问:
Exercise $2.3.2$ of Grafakos, Classical Fourier Analysis.
I am trying to prove the exercise $2.3.2$ of "Grafakos-classical Fourier Analysis".

Exercise $2.3.2.$ Let $\varphi,\ f\in \mathcal{S}(\mathbb{R}^n)$, and for $\epsilon>0$ let $\varphi_{\epsilon}(x)=\epsilon^{-n}\varphi(\epsilon^{-1}x)$. Prove that $\varphi_{\epsilon}\ast f\to bf$ in $S$ (schwartz space), where $b$ is the integral of $\varphi$.

The definition of $f_k\to f$ in $\mathcal{S}$ in Grafakos is:

$\lim_{k\to\infty} \sup_{x\in\mathbb{R}^n}\left|x^{\alpha} \partial^{\beta}(f_k-f)(x)\right|=0$, for all $\alpha,\beta$ multi-indices.

What would be an "efficient" way to start this exercise?

Actualization 1:
\begin{align}\left|\varphi_\varepsilon \ast f(x) - b f(x)\right| &= \left|\int_{\mathbb{R}^n} \varepsilon^{-n} \varphi(y/\varepsilon) f(x-y) dy - \int_{\mathbb{R}^n} \varepsilon^{-n} \varphi(y/\varepsilon) f(y) dy \right|\\
&= \left|\int_{\mathbb{R}^n}  \varepsilon^{-n} \varphi(y/\varepsilon) [ f(x-y) - f(y)] dy\right|\\
&\leq \left|\int_{\mathbb{R}^n}  \varepsilon^{-n} \varphi(y/\varepsilon)  |f(x-y) - f(y)| dy\right|
\end{align}

$f\in S$ then $f$ is uniformly continuos therefore, for any $|y|\leq \delta$ some $\delta>0$ implies $|f(x-y)-f(y)|<\epsilon$

therefore

\begin{align}
\sup_{|x|\leq \delta} \left| x^{\alpha}\partial^{\beta}[ \varphi_{\epsilon}*f(x)-bf(x)]\right|
&=\sup_{|x|\leq \delta}\left| x^{\alpha} [\varphi_{\epsilon}*\partial^{\beta}f(x)-b\partial^{\beta}f(x)]\right|\\
&=\sup_{|x|\leq \delta}\left| x^{\alpha} [\varphi_{\epsilon}*g(x)-bg(x)]\right|\\
&\leq \sup_{|x|\leq \delta}\left| x^{\alpha} \int_{R^n}\epsilon^{-n}\varphi({y/\epsilon})|[g(x-y)-g(y)]|dy\right|\\
&\leq \delta^n\int_{R^n}\epsilon^{-n}\varphi(y/\epsilon)\epsilon dy\\
&=\delta^n\left\|\varphi\right\|_{1} \epsilon\to 0
\end{align}
where $g(x)=\partial^{\beta}f(x),\ g\in S$

How proves for $|x|>\delta$?

Actualization 2.
I fixed the above.

\begin{align} &\sup_{x\in R^n} \left|x^{\alpha} \partial^{\beta}(\varphi_{\epsilon}*f-bf)(x)\right|\\
&=\sup_{x\in R^n} \left|x^{\alpha} (\varphi_{\epsilon}*\partial_{x}^{\beta}f-b\partial_{x}^{\beta}f)(x)\right|\\
&=\sup_{x\in R^n}\left|x^{\alpha}(\varphi_{\epsilon}*g-bg)(x)\right|,\quad g=\partial_{x}^{\beta}f\in S\\
&\leq \sup_{x\in R^n} \left|x^{\alpha}\int_{R^n}\varphi_{\epsilon}(y)|g(x-y)-g(x)|dy\right|\\
&=\sup_{x\in R^n}\left|x^{\alpha}\left[\int_{|y|>\delta} \varphi_{\epsilon}(y)|g(x-y)-g(x)|dy+\int_{|y|\leq \delta} \varphi_{\epsilon}(y)|g(x-y)-g(x)|dy\right]\right|
\end{align}
Here I don't know how to continue.
From what I've seen, I should use the following facts:
$\lim_{\epsilon\to 0}\int_{|x|>\delta}\varphi_{\epsilon}(x)dx=0$ for any $\delta>0$
and $f\in S\rightarrow g:=\partial_{x}^{\beta}f\in S$ then $g$ uniformly continuous.
But I don't know how to "kill" the $x^{\alpha}$

Actualization 3.
\begin{align}
&\sup_{x\in R^n}\left|x^{\alpha}\left[\int_{|y|>\delta} \varphi_{\epsilon}(y)|g(x-y)-g(x)|dy\right]\right|\\
&\leq \sup_{x\in R^n}\left|x^{\alpha}\left[\int_{|y|>\delta} \varphi_{\epsilon}(y)(|g(x-y)|+|g(x)|)dy\right]\right|\\
&\sup_{x\in R^n}\left|\int_{|y|>\delta} x^{\alpha}|g(x-y)|\varphi_{\epsilon}(y)dy+\int_{|y|>\delta} x^{\alpha} |g(x)|\varphi_{\epsilon}dy\right|\\
&\leq \left|\int_{|y|>\delta} C(\alpha)\varphi_{\epsilon}dy+\int_{|y|>\delta} C(\alpha)\varphi_{\epsilon}dy\right|
\to 0 \quad (|y|>\delta  \text{ and } \epsilon\to 0)
\end{align}
because $|g|$ and $|\tau_{y}g|$  are Schwartz functions because $g$ is Schwartz.

In $x^{\alpha}\int_{|y|\leq \delta} \varphi_{\epsilon}(y)|g(x-y)-g(x)|dy$ I am lost by the $x^{\alpha}$.

答:
I was working on the same Grafakos problem and found my way here.  I had some of the same issues as you, and perhaps found the fix you needed.  It seems your only issue is the case $|x| \geq \delta$ and $|y| < \delta$.  My first step is to do
$$
x^\alpha \int_{|y| < \delta} \epsilon^{-n}\varphi(y/\epsilon)(\partial^\beta \tau^y f(x) - \partial^\beta f(x))dy = \frac{\epsilon^{-n}}{x^\alpha}\int_{|y|<\delta} x^{2\alpha}\varphi(y/\epsilon)(\partial^\beta \tau^y f(x) - \partial^\beta f(x))dy.
$$
Since $\varphi$ is a Schwartz function then $\sup_{x \in \mathbb{R}^n}|x^{2\alpha}f(y/\epsilon)| < C_2 < \infty$ for some fixed $C_2$ and for all $|y| < \delta$.  Thus we have
$$
\begin{align*}
\sup_{|x| \geq \delta} \left| \frac{\epsilon^{-n}}{x^\alpha}\int_{|y|<\delta} x^{2\alpha}\varphi(y/\epsilon)(\partial^\beta \tau^y f(x) - \partial^\beta f(x))dy \right|
<{}& \epsilon^{-n}\delta^{1 + |\alpha|}v_n C_2 \|\partial^\beta \tau^y f - \partial^\beta f\|_\infty,
\end{align*}
$$
where $v_n$ is the volume of the unit ball in $\mathbb{R}^n$.  As you have mentioned, by making $\delta$ small enough you can let $\|\partial^\beta \tau^y f - \partial^\beta f\|_\infty \rightarrow 0$.


问:
Version of the Riemann-Lebesgue lemma
Let $f \in L_{\mathbb{R}^+}^1(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda).$ We want to prove the following version of the Riemann-Lebesgue lemma: $$\lim_{y \to +\infty}\int_{\mathbb{R}}f(x)\sin(yx)dx=0.$$

Hint: consider the sequence of functions $f_n=\min(f,n)1_{[-n\pi;n\pi]}.$

Let $\phi_n=\min(f,n).$

For all $n \in \mathbb{N},y>0$
$$|\int_{\mathbb{R}}f(x)\sin(yx)|dx \leq \int_{\mathbb{R}}|f_n(x)-f(x)|dx+|\int_{-n\pi}^{n\pi}\phi_n\sin(yx)dx|$$
then we take $\limsup_{y}$ and then we let $n \to +\infty,$ after noticing that $\lim_n\int_{\mathbb{R}}|f_n(x)-f(x)|dx=0$ from the dominated convergence theorem.

So it remains to prove that for a fixed $n \in \mathbb{N},$ $\lim_{y \to+\infty}\int_{-n\pi}^{n\pi}\phi_n(x)\sin(yx)dx=0$

The following argument, is it true?
$$\int_{-n\pi}^{n\pi}\phi_n(x)\sin(yx)dx=\frac{1}{y}\int_{-n \pi y}^{n\pi y}\phi_n(\frac{x}{y})\sin(x)dx$$
So, $$|\int_{-n\pi}^{n\pi}\phi_n(x)\sin(yx)dx| \leq \frac{1}{y} (\int_{-n \pi y}^{-[ny]\pi}ndx +\int_{[ny] \pi}^{ny \pi}ndx+|\int_{-[n y] \pi}^{[ny] \pi}\phi_n(\frac{x}{y})\sin(x)dx|)$$

which implies

$|\int_{-n\pi}^{n\pi}\phi_n(x)\sin(yx)dx| \leq \frac{2n \pi}{y}+\frac{1}{y}|\sum_{k=0}^{[ny]-1}(-1)^k\int_0^{\pi}\sin(x)(\phi_n(\frac{x-k\pi}{y})-\phi_n(-\frac{x-k \pi}{y}))dx|,$

we conclude,

$|\int_{-n\pi}^{n\pi}\phi_n(x)\sin(yx)dx| \leq \frac{2 \pi n}{y}+\frac{2 \pi n}{y}|\sum_{k=0}^{[ny]-1}(-1)^k| \leq \frac{4 \pi n}{y} \rightarrow_{y \to +\infty}0.$

答:
First consider case when f is constant on an interval [a,b] and 0 outside of it.
This is an easy computation.
Then consider a case when f is a finite sum of function constant on disjoint intervals.
It just directly follows from the first case and property of limits.
Next, consider a case of f continuous with compact support.
This function can be uniformly approximated by functions from step two.
Lastly, integrable functions can be approximated (in L1 norm) by continuous functions with compact support.</p>

ContentLicense: CC BY-SA 4.0

Title: Version of the Riemann-Lebesgue lemma


问:
Binomial distribution as sum of Bernoulli distributions
Let's give a generic discrete probability space $(\Omega,P)$ where $|\Omega| \le \aleph_0$.

A random variable $X\colon \Omega \to \{0,\dots,n\}$ is a Binomial one if, by definition, $p_X(k):=P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$ where $k \in \{0,\dots,n\}$ and $p \in [0,1]$ (Is this a correct definition? Is there a more general one?).

I know that if we have $X_1,\dots,X_n\colon\Omega \to \{0,1\}$ independent Bernoulli random variables, then $X:=X_1+\dots+X_n\colon \Omega \to \{0,\dots,n\}$ is a Binomial random variable.

I'm wondering if the converse holds. Namely, if $X\colon \Omega \to \{0,\dots,n\}$ is a Binomial random variable, then can it be written as the sum of independent Bernoulli random variables?

Thank you for your time!

答:
The answer is no.

Consider the probability space $(\Omega, \mathcal{F}, \mathbb{P})$, where $\Omega = \{0,\ldots,n\}$, $\mathcal{F} = \mathcal{P}(\Omega)$ and $\mathbb{P}: \mathcal{F} \to [0,1]$ defined as $\mathbb{P}(A) = \sum\limits_{i \in A} \binom ni p^i(1-p)^{n-i}$ for every $A \in \mathcal{F}$.

Consider now $X: \Omega \to \{0,\ldots, n\}$ as $X(i)=i$ (the identity function).

The distribution of $X$ is clearly binomial.

Suppose now that $X = X_1 + \ldots + X_n$ with $X_i$ iid with law Bernoulli of parameter $p$, all defined on $(\Omega, \mathcal{F}, \mathbb{P})$.

This would mean that there is $A \in \mathcal{F}$ such that $\mathbb{P}(A) = p$.

And this must be true for every $p \in (0,1)$.

Consider the map $F: (0,1) \to \mathcal{P}(\{0,\ldots,n\})$ that for a given $p$ returns such a set $A$. Since the domain is infinite and the codomain is finite there must be a given $A \subseteq \{0,\ldots,n\}$ such that for infinite many $p$ we have $$\sum\limits_{i \in A} \binom ni p^i(1-p)^{n-i} = p$$
Therefore we have $\sum\limits_{i \in A} \binom ni x^i(1-x)^{n-i} = x$ for every real $x$ (polynomial identity).

And now, for example with $n=2$, we obtain that the LHS can be equal to $0,1,x^2,(1-x)^2,2x(1-x),1-x^2,1-(1-x)^2,1-2x(1-x)$, none of which is equal to $x$.


问:
Finite straight line question, will this formula create a straight line between two end points
I created a formula that I hope creates a finite straight line between any two points, I wanted to know if my math logic is correct or if I have an error in my formula. The idea is to create a straight line at y=1 between two points and then use that intersection to multiply by any other formula to get the part of the formula that fits within the bounds of my defined limit.

Here is the formula that I created:
$$\ln\left(\sqrt{9-x^2}*2\right)-\ln\left(\exp\left(\frac{\ln\left(\left(9-x^2\right)^2\right)}{4}\right)*2\right)+1$$

please note that the 9 can be any number and the square root of that number will be the endpoints of the line, so in the example above I am hoping to get a line between -3 and 3 exclusive.

Can you please tell me if my logic is correct and if this will always return just a positive line at y=1 between two defined endpoints. if not can you please tell me where the issue is.

Thank you for your help!

Thank you all for your feedback. just to clarify why I am doing this, I would like to, using a standard math formula get any section of another formula without the need to to use absolute value, or other math functions like MOD. I would like to be able to create a standard math function that I can insert into any other formula and get a subsection of the formula using standard math.

Thank you and sorry if it is confusing or if I am making things more complicated than they have to be.

My idea is only the first part will be invalid if $9-x^2$ is negative, and the rest of the logic will always be the exact same value without the value returning square root of a negative number so that it does not cancel out. and finally add a one so that if x is within the limit then the value is always 1 to later be able to multiply by any other formula.

My apologies again if I am making things way too complicated for my self.

I now understand what the feedback meant about it being too complicated. her is what I think the formula should be;
$$\ln\left(\sqrt{9-x^2}\right)-\frac{\ln\left(\left(9-x^2\right)^2\right)}{4}+1$$

I believe this will be simpler and since $\ln\left(\sqrt{9-x^2}\right)$ does not allow x  > 3 but $\frac{\ln\left(\left(9-x^2\right)^2\right)}{4}$ does allow x > 3 then this simpler function can be used to get any slice of a graph.

Please let me know if this is not correct.

Thank you, all again.

答:
This is absurdly complicated. The simple way would be to write:
$$y=1,\quad -3<x<3.$$

But if your goal is just to find a function that implies the domain rather than explicitly writing out $-3<x<3$, then you could do something like
$$y = \frac{\sqrt{9-x^2}}{\sqrt{9-x^2}},$$
which is much simpler than what you have.


问:
testing if the infinite series are converges or diverges
here are two problems on testing if the infinite series are conv. or div.

$\sum_{n=1}^\infty \frac{{{7\sqrt[6]{n^{13}}+2n}}}{{\sqrt[3]{27n^{9}-10n+16}}}$

$\sum_{n=2}^\infty \frac{1}{\ln(n!)}$

the first one I couldn't do :(

but the second one here is what I did using the ratio test I got that

$$\frac{\ln(n!)}{\ln((n+1)!)}=\frac{\ln(n)+\cdots+ \ln(2)}{\ln(n+1)+\cdots +\ln(2)}<1$$

which I thought would be effective but it isn't :(

答:
In the first one, you have
\begin{align}
& \frac{n^{13/6} + \text{comparatively negligible things}}{n^3 + \text{comparatively negligible things}} \\[12pt]
= {} & \frac 1 {n^{5/6} + \text{comparatively negligible terms}} \ge \frac {\text{some constant}} {n^{5/6}} \\[6pt]
& \text{(and “constant” means not changing as $n$ changes)}
\end{align}
Here I would work on trying to prove that the $n$ term of the series is greater than or equal to some constant multiple of $n^{-5/6}.$

\begin{align}
& \frac{7\sqrt[6]{n^{13}}+2n}{\sqrt[3]{27n^9-10n+16}} = \frac{7 + \frac 2 {n^{7/6}}}{\sqrt[3]{27n^{5/2} + \frac {10} {n^{11/2}} + \frac{16}{n^{13/2}}}} \\[12pt]
\ge {} & \frac 7 {\sqrt[3]{27n^{5/2} + 10 n^{5/2} + 16n^{5/2}}} = \frac 7 {n^{5/6} \cdot\sqrt[3]{27+10+16}}
\end{align}

And $\displaystyle \sum_{n=1}^\infty \frac 1 {n^{5/6}} =  +\infty.$


问:
Plotting parametric equations in Maple
I have the following problem. Let $\alpha:=\alpha(v)$ and $\beta:=\beta(v)$, where $v\in\mathbb{R}$. Using MAPLE, I would like to plot $\alpha$ and $\beta$ in the $(\alpha,\beta)$ space as $v$ varies continuously. I suspect that this will involve creating lists in Maple. I need some help! Here is what I am thinking at this time:

```<code>restart:

alpha:=v-> some function of v:
beta:=v-> some function of v:

for v from 0.01 to 10 by 0.001 do
  alpha:=alpha(v):
  beta:=beta(v):
  ? put the alpha and beta into a list?
od:

?plot the sequence of points in the list?
</code>```

I need some direction/help from Maple users out there, please.

答:
```<code>plot([alpha(v),beta(v),v=0..10]);
</code>```

No manual lists required.

ContentLicense: CC BY-SA 3.0

Title: Plotting parametric equations in Maple

Body:
plot([alpha(v),beta(v),v=0..10]);

No manual lists required.


问:
prove the following inequality for integral
Let $\eta$ be the smooth function supported in $B_1(0)$.such that $\int\eta = 1$

Let $u$ be a smooth function  defined on an open set $V\subset \Bbb{R}^n$ that is $u\in C_c^\infty(V)$.

Prove the following inequality holds：

$$\int_{B_1(0)}\eta(y)\int_0^1\int_V|Du(x-\epsilon ty)|dxdtdy\le \int_V|Du(z)|dz$$

I do it as follows extend the integral domain $V\to \Bbb{R^n}$ then the inequality $\int_V|Du(x-\epsilon ty)dx\le \int_\mathbb{R^n}|Du(x-\epsilon ty)|dx = \int_V|Du(z)|dz$.Is my proof correct?If not how to do it?

答:
$$
\begin{aligned}
\int_{B(0,1)} \eta(y) \int_{0}^{1} \int_{V}\left|D u_{m}(x-\epsilon t y)\right| d x d t d y &=\int_{B(0,1)} \eta(y) \int_{0}^{1} \int_{V}\left|D u_{m}(x)\right| d x d t d y \\
&=\int_{V}\left|D u_{m}(x)\right| d x \int_{B(0,1)} \eta(y) d y \int_{0}^{1} d t \\
&=\int_{V}\left|D u_{m}(x)\right| d x
\end{aligned}
$$


答:
One helpful clarification you could make is to describe the graph you want to create as a line segment, not just a line. A line, strictly speaking, has no endpoints, but a line segment does.

The formula you gave does indeed create a line segment at $y=1$ between $x=-\sqrt{a}$ and $x=\sqrt{a}$, so your logic is correct. The way you've written the formula is slightly redundant, though. For example, the $\ln$ and the $e$ in the second term cancel each other out. With a couple other simplifications, your formula can be reduced to $$\ln\left(\sqrt{a-x^2}\right)-\ln\left(\sqrt{a-x^2}\right)+1$$

As you might have already figured out, the fact that this function isn't defined for $-\sqrt{a}<x<\sqrt{a}$ means that even though the first two terms look like they should cancel each other out completely, they still affect the final result. That's pretty cool!

For a more general approach to plotting a function with restricted domain, check out this question and my answer to it. Essentially, for any function $f(x)$, you can plot a modified version (call it $g(x)$) that looks like this: $$g(x)=f(x)\sqrt{\frac{||x-b|-a|}{|x-b|-a}}$$ This will plot the portion of $f(x)$ centered at $x=b$ that is $a$ units wide in either direction.

Title: Finite straight line question, will this formula create a straight line between two end points

Body:
I created a formula that I hope creates a finite straight line between any two points, I wanted to know if my math logic is correct or if I have an error in my formula. The idea is to create a straight line at y=1 between two points and then use that intersection to multiply by any other formula to get the part of the formula that fits within the bounds of my defined limit.

Here is the formula that I created:
$$\ln\left(\sqrt{9-x^2}*2\right)-\ln\left(\exp\left(\frac{\ln\left(\left(9-x^2\right)^2\right)}{4}\right)*2\right)+1$$

please note that the 9 can be any number and the square root of that number will be the endpoints of the line, so in the example above I am hoping to get a line between -3 and 3 exclusive.

Can you please tell me if my logic is correct and if this will always return just a positive line at y=1 between two defined endpoints. if not can you please tell me where the issue is.

Thank you for your help!

Thank you all for your feedback. just to clarify why I am doing this, I would like to, using a standard math formula get any section of another formula without the need to to use absolute value, or other math functions like MOD. I would like to be able to create a standard math function that I can insert into any other formula and get a subsection of the formula using standard math.

Thank you and sorry if it is confusing or if I am making things more complicated than they have to be.

My idea is only the first part will be invalid if $9-x^2$ is negative, and the rest of the logic will always be the exact same value without the value returning square root of a negative number so that it does not cancel out. and finally add a one so that if x is within the limit then the value is always 1 to later be able to multiply by any other formula.

My apologies again if I am making things way too complicated for my self.

I now understand what the feedback meant about it being too complicated. her is what I think the formula should be;
$$\ln\left(\sqrt{9-x^2}\right)-\frac{\ln\left(\left(9-x^2\right)^2\right)}{4}+1$$

I believe this will be simpler and since $\ln\left(\sqrt{9-x^2}\right)$ does not allow x  > 3 but $\frac{\ln\left(\left(9-x^2\right)^2\right)}{4}$ does allow x > 3 then this simpler function can be used to get any slice of a graph.

Please let me know if this is not correct.

Thank you, all again.


问:
Efficiency physics problem
84 An internal combustion engine consumes 1 gallon of diesel fuel per hour. It provides a
constant 10 hp. One gallon of diesel fuel has potential energy of about 115,000 Btu. Remember
that 1 hp × 42.44 = 1 Btu/min. What is the efficiency of the system?

A.  22% Answer
B.  28%
C.  31%
D.  35%

How I approach this Since 1 gallon an hour is 115,000 then 10hp constant an hour is 424.4
There I use output/input x100. So $424.4/100000 * 100$ I got .42 I am confused in where I went wrong?

答:
First: read carefully the following statement from the problem:

Remember that $1$ hp × $42.44$ = 1 Btu/${\bf min.}\;$ [Bold-face mine.]

There are $60$ minutes in an hour.

Second: where are you making use of the information regarding potential energy of $\;115,000\;$ BTU?

Title: Efficiency physics problem


问:
$\mathbb{R}P^3$ is orientable, but $\mathbb{R}P^2$ is not

Def: Let $U, V \subset \mathbb{R}^n$ be open and $f: U \to V$ a
homeomorphism. We say $f$ is orientation preserving if for all $x \in
 U$ the composite $$ H_n(\mathbb{R}^n, \mathbb{R}^n \setminus 0)
 \xrightarrow{tr} H_n(\mathbb{R}^n, \mathbb{R}^n \setminus x)
 \xrightarrow{exc} H_n(U, U \setminus x) \xrightarrow{f_*}H_n(V,
 V\setminus f(x)) \xrightarrow{exc} H_n(\mathbb{R}^n, \mathbb{R}^n
 \setminus f(x)) \xrightarrow{tr} H_n(\mathbb{R}^n, \mathbb{R}^n
 \setminus 0)$$ is the identity map. (Here tr denotes the map induced
by translation of $\mathbb{R}^n$ and exc denotes the excision
isomorphism.)

Lemma: A manifold $M$ is orientable iff it admits an atlas with transition maps orientation preserving homeomorphisms of open subsets of $\mathbb{R}^n$.

I am trying to show that $\mathbb{R}P^3$ is orientable using this lemma. So I constructed an atlas with four charts
$U_1 = \{  [1:y:z:w] \in \mathbb{R}P^3   \}  \xrightarrow{\phi_1} \mathbb{R}^3$ via $[1:y:z:w] \mapsto (y,z,w)$,

$U_2 = \{  [x:1:z:w] \in \mathbb{R}P^3   \} \xrightarrow{\phi_2} \mathbb{R}^3$ via $[x:1:z:w] \mapsto (x,z,w)$

and similarly for $U_3$ and $U_4$.

Now we would like the transition maps $\phi_i \phi_j^{-1}$ to be orientation preserving. This is where the problems start for me.

Let's check that $f=\phi_2 \phi_1^{-1}$ is orientation preserving. Let $(y,z,w) \in \mathbb{R}^3$ be arbitrary. Then $f(y, z, w)=(1/y, z/y, w/y)$.

However, the same proof should break down for $\mathbb{R}P^2$, but I can't see where this happens, so I know something is wrong with this "proof".

答:
You have to show that the maps $\phi_i \phi_j^{-1}$ are orientation preserving. For $i=2$,$j=1$ and your notation, this is:
$$(y,z,w) \mapsto (\frac{1}{y},\frac{z}{y},\frac{w}{y})$$
where we have to restrict to the set where $y\neq 0$. This changes the orientation of $\mathbb{R}^3\setminus \{ y=0\} $ (in particular, it is not the identity!).

One thing you could try to construct an atlas would be the following: Pick an atlas of orientation preserving charts of $S^3$ such that all domains of the charts are contained in a hemisphere. Since the antipodal map on $S^3$ is orientation-preserving, this will give you an atlas of charts such that on the intersection of two charts the transition diffeomorphisms are orientation-preserving.


答:
As @Yuval showed, the first one diverges.

Since the integral test is not allowed, you can use direct comparison test:

Comparing your series to the harmonic series $\sum_{k=1}^{\infty}\frac{1}{k}$, which diverges, gives us

$\frac{7}{3}\frac{1}{k^{\frac{5}{6}}}>\frac{1}{k} \Rightarrow \sum_{k=1}^{\infty}\frac{7}{3}\frac{1}{k^{\frac{5}{6}}}>\sum_{k=1}^{\infty}\frac{1}{k}$.

Now, by the direct comparison test, the series diverges.

Remarks:

(1) I used the asymptotic order of your series, but it works similar with the original term.

(2) Divergence of the harmonic series can be proven by direct comparison test as well. (https://en.wikipedia.org/wiki/Harmonic_series_(mathematics))

Title: testing if the infinite series are converges or diverges

Body:
here are two problems on testing if the infinite series are conv. or div.

$\sum_{n=1}^\infty \frac{{{7\sqrt[6]{n^{13}}+2n}}}{{\sqrt[3]{27n^{9}-10n+16}}}$

$\sum_{n=2}^\infty \frac{1}{\ln(n!)}$

the first one I couldn't do :(

but the second one here is what I did using the ratio test I got that

$$\frac{\ln(n!)}{\ln((n+1)!)}=\frac{\ln(n)+\cdots+ \ln(2)}{\ln(n+1)+\cdots +\ln(2)}<1$$

which I thought would be effective but it isn't :(


问:
Proving $T$ and $U$ are equal if I know $(X,T)$ and $(X,U)$ are hausdorff and compact
This question is from Wayne Patty Exercise 4.2

Let $T$ and $U$ are Topologies on set X. Suppose $ (X,T) $ and $(X,U)$ are compact Haurdsorff Spaces. Prove that $T=U$ or $T$ is not a subset of $U$ or $U$ is not a subset of T.

I thought of proving it by this way: Assuming that if $T$ is a subset of $U$ or $U$ is a subset of $ T$ then $T=U$ must be true.
But I was unable to think on how to use compactness or hausdorff property and was unable to proceed.

Well, also this question was asked before on this website. But I am unable to understand how the solution will work despite trying to understand it for a fair amount of time as it was just outlined.

So, I am looking for a detailed solution.

答:
Recall the fact that a continuous bijection $f: A \rightarrow B$ where the domain $A$ is compact Hausdorff and the codomain $B$ is hausdorff is a homeomorphism.

If you take the finer topology in the domain and the coarser on the codomain, the identity function is the continuous bijection we need. That is take $$i: (X, \tau_1) \rightarrow (X,\tau_2)$$ with $\tau_2 \subseteq \tau_1$. Thus the identity a homeomorphism, from which you conclude that, if comparable, the two topologies are the same.


问:
Property of a divergent sequence $\{a_n\}$ of positive real numbers satisfying $\lim_{n\to\infty}(a_{n+1}-a_n)=0$
Let $\{a_n\}$ be an unbounded sequence of positive real numbers s. t. $\lim\limits_{n\to\infty}(a_{n+1}-a_n)=0$.

Question:

Is $a_n$ necessarily equal to a partial sum $\sum_{k=1}^n b_k$ of some sequence $\{b_k\}$ s. t. $\lim\limits_{k\to\infty}b_k=0$, but $\sum_{k=1}^\infty b_k=\infty$?

Thoughts:

Since $\{a_n\}$ is an unbounded sequence of positive real numbers, I thought it must diverge.

On the other hand, $\lim\limits_{n\to\infty}(a_{n+1}-a_n)=0$ reminded me of the limits of the form $\lim\limits_{n\to\infty}\frac{x_n}{y_n}$, where also $x_n\to\infty$, which could be evaluated via Stolz-Cesaro, but the only examples of divergent sequences $\{a_n\}$ satisfying $\lim\limits_{n\to\infty}(a_{n+1}-a_n)=0$ I could think of were the partial sums $\sum_{k=1}^nb_k$ of a sequence $b_k$ converging to $0$.

However, my statement seems fragile and I'm pretty sceptical.

Is there any counter-example?

I also took into account sequences that contain infinite products, but, if the logarithms are involved, we could transform the product into a sum.

Thank you in advance!

答:
I later saw a bit trivial example of a divergent positive sequence $\{a_n\}$ with the property $\lim\limits_{n\to\infty}(a_{n+1}-a_n)$ worth mentioning: $a_n=\sqrt n$.
$$\lim_{n\to\infty}a_n=+\infty\quad\land\quad \lim_{n\to\infty}(a_{n+1}-a_n)=\lim_{n\to\infty}(\sqrt{n+1}-\sqrt n)=\lim_{n\to\infty}\frac1{\sqrt{n+1}+\sqrt n}=0$$


问:
Stuck on a step for finding the closed formula for catalan numbers from generating function!
I am looking at the Frazer Jarvis paper, entitled Catalan Numbers.
http://www.afjarvis.staff.shef.ac.uk/maths/jarvisspec01.pdf

In this paper, he derives the closed formula from the generating function.

The $nth$ Catalan number, denoted $C_n$, is the number of ways
of multiplying together $n$ symbols.

I am stuck on page 7, particularly the following steps:

$\begin{equation}C_n=-\frac{1}{2}\bigg\{\frac{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})...(-\frac{2n-3}{2})}{n!}(-4)^n\bigg\}\end{equation}$

$C_n=\frac{1}{2}\bigg\{\frac{\frac{1}{2}(\frac{1}{2})(\frac{3}{2})...(\frac{2n-3}{2})(n-1)!}{n!(n-1)!}(2^2)^n)\bigg\}$

$C_n=\frac{1}{2}\bigg\{\frac{\frac{1}{2}[\frac{1}{2}\cdot 1 \cdot\frac{3}{2} \cdot 2...(n-2)(\frac{2n-3}{2})(n-1)]}{n!(n-1)!}2^{2n}\bigg\}$

$C_n=\frac{1}{2}\bigg\{\frac{\frac{1}{2}(1 \cdot 2 \cdot 3 \cdot 4...(2n-4)(2n-3)(2n-2)}{n!(n-1)!}2^{2}\bigg\}$

Particularly, from the 2nd to the 4th line, I am struggling to understand the steps taken here.
I am confused with what he is doing with the $\frac{1}{2}$. Is he pulling it out in the 3rd step? And most of all, I am very confused with how the $2^n$ went away in the 4th step. If anyone could please help me understand some of the steps that he skipped here, I would really appreciate it! Thank you so much!!!

Body:
Start from the second line,
$$
\begin{aligned}
C_n=&\frac{1}{2}\bigg\{\frac{\frac{1}{2}(\frac{1}{2})(\frac{3}{2})...(\frac{2n-3}{2})(n-1)!}{n!(n-1)!}(2^2)^n)\bigg\}\\
\text{line 3     } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
=&\frac{1}{2}\bigg\{\frac{\frac{1}{2}[\frac{1}{2}\cdot 1 \cdot\frac{3}{2} \cdot 2...(n-2)(\frac{2n-3}{2})(n-1)]}{n!(n-1)!}2^{2n}\bigg\}\\
=&\frac{1}{2}\bigg\{\frac{\frac{1}{2}[\frac{1}{2}\cdot \frac 22 \cdot\frac{3}{2} \cdot \frac 42...(\frac{2n-4}{2})(\frac{2n-3}{2})(\frac{2n-2}{2})]}{n!(n-1)!}2^{2n}\bigg\}\\
=&\frac{1}{2}\bigg\{\frac{1}{2^{2n-1}}\frac{(2n-2)!}{n!(n-1)!}2^{2n}\bigg\}\\
=& \frac{(2n-2)!}{n!(n-1)!}=\frac 1n \binom{2n-2}{n-1}.
\end{aligned}$$


问:
Calculating a limit to infinity on e

$$\lim_{x\to0^+}(1+3x)^\frac{-2}{3x^3}$$

Okay so I got to the point where I have
$$\lim _{t\to \infty }\left(\left(1+\frac{1}{t}\right)^{t}\right)^{-18t^{2}}$$

Now, my instructors don't allow us to use L'Hopital and we haven't proven anything about situations such that
$$\lim _{x\to \infty }\left(f\left(x\right)^{g\left(x\right)}\right)$$

so is my only option is to use squeeze rule? since I know the expression is basically $e^{-18t^{2}}$
and we know that $2 < e < 3$..
is there a simpler way to solve this? if so then how? if not then how do I prove $2^{-18t^{2}} < e^{-18t^{2}} < 3^{-18t^{2}}$?

答:
[Hint] : $$(1+3x)^{-2/3x^{3}}=e^{\displaystyle -\frac{2}{3x^{3}}\ln(1+3x)}$$ this is assuming $x>-\frac{1}{3}$
Now let $u=3x$, we get :
$$
\lim_{t\to 0^{+}}\displaystyle e^\frac{\displaystyle-2\ln(1+t)}{\displaystyle t^3}=\lim_{t\to 0^{+}}\displaystyle e^{\frac{\displaystyle-2\ln(1+t)}{\displaystyle t}\cdot\displaystyle\frac{1}{t^{2}}}
$$

Title: Calculating a limit to infinity on e


问:
Existence of flat connections and the application of the Frobenius theorem
I know that this question has been asked before but there are some doubts about it that I need to clarify, mainly what is the Frobenius theorem really saying.

The main goal is to prove the following :

If $\nabla$ is a flat connection on the vector bundle $(E,M,p)$ then around any point $p\in M$ we can find a basis of local flat sections.

So we start with an arbitrary basis of local sections $\{s_i\}_{i=1}^{r}$ with connection form $\omega$, and we want to find $s'=sA$ such that $\omega'=0$, where $\omega$ and $\omega'$ are the corresponding connection forms.  Now basically we need to find $A\in C^{\infty}(U,GL(r))$. We know that $\omega'=A^{-1}\omega A+A^{-1}dA$ and so we would want that $\omega A+dA=0$.

Now this is where I am getting some trouble. Now we want to define an integrable distribution on $U\times GL(r)$ and then prove that it's involutive and use the Frobenius theorem . First I though we wanted to find $A$ and that is to solve the system of differential equations $\omega A+ dA=0$, and I don't quite see how the frobenius theorem helps. I know that it has something related to it , I saw this reading the Wikipedia page,  but I don't see how it helps, what I mean is, this is the formulation of the frobenius theorem that I am aware of

Let $D$ be a smooth involute distribution in $M$ that is involutive then $D$ is integrable.

By integrable we mean that there exists a foliation $\mathcal{F}$ such that $T\mathcal{F}=D.$  And I don't see how we will get the $A$ we want .

Any help is aprecciated, thanks in advance.

答: This is a method going back to Élie Cartan and appears in various textbooks. The goal is to find such a function $A$ on $U$ satisfying that differential equation by finding its graph as an integral manifold of the differential system on $U\times GL(r)$. Flatness gives the integrability condition, and a little bit of thought will show you that locally an integral manifold projects diffeomorphically to $U$ and hence is the graph of a smooth map. You will, of course, want to use the differential forms formulation of integrability: If you define the $\mathfrak{gl}(r)$-valued $1$-form $\phi = \omega A + dA$ on $U\times GL(r)$, then note that $d\phi = -\omega\wedge\phi$, which tells us that the differential system $\phi=0$ is integrable. (You can also get it quite prettily from the original $\eta = A^{-1}\omega A + A^{-1}\,dA$; then $d\eta = -\eta\wedge\eta$.)


答:
We add the missing even terms of $(2n-2)!$ from the $(n-1)!$ and one portion of $2^{n-1}$.

The other $2^{n-1}$ is cancelled by the $n-1$ pieces of $\frac12$.

Title: Stuck on a step for finding the closed formula for catalan numbers from generating function!

Body:
I am looking at the Frazer Jarvis paper, entitled Catalan Numbers.
http://www.afjarvis.staff.shef.ac.uk/maths/jarvisspec01.pdf

In this paper, he derives the closed formula from the generating function.

The $nth$ Catalan number, denoted $C_n$, is the number of ways
of multiplying together $n$ symbols.

I am stuck on page 7, particularly the following steps:

$\begin{equation}C_n=-\frac{1}{2}\bigg\{\frac{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})...(-\frac{2n-3}{2})}{n!}(-4)^n\bigg\}\end{equation}$

$C_n=\frac{1}{2}\bigg\{\frac{\frac{1}{2}(\frac{1}{2})(\frac{3}{2})...(\frac{2n-3}{2})(n-1)!}{n!(n-1)!}(2^2)^n)\bigg\}$

$C_n=\frac{1}{2}\bigg\{\frac{\frac{1}{2}[\frac{1}{2}\cdot 1 \cdot\frac{3}{2} \cdot 2...(n-2)(\frac{2n-3}{2})(n-1)]}{n!(n-1)!}2^{2n}\bigg\}$

$C_n=\frac{1}{2}\bigg\{\frac{\frac{1}{2}(1 \cdot 2 \cdot 3 \cdot 4...(2n-4)(2n-3)(2n-2)}{n!(n-1)!}2^{2}\bigg\}$

Particularly, from the 2nd to the 4th line, I am struggling to understand the steps taken here.
I am confused with what he is doing with the $\frac{1}{2}$. Is he pulling it out in the 3rd step? And most of all, I am very confused with how the $2^n$ went away in the 4th step. If anyone could please help me understand some of the steps that he skipped here, I would really appreciate it! Thank you so much!!!


答:
Alternative resolution

$$L=\underset{x\to 0^+}{\text{lim}}(3 x+1)^{-\frac{2}{3 x^3}}$$
consider
$$\log L = \log \left(\underset{x\to 0^+}{\text{lim}}(3 x+1)^{-\frac{2}{3 x^3}}\right)=$$
$$=\underset{x\to 0^+}{\text{lim}}\left(-\frac{2}{3 x^3}\log(1+3 x)\right)$$
As $\log (1+3x)\sim 3x $ as $x\to 0^+$ we have
$$ \log L=\underset{x\to 0^+}{\text{lim}}\left(-\frac{2\cdot 3x}{3 x^3}\right)=-\infty$$
if $\log L\to -\infty$ then $L=0$.

Hope this is useful

Title: Calculating a limit to infinity on e

Body:
$$\lim_{x\to0^+}(1+3x)^\frac{-2}{3x^3}$$

Okay so I got to the point where I have
$$\lim _{t\to \infty }\left(\left(1+\frac{1}{t}\right)^{t}\right)^{-18t^{2}}$$

Now, my instructors don't allow us to use L'Hopital and we haven't proven anything about situations such that
$$\lim _{x\to \infty }\left(f\left(x\right)^{g\left(x\right)}\right)$$

so is my only option is to use squeeze rule? since I know the expression is basically $e^{-18t^{2}}$
and we know that $2 < e < 3$..
is there a simpler way to solve this? if so then how? if not then how do I prove $2^{-18t^{2}} < e^{-18t^{2}} < 3^{-18t^{2}}$?


答:
In the first step he observed that there are $2n$ minus signs, so they cancel out, and he multiplied top and bottom by $(n-1)!$. In the next step he rearranged the factors in the numerator:

$$\begin{align*}
&\frac12\cdot\frac32\cdot\ldots\cdot\frac{2n-3}2(n-1)!\\
&\qquad=\frac12\cdot\frac32\cdot\ldots\cdot\frac{2n-3}2\cdot\color{red}1\cdot\color{red}2\cdot\ldots\cdot\color{red}{(n-1)}\\
&\qquad=\frac12\cdot\color{red}1\cdot\frac32\cdot\color{red}2\cdot\frac52\cdot\color{red}3\cdot\ldots\cdot\color{red}{(n-2)}\cdot\frac{2n-3}2\cdot\color{red}{(n-1)}\\
&\qquad=\frac12\cdot\color{red}{\frac22}\cdot\frac32\cdot\color{red}{\frac42}\cdot\frac52\cdot\color{red}{\frac62}\cdot\ldots\color{red}{\frac{2n-4}2}\cdot\frac{2n-3}2\cdot\color{red}{\frac{2n-2}2}\\
&\qquad=\frac{(2n-2)!}{2^{2n-2}}\,.
\end{align*}$$

Thus, the whole thing is

$$\begin{align*}
\frac12\cdot\frac{\frac12(2n-2)!}{2^{2n-2}n!(n-1)!}\cdot2^{2n}&=\frac14\cdot\frac{(2n-2)!}{n!(n-1)!}\cdot2^{2n-(2n-2)}\\
&=2^{-2}\cdot\frac{(2n-2)!}{n!(n-1)!}\cdot 2^2\\
&=\frac{(2n-2)!}{n!(n-1)!}\\
&=\frac1n\binom{2n-2}{n-1}\,.
\end{align*}$$

I wrote up essentially the same derivation in this answer, if you’d like to see a slightly different take on it.


问:
"Set of all formal products" - what does this mean?
List the set of all formal products of $(1+x^2+x^4)^2(1+x+x^2)^2$ with exponents summing to $4$.

What is this question asking exactly? What is a "formal product"? Does it have anything to do with formal power series? How can be different products of the same expression? What counts as a valid "formal" product of an expression? Obviously if I just simplify the expression I get one thing, not a set. The book seems to not define "formal product", only "product of formal series" which yields a simplified form that I don't think is a set of all possible products.

答: The only difference between a formal product and a product is that in the former we are not concerned about whether it represents an actual number; it's just moving symbols and numbers around formally, i.e. by the rules of arithmetic.  For example $\sum_{i\ge 0}i!x^i=0!+1!x+2!x^2+3!x^3+\cdots$ is a series that has a radius of convergence 0, but formally we don't worry about which $x$, if any, lead to convergence.


答:
This is rather extremely complicated almost to a ludicrous extreme.

$\ln$ and $\exp$ are inverses so you can simplify this to:

$\ln\left(\sqrt{9-x^2}\right)-\ln\left(\exp\left(\frac{\ln\left(\left(9-x^2\right)^2\right)}{4}\right)\right)+1$

$\ln\left(\sqrt{9-x^2}\right)-\left(\frac{\ln\left(\left(9-x^2\right)^2\right)}{4}\right)+1=$

$\frac 12 \ln(9-x^2) - \frac {2\ln (9-x^2)}4 + 1 =$

$\frac 12 \ln (9-x^2) - \frac 12 \ln (9-x^2) + 1=$

$1$.

....

From what I can figure you have the equation of the line to begin with:  $y=1$.  So the line is simply all the points $(x, 1)$.

To get the equation from any two endpoints $(-3,1)$ and $(3,1)$ you really should just use the simple equation.

Endpoint 1: is $(x_1, y_1) = (3, 1)$.  Endpoint 2: is $(x_2, y_2) = (-3,1)$.  Slope is $m = \frac {y_2 - y_1}{x_2 - x_1} = \frac {1-1}{3-(-3)} = \frac 0 6 = 0$.

And the formula is $y - y_1 = m(x - x_1)$ or

$y - 1 = 0(x-3)$ or

$y - 1= 0$ or

$y = 1$.

That's it that's the equation of the line.  Which you knew from the very beginning.


问:
bad primes and torsion primes
Let $Φ$ be a root system of type $G_2$, with base ${α, β}$
where $α$ is short. Then $Ψ_1 :=$ ±{$α, 3α + 2β$} is a closed
subsystem of type $A_1A_1$, and clearly $|\mathbb{Z}Φ/\mathbb{Z}Ψ_1| = 2$; furthermore, $Ψ_2 =
±${$β, 3α + β, 3α + 2β$} is a closed subsystem of type $A_2$ with $|\mathbb{Z}Φ/\mathbb{Z}Ψ_2| = 3$, so $2$ and $3$ are certainly bad primes for $G_2$.

I can't get this: $|\mathbb{Z}Φ/\mathbb{Z}Ψ_2| = 3$. It's supposed to be the number in $\mathbb{Z}Φ$ which cannot be expressed as a linear combination of $\mathbb{Z}Ψ_2$, if I am correct? I only see $α,α + β$ .

答:
Hint 1: Via the isomorphism
$$\mathbb Z \Phi \simeq \mathbb Z \oplus \mathbb Z$$
$$n \alpha + m \beta \mapsto (n,m)$$

you are looking at the quotients of $\mathbb Z \oplus \mathbb Z$ with respect to the subgroups generated by $\{(1,0),(3,2)\}$ (for $\Psi_1$) and by $\{(0,1), (3,1),(3,2)\}$ (for $\Psi_2$), respectively. (Note also that the last subroup is already generated by $(0,1)$ and $(3,1)$ alone.)

Hint 2: The two elements of $\mathbb Z \Phi/ \mathbb Z\Psi_1$ are represented by $0$ and $\beta$. In $\mathbb Z \Phi/ \mathbb Z\Psi_1$, we have $2\beta \equiv -3\alpha \equiv 0$.

Hint 3: In $\mathbb Z \Phi/ \mathbb Z\Psi_2$ we actually have $\alpha \equiv \alpha+ \beta$ because $\beta \equiv 0$, so the two representatives you write in the question actually give the same element. Rather, since $\beta \equiv 0$, look at multiples of $\alpha$ alone and check how many there are up to equivalence in that quotient.

Title: bad primes and torsion primes


问:
Prove that idempotent operator $E$ is self-adjoint if and only if $EE^∗$ = $E^∗E$
Let $V$ be a finite-dimensional inner product space, and let $E$ be an idempotent
linear operator on $V$, i.e., $E^2 = E$. Prove that E is self-adjoint if and only if $EE^* = E^*E$.

Are there any simpler answers to the question that the answers provided here Normal, idempotent operator implies self-adjointness. . Both answers seem to be correct but contain logical steps that I can't comprehend e.g $(I−E)Ex=0 \Rightarrow (I−E^∗)Ex=0 $ and $v^\ast E^\ast Ev=0 \Rightarrow Ev=0$

答: Suppose $E$ is self-adjoint then $E=E^*$, So $EE^*=E^2=E^*E$. Now suppose conversely that $EE^*=E^*E$ then, $<EE^*(v),E^*E(v)> = <EE^*(v),E(v)>$ and also $$<E^*E(v),EE^*(v)> = <E(v),EE^*(v)> = <E^*E(v), E^*(v)>$$ and hence from these two equations that are true for all $v$ in $V$. We get $E^*(v)=E(v)$ for all $v$ belong to $V$ that is, $E$ is self adjoint.


问:
Can continuity be characterized by nets?
Let $X$ and $Y$ be topological spaces  (not necessarily assumed to be  Hausdorff or to have any additional property) and let $f:X\to Y$ be a given function.  Is it true that $f$ is continuous iff
for every $x\in  X$, and every net $\{x_i\}_i$ converging to $x$, one has that
$f(x_i)\to f(x)$.

PS: I have searched for this specific question in MSE and, although I found several posts discussing it (this,
this, and this) with varying degress of objectivity, and using various additional hypothesis (Hausdorff,
completely regular, first countable) I do not believe it can be found in the exact terms above.

I therefore thought it would be nice to register it here.

I am also providing an answer which I hope  is similar to the one in The Book $\ddot \smile$

答:
The proof of the "only if" part is well known, so let us prove the "if" part.

Assuming by contradiction that $f$ is discontinuous at a point $x$ in $X$, we will build a net $\{x_i\}_{i\in I}$ converging
to $x$, such that $\{f(x_i)\}_{i\in I}$ does not converge  to $f(x)$.  The set of indices for our net will be the set $N_x$
formed by  all neighborhoods of
$x$, and viewed as a directed set with order given by reverse inclusion,  namely,
$$
  U\geq V \Leftrightarrow   U\subseteq  V.
  $$

Since $f$ is discontinuous at $x$, there exists a neighborhood $U$ of $f(x)$ such that $f(V)\not\subseteq U$, for all
$V\in N_x$. Therefore,   for any such $V$ we may  choose some $x_V\in V$ such that $f(x_V)\not\in U$.

It is then evident that the net $\{x_V\}_{V\in  N_x}$ converges to $x$, while $\{f(x_V)\}_{V\in  N_x}$ does not converge to $f(x)$.

PS:  The same argument above shows that $f$ is continuous at a given point $x_0$,
if and only  if,  for every net $\{x_i\}_{i\in I}$,  converging to $x_0$,  one has that
$\{f(x_i)\}_{i\in I}$ converges to $f(x_0)$.


问:
Fixed Charge Transportation Problem, only fixed cost (constant), non-minimal cost solution acceptable
I want to solve the Fixed Charge Transportation Problem, although there is only a fixed cost for opening the route, no cost proportional to the edge weight. Each edge will have a maximum possible weight defined from the beginning.

The fixed cost d is 1 for every i,j and, at first, any solution, not only the one with minimal cost is acceptable, as it is possible to take cells forming a rectangle (in the matrix formulation) and redistribute quantities until one of them becomes zero, minimizing the cost. Getting the minimum cost is a hard problem in general and I may go without it.

What is the fastest algorithm to find a solution given that any cost is acceptable?

答:
You can solve the original problem via mixed integer linear programming as follows.  For each arc $(i,j)\in A$, let $u_{i,j}$ be the capacity of that arc, let nonnegative variable $x_{i,j}$ be the flow across that arc, and let binary variable $y_{i,j}$ indicate whether that arc is used.  The problem is to minimize $\sum_{(i,j)\in A} y_{i,j}$ subject to
\begin{align}
\sum_{(i,j)\in A} x_{i,j} &= s_i &&\text{for all $i$}\\
\sum_{(i,j)\in A} x_{i,j} &= d_j &&\text{for all $j$}\\
x_{i,j} &\le u_{i,j} y_{i,j} &&\text{for all $(i,j)$}
\end{align}
To find a feasible solution, you can fix $y\equiv 1$ and solve the resulting transportation problem, using either a linear programming solver or some specialized network algorithm.  You can then post-process to set $y_{i,j}=1$ if $x_{i,j}>0$.


答:
Step 1 (in the question): $\begin{equation}C_n=-\frac{1}{2}\bigg\{\frac{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})...(-\frac{2n-3}{2})}{n!}(-4)^n\bigg\}\end{equation}$

Step 2: $C_n=\frac{1}{2}\bigg\{\frac{\frac{1}{2}(\frac{1}{2})(\frac{3}{2})...(\frac{2n-3}{2})(n-1)!}{n!(n-1)!}(2^2)^n)\bigg\}$

In step 1, $-\frac{1}{2}, -\frac{3}{2}... -\frac{2n-3}{2}$ has $(n-1)$ terms with negative signs as $2n - 3 = 1 + 2 (n-2)$ and there is $-\frac{1}{2}$ outside curly brackets too. So the negative sign is taken out in step 2 as $(-1)^n \times (-1)^n = (-1)^{2n} = 1$.

Also note there are $n$ $2's$ in the denominator of $\frac{1}{2} (-\frac{1}{2}), (-\frac{3}{2})... (-\frac{2n-3}{2})$

Step 3: $C_n=\frac{1}{2}\bigg\{\frac{\frac{1}{2}[\frac{1}{2}\cdot 1 \cdot\frac{3}{2} \cdot 2...(n-2)(\frac{2n-3}{2})(n-1)]}{n!(n-1)!}2^{2n}\bigg\}$

i) Writing $4^n$ as $2^{2n}$, ii) In step 2, we multiplied numerator and denominator by $(n-1)!$. We now expand $(n-1)!$ in the numerator as $1 \times 2 \times 3 ...\times (n-1)$ and write in order.

Step 4: $C_n=\frac{1}{2}\bigg\{\frac{\frac{1}{2}(1 \cdot 2 \cdot 3 \cdot 4...(2n-4)(2n-3)(2n-2)}{n!(n-1)!}2^{2}\bigg\}$

In this step we cancel $2^{2n-2}$ in the numerator with $2^{n-1}$ in denominator and we also use $2^{n-1}$ to double each term of $(n-1)!$ from $1 \times 2 \times 3 ..(n-1)$ to $2 \times 4 \times 6 ... \times (2n-2)$ which will lead to numerator as $(2n-2)!$


问:
Show the differential operator is bounded
I need to show the differential operator is bounded. I believe it is bounded on $C[0,1]$ with the sup norm by 1, but I am struggling to show this. What I have so far:
NTS: $\|Tf\|\leq\|T\|\|f\|$ Where:

\begin{equation} \|Tf\|=\sup_{\|f\|=1}\|Tf\|=\sup_{\|f\|=1} \sup_{x\in [0,1]} |f'(x)|=\sup_{\|f\|=1} \sup_{x\in [0,1]} \left|\frac{\lim_{h\rightarrow 0} f(a+h)-f(a)}{h}\right|
\end{equation}

Not entirely sure if I should have written the derivative like that in the last equality above. I need a hint to see where to go next.

Thanks in advance!

答:
What about the effect of differentiating a sequence f sub n = sin(2pi* n* x) where n is integer > 0?

You get 2pi *n *cosine, whose supremum is infinite


问:
Prove that integral converges
How can I prove that this integral converges?

$$\int\limits_1^{\infty}\frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}\ dx$$

I tried to show that function $0 < \frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}$ is decreasing and is continious for $x$ in $(1,\infty)$. Furthermore the above integral converges when series $\sum_{1}^{\infty}\frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}$ converges.

Are there better methods?

答:
Near $x=1$, since $\ln x=\ln(1+(x-1))\sim x-1$, and $\sqrt{x^2-1}\sim\sqrt{x-1}$, you can compare your integral with $\frac{1}{\sqrt{x-1}}$.

Near $x=\infty$ (let me express in this way), since $\ln x \ll x^\alpha$ for every $\alpha>0$, and $(x-1)\sqrt{x^2-1}\sim x^{3/2}$, you can compare your integral with $\frac{1}{x^{3/2-\alpha}}$ for any $\alpha>0$ such that $\frac{3}{2}-\alpha>1$ (take $\alpha=1/4$ for instance).

Title: Prove that integral converges


问:
Maximal domain of definition of rational morphism induced by linear system.
Let $L$ be a invertible sheaf on $X$ and $s_0, \dots , s_n \in \Gamma(X,L)$ linearly independent. Then $\{s_i\}$ induces a rational morphism from $X$ to $\mathbb{P}^n$.

It is well know that if $s_i(x)$ all vanish, then the rational morphism cannot be defined at $x$. But what if $s_i = f s_i'$ for all $s_i$ locally and $s_i'$ not all vanishing at $x$. Then We can extend this morphism to $x$.

This may be impossible for $X$ a (projective) variety. But I don't know how to show it.

QUESTION: Whether or not, the maximal domain of definition of rational morphism induced by linear system is the set away from the base.

答: This isn't true. Consider $X=\Bbb P^1$, $\mathcal{L}=\mathcal{O}(2)$, and $s_0=x^2,s_1=xy$. Then this defines a nonconstant rational map $\Bbb P^1\to\Bbb P^1$, which extends to a morphism - any rational map from a regular curve to a projective variety extends to an honest morphism (ref).


问:
Where does the cos(phi) term come from that often shows up in spherical integrals?
When integrating functions defined on the surface of a unit sphere, you often get an integral that looks something like:

$\int{d\theta}\int{d\phi}\cos\phi S(\theta,\phi)$

Where $S(\theta,\phi)$ is your function defined on the surface of the sphere, and $\theta$ and $\phi$ are the azimuth and elevation angles, respectively (or longitude and latitude angles, if you prefer).

I've most often seen this in situations where you're trying to find the amount of total energy falling on a point (i.e. by summing in all incoming directions to that point), so you see it a lot in rendering equations (especially radiosity), computational acoustics simulations using plane waves, flux, etc...

My question though, is where does the $\cos\phi$ or, depending on your coordinate system, sometimes $\sin\phi$, term come from?

My guess is this is accounting for the fact that as you near the poles on the sphere, the elevation angles become closer together, and this accounts for the change in infinitesimal area that you're breaking the integral up into. Is that the correct interpretation?

答:
In the above spherical system, $r$ is the radius, $\theta$ is the azimuth angle and $\phi$ is the elevation angle.

Here $\theta$ is the angle between $x$ axis (in the Cartesian system ) and $r\sin\phi$.

Considering $r$ as radius, infinitesimal change in arc length for infinitesimal change in $\phi $ is $dl = r \ d\phi$.

Similarly you can get $(r\sin\phi)\ d\theta$.

We can vary any $2$ of these $3$ quantities infinitesimally keeping the other  quantity constant and get elemental areas,

$dS_{r = \text{const}} = r\sin\phi \ d\theta \cdot rd\phi = r^2\sin\phi \ d\theta d\phi$

$dS_{\phi = \text{const}} = dr \cdot r\sin\phi \ d\theta = r\sin\phi \ dr d\theta$

$dS_{\theta= \text{const}} = dr \cdot rd\phi = r \ dr d\phi$

If you integrate over the sphere keeping $r = 1$ as constant, you need to consider the 1st case,

$dS_{r = 1} =\sin\phi \ d\theta d\phi$

Based on your definition of the angles, you may get either $\sin\phi$ or $\cos\phi$


Title: Prove that integral converges

Body:
How can I prove that this integral converges?

$$\int\limits_1^{\infty}\frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}\ dx$$

I tried to show that function $0 < \frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}$ is decreasing and is continious for $x$ in $(1,\infty)$. Furthermore the above integral converges when series $\sum_{1}^{\infty}\frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}$ converges.

Are there better methods?

Body:
Hint:

Observe that $\;(x-1)\sqrt{x^2-1}=(x-1)^{3/2}(x+1)^{1/2}\;$ , and thus

$$\frac{\log x}{(x-1)^{3/2}\sqrt{x+1}}\le\frac{x^{1/4}}{(x-1)^{3/2}}\le\frac1{x^{9/8}}$$


问:
Numerical analysis objective question.
The equation $e^x-4x^2=0$ has a root between $4$ and $5$. Fixed point iteration with iteration function $\frac{1}{2}e^{\frac{x}{2}}$ is

$1.$ diverges .

$2.$ converges.

$3.$ oscillates.

$4.$ converges montotonically.

The same question is asked before FIxed Point Iteration (numerical analysis) but for me it seems to be convergent .

I am thinking it like if I take any initial point in between $4$ and $5$ then sequence start to decreasing and converges to the fixed point of iteration function that is before $4$. Like the picture added below So according to me answer is $D$. Am I right . Please comment. Thank you .

答:
Diverges as $f(x)=\frac{e^{x/2}}{2}$ is not Lipschitz in $(4,5)$

Indeed derivative is $f'(x)=\frac{e^{x/2}}{4}$ and $|f'(x)|>1$ in the interval $(4,5)$

Out of curiosity, $f(x)=x$ has another solution in $(0,1)$ which can be found with fixed point method.

In the picture below we can see $f(x)$ and $y=x$. Should be clear the reason why $|f'(x)|<1$ is a requirement for the convergence of the fixed point method.

$$...$$

Title: Numerical analysis objective question.


问:
How to sum numerator and denominator of a fraction?
I want to do sum over this.  Can apply the summation to top and bottom separately?

$$\sum\limits_{i=1}^{n} \frac{-a(x_i-\mu)^2}{x_i}$$
$$=\frac{\sum\limits_{i=1}^{n}-a(x_i-\mu)^2}{\sum\limits_{i=1}^{n}x_i}$$

Is this correct?

Where can I find the rules to summations of a division, product, or addition?

Edit:

I want to solve for y,
$$\frac{5}{y}=\sum\limits_{i=1}^{n} \frac{-a(x_i-\mu)^2}{x_i}$$
$$\frac{y}{5}=\sum\limits_{i=1}^{n} \frac{x_i}{-a(x_i-\mu)^2}$$

Is this correct?

答: One thing you can do which might be helpful is, assuming $x_i>0$ for all $i$,  $$\frac{-a(x_i-\mu)^2}{x_i}=\frac{-a(x_i-\mu)^2}{(\sqrt{x_i})^2}=-a\left(\sqrt{x_i}-\frac{\mu}{\sqrt{x_i}}\right)^2$$


问:
AF C*-Algebras and Continuous Functions on Totally Disconnected Sets
Why do the continuous functions on a totally disconnected set, such as the Cantor set, form an AF C$^*$-algebra?  Conversely, why do commutative AF C$^*$-algebras consist of continuous complex functions on a totally disconnected compact metrizable space?  I am new to operator theory and have only begun reading papers in the subject.  Both claims appear in different papers and I have not been able to find references for these results.</p>

答:
Let $\mathcal{A}$ be an unital commutative $C^{*}$-algebra. Suppose that $\mathcal{A}$ is Approximately Finite-dimensional. By the Gelfand Theorem, we can suppose that $\mathcal{A}=C(X,\mathbb{C})$, where $X$ is a compact Hausdorff topological space.

A commutative finite-dimensional $C^{*}$-álgebra is generated by a finite number of projections (idempotents and self-adjoints elements). Since our $C^{*}$-algebra $C(X,\mathbb{C})$ is AF, it must be generated by a sequence of projections, that is, there exists a sequence $(p_{n})$ of projections of $C(X,\mathbb{C})$ such that the closed linear span generated by the $p_{n}$'s coincides with $C(X,\mathbb{C})$.

Now, if $f\in C(X,\mathbb{C})$ is a projection, we have $f^{2}=f$, and so, $f=\chi_{A}$ (the characteristic funcion of $A$), where $A=f^{-1}(\{1\})$. Since $f$ is continuous, the sets $A=f^{-1}(\{1\})$ and $X\setminus A=f^{-1}(\{0\})$ are closed subsets of $X$, that is, $A$ is a clopen (i.e, open and closed) subset of $X$. Conversely, the characteristic functions of clopen subsets of $X$ are continuous functions on $X$. Therefore, the projections of the $C^{*}$-algebra $C(X,\mathbb{C})$ are the characteristic functions of the clopen subsets of $X$

Therefore, our $C^{*}$-algebra $C(X,\mathbb{C})$ coincides with the closed linear span  of a family of characteristic functions $\{\chi_{A_{n}}:n\in\mathbb{N}\}$, where $A_{n}$ is a clopen subset of $X$ for each $n\in\mathbb{N}$. In particular, the function constantly equal to $1$ on $X$ can be uniformly approximated by linear combinations of the $\chi_{A_{n}}$'s, and then, $X=\bigcup\limits_{n}A_{n}$ (if there existed an element $x\in X$ that does not belong to any $A_{n}$, any linear combination of the $\chi_{A_{n}}$'s would take the value $0$ in $x$, and then, we could not approximate the function constantly equal to $1$ on $X$)

Now, let us see that $X$ is totally disconnected.

We will use the following fact: a compact Hausdorff space $Y$ is totally disconnected if and only if for all $x, y\in Y$ with $x\neq y$, there exists a clopen subset $A$ of $Y$ such that $x\in A$ and $y\notin A$ (the reverse implication is almost immediate. For the other implication, see this Any two points in a Stone space can be disconnected by clopen sets)

We proceed by contradiction. Let us suppose that $X$ is not totally disconnected. Then, there exists $x, y\in X$, $x\neq y$  such that every clopen subset of $X$ that contains $x$, also contains $y$. Since the $A_{n}$'s are a covering of $X$, there exists a minimum $n_{0}\in\mathbb{N}$ such that $x\in A_{n_{0}}$, and then, $A_{n_{0}}$ also contains $y$. Since the singletons $\{x\}$ and $\{y\}$ are closed and $X$ is a compact Hausdorff space (in particular, it is a normal topological space), by the Urysohn's Lemma, there exists a continuous function $\varphi:X\longrightarrow [0,1]$ such that $\varphi(x)=1$ and $\varphi(y)=0$. Since $C(X,\mathbb{C})$ coincides with the closed linear span of the $\chi_{A_{n}}$'s, there exists $N\in\mathbb{N}$ and scalars $a_{1},\dots,a_{N}$ such that $\|\varphi-\sum_{i=1}^{N}a_{i}\chi_{A_{i}}\|_{\infty}<1/2$.

Denote by $J=\{1\leq m\leq N: x,y\in A_{m}\}$. We know that $J=\{1\leq m\leq N: x\in A_{m}\}$. Also, $J=\{1\leq m\leq N: y\in A_{m}\}$ (if we had $1\leq m\leq N$ such that $y\in A_{m}$ and $x\notin A_{m}$, then $X\setminus A_{m}$ is a clopen subset which contains $x$, but it does not contain $y$, and we would have a contradiction).

Evaluating at $x$ and $y$, we obtain respectively $|1-\sum_{i\in J}a_{i}|<1/2$ and $|\sum_{i\in J}a_{i}|<1/2$, which is absurd.

Therefore, $X$ is totally disconnected


问:
Atiyah-Macdonald: Exercise 3.5
Let $A$ be a ring. Suppose that, for each prime ideal $\mathfrak{p}$, the local ring $A_{\mathfrak{p}}$ has no nilpotent elements $\neq 0$. Show that $A$ has no nilpotent element $\neq 0$. If each $A_{\mathfrak{p}}$ is an integral domain, is $A$ necessarily an integral domain?

I have found an answer to the second question. Consider $A=\mathbb{Z_{4}}$, the only prime ideal of $A$ is $\{\bar{0},\bar{2}\}$ then for $S:=A\setminus\mathfrak{p}$ we have $S^{-1}A=\{\frac{0}{1},\frac{1}{1},\frac{2}{1},\frac{1}{3}\}$ which is an integral domain, but $A$ is not. (The property of being Integral Domain is not local.)

For the first part, I tried to proceed as follows: Let there exist $(0\neq)a \in A$ such that $a^n=0$, again $1 \in S$ $\implies$ $\frac{a}{1} \in S^{-1}A$ such that $(\frac{a}{1})^n=\frac{a^n}{1}=\frac{0}{1}$. To arrive at a contradiction it remains to show that $\frac{a}{1}\neq 0.$ If $\frac{a}{1}=\frac{0}{1}$ then there exist $t \in A\setminus \mathfrak{p}$ such that $ta=0$.

I cannot think of any way out from here. Please help.

Is the converse true? (i.e Is "the existence of a nonzero nilpotent element in $A$" a local property?)

答: You are on the right track, and you're correct that you're done if you can show $\frac{a}{1}\ne 0,$ i.e. there is no $t\in A\setminus\mathfrak p$ such that $ta=0.$ But keep in mind you only need to find one prime ideal $\mathfrak p$ for which this holds. So you can take $\mathfrak p$ to be a maximal extension of the ideal $\{b\in A: ba=0\}.$

Title: Atiyah-Macdonald: Exercise 3.5


问:
Solution to an inexact differential equation with the difference between partial derivatives not single variable
How can I find the general solution to the equation $$\left(x^2+xy+\frac{y^2}{x}\right)dx+(x^2+xy-y)dy=0$$
Note that, by using exact differentials, I could reduce it to $(x+y)d(x+y)=yd\left(\frac{y}{x}\right)$, but could not solve it. The difference between $M_y-N_x=\frac{2y}{x}-x-y$, where $M=x^2+xy+\frac{y^2}{x}$, $N=x^2+xy-y$. Any hints? Thanks beforehand.

答:
You already reduced it to $(x+y) \,d(x+y) = y \,d(\frac{y}{x})$.

Now you can just use change of variable -

$\displaystyle x + y = u, \frac{y}{x} = v$

$ \displaystyle \implies x = \frac{u}{v+1}, y = \frac{uv}{v+1}$

So $d(x+y) = \frac{y}{(x+y)} \,d(\frac{y}{x})$ becomes

$\displaystyle du = \frac{v}{v+1} dv \,$.


问:
Showing that a rational number exists between any two reals
The following is from Real Analysis by N. L. Carothers.

Theorem: If $a$ and $b$ are real numbers with $a < b$, then there is a rational number $r$ $\in \mathbb{Q}$ with $a < r < b$.

Proof: Since $b - a > 0$, we may apply the Archimedean property to get a positive integer $q$ such that $q(b - a) > 1$. But if $qa$ and $qb$ differ by more than one, then there must be some integer in between.$\\$I.e. $\;qa < p < qb\implies a < \dfrac{p}{q}< b\;$, $\;\dfrac{p}{q}\in \mathbb{Q}\;$.

What I do not understand is the "must" part. Why does there exist an integer between the two reals? Is this tied to some general result that shows the existence of some $p$ quantity of something when two objects (= numbers) differ by some $c$?

答:
If $x,y\in\Bbb R$ are such that $\color{blue}{y-x>1}$, then $\lfloor x\rfloor+1$ is an integer and, since $\color{blue}{x<\lfloor x\rfloor+1}$ and$$\color{blue}{\bigl(\lfloor x\rfloor+1\bigr)-x}=1-\bigl(x-\lfloor x\rfloor\bigr)\color{blue}{\leqslant1},$$the integer$\lfloor x\rfloor+1$ belongs to $(x,y)$.

Indeed,

$$x<\lfloor x\rfloor+1=\bigl[\bigl(\lfloor x\rfloor+1\bigr)-x\bigr]+x\leqslant1+x<y\;.$$

Note: $\lfloor x\rfloor$ is the greatest integer smaller than or equal to $x$.


问:
Chain rule for partial derivates
It's given that  $P \equiv P(V,T)$ and $E \equiv E(V,T)$. These are known expressions.

If $S=-\frac{\partial F}{\partial T}$, $\boldsymbol{P}=-\frac{\partial F}{\partial V}$ and $F(V,T)= E(V,T)- TS(V,T)$, what will be an expression for $S$ using chain rules of partial differentiation?

答: $$-S= \frac{\partial (E-TS)}{\partial T} =\frac{\partial E}{\partial T} -\frac{\partial(TS)}{\partial T}=\frac{\partial{E}}{\partial T}-\left[T\frac{\partial S}{\partial T}+S\right]$$ using the usual product rule. Simplifying yields $$\frac{\partial S}{\partial T} = \frac 1T \frac{\partial E}{\partial T} \\ S=\int \frac 1 T \frac{\partial E}{\partial T} \ \partial T $$

Title: Chain rule for partial derivates


问:
Reference for a proof of the Hahn-Mazurkiewicz theorem
The Hahn-Mazurkiewicz theorem states that a space $X$ is a Peano Space if and only if $X$ is compact, connected, locally connected, and metrizable. If anybody knows a book with a proof, please let me know. Thanks.

P.S. (added by t.b.) A Peano space is a topological space which is the continuous image of the unit interval.

答:
Hocking and Young, Topology
also has a proof, theorem 3-30, page 129. Visible in Google Books.</p>


问:
The level surfaces of the equation $f(x,y)=y(x^2-y^2)$
So as the title suggests that I have to find the level curves of the equation $f(x,y)=y(x^2-y^2)$ for f(x,y)= 1,0,-1. I trivially found that for 0 the equation simply becomes $x^2=y^2$ which is a cross. However, for 1 and -1 I can't wrap my head around it. I see that we define the equation as $Y(x^2-y^2)=C $ and we see that there is the term $x^2-y^2$ that can be converted to the hyperboloid. The problem with this equation is the Y outside of the brackets, due to this I have no clue how to convert it to the equation. $x^2/C+Y^2/C$
If somebody could provide me with help, it would greatly be appreciated!

答:
The level curves are cubics $x^2y-y^3=c$

I draw some of them in the picture below.

Hope this helps

$$...$$


问:
Proving $T$ and $U$ are equal if I know $(X,T)$ and $(X,U)$ are hausdorff and compact
This question is from Wayne Patty Exercise 4.2

Let $T$ and $U$ are Topologies on set X. Suppose $ (X,T) $ and $(X,U)$ are compact Haurdsorff Spaces. Prove that $T=U$ or $T$ is not a subset of $U$ or $U$ is not a subset of T.

I thought of proving it by this way: Assuming that if $T$ is a subset of $U$ or $U$ is a subset of $ T$ then $T=U$ must be true.
But I was unable to think on how to use compactness or hausdorff property and was unable to proceed.

Well, also this question was asked before on this website. But I am unable to understand how the solution will work despite trying to understand it for a fair amount of time as it was just outlined.

So, I am looking for a detailed solution.

答:
Suppose that $\mathscr{T}\subsetneqq\mathscr{U}$, and let $U\in\mathscr{U}\setminus\mathscr{T}$. Let $K=X\setminus U$; $K$ is not closed in $\langle X,\mathscr{T}\rangle$, but $K$ is closed in $\langle X,\mathscr{U}\rangle$, which is a compact space, so $K$ is compact in $\langle X,\mathscr{U}\rangle$: every cover of $K$ by $\mathscr{U}$-open sets has a finite subcover.

Let $\mathscr{V}$ be a cover of $K$ by $\mathscr{T}$-open sets. (I.e., $K\subseteq\bigcup\mathscr{V}$, and $\mathscr{V}\subseteq\mathscr{T}$.) $\mathscr{T}\subseteq\mathscr{U}$, so $\mathscr{V}$ is automatically also a $\mathscr{U}$-open cover of $K$, and as such it has a finite subcover $\mathscr{V}_0$. Thus, every cover of $K$ by $\mathscr{T}$-open sets has a finite subcover, and $K$ is therefore compact in $\langle X,\mathscr{T}\rangle$. But every compact subset of a Hausdorff space is closed, and $\langle X,\mathscr{T}\rangle$ is Hausdorff, so $K$ is closed in $\langle X,\mathscr{T}\rangle$. This contradiction shows that $\mathscr{T}$ cannot be a proper subset of $\mathscr{U}$.

The same argument, with the rôles of $\mathscr{T}$ and $\mathscr{U}$ reversed, shows that $\mathscr{U}$ cannot be a proper subset of $\mathscr{T}$. Thus, either $\mathscr{T}=\mathscr{U}$, or $\mathscr{T}$ and $\mathscr{U}$ are not related by inclusion at all.


问:
detail in proof of the mean value theorem for integrals
I am trying to prove the following result

If $f$ is a continuous function on a closed interval $[a,b]$, then
there exists a $c \in (a,b)$ such that,

$$f(c)=\frac{1}{b-a}\int_a^bf(x)\;\mathrm{d}x$$

I have worked out the following proof

Because $f$ is continuous on the interval, it reaches a minimum $m$, and a maximum $M$ so that we have
$$\begin{aligned} m&\leq f(x)\leq M \\ m(b-a)\leq  & \int_a^bf(x)\;\mathrm{d}x\leq M(b-a) \\  m\leq &\frac{1}{b-a}\int_a^bf(x)\;\mathrm{d}x\leq M\end{aligned}$$
and the IVT guarantees the exxistence of $c$ as above.

How do I prove that $c \neq a\ \& \ c\neq b$?

Thanks for the help !

答:
Sometimes a figure is worth 1000 words:

What is the area in blue?

$$\int\limits_a^b f(x)\ dx$$

What is the area in orange?

$$f(c) \cdot (b-a)$$

QED.


答:
Observe that for $x\geq 2$ we have $\frac{x}{2}\geq 1$ and $\frac{x^2}{2}\geq 2\geq 1$ . Hence,
\begin{align}
(x-1)\cdot \sqrt{x^2-1}&\geq (x-\frac{x}{2})\cdot \sqrt{x^2-\frac{x^2}{2}}\\
&=\frac{x}{2}\cdot \sqrt{\frac{x^2}{2}}\\
&=\frac{x^2}{2\sqrt{2}}
\end{align}
Hence for every $x\geq 2$
$$\frac{1}{(x-1)\cdot \sqrt{x^2-1}}\leq \frac{2\sqrt{2}}{x^2}$$
This shows that
\begin{align}
\int_{2}^{\infty}\frac{\ln x}{(x-1)\cdot \sqrt{x^2-1}}\,dx&\leq \int_{2}^\infty \frac{2\sqrt{2}\ln x}{x^2}\,dx\\
&=2\sqrt{2}\int_{2}^\infty \biggl(\frac{-1}{x}\biggr)'\cdot \ln x\,dx\\
&=2\sqrt{2}\biggl(-\frac{1}{x}\ln x \biggr|_{2}^{\infty}\biggr)+2\sqrt{2}\int_{2}^\infty\frac{1}{x^2}\,dx\\
&=\sqrt{2}\ln 2+2\sqrt{2}\biggl(-\frac{1}{x}\biggr|_{2}^\infty\biggr)\\
&=\sqrt{2}\ln 2+\sqrt{2}
\end{align}

Now, to treat the case where $x>1$ and $x$ being close to $1$, observe that
$$(x-1)\leq (x-1)\cdot \sqrt{x^2-1}\leq x\cdot (x-1)$$
for all $x>1$. Hence, for $x>1$
$$\frac{\ln x}{x\cdot (x-1)}\leq \frac{\ln x}{(x-1)\cdot \sqrt{x^2-1}}\leq \frac{\ln x}{x-1}$$
Using DLH to evaluate the limits of the right and left hand side of the above inequality we obtain
$$\lim_{x\to 1^+}\frac{\ln x}{(x-1)\cdot \sqrt{x^2-1}}=1$$
This shows also that
$$\int_{1}^2 \frac{\ln x}{(x-1)\cdot \sqrt{x^2-1}}\,dx<\infty$$

Title: Prove that integral converges


问:
Making $-\log(2-4x)\log(x-5)+\log(2x-1)=\log(3-4x)$ into a quadratic equation
How we can work with below equation to get  quadratic equation?

$$-\log(2-4x)\log(x-5)+\log(2x-1)=\log(3-4x)$$

No need to get the variable x

答:
we know that $\log x$ is defined for $x>0$ so

$\log(2-4x) \Rightarrow 2-4x>0 ,\  \frac{1}{2}>x$

$\log(2x-1) \Rightarrow 2x-1 >0 ,\ x> \frac{1}{2}$

that is a contradiction which means there is not a solution


问:
Prove that $[\mathbb{Q}(\sqrt{ 3+4i}+\sqrt{ 3-4i}):\mathbb{Q} ] =1 $
let $\sqrt{ 3+  4i}$   denote the square root of the complex number $3 + 4i$ that lies in the first quadrant and let    $\sqrt{ 3-  4i}$   denote the square root of $3 - 4i$ that lies in the fourth quadrant. Prove that $[\mathbb{Q}(\sqrt{ 3+4i}+\sqrt{ 3-4i}):\mathbb{Q} ] =1    $

My attempt:         $\sqrt{ 3+  4i}=\sqrt{ 3+  \sqrt{(-16)}}$

$\sqrt{ 3-  4i}=\sqrt{ 3-  \sqrt{(-16)}}$

take       $x=\sqrt{ 3+  \sqrt{(-16)}}+\sqrt{ 3-  \sqrt{(-16)}}$

taking square on both sides,we get

$x^2=3+  \sqrt{(-16)}+3-  \sqrt{(-16)}+2\sqrt{ 3+  \sqrt{(-16)}}\sqrt{ 3-  \sqrt{(-16)}}$

$x^2=6+10=16$ this implies $x=4$

degree of $4$ is $1$

Therefore $ [\mathbb{Q(4):\mathbb{Q}}]=1$

Is its correct ?

答:
It is enough to verify that $\sqrt{3+4i}=2+i$ and $\sqrt{3-4i}=2-i$ from which

$$\sqrt{ 3+4i}+\sqrt{ 3-4i}=4\in\mathbb Q$$


问:
Free abelian group $F_i$ with free basis $B_i$
I have to prove this statement:

Let free abelian group $F_i$ with free basis $B_i$ for $i=1,2$ then $F_1\cong F_2$ iff $\lvert B_1 \rvert = \lvert B_2 \rvert$.

I prove it by the idea that element of basis is linear independent and they generate whole group. So we have injection $F_1= \langle B_1 \rangle \rightarrow \langle B_2\rangle=F_2$, WLOG $\lvert B_1 \rvert \leq \lvert B_2 \rvert$.

So now is trivial both implication.

My question is can I do use this idea for proving?

答:
I believe this can be shown using the universal property of free objects. I am assuming that all bases of a free abelian group have the same cardinality.

Let $\phi: F_1 \to F_2$ be an isomorphism. Then, I claim that $\phi(B_1)$ is a basis for $F_2$. To show this, let $G$ be an abelian group and $f: \phi(B_1) \to G$ be a function. Then, $f \circ \phi: B_1 \to G$ is a function. By the universal property of free objects, there is some unique $g: F_1 \to G$ so that $g$ restricts to $f \circ \phi$ on $B_1$. Then, $\phi \circ g: F_2 \to G$ is the unique homomomorphism which restricts to $f$ on $\phi(B_1)$, proving that $\phi(B_1)$ is a basis of $F_1$. Since $\phi$ is an isomorphism, it is a bijection. Then, since all bases of free abelian groups have the same cardinality, $$|B_1| = |\phi(B_1)| = |B_2|.$$

EDIT: Here is a proof that bases of free modules over sufficiently nice rings have equal cardinality. This applies here since $\mathbb{Z}$ is a commutative ring with 1.


问:
Stiefel-Whitney numbers for disjoint union of a Manifold
I am reading the book by Milnor & Stasheff on Characteristics Classes, In the 4th Chapter There's is this beautiful result by Thom" If all of the Stiefel-Whitney numbers
of M are zero, then M can be realized as the boundary of some
smooth, compact manifold."

Just after which as an example It is written that "A disjoint union of two copies of a manifold M has all Stiefel-Whitney no. zero."

Why is that true? It can easily be seen by applying the converse of above as the disjoint union can be regarded as the boundary of M x [0,1].
Is there a way to directly prove it?

Any hints are appreciated.

答:
It is clear from the proof of additivity of cohomology under disjoint union that $H^*(M \sqcup M) = H^*(M) \oplus H^*(M)$ as rings.

It is also clear from the definition of the Stiefel-Whitney classes (for instance, naturality) that $w_i(M \sqcup M) = (w_i(M), w_i(M))$ above.

For a partition $p$ of $n = \dim M$ (a sequence of integers $1 \leq i_1 \leq \cdots \leq i_j \leq n$ with $i_1 + \cdots + i_j = n$) there is an associated product $$w_p(M) = w_{i_1}(M) \cdots w_{i_j}(M) \in H^n(M).$$

In your case, the first two sentences give you that $w_p(M \sqcup M) = (w_p(M), w_p(M)) \in H^n(M \sqcup M) = H^n(M) \oplus H^n(M)$.

Now the Stiefel-Whitney numbers of $M$ are obtained by pairing with the fundamental class; I write $s_p(M) = \langle w_p(M), [M]\rangle$ for this integer mod 2.

Then we have $$s_p(M \sqcup M) = \langle w_p(M \sqcup M), [M \sqcup M]\rangle = \langle (w_p(M), w_p(M)), ([M], [M])\rangle = \langle w_p(M), [M]\rangle + \langle w_p(M), [M]\rangle = 2s_p(M) = 0.$$


问:
Cantor set prove question solution verfication
Question:
Let C be the Cantor set contained in $[0, 1]$, as constructed in class. Let A be any (nonempty) connected subset of C. Prove that A consists of exactly one point.
You may use the key theorems that we have proved about connected sets, and the definition of the Cantor set as constructed (it is the intersection of the sets Cn, which are unions of $2^n$ separated intervals each of width $1/3^n$).

My proof:
Let $C⊆[0,1]$ be the Cantor set. For every x,y∈C with x≠y there exists by the construction of C some z∈[0,1] with z∉C and $x<z<y$. Then both $(−1,z)∩C=[−1,z]∩C $and $(z,2)∩C=[z,2]∩C$ are clopen in C with respect to the subspace topology.

So every two points of C are respectively contained in two disjoint clopen subsets of C. Because every connected component of C is contained in a clopen subset of C it follows that no two distinct points of C are contained in the same connected component. So C is totally disconnected, i.e. all connected components are singletons.

Can anyone help me check my proof is right or wrong? thank you so much

答:
You are assuming that the construction of $C$ ensures that between any two points of $C$ there is a point not in $C$. If you actually proved that fact during the construction of $C$, then you can use it as you did, the problem is trivial, and your proof is correct. If not, however, you are essentially assuming the result that you’re trying to prove, and you have more work to do.

HINT: Suppose that $x,y\in C$, and $x<y$. For each $n\in\Bbb N$ let $I_n(x)$ be the interval of $C_n$ containing $x$, and let $I_n(y)$ be the interval of $C_n$ containing $y$.

Show that $\bigcap_{n\in\Bbb N}I_n(x)=\{x\}$ and $\bigcap_{n\in\Bbb N}I_n(y)=\{y\}$.

Conclude that there is an $n\in\Bbb N$ such that $I_n(x)\cap I_n(y)=\varnothing$.

Let $I_n(x)=[a,b]$ and $I_n(y)=[c,d]$. Show that $b<c$.

Now you can pick any $z\in(b,c)$ and complete the argument just as you did originally.


答:
Let $g(x) = \frac{1}{2}e^{x/2}$. Then $g'(x) = \frac{1}{4}e^{x/2}$ so $g'(4) = 1.85$, $g'(5) = 3.05$. Let $c$ be the solution between $4$ and $5$ Based on the equation if you define a sequence by $x_{n + 1} = g(x_{n})$ then by Taylor's theorem and using $g(c) = c$,
$$x_{n + 1} - c = g'(\xi(c, x_n))(x_n - c).$$
where $\xi(c, x_n)$ is between $c$ and $x_n$.
Since $|g'(c)| > 1$, the iteration moves away from the fixed point when you are close to it, so the iteration likely won't converge to $c$ unless you start exactly at $x_0 = c$. For example, suppose $x_0 > c$. Then $g'(\xi(c, x_n)) > 1$ for every $n$ so $x_n > c$ for every $n$ and
$$x_{n + 1} - c > g'(c)(x_n - c) > g'(c)^{n + 1}(x_0 - c) \to \infty$$
If $x_0 < c$, then the iteration won't converge to $c$, but may converge to some other value less than $c$. Simulating it demonstrates that it converges to a solution at $0.7148059$. When in doubt, simulate it.

Title: Numerical analysis objective question.

Body:
The equation $e^x-4x^2=0$ has a root between $4$ and $5$. Fixed point iteration with iteration function $\frac{1}{2}e^{\frac{x}{2}}$ is

$1.$ diverges .

$2.$ converges.

$3.$ oscillates.

$4.$ converges montotonically.

The same question is asked before FIxed Point Iteration (numerical analysis) but for me it seems to be convergent .

I am thinking it like if I take any initial point in between $4$ and $5$ then sequence start to decreasing and converges to the fixed point of iteration function that is before $4$. Like the picture added below So according to me answer is $D$. Am I right . Please comment. Thank you .


问:
About the correct definition of a binomial random variable
I'm having trouble with how to exactly define a binomial random variable.

Let's fix a discrete probability space $(\Omega,P)$, where $|\Omega| \le \aleph_0$.

Then, if $X\colon \Omega \to \mathbb{R}$ is a real random variable, when exactly can we say that it is binomial ($X \sim Bi(n,p)$)?

I found two possible (non equivalent) definitions:

$X$ is the sum of $n$ independent Bernoulli random variables $X_1\colon \Omega \to \{0,1\},\dots,X_n\colon \Omega \to \{0,1\}$, where $X_1 \sim Be(p),\dots,X_n \sim Be(p)$;

The probability distribution of $X$ is given by $p_X(k)=P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$ when $k \in \{0,\dots,n\}$, while $p_X(x)=0$ if $x \in \mathbb{R} \setminus \{0,\dots,n\}$.

I know that 1) implies 2), but in general it is not true the converse, so that 1) is not equivalent to 2).

So maybe I should use 1) as a definition. Right?

I apologize for the low level of my question.

Thank you!

答:
It's not a natural thing to do in probability to fix the sample space $\Omega$. The important thing is the random variable(s) and you can and should give yourself the freedom to modify the sample space as necessary to incorporate any new random variable(s) you want to discuss. This is discussed clearly and in detail in, for example, Terence Tao's 275A, Notes 0: Foundations of probability theory.

With the freedom to modify the sample space in mind, the two definitions are equivalent.

Said another way, the question is: what does it mean for two random variables to be "equivalent"? To my mind the cleanest answer is that they have the same distribution (not that they are literally the same function on the same sample space $\Omega$), or equivalently that they induce the same pushforward probability measures on $\mathbb{R}$. So when we define binomial random variables (or Gaussians or whatever else) we really only care about defining random variables up to this notion of equivalence, which does not depend on a choice of probability space. From this point of view the sense in which definitions 1) and 2) are equivalent is that they define the same probability measure on $\mathbb{R}$.

If you want to be really pedantic you can use definition 2) but I can't imagine a situation where the distinction would really matter anyway.


答:
As the example in the comment shows nicely, no - Another simple example shows why this can't work: $$\frac 12 + \frac 12 \neq \frac 24 = \frac 12$$

Next: You might want to visit Wikipedia's entry for Summation, which includes a nice list of valid manipulations that hold for sums.

Title: How to sum numerator and denominator of a fraction?

Body:
I want to do sum over this.  Can apply the summation to top and bottom separately?

$$\sum\limits_{i=1}^{n} \frac{-a(x_i-\mu)^2}{x_i}$$
$$=\frac{\sum\limits_{i=1}^{n}-a(x_i-\mu)^2}{\sum\limits_{i=1}^{n}x_i}$$

Is this correct?

Where can I find the rules to summations of a division, product, or addition?

Edit:

I want to solve for y,
$$\frac{5}{y}=\sum\limits_{i=1}^{n} \frac{-a(x_i-\mu)^2}{x_i}$$
$$\frac{y}{5}=\sum\limits_{i=1}^{n} \frac{x_i}{-a(x_i-\mu)^2}$$

Is this correct?


问:
Solution verification: check if $H_t(x)= \frac{1}{2 \sqrt{\pi t}}e^{-\frac{x^2}{4t}}$ is approximate identity
Let $H_t(x)= \frac{1}{2 \sqrt{\pi t}}e^{-\frac{x^2}{4t}}$, $t>0$. Show that $H_t(x)$ is an approximate identity on $\mathbb{R}$ when $t\rightarrow 0$.

I need to check the third property of the approximate identity. Here is my solution, can someone just check if this is correct:

$$I = \frac{1}{2 \sqrt{\pi t}} \int_{|x|> \delta} e^{-\frac{x^2}{4t}} dx  =\frac{1}{2 \sqrt{\pi t}} \int_{|x|> \delta} e^{-(\frac{x}{2 \sqrt{t}})^2} dx .$$

Substitute: $u= \frac{x}{2 \sqrt{t}} $, $ du= \frac{dx}{2\sqrt{t}} $:

$$I= \frac{1}{ \sqrt{\pi }} \int_{|u|> \frac{\delta}{2\sqrt{t}}} e^{-u^2}  du .$$
Then since $t \rightarrow 0 $ we have $ \frac{\delta}{2\sqrt{t}}  \rightarrow  \infty$. Function  $ e^{-u^2} $ is integrable, so  $I \rightarrow 0$.

Thanks a lot in advance.

答: It's the nascent delta $\tfrac{1}{\epsilon}\eta(\tfrac{x}{\epsilon})$ with $\epsilon:=\sqrt{t}$ and $\eta(z):=\frac{1}{2\sqrt{\pi}}\exp(-z^2/4)$ a PDF for which $\eta(0)>0$, namely that of a Normal distribution with $\mu=0,\,\sigma^2=2$.


问:
Where does the cos(phi) term come from that often shows up in spherical integrals?
When integrating functions defined on the surface of a unit sphere, you often get an integral that looks something like:

$\int{d\theta}\int{d\phi}\cos\phi S(\theta,\phi)$

Where $S(\theta,\phi)$ is your function defined on the surface of the sphere, and $\theta$ and $\phi$ are the azimuth and elevation angles, respectively (or longitude and latitude angles, if you prefer).

I've most often seen this in situations where you're trying to find the amount of total energy falling on a point (i.e. by summing in all incoming directions to that point), so you see it a lot in rendering equations (especially radiosity), computational acoustics simulations using plane waves, flux, etc...

My question though, is where does the $\cos\phi$ or, depending on your coordinate system, sometimes $\sin\phi$, term come from?

My guess is this is accounting for the fact that as you near the poles on the sphere, the elevation angles become closer together, and this accounts for the change in infinitesimal area that you're breaking the integral up into. Is that the correct interpretation?

答:
In performing a change of variables to spherical coordinates, $(\rho, \phi, \theta) \mapsto (\rho \sin\phi \cos\theta, \rho \sin\phi \sin\theta, \rho \cos\phi)$, you need to take into account the Jacobian determinant
$$ \left| \begin{matrix} \sin\phi\cos\theta & \rho\cos\phi\cos\theta & -\rho\sin\phi\sin\theta \\
\sin\phi\sin\theta & \rho\cos\phi\sin\theta & \rho\sin\phi\cos\theta \\
\cos\phi & -\rho\sin\phi & 0 \end{matrix} \right| = \rho^2 \sin\phi. $$


问:
Mathematics puzzles
I have a question about this math puzzle and the question offers the numbers 1, 3, 4, 5, 6, 7, 8, 9, 10:

I already solved it:

A=4,
B=9,
C=10,
D=8,
E=7,
F=3,
G=1,
H=6,
J=5

However, it took me a while because I was trying the possibilities without a specific rule, so is there a rule to solve it?

Thanks

答:
Immediately we see that $D=8$.

Then, I think the most logical step is to try to place the $10$. It is trivially not in places $A, E, F, G, H, J$ because the minimum row and column sums fail in those cases. Thus, either $B$ or $C$ contains the $10$.

Now, $B=10 \implies H=5\implies (G,J)\in \{(1,6),(6,1),(3,4),(4,3)\}$

Using that $C+F+J=18$, $A+E+G=12$, $A+J=9$, $C+G=11$ we can show that each of these pairings break somewhere:

If $G=1$ then $C=10$ but $B=10$.

If $J=1$ then $C+F=17 \implies C,F=8,9$ but $D=8$

If $J=4$ then $C+F=14\implies C,F = 6,8$ or $9,5$ but $D=8$ and $H=5$ (also $J=4\implies A=5$)

If $G=4$ then $A+E=8\implies A,E=1,7; 2,6; 3,5$ but $C=7$; $2$ is already placed and $H=5$

So $B\neq 10 \implies C=10\implies G=1$

We can now see that:
$$A+E=11\implies A\in\{4,5,6,7\}=A_1$$
$$A+B=13\implies A\in\{4,6,7,9\}=A_2$$
$$A+J=9\implies A\in\{3,4,5,6\}=A_3; J\in \{3,4,5,6\}=J_1$$
$$F+J=8\implies J\in\{3,5\}=J_2$$
$$H+J=11\implies J\in\{4,5,6,7\}=J_3$$

I've achieved this by listing pairs and discarding any that contain $\{1,2,8,10\}$ that we have already placed.

Now: $$J_1\cap J_2\cap J_3 =\{5\}\implies J=5\implies A=4$$

and the rest fall out from there.

Title: Mathematics puzzles


问:
Model categories: Any two solutions to a lifting problem are homotopic
I am reading the paper "Model Categories and Simplicial Methods" by Goerss and Schemmerhorn  and I am struggling to prove Lemma 2.11, which states the following.

Suppose that there is a lifting problem in a model category $\mathcal{C}$ (shown below), where $j$ is a cofibration, $q$ is a fibration, and one of $j$ or $q$ is a weak equivalence. Then $f$ exists and is unique up to left homotopy under $A$ and over $Y$.

The paper says that this Lemma follows from the model category axioms, but I don't see how.

My attempt

I don't have any very promising angles, but I will list a few things that I have tried.

By the model category axioms, the lifting problem admits a solution (i.e. at least one $f$ exists), so the difficulty is proving that for any two such solutions $f, f'$, we have a (left) homotopy under $A$ and over $Y$ from $f$ to $f'$. I am not very confident with the notions of homotopies over and under objects, so I have just been trying to prove that there is a left homotopy from $f$ to $f'$.

Any homotopy is a factoring of the map
$$
f\coprod f':B\coprod B \to X.
$$
The model category axioms tell us that there is a factorisation
$$
A\coprod A \overset{\alpha}{\to} Z \overset{\beta}{\to} X,
$$
where $\alpha$ is a cofibration and $\beta$ is an acyclic fibration. This factorisation looks a bit like a homotopy, particularly because $i$ is a cofibration. Therefore it would suffice for $Z$ to be a cylinder object of $A$. However, I do not see any reason why $Z$ should be a cylinder object, so this seems to be a dead end.

答:
Suppose that we have two lifts $f,f'$ $q$ is a weak equivalence (the two cases are dual), so we hava a diagram as below

Since this diagrams commutes, we have a lift $\eta$ which is a left homotopy between  $f$ and $f'$.  Also, if all these objects were fibrant and cofibrant (which most of the time we require them to be), we would have that $f \circ j = f' \circ j$,  so that $[f \circ j]=[f] \circ [j] = [f'] \circ [j]$, and since $j$ is a weak equivalence, [j] is an isomorphism, and so we could've canceled out $[j]$ and arrive at $[f]=[f']$.


答:
Assume $w<z, w,z\in [a,b]$ with $f(w) = m, f(z) = M$

If $m=M \implies f$ is a constant function then all $c\in(a,b)$ works.

Suppose $m<M$

We have that for $c\in [w,z]$, $f(c) = \frac{1}{b-a}\int_{a}^{b}f(x)dx$

Suppose $c = a$, then $a = w$, because $a\leq w \leq c \leq  z \leq b$

Then $f(a) = m$ is the  minimum of the function

$f(z) =M>m \implies$ there exist some $\delta >0$ such that $f(x)>m+\epsilon$ for some $\epsilon>0$ for all $x\in (z-\delta,z]$

Thus we have that $\begin{aligned}\int_{a}^{b}f(x)dx & = \underbrace{\int_{a}^{z-\delta}f(x)dx}_{\geq f(a)\cdot(z-\delta -a)} +\underbrace{\int_{z-\delta}^{z}f(x)dx}_{\geq (f(a) + \epsilon)\cdot\delta>f(a)\cdot \delta}+\underbrace{\int_{z}^{b}f(x)dx}_{\geq f(a)\cdot(b-z)}>f(a)\cdot(b-a)\\  &\underbrace{\frac{1}{b-a}\int_{a}^{b}f(x)dx}_{f(c)} > f(a)\end{aligned}$

that is a contradiction. Then $c \neq a$

I hope it help you,
I think the other cases will be analogous

(for example, apply this reasoning to $-f$)

Title: detail in proof of the mean value theorem for integrals

Body:
I am trying to prove the following result

If $f$ is a continuous function on a closed interval $[a,b]$, then
there exists a $c \in (a,b)$ such that,

$$f(c)=\frac{1}{b-a}\int_a^bf(x)\;\mathrm{d}x$$

I have worked out the following proof

Because $f$ is continuous on the interval, it reaches a minimum $m$, and a maximum $M$ so that we have
$$\begin{aligned} m&\leq f(x)\leq M \\ m(b-a)\leq  & \int_a^bf(x)\;\mathrm{d}x\leq M(b-a) \\  m\leq &\frac{1}{b-a}\int_a^bf(x)\;\mathrm{d}x\leq M\end{aligned}$$
and the IVT guarantees the exxistence of $c$ as above.

How do I prove that $c \neq a\ \& \ c\neq b$?

Thanks for the help !


问:
How to solve $\int \frac{x^3}{(1+x)^5}\ dx$?
I am stuck at this integral. Can anybody help or give me a hint what to do with the denominator?

$$\int \frac{x^3}{(1+x)^5} \, dx$$

答:
Try to integrate by parts:

$$
\int x^2\cdot (1+x)^{-5}\ dx=x^2\cdot \left(-\frac{1}{4}(1+x)^{-4}\right)+\int 2x\cdot \frac{1}{4}(1+x)^{-4}\ dx
$$

and then apply again the rule:

$$
\int 2x\cdot \frac{1}{4}(1+x)^{-4}\ dx= \frac{1}{2}x\left(-\frac{1}{3}(1+x)^{-3}\right)+\int \frac{1}{2}\cdot \frac{1}{3}(1+x)^{-3}\ dx
$$

and the last integral is immediate

$$
\int \frac{1}{2}\cdot \frac{1}{3}(1+x)^{-3}\ dx=-\frac{1}{12}(1+x)^{-2}+cost
$$

so, summing up all the terms, you get

$$
\int \frac{x^2}{(1+x)^5}\ dx=-\frac{6x^2+4x+1}{12(1+x)^4}+cost.
$$


问:
Does the intersection of two concave polygons without holes result in a set of several polygons without holes?
I did not find a proof anywhere, so here is my assumption :

If I have 2 concave and simple polygons without holes, then the intersection yields a set of different polygons without any holes.

If it can result to a polygon with a hole inside, could you provide an example?

Edit:
the polygons are simple. edges of one polygone don't intersect, except consecutive edges, which intersect in their common vertex. The polygons in input do not have holes. I consider the intersection of the interiors of the polygons. The resulting set is all the points that are included both in interior of polygon P1 and the interior of the polygon P2.

Thanks.

答:
Call the two polygons $P_1$ and $P_2$. Suppose that some point $x \not \in P_1 \cap P_2$, but there is a curve inside $P_1 \cap P_2$ encircling $x$.
Since $x \not \in P_1 \cap P_2$ then $x \not \in P_1$ wlog. But there is a curve encircling $x$ which is inside $P_1$ and therefore $P_1$ has a hole.


Title: How to solve $\int \frac{x^3}{(1+x)^5}\ dx$?

Body:
I am stuck at this integral. Can anybody help or give me a hint what to do with the denominator?

$$\int \frac{x^3}{(1+x)^5} \, dx$$

Body: Note that\begin{align}\int\frac{x^3}{(1+x)^5}\,\mathrm dx&=\int\frac{((1+x)-1)^3}{(1+x)^5}\,\mathrm dx\\&=\int(1+x)^{-2}\,\mathrm dx-3\int(1+x)^{-3}\,\mathrm dx+\\&\phantom{=}+3\int(1+x)^{-4}\,\mathrm dx-\int(1+x)^{-5}\,\mathrm dx.\end{align}Besides$$n\ne-1\implies\int(1+x)^n\,\mathrm dx=\frac1{n+1}(1+x)^{n+1}.$$


问:
Generating points in rectangle
I am trying to generate $N$ points randomly and uniformly distributed in an $m \times n$ rectangle. How can this be done? I have tried to split the initial rectangle into as many rectangles i could, but this didn't work.</p>

答: For each point, independently generate two numbers: $l_i\sim U(0,m)$ and $h_i\sim U(0,n)$. The resulting point $(l_i,h_i)$ is a point in the rectangle and the joint distribution is uniform over the entire rectangle. </p>

Title: Generating points in rectangle


问:
Exchanging limit and integral
If $\sup_{x \in \mathbb{R}^n} |f_m(x)|\rightarrow 0$ for $f$ a Schwartz function (i.e. $f$ is smooth and decays rapidly), is it true that $\|f_m\|_{L^1}=\int_{\mathbb{R}^n} |f_m(x)| dx \rightarrow 0$ as $m \rightarrow \infty$?

This seems to be true but I am not sure how to justify it.

答:
Not true.
Pick $f_m = \frac{1}{m} 1_{\lbrack 0, m \rbrack }$.

Title: Exchanging limit and integral


问:
Prove $A, B, C, D$ are $N(0,1)$ random variables
$A$ and $B$ are i.i.d RVs, assume $\mathbb{E}[A^2]$ exists, $\mathrm{var}{A}=1$, and that MGF of $A$ exists near zero. $C=(A+B)/\sqrt{2}$ and $D=(A-B)/\sqrt{2}$. Now prove that if. $C$ and $D$ are i.i.d., $A, B, C, D$ are $N(0,1)$ random variables.

Basic computation got me to figure out that the expectation and variance of all four RVs are 0 and 1 respectively, and I also know that RVs that have an MGF such that $M(2s)=(M(s))^4$ have a normal distribution, but I am not sure how to proceed from there.

Body:
Put $\phi(t) = Ee^{itA} = Ee^{itB}$.  We have

\begin{gather}
\phi(\frac{t+s}{\sqrt{2}}) \phi(\frac{t-s}{\sqrt{2}}) = Ee^{i A ( \frac{t+s}{\sqrt{2}})} Ee^{i B (\frac{t-s}{\sqrt{2}})} = Ee^{i( A ( \frac{t+s}{\sqrt{2}}) +   B (\frac{t-s}{\sqrt{2}}))}=
Ee^{i(t \frac{A+B}{\sqrt{2}} + s \frac{A-B}{\sqrt{2}} )} = Ee^{i(t \frac{A+B}{\sqrt{2}} )} Ee^{i(s \frac{A-B}{\sqrt{2}} )} =\\
= \phi(\frac{t}{\sqrt{2}}) \phi(\frac{t}{\sqrt{2}}) \phi(\frac{s}{\sqrt{2}}) \phi(-\frac{s}{\sqrt{2}})  \text{  ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ (1)}
\end{gather}
Moreover, $Ee^{i(t \frac{A+B}{\sqrt{2}} )} =  Ee^{i(t \frac{A-B}{\sqrt{2}} )} $ and thus
\begin{gather}
 \phi(\frac{t}{\sqrt{2}}) \phi(\frac{t}{\sqrt{2}}) = \phi(\frac{t}{\sqrt{2}}) \phi(-\frac{t}{\sqrt{2}})   \text{  ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ (2)}
\end{gather}
Taking $\sqrt{2}t$ and $\sqrt{2}s$ instead of $t$ and $s$ we get from (1) and (2) that
$$\phi(t+s) \phi(t-s) =  \phi^2(t)\phi(s)\phi(-s),  \text{  ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ (3)}$$
$$\phi^2(t) = \phi(t) \phi(-t).  \text{ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ (4)}$$
Hence $$\phi(t+s) \phi(t-s) =  \phi^2(t)\phi^2(s),  \text{  ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ (5)}.$$
Taking $-t$ instead of $t$ in (4) we get
$$\phi^2(-t) = \phi(t) \phi(-t) \text{  ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ (6)}$$
It follows from (4) that either $\phi(t) = 0$ or $\phi(t) = \phi(-t)$. It follows from (6) that either $\phi(-t) = 0$ or $\phi(t) = \phi(-t)$. Hence we have $\phi(t) = \phi(-t)$ for all $t$.  Thus $\overline{\phi(t)} = \overline{Ee^{itA}} = Ee^{-itA} = \phi(-t) = \phi(t)$ . We got that $\phi(t) \in \mathbb{R}$.

Further, MGF exists near $0$, so $\exists EA^n$ for all $n \ge 1$ and hence $\phi(t) \in C^{\infty}(\mathbb{R})$. But $\phi(0)=1$, thus $\phi(t) \ne 0$ near zero. Put $T = \inf(t>0: \phi(t) = 0)$. If $T < \infty$, then let $t \to T-$ and $s \in (0, \frac{T}2)$. We have $\phi(t), \phi(s) \ne 0$ and it follows from (5) that $\phi(t+s) \ne 0$. We got a contradiction. Hence, $T = \infty$ and $\phi(t) \ne 0$  for all $t \ge 0$. Similarly we get that $\phi(t) \ne 0$ for all $t$. We also know that $\phi(t) \in \mathbb{R}$ is continuous and $\phi(0)=1$. Thus $\phi(t) > 0$.

Put $g(t) = \ln \phi(t)$. Then $g(t) \in C^{\infty}(\mathbb{R})$. It follows from (5) that
$$g(t+s) + g(t-s) = 2(g(t) + g(s)).\text{  ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ (7)}$$

Fix $t \ne 0$. We know that
$$g((k+1)t) + g((k-1)t) = 2(g(kt) + g(t))$$
for all integer $k$ and that $\phi(0)=0$. Taking $k=1$ we get that $g(2t) = 4g(t)$ and by induction we get that $g(pt) = p^2g(t)$ for integer $p \ge 0$. Moreover, we may get similarly that $g(pt) = p^2g(t)$ for all integer $p$.
Put $u = tq$ for some integer $q$. Thus
$$g(\frac{p}{q}u) = g(pt) = p^2g(t) = p^2g(\frac{u}{q}) = \frac{p^2}{q^2} f(u).$$
Hence $g(\frac{p}{q}) = \frac{p^2}{q^2} g(1)$ and by continuity of $g$ we get that $g(x) = cx^2$.

We got that $$Ee^{iAt} = e^{g(t)} = e^{ct^2}$$
for some $c \in \mathbb{R}$. We know that $|Ee^{iAt}| \le E|e^{iAt}| = 1$, hence $c \le 0$. If $c=0$ then $A=0$, $DA = 0$ and we got a contradiction. Hence $c < 0$. Thus $A \sim N(0, \sigma^2)$. But $DA =1$. Hence $A \sim N(0,1)$ and $B \sim N(0,1)$ also. Now it's easy to see that $C \sim N(0,1)$  and $D \sim N(0,1)$ , q.e.d.


问:
the cofinite topology in $Y$ and the topology of $Y$-subspace induced by the topology of $X$
Be $X$ a set of cofinite topology and $Y \subseteq X$. Are $\tau_{c}$ and $\tau_{s}$, respectively, the cofinite topology in $Y$ and the topology of $Y$-subspace induced by the topology of $X$. Compare $\tau_{c}$ and $\tau_{s}$ with respect inclusion.

Definition 1. Be $X$ an infinite set and set $\tau=\{\emptyset\}\cup \{V\subseteq X:X-V$ is finite$\}$.

Definition 2. Be $X$ a set, $\tau$ a topology, and $Y\subseteq X$.  Then the $X$-induced topology is given by: take the $X$ open and intersect with $Y$.

Can anyone help me with this? I think the only difference from $\tau_{c}$ topology and topology is that in $\tau_{c}$ topology:
\begin{equation}
\tau_{c}=\{\emptyset\}\cup \{V\subseteq Y: Y-V \text{ is finite } \}
\end{equation}

while whereas

\begin{equation}
\tau_{s}=\{\emptyset\}\cup \{V\subseteq X: X-V \text{ is finite } \}\cap Y
\end{equation}

in my view it doesn't change anything. Change anything, which is the finest?

答: Take a finite set $V\subset Y$, this means $Y- V$ is open in the cofinite topology of $Y$, now $X-V$ is also open in the cofinite topology of $X$ and $Y\cap (X-V)=Y-V$, so $Y- V$ is also open in the topology inherited by $X$. It is pretty obvious that $Y\cap (X- V)$ is open in the cofinite topology of $Y$ for any finite $V\subset X$, so you have the double incusion.


答:
If $V\in\tau_c$ let $F=Y\setminus V$, and let $U=X\setminus F$. $F$ is a finite subset of $X$, so $U$ is open in $X$, and clearly

$$U\cap Y=(X\setminus F)\cap Y=(X\cap Y)\setminus F=Y\setminus F=V\,,$$

so $V\in\tau_s$. This shows that $\tau_c\subseteq\tau_s$; now see if you can show that $\tau_s\subseteq\tau_c$ to conclude that $\tau_s=\tau_c$.

Body:
Be $X$ a set of cofinite topology and $Y \subseteq X$. Are $\tau_{c}$ and $\tau_{s}$, respectively, the cofinite topology in $Y$ and the topology of $Y$-subspace induced by the topology of $X$. Compare $\tau_{c}$ and $\tau_{s}$ with respect inclusion.

Definition 1. Be $X$ an infinite set and set $\tau=\{\emptyset\}\cup \{V\subseteq X:X-V$ is finite$\}$.

Definition 2. Be $X$ a set, $\tau$ a topology, and $Y\subseteq X$.  Then the $X$-induced topology is given by: take the $X$ open and intersect with $Y$.

Can anyone help me with this? I think the only difference from $\tau_{c}$ topology and topology is that in $\tau_{c}$ topology:
\begin{equation}
\tau_{c}=\{\emptyset\}\cup \{V\subseteq Y: Y-V \text{ is finite } \}
\end{equation}

while whereas

\begin{equation}
\tau_{s}=\{\emptyset\}\cup \{V\subseteq X: X-V \text{ is finite } \}\cap Y
\end{equation}

in my view it doesn't change anything. Change anything, which is the finest?


问:
Proof of Eakin-Nagata theorem, Matsumura $thm2.6$
Eakin-Nagata theorem,in Matsumura $thm3.6$,I have question.

Let $A$ be a commutative ring and $M$ is a faithful finitely generated module over it.If the ascending condition holds on the submodules of the form $IM$ for ideals $I⊂A$, the module $I/IM$ is noetherian ring.

Proof

It is enough to show that $M$ is a Noetherian module since, in
general, strong texta ring admitting a faithful Noetherian module over
it is a Noetherian ring. Suppose otherwise. By assumption, the set of
all $IM$, where $I$ is an ideal of $A$ such that $M/I$ is not
Noetherian has a maximal element, $J/M$  .
Replacing $M$ and $A$
by$M/JM$ and $A/Ann（M/JM）$, we can assume, for each nonzero submodule
$I⊂A$,the module $M/IM$ is noetherian.

Why can we replace $M$ and $A$ to $M/JM$ and $A/Ann（M/JM）$?
I think the latter two satisfies the hypothesis of the theorem,
but I believe we cannot say every $M$ and $A$ can be denoted like
$M/JM$ and $A/Ann（M/JM）$.

Thank you for your kind help.

答:
The argument is a proof by contradiction. Here is a rough outline of the strategy: Suppose there is a counterexample $M$ over a ring $A$ (meaning that $M$ satisfies the hypotheses of the theorem but is non-Noetherian). Then we can construct from this one counterexample $M$ another counterexample $M'$ over another ring $A'$ that has an additional property (that $M'/IM'$ is Noetherian for any non-zero ideal $I$ of $A'$). The remainder of the proof then derives a contradiction using this additional property.

So, what this shows is that there's no counterexample that has this additional property. But the proof starts by showing: if there is any counterexample whatsoever to the theorem, then there is a counterexample with that additional property. So the theorem can't have any counterexample, which completes the proof by contradiction.

Of course, if this were a direct proof, then the strategy wouldn't make sense, because you'd only be proving something about modules of that particular form. But since it's a proof by contradiction, we just need to start by assuming there's a counterexample and derive a contradiction from this, and it's perfectly fine if one of the intermediate steps is "show that the existence of a counterexample implies the existence of another counterexample, and consider this other counterexample instead".


问:
The minimum distance of the $(u | u+v)$ construction.
For some $\mathcal{C}_i = [n, k_i, d_i]$ in $\mathbb{F}_2$ where $i\in\{1,2\}$ the $(u | u+v)$ construction is defined as
\begin{align*}
    \mathcal{C} = \{(u\ |\ u+v)\ |\ u\in\mathcal{C}_1,\ v\in\mathcal{C}_2\}
\end{align*}

I have to derive the deriving the minimum distance of this code, and the value that all other derivations* give is $d_{\min} = \min(2d_1, d_2)$. However in all these derivations the case where $v=u\neq 0$ is not considered. In this case you would have
\begin{align*}
    ||(u\ |\ u+v)|| = ||u\ |\ 2u|| = ||u\ |\ 0|| \geq d_1
\end{align*}
So in the cases where $u$ and $v$ can be equal, $d_{\min} = \min(2d_1, d_2)$ would not hold. Instead it should be $d_{\min} = \min(d_1, d_2)$.

Is there anything wrong with the counterproof I made, or is there a particular reason that they do not consider the case where $v=u\neq 0$?

*Like the derivation on page 28 here, or the derivation from page 4 onward here.

答:
If $u=v$, then $u$ is a word of the code $\mathcal{C}_2$. Hence $(u|0)=(v|0)$ has weight at least $d_2\ge\min\{2d_1,d_2\}$, so the claim still holds.

My experience may be a bit more limited than it should, but I'm under the impression that the $(u|u+v)$ construction is typically used when $d_2>d_1$ and $\mathcal{C}_2\subset \mathcal{C}_1$ — the Reed-Muller codes being a well known case. I recall being a bit surprised to learn that the bound holds more generally.