问:
Integral involving ratios of polynomial and square roots
Is there a simple way to compute the following integral

$$I(y,\beta,\gamma)=\int_1^y \frac{x^2}{\sqrt{\left(x^2-1\right)\left(x^2-\gamma ^2\right) \left(x^2+\beta ^2\right) }}\mathrm{d}x$$

with $y>1$, $0<\gamma<1$ and $\beta>0$?

答:
Define the function $\mathcal{I}:\left(0,1\right)\times\mathbb{R}_{>0}\times\left(1,\infty\right)\rightarrow\mathbb{R}$ via the improper integral

$$\mathcal{I}{\left(p,q,y\right)}:=\int_{1}^{y}\mathrm{d}x\,\frac{2x^{2}}{\sqrt{\left(x^{2}-1\right)\left(x^{2}-p^{2}\right)\left(x^{2}+q^{2}\right)}}.$$

Suppose $\left(p,q,y\right)\in\left(0,1\right)\times\mathbb{R}_{>0}\times\left(1,\infty\right)$, and set $z:=y^{2}\land a:=1\land b:=p^{2}\land c:=0\land d:=-q^{2}$.

Note that $z>a>b>c>d$, and then also

$$0<\frac{\left(a-d\right)\left(b-c\right)}{\left(a-c\right)\left(b-d\right)}<1\land0<\frac{\left(b-d\right)\left(z-a\right)}{\left(a-d\right)\left(z-b\right)}<1.$$

The integral $\mathcal{I}$ can be transformed into an elliptic integral by means of a simple power substitution:

$$\begin{align}
\mathcal{I}{\left(p,q,y\right)}
&=\int_{1}^{y}\mathrm{d}x\,\frac{2x^{2}}{\sqrt{\left(x^{2}-1\right)\left(x^{2}-p^{2}\right)\left(x^{2}+q^{2}\right)}}\\
&=\int_{1}^{y^{2}}\mathrm{d}t\,\frac{t}{\sqrt{t}\sqrt{\left(t-1\right)\left(t-p^{2}\right)\left(t+q^{2}\right)}};~~~\small{\left[x=\sqrt{t}\right]}\\
&=\int_{a}^{z}\mathrm{d}t\,\frac{t}{\sqrt{\left(t-a\right)\left(t-b\right)\left(t-c\right)\left(t-d\right)}}.\\
\end{align}$$

Next, consider the linear fractional transformation given implicitly by the relation

$$\left(t-b\right)\left(1-u\right)=\left(a-b\right).$$

Then,

$$t=\frac{a-bu}{1-u}\implies dt=du\,\frac{\left(a-b\right)}{\left(1-u\right)^{2}},$$

and

$$\frac{t-a}{t-b}=u\land\frac{t-c}{t-b}=\frac{a-c}{a-b}-\frac{b-c}{a-b}u\land\frac{t-d}{t-b}=\frac{a-d}{a-b}-\frac{b-d}{a-b}u.$$

Applying this LFT to $\mathcal{I}$, we obtain

$$\begin{align}
\mathcal{I}{\left(p,q,y\right)}
&=\int_{a}^{z}\mathrm{d}t\,\frac{t}{\sqrt{\left(t-a\right)\left(t-b\right)\left(t-c\right)\left(t-d\right)}}\\
&=\int_{a}^{z}\mathrm{d}t\,\frac{t}{\left(t-b\right)^{2}\sqrt{\left(\frac{t-a}{t-b}\right)\left(\frac{t-c}{t-b}\right)\left(\frac{t-d}{t-b}\right)}}\\
&=\int_{0}^{\frac{z-a}{z-b}}\mathrm{d}u\,\frac{\left(a-b\right)}{\left(1-u\right)^{2}}\cdot\frac{a-bu}{1-u}\cdot\frac{\left(1-u\right)^{2}}{\left(a-b\right)^{2}}\\
&~~~~~\times\frac{1}{\sqrt{u\left(\frac{a-c}{a-b}-\frac{b-c}{a-b}u\right)\left(\frac{a-d}{a-b}-\frac{b-d}{a-b}u\right)}};~~~\small{\left[t=\frac{a-bu}{1-u}\right]}\\
&=\int_{0}^{\frac{z-a}{z-b}}\mathrm{d}u\,\frac{a-bu}{1-u}\cdot\frac{1}{\sqrt{u\left[\left(a-d\right)-\left(b-d\right)u\right]\left[\left(a-c\right)-\left(b-c\right)u\right]}}\\
&=\int_{0}^{\frac{\left(b-d\right)\left(z-a\right)}{\left(a-d\right)\left(z-b\right)}}\mathrm{d}v\,\frac{a-b\left(\frac{a-d}{b-d}\right)v}{1-\left(\frac{a-d}{b-d}\right)v}\\
&~~~~~\times\frac{1}{\sqrt{v\left(1-v\right)\left[\left(a-c\right)\left(b-d\right)-\left(b-c\right)\left(a-d\right)v\right]}};~~~\small{\left[u=\left(\frac{a-d}{b-d}\right)v\right]}\\
&=\frac{1}{\sqrt{\left(a-c\right)\left(b-d\right)}}\int_{0}^{\frac{\left(b-d\right)\left(z-a\right)}{\left(a-d\right)\left(z-b\right)}}\mathrm{d}v\,\left[\frac{a-b}{1-\left(\frac{a-d}{b-d}\right)v}+b\right]\frac{1}{\sqrt{v\left(1-v\right)\left[1-\frac{\left(a-d\right)\left(b-c\right)}{\left(a-c\right)\left(b-d\right)}v\right]}}\\
&=\frac{1}{\sqrt{\left(a-c\right)\left(b-d\right)}}\int_{0}^{\sin^{2}{\left(\varphi\right)}}\mathrm{d}v\,\left[\frac{a-b}{1-\eta\,v}+b\right]\frac{1}{\sqrt{v\left(1-v\right)\left(1-\kappa^{2}v\right)}}\\
&=\frac{2}{\sqrt{\left(a-c\right)\left(b-d\right)}}\int_{0}^{\sin{\left(\varphi\right)}}\mathrm{d}x\,\left[\frac{a-b}{1-\eta\,x^{2}}+b\right]\frac{1}{\sqrt{\left(1-x^{2}\right)\left(1-\kappa^{2}x^{2}\right)}};~~~\small{\left[v=x^{2}\right]}\\
&=\frac{2}{\sqrt{\left(a-c\right)\left(b-d\right)}}\left[\left(a-b\right)\Pi{\left(\varphi,\eta,\kappa\right)}+bF{\left(\varphi,\kappa\right)}\right],\\
\end{align}$$

where we've introduced the auxiliary parameters

$$\kappa:=\sqrt{\frac{\left(a-d\right)\left(b-c\right)}{\left(a-c\right)\left(b-d\right)}}\in\left(0,1\right),$$

$$\varphi:=\arcsin{\left(\sqrt{\frac{\left(b-d\right)\left(z-a\right)}{\left(a-d\right)\left(z-b\right)}}\right)}\in\left(0,\frac{\pi}{2}\right),$$

$$\eta:=\frac{a-d}{b-d}\in\left(1,\infty\right).$$

Note: even though $\eta>1$, we can still define $\Pi$ in the usual way without worrying about Cauchy principal values since $0<1-\eta\sin^{2}{\left(\varphi\right)}$.


问:
How to study mathematics (Graduate level)
I know that normally, after attending class and understand the theorem and proofs, we will go directly to the problem and solve a couple of them, strengthening the concepts. Clearly, this is the best way to study and this is why in the US they always give a homework assignments to keep track of the study.

But it seems this way doesn't really work for me. The main problem is always, I found it's hard for me to do the problem directly after the class as I need more time to absorb the materials (In some sense, I might need to prove a theorem 5 times on my own to really understand what that theorem is all about.) So I don't usually do any homework on my own, instead, I focus on understanding the material first and do extra homework 2 weeks before exams.

This causes a big problem, that I don't have enough time to think about the homework assignment and end up "discussing problems" (mimicking solutions) with the others during the week. Also, I feel more comfortable just having a solid understanding of the whole course before moving on to doing questions, say two weeks before final exams.

Is there any way to solve this problem? Due to US system, there are always homework assignment and mid term and I cannot avoid them.

答:
You don't really need to 100% understand the proof of the theorem (or the theory itself) when solving problems. I don't know how hard are your homeworks but sometimes it is good to go to the problem headfirst. Take a theorem(s) as blackbox into which you put some assumptions and you get some consequences and try to proceed with the homework.

Of course, you usually won't solve the whole homework this way but you will put the theory in use. And it is easier to spot the places where you need a deeper understanding of theory.

I personally find it more useful for me to get over a problem set multiple times then to go over the proofs of theorems in theory multiple times. Also some theory is not to be understood immediately. That's just how math works. But seeing it in "practice" (on homework problems) helps you realize the connections.

ContentLicense: CC BY-SA 4.0


问:
Integration by Trigonometric Substitution vs Table of Integral Solution
I'm not sure how to phrase this question, but I find myself confused over the correctness of a particular solution in the table of integral, specifically:
$$\int\frac{dx}{\sqrt{x^2-a^2}}=\ln|\sqrt{x^2-a^2}+x|+C$$

I find that if you try to solve it with trigonometric substitution, you get a different answer:

So, for the example, since the form $\sqrt{x^2-a^2}$ is present, we use the substitution $x=asec \theta$, and $dx=asec\theta tan \theta d\theta$. Doing so:

$$\int\frac{asec\theta tan \theta d\theta}{\sqrt{(asec \theta)^2-a^2}}$$
$$\int\frac{asec\theta tan \theta d\theta}{\sqrt{a^2sec^2 \theta-a^2}}$$
$$\int\frac{asec\theta tan \theta d\theta}{a\sqrt{sec^2 \theta-1}}$$
Using the trigonometric identity $tan^2 \theta=sec^2 \theta -1$
$$\int\frac{asec\theta tan \theta d\theta}{atan \theta}$$

Cancelling like terms, we get $\int sec \theta$, which finally integrates to $ln|sec \theta + tan \theta|+C$

Undoing the substitution, using this triangle:

https://i.stack.imgur.com/6tkR7.png

With $sec \theta = \frac{x}{a}$ and $tan \theta = \frac{\sqrt{x^2-a^2}}{a}$, the final answer then is:

$$ln\biggl|\frac{x}{a} + \frac{\sqrt{x^2-a^2}}{a}\biggl|+C$$

Which is evidently different from the one from the table of integrals. Is my solution wrong or something?

答:
You're right. Indefinite integrals represent a family of curves(functions) which vary by some constant.

$\ln \left|\dfrac{x+\sqrt{x^2-a^2}}{a}\right| +c= \ln |x+\sqrt{x^2-a^2}| - \ln|a| +c  = \ln(x+\sqrt{x^2-a^2}) +k$

$k = c-\ln |a| = $ constant


问:
Show that $\{\alpha_1,\alpha_2,\alpha_3\}$ is a basis for V
$V=\mathbb{R}^3$

Let,

$\alpha_1=\begin{bmatrix}1\\0\\1\end{bmatrix},~\alpha_2=\begin{bmatrix}0\\1\\1\end{bmatrix},~\alpha_3=\begin{bmatrix}1\\-1\\2\end{bmatrix}$

Show that $\{\alpha_1,\alpha_2,\alpha_3\}$ is a basis for V and find its dual basis in $\mathbb{R}_3$.

I think we should show that elements are linear independent. (But is this enough for being basis?) I couldnt do the rest.

答:
$\text{Solution sketch:}$ Since the dimension of the vector space is $3$, if $\alpha_1,\alpha_2$ and $\alpha_3$ are linearly independent, then $\mathcal A=\{\alpha_1,\alpha_2,\alpha_3\}$ will be a basis for $V=\mathbb R^3$. Observe that
$$A=\begin{pmatrix}1&0&1\\0&1&-1\\1&1&2 \end{pmatrix}\overset{R_3-R_1}\sim \begin{pmatrix}1&0&1\\0&1&-1\\0&1&1 \end{pmatrix}\overset{R_3-R_2}\sim \begin{pmatrix}1&0&1\\0&1&-1\\0&0&2 \end{pmatrix}\implies\text{rk}(A)=3$$
This shows that $\mathbb R^3=\langle \alpha_1,\alpha_2,\alpha_3\rangle$.

If we want the dual basis for $V^*$, we have to find a set of linear maps $\mathcal A^*=\{\alpha_1^*,\alpha_2^*,\alpha_3^*\}=\{f_1,f_2,f_3\}$ such that $f_i(\alpha_j)=\begin{cases}1&\text{ if }i=j\\0&\text{ otherwise} \end{cases}$.

For the first function $f_1$ you have to impose $f_1(\alpha_1)=1$ and $f_2(\alpha_1)=f_1(\alpha_3)=0$ and solve the system. You do the same thing for the other maps.


问:
Understanding the definition of the direct sum of subspaces of a vector space
I have a question regarding the definition of direct sum of a vector space in relation to subspaces.

Definition: A vector space $V$ is called the direct sum of $W_1$ and $W_2$ if $W_1$ and $W_2$ are subspaces of $V$ such that $W_1\cap W_2 = \{0\}$ and $W_1 + W_2 = V$. We denote that $V$ is the direct sum of $W_1$ and $W_2$ by writing $V = W_1\oplus W_2$.

Is this definition saying that any vector in $V$ can be written as a linear combination of the vectors in the set $W_1 + W_2$?

Thanks!

答:
$W_1\cap W_2=\{0\}$ if and only if
\begin{align}
\phi\colon W_1\times W_2&\to V\\
(v,w)&\mapsto v+w
\end{align}
is injective and

$V=W_1+W_2$ if and only if $\phi$ is surjective.

In summary:
\begin{equation}
V=W_1\oplus W_2\Leftrightarrow \phi\text{ is a vector space isomorphism.}
\end{equation}

In other words: $V=W_1\oplus W_2$ if and only if for all $v\in V$, there is exactly one $(w_1,w_2)\in W_1\times W_2$ such that $v=w_1+w_2$.

Proof of 1.:

(Reminder: If $W_1$ and $W_2$ are vector subspaces of $V$, $W_1\cap W_2$ is also a vector subspace of $V$.)

If $v\in W_1\cap W_2$, $\phi(v,0)=v+0=v=0+v=\phi(0,v)$. $\Rightarrow v=0$ if $\phi$ is injective.

Conversely, if $W_1\cap W_2=\{0\}$ and $(v,w),(v',w')\in W_1\times W_2$, then
\begin{equation}
\phi(v,w)=\phi(v',w')\Leftrightarrow v+w=v'+w'\Leftrightarrow v-v'=w'-w=:u.
\end{equation}
Since $v-v'\in W_1$ and $w'-w\in W_2$, $u\in W_1\cap W_2=\{0\}$. Thus, $v=v'$ and $w=w'$.

ContentLicense: CC BY-SA 3.0


问:
Question about a step in Prop 7.50 in Milne's Algebraic Number Theory

This question talks about the correspondence between finite unramified extensions of $K$, a complete field with a discrete, nonarchimedean absolute value, that are contained in an algebraic extension $L$ of $K$ and finite subextensions of the tower of residue fields $k \subset l$.

So Milne starts by considering a finite subextension $k' \supset k$ of the residue field of $k$ and writes it as $k[a]$ for some $a$. Now he uses Newton's Lemma (which was proved along with Hensel's Lemma) to argue that there should be an $\alpha \in L$ such that $f(\alpha)=0$ and $\alpha \equiv a \mod \mathfrak{p}$.

My issue is this: why should such an $\alpha$ exist in L? Since L is an arbitrary algebraic extension (not necessarily finite), it doesn't have to be complete under the extension of the absolute value and so I'm not sure why Newton's Lemma applies. Also, since L is not necessarily algebraically closed, why should the extension $K[x]/(f)$ embed into L (where f is a lift of the minimal polynomial of the primitive generator k')?

Am I missing something obvious or is there a mistake here?

You can look at the proposition and its proof in page 127 in Milne's notes.

EDIT:

I also took a look at these notes from 18.785 (page 1-2) where they consider $K[X]/(f)$ but they say that by construction it is true that the valuation ring of this is $A[\alpha]$. I don't see why this should be so. Could someone help with this as well?

答:
There is nothing wrong here when $L$ is infinite. You just need to think more carefully by focusing on the actual finite extensions you use to create the lifting of a root. An infinite algebraic extension of $K$ is a union of finite extensions of $K$, and each finite extension of $K$ is complete.

Even if $L/K$ is not finite, each particular element of $L$ generates a finite extension of $K$, and finite extensions of $K$ are complete. So if $f(x) \in A[x]$ and $f(x) \bmod \mathfrak p$ is separable, then if you lift a root of $f(x) \bmod \mathfrak p \in k[x]$ to $A$, it is a number $\gamma \in A$ such that $|f(\gamma)| < 1$ and $|f'(\gamma)| = 1$.  The field $K(\gamma)$ in $L$ has to be complete since it is finite over $K$. Apply Hensel's lemma to the complete field $K(\gamma)$ to see that $f(x)$ has a root $\alpha$ in the integers of $K(\gamma)$ such that $f(\alpha) = 0$ and $|\alpha - \gamma| < 1$.

Suppose $f(x)$ has two roots $\alpha$ and $\beta$ in $L$ such that $|\alpha - \gamma| < 1$ and $|\beta - \gamma| < 1$. Even if $L$ is not complete, the field $K(\alpha,\beta)$ inside of $L$ is a finite extension of $K$, and thus is complete. The numbers $\alpha$ and $\beta$ are in the integers of $K(\alpha,\beta)$ and they reduce to the same root of $f(x) \bmod \mathfrak p$ in the residue field of $K(\alpha,\beta)$. By the uniqueness of the lifting of a root in Hensel's lemma, we get $\alpha = \beta$. Thus in $L$ there is only one lifting of a root of $f(x) \bmod \mathfrak p$ in the residue field of $L$.


答:
One way of determining whether a set of vectors is linearly independent is to stick them in a determinant. If the determinant is not $0$ then you're done. So calculate
$$
\begin{array}{|ccc|}
\ 1 & 0 & 1 \ \\
\ 0 & 1 & 1 \ \\
\ 1 & -1 & 2 \
\end{array}
$$
and it shold give you a non-zero answer.

For your dual basis point, can you define it?

Body:
$V=\mathbb{R}^3$

Let,

$\alpha_1=\begin{bmatrix}1\\0\\1\end{bmatrix},~\alpha_2=\begin{bmatrix}0\\1\\1\end{bmatrix},~\alpha_3=\begin{bmatrix}1\\-1\\2\end{bmatrix}$

Show that $\{\alpha_1,\alpha_2,\alpha_3\}$ is a basis for V and find its dual basis in $\mathbb{R}_3$.

I think we should show that elements are linear independent. (But is this enough for being basis?) I couldnt do the rest.


问:
What's wrong with this procedure?
Consider the problem of evaluating the $$\lim_{x\to 0}{\frac{x^2+x+1}{2x^2-3}},$$ for example. Then the value is $-1/3.$ But suppose for $x\ne 0$ we rewrite the expression as $$\frac{1+\frac 1x+\frac{1}{x^2}}{2-\frac{3}{x^2}}=g(x).$$ Then the original problem is equivalent to evaluating $$\lim_{x\to\infty}{g(x)},$$ which is equal to $1/2,$ as opposed to $-1/3.$

Clearly the problem is with the second method since the rational function is continuous at $x=0.$ My question is, where exactly has a procedural error been made in the second method? Thank you.

答:
The error is in the step "Then the original problem is equivalent to evaluating ...". It is not. To be precise,
$$
\lim_{x\to 0}\frac{x^2+x+1}{2x^2-3}
=\lim_{x\to \color{red}{0}}\frac{1+\frac1x+\frac1{x^2}}{2-\frac3{x^2}}
\color{red}{\ne}
\lim_{x\to \color{red}{\infty}}\frac{1+\frac1x+\frac1{x^2}}{2-\frac3{x^2}}
$$

Title: What's wrong with this procedure?


问:
Find RV distribution and prove its convergence to zero
Let $X_n$ be a sequence of independent and uniformly distributed over $[0; 1]$ random variables. The task is to find distribution of the following random variable: $m_n=\min\{X_1, X_2, \ldots, X_n\}$ and also proof that it converges almost surely to zero, in other words: $$\Pr(\lim_{n \rightarrow \infty}m_n=0)=1$$

答:
For any $x\in[0,1]$ we have
$$
\left\{\min_{1\leqslant i\leqslant n} X_i \geqslant x \right\} = \bigcap_{i=1}^n \{X_i\geqslant x\}.
$$
It follows that
\begin{align}
\mathbb P(m_n\geqslant x) &= \mathbb P\left(\bigcap_{i=1}^n \{X_i\geqslant x\} \right)\\
&= \prod_{i=1}^n \mathbb P(X_i\geqslant x)\\
&= \mathbb P(X_1\geqslant x)^n\\
&= (1-x)^n,
\end{align}
and so the distribution function of $m_n$ is
$$
F_n = (1 - (1-x)^n)\cdot\mathsf 1_{[0,1]}(x) + \mathsf 1_{(1,\infty)}(x)
$$
and the density
$$
f_n = n(1-x)^{n-1}\cdot\mathsf 1_{[0,1]}(x).
$$
Recall that $Z_n\to Z$ almost surely if and only if
$$
\mathbb P\left(\limsup_{n\to\infty} \{|Z_n-Z|>\varepsilon\}\right) = 0
$$
for all $\varepsilon>0$. Let $\varepsilon\in(0,1)$, then
$$
\sum_{n=1}^\infty \mathbb P(m_n>\varepsilon) = \sum_{n=1}^\infty (1-\varepsilon)^n = \frac{1-\varepsilon}\varepsilon<\infty,
$$
so by the Borel-Cantelli lemma
$$
\mathbb P\left(\limsup_{n\to\infty} \{X_n>\varepsilon\}\right)=0,
$$
and hence $m_n\to 0$ almost surely.


答: In your second form, although equivalent, you are in fact dividing by $0$ which you absolutely do not want to do! So you cannot accurately compute the limit as $x\rightarrow 0$ of $\frac{k}{x}$, as you have done. If you were calculating an infinite limit, however, this would be ok (as $\frac{k}{x}\rightarrow 0$).

Title: What's wrong with this procedure?

Body:
Consider the problem of evaluating the $$\lim_{x\to 0}{\frac{x^2+x+1}{2x^2-3}},$$ for example. Then the value is $-1/3.$ But suppose for $x\ne 0$ we rewrite the expression as $$\frac{1+\frac 1x+\frac{1}{x^2}}{2-\frac{3}{x^2}}=g(x).$$ Then the original problem is equivalent to evaluating $$\lim_{x\to\infty}{g(x)},$$ which is equal to $1/2,$ as opposed to $-1/3.$

Clearly the problem is with the second method since the rational function is continuous at $x=0.$ My question is, where exactly has a procedural error been made in the second method? Thank you.


问:
Why does the compression criterion require homotopy relative to the boundary?
The Compression Criterion states the following:

Let $(X,A,x)$ be a nested space triple and $$f: (D^n, \partial D^n, z) \longrightarrow (X,A,x)$$ a continuous map of space triples. Then the following are equivalent:

$f$ is triple homotopic to $const_x$.

$f$ is triple homotopic, relative $\partial D^n$, to a map with image in A.

The direction 2. $\Rightarrow$ 1. works by concatenating the given homotopy with one that contracts $D^n$ to $z$.

The direction 1. $\Rightarrow$ 2. is the interesting one: Here one can picture a cylinder standing upright with a horizontal slice corresponding to the homotopy at a point in time. The bottom face is $f$, the top face is $const_x$. The boundary maps to $A$. By blowing the bottom face into the cylinder as if it was made from rubber while fixing $\partial D^2 \times \{0\}$, and doing this all the way until the former bottom faces is contained in the cylinder's boundary again, we can find a homotopy from $f$ to a map with image in $A$, relative to the boundary.

A practically identical proof should show that:

$f$ is triple homotopic to a map with image in $A$.

implies 2. The picture is identical except that the top face maps into $A$, instead of being $const_x$.
In particular a stronger version of the compression criterion would be that the following are equivalent:

$f$ is triple homotopic to $const_x$.

$f$ is triple homotopic, relative $\partial D^n$, to a map with image in A.

$f$ is triple homotopic to a map with image in $A$.

My question is the following: How come I have never seen this version of the compression criterion before, especially since for me, personally, this really helps with understanding relative homotopy? Or is this result not correct? In this case I must really be overlooking something.

Any help is greatly appreciated.
Note that this question is a duplicate of Compression Criterion for $\pi_n(X,A,x_0)$. Why do we need homotopies $\text{rel} \ S^{n-1}$?, but since this is a 5 year old question with no answer and it seems like quite a relevant question, I am okay with possibly having this question deleted to get an answer to the original one.

答:
The implications $2. \Rightarrow 3. \Rightarrow 1.$ are obvious. Let us prove $1. \Rightarrow 2.$

Let $H : (D^n,S^{n-1}) \times I \to (X,A)$ be a homotopy of triples from $f$ to a constant map $c : (D^n,S^{n-1})\to (X,A)$ with value $a_0 \in A$ (for our argument it is irrelevant wether or not $H$ is stationary on the basepoint $y$).

Define $M = \{(x,t) \in D^n \times I \mid  \lVert x \rVert \ge \frac{2-t}{2} \}$ and $M' = \{(x,t) \in D^n \times I \mid \lVert x \rVert \le \frac{2-t}{2}\}$. These are closed subsets which cover $D^n \times I$.
Define
$$\bar H : D^n \times I \to X, \bar H(x,t) =\begin{cases} H(\frac{x}{\lVert x \rVert},2 - 2\lVert x \rVert) & (x,t) \in M \\ H(\frac{2x}{2-t},t) & (x,t) \in M'   \end{cases}$$

$\bar H$ is well-defined and continuous since the definitions in line 1 and 2 agree on $M \cap M'$.

We have $S^{n-1} \times I \subset M$, thus for all $x \in S^{n-1}$ we get $\bar H(x,t) = H(x,0) = f(x)$.

For $x \in D^n$ we have $\bar H(x,1) \in H(S^{n-1} \times I) \subset A$ if $(x,1) \in M$ and $\bar H(x,1) \in H(D^n \times \{1\}) = \{a_0\} \subset S^{n-1}$ if $(x,1) \in M'$. Hence $\bar H(D^n \times \{1\}) \subset S^{n-1}$.

This shows that $f$ is homotopic relative $S^{n-1}$ to a map with image in $A$ (via the "compressing homotopy" $\bar H$). Clearly $\bar H$ is a homotopy of triples because it is stationary on $S^{n-1}$.

Let us finally come to the intuition to find $\bar H$. Look at the level $t$. The level set $D^n \times \{t\}$ can be decomposed in a closed ball $D^n_t$ with center $(0,t)$ and radius $\frac{2 -t}{2}$ and a closed spherical shell $A^n_t$ with center $(0,t)$ , inner radius $\frac{2 -t}{2}$ and outer radius $1$. Note that $D^n_0 = D^n \times \{0\}$ and $A^n_0 = S^{n-1} \times \{0\}$. We  can identity $D^n_t$ with $D^n$ via "radial stretching", i.e. $(x,t) \mapsto \phi(x,t) = \frac{2x}{2-t}$. Define $\bar H(x,t) = H(\phi(x,t),t)$ on $D_t$. Similarly we can identify $A_t$ with $S^{n-1} \times [0,t]$ via $\psi(x) = (\frac{x}{\lVert x \rVert},2 - 2 \lVert x \rVert)$. Define $\bar H(x,t) = H(\psi(x,t))$ on $A_t$. Then on the common boundary sphere $C_t$ of $A_t$ and $D_t$ the map $\bar H$ is nothing else than $H_t \mid_{S^{n-1}}$ "compressed" to $C_t$.


答:
You have made an error of not converting the limit value when changing the function. Have a look at the equation below to get a better understanding of the flaw in your logic.

$ Y = \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{x^2+x+1}{2x^2-3} = \lim_{x \to 0} \frac{1+1/x+1/x^2}{2-3/x^2} = \lim_{y \to \inf} \frac{1+y+y^2}{2-3y^2} \neq \lim_{y \to 0} \frac{1+y+y^2}{2-3y^2} $

If it's still not clear, comment and I will elaborate.

Title: What's wrong with this procedure?


问:
Why is $\lvert \tanh x\rvert < 1$?
In the solutions to my exercises, it says that it is easy to see that $\lvert \tanh x\rvert < 1$. Sadly, this doesn't really help me. I know that $$\lvert \tanh x\rvert = \left\lvert \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} \right\rvert = \left\lvert \frac{1-e^{-2x}}{1 + e^{-2x}} \right\rvert$$
However, I don't see how this would imply or even help me show $\lvert \tanh x\rvert < 1$.

答:
$$\tanh(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}.$$

As $x\rightarrow-\infty,\; e^{2x}\rightarrow0$ so
$\displaystyle\tanh(x)=\frac{e^{2x}-1}{e^{2x}+1}\rightarrow-1.$

As $x$ increases, $1-e^{-2x}$ increases while $1+e^{-2x}$ decreases;
so, $\tanh(x)=\displaystyle\frac{1-e^{-2x}}{1+e^{-2x}}$ is an
increasing function.

As $x\rightarrow\infty,\; e^{-2x}\rightarrow0$ so
$\displaystyle\tanh(x)=\frac{1-e^{-2x}}{1+e^{-2x}}\rightarrow1.$

Therefore $\displaystyle-1<\tanh(x)<1.$

Title: Why is $\lvert \tanh x\rvert < 1$?


问:
How to do integration over a unit sphere with 3 independent vectors?
I got this equation I want to integrate over a sphere:

$min( max(-vSurfaceToCameraDir \cdot (vLight + vNormal * A), 0), 1 )^B$

A & B are constants and vSurfaceToCameraDir, vLight & vNormal are unit vectors that can point in any direction. If A=0 this could be expressed as $\int_{0}^{2π}{\int_{0}^{π/2} cosθ^Bsinθ \,dθ}\,dφ$ which would add up to π/(B + 1). However with vNormal there's a 3rd independent vector and I'm not sure how to proceed.

Is there some analytical way to compute this? If not I suppose I can do a numeric integration by making lots of random samples but I thought there might be some more elegant & faster way.

答:
Let $T \left(s\right) = \min  \left(1 , \max  \left(0 , s\right)\right)$ and $\overrightarrow{w} =-\left({\overrightarrow{v}}_{\text{light}}+A {\overrightarrow{v}}_{\text{normal}}\right)$. Assuming we want the integral

\begin{equation}
I = \int_{{S}^{2}}^{}T \left({\left(\overrightarrow{{\xi}} \cdot  \overrightarrow{w}\right)}^{B}\right) d \sigma(\overrightarrow{{\xi}}) = 2 {\pi} \int_{0}^{1}T \left({\left|w\right|}^{B} {s}^{B}\right) d s
\end{equation}

If $\left|w\right|  \leqslant  1$ we have

\begin{equation}
I = 2 {\pi} {\left|w\right|}^{B} \int_{0}^{1}{s}^{B} d s = \frac{2 {\pi} {\left|w\right|}^{B}}{B+1}
\end{equation}

If $\left|w\right|  \geqslant  1$ we have

\begin{equation}
\renewcommand{\arraystretch}{1.5}  \begin{array}{rcl}I&=&\displaystyle  2 {\pi} {\left|w\right|}^{B} \int_{0}^{1/\left|w\right|}{s}^{B} d s+2 {\pi} \int_{1/\left|w\right|}^{1}d s\\
&=&\displaystyle  \frac{2 {\pi}}{\left(B+1\right) \left|w\right|}+2 {\pi} \left(1-\frac{1}{\left|w\right|}\right)
\end{array}
\end{equation}


问:
Is $V=S\oplus T$ if and only if $S\times T\ni(v,w)\mapsto v+w\in V$ is a vector space isomorphism?
Definition (direct sum): If $S$ and $T$ are vector subspaces of $V$, we write $V=S\oplus T$ if $S\cap T=\{0\}$ and $\{v+w:v\in S,w\in T\}=V$.

Thus, if $V=S\oplus T$,
\begin{align}
\phi\colon S\times T&\to V\\
(v,w)&\mapsto v+w
\end{align}
is a vector space isomorphism.

Conversely, can we show that $V=S\oplus T$ if $\phi$ is a vector space isomorphism? That is, can we prove $S\cap T=\{0\}$, possibly without assuming that $V$ is finite dimensional?

答:
Yes, $S\cap T=\{0\}$ if and only if
\begin{align}
\phi\colon S\times T&\to V\\
(v,w)&\mapsto v+w
\end{align}
is injective. (Reminder: $S\cap T$ is a vector subspace of $V$, because the intersection of vector subspaces is a vector subspace.)

Proof.

If $v\in S\cap T$, $\phi(v,0)=\phi(0,v)$ and if, in addition, $\phi$ is injective,
\begin{equation}
\phi(v,0)=\phi(0,v)\Leftrightarrow (v,0)=(0,v)\Leftrightarrow v=0.
\end{equation}

Conversely, if $S\cap T=\{0\}$ and $(v,w),(v',w')\in S\times T$, then
\begin{equation}
\phi(v,w)=\phi(v',w')\Leftrightarrow v+w=v'+w'\Leftrightarrow v-v'=w'-w=:u.
\end{equation}
Since $v-v'\in S$ and $w'-w\in T$, $u\in S\cap T=\{0\}$. Thus, $v=v'$ and $w=w'$.


问:
How can I develop this?
$$\operatorname{var}(X - Y)^2 $$
How can I develop this? I think I should arrive at a covariancy, but I can't think of any property.

答: Assuming you mean var($ (X-Y)^2 $), that is var ($X^2 - 2XY + Y^2$).  Repeated use of V(A+B) = V(A) + 2 Cov(A,B) + V(B) as well as V(aX) = $a^2$V(X) gives me V($X^2$) -4 Cov($X^2, XY$) + 2Cov($X^2, Y^2$) + 4V(XY) - 4Cov($XY,Y^2$) + V($Y^2$). There is no simple relationship between V(X) and V($X^2$) etc. For example if X were normal, $X^2$ would be a $\chi ^2$ random variable.

Title: How can I develop this?


问:
Show that $|f(p_n)|<10^{-3}$ whenever $n>1$ but that $|p-p_n|<10^{-3}$ requires that $n>1000$.
Let $f(x)=(x-1)^{10}$.

The root of the equation , $p=1$.

The approximates of the root, $p_n=1+\frac{1}{n}$

Show that $|f(p_n)|<10^{-3}$ whenever $n>1$ but that $|p-p_n|<10^{-3}$ requires that $n>1000$.

I started to solve it by the following way:

$|p-p_n|<10^{-3}\Rightarrow |1-(1+\frac{1}{n})|<10^{-3}\Rightarrow |\frac{1}{n}|<10^{-3}\Rightarrow |n|>1000\Rightarrow n>1000$

And

$f(p_n)=[(1+\frac{1}{n})-1]=\frac{1}{n}$; whenever $n>1$

But how to show

$|f(p_n)|<10^{-3}$ ?

Body:
Show that $|f(pn)|<10^{-3}$ whenever $n>1$
$F(x)=(x-1)^{10}$ then $f(pn) =((1+1/n)-1)^{10} => |f(pn)| =(1/n)^{10}$
And $n>1$ then $|f(pn)| < = (1/2)^{10} = 1/1024$ and $1/1024 <10^{-3 }$


问:
Trigonometric definite integral $\int ^\frac{\pi}{2} _0 \cos^4(s)\mathrm{ds}-\int^\frac{\pi}{2}_0 \cos^6(s)\mathrm{ds} $
$$\int ^\frac{\pi}{2} _0 \cos^4(s)\mathrm{ds}-\int^\frac{\pi}{2}_0 \cos^6(s)\mathrm{ds}  $$

the answer is $\dfrac{3\cdot1}{4\cdot2}-\dfrac{5\cdot3\cdot1}{6\cdot4\cdot2}$

I really don't know why

There may be some tricks in this, but I have not seen it in the textbook

Body:
Define
$$I_n=\int_0^{\pi/2}\cos^{2n}(t)dt$$

Integrate by parts
$$I_n=\int_0^{\pi/2}\cos^{2n-1}(t)\cos(t)dt=\color{blue}{\cos^{2n-1}(t)\sin(t)\Big|_0^{\pi/2}}+(2n-1)\int_0^{\pi/2}\cos^{2n-2}(t)\sin^2(t)dt$$
and observe that the blue part is equal to zero.

Use $\cos^2(t)+\sin^2(t)=1$
$$(2n-1)\int_0^{\pi/2}\cos^{2n-2}(t)\sin^2(t)dt=(2n-1)\Big(\int_0^{\pi/2}\cos^{2(n-1)}(t)dt-\int_0^{\pi/2}\cos^{2n}(t)dt\Big)$$

Collect all results
$$I_n=(2n-1)(I_{n-1}-I_n)$$

Isolate $I_n$
$$I_n=\frac{2n-1}{2n}I_{n-1}$$

Compute $I_0, I_1, I_2, I_3$
$$I_0=\frac{\pi}{2},\quad I_1=\frac{1}{2}\frac{\pi}{2},\quad I_2=\frac{3\cdot 1}{4\cdot 2}\frac{\pi}{2},\quad I_3=\frac{5\cdot 3\cdot 1}{6\cdot 4\cdot 2}\frac{\pi}{2}$$

Compute $I_2-I_3$


问:
Normal subgroup if conjugate subgroup is subset
I find this explanation in Isaacs' Algebra:

Lemma. Let $H\subseteq G$ be a subgroup. Then $H$ is a normal subgroup if $H^g\subseteq H$ for all $g\in G$.

The reader should be warned that this lemma does not state that $H^g=H$ whenever $H^g\subseteq H$. Since the inner automorphism induced by the element $g$ is a bijection, it is certainly true that $|H^g|=|H|$, and if $H$ is finite, this equality of orders together with the containment $H^g\subseteq H$ certainly does imply that $H^g=H$. For infinite subgroups, however, this does not follow and is not generally true.

How does the lemma not state that? If $H^g\subseteq H$, then the lemma says that $H$ is a normal subgroup, which by definition means $H^g=H$. What am I missing?

答:
If $H^g\subseteq H$, then the lemma says that $H\trianglelefteq G$, which by definition means $H^g=H$.

With $g$ fixed, the condition $H^g\subseteq H$ does not imply $H^g=H$. It is true that if $H$ is normal, then the condition $H^g\subseteq H$ does imply $H^g=H$ (again, $g$ is some fixed element here), but it is arguable if it "follows from definition," specifically the definition the text gives, because the text's definition does not explicitly say anything about equality. Here is the proof:

Suppose $H\trianglelefteq G$ is a normal subgroup and $H^g\subseteq H$ for some $g\in G$. Since $H$ is normal, we have $H^{g^{-1}}\subseteq H$; conjugate both sides by $g$ to obtain $H\subseteq H^g$. Since $H^g\subseteq H$ and $H\subseteq H^g$, we have obtained equality $H^g=H$.

Also note that, again with $g$ fixed, the lemma does not say that $H^g\subseteq H$ implies $H\trianglelefteq G$. Rather, the lemma says that if $H^g\subseteq H$ holds true for all $g\in G$, then $H$ is normal. The inclusion $H^g\subseteq H$ holding for some isolated individual element $g\in G$ is not enough for $H$ to be normal.

But here is what the text is getting at. In general (without a priori knowledge about normality), the condition $H^g\subseteq H$ does not imply $H^g=H$. That is, there exist situations where you have a subgroup $H\le G$ that strictly contains one of its conjugates $H^g$, i.e. $H^g\subset H$. In this case the $H$ is not normal, since $H^g\subset H\implies H\subset H^{g^{-1}}\implies H^{g^{-1}}\not\subseteq H$.

Let's look at an example. Qiaochu gives a matrix example. Here is a semidirect product example. Take $\bf Q$ and $\bf Z$ to be additive groups, and define ${\bf Z}\to{\rm Aut}({\bf Q})$ so that $1\in{\bf Z}$ acts as e.g. the map $x\mapsto 2x$ on $\bf Q$. Then in $G={\bf Q}\rtimes{\bf Z}$, the subgroup $H=\langle1\rangle\subset \bf Q\hookrightarrow G$ conjugated by $1\in\bf Z$ is properly contained in $H$ - specifically, it is $2{\bf Z}\subsetneq H$.


问:
How to do the PFE of a function whose polynomial does not easily expand?
I want to do a PFE of

$$ y(x) = \frac{1}{ x^2 + \sqrt{2}x +1} $$

when I try to expand the polynomial I end up with

$$ y(x) = \frac{1}{(x + \sqrt{\frac{1}{2}})( x + \sqrt{\frac{1}{2}}) + \frac{1}{2}} $$

I would know how to proceed if the $+\frac{1}{2}$ would not be there. That is what is confusing me. I tried

$$ y(x) = \frac{A}{x + \sqrt{\frac{1}{2}}} + \frac{B}{x + \sqrt{\frac{1}{2}} + \frac{1}{2}} $$

but it is not equivalent.

答:
What you have is correct.  If you are trying to integrate, the next step would be to substitute u = x + $1/\sqrt(2)$, then substitute again w = $u\sqrt(2)$ to get an expression of the form $\frac{1}{w^2 + 1}$ which integrates to arctangent w.  If instead you are fine with complex numbers in your terms you can just use the quadratic formula as @Calvin Khor suggested.
EDIT TO ADD: the Laplace transform of $ \sin kt = \frac {k}{s^2 + k^2}$ so adjust constants, using shift theorem as needed: the Laplace transform of $ e^{at} \sin kt = \frac{k}{(s-a)^2 + k^2} $


问:
Definite Integration explanation
We know that definite integration provides us the area under the curve but I still don't understand one simple thing.

If we take
$\int f(x)dx = F(x)+ c$

Then we say $\int_a^b f(x)dx= F(b)-F(a)$

$i.e$ how do we know that the area under the curve is difference of values of the integrated function

答:
I think that instead of viewing it as "the area under the curve", you should view it as "an infinite sum of infinitesimal values".  Using it to find the area under the curve is just one usage.  Thinking of it as an infinite sum of infinitesimal values makes the process make more sense.

Let's start with a function:

$$ y = F(x)$$

Now, let's differentiate it without dividing by $dx$:

$$ dy = f(x)\,dx $$

$dy$ is the infinitesimal change in $y$.  Now, if our graph is smooth and continuous, then the total change in $y$ between two $x$ values is just going to be the sum of all of the infinitesimal changes between those two places.

So, we can write:

$$ \text{total change in }y = \int_{x = a}^{x = b} dy = \int_{x = a}^{x = b} f(x)\,dx $$

Now, interestingly, the "total change in $y$" is actually the same as $y$, but possibly offset by a constant.  That is, if we graphed the "change" in $y$ from a given starting point, it would give us the same graph as $y$ itself, just with a possible vertical offset.

Therefore:

$$ \text{total change in }y = y + C = \int dy = \int f(x)\,dx $$

So, if we just pick two pieces of this puzzle that we are interested in:

$$ y + C = \int f(x)\,dx $$

The big question is, why does the value on the right give the area under the curve of $f(x)$?  The answer is that if we break the graph up into rectangles, then each one will be $f(x)$ high, and $dx$ wide. So $f(x)\cdot dx$ is the area of each rectangle, and so $ \int f(x)\,dx $ is simply the sum of all of them.  The ones we are interested in are those from $x = a$ to $x = b$.

So why the difference between $F(a)$ and $F(b)$?  Well, since $y + c$ is equal to the "total change in $y$", then the difference between two points in $y$ is going to give us that total change.  Since we have a formula for $y$ based on points in $a$ and $b$, ($F(x)$), we can use that formula.  I.e., since $y = F(x)$ (our original given),

$$ \text{total change in }y_{x = a}^{x = b} = (y + C) |_{x = a}^{x = b} = \int_{x = a}^{x = b} dy = \int_{x = a}^{x = b} f(x)\,dx \\
F(b) - F(a) = \int_{x = a}^{x = b} f(x)\,dx
$$

Title: Definite Integration explanation


问:
Why is $\left(\sum_{n=0}^{+\infty} \frac{\left(-x^{4}\right)^{n}}{n !}\right) 2 x=\sum_{n=0}^{+\infty}(-1)^{n} \frac{2 \cdot x^{4 n+1}}{n !}$?
How do we integrate $2x$ on the left into the sum such that $\left(\sum_{n=0}^{+\infty} \frac{\left(-x^{4}\right)^{n}}{n !}\right) 2 x=\sum_{n=0}^{+\infty}(-1)^{n} \frac{2 \cdot x^{4 n+1}}{n !}$?

答:
For every $N\in\mathbb{N}$ You have by the distributive law:
$$\left(\sum_{n=0}^{N}\frac{(-x^4)^n}{n!}\right)2x=\sum_{n=0}^{N}\frac{(-x^4)^n2x}{n!}=\sum_{n=0}^{N}(-1)^n\frac{2x^{4n+1}}{n!}.$$


问:
Given $P(D)(x)=\sin(t)$ Prove that the equation has unique periodic solution.

Given $P(D)(x)=\sin(t)$ where $P(\lambda)$ is a real value polynom and it is known that all the roots of the polynom has a negative real part $Re(\lambda_i)<0$.

Prove that the equation has unique periodic solution.

My try if I understood the questions right.

First solving the homog. part and getting the solution of the form $C_ie^{\lambda_i t}$ and We wish that all the eigenvalues are complex to get homog. solution of the form $\sum_i e^{\alpha t}(c_{1i}\cos(\beta t)+c_{2i}\sin(\beta t))$ and from the undetermined coeff because $\pm i$ is not a root a particular solution is of the form $A\cos(t)+B\sin(t)$ but I don't see how can I say that the solution is periodic if I have a real part which is not zero in the eigenvalues

答:
You are already at the finish line. Finding that $A\cos(t)+B\sin(t)$ for some $A,B\in \Bbb R$ is a solution already provides the demanded periodic solution. The claim is not that every solution is periodic. Any other solution has a complementary part added to the particular solution, and the complementary solutions are, as you observed, not periodic.

You should know that $P(D)[e^{λt}]=P(λ)e^{λt}$. Then you could also obtain the particular and periodic solution from $\sin(t)=Im(e^{it})$ as
$$
y_p(t)=Im\left(\frac{e^{it}}{P(i)}\right)
$$


问:
Closed subspace consisting of continuous functions in $L^{2}([0,1])$ is finite-dimensional
I wish to show that any closed subspace $M$ of $L^{2}[0,1]$ consisting of continuous real-valued functions is finite-dimensional (which amazes me quite a bit, I must admit).

My approach proceeds as follows:

Step 0. Note that $L^{2}[0,1]$ is a separable Hilbert space with the usual scalar product on $L^2$. Moreover, since $M$ is closed, it is separable Hilbert, too, with inner product of $L^2([0,1])$ restricted to $M$.

Step 1. Let $I: (M, \| \cdot \|_{\infty}) \to (M, \| \cdot \|_{2})$ be the identity operator on $M$. Since $I$ is bijective, by the open mapping theorem, $I^{-1}$ exists in the space of bounded linear operators on $M$. Hence, for any $f \in M$, we have $$ \| f\|_{\infty} = \| I^{-1} (f)\|_{\infty} \leq C \|f \|_{2} \quad \forall f \in M,$$
where $C := \| I^{-1} \|.$

Step 2. Now, let $t \in [0,1]$ and consider the linear functional $\ell_{t}: M \to \mathbb{R}, f \mapsto f(t).$ As $M$ is Hilbert, we may apply Riesz representation theorem to deduce that $\ell_{t} = \langle \cdot, g_t \rangle =: \ell_{g_t}$ for some $g_t \in M$ and $\|\ell_{g_t}\| = \| g_t\|_{2}$. Moreover, $$ f(t) = l_t (f) =  \langle f, g_t \rangle \quad \forall t \in [0,1].$$
Using step 1, we may further conclude that $\| g_t \|_{2} \leq C$.

Step 3. By seperability of $M$, we may suppose that $S = \{h_{n}: n \in \mathbb{N}\}$ is a orthonormal basis of $M$. Then, Parseval's identity yields

$$
\infty > \|g_t\|_{2} = \sum_{h \in S} | \langle h, g_t \rangle |^{2} = \sum_{h \in S} |h(t)|^{2}
$$

I am stuck in the last line. I want to argue that this sum has to be finite. Can we bound $|h(t)|$ from below? I guess, this suffices to conclude, right? Does this argument work?

答:
In fact, if $M$ is a closed subspace of $L^2(X)$ for some finite measure space $X$, and $M\subset L^{\infty}(X)$, then $M$ is finite dimensional.

First note that step $1$ doesn't change.: we still have $\|f\|_2\leq \mu(X)^{1/2}\|f\|_{\infty}$ for $f\in L^{\infty}(X)$ which 1) guarantees that $M$ is Banach with the $\|\cdot\|_\infty$ norm and 2) that the identity map $(M,\|\cdot\|_\infty)\rightarrow (M,\|\cdot\|_2)$ is continuous. So the bounded inverse  theorem gives a constant $C$ such that $\|f\|_\infty\leq C\|f\|_2$, $f\in M$.

For the remainder of the proof we have to do somehting different. Take an orthonormal system $\{f_k\}_{k=1}^n$ in $M$. Then for any $c_1, \ldots, c_n \in \mathbb{C}$ the inequality $\|f\|_\infty\leq C\|f\|_2$ for the function $f=c_1 f_1+ c_2f_2+\ldots +c_n f_n$ means that
$$|c_1f_1(x)+\ldots+c_nf_n(x)|^2\leq C^2(|c_1|^2+\ldots +|c_n|^2)\; \; \; \; \; \;(1)$$
for a.e. $x$. Choosing a countable dense subset $\Omega\subset \mathbb{C}^n$ we can find a measurable set $Y$ with $\mu(X\backslash Y)=0$ such that $(1)$ holds for all $x\in Y$ and $(c_1, \ldots, c_n)\in \Omega$. By continuity $(1)$ then holds for all $y\in Y$ and any scalars $c_1, \ldots, c_n \in \mathbb{C}$. In particular we can set $c_k=\overline{f_k(x)}$ to get
$$|f_1(x)|^2+\ldots +|f_n(x)|^2\leq C^2$$
for all $x\in Y$. Now we simply integrate both sides over $X$ (or, equivalently, over $Y$) to get $n\leq C^2\mu(X)$. Hence we have a uniform bound on the size of any orthonormal system in $M$ which implies that $M$ is finite dimensional.


问:
Non-Metrizability of Compact-Open Topology
Let $X$ be a compact metric space and let $Y$ be a non-metrizable topological space.  How can we show that $C(X,Y)$ with compact-open topology is non-metrizable?

I was thinking of embedding $Y$ into $C(X,Y)$ via $y\mapsto E_y$ where $E_y$ is the evaluation map $f\mapsto f(y)$.  Then arguing by contradiction.  However, that's as far as I've gotten.

答: Instead of the evaluation map, which maps $C(X,Y)$ to $Y$, use the map $c$ which maps $y \in Y$ to the constant map $c(y):X\to Y$, with value $y$. This is an embedding of $Y$ into $C(X, Y)$. It’s well-defined as constant maps are continuous and $c$ is clearly 1-1 as well. Just check continuity of $c$ and its openness as a map from $Y$ onto its image. ($c^{-1}[[K,O]] = O$ and $c[O] = [X,O] \cap c[Y]$ e.g.)


问:
Prove the existence of $\Psi\in(\ell_\infty(G))^\ast$ satisfying the amenability conditions
Let $G$ be a finitely generated group satisfying the Folner condition and let $S\subset G$ be a finite generating set of $G$. Denote by $\rho=\{\rho_g\}_{g\in G}$ the right translation action of $G$ on $\ell_{\infty}(G)$ defined by
$$[\rho_g(\zeta)](k)=\zeta(kg), \ \ \zeta\in\ell_\infty(G),k\in G.$$
Denote by $\lambda$ the left translation action of $G$ on $\ell_\infty(G)$. Prove that there exists $\Psi\in(\ell_\infty(G))^\ast$ satisfying the following conditions:

$\Psi(1)=1$, where $1$ on the left hand side is the constant function $1$ on $G$,

$\left\|\Psi\right\|=1$,

$0\leq\Psi(\zeta)$ for every non-negative real-valued bounded function $\zeta$ on $G$,

$\Psi(\lambda_g\zeta)=\Psi(\zeta)$ for every $g\in G$ and every $\zeta\in\ell_\infty(G)$,

$\Psi(\rho_g\zeta)=\Psi(\zeta)$ for every $g\in G$ and every $\zeta\in\ell_\infty(G)$.

Attempt: Already proved that there exists $\Phi\in(\ell_\infty(G))^\ast$ satisfying $1,2,3, 4$.

Problem: I want to show that for an arbitrary non-empty finite subset $F\subset G$, the linear functional $\Psi_F\in(\ell_\infty(G))^\ast$ defined by
$$\Psi_F(\zeta)=\frac{1}{\#F}\sum_{h\in F}\Phi(\rho_h\zeta), \ \ \zeta\in\ell_\infty(G)$$
also satisfies 1,2,3,4. If $F$ is an $(\varepsilon,S)$-Folner set, then $\Psi_F$ satisfies an inquality. By Banach-Alaoglu theorem, we can obtain something that is essential for the proof.

答:
Here is one way to prove (4):

$$
  \Psi _F(\lambda _g\zeta ) =
  \frac{1}{\#F}\sum _{h\in F}\Phi (\rho _h\lambda _g\zeta ) {\ \buildrel {(i)} \over {=}\ }
  \frac{1}{\#F}\sum _{h\in F}\Phi (\lambda _g\rho _h\zeta ) {\ \buildrel {(ii)} \over {=}\ }
  \frac{1}{\#F}\sum _{h\in F}\Phi (\rho _h\zeta ) =
  \Psi _F(\zeta ).
  $$

Notes:

(i) Use that $\rho _h$ and $\lambda _g$ commute.  This is a simple consequence of the associativity axiom of $G$.

(ii) Use that $\Phi $ is invariant under $\lambda _g$.


问:
How to show that $r_1, r_2,.., r_k$ are linearly independent
Prove that if $R$ is a matrix in echelon form , then a basis  for $\mathcal{R}(R)$ consists of the non zero rows  of $R$

My attempt  : For basis we must show the span and linearly independent properties

let $r_1,r_2 ,..., r_k$ be the non zero rows of $R$ starting from the $1$st row to the  $k-$th row

now we will prove  that $\mathcal{R}(A)= \operatorname{span}  \{ r_1,r_2,....., r_k\}$

for any vector  $v \in \mathcal{R}(A)$, we can write

$ v= c_1r_1 +....+c_kr_k +c_{k+1}0 +......+c_m0= c_1r_1 +....+c_kr_k$

so $v \in \operatorname{span}\{r_1,r_2,........,r_k\}$

Therefore $\mathcal{R}(R) \le\operatorname{span} \{r_1,....,r_k\}$

now trivially we also have $\operatorname{span}\{r_1,....,r_k\} \le  \operatorname{span}\{r_1,r_2,...,r_k,0,0,...,0\} =\mathcal{R}(R)$

This implies  that $\operatorname{span}\{r_1,....,r_k\}=\mathcal{R}(R)$

My doubts : How to show that  $r_1, r_2,.., r_k$ are linearly independent

答:
linear independence By echelon form, you have ordered the rows by the number of zeros on the left. This number of zeros is strictly increasing. Take they order from the maximum number of zeros to the minimum (so in your row-echelon matrix you go from the lowest one up to the first row).

Let say you have $r_1,\dots, r_m$ non-zero rows. Now $r_1\neq 0$ so it is lineary independ. Now you add $r_2$, this can't by scalar multiple of $r_2$ because it have non-zero on the position where the $r_1$ has zero.

Now you have proved that $r_1,\dots r_k$ are independ. $r_{k+1}$ has non-zero on the position where all $r_1,\dots r_k$ have zero so it cant be spanned by them.


问:
What is Inverse Fourier Transform of $2^2/(^2−^2)$?
What is the inverse Fourier transform of  $F(w) = \frac{4}{4+(j2\pi f)^2} $?
 I have two suggested solutions:

Assume $j2\pi f = \omega$ and use the standard transform
$e^{-\alpha |\tau |} = \frac{2\alpha}{\alpha ^2 + \omega ^2}$ to obtain $\frac{1}{4}e^{-2|\tau|}$

Calculate $f(\tau) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2\alpha}{\alpha^2-\omega^2}e^{i\omega\tau}d\omega$

For (2), how can I solve the integral in order to get the inverse Fourier transform $f(\tau)$?

答:
$\newcommand{\eqd}{\triangleq}$
$\newcommand{\eqa}{\approx}$
$\newcommand{\abs}[1]{{\left\lvert #1 \right\rvert}}$
$\newcommand{\brp}[1]{{\left(#1\right)}}$
$\newcommand{\brs}[1]{{\left[#1\right]}}$
$\newcommand{\brlr}[1]{\left.#1\right|}$
$\newcommand{\deriv} [2]   {{\frac{\mathrm{d}#1}{\mathrm{d}#2} }}$
$\newcommand{\R}{\Bbb{R}}$
$\newcommand{\intcc} [2]  {{\left[#1:#2\right]}}$
$\newcommand{\intoo} [2]  {{\left(#1:#2\right)}}$
$\newcommand{\intoc} [2]  {{\left(#1:#2\right]}}$
$\newcommand{\intco} [2]  {{\left[#1:#2\right)}}$
$\newcommand{\ff}{\mathrm{f}}$
$\newcommand{\fF}{\mathrm{F}}$
$\newcommand{\fG}{\mathrm{G}}$
$\newcommand{\fg}{\mathrm{g}}$
$\newcommand{\fphi}{\mathrm{\phi}}$
$\newcommand{\dx}{\mathrm{dx}}$
$\newcommand{\dtau}{\mathrm{d\tau}}$
$\newcommand{\du}{\mathrm{du}}$
$\newcommand{\dv}{\mathrm{dv}}$

It looks like there is some problem with the answer in "1."
First note that it is indeed true that the Fourier Transform $\fG(\omega)$ of
the function $\fg(\tau)\eqd e^{-\alpha\abs{\tau}}$ is
$\fG(\omega)\eqd\frac{2\alpha}{\alpha^2+\omega^2}$.
Assuming the definition of the Fourier Transform used is
$\fG(\omega)\eqd \int_{-\infty}^{\infty}\fg(\tau)e^{-j\omega\tau}\dtau$,
we can check the relation as follows:
\begin{align}
  \fG(\omega)
    &\eqd \int_{-\infty}^{\infty}\fg(\tau)e^{-j\omega\tau}\dtau
    && \text{by definition Fourier Transform}
  \\&\eqd \int_{-\infty}^{\infty} e^{-\alpha\abs{\tau}} e^{-j\omega\tau}\dtau
    && \text{by definition of $\fg(\tau)$}
  \\&= \int_{-\infty}^{0} e^{-\alpha(-\tau)} e^{-j\omega\tau}\dtau
     + \int_{0}^{\infty} e^{-\alpha(\tau)} e^{-j\omega\tau}\dtau
  \\&= \int_{-\infty}^{0} e^{\tau(\alpha-j\omega)}\dtau
     + \int_{0}^{\infty} e^{\tau(-\alpha-j\omega)}\dtau
  \\&= \brlr{\frac{e^{\tau(\alpha-j\omega)}}{\alpha-j\omega}}_{-\infty}^{0}
     + \brlr{\frac{e^{\tau(-\alpha-j\omega)}}{-\alpha-j\omega}}_{0}^{\infty}
    && \text{by Fundamental Theorem of Calculus}
  \\&= \brs{\frac{1}{\alpha-j\omega} - 0}
     + \brs{0- \frac{1}{-\alpha-j\omega}}
  \\&= \brs{\frac{1}{\alpha-j\omega}}\brs{\frac{\alpha+j\omega}{\alpha+j\omega}}
     + \brs{\frac{1}{\alpha+j\omega}}\brs{\frac{\alpha-j\omega}{\alpha-j\omega}}
    && \text{(to rationalize the denominator)}
  \\&= \frac{\alpha+j\omega}{\alpha^2+\omega^2}
     + \frac{\alpha-j\omega}{\alpha^2+\omega^2}
  \\&= \boxed{\frac{2\alpha}{\alpha^2+\omega^2}}
\end{align}

And so the Inverse Fourier Transform (IFT) of $\frac{2\alpha}{\alpha^2+\omega^2}$
is $e^{-\alpha\abs{\tau}}$.

However, your question does not ask for the IFT of
$\frac{2\alpha}{\alpha^2+\omega^2}=\frac{2\alpha}{\alpha^2+(2\pi f)^2}$,
but rather for the IFT of
$$
  \frac{2\alpha}{\alpha^2+(j2\pi f)^2}
    = \frac{2\alpha}{\alpha^2+(j\omega)^2}
    =\frac{2\alpha}{\alpha^2-\omega^2}
$$


问:
Expressing a projective conic in standard form
I was given the task to find a projective transformation $\phi: \mathbb{P}_\mathbb{C} \rightarrow \mathbb{P}_\mathbb{C}$ that transforms the following conic:

$$2\,{x}^{2}+2\,xy-3\,xz+4\,yz+{z}^{2}=0$$

into the form:

$$X^2 + Y^2 + Z^2 = 0$$

Then my attempt to do so was to first find the matrix of the symmetric bilinear form which is given by:

$$ \left[ \begin {array}{ccc} 2&1&-3/2\\ 1&0&2
\\ -3/2&2&1\end {array} \right]
$$

Then I wanted to orthogonally diagonalise this matrix to obtain an expression that doesn't have any cross terms in it. However the eigenvectors of this matrix are not "nice" (I've checked with Maple and they involve a lot of square roots).

In the solution given to me, the matrix that expresses this conic in standard form was given by the following matrix:

$$  \left[ \begin {array}{ccc} 1/2\,\sqrt {2}&1/10\,\sqrt {10}&-1/3\,
\sqrt {3}\\ 0&1/10\,\sqrt {10}&-i/6\sqrt {3}
\\ 0&1/5\,\sqrt {10}&i/3\sqrt {3}\end {array}
 \right]
$$

I can see that this matrix is orthogonal, but how does it simplify the expression of the conic above? And how do I find it from scratch? And how do this in general?

答:
We can compute the Jordan decomposition:
$$
\begin{align}
M
&=\begin{bmatrix}
2&1&-\frac32\\
1&0&2\\
-\frac32&2&1
\end{bmatrix}\tag{1a}\\[9pt]
&=PDP^{-1}\tag{1b}
\end{align}
$$
where
$$
D=\begin{bmatrix}
 -2.25995870076 & 0 & 0 \\
 0 & 2.10129990875 & 0 \\
 0 & 0 & 3.15865879201
\end{bmatrix}\tag2
$$
and
$$
P=\begin{bmatrix}
 0.377481248580 & 0.604631637473 & -0.701376140125 \\
 -0.701739221400 & 0.680975120463 & 0.209367978594 \\
 0.604210205243 & 0.413150660507 & 0.681349073238
\end{bmatrix}\tag3
$$
We compute the Jordan decomposition by computing the eigenvalues and eigenvectors of $M$; note that $\det(M-I\lambda)=-\frac14\left(4\lambda^3-12\lambda^2-21\lambda+60\right)$. Since $MP=PD$, the columns of $P$ are eigenvectors of $M$ (so they are orthogonal) and the eigenvalues are the diagonal elements of $D$. Normalize the columns of $P$ (that is, pick unit eigenvectors) so that $P^TP=I$.

Set
$$
\begin{align}
\begin{bmatrix}
u\\v\\w
\end{bmatrix}
&=P^T
\begin{bmatrix}
x\\y\\z
\end{bmatrix}\tag{4a}\\
&=\begin{bmatrix}
 0.377481248580 & -0.701739221400 & 0.604210205243 \\
 0.604631637473 & 0.680975120463 & 0.413150660507 \\
 -0.701376140125 & 0.209367978594 & 0.681349073238
\end{bmatrix}
\begin{bmatrix}
x\\y\\z
\end{bmatrix}\tag{4b}
\end{align}
$$
Then
$$
\begin{align}
\begin{bmatrix}
u\\v\\w
\end{bmatrix}^T
D
\begin{bmatrix}
u\\v\\w
\end{bmatrix}
&=
\begin{bmatrix}
x\\y\\z
\end{bmatrix}^T
PDP^T
\begin{bmatrix}
x\\y\\z
\end{bmatrix}\tag{5a}\\
&=
\begin{bmatrix}
x\\y\\z
\end{bmatrix}^T
M
\begin{bmatrix}
x\\y\\z
\end{bmatrix}\tag{5b}
\end{align}
$$
That is,
$$
\begin{align}
&-2.25995870076u^2+2.10129990875v^2+3.15865879201w^2\\
&=2x^2+2xy-3xz+4yz+z^2\tag6
\end{align}
$$


问:
find the number of injections/surjections
I've been banging my head on this problem for some time now, and could really use help. Bear in mind I'm not very good at this sort of thing and am struggling to get by in class.

Problem:

Given $ \alpha : A \to B       \;\;\;\;\;   |A|=p \;\;\;\;|B|=q $

Find:

Number of Injections: $A \to B$

Number of Surjections: $ A \to B$

Prove that ∃ a bijection $A \to B$

$\Rightarrow$ p = q

find number of bijections: $A \to B$

I apologize for dropping such a big problem on you all, but I just can't get anywhere with this. Help?

EDIT:
These answers are really helpful and I believe I've got down the number of Injections, but  we haven't gone through permutations or combinatorics in class so my understanding of it is really hamstrung. I'm unclear on the Stirling theorem as discussed here: http://www.ma.utexas.edu/users/kbi/COURSES/TERM/11S/325K/L17.pdf, specifically the part where inclusion-exclusion comes into play. I'm not sure how to apply that, since right now I'm letting p = 4 and q = 3, so I don't know where inclusion-exclusion would be used (if at all!) since I don't know how elements would be missing.

答:
The number of bijections is given by $n!$, in which $n$ denotes the common cardinality of your sets.  To see this, note that any bijection can be written as a permutation followed by a given bijection.

An injection is a bijection onto its image.  Thus you can find the number of bijections by counting the possible images and multiplying by the number of bijections to said image.  In your notation, this number is
$$\binom{q}{p} \cdot p!$$

As others have mentioned, surjections are far harder to calculate.  They satisfy certain recursions, e.g. given here.  The number of surjections is also related to the Stirling numbers (of the second kind.


问:
Prove that two subspaces of a vector space intersect only at 0
Let $V \subset \mathbb{R}^n$ & $W \subset \mathbb{R}^m$ with set of basis $S_V=\{v_1,v_2,...,v_n\}$ and $S_W=\{w_1,w_2,...,w_m\}$.
The vector space spanned by these basis vectors is the direct sum of $V$ and $W$ with a dimension of $n+m$.

How do I prove that $V \cap W=\{0\}$?

I understand how to draw it conceptually because the only intersection that both subsets share is the zero vector. But I don't quite know how to do the proof mathematically.

答:
Since $V$ and $W$ are subspaces of different vector spaces, so $V\cap W=\emptyset$. But we can try to make the question meaningful by introducing the maps
\begin{align}
F\colon V\subset\mathbf{R}^m&\to\mathbf{R}^{m+n}\\
v&\mapsto(v,\underbrace{0,\ldots,0}_{n\text{ copies}})
\end{align}
and
\begin{align}
G\colon W\subset\mathbf{R}^n&\to\mathbf{R}^{m+n}\\
w&\mapsto(\underbrace{0,\ldots,0}_{m\text{ copies}},w)
\end{align}
$F$ and $G$ are injective and if we define $V':=F(V)\subset\mathbf{R}^{n+m}$ and $W':=G(W)\subset\mathbf{R}^{n+m}$, the maps $V\ni v\mapsto F(v)\in V'$ and $W\ni w\mapsto G(w)\in W'$ are vector space isomorphisms. You can easily prove that $V'\cap W'=\{0\}$ and if $V$ and $W$ span $\mathbf{R}^m$ and $\mathbf{R}^n$, respectively, then $\mathbf{R}^{n+m}=V'\oplus W'$.


问:
,For $y=x^2+1$ and $x=y^2+1$, if PQ is the shortest distance and R,S are points of contact of common tangent, find area PQRS.
From equation, it is evident that curves are symmetric about $y=x$

So I tried writing the equations of tangents
$$k=mh +\frac{1}{4m}$$ and $$h=mk-\frac{m^2}{4}$$

Where $(h,k)$ is a point on curve 1

But on plugging $m=1$ the lines turn out to be coincident, and not two separate parallel lines. I was hoping to find $(h,k)$ by solving these two equations but that clearly isn’t working. What should I do?

答:
Per the symmetry with respect to $y=x$, the tangent is $1$ at $P $ and $-1$ at $R$. Then, set $y’=(x^2+1)’=2x  =\pm 1$ to obtain $P(\frac12,\frac54)$ and $R(-\frac12,\frac54)$. As a result, $PR=1$, their distances to the symmetry line $y=x$ are respectively $\frac3{4\sqrt2}$ and $\frac7{4\sqrt2}$, and
the height of the trapezoid $PQRS $ is $\frac{PR}{\sqrt2} $.  Thus, the area is
$$ 2\cdot \frac12(\frac3{4\sqrt2}+ \frac7{4\sqrt2}) \frac1{\sqrt2}=\frac54
$$


问:
Prove that the minimum of a functional doesn't exist
Prove that there is no smooth solution ho the minimization problem:

$$\mathcal{L} (u)= \int_{0}^1 e^{-u'}+u^2 dx$$

Where the admissible space is $X =\{ u \in \mathcal{c}^2 [0,1] | u(0)=0,  u(1)=1
       \} $

UPDATE:

GOAL: I am trying to define a sequence of functions $u_n(x)$ in $X$ s.t their integrals $\mathcal{L}(u_n) \to 0$

What I have done so far:

Define $u_n(x)$ in the following manner:
$$u_n (x)= \begin{cases}
0, \quad 0 \leq x\leq 1-\frac{1}{n} \\[2ex]
ax^2 +bx+ c, \quad 1-\frac{1}{n}<x \leq 1
\end{cases}$$

where $n=1,2,3,..$. Require that $u_n$ satisfy:
\begin{align}
 u_n(1-\frac{1}{n})=0\\
 u_n(1)=1\\
 u_n'(1-\frac{1}{n})=0\\
\end{align}

We then have:
\begin{align}
a+b+c=1\\
a(1-\frac{1}{n})^2+b(1-\frac{1}{n})+c=0\\
2a(1-\frac{1}{n})+b=0
\end{align}

Find $a,b,c$ then:

$$u_n (x)= \begin{cases}
0, \quad 0 \leq x\leq 1-\frac{1}{n} \\[2ex]
n^2x^2 -2n(n-1)x+ (n-1)^2, \quad 1-\frac{1}{n}<x \leq 1
\end{cases}$$

\begin{align}
\int_0^1 e^{-u_n(x)'} + u_n^2(x) dx &= \int_0^{1-\frac{1}{n}} e^{0} + 0 dx +  \int_{1-\frac{1}{n}}^1 e^{-(2n^2 x -2n(n-1))} + (n^2x^2 -2n(n-1)x+ (n-1)^2)^2 dx\\& = 1- \dfrac{1}{n}+ \dfrac{\mathrm{e}^{-2n}\left(\left(2n+5\right)\mathrm{e}^{2n}-5\right)}{10n^2}
\end{align}

$$\dfrac{\mathrm{e}^{-2n}\left(\left(2n+5\right)\mathrm{e}^{2n}-5\right)}{10n^2} \to 0$$ and $$1-\dfrac{1}{n} \to 1$$

答:
We can more or less use the same proof technique as in my Math.SE answer here.

Define function $$f(a)~:=~(a+1)e^{-a}.\tag{1}$$
Now it is a fact that
$$ \exists ! a_0 <0:~~ f(a_0)~=~0. \tag{2}$$
$$ \forall a> a_0: ~~ 0~<~f(a)~\leq~1.\tag{3}$$
$$ \forall a< a_0: ~~ f(a)~<~0.\tag{4}$$
Here $a_0=-1$. There is a first integral to OP's EL equation. Together with OP's boundary conditions, it becomes $$  f(u^{\prime}(0))~=~ f(u^{\prime}) + u^2
~=~f(u^{\prime}(1))+1 .\tag{5} $$
In light of eqs. (2)-(4) we see that eq. (5) is impossible to satisfy for $u^{\prime}(1)>a_0$ because then $f(u^{\prime}(1))>0$ while $f(u^{\prime}(0))\leq 1$.

The EL equation reads
$$ \left(-e^{-u^{\prime}}\right)^{\prime}~=~2u.\tag{6} $$
If in the interval $[x_0,1]$ the function $u > 0$ is positive, then $-e^{-u^{\prime}}$ (and thereby $u^{\prime}$) is increasing, and hence $u^{\prime}\geq 0$. Then
$$u^{\prime}(1)~\geq~ 0,\tag{7}$$
which according to section 1 leads to a contradiction. $\Box$


问:
Tools for Solving Homogeneous Diophantine Equations of Degree 2
Wikipedia describes only one way to prove that a homogeneous diophantine equation of degree 2 has no trivial solution (namely by reducing the equation modulo p).

My toolbox for solving homogeneous diophantine equations of degree 2 currently breaks down to the following methods:

Try out some values for x,y,z. Hope that x,y,z is a non-trivial solution.

Try out some values for p. Reduce the equation modulo p and hope that you can construct a prove similiar to the corresponding wikipedia page.

These approaches are obviously depending on luck. What other tools do exist?

答:
It is a theorem (the so called Hasse Principle for genus 0 curves) that if $Q(x,y,z) \in \mathbb{Q}[x,y,z]$ is a homogeneous degree $2$ polynomial, then $Q(x,y,z) = 0$ has a solution in $\mathbb{Q}$ if and only if it has a solution in $\mathbb{Q}_p$ for all primes $p$ (roughly speaking (by Hensel's lemma) this is the same as saying there are solutions mod $p$), and a solution in $\mathbb{R}$.

That is to say, this is the only tool one needs when trying to investigate such curves is reduction mod $p$, and the information from $\mathbb{R}$.

Note that in general we do not get nearly as lucky, and a lot of modern number theory is concerned with investigating the Hasse Principle for examples of varieties.


问:
Probability on product spaces
I am having some trouble, more of an argument with someone else, about a simple question regarding product spaces.

Let $X_1,X_2,\dots,X_n$ a set of independent and identically distributed random variables from a population $P\in\mathcal{P}$, where $\mathcal{P}$ is a family of probability measures (non explicitly parameterized). The generic random variable $X$ is a measurable function from a probability space $\left(\Omega,\mathcal{F},P'\right)$ to $(\mathbb{R},\mathcal{B}(\mathbb{R}))$.

In order to build the random vector containing the elements of the sample, $\tilde{X}=[X_i]_{n\times 1}$, and be able to calculate probability measures of events like $\{X_1<X_2\}$ or $\{X_1=X_2\}$, I decided to build a product space $(\Omega^n=\Omega\times\Omega\dots\times\Omega,\sigma(\mathcal{F}^ n)=\sigma(\mathcal{F}\times\mathcal{F} \dots,\times\mathcal{F}),P=P'\times P'\times\dots\times P')$, and let the vector function $\tilde{X}$ go from this product space to $(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n))$.

Therefore, the events

$\{X_1=X_2\}=\{(w_1,w_2,\dots,w_n)\in\Omega^n:X_1(\omega_1)=X_2(\omega_2)\}$
and

$\{X_1<X_2\}=\{(w_1,w_2,\dots,w_n)\in\Omega^n:X_1(\omega_1)<X_2(\omega_2)\}$

have a clear meaning.

However, the person I am having the argument with argues that

$\tilde{X}$ goes from the original probability space $\left(\Omega,\mathcal{F},P'\right)$ to $(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n))$, and

If I keep doing it the way of product spaces it is a lot harder to calculate the probabilities I want.

If my friend's argument is true, I am having trouble finding the meaning of, and calculating the probabilities of the events above.

What am I missing?

Best regards,

JM

答:
You're on the right track. The sample space you want is the product probability space $\prod_1^n (\mathbb{R}, \mu_n)$ where $\mu_n$ is the push-forward measure on $\mathbb{R}$ given by $X_n$. The random variable $X_m$ is then the $m$-th coordinate projection from  $\prod_1^n (\mathbb{R}, \mu_n)$.

The generalization of this construction to arbitrary family of random variables is due to Kolmogorov, I believe.

Title: Probability on product spaces


问:
What is the smallest extension field of $Q$ that contains this radical?
for a radical like $\frac{2+\sqrt[11]{1+\sqrt[5]{3}}}{3}$, what is the smallest extension field of $Q$ that contains it?

I'm not sure how to deal with radicals inside radicals

答:
If this is $\alpha$, then teh smallest extension field is clearly just
$$\Bbb Q[\alpha].$$
By transforming the defining expression,
$$ \bigl((3\alpha-2)^{11}-1\bigr)^5=3,$$
i.e., $\alpha$ is a root of some complicated degree 55 polynomial.
Once we note that $\beta:=3\alpha-2\in\Bbb Q[\alpha]$ and $\alpha\in\Bbb Q[\beta]$, we can look for $\beta$ instead of $\alpha$. As
$$(\beta^{11}-1)^5=3, $$
its polynomial is less horrible:
$$  X^{55} - 5X^{44} + 10X^{33} - 10X^{22} + 5X^{11} - 4$$
No show that is polynomial is irreducible and conclude that
$$ \Bbb Q[\alpha]\cong \Bbb Q[X]/(X^{55} - 5X^{44} + 10X^{33} - 10X^{22} + 5X^{11} - 4).$$


问:
compute $\iint_Y F.N\ dS $ with Gauss
The question is:
$$ \iint_Y F.N\ dS \quad Y=(x-z)^2+(y-z)^2=1+z^2,\ \  0\leq z\leq 1 \quad N=\text{pointing outward}
$$ $$ F=(y,x,1+x^2z)$$
Here is how I have tried to solve it: I tried to solve it with Gauss theorem like this, $\gamma=\text{upper lid}$, $\sigma=\text{lower lid} $
$$ \iint_{Y+\sigma+\gamma}F.N\ dS=\iiint_K \text{div}f\  dA \iff \iint_Y F.N\ dS=\iiint_k-\iint_\sigma-\iint_\gamma$$
$$ \iint_\sigma=-\pi, \iint_\gamma=2+\pi$$
But I have trouble to integrate this triple integral
$$ \iiint_kx^2 dxdydz$$
I don't know if I have been  calculating right so far, and how should I proceed? Any suggestion would be great, thanks

答:
Surface is $(x-z)^2+(y-z)^2 = 1 + z^2$

$\vec F = (y,x,1+x^2z)$

To apply divergence theorem, we close the surface with discs at $z = 0$ and at $z = 1$.

a) Flux through disc at $z = 1$,

Plugging in $z = 1$ in the equation of hyperboloid, we get the equation of the disc as

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

In polar coordinates, $x = 1 + \rho \cos\theta, y = 1+\rho \sin\theta, z = 1 \ (0 \leq \rho \leq \sqrt2, 0 \leq \theta \leq 2\pi)$

Outward normal vector is $(0, 0, 1)$.

So the integral to find flux becomes

$\displaystyle \int_0^{2\pi} \int_0^{\sqrt2} [1+(1+ \rho \cos\theta)^2] \  \rho \ d\rho \ d\theta = 5 \pi$

b) Flux through the disc at $z = 0$ is simply the area of the disc with negative sign as $\vec F \cdot (0,0,-1) = -1$. So it is $-\pi$ as you correctly calculated.

c) Now to find flux through the closed surface, apply divergence theorem.

$\nabla \cdot \vec F = x^2$ as you mentioned.

Now the region for triple integral is a hyperboloid which is difficult to calculate without a Jacobian. So as the other answer suggests,

use change of variable $x = u + w, y = v + w, z = w$. Its Jacobian is $1$.

So the transformed region is $u^2 + v^2 - w^2 = 1$. In cylindrical coordinates,

$u = \rho \cos \theta, v = \rho \sin \theta, w$

$0 \leq \rho \leq \sqrt{1+w^2}, 0 \leq w \leq 1, 0 \leq \theta \leq 2\pi $

Integrand $x^2 = (u+w)^2 = (\rho \cos \theta + w)^2$

So integral to find flux,

$\displaystyle \int_0^{2\pi} \int_{0}^{1} \int_0^{\sqrt{1+w^2}} (\rho \cos \theta + w)^2 \ \rho \ d\rho \ dw \ d\theta = \pi$.

d) So flux through hyperboloid surface $ = \pi - 5 \pi - (-\pi) = - 3 \pi$

Title: compute $\iint_Y F.N\ dS $ with Gauss


答:
You did a great job in finding $g(\alpha)$ and $h(\alpha)$. Yet in my opinion there is no way to find a general expression for $f(\alpha)$.

My doubts are explained through a simple analysis of the first six pure coefficients:

$$3 ~~~~~ \frac{25}{16} ~~~~~ \frac{227}{216} ~~~~~ \frac{5443}{6912} ~~~~~ \frac{677309}{1080000} ~~~~~ \frac{673333}{1296000}$$

Writing them in terms of primes we have ($p$ denotes a pure prime):

$$p ~~~~~ \frac{5^2}{2^4} ~~~~~ \frac{p}{2^3 3^3} ~~~~~ \frac{p}{2^8 3^3} ~~~~~\frac{p}{2^6 3^3 5^4} ~~~~~ \frac{419 \times 1607}{2^7 3^4 5^3}$$

It's a rather messy series, where second and sixth numerators are not primes, whilst others are.

Analysing the denominators, we do not recognise a pattern in terms of the exponents concerning the first three primes.

The sixth numerator is relatively close to the previous one, rather irregular for a pattern.

So this is my opinion: there is no general expression for $f(\alpha)$.

However I could be wrong, and I could have missed something very subtle someone else might notice!
Maybe my very rude and simple analysis can lead someone to something else (better, new, more argumentative), so who knows!

That's the beauty of maths and numbers.

Title: General formula for $\int^1_0 x^\alpha \log(1-x)\operatorname{Li}_2 (x)\, \mathrm dx$

Body:
Consider a following definite integral

$$I(\alpha)  = \int^1_0 x^\alpha \log(1-x)\operatorname{Li}_2 (x)\, \mathrm dx$$

Mathematica is able to provide a result for many $\alpha$ integers. See table here:

$$\begin{array}{c}
 I(0)=-2 \zeta (3)+3-\frac{\pi ^2}{6} \\
 I(1)=-\zeta (3)+\frac{25}{16}-\frac{\pi ^2}{8} \\
 I(2)=-\frac{2 \zeta (3)}{3}+\frac{227}{216}-\frac{11 \pi ^2}{108} \\
 I(3)=-\frac{\zeta (3)}{2}+\frac{5443}{6912}-\frac{25 \pi ^2}{288} \\
 I(4)=-\frac{2 \zeta (3)}{5}+\frac{677309}{1080000}-\frac{137 \pi ^2}{1800} \\
 I(5)=-\frac{\zeta (3)}{3}+\frac{673333}{1296000}-\frac{49 \pi ^2}{720} \\
 I(6)=-\frac{2 \zeta (3)}{7}+\frac{229495199}{518616000}-\frac{121 \pi ^2}{1960} \\
 I(7)=-\frac{\zeta (3)}{4}+\frac{1824315697}{4741632000}-\frac{761 \pi ^2}{13440} \\
 I(8)=-\frac{2 \zeta (3)}{9}+\frac{16317321883}{48009024000}-\frac{7129 \pi ^2}{136080} \\
 I(9)=-\frac{\zeta (3)}{5}+\frac{1946452513}{6401203200}-\frac{7381 \pi ^2}{151200} \\
 I(10)=-\frac{2 \zeta (3)}{11}+\frac{64401732356723}{234300040128000}-\frac{83711 \pi ^2}{1829520} \\
\end{array}$$

Is there a closed form expression for $f,g,h$ in $I(\alpha) = f(\alpha) + g(\alpha)\pi^2 + h(\alpha)\zeta(3)$?


问:
Intuition of the meaning of homology groups
I am studying homology groups and I am looking to try and develop, if possible, a little more intuition about what they actually mean. I've only been studying homology for a short while, so if possible I would prefer it if this could be kept relatively simple, but I imagine it is entirely possible there is no real answer to my query anyway.

As I said above, I want to gain a little deeper understanding of what the n-th homology group actually means: I can happily calculate away using Mayer-Vietoris but it doesn't really give me a great deal of intuition about what the n-th homology group actually means. For example, with homotopy groups, the fundamental group is in some sense a description of how loops behave on the object in question, and it is obvious to me why that is what it is for say, the torus or the circle. However, I have no idea what, if anything, I am actually saying about a triangulable object when I talk about it having 0-th homology group this or 1st homology group that.

The best I have been able to find online or in my limited book selection is the brief description "intuitively, the zeroth homology group counts how many disjoint pieces make up the shape and gives that many copies of $\Bbb Z$, while the other homology groups count different types of holes". What 'different types of holes' are there, roughly speaking? I appreciate that it may often be completely non-obvious what the low-order homology groups are for some complicated construction, but perhaps in simpler examples it might be more explicable. Are there (simple) cases where I could say, just from looking something like e.g. the torus, what its zero-th or first or second etc. homology group was based on the nature of the object? I guess in the zero-th case it is, as my source (http://teamikaria.com/hddb/wiki/Homology_groups) above says, related to the number of disjoint pieces. Can we delve deeper than this for the other homology groups?

Any book/website suggestions would be welcomed (preferably websites as I am nowhere near a library!) - I have Hatcher but not a great deal else, and I haven't gleaned as much as I wish to from that alone. Of course I know that there is a great deal we don't know about homology groups even today, so I don't expect some magical all-encompassing answer, but any thoughts you could provide would be appreciated. I hope this question is appropriate for SE Mathematics, apologies if not! -M

答:
To a certain extent, there is no great answer that works in complete generality, or at least in a way where you can have a reasonable intuition about it, enough to see at a glance what the homology of high dimensional spaces will be.  Part of this is because of the difficulty of picturing higher dimensional spaces.  However, we can get a glimpse of what the homology records by considering some examples.

First, understanding the finer details of homology is hard, so we will look at a simpler invariant, the betti numbers $b_i=\dim_{\mathbb{Q}} H_i(X;\mathbb{Q})$.  This disregards all sorts of subtle information which is safe to ignore for starting to build your intuition.

$b_0$ is the number of connected components.  $H_1(X)$ is the abelianization of $\pi_1(X)$, and so $b_1$ is the number of "independent loops" in $X$, or the number of holes .  The best example of this looking at surfaces.  Take the $3$-dimensional sphere, and attach $g$ handles to it.  Now, take the surface which is the boundary of this three dimensional object.  For each handle, we have a loop that goes around the handle (in the way you might pick up a coffee mug), and a loop that goes from the body, into the handle and back again.  For this, imagine that inside the $3$-dimensional space, you took a rubber band that went through the handle, and you shrunk the band until it was taut against the surface.

So we have $b_0=1$, $b_1=2g$, and $b_2=1$ (In general, with a compact orientable $n$-manifold, $b_n=1$). However, instead of viewing the surface as coming from attaching handles to the $3$-sphere, we can view it as punching $g$ holes through the sphere.  The fact that $b_1=2g$ and not $g$ might be a little disconcerting, as we have $g$ holes.  This is remedied somewhat by the fact that the handle-body (the object before we took the surface) is homotopy equivalent to a wedge of $g$ circles, and so it does have $b_1=g$.  However, this shows how subtle it is to say that $b_1$ is "the number of holes", as it makes it difficult to say what exactly a hole is.

To get a grasp about what's going on in the other directly, let's consider $\mathbb{R}^n\setminus \{0\}$.  This is putting a hole in the space.  Let's try to detect the hole by using spheres.  We can't encase the hole inside of a $S^k$ with $k<n-1$, because there is enough ambient space that we can move the sphere past the hole.  However, we can encase it inside of an $n-1$ sphere.  You can build some more intuition about detecting holes with spheres by looking at $\mathbb{R}^n\setminus \mathbb{R}^m$, and thinking about when we can encase the "hole" inside a sphere.

Unfortunately, this has a lot more to do with higher homotopy groups.  There are relations between homology and homotopy groups, but the connection is subtle.  For example, while the Hurewicz isomorphism theorem gives a close connection in one dimension (for any particular space), there is a Hurewicz map in all dimensions, but it is not in general easy to understand what it does.

One good way of understanding homology of CW complexes is with cellular homology.  In general, this is still hard to understand, but if you have a buffer between the dimensions of the spheres used to construct the space (e.g. complex projective spaces, which are built up out of even-dimensional spheres), homology is counting the spheres used to construct the space.

Beyond these things, I can only recommend that you let your intuition be built up from examples.  Sadly, there is no magic bullet that will give you a great understanding.


问:
find $a,b,c$ such that $e^{2\pi A}=I$

Given $A$ is real valued matrix of the form $\begin{pmatrix}a&b\\c&0\end{pmatrix}$ find all $a,b,c$ such that $e^{2\pi A}=I$

My attempt is the following, Calculating the eigenvalues I get $\frac{a\pm\sqrt{a^2+4bc}}{2}$ and I said that $a=0$ otherwise we won't get the desired result and $bc<0$ to get the right form of diagonlize matrix of $A$ and afterwards that $bc\in\mathbb{Z}$ but not sure if I've done this right

答:
The eigenvalues should be in the form $\{ki|k\in\Bbb{Z}\}$. $a$ can't be imaginary, so it must be 0. So the eigenvalues are $\pm\sqrt{bc}$. So $\sqrt{bc} \in \{ki|k\in\Bbb{Z}\}$. You are right that it works for $\sqrt{bc} < 0$ but it works also for $\sqrt{bc}=0$. Therefore $bc \in \{-k^2|k \in \Bbb{Z}\}$.

There is just one exception for $bc=0$. If only $b$ or only $c$ is zero, there will be a 2×2 Jordan block, so the exponential of this matrix wouldn't be the identity matrix. Only if both $b$ and $c$ are zero, it works.

Title: find $a,b,c$ such that $e^{2\pi A}=I$


问:
Smallest possible area for triangle
I'm solving the following question:

The two legs of a right triangle lie along the positive x and y axes.
The hypotenuse is tangent to the ellipse $2x^2 + y^2 = 1$. What is the
smallest possible area for such a triangle ?

This is my attempt at solving the problem:

Let $a$ be the length of the x axis and $b$ be the length in it's y axis if the triangle. So it's area is $1/2 a.b$

Let's find the tangent of the ellipse:

$2x^2 + y^2 = 1$

$4x + 2y \dfrac{dy}{dx} = 0$

$\dfrac{dy}{dx} = \dfrac{-2x}{y}$

Since the two points on the triangle are $(0,b)$ and $(a,0)$, we can
find the slope of the line:

$m = \dfrac{b-0}{0-a} = \dfrac{-a}{b}$

So the equation of hypotonuse is $y = \dfrac{-a}{b}x + c$

Since $(0,b)$ is one of the point in the line, we can substitute it in
the above line to find that $b = c$. Solving it with the other
co-ordinate, we can find the following fact $b = a$.

So now area is $1/2 a.b = 1/2 a^2$

Now I used the first derivative test to find that the function attains
it's local minima at $0$. So the smallest possible area for such a
triangle is zero.

I have missed lots of steps in my above solution for brevity,
but I can expand if needed. My question: is zero
the right answer ?  Unfortunately, the book I'm using doesn't provide
the answer to this question.

答:
You have made a small error. You wrote that
$$m = \dfrac{b-0}{0-a}$$ but then said that this is equal to
$$ \dfrac{-a}{b}$$
This is clearly wrong. In fact,
$$m=-\frac{b}{a}$$

The Solution
We have $y=-\frac{b}{a}x+b=\frac{ab-bx}{a}$. Substituting this into $2x^2+y^2=1$ yields
$$\begin{align}
&2x^2+\left(\frac{ab-bx}{a}\right)^2=1\\
\implies& 2a^2x^2+b^2x^2-2ab^2x+a^2b^2=a^2\\
\implies& x^2(2a^2+b^2)+x(-2ab^2)+(a^2b^2-a^2)=0\\
\end{align}$$
We could then go on to try to solve this quadratic, but in fact we can do better. Note that the tangent line will intersect the ellipse once, and only once. Hence, the discriminant of this quadratic in $x$ is equal to $0$:
$$4a^2b^4-4(2a^2+b^2)(a^2b^2-a^2)=0$$
We can divide through by $a^2$, as we don't have a degenerate triangle with $a=0$, simplifying to
$$\begin{align}&4b^4-4(2a^2+b^2)(b^2-1)=0\\
\implies&4b^4-4(2a^2b^2-2a^2+b^4-b^2)=0\\
\implies & a^2(8b^2-8)=4b^2\\
\implies & a^2=\frac{b^2}{2(b^2-1)}
\end{align}$$
Hence, we have, assuming without loss of generality that $a,b\geqslant 0$, that
$$a=\frac{b}{\sqrt{2(b^2-1)}}$$
So we find that the area of the triangle, which we know is equal to $\frac{1}{2}ab$, is in fact equal to
$$\frac{b^2}{2\sqrt{2(b^2-1)}}$$
So we now have an expression for the area, $A$, in terms of one variable only! This means that to find the minimum value of $A$, we can just differentiate with respect to $b$, solve for stationary points, check it is a minimum, and we are basically done!

I hope that was helpful. Thank you for giving me the opportunity to attack this interesting question; I'd never come across a similar one before. What I find particularly pleasing about this solution is that we have no need to look at the derivative of the ellipse, which would've made things rather complicated. If you have any questions please don't hesitate to ask!

Title: Smallest possible area for triangle


问:
Show the limit of parametric integral
Let's consider the parametric integral:

$F:\mathbb{R}\to\mathbb{R}$, where $F(x):=\int\limits_0^1 \frac{1}{x}\left(e^{-x^2(1+t)}(1+t)^{-1}-(1+t)^{-1}\right)dt$.

What is the value of the limit: $$\lim\limits_{x\to 0}F(x):=\lim\limits_{x\to 0}\int\limits_0^1 \frac{1}{x}\left(e^{-x^2(1+t)}(1+t)^{-1}-(1+t)^{-1}\right)dt?$$

Splitting up the integral into two summands and applying integration by parts at the first one doesn't get me anyhwere?! Any thoughts?

Title: Show the limit of parametric integral

Body:
$$\lim_{x\to 0}\frac{\int_0^1 \left(\frac{e^{-x^2(t+1)}}{t+1}-\frac{1}{t+1}\right)\, dt}{x}=$$
$$\int_0^1\frac{dt}{t+1}=\log 2$$
Numerator becomes
$$\int_0^1 \frac{e^{-x^2(t+1)}}{t+1}\, dt-\log 2$$
by  Leibniz integral rule
$$\frac{d}{dx}\int_0^1 \frac{e^{-x^2(t+1)}}{t+1}\, dt=\int_0^1 \frac{-2 (t+1) x e^{(-t-1) x^2}}{t+1} \, dt=\frac{2 e^{-2 x^2}}{x}-\frac{2 e^{-x^2}}{x}$$

Apply L'Hopital rule again:

$$=\lim_{x\to 0}\left(\frac{2 e^{-2 x^2}}{x}-\frac{2 e^{-x^2}}{x}\right)=\lim_{x\to 0}\left(4 e^{-x^2} x-8 e^{-2 x^2} x\right)=0$$


问:
invertible elements of polynomial ring
what is invertible elements of polynomial ring $k[x_1,...,x_n]$  ? for case  $n=1$ we have $k[x_1]$ and $$
p(x)=a_{0}+a_{1} x+\ldots a_{n} x^{n}
$$
then $p(x)$ is invertible element such that $a_{0}$ is a unit in $R,$ and the remaining coefficients $a_{1}, \ldots, a_{n}$ are nilpotent elements. ( $k$ is a field)

答:
If $k$ is a field then for nonzero polynomials in $k[x_1]$, say $p$ and $q$ You have
$$\deg(pq)=\deg(p)+\deg(q)$$
and since $\deg(1)=0$, where $1$ is the constant polynomial with coefficient $1$, You see that the image of $k^*=k\backslash\{0\}$ in $k[x_1]$ is the set of units in $k[x_1].$

Title: invertible elements of polynomial ring


问:
How should one think to get the radius of the resulting curve?
For example, the curve C is given as the intersection between
$$
C: x²+y²+z²=1, x+y+z=0
$$
Radius: 1

Another one:
$$
C: x²+y²+z²=1, x+y=1
$$

Radius:

$$
\frac{1}{\sqrt{2}}
$$

How should I think to get these? I know they're ellipses, but substituting them into each other yields "hard" expressions.

答: A plane at distance $d$ from the center of a sphere of radius $R$ intersects the sphere by a circle of radius $\sqrt{R^2-d^2}$ (This is just the old good Pythagorean theorem). </p>


答:
Since you've already gotten to
$$
A=\begin{bmatrix}0&b\\c&0\end{bmatrix}\tag1
$$
where $bc=-\lambda^2\lt0$, let's compute...
$$
\begin{align}
A^2
&=\begin{bmatrix}bc&0\\0&bc\end{bmatrix}\tag{2a}\\[6pt]
&=-\lambda^2I\tag{2b}
\end{align}
$$
Thus, $A^{2n}=(-1)^n\lambda^{2n}I$ and $A^{2n+1}=(-1)^n\lambda^{2n}A$. Therefore,
$$
\begin{align}
e^{2\pi A}
&=\sum_{n=0}^\infty\left(\frac{(2\pi A)^{2n}}{(2n)!}+\frac{(2\pi A)^{2n+1}}{(2n+1)!}\right)\tag{3a}\\
&=\sum_{n=0}^\infty\left(\frac{(-1)^n(2\pi\lambda)^{2n}}{(2n)!}I+\frac{(-1)^n(2\pi\lambda)^{2n+1}}{\lambda(2n+1)!}A\right)\tag{3b}\\
&=\cos(2\pi\lambda)I+\frac1\lambda\sin(2\pi\lambda)A\tag{3c}
\end{align}
$$
So $bc=-\lambda^2$ where $\lambda\in\mathbb{Z}$, except for $\lambda=0$ because of the division by $0$ (which actually gives $I+2\pi A$, which works if $A=0$).

Title: find $a,b,c$ such that $e^{2\pi A}=I$

Body:
Given $A$ is real valued matrix of the form $\begin{pmatrix}a&b\\c&0\end{pmatrix}$ find all $a,b,c$ such that $e^{2\pi A}=I$

My attempt is the following, Calculating the eigenvalues I get $\frac{a\pm\sqrt{a^2+4bc}}{2}$ and I said that $a=0$ otherwise we won't get the desired result and $bc<0$ to get the right form of diagonlize matrix of $A$ and afterwards that $bc\in\mathbb{Z}$ but not sure if I've done this right


问:
Is it possible that the tangent can look like this
Consider this:

$$ K(v) = \frac{v}{v^2+9} $$

Approximate the function for v = 1 by a tangent. I first did the derivitave of the function.

$$ K'(v) = \frac{-v^2+9}{(v^2+9)^2} $$ And now the tangent formula

$$ t(v) = f'(v) * (v - v0) + f(v) $$
Now if I input 1 in both K and K' I get this

$$K'(v) = \frac{-1^2+9}{(1^2+9)^2} $$ That should be 1/10 for K we also get 1/10. Now if I put that in the tangent formula

$$ t(v) = \frac{1}{10} * (v - 1) + \frac{1}{10} $$ Now when I multiply this I only have 1/10v left since 1/10 - 1/10 is 0. Is it possible to have a tangent looking like this,or did I make an error somwhere along the lines.

Thank you!

答:
hint

near $ v_0=1 $, we have the approximation

$$K(1+\epsilon)\approx t(1+\epsilon)$$

or

$$K(1+\epsilon)\approx K'(1)(1+\epsilon-1)+K(1)$$
and

$$K(1+\epsilon)\approx \frac{\epsilon}{25}+\frac{1}{10}$$


答:
Outline : We will write down the equation of tangent at arbitrary point $P$ of ellipse. Let it intersect the coordinate axes in $A$ and $B$. We identify the legs of the right triangle formed can be identified as $x$- and $y$- intercepts of the tangent line. So we find the smallest area of triangle minimizing the product of the intercepts.

Let $P=(h,k)$. Then tangent at $P$ is $2xh+ky=1$ which can be written as
$$\frac{x}{1/(2h)}+\frac{y}{1/k}=1$$

Area $=\frac{1}{2}\cdot \frac{1}{(2h)}\cdot \frac{1}{k}\cdot$ where $2h^2+k^2=1$

By AM-GM,
$$\frac{2h^2+k^2}{2} \ge \sqrt{2h^2\cdot k^2}$$
$$\Rightarrow \frac{1}{hk} \ge \frac{2\sqrt{2}}{2h^2+k^2}=2\sqrt{2}$$
$$\therefore A_{\text{min}}=\frac{1}{4hk}=\frac{2\sqrt{2}}{4}=\frac{1}{\sqrt{2}}$$

Equality occurs at $2h^2=k^2$ which can be used to determine coordinates of optimal $P$.

Title: Smallest possible area for triangle

Body:
I'm solving the following question:

The two legs of a right triangle lie along the positive x and y axes.
The hypotenuse is tangent to the ellipse $2x^2 + y^2 = 1$. What is the
smallest possible area for such a triangle ?

This is my attempt at solving the problem:

Let $a$ be the length of the x axis and $b$ be the length in it's y axis if the triangle. So it's area is $1/2 a.b$

Let's find the tangent of the ellipse:

$2x^2 + y^2 = 1$

$4x + 2y \dfrac{dy}{dx} = 0$

$\dfrac{dy}{dx} = \dfrac{-2x}{y}$

Since the two points on the triangle are $(0,b)$ and $(a,0)$, we can
find the slope of the line:

$m = \dfrac{b-0}{0-a} = \dfrac{-a}{b}$

So the equation of hypotonuse is $y = \dfrac{-a}{b}x + c$

Since $(0,b)$ is one of the point in the line, we can substitute it in
the above line to find that $b = c$. Solving it with the other
co-ordinate, we can find the following fact $b = a$.

So now area is $1/2 a.b = 1/2 a^2$

Now I used the first derivative test to find that the function attains
it's local minima at $0$. So the smallest possible area for such a
triangle is zero.

I have missed lots of steps in my above solution for brevity,
but I can expand if needed. My question: is zero
the right answer ?  Unfortunately, the book I'm using doesn't provide
the answer to this question.


问:
A problem V.I. Arnold solved as a primary school student
According to a 1995 interview that Vladimir I. Arnold gave to the Notices of the AMS,
his primary school teacher I.V. Morozkin gave in 1949 (when Arnold was 12 years old) to a Soviet classroom, most likely the 6th grade, the following question

Two women started at sunrise and each walked at a constant velocity. One went from $A$
  to $B$ and the other from $B$ to $A$.  They met at noon and, continuing with no stop,
  arrived respectively at $B$ at 4 p.m. and at $A$ at 9 p.m. At what time was the sunrise
  that day?

The question is not how to solve this problem, but rather How to solve this problem using what 12 year old kids know (or knew during the Soviet era).

UPDATE. In the same interview, Arnold also said:

I spent a whole day thinking on this
  oldie, and the solution (based on what
  is now called scaling arguments,
  dimensional analysis, or toric variety
  theory, depending on your taste) came
  as a revelation.

It would be of interest to know how, in Arnold's mind, dimensional analysis and toric varieties are related to this problem.

答:
A trivial solution: system of equations

Call $|AB|=L$, speed of first lady $v_1$, speed of second lady $v_2$ and time passes from dawn to $4pm$ as $t$.

Then we have $4$ equation with $4$ unknown.

$$4v_1+9v_2=L \tag1$$$$ (v_1+v_2)(t-4)=L \tag2$$$$ v_1 t=L\tag 3$$$$ v_2 (t+5)=L\tag 4$$


问:
Apply Ito's formula to $d(X_t)$, where $X_t = e^{\int_0^t(X_s-1/2)ds+W_t}$
I'd like to find $dX_t$, where $X_t = e^{\int_0^t(X_s-1/2)ds+W_t}$. I have no idea how to apply Ito's formula here.

答:
Let's denote $Y_t = \int_0^t(X_s-1/2)ds+W_t$, so we have $X_t=e^{Y_t}$.

Applying the Ito's lemma on $X_t=e^{Y_t}$
$$dX_t = e^{Y_t}dY_t+\frac{1}{2}e^{Y_t}<dY_t,dY_t> \tag{1}$$

But from $Y_t = \int_0^t(X_s-1/2)ds+W_t$, we have
$$dY_t = (X_t-\frac{1}{2})dt+dW_t \tag{2}$$

Replacing (2) to (1), by knowing that $<dY_t,dY_t>  =  <dW_t,dW_t>  =  dt$:
\begin{align}
dX_t &= e^{Y_t}dY_t+\frac{1}{2}e^{Y_t}<dY_t,dY_t> \\
 &= X_t ((X_t-\frac{1}{2})dt+dW_t)+\frac{1}{2}X_tdt \\
 &= X_t (X_tdt+dW_t) \\
 &= X_t^2  dt + X_tdW_t
\end{align}


问:
How to evaluate $\lim_{x\to 0} (1+2x)^{1/x}$
Good night guys! I'm having some trouble with this:

$$\lim_{x\to 0} (1+2x)^{1/x}$$

I know that $\lim_{x\to\infty} (1 + 1/x)^x = e$
but I don't know if i should take $h=1/(2x)$ or $h=1/x$

Can someone please help me? Thanks!

答:
We first find the limit as $x$ approaches $0$ from the right.

Let $y=\frac{1}{2x}$. Then $2x=\frac{1}{y}$ and $\frac{1}{x}=2y$. We want
$$\lim_{y\to\infty}\left(1+\frac{1}{y}\right)^{2y}.$$
This is the square of the familiar
$$\lim_{y\to\infty}\left(1+\frac{1}{y}\right)^{y}.$$
We conclude that
$$\lim_{x\to 0^+}\left(1+2x\right)^{1/x}=e^2.$$

Limit from the left is more troublesome with this approach. Let $1+2x=\frac{t}{1+t}$. Then $\frac{1}{x}=-2\frac{1+t}{t}$. So we are looking for
$$\lim_{t\to 0^+}\frac{1}{\left(1+t\right)^{-2(1+t)/t}},$$
which is
$$ \lim_{t\to 0^+}\left(1+t\right)^{2(1+t)/t}.$$
Now an argument much like the one before works.   The limit from the left is also $e^2$.


问:
How to use uniform boundedness principle
If $f_n$ be orthogonal sequence in Hilbert space and $\sum \langle f_n,g\rangle$ converges for all $g \in H$. Then show that $\sum f_n$ converges in Hilbert space.

I want to use the uniform boundedness principle to solve this but I have one problem with that. I think we should define $T_n = \sum\langle f_n,g\rangle f_n$ and then use UBC but I think if we take $T_n = \sum \langle f_n,g\rangle$ is also should work. So my problem is that which one is correct or both of them is working in this question?

答:
Define $$T_n: \mathcal{H} \to \mathbb{C}$$ by $$T_n (g) =\sum_{k=1}^{n} \left< f_k , g\right>$$
Let $$u_n =\frac{\sum_{k=1}^n f_k}{\left|\left| \sum_{k=1}^n f_k\right|\right|}$$ then $$||u_n|| =1$$
and $$|T_n (u_n ) |=\left|\left| \sum_{k=1}^n f_k\right|\right|$$
the above with some obvious observations implies that$$||T_n || =\left|\left| \sum_{k=1}^n f_k\right|\right|.$$
Now we know that $T_n $ converges for every $g$ therefore by Banach - Steinhaus theorem the sequence $$||T_n ||$$ is bounded.


问:
More understandable explanation for Cat-Rabbit-Dog quiz
This is a quiz from Brilliant.org Algebra Fundamentals courses.

Quiz:

At Step 40, how many cats will there be in the 7th row?

Answer: 14

Explanation given by Brilliant.org is not clear to me so if someone can give a better explanation it will be helpful.

Explanation by Brilliant.org

In rows where dividing by 3 has a remainder of 1, the pattern is
cat-rabbit-dog.

7 when divided by 3 has a remainder of 1, so cats occur at 1, 4, 7,
etc., i.e. at points where dividing by 3 has a remainder of 1.

In step 40, the grid is 40 columns wide. The full cat-rabbit-dog
pattern happens for $39\div 3=13$ times, but since the pattern
at row 7 starts with a cat, column 40 has one extra cat. So there are
$13 + 1 = 14$ cats.

Note: A more compact way of saying "dividing $x$ by 3 has a remainder
of 1" is $x \mod 3 = 1$.

答:
In Step 40, there will be $40$ rows each containing $40$ animals. In each row, each group of three consecutive animals consists of one of each type. You can separate the seventh row into the first animal followed by $13$ groups of three. The $13$ groups of three will contain $13$ cats, so the answer will be $14$ if the first animal in the seventh row is another cat, and $13$ otherwise.

Counting down the first column, we see that the first, fourth, seventh, etc. (going up by $3$ each time) are cats. So the answer is $14$, because there is an extra cat at the start of the $7$th row.

What the explanation is trying to add is that you would get the same answer if the row was changed to any other number in the sequence $1,4,7,...$ (but no bigger than $40$). These numbers are the numbers that leave a remainder $1$ when divided by $3$. Other rows - those whose position divides exactly by $3$ or leaves remainder $2$ - will not have a cat at the start, so would have $13$ cats.


答:
Let $(h,k)$ be the point on ellipse from where you draw the tangent, then,
$$2h^2+k^2=1$$

You already know the slope of tangent is $$\frac{dy}{dx}=\frac{-2h}{k}$$

Constructing the equation of hypotenuse from this,

$$\frac{y-k}{x-h}=\frac{-2h}{k}$$

Putting $y=0$ , you get $x=a$,

$$a=\frac{1}{2h}$$
Similarly $$b=\frac{1}{k}$$

You need to minimise the area, i.e. , $$=\frac{ab}{2}$$

$$=\frac{1}{4hk}$$

$$=\frac{1}{4\cdot{k}\cdot{}\sqrt{\frac{1-k^2}{2}}}$$

$$=\frac{1}{2\sqrt{2}\cdot{k}\cdot{}\sqrt{1-k^2}}$$
Can you continue?

Title: Smallest possible area for triangle


答: For problems like these, try taking the natural log of the expression. More precisely, compute $\ln(\lim_{x\rightarrow 0} (1+2x)^{1/x}) = \lim_{x\rightarrow 0} \ln((1+2x)^{1/x}) = \lim_{x\rightarrow 0} \frac{\ln((1+2x))}{x}$. Now it is a form that allows you to apply L'Hopital. Don't for get to exponentiate at the end!</p>

Title: How to evaluate $\lim_{x\to 0} (1+2x)^{1/x}$

Body:
Good night guys! I'm having some trouble with this:

$$\lim_{x\to 0} (1+2x)^{1/x}$$

I know that $\lim_{x\to\infty} (1 + 1/x)^x = e$
but I don't know if i should take $h=1/(2x)$ or $h=1/x$

Can someone please help me? Thanks!


答: Apparently, in step $n$, there is an animal in position $(r,c)$ iff $1\le r,c\le n$, and that animal is a cat iff $r+c\equiv 2\pmod 3$. So for $n=40$ and $r=7$, we want to find the number of $c$ with $1\le c\le 40$ and $7+c\equiv 2\pmod 3$. Those $c$ have the form $c=1+3k$ where $k$ is allowed to range from $0$ (making $c=1$) up to $13$ (making $c=40$).  So there are $$14$$ valid choices for $k$.

Title: More understandable explanation for Cat-Rabbit-Dog quiz

Body:
This is a quiz from Brilliant.org Algebra Fundamentals courses.

Quiz:

At Step 40, how many cats will there be in the 7th row?

Answer: 14

Explanation given by Brilliant.org is not clear to me so if someone can give a better explanation it will be helpful.

Explanation by Brilliant.org

In rows where dividing by 3 has a remainder of 1, the pattern is
cat-rabbit-dog.

7 when divided by 3 has a remainder of 1, so cats occur at 1, 4, 7,
etc., i.e. at points where dividing by 3 has a remainder of 1.

In step 40, the grid is 40 columns wide. The full cat-rabbit-dog
pattern happens for $39\div 3=13$ times, but since the pattern
at row 7 starts with a cat, column 40 has one extra cat. So there are
$13 + 1 = 14$ cats.

Note: A more compact way of saying "dividing $x$ by 3 has a remainder
of 1" is $x \mod 3 = 1$.


问:
sum up $\sum\limits_{i=0}^n\ln(i^2+a)b^i$
How to sum up  $\sum\limits_{i=0}^n \ln(i^2+a) b^i$, getting a simple analytical formula, where $a>0, 0<b<1$.

I guess it has something to do with $\arctan$ for the same form they have. Here I show the curve of the sum function when $a=1$ and $b=0.99$.

In addition, the series expansions of $\ln(x)$ and $\arctan(x)$ are similar, which may also indicate that these two functions are related. But I only found such a relationship between $\ln(x)$ and $\arctan(x)$, $\arctan(x)=\frac{1}{2i}\ln(\frac{x-i}{x+i})+q$, and there are imaginary numbers $i$.

Of course, it may be not right to try to simplify this summation formula to arctan function...

Can anyone give some advice?

答:
Let

$$
S(a)=\sum_{n=0}^N b^n \ln(a^2+n^2)=\sum_{n=0}^N b^n \ln \left(a^2(1+(n/a)^2 \right) \\
S(a)=\sum_{n=0}^N b^n \ln(a^2)+\sum_{n=0}^N b^n \ln \left (1+(n/a)^2 \right)
$$

The first sum on the right is a geometric sum. We will first investigate the sum to infinity.

$$
S(a)=\frac{\ln(a^2)}{1-b}+\sum_{n=0}^\infty b^n \ln \left (1+(n/a)^2 \right)
$$

Let $t_n=n/a$

$$
\sum_{n=0}^\infty b^n \ln \left (1+(n/a)^2 \right)=a\sum_{n=0}^\infty b^{at} \ln(1+t^2) \frac{1}{a}
$$

Since $1/a=t_{n+1}-t_n$, we now have a Riemann sum that may be written as an integral for $a \to \infty$.

$$
T(a)=a\int\limits_0^{\infty} dt \ b^{at} \ln(1+t^2)
$$

With the assistance of Mathematica we find

$$
T(a)=\frac{1}{\ln(b)} \left\{ 2 \cos(a \ln b) \operatorname{ci}(-a\ln b) + \sin(a \ln b) \left[ \pi +2 \operatorname{si}(a\ln b) \right] \right\}
$$

Where $\operatorname{si}$ and $\operatorname{ci}$ are the sine and cosine integrals.  Thus we have, for fixed $b$

$$
S(a) \sim \frac{\ln(a^2)}{1-b}+T(a) \ \ , \ \ a \to \infty
$$

Here is a plot of the approximation versus the exact finite sum, with $a=2$, $b=0.6$:

Here is a plot of the approximation and exact sum to infinity$^\dagger$, with $b=0.999$, $b=0.001$, and for 'large' values of $a$ up to unity:

Notice that if $b \to 0, T \to 0$ while $\frac{\ln(a^2)}{1-b} \to \ln(a^2)$, thus we also have

$$
S(a) \sim \frac{\ln(a^2)}{1-b} \ \ , \ \ a \to \infty \ \ , \ \ b \to 0
$$

It turns out this is good when either $a \to \infty$ or $b \to 0$. I think the error here is $O(1/a(\ln b)^2)$. Here is a plot of this leading order term as a function of $b$:

If you want only elementary functions, we can expand $T(a)$ in terms of the large parameter $(-a \ln b)$ to find the next term

$$
S(a) \sim \frac{\ln(a^2)}{1-b}- \frac{2}{(a \ln b)^3} \ \ , \ \ a \to \infty \ \ , \ \ a \ln b \to -\infty
$$

The finite sum may be performed similarly. I find

$$
S(a) \sim \ln(a^2) \frac{1-b^{N+1}}{1-b}+   \frac{b^{-ia}}{\ln(b)} \left\{ \operatorname{Ei} (\small{ ia\ln b})+b^{ai+N} \ln(1+(a/N)^2) \\ +b^{2ia}\left[ \operatorname{Ei}( \small{-ia \ln b} )- \operatorname{Ei}( \small{ (N-ia )  \ln b}) \right] - \operatorname{Ei}( \small{(ai+N) \ln b}) \right\} \ \ \ , \ \ \ a \to \infty
$$

Where $\operatorname{Ei}$ is the exponential integral, and $b$ is fixed. Using just the first term on the right:

$\dagger$ Mathematica evaluates the infinite sum as derivatives of the Lerch Phi function

$$
\sum_{n=0}^\infty b^n \ln(a^2+n^2)= -\Phi^{0,1,0}(b,0,-ia)-\Phi^{0,1,0}(b,0,ia)
$$

Beware! The expression on the right has a nonzero imaginary part. The real part is equal to the sum on the left, and is what I'm using in plots for the 'exact sum to infinity'.


答:
If you expand the exponential into its Taylor series and simplify the result, the integrand becomes
$$
\frac{1}{x}\left( {e^{ - x^2 (1 + t)} (1 + t)^{ - 1}  - (1 + t)^{ - 1} } \right) = \sum\limits_{n = 0}^\infty  {( - 1)^{n + 1} \frac{{x^{2n + 1} }}{{(n + 1)!}}(1 + t)^n } .
$$
Integrating term-by-term (the series converges absolutely and uniformly in $t\in [0,1]$) yields
$$
F(x) = \sum\limits_{n = 0}^\infty  {( - 1)^{n + 1} \frac{{x^{2n + 1} }}{{(n + 1)!}}\frac{{2^{n + 1}  - 1}}{{n + 1}}} .
$$
This clearly shows that $\lim _{x \to 0} F(x) = 0$.

Title: Show the limit of parametric integral

Body:
Let's consider the parametric integral:

$F:\mathbb{R}\to\mathbb{R}$, where $F(x):=\int\limits_0^1 \frac{1}{x}\left(e^{-x^2(1+t)}(1+t)^{-1}-(1+t)^{-1}\right)dt$.

What is the value of the limit: $$\lim\limits_{x\to 0}F(x):=\lim\limits_{x\to 0}\int\limits_0^1 \frac{1}{x}\left(e^{-x^2(1+t)}(1+t)^{-1}-(1+t)^{-1}\right)dt?$$

Splitting up the integral into two summands and applying integration by parts at the first one doesn't get me anyhwere?! Any thoughts?