\frac 2{1-t^2}
\\
&=
\Re
\int_0^{\pi/4}
-\frac 14\cdot\underbrace{\log^2(i\tan s)}_{=\left(i\frac \pi2+\log\tan s\right)^2}\cdot i\;\left(\frac \pi 2-2s\right)\cdot\frac2{\cos 2s}\; ds
\\
&=
\int_0^{\pi/4}
\frac 14\cdot2\cdot\frac \pi 2\cdot\log\tan s
\cdot\left(\frac \pi 2-2s\right)\cdot\frac2{\cos 2s}\; ds\qquad(t=\tan s)
\\
&=
\frac \pi 2
\int_0^1
\log t
\cdot\left(\frac \pi 2-2\arctan t\right)\cdot\frac{1+t^2}{1-t^2}\; \frac 1{1+t^2}\;dt
\\
&=
\frac{\pi^2}4
\cdot\frac 12
\int_0^1
\log t\left(\frac1{1-t}+\frac 1{1+t}\right)\; \;dt
-
\frac \pi
\int_0^1
\log t\cdot \arctan t\cdot\frac 1{1-t^2}\;dt
\ .
\end{aligned}
$$
For the first integral above, note that
$\partial_t\operatorname{Li}_2(1-t)
=\frac 1{1-t}\log t$, and
$\partial_t\operatorname{Li}_2(1+t)
=-\frac 1{1+t}\log (-t)$.
So its value is
$$
\begin{aligned}
\Big[
\operatorname{Li}_2(1-t)
\Big]_0^1
-
\Re
\Big[
\operatorname{Li}_2(1+t)
\Big]_0^1
&=
(\operatorname{Li}_2(0)-\operatorname{Li}_2(1))
-
\Re
(\operatorname{Li}_2(2)-\operatorname{Li}_2(1))
\\
&=
-\Re\operatorname{Li}_2(2)
=-\frac {\pi^2}4\ .
\end{aligned}
$$
The second integral above was already computed as part of $(15)$, it is related to
$J_{1211a}$ and $J_{1211b}$. This gives:
$$
\begin{aligned}
K &= K_1+K_2 = \frac{\pi^4}{64} + K_2
\\
&=\frac{\pi^4}{64} +
\frac{\pi^2}8\cdot\left(-\frac {\pi^2}4\right)
-\frac\pi 2(J_{1211a}+ J_{1211b}) =\dots
\end{aligned}
$$
and so on.
We are in position to put all together. First of all
$$
\tag{18}
$$
$$
\begin{aligned}
J_{111}
&=
8(J_{111u1}-\underbrace{J_{111u2}}_{K})
-4\pi(J_{111t}-J_{1211a}-J_{1211b}) +(\pi^2-4)J_{111s}
\\
&=
-\frac{\pi^4}4 - \frac{\pi^2}4\log^2 2 +8 \pi G\log 2
+16\pi \operatorname{Li}_3\left(\frac {1+i}2\right)
+4\pi - \pi^2 -\frac 13\pi^3 -4\pi\log 2
\ .
\end{aligned}
$$
From $(18)$ and $(15)$ we get:
$$
\begin{aligned}
J
&= 2\underbrace{J_1}_{\frac 12J_{11}-\frac 14 J_{12}}
+ 2\pi^2\cdot \underbrace{J_2}_{\frac{\pi^2}2-\pi}
\\
&={\color{blue}{J_{11}}} -\frac 12{\color{brown}{J_{12}}} +\pi^4-2\pi^3
\\
&=
{\color{blue}
{\left(\frac {\pi^3}2 -8\pi\log 2 +4\pi + 2J_{111}\right)}}
-\frac 12
{\color{brown}
{\left(-\frac {\pi^3}3 -8\pi\log 2 +8\pi -4\pi J_{121}\right)}}
+\pi^4-2\pi^3
\\
&=
\pi^4-\frac 43\pi^3-4\pi\log2 + 2J_{111}
+
2\pi\cdot J_{121}
\\
&=
\pi^4-\frac 43\pi^3-4\pi\log2
\\
&\qquad
+\frac 18\left[
-4\pi^4 - 8\pi^2\log^2 2 +128 \pi G\log 2 +256\pi \operatorname{Li}_3\left(\frac {1+i}2\right)
 -16 \pi^2
\right]
-\frac 23\pi^3
-8\pi\log 2 + 8\pi
\\
&\qquad
+
\frac 18\left[\ 5\pi^4+4\pi^2\log^2 2  -64\pi G\log 2 - 128\pi\Im\operatorname{Li}_3\left(\frac {1+i}2\right)\ \right]
+4\pi\log 2-8\pi
\\
&=
\frac 98\pi^4-2\pi^3-2\pi^2
-\frac 12\pi^2\log^2 2
+8\pi G\log 2
+ 16\pi\Im\operatorname{Li}_3\left(\frac {1+i}2\right)
-8\pi\log 2
\ .
\end{aligned}
$$

$\square$

Title: Challenging problem: Calculate $\int_0^{2\pi}x^2 \cos(x)\operatorname{Li}_2(\cos(x))dx$

Body:
The following problem is proposed by a friend:

$$\int_0^{2\pi}x^2 \cos(x)\operatorname{Li}_2(\cos(x))dx$$

$$=\frac{9}{8}\pi^4-2\pi^3-2\pi^2-8\ln(2)\pi-\frac12\ln^2(2)\pi^2+8\ln(2)\pi G+16\pi\Im\left\{\operatorname{Li}_3\left(\frac{1+i}{2}\right)\right\}$$

My only try is writing

$$\operatorname{Li}_2(\cos(x))=-\int_0^1\frac{\cos(x)\ln(y)}{1-\cos(x)y} \, dy$$

and have no idea how to continue with the double integral. I also tried $\cos(x)=u$, didn't do much.

Any help would be much appreciated.

Body:
Part II

Please look around for the first part, if this is incidentally first.

Finally, the most complicated integral, $J_{11}$. We have:
$$
\begin{aligned}
J_{11}
&=
\int_0^{\pi}x^2\;\cos x\; \Big(
  \operatorname{Li}_2(\cos x)
- \operatorname{Li}_2(-\cos x)
\Big)
\;dx
\\
&=
\int_0^{\pi/2}x^2\;\cos x\; \Big(
  \operatorname{Li}_2(\cos x)
- \operatorname{Li}_2(-\cos x)
\Big)
\;dx
\\
&\qquad\qquad
+
\int_0^{\pi/2}(\pi-x)^2\;\cos (\pi-x)\; \Big(
  \operatorname{Li}_2(\cos (\pi-x))
- \operatorname{Li}_2(-\cos (\pi-x))
\Big)
\;dx
\\
&=
\int_0^{\pi/2}(x^2+(\pi-x)^2)\;\cos x\; \Big(
  \operatorname{Li}_2(\cos x)
- \operatorname{Li}_2(-\cos x)
\Big)
\;dx
\\
&=
\int_0^{\pi/2}\Big( \ (4x-2\pi)\cos x + (2x^2-2\pi x +\pi^2-4) \sin x\ \Big)'\;
\\
&\qquad\qquad\cdot
\Big(
  \operatorname{Li}_2(\cos x)
- \operatorname{Li}_2(-\cos x)
\Big)
\;dx
\\
&=
\frac 12\pi^3
-
\int_0^{\pi/2}
\Big( \ (4x-2\pi)\cos x + (2x^2-2\pi x +\pi^2-4) \sin x\ \Big)
\\
&\qquad\qquad\cdot
\left(
\frac 1{\cos x}\operatorname{Li}_1(\cos x)\cdot (-\sin x)
-
\frac 1{-\cos x}\operatorname{Li}_1(-\cos x)\cdot \sin x
\right)\; dx
\\
&=
\frac 12\pi^3
-
\int_0^{\pi/2}
\Big( \ (4x-2\pi)\cos x + (2x^2-2\pi x +\pi^2-4) \sin x\ \Big)\;
\frac {\sin x}{\cos x}\log\frac{1-\cos x}{1+\cos x}
\; dx
\\
&=
\frac 12\pi^3
-
\int_0^{\pi/2}
(4x-2\pi)\;
\sin x\;\log\frac{1-\cos x}{1+\cos x}
\; dx
\\
&\qquad\qquad
-
\int_0^{\pi/2}
(2x^2-2\pi x +\pi^2-4) \;
\frac {\sin^2 x}{\cos x}\log\frac{1-\cos x}{1+\cos x}
\; dx
\\
&=
\frac 12\pi^3
-
\int_0^1
(8\arctan t-2\pi)\;
\frac{2t}{1+t^2}\;\log t^2\;\frac2{1+t^2}
\; dt
\\
&\qquad\qquad
-
\int_0^1
(8\arctan^2 t-4\pi \arctan t +\pi^2-4) \;
\frac {4t^2}{(1+t^2)(1-t^2)}\log t^2\;\frac2{1+t^2}
\; dt
\\
&=
\frac 12\pi^3
-
\underbrace{
\int_0^1
(8\arctan t-2\pi)\;
\left(
\frac{4t^2}{1+t^2}\log t - 2\log(1+t^2)
\right)'
\; dt}
_{=8\pi\log 2- 4\pi}
\\
&\qquad\qquad
+2
\underbrace{
\int_0^1
(8\arctan^2 t-4\pi \arctan t +\pi^2-4) \;
\left(
\frac{2t}{1+t^2} + \log\frac{1-t}{1+t}
\right)'
\;\log t
\; dt}_{=J_{111}}
\ .
\end{aligned}
$$
Some words on the value of the known integral above, the one with the value
$8\pi\log 2-4\pi$. We use partial integration. Then the piece in $(8\arctan t-2\pi)'=8/(1+t^2)$ is rational, we use partial fraction decomposition over $\Bbb C$, and thus the integral is broken into pieces, each piece
being of the form $\frac 1{t-b}\log(t-a)$ or $\frac 1{(t-b)^2}\log(t-a)$ with $a,b$ among $0,\pm i$. These integrals can be handled.
Dilogarithms appear, we use than the dilogarithm identities.

A numerical check:

? 8*Pi*log(2) - 4*Pi
%41 = 4.8543181080696440901549376527829000419
? intnum( t=0, 1, (8*atan(t) - 2*Pi) * 4*t/(1+t^2)^2 * 2*log(t) )
%42 = 4.8543181080696440901549376527829000419
? intnum( x=0, Pi/2, (4*x - 2*Pi) * sin(x) * log( (1-cos(x)) / (1+cos(x)) ) )
%43 = 4.8543181080696440901549376527829000419

Now we look closer to the remained integral, $J_{111}$. Parts of it are rather easy. We have
$$
\begin{aligned}
J_{111s}
&:=
\int_0^1
\left(
\frac{2t}{1+t^2} + \log\frac{1-t}{1+t}
\right)'
\;\log t\; dt
\\
&=
\underbrace{\int_0^1
\left(
\frac{2t}{1+t^2}\right)'
\;\log t\; dt}_{=-2\arctan 1=-\pi/2}
-
\underbrace{
\int_0^1
\left(
\frac 1{1-t}+\frac 1{1+t}
\right)
\;\log t\; dt}_{=-\pi^2/4}
\ .
\end{aligned}
$$
The integral with value $-2\arctan 1$ is done by partial integration, the other one using the (real part of the) primitive $\operatorname{Li}_2(1\pm t)$.
The dilog vanishes in $0$, the two contributions in $t=0$ of $\operatorname{Li}_2(1\pm t)=\operatorname{Li}_2(1\pm t)$ cancel each other (different signs),
and there remains $-\Re\operatorname{Li}_2(1+1)=-\frac 14\pi^2$.

The part in $\arctan t\cdot \log t$ from $J_{111}$ can also be computed. The parts derived from
$\left(
\frac 1{1-t}+\frac 1{1+t}
\right)
$ are considered in $J_{1211a}$, $J_{1211b}$. The remained part in $\arctan t\cdot \log t$ is
$$
\begin{aligned}
J_{111t}
&:=
\int_0^1
\arctan t\;
\left(
\frac{2t}{1+t^2}
\right)'
\;\log t
\; dt
\\
&=
-
\int_0^1
\frac{2t}{1+t^2}
\left(
\frac 1{1+t^2}\cdot\log t
+
\arctan t\cdot\frac 1t
\right)
\; dt
\\
&=
\int_0^1
\left(
\frac 1{1+t^2}-1
\right)'
\cdot\log t
\; dt
-
\int_0^1
\frac{2}{1+t^2}
\arctan t
\; dt
\\
&=\frac 12\log 2-\frac 1{16}\pi^2\ .
\end{aligned}
$$
So the main issue is
$$
\tag{16}
$$
$$
\begin{aligned}
J_{111u}
&:= \int_0^1
\arctan^2 t\;
\left(
\frac{2t}{1+t^2}
+
\log\frac{1-t}{1+t}
\right)'
\;\log t
\; dt
\\
&=
\underbrace{
\int_0^1
\arctan^2 t\;
\left(
\frac{2t}{1+t^2}
\right)'
\;\log t
\; dt}_{J_{111u1}}
-
\underbrace{
\int_0^1
\arctan^2 t\;
\log t\;
\left(\frac 1{1-t} +\frac 1{1+t}\right)
\; dt}_{J_{111u2}}
\ .
\end{aligned}
$$
The term listed first is simpler.
$$
\begin{aligned}
J_{111u1}
&=
-
\int_0^1
\frac{2t}{1+t^2}
\left(
\arctan^2 t\cdot \frac 1t
+
2\arctan t\cdot\log t\cdot\frac 1{1+t^2}
\right)
\; dt
\\
&=
-
\left[\frac 23\arctan^3 t\right]_0^1
+
2\int_0^1
\left(\frac 1{1+t^2}\right)'\cdot\arctan t\cdot \log t\; dt
\\
&=
-\frac 23\cdot\frac 1{4^3}\pi^3
-
2\int_0^1
\frac 1{1+t^2}\;
\left(\frac 1{1+t^2}\cdot \log t + \arctan t\cdot\frac 1t\right)\; dt
\\
&=
-\frac 1{96}\pi^3
-2\underbrace{\int_0^1\frac{\log t}{(1+t^2)^2}\; dt}_{-(\pi+4G)/8}
-2\underbrace{\int_0^1 \frac 1t\;\arctan t\; dt}_{G}
+\underbrace{\int_0^1 \frac {2t}{1+t^2}\;\arctan t\; dt}_{-\frac 14\pi\log 2+G}
\\
&=
-\frac 1{96}\pi^3 +\frac 14\pi -\frac 14\pi\log 2
\ .
\end{aligned}
$$
For the integral of $\frac 1t\arctan t$ use maybe the Taylor expansion.
For the integral with the $\log t$ use the partial fraction decomposition over $\Bbb C$ for
$\frac 1{(t^2-a^2)^2}=\frac 1{4a^3}\left(\frac 1{t+a}-\frac 1{t-a}\right)+\frac 1{4a^2}\left(\frac 1{(t-a)^2}+\frac 1{(t+a)^2}\right)$,
where $a=i$.  For the remained integral, using partial integration we compute instead
$\int_0^1\frac{\log(1+t^2)}{1+t^2}\; dt$. We split again in $\log$, and in partial fractions, using $(t^2+1)=(t-i)(t+i)$.
Then $\int_0^1\frac{\log(t+i)}{t+i}\; dt=\int_0^1\log(t+i)\cdot(\log(t+i))'\; dt$, so we can integrate. And
$\int_0^1\frac{\log(t+i)}{t-i}\; dt$ leads to a dilog value, $\operatorname{Li}_2((1-i)/2)$, and we need finally from it
$\Im\operatorname{Li}_2((1-i)/2)=-G+\frac 18\pi\log 2$.

So up to mentioned white noise integrals, that can be handled, we are in position to start the solution.
We show, using $K$ for a shorter name:
$$
\tag{17}
$$
$$
\begin{aligned}
K:=J_{111u2}
&:=
\int_0^1
\arctan^2 t\cdot\log t\;
\left(
\frac 1{1-t}
+
\frac 1{1+t}
\right)
\; dt
\\
&=
\frac 3{128}\pi^4
+
\frac 1{32}\pi^2\log^2 2
-
\frac 12\pi G\log 2
-
\pi\Im\operatorname{Li}_3\left(\frac{1+i}2\right)\ .
\end{aligned}
$$
(This relation resisted to all standard attacks. I tried several ideas and tricks like
partial integration, and the substitutions $s=1/t$ and $t=(1-u)/(1+u)$, and the deformation
of the $\arctan t$ using the parameter $a$, thus replacing
$\arctan t=\int_0^1\frac {t\; da}{1+a^2t^2}$ and/or replacing
$\arctan^2 t=\iint_{[0,1]^2}\frac {t\; da}{1+a^2t^2}\cdot \frac {t\; db}{1+b^2t^2}$, and so on.
The complexity
of the resulted expressions was not reduced. Some of these lines are mentioned after the solution below,
which is something experimentally found and new, that i never saw before.
Well, when ideas and tricks do not work, only madness can help us...)

We will work using complex analysis and the first step is to write:
$$
\int_0^1=\int_0^i+\int_i^1\ .
$$
Since $K\in \Bbb R$, we compute only the real part of the integrals in the R.H.S. above. Using
$$
\arctan t=\frac 1{2i}\log\frac{1+it}{1-it}
$$
around zero we can compute by the parametrization $t=iu$, $u\in[0,1)$:
$$
\begin{aligned}
K_1&:=
\Re \int_0^i
\arctan^2 t\cdot\log t
\cdot\frac 2{1-t^2}
\\
&=
\Re\int_0^1
-\frac 14\cdot\log^2\frac{1+i\cdot iu}{1-i\cdot iu}\cdot\log(iu)\cdot\frac2{1+u^2}\; i\; du
\\
&=
\Re
\int_0^1
-\frac 14\cdot\log^2\frac{1-u}{1+u}\cdot\left(i\frac \pi 2+\log u\right)\cdot\frac2{1+u^2}\; i\; du
\\
&=
\frac \pi 4
\int_0^1
\log^2\frac{1-u}{1+u}\cdot\frac 1{1+u^2}\; du
\\
&=
\frac \pi 4
\int_0^1
\log^2 s\cdot\frac 1{1+\left(\frac{1-s}{1+s}\right)^2}\; \frac 2{(1+s)^2}\;ds
=
\frac \pi 4
\int_0^1
\log^2 s\cdot\frac 2{(1+s)^2 + (1-s)^2}\; ds
\\
&=
\frac \pi 4
\int_0^1
\log^2 s\cdot\frac 1{2i}\left(\frac 1{s-i}-\frac 1{s+i}\right)\; ds
\\
&=
\frac \pi 4
\cdot\frac 1{2i}
\left(-2\operatorname{Li}_3\left(\frac 1i\right)
+2\operatorname{Li}_3\left(\frac 1{-i}\right)
\right)
=\frac \pi 4\cdot 2\cdot\frac{\pi^3}{32}
\\
&=\frac{\pi^4}{64}\ .
\end{aligned}
$$
(Similarly to the $\log(1-t)$ factor from the original integral, that may show some integration problem in $u=1$,
but there isn't any, there also no issue for $\log(1+iu)$ in $i$.)
Numerical check, pari/gp can do it also under such unusual circumstances:

? real(intnum( t=0, i, atan(t)^2 * log(t) *2/(1-t^2) ))
%290 = 1.5220170474062880818193801982610173632769935261357097139291853029682946165
? pi^4 / 64
%291 = 1.5220170474062880818193801982610173632769935261357097139291853029682946165

The other integral turns out to be also feasible by using a path from $i$ to $1$ on the unit circle,
explicitly we use the parametrization "$t=t(s)$" with
$$ t(s) = \sin 2x + i\cos 2s=i\cdot e^{-2is}\ ,\qquad s\in[0,\ \pi/4]\ .$$
Then
$$
\begin{aligned}
\frac
{1+i\; t(s)}
{1-i\; t(s)}
&=
\frac
{1-e^{-2is}}
{1+e^{-2is}}
=
\frac
{(1-e^{-2is})(1+e^{+2is})}
{(1+e^{-2is})(1+e^{+2is})}
=
\frac
{2i\; \sin 2s}
{2+2\cos 2s}
\\
&=i\cdot\frac{2\sin s\cos s}{2\cos ^2 s}
=i\;\tan s\ .
\end{aligned}
$$
The other factors in the integrand can be reshaped in a similar way. So...
$$
\begin{aligned}
K_2&:=
\Re
\int_i^1
\arctan^2 t\cdot\log t
\cdot\frac 2{1-t^2}
\\
&=
\Re
\int_0^{\pi/4}
-\frac 14\cdot\underbrace{\log^2(i\tan s)}_{=\left(i\frac \pi2+\log\tan s\right)^2}\cdot i\;\left(\frac \pi 2-2s\right)\cdot\frac2{\cos 2s}\; ds
\\
&=
\int_0^{\pi/4}
\frac 14\cdot2\cdot\frac \pi 2\cdot\log\tan s
\cdot\left(\frac \pi 2-2s\right)\cdot\frac2{\cos 2s}\; ds\qquad(t=\tan s)
\\
&=
\frac \pi 2
\int_0^1
\log t
\cdot\left(\frac \pi 2-2\arctan t\right)\cdot\frac{1+t^2}{1-t^2}\; \frac 1{1+t^2}\;dt
\\
&=
\frac{\pi^2}4
\cdot\frac 12
\int_0^1
\log t\left(\frac1{1-t}+\frac 1{1+t}\right)\; \;dt
-
\frac \pi
\int_0^1
\log t\cdot \arctan t\cdot\frac 1{1-t^2}\;dt
\ .
\end{aligned}
$$
For the first integral above, note that
$\partial_t\operatorname{Li}_2(1-t)
=\frac 1{1-t}\log t$, and
$\partial_t\operatorname{Li}_2(1+t)
=-\frac 1{1+t}\log (-t)$.
So its value is
$$
\begin{aligned}
\Big[
\operatorname{Li}_2(1-t)
\Big]_0^1
-
\Re
\Big[
\operatorname{Li}_2(1+t)
\Big]_0^1
&=
(\operatorname{Li}_2(0)-\operatorname{Li}_2(1))
-
\Re
(\operatorname{Li}_2(2)-\operatorname{Li}_2(1))
\\
&=
-\Re\operatorname{Li}_2(2)
=-\frac {\pi^2}4\ .
\end{aligned}
$$
The second integral above was already computed as part of $(15)$, it is related to
$J_{1211a}$ and $J_{1211b}$. This gives:
$$
\begin{aligned}
K &= K_1+K_2 = \frac{\pi^4}{64} + K_2
\\
&=\frac{\pi^4}{64} +
\frac{\pi^2}8\cdot\left(-\frac {\pi^2}4\right)
-\frac\pi 2(J_{1211a}+ J_{1211b}) =\dots
\end{aligned}
$$
and so on.
We are in position to put all together. First of all
$$
\tag{18}
$$
$$
\begin{aligned}
J_{111}
&=
8(J_{111u1}-\underbrace{J_{111u2}}_{K})
-4\pi(J_{111t}-J_{1211a}-J_{1211b}) +(\pi^2-4)J_{111s}
\\
&=
-\frac{\pi^4}4 - \frac{\pi^2}4\log^2 2 +8 \pi G\log 2
+16\pi \operatorname{Li}_3\left(\frac {1+i}2\right)
+4\pi - \pi^2 -\frac 13\pi^3 -4\pi\log 2
\ .
\end{aligned}
$$
From $(18)$ and $(15)$ we get:
$$
\begin{aligned}
J
&= 2\underbrace{J_1}_{\frac 12J_{11}-\frac 14 J_{12}}
+ 2\pi^2\cdot \underbrace{J_2}_{\frac{\pi^2}2-\pi}
\\
&={\color{blue}{J_{11}}} -\frac 12{\color{brown}{J_{12}}} +\pi^4-2\pi^3
\\
&=
{\color{blue}
{\left(\frac {\pi^3}2 -8\pi\log 2 +4\pi + 2J_{111}\right)}}
-\frac 12
{\color{brown}
{\left(-\frac {\pi^3}3 -8\pi\log 2 +8\pi -4\pi J_{121}\right)}}
+\pi^4-2\pi^3
\\
&=
\pi^4-\frac 43\pi^3-4\pi\log2 + 2J_{111}
+
2\pi\cdot J_{121}
\\
&=
\pi^4-\frac 43\pi^3-4\pi\log2
\\
&\qquad
+\frac 18\left[
-4\pi^4 - 8\pi^2\log^2 2 +128 \pi G\log 2 +256\pi \operatorname{Li}_3\left(\frac {1+i}2\right)
 -16 \pi^2
\right]
-\frac 23\pi^3
-8\pi\log 2 + 8\pi
\\
&\qquad
+
\frac 18\left[\ 5\pi^4+4\pi^2\log^2 2  -64\pi G\log 2 - 128\pi\Im\operatorname{Li}_3\left(\frac {1+i}2\right)\ \right]
+4\pi\log 2-8\pi
\\
&=
\frac 98\pi^4-2\pi^3-2\pi^2
-\frac 12\pi^2\log^2 2
+8\pi G\log 2
+ 16\pi\Im\operatorname{Li}_3\left(\frac {1+i}2\right)
-8\pi\log 2
\ .
\end{aligned}
$$

$\square$


问:
$\sum_{i_{K+1},...,i_N=1}^N ε^{i_1,...,i_K,i_{K+1},...,i_N}δ^{j_{K+1},...j_N}_{i_{K+1},...,i_N}=(N-K)!ε^{i_1,...,i_K,j_{K+1},...,j_N}$
What shown below is a reference from Introduction to vectors and tensors by Ray M. Bowen and C.C Wang. Precisely it is de definition of the generalised Kroneker delta thus if you know it you can even not read what is written in the image and so you can directly read what I ask to prove.

So using this definition I ask to prove that

$$
\sum_{i_{K+1},...,i_N=1}^N ε^{i_1,...,i_K,i_{K+1},...,i_N}δ^{j_{K+1},...j_N}_{i_{K+1},...,i_N}=(N-K)!ε^{i_1,...,i_K,j_{K+1},...,j_N}
$$

So could someone help me, please?

答:
So we let's start to analyse the summation to the left hand of the equality $(1)$. So if
$$
ε^{i_1,...,i_K,i_{K+1},...,i_N}δ^{j_{K+1},...j_N}_{i_{K+1},...,i_N}\neq 0
$$
then
$$ε^{i_1,...,i_K,i_{K+1},...,i_N}\neq 0\,\,\,\text{and}\,\,\,δ^{j_{K+1},...j_N}_{i_{K+1},...,i_N}\neq 0
$$
and so
$$
\{i_{K+1},...,i_N\}\subseteq\{1,...,N\}\setminus\{i_1,...,i_K\}
$$
and thus $I:=\{i_1,...,i_K\}$ and $J:=\{j_{K+1},...,j_N\}$ determine a partition of $\{1,...,N\}$ because $δ^{j_{K+1},...j_N}_{i_{K+1},...,i_N}\neq 0$ and this implies $\{j_{K+1},...j_N\}=\{i_{K+1},...,i_N\}$. So if the sets $I$ and $J$ do not determine a partition of $\{1,...,N\}$ then clearly any term of the summation to the left hand is equal to zero and the same is for the right hand so that the equality trivially holds. So now we suppose that the sets $I$ and $J$ determine a partition of $\{1,...,N\}$ and we analyse which terms to the summation are not zero. So by the previous observation we know that the only indices $i_{K+1},...,i_N$ such that the corresponding term of the summation to the left hand of the equation $(1)$ is not zero are only that whose indices determine a permutation of $J$ so that we have $(N-K)!$ not zero terms. So if we now observe that
$$
ε^{i_1,...,i_K,i_{K+1},...,i_N}δ^{j_{K+1},...j_N}_{i_{K+1},...,i_N}=
\\ε^{i_1,...,i_K,i_{K+1},...,i_N}δ^{i_1,...,i_K,j_{K+1},...j_N}_{i_1,...,i_K,i_{K+1},...,i_N}=
\\
\\ε^{i_1,...,i_K,i_{K+1},...,i_N}ε^{i_1,...,i_K,j_{K+1},...j_N}ε_{i_1,...,i_K,i_{K+1},...,i_N}=
\\ε^{i_1,...,i_K,i_{K+1},...,i_N}ε_{i_1,...,i_K,i_{K+1},...,i_N}ε^{i_1,...,i_K,j_{K+1},...j_N}=
\\δ^{i_1,...,i_K,i_{K+1},...,i_N}_{i_1,...,i_K,i_{K+1},...,i_N}ε^{i_1,...,i_K,j_{K+1},...j_N}=
\\ε^{i_1,...,i_K,j_{K+1},...j_N}
$$
and thus the equation follows immediately.

ContentLicense: CC BY-SA 4.0


问:
Basis and dimension with finite vectors
I am trying to solve the following problems:

Considering the linear independet vectors $\vec T_1, ..., \vec T_s \in \mathbb{R}^s$ and the numbers $n_1, ... n_s \in \mathbb{R} - \{0\}$,

How can I check the vectors

$n_s \vec T_s, n_s \vec T_s + n_{s-1} \vec T_{s-1}, ...,  n_s \vec
   T_s + n_{s-1} \vec T_{s-1} + ... + n_1 \vec T_1$

compose a $\mathbb{R}^s$ basis?

Considering $1 \le p \le s$ and the subspace

$\begin{equation} W = \langle n_s    \vec T_s, n_s \vec T_s + n_{s-1}
   \vec T_{s-1}, ...,  n_s \vec       T_s + n_{s-1} \vec T_{s-1}    +
   ... +    n_p \vec T_p \rangle    \end{equation}$,

how can I calculate $W$ dimension?

答:
Let $\{v_1,v_2,\dots,v_n\}$ be a basis of the vector space $V$.

First fact. The set
$$
\{v_1,v_1+v_2,\dots,v_1+\dots+v_{n-1},v_1+\dots+v_n\}
$$
is a basis of $V$.

Proof. Setting $w_k=v_1+\dots+v_k$, we have
$$
v_1=w_1,\quad v_2=w_2-w_1,\quad v_3=w_3-w_2,\quad \dots,\quad v_n=w_n-w_{n-1}
$$
Hence the set $\{w_1,w_2,\dots,w_n\}$ spans $V$. QED

Second fact. If $c_1,c_2,\dots,c_n$ are nonzero scalars, the set
$$
\{c_1v_1,c_2v_2,\dots,c_nv_n\}
$$
is a basis of $V$.

Proof. We have $v_k=c_k^{-1}(c_kv_k)$, so the set spans $V$. QED

Third fact. For $1\le k\le n$, the set $\{v_1,v_1+v_2,\dots,v_1+\dots+v_k\}$ is linearly independent.

Proof. It is a subset of a basis. QED

Now translate this in your notation and you're done.

Title: Basis and dimension with finite vectors


问:
How to find $R$ as a set of ordered pairs?
Suppose I am given a set $A = \{0,1,2,3,4,5,6,7,8,9\}$ and a relation $R$ on $A$ which is given by $x R y \iff y=2x+3.$

In order to write $R,$ how would I go answering this question? Would I just input all the values of the set $A$ as $x$ giving me $\{(0,3),(1,5),(2,7),(3,9),(4,11),(5,13),(6,15),(7,17),(8,19),(9,21)\}?$ Or perhaps do I only input numbers in $y=2x+3$ that would output numbers only between $0-9?$

Thanks in advance!

答:
Let $A$ be the following set

$$A = \{0,1,2,3,4,5,6,7,8,9\}$$

and $R$ the relation on $A$ given by the following condition

$$xRy \iff y = 2x+3.$$

What you are trying to determine is the relation $R$ itself (by defining it in extension). I.e., the relation $R$ will be

$$R = \{(x,y) \in A \times A \mid y = 2x + 3\}.$$

Now, doing the arithmetic, you observe that

$$\begin{align}
3 = 2 \cdot 0 +3 && \implies (0,3) \in R\\
5 = 2 \cdot 1 +3 && \implies (1,5) \in R\\
7 = 2 \cdot 2 +3 && \implies (2,7) \in R\\
9 = 2 \cdot 3 +3 && \implies (3,9) \in R
\end{align}$$

and also that

$$\begin{align}\forall x \in A, x \geq 4 & \implies y \geq 11\\
& \implies y \notin A\\
& \implies (x,y) \notin R.\end{align}$$

This means that there is no more ordered pairs of $A \times A$ that belong to R.

Therefore

$$R = \{(0,3),(1,5),(2,7),(3,9)\}.$$


问:
Problem on temperate distribution theory

Problem: Give an example of an infinitely differentiable function f on $\mathbf{R}$ such that (1): there does not exist a polynomial P on $\mathbf{R}$ such that $|f(x)| \le |P(x)|, x \in \mathbf{R}$ and (2): the mapping $\mathbf{S}(\mathbf{R}) \ni \varphi \rightarrow \int f(x) \varphi(x)dx$ defines a temperate distribution.

Denote by $\mathbf{S}$ the Schwartz space of infinitely differentiable functions $\phi(x)$ such that $|\frac{\partial^k \varphi(x)}{\partial x^k}| \le C_{m,k}(1+|x|)^{-m}$ for any $k$ and any positive integer $m$. Introduce the norms $|[\varphi]|_{m,S}=\textbf{sup}_{x \in \mathcal{R}^n} (1+|x|)^m \sum_{|k|=0}^m |\frac{\partial^k \varphi(x)}{\partial x^k}|$. We say that $\varphi_n \rightarrow \varphi in S$ if $|[\varphi_n-\varphi]|_{m,S} \rightarrow 0$ as $n \rightarrow \infty$ for all $m=1,2,...,$

Definition: Linear continuous functional on S is called a tempered distribution. The linear space of tempered distributions is denoted by $S'$.

Theorem 9.1: For any tmepered distrubition f, there exist an integer m and a constant C such that $|f(\varphi)| \le C |[\varphi]|_{m,S} \forall \varphi \in S$.

Theorem 9.2: Any $f \in S'$ can be represented in the following form: $f=\sum_{|k|=0}^{m_1} \frac{\partial^k \varphi(x)}{\partial x^k} f_k$, where $f_k$ are regular functions in $S'$ corresponding to continuous functions $f_k(x)$ satisfying the estimates $|f_k(x)| \le C_k(1+|x|)^{m_2}$. Here $m_1,m_2$ are integers.

Question: I have no clue how to approach this problem. Any help is appreciated.

答: Let $F(x)=\sin e^x$. This is clearly both continuous and bounded, so it defines a tempered distribution. And if $F$ defines a tempered distribution, then so does its derivative! Thus, $f(x)=F'(x)=e^x \cos e^x$ defines a tempered distribution. But $|f(x)|$ grows exponentially and is therefore not bounded by any polynomial.


问:
Conditional expectation on Cross product von Neumann algebra
Let $M$ be a von Neumann algebra on a Hilbert space $\mathcal{H}$. Let $(M,G,\alpha)$ be a $W^*$-dynamical system. Let $G$ be discrete. Then the cross product von Neumann algebra $M\rtimes_\alpha G$ is defined in the Hilbert space $l^2(G,\mathcal{H})$. Now since $G$ is discrete we can write $l^2(G,\mathcal{H})=\oplus_{g \in G} \mathcal{H}\otimes \epsilon_g $ where $\epsilon_g$ is defined as $\epsilon_g(h)=\delta_{h,g}$. Using this identification we can write any element of $\mathcal{B}(l^2(G,\mathcal{H}))$ as matrix of operators on $H$.

Now let $P_g$ be the orthogonal projection onto the subspace $H\otimes \epsilon_g$ of  $l^2(G,\mathcal{H})$. Then I saw a result (link given below), that the map $\phi: M\rtimes G \rightarrow M$ defined by $\phi(T)=\sum_gP_gTP_g$ where the sum is in the SOT topology is a conditional expectation.  Interms of matrix representation of $M\rtimes G \subset \mathcal{B}(l^2(G,\mathcal{H}))$, I guess that the map $\phi$ is mapping a matrix to its diagonal? Is it true? How to prove it? Why does the sum infact belong to $M$?

https://www.jstor.org/stable/pdf/2373237.pdf

https://projecteuclid.org/download/pdf_1/euclid.tmj/1178241528

答:
You never mention how you define $M\rtimes_\alpha G$. In the direct sum picture, one defines, using the matrix units $\{E_{gh}:\ g,h\in G\}$,
$$
M\rtimes_\alpha G=\Bigg[\Big\{\sum_{g\in G}\alpha_g(x)\,E_{gg}:\ x\in M\Big\}\cup\Big\{\sum_{h\in G}E_{h,g^{-1}h}:\ g\in G\Big\}\Bigg]'',
$$
where $\hat x=\sum_{g\in G}\alpha_g(x)\,E_{gg}$ is how we see $M$ inside $M\rtimes_\alpha G$. The elements $u_g=\sum_{h\in G}E_{h,g^{-1}h}$ are unitaries, and the algebra $\hat M$ of elements $\sum_{g\in G}\alpha_g(x)\,E_{gg}$ is invariant under conjugation by each $u_g$.

The expectation is indeed "compression to the diagonal" in a block sense. And the "diagonal"  is $\hat M$. In this notation, the projections are $E_{gg}$. Then
$$
\sum_{g\in G} E_{gg}\Big(\sum_{s\in G}\alpha_s(x)\,E_{ss}+\sum_{r\in G\setminus\{e\}} c_{r}\sum_{t\in G}E_{r^{-1}t,t}\Big)E_{gg}=\sum_{s\in G}\alpha_s(x)\,E_{ss}\in\hat M.
$$


问:
Derivation of a classical fact in the theory of Markov chains

My question comes from the book "Mathematical Aspects of Mixing Times in Markov Chains". Specifically, the authors stated Lemma 1.3 in the figure above but does not include a proof for this lemma. I am wondering how can this lemma be justified in a rigorous manner. I feel like it is just a Fokker-Planck equation, but I am really stuck at the rigorous proof of the fact that $h_t = H^*_t h_0$. Thanks for any help!

答: I end up writing my own answer. The proof is very similar to the corresponding case in the discrete time. Fix $x \in \Omega$, then for any $t \geq 0$ we have that $$ (H^*_th_0)(x) = \sum_y H^*_t(x,y)\cdot \frac{H_0(\cdot,y)}{\pi(y)} = \frac{1}{\pi(x)}\sum_y H_t(y,x)\cdot H_0(\cdot,y) = \frac{H_t(\cdot,x)}{\pi(x)} = h_t(x), $$ which leads to the desired claim that $h_t = H^*_th_0$.


问:
How to prove recursive sequence converges? - $a_{n+1}=1/(a_n+1)$
I have a recursive sequence defined by $$a_{n+1}=\frac{1}{a_n+1}$$, where $a_0=7$.

The sequence is bounded, but it's not monotonic. So how does one show, that the sequence converges?
I already found it's limit under the assumption, that it is converging. Which is $a=\frac{\sqrt5-1}{2}$.

Note, that I'm not allowed to use epsilon proofs or proof by induction.

答:
Write $a_n=L+u_n$, with $L=(\sqrt5-1)/2$, noting that $L^2+L-1=0$.  From the recursion we get

$$u_{n+1}={1\over1+L+u_n}-L={1-L-L^2-Lu_n\over1+L+u_n}={-Lu_n\over1+L+u_n}$$

Since $L\approx0.618\lt1$ and $L+u=a_n\gt0$ for all $n$ (by induction: $a_0=7\gt0$, and $a_n\gt0$ implies $a_{n+1}=1/(a_n+1)\gt0$), it follows that

$$|u_{n+1}|={L|u_n|\over|1+L+u_n|}\lt{|u_n|\over|1+L+u_n|}\lt{|u_n|\over|1+0|}=|u_n|$$

Thus $|u_n|$ is a decreasing sequence, hence has a limit (being bounded below by $0$). If that limit were anything other than $0$, we would have

$$\lim_{n\to\infty}|1+L+u_n|=L\lim_{n\to\infty}{|u_n|\over|u_{n+1}|}=L$$

But this is a contradiction, since $|1+L+u_n|=|1+a_n|\gt1\gt L$ for all $n$. Thus $u_n\to0$, so $a_n\to L$ as $n\to\infty$.

Remark: It's easy, but unnecessary, to show that $u_n$ is alternately positive and negative.


问:
How to compute the homology group $H_q(\mathbb{R}^n-e^r)$?
Let $e^r$ be a homeomorphic copy of $I^r$ in $\mathbb{R}^n$($I=[0,1]$).How to compute the homology group $H_q(\mathbb{R}^n-e^r)$?($r,n,q$ are non-negative integers)</p>

答:
I applied the Mayer-Vietoris sequence in $\mathbb{S}^n -Y$ and the following result:

$``$Let $Y$ a subset of the $\mathbb{S}^n$ homeomorphic of $I^r$. Then the reduced homology groups is
$$
\tilde{H}_j(\mathbb{S}^n -Y)=0 \quad \forall j."
$$
The subsets were used in sequence:
$$
U = \mathbb{S}^n -(Y \cup \{pto\}), \quad V \cong \mathbb{D}^n \quad \text{and} \quad U \cap V \cong \mathbb{D}^n - \{pto\}.
$$
Comments: The $\mathbb{R}^n$ is homeomorphic of $\mathbb{S}^n-\{pto\}$, $\{pto\} \subset int (V)$ and $V \cap Y = \varnothing$.

ContentLicense: CC BY-SA 3.0


问:
Given an equation, I'm having trouble finding a solution to another equation. Help proving whether or not it can be solved? (Differential Equations)
This is a problem I've come up with, not a homework problem. Ideally I'd like to find a solution, but if I can prove that it can't be solved, that's okay too. The problem is this:

Given $n'' = hn$, find $\frac{n'''n'}{(n'')^2}$ in terms of $h$ and $z$ if necessary. This is using the standard notation of each $'$ being a derivative. h and n are both functions of the variable z, that is h -> h(z) and n -> n(z). All derivatives are with respect to z.

I've done so many different things to solve this and just can't figure it out. If you have any ideas on how to approach this, I'd be really grateful.

Edit: For context, I'm very familiar with the rules of calculus. I haven't posted any attempts because I've made so many and they all just become crazy and insane. I've been working on this problem for ~ 2.5 years, and only just thought now I'd ask for help because I'm beginning to think that it isn't possible.

答:
You have a single initial relation

$$n''=hn$$ and you want to establish a relation that involves $n',n'',n'''$. Hence you must differentiate once,

$$n'''=h'n+hn'.$$

There is no point differentiating again, as this would introduce $n''''$, which we don't want.

We must also eliminate $n$, which we can do using the first relation and we end up with the sole relation

$$hn'''=h'n''+h^2n'.$$

This equation does not allow you to express $\dfrac{n'''n'}{n''^2}$ in terms of $h$ and derivatives, so I don't think that the question has an answer.


问:
Conflict in expression for Lie derivative of a metric
Let $S \subset M$ be an embedded submanifold. Here, $(M, g)$ is a (pseudo)Riemannian manifold. One can choose a vector field defined on $S$ (not necessarily normal or tangent to $S$), say, $X$. We can decompose this into its normal/tangental components $X = \lambda \nu + X^T$ where $X^T$ is tangent to $S$ and $\nu$ is the normal field to $S$. Here $\lambda$ is a scalar function. Computation shows that
$$
L_X g = 2\lambda\theta + \pi,
$$
where $\pi_{ab} = \nabla_a X^T_b + \nabla_b X^T_a$ is the deformation tensor of $X^T$ and $\theta$ is the second fundamental form of $S$ in $M$, defined by $\theta(X, Y) = g(\nabla_X \nu, Y)$. Edit: Here, $S$ is codimension 1 and $\nabla$ is the connection on $M$.

My question: The Lie derivative is known to depend not just on the vector $X$ at a point, but the 1-jets; which, in this case, may leave $S$ (they do, as long as $\lambda \neq 0$). However, the right hand side of this equation, to the best of my knowledge, depends only on the values of $X$ on the submanifold $X$.

A possible explanation: At least, this is definitely true for $\theta$ and $\lambda$, as $\theta$ doesn't depend on $X$ at all, and $\lambda$ was defined in a way that only depends on $X|_S$. So perhaps $\pi$ depends on $X$'s values outside of $S$. However, if we restrict the above to tensor fields on $S$ (so they take $V, W$ in $T_p S$ for some $p \in S$), I believe this is false:
$$
\pi(V, W) = g(\nabla_V X^T, W) + g(\nabla_W X^T, V)
$$
and since $V, W$ are tangent to $S$, this only depends on the values of $X$ (in particular, the values of $X^T$) on $S$. So, some follow-up questions are:

Does $L_X g(V, W)$ not depend on the 1-jets of $X$, but only the value at a point, if $V, W$ are in $S$? Or perhaps it still depends on 1-jets of $X$, but only the tangential components? However, this goes against what I know about Lie derivatives.

Does $\pi(V, W)$ in general depend on the values of $X$ outside $S$, if the vectors $V, W$ are not assumed to be tangent to $S$?

答:
Let $X = X^{T} + \lambda \nu$ be the above decomposition. Let $U$ and $V$ be vector fields. Then,
\begin{align}
\left(L_Xg\right)(U,V) &= X\left(g(U,V) \right) - g\left([X,U],V \right) - g\left(U,[X,V]\right) \\
&= g \left(\nabla_XU,V \right) + g\left(U,\nabla_XV \right) -g\left([X,U],V \right) - g\left(U,[X,V]\right) \\
&= g\left(\nabla_XU - [X,U],V \right) + g\left(U,\nabla_XV-[X,V] \right)\\
&= g\left(\nabla_U X,V \right) + g\left(U,\nabla_V X \right) \\
&= g\left(\nabla_U\left(X^T\right),V \right) + g\left(U, \nabla_V\left(X^T \right)\right) \\
&~~~+ g\left(\nabla_U\left(\lambda\nu\right),V \right) + g \left(U,\nabla_V(\lambda \nu) \right) \\
&= g\left(\nabla_U\left(X^T\right),V \right) + g\left(U, \nabla_V\left(X^T\right) \right) \\
&~~~ + g\left( \mathrm{d}\lambda(U)\nu,V \right) + g\left(\lambda\nabla_U\nu,V \right)\\
&~~~ + g\left(U, \mathrm{d}\lambda(V)\nu\right) + g\left(U,\lambda \nabla_V\nu\right)\\
&= g\left(\nabla_U\left(X^T\right),V \right) + g\left(U, \nabla_V\left(X^T \right)\right) \\
&~~~ + g\left(\mathrm{d}\lambda(U)\nu,V \right) + g\left(U, \mathrm{d}\lambda(V)\nu\right) \\
&~~~+ 2\lambda \theta\left(U,V \right)
\end{align}
So it seems your formula is valid only if $U$ and $V$ are tangent to $S$, i.e orthogonal to $\nu$. Otherwise, you would have to consider the term $g\left((U\cdot \lambda)\nu,V \right) + g\left(U, (V\cdot \lambda)\nu\right)$, and $\mathrm{d}\lambda$ is a component of the $1$-jet of $X$. Moreover, $\nabla_UX^T$ and $\nabla_V X^T$ depends too on the $1$-jet of $X$, not just on the value of $X$ at a point.

More generally, $\left(\nabla_AB\right)_p$ depends on the value $A_p$ and on the value and the $1$-jet of $B$ at $p$.

So to answer directly your questions:

$(L_Xg)(U,V)$ depends on the $1$-jet of $X$ by $\mathrm{d}\lambda$ and $\nabla \left(X^T\right)$.

if $U$ and $V$ are tangent to $S$ (thus, orthogonal to $\nu$), $\lambda$ affects the result only at the $0$-th order. Otherwise, its infinitesimal growth affects the result, thus the growth of $X$ along the integral curves of $\nu$ affects the results too.

Edit If $U$ and $V$ are tangent to $S$, the calculation shows that $(L_Xg)(U,V)$ depends on $X$ this way: it depends on the $1$-jet of $X^T$ and on the $0$-jet of $\lambda$. This can be interpreted like this: $U$ and $V$ are intrinsic object of $S$. If $\lambda = 0$, $X$ is also an intrinsic object of $S$, so is $(L_Xg)(U,V)$. So the result should not depend on how you expand $X$ outside of $S$. As you can expand $X$ by many ways and at many growth outside of $S$ (with any sort of function $\lambda)$, this should only depend on the fact that $\lambda = 0$ on $S$ and not on its derivative.


问:
Show that if $I$ and $I'$ are initial objects there exists a unique isomorphism between them.
I am new to set theory and I have come across this exercise. I think the proof I have come up with is too simple to be right. would greatly appreciate some tips.

If $I$ is an initial set, then for $\forall B$ in a category C, $\exists !$ an arrow $I \to B$.
Since $I'$ is a set in $C$, $\exists !$ arrow to A such that $I \to I'$. The same theory can be applied to prove that there is an arrow $B$ such that $I' \to I$. Therefore, there is an isomorphism between $I$ and $I'$.

答: You're halfway through, because you didn't show these arrows going in opposite directions are actually inverses. Assume $I$ and $I'$ are initial. Then since $I$ is initial, there is a unique arrow $f:I \to I'$. Since $I'$ is initial, there is a unique arrow $g:I' \to I$. Now, both $g \circ f$ and $1_I$ are arrows $I \to I$, but since $I$ is initial, there can be only one: we conclude that $g\circ f = 1_I$. Similarly, $f \circ g = 1_{I'}$. Now you can say that $I \cong I'$.


问:
The relation $\sim$ in $\mathbb{R}$ is defined as: $x \sim y \iff x − y \in \mathbb{Z}$
$(a) \quad$ Show that $\sim$ is an equivalent relation.

$(b) \quad$ Give $2$ distinct equivalent classes (must show they are distinct).

$(c) \quad$  $[0, 1) = \{x \in \mathbb{R} \mid 0 \leq x < 1\}.$ Show that $^{\mathbb{R}}\big/_{\sim} \cong [0, 1).$

I was able to prove that the relation is reflexive, symmetric and transitive. So, it is equivalent.

For $(b)$ one class I could find was

$$\{y \in \mathbb{R} \mid x \sim y\} = \{y \in \mathbb{R} \mid \exists z \in \mathbb{Z}, x - y = z\} = \{y \in \mathbb{R} \mid \exists z \in \mathbb{Z}, x = z + y\}.$$

But I cannot find another one which is distinct!

And for $(c)$ I don't even know how to start this proof. Any ideas? Thanks

答:
(a) It's called equivalence relation

(b) The set you provided here is the equivalence class of a fixed element $x$. For example what is the equivalence class of $0$, or of $1/2$?

(c) Consider a specific function $f:\Bbb R\to [0,1)$ which satisfies $f(x)=f(y)\iff x\sim y$.


问:
Gradient of a convex function is continuous on the interior of its domain
Given a convex, lower semicontinuous, and proper function $f:\mathbb{R}^n\to \mathbb{R}$ which is differentiable on its domain, is it true that its gradient $\nabla f$ is continuous on the interior of the domain of $f$? Here I am taking $\text{dom}f = \{x\in\mathbb{R}^n: f(x)<\infty\}$. What I came up with was that for such a function $f$, it must be true that $f$ is locally Lipschitz continuous on its domain and then by Rademacher's theorem it is locally differentiable a.e.. This doesn't get what I want, however. Anyone have a proof or counter example?

Edit: this is corollary 9.20 in Rockafellar and Wets, as it turns out.

答: I am updating this post with the follow-up question: If $f$ is a convex function defined on some convex set $E\subseteq \mathbb R^n$ and if it is differentiable on $E$, is it true that its gradient must be continuous on $E$ (and not only in the interior) ?


问:
If a power of 2 divides a number, under what conditions does it divide a binomial coefficient involving the number that it divides?
We have had many questions here about the divisibility of $\binom{n}{k}$, many of them dealing with divisibility by powers of primes, or expressions involving the $\textrm{gcd}(n,k)$ (I originally gave several more examples but took TheSimpliFire's advice to shorten the list, and many other examples can still be found by looking at this question's edit history):

How can we prove that a binomial coefficient (n choose k) is divisible by the ratio of n and the gcd(n,m)?

Proving prime $p$ divides $\binom{p}{k}$ for $k\in\{1,\ldots,p-1\}$

The topic has also lead to some interesting research papers recently:

In 2014: Some divisibility properties of binomial and $q$-binomial coefficients.

In 2017: Divisibility of binomial coefficients and generation of alternating groups.

In 2019: On the divisibility of binomial coefficients.

There's also many associated theorems:

Kummer's theorem

Lucas's theorem

Granville's theorem

Désarménien's theorem

However I've come across a related problem which is expressed completely in the title, and is remarkably not covered by the above extensive body of literature. In MathJax, if $s>0$ is an integer (let's also make it the largest one for which $2^s$ divides $n$), under what conditions of $k$ do we have the following:

$$
2^s \mid n \implies 2^s \left\vert \binom{n}{k} \right. . \tag{1}
$$

Since $\binom{n}{k} = \frac{n}{k} \binom{n-1}{k-1}$ and $2^s \mid n$ we are left with determining when $\frac{q}{k} \binom{n-1}{k-1}$ is an integer ($q$ being the result when $n$ is divided by $2^s$).

The implication works for a remarkable number of cases, but not always (for example if $\binom{n}{k}$ is odd).

答:
With the very helpful tip from user BillyJoe, we can say that for any prime $p$,

$$\tag{1}
\binom{n}{k} \equiv  \left\langle {n_{s-1}\cdots n_0 \atop k_{s-1}\cdots k_0} \right\rangle \prod_{j=1}^{r-s+1}\left\langle {n_{j+s-1}\cdots n_j \atop k_{j+s-1}\cdots k_j} \right\rangle\left\langle {n_{j+s-2}\cdots n_j \atop k_{j+s-2}\cdots _j} \right\rangle^{-1} \left( \textrm{mod} ~ p^s \right),
$$

where the indices 0 and $r$ represent the terms corresponding to the lowest and highest powers (respectively) in the $p$-adic expansions of $n$ and $k$, and the angular brackets correspond to a slight modification (described in more detail here) of the ordinary binomial coefficient symbols.

We then have that:

$$
\left\langle {n_{s-1}\cdots n_0 \atop k_{s-1}\cdots k_0} \right\rangle \prod_{j=1}^{r-s+1}\left\langle {n_{j+s-1}\cdots n_j \atop k_{j+s-1}\cdots k_j} \right\rangle\left\langle {n_{j+s-2}\cdots n_j \atop k_{j+s-2}\cdots _j} \right\rangle^{-1} \equiv 0 \left( \textrm{mod} ~ p^s \right),\tag{2}
$$

if and only if $p^s \mid \binom{n}{k}$.

We therefore have a general condition for when $p^s$ divides $\binom{n}{k}$, whether or not $p^s \mid n$.

For the special case where $p=2$ and $2^s \mid n$, which was the premise of the original question, we may be able to get a cleaner expression than what I have above in Eq. 2.


问:
How to compute $\int_{0}^{+\infty}\frac{x^2\mathrm{d}x}{e^{x}-1}$ analytically?
Does anyone know how to compute analytically the following integral:

$$\int\limits_{0}^{+\infty}\dfrac{x^2\mathrm{d}x}{e^{x}-1}$$

It should be equal to $2\zeta(3)$ according to Maple. I tried the following using the binomial theorem for negative integer exponents:

$$I = \int\limits^{+\infty}_{0} e^{-x}(1-e^{-x})^{-1}x^2\mathrm{d}x = \int\limits^{+\infty}_{0}\left[\sum_{k=0}^{+\infty}(-1)^k(-1)^ke^{-(k+1)x}\right]x^2\mathrm{d}x=\int\limits^{+\infty}_{0}\left[\sum_{k=0}^{+\infty}(-1)^{2k}e^{-(k+1)x}\right]x^2\mathrm{d}x$$

After another change of variables, $y=(k+1)x$:

$$I = \sum_{k=0}^{+\infty}(-1)^{2k} \frac{1}{(k+1)^3}\int\limits_0^{+\infty} y^2e^{-y}\mathrm{d}y$$

The keen eye might recognize $\int\limits_0^{+\infty} y^2e^{-y}\mathrm{d}y$ as the gamma function, $\Gamma(3)=(3-1)!=2$. This, together with a slight nudge to the bottom limit of the summation we can rewrite things as:

$$I = \Gamma(3)\sum_{k=1}^{+\infty} \dfrac{(-1)^{2k}}{k^3}$$

And i see immediately (since the beginning in fact...) an infinite sum that makes me troubles and i can't get rid of. I tried to found if i did any trivial error but i'm focusing since to many hours to found it. That's why I need an external view to point me out my obvious error.

Thanks in advance for your help

Body:
You're actually almost there! Note that

$$(-1)^{2k} = \Big( (-1)^2 \Big)^k = 1$$

Thus,

$$\sum_{k=1}^\infty \frac{(-1)^{2k}}{k^3} = \sum_{k=1}^\infty \frac{1}{k^3} = \zeta(3)$$


问:
Deducing Identity from Stirling Formula
Stirling's Formula states that $\Gamma(z+1) \sim \sqrt{2 \pi z} (\frac{z}{\mathbb{e}})^{z}$ as $z \rightarrow \infty$. I need to prove the following identity using Stirling's formula:

$$ (2n)! \sim \frac{2^{2n} (n!)^{2}}{\sqrt{\pi n}}     $$ as $n \rightarrow \infty$.

In Stirling's formula I plugged in $z = 2n$ to get:

$$ \Gamma(2n+1) = 2n\Gamma(2n) \sim \sqrt{4n \pi} (\frac{2n}{\mathbb{e}})^{2n}  $$
where the first equality follows from the functional equation for the gamma function. Simplifying a little bit more I achieve:

$$ (2n)! \sim \frac{\sqrt{\pi n} 2^{2n} n^{2n-1}}{\mathbb{e}^{2n}}    $$

But, I don't know how to prove the identity from here. Can someone give me some hints on how to move on from this step?

答:
Using Stirling formula, we get
$$ \frac{(2n)!}{(n!)^2}\underset{n\rightarrow +\infty}{\sim}\frac{\sqrt{4\pi n}\left(\frac{2n}{e}\right)^{2n}}{2\pi n\left(\frac{n}{e}\right)^{2n}}=\frac{2^{2n}}{\sqrt{\pi n}} $$

Title: Deducing Identity from Stirling Formula


问:
Hartshorne's Exercise II.4.5(c). A third time.
It seems that I am the third one to pose this question. I think the proof in this question has a gap. He have not shown the lifting he constructed is compatible with $\mathrm{Spec}k(p) \rightarrow X$. And the proof given in wikiproof have not shown why Z satisfies the property as described in the exercise, which in my opinion is the most key part.

Any way to fill the gap is admired.

答:
$\newcommand{\Spec}{\operatorname{Spec}} \newcommand{\cO}{\mathcal{O}} \newcommand{\k}{\kappa} \newcommand{\Frac}{\operatorname{Frac}} \newcommand{\ol}[1]{\overline{#1}}$ Here is a proof which does not use Chow's lemma in contrast to EGA II 7.3.10. The main idea is that if $X\to Y$ is a proper dominant morphism of integral schemes over $k$, then $X$ satisfies the condition on valuations iff $Y$ does. From there, we can replace $X$ by its normalization $X'$ and show that every closed integral subscheme of $X'$ of codimension one satisfies the condition on valuations, implying every closed integral subscheme of $X$ of codimension one satisfies the condition on valuations by the lemma. By downwards induction, this implies that every closed integral subscheme of $X$ satisfies the condition on valuation rings. This shows that $X$ satisfies the valuative criteria, finishing the problem.

Warning: this post is somewhat long. If you have ideas on how to improve it or shorten it up while staying relatively true to the material presented in Hartshorne, please do not hesitate to say something in the comments.

First, a preliminary about valuation rings:

Lemma. If $A\subset K$ is a subring of a field, then $A$ is a valuation ring iff for every nonzero $x\in K$, at least one of $x$ and $x^{-1}$ belongs to $A$.

Proof. The forward direction is clear: $x\cdot x^{-1}=1$, so $v(x)+v(x^{-1})=0$ in the value group and therefore at least one of $v(x),v(x^{-1})$ is non-negative.
For the reverse direction, if $A\neq K$, then $A$ has a nonzero maximal ideal $\mathfrak{m}$.
If there's another maximal ideal $\mathfrak{m}'$, then we can find $x\in\mathfrak{m}$, $y\in\mathfrak{m}'$ with $x\notin\mathfrak{m}'$ and $y\notin\mathfrak{m}$.
Then neither $x/y$ or $y/x$ can be in $A$, contradicting the asumption, so $A$ has a unique maximal ideal.
Now suppose $A'$ is a local ring dominating $A$, and suppose $x\in A'$ - we have to show $x\in A$.
If not, then $x^{-1}\in A$ and in fact $x^{-1}\in\mathfrak{m}$, so $x,x^{-1}\in A'$.
But this means that $x^{-1}$ goes from being in the maximal ideal of $A$ to being a unit in $A'$, which is impossible because $A'$ dominates $A$.
The claim is proven. $\blacksquare$

To solve the problem, we'll show that the conditions about centers of valuations on $k(X)/k$ give the appropriate conditions for the valuative criteria for separatedness and properness.
We start by explaining how to connect the left side of the diagram in the valuative criteria to valuations of $k(X)/k$.
Let $R$ be a valuation ring with field of fractions $L$ and suppose we have the standard commutative diagram:

$\require{AMScd}$
\begin{CD}
\Spec L @>>> X\\
@VVV  @VVV\\
\Spec R @>>> \Spec k
\end{CD}

Let $z$ denote the unique point in the image of $\Spec L\to X$, and let $Z$ be the closure of $z$ equipped with the reduced induced subscheme structure.
Then $Z$ is an integral subscheme of $X$ with generic point $z$ and function field $k(Z)=\k(z)$, the residue field at $z$.
The map $\Spec L \to X$ gives us an inclusion $\k(z)\subset L$, and we let $S=\k(z)\cap R$.
It is straightforward to check that $S$ is a valuation ring: for any element $a\in \k(z)=\Frac(S)$, we have that considering $a$ as an element of $L$, we have that either $a\in R$ or $a^{-1}\in R$, so either $a\in S$ or $a^{-1}\in S$.
So our diagram can be rewritten as follows:

$\require{AMScd}$
\begin{CD}
\Spec L @>>> \Spec \k(z) @>>> X\\
@VVV @VVV  @VVV\\
\Spec R @>>> \Spec S @>>> \Spec k
\end{CD}

and therefore it's enough to check the valuative criteria on valuation rings $R$ with fields of fractions $L$ the residue field of a point $z$ in $X$.

In the case where $z$ is the generic point of $X$, this gives the answer immediately: $R$ is a valuation ring of $k(X)$, so by assumption it has at most one (respectively, a unique) center $x\in X$, which means that there exists at most one (respectively, a unique) lifting $\Spec R\to X$ making the relevant diagram commute by lemma II.4.4.
To show the result in general, we will prove that if $X$ is an integral scheme of finite type over a field so that every valuation on $k(X)/k$ has at most one (respectively, a unique) center on $X$, then the same is true for every integral closed subscheme $Z\subset X$.
We can reduce this further to proving that if $X$ satisfies the condition on valuations, then every closed integral subscheme $Z\subset X$ of codimension one does as well by downward induction.
To show this, we start with a lemma.

Lemma. Let $f:X\to Y$ be a proper dominant (equivalently, proper surjective) morphism of integral schemes over $k$.
Every valuation on $k(X)/k$ has at most one (respectively, a unique) center on $X$ iff the same is true for valuations of $k(Y)/k$ and $Y$.

Proof.

We have four things to prove:

(i) If every valuation on $k(X)/k$ has at most one center, then every valuation on $k(Y)/k$ has at most one center;

(ii) If every valuation on $k(X)/k$ has a center, then every valuation on $k(Y)/k$ has a center;

(iii) If every valuation on $k(Y)/k$ has at most one center, then every valuation on $k(X)/k$ has at most one center;

(iv) If every valuation on $k(Y)/k$ has a center, then every valuation on $k(X)/k$ has a center.

Let $R$ be a valuation ring for $k(Y)/k$.
As $f:X\to Y$ is dominant, it maps the generic point of $X$ to the generic point of $Y$ and thus induces an injection of fields $k(Y)\hookrightarrow k(X)$ which we may assume to be an inclusion.
Let $R'$ be a valuation ring of $k(X)$ dominating $R\subset k(X)$.
(We note that this also implies that $R'\cap k(Y)=R$: if $k(Y)\cap R'$ contained an element $e$ not in $R$, then $e^{-1}$ is in $R$, therefore $\mathfrak{m}_{R'}\cap R\neq \mathfrak{m}_R$ which contradicts the fact that $R'$ dominates $R$.)
By lemma II.4.4, a center $y$ for $R$ on $Y$ is equivalent to $R$ dominating $\cO_{Y,y}$, which implies $R'$ dominates $\cO_{Y,y}$ as subrings of $k(X)$.
This means we have a valuative diagram

$\require{AMScd}$
\begin{CD}
\Spec k(X) @>>> X\\
@VVV  @VVV\\
\Spec R' @>>> Y
\end{CD}

and by the valuative criteria for properness, we have a unique lifting $\Spec R'\to X$.
Thus for every center $y\in Y$ of $R$, we get a unique center $x\in X$ of $R'$.
So if there is at most one center in $X$ for all valuation rings $R'$ of $k(X)/k$, then there is at most one center in $Y$ for any valuation ring $R$ on $Y$, and (i) is proven.

To show that the existence of a center for all valuation rings $R'$ of $k(X)/k$ implies the existence of a center for all valuation rings of $k(Y)/k$, let $R$ and $R'$ be as in the previous paragraph.
As $R'$ has a center on $X$ by assumption, we get a map $\Spec R'\to X$ by lemma II.4.4, and composing with the map $X\to Y$, we obtain a map $\Spec R'\to Y$.
I claim that $\Spec R'\to Y$ factors through $\Spec R\to Y$.
This can be seen from examining the maps on local rings: letting $x\in X$ be the image of the closed point of $\Spec R'$ and $y\in Y$ the image of $x$, we have a sequence of local maps of local rings $\cO_{Y,y}\to \cO_{X,x}\to R'$ which are all injections because the maps on function fields are.
But $\cO_{Y,y}$ lands inside $k(Y)\subset k(X)$, which means it lands inside $k(Y)\cap R'$, which is exactly $R$.
Therefore $R$ dominates $\cO_{Y,y}$, so $y$ is a center for $R$ and (ii) is proven.

Now suppose $R'$ is a valuation ring for $k(X)/k$ with two centers $x,x'$.
Then by the logic above, the images of $x$ and $x'$ must be centers for $R=R'\cap k(Y)$, and $x$ and $x'$ cannot map to the same point, otherwise this would violate our result that every center for $R$ uniquely determines a center for $R'$.
Therefore we've proven (iii) by contrapositive.

Finally, if $R'$ is a valuation ring for $k(X)/k$, then $R=R'\cap k(Y)$ is a valuation ring for $k(Y)/k$, and so if $R$ has a center, we get a valuative diagram as above.
As $X\to Y$ is proper, we get a lifting $\Spec R\to X$, and considering the composition $\Spec R'\to \Spec R\to X$, we see that $R'$ has a center on $X$ by lemma II.4.4, so we've proven (iv) and we're done. $\blacksquare$

Now recall the construction of the normalization from exercise II.3.8: for a $X$ an integral $k$-scheme of finite type, we let $X'$ denote the normalization, which comes with a natural dominant map $\nu:X'\to X$ which is finite in our case.
By exercise II.4.1, a finite map is proper, so the map $\nu:X'\to X$ verifies the conditions of the lemma (we also note $\nu$ is surjective).

Let $Z\subset X$ be a codimension one integral closed subvariety.
As finite and surjective morphisms are stable under base change, we have that $Z\times_X X'\to Z$ is finite and surjective.
Since finite morphisms are closed, each irreducible component of $Z\times_X X'$ must map to a closed irreducible subset of $Z$, and therefore by surjectivity there must be an irreducible component of $Z\times_X X'$ which surjects on to $Z$.
Let $Z'\subset X'$ be such an irreducible component equipped with the reduced induced subscheme structure.
As $Z'\to Z\times_X X'$ is a closed immersion, it is finite, so the composite $Z'\to Z$ is a finite surjective morphism, and as finite morphisms are proper by exercise II.4.1, this satisfies the conditions of our lemma.
So it suffices to show that if $Z$ is a closed codimension one integral subscheme of a normal integral scheme $X$ of finite type over a field, then the condition about valuations on $X$ implies the condition about valuations on $Z$.

Let $Z\subset X$ as described in the previous sentence, and let $z\in Z$ be the generic point of $Z$.
Note that $\cO_{X,z}$ is a DVR by theorem I.6.2A: it's a noetherian local domain of dimension one which is integrally closed.
Suppose $R\subset k(Z)=\k(z)$ is a valuation ring trivial on $k$, and let $q:\cO_{X,z}\to \cO_{Z,z}=\k(z)$ be the natural quotient map.
Define $S=q^{-1}(R)\subset\cO_{X,z}$.
I claim $S$ is a valuation ring.
Let $e\in k(X)$ be an arbitrary nonzero element.
As at least one of $e,e^{-1}$ are in $\cO_{X,z}$, we may assume $e$ is actually in $\cO_{X,z}$.
If $e\in\mathfrak{m}_z$, then $e\in S$.
If $e\notin\mathfrak{m}_z$, then $e,e^{-1}\in\cO_{X,z}$ and so at least one of $q(e)$ or $q(e^{-1})=q(e)^{-1}$ is in $R$, which implies that one of $e$ or $e^{-1}$ is in $S$, which implies it is a valuation ring.

Suppose $z'\in Z$ is a center for $R$ on $Z$.
Then $\cO_{X,z'}=q^{-1}(\cO_{Z,z'})$ is dominated by $S$, so $z'$ is a center for $S$ on $X$.
As $Z\to X$ is injective on underlying sets, this implies that if $R$ has two distinct centers on $Z$, $S$ must have two distinct centers on $X$.
This shows that if every valuation on $k(X)/k$ has at most one center on $X$, then every valuation on $k(Z)/k$ has at most one center on $Z$.
Now suppose $z'\in X$ is a center for $S$: then $\cO_{X,z'}\subset S\subset \cO_{X,z}$, so $z'\in \ol{z}=Z$ and taking the quotient by the maximal ideal of $\cO_{X,z}$ we see that $\cO_{Z,z'}$ is dominated by $R$.
We are done.


问:
$\int_0^1\frac{(f(x)-1)^2 -4x^2}{x^{3.5}}\,dx$ exists. Calculate $f(0)$ and $f'(0)$
I've tried somehow using Taylor to try and figure this one out.

Unfortunately, I couldn't seem to get a solid answer.
Thank you very much for your help!

Let f be a continuos function,
$$f:R\rightarrow R$$
$$f(x), \  f'(x)  \ , \ f''(x) \ \text{are continuous} $$

and let the integral
$$\int_0^1\frac{(f(x)-1)^2 -4x^2}{x^{3.5}}\,dx \space \text{exist and be finite}$$

Find the value of
$$f(0)\ \text{and}\ |f'(0)|$$

答:
Let $a=f(0)-1, b=f'(0)$. Then continuity of the second derivative gives $f(x)= 1+a+ bx+O(x^2)$ and hence the integrand is
$$\frac{(a+bx+O(x^2))^2-4x^2}{x^{3.5}} = \frac{a^2+2abx+(b^2+a\times\cdots-4)x^2+O(x^3)}{x^{3.5}}$$
A function like $x^c$ is integrable at the origin only if $c>-1$ (check) and hence the first term in the numerator which may be nonzero is the cubic term.

You should be able to figure out what $a,|b|$ are now.


答:
You're on the right track:

$$(2n)! \sim \sqrt{4\pi n}\left(\frac{2n}{e}\right)^{2n} as \ n \to \infty$$
$$(2n)! \sim 2\sqrt{\pi n}\left(\frac{2n}{e}\right)^{2n} =\frac{2\sqrt{\pi n}\sqrt{\pi n}}{\sqrt{\pi n}}\left(\frac{2n}{e}\right)^{2n} $$
$$(2n)! \sim \frac{2^{2n}}{\sqrt{n\pi}}2n\pi \left(\frac{n}{e}\right)^{2n} $$

Observe from definition that $(n!)^2 \sim 2n\pi \left(\dfrac{n}{e}\right)^{2n} $

I think you can finish off from here

Title: Deducing Identity from Stirling Formula

Body:
Stirling's Formula states that $\Gamma(z+1) \sim \sqrt{2 \pi z} (\frac{z}{\mathbb{e}})^{z}$ as $z \rightarrow \infty$. I need to prove the following identity using Stirling's formula:

$$ (2n)! \sim \frac{2^{2n} (n!)^{2}}{\sqrt{\pi n}}     $$ as $n \rightarrow \infty$.

In Stirling's formula I plugged in $z = 2n$ to get:

$$ \Gamma(2n+1) = 2n\Gamma(2n) \sim \sqrt{4n \pi} (\frac{2n}{\mathbb{e}})^{2n}  $$
where the first equality follows from the functional equation for the gamma function. Simplifying a little bit more I achieve:

$$ (2n)! \sim \frac{\sqrt{\pi n} 2^{2n} n^{2n-1}}{\mathbb{e}^{2n}}    $$

But, I don't know how to prove the identity from here. Can someone give me some hints on how to move on from this step?


问:
Let $G$ be a group containing a normal subgroup $H$ isomorphic to $D_8$. Prove that $G$ must have a nontrivial center.
Let $G$ be a group containing a normal subgroup $H$ isomorphic to $D_8$. Prove that $G$ must have a nontrivial center.

I proved that the $|Z(H)|=2$. I tried to solve further by class equation but not able to reach any conclusion.

Is there any relation between $|Z(G)|$ and $|Z(H)|$?

答:
As you noted, $D_8$ has nontrivial center, with order two.

By normality, conjugation by any element of $G$ defines an automorphism of $H$. Now, the center is always a characteristic subgroup.

Thus if we let $h\ne e\in Z(H)$, then for any $g\in G$, $ghg^{-1}=h$.  Thus $h\in Z(G)$.  Thus $Z(H)\le Z(G)$.


问:
How to find equation of both parallel planes
I need help with the following problem:

Planes $S_1$ and $S_2$ are parallel to each other and the distance between them is $2$.
Plane $S_1$ passes through the points $A=(2,0,3)$ and $B=(0,0,6)$. Plane $S_2$ passes through the point $C=(-2,0,2)$.

I Need to find the equations of both planes.

I tried to write a few equations and solve without success.

答:
Assume the equation of $S_1$ and $S_2$ are $ax + by + cz + d_1 = 0$ and $ax + by + cz + d_2 = 0$ respectively.

Given: $2a + 3c + d_1 = 0$, $6c + d_1 = 0$ and $-2a + 2c + d_2 = 0$

Also $$\frac{|d_2 - d_1|}{\sqrt{a^2 + b^2 + c^2}} = 2$$

From the first three equations, $c = \frac{2a}{3}$, $d_1 = -4a$ and $d_2 = \frac{2a}{3}$

From the last equation, $b = \pm 2a$

Hence $S_1$ is $x \pm 2y + \frac{2}{3}z - 4 = 0$ and

$S_2$ is $x \pm 2y + \frac{2}{3}z + \frac{2}{3} = 0$


问:
Factorial of infinity
So, I've read in this article that:

$$\zeta'(0) = \log\sqrt\frac 1 {2\pi}$$

And that:
$$e^{-\zeta'(0)} = 1\cdot2\cdot3\cdots \infty = \infty! = \sqrt{2\pi}$$

I found this result very strange and tried to simplify if but i couldn't. Can somebody help me?

Thanks!

答:
One way to crosscheck (not prove/disprove (!)) that results approximative and numerically is to use the alternating zeta at the same parameter, which can be summed using simple summation procedures for divergent summation. If we write
$$ \eta(s) = \frac 1{1^s} - \frac 1{2^s} + \frac 1{3^s} - \cdots $$
and furtherly
$$ \eta(s) = \exp(-s\ln 1) - \exp(- s\ln2) + \exp(-s\ln 3) - \cdots $$

then we can write the derivative as
$$ \eta(s)' = (-\ln1)\cdot \exp(-s\ln 1) - (-\ln2)\cdot \exp(- s\ln2) + (-\ln3)\cdot \exp(-s\ln 3) - \cdots $$
and letting $\lim s \to 0$ this gives
$$ \begin{eqnarray}
\eta(0)' &=& (-\ln1) - (-\ln2) + (-\ln3) - \cdots  \\
 \eta(0)' &=& - (\ln1 - \ln2 + \ln3 - \cdots) \tag 1 \end{eqnarray}$$
The latter can be handled by a simple procedure for divergent summation; for instance Euler-summation $ \mathfrak E $ which gives then
$$ \eta(0)'  \underset{\mathfrak E}{=} 0.225791352645\ldots $$
Now in (1) we see the alternating signs but for the equivalent value of the zeta-function this series is non-alternating. So
$$ \zeta(0)' = - (\ln1 + \ln2 + \ln3 + \cdots) \tag 2$$
We assume now, that the value $ \zeta(0)' = s$ exists and that we can handle it into the divergent sum. Here we allude to Euler's trick of the transformation-formula between zeta() and eta() and exploit that the logs at even arguments are negative in the eta-formula. But $ \ln(2m) = \ln2 + \ln m$ and thus we might write
$$ \begin{eqnarray}
 s &=& - ( &(\ln 1 + \ln 2 + \ln 3 + \ln 4 + \ln 5 + \ln 6 + \ln 7\cdots) \\
 s &=& - ( &(\ln 1 - \ln 2 + \ln 3 - \ln 4 + \ln 5 - \ln 6 + \ln 7\cdots) \\
         & & &+2 \cdot ( \ln 2 + \ln 4 + \ln 6 + \ln 8 + \cdots) ) \\
 s&=& &\eta(0)' - 2 \cdot ( (\ln2 + \ln1) + (\ln2 + \ln 2) + (\ln 2 + \ln 3) + \cdots)) \\
 s&=& &\eta(0)' - 2 \cdot ( \zeta(0) \cdot \ln 2 - s ) \tag 3\\
\end{eqnarray} $$
where we have accumulated the constant $\ln 2$ which occurs as summand at each consecutive logarithm in the last equation (3) as $\zeta(0) \cdot \ln 2$. Then we have a determination for $s$ by
$$ \begin{eqnarray} &s&=& \eta(0)' - 2 \cdot ( \zeta(0) \cdot \ln 2 - s ) \\
 &-s&=& \eta(0)' - 2 \cdot \zeta(0) \cdot \ln 2  \\
  \zeta(0)'&=s&=& -\eta(0)' - \ln 2  \tag 4 \end{eqnarray}$$
$$\begin{eqnarray} &\zeta(0)' &\sim - 0.225791352645\ldots - 0.693147180560 \\
 && \sim -0.918938533205\ldots \tag 5 \\
 \end{eqnarray}$$
And indeed, numerically we get using Pari/GP

```<code>log(1/sqrt(2*Pi))
 %393 = -0.918938533205
</code>```

and also

```<code>h=1e-12;(zeta(0+h/2)-zeta(0-h/2))/h
 %394 = -0.918938533205
</code>```

This all required assumptions about the general summability of the divergent terms. But there is a fairly well developed theory of divergent summation, from which one may take the confirmation about the acceptability of the above transformations. They are intended to show one path to obtain the result; perhaps/likely there are others more elegant or even rigid ones.

Appendix: The Pari/GP code for the numerical evaluation of $\eta(0)'$ in (3) is

```<code>-sumalt(k=0,(-1)^k*log(1+k))
 %387 = 0.225791352645
</code>```

Title: Factorial of infinity

Body:
One way to crosscheck (not prove/disprove (!)) that results approximative and numerically is to use the alternating zeta at the same parameter, which can be summed using simple summation procedures for divergent summation. If we write
$$ \eta(s) = \frac 1{1^s} - \frac 1{2^s} + \frac 1{3^s} - \cdots $$
and furtherly
$$ \eta(s) = \exp(-s\ln 1) - \exp(- s\ln2) + \exp(-s\ln 3) - \cdots $$

then we can write the derivative as
$$ \eta(s)' = (-\ln1)\cdot \exp(-s\ln 1) - (-\ln2)\cdot \exp(- s\ln2) + (-\ln3)\cdot \exp(-s\ln 3) - \cdots $$
and letting $\lim s \to 0$ this gives
$$ \begin{eqnarray}
\eta(0)' &=& (-\ln1) - (-\ln2) + (-\ln3) - \cdots  \\
 \eta(0)' &=& - (\ln1 - \ln2 + \ln3 - \cdots) \tag 1 \end{eqnarray}$$
The latter can be handled by a simple procedure for divergent summation; for instance Euler-summation $ \mathfrak E $ which gives then
$$ \eta(0)'  \underset{\mathfrak E}{=} 0.225791352645\ldots $$
Now in (1) we see the alternating signs but for the equivalent value of the zeta-function this series is non-alternating. So
$$ \zeta(0)' = - (\ln1 + \ln2 + \ln3 + \cdots) \tag 2$$
We assume now, that the value $ \zeta(0)' = s$ exists and that we can handle it into the divergent sum. Here we allude to Euler's trick of the transformation-formula between zeta() and eta() and exploit that the logs at even arguments are negative in the eta-formula. But $ \ln(2m) = \ln2 + \ln m$ and thus we might write
$$ \begin{eqnarray}
 s &=& - ( &(\ln 1 + \ln 2 + \ln 3 + \ln 4 + \ln 5 + \ln 6 + \ln 7\cdots) \\
 s &=& - ( &(\ln 1 - \ln 2 + \ln 3 - \ln 4 + \ln 5 - \ln 6 + \ln 7\cdots) \\
         & & &+2 \cdot ( \ln 2 + \ln 4 + \ln 6 + \ln 8 + \cdots) ) \\
 s&=& &\eta(0)' - 2 \cdot ( (\ln2 + \ln1) + (\ln2 + \ln 2) + (\ln 2 + \ln 3) + \cdots)) \\
 s&=& &\eta(0)' - 2 \cdot ( \zeta(0) \cdot \ln 2 - s ) \tag 3\\
\end{eqnarray} $$
where we have accumulated the constant $\ln 2$ which occurs as summand at each consecutive logarithm in the last equation (3) as $\zeta(0) \cdot \ln 2$. Then we have a determination for $s$ by
$$ \begin{eqnarray} &s&=& \eta(0)' - 2 \cdot ( \zeta(0) \cdot \ln 2 - s ) \\
 &-s&=& \eta(0)' - 2 \cdot \zeta(0) \cdot \ln 2  \\
  \zeta(0)'&=s&=& -\eta(0)' - \ln 2  \tag 4 \end{eqnarray}$$
$$\begin{eqnarray} &\zeta(0)' &\sim - 0.225791352645\ldots - 0.693147180560 \\
 && \sim -0.918938533205\ldots \tag 5 \\
 \end{eqnarray}$$
And indeed, numerically we get using Pari/GP

log(1/sqrt(2*Pi))
 %393 = -0.918938533205

and also

h=1e-12;(zeta(0+h/2)-zeta(0-h/2))/h
 %394 = -0.918938533205

This all required assumptions about the general summability of the divergent terms. But there is a fairly well developed theory of divergent summation, from which one may take the confirmation about the acceptability of the above transformations. They are intended to show one path to obtain the result; perhaps/likely there are others more elegant or even rigid ones.

Appendix: The Pari/GP code for the numerical evaluation of $\eta(0)'$ in (3) is

-sumalt(k=0,(-1)^k*log(1+k))
 %387 = 0.225791352645


答:
This is a fun question, and lets us showcase a really powerful idea in group theory (and math more broadly):

If something is unique with a given property, it must be a fixed point for anything preserving that property

We see this idea with, for instance, $P$-sylow subgroups. If they are unique, they must be fixed by conjugation. Thus they're normal.

There are lots of places where, by showing something is unique, we can then conclude extra information about it. This is one of those places!

Say $x$ is the (unique!) nontrivial central element of $H = D_{8}$. We want to show that $xg = gx$ for every $g \in G$... Of course, that's the same thing as saying $g^{-1}xg = x$. And so we're trying to show that $x$ is fixed by the conjugation operation.

We know that $g^{-1}Hg = H$, since $H$ is normal. So at the very least, $g^{-1}xg \in H$. But we know more. Since conjugation by $g$ is an automorphism, it must preserve any equations that exist. In particular, if $xh = hx$, then that equation stays true after conjugating by $g$!

So if $x$ started in the center of $H$, $g^{-1}xg$ must end up in the center too. But, since $x$ is the unique nontrivial element of the center, we have no choice in the matter: $g^{-1}xg = x$. So $xg = gx$, and $x \in Z(G)$ too.

I hope this helps ^_^

Body:
Let $G$ be a group containing a normal subgroup $H$ isomorphic to $D_8$. Prove that $G$ must have a nontrivial center.

I proved that the $|Z(H)|=2$. I tried to solve further by class equation but not able to reach any conclusion.

Is there any relation between $|Z(G)|$ and $|Z(H)|$?


问:
Second Course on ODEs - Self Study - Book Recommendation
I'm looking for a book that would serve as a a second course on Ordinary Differential Equations. The book should be theoretically-oriented and for self study. I just finished the book by George Simmons - Differential Equations with Historical notes and Applications and in all honestly didn't really like it, finished it for the sake of completeness but it isn't what I was looking for.
I would like the book to contain a decent number of exercises after each section/chapter and, I stress, to be a theoretical approach to the topic. I originally started with Birkhoff and Rota's ODEs but one of the requierements was basic knowledge with complex functions, which I haven't studied yet.
I have knowledge of Calculus I (Spivak and Apostol), Calculus II (Apostol), Linear Algebra (Friedberg), Abstract Algebra (Herstein). Also, I will be finishing Introduction to Analysis (Mattuck) this week and will be starting with Rudin's Principles of Mathematical Analysis afterwards.
I have added the authors of the books, all of which I've liked, I used to study the previous topics for reference.
I'm looking for a book that I will be able to complete with the current knowledge that I have. I don't know if learning complex functions first, which I beleive I can take after finishing Rudin's Analysis, and then going back to Birkhoff's would be an option, but if it is I'm open to it too.

Thanks in advance

答: I would like to recommend L. Perko's book: Differential Equations and Dynamical Systems, 3rd ed to you. This is because it contains more modern and hot topics in dynamical system. P. Smith's book Nonlinear Ordinary Differential Equations is also a good book for studying nonlinear ODEs.</p>


问:
Find the second term of a sequence
I am posting this question here only because I find it so elegant. I found it on the French forum les-mathematiques.net, where somebody posted it, and it comes from the oral entrance exam of Ecole Polytechnique in 1987.

Let $(a_n)_{n\geq 0}$ be a nondecreasing sequence satisfying, for all integers $n\geq 2$, $(n-1)a_n=na_{n-2}$ and $a_0=2$. Find $a_1$.

答:
Define a sequence by the initial values $\,a_0=2\,$ and
$\,a_1=x\,$ with
$$ (n-1)a_n = n\,a_{n-2}\quad \text{ for } \quad
n\geq 2. \tag{1}$$
Also, suppose that $\,x\,$ is such that
$$ \,a_n\ge a_{n-1}\quad  \text { for } \quad
n\geq 1. \tag{2}$$
We want to find the unique value of $\,x.$

For the even values of $\,n\,$ the sequence begins
$$ 2,\,4,\,16/3,\,32/5,\,256/35,\,512/63,\,\dots
\tag{3}$$
with powers of $2$ in the numerator. In fact, for even $\,n\,$
$$ 2^n/a_n ={n-1\choose n/2}\sim 2^n/\sqrt{\pi n/2}
\tag{4}$$
using induction and Stirling's formula.
A similar result hold for odd $\,n.\,$ Since we are
given that the sequence is non-decreasing we must have
the inequalities
$$ x\ge 2,\,x\le 4,\,x\ge 8/3,\,
  x\le 32/9,\,x\ge 128/45,\,\dots . \tag{5}$$
The upper and lower bounds converge to a common limit
$\,x=\pi.$

Title: Find the second term of a sequence


答:
In your answer for b) you have to declare what $x$ is.

For instance if $x = \pi$ then we can define the equivalence class of pi is

$\{y\in \mathbb R: \pi\sim y\} = \{y\in \mathbb R|\exists z \in \mathbb Z; \pi-y = z\}=\{y\in \mathbb R| y=-z-\pi;z\in \mathbb Z\} = \{k-\pi|k\in \mathbb  Z\}=$

$\{..., -3-\pi, -2-\pi, -1, -\pi, 1-\pi, 2-\pi, 3-\pi, ....\}$.

You can do this for any real number $x$.

Do it for $x = \sqrt 2$ and get $\{y\in \mathbb R: \sqrt 2 \sim y\} = \{k-\sqrt 2|k\in \mathbb Z\} = \{...., -1 -\sqrt 2, -\sqrt 2, 1-\sqrt 2, 2-\sqrt 2,....\}$

Or eve for $x = 3$.  The equivalence class for $3$ is $\{k-3|k\in \mathbb Z\} = \mathbb Z$.

To show $\{..., -3-\pi, -2-\pi, -1, -\pi, 1-\pi, 2-\pi, 3-\pi, ....\}$ and  $\{...., -1 -\sqrt 2, -\sqrt 2, 1-\sqrt 2, 2-\sqrt 2,....\}$ are distinct, is a matter of saying

Let  $x \in \{y\in \mathbb R: \sqrt 2 \sim y\}\cap \{y\in \mathbb R: \pi\sim y\}$.  let $\pi \sim x$ and $\sqrt 2 \sim x$.  So $\pi -x = k\in \mathbb Z$ and $\sqrt 2-x= m\in \mathbb Z$ for some $k,m$.  But that means $x= \pi - k = \sqrt 2-m$ and that means $\pi = \sqrt 2 +(k -m)$ or that $\pi \sim 2$ (which follows by symmetry and transitivity).

And that's nonsense.  SO the two classes are distinct.

Title: The relation $\sim$ in $\mathbb{R}$ is defined as: $x \sim y \iff x − y \in \mathbb{Z}$

Body:
$(a) \quad$ Show that $\sim$ is an equivalent relation.

$(b) \quad$ Give $2$ distinct equivalent classes (must show they are distinct).

$(c) \quad$  $[0, 1) = \{x \in \mathbb{R} \mid 0 \leq x < 1\}.$ Show that $^{\mathbb{R}}\big/_{\sim} \cong [0, 1).$

I was able to prove that the relation is reflexive, symmetric and transitive. So, it is equivalent.

For $(b)$ one class I could find was

$$\{y \in \mathbb{R} \mid x \sim y\} = \{y \in \mathbb{R} \mid \exists z \in \mathbb{Z}, x - y = z\} = \{y \in \mathbb{R} \mid \exists z \in \mathbb{Z}, x = z + y\}.$$

But I cannot find another one which is distinct!

And for $(c)$ I don't even know how to start this proof. Any ideas? Thanks


问:
solve $|z+2|z=-i$ in $\mathbb{C}$
Before I wrote this question I've tried to solve this by using the definition of $|z|= \sqrt{a^{2}+b^{2}}$ so in this case should be $|z+2|= \sqrt{(a+2)^{2}+b^{2}}$ but this didn't help. I' tried also to use the fact $a=0$ because in $|z+2|z=-i$, there isn't the real part but I don't think this is legit. I think could be useful try the polar form, but I don't really know how to take advantages of this form in this case.
Thank you in advance

答:
If $z=a+ib$, $$\left|a+ib+2\right| (a+ib)=-i$$
hence
$$\left(\sqrt{(a+2)^{2}+b^{2}}\right)(a+ib)=-i \tag 1$$

Expand the $(1)$,
$$a\sqrt{a^2+4a+4+b^2}+ib\sqrt{a^2+4a+4+b^2}=0-i$$

and equalize the real part with the real part and the complex part with the complex part. In other word you will calculate two irrational equations.

\begin{cases}
a\sqrt{a^2+4a+4+b^2}=0 & (\star)\\
b\sqrt{a^2+4a+4+b^2}=-1 & (\star\star)
\end{cases}

The $(\star)$ give you $a=0$ being $\sqrt{a^2+4a+4+b^2} \neq 0$ ....and then you can continue....having

$$b\sqrt{4+b^2}=-1$$ with $b\neq 0$.

Title: solve $|z+2|z=-i$ in $\mathbb{C}$


问:
Use of epsilon in proofs about diameters
I have been working on the following question:

Assume that $K$ is a compact metric space. Prove that there are two points $x,y \in K$ such that $d(x,y)=$diam $K$. (Recall that diam $K = sup${$d(x,y) | x,y \in K$}.

Here is an attempted proof (imitating Baby Rudin proof 3.10(a)).

By definition of sup, $\forall \epsilon >0$, $\exists x,y \in K$ s.t. $d(x,y)+\epsilon ≥ $ diam $K$.

We also have that $d(x,y) ≤ $ diam $K$ $\forall x,y \in K$.

Since $\epsilon$ was arbitrary, it follows that $d(x,y)=$diam $K$.

This proof can't be right since it doesn't require compactness of K (and it's easy to come up with examples where the statement doesn't hold for K not compact). Can you help me understand where it's going wrong? Thanks!

答:
You say in point three, that there is a $x,y$ such that $d(x,y)=\operatorname{diam} K$.
You can't conclude that one. But maybe you can show that $d(x,y)$ is continuous and what do you know about continuous functions and compact sets?</p>


问:
If the gradient descent flow is a monotone decreasing function, why isn't my training loss monotonic?
The gradient descent flow has the property that the loss is a monotone decreasing function. My training loss increases and decreases along the trajectory. Why does this happen?

答: Gradient flow is gradient descent with "infinitesimal step sizes." When you actually perform gradient descent you of course use steps which are some non-infinitesimal size, and these steps can "overshoot" what gradient flow would theoretically do and end up increasing the loss function. You can try to compensate for this by making your step sizes smaller but this comes at the cost of training taking longer.</p>


问:
Fault in this proof of $\overline{\lim} (a_n+b_n) = \overline{\lim} a_n + \overline{\lim} b_n $ for bounded sequences $(a_n)$ and $(b_n)$
The statement

$\overline{\lim} (a_n+b_n) = \overline{\lim} a_n + \overline{\lim} b_n $, for bounded sequences $(a_n)$ and $(b_n)$

is false for arbitrary bounded sequenecs. What is true generally however is subadditivity:

$\overline{\lim} (a_n+b_n) \leq \overline{\lim} a_n + \overline{\lim} b_n $, for bounded sequences $(a_n)$ and $(b_n)$.

Knowing that, I am not able to spot the error in the arguments I've written up below that seemingly aver the statement of equality. Can someone see where exactly the error is present?

The definition of limit superior that I'm working with here is:

$$ \overline{\lim} (a_n + b_n) = \sup A = \sup \{ x  \in \mathbb{R}: x < a_n + b_n \ \text{ infinitely many times}  \} $$

Putative proof:

We know there exist subsequences $(a_{n_k})$ and $(b_{n_k})$ that converge to $\overline{\lim} a_n$ and $\overline{\lim} b_n$ respectively.

That means:

For each real $\varepsilon > 0$ there is a positive integer $N_1$ such that $k \geq N_1$ implies

$ \overline{\lim} a_n - \varepsilon < a_{n_k} < \overline{\lim} a_n + \varepsilon $

And similarly for the other sequence, there is a positive integer $N_2$ such that $k \geq N_2$ implies

$ \overline{\lim} b_n - \varepsilon < b_{n_k} < \overline{\lim} b_n + \varepsilon $

Consequently, for each integer $k \geq \max(N_1,N_2)$ we have

$ \overline{\lim} a_n + \overline{\lim} b_n - 2\varepsilon < a_{n_k} + b_{n_k} < \overline{\lim} a_n + \overline{\lim} b_n + 2\varepsilon    $

This means that the number $\overline{\lim} a_n + \overline{\lim} b_n - 2 \varepsilon $ has above it infinitely many terms from the sequence $(a_n + b_n)_{n\geq 1}$.

Therefore it is an element of the set $A$ and bounded above by the supremum $\overline{\lim} (a_n + b_n)$, that is

$ \overline{\lim} a_n + \overline{\lim} b_n - 2\varepsilon \leq \overline{\lim} (a_n+b_n) $

If $x \in A$, then $x < a_n + b_n$ infinitely many times. What does that mean, "infinitely many times"? Well it must mean that given any positive integer $m$, there exists some other positive integer $p > m$ such that $x < a_p + b_p$ (otherwise there would be finitely many terms above $x$, which would be false).

So there must exist some integer $m > \max(N_1, N_2)$ such that $x < a_m + b_m$. But then since $m > \max (N_1, N_2)$, this means $x < a_m + b_m < \overline{\lim} a_n + \overline{\lim} b_n + 2 \varepsilon$.

Consequently $A$ is bounded above by the number $\overline{\lim} a_n + \overline{\lim} b_n + 2 \varepsilon$, so this number must greater or equal to $\sup A$. That is,

$ \overline{\lim} ( a_n + b_n ) \leq \overline{\lim} a_n + \overline{\lim} b_n + 2 \varepsilon $

All together we have shown for each real $\varepsilon > 0$ that

$ \overline{\lim} a_n + \overline{\lim} b_n - 2 \varepsilon \leq  \overline{\lim} ( a_n + b_n ) \leq \overline{\lim} a_n + \overline{\lim} b_n + 2 \varepsilon  $

This implies $\overline {\lim} (a_n + b_n) = \overline{\lim} a_n + \overline {\lim} b_n$. $\blacksquare$

But of course, that is false! Yet I can't manage to find the error in my arguments above.

答:
The problem is that the sequences $a_{n_k}$ and $b_{n_k}$ don't necessarily follow the same subsequences. Instead, replace $b_{n_k}$ with, say, $b_{m_k}$.

Eventually, this part:

$\overline{\lim} a_n + \overline{\lim} b_n - 2\varepsilon < a_{n_k} + b_{n_k} < \overline{\lim} a_n + \overline{\lim} b_n + 2\varepsilon$

should be

$\overline{\lim} a_n + \overline{\lim} b_n - 2\varepsilon < a_{n_k} + b_{m_k} < \overline{\lim} a_n + \overline{\lim} b_n + 2\varepsilon$

and this fact does not guarantee that

"This means that the number $\overline{\lim} a_n + \overline{\lim} b_n - 2 \varepsilon$ has above it infinitely many terms from the sequence $(a_n + b_n)_{n\geq 1}$."

Even as just a trivial example, suppose $n_k$ are odd integers and $m_k$ are even integers. Then actually the subsequences form a partition.


问:
Show that if a nonsingular linear system is altered by multiplying its j-th column by c $\not =$ 0, the solution is altered only in the jth component
How can I show that if the nonsingular linear system Ax = b is altered by multiplication of its j-th column by c $\not =$ 0, then the solution is altered only in the jth component, which is multiplied by 1/c ?

答:
It's working simply because of the definition of matrix product:

Let $A'$ be the modified matrix: column $j$ is multiplied by $c$.

For any vector $x$, let $x'$ be the modified vector: coordinate $j$ is divided by $c$.

Show that $Ax=A'x'$.


答:
Since $|z+2|$ is real, we see that $z=-i/|z+2|$ has no real part, i.e., $z=iy$ for some $y\in\mathbb{R}$. But we now have $|2+iy|iy=-i$, which implies $y\sqrt{4+y^2}=-1$. Squaring this gives $y^2(4+y^2)=1$, or $(y^2)^2+4(y^2)-1=0$, for which the quadratic formula tells us $y^2=-2+\sqrt5$ (rejecting $-2-\sqrt5$ as not the square of a real number). Finally, noting that $y\sqrt{4+y^2}=-1\lt0$ implies $y\lt0$, we have

$$z=iy=-i\sqrt{\sqrt5-2}$$

as the only solution.

Title: solve $|z+2|z=-i$ in $\mathbb{C}$

Body:
Before I wrote this question I've tried to solve this by using the definition of $|z|= \sqrt{a^{2}+b^{2}}$ so in this case should be $|z+2|= \sqrt{(a+2)^{2}+b^{2}}$ but this didn't help. I' tried also to use the fact $a=0$ because in $|z+2|z=-i$, there isn't the real part but I don't think this is legit. I think could be useful try the polar form, but I don't really know how to take advantages of this form in this case.
Thank you in advance


问:
If $f(nx)=f(x)^n$ then $f(x+y)=f(x)f(y)$
If $f:\mathbb{R}\to\mathbb{R}_{+}$ is injective monotone function and if, $f(nx)=f(x)^n$ for all $n\in\mathbb{Z}$ and for all $x\in\mathbb{R}$, then,
$$f(x+y)=f(x)f(y),\text{ for all }x,y\in\mathbb{R}$$

It is easy to check the equality if $y$ is a multiple of $x$.

I thought about proving the inequalities, $f(x+y)\le f(x)f(y)$ and $f(x+y)\ge f(x)f(y)$, but I couldn't get to them.

Assuming that $f$ is increasing and $x<y$, then $y=kx+r$, for some $k\in\mathbb{Z}$ and $0< r\le \lvert x \rvert$, so

$$f(x+y)=f(x+kx+r)=f(x(k+1)+r)\ge f(x(k+1))=f(x)^{k+1}=f(x)f(x)^k=f(x)f(kx)$$

That is, $f(x+y)\ge f(x)f(kx)$, but the problem is that $f(kx)<f(y)$ so I can't continue the demonstration.

答:
If $x,y \in \mathbb{Z}$ then
$$
f(x+y) = f((x+y)1) = f(1)^{x+y} = f(1)^x f(1)^y = f(x1) f(y1) = f(x)f(y).
$$

If $x,y \in \mathbb{Q}$ then there is $n\in\mathbb{N}$ such that $nx,ny\in\mathbb{Z}.$ Then
$$
f(x+y)^n = f(n(x+y)) = f(nx+ny) = f(nx) f(ny) = f(x)^n f(y)^n = (f(x)f(y))^n,
$$
so $f(x+y)=f(x)f(y).$

If $x,y\in\mathbb{R}$ then take sequences $x_k^+, x_k^-, y_k^+, y_k^- \in \mathbb{Q}$ such that $x_k^+\to x+,\ x_k^-\to x-,\ y_k^+\to y+,\ y_k^-\to y-$. The result then follows from density of $\mathbb{Q}$ in $\mathbb{R}$ and from monotonicity. Try to complete this step yourself.


答:
There's something missing. For each $\epsilon>0$ (for example $\epsilon=\frac1n$) you may find different $x,y$, that is you actually get sequences $x_n,y_n$ with $d(x_n,y_n)\to \operatorname{diam}(K)$. By compactness you can find a subsequence that converges to som $(x,y)$.

A nicer proof is: If $K$ is compact, then so is $K\times K$, hence the continuos function $K\times K\to\mathbb R$, $(x,y)\mapsto d(x,y)$ assumes its maximum.

Title: Use of epsilon in proofs about diameters

Body:
I have been working on the following question:

Assume that $K$ is a compact metric space. Prove that there are two points $x,y \in K$ such that $d(x,y)=$diam $K$. (Recall that diam $K = sup${$d(x,y) | x,y \in K$}.

Here is an attempted proof (imitating Baby Rudin proof 3.10(a)).

By definition of sup, $\forall \epsilon >0$, $\exists x,y \in K$ s.t. $d(x,y)+\epsilon ≥ $ diam $K$.

We also have that $d(x,y) ≤ $ diam $K$ $\forall x,y \in K$.

Since $\epsilon$ was arbitrary, it follows that $d(x,y)=$diam $K$.

This proof can't be right since it doesn't require compactness of K (and it's easy to come up with examples where the statement doesn't hold for K not compact). Can you help me understand where it's going wrong? Thanks!


问:
Help me solve a differential equation?
Help me please with this one: $$y'=\arcsin(y+2x-1)$$

答:
Hint: Let $u=2x+y-1$ and then
$$ u'=2+\arcsin(u)$$
which is separable.

Title: Help me solve a differential equation?


答:
In such examples, you have something of type $y' = f(y+2x-1)$ for some function $f$. Those terms under $f$ are a bit challening, so why not replace it "outside" $f$, where we can somehow handle it. By that I mean, let $u(x) = y(x)+2x-1$. Then $u'(x)=y'(x) + 2$ so that $y'(x) = u'(x) - 2$.

After substituting we get an equation of type $$ u'(x)  = \arcsin(u(x)) + 2$$
Can you proceed?

Title: Help me solve a differential equation?

Body: Help me please with this one: $$y'=\arcsin(y+2x-1)$$


问:
Question on Definition of a Dual Map
Let L: V --> W be a linear map where V and W are vector spaces over a common field K. Why does the definition of the dual map of L not require L to be onto when the definition involves a composition.

My understanding of the definition follows. Let W* and V* be the dual spaces of V and W respectively. The dual map of L is the map L*: W* --> V* given by L*(M) = ML for all M of W*. But the composition ML exists only if L is onto.

If L is not onto when written as L: V --> W, then write L as L:V --> V(L), so L is indeed onto. But then there exists vectors of W which are excluded from the image V(L), so M cannot map every vector of W into K and M cannot be an element of W*.

What am I missing? My apologies ahead of time for not using the cool formatting tools.

答:
If $M \in W^*$, then it is able to map any vector of $W$ to some vector in $W$.

The composition $ML : V \to W$ exists no matter what the image of $L$ is. It does not matter that $M$ doesn't have the opportunity to see vector of $W$ outside the image of $L$; the composition is still a map $V \to W$ since it takes a vector in $V$ and sends it to a vector in $W$.

Title: Question on Definition of a Dual Map


问:
Prove that there are at least $n$ pairwise disjoint sets in the same class.
Let $S$ be a set containing $n^{2}+n-1$ elements, for some positive integer $n$. Suppose that the $n$-element subsets of $S$ are partitioned into two classes. Prove that there are at least $n$ pairwise disjoint sets in the same class.

I was thinking to use induction on $|S|$, with base $n=1$, and it works for $n=1$. However I am not able to proceed.

答:
(Not a solution, but a response to OP's comment. Too long to be a comment.)

There can be several motivating reasons for the induction on $k$, independent of knowing the solution.

Knowing that olympiad problems that seem inaccessible often have a "nice" solution after making an insightful observation.

Playing with $n^2 + n -1$ early on quickly gets out of hand. If we wanted to get a stronger handle on the problem, we should look for "Find the smallest $N(n,k)$ such that when we partition ....".

It is not uncommon for olympiad problems to be a unique case of an inductive process. This allows them to hide suggestions that would have made the problem much easier. E.g. Prove that a $8 \times 8$ chessboard with 1 square removed can be tiled by $L-$shaped triminos.

Study the counterexample when we have $n^2+n - 2$ elements and no $n$ pairwise disjoint sets: Let set $A$ have $n^2 - 1$ elements and set $B$ have the other $n-1$ elements. Let the first class be any n-element subset of $A$, and the second class be the rest of the n-element subsets (which clearly have a non-zero intersection with $B$). Then clearly we can find at most $n-1$ disjoint sets in a class, with a suggested indexing via a bijection with $B$. This suggests a generalized counterexample that $k$ subsets of size $n$ are not possible if we have a $B$ with $k-1$ elements, and set $A$ with $ kn-1$ elements which gives us $(k+1)n - 2 $ elements, and suggests (wishful thinking) that with $N(n,k) = (k+1)n - 1 $ elements we can find $k$ disjoint subsets.

(weak) Your suggested induction approach doesn't seem to yield a nice inductive hypothesis step, so if we really wanted to use induction we need to figure out a valid variable $k$ to induct on.


问:
Square root in $\mathbb{Q}_p$.
Problem: Suppose that $p > 2$. I want to show that $x^2  \equiv 2($mod $p$) has a solution iff 2 has a square root in $\mathbb{Q}_p$.

My attempt: Suppose that $x^2  \equiv 2($mod $p$) has a solution, say $x_1$, then take $x_2 = x_1 + \delta_1 p$ so that $x_2^2 \equiv 2($mod $p^2$). Similarly, take $x_3 = x_2 + \delta_2p^2$ so that $x_3^2 \equiv 2 ($mod $p^2$). Repeat this process, we obtain a sequence $(x_n)$ such that $x_n^2 \equiv 2 ($mod $p^n$). So $\lim\limits_{n \to \infty} |x_n^2-2|_2 \leq \lim\limits_{n \to \infty} p^{-n} = 0$. Thus $x_n^2 \to 2$ and hence $x_n \to \sqrt{2}$. This shows that 2 has a square root in $\mathbb{Q}_p$.

Is this correct? I've only been introduced to the concept of $\mathbb{Q}_p$ in the context of metric completion. Is there any hint or suggestion for the backward direction?

答:
Your approach works, but you need to justify that $x^2 \equiv 2 \pmod{p^n}$ has a solution if we only assume that $x^2 \equiv 2 \pmod{p}$ has a solution. Take a look at Hensel's lemma for this.

For the backwards direction, note that $1=|2|_p=|\sqrt{2}|_p^2$ implies $|\sqrt{2}|_p=1$, so $\sqrt{2} \in \Bbb Z_p$. Now reduce mod $p$ and use that $\Bbb Z_p/p\Bbb Z_p \cong \Bbb Z/p\Bbb Z$.

Title: Square root in $\mathbb{Q}_p$.


问:
Are all functions that take the form $ x ^ 2$ convex (with respect to some parameters)?
I know that $f(x) = x ^ 2$ is a convex function. I can see  that from plotting $f(x)$ on 2D graph with respect to $x$.

Now, lets say I have $f(\ldots) = (g(\ldots))^2$ . Let assume $g(\ldots)$ could be a simple linear function or a polynomial or the kind of function whose formula we can't even write it down (eg; type of 'learnt' function with billions of parameter commonly found in deep neural network).

I assume function "$f$" is still convex when plotted with respect to '$g$'?

But will '$f$' still be considered convex when plotted with respect to some (let's say 2 parameters) of a billion parameters of '$g$'?

答: No, consider the function $g(x) = x^{1/4}$ which when squared is concave.


问:
Markov Chain: Using one-step analysis to calculate the probability to reach state $5$ before reaching state $0$ when starting in state$ 3.$
I have an alternating random walk question where the stochastic process ${X_n : n = 0, 1, 2, . . .}$ is a Markov chain on the integers with
$X_0 = 0$. If $|X_n|$ is even, the conditional probabilities $P(X{n+1} = i + 1|X_n = i) = 3/4$ and
$P(X_{n+1} = i − 1|X_n = i) = 1/4 $ are given, whereas if $|X_n|$ is odd, the conditional
probabilities are$ P(X_n+1 = i + 1|X_n = i) = 1/4$ and $P(X_{n+1}= i − 1|X_n = i) = 3/4.$ I have been asked to use a one-step analysis to calculate the probability to reach state $5$ before reaching state $0$ when starting in state $3$.

I thought of this question as a Gambler's Ruin problem, but i'm not sure if i'm correct. I defined $A_j$ as the event of reaching state $5$ starting in state $j=3$ before reaching state $N=0$. So there are two absorbing states $5$ and $0$. So $P(A_3)= \theta_3$ with $\theta_5 =1$ (since probability of reaching 5 when starting at 5 is equal to 1) and using one step analysis i have that $\theta_3 = 3/4\theta_4 +1/4\theta_2 $ from  $\theta_j = p\theta_{j+1} +q\theta_{j-1} $  since they're even and from the question i have $p=3/4$ and $q=1/4$. How should i find $\theta_2, \theta_4$? Would greatly appreciate the help

答: You would find $\theta_2$ and $\theta_4$ in the same way as you found $\theta_3$, so in this case $\theta_2 = 3/4 \theta_3 + 1/4 \theta_1$ and $\theta_4 = 3/4 \theta_5 + 1/4 \theta_1$.  This will give you 4 equations in the 4 unknowns $\theta_1, \theta_2, \theta_3, \theta_4$, which you can then solve with linear algebra.