问:
Contour integral with power of sine
I am trying to show that, for $\alpha > 1$,

$$\int_{0}^{\infty} \sin(x^{\alpha}) dx = \sin \left ( \frac{\pi}{2\alpha} \right ) \int_{0}^{\infty} \exp(-x^{\alpha}) dx$$

The approach I've used has been to take the imaginary part of the integral of $\exp(i x^{\alpha})$ over the line. Then I'm trying to compute this with a wedge contour with angle $\pi/2\alpha$. The horizontal segment and the diagonal segment are fine, but I'm having trouble showing that the arc integral goes to zero, i.e.

$$\lim_{R \to \infty} \int_{0}^{\pi/2\alpha} \exp(i R^\alpha \exp(i \alpha \theta)) R i \exp(i \theta) d\theta = 0$$

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{\left.\int_{0}^{\infty}\sin\pars{x^{\alpha}}\dd x
\,\right\vert_{\,\alpha\ >\ 1} =
\sin\pars{\pi \over 2\alpha}
\int_{0}^{\infty} \exp(-x^{\alpha})\,\dd x}:\ {\Large ?}}$.

\begin{align}
&\bbox[5px,#ffd]{\left.\int_{0}^{\infty}\sin\pars{x^{\alpha}}
\dd x\,\right\vert_{\,\alpha\ >\ 1}}
\,\,\,\stackrel{x^{\alpha}\ \mapsto\ x}{=}\,\,\,
{1 \over \alpha}\,\Im\int_{0}^{\infty}
x^{1/\alpha - 1}\,\,\expo{\ic x}\dd x
\\[5mm] = &\
{1 \over \alpha}\,\Im\int_{0}^{\infty}
\ic^{1/\alpha - 1}\,\,\,y^{1/\alpha - 1}\,\,
\expo{-y}\,\ic\,\dd y\quad
\pars{\substack{\mbox{I'll "close" the integral}
\\[0.5mm] \mbox{ along a quarter circle}
\\[0.5mm] \mbox{ in the complex plane}
\\[0,5mm] \mbox{first quadrant.}}}
\\[5mm] = &\
{1 \over \alpha}\sin\pars{\pi \over 2\alpha}
\int_{0}^{\infty}y^{1/\alpha - 1}\,\,\expo{-y}\,\dd y
\\[5mm] \stackrel{y\ =\ x^{\alpha}}{=}\,\,\,&
{1 \over \alpha}\sin\pars{\pi \over 2\alpha}
\int_{0}^{\infty}\pars{x^{\alpha}}^{1/\alpha - 1}\,\,\expo{-x^{\alpha}}\pars{\alpha x^{\alpha - 1}}\dd x
\\[5mm] = &\
\bbx{\sin\pars{\pi \over 2\alpha}\int_{0}^{\infty}
\expo{-x^{\alpha}}\dd x} \\ &
\end{align}

Title: Contour integral with power of sine


问:
A question related to box topology ( unable to understand the contradiction)
This question is from take home assignments and some examples preceed them.

Consider these 2 images:

In the last line of 2nd image, I am unable to see what will be the contradiction?

Rest all analysis is clear to me.

Please help with it.

答:
Directly:

If $x\in \Bbb R$ then $f(x)=(x_n)_{n\in \Bbb N}\in\Bbb R^{\omega}$ where $x_n=x$ for every $n.$ So for any $x$ we have $$x\in f^{-1}B\implies f(x)\in B\implies \forall n\in \Bbb N\,(  x=x_n\in (-1/n, 1/n) \,)\implies$$ $$\implies  x\in\cap_{n\in \Bbb N}(-1/n,1/n)=\{0\}.$$

So $f^{-1}B\subseteq \{0\}.$ And (obviously) $0\in f^{-1}B.$ So $f^{-1}B=\{0\},$ which is not open in $\Bbb R.$

ContentLicense: CC BY-SA 4.0


问:
Ramanujan summation of alternating factorials: why is it $0.596...$?
I was reading Ramanujan's letter to Hardy and came across this identity:
$$0!-1!+2!-3!+...=0.596...$$
Ramanujan didn't give the exact value of that constant, but it is the Euler-Gompertz constant, as stated in the comments.
This seems nonsensical, but in fact, this makes sense, if we study some methods of "forcing divergents to converge". For example, $1+2+3+....$ is $-1/12$ because $\zeta(-1)$ is $-1/12$. But I can't prove why this is true. So my question is:

How can this be proved?

Sorry if this question is stupid.

答:
The answer is a little bit buried in one of the formulas in the Wikipedia article for the Euler-Gompertz constant:

$$G = \int_0^{\infty} \frac{e^{-x}}{1 + x} \, dx.$$

This integral can be expanded formally into a divergent series as follows. We have

$$\int_0^{\infty} \frac{e^{-x}}{1 + x} \, dx = \int_0^{\infty} \sum_{n \ge 0} (-1)^n x^n e^{-x} \, dx "=" \sum_{n \ge 0} (-1)^n \int_0^{\infty} x^n e^{-x} \, dx$$

(the equal sign being in quotes because the exchange of sum and integral here is not actually justified) and the integral defining the Gamma function gives

$$\int_0^{\infty} x^n e^{-x} \, dx = n!.$$

This can be proven pretty straightforwardly by induction on $n$ using integration by parts, and there is also a cute generating function proof. So, formally, we get $G "=" \sum_{n \ge 0} (-1)^n n!$. (I consider it quite misleading to use ordinary equality to talk about sums of divergent series.)


问:
What is the Kernel of $\omega$?
I'm totally confused about the concept of kernel, could someone help me solve this?

Suppose that $\omega(\begin{bmatrix}j&k\\l&m\end{bmatrix})=(j+k)+(j+l)x+(j+m)x^2$. What is the Kernel of $\omega$?

答:
It seems $\omega$ is a map $\omega: \mathrm{Mat}_{2\times 2}\to P^2$, so a $2\times 2$ matrix is mapped on a polynomial of (at most) degree 2. To find the kernel, you have to find all such matrices which are mapped onto the null polynomial. So every coefficient has to be zero.

So $j+k=0$ and $j+l=0$ and $j+m=0$, which gives $j=-k=-l=-m$.

So every matrix of the form $\begin{pmatrix}j&-j\\-j&-j\end{pmatrix}$ is an element of the kernel of $\omega$.

Title: What is the Kernel of $\omega$?


问:
Non-prime-generating polynomial
Is there a polynomial, which generates only composites and never primes? I only need a theoretic proof, no need for the actual polynomial. It can have non-integer coefficients.</p>

答: If $x$ is an integer, then $4x^2+4$ is a positive multiple of $4$.

Title: Non-prime-generating polynomial


问:
How to determince convergence and divergence of
How to determine convergence and divergence of $a_n =\sqrt[\Large^n]c$ ?

答: Assume $c > 0$. Further assume $c > 1$ because if $c < 1$ we can put $c = \dfrac{1}{b} < 1, b > 1$. Thus put $c = a+1, a > 0$, then $ 1 < a_n = (1+a)^{\frac{1}{n}}= \sqrt[n]{1\cdot 1\cdot 1\cdots (1+a)} < \dfrac{1+1+\cdots + 1+(1+a)}{n}= \dfrac{n+a}{n}= 1+\dfrac{a}{n}$. Thus $a_n \to 1$ by sandwich theorem. If $c = 1$, the limit is $1$.


问:
How can I find all possible solutions of a trigonometric equation?
I know the modus operandi,

given $\sin x + \sin 2x = 0$, (range of x = $[0, \pi]$)

1st, rewrite the equation: $\sin x = -\sin 2x$.

2nd, on the cartesian plane, plot the first sine out, and do the same thing for the other one.

3rd, find the intercepts between the first and the second function.

I've followed this algorithm, and I've figured out that this equation has at most two solutions (0 and $\pi$), but the multiple choice question said otherwise.

(1) it has exactly one solution.

(2)  it has exactly 3 solutions.

(3) it has infinite solutions.

(4) it has exactly 2 solutions.

the correct answer (according to exercise) is (2).

EDIT (correct solution):

the modus operandi is correct, the problem is that I didn't consider ($\pi/4
\space and \space 3/2\space\pi)$. Meaning, half of $\pi/2$ and $\pi/2 + \space half$. Personally, I prefer using this method over the others.

答:
$\sin(2x) = 2\sin(x)\cos(x).$

Therefore, the equation is equivalent to

$\sin(x) + 2\sin(x)\cos(x) = 0.$

Therefore $[\sin(x)][1 + 2\cos(x)] = 0.$


答: $-\sin2x=\sin(-2x)$ and so we have $\sin x=\sin(-2x)$. Use the identity $\sin\theta=\sin(\pi-\theta)$ to solve your equation. We get two branches: $x=-2x+2k\pi$ and $x=\pi-(-2x)+2k\pi$. This will give you the general solution set. Then you can apply the required interval. Can you take it from here?

Title: How can I find all possible solutions of a trigonometric equation?

Body:
I know the modus operandi,

given $\sin x + \sin 2x = 0$, (range of x = $[0, \pi]$)

1st, rewrite the equation: $\sin x = -\sin 2x$.

2nd, on the cartesian plane, plot the first sine out, and do the same thing for the other one.

3rd, find the intercepts between the first and the second function.

I've followed this algorithm, and I've figured out that this equation has at most two solutions (0 and $\pi$), but the multiple choice question said otherwise.

(1) it has exactly one solution.

(2)  it has exactly 3 solutions.

(3) it has infinite solutions.

(4) it has exactly 2 solutions.

the correct answer (according to exercise) is (2).

EDIT (correct solution):

the modus operandi is correct, the problem is that I didn't consider ($\pi/4
\space and \space 3/2\space\pi)$. Meaning, half of $\pi/2$ and $\pi/2 + \space half$. Personally, I prefer using this method over the others.


问:
Volume integral - finding the region
I need to find the volume of an object set with the following function:

$$x + y + z = 1$$

And all three axis.

So I converted the function into $z = -x - y + 1$, and it gave me kind of a clepsydra, crossing x,y plane in x = y = 1.

So the region seems to be the circle $x^{2}+y^{2}=1$.

However, changing x to r, and y to $\phi$ gives me wrong result - $\pi$, and it should be 1/6.

Is the region I've came up with correct?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
{\cal V} & \equiv\bbox[5px,#ffd]{\iiint_{\pars{0,1}^{\,3}}\,
\bracks{x + y + z < 1}\dd x\,\dd y\,\dd z}
\\[5mm] = &\
\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}
\bracks{x < 1 - y - z}\dd x\,\dd y\,\dd z
\\[5mm] = &\
\int_{0}^{1}\int_{0}^{1}
\bracks{0 < 1 - y - z < 1}\int_{0}^{1 - y - z}
\dd x\dd y\,\dd z
\\[5mm] = &\
\int_{0}^{1}\int_{0}^{1}
\bracks{-z < y < 1 - z}\pars{1 - y - z}\dd y
\\[5mm] = &\
\int_{0}^{1}\int_{0}^{1}
\bracks{y < 1 - z}\pars{1 - y - z}\dd y\,\dd z
\\[5mm] = &\
\int_{0}^{1}\int_{0}^{1 - z}\pars{1 - y - z}\dd y\dd z
\\[5mm] = &\
{1 \over 2}\int_{0}^{1}\pars{1 - z}^{2}\dd z = \bbx{1 \over 6} \\ &
\end{align}

ContentLicense: CC BY-SA 3.0

Title: Volume integral - finding the region


问:
An urn contains ten balls numbered 1 to 10. Using the Law of total probability, What is the probability that the 3rd ball drawn is numbered 3
I know the answer to this question is 1/10 because we know nothing about the previous two balls drawn before we draw the third ball, however Im unsure how to prove it using the law of total probability. I am able to prove it if it was the second ball drawn ie:

P(Draw2 is Ball3) = P(Draw2 is Ball3 | Draw1 is Ball3)P(Draw1 is Ball3) + P(Draw2 is Ball3|Draw1 not Ball3)P(Draw1 not Ball3)

The numbers end up being:

P(Draw2 is Ball3) = 0*(1/10) + 1/9*(9/10) = 1/10

Im not quite sure how to show it for the third draw using the same logic.

答:
If we call the event "ball $i$ is $3$" $E_i(3)$, then the probability we want is

\begin{align*}
P(E_3(3)) &= P(E_3(3) \cap E_2(3)^c \cap E_1(3)^c) \\
 &= P(E_3(3) | E_2(3)^c \cap E_1(3)^c) \\
& \times P(E_2(3)^c | E_1(3)^c) \\
 & \times P(E_1(3)^c). \
\end{align*}

From here, it's not hard to calculate the individual conditional probabilities -- for instance, $P(E_2(3)^c | E_1(3)^c) = 8/9$. Can you finish?


答:
$R$ is the event that ball 3 drawn on either of the first two draws.

$S$ is the event that ball 3 drawn on the 3rd draw.

Then,

$$p(S) = [p(S|R) \times p(R)] + [p(S|\neg R) \times p(\neg R)].$$

You have already used the Law of Total Probability to compute the probability that Ball 3 was drawn on the 2nd draw.  You can simply add that to the probability of Ball 3 being drawn on the 1st draw.  This gives you $p(R).$

Title: An urn contains ten balls numbered 1 to 10. Using the Law of total probability, What is the probability that the 3rd ball drawn is numbered 3

Body:
I know the answer to this question is 1/10 because we know nothing about the previous two balls drawn before we draw the third ball, however Im unsure how to prove it using the law of total probability. I am able to prove it if it was the second ball drawn ie:

P(Draw2 is Ball3) = P(Draw2 is Ball3 | Draw1 is Ball3)P(Draw1 is Ball3) + P(Draw2 is Ball3|Draw1 not Ball3)P(Draw1 not Ball3)

The numbers end up being:

P(Draw2 is Ball3) = 0*(1/10) + 1/9*(9/10) = 1/10

Im not quite sure how to show it for the third draw using the same logic.


问:
Mean-type series convergence if a sequence converges
Reading a paper i have found that they use the following proposition:

If  a convergent sequence $\{c_k\}$ with limit $c$, then
$$\sum_{k=1}^N \frac{k^2 c_k}{N^3} \rightarrow \frac{c}{3}.$$
It's similar to the well-known arithmetic mean convergence, i.e., under the same assumption we have
$$\sum_{k=1}^N  \frac{c_k}{N} \rightarrow c.$$

I've alredy tried to use a similar approach to arithmetic mean convergence
\begin{align*}
|\sum_{k=1}^N \frac{k^2 c_k}{N^3} -\frac{c}{3}|&\leq \frac{1}{3N^3} \sum_{k=1}^N |3k^2 c_k -cN^3|\\
&\leq \frac{1}{3N^3} \sum_{k=1}^N |3k^2 c_k -3k^2 c|+\frac{1}{3N^3} \sum_{k=1}^N |3k^2 c -cN^3|\\
&=\frac{1}{N} \sum_{k=1}^N | c_k - c|+c\sum_{k=1}^N (\frac{1}{3}-\frac{k^2}{N^3})
\end{align*}
So, the first term of the last inequality goes to zero, as $N\rightarrow \infty$, but the second term even doesn't converge. And i have the same problem using different "zeros" in the second step.

Thanks for helping.

答:
Since $\sum_{k=1}^N k^2=1+2^2+...+N^2=\frac{N(N+1)(2N+1)}{6}$, we have
$$
\frac c3 =\lim_{N\to\infty}
\sum_{k=1}^N {k^2c}{N^3}.
$$
Then along your original approach,
$$
\Big(\lim_{N\to\infty}
\sum_{k=1}^N \frac{k^2c_k}{N^3}\Big)-\frac c3=\lim_{N\to\infty}
\sum_{k=1}^N \frac{k^2c_k}{N^3}-\lim_{N\to\infty}
\sum_{k=1}^N \frac{k^2c}{N^3}=\lim_{N\to\infty}
\sum_{k=1}^N \frac{k^2(c_k-c)}{N^3}.
$$
The limit on the right is $0$ as you observed.


答:
There are infinitely many solutions. For example, the polynomial
$$a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_1x^1$$
is always composite if $x>1$ is an integer an at least two of the sequence $a_k$ are nonzero. Also, if the elements of $a_k$ share a common factor other than $1$, the polynomial $a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_1x^1+a_0$
is composite.

Title: Non-prime-generating polynomial

Body: Is there a polynomial, which generates only composites and never primes? I only need a theoretic proof, no need for the actual polynomial. It can have non-integer coefficients.</p>


问:
How to determine a function of 2 variables from its derivative?
Please even the slightest advice would help!

If I have a function $V$ made of 2 variables $x_1$ and $x_2$,
 and its derivative $$\frac{dV}{dt} = \frac{dV}{dx_1}\frac{dx_1}{dt} + \frac{dV}{dx_2}\frac{dx_2}{dt},$$
 how do I find $V$? I don't know how to integrate $dV/dt$. Both $dx_1/dt$ and $dx_2/dt$ are given.

Here is an example:
 $$dx_1/dt = -10x_2$$
 $$dx2/dt = 2x_1 - 20x_2 $$
 $$dV/dt = -2x_1^2 - 2x_1x_2 - x_2^2$$
 What is $V$?

答:
To start you can make the following substitutions to find $x_1$ and $x_2$ :
from second equation we got : $x_1=1/2(dx_2/dt)+10x_2$ now differentiate $x_1$ with respect to
$t$ : $dx_1/dt=1/2(d^2x_2/dt^2) + 10dx_2/dt$ now putting it in the first equation you have a
differential equation in one variable. $10x_2=-1/2(d^2x_2/dt)-10dx_2/dt$ this is a simple diff. equation that you can solve with the approach: $a\exp(\lambda t)$. From here on the problem gets a bit simpler...</p>


问:
Evaluating $\int_0^\infty \frac{e^{-kx}\sin x}x\,\mathrm dx$
How to evaluate the following integral?

$$\int_0^\infty \frac{e^{-kx}\sin x}x\,\mathrm dx$$

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
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 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}
{\expo{-kx}\sin\pars{x} \over x}\,\dd x} =
\int_{0}^{\infty}\expo{-kx}
\pars{{1 \over 2}\int_{-1}^{1}\expo{-\ic qx}\,\dd q}\,\dd x
\\[5mm] = &\
{1 \over 2}\int_{-1}^{1}\int_{0}^{\infty}
\expo{-\pars{k + \ic q}x}\,\,\,\dd x\,\dd q =
{1 \over 2}\int_{-1}^{1}{\dd q \over k + \ic q} =
{1 \over 2}\int_{-1}^{1}{k - \ic q \over q^{2} + k^{2}}\,\dd q
\\[5mm] = &\
\int_{0}^{1/k}{\dd q \over q^{2} + 1} =
\bbx{\arctan\pars{1 \over k}} \\ &
\end{align}


问:
How to show $\mu$ is measure when $E_n^c$ is countable?

Let $X$ be an uncountable set, let $\mathfrak{M}$ be the collection of all sets $E \subset X$ such that either $E$ or $E^c$ is at most countable ( i.e, finite or countable), and define $\mu(E)=0$ in the first case, $\mu(E)=1$ in the second. Prove that $\mu$ is  measure on $\mathfrak{M}?$

My attempt : Here to prove $\mu$  is measure mean that $\mu$ is countably additive on $\mathfrak{M}$

we have two cases

$1.$when $E_n$ is  countable

$2.$ when $E_n^c$ is  countable

Case $1.$ : let $E_1, E_2...$ be  a countable sequence  of pairwise disjoint element of $\mathfrak{M}$

then their union  is also countable

so we have the following inequality

$\mu(\bigcup E_n)=\sum_{n=1}^{\infty}\mu(E_n)=0$

My doubt : How to show $\mu$  is measure when $E_n^c$ is countable ?

答:
Let $F$ be a countable subset of $M.$

If every $f\in F$ is countable then $\cup F$ is countable so $$\cup F\in M \;\land\;\mu(\cup F)=0=\sum_{f\in F}0=\sum_{f\in F}\mu(f).$$ If there exists an uncountable $f_0\in F$ then

(i) since $f_0\in M,$ we have $\cup F\in M$ because $(\cup F)^c$ is a subset of the countable set $(f_0)^c,$ and

(ii) since $(\cup F)^c$ is countable by (i), we have $$\mu(\cup F)=\infty=\mu(f_0)\le\sum_{f\in F}\mu(f)\le \infty.$$


答:
Let
$
a_n := \sum_{k=1}^n k^2c_k
$
and $b_n := \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ (which defines an increasing sequence divergence to $\infty$). One can check that
$$
\frac{a_{n+1}-a_n}{b_{n+1}-b_n}
= \frac{(n+1)^2 c_{n+1}}{(n+1)^2} = c_{n+1} \xrightarrow[n\to\infty]{} c
$$
so, by the Stolz—Cèsaro theorem,
$$
\frac{a_n}{b_n} \xrightarrow[n\to\infty]{} c
$$
as well.
Now, note that
$$
\frac{a_n}{b_n} = \frac{6n^3}{n(n+1)(2n+1)}\sum_{k=1}^n \frac{k^2}{n^3} c_k
$$
to conclude, as $\frac{6n^3}{n(n+1)(2n+1)} \xrightarrow[n\to\infty]{} 3$.

Title: Mean-type series convergence if a sequence converges

Body:
Reading a paper i have found that they use the following proposition:

If  a convergent sequence $\{c_k\}$ with limit $c$, then
$$\sum_{k=1}^N \frac{k^2 c_k}{N^3} \rightarrow \frac{c}{3}.$$
It's similar to the well-known arithmetic mean convergence, i.e., under the same assumption we have
$$\sum_{k=1}^N  \frac{c_k}{N} \rightarrow c.$$

I've alredy tried to use a similar approach to arithmetic mean convergence
\begin{align*}
|\sum_{k=1}^N \frac{k^2 c_k}{N^3} -\frac{c}{3}|&\leq \frac{1}{3N^3} \sum_{k=1}^N |3k^2 c_k -cN^3|\\
&\leq \frac{1}{3N^3} \sum_{k=1}^N |3k^2 c_k -3k^2 c|+\frac{1}{3N^3} \sum_{k=1}^N |3k^2 c -cN^3|\\
&=\frac{1}{N} \sum_{k=1}^N | c_k - c|+c\sum_{k=1}^N (\frac{1}{3}-\frac{k^2}{N^3})
\end{align*}
So, the first term of the last inequality goes to zero, as $N\rightarrow \infty$, but the second term even doesn't converge. And i have the same problem using different "zeros" in the second step.

Thanks for helping.


问:
Problem's solving the integral $\int_0^\infty \frac{1} {x^{1/3}(x+a)} dx$ where a is a constant
I found this integral

$$ \int_{0}^{\infty} \frac{1} {x^{1/3}(x+a)} \, \mathrm{d}x $$

as part of an example of differentiation under the integral sign in the book Advanced Calculus Explored, by Hamza E. Asamraee. There is only the hint that you can use the substitution:

$$y= x^{1/3}$$

so in the end:

$$ \mathrm{d}x = 3y^{2}\mathrm{d}y$$

And after performing the substitution the integral becomes:

$$ \int_{0}^{\infty} \frac{3y^2} {y(y^3+a)} \, \mathrm{d}y = \int_{0}^{\infty} \frac{3y} {(y^3+a)} \, \mathrm{d}y $$

After this you can factorize the denominator in the form

$$y^3+a = (y+a^{1/3})(y^2-a^{1/3}y+a^{2/3})$$

and use partial fractions so that the integral becomes:

$$ -\frac{1}{a^{1/3}} \int_{0}^{\infty} \frac{\mathrm{d}y}{y+a^{1/3}} + \int_{0}^{\infty} \frac{y+a}{y^2-a^{1/3}y+a^{2/3}} \, \mathrm{d}y $$

And the problem is that these two integrals diverge but in the book or even in wolfram alpha the result is:

$$ \frac{2\pi}{a^{1/3}\sqrt{3}} $$

But I don't know if I made a mistake in the substitution or in which part, so I request for help :c

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
This is a simple application of
Ramanujan's Master Theorem:
\begin{align}
&\bbox[5px,#ffd]{\left.\int_{0}^{\infty}
{\dd x \over x^{1/3}\pars{x + a}}\,\right\vert_{\,a\ >\ 0}} =
{1 \over a^{1/3}}\int_{0}^{\infty}
{x^{\color{red}{2/3} - 1} \over x^{1/3}\pars{1 + x}}\,\dd x
\end{align}
Note that
$\ds{{1 \over 1 + x} = \sum_{k = 0}^{\infty}
\color{red}{\Gamma\pars{1 + k}}{\pars{-x}^{k} \over k!}}$. Then,
\begin{align}
&\bbox[5px,#ffd]{\left.\int_{0}^{\infty}
{\dd x \over x^{1/3}\pars{x + a}}\,\right\vert_{\,a\ >\ 0}} =
{1 \over a^{1/3}}\,\Gamma\pars{\color{red}{2 \over 3}}
\Gamma\pars{1 - \color{red}{2 \over 3}}
\\[5mm] = &\
{1 \over a^{1/3}}\,{\pi \over \sin\pars{\pi/3}} =
{1 \over a^{1/3}}\,{\pi \over \root{3}/2} =
\bbx{{2\root{3} \over 3}\,\pi\,a^{-1/3}} \\ &
\end{align}


问:
How to show that there does not exist any integer $b$ with $f(b)=14.$

Let $f(x)$ be a polynomial with integer coefficients. Suppose that there exist distinct integers $a_1,a_2,a_3,a_4,$ such that $f(a_1)=f(a_2)=f(a_3)=f(a_4)=3.$ Then show that there does not exist any integer $b$ with $f(b)=14.$

答: We have $f(x)=3+(x-a_1)(x-a_2)(x-a_3)(x-a_4)g(x)$ for some polynomial $g$ with integer coefficients. But then $(b-a_1)(b-a_2)(b-a_3)(b-a_4)g(b)=11$. Use the primality of $11$ to get a contradiction.</p>


问:
$p$ smallest prime dividing $G$ subgroup with index $p$ normal in $G$
I am struggling to understand this proof.

Suppose that $p$ is the smallest prime that divides $|G|$ show that any subgroup of index $p$ is normal in $G$.

Proof

Let $\phi \rightarrow S_p$ be the homomorphism given by the generalized Cayley theorem. We claim $ker(\phi)=H$ which implies $H$ is normal. By first isomorphism theorem $|G/Ker \phi|$ divides $|S_p|=p!$ON the other hand $|G/Ker \phi|$ divides $|G|$ by lagrange's theorem. Since $ker\phi \leq H$ and $p$ is the smallest prime factor of $|G|$, $ker\phi=H$. I am struggling to understand this if anyone could help explain this?

答:
You don't actually detail your difficulties, so here's a description based on a guess...

$|G / \ker \phi|$ divides $|S_p| = p!$

means the factorization of $|G / \ker \phi|$ uses only powers of primes in the list $1, 2, \dots, p$.

$|G / \ker \phi|$ divides $|G|$

means the factorization of $|G / \ker \phi|$ uses only power of primes listed in the factorization of $|G|$, the smallest of which is $p$.

The only prime on both lists is $p$.

Going back to the first quote, this says the index of $H$ in $G$ is a multiple of $p$.  Going back to the second quote, this says the index of $H$ in $G$ is a divisor of $p$.  So the index of $H$ in $G$ is $p$.


答: If $x$ is an integer, $x^2$ is never a prime.

Title: Non-prime-generating polynomial


问:
proving product topology induced by Discrete Topology and Usual topology equal to this topology
While solving assignment questions of my topology course , I am unable to solve this particular question and so posting it here.

Let $\leq_R$ denote the usual order on $\mathbb{R}$ , let $\leq$ denote the dictionary order on $\mathbb{R}^2$ determined by $\leq_R$ and let U denote the order topology on $\mathbb{R}^2$ determined by $\leq$. Furthermore ,$T_1$ denote the discrete topology on $\mathbb{R}$ , let $T_2$ denote the usual topology on $\mathbb{R}$, and let T denote the product  topology on $\mathbb{R}^2$ determined by $T_1$ and $T_2$. Prove that U=T.

First I have to write T ,which will be generated by { r,(a,b) | r,a,b $\in \mathbb{R}$ }.

Let X$\in $U.So, X is of the form $(x_1, x_2)$ and I have to prove that X $\in $ T.but the elements in T are given by  { r,(a,b) | r,a,b $\in \mathbb{R}$ } and we can see that 2nd element there is an interval so, do I have to prove that $x_2$ lies in that interval?

It is my humble request that can you please prove 1 side of the proof that If X$\in $U then how can I show that X $\in $T.

I would do 2nd part by myself.

答:
$T$ has a base $\mathscr{B}$ consisting of all sets of the form $B(a,b,c)=\{a\}\times(b,c)$, where $a,b,c\in\Bbb R$, and $b<c$. Each of these sets belongs to $U$, since

$$B(a,b,c)=\left\{\langle x,y\rangle\in\Bbb R^2:\langle a,b\rangle<\langle x,y\rangle<\langle a,c\rangle\right\}$$

is an open interval in the order $\le$. Thus, $T\subseteq U$.

One way to show that $U\subseteq T$ is to show that the sets $B(a,b,c)$ are also a base for $U$. You can do it by showing that every open interval in the order $\le$ is a union of such sets. Consider the open interval $V$ with endpoints $\langle a,b\rangle$ and $\langle c,d\rangle$, with $\langle a,b\rangle\le\langle c,d\rangle$. There are two possibilities.

If $a=c$, check that $V=B(a,b,d)$.

If $a<c$, verify that $$V=\big(\{a\}\times(b,\to)\big)\cup\bigcup_{a<x<c}\big(\{x\}\times\Bbb R\big)\cup\big(\{c\}\times(\leftarrow,d)\big)\,,$$ and show how to write this set as a union of members of $\mathscr{B}$.

Note: The notations $(b,\to)$ and $(\leftarrow,d)$ are a less common standard notation equivalent to the more familiar notations $(b,\infty)$ and $(-\infty,d)$.


答:
Note that the integral is

$$\mathcal{L}\bigg(\frac {\sin(t)}t\bigg)=F(k)$$
Where $\mathcal{L}$
is the Laplace transform operator

You can find the transformation using power series, which ultimately gives
$$F(k)=\arctan(\frac 1k)$$

Title: Evaluating $\int_0^\infty \frac{e^{-kx}\sin x}x\,\mathrm dx$

Body:
How to evaluate the following integral?

$$\int_0^\infty \frac{e^{-kx}\sin x}x\,\mathrm dx$$


问:
If $u$ is harmonic in $B(0,1)\subset\mathbb R^2$ with $\vert u\vert\leq1,$ then $\vert\nabla u(0)\vert\leq4/\pi$
Let $u$ be harmonic in $B(0,1)\subset\mathbb R^2$ and $\vert u\vert\leq1,$ then $\vert\nabla u(0,0)\vert\leq4/\pi.$

Since $\vert\nabla u(0,0)\vert=\sqrt{u_{x}(0,0)^2+u_{y}(0,0)^2}$, we need to estimate the previous two partial derivatives at the origin. Using the mean value formula for harmonic functions we find that
$$\vert u_x(0,0)\vert=\left\vert\frac{1}{\vert B(0,\pi/4)\vert}\int_{B(0,\pi/4)}u_x(x,y)\,dx\,dy\right\vert=\frac{2^2}{\alpha(2)(\pi/2)^2}\left\vert\int_{\partial B(0,\pi/4)}u\nu_x\,dS\right\vert\leq\frac{4\cdot2\cdot2}{\pi},$$
however, this does not yield the bound we are after, and I see no way to improve on the calculation above.

Does anyone have any suggestions on how to improve this bound?

答:
One can proceed similar as in Estimate $|f’(0)|$ by $Re(f(z))$.

If we identify $\Bbb R^2$ with $\Bbb C$ then $u$ is the real part of a holomorphic function $f$ in the complex unit disk $\Bbb D$. The goal is to show that $|\nabla u(0)| = |f'(0)| \le 4/\pi$.

We can assume that the strict inequality $|\operatorname{Re}(z)| < 1$ holds in $\Bbb D$ because $f$ is constant otherwise. Then
$$
g(z) = \tan\left(\frac \pi 4 f(z)\right)
$$
maps the unit disk into itself, so that the Schwarz–Pick theorem can be applied to $g$ at $z=0$:
$$
 1 - |g(0)|^2 \ge |g'(0)| = \left| \frac{\frac \pi 4 f'(0)}{1 + \tan^2\left(\frac \pi 4 f(0)\right)} \right| = \frac \pi 4 |f'(0)| \, .
$$
i.e. $ |f'(0)| \le \frac 4 \pi (1 - |g(0)|^2) \le \frac 4 \pi$.


答: Hint: Consider the polynomial $g(x) = f(x) - 3$.  Then $g$ has four known linear factors $(x - a_1), (x - a_2), (x - a_3), (x - a_4)$.  What can these four factors evaluate to if $f(b) = 14$?</p>

Title: How to show that there does not exist any integer $b$ with $f(b)=14.$

Body: Let $f(x)$ be a polynomial with integer coefficients. Suppose that there exist distinct integers $a_1,a_2,a_3,a_4,$ such that $f(a_1)=f(a_2)=f(a_3)=f(a_4)=3.$ Then show that there does not exist any integer $b$ with $f(b)=14.$


问:
To prove an apparently obvious statement: if $A_1\subseteq A_2 \subseteq ... \subseteq A_n$, then $\bigcup_{i=1}^n A_i = A_n$

To prove:$$\text{If } A_1\subseteq A_2 \subseteq ... \subseteq A_n\text{ ,then } \bigcup_{i=1}^n A_i = A_n$$ using the axioms of ZFC Set Theory.

Honestly, this statement is very obvious, but I do not want to take that for granted. How can I prove this from the axioms, especially the axiom of unions?

I'm fairly new to ZFC Set Theory, so I don't know where to start. I do know that the Axiom of Unions (and Specification) gives rise to the definition of $\bigcup_{i=1}^n A_i$ for sets $A_1,...,A_n$ though. The uniqueness follows from the Axiom of Extensionality.

My thoughts:

We already know that the union exists, and it is defined such that for every element $x\in \bigcup_{i=1}^n A_i$, there exists at least one set $A_j$ such that $x\in A_j$. What's next?

Thank you!

Edit:

Induction works if the chain is finite or countably infinite. What if the chain length is uncountable? Does this result still hold, and if yes, how do we prove it?

Edit 2:

So now I wish to prove (or disprove) the following using the axioms of ZFC Set Theory:

Consider an infinite set $A$, and an index set $I$ such that $$A_1\subseteq A_2\subseteq ... \subseteq A_i \subseteq ... \subseteq A$$
where $i\in I$. The index set $I$ is uncountable. We have $$\bigcup_{i\in I}A_i = A$$

答:
This is regarding your second edit. First of all denoting $A_1 \subseteq A_2 .... \subseteq A$ is not correct when the index set $I$ is uncountable and moreover you don't know the elements of $I$.

Take $A_i =(0,i)$ where $i\in (0,1)$ and $A = [0,1]$. These are nested in the sense that $i \leq j$ implies that $A_i \subseteq A_j$ and observe that for $i$ in index set $A_i \subset A$. Clearly union of all $A_i$ is the interval $(0,1)$ which is not $A$.


问:
Find $\oint_{| z|=1}\frac{\log \mid \zeta(\frac{z}{z-1})\mid}{z}dz$
Evaluate
$$
\oint_{\left\vert z\right\vert\ =\ 1}
\log\left(\left\vert
\zeta\left(\frac{z}{z - 1}\right)\right\vert\right)
\,{\mathrm{d}z \over z}
$$
where $\zeta(s)$ is the analytic continuation of Riemann zeta function which is analytic except for a simple pole at $s = 1$.

MY TRY-
Define $f(z)=\frac{\log \left| \zeta(\frac{z}{z-1})\right|}{z}$

Poles of $f(z)$ are $z=0$.

R=Residue of $f(z)$ at $z=0$.

R$=\lim_{z\rightarrow 0} (z-0) f(z)$

R= $\lim_{z\rightarrow 0}  \log \left|\zeta(\frac{z}{z-1})\right|  $

$\zeta(0)=-1/2 $ gives,

R$=\log\left(\left| \frac{-1}{2}\right|\right)=-\log \ 2$

By Cauchy's Residue Theorem,

$$\oint_{|z|=1}\frac{\log \left| \zeta(\frac{z}{z-1})\right|}{z}dz=-2\pi i \log \ 2$$

Is it correct?

答:
$g(z)=\frac{z}{z-1}$ is biholomorphic $\Bbb{C}-1$ to $\Bbb{C}-1$ and $|z|<1$ to $\Re(s)<1/2$.

$g(g(z))=z$ thus $\zeta( g(z))$ is analytic on $|z|<1$ with some zeros at $g(-2n)$ and at the $g(\rho),\Re(\rho)<1/2$.

$\log \zeta(g(z))$ has some logarithmic branch points there.

Then apply Jensen's formula.

For $r<1$

$$\frac1{2i\pi}\int_{|z|=r} \log |\zeta(g(z))| \frac{dz}z=  \log |\zeta(0)|-\sum_{g(-2n)<r} \log \frac{g(-2n)}{r} - \sum_{g(\rho)<r} \log \frac{|g(\rho)|}{r}$$
so that
$$\lim_{r\to 1^-}\frac1{2i\pi}\int_{|z|=r} \log |\zeta(g(z))| \frac{dz}z= \infty$$


问:
Mistake converting $2.3151515...$ to a fraction?
So I think I made a mistake but I just want to confirm I'm getting this right:

In the example, we had the number $2.3151515...$ and were supposed to convert this number into a fraction.
It was then shown to be equal to the sequence $2.3 + \frac{15}{10^3}+\frac{15}{10^5}+\frac{15}{10^7}$ etc.
As such the common ratio is $r = \frac{1}{100}$

Thus we get the sum of the sequence equalling $(2.3)\frac{1}{1-\frac{1}{100}} = \frac{230}{99}$
The problem is, when I checked my solution $\frac{230}{99}$, it's rougly $2.32323232...$ yet the right solution would actually be $\frac{382}{165}$

Does the problem lie within the simplification of the sequence or am I doing something wrong?

Thanks in advance.

答:
Here is the standard technique. We have $$2.3\overline{15}=\frac{2315-23}{990}=\frac{2292}{990}$$ The overline / bar denotes the group of repeating digits. Just form the integer without any decimals to get $2315$ and subtract the number formed by non-repeating digits $23$ and you get numerator $2315-23=2292$.

For denominator just check the digits after decimal point. It consists of as many 9's as there are repeating digits followed by as many zeroes as there are non-repeating digits after decimal point. Thus we get $990$ as denominator.

The above law can be proved using the standard formula for sum of an infinite geometric progression.

Your mistake is in ignoring the first term $a$ in formula for sum of infinite GP $$a+ar+ar^2+\dots=\frac{a} {1-r}$$ and treating the sum as just $1/(1-r)$.


问:
What are the equations of the three straight lines represented by $x^3+bx^2y+cxy^2+y^3=0$ when $b+c=-2$?
I am given that $$x^3 + bx^2y + cxy^2 + y^3 = 0$$ represents three straight lines if $b + c = -2$.

Is there a way to find the equations of the three lines separately?

I tried factorizing the equation but wasn't able to get anywhere. Is there a way to get the separate equations using partial differentiation?

答:
If $b+c=-2$, you can write $b=-k$ and $c=k-2$, so

$$\begin{align}x^3+bx^2y+cxy^2+y^3&=x^3-kx^2y+(k-2)xy^2+y^3\\&=x^3-kx^2y+kxy^2-2xy^2+y^3\\&=(x^3-xy^2)-(kx^2y-kxy^2)-(xy^2-y^3)\\&=(x-y)(x(x+y)-kxy-y^2)\\&=(x-y)(x^2-(k-1)xy-y^2)\end{align}$$

You can factor the latter part by completing the square but the computation is a bit ugly.

The three lines will be:

$$x=y\\ x=y\left(\frac{k-1}2+\sqrt{1+\left(\frac{k-1}2\right)^2}\right)\\ x=y\left(\frac{k-1}2-\sqrt{1+\left(\frac{k-1}2\right)^2}\right)$$

where $k$ is just $-b$, a constant, so the entire expression in the brackets is a constant, the reciprocal of their slopes.


问:
Find intersection of three planes?
Find the intersection of

$2x_1+x_2+x_3-3=0$, $2x_1+x_2+4x_3-6=0$, $2x_1+x_2-2=0$.

So I calculated the determinant of the three planes and it's equal to zero, then I calculated the cross product between the two first planes and it gave me $3x_1-6x_2$

The thorem in the course book says that if the determinant of the three planes is equal to zero, and cross product it's not zero, then there's two posibilities if $r_4-up_4-vq_4=0$ then the three planes intersect in a line $l$ and $l=\{(x_1,x_2,x_3):SX=up×q\}$ and S is a point in the line.

What is $r_4$, $up_4$ and $vq_4$ ? Thanks for your help.

答:
These three planes are pairwise non-identical and non-parallel so they will meet in a point, or a line or they will form an open prism.

If you write it like  $Ax=B$ and if $\det|A|=0$, the planes will either meet in a line or the will form an open prism. If you find the edges of pairwise planes by finding cross products of their normals $\vec n_1, \vec n_2, \vec n_3$. If edges are identical/coincident the planes will meet in a line. If edges are non coplanar planes will form a prism.

But more simply here if you put (3) in (1) and (2), you get $x_3=1$  and $x_3=2$ contradiction (an inconsistency). So these three Eqs. are inconsistent and hence they form a prism. In a prism two plane meet leaving out one, so no solution (inconsistency).

Title: Find intersection of three planes?


问:
Infinite sets with bijection are strongly minimal
I'm working through Tent and Ziegler "A Course in Model Theory" and I got stuck on the Exercise 5.7.2:

Show that the theory of an infinite set equipped with a
bijection without finite cycles is strongly minimal and that the associated
geometry is trivial.

I think I understand how to do the trivial geometry part. For the strong minimality, however, I do not know how to get quantifier elimination, which seems to be the key to the problem. Any hints would be appreciated!

答:
Let $\mathcal{L}=\{f\}$, the language of a unary function, and let $T$ be the theory of a bijection without finite cycles. We first show that $T$ eliminates quantifiers; by, eg, lemma 3.2.4. of Tent and Ziegler, it suffices to consider primitive existential formulas.  There are no constant symbols in our language, so the only terms are of the form $f^n(v)$ for some $n\in\omega$ and some variable $v$, where we use the notation indicated by Alex in his comments. Additionally, the only relation symbol in our language is equality, so the only atomic formulas are of the form $f^m(v)=f^n(w)$. Thus, any conjunction of atomic formulas and their negations will be of the form $$\phi(\overline{v})\wedge\left(\bigwedge_{l=1}^af^{m'_l}(w)=f^{m_l}(v_{i_l})\right)\wedge\left(\bigwedge_{l=1}^{b}f^{n'_l}(w)\neq f^{n_l}(v_{j_l})\right)\wedge\psi(w)$$ where $\psi$ and $\phi$ are themselves quantifier-free and of the same form as the two conjuncts in the middle. Now, denoting by $\theta(v_1,\dots,v_k,w)$ the formula above, we wish to show that $\exists w\theta$ is equivalent to a quantifier-free formula, modulo $T$.

First we claim that we can ignore $\psi(w)$; indeed, if $\psi(w)$ contains any conjunct of the shape $f^m(w)\neq f^m(w)$, then $\exists w\theta$ will of course be unsatisfiable and so equivalent to $\bot$. On the other hand, consider the case where $\psi(w)$ contains any conjunct of the shape $f^m(w)=f^n(w)$ for $m\neq n$. Any value of $w$ satisfying this formula would be part of a finite cycle, of cycle length dividing $m-n$. Since models of $T$ have no finite cycles, in this case $\exists w\theta$ is again equivalent to $\bot$. Conversely, if $\psi$ does not contain any conjuncts of the above two forms, then every element of a model of $T$ will satisfy it, and so we may safely omit it from considerations. Indeed, in this case every conjunct of $\psi$ will be of form $f^m(w)=f^m(w)$ (clearly satisfied by any element of a model of $T$) or $f^m(w)\neq f^n(w)$ for $m\neq n$ (again satisfied by any element of a model of $T$, since such models have no finite $f$-cycles).

Thus it suffices to consider the case where $\theta$ has an empty $\psi(w)$ term. Since $f$ is a bijection, we may replace each $m_l$ in $\theta$ with $m_l-m_l'$ and each $n_l$ in $\theta$ with $n_l-n_l'$ and hence assume that each $m'_l=0$ and $n'_l=0$. (Note, however, that this means $m_l$ and $n_l$ may be negative, so the corresponding expression may not be an $\mathcal{L}$-formula.)

Now, because any model of $T$ has infinitely many elements, if $a=0$ then $\exists w\theta$ will be equivalent to $\phi(\overline{v})$. Thus we may assume $a>0$, and then we have that $\exists w\theta$ is equivalent to $$\phi(\overline{v})\wedge\left(f^{m_1}(v_{i_1})=\dots=f^{m_a}(v_{i_a})\right)\wedge\left(\bigwedge_{l=1}^b f^{m_1}(v_{i_1})\neq f^{n_l}(v_{j_i})\right).$$ Now, since there may be negative values of $m_l$ and $n_l$, this will not necessarily be an $\mathcal{L}$-formula. However, if there are negative values of $m_l$ and $n_l$, then we may simply apply $f$ a total of $|\min\{m_1,\dots,m_a,n_1,\dots,n_b\}|$ times to every term and thence obtain an $\mathcal{L}$-formula. Since $f$ is a bijection, the resulting formula will be logically equivalent to the original.

It follows immediately that $T$ is strongly minimal. Indeed, if $M\models T$, then by quantifier elimination (and the fact that $\mathcal{L}$ has no relation symbols) every $\mathcal{L}_M$-formula in one free variable will cut out a boolean combination of finite sets, and thus be either finite or cofinite, as desired.


问:
Is T discrete topology on X or not?
I want to verify my argument  from  question of my assignments in topology.

For each n$\in  \mathbb{N}$ , let $X_n =\mathbb{R}$ and let $T_n$ be discrete topology on $X_n$. Let T be product topology on X=$\prod_{n \in \mathbb{N}} X_n$ . Is T discrete topology on X or not?

I think It will be discrete topology as every element of  $P(\prod_{\mathbb{N}} X_n)$, where P(X) means power set of X are in X.

Am I right?

答:
The box topology would be the discrete topology; but the product topology is smaller.  One way to see it is that the product topology is the coarsest making the projections continuous.

However, every function from a discrete space is continuous.  That's it's the finest topology, period.

The product topology is in fact strictly coarser, because it doesn't include the entire power set.  It only includes finite products of non-empty open sets.

Title: Is T discrete topology on X or not?


问:
How do I create symmetrical bands on both sides of a curve?
I am playing with Amazon's DeepRacer, which is a sort of intro into machine learning where you train a car to drive around a track.

An input to the program is an ordered list of waypoints that form the track. Another input is the width of the track. Using these inputs, I'd like to be able to generate the inner and outer bounds of the track, but I am struggling.

An example of a track, showing data points for the center, inner bound, and outer bound:

A zoomed in view of the chicane on the top of the track:

Given just the dataset represented by the blue dots, how can I generate the data represented in green and orange? I have tried a couple of different methods that just are not working out.

Specifically, I have tried using the arctangent to determine the angles from each waypoint to the next, and then trying to place points on both sides of a line that bisects that angle. That seems like a pretty simple approach, but I just can not seem to get it right. How would you approach this problem?

答:
Given a (blue) curve $\mathbf{r}(t)=(x(t),y(t))$, then $(x'(t),y'(t))$ is its tangent and $(y'(t),-x'(t))$ is a normal vector, made unit as $\mathbf{n}=\frac{(y'(t),-x'(t))}{\sqrt{x'(t)^2+y'(t)^2}}$. Hence the green and yellow curves are given by $$\mathbf{u}_{\pm}(t):=\mathbf{r}(t)\pm \epsilon\,\mathbf{n}(t)$$ However, if you plot these curves directly, the dots will not be equally spaced. So let $s_\pm(t):=\int^t\sqrt{x'(t)^2+y'(t)^2}\,dt$ be the arclength function of $\mathbf{u}_\pm(t)$ and reparametrize $\mathbf{u}$ in terms of $s$, then plot $\mathbf{u}_\pm(ns_0)$.

Here's some pseudo-code to achieve this ($x(t),y(t),x'(t),y'(t),x''(t),y''(t)$ should be given or calculated before-hand):

```<code>Parameters:
e (= distance of green/yellow curves from blue curve),
T0, T1, dt (= starting/ending values of t, with increment dt),
s0 (= distance between points)

sum[0]=0; sum[1]=0; sum[2]=0;
for t=T0 to T1 step dt:
   x=x(t); y=y(t); xt=x'(t); yt=y'(t); xtt=x''(t); ytt=y''(t);
   s = sqrt(xt*xt + yt*yt);

   x1 = x + e*yt/s; y1 = y - e*xt/s;
   xt1 = xt + e*(ytt/s-yt/(2*s*s*s)); yt1 = yt + e*(-xtt/s + xt/(2*s*s*s));
   s1 = sqrt(xt1*xt1 + yt1*yt1);

   x2 = x - e*yt/s; y1 = y + e*xt/s;
   xt2 = xt - e*(ytt/s-yt/(2*s*s*s)); yt2 = yt - e*(-xtt/s + xt/(2*s*s*s));
   s2 = sqrt(xt2*xt2 + yt2*yt2);

   sum[0] += s*dt; sum[1] += s1*dt; sum[2] += s2*dt;
   if(sum[0]>=s0): plot blue point(x,y); sum[0]=0;
   if(sum[1]>=s0): plot green point(x1,y1); sum[1]=0;
   if(sum[2]>=s0): plot yellow point(x2,y2); sum[2]=0;
</code>```

Body:
Given a (blue) curve $\mathbf{r}(t)=(x(t),y(t))$, then $(x'(t),y'(t))$ is its tangent and $(y'(t),-x'(t))$ is a normal vector, made unit as $\mathbf{n}=\frac{(y'(t),-x'(t))}{\sqrt{x'(t)^2+y'(t)^2}}$. Hence the green and yellow curves are given by $$\mathbf{u}_{\pm}(t):=\mathbf{r}(t)\pm \epsilon\,\mathbf{n}(t)$$ However, if you plot these curves directly, the dots will not be equally spaced. So let $s_\pm(t):=\int^t\sqrt{x'(t)^2+y'(t)^2}\,dt$ be the arclength function of $\mathbf{u}_\pm(t)$ and reparametrize $\mathbf{u}$ in terms of $s$, then plot $\mathbf{u}_\pm(ns_0)$.

Here's some pseudo-code to achieve this ($x(t),y(t),x'(t),y'(t),x''(t),y''(t)$ should be given or calculated before-hand):

Parameters:
e (= distance of green/yellow curves from blue curve),
T0, T1, dt (= starting/ending values of t, with increment dt),
s0 (= distance between points)

sum[0]=0; sum[1]=0; sum[2]=0;
for t=T0 to T1 step dt:
   x=x(t); y=y(t); xt=x'(t); yt=y'(t); xtt=x''(t); ytt=y''(t);
   s = sqrt(xt*xt + yt*yt);

   x1 = x + e*yt/s; y1 = y - e*xt/s;
   xt1 = xt + e*(ytt/s-yt/(2*s*s*s)); yt1 = yt + e*(-xtt/s + xt/(2*s*s*s));
   s1 = sqrt(xt1*xt1 + yt1*yt1);

   x2 = x - e*yt/s; y1 = y + e*xt/s;
   xt2 = xt - e*(ytt/s-yt/(2*s*s*s)); yt2 = yt - e*(-xtt/s + xt/(2*s*s*s));
   s2 = sqrt(xt2*xt2 + yt2*yt2);

   sum[0] += s*dt; sum[1] += s1*dt; sum[2] += s2*dt;
   if(sum[0]>=s0): plot blue point(x,y); sum[0]=0;
   if(sum[1]>=s0): plot green point(x1,y1); sum[1]=0;
   if(sum[2]>=s0): plot yellow point(x2,y2); sum[2]=0;


问:
Find multiplicative inverses for elements in U(20)
I am self-learning abstract algebra using Abstract Algebra: An Introductory Course by Gregory Lee.

I come across this group:

$$U(20)=\{1,3,7,9,11,13,17,19\}$$

I have managed to find two pairs of multiplicative inverses, namely:

$$3\cdot 7=1$$
$$13\cdot 17=1$$

I am wondering how to find the multiplicative inverses for $9$, $11$, and $19$.

For example, for $19$, the closest I can get is:

$$9\cdot 19=11$$

Also, must inverse be unique in a group? In the other words, can two elements have the same inverse?

Thank you.

答: In groups inverses are unique. $9$ and $11$ are their own inverses.  For $19$, it is also, since $19\equiv-1\bmod20$.


问:
question from hatcher basic 3 manifolds
The question is: why should a homologically trivial embedded sphere in a simply connected  (not necessarily compact) 3 manifold M bound a compact 3 manifold embedded in M?

I had this problem reading the first part of the proof of proposition 3.10 in "hatcher on 3 manifolds topology", available here:
http://www.math.cornell.edu/~hatcher/3M/3M.pdf page 52

Warning! I don't know if the simple connectedness is necessary. Better is read directly from the link!

答:
Here's an argument that I can see. There may be an easier method. There is an intersection pairing on compact oriented manifolds:
$$H_p(M)\otimes H_{n-p}(M,\partial M)\to \mathbb Z$$
and a similar pairing for unoriented manifolds:
$$H_p(M;\mathbb Z_2)\otimes H_{n-p}(M,\partial M;\mathbb Z_2)\to \mathbb Z_2.$$
These can be described by looking at the algebraic intersection number of representatives of the homology classes. In particular, if you can ever find two manifolds that intersect transversely in a single point, then both represent nontrivial homology classes. Therefore, assuming $M$ is a compact manifold for the time being, if there is a homologically trivial embedded sphere, it must separate $M$ into two pieces (else it meets a circle in a point) and it cannot separate boundary components (else it meets an arc from the boundary to itself in one point.) So it is the boundary of the part of $M$ that it separates off with no boundary components. Okay, so this proves the case for $M$ compact with boundary. Suppose now that $M$ is not compact. Then the homologically zero sphere bounds some singular chain in $M$, which is compact. Now take a tame neighborhood of this singular chain. (That is, take a small neighborhood that has a well-defined 2-dimensional boundary.) Now apply the previous argument to this new manifold.</p>

Title: question from hatcher basic 3 manifolds


问:
Question 2.3.5 Wayne Patty ( Closed sets in Product Topology)
This question is from Foundations of topology by Wayne Patty(Pg 85).

Let $X_n=\mathbb{R}$ for all n$\in \mathbb{N}$ and $T_n$ be lower limit topology on $X_n$ and T be the product topology on $X=\prod_{\mathbb{N}}X_n$. Prove that {x} , where $x_n$ =1 for all $n\in \mathbb{N}$ is closed subset of (X,T).

I am thinking by proving that complement of sequence is open and using the result that if $U_i$ is open for all $n \in \mathbb{N}$ then $\prod_{n\in \mathbb{N}}X_n $ is open in X.

complement of 1 is $(-\infty , 1) \bigcup (1, \infty)$ . but In lower limit topology $ (1,\infty)$ is not open.So, complement of 1 will not be open.

What mistake I am making ?

Can you please tell.

thanks!

答:
There are quite a few mistakes.

It is not in general true that $\prod_{n\in\Bbb N}U_n$ is open $X$ if each $U_n$ is open in $X_n$: this product set is open if and only if $\{n\in\Bbb N:U_n\ne X_n\}$ is finite.

The complement of $\{1\}$ is indeed $(\leftarrow,1)\cup(1,\to)$, but the complement of $\{x\}$ is not $\prod_{n\in\Bbb N}U_n$, where $U_n=(\leftarrow,1)\cup(1,\to)$ for each $n\in\Bbb N$.

$(1,\to)$ is open in the lower limit topology: it’s the union of the sets $[x,\to)$ for $x>1$, and those sets are clearly open in the lower limit topology.

There are at least two reasonable ways to prove that $\{x\}$ is closed in the product topology. One is indeed to show that it’s complement is open, and you can use the sets $U_n=(\leftarrow,1)\cup(1,\to)$ to do it. HINT: For $n\in\Bbb N$ let

$$B_n=\prod_{k\in\Bbb N}V_k\,,$$

where

$$V_k=\begin{cases}
U_n,&\text{if }k=n\\
X_k,&\text{otherwise.}
\end{cases}$$

These sets $B_n$ are open in $X$. (Why?) Now show that $$X\setminus\{x\}=\bigcup_{n\in\Bbb N}B_n\,.$$

Another approach uses the following facts.

The lower limit topology is $T_1$.

The product of $T_1$ spaces is $T_1$.

A space is $T_1$ if and only if every singleton (one-point subset) in it is closed.

(Of course if you’ve not yet proved these facts, you’d have to prove them in order to use them.)


答:
We are told that $\{1\}$ is closed in the lower limit topology, so its complement is open; that one of the sets making up that union is not open does not imply that the union isn't open. (For example, in $\mathbb{R}$ with the usual topology, the set $(0, 1)$ is open and can be written as $(0, 1) = (0, 1/2) \cup [1/2, 1)$, but $[1/2, 1)$ is not open).

One way to prove that $\{x\}$ is closed: the singleton set $\{1\}$ is compact in the lower limit topology; therefore $\{x\}$ is compact in the product topology by Tychonoff's theorem. So $\{x\}$ is closed (since the product topology of Hausdorff spaces is Hausdorff, compact sets are closed in this product space).

Title: Question 2.3.5 Wayne Patty ( Closed sets in Product Topology)

Body:
This question is from Foundations of topology by Wayne Patty(Pg 85).

Let $X_n=\mathbb{R}$ for all n$\in \mathbb{N}$ and $T_n$ be lower limit topology on $X_n$ and T be the product topology on $X=\prod_{\mathbb{N}}X_n$. Prove that {x} , where $x_n$ =1 for all $n\in \mathbb{N}$ is closed subset of (X,T).

I am thinking by proving that complement of sequence is open and using the result that if $U_i$ is open for all $n \in \mathbb{N}$ then $\prod_{n\in \mathbb{N}}X_n $ is open in X.

complement of 1 is $(-\infty , 1) \bigcup (1, \infty)$ . but In lower limit topology $ (1,\infty)$ is not open.So, complement of 1 will not be open.

What mistake I am making ?

Can you please tell.

thanks!


问:
Prove $\int_{0}^{\frac{\pi}{2}} x\mathrm{sin}(x) \;\mathrm{d}x \leq \frac{\pi^{2}}{8}$
An answer to this question is:

Since $x\mathrm{sin}x \leq x$, by the comparison property...
\begin{align}
\int_{0}^{\frac{\pi}{2}} x\mathrm{sin}(x) \;\mathrm{d}x &\leq \int_{0}^{\frac{\pi}{2}} x \;\mathrm{d}x\\
\int_{0}^{\frac{\pi}{2}} x\mathrm{sin}(x) \;\mathrm{d}x &\leq \frac{\pi^{2}}{8}
\end{align}

Hence proved.

But what I am really having trouble understanding is how do you recognize, in this case, that the integral to compare is $\int_{0}^{\frac{\pi}{2}} x \;\mathrm{d}x$? I understand that the reason this integral was chosen was because it equals $\frac{\pi^{2}}{8}$; In a more general case, however, how does one recognize what would be a good comparison to make? There's also the issue of somehow recognizing, and proving that $x\mathrm{sin}(x)
 \leq x$. Is this something that just comes with practice to somehow apply known patterns? I can't imagine that simple pattern recognition is the solution in all scenarios...

Are there any tricks, or tactics that you can recommend to help the process of narrowing down the possibilities for these types of questions in math? They seem to pop up rather often.

答:
In a more general case, however, how does one recognize what would be
a good comparison to make?

There is no general rule. The answer to your question is "with experience".


问:
How can I solve this limit? $\lim _{x \to a}\left(\frac{\sin(x)}{\sin(a)}\right)^{\frac{1}{x-a}}$
$\lim _{x \to a}\left(\frac{\sin(x)}{\sin(a)}\right)^{\frac{1}{x-a}}$
Where $a$ can be any number.

I started by using $e$ like this:

$\lim _{x \to a}e^{\ln\left(\left(\frac{\sin(x)}{\sin(a)}\right)^{\frac{1}{x-a}}\right)} \to \lim _{x \to a}e^{\ln\left(\frac{\sin(x)}{\sin(a)}\right)\cdot\frac{1}{x-a}} \to \lim _{x \to a}e^{\frac{\ln\left(\sin(x)\right)-\ln\left(\sin(a)\right)}{x-a}}$

The answer should be $e^{\cot{a}}$

答:
To make life easier, let $x=y+a$ in order to work around $y=0$. So

$$L=\lim _{x \to a}\left(\frac{\sin(x)}{\sin(a)}\right)^{\frac{1}{x-a}}=\lim _{y \to 0}\left(\frac{\sin(y+a)}{\sin(a)}\right)^{\frac{1}{y}}$$

$$\log(L)=\lim _{y \to 0}\frac 1y \log\left(\frac{\sin(y+a)}{\sin(a)}\right)$$
Expand the sine
$$\frac{\sin(y+a)}{\sin(a)}=\frac{\sin (a) \cos (y)+\cos (a) \sin (y)}{\sin(a)}=\cot (a) \sin (y)+\cos (y)$$

A bit of Taylor series
$$\cot (a) \sin (y)+\cos (y)=1+y \cot (a)-\frac{y^2}{2}+O\left(y^3\right)$$
$$\frac 1y\log\left(\frac{\sin(y+a)}{\sin(a)}\right)= \cot (a)-\frac{1}{2} \left(\cot ^2(a)+1\right)y+O\left(y^2\right)$$

Now, using $t=e^{\log(t)}$,
$$\left(\frac{\sin(y+a)}{\sin(a)}\right)^{\frac{1}{y}}=e^{\cot (a)}\left( 1-\frac{1}{2} \left(\cot ^2(a)+1\right)y+O\left(y^2\right)\right)$$ which shows the limit and also how it is approached.


答:
Observe that all three straight lines pass through the origin $(0, 0)$

Let the three straight lines are

$y = m_1x$, $y = m_2x$ and $y = m_3x$

Then
$$ (y -m_1x)(y -m_2x)(y -m_3x) \equiv x^3 + bx^2y + cxy^2 + y^3$$

Comparing the coefficients, we have

$m_1 m_2 m_3 = -1$

$m_1 m_2 + m_1 m_3 + m_2 m_3 = b$

$m_1 + m_2 + m_3 = -c$

These are the three roots of the equation
$m^3 + cm^2 + bm +1 = 0$

If $b + c = -2$, then $m=1$ must be a root of the above equation. The other quadratic factor is $m^2 - bm - m - 1$ and has the roots $\frac{1}{2} (1 + b \pm \sqrt{b^2 + 2 b + 5})$

Finally the three straight lines are

$y = x$

$y = \frac{x}{2} (1 + b + \sqrt{b^2 + 2 b + 5})$

and

$y = \frac{x}{2} (1 + b - \sqrt{b^2 + 2 b + 5})$

Title: What are the equations of the three straight lines represented by $x^3+bx^2y+cxy^2+y^3=0$ when $b+c=-2$?

Body:
I am given that $$x^3 + bx^2y + cxy^2 + y^3 = 0$$ represents three straight lines if $b + c = -2$.

Is there a way to find the equations of the three lines separately?

I tried factorizing the equation but wasn't able to get anywhere. Is there a way to get the separate equations using partial differentiation?


问:
Inequality of exponential
Trying to show $$\frac{e^{-x}-e^{-y}}{y-x}\leq \frac{1}{2}(e^{-x}+e^{-y})$$ I tried working with $y>x$, and the case $y-x\geq2$ was trivial. But I guess this isn't useful. Using the mean value theorem and writing the problem as $\frac{e^y-e^x}{y-x}\leq\frac{1}{2}(e^y+e^x)$ I get that exists a $c\in(x,y)$ s.t. $\frac{e^y-e^x}{y-x}=e^c$ but I can't get the inequality. In other hand, $e^x$ is convex, so $\frac{1}{2}(e^{x}+e^{y})\geq e^{\frac{x+y}{2}}$. Ought I try to show $e^{\frac{x+y}{2}}\geq e^c$ ?

答:
As

$$F(x,y) = -\frac{e^{-x}-e^{-y}}{y-x}+\frac{1}{2}(e^{-x}+x^{-y})$$ is symmetric in $x,y$, we can suppose $y\ge x$. We have to prove that

$$(y-x)e^x F(x,y)=-(1-e^{-(y-x)})+\frac{y-x}{2}(1+e^{-(y-x)})$$ is non negative for $y \ge x$.

This is equivalent to prove that

$$f(x)=-(e^x-1)+\frac{x}{2}(e^x +1)$$ is non negative for $x \ge 0$.

$$f^\prime(x)=\frac{1}{2}\left(1-e^x+xe^x\right), \, f^{\prime\prime}(x)=\frac{xe^x}{2}\ge 0.$$

As $f^\prime(0)=0$, $f^\prime$ is non negative, $f $ non decreasing and as $f(0)=0$ we get the desired conclusion.

Title: Inequality of exponential


问:
Solving $x^{\log(x)}=\frac{x^3}{100}$
How do I find the solution to:

$$x^{\log(x)}=\frac{x^3}{100}$$

So I multiplied 100 both sides getting:

$$100x^{\log(x)}=x^3$$

Now what should I do?

答:
I am entering an answer just to point out a peculiarity of this equation, and the importance of interpreting the exponent "correctly". (This should also clarify  Charles' answer.)  I have graphed the quadratic functions $ \ (\ln x)^2 - (3 \ln x) + (\ln 100) \ $ in blue and $ \ (\log x)^2 - (3 \log x) + 2 \ $ in red.  Notice that the blue curve has no x-intercepts; the quadratic equation for natural logarithms yields a negative discriminant since $ \ (-3)^2 < 4 \ln 100 \ $.  So common logarithms are intended in this problem.

I had to use two graphs because the graph of the common logarithmic function is very shallow, but it does cross the x-axis at 100.

Title: Solving $x^{\log(x)}=\frac{x^3}{100}$


问:
how to find the volume of the intersection of this cylinder and sphere?
I need to find the volume of $x^2+y^2=y$ bounded by $x^2+y^2+z^2=2$. I am capable of doing the integral, but I'm not sure how do I find the bounds for the integral. I know it's a triple integral. Should I just convert into cylindrical coordinates and calculate?

Any help?

答: Note that $x^2+y^2\leqslant y\iff x^2+\left(y-\frac12\right)^2\leqslant\frac14$. So, $0\leqslant y\leqslant1$ and therefore if you express the volume that you want to compute in cylindrical coordinates, you will have $0\leqslant\theta\leqslant\pi$. On the other hand,$$x^2+y^2\leqslant y\iff r^2\leqslant r\sin(\theta)\iff r<\sin(\theta)$$and$$x^2+y^2+z^2\leqslant2\iff z^2\leqslant2-r^2.$$So, compute the integral$$\int_0^\pi\int_0^{\sin(\theta)}\int_{-\sqrt{2-r^2}}^{\sqrt{2-r^2}}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta.$$


问:
Give an example to show that if T is the product topology on X , then $\prod_{B}$ is need not be closed( Wayne patty 2.3.8)
This question is from page85 of Wayne patty's topology.

Let {$(X_\alpha ,T_\alpha): \alpha \in \Lambda$} be an indexed family of topological spaces, let $X= \prod_{\alpha \in \Lambda}X_\alpha $ , let T be a topology  on X, and for each $\beta \in \Lambda$, let $\pi_{\beta} : X \to X_{\beta}$ be the projective map. Then, Give an example to show that if T is the product topology on X, then $\pi_\beta$ need not be closed.

I am unable to construct such an example and so requesting you to help me.

答: The  projection onto the first coordinate on  $\mathbb R^{2}$ is not a closed map: $\{(a,b):ab=1\}$ is closed but its projection is not. The projection is $\mathbb R \setminus \{0\}$.


答: Since $e^{-x}$ is convex, you have $$e^{-tx-(1-t)y}\le t\,e^{-x}+(1-t)\,e^{-y}$$ for $t\in[0,1]$. Now integrate both sides over $t$ from $0$ to $1$, and you get exactly your inequality.

Title: Inequality of exponential

Body: Trying to show $$\frac{e^{-x}-e^{-y}}{y-x}\leq \frac{1}{2}(e^{-x}+e^{-y})$$ I tried working with $y>x$, and the case $y-x\geq2$ was trivial. But I guess this isn't useful. Using the mean value theorem and writing the problem as $\frac{e^y-e^x}{y-x}\leq\frac{1}{2}(e^y+e^x)$ I get that exists a $c\in(x,y)$ s.t. $\frac{e^y-e^x}{y-x}=e^c$ but I can't get the inequality. In other hand, $e^x$ is convex, so $\frac{1}{2}(e^{x}+e^{y})\geq e^{\frac{x+y}{2}}$. Ought I try to show $e^{\frac{x+y}{2}}\geq e^c$ ?


问:
Using linear least-squares to solve this system
I have $\frac{n(n-1)}{2}$ equations of the form:

$$2(x_j-x_i)X+2(y_j-y_i)Y+2(z_j-z_i)Z+2v^2\tau(t_i-t_j)=2v^2(t_i^2-t_j^2)+(x_j^2+y_j^2+z_j^2)-(x_i^2+y_i^2+z_i^2)$$

With $n\geq 4$, and where everything except $X,Y,Z$ and $\tau$ are known

I understand that this equation (actually, $\frac{n(n-1)}{2}$ equations/datapoints) can be solved for $X,Y,Z,\tau$ (values which I'll use as initial estimates for a more ``precise" solver) with a linear least-squares fit. How, in practice, is this done? I know not too much about the topic. Would I want a multivariate least-squares fit? A multiple least-squares fit? "Standard" least-squares?

My recollection is that I'd wan't a multiple least-squares fit, however I'm not sure.

Note that I specifically want to use least-squares here, and the goal is to obtain estimates for $X,Y,Z,\tau$.

Thanks in advance.

答:
Your equations are of the form

$Xv_1+Yv_2+Zv_3+\tau v_4 = b$

where $v_1,v_2,v_3, v_4$ and $b$ are vectors. Write A = {the matrix whose columns are $v_1,v_2,v_3,v_4$} then the 4 vector $w$ of numbers you want is the solution of the normal equation

$A^TAw = A^Tb$

which is your least squares solution.


问:
GCD in polynomial rings with coefficients in a field extension
Let $E/F $ be a field extension and $f,g$ $\in$ $F[x]$ (the polynomial ring with coefficients in $F$ ). Let's denote with $\gcd_F(f,g)$ the greatest common divisor of $f$ and $g$ in $F[x]$. Is it true that $\gcd_F(f,g)$ $=$ $\gcd_E(f,g)$ ? </p>

Body: Let $d=\gcd_F(f,g)$. Then $$dE[x]=(dF[x])E[x]=((f,g)F[x])E[x]=(f,g)E[x].$$Thus $d=\gcd_E(f,g).$


问:
Weird limit with/without derivation
I have no idea how can I solve the following:

$$ x^2\left(1+\frac{1}{x}\right)^{x}-e\,x^{3}\ln\left(1+\frac{1}{x}\right)$$

with $x$ tends to infinity. Thank you.

答:
We could go beyound the limit itself and get the asymptotics composing Taylor series one piece at the time.
$$y=x^2\left(1+\frac{1}{x}\right)^{x}-e\,x^{3}\ln\left(1+\frac{1}{x}\right)$$

$$a=\left(1+\frac{1}{x}\right)^{x}\implies \log(a)=x \log\left(1+\frac{1}{x}\right)$$
$$\log(a)=x \left(\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{3
   x^3}-\frac{1}{4
   x^4}+O\left(\frac{1}{x^5}\right) \right)$$

$$\log(a)=1-\frac{1}{2 x}+\frac{1}{3 x^2}-\frac{1}{4
   x^3}+O\left(\frac{1}{x^4}\right)$$
$$a=e^{\log(a)}=e-\frac{e}{2 x}+\frac{11 e}{24 x^2}-\frac{7e}{16
   x^3}+O\left(\frac{1}{x^4}\right)$$ Back to $y$
$$y=x^2\left(e-\frac{e}{2 x}+\frac{11 e}{24 x^2}-\frac{7e}{16
   x^3}+O\left(\frac{1}{x^4}\right) \right)-
 e x^3\left(\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{3
   x^3}-\frac{1}{4
   x^4}+O\left(\frac{1}{x^5}\right) \right)$$

$$y=\frac{e}{8}-\frac{3 e}{16 x}+O\left(\frac{1}{x^2}\right) $$

Title: Weird limit with/without derivation


问:
Use the Divergence Theorem to compute the flux of F across
Consider the vector field $F=<0,0,x+z>$,and the surface S which is the part of the plane $x+y+2z=4$  inside the first octant. Assume the unit normal vector n to S has positive third component.Use the Divergence Theorem to compute the flux of F across S by finding instead the flux of F across a different surface and an appropriate triple integral.

Here is my work:
$$div(\mathbf F) = \frac{\partial 0}{\partial x} + \frac{\partial 0}{\partial y}+\frac{\partial}{\partial z}(x+z) = 1 $$
By divergence theorem, $$\int\int\int_Ddiv\mathbf FdV=\int\int_S(F\cdot n)d\sigma$$
$$\int\int\int_D1dV = the\;volume\;under\;"x+y+2z=4"$$
so I got $\frac {16}3$, but the correct answer is 16.
Does anyone know where I did it wrong?

答:
That is because applying Divergence Theorem will give you flux through the entire closed region which is a tetrahedron with $4$ surfaces enclosed by $x + y + 2z = 4$ in the first octant and the coordinate planes, including the XY plane between $(0, 0, 0), (4, 0, 0), (0, 4, 0)$ and similarly part of $YZ$ and $XZ$ planes.

But I think your question asks you to find flux only through the surface $x + y + 2z = 4$. So you need to subtract the flux through the other $3$ surfaces.

Given our vector field is $(0, 0, x + z)$, the flux through $XZ$ plane and $YZ$ planes will be zero as the unit normal vectors to these planes will have no $z$ component and dot product $\vec{F} \cdot \hat{n}$ will be zero. But that is not the case for $XY$ plane.

The outward pointing unit normal vector through the surface which is $XY$ plane is $(0, 0, -1)$.

Flux through this surface will be $\displaystyle \int_0^4 \int_0^{4-y} (0, 0, x + 0) \cdot (0, 0, -1) \, dx \, dy = -\frac{32}{3}$.

So the flux through surface ($S, x + y + 2z = 4$) will be the flux through the entire closed region minus the flux through other $3$ surfaces.

$$ = \frac{16}{3} - (-\frac{32}{3}) = 16$$.


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

答:
After having needed to spend a bit of time in order for things to come to light, may I say that this is quite a lovely problem and may I convey my regards to your professor for having included it in his textbook.

Let us assume a left Artinian, non-commutative ring $A$ such that for every nonzero bilateral (I dislike the English terminology of "two-sided") ideal we have $A=[I \cup \{1_A\}]_{\mathrm{Ann}}$, where for arbitrary subset $X \subseteq A$ I use the notation $[X]_{\mathrm{Ann}}$ to refer to the subring generated by $X$ in $A$. Let us remark that for any bilateral ideal $I \leqslant_{\mathrm{b}} A$ we have the relation $[I \cup \{1_A\}]_{\mathrm{Ann}}=I+\mathbb{Z}_A$, where I have used the abbreviation $\mathbb{Z}_A\colon=\mathbb{Z}1_A\colon=\{n1_A\}_{n \in \mathbb{Z}}$ (on the same note let us also introduce $n_A\colon=n1_A$ for arbitrary integer $n \in \mathbb{Z}$).

We seek to prove that $A$ is simple under these conditions and we shall proceed in the following sequence:

Introducing the abbreviation $C\colon=\mathrm{Z}(A)$ for the centre of $A$, we first argue that $C$ is a field.

Since $A$ is non-commutative it is in particular nonzero, which means $0_A \neq 1_A$ and therefore $C$ is also nonzero. Given arbitrary ring $R$ we shall write $\mathrm{U}(R)$ for the subset of units of $R$ and similarly $R^{\times}\colon= R \setminus \{0_R\}$. For arbitrary subset $X \subseteq R$ we recall the left and right annihilators $\mathrm{Ann}_{\mathrm{s}}(X)\colon=\{t \in A \mid tX \subseteq \{0_R\}\}$ respectively $\mathrm{Ann}_{\mathrm{d}}(X)=\{t \in A \mid Xt \subseteq \{0_R\}\}$, remarking that left (right) annihilators are always left (right) ideals: $\mathrm{Ann}_{\mathrm{s}}(X) \leqslant_{\mathrm{s}} R$ respectively $\mathrm{Ann}_{\mathrm{d}}(X) \leqslant_{\mathrm{d}} R$. In the particular case when $X \subseteq \mathrm{Z}(R)$ is a central subset it is easily seen that the two lateral annihilators coincide and are both a bilateral ideal of $R$: $\mathrm{Ann}_{\mathrm{s}}(X)=\mathrm{Ann}_{\mathrm{d}}(X) \leqslant_{\mathrm{b}} R$.

We first establish the following characterisation: $z \in \mathrm{U}(C) \Leftrightarrow \mathrm{Ann}_{\mathrm{s}}(z)=\{0_A\}$ (a central element is invertible if and only if it is cancellable). Indeed, the implication $z \in \mathrm{U}(C) \Rightarrow \mathrm{Ann}_{\mathrm{s}}(z)=\{0_A\}$ is a particular case of the more general implication $x \in \mathrm{U}(A) \Rightarrow \mathrm{Ann}_{\mathrm{s}}(x)=\mathrm{Ann}_{\mathrm{d}}(x)=\{0_A\}$, relation which is valid in any ring without imposing any supplementary hypotheses.

As for the converse implication, we turn to the left Artinian nature of $A$ by considering the decreasing sequence of principal left (they are actually bilateral) ideals $\left(Az^n\right)_{n \in \mathbb{Z}}$ and inferring the existence of $k \in \mathbb{N}$ such that $Az^k=Az^{k+1}$, which further entails the existence of $x \in A$ such that $z^k=z^{k+1}x$ or equivalently $(1_A-zx)z^k=0_A$ and thus $1_A-zx \in \mathrm{Ann}_{\mathrm{s}}\left(z^k\right)$. In general, for arbitrary $y \in A$ it is elementary that $\mathrm{Ann}_{\mathrm{s}}(y)=\{0_A\}$ entails $\mathrm{Ann}_{\mathrm{s}}\left(y^n\right)=\{0_A\}$ for any $n \in \mathbb{N}$ (if $y$ has a zero left annihilator then the group endomorphism of $(A, +)$ given by right multiplication with $y$ is injective, property which is preserved by all the powers of this endomorphism in the ring $\mathrm{End}_{\mathbf{Gr}}((A, +))$, powers which are none other than the endomorphisms given by right multiplication with $y^n$ for generic $n \in \mathbb{N}$). We thus infer that $zx=xz=1_A$, which means that $z \in \mathrm{U}(A) \cap \mathrm{Z}(A)=\mathrm{U}(C)$ (recall that in general for any monoid $M$ the equality $\mathrm{U}(\mathrm{Z}(M))=\mathrm{U}(M) \cap \mathrm{Z}(M)$ holds, since indeed any central element of $M$ which is a unit of $M$ will necessarily have a central inverse, rendering it into a unit of the centre).

Having obtained the above characterisation, let us further establish the relation $z=0_A \Leftrightarrow z \in C \wedge \mathrm{Ann}_{\mathrm{s}}(z) \neq \{0_A\}$. The $\Rightarrow$ direction is clear, so we focus on the converse, considering an arbitrary central element $z \in C$ whose (left...) annihilator $I\colon=\mathrm{Ann}_{\mathrm{s}}(z)$ is nonzero. As we have already remarked, $I \leqslant_{\mathrm{b}} A$ is in this case a bilateral ideal and as it is nonzero we gather that $A=I+\mathbb{Z}_A$ by virtue of our very particular assumption. This immediately entails the description $Az=\left(I+\mathbb{Z}_A\right)z=Iz+\mathbb{Z}z=\mathbb{Z}z$, which means that any element in the principal (left...) ideal generated by $z$ is actually an integer multiple of $z$. Let us now additionally assume that $z \neq 0_A$ and derive from this a contradiction. If indeed $z \neq 0_A$ then $z \in Az \leqslant_{\mathrm{b}} A$, in other words $Az$ is a nonzero bilateral ideal (left ideals generated by central subsets are automatically bilateral). The particular assumption of our problem once again applies to the conclusion that $A=Az+\mathbb{Z}_A=\mathbb{Z}z+\mathbb{Z}_A$, which further entails the commutativity of $A$ thus contradicting the hypothesis. It follows that with necessity $z=0_A$, under the above assumptions.

Hence, an arbitrary central element $z \in C$ either has nontrivial (left...) annihilator - in which case it is $0_A$ - or a trivial one, in which case it is a unit of $C$. This establishes $C$ as a field.

We now argue that $A$ is a simple ring, for which it will suffice to show that its Jacobson radical $R\colon=\mathrm{Rad}(A)$ is null.

Indeed, having established the fact that $R=\{0_A\}$ we recall the general property of any left Artinian ring of having a semisimple quotient by its radical. It follows from one of the Wedderburn structure theorems that $A/\{0_A\} \approx A \ (\mathbf{Ann})$ is semisimple and thus isomorphic to the direct product of a finite family $S$ - indexed by $\Lambda$ - of nonzero simple rings. This entails that the centre $C$ is isomorphic to the direct product of the family $(\mathrm{Z}(S_{\lambda}))_{\lambda \in \Lambda}$ of centres. By virtue of another one of the Wedderburn structure theorems, it can be easily inferred that the centre of a nonzero simple ring is necessarily a field. Combining this with the fact that $C$ is itself a field and hence a nonzero domain, we gather that $|\Lambda|=1$, in other words the direct product isomorphically expressing $A$ must consist of precisely one factor (the case $|\Lambda| \geqslant 2$ would contradict the integrity of $C$ whereas $\Lambda=\varnothing$ is prohibited by our previous observation that since $A$ is noncommutative it is in particular nonzero).

Let us proceed by Reductio ad Absurdum and assume that $R \neq \{0_A\}$. We first remark that under this hypothesis it is impossible that $\mathrm{char}\ C=0$. Indeed, if this were the case then $2_A \neq 0_A$. Since $R \leqslant_{\mathrm{b}} A$ is nonzero we gather that $A=R+\mathbb{Z}_A$ and as $C$ is a field we can claim that $\frac{1}{2_A} \in A$, whence there must exist $a \in R$ and $k \in \mathbb{Z}$ such that $\frac{1}{2_A}=a+k_A$. This equation leads to $a=-k_A+\frac{1}{2_A} \in C^{\times}=\mathrm{U}(C)$ and signifies that the proper ideal $R$ would contain a unit, clearly an absurd situation (the hypothesis that $C$ is of characteristic $0$ is the one which justifies the assertion that $k_A \neq \frac{1}{2_A}$, equivalent to $(2k-1)_A \neq 0_A$).

Since $C$ is a field, we must have $\mathrm{char}\ C=p$ for a certain (strictly positive) prime number $p$. This means that $A$ has a canonical (associative and unitary) $\mathbb{Z}_p$ - algebra structure, where I am using the notation $\mathbb{Z}_r\colon=\mathbb{Z}/r\mathbb{Z}$ for arbitrary $r \in \mathbb{N}$. From the particular assumption in the hypothesis we gather that given any proper nonzero bilateral ideal $I \leqslant_{\mathrm{b}} A$, we have $A=I+(\mathbb{Z}_p)1_A$, where the sum on the right-hand side is direct: indeed, $I \cap \left(\mathbb{Z}^{\times}_p\right)1_A \subseteq I \cap \mathrm{U}(A)=\varnothing$ whence it follows that $I \cap (\mathbb{Z}_p)1_A=\{0_A\}$. From this we infer that $\mathrm{codim}^{A}_{\mathbb{Z}_p}I=\mathrm{dim}_{\mathbb{Z}_p}(A/I)=1$, in other words any nonzero proper bilateral ideal of $A$ is a hyperplane in the $\mathbb{Z}_p$-vector space structure subjacent to the algebra structure introduced above.

Let $\mathscr{Id}_{\mathrm{b}}(A)$ denote the set of all bilateral ideals of $A$, $\mathscr{I}\colon=\mathscr{Id}_{\mathrm{b}}(A) \setminus \{A\}$ and $\mathscr{I}’\colon=\mathscr{I} \setminus \{\{0_A\}\}$. As $A$ is nonzero we clearly have $\mathscr{I} \neq \varnothing$. Bearing in mind the general relation $R \subseteq \displaystyle \bigcap \mathrm{Max}\left(\mathscr{I}\right)$ ($\mathscr{I}$ is considered implicitly ordered by inclusion and the syntax $\mathrm{Max}$ refers to the collection of maximal elements with respect to this ordered set, in other words the collection of all maximal bilateral ideals), we infer that $\mathscr{I}’ \neq \varnothing$ (otherwise we would have $\mathscr{I}=\{\{0_A\}\}$, $\mathrm{Max}\left(\mathscr{I}\right)=\{0_A\}$ and eventually $R \subseteq \{0_A\}$).

It is also an elementary fact that the ordered set $\mathscr{I}$ is inductive and therefore admits at least a maximal element, say $M \in \mathrm{Max}\left(\mathscr{I}\right)$. It is clear that $M \neq \{0_A\}$, for assuming the contrary would once again entail $\mathscr{I}=\{\{0_A\}\}$, situation which we have already ruled out as impossible under the given assumptions. Since $M, R \in \mathscr{I}’$ we have that $M$ and $R$ are both $\mathbb{Z}_p$ - hyperplanes of $A$, according to one of the previous observations. Since $R \subseteq M$, we thus gather that $R=M$, meaning that the Jacobson radical $R$ is a maximal bilateral ideal. Since $R$ is at the same time included in every maximal bilateral ideal, we infer that $\mathrm{Max}\left(\mathscr{I}\right)=\{R\}$.

Since any proper bilateral ideal is included in a certain maximal bilateral ideal (by virtue of the same token of inductivity of $\mathscr{I}$ ordered by inclusion), the existence of just one maximal bilateral ideal means that $A$ actually has $R$ as a maximum proper bilateral ideal, symbolically $R=\max \mathscr{I}$. Thus, if we furthermore consider an arbitrary $I \in \mathscr{I}’$ we gather that $I \subseteq R$ and $I$, $R$ are both $\mathbb{Z}_p$ - hyperplanes of $A$. Once again, this entails $I=R$ and signifies that we actually have the relation $\mathscr{I}=\{\{0_A\}, R\}$.

Let us however keep in mind another general property of left Artin rings, namely that they have nilpotent Jacobson radicals. I will use the notation $I^{.n}$ to refer to the power of exponent $n \in \mathbb{N}$ of the bilateral ideal $I$ within the multiplicative monoid $(\mathscr{Id}_{\mathrm{b}}(A), \bullet)$. In our case, since $R^{.2} \in \mathscr{I}$ we have either $R^{.2}=\{0_A\}$ or $R^{.2}=R$. The latter option would mean that $R$ is at the same time a nonzero idempotent as well as a nilpotent element, which constitutes a contradiction. This leaves the only possible option $R^{.2}=\{0_A\}$.

The above finding means in particular that $R^{(2)}=RR=\{0_A\}$ and - together with the relation $A=R+\mathbb{Z}_A$ - allows us to infer that $R \subseteq \mathrm{C}_A(R)$, in other words $R$ centralises itself. As it is also obvious that $R \subseteq \mathrm{C}_A(\mathbb{Z}_A)$, taking into account the relation $A=R+\mathbb{Z}_A$ we easily gather that $A \subseteq \mathrm{C}_A(A)=\mathrm{Z}(A)$, contradicting the restriction that $A$ is not commutative. This contradiction is caused by the assumption $R\neq \{0_A\}$ and signifies that indeed $R=\{0_A\}$, as we had set out to prove.

Our argument thus comes to an end.

Title: Why this Artin ring is simple?


问:
explicit formula of $(x \frac{\partial}{\partial x})^n f(x)$
Is there any explicit expression of

$$
(x \frac{\partial}{\partial x})^n f(x)
$$

as function of $x$ and $\frac{\partial^{k}f}{\partial x^k} $$, $$ 1\leq k \leq n$.

Any idea

Thanks

答:
I focused on it.

Let's suppose that it takes the following polynomial formula

$$
(x\frac{\partial}{\partial x})^n = \sum_{k=0}^{n} a_{k}^{n} x^k \frac{\partial^k}{\partial x^k}
$$

when making calculation for the first values of n, $n=1, n=2, n=3 ...$
we notice that

$a_0^{n} = 0$

$a_1^{n} = 1$

and   $\ a_n^n = 1$

For the other coefficients $2 \leq k \leq n-1$ we get

$a_2^n = \frac{1}{2}  (2^n - 2)$

$a_3^n = \frac{1}{6} (3^n - 3 \ 2^n + 3)$

$a_4^n = \frac{1}{24} (4^n - 4 \ 3^n + 6 \ 2^n - 4)$

$a_5^n = \frac{1}{120} (5^n - 5 \ 4^n + 10 \ 3^n - 10 \ 2^n + 5)$

This is likely to be the following general formula
$$
a_k^n = \frac{(-1)^k}{k!} \sum_{p=1}^{k}C_{k}^{p} (-1)^p p^n$$

Using the recurrence relation for $a_k^n$ coefficients $a_k^{n+1} = k a_k^n + a_{k-1}^n$, we can check the general formula.

Then we get,

$$
(x\frac{\partial}{\partial x})^n f = \sum_{k=1}^{n-1} \frac{(-1)^k}{k!} \sum_{p=1}^{k}C_{k}^{p} (-1)^p p^n x^k \frac{\partial^k f}{\partial x^k} + x^{n} \frac{\partial^n f}{\partial x^n}
$$


问:
Determine wether the function $f(x,y)=\frac{x^2+y^2}{\sin \sqrt{x^2+y^2}}$ is defferentiable at $(0,0)$.

Let $D=\{(x,y)\in\Bbb R^2:\sqrt{x^2+y^2}<\pi\}$.
Define $f(x,y)=\begin{cases}
0,  & \text{if $x= 0$ } \\
\frac{x^2+y^2}{\sin \sqrt{x^2+y^2}}, & \text{if $x\in D-(0,0)$}
\end{cases}$. Is $f$ différentiable at $(x,y)=(0,0)$?

I have computed $\frac{\partial f}{\partial x}(0,0)=1$, and $\frac{\partial f}{\partial y}(0,0)=1$. I'm stuck at $$\lim \limits_{(h,k)\to (0,0)}\frac{\frac{h^2+k^2}{\sin \sqrt{h^2+k^2}}-h-k}{\sqrt{h^2+k^2}}$$

What should I do next?

答:
Let us see if $f$ is partially derivable w.r.t. $x$ at $(0,0)$. Since
$$
\frac{f(x, 0) - f(0,0)}{x}
= \frac{|x|}{x} \cdot \frac{|x|}{\sin |x|}\,,
$$
we conclude that the limit as $x\to 0$ of the above quantity does not exist.


问:
Largest disc around which this complex function is one-to-one?
How would you determine explicitly the largest disc round the origin on which the function $f(z) = z^2 +z$ is one-to-one?

Is there a general method to do this for functions of this type?

答:
0) "Disk" shall mean disk centered at $0\in \mathbb C$.

1) a) The equation $z^2+z=w^2+w$ is equivalent to $(z-w)(z+w+1)=0$.

So two different points $z,w\in \mathbb C $ will have the same image under $f$ iff $w=-z-1$.

b)In every open disk $D$  of radius $\gt \frac {1}{2}$ there exists $\epsilon $ such that the points  $z=-\frac {1}{2}+\epsilon $ and $w=-\frac {1}{2}-\epsilon $  belong to $D$.

 Since $z$ and $w$ satisfy $z+w+1=0$ we have $f(z)=f(w)$,  so that $f$ is not injective on $D$.

Hence the required radius must be  $\leq \frac {1}{2}$ .

c) On the other hand  if $z'$ is in the  disk centered at $0$ and of radius $ \frac {1}{2}$, then $z'-1$ is not in the disk.[Draw  a picture !].

Applying this to $z'=-z$, we conclude that  $f$ is is injective on that disk .

d) Conclusion: the required radius is $\frac {1}{2}$

2) One could  make a change of variable $\zeta=z+\frac {1}{2}$,  and reduce the question to the radius of injectivity of $\zeta\mapsto \zeta ^2$ around $\zeta=\frac {1}{2}$.

This allows one to use that  the squaring function is injective on $\text {Re}  (z)\gt 0$.

3) More advanced users will notice that this example comes from a degree two covering of $\mathbb P^1$ ramified only at $z=\infty$ and $z=-\frac {1}{2}$

4) The injectivity of a holomorphic function on a disk is a very subtle problem.

Louis de Branges de Bourcia's solution of the Bieberbach conjecture on holomorphic functions injective on the unit disk is one of the most profound results in complex analysis proved in the twentieth century.

You can find a photo of the solver  sporting a fine Basque beret here.


答: You made a mistake computing the partial derivatives, they actually don't exist at the origin. For example, if you set $y$ to zero, and use the approximation $\sin x = x$ near the origin, you can see that your function reduces to $|x|$, which is not differentiable.

Title: Determine wether the function $f(x,y)=\frac{x^2+y^2}{\sin \sqrt{x^2+y^2}}$ is defferentiable at $(0,0)$.

Body:
Let $D=\{(x,y)\in\Bbb R^2:\sqrt{x^2+y^2}<\pi\}$.
Define $f(x,y)=\begin{cases}
0,  & \text{if $x= 0$ } \\
\frac{x^2+y^2}{\sin \sqrt{x^2+y^2}}, & \text{if $x\in D-(0,0)$}
\end{cases}$. Is $f$ différentiable at $(x,y)=(0,0)$?

I have computed $\frac{\partial f}{\partial x}(0,0)=1$, and $\frac{\partial f}{\partial y}(0,0)=1$. I'm stuck at $$\lim \limits_{(h,k)\to (0,0)}\frac{\frac{h^2+k^2}{\sin \sqrt{h^2+k^2}}-h-k}{\sqrt{h^2+k^2}}$$

What should I do next?


问:
Can it be proved using something like functional calculus?
I have a question about a part of some proof.

Let $\lambda$ be any number on the unit circle in the complex plane.

Then, we know that there exists a real bounded Borel function $g(\lambda)$ such that $\lambda=\exp(ig(\lambda))$, where $g(\lambda)\in(-\pi,\pi]$.

Now, let $H=M_g$ be the multiplication operator defined by $g$, which is a self-adjoint operator on some Hilbert space, the detail of which I omit for simplicity.

Then, could we prove easily exp$(iH)=M_\lambda$ using something like Borel functional calculus or anything else?

答:
Borel functional calculus can be defined (or otherwise characterized) in such a way that if a normal operator $T$ is unitarily equivalent to the multiplication operator $M_\phi$ and $f$ is a Borel function, then $f(T)$ is unitarily equivalent to $M_{f\circ\phi}$ (through the same unitary).

In your case $T=M_g$ and $f=\exp(i\,\cdot\,)$, hence $f(T)=M_{f\circ g}=M_{\mathrm{id}}$.

However, the exponential of a bounded operator (normal or not) can also be defined via the exponential series. One can show that both definitions coincide for normal operators. Using that $g\mapsto M_g$ is a $\ast$-homomorphism, this definition also implies $\exp(iM_g)=M_{\mathrm{id}}$ immediately.


问:
The boundary of a closed Mobius band
The Mobius band is defined by the quotient: $r:I\times I\rightarrow M$, with the equivalence relation $(0,x)\sim (1,1-x)$ for all $x\in I$.

The boundary of Mobius band $M$ is defined as the set of points that have an open neighbourhood which is homeomorphic to the closed half space.

I know its boundary is $r(B)$ where $B=\{(x,y)\in I\times I: y\in\{0,1\}, x\in I\}$. We need to find an open neighbourhood for each point in $r(B)$ s.t. homeomorphic to the closed half space $H=\{(x,y)\in\mathbb{R}^2:y\geq 0\}$.

It is easy to prove $r((0,1)\times (0,1))$ is not in the boundary $\partial M$, and $r((0,1)\times\{0,1\})\subset\partial M$.

But how to prove $r(\{0,1\}\times (0,1))$ is not in the boundary, and $r(0,0)$ $r(1,0)$ are in the boundary?

答:
But how to prove $r(\{0,1\}\times (0,1))$ is not in the boundary

Consider $[x,y]\in r(\{0,1\}\times(0,1))$. Note that $[x,y]=\{(0,t),(1,1-t)\}$ for some $t\in(0,1)$, as an equivalence class. Pick $0<r<\min(t,1-t)$ and consider two half open disc:

$$D_1=\big\{v\in I^2\ \big|\ \lVert v-(0,t)\rVert < r\big\}$$
$$D_2=\big\{v\in I^2\ \big|\ \lVert v-(1,1-t)\rVert < r\big\}$$

These are two disjoint and open subsets of $I^2$. They are not standard disks however we can define a continuous function

$$h:D_1\cup D_2\to \mathbb{R}^2$$
$$h(x,y)=\begin{cases}
(x,y) &\text{if }x\leq 1/2 \\
(x-1, y-1) &\text{otherwise}
\end{cases}$$

and note that the image of $h$ is precisely the open ball $B$ in $\mathbb{R}^2$ around $(0,t)$  of radius $r$. Of course $h$ is not a homeomorphism, because it is not injective, however it induces an injective map $H:(D_1\cup D_2)/\sim\to B$ on the quotient space, and thus $H$ is a homeomorphism. Note that $(D_1\cup D_2)/\sim$ is an open neighbourhood of $[x,y]$.

In particular $[x,y]$ has an open neighbourhood homeomorphic to $\mathbb{R}^2$ and thus it cannot be a boundary point.

and $r(0,0)$ $r(1,0)$ are in the boundary?

For that you can utilize the fact that if $v\in M$ is such that any open neighbourhood $U$ of $v$ satisfies $U\cap\partial M\neq \emptyset$ then $v\in\partial M$. This is a simple consequence of the fact that $\partial M$ is a closed subset of $M$. Which follows (as earlier) from the observation that $x\in M\backslash\partial M$ if and only if $x$ has an open neighbourhood homeomorphic to $\mathbb{R}^n$.

Title: The boundary of a closed Mobius band


问:
Why does Totally bounded need Complete in order to imply Compact?
Why does "Totally bounded" need "Complete" in order to imply "Compact"?
Shouldn't the definition of totally bounded imply the existence of a convergent subsequence of every sequence?</p>

答:
The confusion is understandable. We restate the two definitions:

Total boundedness says that every uniform open cover has a finite subcover.

Covering compactness says that every open cover, including non-uniform covers, have a finite subcover.

(In metric spaces, open balls can be used in place of open sets/covers. This is because they form a base of the topology: that is, every open set can be decomposed into a union of open balls, implying that every open cover, being a union of open sets, can as well.)

A more illustrative counter-example would be $(0, 1)$, as user Daniel Fischer mentioned in a comment. It is not compact because I can find an open cover that has no finite subcover, namely

$$B\left(1, {1 \over 2}\right) \cup B\left({1 \over 2}, {1 \over 3}\right) \cup \cdots = \bigcup_{n = 1}^{+ \infty} B\left({1 \over n}, {1 \over n + 1}\right)$$

If we stop the construction of the union at any finite $n$, the interval $(0, {1 \over n} - {1 \over n + 1}]$ won't be covered.

It is totally bounded, though. There exists a finite, positive real number, namely $1$, such that, $\forall x, y \in (0, 1)$, $\ d(x, y) < 1$, making $(0, 1)$ bounded. On the real line, a bounded interval is totally bounded (given $\varepsilon > 0$ and an interval $(a, b)$, subdivide your interval into ${(b - a)\over\varepsilon} + 1$ open pieces, then union these pieces with open balls of radius $\varepsilon$ centered at the partition points).


问:
Let $G$ be a domain and $f$ be holomorphic in $G$. Show that, if $|f|^2$ is harmonic in $G$, then $f$ is constant.
Complex Analysis

Let $G$ be a domain and $f$ be holomorphic in $G$. Show that, if $|f|^2$ is harmonic in $G$, then $f$ is constant.

My attempt at a solution so far:

Write $|f|^2=|u+iv|^2=u^2+v^2$

If $|f|^2$ is harmonic, then $\nabla^2|f|^2=$$ \frac{∂^2|f|^2 }{∂x∂x}+\frac{∂^2|f|^2 }{∂y∂y}=0$

Expanding on this:

$$\frac{∂^2|f|^2 }{∂x∂x}+\frac{∂^2|f|^2 }{∂y∂y}=\frac{∂^2 }{∂x∂x}(u^2+v^2)+\frac{∂^2 }{∂y∂y}(u^2+v^2)=\frac{∂ }{∂x}(2u\frac{∂u }{∂x}+2v\frac{∂v }{∂x})+\frac{∂ }{∂y}(2u\frac{∂u }{∂y}+2v\frac{∂v }{∂y})=(2\frac{∂u }{∂x}\frac{∂u }{∂x}+2u\frac{∂^2u }{∂x∂x}+2\frac{∂v }{∂x}\frac{∂v }{∂x}+2v\frac{∂^2v }{∂x∂x})+(2\frac{∂u }{∂y}\frac{∂u }{∂y}+2u\frac{∂^2u }{∂y∂y}+2\frac{∂v }{∂y}\frac{∂v }{∂y}+2v\frac{∂^2v }{∂y∂y})=0$$$($take this as equation $(1))$

At this point I'm unsure (even more), but some factoring and rearrangement is certainly possible.
Does this approach lead anywhere? Somewhere down the line I would need $$\frac{∂u }{∂x}=\frac{∂u }{∂y}=\frac{∂v }{∂x}=\frac{∂v }{∂y}=0$$ for $f$ to be constant. Any ideas?

答:
Your expansion $(1)$ can be written in the form
$$u_x^2+v_x^2+u_y^2+v_y^2+u(u_{xx}+u_{yy})+v(v_{xx}+v_{yy})=0\tag2$$
Given that $f=u+iv$ be holomorphic, so it satisfies Cauchy-Riemann equation and also the real and imaginary parts of the holomorphic function $f$ i.e., $u$ and $v$ are harmonic. Therefore
$$u_x=v_y~,~~~~~u_y=-v_x~,~~~~~\text{and}~~~~u_{xx}+u_{yy}=0~,~~~~~v_{xx}+v_{yy}=0~.$$
Hence from $(2)$ we have$$u_x^2+v_x^2+u_y^2+v_y^2=0$$
$$\implies u_x=v_x=u_y=v_y=0$$
$$\implies u~~\text{and}~~v~~\text{are constants.}$$
Therefore $~f~$ is constant.


问:
Double integral of $\frac{y}{\sqrt{16+x^8}}$
I need to evaluate the following integration problem: $\int_{0}^8 \int_{\sqrt[3]{y}}^2 \frac{y}{\sqrt{16+x^8}} dx dy$.

Since it cannot be evaluated in dx, the order of integeration must be changed as follow: $\int_{0}^2 \int_{0}^{x^3} \frac{y}{\sqrt{16+x^8}} dy dx$, and that can be reduced to: $\frac{1}{2}\int_{0}^2 \frac{x^6}{\sqrt{16+x^8}}dx$.

I've reached that point and couldn't move further. Is there a mistake in the problem itself (like the $x^8$ in the denumerator should be $x^7$ instead?)

答:
tl;dr Given that the problem was found on a past exam, there was almost certainly a mistake in how the problem was printed; if the integrand had been $\frac{\mathbf 1}{\sqrt{16+x^8}}$, after swapping integration order and solving the easy first integral we get $\int_0^2\frac{x^3}{\sqrt{16+x^8}}\,dx$, which is elementary by the substitution $u=x^4$; the answer is $\frac{\sinh^{-1}4}4$. But for the answer to the question in the title, read on…

First substitute $u\sqrt2=x$ to transform the remaining single integral to
$$I_0=\sqrt2\int_0^{\sqrt2}\frac{u^6}{\sqrt{1+u^8}}\,du$$
Now substitute $t=u^2+\frac1{u^2}$. By Byrd and Friedman 584.04 and .05 this reduces $I_0$ to elliptic integrals:
$$I_0=\frac{\sqrt2}4\left(\int_2^\infty\frac{t^2-t-1}{\sqrt{(t-2)(t^2-2)}}\,dt+\int_2^{5/2}\frac{t^2-t-1}{\sqrt{(t-2)(t^2-2)}}\,dt+\int_{5/2}^\infty\frac{-t^2-t+1}{\sqrt{(t+2)(t^2-2)}}\,dt\right)$$
The first two integrals above are solved by B&F 237.20 followed by 313.00, .02, .04:
$$I_1(y)=\int_2^y\frac{t^2-t-1}{\sqrt{(t-2)(t^2-2)}}\,dt=\frac{\sqrt{4-2\sqrt2}}3\left[F(\varphi_1,m_1)-(2+\sqrt2)E(\varphi_1,m_1) + \sqrt{2+\sqrt2}\sqrt{\frac{(y-2)(y+\sqrt2)}{y-\sqrt2}}(y-\sqrt2+1)\right]$$
$$\varphi_1=\sin^{-1}\sqrt{\frac{y-2}{y-\sqrt2}},m_1=2\sqrt2-2$$
($m$ is the parameter.) $I_1(\infty)$ has an infinite part of $\frac23\sqrt{\frac{(y-2)(y+\sqrt2)}{y-\sqrt2}}(y-\sqrt2+1)$; remember this for later. The last integral is solved by B&F 238.18 followed by 311.00, .02, .04:
$$I_2(y)=\int_y^\infty\frac{-t^2-t+1}{\sqrt{(t+2)(t^2-2)}}\,dt=\frac{\sqrt{4-2\sqrt2}}3\left[(1+\sqrt2)F(\varphi_2,m_2)-(2+\sqrt2)E(\varphi_2,m_2) + \sqrt{2-\sqrt2}\sqrt{\frac{y^2-2}{y+2}}(1+\sqrt2)(y+1)\right]$$
$$\varphi_2=\sin^{-1}\sqrt{\frac{\sqrt2+2}{y+2}},m_2=3-2\sqrt2$$
$$I_2'(y)=\int_{5/2}^y\frac{-t^2-t+1}{\sqrt{(t+2)(t^2-2)}}\,dt=I_2(5/2)-I_2(y)$$
Note that $I_2$ has no meaning by itself, but only as a part of $I_2'$. The infinite part in $I_2'$ works out to $-\frac23\sqrt{\frac{y^2-2}{y+2}}(1+\sqrt2)\sqrt{3-2\sqrt2}(y+1)$; when combined with the infinite part of $I_1$, their sum tends to zero as $y\to\infty$. Now
$$I_0=\frac{\sqrt2}4(I_1(\infty)+I_1(5/2)+I_2(5/2)-I_2(\infty))$$
$$=\frac{\sqrt2}4\cdot\frac{\sqrt{4-2\sqrt2}}3\left(
\begin{align}
&K(m_1)-(2+\sqrt2)E(m_1)\\
+&F(\varphi_1|_{y=5/2},m_1)-(2+\sqrt2)E(\varphi_1|_{y=5/2},m_1)+\sqrt{\frac{1193\sqrt2+1954}{136}}\\
+&(1+\sqrt2)F(\varphi_2|_{y=5/2},m_2)-(2+\sqrt2)E(\varphi_2|_{y=5/2},m_2) + 7\sqrt{\frac{17\sqrt2+34}{72}}\end{align}\right)$$
$$=\frac{\sqrt{2-\sqrt2}}6(K(m_1)-(2+\sqrt2)E(m_1)+F(\psi_1,m_1)-(2+\sqrt2)E(\psi_1,m_1)+(1+\sqrt2)F(\psi_2,m_2)-(2+\sqrt2)E(\psi_2,m_2))+\sqrt{\frac{2518+300\sqrt2}{1377}}$$
$$\psi_1=\sin^{-1}\sqrt{\frac{5+2\sqrt2}{17}}\qquad\psi_2=\sin^{-1}\frac{\sqrt{2\sqrt2+4}}3$$


问:
Test for convergence for improper integral $1/x^x$
I am having trouble determining if this is convergence or divergence

$$\int^1_0 1/x^x dx$$

答: There is another way which is based on Caparison Test again. In fact, the following limit says that the improper integral is convergent: $$\lim_{x\to 0^+}~~x^{1/2}\times\frac{1}{x^x}=0<\infty$$ </p>


问:
The set of all self-inverting functions in $\mathbb{R}$
I had asked a question about self-induced functions a few days ago and would like to discuss again in the topic a bit. We found out that if $f(x) = f^{-1}(x)$, then that for an invertible function $g(x)$, the function
\begin{align}
f_g(x) = g^{-1}(f(g(x)))
\end{align}
is also a self-inverting function again. What I am wondering now is whether with the function
\begin{align}
f(x) = \frac{ax+b}{cx-a}
\end{align}
all other self-invertible functions in the space $\,\mathbb{R} \,$ follow, so the set of self-inverting functions is thus completely described.

答: If $f$ is self-inverse, i.e. $f(f(x))=x$ on $\mathbb{R}$, you might consider $g(x)=x-f(x)$. Then, $$g(f(x))=f(x)-f(f(x))=-g(x),$$ i.e. $$f(x)=g^{-1}(-g(x)).$$ Wouldn't that be a description of all self-inverse functions using just one, $f(x)=-x$?


问:
the boundary of Mobius band is homeomorphic to a circle
It is know that $\partial M\approx S^1$, where $\partial M$ is the boundary of the mobius band. But how do we prove that? I am trying to define the homeomorphism explicitly, but it is messy.

The Mobius band $M$ is defined by the quotient $q:I\times I\rightarrow M$ with the equivalence relation $(0,x)\sim(1,1-x)$.

$\partial M=q(I\times\{0,1\})$.

答:
So first of all you need to know that $\partial M=(I\times\{0,1\})/\sim$. Then consider

$$f:I\times\{0,1\}\to S^1$$
$$f(x,y)=\exp(\pi i (x+y))$$

This function maps $I\times\{0\}$ onto the first half circle and $I\times\{1\}$ onto the second half circle. It is not a homeomorphism, because it is not injective. However it induces a well defined continuous bijection

$$F:\partial M\to S^1$$
$$F([x,y])=f(x,y)$$

on the quotient and thus a homeomorphism since $\partial M$ is compact.


问:
Determining the number of solutions of a system of linear inequalities.
I want to determine the number of solutions of a system of linear inequalities, and I was wondering if there was a simple way to to that. I know that linear programming is often used to check whether there are a zero or non-zero number of solutions, i.e. if the system/bounds is/are feasible, but is it possible to distinguish between there being a finite amount of unique solutions or infinitely many solutions? For instance, the system
$$
\begin{array}{lcl}
x & \leq & y \\
x & \geq & y \\
x + y & \leq & 1 \\
x + y & \geq & 1
\end{array}
$$

has 1 unique solution, namely $x=y=\frac{1}{2}$, while the system

$$
\begin{array}{lcl}
x & \geq & y \\
x + y & \leq & 1 \\
x + y & \geq & 1
\end{array}
$$

has infinitely many solutions. Is there a away to find out how many solutions a system of linear inequalities has, if any?

答:
As @Milten noted in the comments, the number of solutions to a system of linear inequalities (over $\mathbb{R}$) may only be $0,1$ or infinite. This is because of convexity: if $v,w$ are two solutions, then $\alpha v + (1-\alpha)w$ is a solution for any $0 \leq \alpha \leq 1$, and for $v \neq w$ this gives an infinite number of solutions. This means that you only need to find two distinct solutions to decide whether there is an infinite number of them.

As you say, linear programming can be used to decide whether there is any solution, but with a bit more work it can also tell you whether there are multiple solutions or just one. Heuristically, you can use linear programming to maximize a random objective function $c\cdot x$ over the feasible region; one would expect that if the feasible region has more than one points then with high probability you would obtain multiple solutions. Granted, this is just a heuristic, but for practical purposes it should be good enough. (See also this answer.)

For an honest polynomial-time algorithm to decide whether the solution is unique you may want to take a look at this article, in which the authors reduce the problem of deciding uniqueness to finding the solution to another linear program. The article also contains a survey of previous results on this problem. Sadly, all the (legal) links I could find to the article are paywalled. If you would like to, I can describe their solution in more detail.

Finally, I would just like to note that in general you "can't avoid" linear programming in the sense that deciding whether there is at least one solution to a linear program is almost as difficult as finding an optimal solution. This is true in the sense that if you can decide whether there is a solution, then you can use a binary search-like algorithm to find an optimal solution in reasonable time.