问:
How to Show Representations of an Infinite Group are not Equivalent?
I have unitary representations of an infinite group (but finitely generated) on an infinite dimensional Hilbert space. I would like to show that two representations are not equivalent. I am guessing I would need to find some invariant property of an representation and then show the two representations do not share the same invariant property. However after quite a bit of searching on the internet, I could not find any invariants that apply to representations on an infinite group. All I could find were invariants of representations of finite groups.

Are there are any invariants of representations of infinite groups? Are there any other methods of proving two representations are not equivalent? Also in general, I would like to learn more about representations of infinite groups. Could you recommend some resources I could look at?

答:
To fix notation let $u_1$ and $u_2$ be two unitary representations of the group $G$ on Hilbert spaces $H_1$ and $H_2$.

One of the most elementary reasons why $u_1$ and $u_2$ might fail to be unitarily equivalent is spectra:  if there is an element
$g$ in $G$ such that the spectrum of $u_1(g)$ differs from the spectrum of $u_2(g)$, then clearly $u_1$ and $u_2$ are
not unitarily equivalent.

This is not such a fine invariant, though, since representations if infinte groups often involve only  unitary operators
with full spectrum, but of course this is a start.

The representation theory of $G$ is equivalent to the representation theory of the Banach $^*$-algebra $\ell ^1(G)$, in the
sense that each $^*$-representation of the latter restricts to a unitary representation of $G$ and,  conversely,  every
representation $u$ of $G$ integrates to a representation $\pi $ of $\ell ^1(G)$ according to the formula
$$
  \pi (f) = \sum_{g\in G}f(g) u_g, \quad \forall f \in  \ell ^1(G).
  $$

The advantage of working with $\ell ^1(G)$ is the availability of a linear structure.  For example,  the kernel of the
integrated representation $\pi $ is a two-sided ideal in $\ell ^1(G)$,  and is thus an invariant for the representation $u$
from which $\pi $ has been integrated.

The envelopping C$^*$-algebra of $\ell ^1(G)$,  called $C^*(G)$,  is also a very useful gadget because, like $\ell ^1(G)$, its representation theory is
also equivalent to the representation theory of $G$. However   it  has several  advantages due to the fact that its
representation theory is highly developped.

To be a bit more precise,
given a unitary representation $u$ of $G$, the integrated representation $\pi $ on $\ell ^1(G)$ extends to
$C^*(G)$ due to the universal properties of the envelopping C$^*$-algebra, and if $\rho $ denotes this
extension, then again the kernel of $\rho $ is an invariant of $u$.

Given the finer structure of $C^*(G)$, the kernel of
$\rho $ carries much more information about $u$.  For instance, if two unitary representations $u_1$ and $u_2$ give rise to
representations $\rho _1$ and $\rho _2$ of $C^*(G)$ with the same kernel, then $u_1$ and $u_2$ are mutually weakly contained in
each other in the sense that, for every $\xi $ in $H_1$, the  positive type function on $G$ given by
$$
  g\in   G\mapsto \langle u _1(g)\xi ,\xi \rangle \in {\mathbb C}
  $$
can be approximated by positive type functions coming from vectors in the representation space of $u_2$, and vice-versa.

Another very useful algebraic invariant is the commutant of a unitary representation $u$,  defined to be the set $u(G)'$
formed by all
bounded, linear
operators $T$ commuting with every $u_g$.

$u(G)'$ is  a von Neumann algebra and it carries a lot of information about multiplicities of sub-representations of
$u$.  For instance, $u(G)'$ is isomorphic to ${\mathbb C}$ iff $u$ is irreducible while $u(G)'$ is abelian iff $u$ is multiplicity
free.

The von Neumann algebra generated by $u(G)$, namely the  commutant of $u(G)'$, is a further very important invariant.
Given the rich theory of classification of von Neumann algebras  in types, many invariants can be extracted from $u(G)''$.


问:
Why does $a^n \mod p$ always result in a number with Legendre symbol as 1?
I noticed that the following expression

$a^n\mod p$ where p is a prime and $n >=1$ and $n <= p$

always results in a number with Legendre Symbol (with p as the prime) as 1.

I tested it out with a couple of a & p combinations.
$a = 15, p = 71$

$a = 288260533169915, p = 1007621497415251$

Generated random n's & tried it out & I got only Legendre symbols as 1

What is the reason for this?

答:
Let $P$ signify a quadratic residue and $N$ signify a quadratic non-residue.

Then, $NP=N$ (i.e. non-residue $\times$ residue $=$ non-residue), $NN=P$, and so $NNN=N$.

This means that if $a$ is any quadratic non-residue $\mod{p}$, so are $a_3, a_5, \cdots.$

On the other hand, if $a$ is a quadratic residue, then so are $a_2,a_3,a_4,a_5,\cdots$.

ContentLicense: CC BY-SA 4.0


问:
Big O for error terms
The following link from wikipedia explains the Big O notation really good. I have only one problem, which is to formalize the usage of Big O notation for error terms in polynomials. In the  example give here  we have
$$
e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...=1+x+\frac{x^2}{2!}+\mathcal{O}(x^3)=1+x+\mathcal{O}(x^2)
$$
as $x\rightarrow 0$. Now we find a similar notation also for the error terms in taylor polynomials. I would like to understand why this is right just formaly.

1.) Why is $\frac{x^3}{3!}+\frac{x^4}{4!}+... = \mathcal{O}(x^3)$

Is this because
$$
\frac{x^3}{3!}\frac{1}{x^3}+\frac{x^4}{4!}\frac{1}{x^3}+\frac{x^5}{5!}\frac{1}{x^3}+...=\frac{1}{3!}+\frac{x}{4!}+\frac{x^2}{5!}+...\leq M, \quad x\rightarrow 0
$$
for some $M$ that has to be bigger than $\frac{1}{3!}$ or can one show this in a different way formaly?

2.) Why is $\frac{x^2}{2!}+\mathcal{O}(x^3) = \mathcal{O}(x^2)$?

I appreciate your help! :)

答:
Always remember that $\mathcal{O}(g)$ is a set. So it is more appropriate to say that when $x\to a, f\in \mathcal{O}(g)$ which means that there exists an $\epsilon\gt 0 $ for which you can find a $\delta\gt 0$ such that $0\le |f(x)|\le \epsilon |g(x)|$ for all $0\lt|x-a|\lt \delta$.

Your question 1): Why $\frac{x^3}{3!}+\frac{x^4}{4!}+... = \mathcal{O}(x^3)$

Answer:

Note that $\lim_{x\to 0}\Big|\frac{\frac{x^3}{3!}+\frac{x^4}{4!}+...}{x^3}\Big|=\frac 16$ hence by limit definition there exists a $\delta\gt 0$ such that

$\begin{align}
0\lt|x|\lt \delta& \implies \Big|\Big|\frac{\frac{x^3}{3!}+\frac{x^4}{4!}+...}{x^3}\Big|-\frac 16\Big|\lt 1\\
&\implies \Big|\frac{\frac{x^3}{3!}+\frac{x^4}{4!}+...}{x^3}\Big|\lt \frac 76\\
& \implies \Big|\frac{x^3}{3!}+\frac{x^4}{4!}+...\Big|\le \frac 76 |x^3|\\
&\implies \frac{x^3}{3!}+\frac{x^4}{4!}+...\in \mathcal{O}(x^3) \end{align} $

Similarly, you can prove the second one.

Title: Big O for error terms


问:
Vertex coloring complete minors
Say you have a graph G and with a $K_t$ minor. Let H be the smallest subgraph of G that has $K_t$ as a minor (for all $v\in V(H)$, $H\setminus v$ does not contain $K_t$ as a minor). Then we can partition H into sets $V_1, V_2, \ldots V_t$ such that $H\left[V_i\right]$ is connected for all $i\in \{1, 2, \ldots, t\}$.  If $\chi(H\left[V_i\right]) < t$ for all $i\in \{1,2,\ldots, t\}$ and the clique number of H is less than t can we say $\chi(H) \leq t$?

答:
Isn't any odd cycle other than $K_3$ as $H$ a counterexample already?

$\chi(H)=3$ and $\chi(H[V_i])\le2$ as any proper subgraph of $H$ is just a path.

Title: Vertex coloring complete minors


问:
Is $\frac{\sin x^2}{\sin^2 x}$ uniformly continuous on $(0,1)$? Is my proof correct?
I'm said to check if $\dfrac{\sin x^2}{\sin^2 x}$ is uniformly continuous on $(0,1)$

I can see that $\dfrac{\sin x^2}{\sin^2 x}$ is continuous on $(0,1].$ So $\displaystyle\lim_{x\to1-}\dfrac{\sin x^2}{\sin^2 x}$ is finite.

I need to check if $\displaystyle\lim_{x\to0+}\dfrac{\sin x^2}{\sin^2 x}$ is finite.

$\displaystyle\lim_{x\to0+}\dfrac{\sin x^2}{\sin^2 x}\\=\displaystyle\lim_{x\to0+}\dfrac{x\cos x^2}{\sin x\cos x}\\=\displaystyle\lim_{x\to0+}\dfrac{\cos x^2-2x^2\sin x^2}{\cos^2 x-\sin^2 x}=1$

So uniformly continuous. Am I correct?

答: You are using that if a function is continuous on $(a,b]$ and the limit $\lim_{x\rightarrow a+}f(x)$ exists, then $f$ (or to be a little pedantic, its extension) is uniformly continuous on $[a,b]$. Can you prove this result? If you can, then you are absolutely right in whatever you have done above.</p>

ContentLicense: CC BY-SA 3.0


问:
Can you solve $\int 1/(x^2+1)^2\, dx$
I know that the integral looks like like the anti-derivative of $\arctan$, but i don't know how to use this fact so I tried to use a fraction decomposition to: $$\frac{1}{(x^2+1)^2}$$
to solve the anti-derivative : $$\int \frac{1}{(x^2+1)^2}\, dx$$
so :$$\frac{1}{(x^2+1)^2}=\frac{A}{x^2+1}+\frac{B}{x^2+1}\iff \frac{1}{x^2+1}=A+B$$
$$1+0x^2=(A+B)x^2+(A+B)1$$
Thus I got this system of equations:
$$\cases{A+B=1 \\ A+B=0}$$ Which is impossible. I also tried to use substitution but it didn't help me.

答:
There's multiple ways to do this integral including IBP, substitution... Here's one that uses Feynman's trick (differentiation under the integral sign)

We know the following antiderivative

$$I(a)=\int^x\frac{ dx'}{x'^2+a^2}=\frac{1}{a}\arctan(x/a)$$

You can see we introduced a parameter that did not exist before. We differentiate under the integral sign to obtain the desired integral form

$$J(a)=\int^x\frac{dx'}{(x'^2+a^2)^2}=-\frac{1}{2a}\frac{\partial I}{\partial a}=\frac{1}{2a^3}\left(\arctan(x/a)+\frac{a x}{x^2+a^2}\right)$$

$J(1)$ is the indefinite integral we seek.

EDIT: Here's a way that utilizes IBP.

Rewrite by adding and subtracting $x^2$ in the numerator

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

However note that

$$\int xdx\frac{x}{(x^2+1)^2}=-\frac{1}{2}\int dx~x \frac{d}{dx}\left(\frac{1}{x^2+1}\right)=-\frac{x}{2(x^2+1)}+\frac{1}{2}\arctan x$$

which yields the desired result

$$\int\frac{dx}{(x^2+1)^2}=\frac{1}{2}\arctan x+\frac{1}{2}\frac{x}{x^2+1}$$

Title: Can you solve $\int 1/(x^2+1)^2\, dx$


问:
Exact Deformation of Lagrangian Submanifolds
Let $j_{t}:L \rightarrow P$ be a family of Lagrangian submanifolds. I'm trying to show that the form $j_{t}^{*}(i(X_{t})\omega)$, $X_{t}(j_{t}(x)):=\frac{dj_{t}(x)}{dt}$ is exact for all $t$ if and only if there is a family of Hamiltonian diffeomorphisms $\psi_{t}^{H}:P \rightarrow P$ such that $j_{t}(L)=\psi_{t}^{H}(L)$.

What I have so far for one implication is that if the form is exact, it is of the form $j_{t}^{*}(i(X_{t})\omega)=dh_{t}$. The idea I have is to extend $h_{t}$ to the family of Hamiltonians $H_{t}$ I need using the canonical almost complex structure on P, but I'm struggling with the technical details. Thank you for your help in advance!

答:
Let $\psi^{H}_t : P \to P$ be a Hamiltonian isotopy, i.e. there exists a smooth time-dependent function $H_t : P \to \mathbb{R}$ satisfying $i(X_t)\omega = dH_t$ where $X_t$ is defined via $X_t(\psi^{H}_t(x)) = d(\psi^{H}_t(x))/dt$ for $x \in P$, such that $j_t = \psi^{H}_t \circ j_0 : L \to P$. Then $j_t^*(i(X_t)\omega) = j_t^*(dH_t) = d(j_t^*H_t)$ is exact on $L$.

Conversely, let $j_t : L \to P$ be a smooth family of Lagrangian submanifolds satisfying $j_t^*(i(X_t)\omega) = dh_t$ where $X_t \in C^{\infty}(j_t^* TP)$ is defined via $X_t(j_t(x)) = d(j_t(x))/dt$ for $x \in L$ and $h_t : L \to \mathbb{R}$ is a smooth time-dependent function. (Here, $j_t^*TP$ is the pullback tangent bundle, namely the vector bundle over $L$ whose fiber over $x \in L$ is the tangent space $T_{j_t(x)}P$.) Note that it makes sense to let the 1-form $A_t := i(X_t)\omega$ act on any element in $j_t^*TP$, and the equation $j_t^*(i(X_t)\omega) = dh_t$ is then a constraint on $A_t$: it means that the restriction of $A_t$ to the subbundle $j_t^* TL_t = j_t^* (j_t)_* TL$ is exact. Because of this constraint, the circle of ideas known as "Whitney extension theorem" shows that we can extend $h_t$ (thought of as a function on the submanifold $L_t$) to a function $H_t$ on $P$ such that $dH_t$ is given along $L_t$ by $A_t$. We can therefore define the vector field $X_t$ on the whole of $P$ via $i(X_t) \omega = dH_t$ and define $\psi^{H}_t$ to be the isotopy generated by $X_t$. Since this $X_t$ clearly extends the vector field $X_t$ which was already defined along each $L_t$, it follows that $j_t = \psi^{H}_t \circ j_0$.


答:
If we use the substitution $x = \tan \theta$, $dx = \sec^2 \theta \,d\theta$, the integral becomes \begin{align*}
\int \frac{1}{(x^2+1)^2}\, dx &= \int \frac{\sec^2 \theta}{(\tan^2\theta + 1)^2}\,d\theta\\
&= \int \frac{\sec^2\theta}{\sec^4 \theta}\,d\theta\\
&= \int\cos^2\theta \,d\theta\\
&= \frac{1}{2}(\theta + \sin\theta\cos\theta) + C.
\end{align*}  I will leave that last integral to you, but we can use the double angle formula for cosine.  So, we just need to undo the substitution.  Given that $x = \tan\theta$, we can conclude that $\theta = \arctan x$ and  $$\sin\theta = \frac{x}{\sqrt{1+x^2}}\quad\text{and}\quad \cos\theta = \frac{1}{\sqrt{1+x^2}}.$$
This follows from the fact that $$x^2 + 1 = \tan^2\theta + 1 = \sec^2\theta \implies \cos^2\theta = \frac{1}{1+x^2}.$$

This gives us
$$\int \frac{1}{(x^2+1)^2}\, dx = \frac{1}{2}(\theta + \sin\theta\cos\theta) +C= \frac{1}{2}\left(\arctan x + \frac{x}{1+x^2}\right) +C.$$

Title: Can you solve $\int 1/(x^2+1)^2\, dx$

Body:
I know that the integral looks like like the anti-derivative of $\arctan$, but i don't know how to use this fact so I tried to use a fraction decomposition to: $$\frac{1}{(x^2+1)^2}$$
to solve the anti-derivative : $$\int \frac{1}{(x^2+1)^2}\, dx$$
so :$$\frac{1}{(x^2+1)^2}=\frac{A}{x^2+1}+\frac{B}{x^2+1}\iff \frac{1}{x^2+1}=A+B$$
$$1+0x^2=(A+B)x^2+(A+B)1$$
Thus I got this system of equations:
$$\cases{A+B=1 \\ A+B=0}$$ Which is impossible. I also tried to use substitution but it didn't help me.


答:
For $0\le x\le1$,

$\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}+\cdots\le x^3\left(\dfrac1{3!}+\dfrac1{4!}+\dfrac1{5!}+\cdots\right)$ and the sum inside the parenthesis is finite.

Similarly, $\dfrac{x^2}2+\mathcal O(x^3)\le \dfrac{x^2}2+cx^3\le x^2\left(\dfrac12+c\right).$

Title: Big O for error terms

Body:
The following link from wikipedia explains the Big O notation really good. I have only one problem, which is to formalize the usage of Big O notation for error terms in polynomials. In the  example give here  we have
$$
e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...=1+x+\frac{x^2}{2!}+\mathcal{O}(x^3)=1+x+\mathcal{O}(x^2)
$$
as $x\rightarrow 0$. Now we find a similar notation also for the error terms in taylor polynomials. I would like to understand why this is right just formaly.

1.) Why is $\frac{x^3}{3!}+\frac{x^4}{4!}+... = \mathcal{O}(x^3)$

Is this because
$$
\frac{x^3}{3!}\frac{1}{x^3}+\frac{x^4}{4!}\frac{1}{x^3}+\frac{x^5}{5!}\frac{1}{x^3}+...=\frac{1}{3!}+\frac{x}{4!}+\frac{x^2}{5!}+...\leq M, \quad x\rightarrow 0
$$
for some $M$ that has to be bigger than $\frac{1}{3!}$ or can one show this in a different way formaly?

2.) Why is $\frac{x^2}{2!}+\mathcal{O}(x^3) = \mathcal{O}(x^2)$?

I appreciate your help! :)


问:
MIP modelling of piecewise linear function
Suppose we have the following piecewise linear function $f(x)$:

$3x + 8,  x\in [0,5]$

$33 - 2x,    x\in [5,10]$

$3 + x,    x\in [10,20]$

How to model the relations between $f(x)$ and  $x$ using integer variables and linear constraints. How to specify the values of M if we model it using big-M method.

答:
Introduce a binary variable $z_i$ for each segment and impose the following linear constraints:
\begin{align}
z_1 + z_2 + z_3 &= 1 \\
0z_1 + 5z_2 + 10z_3 \le x &\le 5z_1 + 10z_2 + 20z_3 \\
L_1(1-z_1) \le y - (3x+8) &\le U_1(1-z_1) \\
L_2(1-z_2) \le y - (33-2x) &\le U_2(1-z_2) \\
L_3(1-z_3) \le y - (3+x) &\le U_3(1-z_3)
\end{align}
I'll leave the computations of the big-M values $L_i$ and $U_i$ to you.


问:
Differentiability of sum of eigenvalues.
Let $A(x)=(a_{ij}(x))_{i,j}$ be a smooth map of symmetric matrices for $x\in\Omega\subseteq\mathbb R^n$. (i.e., every $a_{ij}(x)=a_{ji}(x)$ is smooth.) Let $k_1\le\cdots\le k_n$ be its eigenvalues. If we know that at a point $p,$ $k_1=k_2<0$ and $k_1=k_2<k_3,$ can we derive that $k_1+k_2$ is also smooth?

When $n=3,$ since $k_3$ has multiplicity $1$ and $k_1+k_2=\text{tr} A-k_3,$ we know this statement is correct. However, in general, I am not sure how to see this.

Any ideas or comments are appreciated!

答:
Assume as in your question that $p\in\Omega$ is a point such that the eigenvalues of $A(x=p)$ are $k:=k_1=k_2<k_3\leq \cdots \leq k_n$.

Let us write the characteristic polynomial of $A(x)$ in a neighborhood of $x=p$ as follows
$$
\det(y-A(x))=(y^2-u(x)y+v(x))(y^{n-2}+c_{n-3}(x)y^{n-3}+\cdots+c_1(x) y+c_0(x)),
$$
where: (1) $u,v,c_0,...,c_{n-3}$ are smooth functions of $x$ in a neighborhood of $x=p$, (2) the values of $u,v,c_0,...,c_{n-3}$ at $x=p$ are uniquely determined by $u(x=p)=2k_1$ and $v(x=p)=k_1^2$.

We want to do this by the inverse function theorem. Consider the usual expansion
$$
\det(y-A(x))=y^n+d_{n-1}(x)y^{n-1}+\dots+d_1(x)y+d_0(x).
$$
Since $d_i(x)$ are by assumption smooth in $x\in\Omega$, so we only need to invert the map
$$
(u,v,c_0,...,c_{n-3})\mapsto(d_0,\dots,d_{n-1})
$$
defined by $d_k=c_{k-2}-uc_{k-1}+c_kv$ with the conventions: $c_{-1}=c_{-2}=0=c_{n-1}$ and $c_{n-2}=1$. The Jacobian matrix of such transformation is
$$
J=\begin{pmatrix}
0 & -c_0 & -c_1 & -c_2 &\cdots &-c_{n-3} & -1
\\
c_0 & c_1 & c_2 & c_3 &\cdots & 1 & 0
\\
v & -u & 1 & 0 & \cdots & 0 & 0
\\
0 & v & -u & 1 & \cdots & 0 & 0
\\
0 & 0 &v & -u &  \cdots & 0 & 0
\\
\vdots & \ddots & \ddots & \ddots & \ddots & \ddots &\vdots
\\
0 & 0 &0 &\cdots & v & -u & 1
\end{pmatrix}
$$
It is a known fact about Sylvester matrices (https://en.wikipedia.org/wiki/Sylvester_matrix) that this matrix has zero determinant if and only if the polynomials $y^2-uy+v$ and $y^{n-2}+c_{n-3}y^{n-3}+\dots+c_1y+c_0$ share a common nontrivial factor. Thus whenever they do not share a common factor, the map can be inverted obtaining $u(\vec d),v(\vec d),c(\vec d)$ as smooth functions of $\vec d=(d_0,...,d_{n-1})$ and we may express in particular $u$ as a smooth function of $x$ in a neighborhood of $x=p$.


问:
How to show that the intersection of two elementary classes is a elementary class
I am wondering how I can show that the intersection between two elementary classes is a elementary class.

I have two elementary classes A and B, $A,B = (M|M\models\Gamma)$ and I want to give a proof that the intersection $A\cap B$ is a new elementary class C which is the class of the models found in both A and B.

I see in this situation two approaches, either assume that this is true and prove it or assume that C is not elementary and derive a contradiction, am I on the right track with this? Or what is the best approach?

Thanks in advance!

答:
I suspect you are overthinking this. Suppose $A=\{\mathcal{M}:\mathcal{M}\models\Gamma\}$ and $B=\{\mathcal{N}:\mathcal{N}\models\Theta\}$ for some sets of sentences $\Gamma,\Theta$. Elements of $A\cap B$ are exactly structures which satisfy both $\Gamma$ and $\Theta$ at once; do you see a set of sentences $\Delta$ whose models are exactly the models of both $\Gamma$ and $\Theta$?

HINT: don't work too hard ...

 Just consider the union of $\Gamma$ and $\Theta$: $$\Delta=\Gamma\cup\Theta.$$

A more interesting example concerns unions of elementary classes. For a toy example, suppose $A=\{\mathcal{M}:\mathcal{M}\models\gamma\}$ and $B=\{\mathcal{N}:\mathcal{N}\models\theta\}$ for some single sentences $\gamma$ and $\theta$. Then $A\cup B=\{\mathcal{A}: \mathcal{A}\models\gamma\vee\theta\}$. Do you see how to extend this to sets of sentences in place of infinite sentences?


Title: Can you solve $\int 1/(x^2+1)^2\, dx$

Body:
Integrate as follows

\begin{align}
\int {\dfrac{1}{(x^2+1)^2}}\,dx
&= \int {\dfrac{1}{x^2+1}}\,dx - \int {\dfrac{x^2}{(x^2+1)^2}}\,dx\\
 &= \int {\dfrac{1}{x^2+1}}\,dx + \frac1{2}\int x\>d\left({\dfrac{1}{x^2+1}}\right)\\
 &= \frac1{2} \frac x{x^2+1} + \frac1{2}\int {\dfrac{1}{x^2+1}}\,dx\\
 &= \frac1{2} \frac x{x^2+1} + \frac1{2}\tan^{-1} x+C
\end{align}


问:
Find where the cut wire theorem fails
First, to add more context, let me write the theorem in question:

Cut wire theorem:
If $(X, \tau)$ is a compact Hausdorff space and $A, B \subseteq X$
are two different closed sets such that every connected component of
$X$ intersects at most one of them, then there exists $ U, V \in \tau$
disjoint, such that $A \subseteq U,\,  B \subseteq V,\,\text{ and } U
 \cup V = X$

I'm trying to find some spaces where this doesn't hold. One of them Hausdorff but not compact and the other one compact but not Hausdorff. I would appreciate any help to find spaces like that. My only idea about this is that such space must be disconnected wich is not quite helpful. Thanks

答:
A standard example for the first case is the subspace of the plane consisting of the lines $[0,1] \times \{ 1/n \}$, $n \in \mathbb N$, and the singletons $\{(0,0)\}$ and $\{(1,0)\}$. The two singletons are connected components but lie in the same quasi-component.

For a non-Hausdorff example, take the quotient of above formed by shrinking each of the lines to a point. (A sequence with two limit points.)

Title: Find where the cut wire theorem fails


问:
Uniqueness in spectral theorem for commutative (non-unital) Banach $*$-algebras
$\require{AMScd}$I'm struggling to understand the uniqueness part of Theorem 1.54 in Folland's book on Harmonic analysis.

Theorem 1.54: Let $\mathcal{A}$ be a commutative Banach $*$-algebra (non necessarily unital) and let $\phi$ be a nondegenerate $*$-representation $\phi:\mathcal{A}\to \mathcal{L}(H)$ (the codomain is the set of bounded operators on a Hilbert space). There is a unique projection-valued measure $P$ on $\sigma(\mathcal{A})$ such that $\phi(x) = \int G_{\mathcal{A}}(x)dP
$ for all $x\in\mathcal{A}$. Here $G_{\mathcal{A}}(x)$ is the continuous function on the spectrum $\sigma(\mathcal{A})$ which is given by evaluation on $x$.

Attempt: (Uniqueness) Given two $P, Q$ satisfying the condition, we would have
$$
\int G_{\mathcal{A}}(x) dP_{u,v} = \int G_{\mathcal{A}}(x) dQ_{u,v}
$$
for all $u, v \in H$. In the case of (non-unital) $C^*$-algebras, one could use the fact that the $G_A(x)$ exhaust $C_0(\sigma(\mathcal{A}))$ and conclude equality of $P, Q$ by Riesz representation. However, this case is different since the set $\{G_A(x) : x\in \mathcal{A}\}$ might not be big. So how do I go about proving uniqueness?

I include the existence part of the theorem here, in case it helps. Let $\mathcal{B}$ denote the norm closure of $\phi(\mathcal{A})$, and let $\mathcal{\tilde{A}},\ \mathcal{\tilde{B}}$ denote the algebras with units attached. $\mathcal{\tilde{B}}$ is then $\mathcal{B}\oplus \mathbb{C}I$ and is a commutative unital $C^*$-algebra.
Moreover, the $*$-representatiom $\phi$ extends to give the diagram

$$
\begin{CD}
\mathcal{A} @>{\phi}>> \mathcal{B}  \\
@V{}VV @V{}VV \\
 \mathcal{\tilde{A}} @>{\phi}>> \mathcal{\tilde{B}}
\end{CD}  \qquad\quad (1)
$$
On identifying the spectrum of $\mathcal{\tilde{A}}$ with a subset of bounded linear functionals on $\mathcal{A}$, we can write $\sigma(\mathcal{\tilde{A}}) = \sigma(\mathcal{A})\cup \{0\}$ which,  if $\sigma(\mathcal{A})$ is noncompact, is the one-point compactification of $\sigma(\mathcal{A})$. Similarly, this holds for $\mathcal{B}$. Thus $(1)$ gives a dual diagram (the vertical arrows are inclusions and not the dual)
$$
\begin{CD}
\sigma(\mathcal{A}) @<{\phi^*}<< \sigma(\mathcal{B})  \\
@V{}VV @V{}VV \\
 \sigma(\mathcal{A})\cup \{0\} @<{\phi^*}<< \sigma(\mathcal{B})\cup \{0\}
\end{CD}  \qquad\quad (2)
$$
It follows from the definitions that the arrows are continuous injections. The bottom row is a homeomorphism onto the image. Taking duals again, we have the following commutative diagram of Gelfand transforms

$$
\begin{CD}
\mathcal{A} @>{\phi}>> \mathcal{B}  \\
@V{}VV @V{}VV \\
 \mathcal{\tilde{A}} @>{\phi}>> \mathcal{\tilde{B}}
\end{CD}  \qquad \overset{G_\cdot}{\longrightarrow}
\begin{CD}
C_0(\sigma(\mathcal{A})) @>{(\phi^*)\check{}}>> C_0(\sigma(\mathcal{B}))  \\
@V{}VV @V{}VV \\
 C(\sigma(\mathcal{A})\cup \{0\}) @>{(\phi^*)\check{}}>> C(\sigma(\mathcal{B})\cup \{0\})
\end{CD}
$$
Now the spectral theorem applies to $\mathcal{\tilde{B}}$ to give a unique projection-valued measure on Borel sets $Q:\text{Borel}(\sigma(\mathcal{B})\cup\{0\}) \to \mathcal{L}(H)$ such that for every $T\in \mathcal{\tilde{B}}$,
$$
T = \int G_\mathcal{\tilde{B}}(T) dQ.
$$
Now, we claim $Q(\{0\})=0$. Put $E=\{0\}$. If $u \in H$ is in the image of $Q(E)$, then we have, for any $S \in \mathcal{B}$ and $v\in H$, by the spectral theorem
$$(Su,v) = (SQ(E)u,v) = \int_{h\in E} G_{\mathcal{\tilde{B}}}(S)(h) dQ_{u,v}(h)=0.
$$
Since $\phi$ is non-degenerate, $Q(E)$ must be the zero operator.
Thus we have a projection-valued measure $R$ on $\sigma(\mathcal{B})$ uniquely defined by the property that, for every $T\in \mathcal{B}$ we have $$
T = \int G_{\mathcal{B}}(T) dR. \qquad (3)
$$
I guess the uniqueness here follows by the fact that $\{G_\mathcal{B}(T) : T\in \mathcal{B}\}=C_0(\sigma(\mathcal{B}))$.

Finally, we define a projection-valued measure $P$ on $\sigma(\mathcal{A})$ using the diagram
$$
\text{Borel}(\sigma(\mathcal{A})) \overset{(\phi^*)\check{}}{\longrightarrow} \text{Borel}(\sigma(\mathcal{B})) \overset{R}{\longrightarrow} \mathcal{L}(H)
$$
With this definition, it is easy to see that $P$ is the pushforward measure from $R$ by $\phi^*$. That is, $P_{u,v} = (\phi^*)_*R_{u,v}$.
To see that $P$ has the required property, take any $x\in\mathcal{A}, u, v \in H$ and observe that, by $(3)$, we have
\begin{equation}\begin{split}
(\phi(x)u,v) &= \int_{\sigma(\mathcal{B})}G_\mathcal{B}(\phi(x))(h)dR_{u,v}(h)\\
&= \int_{\sigma(\mathcal{B})} h(\phi(x)) dR_{u,v}(h)\\
&= \int_{\sigma(\mathcal{B})} \phi^*(h)(x) dR_{u,v}(h)\\
&= \int_{\sigma(\mathcal{B})} G_\mathcal{A}(x)(\phi^*(h)) dR_{u,v}(h)\\
&= \int_{\sigma(\mathcal{A})} G_\mathcal{A}(x)(l) d\left((\phi^*)_*R_{u,v}\right)(l)\\
&= \int_{\sigma(\mathcal{A})} G_\mathcal{A}(x)(l) dP_{u,v}(l)
\end{split}\end{equation}
Again, I don't see why such a measure should be unique. Please send help.

答: The set of functions of the form $G_{\mathcal{A}}(x)$ is a $*$-subalgebra of $C_0(\sigma(\mathcal{A})$ that separates points from each other and from $0$.  So by Stone-Weierstrass, it is dense in $C_0(\sigma(\mathcal{A}))$.  It follows that the bounded functionals of the form $G_{\mathcal{A}}(x)\mapsto \int G_{\mathcal{A}}(x) dP_{u,v}$ extend uniquely to bounded functionals on all of $C_0(\sigma(\mathcal{A}))$, and you can apply the Riesz representation theorem.


问:
Mittag Leffler-like expansion of $\frac{s}{e^{sx}-1}-\frac{1}{x}+\frac{s}{2}$
Let $x\in \mathbb{R^{+}}$ and $s\in\mathbb{C}\;\;\Re(s)>0$. The function in question has the well-known Mittag Leffler expansion:
$$\frac{s}{e^{sx}-1}-\frac{1}{x}+\frac{s}{2}=2xs^{2}\sum_{n=1}^{\infty}\frac{1}{4\pi^{2}n^{2}+(sx)^{2}}$$
But is it possible for our function to have an expansion of the form :
$$\frac{s}{e^{sx}-1}-\frac{1}{x}+\frac{s}{2}=F(s,x)+\sum_{n=0}^{\infty}\frac{A_{n}(s)}{e^{x}-Z_{n}(s)}$$
Where $A_{n}(s)$, $Z_{n}(s)$ are appropriate functions in $s$, and $F(s,x)$ is entire in $x$.

答:
Why would you want such an expansion?

I doubt your condition $x>0$ makes much sense, the continuation to $x$ complex should be implied by your expansions.

In that case fix $s=\pi$.

$F(s,x)$ analytic in $x$ and the $2i\pi$-periodicity in $x$ of $\sum_{n=0}^{\infty}\frac{A_{n}(s)}{e^{x}-Z_{n}(s)}$ would give that

$$\frac{s}{e^{sx}-1}-\frac{1}{x}-\frac{s}{e^{s(x+2i\pi)}-1}+\frac{1}{x+2i\pi}$$ is analytic in $x$ which is absurd.


问:
A matrix calculus problem with vectorization
I'm trying to find out the derivative of $vec(AA^T)$ w.r.t to $vec(A)$, where $A$ is a $m$ by $n$ matrix

I use a simple $3$ by $2$ case and find it entry-wisely.

I guess the answer would be $A^T\otimes I_m+[(I_m\otimes a_1),\ (I_m\otimes a_2),\ \ldots,\  (I_m\otimes a_n)]^T$ where $a_i$'s is columns of $A$ (not sure it is correct)

Is there any systematical way to find out such an expression?

My intuitive thought is to express $vec(AA^T)$ as some differentiable function of $vec(A)$ and utilize some simple matrix calculus rule.

Any tips would be appreciated!

答:
Define a vector function, then calculate its differential and gradient.
$$\eqalign{
 f &= {\rm vec}(AA^T) \\
df &= {\rm vec}\Big(I_m\,dA\,A^T+A\,dA^TI_m\Big) \\
   &= (A\otimes I_m)\,da + (I_m\otimes A)K\,da \\
   &= \Big((A\otimes I_m) + (I_m\otimes A)K\Big)\,da \\
\frac{\partial f}{\partial a}
   &= (A\otimes I_m) + (I_m\otimes A)K \\
}$$
where $K$ is the Commutation Matrix associated with vectorizing a matrix transpose via the Kronecker product.

For a given $(m,n)$ one can readily calculate $K$, e.g. in Julia

```<code>i = collect(1:m*n)
j = vec(reshape(i,m,n)')
K = sparse(i,j,1)
</code>```

Body:
Define a vector function, then calculate its differential and gradient.
$$\eqalign{
 f &= {\rm vec}(AA^T) \\
df &= {\rm vec}\Big(I_m\,dA\,A^T+A\,dA^TI_m\Big) \\
   &= (A\otimes I_m)\,da + (I_m\otimes A)K\,da \\
   &= \Big((A\otimes I_m) + (I_m\otimes A)K\Big)\,da \\
\frac{\partial f}{\partial a}
   &= (A\otimes I_m) + (I_m\otimes A)K \\
}$$
where $K$ is the Commutation Matrix associated with vectorizing a matrix transpose via the Kronecker product.

For a given $(m,n)$ one can readily calculate $K$, e.g. in Julia

i = collect(1:m*n)
j = vec(reshape(i,m,n)')
K = sparse(i,j,1)


问:
Proof: continuous function is bounded
I've seen some proofs for this Lemma and wanted to know what's wrong with my take.

Lemma. Let $a < b$ be real numbers, and let $f : [a, b] → R$ be a function continuous on $[a, b]$. Then $f$ is a bounded function.

Proof:
Because we know $f$ is continuous, for a given $\epsilon$ we can find $\delta$ such as $|x-x_0|< \delta \Rightarrow |f(x) - f(x_0)| < \epsilon$.
I proceeded to use triangle inequality so: $|f(x)| - |f(x_0)| \le |f(x) - f(x_0)| < \epsilon$.
Then $|f(x)| - |f(x_0)| < \epsilon$. And finally assert that $\epsilon + |f(x_0)|= M$ so that $|f(x)|< M$

Hope I am using MathJax the right way.
Thanks :)

Edit: I forgot to add that $x_0 \in [a,b]$

答:
You don't say what $x_0$ is. If $x_0$ is an arbitrary point of $[a,b]$, then what you have proved is that near $x_0$, $f$ is bounded. It is not automatic that it follows from that that $f$ is bounded on $[a,b]$.

Besides, the same argument who work on any subset of $\Bbb R$, but, say, $\frac1x$ is unbounded on $(0,1]$.

Title: Proof: continuous function is bounded


问:
If an exact form vanishes on a submanifold, can I find a primitive that also does?
Let $M$ be a closed smooth manifold and $Q\subset M$ a closed embedded submanifold.
Furthermore, let $\omega$ be an exact differential form $\omega\in\Omega^k(M)$ and vanishing identically on $Q$ (i.e. $\omega_q=0$ for any $q\in\ Q$).

Can we always find a primitive $\alpha\in\Omega^{k-1}(M)$ (i.e. such that $d\alpha=\omega$), whose restriction to $Q$ also vanishes?

This question came up during class and maybe it is obvious but I can't even seem to convince myself whether it is true or not, so any help is greatly appreciated.

Edit: counterexamples are given in the comments for $k=\dim Q+1$ and for the case of $\omega$ being $1$-form with $Q$ disconnected. In the context of the class, we were specifically considering $\omega$ to be a $2$-form, but I am also interested in the general case.

Edit2: There is an answer dealing with the condition of the pullback of $\omega$ to $Q$ being $0$. However, I meant that $\omega$ itself vanishes identically in points that belong to the submanifold $Q$.

答:
Proposition 6.8 of Cannas da Silva's "Lectures on Symplectic Geometry" says:

Proposition: Let $U$ be a tubular neighborhood of a compact submanifold $Q \subset M$, and $i: Q\to U$ the inclusion. If $\tau$ is a closed $k$-form on $U$ such that $i^*\tau = 0$, then $\tau$ is exact. Moreover, there is a $k-1$ form $\alpha$ with $d\alpha = \tau$ on $U$, such that $\alpha|_{T_QM} = 0$.

In the context of Moser's trick, this is applied to $\tau = \omega_1 - \omega_0 $, the difference of two symplectic forms that have been assumed or arranged to agree on $T_QM$, the tangent bundle of $M$ restricted over $Q$.

By the example in Ted Shifrin's answer, it is necessary to restrict to a tubular neighborhood: even if $\tau$ is globally exact, there need not exist a global primitive $\alpha$ that also vanishes on $Q$.


问:
What does square brackets around a polynomial mean?
EDIT: The first paragraph has been indicated as inaccurate, please see this answer.

When we have something like $\mathbb{Z}_2[x]/(x^2 + x + 1)$, we
understand that this means the set of polynomials over indeterminate
$x$ where the coefficients are drawn from $\mathbb{Z}_2 = \{0,1\}$ with
degree less than $\text{deg} (x^2 + x + 1)$.

So, in this set we can include polynomials like $x$, $x+1$ and
even $1$, but not $x^2+1$ or $2x+1$. This is okay so far with me.

But in some lecture note (Rings and fields,
Sergei Silvestrov,
Spring term 2011, Lecture 5; page 3, Example 1
and subsequent ones) I find representations like,
$\mathbb{Z}_2[x]/(x^2 + x + 1) = \{[0], [1], [x], [x + 1]\}$.

I fail to understand why the square brackets have been put around
the polynomials.

Answer to this question says that this depends on the context and
may be same as parentheses or may mean braces to represent sets.

Your help will be appreciated.

答:
Your first three lines aren't accurate: that symbol is for a field , represented by that quotient ring, the elements of which can be expressed as polynomials of degree zero or degree one in one of the roots of the polynomial $\;x^2+x+1\;$ over $\,\Bbb F_2\,$.

Thus, the elements in that "set" are not polynomials but expressions of the form $\,w, w+1,$ , etc., where $\, w^2+w+1=0\pmod 2\,$ , and precisely in order to avoid this confusion is that Silvestrov uses the symbols he does...


问:
integer ring and valuation ring of local fields
Let $K$ be a local field, that is, complete discrete valuation field, with finite residue field.
Then, integer ring of $K$ and valuation ring of $K$ corresponds?

And to what extent can I extend to another field?

In trivial case, integer ring of $\Bbb Q_p$ is $\Bbb Z_p$, and valuation ring is also$\Bbb Z_p$ .

Thank you in advance.

答:
For a discretely valued field "the ring of integers" and "the valuation ring" means the same thing: $O_K=\{ a\in K,v(a)\ge 0\}$.

This is different to the ring of integers of a number field understood as the integral closure of $\Bbb{Z}$.

Complete discretely valued field with finite residue field (a local field) gives that $K$ is a finite extension of $\Bbb{Q}_p$ or $\Bbb{F}_p((t))$.

For any finite extension $K/F$ then $F$ is discretely valued and complete and $O_K$ is the integral closure of $O_F$ in $K$.

(take the normal closure $L$ of $K/F$, there is a standard proof that the discrete valuation on $L$ is unique because it is recovered from algebraic properties such as "the elements having a $n$-th roots in $L$ for all $p\nmid n$", whence $Aut(L/F)$ acts continuously from which its fixed subfield $E$ is closed (complete), if $L/F$ is not separable then $F=E^{p^r}$ is complete too, so we are in the standard case of $K/F$ a finite extension of local field...)

The same should hold for any complete discretely valued field. It doesn't hold for non-complete ones: try with $O_K=\Bbb{Z}[i]_{(2+i)}$, the integral closure of $O_F=\Bbb{Z}_{(5)}$ is $\Bbb{Z}[i]_{(2+i)}\cap \Bbb{Z}[i]_{(2-i)}$.


问:
How to correctly understand a Corollary of the Metatheorem on Constants in predicate logic
In Lectures in Logic and Set Theory(Vol.2) by George Tourlakis, there is a metatheorem as follows:

I cannot see intuitively why the $x_1$ that occurs free in $A[e_1,...,e_n]$ does not matter. For example, let $A[e_1,...,e_n] \equiv x_1=e_1\rightarrow P(e_1, e_2)$, then $A[y_1,...,y_n] \equiv x_1=y_1\rightarrow P(y_1, y_2)$ and $A[x_1,...,x_n] \equiv x_1=x_1\rightarrow P(x_1, x_2)$. Doesn't the $x_1$ introduced in the final step interfere with the original one?

答: The last boils down to $P(x_1,x_2)$, which looks like being too strong. However, we have to start from $\vdash_{\mathfrak T'}\mathscr A[e_1,e_2]$, but as $\Gamma'=\Gamma$, we cannot have used any special properties of the new names $e_1,e_2$ in deriving $\mathscr A[e_1,e_2]$. Put differently, when $x_1,x_2$ are free, we certainly have $\Gamma'\cup\{x_1=e_1,x_2=e_2\}\vdash_{\mathfrak T'}\mathscr A[x_1,x_2]$ and can translate any derivation of that into $\Gamma\vdash_{\mathfrak T}\mathscr A[x_1,x_2]$.


问:
Maximize (but not minimize?) entropy for probability distribution
Let $p$ be an arbitrary discrete probability distribution on $\{1, \dots, 10\}$. We can represent such distribution by a vector indexed by by $i$ subject to the constraints $p_i \ge 0$ and $\sum^{10}_{i=1}p_i = 1$. I want to find analytically the $p$ with maximum entropy which is given by

$$
H(p) = - \sum_{i=1}^{10} p_i \log(p_i)
$$

My question consists of two parts:

How do I show that the optimal $p$ is given by the uniform distribution using the Lagrange method.

Why can't the Lagrange method be used to find $p$ with minimal entropy (which I suppose is given by the dirac distribution)?

For 1. I know that the Lagrangian function associated with this constrained problem is given by

$$
\mathcal{L}(p,\lambda) = - \sum_{i=1}^{10} p_i \log(p_i) + \lambda(1 - \sum^{10}_{i=1}p_i)
$$

and the respective gradient by

$$
\frac{\partial \mathcal{L}}{\partial p_i} = - \log(p_i) - 1 - \lambda
$$

and

$$
\frac{\partial \mathcal{L}}{\partial \lambda} = 1 - \sum^{10}_{i=1}p_i
$$

but I don't know how to continue from there.

答:
When applying the Lagrange method, you need to cancel the gradient, which yields $ - \log(p_i) - 1 - \lambda = 0$ and $1 - \sum^{10}_{i=1}p_i = 0$. The first equality yields $p_i =e^{-1-\lambda}$ and the second one yields $\sum^{10}_{i=1}p_i = 1$. So we want to have $\sum^{10}_{i=1} e^{-1-\lambda} = 1$, or, in other words, $10 e^{-1-\lambda} = 1$. Thus, substituting $e^{-1-\lambda} = \frac 1 {10}$ into the expression $p_i = e^{-1-\lambda}$ gives $p_i = \frac 1 {10}$, as we expect.

You cannot use your Lagrangian to find the minimum, because the minima (which are indeed all 10 Dirac distributions) lie at the "boundary" of the constraints $p_i \geq 0$ and $p_i \leq 1$, and your Lagrangian does not know anything about these constraints.

Update 1: the Lagrangian actually knows about the constraints $p_i \geq 0$ because otherwise it is not defined, but it does not know about the constraints $p_i \leq 1$. In fact the Lagrangian method here would minimize over the set $\{(p_1, \dots, p_n)\in {\mathbb R_+^{10}}, \sum^{10}_{i=1}p_i = 1\}$, which is larger than $\{(p_1, \dots, p_n)\in [0,1]^{10}, \sum^{10}_{i=1}p_i = 1\}$, and the Dirac measures are not minimizers over the larger space.

Update 2 (following LinAlg's comment and answer): you could try to change your Lagrangian to take the inequalities into account, see for example these notes, but then you would run into trouble due to the fact that $x\mapsto x\log x$ is not differentiable at 0.


问:
Regular and irregular points on the boundary for Brownian motion
If we take unit ball which is punctured at the origin and consider the origin as the boundary point, then why is it so that the origin is a irregular point if d>1 but regular for d=1?</p>

答:
You know that the zero set of Brownian Motion intersects $[0,t]$ at infinitely many points (this follows, for instance, by the law of the iterated logarithm). This proves that $(-a,b)\setminus \{0\}$ is a regular set for one-dimensional Brownian Motion.

For two dimensions or higher, you can simply pick a sequence $t_n\to 0$ and note that $\mathbb{P}^0(X_t\neq 0\;\forall t>0)=\lim_{n\to\infty}\mathbb{P}^0(X_t\neq 0\;\forall t>t_n)$. By the Markov Property, we have
$$
\mathbb{P}^0(X_t\neq 0\;\forall t>t_n)=\mathbb{E}^0 [\mathbb{P}^{X_{t_n}}[X_t\neq 0\; \forall t>0]]
$$
Now, by Brownian scaling and rotation invariance, we have
$$
\mathbb{P}^{x}[X_t\neq 0\; \forall t>0]=\mathbb{P}^{e_1}[X_t\neq 0\; \forall t>0]
$$
for all $x\neq 0$. Hence, since $X_{t_n}\neq 0$ almost surely, we get that $ \mathbb{P}^0(X_t\neq 0\;\forall t>t_n)=\mathbb{P}^{e_1}[X_t\neq 0\; \forall t>0]=1$, where the last equality follows from the fact that we're in dimension two or greater. If you don't know this result, you can see the following for a solution in three dimensions or higher (Modulus 3-dim. Brownian motion convergence). The two dimensional case is similar but the harmonic function involved is a logarithm.

Hence, if $\tau_{B'}$ and $\tau_B$ denote the first exit times from the punctured and the  unpunctured balls respectively, we get
$$
\mathbb{P}^0[\tau_{B'}=\tau_B]=1
$$
and $\mathbb{P}^0[\tau_B=0]=0$. Since $0$ is on the boundary of the punctured ball, this implies that $B'$ is not regular for Brownian Motion.


问:
If a function is the limit in $L^1$ of simple functions, then that function is integrable
I'm trying to prove that the completion of the integrable simple functions on $ (X,\Sigma,\mu) $ is $L^1(X,\Sigma,\mu)$. Suppose we have:

$lim_{n\to \infty}||f-s_n||_1=0$ for some sequence $(s_n)_n$ of simple integrable functions. How can I prove that $\int{|f(x)|}d\mu < \infty$?

答:
Write
$$\int |f| \ d\mu \leq \int |f - s_n| \ d\mu + \int |s_n| \ d\mu,$$
where $n$ is taken to be sufficiently large.


问:
Can you solve $\int 1/(x^2+1)^2\, dx$
I know that the integral looks like like the anti-derivative of $\arctan$, but i don't know how to use this fact so I tried to use a fraction decomposition to: $$\frac{1}{(x^2+1)^2}$$
to solve the anti-derivative : $$\int \frac{1}{(x^2+1)^2}\, dx$$
so :$$\frac{1}{(x^2+1)^2}=\frac{A}{x^2+1}+\frac{B}{x^2+1}\iff \frac{1}{x^2+1}=A+B$$
$$1+0x^2=(A+B)x^2+(A+B)1$$
Thus I got this system of equations:
$$\cases{A+B=1 \\ A+B=0}$$ Which is impossible. I also tried to use substitution but it didn't help me.

Title: Can you solve $\int 1/(x^2+1)^2\, dx$

Body:
$I=\int \frac{dx}{(x^2+1)^2}=\frac{1}{2} \int \frac{1+x^2+1-x^2}{x^4+2x^2+1}.dx$

$\implies I= \frac{1}{2} \left ( \int \frac{1+x^2}{x^4+2x^2+1}.dx +\int \frac{1-x^2}{x^4+2x^2+1}.dx\right).$

$I_1=\int \frac{1+x^2}{x^4+2x^2+1}.dx=\int \frac{1+\frac{1}{x^2}}{x^2+2+\frac{1}{x^2}}.dx=\int \frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})^2+4}.dx$

Now $t=x-\frac 1 x \implies dt=\left(1+\frac{1}{x^2} \right)dx$

$I_1=\int\frac{dt}{t^2+4}=\frac{1}{2}\tan^{-1}\left(\frac{t}{2}\right)$.

$I_2=\int \frac{1-x^2}{x^4+2x^2+1}.dx=-\int \frac{1-\frac{1}{x^2}}{x^2+2+\frac{1}{x^2}}.dx=-\int \frac{1-\frac{1}{x^2}}{(x+\frac{1}{x})^2+1}.dx$

Now $p=x+\frac 1 x \implies dp=\left(1-\frac{1}{x^2} \right)dx$

$I_2=\int\frac{dp}{p^2+1}=\tan^{-1}(p)$

$I=I_1+I_2=\frac{1}{2}\tan^{-1}\left(\frac{x-\frac 1 x}{2}\right)+\tan^{-1}(x+\frac 1 x).$


答:
You cant simplify with PFD on something like this, you do have a few choices though:

Trigonometric substitution

manipulate the expression

I think the Trig is fairly well covered so I will do how to manipulate the expression:
$$\frac{1}{(1+x^2)^2}=\frac{1+x^2-x^2}{(1+x^2)^2}=\frac{1+x^2}{(1+x^2)^2}-\frac{x^2}{(1+x^2)^2}$$
$$=\frac{1}{1+x^2}-\frac{x^2}{(1+x^2)^2}=\frac1{1+x^2}-x\frac{x}{(1+x^2)^2}$$
Now the first part of this is a standard integral and the second can be done using integration by parts, as shown well by @Quanto now that I check.

Title: Can you solve $\int 1/(x^2+1)^2\, dx$


问:
Is $\exists x \, \forall y : P(x,y)$ equivalent to $\exists x : P(x,y) \; \forall y$?
Are the two statements
\begin{align*}
\exists x \, \forall y : P(x,y) \qquad \text{and} \qquad  \exists x : P(x,y) \; \forall y
\end{align*}
equivalent?

Or is this more of a matter of what one understands by the second statement, which I think of as really $\exists x : (P(x,y) \, \forall y)$?

答:
Let $P(x,y)$ be some property on two variables $x$ and $y$.

Note that when one writes $\exists x \, \forall y \, \colon P(x,y)$, it means that “There is an $x$, such that for all $y$, $P(x,y)$ holds.”. On the other hand, when one writes $\exists x \, \colon P(x,y) \, \forall y$, it means “There is an $x$ such that $P(x,y)$ holds for all $y$.”.

They are not strictly identical (they are different expressions), but they are equivalente, in the sense of being two ways of expressing the same idea.

Although, most people don’t write $\exists x \colon P(x,y) \, \forall y$, in mathematical notation.


答:
The smoothness of the "eigenvalue group" is a standard fact which can be found in
T. Kato's perturbation theory of linear operators (first chapter, I think page 71).</p>

Title: Differentiability of sum of eigenvalues.

Body:
Let $A(x)=(a_{ij}(x))_{i,j}$ be a smooth map of symmetric matrices for $x\in\Omega\subseteq\mathbb R^n$. (i.e., every $a_{ij}(x)=a_{ji}(x)$ is smooth.) Let $k_1\le\cdots\le k_n$ be its eigenvalues. If we know that at a point $p,$ $k_1=k_2<0$ and $k_1=k_2<k_3,$ can we derive that $k_1+k_2$ is also smooth?

When $n=3,$ since $k_3$ has multiplicity $1$ and $k_1+k_2=\text{tr} A-k_3,$ we know this statement is correct. However, in general, I am not sure how to see this.

Any ideas or comments are appreciated!


问:
For all real numbers satisfying $a < b$, there exists an $n \in \mathbb{N}$ such that $a + 1/n < b.$
From Stephen Abbott's Understanding Analysis 1.2.11:

For all real numbers satisfying $a < b$, there exists an $n \in \mathbb{N}$ such that $a + 1/n < b.$

My try:

$$\forall a\in \Bbb R, \forall b\in \Bbb R, \exists n \in \Bbb N \space \text{that satisfies}: \space a<b \Rightarrow a+\frac 1n<b$$

$\because a<b $

$\therefore\exists m\in\Bbb R, m>0$ such that $b=a+m$.

So the statement we are proving becomes (substitute $b$ by $a+m$):

$$\forall a\in \Bbb R, m \in \Bbb R, \exists n\in\Bbb N \space \text{that satisfies}: a<a+m \Rightarrow a+\frac 1n < a+m$$

Which is (take $a$ from both sides):
$$\forall m\in \Bbb R, \exists n\in\Bbb N \space \text{that satisfies}:0<m \Rightarrow \frac 1 n<m$$
We write this as: $$\forall m\in \Bbb R, 0<m, \exists n\in\Bbb N \space \Rightarrow \frac 1 n<m$$

We define set $M = \{m\mid M=\Bbb R \cap m>0\}$

And set $N = \{\frac 1n \mid n \in \Bbb N\}$

$\because\forall q \in \Bbb Z, p\in\Bbb Z, \frac p q\in\Bbb Q$

$\space\space\space n\in\Bbb N\subset\Bbb Z$

$\therefore\frac 1n\in\Bbb Q$

Mention that $m\in\Bbb R$

$\because\Bbb Q\subsetneq\Bbb R$

$\therefore\forall n\in\Bbb N, \exists m \in \Bbb R \Rightarrow \frac 1n \geqslant m$

This statement is equivalent to:
$$\forall m\in \Bbb R, 0<m, \nexists n\in\Bbb N \space \Rightarrow \frac 1 n<m$$

Which disproves the statement, which is wrong - What is wrong with my disproof?

Also, I am learning how to use mathematical notations properly (because I am only in year 11 but anyway). Please tell me if there is any error in my expressions.

Thanks a lot!

答:
In your eagerness to make things formal (very formal), you've ended up juggling a bit too many entities. :)

So if I understand correctly, you're saying that the statement from Abbott's textbook should be true, whereas you found a proof that it isn't. Informally, your argument seems to boil down to something like this:

I have to show that, given $m\in\mathbb{R}$, $m>0$, there exists $n\in\mathbb{N}$, such that $\frac{1}{n}< m$. But it also holds that, for any $n\in\mathbb{N}$, I can always find an $m\in\mathbb{R}$ such that $\frac{1}{n}\geq m$. So Abbott's statement is false.

I guess the misunderstanding comes from using the same symbol '$m$' to refer to what are two distinct numbers.

First off, let's check that the original statement is indeed true. As Ethan mentions, your chain of reasoning (with the $m$ and the sets $M$ and $N$) is a bit overkill. More simply, just notice that the inequality you start with can be rewritten as $n>\frac{1}{b-a}$. In other words, you have to show that for any $a,b\in\mathbb{R}$ such that $a<b$, there exists an $n\in \mathbb{N}$ such that $n>\frac{1}{b-a}$.

Since $b-a>0$, it is not difficult to see that this, in general, true. As an example, suppose $a=0.02$ and $b = 0.34$ (I chose these numbers at random). Then $\frac{1}{b-a} = 3.125$. Can we find a natural number greater than $3.125$? Yes: $4$ works; $5$ works; anything greater than $5$ also works. Broadly speaking, no matter what $a$ and $b$ you start with, the bag of natural numbers is large enough that you can always find some $n$ that is large enough and that fits your needs.

This is just to give some intuition for why the original statement is true.

Now coming back to your proposed answer, I would say the last inference is a mistake. Indeed, for any $n\in\mathbb{N}$ there exists $m\in\mathbb{R}$ such that $\frac{1}{n}\geq m$. That is to say, given $n\in\mathbb{N}$ you can always find some very small $m$, smaller than $\frac{1}{n}$. But that doesn't mean that you've disproven the original statement, because this $m$ is not the same as the $m$ you've started with! It's a different $m$, which, to avoid confusion, should probably not even be denoted $m$ in the first place. Perhaps if you give it a different name, say $r$, it becomes clearer that from the last statement you introduce it does not follow that you've disproven the original claim.


问:
What does it mean when an integral contains dx'?
I am trying to use the following mathematical model to model some Physical properties of materials. Alpha and Beta are constants but I have a question about the integration. Here is the equation:

Equation from McIntyre (1966) that I am trying to solve

My question is: there is a dx' (and dx'') in the integrals. Am I supposed to evaluate this integral any differently than normal? Additionally, there is an x' bound. How would this be handled?

答:
as there is nothing else relating to $x$ present in each of the integrals, you can treat $x'$ as a variable in its own right, that is to say:
$$\int dx'=x'+C$$
if there were an expression of $x$ present it would be much more difficult although $x$ could also represent a function i.e. $x=x(t)$ which would give you:
$$dx'=d\left[\frac{dx}{dt}\right]=\frac{d^2x}{dt^2}dt$$


答:
Unfortunately, sometimes people write:

there exists $x$,  $y=x^2$, for all $y$

but it is ambiguous.  It might mean

there exists $x$, for all $y>0$, $y=x^2$

which is false,

or it might mean

for all $y>0$ there exists $x$, $y=x^2$

which is true.

Title: Is $\exists x \, \forall y : P(x,y)$ equivalent to $\exists x : P(x,y) \; \forall y$?

Body:
Are the two statements
\begin{align*}
\exists x \, \forall y : P(x,y) \qquad \text{and} \qquad  \exists x : P(x,y) \; \forall y
\end{align*}
equivalent?

Or is this more of a matter of what one understands by the second statement, which I think of as really $\exists x : (P(x,y) \, \forall y)$?


问:
What is wrong with this fake proof that $\lim\limits_{n\rightarrow \infty}\sqrt[n]{n!} = 1$?
$$\lim_{n\rightarrow \infty}\sqrt[n]{n!}=\lim_{n\rightarrow \infty}\sqrt[n]{1}*\sqrt[n]{2}\cdots\cdot\sqrt[n]{n}=1\cdot1\cdot\ldots\cdot1=1$$
I already know that this is incorrect but I am wondering why. It probably has something to do with the fact that multiplication in $n!$ is done infinite number of times.</p>

答:
Start by figuring out a simpler example:
$$1 = \lim_{n\to\infty} \frac n n = \lim_{n\to\infty} \frac {1+1+\ldots+1} n = \lim_{n\to\infty} \frac 1 n + \frac 1 n + \ldots + \frac 1 n = 0 + 0 + \ldots + 0 = 0$$

Indeed, you cannot exchange sum (or product) and limit if the amount of terms in the sum or product depend on the limiting variable.


问:
Plane parallel to the base of a triangular pyramid
A triangular pyramid $ABCD$ is given. A plane parallel to the plane $(ABC)$ intersects $DA,DB$ and $DC$ at $A_1,B_1$ and $C_1,$ respectively. Show that $\dfrac{DA_1}{DA}=\dfrac{DB_1}{DB}=\dfrac{DC_1}{DC}$.

I think I am supposed to use Thales theorem for the given equality but I am not sure how to begin the solution. Thank you in advance!

答:
Consider a side of the pyramid, say $\triangle ABD$. $\frac{DA_1}{DA}=\frac{DB_1}{DB}$ then follows from intercept theorems.

$\hspace{6cm}$


问:
Question about continuity of stopping time probability in proof of Dirichlet problem in Stein and Shakarchi
In Chapter 6 in the book Functional analysis by Stein and Shakarchi the following theorem (Dirichlet problem) is proved:

$\mathcal{R}$ denotes a bounded open set and a point $y$ is called regular if
$$\mathbb{P}(\tau_\ast^y=0) = 1$$
where $\tau^x = \inf\{t\geq 0: x+B_t\notin \mathcal{R}\}$ and $\tau_\ast^x = \inf\{t>0: x+B_t\notin \mathcal{R}\}$ and $B_t$ is a Brownian motion started at $0$ and $B_t^x = x+B_t$.

Now my question concerns a particular step in the proof of (b).

I find the fact that $x\mapsto 1_{\{\omega':x+B_t(\omega')\in \mathcal{R}, \, \forall \epsilon\leq t \leq \delta\}}(\omega)$ should be continuous, hard to believe. Assume that $\mathcal{R} = \mathbb{D}\setminus \{0\}$, the unit disk with the $0$ removed. Then if $B_{t_0}^y(\omega) = 0$ for precisely one $t_0\in [\epsilon,\delta]$ it is not necessarily so that $B_t^x(\omega) = 0$ for some $t\in [\epsilon,\delta]$ regardless of how close $x$ is to $y$. Could someone clarify this for me? Is this an error in the book?

答:
DISCLAIMER: THIS ANSWER IS INCOMPLETE

I think the key point is that $\mathcal{R}$ is an open set. Fix an $\omega$ such that $B_t$ is continuous. First note that $g(y,t)= B_t^y =y+B_t$ is jointly continuous.

Let us first show that $A:=\{y:y+B_t \in \mathcal{R}, \text{  all }\varepsilon\leq t \leq \delta\}$ is open. Suppsoe that $A$ is not open. Then there is a point $y \in A$ such that for every $n\geq 1$, you will get a $y_n \in (y-\frac{1}{n}, y+\frac{1}{n})$ such that $y_n \not\in A$. This means $y_n+B_{t_n} \not\in \mathcal{R}$ for some $t_n \in[\varepsilon,\delta]$. Now take a subsequence $y_{a_n}$ of $y_n$ such that for $y_{a_n}+B_{t_{a_n}}\not\in \mathcal{R}$, $t_{a_n} \in [\varepsilon,\delta]$, and $t_{a_n}\to t'$ for some $t'\in [\varepsilon,\delta]$ (you can always choose such a subsequence since $[\varepsilon,\delta]$ is compact). Now note that by construction $(y_{a_n},t_{a_n})\to(y,t')$ and therefore $y_{a_n}+B_{t_{a_n}} \to y+B_{t'} \in \mathcal{R}$ (using joint continuity of $B_t^y$). On the other hand $y_{a_n}+B_{t_{a_n}} \in \mathcal{R}^c$ which is closed, therefore the limit must be in $\mathcal{R}^c$ which is a contradiction. Therefore $A$ is open.

Coming to your problem, let $$f(y)=1\{B_t^y \in \mathcal{R}, \text{  all }\varepsilon\leq t \leq \delta\}= 1\{y \in A\}$$,
and we showed $A$ is open. Now you can conclude $f(y)$ is continuous on $A$ from here.

It still does not show that $f$ is continuous everywhere.


问:
Definition of Modular Arithmetic and Congruence
Definition Modular Arithmetic: Based on the definition of modular arithmetic, $a \equiv b \pmod{m}$ is meant to say that $a$ will leave the same remainder as $b$ when either $a$ or $b$ are divided by $m$. So the remainder of $\frac{a}{m}$ would be the same as $\frac{b}{m}$.

In a question, it was asked to find the integer $a$ such that $a \equiv 99 \pmod{41}$ and $100 \le a \le 140$.

Question 1: A solution to that was, "It's also good to note that for $x \equiv y \mod z$, $x$ is just the remainder of integer division of $y$ by $z$. For example, $35/14$ is equal to $2$ remainder $7$, and $7 \equiv 35 \mod 14$." But is this is not the same as the definition provided above, is not it?

答:
Yes, you are right. They meant to say, (as I guess):

Note that if $x$ is the remainder when $y$ is divided by $z$, then $x\equiv y\ (\text{mod}\ z)$.

Hope this helps. Ask anything if not clear :)


问:
A divisible abelian group is the direct sum of torsion subgroup and a torsion-free divisible subgroup.
Let $D$ be a divisible abelian group. It is known that $D \cong D_t \oplus D/D_t$, where $D_t$ is the torsion subgroup and $D/D_t$ is torsion-free and divisible. I want to find a torsion-free divisible subgroup of $D$, say $G$ such that $D = D_t \oplus G$. I can’t find an intuitive subgroup that satisfies this condition. My intuition told me that $G \cong D/D_t$, but I can’t find any intuitive isomorphism. Any help will be appreciated.

答:
There won't be any "intuitive" such subgroup, since there is not any canonical choice of such a $G$, and in general, I'm pretty sure the axiom of choice is needed to prove that one exists.  If you're familiar with injective modules, the quick proof is that $D_t$ is divisible and hence injective (by Baer's criterion), so whenever $D_t$ is a subgroup of another abelian group it is a direct summand.

Or, here is a more direct (but basically equivalent) argument.  By Zorn's lemma, there exists a subgroup $G$ of $D$ that is maximal with the property that $G\cap D_t=0$.  Since $D$ is divisible, $G$ must be divisible by maximality (if some $g\in G$ was not divisible by $n$ in $G$, you could adjoin an element $h\in D$ such that $nh=g$ to $G$ to contradict maximality of $G$).  Now suppose there were some $x\in D$ that is not in $G+D_t$.  Let $G'$ be the subgroup generated by $G$ and $x$.  By maximality of $G$, the intersection $G'\cap D_t$ must be nontrivial, which means that there is some $n\in\mathbb{Z}$ and $g\in G$ such that $nx+g$ is a nonzero torsion element of $D$.  Then $n$ must be nonzero, since $G\cap D_t=0$.  Since $nx+g$ is torsion, there is a nonzero $m\in\mathbb{Z}$ such that $mnx+mg=0$, so $mnx\in G$.  But now since $G$ is divisible, there exists $h\in G$ such that $mnh=mnx$.  Then $x-h$ is torsion so $x=h+(x-h)\in G+D_t$, a contradiction.  Thus $G+D_t$ is all of $D$, and $D=G\oplus D_t$, as desired.


问:
Limit $\lim_{(x,y)\to(0,0)}\frac{-16x^3y^3}{(x^4+2y^2)^2}$
Find the limit

$$\displaystyle\lim_{(x,y)\to(0,0)}\frac{-16x^3y^3}{(x^4+2y^2)^2}$$

Let $x=r\sin \theta$ and $y=r\cos \theta$.

$$
\begin{align}
\lim_{(x,y)\to(0,0)}\frac{-16x^3y^3}{(x^4+2y^2)^2}&=\lim_{r\to0}\frac{-16r^3\cos ^3 \theta r^3\sin ^3\theta}{(r^4\cos^4\theta+2r^2\sin^2\theta)^2} \\
&=\lim_{r\to0}\frac{-16r^3\cos ^3 \theta r^3\sin ^3\theta}{r^4(r^2\cos^4\theta+2\sin^2\theta)^2}\\
&=\lim_{r\to0}\frac{-16r^2\cos ^3 \theta \sin ^3\theta}{(r^2\cos^4\theta+2\sin^2\theta)^2}\\
&=\frac{-16(0)^2\cos ^3 \theta \sin ^3\theta}{((0)^2\cos^4\theta+2\sin^2\theta)^2}\\
&=\frac{0}{(0+2\sin^2\theta)^2}\\
&=0
\end{align}
$$

Does this logic follow? Can I ignore the case where $2\sin^2\theta=0$ meaning it is of the form $\frac{0}{0}$?
How else would I approach this limit.

答:
These things get confusing, I like to work out the extreme values using Lagrange multipliers. Here, I demand $x > 0, y > 0,$   maximum value of
$$ \frac{x^3 y^3}{(x^4+2y^2)^2}  $$
when the denominator is thought of as the constraint. Go through the gradients being parallel, the largest value occurs when
$$  24 x^2 y^2 (x^4 + 2 y^2)(x^4 - y^2) = 0 \; . \;  $$
The letters are real and positive, so $y=x^2$ gives the maximum for fixed denominator. This is then
$$ \frac{x^9 }{(x^4+2x^4)^2} = \frac{x^9 }{9x^8} = \frac{x }{9} $$
That is the biggest (my) fraction can be, in the first quadrant.

I was disappointed to find that the apparent inequality suggested  does not hold in general;safer to say that, when the denominator is $C>0,$  we know $x < C^{1/8}$ in the first quadrant. Then, let's see, the ratio is less than $$\frac{ C^{1/8}}{9}$$


问:
Is there an automaton with the given properties?
Let us assume that automaton A has the property that at most one transition enters each state. I wonder if such an automaton can recognize a language containing arbitrarily long words?</p>

答: The answer is yes. Consider the automaton $1 \xrightarrow{a} 1$, where the alphabet is $\{a\}$ and the unique state $1$ is both initial and final. This automaton recognizes the language $a^*$ which contains exactly one word of every length.


问:
Simplicial set map from $\Delta[n]$ to $K$ induced by an $n$-simplex $\sigma \in K_n.$
I am reading "rational homotopy theory" by Yves Felix, Stephen Halperin and
Jean-Claude Thomas on simplicial sets and simplicial cochain algebras. There is a lemma 10.3 page 118 which states that

If $K$ is any simplicial set then any $\sigma\in K_n$ determines a unique simplicial set map $\sigma_*:\Delta[n]\to K$ such that $\sigma_*(c_n)=\sigma$.

Here $c_n$ is the fundamental simplex of $\Delta[n]$.

What I tried: In order to get a map $\sigma_*:\Delta[n]\to K$, we need to construct a sequence of maps $(\sigma_*)_k: \Delta[n]_k \to K_k$ for all $k\geq 0$ such that $ (\sigma_*)_k \circ\tilde{\partial}_i^{k+1} = \partial_i^{k+1} \circ (\sigma_*)_{k+1}$ and $s_i^{k+1}\circ(\sigma_*)_k =  (\sigma_*)_{k+1}\circ\tilde{s}_i^{k+1} $ for all $k\geq 0$, where $\tilde{\partial}_i^k , \partial_i^k $ are face maps, $\tilde{s}_i^k, s_i^k $ are degeneracy maps  of $\Delta[n]$ and $K$ respectively, for all $k\geq 0$. That is the following two ladders are commutative:
$\require{AMScd}$
\begin{CD} \Delta[n] : \cdots @>>>\Delta[n]_n @>{\tilde{\partial}_i^n}>> \cdots @>>> \Delta[n]_{k+1} @>{\widetilde{\partial}^{k+1}_i}>> \Delta[n]_k @>{\widetilde{\partial}^{k}_i}>> \cdots @>>> \Delta[n]_1 @>{\widetilde{\partial}^1_i}>>\Delta[n]_0 \\
 @V{\sigma_*}VV@V{(\sigma_*)_n}VV@.@V{(\sigma_*)_{k+1}}VV@V{(\sigma_*)_k}VV@.@V{(\sigma_*)_1}VV@V{(\sigma_*)_0}VV\\
K: \cdots @>>> K_n @>{\partial^n_i}>>\cdots @>>>K_{k+1} @>{\partial^{k+1}_i}>> K_k @>{\partial^k_i}>> \cdots @>>> K_1 @>{\partial^1_i}>>K_0
\end{CD}

\begin{CD} \Delta[n] : \cdots @<<<\Delta[n]_n @<{\tilde{s}_i^n}<< \cdots @<<< \Delta[n]_{k+1} @<{\widetilde{s}^{k+1}_i}<< \Delta[n]_k @<{\widetilde{s}^{k}_i}<< \cdots @<<< \Delta[n]_1 @<{\widetilde{s}^1_i}<<\Delta[n]_0 \\
 @V{\sigma_*}VV@V{(\sigma_*)_n}VV@.@V{(\sigma_*)_{k+1}}VV@V{(\sigma_*)_k}VV@.@V{(\sigma_*)_1}VV@V{(\sigma_*)_0}VV\\
K: \cdots @<<< K_n @<{s^n_i}<<\cdots @<<<K_{k+1} @<{s^{k+1}_i}<< K_k @<{s^k_i}<<\cdots @<<< K_1 @<{s^1_i}<<K_0
\end{CD}

First I define $(\sigma_*)_0:\Delta[n]_0 \to K_0$ by $(\sigma_*)_0(<e_j>) = \partial_0^1 \circ \cdots \circ \partial_0^n (\sigma)$, and for any $k>0$ define $(\sigma_*)_k:\Delta[n]_k\to K_k$ by $(\sigma_*)_k(<e_{j_0}, \ldots, e_{j_k}>) = s_i^k \circ (\sigma_*)_{k-1} \circ \tilde{\partial}_j^k(<e_{j_0}, \ldots, e_{j_k}>)$ for any choice of $i$ and $j$. I manage to show that the $(\sigma_*)_k$ defined as above are well defined irrespective of the choice of $i$ and $j$ in the definition and that both of the two ladders are commutative. That is $\sigma_*$ is a simplicial set map, but I cannot prove that $(\sigma_*)_n (c_n) =\sigma.$

Question: Is my definition of $\sigma_*$ correct? if this is not the right way to define $\sigma_*$, then how can I define? secondly if the definition of $\sigma_*$ is correct can we show that $(\sigma_*)_n(c_n)=\sigma$.

Any help is appreciated.

答:
I have not checked your definition of $\sigma_*$ carefully, but I suspect it is not correct. (For example, your definition of $(\sigma_*)_0(\langle e_j \rangle)$ does not depend on $j$, which is a bit suspicious.) I will sketch a slightly different approach, and leave the details for you.

The key point is that any element $\alpha \in \Delta[n]_k$ can be written as $s_{j_0} \cdots s_{j_a} \partial_{\ell_0} \cdots \partial_{\ell_b}(c_n)$ for some sequence of $j$s and $\ell$s. The idea here is that, if $\alpha = \langle e_{i_0}, \ldots, e_{i_k} \rangle$, then we can use the $\partial$s to "get rid of any unused elements in the sequence $\langle e_0, \ldots, e_n \rangle$", and then use the $s$s to " introduce any necessary repetitions". I encourage you to work out a few examples to see what I mean.

Now, to define $(\sigma_*)_k$, we proceed as follows. For $\alpha \in \Delta[n]_k$, we write $\alpha = s_{j_0} \cdots s_{j_a} \partial_{\ell_0} \cdots \partial_{\ell_b}(c_n)$. Then $(\sigma_*)_k(\alpha) = s_{j_0} \cdots s_{j_a} \partial_{\ell_0} \cdots \partial_{\ell_b}(\sigma)$.

I will leave it to you to check all the necessary details for this definition. I will also mention that there is another (essentially equivalent) approach to the theory of simplicial sets (via presheaves on the simplex category) that makes this problem much easier to deal with.


问:
Proving the gamma function integral converges.
I am trying to prove that the integral $\int_0^{\infty}x^{t-1}e^{-x}dx$ converges for all $t>0$.

My first thought is to bound the function above by some other function which converges, however I could not think up of a function which does so.

I know the function has a local maximum on $[0,\infty)$ at $x = t-1$, so I thought about separating the integral about that point and then bounding each one separately. However I could not, again, think of a function which bounds the curve for any $t$.

My final approach was to use the power series of $e$ and then integrate power by power.

Are any of my approaches correct or am I completely on the wrong path?

答:
Indeed, split it up into an integral away from $0$ and an integral near $0$, say $(1,\infty)$ and $[0,1]$. First consider $(1,\infty)$.
In these situations it is common to "borrow" from the dominating term as follows.
Write $x^{t-1} e^{-x}$ as
$$
\frac{x^{t-1}}{e^{\epsilon x}} e^{-(1-\epsilon)x}
$$
for $\epsilon >0$ very small ($<1$ is enough).
Next you can show using limits that the term on the left is bounded as $x \to \infty$. Thus an integral away from $0$ will converge because $\int_1^\infty e^{-(1-\epsilon)x} dx$ does. For the part close to $0$, the $e^{-x}$ term is bounded and the integral converges because $\int_0^1 x^{t-1} dx$ converges.


答:
According to subadditivity of definite integrals, we split the integral into two as follows
$$\int_0^{\infty}x^{t-1}e^{-x}dx = \underbrace{\int_0^1 x^{t-1}e^{-x}dx}_{(1)} + \underbrace{\int_1^\infty x^{t-1}e^{-x}dx}_{(2)}$$

For the first integral, we note that since $t > 0$, we can find a segment $[a, A]$ such that $t \in [a,A]$. Then, for $t \in [a,A]$ and $x \in [0, +\infty)$ we have
$$x^{t-1}e^{-x} = \frac{1}{x^{1-t}e^{x}} \le \frac{1}{x^{1-A}\cdot 1}$$
By comparison test, we observe that the integral $(1)$ converges.

For the second integral we note that
$$\lim_{x\to\infty}\frac{x^{t-1}e^{-x}}{\frac{1}{x^2}} = \lim_{x\to\infty}x^{t+1}e^{-x} = 0$$
and again by comparison test, the integral $(2)$ converges.

Title: Proving the gamma function integral converges.

Body:
I am trying to prove that the integral $\int_0^{\infty}x^{t-1}e^{-x}dx$ converges for all $t>0$.

My first thought is to bound the function above by some other function which converges, however I could not think up of a function which does so.

I know the function has a local maximum on $[0,\infty)$ at $x = t-1$, so I thought about separating the integral about that point and then bounding each one separately. However I could not, again, think of a function which bounds the curve for any $t$.

My final approach was to use the power series of $e$ and then integrate power by power.

Are any of my approaches correct or am I completely on the wrong path?


问:
What's the difference between $(\mathbb Z_n,+)$ and $(\mathbb Z_n,*)$?
I noticed that $(\mathbb Z_n,+)$ and $(\mathbb Z_n,*)$ are not the same thing. For example 2 is not invertible in $(\mathbb Z_6,*)$. However, since $\bar2+\bar4=\bar0$, thus it is invertible in  $(\mathbb Z_6,+)$.

I found the following problems:

Am I right? If so, why does any book not mention this difference which is very important?

When we study groups, what's the standard operation of $\mathbb Z_n$ when the operation is omitted?

When we study rings, do we generally use $(\mathbb Z_n,+)$ to be the abelian group of the ring $(\mathbb Z_n,+,*)$?

If I work with $(\mathbb Z_n,+)$ I would lose these theorems which I proved for $(\mathbb Z_n,*)$?

An element $a$ in $(\mathbb Z_n,*)$ is an unit iff $a$ and $n$ are  coprimes

An element $a$ in $(\mathbb Z_n,*)$ is an zero divisor iff $a$ and
  $n$ aren't coprimes

I'm sure there are a lot of student with this same doubt.

I really need help.

Thanks a lot.

答: Well, $(Z_{n}, *)$ is just the set $Z_{n}$ and a binary operation multiplication modulo $n$. Yes, you are right, this is entirely different from $(Z_{n}, +)$. The reason the books will not mention it because the former is not a group and hence not discussed, but nobody will say that they are the same things.</p>


问:
Is the difference between stable nonlinear systems Lyapunov stable?
If there are two nonlinear systems with stable equilibria $x_1 = x_2 = 0$
$$\dot x_1 = f(x_1, u) \qquad \dot x_2 = g(x_2, u)$$
with identical inputs $u$, is the difference between the system states $x_1, x_2 \in \mathbb{R}^n$
$$e = x_1 - x_2$$
stable?

Trivial case

If the systems are LTI with identical system and input matrices $A$ and $B$
$$f(x_1 = x, u) = g(x_2 = x, u) = Ax + Bu,$$
the system dynamics is
$$\dot e = Ae.$$
A Lyapunov function
$$V(e) = e^T P e$$
results in
$$\dot V(e) = \dot e P e^T + e^T P \dot e = e^T (A^T P + P A) e.$$
If $A$ is Hurwitz, then $A^T P + P A = -Q$ with a positive definite real symmetric $P$ and a positive definite $Q$. As $\dot V(e) = -e^TQe < 0\ \forall x \neq 0$, the difference between the two asymptotically stable LTI systems is asymptotically stable.

My Question

Can this be generalized to arbitrary nonlinear globally/locally/asymptotically/... stable systems,

where $f = g$? (1)

where $f \neq g$? (2)

Thought experiment: An equilibrium is stable if for each $\epsilon > 0$, there is a $\delta$ such that with $|| x || < \delta$ at $t = t_0$ the state remains within $|| x || < \epsilon$. Wouldn't that imply that the vector $e$ connecting two $x_1, x_2$ remains within a hypersphere $\mathcal{B}$ that includes both $|| x_1 || < \epsilon_1$ and $|| x_2 || < \epsilon_2$? Then for $|| e || < \min(\delta_1, \delta_2)$ at $t = t_0$, it would hold $|| e || < 2r_\mathcal{B}$ with $r_\mathcal{B}$ radius of $\mathcal{B}$?

At the same time: Starting from
$$V(e) = \frac{1}{2} e^T e,$$
I quickly get stuck at
$$\dot V(e) = e^T \dot e = e^T \left(f\left(x_1, u\right) - g\left(x_2, u\right)\right) =\ ...?$$

Update

Arastas answer has a counterexample for $f \neq g$ (loosely quoting):

$\dot x_1 = u$ and $\dot x_2 = -u$ leads to $\dot e = 2u$. With $u=1$, we get $e(t) \rightarrow \infty$ for $t \rightarrow \infty$.

In this SISO example, the derivatives of $f$ and $g$ w.r.t. $u$ differ:
$$\frac{\partial}{\partial u} f(x_1, u) = 1 \neq -1 = \frac{\partial}{\partial u} g(x_2, u).$$

Speaking more generally, the (transposed) gradient matrices of $f$ and $g$ w.r.t. $u$ differ:
$$(\nabla_u f)^T = \frac{\partial}{\partial u} f(x_1, u) \neq \frac{\partial}{\partial u} g(x_2, u) = (\nabla_u g)^T.$$

What can we say about the stability of $e$ for arbitrary $f, g$, but allowing for restrictions on the gradients $\nabla_u f$, $\nabla_u g$? (3)

If the restriction were made $\nabla_u f = \nabla_u g$, (3) would capture case (1).

答:
I will assume that although $u$ is the same in both, that it is not a-priori fixed. That is you are considering that you can make the error asymptotically stable with the appropriate choice of $u.$ My interpretation of your question is:

When is the parallel interconnection of two IO stable systems IO stable? (Where IO Stable can mean whatever type of stability you want it to mean)

For all practical purposes this is what you are asking since we can always absorb a sign into one of the system's outputs and ask that question for $e' = y_1 + y_2.$ Now, this is still not obviously true. For example, Arastas' combined system is not controllable. In particular, note that the stacked system is

$$\begin{aligned}\begin{pmatrix} \dot{x}_1 \\ \dot{x}_2 \end{pmatrix} &= \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} {x}_1 \\ {x}_2 \end{pmatrix} + \begin{pmatrix} 1 \\ -1 \end{pmatrix} u, \\ y &= x_1 - x_2\end{aligned}$$

Apply the Kalman Controllability Conditions and verify a lack of controllability. However, it is stabilizable! The eigenvalues of the system matrix remain negative. In fact, this is true for any stacked systems that are otherwise decoupled. The eigenvalues of the combined system are necessarily the union of eigenvalues. So if you have two systems whose system matrix is Hurwitz, then the stacked state also has a Hurwitz system matrix. Moreover, if both systems are controllable, you at the minimum have stabilizability. And, of course, if all the states decay, clearly the output $y := x_1 - x_2 \to 0$ as well.

Of course, by stacking the system, it should be made clear that the feedback used to stabilize the system may be different than the feedback used to stabilize the individual subsystems. As does the nature of the stability. That shouldn't be surprising. However, my interpretation of your question is that $u$ is the same in both, but not a-priori fixed. So under that interpretation this is reasonable.

Now, what about in the nonlinear case. Clearly, even in the linear case, you are going to lose modes of controllability and so that is likely to highly constrain the nonlinear case (and being nonlinear the problem gets harder). First off, recognize that if both systems can be linearized, have Hurwitz jacobians and are controllable we can just defer to the above result. So let us consider the other cases. This is where it gets tricky. One place to look is passivity and that is not a bad place to look since you are already considering Lyapunov stability. Assuming you absorb a sign in the right place and prove that the sign flipped system is passive (i.e. you proved the map $u_2\mapsto -y_2$ is passive) you can consider

When is the parallel interconnection of two passive systems, passive?

Always. Once the combined system is passive, you can stabilize it with a strictly passive feedback to get $L_2$ stability ($L_2$ inputs produce $L_2$ outputs).

You can also consider feedback interconnections and cleverly organize your inputs and outputs to get some input to the error is passive probably. However all of these considerations depend highly on the system you are considering as they are quite restrictive! There are a wide variety of tools available that, once you have determined a system and specific category of stability, could be leveraged.


答:
First note that
$$\int_0^1x^{t-1}dx$$
converges and then use that $|x^{t-1}e^{-x}|\leq x^{t-1}$ for $x\in[0,1]$.
Then see that
$$\int_1^{\infty}x^{-2}dx$$
converges and use that
$$\frac{x^{t-1}e^{-x}}{x^{-2}}=\frac{x^{t+1}}{e^x}\rightarrow 0,x\rightarrow \infty.$$


问:
Finding the smallest chord that passes through a point
I'm having some trouble with the following exercise:

Let $A$ be a point inside a circumference. Determine the smallest chord that passes through $A$.

So my guess was that the smallest chord that passes through $A$ is the one that is perpendicular to the segment $AA'$ in the following picture ($[MN]$):

I think that this is the correct answer but I'm having some trouble proving this.

Is this correct? If so, how can I prove this?

答:
Draw any chord through $A$. Let its mid-point be $P$.

The length of the chord is equal to $2\sqrt{r^2 - OP^2}$.

We know that $AP \perp OP$. We have $OP^2 + AP^2 = OA^2$.

Hence the length of the chord is equal to $2\sqrt{r^2 - OA^2 +AP^2}$.

Since both $r$ and $OA$ are fixed, to minimize the length is to minimize $AP$.

When $AP = 0$, $A$ is the mid-point of the chord, and $OA$ is perpendicular to the chord.


答:
There are two related notions, and similar notation; which may be confusing.

"Binary mod operator": given an integer $a$ and a positive integer $m$, the binary mod operator $a\bmod m$ (pronounced "$a$ mod $m$" or "$a$ modulo $m$") takes as value the remainder when dividing $a$ by $m$ (integer division with remainder); this remainder is assumed to satisfy $0\leq a\lt m$.  (This may be different in some computer languages, where the value is taken to have least possible absolute value; so for example, $a\bmod 3$ could be $-1$, $0$, or $1$.)  This is an operation: it takes to arguments ($a$ and $m$), and returns a value (the remainder).

The relation "mod $m$". We say that $a\equiv b\pmod{m}$ (pronounced "$a$ is congruent to $b$ modulo $m$) if and only if $m|a-b$. This is a binary relation on the integers. You take two integers, and either they are in the relation, or they are not. (In fact, it is an equivalence relation).

The connection between the two notions is the following:

Theorem. Let $m\gt 0$. For integers $a$ and $b$,
$$a\equiv b\pmod{m} \iff a\bmod m = b\bmod m.$$
That is: $a$ is congruent to $b$ modulo $m$ if and only if $a$ and $b$ leave the same remainder when you divide them by $m$.

Proof. Write $a=qm+r$, $0\leq r\lt m$, and $b=pm+s$, $0\leq s\lt m$. Without loss of generality, we may assume $s\leq r$.

Since $a-b = (q-p)m + (r-s)$, with $0\leq r-s\lt m$, we have that $m|a-b$ if and only if it divides $r-s$, if and only if $r-s=0$, if and only if $a\bmod m = r = s =b\bmod m$. $\Box$

So you could define "$a\equiv b\pmod{m}$ if and only if $a$ and $b$ have the same remainder when divided by $m$".

Note that you use $\equiv$ when you work with the equivalence relation, but you use $=$ when you work with the operator. Your final paragraph should really say that

if $x= y\bmod z$, then $x$ is the remainder of integer division of $y$ by $z$

But note the use of the "binary mod" (MathJax/$\LaTeX$ code \bmod) and the equal sign. If you say "$x\equiv y \bmod{z}$ means that $x$ is the remainder of integer division of $y$ by $z$", then that's not technically correct, since you are mixing notations, and in addition, for instance $5\equiv 7\pmod{2}$, and $5$ is not the remainder when you divide $7$ by $2$; the remainder is $1$.

So the answer to your question 1 is that (i) as written it is false because you are mixing the relational and the operational mods, and they are not the same. But (ii) written correctly it asserts something about the mod operator, which is not the mod relation which is what the definition is about.

Title: Definition of Modular Arithmetic and Congruence

Body:
Definition Modular Arithmetic: Based on the definition of modular arithmetic, $a \equiv b \pmod{m}$ is meant to say that $a$ will leave the same remainder as $b$ when either $a$ or $b$ are divided by $m$. So the remainder of $\frac{a}{m}$ would be the same as $\frac{b}{m}$.

In a question, it was asked to find the integer $a$ such that $a \equiv 99 \pmod{41}$ and $100 \le a \le 140$.

Question 1: A solution to that was, "It's also good to note that for $x \equiv y \mod z$, $x$ is just the remainder of integer division of $y$ by $z$. For example, $35/14$ is equal to $2$ remainder $7$, and $7 \equiv 35 \mod 14$." But is this is not the same as the definition provided above, is not it?


问:
Maximize (but not minimize?) entropy for probability distribution
Let $p$ be an arbitrary discrete probability distribution on $\{1, \dots, 10\}$. We can represent such distribution by a vector indexed by by $i$ subject to the constraints $p_i \ge 0$ and $\sum^{10}_{i=1}p_i = 1$. I want to find analytically the $p$ with maximum entropy which is given by

$$
H(p) = - \sum_{i=1}^{10} p_i \log(p_i)
$$

My question consists of two parts:

How do I show that the optimal $p$ is given by the uniform distribution using the Lagrange method.

Why can't the Lagrange method be used to find $p$ with minimal entropy (which I suppose is given by the dirac distribution)?

For 1. I know that the Lagrangian function associated with this constrained problem is given by

$$
\mathcal{L}(p,\lambda) = - \sum_{i=1}^{10} p_i \log(p_i) + \lambda(1 - \sum^{10}_{i=1}p_i)
$$

and the respective gradient by

$$
\frac{\partial \mathcal{L}}{\partial p_i} = - \log(p_i) - 1 - \lambda
$$

and

$$
\frac{\partial \mathcal{L}}{\partial \lambda} = 1 - \sum^{10}_{i=1}p_i
$$

but I don't know how to continue from there.

答:
Why can't the Lagrange method be used to find $p$ with minimal entropy (which I suppose is given by the dirac distribution)?

The KKT conditions, which can incorporate the nonnegativity constraints, are not necessary for optimality when the objective function is not continuously differentiable. In this case the objective function is not differentiable at certain points, so those points can never be found by this method (the stationarity condition is $\log(p_i)=\mu_i-\lambda-1$ which cannot have $p_i=0$ (for some $i$) as a solution).

You can use that the minimum of a concave function over a convex set is attained at an extreme point of the convex set. This is known as the maximum principle. The extreme points are the vectors with $p_i=1$ for some $i$ and $p_i=0$ elsewhere. By comparing the function value at those points, you find that they are all globally optimal.

How do I show that the optimal p is given by the uniform distribution using the Lagrange method.

You should do what Nown did to find potential local optima. Then you should compare the function value with the function values at all points for which the objective function is not differentiable (or the derivative is not continuous), because the Lagrange method may not find those points. That's a lot of work here because the derivative does not exist when $p_i=0$ for at least one $i$. I would argue that the Lagrange method is just not suitable to solve this problem.

Maximizing a symmetric concave function over the standard simplex is an easy task if you follow this answer.


问:
Finitely generated module over $A[[x]]$
I have a question about module over power series ring.

Let $A$ is a local ring with maximal ideal $\mathfrak m$, I'm more interested in the case that $A=\mathbb Z_p/p^n\mathbb Z_p$

If $M\neq0$ is a finitely generated torsion module over $A[[x]]$, then is $M$ finitely generated over $A$?

I think it suffices to assume that $M$ is generated by one element $m$ and $m$ is killed by some series $f(x)=\sum_{i=0}a_ix^i$. But if $a_0\in\mathfrak m$, I can't deduce anything.

Thanks.

答:
$p$ is nilpotent. For any $f\in pA[[x]], g\in x^m A[[x]]^\times$,

$f-g$ is not a zero-divisor and any non-zero divisor is of this form.

let $$h=\prod_{m=0}^{n-1}(f^{2^m}+g^{2^m})$$

we have that $$(f-g) (h)=(f^{2^n}-g^{2^n})=(g^{2^n})=(g)^{2^n}=(x^m)^{2^n}=(x^{m 2^n})$$

When including non-zero divisor in the definition of torsion module you get that
$M$ is a finitely generated $A[[x]]$-module killed by such a non-zero divisor $f-g\in A[[x]]$, whence $M$ is killed by some $x^{m2^n}$,  it is a finitely generated $A[[x]]/(x^{m 2^n})$ module.

Therefore it is also a finitely generated $A$-module.

Title: Finitely generated module over $A[[x]]$


问:
Factors of a stochastic matrix
If $A B = C$ where matrix $C$ is stochastic (all entries are positive and all rows add to $1$) then is it necessary that both $A$ and $B$ be stochastic?

答: $$A=\begin{pmatrix} 2 & 0 \\ 0 & 2\end{pmatrix}, B = \begin{pmatrix}\frac12 & 0\\0 & \frac12\end{pmatrix}.$$

Title: Factors of a stochastic matrix


问:
Proof functions in involution for R matrix Lie-Poisson bracket
I'm reading the well-known article of Reyman and Semenov-Tian-Shansky called "Group-Theoretical Methods in the Theory of Finite-Dimensional Integrable Systems".

Suppose $\mathfrak{g}$ is a Lie algebra, and $R$ is an $R$-matrix (so $[x,y]_R = [Rx,y] + [x,Ry]$ is a Lie bracket on $\mathfrak{g}$). So $\mathfrak{g}$ has two Lie brackets (the usual one, and $[-,-]_R$), and thus $\mathfrak{g}^*$ has two Lie-Poisson structures. I'm reading the proof of the fact that if $f$ and $g$ are coadjoint-invariant functions on $\mathfrak{g}^*$, then $\{f,g\}_R = 0$.

The authors say that

"The invariance of $f$ is equivalent to the relation $\left<\lambda, [d_\lambda f, x] \right> = 0$ for all $x \in \mathfrak{g}$"

Here, $df$ (at a point $\lambda \in \mathfrak{g}^*$) is identified with an element of $\mathfrak{g}$, and $\left<-,-\right>$ is the dual pairing of $\mathfrak{g}^*$ and $\mathfrak{g}$.

This is their entire proof of this fact. I see why the result follows from this, but I do not see why this statement is true. Why is the fact that $f$ is coadjoint-invariant equivalent to this relation?

答:
Let $G$ be a simply connected Lie group with Lie algebra $\mathfrak{g}$. Note that $f$ being coadjoint invariant means that
\begin{align*} f(\mathrm{Ad}^*_g(\lambda))= f(\lambda), \quad \forall \lambda \in \mathfrak{g}^*,g\in G  \end{align*}
which for $g_t=e^{tx}$ with $x\in \mathfrak{g}$ implies
\begin{align*}
\frac{d}{dt}|_{t=0}f(\mathrm{Ad}^*_{e^tx}(\lambda))= \frac{d}{dt}|_{t=0} f(\lambda) =0
\end{align*}
but also
\begin{align*}
\frac{d}{dt}|_{t=0}f(\mathrm{Ad}^*_{e^tx}(\lambda))= \mathrm{d}_{\lambda}f(\mathrm{ad}^*_x(\lambda)) =\lambda ( [\mathrm{d}_{\lambda}f,x]).
\end{align*}


问:
is number of topologies can be defined on a finite set with $n$ element finite?
Given a set $A$ with $n$ elements, the finest topology is powerset $2^A$ and the coarsest is the trivial topology $\{ \emptyset, A\}$. And it seems to count the exact number of different topologies in between is an open problem.

But maybe to fulfill my curiously, do we know if the number of different topologies is finite?

答:
Yes, the number of different topologies on a set of cardinality $n$ is at most $2^{2^n}$, the number of sets of subsets of a set of cardinality $n$ (since every topology is a set of subsets of the set of points).

Note, incidentally, that there are two possible ways to interpret "the number of topologies on a set with $n$ elements." We could count the literal number of topologies on such a set, or the number of topologies up to homeomorphism. The latter is vastly smaller than the former in general; for example, there are $355$ different topologies on a $4$-element set but only $33$ different homeomorphism types. The former is, to my understanding, significantly easier to count than the latter; for example, there is a simple formula for the "log-asymptotic" behavior of the former, but to the best of my knowledge no such result is known for the latter.


答:
For any chord $PQ$ which is not perpendicular to $OA$, $OA \geq OS$ where $OS$ is perp to $PQ$. In other words, $OA$ is the maximum perpendicular distance from center to any chord passing through $A$.

If $r$ is radius of the circle, $l$ is chord length and $d$ is perpendicular distance from center to the chord,

$l^2 = 4 (r^2-d^2)$

$l$ is minimum when $d$ is maximum, which is $OA$.

Title: Finding the smallest chord that passes through a point

Body:
I'm having some trouble with the following exercise:

Let $A$ be a point inside a circumference. Determine the smallest chord that passes through $A$.

So my guess was that the smallest chord that passes through $A$ is the one that is perpendicular to the segment $AA'$ in the following picture ($[MN]$):

I think that this is the correct answer but I'm having some trouble proving this.

Is this correct? If so, how can I prove this?


问:
Calculation of eigenvalues
I want to calculate the eigenvalues of the matrix

$$\begin{pmatrix} \frac{apb}{\delta} - c & \frac{dpb}{\delta} \\ \frac{a(1 - p)b}{\delta} & \frac{d(1 - p)b}{\delta} - t \end{pmatrix}$$

I tried to use the characteristic polynomial but I can't factor it to find the polynomial root. Can you please help me?

答:
If $A = \begin{pmatrix} \frac{apb}{\delta} - c & \frac{dpb}{\delta} \\ \frac{a(1 - p)b}{\delta} & \frac{d(1 - p)b}{\delta} - t \end{pmatrix}$, then finding the eigenvalue of $A$ is solving the quadratic equation that one obtains from

\begin{align*}
\det{(A - x I_2)} = 0
\end{align*}

in order to $x.$ Then, $x$ would be an eigenvalue of $A$.

Computing the determinant of the LHS of the equation, you get

\begin{align*}
(c+x)(t+x) + \frac{b(cd(p-1) + d(p-1)x - ap(t+x))}{\delta} = 0.
\end{align*}

By solving this quadratic equation (like you usually do) you get

\begin{align*}
x = \frac{b(d+ap-dp) - \delta\Big(c + \sqrt{\frac{b^2(d+ap-dp)^2}{\delta^2} - \frac{2b(d(p-1)+ap)(c-t)}{\delta} + (c-t)^2 } + t\Big)}{2\delta}
\end{align*}

and

\begin{align*}
x = \frac{b(d+ap-dp) + \delta\Big(\sqrt{\frac{b^2(d+ap-dp)^2}{\delta^2} - \frac{2b(d(p-1)+ap)(c-t)}{\delta} + (c-t)^2 }\Big) + -\delta(c+t)}{2\delta}
\end{align*}

Title: Calculation of eigenvalues


问:
Quadratic form problem optimization
If
$$
f(\mathbf{x})=\frac{1}{2} \mathbf{x}^{T} \mathbf{Q} \mathbf{x}-\mathbf{b}^{T} \mathbf{x}
$$

where $\mathbf{Q} \in \mathbb{R}^{n \times n}$ is symmetric and positive definite; $\mathbf{x}, \mathbf{b} \in \mathbb{R}^{n} .$
Given an initial point $ \mathbf{x}_{0} $ and the following iterative scheme

$$\mathbf{x}_{k+1}=\mathbf{x}_{k}+\alpha_{k} \mathbf{d}_{k} $$

With exact step size

$$
\alpha_{k}=\arg \min_{\alpha>0} f\left(\mathbf{x}_{k}+\alpha \mathbf{d}_{k}\right)
$$

with $\mathbf{g}_{k+1}=\mathbf{Q} \mathbf{x}_{k+1}-\mathbf{b}$ and direction $\mathrm{d}_{k}$ defined as follows

$$
\mathbf{d}_{k+1} =-\gamma_{k+1} \mathbf{g}_{k+1}+\mathbf{d}_{k} $$

for $k=-1,0,1, \ldots$, with $\mathbf{d}_{-1}= 0$, $\mathbf{d}_{k+1}^{T} \mathbf{Q} \mathbf{d}_{k}=0$ , $\gamma_{0}=1$

a) Calculate  $\gamma_{k+1}$ based on $\mathbf{g}_{k+1}, \mathbf{d}_{k}$, $\mathbf{Q}$.

b) Verify that $\mathbf{g}_{k+1}=\mathbf{g}_{k}+\alpha_{k} \mathbf{Q} \mathbf{d}_{k}$

c) Show that $\mathbf{g}_{k+1}^{T} \mathbf{d}_{k-1}=0$ and $\mathbf{g}_{k+1}^{T} \mathbf{g}_{k}=0$

d) Show that $\mathbf{d}_{k+1}$ is a descending direction, for $k>-1$.

I have already proven parts a) and b), but I can't solve parts c) and d), for the part a) I get
$$ \gamma_{k+1}=\frac{\mathbf{d}_k^T \mathbf{Q} \mathbf{d}_k }{\mathbf{g}_{k+1}^{T} \mathbf{Q} \mathbf{d}_k} $$

For part c) and d) I tried substituting the given expressions, but I get something recursive, which makes me think that I should proceed by induction over $k$.

答:
The proof is a bit longer...

Let us minimize function $f(\mathbf{x})=\frac{1}{2}\mathbf{x}^{\text{T}}\mathbf{Q}\mathbf{x}-\mathbf{x}^{\text{T}}\mathbf{b}$, where $\mathbf{Q}$ is positive definite. The gradient yields $\mathbf{g}\equiv\nabla f(\mathbf{x})=\mathbf{Q}\mathbf{x}-\mathbf{b}$.

Theorem (algorithm of the conjugate gradient method):
Let starting point $\mathbf{x}_0$ be given and the initial direction is chosen as a steepest descent:
\begin{align}
 \mathbf{d}_0=-\mathbf{g}_0~~~~~~~~~~~~~(0)
\end{align}
Let us calculate
\begin{align}
 \mathbf{x}_{k+1}&=\mathbf{x}_k+\alpha_k\mathbf{d}_k,~\alpha_k=\frac{\mathbf{g}_k^{\text{T}}\mathbf{g}_k}{\mathbf{d}_k^{\text{T}}\mathbf{Q}\mathbf{d}_k}~~~~~~~~~~~~~(1)\\
 \mathbf{d}_{k+1}&=-\mathbf{g}_{k+1}+\beta_k\mathbf{d}_k,~\beta_k=\frac{\mathbf{g}_{k+1}^{\text{T}}\mathbf{g}_{k+1}}{\mathbf{g}_k^{\text{T}}\mathbf{g}_k}~~~~~~~~~~~~~(2)
\end{align}
where
\begin{align}
\mathbf{g}_k=\mathbf{Q}\mathbf{x}_k-\mathbf{b}~~~~~~~~~~~~~(3)
\end{align}
for $k=1,2,\dots$. Then
\begin{align}
\mathbf{g}_i^{\text{T}}\mathbf{g}_j&=0~~\forall i\neq j, ~~(\text{symmetric in }i,j)~~~~~~~~~~~~~(4)\\
\mathbf{g}_i^{\text{T}}\mathbf{d}_j&=0~~\forall i>j~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(5)\\
\mathbf{d}_i^{\text{T}}\mathbf{Q}\mathbf{d}_j&=0~~\forall i\neq j ~~(\text{symmetric in }i,j)~~~~~~~~~~~~~~~~(6)
\end{align}
Moreover is $\mathbf{d}_{k+1}$ the descending direction, i.e. $\mathbf{d}_{k+1}^{\text{T}}(-\mathbf{g}_{k+1})\ge0$.

Proof: We start with
\begin{align}
 \mathbf{g}_{k+1}=_{(3)}\mathbf{Q}\mathbf{x}_{k+1}-\mathbf{b}=_{(1)} \mathbf{Q}\mathbf{x}_{k}-\mathbf{b}+\alpha_k\mathbf{Q}\mathbf{d}_k=_{(3)}\mathbf{g}_k+\alpha_k\mathbf{Q}\mathbf{d}_k. ~~~~~~~~~~~~~(7)
\end{align}
Let us prove Eqs.(4,5,6) by induction. We prove only inequalities $j<i$, the case $j>i$ for Eqs.(4,6) is a consequence of symmetry. In the first step we show:
\begin{align}
\mathbf{g}_1^{\text{T}}\mathbf{g}_0&=_{(7)}(\mathbf{g}_0+\alpha_0\mathbf{Q}\mathbf{d}_0)^{\text{T}}\mathbf{g}_0=_{(0,1.2)}\mathbf{g}_0^{\text{T}}\mathbf{g}_0+\frac{\mathbf{g}_0^{\text{T}}\mathbf{g}_0}{\mathbf{d}_0^{\text{T}}\mathbf{Q}\mathbf{d}_0}\mathbf{d}_0^{\text{T}}\mathbf{Q}(-\mathbf{d}_0)=0~~~~~~~~~~~~~(8)\\
\mathbf{g}_1^{\text{T}}\mathbf{d}_0&=_{(0)}-\mathbf{g}_1^{\text{T}}\mathbf{g}_0=_{(8)}0~~~~~~~~~~~~~(9)\\
\mathbf{d}_1^{\text{T}}\mathbf{Q}\mathbf{d}_0&=_{(2)}(-\mathbf{g}_1^{\text{T}}+\beta_0\mathbf{d}_0^{\text{T}})\mathbf{Q}\mathbf{d}_0=_{(2.2,7)}\left(-\mathbf{g}_1^{\text{T}}+\frac{\mathbf{g}_1^{\text{T}}\mathbf{g}_1}{\mathbf{g}_0^{\text{T}}\mathbf{g}_0}\mathbf{d}_0^{\text{T}}\right)\frac{\mathbf{g}_1-\mathbf{g}_0}{\alpha_0}=_{(0,8)}\frac{-\mathbf{g}_1^{\text{T}}\mathbf{g}_1+\mathbf{g}_1^{\text{T}}\mathbf{g}_1}{\alpha_0}=0
\end{align}
Let us do the second induction step: let us suppose, that Eqs. (4-6) are valid for $\forall~j<i$ for fixed index $i$ and we want to prove it for $\forall~j<i+1$:
\begin{align}
\mathbf{g}^{\text{T}}_{i+1}\mathbf{g}_j&=_{(7)}(\mathbf{g}_i+\alpha_i\mathbf{Q}\mathbf{d}_i)^{\text{T}}\mathbf{g}_j=_{(2)}\underbrace{\mathbf{g}_i^{\text{T}}\mathbf{g}_j}_{=0}+\alpha_i\underbrace{\mathbf{d}_i^{\text{T}}\mathbf{Q}(-\mathbf{d}_j+\beta_{j-1}\mathbf{d}_{j-1})}_{=0}=0~~~~~~~~~~~~~(11)\\
\mathbf{g}^{\text{T}}_{i+1}\mathbf{d}_j&=_{(2)}\mathbf{g}^{\text{T}}_{i+1}(-\mathbf{g}_j+
\beta_{j-1}\mathbf{d}_{j-1})=_{(7)}-\underbrace{\mathbf{g}^{\text{T}}_{i+1}\mathbf{g}_j}_{=0~(11)}+\beta_{j-1}\underbrace{(\mathbf{g}_i^{\text{T}}+\alpha_i\mathbf{d}_i^{\text{T}}\mathbf{Q})\mathbf{d}_{j-1}}_{=0}=0 \\
\mathbf{d}_{i+1}^{\text{T}}\mathbf{Q}\mathbf{d}_j&=_{(2)}(-\mathbf{g}_{i+1}^{\text{T}}+\beta_i\mathbf{d}_i^{\text{T}})\mathbf{Q}\mathbf{d}_j=_{(7)}-\mathbf{g}_{i+1}^{\text{T}}\frac{\mathbf{g}_{j+1}-\mathbf{g}_j}{\alpha_j}+\beta_i\mathbf{d}_i^{\text{T}}\mathbf{Q}\mathbf{d}_j=_{(1.2,11)}\nonumber\\
&-\frac{\mathbf{g}_{i+1}^{\text{T}}\mathbf{g}_{j+1}}{\mathbf{g}_j^{\text{T}}\mathbf{g}_j}\mathbf{d}_j^{\text{T}}\mathbf{Q}\mathbf{d}_j+\beta_i\mathbf{d}_i^{\text{T}}\mathbf{Q}\mathbf{d}_j
\end{align}
which in case $j=i$ identically equals to zero and in case $j<i$ (i.e. $j+1<i+1$) it is zero due to Eq.(11) and assumption due to the induction step. This finishes the proof by induction.

Let us finally show, that $\mathbf{d}_{k+1}$ is the descending direction:
\begin{align}
 \mathbf{d}_{k+1}^{\text{T}}(-\mathbf{g}_{k+1})=_{(2)}||\mathbf{g}_{k+1}||^2-\beta_k\underbrace{\mathbf{d}_{k}^{\text{T}}\mathbf{g}_{k+1}}_{=0}\ge0.~~\spadesuit
\end{align}

Title: Quadratic form problem optimization


答:
We show that

a chord bisected at a given point in a circle, is the shortest chord through that point.

Let the chord bisected have length $x,x$ on either side of the given point. Draw any other chord through that point which has segments $a,b$, as divided by the point and $a \neq b$.

We know by intersecting chords theorem,
$$x\cdot x = a\cdot b$$
$$\Rightarrow x=\sqrt{x^2}=\sqrt{ab}\lt \frac{a+b}{2}$$
$$\Rightarrow 2x \lt a+b$$

using AM-GM inequality. That is, $2x$ is shorter than length of any other chord through the point.


问:
Let R be the relation on Z, defined as follows: ∀m, n∈Z, mRn⇐⇒2|(m−n) Determine whether the relation is reflextive, symmetric, and/or transitive
Let R be the relation on Z, defined as follows:
∀m,n∈Z, mRn⇐⇒2|(m−n)

Determine whether the relation is reflextive, symmetric, and/or transitive

答:
$2/m-n$  so $m-n=2k$ that mean $m=n[2]$

And we know $m=n[2]$$\Leftrightarrow $$n=m[2]$ so this relation is symmetry

and also we know  $m=m[2]$ so it's reflexive

And we know if $m=n[2] $ and $n=j[2]$ that's implies $ m=j[2]$ So it's transitive

And after this(transitivity, reflexivity, symmetry) we can say this relation is  an equivalence relation


答:
No, it is not necessary.

It may be that $A$ and $B$ are both stochastic:

$$ A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, B = \begin{pmatrix} 0.5 & 0.5 \\ 0.5 & 0.5 \end{pmatrix} \implies C = AB = \begin{pmatrix} 0.5 & 0.5 \\ 0.5 & 0.5 \end{pmatrix} . $$

It may be that $A$ and $B$ are not stochastic:

$$ A = \begin{pmatrix} 0 & 2 \\ 2 & 0 \end{pmatrix}, B = \begin{pmatrix} 0 & 0.5 \\ 0.5 & 0 \end{pmatrix} \implies C = AB = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} . $$

Title: Factors of a stochastic matrix

Body: If $A B = C$ where matrix $C$ is stochastic (all entries are positive and all rows add to $1$) then is it necessary that both $A$ and $B$ be stochastic?