问:
Contour integration to evaluate a real-valued integral
I am evaluating this integral:

$$\int_{-\infty}^{\infty} \frac{\sin x}{x(x^2+1)^2}\,dx$$

with the formula

$$\int_{-\infty}^{\infty} f(x) \sin(sx) dx = 2\pi \sum\text{Re } \text{Res}[f(z) e^{isz}]$$

where the sum is over the residues in upper half plane.

So since the only two singularities that are inside the upper half plane are at $z = 0$ and $z=i$, I found that

$$\begin{align}
2\pi \sum\text{Re } \text{Res}\left(f(z) e^{isz}\right) &= 2 \pi \left(\text{Re }  \text{Res}_{z= 0}\left[\frac{1}{z (z^2+1)^2} e^{i z}\right]
+ \text{Re } \text{Res}_{z= i}\left[\frac{1}{z (z^2+1)^2} e^{i z}\right]\right) \\\\
  &=2 \pi \left(1 + \frac{-3}{4e}\right)
\end{align}$$

I am pretty sure that I calculated the two residues correctly, since in mathematica

```<code>Residue[E^(I z)/(z (z^2 + 1)^2), {z, 0}]
</code>```

is $1$ and

```<code>Residue[E^(I z)/(z (z^2 + 1)^2), {z, I}]
</code>```

is $\frac{-3}{4e}$

But evaluating the integral

```<code>Integrate[Sin[x]/(x (x^2 + 1)^2), {x, -Infinity, Infinity}]
</code>```

mathematica gives $\pi - \frac{3 \pi}{2e}$.

I am wondering if this is because I did something wrong somewhere or if it is because mathematica gives the wrong answer.

Thank you!

答: Since $z=0$ is a single pole of $f(z)$ on the boundary of the upper semicircle, it should be multiplied in $\pi$ rather than $2\pi$.


问:
Is this approach to the limit solution kosher?$ \lim_{n\rightarrow\infty}\frac{(n!)^{\tfrac{1}{n}}}{n}=\frac{1}{e}$
Here is a non-rigourous approach to a limit:
$$ \lim_{n\rightarrow\infty}\frac{(n!)^{\tfrac{1}{n}}}{n}=\frac{1}{e}$$
I know the solution is correct. I'm suspicious of my methods.

Rewrite the expression:

$$ \lim_{n\rightarrow\infty}\left(\frac{n!}{n^n}\right)^{\frac{1}{n}}$$

I know the $\lim_{n\rightarrow\infty}\left(\frac{n!}{n^n}\right)=0$. So bear with me.

$$\frac{n!}{n^n}=\frac{n(n-1)(n-2)...[n-(n-2)][n-(n-1)]}{n^n}$$

Written this way, the numerator is a polynomial of degree $n$. For large $n$, the leading term will prevail. I will write that polynomial as a sum of $n^n$ and a polynomial of degree $n-1$.

$$\frac{n!}{n^n}\Rightarrow\frac{n^n+[\text{polynomial}]^{n-1}}{n^n}\Rightarrow 1+\frac{1}{n}$$

As noted in Step 2, this is not the expected result for this limit. But watch:

$$ \lim_{n\rightarrow\infty}\frac{(n!)^{\tfrac{1}{n}}}{n}=\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{\tfrac{1}{n}}=\lim_{n\rightarrow\infty}\frac{1}{\left(1+\frac{1}{n}\right)^n}=\frac{1}{e}$$

This is the correct solution. But Step 2 is sketchy. Did I make another error which undid this error? Can this derivation be salvaged, or is my proof "not even wrong"?

Please be gentle with me.

答:
This can be considered an heuristic way to proceed, to be rigorous we can use the ratio-root criterion as explained here

Why does $\;\lim_{n\to \infty }\frac{n}{n!^{1/n}}=e$?

As noticed there is a mistake in your heuristic derivation, we could try to proceed as follows

$$\left(\frac{n!}{n^n}\right)^n =1 \cdot \left(1-\frac1n\right)^n\cdot \ldots \cdot\left(1-\frac{n-1}n\right)^n \sim \frac1{e^0}\cdot\frac1{e^1}\cdot \ldots \cdot\frac1{e^{n-1}}=\frac1{e^{\frac{n(n-1)}2}}$$

but we would conclude that

$$\left(\frac{n!}{n^n}\right)^\frac1n \sim \frac1{e^{\frac{n(n-1)}{2n^2}}} \implies \lim_{n\rightarrow\infty}\frac{(n!)^{\tfrac{1}{n}}}{n}=\frac1{\sqrt e}$$

ContentLicense: CC BY-SA 4.0


答:
The inclusion of the residue at $z=0$ is not correct.  Rather, we begin by writing

$$\int_{-\infty}^\infty \frac{\sin(x)}{x(x^2+1)^2}\,dx=\text{Im}\left(\text{PV}\int_{-\infty}^\infty \frac{e^{ix}}{x(x^2+1)^2}\,dx\right)$$

where the Cauchy Principal Value is given by

$$\text{PV}\int_{-\infty}^\infty \frac{e^{ix}}{x(x^2+1)^2}\,dx=\lim_{\varepsilon\to 0^+}\int_{|x|>\varepsilon}\frac{e^{ix}}{x(x^2+1)^2}\,dx$$

Next, we move to the complex plane.  Ler $R>1$, $\varepsilon>0$, and $C$ be the contour in the upper-half plane that is comprised as $(i)$ the straight line paths from $-R$ to $-\varepsilon$ and from $\varepsilon$ to $R$, $(ii)$ the semi-circular arc centered at $z=0$ with radius $\varepsilon$ from $-\varepsilon$ to $\varepsilon$, and $(iii)$ the semi-circular arc centered at $z=0$ with radius $R$ from $R$ to $-R$.  Note that $z=0$ is excluded from the interior region bounded by $C$.

Then, we have can write

$$\begin{align}
\oint_{C}\frac{e^{iz}}{z(z^2+1)^2}\,dz&=\text{PV}\int_{-\infty}^\infty \frac{e^{ix}}{x(x^2+1)^2}\,dx\\\\
&+\int_{\pi}^0 \frac{e^{i\varepsilon e^{i\phi}}}{\varepsilon e^{i\phi}((\varepsilon e^{i\phi})^2+1)^2}\,i\varepsilon e^{i\phi}\,d\phi\\\\
&+\int_0^{\pi} \frac{e^{iR e^{i\phi}}}{R e^{i\phi}((R e^{i\phi})^2+1)^2}\,iR e^{i\phi}\,d\phi\tag1
\end{align}$$

As $R\to \infty$, the last integral on the right-hand side of $(1)$ approaches $0$.

As $\varepsilon\to0^+$, the second integral on the right-hand side of $(1)$ approaches $-i\pi$.

Since $C$ has excluded the $z=0$, the only residue implicated is at $z=i$.  Hence, we find

$$\text{PV}\int_{-\infty}^\infty \frac{e^{ix}}{x(x^2+1)^2}\,dx=i\pi +\text{Res}\left(\frac{e^{iz}}{z(z^2+1)^2}, z=i\right)\tag2$$

Now, calculate the residue at $z=i$ and take the imaginary part of both sides of $(2)$.  Can you finish now?

Title: Contour integration to evaluate a real-valued integral

Body:
I am evaluating this integral:

$$\int_{-\infty}^{\infty} \frac{\sin x}{x(x^2+1)^2}\,dx$$

with the formula

$$\int_{-\infty}^{\infty} f(x) \sin(sx) dx = 2\pi \sum\text{Re } \text{Res}[f(z) e^{isz}]$$

where the sum is over the residues in upper half plane.

So since the only two singularities that are inside the upper half plane are at $z = 0$ and $z=i$, I found that

$$\begin{align}
2\pi \sum\text{Re } \text{Res}\left(f(z) e^{isz}\right) &= 2 \pi \left(\text{Re }  \text{Res}_{z= 0}\left[\frac{1}{z (z^2+1)^2} e^{i z}\right]
+ \text{Re } \text{Res}_{z= i}\left[\frac{1}{z (z^2+1)^2} e^{i z}\right]\right) \\\\
  &=2 \pi \left(1 + \frac{-3}{4e}\right)
\end{align}$$

I am pretty sure that I calculated the two residues correctly, since in mathematica

```<code>Residue[E^(I z)/(z (z^2 + 1)^2), {z, 0}]
</code>```

is $1$ and

```<code>Residue[E^(I z)/(z (z^2 + 1)^2), {z, I}]
</code>```

is $\frac{-3}{4e}$

But evaluating the integral

```<code>Integrate[Sin[x]/(x (x^2 + 1)^2), {x, -Infinity, Infinity}]
</code>```

mathematica gives $\pi - \frac{3 \pi}{2e}$.

I am wondering if this is because I did something wrong somewhere or if it is because mathematica gives the wrong answer.

Thank you!


问:
Euler characteristic of a compact surface
Determine the Euler characteristic of the surface

$$ M=\left\{(x,y,z); \sqrt{x^2+y^2}=1+z^{2n}, 0< z< 1\right\} $$

答:
For any $n\in \mathbb{R}$, your surface is a cylinder, and homotopic to the circle.

(I don't see why Agusti Roig gets the disjoint union of two cylinders)

OwnerDisplayName: swallenberg

ContentLicense: CC BY-SA 2.5

ContentLicense: CC BY-SA 3.0


问:
Connection between Gauss and Poisson distribution?
I know that these distributions are connected by the central limit theorem.

But as it is written here: The central limit theorem says that the distribution of the mean of N draws from a probability distribution approaches a Gaussian of width $\frac{\sigma}{\sqrt{N}}$ for large N.

In the same script, in Eq(70) he says the standard deviation of a poisson distribution is $\sigma = \sqrt{\lambda t}$

Below Eq(79) he says the standard deviation of the gaussian (as limit of the poisson) is also  $\sigma = \sqrt{\lambda t}$.

For me this is confusing, where is with the $\frac{1}{\sqrt{N}}$ ?

In our case, the central limit theorem says that the distribution of the mean of N draws from the poisson distribution approaches a Gaussian.

As in the example of the radioactive decay:

I measure N times, the number of decays during the time t (this is the mean) and consider the distribution. This should be a Gaussian? But in this script, he do not sample over anything, he just says that he consider m >> 1 with m as the number of events during t. Is this right? I mean it should be right - he is professor in Harvard - but I don't get it.

答:
The answer on my question is in the same script:

Sometimes we add the values of the draws from a distribution instead of averaging them. In
this case, the mean $\bar{x}$ grows as  $N\bar{x}$ and the standard deviation $\sigma$ grows like $\sqrt{N}\sigma$.

Where N is somehow the time t we measure (for example the radioactive decay)


问:
How can I prove that W is a subspace of $\Bbb{R}^n$
My question is if $v_0 \in \Bbb{R}^n$ is a fixed vector and we define $W\subset \Bbb{R}^n$ to be the set of all vectors in $\Bbb{R}^n$ that are othogonal to the particular vector $v_0$. Then using the subspace test, and abstract properties of dot products how does one prove that $W$ is a subspace of $\Bbb{R}^n$?

答: The subspace test requires showing that $W$ (all vectors orthogonal to $v_0$, i.e. all vectors $w$ such that $w\cdot v_0=0$) is closed under addition and scalar multiplication. Take two arbitrary vectors in $W$, call them $w_1$ and $w_2$. All we start with is that $w_1\cdot v_0= 0$  and $w_2\cdot v_0= 0$. To show closure under addition we need that $(w_1+w_2)\in W$, which means we need to show that $(w_1+w_2)\cdot v_0 =0$. Which property of dot product give us this? Similarly we need to show that for an arbitrary vector $w\in W$ that $cw\in W$ for any scalar $c$.


问:
Derivative of convolution proof
Using the definition of convolution: $$ (f*g)(x) = \frac {1}{2\pi} \int_{-\pi}^{\pi} f(x-t)g(t)dt$$ I need to prove the following:

Let $f$ be a continuously differentiable function. Then $f*g$ is also differentiable and it's derivative at point $t$ is $((f')*g)(t)$.

Any help would be appreciated.

答:
Hint

Use the definition of derivative and swap the convolution integral with the $\lim$ in the definition of derivative.

Title: Derivative of convolution proof


问:
For vectors $\vec{a}\times\vec{b}$, $\vec{b}\times\vec{c}$, $\vec{c}\times\vec{a}$, which are linearly dependent, show that they must be parallel.

For vectors $\vec{a}, \vec{b}, \vec{c} \in \mathbb{R}^3$ show: If vectors  $$\vec{a} \times \vec{b}, \;\vec{b} \times \vec{c},\; \vec{c} \times \vec{a}$$
are linearly dependent, then they must be parallel.

How would you prove that?

答:
If they are linearly dependent, then we can find real numbers $x,y,z$ such that
$$
 x(a \times b) + y (b \times c) + z(c \times a) = 0.
$$
To get the idea across, assume that $x,y,z \not= 0$. Otherwise you just have to do case work.

Note that $c \cdot (c \times \text{anything}) = 0$. If we dot product the left and right hand sides by $c$, we see that
that$$
 x(a \times b) \cdot c = 0.
$$
Since $x \not= 0$, this shows that all three $a \times b$, $b \times c$, and $c \times a$ are orthogonal to $c$. Repeating this for $a,b$, we see that all three cross products are orthogonal to $a,b,c$. From here, you can deduce that they are all parallel. (Again, more case works)


答:
We need to prove that $W$ is not empty which it is indeed because $0\in W$; that it is stable by addition and multiplication by a scalar.

Two vectors $w_1,w_2\in W$, are such that $w_1\cdot v_0= w_2\cdot v_0=0$. So $(w_1+w_2)\cdot v_0=\lambda w_1\cdot v_0=0$

And we’re done.

Title: How can I prove that W is a subspace of $\Bbb{R}^n$

Body: My question is if $v_0 \in \Bbb{R}^n$ is a fixed vector and we define $W\subset \Bbb{R}^n$ to be the set of all vectors in $\Bbb{R}^n$ that are othogonal to the particular vector $v_0$. Then using the subspace test, and abstract properties of dot products how does one prove that $W$ is a subspace of $\Bbb{R}^n$?


问:
Showing $\frac{\sin(x+a)-\sin(x)}{a}= \frac{\cos(x+\frac{a}{2})\sin\frac{a}{2}}{a/2}$
In my mathematics textbook, when deriving the differential of sin(x), it goes from the first expression to the second one without any explanation as to why.

$$\frac{\sin(x+a)-\sin(x)}{a}= \frac{\cos(x+\frac{a}{2})\sin\frac{a}{2}}{a/2}$$

Myself and my flatmates have been attempting to rearrange it for about an hour now with no luck.

Any help would be greatly appreciated.

答:
Rewrite

$$ \sin(x+a)-\sin(a) $$

as

$$\sin\left(\left(x+\frac{a}{2}\right)+\frac{a}{2}\right)-\sin\left(\left(x+\frac{a}{2}\right)-\frac{a}{2}\right)$$

then apply the sum and difference formulas for $\sin$ and simplify.


答:
They simply use one of the four factorisation formulæ, namely
$$\sin p-\sin q=2\sin\Bigl (\frac{p-q}2\Bigr)\cos\Bigl (\frac{p+q}2\Bigr),$$
which is a consequence of this corollary of the addition formulæ:
$$\sin(\alpha+\beta)-\sin(\alpha-\beta)=2\cos\alpha\sin\beta.$$

Title: Showing $\frac{\sin(x+a)-\sin(x)}{a}= \frac{\cos(x+\frac{a}{2})\sin\frac{a}{2}}{a/2}$

Body:
In my mathematics textbook, when deriving the differential of sin(x), it goes from the first expression to the second one without any explanation as to why.

$$\frac{\sin(x+a)-\sin(x)}{a}= \frac{\cos(x+\frac{a}{2})\sin\frac{a}{2}}{a/2}$$

Myself and my flatmates have been attempting to rearrange it for about an hour now with no luck.

Any help would be greatly appreciated.


问:
Write a maximum as integral of indicator function (clarification)
It is known that a random variable $X$ can be written in terms of positive and negative parts $X=X^+-X^-$.
Also, $X^+=\max(0,X)=\int_0^\infty I(X>x)dx$. I don't know how to show it.

Say that $a\in\mathbb{R}$. I can see that $\max(0,a)=aI(a>0)$, and that $a=\int_0^adx$. It indicates me that $\max(0,a)=\int_0^{aI(a>0)}dx$, but I can't relate this with $\max(0,a)=\int_0^\infty I(a>x)dx$.

Can you help me with it?

答:
As you noticed, a real number $x\ge 0$ can be written as $\int_0^x 1\,dz$. Thus,
$$
X\vee 0=\int_0^{\infty} 1\{X\vee 0>z\}\,dz=\int_0^{\infty} 1\{X>z\}\,dz.
$$


答: The bilinearity of the dot product gaurantees $\psi:v\mapsto v\cdot v_0$ defines a linear map on $\Bbb R^n$ and therefore $\ker(\psi)$ is an $\Bbb R^n$ subspace where $\ker(\psi)=\{v\in\Bbb R^n:v\cdot v_0=0\}$.


问:
using diagonal argument to show existence of uncountable antichain
Consider the poset $$(\mathcal{P}(\mathbb{N}),\subseteq)$$

i.e. subsets of the naturals with partial ordering given by set inclusion.
An antichain in this poset is a family of sets $\mathcal{F}\subseteq \mathcal{P}(\mathbb{N})$ such that for any distinct $A,B\in \mathcal{F}$, we have $A\not\subseteq B$ and $B\not\subseteq A$.

The objective is to construct an uncountable antichain in this poset.
We can represent subsets $A\subseteq \mathbb{N}$ by their indicator sequence $(a_i)_{i\in\mathbb{N}}$, with $a_i=1$ if $i\in A$ and $a_i=0$ if $i\notin A$.
Using this representation we can apply a variant of Cantor's diagonal argument to show that given a countably infinite antichain (modulo some conditions), we can always construct a new element that can be added to the original antichain to get a larger antichain.

My question is: how do we use this insight to show that we can get an uncountable antichain?

I've been told to consider a 'maximal' antichain, say $\mathcal{A}$.
Then if $\mathcal{A}$ is still countably infinite, we can use the diagonal argument to find an additional element to append to it, which is a contradiction.
However I'm unsure 1) what 'maximality' should mean here exactly, and/or 2) why such a 'maximal' antichain should exist.

My guess is that 'maximal' = 'no elements can be appended without losing the antichain property' - but then I am unsure as to why there must exist a maximal antichain.

EDIT:
I am aware of other ways to show the existence of such uncountable antichains in $\mathcal{P}(\mathbb{N})$, but I am interested here in understanding this specific approach.

答:
Your understanding of maximal is correct. The existence of a maximal antichain is a straightforward consequence of Zorn’s lemma applied to the set of all antichains in $\wp(\Bbb N)$.

Let $\mathfrak{A}$ be the set of all antichains in $\wp(\Bbb N)$, ordered by inclusion. Let $\mathfrak{C}$ be a chain in $\mathfrak{A}$. That is, $\mathfrak{C}\subseteq\mathfrak{A}$, and for any antichains $\mathscr{A}_0,\mathscr{A}_1\in\mathfrak{C}$, either $\mathscr{A}_0\subseteq\mathscr{A}_1$, or $\mathscr{A}_1\subseteq\mathscr{A}_0$.

Show that $\bigcup\mathfrak{C}\in\mathfrak{A}$, i.e., that $\bigcup\mathfrak{C}$ is an antichain in $\wp(\Bbb N)$.

Clearly $\bigcup\mathfrak{C}$ is an upper bound for $\mathfrak{C}$ in the partial order $\subseteq$ on $\mathfrak{A}$, so this shows that every chain in $\mathfrak{A}$ has an upper bound in $\mathfrak{A}$. Zorn’s lemma then says that $\mathfrak{A}$ has a maximal element $\mathscr{M}$. $\mathscr{M}$ is an antichain in $\wp(\Bbb N)$, and it is maximal among all such antichains, meaning that if $\mathscr{M}\subseteq\mathscr{A}\in\mathfrak{A}$, then $\mathscr{M}=\mathscr{A}$: $\wp(\Bbb N)$ has no antichain that properly contains $\mathscr{M}$.

And once you have $\mathscr{M}$, the diagonal argument that you sketched will show that $\mathscr{M}$ must be uncountable.


问:
Joint Discrete distribution and continuous distribution
$X$ follows a continuous uniform distribution on $[0,3]$.$Y$ follows a discrete uniform distribution on $\{1,2,3\}$
Calculate the probability that $X<Y$.

Can we compare probability a continuous random variable with a discrete random variable? Do they have joint distribution? I tried to use the concept of joint distribution to solve this problem, Like drawing a pic and calculate the area.

What kind concept that I should use to solve this problem?

Thanks for the helps!

答: Try using the law of total probability: $P(X<Y) = \sum_{i=1}^3 P(X<Y\cap Y=i) =  \sum_{i=1}^3 P(X<Y| Y=i)P(Y=i)$.


问:
Is every subgroup of $S_n$ cyclic?
I've just showed that every subgroup of $S_3$ is cyclic.

I'm not familiar with permutation groups yet and I'm curious to know whether every subgroup of $S_n$ is cyclic.

As a counterexample might be $S_4$ probably, but I'm not sure at all.

Thanks in advance for explanation.

答: The Klein subgroup of $S_4$ is a non cyclic proper subgroup of order $4$. All its elements have order $2$.

Title: Is every subgroup of $S_n$ cyclic?


问:
A proof on the Jacobian and error to prove differentiability
Let $F : \mathbb{R}^n \to \mathbb{R}^m$. If there exists an $m \times n$ matrix $A$ and $\rho : \mathbb{R}^n \to \mathbb{R}^m$ such that $\forall h \in \mathbb{R}^n$, $F(a+h) = F(a) + Ah + ||h||\rho(h)$, where $\rho(h) \to 0$ as $h \to 0$, then $F$ is differentiable at a.

My attempt:
Rearranging yields
$\dfrac{F(a+h) - F(a) - Ah}{||h||} = \rho(h)$, and $\rho(h) \to 0$ means the left hand goes to $0$, and that's the definition of differentiability (if $A$ is the Jacobian).

This almost seems too easy but I feel like I'm missing many details: could someone help me out in filling the missing assumptions?

答: The only think is that you also need to prove that $h \mapsto Ah$ is continuous. Which is the case as a linear map defined on a linear space of finite dimension is always continuous.


问:
Find radius of convergence of the power series $\sum_{n=0}^{\infty}a_n(z-z_0)^n$ when $a_n = \frac{n^2}{4^n+3n}$.

Find radius of convergence of the power series $\sum_{n=0}^{\infty}a_n(z-z_0)^n$ when $a_n = \frac{n^2}{4^n+3n}$.

I am posting this question to verify if my work and reasoning is correct.
Using the divergence test, we have

\begin{align}
\lim_{n\to\infty}|a_n| &= \lim_{n\to\infty} \frac{n^2}{4^n+3_n} \\
&= \lim_{n\to\infty} \frac{\frac{n^2}{4^n}}{1+\frac{3n}{4^n}} \\
&= \frac{\lim_{n\to\infty}\bigg(\frac{n^2}{4^n}\bigg)}{\lim_{n\to\infty}\bigg(1+\frac{3n}{4^n}\bigg)} \\
&= \frac{0}{1} = 0
\end{align}

Since the limit converges to 0, then the radius of convergence is $\infty$.

Body:
hint

$$|a_n|=\frac{n^2}{4^n(1+\frac{3n}{4^n})}\sim \frac{n^2}{4^n}=b_n$$

$$R=\lim_{n\to\infty}\frac{b_n}{b_{n+1}}$$

$$=\lim_{n\to\infty}\frac{n^2}{(n+1)^2}\frac{4^{n+1}}{4^n}$$
$$=4$$


答: You actually want, by Cauchy-Hadamard, $r=\limsup_{n\to\infty}\dfrac1{\sqrt[n]{\dfrac{n^2}{4^n+3n}}}=4$.

Body:
Find radius of convergence of the power series $\sum_{n=0}^{\infty}a_n(z-z_0)^n$ when $a_n = \frac{n^2}{4^n+3n}$.

I am posting this question to verify if my work and reasoning is correct.
Using the divergence test, we have

\begin{align}
\lim_{n\to\infty}|a_n| &= \lim_{n\to\infty} \frac{n^2}{4^n+3_n} \\
&= \lim_{n\to\infty} \frac{\frac{n^2}{4^n}}{1+\frac{3n}{4^n}} \\
&= \frac{\lim_{n\to\infty}\bigg(\frac{n^2}{4^n}\bigg)}{\lim_{n\to\infty}\bigg(1+\frac{3n}{4^n}\bigg)} \\
&= \frac{0}{1} = 0
\end{align}

Since the limit converges to 0, then the radius of convergence is $\infty$.


答: If you mean that all proper subgroups of $S_n$ are cyclic, it  would imply that all proper subgroups are abelian. Now  it is quite easy to show that $A_n$ is not abelian for $n\ge 4$.

Title: Is every subgroup of $S_n$ cyclic?

Body:
I've just showed that every subgroup of $S_3$ is cyclic.

I'm not familiar with permutation groups yet and I'm curious to know whether every subgroup of $S_n$ is cyclic.

As a counterexample might be $S_4$ probably, but I'm not sure at all.

Thanks in advance for explanation.


答:
In step 2 you don't consider that
$\frac{n^n+[\text{polynomial}]^{n-1}}{n^n}$ has infinite polynomial degree in the limit, so is false
$\frac{n^n+[\text{polynomial}]^{n-1}}{n^n}\rightarrow 1+\frac{1}{n}$.

I recommend using Stirling's formula.

Title: Is this approach to the limit solution kosher?$ \lim_{n\rightarrow\infty}\frac{(n!)^{\tfrac{1}{n}}}{n}=\frac{1}{e}$

Body:
Here is a non-rigourous approach to a limit:
$$ \lim_{n\rightarrow\infty}\frac{(n!)^{\tfrac{1}{n}}}{n}=\frac{1}{e}$$
I know the solution is correct. I'm suspicious of my methods.

Rewrite the expression:

$$ \lim_{n\rightarrow\infty}\left(\frac{n!}{n^n}\right)^{\frac{1}{n}}$$

I know the $\lim_{n\rightarrow\infty}\left(\frac{n!}{n^n}\right)=0$. So bear with me.

$$\frac{n!}{n^n}=\frac{n(n-1)(n-2)...[n-(n-2)][n-(n-1)]}{n^n}$$

Written this way, the numerator is a polynomial of degree $n$. For large $n$, the leading term will prevail. I will write that polynomial as a sum of $n^n$ and a polynomial of degree $n-1$.

$$\frac{n!}{n^n}\Rightarrow\frac{n^n+[\text{polynomial}]^{n-1}}{n^n}\Rightarrow 1+\frac{1}{n}$$

As noted in Step 2, this is not the expected result for this limit. But watch:

$$ \lim_{n\rightarrow\infty}\frac{(n!)^{\tfrac{1}{n}}}{n}=\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{\tfrac{1}{n}}=\lim_{n\rightarrow\infty}\frac{1}{\left(1+\frac{1}{n}\right)^n}=\frac{1}{e}$$

This is the correct solution. But Step 2 is sketchy. Did I make another error which undid this error? Can this derivation be salvaged, or is my proof "not even wrong"?

Please be gentle with me.


答:
No.

By Cayley's Theorem, every finite group is isomorphic to a subgroup of a symmetric group $S_n$ for some $n$ (dependent upon the finite group).

Title: Is every subgroup of $S_n$ cyclic?


问:
Finding the horizontal and vertical tangents of $r^2=\sin(2\theta)$
Finding the horizontal and vertical tangents of $r^2=\sin(2\theta)$ where we are using polar coordinates.

I was wondering if you guys could show me multiple ways to do this problem. I did it by using implicit differentiation, but whenever I ask questions here I get a bunch of brilliant solutions so I thought I'd ask. Thanks!

答:
$$r^2=\sin(2\theta)$$
$$r^2=2\sin\theta\cos\theta\to r^4=2(r\sin\theta)(r\cos\theta)$$
as $r\cos\theta=x;\;r\sin\theta=y$ we get the cartesian equation
$$(x^2+y^2)^2=2xy\to f(x,y)=(x^2+y^2)^2-2xy=0$$
I use the formula for implicit diferentiation
$$\frac{dy}{dx}=-\frac{\partial f/\partial x}{\partial f/\partial y}$$
So we have
$$\frac{dy}{dx}=\frac{2 y-4 x \left(x^2+y^2\right)}{4 y \left(x^2+y^2\right)-2 x}$$
Horizontal tangents are where derivative is zero.
$$2 y-4 x \left(x^2+y^2\right)=0\to x^2+y^2=\frac{y}{2x}$$
Plug in the equation of the curve
$$\left(\frac{y}{2x}\right)^2=2xy\to y=0;\;y=8x^3$$
The first solution is then $(0,0)$

Plugging $y=8x^3$ in the equation of the curve we get
$$(x^2+64x^6)^2=16x^4\to x^4 \left(64 x^4-3\right) \left(64 x^4+5\right)=0$$
$x=\pm\sqrt[4]{\frac{3}{64}}\approx \pm 0.4653$ and $y\approx \pm 0.806$.

The points where the tangents are vertical are the symmetric wrt line $y=x$ bisector of the first quadrant.

$$
.
$$


问:
Largest prime factor of 600851475143
I'm trying to use a program to find the largest prime factor of 600851475143.
This is for Project Euler here: http://projecteuler.net/problem=3

I first attempted this with the code that goes through every number up to 600851475143, tests its divisibility, and adds it to an array of prime factors, printing out the largest.

This is great for small numbers, but for large numbers it would take a VERY long time (and a lot of memory).

Now I took university calculus a while ago, but I'm pretty rusty and haven't kept up on my math since.

I don't want a straight up answer, but I'd like to be pointed toward resources or told what I need to learn to implement some of the algorithms I've seen around in my program.

答:
The number looks small enough to be brute-forced on a computer. Just try every possible factor, starting with 2, 3, 4, ... and keep dividing them out as long as the division comes out even. Then continue looking for factors of the quotient. You don't even need to explicitly restrict to primes, because any composite number you try simply won't divide the quotient so far (why?).

If you reach a trial factor that's less than the square root of the number you're dividing into, you know that the larger number must be the last prime factor. So there cannot possibly be more than a few million trial divisions to do.

Title: Largest prime factor of 600851475143


问:
Relative homology and homology of subset
Let $X$ be homotopy equivalent to a point and let $A \subset X$.

 I have shown that $H_n(X,A) \cong H_{n-1} (A)$ for $n \geq 2$. I did this using the long exact sequence property in relative homology.

 Does it also hold for $n=1$. If not, what would a counterexample be?

答:
If you work with reduced homology, then by the long exact sequence you get that $H_1(X,A)\cong H_0^{\text{#}}(A) $. But we know that $H_0^{\text{#}}(A)$ is never isomorphic to $H_0(A)$ as they have different ranks as Abelian groups (If $A$ is nonempty subspace with finitely many components and assuming we are working over the integers).

Recall that the rank of $H_0^{\text{#}}(A)$ is number of path components of $A$ minus 1 and that the rank of $H_0(A)$ is equal to the number of path components of $A$.


问:
Upper Bound Theorem for Simplicial Spheres
I am recently approaching combinatorial commutative algebra and I am studing Upper Bound Theorem for Simplicial Spheres (Stanley 1975).
My question is so a bit general and maybe ingenuous...

Momentum curve is the algebraic curve $M \subset \mathbb{R}^d$ defined parametrically by
$x(\tau) = (\tau, \tau^2, ..., \tau^d) $.
A cyclic polytope $C(n, d)$ is the convex hull of any $n$ distinct points on $M$.

There is a first Proposition I have seen,that says any cyclic polytope is a symplicial $d$- polytope (since you can prove any $d+1$ dinstinct points on $M$ are affinely independent).

Question 1) Are there other algebraic curves with this property?

More over you can prove the combinatorial type of a cyclic polytope $C(n, d)$ depends only upon $n$ and $d$ and not on the particular vertex set $V \subset M$.

Question 2) Are there other algebraic curves with this property?

So the point is: I do not understand the peculiarity of momentum curve.
Maybe there is another curve $N$ such that if you take the convex hull of $n$ distinct points, call it $N(n, d)$, you can reformulate upper bound theorem for simplicial spheres in this way:

Given $P$ a simplicial complex whose geometric realization is topologically a sphere, with n vertices, the following inequality holds: $f_j(P) \le f_j(N(n, d))$, where $f_j$ is any component of the $f$-vector.

If such a curve does not exist, why?
I think McMullen's formulated the conjecture, so how did he find the momentum curve?

Thanks

答:
To your last (unnumbered) questions:

Suppose that such a function $N(n,d)$ exists with $f_i(P)\le f_i(N(n,d))$ for all simplicial spheres $P$.
This must in particular hold for the boundary complex of the cyclic polytope.
Therefore

$$f_i(C(n,d))\le f_i(N(n,d)).$$

But by the upper bound theorem for simplicial spheres (proven by Richard Stanley), we also have

$$f_i(N(n,d))\le f_i(C(n,d))$$

as well, establishing identity, and so $N(n,d)$ must be cyclic.


问:
Changing of Basis Questions

Hello there, I am currently tackling this Linear Algebra question and I'm struggling to get a grasp on it.

So far I have figured out a) using a theorem from my notes that says that addition is associate and multiplication is distributive and hence I can split:

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

And achieve through substitution:

$$T(x^2 - 2x +1)= 3x^2 - 5x - 1$$

And here is where I start to struggle.

Any pointers/help would be greately appreciated.

答:
In part b, you've been given the map in functional notation i.e. $T(v)=\text{something}$, and are asked for a matrix. The matrix "associated" with a linear map is the one such that left-multiplying by that particular matrix is equivalent to applying that map, so you need to work backwards from the map outputs you've been given, along with the rules for matrix multiplication, to find the matrix elements. Notice also that the elements of the set $B$ are the same as the map inputs you've been given. Another hint is that you don't have to compute anything new to answer this - only arrange expressions (or pieces of expressions) you already have in the right order.

This is a question of the conditions that make a set a basis for a given vector space - are there are enough elements for the full dimension of the space (in this case, $3$), and are all of the elements of the set linearly independent (i.e. not scalar multiples of) all of the others?

$[C]_B$ is (if I'm reading right) the matrix that takes the components of some vector expressed in the standard basis $B$ and outputs the components necessary to express that the same "actual" function in terms of elements of $C$, e.g. the function $x^2-x$ as a vector is written $(0, -1, 1)$ in the standard basis, and needs to be transformed into $(1,0,0)$ by the map you calculate. $[B]_C$ is the matrix that does the reverse. This also implies that $[C]_B [B]_C = I$, so they're inverses of each other. These should be relatively easy to calculate if you take advantage of the reasoning from part B to split the problem of computing the matrix's $9$ elements into smaller independent subsets.

$[T]_C$ is (AIUI) a transform that "has the same effect" as the original $T$, but written in terms of coordinates from the basis $C$ rather than the standard $B$. But you already know how to transform one basis into another from the previous question - simply glue that transform in front (to the right of) the original $T$, and the product of the two is the answer. I'm assuming that this has some sort of nice form because the $([T]_C)^N$ part is asking you to write what this matrix looks like multiplied by itself repeatedly.

This is similarly glueing together existing pieces. You can answer the first part by taking the $([T]_C)^N$ from the previous question and transforming it back to the standard basis. (You already have the function you have to apply to do that) You can then take this combined result and use it to transform the function $x^2+1=(1,0,1)$

Hope that helps!

Title: Changing of Basis Questions


问:
finding $x^6+y^6$ given $x+y$ and $xy$
Here is the question

If $x + y = 4$ and $xy = 2$, then find $x^6+ y^6$.

This is from a previous timed competition, so fastest answers are the best answers.

I've tried using sum of cubes, but I dont know what to do after $(x^2+y^2)(x^4-x^2y^2+y^4)$
. The only other way I can think of is solving for x and y, but that wouldn't be too quick. Any help?

Title: finding $x^6+y^6$ given $x+y$ and $xy$

Body:
$x^2+y^2=(x+y)^2-2xy=16-4=12.$

$x^4+y^4=(x^2+y^2)^2-2(xy)^2=12^2-2\cdot 2^2=144-8=136$.

Now you can use your factorization:

$x^6+y^6=(x^2+y^2)[(x^4+y^4)-(xy)^2]$.


问:
qq, can I solve equations with complex numbers like that?
So I have 2 examples:

$\frac{27|z|}{z}=\frac{1}{3}\bar z *z^3 \implies 27|z|=\frac{1}{3}\bar z *z^4 \implies 27|z|=\frac{1}{3} |z|^2*z^3 \implies \frac{27}{|z|}=\frac{1}{3} z^3 \implies \frac{27}{ \sqrt{a^2 + b^2}}=\frac{1}{3} (a+bi)^3 $

On the left hand side I see no imaginary part, therefore I know that: $ b = 0$ nad $ \frac{27}{ a}=\frac{1}{3} a^3 $

$8z(\bar z) = \bar z^5 \implies 8z|z|= \bar z ^5 \implies 8\frac{|z|^2}{\bar z}|z|= \bar z ^5 \implies 8|z|^3= \bar z ^6 \implies 8(\sqrt{a^2 + b^2})^3 = (a-bi)^6$

Same thing - so I know that: $ b = 0 $ , so:   $ 8a^3 = a^6$

Is that ok?

答:
hint

Approach by polar expression with
$$z=re^{it}=r(\cos(t)+i\sin(t))$$

the equation becomes
$$27\frac{r}{re^{it}}=\frac 13.re^{-it}.r^3e^{3it}$$

or, after simplification

$$81=r^4e^{3it}=3^4e^{2ik\pi}$$

So, the solution of your equation is
$$z=3e^{i\frac{2k\pi}{3}}\; where\;\; k=0,1,2$$


问:
Number of functions from A to B, where C $\subseteq$ B
Be A, B, C finite sets with C $\subseteq$ B, |A| = m, |B| = n, |C| = r; where m, n$\geq$ r. Find the number of functions f from A to B with the property C $\subseteq$ f(A).

Hello! I found the following result for this problem: number of functions is $n^m - (n - r)^m$. Could someone tell me please if this is the right answer? Thank you in advance!

答:
$(n-r)^m$ is the number of functions from $A$ to $B\setminus C$, so $n^m-(n-r)^m$ is the number of functions from $A$ to $B$ such that $f[A]\nsubseteq B\setminus C$, i.e., the number of functions from $A$ to $B$ such that $f[A]\cap C\ne\varnothing$. This isn’t what you want: you want $f[A]$ to include every element of $C$, not just at least one element of $C$.

You can solve this problem with an inclusion-exclusion argument. For that we want to count the ‘bad’ functions from $A$ to $B$, the ones whose ranges don’t include all of $C$. For each $c\in C$ let $F_c$ be the set of $f:A\to B$ such that $c\notin f[A]$; $\bigcup_{c\in C}F_c$ is the set of ‘bad’ functions, so we want $\left|\bigcup_{c\in C}F_c\right|$.

For any non-empty $I\subseteq C$ there are $(n-|I|)^m$ functions from $A$ to $B\setminus I$, so the inclusion-exclusion principle says that

$$\begin{align*}
\left|\bigcup_{c\in C}F_c\right|&=\sum_{\varnothing\ne I\subseteq C}(-1)^{|I|+1}\left|\bigcap_{c\in I}F_c\right|\\
&=\sum_{\varnothing\ne I\subseteq C}(-1)^{|I|+1}(n-|I|)^m\\
&=\sum_{k=1}^r(-1)^{k+1}\binom{r}k(n-k)^m\,,
\end{align*}$$

and subtracting this from $n^m$, the total number of functions from $A$ to $B$, will give us the desired result:

$$\begin{align*}
n^m-\sum_{k=1}^r(-1)^{k+1}\binom{r}k(n-k)^m&=n^m+\sum_{k=1}^r(-1)^k\binom{r}k(n-k)^m\\
&=\sum_{k=0}^r(-1)^k\binom{r}k(n-k)^m\,.
\end{align*}$$


问:
MLE Normal Variance Changing Means
I have $n$ draws from a 2d normal distribution with a different mean vector every draw. The variance stays constant at $\tau I_2$. How can I get the MLE estimator for $\tau$? Since the mean is changing, it seems 0 or a very small $\epsilon$ is the best, but this doesn’t make much sense in a normal setup.

答:
The PDF of a multivariate normal distribution of dimension $k$ and mean vector $\mu$ and covariance matrix $\tau I$ is
$$f_x(x | \mu, \tau) = (2 \pi)^{-k} \tau^{-k/2} \exp(-\frac{1}{2\tau}(x-\mu)^T(x-\mu))$$
Given $N$ observations of the distribution, the likelihood function could be written as
$$L(\mu,\tau) = f_x(x_1,x_2\ldots x_N \vert \mu,\tau) $$
If the observations are independent then we get the following product
$$L(\mu,\tau) = \Pi_{n=1}^N f_x(x_n\vert \mu,\tau) $$
which is
$$L(\mu,\tau) =  (2 \pi)^{-Nk} \tau^{-Nk/2} \exp(-\frac{1}{2\tau}\sum\limits_{n=1}^N (x_n-\mu)^T(x_n-\mu)) $$
The maximum likelihood estimate of $\tau$ maximizes the above or any function that is increasing of the above, one of which is the log-likelihood, which is
$$\ell (\mu,\tau) =  \log L(\mu,\tau) = \log (2 \pi)^{-Nk} \tau^{-Nk/2} \exp(-\frac{1}{2\tau}\sum\limits_{n=1}^N (x_n-\mu)^T(x_n-\mu)) $$
which is also
$$\ell (\mu,\tau)  = \log (2 \pi)^{-Nk} + \log  \tau^{-Nk/2} + \log \exp(-\frac{1}{2\tau}\sum\limits_{n=1}^N (x_n-\mu)^T(x_n-\mu)) $$
Using $\log \exp(x) = x$ we get
$$\ell (\mu,\tau)  = \log (2 \pi)^{-Nk} + \log  \tau^{-Nk/2} -\frac{1}{2\tau}\sum\limits_{n=1}^N (x_n-\mu)^T(x_n-\mu) $$
Using $\log x^a = a \log x$, we get
$$\ell (\mu,\tau)  = -Nk\log (2 \pi) - \frac{Nk}{2} \log  \tau -\frac{1}{2\tau}\sum\limits_{n=1}^N (x_n-\mu)^T(x_n-\mu) $$
Deriving with respect to $\tau$,
$$\frac{\partial}{\partial \tau}\ell (\mu,\tau) = - \frac{Nk}{2\tau} + \frac{1}{2\tau^2}\sum\limits_{n=1}^N (x_n-\mu)^T(x_n-\mu) = 0 $$
which gives you
$$\hat{\tau} = \frac{1}{Nk}\sum\limits_{n=1}^N (x_n-\mu)^T(x_n-\mu) \tag{1}$$
But we do not know the mean $\mu$, and hence it should be estimated using the same process, i.e.
$$\frac{\partial}{\partial \mu}\ell (\mu,\tau) = - \frac{1}{2\tau^2}\sum\limits_{n=1}^N (-2 x_n + 2 \mu ) = 0 $$
which gives us
$$\hat{\mu} = \frac{1}{N}\sum\limits_{n=1}^N x_n \tag{2} $$
Replacing (2) in (1) we get
$$\hat{\tau} = \frac{1}{Nk}\sum\limits_{n=1}^N (x_n-\frac{1}{N}\sum\limits_{l=1}^N x_l)^T(x_n-\frac{1}{N}\sum\limits_{l=1}^N x_l)$$

Title: MLE Normal Variance Changing Means


问:
How do we define the $L^p$ norm of a tempered distribution?
I am finishing up a class on function theory and I am trying to reconcile
a few statements that I have seen.

Let us define $L^p(\mathbb R^n)$ to be the set of measurable functions $f$ so that
$\int_{\mathbb R^n} |f|^p dx < \infty$, with the norm $||f||_p = \Big(\int_{\mathbb R^n
} |f|^p dx\Big)^{1/p}$.

I have also seen it written that $L^p(\mathbb R^n)$ is the set of all tempered distributions
$f \in S'$ satisfying $ ||f||_p < \infty$. Here $S$ is the collection of rapidly decreasing complex valued functions on $\mathbb R^n$, and $S'$ is the space of continuous linear functionals on $S$ (tempered distributions).

I would like to show that these two definitions "are the same". However, I am at a disadvantage, since I don't even see how to define the $||\cdot||_p$ norm on
the space of tempered distributions.

Thus, how do we define the $L_p$ norm of a tempered distribution?

答:
Distributions act on the space $\mathcal{D}$ of test functions (infinitely differentiable with compact support). For all $p$ it holds that $\mathcal{D}\subset L^p$ and hence, it makes sense to take the $p$-norm test functions for any $p$. Now take some $p$ and its conjugate $q$ (i.e. $p+q=pq$). For any distribution $T$ we can define
$$
\|T\|_p = \sup\{ T(\phi)\ :\ \|\phi\|_q\leq 1\}
$$
which may or may not be finite.

Note that this produces the $p$ norm of a function $f$ when applied to the distribution induced by $f$.


Title: finding $x^6+y^6$ given $x+y$ and $xy$

Body:
Here is the question

If $x + y = 4$ and $xy = 2$, then find $x^6+ y^6$.

This is from a previous timed competition, so fastest answers are the best answers.

I've tried using sum of cubes, but I dont know what to do after $(x^2+y^2)(x^4-x^2y^2+y^4)$
. The only other way I can think of is solving for x and y, but that wouldn't be too quick. Any help?

Body:
Another way is to write out

$$ (x+y)^3=(x^3+y^3)+3xy(x+y), $$

$$ (x^3+y^3)^2=(x^6+y^6)+2(xy)^3. $$


Title: Changing of Basis Questions

Body:
Hello there, I am currently tackling this Linear Algebra question and I'm struggling to get a grasp on it.

So far I have figured out a) using a theorem from my notes that says that addition is associate and multiplication is distributive and hence I can split:

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

And achieve through substitution:

$$T(x^2 - 2x +1)= 3x^2 - 5x - 1$$

And here is where I start to struggle.

Any pointers/help would be greately appreciated.

Body:
Let $B=\{1,x,x^2\}$ then $\forall p \in P_2(R)$ $\exists c_0,c_1,c_2$: $p=c_0+c_1x+c_2x^2$

$T(p)=c_0T(1)+c_1T(x)+c_2T(x^2)$ $\Rightarrow$ $[T(p)]_B=A\begin{pmatrix} c_0 \\ c_1 \\ c_2 \end{pmatrix}$

$A_1=[T(1)]$ ,$A_2=[T(x)]$,$A_3=[T(x^2)]$  where $A_1,A_2,A_3$ are column vectors of $A$

$A=\begin{pmatrix} -1 & 0 & 0\\ -2 & 2 & 1 \\ 2 & 0 & 1 \end{pmatrix}$

We know that dimension of space is 3.All we have to do to prove (c) is show that they are linearly independent

$c_0(x^2-x)+c_1x+c_2(-x^2+x+1)=0$ $\Rightarrow$ $(c_2-c_0+c_1)x+c_2+(c_0-c_2)x^2=0$

$(c_2-c_0+c_1)x+c_2+(c_0-c_2)x^2=0$ $\Rightarrow$ $c_2-c_0+c_1=0$ ,$c_2=0$,$c_0-c_2=0$

which means $c_0=c_1=c_2=0$ so they are linearly independent

Let $C=\{x_2-x,x,-x^2+x+1\}$ and $p=c_0(x^2-x)+c_1x+c_2(-x^2+x+1)=n_0+n_1x+n_2x^2$  then

$c_2-c_0+c_1=n_1$ , $c_2=n_0$,$c_0-c_2=n_2$ so

$\begin{pmatrix} 0 & 0 & 1 \\ 1 & -1 & 1 \\ 1 & 0 &-2\end{pmatrix}$$[p]_B=[p]_C$


Title: finding $x^6+y^6$ given $x+y$ and $xy$

Body: Your sum is$$\begin{align}\sum_\pm(2\pm\sqrt{2})^6&=2(2^6+\sqrt{2}^6+15(2^4\sqrt{2}^2+2^2\sqrt{2}^4))\\&=2(64+8+15(2\sqrt{2})^2(2^2+2))\\&=2(72+15\cdot8\cdot6)\\&=1584.\end{align}$$


问:
infinite sum of inverse binomial coefficient encountered in Bayesian treatment of the German tank problem
in the Bayesian treatment of the German tank problem in Wikipedia here, they use:

$\displaystyle \sum_{n=m}^\infty \dfrac{1}{\binom{n}{k}}=\dfrac{k}{k-1}\dfrac{1}{\binom{m-1}{k-1}}$

how can I prove this in a clever combinatorics fashion?

I found this paper, see eqn. (9), which uses Gauss' hypergeometric function-- a bit beyond me.

there must be some way via a recursion relation, like I found in this old paper. theorem 1 in that reference has a similar infinite sum of an inverse binomial coefficient.

答:
we can write this as a telescoping sum using the identity:

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

write the left side as factorials and get a common denominator to see :)

then we can write the sum as:

\begin{equation}
\lim_{\Omega\rightarrow\infty}\dfrac{k}{k-1}\displaystyle\sum_{n=m}^\Omega [a_{n-1}-a_n]=\dfrac{k}{k-1} [a_{m-1}-a_\Omega]
\end{equation}

with:

\begin{equation}
   a_n:=\frac{1}{\binom{n}{k-1}}
\end{equation}

the right-most equality follows from the telescoping property of the sum.

as $\Omega \rightarrow \infty$, $a_\Omega \rightarrow 0$. thus:

$\displaystyle\sum_{n=m}^\infty \dfrac{1}{\binom{n}{k}}=\dfrac{k}{k-1}\dfrac{1}{\binom{m-1}{k-1}}$


问:
Why does $\lim\limits_{\theta\to\infty}n\sin\frac {\theta}n=\theta$

Questions:

Why do we have$$\lim\limits_{\theta\to\infty}n\sin\dfrac {\theta}n=\theta\tag1$$

How do we prove $(1)$?

I started off with the well known limit:$$\lim\limits_{\theta\to 0}\dfrac {\sin\theta}{\theta}=1\tag2$$
And substituted $\theta:=\dfrac \theta n$ into $(1)$ to get$$\lim\limits_{\frac \theta n\to 0}\dfrac {\sin\dfrac \theta n}{\dfrac \theta n}=\lim\limits_{\frac \theta n\to0}\dfrac {n\sin\frac \theta n}{\theta}\tag3$$
However after that, I'm not sure what to do. Apparently, the RHS of $(1)$ has a limit of $1$, so that the limit of $n\sin\frac \theta n=\theta$.

答: The original expression does not seem correct to me, let's just substitute $n$ for $\theta$ and rewrite $$\lim_{n\to \infty}n \sin\frac{\theta}n$$ At this point you can easily deduce that $$\lim_{n\to \infty}\Biggl(\frac{\sin\frac{\theta}n}{\frac1n}\Biggl)\space=\space\lim_{n\to \infty}\Biggl(\frac{\sin \frac{\theta}{n}}{\frac{\theta}{n}}\theta\Biggl)$$ Since we know that argument of sinus function will be ridiculously close to zero, we can use below expression to make things easier $$\sin \frac{\theta}{n}\space\approx\space\frac{\theta}{n}$$ If we make required substitutions, we'll eventually get $$\lim_{n\to \infty}\frac{\frac{\theta}{n}}{\frac{\theta}{n}}\theta\space=\space\theta$$


问:
Evaluating $\lim_{x \to 3} \frac{\log(1+x) - \log(4)}{x^2 - 9}$ with L'Hopital's rule gives me a different answer than what I expect
I'm trying to evaluate
$$\lim_{x \to 3} \frac{\log(1+x) - \log(4)}{x^2 - 9}$$
Since $$\lim_{x \to 3} \log(1+x) - \log(4)=0$$
and $$\lim_{x \to 3} x^2 - 9 = 0$$
we should be able to use L'Hopital's rule to see that
$$\lim_{x \to 3} \frac{\log(1+x) - \log(4)}{x^2 - 9} = \lim_{x \to 3} \frac{1/(1+x)}{2x} = \frac{0.25}{6} = 0.042$$
But when I graph the equation I see that the limit is $0.018$.

Why is this the case?

答:
As noticed in the comments there is an issue with the base for logarithm and your evaluation is right, as an alternative without l'Hospital we can refer to the definition of derivative by $f(x)=\log(1+x)$

$$\lim_{x \to 3} \frac{\log(1+x) - \log(4)}{x^2 - 9}=\lim_{x \to 3} \frac{1}{x+3}\cdot \lim_{x \to 3} \frac{\log(1+x) - \log(4)}{x-3}=\frac16 \cdot f'(3)$$

or also by standard limits with $h=x-3 \to 0$

$$\lim_{x \to 3} \frac{\log(1+x) - \log(4)}{x^2 - 9}=\lim_{h \to 0} \frac{\log(4+h) - \log(4)}{h(h+6)}=\lim_{h \to 0} \frac{1}{4(h+6)}\frac{\log\left(1+\frac h4\right)}{\frac h4} = \frac1{24}$$


答: $S_4$ has subgroups of orders 1,2,3,4,6,8,12 and 24. Except subgroups of orders 1,2 and 3 (these groups are always cyclic) and some subgroups of order 4, all other subgroups are non cyclic.

Title: Is every subgroup of $S_n$ cyclic?


问:
Simple model-theoretic arguments in set theory
Consider the following simple proof that PA has models with nonstandard numbers.

The type $\{ x > n \mid n \in \mathbb{N} \}$ is finitely realizable, so there is an $M \models \mathrm{PA}$ which fully realizes it. Any realization of this type is larger than all the standard natural numbers. QED

Arguments of this kind, and analogous ones via omitting types instead of realizing them, are the bread and butter of model theory. My question is this. Are there any obvious reasons why simple arguments of this kind can't prove the basic results of undecidability?

Naively, "PA doesn't prove its own consistency" is equivalent to "there is $M \models \mathrm{PA}$ which contains a proof of 0=1". (Of course, such a proof would be nonstandard.) I tried for a bit to prove this by a type-realization argument like the one above, but couldn't. Are there known obstacles to such a strategy?

Relatedly, could there be a simple type realizing/omitting argument to construct an $M \models \mathrm{PA}$ which contains some sentence $\varphi$ such that, in M, there are no proofs of $\varphi$ nor $\lnot\varphi$? This wouldn't quite show incompleteness of PA (as $\varphi$ could be nonstandard)  but it would be interesting to me.

In general, I guess I'm looking for a bit more of a connection between basic computability and basic model theory -- even if the connection is, "here are good reasons why simple methods from the latter don't work in the former".

答:
Here's one explanation, not too far from the lines Noah is drawing in his answer.

How do you prove your result? Well, first you say, add a constant $c$ and axioms stating that it is realising this type. These are finitely consistent, using $\Bbb N$ itself as a model. So by compactness, there is a model where this type is realised. But how do you prove this compactness result? Using ultraproducts and Łoś's theorem, usually,1 and therefore in this case the model where you realise the type is elementarily equivalent to $\Bbb N$. If we want a countable model, we use the Löwenheim–Skolem theorem, which also preserve elementarity.

So all of the steps we took to get to the model do not change the theory, and in fact, do not change the theory of the language expanded with constants for the original model (in this case this is unnecessary since $\Bbb N$ is very nice: every element has a closed term which defines it).

Therefore, we did not produce a model of $\sf PA$ whose theory is different from that of $\Bbb N$. Now you can argue that "in principle, we could have!" but to do that you'd have to use other models of $\sf PA$ first, and you don't yet know they exist (in principle). Only after we have proved the incompleteness of $\sf PA$ you can use the fact that there models with different theories.

1. Once I was discussing choice-related stuff with a model theorist, and they were quite surprised to find out that compactness does not follow from Łoś's theorem, as a general principle, and it is just a rather convenient proof we get in $\sf ZFC$. The point is that in $\sf ZF$ the compactness theorem is independent from Łoś's theorem.


问:
Proof that the method of Gauss/Jordan yields the inverse of a matrix
I have trouble in solving the following exercise: let A be an invertible matrix. Consider the matrix A|I where I is the identity matrix.
Prove that the matrix obtained by transforming A|I in it's row-echelon form is equal to I|B where B is the inverse matrix of A. Any help would be appreciated.</p>

答:
In the matrix $A|I$ imagine we place in the middle the matrix inverse $B$ and we have
$$AB|I$$
which means obviously that $AB=I$ and an elementary row  operation on A is multiplying A from the left  by an elementary matrix $E_j $ until we find the matrix $I$ so we find
$$\underbrace{E_rE_{r-1}\cdots E_1A}_{=I}B|E_rE_{r-1}\cdots E_1I$$
hence we find that
$$B=E_r E_{r-1}\cdots E_1$$
and we can do the same method for the elementary column operations.</p>


问:
Is this approach to the limit solution kosher?$ \lim_{n\rightarrow\infty}\frac{(n!)^{\tfrac{1}{n}}}{n}=\frac{1}{e}$
Here is a non-rigourous approach to a limit:
$$ \lim_{n\rightarrow\infty}\frac{(n!)^{\tfrac{1}{n}}}{n}=\frac{1}{e}$$
I know the solution is correct. I'm suspicious of my methods.

Rewrite the expression:

$$ \lim_{n\rightarrow\infty}\left(\frac{n!}{n^n}\right)^{\frac{1}{n}}$$

I know the $\lim_{n\rightarrow\infty}\left(\frac{n!}{n^n}\right)=0$. So bear with me.

$$\frac{n!}{n^n}=\frac{n(n-1)(n-2)...[n-(n-2)][n-(n-1)]}{n^n}$$

Written this way, the numerator is a polynomial of degree $n$. For large $n$, the leading term will prevail. I will write that polynomial as a sum of $n^n$ and a polynomial of degree $n-1$.

$$\frac{n!}{n^n}\Rightarrow\frac{n^n+[\text{polynomial}]^{n-1}}{n^n}\Rightarrow 1+\frac{1}{n}$$

As noted in Step 2, this is not the expected result for this limit. But watch:

$$ \lim_{n\rightarrow\infty}\frac{(n!)^{\tfrac{1}{n}}}{n}=\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{\tfrac{1}{n}}=\lim_{n\rightarrow\infty}\frac{1}{\left(1+\frac{1}{n}\right)^n}=\frac{1}{e}$$

This is the correct solution. But Step 2 is sketchy. Did I make another error which undid this error? Can this derivation be salvaged, or is my proof "not even wrong"?

Please be gentle with me.

答:
Another very quick way is to use one of the asymptotic Stirling approximations for the factorial function. That is,
$$
n!\sim \frac{n^n}{e^n}
$$
So
$$
\lim_{n\rightarrow\infty}\frac{(n!)^{\frac{1}{n}}}{n}=\lim_{n\rightarrow\infty}\frac{(\frac{n^n}{e^n})^{\frac{1}{n}}}{n}=\lim_{n\rightarrow\infty}\frac{(n)}{e(n)}=\frac{1}{e}
$$
Hence
$$
\lim_{n\rightarrow\infty}\frac{(n!)^{\frac{1}{n}}}{n}=\frac{1}{e}
$$


问:
Is my approach correct in solving for subspace and finding dimension?
Let $V$ be a vector space over $\mathbb{R}$ and let $v_{1},v_{2},...,v_{m} \in V$; Let $U$={$c_{1}v_{1}+,...+c_{m}v_{m}:c_{1},...c_{m}\in \mathbb{R}$ and $c_{1}+..+c_{m}$=0}. Is $U$ a subspace of $V$. What is the dimension of $U$?

my attempt for finding dimension of $U$ :: $c_{1}$ can be written in terms of $c_{2},..c_{m}$ and substituting $c_{1}$ in the above equation we can express $U$ as linear combination of $v_{2}-v{1},...,v_{m}-v_{1}$ so dimension of $U$ is $m-1$ is this right or is there a case where $dim(U)<m-1$? And also what happens if $c_{1}+..+c_{m}=1$, Is $U$ even a subspace for this case?
EDIT In the above case i assumed $v_{1},v_{2},...,v_{m}$ are linearly independent.

答:
In case your interested:

Theorem: Let be $\;\phi:V\to\Bbb F\;$ a linear functional from a linear space $\;V\;$ to the definition field $\;\Bbb F\;$ . If $\;\phi\neq0\;$ , then $\;\ker\phi\;$ is a maximal subspace of $\;V\;$ . In the finite dimensional case, say $\;\dim V=n<\infty\;$ , this means $\;\dim\ker\phi=n-1\;$ .

The above is not so hard to prove with Zorn's Lemma (or the dimension theorem in the finite dimensional case).

Now in your question: the map $\;\phi:\text{Span}\,\{v_1,...,v_m\}\to\Bbb F\;$ defined by

$$\phi\left(\sum_{k=1}^n a_kv_k\right)=\sum_{k=1}^m a_k$$

is a linear map and thus a linear functional, so $\;U=\ker\phi\;$ is always a subspace of $\;V\;$ . In case $\;\{v_1,...,v_n\}\;$ are linearly independent then $\;\dim\ker\phi=m-1\;$ , and in case $\;\{v_1,...,v_m\}\;$ is a basis of $\;V\;$ then $\;U=\ker\phi\;$ is a maximal subspace of $\;V\;$ .


问:
Context-Free Grammar construction when order isn't specified
I'm having trouble constructing a grammar for

L={wϵ{a,b}|na(w)≠nb(w)}

So I need to construct a grammar for the language wherein the number of a's does not equal the number of b's. The farthest I've attempted is splitting the language into

L={wϵ{a,b}*|na(w)>nb(w)}∪{wϵ{a,b}*|na(w)<nb(w)}

but I don't know where to go from there. What makes this problem challenging is that no order is specified so I don't even know where to begin.

Thus, the second part of my questions is if there is any general rule I should follow when dealing with languages that don't stipulate an order.

答: Hint: The first step in your grammar should decide whether $n_{\mathtt{a}} > n_{\mathtt{b}}$ or $n_{\mathtt{a}} < n_{\mathtt{b}}$, and then the two "halves" of your grammar will then construct appropriate strings.  This related question might help you.


问:
Link projective resolution and length
I am wondering if there is a link between a projective resolution of an R-module M and its composition serie. If we have M of finite length can we conclude something ? If we know one of the composition series can we construct a projective resolution ?

Thank you really much.

答:
The answer is very much no.  For instance, two non-isomorphic modules with the same composition factors can have very different projective resolutions.

Here's an example: take $k$ a field, and $R = k[x]/(x^2)$.  Let $S = k[x]/(x)$ be the simple $R$-module.  Then the $R$-modules $R$ and $S^2$ have the same composition factors, but their projective resolutions are
$$
 0 \to R \xrightarrow{1}R \to 0
$$
and
$$
 \ldots \xrightarrow{x} R\xrightarrow{x} R\xrightarrow{x} R \to S \to 0,
$$
respectively.  Thus $R$ has projective dimension $0$, while $S$ has infinite projective dimension.

Title: Link projective resolution and length


问:
Proof of divisibility with combinatorics
Prove that, for every positive integer $\,k, {2k\choose k}\,$ is divisible by $k+1$. Hint: $\frac{2k+1}{k+1}{2k\choose k}$

What I have done so far

${2k\choose k} = 2k!/k!k!=((k+k)(k+k+1)(k+k+2)\cdots k!)/k!k!= ((k+1)(2k!/(k+1))/k!k!$

Is this proof correct or is there a better way to prove it?

答:
You are correct that $\binom{2k}{k} = \frac{(k+1) (2k)! / (k+1)}{k! k!}$, but this doesn't show that $\binom{2k}{k}$ is divisible by $k+1$. You can take any number $n$ and write $n = n(k+1)/(k+1)$ even though $n$ is not divisible by $k+1$.

The purpose of the hint is for you to recognize that $\frac{2k+1}{k+1} \binom{2k}{k} = \binom{2k+1}{k}$ is an integer. This means $(2k+1) \binom{2k}{k}$ is divisible by $k+1$. To finish the proof, think about the common factors of $2k+1$ and $k+1$.


问:
Relation p^3 on set
For the relation p = {(1, 4), (2, 1), (2, 3), (4, 2)} on the set {1, 2, 3, 4}, determine the relation p^3.

I get that p^3 is p°p°p but how do I apply this to the exercise? Could p be described as = {(a, d), (b, c), (b, c), (d, b)}, and the set as {a, b, c, d} and then I just have to apply this three times?

答: Given a relation $R$ over some set $A$, we define the composition of $R$ with itself as $$R\circ R := \{(a,c)\in A\times A\phantom{i} | \phantom{i} \exists b : aRb \phantom{i} \& \phantom{i}bRc\}.$$ Given your set $p$, we could quite easily manually check which elements satisfy this definition. As a matter of fact, this turns out to be the set $$p^3 := p\circ p\circ p = \{(1,1), (1,2), (1,3), (2,2), (4,4)\}.$$

Title: Relation p^3 on set


问:
Is Synge's world function redundant?
Consider set $\mathcal S$ on which Synge's world function is defined, $\sigma : \mathcal S^2 \rightarrow \mathbb R$.

Set $\mathcal S$ is required to be convex wrt. function $\sigma$. Expressing this rather informally but suggestively: for each pair of (distinct) points $(X, Y) \in \mathcal S^2$ there is a unique sraight line segment connecting $X$ and $Y$, completely contained in set $\mathcal S$. Expressing this explicitly instead, in terms of values of function $\sigma$, where (in application of Heron's formula) three distinct points $J, K, Q \in \mathcal S$ are called "straight wrt. each other" iff
$$(\sigma[ \, J, K \, ])^2 + (\sigma[ \, J, Q \, ])^2 + (\sigma[ \, K, Q \, ])^2 = \\
2 \, \sigma[ \, J, K \, ] \, \sigma[ \, J, Q \, ] + 2 \, \sigma[ \, J, K \, ] \, \sigma[ \, K, Q \, ] + 2 \, \sigma[ \, J, Q \, ] \, \sigma[ \, K, Q \, ],$$
is surely possible, albeit cumbersome.

Considering further some particuar (generally interior) point $P \in \mathcal S$, all pairs of (not necessarily distinct) points $(X, Y) \in \mathcal S^2$ may be distinguished into

those pairs for which $P$ belongs to the unique straight line segment connecting $X$ and $Y$,

namely set $\mathcal G_P \subset \mathcal S^2$ (where specificly also $\forall \, X \in \mathcal S : (X, P) \in \mathcal G_P \text{ as well as } (P, X) \in \mathcal G_P )$;

and all other pairs, namely set $(\mathcal S^2 \setminus \mathcal G_P)$.

My question: Given only the function $$\hat\sigma : (\mathcal S^2 \setminus \mathcal G_P) \rightarrow \mathbb R, \qquad \hat\sigma[ \, A, B \, ] \mapsto \sigma[ \, A, B \, ]$$
is it possible to calculate the remaining values $\sigma[ \, X, Y \, ]$ for any or all argument pairs $(X, Y) \in \mathcal G_P$ ?

For completeness, here the explicit definition:

$\mathcal G_P \equiv \{ (X, Y) \in \mathcal S^2 :$

either: $\sigma[ \, X, Y \, ] \ne 0$ and
$$\text{Sgn}[ \, \sigma[ \, X, Y \, ] \, ] \, \sqrt{ \text{Sgn}[ \, \sigma[ \, X, Y \, ] \, ] \, \sigma[ \, X, Y \, ] } = \\ \text{Sgn}[ \, \sigma[ \, X, P \, ] \, ] \, \sqrt{ \text{Sgn}[ \, \sigma[ \, X, P \, ] \, ] \, \sigma[ \, X, P \, ] } +
\text{Sgn}[ \, \sigma[ \, P, Y \, ] \, ] \, \sqrt{ \text{Sgn}[ \, \sigma[ \, P, Y \, ] \, ] \, \sigma[ \, P, Y \, ] },$$

or: $\sigma[ \, X, Y \, ] = \sigma[ \, X, P \, ] = \sigma[ \, P, Y \, ] = 0$ and
$$\exists \, Q \in \mathcal S \, \mid \, (\sigma[ \, P, Q \, ] = 0) \text{ and } (\sigma[ \, X, P \, ] \, \sigma[ \, P, Y \, ] \lt 0)$$
$ \}$.

答:
1. Determining the signature of $\hat\sigma$

If any four points $A, B, J, K \in (\mathcal S^2 \setminus \mathcal G_P)$ can be found such that

$$\hat\sigma[ \, A, J \, ] \, \hat\sigma[ \, A, B \, ] \gt 0, \qquad \hat\sigma[ \, J, B \, ] \, \hat\sigma[ \, A, B \, ] \gt 0, \\ \hat\sigma[ \, A, K \, ] \, \hat\sigma[ \, A, B \, ] \gt 0, \qquad \hat\sigma[ \, K, B \, ] \, \hat\sigma[ \, A, B \, ] \gt 0, $$

and
$$\sqrt{\text{Sgn}[ \, \hat\sigma[ \, A, J \, ] \, ] \, \hat\sigma[ \, A, J \, ]} +  \sqrt{\text{Sgn}[ \, \hat\sigma[ \, J, B \, ] \, ] \, \hat\sigma[ \, J, B \, ]} \lt \qquad \qquad \qquad \qquad \qquad \qquad \\ \sqrt{\text{Sgn}[ \, \hat\sigma[ \, A, B \, ] \, ] \, \hat\sigma[ \, A, B \, ]} \lt \\ \qquad \qquad \qquad \qquad \qquad \qquad \sqrt{\text{Sgn}[ \, \hat\sigma[ \, A, K \, ] \, ] \, \hat\sigma[ \, A, K \, ]} + \sqrt{\text{Sgn}[ \, \hat\sigma[ \, K, B \, ] \, ] \, \hat\sigma[ \, K, B \, ]}$$

then any pair of points $(U, V) \in (\mathcal S^2 \setminus \mathcal G_P)$ for which $\hat\sigma[ \, U, V \, ] \, \hat\sigma[ \, A, B \, ] \gt 0$ will in the following be called spacelike separated;

any pair $M, N \in (\mathcal S^2 \setminus \mathcal G_P)$ for which $\hat\sigma[ \, M, N \, ] = 0$ will be called lightlike separated;

and any pair $Q, Z \in (\mathcal S^2 \setminus \mathcal G_P)$ remaining will be called timelike separated.

For spacelike separated points $A, B$ and timelike separated points $Q, Z$ therefore $\hat\sigma[ \, A, B \, ] \, \hat\sigma[ \, Q, Z \, ] \lt 0$, of course.

2. Determining the signature of $(X, Y) \in \mathcal G_P$

Consider any and all (simple, invertible) curves $\gamma : [0 \ldots 1] \rightarrow (\mathcal S \setminus \{ P \}), \qquad \gamma[ \, 0 \, ] \mapsto X, \qquad \gamma[ \, 1 \, ] \mapsto Y$.

If among them there exist curves $\overline\gamma$ such that

$\forall \, r \in \mathbb R \, \mid \, 0 \lt r \lt 1 : $ the pair $(X, \overline\gamma[ \, r \, ]) \in (\mathcal S^2 \setminus \mathcal G_P)$ and timelike separated, and

$\forall \, s \in \mathbb R \, \mid \, 0 \lt s \lt 1 : $ the pair $(\overline\gamma[ \, s \, ], Y) \in (\mathcal S^2 \setminus \mathcal G_P)$ and timelike separated, and

$\forall \, r, s \in \mathbb R \, \mid \, 0 \lt r \lt s \lt 1 : $ the pair $(\overline\gamma[ \, r \, ], \overline\gamma[ \, s \, ]) \in (\mathcal S^2 \setminus \mathcal G_P)$ and timelike separated,

then the pair $(X, Y)$ is called timelike separated as well.

Vice versa, if there exists a curve $\overline\gamma$ whose points, with exception of the pair $(X, Y)$ of endpoints itself, are all pairwise spacelike separated, then the pair $(X, Y)$ is called spacelike separated, too.

All remaining pairs $(X, Y) \in \mathcal G_P$ are called lightlike separated.

3. Determining $\sigma[ \, X, Y \, ]$ of points $(X, Y) \in \mathcal G_P$

3.1 $(X, Y)$ lightlike separated:

$$\sigma[ \, X, Y \, ] := 0.$$

3.2 $(X, Y)$ timelike separated:

For each of the completely timelike curves $\overline\gamma \in \overline \Gamma,$

$\overline\gamma : [0 \ldots 1] \rightarrow (\mathcal S \setminus \{ P \}), \qquad \overline\gamma[ \, 0 \, ] \mapsto X, \qquad \overline\gamma[ \, 1 \, ] \mapsto Y$ consider all its finite partitionings $t \in \mathcal T$, with $t_k \in [0 \ldots 1],$ integer indices $k \in [0, 1, \ldots n] \qquad t_0 = 0, \qquad t_n = 1, \qquad (j \lt k) \implies (t_j \lt t_k)$,
with $n \ge 2$ and such that $\forall k \in [0, \ldots (n - 1)] : (\overline\gamma[ \, t_k \, ], \overline\gamma[ \, t_{(k + 1)} \, ]) \in (\mathcal S^2 \setminus \mathcal G_P)$. Then

$$\sigma[ \, X, Y \, ] := -\text{Sgn}[ \, \hat\sigma[ \, A, B \, ] \, ]  \left( \underset{\overline\gamma \in \overline\Gamma}{\text{Sup}} \! \! \left[ \, \underset{t \in \mathcal T}{\text{Sup}} \! \! \left[ \, \sum_{k = 0}^{(n[t] - 1)}\left[ \, \sqrt{ -\text{Sgn}[ \, \hat\sigma[ \, A, B \, ] \, ] \, \hat\sigma[ \, \overline\gamma[ \, t_k \, ], \overline\gamma[ \, t_{(k + 1)} \, ] \, ] } \, \right] \, \right] \, \right] \right)^{\! \! 2}.$$

3.3 $(X, Y)$ spacelike separated:

For each of the completely spacelike curves $\overline\gamma \in \overline\Gamma,$

$\overline\gamma : [0 \ldots 1] \rightarrow (\mathcal S \setminus \{ P \}), \qquad \overline\gamma[ \, 0 \, ] \mapsto X, \qquad \overline\gamma[ \, 1 \, ] \mapsto Y$ consider all its finite partitionings $t \in \mathcal T$, with $t_k \in [0 \ldots 1],$
integer indices $k \in [0, 1, \ldots n] \qquad t_0 = 0, \qquad t_n = 1, \qquad (j \lt k) \implies (t_j \lt t_k)$,
with $n \ge 2$ and such that $\forall k \in [0, \ldots (n - 1)] : (\overline\gamma[ \, t_k \, ], \overline\gamma[ \, t_{(k + 1)} \, ]) \in (\mathcal S^2 \setminus \mathcal G_P)$. Then

$$\sigma[ \, X, Y \, ] := \text{Sgn}[ \, \hat\sigma[ \, A, B \, ] \, ] \, \left( \underset{\overline\gamma \in \overline\Gamma}{\text{Inf}}\left[ \, \underset{t \in \mathcal T}{\text{Sup}}\left[ \, \sum_{k = 0}^{(n[t] - 1)}\left[ \, \sqrt{ \text{Sgn}[ \, \hat\sigma[ \, A, B \, ] \, ] \, \hat\sigma[ \, \overline\gamma[ \, t_k \, ], \overline\gamma[ \, t_{(k + 1)} \, ] \, ] } \, \right] \, \right] \, \right] \right)^{\! \! 2}.$$

4. Determining $\sigma[ \, X, P \, ] = \sigma[ \, P, X \, ]$

4.1 $X \equiv P$:

$$\sigma[ \, P, P \, ] = 0.$$

4.2 $\sigma[ \, X, P \, ]$ for $(X, Y) \in \mathcal G_P$ lightlike separated:

$$\sigma[ \, X, P \, ] = \sigma[ \, P, X \, ] = 0.$$

4.3 $\sigma[ \, X, P \, ]$ for $(X, Y) \in \mathcal G_P$:

Define set $\mathcal H_X \equiv \{ H \in (\mathcal S \setminus \{ P \}) \, \mid \, ((X, H) \in (\mathcal S^2 \setminus \mathcal G_P) \text{ and } $
$(\text{Sgn}[ \, \sigma[ \, X, Y \, ] \, ] \, \sqrt{ \text{Sgn}[ \, \sigma[ \, X, Y \, ] \, ] \, \sigma[ \, X, Y \, ] } = $

$\text{Sgn}[ \, \sigma[ \, X, H \, ] \, ] \, \sqrt{ \text{Sgn}[ \, \sigma[ \, X, H \, ] \, ] \, \sigma[ \, X, H \, ] } +
\text{Sgn}[ \, \sigma[ \, H, Y \, ] \, ] \, \sqrt{ \text{Sgn}[ \, \sigma[ \, H, Y \, ] \, ] \, \sigma[ \, H, Y \, ] })$

$\}.$

In terms of this:

$$\sigma[ \, X, P \, ] := \text{Sgn}[ \, \sigma[ \, X, Y \, ] \, ] \left( \underset{H \in \mathcal H_X}{\text{Sup}}\left[ \, \sqrt{ \text{Sgn}[ \, \sigma[ \, X, H \, ] \, ] \, \sigma[ \, X, H \, ] } \, \right] \right).$$

Title: Is Synge's world function redundant?


问:
Automorphism group of quaternions
Let $\mathbb{H}$ be the algebra of quaternions. How to prove that the automorphism group $Aut(\mathbb{H})$ is isomorphic to the group $SO_3$?

I found that all derivations on algebra $\mathbb{H}$ are inner.  Moreover, they are isomorphic to the Lie algebra $so_3$. It remains to understand whether it is simply connected or not. But I don’t know how to show that the group $Aut(\mathbb{H})$ is not simply connected.

答:
An automorphism $\alpha$ must preserve $1$, so by $\mathbb{R}$-linearity it preserves the real line.

Also $\alpha$ must preserve the set of square roots of $-1$, which is the unit sphere $S^2$, and all of their real multiples, which forms the pure imaginary subspace $\mathbb{R}^3$ of 3D vectors.

If $\mathbf{u}$ and $\mathbf{v}$ are vectors, the real part of $\mathbf{uv}$ is minus the dot product, i.e. $-\mathbf{u}\cdot\mathbf{v}$. Since $\alpha$ preserves real parts, this means it preserves dot products, so it acts as a linear isometry of $\mathbb{R}^3$. So $\alpha\in\mathrm{O}(3)$.

Similarly the imaginary part of $\mathbf{uv}$ is the cross product $\mathbf{u}\times\mathbf{v}$. See if you can show $\alpha$ preserves cross products, and this will further mean $\alpha\in\mathrm{SO}(3)$.

Conversely, any element of $\mathrm{SO}(3)$ preserves dot products and cross products, so preserves the quaternion multiplication (which is built out of them). Because of $S^3\to\mathrm{SO}(3)$, this means all automorphisms of $\mathbb{H}$ are inner.

Title: Automorphism group of quaternions


问:
Is this map an homeomorphism with its image?
Suppose $\varphi:\mathbb{R}^2 \to \mathbb{R}^3$ is defined as:

$$\varphi(u,v)=(u-\frac{u^3}{3}+uv^2,v-\frac{v^3}{3}+vu^2,u^2-v^2)$$

Can I conclude that $\varphi$ is an homeomorphism with its image?

I tried defining $x=u-\frac{u^3}{3}+uv^2$, $y=v-\frac{v^3}{3}+vu^2$ and $z=u^2-v^2$ finding $u=\frac{3x}{3-z}$ and $v=\frac{3y}{1+z}$ but I do not know if this is enough to conclude.

答: No. Your function is not injective, since$$\varphi\left(\sqrt3,0\right)=\varphi\left(-\sqrt3,0\right)=(0,0,3).$$


问:
Find $\sum_{\rho}\frac{1}{\rho^3}$ where $\rho$ is a non trivial zero of zeta function
I am reading the Equivalents of Riemann Hypothesis: Arithmetic Equivalents pg. 35 Lemma 2.10. If $\rho$ is a non trivial zero of the Riemann Zeta function then,
$\sum_{\rho}\frac{1}{\rho}=1+\frac{\gamma}{2}-\frac{1}{2}
     log  \ 4\pi$

I want to know what is $\sum_{\rho}\frac{1}{\rho^3}$?. Please provide a reference if possible..

答:
Edit I thought it was $\sum \rho^{-2}$, for $\sum \rho^{-3}$ it works the same way

$$\lim_{a\to \infty}\int_{a-i\infty}^{a+i\infty} \frac{\zeta'(s)}{s^3\zeta(s)}ds=0$$
This is also equal to

$$\int_{2-i\infty\to 2+i\infty\to -\infty\to 2-i\infty} \frac{\zeta'(s)}{s^3\zeta(s)}ds$$
Then apply the residue theorem, you'll get your series plus the residue at $1$, the residue at $0$ and the sum of the residues at the trivial zeros which is $\sum_{n=1}^\infty \frac1{(-2n)^3}$

The residue at $0$ depends on known constants plus $\zeta'(0),\zeta''(0),\zeta'''(0)$, by the functional equation those relate to the Laurent coefficients of $\zeta(s)$ at $s=1$


问:
Are these statements on open sets and limit points true or false?

If $D \subseteq \mathbb{R}^n$ contains an isolated point then $D$ is not open.

Let $D \subseteq \mathbb{R}^n$ and $a \in D$. If $a$ is a limit point of $D$ then there is some $\epsilon > 0$ such that $B_{\epsilon}(a) \subseteq D$.

For the first, by definition of isolated point, it must be in $D$ but not be a limit point of $D$. I have a feeling if $D = \mathbb{R}^n$ this could be a counter example, as it is a clopen set.

For the second, the consequent is the definition of open if I'm not mistaken, but I'm not sure how this relates to $a$ being a limit point of $D$.

Any help would be appreciated?

答:
If $x$ is an isolated point of $D$, there is an $\epsilon_0>0$ such that $B(x,\epsilon_0)\cap D=\{x\}$. If $D$ were open, there would be an $\epsilon_1>0$ such that $B(x,\epsilon_1)\subseteq D$. Let $\epsilon=\min\{\epsilon_0,\epsilon_1)$; then on the one hand $$x\in B(x,\epsilon)\cap D\subseteq B(x,\epsilon_0)\cap D=\{x\}\,,$$ so $B(x,\epsilon)\cap D=\{x\}$, but on the other hand

$$B(x,\epsilon)\subseteq B(x,\epsilon_1)\subseteq D\,,$$

so $B(x,\epsilon)\cap D=B(x,\epsilon)$. Thus, $B(x,\epsilon)=\{x\}$, which is false in $\Bbb R^n$. This shows that $D$ cannot be open.

For the second question, let $n=1$, and let $D=(0,1)$; $0$ is a limit point of $D$, but there is no $\epsilon>0$ such that $(-\epsilon,\epsilon)\subseteq(0,1)$.


问:
Find a function for well form brackets using generating functions
How to find a function for calculating the number of well form brackets (for "n" pairs of brackets) using generating function? This is a probably a routine problem for some people, but I haven't got any resource on generating function.</p>

答: If Hans Lundmark is right, you could see Wikipedia on the Catalan numbers, where the generating function is given under "Proof of the Formula"  There are also many references in OEIS</p>


问:
finding $x^6+y^6$ given $x+y$ and $xy$
Here is the question

If $x + y = 4$ and $xy = 2$, then find $x^6+ y^6$.

This is from a previous timed competition, so fastest answers are the best answers.

I've tried using sum of cubes, but I dont know what to do after $(x^2+y^2)(x^4-x^2y^2+y^4)$
. The only other way I can think of is solving for x and y, but that wouldn't be too quick. Any help?

答:
Since $(x,y)$ are roots of $\ x^2-sx+p=0$

Then consider this equation as the characteristic equation of the linear induction relation $$U_{n+1}=sU_u-pU_{n-1}$$

Which solution is $\ U_n=x^n+y^n\ $ with initial values $\begin{cases}U_0=2\\U_1=4\end{cases}$

Then go on calculating successive values:

$U_2=4U_1-2U_0=12$

$U_3=4U_2-2U_1=40$

$U_4=4U_3-2U_2=136$

$U_5=4U_4-2U_3=464$

$U_6=4U_5-2U_4=1584$

The advantage of this method, is that it is a no-brainer even if you are asked to get $x^{17}+y^{17}$ for instance, it is quite easy to calculate more terms.

Title: finding $x^6+y^6$ given $x+y$ and $xy$


问:
Closed-form of $\mathbb E(\|G\|_\infty)$ where $G\sim\mathcal N(0,\mathbf{Id}_n)$.
Let $I_n = \mathbb E(\|G\|_\infty)$, i.e.
$$I_n = (2\pi)^{-\frac{n}{2}}\int_{x\in\mathbb R^n}\|x\|_\infty e^{-\frac{1}{2}\|x\|_2^2}\,dx.$$
I wonder if I can get its closed-form. By symmetry I got
$$I_n = 2n\sqrt{\frac{2}{\pi}}\int_0^\infty xe^{-x^2}\operatorname{erf}(x)^{n-1}\,dx,$$
and then by integration by parts, for $n\ge2$,
$$I_n = \frac{2\sqrt2}{\pi}n(n-1)\int_0^\infty e^{-2x^2}\operatorname{erf}(x)^{n-2}\,dx,$$
where $\operatorname{erf}$ is the error function.

These two formulas give me

$$I_1 = \sqrt{\frac{2}{\pi}},\quad I_2 = 2\sqrt{\frac{1}{\pi}},\quad I_3 = \frac{12}{\pi\sqrt\pi}\arctan\frac{\sqrt2}{2}.$$

In this step, I think a general closed-form is almost impossible, so I post here to see if anyone has a better approach (at least for $I_4$).

Update

Series expansion of $I_4$:
$$I_4 = \frac{8\sqrt2}{\pi^2}\sum_{n=0}^{\infty}\left(\frac43\right)^n\frac{n!}{(2n+1)!}\,\Gamma(n+3/2)\,{}_2F_1(1/2,-n;3/2;1/4).$$

By the way
$$I_n = \sqrt2n\int_0^1t^{n-1}\operatorname{erf}^{-1}(t)\,dt \,=\!\!\!?\; \sqrt2n\sum_{k=0}^\infty a_k \left(\frac{\sqrt\pi}{2}\right)^{2k+1}\frac1{2k+n+1},$$
where $a_k$ is the $k$-th coefficient of the Maclaurin series of $\operatorname{erf}^{-1}(2x/\sqrt\pi)$ (see InverseErf).

Well, I don't really know the behavior of $(a_k)$, but numerically the series does converge. I don't think this will lead to anything though.

Let me explain a little about this problem.

Imagine we have $n$ points to throw at 0 at the real axis, and the resulted position of one point is determined by $\mathcal N(0,1)$. We want to study the behavior of the farthest distance from 0.

This distance $D = \|G\|_\infty$ is determined by the density function defined below

$$f:x \mapsto n\sqrt{\frac2\pi}\,\exp\left(-\frac{x^2}2\right) \operatorname{erf}^{n-1}\frac{x}{\sqrt2} \mathbb1_{x\ge0}.$$

(For fun one can check that $\int_0^\infty f(x)\,dx=1$.)

And now, what we want to know is, how to calculate $\mathbb E(D)$ (at least when $n=4$)?

@YuriNegometyanov has given a formula for $\mathbb E(\|G\|_2)$. Even though it's not quite the topic, let's write it down as well:

$$\mathbb E(\|G\|_2) =\sqrt2\,\frac{\Gamma\left(\dfrac{n+1}2\right)}{\Gamma\left(\dfrac n2\right)}.$$

A jupyter notebook to calculate numerical results.

So from the series expansion of $I_4$ mentioned above (and tons of calculation), I got:
$$I_4 = \frac{24}{\pi\sqrt\pi}\arctan\frac{1}{2\sqrt2}.$$
This is kind of interesting since the form is similar to $I_3$. Maybe a general closed-form is in fact possible?

答:
Let $J(a,n) = \int_0^\infty e^{-at}\operatorname{erf}^n\sqrt t\,dt$ for $a>0$.

Let $J_n=J(1,n)$, we have then $I_n = n\sqrt{\frac{2}{\pi}}J_{n-1}$.

By some equalities, we have the recurrence relation below:

$$J(a,0)=\frac1a,\quad J(a,1)=\frac1{a\sqrt{a+1}},$$
$$J(a,n)=J(a,n-2)-\frac{4}{\pi}\int_0^1\frac{1}{1+s^2}J\left(1+s^2+a,n-2\right)\,ds.$$

Then by some calculation, we have

$$J(a,2)=\frac{4}{\pi}\frac1{a\sqrt{a+1}}\arctan\frac1{\sqrt{a+1}},$$
and
$$J(a,3)=\frac{4}{\pi}\frac1{a\sqrt{a+1}}\arctan\frac{1-b}{1+b},\text{ where }b=\frac{a}{a+4}\sqrt{\frac{a+3}{a+1}}.$$

(By the way, for fun one can prove that
$$2\arctan\frac{5-\sqrt2}{5+\sqrt2}=3\arctan\frac{1}{2\sqrt2},$$
which shows up in $I_4$.)

We can also give an expression of $J_4$ (which gives $I_5$):
\begin{align}
J_4&=J_2-\frac{4}{\pi}\int_0^1\frac{1}{1+s^2}J\left(2+s^2,2\right)\,ds\\
&=J_2-\left(\frac{4}{\pi}\right)^2\int_0^1\frac{1}{1+s^2}\frac{1}{2+s^2}\frac{1}{\sqrt{3+s^2}}\arctan\frac{1}{\sqrt{3+s^2}}\,ds.
\end{align}

As you can see, these become more and more complicated. I really don't think there's a closed-form for $I_n$ when $n\ge5$.

Alternatively, we also have
$$J_n=\sum_{k=0}^n\left(-1\right)^k\binom{n}{k}C_k,$$
where
$$C_k=\mathbb E\left[\phi(U)\right]=\pi^{-k}\int_{u\in\mathbb R^k}\phi(u)\prod_{i=1}^k\frac{1}{1+u_i^2}\,du,$$
$$\phi(u)=\frac{1}{1+\sum_{i=1}^k\left(1+u_i^2\right)},$$
and $U=(U_i)_{1\le i\le k}$ is a random vector of independent $\operatorname{Cauchy}(0,1)$ variables.

This might give us a global view of what happens in that recurrence relation (which I believe is unhelpful for a general closed-form).

(By the way, the formula of $\mathbb E(\|G\|_2)$ given by @YuriNegometyanov can be easily found using $\chi^2$-distribution.)

A little simplification (see here).
$$J(a,3) = \frac{12}{\pi}\frac1{a\sqrt{a+1}}\left(\arctan\sqrt{\frac{a+3}{a+1}}-\frac\pi4\right).$$


问:
Derivation of Quantifier Negation: $(\exists x)Px \vdash -(\forall x) -Px$
I am trying to figure out the derivation of: $(\exists x)Px \vdash -(\forall x)-Px$

This is my thought process so far is to establish a contradiction on Px and -Px, but i'm struggling with an assumption to contradict to get.

Here's my though process so far:

$
(\exists x) Px\\
-Px\\
Px \wedge -Px\\
contradiction?
$

But Im not quite sure how to go from a contradiction on that to $-(\forall x)-Px$, I feel like it would just get me back to $--Px$.

Can anyone provide any suggestions or assistance on how to proceed?

答:
$\def\fitch#1#2{~~~~\begin{array}{|l} #1 \\\hline #2\end{array}}$ That is almost correct, but a negation introduction subproof require assumption of the position you seek to negate.

That is: assume $\forall x~\lnot P(x)$.

 $$\fitch{~~1.~\exists x~P(x)}{\fitch{~~2.~\forall x~\lnot P(x)}{\fitch{~~3.~\boxed a~P(a)}{~~4.~\lnot P(a)\hspace{10ex}\forall\mathsf E~2\\~~5.~\bot\hspace{14ex}\neg\mathsf E~3,4}\\~~6.~\bot\hspace{18ex}\exists\mathsf E~1,3{-}5}\\~~7.~\lnot\forall x~\lnot P(x)\hspace{12ex}\neg\mathsf I~2{-}6}$$


问:
Cyclic subgroups of $GL_2(\mathbb{Z}/p\mathbb{Z})$?
What are the cyclic subgroups of $GL_2(\mathbb{Z}/p\mathbb{Z})$, the general linear group over the finite field of order $p$, where $p$ is prime?

Obviously, each cyclic subgroup is generated by some element $g \in  GL_2(\mathbb{Z}/p\mathbb{Z})$.  What are the possible orders of these cyclic subgroups, and how do I explicitly find a $g$ that generates each (i.e., explicitly find the elements of the $2\times 2$ matrix that corresponds to $g$)?

I can see that their order must divide $(p+1)p(p-1)^2$, but I can't immediately see exactly which orders are possible or how to construct them.  I can see how to generate a cyclic subgroup of order $p-1$ (use the matrix $\left(\begin{smallmatrix}x &0\\0 &x\end{smallmatrix}\right)$, where $x \in (\mathbb{Z}/p\mathbb{Z})^*$ has order $p-1$) or of order $(p-1)p$ (use $\left(\begin{smallmatrix}x &1\\0 &x\end{smallmatrix}\right)$), from which it follows how to find a group whose order is any divisor of $(p-1)p$. However I'm having trouble seeing whether there's a cyclic subgroup of order $p+1$ (for example) nor how to construct it.

答:
You can think of $\mathbb{F}_{p^2}$ as a 2-dimensional vector space over $\mathbb{F}_p$. Multiplication by any non-zero element of $\mathbb{F}_{p^2}$ is then an $\mathbb{F}_p$-linear map on $\mathbb{F}_{p^2}$, and so in this way $\mathbb{F}_{p^2}^\times$ embeds into $GL_2(\mathbb{Z}/p\mathbb{Z})$. Since the multiplcative group of a finite field is cyclic, this tells you that there is an element of $GL_2(\mathbb{Z}/p\mathbb{Z})$ of order $p^2-1$.

In general, the conjugacy class of an element of $GL_2(\mathbb{Z}/p\mathbb{Z})$ is almost determined by its characteristic polynomial. It's a nice exercise to go through the possibilities for the characteristic polynomial (irreducible over $\mathbb{F}_p$, reducible with distinct roots, reducible with double root) and try to see what elements have such characteristic polys. The only case where you have two conjugacy classes with the same characteristic polynomial is if the polynomial has a double root.


问:
Prove that if $\gcd( a, b ) = 1$ then $\gcd( ac, b ) = \gcd( c, b ) $
I know it might be too easy for you guys here. I'm practicing some problems in the textbook, but this one really drove me crazy.

From $\gcd( a, b ) = 1$, I have $ax + by = 1$, where should I go from here? The extra $ac$ is so annoying. Any hint?

Thanks,

Chan

答:
Use the GCD in prime factorization form.

If $m = \prod p_i ^ {m_i}, n = \prod p_i^{n_i}$, then $ d = \prod p_i^{\min(m_i, n_i)}$.

The result follows by showing that

$$ \min (m_i, n_i + k_i) = \min (m_i, \min (m_i, n_i) + k_i)$$

 A direct way of showing this is to condition on $ m_i \geq n_i$ (so $\min(m_i, n_i) = n_i$) and $ m_i < n_i$ (so both sides equal $m_i$).


问:
I have to prove that the following is true using the mean value theorem
If $f'(x)=\sqrt{x^3+1}$ for all $x>0$ and $f(2)=10$, then $f(5) > 16$.

I have what is below so far, but I am not sure how to show f(5) > 16.

Assume $f'(x) = \sqrt{x^{3} + 1}$ for all $x > 0$ and $f(2) = 10$.Since $f'(x)$ is differentiable for all $x > 0$ then $f$ is differentiable on $(2, 5)$ and $f$ is continuous on [$2, 5$].

答:
Hint: Applying the MVT on the interval $[2,5]$ implies the existence of some $c \in [2,5]$ such that
$$\frac{f(5)-f(2)}{5-2} = f'(c) = \sqrt{c^3 + 1}$$
i.e.
$$f(5) = f(2) + (5-2) \sqrt{c^3 + 1}.$$


问:
How to prove that arithmetic progression contains non coprime pairs
There is a progression $a_n=1+bn$. Prove that there is at least one pair $(a_i,a_j)$ such as $\gcd(a_i, a_j) \neq 1$.

When the $a_0$ is not |1|, it quite easy to prove:
$$a_{|a|} = a_0 + |a_0|b$$
$$\gcd(a_0, a_0 + |a_0|b) \geq |a_0|$$
But when the $a_0$ is |1|, I don't know how to play with it.

答:
$a_1=1+b$ and $a_{b+2}=1+b(b+2)=1 +b^2+2b=(1+b)^2$

$\text{GCD}(a_1,a_{b+2})=b+1$ therefore there is at least a pair of non coprime elements of the arithmetic progression.


问:
Why is it not sufficient to only check the third condition when verifying equality of functions?
I have been told that two functions $f$ and $g$ are equal if and only if the domain and the subset of the cartesian product of the two functions is the same. My question is, given that a function is a special case of a relation, both $f$ and $g$ are sets, why is it not sufficient to just verify if the cartesian product is the same between the two functions?. In other words, doesn't the third condition implies that both domains are equal?

答: A function $f:X\to Y$ consists of three pieces of information: the domain $X$, the codomain $Y$, and the graph $G_f\subseteq X\times Y$. So formally, it makes sense to define a function not just as its graph, but as the tuple $(X,Y,G_f)$, and two functions $f=(X,Y,G_f)$ and $g=(V,W,G_g)$ are equal iff $X=V$, $Y=W$, and $G_f=G_g$. In words, two functions are equal iff their domains, codomains and graphs are equal. But we get the domain for free by checking the graphs, since the graphs contain a pair $(x,f(x))$ for every $x\in X$, so we can extract the domain from the graph. So we only need to check the graph and the codomain, but not the domain.


答:
$$Hint**:

The Mean value theorem  asserts that
$$\frac{f(5)-f(2)}{5-2}=f'(c),\quad c\in(2,5)$$
and $f'(c)>\dots$

Title: I have to prove that the following is true using the mean value theorem

Body:
If $f'(x)=\sqrt{x^3+1}$ for all $x>0$ and $f(2)=10$, then $f(5) > 16$.

I have what is below so far, but I am not sure how to show f(5) > 16.

Assume $f'(x) = \sqrt{x^{3} + 1}$ for all $x > 0$ and $f(2) = 10$.Since $f'(x)$ is differentiable for all $x > 0$ then $f$ is differentiable on $(2, 5)$ and $f$ is continuous on [$2, 5$].


问:
A Bernoulli trials problem
Two parents decide to have children until they have 3 children of the same gender one after another (3 in a row). If p(boy)=p(girl)=1/2, how many children are they expected to have?

I have tried to draw a tree diagram to analyse the problem, but i can't generate a useful expression for the probability that a 'success' will happen on the nth trial.

So far P(3)=1/4, P(4)=1/8, P(5)=1/8, P(6)=3/32, P(7)=3/64, P(8)=5/128 and P(9)=1/32.

I have read a similar article on this forum about finding E(x) and Var(x) for 2 consecutive successes, but generating an expression here is far more difficult. In the case of first 2 consecutive successes in k trials, every success must be followed by a failure until the (k-1)th trial. Here not every success needs to be followed by a failure, though 2 consecutive successes must be followed by a failure until the (k-2)th trial.

many thanks,
Yun Fei

答:
Let $t_i$ denote the expected number of children to be born for 3 of the same sex to be born consecutively, after $i$ children of the same sex were born. Thus, $t_3=0$ and one is looking for $t_0$. Conditioning on the sex of the next child, one gets $t_0=1+t_1$, $t_1=1+\frac12t_1+\frac12t_2$ and $t_2=1+\frac12t_1$. Hence $t_0=7$.

For example, assume the last children born are ...MFMFF. Then $i=2$. If the next child is F, one goes to $j=3$ and the mean time to wait after this step is $0$ since the last children born are ...MFMFFF. If the next child is M, one goes back to $j=1$ and the mean time to wait after this step is $t_1$ since the last children born are ...MFMFFM. Thus, $t_2=1+\frac12\cdot0+\frac12t_1$. Likewise for $i=0$ (always goes to $1$) and for $i=1$ (goes to $2$ or to $1$).

Title: A Bernoulli trials problem


问:
small details in Hahn-Banach Separation Theorem (Real Case)
In the real case, of Hahn-Banach Separation Theorem, where $A$ and $B$ are nonempty disjoint convex subsets of $X$ with $A$ open.

Let $a\in A, b\in B$.

Let $C=A-B-a+b$.

I see that $C=\bigcup_{b_0\in B} A-b_0 -a+b$. But why is $C$ open? Is it because it's an union of open sets? But why is each $A-b_0 -a+b$ open?

Also, I don't see why $0 \in C$.

Thanks in advance!

答:
First question: Because A is open.

Second question: Because $0 \in A-a$ and $0 \in B-b$


答:
If $U$ is open then so is $U+x$ for any $x$. This is because $y \to y+x$ is a homeomorhpism. Hence $A-b_0-a+b$ is open. Union of open sets is open.

$0=a-b-a+b \in A-b-a+b \subseteq \bigcup_{b_0 \in B} (A-b_0-a+b)$

Title: small details in Hahn-Banach Separation Theorem (Real Case)

Body:
In the real case, of Hahn-Banach Separation Theorem, where $A$ and $B$ are nonempty disjoint convex subsets of $X$ with $A$ open.

Let $a\in A, b\in B$.

Let $C=A-B-a+b$.

I see that $C=\bigcup_{b_0\in B} A-b_0 -a+b$. But why is $C$ open? Is it because it's an union of open sets? But why is each $A-b_0 -a+b$ open?

Also, I don't see why $0 \in C$.

Thanks in advance!


问:
Is this function one-to-one? Is it onto?
The map $$T:V(\mathbb{R}) \rightarrow S$$ from the vector space of real-valued functions defined on $\mathbb{R}$ to the space of ssignals defined by sending each function to the signal obtained by evaluating at integers is linear (i.e. $T(f_ = {u_k}$) where $u_k = f(k)$).

Is it one-to-one? Is it onto?

I'm honestly not really sure where to start with this proof. I'm guessing that in order to prove that the mapping is onto/one-to-one, we'll have to manipulate T. I also know that functions in $V(\mathbb{R})$ are just arbitrary functions f: $\mathbb{R} \rightarrow \mathbb{R}$.

答:
Is it one-to-one?

Can you find two different functions $f$ and $g$ such that $f(k)=g(k)$ for all integers?

Is it onto?

If you have some sequence of real numbers $(u_k)_{k \in \mathbb{Z}}$, can you find a function $f$ such that $f(k) = u_k$ for all $k$?


问:
Is every geodesic-preserving diffeomorphism an isometry?
Let $M$ be a closed $n$-dimensional Riemannian manifold.

Let $f:M \to M$ be a diffeomorphism and suppose that for every (parametrized) geodesic $\gamma$, $f \circ \gamma$ is also a (parametrized) geodesic.

Must $f$ be an isometry?

An equivalent condition on $f$ is that $\nabla df=0$ where $\nabla=\nabla^{T^*M} $ $ \otimes \nabla^{f^*TM}$ is the relevant tensor product connection.

Note that this equivalent assumption implies that $df$ has constant singular values, and in particular that the Jacobian $\det(df)$ is constant, hence it must be $1$ (since $f$ was assumed to be a diffeomorphism.) Thus $f$ is volume-preserving.

This can certainly be false for manifolds with nonempty boundary in general, as the following example shows:

Let $0<a<b$, and set
$
M=D_{a,b}=\biggl\{(x,y) \,\biggm | \, \frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1 \biggr\}
$
to be the ellipse with diameters $a,b$, endowed with the standard Euclidean metric (induced by $\mathbb{R}^2$). Then there exists $A \in \operatorname{SL}_2(\mathbb R) \setminus \operatorname{SO}(2)$ such that $AD_{a,b}=D_{a,b}$, and $A$ clearly preserves geodesics (it maps straight lines to straight lines.)

Indeed one can take $A$ to be of the form
$$
A =A_{\theta}:=  \begin{pmatrix} a& 0 \\ 0 & b \end{pmatrix} \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos \theta \end{pmatrix}\begin{pmatrix} 1/a& 0 \\ 0 & 1/b \end{pmatrix}=
 \begin{pmatrix} \cos\theta & -\frac ab \sin\theta \\ \frac ba \sin\theta & \cos \theta \end{pmatrix}.
$$

For $M=\mathbb{S}^n $ the answer is positive, by this answer here.

答:
No. As a counterexample, take the torus $\mathbb{T}^n=\mathbb{R}^n/\mathbb{Z}^n$ where $\mathbb{R}^n$ is endowed with the Euclidean metric. Since linear maps preserve geodesics in $\mathbb{R}^n$, $GL(n,\mathbb{Z})$ acts by geodesic-preserving diffeomorphisms on $\mathbb{T}^n$, and many of these are not isometries. For example, consider the map $\mathbb{T}^2\to\mathbb{T}^2$ defined by $(x,y)\mapsto(x,x+y)$ modulo $1$.

Qualitatively speaking, the requirements on $f$ do not constrain the shear components of $df$, though I suppose there may be global or even local obstructions to the existence of such shear maps.


问:
Closed form for $\prod_{n=1}^\infty\sqrt[2^n]{\frac{\Gamma(2^n+\frac{1}{2})}{\Gamma(2^n)}}$
Is there a closed form for the following infinite product?
$$\prod_{n=1}^\infty\sqrt[2^n]{\frac{\Gamma(2^n+\frac{1}{2})}{\Gamma(2^n)}}$$</p>

答:
Firstly, $\Gamma\left(m+\frac{1}{2}\right) = \frac{(2m)!}{4^mm!}\sqrt{\pi}$ when $m\in\mathbb{Z}$, so if we let $m=2^n$, what you have can be written as

$$ \frac{\Gamma\left(m+\frac{1}{2}\right)}{\Gamma\left(m\right)} = \frac{\frac{(2m)!}{4^mm!}\sqrt{\pi}}{(m-1)!} = \frac{(2^{n+1})!\sqrt{\pi}}{4^{2^n}2^n(2^n-1)!^2}. $$

Perhaps someone else can see how to simplify this further because I sure can't.


问:
Doubts about the general solution of $y'' +7y' = e^{-7x}$
I have some doubts about how to find the general integral of this ODE:

$$y'' +7y' = e^{-7x}$$

I know that first we find the solution to the homogeneous ODE and then the particular solution. When I was solving the homogeneous  I got to:

$$y_h= K_1 e^0 + K_2 e^{-7x}$$

My doubt is that is it correct to put $y_h= K_1 + K_2 e^{-7x}$? Or is the $e^0$ some kind of special case I need to know and do something else. Also, to find the particular solution I'm doing

$$ y_p = A+Be^{-7x}$$

and then doing the substitutions and equalling them to the LHS of the original ODE.

Thanks.

答:
HINT

\begin{align*}
y'' + 7y' = e^{-7x} \Longleftrightarrow y' + 7y = -\frac{e^{-7x}}{7} + k \Longleftrightarrow (ye^{7x})' = -\frac{1}{7} + ke^{7x}
\end{align*}

Can you take it from here?


答:
I take $y'=p$. The equation becomes
$$p'+7p=e^{-7 x}$$
which gives the solution
$$p=y'=c_1 e^{-7 x}+e^{-7 x} x$$
Integrating again
$$y= -\frac{1}{49} e^{-7 x} ( c_1+7 x)+c_2$$

Title: Doubts about the general solution of $y'' +7y' = e^{-7x}$

Body:
I have some doubts about how to find the general integral of this ODE:

$$y'' +7y' = e^{-7x}$$

I know that first we find the solution to the homogeneous ODE and then the particular solution. When I was solving the homogeneous  I got to:

$$y_h= K_1 e^0 + K_2 e^{-7x}$$

My doubt is that is it correct to put $y_h= K_1 + K_2 e^{-7x}$? Or is the $e^0$ some kind of special case I need to know and do something else. Also, to find the particular solution I'm doing

$$ y_p = A+Be^{-7x}$$

and then doing the substitutions and equalling them to the LHS of the original ODE.

Thanks.


问:
Show that f is a polynomial

Suppose $f$ is an entire function on $\mathbb{C}^n$ that satisfies for every $\epsilon>0$ a growth-condition $$|f(z)|\leq C_{\epsilon}(1+|z|)^{N_{\epsilon}}e^{\epsilon |
\text{Im}\,z|}$$

Show that $f$ is a polynomial. (Hint: study $\hat{f} = \mathcal{F}(f)$ the Fourier-transform).

I know I'm supposed to apply the Paley-Wiener-Schwartz Theorem, but not sure how.;.

(See http://en.wikipedia.org/wiki/Paley%E2%80%93Wiener_theorem below).

Any suggestions and/or tips are greatly appreciated. Thnx.

答:
Well, I think it should be clear as suggested by Aaron that the Schwartz's Theorem in your link implies that for every $ \epsilon >0 $ you have a distribution $ v $ supported in the closed ball $ B(0,\epsilon) $, and $ f = \hat{v} $. So we have the distribution $v$ with $ f = \hat{v} $ and $ supp(v) = \{0\} $. Hence it is a well known result that for some $k $
$$ v = \sum_{|\alpha | = k}C_\alpha D^\alpha \delta $$
Hence $f(\xi) = \hat{v}(\xi) = \sum_{|\alpha | = k}C_\alpha \hat{(D^\alpha \delta)}(\xi)= \sum_{|\alpha | = k}C_\alpha (2\pi i \xi)^\alpha  $ is indeed a polynomial.</p>

Title: Show that f is a polynomial


问:
Lebesgue integral or Lebesgue–Stieltjes integration?
I was given I tricky problem where I am really confused with the possible solution.

let $\mu$ be a measure defined on $[-1,2]$ as follow: on the set $[-1,2]/\{-1,0,1,2\}, \mu(A) = \int_{A}x^2 dx.$  On the set $B = \{-1,0,1,2\}, \mu(x) = \frac{1}{4}, \forall x\in B$.

The question is asking me to find $\int_{[-1,2]}xd\mu(x)$.

It is pretty clear to me if $\mu(x)$ is just an ordinal Lebesgue measure. However, with this non-standard form of measure, I am not sure how to start. Is this some form of Lebesgue–Stieltjes integration?

Please help! Thank you so very much!

答: $\mu =\mu_1+\frac 1  4 \delta_{-1}+\frac 1  4 \delta_{0}+\frac 1  4 \delta_{1}+\frac 1  4 \delta_{2}$ where $\delta_x$ denotes the measure defined by $\delta_x(E)=1$ or $0$ according as $x\in E$ or not and $\mu_1(E)=\int_E x^{2}dx$. For any $\mu -$ intergrable function $f$ we have $\int fd\mu=\int x^{2}f(x)dx+\frac 1  4 f(-1)+\frac 1  4 f(0)+\frac 1  4 f(1)+\frac 1  4 f(2)$.


问:
Prove the identity $\frac{1}{1+x^{1}}+\cdots+\frac{2^{n}}{1+x^{2^{n}}}=\frac{1}{x-1}+\frac{2^{n+1}}{1+x^{2^{n+1}}}$
Show that for all non-negatives integers $n$, it is true that
$$\frac{1}{1+x^{1}}+\frac{2}{1+x^{2}}+\frac{4}{1+x^{4}}+\frac{8}{1+x^{8}}+\cdots+\frac{2^{n}}{1+x^{2^{n}}}=\frac{1}{x-1}+\frac{2^{n+1}}{1+x^{2^{n+1}}}$$
For $x \ne 1$, $x \ne -1$
I tried to solved it by appliying geometric series but I got the following
after doing the $2f\left(x\right)-f\left(x\right)$ step, where $f\left(x\right)=\frac{1}{1+x^{1}}+\frac{2}{1+x^{2}}+\frac{4}{1+x^{4}}+\frac{8}{1+x^{8}}+\cdots+\frac{2^{n}}{1+x^{2^{n}}}$

$$f\left(x\right)=\frac{1}{x-1}+\frac{2^{n+1}}{1+x^{2^{n}}}$$
Not sure if Fermat's theorem is needed in order to solve it

Body:
Subtract $1/(x-1)$ from LHS and note that,

$$\frac 1{x+1}-\frac 1{x-1}=-\frac 2{x^2-1} \\ \frac 2{x^2+1}-\frac 2{x^2-1}=-\frac 4{x^4-1}\\ \frac 4{x^4+1}-\frac 4{x^4-1}=-\frac 8{x^8-1}$$

and so on. Do you see the telescoping pattern?

To finish, note that,

$$\frac {2^n}{x^{2^n}+1}-\frac {2^n}{x^{2^n}-1}=-\frac{2^{n+1}}{x^{2^{n+1}}-1}=\frac{2^{n+1}}{1-x^{2^{n+1}}}$$


答:
For the particular solution, note the r.h.s. is one of the basic solutions of the homogeneous equation, so we  search for a particular solution of the form
$$y(x)=Cx\mathrm e^{-7x}.$$


问:
Maximum Likelihood using Gradient Descent or Coordinate Descent for Normal Distribution with unknown variance

Context

The maximum likelihood estimators for a Normal distribution with unknown mean and unknown variance are
$$
\widehat{\mu} = \frac{1}{n}\sum_{i=1}^n x_i \qquad \text{and} \qquad \widehat{\sigma}^2 = \frac{1}{n}\sum_{i=1}^n (x_i - \mu)^2
$$
These can be found (for example) by taking derivatives of the average log-likelihood
$$
\frac{1}{n}\sum_{i=1}^n \log p(x_i) = -\frac{1}{2}\log(2\pi) - \frac{1}{2n\sigma^2}\sum^n_{i=1} (x^{(i)} - \mu)^2 - \log \sigma
$$

Question: What if I want to use a gradient-based method?

Yes, I know I can just use the estimators found above. However, I want to find such estimators using a gradient-based method such as coordinate descent or gradient descent. These are the gradients with respect to $\mu$ and with respect to $\sigma$ (which you can set equal to zero to find the estimators above)

$$
\begin{align}
\frac{\partial}{\partial \mu} \frac{1}{n} \sum^n_{i=1} \log p(x^{(i)})
&= \frac{\overline{x}}{\sigma^2} - \frac{\mu}{\sigma^2} \\
\frac{\partial}{\partial \sigma} \frac{1}{n}\sum^n_{i=1} \log p(x^{(i)}) &= \frac{1}{n\sigma^3}\sum^n_{i=1}(x^{(i)} - \mu)^2 - \frac{1}{\sigma}
\end{align}
$$
I tried using them in gradient descent
$$
\begin{align}
\mu_{t+1} &\longleftarrow \mu_t + \gamma \left(\frac{\overline{x}}{\sigma^2_t} - \frac{\mu_t}{\sigma^2_t}\right) \\
\sigma_{t+1} &\longleftarrow \sigma_t + \gamma\left(\frac{1}{n\sigma^3_t}\sum^n_{i=1}(x^{(i)} - \mu_{t+1})^2 - \frac{1}{\sigma_t}\right)
\end{align}
$$
or in coordinate ascent (where I would keep, say $\sigma_t$ fixed and optimize $\mu_t$ for $n_{\text{inner}}$ times and then switch: keep $\mu_t$ fixed and optimize $\sigma_t$ for $n_{\text{inner}}$ times. All this for $n_{\text{outer}}$ times. However it seems to blow up for some reason and not give me the obvious answer. You can run the code here.

What am I doing wrong?

答:
The gradient with respect to $\sigma$ is very large when $\sigma$ is less than than the maximizer, and decreases gradually as $\sigma$ grows when it is larger than the maximizer. (See this plot.) The behavior you are seeing is likely due to a large step size; gradient descent creeps down to the maximizer, but if it overshoots and ends below  the maximizer, it will use the huge gradient to take a giant step upward, erasing all the work you have done.

You have to be careful with gradient methods when the gradient is unbounded or varies sharply like this.


问:
Extending nullhomotopic function to boundary of sphere question
Suppose $f:\mathbb{S}^n\rightarrow Y$ is a continuous map null homotopic to a constant map $c$. In other words: $F: f\simeq c$ , where $c(x)=y$

Now, we may extend $f$ to a continuous map $g: D^{n+1}\rightarrow Y$ by defining

$g(x)= y$ if $0\leq ||x||\leq \frac{1}{2}$ and $F(\frac{x}{||x||},2-2||x||)$ if $\frac{1}{2}\leq ||x||\leq  \frac{1}{2}$

Now, on $||x||=\frac{1}{2}$, $F(\frac{x}{||x||},2-2||x||)=F(\frac{x}{||x||},1)=c=y$.

Hence $F$ is continuous by the gluing lemma.

I was wondering as to what the intuition is for constructing such a function, what geometrical clues allow one to define such a function $F$

答:
Your definition of $g$ is unnecessarily complicated, and this this gives rise to asking "what the intuition is for constructing such a function".

It is much easier to proceed as follows.

You have a homotopy $F : f \simeq c$. Let $G : c \simeq f$ be the "inverse" homotopy $G(x,t) = F(x,1-t)$. Define

$$g : D^{n+1} \to Y, g(x) = \begin{cases} G(\frac{x}{\lVert x \rVert}, \lVert x \rVert) & x \ne 0 \\ y  & x = 0 \end{cases}$$
This is an extension of $f$ since for $x \in S^n$ we have $\lVert x \rVert = 1$ and therefore $g(x) = G(x,1) = f(x)$.

The "problem" here is to show that $g$ is continuous in $0$. I shall not give a proof, it is an easy exercise. It also follows from the "formal part" below. The benefit of your (seemingly strange) definition is that we do not need a special argument to prove continuity.

Anyway, now you see what Tyrone's and Tom Ariel's comments want to say you. The intuition for constructing $g$ is that $D^{n+1}$ is the disjoint union of the spheres $S^n_t = \{ x \mid \lVert x \rVert = t \}$ with $t \in (0,1]$ and the one-point space $S^n_0 = \{0\}$. On $S^n_t$ with $t \in (0,1]$ we take the map $G_t : S^n \to Y$, on $S^n_0$ we take the (constant) map $0 \mapsto y$.

To be formal, the map $G : S^n \times [0,1] \to Y$ has the property $G(S^n \times \{0\}) = \{y\}$, thus induces $\tilde G : (S^n \times [0,1])/(S^n \times \{0\}) \to Y$. But $Q  = (S^n \times [0,1])/(S^n \times \{0\})$ can be naturally identified with $D^{n+1}$. In fact, the map $p : S^n \times [0,1] \to D^{n+1}, p(x,t) = t\frac{x}{\lVert x \rVert}$, is continuous and collapses $S^n \times \{0\}$ to $0$. Since $S^n \times [0,1]$ is compact and $D^{n+1}$ is Hausdorff, $p$ is a closed map, thus a quotient map. This gives us the identification of $Q$ with $D^{n+1}$.