问:
Is the splitting $\pi_{k}(X,A)\simeq\pi_{k}(X)\times \pi_{k-1}(A)$ a $\pi_1(A)$-modules isomorphism?
Let $(X,A)$ be a pair of topological spaces with $A\subset X$. Fix a basepoint $x_0$ of $X$ which lies in $A$. Assume that the inclusion $(A,x_0)\to (X,x_0)$ is homotopic to a constant map relatively to the basepoint, that is there is a homotopy $H:A\times [0,1]\to X$  with
$H(x,0)=x$ and $H(x_0,t)=H(x,1)=x_0$. Then, from a map
$f:(D^k,S^{k-1},s_0)\to(X,A,x_0)$
with $s_0 \in S^{k-1}$ one can construct an extension
$$\tilde{f}:(D^k,S^{k-1})\to(A,x_0)$$
by "gluing" a cylinder $S^{k-1}\times [0,1]$ to the disk $D^k$ and by following the homotopy radially from the boundary. After verification, this gives a well defined morphism

$$r:\pi_k(X,A,x_0)\to\pi_k(A,x_0)$$

given by $r([f])=[\tilde{f}]$. Since the inclusion $A\to X$ is homotopically trivial, the long exact sequence of the pair $(X,A)$ becomes, for all $k\geq 2$,

$$0\to\pi_{k}(X,x_0)\overset{j_*}{\longrightarrow}\pi_{k}(X,A,x_0)\overset{\partial_*}{\longrightarrow} \pi_{k-1}(A,x_0)\to0.$$

By construction we have $r\circ j_*=id$, so $r$ is a retraction of the above short exact sequence and it gives a splitting

$$\boxed {\pi_{k}(X,A,x_0)\simeq\pi_{k}(X,x_0)\times \pi_{k-1}(A,x_0)}.$$

Also the three groups above are $\pi_1(A,x_0)$-modules and the maps $j_*$ and $\partial_*$ are both $\pi_1(A,x_0)$-equivariant.

Is the above splinting an isomorphism of $\pi_1(A,x_0)$-modules ?

This amounts to proving that the retraction $r$ is $\pi_1(A,x_0)$-equivariant. Basically I tried to do drawings and interchange the action of $\pi_1(A,x_0)$ and the operation of gluing $H$ but it failed miserably. Maybe someone has a more straightforward approach or a counterexample ? Any help will be greatly appreciated.

答:
Recall that then $F\to X\to Y$ is a fibration sequence, you have an induced long fibration sequence $$\cdots\to\Omega F\to\Omega X\to\Omega Y\to F\to X\to Y$$
meaning that each series of three consecutive spaces forms a fibration sequence, called "exact Puppe sequence".

For a fibration $f:E\to B$ with fiber $F$ you have $\pi_k(E,F,*)\cong\pi_k(B,*)$ so that if $i:A\to X$ is an inclusion with homotopy fiber $F$, there is an isomorphism
$$\pi_k(\Omega X,\Omega A)\cong\pi_k(F)$$
induced from the fibration sequence $\Omega A\to\Omega X\to F$. Of course you also have
$$\pi_k(\Omega X,\Omega A)=\pi_{k+1}(X,A)$$
because the adjunction $(\Sigma,\Omega)$ induces an adjunction on the homotopy category of pairs.

In short, for any pair $(X,A)$, there is an isomorphism $\pi_{k+1}(X,A)\cong \pi_k(F)$ where $F$ is the homotopy fiber of the inclusion.

Unraveling the different isomorphisms, one can make the latter explicit : take a representative $f\colon (D^{k+1},S^k,*)\to (X,A,*)$ of an element of $\pi_{k+1}(X,A,*)$, then the associated element of $\pi_k(F)$ is represented by the map $f^\prime\colon (S^k,*)\to (F,*)$
$$f^\prime(z)=(a(z),\gamma(z))$$
with $a(z)=f(z)$ and $\gamma(z)(t)=f(z,t)$ under the identification $D^{k+1}=CS^k$.

We should make sure that is is $\pi_1(A)$-equivariant, I will just accept it, Tyrone claimed it in his comment. The action of $\pi_1(A)$ on $\pi_k(F)$ is the action of the fundamental group of the total space on the homotopy groups of the fiber: there is a monodrony action $\Omega X\times F\to F$ which is not basepoint preserving, but precomposing with $A\to X$ gives a well defined pointed map
$$\Omega A\times F\to F$$
inducing an action on homotopy groups.

Now if the inclusion $A\subseteq X$ is nulhomotopic, i.e. if it is homotopic to the trivial map $*\colon A\to X$, then the homotopy fibers of the inclusion and of the trivial map are weakly equivalent (by naturality of the long exact sequence of homotopy groups of a fibration, and the five lemma). But the homotopy fiber of the trivial map is $$A\times \Omega X$$ Also, the long exact sequences are $\pi_1(A,*)$-equivariant as well as the morphism between them. For a trivial map $*\colon A\to X$, the additional piece of data we have is that the fiber sequence
$$\Omega X\to F\to A$$
is such that $\Omega X\to F$ has a retract. So, the splitting is equivariant in this case, and hence is always equivariant.

Little check: we used that a commutative square of fibrations induces a map of long exact sequences, which is true in the strict setting. It is also true in the case of a diagram which only commutes up to homotopy: if $f\colon A\to X$ and $g\colon B\to C$ are fibrations and we have maps $k\colon A\to C$, $l\colon B\to D$ making the square commute up to homotopy, take a homotopy
$$H:A\times[0,1]\to D$$
and its dual
$$K:A\to D^{[0,1]}$$
You can form the following two strictly commutative squares:

$$
\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
%
\begin{array}{ccc}
A & \ra{f} & B \\
\da{a\mapsto (a,0)} & & \da{l}\\
A\times[0,1] & \ra{H} & D
\end{array}
$$

and

$$
\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
%
\begin{array}{ccc}
A & \ra{K} & D^{[0,1]} \\
\da{k} & & \da{\mathrm{ev}_1}\\
C & \ra{g} & D
\end{array}
$$

They induce long exact sequence in homotopy, and the lower long sequence of the first diagram is exactly the same as the upper long sequence of the second one (aka the long exact sequences of the maps $l\circ f\simeq k\circ g$), so they can be glued forming the exact sequence we were looking for.


问:
Notation for newton-like expansion
Is there a compact way of referring to the expression $$a^n + a^{n - 1}b + a^{n - 2}b^2 + \cdots + b^n\:?$$ Maybe some notation I do not know about it.

Thanks!

答:
Using sigma notation, you can write it as
$$
\sum_{k=0}^{n}a^{n-k}b^k \, .
$$

ContentLicense: CC BY-SA 4.0

Title: Notation for newton-like expansion


问:
Floor function proof :
I am new to floor functions and I can't see a way to find this exact inequality. I have studied the cases but I'm more interested in being able to find this particular inequality.

$$  -2 \leq 3 \left \lfloor 2x \right \rfloor -2\left \lfloor 3x \right \rfloor \leq 1$$

答:
$2x-1<\lfloor 2x\rfloor\le2x\quad\color{blue}{(*)}$

$3x-1<\lfloor 3x\rfloor\le3x\quad\color{blue}{(**)}$

By multiplying $(*)$ by $3$ and by multiplying $(**)$ by $-2$, it follows that

$6x-3<3\lfloor 2x\rfloor\le6x$

$-6x\le-2\lfloor 3x\rfloor<-6x+2$

and, by adding the last two inequalities, we get that

$-3<3\lfloor 2x\rfloor -2\lfloor 3x\rfloor<2\;.$

Since $\;3\lfloor 2x\rfloor -2\lfloor 3x\rfloor\;$ is an integer, it follows that

$-2\le3\lfloor 2x\rfloor -2\lfloor 3x\rfloor\le1\;.$

Title: Floor function proof :


问:
Surprisingly elementary and direct proofs
What are some examples of theorems, whose first proof was quite hard and sophisticated, perhaps using some other deep theorems of some theory, before years later surprisingly a quite elementary, direct, perhaps even short proof has been found?

A related question is MO/24913, which deals with hard theorems whose proofs were simplified by the development of more sophisticated theories. But I would like to see examples where this wasn't necessary, but rather the theory turned out to be superfluous as for the proof of the theorem. I expect that this didn't happen so often. [Ok after reading all the answers, it obviously happened all the time!]

答:
Perhaps the most exciting and dramatic of the difficult inequalities is Arhangel'skii's theorem that $|X|\le \exp(L(X)\chi(X))$ for every Hausdorff space. The countable version of this result, namely that every Lindelöf, first countable, Hausdorff space has cardinality at most $\mathfrak c$, answer the following fifty-year old question of Alexandroff and Urysohn. Does there exists a compact, first countable space having cardinality greater that the continuum?

As one might guess, Arhangel'skii's original proof was quite difficult. The argument given in Set theoretic Topology Page 19 is due to Pol. It is not difficult for one to undestand. The countable version of this proof should be within the reach of any first-year graduate student in mathematics. The theorem is sufficently important to be included in any introductory graduate course in set-theoretic topology, and provides exposure to modern topology at an early level of mathematical training.

ContentLicense: CC BY-SA 3.0


问:
Problem about find the extreme of a function (Multipliers of Lagrange)
Good morning. I have a problem with this:

Find the maximum and minimum distances from the origin to the curve* $$g\left(x,y\right)=5x^{2}+6xy+5y^{2}$$

I have done this:

Function to optimize:$f\left(x,y\right)=x^{2}+y^{2}$

Restriction: $g\left(x,y\right)=5x^{2}+6xy+5y^{2}=8
 $

Applying Lagrange multipliers:
$\nabla f\left(x,y\right)=\lambda\nabla g\left(x,y\right)$

Then,
$\nabla f\left(x,y\right)=2x\hat{i}+2y\hat{j}$ and   $\lambda\nabla g(x,y)=\lambda(2x+\frac{6}{5}y)\hat{i}+\lambda\left(2y+\frac{6}{5}x\right)\hat{j}
 $

Making the ecuation system:

$\begin{cases}
2x=(2x+\frac{6}{5}y)\lambda\\
2y=(2y+\frac{6}{5}x)\lambda\\
x^{2}+\frac{6}{5}xy+y^{2}=8
\end{cases}$

But I have serious problem solving the system. Any suggestions?

答:
It turns out that this system of Lagrange equations doesn't really warrant the use of Mathematica.  We are faced with

$$ \ 2x \ = \ (2x \ + \ \frac{6}{5}y)\lambda \ \ , \ \ 2y \ = \ (2y \ + \ \frac{6}{5}x)\lambda \ \ . $$

Since bringing all the terms to one side in each equation won't help us to factor the expressions, we might instead solve each equation for $ \ \lambda \ $ to obtain

$$ \lambda \ \ = \ \ \frac{x}{x \ + \ \frac{3}{5} y} \ \ = \ \ \frac{y}{y \ + \ \frac{3}{5} x} \ \ . $$

Assuming for the moment that neither denominator is equal to zero (it will turn out that they are not), we can "cross-multiply" the ratios to produce

$$ xy \ + \ \frac{3}{5} x^2 \ \ = \ \ xy \ + \ \frac{3}{5} y^2 \ \ \Rightarrow \ \  x^2 \  = \  y^2 \ \ \Rightarrow \ \ y \ = \ \pm x \ \ . $$

We have two cases now for insertion into the curve equation:

$$ \mathbf{y = x :} \quad 5x^2 \ + \ 6xy \ + \ 5y^2 \ = \ 8 \ \ \rightarrow \ \ 5x^2 \ + \ 6·x·x \ + \ 5x^2 \ = \ 8 \ \ \Rightarrow \ \ 16x^2 \ = \ 8 $$ $$ \Rightarrow \ \ x^2 \ = \ \frac{1}{2} \ \ ; $$
$$ \mathbf{y = -x :} \quad \quad \quad \rightarrow \ \ 5x^2 \ + \ 6·x·(-x) \ + \ 5x^2 \ = \ 8 \ \ \Rightarrow \ \ 4x^2 \ = \ 8 \ \ \Rightarrow \ \ x^2 \ = \ 2 \ \ . $$

If we only need the "distance-squared" from the origin for the extremal points, then the cases are $ \ \mathbf{y = x :} \ \ x^2 \ + \ y^2 \ = \ \frac{1}{2} \ + \ \frac{1}{2} \ = \ 1 \ \ $ and $ \ \ \mathbf{y = -x :} \ \ x^2 \ + \ y^2 \ = \ 2 \ + \ 2 \ = \ 4  \  ,  $ making the extremal distances from the origin $ \ 1 \ $ and $ \ 2 \ . $  [The points at these distances are $ \ \left( \pm  \frac{1}{\sqrt{2}} \ , \ \pm  \frac{1}{\sqrt{2}} \right) \ $ and $ \ \left( \pm   \sqrt{2} \ , \ \mp   \sqrt{2} \right) \ , $ respectively.

The constraint curve $ \ 5x^2 \ + \ 6xy \ + \ 5y^2 \ = \ 8 \ $ has symmetry about the origin, so it is reasonably to expect that there should be a critical point in each quadrant (with extremal points of the same type in opposite quadrants).  The curve is a "rotated ellipse" for which we have found the lengths of its semi-major and semi-minor axes and the locations of their endpoints.


问:
Does $\log\sqrt x$ grow as fast as $4\log x$ or slower?
Does $\log\sqrt x$ grow as fast as $4\log x$ or slower? Based on their function diagram, I would guess that $\log\sqrt x$ is slower. However
$$\lim_{ x\to \infty} \frac {\log (\sqrt x)}{\log x} = 1/8$$

So we can conclude that $\log\sqrt x$ grows as fast as $4\log x$. How can I solve this problem?

答: $$\log \sqrt{x} = \log x^{1/2} = \frac{1}{2} \log x.$$  So the ratio $$\frac{4 \log x}{\log \sqrt{x}} = \frac{4}{1/2} = 8.$$  So they are both $O(\log x)$ but obviously one is a constant multiple of the other.


答: As @MikeEarnest notes, it's a finite geometric progression with $n+1$ terms, of sum $\frac{a^{n+1}-b^{n+1}}{a-b}$. With classical $q$-analogs it can be written as $a^n[n+1]_{b/a}$, or $b^n[n+1]_{a/b}$.

Title: Notation for newton-like expansion

Body:
Is there a compact way of referring to the expression $$a^n + a^{n - 1}b + a^{n - 2}b^2 + \cdots + b^n\:?$$ Maybe some notation I do not know about it.

Thanks!


问:
Wedge sum of contractible spaces
Let $\vee_i U_i$ be the wedge sum of a set of based spaces $(U_i,x_i)$. Suppose each $U_i$ is a contractible neighborhood of $x_i$, then I want to prove that $\vee_i U_i$ is also contractible. (J. P. May mentions this statement in Section 2.8 of his book A Concise Course in Algebraic Topology)

My attempt: Suppose $F_i: c_{x_i} \simeq id_{U_i}$ are homotopies, where $c_{x_i}: U_i \to U_i$ is the constant map with value $x_i$. I tried to construct a homotpy $F$ between $c_x:\vee_i U_i \to \vee_i U_i$ and $id_{\vee_i U_i}$ via $F|_{U_i}=F_i$, where $x$ is the homotopy class of $x_i$. But the problem is that for fixed $t \in [0,1]$, $F_i(x_i, t)$ are different in general, so $F$ is not well-defined at $x$.

Q: How to explicitly contruct a homotopy between $c_x$ and $id_{\vee_i U_i}$?

答: The statement is wrong, the Griffiths Twin Cone is a counterexample, as suggested by JHK in the comment above. So when May says $U_i$ is a contractible neighborhood of $x_i$, I guess he means $x_i$ is a strong deformation retraction of its open neighborhood $U_i$.

Title: Wedge sum of contractible spaces


问:
How to find the second time when a piezoelectric crystal vibrates given a cubic trigonometric equation?
The problem is as follows:

In an electronics factory a quartz crystal is analyzed to get its
vibration so this frequency can be used to adjust their components.

The length that it expands from a certain charge is given by:

$6\tan\left(\frac{\pi t}{24}\right)\cdot\tan\left(\frac{\pi
 t}{24}+\frac{\pi}{3}\right)\cdot\tan\left(\frac{\pi
 t}{24}-\frac{\pi}{3}\right)$ micrometers.

Assuming this interval $0\leq t \leq 12$. Where $t$ is measured in
seconds. Find on which second the crystal expands $6$ micrometers?

I'm not sure how to solve this question because I'm ending with a cubic equation and I don't really know how to solve that using precalculus tools.

The thing is that all that equation is:

$6\tan\left(\frac{\pi t}{24}\right)\cdot\tan\left(\frac{\pi t}{24}+\frac{\pi}{3}\right)\cdot\tan\left(\frac{\pi t}{24}-\frac{\pi}{3}\right)=6$

Hence:

$\tan\left(\frac{\pi t}{24}\right)\cdot\tan\left(\frac{\pi t}{24}+\frac{\pi}{3}\right)\cdot\tan\left(\frac{\pi t}{24}-\frac{\pi}{3}\right)=1$

$\tan\left(\frac{\pi t}{24}\right)\cdot\left[\frac{\tan\frac{\pi t}{24}+\sqrt{3}}{1-\sqrt{3}\tan\frac{\pi t}{24}}\right]\cdot\left[\frac{\tan\frac{\pi t}{24}-\sqrt{3}}{1+\sqrt{3}\tan\frac{\pi t}{24}}\right]=1$

Then:

$\tan\left(\frac{\pi t}{24}\right)\cdot\left[\frac{\tan \frac{\pi t}{24}-3}{1-3\tan^2\frac{\pi t}{24}}\right]=1$

$\tan\left(\frac{\pi t}{24}\right)\cdot\left[\tan \frac{\pi t}{24}-3\right]=1-3\tan^2\frac{\pi t}{24}$

$\tan^3 \left(\frac{\pi t}{24}\right)-3 \tan \left(\frac{\pi t}{24}\right)=1-3\tan^2\left(\frac{\pi t}{24}\right)$

$\tan^3 \left(\frac{\pi t}{24}\right) +3\tan^2\left(\frac{\pi t}{24}\right) -3 \tan \left(\frac{\pi t}{24}\right) -1 =0$

Then I landed here, now what?

Because I'm not familiar with solving this sort of equation it would help me alot someone could guide on me what should be done next.

The key from what I could notice is that if I could get that time, I would find when it gets its second value, hence the second time. But as I mentioned. I don't know how to do that.

答:
You made the same mistake a few times: you neglected the $t$ in $\tan\frac{\pi t}{24}$ in some of your equations, which may be the cause of your confusion. We have the equation
$$x^3+3x^2-3x-1=0$$ where $x=\tan\frac{\pi t}{24}$. Simply by inspection, after trying a few different values for $x$ we see that $x=1$ is a solution; hence by the factor theorem
$$\begin{align}
x^3+3x^2-3x-1&\equiv(x-1)(Ax^2+Bx+C)\\
&\equiv Ax^3+(B-A)x^2+(C-B)x-C
\end{align}$$
We are now in a position to compare coefficients. We can first see that $A=1$. Also, $B-A=3$, but since $A=1$ this means that $B=4$. Similarly, we can see that the constant terms on both sides must be equal so finally we have $C=1$, and thus
$$
x^3+3x^2-3x-1\equiv(x-1)(x^2+4x+1)$$
This is now something that we can solve, as we have two factors, one of which is simply linear and hence very easy to deal with, and the other quadratic which can be dealt with by employing the quadratic formula.

I think you must have made an error: the quadratic factor is factorisable over the reals, as shown below.
$$x^2+4x+1=(x-(-2+\sqrt 3))(x-(-2-\sqrt 3))$$
I got these values by using the quadratic formula- I didn't work them out just by looking at the equation :) These are actually very nice factors in our case, as... drumroll...
$$\tan\left(\frac{7\pi}{12}\right)=-2-\sqrt3~~~~\tan\left(\frac{11\pi}{12}\right)=-2+\sqrt3$$

I hope that was helpful. If you have any questions please don't hesitate to ask :)


问:
Proof that every quadratic residue has two roots, modulo a prime
Can someone provide a proof that every quadratic residue, when working in $\mathbb Z_p$, where $p$ is a prime, has exactly two roots? Indeed, there cannot be only one root as for any $a^2$, we know $a$ and $-a$ are both roots. So then we need only show that there cannot be more than two- does this follow from some argument for the number of roots a quadratic can have modulo a prime? I'm not well-acquainted with the theorems pertaining to prime moduli.  </p>

答: The definition of an integer $x$ being a quadratic residue (mod p) is equivalent to the statement that there exists an integer $a$ so that $x$=$a$(mod p). But this in itself is equivalent to saying that $x^2$-$a$=$0$(mod p). So now one can conisder working over the field $\mathbb Z$/$p$$\mathbb Z$[$x$]. But by the fundamental theorem of algebra, such a polynomial must have at most $2$ roots in $\mathbb Z$/$p$$\mathbb Z$[$x$].


问:
About the Taylor polynomial of the antiderivative of a function
If $P_{n, g} (x)$ denotes the $n$-order Taylor polynomial of a function $g$ around $x = a$, and if $F$ is an antiderivative of $f$ ($f \in C_{\mathbb{R}}^\infty$).

Does  $P_{n+1, F} (x) = \int_{a}^{x}P_{n,f} (x)dx$ ?
I think the statement is true, but I'm stuck trying to show it.

The LHS should be:
$$ P_{n+1, F}(x) = \sum_{k=0}^{n+1} \frac{F^{(k)(} (a)}{k!} (x-a)^k$$

But I get the following in the RHS:

$$ \int_{a}^{x} P_{n, f} (x) dx = \int_{a}^{x} (\sum_{k=0}^{n} \frac{f^{(k)} (a)}{k!} (x-a)^k )dx
= \sum_{k=0}^{n} \frac{f^{(k)}(a)}{(k+1)! } (x-a)^{(k+1)} \\
\int_{a}^{x} P_{n, f} (x) dx= \sum_{k=0}^{n} \frac{F^{(k+1)}(a)}{(k+1)! } (x-a)^{(k+1)} =
\sum_{k=1}^{n+1} \frac{F^{(k)}(a)}{(k)! } (x-a)^{(k)}\\
\int_{a}^{x} P_{n, f} (x) dx=\sum_{k=0}^{n+1} \frac{F^{(k)}(a)}{(k)! } (x-a)^{(k)} - F(a)$$

Am I doing something wrong ?

答:
The correct answer is
$$
P_{n+1, F} (x) =  F(a) + \int_{a}^{x}P_{n,f} (t) \, dt
$$
which is what you got in the second calculation. It satisfies $P_{n+1, F} (a) = F(a)$ and
$$
P_{n+1, F}^{(k)} (a) = P_{n,f}^{(k-1)}(a) = f^{(k-1)}(a) = F^{(k)}(a)
$$
for $1 \le k \le n+1$.

It is correct that $P_{n+1, F}$ is an antiderivative of $P_{n,f}$, but the constant of integration must be chosen such that $P_{n+1, F} (a) = F(a)$.


问:
$f$ can be extended iff $\partial f = 0$
If

$0\rightarrow{A'}\rightarrow{A}\rightarrow{A''}\rightarrow{0}$

is an exact sequence of modules, then there exists an exact secuence

$0\rightarrow{}Hom(A'',B)\rightarrow{}Hom(A,B)\rightarrow{}Hom(A',B)\xrightarrow \partial{Ext}^1(A'',B)\rightarrow ...$

Suppose $A'\subseteq A$ and $f:A'\to B$. Prove that $f$ can be extended to $A$ if and only if $\partial f = 0$.

Any hint? Thanks!

答:
Let $i: A' \to A$ be the inclusion. If $f$ can be extended, then it's in the image of what map? What do you know about the composition of two consecutive maps in a complex?

Conversely, say $\partial f$=0. Then $f$ is in the kernel of $\partial$, but that sequence is exact, so...


问:
Probability problem where I remove 4 balls without reposition
One box has 7 white balls, 3 black balls and 3 red balls. We remove four balls without replacement.

What is the probability that there are two white, one black and one red.

答:
Here are some hints:

How many ways can you pick out two white balls from the seven white ones?

How many ways to pick out one black ball from the three black ones? (Same for the red balls)

How many ways can you pick out four balls from the whole thing, regardless of color?

Can you take it from here?


问:
Examples of measures that assign positive measure to countable sets of $\mathbb{R}^n$?
Let $A\subset\mathbb{R}^n$ be a countable (and finite) set of Real-valued $N$-dimensional vectors. The Lebesgue measure of this set $A$, denoted $\lambda(A)$, is just $\lambda(A)=0$ because the Lebesgue measure assigns measure zero to all countable subsets. The Dirac measure of this set $A$, denoted $\delta(A)$, may or may not be zero, since:
\begin{gather}
\delta_x(A)=
\begin{cases}
1&\text{if }x\in A\\
0&\text{if }x\notin A
\end{cases}
\end{gather}
for some $x\in\mathbb{R}^n$. My first question is this: in case they exist, can you provide examples of other measures that are not the Dirac one, that can be positive for some countable set $A\subset\mathbb{R}^n$?

BONUS QUESTION: consider some (possibly uncountable and infinite) set $B\subset\mathbb{R}^N$ and consider this construction:
\begin{gather}
\delta_B'(A)=
\begin{cases}
1&\text{if }A\cap B\neq\emptyset\\
0&\text{if }A\cap B=\emptyset
\end{cases}
\end{gather}
My bonus question is as follows: is the construction $\delta_B'(A)$ a measure? Why or why not?

EDIT: While $\delta_B'$ satisfies the properties of 'non-negativity' and 'null empty set', I think it fails to satisfy 'countable additivity'. Consider two sets $A,A'\subset\mathbb{R}^N$ such that $A\cap A'=\emptyset$ yet $A\cap B\neq\emptyset$ and $A'\cap B\neq\emptyset$. Then, $\delta_B'(A)=1$, $\delta_B'(A')=1$ and $\delta_B'(A\cup A')=1$. However, $\delta_B'(A)+\delta_B'(A')=2$, thus implying $\delta_B'(A\cup A')\neq\delta_B'(A)+\delta_B'(A')$ and therefore countable additivity fails to be satisfied. Is this reasoning correct?

答:
The questions is: in case they exist, can you provide examples of other measures that are not the Dirac one, that can be positive for some countable set $A\subset\mathbb{R}^n$?

Answer:

Using the Borel (or the Lebesgue) $\sigma$-algebra on $\Bbb R^n$, let $\nu_1$ be any measure (not signed measure) such that $\nu_1$ is not identically zero and, for all $x\in \Bbb R^n$, $\nu_1(\{x\})=0$.

Let $C$ be any non-empty countable subset of $\Bbb R^n$ and let $f: C \rightarrow (0,+\infty]$ be a function. For any measurable set $A$, define $\nu$ by
$$ \nu(A)= \nu_1(A) + \sum_{x\in A\cap C} f(x) $$

Then $\nu$ is a measure. For all measurable set $A$, $\nu(A)\geqslant 0$ and, for all $E \subseteq C$, if $E \neq \emptyset $ then $E$ is countable and $\nu(E) >0$.  However $\nu$ is not a Dirac measure.

Remark 1:  The construction of the measure $\nu$ is a general construction. Any measure constructed in this way will a measure positive for some countable subset of $\Bbb R^n$, but not a Dirac measure.

Remark 2: We can generalize the answer even further:

Assume your $\sigma$-algebra is such that all one-point sets are measurable.

Let $\nu_1$ be any measure (not signed measure) such that, for all $x\in \Bbb R^n$, $\nu_1(\{x\})=0$, $C$ be any non-empty countable subset of $\Bbb R^n$ and $f: C \rightarrow (0,+\infty]$ be a function. Then $\nu$ defined by, for all $A$ measurable,

$$ \nu(A)= \nu_1(A) + \sum_{x\in A\cap C} f(x) $$

is a measure that is positive on a countable set. The Dirac measures are just special cases where $\nu_1$ is identically zero and $C$ is an one-point set.

Moreover, in the "reverse direction", given  any any measure $\nu$, such that $\nu$ is positive on a countable set, and  for all $x\in \Bbb R^n$, $\nu(\{x\})<+\infty$, then $\nu$ can be obtained from the above construction with $f: C \rightarrow (0,+\infty)$.


问:
Solution of a simple differential equation, but now I forget it
$\frac{dp}{dr}+\frac{p}{r}$=0

what's p?

It was originally a simple equation......

I have learned this part of knowledge before, but now I forget it......thanks

答:
$$\dfrac{dp}{dr}r+p=0$$

Or more generally

$$\dfrac{dp}{dr}F(r)+G(p)=0$$

Is a linear first order homogeneous ODE.
As suggested in the comments can be solved by separation of variables or integrating factor.


问:
Can an odd perfect number be divisible by $165$?
I know that an odd perfect number cannot be divisible by $105$ or $825$. I wonder if that's also the case for $165$ (this is actually a stronger statement).</p>

答:
An odd perfect number, if one exists, could in principle be divisible by 165.

Let $\sigma(n)$ denote the sum of the positive divisors of $n$, and let the "abundancy" be $a(n)=\sigma(n)/n$. A perfect number is then defined by $\sigma(n)=2n$, or equivalently $a(n)=2$.

Let $$n = \prod_{i=1}^k q_i^{\alpha_i}$$
be the prime factorization of $n$, where the $q_i$ are distinct primes. Then
\begin{align}
\sigma(n) &= \sum_{d|n} d\\
          &= \prod_{i=1}^k \sum_{\alpha=0}^{ \alpha_i } q_i^{\alpha}\\
          &= \prod_{i=1}^k \frac{q_i^{\alpha_i+1}-1}{q_i-1}
\end{align}
So
\begin{align}
a(n) &= \prod_{i=1}^k \frac{q_i^{\alpha_i+1}-1}{ q_i^{\alpha_i}(q_i-1)}\\
     &= \prod_{i=1}^k h(q_i, \alpha_i),
\end{align}
where
$$
h(q, \alpha) = \frac{q^{\alpha+1}-1}{ q^{\alpha}(q-1)}.
$$

We note that $h(q,\alpha)>1\ \forall\ q\geq2\textrm{ and }\alpha>0$, and that $h(q,\alpha)$ is an increasing function of $\alpha$ for fixed $q$.

Suppose $n$ is an odd perfect number. Then $n = p^e m^2$ for some prime $p$ with $p\equiv e \equiv 1\ (\operatorname{mod} 4)$ and $\operatorname{gcd}(p,m)=1$. (This is a well-known theorem due to Euler.) Thus, if $105 | n$ then $3^2\cdot5\cdot7^2\ |\ n$, which would imply $a(n) > h(3,2)h(5,1)h(7,2) = 2.01\ldots$, a contradiciton. This is the reason that no odd perfect number can be divisible by $105$: any odd perfect number $n$ divisible by $105$ would, by Euler, be divisible by $3^2\cdot5\cdot7^2$ and would therefore be "abundant", i.e. $a(n)>2$ for such $n$, a contradiction. The same argument gives $a(n)>1.905$ for $n$ divisible by $165$. There is nothing wrong with this statement; it does not lead to a contradiction.

It's difficult to find a reference that something has not been proven to be impossible. However, in the normal machinations to show that a prospective odd perfect number can't have a certain divisibility property, there's nothing obvious to say that one couldn't be divisible by $165$.


问:
Probability that a given pair of people sits next to each other in a round table.
There's a round table with N chairs. $N$ people are seated randomly. What is the probability that a given pair of people will sit next to each other.

If you fix the first person to sit at the i-th chair, then it seems the problem reduces down to the probability of the second person sitting in the i+1 or i-1 chair locations, which is simply just $\frac{2}{N-1}$. Is this correct?

When I try to prob this with monte carlo, I seem to be getting a consistently smaller probability. The way my monte carlo algorithm works is I sample 2 numbers without replacement from $\sim U(1, N)$. Then I check that the absolute difference between the numbers is 1. If it is, then they're deemed to be sitting next to each other. Then I divide this amount by the number of trials.

答:
The first part is right.

For the simulation, are you also counting the case where the chosen numbers are $1$ and $N$?


问:
Calculating the Lie bracket on the Heisenberg algebra of $H=Z\times S^1$
I'm working through Mechanics and Symmetry by Marsden and Ratiu. Let $(Z,\Omega)$ be a symplectic vector space and define on $H:=Z\times S^1$ the operation $$(u,\exp i\phi)(v,\exp i\psi)=(u+v,\exp i[\phi+\psi+\hbar^{-1} \Omega(u,v)])$$

I need to show that on the Lie algebra of $H$, $\mathfrak{h}=Z\times\mathbb{R}$, the Lie bracket is defined as follows: $$[(u,\phi),(v,\phi)]=(0,2\hbar^{-1}\Omega(u,v))$$

I'm using the method the book recommends to calculate this bracket, the steps are:

Calculate the inner automorphisms: $$I_g:H\to H,\quad\text{where}\quad I_g(h)=ghg^{-1}$$

Differentiate $I_g(h)$ with respect to $h$ at $h=e$ to product the adjoint operators $$\textrm{Ad}_g:\mathfrak{h}\to\mathfrak{h};\quad \textrm{Ad}_g\eta=T_eI_g(\eta)$$

Then Differentiating $\textrm{Ad}_g\eta$ with respect to $g$ at the identity $e$ in the direction of $\xi\in\mathfrak{h}$, since $$T_e(\textrm{Ad}_g\eta)\cdot\xi=[\xi,\eta]$$

So I started by calculating the inner automorphisms. Given $g=(v,\exp(i\psi))$ we have $g^{-1}=(-v,\exp(-i\psi))$. If we take $h=(u,\exp i\phi)$, we have \begin{align}
I_g(h)&=(v+u,\exp i(\phi+\psi+\hbar^{-1}\Omega(v,u))(-v,\exp i(-\psi))\\
&=(v+u-v,\exp i(\phi+\psi+-\psi+\hbar^{-1}\Omega(v,u)+\hbar^{-1}\Omega(v+u,-v))\\
&=(u,\exp i(\phi+2\hbar^{-1}\Omega(v,u))
\end{align}
After this, I am confused as to how to take the derivative of this, I believe it should be something like: $$T_eI_g(\eta)=(0, 2\hbar^{-1}\Omega(v,u))$$
But I am unsure if this is correct, or how I could justify this step. How am I supposed to compute this derivative?

答:
To be consistent, it would make more sense to take $g=(u,\exp(i\phi))$ and $h=(v,\exp(i\psi))$, so
$$
I_g(h) = (v,\exp i(\psi+2\hbar^{-1}\Omega(u,v))).
$$
Now one way to calculate $T_eI_g(\eta)$ is to calculate $\frac{d}{dt}\big\vert_{t=0}I_g(h(t))$, where $h(t)$ is a smooth curve such that $h(0) = e=(0,1)$, the identity of the Heisenberg group, and $h'(0) = \eta$. Taking $\eta = (v,\psi)$, a corresponding curve is $h(t) = (tv,\exp(it\psi))$. Then
$$
I_g(h(t)) = (tv,\exp i(t\psi + 2\hbar^{-1}\Omega(u,tv))) \implies T_eI_g(\eta) = \frac{d}{dt}\big\vert_{t=0} I_g(h(t)) = (v,\psi+2\hbar^{-1}\Omega(u,v)).
$$
Now repeat this process with $g(t) = (tu,\exp it\phi)$.


问:
For which $n$ the first player has a strategy so that he wins no matter how the other player plays?

The Question

From Junior O-level Tournament of the Towns paper, Fall 2020:

There are $n$ stones in a heap. Two players play the game by
alternatively taking either $1$ stone from the heap or a prime number
of stones which divides the current number of stones in the heap. The
player who takes the last stone wins. For which $n$ the first player has
a strategy so that he wins no matter how the other player plays?

My Understanding

Let player $A$ play first and player $B$ second. Also, let a heap with $n$ stones be an $n$-heap.

My initial observations:

$A$ wins any $p$-heap for a prime number $p$.

If an $n$-heap is
winning for $B$, then the $(n + 1)$-heap is winning for $A$, since
$A$ can remove $1$ stone and then use $B$'s winning strategy for the
$n$-heap (since for $n$-heap $A$ moves second).

I couldn't think of anything else so I tried finding a pattern with some examples.

\begin{array}{|c|c|c|c|}
\hline
n& \text{winner} \\ \hline
 1 &A \\ \hline
 2  & A\\ \hline
 3  & A\\ \hline
 4 & B\\ \hline
 5  & A\\ \hline
 6  & A\\ \hline
 7 & A\\ \hline
 8  & B\\ \hline
 9  & A\\ \hline
 10  & A\\ \hline
 11 & A\\ \hline
 12  & B\\ \hline
\end{array}

Since it seems $B$ wins exactly when $n = 4k$, I tried to prove this by induction on $k$.

Basis: the claim holds for $k \leq 3$ as shown in the examples.

Inductive hypothesis: Suppose that $B$ wins exactly when $n = 4k$ for arbitrary $k$.

Now I try to prove that $4k + 1$, $4k + 2$, $4k + 3$ have winning strategies for $A$ and that $4k + 4$ has a winning strategy for $B$.

$\textbf{(4k + 1)-heap:}$ $A$ wins by observation 2.

$\textbf{(4k + 2)-heap:}$ $4k + 2 = 2(2k + 1)$ which is even. $A$ can remove 2 stones to get a $4k$-heap which is winning for $A$ since now $B$ moves first.

Now I am stuck for the $(4k + 3)$-heap and $(4k + 4)$-heap. I don't want a complete solution, instead I just want a hint and some confirmation that this is a way to solve this problem. Also if you have an easier way, I would like a hint to discover it!

答:
Yes, you've found a good way of attacking the problem, and you've already got most of the way there.

Hints:

The number of stones that can be removed from a pile of size $\ 4k+4\ $ can only be $\ 1,2\ $ or an odd prime.

A number of the form $\ 4k+3\ $ must have at least one prime factor $\ p\equiv3\pmod{4}\ $.


问:
Express $\operatorname{sech}^{-1}(x)$ in terms of logarithms
I'm trying to express the following  $\operatorname{sech}^{-1}(x)$ in terms of logarithms, and would warmly appreciate feedback towards my approach. The solution should be :

$$\ln\left(\dfrac{1+\sqrt{(1-x^2)}}{x}\right)$$ when $x > 0.$ However, I cannot seem to get this. My working out:

$y = \operatorname{sech}^{-1}(x)\implies \operatorname{sech}(y) = x$

$$-\operatorname{sech}(y)\operatorname{tan}(y) = x; $$

Once I got to here, I decided to transform the $\operatorname{sech}$ and $\tan$ into their hyperbolic identities.

$$-\dfrac{\operatorname{sinh}(y)}{\operatorname{cosh}^2(y)}=\dfrac{e^y-e^{-y}}{2}(\dfrac{e^y+e^{-y}}{2})^2 = x$$

Though I've tried reworking this into a quadratic form $\dfrac{b \space \pm \space \sqrt{b^2-4ac}}{2a}$, however, I couldn't manage to get the right form. I would greatly appreciate some help on the next steps towards this.

答:
Forming a quadratic as outlined in the above answers is a fine approach, and one I recommend above this one. Here's a different one, which relies on knowing beforehand that
$$\cosh^{-1}x=\ln(x+\sqrt{x^2-1})$$
Let $y=\operatorname{sech} ^{-1} x$. Then
$$\begin{align}
\operatorname{sech}y=x&\iff \cosh y=\frac{1}{x}\\
&\iff y=\operatorname{sech}^{-1}x=\cosh^{-1}\frac{1}{x}=\ln\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)
\end{align}$$
Simplifying yields the correct answer. I hope that was helpful. If you have any questions please don't hesitate to ask :)


问:
Is an orthocomplemented lattice automatically complete?
I am reading a paper where I find the following definition:

Definition: Let $L$ be an orthocomplemented lattice. A function $\mu: L \rightarrow [0,1]$ is called a probability measure on $L$ if

$\mu(0) = 0$, $\mu(1) = 1$,

$\mu(a_1 \lor a_2 \lor \ldots) = \sum_{k=1}^\infty \mu(a_k)$ for mutually disjoint $a_i$.

A function $\nu: B(\mathbb{R}) \rightarrow L$, where $B(\mathbb{R})$ is the family of Borel sets on $\mathbb{R}$, is called a $L$-valued measure on $B(\mathbb{R})$ if

$E \cap F = \emptyset$ implies $\nu(E) \perp \nu(F)$ for any $E, F \in B(\mathbb{R})$,

$\nu(E_1 \cup E_2 \cup \ldots) = \nu(E_1) \lor \nu(E_2) \lor \ldots$ for any family $E_i \in B(\mathbb{R})$ such that $E_i \cap E_j = \emptyset$ for $i \neq j$,

$\nu(\emptyset) = 0, \nu(\mathbb{R}) = 1$.

There is one thing that I don't understand in this definition. On both lists, in item 2, there appears a join of potentially infinitely many elements of $L$. Is this well-defined? I thought that for such a thing to be well-defined, the lattice needs to be complete. Is an orthocomplemented lattice automatically complete?

答:
I'd assume they meant that the identity in item 2 should hold when $a_1 \vee a_2 \vee ...$ is defined.

I haven't really studied orthocomplemented lattices before, so I'm just working from the Wikipedia definition, but how about the following for a non-complete orthocomplemented lattice (apologies for the strange notation):

$0<1<2<...< \infty -2 < \infty - 1 < \infty$,

$0<1'<2'<...< (\infty -2)' < (\infty - 1)' < \infty$,

with anything exclusively from line 1 neither bigger nor smaller than anything exclusively from line 2.

This is a bounded lattice (both strands are totally ordered), with lower bound $0$ and upper bound $\infty$, and for $n \geq 1$ $(\infty - n)'$ can be chosen as the orthocomplement for $n$, $n'$ as the orthocomplement for $(\infty - n)$, $(\infty - n)$ as the orthocomplement for $n'$, and $n$ as the orthocomplement for $(\infty - n)'$. $\{0,1,2...\}$ has no least upper bound in this lattice.


问:
Inverse of g(a, b) = (a + b, a - b)?
As the question title suggests, does $g(a, b) = (a + b, a - b)$ have an inverse? If so, what is it?

答:
As suggested in a comment by Martin R, we have

$$u=a+b\tag1$$

and $$v=a-b\tag2.$$

Adding $(1)$ and $(2)$, we get $u+v=2a$, and subtracting $(2)$ from $(1)$, we get $u-v=2b$.

If you solve these for $a$ and $b$, you will get $g^{-1}(u,v)$.

Title: Inverse of g(a, b) = (a + b, a - b)?


问:
How can we prove the following product-sum inequality involving e?
Take any sequence $x_1, \dots, x_k$ of positive reals such that $\sum_{i=1}^{k-1} x_i < 1$ and $\sum_{i=1}^k x_i \ge 1$. I suspect that the following inequality holds:
$$
\prod_{i=1}^k (1 + x_i) \le e \sum_{i=1}^k x_k.
$$

This didn't seem very hard at first, but I can't quite get the desired inequality.  Here's my best attempt so far. By AM-GM inequality, we have
$$
\prod_{i=1}^k (1 + x_i) = (1 + x_k) \prod_{i=1}^{k-1} (1 + x_i) \le (1 + x_k) \left( 1 + \frac{1}{k-1} \sum_{i=1}^{k-1} x_i \right)^{k-1}.
$$
By a standard exponential function inequality, we then have
$$
(1 + x_k) \left( 1 + \frac{1}{k-1} \sum_{i=1}^{k-1} x_i \right)^{k-1} < (1 + x_k) \exp \left( \sum_{i=1}^{k-1} x_i \right).
$$
Finally, by the first constraint on the sequence, we get
$$
(1 + x_k) \exp \left( \sum_{i=1}^{k-1} x_i \right) \le e (1 + x_k).
$$

This is very close to what I want, but not quite there. I'm assuming there's a trick somewhere that I'm missing. Any help would be appreciated.

答:
If $a\ne b$,
$$
\begin{align}
\left(1+\frac{a+b}2\right)^2
&=(1+a)(1+b)+\left(\frac{a-b}2\right)^2\tag{1a}\\
&\gt(1+a)(1+b)\tag{1b}
\end{align}
$$
Induction and $(1)$ shows that for a given sum of $x_k$, the product of $1+x_k$ is maximized when all of the $x_k$ are equal (if two $x_k$ are different, replace them with their average to get greater product while maintaining the sum). That is,
$$
\prod_{k=1}^n(1+x_n)\le\left(1+\frac1n\sum_{k=1}^nx_n\right)^n\tag2
$$

Suppose that
$$
\sum_{k=1}^{n-1}x_k=\alpha\lt1\tag3
$$
and
$$
\alpha+x_n\ge1\tag4
$$
then $(2)$ and $(3)$ imply that the greatest product of $1+x_k$ is when all the $x_k$ are equal; that is,
$$
\begin{align}
\prod_{k=1}^n(1+x_k)
&=\left(\prod_{k=1}^{n-1}(1+x_k)\right)(1+x_n)\tag{5a}\\
&\le\left(1+\frac\alpha{n-1}\right)^{n-1}(1+x_n)\tag{5b}
\end{align}
$$
Consider
$$
f(\alpha)=e^\alpha(1+x_n)-e(\alpha+x_n)\tag6
$$
Since $f''(\alpha)=e^\alpha(1+x_n)\ge0$, $f$ is convex. Furthermore, $(3)$ and $(4)$ constrain $1-x_n\le\alpha\lt1$.

Since $x_n\gt0$,
$$
\begin{align}
f(1-x_n)
&=e^{1-x_n}\overbrace{(1+x_n)}^{\large\lt e^{x_n}}-e(1-x_n+x_n)\tag{7a}\\
&\lt0\tag{7b}
\end{align}
$$
and
$$
\begin{align}
f(1)
&=e(1+x_n)-e(1+x_n)\tag{8a}\\
&=0\tag{8b}
\end{align}
$$
for $1-x_n\le\alpha\lt1$, we have
$$
\begin{align}
\prod_{k=1}^n(1+x_k)-e\sum_{k=1}^nx_k
&\le\left(1+\frac\alpha{n-1}\right)^{n-1}(1+x_n)-e(\alpha+x_n)\tag{9a}\\
&\le e^\alpha(1+x_n)-e(\alpha+x_n)\tag{9b}\\[9pt]
&=f(\alpha)\tag{9c}\\[9pt]
&\lt0\tag{9d}
\end{align}
$$
Therefore,
$$
\prod_{k=1}^n(1+x_k)\lt e\sum_{k=1}^nx_k\tag{10}
$$


问:
Addition table for a finit field with 3 elements $\mathbb{F}_{3}=\{0,1,-1\}$
just a quick question :-)

I need to make an addition table for a finit field with 3 elements $\mathbb{F}_{3}=\{0,1,-1\}$

Is this table correct?

$ \begin{matrix}
\boldsymbol{\textbf{+}} & \mathbf{0} & \textbf{1} & \textbf{-1} \\
\boldsymbol{\textbf{0}} & 0 & 1 & -1\\
\boldsymbol{\textbf{1}} & 1 & 0 & 1 \\
\boldsymbol{\textbf{-1}} & -1 & 1 & 0 \\
\end{matrix}$

@ZeroTheHero:

$ \begin{matrix}
\boldsymbol{\textbf{+}} & \mathbf{0} & \textbf{1} & \textbf{2} \\
\boldsymbol{\textbf{0}} & 0 & 1 & 2 \\
\boldsymbol{\textbf{1}} & 1 & 2 & 0 \\
\boldsymbol{\textbf{2}} & 2 & 0 & 1 \\
\end{matrix}$

答:
You can't say that 1+(-1) = 0 = 1
So this is wrong because this 'field' now only has 2 elements, and also, a field requires that 1!=0. The alternative is that the -1 is an abuse of notation is not referring to additive inverse of 1, in which case this field is stupid lol. Theres only one possible way to define F_3 I'm pretty sure. Show That There Is One and (Essentially) Only One Field With 3 Elements</p>


问:
About the exactness of $n$-forms
I have the following statement in my Analysis in $\mathbb{R}^n$ book:

If $\omega$ is an $n$-form on a orientable compact $n$-manifold $M$ without boundary, then $\omega $ is exact if and only if $\int\limits_{M}\omega=0$.

In the comment on this post, it is mentioned that it is necessary that $M$ be connected. But in my book there is no comment on the need to be connected, can this be due to the definition of manifold?

I am particularly interested in the volume form, I have that the volume form cannot be exact, the problem comes from using Stokes, the volume form would be zero (because the boundary is empty), can I still state this if the manifold is disconnected?

A manifold of dimension $m$ and class $C^k$ in $\mathbb{R}^n$ is a set $M\subset R^n$ that can be covered by a collection of open $U\subset\mathbb{R}^n$ such that $V=U\cup M$ admits $C^k$ parameterization $\phi: V_0 \to V$ defined in an open $V_0\subset\mathbb{R}^m$

If I don't need connectivity because of my definition, what would be the definition that requires connectivity?

答:
The "only if" is fine, but the "if" requires connectedness. For consider the disjoint union of two spheres, and let $\omega$ be a form that integrates to $1$ on the first sphere and to $-1$ on the second. Obviously, such an $\omega$ cannot be exact. (Apply Stokes's Theorem on each sphere separately.)

Indeed, the argument I just gave answers your question about the volume form. It follows directly from the contrapositive of the "only if."

Title: About the exactness of $n$-forms


问:
Gauss Map and Geodesic Flow
I was reading chpater (9) of the "Ergodic Theory with a view towards Number Theory" book by Manfred Einsiedler and Thomas Ward.

To be more precise, I was trying to understand the connection between the Gauss Map and the Geodesic Flow as it is illustrated in the Section 6 of the chpater (9.6 Ergodicity of the Gauss Map).

To be honest, the idea was not too much clear to me (I did not say that it is not clear at all). Therefore, I am now looking for references that illustrate the connection between the Gauss Map and the Geodesic Flow is a way that is easier than the one in this chapter.

I would be very grateful if one could suggest some good references.

See pages 317, 318 and 319.

One of the questions is:

What is the "first" visit of the geodesic flow to the set $\pi(C)$ to consider the next visit?

答:
I do not have time to check the calculations, but the authors define a certain closed subset $C$ of the unit tangent bundle $X=UT(M)$ (of the modular orbifold). Given a point $v\in C$, the geodesic $c_v$ in $X$ defined by $v$ (so that $c_v(0)=v$) first leaves $C$ at time $t_0\ge 0$, which can be regarded as the end of the  "first visit" of $C$. Formally speaking, $t_0$ is the supremum of all $t\ge 0$ such that $c_v(t)\in C$.

Then there is the smallest $t_1^- >t_0$ such that $c_v(t)\in C$. This is the "next time" $c_v$ visits $C$, more precisely, the first time $c_v$ reenters $C$.

You did not ask, but there will be $t_1^+$, the "end of the 2nd visit," where $t_1^+$ is the supremum of all $\{t \ge t_1^-: c_v(t)\in C\}$. This is the end time of the 2nd visit. And so on.

Title: Gauss Map and Geodesic Flow


问:
Need help on understand a question on inverse functions
I need help understanding what this question is asking and I am not sure what to do. The question is given below

For each number y find the maximum value of $yx - 2x^4$. This maximum is a function $G(y)$. Verify that the derivative $G(y)$ and $2x^4$ are inverse functions

Please do not give away the answer but guide me in the right direction.

Many thanks and stay safe!!!

答:
For the first part, it wants you to treat $y$ as a constant. For instance, if $y=2$, you would need to find the maximum of the function $2x-2x^4$ with respect to $x$. Finding the maximum for general $y$ gives you a function $G(y)$ such that $yG(y)-2(G(y))^4\geq yx-2x^4$ for any $x$.

As an example, if you wanted to find the maximum of $x-yx^2$ with respect to $x$, you would first find where the derivative is $0$, so
$$0=\frac{d}{dx}x-yx^2=1-2yx$$
Solving for $x$, you get the function $\frac{1}{2y}$, which is what they mean by $G(y)$ in the problem (you still need to verify this is actually a maximum of course).

The second part is worded a little strangely, but I think they want you to verify that the inverse function of $G$ is equal to the derivative of $2x^4$, although I’m not sure why they would ask for this without more context, so I might be misreading it.


问:
There are $7$ empty seats on a bus and four people get on. How many different ways can they be seated?
There are $7$ empty seats on a bus and four people get on. How many different ways can they be seated?

Would it be ${}^7C_{4}$ or ${}^7P_{4}$?

答: I think the answer is $$ 4! C_7^4 = 7 * 6 * 5 *4 $$ Where the C is a notation for the combination of 4 out of 7. Explanation: first you pick 4 out of 7 seats which will be taken and then you determine the person for each seat.</p>


问:
Why are all finite state Markov Chains positive recurrent?
This makes sense to me intuitively, since infinite return time if null recurrent (or transient i.e positive probability of no return) doesn't seem to be possible if there are finite states.

I found a convincing argument here, but I'm not able to prove this mathematically. Could someone please provide hints/outline for a proof?

Thanks in advance

答:
for an algebraic proof, note that if your chain had no positive recurrent class then that means it decomposes purely into null recurrent and transient classes.

This implies  $P^k\to \mathbf 0$

using the Frobenius norm for convenience, this is equivalent to

$\big\Vert P^k-\mathbf 0\big \Vert_F \lt \epsilon$

for any $\epsilon \gt 0$ for $k$ large enough.

but $P\mathbf 1 = \mathbf 1$ so $\lambda_1=1$ is an eigenvalue of $P$.  Applying Schur Triangularization to $P$ we get

$1 =\lambda_1^k \leq \big\Vert P^k-\mathbf 0\big \Vert_F $

for all $k$ so $P^k\to \mathbf 0$ is impossible and $P$ must contain a positive recurrent class.

wordsmithing note to OP: it isn't actually true that all finite state markov chains 'are' positive recurrent, it is true that they must contain a positive recurrent class.  (To falsify the former, e.g. construct a simple absorbing state chain where absorbtion happens WP1.) Alternatively you could say that all irreducible finite state markov chains are positive recurrent.


问:
Integral of square of inverse of CDF
Suppose $X$ is some non-negative random variable with $\mathbb{E} X^2<\infty$ and that $F$ is its cumulative distribution function, with generalized inverse $F^{-1}$. I'm wondering how we might be able to bound the integral $\int_0^{1/2}(F^{-1}(1-x))^2dx.$

If we have a $k$-th moment for $k>2$, then I can use Markov's inequality to show that
$$F^{-1}(1-x)\leq ||X||_k/x^{1/k}$$
and hence
$$\int_0^{1/2}(F^{-1}(1-x))^2dx \leq ||X||_k^2 \int_0^{1/2} 1/x^{2/k}dx<\infty.$$
The problem with filling in $k=2$ is that the final integral in the bound is no longer finite.

How could I solve this problem (if it is even possible)?

答:
I will assume that $F$ is continuous and strictly increasing to simplify things. If you need a result for more general $F$, you will need to figure that out on your own.

Using this trick for integrating inverse functions, you get
$$
\int_{0}^{1/2}(F^{-1}(1-x))^2\,dx
=\frac12 F^{-1}(1/2)^2+\int_{F^{-1}(1/2)^2}^\infty (1-F(x^{1/2}))\,dx
$$
Using the substitution $x=u^2,dx=2u\,du$, this becomes
$$
\frac12 F^{-1}(1/2)^2+\int_{F^{-1}(1/2)}^\infty 2u(1-F(u))\,du
$$
Write $m=F^{-1}(1/2)$, which is the median of $X$. Using the same Fubini trick which proves $E[X^2]=\int_0^\infty 2tP(X>t)\,dt$,
$$
\begin{align}
\frac12 m^2+\int_{m}^\infty 2uP(X>u)\,du
&=\frac12m^2+\int_m^\infty 2uE[{\bf 1}(X>u)]\,du
\\&=\frac12m^2+E\left[\int_m^X2u\,du\right]
\\&=\frac12m^2+E[(X^2-m^2)\cdot {\bf 1}(X>m)]
\\&=E[X^2\cdot {\bf 1}(X> m)]
\end{align}
$$
This is an exact expression for your integral, which also comes with the simple bounds
$$
E[X^2]-\frac12m^2\le E[X^2\cdot {\bf 1}(X> m)]\le E[X^2]
$$

Title: Integral of square of inverse of CDF


问:
Determining the limit: What does it mean when one obtains zero in the calculation?
Consider the sequence $a_n = \sqrt{n+\sqrt{n}}-\sqrt{n-\sqrt{n}}\\$. To determine the limit I did the following:
\begin{aligned}
a_{n} &=\left(\sqrt{n+\sqrt{n}}-\sqrt{n-\sqrt{n}}\right) \frac{\sqrt{n+\sqrt{n}}+\sqrt{n-\sqrt{n}}}{\sqrt{n+\sqrt{n}}+\sqrt{n-\sqrt{n}}} \\[10pt]
&=\frac{2 \sqrt{n}}{\sqrt{n+\sqrt{n}}+\sqrt{n-\sqrt{n}}}=\frac{2 \sqrt{n}}{\sqrt{n} \sqrt{1+\frac{1}{\sqrt{n}}}+\sqrt{n} \sqrt{1-\frac{1}{\sqrt{n}}}} \\[10pt]
&=\frac{2}{\sqrt{1+\underbrace{\frac{1}{\sqrt{n}}}_{\rightarrow0 \text{ for }n \rightarrow \infty}}+\sqrt{1-\underbrace{\frac{1}{\sqrt{n}}}_{\rightarrow0 \text{ for }n \rightarrow \infty}}} \\[10pt] &= \dfrac{2}{2} = 1.
\end{aligned}

However, my first thought was $a_n = \sqrt{n+\sqrt{n}}-\sqrt{n-\sqrt{n}} = \sqrt{n}\left(\sqrt{1 + \dfrac{1}{\sqrt{n}}}\space - \sqrt{1 - \dfrac{1}{\sqrt{n}}}\right) = \sqrt{n}\space (1-1) = 0$

with the same argument as above. Is it correct that one can't make a statement about the convergence in the latter calculation because we have $\infty \cdot 0$ ? If yes, why exactly is this the case?

Edit: To see that $\lim\limits_{n\to \infty} \dfrac{1}{\sqrt{n}} = 0$ pick some arbitrary $\epsilon > 0$. We want to find a $N$ s.t. $\forall n\geq N\colon |\dfrac{1}{\sqrt{n}} - 0|< \epsilon \Longleftrightarrow \dfrac{1}{\sqrt{n}} < \epsilon \Longleftrightarrow n >\dfrac{1}{\epsilon^2}$ meaning we can choose $N = \dfrac{1}{\epsilon^2} + 1$ for example.

答:
When one factor of a product goes to $\infty$ and the other goes to 0, you can't immediately conclude what the limit is. The easiest way to see this is to consider $\lim_{n \to \infty} n \cdot \frac{1}{n}$.  Now replace the 1 by any other number.

The error in your calculation is that when you evaluate a limit by substitution, you have to substitute all instances of $n$ at the same time.  In your calculation, in the second to last = sign, you substitute some instances to get $1-1$ while leaving the $\sqrt{n}$ unsubstituted.  That is wrong because when you take a limit, all instances of $n$ are tending to the limit at the same time.


问:
Tensor product of two spaces one of them is free module
Let $V$ be a real vector space and let $S(V^*)$ be the symmetric algebra of $V^*$.
Suppose $L$ and $N$ are two $S(V^*)$-modules.

How to prove the statement in corollary 16 in the article by Kumar and Vergne on equivariant cohomology, which says that

If $L$ is a free $S(V^*)$-module, then the tensor product space $L \otimes_{\mathbb R} N$ is also  free.

答: According to Lemma 15, whenever you consider a tensor product of the form $S(V^*)\otimes N$ where $S(V^*)$ is the regular module, this is isomorphic to $S(V^*)\otimes N'$ where $N'$ has the trivial module structure. If $M$ is free, it is of the form $S(V^*)\otimes W$ and their argument in Lemma 15 shows that $M\otimes N$ is isomorphic to $M\otimes N'$, which is ---unless I am missing something obvious--- free over $W\otimes N'$.


问:
The Laplace operator.
What will be the value of $\Delta\left(\frac 1{r^2}\right)$ if $r=|x|=\sqrt{x_1^2+x_2^2+x_3^2}$?
May be this can be determined using Green's formula.</p>

答:
This is not a PDE. This is just the Laplacian. A PDE is an equation. See https://en.wikipedia.org/wiki/Laplace_operator#Coordinate_expressions.
$$
\Delta\left[\frac{1}{r^{2}}\right]=\frac{1}{r^{2}}\frac{\partial}{\partial r}\left[r^{2}\frac{\partial}{\partial r}\left[\frac{1}{r^{2}}\right]\right]=\frac{2}{r^{4}}
$$</p>

Title: The Laplace operator.


问:
Inequality $\sum_{k=1}^n\frac{k^2}{n^3+n^2}\le\sum_{k=1}^n\frac{k^2}{n^3+k^2}\le\sum_{k=1}^n\frac{k^2}{n^3+1}$
From this question How do you calculate this limit? $\lim_{n\to\infty}( \frac{1^2}{n^3+1^2} + \frac{2^2}{n^3+2^2} ... + \frac{n^2}{n^3+n^2})=?$

I have not understood the reason of the @xpaul's answer for the $(1)$,

$$\sum_{k=1}^n\frac{k^2}{n^3+n^2}\le\sum_{k=1}^n\frac{k^2}{n^3+k^2}\le\sum_{k=1}^n\frac{k^2}{n^3+1} \tag 1$$

答: Since the sum ranges from $k=1$ to $k=n$, we have $k\geq 1$ and $k\leq n$; thus $n^2\geq k^2\geq 1$. Translate by $n^3$ to get $n^3+n^2\geq n^3+k^2\geq n^3+1$, then invert (changing the direction of the inequalities) to get $\dfrac1{n^3+n^2}\leq \dfrac1{n^3+k^2}\leq \dfrac1{n^3+1}$. Multiply by $k^2$ to get $\dfrac{k^2}{n^3+n^2}\leq\dfrac{k^2}{n^3+k^2}\leq\dfrac{k^2}{n^3+1}$, and then just sum termwise over $k$.


问:
Is every abstract simplicial complex the independence complex of a simple graph?
Given a simple (no loops, no multi-edges) undirected graph $G$ on $n$-vertices, one can assign an abstract simplicial complex known as the independence complex (https://en.m.wikipedia.org/wiki/Independence_complex) on $n$-vertices whose faces are exactly the independent sets of the graph $G$.

My question is: Is every abstract simplicial complex the independence complex of a simple graph?

答: As @AndreasBlass said, this is not true because whenever all the 2-faces of a simplex are present in the complex, the the simplex itself is present. This is a characterization of the clique complexes. And actually, the independence complex of a graph $G$ is the clique complex of the complement of $G$, so this makes sense.


答:
Your second approach is wrong.

To save writing let $x=\sqrt{n}$, and expand square roots in power series in $\frac{1}{x}$, so expression becomes $x(\sqrt{1+\frac{1}{x}}-\sqrt{1-\frac{1}{x}})$

$=x(1+\frac{1}{2x}+O(\frac{1}{x^2})-1+\frac{1}{2x}+O(\frac{1}{x^2}))$

$=1+O(\frac{1}{x})\to 1$ as $x\to \infty$

Title: Determining the limit: What does it mean when one obtains zero in the calculation?

Body:
Consider the sequence $a_n = \sqrt{n+\sqrt{n}}-\sqrt{n-\sqrt{n}}\\$. To determine the limit I did the following:
\begin{aligned}
a_{n} &=\left(\sqrt{n+\sqrt{n}}-\sqrt{n-\sqrt{n}}\right) \frac{\sqrt{n+\sqrt{n}}+\sqrt{n-\sqrt{n}}}{\sqrt{n+\sqrt{n}}+\sqrt{n-\sqrt{n}}} \\[10pt]
&=\frac{2 \sqrt{n}}{\sqrt{n+\sqrt{n}}+\sqrt{n-\sqrt{n}}}=\frac{2 \sqrt{n}}{\sqrt{n} \sqrt{1+\frac{1}{\sqrt{n}}}+\sqrt{n} \sqrt{1-\frac{1}{\sqrt{n}}}} \\[10pt]
&=\frac{2}{\sqrt{1+\underbrace{\frac{1}{\sqrt{n}}}_{\rightarrow0 \text{ for }n \rightarrow \infty}}+\sqrt{1-\underbrace{\frac{1}{\sqrt{n}}}_{\rightarrow0 \text{ for }n \rightarrow \infty}}} \\[10pt] &= \dfrac{2}{2} = 1.
\end{aligned}

However, my first thought was $a_n = \sqrt{n+\sqrt{n}}-\sqrt{n-\sqrt{n}} = \sqrt{n}\left(\sqrt{1 + \dfrac{1}{\sqrt{n}}}\space - \sqrt{1 - \dfrac{1}{\sqrt{n}}}\right) = \sqrt{n}\space (1-1) = 0$

with the same argument as above. Is it correct that one can't make a statement about the convergence in the latter calculation because we have $\infty \cdot 0$ ? If yes, why exactly is this the case?

Edit: To see that $\lim\limits_{n\to \infty} \dfrac{1}{\sqrt{n}} = 0$ pick some arbitrary $\epsilon > 0$. We want to find a $N$ s.t. $\forall n\geq N\colon |\dfrac{1}{\sqrt{n}} - 0|< \epsilon \Longleftrightarrow \dfrac{1}{\sqrt{n}} < \epsilon \Longleftrightarrow n >\dfrac{1}{\epsilon^2}$ meaning we can choose $N = \dfrac{1}{\epsilon^2} + 1$ for example.


问:
Double integral variable substitution

Let $(a,b)\in (0,1)$ and let $T>0$. Consider the following integral:
$$
\iint_{\Gamma }f(x)g(x+t)\,dt\,dx,
$$
where
$
\Gamma =\left\{ (t,x)\in (0,T)\times (0,1):t+x\in (0,1)\right\} .
$

Let $s=x+t$, then $s\in (0,1)$ and $s-t\in (0,1)$, $t\in (0,T)$.
Am I right?
$$
\iint_{\Gamma }f(x)g(x+t)\,dt\,dx={\int_{0}^{1}\int_{0}^{\min(1-x,T)}}f(x)g(t+x)\,dt\,dx={\int_{0}^{1}\int_{0}^{\min(s,T)}}f(s-t)g(s)\,dt\,ds,
$$
Thank you.

答:
The substitution is fine. Here is a variation with an intermediate step.

We obtain
\begin{align*}
\color{blue}{\int_{x=0}^1}&\color{blue}{\int_{t=0}^{\min(1-x,T)}f(x)g(t+x)\,dt\,dx}\\
&=\int_{t=0}^{\min(T,1)}\int_{x=0}^{1-t}f(x)g(t+x)\,dx\,dt\tag{1}\\
&=\int_{t=0}^{\min{(T,1)}}\int_{s=t}^{1}f(s-t)g(s)\,ds\,dt\tag{2}\\
&\,\,\color{blue}{=\int_{s=0}^{1}\int_{t=0}^{\min{(s,T)}}f(s-t)g(s)\,dt\,ds}\tag{3}
\end{align*}
in accordance with OPs calculation.

Comment:

In (1) we change the order of integration.

In (2) we substitute $s=x+t, ds=dx$ and set upper and lower limit of the integral accordingly.

In (3) we change the order of integration once more.

Title: Double integral variable substitution


问:
When is commutative reduced ring a finite direct product of domains?
All rings considered are unital and commutative.

Intro Direct products of domains are reduced rings. The opposite is not true but I must admit I have troubles finding counter-examples. It holds that reduced rings are subdirect products of domains. But this is not that useful as many interesting ring-theoretic properties are not closed under subalgebras (e.g. being Artinian).

Question What are some general, ring-theoretic properties that characterize when a reduced ring is a finite direct product of domains? Are there at least some special cases (rings of small dimension, with zero socle, zero radical...) for which this has some nice characterization?

I do not need this for a specific purpose, and I am not sure whether this has some good, clear answer. It is just a question that is intrinsically interesting to me.

I am not sure how to start. An obvious necessary condition is that socle of the ring is zero (because domains that are not fields have zero socle). But I can't prove that this is sufficient, neither can I find a counterexample.

EDIT: Thanks Eric for pointing out that direct sum of fields is counterexample.  As a field is (the only) example of domain that has non-zero socle.

答:
A ring $R$ is a finite product of domains iff $R$ has finitely many minimal primes and for each maximal ideal $m\subset R$, the localization $R_m$ is a domain.  It is easy to see these properties are necessary (if $R\cong\prod R_i$ where each $R_i$ is a domain, the minimal primes are the kernels of the projections and every localization at a maximal ideal coincides with a localization of some $R_i$).

Conversely, suppose $R$ has finitely many minimal primes and its localization at each maximal ideal is a domain.  Suppose $p,q\subset R$ are two distinct minimal primes and suppose $p+q$ is a proper ideal.  Then we can extend $p+q$ to a maximal ideal $m$.  Since $p$ and $q$ are both contained in $m$, they give two distinct minimal prime ideals in the localization $R_m$.  But this is a contradiction, since $R_m$ is a domain.

Thus the minimal primes of $R$ are pairwise comaximal.  By the Chinese remainder theorem, this gives an isomorphism $R/\sqrt{0}\to \prod R/p_i$ where $p_i$ ranges over the minimal primes of $R$.  But the nilradical $\sqrt{0}$ is trivial, since $R$ is locally a domain and thus reduced.

(This is a basic and well-known result in scheme theory, though it is typically stated with "Noetherian" as a hypothesis rather than "finitely many minimal primes".  In geometric terms, if $\operatorname{Spec} R$ has finitely many irreducible components, then as long as those irreducible components are disjoint, $\operatorname{Spec} R$ will be the coproduct of the irreducible components, and so this gives a finite direct product decomposition of $R$.  But if two irreducible components intersect, then $R$ will not be locally integral at any point in the intersection.)


答:
Here is how you can distinguish between permutations and combinations. Suppose four people sit down on the bus in one way, then those same four people get up and change seats. Does this count as a different seating for you? If yes, you should use permutations (7P4). If not, you should use combinations (7C4).

This idea carries over to similar problems. If you are placing the items on a shelf (i.e. the order in which they appear matters), then permutations are appropriate. If you are placing the items in a bucket (i.e. the order in which they appear does not matter), then combinations are appropriate.

Body:
There are $7$ empty seats on a bus and four people get on. How many different ways can they be seated?

Would it be ${}^7C_{4}$ or ${}^7P_{4}$?


问:
Linear maps from $\mathbb{R}^{2}$ to $\mathbb{R}^{3}$ and cross products.
Let $S: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3}$ be a linear map, and let $T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ be an orthogonal map. Let $\{ e_{1}, e_{2} \}$ be the standard basis of $\mathbb{R}^{2}$.
Is it true that
\begin{equation*}
 \| S(e_{1}) \times S(e_{2}) \| = \| ST(e_{1}) \times ST(e_{2}) \|?
\end{equation*}
If the rank of $S$ is $0$ or $1$, then both sides of the equation are equal to zero. So that leaves the case that the rank of $S$ is $2$, but I got pretty much stuck there.
I feel like this should be true, because it feels like the quantity I'm trying to calculate is somehow a property of the linear map $S$, that shouldn't depend on the choice of basis, but maybe that's just wrong.

答: Yes, it is true. Suppose that the matrix of $S$ with respect to the canonical basis is$$\begin{bmatrix}a&b\\c&d\\e&f\end{bmatrix}.$$Then$$S(e_1)\times S(e_2)=(a,c,e)\times(b,d,f).$$On the other hand, if the matrix of $T$ with respect to the canonical basis is$$\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix},$$then the matrix of $S\circ T$ with respect to the canonical basis is$$\begin{bmatrix}a \cos (\theta )+b \sin (\theta ) & b \cos (\theta )-a \sin   (\theta ) \\ c \cos (\theta )+d \sin (\theta ) & d \cos (\theta )-c \sin   (\theta ) \\ e \cos (\theta )+f \sin (\theta ) & f \cos (\theta )-e \sin   (\theta )\end{bmatrix}.$$So,\begin{multline}\bigl\|S\bigl(T(e_1)\bigr)\times S\bigl(T(e_2)\bigr)\bigr\|=\\=\bigl((a,c,e)\cos(\theta)+(b,d,f)\sin(\theta)\bigr)\times\bigl(-(a,c,e)\sin(\theta)+(b,d,f)\cos(\theta)\bigr)=\\=(\cos^2\theta+\sin^2\theta)(a,c,e)\times(b,d,f)=(a,c,e)\times(b,d,f).\end{multline}If the matrix of $T$ with respect to the canonical basis is$$\begin{bmatrix}\cos\theta&\sin\theta\\\sin\theta&-\cos\theta\end{bmatrix},$$the computations are similar.


问:
Let $c$, $d$ $\in$ $\mathbb{N}^+$ be positive integers which are coprime. Show that $cd \mid a$, for an integer $a$.
I wrote a proof for this, not sure if it's correct though.

Let $c$, $d$ $\in$ $\mathbb{N}^+$ be positive integers which are coprime. Show that $cd \mid a$, for an integer $a$.

Assume $cd \mid a$

By the assumption, $a$ is some multiple of $cd$

So $a$ can be written as $cda$

Hence $cd \mid a$.

I'm not sure if this is correct because the question mentions that $c$ and $d$ are coprime. Any help appreciated.

答:
In a proof like this, you have to quantify what $a$ is, in terms of $c$ and $d$. In other words, you want to find an $a(c,d)\in\Bbb N$ such that for each $c,d\in \Bbb N$, it follows that $cd|a(c,d)$. Can you find this?

Extra Hint: Use the property that $$\{c|a \cap d|a\}\implies cd|a$$


问:
Real Analysis: Affine Maps and Closures of Sets
my task is to show that for a set $G \subset \mathbb R^n$, we have

$T(cl(conv(G))\subseteq cl(conv(T(G))$
where
$T$ is an affine map and $cl$ denotes closure of a set and $conv$ denotes the convex hull of a set.

Note that convex hull of a set is the set of all convex combinations of points from that set.
This seems easy on the surface but for some reason I get stuck.

Attempt at Proof:

Let $x \in T(cl(conv(G))$. Then $x=T(y)$ where $y \in cl(conv(G))$. This implies that $y=\lim_{k \rightarrow \infty} \alpha x_{k}+(1-\alpha)y_{k}$ where $x_{k} \subseteq G$ and $y_{k} \subseteq G$.

Hence, let $\alpha \in [0,1]$:
\begin{align*}
x&=T(\lim_{k \rightarrow \infty} \alpha x_{k}+(1-\alpha)y_{k})\\
&=\alpha T(\lim_{k \rightarrow \infty} x_{k})+(1-\alpha)T(\lim_{k \rightarrow \infty} y_{k})\\
\end{align*}

But then here I get stuck since I believe I am supposed to show $\lim_{k \rightarrow \infty}x_{k} \in G$ and $\lim_{k \rightarrow \infty}y_{k} \in G$.

But this isn't necessarily true since $G$ isn't necessarily closed so I'm confused. What mistake am I making? Thank you.

答:
Step 1 $T(cl(E)) \subseteq cl(T(E))$ for every set $E \subseteq \mathbb{R}^n$. If $y \in T(cl(E))$, let $y = T(\lim x_n)$. Since $T$ is continuous, $y = \lim T(x_n) \in cl(T(E))$.

Step 2 $T(conv(G)) \subseteq conv(T(G))$. Let $y \in T(conv(G))$, and $T(x) = a + Bx$ the affine map. Then $y = T(\alpha x + (1-\alpha)w) = a + \alpha B x + (1-\alpha)Bw = \alpha T(x) + (1-\alpha)Tw \in conv(T(G))$ with $x,w,\in G$.

Finally, apply the above with $E = conv(G)$ and use the fact that $cl(E) \subseteq cl(F)$ if $E \subseteq F$.


问:
Inverse of g(a, b) = (a + b, a - b)?
As the question title suggests, does $g(a, b) = (a + b, a - b)$ have an inverse? If so, what is it?

Title: Inverse of g(a, b) = (a + b, a - b)?

Body:
If $h(a,b) = (u(a,b), v(a,b))$ and we need $g(h(a,b)) = g(u(a,b), v(a,b)) = (u(a,b) + v(a,b), u(a,b) - v(a,b) = (a,b)$ and $h(g(a,b)) = h(a+b, a-b) = (u(a+b,a-b),v(a+b,a-b)) = (a,b) $

So we nut $u(a,b) + v(a,b) = a$ while $u(a,b) -v(a,b) = b$.  And we need $u(a+b, a-b) = a$ and $v(a+b, a-b)) = b$>

$u(a,b)+v(a,b) = a$ and $u(a,b) - v(a,b) = b$ so $[u(a,b)+v(a,b)] + [u(a,b) - v(a,b) ] = 2u(a,b) = a+b$.  So $u(a,b) = \frac {a+b}2$.

And  $[u(a,b)+v(a,b)] -[u(a,b) - v(a,b) ] = 2v(a,b) = a-b$.  So $v(a,b) = \frac {a-b}2$.

So $h(a,b) = g^{-1}(a,b) = (\frac {a+b}2, \frac {a-b}2)$.

We can verify that.

$g(\frac {a+b}2, \frac {a-b}2 ) = (\frac {a+b}2 + \frac {a-b}2, \frac {a+b}2 - \frac {a-b} 2) = (a,b)$.

And $g^{-1}(g(a,b)) = g^{-1}(a+b, a-b) = (\frac {(a+b) + (a-b)}2, \frac {(a+b)-(a-b)}2)= (\frac {2a}2, \frac {2b}2) = (a,b)$.

It holds.


问:
Proof by contradiction and invalid reasoning
I am very aware of the several posts that are in this forum, but to be honest I am not quite satisfied with the provided answers. Also, I can tell that the best post is the following:

Proof by Contradiction, Circular Reasoning?

I would love to continue the discussion in that post, however my reputation level does not allow me to add any comments, and I still have some questions. Therefore, I want to continue discussing more about proof by contradiction and circular reasoning.

Suppose we wish to prove $(P \wedge Q) \rightarrow R$. We assume P and Q. Proof by contradiction begins by assuming $\neg R$, and from these two assumptions, a "contradiction" is derived, and sometimes that contradiction is that $\neg P$ or $\neg Q$ (our initial valid assumptions). Therefore, since there is a contradiction with our assumptions, we conclude that it cannot happen that $\neg R$, so it must be the case that $R$.

This proof strategy bothers me, because I have the feeling that there is some kind of circular (or invalid) reasoning going on there i.e. $(P \wedge Q) \rightarrow R$ is equivalent to $\neg R \rightarrow \neg(P \wedge Q)$, so if we already know that the consequent is false, in order for the statement to be true, $\neg R$ has to be false, so R is true, but R is part of the consequent part of the conditional statement (isn't doing this like assuming there is a biconditional i.e. because R is true, then $P \wedge Q$ is true?). I probably don't know how to explain it better, but how/why can we say that this is indeed a proof? Hence, my main question is why it is valid to assume $\neg R$ if our conclusion involves R? I have one explanation that kind of makes sense to me (still a little iffy though):

In more simple terms, where we want to prove that $P \rightarrow Q$, we are trying to do the following: $[(P \wedge \neg Q) \rightarrow Contradiction] \rightarrow (P \rightarrow Q)$ if we can prove that $(P \wedge \neg Q)$ leads to a Contradiction, then we are indeed proving that $P \rightarrow Q$, which was our original goal. Given that this is a tautology, whenever we follow this proof strategy structure, we are always proving $P \rightarrow Q$. However, saying that this is a tautology kind of bothers me for some reason.

I would sincerely appreciate your insights, and if there is anything you would improve in my post, please feel free to do so. I'm starting my formal journey in the proof realm (currently reading How To Prove It), and I have always heard that proofs by contradiction were the easiest strategy. However, I have learned that this strategy seems to be difficult at the beginning for all those who are trying to learn proof strategies.

答:
There is nothing circular about the method of proof by contradiction, but the method does assume a principle called the law of the excluded middle (LEM) that is not accepted by mathematicians and logicians who adopt the position of intuitionism rather than classical logic. LEM asserts that for any proposition $A$, $A \lor \lnot A$ is true.

In classical logic, if you are trying to prove some proposition $R$, then as you "know" that $R \lor \lnot R$ is true, it is valid to assume that $\lnot R$ is true (otherwise, for $R \lor \lnot R$ to be true, we must have that $R$ is true already). Intuitionistic logic does not consider this line of reasoning to be valid, unless we can prove for this specific $R$ that $R \lor \lnot R$ is true. For an intuitionist, LEM does not apply to a proposition like the abc conjecture that has neither been proved nor disproved.


问:
Return tuple from intersection between tuple and set
I'm interested in the problem from this post, only I would like to return a tuple instead of a set.

Specifically, if $\theta = (3, 4, 2)$ and $E = \{1, 2, 3\}$, then I'd like to get $(3, 2)$ where the order of the intersecting elements is preserved from the original tuple.

Based on the linked post above, I feel like

$(3, 2) = (\theta_i|\theta_i \in \theta, \theta_i \in E)$

wouldn't work. Is there a nice notation that would allow me to achieve what I'm looking for? I also have seen this answer, but I'm afraid of putting too much reliance on my audience knowing the details of set notation.

答:
I think that what you suggest would work, but recommend not relying on anything like it. Don't make the reader work harder than necessary. Invent your own notation, perhaps
$$
\theta \sqcap E
$$
along with a definition (mostly in words) and an example like the one you provided here.


问:
Why a sequence of functions cannot be dominated by an integrable function
I have $$f_n(x)=\begin{cases}
      \frac{1}{n} & |x|\leq n, \\
      0 & |x|>n .
   \end{cases}$$

Why cannot be dominated by an integrable function $g$ by the Dominated Convergence Theorem? I am also wondering what exactly it means for a function $g$ to dominate a sequence of functions, since I believe my definition of this is what I am understanding incorrectly.

答:
Well, say $g(x)\geq |f_n(x)|$ for every $x$ and $n$. Then,
$$
g(x)\geq \sup_n|f_n(x)|=\begin{cases} \frac{1}{n} & |x|\in (n-1,n] \\ 0 & x=0 \end{cases}
$$
and thus,
$$
\int_{\mathbb{R}}g(x)\textrm{d}x=2\sum_{n=1}^{\infty} \frac{1}{n}=\infty
$$
and hence, $g$ cannot be integrable.


答: $f_n\to 0$ on $\mathbb{R}$, so if the $f_n$ were dominated by an integrable function, DCT would give $$\lim_{n\to\infty}\int_{\mathbb{R}}f_n(x)\,\mathrm{d}x=0$$ whereas it's obvious that the limit is $1$.

Title: Why a sequence of functions cannot be dominated by an integrable function

Body:
I have $$f_n(x)=\begin{cases}
      \frac{1}{n} & |x|\leq n, \\
      0 & |x|>n .
   \end{cases}$$

Why cannot be dominated by an integrable function $g$ by the Dominated Convergence Theorem? I am also wondering what exactly it means for a function $g$ to dominate a sequence of functions, since I believe my definition of this is what I am understanding incorrectly.


答: In order to dominate a sequence of functions, it has to be larger than ALL of the functions in the entire sequence.  There is no such $g$ for your sequence of functions, because no matter how large you set $g$ near $0$, it will eventually be exceeded by all the terms in the sequence beyond some point.


问:
Is the "derivative limit theorem" true  in the higher dimension?
<s>I am wondering if the following statement is a canonical theorem in real analysis. Does anybody here know the exact reference for it? It maybe a corollary of the mean value theorem, I think. Motivated by the answer to this question, I curious about that if this statement is still true in the higher dimension case.</s>

The following statement is a corollary of the Mean Value Theorem.

Let $x_0\in{\bf R}$ and $U(x_0)$ be
  the neighborhood of $x_0$.

$$f:U(x_0)\to{\bf R}$$

is continuous on $U(x_0)$ and
  differentiable on $U(x_0)\setminus\{x_0\}$.
   If the limit

$$\lim_{x\to x_0;x\in
U(x_0)\setminus\{x_0\}}f'(x)$$

exists, then $f$ is differentiable at
  $x_0$ and

$$f'(x_0)=\lim_{x\to x_0;x\in
U(x_0)\setminus\{x_0\}}f'(x)$$

Here is my question:

Is this statement still true for the higher dimension if
$$f:U(x_0)\subset{\mathbb R}^n\to{\mathbb R}^m ?$$

Since I don't know any generalization of the MVT in the higher dimension, I think one may need to approach it directly by definition. On the other hand, if one can deal with it "component-wise", one may be able to reduce it to the one dimension case.

答:
It is, in fact, a consequence of the mean value theorem ; supposing your neighborhood contains an open interval centered on $x_0$, call the limit of $f'(c)$ to be $L$, take $x$ in this interval ; hence there exists $c$ such that
$$
f(x)-f(x_0) = f'(c)(x-x_0)  \quad \Rightarrow \quad \frac{f(x) - f(x_0)}{x-x_0} = f'(c) \to L(x_0)
$$
because $c \to x_0$ when $x \to x_0$ and  $f'(c) \to L(x_0) = f'(x_0)$. There is a way of seeing this in the higher dimensional case ; in $\mathbb R^n$, use the same way of thinking, but instead of calling it the mean value theorem, use Taylor's Theorem which says that there exists a $c \in [x_0, x] \overset{def}{=} \{ y \in \mathbb R^n \, | \, y = \lambda x_0 + (1-\lambda) x_0, \, \lambda \in [0,1] \}$ such that
\begin{align*}
f(x) - f(x_0) &= f'(c)(x-x_0)  \quad \Longrightarrow \quad \frac{f(x) - f(x_0) - L(x_0)(x-x_0)}{\| x-x_0 \|} \\ &= (f'(c) - L(x_0)) \left( \frac{x-x_0}{\| x-x_0 \| } \right) \to 0.
\end{align*}
My arguments are a little sketchy ; I'm just giving you an idea of what it looks like. You need to watch out what happens in higher dimension to what happens when you evaluate that the last term goes to $0$.</p>


问:
Surjective map between modules and generators
Suppose that $ \phi : N \to M$ is a surjective $R$-linear map. Show that if $N$ is finitely generated then so is $M$; and on the other hand if $M$ and ker $\phi$ are finitely generated then $N$ is finitely generated.

Here we are taking $M$ and $N$ to be $R$-modules. My initial thought about the first part of this problem is to consider the $(n_i)_{i \in I}$ (for a finite indexing set I) that generates $N$ and to use surjectivity to get the set $(m_i)$ that generate $M$ but I don't know if this is the right way to go or how to formalise this. For the second part, could the first isomorphism theorem be used and if so, how?

答:
Your idea is good. Suppose $\{n_1,\dots,n_k\}$ is a finite set of generators of $N$. If $y\in M$, there is $x\in N$ such that $y=\phi(x)$.

By assumption, $x=r_1n_1+\dots+r_kn_k$ and therefore
$$
y=\phi(x)=r_1\phi(n_1)+\dots+r_k\phi(n_k)
$$
which proves that $\{\phi(n_1),\dots,\phi(n_k)\}$ is a set of generators for $M$.

What about the other direction? Consider $\{n_1,\dots,n_k\}$ in $N$ such that $\{\phi(n_1),\dots,\phi(n_k)\}$ is a set of generators for $M$. This is possible because $M$ is, by assumption, finitely generated and $\phi$ is surjective. Take also a set of generators $\{v_1,\dots,v_l\}$ of $\ker\phi$.

Let $x\in N$. Then
$$
\phi(x)=r_1\phi(n_1)+\dots+r_k\phi(n_k)
$$
Consider $x'=r_1n_1+\dots+r_kn_k$. Observe that $x-x'\in\ker\phi$ and finish up.


问:
Surprisingly elementary and direct proofs
What are some examples of theorems, whose first proof was quite hard and sophisticated, perhaps using some other deep theorems of some theory, before years later surprisingly a quite elementary, direct, perhaps even short proof has been found?

A related question is MO/24913, which deals with hard theorems whose proofs were simplified by the development of more sophisticated theories. But I would like to see examples where this wasn't necessary, but rather the theory turned out to be superfluous as for the proof of the theorem. I expect that this didn't happen so often. [Ok after reading all the answers, it obviously happened all the time!]

答:
Gauss's initial proof of quadratic reciprocity in Disquisitiones was exceptionally long spanning I believe 30+ pages, and in my opinion it is incredibly nasty. Gauss used induction and then reduced to a something like eight cases. While not necessarily relying on big theorems this proof seems overly complicated especially when compared to some of the proofs that followed.

For example, Eisenstein's proof is quite short and has a very nice geometric component that I think is very nice, and it is does not require a great deal more machinery than Gauss's original.

There are in fact may other nice proofs of quadratic reciprocity.  Here's a nice list of some of them


问:
Determine the value of each of the constants $a$, $b$, $c$, and $d$ in the identity $a(x+b)^3+c\equiv4x^3-24x^2+48x+d$

Determine the value of each of the constants $a$, $b$, $c$, and $d$ in the identity

$$a(x+b)^3+c\equiv4x^3-24x^2+48x+d$$

I have already found $a=4$ and $b=-2$ but I'm struggling to find $c$ and $d$.

The answers for $c$ and $d$ are $c=29$, $d=-3$.

答: $4x^3 - 24x^2+48x + d = 4(x^3 - 6x^2 + 12x - 8) + 32+ d = 4(x-2)^3 + 32+ d\implies a = 4, b = -2, c = 32+d$. Note that there are more than one right answer for this. You can take for example $c = 3, d = -29$ or $c = 0, d = -32$ as they are just two answers among many !


问:
Eigenvalues of a high dimensional sample covariance matrix
I am not a math/stat major, but I encountered high dimensional matrix in my research. Particularly, consider a sample covariance matrix of the form
$S_{X}=\frac{1}{T}\sum_{t=1}^{T}\mathbf{X}_{t}\mathbf{X}_{t}^{T},$ where $\mathbf{X}_{t}=\left( x_{1t},\ldots ,x_{Nt}\right) ^T.$

Assuming both $N$ and $T$ are large, then $S_{X}$ is a $N\times N$ high
dimensional sample covariance matrix. Since the elements in $\mathbf{X}_{t}$
could be correlated, $S_{X}$ may not be a diagonal matrix. My question is
what's the eigenvalues of $S_{X}$? Are there any references to find the
properties of the eigenvalues of $S_{X}?$ The results I need is that the
maximum eigenvalues of $S_{X}$ is bounded for large $N$ and $T.$ Thanks!

答:
With normal 0 mean assumptions, $TS_x$ is Wishart distributed and the joint distribution of eigenvalues is known in some case (see wiki). Some result about the dominant eigenvalue distribution was derived in

Khatri, C. G. "Non-central distributions of i th largest characteristic roots of three matrices concerning complex multivariate normal populations." Annals of the Institute of Statistical Mathematics 21.1 (1969): 23-32.

Also this

Chiani, Marco. "Distribution of the largest eigenvalue for real Wishart and Gaussian random matrices and a simple approximation for the Tracy–Widom distribution." Journal of Multivariate Analysis 129 (2014): 69-81..

You might start here.


问:
How to find the periodicity of a cycle through iterative assignments?
First, I'm not a mathematician by trade so if the terminology I'm using can be made more precise, please don't hesitate to improve this post.

Now that that's out of the way, this question looks like a more specific form of Iterated function 'periodicity'. Specifically, if I have some function of the form $f(x)=(ax+c) \space mod \space m$ for some $a$, $c$, and $m$, how would I go about finding how many iterated compositions of the function (ie $n$ in $f^n$ where $f^n$ denotes the iteration of composition, e.g. $f^3(x)=(f\circ f\circ f)(x)$) will be the same as $f(x)$? Even more specifically, how would I know if $n=m$ (especially if $m$ is not prime)?

Context: I'm trying to generate a sequence of numbers such that all integers within $[0, m)$ are used and would prefer the given sequence not be easily predictable (although this doesn't have to be cryptographically secure).

答:
These are called linear congruential generators, so you'd probably be able to find a lot of information using that term. You won't get every integer from $0$ to $m-1$ this way, but you can get most of them (if $m$ is a prime then you'll get all but one such integer, for example, except in the case where $a\equiv1\pmod m$ which is a horrible random-number generator).

For the general situation of having an iterated function and wanting to find its period length on a given initial value: a standard observation is that it suffices to find a positive integer $k$ such that the $k$th iterate equals the $2k$th iterate (for then the period will be a divisor of $k$). This observation, which I learned in the context of the "Pollard rho algorithm" for factoring integers, is a lot easier to do in practice than comparing the $k$th iterate against all of the 1st, 2nd, ..., $(k-1)$st iterates every time.


问:
Proving the inequality $\limsup (a_n+b_n)\le \limsup(a_n) +\limsup(b_n)$.
Definition: $\limsup x_n$= supremum of all subsequential limits of sequence $(x_n)=\sup E_x$, where $E_x$ is set of all subsequential limits of sequence $(x_n)$. Let $x^*=\limsup x_n$.

Given any two sequences $(a_n)$ and $(b_n)$, it is to be proven that: $\limsup (a_n+b_n)\le \limsup(a_n) +\limsup(b_n)$, provided sum on right side is not of the form $\infty -\infty$. $\tag 1$

I tried to prove it as follows:

Let $c_n=a_n+b_n$ for all $n\in \mathbb N$

Case $1: a^*, b^*, c^* \in \mathbb R$

It follows that the sequences $(a_n),(b_n),(c_n)$ are bounded. And since set of all subsequential limits of a sequence is closed, it follows that $a^*\in E_a, b^*\in E_b, c^*\in E_c\implies$ There exist subsequences $(a_{n_k}), (b_{n_l})$ and $(c_{n_m})$ such that $a_{n_k}\to a^*, b_{n_l}\to b^*, c_{n_m}\to c^* $.

For any $\epsilon \gt 0, \exists s_\epsilon\in \mathbb N$ such that $s\ge s_\epsilon \implies |a_{n_s}-a^*|\lt \epsilon/4 \space, |b_{n_s}-b^*|\lt \epsilon/4 \implies |a_{n_s}+b_{n_s}-a^*-b^*|\lt \epsilon/2$.

Now suppose on the contrary that $c^*\gt a^*+b^*$. We have: for any $\epsilon \gt 0$

$\exists M_\epsilon \in \mathbb N: m\ge M_\epsilon\implies |c_{n_m}-c^*|\lt \epsilon/2.$

Hence for $r\in \mathbb N: r\ge \sup(M_\epsilon, s_\epsilon),$ we have:

$0\lt c^*-a^*-b^*=|c^*-(a_{n_r}+b_{n_r})+ (a_{n_r}+b_{n_r}-a^*-b^*)|\le |c^*-(a_{n_r}+b_{n_r})|+ |(a_{n_r}+b_{n_r}-a^*-b^*)|\lt\epsilon \implies c^*=a^*+b^* $, which is contradiction. Therefore, $c^*\le a^*+b^*$.

Case $2$: $a^*=\infty, b^*\in \mathbb R\cup \{\infty\},$ then the result is trivial.

Case $3$: $b^*=\infty, a^*\in \mathbb R\cup \{\infty\}$, then as in Case $3$, the result is true.

Case $4$: $a^*\in \mathbb R, b^*=-\infty .$

$E_b=\{-\infty\}$ and $ (a_n)$ is bounded by some $M\gt 0$. So, $(c_n)$ is not bounded below as $(b_n)$ is not bounded below.
 So for some $M\in \mathbb N$, $n\ge M\implies c_n\lt a^*+b_n\implies c^*\le a^*+b^* $.

Case $5$: $b^*\in \mathbb R, a^*=-\infty .$

Same as case $4$.

So proved.

My question is what happens when right hand side of $(1)$ is $\infty -\infty $?

Consider for example: $(a_n)=(n), (b_n)=(-n)$, clearly $c_n=0\to \limsup c_n=0$. But can we claim that $0\gt \infty -\infty $ (I don't think so as $\infty -\infty $ is undefined.)?. If not, then $(1)$ always holds?

Please help. Thanks.

答:
I am posting an answer based on inputs by Mr. Martin R:

If exactly one of $a^*$ or $b^*$ is infinite, we are done. If both $a^*$, $b^*$ are infinity of same sign, then also we are done.

We consider the case, when $a^*, b^*$ and $c^*$ are finite. Since $c^*\in E_c, \exists$ subsequence $c_{n_k}\to c^*$. That is $c_{n_k}=a_{n_k}+b_{n_k} \to c^* $.

$(a_{n_k})$ is bounded and therefore has convergent subsequence $(a_{n_{k_l}}): (a_{n_{k_l}})\to \alpha$. Therefore, $b_{n_{k_l}}=(a_{n_{k_l}}+b_{n_{k_l}})-(a_{n_{k_l}})\to c^*-\alpha$.

Now, $a_{n_{k_l}}+b_{n_{k_l}}\to (\alpha)+(c^*-\alpha)\le a^*+b^*\implies c^*\le a^*+b^*$.
 Proved.


问:
Where digit sum matches digit sum of the square - formal name?
Just came across this.

The UK equivalent is 999, for which the square is 998,001 - which interestingly shares the same digit sum.  I just wonder if this has a name and/or any literature?  I note there are some applications of digit sum.

答:
[Made from comment to answer at OP's request]

See OEIS sequence A058369 and Hare, Laishram & Stoll reference there.


问:
Expected value of the integral of a Brownian excursion
Let $\mathrm{B}_e(t)$ be a Brownian excursion on $[0,1]$. I would like to find
$$ \mathbb{E}\left[ \int_0^1 \mathrm{B}_e(t) dt \right]$$
In this post it is explained that the integral of a Brownian motion is Gaussian. Following the answers it states that if $\mathrm{B}(t)$ is a Brownian motion, then
$$ \int_0^\tau \mathrm{B}(t) dt \sim N\left(0, \int_{0}^{\tau}(\tau-s)^{2} d s\right) $$
That would be $\sim N\left(0, \frac{1}{3}\right)$ on $[0,1]$.

But here we are dealing with an excursion, intuitively the integral will not be $0$ as it is positive everywhere on $(0,1)$. So i.e. my question is if such a relation exists for a Brownian excursion.

答:
Let $\def\B{\mathrm{B}}\B(t)$ be usual Brownian motion, and let $\B_b(t)$ be the Brownian bridge. Explcitly, $$\B_b(t)=\B(t)-tB(t).$$

It can be shown (see A Relation between Brownian Bridge and Brownian Excursion) that
$$
\B_e(t)\stackrel{D}= B_b(\tau_m+t\!\!\mod 1)-B_b(\tau_m),
$$
where $\tau_m$ is the time when $\B_b$ attains its minimum. It follows that
$$
E\left[\int_0^1\B_e(t)\,dt\right]=E\left[\int_0^1\B_b(t)\,dt\right]-E\left[\B_b(\tau_m)]=0-E[\B_b(\tau_m)\right],
$$
Therefore, the expectation you want is the opposite of the expected minimum of a Brownian bridge, or equivalently the expected maximum of a Brownian bridge.

Let $M=\sup_{0\le t\le 1}\B_b(t)$ be this maximum. Using the reflection principle, you can show that
$$
P(M> m)=P(\B_1=2m\mid \B_1=0)=\frac{
\frac1{\sqrt{2\pi}}\exp(-(2m)^2/2)}{\frac1{\sqrt{2\pi}}\exp(-0^2/2)}=\exp(-2m^2).
$$
Namely, if you take a Brownian bridge and reflect it after the time it reaches $+m$, you get a Brownian motion path which ends at $2m$. Conversely, any path with reaches $2m$ reflects to a bridge which attains $m$. Of course, conditioning on a zero probability event is not allowed, but the above is the intuition which can be made rigorous.

Putting this altogether,
$$
E\left[\int_0^1\B_e(t)\,dt\right]=E[M]=\int_0^\infty P(M>m)=\int_0^\infty \exp(-2m^2)\,dm=\sqrt{\pi/8}.
$$


问:
$f$ continuous and bounded on $\epsilon$ cover of $S$ implies $f$ continuous and bounded on $S$
I am trying to prove this statement:

Let:

$X,Y$ be compact sets

$d$ be a distance metric on $X$

$S\subset X$ be closed and bounded

$f:X\rightarrow Y$ be a function

$B_{\delta}(x)$ denote the open $\delta$-ball around $x\in X$, defined by $B_{\delta}(x)=\{x':d(x,x')<\delta\}$.

If;

$ \forall s\in S, \exists \epsilon_s>0$ such that $f$ is continuous and bounded on $B_{\epsilon_s}(s)$

Then;

$f$ is continuous and bounded on $S$.

I was given this in an exercise from some old lecture notes (I don't think I am allowed to share them on here as they were someones private notes). I think it needs an $\epsilon-\delta$ argument as it must use the fact that the balls are open and therefore they must overlap.

I thought about using some sort of boundedness theorem and saying a union of the balls contains a closed region over which $f$ must be continous and therefore bounded. But I am confused and don't think this works. I have spent a lot of time on this already and feel like I am now going back and forth on the argument.

答:
For the continuity there is no problem, as it is a local feature. More precisely, let $s \in S$. For every $V$ neighborhood of $f(s)$ there exists $U$ neighborhood of $s$ inside $B_{\epsilon_s}(s)$ such that $f(U) \subset V$. Then $U$ is also a neighborhood of $s$ in $S$, so $f : S\to Y$ is continuous.

For every $s \in S$ let $\epsilon_s$ as in the hypothesis. Then $\{B_{\epsilon_s}\}_{s \in S}$ is an open cover of $S$, and we can select $s_1, \dotsc, s_n$ finitely many points such that $S \subset B_{\epsilon_{s_1}} \cup \dotsb \cup B_{\epsilon_{s_n}}$. Hence, $f$ will be continuous and bounded on $B_{\epsilon_{s_k}}$ for every $k = 1, \dotsc, n$.

Let $M_k = \max \{|f(x)| : x \in B_{\epsilon_{s_k}}\}$ and $M = \max \{M_1, \dotsc, M_n\}$. Then $|f(x)| \leq M$ for every $x \in S$, since every $x$ belongs to one of the selected balls.


问:
Let $H$ and $K$ be subgroups of $G$. Prove that if $G = HK$, then for any $x, y \in G$ $xH \cap yK \neq \emptyset$.
It's easy to show that if $G = HK$, $HK = KH$. Thus $xH = kH$ for some $k \in K$ and $yK = hK$ for some $h \in H$. So the question is narrowed to proving that $kH \cap hK \neq \emptyset$. Unfortunately, I can't go any further.

答: Hint: you are very close. If $G = HK$, then $G$ is the union of the cosets $hK$ for $h\in H$, so any coset $zK$ is $hK$ for some $h \in H$. But then $xH \cap yK = x(H \cap (x^{-1}y)K) = x(H \cap h K)$ for some $h \in H$ and so, as $h = he \in H \cap hK$, we have $xh \in xH \cap yK$. So $xH \cap yK \neq \emptyset$


问:
Span of an empty list
Its a simple question . But still I thought of asking it coz I don't see the logic behind it. Recently I read a statement "the Span of empty list ( ) equals {0}" .What we know is an empty list has no elements in it,so how can we even talk of a span of it as span is a linear combination of all the vectors in a list and in empty list no vector is present at all. So, is it kind of just a declaration that we make just like that?? i guess there is no logic behind it and we just assume it like that. But why? What's the reason of doing so? And why do we need to declare a span of an empty list anyway?? Sorry I am not getting the logic behind it and that's why I thought of asking you guys. Thanks.</p>

答:
The singleton $V=\{0\}$ satisfies all of the axioms of a vector space over any field, when addition and scalar multiplication are defined in the only way they can be.  What are the natural answers to the following questions?

Is $V$ finite-dimensional?  If so, what is its dimension?

Since $V$ is a subspace of all vector spaces (including one-dimensional vector spaces), it makes sense to answer that $V$ is indeed finite-dimensional and that its dimension is $0$.  Next, if the dimension of a vector space is equal to the cardinality of a basis, we are only left with the option of declaring the set $\varnothing$ to be a basis for $V$.  Since a basis is a spanning set, it follows that we have $\operatorname{Span}\varnothing=\{0\}$.

Title: Span of an empty list


答: Set $a=4$, $b=-2$, $c=32+t$ and $d=t$ and you'll get that the identity is true for all $t$ which means the solution is not unique, in other words, there's no enough information to get the supposed answer you were given.

Title: Determine the value of each of the constants $a$, $b$, $c$, and $d$ in the identity $a(x+b)^3+c\equiv4x^3-24x^2+48x+d$

Body:
Determine the value of each of the constants $a$, $b$, $c$, and $d$ in the identity

$$a(x+b)^3+c\equiv4x^3-24x^2+48x+d$$

I have already found $a=4$ and $b=-2$ but I'm struggling to find $c$ and $d$.

The answers for $c$ and $d$ are $c=29$, $d=-3$.


问:
Constructing convergent sequences in p-adic numbers
I am trying to find a prime p such that there exists a sequence of integers S_n where (S_n)^2 converges to 21 under the p-adic norm.

I'm familiar with the definition of p-adic distance and with the notion of convergence that applies. One idea I've had is that we're looking for primes p where 21 is a quadratic residue (mod p). I thought of this because we're looking at the squared terms in the sequence, but I'm not sure if it's helpful.

I can intuitively understand the idea of convergence to 0 in the p-adic norm (i.e. 5^x converges to 0 in 5-adic distance as x goes to infinity), but I'm having trouble extending this thinking to a non-zero convergence. Hints would be highly appreciated.

答:
For any prime $p$ integer $x^2$, $x^2$ is close to $21$ in $p$-adic norm if $x^2 - 21$ is divisible by a large power of $p$.

Well, let's try $p=5$.

$21 - 1^2$ is divisible by $5^1$.

$21 - (5 k + 1)^2 \equiv 15 k + 20 \mod 5^2$; this is $0$ if $k = 2$, so $5k+1 = 11$.

$21 - (5^2 k + 11)^2 \equiv 75 k + 25 \mod 5^3$; this is $0$ if $k = 3$, so $5^2 k + 11 = 86$.

Can you see how to continue?  Now look up Hensel lifting.


问:
Find the CDF and PDF of X
We produce a real number X through a two-stage experiment.

First roll a fair die to get an outcome $Y\in\{1,2,...,6\}$.

Then if $Y=k$, choose a point uniformly at random in $[0,k]$, denote it X.

What is the c.d.f. of X? What is the p.d.f. of X?

My Attempt:

The c.d.f. is defined as $F(t)=\mathbb{P}(X\leq t)$, so

\begin{equation}
F(t)=
\begin{cases}
0&\text{if } t<0\\
\frac{1}{6}\cdot\frac{1}{k-0}&\text{if }t\leq k\\
1&\text{if } ???
\end{cases}
\end{equation}

My thinking on how to get the probability is that there is a $\frac{1}{6}$ chance of rolling a number Y and then there is a $\frac{1}{k-0}$ chance of choosing a random number in $[0,k]$ because it is uniformly distributed. I am not sure what conditions of $t$ would cause a probability of $1$.

The p.d.f can be found by differentiating the c.d.f., but since there would be 6 different functions for $k\in\{1,2,...,6\}$, I am unsure how to combine them to differentiate. Should I sum them?

答:
Your computation of $F$ is wrong. There should not be  any $k$ in $F(t)$.

$P(X \leq t)=\sum_{\{k: t \leq k\}} P(Y=k) \frac t k=\frac 1  6 \sum_{\{k: t \leq k\}}  \frac t k$. [The sum is taken over $k$ such that $t \leq k$].

Now split this into the cases $0 \leq t \leq 1, 1<t\leq 2,...,5 <t \leq 6$. In the first case you get $f(t)=\frac  1 6  \sum\limits_{k=1}^{n} \frac t k$. In the second case you get $f(t)=\frac  1 6  \sum\limits_{k=2}^{n} \frac t k$, and so on. Now differentiate w.r.t. $t$ to find the density function of $X$.

Title: Find the CDF and PDF of X


问:
Two-dimensional recurrence relation similar to that of Bessel numbers
I am trying to determine the solution for the two-dimensional recurrence relation
$$
C_{n+1,k}=C_{n,k}+nC_{n-1,k-1}
$$
I've noticed from this paper (1.3) that this is very close to that of the Bessel numbers

$$
B_{n+1,k}=B_{n,k-1}+nB_{n-1,k-1}
$$
with the difference being the second index in the $n$ term.

Is there a resource which gives the solution to the first recurrence relation? Are they known by any name?

EDIT:
The initial values are $C_{n,0}=1$ and $C_{n,k}=0$ for $2k>n$.

答:
I plugged the numbers determined by your sequence, row by row, into OEIS, and found https://oeis.org/A100861, which gives the formula
$$
C_{n,k}=\frac{n!}{k!(n-2k)!2^k}
$$
They are also called the Bessel numbers, and indeed each row is just a shift of the rows for $B_{n,k}$.


问:
What does it mean when one says "$B$ has no limit point in $A$"?
If $B$ is a subset of a set $A$, what does the sentence "$B$ has no limit points in $A$" mean? I am aware that $x$ is a limit point of $A$, if for every neighbourhood $U$ of $x$, $(U-\{x\})\cap A$ is non-empty.

Please let me know. Thank you.

答:
Giving an example may be helpful for you.

Example: Let $X=\mathbb R$ with usual topology. $A=\{x\in \mathbb R: x > 0\} \subseteq X$ and $B=\{1, \frac12, \frac13,... \frac1n,...\} \subset A$. It is not difficult to see that $B$ has no limit points in $A$ since the unique limit point 0 of $B$ is not in $A$.


问:
Combo problem, complementary counting
You have 5 blue nails in a column and 3 red nails in another column. You can attach a string between any red nail and blue nail. How many ways can we attach these strings such that every nail has at least 1 string attached to it?

For example, drawing all possible strings would work. Or connecting the first 3 blue with first 3 red, then connecting the last 2 blue to any red would work.

I approached this using Principle of Inclusion-Exclusion (if there is a simpler solution, let me know), but I am stuck. All I have is

$(2^5-1)^3$-something

$(2^5-1)^3$ is the number of non-empty subsets of blue nails for the red nails to connect to, cubed because 3 red nails. I don't know what to subtract from this though

答:
To figure out what to subtract, figure out what you've counted that you don't want to count.

You've counted every configuration that uses all 3 red nails, but you've included configurations that don't use all the blue nails. So count those (e.g. the ones that use only four blue nails) and subtract them. Using the inclusion-exclusion principle, you can correctly account for the configurations that use 3 blue nails, etc.

Note that this will be easier if you do it in the other direction, with cases based on the number of red nails used.

Title: Combo problem, complementary counting


问:
Find the sum of the following infinite series. Don't know how to separate the denominators out.
Find the sum of the following infinite series:
$$\frac{1}{6}+\frac{13}{6^2}+\frac{19}{6^3}+\frac{97}{6^4}+\cdots+\frac{3^n+(-2)^n}{6^n}+\cdots$$

答: Hint: $\dfrac{3^n+(-2)^n}{6^n} = \left(\dfrac{3}{6}\right)^n + \left(\dfrac{-2}{6}\right)^n= \left(\dfrac{1}{2}\right)^n+\left(\dfrac{-1}{3}\right)^n$. Are you able to take it from here?


答:
Too long for a comment.

I think Bill Johnson's answer (Lomonosov 1973) on MO applies to your question as well. Some superfluous theory that was used for the first proof of a weaker result (Bernstein-Robinson 1966) was nonstandard analysis. It was Halmos who immediately removed the nonstandard analysis from the argument.

These things were certainly not easy to figure out because von Neumann himself only proved the existence of nontrivial invariant subspaces for compact operators on Hilbert space ($\dim \geq 2$). Lomonosov entails hyperinvariant ones on general Banach spaces.

See this wikipedia page on the invariant subspace problem for statements and a chronology of related results. Maybe that's a good opportunity to recall that it is still not known whether every bounded operator on an infinite-dimensional separable Hilbert space admits a nontrivial invariant subspace.

Title: Surprisingly elementary and direct proofs

Body:
What are some examples of theorems, whose first proof was quite hard and sophisticated, perhaps using some other deep theorems of some theory, before years later surprisingly a quite elementary, direct, perhaps even short proof has been found?

A related question is MO/24913, which deals with hard theorems whose proofs were simplified by the development of more sophisticated theories. But I would like to see examples where this wasn't necessary, but rather the theory turned out to be superfluous as for the proof of the theorem. I expect that this didn't happen so often. [Ok after reading all the answers, it obviously happened all the time!]