问:
Can't find the inverse of Laplace Transform.
Find $\mathcal{L}^{-1}{(F(s))}$, if given $F(s)=\dfrac{2}{s(s^2+4)}$.

I have tried as below.

To find inverse of Laplace transform, I want to make partial fraction as below.
\begin{align*}
\dfrac{2}{s(s^2+4)}=\dfrac{A}{s}+\dfrac{Bs+C}{s^2+4}=\dfrac{(A+B)s^2+Cs+4A}{s(s^2+4)}.
\end{align*}
After that, we have system of linear equation
\begin{align*}
A+B&=0\\
C&=0\\
4A&=2.
\end{align*}
Thus we have $A=2$, $B=-2$, and $C=0$.
Now, substituting $A, B, C$ and we have
\begin{align*}
\dfrac{2}{s(s^2+4)}=\dfrac{2}{s}+\dfrac{-2s}{s^2+4}.
\end{align*}
But the fact is
\begin{align*}
\dfrac{2}{s(s^2+4)}\neq \dfrac{2}{s}+\dfrac{-2s}{s^2+4} = \dfrac{8}{s^2+4}.
\end{align*}

I'm stuck here. I can't make a partial fraction for $F(s)$ and I can't find inverse of Laplace transform for $F(s)$.

Anyone can give me hint to give me hint for this problem?

答:
$$F(s)=\dfrac{2}{s(s^2+4)}$$
$$F(s)=\dfrac{1}s\dfrac 2{(s^2+4)}= \mathcal {L}(1) \mathcal{L} ( \sin (2t)$$
You can also use the theorem of convolution :
$$f(t)=1* \sin (2t)=\int_0^t 1 \times \sin(2\tau) d\tau$$
$$f(t)= \dfrac {-\cos (2 \tau)}{2} \bigg |_0^t$$
$$f(t)= \dfrac 12 -\dfrac {\cos (2 t)}{2} $$


问:
Does the "coordinate-to-basis-vector" functor have an adjoint?
Consider global coordinate systems on $\mathbb R^n$. Such a coordinate system is defined by a list of $n$ smooth functions
$$x^i: \mathbb R^n \to \mathbb R, \quad i = 1,... n$$
such that their gradients at every point make up a basis (that is, their contour surfaces are fully transverse at every point).

Also consider global frame systems on $\mathbb R^n$. Such a frame system is defined by a list of $n$ smooth vector fields
$$v^i: \mathbb R^n \to \mathbb R^n, \quad i = 1,... n$$
such that they make up a basis at every point.

The global coordinate systems make up a category $\mathcal C$, with coordinate transforms being the arrows. Similarly, the global frame systems make up a category $\mathcal F$, with smooth pointwise linear transforms being the arrows.

Then, we have a contravariant functor $V: \mathcal C \to \mathcal F$, which sends $\{x^1, ... x^n\}$ to $\{\partial_{x^1}, ... \partial_{x^n}\}$, and sends coordinate transform $y^i = f^i(x^1, ... x^n)$ to linear transform $\partial_{x^i} = (\partial_{x^i}f^j)\partial_{y^j}$.

There are strictly more frame systems than coordinate systems, because given the frame system corresponding to a coordinate system, we can integrate it to recover the coordinate system. However, there are many frame systems that cannot be integrated.

 For example, in $\mathbb R^2$, consider this frame system:
 $$\{(1 + a^2 a^2) \hat e_1, \hat e_2 \}\quad\text{at point } (a^1, a^2)$$
 There is no way to consistently assign a coordinate to the point $(1, 1)$, since if you follow the path $(0, 0) \to (0, 1) \to (1, 1)$, you would move by $(1, 1)$. But if you follow the path $(0, 0) \to (1, 0) \to (1, 1)$, you would move by $(0.5, 1)$.

So $V$ is an injective functor, like free functors. Free functors often have right adjoints called forgetful functors.

So the question is: does $V$ have a right adjoint $W: \mathcal F \to \mathcal C$, a forgetful functor that "forgets" a frame system into a coordinate system?

(Of course, $V$ is contravariant, so it would have to actually be a functor $W: \mathcal F^{op} \to \mathcal C$, but you get the point.)

答:
Yes, and the right adjoint is whatever functor you want :D However, this is due to the particular nature of involved categories more than the free-forgetting yoga.

Firstly we want to show that for all $z, w \in \mathcal{F}$ it holds

$$ Hom_{\mathcal{F}}(z, w) = \{*\}$$

i. e. it has cardinality one. Indeed, recall that a global frame $z$ is given by vectors $z_1(x), \ldots, z_n(x) $ that spans $\mathbb{R}^n$ smoothly varying with $x$. We denote by $Z(x) $ the matrix obtained by putting the columns $z_1(x), \ldots, z_n(x) $ one after another. A morphism from z to w is a smoothly varying $L(x) $ such that for every $i$

$$L(x) z_i(x) = w_i(x) $$

Which can be rewritten as

$$ L(x) Z(x) = W(x) $$

Since $Z(x) $ is invertible, this implies

$$ L(x) = W(x) Z(x) ^{-1}$$

Also, notice that the inverse is a smooth function, since it can be written as the adjoint matrix (polynomial) multiplied by 1 over the determinant (smooth where different by zero). This gives us a unique morphism.

Now we notice that this extends to $\mathcal{C}$. As you noticed, the two functors $V, I$ exhibits $Hom_{\mathcal{C}}(z, w) $ as a non empty subset of $Hom_{\mathcal{F}}(Vz, Vw) $; thus it has cardinality one, too.

Let's pass to the adjointness part. We will show that any functor from $\mathcal{F}$ to $\mathcal{C}$ is adjoint to $V$. Take any functor $F$, and define the natural transformations as the only morphisms $1 \to FV, VF \to 1$. Commutation rules are always satisfied, because when you have to verify if two morphism are equal, you fall into an hom set of cardinality one, so they must be the same.

My suggestion: to get something more meaningful, you should relax a bit the conditions on morphisms. In particular, it would be nice if naturality was more demanding to get, and if global frames morphisms were not all isomorphisms. For example, if you allow "singular frames" to exist, you could have non invertible transformations too.

Hope I didn't mislead what you meant!

ContentLicense: CC BY-SA 4.0


问:
Extremal distribution of random variable that averages to a given value
Denote by $\mathcal M$ the set of probability measures over $[0,1]$ that average to $p$. This is a convex set and I am trying to characterize its extreme points. For any $x$, $y$ such that $0\leq x< p< y\leq 1$ define
\begin{align*}
\lambda_x^y = \frac{p-y}{x-y}\cdot\delta_x+\frac{x-p}{x-y}\cdot\delta_y
\end{align*}
where for any measurable $A\subseteq [0,1]$ and $a\in[0,1]$, $\delta_a(A)=\mathbf 1(a\in A)$. Since $\delta_x$ (resp. $\delta_y$) averages to $x$ (resp. $y$) we get that $\lambda_x^y$ averages to $\frac{p-y}{x-y}\cdot x+\frac{x-p}{x-y}\cdot y=p$ and so is in $\mathcal M$.

I am very tempted to say that the extreme points of $\mathcal M$ is the set $\{ \lambda_x^y : 0\leq x< p < y \leq 1 \}\cup \{ \delta_p \}$ without being able to prove it. If I am not mistaking I am supposed to prove that (ignoring the $\delta_x$ component)

If for some $x$ and $y$ and $a\in[0,1]$, $\lambda_x^y=a \mu+(1-a) \nu$ then $\mu=\nu=\lambda_x^y$.

For any $\mu\in\mathcal M$ there is a probability measure $\kappa$ over $[0,p]\times[p,1]$ such that $\mu(A)=\int \lambda_x^y(A) d\kappa(x,y)+\mu(\{ p \})\delta_p(A)$ for all measurable $A\subseteq [0,1]$.

For the first point, it is quite clear by positivity of the measures $\mu$ and $\nu$ that their support $\{ x,y \}$, then there is only one probability measure with such a support that averages to $p$ and it is $\mu_x^y$.

For the second statement however it is less clear to me, Maybe one way to proceed is computing the integral
\begin{align*}
\int \lambda_x^y(A) d\kappa(x,y)&=\int \left( \frac{p-y}{x-y}\cdot\delta_x(A)+\frac{x-p}{x-y}\cdot\delta_y(A) \right) d\kappa(x,y)\\
&=\int_{\left(A\cap [0,p]\right)\times [p,1]}\frac{p-y}{x-y} d\kappa(x,y)+\int_{[0,p]\times\left(A\cap [p,1]\right)}\frac{x-p}{x-y}d\kappa(x,y)
\end{align*}
It now feel easier to separate cases $A\subseteq [0,p]$ and $A\subseteq [p,1]$ to set one of the two integral to $0$ but I still don't know how to proceed.

Actually thinking about how I could use it in my research it would be much more powerful to not enforce $x\leq p \leq y$ but just that $p$ is in the convex hull of $\{ x,y \}$ and then we get repetition $\lambda_x^y=\lambda_y^x$ but the of extreme points is the same. This makes $\kappa$ a probability measure over $[0,1]^2$ where we have to be careful assigning $0$ probability to sets $A\times B\subseteq [0,1]^2$ where the convex hull of $A\cup B$ does not contain $p$.

Here is the last thing I tried :
Denote $a_x^y=\frac{p-y}{x-y}$, for $\mu \in\mathcal M$ and $A$ measurable
\begin{align*}
\mu(A)&=\int_{[0,p)} \delta_x(A) d\mu(x)+\int_{(p,1]} \delta_y(A) d\mu(y)+\mu(\{ p \})\delta_p(A)\\
&=\int_{[0,p)\times (p,1]} \left(\frac{a_x^y}{a_x^y}\delta_x(A) + \frac{1-a_x^y}{1-a_x^y}\delta_y(A) \right)d\mu\otimes\mu(x,y)+\mu(\{ p \})\delta_p(A)\\
\end{align*}
It feels like we can define $\kappa$ as a function of $\mu\otimes \mu$ and $a_x^y$, I cannot finish the argument though.

答: Your identificaton of the extreme points of $\mathcal M$ is correct.  A proof of the difficult part can be based on a theorem of R.G. Douglas. See my recent answer to Two term martingales and their extreme points, with a reference to a paper of A.F. Karr containing such a proof.


问:
Sequences and integration problem
Is it true that 2 sequnces of real numbers $a_n$ and $b_n$ converge to 0 if   $\lim\limits_{n\to\infty}\int_{0}^{1}|a_nx+b_n|dx=0$ ?

I tried to relate to the integral to see if I get to some result, but it didn't worked out. I learned integrals and also limits and the problem it is from a book I'm working from.

答:
Yes! Notice that $|a_n + b_n|$ needs to converge to $0$. If it doesnt, then ther would need to be some $\varepsilon > 0$ such that $|a_n + b_n| > \varepsilon$ for some $n_0\in\mathbb{N}$ and all $n>n_0$. Since the function $|a_nx + b_n|$ is continuous  there exists some $\delta > 0$ such that $|a_nx + b_n| > \varepsilon/2$ for all $x\in(1-\delta,1]$ This means that

$$
\int_0^1 |a_nx + b_n|dx \geq \int_{1-\delta}^1 |a_nx + b_n|dx \geq\frac{\varepsilon\delta}{2} > 0
$$
Which would imply that the sequence of integrals cant converge to $0$. This means that
$$
\lim_{n\to\infty}(a_n+b_n) = \lim_{n\to\infty}|a_n+b_n|=0
$$
Using a similar approach you can show that the sequence $|\frac{a_n}{2}+b_n|$ also needs to converge to zero which implies that $\lim_{n\to\infty}(a_n/2+b_n)=0$. Multiplying both sides by $2$ you get.
$$
\lim_{n\to\infty}(a_n+2b_n)=0
$$
Subtracting the bottom equality from the top one you get
$$
\lim_{n\to\infty}b_n = \lim_{n\to\infty}(a_n + 2b_n) - \lim_{n\to\infty}(a_n + b_n) = 0
$$
from which you can also deduce that $a_n$ has to go to $0$ aswell.

Title: Sequences and integration problem


问:
Finding a power series solution
It's a problem on Qing Han's book. Given:

$$ u_{tt}-u_{xx}-u=0 \in \mathbb{R}\times (0,\infty)\\
u(x,0)=x, u_t(x,0)=-x$$

Then, there is a power series solution in $(0,0)$ and find this solution.

The existence can be verified by Cauchy-Kovalevskaya, because all the coefficients are analytic.
Now, using the initial condition, can we deduce that the power series solution has order 2? How can I show this explicitly?

答:
I am not sure what your definition of the order of a power series is, but I solved this exercise simply by making the ansatz
\begin{align}
u(t,x)=\sum_{n,k}c_{n,k}x^nt^k
\end{align}

You can plug this expression into the PDE and set the coefficients of $x^nt^k$ equal to zero. Using the initial conditions it's not very difficult to work out the coefficients $c_{n,k}$.

The solution is given by

 \begin{align} u(t,x)=xe^{-t} \end{align}

Title: Finding a power series solution


问:
Example of Normed Spaces $X$ and $Y$ and a linear operator $T$, so that $\ker (T)$ is not closed.
So first, I thought about the zero transformation, however I saw it would not work. After that, I thought about the differentiation transformation, being $X$ and $Y$ the set of all infinitely differentiable functions, however $\ker(T)$ would be closed, since it would be the set of all constant functions, wouldn't it?

Can anyone help or just give a hint of something I am not thinking about? Thanks in advance.

答:
Hint:

For a continuous operator $T$ this can never work,
because then the kernel is always closed.

Choose a linear operator $T$ which is not continuous
and which has a non-trivial kernel!
Then the chances are good that the kernel is not closed.


问:
Factoring third degree polynomials using long division
I am sure there is a better strategy that someone smarter than me would use, but I am not that person.

I am trying to factor

$$x^2 - 3x^2 - 4x + 12$$

I don't know how so I attempt to guess with long division. I cheat and look at the answer so find out one of the factors to save myself time, so I try x - 2

I am not sure how to type out long division but I get

$$ x - 2  |  x^3 - 3x^2 - 4x + 12$$

so I know I can have an $x^2$ for how many times x goes into the leading term.

Subract it all and I am left with

$$ x - 2  |  -x^2 - 4x + 12$$

I know that the leading term goes into the inner leading term $-x$ times or however you say that.

$$ x - 2  |   - 2x + 12$$

Now -2

$$ x - 2  |   8$$

Now I don't know what to do, how did this go so wrong? I have $x^2 - x -2$ ontop and I have a remainder of 8. This can't be right, I cheated so I know that this should be a factor.

答:
Your first term, upon division: $x^2$ is correct $\large\checkmark$ (in the quotient), leaving

$$ x - 2  |  -x^2 - 4x + 12\quad\large\checkmark$$

I know that the leading term goes into the inner leading term $-x$ times or however you say that.

$$ x - 2  |   - 2x + 12 \quad \longleftarrow \text{error}$$

Here you subtracted incorrectly: We should have $x^2 - x$ in the quotient, that's correct, but multiplying $-x(x - 2) = -x^2 + 2x$

So when we subtract, we subtract, from $(-x^2 - 4x + 12) - (-x^2 + 2x) = -6x + 12$.

Now, we have $$ x -2 \mid -6x + 12$$ and so our ongoing quotient becomes $x^2 - x  {\bf - 6}$

which, leaves a zero remainder since $-6(x-2) = -6x + 12$, as desired.

So...we have that $$\frac{x^3 - 3x^2 -4x + 12}{x - 2} = x^2 - x - 6$$

Or, that is,  $$x^3 - 3x^2 - 4x + 12 = (x -2)(x^2 - x - 6) = (x-2)(x +2)(x - 3)$$

ContentLicense: CC BY-SA 3.0


问:
How do this question?
A discrete random variable $X$ can only take the value of 1, 2, 3 and 4, with the probabilities $P(X=1)=P(X=3)$ and $P(X=2)=P(X=4)=2P(X=1)$.

A random sample of size 3 is chosen from $X$ with replacement.

(a) given that $X_1,\ X_2,\text{ and }X_3$ are the three independent random values of $X$, find the probability that exactly one of these values is more than 3. $\left[ans:\dfrac{4}{9}\right]$

(b) find the probability that the sample mean is greater than  $\dfrac{4}{3}$. $\left[ans:\dfrac{209}{216}\right]$

I do part (a) like this,

$\Sigma P(X=x)=1$

$P(X=1)+P(X=2)+P(X=3)+P(X=4)=1$

$P(X=1)+2P(X=1)+P(X=1)+2P(X=1)=1$

$6P(X=1)=1$

$P(X=1)=\dfrac{1}{6}$

$X$
$1$
$2$
$3$
$4$

$P(X=x)$
$\dfrac{1}{6}$
$\dfrac{2}{6}$
$\dfrac{1}{6}$
$\dfrac{2}{6}$

$\dfrac{\left(\dfrac{1}{6}+\dfrac{2}{6}+\dfrac{1}{6}\right)^2 \cdot \dfrac{2}{6} \cdot 3!}{2!}=\dfrac{4}{9}$

My part (b),

$E(X)=\Sigma xP(X=x)=\dfrac{8}{3}$

$E(X^2)=\Sigma x^2P(X=x)=\dfrac{25}{3}$

$Var(X)=\sigma^2=E(X^2)-[E(X)]^2=\dfrac{17}{3}$

$\overline X \sim \mathcal{N} \left(\dfrac{8}{3},\dfrac{\dfrac{17}{3}}{3}\right)$

$\overline X \sim \mathcal{N} \left(\dfrac{8}{3},\dfrac{17}{9}\right)$

$P(X>\dfrac{4}{3})=P\left(\dfrac{\overline X-\dfrac{8}{3}}{\sqrt{\dfrac{17}{3}}}>\dfrac{\dfrac{4}{3}-\dfrac{8}{3}}{\sqrt{\dfrac{17}{3}}}\right)$

$=P(Z>-0.5601) $

$=1-P(Z<-0.5601) $

$=1-0.2877 $

$=0.7123 \neq \dfrac{209}{216}$

Can anyone check my work?

答:
Your calculations for (a) is correct.  You choosing three events from

event
probability

$\{X \le 3 \}$
$2/3$

$\{X > 3 \}$
$1/3$

You need $2!$ in the denominator as events with two '$\le3$' like $\{X_1\le3, X_2\le3, X_3>3\}$ are double-counted by $3!$.

Your error in (b) comes from $\bar{X} \sim \mathcal{N}(E[X], \frac13 \mathrm{Var}(X))$.  Why does the sample mean $\bar{X}$ follow the normal distribution?

As @sudeep5221 points out, the required probability is

$$\begin{aligned}
P(\bar{X} > 4/3) &= P(X_1+X_2+X_3 > 4) \\
&= 1 - P(X_1+X_2+X_3 \le 4) \\
&= 1 - P(X_1=X_2=X_3=1) - P(X_1+X_2+X_3 = 4) \\
&= 1 - (1/6)^3 - 3 \cdot (1/6)^2 (2/6) \\
&= \frac{6^3 - 1 - 3 \cdot 1 \cdot 2}{6^3} \\
&= \frac{209}{216}
\end{aligned}$$

Title: How do this question?


问:
Prove that $x \mapsto \boldsymbol{\mathrm P} \left[ \mathrm{B}(\boldsymbol\cdot + x ) \in A \right]$ is measurable for a Brownian motion $\mathrm B$

Given a (one-dimensional) Brownian motion $\mathrm{B}$ and any $A \in \mathcal B\left(\mathbb R^{[0,1)}\right)$, how may one prove that the map, from $\mathbb R$ to $[0,1]$, given by
$$
x \mapsto \boldsymbol{\mathrm P} \left[ \mathrm{B}(\boldsymbol\cdot + x ) \in A \right]
$$
is measurable?

The notation
$$
\boldsymbol{\mathrm P} \left[ \mathrm{B}(\boldsymbol\cdot + x) \in A \right]
$$
means that we consider the whole path; or in other words, if the function
$$
t \mapsto \mathrm{B}(t + x)
$$
is in $A$.

Most grateful for any advice provided!

答: A start: If $A$ is of the form $\cap_{k=1}^n\{B(t_k)\in G_k\}$ where $0<t_1<t_2<\cdots <t_n\le 1$ and the $G_k$ are open, then $x\mapsto{\bf P}[B(\cdot+x)\in A]$ is lower semi-continuous (hence Borel measurable) because $B$ is continuous in $t$. Events of this type form a $\pi$-system that generates $\mathcal B(\Bbb R^{[0,1]})$. Now use the Monotone Class Theorem.


问:
Trying to calculate matches in a tournament using combinations, in games that have multiple players
I am trying to calculate opponents in matches for my app game.

In it I will have a league of a varied number (between 5-14) of players and I want to know, given the number of players in a league, what's the smallest number of games each person can play so they've played all other people in the league and equal number of times. I could obviously play out every combination, but I want to play less games if possible. Each match will have either 3 or 4 players in it (including yourself)

Some examples to help better explain...

I know that in a 9 player league, if each match has 3 opponents only, then each player can play only 8 games, resulting in playing each opponent twice. This is fewer than the full 84 possible combinations.

In an 8 player league with 4 people then it only needs 7 games per person playing each opponent 3 times.

I THINK I have worked out the maths on how to do get the number of matches and number of times to play each opponent, but I can't work out how to come up with the combinations of players in order to satisfy this.

I can add the maths I've worked out, but it would be quite long so I'll only do that if people need it. As that's not actually the bit I need solving. (though it's part of the solution)

I've tried both on pen and paper, and writing a program to do it. The program has had the most success, working on a variety of combinations of league and match size, but it doesn't work on all of them.

I'm at a loss right now, and looking for any sort of help.

答:
Organize it in some way and then look for a systematic method.

Nine players with three in each game:

```
1:2:3 1:4:5 1:6:7 1:8:9
2:4:6 2:5:8 2:7:9
3:4:9 3:5:7 3:6:8
4:7:8
5:6:9
```

Body:
Organize it in some way and then look for a systematic method.

Nine players with three in each game:

1:2:3 1:4:5 1:6:7 1:8:9
2:4:6 2:5:8 2:7:9
3:4:9 3:5:7 3:6:8
4:7:8
5:6:9


问:
How many different ways can the kinds of milk to buy be selected?
This combinatorics problem is giving me a hard time:

A grocery store stocks 1-gallon cartons of skim milk, 1% milk, 2% milk, and whole milk. A customer is asked to buy 10 gallons of milk. The customer needs to buy at least one carton of each type of milk.

How many different ways can the kinds of milk to buy be selected?

The answer is supposed to be 84 however I am not sure how to arrive at that
answer. I saw a similar question with the stars and bars method used however
I don't fully understand how to implement that method into the context of this problem.

答:
Ans is $4^{6}$

Explanation:
After getting at least one gallon of each type, There are 4 ways for each next six gallon can be bought.


问:
Why do we need invariance of domain to conclude that the neighborhood in the definition of a manifold is open?
I'm reading Spivak's first differential geometry text. Spivak claims that in the definition of a manifold, the neighborhood $U\cong\mathbb{R}^n$ of a point $x$ must be an open set, and that this follows from invariance of domain. I'm almost certainly missing something obvious, but I don't see why we need invariance of domain for this. Here's what I was thinking:

Let $U$ be a neighborhood of $x$ which is homeomorphic to $\mathbb{R}^n$. Then there exists a continuous map $\phi:U\to\mathbb{R}^n$ such that $\phi(U)=\mathbb{R}^n$ (since $\phi$ is surjective). It follows that $\phi^{-1}(\mathbb{R}^n)=U$ is open, since the preimage of the open set $\mathbb{R}^n$ is open.

What's wrong with this argument, which doesn't use invariance of domain?

答: The issue with this argument is that all you've shown is that $U$ is open in $U$ in the subspace topology, which is true of any space. If it had a neighborhood isomorphic to the Cantor set instead of $\mathbb R^n$, that neighborhood would be open, at least in itself.


问:
Cantor Set Looking for what is Not remaining?
I have the Cantor Set parsed out into as follows:
\begin{align}\mathrm{Segment\ 1}&=[0,\frac13],[\frac23,1]\\ S_1&=\frac13\\ \mathrm{Segment\ 2}&= [0,\frac19],[\frac29,\frac13],[\frac23,\frac79],[\frac89,1]\\ S_2&=\frac13+\frac29\end{align}
My question are my $S_n$s what is not remaining? Because I have to construct a series of what is not remaining so is that right because then so far I have the following. \begin{equation}\sum_0^\infty \frac{1}{3}(\frac23)^n=\frac13+\frac29+\frac{4}{27}+... \end{equation}

答: I'm not sure if I understood you correctly. At each step you're calculating a measure of the complement of the Cantor set (in the unit interval). Since the Cantor set is the subset of $[0,1]$ then $\mu(\mathcal{C})=\mu([0,1])-\mu([0,1]\setminus\mathcal{C})$. Now just observe that at step $n$ you've removed total length of $\sum_{k=1}^n\frac{2^{n-1}}{3^n}$ which is geometric series that converges to $1$.


问:
A subbundle of $(0,1) \times \mathbb{R}^m$ over $(0,1)$ is trivial if derivatives of all sections are in it
Let $K \subseteq (0,1) \times \mathbb{R}^m$ be a subbundle of rank $k$ of the trivial bundle $(0,1) \times \mathbb{R}^m$ over $(0,1)$, that is, $K$ is a differentiable submanifold of $(0,1) \times \mathbb{R}^m$ and $K_{t}:= \left\{ v \in \mathbb{R}^m: (t,v) \in K \right\} $ is a vector subspace of $\mathbb{R}^m$ of dim $k$ for any $t \in (0,1)$. I am trying to prove that if for any section $\sigma \in \Gamma(K)$ we have that $\frac{d \sigma} {d t} \in K$, then $K = (0,1) \times \mathbb{R}^k$.

Firstly, to make sense of $\frac{d \sigma} {d t}$, I am seeing it as a shorthand for $d_{t_0} \sigma ( \left. \frac{d}{dt} \right\vert_{t=t_0} ) \in T_{\sigma(t_0)} K$; thus the condition in the hypothesis would read out be that $d_{t_0} \sigma ( \left. \frac{d}{dt} \right\vert_{t=t_0} ) \in K_{t_0}$. Is this correct? If so, can one see that $T_{\sigma(t_0)}K \simeq K_{t_0}$ (this seems like it could be useful, if true)?

And of course, I am lost about how to prove the result. I have the hint that the composition of the maps
$$
K \hookrightarrow (0,1) \times \mathbb{R}^m \xrightarrow{pr_2} \mathbb{R}^m
$$
is supposed to be of rank $k$, but I don't know how to show this i.e. that the composition $f = pr_2 \circ i$ has $\dim (d_{(t,v)}f(T_{(t,v)}K)) = k$ and also I am confused as to why that would imply that $K_t$ doesn't depend on $t$ - assuming it's true that $T_{(t,v)} \simeq K_t$ for any $(t,v) \in K$, we'd have that $K_t$ is a $k$-dimensional subspace of $T_v \mathbb{R}^m \simeq \mathbb{R}^m$, but it's still unclear why $K_t$ would have to be the same subspace for all $t$.

答:
can one see that $T_{\sigma(t_0)}K \simeq K_{t_0}$ (this seems like it could be useful, if true)?

$T_{\sigma(t_0)}K$ has one more dimension than $K_{t_0}$ - this last one captures all the fiber directions, but there is also the base direction.

Firstly, to make sense of $\frac{d \sigma} {d t}$, I am seeing it as a shorthand for $d_{t_0} \sigma ( \left. \frac{d}{dt} \right\vert_{t=t_0} ) \in T_{\sigma(t_0)} K$; thus the condition in the hypothesis would read out be that $d_{t_0} \sigma ( \left. \frac{d}{dt} \right\vert_{t=t_0} ) \in K_{t_0}$. Is this correct?

The $d_{t_0} \sigma ( \left. \frac{d}{dt} \right\vert_{t=t_0})$  will be a vector wich pushes forward by the projection to $\frac{d}{dt}$, but we want the "other part". Luckily the $T[(0,1) \times \mathbb{R}^m]=T[(0,1)]\times T[\mathbb{R}^m]$  so we can just throw away the first part -- i.e. project to $T[\mathbb{R}^m]$. Another way to get the same answer would be just to take $\lim_{t\to t_0} \frac{pr_2(\sigma(t))-pr_2(\sigma(t_0)}{t-t_0}\in \mathbb{R}^m$ and think of it as a vector in the fiber $\{t_0\}\times \mathbb{R}^m$.

And of course, I am lost about how to prove the result.

Well, it's not entirely clear what $K=(0,1)\times \mathbb{R}^k$ means; unless we know one of the fibers is $\mathbb{R}^k\subset \mathbb{R}^k$ (this is abuse of notation: $k$ tuples of real numbers are not $m$ tuples of real numbers; but you know what I mean -- we extend by zeroes), it is in fact not true that $K=(0,1) \times \mathbb{R}^k$ as a set (and even then -- see above about abuse of notation).

But you are right that we can conclude that there is a $k$-dimensional subspace $V$ of $\mathbb{R}^n$  such that $K=(0,1)\times V$ (as a set).

The idea is that in order for $K_t$ to "move" with $t$ some point of it would have to escape $V$, and we should be able to detect it by taking the path of this point and seeing that at some point its velocity was not in $V$. Making this into a proof requires some technical manipulations. Perhaps you can come up with better ones, but here is an option:

I will habitually identify points in a fiber $K_t$ and their projections to $\mathbb{R}^m$.

Take $T_0\in (0,1)$ and let $V=K_{T_0}$ thought of as a subspace of $\mathbb{R}^n$. Let $V$ be given by $n-k$ (linear, but smooth would do just as well) equations $f_1(x)=0, \ldots, f_{n-k}(x)=0$ (here $x\in \mathbb{R}^n$ and $f_j:\mathbb{R}^n\to \mathbb{R}$).

Now we want to show that for every $t$, and any $x_t\in K_t\subset \mathbb{R}^n$ we also have $f_j(x_t)=0$ - so that $K_t=V$. The plan is to show that this is true "locally" using the derivative condition, and then deduce that it is true globally. Technically, we can set it up using connectedness if $(0,1)$.

For this, consider the set of all $t$ for which the above is true. It is non-empty (since it includes $T_0$) and we want to show it is open and closed. Then it would be all of $(0,1)$.

It is closed for general reasons: suppose not, then there is $t$ and $t_i\to t$ such that $K_t$ is not in zero set of some $f_j$ but all $K_{t_i}$s are; but then there is a sequence of $x_i\in K_{t_i}$ converging to $x\in K_t$ with $f_j(x_i)=0$ and $f_j(x)\neq 0$, which is impossible since $f_j$ is continuous.

Let's show that it is open.

Suppose $t_0$ is in this set. We know that $K$ is locally trivial, so there is $(a,b)\subset(0,1)$ containing $t_0$ such that over $(a,b)$ we have a trivialization. We use this to get a frame over $(a,b)$ - a $k$-tuple of vectors $v_1(t), \ldots, v_k(t)$ in $\mathbb{R}^m$ smoothly depending on $t$ and forming (for each $t\in (a,b)$) a basis for $K_t$.

First, let's treat the case when $k=1$. Then we have $v'(t)=\lambda(t)v(t)$ for some smooth function $\lambda:(a,b)\to \mathbb{R}$. Now there are at least two ways to go:

We can observe that each of the $m$ components of $v$ satisfy the same linear ODE $f'(t)=\lambda(t)f(t)$ (with possibly different initial conditions). By uniqueness of solutions and linearity,  all solutions are proportional (as functions of $t$) -- which means precisely that $v(t)$ is a rescaling of $v(t_0)$.

Let $f(t)$ be the solution of the ODE $f'(t)=\lambda(t)f(t)$ with initial value $f(t_0)=1$. Observe that since the ODE has solution $f(t)=0$ and by uniqueness of solutions two distinct solutions never cross, $f(t)>0$ for all $t\in(a,b)$. Now let $u(t)=v(t)/f(t)$. We compute:$$u'=\frac{v'f-f'v}{f^2}=\frac{\lambda v f-\lambda f v}{f^2}=0$$Hence $u$ is constant, $u(t)=u(t_0)$ and $v$ is always a multiple of $v(t_0)$.

This completes the proof for the $k=1$ case.

For the general $k$ we again have at least two ways to go:

Consider the alternating form $\omega(t)=v_1^*(t)\wedge v_2^*(t)\wedge \ldots \wedge v_k^*(t)$. It is now a vector in the vector space of alternating forms, and since each $\frac{d v_i(t)}{dt}$ is a linear combination of $v_1(t), \ldots, v_k(t)$, it has $\frac{d \omega(t)}{dt}=\lambda(t)\omega(t)$. Hence by the same arguments as in the $k=1$ case it is constant up to rescaling, which exactly means that the vector space $K_t$ spanned by $(v_1, \ldots, v_k)$ is constant.

To do what is essentially the same thing, but in simpler language, we form an $m\times k$ matrix $V(t)$ with columns $v_i(t)$, and the condition that all $v'_i$ are in the span of the $v_i$ becomes $V'(t)= V(t)\Lambda(t)$, where now $\lambda(t)$ is a $k\times k$ matrix which tells us how to mix $v_i$s to get $v'_j$s. We can let $F(t)$ be the $k\times k$ matrix solution to matrix ODE $F'=F\Lambda$ with $F(t_0)=Id_k$. Then $f(t)=\det F(t)$ satisfies $f'=tr(\Lambda)f$ and so is non-zero, meaning $F(t)$ is invertible, and from $0=(FF^{-1})'=F'F^{-1}+F(F^{-1})'$ we have $(F^{-1})'=-F^{-1}F'F^{-1}=\Lambda F^{-1}$, so that
$$(VF^{-1})'=V'F^{-1}-V(F^{-1})'= V\Lambda F^{-1}-V\Lambda F^{-1}=0$$
So $V(t)F^{-1}(t)$ is constant, equal to $V(t_0)$ and $V(t)=V(t_0)F(t)$, meaning that $K_t$, the span of $v_i(t)$s is $K_{t_0}$.

This completes the proof.


问:
Integrate $\int {{{\left( {\cot x - \tan x} \right)}^2}dx} $
$\eqalign{
  & \int {{{\left( {\cot x - \tan x} \right)}^2}dx}   \cr
  &  = {\int {\left( {{{\cos x} \over {\sin x}} - {{\sin x} \over {\cos x}}} \right)} ^2}dx  \cr
  &  = {\int {\left( {{{{{\cos }^2}x - {{\sin }^2}x} \over {\sin x\cos x}}} \right)} ^2}dx  \cr
  &  = \int {{{\left( {{{\cos 2x} \over {{1 \over 2}\sin 2x}}} \right)}^2}dx}   \cr
  &  = \int {{{\left( {2\cot 2x} \right)}^2}}   \cr
  &  = \int {4{{\cot }^2}2xdx}   \cr
  &  = \int {4\left( {{{\csc }^2}2x - 1} \right)dx}   \cr
  &  = \int {\left(4{{\csc }^2}2x - 4\right)dx}   \cr
  &  = 4 \times {{ - 1} \over 2}\cot 2x - 4x + C  \cr
  &  =  - 2\cot 2x - 4x + C \cr} $

Where have I gone wrong? I've tried to spot an error so many times yet I can't find it, I need another pair of eyes.. Thanks.

答:
A cleaner way could be

$$\int (\cot x-\tan x)^2dx=\int(\cot^2x+\tan^2x-2\cot x\tan x)dx$$

$$=\int (\csc^2x-1+\sec^2x-1-2)dx$$

$$=\int \csc^2x dx+\int \sec^2x dx-4\int dx$$

$$=-(\cot x-\tan x)-4x+C$$

From another answer $\cot x-\tan x=2\cot2x$


问:
General approach to find a function
Is there a particular way to find a function using given conditions

If f(x) is a differentiable function satisfying $f^2
(x) + f^2
(y) + 2(xy – 1) = f^2
(x + y) $
, Also
f(x) > 0
$f  (\sqrt2)= 2$

I can only see a symmetry in $f^2
(x) + f^2
(y) + 2(xy)$.......$€(a+b)^2$
But I dont know how exactly should I approach such problems

答:
Let $g(t)=f^2(t)$, which is also differentiable.
Plug in $x=y=0$ to find $2g(0)-2=g(0)$, i.e.,
$$g(0)=2. $$
Now shift terms around
$$ g(x+y)-g(x)=g(y)+2(xy-1)=g(y)-g(0)+2xy$$
and divide by $y$ (assuming $y\ne 0$):
$$ \frac{g(x+y)-g(x)}y = \frac{g(y)-g(0)}y+2x.$$
Taking the limit as $y\to 0$, we recognize derivatives:
$$ g'(x) = g'(0) +2x.$$
From this, we infer by integration
$$g(x)=xg'(0)+x^2+C$$
for some constant $C$.
We know $g(0)=2$ and $g(\sqrt 2)=4$, so must have $C=2$ and $g'(0)=0$.
Hence ultimately,
$$ f(x)=+\sqrt{g(x)}=\sqrt{x^2+2}.$$

Remark: We did not really need the given property $f(x)>0$.

Title: General approach to find a function


问:
groups of conics
Let $\mathcal{C}$ be a conic on the projective plane $PG(2,\mathbb{F})$ where $\mathrm{char}\mathbb{F}\neq 2$. Let $\ell$ be a line and let $N$ be a point on $\mathcal{C}\setminus \ell$. For $A,B\in \mathcal{C}$, let $L_{AB}=AB\cap \ell$. Prove that the following operation defines an abelian group $G$ on $\mathcal{C}\setminus\ell$:

$$A \circ B =\begin{cases}
      N, & \text{ if } L_{AB}N \text{ is the tangent at } N \text{ on  }\mathcal{C} \\
      M, & \text{ if } L_{AB}N = \{N,M\}
   \end{cases}.$$

Further show that if

$\ell$ intersects the conic in one point (tangent) then $G$ is the additive group of $\mathbb{F}$

$\ell$ intersects the conic in two points (secant) then $G$ is the multiplicative group of $\mathbb{F}$

Proof. Here is a figure where $A\circ B$ equals a point $M$ and $A\circ C$ equals the point $N$.

For the first part I can easily show closure, abelianity, identity (which is $N$) and inverses. I am a bit stuck on the associative part $(A\circ B)\circ C = A\circ (B\circ C)$. Let $A\circ B = M_1$ and $B\circ C = M_2$. I thought we can use Pascal's theorem to show that $M_1C$ is parallel to $AM_2$ but I am not sure how to continue.
The second part of the exercise is not clear to me at all. Has anyone seen this type of group construction before? I've been looking at literature and cannot find anything else.

Update: The group part is done, and now we have to consider the cases when $\ell$ is a tangent and when $\ell$ is a secant. I think we can use the parabola and hyperbola for these cases and then use the fact that we can map conics to parabola and hyperbola - we usually do things like this but not sure how to continue.

答:
Consider hexagon $ABCM_1NM_2$ inscribed in the conic. By construction, opposite sides $AB$, $M_1N$ meet on $\ell$, and so do opposite sides $BC$, $M_2N$. It follows by Pascal's Theorem that opposite sides $M_1C$ and $AM_2$ also meet on $\ell$, which implies $(A\circ B)\circ C=A\circ(B\circ C)$.

EDIT.

See this answer to a similar question for a proof that
$G$ is the additive or multiplicative group of $$.

Title: groups of conics


问:
Laplacian Graph Embeddings with Weighted Graphs
In "A Simple Baseline Algorithm for Graph Classification" it says your graph $G$ must be "undirected and unweighted". Why must it be unweighted? Looking through the equations it is not obvious why this must be the case. Is there a workaround to allow me to use weighted adjacency matrices?

答:
This question is a bit open-ended. The authors may only deal with unweighted graphs for simplicity. In an unweighted graph, the unweighted adjacency matrix $A$ is equal to the weighted adjacency matrix $W$. This allows the authors to define the symmetric normalized Laplacian matrix:

$$L=I-D^{-1/2}AD^{-1/2}.$$
where $L$ denotes the Laplacian matrix, $D$ the degree matrix, and $A$ the adjacency matrix.

Using the symmetric normalized Laplacian matrix has many advantages; see https://www.math.ucsd.edu/~fan/research/cb/ch1.pdf.

You can still use the symmetric normalized Laplacian for a weighted graph, but the eigenvalue theorems cited in the paper you linked may no longer apply.


答:
Milk of the same type is indistinguishable.  So all gallons of 2% milk (for example) look alike.

To understand this answer, please read Stars and Bars closely.

If there was no constraint re at least one of each type of milk has to be bought, then the answer would be

$$\binom{10 + [4-1]}{[4-1]}.$$

However, since at least one of each type has to be bought, when considering the number of integer solutions to

$$x_1 + x_2 + x_3 + x_4 = 10$$

The lower bound for each variable is $1$, rather than $0$.

This bijects to the number of non-negative integer solutions to

$$x_1 + x_2 + x_3 + x_4 = 6$$

which will have

$$\binom{6 + [4-1]}{[4-1]} = \binom{9}{3} ~\text{solutions}.$$

Body:
This combinatorics problem is giving me a hard time:

A grocery store stocks 1-gallon cartons of skim milk, 1% milk, 2% milk, and whole milk. A customer is asked to buy 10 gallons of milk. The customer needs to buy at least one carton of each type of milk.

How many different ways can the kinds of milk to buy be selected?

The answer is supposed to be 84 however I am not sure how to arrive at that
answer. I saw a similar question with the stars and bars method used however
I don't fully understand how to implement that method into the context of this problem.


问:
Classical number theoretic applications of the $p$-adic numbers
I am sure we can all agree that the $p$-adic numbers are highly fascinating objects in their own right - just as the closely related theory of valuations.

Having independently read up on the $p$-adic numbers for a few weeks now, I have so far only seen one application of them to what I would call classical number theory - namely the proof given in Serre's Cours d'arithmétique that a natural number is expressible as the sum of $\leq 3$ squares if and only if it is not of the form $4^a(8b-1)$ for some $a,b \in \mathbb{N}$.

Since I have a tendency to appreciate the value of the higher theories of mathematics in proportion to their applications to elementary number theory, I immediately found myself wondering if there are any other applications.

So my question to the community is: What are the most delightful applications of the $p$-adic numbers and the theory of valuations to elementary number theory?

Many thanks.

P.s.: I am aware that there are already several posts on the forum about the applications of the $p$-adic numbers, but none that refers to elementary number theory specifically.

Edit: I agree that I have been too vague in what I mean by "elementary number theory", so I will try to be a little more specific: By a classical "elementary" number theoretic proposition, I mean a number theoretic proposition that Fermat might have come up with. Thus, the above proposition about the sum of three squares is an elementary number theoretic proposition, as is e.g. Fermat's Last Theorem and the Twin Prime Conjecture, while e.g. the BSD Conjecture or the Class Number Problem are not.

Edit 2: Thank you for all the answers below - they are all excellent! In case anyone should come up with another one, I should like to say that bonus-points are given for results that have so far only been proven using the theory of $p$-adic numbers, or whose proof using $p$-adic numbers is far more conceptual and insightful than the original / more elementary one.

答:
One of my favourite classical results using $p$-adic methods in elementary number theory is the theorem of Skolem-Mahler-Lech:

This is a theorem about linear recurrence sequences, which are sequences of integers where each term is a fixed linear combination of $n$ previous ones.
So fixing $n$ the sequence $s_i$ is defined by choosing the first $n$ terms
$$s_0,\ldots, s_{n-1}\in \mathbf Z$$
and a relation for all $k$
$$s_{k + n} = \sum_{i=0}^{n-1} a_i s_{k+i}$$
for fixed $a_i$.

Some examples are the Fibonacci sequence ($n = 2$,$s_0 = 0, s_1 = 1$, $a_0=a_1= 1$), and simpler things like any eventually periodic sequence, or the sequence $s_k = k$ (here $n=2$, $s_0 = 0, s_1=1$, $a_0 = -1, a_1= 2$).
We can make other such sequences easily by noting that  that the sum of any two linear recurrence sequences is also a linear recurrence sequence.

An important fact about such sequences are that their generating functions
$$f_s = \sum_{k= 0}^\infty s_k x^k$$
are always rational functions of  the variable $x$ (one polynomial divided by another), where the numerator defines the initial terms $s_0, \ldots, s_{n-1}$ and the denominator defines the recurrence relation.

Of the examples I mentioned above, the fibonacci sequence grows (exponentially), eventually periodic sequences are bounded, and the sequence $s_k=k$ also grows, just less quickly than fibonacci.

One question that one might then ask is:

What is the set of $k$ for which $s_k = 0$?

from these examples (and others) we might conjecture that this set is periodic, except for finitely many exceptions (after all we can always change finitely many terms of any linear recurrence sequence to make a sequence with the same behaviour eventually but with zeroes wherever we want at the start).

How might one go about proving this?
The first step of the proof is to use the rational generating function $f_s$ and write out its partial fraction decomposition over an algebraically closed field (like  $\overline {\mathbf Q}$), this will be of the form

$$f_s = \sum_{i=1}^{\ell} \sum_{j = 1}^{r_i} \frac{\alpha_{ij}}{(x - \beta_{i})^j}  $$

for some fixed roots $\beta_j$ of the original denominator of $f_s$.

Now using this decomposition
we have
$$f_s = \sum_{i=1}^{\ell} \sum_{j = 1}^{r_i} \alpha_{ij}{\left(\sum_{n=0}^\infty \beta_i^n x^n\right)^j}  $$

what this gives is that
$$s_n = \text{some polynomial expression involving terms }\beta^n  $$

For example for the fibonacci sequence this recovers Binet's formula
$$s_n = \frac{1}{\sqrt 5} \left(\frac{1+ \sqrt 5}2\right)^n-\frac{1}{\sqrt 5} \left(\frac{1- \sqrt 5}2\right)^n.$$
Or for the periodic sequence $0,1,0,1,0,1,\ldots$ this is
$$ s_n = 1^ n - (-1)^n$$

So we've written $s_n$ as a sum of exponential-type functions in $n$ with different bases, that we want to describe the zeroes of this function for $n \in \mathbf N$.

Now the magic part: the function $e^x$ is an analytic function, and on a bounded domain analytic functions have only finitely many zeroes (unless they are zero everywhere). This would give us a lot of control over the zeroes of $s_n$ if the naturals were bounded.
Which leads to the slightly strange question:

What if the natural numbers were bounded? And the functions $\beta^n$ were still analytic in some way?

Of course using the usual absolute value and metric on $\mathbf Q$ and $\mathbf C$ this is totally false.

But in the $p$-adic numbers this is true! The integers are all bounded ($p$-adically) by norm $\le 1$. So lets treat these functions as $p$-adic functions and control the zero sets in some way.

How does this prove the result? The functions $\beta^n$ are not $p$-adic analytic functions of $n$ on their own, but they are so on small enough $p$-adic disks though, what ends up happening is we get distinction between congruence classes of $n$ mod  $p-1$ for some well-chosen $p$ such that in each congruence class there are only finitely many zeroes of $s_n$ or the function $s_n$ is identically zero on that congruence class.
This gives us the theorem mentioned above, that the zeroes of $s_n$ are periodic, except for finitely many exceptions.


问:
Tensor product of fields from Romyar Sharifi's lecture notes. $K[X]/(f) \otimes_K M$.
I'm going through Romyar Sharifi's lecture notes on ANT.  In the first section, titled "Tensor product of fields", there's the isomorphism:

$$
f: K[X]/(f) \otimes_K M \simeq M[x]/(f)
$$

an isomorphism of $M$-algebras, where $f \in K[X]$ is monic and irreducible, and $M \supset K$ is an extension of fields.

$K[X]/(f) \otimes_K M$ is generated by elements of the form $\bar{g} \otimes m = (g + (f))\otimes m$ where $g \in K[X]$ and $m \in M$.  Was wondering how to interpret $\overline{g} \otimes m$?  Is it defined to be something?

答:
The map is fairly easy to define:  map the element $\bar{g}\otimes m$ to $m\cdot \bar{g}$, where that second product is just scalar multiplication ( i.e. multiply all the coefficients by $m$ ).

There are several abuses of notation in there, and a big caveat that we have to check the map is well-defined, so let me be more verbose.

$\bar{g}\in K[X]/(f)$ is represented by $g\in K[X]$.  $K$ is a subfield of $M$, given by an inclusion map $i:K\to M$.  We can use $i$ to define a map from $K[X]$ to $M[x]$ by mapping all the coefficients using $i$ and mapping $X$ to $x$.  Call this map $\phi: K[X]\to M[x]$.  This map is injective.  When we write $M[x]/(f)$, that's an abuse of notation that means $M[x]/(\phi(f))$.  And when I wrote "$\bar{g}$" in the expression $m\cdot \bar{g}$ I was using $\bar{g}$ to reference an element of $M[X]/(\phi(f))$.  That is another abuse of notation.  It should be $\bar{\phi(g)}$, where the bar over the top in this case means the coset $(\phi(g)+(\phi(f))$.  I'm pretty sure that's all the abuses of notation. The abuses of notation here are really not that bad; we're just taking the "same" polynomials and thinking of the coefficients as belonging to a larger field.

So now there's a couple of things left to show.  The biggest is probably that this map is well-defined; I defined it by referencing a representative of the coset $\bar{g}$, so you have to check that the image does not depend on the choice of representative.  You also have to check that the map is a ring-homomorphism and $M$-linear.

Finally, you have to check that it's bijective.  I believe that is again straightforward.

I spent most of the time above explaining the abuses of notation, because I believe that's what's tripping you up, but let me know if you're still confused ( or more confused).


问:
How to solve Lorenz equations numerically
I wrote a swift program to solve the Lorenz equations numerically with the classical Runge-Kutta method of order $4$.

I used the start coordinate $(1, 1, 1)$ and the step $h = 0.003$.

After about $45$ million steps, the solution converges to the "numerical fixpoint" $(x:  -8.48528137423224, y:  -8.48528137422238, z:  27.0000000000021$) and the game is over.

Could somebody link me another numerical method to solve the equations, which is not too difficult to implement and is better than the classical Runge-Kutta method ?

Many different suggestions are welcome. Thanks.

Code used

Here my swift code, which is easy to check:

```<code>import Foundation

typealias coordinata3D    = ( x: Double, y: Double, z: Double )

let sigma : Double        = 10
let r     : Double        = 28
let b     : Double        = 8/3
let dt    : Double        = 0.003
let max   : Int           = 45_000_000
let x0    : coordinata3D  = ( x: 1, y: 1, z: 1 )

func f ( p : coordinata3D ) -> coordinata3D {
    return ( x: sigma * (p.y - p.x),
             y: r * p.x - p.y - p.x * p.z,
             z: p.x * p.y - b * p.z )
}

func nextStepRungeKuttaMethod ( X : coordinata3D ) -> coordinata3D {
    var k1: coordinata3D = f( p: X )
    k1.x = k1.x * dt
    k1.y = k1.y * dt
    k1.z = k1.z * dt

    var k2: coordinata3D = f( p: ( x: X.x + k1.x/2,
                                   y: X.y + k1.y/2,
                                   z: X.z + k1.z/2 ) )
    k2.x = k2.x * dt
    k2.y = k2.y * dt
    k2.z = k2.z * dt

    var k3: coordinata3D = f( p: ( x: X.x + k2.x/2,
                                   y: X.y + k2.y/2,
                                   z: X.z + k2.z/2 ) )
    k3.x = k3.x * dt
    k3.y = k3.y * dt
    k3.z = k3.z * dt

    var k4: coordinata3D = f( p: ( x: X.x + k3.x,
                                   y: X.y + k3.y,
                                   z: X.z + k3.z ) )
    k4.x = k4.x * dt
    k4.y = k4.y * dt
    k4.z = k4.z * dt

    let dtSesto : Double   = dt / 6

    return ( x: X.x + dtSesto * ( k1.x + 2 * ( k2.x + k3.x )  + k4.x ),
             y: X.y + dtSesto * ( k1.y + 2 * ( k2.y + k3.y )  + k4.y ),
             z: X.z + dtSesto * ( k1.z + 2 * ( k2.z + k3.z )  + k4.z ) )

}

var X : coordinata3D = x0

for _ in 1...max {
    X = nextStepRungeKuttaMethod( X: X )
}
Swift.print("x:  \(X.x)  y:  \(X.y)  z:  \(X.z)")

X = nextStepRungeKuttaMethod( X: X )
Swift.print("x:  \(X.x)  y:  \(X.y)  z:  \(X.z)")

X = nextStepRungeKuttaMethod( X: X )
Swift.print("x:  \(X.x)  y:  \(X.y)  z:  \(X.z)")
</code>```

and here the result:

```<code>x:  -8.48528137423224  y:  -8.48528137422238  z:  27.0000000000021
x:  -8.48528137423224  y:  -8.48528137422238  z:  27.0000000000021
x:  -8.48528137423224  y:  -8.48528137422238  z:  27.0000000000021
</code>```

Images of the solution

the lorenz-attractor for x0 = (8, 8, 8)
detail around limit point

original: lorenz
original: detail

it is quite strange....

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
I used an Euler iteration which is rescaled at each step to enforce the smoothness of the spatial performance.
\begin{align}
&\mbox{Write}\ \Delta\vec{r} = \vec{v}\pars{\vec{r}}\Delta t\
\mbox{as}\  \Delta\vec{r} = {\vec{v}\pars{\vec{r}} \over \verts{\vec{v}\pars{\vec{r}}}}\Delta\ell\,,\quad
\vec{v}\pars{\vec{r}} \not= \vec{0}
\end{align}
where $\ds{\Delta\ell \equiv \verts{\vec{v}\pars{\vec{r}}}\Delta t}$ is a fixed quantity.

In this way, you enforce
$\ds{\verts{\Delta\vec{r}} = \Delta\ell}$. It means that at each step, $\ds{\Delta t}$ is changed to enforce the data smoothness ( by the way, in a picture we see $\ds{\vec{r}}$ instead of
$\ds{t}$ ). In addition, that avoids the cumbersome
Runge-Kutta.


问:
Does a singular Jacobian matrix imply functional dependence?
I often find the following "fact" mentioned in engineering mathematics texts and videos:

If $f_1,...,f_n$ are $C^1$ functions from $\mathbb{R}^n$ to $\mathbb{R}$ such that the Jacobian matrix of the map $(f_1,...,f_n)$ vanishes everywhere, then the $f_i$'s must be functionally related.

They seem to use the following definition of functionally related: There is some $C^1$ function $\phi:\mathbb{R}^n \rightarrow\mathbb{R}$ such that $\nabla\phi$ is nonzero everywhere, and
$\phi(f_1,...,f_n)=0$ everywhere.

It is easy to see (using chain rule and some elementary linear algebra) why this definition of functionally related would lead to the Jacobian being singular everywhere. But what intrigues me is the converse claim. I am unable to come up with a rigorous proof or a counterexample.  All the multivariable calculus books I have seen so far are silent on this point.

Any helpful pointer would be greatly appreciated.

答:
Let $f_1$, $\ldots$, $f_{k+1}$ such that the jacobian of $(f_1, \ldots, f_k)$ has maximal rank $k$, but the jacobian of $(f_1, \ldots,f_{k+1})$ has constant rank $k$. Then $f_{k+1}$ is a function of $f_1$, $\ldots$, $f_k$.

Indeed, any locus $(f_1,\ldots, f_k)=$ const is a manifold of codimension $k$. Because of the condition on the jacobian of $(f_1, \ldots,f_{k+1})$ the function $f_{k+1}$ is constant on such a locus. It follows that $f_{k+1}= F(f_1, \ldots, f_k)$, for $F$ function of $k$ variables. This function will be smooth, since the map $(f_1, \ldots, f_k)$ is a submersion.

$\bf{Added:}$ Let's explain why if $f_1$, $\ldots$, $f_k$ have jacobian of maximal rank $k$, and $\operatorname{grad}f$ is linearly dependent on $\operatorname{grad}f_1$, $\ldots$, $\operatorname{grad}f_k$, then $f$ is locally constant on any submanifold of the form $f_i = c_i$, $i=1, k$. Indeed, consider a smooth path $t \mapsto \phi(t)$ on such a submanifold. Then at every $t$ we have
$$\langle \operatorname{grad}f_i\ _ {\phi(t)}, \phi'(t)\rangle = 0$$ and by the linear dependence of the gradients we get
$$\langle \operatorname{grad}f _ {\phi(t)}, \phi'(t)\rangle = 0$$
and so the function $f(\phi(t))$ is constant in $t$.


问:
Calculation of integral around circle
I had the equation$$\int(xdy-ydx)$$around the circle
$$x^2+y^2=1$$
I tried this:
$$\begin{align}x&=(1-y^2)^{1/2}\\y&=(1-x^2)^{1/2}\end{align}$$
$$\int(1-y^2)^{1/2}dy-\int(1-x^2)^{1/2}dx=0$$
The limits of both $x$ and $y$ are same so both should cancel out and give ans$= 0$
but the answer is $2\pi$.

答: I don't know whether you took the correct Cartesian limits, but it's more convenient to work in polar coordinates. Since the circle is $r=1$,$$xdy-ydx=\cos\theta\cdot\cos\theta d\theta-\sin\theta(-\sin\theta)d\theta=d\theta.$$Now it's obvious the answer is $2\pi$.

Title: Calculation of integral around circle


问:
On the proof of The Bounded Convergence Theorem
From the text Real Analysis by Royden, he gives the following proposition

Proposition 8: Let $\{f_n\}$ be a sequence of bounded measurable functions on a set of finite
measure $E$.

$$\text{If } \{f_n\}\to f \text{ uniformly on $E$, then } \lim_{n\to\infty}\int_{E}f_n = \int_{E}f.$$

The proof is given as:

Proof: Since the convergence is uniform and each $f_n$ is bounded, the limit function $f$ is
bounded. The function $f$ is measurable since it is the pointwise limit of a sequence of
measurable functions. Let $\epsilon>0$. Choose an index $N$ for which

$$\left|f-f_n\right|<\epsilon/\operatorname{m}\left(E\right)\text{ on $E$ for all $n\geq N$}.$$

By the linearity and monotonicity of integration and the preceding corollary, for each $n\geq N$,

$$\left|\int_{E}f - \int_{E}f_n\right| = \left|\int_{E}[f - f_n]\right|\leq\int_{E}\left|f-f_n\right|\leq [\epsilon/\operatorname{m}\left(E\right)]\cdot\operatorname{m}\left(E\right) = \epsilon.$$

Therefore $\lim_{n\to\infty}\int_{E}f_n = \int_{E}f$.

My question is: How come if we switch bounded to uniformly bounded, and uniform convergent to pointwise convergent the proof no longer holds? Moreover, how come we can not longer say: Let $\epsilon>0$. Choose an index $N$ for which

$$\left|f-f_n\right|<\epsilon/\operatorname{m}\left(E\right)\text{ on $E$ for all $n\geq N$}.$$

Is this statement no longer true if $\{f_n\}\to f \text{ pointwise on $E$}$?

答:
If all that you assume is that $(f_n)_{n\in\Bbb N}$ converges pointwise to $f$, then all you know is that $\lim_{n\to\infty}f_n(x)=f(x)$ for each individual $x\in E$. But you cannot deduce from this that, given $\varepsilon>0$, you have $\bigl|f(x)-f_n(x)\bigr|$ for every sufficiently large $N$ and for all $x\in E$.

For instance, take $E=[0,1]$. For each $x\in E$ and each $n\in\Bbb N$, defined$$f_n(x)=\begin{cases}n^2x&\text{ if }x<\frac1{2n}\\-n^2\left(x-\frac1n\right)&\text{ if }x\in\left[\frac1{2n},\frac1n\right]\\0&\text{ if }x>\frac1n,\end{cases}$$Then $(f_n)_{n\in\Bbb N}$ converges pointwise to the null function, but, for each $n\in\Bbb N$, $\int_0^1f_n(x)\,\mathrm dx=\frac14$.


问:
What is a uniform convergence counterpart to Bernstein's inequality?
If Pollard's inequality is a uniform convergence counterpart to Hoeffding's inequality, what is a uniform convergence counterpart to Bernstein's inequality?

For completeness, I include all these three concentration inequalities here:

Hoeffding's inequality: If $\{X_i\}_{i=1}^n$ are i.i.d. random variables such that $a_i \leq X_i \leq b_i, \forall i$. Then, for any $\epsilon > 0$, we have
$$
P(\frac{1}{n}\sum_{i=1}^X X_i - \mathbb{E}[X_1] > \epsilon) \leq \exp( \frac{-2 n^2 \epsilon^2}{ \sum_{i=1}^n (a_i - b_i)^2 } )
$$

Pollard's:

where $\mathcal{N}$ is the covering number notation (From here https://jmlr.org/papers/v9/munos08a.html).

Bernstein's inequality:

From here https://en.wikipedia.org/wiki/Bernstein_inequalities_(probability_theory).

What I am looking for is a uniform convergence counterpart to the Bernstein's inequality, something possibly looks like this:

$$
P(\sup_{f \in \mathcal{F}} | \frac{1}{n} \sum_{i=1}^n f(X_i) - \mathbb{E}[f(X_1)] | > \epsilon) \leq \text{covering_number} \cdot \exp( \frac{-0.5 \epsilon^2}{ \sum_{i=1}^n \mathbb{E}[X_i^2] + 1/3 M \epsilon } ).
$$

答: This is exactly Theorem B in http://pages.cs.wisc.edu/~brecht/cs838docs/cuckersmale.pdf :) :


问:
Continuous Function + open set
A mapping $T$ of a metric space $X$ into a metric space $Y$ is continuous iff the inverse image of any open subset $Y$ is open subset of $X$.

Proof:
(a)Suppose that $T$ is continuous. Let $S \subset Y$ be open and $S_0$ the inverse image of $S$. If $S_0 = \emptyset$, it is open. Let $S_0 \neq \emptyset$. For any $x_0\in S_0$ let $y_0=Tx_0$. Since $S$ is open, it contains an $\epsilon$-neighborhood $N$ of $y_0$. Since $T$ is continuous, $x_0$ has a $\delta$-neighborhood $N_0$ which is mapped into $N$. Since $N\subset S$, we have $N_0\subset S_0$ ...

Please You will be able explain me in detailed this part "Since $N\subset S$, we have $N_0\subset S_0$", I don't understand "we have $N_0\subset S_0$".

答:
Inverse images preserve set inclusion. More explicitly:

$x\in N_0 \Rightarrow Tx \in N \Rightarrow Tx \in S \Rightarrow x \in S_0.$

Title: Continuous Function + open set


问:
Find area of a crescent
Find the area of the shaded region that is enclosed by a full circle and a quarter circle inscribed in a square.

My math teacher gave me this question as a challenge. But I can't seem to solve it. Can you guys help? Please tell me the steps to solve it.

Thanks.

答:
Assuming unitary square, the demicircles have equations
$$y=\frac{1}{2} \left(2 \sqrt{x-x^2}+1\right);\;y=\sqrt{1-x^2}$$
Intersection point is at $x_1=\frac{1}{8} \left(5-\sqrt{7}\right)$

Intersections with the diagonal are at $$x_2=\frac{1}{\sqrt{2}};\;x_3=\frac{1}{4} \left(\sqrt{2}+2\right)$$

The requested area is twice the area colored in the picture below that is
$$A/2=\int_{{x_1}}^{{x_2}}\left(\frac{1}{2} \left(2 \sqrt{x-x^2}+1\right)-\sqrt{1-x^2}\right)\,dx+\int_{{x_2}}^{{x_3}}\left(\frac{1}{2} \left(2 \sqrt{x-x^2}+1\right)-x\right)\,dx$$
$$A=\frac{1}{16} \left(2 \sqrt{7}-5 \pi +16 \arcsin \left(\frac{1}{8} \left(5-\sqrt{7}\right)\right)+8 \text{ arcsec}\left(\frac{2 \sqrt{\sqrt{7}+5}}{3}\right)\right)\approx 0.14638$$

$$..$$

Title: Find area of a crescent


问:
What are the extreme points of this polytope?
Let $n, k$ be positive integers with $n\geq k$.

Let $P_{n,k}$ be the set of vectors $x$ in $[0,1]^n$ for which

$$ \sum_{i=1}^n x_i = k $$

$P_{n,k}$ is defined by linear equations, so it is a polytope. What are its corners (extreme points)?

My guess is that the corners are the "$k$-binary" vectors - the vectors with exactly $k$ ones and $n-k$ zeros. To prove this, it is sufficient to prove (I think) that every vector in $P_{n,k}$ is a convex combination of $k$-binary vectors. Is this correct?

Note: when $k=1$, $P_{n,1}$ is just the standard simplex in $\mathbb{R}^n$. Does it have a name when $k > 1$?

答:
We prove that every $x$ in $P_{n,k}$  -- i.e. with  $x_i\in[0,1]$ and $x\cdot \vec{1}=k$ -- is in the convex hull of the integer vectors in $P_{n,k}$, by induction on the number of non-integer components of $x$ (of course all integer components have values $0$ or $1$).

If all $x_i$ are integers we are done. Now suppose some component of $x$ is non-integer. Then since the sum of them is integer, there are at least two non-integer components, $x_i$ and $x_j$.  First, let's decrease $x_i$ and increase $x_j$ until one of them becomes integer. Call the resulting vector $x_l$. Then let's increase $x_i$ and decrease $x_j$ until one of them becomes integer. Call the resulting vector $x_u$. Then $x$ is a convex combination of $x_l$ and $x_u$ (Proof: Denoting by $d$ the vector with $1$ in $i$th coordinate and $-1$ at $j$the coordinate,  $x_u=x+t_ud$ and $x_l=x-dt_l$ for some positive $t_u, t_l$ so $x$ is on the segment between $x_u$ and $x_l$, as wanted). Of course, both $x_l$ and $x_u$ are in $P_{n,k}$.

Now by induction hypothesis, $x_l$ and $x_u$ are both convex combinations of integer points in $P_{n,k}$, and hence so is $x$.


答:
To find its corners, let $\mathbf{x} = (x_i)_{i=1}^n \in P_{n,k}$.  Let the elementary vector $\mathbf{e}_i \in \mathbb{R}^n$ be a vector whose $i$-th component is $1$, and $0$ elsewhere.

It's easy to see that if there exists $i$ such that $x_i \in (0,1)$, then $\mathbf{x}$ isn't a corner, as we also have another index $i' \ne i$ such that $x_{i'} \in (0,1)$.  (Otherwise, if $i$ is the only non-integer index, the sum of components $\sum_i x_i \notin \mathbb{Z}$, contradicting $\sum_i x_i = k \in \mathbb{Z}$.)  It remains to construct the interval containing $\mathbf{x}$.  Choose a sufficiently small $\epsilon > 0$ such that $\mathbf{x}$ is the midpoint of the interval $[\mathbf{x} - \epsilon \mathbf{e}_i + \epsilon \mathbf{e}_{i'}, \mathbf{x} + \epsilon \mathbf{e}_i - \epsilon \mathbf{e}_{i'}] \subseteq P_{n,k}$.  That's possible as $\mathbf{x}$ has finitely many components.

As a result, we can restrict the possible candidates to $\{0,1\}^n$.  The equality constraint $\sum_i x_i = k$ tells us everything, so that the proof is finished.

N.B. If $x_i = 1$, then the $i$-th component of $\mathbf{x} + \epsilon \mathbf{e}_i - \epsilon \mathbf{e}_{i'}$ would be $x_i + \epsilon > 1$, contradicting our choice of $\mathbf{x} \in [0,1]^n$.

Title: What are the extreme points of this polytope?

Body:
Let $n, k$ be positive integers with $n\geq k$.

Let $P_{n,k}$ be the set of vectors $x$ in $[0,1]^n$ for which

$$ \sum_{i=1}^n x_i = k $$

$P_{n,k}$ is defined by linear equations, so it is a polytope. What are its corners (extreme points)?

My guess is that the corners are the "$k$-binary" vectors - the vectors with exactly $k$ ones and $n-k$ zeros. To prove this, it is sufficient to prove (I think) that every vector in $P_{n,k}$ is a convex combination of $k$-binary vectors. Is this correct?

Note: when $k=1$, $P_{n,1}$ is just the standard simplex in $\mathbb{R}^n$. Does it have a name when $k > 1$?


问:
Why this Artin ring is simple?
Suppose we have a non-commutative left Artin ring $R$, such that the subring generated by any nonzero two-sided ideal together with the unity is $R$ itself. I read that $\mathrm{Z}(R)$, which is the center of $R$, forms an integral domain, but why is this true? And how can one proceed to deduce that $R$ is simple? I tried to use DCC on the sets of all two-sided ideals to go further, but don’t see why this implies there can only be extremal two-sided ideals in this case. Any hints on this?

答:
This is what I was working on.  Let $R$ be a right ring such that $1\cdot \mathbb Z+I=R$ for every nonzero ideal $I$, such that $R$ is not commutative.

Lemma 1: under the conditions above, there is no nonzero element $x$ such that $x\mathbb Z$ is an ideal of $R$.

Proof: if $x\mathbb Z$ were a two-sided ideal, $1\cdot\mathbb Z+x\mathbb Z$ would be commutative, contradicting our standing hypotheses.

Lemma 2: under the conditions above, if $I$ is a nonzero ideal, $I^2\neq \{0\}$.

Proof: if $I^2=\{0\}$, then $R=1\cdot \mathbb Z+I$ is commutative, so the square of $I$ must be nonzero.

Proposition:  The center of $R$ is a domain.

Proof: If $xy=0$, and $x,y$ are nonzero elements of the center of $R$, then from $R=1\cdot \mathbb Z+(y)$, we learn $xR=x\mathbb Z$. Since the right hand side is manifestly a nonzero ideal, the Lemma 1 above contradicts this. Therefore there is no such pair $x,y$, and the center is a domain.

Proposition: $R$ is simple Artinian

Proof: it suffices to show that $R$ is prime, i.e. there are no nonzero ideals $A,B$ such that $AB=\{0\}$.  Suppose two such ideals existed.  Since $(BA)^2=\{0\}$, we see by Lemma 2 that $BA=\{0\}$ as well.

Now let $a$ be a nonzero element of $A$.  From $R=1\cdot \mathbb Z +B$ we learn both $aR=a\mathbb Z$ and $Ra=a\mathbb Z$. Therefore $aR=Ra=a\mathbb Z$ is a nonzero ideal of $R$, something prevented by the Lemma 1 above.

Therefore $R$ is a right Artinian prime ring, which is well-known to be a simple Artinian ring.

If one does not recall the proof that a right Artinian prime ring is simple, we could also say that $R$ is right Artinian and semiprime, hence semisimple. It would remain to show that it is simple, and this would follow from a similar argument about existence of two ideals $A,B$ such that $AB=BA=\{0\}$.

Title: Why this Artin ring is simple?


问:
Graph laplacian: second smallest eigenvalue
Let $G$ be a weighted undirected graph with vertex set $V$ and weight matrix $W$.
Given $S\subset V$, we define
$$cut(S,S^c):=\sum_{i\in S,j\in S^c} w_{ij}$$
Let $L$ be the graph Laplacian of $G$. It is well known that the second smallest eigenvalue $\lambda_2(L)$ is linked to the existence of a of a cut in $V$; in particular, it has been proved that
$$\text{min}_{S\subset V} \frac{cut(S,S^c)|V|}{|S^c||S|}\ge \lambda_2(L).$$
Do there exists any result that bounds this eigenvalue from the below?

$$\lambda_2(L)\ge ...$$
(even in some special cases)

答:
The second smallest eigenvalue $\lambda_2$ is closely related to the isoperimetric constant of a graph. A general result is that if a graph $G$ has maximal degree $d$ and isoperimetric constant $iso(G)$, then we have
$$iso(G) \leq \sqrt{2d\lambda_2}.$$

See Theorem 11.15, Bogdan Nica, A Brief Introduction to Spectral Graph Theory. The theorem is originally from this paper: https://www.ams.org/journals/tran/1984-284-02/S0002-9947-1984-0743744-X/S0002-9947-1984-0743744-X.pdf.


问:
What is the universal cover of SL(2,R)?
Wikipedia [1] says,

For $n \geq 2$, the universal cover of the special linear group SL($n, \mathbb{R}$) is not a matrix group (i.e. it has no faithful finite-dimensional representations).

What group is it, then, for example for $n=2$? Does it have a name? What index does SL($2, \mathbb{R}$) have in it?

答:
I'm learning this topic right now, so I'm not expert, but actually, we can say a lot about $\widetilde{SL}(2,\mathbb{R})$. I'll write about three points of view, one group theoretic, one differential, and one Riemannan.

Group theory

First of all there is a short exact sequence of groups [see Scott, p. 464, for a reference about this discussion]
$$0\to\mathbb{Z}\overset{i}{\to}\widetilde{SL}(2,\mathbb{R})\overset{p}{\to} PSL(2,\mathbb{R})\to0$$
and since $PSL(2,\mathbb{R})=SL(2,\mathbb{R})/\{\pm1\}$ is just a two-fold cover, i think we also have a s.e.s.
$$0\to\mathbb{Z}\overset{i}{\to}\widetilde{SL}(2,\mathbb{R})\overset{p'}{\to} SL(2,\mathbb{R})\to0$$
which correspond to the multiplication $\mathbb{Z}\overset{\cdot 2}{\to}\mathbb{Z}$.

Anyways, let $G$ be either $SL(2,\mathbb{R})$ or $PSL(2,\mathbb{R})$, and by a little abuse of notation $p:\widetilde{SL}(2,\mathbb{R})\to G$ any of the two projections.
These are central extensions (even though I don't know how to prove it), so there exist a function (of sets!) $$\phi:G\times G\to \mathbb{Z}$$
such that $\widetilde{SL}(2,\mathbb{R})$ is isomorphic to the set $\mathbb{Z}\times G$ endowed with the product law
$$(a,f)\cdot_\phi(b,g):=(a+b+\phi(f,g),fg).$$
Of course there are too many possible functions $g\times G\to \mathbb{Z}$, and only some of them will give rise to $\widetilde{SL}(2,\mathbb{R})$, but all those that do it are characterised by the fact that, if we choose a section (of sets!) $s:G\to\widetilde{SL}(2,\mathbb{R})$ such that $p\circ s=id_G$, then
$$i(\phi(f,g))=s(f)s(g)s(fg)^{-1}.$$
These last part is just general theory about abelian extensions [see for example Brown, IV.3, or this thread], but I don't know how write explicitly such a function $\phi$.

Differentiable manifolds

Another interesting approach, which I think is related even though I don't know how [see these slides as a sloppy reference], is the following.

The KAN (or Iwasawa) decomposition of $SL(2,\mathbb{R})$ gives us the following isomorphism (of smooth manifolds, but not of groups since, for example, the only normal subgroup of $SL(2,\mathbb{R})$ is $\{\pm1\}$)
$SO(2)\times AN(2,\mathbb{R})\cong SL(2,\mathbb{R})$
where $S^1\cong SO(2)=\Big\{\begin{pmatrix}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\\\end{pmatrix}:\theta\in\mathbb{R}\Big\}$ and $\mathbb{E}^2\cong AN(2,\mathbb{R})=\Big\{\begin{pmatrix}r & s \\ 0 & r^{-1}\\\end{pmatrix}:r,s\in\mathbb{R}\Big\}.$

It follow that, as an analitic manifold, it's universal cover is
$$\widetilde{SL}(2,\mathbb{R})\cong \mathbb{R}\times AN(2,\mathbb{R})$$

Riemannian geometry

I finish pointing out that one might be tempted to write $\widetilde{SL}(2,\mathbb{R})\cong \mathbb{R}\times \mathbb{E}^2$ which for example tells us that $SL(2,\mathbb{R})$ is aspherical, since its universal cover is contractible, but it is in fact even more interesting to write $\widetilde{SL}(2,\mathbb{R})\cong \mathbb{R}\times \mathbb{H}^2$, where $\widetilde{SL}(2,\mathbb{R})$ get the structure of a line bundle over $\mathbb{H}$ of constant curvature $1$. [for this Riemaniann point of view see Thurston, 3.8].

In fact $\widetilde{SL}(2,\mathbb{R})$ is one of the eight $3$-dimensional geometries.

A final note:
I don't know if, and in case how, all this blabbering can be extended to the universal cover of $SL(n,\mathbb{R})$, and I'd be glad if anyone could fill the gaps and questions that I opened up.

Title: What is the universal cover of SL(2,R)?


问:
How to integrate along velocity vector fields?
Let us assume I have sampled a vector field of "velocity" vectors : $\bf x \to {\bf v}({\bf x})$

How can I calculate a vector-field of orts-vectors where I end up after some time $t$?

Let us call say (${\bf x},t) \to {\bf s}_t({\bf x})$

Simple example (unless I am mistaken) :

$${\bf v}({\bf x}) = \begin{bmatrix}0&-1\\1&0\end{bmatrix}{\bf x} = \begin{bmatrix}-x_2\\x_1\end{bmatrix}$$

Should give $${\bf s}_t({\bf x}) = \begin{bmatrix}\cos(t)&\sin(t)\\-\sin(t)&\cos(t)\end{bmatrix}{\bf x}$$

If we "follow" the velocity field of a revolution, we will end up on a circle. Compare the in physics circular motion with constant speed.

My intuition tells me that this should be conceptually similar to integration in some sense. I also have some memory of a professor once telling me there is a connection to abstract algebra and Lie groups and Lie algebras. There was a corresponding "infinitesimal" rotation which by exponentiation was related to this integral. I am sorry if this all sounds very fuzzy.

What I mostly am after is some reference to literature of where to read more on this, although an explanation how to calculate the example is also welcome.

答:
You are correct that this is reminiscent of integration — you are trying to solve a differential equation.  Namely, if you have a relationship $\vec{v}=\vec{F}(\vec{x})$, then, since the definition of velocity is $\vec{v}=\frac{d\vec{x}}{dt}$, one obtains the relation $\frac{d\vec{x}}{dt}=\vec{F}(\vec{x})$.

This ODE does not have a general solution, but the specific example you cite, $$\frac{d\vec{x}}{dt}=\begin{bmatrix}
0&-1\\1&0
\end{bmatrix}\vec{x}\mathrm{,}$$ does.  In particular, let $A$ be any matrix.  Then, for the same reason $x=Ce^{at}$ solves $\frac{dx}{dt}=ax$, the function $$\vec{x}=M\left(\sum_{j=0}^{\infty}{\frac{(At)^j}{j!}}\right)\vec{x}_0$$ (where $M$ is an arbitrary constant matrix, and $\vec{x}_0$ an arbitrary constant vector) also satisfies $\frac{d\vec{x}}{dt}=A\vec{x}$.


答:
If the $(f_n)$ are uniformly bounded and converge pointwise to $f$, the condition
$$
|f(x)-f_n(x)|<\epsilon/m(E)\qquad\forall x\in E\ \forall n\ge N\tag1$$
may never be satisfied.  As a counterexample, on $E:=[0,1]$ consider $f_n$ to be a rectangle of height $1$ over the interval $(0,\frac1n)$. Then for any $\epsilon>0$ there is always a set $A$ of positive measure where $|f-f_n|>\epsilon$ on $A$. So the proof doesn't go through using (1). (The result that $\int f_n\to\int f$ is still true, though, by dominated convergence.)

Title: On the proof of The Bounded Convergence Theorem

Body:
From the text Real Analysis by Royden, he gives the following proposition

Proposition 8: Let $\{f_n\}$ be a sequence of bounded measurable functions on a set of finite
measure $E$.

$$\text{If } \{f_n\}\to f \text{ uniformly on $E$, then } \lim_{n\to\infty}\int_{E}f_n = \int_{E}f.$$

The proof is given as:

Proof: Since the convergence is uniform and each $f_n$ is bounded, the limit function $f$ is
bounded. The function $f$ is measurable since it is the pointwise limit of a sequence of
measurable functions. Let $\epsilon>0$. Choose an index $N$ for which

$$\left|f-f_n\right|<\epsilon/\operatorname{m}\left(E\right)\text{ on $E$ for all $n\geq N$}.$$

By the linearity and monotonicity of integration and the preceding corollary, for each $n\geq N$,

$$\left|\int_{E}f - \int_{E}f_n\right| = \left|\int_{E}[f - f_n]\right|\leq\int_{E}\left|f-f_n\right|\leq [\epsilon/\operatorname{m}\left(E\right)]\cdot\operatorname{m}\left(E\right) = \epsilon.$$

Therefore $\lim_{n\to\infty}\int_{E}f_n = \int_{E}f$.

My question is: How come if we switch bounded to uniformly bounded, and uniform convergent to pointwise convergent the proof no longer holds? Moreover, how come we can not longer say: Let $\epsilon>0$. Choose an index $N$ for which

$$\left|f-f_n\right|<\epsilon/\operatorname{m}\left(E\right)\text{ on $E$ for all $n\geq N$}.$$

Is this statement no longer true if $\{f_n\}\to f \text{ pointwise on $E$}$?


问:
Topology Induced by $b_s$-pseudo metric

Given $s\geq1$, $X$ is nonempty set and $d: X\times X \rightarrow [0,\infty)$ is $b_s$-pseudo metric on $X$ if for $x,y,z\in X$

i. $d(x,x)=0$

ii. $d(x,y)=d(y,x)$

iii. $d(x,y)\leq s[d(x,z)+d(z,y)]$

$F=\{d_n:n\in\mathbb{N}\}$ is a family of $b_s$-pseudo metric.
$X$ is non-empty set.

The pair $(X,\tau(X))$ is called a $b_s$-gauge space where $\tau(X)$ is the topology having subbases in the family
$$B(F)=\{B(x,d_n,\varepsilon);x\in X, d_n\in F,\ and\ \varepsilon>0\}$$

I don't understand the 'topology having subbases in ...' part. What is the subbases definiton here? How do we induced a topology from subbases?

If I have $C([0,\infty),\mathbb{R})$ is a set of all continuous function and the family of $b_2$-pseudo metric $F=\{d_k(x,y):k\in \mathbb{N}\}$ where $$d_k(x,y) = \max_{t\in [0,k]} (x(t)-y(t))^2$$
for every $x,y\in X$ and $k\in\mathbb{N}$, what is the example of the topology induced by $F$?

答:
What it really means is that $B(F)$ is a subbase for $\tau$. $B(F)$ is simply the set of all open $\epsilon$-balls for all of the $b_s$-pseudometrics in $F$.

The topology induced by $B(F)$ is formed in the standard way. First let

$$\mathscr{B}=\left\{\bigcap\mathscr{F}:\mathscr{F}\subseteq B(F)\text{ is finite}\right\}\,,$$

the set of all intersections of finitely many of the open balls in $B(F)$; then $\mathscr{B}$ is a base for $\tau$, i.e., $\tau$ is the set of all unions of members of $\mathscr{B}$.

As for your example, let $x\in C([0,\infty),\Bbb R)$, and let $A\subseteq\Bbb N$ be finite and non-empty. For each $k\in A$ let $\epsilon_k>0$. Then

$$\left\{y\in C([0,\infty),\Bbb R):d_k(x,y)<\epsilon_k\text{ for all }k\in A\right\}\tag{1}$$

is the intersection of finitely many members of $B(F)$ and is therefore a basic open nbhd of $x$. Let $\epsilon=\min\{\epsilon_k:k\in A\}$ and $m=\max A$; then

$$\begin{align*}
B(x,d_m,\epsilon)&=\{y\in C([0,\infty),\Bbb R):d_m(x,y)<\epsilon\}\\
&=\left\{y\in C([0,\infty),\Bbb R):\max_{t\in[0,m]}\big(x(t)-y(t)\big)^2<\epsilon\right\}
\end{align*}$$

is an open nbhd of $x$ contained in the set in $(1)$, so in this case the subbase $B(F)$ is actually a base for the topology.

Title: Topology Induced by $b_s$-pseudo metric


问:
Is it true ZF can't prove there is a choice function on the Turing degrees?
It is possible to show there is no Borel choice function on the Turing degrees. In fact we can even prove there is no Turing invariant Borel injection from $2^\omega$ to $2^\omega$, which sort of says that there is no injection from the Turing degrees into $2^\omega$. The proof relies on the measurability of Borel functions.

Now using some strong large cardinal-like axioms such as "every set is determined", one can show that there is no choice function at all on the Turing degrees. With much lower large cardinal assumptions, Solovay built a model of ZF in which every set is measurable. I'm not entirely sure, but it is probably the case that there is never a choice function on the Turing degrees in such models.

My question is : has anyone succeeded in building a model of ZF in which there is no choice function on the Turing degrees (or maybe on simpler equivalence relations, such as being equal almost everywhere), without using large cardinal assumptions ? or maybe such a thing known to be impossible ?

答:
Here's an easy solution for the $E_0$ relation, i.e. equality mod finite.

Start with your favourite model, say $L$. Next add $\omega_1$ Cohen reals, and consider the permutations that modify each coordinate by "bit flipping" on finitely many natural numbers, independently on each coordinate. Finally, take the filter of subgroups given by countable support (i.e. on a countable set of coordinates we are guaranteed to not change anything).

It is now easy to show that in the symmetric extension:

$\sf ZF$ holds, and for good measure so does $\sf DC$.

The set of $E_0$-classes of the Cohen reals exists, i.e. $([r_\alpha]\mid\alpha<\omega_1)$, where $r_\alpha$ is the Cohen real.

There is no choice function from that set. Therefore no choice function from $2^\omega/E_0$.

Doing something similar with Turing degrees is trickier, I'm guessing, since those are more complicated than just equivalence mod finite, but it's not unthinkable (I suppose something along the lines of bit flipping on a computable set might work, but I never really grokked Turing degrees to the level I can just throw it out there with certainty).


问:
Prove combinatorically - $\sum_{k=0}^{n} 2^{k}\binom{n}{k} = 3^{n}$
$\sum_{k=0}^{n} 2^{k}\binom{n}{k} = 3^{n}$

I have no idea which story to form, I thought about $n$ students assuming n=4

picking 4 students for a committee with 3 roles. since each students has 3 options.

then on the other side.. $2^{0}\binom{4}{0} + 2^{1}\binom{4}{1} +2^{1}\binom{4}{2} + 2^{3}\binom{4}{3}+2^{4}\binom{4}{4}$ = $2^{0}\binom{4}{4} + 2^{1}\binom{4}{3} +2^{1}\binom{4}{2} + 2^{3}\binom{4}{1}+2^{4}\binom{4}{0}$

but im just stuck here...

答:
HINT: You have a pool of $n$ students, and you want to divide them into $3$ groups, say Groups $1,2$, and $3$.

One way is simply to assign each student to one of the groups. How many such assignments are there?

Another way is first to select the students who are not going to be in Group $3$, and then to select a subset of them to be in Group $1$; the rest of the selected students will go in Group $2$. Of course the students who were not selected will be in Group $3$. You can split up this approach according to the number of students selected in the first step.


问:
Non-trivial $2$-torsion point on elliptic curves
If $E$ is an elliptic curve over $\mathbb{F}_p$ where $p\geq5$ and $\#E(\mathbb{F}_p)$ is even, then does $E$ have a non-trivial $2$-torsion point defined over $E(\mathbb{F}_p)$?

In other words, if we write $E$ as $y^2=f(x)$, then is $f$ reducible over $\mathbb{F}_p$?

答:
Yes, this is Cauchy's theorem (in fact its even less than this as $E(\mathbf F_p)$ is an  abelian group). If two divides the order of the group then there is an element of order 2.

Another way to look at it is that the hyperelliptic involution is the map on eliptic curves of the form $y^2 = f(x)$.

$$E(\mathbf F_p) \to E(\mathbf F_p)$$
defined away from infinity as
$$ (x, y) \mapsto (x,-y)$$
and it has the point at infinity as a fixed point, so if the cardinality of $E(\mathbf F_p)$ is even then there must in fact be another fixed point, which will be of the form
$(x_0,0)$, i.e. a root $x_0$ of $f$.


问:
Better method to show $-x^3 + 1 = (-x + 1)(x^2 + x + 1)$
Is there a rule, algorithm, or theorem used for this equality below:

$$-x^3 + 1 = (-x + 1)(x^2 + x + 1)$$

I know the RHS can be established using trial and error distributing the first term through the second, or alternatively same with FOIL.  I'm curious if there is a better method?

Image below from a CAS.

答: The factor theorem states that if $f(a)=0$ for some polynomial $f(x)$ then $(x-a)$ is a factor of $f(x)$. Try applying that to your case by letting $a=1$.


问:
Rainbow covering by rectangles
There are $n$ coverings of the unit square, each of which contains $k$ axes-parallel rectangles of a unique color.
Define a rainbow covering as a covering of the unit square that contains exactly one rectangle of each color.

Given $k$, what is the smallest $n$ for which a rainbow covering always exists?

For $k=2$, a rainbow covering might not exist if $n=2$, since we might have a covering by blue "vertical" rectangles and another covering by red "horizontal" rectangles, and each pair of one blue and one red rectangle does not cover the entire unit square. However, when $n=3$, there are at least two horizontal pairs or two vertical pairs, and taking one rectangle of each of these coverings (plus an arbitrary one from the third covering) yields a rainbow covering of the unit square. So for $k=2$ the answer is $3$.

What is the smallest $n$ for $k\geq 3$?

答:
I am impressed by your problem inventiveness.

For each natural $k$ let $n(k)$ be the smallest $n$ for which a rainbow covering always exists.

Proposition. $n(k)\le k(k+1)/2$.

Consider first a one-dimensional counterpart of $n(k)$.

Lemma.  Given coverings $\mathcal C_1,\dots, \mathcal C_k$ of the unit segment $I$ by $k$ its subsegments, we can choose a segment $I_i\in\mathcal C_i$ for each $i=1,\dots, k$ such that $\bigcup_{i=1}^k I_i=I$.

Proof. We use induction with respect to $k$. The lemma is obvious when $k=1$. Suppose that the lemma is already proved for $k$. Let  we have coverings $\mathcal C_1,\dots, \mathcal C_{k+1}$ of $I$ by $k+1$ its subsegments. For each $i=1,\dots, k+1$ we have $[0,x_i]\in \mathcal C_i$  for some $x_i\in (0,1]$. Let $x_j$ be maximal of the numbers $x_i$. Pick a segment $I_j=[0,x_j]$ of $\mathcal C_j$ and remark that for each $i\ne j$, $\mathcal C_j\setminus [0,x_i]$ is a covering of $[x_j,1]$ by $k$ segments. By the induction hypothesis, we can pick a segment $I_i\in\mathcal C_i$ for each $i\ne j$ providing that $[x_j,1]\subset\bigcup_{i\ne j} I_i$. Then $$I=[0, x_j]\cup [x_j,1]\subset I_j\cup \bigcup_{i\ne j} I_i=\bigcup_{i=1}^{k+1} I_i. \square$$

Proof of Proposition. We use induction with respect to $k$. The proposition is obvious when $k=1$. Suppose that the proposition already proved for $k$. Let $n=(k+1)(k+2)/2$ and we have coverings $\mathcal C_1,\dots, \mathcal C_n$ of a unit square $I^2$ by $k+1$ its subrectangles. For each $i=1,\dots, n$ let $x_i=\min\{y: [0,y]\times [a,b]\in \mathcal C_i$ for some $a<b\}$. Renaming $\mathcal C_i$, if necessarily, we can assure that $x_1\le x_2\le\dots\le x_n$. Put $n’=k(k+1)/2+1$. Lemma implies that we can pick a rectangle $R_i=[0, y_i]\times [a_i, b_i]\in\mathcal C_i$ for each $i=n’,n’+1,\dots n’+k=n$ providing $[0,1]=\bigcup_{i=n’}^n [a_i, b_i]$. Since $y_i\ge x_i\ge x_{n’}$ for each $i\ge n’$, we have $$[0, x_{n’}]\times [0,1]\subset \bigcup_{i=n’}^n [0, x_i]\times [a_i, b_i]\subset \bigcup_{i=n’}^n R_i.$$

For each $i=1,\dots n’-1$, pick $R’_i=[0,x_i]\times [a_i,b_i]\in\mathcal C_i$ and remark that $\mathcal C_i’=\mathcal C_i\setminus R’_i$ is a covering of $[x_{n’},1]\times [0,1]$ by $k$ rectangles. By the induction hypothesis, we can pick a rectangle $R_i\in\mathcal C_i’$ for each $i=1,\dots n’-1$ providing  $[x_{n’},1]\times [0,1]\subset\bigcup_{i=1}^{n’-1} R_i$. Then
$$I^2=[0, x_{n’}]\times [0,1]\cup [x_{n’},1]\times [0,1]\subset \bigcup_{i=n’}^n R_i\cup \bigcup_{i=1}^{n’-1} R_i=\bigcup_{i=1}^{n} R_i.$$

Title: Rainbow covering by rectangles


答:
Define $A_n$ to be the set of strings of length $n$ made of $\{0, 1, 2\}$. It is clear that $|A_n| = 3^n$. We will prove that $|A_n| = \sum_{k=0}^n 2^{k}\binom{n}{k}$.

For $k=0, 1, \dots, n$ partition $A_n$ into $k+1$ disjoint sets $B_{n,k}$ each consisting of strings that contain exactly $k$ $2$s. It is clear that

$$
A_n = \bigcup_{k=0}^n B_{n,k}
$$

and so

$$
|A_n| = \sum_{k=0}^n |B_{n,k}|\tag1.
$$

Each element in $B_{n,k}$ is uniquely determined by the choice of $k$ positions on which to place a $2$ and a binary string on the remaining $n-k$ positions. Therefore, $|B_{n,k}| = 2^{n-k}\binom{n}{k}$. Substituting into $(1)$ and using  $\binom{n}{k} = \binom{n}{n-k}$, we get

$$
|A_n| = \sum_{k=0}^n 2^{n-k}\binom{n}{k} = \sum_{k=0}^n 2^{k}\binom{n}{k}.
$$

Title: Prove combinatorically - $\sum_{k=0}^{n} 2^{k}\binom{n}{k} = 3^{n}$

Body:
$\sum_{k=0}^{n} 2^{k}\binom{n}{k} = 3^{n}$

I have no idea which story to form, I thought about $n$ students assuming n=4

picking 4 students for a committee with 3 roles. since each students has 3 options.

then on the other side.. $2^{0}\binom{4}{0} + 2^{1}\binom{4}{1} +2^{1}\binom{4}{2} + 2^{3}\binom{4}{3}+2^{4}\binom{4}{4}$ = $2^{0}\binom{4}{4} + 2^{1}\binom{4}{3} +2^{1}\binom{4}{2} + 2^{3}\binom{4}{1}+2^{4}\binom{4}{0}$

but im just stuck here...


问:
Solve triple integral and draw graph
I need to solve this triple integral:
$$\iiint y^2\exp(2xy)\, dx\,dy\,dz$$
Where $$V=\{x=0,y=1,z=0,y=x,z=8\}$$

My answer $\exp(2)-1$ was wrong

I adjusted my limits: $x$ from $0$ to $y$, $y$ from $x$ to $1$, $z$ from $0$ to $8$ and got answer:
$$2x^2 - \exp(2x^2)+\exp(2)-2$$But I still have no idea whether it is right or how to draw a proper graph

Here is link to my attempt of solving it
https://fex.net/ru/s/prpd5ep

答:
so, what we have is:
$$0\le z\le8$$
and since this variable is separable we can say:
$$I=\int_0^8dz\iint y^2\exp(2xy)dxdy=8\iint y^2\exp(2xy)dxdy$$

Now the lines $x=0,y=1,y=x$ forms a triangular region which I assume is the domain you want. In this region $0\le x,y\le1$ is true but we need to put the limits on properly:
$$I=8\int_0^1\int_0^yy^2\exp(2xy)dxdy$$
$$I=8\int_0^1y^2\left[\frac{\exp(2xy)}{2y}\right]_{x=0}^ydy=4\int_0^1y\left(e^{2y^2}-1\right)dy$$
$$I=4\int_0^1ye^{2y^2}dy-4\int_0^1ydy$$

Title: Solve triple integral and draw graph


问:
The number of operations appear in the multiplication of two upper triangular matrices
Imagine $A$ and $B$ are both $n\times n$ upper triangular matrices. How many operations exist in the $AB$?

I know that multiplication of two upper triangular matrices is upper triangular and operations that appear in finding the main diagonal is $n$, but I can't find how many operations appear in the $j>i$ matrix elements, can someone help me?

答:
Just a minor comment, initially, on

operations that appear in finding the main diagonal is n

I think the operations in finding the diagonal in fact is 1, because $i$th row of $A$ has zero values from column 1 up to column $i-1$, and $i$th column of $B$ has zeroes from row $i+1$ to $n$. Thus, $c_{ii} = a_{ii} \times b_{ii}$

Or maybe I'm losing something...

Now, following the same reasoning, i.e. taking into account that matrices are upper triangular, a similar comment on

taking the dot product of the  $i^{th}$  row of the first factor with the  $j^{th}$  column of the second, then that operation will require $n−j+1$  multiplications

I think that, since  $i^{th}$  row of the first factor has zero values from column 1 to column $i-1$ and the $j^{th}$ column of the second has zero values from rows $j+1$ to $n$, then the number of multiplications will be $j-i+1$, not $n-j+1$

right?


答: The "binomial theorem" says that $(x+ y)^n= \sum_{i=0}^n \begin{pmatrix}n\\ i\end{pmatrix} x^i y^{n-i}$.  If x= 2 and y= 1 that becomes $(2+ 1)^n= \sum_{i=0}^n\begin{pmatrix}n \\i \end{pmatrix} (2^i)(1^{n-i})$ or $3^n= \sum_{i=0}^n \begin{pmatrix}n \\ i\end{pmatrix} 2^i$.


问:
Tetrahedron circumradius in high dimensions
I guess this answer had already been answered a long time ago, but indeed I cannot find any reference.

What is the circumradius of a $n$-dimensional regular hypertetrahedron? Does it approach the length of the hypertetrahedron edges as $n$ increases?

答:
The standard $n$ simplex is the convex hull in $\mathbb{R}^{n+1}$ of the points $$(1,0,0,\dots,0),\\
(0,1,0,\dots,0)\\
\vdots\\
(0,0,\dots,0,1)$$
That is, the set of all points in $\mathbb{R}^{n+1}$ whose coordinates are nonnegative and add up to $1$.  The edge length is $\sqrt2$ and the center is at $(\frac1{n+1},\frac1{n+1}, \dots,\frac1{n+1})$.  A simple calculation shows that the distance from the center to a vertex approaches $1$ as $n\to\infty,$ so the ratio of the edge length to the circumradius actually goes to $\sqrt2$ as $n\to\infty$.

Thanks to Adam Zalcmann for pointing out the error in my previous calculation.  See his comment below for the detailed calculation.


问:
Is $\pi$ more transcendent than $e$?
I’ve often wondered about this, and I conjecture the affirmative, based mainly on that it is so much easier to prove the transcendence of $e$ than that of $\pi$.

I would be surprised if, just as numbers form a linearly ordered classes ranging from algebraic of degree $n$ and culminating in the class of transcendental numbers, the transcendental numbers themselves are not further divided into a linearly ordered classes, with $\pi$ belonging to the top class, and $e$ belonging to some class below it, so this is really a reference request. Could someone please cite chapter and verse where this is done?

答: You might be interested in Mahler's classification of transcendental numbers, which you can start to read about at http://en.wikipedia.org/wiki/Transcendental_number#Mahler.27s_classification</p>

Title: Is $\pi$ more transcendent than $e$?


问:
Prove that : 1. $\sum_{n=1}^{\infty}\int_{0}^{\infty}|f_n|dx=\infty$ 2. $\sum_{n=1}^{\infty}\int_{0}^{\infty}f_ndx=0$
Given the function sequence $f_n(x)=ae^{-anx}-be^{-bnx}$ where $0<a<b$ Prove that :

$\sum_{n=1}^{\infty}\int_{0}^{\infty}|f_n|dx=\infty$

$\sum_{n=1}^{\infty}\int_{0}^{\infty}f_ndx=0$

The second part I've managed to prove by just integrating and then it's a sum of zeroes($\sum_{n=1}^{\infty}\int_{0}^{\infty}f_ndx=\sum_{1}^{\infty}(-\frac{e^{anx}}{n}+\frac{e^{-bnx}}{n})|_{0}^{\infty}=\sum_{n=1}^{\infty}0=0$ but for the first part I've got no idea. any hint please

答:
find out when $f_n>0$ or $<0$ so:
$$0=f_n=ae^{-anx}-be^{-bnx}$$
$$a(e^{-nx})^a=b(e^{-nx})^b$$
$$\frac ab=(e^{-nx})^{b-a}$$
$$\ln a-\ln b=(b-a)(-nx)$$
so:
$$x=\frac{\ln a-\ln b}{n(a-b)}$$
is when $f_n=0$

$$\int_0^\infty f_ndx=\int_0^\infty\left(ae^{-anx}-be^{-bnx}\right)dx=\left[\frac{e^{-anx}-e^{-bnx}}{-n}\right]_{x=0}^\infty=0$$

$$\int_0^\infty|f_n|dx=-\int_0^{\frac{\ln a-\ln b}{n(a-b)}}(ae^{-anx}-be^{-bnx})dx+\int_{\frac{\ln a-\ln b}{n(a-b)}}^\infty(ae^{-anx}-be^{-bnx})dx$$
Try and find the value of this then plug it into the summation


问:
How to minimize studying using mathematics?
A friend asked me this question earlier today, and it made me wonder how to come up with a general solution (where each variable is an integer):

I have a vocabulary test tomorrow at school. On it, there will be $W$
  terms listed, and I will have to define $X$ of them. However, before
  the test, I am given a list of $Y$ possible terms that could be on the
  list given on the test. How many terms must I be able to define in order to
  assure that I get at least a mark of $Z$% correct on the test?

I've tinkered with some numbers a tad, and came up with some possibilities; i.e. if $W = 10$, $X = 4$, $Y = 19$, and $Z = 100$, then the individual needs to know $13$ terms. I'm having a bit of trouble, however, coming up with a generalized solution.

Any help is appreciated.

答:
To guarantee a particular minimum grade, simply assume that every term you don't study will be on the test.

For the example given, if every term not studied is on the test, to get 100%, you  need to find 4 terms that you've studied, no matter which 10 terms appear.  Hence you can only afford to skip 6 terms, and must therefore study $19-6=13$ terms.

More details, as per request:

To achieve grade of $Z$, you must get at least $\frac{Z}{100}X$.  Hence you can only afford to skip $W-\lfloor \frac{Z}{100} X\rfloor$ terms.  Consequently, you must study at least

$$Y-\left(W-\bigg\lfloor \frac{Z}{100} X\bigg\rfloor\right)=Y-W+\bigg\lfloor \frac{Z}{100} X\bigg\rfloor$$

terms.  Note that $\lfloor \cdot \rfloor$ denotes the floor function.


问:
Double limit question
Let $f: X \times Y \rightarrow \mathbb{R}$ be a continuous function where $X \subseteq \mathbb{R}$ and $Y \subseteq \mathbb{R}$. Suppose that

$(x_{n},y_{n}) \rightarrow (x,y)$ in $\mathbb{R}^{2}$,

$(x_{n},y_{n}) \in  X \times Y$, and

$\lim _{y_{n} \rightarrow y} f(x',y_{n}) = \infty$ for all $x' \in X$.

Is it true that there exists an $N>0$ such that $f(x_{n},y_{n}) > 0 $ for all $n \geq N$?
Seems intuitive but I am a bit stuck.

答:
No, this is not necessarily true. The function $f(x,y)=\frac{1}{y}-\frac{1}{x^2}$ is continuous in $(0,1)^2$, and for $x_n=y_n=\frac{1}{n}$, we have $f(x_n,y_n)<0$ for $n\geq 2$, while for $x'\in(0,1)$, we have $f(x',y_n)=n+1-\frac{1}{x'^2}\to\infty$.

Edit: To have a counterexample with $f$ bounded below, for $(x,y)\in (0,1)^2$, consider the function $$f(x,y)=\left\{\begin{array}{c l}\frac{1}{y}-\frac{1}{x}-1, & 0<y<x<1 \\ -1, &\text{otherwise}, \end{array}\right.$$ and choose  $x_n=y_n=\frac{1}{n}$.

Title: Double limit question


问:
$\sigma$-algebra generated by the function
Suppose function $f:\Omega \rightarrow (\Omega', Y)$ measurable. Define $$X = \sigma(\{f^{-1}(K) : K \in Y\})$$ Prove that for any $A \in X$ and any $w_1, w_2\in \Omega$ with $f(w_1) = f(w_2)$, either $w_1, w_2 \in A$ or $w_1, w_2 \in A^{\complement}$.

I was trying to establish that there exists some $D \in Y$ such that $A = f^{-1}(D)$, then the result is trivial. Is it the case that there is such a set $D$?

答:
As others have shown, the result is very simple if you know that the inverse image of a $\sigma$-algebra is a $\sigma$-algebra.

But I would like to suggest a different approach I find instructive.

For any $\omega_1, \omega_2 \in \Omega$ such that $f(\omega_1) = f(\omega_2)$, let
$$Z = \{ A \subset \Omega \,:\, \{\omega_1,\omega_2\} \subset A \text{ or } \{\omega_1,\omega_2\} \subset A^c\}.$$
It is very easy to check that $Z$ is a $\sigma$-algebra that contains
$$\{ f^{-1}(K) \,:\, K \in Y \}.$$
And $X$ is the smallest $\sigma$-algebra that contains this set.


答:
I think the easiest way to go about this problem is to embed the $n-1$ simplex in a plane in $\Bbb{R}^{n}$. If we do this, it is well known that we can represent the vertices of the $n-1$ simplex in $\Bbb{R}^{n}$ as simply $\{\mathbf{e}_i\}$, the unit vectors. However, one might notice that that the mean position of the vertices is
$$\frac{1}{n}\sum_{i=1}^n\mathbf{e}_i=\left(\frac{1}{n},...,\frac{1}{n}\right)$$
So it might make sense to shift our coordinates system so that the "center of mass" is the origin. That is, we take the vertices to be
$$\mathbf{v}_i=\frac{n-1}{n}\mathbf{e}_i-\frac{1}{n}\sum_{j\in\{1,...,n\}\backslash\{i\}}\mathbf{e}_j$$
However, after doing this one might notice that the side lengths are
$$\Vert\mathbf{v}_i-\mathbf{v}_j\Vert=\sqrt{2\left(\frac{n-1}{n}-\frac{-1}{n}\right)^2}=\sqrt{2}$$
So we scale down by this factor and let the vertices of our simplex be
$$\mathbf{u}_i=\frac{1}{n\sqrt{2}}\left((n-1)\mathbf{e}_i-\sum_{j\in\{1,...,n\}\backslash\{i\}}\mathbf{e}_j\right)$$
Great. So now the circumradius of the $\boldsymbol{n-1}$ simplex is simply
$$R_{n-1}=\Vert\mathbf{u}_i\Vert=\frac{1}{n\sqrt{2}}\sqrt{(n-1)^2+(n-1)}$$
So the circumradius of the $n$ simplex is
$$R_n=\frac{\sqrt{n^2+n}}{(n+1)\sqrt{2}}$$
The first few terms in the sequence are
$$\{0,1/2,1/\sqrt{3},\sqrt{6}/4,\sqrt{2/5},...\}$$
We see that $R_n\to\frac{1}{\sqrt{2}}\text{ as }n\to\infty$.

The upshot is that the circumradius tends to $(\sqrt{2})^{-1}$ times the edge length.

Title: Tetrahedron circumradius in high dimensions

Body:
I guess this answer had already been answered a long time ago, but indeed I cannot find any reference.

What is the circumradius of a $n$-dimensional regular hypertetrahedron? Does it approach the length of the hypertetrahedron edges as $n$ increases?


问:
A question about death (persistent homology)
I've been referring to this set of notes on persistent homology, and am confused with the definition and intuition for the death of a homology class for the persistent homology of a filtration.

Given a filtration $X^0 \subseteq X^1 \subseteq X^2\subseteq \cdots$ of simplicial complexes, the inclusions induce homomorphisms on homology grousp $f^{i, j}_n : H^i_n(X^i) \rightarrow H^j_n(X^j)$, and we define $H^{i, j}_n = im(f^{i,j}_n)$. Standard stuff.

I am confused about this definition about death (pg 3, defn 39):

A $p$-th homology class $[c]$ born at stage $i$ dies entering $X_j$ if it merges with a class born earlier. Formally, if $f^{i, j-1}_p([c]) \not \in H^{i-1, j-1}_p$, but $f^{i, j}([c]) \in H^{i, j-1}_p$

My doubt isn't necessarily about the definition itself, but rather the intuition/reasoning behind the definition. As I understand it, we want to mark the death of a homology class born at stage $i$ when it becomes trivial with respect to some stage $j$. The best example I can think of is conducting a Rips complex sequence on a point cloud of data. Say at stage $\epsilon_i$, we have a $n$-hole in complex, corresponding to some $[c]$ in the $n$-th homology of $X^i$. At some future stage $j$, when $\epsilon_j$ becomes big enough, this $n$-hole gets filled with an $n$-simplex (when all the vertices are <$\epsilon_j$ away from each other). Thus the $n$-hole dies because, well, there no longer is an $n$-hole. So $[c]$ is trivial with respect to $X^j$'s $n$th homology. Is this the wrong interpretation?

I don't see how this matches up with the definition above, which is measuring when the homology class merges with some homology class before. What is our $n$-hole merging with? Isn't it simply disappearing?

A similar example: suppose at some stage $i$ we have a cluster of points $B$, which shows up a single connected component in $0$th homology group. As our distance parameter $\epsilon$ grows bigger, we will eventually at some future stage disappear by overlapping with some other component, and now we're left with one less component; i.e, the homology element associated with $B$ disappears, and thus "dies". I don't see how this squares with the idea of merging with something that was already present previously.

To put it succinctly, in what sense is $[c]$ dying if it is still a (possibly) non trivial element of $H^{i, j-1}_p$?

So my question is: Am I completely wrong in my interpretation? Are the two interpretations equivalent and if so, how?

答:
Your first example is of a class that "merged" with the zero class. That is, it's a smallest $j$ such that $f^{i,j}([c])=0$. Of course it is in $H^{i-1,j}$ then, since it is in the image of the "ancient" class $0$. If it is not in the image of $0$ at any previous stages it may have not died until then - thus making its death time $j$. But it could have died earlier, by merging with another, non-zero class.

This actually happens in your second example. Suppose, for radical simplicity, that we just start with two points, $p_1$ and $p_2$, and at stage $j$ they get connected by a segment. We could say that the class $[p_1-p_2]$ has merged the class $[0]$, thus giving us the death of $[p_1-p_2]$  at time $j$. But what about the classes $[p_1]$ and $[p_2]$? Neither of them got killed, they just merged. As noted in the notes you linked, one can choose to say that the younger of $p_1$ and $p_2$ is killed, and the older survives.  Now, this only works if there is a younger and and older. The notes you linked do say "for simplicity we assume that
we add one simplex at a time into a filtration" - but unfortunately they only say this after the definition of death. However, without this the definition will indeed be deficient -- the two points having been born simultaneously will never die, which is not what we want! But if all the ages are distinct then this is perfectly fine. Of course one could artificially impose the "one at a time" condition, at least for finite complexes -- just insert  artificially some extra timestamps and "split" adding of all the new simplexes at some time $t$ into multiple additions, one by one.