问:
Show that $g$ is constant
Let $f$ be holomorphic in $\{\operatorname{Im}(z) \ge 0\}$, real on the real axis and bounded. Show that $f$ is constant

My attempt : I found the answer here

But i don't want to used schwarz reflection principle

My approach: Define  a function $g : Imz \ge 0 \to \mathbb{C}$  by $g(z)= f(z) - \overline{f(z)} $ where $\overline{f(z)}$ represent  conjugate  of  $f(z)$

Then $g$ is an analytics  function  and all real number are zeroes of $g$

Now if $g$ were non constant , then the set of zeroes  of are  isolated so zeros  will be countable   but here $g$ has uncountably many zeroes and non isolated.

so we get  a  contradiction

Therefore  $g$ must be constant

this implies $f$ is constant

Hence proved

Is my proof/approached is correct or not ?

答: Your proof is not correct because it assumes that $\overline f$ is an analytic function. In general, it is not. The only analytic functions $f$ with connected domain such that $\overline f$ is also analytic are the constant functions.

Title: Show that $g$ is constant


问:
Really confused with unit vectors

Let $\mathbf{u}$ and $\mathbf{v}$ be linearly independent unit vectors. Find the set of all possible values of $\mathbf{u}\cdot\mathbf{v}$. Give your answer in interval notation.

What is the maximum possible length of the vector resulting from the following linear combination?
$$
\frac{1}{\| \mathbf{v_1} \|} \,\mathbf{v_1} + \frac{1}{\|\mathbf{v_2}\|} \,\mathbf{v_2} + \cdots + \frac{1}{\| \mathbf{v_n} \|} \,\mathbf{v_n}
$$

So far I have gotten nowhere with this problem and I cant even look at a similar problem in the textbook. If someone can help me i would really appreciate it.

答: For part 1, linear independence ensures $u\cdot v\ne\pm1$, so the possible values comprise an open interval $(-1,\,1)$ instead of the closed $[-1,\,1]$. For part 2, you've summed $n$ unit vectors, so by the triangle inequality the maximum possible length is $n$, achieved when the vectors are all parallel as opposed to being antiparallel.

ContentLicense: CC BY-SA 4.0

Title: Really confused with unit vectors


问:
Where does the function $f(x) = \frac{2x}{x - 7}$ have an increasing slope?
Where does the function $f(x) = \frac{2x}{x - 7}$ have an increasing slope?

$a. x \le 0, x > 7$

$b. x<7$

$c. x > 7$

$d. x \in \Bbb R, x \neq 7$

This question is from a test of mine in a pre-calculus course (so no calculus allowed in answering the question).I have no idea as to how to solve this problem. I can tell where the function is negative and where it is positive, but that's about it, and I'm fairly sure that's no use here.

Any ideas?

答:
I'd go for $$\frac{2x}{x-7}=\frac{2(x-7)}{x-7} +\frac{14}{x-7}$$

The first term is constant, and the second one is just the graph of $1/x$ shifted and rescaled.

ContentLicense: CC BY-SA 3.0


问:
show this inequality $x^{n+1}+y^{n+1}\ge x^n+y^n$
let $x,y>0$ and $n$ be positive integer.if
$$x^{2n+1}+y^{2n+1}\ge 2$$
show that
$$x^{n+1}+y^{n+1}\ge x^n+y^n$$

maybe use Holder inequality: for example
$$(x^{n+1}+y^{n+1})^n(1+1)\ge (x^n+y^n)^{n+1}$$
so we must prove
$$\dfrac{1}{2}(x^n+y^n)^{n+1}\ge (x^n+y^n)^n$$
or
$$x^n+y^n\ge 2$$
this last inequality seem it is not hold.so How to prove my question.Thanks

答:
A different approach is to consider the equation $x^{2n+1}+y^{2n+1}=k$ and take polar coordinates. This gives $(x,y)=(k^a\cos^{2a}\varphi,k^a\sin^{2a}\varphi)$ where $a=1/(2n+1)$. We wish to show that* if $k\ge2$ then $$\frac{x^{n+1}+y^{n+1}}{x^n+y^n}=k^a\cdot\frac{\cos^{1+a}\varphi+\sin^{1+a}\varphi}{\cos^{1-a}\varphi+\sin^{1-a}\varphi}\ge1$$ for all $a\in(0,1)$ which we can rewrite as $$k^a\ge\sec^{2a}\varphi\cdot\frac{1+\tan^{1-a}\varphi}{1+\tan^{1+a}\varphi}=\frac{(1+s^2)^a(1+s^{1-a})}{1+s^{1+a}}$$ where $s=\tan\varphi\in[0,1]$ as the function is invariant under the transformation $s\mapsto1/s$. This means that $$k^{1-2b}\ge\frac{(1+t)^{1-2b}(1+t^a)}{(1+t^{1-a})}\implies\left(\frac{1+t}k\right)^{1-2b}-\frac{1+t^{1-b}}{1+t^b}\le0$$ where $t\in[0,1]$ and $b=(1-a)/2\in(0,1/2)$. Since the LHS is a decreasing function of $k$ it remains to show that $$\max_t\left\{\left(\frac{1+t}2\right)^{1-2b}-\frac{1+t^{1-b}}{1+t^b}\right\}=0.$$ The easiest way to prove this uses calculus and I don't think there are methods based solely on inequality results to do so.

 *This is actually a stronger claim, that for all real $x,y,n>0$, if $x^{2n+1}+y^{2n+1}\ge2$ then $x^{n+1}+y^{n+1}\ge x^n+y^n$.


问:
Can there be a different Eigen vector for a particular Eigen value?
Please see the photo.

Here, my answer came $k$
$\begin{bmatrix}
                   1\\
                 1\\
                   -1
\end{bmatrix}$

But their answer is given :
$k$
$\begin{bmatrix}
                     1\\
                    -1\\
                     1
\end{bmatrix}$
My answer is correct?? Please tell--

THE ACTUAL MATRIX]

答:
For $\lambda=1$ we have:
$$(A-I)v=0$$
$$\pmatrix{0&1&1\\-1&-2&-1\\0&0&0}\pmatrix {x\\y\\z}=\pmatrix {0\\0\\0}$$
This gives us two equations:
$$y=-z$$
$$-x-2y-z=0 \implies x=z$$
Therefore we have:
$$\implies v=(x,y,z)=x(1,-1,1)$$
You certainly made a sign mistake somewhere.


问:
Argument by symmetry that $\sum_{k=0}^{n-1} e^{2\pi i\frac{k}{n}}=0$
By summing up the geometric series, it can be shown that,

$$\sum_{k=0}^{n-1} e^{2\pi i\frac{k}{n}}=\frac{1- e^{2\pi i\frac{n}{n}}}{1- e^{2\pi i\frac{1}{n}}}=0.$$

However despite having proof of this fact, it feels somewhat magical that the (signed) heights and widths of the right triangles formed by equally partitioning the unit circle always sum to zero.

Consider the case with $n=3$, then we have

It's fairly easy to see in both this example, and in the general case, that the imaginary part always cancels by complex conjugate symmetry.

However even in the case of $n=3$, one does not have pairwise cancellation of the real part. Instead the right triangle widths corresponding to the two non-zero angles together are required to cancel out the third angle's corresponding width, i.e.

$$\cos\frac{2\pi}{3}+\cos\frac{4\pi}{3}=-1,$$

which to me is an entirely non-obvious fact.

So I'm wondering if there exists some argument by symmetry, from which one can deduce that $\sum_{k=0}^{n-1} e^{2\pi i\frac{k}{n}}=0$.

答:
A mechanical view (not an answer)

The equality

$$\sum_{k=0}^{n-1} e^{2\pi i\frac{k}{n}}=0$$ is asserting that the center of mass of a regular polygon is the center of the inscribed circle.

A way to "see it" is by deforming each edge of the polygon to the corresponding arc and to notice that the center of mass of a circle is located at its center.


答:
Let $S = \sum_{k=0}^{n-1} z^k$, where $z = e^{2\pi i/n}$.  You want an argument by symmetry that $S = 0$. Note that $S$ is not $0$ when $n = 1$, so any argument offered has to distinguish $n = 1$ from $n > 1$.

The number $S$ is a sum over all solutions in $\mathbf C$ to $x^n = 1$, and for $n > 1$ the sum $S$ is symmetric in the roles of all of those solutions.  Those solutions are symmetrically placed with respect to the origin and nowhere else, so wouldn't it be strange for $S$ to be located anywhere except $0$? This is not a serious proof, but an intuitive argument that it's reasonable to expect $S = 0$.

Here is a geometric argument by symmetry: the number $S/n$ is the average of the solutions of $x^n = 1$, the "center of mass" of those points.  The solutions of $x^n = 1$ divide up the unit circle into $n$ equal arcs, and for $n > 1$ it is visually evident that their average is the center of the circle, which is $0$.  Thus $S/n = 0$, so $S = 0$.

Here is an algebraic argument by symmetry: let $r$ be a solution of $r^n = 1$ with $r \not= 1$. Then $rS = \sum_{w^n = 1} rw = \sum_{w^n = 1} w = S$, where the second equation is due to multiplication by $r$ being a permutation of the solutions of $x^n = 1$.  Since $rS = S$ and $r \not= 1$, $S = 0$.  This algebraic argument is very important, because it is how a massive generalization of this result, called the orthogonality relation for characters, is proved in more advanced math.

Title: Argument by symmetry that $\sum_{k=0}^{n-1} e^{2\pi i\frac{k}{n}}=0$

Body:
By summing up the geometric series, it can be shown that,

$$\sum_{k=0}^{n-1} e^{2\pi i\frac{k}{n}}=\frac{1- e^{2\pi i\frac{n}{n}}}{1- e^{2\pi i\frac{1}{n}}}=0.$$

However despite having proof of this fact, it feels somewhat magical that the (signed) heights and widths of the right triangles formed by equally partitioning the unit circle always sum to zero.

Consider the case with $n=3$, then we have

It's fairly easy to see in both this example, and in the general case, that the imaginary part always cancels by complex conjugate symmetry.

However even in the case of $n=3$, one does not have pairwise cancellation of the real part. Instead the right triangle widths corresponding to the two non-zero angles together are required to cancel out the third angle's corresponding width, i.e.

$$\cos\frac{2\pi}{3}+\cos\frac{4\pi}{3}=-1,$$

which to me is an entirely non-obvious fact.

So I'm wondering if there exists some argument by symmetry, from which one can deduce that $\sum_{k=0}^{n-1} e^{2\pi i\frac{k}{n}}=0$.


问:
What is the actual geometric meaning of trigonometric operations such as adding cos,sine,tan
$$\sin(\pi/4)+\cos(\pi/4)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}= \frac{2\sqrt{2}}{2}=\sqrt{2}$$

Thinking of trig components (cosine, sine) that I used to produce the result using the mechanics of algebra, makes me wonder what is the geometric representation of
  $$\sin(π/4)+\cos(π/4)$$

The sine function corresponds to the shadow projection on $y$-axis (opposite) and the cosine function to the shadow projection on $x$-axis (adjacent).

At the previous operations I actually added those lines shadowed on the Cartesian axes. In other words, I added those sides of the triangle that form a $45-45-90$.

What is the actual geometric meaning of trigonometric operations such as adding cosine, sine, tangent, etc., or subtracting them? Am I just adding those sides and lines in order to get one new line with length $$\sqrt{2}$$ Is that all?

答:
Well, in the case of adding sines and cosines, you can think of it algebraically:

$$\sin\theta=\frac{y}{r};\cos\theta=\frac{x}{r}$$

Therefore, when you're taking $\sin\theta+\cos\theta$ you're really finding an expression for $\displaystyle\frac{x+y}{r}$. Similarly:

$$\cos\theta+\tan\theta=\frac{x}{r}+\frac{y}{x}=\frac{x^2+ry}{rx}$$
$$\sin\theta+\tan\theta=\frac{y}{r}+\frac{y}{x}=\frac{xy+ry}{rx}$$

In the case of a 45-45-90 triangle, represented by $\theta=\frac{\pi}{4}$, you have $x=y=\sqrt{2}$ and $r=2$, therefore:$$\sin\theta+\cos\theta=\frac{\sqrt{2}+\sqrt{2}}{2}=\sqrt{2}$$

$$\cos\theta+\tan\theta=\frac{2+2\sqrt{2}}{2\sqrt{2}}=\frac{2+\sqrt{2}}{2}$$

And the same for $\sin\theta+\tan\theta$.


问:
Problem of Hahn-Banach theorem
in $(l_\infty)'$ space there is a $f$ functional that
$$\forall x \in c_{00};f(x)=0$$
$$f((1,1,1,...))=1$$
Provide these properties. I find this functional and showed by using Hahn-Banach theorem please help

答:
You can use an approach similar to my answer here.
Since $c_0$ is a closed subspace and $x=(1,1,1,\ldots) \in \ell^\infty \setminus c_0$, we can (by using Hahn-Banach) find a linear functional $\lambda\in (\ell^\infty / c_0)'$ such that
\begin{align*}
\lambda(x+c_0) = \|x+c_0\|_{\ell^\infty/c_0} \neq 0.
\end{align*}
Now since $q\colon \ell^\infty \to \ell^\infty /c_0, y\mapsto y+c_0$ is continuous, we get $g := \lambda\circ q \in (\ell^\infty)'$ with
\begin{align*}
g(x) \neq 0 \qquad \text{and}\qquad g|_{c_0} = 0.
\end{align*}
We choose $f := \frac{1}{g(x)} g \in (\ell^\infty)'$ and thus get the desired functional, because $c_{00}\subset c_0$.

Title: Problem of Hahn-Banach theorem


问:
Does this sequence of operators converge uniformly?
Let $H$ be a Hilbert space and let $\{x_n\}_{n\in\mathbb N}$ and $\{y_n\}_{n\in\mathbb N}$ be orthonormal sets (not necessarily maximal). Consider the operators $T_n$ defined by

$$T_nu=(x_n|u)y_n.$$

Does the series $\sum_{n=1}^\infty T_n$ converge uniformly? I.e., does the sequence of partial sums $S_n=\sum_{i=1}^n T_n$ converge in the operator norm of $\mathfrak B(H)$?

I'm asking because the book I am reading (V. Moretti, Spectral Theory and Quantum Mechanics) casually mentions that if such a series converges uniformly, then the limit is a compact operator [Example 4.16 (2)]. But I am curious whether such a series must always converge uniformly anyway.

答:
Let $k, n$ be natural and $||u|| = 1$. We have
$$\left|\left|\sum_{m=n}^{k+n}(x_m, u)y_n\right|\right|^2 =  \sum_{m=n}^{k+n}|(x_m,u)|^2 \to 0,$$
when $n \to \infty$, because we can extend $\{x_n\}$ to a basis. But this sum has a maximal value $1$ (by the Bessel inequality) for any $n$ and $k$. Moreover, when $u \in \mathop{span}(x_n, \dots, x_{n+k})$ this value is achieved. In other words, $||S_{n+k} - S_{n}|| = 1$ for all $n, k \in \mathbb{N}$.


问:
Probability density of rational function of a normal distribution
Let $X\sim N(\mu,\sigma)$ be a normally distributed random variable. I would like to compute the probability density function of $Y=\frac{1+aX}{1-bX}$.

I know the quotient distribution is
$$
p(y)=\int_{-\infty}^{\infty} |z|p_{A,B}(yz,z) dz
$$
where $p_{A,B}(a,b)$ is the joint distribution of of $1+aX$ and $1-bX$.

However, I am not sure how to compute the latter. I assume $A$ and $B$ cannot be taken as independent !

答:
Thanks to A rural reader's comment, it was actually easier than I thought. The probability density $p(y)$ is obtained without using the quotient formula:

$$
p(y) = \frac{d}{dy}P(Y\leq y) = \frac{d}{dy}P\bigg(\frac{1+aX}{1-bX} \leq y\bigg) =  \frac{d}{dy}P\bigg(X \leq \frac{y-1}{a+by}\bigg) = \frac{d}{dy}F_{\mu,\sigma}\bigg(\frac{y-1}{a+by}\bigg) = \frac{a+b}{(a+by)^2}f_{\mu,\sigma}\bigg(\frac{y-1}{a+by}\bigg)
$$
if $1-bY\ge 0$.

Here $F_{\mu,\sigma}(z)$ is the Normal CDF and $f_{\mu,\sigma}(z)$ is the Normal PDF.


答:
We have
$$a\,\sin\theta+b\,\cos\theta=\sqrt{a^2+b^2}\,\sin(\theta+\alpha)$$
where $\alpha$ is the unique angle such that $\cos\alpha=a/\sqrt{a^2+b^2}$ and $\sin\alpha=b/\sqrt{a^2+b^2}$, in case $a^2+b^2 >0$.

Note that $\alpha$ is the angle of the vector $(a,b)$, measured from the positive half of $x$-axis, and $r:=\sqrt{a^2+b^2}$ is its length.

It means that, $a\,\sin\theta+b\,\cos\theta$ is the $y$ coordinate of the rotation of $(\cos\theta,\,\sin\theta)$ by $\alpha$, multiplied by $r$.

In your example $a=b=1$ so $r=\sqrt2$ and $\alpha=\pi/4$. Then the rotated and stretched vector will have angle $\pi/4+\pi/4=\pi/2$ and length $\sqrt2$. Its $y$ coordinate is indeed $\sqrt2$.

Title: What is the actual geometric meaning of trigonometric operations such as adding cos,sine,tan

Body:
$$\sin(\pi/4)+\cos(\pi/4)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}= \frac{2\sqrt{2}}{2}=\sqrt{2}$$

Thinking of trig components (cosine, sine) that I used to produce the result using the mechanics of algebra, makes me wonder what is the geometric representation of
  $$\sin(π/4)+\cos(π/4)$$

The sine function corresponds to the shadow projection on $y$-axis (opposite) and the cosine function to the shadow projection on $x$-axis (adjacent).

At the previous operations I actually added those lines shadowed on the Cartesian axes. In other words, I added those sides of the triangle that form a $45-45-90$.

What is the actual geometric meaning of trigonometric operations such as adding cosine, sine, tangent, etc., or subtracting them? Am I just adding those sides and lines in order to get one new line with length $$\sqrt{2}$$ Is that all?


答:
Yes, you can prove that by geomery. I  remember this method when I was a high school student.

Let's denote $A_1, A_2,...,A_n$ be the vertex of a regular n-polygon on the unit circle.

We have $S$ the sum of vectors, defined by $S = \sum_{i=1}^n \vec{OA_i}$ is equal to $\vec{0}$.

In deed, suppose by absurdity that the sum $S$ is equal to  $\vec{p} \ne  \vec{0}$. By rotating the regular n-polygon of $\frac{2\pi}{n}$, we obtain the same polygon. But the sum $S = \vec{p}$ is also rotated $\frac{2\pi}{n}$ and is equal to $\vec{p'}$. We have the vector $\vec{p'} = \vec{p}$ which has a different direction. There is only 1 possibility $\vec{p} = \vec{0}$.

Hence, by absurdity, we proved that $S = \sum_{i=1}^n \vec{OA_i} = \vec{0}$.

Now, just make a remark that, for all $\theta$
\begin{align}
S &= \sum_{k=1}^n \vec{OA_i} \\
&= \left(\sum_{k=1}^n \cos \left(\theta + \frac{2\pi k}{n} \right)\right)\vec{e_x} +\left(\sum_{k=1}^n \sin \left(\theta + \frac{2\pi k}{n} \right)\right)\vec{e_x} \\
\end{align}

In particular, for $\theta = 0$, we have
$$A_1 = \sum_{k=1}^n \cos \left(\frac{2\pi k}{n} \right) = 0$$
$$A_2 = \sum_{k=1}^n \sin \left(\frac{2\pi k}{n} \right) = 0$$

So, the sum $A_1 + i A_2 = 0$, we can conclude that
$$\sum_{k=0}^{n-1} e^{2\pi i\frac{k}{n}}=A_1 + i A_2 = 0$$


问:
What does the next $x$% mean in this case?
So I got this homework about statistics and the sentence is quite confusing.

"The result of the math exam in a school is normally distributed with an average of $74$ and a standard deviation of $7.9$. If the top $10\%$ got A in their report and the next $25\%$ got B, what is the lowest possible score of getting B?

So what confusing me is the next thing. Does it mean just $25\%$ of all so the lowest possible score of B is at the top $10\% + 25\% = 35\%$ (or $65\%$ from the bottom), or $25\%$ after A is excluded (or $75\%$ of $90\%$ = $67.5\%$ from the bottom)?

答:
If it resembles "percentiles" like 90th percentile is top 10 percent, then 80th being the next 10 percent we'd have 80th percentile from 80 percent to 90 percent and so on. Then for your example, if they were doing it that way, you'd be looking for the mark m such that exactly 35 percent got m percent  or better.

Note that if they were always going down some percent less than 100 percent of what was so far remaining, then the whole wouldn't get used up in a finite number of steps, which would seem odd (at least to me).


问:
Line along shortest distance between two skew lines
We have points $A(1,-1,1),B(6,-3,1),C(2,-1,5),D(5,1,1)$

We also have a line $p$ that goes through the edges $AD$ and $BC$ perpendicularly.

Find the equation of this line:
My try:

$$AD = (4,2,0)$$
$$BC = (-4,2,4)$$
$$s = (s_1,s_2,s_3)$$

Then we know that(*...scalar product):
$$s*AD = 0$$
$$s*BC = 0$$
From this two equations we get: $$4s_1+2s_2+0s_3 = 0 = -4s_1+2s_2+4s_3 $$
$$ 4s_1+2s_2+0s_3 = 0\\-4s_1+2s_2+4s_3 = 0\\
 \text{We add both equations together }\implies 4s_2 + 4 s_3 = 0 \implies 4s_2 = -4s_3\implies s_2 = -s_3$$
Then we get:
$$4s_1 = 4s_3\implies s_1 = s_3$$
However, when I plug this back into the equation I get zero at the end, which is not correct.

The solution is $p:x - 3 = \frac{y}{-2} = \frac{z-1}{2}$

答:
Another method

line $AD:(1,-1,1)+t (4,2,0)$ generic point on $AD$ is $P=(1+4t,-1+2t,1)$

line $BC:(6,-3,1)+u(-4,2,4)$ point on $BC$ is $Q=(6-4u,-3+2u,1+4u)$

$$PQ^2=f(t,u)=(-2 t+2 u-2)^2+(-4 t-4 u+5)^2+16 u^2$$
minimum distance is when
$$\frac{\partial f}{\partial t}=0;\;\frac{\partial f}{\partial u}=0$$
that is
$$
\begin{cases}
5 t+3 u=4\\
t+3 u=2\\
\end{cases}\to t=u=\frac12
$$
So the points having minimum distance on the two lines are
$$P_0=(3,0,1);\;Q_0=(4, -2, 3)$$
and the line $P_0Q_0$ has parametric equation
$$(x,y,z)=(3, 0, 1) + v (1, -2, 2)$$
or
$$
\begin{cases}
2 x=6-y\\
y=1-z\\
\end{cases}
$$


问:
Working in $S_6$ compute $(134) \cdot (12).$

Working in $S_6$ compute $(134) \cdot (12).$

I know that cycle multiplication is performed from left to right for example $(1324)\cdot(1243) = (142)(3).$ But in this case I'm confused I don't have $3$ or $4$ on the lhs nor $2$ on the rhs. How should I compute this?

答:
Observe that the first cycle is
$$\color{red}{1\to 3, \quad 3 \to 4, \quad 4\to 1}$$
and the second cycle is
$$\color{blue}{1\to 2, \quad 2\to 1}$$
When we compose the permutations, then we get
$$\color{red}{1\to 3, \quad 3 \to 4, \quad 4} \color{green}{\to}\color{blue}{ 2, \quad 2 \to 1}$$
where the green line remark the point when we change the permutation we are working on. Then, the product permutation is $(1 3 4 2)$.


问:
How to solve this optimization by using Lagrangian?
Let $f(x,y)=x(1+y).$

When is the value given by this function maximized subject to $x^2+y^2=1$ ?

How to do it with Lagrange multipliers?

答: The Lagrangian $L=x+xy+\lambda(1-x^2-y^2)$ has $L_x=1+y-2\lambda x,\,L_y=x-2\lambda y$, so$$L_x=L_y=0\iff(\lambda=x=0,\,y=-1)\lor(x=2\lambda/(4\lambda^2-1),\,y=1/(4\lambda^2-1)).$$The first solution gives $x^2+y^2=1,\,L=0$. The second gives$$1=x^2+y^2=(4\lambda^2+1)/(4\lambda^2-1)^2\stackrel{t:=4\lambda^2}{\implies}t(t-3)=0,\,L=\frac{8\lambda^3}{(4\lambda^2-1)^2}.$$The maximum $L$ occurs when$$t=3,\,\lambda=\sqrt{3}/2,\,x=\sqrt{3}/2,\,y=1/2,\,L=3\sqrt{3}/4.$$


问:
What does 16th (the identity matrix) generator of $SL(4,\mathbb{R})$ stand for?
While exploring the group $SL(4,\mathbb{R})$ I have analyzed all its 15 generators.
As an example, they are explicitly given in Elementary constituents of the group SL(4, R),
These generators can be obtained by combining the products of the Dirac matrices and generate the 15 subgroups of $SL(4,\mathbb{R})$.

However, it appears (p.5) that there is also the 16th generator which is the identity matrix
$$
G_{16}=I=\left(\begin{array}{cc}
\sigma_0&0\\
0&\sigma_0
\end{array}\right)
$$
Basically, all 16 generators are the same as the basis of the Clifford algebra $Cl(3,1)$ - which also includes $I$.
On the other hand, $SL(4,\mathbb{R})\cong SO(3,3)$, and the latter is the 15-dimensional group.
Does it imply that the 16th generator being the scalar multiply when added turns $SL(4,\mathbb{R})$ into $GL(4,\mathbb{R})$? Or is it just a dummy element?

What does this generator signify? As I understand, it can be included in the basis for Lie algebra as it is not traceless.

答:
If one considers $SL(4,\mathbb{R})$ as an abstract Lie group, and $\mathfrak{sl}(4,\mathbb{R})$ as an abstract Lie algebra, then there is no $16$th generator: both are $15$-dimensional and a set of generators is just a choice of basis of the Lie algebra.

That said, it is common to instead view $SL(4,\mathbb{R})$ as a subgroup of $GL(4,\mathbb{R})$, and correspondingly view $\mathfrak{sl}(4,\mathbb{R})$ as a subalgebra of $\mathfrak{gl}(4,\mathbb{R})$. This is being done implicitly when one write elements of either as $4\times 4$ matrices. Since $SL(4,\mathbb{R})$ has one less dimension than $GL(4,\mathbb{R})$, a superfluous extra "degree of freedom" will appear in various places.

In this case, that extra degree of freedom is a subgroup in its own right: if we view $\mathbb{R}^*$ as the subgroup of scalar matrices $\{cI:c\neq 0\}\subset GL(4,\mathbb{R})$, we have the decomposition $GL(4,\mathbb{R})\cong SL(4,\mathbb{R})\times\mathbb{R}^*$, and $G_{16}=I$ is one choice of generator of this $\mathbb{R}^*$ subgroup.


问:
Kinematics question, how to find the time when only given average speed, please help.
A particle moves in a straight line, starting from rest at point $P$. It accelerates for $5$ seconds, until it reaches a speed of $16\frac m s$. It maintains this speed for $T$ seconds and then decelerates at $2 \frac{m}{s^2}$ until it comes to a rest at point $Q$. Given that the average speed of the particle on the journey from $P$ to $Q$ is $12 \frac m s$, find the value of $T$.
How would I do this without the distance? I’ve found the time it took to come back to rest, but not very sure where to go from there and what to do with the given average speed.

答:
The distance travelled till the speed is $16$  can be obtained using the relation $v^2-u^2=2as$:

$$ 16^2 -0^2 =2 a s =2\frac{16-0}{5} s =16s \implies s=40$$ where I have further used $a=\frac{v-u}{t} $.

The distance travelled during the $T$ seconds is $16T$, and that while decelerating can be calculated by using $v^2-u^2 =2as$:

$$0^2-16^2 = 2(-2) s \implies s=64 $$ Now that you have the total distance and total time (and the average speed), you can set up a linear equation in $T$.


问:
How to ensure continuity of a piecewise function?
For a function with variable $t$ and constants $k,\delta$, how can I make it continuous?
$$  h=
    \begin{cases}
       t^2 & \quad\text{if}\quad |t|\le k \\
       ln\left(1+e^{\delta t}
 \right)^{\frac{2}{\delta}} - t&\quad\text{else}
     \end{cases}
$$

Clearly this function is discontinuous at the point $k$, since $k^2 \neq ln\left(1+e^{\delta k} \right)^{\frac{2}{\delta}}-k$

What could I do to make it continuous? I am thinking when $|t|\le k$, I could make the function $h=\beta t^2$, and find a $\beta$ which makes these equal? Is this the right way?

By solving $\beta$ in this way I get $$\beta = \left(ln\left(1+e^{\delta t}
 \right)^{\frac{2}{\delta}} - t\right) $$

But if I do this, the shape of the graph seems very different from a quadratic, it is a lot steeper and changes the quadratic function a lot.

答:
This can only be a partial answer, because your question is a bit vague. "What could I do to make it continuous?" Well, plenty of things, eg drop the second part of the definition, so that $h$ is simply the quadratic defined in the first part! But i assume you want a simple modification such as a constant (non-zero) factor.

A potential snag is the first part of the definition for $h$ holds for $|t|\le k$. But the second part depends on $t$, not $|t|$. That might be ok if second part, when simplified, turned out to be a function of $t^2$.

The factor $k/n$ does not depend on $t$, so we have $$\ln\big((1+e^{\delta t})^{2/\delta}\big)-t$$ We have $\ln(a^b)=b\ln a$, so we get: $$(2/\delta)\ln(1+e^{\delta t})-t$$ The power series for $\ln(1+x)$ and $\exp(x)$ are well-known, but a little effort is needed to get the series for $\ln(1+e^t)$, and anyway you are not familiar with the theory. But in practice one just uses a math software program and we find it is: $$\ln2+\frac{t}{2}+\frac{t^2}{2^3}-\frac{t^4}{3\cdot2^6}+\frac{t^6}{2^63^25}-\dots$$

And, to my surprise, it turns out that the 2nd part of the definition is indeed a function of $h^2$! An actual proof of that takes more work, of course.

Anyway, I got that wrong earlier: you can make the $h$ continuous at $t=\pm k$. I was careless and rushed too much. I am sorry!


问:
Another interesting integral $\int{\frac{x^m}{x^{2m}-1}dx}$
After An interesting integral $\int{\dfrac{x^m}{x^{2m}+1}dx}$, I want to generalize a similar integral below
$$\int{\frac{x^m}{x^{2m}-1}dx}$$
for all values of $m\in\mathbb{N}$. Below are my steps:

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

I don't know what my next step should be.

How should I solve this integral?

答:
Fractionalize  partially the integrand
$$\frac{x^n}{x^{2n}-1}=\sum_{k=1}^{2n}\frac{c_k}{x-x_k}$$
with $x_k= e^{i a_k},\>a_k=\frac{k\pi}{n}$ and apply the L’Hoptital’s rule to obtain the coefficients
$$c_k = \lim_{x\to x_k}\frac{x^n(x-x_k)}{x^{2n}-1}=\frac{(-1)^kx_k}{2n}$$

Then
\begin{align}
&\int \frac{x^n}{x^{2n}-1}dx =\frac1{2n}\sum_{k=1}^{2n}\int \frac{(-1)^kx_k}{x-x_k} dx
= \frac1{2n}\sum_{k=1}^{2n} (-1)^kx_k\ln(x-x_k)\\
= &\frac1{2n} \sum_{k=1}^{2n}(-1)^k (\cos a_k +i \sin a_k)
\ln(x-\cos a_k -i\sin a_k)\\
= &\frac1{2n} \sum_{k=1}^{2n} (-1)^k\left(\cos a_k \ln\sqrt{x^2-2x\cos a_k+1}+\sin a_k \tan^{-1}\frac{\sin a_k}{x-\cos a_k} \right)
\end{align}


问:
Venn Diagram symmetric difference

Draw the Venn Diagram for $C \backslash (A \oplus B)$

Attempt at solution

From what I understand the symmetric difference between $A$ and $B$ is the region of everything outside the shared region between the two spheres, which includes the $C$ region. But then $C \backslash$ that region means everything that's in $C$ except that region defined previously (which included $C$), so that would mean the big region of $C$ is not greyed out ? I guess I'm just confused by this.

$\hspace{3cm}$

Can anyone confirm my solution ?

答:
I think you're misinterpreting what the symmetric difference $A \oplus B$ is: it's all the points that are in exactly one of $A$ or $B$, but not both. In particular, this means that the "exclusively $C$" region is not included in $A \oplus B$, because elements exclusively in $C$ are in neither $A$ nor $B$.

Based on this, to fix your diagram you also need to shade the exclusively $C$ region.

Title: Venn Diagram symmetric difference


问:
Distinction between series and sum of infinite terms
Look, we all know that $$\sum_{k=1}^\infty a_k$$ represents an infinity series. Also it represents a sum.

So my question is:

How can one distinguish between the two ? Is there any difference between the ordinary addition of the terms with the value of the series ?

By ordinary addition, I mean the normal  addition what we done in number system, e.g., $2+3=5$ type.

My thoughts: After getting useful information from @JWP_HTX and also searching from various books, I think, the sum (which is used here in a very special sense) of a convergent series is the limit of the sequence of partial sums that is not obtained by ordinary addition. Again the sum gives only a number and it is not capable of being convergent or divergent.

Note: I had already made an edit of my question (on the same date of posting) and there is no other additional information with me. So please help and vote to reopen the question so that I can get the correct explanation of my doubts.

答: I think the answer to your question is yes - there is a difference.  When we consider conditionally convergence series, such as $\sum_{n=1}^{\infty}\frac{(-1)^n}{n}$, we know it converges as written to a finite sum $S$ by Leibnitz' test for convergence.  However, by Riemann's rearrangement theorem, we can actually rearrange the terms of this series to formulate any other finite sum.  So, in a sense, if we define 'normal addition' as adding the series term-by-term in order as written, then we get its sum.  But, the same series summed differently can give us different partial sums.


问:
I need to simplify $\sqrt[4]{x^{12}}$
I need to simplify

$\sqrt[4]{x^{12}}$

my answer is $|x|^3$ because the original radical is $|a| = \sqrt[n]{a^n}$ when n is even.

Am I wrong? Should the answer be $x^3$?

答:
Since $12$ is an even number, $x^{12}$ will always be positive, even if $x$ is negative. Then, taking the fourth root we get the positive value of $x^3$.

Therefore, you are correct that the answer is $|x|^3$. Also as @Martin R suggests, you can set $x=-1$ as a quick check.

Title: I need to simplify $\sqrt[4]{x^{12}}$


问:
Properties of a probability distribution function
Let X be a random variable and let $F_X$ be its probability distribution function.

For each statements state whether it is always true, sometimes true or never true.

a) $F_X$ is right continuous

b) {$F_X(x)$, x $\in$ $\mathbb{R}$ } fully determines the distribution function of the random variable $X$

c) $F_X$ is left continuous

d) $F_X(x)$ = $F_X(x^+)$ $\forall$$x$ $\in$ $\mathbb{R}$

Attempt:

b - always true
d- always true

How can I know if $F_X$ is always/sometimes right and/or left continuous?

答:
How can I know if $F_X$ is always/sometimes right and/or left continuous?

Deduction!

Since this is homework, I'll give you the necessary hints to solve this yourself:

Can it be "always left- and right-continuous"  at the same time? How do we call functions that are both left- and right-continuous? (The answer to this is easy.) Are CDFs always such functions?

Assuming it's only one of right- and left-continuos, try finding an easy example where at least one of these must be wrong. Test your hypothesis by finding the relevant point and apply the definitions of "left-continuous" and "right-continuous" at that specific point (that's easier than it sounds! Ask yourself how $F_X$ is defined in terms of probability!)


问:
Same linear transformation, different basis.

Let $\beta=\{(1,0,0),(0,1,0),(0,0,1\}$be a basis of $\mathbb{R^3}$ and $g: \mathbb{R^3} \to \mathbb {R^3}$ a linear transformation, which matrix is:

$$G=\begin{bmatrix}1 & 0 &-1 \\ 6 & 4 & -2\\ 0 & 1 & -1 \end{bmatrix}$$
Now let $\lambda=\{(0,0,1),(0,1,0),(1,0,0)\}$ be a basis also.

What is the matrix of $M(g;\lambda, \lambda)$?

I started to write each vector of $\beta$ into $\lambda$ coordinates.

$(1,0,0)_{\beta}=(0,0,1)_{\lambda}$

$(0,1,0)_{\beta}=(0,1,0)_{\lambda}$

$(0,0,1)_{\beta}=(1,0,0)_{\lambda}$

So, the change of basis's matrix, from $\beta$ to $\lambda$ is:

$$T=\begin{bmatrix}0 & 0 & 1 \\ 0 & 1 & 0\\ 1 & 0 & 0 \end{bmatrix}$$

This last matrix transforms any vector on $\beta$ coordinates into a vector on $\lambda$ coordinates. Then I wrote this scheme:

$$[v]_{\lambda} \to C \to [v]_{\beta} \to G \to [u]_{\beta} \to T \to [u]_{\lambda}$$

Where C is the change of basis's matrix from $\lambda$ to the canonic basis.

My difficult is how to merge theese matrix in one. Can You help me? thanks.

答:
What you need is nothing else but the matrix product $TGT^{-1}$.

Let's call the standard basis vectors $e_1,\ e_2,\ e_3$. Then we have $\beta=(e_1,e_2,e_3)$ and $\lambda=(e_3,e_2,e_1)$.

Now you have luckily $T^{-1}=T$, as $T$ simply exchanges the basis vectors $e_1$ and $e_3$.

But, for this specific exercise, we can find the answer more directly:

We need $G'=M(g;\lambda,\lambda)$. Because $G\cdot e_i$ gives the $i$th column of the matrix $G$, we have
$$g(e_1)=G\cdot e_1 = e_1+6e_2$$
This means that the third column of $G'$ will be $\pmatrix{0\\6\\1}$, as these are the $\lambda$-coordinates of the $g$-image of the third basis vector of $\lambda$.

Similarly, you can see what happens to the rest of the matrix.


问:
Understanding the geometric realization of a graph
I'm working through the beginnings of the graph theory necessary for pasting diagrams in $2$-categories using Johnson and Yau's 2-Dimensional Categories, and I'm not quite sure what the relation they're using to quotient by for their geometric realization is. For reference:

Definition. A graph $G$ is a triplet $(V_G,E_G,\psi_G)$ where $V_G$ is a finite set called the set of vertices, $E_G$ is a finite set called the set of edges, and $\psi_G:E_G\to V_G\times V_G$ is a function called the incidence function. The geometric realization of $G$, denoted $|G|$, is the topological quotient $$|G|=\Big[\big(\coprod_{v\in V_G}\{v\}\big)\coprod\big(\coprod_{c\in E_G}[0,1]_e\big)\Big]\big/\sim$$ where $\{v\}$ is a one point space indexed by a vertex $v\in V_G$, $[0,1]_e$ is a copy of the unit interval indexed by an edge $e\in E_G$, and the identification $\sim$ is generated by $$u\sim0\in[0,1]_e\ni1\sim v\ \ \ \ \text{if}\ \ \ \ \psi_G(e)=(u,v).$$

What is the equivalence relation $\sim$?

Unless I'm mistaken $\coprod_{v\in V_G}\{v\}$ is homeomorphic to $V_G$ with the discrete topology and $$\coprod_{e\in E_G}[0,1]_e=\{(e,x):x\in[0,1]_e\}$$ with open sets of the form $$U=\coprod_{e\in I}(a_e,b_e)$$ (endpoints $[0$ and $1]$ allowed) for some subset $I\subseteq E_G$, so the coproduct of these sets should be $$X=\big(\coprod_{v\in V_G}\{v\}\big)\coprod\big(\coprod_{c\in E_G}[0,1]_e\big)=\{(n,y):(n=0\wedge y\in\coprod_{v\in V_G}\{v\})\vee(n=1\wedge y\in\coprod_{c\in E_G}[0,1]_e)\}$$ with opens of the form $$V\coprod U$$ with $V\subseteq V_G$ and $U$ as above, so I would expect the relation $\sim$ to be expressible as a subset of $X\times X$ but I can't see exactly how to get there from their definition. Any assistance is appreciated.

答:
The answer was just to think about it some more. The relation $\sim$ is, unless I'm mistaken, the smallest equivalence relation generated by the relation

$$\sim'=\{(u,(e,0)):\pi_0(\psi_G(e))=u\}\cup\{(v,(e,1)):\pi_1(\psi_G(e))=v\}.$$

By taking the coproduct above we are essentially creating a copy of all the vertices $V_G$ and a copy of the unit interval for each edge $e\in E_G$, and the quotient by $\sim$ glues the tails and heads of each edge $e$ (viewed as the first and second coordinate of $\psi_G(e)$ respectively) to the beginnings and ends of the unit intervals indexed by $e$, respectively.

Further, since $\psi_G$ is a function each edge has only two unique vertices as its head/tail so no vertices are related to the same interval endpoints, thus the relational closure $\sim$ of $\sim'$ should only add the diagonal $\Delta X\subseteq X\times X$ and mirror images of existing pairs to $\sim'$. That is, if we let $\langle\pi_1,\pi_0\rangle:X\times X\to X\times X$ be the isomorphism interchanging coordinates, an explicit expression for $\sim$ is $$\sim\ =\ \sim'\ \cup\ \Delta X\ \cup\ \langle\pi_1,\pi_0\rangle(\sim').$$

EDIT: I missed a term for the transitive closure, which is necessary for vertices $u$ which are the tail of one or more edges $\{e_i\}_{0<i\leq m}$ and the head of one or more edges $\{e'_I\}_{0<i\leq m}$ simultaneously. We will then have $$u\sim'(e_0,0)\wedge\dots\wedge u\sim'(e_n)\wedge u\sim'(e'_0,1)\wedge\dots\wedge u\sim'(e'_m,1)$$ so the corresponding beginnings and ends of the unit intervals for these edges will need to be glued together. This adds three more terms to the relational closure corresponding to tails glued together ($TT$), heads glued to tails ($HT$), and heads glued together ($HH$). Explicitly, to the above definition of $\sim$ we need to union in

$$TT=\{((e,0),(e',0)):\pi_0(\psi_G(e))=\pi_0(\psi_G(e'))\}$$

$$HT=\{((e,1),(e',0)):\pi_1(\psi_G(e))=\pi_0(\psi_G(e'))\}$$

$$HH=\{((e,1),(e',1)):\pi_1(\psi_G(e))=\pi_1(\psi_G(e'))\}$$

and $\langle\pi_1,\pi_0\rangle(HT)=TH$ as well since this subset is not necessarily symmetric. This finally yields $$\sim\ =\ \sim'\ \cup\ \Delta X\ \cup\ \langle\pi_1,\pi_0\rangle(\sim')\ \cup HH\ \cup\ HT\ \cup\ TH\cup\ TT.$$


问:
Why does this pattern work: $1 \cdot{1} = 1, 11 \cdot{11} = 121, 111 \cdot{111} = 12321\ldots$
I have recently learned about pattern that goes like this:

$$\begin{align}
1^2 &= 1\\
11^2 &= 121\\
111^2 &= 12,321\\
1,111^2 &= 1,234,321\\
11,111^2 &= 123,454,321.
\end{align}$$

It is a very cool pattern, but after a bit it stops:

$$1,111,111,111^2 = 1,234,567,900,987,654,321$$

My main question:

Why does this pattern work?

A side question that's less important:

Is there an algebraic equation to describe this pattern?

答:
Other answers have already provided some intuition behind the pattern. Here is some terminology:

The numbers being squared, e.g. $111111$, are called repunits.

The resulting products are called Demlo numbers: $\,1, 121, 12321, 1234321,\, \dots$

Repunits can be written as sums of powers of $10$, and from that perspective, base-$10$ repunits are finite geometric series. For example:
\begin{align}
111111 &= 1 + 10 + 100 + 1000 + 10000 + 100000\\
111111&=1 + 10 + 10^2 + 10^3 + 10^4 + 10^5\\\\
111111&=\displaystyle\frac{10^6-1}{10-1}\\\\
111111&=\displaystyle\frac{10^6-1}{9}
\end{align}

The last two lines above use the formula for the sum of a finite geometric series.

Squaring gives:

\begin{align}
111111^2 &= \left(\displaystyle\frac{10^6-1}{9}\right)^2\\
111111^2 &= \displaystyle\frac{10^{12} - 2\cdot 10^6 +1}{81}
\end{align}

In general:

$$\left(\,\underbrace{\,1111\,\dots\,1111\,}_{ n \text{ ones} } \,\right)^2 \, = \,\, \boxed{\displaystyle\frac{10^{2n} - 2\cdot 10^n +1}{81}\,}$$

Demlo numbers are sequence A002477 in the OEIS.


答:
Recall that, in $S_6$, $(134) \cdot (12)=(134)(12)(5)(6)$. Thus we have

$$\begin{align}
1&\xrightarrow{(134)}3\xrightarrow{(12)}3\xrightarrow{(5)}3\xrightarrow{(6)}3,\\
3&\xrightarrow{(134)}4\xrightarrow{(12)}4\xrightarrow{(5)}4\xrightarrow{(6)}4,\\
4&\xrightarrow{(134)}1\xrightarrow{(12)}2\xrightarrow{(5)}2\xrightarrow{(6)}2,\\
2&\xrightarrow{(134)}2\xrightarrow{(12)}1\xrightarrow{(5)}1\xrightarrow{(6)}1,\\
5&\xrightarrow{(134)}5\xrightarrow{(12)}5\xrightarrow{(5)}5\xrightarrow{(6)}5,\\
6&\xrightarrow{(134)}6\xrightarrow{(12)}6\xrightarrow{(5)}6\xrightarrow{(6)}6,
\end{align}$$

so the answer is $(1342)(5)(6)=(1342)$.

Body:
Working in $S_6$ compute $(134) \cdot (12).$

I know that cycle multiplication is performed from left to right for example $(1324)\cdot(1243) = (142)(3).$ But in this case I'm confused I don't have $3$ or $4$ on the lhs nor $2$ on the rhs. How should I compute this?


答:
This is immediately revealed by combining the Cauchy product
$$ \log(1-x)\operatorname{Li}_2(x)=3\sum_{n=1}^{\infty} \frac{x^n}{n^3}-2\sum_{n=1}^{\infty} x^n\frac{H_n}{n^2}- \sum_{n=1}^{\infty} x^n \frac{H_n^{(2)}}{n} ,\ |x|\le1 \land x\neq1$$
and well-known harmonic series in the mathematical literature.
You might find useful results in (Almost) Impossible Integrals, Sums, and Series and in the paper Euler sum involving tail by Moti Levy and C.I. Valean, here.

More precisely, we need slightly adjusted forms (or we let them the way they
are and rearrange the series result obtained after integration) of
$$\sum_{k=1}^{\infty} \frac{H_k}{(k+1)(k+n+1)}=\frac{H_n^2+H_n^{(2)}}{2n}$$
and
$$\sum_{k=1}^{\infty} \frac{H_k^{(2)}}{(k+1)(k+n+1)}= \zeta(2)\frac{H_n}{n}-\frac{1}{n}\sum_{i=1}^n\frac{H_i}{i^2},$$
as they appear in the generalization from Sect. 4.16, page 289 in (Almost) Impossible Integrals, Sums, and Series.

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

Body:
Consider a following definite integral

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

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

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

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


问:
$\int_{0}^{\pi/2} \text{arctanh}(\sin x) \text{arctan}(a \tan(x)) \cos(x) \ dx$
Are you kind to let me know the way? By the way, don't you have a "curiosity" tag?
$$\int_{0}^{\pi/2} \text{arctanh}(\sin x) \text{arctan}(a \tan(x)) \cos(x) \ dx, \quad a>0$$</p>

答:
$$\forall a>0,\quad\int_{0}^{\pi/2}\text{artanh}(\sin x)\,\arctan(a\,\tan(x))\cos(x)\ dx=\\\frac\pi2\Re\left(\ln\left(2\,a^{-1}+2\right)-\frac{\ln\left(2\sqrt{a^{-4}-a^{-2}}+2\,a^{-2}-1\right)}{2 \sqrt{1-a^2}}\right),$$
where $\Re$ denotes the real part.</p>


问:
Argument by symmetry that $\sum_{k=0}^{n-1} e^{2\pi i\frac{k}{n}}=0$
By summing up the geometric series, it can be shown that,

$$\sum_{k=0}^{n-1} e^{2\pi i\frac{k}{n}}=\frac{1- e^{2\pi i\frac{n}{n}}}{1- e^{2\pi i\frac{1}{n}}}=0.$$

However despite having proof of this fact, it feels somewhat magical that the (signed) heights and widths of the right triangles formed by equally partitioning the unit circle always sum to zero.

Consider the case with $n=3$, then we have

It's fairly easy to see in both this example, and in the general case, that the imaginary part always cancels by complex conjugate symmetry.

However even in the case of $n=3$, one does not have pairwise cancellation of the real part. Instead the right triangle widths corresponding to the two non-zero angles together are required to cancel out the third angle's corresponding width, i.e.

$$\cos\frac{2\pi}{3}+\cos\frac{4\pi}{3}=-1,$$

which to me is an entirely non-obvious fact.

So I'm wondering if there exists some argument by symmetry, from which one can deduce that $\sum_{k=0}^{n-1} e^{2\pi i\frac{k}{n}}=0$.

答:
Let
$$ S=\sum_{k=0}^{n-1}e^{\frac{2k\pi}{n}i}. $$
By the invariant of $S$ of rotation through $e^{\frac{2\pi}{n}i}$, one has
$$ e^{\frac{2\pi}{n}i}S=S $$
which implies $S=0$.


问:
Reflection group action on a reflection hyperplane
Let $G$ be a finite reflection group acting on a euclidean vector space $V$ of dimension $n$ and let $H$ be one of the reflection hyperplanes. Then $H$ is stabilized by some (reflection) subgroup of $G$. Which subgroup of $G$ acts on the $n-1$ dimensional subspace $H$? Intuitively, this should be the group $G'$ which is generated by all the roots which are orthogonal to the root corresponding to $H$. But which groups arise in such a way?

答: Any finite reflection group can arise in the way you describe. Take your desired $G'$ acting in a vector space $V'$, and let $V=V'\oplus \mathbb{R}$, and take $G=\langle G',w\rangle$ where $w$ acts as the identity on $V'$ and negation on $\mathbb{R}$.


问:
How to find expectancy of the sum of several random variables
Suppose that $X_{1}, X_{2}, X_{3}... X_{n}$ are random variables independent from each other, all with an unknown distribution, with a standard deviation of 1 and mean of 0.

Let $Y= [\frac{X_{1}+ X_{2}+ ... +X_{n}}{n}]$

Find $E(Y)$ and $\sigma(Y)$

I tried using linearity of expectation  but got nowhere, what can I do to solve this problem?

答:
Using the linearity of the expectation, we get that
$$\mathbb{E}[Y] = \mathbb{E}\left[\frac{X_1 + X_2 + \ldots + X_n}{n}\right] = \frac{1}{n}\mathbb{E}\left[X_1 + X_2 + \ldots + X_n\right] = \frac{1}{n}\sum_{i=1}^n \mathbb{E}\left[X_i\right]$$
Since all the expectations $\mathbb{E}\left[X_i\right]$ are equal by hypothesis (identically distributed), the previous sum is equal to $n\mathbb{E}\left[X_1\right]$ (I wrote the first random variable $X_1$ but it can be the one you choose, since they are identically distributed). Plugging this result in the previous expression, we get
$$\mathbb{E}[Y] = \frac{n}{n} \mathbb{E}\left[X_1\right] = \mathbb{E}\left[X_1\right]$$
Since the mean of all $X_i$ is $0$, then $\mathbb{E}[Y] = 0$.

Can you proceed with the standard deviation using that
$$\mathbb{V}\text{ar}[aX] = a^2\mathbb{V}\text{ar}[X]$$
$$\mathbb{V}\text{ar}\left[X_1 + X_2 + \ldots + X_n\right] = \sum_{i=1}^n\mathbb{V}\text{ar}[X_i] + \sum_{i\neq j}\mathbb{C}\text{ov}[X_i, X_j] $$
and remembering that the random variables $X_1, X_2, \ldots, X_n$ are independent?


问:
Proving that a function has primitives
Let f a differentiable function such that $\lim_{x\to\infty}\frac{f(x)}{x}$=$\lim_{x\to-\infty}\frac{f(x)}{x}=0$ .Prove that the function $g:\Bbb R\to\Bbb R$ admits primitives when g is:
$g(x)=f'(1/x)$, for any $x\neq 0$
and
$g(0)=0$
I tried to integrate the function g and to make the substitution $x:=1/x$ but it did not helped me out.

答:
Let

$$H(x) = \begin{cases}x^2f(1/x),& x \neq 0\\0,& x= 0 \end{cases}$$

Since $\lim_{x \to \pm\infty} \frac{f(x)}{x} = 0$, it follows that $\lim_{x \to 0} xf(1/x) = 0$ and both $H$ and $x \mapsto H(x)/x$ are continuous on $\mathbb{R}$.

Thus, a  primitive of $g$ is

$$G(x) = -H(x) + 2\int_0^x \frac{H(t)}{t} \, dt,$$

Note that for $x \neq 0$, we have

$$G'(x) = -H'(x) + 2\frac{H(x)}{x} = -2xf(1/x) +f'(1/x) + 2xf(1/x) = f'(1/x) = g(x) $$

Try to finish by showing that $G'(0) = g(0) = 0$.

Title: Proving that a function has primitives


问:
Matrix formulas
Let $\mathbb{K}$ be a commutative field, $n ∈ \mathbb{N}^{*}$ fixed. For $A = (a_{i,j})_{1≤i,j≤n} ∈ M_{n}(k)$, we set

$$\tilde{P}_{A}(X) = \det(A + XI_{n} ).$$
where $I_{n}$ is the identity matrix of $M_{n}(k)$. we write

$$\tilde{P}_{A}(X) = P_{0}(A)X^n + P_{1}(A)X^{n-1} + ... + P_{n-1}(A)X + P_{n}(A)$$

and we give that $P_{0}(A) = 1,\;P_{1}(A) = \sum_{i=1}^n a_{ii} = T_{r}(A)$ and $P_{n}(A) = \det(A)$

in the following, we set $\mathbb{K}'= \mathbb{Q}(\mathbb{K}[X])$ the field of the fractions of $\mathbb{K}[X].$ For $R ∈ \mathbb{K}^{'}$ and

$P(X) = \sum_{i=0}^n a_{i}X^{i} ∈ \mathbb{K}[X]$. we naturally denote by $P(R) = \sum_{i=0}^n a_{i}R^{i}$

a)  show that if $A$ is invertible $B = A^{-1}$, then we have

$$\tilde{P}_{B}(X) = \frac{1}{\det(A)}X^{n}\tilde{P}_{A}(\frac{1}{X})$$

b)deduce that when $A$ is invertible we have:

$$tr(A^{-1})\cdot \det(A) = P_{n-1}(A)$$

please can someone help with those two question i thought that if i take in consideration that $tr$ is linear that would help me, but i can't go further, can anyone help me please.

答:
For a) note that:
\begin{align}
\tilde{P}_B(X)&=\det(B+XI_n)\\
&=\det(A^{-1})\det(I_n+XA)\\
&=\det(A)^{-1}\det\left(X\left(\frac{1}{X}I_n+A\right)\right)\\
&=\det(A)^{-1}X^n\det\left(A+\frac{1}{X}I_n\right)\\
&=\frac{1}{\det(A)}X^n\tilde{P}_A\left(\frac{1}{X}\right)
\end{align}
Now b) follows from the formula in a) by comparing the coefficient of $X^{n-1}$ on both sides.

Title: Matrix formulas


问:
Express $169$ as the sum of $1,2,3,4,5$ non-zero squares
I'm trying to solve the following exercise.

Show that $169$ can be expressed as a sum of $1,2,3,4,5$ non-zero squares, and deduce that any $n \ge 169$ is the sum of five non-zero squares.

The latter part of the exercise is clear from Lagrange's four squares theorem, namely that every integer can be expressed as the sum of four integers. (There's also an answer Integers which are the sum of non-zero squares, detailing the exact steps.) I'm having trouble with the first part of the exercise.

For expressing $169$ as a sum of one square, we have $169=13^2$. Now, $13 = 2^2 + 3^2$, so we can use the identity $(a^2+b^2)(c^2+d^2) = (ac+bd)^2 + (ad-bc)^2$ to get $13^2 = (6+6)^2 + (2^2-3^2)^2$. But what about the rest? Is there any way of obtaining $169$ as a sum of $3,4,5$ non-zero squares, except by brute force, perhaps? (Brute force answers are also welcome, especially if they employ intuition or clever tricks! I will upvote them too)

There is one numerical answer to this in the link above, but I'm looking for intuition or a systematic way of doing things, not just a number solution expressing $169$ as a sum of non-zero squares.

答:
I wondered how long this would take me, so I sat down to do it and here was my process.

I thought I would have to do some quick comparisons, so I first wrote the squares from $1$ to $169$ at the top of a sheet of paper. I had expected to repeatedly try a greedy algorithm of taking one large square and seeing what one can do with the remaining small number, but in fact this was only necessary once.

$13^2$ gives the expression as one square.

This amounts to choosing two squares and adding them up. In principle I could have looked at my list of squares and begun to compare, but I think it's pretty common to recognize the Pythagorean triple $5^2 + 12^2 = 13^2$.

For $3$, I think it's easiest to modify the previous triple by re-expressing $5$ in its own Pythagorean triple to get $(3^2 + 4^2) + 12^2 = 13^2$.

Although I'm listing these in order, I actually did $4$ squares last. I didn't do anything clever.

Instead I proceeded greedily: begin with $169$, subtract a large square and see if what remains was obviously a sum of $3$ squares. $25$ is not a sum of three squares, so the initial try of representing $169 - 12^2 = 5^2$ as a sum of three squares doesn't work. But the second guess leads to $169 - 11^2 = 48 = 4^4 + 4^4 + 4^2$ does work. Thus $4^2 + 4^2 + 4^2 + 11^2$ works for $4$.

I know the theorem that every number can be written as a sum of $4$ squares, so you can't really go too wrong here. I thought to examine again the basic Pythagorean triple $5^2 + 12^2 = 13^2$ and to rewrite $5^2$ as a sum of $4$ squares. This can be done with $1^2 + 2^2 + 2^2 + 4^2$ (and all these numbers are small enough to quickly get this with no real work), giving the set $1^2 + 2^2 + 2^2 + 4^2 + 12^2$.


问:
Does this sequence of terms converge or diverge as n approaches infinity?
Does the ratio $$\lim_{n\to\infty}\dfrac{p_{n}}{n}$$
converge to a defined limit?
Here $p_{n}$ denotes the nth prime number.
Just asking out of curiosity.

答:
Hint.

The prime number theorem says that
$
p_n\sim n\log(n)
$; precisely,
$$
\lim_{n\to\infty}\frac{p_n}{n\log (n)}=1
$$
So $\frac{p_n}{n}$ grows like $\log(n)$.


问:
Find the area between two polar curves
I need to find the area between two polar curves, $$r = \frac1{\sqrt{2}}$$ $$r = \sqrt{\cos(θ)}$$

I've found the intersections to be at $\fracπ3$ and $\frac{5π}3$, and I've set up the equation to find the area as

$$\int\limits_{\fracπ3}^{\frac{5π}3} \sqrt{\cos(θ)}^2 - \frac1{\sqrt{2}}^2 \, \mathrm dθ,$$

but whenever I plug it into a calculator, it comes up as undefined, so it can't possibly be correct. Could you help me with this?

答: If you plot both functions, and restrict the domain of r = $\sqrt{cos \theta}$ you see that from $\frac{\pi}{3}$ to $\frac{\pi}{2}$ you get half of the inner function.  So integrate $1/2 - \cos \theta$ between those limits and double your answer to get the full area. I get $\frac{\pi}{6} - 2 + \sqrt{3}$.

Title: Find the area between two polar curves


问:
Shannon entropy of languages
In his paper Prediction and Entropy of Printed English Shannon defines the entropy $H$ of a language as

$$H = \lim_{N \to \infty} F_N$$

where $$F_N = \sum_{i, j} p(b_i, j) \log p(j | b_i)$$

where $b_i$ is a sequence and $j$ is a word.

The interpretation is: $F_N$ computes the surprise of the average continuation $j$ of a sequence $b_i$ (of length $N-1$) weighted by the probability of such a sequence $(b_i, j)$ occurring in the first place.

Shannon further states that

$$F_N = \sum_{i, j} p(b_i, j) \log p(j | b_i) =   \sum_{i, j} p(b_i, j) \log p(b_i, j) + \sum_i p(b_i) \log p(b_i)$$

We arrive at that equivalence as follows:

\begin{align}
\sum_{i, j} p(b_i, j) \log p(j | b_i) & = - \sum_{i, j} p(b_i, j) \log \frac{p(b_i, j)}{p(b_i)}\\
                                      & = - \sum_{i, j} p(b_i, j) ( \log p(b_i, j) - \log p(b_i) ) \\
                                      & = - \sum_{i, j} p(b_i, j) \log p(b_i, j) - p(b_i, j) \log p(b_i) \\
                                      & = - \left(\sum_{i, j} p(b_i, j) \log p(b_i, j) - \sum_{i, j}  p(b_i, j) \log p(b_i)\right) \\
                                      & \text{marginal distribution in second sum}\\
                                      & = - \sum_{i, j} p(b_i, j) \log p(b_i, j) + \sum_{i}  p(b_i) \log p(b_i) \\
\end{align}

答:
My question was about the interpretation of the formula. I answered it myself while writing the question:

The interpretation is: $F_N$ computes the surprise of the average continuation $j$ of a sequence $b_i$ (of length $N-1$) weighted by the probability of such a sequence $(b_i, j)$ occurring in the first place.

But since references to this paper are sparse, and I only found a very badly scanned version that is hard to read, maybe this helps somebody.

Title: Shannon entropy of languages


答: Try running it as two separate integrals, each with their own integrand and dtheta. Evaluate each integrand and then subtract them from each other. However, make sure you have the problem written down exactly the same - the error is not with the formatting but that this equation only has one imaginary answer (~ -2.4616 + 2.3962i). If your problem does not ask for an imaginary anwer for answer for this specific equation, its not defined.</p>

Title: Find the area between two polar curves

Body:
I need to find the area between two polar curves, $$r = \frac1{\sqrt{2}}$$ $$r = \sqrt{\cos(θ)}$$

I've found the intersections to be at $\fracπ3$ and $\frac{5π}3$, and I've set up the equation to find the area as

$$\int\limits_{\fracπ3}^{\frac{5π}3} \sqrt{\cos(θ)}^2 - \frac1{\sqrt{2}}^2 \, \mathrm dθ,$$

but whenever I plug it into a calculator, it comes up as undefined, so it can't possibly be correct. Could you help me with this?


问:
about affine variety
let $ f(x,y,z) \in \mathbb{R}[x, y, z] $ such that :

$$ f(x,y,z) =x^{4}+x^{2} y^{2}+x^{2} z^{2}-x^{3} y-x^{3} z-x^{2} z y+\\  x^{2} y^{4}+y^{6}+y^{4} z^{2}-x y^{5}-x z y^{4}-z y^{5}+\\  x^{2} z^{6}+y^{2} z^{6}+z^{8}-x y z^{6}-x z^{7}-z^{7} y  $$

i want to show that that $Z(f(x,y,z))$ is affine variety in $\mathbb{A^3}_{\mathbb{R}}$ .

we know
$Z(f)=\left\{P \in \mathbb{A^3}_{\mathbb{R}} \mid f(P)=0\right. \}$

now $Z(f)$ is close and it's enough that we show $Z(f)$ is irreducible . on the other hand we know an algebraic set is irreducible if and only if its ideal is a prime ideal.so we have $I(Z(f))=\sqrt{f}$ so we must show that $\sqrt{f}$ is prime . also if we can show $f$ is irreducible then $Z(f)$ is irreducible.

答:
Not a complete solution but I hope it helps.

Generally, when such a task is given to you (like homework), there is good hope there is some significant way how to make it much simpler.

In the first row you can factor out $x^2$ in second $y^4$ and in the third, you can factor out $z^6$. Observe that you have the same remainder in each row.

So we can write: $f(x,y,z)=(x^2+y^4+z^6)(x^2+y^2+z^2-yx-zx-zy).$

So $Z(f)=Z(x^2+y^4+z^6)\cup Z(x^2+y^2+z^2-yx-zx-zy)$.

But $Z(x^2+y^4+z^6)=(0,0,0)$ and that is also an element of $Z(x^2+y^2+z^2-yx-zx-zy)$ so you can simply it all to:

$Z(f)=Z(x^2+y^2+z^2-yx-zx-zy)$

2)Now one can guess that $x=y=z$ are zeroes. So the set contains affine line. If we can prove that no other points our in $Z(f)$ then points from this line, we have a solution, because line is irreducible over infinite field.

Title: about affine variety


问:
Prove or disprove: $A\in \mathbb{C}^{3\times 3}$ and $A^3+4A=0$ then $A$ is not invertible.
Prove or disprove: $A\in \mathbb{C}^{3\times 3}$ and $A^3+4A=0$ then $A$ is not invertible.

My work:

If $A^3+4A=0$ then $A^3=-4A$, taking determinant on both sides we get $|A^3|=|-4A| \Rightarrow |A|^3=-64|A| \Rightarrow |A|^2=-64 \Rightarrow |A|=\sqrt {-64}\ne0$, so $A$ is invertible since $|A|\ne0.$

Would this be a valid proof that $A$ is indeed invertible and the statement is false? or do I need to come up with a counter example? (I'm struggling to do that with the requirments).

Any feedback is appreciated.

答:
Write the given equation as $A(A^2+4)=0$. Then either $A^2+4I=0$ or there exists some vector $u$ so that $v=(A^2+4I)u$ is non-zero and thus a non-trivial kernel vector of $A$.

In the first case, $A$ is invertible, with inverse $-\frac14A$. In the second case a non-trivial kernel exists, so not invertible. With the given assumptions, both cases are possible.


答:
Here's a counterexample:
$$A = 2iI_3.$$

Body:
Prove or disprove: $A\in \mathbb{C}^{3\times 3}$ and $A^3+4A=0$ then $A$ is not invertible.

My work:

If $A^3+4A=0$ then $A^3=-4A$, taking determinant on both sides we get $|A^3|=|-4A| \Rightarrow |A|^3=-64|A| \Rightarrow |A|^2=-64 \Rightarrow |A|=\sqrt {-64}\ne0$, so $A$ is invertible since $|A|\ne0.$

Would this be a valid proof that $A$ is indeed invertible and the statement is false? or do I need to come up with a counter example? (I'm struggling to do that with the requirments).

Any feedback is appreciated.

Body:
Here's a counterexample:
$$A = 2iI_3.$$


答: There is an easier way to do this: Take $A$ to be of the form $A=\lambda I$, where $\lambda$ is nonzero and satisfies $$\lambda^3 +4\lambda =0.$$

Title: Prove or disprove: $A\in \mathbb{C}^{3\times 3}$ and $A^3+4A=0$ then $A$ is not invertible.


答:
The area show in green

is given by the following integral, where $r(\theta)=\sqrt{\cos\theta}$
$$
\int_{\pi/3}^{\pi/2}d\theta\int_0^{r(\theta)}rdr=\frac{1}{2}\int_{\pi/3}^{\pi/2}\cos\theta d\theta=\frac{1}{2}\left(1-\frac{\sqrt{3}}{2}\right)
$$

The rest should be pretty easy.

Title: Find the area between two polar curves


问:
Cauchy-Riemann Equation for Surface Equations
Suppose we are given smooth functions $W,V:\mathbb{R}^2\times\mathbb{C}\to\mathbb{C}^2$ and are asked to determine if the function $Q(W(x,y,z(x,y)),V(x,y,z(x,y))$ solving
$$
Q(W(x,y,z(x,y)),V(x,y,z(x,y)))=0,
$$
is analytic. Following Chapter 10 of Lectures on Random Lozenge  Tilings, Gorin states that the sufficient condition for $Q$ to be analytic is to establish $W(V)$ is analytic or in other words
$$\frac{\partial W}{\partial\bar{V}}=0.
$$
Of course this form of the Cauchy-Riemann equations
$$\frac{\partial f}{\partial \bar{z}}=0
$$
is recognisable when establishing $f(z):\mathbb{C}\to\mathbb{C}$ is analytic, but I don't understand how $\frac{\partial W}{\partial\bar{V}}=0$ implies $Q$ is analytic. What's the basic reason for this? Also, would $\frac{\partial V}{\partial\bar{W}}=0$ do the job, too?

For context, see the latter portion of Chapter 9 from the linked reference which derives the surface equation $Q=0$ from the method of complex characteristics and where there is a concrete example of the functions $W,V$.

答:
I think establishing function $f(z): \mathbb{C} \rightarrow \mathbb{C}$,  mentioned by you, is needed for our consideration to answer your question.

We have, that partial derivatives $u_x$,$u_y$,$v_x$,$v_y$ are continuous(by smoothness in your case), and we have, generally, that your function Q pass circles in circles,particularly, circle with center in $z_0 \in D$ into circle with center in $\overline{z_0} \in D$(we can consider your curves W and V from Gorin's book like bounded Jordan curves, and, within it we can split it on infinite amount of circles, from smoothness and Jordan lemmas).

$\omega - \omega_0 = f_{z_0}(z-z_0) + f_\overline{z_0}(\overline{z}-\overline{z_0})$ - main linear part of that mini-functions, where

$ f_{z_0} = \frac{\partial f}{\partial z}_{z=z_0}$ and
$f_{\overline{z_0}} = \frac{\partial f}{\partial \overline{z}}_{z=z_0}$

and, because of that, $f(z)$ is analytic or anti-analytic in $D$.

What's that got to do with your question? So, if you write down square of module of $\omega - \omega_0$, you will get

$
 \left\lvert  \omega  - \omega_0   \right\rvert^{2} = (\left\lvert f_{z_0}\right\rvert^{2} + \left\lvert f_{\overline{z_0}}\right\rvert^{2})\left\lvert z-z_0   \right\rvert^{2}+2Re [f_{z_0}\overline{f_{\overline{z_0}}}(z-z_0)^{2}]$

thus, circle $\left\lvert z-z_0 \right\rvert = r_1$ goes to $\left\lvert \omega-\omega_0 \right\rvert = r_2$, in last equation we will have
$f_{z_0}\overline{f_{\overline{z_0}}} = 0$, wherein $f_{z_0} = 0$,$f_{\overline{z_0}} \ne 0$ or $f_{z_0} \ne 0$,$f_{\overline{z_0}} = 0$.


问:
Working in $S_6$ compute $(134) \cdot (12).$

Working in $S_6$ compute $(134) \cdot (12).$

I know that cycle multiplication is performed from left to right for example $(1324)\cdot(1243) = (142)(3).$ But in this case I'm confused I don't have $3$ or $4$ on the lhs nor $2$ on the rhs. How should I compute this?

答:
The notation for permutations and cycles is confusing similar, so when you write them out, try to keep the distinction clear:

$$P(123456)C(134)=P(324156)$$

$$P(324156)C(12)=P(314256)$$

To write out in shorthand notation, we get $(3421)(5)(6)=(1342)$.


答:
$A^3-4A = 0$ tells us that A has eigenvalues in $\{2i, -2i, 0\}$

If $0$ is an eigenvalue then $A$ is not invertible.  If $A$ were a real $3\times 3$ matrix, we must choose a real eigenvalue, zero is our only choice, and $A$ is singular.  However, if $A$ has complex entries, there is no reason that 0 must be an eigenvalue of our matrix.

$A = 2iI$ would be a counter-example.


问:
What this parameter in the Riemann surface metric has to do with its complex structure?
As far as I'm aware, a complex manifold $M$ is a topological manifold together with an atlas ${\cal A}$ of charts $(U_i,\varphi_i)\in{\cal A}$ such that the open sets $U_i$ cover $M$, the maps $\varphi_i: U_i\to \mathbb{C}^n$ are homeomorphisms and whenever $U_i\cap U_j\neq\emptyset$ the transition maps $\varphi_i\circ\varphi_j^{-1}:\mathbb{C}^n\to \mathbb{C}^n$ and $\varphi_j\circ\varphi_i^{-1}:\mathbb{C}^n\to \mathbb{C}^n$ are holomorphic. In that scenario, the atlas ${\cal A}$ is called a complex structure.

Now I'm studying string theory and in the book I'm using (Becker, Becker and Schwarz) what the authors call a complex structure seems like something different. Quoting their book, pages 89 and 90:

Here $\int Dh$ means the sum over all Riemann surfaces $(M,h)$. However, this is a gauge theory, since $S$ is invariant under diffeomorphisms and Weyl transformations. So one should really sum over Riemann surfaces modulo diffeomorphisms and Weyl transformations.
Worldsheet diffeomorphism symmetry allows one to choose a conformally flat worldsheet metric $$h_{\alpha\beta}=e^\psi \delta_{\alpha\beta}.\tag{3.109}$$
When this is done, one must add the Faddeev-Popov ghost fields $b(z)$ and $c(z)$ to the worldsheet theory to represent the relevant Jacobian factors in the path integral. Then the local Weyl symmetry $(h_{\alpha\beta}\to \Lambda h_{\alpha\beta}$) allows one to fix $\psi$ (locally) - say to zero. However, this is not possible globally, due to a topological obstruction: $$\psi=0\Longrightarrow R(h)=0\Longrightarrow \chi(M)=0.\tag{3.110}$$
So, such a choice is only possible for worldsheets that admit a flat metric. Among orientable Riemann surfaces without boundary, the only such case is $n_{\rm h}=1$ (the torus). For ach genus $n_{\rm h}$ there are particular $\psi$s compatible with ${\chi}(M)=2-2n_{\rm h}$ that are allowed. A specific choice of such a $\psi$ corresponds to choosing a complex structure for $M$.

Now I'm puzzled with this last statement. This $\psi$ has to do with the metric, while the complex structure has to do with how we construct an atlas of holomorphic charts and therefore with how holomorphic functions are defined. I can't see how these two things are related.

What I imagined was this: we want to consider all possible Riemann surfaces $(M,h)$. Given any such surface the diffeomorphism symmetry allows the metric to be put in the form (3.109) and therefore be classified by $\psi$. Still, given a certain topology, not all metrics can be defined in the surface with that topology. Therefore we partition the space of such surfaces by genus $g$ and for each genus $g$ we have a class of possible $\psi$.

I'm conjecturing that the same happens with complex structure. In the same way that given a topology not all metrics can be defined in the surface with that topology, given a complex structure not all metrics are compatible with it.

Still I don't know if this is correct, and I think a more complete understanding is necessary here. So why does a specific choice of such a $\psi$ corresponds to choosing a complex structure for $M$?

答:
As I said in the comment, what they wrote is simply nonsense. Here is a couple of references where at least you will not find any obvious nonsense:

As a mathematician, I like the first 6 chapters in

Ji, Lizhen; Looijenga, Eduard J. N., Introduction to moduli spaces of Riemann surfaces and tropical curves, Surveys of Modern Mathematics 14. Somerville, MA: International Press; Beijing: Higher Education Press (ISBN 978-1-57146-353-1/pbk). iv, 221 p. (2017). ZBL1375.14002.

The tropical staff is (probably) irrelevant for you and is discussed in Part II of the book.

Also take a look at this, which is aimed to discuss integration over moduli spaces and is probably more relevant for your purposes:

Nag, Subhashis, Mathematics in and out of string theory, Kojima, Sadayoshi (ed.) et al., Topology and Teichmüller spaces. Proceedings of the 37th Taniguchi symposium, Katinkulta, Finland, July, 24–28, 1995. Singapore: World Scientific (ISBN 981-02-2686-1/hbk). 187-220 (1996). ZBL1050.32501.

A free version is available here, but there might be some differences with the published version.


问:
If $y<x$ is there a way to phrase it in terms of $y>$something?
For instance I know $y<x$ implies $-y > -x$ but is there a way to phrase it in terms of $y>$ something (that does not itself contain $y$)?</p>

答:
No.

For each $x\in\Bbb R$ we have that the set $L_x:=\{y\,\mid\,y<x\}$ is downward closed meaning that $u<v\in L_x \,\implies\,u\in L_x$.

The other kind of sets, $G_x:=\{y\,\mid\,y>x\}$ don't have this property, because there exist $u,v\in\Bbb R$ such that $v>x>u$.


问:
Differentiating the maximum?
Consider a function $f(x,y) : \mathbb [0,1]^2 \to \mathbb R$, smooth in both variables (very strong assumption for the purposes). Define $f_{X}(x) = \max_{y \in [0,1]} f(x,y)$ and $f_Y{(y)} = \max_{x \in [0,1]} f(x,y)$ , which are well-defined. Oftentimes, in optimization, it is a useful statement to say that $f_X$ and $f_Y$ have unique maxima/minima. For this purpose, it is convenient to see if they are differentiable, and set the derivative to zero.

Under what sufficient conditions are $f_X$ and $f_Y$ differentiable (they are always continuous, as pointed out below)? Is the condition of "f smooth" acceptable?

How can I study the derivative of $f_X$ and $f_Y$, so that I can look at the maxima/minima of such functions?

EDIT : Since we have come up with a counterexample to non-differentiability, given by $(x,y) \to 1-xy$ on $[-1,1]^2$, I would like sufficient conditions under which this is true. One may use the language of sub-differentials etc. (from convex analysis) if required for this purpose.

答: Sufficient conditions for this to hold are for $f$ to be of the form $f(x,y) = g(x) + h(y)$ with $g$ and $h$ convex functions.

Title: Differentiating the maximum?


问:
About limit point in a topological space.
Let $(\mathbb{N},T)$ be a topological space, where $T$ is the topology generated by $\{\{2n-1,2n\}: n \in \mathbb{N}\}$.

Now, the statement : every infinite subset in $(\mathbb{N},T)$ has a limit point.

Is this statement true or false?

My try, the subset $\{2n:n\in\mathbb{N}\}$ has no limit point, as for every $2n$ , open neighborhood $\{2n-1,2n\}$ intersects nothing other than $2n$.

So, the statement is false.

答:
If $A$ is non-empty let $x \in A$. If $x$ is even, $x-1$ is a limit point of $A$, and if $x$ is odd, $x+1$ is a limit point of $A$.

That $A$ is infinite is irrelevant.


问:
We have a board $15\times15$. Every square we paint in red, blue or green.
We have a board $15\times15$ divided into 225 unit squares. Every square we paint in red, blue or green. Prove that there always exist two rows with the same number of squares of the same color.

答:
HINT

Represent a colouring by as a list of 15 triples: $(R,G,B)$ where $R$ is the number of reds in the row etc. You know that $R=0$ can only appear once in the list. Similarly $R=1$ etc. Now what can you say about the total of each component (ie the sum of the 15 $R$s etc)?


问:
The quotient group of the Heisenberg group and its center is isomorphic to $(\mathbb{R}^2, +)$.

Let $$ H = \left\{ \;\begin{bmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 &
 1 \end{bmatrix} : x,y,z \in \mathbb{R}\right\}, \qquad Z(H) = \left\{
 \begin{bmatrix} 1 & 0 & y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} :
y \in \mathbb{R}\right\}, $$ then show that $H/Z(H) \cong (\mathbb{R}^2, +)$.

Proof: First I show that $Z(H) \cong (\mathbb{R}, +)$ using the isomorphic map $\phi(h) = y$ for $h \in Z(H)_y$. Then I show that $H \cong (\mathbb{R}^3, +)$ using the isomorphic map $\Phi(h) = (x, y, z)$ for $h \in H_{(x,y,z)}$. So we have
$$
H/Z(H) \cong (\mathbb{R}^3, +)/(\mathbb{R}, +).
$$
We can construct a homomorphic map $\varphi\left((x,y,z)\right) = x + y + z$ with $$\text{ker} \varphi = \{ x,y \in \mathbb{R}: z = -x -y\}$$ which is clearly $(\mathbb{R}^2, +)$ so by the first isomorphism theorem we have
$$
H/Z(H) \cong (\mathbb{R}^3, +)/(\mathbb{R}, +) \cong (\mathbb{R}^2, +)
$$
finishing the proof. Does this sort of iterative isomorphic map proof work? These concepts are new to me so I'm unsure.

答:
Your map $\Phi$ is not an isomorphism. For example let $h_1 \in H$ have coordinates $(x_1,y_1,z_1)=(1,0,0)$, and let $h_2$ have coordinates $(x_2,y_2,z_2) = (0,0,1)$. A short computation shows that $h_3 = h_1 h_2 \in H$ has $z_3=1$ whereas $h_4 = h_2 h_1 \in H$ has $z_4 = 0$. Therefore $h_1 h_2 \ne h_2 h_1$, whereas $\Phi(h_1) + \Phi(h_2) = \Phi(h_2)  + \Phi(h_1)$.

What you should do instead is to construct a map $\Phi : H \to \mathbb R^2$ (not $\mathbb R^3$) which you can prove is a homomorphism, is surjective, and has kernel $Z(H)$.


问:
Need help on the Proof of Lemma 7.3.11 in Statistical Inference Casella&Berger (v2)
The Lemma states that if $f(x|\theta)$ satisfies

$\frac{d}{d\theta}\mathbb{E}_\theta\left(\frac{\partial}{\partial\theta} \log(X|\theta)\right) = \int \frac{\partial}{\partial\theta} \left[\left(\frac{\partial}{\partial\theta}\log f(x|\theta)\right) f(x|\theta)\right] dx$

then

$\mathbb{E}_\theta \left( \left(\frac{\partial}{\partial\theta} \log f(X|\theta)\right)^2\right)  = -\mathbb{E}_\theta \left( \frac{\partial^2}{\partial \theta^2}\log f(X|\theta)\right). $

The proof of the Lemma is left as an Exercise (7.39). what I get is

$\mathbb{E}_\theta \left( \frac{\partial^2}{\partial \theta^2}\log f(X|\theta)\right) = \int \left( \frac{\partial^2}{\partial \theta^2}\log f(X|\theta)\right) \log(x|\theta) dx
$

$=\int \frac{\partial}{\partial\theta} \left[\left(\frac{\partial}{\partial\theta}\log f(x|\theta)\right) f(x|\theta)\right] dx - \int\left(\frac{\partial}{\partial\theta}\log f(x|\theta)\right)\frac{\partial}{\partial\theta} f(x|\theta) dx   $

Note that

$\mathbb{E}_\theta \left( \left(\frac{\partial}{\partial\theta} \log f(X|\theta)\right)^2\right) =\int\left(\frac{\partial}{\partial\theta}\log f(x|\theta)\right)\frac{\partial}{\partial\theta} f(x|\theta) dx,$

So basically I need that $\int \frac{\partial}{\partial\theta} \left[\left(\frac{\partial}{\partial\theta}\log f(x|\theta)\right) f(x|\theta)\right] dx=0$,
for which I am unable to derive from the assumption of the Lemma.

Am I missing anything?

Thanks.

答:
It's been a long time since you asked, but it's still useful to have a proper answer I think.

If we do the tedious algebra carefully with the hypothesis we get
\begin{equation}
\begin{aligned}
\frac{d}{d\theta}\int_{-\infty}^{\infty}\frac{d}{d\theta}f_{\theta}(x)dx&=\frac{d}{d\theta}E_{\theta}\left(\frac{d}{d\theta}\log f_{\theta}(X)\right)\\&=\int_{-\infty}^{\infty}\frac{d}{d\theta}\left[\left(\frac{d}{d\theta}\log f_{\theta}(x)\right)f_{\theta}(x)\right]dx\\&=\int_{-\infty}^{\infty}\frac{d^2}{d\theta^2}f_{\theta}(x)dx\text{, }
\end{aligned}
\end{equation}
then, as you said, we have that
\begin{equation}
\begin{aligned}
E_{\theta}\left(\frac{d^2}{d\theta^2}\log f_{\theta}(X)\right)&=\int_{-\infty}^{\infty}\frac{d^2}{d\theta^2}f_{\theta}(x)dx-E_{\theta}\left(\frac{d}{d\theta}\log f_{\theta}(X)\right)^2\\&=\frac{d}{d\theta}\int_{-\infty}^{\infty}\frac{d}{d\theta}f_{\theta}(x)dx-E_{\theta}\left(\frac{d}{d\theta}\log f_{\theta}(X)\right)^2\text{, }
\end{aligned}
\end{equation}
also, we can apply Leibniz's theorem of differentiation under the integral sign, as long as $f_{\theta}(x)$ is integrable for all $\theta$, derivable with respect to $\theta$ for all $x$, and $\frac{d}{d\theta}f_{\theta}$ in integrable for all $\theta$, so
\begin{equation}
0=\frac{d}{d\theta}1=\frac{d}{d\theta}\int_{-\infty}^{\infty}f_{\theta}(x)=\int_{-\infty}^{\infty}\frac{d}{d\theta}f_{\theta}(x)\text{, }
\end{equation}
then, substituting this last result, we have the lemma proved.

You can also check the book solutions for an alternative.


问:
Show the following function is differentiable
$$f(x) = \int_{-\sqrt{x}}^{\sqrt{x}} e^{-\sqrt{1 + t^2}} dt$$

Explain why this function is differentiable at each $x \in [0,\infty)$ and compute $f'(x)$.

I know that $f'(x) = e^{-\sqrt{1 + \sqrt{x}^2}} - e^{-\sqrt{1 + (-\sqrt{x})}^2} = e^{-\sqrt{1 + x}} - e^{-\sqrt{1 + x}} = 0$. However, I am having a hard time explaining why it is differentiable at each $x$.

I tried to apply the limit definition $f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$. However, it becomes pretty nasty since it has integrals within the limit. I'm not convinced this is the way to show it just because it doesn't really get me anywhere.

My next idea seemed golden to me.

I can state the Fundamenetal Theorem of Calculus as follows:

Let $f: [a,b] \to \mathbb{R}$ be integrable and define
$$F(x) = \int_{a}^{x} g(t) dt$$
$\forall x \in [a,b]$. Then, $F$ is continuous on $[a,b]$. If $f$ is continuous at some point $c \in [a,b]$, then $F$ is differeitable at $c$ and $F'(c) = f(c)$.

Our function $f$ has no discontinuities, so its set of discontinuities is of measure $0$. So it is integrable. $f$ is continuous $\forall c$ in the domain because it is an exponential, which is always continuous. So, this would mean $F$ is differentiable everywhere.

BUT, our function is $f: [0,\infty) \to \mathbb{R}$. not an interval of the form $[a,b]$. Is this option out of the picture as well then? It seems so close!

Any help is appreciated!

答: If we set the relation such that: $-\sqrt{x}\lt a \lt \sqrt{x}$, we can rewrite the integral function in this form: $$f(x)=\int_{-\sqrt{x}}^{a}e^{-\sqrt{1+t^2}}dt \space +\space \int_{a}^{\sqrt{x}}e^{-\sqrt{1+t^2}}dt$$ Then we can take the derivative of it by applying the chain rule: $$f'(x)=0-\Biggl(\frac{e^{-\sqrt{1+x}}}{-2\sqrt{x}}\Biggl)\space + \space \frac{e^{-\sqrt{1+x}}}{2\sqrt{x}}-0=\frac{e^{-\sqrt{1+x}}}{\sqrt{x}}$$ The function is differentiable on the open interval $(0,\infty).$


问:
Algebra Equation with equals on left hand side
$6 = 4 - 2x =$

Show the answer with the mechanics of working out.

答:
Equality: "$=$" is a symmetric relation, as are all equivalence relations: $$a = b \quad \overbrace{\iff}^{\text{equivalent}}\quad b = a$$ That is, it doesn't matter if we replace $\;$ "$a$ is equal to $b$" $\;$ with $\;$ "$b$ is equal to $a$:" $\;$ both expressions tell us precisely the same thing.

So, the following two equations (left hand side & right hand side of the arrow) are equivalent (meaning, really, that they give the same identical information:
$$6 = 4 - 2x \iff 4 - 2x = 6$$

Now, I am assuming that you need to solve for $x$. Let's just go ahead and use the form of the equation you were given:

$(1)$ add $+ 2x$ to each side of the equation.$$\begin{align} 6 & = 4 - 2x \\ \\ 2x + 6 & = 4 - 2x + 2x\end{align}$$ Simplify.

$(2)$ Subtract $6$ from each side of the equation. Simplify.

$(3)$ Divide each side of the equation by $2$. (Put differently, you can multiply each side of the equation by $\frac 12).$


问:
Tangent integration $\int \tan (2x)\tan (3x)\mathrm{d}x$
I know how to calculate $\displaystyle\int \tan (x) \tan (2x)\tan (3x)\mathrm{d}x$ with the tangent sum formula
$\tan 3x\tan 2x\tan x =\tan3x -\tan 2x -\tan x$. But I don't know how I can solve

$$\displaystyle\int \tan (2x)\tan (3x)\mathrm{d}x$$

I tried u sub, trig sub, integration by parts. But I couldn't make it.

Thanks.

答:
Proceed as follows

\begin{align}
\int \tan 2x\tan 3x {d}x
&=\int \frac{\sin 2x\sin 3x }{\cos 2x\cos 3x }dx\\
 &=\int \frac{(2\sin x\cos x)[\sin x(4\cos^2x-1)] }{(2\cos^2x-1)[\cos x(4\cos^2x-3) ]}dx\\
 &=2\int \frac{(1-\cos^2x)(4\cos^2x-1)}{(2\cos^2x-1)(4\cos^2x-3)}dx\\
 &=\int \left( -1 -\frac1{2\cos^2x-1} + \frac2{4\cos^2x-3}\right)dx\\
 &=-x -\int \frac{d(\tan x)}{1-\tan^2x} +2\int \frac{d(\tan x)}{1-3\tan^2x}\\
 &=-x -\tanh^{-1}(\tan x) +\frac2{\sqrt3}\tanh^{-1}(\sqrt3\tan x)
\end{align}


问:
Prove: $\phi$ is an isomorphism of vector spaces if and only if $\det(M_{C,B}(\phi))\neq 0$
Let $V, W$ be finite-dimensional $F$-vector spaces of dimension $n> 0$ and let $\phi: V \to W$ be a linear mapping

Prove:
For every ordered basis $B$ of $V$ and every ordered basis $C$ of $W$: $ϕ$ is an isomorphism of vector spaces if and only if
$$\det(M_{C,B}(\phi))\neq 0.$$

I understand this question but I can't get anywhere with the proof. I would appreciate any help!

答:
So $\phi$ is linear mapping, in other words, it is a homomorphism. So what you need to do is just prove that it is bijection.

If $\phi$ is isomorphism then $\phi$ is bijection, so that transformation matrix is regular. So its determinant is non-zero.

Now for the other implication, we know that matrix is regular, hence our mapping is bijection.


问:
A net clusters at $x$ iff the "filter generated by the net" clusters at $x$
Let $\langle x_\lambda \rangle_{\lambda\in\Lambda}$ be a net. Let $\mathscr{F}_{x_\lambda}$ be the filter generated by the net $\langle x_\lambda \rangle$, i.e. the generated by the filter base
$\mathscr{C}_{x_\lambda} = \{\{x_{\lambda }:\lambda\geq\mu\}\}_{\mu\in\Lambda}$.

What I would like to show is,

$\langle x_\lambda \rangle_{\lambda\in\Lambda}$ clusters at $x$ iff $\mathscr{F}_{x_\lambda}$ clusters at $x$.

Proof:

($\Rightarrow$)Note,

$$\left(\forall U\in\mathscr{U}_x,\exists C\in\mathscr{C}_{x_\lambda}: C\subseteq U\right)\iff\left(\mathscr{C}_{x_\lambda}\to x\right)\iff\left(\mathscr{F}_{x_\lambda}\to x\right) \implies \left(\mathscr{F}_{x_\lambda}\text{ clusters at } x\right).$$

Therefore, it is enough to show that $\forall U\in\mathscr{U}_x,\exists C\in\mathscr{C}_{x_\lambda}: C\subseteq U$.

So, let $U\in\mathscr{U}_{x}$. By definition of $\langle x_\lambda \rangle_{\lambda\in\Lambda}$ clustering at $x$ and $U$ and open nbhd of $x$,

$$\forall\mu\in\Lambda, \exists \lambda\in\Lambda: \lambda\geq\mu \text{ and } x_{\lambda}\in U.$$

This implies that $C:=\{x_{\lambda}:\lambda\geq\mu\}\subseteq U$. As $U$ was arbitrary and $C\in\mathscr{C}_{x_\lambda}: C\subseteq U$, this direction is done.

($\Leftarrow$) Suppose $\mathscr{F}_{x_{\lambda}}$ clusters at $x$. Then for $U\in\mathscr{U}_x$ and $F\in\mathscr{F}_{x_{\lambda}}$, we have $U\cap F\neq\emptyset$. By def. of $F\in\mathscr{F}_{x_{\lambda}}$, there exists $C\in\mathscr{C}_{x_{\lambda}}: C\subseteq F$ hence $U\cap C\neq\emptyset$. By def. of $\mathscr{C}_{x_{\lambda}}$, this means that there exists $\{x_{\lambda}:\lambda\geq\mu\}$

$$U\cap\{x_{\lambda}:\lambda\geq\mu\}\neq\emptyset.$$

However, my question is, how does this imply "for each $\mu$, there is a $\lambda$ such that $\lambda\geq\mu$ and $x_{\lambda}\in U$"? More specifically, is appears to me that this implies an existence of $\mu$ rather than for each $\mu$.

答:
Slowly unwrap the (correct!) definitions:

Suppose $(x_\lambda)_{\lambda \in \Lambda}$ cluster at $x$. Then $\mathcal{F}$ (I leave off the subscript) clusters at $x$ too, which means that every neighbourhood of $x$ intersects every set from the filterbase.

So let $O$ be an open neighbourhood of $x$, and let $C_\mu:=\{x_\lambda: \lambda \ge \mu\}$ be a set from the filterbase for some $\mu \in \Lambda$.

Now apply the net-cluster definiton at $x$ to $U$ and $\mu$, to get $\mu' \ge \mu$ such that $x_{\mu'} \in U$ and then $x_{\mu'} \in U \cap C_\mu$ as required. So we're done.

The reverse is very similar.


问:
Function generation by input $y$ and $x$ values
I wonder if there are such tools, that can output function formulas that match input conditions. Lets say I will make input like that:

$y=0, x=0$

$y=1, x=1$

$y=2, x=4$

and tool should generate for me function formula y=x^2. I am aware its is not possible to find out exact function, but it would be great to get some possibilities. I'm game developer and i need to code some behaviours in game mechanics, that aren't simply linear, sometimes I need for example arcus tangens, when i want my value to increase slower and slower for higher arguments. Problem is that I finished my school long time ago and simply I don't remember how does many functions looks like and such tool would be great to quickly find out what i need.

答:
Simple "ramp" or "easing" functions that describe position as a function of time are common in games and other types of animation. Often, you want the speed of the object to vary, which means that the function graph won't be a straight line.

A couple of simple cases are described in this answer and in this answer.

You can usually get what you need just by using a simple low-degree polynomial (quadratic or cubic). If you tell us what kind of behaviour/properties you want, it should be easy to construct the function. It's best if you can provide a graph of the desired function.


问:
$e^{5 \frac{2 \pi i}{6}}+e^{\frac{2 \pi i}{6}}=1$
Why is $e^{5 \frac{2 \pi i}{6}}+e^{\frac{2 \pi i}{6}}=1$?

I know $e^{5 \frac{2 \pi i}{6}} = e^{-\frac{\pi i}{3}}=\overline{e^{\frac{\pi i}{3}}}=\overline{e^{\frac{2 \pi i}{6}}}$, but I still don't see why the equation is true.

答:
Hint.

Note that for any complex number $z$,
$
z+\overline{z}=2\textrm{Re}(z)
$.
Also $\textrm{Re}(e^{i\theta})=\cos(\theta)$ for $\theta\in\mathbb{R}$.

Now observe that $\cos(2\pi/6)=\frac12$.


问:
How to express by a single equation a region delimited by confocal ellipse and hyperbola
I am trying to find the equation of a shape formed by a conic (ellipse) that excludes its intersection with another conic (hyperbola). The hyperbola and ellipse are assumed to share common foci. But they are general conics  in 2D which means they can be rotated about the z-axis and translated along x and y axes. As it can be seen in the figure, the conics are rotated by an angle theta and not centered at the origin.

In the figure, I want the equation of resulting shape shown in grey shade (formed by the two conics). This new shape can be shown as follows:

Please let me know if it is possible to get an equality representing this new shape in the quadratic form as follows:

$ Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$

答:
Let $a$ and $b$ respectively be half the lengths of the major and minor axes of the ellipse.
Let $A$ and $B$ respectively be half the lengths of the transverse and conjugate axes of the hyperbola.
Let $f$ be half the distance between the foci, which are the same points for both the ellipse and the hyperbola. Then
\begin{align}
a^2 &= f^2 + b^2, \\
A^2 &= f^2 - B^2,
\end{align}
which implies that $a > f$ and $B < f.$
(All distance measurements are non-negative.)

The equation of the ellipse is
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} - 1 = 0  \tag1 $$
and the equation of the hyperbola is
$$ \frac{x^2}{A^2} - \frac{y^2}{B^2} - 1 = 0.  \tag2 $$

The desired curve satisfies Equation $(1)$ when
$\frac{x^2}{A^2} - \frac{y^2}{B^2} < 1$
and satisfies Equation $(2)$ when $\frac{x^2}{a^2} + \frac{y^2}{b^2} < 1.$

Inspired by the clever solution in
another answer,
we may try to unify the two equations by first setting the $x^2$ terms equal.
For the ellipse we get
$$ x^2 + \frac{a^2}{b^2} y^2 - a^2 = 0 \tag3$$
and for the hyperbola
$$ x^2 - \frac{A^2}{B^2} y^2 - A^2 = 0.  \tag4$$
The "average" of these equations (taking half the sum of the left-hand sides of $(3)$ and $(4)$) is
$$ x^2 + \frac12\left(\frac{a^2}{b^2} - \frac{A^2}{B^2}\right) y^2
    - \frac12 (a^2 + A^2) = 0 \tag5$$
and "half the difference" of these equations (taking half the difference of the left-hand sides of $(3)$ and $(4)$) is
$$ \frac12\left(\frac{a^2}{b^2} + \frac{A^2}{B^2}\right) y^2
    - \frac12 (a^2 - A^2) = 0. \tag6$$

Equation $(3)$ is then the "sum" of Equations $(5)$ and $(6)$ and Equation $(4)$ is the "difference", subtracting $(6)$ from $(5).$

But if we add the absolute value of the left side of $(6)$ to the left side of $(5)$, we get
\begin{multline}
\qquad
x^2 + \frac12\left(\frac{a^2}{b^2} - \frac{A^2}{B^2}\right) y^2
    - \frac12 (a^2 + A^2) \\
   + \left\lvert \frac12\left(\frac{a^2}{b^2} + \frac{A^2}{B^2}\right) y^2
                  - \frac12 (a^2 - A^2) \right\rvert = 0 \qquad \tag7
\end{multline}

Observe that when
$\left(\frac{a^2}{b^2} + \frac{A^2}{B^2}\right) y^2 \geq (a^2 - A^2)$,
Equation $(7)$ matches Equation $(5)$ and therefore follows the path of the ellipse, but when
$\left(\frac{a^2}{b^2} + \frac{A^2}{B^2}\right) y^2 \leq (a^2 - A^2)$,
Equation $(7)$ matches Equation $(6)$ and therefore follows the path of the hyperbola.
When $\left(\frac{a^2}{b^2} + \frac{A^2}{B^2}\right) y^2 = (a^2 - A^2)$,
Equation $(7)$ matches both $(5)$ and $(6)$ simultaneously and it determines
the intersections of the ellipse and hyperbola.
That is, this is exactly the equation we want when the two foci are on the $x$ axis and are equidistant from the origin.

Note that since $a$ is a function of $f$ and $b$ and since $A$ is a function of $f$ and $B$, we can express the two equations using just three variables
(and should do so, in order to ensure the same foci):
$f$, $a$, and $A$; $f$, $b$, and $B$; or some other combination of three of the five variables. As long as we choose three mutually independent variables we can set them to whichever values we like (under the constraints $a > f$ and $B < f$) and derive the other two.
For example, we could choose the length and width of the ellipse ($2a$ and $2b$) and the distance $2A$ between the vertices of the hyperbola,
under the constraint that $A^2 < a^2 - b^2.$

If we choose $f,$ $b,$ and $B$ such that $b^2 = 1 + \sqrt{f^2 + 1}$ and
$B^2 = b^2 - 2,$ then Equation $(7)$ becomes
$$
x^2 + \left(\sqrt{f^2 +1}\right) \left\lvert y^2 - 1\right\rvert - (f^2 + 1) = 0,
$$
the same equation found in
the answer that inspired this one.

For a curve at any other location and orientation in the plane, we can replace $x$ and $y$ with linear combinations of $x$ and $y$ that translate and rotate the figure as required.
We then get an equation in which the sum of one quadratic polynomial in $x$ and $y$
and the absolute value of another quadratic polynomial in $x$ and $y$ is equal to zero.


问:
[Proof Verification]Is my proof for the following statement on closed sets correct?

Let $A \subseteq \Bbb{R}$. Let $L$ be the set of all limit points of $A$. Then $L$ is closed.

Proof: We need to show that $L$ is closed, ie. it contains all its limit points. Let $x$ be a limit point of $L$. Then by definition $x \notin V_{\epsilon}(x) \cap L$ where $V_{\epsilon}(x)$ is the $\epsilon$-neighborhood of $x$. We now need to show that $x \in L$. Suppose $x \notin L$. Then choose an $\epsilon > 0$ such that $y \in V_{\epsilon}(x)\cap A$ and $y \neq x$. Then $x$ is a limit point of $A$, but we have arrived at a contradiction, so it must be that $x \in L$ and hence $L$ is closed.

答:
I think there is something odd where you say "Then by definition $x\not \in V_{\epsilon}(x)\cap L$ where $V_{\epsilon}(x)$ is the $ϵ$-neighborhood of $x$".

First, by definition, $x$ is a limit point of $L$ if and only if for every $\epsilon>0$, the intersection $V_{\epsilon}(x)\cap L$ is non-empty. When you say the $ϵ$-neighborhood it seems as if you were fixing an $\epsilon$ and then saying stuff about it. With this being understood, the relation $x\not \in V_{\epsilon}(x)\cap L$ doesn't have anything to do with the definition of being a limit point of $L$. Even more, we can deduce from $x\not \in V_{\epsilon}(x)\cap L$ that $x\not\in L$, and that's exactly the opposite of what you want.

Whenever you try to show that something is closed, it is better almost every time to prove that the complement is open. Having this in mind, take a point $x\not\in L$ (i.e. $x\in \mathbb{R}\setminus L$). By definition, there exists $\epsilon>0$ in such a way that $V_\epsilon(x)\cap L=\emptyset$. Then, we infer that $V_\epsilon(x)$ is an open neighborhood of $x$ that is contained in $\mathbb{R}\setminus L$. Thus, since $x$ was an arbitrary point in the complement $\mathbb{R}\setminus L$, the set $\mathbb{R}\setminus L$ is open and so, $L$ is closed.

Edit (proof with accumulation points): Take a point $x\not\in L$ (i.e. $x\in \mathbb{R}\setminus L$). By definition, there exists $\epsilon>0$ in such a way that $\left( V_\epsilon(x)\setminus \{x\} \cap L\right)=\emptyset$. Then, since $x\not\in L$, we get that $V_\epsilon(x)\cap L=\emptyset$ and so, $V_\epsilon(x)$ is an open neighborhood of $x$ that is contained in $\mathbb{R}\setminus L$. Thus, since $x$ was an arbitrary point in the complement $\mathbb{R}\setminus L$, the set $\mathbb{R}\setminus L$ is open and so, $L$ is closed.


问:
For $dX = Adx+Bdy$, how to compute $\frac{\partial x}{\partial X}$?
Suppose I know

\begin{align}
dX = A dx + B dy
\end{align}
Then I know
\begin{align}
\frac{\partial X}{\partial x} =A, \quad \frac{\partial X}{\partial y} =B
\end{align}
Then is there any smart way to compute
\begin{align}
\frac{\partial x}{\partial X}, \quad \frac{\partial y}{\partial X}
\end{align}

What I know :

Take simple example "two dimensional polar system"
\begin{align}
&x= r\cos(\theta), \quad y = r\sin(\theta)
\end{align}
\begin{align}
  &dx = \cos(\theta) dr - r \sin(\theta) d\theta \\
  &dy = \sin(\theta) dr + r\cos(\theta) d\theta
\end{align} in terms of $dr, d\theta$, by multiplying inverse metric for coefficients,  I have
\begin{align}
  &dx = \cos(\theta) dr - r \sin(\theta) d\theta \\
  &dy = \sin(\theta) dr + r\cos(\theta) d\theta
\end{align}
Then Finally
\begin{align}
&\frac{\partial x}{\partial r} = \frac{x}{r}=\cos(\theta), \quad \frac{\partial r}{\partial x} =\cos(\theta) \\
&\frac{\partial y}{\partial r} =\frac{y}{r} =\sin(\theta),
\quad \frac{\partial r}{\partial y} =\sin(\theta) \\
& \frac{\partial x}{\partial \theta} =-y = -r\sin(\theta), \quad \frac{\partial \theta}{\partial x} =-\frac{1}{r} \sin(\theta) \\
& \frac{\partial y}{\partial \theta} = x = r\cos(\theta), \quad \frac{\partial \theta}{\partial y} = \frac{\cos(\theta)}{r}
\end{align}
It seems they related somehow but how?

答:
$\def\o{\tt1}\def\m#1{\left[\begin{array}{c}#1\end{array}\right]}\def\p#1#2{\frac{\partial #1}{\partial #2}}$The
problem is that single equation does not provide enough information. The desired calculation requires the full system of equations:
$$\eqalign{
dX &= A\,dx + B\,dy \\
dY &= C\,dx + D\,dy \\
}$$
Now calculate the inverse of the coefficient matrix
$$\eqalign{
M &= \m{A&B\\C&D}
\quad\implies M^{-1} &= \m{E&F\\G&H} \implies
\m{dx = E\,dX + F\,dY \\dy = G\,dX + H\,dY} \\
}$$
and extract the desired gradient
$$\eqalign{
\p{x}{X} &= E \\
}$$
These block matrix inversion formulas can be used to express $E$ in terms of $(A,B,C,D)$ by treating them as $(\o\times\o)$ matrices.


问:
Explore convergence of improper integral $\int_{0}^{\infty} \frac{e^{-ax}}{(1+x)^2}dx$
First I thought of Dirichlet's principle, but we don't have a bounded function. Then, I assumed, we can use it after decomposition in parts: $$\int_{0}^{\infty} \frac{e^{-ax}}{(1+x)^2}dx = \lim_{t\to\infty } e^{-ax}  \arctan(x)\bigg|_0^t + \frac{1}{a}  \lim_{t\to\infty } \int_{0}^{t}e^{-ax}  \arctan(x)dx,$$but it seems pointless to me. What should be done?

Body:
$$\int_0^{\infty } \frac{e^{-a x}}{(x+1)^2} \, dx=\int_0^1 \frac{e^{-a x}}{(x+1)^2} \, dx+\int_1^{\infty } \frac{e^{-a x}}{(x+1)^2} \, dx=$$
$$=\left(1-\frac{e^{-a}}{2}-a e^a \Gamma (0,a)+a e^a \Gamma (0,2 a)\right)+\left(\frac{e^{-a}}{2}-a e^a \Gamma (0,2 a)\right)=1-a e^a \Gamma (0,a)$$
which exists only if $a>0$


问:
Formal notation related to Cartesian product
I have defined a set L and another set R, and a set S = L x R for the Cartesian product of the two sets.

My question is quite naive. Given an element $s \in S$, is there a formal way to express its left part $l$ and right part $r$, where $s = (l,r)$?

Another question is... given a set $S' \subseteq S$, is there a formal way to express its left part $L'$ and right part $R'$, where $S'=L' \times R'$?

Thank you very much!

答:
Yes, you can use the projection maps.

For example, define a map $\pi_L: S \rightarrow L$ that takes $s=(l,r)$ to $l$, and similarly define $\pi_R: S \rightarrow R$. Then, for $S' \subseteq S$, $L' = \pi_L(S')$ and $R' = \pi_R(S')$. Note this does not guarantee $S' = L' \times R'$. Consider $S' = \{(1,1),(2,1),(1,2)\}$. Then $\pi_L(S') = \{1,2\}$ and $\pi_R(S') = \{1,2\}$, but $\pi_L(S') \times \pi_R(S') = \{(1,1),(1,2),(2,1),(2,2)\} \neq S'$.


问:
Singular power cardinals?
I am recently exploring set-theoretical postulates that contradict GCH. One particularly interesting one is the proposition "$2^\kappa$ is singular for each infinite cardinal $\kappa$". However I could not prove that this proposition is at all consistent with ZFC.

It seems that a slightly weakened form "$2^\kappa$ is singular for each infinite regular cardinal $\kappa$" is indeed consistent with ZFC, since the function $\kappa\mapsto\aleph_{\kappa^+}$ satisfies the conditions of Easton's theorem and there therefore exists a model where $2^\kappa=\aleph_{\kappa^+}$ for each regular $\kappa$. Can this result also be extended to all singular cardinals?

答:
Lemma 3.4 in the following paper by Golshani & Hayut implies that this is consistent relative to the existence of a strong cardinal.

Golshani, Mohammad; Hayut, Yair, On Foreman’s maximality principle, J. Symb. Log. 81, No. 4, 1344-1356 (2016). ZBL1387.03037.

Title: Singular power cardinals?


问:
Triple integral with cylinder and sphere intersection domain

Let $f(x,y,z)=z$ and
$$D=\{(x,y,z)\in\mathbb{R}^3:\ y\geq0,\ z\geq0,\ x^2+y^2+z^2\leq2,\ x^2+y^2\leq1\}.$$
Then
$$\iiint_Df(x,y,z)dxdydz$$ is (hint: use cylindrical coordinates)

(A) $\frac{3\pi}{2}\qquad$ (B) $\frac{\pi}{2}\qquad$ (C) $\frac{3\pi}{8}\qquad$ (D) $\frac{5\pi}{6}\qquad$ (E) $\frac{3\pi}{4}$

I need to calculate that integral. We are suggested to use cylindrical coordinates. After a lot of thought I know that the integration domain is the intersection of a radius $\sqrt2$ sphere centered on $(0,0,0)$ and a half cylinder of radius $1$, also centered on $(0,0,0)$.

I noticed that the radius of my cylinder coordinates started to change after $z=1$, so I divided the integral in those two:

$\int_{0}^{1}\int_{0}^{\pi}\int_{0}^{1}z\rho d\rho d\theta dz$, which is the half cylinder part of the domain, and

$\int_{1}^{\sqrt2}\int_{0}^{\pi}\int_{0}^{\sqrt{2-z^2}}z\rho d\rho d\theta dz$, wich is the curved top.

I see it this way:

But after computing everything, it gives me a wrong result. It's my reasoning flawed or I'm computing something wrong? Thank you very much.

答:
An idea for you: in cylindric coordinates, you'll have

$$0\le \theta\le\pi\;\; \text {(just the part for}\;y\ge0)\;,\;\;0\le r\le 1\;\;\text{(within the cylinder)}\,,\;\;0\le \sqrt{2-r^2}$$

the last part being from "the floor" $\;z=0\;$ to the top $\;z=\sqrt{2-x^2-y^2}=\sqrt{2-r^2}\;$ , so your integral becomes

$$\int_0^\pi\int_0^1\int_0^{\sqrt{2-r^2}}\,r\,z\,dz\,dr\,dz$$

with $\;r=\;$ Jacobian. The answer is (c)


问:
Where does the function $f(x) = \frac{2x}{x - 7}$ have an increasing slope?
Where does the function $f(x) = \frac{2x}{x - 7}$ have an increasing slope?

$a. x \le 0, x > 7$

$b. x<7$

$c. x > 7$

$d. x \in \Bbb R, x \neq 7$

This question is from a test of mine in a pre-calculus course (so no calculus allowed in answering the question).I have no idea as to how to solve this problem. I can tell where the function is negative and where it is positive, but that's about it, and I'm fairly sure that's no use here.

Any ideas?

答:
Differentiating gives f'(x) = -14/(x-7)^2 which is always negative. Drawing the curve of y = 2 + 14/(x-7) as suggested by newbie one gets a rectangular hyperbola with asymptotes x = 7, y = 2 showing negative gradient at all times.I'd be happier if the problem stated 'decreasing' rather than 'increasing' slope, in which case the answer would be d.

CORRECTION. I've just read the solution above, and must have misunderstood the phrase 'increasing slope' to mean 'positive slope". The slope is always negative, but changes as x changes, so can of course be increasing negatively.