问:
On understanding $\mathrm{ad}^0 \overline{\rho}$ and $\mathrm{ad}^0 \overline{\rho}(1)$ in Taylor-Wiles method
I'm currently learning Taylor-Wiles method and modularity lifting and comming up with following difficulties, which I think is based on understanding how (global and local) Galois groups act on $\mathrm{ad}^0 \overline{\rho}$ and $\mathrm{ad}^0 \overline{\rho}(1)$. Here:

$\overline{\rho}: G_{F,S} \rightarrow \mathrm{GL}_2(\mathbb{F})$ is the residue representation. Here $S$ is a finite set of finite places of $F$ containing $\{v | p\}$, where $F$ is a number field (a finite extension over $\mathbb{Q}$). Here $G_{F,S}$ is the Galois group of the maximum extension over $F$ which is unramified outside $S \sqcup \{ v | \infty \}$.

$\mathrm{ad}^0 \overline{\rho}$ consists of traceless matrices $M \in M_n(\mathbb{F})$ with $G_{F,S}$-conjugation action $\sigma \cdot X := \overline{\rho}(\sigma) X \overline{\rho}(\sigma)^{-1}$.

$\mathrm{ad}^0 \overline{\rho}(1)$ is its Tate twist $\mathrm{ad}^0 \overline{\rho} \otimes \mathbb{Z}_p(1)$.

Via the $G_{F,S}$-action on $\mathrm{ad}^0 \overline{\rho}$ and $\mathrm{ad}^0 \overline{\rho}(1)$, we can consider the $G_{F,S}$-cohomology with coefficients in $\mathrm{ad}^0 \overline{\rho}$ and $\mathrm{ad}^0 \overline{\rho}(1)$ accordingly.

My question 1: How to prove the following claim:

If $\overline{\rho}|_{G_{F(\zeta_p)}}$ is absolutely irreducible, then $H_0(G_{F,S}, \mathrm{ad}^0 \overline{\rho}(1)) = 0$. (Where $G_{F(\zeta_p)}$ is the absolute Galois group of $F(\zeta_p)$).

My attempt: The zeroth cohomology group $H_0(G_{F,S}, \mathrm{ad}^0 \overline{\rho}(1))$ consists of exactly the elements in $\mathrm{ad}^0 \overline{\rho}(1)$ fixed by $G_{F,S}$, so it is essential to figure out the action of $G_{F,S}$ on $\mathrm{ad}^0 \overline{\rho}(1)$. Then I have found someone claimed that

Elements in $H_0(G_{F,S}, \mathrm{ad}^0 \overline{\rho}(1))$ correspond to the intertwining operators $\overline{\rho} \rightarrow \overline{\rho}(1)$ between $G_{F(\zeta_p)}$-modules.

But how to prove this claim?. After proving this claim, Schur's lemma implies that $H_0(G_{F,S}, \mathrm{ad}^0 \overline{\rho}(1)) = 0$ and we are done.

My question 2: Let $v$ be a Taylor-Wiles prime of $F$. (i.e. $v$ is a finite place of $F$ satisfying the two conditions: the norm of $v$ (denoted by $q_v$) satisfies $q_v \equiv 1 \pmod{p}$; and $\overline{\rho}(\mathrm{Frob}_v) \in \mathrm{GL}_2(\mathbb{F})$ has distinct eigenvalues.) Let $F_v$ be the completion of $F$ wrt $v$ and $G_v$ be its (local) absolute Galois group. Via
$$
G_v \hookrightarrow G_F \twoheadrightarrow G_{F,S} \xrightarrow{\overline{\rho}} \mathrm{GL}_2(\mathbb{F}),
$$
we can consider the $G_v$ action on $\mathrm{ad}^0 \overline{\rho}(1)$. Then how to prove the following claim:

$H_0(G_{v}, \mathrm{ad}^0 \overline{\rho}(1)) = H_0(G_{v}, \mathrm{ad}^0 \overline{\rho}).$

So again the essential point is to understand how $G_v$ acts on $\mathrm{ad}^0 \overline{\rho}(1)$ and I have difficulty on this. I even guess

The actions of $G_{v}$ on $\mathrm{ad}^0 \overline{\rho}(1)$ and $\overline{\rho}$ are the same.

But I don't know how to prove this. (I feel like this is due to the condition that $q_v \equiv 1 \pmod{p}$.)

答:
I think a lot of this is easier to see if we don't fix a basis. Consider $$\overline\rho\colon G_{F, S}\to \mathrm{GL}(V)$$
where $V$ is a two-dimensional $\mathbb F$-vector space (although the arguments here work in higher dimensions too). Then $\mathrm{ad}(\overline\rho)$ is the representation
$$\mathrm{ad}(\overline\rho)\colon G_{F, S}\to \mathrm{GL(End}(V))$$
where $g\in G_{F, S}$ acts on $\phi\in \mathrm{End}(V)$ by sending it to the map $v\mapsto \overline\rho(g)\cdot\phi(\overline\rho(g)^{-1}\cdot v)$. As a representation, $\mathrm{End}(V)$ is canonically isomorphic to $V\otimes V^*$, which decomposes as $\mathbb 1 \oplus \mathrm{ad}^0(\overline\rho)$, where $\mathbb 1$ is the trivial representation.

In fact, in the two-dimensional case, $V^*\cong V\otimes \det(V)^{-1}$, so
$$V\otimes V^* \cong (V\otimes V\otimes \det(V)^{-1}) \cong \mathrm{Sym}^2(\overline\rho)\otimes \det(\overline\rho)^{-1} \oplus\mathbb 1$$
and $\mathrm{ad}^0(\overline\rho) \cong \mathrm{Sym}^2(\overline\rho)\otimes \det(\overline\rho)^{-1}$.

But regardless, $\mathrm{ad}^0(\overline\rho)(1)$ is a subrepresentation of $\mathrm{ad}(\overline\rho)(1) \cong V\otimes\epsilon\otimes V^*\cong \mathrm{Hom}(V, V(1))$, where $\epsilon$ is the mod $p$ cyclotomic character. Hence, elements of $H^0(G_{F, S}, \mathrm{ad}^0(\overline\rho))\cong \mathrm{Hom}_{G_{F, S}}(V, V(1))$ naturally corresponds to intertwining operators between $\rho$ and $\rho(1)$ as $G_{F, S}$-modules.

If $V$ is irreducible as a $G_{F, S}$-module, then by Schur's lemma, $\mathrm{Hom}_{G_{F, S}}(V, V(1)) = 0$ unless $V(1)\cong V$ as $G_{F, S}$-modules, which (by Clifford theory) will happen if and only if $\overline{\rho}|_{G_{F(\zeta_p)}}$ is reducible. (Note that $G_{F(\zeta_p)}$ is the kernel of the mod $p$ cyclotomic character.)

For your second question, note that, if $q_v \equiv 1\pmod p$, then $\epsilon(\mathrm{Frob}_v) = q_v\equiv 1\pmod p$, so $\epsilon|_{G_v}$ is the trivial character.


问:
What is the probability that the area of the triangle with vertices $(0,0)$, $(3,0)$ and $P$ is greater than 2?
Let $P$ be a point chosen at random on the line segment between the points $(0,1)$ and $(3,4)$ on the coordinate plane. What is the probability that the area of the triangle with vertices $(0,0)$, $(3,0)$ and $P$ is greater than 2?

I started off by noticing that the base of the triangle is 3, so the height must be greater than $\frac{4}{3}$ if the area is greater than 2 because $\frac{1}{2}(3)(h)>2 \rightarrow h >\frac{4}{3}.$

Since $y>\frac{4}{3}, x > \frac{1}{3}$ because the line is $y=x+1$.

So for the area of the triangle to be greater than 2, $P$ can be any point from $(\frac{1}{3}, \frac{4}{3})$ to $(3,4)$ on the line $y=x+1$.

Therefore the probability of the triangle area being greater than 2 is  $\dfrac{\text{length of segment (1/3, 4/3) to (3,4)}}{\text{length of segment (0,1) to (3,4)}}=\dfrac{2/3\sqrt{17}}{3\sqrt{2}}=\dfrac{\sqrt{34}}{9}.$

However the solution says the answer is 8/9, so what am I doing wrong here? I'm also pretty sure I overcomplicated a lot of things, and I was also wondering if there's a simpler solution. Thanks!

答:
The area of the triangle is given by the function $\displaystyle A(X) = \frac{3}{2}(X+1)$, where X is uniformly distributed in $[0,3]$, so
\begin{equation}
\mathbb{P}(A(X) \geq 2) = \mathbb{P}\left(X\geq\frac{1}{3} \right) = \frac{1}{3} \left(3 - \frac{1}{3} \right) = \frac{8}{9}
\end{equation}

ContentLicense: CC BY-SA 4.0


答:
The point, $Z$ on the dark green line has coordinates
$x = x(P)$ and $y=area(\triangle PQR)$

So the ratio you are looking for is

$\dfrac{6-2}{6-1.5}=\dfrac{4}{4.5}=\dfrac 89$

Body:
Let $P$ be a point chosen at random on the line segment between the points $(0,1)$ and $(3,4)$ on the coordinate plane. What is the probability that the area of the triangle with vertices $(0,0)$, $(3,0)$ and $P$ is greater than 2?

I started off by noticing that the base of the triangle is 3, so the height must be greater than $\frac{4}{3}$ if the area is greater than 2 because $\frac{1}{2}(3)(h)>2 \rightarrow h >\frac{4}{3}.$

Since $y>\frac{4}{3}, x > \frac{1}{3}$ because the line is $y=x+1$.

So for the area of the triangle to be greater than 2, $P$ can be any point from $(\frac{1}{3}, \frac{4}{3})$ to $(3,4)$ on the line $y=x+1$.

Therefore the probability of the triangle area being greater than 2 is  $\dfrac{\text{length of segment (1/3, 4/3) to (3,4)}}{\text{length of segment (0,1) to (3,4)}}=\dfrac{2/3\sqrt{17}}{3\sqrt{2}}=\dfrac{\sqrt{34}}{9}.$

However the solution says the answer is 8/9, so what am I doing wrong here? I'm also pretty sure I overcomplicated a lot of things, and I was also wondering if there's a simpler solution. Thanks!


问:
I need a hint to combine the two sums into only one
Would it be possible for me to write and if yes could somebody please help me out:

$$\sum_{i=1}^{\lfloor n\{x\}\rfloor}\frac{1}{\lfloor x+1 \rfloor^k} +\sum_{i=\lfloor n\{x\}\rfloor+1}^{n}\frac{1}{\lfloor x \rfloor^k}=\sum_{i=1}^{n}\frac{1}{\lfloor x+ \frac{i-1}{n}\rfloor^k}$$

My main problem is to include $i$ and $n$ in the sum. I would be really thankful for help.

P.S Please, I only want hints, not solutions. I would really like to solve it on my own, it's just that in this specific point i am stuck for days now.

Edit: $\{x\}=x-\lfloor x \rfloor$

答:
Hint:
$$1 \le i \le n \implies 0 \le \frac{i-1}{n} \le \frac{n-1}{n} < 1,$$
so
$$x \le x+\frac{i-1}{n} \le x+\frac{n-1}{n} < x+1$$
and
$$\lfloor x \rfloor
\le \left\lfloor x+\frac{i-1}{n} \right\rfloor
\le \left\lfloor x+\frac{n-1}{n} \right\rfloor
\le \lfloor x+1 \rfloor,$$
hence $$\left\lfloor x+\frac{i-1}{n} \right\rfloor \in \{\lfloor x \rfloor, \lfloor x+1 \rfloor\}.$$

By the way, simplifying the sum of the floor powers (rather than their reciprocals) was the first part of Mathematics Magazine Problem 1785 (December 2007), and the same technique applies here.


问:
Why in scalar product I can multiply two vectors $\overrightarrow{a}_{1x3} * \overrightarrow{b}_{1x3} $ , whereas in matrices products I couldn't?
If a vector is a matrix of one dimention, Why in scalar product I can multiply  two vectors $\overrightarrow{a}_{1x3} * \overrightarrow{b}_{1x3} $ , whereas in matrices products I'd need $a_{1x3} * b_{3x1} $?

For example : scalar product of two vectors :

$$
\overrightarrow{v}*\overrightarrow{w} = \begin{bmatrix}1&2&3 \end{bmatrix}*
\begin{bmatrix}0&1&-1\end{bmatrix}
\\
=(1*0 )+(2*1)+(3*(-1))
\\
\overrightarrow{v}*\overrightarrow{w}=-1
$$

But in matrix product I'd need the transpose of $w$:

$$
v*w^T = \begin{bmatrix} 1&2&3 \end{bmatrix} * \begin{bmatrix}0\\1\\-1 \end{bmatrix}
\\
=(1*0 )+(2*1)+(3*(-1))
\\
v*w^T = -1
$$

So vectors are not matrices of one dimention ?  I'm a little confused.

答: In reality you transpose one vector for scalar multiplication. What is loosely written as $u\cdot v$ is in fact $u^Tv$ assuming that both vectors are columns. If the vectors are complex the scalar product requires not only to transpose the vector but also perform its complex conjugation.


问:
Set-up for the Paris-Harrington Theorem
In his book "Models of Peano Arithmetic" Kaye proves the Paris-Harrington Theorem. He starts off by introducing a "simplification" then proves the short Lemma 14.11 about it (see picture). I don't understand any of this. Specifically:

What is the rational of the "Truth definition for $I$" (just above Lemma 14.11 in the picture)?

Why does "$I \subset_e M$ is a structure of the language $\mathcal{L}_A$ of arithmetic" imply that $I \models \text{PA}^{-}$?

I don't understand at all why the proof shows that $I \models \text{PA}$. Why does the proof imply that $I$ models all induction axioms?

Basically, can someone explain the proof and its background, because I'm completely lost. I would greatly appreciate any help!

答:
Re: (2), first remember that all but one of the axioms of $\mathsf{PA^-}$ are universal, hence downwards-absolute between structures: since $I$ is a substructure of a model of $\mathsf{PA^-}$ it satisfies all of these sentences too.

So this just leaves the thirteenth axiom: $$\mathsf{Ax13}:\quad\forall x,y(x<y\rightarrow \exists z(x+z=y)).$$

This is where we use the fact that $I\subset_eM$, or more intuitively that $I$ is downwards closed. Let $x,y\in I$ with $x<y$. Applying $\mathsf{Ax13}$ in $M$ we get in $M$ some $z$ such that $x+z=y$. In $M$ we have $z\le y$; since $I$ is downwards closed this means $z\in M$. Now the property $x+z=y$ is absolute between $I$ and $M$, so $I\models x+z=y$. And this gives $\mathsf{Ax13}$ in $I$ as desired.

Note that we needed end extension as opposed to mere superstructurehood, in verifying axiom $13$. It's a good exercise to check that every nonstandard model of $\mathsf{PA}$ has a (necessarily non-downwards-closed) substructure not satisfying axiom $13$. On the other hand, if we included "symmetric subtraction" $(x,y)\mapsto \vert x-y\vert$ as a primitive operation, then $\mathsf{Ax13}$ can be rephrased as a universal and after such rephrasing every substructure of a model of $\mathsf{PA^-}$ would again satisfy $\mathsf{PA}^-$.

Title: Set-up for the Paris-Harrington Theorem


答:
The base is of length $3;$ the height is $h = 1 + U,$ where $U \sim \mathsf{Unif}(1,4).$ So the following R code finds areas of a million triangles and finds the proportion with areas exceeding 2.
With a million such simulated triangles the desired probability
$P(A > 2)$ should be estimated to two decimal places.

```<code>set.seed(2021)
a = .5*3*runif(10^6, 1, 4)
mean(a > 2)
[1] 0.889393   # aprx P(A > 2)
8/9
[1] 0.8888889  # exact
</code>```

Congratulations on asking a popular questions. Lots of answers are appearing almost simultaneously. My simulation
is essentially the same idea as in @W's Answer (+1)---and also $Zarrax' (+1).

Title: What is the probability that the area of the triangle with vertices $(0,0)$, $(3,0)$ and $P$ is greater than 2?

Body:
The base is of length $3;$ the height is $h = 1 + U,$ where $U \sim \mathsf{Unif}(1,4).$ So the following R code finds areas of a million triangles and finds the proportion with areas exceeding 2.
With a million such simulated triangles the desired probability
$P(A > 2)$ should be estimated to two decimal places.

set.seed(2021)
a = .5*3*runif(10^6, 1, 4)
mean(a > 2)
[1] 0.889393   # aprx P(A > 2)
8/9
[1] 0.8888889  # exact

Congratulations on asking a popular questions. Lots of answers are appearing almost simultaneously. My simulation
is essentially the same idea as in @W's Answer (+1)---and also $Zarrax' (+1).


问:
How to sove this inequality: $2 \cdot \sqrt{x^2-x-2} -2x +1 > 0$. I know it looks obvious, but there something into it
I get $-8>1$, which mean that $2\sqrt{x^2-x-2} -2x +1 > 0$ is always false , but actually it isn't.

Edit: here is how I got it.

$2\sqrt{x^2-x-2} -2x +1 > 0  $

$2\sqrt{x^2-x-2} > 2x -1 $

$(2\sqrt{x^2-x-2})^2 > (2x -1)^2 $

$4x^2 -4x -8 > 4x^2 -4x +1 $

$4x^2 -4x -8 - 4x^2 + 4x>   +1 $

$ -8 > +1 $

答:
First you should check the domain of this inequality. Since the quantity under the radical must not be negative, the domain is the solution set of
$$x^2 - x - 2 \ge 0
$$
which is
$$x \in (-\infty,-1] \cup [2,+\infty)
$$
Next let's check each of the two pieces of this domain.

On the piece $x \in (-\infty,-1]$, the left hand side of the inequality is $\ge 0$ (because square roots are always $\ge 0$), and the right hand side is negative. So the inequality is true for all $x \in (-\infty,-1]$.

On the piece $x \in [2,\infty)$, both sides of the inequality are positive. Your method therefore applies, to prove that your inequality is equivalent to the false inequality $-8>1$ (you need to carefully justify the step where you squared both sides: but squaring both sides of an inequality of positive numbers does produce an equivalent inequality). So the inequality is false for all $x \in [2,\infty)$.

The solution set is therefore $(-\infty,-1]$.


问:
What does 'i-th' mean?
I have seen a problem set for the tower of hanoi algorithm that states:

Each integer in the second line is in the range 1 to K where the i-th
  integer denotes the peg to which disc of radius i is present in the
  initial configuration.

What does i-th actually mean? If i have a line conforming to the format above such as:

4 2 4 3 1 1

How exactly does this translate to the format described above?

答:
The statement that

the $i$-th integer denotes the peg on which the disk of radius $i$ is present in the initial configuration

is shorthand for this statement:

the first integer in the list is the number of the peg on which the disk of radius $1$ is initially placed; the second integer in the list is the number of the peg on which the disk of radius $2$ is initially placed; the third integer in the list is the number of the peg on which the disk of radius $3$ is initially placed; and so on, until the $K$-th and last integer in the list is the number of the peg on which the disk of radius $K$ is initially placed.

The list $4,2,4,3,1,1$ indicates that the disk of radius $1$ starts on peg $4$, the disk of radius $2$ starts on peg $2$, the disk of radius $3$ starts on peg $4$, the disk of radius $4$ starts on peg $3$, and the disks of radius $5$ and $6$ start on peg $1$. The configuration is therefore this one, where the numbers in the array are the radii of the disks.

$$\begin{array}{cc}
\text{Peg }1&\text{Peg }2&\text{Peg }3&\text{Peg }4\\ \hline
5&&&1\\
6&2&4&3
\end{array}$$

ContentLicense: CC BY-SA 3.0

Title: What does 'i-th' mean?


问:
Intuition behind the maximum principle for the heat equation
The maximum principle for the heat equation say that if $u$ solves the heat equation on $\Omega_T=\Omega \times (0,T]$, then it will take its maximum on the parabolic boundary $\Gamma_T=\overline{\Omega_T}-\Omega_T$, and if the maximum is attained at some $(x_0,t_0)$ point inside, then $u$ is constant on $\bar{\Omega}$ for all $t\leq t_0$.

I wonder what is the physical intuition behind this theorem. Any help would be greatly appreciated.

答:
The heat operator is "smoothing" in the sense that the temperature at any point is the average of the temperature in a ball around that point at a previous time, i.e.
$$u(x_0,t_0) = \text{avg} \; u(B_r(x_0), s_0) \;\;  \text{ for } s_0 < t_0.$$

This is intuitive if you think about what happens to the temperature of an object (a wire in 1D, a thin surface in 2D, etc.).

The maximum principle says that the maximum temperature occurs either at the initial condition (temporal boundary), or on  $\bar{\Omega} - \Omega$ (spatial boundary). Intuitively, this says that any "spikes" in temperature diffuse out over time, and that if at some time $t_0$ there is a maximum at $x_0$ in the interior of $\Omega$, then it cannot be a (new) spike.

This second point makes sense because because of the property $$u(x_0,t_0) = \text{avg} \; u(B_r(x_0), s_0) \;\;  \text{ for } s_0 < t_0.$$

Since we're assuming $u(x_0,t_0) = \max_{\bar{\Omega}_T} u$, this means that $u(B_r(x_0),s_0) \equiv u(x_0,t_0)$ (i.e. the temperature throughout the ball is the same as the max temperature, since any point in the ball is not allowed to be hotter than $u(x_0,t_0)$, and yet the average needs to be equal). If you extend this argument to previous times, expanding the ball, we can see that the temperature must have been constant over all $\bar{\Omega}$ up to time $t_0$!


问:
A very small disc winds around a larger circular disc of radius $R$ connected to it by a string. How long is the spiral it travels?
A very small disc winds around a larger circular disc of radius $R$. It is connected to it by a string of length l that remains tight. What distance does it travel before it hits the larger disc of radius $R$ ? I've estimated that it is $2\pi(l-R)$ based on assuming a spiral is a little like a series of concentric circles.

答:
This has been updated with corrections given in the accepted answer.

Assuming $L>>R$.

Path length approximated assuming that the spiral is approximately a sum of circles with circumference $2\pi R$ for each revolution of the spiral.

Assume $N=\frac{L}{2\pi R}$ is an integer in suitable units for the particular problem.

$\text{sum of diminishing circle circumferences}=$

$(L-2\pi R)2\pi + (L-2\pi R -2\pi R)2\pi +...$

$=2\pi NL - 4\pi^2 R (1+2 +3+...N)=2\pi NL - 4\pi^2 R(N(N+1))/2$

$=\frac{L^2}{R} - 4\pi^2 R\frac{(\frac{L^2}{(2\pi R)^2}+\frac{L}{2\pi R}))}{2}$

$=\frac{L^2}{R} - (\frac{L^2}{2R}+\pi L))$

$=\frac{L^2}{2R}-\pi L$


问:
Need help to understand different formula for Trapezoidal's rule
My book (Numerical Analysis by Douglas) said the formula for Trapezoidal's rule is

$$\int_a^b f(x)\, \Bbb dx \approx \frac h2 \left(f(a) + f(b) + 2\sum_{k=1}^{n-1} f(x_k)\right)$$

Here is the screenshot from the book. (i exclude the error)

And also from wolfram documentation i found the different formula like this:

$$\int_a^b f(x)\, \Bbb dx \approx \frac h2 \left(\sum_{k=0}^{n-1} (f(hk)+ f(h(1+k)))\right)$$

Here is the screenshot of the formula on my android app "Wolfram Calculus" when i was evaluating the $\sin(x)$:

I've tested those formula for finding the integral of $\sin(x)$ from $0$ to $\pi$ and got the same result, the same absolute error, and the same relative error. Which means there's a possibility those formulas are supposed to be equivalent.

So, i realize there is a property like this:
$$\sum_{k=0}^{n-1} i = \sum_{k=1}^{n} i-1$$
But the upper bound on the wolfram documentation doesn't change. I'm really confuse now. Can you explain why those are equivalent? I don't get why $f(a)+f(b)$ dissapeared and $f(x_k)$ is replaced by $(f(hk)+ f(h(1+k)))$. And the constant 2 is missing too.

I included the algebra precalculus as a tag because my question is more about algebraic manipulation rather than numerical analysis.

Hope you can explain it to me. Thank you so much.

答:
It helps if you write out the first couple of terms and the last couple of terms.

The formula you got from "Wolfram documentation" has an error, in that it assumes that $a = 0$.  The correct formula is a bit messier:
$$\int_{a}^{b} f(x) dx ≈ \frac{h}{2}\cdot \sum_{k=0}^{n-1}\big(f(a + hk) +f(a + h(k+1))\big)$$
Now if you put in $k=0$, the first term becomes $$\frac{h}{2}\cdot\big(f(a) + f(a + h)\big)$$ and the second term becomes $$\frac{h}{2}\cdot\big(f(a + h) + f(a + 2h)\big)$$ while the last two terms become
$$\frac{h}{2}\cdot\big(f(a + (n-2)h) + f(a + (n-1)h)\big)$$ and $$\frac{h}{2}\cdot\big(f(a + (n-1)h) + f(a + nh)\big)$$ so the last term in the last part is $f(b)$ because $a + nh = b$.
So you can see that the $f(a)$ and $f(b)$ terms show up once, and all the terms in the middle show up twice.


问:
How to solve $e^{ix} = i$
I am taking an on-line course and the following homework problem was posed: $$e^{ix} = i$$ I have no idea how to solve this problem. I have never dealt with solving equations that have imaginary parts. What are the steps to solving such equations? I am familiar with Taylor series and the Euler formula if that is any help.

Thanks.

答:
$$ e^{ix} = i \implies ix =\ln(i) \implies ix= \ln(|i|)-i \left(\frac{\pi}{2}+2k\pi\right) $$

$$ \implies x = \frac{\pi}{2}+2k\pi,\quad k\in \mathbb{Z}. $$

Title: How to solve $e^{ix} = i$


问:
If the distances between every pair of airports is distinct, prove that more than $5$ planes can't land at any airport

The distances between any two airports of a country are distinct. From every airport one airplane takes off and lands at the closest airport. Prove that there doesn't exist an airport where more than 5 planes will land.

I proved it in the following way:

In order for a plane to take off from a certain airport and land at another, it means that the other airport is the closest to it at a distance $r$. Hence it has a circle with radius r with center the airport which it took off from and with a point on its circumference, the airport at which it landed. If there are more than $5$ points which land at a certain airport, then you get a polygon $A_1A_2A_3...A_n$ with a point inside it $O$. If $n\gt 5$ then we have that at least one $\angle OA_iA_{i+1}\lt60^o$ which is indeed a contradiction.

This is what I managed to do for this question, but I am not completely certain I am completely correct. Could you please verify my solution and explain any other nice approaches and also explain to me why my last statement that $\angle OA_iA_{i+1}\lt60^o$ indeed is a contradiction as I worked it out only intuitively?

答:
Your proof is pretty much complete. Here are the gory details.

The only facts you need about triangles are:

For any two vertices $A$ and $B$, $m(\angle A)\ge m(\angle B)$ implies the side opposite $A$ is at least as long as the side opposite $B$. This follows from the law of sines.

$m(\angle A)+m(\angle B)+m(\angle C)=180^\circ$

Suppose planes at airports $A$ and $B$ both fly to the airport at $C$. I claim this implies $m(\angle ACB)>60^\circ$.

Proof: Since $A$ flew to $C$ instead of $B$, the distance $AC$ must be strictly less than $AB$, implying $m(\angle B)<m(\angle C)$. Similarly, since $B$ flew to $C$ instead of $A$, it must be true that $m(\angle A)<m(\angle C)$. Finally, if we had $m(\angle C)\le 60^\circ$, we would have $$180^\circ=m(\angle A)+m(\angle B)+m(\angle C)<60^\circ+60^\circ+60^\circ=180^\circ,$$ a contradiction. We conclude $m(\angle C)>60^\circ$.

Now, suppose that planes at $A_1,\dots,A_n$ all fly to point $O$, where $A_i$ are listed in clockwise order around $O$. On the one hand,  $m(\angle A_1OA_2)+m(\angle A_2OA_3)+\dots+m(\angle A_nOA_1)=360^\circ$, since the angle form a complete circle. On the other hand, we just proved that each $m(\angle A_iOA_{i+1})>60$. This implies that $n\cdot 60<360$, so that $n<6$.


问:
limit of sequence $U_{n}=\sum_{j=1}^{n}\frac{an+1}{cn^{k}+j}$
How to find the limit of sequences of this general form:
$$
U_{n}=\sum_{j=1}^{n}\frac{an+1}{cn^{k}+j}
$$

$$
k\in\Bbb{N}-\{0, 1\}
$$
$$
(a, b)\in\Bbb{R}^{+}_*\times\Bbb{R}^{+}_*
$$

答:
Assuming $a,c >0$, the function $\displaystyle f_n(x) = \frac{an+1}{cn^k+x}$ is decreasing in $[0, \infty)$, so

\begin{equation}
U_n \leq \int_0^n \frac{an+1}{cn^k+x} \mathrm{d}x
\end{equation}

and

\begin{equation}
\lim_{n\to \infty} U_n \leq \lim_{n \to \infty} (an+1)\log\left(1+\frac{1}{cn^{k-1}}\right)
\end{equation}

So for $k>2$ we have that $U_n \to 0$ if $n \to \infty$. If $k=2$ we get the upper bound of $a/c$. Now use the fact that

\begin{equation}
U_n \geq \int_1^{n+1} f_n(x) \mathrm{d}x
\end{equation}

to conclude that the limit is equal to $a/c$ if $k=2$.


问:
Covariant derivative and $\nabla X$
If I consider the covariant derivative of a vector field $Y$ with respect to $X$ this is usually indicated as $\nabla_X Y$, but what's about $\nabla X$? This is the covariant derivative of $X$ and I have read it is  tensor field of type (1,1) such that if $Y\in \chi(\cal{M})$ (vector field) and $w\in \chi^*(\cal{M})$ (covector)
$$\nabla X(Y,w)=w(\nabla_XY)$$
Why it is true? I only know that given a smooth function $f$ then $\nabla_{f X} Y=f\nabla_X Y$, but really I can't understand the relationship between $\nabla X$ and $\nabla_X Y$ and so why the expression above holds.

Can you help me please?

答:
First off, it should be
\begin{equation}
\nabla X(Y,\omega)=\omega(\nabla_YX)
\end{equation}
Over a general vector bundle $E\rightarrow M$ we have that the connection is a map
$$
\Gamma(E)\xrightarrow{\nabla}\Gamma(T^*M\otimes E)
$$
where $\Gamma(-)$ denotes smooth sections. In your case $E=TM$:
$$
\Gamma(TM)\xrightarrow{\nabla}\Gamma(T^*M\otimes TM)
$$
i.e. $\nabla$ eats a vector field and spits out a (1,1) tensor field. We'll work in coordinates for these calculations. So if $\omega$ is a covector field and $Y$ a vector field, then
\begin{align}
\nabla X(Y,\omega)&=\nabla(X^i\partial_i)(Y,\omega) \\
&=(\nabla X^i\otimes\partial_i)(Y,\omega)+(X^i\nabla\partial_i)(Y,\omega) \\
\end{align}
This is since $\nabla$ satisfies a sort of Leibnitz rule. Continuing:
\begin{align}
&=(d(X^i)\otimes\partial_i)(Y,\omega)+(X^i\nabla\partial_i)(Y,\omega)
\end{align}
Since by definition, $\nabla(f)=df$ for any smooth function. Then:
\begin{align}
&=(\partial_m X^idx^m\otimes\partial_i)(Y,\omega)+(X^i\Gamma^{l}_{mi}dx^m\otimes\partial_l)(Y,\omega) \\
\end{align}
Where the $\Gamma$'s are the usual connection coefficients: $\nabla\partial_i=\Gamma^l_{mi}dx^m\otimes\partial_l$. Hence:
\begin{align}
&=(\partial_mX^i)Y^p\omega_\alpha dx^m(\partial_p)\partial_i(dx^\alpha)+X^i\Gamma^l_{mi}Y^p\omega_\alpha dx^m(\partial_p)\partial_l(dx^\alpha) \\
&=(\partial_m X^i)Y^m\omega_i+X^i\Gamma^l_{mi}Y^m\omega_l
\end{align}
On the other hand,
\begin{align}
\omega(\nabla_YX)&=\omega(\nabla_{Y^m\partial_m}X^i\partial_i) \\
&=\omega[Y^m((\partial_m X^i)\partial_i+X^i \nabla_{\partial_m}\partial_i)] \\
&=\omega[Y^m((\partial_m X^i)\partial_i+X^i\Gamma^s_{mi}\partial_s)] \\
&=\omega_ldx^l[Y^m((\partial_m X^s)+X^i\Gamma^s_{mi})\partial_s] \\
&=\omega_l[(\partial_mX^s)Y^m+X^i\Gamma^s_{mi}Y^m]dx^l(\partial_s) \\
&=(\partial_mX^s)Y^m\omega_s+X^i\Gamma^s_{mi}Y^m\omega_s
\end{align}
where I leave the justifications to you. And then replacing dummy indices yields the result.

Title: Covariant derivative and $\nabla X$


答: This is really not the right way to look at it. $\nabla X$ is a vector-field valued $1$-form $\omega$ whose value  on $Y$ is $\omega(Y)=\nabla_Y X$. That is, it's giving you a (tensorial) map from vector fields to vector fields. It's a section of $T^*M\otimes TM$, and as such is a $(1,1)$-tensor.

Title: Covariant derivative and $\nabla X$

Body:
If I consider the covariant derivative of a vector field $Y$ with respect to $X$ this is usually indicated as $\nabla_X Y$, but what's about $\nabla X$? This is the covariant derivative of $X$ and I have read it is  tensor field of type (1,1) such that if $Y\in \chi(\cal{M})$ (vector field) and $w\in \chi^*(\cal{M})$ (covector)
$$\nabla X(Y,w)=w(\nabla_XY)$$
Why it is true? I only know that given a smooth function $f$ then $\nabla_{f X} Y=f\nabla_X Y$, but really I can't understand the relationship between $\nabla X$ and $\nabla_X Y$ and so why the expression above holds.

Can you help me please?


问:
Help on probability distribution question
Let a random variable $X$ take its values in the set $\{-2, -1, 1, 2, 3\}$. Assume
that its pmf function can be written in the form:
$$
P(X = x_i)=
\begin{cases}
    \frac{x_i^2}{a},& \text{if } x_i\in\{-2, -1, 1, 2, 3\}\\
    0,              & \text{otherwise}
\end{cases}
$$

Find the constant $a$.

I'm not sure where to even start.

答: The sum of $x^2/a$ over your set is $19/a$. For this to be a pmf the sum must equal $1$, so $a=19$.


问:
Given a normed vector space $X$, can you always define a bounded linear functional $f$ which is bounded above and below by the norm?
Let $X$ be a (possibly infinite-dimensional) normed vector space over $\mathbb{R}$, whose norm is denoted $|| \cdot ||$.

Given two non-negative scalars, say $M, N \in \mathbb{R}$ with $M < N$, can you always find a bounded linear functional $f: X \to \mathbb{R}$ such that $M||x|| \leq | f(x) | \leq N || x||$ for all $x \in X$?

Obviously, if $f$ is bounded then $|f(x)| \leq ||f||_{op} || x|| \leq K || x||$ for some non-negative $K \in \mathbb{R}$. But this fact is kind of going in the "wrong direction" so I'm not too sure if this is helpful.

It does seem like there are several corollaries to the Hahn-Banach theorem that might also be related, but again, I'm not sure if there are additional assumptions that I would need to make to force this question to have a positive answer.

答: If $M=0$ then the answer is clearly YES. If $M >0$ then the condition $M\|x\|\leq |f(x)|$ implies thast $f$ is one-to-one. This forces your space $X$ to be one-dimensional.


答: This is never the case if the dimension is bigger than $1$ and $M>0$, simply because you will have a nontrivial kernel. On the other hand if $M\leq 0$, then that part of the condition is automatically satisfied.

Title: Given a normed vector space $X$, can you always define a bounded linear functional $f$ which is bounded above and below by the norm?

Body:
Let $X$ be a (possibly infinite-dimensional) normed vector space over $\mathbb{R}$, whose norm is denoted $|| \cdot ||$.

Given two non-negative scalars, say $M, N \in \mathbb{R}$ with $M < N$, can you always find a bounded linear functional $f: X \to \mathbb{R}$ such that $M||x|| \leq | f(x) | \leq N || x||$ for all $x \in X$?

Obviously, if $f$ is bounded then $|f(x)| \leq ||f||_{op} || x|| \leq K || x||$ for some non-negative $K \in \mathbb{R}$. But this fact is kind of going in the "wrong direction" so I'm not too sure if this is helpful.

It does seem like there are several corollaries to the Hahn-Banach theorem that might also be related, but again, I'm not sure if there are additional assumptions that I would need to make to force this question to have a positive answer.


问:
Find all triples of natural numbers $x$, $y$, $z$ such that $x+2y+2z=xyz$.

Find all triples of natural numbers $x$, $y$, $z$ such that $x+2y+2z=xyz$.

I found a solution, but it involves a lot of case work. Can someone help me find a solution which doesn't involve a lot of case work?

答: We have: $y = \dfrac{x+2z}{xz-2}\ge 1\implies x+2z\ge xz-2\implies 2z+2\ge xz-x = x(z-1)\implies x \le \dfrac{2z+2}{z-1}$. Of course the case $z = 1$ warrants a separate consideration and it's not hard to solve for this case as you have down to $2$-variable equation in $x,y$ only and you can isolate either $x$ or $y$ and take care of the case when a fraction is an integer. So for $z > 1$, then $x \le \dfrac{2z+2}{z-1} \le 3$ when $z \ge 5$. Thus you have to consider cases $z = 1,2,3,4$ first as well. And when $z \ge 5$, you have $x = 1,2,3$ and treat each case again. So in brief, it seems that's what amounts to and you have to be willing to work hard this time to treat about $10$ cases !


问:
Intuition behind the coupon collector problem. Is there inclusion-exclusion principle in play?
As well-known the expected number of coupons $N$ required to collect the complete set of $n$ coupons in the general case of non-uniform probability distribution can be computed as:
$$
\mathbb E (N)=\sum_i\frac1{p_i}-\sum_{(i,j)}\frac1{p_i+p_j}+\sum_{(i,j,k)}\frac1{p_i+p_j+p_k}-\cdots-\frac{(-1)^n}{\underbrace{p_1+p_2+\cdots+p_n}_{=1}},\tag1
$$
where $p_i$ is the probability to obtain $i$-th coupon.

My question is: can this pattern be explained on the basis of the inclusion-exclusion principle? It really strikes that the first term $\frac1{p_i}$ is nothing else as the expected number of coupons till obtaining the $i$-th coupon, next term $\frac1{p_i+p_j}$ is the expected number of coupons till obtaining the $i$-th or the $j$-th coupon and so on.

I am looking for intuitively understandable explanation which would be consistent with the above interpretation of the separate terms.

I would appreciate any hint.

答:
Yes, it is exactly inclusion exclusion. For each $t\in \mathbb N_0$ and $k\in [n]$, let

$E_t$ be the event that not all the coupons have been collected after $t$ coupons have been drawn,

$E_{k,t}$ be the event that coupon $k$ does not appear in the first $t$ coupons.

Then $E_t=\bigcup_{k=1}^n E_{k,t}$, so inclusion-exclusion applies. It all comes together with these three pieces:
$$
E[N]=\sum_{t\ge 0}P(E_t),\\
P(E_{t})=\sum_{S\subseteq [n]} (-1)^{|S|+1}P\left(\bigcap_{k\in S} E_{k,t}\right),\\
P\left(\bigcap_{k\in S} E_{k,t}\right)=\left(1-\sum_{k\in S}p_k\right)^t.
$$
To be clear, $S$ ranges over only nonempty subsets of $[n]$.


问:
Kiselev's geometry Problem 67: In an isosceles triangle, two medians/bisectors/altitudes are congruent
Problem 67 from Kiselev's Geometry,

Prove that in an isosceles triangle, two medians are congruent, two angle bisectors are congruent, two altitudes are congruent.

Here is the content page in the book. The problem belongs to Chapter 1.5.

The theorems covered in chapter 1.5 are,

(i) In an isosceles triangle, the bisector of the angle at the vertex is at the same time the median and the altitude.

(ii) In an isosceles triangle, the angles at the base are congruent.

(iii) In an isosceles triangle, the bisector of the angle at the vertex is an axis of symmetry of the triangle.

I have understood the proof for these theorems and I have tried to use these theorems and others mentioned in previous chapters to prove the theorem in problem 67, but I can't prove it.

Please help. Also, please avoid using theorems that are covered in the later part of the book.   :)

答:
This problem comes before the theorems on triangle congruence are introduced. An earlier commenter says, "you could make an argument based on the axis of symmetry", and in fact this is exactly what Kiselev intends: half the chapter is devoted to proving various properties of isosceles triangles, and the other half is devoted to explaining the concept of 'axial symmetry' and showing that (in Kiselev's words):

in an isosceles triangle, the bisector of the angle at the vertex is
an axis of symmetry of the triangle.

Following this, Kiselev remarks:

Two symmetric figures can be superimposed by rotating one of them in
space about the axis of symmetry until the rotated figure falls into
the original plane again.

Remembering that congruence is defined in terms of superposition, it is easy to understand the proofs Kiselev was trying to lead his reader towards:  By rotating an isosceles triangle around its axis of symmetry, we carry one median onto another (because the sides are congruent), one angle bisector onto another (because the base angles are congruent), and one altitude onto another (because the perpendicular drawn from a point to a line is unique (§24)).


问:
Uniform convergence, differentiability theorem's converse
Suppose $\{f_n \}$ is a sequence of differentiable functions on $[a,b]$ such that $f_n \to f$ uniformly, and $f'(x) = \lim_{n \to \infty} f_n'(x)$ (pointwise). Then is it true that $f_n'$ converges uniformly?

My initial reaction is that the answer is yes, because all the differentiable $f_n$'s will be able to fit inside an "$\epsilon \text{-thick}$ tube" around $f$, but intuition for sequences of functions rarely seem to be correct. I'd appreciate some help with this; thank you!

答: Let $f_n(x)=\frac {x^{n}} n$ on $[0,1]$ Then $f_n \to 0$ uniformly and $f_n'(x) \to 0$ for $x <1$, $f_n'(x) \to 1$ for $x=1$. Since the limit is not continuous it follows that $f_n'$ does not converge uniformly.


问:
Polynomial with real coefficient which has root $\sqrt{5}+\sqrt{7}$
Is it possible to find a polynomial with real coefficients with a root of $\sqrt{5}+\sqrt{7}$ ?

答:
If all you want are "real coefficients", then $x-(\sqrt 5 + \sqrt 7) =0$ will suffice. Just a simple linear polynomial.

But this sort of question usually involves finding polynomials with integer coefficients.

The usual way is to start by writing $x = \sqrt 5 + \sqrt 7$

Then square both sides. Use the binomial theorem to expand the RHS.

You'll have $x^2 = m +2\sqrt n$, where $m, n$ are integers you can find (you have to do this work yourself).

Then rearrange to $x^2 - m = 2\sqrt n$ (keep the surd terms on one side in every step) and square again.

You'll end up with a quartic (degree 4) with integer coefficients, one root of which is guaranteed to be the surd you started with.


问:
what's the normal vector of plane doesn't cross origin
Given two vectors $x, y \in \mathbb{R}^3$, one might find their normal vector $n:=x \times y$ via a cross product.
But then the cross product is only applicable to planes that cross the origin. What about planes that don't cross the origin, how can one find the normal vectors? For example, we translate the $x, y$ plane via adding another vector $z \in \mathbb{R}^3$. Would the resulting normal vector just be $z + x \times y$ ?

答: You might want the foot of the vector to be at $z$ (so it is a free vector), but the direction will still be $x \times y$.


问:
Covariance of Brownian motion intuition
So the covariance between two time-instances of Brownian motion is $$\text{Cov}(B_s, B_t) = \min(s,t).$$

This post gives a derivation of this fact, but I'm lacking intuition.

Suppose $\epsilon \ll 1 $ is small and $t$ is larger. Then
$$ \text{Cov}(B_\epsilon, B_t) = \epsilon,$$
$$\text{Cov}(B_{t+\epsilon}, B_{2t}) = t+\epsilon.$$

In both cases the time step is $t-\epsilon$, but the covariances are drastically different.

Brownian motion is stationary, so why is this true? (And if not, what am I doing wrong?)

答: Your computation is correct. However, Brownian motion itself is not a stationary process. It's increments are though. This means that for for any $\epsilon>0$, the process $\{B_{t+\epsilon}-B_t\}_{t\geq 0}$ is stationary, ie has the same distribution. In particular by setting $t=0$ we get that $B_{t+\epsilon}-B_t$ is equal in distribution to $B_\epsilon-B_0$ and hence to $B_\epsilon$.

Title: Covariance of Brownian motion intuition


答:
If a plane does not pass through the origin, you can translate it so that it does. The normal vectors to the two planes are the same. Notice that if you have the plane $x+y+2z=0$, its normal vector is $\vec n = (1,1,2)$. If I take the plane $x+y+2z=3$, it is parallel, and so the normal must be in the same direction. This does not mean that the normal vector is orthogonal to the vectors from the origin to points in the plane; it means that it is orthogonal to vectors lying in the plane. This means that if we take $(x_0,y_0,z_0)$ (for example, $(0,1,1)$) in the plane, then
$$\vec n\cdot [(x,y,z)-(x_0,y_0,z_0)] = 0,$$
which is the same as saying
$$\vec n\cdot (x,y,z) = \vec n\cdot (x_0,y_0,z_0) = 3,$$
in our case.

Title: what's the normal vector of plane doesn't cross origin

Body:
Given two vectors $x, y \in \mathbb{R}^3$, one might find their normal vector $n:=x \times y$ via a cross product.
But then the cross product is only applicable to planes that cross the origin. What about planes that don't cross the origin, how can one find the normal vectors? For example, we translate the $x, y$ plane via adding another vector $z \in \mathbb{R}^3$. Would the resulting normal vector just be $z + x \times y$ ?


答: Per your question with a "real coefficient", all you have to do is take $P(x) = x - \sqrt{5}-\sqrt{7}$ and it does it. But if you want a polynomial with "integer" coefficients, then you need a bit more work. You can take advantage of "conjugate" and form your answer quickly as follows: Take $P(x) = (x-\sqrt{5}-\sqrt{7})(x-\sqrt{5}+\sqrt{7})((x+\sqrt{5})^2-7)$. You need not simplify this and still know it's an integer coefficient polynomial with a root $x = \sqrt{5}+\sqrt{7}$.

Title: Polynomial with real coefficient which has root $\sqrt{5}+\sqrt{7}$

Body: Is it possible to find a polynomial with real coefficients with a root of $\sqrt{5}+\sqrt{7}$ ?


问:
If $g : S \rightarrow \mathbb{R}$ is a continuous function which $\forall s\in S:g(s)\ge 0$ and $\exists s\in S:g(s)>0$, does $\int_{S}g>0$?
I'm trying to prove that if a continuous function $g : S \rightarrow \mathbb{R}$ verifies that $\forall s \in S : g(s) \ge 0$ and $\exists s \in S :g(s) > 0$ where S is an open set that has volume $\Rightarrow\int_{S} g > 0$.

I think it might be useful to find a $B(s,r) \subset A $ which verifies $\forall k \in B(s,r): f(k) > 0$
but I'm not sure how I could do it.

I'm stuck on this proof, any hints or suggestions?

答: Hint: use $\varepsilon := g(s)/2$ in the $\varepsilon$-$\delta$ definition of continuity where $s \in S$ such that $g(s) > 0$.


答: $\{s': g(s') >\frac {g(s)} 2\}$ is an open set containing $s$ so there exists $r>0$ such that $B(s,r) \subset S$ and $g(s') > \frac {g(s)} 2$ for all $s' \in B(s,r) $.

Title: If $g : S \rightarrow \mathbb{R}$ is a continuous function which $\forall s\in S:g(s)\ge 0$ and $\exists s\in S:g(s)>0$, does $\int_{S}g>0$?

Body:
I'm trying to prove that if a continuous function $g : S \rightarrow \mathbb{R}$ verifies that $\forall s \in S : g(s) \ge 0$ and $\exists s \in S :g(s) > 0$ where S is an open set that has volume $\Rightarrow\int_{S} g > 0$.

I think it might be useful to find a $B(s,r) \subset A $ which verifies $\forall k \in B(s,r): f(k) > 0$
but I'm not sure how I could do it.

I'm stuck on this proof, any hints or suggestions?


问:
Does the Borsuk-Ulam theorem work for the real projective space?
The Borsuk-Ulam theorem states that if $f$ is a continuous function from the $n$-dimensional sphere $S^n$ to the $n$-dimensional euclidean space $\mathbb{R}^n$, then there are two antipodal points of $S^n$ which map to the same point.

I would like to apply this to a function $f$ which maps some points of $S^n$ to points at infinity.
Is the theorem still true if we replace $\mathbb{R}^n$ with the real projective space $RP^n$?

I tried searching for "Borsuk-Ulam real projective space" online but only came across things which go way over my head. As I am not a mathematician (nor math student) I'd prefer answers which stick to the basics and do not go to deep into stuff which I'm likely not to understand with my current level of mathematical sophistication (cohomology, fiber bundles, CW-complex, etc.) I would also be happy with references to introductory material which could answer my question (or help me answer it for myself).

答:
This is not true. A projective line is homeomorphic to a circle.

Map points $f: S^1 \rightarrow RP^1: (\cos(\theta), \sin(\theta) ) \mapsto [\cos(\frac{\theta}{2}): \sin(\frac{\theta}{2})]$
This is a bijection since angles that differ by $\pi$ represent the same point in $RP^1$.
It is also a homeomorphism with a continuous inverse.

Clearly this disproves the idea that there would be antipodal points mapped to the same point for any continuous map.


问:
Estimate limit without using L'Hopital
It is possible to estimate the following limit without using L'Hopital Rule?

$$\lim_{x\to \frac{\pi }{3}}\left(\frac{\sin\left(x-\frac{\pi}{3}\right)}{1-2\cos\left(x\right)}\right)$$

I will be happy to get any tip!

答:
First, recenter around $0$, because $0$ is nicer to think about, and it's easier to recognize identities around it:
$$
\lim_{x\to \frac{\pi }{3}} \frac{\sin\left(x-\frac{\pi}{3}\right)}{1-2\cos x}
= \lim_{x\to 0} \frac{\sin x}{1-2\cos\left(x+\frac{\pi }{3}\right)}
$$

Next, using the identity $\cos(a+b)=\cos a \cos b-\sin a \sin b$, you get
$$
1-2\cos\left(x+\frac{\pi }{3}\right)
= 1-\cos x + \sqrt{3}\sin x
$$

At this point, you may "expect" $\sin x \approx x$ and $1-\cos x \approx x^2/2$ around $0$ (by a Taylor series expansion), which tells you the limit should be $1/\sqrt{3}$. If you don't have that intuition, no problem! It's not necessary. What we want to use are standard limits around $0$, like $\lim_{x\to 0}\frac{\sin x}{x} = 1$.

You can rewrite
$$
\frac{\sin x}{1-2\cos\left(x+\frac{\pi }{3}\right)}
= \frac{\sin x}{1-\cos x + \sqrt{3}\sin x}
= \frac{1}{\frac{1-\cos x}{x}\cdot \frac{x}{{\sin x}} + \sqrt{3}}
$$
and use the facts that
$$
\lim_{x\to0} \frac{1-\cos x}{x} = 0, \qquad \lim_{x\to0} \frac{\sin x}{x} = 1
$$
(can you see why?) to conclude that
$$
\lim_{x\to\frac{\pi}{3}}\frac{\sin x}{1-2\cos\left(x+\frac{\pi }{3}\right)}
= \frac{1}{0\cdot 1 + \sqrt{3}} = \frac{1}{\sqrt{3}}
$$

Title: Estimate limit without using L'Hopital


问:
Intersection of sets is indeed a set
My question is pretty straight forward - how do I use ZF axioms to prove that if $\{A\}_{i\in\Bbb{N}}$ is a family of sets, then $\bigcap_{i}A_i$ is a set?

I feel like I should be using the axiom of union and the axiom of separation, but I'm not quite sure how.

答:
To be precise, we must prove the following statement:

For all non-empty $I$, for all index families $\{A_i\}_{i \in I}$, there exists a unique set $S$ such that for all $x$, $x \in S \iff \forall i, x \in A_i$.

To prove this, suppose we such an $I$ and a family $A$. Then let $j$ be an element of $I$. Then by the axiom of separation, there exists $S$ such that for all $x$, $x \in S \iff x \in A_j \land \forall i \in I, x \in A_i$. We then immediately see that for all $x$, $x \in S \iff \forall i \in I, x \in A_i$. Thus, we have shown existence.

For uniqueness, we note that if we have two such sets $S$ and $S'$, then for all $x$, $x \in S \iff (\forall i \in I, x \in A_i) \iff x \in S'$ and hence $S = S'$ by the axiom of extensionality.

Title: Intersection of sets is indeed a set


问:
Evaluate conditional expectation $E[e^X | Y=1]$
Given $X,Y \sim N(0,1)$, X,Y follows a joint normal distribution and $corr(X,Y) = \frac{1}{2}$.

Evaluate $E[e^X|Y=1]$.

I'm not sure where to start on this one. I did a change of variable $Z = e^X$ and use the definition of conditional expectaion: $E[Z|Y=1] = \sum_{z} zp_{Z|Y}(z|y)$ but it isn't helpful.

答:
Since the law of a Gaussian vector of mean zero is determined by its covariance matrix,
you can write $X=Y/2+t Z$ where $Y,Z$ are i.i.d. $N(0,1)$ variables and $t=\frac{\sqrt 3}{2}$ (In the sense that this pair $(X,Y)$ has the same joint law as the given variables.) Therefore
$$E[e^X|Y=1]=e^{1/2} E[e^{tZ}]= e^{1/2} e^{t^2/2}= e^{5/4} \,$$

https://en.wikipedia.org/wiki/Normal_distribution#Moment_and_cumulant_generating_functions


问:
Necessary and sufficient condition for a degree 4 polynomial to be a sum of 4th powers
This is a problem I encountered in Hall and Knight's Higher Algebra. Suppose we have a polynomial $p(x,y)=a_0x^4+4a_1x^3y+6a_2x^2y^2+4a_3xy^3+a_4y^4.$
I want to find a necessary and sufficient condition on the coefficients such that $p(x,y)$ is equal to the sum of the fourth powers of 2 polynomials linear in $x$ and $y$.

I can find a necessary condition as follows:
Suppose $p(x,y)=(ax+by)^4+(cx+dy)^4$. Then, comparing coefficients,
$$a^4+c^4=a_0,$$
$$a^3b+c^3d=a_1,$$
$$a^2b^2+c^2d^2=a_2,$$
$$ab^3+cd^3=a_3,$$
$$b^4+d^4=a_4.$$

Then, doing some simple computations:
$$a_0a_2-a_1^2=a^2c^2(ad-bc)^2,$$
$$a_1a_3-a_2^2=abcd(ad-bc)^2,$$
$$a_2a_4-a_3^2=b^2d^2(ad-bc)^2.$$
Hence,
$$(a_0a_2-a_1^2)(a_2a_4-a_3^2)=(a_1a_3-a_2)^2.$$

Therefore, this condition is definitely necessary. I have a hunch that this should also be sufficient, but I don't know how to proceed with that. Are there any more conditions needed? If not, can we prove that if this condition is satisfied, one can solve for $a,b,c,d$? Any hints/suggestions would be appreciated.

答:
Not complete answer but some observations

Eigenvalues

\begin{align}
  \lambda_1 \lambda_2 &= \det
  \begin{pmatrix}
    a_0 & a_1 \\
    a_1 & a_2
  \end{pmatrix} \\
  \mu_1 \mu_2 &= \det
  \begin{pmatrix}
    a_1 & a_2 \\
    a_2 & a_3
  \end{pmatrix} \\
  \nu_1 \nu_2 &= \det
  \begin{pmatrix}
    a_2 & a_3 \\
    a_3 & a_4
  \end{pmatrix} \\
\end{align}

Necessary condition

$$\lambda_1 \lambda_2 \nu_1 \nu_2 =(\mu_1 \mu_2)^2 \iff a_2 \det
\begin{pmatrix}
  a_0 & a_1 & a_2 \\
  a_1 & a_2 & a_3 \\
  a_2 & a_3 & a_4
\end{pmatrix}=0$$

(Binary) quartic form:

$$p(x,y)=
\begin{pmatrix}
  x^2 & 2xy & y^2
\end{pmatrix}
\begin{pmatrix}
  a_0 & a_1 & a_2 \\
  a_1 & a_2 & a_3 \\
  a_2 & a_3 & a_4
\end{pmatrix}
\begin{pmatrix}
  x^2 \\ 2xy \\ y^2
\end{pmatrix}$$


问:
Is knowing two matrices have the same eigenvalues sufficient evidence to prove similarity?
I feel like a lot of resources imply that

Matrixes A and B are similar if and only if tr(A) = tr(B) and det(A) = det(B) and the set of eigenvalues for A and the set of eigenvalues for B are the same.

but I haven't been able to confirm this with Google.

答:
No, this is false. Consider the matrices

$M = \begin{pmatrix}
  1 & 0 \\
  1 & 1 \end{pmatrix}$

and

$I = \begin{pmatrix}
  1 & 0 \\
  0 & 1 \end{pmatrix}$

The two matrices both have the same determinant, trace, and set of eigenvalues. But they are clearly not similar. Because if $M = K^{-1} I K$, then we would have $M = I$.

If you require that both matrices have a basis of eigenvectors and the same eigenvalues (with the same multiplicities) then the matrices are similar - just change basis from one eigenbasis to the other.


答:
They way I learned it (I believe this is how Halmos gives it, but he isn’t writing a fully formal presentation of $\mathsf{ZF}$), the Axiom of Union guarantees the existence of the unary union. We know that for all $X$ there exists $A$ such that
$$\forall z\Bigl(\bigl(\exists y (y\in X\wedge z\in y)\bigr)\longrightarrow z\in A\Bigr).$$
and then with Separation we can define
$$\bigcup X = \{z\mid \exists y(y\in X\wedge z\in y)\}.$$
Once we have defined the (unary) union, we define the unary intersection by
$$\bigcap X = \left\{z\in \bigcup X\Bigm| \forall y (y\in X\longrightarrow z\in y)\right\}$$
which is a set by the Axiom of Separation.

Note that with this definition, you would get that the empty intersection is empty; some people don’t like that (and there are good reasons for avoiding it, and good reasons for not minding or even wishing that). In any case, you can use the latter definition to conclude that $\bigcap X$ is a set in $\mathsf{ZF}$ for any particular nonempty $X$.

Title: Intersection of sets is indeed a set

Body:
My question is pretty straight forward - how do I use ZF axioms to prove that if $\{A\}_{i\in\Bbb{N}}$ is a family of sets, then $\bigcap_{i}A_i$ is a set?

I feel like I should be using the axiom of union and the axiom of separation, but I'm not quite sure how.


问:
On existence of an hamiltonian path in cartesian power of directed cycle graph
Is it true, and if so, how to show it, that there is a Hamiltonian path in the cartesian power of a directed cycle graph (i.e. the iterated cartesian/box product $\square$) $C_n^{\square r}$ where $n, r \geq 2$?

I found some articles dealing with this topic, like this one: http://people.uleth.ca/~dave.morris/papers/HPathCartPower.pdf which seems to show a stronger result and is of a very high level for me, which is the characterization of the extremities of the Hamiltonian path according to $n$. Perhaps it is possible from these results to obtain that there is always such a path, but I have the impression (perhaps wrong) that it could exist simpler, as a combinatorial argument.

I'm sorry if it's obvious, but it's not for me as a beginner in graph theory.

答:
If $G$ and $H$ both have a Hamiltonian path starting at any vertex, then the same thing is true of $G \mathbin{\square} H$. From this, we can find a Hamiltonian path in $C_n^{\square r}$ by induction on $r$.

To prove the claim, pick an arbitrary vertex $(v_1,w_1)$ in $G \mathbin{\square} H$. Use our assumption on $G$ and $H$ to find:

A Hamiltonian path $v_1, v_2, \dots, v_n$ in $G$.

A sequence of $n$ Hamiltonian paths $P_1, P_2, \dots, P_n$ in $H$, where $P_1$ begins at $w_1$ and $P_i$ begins at the last vertex of $P_{i-1}$ for all $i>1$.

Let $P_i'$ be the path in $G \mathbin{\square} H$ where every vertex of $P_i$ is prefixed by $v_i$. Then concatenating $P_1', P_2', \dots, P_n'$ gives a Hamiltonian path in $G \mathbin{\square} H$ starting at $(v_1, w_1)$.


答:
If the underlying vector space is one-dimensional, then yes. Otherwise, no.

Counter-example in the latter case is the zero matrix and any non-zero nilpotent matrix, i.e. a matrix $N$ such that $N^k=0$ for a positive integer $k > 1$. Both have only zero as the eigenvalue and consequently have zero trace and zero determinant, but are not similar since the only matrix similar to the zero matrix is the zero matrix itself.

Title: Is knowing two matrices have the same eigenvalues sufficient evidence to prove similarity?

Body:
I feel like a lot of resources imply that

Matrixes A and B are similar if and only if tr(A) = tr(B) and det(A) = det(B) and the set of eigenvalues for A and the set of eigenvalues for B are the same.

but I haven't been able to confirm this with Google.


问:
Expected number of elements before 1 out of 2 hash tables are full?
From a previous asked question it became clear to me that the expected number of elements to fill up one hash table (assuming ideal randomness) is just the Harmonic number: $mH_m$ where $m$ is the size of the table (collisions could just be solved with chaining or whatever. Don't care.)

However, a more tricky question is:

What is the expected number of elements we need before one of the hash tables are full? I.e. I want to know how fast it takes before one of the tables are full. We can assume that choosing either table $a$ or $b$ has probability $P[a|b]= \frac{1}{2}$ (see figure below).

答:
At the prompting of @user, I wanted to give an intuitive explanation of the formula found the in other answer.

Let $T_1$ be the time it takes to fill the first hash table, and $T_2$ be the time it takes to fill the second. You want $E[\min(T_1,T_2)]$. Since $\min(T_1,T_2)=T_1+T_2-\max(T_1,T_2)$, this is the same as
$$
E[T_1]+E[T_2]-E[\max(T_1,T_2)]
$$
Note that $\max(T_1,T_2)$ is the time it takes to fill out both hash tables, which is just like filling out a single larger hash table with size $2m$. Therefore, $E[\max(T_1,T_2)]=2m H_{2m}$, using what you already knew.

Symmetry implies $E[T_1]=E[T_2]$, but what is $E[T_1]$? It is not the same as the single hash table problem, since only about half of the hashes contribute to the first table. If every hash went to the first table, the expected time would be $mH_m$, so it stands to reason the expected time when you are going at half-speed should be $2mH_m$. This can be made rigorous using Wald's equation and an appropriate stochastic process, but hopefully this is at least believable.

Putting this altogether, you get $E[\min(T_1,T_2)]=2mH_m+2mH_m-2mH_{2m}$, as in the other answer.


问:
Regarding limits and $1^\infty$

Possible Duplicate:

  Why is $1^{\infty}$ considered to be an indeterminate form

I have some questions about limits and the undefinability of $1^\infty$.

For example, is $\lim_{x\to\infty}1^x$ indefinite? Why is it not $1$? Or do mathematicians, when saying that $1^\infty$ is indefinite, actually refer to cases such as $lim_{x\to\infty} \left(1 + \frac{a}{x}\right)^x$ where even though at a first glace the result is $1$, this is actually a special case and it is equal to $e^a$?

答: By itself, $1^{\infty}$ doesn't mean anything, except the implication that you had a quantity of the form $a^b$, and $\lim_{x\rightarrow\infty}a=1$ and $\lim_{x\rightarrow\infty}b=\infty$.  Depending on how quickly and from what direction $a$ approaches 1, and how quickly $b$ approaches $\infty$, $a^b$ could approach any non-negative number, infinity, or nothing in particular at all.</p>

Title: Regarding limits and $1^\infty$


问:
Subtracting roots of unity. Specifically $\omega^3 - \omega^2$
This is question that came up in one of the past papers I have been doing for my exams. Its says that if $\omega=\cos(\pi/5)+i\sin(\pi/5)$. What is $\omega^3-\omega^2$. I can find $\omega^3$ and $\omega^2$ by De Moivre's Theorem. But I cant make much headway into how to subtract these?
The answer they give is $2cos(\dfrac{3\pi}{5})$

Please, some help will be greatly appreciated. I cant seem to find help over the internet either.

答:
Hint: $\cos(3\pi/5)$ and $\cos(2\pi/5)$ are closely related, as are $\sin(3\pi/5)$ and $\sin(2\pi/5)$.

Remark If you are familiar with the complex exponential, note that $\omega^2\omega^3=e^{i\pi}=-1$. So we are looking at $e^{3\pi i/5}+e^{-3\pi i/5}$, which is $2\cos (3\pi/5)$.


问:
Why doesn't the lack of pointwise convergence of $f_n(x)=x^n$ contradict the fact that $C[0,1]$ is Banach?
On this answer, the function $f_n(x)=x^n$ in the interval $[0,1]$ is given as a pathologic example with pointwise convergence.

Can I say that this Cauchy sequence does not (pointwise) converge because the limit of the sequence is a function like this (not continuous):

without specifying any particular norm? I read that pointwise convergence, doesn't imply $d_\infty$ (uniform) convergence, and that uniform convergence implies pointwise convergence. But does lack of pointwise convergence negate uniform convergence?

Does this contradict in any way (or under certain norms) the fact that $C[a,b]$ with respect to $\Vert f \Vert_{\infty}$ is a Banach space? In other words, why is not an example of a Cauchy sequence that does not converge to some $f\in C[0,1]$?

答:
The notion of a sequence being "Cauchy" is inherently a metric concept. When you talk about a Cauchy sequence, you need to be talking about convergence in a metric. Pointwise convergence does not correspond to any metric (this can be proven). If you use the $\|f\|_{\infty}$ metric, the sequence isn't Cauchy. For any finite $m$, there is some point $x_m$ sufficiently close to $1$ for which $x_{m}^{m} \approx 1$. However, if you take $n$ sufficiently large, $x_{m}^{n} \approx 0$, so that $\|x^{n} - x^{m}\|_{\infty} \approx 1$. Since the sequence isn't Cauchy, it won't converge in our metric.

As for the question: if the pointwise limit is not continuous, can the sequence converge in the $\|f\|_{\infty}$ metric? The answer here is no: if it did converge in $\|f\|_{\infty}$, the limit would be continuous (because our space is Banach, so complete) , and since the pointwise limit has to be the same as the uniform limit, we would arrive at a contradiction.


问:
Limits using Cramer's Rule as determinant approaches 0
I'm in Linear Algebra 1, and having just covered Cramer's Rule, the prof showed this interesting case that I have a further question about the significance of.

Say we have a matrix containing a constant that can be adjusted, for instance, the system of equations

2cx+3y=6

4x+(c-1)y=4

giving

$\begin{pmatrix}2c&3\\4&c-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}6\\4 \end{pmatrix}$

Since Cramer's rule only holds in cases where the determinant is nonzero, a typical question would be to find the values of c for which that is true. In this case, det(A)=0 when c=-2 or c=3. At c=-2 there are no solutions to the system, and at c=3 there are infinitely many solutions.

In the case of c=3, we cannot simply apply Cramer's rule, because the denominator of x=$\frac{detA(1)}{detA}$ and y=$\frac{detA(2)}{detA}$ are both detA=0.

However, what we can do is go back to the original system, leaving the variable c in the matrix, and calculate the values of detA, detA(1) and detA(2) in relation to c.

If I do that, and completely factor, I get:

detA=2(c+2)(c-3)

detA(1)=6(c-3)

detA(2)=8(c-3)

Now I can use limits to get an answer from the formulation of Cramer's rule in the case of c=3.

x= $\lim\limits_{c \to 3}\frac{6(c-3)}{2(c+2)(c-3)}$

y= $\lim\limits_{c \to 3}\frac{8(c-3)}{2(c+2)(c-3)}$

From which we can easily get the values of x=$\frac{3}{5}$ and y=$\frac{4}{5}$, which is a valid solution.

So Cramer's rule, despite its initial misgivings, has provided a solution to a system with a determinant of 0. My question (which my prof couldn't answer on the spot, which is why I'm bringing it here) is, which solution? What is special about these numbers, that they're the ones that happen to be spat out using this method? My first thought was that perhaps it's the particular solution, but it's not: the solution in parameterized form of the matrix when c=3 is

$\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}1\\0 \end{pmatrix}+s\begin{pmatrix}-1/2\\1 \end{pmatrix}$

So I am at a loss as to what these numbers "are," if they "are" anything in particular. Surely they're not just random?

答:
Maybe it helps to visualize the two equations as lines in the plane. As you vary $c,$ the lines move around in the plane in a certain way. Most of the time, they will have exactly one point of intersection. But if $c=-2,$ they are parallel and distinct, and if $c=3,$ they are identical and can both be described by the equation $2x+y=2$.

If you vary $c$ continuously, the point of intersection will also move around continuously in the plane. When $c$ approaches $3,$ the point of intersection will approach one particular point on the line described by the equation $2x+y=2,$ namely $\left(\frac 35,\frac 45\right).$

The way in which $c$ influences the matrix elements also determines the way the lines move in the plane when you vary $c$ and hence determines the point approached by the intersection of the lines. You can construct other systems of linear equations that also have infinitely many solutions of the form $2x+y=2$ for a particular value of a certain parameter, but the point, which makes the curve of the intersection points continuous, is somewhere else on that line.

As the location of that point is strongly coupled to the particular way your parameter modifies the system of equations, you cannot make more general statements about this point. This point is special only for your particular system of equations and your particular way of parameterizing it.

So all you can say is: If you want to get exactly one particular solution of the system of equations for each $c$ for which at least one solution exists, and if you want this solution to depend continuously on the parameter $c$ where continuity is possible, then you have to choose $(x,y) = \left(\frac 35,\frac 45\right)$ for $c=3.$


问:
Show a geometry can be embedded in 3D euclidean space
Consider the 2D space with line element $ds^2=\frac{dr^2}{1-\frac{2\mu}{r}}+r^2d\phi^2$. How can I prove that this geometry can be embedded in a 3D Euclidean space and also find the equations for the corresponding 2-dimensional surface?

On a first attempt, I tried using the induced metric functions equation: $h_{ij}=g_{ab}\frac{\partial x^a}{\partial u^i}\frac{\partial x^b}{\partial u^j}$, so that parametrizing the surface in terms of $r$ and $\theta$, I could try to satisfy the relation between the euclidean metric tensor $g_{ab}=\mathbb{I}$ and the induced metric functions, which I found to be: $h_{11}=\frac{1}{1-\frac{2\mu}{r}}$ and $h_{22}=r^2$.

Anyway, I couldn't find any way forward, so any help will be much appreciated!

答:
Using @Arctic Char' suggestion: it make sence to use the rotational symmetry.

If we embed the required surface into 3-dymensional Euclidean space, we can require the rotational symmetry over axis Z (the coordinate system in 3D space - this is our choice). We can also use the angle $\phi$ for parametrization in 3D, so this angle will be the same for both surface and 3D space.
To define the surface it would be enough to draw its cross-section (a curve), for instance, in X-Z plane.

$ds^2=\frac{dr^2}{1-\frac{r_s}{r}}+r^2d\phi^2$, where $r_s$ is the "Schwarzschild radius".

In 3D space we introduce $z$ and $\rho$ coordinates: $ds^2=dz^2+d\rho^2+\rho^2d\phi^2$. On the other hand, if we stay in the X-Z plane, $d\phi=0$, so

$ds^2=\frac{dr^2}{1-\frac{r_s}{r}}\Rightarrow dz^2+d\rho^2=\frac{1}{1-\frac{r_s}{r}}dr^2$

$(\frac{dz}{dr})^2+(\frac{d\rho}{dr})^2=\frac{r}{r-r_s}$ - the parameterized equation for $z(r)$ and $\rho(r)$

Considering $z(r)$ as a function $z(\rho(r))$ we get

$\Bigr((\frac{dz}{d\rho})^2+1\Bigl)(\frac{d\rho}{dr})^2=\frac{r}{r-r_s}$, or $\Bigr((\frac{dz}{d\rho})^2+1\Bigl)=\frac{r}{r-r_s}(\frac{dr}{d\rho})^2$.

But we are allowed to transform the variable , switching from $r$ to $r(\rho)$. We can choose simply $r=\rho$.

$(\frac{dz}{dr})^2+1=\frac{r}{r-r_s}\Rightarrow \frac{dz}{dr}=\pm\sqrt{\frac{r_s}{r-r_s}}$

$z(r)=\pm2\sqrt{r_s(r-r_s)}+const$. A constant here defines the origine for the variable $z$ and can be chosen zero.

We got a parabola lying "on the side" with the beginning at the point $r=r_s$. Rotating this parabola around axis Z we get the required surface.

So, in 3D space we got the surface defined by the coordinates

Polar angle $\phi$

Radius r

$z$ coordinate depending on $r$: $z(r)=\pm2\sqrt{r_s(r-r_s)}$

At $r>>r_s$ $z(r)\approx2\sqrt{r_sr}<<r$, and $ds=\sqrt{1+(\frac{dz}{dr})^2} $ $dr\approx(1+\frac{r_s}{2r})$ $ dr\approx dr$, so at $r\to\infty$ the surface has two-dimension Euclidean metrics.

At $r\to{r}_s$ the surface is curved and $ds\to\infty$, but $S=\int_{r_1}^{r_s}ds$ is finite, since the singularity at $r=r_s$ is integrable.

If we suppose that the time in this system is absolute, a two-dimensional creature living on the surface and moving along the radius $r$  with a constant velocity $v$ will reach $r_s$ ($r_1>r_s$) in finite time

$$T=-\frac{1}{v}\int_{r_1}^{r_s}\sqrt{\frac{r}{r-r_s}}dr=\frac{1}{v}\Bigl(\sqrt{r_1(r_1-r_s)}+r_s\log\frac{\sqrt{r_1}+\sqrt{r_1-r_s}}{\sqrt{r_s}}\Bigr)$$

At $r_1>>r_s$ $$ T=\frac{r_1}{v}(1-\frac{r_s}{2r_1})+\frac{r_s}{v}\log(2\sqrt{\frac{r_1}{r_s}})\geqslant\frac{r_1}{v}$$

At $(r_1-r_s)<<r_s$ $$ T=2\frac{\sqrt{r_s}}{v}\sqrt{r_1-r_s}>>\frac{r_1-r_s}{v}$$


答: Working over the complex numbers, your two matrices will be similar if and only if they have the same Jordan canonical form. (This is a bit more subtle than saying the algebraic and geometric multiplicities of each eigenvalue have to agree, although in some cases that will be enough.)</p>


问:
A new stopping time built from a stopping time
Let $T$ be a stopping time for the filtration $(\mathcal{F_n})_{n \in \mathbb{N}}.$ For all $n \in \mathbb{N}  \cup \left\{+\infty \right\},$ we set $\phi(n)=\inf\left\{k \in \mathbb{N};\left\{T=n \right\} \in \mathcal{F_k}\right\}$.

Show that $\phi(T)$ is a stopping time and that $\phi(T) \leq T.$

I wrote $\left\{w;\phi(T(w))=n \right\}=\left\{w;T^{-1} (\left\{T(w) \right\}) \in \mathcal{F_n} \ and \ T^{-1}(\left\{T(w) \right\}) \notin \bigcup_{k=0}^{n-1}{\mathcal{F}_k} \right\} $ by I don't see how this will help us to prove that $\left\{w;\phi(T(w))=n \right\} \in \mathcal{F}_n$.

答: for $k \in \mathbb{N},\{\phi(T)\leq k \} =\bigcup_{n \in \mathbb{N}}(\{T=n\}\cap \{\phi(T) \leq k\})$ and for all $n \in \mathbb{N},\{T=n\}\cap \{\phi(T) \leq k\}$ is either $\emptyset$ (if $\phi(n)>k$) or $\{T=n\} \in \mathcal{F}_{\phi(n)} \subset \mathcal{F}_k$ (if $\phi(n) \leq k$), so $\{\phi(T) \leq k\} \in \mathcal{F}_k,$ which means that $\phi(T)$ is a stopping time. Since, for a fixed $w,T^{-1}(\{T(w)\}) \in \mathcal{F}_{T(w)}$ this means that $\phi(T(w)) \leq T(w)$


问:
Why my this formula doesn't satisfy the formula of volume of sphere?
I've defined the function which returns a volume of sphere as a radius value given.

$f(a):=2\int_{0}^{a}\pi*r^{2}*dr$

I thought that above function is correct but actually it is not correct.

The return value of the above function is $\frac{2\pi a^{3}}{3}$

The $2$ at the leftmost of the function represents of a double of volume of hemisphere.

Can anyone tell my why this function is not working?

答:
Given a radius $r$, consider the graph of the semicircle given by $y = \sqrt{r^2 - x^2}$ for $x \in [-r, r]$.
Rotate this graph about the $x$-axis, and you will find the volume of the resulting sphere as an integral.


问:
Semi-algebraic sets
The Astroid is a semi-algebraic curve. Since the points $(x,y)$ of the curve fulfill the equation $x^{2/3} + y^{2/3} = 1 $, and since they are points of said polynomial it is semi-algebraic. But my question is that enough? Since the coefficients of the polynomial are rational and I understand that a set is semi-algebraic if it is the finite union of finite intersections of polynomial systems with integer exponents.

答: It's possible to write down the explicit polynomial system if you want, but it's easier to use the fact that the projection of a semi-algebraic set is again semi-algebraic by Tarski-Seidenberg. Consider $\Bbb R^4$ with coordinates $x,y,z,w$ and the set $S$ cut out by the constraints $x^3=z^2$, $y^3=w^2$, and $z+w=1$. Then the projection of $S$ to the copy of $\Bbb R^2$ with coordinates $x,y$ exactly recovers your asteroid: the fiber of $S$ over $(x,y)$ is a single point when $x^{2/3}+y^{2/3}=1$ and empty otherwise.

Title: Semi-algebraic sets


问:
How to approximate a sharply peaked function with deltas?
I have the function

$$
f(x)=\cases{\frac{1}{\sqrt{\sinh^2(a)-\sinh^2(ax)}} & $-1 \leq x \leq1$ \\ 0 & otherwise}
$$

Where $a \in \mathbb{R}^+$. Here is a plot of $f$:

It would be very convenient if I could justify (in any way: maybe distributionally?) that

$$
f(x)=A\left[ \delta(x-1)+\delta(x+1) \right]+B+\mathcal{O}(?) \quad ; \quad a \to \infty \tag{1}
$$

For some $x$ independent constants $A$ and $B$. I am uncertain of how to do this, or how to estimate the error.

What I've tried

My thought was to look at the Fourier series of $f$, and associate terms with the Fourier series of (2). By expanding (1) around $x= \pm1$, I can find a Laurent series in $(x \pm 1)$, and perform the Fourier integral term by term. The result is messy, and worse: does not allow easy comparison to the Fourier series of (2). This may not be a fruitful approach.

By looking at the wiki page about mollifiers, it seems I need to claim something like: $a f(x) \to \delta (x \pm1), \ a \to \infty$. Superficially, this appears to work, but I'm not sure how to make it concrete, or if there should be any other factors next to $af(x)$ (other than a factor to ensure normalization to unity).

Questions

How can I estimate the error and find $A$ and $B$ in (1)?

If $A$ and $B$ cannot be found, how can I find the ratio $A/B$?

Background

$f(x)$ will eventually appear beneath an integral with a Green's function $G$. I'd like to formulate this for arbitrary $G$. We can take some liberties about the 'niceness' of $G$, as necessary. In particular, it has a well behaved Fourier transform.

答:
@Sal: Not the answer you're looking for I'm sure, and I anticipate the MSE gods will delete this in short order, but perhaps it helps you in the meantime as you continue your research.

After messing about with things a bit, analytically and numerically, a similar function has the delta-like property. Take $\phi_a(x) = \frac{2}{\pi}\frac{a\sinh(ax)}{(\sinh^2 a - \sinh^2(ax))^{1/2}}$ on $0 < x < 1$. Then completely experimentally and without proof (or perhaps fodder for another post!) it appears $\int_0^1\, g(x)\phi_a(x)\, dx \rightarrow g(1)$ as $a \rightarrow \infty$ for a smooth function $g$.

== ADDED ==

If you want to approximate $f$ by $A\delta(x + 1) + A\delta(x - 1) + B$ in some sense, you'll probably want something like
\begin{equation*}
\int_{-1}^{1}\, g(x)f(x; a)\, dx - A g(-1) - Ag(1) - B\int_{-1}^1\, g(x)\, dx \rightarrow 0
\end{equation*}
as $a \rightarrow \infty$ to hold. Use a few examples, $g(x) = 1, x^2, x^4, \dots$ and work out what $A$ and $B$ would need to look like. For example with $g = 1$ I get $\int_{-1}^1\, f(x; 20)\, dx = 4.12252\times 10^{-9}$ and with $g(x) = x^2$, $\int_{-1}^1\, x^2 f(x; 20)\, dx = 1.37424 \times 10^{-9}$. To me this path doesn't look promising. The area under the singular portions simply isn't big enough to mimic a delta function.


问:
The difference of having another layer of {} in sets
Say we have a set X

X = {∅,{a},{a,b},{a,b,c}}

And there are three different statements

{{a}} ∈ X

{a} ⊆ X

{{a}} ⊆ X

One or more of these statements is false, and I can't figure out which one

What is the difference between {a} and {{a}} ?

答:
Can you tell the difference between "a ball inside a box" and "a ball inside a box inside a bigger box"?

Then you should be able to tell the difference between $\{a\}$ and $\{\{a\}\}$.


问:
Need help with integral and change the variable
I have

$$\displaystyle \int_0^\pi \sqrt{x} ~\cos x ~dx$$

and I need to make change of the variable $u = \sin x$.


问:
Simple Double Summation
I've seen how nesting works with a simple $(i+j)$ but this problem below is tripping me up. It's either because of the multipliers or because they each start at zero but I get 60, and the answer I believe is 78. Not sure where I'm missing the last 18.

$$\sum_{i=0}^2\sum_{j=0}^3(2i+3j)$$

For the inner sum, I come up with $18$ $((0+0) + (0+3) + (0+6) + (0+9))$

I plug that into the outer sum $((0+18) + (2+18) + (4+18))$

What am I doing wrong? This discrete math book is horribly short in it's explanation (it only speaks of double sums in product form) and it throws curve balls right off the bat. It doesn't do much to build confidence in the material.

Thanks for the help and the place to vent!

答:
I hope maybe a related presentation familiar to programmers helps readers "see" the double summation better. With double summation, I imagine a 2d matrix with i=rows 0,1,2 and j = columns 0,1,2,3:

\begin{matrix}
 & j=0 & j=1 & j=2 & j=3 \\
i=0 & 0 & 3 & 6 & 9 \\
i=1 & 2 & 5 & 8 & 11 \\
i=2 & 4 & 7 & 10 & 13
\end{matrix}

With each row, the value of i is fixed, and j varies. Then our problem is just, how do we reduce this to a single number? Well, we sum it... twice! One pass in each dimension. First we can sum across each row's columns to "smoosh" into one column (you could equally add down each column instead to produce [6,15,24,33]):

\begin{matrix}
0+3+6+9&=18\\
2+5+8+11&=26\\
4+7+10+13&=34
\end{matrix}

Finally we sum down that column to get to the shape we expect, a single number: 78

Title: Simple Double Summation


问:
How to find the minimum perimeter of this triangle within $30^o$ angle?
$\angle AOB = 30^o $, C is an arbitrary point inside the angle and  OC =$1$. If we  choose two points D and E on OA and OB  to form $\Delta$CDE, what is its minimum perimeter? With very little information, I tried to draw perpendicular lines CD $\perp$ OA, and CE $\perp$ to OB, but it does not seem to work.  Since C is an arbirtary point, need to find some symmetry somewhere?

答:
Edited: Note that I personally prefer Player3236's answer.

Consider $OC$ your baseline.  Then $\angle DOC = 15^\circ$ and $\angle COE = 15^\circ$, symmetrically above and below $OC$.  Due to symmetries, you can assume $DE \perp OC$.

In a coordinate system where $O = (0,0)$, $C = (1,0)$, $D = (x(t), y(t))$, and $E = (x(t), - y(t))$, the perimeter $p$ is
$$p(t) = 2 y(t) + 2 \sqrt{ y(t)^2 + (x(t) - 1)^2 }$$
This reaches a minimum where its derivative reaches zero,
$$\frac{d p(t)}{d t} = 2 \frac{d y(t)}{d t} + \frac{2 y(t)\left(\displaystyle \frac{d y(t)}{d t}\right) + 2 \left(x(t) - 1\right)\left(\displaystyle \frac{d x(t)}{d t}\right)}{\sqrt{y(t)^2 + (x(t)-1)^2}} = 0$$
You can use either
$$\left\lbrace \begin{aligned}
x(t) &= t \cos 15^\circ = t \displaystyle \frac{\sqrt{3} + 1}{2 \sqrt{2}} \\
y(t) &= t \sin 15^\circ = t \displaystyle \frac{\sqrt{3} - 1}{2 \sqrt{2}} \\
\end{aligned} \right. \Rightarrow \left\lbrace \begin{aligned}
\displaystyle \frac{d x(t)}{d t} &= \cos 15^\circ = \displaystyle \frac{\sqrt{3} + 1}{2 \sqrt{2}} \\
\displaystyle \frac{d y(t)}{d t} &= \sin 15^\circ = \displaystyle \frac{\sqrt{3} - 1}{2 \sqrt{2}} \\
\end{aligned} \right. $$
or
$$\left\lbrace \begin{aligned}
x(t) &= t \\
y(t) &= t \tan 15^\circ = t \left( 2 - \sqrt{3} \right) \\
\end{aligned} \right. \Rightarrow \left\lbrace \begin{aligned}
\displaystyle \frac{d x(t)}{d t} &= 1 \\
\displaystyle \frac{d y(t)}{d t} &= \tan 15^\circ = 2 - \sqrt{3} \\
\end{aligned} \right. $$
The derivative has two minima (just expand and simplify, noting that whenever you know both sides are nonnegative, you can square both sides), one of which is the true minimum, for which $p = 4 ( \cos 15^\circ ) ( \sin 15^\circ ) = 2 \sin 30^\circ = 1$, $x = \sqrt{3/4} \approx 0.866$, $y = \sqrt{3/4}(2 - \sqrt{3}) \approx 0.232$

Without symmetries, we can define $E = (x_1, 0)$, $C = (\cos \theta, \sin\theta)$, and $D = (x_2, x_2 \tan 30^\circ) = (x_2, x_2 / \sqrt{3})$.  The perimeter length is then
$$\begin{aligned}
p & = \lVert ED \rVert + \lVert DC \rVert + \lVert CE \rVert \\
~ & = ~ \sqrt{ (x_2 - x_1)^2 + \left(\frac{x_2}{\sqrt{3}} - 0\right)^2 } \\
~ & + ~ \sqrt{ (\cos\theta - x_2)^2 + \left(\sin\theta - \frac{x_2}{\sqrt{3}}\right)^2 } \\
~ & + ~ \sqrt{ (x_1 - \cos\theta)^2 + (0 - \sin\theta)^2} \\
\end{aligned}$$
i.e.
$$\begin{aligned}
p & = ~ \sqrt{ \frac{4}{3} x_2^2 + x_1^2 - 2 x_1 x_2 } \\
~ & + ~ \sqrt{ \frac{4}{3} x_2^2 - 2 x_2 \left( \frac{\sin\theta}{\sqrt{3}} + \cos\theta \right) + 1 } \\
~ & + ~ \sqrt{ x_1^2 - 2 x_1 \cos\theta + 1 } \\
\end{aligned}$$
I don't know of a good trivariate minimization for this, but brute force numerical evaluation does trivially find the solution $p = 1$, $\theta = 15^\circ$, $x_1 \approx 0.89657548$, $x_2 \approx 0.77645714$.

The symmetry is not obvious here from the numbers, but $E = (0.89657548, 0)$ and $D = (0.77645714, 0.44828774)$ are symmetrically positioned around $OC$, $O = (0,0)$, $C = (0.96592583, 0.25881905)$.

Personally, I find this a good example of the situation when finding the easiest way to solve a problem is the key.


问:
Complex vector/matrix multiplication
I am currently studying a textbook and I have a limited background in complex linear algebra and unfortunately the textbook does not provide background knowledge for this either. As a result I am a bit stuck with something that should be fairly simple. The problem is as follows:

Given known $3\times1$ vectors $\mathbf{h_1}$ and $\mathbf{h_2}$ we want to solve for an unknown matrix $C$ of shape $3\times3$ consisting of real entries. Suppose that we form a constraint with them such that $$(\mathbf{h_1} \pm i\mathbf{h_2})^TC(\mathbf{h_1} \pm i\mathbf{h_2})=0$$
Where $i\mathbf{h_2}$ denotes that vector $\mathbf{h_2}$ is an imaginary component. Then such constraint can be rewritten as two constraints:
$$\mathbf{h_1}^TC\mathbf{h_2}=0$$
$$\mathbf{h_1}^TC\mathbf{h_1}=\mathbf{h_2}^TC\mathbf{h_2}$$

Can anyone provide a thorough derivation of this? I suspect it is something to do with splitting up the real and imaginary parts or the plus minus sign?

答:
The second constraint can be easily shown by the separation of the real and imaginary parts (as you actually noted). I can prove the first constraint similarly but only for a symmetric C matrix (it is not clearly stated but I suppose it is symmetric from your comment).

I start by expanding the left-hand side:
$$ (\mathbf{h_1} \pm i\mathbf{h_2})^TC(\mathbf{h_1} \pm i\mathbf{h_2})=(\mathbf{h_1}^T \pm i\mathbf{h_2}^T)C(\mathbf{h_1} \pm i\mathbf{h_2}) = \mathbf{h_1}^TC\mathbf{h_1} \pm i\mathbf{h_1}^TC\mathbf{h_2} \pm i\mathbf{h_2}^TC\mathbf{h_1} + i^2\mathbf{h_2}^TC\mathbf{h_2} =  \mathbf{h_1}^TC\mathbf{h_1} \pm i\mathbf{h_1}^TC\mathbf{h_2} \pm i\mathbf{h_2}^TC\mathbf{h_1} - \mathbf{h_2}^TC\mathbf{h_2} = 0 $$

You need to realize that $ 0=0+0i $ and you can balance both real and imaginary part separately.
Balancing the real part gives you directly the second constraint:
$$ \mathbf{h_1}^TC\mathbf{h_1} - \mathbf{h_2}^TC\mathbf{h_2} = 0 \implies \mathbf{h_1}^TC\mathbf{h_1} = \mathbf{h_2}^TC\mathbf{h_2} $$

Balancing the imaginary part is a little bit more tricky:
$$ \pm \mathbf{h_1}^TC\mathbf{h_2} \pm \mathbf{h_2}^TC\mathbf{h_1} = 0 $$
By using the property of transpose operation that $ (AB)^T=B^TA^T $ I get:
$$ \pm \mathbf{h_1}^TC\mathbf{h_2} \pm \mathbf{h_2}^TC\mathbf{h_1} = \pm \mathbf{h_1}^TC\mathbf{h_2} \pm (\mathbf{h_1}^TC^T\mathbf{h_2})^T = 0 $$
Both terms are scalar, which can be proved simply by dimensionality analysis, and transpose of a scalar is the scalar itself:
$$ \pm \mathbf{h_1}^TC\mathbf{h_2} \pm (\mathbf{h_1}^TC^T\mathbf{h_2})^T = \pm \mathbf{h_1}^TC\mathbf{h_2} \pm \mathbf{h_1}^TC^T\mathbf{h_2} = 0 $$
Now, you get finally your other constraint if $C$ is symmetric (i.e. $ C=C^T $):
$$ \pm \mathbf{h_1}^TC\mathbf{h_2} \pm \mathbf{h_1}^TC^T\mathbf{h_2} = \pm 2 \mathbf{h_1}^TC\mathbf{h_2} = 0 \implies \mathbf{h_1}^TC\mathbf{h_2} = 0 $$

Title: Complex vector/matrix multiplication


问:
How do I prove Binet's Formula?
My initial prompt is as follows:

For $F_{0}=1$, $F_{1}=1$, and for $n\geq 1$, $F_{n+1}=F_{n}+F_{n-1}$.
  Prove for all $n\in \mathbb{N}$:

$$F_{n-1}=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt5}{2}\right)^n-\left(\frac{1-\sqrt5}{2}\right)^n\right)$$

Which, to my understanding, is Binet's Formula.

I came up with a proof strategy similar to that described here. But as you can see, it is considering the case of $F_n$, not $F_{n-1}$.

So, my question is: does the same strategy hold for $F_{n-1}$? If not, how is it different?

答: It's impossible to tell without reading your proof, but induction (as a principle) will work regardless of whether you consider $n$ or $n-1$.  The only difference is the indexing of the base case, but with the shift $n \mapsto n-1$ you needn't worry about introducing a "gap" at all.</p>

Title: How do I prove Binet's Formula?


问:
Does exist a Banach-Tarski paradoxical decomposition of the unit cube in which each piece has the Baire Property?
The Banach-Tarski Paradox for the cube states that the open unit cube in three dimensions can be decomposed into finitely many pieces, which can then, by rotation and translation, be re-assembled into two copies of the unit cube. That is, there exist disjoint sets $\left(A_{1}, A_{2}, \ldots, A_{n}, B_{1}, \ldots, B_{m}\right)$ whose union is $(0,1)^{3}$ and there exist Euclidean isometries $\left(\rho_{1}, \rho_{2}, \ldots, \rho_{n}, \sigma_{1}, \ldots, \sigma_{m}\right)$ such that
$$
(0,1)^{3}=\bigcup \rho_{j}\left(A_{j}\right)=\bigcup \sigma_{i}\left(B_{i}\right)
$$
Clearly the pieces $\left(A_{1}, \ldots, A_{n}, B_{1}, \ldots, B_{m}\right)$ cannot all be Lebesgue measurable, because Lebesgue measure is preserved by rotation and translation.

The problem of Marczewski for the cube (which is some sixty years old) asks: Does there exist a Banach-Tarski paradoxical decomposition of the unit cube, in which each piece has the Baire Property?

Although little is known about this question, it seems reasonable to conjecture that the answer is no. First of all, it is unlikely that the proof of the Banach-Tarski Paradox, which uses the Axiom of Choice to define the pieces, could be modified so that the pieces have the Property of Baire. Furthermore, a positive answer to Marczewski's Problem would imply the following, seemingly implausible conclusion. Any cube, no matter how small, contains pairwise disjoint open subsets $U_{1}, \ldots, U_{n}$ such that, for suitable isometries $\rho_{i}$,
$\bigcup \rho_{i} U_{i}$ is dense in the unit cube.
Every hyperfinite von Neumann algebra is an injective $\mathrm{C}^{*}$ -algebra. It is very plausible to conjecture that each hyperfinite monotone complete $\mathrm{C}^{*}$ -algebra is injective. However, seems that that if this is true then there exists a paradoxical decomposition of the open unit cube in $\mathbb{R}^{n}(n \geqslant 3)$ into pieces with the Baire Property.

答:
Marczewski's question was answered affirmatively (!) for the unit ball in:

Randall Dougherty and Matthew Foreman.
Banach—Tarski decompositions using sets with the property of Baire.
Journal of the American Mathematical Society, vol. 7 (1994), pp. 75–124.

The result can surely be transferred to the cube.

Edit: In the last sentence, "surely" (which usually means "I'm confident but haven't checked") can be upgraded: Dougherty and Foreman state the expected extension, to arbitrary bounded, open, nonempty subsets of $\mathbb R^n$ for any $n\geq 3$, on the first page of their paper.


问:
Adjoing $p$th roots to rings: as quotients of polynomials
Motivation/Setup: Suppose $R$ is a domain. I'd want to "adjoin $p$th roots of an element $a \in R$." where $p$ is a prime (which i am mostly interested in $p$-adic) This seems to be a cmomon operationo in $p$-adic goemetry. See the rings of interst below.

Goal: I would like to express these objects as a quotient of polynomial.

Let $K$ be the algebraic closure of fraction field of $R$, and make a choice of $a^{1/p}$ for $a \ inR$

Suppose $a^{1/p} \notin R$.

There is a map
$$ R[x]/(x^{p}-a) \rightarrow R[a^{1/p}] \hookrightarrow K$$
$$x \mapsto a^{1/p}$$

Question: Are there simple criteria to test is if first map is iso? For instance, i'd like to adress the three cases below.  I could explain 1. in example.

Examples of interest:

$R=\Bbb F_p((t)), a=t$. This is adressed below combining the two observations below in my thoughts. In otherwords, I bellieve I have proven :
$$F_p((t))[t^{1/p}] \simeq F_p((t))[y]/(y^p-t)$$

$R=\Bbb Z_p, a=p$. I would like to adjoint $p$th roots of unit as

$$\Bbb Z_p[\xi_p] \simeq \Bbb Z_p[y]/(y^{p-1}+\cdots +1)$$

$R=S[x]$, $a=x$, where $S$ is some domain like $\Bbb Z, \Bbb Z_p, \Bbb Q_p$. I would like to state something as

$$\Bbb Z_p[x][x^{1/p}] \simeq \Bbb Z_p[x][y]/(y^p-x)$$

Thoughts:

The map is clearly surjectiev. So we are down to the case of innjectivitiy.
Let $I:= \ker(R[t] \rightarrow R[a^{1/p}])$. This is a prime ideal.

Condition 1: When all prmies of $R[t]$ are principal/$R$ is a field These are equivalent condition.

Condition 2: $R$ is of charateristic $p$ The question is equivalent to finding minimal polynomial of $a^{1/p}$ over $R$. Over $K$, $x^p-a=(x- a^{1/p})^p$. Thus, if in $R[t]$, it is reducible, we must have $a^{k/p} \in R$ for some $1 \le k <p$, and $(k,p)=1$ would imply $$(a^{lk/p}a^b)^p=a$$ for some integers $b,l\in \Bbb Z$, this shows we have found a $p$th root.

答:
For $p$ prime $x^p-a\in K[x]$ is irreducible iff it has no root in $K$.

Let $b$ be a root in the algebraic closure and $f$ its minimal polynomial, of degree $m$.

In characteristic $p$: $x^p-a=(x-b)^p$ so $f=(x-b)^m, f(0)=(-b)^m\in K$, if $m<p$ then $b\in K$.

In characteristic not $p$:

$f=\prod_{j=1}^m (x-\zeta_p^{c_j} b)$ so $f(0)=(-1)^m \zeta_p^r b^m\in K$. If $m<p$ then $\zeta_p^{rs} b\in K$ where $sm=1\bmod p$.

And hence $x^p-a$ has a root $\zeta_p^{rs} b\in K$.

Conversely if $x^p-a$ has no root in $K$ then $m=p$ ie. $x^p-a$ is irreducible and for any subring $R\subset K$ containing $a$,  $R[b]\cong R[x]/(x^p-a)$.


问:
proof $\Vert a\Vert^2\Vert b\Vert^2=\Vert a \times b \Vert^2+\vert a\cdot b\vert^2$
$\Vert a\Vert^2\Vert b\Vert^2=\Vert a \times b \Vert^2+\vert a\cdot b\vert^2$, where a and b are vectors

答:
Hint
$$
\| a \times b\|=\|a\| \cdot \| b\| \cdot \sin(\theta) \\
a \cdot b = \|a \|\cdot \| b\| \cdot \cos(\theta)$$


问:
Covariance of Brownian motion intuition
So the covariance between two time-instances of Brownian motion is $$\text{Cov}(B_s, B_t) = \min(s,t).$$

This post gives a derivation of this fact, but I'm lacking intuition.

Suppose $\epsilon \ll 1 $ is small and $t$ is larger. Then
$$ \text{Cov}(B_\epsilon, B_t) = \epsilon,$$
$$\text{Cov}(B_{t+\epsilon}, B_{2t}) = t+\epsilon.$$

In both cases the time step is $t-\epsilon$, but the covariances are drastically different.

Brownian motion is stationary, so why is this true? (And if not, what am I doing wrong?)

答: Basically you are saying $X-Y$ and $Z-W$ has the same distribution, why $\mathrm{Cov}(X,Y)$ can be very different from $\mathrm{Cov}(Z,W)$? Think of a baby example: $X$ has some distribution with large variance, take $Y = X+ c$, where $c$ is a constant. Now say $Z$ has some distribution with small variance, take $W = Z+ c$. Can you see that $\mathrm{Cov}(X,Y)$ can be much larger than $\mathrm{Cov}(Z,W)$?

Title: Covariance of Brownian motion intuition


答: $\|a \times b\| = |a| |b| \sin \theta,$ where $\theta$ is the angle between $a, b.$

Title: proof $\Vert a\Vert^2\Vert b\Vert^2=\Vert a \times b \Vert^2+\vert a\cdot b\vert^2$

Body: $\Vert a\Vert^2\Vert b\Vert^2=\Vert a \times b \Vert^2+\vert a\cdot b\vert^2$, where a and b are vectors


问:
$ab$-plane phase portraits
I have the following system of equations:

$x'_1=ax_1+bx_3$

$x'_2=bx_2$

$x'_3=-bx_1+ax_3$

I found the solution:

$x_1=c_1e^{at} cos(bt)+c_3 e^{at}sin(bt)$

$x_2=c_2e^{bt}$

$x_3=-c_1e^{at} sin(bt)+c_3 e^{at}cos(bt)$

Now I'm trying to answer this:
Sketch the region in the $ab$–plane where this system has different
types of phase portraits.
I don't even understand what the question is asking. Am I supposed to play around plugging in $0$ for $a$ and $b$?

答:
The system is given as
$$\begin{align}x'_1&=ax_1+bx_3 \\ x'_2&=bx_2 \\ x'_3&=-bx_1+ax_3\end{align}$$

The eigenvalues are given by

$$(b-\lambda ) \left(a^2-2 a \lambda +b^2+\lambda ^2\right) = 0 \implies \lambda_{1, 2, 3} = a-i~ b, a+i~ b, b$$

The solution is given by

$$\begin{align}x_1(t)&= c_1 e^{a t} \cos (b t)+c_3 e^{a t} \sin (b t)\\y_1(t) &= c_2 e^{b t}\\z_1(t) &=  -c_1 e^{a t} \sin (b t) + c_3 e^{a t} \cos (b t)\end{align}$$

Lets investigate system stability using eigenvalues in the different quadrants of the $ab-$plane

Case 1: $a = b = 0$, solution is a constant

Case 2: $b = 0, a > 0$, solution lies on a line and is unstable

Case 3: $b = 0, a < 0$, solution lies on a line and is stable

Case 4: $a <0, b \ne 0$, system is stable

Case 5: $a > 0, b \ne 0$, system is unstable

Try other cases to see if anything interesting shakes out to complete the analysis, like $a = \pm b$.

For case $1$, we have the solution as

$$X(t) = \begin{pmatrix} x(t) \\ y(t) \\ z(t) \end{pmatrix} = c_1 \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} + c_2 \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + c_3 \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} $$

For case $2$, we have

For case $3$, we have

For case $4$, we have

For case $5$, we have

Note: Because this system is decoupled, you could have also just stated what $y(t)$ does and then analyzed $x(t), z(t)$ using a $2D$, $ab-$plane, similar to these notes, which is preferred. These notes are also very helpful for bifurcations in general.

Title: $ab$-plane phase portraits


问:
Does $(0, \infty)$ inherit its usual topology as a subspace of $\mathbb{R}_K$?
Recall that $\mathbb{R}_K$ denotes the real line with the $K$ topology. In other words, its basis elements are of the form $(a, b) - K$, where $K = \{1/n: n \in \mathbb{Z}_+\}$. On page 178 of Munkres's 'Topology' book, the author states that $(0, \infty)$ inherits its usual topology as a subspace of $\mathbb{R}_K$. But I disagree because $(0,1) - K$ is open in $\mathbb{R}_K$ and so its intersection with $(0,\infty)$ is open in $(0, \infty)$. In other words, $((0,1) - K) \cap (0, \infty) = (0,1) - K$ is open in $(0, \infty)$, but the standard topology on $(0, \infty)$ does NOT have $(0,1) - K$ as one of its open sets. So where am I wrong in my argument here? I am assuming that Munkres is correct and I am not. Thanks!

答: On the contrary, $(0,1)\setminus K$ is open in the standard topology: it is equal to the intersection of the open sets $(0,1)$ and $\Bbb R\setminus\big(\{0\}\cup K\big)$. The latter is open because $\{0\}\cup K$ is closed.


答:
The problem that you’ve been given uses a less-standard indexing of the Fibonacci numbers. The usual definition has initial values $F_0=0$ and $F_1=1$; your $F_n$ is the usual $F_{n+1}$, and the usual $F_n$ is your $F_{n-1}$. Thus, where your link has $$\varphi^a=F(a)\varphi+F(a-1)\;,$$ you’ll need to write $$\varphi^a=F_{a-1}\varphi+F_{a-2}\;.$$ As a quick check, when $a=2$ that gives you $\varphi^2=F_1\varphi+F_0=\varphi+1$, which you can see from the link is correct.

(I’m assuming here that your proof really does follow pretty much the pattern in the link. You can also just do a straight induction: the induction step is the same whichever indexing you use.)

Title: How do I prove Binet's Formula?

Body:
My initial prompt is as follows:

For $F_{0}=1$, $F_{1}=1$, and for $n\geq 1$, $F_{n+1}=F_{n}+F_{n-1}$.
  Prove for all $n\in \mathbb{N}$:

$$F_{n-1}=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt5}{2}\right)^n-\left(\frac{1-\sqrt5}{2}\right)^n\right)$$

Which, to my understanding, is Binet's Formula.

I came up with a proof strategy similar to that described here. But as you can see, it is considering the case of $F_n$, not $F_{n-1}$.

So, my question is: does the same strategy hold for $F_{n-1}$? If not, how is it different?


答:
Hint: Reflect $C$ across $OA$ and $OB$ as shown.

The perimeter $CD+DE+EC = C'D + DE + EC'_1$. When does the latter reach its minimum? It should then follow trivially that the minimal perimeter is $1$.

Title: How to find the minimum perimeter of this triangle within $30^o$ angle?

Body: $\angle AOB = 30^o $, C is an arbitrary point inside the angle and  OC =$1$. If we  choose two points D and E on OA and OB  to form $\Delta$CDE, what is its minimum perimeter? With very little information, I tried to draw perpendicular lines CD $\perp$ OA, and CE $\perp$ to OB, but it does not seem to work.  Since C is an arbirtary point, need to find some symmetry somewhere?


问:
Given a triangle with sides $a, b, c$ prove that: $a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc$
Given a triangle with sides $a, b, c$ prove that: $a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc$

I attempted to solve this question as follows:

$a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc$

$a^2(b+c)+b^2(c+a)+c^2(a+b)-(a^3+b^3+c^3)\le 3abc$

$(a+b)(b+c)(c+a)-2abc-3abc\le 3abc$

$(a+b)(b+c)(c+a)-5abc\le 3abc$

$(a+b)(b+c)(c+a)\le 8abc$

And I got stuck here. Could you please explain to me how to finish it off?

答:
$a=y+z,b=x+z,c=x+y$ then we have to prove $$\sum 2z{(y+z)}^2\le 3(x+y)(y+z)(z+x) \tag 1$$ Now note that $$\sum (2x{(z+y)}^2-8xyz)=\sum 2x{(y-z)}^2$$ and $$(x+y)(y+z)(x+z)-8zxy=\sum z{(x-y)}^2$$ therfore  after subtracting $24xyz$ from both sides of $(1)$ it remains to rpove $$\sum z{(x-y)}^2\ge 0$$ which is true

Note that one could actually multiply out inequality $(1)$ to get $$x^2(y+z)+y^2(z+x)+z^2(x+y)\ge 6xyz$$ which is true by AM_GM but i felt writing it as an SOS is more nice


问:
Meaning of $\int_\Omega\langle\phi,Du(x)\rangle$, in the definition of BV space
I was going through the definition of bounded variation functions on Wikipedia : https://en.wikipedia.org/wiki/Bounded_variation

Definition. Let $\Omega$ be an open subset of $\mathbb{R}^n$. A function $u\in L^1(\Omega)$ is said of bounded variation if there exists a finite vector Radon measure $Du\in \mathcal{M}(\Omega,\mathbb{R}^n)$ such that the following equality holds : $$\int_\Omega u(x)\text{div}\phi(x)dx=-\int_\Omega\langle\phi,Du(x)\rangle\quad\forall \phi\in C_c^1(\Omega,\mathbb{R}^n).$$

I am very confused with this definition, since I am not sure what $\int_\Omega\langle\phi,Du(x)\rangle$ is. I never saw such a notation, and I could not find any reference to this 'vector measure' and corresponding Lebesgue integral. It will be very appreciated if anyone could give me some explanation on this.

答:
So, since there is no answer written except on the comment section, and now I know a bit more, for the sake of the completness of the question, I will write my own answer.

In the above definition, $Du=(\partial_iu)_{i}$ is a distributional derivative (it always exists for $u\in L_\text{loc}^1$) of $u$, and $Du\in \mathcal{M}(\Omega,\mathbb{R}^n):=(C_c(\Omega,\mathbb{R}^n))^*$, i.e. is an element in the dual space of compactly supported uniformly continuous functions, and $\int_\Omega\langle\phi,Du\rangle$ simply means $Du(\phi)=(\partial_iu(\phi_i))_i$.

The definition does make sense in the context that, if $u\in C^1(\Omega)$, then $Du\in (C^0(\Omega))^n$, and
$$-\int_\Omega\langle \phi,Du\rangle=-\sum_i\int_\Omega\phi_i\partial_iu=\sum_i\int_\Omega(\partial_i\phi)u=\int_\Omega u\,\text{div}\phi.$$
And this condition implies "bounded variation" because, if we can find such $Du$ in $\mathcal{M}(\Omega,\mathbb{R}^n)$, then $Du$ is not "too wild". For example, consider non-measure distributions such as $Du=\sum_{n\in\mathbb{N}}(-1)^n\delta_{1/n}$ where $\delta_x$ is a dirac-delta at $x\in\mathbb{R}$. We can come up with a function $u\in L^1$ with this distributional derivative, e.g. $u=1$ on $[1/(2n),1/(2n-1)]$ and $0$ otherwise. However this function will not have a "bounded variation" since it fluctuates super hard ($``\int_{1/N}^1|Du|=\sum_{n\leq N}\delta_{1/n}"$ goes infinity as $N\rightarrow \infty$). In contrast, if we have $Du$ as a finite Radon measure, then such phenomenon cannot occur, due to the fact that it should be finite.

If fact, there is the another equivalent formulation of BV space: For any $u\in L^1(\Omega)$, the total variation of $u$ in $\Omega$ is $$V(u,\Omega):=\sup_{\substack{\phi\in C_0^1(\Omega:\mathbb{R}^n)\\\|\phi\|_\infty\leq 1}}\int_\Omega u\,{\rm div}\,\phi.$$ Here $\text{div}\,\phi=\nabla\cdot \phi=\sum_{i=1}^n\partial_i\phi_i$ denotes the divergence. If the total variation of $u$ in $\Omega$ is finite, then we say $u\in BV(\Omega)$. The space of functions with bounded variation, $BV(\Omega)$, is Banach space, with the norm called called bounded variation $$\|u\|_{BV(\Omega)}:=\|u\|_{L^1(\Omega)}+V(u,\Omega).$$ When it is clear we will sometimes write $\|\cdot\|_{BV}$.