问:
Moving 'eye' along a vector while focused on a target
I am struggling a little bit with my trigonometry. I am writing a little 3D simulation in which I have a character (the 'eye') which is always fixed on a target.

I want to move along a vector from any given position left or right along a vector. This would be the equivalent of walking left or right in the real world, but while always facing the target.

I have drawn a picture that I hope will help.

I think what I need to do is figure out the angle my eye to the target with respect to the x and y axis (in 2D), but am stumbling to figure out how that fits into moving along the 'blue line' in my diagram.

Thank you for any hints or guidance!

答:
First, project the target onto the ground plane if it's not already there (i.e., just ignore its $z$-component). Let the vector from the eye to the (ground) target be $\Bbb v = (\Delta x, \Delta y)$. Normalize this to a unit vector $$\Bbb u = \frac{\Bbb v}{|\Bbb v|}.$$ Then there are two unit vectors perpendicular to this on the ground plane, corresponding to stepping left or right:
$$\Bbb p_L = (-u_y, u_x)$$
$$\Bbb p_R = (u_y, -u_x).$$


问:
Evaluating $\lim_{x\to0} \frac{\cos x - \cos 3x}{\sin 3x^2 - \sin x^2}$

$$ \lim_{x\to0}\frac{\cos x-\cos (3x)}{\sin (3x^2)-\sin (x^2)} $$

Is there a simple way of finding the limit?

I know the long one: rewrite it as
$$ -\lim_{x\to 0}\frac{\cos x-\cos(3x)}{\sin(3x^2)}\cdot\frac{1}{1-\dfrac{\sin(3x^2)}{\sin(x^2)}} $$
and then find both limits in separately applying L'Hospital's rule several times. The answer is $2$.

答: Start with dividing both the denominator and nominator by $x^2$ and applying quotitent rule for limits, you'll get $$\lim_{x\to 0}\Biggl(\frac{\frac{\cos x-\cos 3x}{x^2}}{\frac{\sin 3x^2-\sin x^2}{x^2}}\Biggl)\space=\space\frac{\lim_{x\to 0}\frac{\cos x}{x^2}-\lim_{x\to 0}\frac{\cos 3x}{x^2}}{\lim_{x\to 0}\frac{\sin 3x^2}{x^2}-\lim_{x\to 0}\frac{\sin x^2}{x^2}}$$ Now remember the particular limits $$\lim_{x\to 0}\frac{\sin x}{x}=1$$ $$\lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}$$ Then we can manipulate the first expression $$\Biggl(\lim_{x\to 0}\frac{\cos x -1}{x^2}\space+\space\lim_{x\to 0}\frac{1-\cos 3x}{x^2}\Biggl)\space:\space\Biggl(\lim_{x\to 0}\frac{\sin 3x^2}{x^2}\space-\lim_{x\to 0}\frac{\sin x^2}{x^2}\Biggl)\space=\space\Biggl(-\frac{1}{2}\space+\space\frac{9}{2}\Biggl)\space:\space(3-1)\space=\space\frac{8}{2}\space:\space2\space=\space2  $$

ContentLicense: CC BY-SA 4.0


问:
The diagram shows an equilateral triangle $ADE$ inside a square $ABCD$ . What is the value of $\frac{[\Delta ADE]}{[\Delta DEC]}$ .

The diagram shows an equilateral triangle $ADE$ inside a square $ABCD$ . What is the value of $\frac{[\Delta ADE]}{[\Delta DEC]}$ .

What I Tried: Here is the diagram :-

You can see I marked the angles as required. Now let $AB = x$ . We then have :-
$$[\Delta ADE] = \frac{\sqrt{3}}{4}x^2$$
Now from here :- https://www.quora.com/What-is-the-ratio-of-sides-of-a-30-75-75-angle-triangle-without-sine-rule , I could understand and show that :- $$ED : DC : CE = \bigg(\frac{\sqrt{3} + 1}{2} : \frac{\sqrt{3} + 1}{2} : 1\bigg)$$

So let $EC = k$ , $CD = DE =  \frac{(\sqrt{3} + 1)k}{2}$ .

From here :- $$x = \frac{(\sqrt{3} + 1)k}{2}$$
$$\rightarrow k = EC = \frac{2x}{(\sqrt{3} + 1)}$$

Now, we can find area by Heron's Formula. We have :- $$s = x + \frac{x}{(\sqrt{3} + 1)}$$
$$\rightarrow s = \frac{x\sqrt{3} + 2x}{(\sqrt{3} + 1)}$$
So :- $[\Delta DEC] = \sqrt{s(s-a)(s-b)(s-c)}$

$$\rightarrow \sqrt{\Bigg(\frac{(x\sqrt{3} + 2x)}{(\sqrt{3} + 1)}\Bigg)\Bigg(\frac{x}{(\sqrt{3} + 1)}\Bigg)\Bigg(\frac{x}{(\sqrt{3} + 1)}\Bigg)\Bigg(\frac{(x\sqrt{3})}{(\sqrt{3} + 1)}\Bigg)}$$

This looks like really a complicated expression, and I really am not going to attempt to simplify this. So can anyone give me a different solution?

Thank You.

答:
If $a,b$ are sides of a triangle and $x$ the mesure of angle between them, then the area of it is $${a\cdot b \cdot \sin x\over 2}$$

We use that formula here.

We have $AD = AE =DE =DC=a $ so$$\frac{[\Delta ADE]}{[\Delta DEC]} = {{a^2\sin 60 \over 2}\over {a^2 \sin 30 \over 2}} = \sqrt{3}$$


问:
Does every event map to a Borel set?
Given a (real-valued) random variable $X$ in the probability space $(\Omega,\mathcal{F},P)$. Let $\mathcal{B}$ be the Borel sigma algebra on $\mathbb{R}$.

We know that given any $B \in \mathcal{B}$, $X^{-1}(B) \in F$. But is $X(A) \in \mathcal{B}$ for any $A \in F$?

答: No. Take your favorite example of a non-Borel set $S\subseteq \mathbb{R}$. Consider the probability space $(S,\mathcal{P}(S),P)$, where $P$ is any probability measure on $(S,\mathcal{P}(S))$ (it exists, take $P=\delta_x$ for some fixed $x\in S$). Then the function $f:S\to \mathbb{R}$ defined by $f(s)=s$ is measurable but $f(S)\notin\mathcal{B}$.

Title: Does every event map to a Borel set?


问:
How to prove $(\frac{n+1}{e})^n<n!<e(\frac{n+1}{e})^{n+1}$ without integrating method?
How to prove $$\left(\frac{n+1}{e}\right)^n<n!<e\left(\frac{n+1}{e}\right)^{n+1}$$ without integrating method? In fact we could prove this by noticing that
$$i<x<i+1\Rightarrow \ln i<\ln x<\ln (i+1),$$ then integrating it.</p>

ContentLicense: CC BY-SA 3.0

Body:
I'll enhance an earlier answer of mine
and show that
$e(n/e)^n < n! < e^{3/2}\sqrt{n}(n/e)^n$.

This is better than OP's request
since
$((n+1)/e)^n  < e(n/e)^n$
is the same as
$(1+1/n)^n < e$
and
$e^{3/2}\sqrt{n}(n/e)^n
<e((n+1)/e)^{n+1}$
is the same as
$e\sqrt{e/n}<(1+1/n)^{n+1}$
which is more than true since
$(1+1/n)^{n+1} > e$
(proofs available upon request).

The only result from analysis we need is
 $z-z^2/2 < \ln(1+z) < z$
for $0 < z < 1$.

We first bound
$H_n = \sum_{k=1}^{n} 1/k$.

$\begin{align}
H_n &= \sum_{k=1}^{n} 1/k\\
&> \sum_{k=1}^{n} (\ln(1+1/k))\\
&= \sum_{k=1}^{n} (\ln(k+1)-\ln(k))\\
&= \ln(n+1)\\
\end{align}
$

and

$\begin{align}
H_n &= \sum_{k=1}^{n} 1/k\\
&< \sum_{k=1}^{n} (\ln(1+1/k) -1/(2k^2))\\
&= \sum_{k=1}^{n} (\ln(k+1)-\ln(k)) - (1/2)\sum_{k=1}^{n} 1/k^2\\
&= \ln(n+1)- (1/2)\sum_{k=1}^{n} 1/k^2\\
\end{align}
$.

To bound the right-hand sum

$\begin{align}
\sum_{k=1}^{n} 1/k^2
&=1+\sum_{k=2}^{n} 1/k^2\\
&<1+\sum_{k=2}^{n} 1/(k(k-1))\\
&=1+\sum_{k=2}^{n} (1/(k-1)-1/k)\\
&=1+1-1/n\\
&< 2\\
\end{align}
$

so
$\ln(n+1)< H_n< \ln(n+1)+1$.

We start with
$\ln(n!) =
\sum_{k=1}^n \ln k$
and estimate $\ln k$.

$\begin{align}
(x+1)\ln(x+1)-x \ln x
&= x(\ln(x+1)-\ln(x))+\ln(x+1)\\
&=x\ln(1+1/x)+\ln(x+1)\\
\end{align}
$

so $ (x+1)\ln(x+1)-x \ln x -\ln(x+1)=x\ln(1+1/x)$.

Using this,
$x\ln(1+1/x)
<x(1/x) = 1
$
and
$x\ln(1+1/x)
>x(1/x-1/(2x^2)
=1-1/(2x)
$
so
$1-1/(2x) < (x+1)\ln(x+1)-x \ln x -\ln(x+1) < 1$.
Note that this is just an approximate form of
$\int \ln x\,dx = x \ln x - x$
or $\ln x = (x \ln x)' - 1$.

Summing for $x$ from 1 to $n-1$,
$n-1-\sum_{k=1}^{n-1} 1/(2k) < \sum_{x=1}^{n-1}
\big((x+1)\ln(x+1)-x \ln x -\ln(x+1)\big) < n-1
$
or
$n-1-H_{n-1}/2 < \sum_{x=1}^{n-1}
\big((x+1)\ln(x+1)-x \ln x -\ln(x+1)\big) < n-1
$

Since the left part of the sum is telescoping
and the right part gives $\ln(n!)$,
$n-1-H_{n-1}/2 < n \ln n -\ln(n!) < n-1$.

Using the right side,
$\ln(n!) > n \ln n-n+1$
or
$n! > e(n/e)^n$.

Using the left side,
$\ln(n!)
< n \ln n-n+1+H_{n-1}/2
< n \ln n-n+1+(\ln(n)+1)/2
$
or
$n! < e^{3/2}\sqrt{n}(n/e)^n$.

By taking more terms of
the expansion of $\ln(1+1/z)$,
we can get the
asymptotic series for $\ln(n!)$,
which is Stirling's formula without the constant.


答:
By Area of triangle using trigonometry

Area (DEC) $ = \frac12 (CD)(DE) \sin (30^\circ) = \frac 14 x^2$

Body:
The diagram shows an equilateral triangle $ADE$ inside a square $ABCD$ . What is the value of $\frac{[\Delta ADE]}{[\Delta DEC]}$ .

What I Tried: Here is the diagram :-

You can see I marked the angles as required. Now let $AB = x$ . We then have :-
$$[\Delta ADE] = \frac{\sqrt{3}}{4}x^2$$
Now from here :- https://www.quora.com/What-is-the-ratio-of-sides-of-a-30-75-75-angle-triangle-without-sine-rule , I could understand and show that :- $$ED : DC : CE = \bigg(\frac{\sqrt{3} + 1}{2} : \frac{\sqrt{3} + 1}{2} : 1\bigg)$$

So let $EC = k$ , $CD = DE =  \frac{(\sqrt{3} + 1)k}{2}$ .

From here :- $$x = \frac{(\sqrt{3} + 1)k}{2}$$
$$\rightarrow k = EC = \frac{2x}{(\sqrt{3} + 1)}$$

Now, we can find area by Heron's Formula. We have :- $$s = x + \frac{x}{(\sqrt{3} + 1)}$$
$$\rightarrow s = \frac{x\sqrt{3} + 2x}{(\sqrt{3} + 1)}$$
So :- $[\Delta DEC] = \sqrt{s(s-a)(s-b)(s-c)}$

$$\rightarrow \sqrt{\Bigg(\frac{(x\sqrt{3} + 2x)}{(\sqrt{3} + 1)}\Bigg)\Bigg(\frac{x}{(\sqrt{3} + 1)}\Bigg)\Bigg(\frac{x}{(\sqrt{3} + 1)}\Bigg)\Bigg(\frac{(x\sqrt{3})}{(\sqrt{3} + 1)}\Bigg)}$$

This looks like really a complicated expression, and I really am not going to attempt to simplify this. So can anyone give me a different solution?

Thank You.


问:
A question on a $6 \times 6 $ table(Olympiad question)
There are some mines in some cells of a $6 \times 6$ table. Among all $2 \times 2 $ squares, exactly $n$ squares contain an odd number of mines and the rest contain an even number of mines. What is the possible values of $n$ ?

Here is my try to solve this question :
For convenience I did it for a $ 4 \times 4 $ table which contains nine, $2\times 2$ squares. In this case suppose the following coordinates are mines : $(1,1) , (1,3) , (3,1) , (3,3)$, in this case all $9$ squares contain a mine, now we could reduce their number one by one forexample at first we add a mine in the $(4,4)$ square, so the number of $2\times 2$ square in which there is the number of odd mines is $8$. In the next step we'll put a mine in the square with $(4,3)$ coordinate in this case two of squares contain the even number of mines, and seven squares contain the number of odd mines.

I think we could continue above solution for solving this question in general, but I'm a little confused. I hope you can help me.

答:
Divide the table into four $3\times3$ squares. Each contains four $2\times2$ squares. It is easy to place mines such that precisely $0$, $1$, $2$, $3$ or $4$ of these squares contain an odd number of mines, and none of the remaining $2\times2$ squares contain a mine. This way you get all $n\leq16$.

For $n\geq16$ place mines at the squares that have both coordinates even. This places $9$ mines in a square grid in such a way that every $2\times2$ square contains precisely one mine; omit some of the corner mines to get any number $n\geq16$.


问:
Function depending on the empirical measure
Let $X_1, \ldots, X_n$ be $\mathbb{R}^d$-valued random variables on the probability space $(\Omega, \mathcal{F}, P)$ and let $\delta_c$, $c \in \mathbb{R}^d$, denote the Dirac measure. Consider the emirical measure defined by
$$
\mu := \frac{1}{n} \sum_{k=1}^n \delta_{X_{k}},
$$
so that for every fixed $\omega \in \Omega$ and a measurable set $A$ we get
$$
\mu(A) = \frac{1}{n} \sum_{k=1}^n \delta_{X_k(\omega)} ( A ) = \frac{1}{n} \sum_{k=1}^n 1_A (X_k(\omega)).
$$
Now assume that an $\mathbb{R}^d$-valued function $G$ is known to depend on $\mu$ which allows for the notation $G(\mu)$.

Why is it justified to have $$G(\mu)= \frac{1}{n} \sum_{k=1}^n X_k, \tag{1}$$ i.e. can $(1)$ be expressed via the measure $\mu$ (for example, through some integrals w.r.t. $\mu$ for every fixed $\omega \in \Omega$)?

If $d=1$, then for every fixed $\omega$ we can write
$$
\int_{\mathbb{R}} x \, d \left( \frac{1}{n} \sum_{k=1}^n \delta_{X_k(\omega)} \right) = \frac{1}{n} \sum_{k=1}^n \int_{\mathbb{R}} x \, \delta_{X_k(\omega)} = \frac{1}{n} \sum_{k=1}^n X_k(\omega) = G(\mu(\omega)).
$$
Can something similar be done for general $d$? Is there maybe an alternative representation?

答:
Well, I'm rather confused by your notation. So I assume that you want to know why can we find  (roughly speaking) a function $G$ defined on space of measures such that $G(\mu) = \frac{1}{N} \left( \sum_{n} X_n \right)$.

And to be honest, it is not anything mysterious and I guess you also knew it. It is just linear functional on space of measures.

To be more precise, for any bounded function $g$ , we define:

$G: \mathcal{M}_f\left( \mathbb{R}^d \right) \longrightarrow \mathbb{R}$

$ \mu \mapsto \int g d\mu$

where $\mathcal{M}_f\left( \mathbb{R}^d \right)$ is the space of all finite measures on $\mathbb{R}^d$

Remark: There are some useful properties of $G$ that you can see. $G$ is clearly continuous wrt to strong norm in $\mathcal{M}_f$, and if $g$ is continuous, $G$ is also continuous wrt weak topology on $M_f$.etc.

If $g$ is not bounded, $G$ is just a linear operator defined only on some subspace of $\mathcal{M}_f$, but you can always adjust $g$ for your needs.


问:
Convergence of maximum points on compact sets
Suppose $f$ and $f_n$ ($n=1,2,\dots$) are continuous functions $[0,1]\rightarrow\mathbb{R}$, achieving their maxima at unique points $x^*$ and $x^*_n$ respectively. If $f_n(x)\rightarrow f(x)$ for all $x\in [0,1]$, do we also have $x^*_n\rightarrow x^*$?

Here are a few thoughts on the problem:

One can ask a similar question when $[0,1]$ is replaced by $[0,\infty).$ In that case the answer is negative, as the maximum may escape to infinity. This behaviour is prohibited on $[0,1]$ due to compactness, but of course something else might go wrong.

If $f_n\rightarrow f$ uniformly, then the answer is positive; indeed, say (after selecting a subsequence) $x_n^*\rightarrow y \neq x^*$. Then $\vert f_n(x_n^*) - f(y) \vert \le \vert f_n(x_n^*)-f(x_n^*)\vert + \vert f(x_n^*) - f(y) \vert \le \Vert f - f_n \Vert_{L^\infty[0,1]} + \vert f(x_n^*) - f(y) \vert \rightarrow 0$ and thus taking limits in the inequality $f_n(x^*) \le f_n(x_n^*)$ yields $f(x^*) \le f(y)$. This is a contradiction, as $f$ is assumed to achieve its maximum at a unique point.

答:
Hint

What about $f(x) = x$ and $f_n = \sup(f, 2g_n)$ where

$$g_n(x)=
\begin{cases}
nx & 0 \le x \le 1/n\\
2-nx & 1/n \le x \le 2/n\\
0 & x \ge 2/n
\end{cases}$$


答:
$DC=x$ and the altitude of $\Delta DEC$ from $E$ is $\frac{x}{2}$. This gives us
$$[\Delta DEC] = \frac{x^2}{4}$$
And the result
$$\frac{[\Delta ADE]}{[\Delta DEC]}=\sqrt{3}$$

Title: The diagram shows an equilateral triangle $ADE$ inside a square $ABCD$ . What is the value of $\frac{[\Delta ADE]}{[\Delta DEC]}$ .


问:
Deriving A implies B from Not A
My logic textbook has the following example showing how to derive $A \to B$ from $\neg A$:

First we assume $A$ and use the conjunction introduction rule which results in a contradiction $[A] \land \neg A$. Now we can use the negation introduction rule to get $\neg \neg B$. With the negation elimination rule we get $B$ and after using the implication introduction rule we have $A \to B$.

However, I'm confused when it comes to using the negation introduction rule to get $\neg \neg B$ from $A \land \neg A$. While $A \land \neg A$ is clearly a contradiction resulting from our (temporary) assumption $A$, shouldn't this result the opposite of our assumption, that is $\neg A$, being true? Where does the $\neg \neg B$ come from? Quoting my textbook: "When we have derived a contradiction $B \land \neg B$ from the temporary assumption $A$, we consider $\neg A$ proved."

答:
A negation introduction rule often means that from a contradiction we can infer the negation of the proposition that we've assumed.  So, I think your textbook aims at some proof like this (note the difference with amWhy's proof):

1 $\lnot$ A assumption

2 |  A assumption

3 || $\lnot$ B assumption

4 || (A $\land$ $\lnot$ A) 1, 2 conjunction introduction

5 | $\lnot$ $\lnot$ B 3-4 negation introduction

6 | B 5 negation elimination

7 (A $\implies$ B) 3-6 conditional introduction

The $\lnot$ $\lnot$ B comes from assuming $\lnot$ B, deriving a contradiction, and then using negation introduction.

Title: Deriving A implies B from Not A


问:
Convergent sequence in Range has convergent premimage for a surjective bounded operator
Let $T:X\rightarrow Y$ be a bounded linear surjective operator where $X$ and $Y$ are Banach spaces and $y\in Y$. Then for $\{y_n\}$ converging to $y$,  $\exists \{x_n\}$ converging to $x$ s.t. $T(x)=y$.

Edit: I meant there exists a convergent $\{x_n\}$ such that $T(x_n)=y_n$ and $x_n\rightarrow x$ as $n \rightarrow \infty$.

答:
Its an application of the open mapping theorem. In particular a useful proposition is the following:

If $T:X\to Y$ is a surjection bounded linear operator between the two Banach space $X,Y$, then, there is exists a constant $M>0$ such that for every $y\in Y$ there is some $x\in X$ with $Tx=y$ and $||x||\leq M||y||$.

To prove this, from the open mapping theorem there is some $M'>0$ such that $M'B_Y\subseteq T(B_X)$ (i.e. $0$ belongs to the interior of $T(B_X)$), where $B_X=\{x\in X:\ ||x||\leq 1\}$ denotes the closed unit ball of $X$. Equivalently, $$\tag{1}B_Y\subseteq \frac{1}{M'}T(B_X)$$

Now pick $y\neq 0$ in $Y$. Then $y/||y||\in B_Y$, so from $(1)$ there is some $x'\in B_X$ such that $$\frac{1}{M'}Tx'=\frac{y}{||y||}$$

Then $T(\frac{||y||x'}{M'})=y$. Setting $x=\frac{||y||x'}{M'}$ we have $Tx=y$ and $||x||\leq \frac{1}{M'}||y||$. So if we pick $M=1/M'$ we have proved the above proposition. Now its easy to prove what you want since if $y_n\to y$. Then, fix some $x\in X$ with $Tx=y$.

Then use the proposition for the element $y_n-y$ and find $z_n\in X$ such that $Tz_n=y_n-y$ and $$\tag{2}||z_n||\leq M||y_n-y||$$

Then by $(2)$ it follows that $z_n\to 0$. Take $x_n=z_n+x$.


问:
Definition of finite dimensional representation of a compact group
Consider the following fragment:

One asks that the map $\pi: G \to GL(V)$ is continuous. What does this mean? If $V$ is finite-dimensional, then we can view $GL(V)$ as matrices and then I guess it means that all the matrix entries are continuous?

What does continuity mean when $V$ is a Banach space (not necessarily finite-dimensional)? Which topology is put on $GL(V)$? Is it pointwise convergence topology?

答:
To actually answer the question "What does continuity mean when $V$ is a Banach space" we need something stronger, that your text does not seem to mention:

We require that the group action of $G$ on $V$ defined by

$$G \times V \to V: (g, v) \mapsto \pi(g)v$$ is continuous. Here, $G \times V$ has the product topology.


问:
Show that $T:=\{(\cos(t),\sin(t)):t\in[0,2\pi)\}$ is Lebesgue measure zero
Show that $$T:=\{(\cos(t),\sin(t)):t\in[0,2\pi)\}$$ is Lebesgue measure zero.

I've got this problem and I know that i can solve this by showing the measure of 2 circle of radius $1
+\epsilon,1-\epsilon$ but I don't know how to prove this by taking a bounding countable sets and bounded countable sets.

I had an idea of taking the interval $[-1,1]\times[-1,1]$ and divide it by $4n^2$ equal squares with $1/n$ segment. and then Showing the amount of squares which covers the sphere will be some constant times $n$.

Any help will be appreciated

答:
Hint: We know that:

If $f: [a,b] \to \mathbb{R}$ is continuous function, so its graph has measure zero.

If you have problem with prove that, you can say me and I will write the prove about that fact.

And, now you can divide the circle into two genuine functions, and since the union of two sets of Lebesgue measure zero still has Lebesgue measure zero, so you can prove what $T$ is Lebesgue measure zero.


问:
No proper dense open subsets of $\mathbb{R}^n$?
Let $U$ be a dense open subset of $\mathbb{R}^n$, a positive integer, and suppose that $X:=\mathbb{R}^n-U$ is $0$-dimensional; in the sense of the Lebesgue covering dimension.  Then is $U$ homeomorphic to $\mathbb{R}^n$ if and only if $U=\mathbb{R}^n$?

答:
You are right.
Clearly, if $U=\mathbb{R}^n$, $U$ is a dense open subset of itself.

For the other implication, let us assume $\exists U\subsetneq \mathbb{R}^n$, a dense open subset of $\mathbb{R}^n$, with $V:=\mathbb{R}-U$ $0$-dimensional. Since $V$ is $0$-dimensional, for any $\mathbf{y}\in V$, $B^n_{\epsilon}(\mathbf{y})\cap U\neq \emptyset,\;\forall \epsilon>0$ (Where $B^n_\epsilon(\mathbf{y})$ is $n$-dimensional ball, centered in $\mathbf{y}$ and radius $\epsilon$,).

Let, $$x_0\in \bigcap_{\epsilon >0} B^n_{\epsilon}(\mathbf{y})\cap U,$$
Then, since $U$ is an open set, the  $\exists\delta >0\;|\;B_{\delta}^n(\mathbf{x}_0)$ is contained in $U$. But, by construction, $\mathbf{y}\in B_{\delta}^n(\mathbf{x}_0)$, for any $\delta >0$. This is a contradiction, since we assumed $\mathbf{y}\notin U$.


问:
Question about isomorphisms
Let $A=\operatorname{span}\big(\begin{bmatrix} 1 \\ 0\end{bmatrix}, \begin{bmatrix} 0 \\ 1\end{bmatrix}, \begin{bmatrix} h \\ 0\end{bmatrix}\big)$ and $B=\operatorname{span}\big(\begin{bmatrix} 1 \\ 0\end{bmatrix}, \begin{bmatrix} 0 \\ 1\end{bmatrix}\big)$, then are $A$ and $B$ isomorphic?

I think since $\dim(A)=2$ and $\dim(B)=2$, then $A$ and $B$ are isomorphic. I don't know if this is true for $h=0$ because if we have the linear combination $a_1\begin{bmatrix} 1 \\ 0\end{bmatrix}+a_2\begin{bmatrix} 0 \\ 1\end{bmatrix}+a_3\begin{bmatrix} h \\ 0\end{bmatrix}=\begin{bmatrix} 0 \\ 0\end{bmatrix}$, then $a_3$ is not necessarily zero if $h=0$.

So $A$ and $B$ are isomorphic for $h\neq 0$?

Please give some hint or ideas.

答: Yes, they are isomorphic since, for any $h$, $\left[\begin{smallmatrix}h\\0\end{smallmatrix}\right]$ is a linear combination of the other two vectors.

Title: Question about isomorphisms


问:
How to prove that the expectation of Cox Ingersoll Ross interest rate model is $r_0e^{-at} + b(1-e^{-at})$?
The Cox Ingersoll Ross is given in the form:

$$dr_t = a(b-r_t)dt + \sigma \sqrt{r_t}dW_t$$

According to Wikipedia, the expected value of it i.e $E[r_t|r_0]$ is

$$r_0e^{-at} + b(1-e^{-at})$$

but I don't really understand how it's derived. Can you give some hints?

答:
Hint

Remark that

$$\mathbb E[r_t]=\mathbb E[r_0]-a \int_0^t\mathbb E[r_s]\,\mathrm ds+abt$$

and use Gronwall inequality.


问:
A mathematical model for rotations of a die
I have a normal 6-sided die and I would like to find a model for how it transforms under rotatins (north, south, east, west), that can help me determine which side is up. Just to make sure we are using the same die, here's how mine is numbered:

To give a few examples, when the die is in that position (1 facing up, 2 facing at me), if I rotate it south it will end up with 5 facing up and 1 facing me. Similarly, a rotation to the west gives 3 facing up and 2 facing at me.

I tried representing the die and its rotations as a graph where each node is one of the states the die can be in and there are four transitions for every node (corresponding to each of the four rotations). The problem is that there is a total of $24$ states and $4 * 24 = 96$ transitions and writing down all of them is laborious and very error prone.

The reason there are $24$ states is because when 1 is facing up and 2 is facing at me is quite different from when 1 is facing up and 3 is facing me. So a state of the die is a 2-tuple of the form $(\text{number facing up},\ \text{number facing me})$. (There are $6$ numbers that can face up and for each of them there are are $4$ numbers that can face me, so $24$.)

I need some help in finding a nicer mathematical model for determining what face is up, given a starting position and a sequence of rotations.

答: Look at section 2 of these lecture notes.  What you are looking for is a faithful group action of $S_4$ on the cube.</p>


问:
What are the equations for the modeling?
We have a system of spring-mass-spring-mass-spring-mass and required to write the motion equation of the described system.

If I apply hooke's law to the system I got the following :

Are these the correct equation form?

答:
if you apply Newton's second law on $x_1,x_2,x_3$, you should get (with $l_0$ the caracterisitc length of the spring) (and neglecting gravity)  :

$$m\ddot{x_1}=-k(2x_1-x_2)\\m\ddot{x_2}=-k(2x_2-x_3)\\m\ddot{x_3}=-k(x_3-x_2-l_0) $$


答:
Note that a basis for $B$ is $$\beta_{B}=\left\{\begin{pmatrix} 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \end{pmatrix} \right\} \implies \dim(B)=2$$

also you can see that the set $$\left\{\begin{pmatrix} 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \end{pmatrix} , \begin{pmatrix} h \\ 0 \end{pmatrix} \right\}$$ is dependent linearly so, the set $\beta_{A}$ is a basis for $A$ where $$\beta_{A}=\left\{\begin{pmatrix} 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \end{pmatrix} \right\} \implies \dim(A)=2$$
Therefore, $A$ is isomorphic to $B$ forall $h \in \mathbb{R}$.

Note: You can remove (delete) linearly dependent vectors from a span, and the set remains the same as before removing the linearly dependent vector. This can be formally proven.

Title: Question about isomorphisms

Body:
Let $A=\operatorname{span}\big(\begin{bmatrix} 1 \\ 0\end{bmatrix}, \begin{bmatrix} 0 \\ 1\end{bmatrix}, \begin{bmatrix} h \\ 0\end{bmatrix}\big)$ and $B=\operatorname{span}\big(\begin{bmatrix} 1 \\ 0\end{bmatrix}, \begin{bmatrix} 0 \\ 1\end{bmatrix}\big)$, then are $A$ and $B$ isomorphic?

I think since $\dim(A)=2$ and $\dim(B)=2$, then $A$ and $B$ are isomorphic. I don't know if this is true for $h=0$ because if we have the linear combination $a_1\begin{bmatrix} 1 \\ 0\end{bmatrix}+a_2\begin{bmatrix} 0 \\ 1\end{bmatrix}+a_3\begin{bmatrix} h \\ 0\end{bmatrix}=\begin{bmatrix} 0 \\ 0\end{bmatrix}$, then $a_3$ is not necessarily zero if $h=0$.

So $A$ and $B$ are isomorphic for $h\neq 0$?

Please give some hint or ideas.


问:
Conditional expectation and random variables
Let $X, Y$ be independent Bernoulli random variables with parameter $p$. Let $A=\{\omega \in \Omega: X(\omega)+Y(\omega) = 0\}$. How does one compute $E[X\cdot\mathbb{1}_{A^{c}}]$? My guess is that since $X$ can be both $0$ and $1$ and the sum could still be bigger than $0$, the result is simply $0\cdot(1-p)+1\cdot p=p$. Is this correct?

答:
Hint

$$\mathbb E[X\boldsymbol 1_{A^c}]=\mathbb E[X]-\mathbb E[X\boldsymbol 1_A]$$

and $$\mathbb E[X\boldsymbol 1_A]=\mathbb P\{X=1,X+Y=0\}=0.$$


问:
What is the probability of randomly picking x amount items out of 2 sets, and the x items being of only one set.
Suppose you are choosing 4 people for a job. There are 5 males and 10 females. How would one go about calculating the probability that all 4 people are of the same gender. I started with the basic Combination without repeats formula but quickly understood that this wont work. I assume that combinations are the way to go here?</p>

答:
You can still use combination, but in a proper sort of usage. Obviously choosing 4 people out of 15 is possible in $\binom{15}{4}$ ways. If they are all from same gender, they should be all males or females, totally in
$\binom{10}{4}+\binom{5}{4}$ different ways. The correct answer would be
$$
\binom{10}{4}+\binom{5}{4}\over \binom{15}{4}
$$


问:
Prove that the number of prime numbers is infinite.
Prove that the number of prime numbers is infinite

I want to solve it with this theorem :

For n, a positive integer, and integers i,j with 1≤i<j≤n, we know that gcd ($n!$ $\times$ $i+1$ , $n!$$\times$ $j+1$) $ =1$.

.

The theorem has been proven.But I don't know how to use this theorem to prove my question.

答:
Let us use a proof by contradiction. Let us assume $P$ is finite, let $n=\max(P)$, you can see that $n!+1$ is prime. In fact, you can see that each prime number can't divide $n!+1$ (because they divide $n!$, if one would divide $n!+1$, it would divide $n!+1-n!=1$.

So $n!+1$ is prime and $n!+1>n $, so here you have your contradiction


问:
How to split large numbers easily for factoring the middle term?
I was doing a problem when I got suddenly stuck with a middle term factorisation.

The equation is as follows:

$$x^4-289x^2+14400=0.$$

I managed to somehow solve it, but it took a lot of time.

Can you please help me with a fast way if any, or can you at least tell me what is the fastest method you know?

Thank you.

答:
A direct method would be to let $y=x^2$,

and solve $y^2-289y+14400=0$ using the quadratic formula.


答:
The function $X \cdot 1_{A^c}$ takes on values in $\{0,1\}$, so $\mathbb{E}X1_{A^c} = \mathbb{P}[ X1_{A^c} = 1 ]$. The latter event holds iff $X=1$ and $X+Y > 0$. If $X=1$ then $X+Y >0$ always, so that
$$ \mathbb{P}[ X1_{A^c} = 1 ] = \mathbb{P}[ X = 1 \wedge X+Y >0 ] = \mathbb{P}[X+Y>0 | X=1] \cdot \mathbb{P}[X=1] = \mathbb{P}[X=1] =p.  $$

Title: Conditional expectation and random variables

Body: Let $X, Y$ be independent Bernoulli random variables with parameter $p$. Let $A=\{\omega \in \Omega: X(\omega)+Y(\omega) = 0\}$. How does one compute $E[X\cdot\mathbb{1}_{A^{c}}]$? My guess is that since $X$ can be both $0$ and $1$ and the sum could still be bigger than $0$, the result is simply $0\cdot(1-p)+1\cdot p=p$. Is this correct?


问:
Are weak homotopy equivalences always homotopic?
I was wondering whether two weak equivalences between spaces $X$ and $Y$, are homotopic, or not. If yes, a proof / reference would be very welcome. If not, a counterexample would be interesting!
More generally, I am interested in the case of any category with a notion of weak equivalences / homotopies between weak equivalences, but I think the archetypal case of topological spaces is a good way to start.

Have tried a few things so far, but nothing worth mentioning yet.

答: For a very simple counterexample, take $X$ and $Y$ to be discrete spaces of the same cardinality.  Then any bijection between them is a homotopy equivalence (indeed, a homeomorphism), but there are no nontrivial homotopies of maps $X\to Y$.


问:
A combinatorial sum

Any idea to evaluate the sum
$$
\sum_{j=m}^{k}\frac{\binom{m}{2m - j\,\,}}{\binom{k}{j}}
\quad\mbox{with}\quad m \leq k < 2m - 1.
$$

I have found the sum for $k=2m-1$. In fact, it is verified that
$$
\sum_{j = m}^{2m - 1}\ \frac{\binom{m}{2m-j\,\,}}
{\binom{2m - 1\,\,}{j}} = 2m\left(H_{2m} - H_{m}\right),
$$
where $H_{j}$ is the $j$-th harmonic number.

答: Here maxima's Zeilberger package (take a peek at Petkovsek, Wilf, Zeilberger "A = B" for the gory details) says your sum isn't Gosper summable. That means there is no simple closed form for it.</p>

Title: A combinatorial sum


问:
Introduction to Abstract Harmonic Analysis for undergraduate background
I'm looking for a good starting book on the subject which only assumes standard undergraduate background.

In particular, I need to gain some confidence working with properties of Haar measures, so I can better understand the spaces $L^{p}(G)$ for a locally compact group $G$.

For some perspective on what I currently know, I can't yet solve this problem:  Verifying Convolution Identities

Something with plenty of exercises would be ideal.

答:
For a historical view, see

Mackey, George W. Harmonic analysis as the exploitation of symmetry—a historical survey. Bull. Amer. Math. Soc. (N.S.) 3 (1980), no. 1, part 1, 543–698.


答:
$\gcd(n!+1, n!\times 2 + 1)=1$  but neither $n!+1$ nor $n!\times 2 + 1$ is divisible by any prime less than or equal to $n$.

so there must always be two primes larger than any $n$-- one that divides $n!+1$ and another to divide $n!\times 2 + 1$.

ANd as $n$ is unbounded so are the primes larger than $n$.

.....

But you didn't really need the awkward theorem.

Its enough to point out that $n!+1$ is not divisible by any prime less than or equal to $n$

......

But even that is overkill.  The classic way, the way Euclid did it, is that if you have any finite list of primes, $A$, then $1+\prod_{p\in A}p $ is not divisible by any prime in the list so there must be primes not on the list so no finite list is complete.

Sometimes the classics are best.

Title: Prove that the number of prime numbers is infinite.

Body:
Prove that the number of prime numbers is infinite

I want to solve it with this theorem :

For n, a positive integer, and integers i,j with 1≤i<j≤n, we know that gcd ($n!$ $\times$ $i+1$ , $n!$$\times$ $j+1$) $ =1$.

.

The theorem has been proven.But I don't know how to use this theorem to prove my question.


问:
How to prove a GCD question?
Prove for all positive integers d, m, n. If d=gcd(m, n) then gcd(m, nk)=gcd(m, dk).

So I messed around with it, and I started from the conclusion. I made gcd(m, nk)=x, gcd(m, dk)=y. Then by Bezout's Lemma ms+nkt=x, ms+dkt=y, s and t are integers. I think I can make these 2 equations and change it to the form of d=gcd(m, n). Maybe by dividing? But If I divide it by d, then ms/d+nkt/d=x/d, then this equation might not be integers anymore. I don't know that the next step is. Thank you for looking this over for me.

答:
Hint: Use the GCD in prime factorization form.

If $m = \prod p_i ^ {m_i}, n = \prod p_i^{n_i}$, then $ d = \prod p_i^{\min(m_i, n_i)}$.

The result follows by showing that

$$ \min (m_i, n_i + k_i) = \min (m_i, \min (m_i, n_i) + k_i)$$

 A direct way of showing this is to condition on $ m_i \geq n_i$ (so $\min(m_i, n_i) = n_i$) and $ m_i < n_i$ (so both sides equal $m_i$).

Title: How to prove a GCD question?


问:
Number Theory : Find the group of A such that $A=\{x \in \mathbb{Z}:\frac{x^3-3x+2}{2x+1}\in \mathbb{Z}\}$
Find the group A such that  $A=\{x \in \mathbb{Z}:\frac{x^3-3x+2}{2x+1}\in \mathbb{Z}\}$ ?

Polynomial Long Division we get
$\frac{x^{2}}{2}-\frac{x}{4}-\frac{11}{8}+\frac{27}{8\left(2x+1\right)}$

but how i can from here find all x such that $x \in \mathbb{Z}$ ??

i did use the hint for $y = 2x+1$ so that  $x =\frac{y-1}{2}$

made the equation to be $\frac{(y-1)^3}{8}-\frac{3(y-1)}{2}+2=\frac{y^3-3y^2-9y+27}{8}$

$\frac{\frac{y^3-3y^2-9y+27}{8}}{y}=\frac{y^3-3y^2-9y+27}{8y}$

I can not understand with the help of the clues, can some 1 some formal proof ?

答: Hint: Since $2x+1$ is odd for all integers $x$, we have that $\frac{x^3-3x+2}{2x+1}\in\mathbb Z$ if and only if $\frac{2(x^3-3x+2)}{2x+1}\in\mathbb Z$.


问:
Show that the determinant of the matrix $E$ is $0$
Let
$E=\begin{pmatrix}
(1-n)e_1 & e_2 & ... & e_n\\
e_1 & (1-n)e_2 & ... &e_n\\
. & . & . &.\\
. & . & . &.\\
e_1 & e_2 & ... & (1-n)e_n
\end{pmatrix}$
an $n \times n$ matrix. I need to prove that the determinant is zero. I thought about using guassian elimination, and somehow $e_{nn}$ (position) will lead to zero. So, as a triangular matrix with a zero in the diagonal means $Det(E)=0$, then it is solved.

However, I would like to know if there's a better way to prove it, or a hint, and if what I thought is correct.

答: HINT: If you sum all the rows, which is a non-trivial linear combination of them, you get... And this would imply the determinant is zero because any non-trivial linear combination of the rows of an invertible (i.e. non-zero determinant) matrix cannot be this value.


答:
Dym and McKean's book Fourier Series and Integrals is very well-written, begins with standard undergraduate harmonic analysis on $\mathbb{S}^1$ and $\mathbb{R}$, and devotes the latter half to Fourier analysis on finite groups and some good examples of compact groups ($SO(3)$ in particular).

Since I have not yet read it (it was recommended to me by my advisor, so I am in the process of acquisition), I am not sure at what level Helgason's book Topics in Harmonic Analysis on Homogeneous Spaces is written. You might consider checking it out to see if you can follow his arguments, or at least understand statements of results.

Title: Introduction to Abstract Harmonic Analysis for undergraduate background

Body:
I'm looking for a good starting book on the subject which only assumes standard undergraduate background.

In particular, I need to gain some confidence working with properties of Haar measures, so I can better understand the spaces $L^{p}(G)$ for a locally compact group $G$.

For some perspective on what I currently know, I can't yet solve this problem:  Verifying Convolution Identities

Something with plenty of exercises would be ideal.


问:
Relation between (the $2 \pi$ in) Gauss-Bonnet and in Cauchy's differentation formula
When I 1st saw Gauss-Bonnet, I was wondering if this $2 \pi$ had any relation with the $2 \pi$ in Cauchy's differentiation formula. Maybe a better question is to ask about the relation between Gauss-Bonnet and Cauchy's differentiation formula. I recall the $2 \pi$ (or $\tau$) in Cauchy's differentiation formula is to do with homotopy with a circle in the original Cauchy's integral formula. I guess Gauss-Bonnet will have something to do with circles or spheres. So far I know only some basic version of Gauss-Bonnet and not yet its generalisations.

Okay so not exactly any question yet ummm...

Where does the $2 \pi$ come from in Gauss-Bonnet? Is it something to do with circles or spheres (Or $S^n$)?

Is the $2 \pi$ in Gauss-Bonnet related to the $2 \pi$ in Cauchy's differentiation formula?

What's the relation of Gauss-Bonnet with Cauchy's differentiation formula?

答:
The Gauss-Bonnet theorem can be seen as a limiting continuous case of a simpler theorem, to do with surfaces made out glued-together polygons. To each point $p$ on such a surface, we can measure the angle $\theta(p)$ "visible" around $p$. If $p$ lies on a face or an edge we have $\theta(p) = 2\pi$, but on the corners things are different: if $p$ lies on the corner of a cube then $\theta(p) = 3 \pi / 2$ for example, while a point on the corner of a tetrahedron would have the visible angles adding to $\theta(p) = \pi$. We could also imagine gluing together "too many" polygons along their edges, to get a point with $\theta(p) > 2 \pi$.

Define the angle defecit at a point to be $\delta(p) = 2\pi - \theta(p)$, so $\delta(p) = 0$ for everything except corner points, where it measures "deviation" from being flat. Then we have a discrete analogue of the Gauss-Bonnet theorem: for any polygonal surface $S$,
$$ \sum_{p \in S} \delta(p) = 2 \pi \chi(S),$$
where $\chi(S)$ is the Euler characteristic of $S$. We can quickly check this for the tetrahedron for example: $\delta(p) = \pi$ at the four corner points, so we have $\sum_{p \in S} \delta(p) = 4 \pi$. On the other hand, the Euler characteristic of a tetrahedron is $2$, since it is homeomorphic to a sphere. Of course the wonderful thing is that no matter how you assemble polygons to make a (orientable) surface, the sum of the angle defecits only depends on the underlying topology. You can read more about this in the wonderful book Mostly Surfaces by Richard Evan Schwartz.

The differential geometry version of Gauss-Bonnet is some kind of limiting version of the above: the Gaussian curvature $K$ is the infinitesimal version of the angle defecit, and we have
$$ \int_{S} K \, dA = 2 \pi \chi(S).$$
The interpretation of the $2 \pi$ here being an angle also makes sense from dimensional analysis: area has units of $[\mathrm{length}]^2$ while Gaussian curvature has units of $[\mathrm{length}]^{-2}$, so the integral of one over the other should be dimensionless. (A good rule of thumb is to always assume that a $2 \pi$ appearing in a dimensionless quantity is an angle).

As for Cauchy's integral formula, I feel that the spirit of the question is the following: why is it true that $\oint_C \frac{dz}{z} = 2 \pi i$ for any closed curve $C$ encircling the origin counter-clockwise, and where does the $2 \pi$ come from? Again the $2 \pi$ appearing is an angle, and the quantity being counted is called the winding number: if $C$ were to wrap around the origin twice, we would get $4 \pi i$. Imagine standing at the origin and watching the curve $C$ being traced out: when it closes in on itself, you may have turned around a net $n$ times, and the result of the integral will be $2 \pi n i$.

I cannot myself draw a direct connection between them further than this: The $2 \pi$ appearing in each is an angle, and they are both theorems relating complicated things (differential geometry, or contour integration) to simpler topological things (Euler characteristic, or winding number).


问:
Lebesgue measure of complement of this set
This question was part of  my real analysis quiz and I need help in correctly proving it.

For each a $\in \mathbb{R}$ , let $ S_a$ ={$(x,y,z)\in {\mathbb{R}}^3 | x^2+y^2+z^2= a^2$} Let $E=\bigcup_{a\in \mathbb{R}/ \mathbb{Q}} S_a$ Then does every open set containing $E^C$ has finite Lebesgue measure ?

I can't think why measure must be finite , even intutively I think that measure must be infinite. I am looking for a rigorious argument why measure must be finite. I am not really good in Lebesgue measure but really want to learn.

答:
(edit) I assume you meant to say that there is not open set contained in E, with positive measure, so I answered that question.

E is $\mathbb{R}^3$, except for the points with rational norm ( $||\mathbf{x}||\in \mathbb{Q}\Rightarrow \mathbf{x}\notin E$).
The idea here is that, $F:=\mathbb{R}^3-E$ is dense in $\mathbb{R}^3$ (and $\overline{F}=\mathbb{R}^3$)
Then, you can show that, there are no open sets in $E$, with non cero messure, since they would intersect with $F$.

Let me know if you need more details.


Title: Show that the determinant of the matrix $E$ is $0$

Body:
Let
$E=\begin{pmatrix}
(1-n)e_1 & e_2 & ... & e_n\\
e_1 & (1-n)e_2 & ... &e_n\\
. & . & . &.\\
. & . & . &.\\
e_1 & e_2 & ... & (1-n)e_n
\end{pmatrix}$
an $n \times n$ matrix. I need to prove that the determinant is zero. I thought about using guassian elimination, and somehow $e_{nn}$ (position) will lead to zero. So, as a triangular matrix with a zero in the diagonal means $Det(E)=0$, then it is solved.

However, I would like to know if there's a better way to prove it, or a hint, and if what I thought is correct.

Body:
First $(1)$, factor out the coefficients. Second $(2)$, add all rows to the first row:
$$\begin{vmatrix}
(1-n)e_1 & e_2 & ... & e_n\\
e_1 & (1-n)e_2 & ... &e_n\\
. & . & . &.\\
. & . & . &.\\
e_1 & e_2 & ... & (1-n)e_n
\end{vmatrix} $$$$
\overset 1=(e_1\cdots e_n)\begin{vmatrix}
(1-n) & 1 & ... & 1\\
1 & (1-n) & ... &1\\
. & . & . &.\\
. & . & . &.\\
1 & 1 & ... & (1-n)
\end{vmatrix} $$$$\overset 2 =(e_1\cdots e_n)\begin{vmatrix}
(1-n)+(n-1) & (1-n)+(n-1) & ... & (1-n)+(n-1)\\
1 & (1-n) & ... &1\\
. & . & . &.\\
. & . & . &.\\
1 & 1 & ... & (1-n)
\end{vmatrix} \\= (e_1\cdots e_n)\begin{vmatrix}
0 & 0 & ... & 0\\
1 & (1-n) & ... &1\\
. & . & . &.\\
. & . & . &.\\
1 & 1 & ... & (1-n)
\end{vmatrix}$$$$=0$$


问:
Proof of Idempotency for Matrices
I'm currently looking at achieving some proof that an $n \times n$ matrix (let's say $Y$) is idempotent if and only if $I_n-Y$ is an idempotent matrix, (with $I_n$ being the identity matrix).

I have so far proved that $I_n-Y$ is idempotent if $Y$ is idempotent as:

$(I_n-Y)^2 = I_n - 2Y + Y = I_n - Y$ hence idempotent.

But I don't know if this helps me at all and how to flip it so that if $I_n - Y$ is idempotent then $Y$ must be also idempotent.

Any help would be great!

答: $(I-Y)^2=I-Y$ $\Rightarrow$ $Y^2-2Y=-Y$ $\Rightarrow$ $Y^2=Y$ hence it is idempotent

Title: Proof of Idempotency for Matrices


答:
The situation is of course far more ridiculuous than you could possibly hope for.

For instance take the paper Every finite group is the group of self-homotopy equivalences of an elliptic space, Acta Math. 213, 1, (2014), 49-62, by C. Costoya, A. Viruel. If you have not yet guessed its content:

Every finite group can be realised as the group of self-homotopy equivalences of a simply-connected CW complex.

In fact the authors show that for any given finite group there are infinitely many such distinct complexes with the stated property, and moreover that each can be taken as the rationalisation of a simply-connected manifold. Even more details are included in the paper, and anyone interested in rational homotopy theory should enjoy skimming through it.

To include some details for the beginner, the 'group of self-homotopy equivalences' of $X$ is the set of all homotopy classes of homotopy self-equivalences $X\xrightarrow\simeq X$ given the composition product. Distinct elements of this group are distinct homotopy classes of maps. If $X$ is (homotopy equivalent to) a CW complex then there is no distinction between homotopy self-equivalence and weak homotopy self-equivalences.

Title: Are weak homotopy equivalences always homotopic?

Body:
I was wondering whether two weak equivalences between spaces $X$ and $Y$, are homotopic, or not. If yes, a proof / reference would be very welcome. If not, a counterexample would be interesting!
More generally, I am interested in the case of any category with a notion of weak equivalences / homotopies between weak equivalences, but I think the archetypal case of topological spaces is a good way to start.

Have tried a few things so far, but nothing worth mentioning yet.


问:
Tensor product in an exponent
I stumbled over the expression $\mathcal{H}=(\mathbb{C}^d)^{\otimes k}$. But I don't understand what this this means in context to the dimension of the Hilbert space $\mathcal{H}$.

答:
If $V$ is a vector space, $V^{\otimes k}$ denotes the tensor product of $k$ copies of $V$. For example
$$V^{\otimes 2}=V\otimes V$$
and
$$V^{\otimes 3}=V\otimes V\otimes V$$

Since the dimension of a tensor product is the product of the dimensions of the components
$$\text{dim }V^{\otimes k}=(\text{dim }V)^k$$

In your case, the dimension of $\mathcal{H}$ will be $d^k$.

Title: Tensor product in an exponent


问:
Repeating u-substitution
A question about the integration technique of u-substitution:

Is it allowed to apply u-substitution over and over again, to reduce the integral to a more manageable form?

答:
Most certainly.

However, I do suggest that you do not keep using the letter $u$ for your substitutions to avoid confusion; I typically use $w$ for my second substitution, and $t$ for a third, etc.  Situations in which you need more than $3$ substitutions are fairly rare... (in my experience)

Also, these situations oftentimes arise with applying $u$ substitution, followed by some other technique, then another substitution, etc.  Back-to-back substitutions can be handled by one (bigger) substitution.

Title: Repeating u-substitution


答:
Consider the vector $v = [1,1,\ldots, 1]^T$. You can calculate component wise that
$E^Tv = 0$ which implies that $E$ is singular so $\det(E^T) = 0$ and since $\det(E)=\det(E^T)$ we have that
$$ \det(E)=\det(E^T) = 0$$
as desired.


答:
In some sense, the answer is yes, there is a relation, passing through Riemann-Roch theorem (two corollaries of it, actually). Indeed, one can prove Gauss-Bonnet (at least for Riemann surfaces$^1$) with Cauchy and RR.

Proof:
On a Riemannian surface we can always find local coordinates, in which the metric tensor is conformal and the Gaussian curvature take the particularly easy form

$$\mathbf{g}=\frac12\lambda^2(z,\overline{z}) (dz\otimes d\overline{z}+d\overline{z}\otimes dz)=\lambda^2(dx\otimes dx+dy\otimes dy)\\
K=-\frac1{\lambda^2}\Delta\log(\lambda)$$

By partition of unity, we can glue them to a global construction. The volume form $dS$ takes the form $dS=\lambda^2 dx\wedge dy$, and thus we obtain

$$KdS=-\Delta \log(\lambda)dx\wedge dy$$

Using Dolbeaut's operators we can write it as
$$KdS=2id(\partial \log(\lambda))$$
Now, given a differential meromorphic $1-$form $\omega$ (which exists thanks to RR), one can prove that, writing locally $\omega=fdz$, the expression $$\varphi=\frac{\lambda}{|f|}$$
defines a meromorphic function. Since in particular $\ln(|f|)$ is harmonic, we can write
$$KdS=2id(\partial\log(\varphi))$$

Now, to the Gauss-Bonnet theorem: writing $X_\varepsilon$ as the riemann surface minus small "discs" $D_{k,\varepsilon}$ around the singularities $z_k$ of $\varphi$, we have

$$\int_X KdS=\lim_{\varepsilon\to 0}\int_{X_\varepsilon}2id(\partial \log(\varphi))=\lim_{\varepsilon\to 0}
2i\int_{\partial X_\varepsilon}\partial \log(\varphi)=\lim_{\varepsilon\to 0}\sum _k\int_{\partial D_{k,\varepsilon}}\partial \log(\varphi)
$$

Now, near the singularities of $\varphi$, i.e. the poles or zeros of $f$, we can write $\varphi$ as $\frac{\psi}{|z|^m}$. Thus, by Cauchy's integral formula
$$\lim_{\varepsilon\to 0}\int_{\partial D_{k,\varepsilon}}\partial \log(\varphi)=\lim_{\varepsilon\to 0}-m_k\int_{\partial D_{k,\varepsilon}}\partial \log(|z|)=\lim_{\varepsilon\to 0}-\frac {m_k}{2}\int_{\partial D_{k,\varepsilon}}\frac{dz}{z}=-\pi im_k$$

Since, by RR, the degree of the canonical divisor is $2g-2$ and $\sum m_k=-\text{deg}(\omega)$
Thus we get

$$\int_X KdS=-2\pi \text{deg}(\omega)=2\pi (2-2g)=2\pi \chi(X)$$

$^1:$ One can actually prove, though it is much harder, that every orientable surface admits such a local coordinates system, called isothermal coordinates. The usual proofs relies on the solution of the Beltrami equation

Title: Relation between (the $2 \pi$ in) Gauss-Bonnet and in Cauchy's differentation formula

Body:
When I 1st saw Gauss-Bonnet, I was wondering if this $2 \pi$ had any relation with the $2 \pi$ in Cauchy's differentiation formula. Maybe a better question is to ask about the relation between Gauss-Bonnet and Cauchy's differentiation formula. I recall the $2 \pi$ (or $\tau$) in Cauchy's differentiation formula is to do with homotopy with a circle in the original Cauchy's integral formula. I guess Gauss-Bonnet will have something to do with circles or spheres. So far I know only some basic version of Gauss-Bonnet and not yet its generalisations.

Okay so not exactly any question yet ummm...

Where does the $2 \pi$ come from in Gauss-Bonnet? Is it something to do with circles or spheres (Or $S^n$)?

Is the $2 \pi$ in Gauss-Bonnet related to the $2 \pi$ in Cauchy's differentiation formula?

What's the relation of Gauss-Bonnet with Cauchy's differentiation formula?


问:
Expectation conditional on a linear combination of i.i.d random variables
Suppose $X_1,X_2,...,X_N$ are i.i.d random variables. I want to calculate the conditional expectation
$$
\mathbb{E}[X_i|\sum_{k=1}^N\alpha_k X_k].
$$

For now, my idea is as the following:

Denote $Y_k=\alpha_kX_k$ and let $\pi=(\pi_1,...,\pi_N)$ be any permutation of $(1,...,N)$. It follows from the pairwise independence of $(Y_1,...,Y_N)$ that $Y_1,...,Y_N$ are exchangeable, i.e., for any two permutations $\pi$ and $\pi'$, $(Y_{\pi_1},...,Y_{\pi_N})$ and $(Y_{\pi_1'},...,Y_{\pi_N'})$ share the same joint distribution. Thus, function $f(Y_{\pi_1},...,Y_{\pi_N})$ and $f(Y_{\pi_1'},...,Y_{\pi_N'})$ have the same distribution. Applying this fact to function $f(Y_{\pi_1},...,Y_{\pi_N})=(Y_{\pi_1},\sum_{k=1}^NY_k)$ implies $(Y_i,\sum_{k=1}^NY_k)$ and $(Y_j,\sum_{k=1}^NY_k)$ share the same joint distribution.

By definition, the conditional expectation $\mathbb{E}[X_i|\sum_{k=1}^NY_k]$ is given by
$$
\mathbb{E}[X_iZ]=\mathbb{E}\left[Z \mathbb{E}[X_i|\sum_{k=1}^NY_k]\right],\ Z\in\mathbf{M}\left(\sum_{k=1}^NY_k\right)
$$
where $\mathbf{M}\left(\sum_{k=1}^NY_k\right)$ is the closed subspace of $L^2$ consists of all Borel function of $\sum_{k=1}^NY_k$. Multiplying both sides of the above equation by $\alpha_i$, we can also write
$$
\mathbb{E}[Y_iZ]=\mathbb{E}\left[Z \mathbb{E}[Y_i|\sum_{k=1}^NY_k]\right].
$$
Then we have
$$
\mathbb{E}[Y_iZ]=\mathbb{E}[Y_jZ]=\mathbb{E}\left[Z \mathbb{E}[Y_j|\sum_{k=1}^NY_k]\right]
\Longrightarrow
\mathbb{E}[Y_i|\sum_{k=1}^NY_k] = \mathbb{E}[Y_j|\sum_{k=1}^NY_k],
$$
since $(Y_i,\sum_{k=1}^NY_k)$ and $(Y_j,\sum_{k=1}^NY_k)$ share the same joint distribution and $Z$ is a function of $\sum_{k=1}^NY_k$. Thus,
$$
\alpha_i\mathbb{E}[X_i|\sum_{k=1}^NY_k]=\alpha_j\mathbb{E}[X_j|\sum_{k=1}^NY_k].
$$
Summing up both sides with respect to $j$ gives
$$
N\alpha_i\mathbb{E}[X_i|\sum_{k=1}^NY_k]=\mathbb{E}[\sum_{j=1}^N\alpha_jX_i|\sum_{k=1}^NY_k]=\sum_{j=1}^N\alpha_jX_i.
$$
Therefore, we have
$$
\mathbb{E}[X_i|\sum_{k=1}^NY_k]=\frac{1}{N\alpha_i}\sum_{j=1}^N\alpha_jX_i,\ \alpha_i\ne0.
$$
For $\alpha_i=0$, it follows from the independence that $\mathbb{E}[X_i|\sum_{k=1}^N\alpha_kX_k]=\mathbb{E}[X_i|\sum_{k\ne i}\alpha_kX_k]=0$.

Implications:

If $\alpha_1=...=\alpha_N=\alpha$, then $\mathbb{E}[X_i|\sum_{k=1}^N\alpha_kX_k]=\mathbb{E}[X_j|\sum_{k=1}^N\alpha_kX_k]=\frac{1}{N\alpha}\sum_{k=1}^NX_k,\ \forall i,j=1,...,N$;

In general, for $\alpha_i\ne \alpha_j$ with $\alpha_i\alpha_j\ne 0$, $\mathbb{E}[X_i|\sum_{k=1}^N\alpha_kX_k]\ne\mathbb{E}[X_j|\sum_{k=1}^N\alpha_kX_k]$, although $X_i$ and $X_j$ are i.i.d.

Does it make sense? Any hint is appreciated!

EDIT 1: Thank @Leander Tilsted Kristensen for pointing out the mistake in the comment. Now I know $(Y_{\pi_1},...,Y_{\pi_N})$ and $(Y_{\pi_1'},...,Y_{\pi_N'})$ do not necessarily have the same distribution. Then, if $\alpha_i$s are not identical, can I make a statement that $\mathbb{E}[X_i|\sum_{k=1}^N\alpha_kX_k]\ne\mathbb{E}[X_j|\sum_{k=1}^N\alpha_kX_k]$ in general?

EDIT 2: Thank @Paresseux Nguyen for the examples and comments. It turns out that in the EDIT 1, "not equal" is still a strong conclusion and we should say the two conditional expectation are "not necessarily" equal.

答:
As showed by Kristensen, you kicked off your approach with a not true observation.

In fact, I don't think you can do anything for it in general. I think I can dive a little bit in analysis of this problem to show you where things get complicated.
 But I guess it'll be better if I give some examples.

Case 1

Let's say $X=\alpha_1 X_1+\alpha_2X_2+...+\alpha_n X_n$

If $\alpha_1=\alpha_2=...=\alpha_n=1$,

$\mathbb{E}( X_1|X)= \frac{X}{n}$

Case 2

If $(X_k, k \in [n])$ are all standard normal distribution, then

$\mathbb{E}( X_1|X)= \frac{\alpha_1X}{\sum_{k=1}^n \alpha_k^2}$

Case 3

If $ (X_k, k \in [n])$ are all Bernoulli variable $\mathcal{B}(p) (0<p<1)$, $a_k=2^{-k}$, then:

$\mathbb{E}( X_1|X)= [2X]$

$\mathbb{E}( X_2|X)= [4X]-2[2X] $

etc.

**Disclaimer **: I may have mistaken somewhere but the generality stays the same.


问:
Define *Formally* a set $T$ which includes all sets
How may I define formally a set $T$ which includes all sets of sets which their members' size is finite?

For example $\{\{1\},\{2\},\{3\},\{4\}, \cdots\}$ is an element of $T$ since each set size inside of it is finite and equal to $1$.

However, the set $\{\{1,2,3,4, \cdots\}\}$ isn't in $T$, because its only member has infinitely many elements.

Note: By formally I mean this is how we define a set which includes all even numbers: $T=\{x \in R\mid x\text{ is even}\}$.

Another example from wikipedia:

$$F=\{n\mid n{\text{ is an integer, and }}0\leq n\leq 19\}.$$

答:
Basically, like this:

$$\{A\ |\ \forall a\in A, |a|<\aleph_0 \}$$

The condition $|a|<\aleph_0$ means "the cardinality of $a$ is less than $\aleph_0$", $\aleph_0$ being the smallest infinite cardinal number. You could replace this with many other equivalent ways of saying "$a$ is finite", such as "there is no injection from $a$ into itself".

I think, given the examples you've listed, this is basically what you mean by "formal". Of course, strictly speaking in formal set theory this is not correct, but I think those sorts of concerns might be a bit beyond the scope of the question as you intended it.

The reason it isn't correct is basically because we haven't specified the domain of the variable $A$ on the left hand side of the $|$ sign. For example, the following would be perfectly fine in formal set theory:

$$\{A\in\mathcal P(\mathbb N) \mid \forall a\in A, a \text{ is finite} \}$$

...where again, we replace "$a$ is finite" via some appropriate formalization. However, this only gives you the set of all finite sets of positive integers.


问:
Homomorphism from $S_3$ to ($\mathbb{Q},+)$
I am solving exercise in abstract algebra and could not solve this 1 correctly.

Does there exists a homomorphism from $S_3$ to the additive group ($\mathbb{Q},+)$ of rational numbers?

I think it exists. Map $A_3$ to $1$ and remaining elements to $-1$. But answer is no.

So, what mistake I am making? Please tell.

答:
Your map is a homomorphism to the multiplicative group of non-zero rational numbers.

You could map every element of $S_3$ to $0$ in $\mathbb Q$

to obtain a homomorphism to the additive group of rational numbers.


问:
Solving the non-homogeneous two dimensional heat equation
I am currently interested in solving the inhomogeneous heat equation for a two dimensional flow for a bounded domain. The PDE is given as:

$\frac{\partial u_1'}{\partial t}-\nu(\frac{\partial^2 u_1'}{\partial x_1^2}+\frac{\partial^2 u_1'}{\partial x_2^2})=f(x_1,x_2,t)$

with $f(x_1,x_2,t)=cos(\sqrt{a}x_1)cos(\sqrt{b}x_2)e^{-\nu(a+b)t}$.

I was thinking about solving this by using the Fundamental solution for two dimensions, given as:

$\Phi ={\frac {1}{\sqrt {(4\pi \nu t)^{n}}}}\exp \left(-{\frac {\mathbf {x} \cdot \mathbf {x} }{4 \nu t}}\right)$

The solution would then be given (atleast for the 1D case) as:

$u_1^{'p}=\int_0^t \int_{-\infty}^{\infty}\Phi(\vec{x}-\vec{y},t-s)f(\vec{y},s)d\vec{y}ds$

However since the inhomogeneous function is given as a rather "complicated" structure in terms of integration, I struggle quiete a lot there.

Do you have any ideas or experience in how to solve suche an inhomogeneous heat equation?

I would also appreciate any guidance in how to solve the inhomogeneous heat equation in genereal for a two dimensional flow, because I am not quite sure if this approach is the correct one.

I am very thankful for any advice

答:
The PDE is linear so the general solution is the sum of a homogeneous and particular solution, $u = u_h + u_p$.

The particular solution satisfies

$$ \frac{\partial u_p}{\partial t} - \nu \left(\frac{\partial^2 u_p}{\partial x_1^2}  + \frac{\partial^2 u_p}{\partial x_2^2}\right) = f $$

Making the ansatz $u_p(x_1,x_2,t) = T(t)\cos (\sqrt{a}x_1)\cos (\sqrt{b}x_2)$ and subsituting into the PDE, we get

$$T'(t)\cos (\sqrt{a}x_1)\cos (\sqrt{b}x_2) + \nu T(t)(a+b) \cos (\sqrt{a}x_1)\cos (\sqrt{b}x_2) = e^{-\nu(a+b) t}\cos (\sqrt{a}x_1)\cos (\sqrt{b}x_2), $$

and a solution is obtained when

$$T'(t) + \nu(a+b)T(t) = e^{-\nu(a+b) t}$$

Solving the ODE with an integrating factor we get  $T(t) = (C+t)e^{-\nu (a+b)t}$, and the particular solution is

$$u_p = (C+t)e^{-\nu (a+b)t}\cos (\sqrt{a}x_1)\cos (\sqrt{b}x_2)$$

As you have not specified an initial condition, nothing can be said yet about the constant $C$ appearing in the solution.

It remains to solve the homogeneous part,

$$ \frac{\partial u_h}{\partial t} - \nu \left(\frac{\partial^2 u_h}{\partial x_1^2}  + \frac{\partial^2 u_h}{\partial x_2^2}\right) = 0$$

As you want to solve on a "bounded domain", once you supply the initial and boundary conditions you can apply separation of variables seeking a solution of the form

$$u_h(x_1,x_2,t) = X_1(x_1)X_2(x_2)T(t)$$


问:
prove that the unit circle $x^2+y^2=1$ is a closed set
I need to prove that the unit circle $x^2+y^2=1$ is a closed set in $\mathbb{R}^2$ is closed using convergent sequence method.</p>

答:
HINT: Work in polar coordinates: then a sequence in $S^1$ is of the form $\big\langle\langle 1,\theta_n\rangle:n\in\Bbb N\big\rangle$, and you can think about it in terms of the sequence $\langle\theta_n:n\in\Bbb N\rangle$, which you can take to be in $[0,2\pi]$. (This isn’t the only way, but it makes life pretty easy.)

Added: Note that if the sequence converges to the point whose rectangular coordinates are $\langle 1,0\rangle$, this won’t quite work, since you may have infinitely many angles a little larger than $0$ and infinitely many a little less than $2\pi$. In that case use angles in the range $[-\pi,\pi]$ instead.


问:
Which notion of equivalence is being used here?
I'm reading Homotopy associativiy of H-spaces I by Stasheff and in the proof of Proposition 2 he says that any map is equivalent to a fibring (in modern day language I think this is either a fibration or a fiber bundle). He refers to Lemma 13 of this paper by Nomura, where he states that any map $f:X\to Y$ is equivalent to a fibring. The fibring is of the form $p:Z_f\to Y$ where $Z_f$ is constructed in the proof.

I would like to know in what sense these maps are equivalents. The map $f$ need not be a fibring  so this is not an equivalence of fibrings. A homotopy equivalence wouldn't make sense either because the domains are different. I haven't found in any of the linked papers any definition of this equivalence, so does anyone knows what they're talking about?

According to Stashef there should be a map $X\to Z_f$ commuting with $f$ and $p$, so maybe they mean that $f$ is homotopy equivalent to the composition of $p$ with this map.

答:
By fibring Stasheff means Hurewicz fibration. That is a map $p:E\rightarrow B$ which has the homotopy lifting property with respect to all spaces.

By equivalent to Stasheff means the following. Given any map $f:X\rightarrow Y$ there is a Hurewicz fibration $p_f:W_f\rightarrow Y$ and inverse homotopy equivalences
$$X\xrightarrow{i} W_f\xrightarrow{q} X$$
such that
$$p_f\circ i=f,\qquad \qquad f\circ q\simeq p_f.$$
It is in the sense of the first equation that $f$ is equivalent to $p_f$.

As a model for the space it is standard to take
$$W_f=\{(x,l)\in X\times Y^I\mid f(x)=l(0)\}$$
with
$$p_f(x,l)=l(1).$$
Here $Y^I$ is the space of all maps $I\rightarrow Y$ in the compact-open topology. The map $i$ sends $x$ to the pair $(x,c_{f(x)})$, where $c_{f(x)}$ is the constant path at $f(x)$. The map $q$ is the projection $(x,l)\mapsto x$.

See, eg.  M. Arkowitz's book Introduction to Homotopy Theory $\S$3.3 for details.


问:
The complex conjugate of the character of an irreducible representation is the character of another irreducible representation
Here the group is finite and the base field is $\mathbb{C}$.

I googled this phrase but only saw some exercises on this problem. Any reference or hint would be helpful. I remember there is a way of going to dual space but didn't manage to redo that. I am also interested in seeing different proofs or different ways of understanding this. Many thanks in advance.

答:
If $\rho \colon G \to \operatorname{GL}(V)$ is a representation of $G$ on the finite-dimensional vector space $V$, then there is a dual representation $\rho^* \colon G \to GL(V^*)$ defined by $\rho^*(g) = \rho(g^{-1})^*$. It is important here that the representation on $V^*$ involves both taking the dual operator and inverting the group element, so that we have
$$ \rho^*(g) \rho^*(h) = \rho(g^{-1})^* \rho(h^{-1})^* = (\rho(h^{-1}) \rho(g^{-1})^* = \rho((gh)^{-1})^* = \rho^*(gh).$$

Since the eigenvalues of a transpose operator are the same as the eigenvalues of the original, this means that the eigenvalues of $\rho^*(g)$ are the same as the eigenvalues of $\rho(g^{-1})$. Now,

Each entry in the character table is a trace (sum of eigenvalues),

Each eigenvalue in a group representation must be a root of unity.

Can you take it from here?


答:
Let $n \in \mathbb{N}$ and let $Y \in \cal{M}_{n \times n}\mathbb{(K)}$ where $\mathbb{K}$ is some field (with characteristic zero).

$\implies.$ Suppose that $Y$ is idempotent. Then, we know that $Y^2=Y.$ Hence

$$\begin{align}(I_n - Y)^2 & = (I_n - Y)(I_n - Y)\\
& = I_n^2 - I_nY -YI_n + Y^2\\
& = I_n -Y - Y + Y\\
& = I_n - Y
\end{align}$$

Therefore, $Y - I_n$ is also idempotent.

$\Longleftarrow.$ Supposee that $I_n - Y$ is idempotent. Then, we know that $(I_n - Y)^2 = I_n - Y.$ Hence

$$\begin{align}
(I_n - Y)^2 = I_n - Y & \implies I_n^2 - I_nY -YI_n + Y^2 = I_n - Y\\
& \implies I_n -2Y + Y^2 = I_n - Y\\
& \implies Y^2 = Y
\end{align}$$

Therefore, $Y$ is also idempotent. $\square$

Title: Proof of Idempotency for Matrices

Body:
I'm currently looking at achieving some proof that an $n \times n$ matrix (let's say $Y$) is idempotent if and only if $I_n-Y$ is an idempotent matrix, (with $I_n$ being the identity matrix).

I have so far proved that $I_n-Y$ is idempotent if $Y$ is idempotent as:

$(I_n-Y)^2 = I_n - 2Y + Y = I_n - Y$ hence idempotent.

But I don't know if this helps me at all and how to flip it so that if $I_n - Y$ is idempotent then $Y$ must be also idempotent.

Any help would be great!


问:
The locus of points from which tangents to a parabola make a constant angle is a hyperbola with the same focus as the parabola
It is well known that when two tangents to a parabola are perpendicular to each other, they intersect on the directrix. In other words, the intersection point of the two tangents make a straight line, in this case, the directrix,

However, when the two tangents to a parabola intersect at another angle, it seems that the intersection point of the two tangents always forms a hyperbola, regardless of the angle. (The original problem I encountered asked the trace of the intersection point when the angle was 45degrees)

And surprisingly, the hyperbola has the same focus with the parabola.

This can be proven by brute-force algebra, with the same method used to prove the directrix property.

The question I'd like to ask is :
Is there a geometric, or maybe an intuitive proof to why the trace of the intersection point forms a hyperbola, and why the focus of the hyperobla and parabola are the same?

It can't be a coincidence that these two conic sections have the same focus, and I assumed that there would be a simple geometric observation that can be made to prove this property without algebra, as many other properties of the conic sections do, but I just can't seem to find it. My math teachers also seem to be stumped with this problem.

Can anyone give me help?
I couldn't find any proof regarding this property on google either.

(Sorry for my bad English; it is my first time writing mathematical topics in English, please correct me if anything is wrong)

答:
Not only do the parabola and hyperbola have the same focus, they also have the same directrix.

Recall that a conic is the set of points for which the distance to the focus, divided by the distance to the directrix, equals the eccentricity
$e$.

Now let's say a point $C$ lies outside a parabola with focus $F$, so the two tangents from $C$ to the parabola meet at an angle of $\theta$.

As shown in the diagram, let $A$ and $B$ be the points of tangency, and let $P_A$, $P_B$, and $P_C$ be the projections of these points onto the directrix.

Looking at tangent point $A$, we see that since it lies on the parabola, it is equidistant from the focus $F$ and from the directrix,
so triangle $FAP_A$ is isosceles.

Furthermore, this triangle is cut in half by the line of tangency, $AC$, because it bisects angle $FAP_A$.
(This is the same angle property as the "radio dish" property: The line $FA$, reflected off the tangent at $A$, is reflected straight upwards.)
So $AC$ is the perpendicular bisector of $FP_A$.
So $CF=CP_A$.

Similarly, $BC$ is the perpendicular bisector of $FP_B$, and $CF=CP_B$.

Since $CP_A=CF=CP_B$, we can draw a circle centered at $C$, passing through $F$, $P_A$, and $P_B$.

This circle lets us see that angle $\angle P_A F P_B = \frac{1}{2} \angle P_A C P_B = \angle P_C C P_B$.

And the angle between lines $FP_A$ and $FP_B$ must be the same as the angle between their perpendicular bisectors, which is $\theta$ (or $\varphi=\pi-\theta$, depending on which angle you want). In our case, we have $\angle P_A F P_B = \varphi$, so $\angle P_C C P_B = \varphi$.

So $\cos(\varphi) = \frac{CP_C}{CP_B} = \frac{CP_C}{CF}$.

Thus $C$ lies on the hyperbola having the same focus and directrix as the parabola, but having eccentricity $\frac{1}{\cos(\varphi)} = \frac{-1}{\cos(\theta)}$.

Comments:
If we consider all points $C$ for which the parabola spans a given angle $\theta$, we get one branch of a hyperbola.  The parabola and directrix are both limiting cases of such a hyperbola branch.  If $\theta<\frac{\pi}{2}$, the "negative eccentricity" just means the hyperbola branch lies below the directrix: We can think of the signed distance to the directrix being negative there.


问:
Proving differentiability at 0
Let $h:\mathbb{R}\to\mathbb{R}$  be defined by

\begin{align*}
h(x) =
\begin{cases}
\sin(x^2)\cos(1/x) & x\neq 0 \\
0 &  x=0
\end{cases}
\end{align*}
Prove that $h$ is differentiable at $0$ and find $h'(0)$.

So far I have used
\begin{align*}
h'(0) = \lim_{x\to 0}\frac{h(x) - h(0)}{x - 0}
\end{align*}

答:
HINT

Start with the definition of derivative:
\begin{align*}
\lim_{x\to0}\frac{h(x) - h(0)}{x - 0} = \lim_{x\to0}\frac{\sin(x^{2})\cos(1/x)}{x}
\end{align*}

Now observe that
\begin{align*}
\lim_{x\to0}\frac{\sin\left(x^{2}\right)}{x^{2}} = 1\quad\text{and}\quad\lim_{x\to0}x\cos\left(\frac{1}{x}\right) = 0
\end{align*}

Can you take it from here?

Title: Proving differentiability at 0


答:
Yes, it is. Example: First substitution $t=x^{1/3}$

$$\begin{equation*}
I=\int \frac{x^{1/2}+3}{2+x^{1/3}}dx=3\int \frac{t^{2}\left(
t^{3/2}+3\right) }{2+t}\,dt.
\end{equation*}$$
You could use the additional substitution $u=t^{1/2}=x^{1/6}$ to obtain
$$
\begin{equation*}
I=6\int \frac{u^{8}+3u^{5}}{u^{2}+2}\, du.
\end{equation*}
$$

Title: Repeating u-substitution

Body:
A question about the integration technique of u-substitution:

Is it allowed to apply u-substitution over and over again, to reduce the integral to a more manageable form?


问:
Find matrix $R$ such that $(R^TPR-P)x=y$ or that $PRx = y$
Consider an invertible matrix $P\in\mathbb{R}^{n\times n}$ and two vectors $x,y\in\mathbb{R}^n$ ($P,x,y$ are given). Now consider the following two problem.

Problem 1: Find a matrix $R\in\mathbb{R}^{n\times n}$ (if it exists) such that $(R^TPR-P)x=y$.

Problem 2: Find a matrix $R\in\mathbb{R}^{n\times n}$ (if it exists) such that $PRx = y$

Lately I've been encountering problems of this type. I've been trying to use some vectorization tricks without any success. I wanted to see how would you solve these equations so I can make a general idea of what would be the general strategy if it exists.

答:
For Problem $1$

This is not a rigorous approach, but shows that $R$ exists under certain conditions,
$$(R^TPR-P)x=y$$
Let the eigenvalue decomposition of $P = UDU^{-1}$ and that of $R = VSV^{-1}$. Choose the eigenvectors of $V$ to be that of $P$, i.e. $V= U$ we get
$$(USU^{-1} UDU^{-1}USV^{-1}- UDU^{-1})x=y$$
or
$$(USDSU^{-1}- UDU^{-1})x=y$$
or
$$U(SDS- D)U^{-1}x=y$$
or
$$(SDS- D)U^{-1}x=U^{-1}y$$
Since $S$ and $D$ are diagonal then we can exchange them as
$$(S^2 D- D)U^{-1}x=U^{-1}y$$
which is
$$(S^2 - I)DU^{-1}x=U^{-1}y$$
Let $a = DU^{-1}x$ and $b = U^{-1}y$ so
$$(S^2 - I)a=b$$
If the $k^{th}$ diagonal of $S$ is denoted by $s_k$, also if the $k^{th}$ element of vector $a,b$ is denoted by $a_k,b_k$ then the $k^{th}$ row of the above equation is nothing other than
$$(s_k^2 - 1)a_k = b_k$$
If $a_k \neq 0 $ and $\frac{b_k}{a_k} > -1$, then $s_k = \pm \sqrt{\frac{b_k}{a_k} + 1}$.
So your matrix $R = U S U^{-1}$ where $s_k = \pm \sqrt{\frac{b_k}{a_k} + 1}$ where $a = DU^{-1}x$ and $b = U^{-1}y$ given that all entries of $a$ are non-zero and all $\frac{b_k}{a_k} > -1$.

For Problem $2$

If $P$ is invertible, then $Rx = P^{-1}y$, you can now choose $R$ as $R = \frac{1}{x^T x} P^{-1}yx^T$. Verification:
$$PRx = P(\frac{1}{x^T x} P^{-1}yx^T)x = \frac{1}{x^T x}PP^{-1}yx^Tx =  \frac{x^Tx }{x^T x}y= y$$


问:
$\operatorname{Ann}(G)=$?, where $G$ is a module over $\mathbb{Z}$.
As every Abelian Group $G$ is a $\mathbb{Z}$-module. It is easy to see that if $|G|=n$, then $n\mathbb{Z} \subset \operatorname{Ann}(G)$.
Can we tell that for every group $G$, $n\mathbb{Z}=\operatorname{Ann}(G)$?

If $G=\mathbb{Z}_n$ then it is true. Is this true in general?

答:
You have ${\rm Ann}(G)=e\mathbb{Z}$ where $e$ is the exponent of $G$, i.e. the lcm of the orders of the elements of $G$.

For instance
$${\rm Ann}((\mathbb{Z}/2\mathbb{Z})^r)=2\mathbb{Z}
$$
whatever $r$.


问:
A question about Lebesgue-Stielties measure
Let $E\subset R$.

($A_{\mu}$ is the domain of the Lebesgue-stielties measure $\mu$ ($A_{\mu}$ is a sigma-algebra that contains the borel sigma algebra).

A. Prove that if $E=H\cup N$ where H is a countable union of closed sets and $\mu(N)=0$ then $E\in A_{\mu}$.

B. prove that if $E\in A_{\mu}$ and $\mu(E) < \infty$, then $E=H\cup N$ where H is a countable union of closed sets and $\mu(N)=0$.

My try:

A.

$H\subset B_{R}\in A_{\sigma}$.
$N\in A_{\mu}$ as $\mu(N)=0$, so $E=H\cup N\in A_{\mu}$ as an union of two sets in $A_{\mu}$.

B.

$\forall n\in N$ there's $K_n\subset E\subset R$  such that $\mu(E)\leq \mu(K_n)+\frac{1}{n}$.

Notice that since $K_n$ is compact in $R$ so by a theory in topology ee get that it's closed and bounded.
So, lets define $H=\cup_{n=1}^{\infty} K_n$ so H is a countable union of closed sets. And $H\subset B_R\in A_{\mu}$.
Now, let's look at $E\setminus H$.
since $K_n\subset H$ therfore $E\setminus H\subset E\setminus K_n$.
And $\mu(E\setminus H)\leq \mu(E\setminus K_n)=\mu(E)-\mu(K_n)≤ 1/n$ (notice that $\mu(K_n)$ is finite since $K_n\subset E$). This inequality holds for every $n\in N$, so $\mu(E\H)=0$, and finally define $N=E\setminus H$.
Getting, $N\cup H= (E\setminus H) \cup H= E$.

Can you help in this..

答:
(a) is obvious because $H$ is a Borel set and $\mu$ is complete. As for (b) you need to use the fact that $\mu$ is regular, i.e., for $E\in\mathcal{A}_{\mu}$,
\begin{align}
\mu(E)&=\inf\{\mu(O):O\supset E, O\text{ is open}\} \\
&=\sup\{\mu(K):K\subset E, O\text{ is compact}\}.
\end{align}
For each $n\ge 1$, choose a compact set $K_n\subset E$  s.t.
$$
\mu(E)\le \mu(K_n)+n^{-1}.
$$
Let $K=\bigcup_{n\ge 1}K_n\subset E$. It follows that $\mu(K)=\mu(E)<\infty$ and $\mu(E\setminus K)=0$.


问:
Is there a condition for the closure of an open subset to be a manifold with boundary?
A necessary condition for a manifold with non-empty manifold boundary embedded in $\mathbb{R}^n$ to have its topological boundary coincide with its manifold boundary is that its manifold interior is an open subset of $\mathbb{R}^n$.

Questions:

Is this also a sufficient condition? (Note that I am assuming a prior that the manifold boundary is non-empty -- this does not hold otherwise.)

What conditions are necessary for a sort of "converse" to this to hold? I.e., given a proper open subset of $\mathbb{R}^n$, which is always a smooth embedded submanifold, when is its topological closure an embedded smooth submanifold with (non-empty) boundary?

Attempt: The second question is not as straightforward as I first thought: consider the open hypercube, which is clearly a proper open subset of $\mathbb{R}^n$ -- however, unless $n=1$, its topological closure is a manifold with corners, but not a manifold with boundary.

Since the open hypercube is homeomorphic (diffeomorphic too, right?) to any open ball, whatever conditions need to be placed on such a proper open subset of $\mathbb{R}^n$ for this "converse" to hold must be "more than topological" (I don't know how to phrase that precisely).

Note: this is not a duplicate of this question, it is a follow-up to it.

Context:
This answer established that, if an $m$-dimensional manifold with boundary $M$ can be embedded into a Euclidean space $\mathbb{R}^n$ such that its topological and manifold boundaries coincide, then necessarily one has that $m=n$.

This implies in particular that for $n$-dimensional manifolds with or without boundary which cannot be embedded into $\mathbb{R}^n$, the topological and manifold boundaries never coincide when embedded in any Euclidean space. (If the manifold with boundary has empty manifold boundary then of course the topological and manifold boundaries coincide when it is embedded in itself.)

The interior of any such embedded manifold with boundary is always an embedded manifold (without boundary). We also have that the embedded submanifolds of codimension $0$ of any manifold are exactly the open subsets of that manifold (e.g. Proposition 5.1. p.99 of Lee's Introduction to Smooth Manifolds).

Thus we have a necessary condition for the topological and manifold boundaries to coincide, at least when the embedding is inside of $\mathbb{R}^n$ (although I suspect that this holds for embeddings in an arbitrary $n$-dimensional manifold): the manifold interior of $M$ is an open subset of $\mathbb{R}^n$.

However, this is can not be a sufficient condition: if the manifold boundary is empty then this fails because the topological boundary is non-empty (unless the manifold is all of $\mathbb{R}^n$ itself -- e.g. see this answer). It is unclear to me whether this is a sufficient condition if we assume a priori that the manifold boundary is non-empty.

答:
A partial answer to the second question: if $W$ is an open set in an affine space and $H=\overline W\setminus W$ is a smooth hypersurface, then $\overline W$ is a smooth manifold, with or without boundary (think of $W=\mathbb R^2\setminus\{x=0\}$). Notice that if $\overline W$ is a manifold, its boundary is a hypersurface contained in $H$. In fact the argument below works to see that that boundary must be a union of connected components of $H$.

The argument. For points $x\in W$ there is nothing to do. If $H$ is a hypersurface, every point $x\in H$ has an open neighborhood $U$ such that $U\setminus H$ has two connected components $A,B$, and $\overline A=U\setminus B$ is a manifold with boundary $U\cap H$, and viceversa. Since $$D:=W\cap(U\setminus H)=\overline W\cap(U\setminus H),$$ the non-empty set $D$ is open and closed in $W\cap(U\setminus H)$. It follows either $D=A$, or $D=B$, or $D=A\cup B$. Hence either
$\overline D\cap U= U\setminus B$ or $\overline D\cap U= U\setminus A$, which are both manifolds with boundary $U\cap H$, or $\overline D\cap U$, which is a manifold without boundary.

Hope this is not redundant!


问:
Expand: $e^{\sin z}$ at $z = 0$
I need hints to find the Laurent expansion of $\displaystyle e^{\sin z}$ at $z = 0$ in a simpler way. I am getting double series which I can't simplify.

ADDED:: Can it be simpler that the below? I also need to find the radius of convergence.
$$\Large e^{\sin z} = e^{\sum_{k = 0}^\infty \frac{(-1)^nz^{2n+1}}{(2n+1)!}} = \sum_{n = 0}^\infty \frac{\small{B(1, -1/3!, \dots , (-1)^n/(2n+1)!)} z^n}{n!}$$

答:
Note that $\frac{\mathrm{d}}{\mathrm{d}x}e^{\sin(x)}=\cos(x)e^{\sin(x)}$. Equating the coefficients for the series on both sides, that is
$$
\sum_{k=0}^\infty(k+1)a_{k+1}x^k=\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}x^{2k} \sum_{k=0}^\infty a_kx^k
$$
we get the following recursion for the coefficients:
$$
a_{n+1}=\frac1{n+1}\sum_{k=0}^{\lfloor n/2\rfloor}\frac{(-1)^k}{(2k)!}a_{n-2k}
$$
Since $a_0=1$, we get
$$
e^{\sin(x)}=1+x+\frac12x^2+0x^3-\frac18x^4-\frac1{15}x^5-\frac1{240}x^6+\frac1{90}x^7+\dots
$$
Not working from the sequence of coefficients, but just knowing that the radii of convergence of both $\sin(x)$ and $e^x$ are infinite, the radius of convergence for the composition will also be infinite.

Generalization on the Same Idea

To get a result similar to the Exponential formula cited by Michael Hardy, we can use the preceding idea to get a recursion for general $f$. Assuming that $f(0)=0$, and $\displaystyle f(x)=\sum_{k=1}^\infty f_kx^k$
we get the following recursion for $\displaystyle e^{f(x)}=\sum_{n=0}^\infty a_nx^n$:
$$
a_0=1\quad\text{and}\quad a_{n+1}=\frac1{n+1}\sum_{k=0}^n(k+1)f_{k+1}a_{n-k}
$$

Title: Expand: $e^{\sin z}$ at $z = 0$


问:
How can I find the optimal solution for combining several people from different (unequal) groups?
I have a (probably) simple problem, but due to my dusty knowledge, I don't even know where to start reading or how to name my problem.

Our team has grown a lot in the last few weeks and it is very difficult to get to know each other. Therefore I want to make a list with meetings of 3 people each to establish contact. Now I would like to depict all possible encounters under the following conditions:

Number of persons/meeting = 3

Only people from different sub-teams should meet in each meeting
(i.e. max. 1 member from three different sub-teams each).

Team size = 21 people in 5 sub-teams

Size of the sub-teams varies (4,7,5,3,2 members)

How can I solve the problem? What are such problems called? (Bonus: Where can I read in on this?)

Posted also on Cross Validated

答:
First, let's see if we can calculate how many meetings of three people are possible, with at most one member from each sub-team.  Consider the polynomial
$$f(x) = (1+4x)(1+7x)(1+5x)(1+3x)(1+2x)$$
and expand, with result
$$f(x)=1+21 x+169 x^2+651 x^3+1198 x^4+840 x^5$$
The coefficient of $x^n$ in this result is the number of meetings that can be formed with $n$ people, at most one from each sub-team.  In particular, the coefficient of $x^3$ is the number of possible meetings with three people: $651$. This seems too long a list to be useful, but maybe we can say something about ways to pick people for meetings.

One way to proceed is to pick three distinct sub-teams.  This can be done in any one of $\binom{5}{3} = 10$ ways.  Then pick one person from each of the three sub-teams.  If you were to pick the first three sub-teams, for example, then three people, one from each sub-team, can be chosen in $4 \times 7 \times 5 = 140$ ways; so you have plenty of choices. The $10$ ways to pick three sub-teams, if the sub-teams are numbered $1$ through $5$, are
$$(1, 2, 3),
(1, 2, 4),
(1, 2, 5),
(1, 3, 4),
(1, 3, 5), \\
(1, 4, 5),
(2, 3, 4),
(2, 3, 5),
(2, 4, 5), \text{ and } (3, 4, 5)$$

I suppose this is obvious, but I don't know what more to say.

($f(x)$, above, is an example of a generating function.  Readers interested in learning about generating functions may find many resources in the answers to this question: How can I learn about generating functions?)


问:
Show that $x \leq f(x)$ where $<$ is a well ordering
I'm stuck on this question: let $<$ be a well ordering on a set $A$ and let $f$ be an operator on $A$ satisfying $x < y \implies f(x) < f(y)$. I need to prove $x \leq f(x)$ for each $x \in A$. A suggestion is to consider $f(f(x))$, and I wanted to use a contradicton argument but I see no leads. Is there any more suggestions?

答: Assume the set $A'=\{a\in A\mid f(a)<a\}$ is non-empty. Let $x$ be the smallest element of $A'$. Then we have $f(x)<x$, and therefore $f(f(x))<f(x)$, so $f(x)\in A'$. But this contradicts $x$ being the smallest element of $A'$.


问:
Homotopy cardinality of weak quotients
Let $G$ be a group (regarded as a one-object groupoid) acting on a groupoid $X$ (i.e.: a functor $F: G \to \mathsf{Groupoids}$ sending the unique object of $G$ to $X$). Denote with $X//G$ the “weak quotient” of such an action, i.e. the Grothendieck construction of the functor $F$.

It seems to be well-known that the homotopy cardinality of the weak quotient $X//G$ is given by the formula
$$
|X//G| = \frac{|X|}{o(G)}\text{,}
$$
where $|X|$ is the homotopy cardinality of $X$ and $o(G)$ is the order of $G$ (of course this makes sense only when $|X|$ and $o(G)$ are finite). However, I wasn't able to find any recorded proof of this fact (maybe it's very simple and I can't see it). Can someone sketch a proof or give some reference where I can find one?

答: I think the key point is that we know the cardinality of a semi direct product, which is the group theoretic background of the homotopy quotient. For instance, suppose $X$ is skeletal, so that isomorphic objects are equal, and that the action is transitive. Then all objects of $X$ have the same automorphism group $H$. By the usual orbit-stabilizer theorem, the stabilizer $K$ of some $x\in X$ has cardinality $|G|/|\mathrm{ob} X|$. Then $X//G$ has one isomorphism class of object, whose automorphism group $H\rtimes K$. The homotopy cardinality of $X$ is $|G|/(|K||H|)$ and of $X//G$ is $1/(|H||K|)$, as desired.

Title: Homotopy cardinality of weak quotients


问:
Number Theory : Find the group of A such that $A=\{x \in \mathbb{Z}:\frac{x^3-3x+2}{2x+1}\in \mathbb{Z}\}$
Find the group A such that  $A=\{x \in \mathbb{Z}:\frac{x^3-3x+2}{2x+1}\in \mathbb{Z}\}$ ?

Polynomial Long Division we get
$\frac{x^{2}}{2}-\frac{x}{4}-\frac{11}{8}+\frac{27}{8\left(2x+1\right)}$

but how i can from here find all x such that $x \in \mathbb{Z}$ ??

i did use the hint for $y = 2x+1$ so that  $x =\frac{y-1}{2}$

made the equation to be $\frac{(y-1)^3}{8}-\frac{3(y-1)}{2}+2=\frac{y^3-3y^2-9y+27}{8}$

$\frac{\frac{y^3-3y^2-9y+27}{8}}{y}=\frac{y^3-3y^2-9y+27}{8y}$

I can not understand with the help of the clues, can some 1 some formal proof ?

Body:
Hint:

If $2x+1=y$

$$x^3-3x+2=(y-1)^3/8-3(y-1)/2+2=\dfrac{y^3-3y^2-9y+27}8$$ which will be divisible by $y$

$\iff y$ divides $y^3-3y^2-9y+27$ as $y$ is odd


问:
How can I use for the generalized logistic equation a series approximation?
The question is how can I show the following approximation stated in this paper.

I have tried to follow , but I am not pretty sure if the following results are correct

Generalized logistic growth modeling of the COVID-19 outbreak: comparing the dynamics in the 29 provinces
in China and in the rest of the world

https://link.springer.com/article/10.1007/s11071-020-05862-6

Eq 5, 6

They have written an equation similar to

\begin{align}
\frac{1}{Y}\frac{dY}{dt}=\frac{   A  e^{-  b t}}{ 1 -  A  e^{- b  t}}
\end{align}
following the definition of the serie geometric

\begin{align}
 \sum_{n=0}^{\infty} Ae^{-n b t} = 1 +   e^{- b  t}+
  e^{-2 b  t} +  e^{- 3b  t} + \ldots
\end{align}

\begin{align}
\frac{1}{Y}\frac{dY}{dt}=
 \frac{   A  e^{-  b t}}{ 1 -  A  e^{- b  t}}  = A e^{-  b t} \left(  1 +   e^{- b  t}+
 e^{-2 b  t} +  e^{- 3b  t} + \ldots\right)
\end{align}

\begin{align}
\frac{1}{Y}\frac{dY}{dt}=
 \frac{   A  e^{-  b t}}{ 1 -  A  e^{- b  t}}  = A   e^{-  b t} +   e^{- 2b  t}+
 e^{-3 b  t} +  e^{- 4 b  t} + \ldots
\end{align}
for long term, it can be written after substitution the approximation.

\begin{align}
\frac{1}{Y}\frac{dY}{dt}= \frac{   A  e^{-  b t}}{ 1 -  A  e^{- b  t}}  = A   e^{-  b t}
\end{align}

Any suggestion

答:
the leading behavior of the growth rate at long times

$$\gamma  \sum _{k=1}^{\infty } \left(\epsilon_0  e^{-\gamma  t}\right)^k=\frac{\gamma  \epsilon_0  e^{-\gamma t}}{1-\epsilon_0  e^{-\gamma t}}$$
In the sum on the left all ther terms with exponent $k$ larger that  $1$ can be neglected for large $t$.

And it can be approximated by $\epsilon_0  e^{-\gamma  t}$.

Just to see what happens, suppose $\epsilon_0 =\gamma=1$ and $t=5$, and let's see the values when exponent $k$ increases

$$
\begin{array}{r|r}
 k & e^{-5k}\\
\hline
 1 & 0.006738 \\
 2 & 0.000045 \\
 3 & 3.06\cdot 10^{-7} \\
 4 & 2.06\cdot 10^{-9}  \\
 5 & 1.39\cdot 10^{-11}  \\
\end{array}
$$


问:
How can I show if $AA^T+A^TA=A+A^T$, then $Col(A)=Col(A^T)$?
Question: For a square matrix $A$, if $$AA^T+A^TA=A+A^T,$$ prove that $$\text{Col}(A)=\text{Col}(A^T).$$

I want to prove this statement, but it's quite difficult...

I think I should use "$Null(A)$ is orthogonal to $Col(A^T)$"

Any good ideas?

答:
Using your observation that the null space is the orthogonal complement to the row space, it suffices to show that $\operatorname{Null}(A) = \operatorname{Null}(A^T)$. Now, decompose $A$ into the sum $A = S + K$, where $S$ is symmetric and $K$ is skew-symmetric. In fact, this implies that $S = \frac 12(A + A^T)$ and $K = \frac 12 (A - A^T)$.

Suppose that $x \in \Bbb R^n$ is such that $Ax = 0$. We have
$$
Ax = 0 \implies (S + K)x = 0 \implies Sx = -Kx.
$$
It follows that $x^TSx = -x^TKx$. Because $K$ is skew-symmetric, we must have $x^TKx = 0$ (why?). Thus, we have $x^TSx = 0$. That is,
$$
\begin{align}
0 = x^TSx &= \frac 12 x^T[AA^T + A^TA]x =
\frac 12 [x^TAA^Tx + x^TA^TAx]
\\ & =
\frac 12 [(A^Tx)^T(A^Tx) + (Ax)^T(Ax)] =
\frac 12 [\|A^Tx\|^2 + \|Ax\|^2].
\end{align}
$$
Thus, we must have $\|A^Tx\| = 0$, so $A^Tx = 0$.

That is, we have $Ax = 0 \implies A^Tx = 0$, so that $\operatorname{Null}(A) \subseteq \operatorname{Null}(A^T)$. Symmetrically, we can conclude that $\operatorname{Null}(A^T) \subseteq\operatorname{Null}(A)$. So, we have $\operatorname{Null}(A) = \operatorname{Null}(A^T)$ as desired.


问:
Root of a quadratic equation that has modulus $1$
Let us suppose $\alpha \in \mathbb C$ and $|\alpha|=1$ and $\alpha$ satisfies a monic quadratic equation. Then prove that $\alpha^{12} =1$.

Show me the right way to solve this. Thanks in advance.

答:
Suppose $\alpha\in \bf C$ is a root of some quadratic polynomial $x^2+ax+b\in{\bf Q}[x]$ and $|\alpha|=1$.

Then $\displaystyle\alpha=\frac{-a\pm\sqrt{a^2-4b}}{2}$. If $\alpha\ne\pm1$ then $a^2<4b$, and

$$1=\alpha\overline{\alpha}=\frac{a^2-(a^2-4b)}{4}=b,\quad -2<a<+2.$$

It is false that $\alpha$ must be a $12$th root of unity; check either root of e.g. $x^2+\frac{1}{2}x+1$. However if we restrict $a,b\in\bf Z$ to be integers, then $-2<a<2\iff a\in\{-1,0,+1\}$, and it can be checked that

the values $\alpha=\pm1$ are second roots of unity,

roots of $x^2+1$ are primitive fourth roots of unity, $\pm i$,

roots of $x^2-x+1$ are primitive sixth roots of unity,

roots of $x^2+x+1$ are primitive cube roots of unity,

and all of the above are twelfth roots of unity. However, observe that none of these is a primitive twelfth root; the possible values for $\alpha$ are precisely the nonprimitive twelfth roots of unity.

The last two bullet points above are covered in the theory of cyclotomic polynomials, as the two polynomials listed are in fact $\Phi_6(x)$ and $\Phi_3(x)$.

If $x^2+x+1=0$ then multiply by $x-1$ to obtain the relation $x^3-1=0$, but $x=1$ is impossible so $x$ must be a primitive cube root.

If $x^2-x+1=0$ then multiply by $(x^2+x+1)(x^2-1)$ to obtain $x^6-1=0$, but $x$ cannot be a root of $x^2-1$ or of $x^2+x+1$ at the same time, so it is not a second or third root of unity while it is a sixth root, hence it is a primitive sixth root of unity.


问:
Prove that a functional on Sobolev space type is convex
Let $a\in(-1,1)$, let:
$$ W_a=\biggl\{ g\in W^{1,1}_{\text{loc}}(0,\infty): \int_0^\infty t^a|g(t)|^2\,dt<\infty,\text{ and } \int_0^\infty t^a|g'(t)|^2\,dt<\infty  \biggr\},$$
let $\lambda>0$, i have to prove that the following functional:
$$ G_\lambda(g)=\int_0^\infty t^a(\lambda|g(t)|^2+|g'(t)|^2)\,dt,\quad\forall g\in W_a,$$
is convex. So I think that I have to prove that for all $\alpha\in[0,1]$ and $f,g\in W_a$:
$$ G_\lambda(\alpha g+(1-\alpha)f)\leq \alpha G_\lambda(g)+(1-\alpha)G_\lambda(f). $$
But I don't know how to go on. It seems to me more difficult than it seems. Any help would be appreciated.

答: My assert follow immediately by the convexity of the function $x\mapsto x^2$.


问:
Prove $|(1+|u|^2)v-(1+|v|^2)u|>|u\bar{v}-\bar{u}v|$
Show that for $u,v \in \mathbb{C}$ with $|u|<1, |v|<1$, and $\bar{u}v\neq u\bar{v}$, we always have
$$\left|\left(1+|u|^2\right)v-\left(1+|v|^2\right)u\right|>\left|u\bar{v}-\bar{u}v\right|.$$</p>

答:
I have answered here as this question which shows context has been closed as a duplicate.

Let $u=re^{i\theta}$ and $v=\rho e^{i\phi}$. Then the inequality is equivalent to $$\left|(1+r^2)\rho e^{i\phi}-(1+\rho^2)re^{i\theta}\right|>r\rho\left|e^{i(\theta-\phi)}-e^{-i(\theta-\phi)}\right|=2r\rho|\sin(\theta-\phi)|$$ and dividing both sides by $r\rho$ yields $$\left|r+\frac 1r-\left(\rho+\frac1\rho\right)e^{i(\theta-\phi)}\right|>2|\sin(\theta-\phi)|.$$ Denoting $s=r+1/r$ and $t=\rho+1/\rho$, we have $s,t>2$. Therefore, it suffices to show that $$(s-t\cos(\theta-\phi))^2+t^2\sin^2(\theta-\phi)^2>4\sin^2(\theta-\phi)$$ on squaring both sides. Denoting $\varphi=\theta-\phi$, the inequality simplifies to $$s^2-2(t\cos\varphi)s+(t^2-4\sin^2\varphi)>0.$$ When $\cos\varphi\le0$ the LHS is minimised at $s=2$ so that $$s^2-2(t\cos\varphi)s+(t^2-4\sin^2\varphi)\ge4-4t\cos\varphi+(t^2-4+4\cos^2\varphi)=(t-2\cos\varphi)^2>0.$$ When $\cos\varphi>0$ the LHS is minimised at $s=t\cos\varphi$ so that $$s^2-2(t\cos\varphi)s+(t^2-4\sin^2\varphi)\ge-t^2\cos^2\varphi+t^2-4\sin^2\varphi=(t^2-4)\sin^2\varphi>0$$ which proves $\vert (1+\vert u \vert^2)v - (1+\vert v \vert^2)u  \vert > \vert u\overline{v}-\overline{u}v  \vert$.


问:
Convergent series $\iff$ absolutely convergent
Consider the measure space $(\mathbb{N}, 2^{\mathbb{N}},\#)$, where $\#$ is the counting measure.

It is easy to show that for $f: \mathbb{N} \rightarrow \mathbb{R}_{\geq 0}$ we have $\int_{\mathbb{N}}f d\#=\sum_nf(n)$.

Now, suppose $f: \mathbb{N} \rightarrow \mathbb{R}$. Then $f \in L^1$ if and only if $\int_{\mathbb{N}}f^+ d\#$ and $\int_{\mathbb{N}}f^- d\#$ are finite. So, if $f\in L^1$ we can write $\int_{\mathbb{N}}f d\#=\sum_nf^+(n)-\sum_nf^-(n)=\sum_nf(n)$, for the properties of convergent series.

This, combined with $f \in L^1 \iff |f| \in L^1$, yields that convergence is equivalent to absolute convergence, for series.

Can you help me spot the mistake?

答: $f$ is defined to be in $L^1$ if $\int_{\mathbb{N}} f^+ \,d\#$ and $\int_{\mathbb{N}} f^- \,d\#$ are both finite.  If this happens, then $\int_{\mathbb{N}} f \,d\# = \sum_{n} f(n)$ is defined to be $\int_{\mathbb{N}} f^+ \,d\#$ - $\int_{\mathbb{N}} f^- \,d\#$.  So $f \in L^1$ is the same as saying $(f(n))_{n}$ is absolutely convergent.  Conditionally convergent sequences are those for which $\int_{\mathbb{N}} f^+ \,d\#$ and $\int_{\mathbb{N}} f^- \,d\#$ are both infinite, but somehow they "collude" to make $\int_{\mathbb{N}} f^+ \,d\#$ - $\int_{\mathbb{N}} f^- \,d\#$ finite (when you consider this integral in the sense of summing from $n=1$ to $\infty$).  These sequences aren't considered to be part of $L^1$.


问:
Proof using derivative definition?

Let $\varphi(h)$ be a function such that $\varphi (h)\neq 0$ for all
$h \neq 0$ and $\lim_{h \rightarrow 0} \varphi(h)=0$. Prove that for
any differentiable function $f(x)$ that

$$ \lim_{h \rightarrow 0}\frac{f(x+\varphi(h))-f(x)}{\varphi(h)} =
f'(x) $$

This seems obviously true to me but I'm having trouble writing the proof. I tried using sequence criteria like so:

$$ Let \space h = \frac{1}{n} $$

$$ f'(x) = \lim_{n \rightarrow
 \infty}\frac{f(x+\frac{1}{n})-f(x)}{\frac{1}{n}} $$

$$ f'(x) = \lim_{n \rightarrow
 \infty}\frac{f(x+\varphi(\frac{1}{n}))-f(x)}{\varphi(\frac{1}{n})} $$

$$ f'(x) = \lim_{h \rightarrow
 0}\frac{f(x+\varphi(h))-f(x)}{\varphi(h)} $$

But this isn't convincing and I'm not sure if it's accurate. I would appreciate a nudge in the right direction.

答:
We have that by differentiability

$$f(x+h)=f(x_0)+f'(x)h+o(h) \quad \frac{o(h)}h \to 0$$

therefore

$$f(x+\varphi(h))=f(x_0)+f'(x)\varphi(h)+o(\varphi(h))$$

and then

$$\frac{f(x+\varphi(h))-f(x)}{\varphi(h)}=\frac{f'(x)\varphi(h)+o(\varphi(h))}{\varphi(h)} = f'(x)+\frac{o(\varphi(h))}{\varphi(h)} \to f'(x)$$

Title: Proof using derivative definition?


问:
Largest prime factor of 600851475143
I'm trying to use a program to find the largest prime factor of 600851475143.
This is for Project Euler here: http://projecteuler.net/problem=3

I first attempted this with the code that goes through every number up to 600851475143, tests its divisibility, and adds it to an array of prime factors, printing out the largest.

This is great for small numbers, but for large numbers it would take a VERY long time (and a lot of memory).

Now I took university calculus a while ago, but I'm pretty rusty and haven't kept up on my math since.

I don't want a straight up answer, but I'd like to be pointed toward resources or told what I need to learn to implement some of the algorithms I've seen around in my program.

答:
Project Euler problems (at least the ones that I have done) tend to deal with a lot of number theory topics.  So, reading an introductory number theory book could be helpful.

With regards to your particular situation, I suggest finding primes first, then testing the primes for divisibility.  That is, to find prime factors of $25$, don't test $1, 2, 3, 4, 5, 6, \ldots$--this would take forever, and you'd also be left with composite factors as well.  Rather, find the primes under $25$, and test those: $2, 3, 5, 7,\ldots$  This will be much faster.

Also, you may want to research prime-finding algorithms.  If you start dealing with elliptic curves, that's probably overkill for Project Euler--there are some simple, yet effective, algorithms out there.

Title: Largest prime factor of 600851475143


问:
Do curves with $(\phi\circ\gamma_1)'(0)=(\phi\circ\gamma_2)'(0)$ for a chart $\phi$ also have same derivative with respect to another chart?
As discussed in this other question, given a manifold $M$ and a point $p\in M$, we can define its tangent vectors in $T_p M$ as the set of equivalence classes $[\gamma'(0)]$ defined so that $\gamma_1,\gamma_2\in[\gamma'(0)]$ iff $(\phi\circ\gamma_1)'(0)=(\phi\circ\gamma_2)'(0)$ for all coordinate charts $\phi:U\to\mathbb R^n$, where $p\in U\subset M$.

In this definition, is it sufficient to ask for the curves to have same derivative with respect to one coordinate chart defined around $p$? In other words, given two charts $\phi,\tilde\phi:U\to\mathbb R^n$ defined on some neighbourhood of $p$, suppose $$(\phi\circ\gamma_1)'(0)=(\phi\circ\gamma_2)'(0).$$
Does this imply that $(\tilde\phi\circ\gamma_1)'(0)=(\tilde\phi\circ\gamma_2)'(0)$?

From the definition of a smooth manifold, I know that $\tilde\phi\circ\phi^{-1}$ is a homeomorphism between $\phi(U)$ and $\tilde\phi(U)$. I would therefore expect that if
$$\phi(\gamma_1(\epsilon))-\phi(\gamma_2(\epsilon)) = o(\epsilon),$$
then the same should hold replacing $\phi\to\tilde\phi$. However, I'm not sure what properties of $\tilde\phi\circ\phi^{-1}$ I could use to show this.

答:
To (hopefully) clarify notation, for a map $F:\mathbb{R}^k\to\mathbb{R}^l$, let $D_xF$ denote the derivative of $F$ at $x$, in the sense of multivariable calculus.

Let $\gamma_1,\gamma_2$ be smooth curves with $\gamma_1(0)=\gamma_2(0)=p$, suppose that $D_0(\varphi\circ\gamma_1)=D_0(\varphi\circ\gamma_2)$ for some chart $\varphi$, and let $\psi$ be any other chart containing $p$. Since $M$ is a smooth manifold, the transition function $\tau=\psi\circ\varphi^{-1}$ is a deffeomorphism between open subsets of $\mathbb{R}^n$. We can compute the coordinate derivative of $\gamma_1$ in the new chart.
\begin{align*}
D_0(\psi\circ\gamma_1)=&D_0(\psi\circ\varphi^{-1}\circ\varphi\circ\gamma_1) \\
=&D_0(\tau\circ(\varphi\circ\gamma_1))
\end{align*}
Applying the chain rule:
\begin{align*}
=&D_{\varphi(p)}\tau\cdot D_0(\varphi\circ\gamma_1)
\end{align*}
We see that $D_0(\psi\circ\gamma_1)$ is related to $D_0(\varphi\circ\gamma_1)$ by the linear map $D_{\varphi(0)}\tau$. The same is true of $\gamma_2$. As a result, of the derivatives are equal in one chart, they are equal in any other:
\begin{align*}
D_0(\varphi\circ\gamma_1)=&D_0(\varphi\circ\gamma_2) \\
\implies D_{\varphi(p)}\tau\cdot D_0(\varphi\circ\gamma_1)=&D_{\varphi(p)}\tau\cdot D_0(\varphi\circ\gamma_2) \\
\implies D_0(\psi\circ\gamma_1)=&D_0(\psi\circ\gamma_2) \\
\end{align*}


问:
Relative homology and homology of subset
Let $X$ be homotopy equivalent to a point and let $A \subset X$.

 I have shown that $H_n(X,A) \cong H_{n-1} (A)$ for $n \geq 2$. I did this using the long exact sequence property in relative homology.

 Does it also hold for $n=1$. If not, what would a counterexample be?

答:
Maybe this is an example.

Let $X=\{x\in\mathbb{R}^2 : |x|\leq 1\}$ and $A=\{(1,0),(-1,0)\}$.
Then $A$ is a strong deformation retract of a suitable compact neighborhood of it, so that $H_n(X,A)=H_n(X/A)$.
The quotient space $X/A$ is homotopy equivalent to $S^1$, so $H_1(X,A)=H_1(X/A)=\mathbb{Z}$.
On the other hand, $H_0(A)=\mathbb{Z}\oplus\mathbb{Z}$ since $A$ has two connected components.