问:
Expectation calculation of hypergeometric distribution with sampling without replacement
Given a population of $n$ black and $m$ green balls, the expected number of black balls in a random sample of $r$ balls can be calculated as follows

We define a random variable $X_k$ taking values 1 or according as the $k$-th element in the sample is black or not. Then $P(X_k = 1) = \frac{n}{n+m}$ and the expected number of black balls in a random sample of $r$ balls is $\frac{nr}{n+m}$.  This is the answer given in Feller's Book (Vol 1). The reason given by Feller is "For reasons of symmetry".

But I am not convinced by this approach. How can we assume a constant probability $P(X_k = 1) = \frac{n}{n+m}$ ?  The sampling process is without replacement and probability of success changes in every trial. I don't understand what is meant by "For reasons of symmetry".

答:
The following seems to be a more rigorous argument.

I think you agree that $P(X_1 = 1) = \frac{n}{n+m}$.

Then \begin{align}P(X_2=1) &= P(X_2=1 \mid X_1=1)P(X_1=1) + P(X_2=1 \mid X_1=0)P(X_1=0) \\
&{= \frac{n-1}{n+m-1} \cdot \frac{n}{n+m} + \frac{n}{n+m-1} \cdot \frac{m}{n+m}} \\
&= \frac{n}{n+m}, \end{align}

if you simplify the expression on the second line.
If you feel like it, you can show that $P(X_k=1) = \frac{n}{n+m}$ for all $1 \leq k \leq r$ similarly, by induction (d.k.o's answer is the induction step).


问:
Simplify the expression $\log(2^x+2^y)$ for large values of $x,y$
Given two numbers $x,y$, I need to calculate $\log(2^x+2^y)$.

Is there a way to simplify this function, in a way that would allow to calculate the complete expression without going through calculating $2^x$ and $2^y$?

Assume that $x,y$ are larger than 1500 and I'm trying to calculate the value using a computer, so it can't really handle $2^{1500}$, but eventually I want to stay in the logarithmic scale so I don't really want to calculate $2^x$.

A (good) approximation is also fine (if exact simplification doesn't exist).

答:
Let $M=\max(x,y)$ and $m=\min(x,y)$. Then
$$
\ln(2^x+2^y)
= \ln(2^M+2^m)
= \ln(2^M(1+2^{-(M-m)})
= \ln(2^M) + \ln(1+2^{-(M-m)})
\\= \ln(2) M + \ln(1+2^{-(M-m)})
.
$$
Now, even if the computer cannot handle $2^M,$ it can handle the above expression.

If you want to reduce the number of calculations of logarithms (hardly needed nowadays) then you can split the calculations into cases: If $M-m$ is bigger than $\sim 10$ then the second term can be approximated with just $2^{-(M-m)}$, and if $M-m$ is bigger than $\sim 100$ then it can be skipped completely. (The limits are qualified guesses; do some tests to see if you need to change them.)

ContentLicense: CC BY-SA 4.0


问:
Spectrum and compactness of $\ell ^1$ operator
Consider complex sequence $(a_n)$, $|a_n| \leq 1$. On $\ell ^1$ space consider the operator
$$ T(x_n) = (a_n x_n).$$
Find the norm. Decide for what sequences is $T$ compact. Find point spectrum and spectrum.

I think the norm is equal to $\lVert a_n \rVert_\infty$. Since $|a_n|\leq 1 $ implies $(a_n) \in \ell ^\infty$. Then
$$\lVert T(x_n) \rVert = \lVert (a_1 x_1, a_2 x_2, \ldots) \rVert =
\sum_{n=1}^{\infty} |a_n x_n| \leq \sup_{n \in \mathbb{N}} |a_n| \sum_{n=1}^{\infty} |x_n| = \lVert a_n \rVert_\infty \lVert x_n \rVert_1.$$
Because of this we have $\lVert T \rVert \leq \lVert a_n \rVert_\infty$. Now we should find $(x_n) \in \ell ^1$ such that $\lVert x_n \rVert_1 = 1$ and $ \lVert T(x_n) \rVert_1 = \lVert a_n \rVert_\infty $. How can we do that?

For point spectrum we need to find nontrivial solutions of equation
$$ (\lambda I - T)(x_n) = 0.$$
We have
$ x_1(\lambda - a_1) = 0$, $x_2(\lambda - a_2) = 0$, $\ldots $

For $|\lambda| > 1 $ we do not have any nontrivial solution. (Since $|a_n| \leq 1$.) For $|\lambda| \leq 1$ we have solutions $\lambda = a_n$. From this the point spectrum $\sigma_p (T) = \{a_n, n \in \mathbb{N} \}$. Is this correct?

To find spectrum $\sigma(T)$ we want to check if the mapping $(\lambda I - T)$ is invertible.
$(\lambda I - T)$ is injective ($\ker T = \{0\}$). We need to check if it is surjective. That is we need to find if this equation always has a solution
$$ (\lambda I - T)(x_n) = (y_n).$$
For $\lambda = 0$ we see that the equation does not have any solution for sequences $(a_n)$ such it contains at least one zero. Let's say $(a_n)$ has a zero in $m$-th position.
For $(y_n) = (0,\ldots, 0,1,0,\ldots) $ where $1$ is in the $m$-th position. For such sequence $(y_n)$ this equation has no solution. Hence $0 \in \sigma(T)$. Which means that $T$ could be compact for such sequences.

How do we proceed from here? Thanks.

答:
$Te_n=a_ne_n$ and this gives $\|T\| \geq |a_n|$ for all $n$

Since $\sigma (T)$ is a closed set it contains the closure of $\{a_n: n \geq 1\}$. To prove the reverse inclusion suppose $\lambda \notin \overline {\{a_n: n \geq 1\}}$. Then there exists $r>0$ such that $|\lambda -a_n| \geq r$ for all $n$. The unique solution of $(T-\lambda I)(x_n)=(y_n)$ is given by $x_n=\frac {y_n} {a_n-\lambda}$.  Note that this sequence $(x_n)$ does belong to $\ell^{1}$.

Also, $\|(x_n)\|\leq \frac1  r \|(y_n)\|$ so $T-\lambda I$ is invertible. Thus $\sigma (T)=\overline {\{a_n: n \geq 1\}}$.

Compactnes: If $T$ is compact then non-zero eigen values cannot have any limit point other than $0$. Hence $a_n \to 0$. Conversely, if $a_n \to 0$ define $T_N(x_n)=(a_1x_1,a_2x_2,...,a_Nx_N,0,0,0...)$. Check that these finite rank operators converge in operator norm to $T$. Hence $T$ is compact.


问:
Is the minimum singular value (amplitude) of $\boldsymbol{AB}$ bounded by the counterparts of $\boldsymbol{A}$ and $\boldsymbol{B}$?
Consider two complex matrices $\boldsymbol{A} \in \mathbb{C}^{M\times N}$ and $\boldsymbol{B} \in \mathbb{C}^{N\times Q}$. Is the minimum singular value's amplitude of $\boldsymbol{AB}$ bounded by the minimum singular value amplitudes of $\boldsymbol{A}$ and $\boldsymbol{B}$? That is,

does
$$
|\sigma_{min}(\boldsymbol{AB})| \leq \min \{|\sigma_{min}(\boldsymbol{A})|,|\sigma_{min}(\boldsymbol{B})|\}
$$
hold?

答:
No, it doesn't even work for scalar matrices.

An inequality of this type should be homogenous, of the correct degree.

The inequality you do have is
$$\sigma_{\min}(A B) \ge \sigma_{\min}(A) \cdot \sigma_{\min}(B)$$


答:
The solution stems from the linearity of expectation, which applies even if the random variables are not independent. At more length,

Let $X_k$ be an indicator random variable that is $1$ if the $k_{th}$ element is black, and $0$ if not

Now black balls (or those of any other color) have no preference for position, thus if you randomly pick up the $k_{th}$ ball,

$\Bbb P(X_k) = \Bbb P(X_1) = \frac {n}{n+m}$

Now the expectation of an indicator random variable is just the probability of the event it indicates, thus

$\Bbb E(X_k ) = \frac{n}{n+m},$

and $\Bbb E(X) = \Bbb E(X_1) + \Bbb E(X_2) + ...+\Bbb E(X_r) = \frac{rn}{n+m}$

Title: Expectation calculation of hypergeometric distribution with sampling without replacement

Body:
Given a population of $n$ black and $m$ green balls, the expected number of black balls in a random sample of $r$ balls can be calculated as follows

We define a random variable $X_k$ taking values 1 or according as the $k$-th element in the sample is black or not. Then $P(X_k = 1) = \frac{n}{n+m}$ and the expected number of black balls in a random sample of $r$ balls is $\frac{nr}{n+m}$.  This is the answer given in Feller's Book (Vol 1). The reason given by Feller is "For reasons of symmetry".

But I am not convinced by this approach. How can we assume a constant probability $P(X_k = 1) = \frac{n}{n+m}$ ?  The sampling process is without replacement and probability of success changes in every trial. I don't understand what is meant by "For reasons of symmetry".


答:
Let $S_k:=\sum_{i=1}^k X_i$. Then for $k<n$,
\begin{align}
\mathsf{P}(X_{k+1}=1)&=\sum_{i=0}^k\mathsf{P}(S_k=i)\mathsf{P}(X_{k+1}=1\mid S_k=i) \\
&=\sum_{i=0}^k\frac{\binom{n}{i}\binom{m}{k-i}}{\binom{n+m}{k}}\times\frac{n-i}{n+m-k} \\
&=\frac{n}{n+m}\sum_{i=0}^k\frac{\binom{n-1}{i}\binom{m}{k-i}}{\binom{n+m-1}{k}}=\frac{n}{n+m}.
\end{align}


问:
Why are smooth exhaustion functions proper maps?
In Lee's Introduction to Smooth manifolds an exhaustion function of a topological space $M$ is defined as a continuous function $f: M \to \mathbb{R}$ such that for all $c \in \mathbb{R}$ we have that $f^{-1}((-\infty,c])$ is compact in $M$. Later in the proof of Whitney's embedding theorem it is used that such a function is proper, i.e. that the pre-image $f^{-1}(K)$ for any compact $K \subseteq \mathbb{R}$ is again compact in $M$. I can't see why every exhaustion function needs to be proper. My approach: Let $K \subseteq \mathbb{R}$ be compact, then $K$ is bounded and thus included in a set $(-\infty, c]$ and therefore $f^{-1}(K)$ is a subset of the compact set $f^{-1}((-\infty,c])$. It remains to show that $f^{-1}(K)$ is closed, but I can't think of an appropriate argument. Since it's not even shown in the book it's probably very easy to see that the claim is true, but I still have some gaps in my knowledge of point-set topology... Any hint/solution is appreciated!

答:
Since $f$ is continuous, $f^{-1}(C) \in M$ is closed whenever $C$ is closed in the codomain.

Your argument may be to take $K$ compact subset of $(-\infty, c]$. Hence it is closed, (since compact sets are closed) and by continuity of $f$, also $f^{-1}(C)$ is closed in $M$. Since $f$ is an exhaustion function, then a closed subset of a compact set is compact and the result follows.


问:
Inverse-Square vector fields have both a divergence and curl of $0$?
Consider an inverse- square vector field

$$ \vec{F} = \frac{x}{r^3}\hat{x} + \frac{y}{r^3}\hat{y} + \frac{z}{r^3}\hat{z} = \frac{\hat{r}}{r^2}$$

where $r = \sqrt{x^2 + y^2 + z^2}$. The curl $\nabla \times \vec{F} = \vec{0}$, therefore we might go looking for a potential $V$. I find that $V = -1/r$ works and therefore one can say that $\vec{F} = \nabla V$ is derivable from a potential function $V$. I'll point out right now that $\vec{F}$ is undefined at the origin.

The divergence $\nabla \cdot \vec{F} = 0$. Therefore, we might go looking for a vector potential $\vec{A}$ such that $\vec{F} = \nabla \times \vec{A}$. One would say that $\vec{F}$ is derivable from a vector potential $\vec{A}$. But I'm having trouble seeing that an inverse-square vector field is derivable from both a vector potential and a scalar potential. So I know we have a trouble point at the origin. Yet this trouble point doesn't really seem to affect the "conservativeness" or path-independence of the vector field. But this trouble point does seem to affect the surface-independence of the vector field. As long as the surface doesn't wrap around the origin, I'd expect the inverse-square vector field to be surface-independent for a given boundary curve.

Can an inverse-square vector field be derivable from both a scalar potential and a separate a vector potential? (Helmholtz theorem comes to mind. But the question I'm asking involves two separate equations. One $\nabla V$ gives $\vec{F}$ and another $\nabla \times \vec{A}$ gives $\vec{F}$ as well).

答:
This is an old question but I find the comments and the other answer are a little evasive. There is a simple, clear answer to your question: NO, your field $F$ does not possess a vector potential everywhere it is defined. It only possesses a vector potential on regions away from the origin, but none of these vector potentials can be extended to all of space minus the origin.

You seem to think $F$ should have a vector potential everywhere because it has zero divergence. But an arbitrary divergence-free field is only guaranteed to have a vector potential on a region with trivial second de Rham cohomology. In this case, the domain of $F$ is space minus the origin, which has nontrivial second de Rham cohomology. As a result, knowing that $F$ has zero divergence is not enough to conclude that it has a vector potential: it could go either way. (There are some divergence-free fields on space minus a point that do have vector potentials; others don’t.) To check, you need to ask whether the integral of $F$ vanishes over every closed surface contained in the domain. If you can find just one closed surface where the integral fails to vanish, then $F$ does not have a vector potential.

That is what happens in this case. (In fact, your $F$ is the classic, canonical example of a divergence-free field without a potential.) You can see that $F$ fails the test I gave in the last paragraph by observing that its integral over the unit sphere is nonzero (this is just Gauss’s law!).


问:
Laplace transform of a solution of differential equation

Consider the differenial equation
$$t\frac{d^2y}{dt^2}+2\frac{dy}{dt}+ty=0\tag{1}$$
$$t>0,y(0+)=1,y'(0+)=0$$
If $Y(s)$ is Laplace transform of $y(t)$, then find the value of $Y(1)$.

My attempt: we know that $\mathcal{L}({tf(t)})=-\frac{d}{ds}F(s)$,where $F(s)$ is the Laplace transform of $f(t)$. Now taking Laplace transform both side in equation $(1)$, we have
$$\begin{align*}&\mathcal{L}[t\frac{d^2y}{dt^2}+2\frac{dy}{dt}+ty]=0
\\\Rightarrow&\mathcal{L}[t\frac{d^2y}{dt^2}]+2\mathcal{L}[\frac{dy}{dt}]+\mathcal{L}[ty]=0
\\\Rightarrow&-\frac{d}{ds}[s^2Y(s)-sy'(0)-y(0)]+2[sY(s)-y(0)]-\frac{d}{ds}Y(s)=0
\\\Rightarrow&-2sY(s)-s^2Y'(s)+2sY(s)-2-Y'(s)=0
\\\Rightarrow&Y'(s)=\frac{-2}{1+s^2}
\\\Rightarrow&Y(s)=-2\tan^{-1}s+C\end{align*}$$

Where $C$ is some constant, therefore $Y(1)=-2\tan^{-1}(1)+C=-\frac{\pi}{2}+C$.

What I need to know is only constant $C$. I tried my best but didn't found the value of this constant.

Any help will be appreciable

Thanks in advance.

答: I spotted one mistake: $\mathcal L[y'']=s^2Y(s)-s\color{red}y(0)-\color{red}{y'}(0)$. Using the correct formula you should get $Y(s)=C-\tan^{-1}s$. To find $C$, simply use the fact that $\lim_{s\to\infty}Y(s)=0$.


问:
Solution for rotation $d\vec{r}/dt = \vec{\omega}\times\vec{r}$.
Let's consider an equation:

\begin{equation}
    \frac{d\vec{r}}{dt}=\vec{\omega}\times\vec{r}.
\end{equation}

This equation  might  be expressed  via "angular velocity" matrix:

\begin{equation}\label{eq:3}\tag{1}
    \frac{d\vec{r}}{dt}= \hat{\Omega} \vec{r},
\end{equation}
where  the "angular velocity" matrix is
\begin{equation}
\Omega =
\begin{pmatrix}
0         & \omega_z  & -\omega_y \\
-\omega_z & 0         & \omega_x  \\
\omega_y  & -\omega_x & 0         \\
\end{pmatrix}.
\end{equation}

Also let's introduce matrixes $\hat{\vec{S}}_{x,y,z}$ ($SO(3)$ generators):

\begin{multline}
\Omega =
\omega_x
\begin{pmatrix}
0  & 0  & 0 \\
0  & 0  & 1 \\
0  & -1 & 0 \\
\end{pmatrix}
+
\omega_y
\begin{pmatrix}
0 &  0  & -1 \\
0 &  0  & 0  \\
1 &  0  & 0  \\
\end{pmatrix}
+
\omega_z
\begin{pmatrix}
0  & 1  & 0 \\
-1 & 0  & 0 \\
0  & 0  & 0 \\
\end{pmatrix}
= \\ =
 i \omega_x
\begin{pmatrix}
0  & 0  & 0  \\
0  & 0  & -i \\
0  & i & 0   \\
\end{pmatrix}
+
i \omega_y
\begin{pmatrix}
0  &  0  & i  \\
0  &  0  & 0  \\
-i &  0  & 0  \\
\end{pmatrix}
+
i\omega_z
\begin{pmatrix}
0  & -i & 0 \\
i  & 0  & 0 \\
0  & 0  & 0 \\
\end{pmatrix}.
\end{multline}

\begin{multline}
\hat{\vec{S}} =
\begin{pmatrix}
0  & 0  & 0  \\
0  & 0  & -i \\
0  & i & 0   \\
\end{pmatrix}
\vec{e}_x
+
\begin{pmatrix}
0  &  0  & i  \\
0  &  0  & 0  \\
-i &  0  & 0  \\
\end{pmatrix}
\vec{e}_y
+
\begin{pmatrix}
0  & -i & 0 \\
i  & 0  & 0 \\
0  & 0  & 0 \\
\end{pmatrix}
\vec{e}_z  =
\hat{S}_x \vec{e}_x +
\hat{S}_y \vec{e}_y +
\hat{S}_z \vec{e}_z
.
\end{multline}

So, we can write a formal solution of~\eqref{eq:3} as:
\begin{equation}\label{eq:solmatrix}\tag{2}
    \vec{r} = e^{i (\vec{\omega} t\cdot\vec{\hat{S}})} \vec{r}_0.
\end{equation}

Now, if I use some convinient axes, I can write \eqref{eq:solmatrix} as:
\begin{equation}\label{eq:solmatrix1}
    \vec{r} = e^{i (\omega t\cdot\hat{S}_{z'})} \vec{r}_0,
\end{equation}
and in explicit form (using Taylor series expansion of $e^M$) it looks like:

\begin{equation}\label{eq:so2}\tag{3}
    \vec{r}(t)=\begin{pmatrix} \cos(\omega t) & -\sin(\omega t) &  0\\ \sin(\omega t) & \cos(\omega t) & 0 \\ 0 & 0 & 1 \end{pmatrix} \vec{r}_0 = \hat{R}_{z'}(\omega t) \vec{v}_0 ,
\end{equation}
where $\hat{R}_{z'}(\omega t)$ is a rotation matrix about $z'$.

Now a question. How can I write the solution \eqref{eq:solmatrix} in explicit form (via rotation matrix like \eqref{eq:so2})? I tried, but there are powers of ($\sum_m i \omega_m \hat{S}_m)^n$ in Taylor series, which is not clear how to simplify. I can't see what they should collapse into.

答:
I'll focus my answer on computing the matrix exponential. Consider components of the unit vector $\hat{\omega}^a  = \omega^a /|\omega| $. Splitting the exponential into odd/ even terms

$$ E=\sum_{n=0}^\infty \frac{\left( i |\omega|\hat{\omega}^a S^a \right)^n}{n!}   = 1+\sum_{n=1}^\infty \frac{(-1)^n |\omega|^{2n}(\hat{\omega}^aS^a)^{2n}}{(2n)!}+i \sum_{n=0}^\infty \frac{(-1)^n |\omega|^{2n+1}(\hat{\omega}^aS^a)^{2n+1}}{(2n+1)!} $$

The components $jk$ of the $\operatorname{so}(3)$ rotation generator $a$ are $S^{ajk}=i\epsilon^{ajk} $. Consider

$$\left[ (\hat{\omega}^aS^a)^2\right]^{jl}=\hat{\omega}^a \hat{\omega}^b S^{ajk}S^{bkl}=\delta^{jl}-\hat{\omega}^j\hat{\omega}^l$$

The last equality follows$^\dagger $ by using the 'contracted epsilon identity' as well as the fact that $\hat{\omega}^a \hat{\omega}^a =1 $. We also note that RHS is a projection operator$^\ddagger$, hence

$$\left[ (\hat{\omega}^aS^a)^{2n}\right]^{jl}=\delta^{jl}-\hat{\omega}^j\hat{\omega}^l$$

$$\left[ (\hat{\omega}^aS^a)^{2n+1}\right]^{jl}=\left[ \hat{\omega}^aS^a \right]^{jl}$$
Substitute back into the sums

$$ E^{jl}=\delta^{jl}+\sum_{n=1}^\infty \frac{(-1)^n |\omega|^{2n}}{(2n)!}(\delta^{jl}-\hat{\omega}^j\hat{\omega}^l) +i \sum_{n=0}^\infty \frac{(-1)^n |\omega|^{2n+1}}{(2n+1)!}\left[ \hat{\omega}^aS^a \right]^{jl} $$

$$E^{jl}=\delta^{jl}+(\delta^{jl}-\hat{\omega}^j\hat{\omega}^l)\left(\cos(|\omega|)-1\right) +i\left[ \hat{\omega}^aS^a \right]^{jl}\sin(|\omega|)$$

$$E^{jl}=\hat{\omega}^j\hat{\omega}^l+(\delta^{jl}-\hat{\omega}^j\hat{\omega}^l)\cos(|\omega|) +i\left[ \hat{\omega}^aS^a \right]^{jl}\sin(|\omega|)$$

Those are the components of the matrix. In matrix form it's probably not going to be very pretty, so I'll leave it there.

$\dagger$ Like this

$$ \hat{\omega}^a \hat{\omega}^b S^{ajk}S^{bkl}=i^2\hat{\omega}^a \hat{\omega}^b \epsilon^{ajk}\epsilon^{bkl}=\hat{\omega}^a \hat{\omega}^b \epsilon^{ajk}\epsilon^{blk}=\hat{\omega}^a \hat{\omega}^b \left( \delta^{ab}\delta^{jl}-\delta^{al}\delta^{bj} \right) \\= \hat{\omega}^a \hat{\omega}^a \delta^{jl}-\hat{\omega}^j \hat{\omega}^l =  \delta^{jl}-\hat{\omega}^j \hat{\omega}^l $$

$\ddagger$ By direct computation
$$  (\delta^{jl}-\hat{\omega}^j \hat{\omega}^l)(\delta^{lm}-\hat{\omega}^l \hat{\omega}^m)= \delta^{jl}\delta^{lm}+\hat{\omega}^j \hat{\omega}^l \hat{\omega}^l \hat{\omega}^m-\delta^{jl}\hat{\omega}^l \hat{\omega}^m-\delta^{lm}\hat{\omega}^j \hat{\omega}^l \\ =\delta^{jm}-\hat{\omega}^j \hat{\omega}^m$$


答:
The spectrum of $T$ contains clearly all the $a_n$'s, since $Te_n=a_ne_n$, and since the spectrum is closed, then it contains all the limit points of $\{a_n\}$. Hence
$$
\overline{\{a_n\}}\subset\sigma(T)
$$
On the other hand, if $\lambda\not\in\overline{\{a_n\}}$, then
$$
r=\mathrm{dist}\big(\lambda,\overline{\{a_n\}}\big)=\inf_{n\in\mathbb N}|\lambda-a_n|>0,
$$
and hence
$$
\frac{1}{|\lambda-a_n|}\ge\frac{1}{r},\quad\text{for all $n$.}
$$
Thus, for every $(x_n)\in \ell^1$, if we set
$$
y_n=\frac{x_n}{\lambda-a_n}
$$
then $(y_n)\in\ell^1$, and in particular, $\|(y_n)\|_{\ell^1}\le\frac{1}{r}\|(x_n)\|_{\ell^1}$, and
$$
(\lambda I-T)(y_n)=(x_n).
$$
Thus, if  $\lambda\not\in\overline{\{a_n\}}$, then $\lambda\not\in\sigma(T)$, and therefore
$$
\sigma(T)\subset\overline{\{a_n\}}.
$$
Altogether
$$
\sigma(T)=\overline{\{a_n\}}.
$$

Title: Spectrum and compactness of $\ell ^1$ operator

Body:
Consider complex sequence $(a_n)$, $|a_n| \leq 1$. On $\ell ^1$ space consider the operator
$$ T(x_n) = (a_n x_n).$$
Find the norm. Decide for what sequences is $T$ compact. Find point spectrum and spectrum.

I think the norm is equal to $\lVert a_n \rVert_\infty$. Since $|a_n|\leq 1 $ implies $(a_n) \in \ell ^\infty$. Then
$$\lVert T(x_n) \rVert = \lVert (a_1 x_1, a_2 x_2, \ldots) \rVert =
\sum_{n=1}^{\infty} |a_n x_n| \leq \sup_{n \in \mathbb{N}} |a_n| \sum_{n=1}^{\infty} |x_n| = \lVert a_n \rVert_\infty \lVert x_n \rVert_1.$$
Because of this we have $\lVert T \rVert \leq \lVert a_n \rVert_\infty$. Now we should find $(x_n) \in \ell ^1$ such that $\lVert x_n \rVert_1 = 1$ and $ \lVert T(x_n) \rVert_1 = \lVert a_n \rVert_\infty $. How can we do that?

For point spectrum we need to find nontrivial solutions of equation
$$ (\lambda I - T)(x_n) = 0.$$
We have
$ x_1(\lambda - a_1) = 0$, $x_2(\lambda - a_2) = 0$, $\ldots $

For $|\lambda| > 1 $ we do not have any nontrivial solution. (Since $|a_n| \leq 1$.) For $|\lambda| \leq 1$ we have solutions $\lambda = a_n$. From this the point spectrum $\sigma_p (T) = \{a_n, n \in \mathbb{N} \}$. Is this correct?

To find spectrum $\sigma(T)$ we want to check if the mapping $(\lambda I - T)$ is invertible.
$(\lambda I - T)$ is injective ($\ker T = \{0\}$). We need to check if it is surjective. That is we need to find if this equation always has a solution
$$ (\lambda I - T)(x_n) = (y_n).$$
For $\lambda = 0$ we see that the equation does not have any solution for sequences $(a_n)$ such it contains at least one zero. Let's say $(a_n)$ has a zero in $m$-th position.
For $(y_n) = (0,\ldots, 0,1,0,\ldots) $ where $1$ is in the $m$-th position. For such sequence $(y_n)$ this equation has no solution. Hence $0 \in \sigma(T)$. Which means that $T$ could be compact for such sequences.

How do we proceed from here? Thanks.


问:
Computing the definite integral $\int_0^1(-1)^{\left\lfloor\frac1x\right\rfloor}\,\mathrm dx$

$$\int_0^1(-1)^{\left\lfloor\frac1x\right\rfloor}\,\mathrm dx$$

For $x\in\left(\frac1{n+1},\frac1n\right)$, $\left\lfloor\frac1x\right\rfloor=n$, and $\frac1n-\frac1{n+1}=\frac1{n(n+1)}$. So the integral is equal to the Riemann sum

$$\int_0^1(-1)^{\left\lfloor\frac1x\right\rfloor}\,\mathrm dx=\lim_{n\to\infty}\sum_{k=1}^n\frac{(-1)^k}{k(k+1)}$$

Now, for $|x|<1$, we have

$$\frac x{1-x}=\sum_{k=1}^\infty x^k\\
\implies -x-\ln(1-x)=C_1+\sum_{k=1}^\infty\frac{x^{k+1}}{k+1}\\
\implies x-\frac{x^2}2+(1-x)\ln(1-x)=C_2+C_1x+\sum_{k=1}^\infty \frac{x^{k+2}}{(k+1)(k+2)}$$

If $x=0$, then we find $C_1=C_2=0$ so that

$$x-\frac{x^2}2+(1-x)\ln(1-x)=\sum_{k=1}^\infty \frac{x^{k+2}}{(k+1)(k+2)}\\
\implies 1-\frac x2+\frac{1-x}x\ln(1-x) = \sum_{k=2}^\infty \frac{x^k}{k(k+1)}\\
\implies 1+\frac{1-x}x\ln(1-x) = \sum_{k=1}^\infty \frac{x^k}{k(k+1)}$$

and when $x=-1$, we have

$$\int_0^1(-1)^{\left\lfloor\frac1x\right\rfloor}\,\mathrm dx=\sum_{k=1}^\infty\frac{(-1)^k}{k(k+1)}=\boxed{1-2\ln2}$$

Is this solution correct?

答:
According to this answer, we get:

$$\mathscr{S}_{-1}:=\int_0^1\left(-1\right)^{\left\lfloor\frac{1}{x}\right\rfloor}\space\text{d}x=\frac{-1+\ln\left(1-\left(-1\right)\right)-\left(-1\right)\ln\left(1-\left(-1\right)\right)}{-1}=$$
$$1-2 \ln (2)\approx-0.386294\tag1$$


问:
Algebraic closure of $\mathbb F_p$
I'm proving that $\overline{\mathbb{F}}_p = \bigcup\limits_{i=1}^{\infty} \mathbb{F}_{p^i}$ is an algebraic closure of $\mathbb{F}_p$ where $p$ is a prime. I think I've gotten down how to prove that $\overline{\mathbb{F}}_p$ is a field and that it is algebraic over $\mathbb{F}_p$.

I have some difficulties with proving that $\overline{\mathbb{F}}_p$ is algebraically closed.

My attempt is as following:

Suppose $f$ is a non-constant polynomial in $\overline{\mathbb{F}}_p [X]$. If $\overline{\mathbb{F}}_p$ contains a root of $f$, then it is algebraiclly closed. Per definition there must exist a $\mathbb{F}_{p^k}$ for a certain positive integer $k$ that contains all the coefficients of $f$. Take a root $\alpha$ of $f$ and consider the extension $\mathbb{F}_{p^k}(\alpha)$. How is this now a field of the form $\mathbb{F}_{p^l}$ for a certain positive integer $l$?

答: Take a polynomial $p(x) \in \overline{\mathbb{F}}_p [x]$. Then, $p(x) \in \mathbb{F}_{p^k}[x]$ for some $k$. The splitting field is a finite extension of characteristic $p$, so it is isomorphic to $\mathbb{F}_{p^l}$ for some $l$ (by characterisation of finite fields). Hence, $p(x)$ must split over $\overline{\mathbb{F}}_p$.

Title: Algebraic closure of $\mathbb F_p$


问:
tail bound of the deviation from sum of functions of random variables to its expectation value
I am struggling at an error propagation recently and I do not know what tools can be used in this problem. Explicitly my problem can be represented as following:

I have expression
$$\varepsilon=\frac{1}{N}\left|\sum_{i=1}^N\left[f(x_i+\epsilon_i)-f(x_i)\right]\right|$$
where $\epsilon_i$ is i.i.d with expectation $\bar{\epsilon_i}=\mu$ and variance $var(\epsilon_i)=\sigma^2$ and $f(\cdot)$ is some kind of function.

Is there any method I can find the tail bound of $\varepsilon$ for universal function $f(\cdot)$, or for some specific function, e.g: $f(x)=e^{-ax}$ or $f(x)=ax^2$?

==========================================

here is some result I found:

for $f(x)=e^{-ax}$
$$\varepsilon=\frac{1}{N}\left|\sum_{i=1}^N\left[f(x_i+\epsilon_i)-f(x_i)\right]\right|\leq \frac{1}{N}\sum_i^N\left|e^{-a\epsilon_i}-1\right|$$

since $\epsilon_i$ is i.i.d, so $X_i=\left|e^{-a\epsilon_i}-1\right|$ is also i.i.d.

use central limit theorem we can find a tail bound of $\frac{1}{N}\sum_i^N\left|e^{-a\epsilon_i}-1\right|$, so we can give a tail bound on $\varepsilon$.

but this result is somehow a little trivial because the tail bound is not zero-centric, which means $\varepsilon$ has a significant probability to be finite.

so is there other result of this problem?

答:
Suppose that $f$ is a  Lipschitz function with constant $c$. For example, $sin (bx), \frac1{1+x^2}$.

In this case $$\varepsilon=\frac{1}{N}\left|\sum_{i=1}^N\left[f(x_i+\epsilon_i)-f(x_i)\right]\right| \le \frac{1}{N}\sum_{i=1}^N|f(x_i+\epsilon_i)-f(x_i)| \le c \frac{\sum_{i=1}^N|\epsilon_i| }{N}.$$

$\frac{\sum_{i=1}^N|\epsilon_i| }{N} \to E|\epsilon_1|$ a.s. and is asymptotically normal.

As $x_i$ are all distinct and $f$ is arbitrary, then we don't have any reason to wait that errors will compensate each other.

In special case $x_1 = x_2 = \ldots$ we may put $Y_i = f(x_i+\epsilon_i)-f(x_i)$ and use CLT for $Y_i$.


问:
Prove Yor's formula not using Itô?
In a book is a exercise to prove Yor's formula for stochastic exponential, i.e.

$$\mathcal{E}(X+Y)\exp{(\langle X,Y\rangle)}=\mathcal{E}(X)\mathcal{E}(Y)$$

where $\mathcal{E}(X)_t=\exp{(X_t-\frac{1}{2}\langle X\rangle_t)}$. Now if we would define the stochastic exponential as $\exp{(X_t-\frac{1}{2}\langle X\rangle)}$, then the above is simple algebra:

$$\mathcal{E}(X)_t\mathcal{E}(Y)_t=\exp{(X_t-\frac{1}{2}\langle X\rangle_t)}\exp{(X_t-\frac{1}{2}\langle X\rangle_t)}=\exp{(X_t+Y_t-\frac{1}{2}(\langle X\rangle_t-\langle Y\rangle_t))}$$

Using that $\langle X,Y\rangle=\frac{1}{2}(\langle X+Y\rangle-\langle X\rangle -\langle Y\rangle$ we would get:

$$\exp{(X_t+Y_t-\frac{1}{2}(\langle X\rangle_t-\langle Y\rangle_t))}=\exp{(X_t+Y_t+\langle X,Y\rangle -\frac{1}{2}\langle X+Y\rangle)}=\exp{(X_t+Y_t-\frac{1}{2}\langle X+Y\rangle)}\exp{(\langle X,Y\rangle)}=\mathcal{E}(X+Y)\exp{(\langle X,Y\rangle)}$$

If I would define the stochastic exponential as the (unique) solution $Z_t$ of the SDE $Z_tdX_t=dZ_t$. Then I guess I have to use Itô, to prove the statement, but how exactly? However, if we define the stochastic exponential as above, then my conclusion would be correct?

答:
The Yor formula above is specialized to continuous processes but the actual Yor formula works for arbitrary semimartingales,
$$\mathscr{E}(X)\mathscr{E}(Y) = \mathscr{E}(X+Y+[X,Y]).$$
To derive it, it is better NOT to use an explicit expression for the stochastic exponential in terms of natural exponential (even though such expression is available also for general semimartingales, see
Doléans-Dade, C., Quelques applications de la formule de changement de variables pour les semi-martingales, Z. Wahrscheinlichkeitstheor. Verw. Geb. 16, 181-194 (1970). ZBL0194.49104.)

One simply writes out the Itô-Meyer formula for the product
$$\mathrm{d}(\mathscr{E}(X)\mathscr{E}(Y)) = \mathscr{E}(Y)_-\mathrm{d}\mathscr{E}(X) + \mathscr{E}(X)_-\mathrm{d}\mathscr{E}(Y) + \mathrm{d}\mathscr{E}(X)\mathrm{d}\mathscr{E}(Y).$$
Now plug in $\mathrm{d}\mathscr{E}(X) = \mathscr{E}(X)_-\mathrm{d}X$ (and likewise for $Y$) to obtain
$$\mathrm{d}(\mathscr{E}(X)\mathscr{E}(Y)) =
\mathscr{E}(X)_-\mathscr{E}(Y)_-(\mathrm{d}X + \mathrm{d}Y + \mathrm{d}X\mathrm{d}Y).$$
This proof can be found, for example, in Proposition 1.18 of  Eberlein and Kallsen (2019).

ContentLicense: CC BY-SA 3.0

Title: Prove Yor's formula not using Itô?


问:
Find the envelope of the family of straight lines $x \cos α + y \sin α = a \sec α$ , where $α$ is the parameter.
Ok, so here's what I have done I have divided both the sides cosα and then simplified it to

$(x-a) + y\tan \alpha - a\tan^2 \alpha = 0$

I am stuck as to what I should do here... Some help needed.

答:
Using ODE (Clairaut)

Write the family of lines as
$$y=-x\cot b+2a\csc 2b~~~(1)$$
D.w.r.t $x$, then $y'=-\cot b$ use this in (1) to eliminate $b$
$$y=xy'-a\frac{1+y'^2}{y'}~~~~(2)$$
D. w. r. $x$
$$y'=y'+xy''+a\frac{y''}{y'^2}-ay''~~~~(3)$$
$$\implies y''=0, y'=\pm \sqrt{\frac{a}{a-x}}~~~~(4)$$
Eq(2) is called Clairaut equation, were $y''=0 \implies y'=C$ will give the family of lines (1) as solution. The fixed solution will come from the other equation in (4), Solve it for $y'$ to get
The fixed curve (the envelope) can be found by solving the ODE in (4) as
$$y=\int\sqrt{\frac{a}{a-x}} dx \implies y^2=4a(a-x).$$
No constant is added in the fixed (singular) solution.


问:
Proof that non-well-orderable sets are not equinumerous to their own cardinals (using Scott's trick)
Azriel Levy's Basic Set Theory (the 2002 Dover edition) contains the following problem (Chapter 3, Exercise 2.24):

Prove that $|x|$ is an ordinal iff $|x| \approx x$, (i.e., iff $|\,|x|\,| = |x|$). If $\mathfrak{a}$ is not a well-ordered cardinal then $|\mathfrak{a}| > \mathfrak{a}$.

Hint. Use the fact that for infinite $\alpha$, $R(\alpha) \times 2 \approx R(\alpha)$, thus $R(\alpha)$ includes two disjoint
copies of itself.

$R(\alpha)$ is defined as the set of sets whose ranks are strictly less than $\alpha$.

The book uses the following definition of cardinal numbers:

The cardinal of $x$, or, synonymously, the cardinality of $x$, which we
shall denote by $|x|$, is

(a) the least ordinal $\alpha$ equinumerous to $x$, if $x$ is well-orderable, and

(b) the set of all sets $y$ of least rank which is equinumerous to $x$, otherwise.

In other words, it uses representative ordinals for well-orderable sets, and Scott's trick for non-well-orderable ones.

The part of this problem I am specifically having trouble with is proving that if $|x|$ is not an ordinal, then $|x| \not\approx x$. The two proof techniques that I can think of which seem relevant are some kind of diagonalization argument, or assuming that $|x| \approx x$ and using the relevant bijection to obtain a well-ordering of $x$. However, after spending several hours thinking about this problem, I cannot come up with a successful way of doing either.

I'm pretty sure that the hint is not relevant here (I found it useful in proving the second sentence of the problem, given the truth of the first, so I assume that's the part it was intended for).

This doesn't seem like a problem that is intended to be this difficult, so I think have probably overlooked something. Any help solving it would be greatly appreciated.

答:
Assume that $|x|$ is not an ordinal. Then the element of $|x|$ is not well-orderable, so $x$ also is. By the definition of $|x|$, every $y\in |x|$ has the same rank, let it call $\alpha$. Then $|x|\subseteq R(\alpha+1)$, and we may assume $x\subseteq R(\alpha)$.

Let $f:R(\alpha)\times R(\alpha)\to R(\alpha)$ be a bijection. Observe that for each $y\in |x|$, $y$ is a subset of $R(\alpha)$. Now consider $f^"[y\times \{a\}]\subseteq R(\alpha)$ for each $a\in R(\alpha)$. Then we have $|R(\alpha)|$ pairwise disjoint copies of elements of $|x|$.
Hence $|R(\alpha)|\le |\,|x|\,|$. Moreover, $x\subseteq R(\alpha)$ proves $|x|\le |R(\alpha)|$.

If $|x|<|R(\alpha)|$, then obviously $|x|<|\,|x|\,|$.
Now assume that $|x|=|R(\alpha)|$. Without loss of generality, assume that $x=R(\alpha)$. I claim that $R(\alpha+1)$ has $|R(\alpha+1)|=2^{|R(\alpha)|}$ copies of $R(\alpha)$. In fact, we can prove

Claim If $|X|=|X|^2$, then $\mathcal{P}(X)$ has $2^{|X|}$ copies of $X$.

Proof. Let $f: X\times X\to X$ be a bijection. Then we have a family of pairwise disjoint sets $X_a:=f^"[\{a\}\times X]$ and a family of bijections $f(a,-): X\to X_a$.

For each $A\subseteq X$, consider
$$Y_A = \bigcup_{a\in A}X_a.$$
Then $|X|\le |Y_A|= |A|\times |X|\le |X|^2=|X|$. (Proving $|Y_A|= |A|\times |X|$ requires the uniform family of bijections.)
Since $A\neq B$ implies $Y_A\neq Y_B$, we are done.

It proves $|\,|R(\alpha)|\,|=2^{|R(\alpha)|}$.


问:
Subset of $GL(n,R)$
I'm trying to understand why the subset of $GL(n,\mathbb{R})$ formed by the block-matrices of the following type:

$$\begin{pmatrix} A & B \\ 0 & C \end{pmatrix}$$ where

$$A \in GL(k,\mathbb{R}),~ C \in GL(n-k,\mathbb{R}),~ B \in M(k,n-k,\mathbb{R}).$$

is closed in $GL(n,\mathbb{R})$. I first tried to see this subset as inverse image of a closed subset of $\mathbb{R}$ via the determinant function, and then it would be closed (since $\det$ is continuous), but I couldn't do it. Can you help me?

答: First, one needs to be specific about what topology you're interested in.  I'm going to assume it is the Zariski topology (although this answer also works for the topology induced by the standard topology on $\mathbb{R}^{n^2}$).  The set $\mathrm{GL}(n,\mathbb{R})$ is considered as a subset of $\mathbb{R}^{n^2}$.  Then note that your set can be defined as the vanishing set of a set of polynomials.  Precisely, it is the vanishing set of the coordinates corresponding to the entries of the lower left block matrix.  (Note that if the lower left block is zero and the matrix is invertible, then the blocks on the diagonal must automatically be invertible, so your other conditions are automatic.)  Since the vanishing set of a set of polynomials is closed, you're done.</p>

Title: Subset of $GL(n,R)$


问:
Multivariable Chain Rule Formula doesn't make sense to me
Consider $z=f(x(t),y(t))$, then its chain rule formula is:

$\frac{dz}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt}$

What I cannot make sense of, are the terms $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$

$\frac{\partial f}{\partial x}$ is defined as the rate of change of $f$ with respect to $x$, when $y$ is held constant.

Now, in our original definition, both $x$ and $y$ depend on $t$. So if we hold $y$ constant, we'll have to hold $t$ constant, and thus $x$ will automatically be held constant. How, then, can we compute $\frac{\partial f}{\partial x}$ if we cannot hold only $y$ as constant?

答: I think the confusion comes from thinking about actually changing the values of the variables $x$ and $y$ when calculating $\frac{df}{dx}$ and $\frac{df}{dy}$. Just think of it as exploring what would happen to the function $f$ if $x$ or $y$ were to change (independently) by a small amount. Likewise for $\frac{dy}{dt}$ and $\frac{dx}{dt}$. Now, $x$ or $y$ cannot ever change independently since they both depend on t. But since calculating the derivatives, we now have information about how a small change in $t$ affects the function $f$.


问:
If rank$(T)\le$rank$(T^3)$, then intersection of range and null space of $T$ is zero
If $T:V\to V$ be linear trannsformation on a vector space $V$ with rank$(T)\le$ rank$(T^3)$. Then, what can be said about the null space and range of $T$. Typically, do they have empty intersection, or, is one the subset of another, or does there exist a subspace $W$ of $V$which equals their intersection.

I know that range of $T^3$ is a subspace of $T$ and null space of $T$ is a subspace of $T^3$. But, am unable to proceed further. Any hints. Thanks beforehand.

答:
Given that, $\operatorname{Rank}(T)\leq \operatorname{Rank}(T^3)\implies n-\operatorname{Rank}(T)\geq n-\operatorname{Rank}(T^3)$

By, Rank-nullity theorem
$$\operatorname N(T)\geq \operatorname N(T^3) \tag{1} \label{eqn1}$$
We know that,
$$\operatorname{Null space}\,(T)\subseteq \operatorname{Null space}\,(T^3) \tag{2} \label{eqn2}$$
$$\implies \operatorname N(T)\leq \operatorname N(T^3) \tag{3} \label{eqn3}$$
Now, from \eqref{eqn1} and \eqref{eqn3} we have
$$\operatorname N(T)=\operatorname N(T^3) \tag{4} \label{eqn4}$$
By \eqref{eqn2} and \eqref{eqn4} we have,
$$\operatorname{Null space}\,(T)=\operatorname{Null space}\,(T^3) \tag{5} \label{eqn5}$$
Now, let $x\in \operatorname{Null space} (T) \cap \operatorname{Range} (T)$ then, $T(x)=0$ and $\exists\,\alpha\in V$ such that, $T(\alpha)=x$
Therefore, $T^2(\alpha)=0\implies T^3(\alpha)=0$.

Now, from \eqref{eqn5} we have $\alpha\in$ Null space $(T)\implies T(\alpha)=0=x$

$\square$


问:
Angle in geometric classification algorithm
I am currently studying the textbook Learning with kernels: support vector machines, regularization, optimization and beyond by Schölkopf and Smola. Chapter 1.2 A Simple Pattern Recognition Algorithm says the following:

We are now in the position to describe a pattern recognition learning algorithm that is arguably one of the simplest possible. We make use of the structure introduced in the previous section; that is, we assume that our data are embedded into a dot product space $\mathcal{H}$. Using the dot product, we can measure distances in this space. The basic idea of the algorithm is to assign a previously unseen pattern to the class with closer mean.

We thus begin by computing the means of the two classes in feature space;
$$\mathbf{c}_+ = \dfrac{1}{m_+} \sum_{\{ i \vert y_i = +1 \}} \mathbf{x}_i, \tag{1.7}$$
$$\mathbf{c}_- = \dfrac{1}{m_-} \sum_{\{ i \vert y_i = -1 \}} \mathbf{x}_i, \tag{1.8}$$
where $m_+$ and $m_-$ are the number of examples with positive and negative labels, respectively. We assume that both classes are non-empty, thus $m_+, m_- > 0$. We assign a new point $\mathbf{x}$ to the class whose mean is closest (Figure 1.1). This geometric construction can be formulated in terms of the dot product $\langle \cdot, \cdot \rangle$. Half way between $\mathbf{c}_+$ and $\mathbf{c}_-$ lies the point $\mathbf{c} := (\mathbf{c}_+ + \mathbf{c}_-)/2$. We compute the class of $\mathbf{x}$ by checking whether the vector $\mathbf{x} - \mathbf{c}$ connecting $\mathbf{c}$ to $\mathbf{x}$ encloses an angle smaller than $\pi / 2$ with the vector $\mathbf{w} := \mathbf{c}_+ - \mathbf{c}_-$ connecting the class means.

It is this part that I'm confused by:

We compute the class of $\mathbf{x}$ by checking whether the vector $\mathbf{x} - \mathbf{c}$ connecting $\mathbf{c}$ to $\mathbf{x}$ encloses an angle smaller than $\pi / 2$ with the vector $\mathbf{w} := \mathbf{c}_+ - \mathbf{c}_-$ connecting the class means.

The angle denoted in Figure 1.1, as far as I can tell, is not the angle "enclosed" by $\mathbf{x} - \mathbf{c}$ and $\mathbf{w}$, and is actually the opposite angle. Am I misunderstanding this, or does this seem to be an error in the image?

The figure is also available in this research paper (Figure 1) by the same authors.

答:
Converting comments to an answer ...

OP has confirmed misinterpreting the angle marked with the dot as the angle described as being "enclosed" by the vectors. However, that dot-mark, an alternative to the "square mark", is meant to indicate that the dashed line/hyperplane is perpendicular to vector $w$. This provides a visual reference for comparing the described angle (determined by $x$ and $w)$ to $\pi/2$.

The "enclosed" angle is in fact the one bounded by the thick arrowheads. Writing $X$, $C$, $C_+$ for the ends of vectors $x$, $c$, $c_+$, this is the angle $\angle XCC_+$.


问:
$3.6$ Example of $0$ Chapter of Do Carmo's Riemannian Geometry
Pictures below are from the Do Carmo's Riemannian Geometry. He want to show $i$ is a embedding. But only at $U$, $i$ is  homeomorphism since $j^{-1}\circ i \circ x=x$. This do not mean $i$ is homeomorphism on $S$. Since in my view, the embedding means that $i:S\rightarrow i(S)$ is homeomorphism. However, only local homeomorphism is proved in picture below. So, this proof don't prove the $i$ is embedding.  How to deal it ?

答:
Stating that the inclusion $i \colon S \longrightarrow \mathbb{R}^3$ is a homeomorphism onto its image is the same as saying that, considering the inclusion now as being the map $i \colon S \longrightarrow i(S) = S \subset \mathbb{R}^3$, we get a homeomorphism. But that is obvious, since now it becomes the identity on $S$ and $S$ is equipped with the same topology (the subspace topology) as both the domain and the image of $i$ .

So, actually, the fact that $i$ is a homeomorphism onto its image does not "follow from a)", but it is trivial.

Hope this helps :).


问:
Exercise of interior of a closed ball
Let $E$ a normed spaces and $a\in E$. How to prove that $$[\overline{B}(a, r)]^{\circ}\subseteq B(a, r)$$ where $\overline{B}(a, r)=\{x\in E: \|x-a\|\leq r\}$ and  $B(a, r)=\{x\in E: \|x-a\|< r\}$ </p>

答: Let $x$ be an interior element of the closed ball. Then there is some $\delta > 0$ so that $B(x, \delta)$ lies in the closed ball.  Every element of $B(x, \delta)$ lies in the open ball. You can use the triangle inequality to see this.</p>

Title: Exercise of interior of a closed ball


问:
Covering space of a topological group is itself a topological group
Let $p:\widetilde{G}\to G$ be a covering, where $G$ is a (locally path-connected and path-connected) topological group with neutral element $e$. Let $\widetilde{e}\in p^{-1}(\{e\})$. I have to show that $\widetilde{G}$ can be made to a topological group with neutral element $\widetilde{e}$ in a unique way.

I allready have show using the lifting property, that there exists a unique continuous map $\widetilde{\mu}:\widetilde{G}\times \widetilde{G}\to \widetilde{G}$ satisfying $\widetilde{\mu}(x,\widetilde{e})=\mu(\widetilde{e},x)=x$ for all $x\in\widetilde{G}$, as well as $p(\widetilde{\mu}(x,y))=\mu (p(x),p(y))$ for all $x,y\in \widetilde{G}$, where $\mu$ denoted the multiplication of $G$. But how to show associativity?

Using the associativity of $\mu$ and the lifting property, I was also able to show that

$$p(\widetilde{\mu}(x,\widetilde{\mu}(y,z)))=p(\widetilde{\mu}(\widetilde{\mu}(x,y),z))$$

for all $x,y,z\in\widetilde{G}$, but how can I conclude associativity?

答: Consider the maps $\alpha,\beta\colon\widetilde G^3\longrightarrow\widetilde G$ defined by$$\alpha(g_1,g_2,g_3)=\widetilde\mu\left(\widetilde\mu(g_1,g_2),g_3\right)\quad\text{and}\quad\beta(g_1,g_2,g_3)=\widetilde\mu\left(g_1,\widetilde\mu(g_2,g_3)\right).$$You want to prove that they are equal. Note that, if $g_1,g_2,g_3\in\widetilde G$, then\begin{align}p\left(\alpha(g_1,g_2,g_3)\right)&=p\left(\widetilde\mu\left(\widetilde\mu(g_1,g_2),g_3\right)\right)\\&=p\left(\widetilde\mu(g_1,g_2)\right)\cdot p(g_3)\\&=p(g_1)\cdot p(g_2)\cdot p(g_3)\\&=p(g_1)\cdot p\left(\widetilde\mu(g_2,g_3)\right)\\&=p\left(\widetilde\mu\left(g_1,\widetilde\mu(g_2,g_3)\right)\right)\\&=p\left(\beta(g_1,g_2,g_3)\right).\end{align}In other words, $p\circ\alpha=p\circ\beta$. Since, furthermore, $\alpha\left(\widetilde e,\widetilde e,\widetilde e\right)=\beta\left(\widetilde e,\widetilde e,\widetilde e\right)=\widetilde e$, $\alpha=\beta$.


问:
Obscuring squares of Rubik's cube
This is a combinatorial question related to Rubik's cube $3\times3\times3$ (and, in the end, its generalizations $n\times n\times n$). I assume that the readers are familiar with this puzzle. Let's recall some basic terminology and conventions to rule out possible ambiguities. In the following, we use the words state and configuration interchangeably, as exact synonyms.

We are only interested in configurations of Rubik's cube where it has the shape of a cube (so, we consider rotations of its layers by an angle of $90^\circ$ as atomic transformations, and no incomplete rotations are allowed). Also, we consider configurations that can be obtained from one another by solid rotations of the cube as a whole (without any relative movements of its parts) as identical.

The cube has $6$ faces, each one of them is a $3\times3$ grid of squares — $9$ squares per face, and $54$ squares in total. Each square is of a solid color. Orientation of every single square is irrelevant (in fact, this rule has significance only for central squares). There are $6$ different colors, and there are $9$ squares of each color. There is a single selected state where each face consists of all squares of the same color — this state is called the solved state. A state that can be reached from the solved state by a sequence of valid rotations of layers is called a valid state (the solved state, of course, is a valid state too).

As a side note, if we allowed disassembling the cube into its parts and reassemling them, then we would get the number of possible states $12$ times larger than the set of valid states (this larger set would be a disjoint union of $12$ equivalence classes under valid rotations, each of the classes called an orbit). We are only interested in valid states in the following.

Usually, it is assumed that colors of all squares on the cube are visible. Let's consider a possibility that some squares may be obscured (e.g. completely covered by an opaque colorless sticker), so that their colors are not visible.

In general, obscuring some squares can result in some states becoming visually indistinguishable. As trivial examples, obscuring all $54$ squares makes all states indistinguishable, but obscuring only $1$ square (no matter which one) does not make any states indistinguishable.

What maximal number of squares in Rubik's cube $3\times3\times3$ can be obscured without making any states indistinguishable?

What maximal number of squares in Rubik's cube $2\times 2\times 2$ can be obscured without making any states indistinguishable?

What maximal number of squares in Rubik's cube $n\times n\times n$ can be obscured without making any states indistinguishable? Can we find a general formula, recurrence relation, etc. to compute it for every $n\in\mathbb N$?

答:
format of terms used in this answer:

removal = obscuring (hiding a sticker)

piece = mini-cube

face = side of a piece =equiv sticker

$n$ = one 1-dimensional face unit

side = whole face of the rubik's cube

cube = entire rubik's cube

state = configuration

type = which form of piece (corner-proper, edge-proper, or edgeless  face)

c = corner_piece (starts with $3C_v$ = three visible faces)

e = edge-without-corner_piece (starts with $2E_v$)

f = face-without-edge_piece (starts with $1F_v$)

$C$ = an individual c face

$C_r$ = a C removed

$C_v$ = a visible C

$C_A$ = all c faces combined

${C_A}_v$ = all visible c faces

$E_A$ = all e faces combined

$F_A$ = all f faces combined

$T_v$ = all visible faces combined ($C_v$+$E_v$+$F_v$)

$X_A$ = all deducible faces

$({X_A})_n$ = arbitrary # deducible faces for $n^3$ cube

(⇒ solve $∀x=r:$ max$[{(X_A)}_n]$ = ${(T_A)}_n$-min$[{(T_v)}_n]$)

This question (well, three questions to the same end) boils down to two considerations: What is minimal stickers required to know what the solved state is supposed to look like, and minimal stickers needed to retain the identities of them all. Considering the first part without the second, as few as two diagonally opposite corner pieces would suffice ($2c$, establishing every color's/side's correct alignment to every other). Of course, this defies the second necessity, since not only the other 6 corner pieces' but all the other pieces' identities would become indistinguishable. Considering the second part (which mostly if not wholly absorbs the first), one can afford one piece of each type getting stripped of its 1, 2, and 3 stickers ($6$ so far). Of course there is a lowerbound of $n^2$ (one whole face, 9n for a 3×3×3 cube), but we can do better than that for the n=3 cube (edit bolded for emphasis). For n≥[some integer ≥4], the general maximal formula will include n^2; this formula might be able to fit for smaller n values despite that the particular algorithms for determining the actual faces to remove from smallish differs from what the formula implies. Consider n=50 cube.

To what extent depends upon which variation of 2^2=4 possible ways to interpret the problem as originally presented: do we know correct color-scheme alignment (medium outside information, reasonable supposition since already know how to play the puzzlegame and have access to internet, and confirmed in response to Vepir's question); do we know a particular pattern to how the faces were removed (heftier outside information, mostly absorbing preceding knowledge); do we know both of these info (heftiest outside information); or de we know none of these info (minimum outside information).

I will approach this problem assuming minimal outside information to start, because considering the difference affordable from there upward seems easier to me, and I am curious how the up to four different general formulas would differ.

Since any valid state with all stickers visible can be deduced as such (by solving it, even if achieving that from a random state might be computationally pricey), they convey the same information; so one sensible way to think of this problem is as subtracting stickers from the solved state. Besides the straightforward description, the response to Vepir's question confirming knowledge of the color scheme may and likely will increase the maximal count by allowing for more removals of non-corner edge pieces. To start, however, I will ignore this fact. Another possible consideration for n>3 (and possibly even n<4) is if you can presuppose knowledge of pattern to which faces you hid. When I get to the part of formulating a general pattern, I will assume initially that the obfuscated cube is to be decipherable by a stranger.

Any piece within a rubik's cube of of side n length 3 or greater can be separated into three disjoint categories: corner [3 different colors], edge without corner [2 different colors], and face without either [1 color]. In the case of 2×2×2 cube, each type shares the same category (is not disjoint). The number of total faces (visible or not) as a function of integer n is equal to 6⋅(n^2). These might be helpful later in constructing a general formula.

Focusing now on the 3×3×3 cube, with all 8⋅3=24 corner stickers removed you could still ascertain what state$_{[solved]}$ should look like by using $c_a$=${(c_a)}_v$ to track a course around the cube. But that would also make those 8 c indistinguishable, so 8⋅2=16 removals allowed. Remove the remaining face from one of the c making it blank and 1 from a single f = 16+2=18 removals allowed (lower bound), since these single "extra" oddball removals from two different types of the three does not by itself prevent deduction of solved state. That leaves the 10 untouched e and 7 one-sticker c pieces to consider. Well, at least one e could be totally removed, so 18+1=19 removals lower_bound for the 3×3×3 cube.

Here is a good place to pause to calculate the formulas for the different types of pieces (by how many original faces contained) as equations based on n.

corners: 8 (constant); $c_n$=8. non-corner edges: as few as 0, for n={1,2}. For every n beyond that, it increases by 10; $e_n$=10(n-2). non-edge/corner faces: all the rest. For n={1,2}, 0. For every n beyond that, it increases by faces times square of two less than n: $f_n$=6(n-2)^2. Any beyond 1 from this category becomes indistinguishable with its only sticker removed, so +1 should be the only effect to the general formula from this type.

Back to optimizing the 3×3×3 cube. None of the remaining 7 C could get removed, since that would make those c indistinguishable. That leaves the nine remaining e to focus on, or possibly all ten if it can ensure a greater maximum (past lowerbound of 20 or alternative removals). Focusing on the 9 is simpler than 10 and likely the correct path so I shall. Every c piece serves for this purpose also as an e piece, but with all of them out for use except for one serving now as an f piece, the remaining e pieces are the only pieces showing more than 1 side/color. The question now is how many of the eight two-side relationships can be removed from the cube before the full identities of the removed (non-corner) ones cannot be deduced. I believe the answer is 4: two on opposite sides of and sharing a side, the other two on reverse side diagonal from the first 2, allowing for an unobstructed route linking all six sides (contiguous, since every side is contiguous). One of these is accounted for already. So long as the other absent face of that piece is considered (rather than preserving it still on) when deciding the remaining 3 to remove, which I believe means choosing the same "obverse" side (as opposed to the opposite "reverse" side) as the first two mentioned (proving this is another task), it should work out. That results in 19+3 = 22 removals from the 3×3×3 cube's 54 faces without knowing the color scheme. If you know the color scheme, then one whole side could be "removed" (possibly better), so long doesn't cost an existing deduction, which I think it wouldn't if you did so unto the remaining "reverse" side: 22+2 = 24 removals from the 3×3×3 with prior knowledge of the color scheme.

This brings up an important consideration for a general formula. While for n<4 a majority of the removals came from corner pieces, as n increases the corner pieces remains constant whereas the edge pieces increases linearly. As such, for higher n values the majority of faux "data-hiding" will be unto non-corner edge pieces, in alternating half-visible fashion. To formulate a general equation, at first I will suppose no prior knowledge of the color scheme nor methodology to the removals. Considering for a moment the particular case of a 4×4×4, there are 20⋅2=40 E, compared to 8⋅3 = 24 C. Since the corner-pieces reveal more information about the whole cube (particularly without prior info), retaining most of them in favor removing all of the non-corner edgepiece stickers likely yields higher value of lossless data-compression. I'm not sure yet how to formulate this into a general equation without if/else-style verbiage included, though it is probably possible. I will come up with a general formula and post it as a comment or update this post, but for now I'll leave with a summary of some relationships to n-width to consider:

corner stickers = ${(C_A)}_n = 3{c_a}  = 3⋅8$

edge stickers = ${(E_A)}_n = 2{(e_a)}_{n-2} = 2[10(n-2)]$

face stickers = ${(F_A)}_n = 1{(f_a)}_{n-2} = 1[6(n-2)^2]$

total stickers = ${(T_A)}_n = 6*{(F_A)}_n = 6⋅n^2$

Considering lower n^3 cubes (to help figure out a general pattern and answer second question):

For n=1, no prior knowledge, 1 face could get removed. With prior knowledge of correct scheme, presumably all except two = 4  of the six could get removed.

For n=2, no prior knowledge, trickier. $T_2$ = ${(C_A)}_2$ = 24. A maximum bound entails linking all sides, from 3 C with 2 c each ⇒ max$[{(X_A)}_2]$ = ${(T_A)}_2$-min$[{(T_v)}_2$] = max$[{(X_A)}_2]$ = $24-(2⋅3)$ = 18. For n=2 with prior knowledge of color scheme: ..

For n=4, no prior knowledge: min$[{(E_v)}_4]$ = min$[{(e_v)}_4]$÷2-1 =20÷2-1=9 ⇒ max$[{(E_r)}_4]$=${(E_A)}_4$-9 = 40-9 =31. max$[{(F_r)}_4]$=1. max$[{C_r}_4]$=8⋅2+1=17. Assuming no contradictions (noting  thatn=4 is even, n=5 is odd,..) max${(X_A)}_4$=${(T_r)}_4$ = max$[{C_r}_4]$+max$[{(E_r)}_4]$+max$[{(F_r)}_n]$ = 17+31+1 = {49 = upperbound}.

$∀n∈N:$upperbound$[{(X_A)}_n]$ = max$[{(C_r)}_n]$+max$[{(E_r)}_n]$+max$[{(F_r)}_n]$,

⇒ max$[{(X_A)}_n] ≤ 17+(½{(E_A)}_n+1)+1$

⇒ max$[{(X_A)}_n] ≤ 17+1+(½(2[10(n-2)])+1)$

⇒ max$[{(X_A)}_n] ≤ (10n-20+1)+18$

⇒ max$[{(X_A)}_n] $?$=$?$ 10n-2$ doesn't fit; too high

but $[{(X_A)}_n]=10n-9$ seems pretty close for n<4

large n value: n=50

lowerbound[max$[{(X_A)}_n]$ = ${f_n}^2+3C_v+2E_v$ = $n^2+3+2$

lowerbound[max$[{(X_A)}_50]$ = $50^2+5$: one whole side removed plus one whole opposite corner and one whole middle edge piece (these latter 5 removals might not all apply, but are guaranteed in a simple quadratic lowerbound). So what other faces can get removed (and still get ascertained by a stranger familiar with the rules of the puzzle but not the color scheme used)?

Title: Obscuring squares of Rubik's cube


问:
how det(A)=0 implies that the solution isn't unique?
Solution of matrix equation Ax=b, where
$$ A=\left(\begin{matrix}
a_1&a_2&\dots&a_n
\end{matrix}\right), \ a_i \in \mathbb{R}^n,$$

is not unique, if vectors $$ a_1, \ a_2, \dots, \ a_n $$
are linearly dependent. Then by properties of determinant, $$ \det A=0. $$
However, does it always follow, that if det A = 0, column vectors of A are linearly dependent? Can someone present a proof?

答:
One possible proof:

Suppose the columns are linearly independent.

Convert the matrix into a column echelon form, starting from the last column and working your way backwards.

You know the number of linearly independent columns is the number of nonzero columns you end up with. However, as you've assumed the columns are independent, there are no zero columns.

In other words, you've ended up with a triangular matrix with all nonzero elements on the diagonal. Its determinant is nonzero.

However, the elementary transformations we use when converting the matrix into a row/column echelon form don't change the property of the diagonal to be zero or nonzero.

Thus, the determinant was nonzero to start with.


问:
Show that $|μ|(C)=\sup_{\substack{f\in C_b\\|f|\le1\\\left.f\right|_{C^c}=0}}\mu f$ for all open $C$
Let $E$ be a metric space and $\mu$ be a finite signed measure on $\mathcal B(E)$.

I would like to show that$^1$ $$|\mu|(C)=\sup_{\substack{f\in C_b(E)\\|f|\le1\\\left.f\right|_{E\setminus C}=0}}\mu f\;\;\;\text{for all open }C\subseteq E\tag1.$$ Moreover, I would like to know whether there is a similar identity for $|\mu|(A)$ and closed $A\subseteq E$.

Note that "$\le$" in $(1)$ is clearly trivial. The other inequality should follow from the following two results:

Lemma 1: If $\nu$ is a finite signed measure on $\mathcal B(E)$, $B\in\mathcal B(E)$ and $\varepsilon>0$, then $$|\nu|(C\setminus A)<\varepsilon\tag2$$ for some closed $A\subseteq E$ and open $C\subseteq E$ with $A\subseteq B\subseteq C$.

Lemma 2: If $\emptyset\subset A\subseteq E$ and $C\subset E$ with $\overline A\subseteq C^\circ$, then there is a $f\in C_b(E)$ with $0\le1 f\le 1$ and \begin{align}\left.f\right|_A&=1\tag{3a}\\\left.f\right|_{E\setminus C}&=0.\tag{3b}\end{align}

Now let $C\subseteq E$ be open and $\varepsilon>0$. Let $E^{\pm}$ be a Hahn decomposition of $E$ wrt $\mu$ and $C^\pm:=C\cap E^\pm$. By Lemma 1, applied to both measures in the Jordan decomposition $\mu^\pm$ of $\mu$ individually, $$\pm\mu(C^\pm\setminus A^\pm)=\mu^\pm(C\setminus A)<\varepsilon\tag4$$ for some closed $A^\pm\subseteq E$ with $A^\pm\subseteq C^\pm\subseteq C$. (We may need to exclude the cases $A^\pm=\emptyset$ and $C^\pm=E$.)

Now, by Lemma 2, there is a $f^\pm\in C_b(E)$ with $0\le f^\pm\le 1$ and \begin{align}\left.f^\pm\right|_{A^\pm}&=1\tag{5a}\\\left.f^\pm\right|_{E\setminus C^\pm}&=0.\tag{5b}\end{align} Let $$g:=f^+-f^-.$$ Note that $g\in C_b(E)$ with $|g|\le1$ and \begin{align}\left.g\right|_{A^\pm}&=\pm1\tag{6a}\\\left.g\right|_{E\setminus C}&=0.\tag{6b}\end{align}

Question 1: We are done if we can show that $\mu g>|\mu|(C)-2\varepsilon$. However, for some reason, I'm not able to obtain this inequality. How do we need to split the integral $$\mu g=\int_Cg\:{\rm d}\mu\tag7$$ in order to obtain this result?

Question 2: Can we find a similar identity for $|\mu|(A)$ and closed $A\subseteq E$?

$^1$ $C_b(E)$ denotes the space of bounded continuous functions and $\mu f:=\int f\:{\rm d}\mu$.

答:
Consider a Hahn decomposition $E = E^{+}\sqcup E^{-}$. For any set, denote $C^{\pm} = C \cap E^{\pm}$. Now, we have
$|\mu|(C)= \mu(C^{+}) - \mu(C^{-})$.  Now take $A^{\pm} \subset C^{\pm}$,
, such that $|\mu|(C^{\pm}\backslash A^{\pm}) < \epsilon$ ( we assume some regularity here). Consider a continuous function $f^{\pm} $ defined on $E$ with values in $[0,1]$, and taking value $1$ on $A^{\pm}$, and $0$ on $A^{\mp}$. Now consider $ g= f^+- f^-$. Multiply it with a function with values in $[0,1]$, taking value $0$ on $E\backslash C$, and $1$ on $A^+\cup A^-$. So may assume from the beginning that $g$ is zero outside $C$. Now, we can see that the integral $\mu(g)$ is close to $|\mu|(C)$.

$\bf{Added:}$ We have
$$||\mu|(C)- \int_C g d\mu |= |\int_{C^+\backslash A} (1-g) d \mu + \int_{C^{-} \backslash B} (-1 - g) d\mu| \le 2 \epsilon + 2 \epsilon = 4\epsilon$$

Note: we can do this also for complex measures, or say for measures with values in some finite dimensional Hilbert space $E$. We then have to consider $\mu(f)$ for $f$ continuous with values in $E$ and $\|f\|\le 1$.  Instead of considering the Hahn decomposition, we consider the vector density $\rho$ (Radon-Nikodym) of $\mu$ with respect to $\|\mu\|$, which is a measurable $E$-valued function of constant absolute value $1$. Now we have to piece-wise interpolate this $\rho$, like above, but with more values.

$\bf{Added:}$

Some calculations about limits of measures
Consider a net of measures $\mu_t$ converging weakly to $\mu$. Then we have
$$\lim\inf_t |\mu_t|(U) \ge |\mu|(U)$$ for every $U$ open.

Assume that we have moreover
$$|\mu_t|(E)|\to |\mu|(E)$$

Consider $A = E\backslash U$ a closed set.

We have
$$\lim\sup_t |\mu_t|(A) = \lim \sup (|\mu_t|(E) - |\mu|(U)) =|\mu|(E) - \lim \sup |\mu_t|(U) \le |\mu|(E) - |\mu|(U)=|\mu|(A)$$

Conversely, if we have the inequality for the closed set $E$, since $E$ is also open, we must have $|\mu_t|(E)\to |\mu|(E)$.

However, we may have a strict inequality for some sequences.

Consider $E=[-1,1]$, $\phi\colon \mathbb{R} \to \mathbb{R}$ support in $[1-,1]$ negative on $(-\infty,0]$, positive on $[0\infty)$ and with $\int_{-1}^1 \phi(x) d x = 1$.  Then the sequence of measures $\mu_n(x) = n \phi_n(nx) dx$ converge to the delta measure $\delta_0$, but $|\mu_n| \to C\cdot \delta_0$, with $C = \int_{-1}^1|\phi(x)|dx> 1$


答:
Since you know that $p(\widetilde{\mu}(x,\widetilde{\mu}(y,z)))=p(\widetilde{\mu}(\widetilde{\mu}(x,y),z))$ holds, the underlying functions differ only by an automorphism of the cover, i. e. $$\widetilde{\mu}(\cdot,\widetilde{\mu}(\cdot,\cdot))=\widetilde{\mu}(\widetilde{\mu}(\cdot,\cdot),\cdot) \circ \phi,$$
where $\phi$ is an automorphism on the cover $\widetilde{G}$.
However since they agree on the identity element $\widetilde{e}$, said automorphism has to be the identity and we get $$\widetilde{\mu}(x,\widetilde{\mu}(y,z))=\widetilde{\mu}(\widetilde{\mu}(x,y),z)$$

Title: Covering space of a topological group is itself a topological group

Body:
Let $p:\widetilde{G}\to G$ be a covering, where $G$ is a (locally path-connected and path-connected) topological group with neutral element $e$. Let $\widetilde{e}\in p^{-1}(\{e\})$. I have to show that $\widetilde{G}$ can be made to a topological group with neutral element $\widetilde{e}$ in a unique way.

I allready have show using the lifting property, that there exists a unique continuous map $\widetilde{\mu}:\widetilde{G}\times \widetilde{G}\to \widetilde{G}$ satisfying $\widetilde{\mu}(x,\widetilde{e})=\mu(\widetilde{e},x)=x$ for all $x\in\widetilde{G}$, as well as $p(\widetilde{\mu}(x,y))=\mu (p(x),p(y))$ for all $x,y\in \widetilde{G}$, where $\mu$ denoted the multiplication of $G$. But how to show associativity?

Using the associativity of $\mu$ and the lifting property, I was also able to show that

$$p(\widetilde{\mu}(x,\widetilde{\mu}(y,z)))=p(\widetilde{\mu}(\widetilde{\mu}(x,y),z))$$

for all $x,y,z\in\widetilde{G}$, but how can I conclude associativity?


问:
How to find the order of the group of rigid motions of platonic solids in $\mathbb{R}^3$?
The following appear as exercises in Dummit and Foote's Algebra (Section $1.2$ - Dihedral Groups):

Let $G$ be the group of rigid motions in $\mathbb{R}^3$ of a tetrahedron. Show that $|G| = 12$

Let $G$ be the group of rigid motions in $\mathbb{R}^3$ of a cube. Show that $|G| = 24$

Let $G$ be the group of rigid motions in $\mathbb{R}^3$ of an octahedron. Show that $|G| = 24$

Let $G$ be the group of rigid motions in $\mathbb{R}^3$ of a dodecahedron. Show that $|G| = 60$

Let $G$ be the group of rigid motions in $\mathbb{R}^3$ of an icosahedron. Show that $|G| = 60$

From this answer I figured that rigid motions are orientation-preserving isometries, i.e. reflections are not allowed.

So, for a tetrahedron, I thought of symmetry axes passing through a vertex and the centroid of the opposite face. There are four such axes (let's call them $A,B,C,D$). Along every axis, we can define $1_i, r_i, r_i^2$ as three rotations with $r_i^3= 1$, the identity element ($i=A,B,C,D$). Since there are four such axes, $|G| = 3\times 4 = 12$. Is this fine, or am I missing something? I'm slightly concerned about the fact that $1_A,1_B,1_C,1_D$ may all possibly be the same (since they are identity transformations), and that I am over-counting?

Minor question (detour): Are identity transformations corresponding to different axes different, or the same?

For the cube, I did the following:

For every pair of opposite faces, we have an axis of symmetry. There are $3$ such pairs, hence $3$ such axes (say $A,B,C,D$). About each axis we define $1,r_i,r_i^2,r_i^3$ with $r_i^4 = 1$ where $i=A,B,C,D$.

There are four body diagonals (say $E,F,G,H$), and about each diagonal (symmetry axis) we define $1,r_j,r_j^2$ with $r_j^3= 1$ where $j=E,F,G,H$.

In view of the above calculations, we have $|G| = 3\times 4 + 4\times 3 = 24$.

Using this method becomes difficult hereafter, for larger solids. It is not easy to identify all the axes of symmetry by hand. Moreover, the only group I have learned about in some detail at this point is $D_{2n}$, so please do not give solutions such as "the required group $G$ is isomorphic to a known and well-studied group $X$, and we know $|X| = ?$ so $|G| = ?$"

I think it boils down to having a good way to count all the distinct rigid motions. Could someone help me with this?

I came across James Ha's solutions here, but I don't understand how the solutions presented in the PDF are equivalent to mine even for the tetrahedron and cube cases. It'd be nice if someone can help me see the equivalence, and also tell me how to proceed with the other platonic solids! Thanks a lot!

答:
For polyhedra in $3$ space you can show that an edge $a$ can be taken to  anoother edge $b$ by  $2$   orientation-preserving transformation of the solid (get one, and then can also rotate around $b$) . If you consider all transformations, then there are $4$ such transformations.transformations.

Therefore, $|G_{+}(S)| = 2 e$, $|G(S)|= 4 e$, where $e$ is the number of edges of $S$.


问:
Is there a Regular Submanifold of $\Bbb R^3$ whose Singular Homology group has non trivial Torsion
Studying Algebraic Topology, I learned that Singular Homology (over de integers) it's supposed to represent the number of "holes" the space has. The main examples are
$$H(S^1,\Bbb Z)=(\Bbb Z,\Bbb Z,0,0,...) \quad \quad H(S^2,\Bbb Z)=(\Bbb Z,0,\Bbb Z,0,0,...) \quad \quad H(S_1 \times S_1)=(\Bbb Z,\Bbb Z^2,\Bbb Z,0,0,...)$$
And they all (kind of) make sense. Now, the first wierd example is the Proyective Space with
$$H(\Bbb P^2(\Bbb R),\Bbb Z)=(\Bbb Z,\, \Bbb Z/2 \Bbb Z \, ,\Bbb Z,0,0,...)$$
Which makes sense, and at the same time it doesn't. What I find "wierd" it's that there is a (differential) embedding of $\Bbb P^2(\Bbb R)$ into $\Bbb R^4$ so there is a reglular submanifold of $\Bbb R^4$ with a...  $\Bbb Z/2 \Bbb Z$-hole?

In my goal of finding intuition of what on earth could that mean, I started wondering if it exists a regular differential submanifold of $\Bbb R^3$ who has an singular homology group with torsion. In that case, I could visualize how a $\Bbb Z/n \Bbb Z$-hole looks like. So the question is, does such a Regular Sub-Manifold of $\Bbb R^3$ exists?

By Regular Sub-Manifold of $\Bbb R^3$, I mean a $C^{\infty}$ real manifold contained in $\Bbb R^3$ with the subspace topology and an inmersive inclusion. The manifold could have boundary and need not to be compact nor orientable.

The Reason I'm interested in $C^{\infty}$ real manifold is because I find them more intuitive and less pathological than an arbitrary Topological subspace of $\Bbb R^3$. I want the Manifold to be a sub-manifold of $\Bbb R^3$ so I can visualize it.

答:
For the first question regarding the intuition behind elements of homology groups that admits finite order. This link is certainly helpful.

For your second question in the comment, I'll answer by parts under the assumption of finitely generated homology groups.

What about compact $2$ (resp. $3$) dimensional manifolds with boundary?

Suppose $X$ is any compact $2$-manifold or $3$-manifold that can be embedded in $\Bbb{R}^3$. Then Alexander Duality tells us that
$$H^2(X)\cong \tilde{H}_0(S^3-X;\Bbb{Z})$$
which must be torsion-free. Then,
by UCT (Universal Coefficient Theorem),
$$ H^2(X)\cong\operatorname{Hom}(H_2(X);\Bbb{Z})\oplus\operatorname{Ext}(H_1(X);\Bbb{Z})\\
\implies H_1(X) \text{ torsion-free}
$$
Besides, one may notice that if $\dim X=2$ and $X$ non-compact, we still have torsion-free $H_1$ by the same argument. (The only difference is that $H_2(X)\cong 0$ in that case, but it doesn't matter)

What about those cases when $X$ is a non-compact $3$ manifold? (This is the intriguing part of your question.)

Since your goal of finding such manifold is to understand the torsion of homology groups, I'll show that "normal" examples don't satisfy your criteria.

For simplicity, assume that $X=S^3-Y$, where $Y$ is a finite simplicial complex, then by Alexander Duality, we can still obtain the usual isomorphism relationship: $H_1(S^3-Y)\cong H^1(Y)$. Then by UCT,
$$H^1(Y)\cong\operatorname{Hom}(H_1(Y);\Bbb{Z})\cong\operatorname{Hom}(\Bbb{Z}^\alpha\oplus T;\Bbb{Z})\cong \Bbb{Z}^\alpha$$
which is torsion-free. This result actually eliminate most common $3$ dimensional non-compact manifolds without boundary. Probably, there are some ugly spaces that admits torsion in their first homology group, but I think they wouldn't be suitable for visualizing elements of finite orders...

Besides the clear answer in the link, I'd like to discuss the cycles with finite order in terms of cellular structure (at least it's imaginable). During the process of constructing $\Bbb{R}P^2$, we often attach a $2$-cell via $f:\partial D^2\to S^1\to\Bbb{R}P^{n-1}$ (fill the circle with a disk and identify antipodes on $S^1$). Alternatively, this can be viewed as gluing the boundary of the disk to a circle via a degree $2$ map, which wraps around $S^1$ twice. By definition, a $1$-cycle must have an image $\partial(1\text{-cycle})=e^0-e^0=0$. In particular, only multiples of $e^1$ ($1$-cell) can be candidates.

For $m$ odd, $\partial(me^1)=0$ and $\not\exists n\in\Bbb{Z}$ s.t. $\partial(ne^2)=me^1$ since going around $S^1$ one time only goes half way around the only $2$-cell, which is not an element of $\Bbb{Z}$.

For $m$ even, then we see that it wraps around the $2$-cell $\frac{m}{2}\in\Bbb{Z}$ times because of the attaching map (degree $2$). Hence it is homologous to $0\implies$ triviality.

You may also connect this to the deformation of closed paths.

I hope this response answers your question. :)


问:
Help for solving first-order differential equations with two unknown functions
I am familiar with first-order differential equations and how to solve them, but only for the case when there is a single unknown function.

For the following case there are two unknown functions and I'm struggling to determine where to even begin. The two equations:

$$ f'(t) = - Af(t)+ Bg(t) \tag{1}$$
$$ g'(t) = - Bg(t)+ Af(t) \tag{2} $$

Where f and g are functions of the variable t, and A & B are constants. The initial conditions are: $f(0) = 0$ and $g(0) = 1$.

I already have knowledge of the answer for f(t), which is:

$$ f(t) = \frac{B}{A+B}(1-e^{-(A+B)t})  \tag{3} $$

I would really like to understand how to tackle questions of this type: where there are two unknown functions. I would attempt to show my working out but I don't know where to start.

Therefore, I would like to ask if anyone could please provide a hint or suggestion for me, something which I can work off. If necessary, I will update my post with original working. Thank you for your time.

答:
You can also sum both equations and get:
$$f'+g'=0 \implies f+g=c$$
With initial conditions we get:
$$c=f(0)+g(0)=1$$
Then use this in the first DE and you have a first order DE:
$$f'=-Af+Bg=-Af+B(1-f)=-(A+1)f+B$$
This equation is separable and can be easily solved.


问:
Meaning of the notation
It should be a very simple question but I can find nowhere being precise on this: what is the meaning of the notation
\begin{equation}
\int_{\mathbb{R}^n}|\nabla f(x)|dx
\end{equation}
(not mentioning the $\int|Df(x)|dx$ notation).

Is it the same as
$$
\int_{\mathbb{R}^n}\sum_{i=1}^n\left|\frac{\partial}{\partial x^i}f(x)\right|dx
$$
I met this when I tried to read some books on total variation, but obviously the authors assumed readers already knew this convention.

答:
The notation $|\nabla f(x)|$ will be "the" norm, one could say the length, of the vector $\nabla f(x) = (\frac{\partial}{\partial x_1}f(x), \dots, \frac{\partial}{\partial x_n}f(x) ) $.

The most common way to define "the" norm of a vector is the Euclidean norm,
that is, $|(y_1, \dots, y_n)| = \sqrt{y_1^2+\dots +y_n^2}$, which corresponds to our usual geometric idea of length,  in your case would mean
$$\sqrt{\sum_{i=1}^n(\frac{\partial}{\partial x_i}f(x))^2}$$
There are however other ways to define the norm of a vector, for example what you wrote would be commonly referred to as the $1$-norm of the vector.

If there is no further commentary, it ought to be what I said initially.

OwnerDisplayName: Matcha

Title: Meaning of the notation


问:
The convergence or divergence of the power series at a point does not determine whether the function can or cannot be continued beyond that point

$f(z)=\frac{1}{1-z}=\sum\limits^{\infty}_{n=0}z^n\text{ for }\mid z\mid<1$.

Although the power series diverges at every point on the unit circle,
$f$ is analytic throughout the punctured plane $z\neq 1$.

How do I show it diverges at every point on unit circle ? How do I see the analyticity? Also it seems to contradict to the definition of "analyticity" because "analytic at a point" means we have a power series at that point with a positive radius of convergence, which is impossible (cause beyond $\mid z\mid=1$, the power series diverges)

$\sum\limits^{\infty}_{n=1}(z^n/n^2)$ converges at all points on the
unit circle; however $g(z)$ cannot be continued analytically to a
domain including $z=1$ since

$g''(z)=\sum\limits^{\infty}_{n=0}\frac{(n+1)z^n}{n+2}\to \infty\text{
> as }z\to 1^-$

I see the convergence on the unit circle. But how do we see the analyticity and why is second derivative relevant? I'd appreciate any insight.

答:
If $|z|=1$, then you don't have $\lim_{n\to\infty}z^n=0$ (since $(\forall n\in\Bbb N):|z^n|=1$) and therefore the series $\sum_{n=0}^\infty z^n$ diverges. And $f$ is analytic since it is the quotient of two analytic functions.

Concerning $g$, if you could expand it to analytically to a domain including $1$, the same would happen to $g''$. But then you would have $\lim_{z\to1}g''(z)=g''(1)$.


问:
Infinite lists in Lambda calculus....
I have been learning Haskell for a number of months now and am now trying to understand the underpinning $\lambda$-calculus,  but I have run into a bit of a mental block! How would I go about writing a $λ$-term corresponding to an infinite list (for example  $[0, 1, 2, \dots]$)?

答:
The infinite list $[0, 1, 2, \dots]$ satisfies the following identity in Haskell (see here):
$$\tag{1}
[0,1,2, \dots] = 0 : \mathtt{map} \, (+1) \, [0,1,2,\dots]
$$
where $(+1)$ is the function computing the successor of an integer, and $:$ is the function that appends an element to the front of a list.
Indeed, $$\mathtt{map} \, (+1) \, [0,1,2,\dots] = [(+1) \, 0, (+1) \, 1, (+1) \, 2, \dots] = [1, 2, 3, \dots]$$ and clearly
$$0 : [1, 2, 3, \dots] = [0,1, 2, \dots]$$
therefore, Identity $(1)$ holds.

Can Identity $(1)$ be expressed in the $\lambda$-calculus? Yes!
Indeed:

The natural number $0$ is represented in the $\lambda$-calculus by the Church numeral $\underline{0} = \lambda f. \lambda x. x$ (see here);

The function $(+1)$ is represented in the $\lambda$-calculus by the $\lambda$-term $\mathsf{succ} = \lambda n. \lambda f. \lambda x. f(n f x)$ (see here);

The function $:$ is represented in the $\lambda$-calculus by the $\lambda$-term $\mathsf{cons} = \lambda x. \lambda l. \lambda c. \lambda n. c x (l c n)$ (see here);

The function $\mathtt{map}$ is represented in the $\lambda$-calculus by the $\lambda$-term $\mathsf{map} = \lambda f. \lambda l. \lambda c. l \, \lambda x. c (f x)$ (see here, unfortunately the author of the question has inexplicably deleted it, I hope he/she will undelete it).

Apparently, the only missing $\lambda$-term to represent $(1)$ in the $\lambda$-calculus is $[0,1,2,\dots]$, that is, the $\lambda$-term that we want to define! Let us denote it by $X$, as an unknown variable in a mathematical equation. Thus, $(1)$ can be rewritten as an equation in the $\lambda$-calculus
$$\tag{2}
X =_\beta \mathsf{cons} \, \underline{0} \, (\mathsf{map} \, \mathsf{succ} \, X)$$
where $=_\beta$ is $\beta$-equivalence, that is (roughly), the unoriented version of $\beta$-reduction.
If we can solve Equation $(2)$ (i.e. we find a $\lambda$-term $X$ such that $(2)$ holds), then such a $\lambda$-term represents the infinite list $[0,1,2,\dots]$.
Since $(\lambda x.N)x \to_\beta N$ for any $\lambda$-term $N$, Equation $(2)$ can be equivalently rewritten as
$$\tag{3}
X =_\beta (\lambda x. \mathsf{cons} \, \underline{0} \, (\mathsf{map} \, \mathsf{succ} \, x)) X
$$
Equation $(3)$ is of interest because it says that the $\lambda$-term $X$ we are looking for is a fixed point of the $\lambda$-term $\lambda x. \mathsf{cons} \, \underline{0} \, (\mathsf{map} \, \mathsf{succ} \, x)$.
This is where the fixpoint combinator $Y$ of the $\lambda$-calculus becomes relevant, as correctly suggested by @MJD in his comment.
Indeed, $Y = \lambda f. (\lambda x. f(xx))(\lambda x. f(xx))$ is a "magic" $\lambda$-term such that, for every $\lambda$-term $M$,
$$\tag{4}
YM =_\beta  M(YM)$$
that is, $YM$ is a fixed point of any $\lambda$-term $M$.
Who cares? From Identity $(4)$, replace the generic $M$ with the $\lambda$-term $\lambda x. \mathsf{cons} \, \underline{0} \, (\mathsf{map} \, \mathsf{succ} \, x)$, so the following identity holds:
$$
Y(\lambda x. \mathsf{cons} \, \underline{0} \, (\mathsf{map} \, \mathsf{succ} \, x)) =_\beta (\lambda x. \mathsf{cons} \, \underline{0} \, (\mathsf{map} \, \mathsf{succ} \, x)) \big( Y (\lambda x. \mathsf{cons} \, \underline{0} \, (\mathsf{map} \, \mathsf{succ} \, x))\big)
$$
This shows that $X = Y (\lambda x. \mathsf{cons} \, \underline{0} \, (\mathsf{map} \, \mathsf{succ} \, x))$ is a solution of Equation $(3)$: just rewrite $(3)$ by replacing $X$ with $Y (\lambda x. \mathsf{cons} \, \underline{0} \, (\mathsf{map} \, \mathsf{succ} \, x))$.
And a solution of Equation $(3)$, and hence of Equation $(2)$, is what we are looking for!

Summing up, the infinite list $[0,1,2,\dots]$ is represented in the $\lambda$-calculus by the $\lambda$-term:
$$
Y (\lambda x. \mathsf{cons} \, \underline{0} \, (\mathsf{map} \, \mathsf{succ} \, x))$$

Digression. There are different ways to define the $\lambda$-terms $\mathsf{cons}$ and $\mathsf{map}$, depending on the way lists are encoded in the $\lambda$-calculus.

According to the encoding of lists presented here, a list $[N_1, \dots, N_k]$ is represented by the $\lambda$-term $$\lambda c. \lambda n.c N_1 (c N_2 (\dots(c N_k n)\dots))$$
and the definitions of $\mathsf{cons}$ and $\mathsf{map}$ above work for this encoding.

An alternative encoding of lists is presented here, it requires other (just slightly different) definitions of $\mathsf{cons}$ and $\mathsf{map}$.
Note that $Y (\lambda x. \mathsf{cons} \, \underline{0} \, (\mathsf{map} \, \mathsf{succ} \, x))$ always represents the infinite list $[0, 1, 2,\dots]$, provided that $\mathsf{cons}$ and $\mathsf{map}$ are suitable for the encoding of lists under consideration.

Title: Infinite lists in Lambda calculus....


问:
Finding all pairs $(m,n)$ of positive integers such that $m(n + 1) + n(m − 1) = 2013$

Find all pairs $(m, n)$ of positive integers such that
$$m(n + 1) + n(m − 1) = 2013$$

What I've done: I tried to make the problem simpler by $2mn+m-n$ and factoring it equals $(2m-1)(2n+1)$

Where do I go now?

答:
We have $$2mn+m-n=2013\implies4mn+2m-2n-1=4025$$
$$\implies(2m-1)(2n+1)=4025=5^2\times7\times23=1^0\times5^2\times7^1\times23^1$$
Hence there are $1\times3\times2\times2 =12$ factors that each bracket could correspond to so you have to check one bracket for each factor of $4025$, and the other bracket for each corresponding factor as you go along.

For example letting
$$2m-1=1, ~~2n+1=4025$$
yields $m=1,~n=2012$ and so on for all the other pairs of factors. Note, you don't have to look for values of $m$ and $n$ for $2n+1=1$ as that would result in a negative value for $n$ which isn't allowed by the question.

If you don't understand something, or you need any more help, please don't hesitate to ask :)


答:
In general we have $\ker T\subset \ker T^3$, so the hypothesis is equivalent to
$\ker T = \ker T^3$, which is equivalent to $\ker T \supset \ker T^3$. This is equivalent to:

if $T^3 v = 0$ then $Tv = 0$, that is : if $T^2 (Tv) = 0$ then $Tv = 0$, so equivalent to
$$\ker T^2 \cap \operatorname{Im} T = (0)$$

Title: If rank$(T)\le$rank$(T^3)$, then intersection of range and null space of $T$ is zero

Body:
If $T:V\to V$ be linear trannsformation on a vector space $V$ with rank$(T)\le$ rank$(T^3)$. Then, what can be said about the null space and range of $T$. Typically, do they have empty intersection, or, is one the subset of another, or does there exist a subspace $W$ of $V$which equals their intersection.

I know that range of $T^3$ is a subspace of $T$ and null space of $T$ is a subspace of $T^3$. But, am unable to proceed further. Any hints. Thanks beforehand.


问:
Minimum degree of $P(P(x)) - Q(Q(x))$
Let $P$ and $Q$ be distinct polynomials of degree $n$, with real coefficients. What is the minimum possible degree of $P(P(x)) - Q(Q(x))$?

(Treat conventionally the degree of the zero polynomial to be $0$.)

For $n$ odd because of $P(x) = x^n$ and $Q(x) = -x^n$ the answer is $0$. By bashing (i.e. writing $P(x) = ax^2 + bx + c$, $Q(x) = mx^2 + nx + p$ and expanding $P(P(x)) - Q(Q(x))$) we see that the degree must be at least $2$. Any idea for an approach for general $n = 2k$, possibly without too much expanding (but more like considering complex roots or something else clever)?

Any help appreciated!

答:
(This is too long for a comment so I submitted an answer instead)

Note 1:

For even $n$:

Let $P(x)=\sum^{n}_{i=0}a_ix^i$ and $Q(x)=\sum^{n}_{i=0}b_ix^i$. Then by Multinomial Theorem:

$$P(P(x))=\sum^{n}_{i=0}a_i\left(\sum^{n}_{j=0}a_jx^j\right)^i=\sum^{n}_{i=0}a_i\sum_{k_1+k_2+\cdots+k_n=i}\binom n{k_1,k_2,\cdots,k_n}\prod^n_{j=1}\left(a_jx^j\right)^{k_j}=\sum^{n}_{i=0}a_i\sum_{k_1+k_2+\cdots+k_n=i}\binom n{k_1,k_2,\cdots,k_n}\prod^n_{j=1}a_j^{k_j}x^{j\cdot k_j}=\sum^{n}_{i=0}a_i\sum_{k_1+k_2+\cdots+k_n=i}\binom n{k_1,k_2,\cdots,k_n}\prod^n_{j=1}\left(a_jx^j\right)^{k_j}=\sum^{n}_{i=0}\sum_{k_1+k_2+\cdots+k_n=i}a_i\binom n{k_1,k_2,\cdots,k_n}\prod^n_{j=1}a_j^{k_j}\cdot x^{\sum^{n}_{j=1}j\cdot k_j}$$

Similarly,

$$Q(Q(x))=\sum^{n}_{i=0}\sum_{k_1+k_2+\cdots+k_n=i}b_i\binom n{k_1,k_2,\cdots,k_n}\prod^n_{j=1}b_j^{k_j}\cdot x^{\sum^{n}_{j=1}j\cdot k_j}$$

Thus:

$$P(P(x))-Q(Q(x))=\sum^{n}_{i=0}\sum_{k_1+k_2+\cdots+k_n=i}a_i\binom n{k_1,k_2,\cdots,k_n}\prod^n_{j=1}a_j^{k_j}\cdot x^{\sum^{n}_{j=1}j\cdot k_j}-\sum^{n}_{i=0}\sum_{k_1+k_2+\cdots+k_n=i}b_i\binom n{k_1,k_2,\cdots,k_n}\prod^n_{j=1}b_j^{k_j}\cdot x^{\sum^{n}_{j=1}j\cdot k_j}$$

Consider the coefficient for $x^{n^2}$ of $P(P(x))-Q(Q(x))$. To first find that, we need to find a combination of nonnegative integers $k_1, k_2,\cdots,k_n$ such that $\sum^{n}_{j=1}j\cdot k_j=n^2$ and $\sum^{n}_{j=1}k_j\leq n$. The only case when this happens is when $k_1=k_2=k_3=\cdots=k_{n-1}=0, k_n=n$. Then $\sum^{n}_{j=1}j\cdot k_j=n^2$, $i=n$ and $a_i\binom n{k_1,k_2,\cdots,k_n}\prod^n_{j=1}a_j^{k_j}\cdot x^{\sum^{n}_{j=1}j\cdot k_j}=a_n\binom n{0,0,\cdots,n}a_n^nx^{n^2}=a_n^{n+1}x^{n^2}$, so the coefficient for $x^{n^2}$ of $P(P(x))$ is $a_n^{n+1}$. Similarly, the coefficient for $x^{n^2}$ of $-Q(Q(x))$ is $-b_n^{n+1}$, so the coefficient for $x^{n^2}$ of $P(P(x))-Q(Q(x))$ is $a_n^{n+1}-b_n^{n+1}$. We want this to be $0$ (which would imply that $P(P(x))-Q(Q(x))$ has degree less than or equal to $n^2-1$), so we want $a_n^{n+1}-b_n^{n+1}=0$. (Presumably) the coefficients of the polynomials must be real, so we have $a_n=b_n$.

Note 2:

Suppose $Q(x)=P(x)+c$ and $P(x)=\sum^{n}_{i=0}a_ix^i$. Then:

$$Q(Q(x))=P(P(x)+c)+c=c+\sum^{n}_{i=0}a_i(P(x)+c)^i=c+\sum^{n}_{i=0}\sum^{i}_{j=0}a_i\binom ijc^jP^{n-j}(x)=c+\sum^{n}_{i=0}P^i(x)c^{n-i}\sum^{i}_{j=0}a_{n-j}\binom {n-j}{n-i}$$

So:

$$P(P(x))-Q(Q(x))=\sum^{n}_{i=0}a_iP^i(x)-c-\sum^{n}_{i=0}P^i(x)c^{n-i}\sum^{i}_{j=0}a_{n-j}\binom {n-j}{n-i}=\sum^{n}_{i=0}\left(a_i-c^{n-i}\sum^{i}_{j=0}a_{n-j}\binom {n-j}{n-i}\right)P^i(x)-c$$

Note that $P^i(x)$ has degree $in$.

Presumably we can find $c$ and $a_1,a_2,\cdots,a_n$ such that $a_i-c^{n-i}\sum^{i}_{j=0}a_{n-j}\binom {n-j}{n-i}=0$ for as many values of $i$ as possible, but I'm not sure.

Title: Minimum degree of $P(P(x)) - Q(Q(x))$


问:
A problem on Angle bisectors of a triangle and perpendiculars.

From $A$ perpendiculars $AX, AY$ are drawn to the bisectors of the exterior angles of $B$ and $C$ of $\Delta ABC$. Prove that $XY || BC$.

Prove that the feet of the four perpendiculars dropped from a vertex of a triangle upon the four bisectors of the other two angle are collinear.

In triangle $\Delta ABC, X $and $Y$ be the feet of perpendiculars from vertex $A$ to the internal angle bisector of $∠B$ and $∠C$ respectively. Line $XY$ meets $AB$ at $P$ and $AC$ at $Q$. If $AB = 7$ cm, $BC = 8$ cm and $CA = 5$ cm then find $PQ$ and $XY$.

These were 3 questions with almost similar configurations; as I wasn't able to solve the first one, I couldn't solve the next two either (I don't know if it's just psychological or something but anyway).

So, could someone please provide the SOLUTION for the first one and just hints for the next two

A solution using just elementary geometry would be ideal but any solution would be really helpful,

Thanks

答:
The dashed lines are angle bisectors. Using angle chasing, congruency and mid-point theorem, bunch of isosceles triangles and various lengths can be found. I'll leave the work to you.

$RS$ contains the midpoints of $AB, AC$. The final results should be :

$$RS = \dfrac{1}{2}A_3A_4 = s$$

$$PQ = \dfrac{1}{2}A_1A_2 = \dfrac{1}{2}(CA_2+BA_1 - BC) = s-a$$


问:
Pairwise independence implies X+Y is independent from Z
Let $X$, $Y$ and $Z$ be three pairwise independent random variables.

Is it true that $(X, Y)$ is independent from $Z$?

Is it true that
$X+Y$ is independent from $Z$?

I know that this holds for mutually independent random variables but can't prove it for the pairwise independent case.

答:
Consider e.g. $(X,Y,Z)$ uniform over $\{(0,0,0),(0,1,1),(1,0,1),(1,1,0)\}$. We can easily see that $(X,Y,Z)$ are pairwise independent.

But knowing $(X,Y)$ or $X+Y$ immediately determines the value of $Z$.

Some motivation for this example: $X$, $Y$ are independent coin flips, and $Z=X\oplus Y$ is XOR. Knowing one of the bits doesn't affect the distribution of any one of the other bits. But knowing two of the bits actually determines the value of the other bit.


问:
Reference request for the constructive algorithm regarding the Kronecker-Weber theorem.
Given an abelian extension $L/\mathbb{Q}$, such that $L=\mathbb{Q}(\alpha)$ for a root $\alpha$ of an irreducible polynomial $f(x)$. By Kronecker-Weber theorem we know that there exists $m \in \mathbb{N}$, such that $L \subseteq \mathbb{Q}(\zeta_m)$. How can we find explicitly the least integer $m \in \mathbb{N}$, with this property? Could you please introduce me to a reference for this?

I know that my question is very short, but I am very curious about this algorithm after this comment. (Also, this question was very informative for me.)

答:
If $K/\Bbb{Q}$ is abelian then $K\subset \Bbb{Q}(\zeta_n)$. Clearly the $p$ ramified in $K$ must divide $n$.

Assuming that $K\subset \Bbb{Q}(\zeta_{rp^l}),p\nmid r$

Since $\Bbb{Q}(\zeta_{rp^l})/\Bbb{Q}(\zeta_r)$ is cyclic and totally ramified at $p$, then $K\subset \Bbb{Q}(\zeta_{rp^{l-1}})$ iff $$e(p,K/\Bbb{Q})=e(p,K(\zeta_r)/\Bbb{Q}(\zeta_r))= [K(\zeta_r):\Bbb{Q}(\zeta_r)] \le [\Bbb{Q}(\zeta_{rp^{l-1}}):\Bbb{Q}(\zeta_r)]= \phi(p^{l-1})$$

(there is a subtlety for $p=2$ here as $\Bbb{Q}(\zeta_8)/\Bbb{Q}$ is not cyclic)

Whence $$K\subset \Bbb{Q}(\zeta_n), n = \prod_p p^{c_p}, \qquad {c_p = \inf \{ l,e(p,K/\Bbb{Q})\ |\ \phi(p^l)\}, \\ c_2=\ldots}$$ is the least possible $n$.

(the given expression for $c_p$ is only valid for $p$ odd, we need a special case for $p=2$ as $\Bbb{Q}(\sqrt2)$ has ramification index $2$ at $p=2$ but is contained in $\Bbb{Q}(\zeta_8)$ not $\Bbb{Q}(i)$.


问:
$a^b = c$, is it possible to express $b$ without logarithms?
$ a^b = c $

is it possible to express b without logarithms?

答: If $e^b=c$, then $b=\ln c$.  Is it possible to express $\ln c$ without logarithms?  Not unless you have a peculiar definition of "without"...</p>


问:
Finding the Degree of a Polynomial, as a function of the Polynomial
Recently, I was wondering about the following question:

Given a polynomial $P(x)$ with real coefficients, express its degree $d$ as a function of only $P(x)$, i.e. $d(P(x))$. Only elementary functions/operations and values of $P$ (or its derivatives/anti-derivatives) at any value may be used.

There must be several creative ways to do this. After some experimentation, I came up with the following:

$$d(P(x)) = \sum_{i=1}^{\infty} \text{sgn}\left(\sum_{j=0}^{\infty} \left| \frac{d^{i+j} P}{dx^{i+j}} (0) \right| \right)$$

Are there simpler functions that work? Any inputs are welcome.

答: You can define the degree of a polynomial $P(x)$ to be the smallest non-negative integer $n$ such that $\lim\limits_{x \to +\infty} \frac{P(x)}{x^n}$ exists.


问:
Can a graph have both a Hamiltonian cycle and path at the same time?
For example, in a complete undirected graph there exists a Hamiltonian cycle, so can we say that this graph has a Hamiltonian path as well?

Or whenever we have a cycle we can't have a path?

Thanks

答: Yes, your first argument is correct. Suppose $G$ is a simple undirected graph with vertex set $V(G) = \{v_1,...,v_n\}$ and WLOG, say $v_1v_2...v_nv_1$ is a Hamiltonian cycle. Then, clearly, $v_1v_2...v_n$ is a Hamiltonian path. So, whenever a graph has a Hamiltonian cycle, it has a Hamiltonian path. Converse may not be true however.


问:
How can $\frac{1+1+1+\ldots}{2+2+2+\ldots} be \frac{(1+1)+(1+1)+\ldots}{2+2+2+\ldots} = 1$?
$\frac{1}{2} = \frac{1+1+1+\ldots}{2+2+2+\ldots} = \frac{(1+1)+(1+1)+\ldots}{2+2+2+\ldots} =  1$ says user Karolis Juodelė.
Why does it get 7 votes up for this comment? I guess that 1/2 is not 1. See: Whats infinity divided by infinity?</p>

答: I upvoted that comment because it appeared to me that it was a very clear and succinct demonstration of exactly why we can't handle expressions like $$\frac{1+1+1+\cdots}{2+2+2+\cdots}$$ the same way we can handle finite expressions.  The original question asked why $\infty\over\infty$ is not simply taken to be 1; Karolis's answer shows why we do not.


问:
Show that a series defines a function continuous at all $x≠2k$
I've been given the following statement to prove:

If $\{a_k\}_{k=1}^{\infty}$ is monotonic and $$\lim_{k\to \infty}{a_k}=0$$ Then the series $$\sum_{k=1}^{\infty}{a_k\sin(kx)}$$ Defines a function continuous for all $x\neq 2\pi k, \hspace{1mm}k\in \mathbb{Z}.$

My strategy would be to use the fact that $$\left|\sum_{k=1}^{n}{\sin(kx)}\right|\leq \frac{2}{|\sin(x/2)|}$$ and given any $x_0\neq2\pi k$, I could construct an intervall $I_{x_0}$ around $x_0$ such that $\frac{2}{|\sin(x/2)|}$ would be bounded on $I_{x_0}$. Then the series would converge uniformly on $I_{x_0}$ by Dirichlet's test and since all functions $a_k\sin(kx)$ are continuous at $x_0$, the series would also be continuous at $x_0$

My main question is if the construction of these intervalls $I_{x_0}$ can legitimately be used to prove what I'm trying to prove? Is it not the case that I need to show that the series converges uniformly on the entire set $\mathbb{R}-\{2\pi k\}_{k\in \mathbb{Z}}$ before I can claim that it is continuous on it? I sense however that it isnt uniformly convergent on that set.

Thanks in advance.

答:
My main question is if the construction of these intervalls $I_{x_0}$ can legitimately be used to prove what I'm trying to prove?

Yes, and that's the natural way to prove it.

Is it not the case that I need to show that the series converges uniformly on the entire set $\mathbb{R}-\{2\pi k\}_{k\in \mathbb{Z}}$ before I can claim that it is continuous on it? I sense however that it isnt uniformly convergent on that set.

You sense rightly in general. Since the terms of the series are continuous functions, if the convergence is uniform on $\mathbb{R} \setminus 2\pi\mathbb{Z}$, then it is in fact uniform on $\mathbb{R}$. Of course this can happen (e.g. if $\sum \lvert a_k\rvert < +\infty$), but typically the sum function will have discontinuities at the points of $2\pi\mathbb{Z}$, so the convergence cannot be uniform on all of $\mathbb{R}$.

However, continuity is a local property, whether a function $f$ is continuous at a point $x_0$ depends only on the values of $f$ on an arbitrarily small neighbourhood of $x_0$. And thus, if a sequence or series of continuous functions converges uniformly on some neighbourhood of $x_0$, however small that neighbourhood is, the limit function is continuous at $x_0$.

That is precisely what your argument with the $I_{x_0}$ gives you, locally uniform convergence on $\mathbb{R}\setminus 2\pi\mathbb{Z}$, and consequently the continuity of
$$\sum_{k = 1}^{\infty} a_k\sin (kx)$$
on that set.


问:
Calculating that determinant is non zero with modulo 2
we have the determinant:

$$
\begin{vmatrix}
6&1&1\\
4&-2&5\\
2&8&7\\
\end{vmatrix}
$$
A stumbled upon this question: Is the following matrix invertible?

If I apply the modulo 2 procedure to this matrix, I get a zero determinant, which is not true. Why doesn't that procedure work in this case?

答:
The modulo $2$ idea does not work because this determinant, albeit nonzero, is divisible by $2$. (In fact, it is $-306$.) If something is nonzero modulo $2$, it is certainly nonzero, but if something is zero modulo $2$, this only means that it is even but doesn't mean it is zero.

You can still apply the same technique, but first divide the first column by $2$: you get the determinant:

$$\begin{vmatrix}3&1&1\\2&-2&5\\1&8&7\end{vmatrix}$$

which is, modulo $2$:

$$\begin{vmatrix}1&1&1\\0&0&1\\1&0&1\end{vmatrix}=1$$

so this new determinant is nonzero, and so the original determinant (which is twice the new determinant) must be nonzero as well.


答: In the given link, we can conclude that the matrix is invertible since its determinant is $1$ modulo $2$, which cannot be $0$, clearly. However, this procedure cannot be used to show that matrix is not invertible, since if it tells you that determinant is $0$ modulo $2$ the determinant still can be an even number other than $0$.

Title: Calculating that determinant is non zero with modulo 2

Body:
we have the determinant:

$$
\begin{vmatrix}
6&1&1\\
4&-2&5\\
2&8&7\\
\end{vmatrix}
$$
A stumbled upon this question: Is the following matrix invertible?

If I apply the modulo 2 procedure to this matrix, I get a zero determinant, which is not true. Why doesn't that procedure work in this case?


问:
Show that a trapezoid is cyclic if and only if it is isosceles

Show that a trapezoid is cyclic if and only if it is isosceles.

I'm new to geometry and only studied the basics, this problem appered in a chapter about Cyclic Quadrilaterals. Could you give me a hint or solution?

答:
A trapezoid always has one pair of parallel sides.

(image not to scale)

For isosceles $\implies$ cyclic, "isosceles" means $\angle C = \angle D$. By the property of co-interior angles, $\angle A + \angle D = 180º, \angle B + \angle C = 180º$. Write each angle in terms of one angle (say angle $D$), and add all the angles up.

For cyclic $\implies$ isosceles, by the definition of "cyclic", $\angle A + \angle C = \angle B + \angle D = 180º$. However, $\angle A + \angle D = \angle B + \angle C = 180º$ again. Now equate the two statements to get $\angle A + \angle C = \angle A + \angle D$, and the conclusion follows.


问:
Is there any way to tell if any givin Number will either blow up or tend towards a Value when squared?
Ok so as a Set Up, i want to do the all time classy, the Mandelbrot set. As far as i know, the equation

$$Z \rightarrow Z²+C$$

Just asks the question "Ok givin any Complex Number, if you Square the Non Complex terme and then Add the Complex compnent, is this new number something REALLY huge or does it tend towards a Value ?

Of course, normaly this Equation is based on Iterations so you do that a bunch of times and bam, either your Value is huge for small.

But, there is sort of a Problem for me. At the moment, my code spits out this:Image

I dont want to be a downer but that doest really look right. My guess is that the Problem is how huge the values get once you square them a few times. And since Computers are pretty dumb, at some point Values might switch from Positive to negative and so on.

So, to the question, please tell me there is a fancy Algorithm that can just tell me if Z will be huge or small D:

Thanks !

答:
The 'normal' Mandelbrot set is derived by plotting the values of $c \in \mathbb C$ such that the sequence $z_{n+1} = z_n^2 + c, z_0 = 0$ remains bounded.

You can verify the sequence diverges when either (a) $|c| > 2$, or, (b), when $|c| \leqslant 2$, and at some stage $|z_n| > 2$.  This allows you to determine when to stop testing the $z_n$.

I don't believe there is any shortcut other than simply trying repeatedly to check on which $c$ led to divergence or not.

Using standard double arithmetic you can create reasonable images as illustrated (1)

or (2)