答: The blue path is the boundary of $S$, denoted $\partial S$. We cannot define an inverse operator $\partial^{-1}$ to apply to this boundary so as to recover $S$, because $S$'s complement $S^\complement$ has the same boundary, i.e. $\partial S=\partial S^\complement$. However, you can disambiguate which set with the given boundary is intended. An important generalization of your example is the interior of a Jordan curve, which has been discussed here before. In one notation, $S=\operatorname{int}(\partial S)$ (see e.g. p.4 here). (As to whether that notation serves your needs, we should address an ambiguity in your question: $\operatorname{int}(\partial S)$ doesn't include $\partial S$ itself, but $\overline{\operatorname{int}}(\partial S):=\partial S\cup \operatorname{int}(\partial S)$ does.)

Title: Is there a mathematical notation for the region enclosed by a set of curves

Body:
I want to define the region $S$ as the area enclosed by the curves $C_1, C_2, C_3, C_4$ as in the picture below, but do not know what notation to use.
Any help would be appreciated.


问:
How to decompose a vector regarding complementary subspaces
How to show that $U_1:=\langle(1,1,1)^T\rangle$ and $U_2:=\langle(1,0,0)^T,(0,0,1)^T\rangle$ are complementary subspaces and how to decompose $(1,2,3)^T$ regarding these ? Best Regards, Thomas</p>

答:
Notice that $U_2$ has zero in the $y$ coordinate, so whatever part is due to $U_1$ must account for the 2 in your vector. Thus $(1,2,3) = (2,2,2)+(-1,0,1)$ is the place to start.

Edit: Whoops! For the first part, the easiest thing is to write the standard basis $(1,0,0) ,(0,1,0),(0,0,1)$ in terms of your vectors since they're so simple.

ContentLicense: CC BY-SA 3.0


问:
Krull Topology on a Galois Group Definition
I want to start study infinite galois theory and I have a problem with a basic definition:

The definition of Krull topology.
So $L/K$ is a Galoisextension, $F$ is a subfield $K\subseteq F\subseteq L$ and
$\mathcal{N}:=\{Gal(L/F)\subseteq Gal(L/K)\text{ }|\text{ }[F:K]<\infty, F/K \text{  galois}\}$.

We define an open neighborhood basis of $\sigma$ as a set of $\sigma\mathcal{N}$.
Now I want to verify that this defines a topology.
We have $U\subseteq Gal(L/K)$ is open if and only if there exists for every $\sigma\in Gal(L/K)$
an Element $S$ in the neighborhood basis such that $S\subseteq U$.

So I don't understand why the empty set is open can somebody explain it please ?

答: This doesn't need to be true for every $\sigma$ in $\mathrm{Gal}(L/K)$, it needs to be true for every $\sigma$ in $U$. If $U$ is the empty set, this is true vacuously - every $\sigma$ in $U$ admits some $S$ such that blah blah blah since there are no elements in $U$ to begin with.

ContentLicense: CC BY-SA 4.0


问:
Prove $ \sum_{n=1}^{\infty} \frac{2}{4n^2-1} = 1$
I want to prove that

$ \sum_{n=1}^{\infty} \frac{2}{4n^2-1} = 1$

My approach is the most logical one, rewrite as follows:

$ \sum_{n=1}^{\infty} \frac{2}{4n^2-1} = \sum_{n=1}^{\infty} \frac{1}{2n-1} - \sum_{n=1}^{\infty} \frac{1}{2n+1} $

But the remaining series are both divergent so I get to a sort of "$ \infty - \infty = 1 $".

答:
Rewriting the term,
$\color{blue}{\left(\sum_{n=1}^{\infty} \frac{2}{4 n^{2} - 1}\right)}=\color{black}{\left(\sum_{n=1}^{\infty} \left(- \frac{1}{2 \left(n + \frac{1}{2}\right)} + \frac{1}{2 \left(n - \frac{1}{2}\right)}\right)\right)}$

This is the telescoping series:

$\sum_{n=1}^{\infty} \left(- \frac{1}{2 \left(n + \frac{1}{2}\right)} + \frac{1}{2 \left(n - \frac{1}{2}\right)}\right)=\left(1-\color{green}{\frac{1}{3}}\right)+\left(\color{green}{\frac{1}{3}}-\color{red}{\frac{1}{5}}\right)+\left(\color{green}{\frac{1}{5}}-\color{red}{\frac{1}{7}}\right)+\left(\color{blue}{\frac{1}{7}}-\color{red}{\frac{1}{9}}\right)+...=1$


问:
Propositional Logic Puzzle
"There is an island with two mythical creatures, the Komos and the Dragos. Komos loves eating visitors to the island while Dragos hates to eat visitors. One fine day, you mysteriously end up on the island and encounter two creatures, Blue and Red. The Blue creature smiles and tell you, “Don’t be afraid! We both are not human eaters”. Determine, if possible, whether the Blue and Red creatures is a Komos or a Dragos."

I found the scenario given is too straight forward, as it doesn't state any possible liar.
What I comprehend from this scenario will be, Let p indicates Blue is Dragos, q indicates Red is Dragos. Since Blue doesn't eat humans, and it doesn't lie, so if p then q, both Blue and Red are Dragos.

Would like to here more thoughts from you all.

答:
Presumably the "we" in question refers to the Blue creature and its interlocutor (i.e., you), as opposed to Blue and Red. Assuming you are human and not a cannibal, then this suggests that Blue is Dragos.

In some languages, there are different words for "we including you", "we not including you", etc., whereas English has no such distinction and relies on context.

This kind of problem is really a language interpretation problem, not so much a mathematics problem. It might be more appropriate to ask it at puzzling.stackexchange.com.

Title: Propositional Logic Puzzle


问:
Is every set that's a union of two sets disconnected?
I'm curious about the definition of disconnected and I'm sure this can't be right. But let's say that $S= A \cup B$, then isn't $S$ automatically disconnected? Since I can re-write It as $(S \cap C) \cup (S \cap D)$ with $C=A$ and $D=B$? This is a circular definition since it basically uses an identity about unions and intersection, but is my understanding of disconnected totally off? Or is this just a trivial case we toss aside? I'm a tad confused here.</p>

答: Disconnectedness requires also that $A,B$ are disjoint.</p>


问:
Express addition of natural numbers ($+$) in structure $(N;=;S;*)$ where $S(n)$ is n + 1
Express addition of natural numbers ($+$) in structure $(N;=;S;\cdot)$ where $S(n)$ $\rightarrow$ $n + 1$

I need to exress operationd of addition in this structure. I tried to: $c = a + b = a * S(0) + b * S(0) $ but stucked. Also tried $a = S(a-1) = S(S(a-2))$, $S(a-1) * S(b-1) = ab $

Please help me a have to send the solution of this problem today.

答:
Definition:

$m + 0 = m$

and

$m + S(n) = S(m+n)$

for all natural numbers $m,n\geq 0$.

Define $1 = S(0), 2=S(1), 3=S(2), \ldots$.

Then $n+1 = n+ S(0) = S(n+0) = S(n)$. This shows that $S$ is the successor function.

Moreover, for instance,

$1+1 = 1 + S(0) = S(1+0) = S(1) = 2$.


问:
$A^* \subset B^* \to B\subset A$
While working with the Hahn-Banach theorem, I came across this question:

Let $ A, B $ be two closed subspaces of the Banach space $ C $. If $ B^* \subset A^* $, do we have $ A \subset B $?

I would say simply yes. In fact, if they were not, we could consider
$
f \in A^*
$
and, if $b \in B\setminus A$, define $f(b)=\infty$, which makes it absurd .
However this reasoning seems too trivial to me...

答:
For your question to make sense you really need to say what you mean by $B^*\subset A^*$.
There is no canonical way of seeing a function $B\to\mathbb C$ as a function $A\to\mathbb C$ if you don't specify any relation between $A$ and $B$.

If you want to say that you see $B^*\subset A^*$ by restriction, you are implying that $A\subset B$ in the definition of your inclusion, so the answer is yes in a trivial sense.

For example, let $$C=\ell^1(\mathbb N)\oplus \ell^2(\mathbb N),\qquad A=\ell^1(\mathbb N)\oplus 0,\qquad B=0\oplus \ell^2(\mathbb N).$$ Then $A^*=\ell^\infty(\mathbb N)$ and $B^*=\ell^2(\mathbb N)$ canonically, so we have a natural inclusion $B^*\subset A^*$. But this tells us nothing about $A$ and $B$ (which have $A\cap B=\{0\}$ in the above example. Or we could have $C=\ell^\infty(\mathbb N)$, $A=\ell^1(\mathbb N)$, $B=\ell^2(\mathbb N)$,  and now we have $A\subset B$ and $B^*\subset A^*$ canonically.

Title: $A^* \subset B^* \to B\subset A$


问:
Solving $\lim_ {a\to \frac{\pi}{4}} f(a)$ where $f$ is a definite integral.
$f:(0,\frac{\pi}{2})\to$$\mathbb{R}$,  $f(a) = \int_0^1\frac{dx}{x^2+2x\tan a + 1}$, $\forall a\in(0,\frac{\pi}{2})$

What is $\lim_ {a\to \frac{\pi}{4}} f(a)$?

I am confused. Function $f$'s domain is given at the beginning but then this  $\forall a\in(0,\frac{\pi}{2})$ is given . Isn't $a$ the variable with which $f$ is defined?

The only thing i can extract from here is $\tan a$ being positive always but aside from that i do not know what to do. I tried writing the denominator by "completing the square" and then i used u-substitution but in the end i get $\frac{1}{2\sqrt {\tan^2{a -1}}}$ multiplied by some logarithms but this is defined only for ${\tan^2a} \gt 1$.

How do i solve it?

答:
We want to find the indefinite integral.

First we complete the square on the denominator($x^2+2x\tan a + 1$).

We get $(x+\tan a)^2 - (\tan^2 a - 1)$

Now this can be easily solved with u substitution with $x + \tan a$ but $(\tan^2 a - 1)$
it going to be a problem under the square root if it is negative in the form $\frac{1}{x^2-c^2}$ where $c$ would be $\sqrt{(\tan^2 a - 1)}$ .

So we take 3 cases :

-$\tan a \gt 1$ we continue with the previous solution

-$\tan a \lt 1$ we write $(\tan^2 a - 1)$ as $-(-\tan^2 a + 1)$ and solve with
elementary integrals

-$\tan a = 1$ we get $(\tan^2 a - 1) = 0$ and solve with elementary integrals as well

After this it is easy.


问:
Show that any graph has an independent set of size at least $\frac{n^2}{2m+n}$
I am trying to show that any graph with n vertices and m edges has an independent set of size at least: $$\frac{n^2}{2m+n}$$

I have already shown that any graph has an independent set of size at least:
$$\sum_{i=0}^n \frac{1}{d(v_i) + 1}$$

where $d(v_i)$ is the degree of vertex $v_i$.

I am trying to deduce the first result from what I have already shown.

I know that $$\sum_{i=0}^n {(d(v_i) + 1)} = 2m + n$$ by the handshaking lemma and I reckon this will be helpful, however I'm struggling to combine it with $\sum_{i=0}^n \frac{1}{d(v_i) + 1}$ to prove the required result.

Any help much appreciated!

答:
Hint

You would be done if you could show that
$$
\sum_i \frac1{d(v_i)+1}\ge \frac{n^2}{2m+n}
$$
Letting $f(x)=\frac1{x+1}$, the LHS is $\sum_i f(d(v_i))$. Use the face that the function $f(x)$ is convex.


答:
Let $A=[0,1]$ and $B=[1,2]$; both sets are connected, and $A\cup B=[0,2]$, which is also connected. Let $A$ be the set of irrational numbers and $B$ the set of rational numbers; then both $A$ and $B$ are totally disconnected, but $A\cup B=\Bbb R$ is connected.

In order for $S$ not to be connected, you must be able to write it as a disjoint union of non-empty sets that are both relatively open in $S$: $S=U\cup V$, say, where $U\ne\varnothing\ne V$, $U\cap V=\varnothing$, and $U\cap\operatorname{cl}V=V\cap\operatorname{cl}U=\varnothing$. Notice the requirements: $U$ and $V$ are non-empty, they’re disjoint, their union is $S$, and each is a relatively open subset of $S$. All of these conditions must be met in order for $U$ and $V$ to be a separation of $S$.

Title: Is every set that's a union of two sets disconnected?

Body: I'm curious about the definition of disconnected and I'm sure this can't be right. But let's say that $S= A \cup B$, then isn't $S$ automatically disconnected? Since I can re-write It as $(S \cap C) \cup (S \cap D)$ with $C=A$ and $D=B$? This is a circular definition since it basically uses an identity about unions and intersection, but is my understanding of disconnected totally off? Or is this just a trivial case we toss aside? I'm a tad confused here.</p>


问:
Dirac function integrated from -infinity to 0
What is the value of integral of $\int_{-\infty}^0 δ(t)dt$ (dirac function)? Wouldn't it make sense for it to be 0.5 since it is an even "function" ? Show proof of whatever the answer is if you know it please.

答: Actually, $\int_{-\infty}^0\delta(t)dt$ is undefined, because $\lim_{a\to0^+}\int_{-\infty}^0\tfrac1a\delta_0(\tfrac{t}{a})dt$ is not the same for all choices of PDFs $\delta_0$ we can use as a nascent delta. For example,$$\begin{align}\delta_0(u)&:=e^{-u}1_{[0,\,\infty)}(u)\\\implies\lim_{a\to0^+}\int_{-\infty}^0\tfrac1a\delta_0(\tfrac{t}{a})dt&=0,\,\\\delta_0(u)&:=e^u1_{(-\infty,\,0]}(u)\\\implies\lim_{a\to0^+}\int_{-\infty}^0\tfrac1a\delta_0(\tfrac{t}{a})dt&=1.\end{align}$$


问:
Solution set of $\frac x{x+2}>0\land\frac{x+1}{x+2}<1$
Okay people, $\Im$ stupid. I have been working on this for literal 2 days but still I can't understand.

Here is the question:
$$\frac x{x+2}>0\land\frac{x+1}{x+2}<1$$
What's the solution set? [Answer is $(0,\infty)$]

I found first inequality means $x > 0$ and second one says $x > -2$ but that doesn't make the solution set given by my lecturer which is $(0,\infty)$.

答:
The best way to solve this kind of inequalities is not to split to different cases but to $\underline{\text{combine the fractions}}$.

For your first inequality: $$\frac{x}{x+2} >0 \iff x(x+2)>0 \iff x \in (-\infty, -2)\cup (0, \infty) \tag 1$$

For your second inequality: $$\frac{x+1}{x+2} < 1 \iff \frac{x+1}{x+2}-1 = - \frac{1}{x+2} < 0 \iff x+2 >0 \tag 2$$

Combine (1) and (2) you get $x>0$.

For another example, see Solving basic inequality


问:
Atoms in a measure space and Sierpinski theorem for non-atomic measures
I have two questions concerning atoms in a measure space.

My first question is about two different definitions I've encountered and I'm not sure if they are equivalent.

Definition 1: Given a measure space $(\Omega, \Sigma, \mu)$ we say that an element $A \in \Sigma$ is an atom of $\mu$ if it satisfies that $\mu(A) >0$ and for every $B \in \Sigma$ such that $B \subset A$ either $\mu(B)=0$ or $\mu(A \setminus B)=0$.

Definition 2: Given a measure space $(\Omega, \Sigma, \mu)$ we say that an element $A \in \Sigma$ is an atom of $\mu$ if it satisfies that $\mu(A) >0$ and for every $B \in \Sigma$ such that $B \subset A$ either $\mu(B)=0$ or $\mu(B)=\mu(A)$.

So I know that if $\mu(\Omega)$ is finite then both definitions are indeed equivalent but, are they still equivalent if the measure is not finite?

Now, concerning atoms of a measure I'm trying to prove the following theorem, apparently due to Sierpinski:

Theorem: If $(\Omega, \Sigma, \mu)$ is a measure space with no atoms, then for every $t \in [0, \mu(\Omega)]$ there exists an $A \in \Sigma$ such that $\mu(A)=t$.

Following Wikipedia's article on atoms, I'm trying to fully proof the sketch of the proof at the end of the page.

So I want to prove there is a funciton $S:[0, \mu(\Omega)] \longrightarrow \Sigma$ satisfying:

$S$ is monotone, that is, if $t \leq t'$ then $S(t) \subset S(t')$

$\mu(S(t))=t$ for every $t \in [0, \mu(\Omega)]$

because this would directly imply the theorem.

To this end, we define
$$\Gamma= \left\{  S:D \longrightarrow \Sigma: D \subset [0, \mu(\Omega)], S \textrm{ is monotone}, \mu(S(t))=t, \forall t \in D \right\}$$

Then $\Gamma$ is not empty beacause we can define $S_0:\{0\} \longrightarrow \Sigma$ given by $S(0)=\emptyset$ and is clearly in $\Gamma$, and we can make it a partially ordered set by establishing that $S_1 \leq S_2$ if, and only if, $\operatorname{graph}(S_1) \subset \operatorname{graph}(S_2)$.

Now, if $\Gamma_c$ is a chain in $\Gamma$ then we define $S_c: D_c \longrightarrow \Sigma$ where

$$D_c=\bigcup_{S \in \Gamma_c} \operatorname{dom}(S)$$

and given $t \in D_c$, we choose any $S \in \Gamma_c$ such that $t \in \operatorname{dom}(S)$, and define $S_c(t)=S(t)$.

First, $S_c$ is well defined because if we have that given $t \in D_c$, there are $S_1, S_2 \in \Gamma_c$ such that $t \in \operatorname{dom}(S_1) \cap \operatorname{dom}(S_2)$, then as $\Gamma_c$ is a chain, we can assume wlog that $S_1 \leq S_2$ and then, $(t,S_1(t)) \in \operatorname{graph}(S_2)$ so $S_2(t)=S_1(t)$ as we wanted to show.

Now, we will prove that $S_c \in \Gamma$.

First, observe that given $t_1, t_2 \in D_c$ there exists $S_1, S_2 \in \Gamma_c$ such that $t_i \in \operatorname{dom}(S_i)$, so as $\Gamma_c$ is a chain, we may assume $S_1 \leq S_2$ and so $\operatorname{dom}(S_1) \subset \operatorname{dom}(S_2)$, getting that $t_1, t_2 \in \operatorname{dom}(S_2)$. So, if $t_1 \leq t_2$ as $S_2 \in \Gamma$ we will have that
$$S_c(t_1)=S_2(t_1) \subset S_2(t_2) = S_c (t_2$$
so $S_c$ is monotone.

On the other hand, it is clear that given $t \in D_c$ if it is $S \in \Gamma_c$ such that $t \in \operatorname{dom}(S)$, then
$$\mu(S_c(t))=\mu(S(t))=t$$

So we have proven that $S_c \in \Gamma$ and it is clear by its construction that it is an upper bound of $\Gamma_c$.

By Zorn's lemma, there exits then some $S: D \longrightarrow \Sigma$ maximal in $\Gamma$, and the claim is that this function is the one we are looking for, and it suffices to prove that $D=[0, \mu(\Omega)]$.

This last part is the one I cannot prove, and if everything I've done is correct I suppose here is where you should use that $\mu$ is non-atomic. If tried to prove that $D$ is closed and open in $[0, \mu(\Omega)]$, and as it is conected, we'll have that $D=[0, \mu(\Omega)]$ as we want.

First, I've shown by using the maximality of $S$ that $\{0, \mu(\Omega)\} \subset D$, for example if we suppose $0 \not \in D$ (the other case is analogue), then defining $\bar S: D \cup \{0\} \longrightarrow \Sigma$ as $\bar S(t)=S(t)$ if $t \in D$ and $\bar S(0)=\emptyset$, it is clear that $\bar S \in \Gamma$ and $S < \bar S$ contradiction with its maximality.

Now, to show it is closed, I've taken a sequence $\{t_n\} \subset D$ converging to some $t_0 \in [0, \mu(\Omega)]$ and I've supposed that $t \not \in D$.

Then, we have that $t_0 \in (0, \mu(\Omega))$ so in particular is finite, and we know then there must exists a monotone subsequence of $\{t_n\}$ converging to $t_0$, so we can suppose that $\{t_n\}$ is monotone.

If it is increasing, defining $A_n=S(t_n)$ and $\bar S: D \cup \{t_0\} \longrightarrow \Sigma$ as $\bar S(t)= S(t)$ if $t \in D$ and $\bar S(t_0)= \cup A_n$ we have:

If $t <t_0$ then there exists some natural number $n$ such that $t \leq t_n$ and so
$$\bar S(t)=s(t) \subset S(t_n) = A_n \subset \bar S(t_0)$$

If $t_0 < t$ then $t_n < t$ for every $n \geq 1$ so
$$A_n=S(t_n) \subset S(t), \forall n \geq 1$$
and then
$$\bar S(t_0) \subset S(t)$$

Since $\{t_n\}$ is increasing, by the monotonity of $S$, $\{S(t_n)\}$ is also increasing and we then have
$$\mu\left(\bar S(t_0)\right)=\mu\left( \bigcup_{n=1}^\infty S(t_n) \right)=\lim_{n \rightarrow \infty} \mu\left( S(t_n)\right) = \lim_{n \rightarrow \infty} t_n = t_0$$

So, this three facts imply that $\bar S \in \Gamma$ but $S < \bar S$ so we have a contradiction with the maximality.

Now, if $\{t_n\}$ is decreasing, we do the same thing but taking $\bar S(t_0)=\cap A_n$ and, as $t$ is finite, we can assume that $\mu(A_1)=\mu(S(t_1))=t_1$ is finite, so we can argue as before arriving at the same contradiction.

If everything I've write here is correct, then to show $D$ is open I must use the fact that $\mu$ is non-atomic because I haven't used it yet, so my doubt concerging this part are:

If the definitions I gave at the beggining are not equivalent, can this last part be shown with both definitions?

In case it is two difficult to show $D$ is open, is there a simple proof that $D=[0,\mu(\Omega)]$? And, in that case, can it be shown with both definitions of atoms if they are not equivalent?

答:
About the definitions:

The definitions $1$ and $2$ are not equivalent in the case of infinite measures.
The definition $1$ is more adequate in some cases, but it is not the most common one.  On the other hand definition $2$ may lead to pathological examples when applied to infinite measures. Consider $(\mathbb{N}, 2^{\mathbb{N}}, \mu)$ where $\mu(\emptyset)=0$ and, for all $E \subseteq \mathbb{N}$ such that $E\neq \emptyset$ , $\mu(E)=\infty$. According to definition $1$,  only the singletons will be atoms. According to definiton $2$, all non-empty subsets of $\mathbb{N}$ (so you have atoms being proper subsets of other atoms, for instance). However definition $2$ is more common and it is more useful when we want to talk about atom-free measures.

About your proof:

Yes, by Zorn's lemma, there exits then some $S: D \longrightarrow \Sigma$ maximal in $\Gamma$. You have already proved that $D$ is closed.
To complete your proof, it is enough to proof that if $t_1, t_2 \in D$ and $t_1<t_2$, there there is $t\in D$ such that  $t_1<t<t_2$.

Proof: Suppose $t_1, t_2 \in D$ and $t_1<t_2$. Suppose  there is no $t\in D$ such that  $t_1<t<t_2$ (in other words, for all $t\in(t_1,t_2), t\notin D$). Let us show that this contradicts the maximality of $S: D \longrightarrow \Sigma$.

Since $t_1<t_2$, $S(t_1)\subseteq S(t_2)$. But clearly $S(t_1)= S(t_2)$ is not possible because $\mu(S(t_1))=t_1$ and $\mu(S(t_2))=t_2$. So $S(t_1)\subsetneq S(t_2)$.

Let $B= S(t_2) \setminus S(t_1)$, since $t_1<t_2$, $t_1$ is finite. So
$\mu(B) = t_2-t_1 >0$. As $\mu$ has no atoms (definition $2$). There is $C\subsetneq B$
such that $0< \mu(C) < \mu(B) = t_2-t_1$. Let $\delta = \mu(C)$.

It is immediate that $t_1<t_1+\delta<t_2$,  $S(t_1)\subsetneq S(t_1) \cup C \subsetneq  S(t_2)$ and $\mu(S(t_1) \cup C) = t_1+\delta$. So  $S: D \longrightarrow \Sigma$ can be extended to include the pair $(t_1+\delta, S(t_1) \cup C)$. Contradiction the maximality of $S: D \longrightarrow \Sigma$.

Since $D$ is closed, $0 \in D$ and $\mu(\Omega) \in D$, we can conclude that  $D=[0,\mu(\Omega)]$.

In fact, suppose there is $x\in [0,\mu(\Omega)]$ such that $x \notin D$ since $D$ is closed, $0 \in D$ and $\mu(\Omega) \in D$, there is an open interval $(a,b)$ such that $x\in (a,b)\subset [0,\mu(\Omega)]$ and $D\cap (a,b) =\emptyset$.

Now note that $0\in \{y\in D: y<a\}$  and $\mu(\Omega) \in \{y\in D: y>b\}$.
So both set are not empty and we can define

$$ a_1 = \sup \{y\in D: y<a\} $$
and
$$ b_1 = \inf  \{y\in D: y>b\}$$

It is easy to see that $(a,b) \subseteq (a_1,b_1)$ and $D \cap (a_1,b_1)=\emptyset$. But $D$ is closed, so $a_1, b_1 \in D$. But then, by what we have proved before, there is $t\in (a_1,b_1)$ such that $t \in D$. Contradiction to  $D \cap (a_1,b_1)=\emptyset$. So for all $x\in [0,\mu(\Omega)]$, $x\in D$. So  $D=[0,\mu(\Omega)]$


答:
$S$ is disconnected if it is the union of two disjoint, open, non-empty subsets $A$ and $B$.

So if $S = A \cup B$ for some $A$ and $B$, it could still be connected because $A$ and $B$ are not disjoint or one of $A$ or $B$ is not open.

Example 1: If $S = (0, 1)$, we can write $S = A \cup B$ where $A = (0, 3/4)$ and $B = (1/4, 1)$. $A$ and $B$ are both open but not disjoint, so this doesn't contradict the fact that $(0,1)$ is connected.

Example 2: Again take $S = (0,1)$, but this time take $A = (0,1/2)$ and $B = [1/2, 1)$. Then $S = A \cup B$ and $A$ and $B$ are disjoint, but the problem here is that $B$ is not open in $(0,1)$.


问:
Partial sum of Stirling numbers of the second kind with falling factorial
Let $S_n = \{ 1, 3, ... \}$ be the set of odd numbers not greater than $n$ and $(x)_n$ be the falling factorial $x(x-1)\cdots(x-n+1)$. I am interested in the following partial sum:
$$
\sum_{i \in S_n} \binom{x}{i} i! {n \brace i} = \sum_{i \in S_n} {n \brace i} (x)_i,
$$
where ${n \brace i}$ is a Stirling numbers of the second kind.

I know that $\sum_{i = 1}^{n} \binom{x}{i} i! {n \brace i} = \sum_{i = 1}^{n} {n \brace i} (x)_i = x^n$ but it is not clear for me how to simplify this partial sum with respect to $S_n$ only. It would be appreciated if someone could give me any hint to simplify or bound this partial sum.

答:
Not an answer, merely a remark.
Let $$O_n(x):=\sum_{i \ odd} {n \brace i} (x)_i.$$ Then by using the basic recurrences for binomial coefficients and Stirling numbers, we see that $O_n$ satisfies the recursion:
$$O_n(x)=x\left(O_{n-1}(x)-2O_{n-1}(x-1)+(x-1)^{n-1}\right)$$ with initial condition $O_1(x)=1$.
Not sure whether it is useful for the question though.

Edit

This recursion can be used to derive the first terms of an expansion for $O_n(x)$:
$$O_n(x)=-\frac{1-(-1)^n}{2}(-x)^n-\frac{n(n-1)}{2}(-x)^{n-1}-\frac{n(n-1)^2(n-2)}{4}(-x)^{n-2}- \cdot\cdot\cdot$$
The calculations become increasingly laborious by this method.

Now, by using the expression of the falling factorial in terms of Stirling numbers of the first kind ${n \brack k}$, one obtains
$$ O_n(x)= -\sum_{h\ge0} P_h(n)(-x)^{n-h} $$
where
$$ P_h(n)= \sum_{i\ge 0} {n \brace 2i+1}{2i+1 \brack n-h} $$
which is indeed a polynomial function of $n$ of degree $2h$. This can be seen when written as
$$ P_h(n)= \sum_{i= 0}^h {n \brace n-i}{n-i \brack n-i-(h-i)} \frac{1-(1-)^{n-i}}{2}.$$
Each summand in the former is the product of a polynomial in $n$ of degree $2i$ and a polynomial in $n$ of degree $2h-2i$.


问:
Conditional Expectation of Brownian motion using projection
Assume that $W_s,W_t$ are Standard Brownian motions with $s<t$. Find the following
$$E[W_s | W_t]=? $$
The hint is to use projection method. If I understand correctly, we have the following property by projection
$$E[W_sZ]=E[YZ] $$ for $Y=E[W_s|W_t]$ and $Z$ is some random variable that is measurable under the same filtration that generates $W_t$. How to use this property and proceed? Any hint is appreciated.

P.S. I do not understand the solution given here and would prefer to understand using projection method (if at all possible).

答:
The main idea is that we're going to guess that $Y=\mathbb{E}[W_s | W_t] = \beta W_t$ where $\beta$ is a (non-random) constant that depends only on $s$ and $t$.  Then the projection method tells us (letting $Z = W_t$) that

\begin{align*}
\mathbb{E}[W_s W_t] &= \mathbb{E}[Y W_t] \\
&= \beta \mathbb{E}[W_t^2].
\end{align*}

Now we know $\mathbb{E}[W_s W_t] = s$ and $\mathbb{E}[W_t^2]=t$, so the equation simplifies to $s = \beta t$ and hence $\beta = \frac st$.  Therefore we conclude that $\mathbb{E}[W_s | W_t] = \beta W_t = \frac st W_t$.


问:
Is there a compact formula for this cross product involving $\nabla$?
Let $\textbf{u}=(u_1(x,y,z),u_2(x,y,z),u_3(x,y,z))$ and $\textbf{v}=(v_1(x,y,z),v_2(x,y,z),v_3(x,y,z))$ be two smooth vector fields in $\mathbb{R}^3$, is there a simplified form of the following difference $$\nabla\times((\textbf{u}\cdot\nabla)\textbf{v})-\nabla\times((\textbf{v}\cdot\nabla)\textbf{u})~~?$$
where $\nabla = \left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right)$ is the del operator and  $(\textbf{u}\cdot\nabla)\textbf{v}=\left(u_1\frac{\partial}{\partial x}+u_2\frac{\partial}{\partial y}+u_3\frac{\partial}{\partial z}\right)\textbf{v}$.

My goal is to simplify above difference in the special case where $\textbf{v} = \nabla\times\textbf{u}$ and $\nabla\cdot\textbf{u}=0$, but it will be nice if there is a universal simplification to the difference when $\textbf{u}$ and $\textbf{v}$ are unrelated, then one could apply that to the special case.

答:
Using the sum convention $$\nabla \times \vec{a} = \epsilon_{ijk} \frac{\partial a_j}{\partial x_i} \vec{e_k} = \epsilon_{ijk} \partial_i a_j \vec{e_k};$$

$$\nabla \times ((\vec{u}\cdot \nabla)\vec{v})−\nabla \times ((\vec{v} \cdot \nabla)\vec{u})= \nabla \times [(\vec{u}\cdot \nabla)\vec{v} - (\vec{v}\cdot \nabla)\vec{u}]$$
$$ = \epsilon_{ijk} \partial_i [(\vec{u}\cdot \nabla)\vec{v} - (\vec{v}\cdot \nabla)\vec{u}]_j \vec{e_k} $$
$$= \epsilon_{ijk} \partial_i [u_m \partial_m v_j - v_m \partial_m u_j]\vec{e_k} $$
It doesn't look like there is much more you can do.

If we assume as well that $\vec{v} = \nabla \times \vec{u}$ then $v_j = \epsilon_{lnj} \partial_{l} u_{n}.$
$$\nabla \times ((\vec{u}\cdot \nabla)\vec{v})−\nabla \times ((\vec{v} \cdot \nabla)\vec{u})= \epsilon_{ijk} \partial_i [u_m \partial_m \epsilon_{lnj} \partial_{l} u_{n} - \epsilon_{lnm} \partial_{l} u_{n} \partial_m u_j]\vec{e_k} $$
$$=\left(\epsilon_{ijk} \epsilon_{lnj} \partial_i [u_m \partial_m \partial_l u_n] - \epsilon_{ijk} \epsilon_{lnm}\partial_i \partial_l[u_n \partial_m u_j]\right)\vec{e_k}$$
$$= \left[\epsilon_{ijk} \epsilon_{lnj}\left(\frac{\partial u_m}{\partial x_i}\frac{\partial^2 u_n}{\partial x_m \partial x_l} + u_m\frac{\partial^3 u_n}{\partial x_m \partial x_i \partial x_l} \right) - \epsilon_{ijk} \epsilon_{lnm}\partial_i \left(\frac{\partial u_n}{\partial x_l}\frac{\partial u_j}{\partial x_m} + u_n \frac{\partial^2 u_j}{\partial x_l \partial x_m}\right)\right]\vec{e_k} $$
$$= \left[\epsilon_{ijk} \epsilon_{lnj}\left(\frac{\partial u_m}{\partial x_i}\frac{\partial^2 u_n}{\partial x_m \partial x_l} + u_m\frac{\partial^3 u_n}{\partial x_m \partial x_i \partial x_l} \right)  - \epsilon_{ijk} \epsilon_{lnm}\left(\frac{\partial^2 u_n}{\partial x_i \partial x_l}\frac{\partial u_j}{\partial x_m} + \frac{\partial u_n}{\partial x_l}\frac{\partial^2 u_j}{\partial x_i \partial x_m} + u_n \frac{\partial^3 u_j}{\partial x_i \partial x_l \partial x_m} + \frac{\partial u_n}{\partial x_i}\frac{\partial^2 u_j}{\partial x_l \partial x_m}\right)\right]\vec{e_k} $$

Which can possibly be simplified.
Kronecker delta and Levi-Civita symbol


问:
For $(x_1,x_2), (y_1,y_2) \in \mathbb{R}^2$, does $⟨(x_1,x_2),(y_1,y_2)⟩={x_1}^2{y_1}^2 + {x_2}^2{y_2}^2$ define an inner product?
For $(x_1,x_2), (y_1,y_2) \in \mathbb{R}^2,$ let $⟨(x_1,x_2),(y_1,y_2)⟩={x_1}^2{y_1}^2 + {x_2}^2{y_2}^2$

Does this define an inner product? I know an inner product has to have linearity, be symmetric and positive definite i.e. $⟨ u, u⟩ > 0$ if and only if u = 0.
It is clearly symmetric and positive definite, however I am struggling with showing if it is linear or not. Any guidance on this would be appreciated.

答: Since $\langle(2,0),(1,0)\rangle=4\ne2\langle(1,0),(1,0)\rangle$, no, it is not linear in the first variable, and therefore $\langle\cdot,\cdot\rangle$ is not an inner product.


问:
What is $\lim_{\leftarrow},\ \prod^{\leftarrow}$
In the page 258 of chapter 8 of Diestel, I found these two notations involving left arrow:
$$\lim_{\leftarrow},\  \prod^{\leftarrow}$$

What's the meaning of this?

答: This is the inverse limit, also known as projective limit.</p>


问:
Solve the Killing equation for a vector field in $\mathbb{R}^2$ with the Euclidean metric
I know that the vector field $$X = a_1\partial_1 + a_2\partial_2$$ where $a_1,a_2 : \mathbb{R}^2 \rightarrow \mathbb{R}$ are smooth,  is a Killing field on $\mathbb{R}^2$ with the Euclidean metric $dx_1^2 + dx_2^2$.

I have to solve the Killing equation $$\mathcal{L}_X(dx_1^2 + dx_2^2) = 0$$ for $a_1$ and $a_2$.

I know I have to use the definition of Lie derivative and find that is equal to zero but I am struggling a bit with the computation, can someone help me ?

I was thiking about using Cartan's formula, is it a good approach ?

答:
Let $U$,$V$ and $X$ be three vector fields and $g$ be the metric tensor field. Then,
\begin{align}
\left(L_Xg\right)(U,V) &= X\cdot g(U,V) - g(L_XU,V) - g(U,L_XV) \\
&= g(\nabla_XU,V) + g(U,\nabla_XV) - g(L_XU,V)-g(U,L_XV) \\
&= g(\nabla_XU - L_XU,V) + g(U,\nabla_XV-L_XV) \\
&= g(\nabla_UX,V) + g(U,\nabla_VX)
\end{align}
Thus, $L_Xg=0$ if and only if for every vector fields $U$ and $V$,
$$
g(\nabla_UX,V) + g(U,\nabla_VX) = 0
$$
that is, if and only if $\nabla X : U \mapsto \nabla_UX$ is a skew-symmetric operator.

In the case $g$ is the euclidean metric of $\mathbb{R}^2$, every vector field $U$ is a smooth combination of $\partial_1$ and $\partial_2$, and
$$
L_Xg = 0 \iff g(\nabla_{\partial_1}X,\partial_1) = 0,~ g(\nabla_{\partial_2}X,\partial_2)=0 \text{ and } g(\nabla_{\partial_1}X,\partial_2) = -g(\partial_1,\nabla_{\partial_2}X)
$$
If $X = a_1\partial_1 + a_2 \partial_2$, recall that $\partial_1$ and $\partial_2$ are parallel for $g$, and:
\begin{align}
\nabla_{\partial_1}X &= \nabla_{\partial_1}\left(a_1\partial_1 + a_2 \partial_2\right) \\
&=(\partial_1a_1) \partial_1 + (\partial_1a_2)\partial_2 \\
\nabla_{\partial_2}X &= \nabla_{\partial_2}\left(a_1\partial_1 + a_2 \partial_2\right) \\
&=(\partial_2a_1) \partial_1 + (\partial_2a_2)\partial_2
\end{align}
Hence, $X$ is a Killing vector field if and only if
\begin{align}
\partial_1a_1 &= 0, & \partial_2a_2&=0, &
\partial_1a_2 &= - \partial_2 a_1
\end{align}
I let you continue the calculations.

Important comment Be carefull with the Cartan magic formula. It says that for a differential form $\omega$, $L_X \omega = (d\circ i_X + i_X\circ d)\omega$. A tensor is not, in general, a differential form. A simple reason why this would not make sense is this: how do you define $dg$ when $g$ is a metric tensor?


问:
How to decompose a vector regarding complementary subspaces
How to show that $U_1:=\langle(1,1,1)^T\rangle$ and $U_2:=\langle(1,0,0)^T,(0,0,1)^T\rangle$ are complementary subspaces and how to decompose $(1,2,3)^T$ regarding these ? Best Regards, Thomas</p>

答:
The best way to do this is to solve the system of equations $Ax=b$, where
$$A=\left[\begin{array}{ccc}1&1&0\\1&0&0\\1&0&1\end{array}\right]\quad\text{and}\quad b= \left[\begin{array}{c} 1\\2\\3\end{array}\right].$$
With the numbers you have, there are shortcuts, as others have suggested.</p>


问:
Find rational $a,b$ with $a\ne b$ such that $x²-bx-\frac{a³-b³}{3b}=0$ has rational solutions
I was working on the generalization of a geometry problem, which resulted in the quadratic equation

$$x^2 - bx - \frac{a^3 - b^3}{3b}= 0$$

Its solutions are

$$-\frac{b}{2} \pm \sqrt{\frac{4a^3 - b^3}{12b}}$$

I wanted to see if there are any rational values for a and b with a not equal to b such that the quadratic equation has rational solutions (in other words: such that $\cfrac{4a^3-b^3}{12b}$ is the square of a rational number). I couldn't find any, so I'm not even sure if there are any.

If there are not, I would want to prove there cannot be such a and b.

答:
By substituting $x=12a/b$ (or rather, $a=bx/12$) and simplifying, the expression becomes
$$\frac{4b^2}{12^4}(x^3-3\cdot 12^2)$$

To make this a rational square, we need to find a rational point on the elliptic curve $y^2=x^3-3\cdot 12^2$. The theory of rational points on elliptic curves is huge and complicated, so it is easiest to just let a computer do it.

According to Magma, the Mordell-Weil group (the group of rational points) on this curve is isomorphic to $Z_3$, so it has only three rational points. Those points are just the ones we already expect, namely two points with $x=12$ (corresponding to $a=b$), and a point at infinity. So there are no other rational points.

You can check this by typing the following into the online Magma calculator:

```<code>E:=EllipticCurve([0,0,0,0,-3*144]);
MordellWeilGroup(E);
RationalPoints(E : Bound:=1000);
</code>```

Body:
By substituting $x=12a/b$ (or rather, $a=bx/12$) and simplifying, the expression becomes
$$\frac{4b^2}{12^4}(x^3-3\cdot 12^2)$$

To make this a rational square, we need to find a rational point on the elliptic curve $y^2=x^3-3\cdot 12^2$. The theory of rational points on elliptic curves is huge and complicated, so it is easiest to just let a computer do it.

According to Magma, the Mordell-Weil group (the group of rational points) on this curve is isomorphic to $Z_3$, so it has only three rational points. Those points are just the ones we already expect, namely two points with $x=12$ (corresponding to $a=b$), and a point at infinity. So there are no other rational points.

You can check this by typing the following into the online Magma calculator:

E:=EllipticCurve([0,0,0,0,-3*144]);
MordellWeilGroup(E);
RationalPoints(E : Bound:=1000);


问:
X is connected iff every application $f:X\to D$ is constant for every discrete space $D$.
How can I prove that a metric space $X$ is connected if and only if every application $f:X\to D$ is constant for every discrete space $D$?

I already know how to prove that $X$ is connected iff every two-valued function on $X$ is constant, however if I try to prove the problem in analogous way to the last one I need that $f$ be continuous which is not an hypthesis of my problem and the open sets formed in the proof of the last one not necessarily "cover" all the space in the part where every function is constant implies $X$ is connected.

答:
This is false as written: you're missing the key assumption of continuity.

A metric space is connected if and only if every continuous map $f\colon X\to D$ is constant, for every discrete space $D$.

One direction is trivial: suppose $X$ is connected and that $f\colon X\to D$ is continuous, with $D$ a discrete space. If $D$ has one element, then $f$ is constant. If $D$ has at least two elements, consider $x\in X$ and note that $A=f^{-1}(\{f(x)\})$ and $B=f^{-1}(D\setminus\{f(x)\})$ are disjoint open sets in $X$ and $A\cup B=X$. Then…

Conversely, if $X$ is disconnected, take two disjoint open sets $A,B$ witnessing it and consider the map $f\colon X\to\{0,1\}$ where $f(a)=0$ for $a\in A$ and $f(b)=1$ for $b\in B$.

The assumption of continuity cannot be lifted off: consider the map $f\colon\mathbb{R}\to\{0,1\}$ defined by $f(x)=0$ if $x$ is rational and $f(x)=1$ if $x$ is irrational.


问:
Finding the volume of surface bounded by planes and $x=y^2$

Find the volume of the surface that's bounded by $x=y^2$ and $x=z, z=0, x=1.$

Looking at the $yz-$plane this seems to form a triangle that's bounded by $y=z$ and $y=1$. I'm not entirely sure if I should form an integral of the form $$\int\int f(y,z) \ dz \ dy$$
or some other way around. From $x=z$ I would have that $z=y^2$ so $f(y,z) = y^2-z$? Any hints would be appreciated.

答:
I had to draw in order to get a better view of the volume. First I handled the parabola given by $x=y^2, \forall z\in \mathbb{R}$ which is cut by the planes $x=z, \forall y \in \mathbb{R}$ and $z=0, \forall (x,y)\in \mathbb{R}^2$. It gave something like this:

In order to get rid of the $z$ variable I did

$$\int_0^x dz = x$$

Next all I have to take care of is in the $xy$ plane where I also used the parabola $x=y^2$ but added $x=1$. This gives this picture:

Now I choose to integrate

$$\int_{y^2}^1 x dx = \frac{1}{2} - \frac{y^4}{2}$$

Finally $y\in [-1,1]$ which gives

$$\int_{-1}^{1} \frac{1}{2} - \frac{y^4}{2} dy = \frac{4}{5}$$

Unless I messed up the reasoning, this is the approach I went with.


问:
Limit of a product with growing number of factors
I'm trying to solve the following limit:
$$L=\lim_{n\to\infty}\frac{(n^2-1)(n^2-2)\cdots(n^2-n)}{(n^2+1)(n^2+3)\cdots(n^2+2n-1)}=\lim_{n\to\infty}\prod_{k=1}^n\frac{n^2-k}{n^2+2k-1}$$
Since originally, the limit yields an $1^\infty$ indeterminate expression, my first idea was taking logarithms:
$$\log L = \lim_{n\to\infty}\log\left(\prod_{k=1}^n\frac{n^2-k}{n^2+2k-1}\right)=\\
=\lim_{n\to\infty}\sum_{k=1}^{n}\log\left(\frac{n^2-k}{n^2+2k-1}\right)=\\=\lim_{n\to\infty}\sum_{k=1}^n\log\left(\frac{1-\frac{k}{n^2}}{1+\frac{2k-1}{n^2}}\right)=\\
=\lim_{n\to\infty}\sum_{k=1}^n\left(\log\left(1-\frac{k}{n^2}\right)-\log\left(1+\frac{2k-1}{n^2}\right)\right)=\\=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(\log\left(1-\frac{k}{n^2}\right)^{n^2}-\log\left(1+\frac{2k-1}{n^2}\right)^{n^2}\right)=\\
=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(\log\left(1+\frac{1}{-\frac{n^2}{k}}\right)^{-\frac{n^2}{k}(-k)}-\log\left(1+\frac{1}{\frac{n^2}{2k-1}}\right)^{\frac{n^2}{2k-1}(2k-1)}\right)$$

Now if I could do the limit of the terms inside the sum I'd have:
$$\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(\log\left(1+\frac{1}{-\frac{n^2}{k}}\right)^{-\frac{n^2}{k}(-k)}-\log\left(1+\frac{1}{\frac{n^2}{2k-1}}\right)^{\frac{n^2}{2k-1}(2k-1)}\right)=\\=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(\log e^{-k}-\log e^{2k-1}\right)=\\
=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(-k-2k+1\right)=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(-3k+1\right)$$
and then the limit would be trivial. The question is: is that allowed? Why? If not, how could I proceed?
Thanks!


问:
$n^\text{th}$ derivative of a function
I'm trying to show that $f^{n+1}(x)=0$. I've calculated the derivatives of $f(x)=2x^{5}+3x^{4}+4$.

$f'(x)=10x^{4}+12x^{3}$

$f''(x)=40x^{3}+36x^{2}$

$f'''(x)=120x^{2}+72$

$f^{4}(x)=240x+72$

$f^{5}(x)=240$

But I'm not sure how to proceed from there?

答:
You are done.

$f^5(x) = 240$ and that is a constant.  The derivative of a constant function is $0$ so $f^6(x) = 0$.  And $0$ is a constant function so it's derivative is $0$ and so by induction.

Base case: $f^6(x) = 0$.

Induction step: If $f^n(x) = 0$, then $f^{n+1}(x) = (0)' = 0$.

Conclusion:  $f^{n} = 0$ for all $n \ge 6$.

Title: $n^\text{th}$ derivative of a function


问:
Prove that $ \liminf_{n}X_{n} $ is a random variable
We define random variable $ X: \varOmega \to \mathbb{R} $ as a function such that for and Borel set $ \forall B\in\mathbb{\mathcal{B}} $ it follows that $ X^{-1}\left(B\right)\in\mathcal{F} $ where $ \mathcal{F} $ is  a sigma-algebra over $ \varOmega $

Now assume that $ \left(X_{n}\right)_{n} $ is a sequence of random variables. Is it true that the function $ Y:\varOmega \to \mathbb{R} $ by :

$ Y\left(\omega\right)=\liminf_{n}X_{n}\left(\omega\right) $ is a random variable?

Here's what I tried.

I want to prove that indeed it is a random variable. Since $ \mathcal{B}=\sigma\left((-\infty,t]:t\in\mathbb{R}\right) $ its enough to show that for any $ t \in \mathbb{R} $ it follows that  $ Y^{-1}\left((-\infty,t]\right)\in\mathcal{F} $

But

$ Y^{-1}\left((-\infty,t]\right)=\left\{ \omega\in\varOmega:\liminf_{n}X_{n}\left(w\right)\in(-\infty,t]\right\}  $

Now if I'll prove that

$ \left\{ \omega\in\varOmega:\liminf_{n}X_{n}\left(w\right)\in(-\infty,t]\right\} =\bigcap_{k}^{\infty}\bigcup_{n\geq k}X_{n}^{-1}\left((-\infty,t]\right) $

Then it will end the proof because the latter is an event in the sigma-algebra. I already proved that $ \left\{ \omega\in\varOmega:\liminf_{n}X_{n}\left(w\right)\in(-\infty,t]\right\} \supseteq\bigcap_{k}^{\infty}\bigcup_{n\geq k}X_{n}^{-1}\left((-\infty,t]\right) $

But Im having trouble to prove that $ \left\{ \omega\in\varOmega:\liminf_{n}X_{n}\left(w\right)\in(-\infty,t]\right\} \subseteq\bigcap_{k}^{\infty}\bigcup_{n\geq k}X_{n}^{-1}\left((-\infty,t]\right) $

Because if I'll take $ \omega $ such that  $ \liminf_{n}X_{n}\left(w\right)\in(-\infty,t] =t $, thenI cannot promise that this $ \omega $ is in $ \bigcap_{k}^{\infty}\bigcup_{n\geq k}X_{n}^{-1}\left((-\infty,t]\right) $.

Any ideas?

答:
We'll instead prove that
$$
\{ \liminf X_n \leq t \} = \bigcap_{r \geq 1} \bigcap_{k \geq 1} \bigcup_{n \geq k} \left\{ X_n < t + \frac1r \right\} =: \bigcap_{r \geq 1} E_r
$$
Adding the wiggle room of "$+1/r$" fixes the problem you were having.

For the $\supseteq$ direction, if $\omega \in E_r$ then $X_n(\omega) < t + 1/r$ infinitely often, which implies $\liminf X_n < t+ 1/r$.  So if this holds for every $r \geq 1$ then $\liminf X_n \leq t$.

For the $\subseteq$ direction, $\liminf X_n(\omega) \leq t$ implies that for every $\epsilon > 0$, $X_n(\omega) < t+\epsilon$ infinitely often, which is equivalent to the statement that $\omega \in E_r$ for every $r$.


问:
Is every set that's a union of two sets disconnected?
I'm curious about the definition of disconnected and I'm sure this can't be right. But let's say that $S= A \cup B$, then isn't $S$ automatically disconnected? Since I can re-write It as $(S \cap C) \cup (S \cap D)$ with $C=A$ and $D=B$? This is a circular definition since it basically uses an identity about unions and intersection, but is my understanding of disconnected totally off? Or is this just a trivial case we toss aside? I'm a tad confused here.</p>

答:
Assuming the topic is topology, I'm interpreting the question is follows: If $X$ is a topological space and $A,B\subseteq X$, is $A\cup B$ always disconnected?

The answer is no, for two reasons. The first is trivial, simply if $A$ and $B$ are not disjoint, then their union may be connected (e.g., take $X=\mathbb R$, A=B=[0,1]). More interestingly, there are famous examples of two disjoint sets whose union is connected, for instance take $X=\mathbb R$, $A=(0,1]$, and $B=\{0\}$. A more interesting example is for instance the topologist's since curve.


问:
How to find details about function in $\mathbb{R}^2$ if $\frac{df} {dx}= \frac{df}{dy}$ for all $(x,y)\in \mathbb{R}^2$
I am trying assignment questions of real analysis and I was unable to solve this question. In fact, I don't have any idea on how to start.

So, I am looking for help here.

Let $f$  be a function on $\mathbb{R}^2$ such that $\frac{df} {dx}= \frac{df}{dy}$ for all $(x,y)\in  \mathbb{R}^2$. Now which of the following options are true:

1 $f$ is constant on all lines parallel to the line $x=-y$.

2 $f(x,y)=0 \quad\forall (x,y) \in \mathbb{R}^2$.

3 $f(x,y)=f(-y,x) \quad\forall (x,y) \in \mathbb{R}^2$.

4 $f(x,y)-f(y,x)=(x-y) \frac {df}{dx}(x*,y*) + (y-x) \frac {df}{dy} (x*,y*) \in \mathbb{R}^2$.

I am unable to get any idea on how to use the hypothesis $\frac{df} {dx}= \frac{df}{dy}$ to deduce or contradict  any of the options.

I studied multivariable calculus from Tom M Apostol in case it is of any help.

Please help me. Kindly just give outlines not complete solution.

答: Since $\displaystyle \frac{\partial f}{\partial x}= \frac{\partial f}{\partial y}$, then we must have $$\nabla f=\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)$$ Therefore $$\nabla f= \frac{\partial f}{\partial x}(1,1)$$ and the gradient can be written as $$\nabla f= \frac{\partial f}{\partial x}\vec{i} + \frac{\partial f}{\partial x}\vec{j}$$ This the direction of steepest ascent is along all lines parallel to the vector $\vec{i}+\vec{j}$ Now the gradient is always normal (perpendicular) to the level curves, so this means all lines parallel to $\vec{i}-\vec{j}$ are level curves, that is $f$ is constant along all lines of the forms $x+y= c$, $c\in \mathbb{R}$. Now use that info to make a determination concerning #$3$


答:
Let's take the first part

$$\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\log\left(1+\frac{1}{-\frac{n^2}{k}}\right)^{-\frac{n^2}{k}(-k)} = -\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n k\log\left(1+\frac{1}{-\frac{n^2}{k}}\right)^{-\frac{n^2}{k}}$$

as example. Note that $-\frac{n^2}{k}$ tends to $\infty$ for any $k=1,2,\cdots,n$, so we can use Taylor's expansion

$$\left(1+\frac{1}{x}\right)^x = e - \frac{e}{2x} + O\left(\frac{1}{x^2}\right) \quad (x \to \infty)$$

to obtain

$$\begin{aligned}
\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\log\left(1+\frac{1}{-\frac{n^2}{k}}\right)^{-\frac{n^2}{k}(-k)}
&= -\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n k\log\left[e+\frac{ek}{2n^2}+O\left(\frac{k^2}{n^4}\right)\right] \\
&= -\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n \left[k + k\log\left(1+\frac{k}{2n^2}+O\left(\frac{k^2}{n^4}\right)\right)\right] \\
&= -\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n \left[k + k\left(\frac{k}{2n^2}+O\left(\frac{k^2}{n^4}\right)\right)\right] \\
&= -\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n k\left[1+O\left(\frac{1}{n}\right)\right],
\end{aligned}$$

where we have used $\frac{k}{n^2}=O\left(\frac{1}{n}\right)$. So the term $O(1/n)$ can be discarded.

Note: An alternative solution is given here:

Lemma. Given $f(0) = 0$ and that finite $f'(0)$ exists, let

$$a_n = \sum_{k=1}^n f\left(\frac{k}{n^2}\right) = f\left(\frac{1}{n^2}\right) + f\left(\frac{2}{n^2}\right) + \cdots + f\left(\frac{n}{n^2}\right),$$

then

$$\lim_{n\to\infty} a_n = \frac{f'(0)}{2}.$$

Proof. First notice that

$$f'(0) = \lim_{x \to 0} \frac{f(x)-f(0)}{x-0} = \lim_{x \to 0} \frac{f(x)}{x},$$

and using the definition of limit, $\forall \varepsilon > 0$, $\exists \delta > 0$, s.t. $\forall 0 < x < \delta$,

$$f'(0) - \varepsilon < \frac{f(x)}{x} < f'(0) + \varepsilon.$$

Particularly for $x > 0$, we have $(f'(0)-\varepsilon)x < f(x) < (f'(0)+\varepsilon)x$.

Now pick $N \in \mathbb{N}$, s.t. $N > \dfrac{1}{\delta}$. For $n > N$,

$$\frac{k}{n^2} \leq \frac{1}{n} < \frac{1}{N} < \delta, \quad k=1,2,\cdots,n,$$

and hence

$$(f'(0) - \varepsilon) \cdot \frac{k}{n^2} < f\left(\frac{k}{n^2}\right) < (f'(0) + \varepsilon) \cdot \frac{k}{n^2}, \quad k=1,2,\cdots,n.$$

Taking summation over $k$ gives

$$\frac{f'(0)-\varepsilon}{2} \cdot \frac{n+1}{n} < \sum_{k=1}^n f\left(\frac{k}{n^2}\right) < \frac{f'(0)+\varepsilon}{2} \cdot \frac{n+1}{n}.$$

Since $\dfrac{n+1}{n} \to 1$ as $n \to \infty$, then $\exists N_1 \in \mathbb{N}$, s.t. $\forall n > N_1$,

$$\frac{f'(0)-\varepsilon}{2} - \frac{\varepsilon}{2} < \sum_{k=1}^n f\left(\frac{k}{n^2}\right) < \frac{f'(0)+\varepsilon}{2} + \frac{\varepsilon}{2},$$

that is, $\dfrac{1}{2}f'(0)-\varepsilon < \sum\limits_{k=1}^n f\left(\frac{k}{n^2}\right) < \dfrac{1}{2}f'(0)+\varepsilon$. Therefore we have

$$\lim_{n\to\infty} a_n = \lim_{n\to\infty} \sum\limits_{k=1}^n f\left(\frac{k}{n^2}\right) = \frac{f'(0)}{2}.$$

Return to your problem, we have

$$I = \lim_{n\to\infty}\prod_{k=1}^n\frac{n^2-k}{n^2+2k-1} = \lim_{n\to\infty}\prod_{k=1}^n\frac{n^2-k}{n^2+2k} = \lim_{n\to\infty}\prod_{k=1}^n \frac{1-\frac{k}{n^2}}{1+2\frac{k}{n^2}} = \lim_{n\to\infty} e^{\sum_{k=1}^n f(k/n^2)},$$

where $f(x) = \ln\left(\frac{1-x}{1+2x}\right)$ and the second equality comes from

$$1 \leftarrow \left(1-\frac{1}{n^2}\right)^n \leq \prod_{k=1}^n \frac{n^2+2k-1}{n^2+2k} = \prod_{k=1}^n \left(1-\frac{1}{n^2+2k}\right) \leq 1.$$

Therefore $I = e^{\frac{f'(0)}{2}} = \boxed{e^{-\frac{3}{2}}}$.

Title: Limit of a product with growing number of factors

Body:
I'm trying to solve the following limit:
$$L=\lim_{n\to\infty}\frac{(n^2-1)(n^2-2)\cdots(n^2-n)}{(n^2+1)(n^2+3)\cdots(n^2+2n-1)}=\lim_{n\to\infty}\prod_{k=1}^n\frac{n^2-k}{n^2+2k-1}$$
Since originally, the limit yields an $1^\infty$ indeterminate expression, my first idea was taking logarithms:
$$\log L = \lim_{n\to\infty}\log\left(\prod_{k=1}^n\frac{n^2-k}{n^2+2k-1}\right)=\\
=\lim_{n\to\infty}\sum_{k=1}^{n}\log\left(\frac{n^2-k}{n^2+2k-1}\right)=\\=\lim_{n\to\infty}\sum_{k=1}^n\log\left(\frac{1-\frac{k}{n^2}}{1+\frac{2k-1}{n^2}}\right)=\\
=\lim_{n\to\infty}\sum_{k=1}^n\left(\log\left(1-\frac{k}{n^2}\right)-\log\left(1+\frac{2k-1}{n^2}\right)\right)=\\=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(\log\left(1-\frac{k}{n^2}\right)^{n^2}-\log\left(1+\frac{2k-1}{n^2}\right)^{n^2}\right)=\\
=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(\log\left(1+\frac{1}{-\frac{n^2}{k}}\right)^{-\frac{n^2}{k}(-k)}-\log\left(1+\frac{1}{\frac{n^2}{2k-1}}\right)^{\frac{n^2}{2k-1}(2k-1)}\right)$$

Now if I could do the limit of the terms inside the sum I'd have:
$$\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(\log\left(1+\frac{1}{-\frac{n^2}{k}}\right)^{-\frac{n^2}{k}(-k)}-\log\left(1+\frac{1}{\frac{n^2}{2k-1}}\right)^{\frac{n^2}{2k-1}(2k-1)}\right)=\\=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(\log e^{-k}-\log e^{2k-1}\right)=\\
=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(-k-2k+1\right)=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(-3k+1\right)$$
and then the limit would be trivial. The question is: is that allowed? Why? If not, how could I proceed?
Thanks!


问:
Prove $S^{-1}A \cong A[x]/(bX-1)$
Let $A$ be a integral domain, $b\in A$ a nonzero element and $S=\{b^n\mid n \in \mathbb{N}\cup\{0\}\}$

Prove: $S^{-1}A \cong A[x]/(bX-1)$

For this we define the homomorphism $\varphi: X\longrightarrow1/b$ and prove that $\ker\varphi=(bX-1)$

Since $\varphi(bX-1)=b·1/b-1=0$ we have that $\ker\varphi \supseteq(bX-1)$

Now we have to see that any $f\in \ker\varphi$ is in the ideal $(bX-1)$, the obvious way to approach this is to divide $f$ by $bX-1$, but we can't apply the division algorithm directly since the main coefficient of $bX-1$ is not necessarily $1$.

Any help to prove $\ker\varphi \subseteq(bX-1)$?

答:
Hint: $S^{-1}A = \{\frac{a}{b^n}\mid a\in A, n\geq 0\}$.

The quotient ring $Q = A[x]/(bX-1)$ contains a zero of $bX-1$, namely the residue class of $x$, which you can denote by $b^{-1}$. This gives the ring extension $A[b^{-1}]$.

Title: Prove $S^{-1}A \cong A[x]/(bX-1)$


问:
Kernel of a Homomorphism.
I am trying to figure out a rather simple and specific concept.

Let's say I have a group $G$ and a subgroup $H<G$.

Define $X=\{xH\mid x\in G\}$ – the set of all the left cosets of $H$

We have $G\curvearrowright X\;$ by $g\cdot xH = (gx)H$.

I know that this operation defines a homomorphism $\phi:G\rightarrow \operatorname{Per}(X)$ (the set of permutations on $X$), and I am really curious about the behaviour of $\operatorname{Ker}(\phi)$ and its relation to $H$.

Do we have $\operatorname{Ker}(\phi) = H$? If not, how do we get that? I'd love to get some insight! thanks

答: $\ker(\theta)=\cap_{x\in G}xHx^{-1}$. This is called the core of $H$ in $G$, denoted by Core$_G(H)$. It is a subgroup of $H$ and a normal subgroup of $G$ (and thus a normal subgroup of $H$ as well). It has some other nice properties, which you can read about here if you want: https://groupprops.subwiki.org/wiki/Normal_core

Title: Kernel of a Homomorphism.


答:
If $f_x=f_y$, then for every $c\in \mathbb R$,
$$
\frac{d}{dt}f(t,c-t)=f_x(t,c-t)-f_y(t,c-t)=0.
$$
Hence $f$ is constant on the set
$$
L_c=\{(t,c-t):t\in\mathbb R\},
$$
which represents the parametric expression of the straight line
$$
x+y=c.
$$
Thus $f$ has to be of the form
$$
f(x,y)=g(x+y).
$$
To understand that, observe first that $\mathbb R^2=\bigcup_{c\in\mathbb R}L_c$, and in particular, note that $\,(x,y)\in L_{x+y}$.

As $f$ is constant on every $L_c$, then for every $c\in\mathbb R$, there exists a function $g$, such that
$$
f(x,y)=g(c), \quad \text{for all $(x,y)\in L_c$}.
$$
But $c=x+y$, for all $(x,y)\in L_c$, and hence $\,f(x,y)=g(c)=g(x+y)$.

Title: How to find details about function in $\mathbb{R}^2$ if $\frac{df} {dx}= \frac{df}{dy}$ for all $(x,y)\in \mathbb{R}^2$

Body:
I am trying assignment questions of real analysis and I was unable to solve this question. In fact, I don't have any idea on how to start.

So, I am looking for help here.

Let $f$  be a function on $\mathbb{R}^2$ such that $\frac{df} {dx}= \frac{df}{dy}$ for all $(x,y)\in  \mathbb{R}^2$. Now which of the following options are true:

1 $f$ is constant on all lines parallel to the line $x=-y$.

2 $f(x,y)=0 \quad\forall (x,y) \in \mathbb{R}^2$.

3 $f(x,y)=f(-y,x) \quad\forall (x,y) \in \mathbb{R}^2$.

4 $f(x,y)-f(y,x)=(x-y) \frac {df}{dx}(x*,y*) + (y-x) \frac {df}{dy} (x*,y*) \in \mathbb{R}^2$.

I am unable to get any idea on how to use the hypothesis $\frac{df} {dx}= \frac{df}{dy}$ to deduce or contradict  any of the options.

I studied multivariable calculus from Tom M Apostol in case it is of any help.

Please help me. Kindly just give outlines not complete solution.


问:
Do not understand where these terms are from in a product of exponential random variable
This example comes from Tsitsiklis' Introduction to Probability.

I don't really quite understand what the product and log are about.

I get that the exponential RV is $\theta e^{-\theta x}$. But then what is the $x_i$ in $\prod^n_{i=1}\theta e^{-\theta x_i}$?

Is the product (ie. $\prod$) essentially the same as $\theta e^{-\theta x_1} \times \theta e^{-\theta x_2} \times ...\theta e^{-\theta x_n}$? (But this makes those indexed $x$ more confusing...what's the difference between them and just $x$?)

Where does the $y_n$ come from in $n \log\theta-\theta y_n$?

I apologise for asking such elementary questions, but I am trying to learn probability with a limited maths background. Thank you so much!

答:
The $x_i$ are the values associated with the $X_i$, i.e. the realisations of the RV $X_i$. A definition is given for the $X_i$ as the difference in arrival times of successive customers. Considering $n$ customers you have $n$ differences in arrival times, with special consideration for $X_1$.

Yes that is what $\prod$ represents, a product. $x$ is a general symbol for a value without regard to a particular observation, $x_i$ is the value of the $i^{th}$ observation.

log of a product is the sum of the logs, i.e.
$$
\log(AB)=\log A+\log B
$$
This extends to any number of terms in the product, in essence for the log-likelihood (ignoring one of the $\theta$ factors, which if included would generate the $n \log \theta$ term)
$$
\log \left( \prod_i^n \exp(-\theta x_i) \right) = \Sigma_i^n  \log(\exp(-\theta x_i))=\Sigma_i^n (-\theta x_i) = -\theta \, \Sigma_i^n x_i = -\theta y_n
$$


问:
Residue field of generic point
I have a question, so let $X\to \operatorname{spec}(K)$ a smooth projective variety over a field K.
Let $\eta$ its generic point.

I think that the residue field of $\eta$ is :
$$ K(\eta) \cong \operatorname{Frac}(\mathcal{O}_{X,\eta}) $$
But I'm looking for a proof.

And what  is the degree of the extension of fields $K(\eta) / K$ ?

Thanks for answers !

答:
Let $X$ be any integral scheme, and let $\eta \in X$ be the generic point. (In particular, this includes any irreducible variety over a field, regardless of whether it's smooth or projective.) Then the residue field $\kappa(\eta)$ at $\eta$ is the function field $K(X)$ of $X$, that is, the field of all rational functions on $X$. This is also equal to the stalk $\mathcal{O}_{X, \eta}$, which is already a field (no need to take the field of fractions).

This is proven in Stacks Project, Lemma 29.49.5 (tag 01RV), though a few of the steps in that proof are unnecessary when there's only one irreducible component. The correspondence between elements of $\mathcal{O}_{X, \eta}$ and rational functions on $X$ essentially amounts to the fact that every nonempty open subset of $X$ contains the generic point $\eta$, which follows from irreducibility. Since $X$ is also reduced, every nonzero rational function on $X$ is invertible on some nonempty open subset, so $\mathcal{O}_{X, \eta}$ is already a field and hence is isomorphic to the residue field $\kappa(\eta)$.

When $X$ is an irreducible variety over a field $k$, the function field of $X$ is an extension of $k$ of transcendence degree equal to the dimension of $X$. This follows from the Noether normalization lemma. In particular, the degree is infinite except in the zero-dimensional case where $X = \operatorname{Spec}(L)$ for some finite extension $L/K$.

Title: Residue field of generic point


答:
Note that kernel of the action is as same as the kernel of the corresponding homomorphism (can you prove this?). So, we can find kernel of the action in order to find $\text{Ker}(\phi)$.

Let $K$ be the kernel of the action you mentioned. Then,
$$K = \{g\in G\ |\ (gx)H = xH, \forall\ xH \in X\} = \{g\in G\ |\ g \in xHx^{-1}, \forall x \in G\} = \bigcap_{x \in G}xHx^{-1} = \text{Ker}(\phi)$$

So, $\text{Ker}(\phi) \subseteq H$ since for $x = e_G$, we have $xHx^{-1} = H$. In order to have $\text{Ker}(\phi) = H$, we must have $H \subseteq xHx^{-1}, \forall x \in G$ since then $H$ will be the smallest set in the intersection $\bigcap_{x \in G}xHx^{-1}$. As a concrete example, if we have $H \unlhd G$, then $xHx^{-1} = H, \forall x \in G$ so we can say that $\text{Ker}(\phi) = H$.

Title: Kernel of a Homomorphism.

Body:
I am trying to figure out a rather simple and specific concept.

Let's say I have a group $G$ and a subgroup $H<G$.

Define $X=\{xH\mid x\in G\}$ – the set of all the left cosets of $H$

We have $G\curvearrowright X\;$ by $g\cdot xH = (gx)H$.

I know that this operation defines a homomorphism $\phi:G\rightarrow \operatorname{Per}(X)$ (the set of permutations on $X$), and I am really curious about the behaviour of $\operatorname{Ker}(\phi)$ and its relation to $H$.

Do we have $\operatorname{Ker}(\phi) = H$? If not, how do we get that? I'd love to get some insight! thanks


问:
Confused about combinatorials
How do I solve 4$\cdot$6 = 8$\cdot$3 by a combinatorial proof? How can I start this proof? I know that I can show a two pictures that represent 24 but I'm not entirely sure how to go about this. Please help!</p>

答:
Completely revised: Since this is in the context of elementary combinatorics, I suspect that you’re intended to apply the multiplication principle. Consider the following problem:

How many three-digit numbers $abc$ are there in which $a\in\{1,2\}$, $b\in\{1,2,3\}$, and $c\in\{1,2,3,4\}$?

Now calculate this in two ways:

First count the ways to choose $a$ and $b$. Then count the ways to combine one of these $ab$ combinations with a choice of $c$.

First count the ways to choose $a$ and $c$. Then count the ways to combine that choice with a choice of $b$.

Title: Confused about combinatorials


问:
Compute $ f(x) = \int_0^{\pi} \frac{ \sin \theta ( \cos \theta - x) d \theta}{ [(\cos \theta - x)^2 + \sin^2 \theta]^{1.5} } x \in \Bbb{R}$
Compute

$$ f(x) = \int_0^{\pi} \frac{ \sin \theta ( \cos \theta - x)}{ [(\cos \theta - x)^2 + \sin^2 \theta]^{1.5} }
\,\,d \theta\,,\quad   x \in \Bbb{R}$$

The denominator makes it a bit annoying. We can evaluate it as

$ [(\cos \theta - x)^2 + \sin^2 \theta = (1 - 2 \cos \theta x + x^2)^{1.5}$ but it doesn't make it any easier.  We can also expand $(1 + x^2 - 2 \cos \theta x )^{1.5}$ using $(1+x)^\alpha = \sum_{n=0}^\infty \binom{\alpha}{n} x^n$ but it's still difficult.

The denominator also has some geometric meaning  (cubed distance from $(x,0)$ to $(\cos \theta, \sin \theta)$ ) but does it help?

答:
Consider the sequence of substitutions:

$$u=\cos\theta-x\implies\mathrm du=-\sin\theta\,\mathrm d\theta$$

$$\implies f(x)=-\int_{1-x}^{-1-x}\frac u{(u^2+1-(u+x)^2)^{\frac32}}\,\mathrm du=\int_{-1-x}^{1-x}\frac u{(1-2ux-x^2)^{\frac32}}\,\mathrm du$$

Then

$$v=1-2ux-x^2\implies\mathrm dv=-2\,\mathrm du$$

$$\implies f(x)=\frac1{4x}\int_{x^2+2x+1}^{-3x^2+2x+1}\frac{v+x^2-1}{v^{\frac32}}\,\mathrm dv$$


问:
Checking if a quotient group is cyclic
Is there an easy way to check if the group $(\mathbb{Z}_9 \times \mathbb{Z}_9)/\langle(3,3)\rangle$ is cyclic? I know I could determine this by writing out each of the elements and checking if one of them is a generator. But there are 27 elements (if my math is correct), which would make that approach very tedious. Is there an easier way?

答: The map $\varphi\colon \Bbb Z_9\times \Bbb Z_9\to \Bbb Z_3\times \Bbb Z_9$, defined by $([m]_9,[n]_9)\mapsto ([m]_3,[m-n]_9)$, is a surjective homomorphism with kernel $\langle(3,3)\rangle$. By the First Homomorphism Theorem, $(\Bbb Z_9\times \Bbb Z_9)/\langle(3,3)\rangle\cong \Bbb Z_3\times \Bbb Z_9$, which is not cyclic, for $\operatorname{gcd}(3,9)=3\ne 1$.

Title: Checking if a quotient group is cyclic


问:
How can i check whether an event is in a sigma-algebra generated by multiple random variables?
In my Probability-Theory course we had the proof of a theorem (Kolmogorov's inequality) and i got stuck at one point...

Let's say for the sake of this question that we consider a simple random walk, i.e. $X_i,1\leq i\leq n$, independent random variables, s.t.

$$P[X_i=1] = P[X_i=-1]=\frac 12$$

We defined events ($u\gt 0$ is a given real number)
$$1 \leq k \leq n :A_k:=\{|S_k|\geq u:|S_j|\lt u \; \forall j\lt k \} = \{first\; time \;j\; at \;which\;|S_j| \ge u\; is \;exactly\; k\}$$
Now in the proof it is claimed without further explanation that  $A_k \in \sigma(X_1,...,X_k)$.

I think that intuitively this makes sense, because it can be determined whether $A_k$ happens or not by just looking at $X_1,...,X_k$. But what is the formal explanation for it?

答:
First, for each $1 \le j \le k$, we have $S_j$ is $\sigma(X_1,...,X_k)$-measurable because it is the sum of $\sigma(X_1,...,X_k)$-measurable random variables, so $|S_j|$ is also $\sigma(X_1,...,X_k)$-measurable for each $1 \le j \le k$.  By the definition of measurability, this implies $\{|S_j| \ge u \} \in \sigma(X_1,...,X_k)$ and $\{|S_j| < u \} \in \sigma(X_1,...,X_k)$ for all $u$ and $1 \le j \le k$.  Then

\begin{align*}
A_k = \{|S_k| \ge u \} \cap \{|S_{k-1}| < u \} \cap \cdots \cap \{|S_1| < u \}
\end{align*}

is a finite intersection of sets in $\sigma(X_1,...,X_k)$ and therefore is itself in $\sigma(X_1,...,X_k)$.


问:
How can I write this sum $ S = z^0 + z^2 + z^4 + ... + z^{2020} $
let z be $ z = e^{i\cdot \pi/2021} $

and  $ S = z^0 + z^2 + z^4 + ... + z^{2020} $

I figured that S could be written in this way $ S = \sum_{k=0}^{1010} z^{2k} $

But I need to write S in this way $ S = 1/(1-z) $

I really got stuck may someone help me?

答:
IF your sum were infinite, then you'd have the right formula.  Since the sum is finite, the formula is

$$S = \frac{1-z^{2022}}{1-z^2}.$$

Since $z^{2021} = 1$, this should be easy to simplify.


问:
$\int \frac{dx}{(x^4 + 1)^2}$
What would be a relatively simple method for computing the indefinite integral below?

$\displaystyle \int \frac{dx}{(x^4+1)^2}$

Furthermore, how would one evaluate the following, possibly by detouring the computation of the indefinite integral?

$\displaystyle\int\nolimits_{-\infty}^\infty \frac{dx}{(x^4+1)^2}$

Note: The resolved value of the definite integral (according to wolfram alpha) is $\displaystyle \frac{3\pi}{4\sqrt{2}}$

答:
Well, the problem is tagged "partial fractions," that seems to me to be the way to go. Of course, first you need to factor $x^4+1$ into irreducible quadratics; that can be done by $$x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-(\sqrt2x)^2$$

It will be messy - do you have any reason to think there is a simpler method?

Edit: for the definite integral, if you've done complex variables, you'll know about contour integration, which ought to handle this example without difficulty. If you haven't done complex variables, you have something truly wonderful to look forward to.

Title: $\int \frac{dx}{(x^4 + 1)^2}$


问:
Why base point makes a huge difference?
While preparing a talk, I was tempted to "prove" the following relationship:

$$\text{Prin}_{G}(X)\cong [X,BG]\cong [B[\pi_{1}(X),BG]\cong [\Omega B[\pi_{1}(X)],\Omega BG]\cong [\pi_{1}(X),G]$$

Here $X$ is a 2 dimensional surface.

The second and third equivalence is right, but the fourth or fifth is wrong. The professor told me I need to think myself to understand why the based loop space does not behave in the way I thought it would, but I have no clue. For example, if $G=\mathbb{S}^{1}$, then $BG\cong \mathbb{Z}$ and the homotopy classes of maps from $X$ to $\mathbb{Z}$ is isomorphic to $\mathbb{Z}$ for any closed surface. But it would be ridiculous if $[\pi_{1}(X),\mathbb{Z}]$ is isomorphic to $\mathbb{Z}$ for any $X$. So something subtle is here and I want to ask for a hint (not a solution).

答:
I attempt to give an answer to address my own confusion.

There are two confusions. One is the homotopy class of maps from $X\cong B(\pi_{1}(X))$ to $BG$ is not behaving nicely under based loop spaces. For any map $f\rightarrow X$, the composition with a map from $X$ to $BG$ can land it to anywhere. Therefore it does not make any sense to talk about applying loopspace operator on both sides.

The other confusion, as others pointed out is equally basic. I confused $B\mathbb{S}^{1}$ with $\mathbb{Z}$, which should be $\mathbb{CP}^{\infty}$ instead. The fact that $[X,\mathbb{CP}^{\infty}]$ is $\mathbb{Z}$ is still true because in this case $\text{Prin}_{\mathbb{S}^{1}}[X]\cong H^{2}(X)=\mathbb{Z}$ using Cech cohomology. So the fact that $\mathbb{Z}\not=\mathbb{Z}^{2g}$ give an example why the above "equivalence" cannot hold. However, I do not feel comfortable in asserting $$\text{Prin}_{\mathbb{S}^{1}}[X]\cong H^{2}(X)=\mathbb{Z}$$
as I read the proof but do not know why intuitively it must be true. So if someone can point it out I would be grateful.

Title: Why base point makes a huge difference?


Title: How can I write this sum $ S = z^0 + z^2 + z^4 + ... + z^{2020} $

Body:
let z be $ z = e^{i\cdot \pi/2021} $

and  $ S = z^0 + z^2 + z^4 + ... + z^{2020} $

I figured that S could be written in this way $ S = \sum_{k=0}^{1010} z^{2k} $

But I need to write S in this way $ S = 1/(1-z) $

I really got stuck may someone help me?

Body:
$$ S = 1 + z^2 + z^4 + ... + z^{2020}= \sum_{r=0}^{1010} z^{2k} = \sum_{r=0}^{1010} (z^2)^{k} =\frac{z^{2022}-1}{z^2-1}=\frac{z.z^{2021}-1}{z^2-1}$$

It's pretty obvious from here


问:
Conditions to assure convergence of MCMC with Metropolis Hastings: ergodicity and detailed balance?
I'm working on an MCMC algorithm for a machine learning project that simulates an adaptive system using a reaction-diffusion SSA framework to build model structure (reactions via Green's rjMCMC) and refine parameters (diffusion via standard MCMC). I've been able to show that the proposal process is irreducible, positive recurrent, and aperiodic (i.e., ergodic) but I don't know how to show that it follows detailed balance. I would like to demonstrate that the approach I'm pursuing will converge to the posterior of interest before I proceed.

I've read elsewhere that ergodicity is a sufficient condition for a proposal for M-H MCMC (e.g., https://jellis18.github.io/post/2018-01-02-mcmc-part1/), but other sites (e.g., https://en.wikipedia.org/wiki/Metropolis%E2%80%93Hastings_algorithm) that provide a "formal derivation" list two requirements: (1) existence and (2) uniqueness of a stationary distribution. Under (2), this site lists ergodicity as sufficient, but under (1) it lists detailed balance as sufficient. My understanding was that ergodicity implied both the existence and uniqueness of a stationary distribution, and moreover that this stationary distribution would be the limiting distribution. Here is my question: why is detailed balance listed separately from ergodicity as a requirement for the M-H MCMC proposal when it would appear that ergodicity would satisfy the two requirements? What am I missing here?

Thank you for your time and thoughts.

答:
I can't give you a nuanced theoretical account of the interplay of the conditions underlying convergence of the Metropolis-Hastings MCMC algorithm, as Casella and Robert have done a better job than I ever could in Chapter 6 of "Monte Carlo Statistical Methods", if you want an account using some machinery in stochastic processes. I'm also certain that other members on this website could illustrate the nuances much more fluently than I could.

To the best of my introductory knowledge on MCMC, and also experience with time-series, the most plausible explanation I have for your query on why the detailed balance conditions are not just redundant in light of ergodicity requirements is that ergodicity can be difficult to assess* (although you have already established it in your context).

From a practitioner's perspective rather than someone concerned with the rigorous technicalities of how it all fits together, I refer you to Kevin Murphy's "Machine Learning: A Probabilistic Perspective", section 17.2.3.4 for the detailed balance conditions in Markov Chains, and on their utility here. And also p375 of David MacKay's "Information Theory, Inference and Learning Algorithms" here. Both those expositions are introductory, whilst the Casella and Robert book is a little more involved.

*That's why in some introductory monographs on time-series for practitioners, ergodicity is often not covered, or given a passing mention at best, and instead we check for stationarity (even though the historical roots of the development of time-series in communications engineering was motivated by ergodic theory and statistical mechanics e.g. here).


问:
Proving identity through combinatorial model
How to prove this identity using combinatorial argument?

$\displaystyle\sum_{r=0}^{n}\binom{n}{r}\binom{p}{r+s}\binom{q+r}{m+n} = \sum_{r=0}^{n}\binom{n}{r}\binom{q}{r+m}\binom{p+r}{n+s}$

Only thing I could think of was to change $\binom{n}{r}$ to $\binom{n}{n-r}$ and then we can probably think of choosing $(n-r+r+s+m+n) = (n+s+m+n)$  people or something but its dead end for me. Please help. Also it seems if I can figure one side, other might trivially follow. Please don't ignore the problem.

答:
Suppose you have a fictional country where there are two big parties. The $\color{red}{red}$ and the $\color{blue}{blue}$ parties. Initially, there are $\color{red}{q}$ and $\color{blue}{p}$ of each party. Suppose you want to pick a political committee (senate?) consisting of $\color{red}{(n+m)}$ and $\color{blue}{(n+s)}$ members. Assume, also, that you have a pool of $n$ candidates that can turn red or blue, they are somewhat neutral and can represent either party.
By law you have to choose at least $\color{red}{m}$ from the original red party and at least $\color{blue}{s}$ from the original blue party.

There are two ways this elections will happen. Either first the red party chooses or the blue party chooses.

For the LHS suppose that first the $\color{blue}{blue}$ party chooses. Assume that $r+s$ is the number of people selected from the original party. Then we need to choose $n-r$ from the pool. This we can do in $\binom{p}{r+s}\binom{n}{n-r}$ ways. Then, the rest of the pool ($r$ of them) will become $\color{red}{red},$ so in the other election you have to choose from $q+r$ candidates, giving you $\binom{q+r}{n+m}$ ways. This gives us a total of $\binom{p}{r+s}\binom{n}{n-r}\binom{q+r}{n+m}$ ways to do this election.

The RHS is the same, but the $\color{red}{red}$ people vote first.


问:
Under what restrictions do piecewise continuous functions in higher dimensions form vector spaces?
Definition 1. A function $ f : [a,b] \to \mathbb R $ is called piecewise continuous if $ [a,b] $ may be broken up into a finite number of subintervals $ [t_i,t_{i+1}] $, $ i = 1,2,\dots, n $, such that $ f $ is continuous on each open subinterval $ (t_i,t_{i+1}) $ and has finite limits at their endpoints.

A natural extension to higher dimensions could be formulated as:

Definition 2. A function $ f : \Omega \subseteq \mathbb R^d \to \mathbb R $ is called piecewise continuous if there is a finite set of open and connected sets $ \omega_1,\omega_2,\dots, \omega_n \subset \Omega $ satisfying $ \bigcup_{i=1}^n\overline{\omega_i} = \Omega $ and $ \omega_i\cap\omega_j=\emptyset $ for $ i\neq j $, and to each set $ \omega_i $ a function $ g_i \in C(\overline{\omega_i}) $ with $ g_i = f $ on $ \omega_i $.

Let us for simplicity assume $ d = 2 $.

Let $ PC(\Omega) $ be the set of all piecewise continuous functions $ f : \Omega \to \mathbb R $, as per the above Definition 2.

Proposition. The set $ V = PC(\mathbb R^2) $ is not a vector space.

Proof: Take for instance
$$ f(x_1,x_2) = \begin{cases} 1 & \text{if $ x_2 \geq \sin(x_1) $} \\ 0 & \text{otherwise},
\end{cases} $$
and
$$ g(x_1,x_2) = \begin{cases} 1 & \text{if $ x_2 \leq -\sin(x_1) $} \\ 0 & \text{otherwise}, \end{cases} $$
each in $ V $. Clearly, $ f+g $ will have the value 2 in an infinite number of disjoint sets, even if both $ f $ and $ g $ were defined by only two sets each. So $ f+g \not\in V $. $ \square $

What can we do to get a vector space? Restricting ourselves to compactly supported functions does not fix the problem.

Proposition. The set $ V = PC(\Omega) $ for a bounded set $ \Omega \subset \mathbb R^2 $ does not form a vector space in general.

Proof: Define
$$ \psi(s) = \begin{cases} 0 & \text{if $ s \leq 0 $} \\ -\sin\left(\frac{s}{1-s^2}\right) & \text{if $ s \in (0,1) $},\end{cases} $$
and consider
$$ f(x_1,x_2) = \begin{cases} 1 & \text{if $ \max(|x_1|,|x_2|) < 1 $ and $ x_2 \geq \psi(x_1) $} \\ 0 & \text{otherwise}, \end{cases} $$
and
$$ g(x_1,x_2) = \begin{cases} 1 & \text{if $ \max(|x_1|,|x_2|) < 1 $ and $ x_2 \leq -\psi(x_1) $} \\ 0 & \text{otherwise}. \end{cases} $$
Clearly, each are compactly supported on in the square domain $ \{(x_1,x_2) : \max(|x_1|,|x_2|) < 1+\epsilon \} $ for any $ \epsilon > 0 $, but again $ f+g $ takes the values 2 on an infinite number of disjoint open sets. So still we don't have a vector space. $ \square $

Question! Is there a simple requirement that immediately yields a vector space. It seems that to get these infinitely interweaving sets on a bounded set must require increasingly sharp turning. <s>Would restricting to compact support along with the boundary of each $ \omega_i $ being Lipschitz continuous be strong enough? I would assume yes, but is there a lesser requirement, which is sufficient?</s>

Edit #1: Clearly, from mihaild's excellent example, which is both Lipschits continuous and infinitely differentiable that assumption was wrong. mr_e_man suggests strictly convex sets $ \omega_i $, but that is incredibly restrictive as very few convex sets can tile together to fill an area. <s>However, it made me think, what if each set $ \omega_i $ was constructible by taking unions and differences of finitely many convex sets. Certainly, that should at least eliminate all these "easy" counterexamples with oscillating boundaries.</s>

Edit #2: Turns out even even my convex set idea is problematic. Consider the set $ \Omega = [-1,1]\times[0,2] $ and let
$$ \theta_i = \pi\left(1-\frac{1}{2^i}\right), i = 0,1,2,\dots,n-1\quad\text{and}\quad\theta_n = \pi, $$
and
$$ \varphi_{i+1} = \theta_i + \frac{\theta_{i+1}-\theta_i}{2}, i = 0,1,2,\dots,n-1,\quad \varphi_0 = 0, \quad\text{and}\quad \varphi_{n+1}=\pi. $$
Consider the point sets
$$ \mathcal P_n = \{(\cos(\theta_i),\sin(\theta_i)) : i = 0,1,\dots,n \} $$
and
$$ \mathcal Q_n = \{(\cos(\varphi_i),\sin(\varphi_i)) : i = 0,1,\dots,n+1 \}. $$
Take $ \mathcal P = \bigcup_{n=1}^\infty\mathcal P_n $ and $ \mathcal Q = \bigcup_{n=1}^\infty \mathcal Q_n $ and let $ \omega_1 $ be the convex hull of $ \mathcal P $ and $ \omega_2 =  \Omega\backslash\omega_1 $ and $ \eta_1 $ the convex hull of $ \mathcal Q $ and $ \eta_2 = \Omega\backslash\eta_1 $. Let
$$ f(\mathbf x) = \begin{cases} 1 & \text{if $ \mathbf x \in \omega_1 $} \\ 0 & \text{if $ \mathbf x \in \omega_2 $} \end{cases} \quad g(\mathbf x)
= \begin{cases} 0 & \text{if $ \mathbf x \in \eta_1 $} \\ 1 &\text{if $\mathbf x \in \eta_2$}. \end{cases}$$
Then the set of $ f $ can be constructed by difference of finitely many (just 2) convex sets and likewise for $ g $ and $ f+g $ will again have infinitely many disjoint sets with value 2.

Visual example here:

Edit #3: More restrictive idea: Each set $ \omega_i $ must be a polygon! That has to do it, right?

答:
Even smooth boundary isn't enough: take
$$f(x, y) = \begin{cases} 0,\ y \leq 0\\ 1\end{cases}$$
and
$$g(x, y) = \begin{cases} 0,\ y \leq \exp(-\frac{1}{x^2})\sin(\frac{1}{x})\\ 1\end{cases}$$

For family $\mathcal F$ of open subsets of $\Omega$, say function $f$ on $\Omega$ is $\mathcal F$-piecewise continuous iff there is a finite set of $\omega_i, \ldots, \omega_n$, $\omega_i \in \mathcal F$ s.t. $\bigcup \overline{\omega_i} = \Omega$ and $f |_{\omega_i}$ can be extended to continuous function on $\overline{\omega_i}$ for each $i$.
Your original definition corresponds to $\mathcal F$ been set of all open connected sets.

Set of all $\mathcal F$-piecewise continuous functions is a vector space iff for $a, b \in \mathcal F$, $a \cap b$ is union of finite number of sets from $\mathcal F$.

If we have a family $\mathcal G$ of sets s.t. $\Omega \in \mathcal G$ and $\mathcal G$ is closed under union and difference (for example, set you described - sets constructible from convex by taing finite unions and differences), then $\mathcal G$ is also closed under intersection: if $a, b \in \mathcal G$ then $\Omega \setminus a$, $\Omega \setminus b$, $(\Omega \setminus a) \cup (\Omega \setminus b)$ and $\Omega \setminus (\Omega \setminus a) \cup (\Omega \setminus b) = a \cap b$ are all in $\mathcal G$. So $\mathcal G$-piecewise continuous functions form a vector space.

UPD: such $\mathcal G$ is actually not exactly what was asked about - the original question was about connected sets from $\mathcal G$ (or equivalently sets with finite number of connected components). And difference of two convex set can have infinitely many connected components, as your example proves.

A bit simple example will be circle and "polygon" with infinitely many non-zero sides inscribed into it (for example, polygon with vertices $(\cos 2\pi - \frac{2\pi}{n}, \sin 2\pi - \frac{2\pi}{n})$).

UPD2: polygons will work. Boundary of intersection is subset of union of boundaries, so boundary of intersection will consist of finite number of line segments (each segment is described by edge it is taken from and at most two edges that give it endpoints) - thus intersection will be finite union of polygons.


问:
$X_i\sim\text{Exp}(1), S_n=\sum_{i=1}^nX_i.$ $\mathbb{E}(\frac{e^{S_{n+1}}(n+1)!}{(1+S_{n+1})^{n+2}}|\mathcal{F_n})=\frac{e^{S_n}n!}{(1+S_n)^{n+1}}$?
$\newcommand{\e}{\operatorname{\mathbb E}}$
We need to show that $ \displaystyle \e \left( \frac{e^{S_{n+1}}(n+1)!}{(1+S_{n+1})^{n+2}} \mid \mathcal{F_n} \right) = \frac{e^{S_n}n!}{(1+S_n)^{n+1}}$.

The $X_i$'s are iid.

My only line of working  so far is:

\begin{align}
& \e \left( \frac{e^{S_{n+1}}(n+1)!}{(1+S_{n+1})^{n+2}} \mid \mathcal{F_n} \right) \\[8pt]
= {} & e^{S_n}(n+1)!\e \left( \frac{e^{X_{n+1}}}{(1+S_n+X_{n+1})^{n+2}}\mid\mathcal{F_n} \right) \\[8pt]
= {} & e^{S_n}(n+1)! \e \left(\frac{e^{X_{n+1}}}{(1+S_n+X_{n+1})^{n+2}}\right)
\end{align}

At this point I'm not sure on how we can proceed? Maybe by using the definition of expectation we could integrate to get the expectation (but this looks long-winded - perhaps there's a neater way)?

Body:
$\newcommand{\e}{\operatorname E}$
\begin{align}
& \e \left(\frac{e^{X_{n+1}}}{(1+S_n+X_{n+1})^{n+2}} \,\big|\, \mathcal F_n \right) \\[8pt]
= {} & \int_0^\infty \frac{e^x}{(1+S_n + x)^{n+2}} \big( e^{-x}\,dx\big) \\[8pt]
= {} & \int_0^\infty \frac{dx}{(1+S_n + x)^{n+2}} = \frac 1 {(n+1)(1+S_n)^{n+1}}.
\end{align}


问:
Do we really need polynomials (In contrast to polynomial functions)?
In the following I'm going to call

a polynomial expression an element of a suitable algebraic structure (for example a ring, since it has an addition and a multiplication)
that has the form $a_{n}x^{n}+\cdots+a_{1}x+a_{0}$, where $x$ is
some fixed element of said structure,

a polynomial function a function of the form $x\mapsto a_{n}x^{n}+\cdots+a_{1}x+a_{0}$ (where the same algebraic considerations as above apply)

a polynomial an element of the form $a_{n}X^{n}+\cdots+a_{1}X+a_{0}$ with indeterminates $X$ (these
can be formalized, if we are in a ring, with strings of $0$s and
a $1$).

Note that when we are very rigorous/formal, polynomial expressions
and polynomials are something different (although in daily life we
often use them synonymously). Polynomial functions and expressions
are also different from each other although in this case the relationship
is a closer one, since every polynomials expression can be interpreted
as a polynomial function evaluated at a certain point (thus "polynomial
functions" are something more general than "polynomial expressions").

My question is: Why do we use polynomials ? It seems to me that every
piece of mathematics I have encountered so far, one could replace
every occurrence of polynomials with polynomial expressions/functions
without any major change in the rest of the proof/theorem/definition
etc.

My thoughts:

The only reasons that I can see to use polynomials are the following
two:

After one makes the idea precise that one can plug "suitable"
elements into polynomials (which may lie a ring containing the ring
in which the coefficients live in), one can save time in certain setting,
by handling the "plugging of suitable elements into polynomials"
more elegantly: For example in case of the theorem of Cayley-Hamilton,
which in its "polynomial function" version would look like:

Let $A$ be an $n\times n$ matrix over $K$, whose characteristic
polynomial (function) is $x\mapsto a_{n}x^{n}+\cdots+a_{1}x+a_{0}$.
Then
$$
a_{n}A^{n}+\cdots+a_{1}A+a_{0}I=0.
$$

whereas the "polynomial" version looks more elegant:

Let $A$ be an $n\times n$ matrix over $K$, whose characteristic
polynomial is $p_{A}\in K\left[X\right]$. Then
$$
p_{A}\left(A\right)=0.
$$

The only thing that polynomials can "do", but algebraic expressions/functions can't, is to be different, when the algebraic expressions/functions are the same (i.e. there's a theorem that tells us that the mapping of polynomials to polynomials expressions/functions
isn't injective, if the field is finite). Maybe this small difference makes a big enough difference
to consider polynomials after all, but as I said, I haven't encountered
any situation in which this difference could manifest itself.

(I'm guessing that maybe cryptography or higher number theory really needs
polynomials and not just polynomial expressions/functions. Since I don't know
anything about these subjects, I would be very happy with an example
of a theorem (whose content isn't merely technical as it is the case with the theorem above) involving polynomials, where these absolutely cannot
be replaced by polynomial expressions/functions. Conversely I would also be happy with an authoritative statement from someone knowledgeable that indeed we could dispense of polynomials, if we wanted to.)

答:
Here are some answers with somewhat different conclusions.  Let us call the form of polynomials with coefficients and indeterminates "abstract" or "algebraic" to distinguish them from polynomial functions.

If you want to work with rational functions, the situation becomes foggy if you do not define everything in terms of abstract polynomials:

Is the function $f(x) = +\frac{1}{x} \ -\frac{1}{x}$ equal to the zero function?

Is $\frac{x}{x}$ equal to $1$ for all $x$, or only for $x \neq 0$?

When composing a chain of linear fractional functions $\frac{ax+b}{cx+d}$ do we exclude more and more points from the domain everywhere that one of the denominators is zero, when the end result as a rational function has at most one one singularity? Over a finite field this can empty the domain of all its points.

For an infinite field you can get around some of this by treating polynomial functions and rational functions as function-equal if they are equal at infinitely many points, and defining rational functions as written in lowest terms (which breaks the analogy with fractions).  Over a finite field, nothing like that can work because ratios like ${p(x)^a}/{p(x)^b}$ are not defined if $p=0$ for all elements in the finite field, and if there were a way to define it based on functions (such as using the lowest-degree representation) it would not depend on the exponents.

In the other direction:

if you want representations of the polynomial to not always be centered at $0$ or to be less coordinate dependent, then you might want polynomial to mean a function killed by differential or difference operators of order $n$ for some $n > 0$.

where polynomials appear by a construction that operates on functions, and cannot be performed on the abstract polynomials, then you have no choice but to work with polynomials as functions.

In algebraic uses of polynomial functions, the arithmetic operations, function composition, differentiation and integration are often sufficient and can be done on the abstract form, and a function is demonstrated to be polynomial by exhibiting an abstract form.  So the evaluation that converts algebraic representation into a function can always be postponed by doing all operations on the algebraic representation.


问:
Showing that $\left(1-x+\frac{x^2}{2}\right)^n-(1-x)^n\leq \frac{x}{2}$ using induction
I was asked to show that

For every $n\in\mathbb{N}$ and $x\in [0,1]$ the following is true:
\begin{equation*}
    \left(1-x+\frac{x^2}{2}\right)^n-(1-x)^n\leq \frac{x}{2}.
\end{equation*}

However, I found a mistake in my proof and now I don't know how to proceed.

This was my attempt:

The case when $n=1$ is true because when $x\in [0,1]$, we have $x^2\leq x$, and so
\begin{equation}
    1-x+\frac{x^2}{2}-1+x=\frac{x^2}{2}\leq \frac{x}{2}.
\end{equation}
Suppose now that the statement holds for some $k\in \mathbb{N}$,
\begin{equation}
    \left(1-x+\frac{x^2}{2}\right)^k-(1-x)^k\leq \frac{x}{2}.
\end{equation}

Knowing that $0\leq x\leq 1$, we can see that $1-1\leq x$, which implies $1-x\leq 1$. Additionally, we can see that $2x\geq x$ and so $x^2\leq x$,
\begin{align}
    x\geq \frac{x}{2}\geq \frac{x^2}{2}&\Rightarrow x\geq \frac{x^2}{2}\\
    &\Rightarrow 0\geq \frac{x^2}{2}-x\\
    &\Rightarrow 1\geq \frac{x^2}{2}-x+1
\end{align}
Thus,
\begin{align}
    { \left(1-x+\frac{x^2}{2}\right)^{k+1}-(1-x)^{k+1}}&=  \left(1-x+\frac{x^2}{2}\right)^k\left(1-x+\frac{x^2}{2}\right)-(1-x)^k(1-x) \\
    &\leq  \left(1-x+\frac{x^2}{2}\right)^k-(1-x)^k\\
    &\leq \frac{x}{2}
\end{align}
The statement holds for $k+1$.

The mistake is the part where I wrongly use
\begin{equation}
    -(1-x)^k(1-x)\leq -(1-x)^k.
\end{equation}

But I don't know how to proceed now.

Any hint or advice is appreciated. Thank you.

答:
Rearrange the induction hypothesis so it reads $\left(1-x+\dfrac {x^2}2\right)^k \le \dfrac x2 + (1-x)^k$.

Now the induction step can go as follows:

\begin{align}\left(1-x+\frac {x^2}2\right)^{k+1} - (1-x)^{k+1} &\le \left(1-x+\frac {x^2}2\right)\left(\frac x2 + (1-x)^k\right)-(1-x)^{k+1}\\
&=\frac x2-\frac {x^2}2+\frac{x^3}4+(1-x)^{k+1}+\frac{x^2}2(1-x)^k-(1-x)^{k+1}\\
&=\frac x2 - \frac{x^2}2\left(1-\frac x2-(1-x)^k\right)\\
&\le \frac x2-\frac{x^2}2\left(1-\frac x2 - (1 - x)\right)\\
&=\frac x2 - \frac{x^3}4 \le \frac x2
\end{align}

The part where we eliminated the exponent $k$ holds since $(1-x)^k \le (1-x)$ for $x \in [0,1]$.


问:
Expectation of sampling integers from a reciprocal distribution without replacement
I have a reciprocal distribution with PDF
$$\frac{1}{x\ ln{N}}$$
I sample $k$ integers from this distribution in the range $[1,N]$ without replacement.
I need to determine the average (expected) number of each integer in $[1,N]$ in my sample.

I came across the multivariate Wallenius' noncentral hypergeometric distribution, which deals with sampling weighted colours of ball from an urn without replacement in sequence. The distribution describes the expected number of each colour $i$ as $\mu_i$ in a vector $\mathbf\mu$, which can be found by solving the following system of non-linear equations
$$\left(1-\frac{\mu_1}{m_1}\right)^\frac{1}{\omega_1}
= \left(1-\frac{\mu_2}{m_2}\right)^\frac{1}{\omega_2}
= \cdots
= \left(1-\frac{\mu_c}{m_c}\right)^\frac{1}{\omega_c}$$
$$\sum_{i=1}^c \mu_i=n$$

For my use case, $c=N$, $m_i=1$, $n=k$ and $\omega_i=\frac{1}{i\ lnN}$, so the equations become
$$\left(1-\mu_1\right)^{lnN}
= \left(1-\mu_2\right)^{2\ lnN}
= \cdots
= \left(1-\mu_N\right)^{N\ lnN} \tag{1}$$
$$\sum_{i=1}^N \mu_i=k \tag{2}$$
The more general Wallenius' mean is normally approximated through e.g. Newton-Raphson, so I'm hoping that the narrowing of the equations makes them directly solvable.
My work so far is as follows:

We can rewrite $(1)$ to put $\mu_i$ in terms of $\mu_j$
$$(1-\mu_i)^{i\ lnN}=(1-\mu_j)^{j\ lnN}$$
Using the identity $a^b=e^{b\ lna}$
$$e^{i\ lnN\ ln(1-\mu_i)}=e^{j\ lnN\ ln(1-\mu_j)}$$
$$i\ lnN\ ln(1-\mu_i)=j\ lnN\ ln(1-\mu_j)$$
$$ln(1-\mu_i)=\frac{j}{i}\ ln(1-\mu_j),\ \ \ N>1$$
$$\mu_i=1-(1-\mu_j)^{j/i}\tag{3}$$
We can then repeatedly substitute $(3)$ into the summation $(2)$ to obtain a formula only in terms of $\mu_i$
$$
\begin{aligned}
k&=\mu_1+\mu_2+\cdots+\mu_N\\
&=(1-(1-\mu_i)^{i/1})+(1-(1-\mu_i)^{i/2})+\cdots+(1-(1-\mu_i)^{i/N})\\
&=N-\sum_{j=1}^N (1-\mu_i)^{i/j}
\end{aligned}
$$
Therefore
$$N-k=\sum_{j=1}^N (1-\mu_i)^{i/j}$$
However, I do not know how to proceed. Can this be rearranged for $\mu_i$? Am I overcomplicating things?

Body:
Since a ball can either be selected once without replacement in $k$ draws, or not at all, the expected number of times it is drawn is the probability of it being drawn.

You said in comments that you did not want to use simulation.  Just to show it is possible, and to allow you to check other calculations you may make, here is an example in R with $n=10$ for all the possible $k$s:

n <- 10
cases <- 10^5
set.seed(2020)
orderedsample <- function(n){unique(sample(n,250*n,p=1/(1:n),replace=TRUE))}
simdat <- replicate(cases, orderedsample(n))
count <- matrix(numeric(n^2), ncol=n, nrow=n)
for (i in 1:n){count[i,] <- table(factor(simdat[1:i,], levels=1:n))}
colnames(count) <- 1:n
rownames(count) <- paste("k=", 1:n, sep="")
count / cases

to give

           1       2       3       4       5       6       7       8       9      10
k=1  0.34305 0.17065 0.11250 0.08565 0.06820 0.05702 0.04878 0.04241 0.03788 0.03386
k=2  0.59220 0.34732 0.24146 0.18280 0.14841 0.12595 0.10735 0.09438 0.08364 0.07649
k=3  0.76348 0.51624 0.37823 0.29484 0.24177 0.20549 0.17734 0.15635 0.13899 0.12727
k=4  0.87436 0.66553 0.51715 0.41541 0.34651 0.30076 0.25883 0.22924 0.20442 0.18779
k=5  0.94059 0.78772 0.65120 0.54302 0.46255 0.40323 0.35356 0.31437 0.28348 0.26028
k=6  0.97631 0.88015 0.77102 0.66755 0.58399 0.51750 0.46126 0.41608 0.37727 0.34887
k=7  0.99167 0.94177 0.86763 0.78511 0.70843 0.64429 0.58418 0.53412 0.48830 0.45450
k=8  0.99803 0.97812 0.94039 0.88480 0.82844 0.77341 0.71887 0.67036 0.62274 0.58484
k=9  0.99976 0.99516 0.98292 0.96018 0.93124 0.89962 0.86237 0.82643 0.78881 0.75351
k=10 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000

This is a simulation so you should not use the later decimal places. But to illustrate how well it does, it is easy to calculate the first row theoretically, which gives

           1       2       3       4       5       6       7       8       9      10
k=1  0.34142 0.17071 0.11381 0.08535 0.06828 0.05690 0.04877 0.04268 0.03794 0.03414


问:
Does this ODE have an exact or well-established approximate analytical solution?
The equation looks like this:
$$\frac{\mathrm{d}y}{\mathrm{d}t} = A + B\sin\omega t - C y^n,$$
where $A$, $B$, $C$ are positive constants, and $n\ge1$ is an integer.  Actually I am mainly concerned with the $n=4$ case.

The $n=1$ case is trivial.  Otherwise the only method I can think of is kind of an iterative approximation, in which a "clean" expression seems not easy to obtain.

Just in case it is too "easy" for the experts, we may generalize it to the form
$$\frac{\mathrm{d}y}{\mathrm{d}t} = f(t) - g(y).$$

答:
You may be interested in the following power series solution for the case $n=4$

$$ y \left( t \right) = y \left( 0 \right) + \left( A-C \left( y\left( 0
 \right)\right) ^{4} \right) t+ \left( \frac{1}{2}\,B\omega-2\,C \left( y
 \left( 0 \right)  \right) ^{3}A+2\,{C}^{2} \left( y \left( 0 \right)
 \right) ^{7} \right) {t}^{2}+O \left( {t}^{3} \right).  $$


问:
Collection of surprising identities and equations.
What are some surprising equations/identities that you have seen, which you would not have expected?

This could be complex numbers, trigonometric identities, combinatorial results, algebraic results, etc.

I'd request to avoid 'standard' / well-known results like $ e^{i \pi} + 1 = 0$.

Please write a single identity (or group of identities) in each answer.

I found this list of Funny identities, in which there is some overlap.

答:
This expression is from Francois Viète: $$\pi = \frac{2}{1} \times \frac{2}{\sqrt2} \times \frac{2}{\sqrt{2+\sqrt2}}\times
 \frac {2}{\sqrt{2+\sqrt{2+\sqrt2}}} \times \cdots$$


答:
Write $r = r(\theta) = \sqrt{1 - 2x\cos\theta + x^2}$. Then we easily check that

$$ \frac{\mathrm{d} r^{\alpha}}{\mathrm{d}\theta} = \alpha r^{\alpha-2} x\sin\theta, \qquad\text{and so},\qquad \int r^{\beta}\sin\theta \, \mathrm{d}\theta = \frac{r^{\beta+2}}{(\beta+2)x}. $$

Now denoting the integral by $f(x)$ and assuming that $x \neq \pm 1$, we have

\begin{align*}
f(x)
&= \int_{0}^{\pi} r^{-3}\sin\theta(\cos\theta - x) \, \mathrm{d}\theta \\
&= \left[ - \frac{r^{-1}(\cos\theta - x)}{x} \right]_{0}^{\pi} - \frac{1}{x} \int_{0}^{\pi} r^{-1}\sin\theta \, \mathrm{d}\theta \\
&= \frac{1}{x}\left( \frac{1-x}{r(0)} + \frac{1+x}{r(\pi)} \right) - \left[ \frac{r}{x^2} \right]_{0}^{\pi}.
\end{align*}

By using $r(\pi) = \left| 1 + x \right|$ and $r(0) = \left| 1 - x \right|$, we get

\begin{align*}
f(x) &= \frac{\operatorname{sgn}(1+x) + \operatorname{sgn}(1-x)}{x} - \frac{\left| 1 + x\right| - \left|1 - x \right|}{x^2} \\
&= \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{\left| 1 + x\right| - \left|1 - x \right|}{x} \right).
\end{align*}

Noting that

$$ \frac{\left| 1 + x\right| - \left|1 - x \right|}{x} = \begin{cases}
2/x, & x > 1, \\
2, & -1 < x < 1, \\
-2/x, & x < -1,
\end{cases} $$

it follows that

$$ f(x) = \begin{cases}
-2/x^2, & x > 1, \\
0, & -1 < x < 1, \\
2/x^2, & x < -1.
\end{cases} $$

Given this form, I suspect that $f$ comes from an electromagnetism problem of computing the electric field arising from a charged metal sphere or something like that.

Remark. It can be checked separately that

$$ f(1)
= -\frac{1}{2\sqrt{2}} \int_{0}^{\pi} \frac{\sin\theta}{\sqrt{1-\cos\theta}} \, \mathrm{d}\theta
= -\frac{1}{\sqrt{2}}  \left[ \sqrt{1-\cos\theta} \right]_{0}^{\pi}
= -1 $$

and

$$ f(-1)
= \frac{1}{2\sqrt{2}} \int_{0}^{\pi} \frac{\sin\theta}{\sqrt{1+\cos\theta}} \, \mathrm{d}\theta
= -\frac{1}{\sqrt{2}}  \left[ \sqrt{1+\cos\theta} \right]_{0}^{\pi}
= 1. $$

Therefore the complete answer is as follows:

$$ f(x) = \begin{cases}
-\frac{2}{x^2}, & x \in (1, \infty), \\
-1, & x = 1, \\
0, & x \in (-1, 1), \\
1, & x = -1, \\
\frac{2}{x^2}, & x \in (-\infty, -1).
\end{cases} $$

Title: Compute $ f(x) = \int_0^{\pi} \frac{ \sin \theta ( \cos \theta - x) d \theta}{ [(\cos \theta - x)^2 + \sin^2 \theta]^{1.5} } x \in \Bbb{R}$

Body:
Compute

$$ f(x) = \int_0^{\pi} \frac{ \sin \theta ( \cos \theta - x)}{ [(\cos \theta - x)^2 + \sin^2 \theta]^{1.5} }
\,\,d \theta\,,\quad   x \in \Bbb{R}$$

The denominator makes it a bit annoying. We can evaluate it as

$ [(\cos \theta - x)^2 + \sin^2 \theta = (1 - 2 \cos \theta x + x^2)^{1.5}$ but it doesn't make it any easier.  We can also expand $(1 + x^2 - 2 \cos \theta x )^{1.5}$ using $(1+x)^\alpha = \sum_{n=0}^\infty \binom{\alpha}{n} x^n$ but it's still difficult.

The denominator also has some geometric meaning  (cubed distance from $(x,0)$ to $(\cos \theta, \sin \theta)$ ) but does it help?


答:
•I will love to add this nested radical from Ramanujan:
$$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\cdots}}}}=3$$ •Also like to add this too:$$1+2+3+4+\cdots=-\frac{1}{12}$$ •This famous Stirling'sapproximation$$n!\approx\sqrt{2\pi n}{\left(\frac{n}{e}\right)}^n$$

Title: Collection of surprising identities and equations.

Body:
What are some surprising equations/identities that you have seen, which you would not have expected?

This could be complex numbers, trigonometric identities, combinatorial results, algebraic results, etc.

I'd request to avoid 'standard' / well-known results like $ e^{i \pi} + 1 = 0$.

Please write a single identity (or group of identities) in each answer.

I found this list of Funny identities, in which there is some overlap.