问:
Lifting to universal covers is a proper map
Let $M$ be an orientable surface equipped with a Riemannian metric, and  $\alpha\in \pi_1(M)$ be a primitive element. Let $f$ be a shortest loop representating $\alpha$ and $p_\alpha:M_\alpha\to M$ be the covering corresponding to the subgroup $\langle \alpha\rangle$ of $\pi_1(M)$. Let $p:\widetilde M\to M$ be the universal cover.

Consider the lift $f_\alpha:\Bbb S^1\to M_\alpha$ be the lift of $f$ w.r.t. the covering $p_\alpha:M_\alpha\to M$. Now, consider the lift $\ell:\Bbb R\to \widetilde M$ of the composite map $\Bbb R\xrightarrow{\exp}\Bbb S^1\xrightarrow{f_\alpha}M_\alpha$.

Problem 1: Show that $\ell:\Bbb R\to \widetilde M$ is a proper map.

Problem 2: Consider the action $\bullet$ of $\pi_1(M)$ on $\widetilde M$ via deck-transformations. For any $g\in \pi_1(M)$, the composition $\Bbb R\xrightarrow{g\bullet \ell}\widetilde M\xrightarrow{\text{universal cover}}M_\alpha$ is a proper map.

$(1)$ I am trying to use use the fact $[f_\alpha\circ \gamma]\bullet \ell(x)=\ell\big([\gamma]\bullet x\big)$ with the fact that the  action of the set of all deck transformations(for the covering $p:\widetilde M\to M$) on $\widetilde M$ is properly discontinuous.

$(2)$ For this I need to use fiber of covering is discrete and $g\bullet\ell$ is an embedding.

I can not complete the proof. Any help will be appreciated. Thanks, in advance.

Below is a screenshot of the book  An Introduction to 3-manifolds by Peter Scott

答:
Suppose that $X$ is a complete (connected) Riemannian manifold, $\Gamma$ is a group of isometries of $X$ acting on $X$ properly discontinuously.

Definition. Let $\gamma\in\Gamma$ be an element of infinite order. Then a complete geodesic $A_\gamma\subset X$ is called an {\em axis} of $\gamma$ if it is invariant under $\gamma$.

(In this case, $\gamma$ acts on $A_\gamma$ as a translation. Note that an axis need not be unique.)

Given an infinite order element $\alpha\in\Gamma$, let $X_\alpha$ denote the quotient Riemannian manifold $X/\langle \alpha\rangle$.

I will be always parameterizing geodesics by the arc-length.

Lemma 1. Suppose that $\alpha, \beta\in\Gamma$ are elements of infinite order so that the subgroups $\langle \alpha\rangle$, $\langle \beta\rangle$ have trivial intersection. Then the projection $A_\beta\to X_\alpha$ is a proper map.

Suppose not. Then there exists a sequence of points $z_i\in A_\beta$ diverging to infinity such that the projections of these points to $X_\alpha$ are within distance $R$ from some point $\bar{x}\in X_\alpha$ (projection of $x\in X$).

Since $\langle \beta\rangle$  acts cocompactly on the axis
$A_\beta$, there is a constant $C$ and a diverging sequence of integers $n_i$ such that the (minimal) distance from $\beta^{n_i}(x)$ to the $\langle\alpha\rangle$-orbit of $x$ is $\le C$. In other words, there is a diverging sequence of integers $(m_i)$ such that
$$
d(\beta^{n_i}(x), \alpha^{-m_i}(x))\le C.
$$
Without loss of generality, we may assume that both sequences consist of distinct integers. We have
$$
d(\alpha^{n_i} \beta^{n_i}(x), x)\le C
$$

By the proper discontinuity of the action of $\Gamma$ on $X$, the set of products
$\alpha^{n_i} \beta^{n_i}$ is finite. Hence, there exist distinct $i, j$ such that
$$
\alpha^{n_i} \beta^{n_i}= \alpha^{n_j} \beta^{n_j},
$$
$$
\alpha^{n_i-n_j}= \beta^{n_j-n_i},
$$
which is a contradiction. qed

Suppose now that $\Gamma$ satisfies the following maximality condition: If $\alpha, \beta$ are primitive elements of $\Gamma$ then the subgroups generated by $\alpha, \beta$ are either equal or have trivial intersection. For instance, fundamental groups of surfaces satisfy this property. (Let me know if you need help proving this property. For noncompact surfaces this follows from the fact that their fundamental groups are free.)

Corollary 1. Suppose that $\alpha\in \Gamma$ is a primitive element. Then for each $\gamma\in\Gamma$ $\alpha, gamma$ either commute or anticommute:
$$
[\alpha,\gamma]=1, \hbox{or} \gamma\alpha\gamma^{-1}=\alpha^{-1},
$$
or the geodesic $\gamma(A_\alpha)$ projects properly to $X_\alpha$.

Proof. Consider
$$
\beta= \gamma\alpha\gamma^{-1}.
$$
Then $A_\beta=\gamma(A_\alpha)$. By Lemma 1, if the projection of $A_\beta$ to $X_\alpha$ is not proper, then the subgroups generated by $\alpha, \beta$ have nontrivial intersection. By the Maximality Property above,  these subgroups have to be equal (note that the primitivity of $\alpha$ implies that of $\beta$ since conjugation is a group automorphism). Thus,
$$
\langle \alpha\rangle= \langle\gamma\alpha\gamma^{-1}\rangle.
$$
Since the infinite cyclic group $\langle \alpha\rangle$ has only two generators, $\alpha^{\pm 1}$, we get

$$
[\alpha,\gamma]=1, \hbox{or} \gamma\alpha\gamma^{-1}=\alpha^{-1},
$$
qed

Suppose now that $X$ is 2-dimensional and $\Gamma$ acts freely. If $\Gamma$ preserves orientation, i.e. $M=X/\Gamma$ is orientable, the anti-commutation is impossible. Furthermore, unless the group $\Gamma$ is isomorphic to ${\mathbb Z}^2$ and the quotient $X/\Gamma$ is diffeomorphic to the torus, the commutation is only possible if $\gamma$ belongs to the subgroup generated by $\alpha$.

Thus, we obtain:

Corollary 2. Suppose that $X$ is 2-dimensional, $M=X/\Gamma$ is orientable and is not diffeomorphic to the torus. Then either $\gamma(A_\alpha)=A_\alpha$ or $\gamma(A_\alpha)$ projects properly to $X_\alpha$.

Note that the torus is a genuine exception: No matter what $\gamma\in \Gamma$ you take, $\gamma(A_\alpha)$ does not project properly to $X_\alpha$ since
$\gamma(A_\alpha)$ is invariant under $\alpha$ and, hence, projects to a compact in $M_\alpha$.


问:
Solve $ab(a^2+b^2)=2c^4$ over $\mathbb Z$
Find all integer solutions to $ab\left(a^2+b^2\right)=2c^4$.

I have found four classes of solutions: $(n,n,n)$, $(n,n,-n)$, $(n,0,0)$ and $(0,n,0)$, for $n\in\mathbb Z$. I think these are all, but I'm not sure.

It suffices to look for positive integer solutions. Let $g=\gcd(a,b)$. Since $g^4$ divides the LHS, it follows that $g$ must divide $c$. So we can divide out by $g$ and look for solutions where $a,b$ are coprime. In this case, $\gcd(ab,a^2+b^2)=1$.

I am stuck from here though.

答:
Multiply by $8$ and rearrange to get
$$(a+b)^4=(2c)^4+(a-b)^4.$$
From here, I have discovered a truly wonderful solution, which math.SE is too small to contain.

ContentLicense: CC BY-SA 4.0


问:
I have to show that the double serie diverges
I have to show that the double serie $$\sum_{k=1}^{\infty} \sum_{i=1}^{\infty} ((\frac{1}{k+1} \cdot (\frac{k}{k+1})^{i}) - (\frac{1}{k+2} \cdot (\frac{k+1}{k+2})^{i}))$$ convergence to $-\frac{1}{2}$.
Wolfram Alpha also says this series convergence to $-\frac{1}{2}$. Unfortunately, I don't know the calculation route.

Body:
i think i am at the end.

$$\sum_{k=1}^{\infty} \sum_{i=1}^{\infty} ((\frac{1}{k+1} \cdot (\frac{k}{k+1})^{i}) - (\frac{1}{k+2} \cdot (\frac{k+1}{k+2})^{i}))$$

=

$$\sum_{k=1}^{\infty} \sum_{i=1}^{\infty} \frac{1}{k+1} \cdot (\frac{k}{k+1})^{i} - \sum_{i=1}^{\infty} \frac{1}{k+2} \cdot (\frac{k+1}{k+2})^{i}$$

=

$$\sum_{k=1}^{\infty} \frac{1}{k+1} \cdot \sum_{i=1}^{\infty} (\frac{k}{k+1})^{i} - \frac{1}{k+2} \cdot \sum_{i=1}^{\infty} (\frac{k+1}{k+2})^{i}$$

=(geometric series)

$$\sum_{k=1}^{\infty} \frac{1}{k+1} \cdot (k) - \frac{1}{k+2} \cdot  (k+1)$$

=

$$\sum_{k=1}^{\infty} -\frac{1}{k^{2} + 3k + 2}$$
Now a little hint is still missing to finish.


问:
Product of smooth varities is smooth
Let $X$ and $Y$ be smooth varieties, then $X\times Y$ is a smooth variety.
Here is my definition of smooth:

It is enough to show that if $X$ is smooth at $p$ and $Y$ is smooth at $q$ then the product is smooth at $(p,q)$. The smoothness of $X$ at $P$ gives us a $U,\phi,(f_1,..,f_a)$ where $a=n-d$ with the Jacobi having maximal rank. Simillary the smoothness of $Y$ gives a $V,\psi,(g_1,...g_b)$ with $b=m-d'$ such that the Jacobi has maximal rank.

By definition the the product $U\times V$ is open in $X\times Y$ and is also a variety.

I think $\phi\times\psi$ should give an isomorphism to $Z(f_1,..,f_a,g_1,..,g_b)$. But this is not obvious to me. It is also not clear why the jacobi of this should be of maximal rank.

答:
$\phi\times\psi$ does indeed give an isomorphism of $U\times V$ with $Z(f_1,\cdots,f_a)\times Z(g_1,\cdots,g_b) \subset \Bbb A^n\times\Bbb A^m$, which is exactly $Z(f_1,\ldots,f_a,g_1,\ldots,g_b)\subset \Bbb A^{n+m}$. We can check the first part by noting that $\phi^{-1}\times\psi^{-1}$ is an inverse to $\phi\times\psi$, and the second part is true because $(x_1,\ldots,x_n)$ satisfies all the $f_i$ and $(y_1,\ldots,y_m)$ satisfies all the $g_j$ iff $(x_1,\ldots,x_n,y_1,\ldots,y_m)$ satisfies all the $f_i$ and the $g_j$.

To check smoothness, it's instructive to write down the Jacobian matrix:

$$\begin{pmatrix}
 \frac{\partial f_1}{\partial x_1} & \ldots & \frac{\partial f_1}{x_n} & \frac{\partial f_1}{\partial y_1} & \ldots & \frac{\partial f_1}{\partial y_m} \\
\vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\
 \frac{\partial f_a}{\partial x_1} & \ldots & \frac{\partial f_a}{x_n} & \frac{\partial f_a}{\partial y_1} & \ldots & \frac{\partial f_a}{\partial y_m} \\
 \frac{\partial g_1}{\partial x_1} & \ldots & \frac{\partial g_1}{x_n} & \frac{\partial g_1}{\partial y_1} & \ldots & \frac{\partial g_1}{\partial y_m} \\
\vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\
 \frac{\partial g_b}{\partial x_1} & \ldots & \frac{\partial g_b}{x_n} & \frac{\partial g_b}{\partial y_1} & \ldots & \frac{\partial g_b}{\partial y_m} \\
\end{pmatrix}$$

but $\frac{\partial f_i}{y_j}=0$ and $\frac{\partial g_i}{x_j}=0$ because no $f$ contains any $y$ and no $g$ contains any $x$. So our Jacobian matrix is of the form $$\begin{pmatrix} J_{f,x} & 0 \\ 0 & J_{g,y} \end{pmatrix}$$ where $J_{f,x}$ is the Jacobian of the $f$s with respect to the $x$s and similarly with $J_{g,y}$. But this exactly means that the big Jacobian is of maximal rank since the small ones are.

Title: Product of smooth varities is smooth


问:
If $p$ is an odd prime (different than $5$) prove that $10 \mid p^{2}-1$ or $10 \mid p^{2}+1$
I was reading some properties on prime numbers in a book and I’ve found this problem.
Using Modular Arithmetic isn’t allowed

Using the division algorithm I have found:

$p \in \{5m+1, 5m+2, 5m+3, 5m+4\}$

I’ve tried to plug those values of $p$ to :$\frac{p^{2}\pm1}{10}$, but I didn’t get an integer.

Thank you for your help

答:
If $p$ is odd and different from 5, then $p=3$ or $p=7$ or $p=9$ or $p=10k+1$ or $p=10k+3$ or $p=10k+5$ or $p=10k+7$, $k\geq 1$. Suppose that $10$ does NOT divide $p^2-1$ nor $p^2+1$.

Then $10$ does not divide the product: $(p^2-1)(p^2+1)$.

You can verify easily this is not true for 3, 7 and 9. Then you can substitute $p=10k+1$ or $p=10k+3$ $p=10k+5$ or $p=10k+7$ for $k\geq1$ to verify this is not true as well and reach an contradiction.

The conclusion is that $10$ must divide $p^2-1$ or $p^2+1$.


问:
Is the class of $k$-letter graphs finitely defined?
I am completely bamboozled by this problem, and although I believe the answer 'should' be yes, I cannot prove it for love nor money! I'll provide a couple of definitions for clarity's sake.

A $k$-letter graph $G$ is a graph defined by a finite word $x_{1}x_{2}...x_{n}$ over an alphabet $X$ of size $k$ together with a subset $S\subseteq{X^2}$ such that:

$V(G)$ = {$1,2,...,n$}

$E(G)$ = {$ij$ : $i\leq{j}$ and $(x_{i},x_{j})\in{S}$}

Note that $V(G)$ and $E(G)$ are the vertex-set and edge-set of $G$, and that $G$ is simple and undirected.

Also a hereditary class of graphs being finitely defined means that the set of minimal induced forbidden subgraphs for this class is finite.

I have a few ideas floating around, but not only are they hard to put into words, they don't appear to help. I spent all my energy trying to prove that the answer is yes, and have given little consideration to the possibility that the answer is no.

Many thanks.

答:
Suppose that $H$ is a minimal forbidden induced subgraph for $k$-letter graphs. If we delete a vertex $v$, the result $H-v$ must be a $k$-letter graph; otherwise, $H-v$ would also be a forbidden induced subgraph, and $H$ would not be minimal.

We can use this to show that $H$ is a $(2k+1)$-letter graph. First, take a description of $H-v$ as a $k$-letter graph with alphabet $X$, word $x_1 x_2 \dots x_n$, and relation $S$. Define a new alphabet $X'$ which has two letters $x^+$ and $x^-$ for every $x \in S$, as well as one extra letter I'll just call $q$ for no reason.

For $i=1, \dots, n$, replace $x_i$ by $x_i^+$ if the $i^{\text{th}}$ vertex of $H-v$ is adjacent to $v$, and $x_i^-$ otherwise, getting a word over $X'$. Then, append $q$ to the end of this word. Define a new relation $S'$ with four pairs $(x^{\pm}, y^{\pm})$ for every pair $(x,y) \in S$, as well as the $k$ pairs $(x^+, q)$ for $x \in X$.  The resulting alphabet, word, and relation define $H$ as a letter graph.

So all minimal forbidden induced subgraphs are $(2k+1)$-letter graphs; now we will show that there are finitely many.

First of all, there are only finitely many relations $S$: only the piddling, insignificant amount of $2^{(2k+1)^2}$. (We assume a fixed alphabet $X$, since it doesn't really matter what the letters are called.) For each choice of $S$, if $H_1$ is a $(2k+1)$-letter graph defined by $S$, $X$, and a word $w$, and $H_2$ is a $(2k+1)$-letter graph defined by $S$, $x$, and a subword of $w$, then $H_2$ is an induced subgraph of $H_1$, so both of them cannot be minimal forbidden subgraphs.

But by Higman's lemma, we can only pick finitely many words over $X$ without including a word and its subword. This means that for every relation $S$, there are finitely many minimal forbidden induced subgraphs defined with that relation.


问:
Let Y be the number of heads and Z the number of tails, find $P \{X \leq x\}$
Three coins are tossed. Let Y be the number of heads and Z the number of tails. For each of the following random variables find $P \{X \leq x\}$

a.) $X=Y-Z$

b.)$X= max\{Y,Z\}$

c.)$X=\left |Y-Z  \right |$

I can't understand what I want, help me with a hint

答:
Well $P\{X\leq x\}$ is cumulative distribution function $F_X(x)$. This function is like

Let's take some $x$. Then what is the probability that the random variable $X$ will be less than $x$.

This $x$ may have any value and is not required $x\in X$.

As Gono said, $Y+Z=3$, so assume $Y=3-Z$.

a) $X=Y-Z=3-Z-Z=3-2Z=\{-3;-1;1;3\}$ - all possible values of $X$ when $Z=3,2,1$ and $0$.

Now let's calculate probabilities of these values:
$$P(-3)=P(3)=0.5^3=0.125\\
P(-1)=P(1)=0.5^3*3=0.375$$
We definitely can say that $P(X\leq -10)=0$ or $P(X\leq -5)=0$ because min value of $X$ is $-3$ and there is no single chance that X will be less than $-10$ or $-5$. Same for $P(X\leq 10)=1$ because max value of $X$ is $3$.

What if $x=-2$? There is only one value exists, that less than $-2$. It's $-3$. And the probability that the random variable $X$ takes a value of $-3$ is $0.125$.

Fine. But what if $x=-0.5$? $X$ might be less or equal than $-0.5$ when it's value is $-3$ or $-1$. Then $P(X\leq -0.5)=P(-3)+P(-1)=0.5$*. Sounds logical, especially if we took $x=0$, because intuition (and expected value $E[X]$) suggests that on average $Y-Z=0$.

*I don't know how "$P(A+B)=P(A)+P(B)$" property will be in English, but this is our case because if $X=-3$ it can't be equal to $-1$ at the same time.

Now, after the examples above, it should become easy to define the entire cumulative distribution function, so that I will omit further reasoning.
$$P\{X\leq x\}=F_X(x)=\begin{cases}
0,  & x\lt -3 \\
P(-3)=0.125, & -3\leq x\lt -1 \\
P(-3)+P(-1)=0.5, & -1\leq x\lt 1 \\
P(-3)+P(-1)+P(1)=0.875, & 1\leq x\lt 3 \\
P(-3)+P(-1)+P(1)+P(3)=1, & 3\leq x
\end{cases}\\$$
b) $$X=max\{Y,Z\}=max\{3-Z,Z\}=\{2,3\}\\
P(2)=0.75\\
P(3)=0.25\\
P\{X\leq x\}=F_X(x)=\begin{cases}
0,  & x\lt 2 \\
0.75, & 2\leq x\lt 3 \\
1, & 3\leq x \\
\end{cases}\\$$
c) $$X=|Y-Z|=|3-Z-Z|=|3-2Z|=\{1,3\}\\
P(1)=0.75\\
P(3)=0.25\\
P\{X\leq x\}=F_X(x)=\begin{cases}
0,  & x\lt 1 \\
0.75, & 1\leq x\lt 3 \\
1, & 3\leq x \\
\end{cases}\\$$
Ask if there are any more questions left.


问:
Distance between a point $x\in H$ and $H\cap \overline{B}(0,1)$, with $H$ a hyperplane
Let $X$ be a normed space. Let $f:X\to\mathbb{R}$ be a linear functional, and consider the hyperplane $H=f^{-1}(r)$, for some $r>0$.

Let  $x\in H$ be such that $1<\left\|{x}\right\|\le 1+\epsilon$, for some $\epsilon>0$.

If $B$ is the unit closed ball and $H\cap B\neq\emptyset$, my question is:

Is it true that $d(x,H\cap B)\le\epsilon$ ?

In any case, I am trying to prove that $d(x,H\cap B)$ has to be "small", so $2\epsilon$ instead of $\epsilon$ would do as well. But I'm still unable to find a suitable $y\in H\cap B$ such that $d(x,y)$ is small. In $\mathbb{R}^2$ (visually speaking), it becomes obvious that such a $y$ has to exist, but I'm stuck when $X$ is any normed space.

答: The answer is no, and here is a counter example in $X=\mathbb R²$, with the norm $\|(x,y)\|=\max\{|x|,|y|\}$.


问:
Find x, y, and z (all of them non-negative) that maximize $ x^\frac14 y^\frac14 z^\frac14 $ subject to the constraint that $ px + y + z ≤ m $
Find x, y, and z (all of them non-negative) that maximize $ x^\frac14 y^\frac14 z^\frac14 $ subject to the constraint that $ px + y + z ≤ m $

Context: undergraduate mathematical economics

I have attempted this problem to the best of my ability but I'm really not sure if I got the mechanics of it right. If somebody would be kind enough to check my work over and point out any potential errors I would be greatly appreciative.

Setting up the Lagrangian:

$ L = x^\frac14 y^\frac14 z^\frac14 - \lambda (px + y + z - m) ≤ 0 $

First Order Conditions:

$ f_x = \frac14x^\frac{-3}4 y^\frac14 z^\frac14 - \lambda p $

$ f_y = x^\frac14 \frac14y^\frac{-3}4 z^\frac14 - \lambda $

$ f_z = x^\frac14 y^\frac14 \frac14z^\frac{-3}4 - \lambda $

$ f_\lambda = -px-y-z+m $

So if I'm not mistaken we'd then get $ px = y = z $ by simplifying the first three equations.

Then, using $ f_\lambda = -px-y-z+m $, we can obtain expressions for x, y, and z:

$ x = \frac m{3p} $
$ y = \frac m3 $
$ z = \frac m3 $

Second part of the problem involves using the envelope theorem to solve for $ \frac {dV}{dm} $. Using the above I calculated this to be:

$ \frac {dV}{dm} = \frac{\partial L}{\partial m} = \lambda $

Taking $ f_z = x^\frac14 y^\frac14 \frac14z^\frac{-3}4 = \lambda $

=> $ \frac{m}{3p}^\frac14 \frac{m}{3}^\frac14 \frac14 \frac{m}{3}^\frac{-3}{4} = \lambda $

$ \lambda = \frac{3^\frac14}{4(m^\frac14)(p^\frac14)} $

Am I on the right track here? Been a while since I've taken a class involving this kind of math so I'm a little rusty.

答:
If $p$ and $m$ are also nonnegative, we might as well assume $px+y+z=m$.  The Lagrange method now directs us to solve the following equations.
$$
\left\{\begin{array}{l}
\frac{\partial}{\partial x}(x^{1/4}y^{1/4}z^{1/4})=\lambda\frac{\partial}{\partial x}(px+y+z)
\\
\frac{\partial}{\partial y}(x^{1/4}y^{1/4}z^{1/4})=\lambda\frac{\partial}{\partial y}(px+y+z)
\\
\frac{\partial}{\partial z}(x^{1/4}y^{1/4}z^{1/4})=\lambda\frac{\partial}{\partial z}(px+y+z)
\\
px+y+z=m\end{array}\right.
$$
Simplifying gives
$$
\left\{\begin{array}{l}
\frac{1}{4}x^{-3/4}y^{1/4}z^{1/4}=\lambda p
\\
\frac{1}{4}x^{1/4}y^{-3/4}z^{1/4}=\lambda
\\
\frac{1}{4}x^{1/4}y^{1/4}z^{-3/4}=\lambda\\
px+y+z=m\end{array}\right.
$$
and hence
$$
\left\{\begin{array}{l}
x^{-3/4}y^{1/4}z^{1/4}=4\lambda p
\\
x^{1/4}y^{-3/4}z^{1/4}=4\lambda
\\
x^{1/4}y^{1/4}z^{-3/4}=4\lambda\\
px+y+z=m\end{array}\right.
$$
Now let's raise the first three equations to the fourth power, to obtain
$$
\left\{\begin{array}{l}
x^{-3}yz=256\lambda^4p^4
\\
xy^{-3}z=256\lambda^4
\\
xyz^{-3}=256\lambda^4\\
px+y+z=m\end{array}\right.
$$
The middle two equations give $xy^{-3}z=xyz^{-3}$ and hence $y=z$, which should be no surprise since both functions are symmetric in $y$ and $z$.  Hence
$$
\left\{\begin{array}{l}
x^{-3}y^2=256\lambda^4p^4
\\
xy^{-2}=256\lambda^4
\\
px+2y=m\end{array}\right.
$$
Making another obvious substitution gives
$$
\left\{\begin{array}{l}
x^{-3}y^2=xy^{-2}p^4
\\
px+2y=m\end{array}\right.
$$
You can take it from here I'm sure.


问:
Boolean Property Simplification
I was wondering if there is a term/property/name to describe this simplification:

$$A + \overline{A}B = A + B$$

thank you

答:
I'll use $\cup$, $\cap$ notation in my proof instead of $+$,$\cdot$, I hope you don't mind (if you aren't familiarized with them, $\cup$ means OR and $\cap$ means AND). Notice that:
$$A\cup(\bar A\cap B)=(A\cup \bar A)\cap(A\cup B),$$
and $(A\cup \bar A)$ is the universe, so
$$(A\cup\bar A)\cap(A\cup B)=A\cup B,$$
to be said;
$$\boxed{A\cup(\bar A\cap B) = A\cup B}$$

Rewriting it using your notation, it goes like this:

$$A+ \bar AB = (A+\bar A)(A+B)=A+B$$

Title: Boolean Property Simplification


问:
Cardinality does not change when substracting a countable subset from an uncountable set
This comes from Exercise 11 of Section 2.1 of Kaplansky's "Set Theory and Metric Spaces".

Let $A$ be an uncountable set. Let $B \subseteq A$ be a countable set and let $B^c$ be its complement in $A$. Prove that there exists a bijection between $A$ and $B^c$.

I've found some related questions here: removing a countable subset from an infinite set doesn't change cardinality, and here: Altering an Infinite Set does not change cardinality, but none seems to be exactly the one I'm looking for.

Suposedly this problem can be solved without the axiom of choice, due to it being proposed before the axiom's introduction in the book.

It is quite easy to show that $B^c$ must be uncountable, but moreover I need to prove that it has the same cardinality as $A$. That doesn't look so simple. Here are some thoughts I've been considering:

Since $B^c$ can be split into some countable $C$ and uncountable $C^c$, and the same for $C^c$, this problem is equivalent to partitioning $A$ into a countable number of countable sets and its complement, which must be uncountable. And if we prove that it has indeed the same cardinality as $A$, we would've proved that any partition of $A$ must include a proper subset. It is a fact that proper subsets exist in $A$ due to it being infinite, but I don't think that's enough to say that it will be present on every partition. ¿Is there a way to prove this?

As a particular case, we obtain from this result that irrationals and trascendents have both the same cardinality as $\mathbb{R}$.

EDIT

So, following Berci's answer, I wrote this proof.

Since $B$ is countable and $A = B \cup B^c$ is uncountable it follows that $B^c$ is uncountable. Then it contains an uncountable infinite subset; call it $B_1$. By the same argument $B^c \backslash B_1$ must be uncountable. On the other hand, we have that $B \cup B_1$ is countable, so there exists a one to one correspondence between $B$, $B_1$ and $B \cup B_1$. And since $A \backslash B \cup B_1$ and $B \backslash B_1$ both have the same elements, it follows that there must exist a bijection between $A$ and $B^c$.

I think that's how it is suposed to be solved. I have some doubts about the conclusion though.

答:
Since $B^\complement$ is uncountable, it contains a countably infinite subset $C$.

Now simply apply a bijection between $B\cup C$ and $C$, and extend it identically on $B^\complement\setminus C$.


问:
Proving union of closures is closure of unions
Is this okay for proving this subset direction, $\text{cl}(A) \cup \text{cl}(B) \supset \text{cl}(A \cup B)$?

Let $x \in \text{cl}(A \cup B)$. Let $U$ be any open set containing $x$. Then $U \cap (A \cup B)=(U \cap A) \cup (U \cap B) \neq \emptyset$.Then $U \cap A \neq \emptyset$ or $U \cap B \neq \emptyset$
If $U \cap A \neq \emptyset \implies x \in \text{cl}(A)$. So $x \in  \text{cl}(A) \cup \text{cl}(B)$.If $U \cap B \neq \emptyset \implies x \in \text{cl}(B)$. So $x \in  \text{cl}(A) \cup \text{cl}(B)$.

答:
$Cl(X)$ is the $\subseteq$-minimum closed set $Y$ such that $Y\supseteq X.$ Equivalently $Cl(X)$ is the intersection of all closed sets that have $X$ as a subset.

$\bullet \;$So if $Y$ is closed and has $X$ as a subset then $Y\supseteq Cl(X).$

$Cl(A)\cup Cl(B)$ is a closed set that has both $A$ and $B$ as subsets.

$\bullet\;$ So $Cl(A)\cup Cl(B)$ is closed and has $A\cup B$ as a subset, so $Cl(A)\cup Cl(B)\supseteq Cl(A\cup B).$

The reverse inclusion follows at once from $[\,Cl(A\cup B)\supseteq Cl(A)\,]\land [\,Cl(A\cup B)\supseteq Cl(B)\,].$


问:
Difficult matrix equation $A(x-x_0) + \alpha\left(\frac{\Sigma x}{\mu^\top x}\right) = 0$
Let $x,x_0,\theta,\mu \in \mathbb{R}^n$ vectors, $\alpha \in\mathbb{R}$ a scalar, and $\Sigma, A\in \mathbb{R}^{n\times n}$ Positive Semi-Definite matrices. Also, $\mu$ is a mean vector and $\Sigma$ is a covariance matrix of a Random Variable $Z$ (I don't know if this information may help). Is there a way to solve in closed form the following matrix equation for $x$?

\begin{equation}
A(x-x_0) + \alpha \left(\frac{\Sigma x}{\mu^\top x}\right) = 0.
\end{equation}

Main difficulty: I find difficulties in handling the denominator. If the equation would have been simply

$$A(x-x_0) + \alpha \Sigma x = 0,$$ the solution would be $x = (A+\alpha\Sigma)^{-1}Ax_0$. I am looking for a similar solution for the first equation but the denominator complicates things and I don't know how to proceed.

What I tried:

\begin{aligned}
& \mu^TxA(x-x_0) + \alpha \Sigma x = 0
\\ &\mu^TxAx - \mu^TxAx_0 - \alpha \Sigma x = 0
\\ &\mu^TxA(x-x_0) = \alpha \Sigma x
\end{aligned}

Is this way of proceeding correct? If yes, how is it possible to solve this matrix equation for $x$? I am not an expert in solving matrix equations and any help would be greatly appreciated. Thanks!

答:
By multiplying with $(\mu^T x)$ we see that the equation is of the form of a system of second degree polynomials. These systems are generally not solvable "in closed form", since (1) higher degree polynomial systems can be reduced to second degree systems in more variables (2) polynomials of degree $\ge 5$ cannot be solved "in closed form". (cf. discussion on mathoverflow).

So you shouldn't expect a "closed form" solution. Depending on your use case, you can try two things:

Try solving particular instantiations of the systems using computer algebra packages such as sympy

Use Numerical methods such as Newton's method or fixed point iteration

At the end of the day, the question you should always ask yourself is whether do you really need a "closed form" solution.


问:
Are the functionals $\omega_{\xi_1, \xi_2}$ norm-dense in $B_0(H)?$
Let $H$ be a Hilbert space and for $\xi_1, \xi_2 \in H$ consider  the bounded functional
$$\omega_{\xi_1, \xi_2}: B_0(H) \to \mathbb{C}: x  \mapsto \langle x \xi_1, \xi_2\rangle$$
Is it true that
$$\overline{\operatorname{span}\{\omega_{\xi_1, \xi_2}: \xi_1, \xi_2 \in H\}}^{\|\cdot \|}= B_0(H)^*$$? I.e. can we approximate every functional on the compact operators by a linear combination of the above functionals? This is used in a proof I'm reading but I don't see why it should hold. It is unclear to me why one would even expect this to be true.

A reference or proof would be appreciated.

答:
If your span were not dense, there would be a non-zero linear functional $\Lambda \in B_0(H)^{**}$ vanishing on every
$\omega_{\xi_1,\xi_2}$.  However, it is well known that
$B_0(H)^{**}=B(H)$, so $\Lambda $ is in fact an operator such that
$$
  \langle \Lambda \xi _1, \xi _2\rangle =0,
  $$
for every $\xi _1$ and $\xi _2$.  Therefore $\Lambda =0$,  a contradiction.


答:
Why on earth would modular arithmetic not be allowed.  As I am of the theory all arithmetic is modular arithmetic that may be hard.

But as $p$ is an odd prime $p^2 -1$ and $p^2 +1$ are both even and $2|p^2 \pm 1$ so it is sufficient to show that $5|p^2 -1$ or $5|p^2 + 1$.  .... Oh, wait, that was modular arithmetic!  Well, too bad.

Now $p$ is odd so $p=2k-1$ for some integer $k$ so $p^2\pm 1 = 4k^2 - 4k+1 \pm 1 =\begin{cases}4k^2 - 4k = &5k^2 - 5k -k^2 + k = &5(k^2-k) -k(k-1)\\4k^2+4k+2 = &5k^2 - 5k - k^2 + k +2=&5(k^2-k) -(k+1)(k-2)\end{cases}$.

Now $k-2,k-1,k,k+1,k+2$ are five consecutive numbers so one of them is divisible by $5$.  (Oh wait... that was modular arithmetic!... Oh, well.)  If the one divisible by $5$ is $k$ or $k-1$ then $p^2 -1 = 4k^2 -4k = 5(k^2-k) -k(k-1)$ is divisible by $5$

If $k+1$ or $k-2$ is divisible by $5$ then $p^2+1 = 4k^2 -4k + 2 = 5(k^2-k) -(k+1)(k-2)$ is divisible by $5$.

And if $k+2$ is divisible by $5$, then there is an integer $m$ so that $5m = k+2$ so $p = 2k -1=2(5m-2)-1=10m-5 = 5(2m-1)$.  But that means $p$ is not a prime other than $5$.  So this is impossible.  $k+2$ is not divisible by $5$ and one of the other four cases must be true.

.....

But that used modular arithmetic.

Title: If $p$ is an odd prime (different than $5$) prove that $10 \mid p^{2}-1$ or $10 \mid p^{2}+1$

Body:
I was reading some properties on prime numbers in a book and I’ve found this problem.
Using Modular Arithmetic isn’t allowed

Using the division algorithm I have found:

$p \in \{5m+1, 5m+2, 5m+3, 5m+4\}$

I’ve tried to plug those values of $p$ to :$\frac{p^{2}\pm1}{10}$, but I didn’t get an integer.

Thank you for your help


问:
Prove with integration the inequality $e(\frac{n}{e})^n < n! < n \times e(\frac{n}{e})^n$
Prove with integration  the inequality, I need some advice about how to start prove it.

I know that if function is Monotonically increasing function so :

$$ f(1)+\int^n_1f(x)dx\leq f(1)+f(2)+....+f(n)\leq f(n)+\int^n_1f(x)dx$$

$$e(\frac{n}{e})^n < n! < n \times e(\frac{n}{e})^n$$

Thanks!

答:
try to show :
$$ n \log n - n + 1 \leq \log (n!) \leq (n + 1)\log n - n + 1 \ldots(1)$$

using
$\log(x)$ over$[1,n]$ taking the integer  partition of $[1,n]$

lower sum of $\int^n_1\log(x)dx=\log2+\log3+\ldots+\log(n-1)=\log(n-1)!$

similarly compute the  upper sum and then

lower sum $\leq \int^n_1\log(x)dx \leq$ upper sum.

Then take exponential of $(1)$ to get the result.

ContentLicense: CC BY-SA 3.0


问:
Is there an $\infty$-category of spaces?
Disclaimer: $\infty = (\infty,1)$.

For some time, I have carried this paradigm that there should be an $\infty$-category of $\text{Top}$ whose

Objects are spaces.

Morphisms are continuous functions.

2-morphisms are homotopies of morphisms.

3-morphisms are homotopies of homotopies, etc.

The homotopy category of $\text{Top}$ would be the ordinary category of spaces modulo homotopies.

I figure one way of capturing this idea by more or less waving hands is by showing that $\infty\text{Gpd}$  is precisely the category of Kan complexes, which in turn correspond to ordinary $\text{Top}$ in a Quillen equivalent way. Then in the homotopical context $\infty\text{Gpd} = \text{Top}$. I'm really shy of this approach because I have no idea how the model structures interact with the $\infty$-language.

Another option (described for instance here), is to take $\text{Top}$ as the nerve of ordinary $\text{Top}$.
The final result, as presented in the link, looks a little like what I conjectured, but I'm not sure.

Is this the correct approach to an $\infty$-category of spaces? Can we write a direct definition, as a quasicategory?

答:
Note: by a space I mean a CW complex.

The $\infty$-category of spaces is usually defined by
$$
\mathbb RN_\Delta(L^H(\text{Kan})),
$$
where

$L^H(\text{Kan})$ is the DK localization of Kan complexes at weak equivalences.

$\mathbb RN_\Delta$ is the right derived functor of the homotopy coherent nerve.

This is indeed a quasi-category from the Joyal-Bergner structures Quillen equivalence.

Also $\text{Sing}:\text{CW}\to\text{Kan}$ will take a continous function $X\to Y$ to an $\infty$-functor between $\infty$-fundamental grupoids $\text{Sing}(X)\to\text{Sing}(Y)$ .


问:
Why negative integers in $(\mathbb{Z}/3\mathbb{Z})[X]/(X^N-1)$?
The NTRU crytposystem makes use of the ring $(\mathbb{Z}/3\mathbb{Z})[X]/(X^N-1)$, which I thought it was the ring of truncated polynomials with degree at most $N-1$, and coefficients reduced modulo $3$, i.e., in the set $\{0,1,2\}$. However, many algorithms for working with these polynomials within the cryptosystem specify that the coefficients are in the set $\{-1,0,1\}$.

Where do these coefficients come from? Is there any relation between the two sets? Maybe I'm not understanding properly the meaning of $(\mathbb{Z}/3\mathbb{Z})[X]/(X^N-1)$.

Also, let $f(X)$ and $g(x)$ be such polynomials with coefficients in the set $\{-1,0,1\}$. What is the correct way of performing the operation
$$
h(X) = f(X) + g(X) \bmod 3
$$
Summing each coefficient and making a modulo operation seems not to be the right way, becuase a computer will always return values in the set $\{0,1,2\}$. I would appreciate any help on this.

答: The set $\Bbb Z{/}3\Bbb Z$ is where the coefficients come from and we can use the representatives $\{0,1,2\}$ for those or $\{-1,0,1\}$ which is the same as $2 \equiv -1 \pmod{3}$. It's more "symmetric" to use the latter one. Modding out by $X^N-1$ only makes some cyclic shifts equivalent (repesenting polynomials as vectors or tuples) and has no bearing on the coefficients of the resulting polynomial classes. Addition is easy for such polynomials as $-1$ and $1$ cancel, $1 + 1$ becomes $-1$ and $-1 + -1$ becomes $1$, and $0$ is the neutral element (power by power of course, or coordinatewise, in tuples).


问:
Shannon limit vs Hamming bound
Under the Binary Symmetric Channel, the Shannon limit can be stated as:

$R \le 1-H(p)$

where $R$ is the rate of a code that can be decoded with high probability, $p$ is the probability of bit-flips on the channel, and $H$ is the binary entropy function. (I believe all of this applies in general, but I'm stating it in terms of binary alphabets for simplicity.) This is suspiciously similar to the Hamming bound:

$R \le 1-H(\delta/2)$

where $\delta$ is the relative distance of the code. And indeed, when $\delta \ge 2p$ then obviously the code will be able to correct all errors with high probability. Naively, the converse feels like it should also be true: That to correct all errors with high probability, $p \le \delta / 2$. And indeed, this is claimed in these lecture notes:

Proposition 2.1. Let $0 ≤ p < \frac{1}{2}$
and $0 < ε ≤ \frac{1}{2} − p$. If an algorithm $A$ can handle $p + ε$ fraction
of worst case errors, then it can be used for reliable communication over $BSC_p$.

... proof snipped ...

Remark 2.2. The converse of Proposition 2.1 is also true. More precisely, if the decoding error
probability is exponentially small for the BSC, then the corresponding code must have constant
relative distance. The proof of this claim is left as an exercise.

However, that leads to a contradiction: The Shannon limit is tight, but there are better upper bounds than the Hamming bound, such as the Elias-Bassalygo bound and the MRRW bound. These bounds rule out the existence of codes that, by Remark 2.2 would be required to achieve the tightness of the Shannon limit.

I know I'm missing something important here, but what is it?

答:
This might help. The Shannon limit places no constraints on computational complexity, nor on the type of coding/decoding. And it is proved by using random coding ideas, which I am sure you know.

Codes are traditionally used for unique decoding while it is possible to do list decoding, i.e., let the decoder output a small list where the correct codeword belongs to one element of the list. See the notes below in the course by Madhu Sudan, for example.

http://people.seas.harvard.edu/~madhusudan/courses/Spring2020/scribe/lect05.pdf

There it is explained how the Elias-Bassalygo bound applies if one allows "small list decoding".

Title: Shannon limit vs Hamming bound


问:
Does $f_1g≃f_2g$ imply $f_1≃f_2$?
Let $f_1,f_2,g$ be continuous map between topological spaces. If we have homotopy $f_1g≃f_2g$, then in what condition of $g$ we have $f_1≃f_2$? For given two continues map $g,h$ be homotopy, I mean there exist continues $f_t$ such that $f_0=g$ and $f_1=h$.

I'm not going to pursue big generality for this question, some sufficient conditions will be fun, just like the one Noel Lundström pointed out in the comment, when $g$ is split, then this can be true.

答:
Here is a partial answer. Since your question is in general quite difficult, I'll hope you find some use for my thoughts. I'll be sketchy with details since this is already long enough, but will be happy to give references for further reading. I'll work in the pointed category. All spaces will be based, all maps and homotopies basepoint-preserving, and all cones and suspensions reduced.

Let $g:X\rightarrow Y$ be a map. First let us study what properties are imposed on $g$ by the assumption:

$(\ast)$ for any pair of maps $\varphi_1,\varphi_2:Y\rightarrow Z$ satisfying $\varphi_1g\simeq\varphi_2g$ it holds that $\varphi_1\simeq\varphi_2$.

We define the mapping cone of $g$ to be the space $C_g$ obtained as the pushout of the span
$$Y\xleftarrow{g}X\xrightarrow{in_0}CX,$$
where $CX=X\wedge I$ is the cone on $X$ ($I$ is given the basepoint at $1$). Let $\delta:Y\rightarrow C_g$ be the inclusion map. Then the composition $\delta g:X\rightarrow C_g$ is canonically null-homotopic, and $(\ast)$ applies to $\delta g\simeq\ast \simeq\ast g$ to give a null-homotopy $\delta\simeq\ast$. In general the mapping cone of $\delta$ is homotopy equivalent to $\Sigma X$. On the other hand, any choice of homotopy $\delta\simeq\ast$ induces a homotopy equivalence $C_\delta\simeq C_\ast\simeq C_g\vee\Sigma Y$. Putting these observations together we get a homotopy equivalence
$$\Sigma X\simeq C_g\vee\Sigma Y.$$

This is a necessary condition for any map satisfying $(\ast)$. Noel Lundström observed in the comments that $g:X\rightarrow Y$ satisfies $(\ast)$ whenever $Y$ is a homotopy retract of $X$. Here we see that even if this stricter condition is not met, $\Sigma Y$ will always be a homotopy retract of $\Sigma X$ whenever $g$ satisfies $(\ast)$.

Now let us drop the assumption $(\ast)$. There isn't much more we can do with the basic ideas outlined above without introducing some more further conditions. Here is a fairly reasonable assumption:

$(\ast\ast)$ $g$ appears in a homotopy cofibration sequence of the form
$$Y'\xrightarrow{w} A\rightarrow X\xrightarrow{g}Y\xrightarrow{\delta} \Sigma A\rightarrow\Sigma X\rightarrow \dots$$

This assumption is still quite strict. For instance it implies that $Y\simeq\Sigma Y'$, so in particular $Y$ is a suspension and has a comultiplication $\nu:Y\rightarrow Y\vee Y$.

Now our assumptions could be weakened a little. I've written them as they are becaues I can't think of any really interesting examples of the more general case that I have in mind. What we need from the assumptions I have given is that the diagram
$\require{AMScd}$
\begin{CD}
Y@>\delta>>\Sigma A\\
  @V\nu VV@VVcV \\
Y\vee Y@>\delta\vee\delta>>\Sigma A\vee \Sigma A
\end{CD}
commutes up to homotopy. This relates the comultiplication on $Y$ with the suspension comultiplication $c$ on $\Sigma A$. That this is implied by the assumptions is clear because they give $\delta\simeq\Sigma w$. Really it is this diagram that we need, and could leave out the existence of $Y'$, assuming instead that the cofibration sequence begins at $A$, and that $Y$ is a co-H-space with comultiplication $\nu$ making the last diagram homotopy commute.

Now, the point is that for any space $Z$ we have a Puppe sequence
$$\dots\leftarrow[A,Z]\leftarrow[X,Z]\xleftarrow{g^*}[Y,Z]\xleftarrow{\delta}[\Sigma A,Z]\leftarrow[\Sigma X,Z]\leftarrow\dots$$
and the assumptions imply that it is exact at $[Y,Z]$ in a fairly strong sense: if $\varphi_1,\varphi_2\in[Y,Z]$ and $g^*\varphi_1=g^*\varphi_2$, then there is $\alpha\in[\Sigma A,Z]$ such that
$$\varphi_1\simeq\varphi_2+\alpha\delta.$$
Suppose that $\delta\simeq\ast$. Then $\alpha\delta\simeq\ast$ and the last equation implies that $\varphi_1\simeq\varphi_2$. In particular $(\ast)$ is satisfied. On the other hand we have seen above that if $g$ satisfies $(\ast)$, then $\delta\simeq\ast$. Thus:

Proposition: Assume $(\ast\ast)$. Then $(\ast)$ holds for $g$ if and only if $\delta\simeq\ast$.

Recall how the necessity here was established and rerun the details after assuming $(\ast\ast)$. This gives us another eqivalent fomulation.

Proposition: Assume $(\ast\ast)$. Then $(\ast)$ holds for $g$ if and only if $\varphi g\simeq\ast$ implies $\varphi\simeq\ast$ for any map $\varphi:Y\rightarrow Z$.

Furthermore, under $(\ast\ast)$ we can refine some of the previous details.

Proposition: Assume $(\ast\ast)$. Then $(\ast)$ holds for $g$ if and only if $\delta$ has a left homotopy inverse if and only if $\Sigma X\simeq \Sigma A\vee \Sigma Y$ under $\Sigma A$.

This gives three equivalent conditions for $(\ast)$ to hold. I'll stress that everthing is true under the more general assumptions I spelled out above.

Here is an example of where this can be put to use. Recall that $S^m\times S^n$ is formed from $S^m\vee S^n$ by attaching an $(m+n)$-cell along the Whitehead map $w:S^{m+n-1}\rightarrow S^m\vee S^n$. The map we will be interested is the collapse map $g:S^m\times S^n\rightarrow S^{m+n}$ which pinches to the top cell.

Then we have a homotopy cofibration sequence
$$S^{m+n-1}\xrightarrow{w} S^m\vee S^n\rightarrow S^m\times S^n\xrightarrow{g}S^{m+n}\xrightarrow{\delta\simeq\Sigma w}S^{m+1}\vee S^{n=1}\rightarrow\dots$$
It's well-known that $\Sigma w\simeq\ast$. So by our results above we have

Proposition: The collapse map $g:S^m\times S^n\rightarrow S^{m+n}$ is a homotopy epimorphism.

Of course we obtain the well-known homotopy equivalence $\Sigma(S^n\times S^n)\simeq S^{m+1}\vee S^{n+1}\vee S^{n+m+1}$, only now we have some insight into where this is coming from.

Title: Does $f_1g≃f_2g$ imply $f_1≃f_2$?


问:
Classify all groups with normal subgroups of order 3 and 4
Suppose a group $G$ has normal subgroups of size 3 and 4. I want to classify all such groups.

The subgroups are clearly $\mathbf Z_3$ and $\mathbf Z_4$ or $K_4=\mathbf Z_2\times\mathbf Z_2$. If $H_4$ denotes one of $\mathbf Z_4$ and $K_4$, then any group of the form
$$\mathbf Z_3\times H_4\times L$$
where $L$ is any group, satisfies the required condition.

I suspect that these are the only groups which satisfy this this condition, but I don't know how to prove it. I am familiar with the following result for obtaining equations of direct products like what I'm after:

If $H,K\leqslant G$, $H\cap K = \{1\}$ and the elements of $H$ and $K$ commute, then $HK\simeq H\times K$.

But I don't believe I can apply it to my situation since I don't know if elements commute.

I appreciate any help with this.

答:
When $G$ is abelian your condition is equivalent to having $12 \mid |G|$.

Necessity stems from Lagrange's theorem, sufficiency by using the following remark for $3$ and $2$ Sylow subgroups,

If $P$ is a group of order $p^n$, with $p$ a prime, it has a normal subgroup of order $p^m$ for all $m \leq n$.

Moreover, in that case such groups are of the form

$$
\Bbb Z_3 \oplus \Bbb Z_2 \oplus \Bbb Z_2 \oplus H, \quad \Bbb Z_3 \oplus \Bbb Z_{2^n} \oplus H \quad (n \geq 2)
$$

with $H$ an abelian group.


问:
Distribution of range of one dimensional simple random walking
The formal question is as follow: Let $X_1,...,X_N,...$ are independent random variables with distribution $\text{Pr}(X_i=1)=1/2$ and $\text{Pr}(X_i=-1)=1/2$,
denote $$Y_N=\text{Max}_{1\leq i\leq N}|\sum_{j=1}^{i}X_j|$$ which is the maximal absolute value achieved in a random process.

I want to estimate the limit of distributions of $\frac{Y_N}{N}$ when $N$ goes to infinity. Especially, I want to get a upper bound on
$$\lim_{N\to\infty}\text{Pr}(\frac{Y_N}{N}\leq \alpha)$$ for $0<\alpha<1$

答:
Let $S_{n}=\sum_{j=1}^{n} X_{j}$.

The number of different walks of n steps where each step is +1 or −1 is $2^{n}$. For the simple random walk, each of these walks is equally likely. In order for $S_{n}$ to be equal to a number k it is necessary and sufficient that the number of +1 in the walk exceeds those of −1 by k. It follows +1 must appear \frac{n + k}{2} times among n steps of a walk, hence the number of walks which satisfy $S_{n} = k$ equals the number of ways of choosing (n + k)/2 elements from an n element set. Notice that n+k must be even since n=j+k+j where j is the number of -1 steps, so $n+k = 2j + 2k$ and we get the formula.

$$P(S_{n}=k) = \frac{n!}{(n-\frac{n+k}{2})!(\frac{n+k}{2})!2^{n}}$$

Using a symmetry we get $P(S_{n}=k)=P(S_{n}=-k)$.

So for n and k even:

$$P(|S_{n}|\leq k) = \sum^{k/2}_{j=0}2 P(S_{n} = 2j) - P(S_{n} = 0) = - \frac{n!}{2^{n}(\frac{n}{2}!)^{2}} + 2\sum^{k/2}_{j=0} \frac{n!}{(n-\frac{n+2j}{2})!(\frac{n+2j}{2})!2^{n}}$$

We can use a bound from https://mathoverflow.net/questions/17202/sum-of-the-first-k-binomial-coefficients-for-fixed-n

To get the bound $$P(|S_{n}|\leq k) \leq \frac{{n}\choose{k/2 +1}}{{n}\choose{n/2}} - \frac{{n}\choose{n/2}}{2^{n}}$$


问:
When unramified maps are smooth coverings?
Let $Y$ be a projective manifold that is also a compact complex manifold. Assume that there is a finite unramified covering $p : \widetilde Y \to Y$. Is it possible to assume that $p$ is a smooth covering map with sheets consisting of a finite number of points?

I must say that I am no algebraic geometer, but as far as I understand, the hypothesis of ''unramified covering map'' on Riemann surfaces implies the result provided if the map $p$ is holomorphic and not constant. So, if $p$ in this case happens to be a topological covering map, it is possible to consider a unique smooth structure on $\widetilde Y$ such that $p$ is a smooth covering map. Is this right?

Concerning the hypothesis of $p$ being finite, I see that this is a very algebraic characterization, can this imply what I asked about the sheets?

答:
This is not true without assuming something extra of the map $\widetilde{Y}\to Y$. Consider $\Bbb P^1\sqcup \Bbb P^0 \to \Bbb P^1$ where the map from $\Bbb P^1$ is the identity and the map from $\Bbb P^0$ is the inclusion of some (closed) point. This satisfies all of your stated assumptions, but is not a covering map.

The missing ingredient is flatness: once you have that, you have an etale morphism. Combining that with your assumption it's surjective and finite, such a morphism is the algebro-geometric interpretation of a finite-sheeted covering map from topology, and everything you ask for is true.

As far as the smooth structure stuff goes, I think there are problems with what you're trying to do. Being able to define holomorphicity of $p$ already gives a smooth structure on $\widetilde{Y}$, and objects in algebraic geometry are generally considered in such a way that the equivalent of a smooth structure is intrinsic to the data of the object.


答:
Let $S_n=X_1+\dots+X_n$ be your position on the random walk.

The key claim is that $\mathbb P(Y_n\geq y)=\mathbb P(\lvert S_n\rvert\geq y)+\mathbb P(\lvert S_n\rvert\geq y+1)$.

First, note that if $s\geq y$, we have $\mathbb P(Y_n\geq y, \lvert S_n\rvert=s)=\mathbb P(\lvert S_n\rvert=s)$. Now if $\lvert S_n\rvert=s<y$, then by the reflection principle,
$$\mathbb P(Y_n\geq y, \lvert S_n\rvert=s)=\mathbb P(Y_n\geq y, \lvert S_n\rvert=2y-s)=\mathbb P(\lvert S_n\rvert=2y-s).$$
So it follows that
\begin{align*}
\mathbb P(Y_n\geq y) &= \sum_{s}\mathbb P(Y_n\geq y, \lvert S_n\rvert=s) \\
&= \sum_{s\leq y-1}\mathbb P(\lvert S_n\rvert=2y-s)+\sum_{s\geq y}\mathbb P(\lvert S_n\rvert=s) \\
&=\mathbb P(\lvert S_n\rvert\geq y+1)+\mathbb P(\lvert S_n\rvert\geq y),
\end{align*}
as claimed. Now we can use the law of large numbers to finish:
$$\lim_{N\to\infty}\mathbb P(Y_N/N<\alpha)=1-2\lim_{N\to\infty}\mathbb P(\lvert S_n\rvert/N\geq\alpha)=1.$$

Title: Distribution of range of one dimensional simple random walking

Body:
The formal question is as follow: Let $X_1,...,X_N,...$ are independent random variables with distribution $\text{Pr}(X_i=1)=1/2$ and $\text{Pr}(X_i=-1)=1/2$,
denote $$Y_N=\text{Max}_{1\leq i\leq N}|\sum_{j=1}^{i}X_j|$$ which is the maximal absolute value achieved in a random process.

I want to estimate the limit of distributions of $\frac{Y_N}{N}$ when $N$ goes to infinity. Especially, I want to get a upper bound on
$$\lim_{N\to\infty}\text{Pr}(\frac{Y_N}{N}\leq \alpha)$$ for $0<\alpha<1$


问:
List of all measures
Let Ω be a four-element set (elements do not matter).

How can I find a list of all measures µ: P(Ω)→ N, for which µ(Ω) = 7, where P(Ω) is Power set of Ω.

答: This is just equivalent to set of all $4$-tuples $(a,b,c,d)$ in $\Bbb N$ such that $a+b+c+d=7$. We only need to assign integer values to the singletons and the measure is determined uniquely. It's quite easy to enumerate these; we will have $\binom{10}{7}$ many of them, by classic combinatiorial facts.

Title: List of all measures


问:
Basis of vector space given the trace equals zero
I have the following question here:

Let $X=\begin{pmatrix}1 &-1\\ 1 & 1 \end{pmatrix}$, and let $U=\{A \in M_2(\mathbb{R})\mid \mathrm{Tr}(XA)=0\}$ , a subspace of $M_2{\mathbb{(R)}}$.

a) Show that the matrices
$$A_1=\begin{pmatrix}1 &1\\ 1 & -1 \end{pmatrix},A_2=\begin{pmatrix}1 &-1\\ -1 & -1 \end{pmatrix},A_3=\begin{pmatrix}1 &-1\\ 1 & 1 \end{pmatrix}$$are all in $U$.

That was easy. Just just the conditions given. $\mathrm{Tr}(XA)=0$ in all cases.

b) Decide whether $\{A_1,A_2,A_3\}$ is a basis for $U$.

I think I have to use the rank nullity theorem here somehow but I am not sure. Can someone help out here? Thanks!

答: Those matrices are linearly independent, because, if $\alpha,\beta,\gamma\in\Bbb R$, then\begin{align}\alpha A_1+\beta A_2+\gamma A_3=\begin{bmatrix}0&0\\0&0\end{bmatrix}&\iff\begin{bmatrix}\alpha+\beta+\gamma&\alpha-\beta-\gamma\\\alpha-\beta+\gamma&-\alpha-\beta+\gamma\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}\\&\iff\left\{\begin{array}{l}\alpha+\beta+\gamma=0\\\alpha-\beta-\gamma=0\\\alpha-\beta+\gamma=0\\-\alpha-\beta+\gamma=0\end{array}\right.\\&\iff\alpha=\beta=\gamma=0.\end{align}If $\{A_1,A_2,A_3\}$ was not a basis of $U$, then $\dim U>3$. But $U$ is a subspace of $M_2(\Bbb R)$, whose dimension is $4$. So, you would have $U=M_2(\Bbb R)$. But this is impossible, since $X^T\notin U$.


问:
Telescoping series: $\sum\limits_{n=1}^{∞}[\tan^{-1}(2n+1)-\tan^{-1}(2n-1)]$
In this question that was asked today the OP wrote that
$$\begin{align}\sum\limits_{n=1}^{∞}[\arctan(2n+1)-\arctan(2n-1)]&=\arctan\infty-\arctan 1\\
&=\frac{\pi}{2}-\frac{\pi}{4}\\&=\frac{\pi}{4}
\end{align}$$

I don't really understand why there is a $\arctan\infty$ term. Surely every single term other than $-\arctan 1$ has been cancelled out? Now I know that I am wrong, as obviously the value of the summation cannot be negative, but I'm not sure where I am wrong.

I have thought of considering the similar finite series, $$\sum\limits_{n=1}^k[\arctan(2n+1)-\arctan(2n-1)]=\arctan(2k+1)-\arctan1$$
and letting $k\to\infty$, so that
$$\lim_{k\to\infty}\sum\limits_{n=1}^k[\arctan(2n+1)-\arctan(2n-1)]=\arctan\infty-\arctan1$$ as required, but I still can't see why all the terms other than $-\arctan1$ don't cancel out, as the upper limit actually is $\infty$.

Thank you for your help.

答:
A few intermediate steps might be helpful.

We obtain
\begin{align*}
\color{blue}{\sum_{n=1}^{\infty}}&\color{blue}{\left[\arctan(2n+1)-\arctan(2n-1)\right]}\\
&=\lim_{N\to\infty}{\sum_{n=1}^{N}\left[\arctan(2n+1)-\arctan(2n-1)\right]}\\
&=\lim_{N\to\infty}\left[\arctan(2N+1)-\arctan(1)\right]\tag{1}\\
&=\lim_{N\to\infty}\arctan(2N+1)-\lim_{N\to\infty}\arctan(1)\tag{2}\\
&=\frac{\pi}{2}-\frac{\pi}{4}\\
&\,\,\color{blue}{=\frac{\pi}{4}}\\
\end{align*}

Comment:

In (1) we apply telescoping.

In (2) we use $\lim_{N\to\infty}\arctan(N)=\frac{\pi}{2}$ and $\lim_{N\to\infty} a=a$.


问:
Find x, y, and z (all of them non-negative) that maximize $ x^\frac14 y^\frac14 z^\frac14 $ subject to the constraint that $ px + y + z ≤ m $
Find x, y, and z (all of them non-negative) that maximize $ x^\frac14 y^\frac14 z^\frac14 $ subject to the constraint that $ px + y + z ≤ m $

Context: undergraduate mathematical economics

I have attempted this problem to the best of my ability but I'm really not sure if I got the mechanics of it right. If somebody would be kind enough to check my work over and point out any potential errors I would be greatly appreciative.

Setting up the Lagrangian:

$ L = x^\frac14 y^\frac14 z^\frac14 - \lambda (px + y + z - m) ≤ 0 $

First Order Conditions:

$ f_x = \frac14x^\frac{-3}4 y^\frac14 z^\frac14 - \lambda p $

$ f_y = x^\frac14 \frac14y^\frac{-3}4 z^\frac14 - \lambda $

$ f_z = x^\frac14 y^\frac14 \frac14z^\frac{-3}4 - \lambda $

$ f_\lambda = -px-y-z+m $

So if I'm not mistaken we'd then get $ px = y = z $ by simplifying the first three equations.

Then, using $ f_\lambda = -px-y-z+m $, we can obtain expressions for x, y, and z:

$ x = \frac m{3p} $
$ y = \frac m3 $
$ z = \frac m3 $

Second part of the problem involves using the envelope theorem to solve for $ \frac {dV}{dm} $. Using the above I calculated this to be:

$ \frac {dV}{dm} = \frac{\partial L}{\partial m} = \lambda $

Taking $ f_z = x^\frac14 y^\frac14 \frac14z^\frac{-3}4 = \lambda $

=> $ \frac{m}{3p}^\frac14 \frac{m}{3}^\frac14 \frac14 \frac{m}{3}^\frac{-3}{4} = \lambda $

$ \lambda = \frac{3^\frac14}{4(m^\frac14)(p^\frac14)} $

Am I on the right track here? Been a while since I've taken a class involving this kind of math so I'm a little rusty.

答:
Another approach is to recognize that $(x y z)^{1/4}$ is maximized when $(p x y z)^{1/3}$ is maximized and apply the AM-GM inequality:
$$(p x y z)^{1/3} \le \frac{p x + y + z}{3} = \frac{m}{3},$$
with equality when $px=y=z$, yielding $$(x,y,z)=\left(\frac{m}{3p},\frac{m}{3},\frac{m}{3}\right).$$


问:
Using the Riemann-Lebesgue lemma to find the integral of a very oscillating function
I'm a bit confused with how the Riemann-Lebesgue lemma is used to show that the integral of a very oscillating function approaches zero as the number of oscillation increases. To be more specific, let $f(x)=\cos(kx)$ for all $x\in [0,1]$ and zero everywhere else, and we want to find
\begin{equation}
\int_{0}^{1} f(x) {\rm d}x
\end{equation}
how does one use the Riemann-Lebesgue lemma to show that this definite integral approaches zero as $k \rightarrow \infty$?

答:
Let $f(x)=\chi_{[0,1]}(x)$. Then the Fourier transform of $f$ is given by
$$\hat{f}(k)=\int_\mathbb{R} e^{-ikx}f(x)\,dx=\int_\mathbb{R}(\cos(-kx)+i\sin(-kx))\chi_{[0,1]}(x)\,dx$$
(perhaps also multiplying by $\pi$ somewhere depending on your convention). Thus, the real part of the Fourier transform of $f$, $\Re \hat{f}(k)$ is just equal to
$$\int_\mathbb{R}\cos(-kx)\chi_{[0,1]}(x)\,dx=\int_0^1\cos(kx)\,dx$$
where we have used that $cos(-kx)=cos(kx)$. The result then follows from the usual Riemann-Lebesgue lemma.

However, I would suggest thinking about this result "by hand" by going back to the proof of the Riemann-Lebesgue lemma. The essence of the lemma is to use oscillation to obtain cancellation and thus decay. This situation occurs often in analysis and the usual way to make this precise is via integration by parts. In this instance, we would have to approximate $f$ by a smooth function close to it in the $L^1$ norm, and then proceed as in the proof of Riemann-Lebesgue.


问:
Simple question about the Brauer group of a variety
If $X$ is a variety over a field $k$ and $\overline{k}$ is an algebraic closure of $k$, then there is a homomorphism $\text{Br}(X) \rightarrow \text{Br}(X \times_k \overline{k})$. Since $\text{Br}$ is a contravariant functor, is the homomorphism from $X_{\overline{k}}$ to $X$ the projection map? I would have thought that the inclusion morphism $X \rightarrow X_{\overline{k}}$ would give a natural homomorphism $\text{Br}(X_{\overline{k}}) \rightarrow \text{Br}(X)$.

答: There is no "inclusion map" $X\to X_{\overline{k}}$. The map $X_{\overline{k}}\to X$ you refer to is indeed the projection map: it's the base change of $\operatorname{Spec} \overline{k}\to \operatorname{Spec} k$ along $X\to \operatorname{Spec} k$, the structure map of $X$.


问:
2-Norm of a Submatrix is $\leq$ 2-Norm of Original Matrix
Say $A$ is a submatrix of $B$. How do I prove that the $\|A\|_2 \leq \|B\|_2$?
I can easily show this for $\|\cdot\|_1, \|\cdot\|_\infty, $ and $\|\cdot\|_F$ and thought maybe the solution lies in relating the inequalities of these other norms to the 2-norm, but this path hasn't proved fruitful.</p>

答:
Let $\boldsymbol{B}=\boldsymbol{R}_{0}\boldsymbol{A}\boldsymbol{C}_{0}$,
where $\boldsymbol{R}_{0},\boldsymbol{C}_{0}$ are the identity matrices
with $1$'s replaced by $0$'s at positions of the removed rows and
columns. We have
\begin{align*}
\|\boldsymbol{B}\|_{2}^{2} & \triangleq\max\limits _{\boldsymbol{x},\,\|\boldsymbol{x}\|=1}\|\boldsymbol{R}_{0}\boldsymbol{A}\boldsymbol{C}_{0}\boldsymbol{x}\|_{2}^{2}=\max\limits _{\boldsymbol{y},\,\|\boldsymbol{y}\|=1}\quad\max\limits _{\boldsymbol{x},\,\|\boldsymbol{x}\|=1}\left(\boldsymbol{y}^{T}\boldsymbol{R}_{0}\boldsymbol{A}\boldsymbol{C}_{0}\boldsymbol{x}\right)^{2}\\
 & =\max\limits _{\boldsymbol{y},\,\|\boldsymbol{y}\|=1,\,\boldsymbol{y}\in\mathcal{C}(\boldsymbol{R}_{0}^{T})}\quad\max\limits _{\boldsymbol{x},\,\|\boldsymbol{x}\|=1,\,\boldsymbol{x}\in\mathcal{C}(\boldsymbol{C}_{0})}\left(\boldsymbol{y}^{T}\boldsymbol{A}\boldsymbol{x}\right)^{2}\\
 & \leq\max\limits _{\boldsymbol{y},\,\|\boldsymbol{y}\|=1}\max\limits _{\boldsymbol{x},\,\|\boldsymbol{x}\|=1}\|\boldsymbol{y}^{T}\boldsymbol{A}\boldsymbol{x}\|_{2}^{2}=\|\boldsymbol{A}\|_{2}^{2},
\end{align*}
where $\mathcal{C}(\boldsymbol{C}_{0})\equiv\mathrm{Im}\boldsymbol{C}_{0}$ denotes the column space of the matrix.

The transition to the second line can be figured out as follows: if $\boldsymbol x$ has a non-zero component in the null-space $\boldsymbol{C}_{0}$, then the objective can be improved at the expense of this component; therefore, an optimum is in $\mathcal{C}(\boldsymbol{C}_{0})$. A similar argument applies to $\boldsymbol{y}$.

The transition to the third line is due that the fact that removing the constraints
$\boldsymbol{x}\in\mathcal{C}(\boldsymbol{C}_{0})$ and $\boldsymbol{y}\in\mathcal{C}(\boldsymbol{R}_{0}^{T})$ can only improve the optimum. As we are maximising here, the inequality sign is $\leq$.

I guess, another way to prove the inequality is by considering the singular value decomposition of the matrix (as a sum of outer products) and making essentially the same argument there.

Update. See also a sketch of a proof relying on submultiplicativity, which holds for induced matrix norms: https://www.physicsforums.com/threads/proof-that-norm-of-submatrix-must-be-less-than-norm-of-matrix-its-embedded-in.436855/


问:
Formulating a problem as a linear program
I have the following problem:

I want to sell five computers. Denote them by $1, 2, 3, 4, 5$. I am allowed to take offers for subsets of the five computers. Three buyers, $A$, $B$, and $C$, want to make an offer. Each buyer has identified a maximum price that they would be willing to pay for various combinations: each individual buyer would only buy at most one parcel combination.

 What is a possible linear programming formulation if I want to maximize the money that I make?

I would appreciate any help in approaching this problem.

答:
Here are a few hints.

Define a binary variable for each combination.  The objective is to maximize revenue.  Two sets of constraints: don't sell the same computer more than once, and don't sell more than one combination to the same buyer.


问:
Compact formula for the $n$th derivative of $f\left(\sqrt{x+1}\right)$?
I'm interested in a general formula for

$$\frac{d^n}{dx^n}\Big[f\left(\sqrt{x+1}\right)\Big].$$

In particular, Fàa di Bruno's formula gives

$$\frac{d^n}{dx^n}\Big[f\left(\sqrt{x+1}\right)\Big]=\sum_{k=1}^n f^{(k)}\left(\sqrt{x+1}\right)B_{n,k}\big(g'(x),g''(x),\dots,g^{(n-k+1)}(x)\big),$$

where $g(x)=\sqrt{x+1}$ and $B_{n,k}$ denote Bell polynomials.

I suspect we can do better than this, using the identities

$$\frac{d^k}{dx^k}\Big[\sqrt{x+1}\Big]=\left(\frac{1}{2}\right)^\underline{k}x^{1/2-k}=-\frac{(2k-3)!!}{(-2)^k}x^{1/2-k}=\frac{\sqrt{\pi}}{2\Gamma(3/2-k)}x^{1/2-k}.$$

In particular, every term in $B_{n,k}$ for fixed $n$ and $k$ will be some numerical factor times $(x+1)^{k/2-n}$ (see notes below), so the result is always of the form

$$\frac{d^n}{dx^n}\Big[f\left(\sqrt{x+1}\right)\Big]
=\sum_{k=1}^n a_{n,k} \frac{f^{(k)}\left(\sqrt{x+1}\right)}{(x+1)^{n-k/2}}$$

for some constants $a_{n,k}.$ Is there an explicit expression for the $a_{n,k}$ above? For example, it seems likely that there is some expression using Stirling numbers, factorials, binomial coefficients, etc., given the combinatorial nature of the problem.

Current progress:

Using the first identity above for derivatives of $\sqrt{x+1}$ [where $(\cdot)^\underline{k}$ denotes the $k$th falling factorial], we have

$$B_{n,k}\big(g'(x),g''(x),\dots,g^{(n-k+1)}(x)\big)
=B_{n,k}\left[\left(\frac{1}{2}\right)^\underline{1}x^{1/2-1},\left(\frac{1}{2}\right)^\underline{2}x^{1/2-2},\dots,\left(\frac{1}{2}\right)^\underline{n-k+1}x^{1/2-(n-k+1)}\right].$$

The definition of the Bell polynomials guarantees that the power of $x$ in every term of $B_{n,k}(\cdots)$ comes out to $x^{k/2-n}$, with pre-factor given by

$$a_{n,k}=B_{n,k}\left[\left(\frac{1}{2}\right)^\underline{1},\left(\frac{1}{2}\right)^\underline{2},\dots,\left(\frac{1}{2}\right)^\underline{n-k+1}\right],$$

so the remaining task is to simplify the above expression, if possible.

答:
Heuristic

By analyzing the matrix of the first values for $a_{n,m}$ it is possible to
deduce that
$$ \bbox[lightyellow] {
a_{\,n,\,m}  =\frac{ \left( { - 1} \right)^{n - m} }{2^n} \left( \matrix{  2n - 1 - m \cr   2n - 2m \cr}  \right)
\left( {2n - 2m - 1} \right)!!\quad \left| {\,0 \le m \le n} \right.
} \tag{1}$$
which can be further simplified in terms of Gamma or other, and which can be
of help for a rigorous demonstration.

Demonstration

We put
$$
{{d^{\,n} } \over {dx^{\,n} }}f\left( {\sqrt {x + 1} } \right)
 = \sum\limits_{0\, \le \,k} {a_{\,n,\,k} {{f^{\,\left( k \right)} \left( {\sqrt {x + 1} } \right)}
 \over {\left( {x + 1} \right)^{\,n - k/2} }}}
$$

The $a_{\,n,\,k} $ are supposed to be constant, independent from $f$.

Then they shall be valid for any infinitely differentiable function, in particular
for analytic functions, and thus in particular for
$$
f(x) = x^{\,2m} \quad \left| {\,0 \le m \in Z} \right.
$$

For this we have
$$
\left\{ \matrix{
  f\left( {\sqrt {x + 1} } \right) = \left( {x + 1} \right)^{\,m}  \hfill \cr
  {{d^{\,n} } \over {dx^{\,n} }}f\left( {\sqrt {x + 1} } \right)
 = m^{\,\underline {\,n\,} } \left( {x + 1} \right)^{\,m - n}  \hfill \cr
  f^{\,\left( n \right)} \left( {\sqrt {x + 1} } \right)
 = \left( {2m} \right)^{\,\underline {\,n\,} } \left( {\sqrt {x + 1} } \right)^{\,2m - n}  \hfill \cr}  \right.
$$
and therefore
$$
\eqalign{
  & m^{\,\underline {\,n\,} } \left( {x + 1} \right)^{\,m - n}  =   \cr
  &  = \sum\limits_{0\, \le \,k} {a_{\,n,\,k} {{\left( {2m} \right)^{\,\underline {\,k\,} }
 \left( {x + 1} \right)^{\,m - k/2} } \over {\left( {x + 1} \right)^{\,n - k/2} }}}
  = \sum\limits_{0\, \le \,k} {a_{\,n,\,k} \left( {2m} \right)^{\,\underline {\,k\,} }
 \left( {x + 1} \right)^{\,m - n} }   \cr
  & \quad \quad  \Downarrow   \cr
  & m^{\,\underline {\,n\,} }
  = \sum\limits_{0\, \le \,k} {a_{\,n,\,k} \left( {2m} \right)^{\,\underline {\,k\,} } }
 \quad  \Leftrightarrow \quad \left( {{m \over 2}} \right)^{\,\underline {\,n\,} }
  = \sum\limits_{0\, \le \,k} {a_{\,n,\,k} m^{\,\underline {\,k\,} } }  \cr}
$$

Since
$$
\eqalign{
  & \left( {{m \over 2}} \right)^{\,\underline {\,n\,} }
  = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n} \right)}
 {\left( { - 1} \right)^{\,n - j} \left[ \matrix{  n \cr   j \cr}  \right]\left( {{m \over 2}} \right)^{\,j} }
  = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n} \right)}
 {{{\left( { - 1} \right)^{\,n - j} } \over {2^{\,j} }}\left[ \matrix{ n \cr  j \cr} \right]m^{\,j} }   \cr
  & \sum\limits_{0\, \le \,k} {a_{\,n,\,k} m^{\,\underline {\,k\,} } }
  = \sum\limits_{0\, \le \,k} {a_{\,n,\,k} \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,k} \right)}
 {\left( { - 1} \right)^{\,k - j} \left[ \matrix{k \cr  j \cr}  \right]m^{\,j} } }  =   \cr
  &  = \sum\limits_{\left( {0\, \le } \right)\,j\,} {\left( {\sum\limits_{0\, \le \,k}
 {\left( { - 1} \right)^{\,k - j} a_{\,n,\,k} \left[ \matrix{ k \cr  j \cr}  \right]} } \right)m^{\,j} }  \cr}
$$
i.e.
$$
{{\left( { - 1} \right)^{\,n - j} } \over {2^{\,j} }}\left[ \matrix{  n \cr  j \cr}  \right]
 = \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k - j} a_{\,n,\,k}
 \left[ \matrix{  k \cr   j \cr}  \right]}
$$
which can be viewed as a matrix relation involving the matrix $A:\; A_{\,n,\,k} = a_{\,n,\,k}$
multiplied by the matrix of the signed Stirling N. of 1st kind, which is well known to be
the inverse of the Stirling N. of 2nd kind, so:
$$ \bbox[lightyellow] {
a_{\,n,\,m}  = \sum\limits_{0\, \le \,k}
 {{{\left( { - 1} \right)^{\,n - k} } \over {2^{\,k} }}
\left[ \matrix{  n \cr   k \cr}  \right]\left\{ \matrix{  k \cr  m \cr}  \right\}}
}\tag{2}$$

The computation of the above checks with the heuristic formula:

it remains to demonstrate this fact analytically.


问:
Calculate the Jordan matrix of a matrix $C$ where $C$ exists of the unity matrix and a matrix $A$ which is diagonalisable.
Let $A \in M^{3 \times 3}(\mathbb{C})$ be diagonalisable matrix with 3 different eigenvalues $\lambda_1, \lambda_2, \lambda_3$ and their corresponding eigenvalues $v_1, v_2, v_3$. Consider the matrix $C \in M^{6 \times 6} (\mathbb{C})$ with
$C = \left( \begin{matrix} A & I \\ 0 & A \end{matrix} \right) $.

Calculate the Jordan Form/Matrix of $C$ and find a matrix P such that $P^{-1}CP = J$.

I know that the characteristic polynomial will be $f_C = (X- \lambda_1)^2(X- \lambda_2)^2(X- \lambda_3)^2$.

To be honest, I don't really know what else I can deduct from the information I have.

I know that in $A$ for each $\lambda_i$ it holds that $\text{dim(Ker}(A-\lambda_i)) = 1$ but I don't know how this translates to $C$.

Can somebody help me?

答:
Let $V$ be the $3\times 3$ matrix with columns $v_1, v_2, v_3$.
Let $L$ be the diagonal matrix with diagonal entries $\lambda_1, \lambda_2, \lambda_3$.
Then, using block matrix computations,
$$
\begin{aligned}
AV
&=A[v_1\ v_2\ v_3]\\
&=[Av_1\ Av_2\ Av_3]\\
&=[v_1\ v_2\ v_3]L\\
&=VL\ .
\end{aligned}
$$
So $V^{-1}AV$ is the diagonal matrix $L$. Similarly:
$$
\begin{aligned}
C
\begin{bmatrix}V & \\ & V\end{bmatrix}
&=
\begin{bmatrix}A & I\\ & A\end{bmatrix}
\begin{bmatrix}V & \\ & V\end{bmatrix}
\\
&=
\begin{bmatrix}AV & V\\ & AV\end{bmatrix}
\\
&=
\begin{bmatrix}VL & V\\ & VL\end{bmatrix}
\\
&=
\begin{bmatrix}V & \\ & V\end{bmatrix}
\begin{bmatrix}L & I\\ & L\end{bmatrix}
\text{ so}\\
\begin{bmatrix}V & \\ & V\end{bmatrix}^{-1}
C
\begin{bmatrix}V & \\ & V\end{bmatrix}
&=
\begin{bmatrix}L & I\\ & L\end{bmatrix}
\ .
\end{aligned}
$$
Using a permutation matrix conjugation, the last $(3+3)\times (3+3)$ matrix is written as a $(2+2+2)\times(2+2+2)$ matrix with diagonal blocks
$$
\begin{bmatrix}\lambda_j & 1\\ & \lambda_j\end{bmatrix}\ ,\qquad j=1,2,3
\ .
$$


问:
How to determine dimensions of a square based prism
asks;

Determine the dimensions of a square based prism box with each volume that requires the least material to make.

$$a) 512 \ \ cm^3$$
$$b) 1000\ \ cm$3$$
$$c) 750\ \  cm^3$$

I attempted the question, sort of understand it. if someone can show me how to do one of them, I can most likely do the rest on my own .

答:
Suppose the prism has side length $a,b,c$ with two of these equal. Then the volume is $v=abc$ and the surface area, which we want to minimize, is $a=2(ab+bc+ac)=2v(\frac 1a+\frac 1b+\frac 1c) = 6v/\text{HM}(a,b,c)$. Now the side length of a cube of the same volume is $\sqrt[3]{abc}$ and the surface area is $a_s=6\left(\sqrt[3]{abc}\right)^2 = 6v/\text{GM}(a,b,c)$. By the inequalities among Pythagorean means, we have $\text{GM}(a,b,c)\geq \text{HM}(a,b,c)$ with equality only when $a=b=c$, so the cube (which is a valid square prism) minimizes surface area for a given volume.

Thus you can calculate the surface area in each case on the basis that the boxes are cubical.


问:
How to show that $\sum_{k=n}^{2n}\frac{k}{(k+1)^{2}}$ converges?
I'm trying to show that the sequence $a_n=\sum_{k=n}^{2n}\frac{k}{(k+1)^{2}}$ converges. I actually proved that it converges by using the Reimann sum showed that it approaches $\log(2)$. But how can I show it using basic theorems for sequences? I'm trying to help a student that didn't study functions/integrals yet. Only plain limits with sequences. Is it possible to do with sequences?

答: Hint: $\frac x{(x+1)^{2}}$ is decreasing on $[1, \infty)$ so  $\frac k{(k+1)^{2}} <\int_{k-1}^{k} \frac x{(x+1)^{2}}dx$ for each $k$. Summing over $k$ we get$\sum\limits_{k=n}^{2n}\frac k{(k+1)^{2}} <\int_{n-1}^{2n} \frac x{(x+1)^{2}}$. Evaluate this integral explicitly by writng the numerator as $(x+1)-1$.


问:
A game consists of rolling a die 8 times. You win if the result is 5 or 6.
A game consists of rolling a die 8 times. You win if the result is 5 or 6. Your numbers of wins will be recorded.

a) Identify the discrete random variable, X
ans: Each time you role a die, if the result is 5 or 6, you win the game

b) State the number of trials, n

ans: As it is stated 8 is number of trials

c) State the probability of success, p, in any trial

ans: 1.52* 10^-4

d) State the probability of failure, q, in any trial.

ans: 0.9999

Hi I was doing prctice questions and saw this question, I answered them but not really sure if the answers are correct. I checked online , it did not help me much.
I wouold appreciate it if you could help me see my answers are correct or not.

答:
Having rolled the die eight times, we are interested in counting how many (if any) of those times a 5 or 6 was rolled.

We give a name to this "counting of how many 5's or 6's"... we call it a random variable and it is customary to name a random variable $X$.

That is to say, $X$ is a way of looking at the results of our eight rolls of dice and being able to say aloud "there were three 5's or 6's" or "there were no 5's and no 6's" etc...

$X$ has "support" $\{0,1,2,3,4,5,6,7,8\}$ as these are the possible answers to the question of "how many 5's or 6's did we roll in our 8 rolls of the dice."

It can be said that our random process consists of several smaller identical independent "trials"... each of which correspond to a single roll of a die.  A "success" in one of these smaller trials correspond to us getting a 5 or 6 on the die.

The probability of success in a specific trial is going to simply be $\dfrac{1}{3}$ as it is assumed that the dice used are standard fair six-sided dice and two of the six sides correspond to a "win" here.

The probability of failure in a specific trial is going to simply be $\dfrac{2}{3}$, the opposite of the probability of success.

The answers as I would have written it:  a) $X$ is the random variable counting the total number of successes across all trials.  b) There are $n=8$ trials.  c) The probability of success for a specific trial is $p=\dfrac{1}{3}$.  d) The probability of failure for a specific trial is $q=1-p=\dfrac{2}{3}$

We call $X$ "binomially distributed" in this case and abbreviate all of the above into one statement, $X\sim\text{Bin}(8,\dfrac{1}{3})$

We can now talk about the probability that $X$ gives a certain value, i.e. that we rolled a certain number of 5's or 6's.  We have that the probability of us rolling $k$ 5's or 6's will be:

$$\Pr(X=k)=\binom{8}{k}\left(\frac{1}{3}\right)^k\left(\frac{2}{3}\right)^{8-k}$$

This information was not needed however for the parts of the question which you asked though it may likely appear in a followup question or parts of the question which you did not yet write.


问:
R infinite space endowed with the infinite norm and fixed point theorem of Schauder
Let $\mathbb{R}^{\infty}$ be the vector space of infinite sequences $(\alpha_1,\alpha_2,\alpha_3,\dots)$ of real numbers as a natural generalization of  $\mathbb{R}^k$, $k\in\mathbb{N}^*$. Let $\boldsymbol{C}$ be the subset of $\mathbb{R}^{\infty}$ defined by,
$$\boldsymbol{C} := \left\{\boldsymbol{p} = (p_1, p_2, p_3, \dots) \in \mathbb{R}^{\infty}\big/ \forall ~k, p_k \in [0, 1] \text{ and } \sum_{k = 1}^{\infty} p_{k} \leq 1\right\}.$$
I make these statements. Tell me if I am wrong.

$\boldsymbol{C}$ is a compact and convex nonempty subset of the infinite dimensional space $\mathbb{R}^{\infty}$ for the the infinite norm defined by,
$$||\boldsymbol{p}||_{\infty} = \max_k p_{k}.$$

Schauder's fixed  point Theorem (generalization of Brouwer's fixed point Theorem to an infinite dimensional space) can be applied: any continuous mapping from $\boldsymbol{C}$ to itself has a fixed point.

答:
Define a sequence $(p^{(n)})_{n \in \mathbb{N}}$ in $C$ by $p^{(n)} = (\delta_{in})_{i \in \mathbb{N}}$.  In other words,
\begin{equation*}
p^{(n)}_{i} = \left\{ \begin{array}{r l}
1, & \text{if} \, \, i = n, \\
0, & \text{otherwise.}
\end{array} \right.
\end{equation*}

$(p^{(n)})_{n \in \mathbb{N}}$ does not have any accumulation points in $C$ with the given norm topology.  The idea is this: $\|p^{(n)} - p^{(k)}\|_{\infty} = 1$ for all $n,k \in \mathbb{N}$ so no subsequence is Cauchy.

Edit: Given $m \in \mathbb{N}$, if we define $C^{m}$ by
\begin{equation*}
C^{m}_{*} = \{x \in \mathbb{R}^{\infty} \, \mid \, \sum_{k = 1}^{\infty} |x_{k}| \leq 1, \, \, |x_{k}| \leq \frac{m}{k} \},
\end{equation*}
then $C^{m}_{*}$ is compact in $\mathbb{R}^{\infty}$ with the norm $\|\cdot\|_{\infty}$ given above.  Note that the set $C^{m}$ defined by Ari.stat below is simply $C^{m} = C^{m}_{*} \cap [0,1]^{\infty}$.  Since $[0,1]^{\infty}$ is a closed subset of $\mathbb{R}^{\infty}$ with respect to $\|\cdot\|_{\infty}$, $C^{m}$ is a closed subset of $C^{m}_{*}$ and, thus, is itself compact.

To see this, first, notice that if $x \in C^{m}_{*}$, then $\|x\|_{\infty} \leq \sum_{k = 1}^{\infty} |x_{k}| \leq 1$.

To see that $C^{m}_{*}$ is compact, I will prove that any sequence in $C^{m}$ has a subsequence that converges.  Suppose that $(x^{(n)})_{n \in \mathbb{N}} \subseteq C^{m}_{*}$.  Since $|x^{(n)}_{k}| \leq 1$ independently of $n$ and $k$, we can employ a diagonalization argument to find a subsequence $(n_{j})_{j \in \mathbb{N}} \subseteq \mathbb{N}$ and a sequence $x \in \mathbb{R}^{\infty}$ so that, for each $k \in \mathbb{N}$, $\lim_{j \to \infty} x_{k}^{(n_{j})} = x_{k}$.  In fact, we can show that $\lim_{j \to \infty} \|x^{(n_{j})} - x\|_{\infty} = 0$ must hold.

Indeed, given $\epsilon > 0$, if we choose $K_{0} \in \mathbb{N}$ such that $\frac{m}{k} \leq \epsilon/2$ for each $k \geq K_{0}$, then the inequality $\max\{|x^{(n)}_{k}|, |x_{k}|\} \leq \frac{m}{k}$ implies
\begin{equation*}
\|x^{(n_{j})} - x\|_{\infty} \leq \max \{ |x^{(n_{j})}_{1} - x_{1}|, \dots, |x^{(n_{j})}_{K_{0}} - x_{K_{0}}|, \epsilon \}.
\end{equation*}
Therefore, by pointwise convergence, $\limsup_{j \to \infty} \|x^{(n_{j})} - x\|_{\infty} \leq \epsilon$.  We conclude by sending $\epsilon \to 0^{+}$.


问:
How many increasing subsequences of length k can be made from an increasing sequence of length n.
For example, say you are given the sequence 1,2,3,4 and you want sequences of length 3, the sequences would be (1,2,3), (1,3,4), (2,3,4).

Is there a way I could compute the number of sequences like this.

答: This is the same problem as the number of $k$-element subsets of a set of size $n$, which is counted by binomial coefficients ${n\choose k}$.


Title: How to show that $\sum_{k=n}^{2n}\frac{k}{(k+1)^{2}}$ converges?

Body: I'm trying to show that the sequence $a_n=\sum_{k=n}^{2n}\frac{k}{(k+1)^{2}}$ converges. I actually proved that it converges by using the Reimann sum showed that it approaches $\log(2)$. But how can I show it using basic theorems for sequences? I'm trying to help a student that didn't study functions/integrals yet. Only plain limits with sequences. Is it possible to do with sequences?

Body:
Note that
$$
\begin{align*}
a_{n+1}-a_n&=\frac{2n+2}{(2n+3)^2}+\frac{2n+1}{(2n+2)^2}-\frac{n}{(n+1)^2}\\
&=\frac{2n+2}{(2n+3)^2}+\frac{2n+1}{(2n+2)^2}-\frac{4n}{(2n+2)^2}\\
&=\frac{2n+2}{(2n+3)^2}-\frac{2n-1}{(2n+2)^2}\\
&=\frac{(2n+2)^3-(2n-1)(2n+3)^2}{(2n+3)^2(2n+2)^2}\\
&=\frac{4n^2 + 18n + 17}{(2n+3)^2(2n+2)^2}\\
&>0.
\end{align*}
$$
So, this sequence is monotone increasing.  So, if we can show that it is bounded above, then it must necessarily converge.

To do that, let's do something really naive: note that for $n\leq k\leq 2n$, we have
$$
\frac{k}{(k+1)^2}\leq\frac{2n}{(n+1)^2}\leq\frac{2n}{n^2}=\frac{2}{n}.
$$
This implies that
$$
a_n=\sum_{k=n}^{2n}\frac{k}{(k+1)^2}\leq\sum_{k=n}^{2n}\frac{2}{n}=\frac{2(n+1)}{n}=2+\frac{1}{n}\leq 3.
$$


问:
Difficult matrix equation $A(x-x_0) + \alpha\left(\frac{\Sigma x}{\mu^\top x}\right) = 0$
Let $x,x_0,\theta,\mu \in \mathbb{R}^n$ vectors, $\alpha \in\mathbb{R}$ a scalar, and $\Sigma, A\in \mathbb{R}^{n\times n}$ Positive Semi-Definite matrices. Also, $\mu$ is a mean vector and $\Sigma$ is a covariance matrix of a Random Variable $Z$ (I don't know if this information may help). Is there a way to solve in closed form the following matrix equation for $x$?

\begin{equation}
A(x-x_0) + \alpha \left(\frac{\Sigma x}{\mu^\top x}\right) = 0.
\end{equation}

Main difficulty: I find difficulties in handling the denominator. If the equation would have been simply

$$A(x-x_0) + \alpha \Sigma x = 0,$$ the solution would be $x = (A+\alpha\Sigma)^{-1}Ax_0$. I am looking for a similar solution for the first equation but the denominator complicates things and I don't know how to proceed.

What I tried:

\begin{aligned}
& \mu^TxA(x-x_0) + \alpha \Sigma x = 0
\\ &\mu^TxAx - \mu^TxAx_0 - \alpha \Sigma x = 0
\\ &\mu^TxA(x-x_0) = \alpha \Sigma x
\end{aligned}

Is this way of proceeding correct? If yes, how is it possible to solve this matrix equation for $x$? I am not an expert in solving matrix equations and any help would be greatly appreciated. Thanks!

答:
I can provide some hints for the case where $A,\Sigma$ are strictly positive definite. This assumption guarantees us that both matrices are invertible; let us set henceforth $\lambda=\frac{\alpha}{\mu^T x}$. By multiplying on the right by $\Sigma^{-1}$ the problem transforms into the following one:

$$(B-\lambda I)x=B x_0~~,~~B=\Sigma^{-1}A$$

Of course $B x_0\neq 0$, since $B$ is itself positive definite. Now it is finally obvious that $B-\lambda I$ can only be singular for $\lambda $ in the set of eigenvalues of the matrix $B$ itself, in which case a solution does not exist. For all other values of the parameter a unique solution is guaranteed; if one uses the eigenvalue decomposition of $B=S\Lambda S^{-1}$ the solution is given by the relation

$$x=S(\Lambda-\lambda I)^{-1}S^{-1}Bx_0$$

This solution is naturally, a function of $\lambda$ itself, which can be written in the concise form
$$x=\sum_{k}\frac{v_k}{\lambda_k-\lambda}$$
where the sum is performed over the distinct eigenvalues of $B$ with the vectors $(v_k)_i=S_{ik}\sum_{lm}S^{-1}_{kl}B_{lm}(x_0)_{m}$ whenever all the eigenvalues have multiplicity 1 (but with trivial modifications if it's greater than 1-just take the limit of coinciding eigenvalues).

We are not done- $\lambda$ depends on $x$ itself and that imposes a constraint on the possible solutions. Dotting from the left with the vector $\mu$, we obtain the rational equation
$$\frac{\alpha}{\lambda}=\sum_{k}\frac{\mu^Tv_k}{\lambda_k-\lambda}$$
which generally has $n$ complex roots. However $\lambda$ is a real parameter. Is a solution guaranteed? It is only guaranteed for odd $n$, since the polynomial equation then is forced to have a real solution. A counterexample to this is given by the function
$$f(\lambda)=-\frac{1}{1-\lambda}+\frac{1}{2-\lambda}-\frac{3}{\lambda}$$
which has no real roots. However, it is true that there may be found a large enough value of $|\alpha|$ for which a real solution exists, regardless of the values of $n$ and the projections $\mu^T v_k$.

EDIT: I believe that this is as close algorithmically as one can get to a "closed form". The problem has certainly been reduced to a simpler one, but the impasse is obvious- we cannot solve arbitrary degree polynomial equations in closed form.


问:
How to partition the numbers 1 to $^2$ into  subsets of equal cardinality and equal sum
I came across this question:

A farmer has 25 cows. These are numbered 1 to 25, and i-th cow gives i litres of milk. The farmer has 5 sons. How should he divide his cows among his sons such that each son gets the same number of cows, as well as the same amount of milk.

The way I found to solve this was to construct a magic square using numbers 1 to 25 to get 2 solutions - the set of rows, and the set of columns.

I was wondering if these 2 are the only solutions, and also what would be a good way to approach a generalisation of this problem, i.e.

How to partition the numbers $1$ to $k^2$ into $k$ subsets of equal cardinality and equal sum.

答:
I'm not familiar with a magic square solution and, if mine could fit into one, I don't know how. The sum of $1$-to-$25$ is $325=5\cdot65$

We can start with groupings that have the same sum of $65$ for each of $5$ sums. These are found by starting with $25$ and adding the next remaining largest number(s) that increase the sum without exceeding $65$.

$$
=25+24+16\qquad
=23+22+20\qquad
=21+19+18+7\\
=17+15+14+13+6\qquad
=12+11+10+9+8+5+4+3+2+1$$

These are not the same cardinality. We need $5$ sums of $65$ with $5$ summands each. The $4^{th}$ sum meets this criteria. Now let's swap $16$ from the $1^{st}$ sum with $10,5,1$ from the $5^{th}$ sum.

$$
=25+24+10+5+1\qquad
=23+22+20\qquad
=21+19+18+7\\
=17+15+14+13+6\qquad
=16+12+11+9+8+4+3+2$$

Now, let's swap $7$ form the $3^{rd}$ sum with $4,3$ from the $5^{th}$ sum.
$$
=25+24+10+5+1\qquad
=23+22+20\qquad
=21+19+18+4+3\\
=17+15+14+13+6\qquad
=16+12+11+9+8+7+2$$
Now, if we swap $20$ from the $2^{nd}$ sum with $11,7,2$ from the $5^{th}$ we get.
$$=25+24+10+5+1\qquad
=23+22+11+7+2\qquad
=21+19+18+4+3\\
=17+15+14+13+6\qquad
=20+16+12+9+8$$

And we are done. All sons have $5$ cows and each gets $65$ gallons of milk a day. There are more solutions (not sure how many) if we simply swap addends between the sums. For instance, let's swap $24,10$ with $23,11$ in the $1^{st}\;2\;$ sums.
$$=25+23+11+5+1\qquad
=24+22+10+7+2\qquad
=21+19+18+4+3\\
=17+15+14+13+6\qquad
=20+16+12+9+8$$

If we now swap $10,7$ with $9,8$ or $15,14$ with $20,9$  (not shown) we would have at least $3$ solutions so the answer to your question is "Yes, there are more than $2$."

Note: The sons do not always get the same proportion of cows from the $1^{st}$ and $2^{nd}$ halves of the herd as suggested in the earlier answer.


答:
Hint

you may start from here
$$
\sum\limits_{k = n}^{2n} {{k \over {\left( {k + 1} \right)^{\,2} }}}
  = \sum\limits_{j = 0}^n {{{n + j} \over {\left( {n + 1 + j} \right)^{\,2} }}}
  = {1 \over n}\sum\limits_{j = 0}^n {{{1 + j/n} \over {\left( {1 + {{1 + j} \over n}} \right)^{\,2} }}}
$$


问:
Values of $a_1$ and $a$ for which $a_{n+1}=a_n^2+(1+2a)a_n+a^2$ is convergent
I need to find the values of $a_1$ and $a \in \Re$ that make convergent  the sequence $(a_n)$ defined as:

$a_1 \in \Re $

$a_{n+1} = a_n^2+(1-2a)a_n+a^2$

I can state easily that the sequence is increasing

$a_{n+1} = a_n + (a_n-a)^2 \ge a_n$ obviously, because $(a_n-a)^2 \ge 0$

Now I know that the sequence is increasing, if it happens to be convergent, the limit has to be $a$ because

$$\lim_{n\to \infty} a_n = \lim_{n\to \infty} a_{n+1} = \lim_{n\to \infty} a_n + (a_n-a)^2$$

$$0 = \lim_{n\to \infty} (a_n-a)^2$$

$$\lim_{n\to \infty} a_n = a$$

Now, the sequence

is increasing

if it converges, it converges at $a$

As a consequence, if $a_1 > a$ then the sequence diverges.

Now I'm wondering how to prove that the sequence is

convergent for $a-1\le a_1 \le a$

divergent for $a_1 \lt a-1$

For the second request, i know it could be something related to $a_2 \gt a$, but I don't know how to proceed from $a_1 \lt a-1$ to $a_2 \gt a$.

Instead, for the first one, I thought it could be convenient to prove by induction that

$$a-1\le a_n \le a \quad \forall n \quad if \quad a-1\le a_1 \le a$$

I can easily get to state that $a_{n+1} \gt a-1$, but I got a few problem with the other hand side.

If you guys could give me some tip or hint, it would be much appreciated.

答:
It makes the proof neater to first simplify the algebraic forms.

Let $x_n=a-a_n$, then $x_{n+1}=x_n(1-x_n)$. If $x_n$ converges, to $ X$ say, then $X=X(1-X)$ and so $X=0$.

If $x_n<0$

Then $x_{n+1}=x_n(1-x_n)<x_n$ and so $x_n$ does not converge if $x_1<0$.

If $x_n>1$

Then $x_{n+1}=x_n(1-x_n)<0$ and we are in the above case. So $x_n$ does not converge if $x_1>1$.

If $1\ge x_n\ge 0$

Then $x_{n+1}=x_n(1-x_n)$ and so $x_n\ge x_{n+1}\ge 0$. Thus we have a non-increasing sequence of terms bounded below. Hence the sequence converges.


问:
Classify all groups with normal subgroups of order 3 and 4
Suppose a group $G$ has normal subgroups of size 3 and 4. I want to classify all such groups.

The subgroups are clearly $\mathbf Z_3$ and $\mathbf Z_4$ or $K_4=\mathbf Z_2\times\mathbf Z_2$. If $H_4$ denotes one of $\mathbf Z_4$ and $K_4$, then any group of the form
$$\mathbf Z_3\times H_4\times L$$
where $L$ is any group, satisfies the required condition.

I suspect that these are the only groups which satisfy this this condition, but I don't know how to prove it. I am familiar with the following result for obtaining equations of direct products like what I'm after:

If $H,K\leqslant G$, $H\cap K = \{1\}$ and the elements of $H$ and $K$ commute, then $HK\simeq H\times K$.

But I don't believe I can apply it to my situation since I don't know if elements commute.

I appreciate any help with this.

答:
I think this question is too broad/vague. It is equivalent to asking for all (finite) groups that have an abelian normal subgroup of order $12$.

Indeed, if $N_1$ is normal of order $3$ and $N_2$ is normal of order $4$, then $N_1N_2$ is normal of order $12$, and has a normal Sylow $3$-subgroup and a normal Sylow $2$-subgroup. But the nonabelian groups of order $12$ are $A_4$ (which does not have a normal Sylow $3$-subgroup), the dihedral group of order $12$, $D_6$ (which does not have a normal Sylow $2$-subgroup), and the dicyclic group $\langle a,s\mid a^6=e, s^2=a^3, sas^{-1}=a^{-1}\rangle$ (which does not have a normal Sylow $2$-subgroup). Thus, $N_1N_2$ is abelian so $G$ has a normal abelian subgroup of order $12$. Conversely, if $G$ has a normal abelian subgroup $N$ of order $12$, then the Sylow subgroups of $N$ are characteristic in $N$, hence normal in $G$, so $G$ has a normal subgroup of order $4$ and one of order $3$.

However, classifying all groups with a normal abelian subgroup of order $12$ (or any given order, for that matter) seems rather broad. The group need not split into $A\times K$ with $A$ the normal abelian group of order $12$; it need not even split into $A\rtimes K$ with $A$ the normal abelian group of order $12$:

$C_{24}$ does not split into $A\times C_2$ with $A$ abelian of order $12$.

$Q_8\times C_3$ has a normal abelian subgroup of order $12$, namely $\{1,-1,i,-i\}\times C_3$, which is cyclic of order $12$ (being the direct product of two cyclic groups of relatively prime order), but it is not a semidirect product $C_{12}\rtimes C_2$, because $Q_8\times C_3$ has three subgroups of order $12$ (depending which subgroup of order $4$ of $Q_8$ you take), but $C_{12}\rtimes C_2$ only has two: if $N\triangleleft C_{12}\rtimes C_2 = \langle x\rangle\rtimes C_2$ has order $12$, the intersection with $\langle x\rangle$ has index at most $2$, hence is either $\langle x\rangle$, so $N=\langle x\rangle$, or else $N\cap\langle x\rangle = \langle x^2\rangle$ and then $N=\langle x^2\rangle\rtimes C_2$.


答:
We have:
$$
\sum_{n\le k\le 2n}\frac k{k^2}
\le
\sum_{n\le k\le 2n}\frac k{(k+1)^2}
\le
\sum_{n\le k\le 2n}\frac {k+1}{(k+1)^2}
\ .
$$
So it is enough to show the existence of the limit of $(b_n)$ given by
$$
\begin{aligned}
b_n
&=
\sum_{n< k\le 2n}\frac k{k^2}
=
\sum_{n< k\le 2n}\frac 1k
=
\sum_{1\le k\le 2n}\frac 1k
-
\sum_{1\le k\le n}\frac 1k
\\
&=
\sum_{1\le k\le 2n}\frac 1k
-
2\sum_{1\le k\le n}\frac 1{2k}
\\
&=
\frac 11-\frac 12+\frac 13-\frac 14+\dots+\frac1{2n-1}-\frac 1{2n}\ .
\end{aligned}
$$
Now use the Leibniz (Alternating series test, or group the terms in pairs
and show that $\displaystyle \frac 1{1\cdot 2}+ \frac 1{3\cdot 4}+\dots
+ \frac 1{(2n-1)\cdot (2n)}$ is increasing and bounded.
Now we may consider the harmonic series, and need for it the logarithmic estimation.

(Alternatively use the logarithmic Euler estimate for the harmonic sequence $(b_n)$ to also have the limit.)


问:
How to write a mathematical expression defining a set with double "such that"?
I have a sentence I am trying to translate as a mathematical expression but I am not sure if the expression matches what I mean.

The mathematical expression I came up with:

What I actually mean:

F is a function that takes an element from set scriptP and set scriptB and returns either 0 or 1.

B_acc is a set depending on set P_s. It is the set of all the elements B belonging to scriptB such that there is at least one element P in the set P_s such that F(P,B) = 1.

P_s* = is the set P_s for which the cardinal of set B_acc is minimum.

I am not sure about how to express the double "such that" in the first equation.

答:
I like the words version $\{B\in\mathcal B\colon \text{there exists } P\in\mathcal P_s \text{ such that } F(P,B)=1\}$ the best.

While I think that's already fine, if you wanted you could write $\mathcal B_{acc}(\mathcal P_s) = \pi_2(F^{-1}(1))$, where $\pi_2$ (or $\pi_{\mathcal B}$) is the projection map from $\mathcal P\times \mathcal B$ to the second coordinate $\mathcal B$. (By the way, it's better to write the domain of $F$ as $\mathcal P\times \mathcal B$ rather than $(\mathcal P,\mathcal B)$.) There's a nuance between $\mathcal P$ and $\mathcal P_s$ that I'm probably missing here, so this might need some tweaking.

Of course, any function to the set $\{0,1\}$ can be naturally identified with a subset of the domain (namely the set of elements sent to $1$). If we write $F$ as a subset of $\mathcal P\times \mathcal B$ rather than a function on $\mathcal P\times \mathcal B$, then it becomes even simpler: $\mathcal B_{acc}(\mathcal P_s) = \pi_2\big( F \cap (\mathcal P_s\times\mathcal B) \big)$ (assuming $\mathcal P_s$ is a subset of $\mathcal P$).


问:
Question about Image and Null space of a linear operator and invariant subspace.
Let $T:V \to V$ a linear operator of finite dimensional vector space $V$. Show that $ImT$ has a complementary $T-$invariant subspace $\iff$ $Im T \cap \ker T = \{0\}$.

This question has asked before here:

Cyclic Decompositions and the Rational form

I got the side $(\Leftarrow)$, but I can't understand the otherside. Let me explain my question. If $W$ is a $T-$invariant subspace of $ImT$, of course that given $w \in W$ then $T(w) \in \ker T$, then $W \subset \ker T$. Ok, why it implies that $\ker T \cap Im T = \{0\}??$

答:
Assume that $V= W \dotplus \operatorname{Im} T$ where $W$ is $T$-invariant. We claim that $W \subseteq \ker T$.

If $w \in W$ then $Tw \in W$ as well, but also $Tw \in \operatorname{Im} T$ and hence $w \in W\cap \operatorname{Im} T=\{0\}$. Therefore $Tw = 0$ so $w \in \ker T$.

Now we have
$$V= W \dotplus \operatorname{Im} T \le \ker T + \operatorname{Im} T$$
so $\ker T + \operatorname{Im} T = V$. It remains to show that the sum is direct, but this follows directly from the rank-nullity theorem:

$$\dim(\ker T \cap \operatorname{Im} T) = \dim\ker T + \dim \operatorname{Im} T - \dim(\ker T + \operatorname{Im} T) = \dim V-\dim V = 0.$$


问:
Proving that if $f'$ has at most $n-1$ zeros, then $f$ has at most $n$ zeros
Is this proof correct?

The problem is the following.

Let $n$ be a natural number. Suppose that the function $f:\mathbb{R}\to\mathbb{R}$ is differentiable and that the following equation has at most $n-1$ solutions:
  $$f'(x)=0, \quad x \in \mathbb{R}.$$
  Prove that the following equation has at most $n$ solutions:
  $$f(x)=0,\quad x \in \mathbb{R}.$$

My proof is:

Let $f'(x)=0$ have solutions $x_1$, $x_2,\ldots ,x_{n-1}$.

since
$$f'(x_1)=\lim\limits_{{x}\to{x_1}}\frac{f(x)-f(x_1)}{x-x_1} = 0$$
and doing so, $f(x)=f(x_{n-1})$
Therefore $f(x)=f(x_1)=f(x_2)=\cdots=f(x_{n-1})$.

Let $x_n$ be one of solutions of $f(x)=0$.
$$0=f(x_n)=f(x_1)=f(x_2)=\cdots=f(x_{n-1}),$$
so $f(x)=0$ has solutions like $x_1$, $x_2,\ldots, x_n$.

If there is wrong part, please let me know.

答:
I'm not that clear on what you're doing, but your solution looks wrong. Just the conclusion that $f$ is zero at the the zeros of $f'$ (you have $n$ of them, by the way) is clearly false in general. You seem to be trying to do some variable point argument with $x$, and using the fact that the limit is zero to conclude that individual terms are zero. </p>

<p>A valid solution is to note that there is a zero of $f'$ between any two zeros of $f$ by the mean value theorem. </p>

Body:
I'm not that clear on what you're doing, but your solution looks wrong. Just the conclusion that $f$ is zero at the the zeros of $f'$ (you have $n$ of them, by the way) is clearly false in general. You seem to be trying to do some variable point argument with $x$, and using the fact that the limit is zero to conclude that individual terms are zero.

A valid solution is to note that there is a zero of $f'$ between any two zeros of $f$ by the mean value theorem.


问:
Question on AP(sequences and series)
Prove that sqrt(2), sqrt(3) and sqrt(5) cannot be terms of an A.P.(not necessarily consecutive)!</p>

答:
If there are terms of an A.P. then $\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{5}-\sqrt{3}}$ is rational (why?). Therefore $$\sqrt{3}(m+n)=\sqrt{5}n+\sqrt{2}m$$
for some $n,m\in\mathbb N.$
Contradiction (Why?).</p>

Title: Question on AP(sequences and series)


问:
can someone explain the difference in uniform convergence and pointwise convergence
In wiki I found this explanation:

for uniform convergence:
$\forall \varepsilon >0\ \underline {\exists N\in \mathbb{N} \ \forall x\in D_{f}}\ \forall n\geq N:\quad \left|f_{n}(x)-f(x)\right|<\varepsilon ,$

for pointwise convergence: $\forall \varepsilon >0\ \underline {\forall x\in D_{f}\ \exists N\in \mathbb{N} }\ \forall n\geq N:\quad \left|f_{n}(x)-f(x)\right|<\varepsilon ,$

I'm trying to understand what the difference in the order of the two underlined parts exactly mean. And why that implies that pointwise convergence follows from uniform convergence.

答:
In the definition of pointwise convergence, the $N$ is chosen after $x$ is chosen, so it may depend on $x$ (there is no guarantee that the same $N$ works for every $x$). The definition of uniform convergence is stronger, because it states that there is a single $N$ that works for all $x$.

Edit. Uniform convergence implies pointwise convergence because, of course, even knowing $x$ we can still choose the same $N$ that the definition of uniform convergence gave us. (We don't need to have it depend on $x$.)


Body:
if we have $f(x) > 0$ but $f'(x) < 0,$  then
$$ \int_a^{b+1} \; f(x) \;  dx \;  < \; \sum_{j=a}^b \;  f(j) \; < \; \int_{a-1}^b \;  f(x) \; dx   $$

so we take $a=n, b =2n$ and $$f(x) = \frac{x}{(1+x)^2}=\frac{1}{1+x}  -\frac{1}{(1+x)^2}$$

$$ \int_n^{2n+1} \; \frac{1}{1+x}  -\frac{1}{(1+x)^2} \;  dx \;  < \; \sum_{j=n}^{2n} \; \frac{j}{(1+j)^2} \; < \; \int_{n-1}^{2n} \; \frac{1}{1+x}  -\frac{1}{(1+x)^2}  \; dx   $$

An antiderivative is $$ \log (1+x) + \frac{1}{1+x}  $$

$$  \log 2 -\frac{1}{2(1+n)}   < \; \sum_{j=n}^{2n} \; \frac{j}{(1+j)^2} \; < \; \log \left( 2 + \frac{1}{n}\right) - \frac{n+1}{2n^2+n} $$


问:
$\phi =\left \{ \cos\frac{2 n \pi t}{T},\sin\frac{2 n \pi t}{T}: \forall n,m \in \mathbb{Z}>0 \right \}$ in L$[0,T]$
MY problem : Show that the real signals of continuous time, defined in $L [0, T]$, with $T ∈ R> 0$, which form the set

$$\phi =\left \{ \cos\frac{2 n \pi t}{T},\sin\frac{2 n \pi t}{T}: \forall n,m \in \mathbb{Z}>0  \right \}$$

they are orthogonal, precisely in the time interval $t ∈ [0, T]$, under the internal product between real signals defined by

$$\left \langle f(t),g(t)) \right \rangle=\int_{0}^{T} f(t)g(t)dt$$

Well, when I resolve the integral don't get zero :c

I do not understand that

答:
Hint:

To calculate the integral $\:\displaystyle\int_0^T\cos\frac{2m\pi t}{T}\sin\frac{2n\pi t}{T}\,\mathrm dt$, use the linearisation formula:

$$ \sin a\cos b=\tfrac12\bigl(\sin(a+b)+\sin(a-b)\bigr). $$


问:
polynomial polynomialfunction unique
polynomialfunction is unambigiously determined by polynomial but polynomial is not unambigiously determined by polynomialfunction.

for example $f:\mathbb{Z}_2\rightarrow\mathbb{Z}_2,f(x)=x^2-x$ is the zerofunction. I.e. we also could have chosen the zeropolynomial. It was used as an example to show the second statement.

I find it difficult to understand the first statement: it says given a polynomial then the function is unambigiously determined.

Why can one polynomial not define two functions?

答:
Consider a commutative ring $R$ (such as $\Bbb Z, \Bbb Q, \Bbb R, \Bbb Z_2$ etc.) and recall that the polynomial ring $R[X]$ is given by the ring of formal polynomials, by which I mean formal linear combinations of powers of an indeterminate $X$ with coefficients in $R$. In other words an element of $R[X]$ hence is a term $a_nX^n + ... + a_1X^1 + a_0X^0$ where all $a_k\in R$. About $X$ we only know that it is not an element of $R$, but it satisfies the distributive law with respect to elements of $R$ and moreover no two distinct powers of $X$ agree.

Now consider a given polynomial $p = a_nX^n + ... + a_1X + a_0 \in R[X]$. Intuitively, given an element $r\in R$ we can substitute it for $X$ and evaluate the expression $a_nr^n + ... a_1r + a_0 \in R$. Let’s denote the resulting element of $R$ by $p(r)$. But as we can do this for any choice of $r\in R$ we have indeed defined a function $\tilde p: R \rightarrow R, r \mapsto p(r)$. As functions are determined by their values, this is the unique function obtained by „evaluating“ the polynomial $p$ at elements of $R$.

It turns out that functions $f:R \rightarrow R$ obtained in this way are quite nice, so let us define that $f$ is a polynomial function, if there exists a polynomial $p \in R[X]$ st. $f=\tilde{p}$. Now two questions should come to mind:

Is every function $R \rightarrow R$ a polynomial function? In general not, since for $R=\Bbb R$ every nonconstant polynomial function is unbounded (tends to $\pm \infty$ as $x$ goes to $\pm infty$), but there sure are nonconstant bounded functions (eg. $\sin x$)

Given a polynomial function $f$, is the polynomial $p\in R[X]$ satisfying $f = \tilde{p}$ unique? Again it won’t be the case in general, as you noted both $0 \in \Bbb Z_2[X]$ and $X^2-X\in \Bbb Z_2[X]$ evaluate at every element $r\in \Bbb Z_2$ to 0 and hence define the constant function at 0. So formally (as linear combinations of powers of $X$) the polynomials are distinct, but their induced polynomial functions behave the same.

Title: polynomial polynomialfunction unique