问:
Infinite subgroups of $\text{PGL}(2,\mathbb C)$ where each element has finite order
I am interested in finite subgroups of $\text{PGL}(2,\mathbb C)\cong\text{Aut}\,\mathbb C(x)$.

Let $G\subset\text{PGL}(2,\mathbb C)$ be finitely generated such that each $g\in G$ has finite order. Is $G$ necessarily finite?

The "finitely generated" hypothesis is necessary here, since $$\left\{\begin{bmatrix}e^{2\pi i t} & 0\\0&1\end{bmatrix}:t\in\mathbb Q\right\}$$ is an infinite subgroup where each element has finite order.

This problem without the restriction of being a subgroup of $\text{PGL}(2,\mathbb C)$ is known as the Burnside problem, for which Golod and Shafarevich came up with a counter-example in 1964. However, my instinct is that $\text{PGL}(2,\mathbb C)$ wouldn't have such weird subgroups.

More generally, is the question true for subgroups of $\text{PGL}(n,\mathbb C)$?

答: The magic words are "Selberg's lemma" (which states that a finitely generated matrix group has a torsion free subgroup of finite index).</p>


问:
Property of bounded sequences in $W_0^{1,p}(\Omega)$
I am a bit embarassed about this question, but I’ll go ahead anyway.

Let $\Omega$ be an open bounded space and consider the Sobolev space $W_0^{1,p}(\Omega)$ endowed with its usual norm $\|\cdot\|$. Let $(u_n)_n\subset  W_0^{1,p}(\Omega)$ be a sequence such that
$$(\| u_n\|)_n \leq \beta,$$
for a positive constant $\beta$ which does not depend on $n$.

My question is: for $\alpha>\beta$, we can say that $n_0\in\mathbb{N}$ exists such that
$$\| u_n\| \leq\alpha\quad\mbox{ for all } n\geq n_0?$$

I am sorry for my probably stupid question, but I hope someone could help.

Thank you in advance!

答:
Of course: if $\|u_n\|\leq \beta$ for all $n\in\mathbb{N}$ and $\alpha>\beta,$ then $$\|u_n\|\leq \beta<\alpha$$ for all $n\in\mathbb{N}$. So, it is true for any $n_0\in\mathbb{N}$.

Saying $(\|u_n\|)_n\leq\beta$ is the same thing as saying $\|u_n\|\leq\beta$ for all $n\in\mathbb{N}.$

ContentLicense: CC BY-SA 4.0


问:
What is a suitable choice of Lyapunov function for this system?
I have the following nonlinear system:

\begin{align}\dot x&=x(2-y^4) \\ \dot y &= -3x(1+y^3) \end{align}

I first verified this system using linearization; however, the eigenvalues are $\lambda =\pm i \sqrt{2} $ which is inconclusive for the stability of $(0,0)$. I tried to find a Lyapunov function to study its stability. I started with classical one $V(x,y) = a x^2 + b y^2$ by adjusting the coefficients $a,b$. However, it takes me hours but still no result. Any hints would be appreciated!! Thanks in advance.

答:
Why the point $(0, 0)$? You have a whole line of equilibria $x=0$, i.e. your whole $y-$axis is a set of equlibrium points. It is very hard to apply linaearization or Lyapunov techniques. The system however is integrable, i.e. there is a first integral, you ca cacluate by integrating
$$I(x, y) = x + \frac{1}{3}\int \frac{2-y^4}{1+y^3}\,dy$$ so you can try to use this as Lyapunov function. But to be clear, you should expect from this system to be unstable around any equilibrium point along $x = 0$.


问:
Notation for ordered pairs of congruence classes
A homework question that I'm trying to solve is

Determine the order of $(\mathbb{Z}_{12} \times \mathbb{Z}_4)/\langle ([3], [2]) \rangle$.

I recognize that that I need to find the order of the direct product $\mathbb{Z}_{12} \times \mathbb{Z}_4$, which is $\vert \mathbb{Z}_{12} \vert \cdot \vert \mathbb{Z}_4 \vert = 12 \cdot 4 = 48$, and that I also need $\vert \langle ([3], [2]) \rangle \vert$. The thing that I'm confused about is what $ ([3],[2]) $ means in this context. I know what $[3]$ and $[2]$ mean individually, that they're congruence classes, but I don't know what $([3], [2])$ means.

答:
Elements of $\mathbb{Z}_{12} \times \mathbb{Z}_4$ are ordered pairs $(a,b)$ where $a \in \mathbb{Z}_{12}$ and $b \in \mathbb{Z}_4$. Does that clarify things? To calculate $\vert \langle ([3], [2]) \rangle \vert$ you'll also need to know that the the product of two ordered pairs is the ordered pair of the products in the components:
$$(a,b)(x,y) = (ax,by)$$


问:
Show that $w(\gamma ,0)=w(\gamma _1 ,0)+w(\gamma _2 ,0)$
$\gamma _1$ and $\gamma _2$ are two closed curves in $\Bbb C\setminus\{0\}$ both with parameter interval $[a,b]$. We define the curve $\gamma(t)=\gamma_1(t)\gamma_2(t)$.

I've just shown that $\gamma$ is closed in $\Bbb C\setminus\{0\}$. Now I have to show that $w(\gamma ,0)=w(\gamma _1 ,0)+w(\gamma _2 ,0)$. I first used a theorem saying that under the given circumstances(or so I thought), $w(\gamma,0)=(1/2\pi i)\int_\gamma (dz/z)$, which easily solved the problem. It turned out, however, that for this to be true, $\gamma$ has to be piecewise $C^1$, which we don't know if it is, so I can't use that theorem. Any ideas what to do instead?

EDIT: My book defines winding number as $w(\gamma,0)=\frac{1}{2\pi}argvar(\gamma)$, and $argvar(\gamma)=\theta(b)-\theta(a)$, where $\theta$ is an arbitrary continuous argument function along $\gamma$.

答:
Here's a more geometric description of what's going on.

Since all the curves avoid the origin, you can parametrize them in polar coordinates using Euler's formula (treating $\mathbb{C} \setminus 0 \cong \mathbb{R}^2 \setminus \{0,0\}$).
$$
\gamma(t) = \begin{pmatrix} r(t) \cos \theta(t) \\ r(t) \sin \theta(t)\end{pmatrix},
$$
where $r$ and $\theta$ are continuous functions from $[a, b] \to \mathbb{R}$, and $r(t) > 0$ for all $t$.  Now, normalize by projecting onto the unit circle:
$$
\begin{array}{rccc}
\epsilon = \frac{\gamma}{||\gamma||}: &[a,b] &\longrightarrow &S^1 \\
& t &\longmapsto &\begin{pmatrix} \cos \theta(t) \\ \sin \theta(t)\end{pmatrix}
\end{array}
$$
This map lifts to a map $\tilde{\epsilon}: [a, b] \to \mathbb{R}$ that is unique once you specify $\tilde{\epsilon}(a)$ (or the image of any one point, for that matter).  In other words,
$$
\epsilon = p \circ \tilde{\epsilon},
$$
where $p$ is the projection $\mathbb{R} \to S^1$, $u \to (\cos u, \sin u)$.

If $\gamma$ is a closed curve, then $\epsilon$ is too, so
$$
\tilde{\epsilon}(b) - \tilde{\epsilon}(a) = 2\pi w
$$
for some $w \in \mathbb{Z}$.  This $w$ is the winding number of $\gamma$ around $0$.

Now, to address your actual question, in which you multiply the curves in the complex plane, first parametrize them as maps $[a, b] \to \mathbb{R}^2 \setminus \{0, 0\}$:
$$
\gamma_1(t) = \begin{pmatrix} r_1(t) \cos \theta_1(t) \\ r_1(t) \sin \theta_1(t)\end{pmatrix}
$$
and
$$
\gamma_2(t) = \begin{pmatrix} r_2(t) \cos \theta_2(t) \\ r_2(t) \sin \theta_2(t)\end{pmatrix}.
$$
Now,
$$
\begin{align}
\gamma(t) &= \gamma_1(t) \gamma_2(t) \\
&= r_1(t) r_2(t) \begin{pmatrix}
\cos \theta_1(t) \cos \theta_2(t) - \sin \theta_1(t) \sin \theta_2(t) \\
\sin \theta_1(t) \cos \theta_2(t) - \cos \theta_1(t) \sin \theta_2(t)
\end{pmatrix} \\
&= r_1(t) r_2(t) \begin{pmatrix}
\cos \big(\theta_1(t) + \theta_2(t) \big) \\
\sin \big(\theta_1(t) + \theta_2(t) \big)
\end{pmatrix}.
\end{align}
$$
From this you can see that the winding number is additive.

ContentLicense: CC BY-SA 3.0


问:
Can you explain this expression
in the book (a text-book of convergence "w.L FERRAR" )

The writer present this ideas

( if y is a fixed number graeter than unity, and k is any fixed positive integer, then the sequence $\frac{n^k}{y^n} $ converges to zero

Before we give the proof of this theorem, we try to explain some of the ideas that lead to the proof

i)...... if  l is a fixed number and n is to be thought of as a very large number, then n-l is about as big as n, further, extending this ideas a little, if k is a fixed number and n is to be thought of as a very large number, then

$n(n-1)(n-2).....(n-k)$

is about as big as $n^{k+1}$

(ii) the y of our Problem exceeds unity and we can write

$y^n=(1+p)^n$ p>0

The right-hand side can be expended, when n is a positive integer, by the binomial theorem, and, since each term of the expansion is positive, so if we take $(n>k+1)$

$y^n=(1+p)^n>\frac{n(n-1)....(n-k)}{(k+1)!}p^{k+1}$)

I can't understand  the last expression، so can you explain it?

And also I need more clarification for the partie (i)

答:
FYI - when people ask for clarification, just edit the post.  No need to keep deleting and redoing it.

To answer your question, if we have $y^n = (1+p)^n$ we can use the binomial theorem, which says that
$$(1+x)^{n} = \sum_{k=0}^{n}\binom{n}{k}x^{k}.$$  In this case, we have
$$y^n = (1+p)^n = \sum_{k=0}^{n}\binom{n}{k}p^{k}.$$  Now, so long as $n > k+1$, and since every term in the sum is positive, we can take just the term with the $k+1$ power and it will be less than the entire sum.  So, we get
\begin{align*}
y^n &= (1+p)^n\\[5pt]
&= \sum_{k=0}^{n}\binom{n}{k}p^{k}\\[5pt]
&=\binom{n}{1}p + \binom{n}{2} + \dotsb + \binom{n}{k+1}p^{k+1} + \dotsb + \binom{n}{n}p^{n}\\[5pt]
&> \binom{n}{k+1}p^{k+1}\\[5pt]
&= \frac{n!}{(k+1)!(n-k-1)!}p^{k+1}\\[5pt]
&= \frac{n(n-1)\dotsb(n-k)}{(k+1)!}p^{k+1}.
\end{align*}

As for part (i), if we think of $n$ as being very large, then $n-1 \approx n$, which means that $n(n-1) \approx n^2$.  Extending this idea, if $k$ is some fixed number, and $n$ is much larger than $k$, then the numbers $$n, n-1, n-2, \dotsc, n-k$$ are all approximately equal to $n$.  So, this gives us $$n(n-1)(n-2)\dotsb(n-k) \approx n^{k+1},$$ because there are $k+1$ terms in the product.

Title: Can you explain this expression


答:
I found the following in Irving Kaplansky's book "Fields and Rings":

Theorem G. Any torsion group of matrices is locally finite.

Here, "locally finite" means every finitely generated subgroup is finite.

Title: Infinite subgroups of $\text{PGL}(2,\mathbb C)$ where each element has finite order

Body:
I am interested in finite subgroups of $\text{PGL}(2,\mathbb C)\cong\text{Aut}\,\mathbb C(x)$.

Let $G\subset\text{PGL}(2,\mathbb C)$ be finitely generated such that each $g\in G$ has finite order. Is $G$ necessarily finite?

The "finitely generated" hypothesis is necessary here, since $$\left\{\begin{bmatrix}e^{2\pi i t} & 0\\0&1\end{bmatrix}:t\in\mathbb Q\right\}$$ is an infinite subgroup where each element has finite order.

This problem without the restriction of being a subgroup of $\text{PGL}(2,\mathbb C)$ is known as the Burnside problem, for which Golod and Shafarevich came up with a counter-example in 1964. However, my instinct is that $\text{PGL}(2,\mathbb C)$ wouldn't have such weird subgroups.

More generally, is the question true for subgroups of $\text{PGL}(n,\mathbb C)$?


答:
The last expression comes from the binomial expansion of $(1+p)^n$.

Remember that $$(1+p)^n=\sum^{n}_0\binom{n}i p^i$$

Now since each term on the RHS is positive, the sum is definitely greater than an individual term.

$$\sum^{n}_0\binom{n}i p^i\ge\binom{n}{k+1} p^{k+1}$$ which is what expands into the last expression.

As for the explanation of part (i) it can be understood as

$$\frac{\sum_{r=0}^{k} (n-r)}{n^{k+1}}=\sum_{r=0}^{k} (1-r/n)\rightarrow 1\ as\ n\rightarrow\infty$$. which justifies the approximation.

Title: Can you explain this expression

Body:
in the book (a text-book of convergence "w.L FERRAR" )

The writer present this ideas

( if y is a fixed number graeter than unity, and k is any fixed positive integer, then the sequence $\frac{n^k}{y^n} $ converges to zero

Before we give the proof of this theorem, we try to explain some of the ideas that lead to the proof

i)...... if  l is a fixed number and n is to be thought of as a very large number, then n-l is about as big as n, further, extending this ideas a little, if k is a fixed number and n is to be thought of as a very large number, then

$n(n-1)(n-2).....(n-k)$

is about as big as $n^{k+1}$

(ii) the y of our Problem exceeds unity and we can write

$y^n=(1+p)^n$ p>0

The right-hand side can be expended, when n is a positive integer, by the binomial theorem, and, since each term of the expansion is positive, so if we take $(n>k+1)$

$y^n=(1+p)^n>\frac{n(n-1)....(n-k)}{(k+1)!}p^{k+1}$)

I can't understand  the last expression، so can you explain it?

And also I need more clarification for the partie (i)


问:
How to Prove That the Variables Come In Plus/Minus Pairs
Consider the numbers, $a_1, a_2, a_3 \cdots a_n$ for some arbitrary $n$.

Given that $\sum_{i=1}^n a_i^k = 0$ for every odd integer $k$, prove that all of the non-zero $a_i$ come in plus-minus pairs.

For example, we could have $a_1 = 2, a_2 = -2, a_3 = 4, a_4 = -4, a_5 = a_6 = \cdots = a_n = 0.$

I tested this out for various values of $n$ and it seems to work, however, I have no idea on how to actually prove this.

Does anyone have ideas?

答:
Hints:

show it is true for $n=1$ and $n=2$

show that you must have $\max_i(x_i)=-\min_i(x_i)$ if this is true for all odd $k$

suppose there is a minimally sizes counterexample, and remove (one of) the maximum value(s) and (one of) the minimum value(s) to get a contradiction


问:
How do we differentiate between $\log_{10}A$, $\log A$, and $\ln A$?
How do we differentiate $\log_{10}A$,  $\log A$, $\ln A$?  I  always thought $\log A= \log_{10}A$, but got stuck in understanding one of the logarithm problems at this site recently. Now, I am aware that quite some folks  interpret $\log A = \ln A$. Just wondering how the majority of us  use and differentiate these logarithm notations to avoid confusions?

答:
I tend to use $\log A = \log_{10}A$ when referring to something in science or engineering. This is because a lot of things in science (scientific notation, pH scale, etc.) tend to use base 10.

Whenever talking about mathematics I use $\log A = \ln A$. This is how I see it done in most math except high school.


问:
Homeomorphism of a Torus
I need to show that the function:

$f(u,v)=((r \cos(u)+a)\cos(v), (r\cos(u)+a)\sin(v), r\sin(u))$, $f:U\to\mathbb{R^3}$, when $U=(0,2\pi)\times (0,2\pi)$ is an homeomorphism to the torus.

I show that $f$ is differentiable and then $f$ is continous, also I show that $f$ is injetive.
I tried to calculate the inverse function...but I'm stucked.

Also...I tried to show that $f$ is an open function...but I don't know how I can start the proof.

答: Well, they're not homeomorphic, as $(0, 2\pi) \times (0, 2\pi)$ has a trivial fundamental group, while $S^1 \times S^1$ does not.

Title: Homeomorphism of a Torus


答: Generally, it depends. From my experience, $\log A=\log_{10} A$. For most fields, this is what the notation means. HOWEVER, in mathematics, $e$ has so many good properties that we define $\log A=\ln A$. That's also why $\ln$ is called the natural log. It's the natural log, the interpreted log. A simple way to remember is What feels natural? The natural log!

Title: How do we differentiate between $\log_{10}A$, $\log A$, and $\ln A$?

Body: How do we differentiate $\log_{10}A$,  $\log A$, $\ln A$?  I  always thought $\log A= \log_{10}A$, but got stuck in understanding one of the logarithm problems at this site recently. Now, I am aware that quite some folks  interpret $\log A = \ln A$. Just wondering how the majority of us  use and differentiate these logarithm notations to avoid confusions?


答:
There are three special cases of the base $b$ that have names/notations (and where they are most often found):

Common logarithm: $\log_{10} x = \log x$ (engineering, seismology, acoustics)

Natural logarithm: $\log_e x = \ln x$ (mathematics)

Binary logarithm: $\log_2 x = \lg x$ (computer science)

Of course because of the change-of-base formula, every logarithm is just a scaled version of every other logarithm:
$$
\log_b x = \frac{\log_c x}{\log_c b}
$$

For example, $\ln x \approx 2.303 \log x$ and $\log x \approx 0.434 \ln x$, and those coefficients are reciprocals.

An issue/ambiguity In many uses of logarithms in higher mathematics, such as when looking at asymptotics of certain sequences or functions, the distinction of this multiple is immaterial, so people just write $\log x$ (and usually mean the natural logarithm), but you probably won't encounter this notation in undergraduate mathematics.

Of course, if the distinction of the base is important, you can always just explicitly state what your notation means. The notation works for you, not the other way around!


问:
Diffeomorphism maps interval into interval in monotone way
Let $I,J$ be two interval(which is connected subset on $\Bbb{R}$).Let $f:I\to J$ be diffeomorphism .Prove $f$ is stricly monotone.

My attempt:

Let two endpoint of $I$ be $x_0,x_1$ ($x_0 < x_1$) (If no endpoint we may extend to it by continuity),$f(x_0) = y_0$ and $f(x_1) = y_1$.Wlog we may assume $y_0<y_1$.  (then the problem becomes to prove that this map is strictly monotone increasing)

Since $f(I)$ is connected which is a interval with two endpoint specified hence $J = [y_0,y_1]$. hence any $w_1,w_2 \in I$  ($w_1<w_2$)exist some $k_1,k_2$ maps to it (by surjective of $f$)

Now we assume $k_1>k_2$ we get contradiction which means that $f$ is monotone increasing,hence complete the proof.

To see the contradiction we just need consider the interval $[x_0,w_1]$ which has range $[y_0,k_1]$ hence exist some point in between  gets the value $k_2$,now we also has $f(w_2) = k_2$  such that $w_2$ different form this point hence contradiction.

Is my proof correct ,is there some better proof?

答:
When you assume that $y_{0}<y_{1}$, I would add something like "with this assumption, the problem becomes proving that the function is increasing. In the other hand, if $y_{1}>y_{0}$, the problem is to show that $f$ is decreasing, and the proof is similar". You also have some typos (e.g., when you say that $w_{1},w_{2} \in J$ it should be $w_{1},w_{2} \in I$).

A more direct way is the following: since $f:I \to J$ is a difeomorfism, its derivative $f'$ is never zero (otherwise $(f^{-1})'=1/f'$ will not be defined). Then, $f>0$ in $I$ or $f<0$ in $I$. In the first case, $f$ is increasing while in the second case $f$ is decreasing.


问:
Bringing an elliptic curve over some finite field in Weierstrass form
I have given the following elliptic curve over $\mathbb{F}_9 = \mathbb{F}_3[\zeta_8]$ (where $\min_{\mathbb{F}_3}(\zeta_8) = x^2 - x - 1$):

$$
E/\mathbb{F}_9: y^2 = \zeta_8 x^3 + x.
$$
Now I would like to know how to bring this equation in Weierstrass form, i.e. in the form $y^2 = x^3 + Ax + B$ for some $A,B \in \mathbb{F}_9$. I thought about doing a substitution $y = \zeta_{16} y'$ for some element $\zeta_{16}$ with $\zeta_{16}^2 = \zeta_8$, so I could divide by $\zeta_8$ on both sides. However, I don't think that $\zeta_{16}$ is in $\mathbb{F}_9$.

Could you please help me here?

答:
Let $\zeta = \zeta_8$. Multiplying both sides of $y^2 = \zeta x^3 + x$ by $\zeta^2$ yields
$$
(\zeta y)^2 = \zeta^2 y^2 = \zeta^3 x^3 + \zeta^2 x = (\zeta x)^3 + \zeta (\zeta x) \, .
$$
Letting $\tilde{x} = \zeta x$ and $\tilde{y} = \zeta y$, then
$$
\tilde{y}^2 = \tilde{x}^3 + \zeta \tilde{x} \, .
$$


问:
Moment of Intertia of hemisphere
Calculate the moment of inertia of a solid uniform hemisphere $x^2+y^2+z^2=a^2$; $z \geq 0$ with mass $m$ about the $z$ – axis.

My attempt: We know that MI of a solid hemisphere is $2Ma^2/5$ where $M$ is mass and a is radius of solid hemisphere. And the MI of a solid hemisphere is $Ma^2/5$. So in above question mass is given $m$ and radius is a so  $MI$ of hemisphere is $2ma^2/5$. Is this correct answer ? Answer given is $15mπa/32$.

答:
Assuming that the mass is m and the radius a, you do not need to integrate in order to find out the MI of a solid hemisphere.

$$MI_{sphere}=\frac{2Ma^2}{5}$$

In the process of making a hemisphere out of it we know that we are eliminating mass symmetrically from the axis, i.e.

$$MI_{cut out}=MI_{hemisphere\ left\ out}$$

and their sum is $\frac{2Ma^2}{5}$.

Thus $$MI_{hemisphere}=\frac{Ma^2}{5}$$.

But remember that M is not the mass of your hemisphere.

$$M=2m$$ where m is the mass of your hemisphere and M that of whole sphere.

Thus $$MI_{hemisphere}=\frac{2ma^2}{5}$$

Title: Moment of Intertia of hemisphere


问:
Expected value of first coordinate of a vector uniformly sampled from a hemisphere
I'm looking for a way to compute the following: Defining $H$ to be the  hemisphere $H=\left\{  x\in S^{d-1} \mid x_1>0 \right\}$, what is $\mathbb{E}[x_1]$ when $x$ is sampled uniformly from $H$?

Another way to look at this that may be easier on the eyes, is to ask what is $\mathbb{E}\left[ \mathbb{1}_{[x_1 > 0]} \cdot x_1\right]$ when $x$ is sampled uniformly from $S^{d-1}$?

Thanks in advance!

答:
Let's first consider when $d=3$, and then we'll generalize. Define the function $f:\mathbb{R}^3\rightarrow \mathbb{R}$ by

$$f(x,y,z) = \left\{ \begin{array}{ll} \frac{1}{2\pi}: \quad (x,y,z)\in H \\ 0: \quad \text{ else} \end{array} \right. $$

Then $f$ corresponds to the pdf for the random vector $(X,Y,Z)\sim \mathcal{U}(H)$. The expected value of $X$ is the surface integral $$\mathbb{E}(X)=\int_{H}xf(x,y,z)\mathrm{d}S$$ We can parameterize $H$ by $$\vec{r}(u,v)=\Big<\sqrt{1-u^2-v^2},u,v\Big>:u^2+v^2<1$$ Notice how $$\mathrm{d}S=\|\vec{r}_u \times  \vec{r}_v\|\mathrm{d}u\mathrm{d}v=\frac{1}{\sqrt{1-u^2-v^2}}$$ So we get $$\mathbb{E}(X)=\int_{\{u^2+v^2<1\}}\frac{\mathrm{d}u\mathrm{d}v}{2\pi}=\frac{\pi}{2\pi}=\frac{1}{2}$$ This generalizes as follows. Re$-$define $f:\mathbb{R}^d\rightarrow \mathbb{R}$ by $$f(\vec{x}) = \left\{ \begin{array}{ll} \frac{\Gamma(d/2)}{\pi^{d/2}}: \quad \vec{x}\in H \\ 0: \quad \text{ else} \end{array} \right. $$ Just as before, $f$ corresponds to a pdf associated with $\vec{X}\sim \mathcal{U}(H)$, and we're tasked to evaluate the surface integral $$\mathbb{E}(X_1)=\int_{H}x_1f(\vec{x})\mathrm{d}S_{d-1}$$ By invoking a similar parameterizaion $$\vec{r}(x_1,\ldots,x_{d-1})=\Big<\sqrt{1-x_1^2-\dots -x_{d-1}^2},x_1,\ldots,x_{d-1}\Big>$$ defined on the domain $$x_1^2+\dots + x_{d-1}^2<1$$ you'll get that \begin{eqnarray*}
\mathbb{E}(X_1) & = & \frac{\Gamma(d/2)}{\pi^{d/2}}\int_{H}x_1\mathrm{d}S_{d-1}
\\ & = & \frac{\Gamma(d/2)}{\pi^{d/2}}\int_{\{x_1^2+\dots +x_{d-1}^2<1\}}\mathrm{d}V_{d-1} \\
 & = & \frac{\Gamma(d/2)}{\pi^{d/2}} \cdot \frac{\pi^{\frac{d-1}{2}}}{\Gamma\big(\frac{d+1}{2}\big)} \\ & = & \frac{\Gamma(d/2)}{\sqrt{\pi}\Gamma\big(\frac{d+1}{2}\big)}
\end{eqnarray*}


问:
Legendre Symbol - Find Prime $p$ Which Divides A Polynomial
I need to find a general form of a prime number $p$ which divides the polynomial $x^2-6$, i.e. $p$ such that $x^2 - 6\equiv 0\text{ (mod }p)$. By Legendre symbol, I actually need to find a prime p such as $\left(\frac{6}{p}\right) = 1$.

I know that $\left(\frac{6}{p}\right) = \left(\frac{3}{p}\right)\left(\frac{2}{p}\right)$, so there are two options at the moment:

Both $\left(\frac{3}{p}\right) = 1$ and $\left(\frac{2}{p}\right) = 1$.

Both $\left(\frac{3}{p}\right) = -1$ and $\left(\frac{2}{p}\right) = -1$.

I'd like to find out how could I find a general form of a prime $p$ which answers the two terms above?

Thanks in advance

答: The second supplement to the law of quadratic reciprocity gives you that $\left(\frac{2}{p}\right)=1$ if $p\equiv\pm 1\text{ (mod }8)$, and $-1$ if $p\equiv\pm 3\text{ (mod }8)$. It can also be shown that $\left(\frac{3}{p}\right)=1$ if $p\equiv\pm 1\text{ (mod }12)$ using quadratic reciprocity. Can you go from here?</p>


问:
Matrix inequality of inverses of sum of two matrices
$A$ and $B$ are two real symmetric non-singular matrices but not necessarily sign definite. Suppose $A+B$ is non-singular. I want to find a $\lambda \in \Bbb R$ such that $$(A+B)^{-1}\leq(A+\lambda I)^{-1}$$ where $A\leq B$ means $A-B$ is negative semi-definite. Does such $\lambda$ exist? How can I find it?

I have tried the identity $(A+B)^{-1}=A^{-1}-A^{-1}(B^{-1}+A^{-1})^{-1}A^{-1}$. But it led me to

$$(A^{-1}+\lambda^{-1} I)^{-1}\leq(A^{-1}+B^{-1})^{-1}$$

Thanks in advance!

答:
I found a way by Schur's complement identity.

It holds that the following three matrices are congruent.
$$\begin{bmatrix}(A+\lambda I)^{-1}&I\\I&A+B\end{bmatrix}\sim\begin{bmatrix}A+B&I\\I&(A+\lambda I)^1-(A+B)^{-1}\end{bmatrix}\sim\begin{bmatrix}(A+\lambda I)^{-1}&I\\I&B-\lambda I\end{bmatrix}$$

Let $n_-(A)$ denote the number of negative eigenvalues of $A$. Then, by Sylvester's law of inertia, $(A+\lambda I)^1\geq(A+B)^{-1}$ iff
\begin{equation}n_-(A+B)=n_-(A+\lambda I)+n_-(B-\lambda I)\end{equation}
But such $\lambda$ does not always exist, as shown by the excellent counterexample by @Rammus. In that example, $n_-(A+B)=0$ but no $\lambda$ can simultaneously make $n_-(A+\lambda I)=0$ and $n_-(B-\lambda I)$.

Anyways, if such $\lambda$ exists, the identity above provides a way to find it.


问:
If a function is not defined at a point then will its inverse be defined at that point?
I have a function $y=\frac{x-2}{x-3}$ clearly the function is not defined at $x= 3$

But for its inverse $ \frac{3x-2}{x-1},$ can we say that it is defined at $x= 3$?

答: Just to fill in the contrapositive which the other questions don't mention. A function whose entire partial function is a bijection from a set onto itself  does have the property that its inverse is defined wherever it is defined.


问:
Orthogonal Complement to arbitrary matrix
Find the orthogonal complement of the subspace

$$S = \{(x,y,z,w)\in\mathbb{R}^{4} \mid x + y = z + w = 0\}$$

I was working through some homework problems, stumbled onto this one and struggled with what to do.

So far, I have created an arbitrary matrix $A$ such that $A = (x,-x,z,-z)$ and I know that the orthogonal complement of $A$ is equals to the null space of $A$ transpose but I am struggling to find the nullspace of $A$ transpose with those arbitrary elements

答:
HINT

You are almost there! Notice that $S = \text{span}\{(1,-1,0,0),(0,0,1,-1)\}$.

Considering that we are dealing with the standard inner product in $\mathbb{R}^{4}$, $v = (a,b,c,d)\in S^{\perp}$ iff
\begin{align*}
\begin{cases}
\langle v,(1,-1,0,0)\rangle = 0\\\\
\langle v,(0,0,1,-1)\rangle = 0
\end{cases}
\end{align*}

Can you take it from here?


问:
Why is the chain rule applied to derivatives of trigonometric functions?
I'm having trouble to understand why is the Chain rule applied to trigonometric functions, like:
$$\frac{d}{dx}\cos 2x=[2x]'*[\cos 2x]'=-2 \sin 2x$$
Why isn't it like in other variable derivatives? Like in:
$$ \frac{d}{dx} 3x^2=[3x^2]'=6x $$
Does it means it is the derivative of the trig function times the derivative of the angle?

Thanks once again.

答:
A more appropriate example would be

$$\frac{d}{dx} (3x)^2=2(3x) \cdot (3x)' = 18x$$


问:
Would the median of union equals to median of medians?
Given a finite set $S \subset \mathbb{Z}$, we can easily obtain its median $median(S)$.
Given a set of sets $T = \{ S_1, S_2, \cdots, S_n \}$. where $S_i \cap S_j = \emptyset, \forall S_i, S_j \in T $

We can define two "median", $M_1 = median(\cup_{S \in T})$ and $M_2 = median(\{ median(S)|S \subset T \})$.

Would $M_1 == M_2$ ?

Body:
The answer is No.

for example, $S_1 = (1, 2, 4)$, $S_2 = (3, 10, 11)$.

$M_1 = 3.5$, while $M_2 = 6$


问:
The equality case of Hölder's inequality at the endpoint $p=1$, $q=\infty$
This is a follow-up question on On the equality case of the Hölder and Minkowski inequalites.

What is the conditions for $$\int \vert fg\vert =\Vert f \Vert_p\Vert g \Vert_q$$ when  $p=1, q= \infty$?

It's seems like that the same proof won't work, as $q = \infty $.

答:
If $\displaystyle \int|fg|=\|f\|_1\,\|g\|_\infty$, this can be written as
$$
\int|fg|=\int|f|\,\|g\|_\infty.
$$
Rewrite as
$$
\int |f|\,(\|g\|_\infty-|g|)=0.
$$
So a necessary and sufficient condition for the equality is that $|g|=\|g\|_\infty$ a.e. whenever $f\ne0$.


答: If the definition of median is as from statistics (no other definition is given), here is a counter example. Let $A = {0,1}$ and $B ={1, 10}$. Then the median of $A$ is $0.5$ and the median of $B$ is $5.5$; the median of these values is $3$. Whereas the median of the union is $1$.

Title: Would the median of union equals to median of medians?

Body:
Given a finite set $S \subset \mathbb{Z}$, we can easily obtain its median $median(S)$.
Given a set of sets $T = \{ S_1, S_2, \cdots, S_n \}$. where $S_i \cap S_j = \emptyset, \forall S_i, S_j \in T $

We can define two "median", $M_1 = median(\cup_{S \in T})$ and $M_2 = median(\{ median(S)|S \subset T \})$.

Would $M_1 == M_2$ ?


问:
Why the sum of the sample autocovariance function of a stationary process is zero?
The sample autocovariance function for a sample $X_1, X_2, \dots, X_n$ from a stationary process is defined as:

$$\widehat{\gamma}(h) = n^{-1}\displaystyle\sum_{t=1}^{n-h}(x_{t+h}-\bar{x})(x_t-\bar{x})$$

with $\widehat{\gamma}(-h) = \widehat{\gamma}(h)\;$ for $\;h = 0,1, ..., n-1$. $\bar{x}$ is the mean.

Show that $\sum_{|h| <n}\widehat{\gamma}(h) = 0$

答:
$\newcommand{\hgamma}{\hat{\gamma}}$
This is Exercise $7.3$ from the classic text Time Series: Theory and Methods. We can prove it by a series of reductions of the original equality.

First note by $\hgamma(h) = \hgamma(-h)$, the equality to be proven is
$$0 = \hgamma(0) + 2\sum_{h = 1}^{n - 1}\hgamma(h). $$

Next note by definition of $\hgamma$, without loss of generality we can assume $\bar{x} = 0$. Therefore, it suffices to prove that under the condition of $x_1 + \cdots + x_n = 0$, it holds that
\begin{align*}
0 = \sum_{t = 1}^n x_t^2 + 2\sum_{h = 1}^{n - 1}\sum_{t = 1}^{n - h}x_{t + h}x_t.
\end{align*}

But this is just the result of expanding the quadratic form
\begin{align*}
0 = \bar{x}^2 = (x_1 + x_2 + \cdots + x_n)^2.
\end{align*}

This completes the proof.


答:
$\log A$ can mean either $\log_{10} A$ or $\ln A = \log_e A$ depending an what book you are using.

Doesn't matter.

$\ln_e x = \log_e x$ is the "natural base" with the natural property that $\frac {de^x}{dx} = e^x$.  And any other base can be derived as $b^x = e^{x \ln b}$.

Now we can use the chain rule that $\frac {d b^x}{dx} = \frac {de^{x\ln b}}{dx} = \frac{d e^{x\ln b}}{d x\ln b} \cdot \frac {dx\ln b}{dx} = e^{x\ln b}\ln b= b^x \ln b$.

So $\frac {d10^x}{dx} = 10^x \ln 10$.

We can use the chain rule to calculat that $x = e^{\ln x}$ so $\frac {dx}{dx} = \frac {de^{\ln x}}{dx} = \frac {de^{\ln x}}{d\ln x}\cdot \frac {d\ln x} {dx}=e^{\ln x} \frac {d\ln x}{dx} = x \frac {d\ln x}{dx}$.

But $\frac {dx}{dx} = 1$ so $\frac {d\ln x}{dx} = \frac 1x$.

$x = b^{\log_b x}$ so $\frac {dx}{dx} = \frac {db^{\log_b x}}{d\log_b x} \cdot \frac {d\log_b x}{dx}$.

So

$1 = \ln b \cdot b^{\log_b x} \cdot \frac {d\log_b x}{dx}$ and

$1 = x\ln b\cdot \frac {d\log_b x}{dx}$ and

$\frac {d\log_b x}{dx} = \frac 1{x\ln b}$

And $\frac {d\log_{10} x}{dx} = \frac 1{x\ln 10}$.

....

That's actually a little backwards as we actually usually define $\ln x = \int_1^x \frac 1t dt$ first and work all the rest out backwards.


问:
Orthogonal complement and subspaces
I have the following multiple response question here.

Suppose $V$ is a subspace of $\mathbb{R}^n$ and $S$ is a set of vectors in $V$ whose span is the subspace $W$ of $V$. Select the correct statements.

The orthogonal complement of $S$ in $V$ is the same as the orthogonal complement $W$ in $V$.

Every vector of $S$ belongs to the orthogonal complement of $S$ in $V$.

If $\mathbf{u}$ is a vector in $V$ which belongs to both $W$ and its orthogonal complement in $V$, then $\mathbf{u}=\mathbf{0}$.

If $\mathbf{u}$ is a vector in $V$ which belongs to both $W$ and its orthogonal complement in $V$, then $\mathbf{u}$ belongs to $S$.

If $\mathbf{u}$ is a vector in $V$ which belongs to both $W$ and its orthogonal complement in $V$, then $|u|=1$.

If $V$ has dimension $10$, and $S$ contains $7$ vectors, then the dimension of $W^\perp$ is at least $3$.

If $V$ has dimension $10$, and $S$ contains $7$ vectors, then the dimension of $W^\perp$ is at most $3$.

If $V$ has dimension $10$, and $S$ contains $7$ vectors, then the dimension of $W^\perp$ is always $3$.

The correct answer should be $1$,$3$ and $6$ but could someone explain why?

I understand $3$ but I don't understand why $1$ and $6$ are true. Could someone explain why the other statements are false?

For instance, I thought $8$ would be right instead of $6$ since I thought by the rank-nullity theorem, the nullity must ALWAYS be $3$.

答:
By definition $S^\perp=\{\bar{x} \in V | \bar{x}\cdot \bar{s} = 0, \forall \bar{s} \in S\}$, where I'm assuming $(\cdot)$ is the regular dot product.

Let $S$ be a set of vectors in $V$.

Then $\bar{x}\in S^\perp$ implies $\bar{x}\cdot\bar{s} =0$ for all $\bar{s}\in S$, and since $W=\text{span}(S)$, $W$ is the set of all linear combinations of the vectors in $S$, thus for $\bar{w} \in W$, $\bar{w} = a_1\bar{s_1} +\cdots + a_n\bar{s_n}$ for $a_1,...,a_n \in \mathbb{R}$, and $\bar{x}\cdot\bar{w} =a_1(\bar{x}\cdot\bar{s_1})+\cdots+a_n(\bar{x}\cdot\bar{s_n}) = a_1(0)+\cdots+a_n(0)=0$.

Similarly $W^\perp = \{\bar{x} \in V|\bar{x}\cdot\bar{w}=0, \forall \bar{w} \in W\}$, thus $\bar{x}\in W^\perp$ because every $\bar{w} \in W$ is a linear combination of the vectors in $S$ and $\bar{x}\cdot\bar{s} =0$ for all $\bar{s}\in S$. $S^\perp \subseteq W^\perp$.

Every vector $\bar{s} \in S$ is an element of $W$ thus if $\bar{x} \in W^\perp$ then in particular $\bar{x}\cdot\bar{s} =0$ for all $\bar{s} \in S$, thus $\bar{x}\in S^\perp$. $W^\perp \subseteq S^\perp$.

In conclusion $S^\perp = W^\perp$. As for $\#6,7,8$ $\dim(S^\perp)+\dim(S) = \dim(V) =10$, however we cannot be certain that $\dim(S)=7$ because the vectors in $S$ may not be linearly independent, thus at best we know that $\dim(S) \leq 7$, and so $\dim(S^\perp)\geq 10-7=3$. For example say $S$ had $7$ vectors but when finding a basis for $S$ we find that the basis only has $5$ vectors, then $\dim(S) =5\leq 7$ and $\dim(S^\perp)=5 \geq 3$.

Title: Orthogonal complement and subspaces


问:
Restriction to minimize vehicles in vehicle routing problem (VRP)
I am currently reading a paper about vehicle routing problem with time windows and when formulating the main goals and restrictions the following is stated:

Objective: Minimize the number of vehicles and the total travel distance.

Assume N is a number of customers (1, 2, ..., n) that need to be serviced.

$c_{ij}$ - the transportation cost from the customer i to j.

$x_{ijk}$ - a variable taking value of 1 if the vehicle k is coming from the customer i to the customer j, and 0 if otherwise.

Objective function:

$$Z = \sum_{k\in V}\sum_{i\in N}\sum_{j\in N} c_{ij}x_{ijk} \to min$$

One of the restrictions is the following:

$$\sum_{j \in N}x_{0jk}=1,\forall k \in V$$

Doesn't the previous restriction imply that every vehicle will move from $0$ (depot) to a customer $j$?

If that was the case the number of vehicles wouldn't be minimized because it would force them all to be in use.

Printscreen of the page here.

答:
Yes, those constraints force every vehicle to be used.

One approach to minimize the number of vehicles is to start with $|V|=1$ and try to solve this problem for fixed $|V|$.  Increment $|V|$ until the problem is feasible.

Another approach is to introduce a binary variable $y_k$, replace the RHS of constraints (4) and (5) with $y_k$, and minimize $\sum_k y_k$.  You can also tighten constraint (3) by replacing the RHS with $q\cdot y_k$.


问:
Why is - $\frac{8}{17}$ taken instead of $\frac{8}{17}$ when trying to find the $cos x$ from $\tan x = \frac{15}{8}$ and $\in [\pi, \dfrac{3\pi}{2}]$?
$\sec^2 (x) =  \tan^2 (x) + 1$

$\sec^2 (x) = \left(\dfrac{15}{8}\right)^2 + 1$

$\sec (x) = \pm \sqrt{\dfrac{289}{64}}$

$\sec (x) = - \dfrac{17}{8}$ because $x \in \left[\pi, \dfrac{3\pi}{2}\right]$ <--- This is where I don't understand why we take the negative.

答:
Further to Deepak's answer: so the mnemonic that I used when growing up was "All Science Teachers are Crazy".

But beyond mnemonics, it is instructive to study the following diagram (noting that angles are positive if and only if measured anti-clockwise)

and understand that the +/- sign of any trigonometric ratio is derived from the +/- signs of the corresponding $x$- and $y$- coordinates.

E.g., $\displaystyle \cot(100°)=\frac{\cos(100°)}{\sin(100°)},\quad$ so $\cot(100°)$ is $\frac-+=$ negative.


问:
If $A + B$ is invertible, do we know if $A$ and $B$ are invertible separately?
If $A$ and $B$ are $n\times n$ matrices and we know that $A + B$ is invertible, can we assume that $A$ and $B$ are invertible by themselves?

答: No. Consider the counterexample $A = 0_{n\times n}$ and $B = I_{n}$.


答:
Clearly, since there are two free variables, $S$ is two-dimensional.  Since $\mathbb R^4$ is four-dimensional, we are looking for a two-dimensional space (since$S\oplus S^\perp\cong\mathbb R^4$).

But, also quite clearly, $(1,1,0,0)$ and $(0,0,1,1)$ are linearly independent elements of $S^\perp$.  Therefore they form a basis.

Title: Orthogonal Complement to arbitrary matrix

Body:
Find the orthogonal complement of the subspace

$$S = \{(x,y,z,w)\in\mathbb{R}^{4} \mid x + y = z + w = 0\}$$

I was working through some homework problems, stumbled onto this one and struggled with what to do.

So far, I have created an arbitrary matrix $A$ such that $A = (x,-x,z,-z)$ and I know that the orthogonal complement of $A$ is equals to the null space of $A$ transpose but I am struggling to find the nullspace of $A$ transpose with those arbitrary elements


问:
Suppose a series $a_n$ is greater than 0 for all positive integer n, and that $\sum \frac {a_n} n$ converges, then does the following also converge?
I was wondering if the following is true.

Suppose a series $a_n$ is greater than 0 for all positive integer n, and that $\sum \frac {a_n}n$ converges, then is $\displaystyle \lim_{m\to \infty}\sum_{n= 1}^m {a_n \over m+n} = 0$?

It seems to be true because if $\sum {a_n\over n}$ converges, then that means that ${a_n\over n }\to 0$ for $n\to \infty$. This means, neglecting $n$, ${a_n \over m+n}$ will also tend to 0, and thus the summation would be equal to 0, but I don't know if this is true.

答:
let $v_n = \frac{a_n}{m+n}$, and $u_n = \frac{a_n}{n}$. Now observe that $\frac{a_n}{m+n} \le \frac{a_n}{2n}$, so by comparison test, we can say that as each term of $v_n$ is smaller or equal to that of $\frac{u_n}{2}$, and as $u_n$ converges, so $v_n$ also converges as $m \to \infty$.

I think this is it.

:)


问:
Continuity problem - bounding $|f(x)|\leq|x|$
Let $f$ be a function such that $|f(x)|\leq|x|$ for all $x\in\mathbb{R}$. Prove that $f$ is continuous at the origin.

So far, I have that in order for $f$ to be continuous at the origin we need a $\delta$ such that $|x|<\delta\Rightarrow |f(x)|<\epsilon$, by the definition of continuity. Surely we can simply take $\delta=\epsilon$ in this case and the claimed result follows immediately?

答:
As is mentioned in the comments, $|f(x)| \leq |x|$ implies $f(0)=0$. Now let $\varepsilon>0$. Notice that

$$|f(x)-f(0)|=|f(x)| \leq |x|=|x-0|.$$

Therefore, if $|x|<\delta=\varepsilon$, we obtain that $|f(x)-f(0)|<\varepsilon$. This prove that $f$ is continuous at $x=0$.


问:
Find the negation of the definition of uniform continuity
We know that:
A function $f$ is uniformly continuous on a set $\mathbb{E}$ $\subseteq$ $\mathbb{R}$ if:
For every $\varepsilon$ $>$ 0, there is a number $\delta$ $>$ 0 such that for all x, x' $\in$ $\mathbb{E}$, $|f(x')-f(x)|$ $<$ $\varepsilon$ whenever | − ′| < .

Written using symbols: ($\forall$ $\varepsilon$ $>$ 0)($\exists$ $\delta$ $>$ 0)($\forall$ x, x' $\in$ $\mathbb{E}$)($|x-x'|$ $<$ $\delta$ $\Longrightarrow$ $|f(x')-f(x)|$ $<$ $\varepsilon$) (is it correct?)

The statement's negation is: A function $f$ is not uniformly continuous on a set $\mathbb{E}$ $\subseteq$ $\mathbb{R}$ if there exists $\varepsilon$ $>$ 0 such that for every $\delta$ $>$ 0 there are x, x' $\in$ $\mathbb{E}$ with $|x-x'|$ $<$ $\delta$ $\Longrightarrow$ $|f(x')-f(x)|$ $<$ $\varepsilon$. (?)

Written using symbol: $|x-x'|$ $<$ $\delta$ $\wedge$ $\sim$ $|f(x')-f(x)|$ $<$ $\varepsilon$ (?)

Please check if the answers are correct. Thank you so much!

答:
You are almost there! You have only forgotten to deny the last statement.

Since $p\to q$ is equivalent to $\neg p\vee q$, the negation of $p\to q$ is $p\wedge\neg q$.

At your case, we say that $f$ is not uniformly continuous on $E\subseteq\mathbb{R}$ if there exists $\varepsilon > 0$ s.t. for every $\delta > 0$ there are $x\in E$ and $y\in E$ with $|x - y| < \delta$ and $|f(x) - f(y)|\geq \varepsilon$.


问:
Why can we think the divergence and the curl as some kind of "derivative"?
My question arises from think about why the divergence theorem and the stoke's theorem are considered generalizations of the fundamental theorem of calculus.</p>

答:
Both of them give information about the way a function is changing locally.

"Locally" means:

knowing the value of the function at a point is not enough to know the value of the "derivative" at that point, but

knowing the value of the function everywhere in some open neighborhood of the point is enough, no matter how small that open neighborhood is.


问:
A proof about a spectrum radius inequality
Let U$_i$ $\in$ $\Bbb R^{n×v_i}$ be a matrix composed of unit orthogonal vectors, U = diag{U$_i$} i $\in$ {1,2,...N}. For instance
\begin{matrix}
U_1=\begin{bmatrix}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{bmatrix} \qquad U_2 =\begin{bmatrix}
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}  \\
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}   \\
0 & 0
\end{bmatrix}
\end{matrix}
then $$
    U =\left [ \begin{matrix}
    0 & 0 & 1 & 0 & 0 \\
    1 & 0 & 0 & 0 & 0 \\
    0 & 1 & 0 & 0 & 0 \\
    0 & 0 & 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\\
    0 & 0 & 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
    0 & 0 & 0 & 0 & 0
   \end{matrix} \right ]
$$

So is that true for any matrix $\phi$ whose spectrum radius is less than 1, the spectrum radius of $U^ \mathrm{T}$ $\phi$ U is less than one. i.e. $\rho$($\phi$) < 1 $\Longrightarrow$ $\rho$($U^ \mathrm{T}$ $\phi$ U) < 1

答:
Here is a counter example:
$$
  U=\pmatrix{1 \cr 0},\quad\text{and}\quad   \phi=\pmatrix{2 &2 \cr -2 & -2}.
  $$
We have  that $\rho(\phi)=0$, but $\rho(U^T\phi U)=2$.


答:
The FTC says (under some conditions) $f(b)-f(a)=\int_a^bf'(t)\,dt$. Or writing it more suggestively, we can say $f(b)-f(a)=\int_{[a,b]}df$. Or even more suggestively, we can write this as
\begin{align}
\int_{[a,b]}df = \int_{\partial [a,b]}f= \int_{\{a,b\}}f
\end{align}
What this is saying is that the integral over a region $[a,b]$ of the derivative $df=f'\,dt$ is somehow related to the "integral" of the original $f$ over the boundary of the interval, namely the two points $\{a,b\}$ (the minus sign has to do with orientation). In this 1-dimensional case, this is pretty overkill, but it sets the stage for higher dimensions.

Morally, this is what Stokes' theorem is about. The general case looks like
\begin{align}
\int_Md\omega = \int_{\partial M}\omega.\tag{$*$}
\end{align}
i.e on the LHS we have the integral of "some kind of derivative" $d\omega$ of an object $\omega$ over a region $M$. Stokes theorem then relates this to the integral of the original thing, $\omega$, but this time the integral is over the boundary $\partial M$.

In the classical vector calculus settings, we have the formulas
\begin{align}
\int_{\Omega}\text{div}(\mathbf{F})\,dV &= \int_{\partial \Omega}\mathbf{F}\cdot \mathbf{n}\,dA \tag{divergence theorem}\\
\int_{S}\text{curl}(\mathbf{F})\cdot \mathbf{n}\,dA &= \int_{\partial S}\mathbf{F} \cdot\mathbf{t}\,dl\tag{classical Stokes}
\end{align}
where $\mathbf{F}$ is a vector field on $\Bbb{R}^3$, $\Omega$ is some region in $\Bbb{R}^3$, and $S$ is some "surface" in $\Bbb{R}^3$. Again, the divergence and curl are given by some formulas involving partial derivatives. Morally speaking, this has the same meaning as above. The LHS is the integral of "some kind of derivative" over a region. The RHS is the integral of the "original" thing over the boundary.

So Stokes theorem roughly says "integration allows us to relate derivatives inside a region to the behavior on the boundary", and this is a direct generalization of the Fundamental theorem of calculus. In fact (*), which is called the generalized Stokes theorem encompasses all these theorems:

The FTC is the special case of a $1$-dimensional region of integration (the interval $[a,b]$) and $0$-dimensional boundary (the two boundary points $a,b$).

The classical Stokes theorem is a special case of $(*)$ with a $2$-dimensional surface with a $1$-dimensional boundary.

The classical divergence theorem is a special case of a $3$-dimensional volume with a $2$-dimensional boundary.

Title: Why can we think the divergence and the curl as some kind of "derivative"?

Body: My question arises from think about why the divergence theorem and the stoke's theorem are considered generalizations of the fundamental theorem of calculus.</p>


问:
Why is the function either maximum or minimum when the constraint is tangential to the function contour?
Given a function $f(x,y)$ and a constraint $g(x,y)=c$, Assuming the fact that the function either maximum or minimum when the constraint is tangential to the function contour, I get why the gradients are proportional, namely $\nabla f = \lambda \nabla g$ . But I don't understand in the first place why the function attains its critical points on the constraint only when they are tangential.

答:
From Wikipedia:

"The method of Lagrange multipliers relies on the intuition that at a maximum, $f(x,y)$ cannot be increasing in the direction of any such neighboring point that also has $g(x,y)=c$. If it were, we could walk along $g(x,y)=c$ to get higher, meaning that the starting point wasn't actually the maximum. Viewed in this way, it is an exact analogue to testing if the derivative of an unconstrained function is $0$, that is, we are verifying that the directional derivative is $0$ in any relevant (viable) direction."

Note that the equations of the Lagrange Multiplier method are necessary but not sufficient conditions for an optimum (on a boundaryless domain). If they have multiple solutions, you will have to compare the solutions further and/or examine their Hessians.


问:
Is the set of points ${i/n},\,n=1,2,3\ldots$ in the complex plane a closed set?
A set is said to be closed if it contains all of its boundary points, where a boundary point of a set in the complex plane is any point such that every neighborhood of the boundary point contains at least one point in the set and at least one point not in the set. At first glance, I assumed that the set $S$ defined by $S=\{ i/n\,\vert \,n\in\mathbb{Z}^+\}$ is closed, since each point $i/n$ is a boundary point of the set and is an element of the set. However, the more I think about it, the more I feel that the complex number $z=0$ is also a boundary point of the set, since any neighborhood of $0$ must contain a complex number that is arbitrarily close to $0$ on the positive imaginary axis (and therefore in $S$) and $0$ itself (which is not in $S$). Since this boundary point $z=0$ is not contained in the set, $S$ does not contain all of its boundary points, and is therefore not closed. I know that this is not a rigorous proof of my claim, but is this reasoning correct?

答:
Yes, your reasoning is correct.

Let $(X,d_{X})$ be a metric space.

We say that a set $E\subseteq X$ is closed iff it contains all of its adherent points (i.e. $E\supseteq\overline{E})$.

Having said that, notice that
\begin{align*}
\lim_{n\to\infty}\frac{i}{n} = 0
\end{align*}
so we can conclude (due the sequential characterization of adherent point) that $0$ is an adherent point of $S$.

On the other hand, $0\not\in S$. Consequently, the given set is not closed.

Hopefully this helps!


问:
Prove $(\cos(2\pi t), \sin(2\pi t))$ is injective for $0\leq t \lt 1$
Prove $(\cos(2\pi t), \sin(2\pi t))$ is injective for $0\leq t \lt 1$

Intuitively it makes sense - this traces out the unit circle - but how we do we show this from the definitions / rigourously?

答: If $(\cos2\pi t,\sin2\pi t)=(\cos2\pi s,\sin2\pi s)$, then we have $2\pi t=2\pi s+3\pi/2\implies t=s+3/4$, and also $2\pi t=2\pi s+\pi/2\implies t=s+1/4$.  Then $s+3/4=s+1/4\implies 1/4=3/4 \Rightarrow \Leftarrow$.


答:
Yes, your reasoning and your conclusion are correct.

Rigorous? Basically. I guess you could add in the detail of showing that any open disk around $0 \in \mathbb{C}$ contains points in your sequence. Such a disk is defined by the inequality $|z| < \varepsilon$ for some $\varepsilon > 0$. Now for any $n > \tfrac{1}{\varepsilon}$,
$$
\biggl\lvert \frac{i}{n} \biggr\rvert = \frac{1}{n} < \varepsilon,
$$
so the point $\tfrac{i}{n}$ in your set is in the neighborhood of $0$.

Title: Is the set of points ${i/n},\,n=1,2,3\ldots$ in the complex plane a closed set?

Body: A set is said to be closed if it contains all of its boundary points, where a boundary point of a set in the complex plane is any point such that every neighborhood of the boundary point contains at least one point in the set and at least one point not in the set. At first glance, I assumed that the set $S$ defined by $S=\{ i/n\,\vert \,n\in\mathbb{Z}^+\}$ is closed, since each point $i/n$ is a boundary point of the set and is an element of the set. However, the more I think about it, the more I feel that the complex number $z=0$ is also a boundary point of the set, since any neighborhood of $0$ must contain a complex number that is arbitrarily close to $0$ on the positive imaginary axis (and therefore in $S$) and $0$ itself (which is not in $S$). Since this boundary point $z=0$ is not contained in the set, $S$ does not contain all of its boundary points, and is therefore not closed. I know that this is not a rigorous proof of my claim, but is this reasoning correct?


答: Just another (counter)example: let $1 \leq k < n$ and define $A : = \operatorname{diag}(1, \dots , \underbrace{1}_{k \text{th entry}} , 0, \dots, 0)$ and $B : = \operatorname{diag}(0, \dots, 0 , \underbrace{1}_{(k +1)\text{st entry}}, \dots, 1)$.

Title: If $A + B$ is invertible, do we know if $A$ and $B$ are invertible separately?

Body: If $A$ and $B$ are $n\times n$ matrices and we know that $A + B$ is invertible, can we assume that $A$ and $B$ are invertible by themselves?


问:
How is the following relation transitive?
Let $A$ be a set and let $\omega$ be a subset of $P(A)$ (power set). Suppose that the elements of $\omega$ are pairwise disjoint. Then, the relation $\sim \omega$ associated with $\omega$ is transitive. I have tried taking example sets and drawing things around but just can't understand the theorem. Could anyone please explain with a simple example? This is what I am learning in high school right now

答:
Suppose $a\sim_\omega b$ and $b\sim_\omega c$, so that there are subsets $S,T\in\omega$ such that $a,b\in S$ and $b,c\in T$. If $S\ne T$, then since $S\cap T\ni b$ is nonempty, this contradicts the assumption that elements of $\omega$ are pairwise disjoint.

Thus, $S=T$ and $a\sim_\omega c$.


问:
Adaptation to Banach–Mazur theorem
I'm trying to prove the following:

For every normed linear space $X$, there exists a isometric ismorphism of $X$ in $C(K)$, where $K$ is a compact space.

I know from Banach–Mazur theorem that every separable Banach space satisfies the problem, so I tried to take a look at the demonstration of this theorem to have any idea how to solve this, but to prove that the function defined on the sketch of the proof of the theorem is in fact a isometry, we have to use Hahn-Banach, since the space $X$ is not necessary complete, I can't use this here and hence I'm stuck.

Thanks for any help!

答:
For the beginning the theorem should be read as follows:

Theorem For every normed space $X$ there is isometric embedding into $C(K)$ for some compact space $K$.

Proof. Consider $K:=\operatorname{Ball}_X(0,1)$ with the weak-$^*$ topology. By the Banach-Alaoglu theorem it is compact. By definition of the weak-$^*$ topology the map
$$
J(x): K\to\mathbb{C}:f\mapsto f(x)
$$
is continuous for each $x\in X$. So we have a well-defined map
$$
J:X\to C(K):x\mapsto J(x)
$$
Note that
$$
\Vert J(x)\Vert=\sup\{|J(x)(f)|:f\in K\}=\sup\{|f(x)|:f\in \operatorname{Ball}_X(0,1)\}=\Vert x\Vert
$$
for each $x\in X$. In the last step we use a corollary of the Hahn-Banach theorem. Thus $J$ is an isometric embedding.

Note that the corollary of the Hahn-Banach theorem does not require completeness, so its usage was valid in the proof above.

Title: Adaptation to Banach–Mazur theorem


问:
Do all rationals $\left ( \mathbb{Q} \right )$ form a perfect set if the metric space is $\mathbb{Q}$?
The following question came to my mind while reading the definition of a perfect set from Walter Rudin's Principles of Mathematical Analysis, 3rd Edition, Page 32, Section 2.18(h):

If the metric space is the set of all rational numbers $\mathbb Q$ (instead of $\mathbb R^1$ -- the set of all real numbers), then does $\mathbb Q$ constitute a perfect set?

In my understanding, the answer is "yes". Because $\mathbb Q$ is closed (since $\mathbb Q$ contains each of its limit points if the underlying metric space is $\mathbb Q$) and because every point of $\mathbb Q$ is a limit point of $\mathbb Q$ (arbitrary proximity). I'm not feeling so much sure somehow, and didn't find a corroboration elsewhere. Therefore this query.

Would appreciate help. Thanks.

答:
Yes, that's exactly right.

Of course, most applications of perfect sets (that I'm aware of anyways) require the ambient space to be "reasonable," usually completely metrizable, and in such spaces perfect sets are very unlike $\mathbb{Q}$ (e.g. uncountable). But if we do take $\mathbb{Q}$ to be our ambient space, then yes, in that context $\mathbb{Q}$ is perfect.

(Note incidentally that perfectness makes sense, not just in the context of metric spaces, but more general topological spaces.)


问:
How can I take the complex logarithm of this equation?
I have
$$
e^{i\beta}=\pm\frac{g_1}{\sqrt{1-g_0^2}}
$$
where $g_0$ is a real value and $g_1$ is complex. I'm trying to solve for $\beta$ in the equation, but I'm not quite sure how I can do that. I tried to take the logarithm so that
$$
\beta =-i \ln \left(\pm\frac{g_1}{\sqrt{1-g_0^2}}\right)
$$
However, the answer looks like
$$
\beta'=-i\log\frac{g_1}{\sqrt{1-g_0^2}}+k\pi,\ \text{where}\ k\in\mathbb{Z}
$$
I'm wondering where does the $\pm$ sign go? Why there's a $+k\pi$ term? Is that relevant to the Riemann surface? Thanks:)

答:
Since $e^{i \theta} = e^{i(\theta + 2k\pi)}$ for any $k \in \mathbb{Z}$, when you solve an equation such as
$$
e^{i \theta} = w
$$
for $\theta \in \mathbb{C}$, you get one value for each integer. (The complex logarithm is multivalued.) However, in your equation, since you have the $\pm$ sign, you get twice as many solutions! To wit,
$$
e^{i(\theta + (2k+1)\pi)} = e^{i(\theta + 2k\pi) + i\pi}
= e^{i(\theta + 2k\pi)} e^{i\pi} = -e^{i(\theta + 2k\pi)}.
$$

Therefore, when you're solving an equation such as $e^{i\theta} = \pm w$, you get both the even and odd multiples of $\pi$, i.e. all the integer multiples of $\pi$.


问:
Given a parabolic shape with maximum height $OC=8m$ and maximum width $AB=20m$. If $M$ is the middle of $OB$, then what is the height $MK$, from $M$?
Given a parabolic shape with maximum height $OC=8m$ and maximum width $AB=20m$. If $M$ is the middle of $OB$, then what is the height $MK$, from $M$?

I attempted to solve the question as follows:

$OB=10m$

Hence, $OM=5m$

$OC=8m$

I state that $O$ is point $(0,0)$ on the Cartesian coordinate system.

$\implies M(5, 0), C(0, 8), B(0, 10), A(-10, 0)$. I then attempted to find the coordinates of K, but I don't know how to. Obviously, $\triangle AKB$ is isosceles, but I couldn't see anything else further than that. Could you please explain to me how to solve this question?

答:
Turning Point form of equation for parabola is $y=a(x-h)^2+k$

$C$ is TP, so $h=0, k=8$.

$A$ is at $(-10,0)$, subst for $x, y$

So $0=a(0-10)^2+8 \rightarrow a=-\frac {2} {25}$

$M$ is at $x=5 \rightarrow  y=-\frac {2} {25}5^2+8$

Height $MK$ is 6m


问:
Find the value of $T=\mathop {\lim }\limits_{n \to \infty } {\left( {1+ \frac{{1+\frac{1}{2}+ \frac{1}{3}+ . +\frac{1}{n}}}{{{n^2}}}} \right)^n}$
I am trying to evaluate

$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n}.$$

My solution is as follow

$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n} \Rightarrow T = {e^{\mathop {\lim }\limits_{n \to \infty } n\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}} - 1} \right)}} = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{n}} \right)}}$$

$$T = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{2n}} + \frac{1}{{3n}} + \cdots + \frac{1}{{{n^2}}}} \right)}} = {e^{\left( {0 + 0 + \cdots + 0} \right)}} = {e^0} = 1$$

The solution is correct  but I presume my approach $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdot + \frac{1}{n}}}{n}} \right) \Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{2n}} + \frac{1}{{3n}} + \cdot + \frac{1}{{{n^2}}}} \right) = 0$ is wrong.

Is there any generalized method

答:
Consider the partial term
$$T_n=\left(1+\frac{H_n}{n^2}\right)^n\implies \log(T_n)=n \log\left(1+\frac{H_n}{n^2}\right)$$
For large $n$
$$H_n=\log (n)+\gamma +\frac{1}{2 n}-\frac{1}{12 n^2}+\frac{1}{120
   n^4}+O\left(\frac{1}{n^6}\right)$$
$$\log\left(1+\frac{H_n}{n^2}\right)=\frac{\log (n)+\gamma }{n^2}+\frac{1}{2 n^3}+O\left(\frac{1}{n^4}\right)$$
$$\log(T_n)=\frac{\log (n)+\gamma }{n}+\frac{1}{2 n^2}+O\left(\frac{1}{n^3}\right)$$
$$T_n=e^{\log(T_n)}=1+\frac{\log (n)+\gamma }{n}+\frac{(\log (n)+\gamma )^2+1}{2
   n^2}+O\left(\frac{1}{n^3}\right)$$ which shows the limit and also how it is approached.

Moreover, this give a shortcut method for the evaluation of $T_n$. Suppose $n=10$. The exact value is
$$T_{10}=\left(\frac{259381}{252000}\right)^{10} \sim 1.334677$$ while the above truncated series gives $\sim 1.334446$ (relative error $=0.017$%).


问:
Power-Law PDFs and Expectations
A random variable $X$ follows the distribution:
$$f_X(x)=\begin{cases}
  Cx^2 & -1\le x\le 2,\\
  0 & \text{otherwise},
  \end{cases}$$

and $Y=X^2$. Calculate

$C$

$P(X\ge0)$

$E[Y]$

$V(Y)$

I know that I need to utilize the Power-law of PDFs but as soon as this went to integrals and derivatives I am lost and don't understand how these equations simplify.  This is my first time looking at derivatives and without explaining each step of them I'm lost immediately.

Any help would be appreciated.  Does it have to do with the equations below:
$$\int_0^1(a+1)x^adx=x^{a+1}|_0^x=1$$

I am at a loss for how to find the $C$ (I assume this means the $constant$.)
But if we take the $\frac{d}{dx}x^2$ that should be $2x^{2-1}=2x$. So the $C=2$?

Does this follow the formula: $$P(X\ge a)=P(X\gt a)=1-F(a)$$
If this is the case than:
$$
\begin{aligned}
F(a)&=\int_{-\infty}^af(a)du=0 \\
F(0)&=\int_{0}^{\infty}f(0^2)dx  \\
&=1-0=1
\end{aligned}
$$

So then $P(X\ge0)=1$?

To find the expected value for x: $$E(X)=\int xf(x)dx$$ correct?

答:
In order to find $C$, we need to utilize this fact:

The integral of the probability distribution function (pdf) over its domain is equal to 1.

Therefore, in mathematical notation, we have-
$$\int_{-\infty}^{\infty} f(x)dx = 1. $$

Now, the function $f$ that we are dealing with has non-zero output (function value) only in the interval $[-1,2]$. Therefore, we split up the integral as follows:
$$ \begin{align}1 &= \int_{-\infty}^{\infty} f(x)dx = \int_{-\infty}^{-1} f(x)dx + \int_{-1}^{2} f(x)dx + \int_{2}^{\infty} f(x)dx\\ &=  \int_{-\infty}^{-1} 0\ dx + \int_{-1}^{2} Cx^2 dx + \int_{2}^{\infty} 0 \ dx \\ &=  0+ C\int_{-1}^{2} x^2 dx + 0 \textrm{, as constants are not affected by integrals}\\&= C\frac{x^3}{3}\bigg|^2_{-1} \textrm{, since the power law says:} \int x^n dx = \frac{x^{n+1}}{n+1} \textrm{ for } n \geq 0 \\&= C \left(\frac{2^3}{3} - \frac{(-1)^3}{3}\right) \\ &= C\left( \frac{8}{3} - \frac{-1}{3} \right)\\ &= C\left( \frac{9}{3}\right) = 3C \end{align}$$
Thus, $3C = 1$, and hence, $C = \frac{1}{3}$.

For the second question, you can use this fact-

$ P(a \leq X \leq b) = \int_a^b f(x) dx$, for every random variable $X$ with pdf $f$.

Here, $P(X\geq 0) = P(0 \leq X \leq \infty)$. Therefore, $ P(X\geq 0) = \int_0^{\infty} f(x) dx.$ So then, we have -
$$ \begin{align} P(X\geq 0) &= \int_0^{\infty} f(x) dx\\ &=\int_0^{2} f(x) dx + \int_2^{\infty} f(x) dx \\&= \int_0^{2} \frac{1}{3}x^2 dx + \int_2^{\infty} 0\ dx.\end{align} $$
I hope you can solve it further using the power law as we did in question 1.

Title: Power-Law PDFs and Expectations


问:
Prove Fibonacci sequence and find partial fraction decomposition
We have a function $z/(1-z-z^2)$ which has simple poles at $z=(-1\pm \sqrt5)/2$.

Furthermore we have the powerseries $\frac{z}{1-z-z^2}=\sum_{n=0}^\infty{F_nz^n}$. I have calculated it's radius of convergence to be $(\sqrt5-1)/2$.

We have to prove that $F_0=0, F_1=1, F_n=F_{n-1}+F_{n-2}, n\ge2$. (The sequence of Fibonacci numbers). I really need some help on this one, I don't really have any idea how to it!

Next, I have to find the partial fraction decomposition of the rational function. From what I understand, you calculate it like this:$$\frac{r(z)}{q(z)}=\sum_{j=1}^k\sum_{l=1}^{m_j}\frac{c_{j,l}}{(z-a_j)^l}$$ where $a_1,...,a_k$ are the zeros of q having multiplicities $m_1,...,m_k$. But then I get $$\frac{z}{1-z-z^2}=\frac{c_1}{z-(\frac{-1+\sqrt5}{2})}+\frac{c_2}{z-(\frac{-1-\sqrt5}{2})}$$, and I can't seem to calculate it in a way that makes sense.

答:
For the first part, let $f(z) = \dfrac{z}{1-z-z^2}$.

Note that $(1-z-z^2)f(z) = z$. Hence

$$\sum_{n=0}^\infty F_n z^n - \sum_{n=1}^\infty F_{n-1} z^{n} - \sum_{n=2}^\infty F_{n-2} z^{n} = z.$$

Equating powers, you get that $F_0 = 0$, $F_1 - F_0 = 1$ and $F_n - F_{n-1} - F_{n-2} = 0$ for $n \ge 2$.

For the second part, see the hints in the comments.


问:
Show that the $p$-series $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent iff $p \gt 1$.
Show that the $p$-series $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent iff $p \gt 1$.

Attempt:
For the right direction:
Let $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent. Then, $\lim\limits_{n\to \infty} \frac{1}{n^p} = 0$.
In particular, $\lim\limits_{n\to \infty} \frac{1}{n^p} = 0$ for $p \gt 0$. But, for $0 \lt p \le 1$, the series $\sum_{n=1}^\infty \frac{1}{n^p}$ is divergent. Hence, we must have $p \gt 1$. $\Box$

Proof that series $\sum_{n=1}^\infty \frac{1}{n^p}$ is divergent:
We know that $n^p \le n$ for all positive integers $n$ and $0 \lt p \le 1$.
Then,
\begin{equation*}
\frac{1}{n} \le \frac{1}{n^p}.
\end{equation*}
Now, since the partial sums of the harmonic series are not bounded, this
inequality shows that the partial sums of the $p-$series are not bounded when $0 \lt p \le 1$. Hence, the $p-$series diverges for theses values of $p. \Box$

For the left direction:
Let $p \gt 1$. Let $f(x) = \frac{1}{x^p}$ for all $x \in [1.\infty)$. Then, $f$ is a positive, continous, and decreasing sequence. Hence, we can apply the Integral Test here. Notice that
\begin{align*}
\int_1^\infty \frac{1}{x^p} &= \lim\limits_{b \to \infty} \int_1^b \frac{1}{x^p} \\
&= \lim\limits_{b\to \infty} \left[\frac{1}{1-p}x^{1-p}\right]_1^b \\
&= \lim\limits_{b \to \infty} \left( \frac{1}{1-p}b^{1-p} - \frac{1}{1-p} \right) \\
&= \frac{1}{p-1} \lim\limits_{b \to \infty} \left(1 - \frac{1}{b^{p-1}} \right) \\
&= \frac{1}{p-1}.
\end{align*}
Thus,
\begin{equation*}
\int_1^\infty \frac{1}{x^p}dx
\end{equation*}
is convergent. Consequently, $\int_1^\infty \frac{1}{n^p}dn$ is convergent. By the Integral Test,
we have that
\begin{equation*}
\sum_{n=1}^\infty \frac{1}{n^p}
\end{equation*}
is convergent. $\Box$

Am I correct, especially for the right direction ?

答:
Here's something you may use:

Consider $p \leq 0$ first. In this case, $\frac{1}{n} = n^{-1} \geq 0$. The limit of this implies that it doesn't converge to $0$ as $n \rightarrow \infty$ and so $\sum_{n = 1}^{\infty} \frac{1}{n^p}$ does not converge.

Now assume $p > 0$. Then $\frac{1}{n^p}$ is a decreasing sequence of positive real numbers. By applying the Dyadic criterion, we have:

$\sum_{n = 1}^{\infty} \frac{1}{n^p}$ converges iff $\sum_{n = 0}^{\infty} 2^n \frac{1}{(2^n)^p}$ converges.

And so $\sum_{n = 0}^{\infty} (2^{1-p})^n$.

Let $q = 2^{1 - p}$ then the sum converges iff $q = 2^{1 - p} < 1$, which is iff $p > 1$.


Body:
Given a parabolic shape with maximum height $OC=8m$ and maximum width $AB=20m$. If $M$ is the middle of $OB$, then what is the height $MK$, from $M$?

I attempted to solve the question as follows:

$OB=10m$

Hence, $OM=5m$

$OC=8m$

I state that $O$ is point $(0,0)$ on the Cartesian coordinate system.

$\implies M(5, 0), C(0, 8), B(0, 10), A(-10, 0)$. I then attempted to find the coordinates of K, but I don't know how to. Obviously, $\triangle AKB$ is isosceles, but I couldn't see anything else further than that. Could you please explain to me how to solve this question?

Body:
From A(-10,0)and B(10,0),  x=-10 and x=10 are two roots:

\begin{align}y(x)=a(x+10)(x-10)\end{align}

From C(0,8), a= -$\frac{8}{100}$

\begin{align}y(x) =-\frac{8}{100}(x+10)(x-10)\end{align}
\begin{align}y(5) =-\frac{8}{100}(5+10)(5-10)=6 \end{align}

K's coordinates (5 cm,6 cm)


问:
Show that a torus is conformally equivalent to a plane
With the metric of the torus given by $ds^2=a^2d\theta^2+(b+asin\theta)^2d\phi^2$, I'm asked to find the conformal transformation which proves that a torus is conformally equivalent to a plane.

I must then find a transformation $\bar{g}_{\mu \nu}=exp(2\Phi)g_{\mu \nu}$ (as far as I know, the exponential is there just to make sure the thing is invertible, but that's the notation my professor is using) such that $\bar{R}_{\lambda \mu \nu \rho}=0$, and the new coordinates satisfy $d\bar s ^2=d\bar\theta^2 + d\bar\phi^2$ (hence rendering the metric equivalent to that of a plane).

I don't know what approach to use. I tried taking the curvature, which I calculated, and finding a variable change that would make $R_{\theta \phi \theta \phi}=\frac{b}{a}sin\theta +sin^2\theta=0$, the only non-null element of the curvature tensor. However, I saw myself wasting a lot of time there and getting me nowhere.

Then I tried using the line element $ds^2=a^2d\theta^2+(b+asin\theta)^2d\phi^2$ and try to fit a transformation to $\bar\theta$ and $\bar\phi$ such that $ds^2$ remained multiplied by some function $\Omega(\theta,\phi)$, so that $d\bar s^2=\Omega(\theta, \phi)ds^2$, which would be consistent with the "locally dilation of the metric"... But I wouldn't know how to find $\Phi$ from there.

Any help on how to work this out will be much appreciated!

答:
HINT: Note that
$$d\theta^2 + f(\theta)^2\,d\phi^2 = f(\theta)^2\left(\left(\frac{d\theta}{f(\theta)}\right)^2 + d\phi^2\right).$$


问:
Simplifying $a(a-2) = b(b+2)$
I reduced a number theory problem to finding all ordered pairs $(a,b)$ that satisfy the equation $a(a-2) = b(b+2)$ in a certain range. After thinking about this for a while, I figured that either $a = b + 2$, $a = -b$, $a = b = 0$ or $a = 2 \text{ and } b = -2$. It is easy to prove that these values for $a$ and $b$ will all satisfy the equation, but how would I solve this equation had I not come up with these solutions? And how do I know I did not miss one? In short, can this equation be solved (more) rigorously?

Edit:
Silly of me I did not recognize that my last two solutions were already covered by my first two.

Title: Simplifying $a(a-2) = b(b+2)$

Body: Put $\, b=x.\, f(x) = x^2\!+\!\color{#c00}2x = a^2\!-\!2a= f(-a)$ has roots $x=-a,\ x'\!=a\!-\!2,$ by $\, x\!+\!x'\!=-\color{#c00}2.$


问:
Show that sample variance is unbiased and a consistent estimator
I am having some trouble to prove that the sample variance is a consistent estimator. I have already proved that sample variance is unbiased. I understand that for point estimates T=Tn to be consistent if Tn converges in probably to theta. However, I am not sure how to approach this besides starting with the equation of the sample variance. Any help would be greatly appreciated. Thank you in advance.</p>

答:
Sample variance is an unbiased estimator of population variance(in iid cases) no matter what distribution of the data.
Now to prove consistency, only need to show variance of sample variance goes to 0 as n goes to infinity, which is true as long as the forth moment of data is finite.</p>


问:
Proving the parallelity of two lines in a convex Pentagon
Can someone show me how to solve this Geometry Problem? I feel like there is a trick behind it that I don’t see

The convex pentagon $ABCDE$ has a circumference and $\overline{AB} = \overline{BD}$ applies. Let the point $P$ be the intersection of the diagonals $AC$ and $BE$. The straight lines $BC$ and $DE$ intersect at point $Q$.
Prove that the line $PQ$ is parallel to the diagonal $AD$.

答:
Observe that,  $\triangle AEP\sim \triangle BEQ$ and hence $\angle APE=\angle BQE$. From here,  it is seen that quadrilateral $PCQE$ is cyclic.

Thus, $\angle CPQ=\angle CEQ=\angle CED=\angle CAD$ and therefore $PQ\parallel AD$.


问:
Cyclic subgroups of Torus
Consider a torus $T_2:= S^1 \times S^1$. Fix integers $a,b$ such that gcd(a,b)=1. We define the circle $S^1(a,b) \subset T_2$ via the following embedding

\begin{align}
S^1 &\hookrightarrow S^1 \times S^1 \\
t &\mapsto (t^a,t^b)
\end{align}

Further we define the finite cyclic group $\mathbb{Z}_n(a,b)$ to be the standard $\mathbb{Z}_n \subset S^1(a,b)$ i.e the cyclic group defined via the embedding

\begin{align}
\mathbb{Z}_n &\hookrightarrow S^1 \times S^1 \\
\zeta &\mapsto (\zeta^a,\zeta^b)
\end{align}

where $\zeta$ is a primitive $\text{n}^{\text{th}}$ root of unity.

Given $\mathbb{Z}_n \subset S^1(a,b)$, under what conditions on $a^\prime, b^\prime$ is $\mathbb{Z}_n \subset S^1(a^\prime, b^\prime) \cap S^1(a,b)$.

I have tried working out several examples with different values for $n,a,b$, and for those values I do get that if $a^\prime \equiv a ~(mod ~n~)$ and $b^\prime \equiv b ~(mod ~n~)$ then
$\mathbb{Z}_n \subset S^1(a^\prime, b^\prime) \cap S^1(a,b)$.

Is $a^\prime \equiv a ~(mod ~n~)$ and $b^\prime \equiv b ~(mod ~n~)$ a necessary and sufficient condition for $\mathbb{Z}_n \subset S^1(a^\prime, b^\prime) \cap S^1(a,b)$?

If this is indeed true then any help in proving it would be appreciated.

答:
The circle subgroup corresponding to the pair $(a,b)$ is an $(a,b)$-torus knot. Say we want to compute the intersection of the $(a,b)$- and $(c,d)$-torus knots. Write

$$ (\zeta^a,\zeta^b)=(\xi^c,\xi^d). $$

With $\zeta=\exp(2\pi i s)$ and $\xi=\exp(2\pi i t)$ (where $s,t\in\mathbb{R}$) this becomes

$$ \begin{cases} as\equiv ct \\ bs\equiv dt \end{cases} \mod1 $$

(that is, as elements of $\mathbb{R}/\mathbb{Z}$). We can do $d$ times the first minus $c$ times the second to obtain $(ad-bc)s\in\mathbb{Z}$, and similarly $a$ times the second minus $b$ times the first for $(ad-bc)t\in\mathbb{Z}$, which implies $\zeta,\xi$ must be $n$st roots of unity, where $n=ad-bc$.

The homomorphisms $(x^a,x^b)$ and $(x^c,x^d)$ have the same image when restricted to the $n$th roots of unity. To see this, it suffices to show we may solve the system

$$ \begin{cases}
am \equiv c \\ bm\equiv d
 \end{cases}
\mod n $$

By Bezout's we may pick $x,y$ so that $ax+by=1$. Then if we add $x$ times the first equation and $y$ times the second we get $m\equiv cx+dy\mod n$. This may be verified a valid solution, e.g.

$$ \begin{array}{ll}
 am & =a(cx+dy) \\
 & = c(ax)+(ad)y \\
 & = c(1-by)+(ad)y \\
 & = c+(ad-bc)y \\
 & \equiv c
\end{array} $$

and similarly $bm\equiv d$ mod $n$. Moreover, if $\zeta$ is a primitive $n$th root then $(\zeta^a,\zeta^b)$ is also a primitive $n$th root, again by Bezout's. Therefore, the intersection is cyclic of order $n=ad-bc$.

This shows $a\equiv c,b\equiv d$ mod $n$ is sufficient for the intersection to contain a cyclic subgroup of order $n$, but it is not necessary, e.g. for $(a,b)=(2,-1)$, $(c,d)=(1,2)$, $n=5$.

Title: Cyclic subgroups of Torus


问:
Unbounded sequence has an divergent subsequence
Show that if $(x_n)$ is unbounded, then there exists a subsequence $(x_{n_k})$ such that $\lim 1/(x_{n_k}) =0.$

I was thinking that $(x_n)$ is a subsequence of itself. WLOG, suppose $(x_n)$ does not have an upper bound. By Algebraic Limit Theorem, $\lim 1/(x_{n_k}) =0.$

Is there any flaws in my proof?

答:
Suppose that you have a subsequence such that

$$a_{n_{1}}>1$$

$$a_{n_{2}}>2$$ and so on such that $$a_{n_{k}}>k$$ which is justified as $a_n$ is unbounded. Thus you get a subsequence which diverges to infinity as $k\rightarrow \infty$.


问:
Centralizers in a monoid of transformations.
This comes from N. Jacobson's Basic Algebra 1, 1.4 #1:

Let $A$ be a monoid, $M(A)$ the monoid of transformations of $A$ into itself, $A_L$ the set of left translations, $A_R$ the right translations. Show that $A_R$ is the centralizer of $A_L$ in $M(A)$ and vice-versa. Further show that $A_L\cap A_R$ are translations by elements in the center of $A$

First I'll include what I have, which was absolutely no trouble at all. $(a_L\cdot b_R)$ is the transformation that maps $k \rightarrow akb$. This is the same as the transformation $(b_R\cdot a_L)$. So the left and right translations commute, which means $A_R \subset C(A_L)$ and $A_L \subset C(A_R)$ Similarly, if $c_L=c_R$ then $ca=ac, \forall a\in A$. So $c$ is in the $A$'s center. The other direction is no harder.

Where I am stuck is showing the other direction of the first two inclusions. I tried to start with the assumption that some transformation $j$ was in the centralizer of $A_L$ with the goal to show that $j$ is a right translation.

I got that if there was such a transformation $j$, it would have the following property by virtue of being in $C(A_L)$: $$\forall a, k \in A, j(ak)=a\cdot j(k)$$ Which makes me think that there may be some sweet choice of $a,k$. However, because we are inside a monoid of transformations, I can't come up with any good ideas for picking $a$ and $k$. $j$ doesn't have to be injective, nor surjective, A is not necessarily finite, ect... The only thing we really have in a monoid is an identity, and letting one or both of $a$ and $k$ be $1$ doesn't seem to lead anywhere.

答: You are almost there. Just take $k = 1$ in your formula to get $j(a) = aj(1)$, which shows that $j = c_R$ with $c = j(1)$.


问:
If $\vec a\in\mathbb{R}^3$ is non-zero, then $\left \{ \vec e_1\times\vec a,\vec e_2\times\vec a,\vec e_3\times\vec a \right \}$ is linearly dependent
Proof: write
$$\;\vec a=(\alpha,\,\beta,\,\gamma)=\alpha e_1+\beta e_2+\gamma e_3$$
then we have
$$\alpha \vec e_1\times\vec a+\beta \vec e_2\times\vec a+\gamma \vec e_3\times\vec a=(\alpha \vec e_1+\beta \vec e_2+\gamma \vec e_3)\times\vec a=\vec a\times \vec a=\vec 0$$
$$as\;(\alpha,\,\beta,\,\gamma)=\vec a\neq\vec 0$$
the set
$$\;\{\vec e_1\times \vec a,\, \vec e_2\times \vec a ,\, \vec e_3\times \vec a\}$$
is linearly dependent.
now we know the proposition is true, but if you look at the contraposition: if the set
$$\{\vec e_1\times \vec a,\, \vec e_2\times \vec a ,\, \vec e_3\times \vec a\}$$ is linearly independent then$$ \vec a=\vec 0$$
which means $\vec e_i \times\vec a=\vec 0,$ the set is linearly dependent, contradiction!

What's wrong with that?

答: All three cross-products are in the subspace  $\rm {span}\{a\}^\perp $, which is two-dimensional.


问:
Let $A, B$ be some sets such that $A\setminus B$ is infinite while $B$ countable or finite. Prove that $A\setminus B\sim A$
The problem is the following :

Let $A, B$ be some sets such that $A\setminus B$ is infinite while $B$ countable or finite. Prove that $A\setminus B\sim A.$

I got stucked in this problem by trying the following :

First, as $A\setminus B$ and $A\setminus B\subseteq A.$ By lemma, I can assume that there is an injection $f$ such $f: A\setminus B\to A.$ Thus, now I need to find an injection from $A\to A\setminus B,$ so I can use the Schröder–Bernstein theorem, that states :

If there's an injection $f$ such $f: A\to B,$ and there's another injection $g$ such $g: B\to A,$ then there is a bijection from $A$ to $B.$

As $A\setminus B$ is inifinite, there must exist a set $C$ countable such $C\subset A\setminus B.$

Now I may write that $A= (A\setminus B)\cup(C)\cup(A\cap B).$ Since $A\cap B\subseteq B,$ and $B$ is countable or finite, then $A\cap B$ is countable or finite as well, that means there exists an injection $A\cap B\to\mathbb{N}.$

But $C$ is countable, then there exists an injection $l$ such that $l: A\cap B\to C.$

That is, I found an injection from $A\cap B\to C,$ but how can I find now an injection $A\to A\setminus B$ in order to prove by the theorem ?

答:
$A\setminus B=A-A\cap B$.

If $A$ is countable , so is $A\setminus B$ as $A\setminus B\subseteq A$ and therefore map every $i$th element of $A\setminus B$ to $i$th element of $A$ to get a bijection.

If $A$ is uncountable then so is $A\setminus B$  because if not then $A=(A\setminus B)\cup (A\cap B)$ is countable which is a contradiction.

Since $A\setminus B$ is uncountable, it contains a countable (infinite) subset $C$.

$A\setminus B=(A\setminus B-C)\cup C$ and $A=(A\setminus B-C)\cup ((A\cap B)\cup C)= (A\setminus B-C)\cup X$, where $X=(A\cap B)\cup C$ is countable.

Define $f:A\setminus B\to A$ by $f(x)=x$, if $x\in A\setminus B-C$ and $f(x_i)=y_i\in X$ for $x_i\in C$. $f$ is one -one.


问:
Upper bound on integrand that converges to zero?
Let $H(x)$ be the cumulative distribution function for a continuous nonnegative random variable $X$ that has a finite mean.
Let $\tilde{H}(x) = 1 - H(x)$ (the tail probability).

For $r \geq 1$, suppose that the integral
$\int_{0}^{\infty} x^r \tilde{H}(x) \, dx$ is convergent. (Remark: This condition arises when the $(r+1)$th moment of $X$ is finite. While the integral is indeed convergent for $r=0$ by assumption of $X$ having a finite mean, this case is not being considered here.)

Does there necessarily exist a function $B(x)$ such that $\tilde{H}(x) \leq \displaystyle\frac{B(x)}{x^{r+1}}$ where $B(x) \to 0$ (monotonically decreasing) as $x \to \infty$?

Remarks

As this is my third attempt to formulate a proposition that can do what I’m seeking, here is what I have tried:

I am aware that I cannot necessarily claim that $x^r \tilde{H}(x) \to 0$ as this does not necessarily hold for improper integrals, even if they converge.  I can make that conclusion if it is established that $x^r \tilde{H}(x) \to \ell$ for some finite limit $\ell$, but the existence of $\ell$ does not necessarily follow solely from $x^r$ being monotone increasing to infinity and $\tilde{H}(x)$ being monotone decreasing to zero.

But if  $\tilde{H}(x) = \displaystyle\frac{B(x)}{x^{r+1}}$ with $B(x) = c>0$ constant then the integral
$\int_{0}^{\infty} x^r \tilde{H}(x) \, dx$ is divergent.

That $\tilde{H}(x)$ decreases monotonically to zero (from being a tail probability) seems to make all the difference. I tried constructing a counterexample for $r = 1$ in which
$$
  \displaystyle
  \int_{0}^{\infty} x \tilde{H}(x) \, dx
  =
  \int_{0}^{\infty} D(x) \, dx
$$
is convergent but $D(x)$ is nondecreasing. I thought to construct
$$
  \tilde{H}(x) = \frac{D(x)}{x}
$$
where $D$ is mostly zero except that at each integer $n$ there is a triangle of height 1 and area $1/n^2$ (so $\int_{0}^{\infty} D(x) \, dx$ converges but $D(x)$ does not decrease to zero). However
$$
  \tilde{H}'\!(x) = \frac{x D'\!(x) - D(x)}{x^2}
=
  \frac{1}{x}\left( D'\!(x) - \tilde{H}(x) \right)
$$
whenever $\tilde{H}(x)$ is differentiable.  Thus as $x$ increases and hence $\tilde{H}(x)$ decreases, the gradient that I can introduce via $D'\!(x)$ must also decrease to preserve $\tilde{H}'\!(x) \leq 0$.  So in particular, if I tried to specify that $D(n) = 1$ for each positive integer $n$ then I would be restricted to $D'\!(x) \leq 1/n$ for the triangle at $n$, which leads to a triangle of area $\geq 1/2n$.  Doing so causes $\int_{0}^{\infty} D(x) \, dx$ to diverge.

(Many thanks again.)

答:
This answer requires that $\tilde{H}(x)$ is differentiable on an interval $(a,\infty)$.  In the following analysis, $\tilde{H}(x)$ has been replaced with $f(x)$ to make the assumptions more explicit and traceable.  I have refactored the logic into small units to increase reusability.

Lemma 1. Let function $f(x)$ be real-valued and nonnegative on $(a,\infty)$, and suppose that for $r  > -1$ the integral
$\int_a^\infty x^r f(x) \, dx$ is convergent.
Write
$$\displaystyle f(x) = \frac{g(x)}{x^{r+1}}$$
Suppose further that $g(x) \to c$ as $x \to \infty$ where $c$ is finite. Then $c = 0$.

Proof of Lemma 1. For $x > a$, $f(x)$ is nonnegative so $g(x)$ is nonnegative hence $c \geq 0$.
Now suppose that $c > 0$ and let $\epsilon = \tfrac{1}{2}c$. Then $\epsilon > 0$, so from $g(x) \to c$:
$$
 \text{there exists}\ x' > a \ \text{such that if}\ x > x' \ \text{then}\ |g(x) - c| < \epsilon
$$
or equivalently
$$
 \text{there exists}\ x' > a \ \text{such that if}\ x > x' \ \text{then}\ c - \epsilon < g(x) < c + \epsilon
$$
so in particular
$$
 \text{there exists}\ x' > a \ \text{such that if}\ x > x' \ \text{then}\ g(x) > c - \epsilon = \tfrac{1}{2}c
$$
Therefore
$$\begin{align*}
  \int_a^\infty x^r f(x) \, dx
&=
  \int_a^\infty \frac{g(x)}{x} \, dx
\\&=
  \int_a^{x'} \frac{g(x)}{x} \, dx
  +
  \int_{x'}^\infty \frac{g(x)}{x} \, dx
\\&\geq
  \int_a^{x'} \frac{g(x)}{x} \, dx
  +
  \int_{x'}^\infty \frac{c}{2x} \, dx
\end{align*}$$
Now the second term on the righthand side diverges to infinity but the lefthand side is a convergent integral by assumption. Contradiction, therefore $c = 0$. $\blacksquare$

Lemma 2. Let function $f(x)$ be real-valued and nonnegative on $(a,\infty)$, and suppose that for $r  > -1$ the integral
$\int_a^\infty x^r f(x) \, dx$ is convergent.
Write
$$\displaystyle f(x) = \frac{g(x)}{x^{r+1}}$$
Suppose further that $f(x)$ is differentiable and nonincreasing on its domain.
Then there exists $c \geq 0$ (finite) such that $g(x) \to c$.

Proof of Lemma 2.

Let $\left\{ x_n \right\}_n$ be any strictly increasing sequence of numbers drawn from $(a,\infty)$. Construct $\left\{ g_n \right\}_n$ by putting $g_n = g(x_n)$ for each $n$. Proceeding in four steps:

It is sufficient to show that there exists $c$ finite such that $g_n \to c$ as $n \to \infty$. The function $f(x)$ is nonincreasing so its limit as $x \to \infty$ is either finite or $-\infty$. Furthermore $f(x)$ is nonnegative so the option of $-\infty$ is excluded and the limit must be nonnegative. Meanwhile the function $d(x) = x^{r+1} \to \infty$ as $x \to \infty$. Now $g(x) = f(x)d(x)$ therefore $g(x)$ either has a finite, nonnegative limit or it diverges to infinity. So suppose there exist sequences $\left\{ x_n \right\}_n$
and $\left\{ x'_m \right\}_m$ such that if $g_n = g(x_n)$ for each $n$ and $g'_m = g(x'_m)$ for each $m$ then $g_n \to c$ as $n \to \infty$ and $g'_m \to c'$ as $m \to \infty$ but $c \neq c'$. Construct sequence $\left\{ x''_k \right\}_k$ by merging $\left\{ x_n \right\}_n$ and $\left\{ x'_m \right\}_m$ in increasing order.  Then as $k \to \infty$, $g(x''_k)$ does not converge. Contradiction hence $c = c'$. Therefore if $g_n \to c$ finite as $n \to \infty$ for any given sequence $\left\{ x_n \right\}_n$ then $g(x) \to c$ as $x \to \infty$.

$\displaystyle g'(x) \leq (r+1)\frac{g(x)}{x}$ for all $x > a$. The function $f(x)$ is differentiable on $(a,\infty)$. Indeed
$$\begin{align*}
  f'(x) &= -\frac{r+1}{x^{r+2}}g(x) + \frac{1}{x^{r+1}}g'(x)
\\&=
  \frac{1}{x^{r+1}}\left(
    g'(x) - (r+1)\frac{g(x)}{x}
  \right)
\end{align*}$$
on that interval. Now for $x > a$, $f(x)$ is nonincreasing so $f'(x) \leq 0$. Meanwhile $1/x^{r+1}$ is nonnegative. Therefore
$\displaystyle g'(x) - (r+1)\frac{g(x)}{x} \leq 0$
for $x > a$ and result follows immediately.

For all $\epsilon > 0$ there exists $t > 0$ such that if $t \leq x_i \leq x_j$ then $g_j - g_i < \epsilon$.  Let $\epsilon > 0$.  The integral
$\int_a^\infty x^r f(x) \, dx$
is convergent so there exists $t$ such that
$$
  \int_{t}^\infty x^r f(x) \, dx < \frac{\epsilon}{r+1}
$$
noting that $r+1 > 0$ by assumption.
Now if $t \leq x_i \leq x_j$ then
$$\begin{align*}
  \int_{t}^\infty x^r f(x) \, dx
&\geq
  \int_{x_i}^{x_j} x^r f(x) \, dx
  & \text{as $f(x)$ is nonnegative}
\\&=
  \int_{x_i}^{x_j} \frac{g(x)}{x} \, dx
\\&\geq
  \int_{x_i}^{x_j} \frac{g'(x)}{r+1} \, dx
  & \text{by step 2}
\\&=
  \frac{g_j - g_i}{r+1}
\end{align*}$$
and result follows immediately.

$\left\{ g_n \right\}_n$ is a Cauchy sequence.  We need to show that for any $\epsilon > 0$ there exists $N$ such that if $m,n > N$ then $|g_n - g_m| < \epsilon$.  So let $\epsilon > 0$.  By step 3, there exists $t > 0$ such that
$$
  \text{if}\ t \leq x_i \leq x_j \ \text{then}\
  g_j - g_i < \frac{\epsilon}{2}
  \tag{1}\label{eqn:boundjumpup}
$$
Let $\ell = \inf\left\{ g_n : x_n \geq t \right\}$.
Now $g(x)$ is bounded below (by zero) so $\ell$ is finite.
Hence by the definition of infimum, there exists $N$ such that $x_N \geq t$ and
$\displaystyle g_N < \ell + \frac{\epsilon}{2}$.
Indeed
$\displaystyle g_N - \frac{\epsilon}{2} < \ell$.
Moreover, to emphasize the construction of $\ell$, if $k \geq N$ then $\ell \leq g_k$. Consequently
$$
  \text{if}\ k \geq N \ \text{then}\
  g_N - \frac{\epsilon}{2} < g_k
  \tag{2}\label{eqn:boundbelowgN}
$$
Meanwhile, applying (\ref{eqn:boundjumpup}) with $i = N$ and $j = k$  yields
$$
  \text{if}\ k \geq N \ \text{then}\
  g_k - g_N < \frac{\epsilon}{2}
$$
or equivalently
$$
  \text{if}\ k \geq N \ \text{then}\
  g_k < g_N + \frac{\epsilon}{2}
  \tag{3}\label{eqn:boundabovegN}
$$
Now suppose $m,n > N$. Then (\ref{eqn:boundbelowgN}) and (\ref{eqn:boundabovegN}) yield
$$\begin{align*}
  g_N - \frac{\epsilon}{2} &< g_m < g_N + \frac{\epsilon}{2} \quad \text{and}
\\
  g_N - \frac{\epsilon}{2} &< g_n < g_N + \frac{\epsilon}{2}
\end{align*}$$
therefore $|g_n - g_m| < \epsilon$ as required.

Having established that $\left\{ g_n \right\}_n$ is a Cauchy sequence of real numbers, it follows immediately that there exists $c$ finite such that $g_n \to c$ as $n \to \infty$. Result follows from step 1.  $\blacksquare$

Proposition. Let function $f(x)$ be real-valued and nonnegative on $(a,\infty)$, and suppose that for $r  > -1$ the integral
$\int_a^\infty x^r f(x) \, dx$ is convergent.
Write
$$\displaystyle f(x) = \frac{g(x)}{x^{r+1}}$$
Suppose further that $f(x)$ is differentiable and nonincreasing on its domain. Then $g(x) \to 0$ as $x \to \infty$.

Proof of Proposition. Apply Lemma 2 and then Lemma 1. $\blacksquare$


问:
Is there another 5 regular connected planar graph?
Fortunately we know the following facts that if $G$ is a planar graph with minimum degree 5 then it has at least 12 vertices.

I consider 5-regular planar graph. We find a graph with 12 vertices that satisfies the conditions. By doing limited copy of following graph, we can get more 5 regular planar graphs. But what I mean is whether it is possible to find more connected 5 regular planar graphs with more than 12 vertices.

Is there any literature on this issue ? Any suggestions are valuable.

This question is related to the question I just asked on the forum.

Can't lower bound be improved on number of light edges in planar graph with minimum degree five?

I see this link seems to be related to my problem.

How many nodes are there in a 5-regular planar graph with diameter 2?

If G is a planar graph with minimum degree 5 then it has at least 12 vertices

答:
The answer is yes.

Pick two copies of your graph, and draw them one next to each other.

Now, on the graph on the left erase the rightmost outside edge. On the graph on the right erase the leftmost outside edge.

Connect the top vertices and the bottom vertices of the two graphs.

It is easy to also produce such graphs from $3,4,..., n$ copies of your graph.


答:
The case where $p \leqslant 0$ is quite simple to establish, so I will look only at the case where $p > 0$.  One way you can establish the convergence/divergence result is by using upper and lower bounds on the summation.  For all values $n \geqslant 1$ and $p>0$ we can easily establish the bounds:

$$\int \limits_{n}^{n+1} \frac{1}{r^p} \ dr \leqslant \frac{1}{n^p} \leqslant \int \limits_{n-1}^n \frac{1}{r^p} \ dr.$$

Applying these bounds to each term in the sum (except for the first term for the upper bound) gives:

$$\int \limits_1^\infty \frac{1}{r^p} \ dr \leqslant \sum_{n=1}^\infty \frac{1}{n^p} \leqslant 1 + \int \limits_1^\infty \frac{1}{r^p} \ dr.$$

Solving the definite integral we get:

$$R(p) \equiv \int \limits_1^\infty \frac{1}{r^p} \ dr
= \begin{cases}
\infty & & \text{if } 0
 1. \\[6pt]
\end{cases}$$

Substituting this form for the integral we therefore obtain:

$$R(p) \leqslant \sum_{n=1}^\infty \frac{1}{n^p} \leqslant 1+R(p).$$

We can now look at the conditions on $p$.  If we have $0 < p \leqslant 1$ then we get:

$$\infty \leqslant \sum_{n=1}^\infty \frac{1}{n^p} \leqslant \infty.$$

If we have $p>1$ then we get:

$$\frac{1}{p-1} \leqslant \sum_{n=1}^\infty \frac{1}{n^p} \leqslant \frac{p}{p-1}.$$

This establishes that the sum is divergent when $0 < p \leqslant 1$ and convergent when $p>1$.

Body:
Show that the $p$-series $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent iff $p \gt 1$.

Attempt:
For the right direction:
Let $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent. Then, $\lim\limits_{n\to \infty} \frac{1}{n^p} = 0$.
In particular, $\lim\limits_{n\to \infty} \frac{1}{n^p} = 0$ for $p \gt 0$. But, for $0 \lt p \le 1$, the series $\sum_{n=1}^\infty \frac{1}{n^p}$ is divergent. Hence, we must have $p \gt 1$. $\Box$

Proof that series $\sum_{n=1}^\infty \frac{1}{n^p}$ is divergent:
We know that $n^p \le n$ for all positive integers $n$ and $0 \lt p \le 1$.
Then,
\begin{equation*}
\frac{1}{n} \le \frac{1}{n^p}.
\end{equation*}
Now, since the partial sums of the harmonic series are not bounded, this
inequality shows that the partial sums of the $p-$series are not bounded when $0 \lt p \le 1$. Hence, the $p-$series diverges for theses values of $p. \Box$

For the left direction:
Let $p \gt 1$. Let $f(x) = \frac{1}{x^p}$ for all $x \in [1.\infty)$. Then, $f$ is a positive, continous, and decreasing sequence. Hence, we can apply the Integral Test here. Notice that
\begin{align*}
\int_1^\infty \frac{1}{x^p} &= \lim\limits_{b \to \infty} \int_1^b \frac{1}{x^p} \\
&= \lim\limits_{b\to \infty} \left[\frac{1}{1-p}x^{1-p}\right]_1^b \\
&= \lim\limits_{b \to \infty} \left( \frac{1}{1-p}b^{1-p} - \frac{1}{1-p} \right) \\
&= \frac{1}{p-1} \lim\limits_{b \to \infty} \left(1 - \frac{1}{b^{p-1}} \right) \\
&= \frac{1}{p-1}.
\end{align*}
Thus,
\begin{equation*}
\int_1^\infty \frac{1}{x^p}dx
\end{equation*}
is convergent. Consequently, $\int_1^\infty \frac{1}{n^p}dn$ is convergent. By the Integral Test,
we have that
\begin{equation*}
\sum_{n=1}^\infty \frac{1}{n^p}
\end{equation*}
is convergent. $\Box$

Am I correct, especially for the right direction ?

Body:
The case where $p \leqslant 0$ is quite simple to establish, so I will look only at the case where $p > 0$.  One way you can establish the convergence/divergence result is by using upper and lower bounds on the summation.  For all values $n \geqslant 1$ and $p>0$ we can easily establish the bounds:

$$\int \limits_{n}^{n+1} \frac{1}{r^p} \ dr \leqslant \frac{1}{n^p} \leqslant \int \limits_{n-1}^n \frac{1}{r^p} \ dr.$$

Applying these bounds to each term in the sum (except for the first term for the upper bound) gives:

$$\int \limits_1^\infty \frac{1}{r^p} \ dr \leqslant \sum_{n=1}^\infty \frac{1}{n^p} \leqslant 1 + \int \limits_1^\infty \frac{1}{r^p} \ dr.$$

Solving the definite integral we get:

$$R(p) \equiv \int \limits_1^\infty \frac{1}{r^p} \ dr
= \begin{cases}
\infty & & \text{if } 0 < p \leqslant 1, \\[8pt]
\frac{1}{p-1} & & \text{if } p > 1. \\[6pt]
\end{cases}$$

Substituting this form for the integral we therefore obtain:

$$R(p) \leqslant \sum_{n=1}^\infty \frac{1}{n^p} \leqslant 1+R(p).$$

We can now look at the conditions on $p$.  If we have $0 < p \leqslant 1$ then we get:

$$\infty \leqslant \sum_{n=1}^\infty \frac{1}{n^p} \leqslant \infty.$$

If we have $p>1$ then we get:

$$\frac{1}{p-1} \leqslant \sum_{n=1}^\infty \frac{1}{n^p} \leqslant \frac{p}{p-1}.$$

This establishes that the sum is divergent when $0 < p \leqslant 1$ and convergent when $p>1$.


问:
Show that $\delta$ is a consistent estimator for $g(\theta)=1\;\;(\theta\ne0)$
Let $X_1,\dots, X_n$ i.i.d. from the $N(\theta,1)$ distribution.
We estimate $g(\theta)=1(\theta\ne0)$, The estimator is $\delta= 1 (|\bar X|\ge\frac{1}{\sqrt[4] n})$.

Show that $\delta$ is a consistent estimator for $g(\theta)$.

Find the rate of convergence of $\delta$.

答:
We want to show that $\delta$ is a consistent estimator for $g(\theta)$, i.e. converges in probability to it as $n\rightarrow \infty$.

Case 1: $\theta=0$

We have $g(\theta)=0$ (as $g$ is just an indicator). Also, $\lim_{n\rightarrow\infty}P\left(|\bar X| \ge\frac 1 {\sqrt[4]{n}}\right)=\lim_{n\rightarrow\infty} P\left(\frac{|\bar X|}{n^{-1/4}}\ge1\right)$. This has the indeterminate form $\frac{0}{0}$ so by L'Hopital's rule we get $\lim_{n\rightarrow\infty}P\left(-\frac{3|\bar X|}{4n^{7/4}}\ge1\right)=P(0\ge 1)=0$, hence $\delta=0$ as $n\rightarrow \infty$. So we have that for all $\epsilon>0$, $\lim_{n\rightarrow\infty}P(|g(\theta)-\delta|<\epsilon)=P(0<\epsilon)=1$.

Case 2: $\theta\ne0$

It can be seen that $g(\theta)=1$ here (as $g$ is just an indicator). Further, we have $|\bar X|\rightarrow |\theta|>0$ while $\frac 1 {\sqrt[4]n}\rightarrow 0$. So $\delta\rightarrow 1(|\theta|>0)=1$. Thus for all $\epsilon>0$, $\lim_{n\rightarrow\infty}P(|g(\theta)-\delta|<\epsilon)=P(|1-1|<\epsilon)=1$.

Hence $\delta\rightarrow^p g(\theta)$.