问:
The relationship between independent and disjoint event
How do we say that disjoint events are not independent? Isn’t the two concepts really different from each other?

According to what I have been taught, disjoint event deals with a single trial and independent event deals with more than one trial

To be specific on my point of doubt

Let’s say A be the event of getting a head in a coin toss and B be the event of getting a tail

In a single trial :
P(A|B)=0

But also isn't P(A and B) [in a single trial] = 0 in this case ?

Like when you toss a coin once you can’t get a head and tail right?

So why do we compare these two concepts when they are really talking about two different things?

I am sorry if I am missing something really obvious.

Thank you

答:
These are two completely different concepts. If $A$ is the event where you get a head from a coin toss and $B$ is the event where you get a tail, then we have $\mathbb P(A\cap B)=0$, meaning that the events are disjoint. However since they are disjoint events, they must be dependent as if we know that $A$ has occurred, then $B$ must not have occured. In other words, $$\mathbb P(A|B)=\frac{\mathbb P(A\cap B)}{\mathbb P(B)}=\frac{0}{1/2}=0\ne \frac{1}{2}=\mathbb P(A)$$
(If events $A$ and $B$ are independent, we have $\mathbb P(A|B)=\mathbb P(A)$)


问:
Why is there a limit for the former and no limit for the latter?
One problem about the limit of the function. It's very confusing to me:

$$\lim_{x \to 0}\frac{f(x)-f(0)}{x} =\lim_{x \to 0}\frac{\cos|x|-1}{x}
=\lim_{x \to 0}\frac{-\frac{1}{2}|x|^2}{x}=0$$ Then the answer say the limit is existed.

But in the other equal:
$$\lim_{x \to 0}\frac{\cos\sqrt{|x|}-1}{x}
= \lim_{x \to 0}\frac{-\frac{1}{2}(\sqrt{|x|})^2}{x}
=-\frac{1}{2}{\lim_{x \to 0} \frac{|x|}{x}}$$

Then the answer say they have no limits!

I really don't know why, and I think these equal is similar, the numerator is always non-negative, only can change plus or minus in the denominator, why the one has limit and the other one didn't?

答: If you plot the two graphs, the difference should be clear.  In one the limit is zero whether $x \to 0$ through positive or negative numbers.  In the other, when $x$ approaches zero only through positive values the limit is $-1/2$ and only through negative numbers the limits is $+1/2$. These not being equal means the limit as $x \to 0$ in an arbitrary manner dos not exist.

ContentLicense: CC BY-SA 4.0


问:
How where these eigenvectors derived?
Excuse me for the very simple question. But I've been looking at this problem for over an hour and I do not understand how they derived vector $\mathbf{v_{1}}$ from the following matrix:

\begin{pmatrix}
1 & 0 & 0 & |& 1 \\
0 & 1 & 1 & | &1 \\
0 & 0 & 0 &| & 0
\end{pmatrix}

This matrix results in:
$$
x_{1} = 1
$$
$$
x_{2}+x_{3} = 1
$$

With $\mathbf{v_{1}} $ as:
\begin{pmatrix}
1 \\
0 \\
1
\end{pmatrix}

And $\mathbf{v_{2}} $ being:
\begin{pmatrix}
1 \\
1 \\
0
\end{pmatrix}

Shouldn't the first vector be $\mathbf{v_{1}} = (1,0,0)^{\mathbf{T}}$ instead? I do not see how they came up with $\mathbf{v_{1}} = (1,0,1)^{\mathbf{T}}$. I left out the rest of the problem which is about finding a generalized eigenvector.

Kind regards,

A confused student

Edit: Thanks to Infinity_hunter I have figured it out! Here is the solution:
$$
\left( \begin{array}{cc}
1 & 0 & 0 & | & 1 \\
1 & 1 & 1 & | & 1 \\
0 & 0 & 0 & | & 0
\end{array} \right)
\rightarrow
\left( \begin{array}{cc}
x_{1} & 0 & 0 \\
0 & x_{2} & x_{3} \\
0 & 0 & 0
\end{array} \right)
=
\left( \begin{array}{cc}
1 \\
1 \\
0
\end{array} \right)
\rightarrow
\left( \begin{array}{cc}
x_{1} \\
x_{2} + x_{3} \\
0
\end{array} \right)
=
\left( \begin{array}{cc}
1 \\
1 \\
0
\end{array} \right)
$$

Then taking for $x_{1} =1$,   $ x_{2} = x_{2}$,   $ x_{3} = 1-x_{2}$, which is written as:

$$
\left( \begin{array}{cc}
1 \\
0 \\
1
\end{array} \right)
+
x_{2}\cdot\left( \begin{array}{cc}
0 \\
1 \\
-1
\end{array} \right)
$$

答: If $v_1 = (x_1,x_2,x_3) ^ T$ then note that $x_1 = 1$ , $ x_2 = 0$ and $x_3 = 1$ are solutions to the given system of equations and similarly $(1,1,0)$. The vector $(1,0,0)$ does not satisfy the given system of equations as sum of second and third entries is not $1$. You can take any $x_1$ , $x_2$ and $x_3$ such that they satisfy the linear equations.

Title: How where these eigenvectors derived?


问:
How to prove that this limit diverges?
Consider the limit
$$\lim_{n\to +\infty} \frac{2^{\sqrt{n}}}{n!}\left(\frac{n}{e}\right)^n.$$

How to show that this limit is $+\infty$? I just tried square criterion but it doesn’t work. Moreover, Stirling’s formula is not allowed (requested in the assignment).

Could anyone please help?

Thank you in advance!

答:
Recall that for $n\ge 1$ the following inequlity holds:
$$
n^ne^{-n}\le n!\le n^{n+1}e^{-n+1}.
$$
You don't need Stirling's approximation here, this inequlaity can be proved by induction on $n$ (for the step the inequality $\left(1+\frac{1}{n}\right)^n\le e\le \left(1+\frac{1}{n}\right)^{n+1}$ is required).

It follows that for all $n\ge 1$
$$
\frac{2^{\sqrt{n}}}{n!}\left(\frac{n}{e}\right)^n\ge\frac{2^{\sqrt{n}}}{n^{n+1}e^{-n+1}}\frac{n^n}{e^n}=\frac{2^{\sqrt{n}}}{en}.
$$
Can you end now?

Title: How to prove that this limit diverges?


问:
Can I avoid using Multiple constraint Lagrange multipliers?

Let $S = \{(x,y) \in \mathbb{R}^n \times \mathbb{R}^n:|x|=|y|=1\}$ and let $F: S \to \mathbb{R}$, $F(x,y) = x \cdot y$.
Use Lagrange multiplier theorem to find $\sup\limits_{(x,y) \in S} F(x,y)$.

Now using the multiple constraint Lagrange multiplier theorem, this should follow rather easily. My question is, can I avoid it and just use the single constraint Lagrange multiplier theorem?

Since $S$ is closed and bounded, and $F$ is continuous, it achieves a max on $S$, say at $(x_0,y_0)$. Now let $S' = \{x \in \mathbb{R}: |x|=1\}$ and define $f: \mathbb{R}^n \to \mathbb{R}$ by $f(x) = x \cdot y_0$. Also define $g:\mathbb{R}^n \to \mathbb{R}$ by $g(x) = |x|^2-1$. Now by the Lagrange multiplier theorem, I have get that $\nabla f(x_0) = \lambda \nabla g(x_o)$, which implies that

$$y_j^0 = 2 \lambda x_j^0,$$

which would imply that $\lambda = \frac{1}{2}$, which would further imply that $x_0 = y_0$. So the maximum of $F$ would be $1$.

Is this a valid approach? Is it obvious that the maximum of $f$ on $S'$ (also closed and bounded) will be at $x_0$?

答:
It is of course obvious that the maximum of $F$ on $S$ is $1$, and that the maximal value is accepted on all diagonal pairs $(z,z)\in S$. This means that we do not have a discrete set of conditionally stationary points $\bigl(x^{(j)},y^{(j)}\bigr)$ $(1\leq j\leq m)$, but a continuum of these.

When we feign not to see this a priori then we have $2$ constraints in a $2n$-dimensional situation, i.e., we should set up the "Lagrangian"
$$\Phi(x,y,\lambda,\mu):=F(x,y)-\lambda\bigl(|x|^2-1\bigr)-\mu\bigl(|y|^2-1\bigr)\ ,$$
then proceed according to the book:
$$\Phi_{.x_i}=y_i-2\lambda x_i\quad(1\leq i\leq n),\qquad \Phi_{.y_k}=x_k-2\lambda y_k\quad(1\leq k\leq n)\ .$$
This leads two times to $x\parallel y$, i.e., $x=y$ or $x=-y$.

In your approach you kept $y_0$ fixed. Normally you would obtain a few conditionally stationary points $x^{(j)}$, but these would belong to the chosen $y_0$. You would then compute the maximal $F$-value $F(x^{(j)},y_0)$ and obtain an auxiliary function $y_0\mapsto \phi(y_0)$, which you then would have to maximize again with respect to $y_0$.


答:
The heart of your question is why it is claimed that the following limit does not exist.

${\lim_{x \to 0} \frac{|x|}{x}}$

If you consider $x>0$ then $\frac{|x|}{x} = \frac{x}{x} = 1$ so the limit as you approach from positive x is 1.

On the other hand, for $x<0$ then $\frac{|x|}{x} = \frac{-x}{x} = -1$ so the limit as you approach from negative x is -1.

As the limit is not the same from the two directions, there is no well-defined limit.

Title: Why is there a limit for the former and no limit for the latter?

Body:
One problem about the limit of the function. It's very confusing to me:

$$\lim_{x \to 0}\frac{f(x)-f(0)}{x} =\lim_{x \to 0}\frac{\cos|x|-1}{x}
=\lim_{x \to 0}\frac{-\frac{1}{2}|x|^2}{x}=0$$ Then the answer say the limit is existed.

But in the other equal:
$$\lim_{x \to 0}\frac{\cos\sqrt{|x|}-1}{x}
= \lim_{x \to 0}\frac{-\frac{1}{2}(\sqrt{|x|})^2}{x}
=-\frac{1}{2}{\lim_{x \to 0} \frac{|x|}{x}}$$

Then the answer say they have no limits!

I really don't know why, and I think these equal is similar, the numerator is always non-negative, only can change plus or minus in the denominator, why the one has limit and the other one didn't?


问:
Prove $f(x)=1$ in almost everywhere $x\in \Omega$.
Let $(\Omega,\Sigma,\mu)$ be a probability space, $f:\Omega \rightarrow \mathbb{C}$ an integrable function such that $\int f \, d\mu=1$ and $|f(x)| \leq 1$ in almost everywhere $x \in \Omega$. Prove $f(x)=1$ in almost everywhere $x\in \Omega$.
Any helps? I don't know how to use the hyphotesis $|f(x)| \leq 1$.

答:
For $n \ge 2$ call
$$\Omega_n= \{ x \in \Omega : f(x) \le  1- (1/n)\}$$
Then
$$1 = \int_{\Omega} f(x) dx = \int_{\Omega_n} f(x) dx + \int_{\Omega \setminus \Omega_n} f(x) dx \le \int_{\Omega_n} (1- (1/n)) dx + \int_{\Omega \setminus \Omega_n} 1 dx$$
Recall that in a probability space $$\mu (\Omega - \Omega_n) = 1- \mu (\Omega_n)$$ Hence
$$1 \le \int_{\Omega_n} (1- (1/n)) dx + \int_{\Omega \setminus \Omega_n} 1 dx = \mbox{computations} =1 - \frac{\mu ( \Omega_n)}{n} \le 1$$
This chain of inequalities show that for all $n$
$$\mu ( \Omega_n)=0$$
To conclude, note that
$$\{ x \in \Omega : f(x) <1\} = \bigcup_n \Omega_n$$
Hence, this set has measure zero.


答:
First note that the Taylor series for $\cos x$ is
$$
\cos x = 1 -x^2/2! + x^/4! +\cdots
$$
Since $ \cos x = \cos (-x)$ it follows that $\cos x = \cos |x|$. Since the Taylor series for $\cos x = 1 -x^2/2! + x^/4! +\cdots $ it is easy to observe that
$$
\frac{ \cos x - 1}{x} = -x/2 + x^3/24 + \cdots
$$ then it follows that the given limit is $0$. In your case since $ |x|^2  = x^2$ so $ \lim_{x \to 0} \frac{|x|^2} {x} = 0$.

In latter case since $|x| = - x$ for negative $x$ we will have $|x|/x = -1$ and for positive $x$ we have $|x|/x = 1$


答:
This is my answer,
We define $h=1-\text{Re}(f)$. Then:
\begin{equation*}
\int hd\mu= \int (1-\text{Re}(f))d\mu= \mu(\Omega)-\text{Re} \int f d\mu = 1-1=0
\end{equation*}
Like $h\geq 0$ (beacuse $\text{Re }(f)\leq |f|\leq 1$), $h=0$ in almost everywhere. So, I conclude $\text{Re}(f)=1$ in almost everywhere. But, how can I conlude $f=1$?

Body:
Let $(\Omega,\Sigma,\mu)$ be a probability space, $f:\Omega \rightarrow \mathbb{C}$ an integrable function such that $\int f \, d\mu=1$ and $|f(x)| \leq 1$ in almost everywhere $x \in \Omega$. Prove $f(x)=1$ in almost everywhere $x\in \Omega$.
Any helps? I don't know how to use the hyphotesis $|f(x)| \leq 1$.


问:
How would one solve the following: $\sin(\frac{\pi x}{30})+\sin(\frac{\pi x}{360})-\cos(\frac{\pi x}{30})-\cos(\frac{\pi x}{360})=0$
I want to use the idea of vectors to solve this problem: "How many times a day do the hour and minute hands of a clock face the opposite of each other?"
We know that the minute hand turns $6^{\circ}$ every minute. If I assume my starting point is 12 o'clock, then my angle (with respect to the positive x-axis, of course), $\theta_m(t)=90^{\circ}-6t$. Similarly, the angle between the hour-hand and the positive x-axis at any given time (in minutes), t, is $\theta_h= 90^{\circ}- \frac{1}{2}t$.
If we think of the hour and minute hands as two dimensional vectors whose tails are on the origin, then the vector of the minute hand can be specified as $\textbf{m}$, where:
\begin{align*}
 \textbf{m} &= \begin{bmatrix}
  \sin(6t) \\
  \cos(6t) \\
 \end{bmatrix}
\end{align*}
In the same way, the vector of the hour hand, $\textbf{h}$= \begin{bmatrix}
 \sin( \frac{1}{2}t) \\
 \cos( \frac{1}{2}t) \\
\end{bmatrix}

We want the instances where:
\begin{align*}
  \textbf{m} &= -\textbf{h} \\
  \textbf{m}+\textbf{h} &=\textbf{0} \\
  \begin{bmatrix}
   \sin(6t)+\sin\left( \frac{1}{2}t\right) \\
   \cos(6t)+\cos\left( \frac{1}{2}t\right) \\
  \end{bmatrix} &= \textbf{0}
 \end{align*}
So we have:
$$\sin(6t)+\sin\left( \frac{1}{2}t\right) -\cos(6t)-\cos\left( \frac{1}{2}t\right)=0$$
In radians:
$$\sin\left( \frac{\pi t}{30}\right)+\sin\left( \frac{\pi t}{360}\right) -\cos\left( \frac{\pi t}{30}\right)-\cos\left( \frac{\pi t}{360}\right)=0$$
My question is, can I find out the roots of that by hand (for my puposes, $0<t \le 1440$)? I plotted it and the period seems to be changing.

Edit: So technically from the vector equation I get this:
$$sin(\frac{\pi t}{30})+sin(\frac{\pi t}{360})=0$$
$$cos(\frac{\pi t}{30})+cos(\frac{\pi t}{360})=0$$

It turns solving these get me the true answers, while solving the one in the title gives me unnecessary ones (I'm saying all of this by looking at the plots). So, how do I solve these 2 sets by hand?

答:
Hint:

Let $\dfrac{\pi x}{360}=t$

$$\sin12t+\sin t=\cos12t+\cos t$$

Use Prosthaphaeresis Formulas  to find $$2\sin\dfrac{13t}2\cos\dfrac{11t}2=2\cos\dfrac{13t}2\cos\dfrac{11t}2$$

So, either $\cos\dfrac{11t}2=0\implies\dfrac{11t}2=(2n+1)\dfrac\pi2$ where $n$ is any integer

or $\sin\dfrac{13t}2=\cos\dfrac{13t}2\iff\tan\dfrac{13t}2=1\implies\dfrac{13t}2=m\pi+\dfrac\pi4$  where $m$ is any integer


问:
Eigenspaces of 2x2 Jacobian matrix with change of variable
Let

$$
A=
\begin{pmatrix}
v & \rho\\
v\left(v+2\rho\right) & \rho\left(2v+\rho\right)
\end{pmatrix}
$$
where $\rho>0$, $v=\alpha-\rho$ with $\alpha\in\mathbb{R}$.

Show that $A$ has two distinct eigenvalues $\lambda_1(\rho,v)<\lambda_2(\rho,v)$ and compute a basis of eigenvectors assoiated to these eigenvalues.

My work
We have
$$\det(\lambda I_2-A)=\lambda^2-(v+\rho(2v+\rho))\lambda+v\rho(v-\rho)$$
The discriminant is
$$\Delta=v^2+4\rho^2v^2+\rho^4+4\rho^3v+6v\rho^2$$

I cannot seem to find a way to show that it is strictly positive. Even if I assume it is, I do not know how to find easily the eigenvectors (apart from resolving a system which will take a very long time). There has to be something that I am missing.

PS: this comes from an hyperbolic PDE where I need to show the system hyperbolicity:
$$
\left\{
\begin{array}{l}
\partial_t \rho+\partial_x(\rho(\alpha-\rho))=0\\
\partial_t(\rho\alpha)+\partial_x(\rho\alpha(\alpha-\rho))=0
\end{array}
\right.
$$
and we chose the change of variables ${\bf u} =(\rho,\alpha\rho)\to {\bf v} = (\rho,v)$ with $v=\alpha-\rho$.

答:
The initial system reads \begin{aligned}
\rho_t + (\rho (\alpha-\rho))_x &= 0 \\
(\rho\alpha)_t + (\rho\alpha (\alpha - \rho))_x &= 0
\end{aligned}
for the conserved variables ${\bf u} = (\rho, \rho\alpha)$.
Now we introduce the new vector of conserved variables $
{\bf v} = (\rho, \alpha-\rho)
$. With $v=\alpha-\rho$, the first equation of our system rewrites as
$$
\rho_t + (\rho v)_x = 0
$$
and the second one becomes $$
(\rho (\rho+v))_t + (\rho v (\rho+v))_x = 0 \, .
$$
Using the first equation and the product rule, we may rewrite the latter as
$$
v_t + (v-\rho) v_x = 0 \, ,
$$
where we have also used $\rho\neq 0$.
Therefore, the Jacobian matrix of the flux such that ${\bf v}_t + A {\bf v}_x = \bf 0 $ reads $$
A = \begin{bmatrix}
v & \rho\\
0 & v-\rho
\end{bmatrix}
$$
which eigenvalues are the diagonal entries. They are distinct provided $\rho>0$, and the system is then strictly hyperbolic.


问:
Show algebraically that $\tan x = \frac{\sin2x - \sin x }{\cos2x - \cos x + 1}$
How can I show this algebraically,

$$\tan x = \frac{\sin2x - \sin x }{\cos2x - \cos x + 1}$$

I don't know where to begin. I simply know, that $\tan x = \dfrac{\sin x}{\cos x}$.

答:
$\sin 2x = 2\sin x \cos x$

$\cos 2x = 2\cos^2x-1$

$\dfrac{\sin2x - \sin x }{\cos2x - \cos x + 1} = \dfrac{2\sin x\cos x - \sin x }{2\cos^2x - \cos x } =\dfrac{\sin x(2\cos x-1)  }{\cos x ( 2\cos x - 1)} = \tan x$


答:
Hint:

Method$\#1:$
$$\sin6t-\cos6t=\cos\dfrac t2-\sin\dfrac t2$$

Using  Solving trigonometric equations of the form $a\sin x + b\cos x = c$ or Solving $a\sin x + b\cos x = c $

$$\sin\left(6t-\dfrac\pi4\right)=\sin\left(\dfrac\pi4-\dfrac t2\right)$$

OR Method$\#2:$

$$\cos6t-\sin6t=-\left(\cos\dfrac t2-\sin\dfrac t2\right)$$

$$\implies\cos\left(6t+\dfrac\pi4\right)=-\cos\left(\dfrac t2+\dfrac\pi4\right)$$

Again, $$-\cos\left(\dfrac t2+\dfrac\pi4\right)=\cos\left(\dfrac t2+\dfrac\pi4-\pi\right)$$

Can you take it from here?

Title: How would one solve the following: $\sin(\frac{\pi x}{30})+\sin(\frac{\pi x}{360})-\cos(\frac{\pi x}{30})-\cos(\frac{\pi x}{360})=0$

Body:
I want to use the idea of vectors to solve this problem: "How many times a day do the hour and minute hands of a clock face the opposite of each other?"
We know that the minute hand turns $6^{\circ}$ every minute. If I assume my starting point is 12 o'clock, then my angle (with respect to the positive x-axis, of course), $\theta_m(t)=90^{\circ}-6t$. Similarly, the angle between the hour-hand and the positive x-axis at any given time (in minutes), t, is $\theta_h= 90^{\circ}- \frac{1}{2}t$.
If we think of the hour and minute hands as two dimensional vectors whose tails are on the origin, then the vector of the minute hand can be specified as $\textbf{m}$, where:
\begin{align*}
 \textbf{m} &= \begin{bmatrix}
  \sin(6t) \\
  \cos(6t) \\
 \end{bmatrix}
\end{align*}
In the same way, the vector of the hour hand, $\textbf{h}$= \begin{bmatrix}
 \sin( \frac{1}{2}t) \\
 \cos( \frac{1}{2}t) \\
\end{bmatrix}

We want the instances where:
\begin{align*}
  \textbf{m} &= -\textbf{h} \\
  \textbf{m}+\textbf{h} &=\textbf{0} \\
  \begin{bmatrix}
   \sin(6t)+\sin\left( \frac{1}{2}t\right) \\
   \cos(6t)+\cos\left( \frac{1}{2}t\right) \\
  \end{bmatrix} &= \textbf{0}
 \end{align*}
So we have:
$$\sin(6t)+\sin\left( \frac{1}{2}t\right) -\cos(6t)-\cos\left( \frac{1}{2}t\right)=0$$
In radians:
$$\sin\left( \frac{\pi t}{30}\right)+\sin\left( \frac{\pi t}{360}\right) -\cos\left( \frac{\pi t}{30}\right)-\cos\left( \frac{\pi t}{360}\right)=0$$
My question is, can I find out the roots of that by hand (for my puposes, $0<t \le 1440$)? I plotted it and the period seems to be changing.

Edit: So technically from the vector equation I get this:
$$sin(\frac{\pi t}{30})+sin(\frac{\pi t}{360})=0$$
$$cos(\frac{\pi t}{30})+cos(\frac{\pi t}{360})=0$$

It turns solving these get me the true answers, while solving the one in the title gives me unnecessary ones (I'm saying all of this by looking at the plots). So, how do I solve these 2 sets by hand?


问:
Action of $S_4$ on $3$ elements
Suppose we have an action $\pi$ of $S_4$ on the set $\{x,y,z\}$, satisfying $\pi_{(1 2)} (x)=y$ and $\pi_{(2 3)} (y)=z$. Find $\pi_{(3 4)} (x)$ and $|Stab(x)|$.

What can I do here? I know that the action is a homomorphism from $S_4$ to the group of symmetries of $\{x,y,z\}$, but how do I say anything about $\pi_{(3 4)} (x)$ when the given $\pi_{(1 2)}$ and $\pi_{(2 3)}$ both fix $4$ when multiplied? Also, I know that the stabilizer is a subgroup of $S_4$, so its order has to divide $24$, and initially I thought that it consists of all permutations that fix $1$, so $6$ in total, but this seems to be wrong. Thanks in advance!

答:
We identify $S_3$ with the permutations of the set $\{x,y,z\}$. (For $S_4$ we use the usual set $\{1,2,3,4\}$.)
Then the action $\pi$ induces a homomorphism $S_4\to S_3$ with $(12)\to(xy)$, $(23)\to(yz)$, this homomorphism is surjective, so its kernel is a normal subgroup of $S_4$ of order $4!/3!=4$. There exists exactly one such subgroup $K$, it is the subgroup with elements

$()$, $(12)(34)$, $(13)(24)$, $(14)(23)$,

so $(34)$ acts like $(12)$, $(24)$ acts like $(13)$, and $(14)$ like $(23)$.

The stabilizer of $x$ in $S_3$ is the subgroup generated by $(yz)$. A preimage of $(yz)$ is $(23)\in S_4$. The group generated by $K$ and $(23)$ has eight elements:
$()$, $(23)$, $(14)$, $(14)(23)$, $(12)(34)$, $(13)(24)$, $(1243)$, $(1342)$.

Title: Action of $S_4$ on $3$ elements


问:
Principal value of $\int_{-\infty}^{\infty} \frac{\cos x}{ a^{2} - x^{2}}dx$ using residue calculus
The principal value has to be calculated for the following integral:

$$\int_{-\infty}^{\infty} \frac{\cos x}{ a^{2} - x^{2}}dx$$

for $a>0$ and $a$ is real. I set the pole to be $z=+a$ or $z=-a$, but I can't seem to get the required answer which is $π\sin(a)/a$. How can I go about solving this?

答:
In general if you have a rational function $R(z)$ that vanishes at $\infty$ and has simple poles $x_1,\dots,x_n$ on the real axis, then for $f(z)=R(z)e^{iz}$ we get $$\lim_{r\to\infty}\mathcal{P}\int_{-r}^r f(x)dx=2\pi i\sum_{\operatorname{Im} a>0}\operatorname{Res}_af+\pi i\sum_{k=1}^n \operatorname{Res}_{x_k}f$$
(exercise: prove this using the residue theorem) Here $\mathcal{P}\int$ denotes the Cauchy principal value.

Now apply this to the case $R(z)=\frac{1}{a^2-z^2}$ and use the fact that $\operatorname{Re}\frac{e^{ix}}{a^2-x^2}=\frac{\cos(x)}{a^2-x^2}$ for real $x$.


问:
If A^T b lies in the column space of A^  then should b not also lie in the column space of A? With application to least squares.
I am trying to get a somewhat intuitive understanding of the Least Mean Square estimator and ran into the following problem.

I start with a matrix A, a target point/vector b and a vector x that takes linear combinations of the columns of A.

Then comes the argument that you have your vector Ax closest to b when the vector $ b - Ax $ is orthogonal to the column space of A (if that is the right entity that the orthogonal vector is orthogonal to).

I.e. $$ A^\intercal (b - Ax) = 0 $$

Leading straightforwardly to
$$ A^\intercal b = A^\intercal Ax $$

Now, $ A^\intercal A $ is the square symmetric matrix which has across each row the inner product of a column vector of A with all column vectors in A, with the self inner products along the diagonal, so:
$$  A^\intercal Ax = \begin{bmatrix}
a_1 \cdot a_1 & a_1\cdot a_2 & ...\\
a_2 \cdot a_1 & a_2 \cdot a_2 & ...\\
... & ... & ...
\end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ ... \end{bmatrix} $$

Each corresponding row element in $ A^\intercal b $ must be equal to the corresponding element in $ A^\intercal A x $, or:
$$ a_1 \cdot a_1 x_1 + a_1 \cdot a_2 x_2+ ... + a_1 \cdot a_n x_n = a_1 \cdot (a_1 x_1 + a_2 x_2+ ... +  a_n x_n) = a_1 \cdot b $$
Therefore
$$ b = (a_1 x_1 + a_2 x_2+ ... +  a_n x_n) $$

Which means that b lies in the column space of A, i.e. it is a linear combination of the columns of A. But I believe LMS works when b does not lie in the column space of A. Was it not allowed to factor out the dot product? What's wrong here?

Thanks.

答:
Your conclusion 'Therefore' simply doesn't follow.

The equations only say that $a_i\cdot b=a_i\cdot (Ax)$ for each column $a_i$ of $A$, and this is not enough to uniquely determine $b$, unless the vectors $a_i$ form a basis.


问:
$\int_0^{\pi} \frac{x \sin x}{2 + \cos x}dx$
Problem : $$\int_0^{\pi} \frac{x \sin x}{2 + \cos x}dx$$

I tried $x=\pi - t$ but this just made integrand more messy
because its denominator isn't $2 + \cos x$.

I want to try this without $\tan(x/2)=t$ (like symmetric or etc.)

Thanks for help.

答:
@Crostul has already done the most difficult part of the question.  I'll show you the tabular method which organizes stuff well.

S
D
I

$+$
$x$
$\sin x / (2 + \cos x)$

$-$
$1$
$-\ln(2 + \cos x)$

\begin{align}
\int_0^{\pi} \frac{x \sin x}{2 + \cos x}\,\mathrm{d}x &= [-x\ln(2+\cos x)]_0^\pi + \int_0^\pi \ln(2 + \cos x)\,\mathrm{d}x \\
&= -\pi\ln(2+\cos\pi) + 0 + \int_0^\pi \ln(2 + \cos x)\,\mathrm{d}x \\
&= 2\pi \ln\frac{1+\sqrt3}2 \tag1 \label1
\end{align}

\eqref{1} comes from Evaluation of $\int_{0}^{\pi}\log(2+\cos x)dx$.


问:
If $A \vDash B$ then $A, C \vDash B$
If $A \vDash B$ then $A, C \vDash B$

I just wanna check if my thinking is correct. What stands above is untrue because if $A$ true, then $B$ true, independent from $C$.

Therefore, premise $C$ can be false while $B$ is true which conflicts the definition of entailment.

答: As correctly remarked by Mauro Allegranza in his comment, $A \models B$ means that every time $A$ is true then $B$ is true as well, i.e. it is impossible that $A$ is true and $B$ is false. As a consequence, if $A \models B$ then $A, C \models B$, because every time $A$ and $C$ are true, in particular $A$ is true and hence (since we are supposing $A \models B$) $B$ is true.

Title: If $A \vDash B$ then $A, C \vDash B$


问:
Notation for "modulo-like indexing"
I have indexed sequence $(A_i)_{i=0,1,...,n,...}$, and want to say that $A_i$ is $A_j$ itself if $i \equiv j\pmod n$. What is the formal mathematical notation for this?

答: The set of equivalence classes modulo $n$ is typically denoted $\mathbb Z/n \mathbb Z$. You can write $(A_i)_{i \in \mathbb Z/n\mathbb Z}$.

Title: Notation for "modulo-like indexing"


问:
Pointwise and uniform convergence of $f_n(x)=n^2 \int_0^x \frac{\sin(x^n t)}{t} dt$
Let $x\in[0,1]$, study the pointwise and uniform convergence of
$$f_n(x)=n^2 \int_0^x \frac{\sin(x^n t)}{t} dt$$
My attempt: if $x=0$ it is $f_n(0)=0$ and so its limit is $f(x)=0$ as well.

If $x\in(0,1)$: since $\sin u \leq u$ for all $u \geq 0$, $x\in(0,1) \implies x^n \in (0,1)$ and $t\in(0,x) \land x\in(0,1) \implies t \in (0,1)$, it is $x^n t \in (0,1)$ and so $0 \leq n^2\frac{\sin(x^n t)}{t} \leq n^2\frac{x^n t}{t}=n^2x^n$; so integrating both sides in the interval $(0,x)$ and letting $n\to\infty$ it follows that
$$0 \leq \lim_{n\to\infty} n^2 \int_0^x \frac{\sin(x^n t)}{t} dt \leq \lim_{n\to\infty} \int_0^x n^2 x^n dt=\lim_{n \to \infty} n^2 x^{n+1}=0$$
Where the latter limit is $0$ because $x\in(0,1) \implies x^{n+1} \in (0,1)$.

If $x=1$, since $\sin u \geq \frac{2u}{\pi}$ for all $u\in \left[0,\frac{\pi}{2}\right]$ and $[0,1] \subset \left[0,\frac{\pi}{2}\right]$, it is
$$f_n(1)=n^2\int_0^1 \frac{\sin t}{t} dt \geq n^2\int_0^1 \frac{2}{\pi}dt \to \infty \ \text{for} \ n \to\infty$$

So $f_n(x)$ converges pointwise to $f(x)=0$ for all $x\in[0,1)$ and diverges to $\infty$ for $x=1$; to study the uniform convergence, let $I_{a}=[0,a)$ with $0 \leq a<1$.

Since, for what said before, it is $f_n(x) \leq n^2 x^{n+1}$ and for the monotonicity of $s^n$ for $s \geq 0$ and $n\in\mathbb{N}$, it is $f_n(x) \leq n^2 x^{n+1} \leq n^2 a^{n+1}$. Since $f_n(x) \geq 0$ for all $x\in[0,1]$ and for all $n\\in\mathbb{N}$, it is $f_n(x)=|f_n(x)|$ and since the right hand side of the inequality $f_n(x) \leq n^2a^{n+1}$ is independent of $x$, the inequality holds for the supremum for $x\in I_a$ as well, hence $\sup_{x \in I_a} |f_n(x)| \leq n^2 a^{n+1}$ and because $a\in[0,1)$ letting $n\to\infty$ in the inequality it is
$$\lim_{n \to \infty} \sup_{x\in I_a} |f_n(x)| \leq \lim_{n \to \infty} n^2a^{n+1}=0$$
This shows that $f_n$ is uniformly convergent to $f(x)=0$ for all the compact sets $I_a$, meaning it is uniformly convergent for all $x\in[0,1)$.

Is this correct? If there are mistakes or imprecisions it would be great to have your feedback to improve myself. Any other solutions are welcome in the answers. Thanks.

答:
Consider a sequence $(x_n)$ of numbers in [0,1) that tends to 1, for example the sequence defined by

$\forall n\in \mathbb{N}^*\setminus\{1\}, x_n=1-\frac{1}{n}$

$f_n(x_n)=n^2\int_0^{x_n} \frac{\sin(x_n^n t)}{t}dt$

$\geq n^2\int_0^{x_n} \frac{2(x_n^n t)}{\pi t}dt$

$\geq \frac{2}{\pi} n^2.x_n^{n+1}$

$\geq \frac{2}{\pi} n^2.\left(1-\frac{1}{n}\right)^{n+1}$

So the limit is $+\infty$


问:
$g(x) = \frac{f'(x)}{f(x)}$ not bounded
Let $f : ]0,1[ \rightarrow ]0,+\infty[$ a differentiable function such that

$\lim_{x\to0+} f(x) = 0$

Show that the function $g:]0,1[ \rightarrow \mathbb{R}$ defined by

$g(x) = \dfrac{f'(x)}{f(x)}$

is not bounded.

I tried a lot of things: I tried the extension by continuity on 0, tried to pass by Mean value theorem and tried to proof by absurd by bounding $g(x)$ but it doesn't lead me anywhere... Thanks for help in advance

答:
Hint: observe that $g(x) = h'(x)$, with $h(x) := \log f(x)$, and that, by assumption, $h(x)  \to -\infty$ as $x\to 0^+$.
Assume by contradiction that $h'$ is bounded. Can $h(x)\to -\infty$ as $x\to 0^+$?

Title: $g(x) = \frac{f'(x)}{f(x)}$ not bounded


问:
Check if the following fields are conservative and solve the integrals
Let $L$ be the straight section connecting the points $A\left(0,-3,-6\right) \rightarrow B\left(3,2,1\right)$

Solve the following integrals (Check if the fields are conservative to avoid unnecessary work)

$\displaystyle \int_{L}
   \left( 9 x^2 y^2 z           \right) \; dx
 + \left( 6 x^3 y z + 12 y^2 z \right) \; dy
 + \left( 3 x^3 y^2 + 4 y^3   \right) \; dz  {}$
$\displaystyle \int_{L}
  \left( 9x^{2}y^{2}z \right) \; dx +
  \left( 3x+6x^{3}yz+12y^{2}z \right) \; dy +
  \left( 3x^{3}y^{2}-3y+4y^{3} \right) \; dz  {}$

What unnecessary work can be avoided if the field is conservative?

I started by calculating the curl to check if the field is conservative. In the first integral the curl is 0. Is this suffiecient to say the field is conservative?

If the field is not conservative what method do I use to solve the integral?

Thank you.

答:
For conservative vector field, the line integral over a smooth curve starting at point $A$ and ending at point $B$ is path independent and can be evaluated as,

$\displaystyle \int_C \vec{F} \cdot dr = f(B) - f(A) \,$

where $\vec{F} = \nabla f(x, y, z)$ i.e. $\vec{F} \, $ is the gradient of the potential function $f(x, y, z)$.

For your first line integral,

$\displaystyle \int_{L}
   \left( 9 x^2 y^2 z           \right) \; dx
 + \left( 6 x^3 y z + 12 y^2 z \right) \; dy
 + \left( 3 x^3 y^2 + 4 y^3   \right) \; dz  {}$

The vector field is $\vec{F} = (9 x^2 y^2 z, 6 x^3 y z + 12 y^2 z, 3 x^3 y^2 + 4 y^3)$

As you have already established that the curl is zero, to find the potential function

$f(x, y ,z) = \int F_x dx + g(y, z) = 3x^3y^2z + g(y, z)$

Now if you look at $F_y$, you see the second term as only a function of $y, z$. Integrating that, $g(y, z) = 4y^3z + h(z)$

As there is no term in $F_z$ which is only a function of $z$,

$f(x, y, z) = 3x^3y^2z + 4y^3z$

Your line integral is simply = $f(3, 2, 1) - f(0, -3, -6)$.

While your second vector field is not conservative, you can easily split it into two and take advantage of the work we already did.

$\displaystyle I_2 = I_1 + 3 \int_{L} x \, dy - y \, dz \,$ (if $I_1$ is the line integral from the first problem).

Specifically on your question on parametrization of line segment $AB$,

$r(t) = (0, -3, -6) + t \, (3-0, 2 - (-3), (1 - (-6))$

$ = (0, -3, -6) + (3, 5, 7) t, 0 \leq t \leq 1$

$x = 3t, y = -3 + 5t, z = -6 + 7t$

$dx = 3 \, dt, dy = 5 \, dt, z = 7 \, dt$


问:
Is learning (theoretical) physics useful/important for a mathematician?
I'm starting to read The Princeton Companion to Mathematics, at the beginning it says:

A proper appreciation of
  pure mathematics requires some knowledge of applied
  mathematics and theoretical physics.

Some of my professors have told me that modern Mathematics require some knowledge about Quantum Mechanics and theoretical Physics.

I attended the second and third day of the José Adem Memorial Lecture Series by Matilde Marcolli at the beginning of this year, they were about Number theory, Quantum statistical Mechanics and Quantum Field theory. I did not understand a single word but somehow the few things I understood had a big impression on me.

Please give me some examples of pure Mathematics that require/use Physics.

答:
I came across this site after reading the question posted here and some of the answers. I really enjoyed it. There are additional links on the page pursuing the discussion, providing examples to consider.

Enjoy!  Chicken-or-egg? Math-Physics?

ContentLicense: CC BY-SA 3.0


问:
Apparently simple integral
I am having trouble solving this apparently simple integral:

$\int\frac{x}{3+\sqrt{x}}dx$

Hints would be preferable than complete answer...

Thanks!

答:
Substitute $x = y^2$ and use long division to simplify the integrand.

Thus, we have $dx = 2 y dy$. Substituting in the original integrand, we have:

$$\int \frac{x}{3+\sqrt{x}} dx = \int \frac{y^2}{3+y} 2y dy$$

i.e.,

$$\int \frac{x}{3+\sqrt{x}} dx  = \int \frac{2y^3}{3+y} dy$$

Title: Apparently simple integral


问:
how to solve trigonometric inequalities?
how does one solve trigonometric inequalities? Is there a method to this or is every solution done ad hoc?

simple equations of the type: $cos3x \leq  0$  when: $0\leq x \leq 2π$

The attempt at a solution: equating $cos 3x = 0$  yields $$ π /6  + 2\frac13πk\leq x \leq 2π -π /6 - πk /3 $$ as a general solution...what happens next?

答:
I use Nghi H Nguyen's method to visually solve trig inequalities by the number unit circle.
Solve: $ F(x) = \cos (3x) < 0$
There are totally 6 end points at: $ \frac{\pi}{6}, \frac{\pi}{2}, 5\frac{\pi}{6}, 7\frac{\pi}{6}, 3\frac{\pi}{2} $ and $11\frac{\pi}{6} $that divide the unit circle into $6$ equal arc lengths.
Use the check point method to find the sign status of arc length $(\frac{\pi}{6}, \frac{\pi}{2})$.

Select $x = \frac{\pi}{3}.$
We get : $F(x) = \cos 3x = \cos \pi = -1. $Then, $f(x) < 0 $ inside this arc length.
Color it blue. Use endpoint's property to easily find other sign status of $F(x)$ for other arc lengths.
We have, starting from end point $\frac{\pi}{6}$ : Blue, red, blue, red, blue, red.
We get the solution set of $F(x) = \cos 3x < 0 $ (blue) that are the 3 open intervals:
$(\frac{\pi}{6}, \frac{\pi}{2})$ and $(5\frac{\pi}{6}, 7\frac{\pi}{6})$ and $(3\frac{\pi}{2}, 11\frac{\pi}{2})$


问:
In how many arrangements of the digits 7,6,5,4,3,2,1 in a row does the consecutive series 247 appear?
My solution is to take the series 247 as one body, then we have 5! permutations

but I could be extremely wrong.

is it correct?

and another extra question is in how many arrangements of the digits 7,6,5,4,3,2,1 in the line appears at least one of the following series: 24, 27 or 47? in which way i can approach this one?

答:
For the second question:

We have $6!$ possibilities for each $24$, $27$ and $47$ as a group. So we have $3\times 6!$ possibilities in total.

But for the first case ($24$ as a group), in certain possibilities, $247$ will occur together. Similarly, for the third case($47$ as a group), $247$ will occur together in certain possibilities. Since $247$ is double-counted, we need to eliminate its set of possibilities once from the total.

Taking $247$ as a group, we have $5!$ possibilities.

So in total, we have

$$6!+6!+6!-5!$$ possibilities.

Thanks to @LightLamps and @JaapScherphuis for pointing out the mistakes I did


问:
The value of $\lim_{x\to +\infty} \dfrac{e^{x^2}}{10^{|x|}}$
I want to find the value of
$$\lim_{x\to +\infty} \dfrac{e^{x^2}}{10^{|x|}}.$$

Since $x \rightarrow + \infty$, I only consider the value of the function for $x \ge 0$, i.e.
$$\lim_{x\to +\infty} \dfrac{e^{x^2}}{10^{x}}.$$

Noticing that the numerator and denominator both approaches $+\infty$ as $x$ approaches $+\infty$, I tried using L'Hôpital's Rule. However, it makes things worse. I did check at Wolfram Alpha, and found out that the value of the limit is $\infty$.

I then noticed that both the numerator and the denominator has powers of some function of $x$. Taking natural logarithm of the numerator and denominator, we have $$\dfrac{\ln e^{x^2}}{\ln 10^{x}}=\dfrac{x^2 \ln e}{x \ln 10}=\dfrac{x}{\ln 10}.$$

This seems really silly, but would I be justified if I say that $$\lim_{x\to +\infty} \dfrac{e^{x^2}}{10^{x}} = \lim_{x\to +\infty}\dfrac{x}{\ln 10}= + \infty?$$

If this is incorrect, kindly give me a hint to tackle this problem.

Thank you for your time.

答:
You can take the natural logarithm of the fraction and use your idea.

$$\lim_{x\to +\infty} \dfrac{e^{x^2}}{10^{x}}=e^{ \lim_{x\to +\infty} \ln\dfrac{e^{x^2}}{10^{x}}}$$

A much simpler solution is to observe that $10 <e^3$ and hence

$$ \dfrac{e^{x^2}}{10^{x}}> \dfrac{e^{x^2}}{e^{3x}}$$

Added: To clarify why I can interchange the exponential and the limit.

Since $\lim_{x\to +\infty} \ln\dfrac{e^{x^2}}{10^{x}} = \infty$, for each $M>0$ there exists some $N>0$ so that $x>N$ implies $f(x) >\ln (M)$. Then for $x >N$ we have

$$e^{f(x)} >e^{\ln(M)} =M \,.$$

Thus, for each $M>0$ there exists some $N>0$ so that $x>N$ implies $e^{f(x)} >e^{\ln(M)} =M$. This shows that

$$\lim_{x\to +\infty}e^{ \ln\dfrac{e^{x^2}}{10^{x}}} = \infty  \,.$$

Had $\lim_{x \to \infty} f(x)$ been finite, to show that

$$\lim_{x \to \infty} e^{f(x)}=e^{\lim_{x \to \infty} f(x)}$$

you'd need to use the continuity of $e^x$ and the following Theorem, which can be typically found in all Calculus/Analysis textbooks:

Theorem If $\lim_{x \to a}f(x)=b$ and $g$ is continuous at $x=b$, then

$$\lim_{x \to a} g(f(x)) = g(b) \,.$$

This Theorem is used to prove that composition of continuous functions is continuous, and it typically be found just before that.


问:
What do the local sections of an nilpotent ideal sheaf look like?
Let $(X, O_X)$ be a scheme. and $I$ an nilptoent ideal sheaf. i.e. $I^n=0$ for some $n$. Would this imply that each $I(U)$ is an nilpotent ideal of $O_X(U)$?

答:
I’m submitting this as an alternative to hm2020’s answer. It’s an elaboration of what I wrote in the comments.

Let $I\subseteq \mathcal{O}_X$ be an ideal sheaf, and let $\mathcal{F}$ be the presheaf assigning to each open subset $U$ of $X$ the ideal $I(U)^n\subseteq \mathcal{O}_X(U)$. You say that $I$ is nilpotent of degree $n$ if $\mathcal{F}^\#$ is zero, where $\#$ is used to denote sheafification. But, since $\mathcal{F}$ is a separated presheaf, being a subsheaf of the sheaf $\mathcal{O}_X$, one has that $\mathcal{F}=0$ if and only if $\mathcal{F}^\#=0$ (e.g. see [1, Tag00WB]). Thus, we deduce the following:

Fact: Let $X$ be a scheme and $I$ an ideal sheaf of $\mathcal{O}_X$. Then, the following are equivalent:

For all open subsets $U$ the ideal $I(U)^n$ is zero.

The sheafification of the presheaf $U\mapsto I(U)^n$ is zero.

[1] Various authors, 2020. Stacks project. https://stacks.math.columbia.edu/


问:
How to evaluate $\int\limits_0^\infty {\frac{{{e^{ - x - \frac{1}{x}}}}}{{1 + x}}dx}$ in form of special function?
Is there anyway to evaluate the following integral in form of special function or maybe an infinite sum ?

$I =\int\limits_0^\infty  {\frac{{{e^{ - x - \frac{1}{x}}}}}{{1 + x}}dx}$

Thank you very much !

Body:
We have, by the change of varaiables $x=e^t$,
\begin{align*}
\int_0^{ + \infty } {\frac{{e^{ - x - \frac{1}{x}} }}{{1 + x}}dx} & = \int_{ - \infty }^{ + \infty } {\frac{{e^{ - 2\cosh t} }}{{1 + e^t }}e^t dt}  = \int_{ - \infty }^{ + \infty } {\frac{{e^{ - 2\cosh t} }}{{\cosh (t/2)}}\frac{{e^{t/2} }}{2}dt}
\\ &
 = \int_0^{ + \infty } {\frac{{e^{ - 2\cosh t} }}{{\cosh (t/2)}}\frac{{e^{t/2} }}{2}dt}  + \int_{ - \infty }^0 {\frac{{e^{ - 2\cosh t} }}{{\cosh (t/2)}}\frac{{e^{t/2} }}{2}dt}
\\ &
 = \int_0^{ + \infty } {\frac{{e^{ - 2\cosh t} }}{{\cosh (t/2)}}\frac{{e^{t/2} }}{2}dt}  + \int_0^{ + \infty } {\frac{{e^{ - 2\cosh t} }}{{\cosh (t/2)}}\frac{{e^{ - t/2} }}{2}dt}
\\ &
 = \int_0^{ + \infty } {e^{ - 2\cosh t} dt}  = K_0 (2)=0.113893872\ldots,
\end{align*}
where $K_\nu(z)$ is the modified Bessel function of the second kind (cf. http://dlmf.nist.gov/10.32.E9).


问:
Triangle inequality in the metric space $(\mathbb R^2 , d)$ where $d$ is the euclidean distance.
$(\mathbb R^2 , d)$ is a metric space where $d((x_1 , y_1) ,(x_2 , y_2))= \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 }$.

How to prove triangle inequality for this metric space?

Can anyone please help me ?

My Attempt : I was trying to prove it using different inequality results.  One of my result was

$\sqrt{(x_1 - x_3)^2 + (y_1 - y_3)^2 } \leq \sqrt{(x_1 - x_2)^2 + (y_2 - y_3)^2 } + \sqrt{(x_2 - x_3)^2 + (y_1 - y_2)^2 } $.

Can anyone please tell me how to proceed from here or else you can tell me a new idea.

答: $\sqrt{(x_1-x_3)^2+(y_1-y_3)^2}=\sqrt{((x_1-x_2)+(x_2-x_3))^2+((y_1-y_2)+(y_2-y_3))^2}$ Applying Minkowski's inequality: $\sqrt{(x_1-x_3)^2+(y_1-y_3)^2}\le \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}+\sqrt{(x_2-x_3)^2+(y_2-y_3)^2}$ $$\Rightarrow d((x_1,y_1),(x_3,y_3))\le d((x_1,y_1),(x_2,y_2))+d((x_2,y_2),(x_3,y_3))$$I would like to prove a more general result, the triangle inequality in Euclidean metric $(\mathbb{R}^n,d)$ using Cauchy-Schwarz inequality.


问:
Show that some tree has an infinite branch
I have a tree with the following properties:

There is a countable number of vertices.

Vertices can have a countable number of "sons".

At each depth, there exists at least a vertex who has a "descendant" at each following depth (those descendants, a priori, are not related in any other way).
Said differently and in formal terms, if we consider that the tree represents the partial order $\leq$, where the root is the minimum - each son being greater than its father and unrelated vertices being uncomparable -, and if we denote by $V_n$ the set of vertices of depth $n$, then:

$\forall n, \exists v \in V_n, \forall m>n, \exists u_m \in V_m, v \leq u_m$

The question is: does the tree possess an infinite branch ?

答:
Edit:

We can define a counterexample. For $n$ a natural number, define $T_n$ to be the tree starting with one vertex $v_n$ in degree $n$, and adding for every $m>n$ a linear branch of length $m-n$. Now let $T$ be the union of all these trees. All these trees are countable, and hence their union is countable. However, there is clearly not an infinite path since every branch is inherently finite.

Wrong answer:

If connections between vertices are only allowed between adjacent degrees, then yes, the tree has an infinite branch, and its proof is related to the one of König's Lemma. We start at degree $n=0$, in particular at the vertex $x_0$ such that there are descendants of $x_0$ for every $m>0$. We now want to decide to which vertex $x_1$ we go. Because $x_0$ has descendants for every $m>0$ there must in particular be some vertex $x_1$ which also has descendants for every $m>0$. We pick this one as the next vertex in our branch. We continue like this and find an infinite path.

If we are also allowed to have paths between vertices which differ more than one in their degree, then we can find a counterexample to the statement. We define $T_n$ to be a tree which has its origin in degree $n$, and has one descendant in every degree $m>n$, which are mutually incomparible. So, for $T_0$ this tree has vertices $v_0,\ldots, v_n,\ldots$ for every $n$, with just the edges $(v_0,v_i)$. Now, we take the union $\bigcup T_n$ of all these trees and obtain a counterexample.

Hope this helps!


问:
Prove $f(x)=1$ in almost everywhere $x\in \Omega$.
Let $(\Omega,\Sigma,\mu)$ be a probability space, $f:\Omega \rightarrow \mathbb{C}$ an integrable function such that $\int f \, d\mu=1$ and $|f(x)| \leq 1$ in almost everywhere $x \in \Omega$. Prove $f(x)=1$ in almost everywhere $x\in \Omega$.
Any helps? I don't know how to use the hyphotesis $|f(x)| \leq 1$.

答:
This answer continues your own answer. Since you where able to prove that $\operatorname{Re}(f)=1$ almost everywhere, it follows from the equatility
$$|f|^2=|\operatorname{Re}(f)|^2 + | \operatorname{Im}(f)|^2,$$
that $\operatorname{Im}(f) = 0$ almost everywhere.


问:
How to show one problem is relaxation of another?
I am new in stackexchange. I am having difficulty in answering the following question in Wolsey's Integer Programming Book.

enter image description here

I know that I can call P2 a relaxation of P1 if the feasible set in the former is larger or the objective function is greater. I also know that LP relaxation creates dual bound. I just don't know how to put together the words to prove it. It seems like the constraint of P2 is in just cx=c format. Any help here would be appreciated.

答:
@LinAlg has outlined the proof.

Observe that \eqref{P1} can be concisely written as

$$\max\{cx \mid Ax = b, x \in B^n\} \tag{$P_1$} \label{P1},$$

where $$A = \begin{bmatrix}
a_{11} & \cdots & a_{1j} & \cdots & a_{1n} \\
\vdots & \ddots & \vdots & \ddots & \vdots \\
a_{i1} & \cdots & a_{ij} & \cdots & a_{in} \\
\vdots & \ddots & \vdots & \ddots & \vdots \\
a_{m1} & \cdots & a_{mj} & \cdots & a_{mn}
\end{bmatrix}.$$

It's more difficult make a similar observation for \eqref{P2}.

$$\max\{cx \mid u^TAx = u^Tb, x \in B^n\} \tag{$P_2$} \label{P2},$$

where $u \in \mathbb{R}^m$.  Once you realize that, it's easily to verify @LinAlg's claim: left multiplication of $Ax = b$ by $u^T$ gives the desired conclusion.

To transform $\sum_j\sum_i u_i a_{ij} x_j$ into $u^TAx$:

order of $\sum_i$ and $\sum_j$ are not important, since it's a finite sum.

imagine the matrix $A = (a_{ij})_{i,j}$ when you see $a_{ij}$.

observe that $j$ is the index that appeared in both $a_{ij}$ and $x_j$.  $j$ runs through the columns of each row of $A$.  Remember that we have $\sum_j$, so this remind us right multiplication of $A$ by $x$: $Ax$.

note that $i$ is the index that appeared in both $a_{ij}$ and $u_i$.  Remember that we have $\sum_i$, and $i$ runs through the rows of each column of $A$, so this remind us right multiplication of $A$ by "$u$": but $uA$ doesn't make sense, as $u$ is a "vertical" vector, so we take transpose and write $u^TA$.

combine the two: $\sum_i u_i (\sum_j a_{ij} x_j) = \sum_i u_i(Ax)_i = u^T Ax$

Remarks: When I need to denote the $i(,j)$-th element of a product of vectors and/or matrices, I personally like the wrapping it with a pair of parenthesis and appending the subscript $i(,j)$, whenever things like "$A$ and $(a_{ij})$" aren't possible.  In the last point, I've done this on $Ax$.  That helps us to understand how the matrix product formula helps to further condense the sum to a matrix product $u^TAx$.


问:
Do there exist examples of integrable functions with these properties?
Question:

Do there exist examples of sequences of integrable functions $(f_n)_n$
and $(g_n)_n$ on $\mathbb{R}$ (with it standard measure structure)
converging pointwise to zero and satisfying:

(i) $\lim_{n \to \infty} \int_{\mathbb{R}} f_n (x) dx = + \infty$

(ii) $g_n \leq 0$, $(g_n)$ converging to zero (from below) and
$\lim_{n \to \infty} \int_{\mathbb{R}} g_n (x) dx = -1. $

If so, provide an example; if not, explain why not.

Attempt: For (i), I think yes. I consider the standard measure space $(\mathbb{R}, \beta, \lambda)$ where $\lambda$ is the Lebesgue measure on the Borel $\sigma$-algebra $\beta$. I let $f_n(x) = \frac{1}{n} \chi_{[0, n^2]} (x)$ where $\chi$ is the indicator function. Then $|f_n(x)| \leq 1/n$ for all $x \in \mathbb{R}$, and so $f_n \to 0$ pointwise (even uniformly), but $$ \lim_{n} \int_{\mathbb{R}} f_n(x) dx = \lim_n \frac{1}{n} n^2 = + \infty. $$

For (ii), I think this is impossible because it contradicts the Dominated Convergence Theorem. In class, we saw this theorem as:

Theorem:

Let $(\Omega, \mathfrak{M}, \mu)$ be a measure space and let $(f_n)_n$
be a sequence in $\mathcal{L}^1 (\Omega, \mathfrak{M}, \mu)$
converging pointwise to a function $f: \Omega \rightarrow C$. Suppose
there exists a positive integrable function $g: \Omega \rightarrow [0,
 + \infty]$ such that $|f_n| \leq g$ for all $n \in \mathbb{N}$. Then $f \in \mathcal{L}^1 (\Omega, \mathfrak{M}, \mu)$ and $$ \lim_n
 \int_{\Omega} | f_n - f| d \mu = 0$$ and $\lim_n \int_{\Omega} f_n d
 \mu = \int_{\Omega} f d\mu. $

So now I was thinking that (ii) would contradict this theorem. If $g_n \leq 0$, then $(g_n)_n$ is dominated by the zero function. So we should have $\lim_n \int_{\mathbb{R}} g_n dx = 0. $ Is this reasoning correct?

Thanks for any help.

答:
This reasonning isn't correct because the there is an absolute value in the domination, so $ g_n\leq 0$ turns into $|g_n|\geq 0 $.

And if you take $g_n=-\frac{1}{n}\chi_{[0,n]} $, you have :

$$g_n\leq 0, g_n\text{ converges pointwise to 0, and : }\lim_{n\to\infty}\int_\Omega g_nd\mu=-1 $$


答:
Here are two simpler examples:

Consider $f_n= \chi_{[n,n^2]}$    and  $g_n=-\chi_{[n,n+1]}$.

They are sequences of integrable functions $(f_n)_n$
and $(g_n)_n$ on $\mathbb{R}$
converging pointwise to zero and satisfying:

(i) $\lim_{n \to \infty} \int_{\mathbb{R}} f_n (x) dx = + \infty$

(ii) $g_n \leq 0$, $(g_n)$ converging to zero (from below, but not monotonically) and
$\lim_{n \to \infty} \int_{\mathbb{R}} g_n (x) dx = -1. $ In fact $ \int_{\mathbb{R}} g_n (x) dx = -1$, for all $n$.

Remark 1: Note that $g_n$ is not dominated by $0$. To be dominated by $0$ means $|g_n|\leqslant 0$, which is not the case.

Remark 2:If you require the $g_n$ converges to zero (from below) monotonically, then (ii) would be impossible.  For all $n$, we would have $|g_n|\leqslant |g_1|$ and since $g_1$ is integrable, so is $|g_1|$ and then by the Dominated Convergence Theorem, we would have:
$\lim_{n \to \infty} \int_{\mathbb{R}} g_n (x) dx = 0$.

Title: Do there exist examples of integrable functions with these properties?

Body:
Question:

Do there exist examples of sequences of integrable functions $(f_n)_n$
and $(g_n)_n$ on $\mathbb{R}$ (with it standard measure structure)
converging pointwise to zero and satisfying:

(i) $\lim_{n \to \infty} \int_{\mathbb{R}} f_n (x) dx = + \infty$

(ii) $g_n \leq 0$, $(g_n)$ converging to zero (from below) and
$\lim_{n \to \infty} \int_{\mathbb{R}} g_n (x) dx = -1. $

If so, provide an example; if not, explain why not.

Attempt: For (i), I think yes. I consider the standard measure space $(\mathbb{R}, \beta, \lambda)$ where $\lambda$ is the Lebesgue measure on the Borel $\sigma$-algebra $\beta$. I let $f_n(x) = \frac{1}{n} \chi_{[0, n^2]} (x)$ where $\chi$ is the indicator function. Then $|f_n(x)| \leq 1/n$ for all $x \in \mathbb{R}$, and so $f_n \to 0$ pointwise (even uniformly), but $$ \lim_{n} \int_{\mathbb{R}} f_n(x) dx = \lim_n \frac{1}{n} n^2 = + \infty. $$

For (ii), I think this is impossible because it contradicts the Dominated Convergence Theorem. In class, we saw this theorem as:

Theorem:

Let $(\Omega, \mathfrak{M}, \mu)$ be a measure space and let $(f_n)_n$
be a sequence in $\mathcal{L}^1 (\Omega, \mathfrak{M}, \mu)$
converging pointwise to a function $f: \Omega \rightarrow C$. Suppose
there exists a positive integrable function $g: \Omega \rightarrow [0,
 + \infty]$ such that $|f_n| \leq g$ for all $n \in \mathbb{N}$. Then $f \in \mathcal{L}^1 (\Omega, \mathfrak{M}, \mu)$ and $$ \lim_n
 \int_{\Omega} | f_n - f| d \mu = 0$$ and $\lim_n \int_{\Omega} f_n d
 \mu = \int_{\Omega} f d\mu. $

So now I was thinking that (ii) would contradict this theorem. If $g_n \leq 0$, then $(g_n)_n$ is dominated by the zero function. So we should have $\lim_n \int_{\mathbb{R}} g_n dx = 0. $ Is this reasoning correct?

Thanks for any help.


问:
Existence of an injection between $\mathbb{Z}$ and $[0,1]$
Is the statement that there is an injection between $\mathbb{Z}$ and $[0,1]$ true or false

$\mathbb{Z}\xrightarrow{(1-1)} [0,1]$

i've build it on the definition of cardinality that is if $\lvert A \lvert \leq \lvert B \lvert$

then there exists an injection $f:A\longrightarrow B$ and one knows that

$\lvert \mathbb{Z} \lvert \leq \lvert [0,1]\lvert$

thus there exists an injection

答:
I see many answers with piecewise functions for positive and negative parts, but you can simply send $\mathbb Z^{-}\hookrightarrow[0,\frac 12]$ and $\mathbb Z^+\hookrightarrow[\frac 12,1]$ with $$\begin{cases}f(n)=\frac 12+\frac 1{2n}\\f(0)=\frac 12\end{cases}$$

The only residual issue is zero, which you still need to set separately though.

Another idea which is common when building injections to send integers to powers of primes, because it ensures injectivity.

The idea is $\mathbb Z\hookrightarrow\mathbb N^2\hookrightarrow [0,1]$
$$\begin{cases}
f(n)=\frac 1{2^n} & n\ge 0\\
f(n)=\frac 1{3^{|n|}}& n\le 0\end{cases}$$

Yet it is also possible to proceed like this $\mathbb Z\hookrightarrow\mathbb R\hookrightarrow[-1,1]\hookrightarrow[0,1]$ and use known bijections from $\mathbb R\mapsto(-1,1)$ like:

$$\dfrac{x}{|x|+1} \text{ or }\tanh(x)\text{ followed by } x\mapsto \frac{x+1}2$$.


问:
A trigonometric limit $ \lim _{x \rightarrow 0^{+}} \frac{\tan 3 x \sqrt{1-\cos 4 x}}{x^{2}+x^{3}} $
I am stuck on evaluating the following limit:
$$
\lim _{x \rightarrow 0^{+}} \frac{\tan 3 x \sqrt{1-\cos 4 x}}{x^{2}+x^{3}}
$$
All I can think of is simplifying it in the form of $\sqrt{2}\lim _{x \rightarrow 0^{+}} \frac{\, \tan3x\, \sin2x}{x^3+x^2}$, but unsure where to go from there. I think one could get to an answer by applying the L'Hôpital's rule twice, but I was wondering whether there is a better approach that I'm missing.

答:
Just do some basic asymptotic analysis:

Near $0,\:$ a polynomial is asymptotically equivalent to its lowest degree term.

$1-\cos u\sim_0 \dfrac{u^2}2$, $\;\tan u\sim_0u$.

Therefore
$$\frac{\tan 3 x \sqrt{1-\cos 4 x}}{x^{2}+x^{3}}\sim_0\frac{3x\sqrt{\frac{16x^2}2}}{x^2}=
\frac{6\sqrt 2x|x|}{|x|^2}=6\sqrt 2\frac x{|x|}=6\sqrt 2\operatorname{sgn}(x).$$
This yields also the limit when $x\to 0^-$.


问:
Describe the level surfaces of the following function: $f(x, y, z) = x^2 − y^2 + z^2$
Question: Describe the level surfaces of the following function: $f(x, y, z) = x^2 − y^2 + z^2$

I'm currently trying find the surface in which the function represents but I can't seem to figure out. I'm leaning into thinking it's a elliptic cone with its format as there's two positives. Anyone have a clue?

答:
There are three level surfaces. The other ones are very similar. If you need a level surface for a specific value, feel free to ask.

The first one is the equation of a cone with symmetry axis the $y$-axis

The second and the third ones are a hyperbolic hyperboloid
$$\Large x^2-y^2+z^2=0$$

$$\Large x^2-y^2+z^2=1$$

$$\Large x^2-y^2+z^2=-1$$


问:
Evaluate integral $\int \frac{(2-x^2)e^x}{(1-x)(\sqrt{1-x^2})}dx $
I tried to solve it by the following method:

\begin{align*}
    \int \frac{(2-x^2)e^x}{(1-x)(\sqrt{1-x^2})}dx&= \int \frac{(1-x^2)e^x+\frac{e^x}{2}+\frac{e^x}{2}}{(1-x)(\sqrt{1-x^2}) }dx \\ &= \int \frac{(1-x^2)e^x+\frac{(1-x)e^x}{2}+\frac{(1+x)e^x}{2}}{(1-x)(\sqrt{1-x^2}) }dx \\ &=\int \frac{\sqrt{(1-x^2)}e^x+\frac{\sqrt{(1-x)}e^x}{2\sqrt{x+1}}+\frac{\sqrt{(1+x)}e^x}{2\sqrt{1-x}}}{(1-x) }dx \\&=
    \int \frac{\sqrt{1-x}(\sqrt{x+1}\cdot e^x+\frac{e^x}{2\sqrt{x+1}})+\sqrt{x+1}\cdot e^x \cdot \frac{1}{2\sqrt{1-x}}}{1-x}dx \\&= \int\left(\frac{\sqrt{x+1}\cdot e^x}{\sqrt{1-x}}\right)'dx \\ &=\left(\frac{\sqrt{x+1}\cdot e^x}{\sqrt{1-x}}\right)+C
\end{align*}
Is this solution right?
if it is right are there other methods to solve it.

Body:
Recognize

$$\frac{2-x^2}{(1-x)(\sqrt{1-x^2})}= \frac{(1-x^2)+1}{(1-x)(\sqrt{1-x^2})}\\=\sqrt{\frac{1+x}{1-x} }+ \frac{1}{(1-x)^{3/2}\sqrt{1+x})}
= \sqrt{\frac{1+x}{1-x} }+ \left( \sqrt{\frac{1+x}{1-x} }\right)’
$$

and then apply to $f(x)= \sqrt{\frac{1+x}{1-x} }$ the general integral result
$$\int e^x(f(x)+f’(x)) dx= e^x f(x)$$
to obtain
$$\int \frac{(2-x^2)e^x}{(1-x)(\sqrt{1-x^2})}dx
=\sqrt{\frac{1+x}{1-x} }\>e^x +C
$$


Title: A trigonometric limit $ \lim _{x \rightarrow 0^{+}} \frac{\tan 3 x \sqrt{1-\cos 4 x}}{x^{2}+x^{3}} $

Body:
I am stuck on evaluating the following limit:
$$
\lim _{x \rightarrow 0^{+}} \frac{\tan 3 x \sqrt{1-\cos 4 x}}{x^{2}+x^{3}}
$$
All I can think of is simplifying it in the form of $\sqrt{2}\lim _{x \rightarrow 0^{+}} \frac{\, \tan3x\, \sin2x}{x^3+x^2}$, but unsure where to go from there. I think one could get to an answer by applying the L'Hôpital's rule twice, but I was wondering whether there is a better approach that I'm missing.

Body:
Note that $\cos(4x)=1-2\sin^2(2x)$, so we have

$$
\lim _{x \rightarrow 0^{+}} \frac{\tan 3 x \sqrt{1-\cos 4 x}}{x^{2}+x^{3}}=\lim _{x \rightarrow 0^{+}}\frac{6\sqrt 2\sin(2x)\sin(3x)}{\cos(3x)(2x)(3x)(1+x)}=6\sqrt{2}
$$


问:
$\sqrt{a}+\sqrt{b}=\sqrt{2+\sqrt{3}}$

Find all positive rational numbers $(a,b)$ so that:

$$\sqrt{a}+\sqrt{b}=\sqrt{2+\sqrt{3}}$$

To solve this problem the first thing comes to my mind is squaring both sides:

$$a+b+2\sqrt{ab}=2+\sqrt{3}$$
I equated $a+b=2$ and $ab=\frac34$ :
$$a,b=\dfrac{2\pm\sqrt{4-3}}{2}$$

$(a,b)\in\{(\frac32,\frac12),(\frac12,\frac32)\}$

Is it true?

答:
Hint:
$$\sqrt{2+\sqrt{3}} = \frac{1+\sqrt{3}}{\sqrt2}$$

Square and multiply for $\sqrt2$ from both sides and you get
$$2(a+b)+4\sqrt{ab} = 4 + 2\sqrt3$$

Here you get a system:
\begin{cases}
a+b = 2 \\
4ab=3
\end{cases}

Title: $\sqrt{a}+\sqrt{b}=\sqrt{2+\sqrt{3}}$


问:
$\mathbb{C}^{2l}$ as a representation of $\mathfrak{sp}(2l)$
I read somewhere that $\mathbb{C}^{2l}$ can be considered as a representation of $\mathfrak{sp}(2l)$, but I don’t understand how the action could be defined. Is it an irreducible representation? If so, what would its highest weight be? I thought of computing the eigenvalues of elements $h$ in the Cartan subalgebra $\mathfrak{h} = span\left\{E_{ii} - E_{l+i, l+i} \right\}_{i=1..l}$, acting on the standard basis of $\mathbb{C}^{2l}$. Does this make any sense?

答:
Taking the standard basis of $\Bbb C^{2l}$ lets us regard elements of $\mathfrak{sp}(2l) := \mathfrak{sp}(\Bbb C^{2l})$ as $(2l) \times (2l)$ matrices in the usual way.

With this matrix realization in hand, the irreducible $(2l)$-dimensional representation of $\mathfrak{sp}(2l)$ is just the set $\Bbb C^{2l}$ of column vectors, where $X \in \mathfrak{sp}(2l)$ acts by multiplication $\mathfrak{sp}(2l) \times \Bbb C^{2l} \to \Bbb C^{2l}$ on the left. This is the standard or defining representation of $\mathfrak{sp}(2l)$.

Remark This representation is isomorphic to its dual via the canonical identification $\Bbb C^{2l} \leftrightarrow (\Bbb C^{2l})^*$, $x \leftrightarrow \omega(x, \,\cdot\,)$, where $\omega : \Bbb C^{2l} \times \Bbb C^{2l} \to \Bbb C$ is the symplectic form on $\Bbb C^{2l}$.

Picking a particular $\omega$ lets us talk about weights more concretely. A typical convention is
$$\omega(x, y) := {}^\top \!x J y, \qquad J := \pmatrix{\cdot&I_l\\-I_l&\cdot} .$$

Like you say, a natural choice for a Cartan subalgebra $\mathfrak h$ with this convention is $$\mathfrak{h} := (H_i), \qquad H_i := E_{ii} - E_{l + i, l + i} .$$

If we denote by $(L_i)$ the basis of $\mathfrak{h}^*$ dual to the basis $(H_i)$ of $\mathfrak{h}$ and by $(e_i, e_{l+i})$ the standard basis of $\Bbb C^{2l}$, we can verify directly that each $e_i$ is an eigenvector of the $h$-action of eigenvalue $L_i$ and each $e_{l+i}$ is an eigenvector of eigenvalue $-L_i$. Explicitly, $H_j \cdot e_i = L_i(H_j) e_i$ and similarly for the elements $e_{l+i}$.

In the usual convention the highest weight is $L_1$, and when we want to emphasize the role of roots or weights we sometimes denote this representation by $\Gamma_{1,0,0}$.

The weights, all of which have multiplicity $1$, form a single orbit of the action of the Weyl group, so the general theory of representations of simple Lie algebras implies that representation is irreducible.

For much more, see Fulton & Harris' Representation Theory: A First Course; $\S\S$16–17 treat the representation theory of the Lie algebras $\mathfrak{sp}(2l)$.


答:
$$
\lim _{x \rightarrow 0^{+}} \frac{\tan 3 x \sqrt{1-\cos 4 x}}{x^{2}+x^{3}}=\lim _{x \rightarrow 0^{+}} \frac{\tan 3 x \sqrt2 \sin(2x)}{x^{2}+x^{3}}=\lim _{x \rightarrow 0^{+}} \frac{\frac{\tan 3 x}{3x} \sqrt2 \frac{\sin(2x)}{2x}\times6}{1+x}\\= 6\sqrt2
$$

I have used the following:

i)$|\sin(2x)|=\sin(2x)$ since $x\to0^+$

ii) $\lim_{t\to0} \frac{\sin t}{t}=\frac{\tan t}{t}=1$


问:
What's the generalisation of right-continuous to non-handed spaces?
A function $f$ is right continuous at a point $c$ if it is defined on an interval $[c, d]$ lying to the right of $c$ and if $\lim_{x→c^+}f(x) =f(c)$.

Moving deliberately into less precise language, I interpret this to mean the function is continuous when approached from the right and it seems reasonable that in the more general setting of topological spaces outside of $\Bbb R$, when approached from the right can be replaced with when approached through some subspace $X$.  Then we might reasonably say the function is $X$-continuous.

Question: Assuming all this is reasonable, what are the proper terminology and definitions for this type of continuity?

As a background only, I have a specific interest in the fact that the function $f(x)=3x+2^{\nu_2(x)}$, which is continuous in the 2-adic metric space, can only approach the number $0$ through the subspace $\{2^i:i\in\Bbb Z\}$, if and only if the Collatz conjecture is true.  Of course, I appreciate that since it is continuous, $f$ is $\{2^i:i\in\Bbb Z\}$-continuous is a vacuous statement but I want to find where to learn about and how to talk about $X$-continuity nevertheless.

答:
I don't think there is an established terminology, but you could just define:

Let $X,Y$ be topological spaces, $S\subseteq X$ and $f:X\to Y$. We call $f$ "continuous at $s\in S$ when approached through $S$" if the restricted function $f\vert_S:S\to Y$ is continuous at $s$ with respect to the subspace topology on $S$.

As Mark S. suggested, "the restriction to $S$ is continuous" is probably clear enough that introducing such nomenclature isn't really needed, though.


问:
derive eulers integration method from taylor series
I'm having a hard time understanding how to derive eulers integration method from tayler series expansion

I have an ODE and an initial value problem

$$
\dfrac{dy}{dt}=f(y,t)
$$

$$
y(0)=y_{0}
$$

How does this taylor series expansion relate to eulers integration method

$$
y(t+h)=y(t)+hy'(t)+\frac{h^2}{2!}y''(t)+\frac{h^3}{3!}y'''....
$$

What exactly is h in this formula? is it that if i know y(0) i can find y(t+h)?

i was also given that if h is small i can write

$$
y(t+h)=y(t)+hy'(t)+\frac{h^2}{2!}y''(t)+\frac{h^3}{3!}y'''+O(h^4)
$$
Why is that? - i do know that it relates to the truncation error

And how do i come from this
$$
y(t+h)=y(t)+hy'(t)+\frac{h^2}{2!}y''(t)+\frac{h^3}{3!}y'''+O(h^4)
$$

to this

$$
y_{k+1}=y_{k}+h\cdot f(y_{k},t_{k})
$$

答:
You did almost everything! As you said,
$$
y(t+h) \approx y(t) + h y'(t),
$$

but, since $y(t)$ is the solution to $y'(t)=f(y,t)$, you can replace $y'(t)$ by $f(y(t),t)$ and obtain
$$
y(t+h) \approx y(t) + hf(y(t),t).
$$

If you prefer to use just integration, you can also start by obtaining an equivalent integral equation:
\begin{align*}
y'(t) = f(y(t),t) & \Rightarrow \int_{t_k}^{t_{k+1}} y'(t) dt = \int_{t_k}^{t_{k+1}}f(y(t),t) dt\\
&  \Leftrightarrow y(t_{k+1}) = y(t_k) + \int_{t_k}^{t_{k+1}}f(y(t),t) dt
\end{align*}

Now, if you use the rectangle rule,

$$
y(t_{k+1}) = y(t_k) + \int_{t_k}^{t_{k+1}}f(y(t),t) dt \approx y(t_k) + (t_{k+1}-t_k) f(y(t_k),t_k)
$$

you also obtain Euler's method.


问:
Arrangements of affine hyperplanes
Fix $n>0$ and $X\subseteq\mathbb{R}^n$. Call a function $f:X\longrightarrow \mathbb{R}$ linear if it is of the form
$$
f(\bar{x})=a_1x_1+\ldots+a_nx_n+b
$$
for some $a_i,b\in\mathbb{R}$.

Now suppose we have linear functions $f_1,\ldots,f_t$ and $g_1,\ldots,g_t$ with the following property:

For all $i\in\{1,\ldots,t\}$ and for all $\bar{x}\in X$,

$$
\textrm{there are }j,k\in\{1,\ldots,t\}\textrm{ such that }f_i(\bar{x})\leq g_j(\bar{x})\textrm{ and }f_k(\bar{x})\leq g_i(\bar{x}).
$$

Is it true that there must be some $i,j\in\{1,\ldots,t\}$ such that $f_i(\bar{x})\leq g_j(\bar{x})$ for all $\bar{x}\in X$?

答:
Let's treat the case $n=1$.

Assume that it is not true that there exist $i,j$ such that $f_i \leq g_j$ on all of $X$.
Then for every pair $i,j$ we can choose $x_{ij} \in X$ such that $f_i(x_{ij}) > g_j(x_{ij})$. Let $a=\min_{i,j}(x_{ij})$ and $b=\max_{i,j}(x_{ij})$.

Upon renumbering the $g_i$ if necessary, we can assume that $g_1(a)$ is the smallest value one of the $g_i$ attains at $a$. By assumption there exists $k$ such that $f_k(a) \leq g_1(a)$, and our assumption on $[a,b]$ above assures that $f_k(b) > g_1(b)$ (as $f_k$ and $g_1$ intersect in $[a,b]$, and lines can intersect at most once). Hence there must exist $j$ such that $f_k(b) \leq g_j(b)$. Again by the construction of $[a,b]$ there exists $\xi \in [a,b)$ such that $f_k(\xi) > g_j(\xi)$. In particular we must have $g_j(a) < f_k(a) \leq g_1(a)$, which contradicts minimality of $g_1(a)$.

Title: Arrangements of affine hyperplanes


答:
This is almost the canonical equation of a hyperboloid, so this is what the surface will be. Some intuition on the three types of hyperboloid is provided below. The TL; DR is that you can rotate a hyperbola either about its major axis or about its conjugate axis (the axis between the two sheets), giving rise to two different kinds of hyperboloid.

The level curves verify
$$x^2-y^2+z^2=A$$
for some $A$, or equivalently
$$y^2-r^2=-A$$
where $r^2=x^2+z^2$. This means that $y$ depends on $x$ and $z$ only through $r$, which is the distance from the point $(x, y, z)$ to the $Y$ axis. Therefore, the surface is rotationally symmetric about the $Y$ axis.

There are three cases depending on the sign of $A$. Nice plots of the three possibilities have been provided by Raffaele in their answer.

If $A$ is positive, let $A=a^2$, and we have
$$\frac{r^2}{a^2}-\frac{y^2}{a^2}=1$$
which is the equation of a hyperbola with asymptotes $r=\pm y$ and vertices at $(r=\pm a, y=0)$. The surface is this hyperbola rotated about the $Y$ axis.

If $A$ is $0$, then we have the degenerate case $y^2= r^2$, or $y=\pm r$, which is a cone along the $Y$ axis.

If $A$ is negative, let $A=-a^2$, and we have
$$\frac{y^2}{a^2}-\frac{r^2}{a^2}=1$$
which is the equation of a hyperbola with asymptotes $r=\pm y$ and vertices at $(r=0, y=\pm a)$. This is the same hyperbola as in the positive $A$ case, but with the $Y$ and $R$ axes swapped. The surface is this hyperbola rotated about the $Y$ axis.

Body:
Question: Describe the level surfaces of the following function: $f(x, y, z) = x^2 − y^2 + z^2$

I'm currently trying find the surface in which the function represents but I can't seem to figure out. I'm leaning into thinking it's a elliptic cone with its format as there's two positives. Anyone have a clue?


答:
You may just apply the denesting formula

$$\sqrt{a+\sqrt{c}}=\sqrt{\frac{a-\sqrt{a^2-c}}2}+ \sqrt{\frac{a+\sqrt{a^2-c}}2}
$$
to obtain$$\sqrt{2+\sqrt3}=\sqrt{\frac12}+\sqrt{\frac32}
$$

Title: $\sqrt{a}+\sqrt{b}=\sqrt{2+\sqrt{3}}$

Body:
Find all positive rational numbers $(a,b)$ so that:

$$\sqrt{a}+\sqrt{b}=\sqrt{2+\sqrt{3}}$$

To solve this problem the first thing comes to my mind is squaring both sides:

$$a+b+2\sqrt{ab}=2+\sqrt{3}$$
I equated $a+b=2$ and $ab=\frac34$ :
$$a,b=\dfrac{2\pm\sqrt{4-3}}{2}$$

$(a,b)\in\{(\frac32,\frac12),(\frac12,\frac32)\}$

Is it true?


答:
Yes it is.
On squaring, you get a+b+2√ab =2+√3

Thus a+b=25

And ab=0.75

Solving the quadratic a(2-a)=0.75, you get the solutions (a,b)=(0.5,1.5) and (1.5,0.5).

Verification:

(√0.5+√1.5)^2

=0.5+1.5 2√(0.5*1.5)

=2+2√0.75

=2+√3

Hence (√0.5+√1.5)=√(2+√3)

Title: $\sqrt{a}+\sqrt{b}=\sqrt{2+\sqrt{3}}$


问:
Minor issue with partial derivatives
Please see the image below for my question concerning partial derivates of functions in polar co-ordinates:

答:
rewrite $r = (x^2+y^2+z^2)^\frac{1}{2}$

now $\frac{\partial r}{\partial x}=\frac{\frac{1}{2}\cdot 2x}{(x^2+y^2+z^2)^\frac{1}{2}} = \frac{x}{r}$ as required

Your $\frac{\partial x}{\partial r}=\frac{x}{r}$ is correct but when you invert it you actually have $r$ as a function of $x,\theta$ and $\phi$. Try to keep strictly to one coordinate system at a time

Title: Minor issue with partial derivatives


问:
Group extension of $\mathbb Z_4$ by $\mathbb Z_2$
Let $f : G →\mathbb Z_2$ be an extension of $\mathbb Z_4$ by $\mathbb Z_2$. Suppose that the induced action $α_f :\mathbb Z_2 →\mathbb Z^{\times}_4$ carries the generator of $\mathbb Z_2$ to $−1$. Then $G$ is isomorphic to either $D_8$ or $Q_8$.

Proof

Let $\mathbb Z_4 = \langle b \rangle$ and $\mathbb Z_2 = \langle a \rangle$. For any $x ∈ G$ with $f(x) = a$, we have $xb^kx^{−1} = α_f (a)(b^k) = b^{−k}$ for all $k$. Also, $f(x^2) = a^2 = e$, so $x^2$ ∈ ker f = b. Thus, $x^2$ = $b^k$ for some k.

Note that x fails to commute with b and $b^{−1}$, so neither of these can equal $x^2$. Thus, either $x^2$ = e or $x^2$ = $b^2$.

If $x^2$ = e, then x has order 2, and hence there is a section of f, specified by s(a)= x. In this case, $G \cong\mathbb  Z_4 \rtimes_{α_f}\mathbb  Z_2$, which, for this value of $α_f$ , is isomorphic to the dihedral group $D_8$.

I'm a bit confused about how they got from $x^2=e$ to $G \cong Z_4 \rtimes_{\alpha_f} Z_2$.

If s(a) = x, then the section $s: Z_2 \rightarrow G$ sends 1 to e and -1 to x, right? So we must have an element of order 2 in G. Additionally, we know that we have a normal subgroup of order 4. However, if we want to construct a semidirect product, we need to make sure that $Z_2$ and $Z_4$ do not have any elements in common, except for the identity, right? But what if $x=b^2$?

The remaining case gives $x^2 = b^2$. But then there is a homomorphism $h : Q_8 → G$
with $h(a) = x$ and $h(b) = b$.

In my notes, it says that any group G is isomorphic to $Q_{2n}$ if the following conditions hold:

$\bullet$ $x^4=e$

$\bullet$ $b^{2n}=e$

$\bullet$ $x^{-1}bx=b^{-1}$

$\bullet$ $x^2 = b^n$

Of course, in this case, all of the first three conditions hold. But what about the 4th one? We know that $x^2 = b^2 \not= b^4$. Then how does that homomorphism exist?

As the image of h contains more than half the elements of G, h must be onto, and hence an isomorphism. (Alternatively, we could use the Five Lemma for Extensions to show that h is an isomorphism.)

I'm not sure if I understand this. The only thing we know about G is that it contains a normal subgroup of order 4, right? So what if we map the group of order 4 in $Q_8$ to the group of order 4 in G? Then we won't ahve mroe than half the elements of G in the image, right?

Also, for the Five Lemma for Extensions... in order to use that theorem, we need an extension for $Q_8$, right? But we're not given any. In other words, we have $\langle b \rangle \rightarrow G \rightarrow \mathbb Z_2$. But For $Q_8$, we have $\langle b' \rangle \rightarrow Q_8 \rightarrow ?$.

$Q_8$, of course, is not a split extension of b by $Z_2$, for two reasons. First, we know that every element of $Q_8$ which is not in b has order 4, and hence it is impossible to define a section from $Q_8/\langle b \rangle$ to Q8. Second, if it were a split extension, then it would be isomorphic to $D_8$, which we know is not so.

Why is it impossible to define a section? Is it not possible to say $s(a)=b^2$. I guess this is related to my previous question.

We have classified the extensions of Z4 by Z2 corresponding to the homomorphism
from $Z_2$ to $Z^{\times}_4$ that takes the generator of $Z_2$ to −1. Since $Z^{\times}_4$ is isomorphic to $Z_2$, there is only one other homomorphism from $Z_2$ to $Z^{\times}_4$ : the trivial homomorphism.

So if we look at all the possibilities of where the generator is being sent, then that would be like classifying all groups of order 8, right?

Thank you in advance

答:
But what if $x=b^2$?

Then $b^x = b^{b^2} = b \neq b^{-1}$. In other words, $x$ and $b$ do not commute, but $b^2$ and $b$ do commute. Equality of $x=b^2$ implies $x$ commutes with $b$, despite our assumption that $x^{-1} b x = b^{-1}$.

In my notes, it says that any group G is isomorphic to $Q_{2n}$ if the following conditions hold:

This should say $Q_{4n}$ not $Q_{2n}$.

I'm not sure if I understand this.

The image of $h$ contains $x$ and $\langle b \rangle$, 5 elements out of 8, so it must contain all 8 by Lagrange's theorem. $x \notin \langle b \rangle$ because $x$ and $b$ do not commute.

…in order to use that theorem, we need an extension for $Q_8$, right?

Yes, we take $1 \to \langle i \rangle \to Q_8 \to C_2 \to 1$ where the surjection takes $\pm j, \pm k$ to the non-identity element of $C_2$ and $\pm1, \pm i$ to the identity of $C_2$. The five lemma probably also requires some homomorphisms from one exact sequence to the other. Map $i\mapsto b$ and $j\mapsto x$. This is basically the same as $h$, but I'm not positive of your notation for $Q_8$.

Why is it impossible to define a section?

$Q_8$ has only one element of order $2$, and that element is central, but the image of $a$ is not central.

So if we look at all the possibilities of where the generator is being sent, then that would be like classifying all groups of order 8, right?

More or less, yes. A group of order 8 contains a group of order 4 and index 2. If that subgroup is cyclic, then the quoted argument classifies all non-abelian possibilities ($D_8$ and $Q_8$). It indicates one could also handle the abelian possibilities ($C_4 \times C_2$ and $C_8$). However, one also has to deal with the possibility that there is no cyclic subgroup of order 4 (then the group has exponent 2 and is a vector space over $\mathbb{Z}/2\mathbb{Z}$ so we get $E_8 = C_2 \times C_2 \times C_2$).  Be careful that if one tries to classify the extensions of $C_2 \times C_2$ by $C_2$ you'll get some repeats, $D_8$ and $C_4 \times C_2$, as well as the desired $E_8$.


问:
Asymptotic estimate of integral $\int_{0}^{r}\frac{dx}{\sum_{k=0}^{l}x^k}$
Let $l\in \mathbb{N}^{+}$ and consider the integral
$$ I(r)=\int_{0}^{r}\frac{dx}{\sum_{k=0}^{l}x^k} $$
I want to know how to estimate the behavior of $I(r)$ for $r\to +\infty$.  By AM-GM inequality, I get
$$ I(r)\approx \int_{0}^{r}\frac{dx}{1+x^l},  $$
then I stuck here. Can someone help me? Thank you very much!

答:
Too long for a comment.

One of my Ph.D. students worked this problem more then $30$ years ago and, playing for sure with the roots of unity, arrived for
$$I_n=\int_{0}^{\infty}\frac{dx}{\sum_{k=0}^{n}x^k}$$ to the simple
$$I_n=\cos \left(\frac{\pi }{n+1}\right)\, \Gamma \left(\frac{n}{n+1}\right)\, \Gamma\left(\frac{n+2}{n+1}\right)-$$
$$\frac{1}{2} \,\cos \left(\frac{2 \pi }{n+1}\right)\,
   \Gamma \left(\frac{n-1}{n+1}\right)\, \Gamma \left(\frac{n+3}{n+1}\right)$$
I have not been able to find the intermediate steps.

From here, the expansions are very simple.

$$I_n=\frac{1}{2}+\frac{\pi ^2}{3 n^2}-\frac{2 \pi ^2}{3 n^3}+\frac{\pi ^2 \left(45+7 \pi ^2\right)}{45 n^4}-\frac{4 \pi ^2 \left(15+7 \pi ^2\right)}{45
   n^5}+O\left(\frac{1}{n^6}\right)$$ Using the above, the relative error is $<1$% for $n>5$,   $<0.1$% for $n>8$,   $<0.01$% for $n>12$.


答: For $n>2$ (the cases $n=1$ and $n=2$ are easy), let $$I_n(r)=\int_0^r\frac{dx}{1+x+\ldots+x^{n-1}}=\int_0^r\frac{1-x}{1-x^n}\,dx.$$ With the help of $f(a):=\int_0^1\frac{y^{a-1}-y^{-a}}{1-y}\,dy=\pi\cot a\pi$ for $0<a<1$ (see e.g. this), $$I_n(+\infty)=\int_0^1\frac{1-x}{1-x^n}\,dx+\int_0^1\frac{1-x^{-1}}{1-x^{-n}}\frac{dx}{x^2}=\frac1n\left[f\Big(\frac1n\Big)-f\Big(\frac2n\Big)\right]=\frac{\pi}{n}\csc\frac{2\pi}{n},$$ and for $r>1$ $$I_n(+\infty)-I_n(r)=\int_r^{+\infty}\sum_{k=1}^\infty(x-1)x^{-nk}\,dx=\sum_{k=1}^\infty\left(\frac{r^{2-nk}}{nk-2}-\frac{r^{1-nk}}{nk-1}\right)$$ (thus, we have an exact expansion in negative powers of $r$, not only an asymptotic one).

Title: Asymptotic estimate of integral $\int_{0}^{r}\frac{dx}{\sum_{k=0}^{l}x^k}$

Body:
Let $l\in \mathbb{N}^{+}$ and consider the integral
$$ I(r)=\int_{0}^{r}\frac{dx}{\sum_{k=0}^{l}x^k} $$
I want to know how to estimate the behavior of $I(r)$ for $r\to +\infty$.  By AM-GM inequality, I get
$$ I(r)\approx \int_{0}^{r}\frac{dx}{1+x^l},  $$
then I stuck here. Can someone help me? Thank you very much!


问:
Showing that a real number is an algebraic integer
For what values of $x,y,z\in\mathbb{Z}$, such that $0\leq x,y,z\leq 2, $ the real number $$\alpha:=\frac{1}{3}\left(x+\sqrt[3]{175} \cdot y+\sqrt[3]{245}\cdot z\right)$$ is an algebraic integer i.e. root of a monic polynomial in $\mathbb{Z}[x]$ ?

I tried to compute the norm but the computation is getting ugly. Also if we can find those $x,y,z $  for which the $\mathbb{Z}$-module generated by $1,\alpha ,\alpha^2,\ldots$ is finitely generated, then those are the required algebraic integers. Can this be done ? Also I would like to know if there are any other simpler approach.

Body: $5 \, \sqrt[3]{245} = ( \sqrt[3]{175})^2$.


答:
While $\frac{\mathrm dx}{\mathrm dy} \cdot \frac{\mathrm dy}{\mathrm dx} = 1$ is true when $x$ is a single-variable function of $y$, this has to fail for partial derivatives of multi-variable functions.

From $x = r \sin \theta \cos \phi$ you have correctly computed that $\frac{\partial x}{\partial r} = \frac{x}{r}$, when $x$ is considered a function of $r,\theta,\phi$. However, the equation $x = r \sin \theta \cos \phi$ can't possibly tell you how $r$ changes with $x$ when $y$ and $z$ are fixed. It only tells you how $r$ changes with $x$ when $\theta$ and $\phi$ are fixed, but that is not what $\frac{\partial r}{\partial x}$ is, when $r$ is considered a function of $x,y,z$.

As a function of $x,y,z$ you have $r = \sqrt{x^2+y^2+z^2}$ which does indeed yield $\frac{\partial r}{\partial x} = \frac{x}{r}$.

Title: Minor issue with partial derivatives

Body: Please see the image below for my question concerning partial derivates of functions in polar co-ordinates:


问:
What is the addition in $U\otimes V$ if $U$ and $V$ are vector space?
Let $U$ a $\mathbb R-$vector space of dimension $m$ and $V$ a $\mathbb R-$vector space of dimension $n$. We define $u\otimes v:=uv^T$ where $u\in U$ and $v\in V$. I have to prove that $$U\otimes V:=\{u\otimes v\mid u\in U,v\in V\}$$
is a vector space of dimension $nm$. My problem is that I even don't know how is defined $u\otimes v+ \tilde u\otimes \tilde v$ where $u,\tilde u\in U$ and $v,\tilde v\in V$. Any idea at this point ? I saw the page on wikipedia, but their construction of $U\otimes V$ is not so intuitive. Their definition is the following one, but I don't see how I can deduce the some of two element in $U\otimes V$.

答:
The tensor product is by definition bilinear, that is, it follows the rules that $u_1\otimes v+u_2\otimes v=(u_1+u_2)\otimes v$ and $u\otimes v_1+u\otimes v_2=u\otimes(v_1+v_2)$, as well as $\lambda(u\otimes v)=(\lambda u)\otimes v=u\otimes(\lambda v)$.

To properly use this to calculate the sum of arbitrary tensor products, you need to choose bases of $U$ and $V$ and decompose $u,\tilde u,v,\tilde v$. For an easy example, let $U,V$ be two-dimensional with bases $\{b_1,b_2\}\subset U,~\{c_1,c_2\}\subset V$. Then we have

$$\begin{align}
u\otimes v+\tilde u\otimes\tilde v&=(u_1b_1+u_2b_2)\otimes(v_1c_1+v_2c_2)+(\tilde u_1b_1+\tilde u_2b_2)\otimes(\tilde v_1c_1+\tilde v_2c_2)\\
&=u_1v_1(b_1\otimes c_1)+u_1v_2(b_1\otimes c_2)+u_2v_1(b_2\otimes c_1)+u_2v_2(b_2\otimes c_2)~+\\
&\color{white}{=}\tilde u_1\tilde v_1(b_1\otimes c_1)+\tilde u_1\tilde v_2(b_1\otimes c_2)+\tilde u_2\tilde v_1(b_2\otimes c_1)+\tilde u_2\tilde v_2(b_2\otimes c_2)\\
&=(u_1v_1+\tilde u_1\tilde v_1)(b_1\otimes c_1)+(u_1v_2+\tilde u_1\tilde v_2)(b_1\otimes c_2)\\
&\color{white}{=}+(u_2v_1+\tilde u_2\tilde v_1)(b_2\otimes c_1)+(u_2v_2+\tilde u_2\tilde v_2)(b_2\otimes c_2)
\end{align}$$

For higher dimensions, it is imperative that you do this using some abbreviating notation. It will come out like this:

$$\left(\sum_{i\in I}x_ib_i\right)\otimes\left(\sum_{j\in J}y_j c_j\right)+\left(\sum_{i\in I}z_ib_i\right)\otimes\left(\sum_{j\in J}w_jc_j\right)=\sum_{i\in I\\j\in J}(x_iy_j+z_iw_j)(b_i\otimes c_j).$$

Note that the sums are always finite even if $I,J$ are not, since $(b_i)_I,(c_j)_J$ are bases of $U$ and $V$, so this seemingly infinite sum is still well-defined.


问:
On self-duality of non-Archimedean local fields
Let $K$ be a non-Archimedean local field. Its additive group $K^+$ is a locally compact Hausdorff abelian group. My question is the following:

Is $K^+$ isomorphic to its Pontryagin dual $\widehat{K^+}$ as a topological group?

Remarks:

1)
Fix a non-trivial unitary character $\psi:K^+ \rightarrow \mathbb{C}^{\times}$. For each $a \in K$, the map $a\psi: K^+ \rightarrow \mathbb{C}^{\times}, \; x \mapsto \psi(ax)$ is a unitary character of $K^+$, and the map $a \mapsto a\psi$ gives an abstract group isomorphism from $K^+$ onto $\widehat{K^+}$ (see e.g. [1, sect. 1.17 Prop.]). Unfortunately, the argument given in [1, sect. 1.17 Prop.] does not prove (as far as I can see) that the previous map is an homeomorphism.

2) Let $p \in \mathbb{N}$ be the characteristic of the residue field of $K$. It is known that as a topological field, $K$ is isomorphic either to a finite extension of $\mathbb{Q}_p$ or to some field of formal Laurent series $\mathbb{F}_q((t))$ where $q$ is a power of $p$. Since the additive group of $\mathbb{Q}_p$ is known to be isomorphic to its Pontryagin dual (see e.g. the introductory paragraphs in [2]), the question can be reduced to the case where $K=\mathbb{F}_q((t))$. But I can't find any references or arguments which hint to a positive answer.

Thank you in advance.

References:

[1]  C. J. BUSHNELL AND G. HENNIART, The Local Langlands Conjecture for GL(2), Springer, 2006.

[2] L. CORWIN, Some remarks on self-dual locally compact abelian groups, Trans. Amer. Math. Soc. 148 (1970), 613-622.

答:
A reference is the paper "Locally compact modules" by Martin D. Levin.

In Levin's terminology, the field $\mathbb{F}_q((t))$ can be seen as the completion of $\mathbb{F}_q(t)$ at an infinite valuation with respect to the ring $\mathbb{F}_q[t^{-1}]$. It then follows from §3, Lemma 1.(i) that $\mathbb{F}_q((t))$ is self-dual. Levin states this without proof, for the proof he refers to Weil's "Basic Number Theory", page 41.

I think the following works as a proof:

If $K$ is a non-archimedean local field, then we can find a compact discrete valuation ring $R \subseteq K$, with uniformizing parameter $\pi$, such that $K$ can be written as a topological union $K = \bigcup_{k} \pi^{-k} R$. The Pontryagin dual of $R = \varprojlim R/\pi^kR$ is given by $\varinjlim\; (\pi^{-k}R)/R = K/R$.

Consider the short exact sequence of locally compact $R$-modules $0 \to R \to K \to K/R \to 0$. Applying Pontryagin duality gives a short exact sequence of $R$-modules $0 \to R \to K^\vee \to K/R \to 0$, where $K^\vee$ is the Pontryagin dual of $K$. So we can see $R$ as a compact $R$-submodule of $K^\vee$. Because $K/R$ is discrete, continuity of the projection $K^\vee \to K/R$ shows that $R \subseteq K^\vee$ is a compact open $R$-submodule.

The inclusion $R \subseteq K^\vee$ extends to a $K$-linear map $f: K \to K^\vee$, which is the identity on $R$.  Because $K$ and $K^\vee$ are both 1-dimensional $K$-vector spaces (by the argument you gave), we know that $f$ is a vector space isomorphism. We claim that it is a homeomorphism. For each $\lambda \in K$, $f$ restricts to a homeomorphism from $\lambda + R$ to $f(\lambda)+R$, because it is a bijection with compact domain. Further, $K$ is the disjoint union of the cosets $\lambda + R$, where $\lambda$ ranges over a set of representatives of $K/R$. Because each $f(\lambda)+R$ is open, $f$ is an homeomorphism from $K$ to an open subset of $K^\vee$. But because $f$ is bijective, $f$ must be an homeomorphism.


问:
Exercise 1.7 from Silverman's AEC
I'm working through Silverman's Arithmetic of Elliptic Curves Exercise 1.7 (3 parts) and I have a question about the last part.

(a) In $\mathbb{P}^1$ at least one of $S$ or $T$ is non-zero, so the map is well-defined at all $[S,T] \in \mathbb{P}^1$. Therefore, $\phi$ is a morphism.

(b) You can easily check that $\psi: [X,Y,Z]\mapsto [Y,X]$ works.

(c) The only problematic point on $\psi$ is $[0,0,1]$. I tried to do a bit of the usual manipulation, but was not able to make $\psi$ well-defined on $[0,0,1]$. For instance:
$$ [Y,X] = [YX^2,X^3] = [YX^2,Y^2Z] = [X^2,YZ] $$
This leads me to believe that $\psi$ is not actually a morphism. However, for $\phi$ to be an isomorphism there just has to exist an inverse morphism $\phi^{-1}$ to $\phi$, and this does not necessarily have to be $\psi$ as I have defined it.

I'm wondering how can I (formally) show that $\phi$ is not an isomorphism?

EDIT: I realize that $Y^2Z=X^3$ has a cusp at $[0,0,1]$. So I guess can someone show me how to prove that an isomorphism of projective varieties must preserve singularities?

答:
If $i: X\rightarrow \mathbb{P}^n_k$ is a projective variety over a field $k$ and if $x\in X$ is a point, you may define the local ring $\mathcal{O}_{X,x}$ at $x$. By definition

L1. $\mathcal{O}_{X,x} \cong lim_{x \in U}\mathcal{O}_{X}(U)$,

The local ring in L1 is invariant wrto choice of embedding $i$, it only depends on the structure sheaf $\mathcal{O}_X$ (the structure sheaf is an "intrinsic" object that does not depend on $i$). The maximal ideal $\mathfrak{m}(x) \subseteq \mathcal{O}_{X,x}$ is also independent of choice of embedding. Hence the notion "non singular" does not depend on choice of embedding or isomorphism.

Title: Exercise 1.7 from Silverman's AEC


问:
What is a Rate in this Wikipedia sentence?
"In mathematics, a rate is the ratio between two related quantities in different units. If the denominator of the ratio is expressed as a single unit of one of these quantities, and if it is assumed that this quantity can be changed systematically (i.e., is an independent variable), then the numerator of the ratio expresses the corresponding rate of change in the other (dependent) variable."

Can someone explain this with an example?
Say using S = vt

答:
For $s=vt$, this can be rearranged to
$$
v=\frac{s}{t} \, .
$$
In other words, your velocity is equal to your displacement divided by time. Here, time is an independent variable—you are measuring your velocity against the amount of time that has been passed. This notion of 'independent variable' becomes even more important in calculus, where the above equation becomes
$$
v=\frac{ds}{dt} \, .
$$
Here, $dt$ can (roughly speaking) be interpreted as infinitesimal change in time. Then, the 'ratio'
$$
\frac{ds}{dt}
$$
an infinitesimal change in displacement divided by the infinitesimal change in time.

Beware: the $d$'s in the notation
$$
\frac{ds}{dt}
$$
cannot be cancelled. They just represent infinitesimal change. So
$$
dt = \text{infinitely small change in time.}
$$


问:
Why $f(x)=\pi$ is a rational function? Is a constant function a polynomial even though the constant is a transcendental?
Wikipedia said

A constant function such as $f(x) = π$ is a rational function since
constants are polynomials. The function itself is rational, even
though the value of $f(x)$ is irrational for all x.

But the definition stated:
A function ${\displaystyle f(x)}$ is called a rational function if and only if it can be written in the form
$${\displaystyle f(x) = {\frac {P(x)}{Q(x)}}}$$
where ${\displaystyle P\,}$ and ${\displaystyle Q\,}$ are polynomial functions of ${\displaystyle x\,}$ and ${\displaystyle Q\,}$ is not the zero function. The domain of ${\displaystyle f\,}$ is the set of all values of ${\displaystyle x\,}$ for which the denominator ${\displaystyle Q(x)\,}$ is not zero.

We can rewrite $f(x)=\pi$ as $f(x)=\frac{\pi}{1}$

I could agree if $1$ is a polynomial since it's a non-zero constant function that is a polynomial of degree 0. But $\pi$ is a trancendental number here. I mean, i never saw a polynomial with a constant of $\pi$. Can you explain it to me?

Thanks.

答:
$\pi$ is a polynomial of degree $0$ of $\mathbb R[x]$. Therefore, the constant real map equal to $\pi$ is a rational function. Which doesn't mean that $\pi$ is a rational number.

The difficulty is that we are (abusively) naming two different objects as $\pi$:

The real number.

A real map.


问:
A trigonometric limit $ \lim _{x \rightarrow 0^{+}} \frac{\tan 3 x \sqrt{1-\cos 4 x}}{x^{2}+x^{3}} $
I am stuck on evaluating the following limit:
$$
\lim _{x \rightarrow 0^{+}} \frac{\tan 3 x \sqrt{1-\cos 4 x}}{x^{2}+x^{3}}
$$
All I can think of is simplifying it in the form of $\sqrt{2}\lim _{x \rightarrow 0^{+}} \frac{\, \tan3x\, \sin2x}{x^3+x^2}$, but unsure where to go from there. I think one could get to an answer by applying the L'Hôpital's rule twice, but I was wondering whether there is a better approach that I'm missing.

答:
Multiply by $\frac{\cos(3x)}{\cos(3x)}$ and note that $\lim_{x \to 0} \cos(x) = 1$ to get $$\sqrt{2} \lim_{x \to 0^+}\frac{1}{x+1}\cdot\frac{\sin\left(3x\right)}{x} \cdot \frac{\sin(2x)}{x}$$

Because $\lim_{x \to 0}\frac{\sin(x)}{x} = 1$, this simplifies to $$\sqrt{2} \cdot \frac{1}{1} \cdot 3 \cdot 2 = 6\sqrt{2}$$


答:
$\pi$ is just a number here, we could calculate in a $\pi$-based number system, it would be a constant still.

Rational function means that it is a fraction of of two polynomials, now whether you take $\mathbb{R}$, $\mathbb{C}$ or even $\mathbb{Q}$ it doesn't matter because the terms rational number and rational function address different issues.

For example let's say you have the function $\pi f(x)$, how do you know this is rational or not?

f(x) could be a constant function or a $\dfrac{x}{\pi}$, the important part is that it satisfies the form
$$a_{n} x^n+a_{n-1} x^{n-1}+...+a_0 \hspace{6mm} \text{where } a_i \in \text{arbitrary } \mathbb{F} \text{ (field)}$$, then it's a polynomial. From which you can constuct rational functions.

Title: Why $f(x)=\pi$ is a rational function? Is a constant function a polynomial even though the constant is a transcendental?

Body:
Wikipedia said

A constant function such as $f(x) = π$ is a rational function since
constants are polynomials. The function itself is rational, even
though the value of $f(x)$ is irrational for all x.

But the definition stated:
A function ${\displaystyle f(x)}$ is called a rational function if and only if it can be written in the form
$${\displaystyle f(x) = {\frac {P(x)}{Q(x)}}}$$
where ${\displaystyle P\,}$ and ${\displaystyle Q\,}$ are polynomial functions of ${\displaystyle x\,}$ and ${\displaystyle Q\,}$ is not the zero function. The domain of ${\displaystyle f\,}$ is the set of all values of ${\displaystyle x\,}$ for which the denominator ${\displaystyle Q(x)\,}$ is not zero.

We can rewrite $f(x)=\pi$ as $f(x)=\frac{\pi}{1}$

I could agree if $1$ is a polynomial since it's a non-zero constant function that is a polynomial of degree 0. But $\pi$ is a trancendental number here. I mean, i never saw a polynomial with a constant of $\pi$. Can you explain it to me?

Thanks.


问:
Principal value of $\int_{-\infty}^{\infty} \frac{\cos x}{ a^{2} - x^{2}}dx$ using residue calculus
The principal value has to be calculated for the following integral:

$$\int_{-\infty}^{\infty} \frac{\cos x}{ a^{2} - x^{2}}dx$$

for $a>0$ and $a$ is real. I set the pole to be $z=+a$ or $z=-a$, but I can't seem to get the required answer which is $π\sin(a)/a$. How can I go about solving this?

答:
$$I=\int_{-\infty }^{\infty } \frac{\cos x}{a^2-x^2} \, dx=2 \pi  i \left(\text{res}\left(\frac{e^{i x}}{2 \left(a^2-x^2\right)},\{x,-a\}\right)+\text{res}\left(\frac{e^{i x}}{2 \left(a^2-x^2\right)},\{x,a\}\right)\right)$$
$$\text{res}\left(\frac{e^{i x}}{2 \left(a^2-x^2\right)},\{x,-a\}\right)=\underset{x\to a}{\text{lim}}\frac{e^{i x} (x-a)}{2 \left(a^2-x^2\right)}=\underset{x\to a}{\text{lim}}\frac{e^{i x} (x-a)}{2 (a+x)(a-x)}=-\frac{e^{i a}}{4 a}$$
Thus
$$I=2\pi i\left(-\frac{e^{i a}}{4 a}+\frac{e^{-i a}}{4 a}\right)= i \pi  \left(\frac{e^{-i a}}{2 a}-\frac{e^{i a}}{2 a}\right)=\frac{\pi}{a}\left(\frac{e^{ia}-e^{-ia}}{2i}\right)=$$
$$=\frac{\pi \sin a}{a}$$


答:
$f(x) = \pi$ is a polynomial function. The coeficients in it do not have to be integers for it to be a polynomial. Yet, the number $\pi$ is irrational.

There are differences between a rational function and a rational number.


问:
The value of $\lim_{x\to +\infty} \dfrac{e^{x^2}}{10^{|x|}}$
I want to find the value of
$$\lim_{x\to +\infty} \dfrac{e^{x^2}}{10^{|x|}}.$$

Since $x \rightarrow + \infty$, I only consider the value of the function for $x \ge 0$, i.e.
$$\lim_{x\to +\infty} \dfrac{e^{x^2}}{10^{x}}.$$

Noticing that the numerator and denominator both approaches $+\infty$ as $x$ approaches $+\infty$, I tried using L'Hôpital's Rule. However, it makes things worse. I did check at Wolfram Alpha, and found out that the value of the limit is $\infty$.

I then noticed that both the numerator and the denominator has powers of some function of $x$. Taking natural logarithm of the numerator and denominator, we have $$\dfrac{\ln e^{x^2}}{\ln 10^{x}}=\dfrac{x^2 \ln e}{x \ln 10}=\dfrac{x}{\ln 10}.$$

This seems really silly, but would I be justified if I say that $$\lim_{x\to +\infty} \dfrac{e^{x^2}}{10^{x}} = \lim_{x\to +\infty}\dfrac{x}{\ln 10}= + \infty?$$

If this is incorrect, kindly give me a hint to tackle this problem.

Thank you for your time.

答: Set $k=ln6$, then $10^x=e^{kx}$. The limit you want is $\lim\limits_{x \to \infty}e^{x^2-kx}$. With this your limits can be easily calculated.  </p>


问:
Picard's successive approximation method for SDE's
As I continue going through stochastic calculus with Krylov, I came across a theorem where he proves that the stochastic differential equation $$ \xi_t=\xi_0+\int_0^t \sigma (s,\xi_s) \ dW_s+\int_o^tb(s,\xi_s) \ ds, $$ has a solution by using the Picard's successive approximation method. Here, $\xi_0$ is an $\mathscr{F}_0$-measurable $\mathbb{R}^{d_1}$-valued random variable, $(W_t,\mathscr{F}_t)$ is a $d$-dimensional Wiener process $(t\geq 0)$, and $b(t,x)$ and $\sigma(t,x)$ are Borel functions defined on $(0,\infty)\times \mathbb{R}^d$. We assume that $b$ is $\mathbb{R}^{d_1}$-valued and $\sigma$ takes values in the set of $d_1\times d$ matrices.

Since the author uses the Picard's method, he first defines for $n\geq0$,
$$ \xi_t(n+1)=\xi_0+\int_0^t \sigma (s,\xi_s(n)) \ dW_s+\int_o^tb(s,\xi_s(n)) \ ds, \ \ \ \xi_t(0)\equiv \xi_0. $$ And at one point we arrive at the following inequality
$$ \mathbb{E}\Big[ \psi_\tau |\xi_\tau(n+1)-\xi_\tau(n)|^2\Big]+c \ \mathbb{E}\Big[ \int_0^\tau\psi_t|\xi_t(n+1)-\xi_t(n)|^2dt\Big]\leq \frac{c}{2} \ \mathbb{E}\Big[\int_0^\tau\psi_t|\xi_t(n)-\xi_t(n-1)|^2dt\Big]$$
for any finite stopping time $\tau$, where $\psi_t=\exp(-N_0t-|\xi_o|)$, $N_0\geq 2$ and $c=N_0-1$ are constants.
Next, by iterating the above inequality, we get $$ \mathbb{E}\Big[ \int_0^\tau\psi_t|\xi_t(n+1)-\xi_t(n)|^2dt\Big]\leq 2^{-n} \ \mathbb{E}\Big[\int_0^\infty\psi_t|\xi_t(1)-\xi_t(0)|^2dt\Big]. $$But I don't see how we can iterate the inequality and obtain this inequality here. Can someone offer me guidance?

Many thanks in advance.

答:
Since $\mathbb{E}[\psi_{\tau} |\xi_{\tau}(n+1) - \xi_{\tau}(n)|^{2}] \geq 0$, the inequality you wrote above (after "And at one point...") implies
\begin{equation*}
c \mathbb{E}\left[ \int_{0}^{\tau} \psi_{t} |\xi_{t}(n+1) - \xi_{t}(n)|^{2} \, dt \right] \leq \frac{c}{2} \mathbb{E} \left[ \int_{0}^{\tau} \psi_{t} |\xi_{t}(n) - \xi_{t}(n-1)|^{2} \, dt \right].
\end{equation*}
Since $c$ is independent of $n$ and $\tau$, we find $a_{n + 1} \leq 2^{-1} a_{n}$, where the sequence $\{a_{n}\}_{n \in \mathbb{N}}$ is given by
\begin{equation*}
a_{n} = \mathbb{E} \left[ \int_{0}^{\tau} \psi_{t} |\xi_{t}(n) - \xi_{t}(n-1)|^{2} \, dt \right].
\end{equation*}

Now notice that if $\{b_{n}\}_{n \in \mathbb{N}}$ is any sequence satisfying $b_{n + 1} \leq 2^{-1} b_{n}$ for all $n \in \mathbb{N} \cup \{0\}$, then $b_{n} \leq 2^{-n} b_{0}$ can readily be proved by induction.

Therefore, setting $\{b_{n}\}_{n \in \mathbb{N}} = \{a_{n}\}_{n \in \mathbb{N}}$ above, we conclude
\begin{align*}
\mathbb{E} \left[ \int_{0}^{\tau} \psi_{t} |\xi_{t}(n) - \xi_{t}(n-1)|^{2} \, dt \right] &\leq 2^{-n} \mathbb{E} \left[ \int_{0}^{\tau} \psi_{t} |\xi_{t}(1) - \xi_{t}(0)|^{2} \, dt \right] \\
&\leq 2^{-n} \mathbb{E} \left[ \int_{0}^{\infty} \psi_{t} |\xi_{t}(1) - \xi_{t}(0)|^{2} \, dt \right].
\end{align*}


问:
Definition of a module in Lang's Algebra
I have a question regarding the definition of a left module in Lang's Algebra. I believe it is missing, but I wanted to confirm here. Here is the definition of a left module in Lang's Algebra. I am quoting it verbatim (I have the third revision edition, and this is on page 117):

Let $A$ be a ring. A left module over $A,$ or a left $A$-module $M$ is an abelian group, usually written additively, together with an operation of $A$ on $M$ (viewing $A$ as a multiplicative monoid by RI 2) such that, for all $a, b \in A$ and $x, y \in M$ we have
$(a+b)x=ax + bx$ and $a(x+y)=ax+ay$.

Now for the sake of completeness, let me write what he means by RI 2, which is one of the three ring axioms when he defines a ring for the first time:

The multiplication is associative, and has a unit element.

In addition to everything he has said in the definition, doesn't he also need to say that $(ab)x=a(bx)?$

答:
This is implied already. We view $A$ as a multiplicative monoid, as suggested. Then an operation of $A$ on $M$ is a monoid homomorphism $\varphi\colon A\rightarrow M^M$. For $a\in A$ and $x\in M$, we denote $\varphi(a)(x)=ax$. Then, for $a,b\in A$ and $x\in X$, $$(ab)x=\varphi(ab)(x)=(\varphi(a)\circ\varphi(b))(x)=\varphi(a)(\varphi(b)(x))=a(bx).$$
The second equality holds, because $\varphi$ is a monoid homomorphism. Note that we also get $1x=\varphi(1)(x)=\operatorname{id}_M(x)=x$ for $x\in M$, where $1$ is the unit in $A$, for the same reason. This recovers the "usual" axioms for a module.


问:
Finding a matrix representation of a linear endomorphism
Consider the following basis of the vector space of $2 \times 2$ matrices with coefficients in $\mathbb{R}$:

$$ e_1 = \left( \begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix} \right) $$

$$ e_2 = \left( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right) $$

$$ e_3 = \left( \begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix} \right) $$

$$ e_4 = \left( \begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix} \right) $$

Now, say we have an endomorphism $T: M(2,\mathbb{R}) \rightarrow M(2,\mathbb{R})$ and we know the image of each basis element. How would I go about construction a matrix representation of $T$? Thank you!

答: Suppose that, say,$$T\left(\begin{bmatrix}a&b\\c&d\end{bmatrix}\right)=\begin{bmatrix}a+b&c+d\\a-d&b-c\end{bmatrix}.$$Then$$T\left(e_1\right)=\begin{bmatrix}1&0\\1&0\end{bmatrix}=e_1+e_3$$and therefore the entries of the first column of your matrix will be $1$, $0$, $1$, and $0$, since $T(e_1)=1\times e_1+0\times e_2+1\times e_3+0\times e_4$. And so on…


问:
Trying to find a basis for the span of 3 vectors
Let $v_1=(1,0,1)$, $v_2=(1,1,2)$, and $v_3=(2,3,5)$. I'm trying to find a basis for $span\{v_1,v_2,v_3\}$. I note that these vectors are not linearly independent because when I put them into a matrix the determinant is zero. What should I do from here? Thank you!

答: Note $c\cdot v_1=(c,0,c)\neq(1,1,2)$ and thus $v_1,v_2$ are linearly independent because $v_2\notin\text{Span}(\{v_1\})$. Therefore $\mathcal B:=\{v_1,v_2\}$ is a Hamel basis for $\text{Span}(\{v_1,v_2,v_3\})$ because $$v_3=-v_1+3v_2\in\text{Span}(\mathcal B)$$ and thus $\text{Span}(\mathcal B)=\text{Span}(\{v_1,v_2,v_3\})$.


答:
You are correct that the three vectors are not linearly independent;

in fact, $(2,3,5)=3(1,1,2)-(1,0,1)$.

But if you take any two of them, you should find that one is not a multiple of the other,

so they are linearly independent, so they are a basis for the span.

Title: Trying to find a basis for the span of 3 vectors

Body: Let $v_1=(1,0,1)$, $v_2=(1,1,2)$, and $v_3=(2,3,5)$. I'm trying to find a basis for $span\{v_1,v_2,v_3\}$. I note that these vectors are not linearly independent because when I put them into a matrix the determinant is zero. What should I do from here? Thank you!