问:
How to check if a transformation $F:\mathbb{R}^2 \to \mathbb{R}^2$ is a contraction mapping?
Let
$$(\forall x_1,x_2 \in \mathbb R) \qquad F \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}
= \begin{bmatrix}
   \frac{1}{4}x_1 + \frac{1}{2}x_2 - 1 \\
   \frac{1}{2}x_1 + \frac{1}{4}x_2 + 2 \end{bmatrix}.$$

How do I check/prove that $F:\mathbb{R}^2 \to \mathbb{R}^2$ is a contraction mapping? How to find the constant point?

I know how to proceed in similar cases when dealing with $\mathbb{R} \to \mathbb{R}$. I check supremum of first derivative. But I don't know what to do here.

If anybody could show me on the above example, that would be great.

答:
You have to compute derivative of $F$, lets denote it by $dF$ and show that its norm, considered as a linear operator from $R^{2} to R^{2}$ is less than $1$.
This is a consequence of, sort of, mean value theorem which states that, locally i.e. for small $||h||$
$$F(x+h)-F(x) = (dF)h + o(||h||)$$
$dF$ is a matrix build out of first derivatives of $F$.
In your case it has this form $$\begin{bmatrix}
   \frac{1}{4},\frac{1}{2} \\
   \frac{1}{2}, \frac{1}{4}
   \end{bmatrix}$$
So, its norm, as a max of its values on vectors $x=(x_{1},x_{2})$  with $x_{1}^{2} + x_{2}^{2} <=1 $ should be less than 1.
In your case, it is easy to see that the norm is $\frac{3}{4}$.


问:
Is the following set empty?
$$
sp\left \{
\begin{pmatrix}
1 \\
-1 \\
1 \\
-1
\end{pmatrix} , \begin{pmatrix}
4\\
-2 \\
4 \\
-2
\end{pmatrix} , \begin{pmatrix}
1\\
1\\
1\\
1
\end{pmatrix}
 \right \} \bigcap \left \{ \begin{pmatrix}
x_1\\
x_2\\
x_3\\
x_4
\end{pmatrix} | \begin{matrix}
x_1 + x_2 = 0\\
x_3 = 2x_4
\end{matrix} \right \}$$

I was asked to find the basis of the following set but the vectors of the LHS set never fit to both requirements on the RHS set , or maybe I'm totally wrong and the binary relation between the requirements is OR?

答:
A group does have a neutral element, hence it can't be empty. But you don't talk about groups at all. You talk about sets (or vector spaces), and this one does have the zero vector in it.

A basis for the Nullvectorspace is the empty set.
Because the empty sum is defined to be the neutral element. And the neutral element of addition in a vectorspace is the null vector.

ContentLicense: CC BY-SA 3.0

Title: Is the following set empty?


问:
Understanding the solution of #14.5.1 in Artin
Here is the question:

Determine a presentation matrix $A$ as an $R-$module for the ideal $ I = (2, 1 + \sqrt{-5})$ where $R = \mathbb Z[\sqrt{-5}].$

Here is a solution of the question:

Exercise 14.5.1. Let $R=\mathbb{Z}[\delta]$ where $\delta=\sqrt{-5}$. Determine a presentation matrix as an $R$-module for the ideal $(2,1+\delta).$

Proof. Let $\varphi:R^2\longrightarrow I$ that sends $(x,y)\rightsquigarrow2x+(1+\delta)y$. Now
$$\ker\varphi=\left\{(x,y)\in R^2\,\middle|\,x=-\frac{1+\delta}{2}y\right\}\leftrightarrow\left\{y\in R^2\,\middle|\,\frac{1+\delta}{2}y\in R\right\}$$
where $\leftrightarrow$ denotes bijection. But $y=2,1-\delta$ are the elements in $R$ with smallest norm satisfying this condition, and moreover the corresponding vectors $(-3,1-\delta),(-1-\delta,2)\in R$ are independent over $R$ since $2r\ne1-\delta$ for any $r\in R$. Thus, $\ker\varphi$ is generated by these two vectors, and by pp. 424-425 we have a presentation matrix
$$A=\left(\begin{array}{cc}
-3\phantom{-} & -1-\delta\\
1-\delta & \phantom{-}2
\end{array}\right).\qquad\square$$

My questions are:

Here is what I know:

If $I = (2, 1 + \delta )$ where $\delta = \sqrt{-5}.$ Define $\varphi: R^2 \rightarrow I$ by $(r,s) \mapsto 2r + (1+ \delta)s.$ Then $\operatorname{ker}\varphi =
\left(\begin{matrix}-\frac 1 2(1+\sqrt5i)s\\s\end{matrix}\right)=\left(\begin{matrix}-\frac 1 2(1+\sqrt5i)\\1\end{matrix}\right)s.$ And hence a generator for $\operatorname{ker}\varphi$  is $\left(\begin{matrix}-\frac 1 2(1+\sqrt5i)\\1\end{matrix}\right).$ But above in a solution of excercise 14.5.1 in Artin book "Algebra"(second edition) the author did not use this generator when the author needed to use a set of generators of the module of relations to be expressed as a linear combination of the set of generators of $I$ to find a presentation matrix as an $R$− module for the ideal $I$, he did not use that generator.
1- Does  $\mathbb Z[\sqrt{-5}]$ make any difference in the set of generators I should find?

2- Also, why we need the $s$ satisfying the condition of  $\operatorname{ker}\varphi$  to be of minimum norm, could someone clarify this to me please?

3-Also, is the solution in the picture above correct?

Note that this question here Determining the presentation matrix for a module helped me a lot in understanding the idea of the presentation matrix.

Note there is no need to answer all questions below as they were already  answered in the comments

1- why in the second braces $y \in R^2$?

2- How do the author get that that $y =2, 1 - \delta $? And how did we know that they have the smallest norm? and what is the importance of having the smallest the norm in the solution?

3- How did the author get the corresponding vectors $(-3, 1- \delta)$ and $(-1-\delta , 2 )$? how these 2 dimensional vectors belong to $R$? Also, I did not uderstand the argument the author used for showing that they are linearly independent, and why we have to show that they are linearly independent?

4- It seems for me until this point the author is trying to find the coefficients of the set of relations because the columns of the presentation matrix are built from them, is that correct?

5- Finally, is that solution correct?

Could anyone clarify these points to me please?

Here is also another solution:

The surjective map is considered as shown below;
\begin{align*}
R^2   & \longrightarrow\varphi(2,1+\delta),\\
(x,y) & \longrightarrow2x+(1+\delta)y.
\end{align*}
As $\ker\varphi$ has two generators. Then, there is the relation that $2(-1-\delta)+(1+\delta)2=0$ and also the relation $2(-3)+(1+\delta)(1-\delta)=0$.

Therefore, two relations cannot be derived from each other. Then $\ker\varphi$ is find by finding $(x,y)\in\mathbb{R}$ such that $2x+(1 +\delta)y=0$.

It means that
$$\ker\varphi=\left\{y:\ y\in\mathbb{R}\ \text{and}\ \left(\frac{1+\delta}{2}\right)y\ \text{is also in}\ \mathbb{R}\right\}.$$

Therefore, the presentation matrix is
$$\left(\begin{array}{cc}
-3\phantom{-} & -1-\delta\\
1-\delta & \phantom{-}2
\end{array}\right).$$

答:
You can proceed directly as following: let $\phi:R^2 \rightarrow I$ be the surjective map $(x,y)\mapsto 2x+(1+\delta)y$. Compute the kernel

$$ v=(x,y) \in \ker(\phi) \iff x=-\frac{1+\delta}{2}y \ , \ x,y \in R \iff v = (-\frac{1+\delta}{2}y,y) \ , \ y \in R $$

By checking directly $\frac{1+\delta}{2}y \in R$ if and only if you can write $y=(a+2b)+a\delta$ where $a,b\in \mathbb{Z}$. So you get that the $v \in \ker(\phi)$ can be written as
$$v=(-\frac{1+\delta}{2}(a+2b+a\delta),a+2b+a\delta) = a\cdot(2-\delta,1+\delta) + b\cdot(-1-\delta,2)$$
for $a,b\in \mathbb{Z}$ and conversely. We conclude that a presentation matrix is
$$ \begin{pmatrix}2-\delta & -1-\delta \\
1+\delta & 2\end{pmatrix}$$
Your equivalent presentation matrix is obtained by multiplying the first column by unit $-1$ and then adding the second column to the first one.

ContentLicense: CC BY-SA 4.0


问:
Binomial distribution closed by linear operations?
I understand that normal distributions are closed by linear operations, but is binomial distribution as well?

In other words, if $X$ is a binomial random variable, does $aX + b$ follow a binomial distribution for any real number $a$ and $b$?

答:
No.  Although if we zoom out on a binomial distribution it looks like a discretized normal distribution, and we even have the normal approximation to the binomial distribution to support this rigorously, the two distributions are not the same especially near the tails.

Translation is especially not obeyed by a binomial distribution: if $X\sim B(n,p)$ is a binomial random variable, $\min(X)=0$.  However, $\min(X+b)=b$, so $X+b$ cannot follow a binomial distribution unless $b=0$.

For scaling, remember the binomial distribution is discrete, so its support (the possible values a binomial random variable can have) is a finite set, in particular the integers from 0 to $n$ inclusive.  However, if we scale a binomial random variable $X$ by $a$, the gaps between the possible values stretch out by a factor of $a$.  If $a$ is not 0 or 1, the support of $aX$ will not be a set of integers from 0 to $an$ inclusive and so $aX$ cannot have binomial distribution.  This problem does not happen for continuous distributions.  (Although not all continuous distributions form a family closed under scaling, and similarly one could create a discrete distribution that was closed under scaling but it would have to either only be closed under scaling up by an integer or allow a discrete but non integer support, both of which are weird.)


答:
Is Y=aX+b binomial for all real a,b? No, if a=2 then the support of Y is 0,2,4,...,2n which is not binomial.

Do the values of Y follow "binomial probabilities"? Yes. If $X\sim Bin(n,p), Y=aX+b$, then the support of Y is $aX+b, X=0,1,...,n$ and the pmf is $f(Y)={n\choose \frac{y-b}{a}}p^{\frac{y-b}{a}}(1-p)^{1-\frac{y-b}{a}}$.

Example

$X\sim Bin(3,.4)$, $Y=2X+1$

$X\begin{cases}0&.6^3\\
1&{3\choose 1}.4.6^2\\
2&{3\choose 2}.4^2.6\\
3&.4^3\end{cases}$

$Y\begin{cases}1&.6^3\\
3&{3\choose 1}.4.6^2\\
5&{3\choose 2}.4^2.6\\
7&.4^3\end{cases}$

Title: Binomial distribution closed by linear operations?

Body:
I understand that normal distributions are closed by linear operations, but is binomial distribution as well?

In other words, if $X$ is a binomial random variable, does $aX + b$ follow a binomial distribution for any real number $a$ and $b$?


问:
proof that any boolean function can be written in canonical form
It's bugging me for a while but although I can vaguely see that when writing canonical forms we kind "build them" in a way specificly to make it be true but I can't grasp exactly why it is possible to do so and if there is any proof of it. I've been having a hard time finding about this online so if anyone can help it would be awesome.</p>

答:
A Boolean function $\varphi$ is specified by its truth table. For the conjunctive normal form, we look at the rows where the function evaluates to $0$. Each row yields a distinct clause, and we take a conjunction of the clauses. Note that $\varphi(x_{1}, \ldots, x_{n}) = 0$ if and only if the specific values for $x_{1}, \ldots, x_{n}$ correspond to one of the rows specified by a given clause. Otherwise, $\varphi(x_{1}, \ldots, x_{n}) = 1$ on those inputs.

To build a clause from a given row, we record $x_{i}$ if $x_{i} = 0$ at that row. Otherwise, we record $\overline{x_{i}}$. We then consider the disjunction (OR) of those literals to obtain the clause.

Section 2.3 of Savage gives a good overview of this in more detail. (http://cs.brown.edu/people/jsavage/book/pdfs/ModelsOfComputation_Chapter2.pdf)


问:
Why is the Nth derivate isolated in many textbook definitions?

My textbook keeps writing definitions in the above form, with the LHS of the equation being the highest order derivative, and the RHS being everything else. Why is this form stressed over a form like this

? Is there any mathematical reason, if it is just a convention then why?

Notice that the second equation has a function of t on the RHS and everything else on the LHS.

答:
As pointed out in a comment, the $\phi^{(n)}(t)$ term has to be there, or else it wouldn't be an $n$th order differential equation, but all of the other stuff could be zero.

I think there is one other reason that it is often useful to write things this way.  It is more conducive to numerical solution of the differential equation.  The simplest way to solve a differential equation numerically is the Euler method, where you basically take the initial conditions for $\phi$ and all its lower derivatives and use this to calculate $\phi^{(n)}(t_0)$.  Then you use this to estimate the function and all of its lower derivatives at a nearby point $t_1$, and then you once again have to calculate $\phi^{(n)}(t_1)$ based on the other information.  The only nontrivial step is calculating $\phi^{(n)}(t)$, so it is nice to have the equation in a form where that is easy to calculate.


问:
Let $a,b$ be integers, and let $m,n$ be positive integers. Show that if $a | (-b+2n)$ then $a \nmid (3b +7m)$ or $a \leq 6n+7m$.
Problem Let $a,b$ be integers, and let $m,n$ be positive integers. Show that if
$a | (-b+2n)$ then $a  \nmid (3b +7m)$ or $a \leq 6n+7m$.

Thoughts:

By contradiction, if $a | (3b+7m)$ and $a>6n+7m$.

If $a>(6n+7m)$ then $a \nmid 6n+7m$ since $m,n$ are positive.

I'm not really sure how to proceed. Hints appreciated.

Example:

Suppose $a=b=1$, and $m=n=2$, then $1$ does divide $3b+7m=3(1)+7(2)=17$ but $a=1 < 6(2)+7(2)=26$, ok this works.

The role of $m$ seems a bit unclear to me. For instance I could try to find a multiple of $(2b+7m)$ added to some multiple of $(6n+7m)$ that is divisible by $a$,  that would somehow contradict that $a$ divides $(-b+2n)$, but not sure how.

答:
Logically Equivalent to show the following: If $a|(-b +2n)$ and $a|(3b+7m)$, then $a \le (6n+7m)$.

So we show this next. To this end note the following: if $a|(-b+2n)$ then $a|3(-b+2n)=-3b+6n$. So $a$ divides both $-3b+6n$ and $3b+7m$, so $a$ divides the sum of these two integers which is $6n+7m$. But then if $a$ divides $6n+7m$ then $a$ must be no larger than $6n+7m$.


答:
Hint

$a|r$ and $a|s$ implies

$a|(3r+s) \implies |a| \leq |(3r+s)|.$

Body:
Problem Let $a,b$ be integers, and let $m,n$ be positive integers. Show that if
$a | (-b+2n)$ then $a  \nmid (3b +7m)$ or $a \leq 6n+7m$.

Thoughts:

By contradiction, if $a | (3b+7m)$ and $a>6n+7m$.

If $a>(6n+7m)$ then $a \nmid 6n+7m$ since $m,n$ are positive.

I'm not really sure how to proceed. Hints appreciated.

Example:

Suppose $a=b=1$, and $m=n=2$, then $1$ does divide $3b+7m=3(1)+7(2)=17$ but $a=1 < 6(2)+7(2)=26$, ok this works.

The role of $m$ seems a bit unclear to me. For instance I could try to find a multiple of $(2b+7m)$ added to some multiple of $(6n+7m)$ that is divisible by $a$,  that would somehow contradict that $a$ divides $(-b+2n)$, but not sure how.


问:
Every Vertical Strip Has A Point With Positive Slope
Consider a differentiable function $f$ on $[x_0, x_1]$.

Suppose $f(x_0) < f(x_1)$.

It seems intuitively obvious that every vertical strip between $f(x_0)$ and $f(x_1)$ contains a point with positive slope.

i.e. a point $x$ with $f'(x) > 0$ and $a < f(x) < b$

Let the lower bound of this vertical strip be $a$.

Let the upper bound of this vertical strip be $b$.

This seems like it should be a very common question, I believe it can be proved formally using the intermediate + mean value theorem.

Does anyone have any idea how to prove this?

答:
First, judging from your drawing, you probably mean horizontal strips not vertical strips.

That aside, this problem is an application of IVT (intermediate value theorem), MVT (mean value theorem) and certain other facts leveraging the continuity of $f$ (e.g. preimages of closed sets are closed).

As $f(x_0) < a < f(x_1)$, by IVT there is some $x_0 < s < x_1$ such that $f(s) = a$. So, the set
$$
S := \{s \in [x_0, x_1] : f(s) = a\} = f^{-1}(\{a\}) \cap [x_0, x_1]
$$ is non-empty, bounded and closed. Thus, $s^\ast := \sup(S)$ is achieved inside $S$ i.e. $f(s^\ast) = a$ and $x_0 \leq s^\ast \leq x_1$ Also, $f(x_1) \neq a$. So in fact $s^\ast < x_1$.

Then, as $a = f(s^\ast) < b < f(x_1)$, yet again by IVT, there is some $s^\ast < t < x_1$ such that $f(t) = b$. So, the set
$$
T := \{t \in [s^\ast, x_1] : f(t) = b\} = f^{-1}(\{b\}) \cap [s^\ast, x_1]
$$ is non-empty, bounded and closed. Thus, $t^\ast := \inf(T)$ is achieved inside $T$ i.e. $f(t^\ast) = b$ and $s^\ast \leq t^\ast \leq x_1$. Also, $f(s^\ast) = a \neq b$. So in fact $s^\ast < t^\ast$.

Finally by MVT, there is some $s^\ast < u < t^\ast$ such that $f'(u) = \frac{f(t^\ast) - f(s^\ast)}{t^\ast - s^\ast} = \frac{b - a}{t^\ast - s^\ast} > 0$.

You must also have $a < f(u) < b$. Otherwise for e.g. if $f(u) \leq a$, then you have the situation $f(u) \leq a < f(x_1)$ so IVT would give you an $u \leq s' < x_1$ such that $f(s') = a$. But then, $s^\ast < u \leq s'$, which is impossible because $s^\ast$ is supposed to be the largest $s \in [x_0, x_1]$ such that $f(s) = a$. A similar argument shows $b \leq f(u)$ is also impossible.


问:
Why the following space is a submanifold of $\mathcal{M}_4(\mathbb{R})$?
Let consider the space $\mathcal{O}:=\{X\in \mathcal{M}_4(\mathbb{R}) \ / \ ^tX DX=D \}$ with $D=diag(1,1,1,-1)$.

Maybe we can apply the submersion's criterion.

We take $f : \mathcal{M}_4(\mathbb{R}) \to \mathcal{S}_4(\mathbb{R})\simeq \mathbb{R}^{10}, \ X\mapsto \ ^tX DX-D$ and we want to prove that $\mathrm{d}f_X$ is a surjective map.

Let $H \in \mathcal{M}_4(\mathbb{R})$ :

$f(X+H) = \ ^t(X+H)D(X+H) - D = f(X) + \ ^tXDH + \ ^tHDX + \text{o}(\Vert H \Vert)$.

So $\mathrm{d}f_X(H)= \ ^tXDH + \ ^tHDX$. Then I do not know how to continue.

Note that the dimension of $\mathcal{O}$ is $6$.

Thanks in advance !

答:
Hint As Daniel Kruse points out in the comments, you do not need $f$ to be a submersion; it's sufficient to show that it is a submersion on $\mathcal O$, i.e., that $$df_X : T_X \mathcal M_4(\Bbb R) \to T_{f(X)} S_4(\Bbb R) \cong S_4(\Bbb R)$$ is surjective for all $X \in \mathcal O$. (Indeed, $df_0$ is the zero map and so is not surjective.)

So, we need to show that for any $X \in \mathcal O$ that any symmetric matrix $S$ can be written in the form $$S = {}^t\! XDH + {}^t\! HDX$$ for some $H$. This equation is complicated by the fact that $D$ is not the identity matrix (which is the case for the orthogonal group of the standard inner product). On the other hand, ${}^t D = D$ and $D^2 = I$, which suggests that we can simplify our condition by writing the unknown $H$ as $DA$ (i.e., setting $A = D^{-1} H = DH$).

Then, the surjectivity condition is that we can find an $A$ such that
$$S = {}^t\! XD(DA) + {}^t (DA)DX,$$
and this expression simplifies.

Incidentally, $\mathcal{O}$ is a group under matrix multiplication; in fact it is the Lorentz group $O(1, 3)$ of linear transformations that fix a light cone in Minkowski space.


问:
Accumulation points of accumulation points of accumulation points
Let $A'$ denote the set of accumulation points of $A$.
Find a subset $A$ of $\Bbb R^2$ such that $A, A', A'', A'''$ are all distinct.

I can find a set $A$ such that $A$ and $A'$ are distinct, but not one where $A,A',A'',A'''$ are all distinct.

答:
Hint: Work backwards. Start with some set, then add a sequence which converges to each point, then add a sequence which converges to each of the points in that set, and so on.

Do this carefully and you can make sure that $A$ is such set.

If you are familiar with the concept of ordinals, then you can also consider the ordinal $\omega^4$ embedded into the real numbers.


问:
If $ a,b,c \gt 0 $ and $ \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = 1 $ Prove $ abc \geq 8 $
If $ a,b,c \gt 0 $  and  $ \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = 1 $

Prove $ abc \geq 8 $

I have tried AM-GM and substituting the 1, without results.

答:
Let
$$x = \frac{1}{a+1}, \quad y = \frac{1}{b+1},  \quad z = \frac{1}{c+1},$$
then $x+y+z=1$ and
$$a = \frac{1-x}{x} = \frac{y+z}{x}.$$
The inequality become
$$\frac{(x+y)(y+z)(z+x)}{xyz} \geqslant 8.$$
Which is true because
$$\frac{(x+y)(y+z)(z+x)}{xyz} \geqslant \frac{2\sqrt{xy} \cdot 2\sqrt{yz} \cdot 2\sqrt{zx}}{xyz} = 8.$$
The proof is completed.


问:
Why is second smallest eigenvalue and the corresponding eigenvector used to partition a graph?

In spectral graph partition theory, the eigenvector $v_2$ (also called Fiedler vector) corresponding to the second smallest eigenvalue $\lambda_2$ (also known as Fiedler eigenvalue, which actually also defines the algebraic connectivity of a graph) of the Laplacian matrix $L$ of a graph $G$, in general, is used to partition the graph.

What is the underlying philosophy behind this? Any reference to any related proofs on this?

答:
Some of this answer repeats a few details in the answers before this response and adds to them.

There is a very readable discussion about graph partitions using eigenvalues of the Laplacian matrix in the paper by
Newman 2006, “Finding Community Structure In Networks Using The Eigenvectors Of Matrices” which can be found as a pre-print at this url:
https://arxiv.org/pdf/physics/0605087.pdf

NOTE: the paper emphasizes that for context such as social graphs, the leading eigenvector of the modularity matrix provides a better partitioning than the leading eigenvector of the Laplacian.

The following is mostly verbatim excerpts from the paper:

[With the goal of partitioning an unlabeled unweighted graph into non-overlapping groups using the eigenvalues of the Laplacian (which is positive and symmetric)] one sees that the smallest eigenvalue = 0, and is always the smallest eigenvalue of the Laplacian.  The corresponding eigenvector is v1 = (1, 1, 1, . . .)/ √ n, and the minimum cut size R, is in the term corresponding to the lowest eigenvalue λ1 = 0, hence R=0. Having a minimum cut size of 0 is equivalent to placing all the vertices in group 1 and none of them in group 2, which is a valid division of the network, but it is not a useful one for most context.  That’s the trivial solution and there are various methods to forbid it to solve whatever graph partition problem one is trying to achieve.
(see page 3 and 4 of the paper).

One goal might be to achieve a partition such that the number of nodes are equal between 2 groups.  This solution works best as an approximation when there is a large spectral gap (= difference between largest absolute values of eigenvalues).

Knowing the definition of R in equation 12, and that the goal is to have n1 = n2, R would be minimized by choosing s proportional to the second smallest eigenvector v2 of the Laplacian, also called the Fiedler vector.
This choice places all of the weight in Eq. (12) in the term involving the second-smallest eigenvalue λ2, also known as the algebraic connectivity. The other terms would automatically be zero, since the eigenvectors are orthogonal. Unfortunately, there is an additional constraint on s imposed by the condition, Eq. (3), that its elements take the values ±1, which means in most cases that s cannot be chosen parallel to v2. This makes the optimization problem much more difficult. Often, however, quite good
approximate solutions can be obtained by choosing s to
be as close to parallel with v2 as possible.
The best solution is achieved by assigning vertices to one of the groups in order of the elements in the Fiedler vector, from most positive to most negative, until the groups have the required sizes.
For groups of different sizes there are two distinct ways of doing this, one in which the smaller group corresponds to the most positive elements of the vector and one in which the larger group does.
We can choose between them by calculating the cut size R for both cases and keeping the one that gives the better result.
This then is the spectral partitioning method in its simplest form. It is not guaranteed to minimize R,  but, particularly in cases where λ2 is well separated from the eigenvalues just larger than it, it often does very well.


问:
Matrix approximation by eigenvalues approximation
We know that given a symmetric $n\times n$ matrix $M$ with rank $r$, by SVD, we have $$M = \sum_{i=1}^r\sigma_i u_iu_i^T,$$
with $\sigma_i >0$ for $i = 1,\dots,r$.

We also know that the rank $k<r$ approximation of $M$, $\hat{M}$,  in the sense that $\|M-\hat{M}\|$ is minimized is $$\hat{M} = \sum_{i=1}^k \sigma_iu_iu_i^T.$$

My question is, can we further approximate $\hat{M}$, $\tilde{M}$, by adjusting singular/eigenvlaues of $M$?

For example, replace $\sigma_1$ by $\tilde{\sigma}_1$ such that for some matrix norm $\|\cdot\|_?$ $$\|\hat{M}-\tilde{M}\|_?$$ is minimized?

Please suggest me some references or advises, thanks!

答:
Vigorously speaking, no. Since you mentioned that $M$ is a symmetric matrix, the SVD you're stating simplifies into the EVD format $M = VDV^T$. Note that the 2-tuple $(V,D)$ are uniquely given by the eigenspace $\mathrm{eig}(M) = \{(\lambda_i, v_i): Mv_i = \lambda_iv_i\}$ of the matrix $M$. This means the matrix is exactly represented by the eigenvalues and the eigenvectors. The rank-$r$ approximation of $M$ follows from the heuristic that supposing $\lambda_1\geq\lambda_2\geq\dots\geq\lambda_n$ and if $\lambda_1 \gg \lambda_{r + 1}$ we can ignore $\lambda_{r+1} \dots \lambda_{n}$. So this time the approximation $\hat{M}$ is uniquely identified by the 2-tuples $\{(\lambda_i, v_i)\}_{i = 1}^r$. This gives $\hat{M} = V_r\mathrm{diag}({\lambda_1}, \dots, {\lambda_r})V_r^T$. So you cannot further approximate a rank-$r$ matrix that gives a ``better" approximation to $M$.

In practice, however, your numerical tools rounds the exact value of $\lambda_i$'s to a certain data space, e.g. 16-bit float or whatever. So one way you may try is to round $\lambda_i$'s up to $\bar{\lambda}_i$'s, say, 2 decimal places and reconstruct the rank-$r$ approximated matrix $\bar{M} = V_r \mathrm{diag}(\bar{\lambda}_1, \dots, \bar{\lambda}_r) V_r^T$ to approximate the matrix $\hat{M}$ using fewer space, while using the same rank.


问:
Combining two DFAs into 1 DFA using concatenation
L= { w : w has exactly two a's and at least two b's}

Picture of DFA's

https://i.stack.imgur.com/4mPHU.png

I have made the two DFA's separately, but I am having trouble combining them specifically the transition. I have attached the two DFA's. Can you help and explain it me please?

Thank you

答:
I like to think of Finite State Automata as memory-less algorithms. Effectively, each state corresponds to a Boolean flag. That is, arriving at a state indicates a true/false condition.

In using a programming analogy, it might be helpful to rename states using descriptive variable names. I might recommend the states:

$q_{0}$: The initial state.

$q_{a}$: We read one $a$ and no $b$'s.

$q_{b}$: We read one $b$ and no $a$'a

$q_{aa}$: We read two $a$'s and no $b$'s

$q_{ab}$: We read one $a$ and one $b$.

$q_{abb}$: We read one $a$ and at least two $b$'s

$q_{aab}$: We read two $a$'s and one $b$.

$q_{aabb}$: We read two $a$'s and at least two $b$'s

$q_{\text{reject}}$: We have determined the input string is not in the language.

Hopefully the names of these states suggest natural transitions to consider.


问:
Symmetric Matrices of $I_{2}$

Find $10$ symmetric matrices $ A = \begin{pmatrix}
a &b \\
 c&d
\end{pmatrix}$ such that $A^{2}=I_{2}$

(I'm going to call matrix A the "square root" of $A^{2}$. If this is the incorrect name for it, may someone please tell me what it is actually called?)

My professor posed this question in class and told us there was an infinite amount of square roots. (Assuming I understood him correctly). However I don't see how there would be many of these, as I was under the impression that a matrix only has one inverse, for $A  A^{-1}=I_{n}$. If someone could tell me if I either misunderstood the professor or if I'm thinking about something incorrectly, please correct me.

My other question is other than blatant guess and check, is there a method to think of these symmetric square roots?

Thanks in advance.

答: Consider in the Euclidean plane the orthogonal reflection in any line; its square is always the identity. If you want linear maps, take a Euclidean vector space and choose the line to go through the origin. That still leaves infinitely many possibilities. In fact, you don't need orthogonal reflections, the reflection in any line parallel to any complementary line will have square the identity. In other words there is plenty of freedom, which is why in this setting one never speaks of the square root if the identity. You can call each of them a square root of the identity, though the term (linear) "involution" is more common.

Title: Symmetric Matrices of $I_{2}$


问:
Calculating the Rolling Variance of a set of numbers
I would like to subscribe to a WebSocket stream which will supply me with many numbers per second. From this data, I would like to calculate the variance of say the last 1000 numbers.

How can I do this in a rolling fashion? That is, I would like some computation comparable to this one for the mean of the last 1000 numbers:

$$\rm{mean}_{i+1} = \rm{mean}_{i} + \frac{1}{1000}\left(x_{i+1}-x_{i-999}\right)$$

Thanks in advance for any help.
Ben

答:
This is basically the same idea as Nameless' answer, but I think it's cleaner if you just keep and update $SX$, the sum of the terms, and $SX^2$, the sum of their squares, update those via

$$\begin{align}
SX_{i+1} &= SX_i + x_{i+1}-x_{i-999}\\
SX^2_{i+1} &= SX^2_i + x_{i+1}^2-x_{i-999}^2\\
\end{align}
$$

and then compute

$$\begin{align}
mean_{i+1} &= SX_{i+1}/1000\\
var_{i+1} &= \frac{1}{1000}SX^2_{i+1} - (\frac{1}{1000}SX_{i+1})^2\\
 &= \frac{1}{1000}SX^2_{i+1} - mean_{i+1}^2\\
\end{align}
$$

(This is the way that old calculators would let you accumulate an arbitrary number of data values and then let you compute mean and variance at the end.)

There is a possible issue in that you have to be able to store a number of magnitude roughly $1000*(\text{typical }x)^2$. Hopefully your values aren't too large.

If you're really worried about numerical instability, @awkward's link to Wellford's online algorithm looks to be the state-of-the-art; I'd check that out if your output values start to go wrong after you've been running for a while.


问:
Extending a pre-Hilbert module operator over a pre-$C^*$-algebra to a Hilbert module over the completion
Let $A_0$ be a pre-$C^*$-algebra, i.e. $A_0$ satisfies all the criteria of a $C^*$-algebra except that it need not be complete. Then we can view $A_0$ as an inner product module over itself, i.e. a $\mathbb{C}$-vector space that is a right $A_0$-module together with an $A_0$-valued inner product, $\langle\,,\,\rangle$, defined by
$$\langle a,b\rangle=a^*b$$
for all $a,b\in A_0$. This structure satisfies the following conditions:

$\lambda(ab)=(\lambda a)b=a(\lambda b)$;

$\langle a,\lambda b+\mu c\rangle=\lambda\langle a,b\rangle+\mu\langle a,c\rangle$;

$\langle a,bc\rangle=\langle a,b\rangle c$;

$\langle b,a\rangle=\langle a,b\rangle^*$;

$\langle a,a\rangle\geq 0$, and if $\langle a,a\rangle=0$ then $a=0$,

for all $\lambda,\mu\in\mathbb{C}$ and $a,b,c\in A_0$.

Now suppose $T$ is a $\mathbb{C}$-linear and $A_0$-linear operator on the inner product module $A_0$. Let $A$ be the $C^*$-algebra completion of $A_0$.

Question: Can $T$ be extended to a densely defined (unbounded) $A$-linear operator on the right Hilbert-$A$-module $A$?

Remark: In general, an unbounded operator $T$ on a Hilbert $A$-module needs to be defined on a dense domain $D(T)$, where $D(T)$ is closed under the action of $A$. In this case, the naive choice $D(T)=A_0$ doesn't work because it is not quite an $A$-module. As a result, it doesn't make sense to say that $T$ is an $A$-linear operator.

I was thinking of getting around this by defining $D(T)=A_0\cdot A^+$, where $A^+$ is the unitization of $A$. This choice of $D(T)$ is dense $A$ and is naturally a right $A$-module, and the action of $T$ on $A_0\cdot A^+$ would then be defined by:

$$T(ab)=\lim_n (Ta)b_n=(Ta)b,$$
where $a\in A_0$ and $b_n$ is a sequence in $A_0$ converging to $b\in A$.

However, it isn't a priori clear that if this is well-defined: one needs to show that if $ab=a'b'$ for some $a'\in A_0$ and $b'\in A$, then $T(ab)=T(a'b')$.

Remark: Perhaps there is standard technique to do this? However, it seems that all of the examples of unbounded operators on Hilbert modules have been of the type where $D(T)$ is already closed under the action of $A$ (e.g. in the setting when the Hilbert module is a Hilbert space and $T$ is an unbounded $\mathbb{C}$-linear operator).

答:
One of the reasons why the theory of bounded operators on Hilbert's space is so advanced is perhaps
because every bounded operator has an adjoint.  The same cannot be said about unbounded operators, and
this is partly the reason why their theory is so hard to develop.  That is, except for self-adjoint
unbounded operators, which, to a large extent, may be studied with a similar level of success as compared
to their bounded cousins.

In other words, the existence of the adjoint is what makes studying operators possible.

Regarding operators on Hilbert modules, even bounded ones sometimes fail to possess an adjoint.  So,
virtually all authors consider only the adjointable ones, namely operators which admit an adjoint.  For
example, if $E$ and $F$ are Hilbert modules over the same C*-algebra, the universally adopted notation
$\mathcal L(E,F)$ refers to the set of all bounded adjointable operators, while there appears to be no
consensus on a notation for the rarely studied (if at all) set of all bounded opertors.

Thus, it is perhaps sensible to assume that the operator $T$ in the question presents a minimum of
adjointness, such as the existence of an opertor $T^*$, defined on a dense subset $D$ (which may or may
not be contained in $A_0$), and such that
$$
  \langle T(a),d\rangle  =   \langle a,T^*(d)\rangle , \quad \forall a\in  A_0,\quad \forall d\in  D.
  $$

With this much,  it is possible to give an affirmative answer,  namely  if $a, a'\in  A_0$,  and $b, b'\in  A$ are
such that $ab=a'b'$,  then $T(a)b = T(a')b'$.

To see why,  choose sequences $\{b_n\}_n$ and $\{b'_n\}_n$ in
$A_0$,  respectively converging to $b$ and $b'$.  Then,  for every $d\in  D$,  one has
$$
  \langle T(a)b,d\rangle  =
  \lim_{n\to\infty } \langle T(a)b_n,d\rangle  =
  \lim_{n\to\infty } \langle T(ab_n),d\rangle  = $$$$ =
  \lim_{n\to\infty } \langle ab_n,T^*(d)\rangle  =
  \langle ab,T^*(d)\rangle  =
  \langle a'b',T^*(d)\rangle  = $$$$ =
  \lim_{n\to\infty } \langle a'b'_n,T^*(d)\rangle  =
  \lim_{n\to\infty } \langle T(a'b'_n),d\rangle  = $$$$ =
  \lim_{n\to\infty } \langle T(a')b'_n,d\rangle  =
  \langle T(a')b',d\rangle .
  $$
Since $D$  is dense, it follows that $T(a)b=T(a')b'$.


答:
Inspired by Mike's answer...

Suppose by contradiction that $a|(-b+2n)$ and $a|(3b+7m)$ and $a > 6n+7m$. Then
$a|3(-b+2n)=-3b+6n$ and $a|(3b+7m)$ and so $a|(-3b+6n+3b+7m)=6n+7m$.

But then $a|6n+7m$ and $a>6n+7m$ where $6n+7m$ is positive. If $a\leq0$ this is impossible since $6n+7m$ is positive.

If $a>0$, then for any $b>c$ with $b,c \in \mathbb{Z}$ and $b,c > 0$ it is true that $b \nmid c$, therefore $a \nmid 6n+7m$, a contradiction, so the desired statement holds.

Title: Let $a,b$ be integers, and let $m,n$ be positive integers. Show that if $a | (-b+2n)$ then $a \nmid (3b +7m)$ or $a \leq 6n+7m$.


问:
How to check if a transformation $F:\mathbb{R}^2 \to \mathbb{R}^2$ is a contraction mapping?
Let
$$(\forall x_1,x_2 \in \mathbb R) \qquad F \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}
= \begin{bmatrix}
   \frac{1}{4}x_1 + \frac{1}{2}x_2 - 1 \\
   \frac{1}{2}x_1 + \frac{1}{4}x_2 + 2 \end{bmatrix}.$$

How do I check/prove that $F:\mathbb{R}^2 \to \mathbb{R}^2$ is a contraction mapping? How to find the constant point?

I know how to proceed in similar cases when dealing with $\mathbb{R} \to \mathbb{R}$. I check supremum of first derivative. But I don't know what to do here.

If anybody could show me on the above example, that would be great.

答:
$F$ is a very simple map (its linear+constant) so  you can just do the computation directly. For the first component, a simple application of Cauchy Schwarz gives
\begin{align}|F_1(x)-F_1(y)|= \left|\frac14(x_1-y_1) + \frac12(x_2-y_2)\right|
&\le \frac14|x_1-y_1| + \frac12|x_2-y_2|
\\&=\binom{\frac14}{\frac12}\cdot \binom{|x_1-y_1|}{|x_2-y_2|}
\\&\le \left|\binom{\frac14}{\frac12} \right|\left|\binom{|x_1-y_1|}{|x_2-y_2|}  \right| \\
&= \frac{\sqrt 5}{4 }|x-y|
\end{align}
basically the same computation also gives $|F_2(x)-F_2(y)|\le \frac{\sqrt 5}{4 }|x-y|$. Then
$$ |F(x)-F(y)|=\sqrt{|F_2(x)-F_2(y)|^2+|F_2(x)-F_2(y)|^2} \le \frac{\sqrt{10}}4|x-y|$$
and $\sqrt{10}/4 = \sqrt{10/16}<1$ so its a contraction (using the Euclidean norm).


问:
Proof that "partition problem in proportion 2:1" is NP-complete
I need to show that problem "partition problem 2:1" is NP-complete.

I know that I need to use $A'$ as certificate to proof that problem is NP.

I know that "partition problem":
given $s: A \rightarrow \mathbb N$, find $A' \subset A : \sum_{a \in A'} s(a) = \sum_{a \in A \backslash A'} s(a)$

is NP-hard.

Using this, I want to show that my problem:

$s: A \rightarrow \mathbb N$, find $A' \subset A : \sum_{a \in A'} s(a) = 2\sum_{a \in A \backslash A'} s(a)$

is also NP-hard.

How can I reduce partition problem to "partition problem 2:1"?

答:
Starting with your definition of the normal or 1:1 partition problem:

given $s: A \rightarrow \mathbb N$, find $A' \subset A : \sum_{a \in A'} s(a) = \sum_{a \in A \backslash A'} s(a)$

Clearly you can multiply each of the members of $A$ by two and not affect the hardness or satisfiability of the problem.  Call this new set of natural numbers $B$.  Compute the sum of all members of $B$ and then divide the result by two.  Call this result $z$ and add it as a new member of $B$.

Now, iff there is a satisfying 2:1 partition of $B$, a 1:1 partition of $A$ can be recovered from this solution by removing $z$ from the 2:1 partition of $B$ and then dividing all the remaining members of $B$'s partition by two to produce a 1:1 partition containing $A$'s values.

This scheme works for the 1:1 partition problem over any $A$ and can be done in polynomial-time.  So this scheme is a polynomial-time reduction from the 1:1 partition problem to the 2:1 partition problem, proving the latter's NP-hardness.


问:
Linear algebra to solve equation for a circle
I am stuck with how to solve this equation:
$$ax^2 + ay^2 + bx + cy + d = 0$$ given three points, $(-2,7), (-4,5)$, and $(4,-3)$.

My steps: 1. Set up three equations,
$$53a-2b+7c+d=0 \\
41a-4b+5c+d=0 \\
25a+4b-3c+d=0$$
2. Created augmented matrix and used my calculator to get rref form (this is what we've been studying and manual row reduction is painful). I get an answer in rref form. Are these the equations for $a, b, c, d$? The answer would be
$$a=x+1/29 \\
b=y-2/29 \\
c=x-4/29 \\
d=d.$$

I don't need the actual answer just confirmation that I'm using the right process, or to be told no and pointed in the right process.

答:
If it's a circle then $a$ cannot be zero, and you can divide through by it:

$$x^2 + y^2 + b'x + c'y + d' = 0$$

where the primes are the variables divided by $a$.

Then you have three equations with three unknowns. You can choose $a$ to be whatever you like.

Does this help?


问:
If a function is continuous at a point and monotone, then does there exist a neighborhood around that point for which the function is continuous?
I am trying to use this property in the context of answering Rudin Exercise 6.1. I will repeat the problem statement and my proof below, and I will embolden the part where I use the statement in the title.

Suppose $\alpha$ increases on $[a,b]$, and $\alpha$ is continuous at some $c \in [a,b]$. If $f(c) = 1$ and $f(x) = 0$ for $x \neq c$, then $f \in \mathcal{R}(\alpha)$ and $\int_a^b f(x) d\alpha(x) = 0$.

We verify the existence and value of the limit directly. Observe that
\begin{align*}
    \underline{\int_a^b}f dx &= \sup L(P,f,\alpha)\\
    &= \sup\left\{\sum_{i=1}^n m_i\Delta \alpha_i\right\} =0
\end{align*}
The above follows from the fact that the infimum of the function on any interval will be $0$ since the function is $0$ everywhere except at $c$.

Meanwhile, for the upper Riemann-Stieltjes integral, we claim that
\begin{align*}
    \overline{\int_a^b} f dx &= \inf U(P,f,\alpha) \\
    &=\inf\left\{\sum_{i=1}^n M_i\Delta \alpha_i\right\} =0
\end{align*}
To prove the equality above, let $\varepsilon>0$ be arbitrary. Since $c \in [a,b]$ it is not an isolated point. Since $\alpha$ is continuous and monotonic, we can find some interval $[x_k, x_{k+1}]$ around $c$ for which $\alpha$ is continuous. Choose points $x_k', x_{k+1}'$ in this interval so that $c \in [x_k', x_{k+1}']$ and so that for any positive integer $n$
$$\alpha(x_{k+1}') - \alpha(x_{k}') = \frac{\alpha(x_{k+1}) - \alpha(x_k)}{n}$$
we can do this by the Intermediate Value Theorem in a similar way to Theorem 6.9 in Rudin.
Let us include $x_k', x_{k+1}'$ in our partition. Then we have
$$\sum_{i=1}^n M_i\Delta\alpha_i = \alpha(x_{k+1}') - \alpha(x_{k}')=  \frac{\alpha(x_{k+1}) - \alpha(x_k)}{n} < \varepsilon$$
for large enough $n \in \mathbb{N}$. Thus, for every $\varepsilon>0$ we can find a partition such that $\sum_{i=1}^n M_i\Delta\alpha_i < \varepsilon$. Therefore,
$$\underline{\int_a^b}f dx = \overline{\int_a^b} f dx = 0$$

答:
No, it is possible for a function to be increasing, continuous at a point $x_0$, and for the function to not be continuous on any neighborhood of $x_0$.

For example, here’s a function that is increasing on all of $\mathbb{R}$, and is discontinuous at exactly all the rationals:

Let $(q_n)_{n\in\mathbb{N}}$ be an enumeration of the rationals. Define $f(x)$ by
$$f(x)  = \sum_{q_n\lt x}2^{-n}.$$
This makes sense, since the series $\sum_{n=1}^{\infty} 2^{-n}$ converges, so $f(x)$ takes real values; it is always positive, and $\lim_{x\to-\infty}f(x) = 0$, $\lim_{x\to\infty}f(x) =2$.

It is also increasing: if $y\gt x$, then $\{n\mid q_n\lt x\}\subset \{n\mid q_n\lt y\}$. Thus,
$$f(y) = \sum_{q_n\lt y}2^{-n} = \sum_{q_n\lt x}2^{-n} + \sum_{x\leq q_n\lt y}2^{-n} \gt \sum_{q_n\lt x}2^{-n} = f(x).$$

It is discontinuous at all rationals: if $a$ is rational, then $a=q_N$ for some $N$. Then for all $y\gt a$ we have
$$f(y) = \sum_{q_n\lt y}2^{-n} = f(a) + 2^{-N} + \sum_{a\lt q_n\lt y} 2^{n} \gt f(a)+2^{-N}.$$
Thus,
$$\lim_{x\to a^+}f(x) \geq f(a)+2^{-N}\gt f(a),$$
and in particular, the limit as $x\to a$ of $f(x)$ cannot equal $f(a)$, so $f$ is not continuous at $a$.

Finally, if $a$ is irrational, let $\epsilon\gt 0$. Then there exists $N\gt 0$ such that $\sum_{n\geq N}2^{-n}\lt \epsilon$. Let $\delta\gt 0$ be such that if $i\lt N$, then $|q_i-a|\gt\delta$ (we can do this, since there are only finitely many rationals we need to avoid). I claim that if $0\lt|x-a|\lt\delta$, then $|f(x)-f(a)|\lt\epsilon$.

Indeed, if $0\lt |x-a|\lt\delta$, then for any $n$, if $q_n$ is between $x$ and $a$, then $n\geq N$. Thus, since the difference between $f(x)$ and $f(a)$ is the sum over all $q_n$ that lie between $x$ and $a$, then
$$|f(x)-f(a)|\leq \sum_{n\geq N}2^{-n}\lt\epsilon,$$
proving that $\lim_{x\to a}f(x) = f(a)$. Thus, $f$ is continuous at all irrationals.

So we have a strictly increasing function $f(x)$, bounded and defined on all reals, with the property that $f$ is continuous at $a$ if and only if $a$ is irrational. In particular, there is no non-degenerate interval in which $f$ is continuous.


问:
Is the image of the intersection of preimage distributive and equal to the intersection of original set in the image?
If $x\in\alpha^{-1}(A)\cap\alpha^{-1}(B)$, then we apply $\alpha$ to both sides and get $\alpha(x)\in\alpha(\alpha^{-1}(A)\cap\alpha^{-1}(B))$. Ok now this is the part where I am a bit unsure of. Can we just distribute $\alpha$ over the intersection and get $\alpha(x)\in A\cap B$?

Because it is true that $f(A)\cap f(B)\subseteq f(A\cap B)$. The intersection of two image is just a subset of the image of intersection so they are not equal, so that means $\alpha(\alpha^{-1}(A))\cap\alpha(\alpha^{-1}(B))\subseteq\alpha(\alpha^{-1}(A)\cap\alpha^{-1}(B))$ right? So if I distribute the $\alpha$ over the intersection, wouldn't I be getting a smaller set?

Edit: Ok the above is not correct. It is the case that $f(A\cap B)\subseteq f(A)\cap f(B)$ if the function is not injective and equal if injective.

答:
The comment at your edit is true.

Suppose that $f(x) = x^{2}$, $A = [-1,0]$ and $B = [0,1]$.

Then we have that $f(A\cap B) = \{0\}$ and $f(A)\cap f(B) = [0,1]$.

Consequently, $f(A)\cap f(B)\not\subseteq f(A\cap B)$.

On the other hand, if $f$ is injective, then we deduce that
\begin{align*}
y\in f(A)\cap f(B) & \Rightarrow (y\in f(A))\wedge(y\in f(B))\\\\
& \Rightarrow (\exists a\in A)(y = f(a))\wedge(\exists b\in B)(y = f(b))\\\\
& \Rightarrow y = f(a) = f(b)\\\\
& \Rightarrow a = b\in A\cap B\\\\
& \Rightarrow y\in f(A\cap B)
\end{align*}

and we are done.

Hopefully this helps.


问:
Prove that there exists $\gamma \in (a,b)$ such that $f \circ f(\gamma)=\gamma$
If $a,b,c \in R$ such that $a<b<c$ and $f:R \to R$ is a continuous function satisfying $f(a)=b,f(b)=c,f(c)=a$, then there exists $\gamma \in (a,b)$ such that $(f \circ f)(\gamma)=\gamma$

My try:

We have $f(a)=b, f(b)=c$ $\implies$ $f \circ f(a)=c$.

But $a=f(c)$, so we get $f \circ f \circ f(c)=c$

Similarly $f \circ f \circ f(b)=b$ and $f \circ f \circ f(a)=a$.

Now since $a<b<c$ we have $f \circ f \circ f(a)<f \circ f \circ f(b)<f \circ f \circ f(c)$

So this means $f \circ f \circ f(x)$ is strictly increasing in $(a,b)\cup[b,c)$ and so it is One-One.

Now since $f \circ f \circ f$ is One-One, that means $f \circ f$ is also One one which means $f$ is One-One.

But how to show that there exists $\gamma \in (a,b)$ such that $f \circ f(\gamma)=\gamma$

答:
I am afraid that your idea does not work.
$$
 f(f(f(a))) < f(f(f(b))) < f(f(f(c)))
$$
for $a < b < c$ does not imply that $f \circ f \circ f$ is strictly increasing on $[a, b]$. Also the conclusion that $f$ is one-to-one cannot be correct because $f(a) < f(b)$ and $f(b) > f(c)$.

Instead use that
$$
 f(f(a)) = c > a \\
 f(f(b)) = a < b
$$
implies that $f(f(x)) - x$ has a zero in $(a, b)$, due to the intermediate value theorem.


问:
Is the category of PL manifolds equivalent to the category of topological manifolds in dimension 2,3?
Balsam and Kirillov write on page 10 in their paper Turaev Viro invariants as an extended TQFT:

"Note that in dimensions 2 and 3, the category of PL (piecewise linear) manifolds is equivalent to the category of topological manifolds."

This seems like a rather non-trivial result. Why does it hold? Sketching a proof or giving a reference would be very much appreciated.

答:
Here is how this equivalence statement should be understood:

Every topological manifold of dimension $\le 3$ admits a PL structure (equivalently, admits a triangulation: In this range of dimensions there is no difference).

Such a PL structure is unique in the following sense:

If $M, N$ are PL manifolds of dimensions $\le 3$ then every homeomorphism $f: M\to N$ is isotopic to a PL homeomorphism. In terms of triangulations: The triangulations can be subdivided so that there exists an isomorphism of the subdivided triangulations isotopic to $f$.

The same holds for manifolds with boundary. Furthermore, the same existence/uniqueness holds for locally flat  (equivalently, tame) submanifolds. For instance, if $M$ is a triangulated 3-dimensional manifold and $S\subset M$ is a properly embedded tame surface, then $S$ is properly isotopic to a simplicial subsurface in $M$.

Proofs are rather nontrivial (even the fact that every topological surface admits a triangulation).

Moise, Edwin E., Geometric topology in dimensions 2 and 3, Graduate Texts in Mathematics. 47. New York - Heidelberg - Berlin: Springer-Verlag. X, 262 p. DM 45.00; $ 19.80 (1977). ZBL0349.57001.

Moise, Edwin E., Affine structures in 3-manifolds. V: The triangulation theorem and Hauptvermutung, Ann. Math. (2) 56, 96-114 (1952). ZBL0048.17102.


问:
Fundamental Theorem of calculus to find a function of the form of two functions
> Find a function $f$ and a positive number $\beta$ such that

$$ \int_{\sqrt{y}}^{\beta} {f(t)\ln (t) dt = \exp(y) -\ln \left( \frac{\sqrt{y}}{\beta} \right) - \pi } $$
for all $y>0$.

Hi, I have no idea how to solve this question, but it seems relatively simple yet I haven’t seen anything with two functions in it before. Please help if you can, thanks in advance!

Edit: I’ve had a great reply but I just don’t understand how taking the derivative of the integral doesn’t just give you the function contained within the integral? Somehow d/dy remains and we have beta and (y)^(1/2) that somehow appear from nowhere? If anyone could explain this would be great! I’ve added a file highlighting what I’m confused about if it helps.

答:
Background on FTC and Derivatives of Definite Integrals

The Fundamental Theorem of Calculus tells us that given a function $g$ and an antiderivative $G$ (so $G'=g$), we have

$$ \int_a^b g(t) dt = G(b) - G(a) . $$

This relationship holds even if the limits $a$ and $b$ are functions of some other variable, say $y$:

$$ \int_{a(y)}^{b(y)} g(t) dt = G\left(b(y)\right) - G\left(a(y)\right) . $$

Now, if we differentiate both sides with respect to $y$, we have

$$ \frac{d}{dy} \int_{a(y)}^{b(y)} g(t) dt = \frac{d}{dy} \left[ G\left(b(y)\right) - G\left(a(y)\right) \right] $$

$$ \frac{d}{dy} \int_{a(y)}^{b(y)} g(t) dt = g\left(b(y)\right)b'(y) - g\left(a(y)\right)a'(y) , $$

where $b'$ and $a'$ appear due to the chain rule being applied when we calculate the derivatives of $G\left(b(y)\right)$ and $G\left(a(y)\right)$, respectively.

Note, if either $a$ or $b$ is a constant function, then its derivative is 0 and that term disappears from the right-hand side.

Solution to Question

In this problem, $g(t) = f(t)\ln(t)$, $a(y) = \sqrt{y}$, and $b(y) = \beta$.

We start by taking the derivative with respect to $y$ on both sides:

$$ \frac{d}{dy}\int_{\sqrt{y}}^{\beta} {f(t) \ln(t) dt} = \frac{d}{dy} \left[ e^y-\ln \left( \frac{\sqrt{y}}{\beta} \right) - \pi \right] $$

Using the FTC on the left, we get

$$ f(\beta)\ln(\beta)\frac{d}{dy}\beta - f\left(\sqrt{y}\right)\ln \left(\sqrt{y}\right)\frac{d}{dy}\sqrt{y} = e^y - \frac{1}{2y} $$

Since $\beta$ is a constant, $\frac{d}{dy}\beta = 0$:

$$ - \frac{f\left(\sqrt{y}\right) \ln \left(\sqrt{y}\right)}{2\sqrt{y}} = e^y - \frac{1}{2y} $$

Let $t = \sqrt{y}$:

$$ - \frac{f(t) \ln (t)}{2t} = e^{t^2} - \frac{1}{2t^2} $$

$$ \Rightarrow f(t) = -\frac{1}{\ln (t)}\left( 2te^{t^2} - \frac{1}{t} \right) $$

Now, to find $\beta$, we plug $f(t)$ back into the integral:

$$ \int_{\sqrt{y}}^{\beta} {- \frac{1}{\ln (t)}\left( 2te^{t^2} - \frac{1}{t} \right) \ln(t) dt} = e^y-\ln \left( \frac{\sqrt{y}}{\beta} \right) - \pi $$

$$ -\int_{\sqrt{y}}^{\beta} { \left( 2te^{t^2} - \frac{1}{t} \right) dt} = e^y-\ln \left( \frac{\sqrt{y}}{\beta} \right) - \pi $$

$$ -\left[e^{t^2} - \ln(t) \right]_\sqrt{y}^{\beta} = e^y-\ln \left( \frac{\sqrt{y}}{\beta} \right) - \pi $$

$$ \left[e^{y} - \ln\left(\sqrt{y}\right) \right] - \left[e^{\beta ^2} - \ln(\beta) \right] = e^y-\ln \left( \frac{\sqrt{y}}{\beta} \right) - \pi $$

$$ e^{y} - \ln\left(\frac{\sqrt{y}}{\beta}\right) - e^{\beta ^2} = e^y - \ln \left( \frac{\sqrt{y}}{\beta} \right) - \pi $$

$$ e^{\beta ^2} = \pi $$

$$ \beta = \sqrt{\ln (\pi)} $$


问:
Is this true? [(∀x)A(x) → (∀x)B(x)] → [(∀x)(A(x) → B(x))]
Is this proposition True?  [(∀x)A(x) → (∀x)B(x)] → [(∀x)(A(x) → B(x))

I cannot think if there is a difference between the first implication and the last one. Therefore, I think the proposition is true, although I  got someone saying that the proposition is false, again, for the proposition to be false, there must be a difference between (∀x)A(x) → (∀x)B(x) and(∀x)(A(x) → B(x).

答: No. The proposition is false. For example, consider the case that the universe of $x$ is $\mathbb{R}$. $A(x):= x>1$ and $B(x):= x>3$.  In this case $(\forall x)A(x)$ is false. Hence $(\forall x)A(x)\rightarrow (\forall x)B(x)$ is true, regardless of $(\forall x)B(x)$. However, $(\forall x)(A(x)\rightarrow B(x))$ is false .


问:
Question on finding maximum of AM GM inequality
How can i solve for minimum value of function $f(x)=\frac{x^2+x+1}{x^2-x+1} $ using AM GM inequality.

I am aware about the AM GM inequality, but am not able to split the terms into the correct form in order to apply the inequality

答:
we prove the answer is $\frac{1}{3}$ which happens at $x=-1$ for proving it we have :
$$\frac{x^2+x+1}{x^2-x+1} \ge \frac{1}{3}$$
if and only if
$$3x^2+3x+3 \ge x^2-x+1$$
if and only if
$$2x^2+4x+2=2(x+1)^2 \ge 0$$
[edit] if you think that it is not AM_GM here is a solution which is the same but the use of AM_GM is more clear on it.
$$\frac{x^2+x+1}{x^2-x+1}=1+\frac{2}{x+\frac{1}{x}-1} \ge 1 +\frac{2}{-2-1}= \frac{1}{3}$$
(we used $\frac{1}{x+\frac{1}{x}-1}\ge \frac{1}{-3}$ for possitive $x$ it is trivially true and for negative ones it is an AM_GM on $-x+\frac{-1}{x} \ge 2$)


问:
Closed linear Lie group
In Knapp's book 'Lie groups beyond introduction', he defined a closed linear group to be a closed subgroup of $GL(n,\mathbb C)=\{\text{all}\ n\times n\ \text{nonsingular complex matrices}\}.$ My question is if this definition can be made independent of the ambient complex linear group. Technically we can not have that $G$ is a subset of both $GL(n,\mathbb C)$ and $GL(m,\mathbb C)$ where $m\neq n.$

答:
You should check the precise statement. Most likely, he means

"A topological group $G$ is said to be closed linear if it admits an embedding in $GL(n,{\mathbb C})$ for some $n$."

In this situation, an "embedding" is equivalent to "a continuous monomorphism with closed image." (An embedding is required to be a homomorphism which is a homeomorphism to its image.)

Thus, $n$ depends on $G$ and is not fixed in advance.

Aside: There are groups which are (complex) linear as abstract groups but do not admit embeddings  with closed image in any $GL(n,{\mathbb C})$.

Title: Closed linear Lie group


问:
How do I rigorously prove Is (R X Z) U ( Z X R) = R X Z false?
I know that $\Bbb{Z}$ is the subset of $\Bbb{R}$, so all the elements in $\Bbb{Z}$ consist in the set $\Bbb{R}$, but elements of $\Bbb{R}$ do not consist in $\Bbb{Z}$. If we do the union of the Cartesian product of $\Bbb{(R \times Z)}$ and $\Bbb{(Z\times R)}$, we will end up getting some elements like $(0.21,0.1)$ which are not the elements of $\Bbb{Z}$, thus the union cannot be equal to $\Bbb{R \times Z}$.

How do I rigorously prove my argument?

Thanks

答:
No need to prove. You have to discard it. One counter example is good enough.

If $x\in \Bbb{R}\times\Bbb{Z}$,
then $x\in (\Bbb{R}\times\Bbb{Z})\cup(\Bbb{Z}\times\Bbb{R}).$
Therefore $\Bbb{R}\times\Bbb{Z}\subset (\Bbb{R}\times\Bbb{Z})\cup(\Bbb{Z}\times\Bbb{R}).$

Clearly, $y=(m,0.a_1a_2...)\in (\Bbb{R}\times\Bbb{Z})\cup (\Bbb{Z}\times\Bbb{R})$

, but $y\notin \Bbb{R}\times\Bbb{Z}$


问:
De re/dicto knowledge
Define $\mathcal{K}$ as a knowledge operator characterized by a S4 modal system. There is a distinction between de re and de dicto expressions of knowledge:

$\exists x\mathcal{K}A(x)$ is a de re expression of knowledge: there exists $x$ such that the agent knows $A(x)$.

$\mathcal{K}\exists xA(x)$ is a de dicto expression of knowledge: the agent knows that there exists $x$ such that $A(x)$.

As is usually assumed in the literature, de re knowledge entails de dicto knowledge (but usually not the other way around):
$$\exists x\mathcal{K}A(x)\rightarrow\mathcal{K}\exists xA(x)\tag{$*$}$$

My question is the following: given the above definitions and $(*)$, can we prove $(**)$?
$$\mathcal{K}(\exists xA(x)\rightarrow\exists yB(y))\rightarrow(\exists x\mathcal{K}A(x)\rightarrow\exists y\mathcal{K}B(y))\tag{$**$}$$

I tried to use the distribution axiom for $\mathcal{K}$ which is available from the S4 system, but I cannot see how to proceed. Can anyone help? Is $(**)$ provable or not?

答:
I tried to use the distribution axiom for K which is available from the S4 system, but I cannot see how to proceed.

No.  It does not seem doable.

$$\def\fitch#1#2{~~~\begin{array}{|l}#1\\\hline#2\end{array}}\def\knows{\mathcal K}\fitch{\exists x~\knows Ax\to\knows\exists x~Ax}{\fitch{\knows(\exists x~Ax\to\exists y~ By)}{\knows\exists x~Ax\to\knows\exists y~By\qquad\textsf{K distibution}\\\fitch{\exists x~\knows Ax}{\knows\exists x~Ax\qquad\to\mathsf E\\\knows\exists y~By\qquad\to\mathsf E\\~~~\vdots\\\exists y~\knows By\qquad\textsf{.... how?}}\\(\exists x~\knows Ax)\to(\exists y~\knows By)}\\(\knows(\exists x~Ax\to\exists y~By ))\to((\exists x~\knows Ax)\to(\exists y~\knows By))}$$

Using Krippke semantics for S4 (reflexive and transitive many worlds), it does not seem that each accessible world having something satisfying $B$ would entail the existence of a single thing that does so in every accessible world.

Title: De re/dicto knowledge


问:
Simplify the difference of Beta functions $B\Big(\alpha, (k-4\alpha)/2\Big) - B\Big(k/2,(k-4\alpha)/2\Big) $
Can we simplify the difference of these two Beta functiona $$B\Big(\alpha, (k-4\alpha)/2\Big) - B\Big(k/2,(k-4\alpha)/2\Big) $$ where $\alpha \in (0,1)$, $k \in (2\alpha,4\alpha)$? In particular, I guess that their difference should be in the form of Beta function(s) with positive arguments instead of negative ones.

答:
The only simplication I can see is through the gamma function
$$B\left(a,\frac{1}{2} (k-4 a)\right)-B\left(\frac{k}{2},\frac{1}{2} (k-4 a)\right)=$$
$$\Gamma \left(\frac{k}{2}-2 a\right) \left(\frac{\Gamma (a)}{\Gamma
   \left(\frac{k}{2}-a\right)}-\frac{\Gamma \left(\frac{k}{2}\right)}{\Gamma (k-2
   a)}\right)$$


问:
Law of Large Numbers, a confusion
According to Law of Large Numbers, if I throw a coin 1000 times approximately 500 will be head and 500 tail. Suppose that I throw the coin 700 times and I got 700 heads. Can I say that in the next 300 throws the probability of getting tails will be higher than probability of getting heads?

Edit: To make an analogy, imagine that you have a box that contains 500 black and 500 white balls. If 700 times (700 exaggerated) you choose black, than it is more probable that for 701 you get white. However if you say me that instead of one box you have 1000 similar box and every time you choose from one box that means the probability of choosing black or white will never change. With independent events you mean this?

Edit2: Imagine that there are billions of people that throwing coins 1000 times. For each person there is an empty box. When he throw the coin, if it is tail he puts a black ball in his box, when the coin is head he puts white ball. So at the end of the experiment there are billions of boxes that each box contains approximately 500 black 500 white ball. So they give me the opportunity to choose one box. The box that I choose represents the one possible coin throwing that I would make. I am asking what is the difference between throwing coins and choosing one box in billions of boxes? If there is no difference, than the first statement holds. For example I pick 400 black balls from my box than it is more probably to choose white ball from remaining 600 hundreds.

答:
Any two coin flips are independent events, and the outcome of one will not affect the outcome of the other.

In other words, the coin has no memory, and doesn't change its composition or shape from toss to toss, so tosses $701-1000$ will not be affected by the outcomes of tosses $1-700$.

Title: Law of Large Numbers, a confusion


问:
how can I find $x,y$ in the following equation and showing linear independence of the result?
Here is the equation I want to solve rigorously:

$2r + (1 + \sqrt{-5})s = 0$

A trial:

1-I know that the solutions intuitively are $(-1 - \sqrt{-5}, 2)$ and $(-3, 1 - \sqrt{-5})$ but I do not know the rigorous way of finding the solution. So, can anyone help me in that direction ?

2- Also, I want to show that $(-1 - \sqrt{-5}, 2),(-3, 1 - \sqrt{-5})$ are linearly independent, could someone show me the details of this please?

答:
$$\begin{split}2r+(1+\sqrt 5i)s&=0\\
2r&=-(1+\sqrt 5i)s\end{split}$$

Set $r=-(1+\sqrt 5i),s=2$ is a solution because you sort of take the thing that's on the other side and set it to be the variable on the side.

But to find all the solutions,

$$r=-\frac 1 2(1+\sqrt 5i)s$$

Then

$$\left(\begin{matrix}-\frac 1 2(1+\sqrt5i)s\\s\end{matrix}\right)=\left(\begin{matrix}-\frac 1 2(1+\sqrt5i)\\1\end{matrix}\right)s$$

is the set of solutions.

There's more unknowns than equations, so the solution isn't unique. You can only have a unique solution if there are at least as many equations as unknowns.

Linear independence

You wanted to show that the two of your vectors are linearly independent, this means they are not scalar multiples of one another.

$$\begin{split}(-1-\sqrt 5 i, 2)&=c(-3,1-\sqrt 5 i)\\
-1-\sqrt 5 i+3c&=0\text { the first component}\\
c&=\frac 1 3(1+\sqrt 5 i)\\
2&=(1-\sqrt 5 i)c\text { the second component}\\
&=(1-\sqrt 5 i)\frac 1 3(1+\sqrt 5 i)\\
&=\frac 1 3(1+5)=2\end{split}$$

Therefore, the first vector is $c=\frac 1 3 (1+\sqrt 5 i)$ times the second vector. This means they are linearly dependent. This makes sense because both of them are solutions to the system, but the solution space has only dimension 1, so any two solutions will line on the same "line".

If you wanted to show that two vectors $v_1, v_2$ were linearly independent, you would show that $v_1\ne cv_2$ for all choices of c.


问:
Show, that $h$ can be continuous when both sets are open or both sets are closed.
In Lipschutz's "outline of theory and problems of general topology" the following problem is posed:

Let $X$ and $Y$ be topological spaces with $X = E \cup F$. Let $f:E \rightarrow Y$ and $g:F \rightarrow Y$ with $f=g$ on $E \cap F$ de continuous functions with respect to the relative topologies. Note that $h = f \cup g$ is a function from $X$ into $Y$. (i) Show by an example that $h$ need not be continuous. (ii) Prove that if $E$ and $F$ are both open then $h$ is continuous. (iii) Prove that if $E$ and $F$ are bith closed then $h$ is continuous.

For (i) i thought about the following example.

$$f:\mathbb{R} \rightarrow \mathbb{R} := \begin{cases}
1, & \text{for } (-\infty, 0)\\
0, & \text{for } [0, \infty)
\end{cases}
$$
Which is not continous at $0$.

I am having trouble proving (ii) and (iii). If both cases were open, for example
$$f:\mathbb{R} \rightarrow \mathbb{R} := \begin{cases}
1, & \text{for } (-\infty, 0)\\
0, & \text{for } (0, \infty)
\end{cases}
$$
i think that it would still be discontinous at 0 and i don't think i could do the same thing for a closed counterpart. Is my approach to this problem right? How do i go about proving (ii) and (iii)? Any help would be appreciated.

答:
Your second function $f$ doesn't make sense: you claim that it is a function from $\Bbb R$ into $\Bbb R$, but it is undefined at $0$.

(ii) Let $A$ be an open subset of $Y$. Then $f^{-1}(A)=(f|_X)^{-1}(A)\cup(f|_Y)^{-1}(A)$, which is the union of two open sets, and therefore it is an open set. So, $f$ is continuous.

(iii) Let $A$ be a closed subset of $Y$. Then $f^{-1}(A)=(f|_X)^{-1}(A)\cup(f|_Y)^{-1}(A)$, which is the union of two closed sets, and therefore it is a closed set. So, $f$ is continuous.


问:
Is $\mathbb{Q}(\sqrt{d})$ always a quadratic field when $d$ is a squarefree integer?
Is $\mathbb{Q}(\sqrt{d})$ always a quadratic field when $d$ is a squarefree integer?

This is the converse of a theorem about quadratic fields, is it true?

答:
Note that $1$ is the smallest square free integer.

So Your argument is false. But other than $d=1$, your argument is true. Indeed, If $d≠ 1  $ square free integer, then $d=p_1p_2...p_n$, and $\sqrt{d}=\sqrt{p_1p_2...p_n}.$


问:
Why is Riemann's Rearrangement Theorem not considered a contradiction?
If we can pick any conditionally convergent series, $\sum_{n=1}^\infty a_n$, and pick any real number, $\ell$, and show that

$$\sum_{n=1}^\infty a_n = \ell,$$

then why is that not considered a contradiction, since we can show that this sum takes on any value, and any real number can be equal to any other real number? In my limited experience in math, usually something like this is considered a contradiction, and I had expected that such a sum would just be said to diverge or DNE.

It's really confusing me today.

答: The theorem does not say that $$\sum_{n=1}^\infty a_n = \ell,$$it says that if $$\sum_{n=1}^\infty a_n$$is conditionally convergent, for every $\ell \in \Bbb R$ there is a bijection $\sigma:\Bbb N \to \Bbb N$ such that $$\sum_{n=1}^\infty a_{\sigma(n)}=\ell.$$It's a different series. For $\sigma = {\rm Id}_{\Bbb N}$ you have the original value of the series. For other values, non-trivial permutations.


问:
Why is the Image sheaf a subsheaf?
Given a morphism of sheaves $\psi:\mathcal{F} \rightarrow \mathcal{G}$, we define the image sheaf Im $\psi$ to be the sheaf associated to the image presheaf $\text{Im}^{\text{pre}} \psi$.

By Hartshorne's construction, the sheaf $\mathcal{H}^+(U)$ associated to a presheaf $\mathcal{H}(U)$ is the set of functions $\{s:U \rightarrow \bigcup \mathcal{H}_p\}$ such that $\forall p \in U, s(p) \in \mathcal{H}_p$ and $\exists V \subset U$ containing p and $t \in \mathcal{H}$ such that $\forall q \in \mathcal{H}(V), s(q) = t_q$.

So my question is that why is the image subsheaf Im $\psi$ a subsheaf of $\mathcal{G}$? Clearly $\text{Im}^{\text{pre }} \psi (U)$ will always be a subgroup/subring of $\mathcal{G}(U)$ by definition, but I don't see why this sheafification of the image presheaf on an open $U$, i.e. $\text{Im}^{\text{pre }} \psi (U)$, needs to be a subgroup/subring of $\mathcal{G}(U)$

Edit: I don't know if Hartshorne mentions this but it is mentioned in Liu's Algebraic geometry and arithmetic curves in Lemma 2.16.

答:
You should look at Proposition-Definition 1.2 of Hartshorne. He defines the sheafification of the presheaf $\mathcal{F}$ to be the sheaf $\mathcal{F}^+$ equipped with a morphism of presheaves $\theta:\mathcal{F}\to \mathcal{F}^+$ satisfying the following universal property. Given a morphism of presheaves $\varphi:\mathcal{F}\to \mathcal{G}$ where $\mathcal{G}$ is a sheaf, there exists a unique morphism of sheaves $\psi:\mathcal{F}^+\to \mathcal{G}$ such that $\varphi=\psi\circ \theta$.

In this case, associated to a morphism of sheaves $\varphi:\mathcal{F}\to \mathcal{G}$ there the image presheaf $\varphi(\mathcal{F})$ which is as you say is a subsheaf of $\mathcal{G}$. We define $\operatorname{im}\varphi$ to be the sheaf associated to the presheaf $\varphi(\mathcal{F})$. As there is an inclusion morphism of presheaves $i:\varphi(\mathcal{F})\to \mathcal{G}$, we get a morphism of sheaves $j:\operatorname{im}\varphi\to \mathcal{G}$ by the universal property.

Lemma: Given an injective map of presheaves $\alpha:\mathcal{E}\to \mathcal{E}'$ the induced map on the associated sheaves $\mathcal{E}^+\to \mathcal{E}'^+$ is injective also.

Hence, as $i:\varphi(\cal{F})\to \mathcal{G}$ is injective, the associated map $j:\operatorname{im}\varphi\to \mathcal{G}$ is also, so that $\operatorname{im}\varphi$ can be identified with a subsheaf of $\mathcal{G}$.

Proof of the Lemma: Examine $\varphi_P^+:\mathcal{F}_P^+\to \mathcal{G}_P^+$ noting that $\mathcal{F}_P^+=\mathcal{F}_P$ and $\mathcal{G}_P^+=\mathcal{G}_P$. Recall, also that $\mathcal{F}\hookrightarrow{} \mathcal{F}_P^+$ as the maps $s:U\to\coprod_{p\in U} \mathcal{F}_P$ given by $s(P)=s_P$. We can check injectivity of a morphism of sheaves at the stalks. If $\varphi_P^+(t,U)=0$, then there exists $W\subseteq U$ containing $P$ such that $\varphi_W^+(t|_W)=0$. But, take an open set $W'\subseteq W$ containing $P$ on which $t|_{W'}=\tau \in \mathcal{F}(W')$. Then $\varphi^+_{W'}|_{\mathcal{F}(W')}$ agrees with $\varphi_{W'}$, and hence is injective. So, $\varphi_{W'}^+(\tau)=0$, and this implies $\tau =0$. So, the germ $(t,U)$ is $0$ at $P$. Therefore, we have injectivity at the stalks, and our map of sheaves is injective, too.

Title: Why is the Image sheaf a subsheaf?


问:
What is the meaning of a big "V" in measure theory?
I am trying to understand random variables and I came across this:

As will become increasing evident as we proceed, infinite families of independent objects possess surprising and beautiful properties. In particular, mutually
independent σ-algebras tend to fill up space in a sense which is made precise by
the following beautiful thought experiment designed by A.N. Kolmogorov. Let
I be any index set, take $F_∅ = \{∅, Ω\},$ and for each nonempty subset $Λ ⊆ I,$ let

$$\mathcal F_\Lambda = \bigvee \mathcal F_i \equiv \sigma \left(\bigcup_{i\in \mathcal I} \mathcal F_i \right)$$

be the $\sigma$-algebra generated by $\displaystyle\bigcup_{i\in \Lambda} \mathcal F_i,$ (i.e., $\mathcal F_\Lambda$ is the smallest $\sigma$-algebra containing $\displaystyle\bigcup_{i\in \Lambda} \mathcal F_i$).

The probability that I understand this anytime soon is zero almost surely, but to begin with, I would like to get what

```<code>\\\$\big vee $:
</code>```

$\bigvee$

means.

答:
I am following up on my comment above. It appears as though the definition is that
$$
\bigvee_{i\in \Lambda}\mathcal{F}_i=\text{The Sigma Algebra Generated by the Sets }\mathcal{F}_i
$$
based on the context.

$\S2.3 $ of the following wikipedia article agrees with this: Wikipedia: $\sigma$-algebra.


问:
How does second partial derivative test be interpreted?
I know that what the formula is used to determine a point whether it has maximum, minimum, or saddle points; but, i do not understand how it is formulated.

$$ H = f_{xx}(x_0,y_0)f_{yy}(x_0, y_0) − f_{xy}(x_0, y_0)^2 $$

The $f_{xx}(x_0,y_0)$ and $f_{yy}(x_0,y_0)$ are used to determine the concavity in x and y direction respectively, but why it is multiplied by each other. I also do not understand why it is subtracted by $f_{xy}(x_0,y_0)^2$ and why it has a power of two?

答: The single-variable condition $f''(x_0)>0$ is replaced by $Hf(x_0,y_0)>0$, where $Hf$ stands for the Hessian of $f$. But this is a matrix, so this ''$>0$'' has to be read carefully: it means that the matrix $Hf(x_0,y_0)$ is positive definite. Since this matrix has order $2$, positive-definiteness is equivalent to the conditions $f_{xx}(x_0,y_0)>0$ and $\det Hf(x_0,y_0)>0$ holding together. This is what you have there. For higher dimensions, look up "Sylvester's Criterion".


答: Coin tosses are independent events- the probability will still be 0.5 for tails (assuming fair coin). Keep in mind that if something is an independent event, that any tosses before it won't change its probability of occurring. The Law of Large Numbers pretty much says that in small number of trials you may not get the true/accurate results but if you increase the number of trials, then the results would be more accurate. That being said, if over 1000 tosses, all were heads, it is very likely that the coin is unfair.</p>

Title: Law of Large Numbers, a confusion

Body:
According to Law of Large Numbers, if I throw a coin 1000 times approximately 500 will be head and 500 tail. Suppose that I throw the coin 700 times and I got 700 heads. Can I say that in the next 300 throws the probability of getting tails will be higher than probability of getting heads?

Edit: To make an analogy, imagine that you have a box that contains 500 black and 500 white balls. If 700 times (700 exaggerated) you choose black, than it is more probable that for 701 you get white. However if you say me that instead of one box you have 1000 similar box and every time you choose from one box that means the probability of choosing black or white will never change. With independent events you mean this?

Edit2: Imagine that there are billions of people that throwing coins 1000 times. For each person there is an empty box. When he throw the coin, if it is tail he puts a black ball in his box, when the coin is head he puts white ball. So at the end of the experiment there are billions of boxes that each box contains approximately 500 black 500 white ball. So they give me the opportunity to choose one box. The box that I choose represents the one possible coin throwing that I would make. I am asking what is the difference between throwing coins and choosing one box in billions of boxes? If there is no difference, than the first statement holds. For example I pick 400 black balls from my box than it is more probably to choose white ball from remaining 600 hundreds.


问:
Combinatorics That Looks Similar to Vandermonde's Identity
How do I simplify:

$$\sum_{r = 0}^{\lfloor \frac{k}{2} \rfloor} \dbinom{k}{2r} \cdot \dbinom{n-k}{k-2r}?$$
Basically, the sum is: $\dbinom{k}{0} \cdot \dbinom{n-k}{k} + \dbinom{k}{2} \cdot \dbinom{n-k}{k-2} + \dbinom{k}{4} \cdot \dbinom{n-k}{k-4} \cdots.$

This seems very similar to Vandermonde's Identity but because we skip two everytime instead of one, I don't see how I can apply it to this problem.

答:
Note that
$$\sum_{r\ge 0} a_{2r} = \sum_{r\ge 0} \frac{1+(-1)^r}{2}a_r.$$
Taking $$a_r=\binom{k}{r}\binom{n-k}{k-r}$$ yields
\begin{align}
\sum_{r\ge0} \binom{k}{2r}\binom{n-k}{k-2r}
&= \sum_{r\ge0} \frac{1+(-1)^r}{2} \binom{k}{r}\binom{n-k}{k-r} \\
&= \frac{1}{2}\sum_{r\ge0} \binom{k}{r}\binom{n-k}{k-r} + \frac{1}{2}\sum_{r\ge0} (-1)^r\binom{k}{r}\binom{n-k}{k-r} \\
&= \frac{1}{2}\binom{n}{k} + \frac{1}{2}\sum_{r\ge0} (-1)^r\binom{k}{r}\binom{n-k}{k-r}.
\end{align}

I don't know whether the remaining sum can be simplified.


问:
How do I get to the coordinates (e,1/e) in this question
This is my working out i know x/x equals to one but from here I don't know how to get the coordinates. I have been stuck on this since yesterday.

This is the question 8.

The equation of a curve is y=lnx/x. Find the exact coordinates of the stationary point on this curve and determined whether it is a maximum or minimum point.

答:
HINT

You have already written that stationary points occur when the first derivative is equal to zero. Equate you final expression to zero and multiply both sides by $x^2$. (You will obtain a famous constant as the x-intercept of your stationary point.)


问:
How can two dependent variables have zero covariance?
The following text is from page 59 of chapter 3 of the book Deep Learning by Ian Goodfellow et al.:

Independence is a stronger requirement than zero covariance, because independence also excludes nonlinear relationships. It is possible for two variables to be dependent but have zero covariance.

For example, suppose we ﬁrst sample a real number $x$ from a uniform distribution over the interval [−1,1]. We next sample a random variable $s$. With probability $\frac{1}{2}$, we choose the value of $s$ to be 1. Otherwise, we choose the value of $s$ to be −1. We can then generate a random variable $y$ by assigning $y=sx$. Clearly, $x$ and $y$ are not independent, because $x$ completely determines the magnitude of $y$. However, $Cov(x, y) = 0$.

What is the difference between sampling a real number and sampling a random variable?
How can I mathematically show that $Cov(x,y)=0$?

答:
It means the same thing. It's just talking about sampling a number from the uniform distribution, call it X, on [-1,1], i.e. from $X\sim Unif[-1,1]$.

To calculate the covariance, you might note the formula $Cov(X,Y)=E(XY)-E(X)E(Y)$. It is known that E(X)=0, beacause X is uniform on [-1,1], so pick the middle value. $E(Y)=E(SX)=E(S)E(X)$ (beacuse they are independent), and their individual expectations are 0, so E(Y)=0. We need to find E(XY) now. Note that $E(XY)=E(SX^2)=E(S)E(X^2)$ again because S and X are independent. So we get $0\cdot E(X^2)=0$. Thus $Cov(X,Y)=0$.

To show that X and Y are dependent events, note that the value of Y is half-determined by the value of X. So Y "depends" on X. To show mathematically that they are dependent, you must find sets A and B such that $X\in A$ and $Y\in B$ are events, where $P(Y\in B|X\in A)\ne P(Y\in B)$. Choose $A=[.25,.75],B=[.25,.75]$. Then $P(Y\in B)=\frac 1 4$ (this is because Y is also uniformly distributed on [-1,1]). But $P(Y\in B|X\in A)=\frac 1 2$. This is because given that $X\in[.25,.75]$, there is equal chance that $Y\in[.25,.75]$ or $Y\in[-.75,-.25]$. Thus we have found an events such that $P(Y\in B|X\in A)\ne P(Y\in B)$, and the random variables are dependent.

Two variables are independent if $P(X\in A\cap Y\in B)=P(X\in A)P(Y\in B)$ for all possible combinations of sets A, B such that $X\in A$ and $Y\in B$. Equivalently, that $P(X\in A|Y\in B)=P(X\in A)$ for all possible combinations of events. This is obtained by dividing the first equation by $P(Y\in B)$.


问:
Introduction to group theory - proof of equivalency of a group $G$ being cyclic and some property of order of $G$
The following is from Czesław Bagiński, "Wstęp do teorii grup". The original is in Polish, the translation is mine.

Remark: "the previous claim" in the last sentence of the quoted proof refers to the claim before theorem 5.10. I don't quote it because it is not relevant for the question.

Theorem 5.10. A finite abelian group of order $n$ is a cyclic group if and only if for every divisor $d$ of the number $n$ there exists exactly $d$ elements satisfying the condition $x^d = e$.

Proof. The implication "=>" is the the theorem 4.4. Let us assume that for every $d$ dividing $n$ there exists exactly $d$ elements whose orders are divisors of $d$. Let $G_d$ be a set of this elements, that is $G_d = \{x \in G: x^d = e\}$. The commutativity of $G$ implies that the subset $G_d$ is a subgroup with $d$ elements. If $G = |p|$ is a prime number then the theorem is obvious. Let us assume that $|G|$ is not a prime and that the theorem is true for groups of the order lesser than $|G|$. If $|G| = p^k$ for some prime $p$ then by inductive hypothesis $h = G_{p^{k-1}}$ is the only cyclic subgroup of the order $p^{k-1}$. Additionally for any $x \in G \setminus H$, $x^{p^{k-1}} \neq e$. At the same time $o(x)|p^k$. Thus $o(x) = p^k$ and by this $G$ is cyclic. If $|G|$ is not a power of a prime then by the fundamental theorem of arithmetic $|G| = rs$ where $r$ and $s$ are relatively prime and $r, s > 1$. By induction hypothesis $G_s$ and $G_r$ are cyclic subgroups of the group $G$ of orders equivalently $s$ and $r$. If $ x = G_s$ and $y = G_r$ then by commutitivity of $G$ and the previous claim $xy$ is an element of the order $rs$ which proves that $G$ is cyclic.

How does the fact that $o(x) = p^k$ imply that $G$ is cyclic?

How does the fact that $xy$ is an elmeent of the order $rs$ proves that $G$ is cyclic?

答:
Thanks for all who commented, I write the full (the full from my point of view) answer for reference in case it is needed in the future.

If we have a finite group $G$ of order $n$ and $o(x) = n$ then $G$ is cyclic because this element raised to 1-th, 2-nd, 3-rd, ..., n-th power must yield all the elements contained in $G$. For if it was not the case, we would have the situation in which at least two elements in such genenerated sequence would be duplicates - but it would mean that generating some initial fragment of this sequence we would obtain $\{x, x^2, ..., x^k, u\}$ where the elements from the set $\{x, x^2, ..., x^k\}$ are different, $k < n$, and $u$ would be equal to one of the previous values and the said sequence (that is $\{x, x^2, ..., x^k, u\}$) is not $\{x, x^2, ..., x^{n-1}, e\}$. Now, there are following possibilities:

$u \neq e$. In this case there is no $e$ in the whole sequence of the length $n$ (because raising $x$ to subsequent powers would just yield the already obtained elements). Contradiction because $o(x) = n$ implies that $x^n = e$ (can also be justified in other ways).

$u = e$. If $k < n - 1$ it would mean that $o(x) < n$ which would contradict the assumption. The case $k = n - 1$ is excluded by the disclaimer.


答:
Covariance is a measure of the linear dependence of two random variables. $Cov(x,y)>0$ means something like the following: if I observe a relatively large $x$, I should also expect to observe a relatively large $y$. However, in the example you gave, if you observe a large $x$, you are equally likely to observe a large $y$ (if $s=1$) as you are to observe a small $y$ (if $s=-1$)! Hence $Cov(x,y)=0$, even though $y$ clearly depends on $x$.

To show $Cov(x,y)=0$ in your problem:

\begin{align}
Cov(x,y) &= Cov(x,sx)\\
&= \mathbb E[x^2 s] - \mathbb E[x]\mathbb E[s x] \\
&= \mathbb E[x^2] \mathbb [s]\\
&= 0
\end{align}

Where we used $\mathbb E[x]=\mathbb E[s]=0$, and the independence of $x$ and $s$, which implies $\mathbb E[x^2 s] = \mathbb E[x^2] \mathbb E[s]$.

In general, independent random variables necessarily have zero covariance, but random variables with zero covariance need not be independent as they may have higher moments which are nonzero.

Title: How can two dependent variables have zero covariance?

Body:
The following text is from page 59 of chapter 3 of the book Deep Learning by Ian Goodfellow et al.:

Independence is a stronger requirement than zero covariance, because independence also excludes nonlinear relationships. It is possible for two variables to be dependent but have zero covariance.

For example, suppose we ﬁrst sample a real number $x$ from a uniform distribution over the interval [−1,1]. We next sample a random variable $s$. With probability $\frac{1}{2}$, we choose the value of $s$ to be 1. Otherwise, we choose the value of $s$ to be −1. We can then generate a random variable $y$ by assigning $y=sx$. Clearly, $x$ and $y$ are not independent, because $x$ completely determines the magnitude of $y$. However, $Cov(x, y) = 0$.

What is the difference between sampling a real number and sampling a random variable?
How can I mathematically show that $Cov(x,y)=0$?


问:
Solve the functional equation $f(x)f(1/x)=f(x+1/x)+1$, $f(1)=2$, where $f(x)$ is a polynomial.
Solve the functional equation
$$f(x)f(1/x)=f(x+1/x)+1,\ f(1)=2,$$
where $f(x)$ is a polynomial.

It is easy to check that $f(x)=x+1$ is a solution. Are there any other solutions? My attempt is described as follows.

Since $f(x)$ is a polynomial, we can write it as $f(x)=\sum_{i=0}^n a_ix^i$, where $a_n\neq 0$. Substituting it into the functional equation, we have
$$
f(x)f(1/x)=\sum_{i=0}^n a_ix^i\sum_{j=0}^n a_jx^{-j}=\sum_{\ell=-n}^n \Bigg(\sum_{i=\max(0,\ell)}^{\min(n,n+\ell)} a_ia_{i-\ell}\Bigg) x^\ell $$
and
\begin{align}
f(x+1/x)+1&=\sum_{i=0}^n a_i(x+1/x)^i+1=\sum_{i=0}^n a_i\sum_{j=0}^i\binom{i}{j} x^{2j-i}+1\\
&=\sum_{\ell=-n}^n \Bigg(\sum_{\max(\ell,0)\leq j\leq \frac{n+\ell}{2}} a_{2j-\ell}\binom{2j-\ell}{j}\Bigg)x^{\ell}+1.
\end{align}
By comparing the coefficients of $x^\ell$ on both sides of the above two equations and noting that $f(1)=\sum_{i=0}^n a_i=2$, we can obtain a very complex system of equations of the coefficients $a_i$, $0\leq i\leq n$. For example, when $n=5$, the system of equations is of the form
\begin{align}
&a_0a_5=a_5,\\
&a_0a_4+a_1a_5=a_4,\\
&a_0a_3+a_1a_4+a_2a_5=a_3+5a_5,\\
&a_0a_2+a_1a_3+a_2a_4+a_3a_5=a_2+4a_4,\\
&a_0a_1+a_1a_2+a_2a_3+a_3a_4+a_4a_5=a_1+3a_3+10a_5,\\
&a_0^2+a_1^2+a_2^2+a_3^2+a_4^2+a_5^2=1+a_0+2a_2+6a_4,\\
&a_0+a_1+a_2+a_3+a_4+a_5=2.
\end{align}
Assuming that $a_5\neq 0$, it is not hard to show that the above system of equations has no solution. But for general case, I have no idea how to solve it or to show it has no solution. I wonder whether there is a clever approach which avoids this tedious computation?

Added. I would like to point out that the condition $f(1)=2$ imposes a strong restriction on the problem. For example, the polynomial $f(x)=1+(1-\sqrt{2})x^2$ satisfies the functional equation but does not satisfy the condition $f(1)=2$. It seems that this condition rules out many solutions.

答:
Here is my solution that was found during an exam :
first let $Deg_p=n>2$ now let :
$$p(x)=\sum_{i=0}^n{a_ix^i}$$
so we have  :
$$p(x)p(\frac{1}{x})=\sum_{i=0}^n{a_ix^i}\sum_{i=0}^n{a_ix^{-i}}=1+\sum_{i=0}^n{a_i(x+\frac{1}{x})^i}$$
first we will look at the constant therm on the both side which is :
$$1+a_0=\sum_{i=0}^n{a_i^2}$$
now we will look at the term $x^n$ at both sides. which gives :
$$a_na_0=a_n$$
since $a_n$ can not be $0$ we must have $a_0=1$ , we rewrite the first condition as :
$$\sum_{i=1}^n{a_i^2}=1$$
now lets calculate the term $x^{n-1}$ on both side we have :
$$a_na_1+a_{n-1}=a_{n-1}$$
which gives $a_1=0$ now we look at the term $n-2$ :
$$a_na_2+a_{n-2}=na_n+a_{n-2}$$
so we have $a_2=n$ but this gives contradiction with the assumption that
$$\sum_{i=1}^n{a_i^2}=1 \ge a_2^2=n^2$$
so we have $n\le 2$ if $n=1$ then we clearly have
$p(1)=2,p(2)=3$ (from $x=1$) so we have $a_1+a_0=2,2a_1+a_0=3$ so $a_1=1,a_0=1$ and $p(x)=x+1$ which clearly works.
if $n=2$ remember we proved $a_1=0$ this also works for $n=2$ so we have
$$p(x)=ax^2+1$$
but since $p(1)=2$ we have $a=1$ bot the polynomial $p(x)=x^2+1$ doesn't work in the statement.
so we are done


答:
I think you're almost done.
Your last expression is

$\frac{dy}{dy} = \frac{1 - \ln(x)}{x^2} = 0$

Since RHS is $0$, numerator of LHS must be zero

$1 - \ln(x) = 0$

$\ln(x) = 1$

$x = e^1$.

Thus, $x = e$.

$y = \frac{\ln(x)}{x} = \frac{1}{e}$

Thus the fixed point is $(e, \frac{1}{e})$

Title: How do I get to the coordinates (e,1/e) in this question

Body:
This is my working out i know x/x equals to one but from here I don't know how to get the coordinates. I have been stuck on this since yesterday.

This is the question 8.

The equation of a curve is y=lnx/x. Find the exact coordinates of the stationary point on this curve and determined whether it is a maximum or minimum point.


答:
Here, $ y = \frac{\ln x}{x} $

For finding stationary points, you need to put $ \frac{dy}{dx} = 0 $

$$ \because \frac{dy}{dx} = \frac{x/x - \ln x}{x^2} $$
We have, $  \frac{x/x - \ln x}{x^2} = 0 \implies  1 - \ln x = 0 \implies x = e $

Hence, Stationary point is $ x = e $ where $ y = \frac{1}{e} $

For finding maxima/minima you need to check the sign of $ \frac{d^2y}{dx^2} $.

If it’s a positive, then it’s a minima and if it’s negative, then it’s a maxima.

Here, you can find $ \frac{d^2y}{dx^2} $ to be negative. That means $ (e, \frac{1}{e}) $ is a maxima.


问:
Let X1 be N(0, 1), and let X2 = −X1, if − 1 ≤ X1 ≤ 1, X1, otherwise. Show that X2 also has an N(0,1) distribution.
Given that X1 follows standard normal distribution, N(0,1). And
define X2 = -X1, if -1 ≤ X1 ≤ 1; X1, otherwise. Then I am asked to prove that X2 follows N(0,1) as well.

My idea is to prove P(X1 ≤ x ) = P(X2 ≤ x) for three cases when x < -1 , when -1 ≤ x ≤ 1, when x > 1.
For x < -1 and x > 1 are easy, I started with P(X1 ≤ x) = P(X2 ≤ x) and P(X2 ≥ x) = P(X1 ≥ x) respectively and show that they have the same CDF.

But I am stucked with the case -1 ≤ x ≤ 1,
I was thinking to start with
when -1 ≤ x ≤ 1,
P(|X1|≤|x|) = P(|X2|≤|x|)
and derive from this equation to get P(X1 ≤ x ) = P(X2 ≤ x), but it does not seem to be the correct way.

Could anyone help me with this proof?

答:
$P(X_2 \leq x)=P(-X_1 \leq x, -1 \leq X_1 \leq 1)+P(X_1 \leq x, | X_1| >1)$. Since $X_1$ has symmetric distribution we can change $X_1$ to $-X_1$ in the first term. Hence, $P(X_2 \leq x)=P(X_1 \leq x, -1 \leq X_1 \leq 1)+P(X_1 \leq x, | X_1| >1)=P(X_1 \leq x)$.

[$P(X_1 \in E)=P(-X_1 \in E)$ for any Borel set $E$].


问:
Accumulation points of accumulation points of accumulation points
Let $A'$ denote the set of accumulation points of $A$.
Find a subset $A$ of $\Bbb R^2$ such that $A, A', A'', A'''$ are all distinct.

I can find a set $A$ such that $A$ and $A'$ are distinct, but not one where $A,A',A'',A'''$ are all distinct.

答:
What might be confusing here is that you won't, of course, get that $\overline{A} \neq \overline{\overline{A}}$ (where $\overline{A}$ denotes the closure of $A$). Thus, the accumulation points of $A'$ will also be accumulation points of $A$, but $A'$ may have fewer accumulation points than $A$ does.

In other words, you'll have that $A' \supset A'' \supset A''' \ldots$. This means that have to start with a set with "enough" accumulation points, otherwise you'll hit the empty set too early for $A',A'',A'''$ to be all different. The easiest way to avoid that is to work backwards, i.e. start with $A'''$, as Asaf suggested.

Let $$
  A''' = \{(0,0)\} \text{.}
$$ Then construct some $A'' \supset A'''$ such that $A''$ has accumulation point $(0,0)$, say $$
  A'' = \{(\frac{1}{n},0) \,:\, n \in \mathbb{N}\} \cup \{(0,0)\} \text{.}
$$
Now continue this idea to find $A'$. $A'$ will have to contain sequences which converge to $(\frac{1}{n},0))$ for every $n \in \mathbb{N}$. Finally, from $A'$, determine $A$.


问:
Does the short exact sequence $0\mapsto \mathbb{Z}^k\mapsto \Gamma\mapsto \mathbb{Z}^\ell\mapsto 0$ split?
I've seen in Nonsplit extension of $\mathbb{Z}$ by itself that every short exact sequence of the form $0\mapsto \mathbb{Z}\mapsto G\mapsto \mathbb{Z}\mapsto 0$ splits.

I wonder if this is still valid if we change the exact sequence by $$0\mapsto \mathbb{Z}^k\mapsto \Gamma\mapsto \mathbb{Z}^\ell\mapsto0$$ where $\Gamma$ is a discrete group.

Thanks in advance

答:
A counterexample is the following short exact sequence:
$0 \rightarrow \mathbb{Z} \rightarrow T \rightarrow \mathbb{Z}^2 \rightarrow 0$. Where $T$ is the group of upper triangular matrices in $\text{GL}_3(\mathbb{R})$ with unit diagonals.

The injection is defined by the formula $z\mapsto \begin{pmatrix} 1 &0&z\\  0 &1 &0 \\ 0 &0&1\end{pmatrix}$.
The surjective morphism is defined by the formula $\begin{pmatrix} 1 &a&c\\0&1&b\\0&0&1\end{pmatrix} \mapsto (a, b)$.

The image of the injection and the kernel of the surjection is the subgroup generated by $\begin{pmatrix} 1 & 0 & 1\\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$. This shows that this is a short exact sequence.

However the sequence is clearly not split because $T$ is not isomorphic to the product of $\mathbb{Z}$ and $\mathbb{Z}^2$ because $T$ is not abelian.


答:
A small point on writing derivatives: when you are calculating derivatives with respect to $x$ , you should write $\frac{d}{dx}(expression)$ . For example, the correct form of the derivative of your problem is:
$$\frac{x (\frac{d}{dx}\ln x) \;- \; (\ln x) \frac{d}{dx}x}{x^2} $$

To find whether the stationary point that you found is a maximum or a minimum, you can calculate the second derivative, then see its sign at the stationary point. A positive second derivative means that the first derivative is increasing, which means that if the first derivative is now zero, it was negative before, and it will be positive after this point. Now, keeping in mind that the first derivative is the slope of the curve, imagine what this looks like: a curve with negative slope (going down) reaches a stationary point and then its slope becomes positive (going up). This is a minimum. Similarly, you can see that a stationary point where the second derivative is negative, is a maximum. Draw a shape for yourself, it always helps!


问:
Find all the functions $f: \mathbb{N} \to \mathbb{N} $ such that $(m+f(n))(n+f(m))$ is a perfect square for all $m,n$
Let $N$ be the set of natural number. Find all functions $f: \mathbb N \to \mathbb N$ , such that the number $(m+f(n))(n+f(m))$ is perfect square for all natural numbers $m$ and $n$.

I was unable to solve this problem so I need a solution to it, I know that the condition that $(m+f(n))(n+f(m))$ is a perfect square would imply that if $n = m$ then $f(n) = f(m)$ but how would that help me in solving the problem?

答:
here is my proof.
first let $T(m,n)=(f(m)+n)(f(n)+m)$ we porve if $p|f(m)-f(n)$ then $p|m-n$ to do so assume :
$$(f(m)+t)(f(t)+m)=g(t,m)^2$$
first assume $V_p(f(m)-f(n))>1$ now let $t=q^{f(m)+f(n)}p-f(n)$ where $q$ is a prime other than $p$ so we have $p|f(m)-f(n)+q^{f(m)+f(n)}p$ but we dont have $p^2|f(m)-f(n)+q^{f(m)+f(n)}p$ so we must have $p|f(q^{f(m)+f(n)}p)+m$ now look at $T(q^{f(m)+f(n)}p-f(n),f(n)$ this gives again $p|f(q^{f(m)+f(n)}p-f(n))+n$ taking the difference gives $p|m-n$.
so this is done now if $V_p(f(m)-f(n))=1$ checking
$$T(p^{f(m)+f(n)}-f(n),m)$$
gives $p|f(p^{f(m)+f(n)}-f(n))+m$ and from
$$T(p^{f(m)+f(n)}-f(n),n)$$
we get $p|f(p^{f(m)+f(n)}-f(n))+n$ so again taking the difference gives $p|n-m$ this means there is no prime such that $p|f(n+1)-f(n)$ so we must have
$$(f(n+1)-f(n))^2=1$$
now we prove the fucntion is injective. assume $f(a)=f(b)$ we have from
$$T(a,m)T(b,m)=(f(a)+m)^2(f(m)+a)(f(m)+b)$$
is a perfect square then so is :
$$(f(m)+a)(f(m)+b)$$
now if we prove $f$ is not bounded we are done. since a polynomial of second degree can not be infinitly many times an square unless the polynomial has $2$ equal roots. which in our case they are assumed to be different . for proving that $f$ is unbounded simply look at
$$T(p-f(1),1)$$
where $p$ is a prime greater than $f(1)$ so we have $p(f(p-f(1))+1)$ is a perfect square so $p|f(p-f(1))+1$ so we must have
$$f(p-f(1)) \ge p-1$$
now taking $p$ big enough we are done.
ass the function is injective we must always have
$$f(n)+1=f(n+1)$$
which gives the answer $f(n)=n+c$ for some constant positive natural $c$ and all these functions work


问:
How to compute $\int_0^{\infty}\sqrt x \exp\left(-x-\frac{1}{x}\right) \, dx$?
How to compute this integral? :

$$\int_0^{\infty}\sqrt x \exp\left(-x-\frac{1}{x}\right) \, dx$$

Wolframalpha gives the answer $\dfrac{3\sqrt{\pi}}{2e^2}$, but how to compute this?

答:
With $t=\sqrt x$

\begin{align}
\int_0^{\infty}\sqrt x e^{-x-\frac{1}{x}} \, dx
&= \int_{-\infty}^{\infty}t^2e^{-t^2-\frac1{t^2}}dt
= \frac1{e^2}\int_{-\infty}^{\infty}t^2e^{-(t-\frac1{t})^2}dt\\
&= \frac1{e^2}\left(\int_{-\infty}^{\infty}\left(t^2 -1\right)e^{-(t-\frac1{t})^2}dt +\int_{-\infty}^{\infty}\underset{t\to \frac1t}{e^{-(t-\frac1{t})^2}}dt\right)\\
&=\frac1{e^2}\int_{-\infty}^{\infty}\left((t-\frac1t)^2 +1 \right)e^{-(t-\frac1{t})^2}dt \\
&= \frac1{e^2}\int_{-\infty}^{\infty}\left(t^2 +1 \right)e^{-t^2}dt
=\frac{3\sqrt\pi}{2e^2}\\
\end{align}
where $\int_{-\infty}^{\infty}f(t-\frac1t)dt = \int_{-\infty}^{\infty}f(t)dt$ is used.


Title: Accumulation points of accumulation points of accumulation points

Body:
Let $A'$ denote the set of accumulation points of $A$.
Find a subset $A$ of $\Bbb R^2$ such that $A, A', A'', A'''$ are all distinct.

I can find a set $A$ such that $A$ and $A'$ are distinct, but not one where $A,A',A'',A'''$ are all distinct.


问:
Does the $\lim_{z \to -1} \sqrt{|z|} \,e^{i\operatorname{Arg}(z)/2}=i$?
I am asked to show whether or not the $\lim_{z \to -1}f(z)$ where $f(z)=\sqrt{|z|} \,e^{i\operatorname{Arg}(z)/2}$ exists.

Simply plugging in $-1$ into the limit, we get, $$\lim_{z \to -1} \sqrt{\left|-1\right|}\: e^{i\operatorname{Arg}(-1)/2}\\=\lim_{z \to -1} e^{i\operatorname{Arg}(-1)/2}$$

The major problem I am facing is figuring out what the $\operatorname{Arg}(-1)$ is going to be. After doing some research, I came across a Wolfram Alpha article which said the $\operatorname{arg}(-1)=\pi$. I am assuming this holds true for the principal argument as well. Solving, we get, $$\lim_{z \to -1} e^{i(\pi/2)}=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}=i$$

Although, I need some help in proving why the $\operatorname{Arg}(-1)=\pi$. Unfortunately, I was not able to find much information about it online, so I would appreciate some assistance.

答:
" I came across a Wolfram Alpha article which said the
$\arg(−1)=\pi$. I am assuming this holds true for the principal
argument as well. "

$\pi$ is the principal argument of $−1$, i.e., Arg$(-1)=\pi.$

But in general, $\arg(−1)=(2k+1)\pi,$ where $k\in\mathbb{Z}.$

" I have always thought about finding $\arg(z)$ by applying the
$\tan^{−1}\left(\frac{y}{x}\right).$ "

This formula doesn't generally work, since
$\tan^{-1}\left(\frac{y}{x}\right) \in
   \left(-\frac{\pi}2,\frac{\pi}2\right),$ which doesn't even span the
principal range $\left(-\pi, \pi\right]$ of $\arg(z)$:
$$\displaystyle\text{Arg}(-1-i)=-\frac34\pi \\\neq \frac{\pi}4
   =\tan^{−1}\left(\frac{-1}{-1}\right).$$

The given limit $$\lim_{z\to-1} \sqrt{|z|} \exp\left(i\frac{\text{Arg}(z)}2\right)$$ doesn't exist because $\displaystyle\lim_{z\to-1}\frac{\text{Arg}(z)}2$ (and in fact, $\displaystyle\lim_{z\to-1}\frac{\arg(z)}2$) has two non-overlapping representations $\displaystyle\pm\frac{\pi}2$ on the Argand diagram, as hinted by F. Tomas in the comments.