问:
Continuous function with irrational period - summation over naturals converges to 0?
A friend gave me this problem as a "Christmas gift":

If $f$ is periodic with irrational period $r$ and $\int_0^r f(x) \mathrm{d}x = 0$, prove or disprove that $\left\{\sum\limits_{i=1}^n f(i)\right\}_{n \in \mathbb{N}}$ is bounded.

One construction that works is $f(x) = \begin{cases} 1 &\mbox{if } x \equiv k \pmod{r}, k \in \mathbb{Z} \\
0 & \mbox{otherwise } \end{cases}$ (I guess this only works if we use the Lebesgue integral); another is $f(x) = \tan(x-\frac{\pi}{2})$.

However, he accidentally forgot to specify that $f$ must be continuous, which invalidates both of these constructions. Since $\mathbb{N} \pmod{r}$ is uniformly distributed in $[0, r)$, I'm pretty sure that we have
$$\lim_{n \to \infty}\frac{1}{n}\sum_{i=1}^{n}f(i) = \int_{0}^{r}f(x)\mathrm{d}x = 0.$$ Unfortunately, this is weaker than the original statement.

My question is, is the original statement true, and if not, what's a counterexample? I'm not sure if this problem can be solved with elementary knowledge, or if it would require some more theory.

答:
No this is not true for continuous functions.
Take $r = 2\pi$. Continuous functions with vanishing integral form Banach space.
Define linear operators from this space: $$T_n (f) = f(1) + f(2) + \dots +f(n)$$
If the claim were true then (by Banach-Steinhaus, a.k.a. uniform boundedenss principle)
$||T_n|| \lt C$ where $C$ is a constant independent from $n$.
In particular, taking for $f_k(t) = \exp(ikt)$ we would get (for any $k \neq 0$)
$$|\exp(ik \cdot 1) + \exp(ik \cdot 2) + \dots + \exp(ik \cdot n)| \lt C$$
But the sum on the left is:
$$\frac{|\exp(ik\cdot n) - 1|}{|\exp(ik) - 1|}$$ and this should be bounded by a constant independent of $k$ and $n$.
Here is where irrationality of $r$ comes into play (that is uniform distribution of $r\cdot k$ and
$r\cdot n\cdot k$).
First take $k$ so that denominator is small and then ($k$ is now fixed and $r\cdot k$ is irrational) take $n$ so that nominator is not small.


问:
Central limit theorem for weighted random variable
Let $(X_n)_n$ be a sequence of i.i.d random variable, $(x_n)_n$ be sequence of $\mathbb{R}^*.$ Let $y^2_n=\sum_{k=1}^nx^2_k.$

Suppose that $E[X_1]=0,E[X_1^2]=1,x_n=o(y_n),y_n \to +\infty.$ Prove that the sequence of the weighted random variable $(x_nX_n)_n$ fulfills the central limit theorem: $$\frac{1}{y_n}\sum_{k=1}^nx_kX_k \implies N(0;1).$$

Remark (optional): more generally, the following holds: $E[X_1^2]<+\infty$ if and only if there exist a sequence of real numbers $(w_n)_n$ such that $\frac{1}{y_n}\sum_{k=1}^nx_kX_k-w_n$ converges in distribution to an arbitrary random variable $Y.$ In this case, $Y$ is normal distributed.

A way to prove that the CLT is fulfilled, is to prove that Lindeberg condition holds. So let $\epsilon>0.$
$$\frac{1}{y_n^2}\sum_{k=1}^nx_k^2E[X_k^2 1_{|x_kX_k|>\epsilon y_n}]=\frac{1}{y_n^2}\sum_{k=1}^nx_k^2E[X_1^2 1_{|x_kX_1|>\epsilon y_n}],$$ can't see how to continue from here, especially how to remove $x_k$ from $1_{|x_kX_1|>\epsilon y_n}?$

Any suggestions are welcomed.

答:
Part 1:

Counterexample.

Consider $X_k$ -i.i.d., $X_k \sim U[-1,1]$, $x_1 = 1$, $x_k  = \frac1{2^{k+2}}$, $k \ge 2$.

$$S_n = \frac{ X_1 + \eta_n}{ \sqrt{(1 + \sum_{k=1}^n  \frac1{4^{k+2}}) }}$$
where $|\eta_n| = |\sum_{k = 2}^n \frac1{2^{k+2}} X_k| \le \sum_{k = 2}^n \frac1{2^{k+2}} \le \frac12$.
As $\eta_n = \sum_{k = 2}^n \frac1{2^{k+2}} X_k $ we have $\exists \eta:$
$\eta_n \to \eta$ a.s. (Kolmogorov's two-series theorem). As$|\eta_n| \le \frac12$ we have $\eta \le \frac12$.

Put $c = \sqrt{1 + \sum_{k=1}^{\infty}  \frac1{4^{k+2}} }$. Hence
$$S_n \to \frac{X_1 + \eta}{c}  $$ a.s. But $|\frac{X_1 + \eta}{c} | \le \frac{2}c$ and hence
$$\lim_n S_n =   \frac{X_1 + \eta}{c} \ne N(0,1).$$

The claim is false. Hence we need some stronger assumption, and assumption $\max_{1 \le k \le n}|x_k| = o(y_n)$ is sufficient, as it's shown below.

Part 2: Suppose that we don't have condition "$X_n$ are i.i.d." - in this case problem is even more interesting.

Claim: instead of condition $x_n = o(y_n)$ we should have more strong condition $\max_{1 \le k \le n}|x_k| = o(y_n)$, otherwise there's no convergence to $N(0,1)$. Let us prove it.

Theorem (from Probability 1 by A.N. Shiryaev): if $\xi_{n1}, \ldots,  \xi_{nn}$ is a sequence of independent r.v., $E \xi_{nk}=0$, $DS_n = 1$, where $S_n = \xi_{n1} + \ldots + \xi_{nn}$ then $S_n \to N(0,1)$ iff $\max_{1 \le k \le n}E\xi_{kn}^2 \to 0$, $n \to \infty$.

Put $\xi_{nk} = \frac{x_k X_k}{y_n}$. We have $E \xi_{nk}=0$, $DS_n = 1$. So $S_n \to N(0,1)$ iff
$$\max_{1 \le k \le n}E\xi_{kn}^2 \to 0, n \to \infty.$$
As $\max_{1 \le k \le n}E\xi_{kn}^2 = \max_{1\le k\le n}\frac{x_k^2}{y_n^2}$ we have that $S_n \to N(0,1)$ iff $max_{1 \le k \le n}|x_k| = o(y_n)$.

For example, we have convergence to $N(0,1)$ for  $x_n$ such that $|x_n|$ is nondecreasing, but not in general case.

Сonclusion: if $X_i$ are not necessary i.i.d., then the condition $\max_{1 \le k \le n}|x_k| = o(y_n)$ is necessary and sufficient  to guarantee the convergence to $N(0,1)$ for all independent $X_i$ with $EX_i = 0$ and $DX_i = 1$.

Let us notice also, that if $X_i \sim N(0,1)$ then $S_n = N(0,1)$  for any $x_k$.

ContentLicense: CC BY-SA 4.0


问:
An example of two functions whose limits don't exist at a but the limit of their product does
Show by example that the $\lim \limits_{x \to a} [f(x) \cdot g(x)]$ may exist even though neither $\lim \limits_{x \to a} [f(x)]$ nor $\lim \limits_{x \to a} [g(x)]$ exist.

答:
There is an example from the old book for this question. Take for example

$$f(x) =
\begin{cases}
0  & \text{if $x\in\mathbb{Q}$} \\
1 & \text{if $x\notin\mathbb{Q}$}
\end{cases} \quad \text{and} \quad g(x) =
\begin{cases}
1  & \text{if $x\in\mathbb{Q}$} \\
0 & \text{if $x\notin\mathbb{Q}$}
\end{cases}$$
and note that $\displaystyle \lim_{x\to a}f(x)$ and $\displaystyle\lim_{x\to a}g(x)$ does not exist but since that $f(x)g(x)$ is constant function, so the $\displaystyle\lim_{x\to a}f(x)g(x)$ there exists.


问:
Proving that $\lim_{n \to \infty} \frac{1}{\sqrt[n]{n!}}=0$
I have to prove this
$$\lim_{n \to \infty} \frac{1}{\sqrt[n]{n!}}=0$$

My proof is the following:

By using $$n!>n^{\frac{n}{2}}$$
$$\frac{1}{n!}<\frac{1}{n^{\frac{n}{2}}}$$
$$\left(\frac{1}{n!}\right)^{\frac{1}{n}}<\left( \frac{1}{n}\right)^{\frac{n}{2}\cdot\frac{1}{n}}=\frac{1}{\sqrt{n}}\underset{n\to \infty}{\to}0$$
Using the Squeeze theorem
$$0<\frac{1}{\sqrt[n]{n!}}<\frac{1}{\sqrt{n}} \implies \frac{1}{\sqrt[n]{n}}\underset{n\to \infty}{\to} 0$$

I think I have done something wrong. Can I even use the first inequality here? and if yes-do I have to prove that too? and is it Ok that I write too little words in my proofs?. Could you please help me?

答:
Another answer has covered how much descriptive detail to include in your proof.  The objective is to ensure your argument is clear, not just to yourself but to another reader who may not be able to follow each step without some help.

The crucial inequality is the first $n! > n^{n/2}$ for sufficiently large $n$.  In fact, for your argument you do need a strict inequality;  your subsequent reasoning would work just as well if $n! \geqslant n^{n/2}$ for $n$ greater than some threshold.  To establish it you can take (natural) logarithms. $n! = \log n+\log(n-1)+\cdots+\log 1$.  Now consider a simple step function,
$$ f(x) = \left\{\array{0 & 0 < x \leqslant 1 \\ \log 2 & 1 < x \leqslant 2 \\ \cdots &\cdots \\ \log n & n-1 < x \leqslant n \\ \cdots & \cdots} \right.$$
This is constructed deliberately so that for $x \geqslant 1$, $f(x)\geqslant \log x$.  Then,
\begin{align}
\log n! &= \int_1^n f(x) ~dx \\ & \geq \int_1^n \log x ~ dx
\\
&= \Big[ x\log x - x \Big]_1^n \\
&= n\log n - n + 1
\end{align}
from which you can deduce
$$\log n! - \frac{n}{2}
\log n \geqslant \frac{n}{2} \Big( \log n - 2 \Big) + 1.$$
The right hand side is clearly positive for all sufficiently large $n$ because $\log n$ is unbonded.  A little calculation can show the threshold is $n \geqslant 5$.  Taking exponentials it follows,
$$n! \geqslant n^{n/2}\quad\text{when}\quad n \geqslant 5.$$
The inequality thus derived gives a sufficient constraint on $n$.  It also shows the inequality is strict for large enough $n$.  A little more calculation can show it also holds when $n=1,2,3$ and $4$.  Our choice of step function was too coarse to identify the inequality remains valid in these cases.


问:
Finding the tangent lines to an implicit curve
Let $x \gt 0$. Find the tangent lines to the curve $x^{\sin y} + y^2 - x = (\pi^2)/(4)$ at the points of the intersection of the given curve and the line $x=1$.
I found this result $y=\pm \pi/2$ but I'm not sure.

答:
$$x^{\sin y} +y^2 -x =\frac{\pi^2}{4} $$ When $x=1$, $$1+y^2-1=\frac{\pi^2}{4} \implies y=\pm \frac{\pi}{2} $$

Now differentiate.  $\bigg [(x ^{\sin y} )’ = ( e^{\sin y \ln x} )’ = x^{\sin y}\left ( \frac{\sin y}{x} + \ln x\cos y \frac{dy}{dx} \right) \bigg]$

$$x^{\sin y}\left ( \frac{\sin y}{x} + \ln x\cos y \frac{dy}{dx} \right) +2y\frac{dy}{dx} -1 =0 $$

to readily obtain $y’(1,\pi/2) = 0$ and $y’(1,-\pi/2) = -\frac{2}{\pi}$ and the equations of the tangents:

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


问:
$T$ is compact iff every closed subspace in $Ran(T)$ is finite dimensional.
Let $T$ be a continuous linear operator on Hilbert space $\mathcal{H}$. Prove that $T$ is compact if and only if every closed subspace contained in the range of $T$ is finite dimensional.

For the left side implies the right side, it is a consequence of open-mapping theorem. But on the other hand I don’t know how to deal with. I tried to use finite rank operator to approach $T$, but field.
Any hint or references? Thanks!

答:
By the polar decomposition one may assume that $T\geq0$.

In this case, for every $n$, let $f_n$ be the characteristic function of $[1/n,\infty)$.
The function
$$
  g_n(t) = f_n(t)/t,
  $$
is then  bounded and $f_n(t)=tg_n(t)$.  Therefore $f_n(T)=Tg_n(T)$,  so we see that the range of $f_n(T)$ is contained
in the range of $T$.  Since $f_n(T)$ is a projection, its range is closed, so by hypothesis we deduce that $f_n(T)$ has
finite rank.

Observing that  $Tf_n(T)$ converges  to $T$, we conclude that  $T$ is compact.


问:
Finding the Extremals of a Functional J.
The functional $J$ is defined on smooth functions $y \colon [a,b] \to \mathbb{R}$ satisfying $y(a) = u$, $y(b) = v$ and is given by

$$J[y]=\int_a^b \sqrt{y} \sqrt{1+(y')^2}\, dx.$$

I have found the Euler-Lagrange equation to be
$\sqrt{y} \sqrt{1+(y')^2} - y' \frac{\sqrt{y}(y')^2}{\sqrt{1+(y')^2}} = k$   for some constand $k$.

solving for $y'$ I got $y' = \sqrt{y/k^2 -1}$

Please could you confirm if this is correct and how to complete the problem from here?
Any help is greatly appreciated!

答:
Try the substitution $u = \frac{y}{k^2} - 1$.

You get that $k^2 \frac{du}{dt} = u^\frac{1}{2}$, which is easy to solve with separation of variables.

ContentLicense: CC BY-SA 3.0


答: A very simple example is $f(x)=\sin(\frac{1}{x-a})$ and $g(x)=\frac{1}{f(x)}$: then neither $\lim_{x\to a} f(x)$ nor $\lim_{x\to a} g(x)$ exist, but clearly $\lim_{x\to a} \frac{f(x)}{g(x)}=1$. If you want an example for $x\to (\pm)\infty$, just take $f(x)=\sin(x)$ and again $g(x)=\frac{1}{f(x)}$.

Title: An example of two functions whose limits don't exist at a but the limit of their product does

Body: Show by example that the $\lim \limits_{x \to a} [f(x) \cdot g(x)]$ may exist even though neither $\lim \limits_{x \to a} [f(x)]$ nor $\lim \limits_{x \to a} [g(x)]$ exist.


问:
Development of a problem of the course on ODE MIT 18.03
From the course MIT 18.03 about ODE, there is that following problem 18.03SC practice problem 2
For the ODE:
$$
\frac{dy}{dx} = x-2y
$$
One of the integral curves seems to be a straight line. What straight line is it? (i.e., for what $m$ and $b$ is $y = mx + b$ a solution?)

answer:

If $y = mx + b$ is a solution, then
$$
m = \frac{dy}{dx} = x-2y = x-2(mx+b)
$$
or, rearranging terms,
$$
m = (1 − 2m)x − 2b\tag{1}
$$
Two polynomials in x can only be the same for all x (over the reals) if they have the same coefficients. Equate coefficients of the powers of x to get two equations in two unknowns that must be satisfied simultaneously.
$$
\begin{cases}
m&=-2b\\
0&=1-2m\tag{2}
\end{cases}
$$

How do we develop step by step to come to (2) from (1):

答:
Although what is stated should be sufficient, you can look at it like this:

The equation must hold for $x=0$, so $$m=-2b$$ Substituting for $b$ gives $$(1-2m)x =0 $$ This equation must hold for $x=1$, so $$1-2m=0$$


问:
How can I integrate $\int_0^1 (1-x^{\frac{1}{q}})^p\, dx$
I start with a substitution:
$$\int_0^1 (1-x^{\frac{1}{q}})^p\, dx $$
we let $u = x^{\frac{1}{q}}$ meaning that $du = \frac{x^{q-1}}{q}dx$:
$$ q\int_0^1 (1-u)^pu^{q(1-q)}du $$
Using integration by parts I get:
$$ \int_0^1 (1-u)^pu^{q(1-q)}du = q\left(\left[\frac{u^{q(1-q)} (1-u)^{p+1}}{p+1} \right]_0^1 -\frac{q(1-q)}{p+1}\int_0^1u^{q(1-q)-1}(1-u)^{p+1}du \right) $$
Simplifying:
$$ \int_0^1 (1-u)^pu^{q(1-q)}du =  -\frac{q^2(1-q)}{p+1}\int_0^1u^{q(1-q)-1}(1-u)^{p+1}du $$
I'm currently a bit lost as I feel like this is just going in circles. What should I do thanks!

答:
$$I=\int_{0}^{1} (1-x^{1/q})^p dx$$
Let $x=\sin^{2q} t$ so
$$I=2p\int_{0}^{\pi/2} \sin^{2q-1} t \cos^{2p+1} t~ dt $$
Use $\beta$ function https://en.wikipedia.org/wiki/Beta_function
to get
$$I=p \frac{\Gamma(p+1) \Gamma(q)}{\Gamma(p+q+1)}$$


问:
The elements of $\mathbb{Z}[\sqrt{-5}]/(2,\sqrt{-5}+1)$
I'm really confused with this one...

How can I determine the elements of the module $\mathbb{Z}[\sqrt{-5}]/(2,\sqrt{-5}+1)$? Or its cardinality?

Does $$\mathbb{Z}[\sqrt{-5}]/(2,\sqrt{-5}+1)=\{\bar{0},\bar{1},\overline{0+\sqrt{-5}}\}?$$

答:
Number theory perspective of this problem is: prime ideal factorization.

The prime ideal $(2)$ in $\mathbb{Z}$ is totally ramified in $\mathbb{Z}[\sqrt{-5}]$

and the ideal $(2, 1+\sqrt{-5})$ is a prime ideal lying over $(2)$

In $\mathbb{Z}[\sqrt{-5}]$, we can write
$$
(2)=(2, 1+\sqrt{-5})^2,$$
and this can be checked by direct calculation.

Thus, the residue field extension
$$\mathbb{Z}/2\mathbb{Z}\subset \mathbb{Z}[\sqrt{-5}]/(2, 1+\sqrt{-5})$$
is a trivial extension of finite field $\mathbb{Z}/2\mathbb{Z}$.

i.e.
$$\mathbb{Z}[\sqrt{-5}]/(2, 1+\sqrt{-5})\simeq \mathbb{Z}/2\mathbb{Z}.$$


问:
How many different combinations are there of three distinct integers between one and thirty (inclusive) such that their sum is a multiple of three?
Now, there are ten numbers of each of the following groups; we shall name one, two and three: $3x, 3x+1$ or $3x+2$.

So you can have 3 of group one, 3 of group two, or one of each from each of the groups. This gives ${10 \choose 3} + {10 \choose 1}^3 + {10 \choose 3} = 1240$.

However, the answers suggest that there is an extra $10C3$ to be added on (making $1360$). Where does this come from?

答: You can have 3 of Group 3: $2+2+2 \equiv 0 \pmod{3}$.


问:
Calculating Differential of multivariable functions the other way.
I never saw this approach before so I beg you to verify:

Is it possible to calculate the differential of, let's say $\partial_xf(x,y)$ at $(1,1)$ by:

\begin{equation} \lim_{x \to 1} \frac{f(x,1)-f(1,1)}{x}\,?\end{equation}
Thank you in advance.

答: No, as that limit is of the form $\frac 01$ and always evaluates to $0$ for continuous $f$. Instead, this would work: $$\lim_{x\to 1} \frac{f(x,1)-f(1,1)}{x-1}$$


答:
If $f$ is continuous, then $$\lim_{x \to 1} \frac{f(x,1)-f(1,1)}{x} = f(1,1)-f(1,1) = 0$$

which is clearly not what we're looking for since you can come up with a continuous function $f$ whose partial derivative at $(1,1)$ is not $0$

Title: Calculating Differential of multivariable functions the other way.

Body:
I never saw this approach before so I beg you to verify:

Is it possible to calculate the differential of, let's say $\partial_xf(x,y)$ at $(1,1)$ by:

\begin{equation} \lim_{x \to 1} \frac{f(x,1)-f(1,1)}{x}\,?\end{equation}
Thank you in advance.


问:
Comparing the value of $\int _0 ^{\frac{\pi}{2}} x^n \sin^nx\,dx$ for different $n$
Let $$I_n = \int _0 ^{\frac{\pi}{2}} x^n \sin^nx \, dx.$$

Could you tell me how to decide which one is true: $I_{2012}>I_{2013} $, or $I_{2013}>I_{2012}$?

答:
Numerically, it would seem that $I_n > I_{n-1}$ for large enough $n$ and that moreover $$\lim_{n \to \infty} \frac{I_n}{I_{n-1}} \sim \frac{\pi}{2}$$ Why is this is so?
Note that as $n \to \infty$ the $\sin^n x$ term is zero almost everywhere execpt for $x=\pi/2$ where it's equal to one. This means that the value of the integral is really just given by the $x^n$ at the end point, so we end up  with the above limit.</p>


问:
Estimation of probability of the distribution of a discrete random variable
I am trying to find the probability of the distribution of a discrete random variable $X$ which is defined $P(X=x)=y(5−)$ where $X\in\{1,2,3,4\}$.

Can anybody help me to find $y$?

答:
A discrete random variable must satisfy the property $\sum f(x)=1$ where $f(x)$ is the probability $P(X=x)$ or in other words: the sum of all probability must equal to $1$.

Let's sum your probabilities to see what we get:
$y(5-1)+y(5-2)+y(5-3)+y(5-4)=4y+3y+2y+y=10y$

so now we know that the sum of our probabilities is equal to $1$ (since they said it is a random variable) and it is equal to $10y$ as well, so we have: $10y=1$ which gives us $y=\frac{1}{10}$

Note that for the continuous case you apply the same method but with the integral.


答: Since $x=u^q$, $dx=qu^{q-1}du$, so the correct calculation is$$\int_0^1(1-u)^pqu^{q-1}dq=q\operatorname{B}(p+1,\,q)=\frac{q\Gamma(p+1)\Gamma(q)}{\Gamma(p+q+1)}.$$

Title: How can I integrate $\int_0^1 (1-x^{\frac{1}{q}})^p\, dx$

Body:
I start with a substitution:
$$\int_0^1 (1-x^{\frac{1}{q}})^p\, dx $$
we let $u = x^{\frac{1}{q}}$ meaning that $du = \frac{x^{q-1}}{q}dx$:
$$ q\int_0^1 (1-u)^pu^{q(1-q)}du $$
Using integration by parts I get:
$$ \int_0^1 (1-u)^pu^{q(1-q)}du = q\left(\left[\frac{u^{q(1-q)} (1-u)^{p+1}}{p+1} \right]_0^1 -\frac{q(1-q)}{p+1}\int_0^1u^{q(1-q)-1}(1-u)^{p+1}du \right) $$
Simplifying:
$$ \int_0^1 (1-u)^pu^{q(1-q)}du =  -\frac{q^2(1-q)}{p+1}\int_0^1u^{q(1-q)-1}(1-u)^{p+1}du $$
I'm currently a bit lost as I feel like this is just going in circles. What should I do thanks!


问:
Simplifying sum of floor functions
Consider
$$S = \sum_{i=0}^{x-2}{\lfloor{a(x-i)}\rfloor}$$

where $x \in \mathbb{N}$, $x \geq 2$, and $a = \frac{p}{10}$, with $p \in \{1, 2, \ldots , 9\}$, is rational.

How can one go about finding a closed form of such summation, if it exists?

Attempt

First I consider the sum for first few values of $x$ expanded out:
$$
\begin{array}{cc}
x & S \\
\hline
2 & {\lfloor 2a \rfloor} \\
3 & {\lfloor 3a \rfloor} + {\lfloor 2a \rfloor} \\
4 & {\lfloor 4a \rfloor} + {\lfloor 3a \rfloor} + {\lfloor 2a \rfloor} \\
5 & \ldots \\
\end{array}
$$
So we can express $S$ as a recursive sum $S(x) = {\lfloor ax \rfloor} + S(x-1)$ with base case $S(2) = {\lfloor 2a \rfloor}$.

But then all I've done is just switched from one type of problem into another.

-- edit --

On a further thought, the constraint $x \geq 2$ isn't really relevant to the problem so we may consider removing it:

$$S = \sum_{i=0}^{x}{\lfloor{a(x-i)}\rfloor}$$

where $x \in \mathbb{N}_{0}$, and $a$ is as above.

答:
A similar sum can be evaluated using simple combinatorics or Pick´s theorem:

$$\sum_{i=1}^n\left\lfloor \frac{p}{n} i\right\rfloor = \frac{pn+p-n + gcd(p,n)}{2}$$

Title: Simplifying sum of floor functions


问:
Symmetric Matrices of $I_{2}$

Find $10$ symmetric matrices $ A = \begin{pmatrix}
a &b \\
 c&d
\end{pmatrix}$ such that $A^{2}=I_{2}$

(I'm going to call matrix A the "square root" of $A^{2}$. If this is the incorrect name for it, may someone please tell me what it is actually called?)

My professor posed this question in class and told us there was an infinite amount of square roots. (Assuming I understood him correctly). However I don't see how there would be many of these, as I was under the impression that a matrix only has one inverse, for $A  A^{-1}=I_{n}$. If someone could tell me if I either misunderstood the professor or if I'm thinking about something incorrectly, please correct me.

My other question is other than blatant guess and check, is there a method to think of these symmetric square roots?

Thanks in advance.

答: For an easy constuction, consider matrices of the form $A=R_\theta^T\pmatrix{1\\ &-1}R_\theta$, where $R_\theta=\pmatrix{\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta}$ is a rotation matrix for angle $\theta\in[0,\pi)$. These $A$s are all distinct (exercise), real symmetric and their squares are equal to $I_2$.</p>

Title: Symmetric Matrices of $I_{2}$


问:
Solving a super basic inequality
Can you please help me find where the problem arises in the solution?

$1 + \frac{1}{x} \geq 0$

My attempt:

$1 + \frac{1}{x} \geq 0$

$1 \geq - \frac{1}{x}$

$x \geq -1$

The answer is $x \in (- \infty, -1] \bigcup (0, \infty)$.

I understand why the answer is so, but I made this mistake while solving a problem and didn't notice, so I want to understand how to approach problems like this.

Thanks!

答:
You cannot just multiply by $x$ since it can be negative too. So, below is the right approach:
$$1 + \frac 1x \ge 0 \tag{$x\ne 0$} \\ \frac{x+1}{x} \ge 0 \iff x(x+1) \ge 0 \\\implies x \in (- \infty, -1] \cup (0, \infty)$$

Title: Solving a super basic inequality


答: If $x$ is negative, when you multiply on both sides of the equality by $x$ you would have to change the direction of the inequality. Avoid doing this when you don't know if $x$ is positive.

Title: Solving a super basic inequality

Body:
Can you please help me find where the problem arises in the solution?

$1 + \frac{1}{x} \geq 0$

My attempt:

$1 + \frac{1}{x} \geq 0$

$1 \geq - \frac{1}{x}$

$x \geq -1$

The answer is $x \in (- \infty, -1] \bigcup (0, \infty)$.

I understand why the answer is so, but I made this mistake while solving a problem and didn't notice, so I want to understand how to approach problems like this.

Thanks!


答:
Multiplying by a negative number reverses the inequality, so you need to consider the cases separately:

$$x\gt 0 \implies x\ge -1 \ \text{(this is always true)}\\ x\lt 0 \implies x\le -1 $$ by your second line and taking the intersection gives $x\gt 0$ or $x\le -1$.

Title: Solving a super basic inequality


问:
Relationship between autocorrelation function and wavelet coefficient
the autocorrelation function can be represented using the spectral density in Fourier space. Is there a similar relationship between the autocorrelation function and the coefficient in the wavelet space?</p>

答:
$\newcommand{\inprod}[2]{\left\langle{#1}\,|\,{#2}\right\rangle}$
$\newcommand{\eqd}{\triangleq}$
$\newcommand{\setn}[1]{{\left\{{#1}\right\}}}$
$\newcommand{\setu}{\cup}$
$\newcommand{\brp}[1]{{\left(#1\right)}}$
$\newcommand{\brbl}[1]{\left\{#1\right.}$
$\newcommand{\brb}[1]{\left\{#1\right\}}$
$\newcommand{\brs}[1]{\left[#1\right]}$
$\newcommand{\setR}{\mathbb{R}}$
$\newcommand{\setC}{\mathbb{C}}$
$\newcommand{\setX}{\mathsf{X}}$
$\newcommand{\opF}{\mathbb{F}}$
$\newcommand{\opW}{\mathbb{W}}$
As you said, the auto-correlation function $R_{xx}(t,u)\eqd E\brs{x(t+u)x^{\ast}(t)}$ is simply a "function" (a mapping from one set $\mathsf{X}$ to another/same set $\mathsf{Y}$). If $x(t)$ is stationary, this definition can be simplified to $R_{xx}(u)\eqd E\brs{x(u)x^{\ast}(0)}$. For a very wide class of functions on $\mathbb{R}\to\mathbb{C}$, we can use the Fourier Transform $\opF$ (transform—a special class of functions often called operators that map from one vector space to another/same vector space). When we use the operator $\opF$ to map the vector $R_{xx}(u)$ to a new vector $S_{xx}(\omega)\eqd \opF R_{xx}(u)$, $S_{xx}(\omega)$ is often called the spectral density of $x$, or the power spectral density (PSD) of $x$.

Of course $\mathbb{F}$ is not the only transform (vector space mapping) that can be applied to $R_{xx}(u)$—any transform that operates over the domain used by $R_{xx}(u)$ could be used. Example transforms include the Laplace Transform (a generalization of the Fourier Transform), B-spline transforms (note on "s": there are an infinite number of B-splines), and of course Wavelet Transforms. Likewise if $R_{xx}$ was discrete (e.g. "$R_{xx}(m)$") rather than continuous, transforms might include the DFT/FFT, Z-transform, and Hadamard Transform.

But to me, your question is not just if the wavelet transform $\opW$ can be applied to $R_{xx}$; it is more like, "Can $\opW$ be applied in like manner as $\opF$ and get a like manner usefulness?" That is, just as the "Power Spectral Density" can be used to calculate the "power" (or something resembling it) in the "spectrum" of $\opF$, can the projection $\opW R_{xx}$ of $R_{xx}$ also be used to calculate power (or something resembling it)?

Here, "spectrum" is defined as the projection coefficients/"Fourier coefficients" $G(\omega)$ of $\mathbb{F}$
$$\brb{\begin{align*}
  G(\omega)
    &\eqd \brs{\opF R_{xx}(t)}(\omega)
     \eqd\inprod{R_{xx}(t)}{\frac{1}{\sqrt{2\pi}}e^{i\omega t}}
     \eqd \frac{1}{\sqrt{2\pi}}\int_{t\in\setR}R_{xx}(t)e^{-i\omega t}dt
  \\&\text{of $R_{xx}$ onto the basis $\setn{\frac{1}{\sqrt{2\pi}}e^{i\omega t} | \omega\in\mathbb{R}}$ and $\opW R_{xx}$}
  \\
  H(a,b)
    &\eqd \brs{\mathbb{W}R_{xx}}(a,b)
     \eqd \inprod{R_{xx}(t)}{\frac{1}{\sqrt{a}}\psi\left(\frac{\omega-b}{a}\right)}
     \eqd \frac{1}{\sqrt{a}}\int_{t\in\setR} R_{xx}(t)\psi\left(\frac{\omega-b}{a}\right)
\end{align*}}$$
is the projection of $R_{xx}$ onto whatever wavelet basis $\{\psi\}$ (e.g. Haar, Daubechies-$p$, Symmlets-$p$) is used in the definition of wavelet transform $\mathbb{W}$.

The answer is "Yes" and is demonstrated by the Plancherel Formula or more generally by Parseval's Identity in any Hilbert Space:
$$\left\lVert R_{xx}(t) \right\rVert^2 = \sum_{n\in\mathbb{z}}\left|\left\langle R_{xx}(t),\phi_n(t)\right\rangle\right|^2$$
where the sequence $\{\phi_n(t)\}$ is any orthonormal basis (Fourier, Wavelet, Hadamard, etc.) for the space.

The term “spectral power” is arguably a bit of an oxymoron because “spectral” deals with leaving the time-domain for an alternate (call it "frequency" if you wish) domain$\ldots$ howbeit the concept of power is solidly founded on the concept of time in that power = energy per time. However, Plancherel/Parseval demonstrate that "power" $\left\lVert R_{xx}(t) \right\rVert^2$ in time can also be calculated from any Hilbert Space transform that is constructed over an orthonormal basis.


问:
For what $x_{0}$ this sequence converges?
$$X_{n+1} = X_{n}^2 - X_{n}/2, \quad n \geq 0$$

For what values of $x_{0}$ this sequence converges? For what value?

I think to solve this problem separating it in three steps:

The series is constant. Ok

The series is crescent:

$X_{n+1} > X_{n}$

And for this case we need to require this:
$X_{n} - X_{n-1} > X_{n+1} - X_{n} $

The series is decrescent:

$X_{n} > X_{n+1}$

And for this case we need to require this:
$X_{n} - X_{n+1} < X_{n-1} - X_{n}$

Now, i think this is not the best approach, it is long and extremely easy to be made an algebraic mistake. There is another way to do that?

答:
Since the fixed points are $0$ and $\frac 32$, one common technique is to subtract them from both sides of the recurrence equation, and you get

$$x_{n+1} -\frac 32 = \left(x_n-\frac 32\right)(x_n+1)$$
$$x_{n+1} - 0 = (x_n-0)\left(x_n-\frac 12\right)$$

Now you need to consider the following intervals for $x_0: (-\infty, -1], (-1, 0], (0, \frac 12], (\frac 12 \frac 32], (\frac 32, \infty)$

Can you start from here?


问:
Neumann problem, stuck on a boundary condition.
I am stuck on a problem that I am trying for exam practice and I would very much appreciate a hint to help me out, here is the section where I am stuck:

A solution is sought to the Neumann problem for $\nabla^2 u = 0$ in the half plane $z > 0$:
  $u = O(|x|^{−a}),
\frac{\partial u}{\partial r}
= O(|x|^{−a−1}) ~~ \mathrm{as} ~~  |x| → ∞,~~ \frac{\partial u}{\partial z} ~
= p(x, y) ~ on ~ z = 0,
\mathrm{where}~ a > 0$. It is assumed that
  $\int_{\infty}^{\infty}\int_{-\infty}^{\infty}
p(x, y) dx dy = 0$. Explain why this condition is
  necessary.

My feeling is that this is to do with Green's third identity and that we need the normal derivative in the $x-y$ plane to be integrable in order to find out solution with a Green's function, am I correct?

EDIT:
the divergence theorem sorts this out.

答: Let $f(z)=\int\int u_z$. Then $f_z=\int\int u_{zz} =\int\int (-u_{xx}-u_{yy})$. Consider this quantity and also $\lim _{z\to\infty} f(z)$. That should give you the result you want.</p>


问:
$\mathbb{C}(x)$ as a representation of $\mathbb{C}$
Let $G$ an algebraic linear group. A representation $V$ of $G$ is a vector space such that the application $\phi : G \to GL(V)$ is defined.

Definition Let $G$ a linear algebraic group. A representation $V$ is locally regular if and only if $\forall W$ vector subspace of $V$ there exists a finite dimensional vector subspace $W'$ such that $W \subseteq W'$ , $W'$ is $G$-stable and $\phi : G \to GL(W')$ is regular (as an application between affine varieties).

Let $G = \mathbb{C}$ and $V = \mathbb{C}(x)$ a representation with the action $\phi: G \to GL(V)$ defined as follows: $\forall g \in G$, $\phi(g) : f(x) \to f(x -g)$, with $f \in V$. I want to prove that $V$ is not locally regular. My idea is to choose $W = \mathbb{C}$ and to find $W'$ finite dimensional vector subspace which is not $G$-stable, but I can't think of any. I tried looking for subfields of $\mathbb{C}(x)$, but they are all infinite dimensional. Any hints?

答:
The $\Bbb{C}$-vector space
$Span(\{ \frac1{x-a},a\in \Bbb{C}\})$ is infinite dimensional.


答:
Notice that for all $x,y$ we have $$x^2+xy+y^2 = {1\over 2}(x^2+(x+y)^2+y^2)\geq 0$$

Now we have:
$$3(a-b) = (b-e)\color{red}{\Big(} \underbrace{b+c+d}_x )^2+\underbrace{(b+c+d)(c+d+e)}_{xy}+(
\underbrace{ c+d+e}_y)^2\color{red}{\Big)} $$

so $${\rm sign} (a-b) = {\rm sign} (b-e) $$ and similary for:

\begin{align} {\rm sign} (b-c) &= {\rm sign} (c-a)\;\;\;\;(2) \\
{\rm sign} (c-d) &= {\rm sign} (d-b)\;\;\;\;(3) \\
{\rm sign} (d-e) &= {\rm sign} (e-c) \;\;\;\;(4)\\
{\rm sign} (e-a) &= {\rm sign} (a-d) \;\;\;\;(5)\\
\end{align}

If $a>b$ then $b>e$ so $a>e$ and from $(5)$ we have $d>a$ so $d>e$ and now from $(4)$ we have $e>c$ and now $d>c$ so from $(2)$ we have $b>d$ so $b>c$ and now from $(2)$ we have $c>a$. So we have $$a>b>d>e>c>a$$ A contradiction.

If $b>a$ we again get a contradiction, so $a=b=c=d=e=:x$ and all we need is to solve $x=9x^3$...

Title: Solving the system $3a=(b+c+d)^3$, $3b=(c+d+e)^3$, ..., $3e=(a+b+c)^3$ for real $a$, $b$, $c$, $d$, $e$

Body:
$$\begin{align}
3a&=(b+c+d)^3 \\
3b&=(c+d+e)^3 \\
3c&=(d+e+a)^3 \\
3d&=(e+a+b)^3 \\
3e&=(a+b+c)^3
\end{align}$$

I tried to use inequality for
$((\sum \alpha)/n)^r\leq \sum \alpha^r/n$ by taking $\alpha=a_1+a_2+a_3$ where $a_1,a_2,a_3\in \{a,b,c,d,e\}$.


问:
Definition of N-commutator
Warnings:

I am not a mathematician and
I am asking this question
out of curiosity.

This question might not have
the correct tags since I'm
not sure about what this
question
is about.

Background:

So I have seen the definition of a commutator which is
$ [A,B]=AB-BA$.

I started wondering that is there a definition for $[A,B,C]$? So I started searching and I found the definition for ternary commutator
(https://en.m.wikipedia.org/wiki/Ternary_commutator)

In the wikipedia page there is a reference to N-commutator but there is no wikipedia page for it.

I am not sure that how the extension from 2-commutator (original commutator) to 3-commutator (ternary commutator) works. So I can't figure out the definition for N-commutator.

Question:

So, I wanna know that how the extension works and what is the definition of N-commutator.

答:
The ternary commutator for an associative binary operator adds together products from all $3!$ permutations of $a,\,b,\,c$, but with a coefficient of $1$ ($-1$) for the even (odd) permutations $abc,\,bca,\,cab$ ($acb,\,bac,\,cba$). The $2$-commutator is the same basic idea, although in that case we'd deem such a description "overkill": the even (odd) permutation $ab$ ($ba$) gets a coefficient $1$ ($-1$). The general $N$-commutator is$$[a_1,\,\cdots,\,a_N]=\sum_{\sigma}\varepsilon_\sigma\prod_{i=1}^Na_{\sigma_i},$$where:

The sum is over all permutations $\sigma$ of $1,\,\cdots,\,N$;

The Levi-Civita symbol $\varepsilon_\sigma$ is $1$ ($-1$) for even (odd) $\sigma$;

The product is left-to-right, e.g. if $\sigma$ is the identity operator $\prod_{i=1}^Na_{\sigma_i}=a_1a_2\cdots a_n$.

The $N$-commutator is multilinear, fully antisymmetric and invariant under cyclic permutations, and satisfies$$[a_1,\,\cdots,\,a_N]=\sum_{i=1}^Na_i[a_{i+1},\,\cdots,\,a_N,\,a_1,\,\cdots,\,a_{i-1}],$$e.g. $[a,\,b,\,c]=a(bc-cb)+b(ca-ac)+c(ab-ba)$.Further,$$[a_{\sigma_1},\,\cdots,\,a_{\sigma_N}]=\varepsilon_\sigma[a_1,\,\cdots,\,a_N].$$

Title: Definition of N-commutator


问:
Round off error in modified Newton-Raphson
Newton's method for approximating roots of $f(x)=0$ is given by $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}, \quad n=0,1,2,\ldots,$$ where $x_n$ denotes the $n$th approximation of the root, and some sufficiently good initial approximation $x_0$ is required for convergence.

In general, this converges quadratically, but if the root, say $x^*$, satisfies $f'(x^*)=0$, the convergence is only linear. I'm aware that quadratic convergence can be restored by using the scheme $$x_{n+1}=x_n-\frac{F(x_n)}{F'(x_n)},$$ where $F(x_n)=f(x_n)/f'(x_n)$.

After substituting the definition of $F(x_n)$, it can be seen that $$x_{n+1}=x_n-\frac{f(x_n)f'(x_n)}{f'(x_n)f'(x_n)-f(x_n)f''(x_n)}.$$

I'm told that this raises another issue, namely that the denominator involves computing the difference of two nearly equal quantities and can lead to large rounding errors.

Questions

Why are the quantities on the denominator nearly equal?

How do we get around this issue?

答:
The quantities in the denominator are not nearly equal.

Subtractive cancellation is not an issue here.

As @SimplyBeatifulArt has already pointed out we have $$ f'(x)f'(x) \approx 2f(x)f''(x)$$
near a double root of $f$. This eliminates the possibility of subtractive cancellation in the calculation of
$$ f'(x)f'(x) -  f(x) f''(x).$$
A general analysis of subtractive cancellation is given in this answer.

In general, the componentwise relative condition number of a subtraction $d = a - b \not = 0$ is given by $$\kappa = \frac{|a|+ |b|}{|a-b|},$$
while the componentwise relative condition number of a division $d = a/b$, $(b \not = 0)$ is given by $$\kappa = 2.$$
In particular, divisions are always well-conditioned in the relative sense.


问:
Does induced topology preserve density?
Let $(X, \tau )$ a topological space and $D \subseteq X$ dense in $(X, \tau)$. Show that if $A \in \tau$, then  $A\cap D$ is dense in subspace $A$.

I have already proven that the intersection of two dense subsets in $X$, one of which is open is dense. Can I enjoy this demonstration?
I feel like we're using the induced topology for subspace A. In this case the induced topology preserves density, compactness as well? Is that what I'm thinking?

答: You have to show that every nonempty set which is open in the subspace $A$ has a nonempty intersection with $D$. But if $A\in \tau$ then the sets which are open in $A$ are exactly the sets of $\tau$ included in $A$ for which, by assumption, you know that (if nonempty) they have a nonempty intersection with $D$.

Title: Does induced topology preserve density?


问:
If given two groups $G_1,G_2$ of order $m$ and $n$ respectively then the direct product $G_{1}\times G_{2}$ has a subgroup of order $m$.
I'm trying to understand if this statement is false or true, I have tried understanding by an example.

For example if $G_1=(\{{\bar0},\bar1\},+_2),G_3=(\{{\bar0},\bar1,\bar2\},+_3)$ then the direct product of these two groups will be

$G_1\times G_2=\{(\bar0,\bar0),(\bar0,\bar1),(\bar0,\bar2),(\bar1,\bar0),(\bar1,\bar1),(\bar1,\bar2)\}$

I see that $o(G_1\times G_2)=o(G_1)\cdot o(G_2)=m\cdot n=6.$ Can I simply say that I take the group $\{{(\bar0,\bar0),(\bar0,\bar1)}\}$ under the addition modulo 2 and 3 respectively and the statement is true for this case ?

Also if this statement is true shouldn't it mean that it will always be true considering that $o(G_1\times G_2)=o(G_1)\cdot o(G_2)$ and there will always be a subgroup of order $m$?

答:
Hint: Consider $$H=\{(g, e_{G_2})\mid g\in G_1\}.$$

 Use the one-step subgroup test. Clearly $H\subseteq G_1\times G_2$. Since $G_1$ is a group, its identity $e_{G_1}\in G_1$, so $(e_{G_1}, e_{G_2})\in H\neq \varnothing$. Let $x=(g_1, e_{G_2})$, $y=(g_2, e_{G_2})\in H.$ Then $$\begin{align}xy^{-1}&=(g_1, e_{G_2})(g_2, e_{G_2})^{-1}\\ &=(g_1, e_{G_2})(g_2^{-1}, e_{G_2})\\ &=(g_1g_2^{-1}, e_{G_2})\\ &\in H\end{align}$$ since $g_1g_2^{-1}\in G_1$ as $G_1$ is a group. Hence $H\le G_1\times G_2$. But clearly $\lvert H\rvert=\lvert G_1\rvert=m$.


问:
A relationship between $a$ and $b$ related to algorithm Rock(1)-Scissor(0)-Paper(2)

This is related to algorithm Rock(1)-Scissor(0)-Paper(2)

I need to find a relationship between $a$ and $b$ so that
$$\begin{matrix}
a & b \\
1 & 0\\
0 & 2\\
2 & 1
\end{matrix}$$
is always true and
$$\begin{matrix}
a & b \\
1 & 2\\
0 & 1\\
2 & 0
\end{matrix}$$
is always false. I had read a book about this involving to binary and computer. I'm quite sure that we must do some analytics among
$$\begin{matrix}
01 & 00 \\
00 & 10
\end{matrix}$$
and
$$\begin{matrix}
00 & 10 \\
10 & 01
\end{matrix}$$
and
$$\begin{matrix}
10 & 01 \\
01 & 00
\end{matrix}$$
What should I do next ?? And I wanna know more about this knowledge, thanks a real lot !

Edit: another solution that is
$$a- b+ 2\equiv 0\mod 3$$

答:
The first two tables you gave are one way of defining the relation. If you want an arithmetic definition, you can take $f(a,b) = ||a-b-1/2|-3/2|$.

Here is how to find $f$. Look at the possible values of $a-b$. When the relation $r$ is true, it is 1 or -2. When $r$ is false, it is -1 or 2. To get the same value for when $r$ is false, make 0 the average of the two outputs then take absolute value: $|a-b-1/2|$. Now the output is 3/2 when $r$ is false, so subtract 3/2: $|a-b-1/2|-3/2$ Now we deal with the output when true, which is $1$ or $-1$. To make this always 1, we take the absolute value: $||a-b-1/2|-3/2|$.


问:
Simplify $\ln(\frac{6x^9}{3x^5})$
I am to simplify $\ln(\frac{6x^9}{3x^5})$. My textbook solution says  $\ln(2x^7)$ whereas I get $\ln(2x^4)$.

I arrived at this by dividing $6x$ by $3x$ to give $2x$. I'm a bit stumped on how to handle the exponents. I thought that in the case of division I am to subtract the denominator power from the numerator power, hence I get 4.

How can I arrive at $\ln(2x^7)$?

答:
$\ln{\frac{6x^9}{3x^5}}=\ln{\frac{2x^9}{x^5}}=\ln{2x^4}$

Your have most likely just found an error in your textbook.

Title: Simplify $\ln(\frac{6x^9}{3x^5})$


答:
The function $x \sin x$ is increasing in the interval $[0, \frac \pi 2]$, and there is a point $x_0 \approx 1.12$ which satisfies $x_0 \sin {x_0} = 1$.

Split the integral to a sum of two integrals: Integral from $0$ to $x_0$ plus integral from $x_0$ to $\frac \pi 2$.

The first one tends to $0$ as $n \to \infty$ (why?). The second one is monotone increasing in $n$ (transitioning from $n$ to $n+1$ we're multiplying by $x \sin x$ which is more than $1$).

Therefore, up to a small error term which is negligible for large $n$, the integral is monotone increasing in $n$.

Title: Comparing the value of $\int _0 ^{\frac{\pi}{2}} x^n \sin^nx\,dx$ for different $n$

Body:
Let $$I_n = \int _0 ^{\frac{\pi}{2}} x^n \sin^nx \, dx.$$

Could you tell me how to decide which one is true: $I_{2012}>I_{2013} $, or $I_{2013}>I_{2012}$?


问:
Induced topology base
Let $X$ be a topological space, $B$ an open base for $X$ and $Y \subseteq X$. Prove that $B_Y = \{U \cap Y: U \in B\}$ is an open base for $Y$ (considering, over $Y$, the subspace topology).

We must show that given an open $A$ relative to the topology of $Y$, we have to show that there is an element of $B_Y$ such that $ A = \cup (U \cap Y) $. Since $ B $ is a base for X, we have that Y can be written as a union of elements of $B$. That is, there is $W \in B$ such that $ Y = \cup W $. Is this demonstration absurd?

答:
Observe that if $B = \{B_i: i \in I \}$ is an open base of $X$ if and only if for any $x$ in open set $G$ there exist $B_i$ in $B$ such that $x \in B_i \subseteq G$.

Now let $A$ be open in $Y$ then  $A = Y \cap G$ where $G$ is open in $X$ and let $x$ be in $A$. Clearly $x \in G$ which implies there is a $B_i$ in $B$ such that $x \in B_i$ and $B_i \subseteq G$. This also implies that $x \in Y \cap B_i$ and clearly $Y \cap B_i \in B_Y$. So we conclude that $B_Y$ is an open base for subspace $Y$.

Title: Induced topology base


问:
What's the use of having both a proper subset and subset definition when we have the of notion equality?
By definition a set $A$ is a proper subset of another set $B$ if we have $A$ is a subset of $B$ and $A\neq B$, I understand the need for such a definition however what I don't understand what's the use of the subset definition when we can say the A is a proper subset of $B$ in case they are not equal and $A=B$ in case they care, what does the subset definition really add to the table here?

答: "$A$ is a subset of $B$" is equivalent to "$A$ is a proper subset of $B$ or $A=B$" and the first is clearly more succint. We often don't care if a subset is proper.


问:
Closure of unitaries in the SOT
Let $H$ be a Hilbert space. I am trying to find the closure of the collection of unitary operators $U(H)$ in $L(H)$ in SOT. I know the strong limit of unitary operators isnt unitary (Unitary shifts on $\ell^2$). I cannot find any information on this in the literature or google.
This isn't a homework problem or an exam.
This looks useful?

答:
The strong closure of the unitaries is the set of all isometries.  To see this suppose that  $\{u_i\}_i$ is a net of unitaries strongly
converging to an operator $v$.  Then, for every $x$ in $H$, one has that
$$
  \|v(x)\| = \lim_i  \|u_n(x)\| = \|x\|,
  $$
so $v$ is an isometry.

Conversely,  given any isometry $v$ in $B(H)$,  let $U$ be a neighborhood of $v$ in the strong
topology.  Then there exists some $\varepsilon >0$,  and vectors $x_1, x_2, \ldots , x_n$ in $H$,  such that
$$
  \{u\in  B(H): \|u(x_i)-v(x_i)\|<\varepsilon \}\subseteq U.
  $$

Let $H_1=\text{span}\{x_1, x_2, \cdots , x_n\}$ and $H_2=\text{span}\{v(x_1), v(x_2), \cdots , v(x_n)\}$.  Then evidently $v$ restricts
to a unitary operator from $H_1$ to $H_2$.  Observing that  $H_1^\perp$ and $H_2^\perp$ have the same dimension, there
exists a unitary operator $v'$ from $H_1^\perp$ to $H_2^\perp$.   The direct sum operator
$$
  u:= v|_{H_1}\oplus v': H_1\oplus H_1^\perp \to H_2\oplus H_2^\perp
  $$
is unitary and it belongs to $U$,  showing that $v$ lies in the closure of the set of unitary operators.

Title: Closure of unitaries in the SOT


答:
Here is another way to see that $G_1 \cong G_1 \times \{e_{G_2}\}$ is a subgroup.

Note that the map
$$\phi: G_1 \to G_1 \times G_2: g \mapsto (g, , e_{G_2})$$
is a group morphism. Hence, $\text{Im}(\phi) = G_1 \times \{e_{G_2}\}$ is a subgroup of $G_1 \times G_2$ and this is the subgroup you are looking for.

Note moreover that $\phi$ is an injection, so one can even say that $G_1 \times G_2$ contains a subgroup isomorphic to $G_1$.

Title: If given two groups $G_1,G_2$ of order $m$ and $n$ respectively then the direct product $G_{1}\times G_{2}$ has a subgroup of order $m$.

Body:
I'm trying to understand if this statement is false or true, I have tried understanding by an example.

For example if $G_1=(\{{\bar0},\bar1\},+_2),G_3=(\{{\bar0},\bar1,\bar2\},+_3)$ then the direct product of these two groups will be

$G_1\times G_2=\{(\bar0,\bar0),(\bar0,\bar1),(\bar0,\bar2),(\bar1,\bar0),(\bar1,\bar1),(\bar1,\bar2)\}$

I see that $o(G_1\times G_2)=o(G_1)\cdot o(G_2)=m\cdot n=6.$ Can I simply say that I take the group $\{{(\bar0,\bar0),(\bar0,\bar1)}\}$ under the addition modulo 2 and 3 respectively and the statement is true for this case ?

Also if this statement is true shouldn't it mean that it will always be true considering that $o(G_1\times G_2)=o(G_1)\cdot o(G_2)$ and there will always be a subgroup of order $m$?


问:
Wrong direction in electric field of a linear charge
I have a big doubt about direction of electric field. When I try to compute a Electric Field over a point at (0,2) from the linear charge in the $x$ axis from (-5,0) to (0,0) with a density linear charge $\sigma=+20 $nC/m`, the Electric Field is $\vec{E}=-55.6i+83.55j$. This $-55.6i$ don't let me fall sleep. Question : Why the "minus" sign?, must the vector $\vec{E}$ be $\vec{E}=+55.6i+83.55j$ because the linear charge is positive?,

$dq = \sigma*ds$

$\overset{\to }{u}=\frac{\left(x\overset{\to }{i}+2\overset{\to }{j}\right)}{\sqrt{x^2+2^2}}$

$\text{dE}=\frac{\text{dq}}{4 r^2 \pi \epsilon _o}\overset{\to }{u}$

$\frac{\sigma }{4 \pi \epsilon _o}\int _{-5}^0\frac{1}{\left(x^2+2^2\right)^{3/2}}dx*\left(x\overset{\to }{i}+2\overset{\to }{j}\right)=-55.6i+83.55j$

答:
Your calculation has a sign error.

If you consider the $x$ component of the electric force on the unit positive charge, it will be given by:

$\displaystyle dF_x = k\frac{e\sigma dx}{4+x^2} \cos \theta = k\frac{e\sigma dx}{4+x^2} \frac {|x|}{\sqrt{4+x^2}} = k\frac{e\sigma dx}{4+x^2} \frac {-x}{\sqrt{4+x^2}} \\= - ke\sigma x(4+x^2)^{-\frac 32}dx$

because, for that range of $x$ values, the absolute value $|x|$ is given by $-x$. Once you do this, you should get the correct answer.

(By the way, $\theta$ is the angle, measured counterclockwise of the electric force vector on the unit charge from the horizontal axis, $e$ is the magnitude of a unit charge (you divide this away for the electric field strength), $\sigma$ is the charge density, a constant, and $k$ is the usual constant of proportionality involving $\pi$ and the permittivity $\varepsilon_0$).


问:
Primitive of $1 / x \sqrt {a x^2 + b x + c}$ for negative discriminant
I'm having problems with this primitive in the real number domain:

Where $x \ne 0$ and $a x^2 + b x + c > 0$:

$\displaystyle \int \frac {\mathrm d x} {x \sqrt {a x^2 + b x + c} }$
for $b^2 - 4 a c < 0$

According to item $14.283$ in Spiegel's "Mathematical Handbook of Formulas and Tables" (Schaum, 1968) it should work out as:

$\dfrac {-1} {\sqrt c} \left({\sinh^{-1} } {\dfrac {b x + 2 c} {|x| \sqrt {4 a c - b^2} } }\right)$

or in an alternative form:

$\dfrac {-1} {\sqrt c} \left({\arcsin^{-1} } {\dfrac {b x + 2 c} {|x| \sqrt {b^2 - 4 a c} } }\right)$

(ignoring the arbitrary constant of integration, to minimise clutter).

Now I can't work out where the absolute value operator comes in on the $x$.

I have got this far:

$x = \dfrac 1 u$

$\leadsto \dfrac {\mathrm d x} {\mathrm d u} = \dfrac {-1} {u^2}$

$\leadsto \displaystyle \int \frac {\mathrm d x} {x \sqrt {a x^2 + b x + c} } = \int {\dfrac 1 {\dfrac 1 u \sqrt {a \left({\frac 1 u}\right)^2 + b \left({\frac 1 u}\right) + c} } } \dfrac {-\, \mathrm d u} {u^2}$

(by integration by substitution)

$= -\displaystyle \int \dfrac {\mathrm d u} {\frac {u^2} u \sqrt {\frac 1 {u^2} (a + b u + c u^2) } }$

$= -\displaystyle \int \dfrac {\mathrm d u} {\sqrt {a + b u + c u^2} }$

as those $u$ factors all cancel.

Then there is a standard result:

$\displaystyle \int \dfrac {\mathrm d u} {\sqrt {a + b u + c u^2} } = \begin {cases}
\dfrac {-1} {\sqrt c} \ln \left\vert {2 \sqrt c \sqrt {a + b u + c u^2} + 2 c u + b}\right\vert + C & : b^2 - 4 a c > 0 \\
\dfrac {-1} {\sqrt c} \sinh^{-1} \left({\dfrac {2 c u + b} {\sqrt{4 a c - b^2} } }\right) + C & : b^2 - 4 a c < 0 \\
\dfrac {-1} {\sqrt c} \ln |2 c u + b| + C & : b^2 - 4 a c = 0
\end {cases}$

I'm only interested in the case where $b^2 - 4 a c < 0$.

Anyway, I substitute back the $u = \dfrac 1 x$ and after algebra I work it out to be:

$= \sinh^{-1} \left({\dfrac {b x + 2 c} {x \sqrt{4 a c - b^2} } }\right)$

Taking a step back I can't see where the $|x|$ would come from in Spiegel's result. Because $\sinh^{-1}$ is a bijection over the reals, an odd function continuous at $0$, there should be no discontinuity in the derivative, so the $|x|$ puzzles me.

答:
The problem is in this place: You have
$$\frac {u^2} u \sqrt {\frac 1 {u^2} (a + b u + c u^2) } =  \sqrt {a + b u + c u^2} $$
It is not true. It should be
$$\frac {u^2} u \sqrt {\frac 1 {u^2} (a + b u + c u^2) } =  \pm\sqrt {a + b u + c u^2}. $$


问:
(NIP Theory) If $I$ is an endless indiscernible sequence and $J$ is a Morley sequence of $\lim(I)$ then $I+J^*$ is indiscernible.
In an NIP theory, show that if $I$ is an endless indiscernible sequence and $J$ is a Morley sequence in the type $\lim(I)$, then $I+J^*$ is indiscernible.

Here, $J^*$ means the sequence $J$ with the order reversed.

As we are working in a NIP theory, the type $p=\lim(I)$ is a global complete type.

I know I can reduce this problem to prove that for any $i_1,\dots,i_{n+1}\in I$ and $j_1,\dots,j_m\in J^*$ there exists $j'\in J^*$ (whose index in $J^*$ is less than the index of $j_1$) such that $$i_1,\dots,i_n,i_{n+1},j_1,\dots,j_m\equiv i_1,\dots,i_n,j',j_1,\dots,j_m.$$
I also know that I can assume that $J$ is $I$-indiscernible.

I don't exactly know how to attack this problem. I think I am missing something. Some hints or references will be appreciated.

答:
You're on the right track with your reduction, but you'll find it easier to "move" elements of $J^*$ into $I$, rather than the reverse. Also, it's easier to work formula-by-formula than to realize the complete type all at once.

First, let's observe the following.

Claim: For any formula $\varphi$, any $i_1<\dots<i_n$ in $I$, and any $j'<j_1<\dots<j_m$ in $J^*$, if $$\varphi(i_1,\dots,i_n,j',j_1,\dots,j_m)$$ then there exists $i'>i_n$ in $I$ such that $$\varphi(i_1,\dots,i_n,i',j_1,\dots,j_m).$$

Proof: By the definition of Morley sequence, $j'$ realizes $\lim(I)$ over $I$ and $j_1,\dots,j_m$. By the definition of $\lim(I)$, any formula in this type is realized by an element of $I$ (in fact, by a whole end segment). $\square$

Ok, now fix some $i_1<\dots<i_n$ in $I$ and $j'<j_1<\dots<j_m$ in $J^*$ and assume $$\varphi(i_1,\dots,i_n,j_1,\dots,j_m).$$ Applying the claim $m$ times, we obtain $i_{n+1},\dots,i_{n+m}$ with $i_1<\dots<i_n < i_{n+1} < \dots < i_{n+m}$ such that $$\varphi(i_1,\dots,i_n,i_{n+1},\dots,i_{n+m}).$$
By indiscernibility of $I$, it follows that for any $i_1'<\dots,i'_{n+m}$ in $I$, $$\varphi(i_1',\dots,i_{n+m}').$$
Since $\varphi$ was an arbitrary formula, we  conclude that for any $i_1'<\dots,i'_{n+m}$ in $I$, $\text{tp}(i_1,\dots,i_n,j_1,\dots,j_m) = \text{tp}(i_1',\dots,i'_{n+m})$. Since $(i_1,\dots,i_n,j_1,\dots,j_m)$ was an arbitrary increasing sequence from $I+J^*$, we conclude that any such sequence has the same type, and hence $I+J^*$ is indiscernible.


问:
Controlling eigenvalues of a 4th order system
Consider the state space system $\dot{z} = Az + Bu$ with $A = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\0 & 0 & 0 & 0\end{bmatrix}$ and $B = \begin{bmatrix} 0 & 0\\ 1 & 0\\ 0 & 0 \\ 0 & 1\end{bmatrix}$. We need to design a control law $u=−Kz$ so that all eigenvalues of $A−BK$ have a real part that is less than $−10$, where $K=\begin{bmatrix} k_{11} & k_{12} & k_{13}& k_{14} \\ k_{21} & k_{22} & k_{23}& k_{24}\end{bmatrix}$.

I am trying to look for a mathematical manual approach as opposed to one involving coding. I am not sure that such a method even exists without any assumptions. For me, I got stuck after computing $\det(\lambda I-(A-BK))=0$, which gives $\lambda(\lambda^2+k_{24}\lambda+k_{23})(k_{11}+k_{12}+\lambda)-(k_{14}\lambda+k_{13})(\lambda k_{22}+k_{21})=0$. This seems quite unapproachable to me from here and I think there should be a nicer approach to solving the problem, since the $A$ matrix has a rank of $2$ which I haven't used anywhere. Any leads and hints would be appreciated.

Thanks

答:
Since $$A-BK=\begin{pmatrix}0&1&0&0\\-k_{11}&-k_{12}&-k_{13}&-k_{14}\\0&0&0&1\\-k_{21}&-k_{22}&-k_{23}&-k_{24}\end{pmatrix}$$ it is enough to take $k_{21}=k_{22}=0$; then the eigenvalues of $A-BK$ are those of the two sub-matrices $$\begin{pmatrix}0&1\\-k_{11}&-k_{12}\end{pmatrix},\qquad\qquad\begin{pmatrix}0&1\\-k_{23}&-k_{24}\end{pmatrix}$$ on the diagonal. The requirement is then that the eigenvalues of each sub-matrix $\begin{pmatrix}0&1\\-a&-b\end{pmatrix}$ (where $a$ represents $k_{11}$, $k_{23}$, and $b$ represents $k_{12}$, $k_{24}$) have real part less than $-10$. Looking at the trace and determinant, this is the case when $b>20$ and $a>100$.

Conclusion: A sufficient condition for all eigenvalues of $A-BK$ having real part less than $-10$ is $$k_{21}=k_{22}=0,\qquad k_{11}, k_{23}>100,\qquad k_{12},k_{24}>20$$


问:
Question about inverse with respect to convolution product.
Let $\mathcal{I}(X)$ be the collection of real valued functions  $f:X\times X\to \mathbb{R}$ with the property that $f(x,y)=0$ when $x>y$. The convolution product $f*g$ for $f,g\in \mathcal{I}(X)$ is defined by $$ f*g(x,y)=\begin{cases}
      \sum_{\lbrace z:x\le y\le z\rbrace}f(x,z)g(z,y) & x\le y \\
      0 & \text{otherwise} \end{cases}$$
From page 14 of http://www.math.ust.hk/~mabfchen/Math232/Inclusion-Exclusion.pdf, I don't follow the last step.

If the latest summ where $\delta$ is the Kronecker Delta, is equal to $\delta(x,y)$, wouldn't that imply the sum representing $g(x,y)$ is 0?

答:
No. It means, if $x<y$ then
$$\sum_{x\le z\le y} g(x,z)\,f(z,y)=0\,.$$
And, also $g(x,x)\,f(x,x)=1$.

For example, if $y$ covers $x\ $ ($y>x$ and there is no element in between), we have
$$g(x,y)=-\frac1{f(y,y)}\big(g(x,x)\,f(x,y)\big)=-\frac{f(x,y)}{f(x,x)\,f(y,y)}\,.$$
Then
$$g(x,x)\,f(x,y)\ +\ g(x,y)\,f(y,y)\ =\ \frac{f(x,y)}{f(x,x)}\ +\ \frac{-f(x,y)}{f(x,x)}\ =\ 0$$
indeed.


问:
Product rule of probabilities and conditioning
I have a probably very basic question regarding the product rule for probabilities. If we have a two probability densities say $p(a|b)$ and $p(b|c,d)$, is $p(a|b)p(b|c,d) = p(a,b | c,d)$? What has me slightly confused is that the second term is conditioned on variables $c$ and $d$ and so I am not sure the product rule applies directly here. I would have thought that the first term $p(a|b)$ would need be conditional on $c$ and $d$ explicitly also.

Would someone mind clarifying this for me please?

答:
$$P(a|b) P(b|c,d ) = P(a,b | c,d) \\ \iff \frac{P(a\cap b)}{P(b)} \times P(b\cap c\cap d) = P(a\cap (b\cap c\cap d))$$

Suppose $a$ and $b$ are independent and $P(b\cap c\cap d)\ne 0$. Then this means $$P(a) P(b\cap c\cap d) = P(b\cap c\cap d) \cdot P(a | b\cap c\cap d) \\ \iff P(a)=P(a|b\cap c\cap d)$$ This is true if and only if $a$ and $b\cap c\cap d$ are also independent, but this is not necessarily true. Therefore, the identity you propose isn’t necessarily true.


问:
Number of lattice points inside a triangle and its area
Consider a triangle that has lattice points as vertices (Lattice points are points with integer coordinates). However it is given that no other lattice point lies inside or on the sides of the triangle. We need to prove that the area of the triangle is $\frac{1}{2}$.

I did it in a way that I didn't like which went on like (first, I saw that it is trivial for triangles in the lattice plane that has one of their side parallel to either the $x$-axis or $y$-axis; the rest of the triangles can be checked exhaustively by taking care of all possible different cases and with the help of Pythagoras theorem).

I wish to know an efficient way to solve this.

Also I was curious whether there is a relation between the area of the polygon and the number of lattice points falling inside or on the sides of a polygon.

答:
You can calculate the amount of latice points inside a triangle using Pick´s theorem.

You find:

Let $$ A (x_1, y_1), B (x_2, y_2), C (x_3, y_3) $$
be three points with integer coordinates not all on one line.

Then the amount of latice points L inside the triangle is given by:

$$ L = \frac{1}{2}A_{ \Delta} + 1 - \frac{1}{2} \sum_{1 \le i \lt j \le 3, i \neq j} {gcd( x_i - x_j, y_i - y_j )} $$

where $$ A_{\Delta} $$ is the area of the triangle ABC.

Like I said, one can proof this easily with Pick´s theorem.


问:
Irreducible polynomials have distinct roots?
I know that irreducible polynomials over fields of zero characteristic have distinct roots in its splitting field.

Theorem 7.3 page 27
seems to show that irreducible polynomials over $\Bbb F_p$ have distinct roots in its splitting field (and all the roots are powers of one root). Is the proof correct? I have never seen this result anywhere else. The proof is very convincing to me.

Does the result hold for $\Bbb F_q$ where $q$ is a power of prime? I don't think it holds because I've heard there are irreducible polynomials with repeated roots?

Please help.

答:
Consider a field $F$ of characteristic $p$.
A polynomial has multiple roots only if it has a root in common with its (formal) derivative; that is, the multiple roots of $f$ are the roots of $\gcd(f,f')$. Since $f$ is irreducible, multiple roots can occur only if the $\gcd$ is $f$ itself, that is $f'$ is a multiple of $f$. And that is only possible if $f'=0$, that is, all monomials in $f$ have degree a multiple of $p$, so $f(x)=g(x^p)$ for some polynomial $g$.

If $F$ is finite, then $\phi\colon a\mapsto a^p$ is an automorphism of $F$ (and also of the splitting field $E$ of our polynomial), and there exists $h(x)$ such that $\phi(h)=g$. Then for $\alpha\in E$ with $f(\alpha)=0$ also $h(\alpha)=0$ (because $\phi(h(\alpha))=\phi(h)(\phi(\alpha))=g(\alpha^p)=f(\alpha)=0$). Since $h$ is of smaller degree than $f$, $f$ is not irreducible.

As this proof shows, one has to look for cases where $\phi$ is not an automorphism to find a counterexample (such as in Andreas Carantis comment).


问:
regarding derivatives
Does the derivative (the slope value) only give us the instantaneous rate change at a point and nothing else? Is this the only significance of that value?

For example I took the function $f(x)=x^2$,lets consider at $x=3$. $f(x)=x^2=9$ and $f'(x)=2x=6$. Does the value $6$ only tells us rate of change at point $x=3$?

Lets consider average rate change, e.g. suppose the average rate of change in y with respect to x over some interval is $7$; that is, for every single unit by which x changes, "y" on average changes by $7$ units. Here the average value "$7$" is related to the y value of the function means y on average changes by $7$ units.

Some of my friends said that the value $f'(x)=6$ in the above example only gives the rate of change at that point, i.e it is just the slope of the tangent at that point and nothing else, it has no effect on the y value ($x^2=9$) of the function. But if it doesn't have to do anything with 'y' value, then why is the slope at point called "the instantaneous rate of change of y with respect to x"?can anyone explain to me this, I really need help.

答:
lets take a function x^2 at x=3, f(x)=9 and here the derivative 2x=6.
the value 6 is called as      "" w.r.t  x (why rate change of y? thats the reason i started thinking if 6 and 9 are connected,and that change in y meant slope causes change in y.)  how does 6 cause rate change of y?how does it change the rate of the function?

Explanation:

There really is no relationship between the 6 in f'(x) = 6 and the 9 in f(x) = 9.

If the rate of change of y wrt x is 6 at that point, then roughly and informally speaking, if x changes by a small amount, then y will change by approximately 6 times that small amount.
For example, if x changes by 0.001, then y will change by approximately 0.006.

More specifically, if x changes from 3 to 3.001, then y will change from 9 to approximately 9.006. i ll explain this below.

Now consider a different x value, which I'll call x' (pronounced "x prime"). This doesn't mean the derivative of x, it just means a different x value. Since it's close to the original x value, we can use it to approximate the value of the function at that new x' value.

Δx = x' - x = 0.001

Δf ≈ f'(x) Δx = (6) (0.001) = 0.006

f(x') ≈ f(x) + Δf = 9 + 0.006 = 9.006 (Approximate value).

So in a way, the 6 is related to the y values near the point we're looking at. If we look at the single point where x=3, then y=9 at that point. If we look at nearby values of x (very close to 3), then the change in y will be approximately 6 times the change in x.

now, when x changes from 3 to 5 i.e x changes by 2 units then y should change 2*6=12, therefore 9+12=21 but at x=5, y=25(x^2) so there is a lot of error. as gap between points increases the error increases.  now if i say x changes from 4.9 to 5 the error will be less i get 24.99 and x^2=25.

so f '(x) does have a relationship with the y values but in a small region around x=3.

my main problem was i used to think if a slope is 6 then the y value(at x=3 which here is 9) should increase by 6,but now i get the meaning of "   ". "with respect to x" the meaning lies in this sentence. i hadnt concentrated on the topic of linearization and differentials that much so i guess i got stuck with this query.
thank you.

Title: regarding derivatives


问:
Can the definition of the Beta function be proved by expanding the integral?
On wikipedia the proof that
$$ B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} $$
puts the whole expression under a double integral.

Would it be possible to prove the fact that
$$\int_0^1 t^{x-1}(1-t)^{y-1}dt = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$
By expanding the integral and having a sum? I have tried to do so but have been unsuccessful many times.

答:
Too long for a comment:

The integral is $\sum_{n\ge0}\frac{(-1)^n(y-1)_n}{n!}\int_0^1t^{n+x-1}dt=\sum_n\frac{(-1)^n(y-1)_n}{n!(n+x)}$ in terms of falling Pochhammer symbols. Since $(y-1)_n=\frac{\Gamma(y)}{\Gamma(y-n)}$ (if $y-n$ is an integer $\le0$ the denominator diverges so the symbol is $0$, or we could just truncate the infinite series), we want to prove $\sum_n\frac{(-1)^n}{n!(n+x)\Gamma(y-n)}=\frac{\Gamma(x)}{\Gamma(x+y)}$. In terms of the Hankel contour, the series is$$\frac{i}{2\pi}\oint_H\sum_n\frac{t^n}{n!(n+x)}(-t)^{-y}e^{-t}dt=\frac{i}{2\pi}\oint_H\Gamma(x,\,0,\,-t)(-t)^{-x-y}e^{-t}dt,$$with the last $\Gamma$ a generalized incomplete Gamma. I expect this can be evaluated to the desired result with the residue theorem.


问:
Primes represented by $x^3-21xy^2+35y^3$.
What do we know about the primes represented by this special binary cubic form $x^3-21xy^2+35y^3$?

I know that my question is very short, but I don't have any idea about it, and I don't know where can I find the answer in the literature.

I searched through the net to find a program to check whether a binary cubic equation $f(x, y)=n$ has a solution or not, but I didn't find anything. In the case of the absence of an answer, or a reference to my question, introducing any program/ engine would be welcome.

Let $\alpha$ be a root of the polynomial $x^3-21x-35=0$, and let $K:=\mathbb{Q}(\alpha)$. Then it is easy to show that $$Norm(x+y\alpha+z\alpha^2)=x^3+35y^3+1225z^3-105xyz-21xy^2+441xz^2+42x^2z-735yz^2.$$ That Binary cubic form is just $Norm(x+y\alpha)$.

Note that the discriminant of $P(x)=x^3-21x-35$ is
$-(4\times(-21)^3+27\times(-35)^2)=3969=3^4\times7^2$, so discriminant of $K$ is a square, so it is a cyclic cubic Galois extension, so we can conclude that $r_1=3$ and $r_2=0$. By Dirichlet's unit theorem, we can conclude that $\mathcal{O}_K^{\times}=\{\pm1\}\times\mathbb{Z}^2$. Also, note that $P(x)=x^3-21x-35$ is $7$-Eisenstein, and $P(x-1)=x^3-3x^2+3x-1-21x+21-35=x^3-3x^2-18x-15$ is $3$-Eisenstein; so we can conclude that $\mathcal{O}_K=\mathbb{Z}[\alpha]=\mathbb{Z}\oplus\mathbb{Z}\alpha\oplus\mathbb{Z}\alpha^2$.

Is the answer to the following question positive?

Assume that $Norm(a+b\alpha+c\alpha^2)=p$. Does there exist a unit $u \in \mathcal{O}_K^{\times}$ such that $(a+b\alpha+c\alpha^2)\times u = A+B\alpha$ for some integers $A, B$? Assume that $a+b\alpha+c\alpha^2$ is given. Can we find a suitable unit, such that after the multiplication, we can write the product as linear combination of $1$, and $\alpha$? without need of $\alpha^2$?

Body:
Not a 'real' answer, but it was too big for a comment. I think that you're looking for a solution without using a calculator or PC but maybe this gives some insight. I did only a quick search with the following bounds: $-50\le x\le50$ and $-50\le y\le50$.

I wrote and ran some Mathematica-code:

In[1]:=Clear["Global`*"];
\[Alpha] = -50;
\[Beta] = 50;
ParallelTable[
  If[TrueQ[PrimeQ[x^3 - 21*x*y^2 + 35*y^3] &&
     x^3 - 21*x*y^2 + 35*y^3 >= 2], {x, y, x^3 - 21*x*y^2 + 35*y^3},
   Nothing], {x, \[Alpha], \[Beta]}, {y, \[Alpha], \[Beta]}] //. {} ->
   Nothing

Running the code gives:

Out[1]={{{-48, 25, 1066283}, {-48, 49, 6427331}}, {{-47, -21,
   7309}, {-47, -15, 127}, {-47, 11, 62189}, {-47, 15, 236377}, {-47,
   21, 655579}, {-47, 26, 1178549}, {-47, 30, 1729477}}, {{-46, -17,
   9883}, {-46, -15, 1889}, {-46, 27, 1295783}, {-46, 33,
   2212433}}, {{-44, -15, 4591}, {-44, 15, 240841}, {-44, 17,
   353807}, {-44, 23, 829457}, {-44, 35, 2547341}}, {{-43, -20,
   1693}, {-43, 15, 241793}, {-43, 34, 2340001}, {-43, 40,
   3605293}, {-43, 45, 4938443}}, {{-41, -18, 5923}, {-41, -15,
   6679}, {-41, 17, 351863}, {-41, 23, 812393}, {-41, 45,
   4863979}, {-41, 48, 5785543}}, {{-39, -17, 5417}, {-39, 25,
   999431}, {-39, 32, 1926217}, {-39, 37, 2834747}, {-39, 43,
   4237757}}, {{-38, -15, 6553}, {-38, 9, 35281}, {-38, 41,
   3698801}}, {{-37, -15, 6047}, {-37, 9, 37799}, {-37, 10,
   62047}, {-37, 16, 291619}, {-37, 21, 616139}, {-37, 39,
   3207329}, {-37, 40, 3432547}}, {{-36, 7, 2393}, {-36, 13,
   158003}, {-36, 35, 2380069}, {-36, 37, 2761163}, {-36, 43,
   4133933}}, {{-34, -15, 3221}, {-34, 7, 7687}, {-34, 27,
   1170107}, {-34, 37, 2711017}, {-34, 43, 4063627}}, {{-33, -14,
   3851}, {-33, 14, 195931}, {-33, 16, 284831}, {-33, 26,
   1047691}, {-33, 34, 2140811}, {-33, 35, 2313613}, {-33, 40,
   3312863}, {-33, 49, 5745671}}, {{-32, -15, 307}}, {{-31, 10,
   70309}, {-31, 12, 124433}, {-31, 15, 234809}, {-31, 22,
   657973}, {-31, 25, 923959}, {-31, 33, 1936943}}, {{-29, -13,
   1637}, {-29, -10, 1511}, {-29, 8, 32507}, {-29, 12, 123787}, {-29,
   15, 230761}, {-29, 17, 323567}, {-29, 20, 499211}, {-29, 23,
   723617}, {-29, 27, 1108477}, {-29, 33, 1896607}, {-29, 38,
   2775527}, {-29, 45, 4398211}, {-29, 50, 5873111}}, {{-27, -11,
   2339}, {-27, -10, 2017}, {-27, 29, 1310779}, {-27, 34,
   2011409}, {-27, 41, 3345679}, {-27, 46, 4586849}, {-27, 50,
   5772817}}, {{-26, 5, 449}, {-26, 27, 1069363}, {-26, 33,
   1834813}, {-26, 35, 2151899}, {-26, 47, 4822343}}, {{-24, 7,
   22877}, {-24, 23, 678637}, {-24, 25, 848051}, {-24, 43,
   3700817}, {-24, 47, 4733317}}, {{-23, 5, 4283}, {-23, 6,
   12781}, {-23, 11, 92861}, {-23, 21, 524971}, {-23, 26,
   929501}, {-23, 29, 1247651}, {-23, 30, 1367533}, {-23, 39,
   2798641}, {-23, 50, 5570333}}, {{-22, -9, 1259}, {-22, 9,
   52289}, {-22, 15, 211427}, {-22, 19, 396199}, {-22, 21,
   517229}, {-22, 25, 824977}, {-22, 45, 4114277}}, {{-19, -8,
   757}, {-19, 7, 24697}, {-19, 10, 68041}, {-19, 18, 326537}, {-19,
   22, 558937}, {-19, 25, 789391}, {-19, 28, 1074277}, {-19, 33,
   1685447}, {-19, 42, 3290057}, {-19, 43, 3513637}, {-19, 48,
   4783157}}, {{-18, 5, 7993}, {-18, 11, 86491}, {-18, 41,
   3041821}}, {{-17, -6, 379}, {-17, 5, 8387}, {-17, 11, 84869}, {-17,
    21, 476659}, {-17, 24, 684559}, {-17, 30, 1261387}, {-17, 35,
   1933037}, {-17, 36, 2090719}, {-17, 44, 3667679}}, {{-16, 7,
   24373}, {-16, 33, 1619603}}, {{-13, -6, 71}, {-13, 10,
   60103}, {-13, 16, 211051}, {-13, 25, 715303}, {-13, 31,
   1302841}, {-13, 34, 1689031}, {-13, 36, 1984571}}, {{-12, -5,
   197}, {-12, 19, 329309}, {-12, 31, 1283129}}, {{-11, 3,
   1693}, {-11, 5, 8819}, {-11, 12, 92413}, {-11, 15, 168769}, {-11,
   20, 371069}, {-11, 30, 1151569}, {-11, 35, 1782269}, {-11, 38,
   2252753}, {-11, 42, 2999233}, {-11, 47, 4142753}}, {{-9, 2,
   307}, {-9, 8, 29287}, {-9, 10, 53171}, {-9, 13, 108107}, {-9, 25,
   664271}, {-9, 32, 1339687}, {-9, 35, 1731421}, {-9, 43,
   3131477}, {-9, 50, 4846771}}, {{-8, 9, 38611}, {-8, 15,
   155413}, {-8, 29, 994391}, {-8, 45, 3529063}}, {{-6, 5, 7309}, {-6,
    13, 97973}, {-6, 25, 625409}, {-6, 43, 3015503}, {-6, 47,
   3911923}}, {{-4, 3, 1637}, {-4, 7, 16057}, {-4, 27, 750077}, {-4,
   33, 1349207}}, {{-3, 1, 71}, {-3, 4, 3221}, {-3, 5, 5923}, {-3, 11,
    54181}, {-3, 19, 262781}, {-3, 40, 2340773}, {-3, 44,
   3103381}, {-3, 46, 3540041}, {-3, 49, 4268951}}, {{-2, 5,
   5417}, {-2, 9, 28909}, {-2, 11, 51659}}, {{-1, 7, 13033}, {-1, 15,
   122849}, {-1, 18, 210923}, {-1, 22, 382843}, {-1, 27, 704213}, {-1,
    30, 963899}, {-1, 40, 2273599}, {-1, 43, 2821573}}, {{1, 2,
   197}, {1, 3, 757}, {1, 5, 3851}, {1, 12, 57457}, {1, 17,
   165887}, {1, 23, 414737}, {1, 35, 1474901}}, {{2, 19, 224911}, {2,
   21, 305621}, {2, 25, 520633}}, {{3, 4, 1259}, {3, 14, 83719}, {3,
   20, 254827}, {3, 26, 572599}, {3, 29, 800659}, {3, 34,
   1302839}, {3, 40, 2139227}, {3, 44, 2859499}}, {{4, 5, 2339}, {4,
   15, 99289}, {4, 17, 147743}, {4, 27, 627733}, {4, 33, 1166383}, {4,
    45, 3019339}}, {{6, 7, 6047}, {6, 13, 55817}, {6, 17, 135757}, {6,
    23, 359407}, {6, 35, 1346491}}, {{8, 1, 379}, {8, 45,
   2849687}, {8, 49, 3714859}}, {{9, 5, 379}, {9, 8, 6553}, {9, 10,
   16829}, {9, 20, 205129}, {9, 22, 281933}, {9, 23, 326593}, {9, 43,
   2434013}}, {{11, -2, 127}, {11, 3, 197}, {11, 7, 2017}, {11, 12,
   28547}, {11, 15, 67481}, {11, 25, 403831}, {11, 45,
   2722931}}, {{12, 1, 1511}, {12, 25, 391103}, {12, 35,
   1193653}, {12, 49, 3514391}}, {{13, -1, 1889}, {13, 11,
   15749}, {13, 14, 44729}, {13, 15, 58897}, {13, 24, 328789}, {13,
   30, 701497}, {13, 35, 1168397}, {13, 36, 1281349}, {13, 45,
   2638747}}, {{16, -3, 127}, {16, 3, 2017}, {16, 5, 71}, {16, 27,
   448057}, {16, 33, 895987}}, {{17, 1, 4591}, {17, 9, 1511}, {17, 19,
    116101}, {17, 24, 283121}, {17, 31, 704521}, {17, 39,
   1538081}, {17, 40, 1673713}, {17, 46, 2656261}}, {{18, -1,
   5419}, {18, 5, 757}, {18, 11, 6679}, {18, 29, 541549}}, {{19, 5,
   1259}, {19, 12, 9883}, {19, 18, 81703}, {19, 30, 592759}, {19, 33,
   830143}, {19, 35, 1018709}, {19, 45, 2388259}}, {{22, -1,
   10151}, {22, 21, 131041}, {22, 29, 475721}, {22, 41,
   1646261}}, {{23, 1, 11719}, {23, 4, 6679}, {23, 6, 2339}, {23, 15,
   21617}, {23, 39, 1353689}, {23, 45, 2223467}}, {{24, 17,
   40123}, {24, 23, 173053}, {24, 35, 897049}, {24, 37,
   1096703}}, {{26, -3, 11717}, {26, 27, 308447}}, {{27, -1,
   19081}, {27, 5, 9883}, {27, 14, 4591}, {27, 16, 17891}, {27, 19,
   55061}, {27, 20, 72883}, {27, 25, 212183}, {27, 31, 517481}, {27,
   35, 825733}}, {{29, -5, 4789}, {29, -2, 21673}, {29, 3,
   19853}, {29, 7, 6553}, {29, 18, 31193}, {29, 25, 190639}, {29, 27,
   269333}}, {{31, 3, 24877}, {31, 5, 17891}, {31, 8, 6047}, {31, 20,
   49391}, {31, 30, 388891}, {31, 32, 510047}, {31, 33, 578647}, {31,
   45, 1900891}}, {{32, -5, 11593}, {32, 9, 3851}, {32, 19,
   30241}, {32, 31, 429661}}, {{33, 1, 35279}, {33, 10, 1637}, {33,
   16, 1889}, {33, 20, 38737}, {33, 29, 306739}, {33, 34,
   610469}, {33, 35, 687637}, {33, 46, 1976309}, {33, 49,
   2489759}, {33, 50, 2678437}}, {{34, 23, 87443}, {34, 33,
   519553}, {34, 35, 665279}, {34, 45, 1782829}}, {{36, 7,
   21617}, {36, 17, 127}, {36, 23, 72577}, {36, 37, 784547}, {36, 43,
   1431557}}, {{37, -6, 15121}, {37, 5, 35603}, {37, 6, 30241}, {37,
   11, 3221}, {37, 20, 19853}, {37, 30, 296353}, {37, 41,
   1156751}}, {{38, 9, 15749}, {38, 31, 330679}}, {{39, -5,
   34469}, {39, -2, 55763}, {39, 7, 31193}, {39, 20, 11719}, {39, 22,
   35603}, {39, 23, 51913}, {39, 28, 185543}}, {{41, 7, 38737}, {41,
   12, 5417}, {41, 13, 307}, {41, 22, 24877}, {41, 43,
   1259677}}, {{43, -6, 39439}, {43, -4, 62819}, {43, -1, 78569}, {43,
    6, 54559}, {43, 11, 16829}, {43, 21, 5419}, {43, 26, 84239}, {43,
   29, 173699}, {43, 39, 782209}, {43, 44, 1312739}}, {{44, -5,
   57709}, {44, 3, 77813}, {44, 7, 51913}, {44, 13, 5923}, {44, 25,
   54559}, {44, 27, 100493}, {44, 37, 593083}, {44, 45,
   1403459}}, {{46, -7, 37997}, {46, -3, 87697}, {46, 33,
   303157}, {46, 35, 414611}}, {{47, 1, 102871}, {47, 4, 90271}, {47,
   9, 49391}, {47, 10, 40123}, {47, 39, 678761}, {47, 40,
   764623}}, {{48, -5, 81017}, {48, 1, 109619}, {48, 5, 89767}, {48,
   35, 376417}, {48, 41, 828379}}}

So, with the bounds $-50\le x\le50$ and $-50\le y\le50$ we find $402$ solutions. In order to find that I used:

In[2]:=Clear["Global`*"];
\[Alpha] = -50;
\[Beta] = 50;
f = Total@*Map[Length];
f[ParallelTable[
   If[TrueQ[
     PrimeQ[x^3 - 21*x*y^2 + 35*y^3] &&
      x^3 - 21*x*y^2 + 35*y^3 >= 2], {x, y, x^3 - 21*x*y^2 + 35*y^3},
    Nothing], {x, \[Alpha], \[Beta]}, {y, \[Alpha], \[Beta]}] //. {} \
-> Nothing]

Out[2]=402

If we extend the bounds to $-10^3\le x\le10^3$ and $-10^3\le y\le10^3$ we find $92522$ solutions. If we extend the bounds, again, to $-10^4\le x\le10^4$ and $-10^4\le y\le10^4$ we find $6950603$ solutions.


问:
Accept reject method to generate random numbers
The method says that having a proposal $g(x)$

Sample $X^* \tilde ~ g(x)$ and $U \tilde ~ Unif(0,1)$

Accept $X = X^*$ if $U ≤ f(X^*) / M g(X^*)$

Moreover, $M$ is constant that satisfies $Mg(x) ≥ f(x) \forall x$

But how should I choose this $g(x)$? And what about the constant $M$?

Say I have the following function $\displaystyle f(x) = \frac{x^3}{e^{x}+1}$ what does $g(x)$ should look like?

Sorry for my no-math-language I'm not used to it.

答:
The main idea is to look at the tail of $f(x)$.  That is how does $f(x)$ behave when $x$ is very large.  Then we need to find something that looks like it, but is easy to integrate.

We want to keep $g$ as close to $f$ as possible If $M$ is high we reject a lot of samples.

In your case the difficult bit is the $\frac 1{1+e^x}$. but for large values of $x$ that's very close to $e^{-x}$ so  $g(x)\sim x^3 e^{-x}$ is the right sort of candidate. It's not too far away from $f(x)$ and we can integrate it to get

$$\int_x^\infty t^3 e^{-t} dt = e^{-x}\left( x^3 + 3x^2 + 6x + 6 \right)$$

We have to be careful because $\int_0^\infty t^3 e^-t dt = 6$ so we should set
$$ g(x) = \frac {x^3}{6 e^{x}}$$
Therefore
$$\mathbb P\left (X^*>x\right) = e^{-x}\frac{ x^3 + 3x^2 + 6x + 6}6.$$
So $X^*$ is easy to simulate.

Now we have to find $\max_{x_\in \mathbb R} \frac{f(x)}{g(x)}$

It's easy to calculate $\dfrac{f(x)}{g(x)} = \dfrac{6e^x}{1+e^x}<6$ So we can set $M = 6$

So for high values of $X^*$ we hardly ever reject, because $\dfrac{f(x)}{Mg(x)}\to 1$  for low values we might reject with probability up to $\frac 12$, which isn't bad.

Choosing a good proposal distribution is more of an art than a science, the main thing is to make sure the tails agree, and that you're don't end up rejecting too many.


答:
For any real $ x>0 $ we have :
$$ \Gamma\left(x\right)=\displaystyle\int_{0}^{+\infty}{z^{x-1}\mathrm{e}^{-z}\,\mathrm{d}z} $$

Thus, if $ u \in \left[0,+\infty\right[ $, and for any $ y>0 $ we have :

\begin{aligned} \Gamma\left(x\right)u^{y-1}\mathrm{e}^{-u}&=u^{y-1}\mathrm{e}^{-u}\displaystyle\int_{0}^{+\infty}{z^{x-1}\mathrm{e}^{- z}\,\mathrm{d}z} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left\lbrace\begin{matrix}z =u t\  \\
\mathrm{d}z =u\,\mathrm{d}t\end{matrix}\right.\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Gamma\left(x\right)u^{y-1}\mathrm{e}^{-u}&=u^{x+y-1}\mathrm{e}^{-u}\displaystyle\int_{0}^{+\infty}{t^{x-1}\mathrm{e}^{-u t}\,\mathrm{d}t} \\ \Longrightarrow\displaystyle\int_{0}^{+\infty}{\Gamma\left(x\right)u^{y-1}\mathrm{e}^{-u}\,\mathrm{d}u}&=\displaystyle\int_{0}^{+\infty}{u^{x+y-1}\mathrm{e}^{-u}\displaystyle\int_{0}^{+\infty}{t^{x-1}\mathrm{e}^{-u t}\,\mathrm{d}t}\,\mathrm{d}u}\\ \iff\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Gamma\left(x\right)\Gamma\left(y\right)&=\displaystyle\int_{0}^{+\infty}{t^{x-1}\displaystyle\int_{0}^{+\infty}{u^{x+y-1}\mathrm{e}^{-\left(1+t\right)u}\,\mathrm{d}u}\,\mathrm{d}t} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left\lbrace\begin{matrix}\tau =\left(1+t\right)u \\
\mathrm{d}u =\frac{\mathrm{d}\tau}{1+t}\ \ \ \ \ \ \ \ \ \ \ \end{matrix}\right.\\ &=\displaystyle\int_{0}^{+\infty}{\displaystyle\frac{t^{x-1}}{\left(1+t\right)^{x+y}}\displaystyle\int_{0}^{+\infty}{\tau^{x+y-1}\mathrm{e}^{-\tau}\,\mathrm{d}\tau}\,\mathrm{d}t}\\ &=\Gamma\left(x+y\right)\displaystyle\int_{0}^{+\infty}{\displaystyle\frac{t^{x-1}}{\left(1+t\right)^{x+y}}\,\mathrm{d}t} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left\lbrace\begin{matrix}\varphi =\frac{1}{1+t}\ \  \\
\mathrm{d}t =-\frac{\mathrm{d}\varphi}{\varphi^{2}}\end{matrix}\right. \\\Gamma\left(x\right)\Gamma\left(y\right)&=\Gamma\left(x+y\right)\displaystyle\int_{0}^{1}{\displaystyle\frac{1}{\varphi^{2}}\left(\displaystyle\frac{1}{\varphi}-1\right)^{x-1}\varphi^{x+y}\,\mathrm{d}\varphi}\\ \iff\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \displaystyle\frac{\Gamma\left(x\right)\Gamma\left(y\right)}{\Gamma\left(x+y\right)}&=\displaystyle\int_{0}^{1}{\varphi^{y-1}\left(1-\varphi\right)^{x-1}\,\mathrm{d}\varphi}\end{aligned}

Title: Can the definition of the Beta function be proved by expanding the integral?

Body:
On wikipedia the proof that
$$ B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} $$
puts the whole expression under a double integral.

Would it be possible to prove the fact that
$$\int_0^1 t^{x-1}(1-t)^{y-1}dt = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$
By expanding the integral and having a sum? I have tried to do so but have been unsuccessful many times.


问:
A really basic inequality
Prove that $5^{44}<4^{53}$

I tried to prove that $5^{44}<4^{52}$

$\implies 5^{11}<4^{13}$

But couldn't

答:
Note that
$
4^{53} = 2^{106} ,
$
so you need to show that $$2^{106} > 5^{44} \; . $$

Now try to find a simpler inequality of that form $2^x > 5^y$:
$$
2^7 = 128 > 125 = 5^3
$$
Now we can use this to prove the original inequality:
$$
2^{106} = 2 \cdot 2^{105} = 2 \cdot \left( 2^7 \right)^{15} > 2 \cdot \left( 5^3 \right)^{15} = 2 \cdot 5^{45} > 5^{45} > 5^{44} \; .
$$

Title: A really basic inequality


问:
How do we find the volume of an Euclidean ball in $\mathbb{R}^n$?
Following up on this post, I am not sure how the volume of a Euclidean ball has been found in the following extract:

Questions:

What does it mean when we say "the mass of a measure"?

What does rotational invariance mean here, and how was it established?

Why do we integrate the function $$x \mapsto \exp \left(-\frac{1}{2}\sum_1^n x_i^2\right)$$ to find $v_n$?

How is the function invariant under rotations?

How did we pull out $nv_n$ from the integral in (1.1)?

I apologize for asking too many questions but I'd really appreciate any help I can get in understanding the discourse above! Thanks a lot.

答:
The order 12435 would have been more helpful, but:

A mass-$1$ measure on the surface is basically a probability density, so the mass of a region therein is the probability of a random point lying therein ("random" in the sense of whatever distribution we choose to sample). A mass-$M$ measure would be $M$ times that.

Rotation about the centre, which is here taken as the origin, doesn't change the surface.

The motive is that $\prod_i\exp(-\tfrac12x_i^2)dx_i=\exp(-\tfrac12\underbrace{\sum_ix_i^2}_{r^2})d^nx=r^{n-1}\exp(-\tfrac12r^2)drd\Omega$, with $d\Omega$ the infinitesimal element of solid angle.

Rotations also don't change sums of the form $\sum_ix_iy_i$ because$$R^TR=I\implies(Rx)^TRy=x^TR^TRy=x^Ty.$$

Integrating over the surface,$$\left(\int_{\Bbb R}\exp(-\tfrac12x^2)dx\right)^n=nv_n\int_0^\infty r^{n-1}\exp(-\tfrac12r^2)dr$$(because, while $v_n$ is the volume, $nv_n$ is the surface $\int_{S^{n-1}}d\Omega$).


问:
$J_n(z)=(z/2)^n\frac{1}{\pi(\frac{1}{2})_n}\int_{0}^{\pi}\cos(z\cos\theta)\sin^{2n}\theta d\theta$
In "A treatise on the theory of Bessel functions, Watson, p.48" he ends up with this relation for the Bessel function:
$$J_n(z)=\frac{(1/2*z)^n}{\Gamma(n+1/2)\Gamma(1/2)}\int_{0}^{1}t^{n-1/2}\left [ \sum_{m=0}^{\infty}\frac{(-1)^mz^{2m}(1-t)^{m-1/2}}{2m!} \right ]dt$$. How can Ι use this to get this identity for Bessel functions?
$J_n(z)=(z/2)^n\frac{1}{\pi(\frac{1}{2})_n}\int_{0}^{\pi}\cos(z\cos\theta)\sin^{2n}\theta d\theta$

答:
Notice that
\begin{align*}
\sum\limits_{m = 0}^\infty  {\frac{{( - 1)^m z^{2m} (1 - t)^{m - 1/2} }}{{(2m)!}}}  & = (1 - t)^{ - 1/2} \sum\limits_{m = 0}^\infty  {\frac{{( - 1)^m (z\sqrt {1 - t} )^{2m} }}{{(2m)!}}} \\ & = (1 - t)^{ - 1/2} \cos (z\sqrt {1 - t} )
\end{align*}
and perform the substitution $t=\sin^2 \theta$, $0<\theta<\frac{\pi}{2}$. This, together with the definition of the Pochhammer symbol, gives
$$
J_n (z) = \left( {\frac{z}{2}} \right)^n \frac{1}{{\pi \left( {\frac{1}{2}} \right)_n }}2\int_0^{\pi /2} {\cos (z\cos \theta )\sin ^{2n} \theta d\theta } .
$$
Finally,
\begin{align*}
\int_0^{\pi /2} {\cos (z\cos \theta )\sin ^{2n} \theta d\theta } & = \int_0^{\pi /2} {\cos (z\cos (\pi  - \theta ))\sin ^{2n} (\pi  - \theta )d\theta }
\\ & = \int_{\pi /2}^\pi  {\cos (z\cos \varphi )\sin ^{2n} \varphi d\varphi }
\end{align*}
shows that
$$
2\int_0^{\pi /2} {\cos (z\cos \theta )\sin ^{2n} \theta d\theta }  = \int_0^\pi  {\cos (z\cos \theta )\sin ^{2n} \theta d\theta } .
$$


问:
Proving or disproving a statement about discontinuity

I'm trying to disprove this statement, but as spend more time on it I'm beginning to think it might actually be true.
How should I approach on proving/disproving this statement?

答:
Im not giving a complete solution but something to start with. Assume $g=fD$ is continous at $x_0$. Then for $\epsilon>0$ there exists $\delta_{\epsilon}>0$, such that
$$|g(x)-g(x_0)|<\epsilon \;\text{for all}\; |x-x_0|<\delta_{\epsilon}\tag{1}$$
Take $r\notin\mathbb{Q}$ with $|r-x_0|<\delta_{\epsilon}$. Plugging $r$ into (1) we get that $\epsilon>|g(r)-g(x_0)|=|g(x_0)|$. Since $\epsilon$ was arbitrary we have that $g(x_0)=0$. From there try to argue that $f(x_0)=0$.

The other direction is also straight forward using epsilon-delta criterion.


问:
If $A=\begin{bmatrix} 1 & 2 \\ 5 & 4 \end{bmatrix}$, find a non singular matrix $P$ such that $P^{-1}AP=\begin{bmatrix} 6 & 0 \\ 0 & -1\end{bmatrix}$.

If $A=\begin{bmatrix} 1 & 2 \\ 5 & 4 \end{bmatrix}$, find a non singular matrix $P$ such that $P^{-1}AP=\begin{bmatrix} 6 & 0 \\ 0 & -1\end{bmatrix}$.

I let $P=\begin{bmatrix}a & b \\ c & d\end{bmatrix}$, so that $P^{-1}=\dfrac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ and I put it in the equation, but as expected, it will get tedious because of two degree terms. I couldn't think of any other method to solve it. I then tried to pre-muliply with $P$ in the equation to get $AP=P\begin{bmatrix} 6 & 0 \\ 0 & -1\end{bmatrix}$ and then letting again $P=\begin{bmatrix}a & b \\ c & d\end{bmatrix}$, but here though I am getting simple linear equations, but they are only two instead of four, namely $5a=2c$ and $b=-d$. So, this too fails.

I wanted to express $P$ solely in terms of $A$, $A^{-1}$ by some kind of manipulation in the original equation but couldn't think of any. I am sure there would exist a simple method for this. So, any hints?

答:
Notice that $A$ has eigenvalues $\lambda_1=6$ and $\lambda_2=-1$. This suggests proceeding by diagonalizing $A$. For a $n\times n$ matrix $A$, if one has eigenvalues $\lambda_1,\lambda_2,\dots,\lambda_n$ (they don't have to be distinct) with $n$ linearly independent eigenvectors, then one has that
$$\begin{bmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n \end{bmatrix}=P^{-1}AP,$$
where $P=\begin{bmatrix} \mathbf{v}_1 & \mathbf{v}_2 & \cdots & \mathbf{v}_n \end{bmatrix}$ is a matrix consisting of eigenvectors $\mathbf{v}_1,\mathbf{v}_2,\dots,\mathbf{v}_n$ corresponding to $\lambda_1, \lambda_2,\dots,\lambda_n$ respectively.

Hence, for our specific problem, we have that (this is one specific choice of eigenvectors, one could also take any non-zero scalar multiple)
$$\mathbf{v}_1=\begin{bmatrix} 2 \\ 5 \end{bmatrix},\quad \mathbf{v}_2=\begin{bmatrix} -1 \\ 1 \end{bmatrix}.$$
Hence a possible matrix $P$ is given by
$$P=\begin{bmatrix} 2 & -1 \\ 5 & 1 \end{bmatrix}.$$

Alternatively, proceeding by your method, you obtained the linear system (which was done correctly)
$$\begin{cases} 5a=2c, \\ b=-d, \end{cases}$$
which has infinitely many solutions (it is a consistent linear system of nullity $2$), but if you fix for instance $c=5$ and $d=1$, you will get the same $P$ I have obtained. In fact, you could pick $P$ to be any matrix of the form (you can obtain this by fixing $c=\alpha$ and $d=\beta$)
$$P=\begin{bmatrix} 2\alpha & -\beta \\ 5\alpha & \beta \end{bmatrix},$$
where $\alpha,\beta\neq 0$.


问:
Geometrical combinatorics with $(n+1)^2$ points inside a square that determine a triangle
Consider a square with the side of length n and $(n+1)^2$ points inside it. Show that we can choose 3 of them to determine a triangle (possibly degenerate) of area at most $\frac{1}{2}$.

I think that I know how to solve the problem for the cases $n=1$ and $n=2$:

For $n=1$ we can easily prove that the triangle formed by those 3 points has the area at most $\frac{1}{2}$,like here:Maximum area of a triangle in a square

For $n=2$ we can divide the square into 4 smaller squares (by uniting the midpoints of the opposite sides),each of them with the length of the side equal to 1.By pigeonhole priciple we know that at least one of those squares contains at least 3 of the 9 points inside it.From here we can use the same proof from the case $n=1$ to prove that those 3 points determine a triangle of area at most $\frac{1}{2}$

This idea doesn't work for the other cases because we would need $2n^2+1$ points instead of $(n+1)^2$

Can you help me please with the other cases?
Source:book:Combinatorics Advances,by Charles J. Colbourn

答:
The idea is to consider the convex hull $C$ of the $(n+1)^2$ points, and split into cases depending on whether $C$ has "many" or "few" points.

If $C$ has at least $4n$ points, then as the perimeter of $C$ is at most $4n$, we can find two adjacent sides whose total length is at most $2$. These two sides determine a triangle of area at most $\frac{1}{2}$.

Else $C$ has $k<4n$ points. Triangulate $C$ into $k-2$ triangles, then iteratively triangulate these triangles via the other $(n+1)^2-k$ points. We have split $C$ into $k-2+2\left[(n+1)^2-k\right]>2n^2$ triangles, so one of them has area at most $\frac{1}{2}$.


问:
Proving that $\lim_{n \to \infty} \frac{1}{\sqrt[n]{n!}}=0$
I have to prove this
$$\lim_{n \to \infty} \frac{1}{\sqrt[n]{n!}}=0$$

My proof is the following:

By using $$n!>n^{\frac{n}{2}}$$
$$\frac{1}{n!}<\frac{1}{n^{\frac{n}{2}}}$$
$$\left(\frac{1}{n!}\right)^{\frac{1}{n}}<\left( \frac{1}{n}\right)^{\frac{n}{2}\cdot\frac{1}{n}}=\frac{1}{\sqrt{n}}\underset{n\to \infty}{\to}0$$
Using the Squeeze theorem
$$0<\frac{1}{\sqrt[n]{n!}}<\frac{1}{\sqrt{n}} \implies \frac{1}{\sqrt[n]{n}}\underset{n\to \infty}{\to} 0$$

I think I have done something wrong. Can I even use the first inequality here? and if yes-do I have to prove that too? and is it Ok that I write too little words in my proofs?. Could you please help me?

答:
Your approach is basically OK, but it really should include a proof of the key inequality $n!\gt n^{n/2}$ (for $n\gt2$). Here's a proof by induction, starting with $2!=2^{2/2}$ and using the inequality $(1+1/n)^n\lt e\lt3$.

If $n!\ge n^{n/2}$, then the strict inequality $(n+1)!\gt(n+1)^{(n+1)/2}$ follows:

$$\begin{align}
(n+1)!
&=(n+1)n!\ge(n+1)n^{n/2}\\
&=(n+1)^{(n+1)/2}\left(n+1\over\left(1+{1\over    n}\right)^n\right)^{1/2}\\
&\gt(n+1)^{(n+1)/2}\left(2+1\over3 \right)^{1/2}\\
&=(n+1)^{(n+1)/2}
\end{align}$$

Note, the assumption $n\ge2$ comes into play in the numerator replacement $n+1\ge2+1$; everything else holds even for $n=1$; in particular the strict inequality $(1+1/n)^n\lt3$ holds for $n\ge1$. Of course, one can (and maybe should) now insist on a proof of $(1+1/n)^n\lt 3$, but I'll take that inequality as known.


问:
Laplace functional of cluster process
Consider the simple cluster process:
$$\sum_n \xi_n \epsilon_{X_n}$$ where $\{X_n\}$ are Poisson points independent of the iid non-negative integer sequence $\{\xi_n\}$. How do I find the Laplace functional? I am a bit confused reading about it online.

This is what I have so far. Let $\psi_N(f)$ be the Laplace functional and suppose $\mu$ is the mean measure of the Poisson random measure defined by $N$.
\begin{align*}
N(A) &= \sum_n \xi_n \epsilon_{X_n}(A) \\
\psi_N(f) &= E[e^{-\sum_n \xi_n \epsilon_{X_n(A_n)}}] \hbox{ where $A_n$ are disjoint} \\
&= E[\prod_{n} e^{-\xi_n N(A_n)}]\\
&= \prod_{n} E[e^{-\xi_n N(A_n)}]\\
&= \prod_{n} \exp(-\int (1-e^{-\xi_n 1_{A_n}(x)}) \mu(dx))
\end{align*}

答:
Let us denote $g$ the probability mass function of $\xi_1$ and $\mu(A)$ the mean measure of the Poisson process on $A$. Then $N(A)$ has the following Laplace functional at point $\lambda$:

$E[e^{-\lambda N(A)}]$ = $E[E[e^{-\lambda \sum_{n=1}^\infty \xi_n\epsilon_{X_n}} | \sum_{n=1}^\infty\epsilon_{X_n} = k]] = \sum_{k=0}^\infty \big[\frac{\mu(A)^ke^{-\mu(A)}}{k!}\big(\sum_{i=0}^\infty e^{-\lambda i} g(i)\big)^k \big]$

$=\exp[-\mu(A)(1-\sum_{i=0}^\infty e^{-\lambda i}g(i))] = \exp[-\mu(A)\sum_{i=0}^\infty (1- e^{-\lambda i})g(i)]$,

where the second equality follows because ${\xi_n}$ and ${X_n}$ are independent and ${\xi_n}$ are iid.

Title: Laplace functional of cluster process


问:
Can complex numbers be visualized

If $z=|x+iy|$ is a complex number satisfying $\;|z-2|-|z-3i|=\sqrt5\;$ then locus of $z$ is $\dots$

On solving this I easily get an equation of part of hyberbola.

My question is: Is there any general result for locus problems like $(x-a)^2-(y-b)^2=c$ is a general equation of hyperbola and can be visualized easily?

Is there any sort of visualization for complex numbers?

答:
Just a little algebra work.
$$
|x-2+iy|-\left| x+\left( y-3 \right) i \right|=\sqrt{5}
\\
\sqrt{\left( x-2 \right) ^2+y^2}-\sqrt{x^2+\left( y-3 \right) ^2}=\sqrt{5}
\\
\left( x-2 \right) ^2+y^2+x^2+\left( y-3 \right) ^2-2\sqrt{\left( x-2 \right) ^2+y^2}\sqrt{x^2+\left( y-3 \right) ^2}=5
\\
\left( x-2 \right) ^2+y^2+x^2+\left( y-3 \right) ^2-5=2\sqrt{\left( x-2 \right) ^2+y^2}\sqrt{x^2+\left( y-3 \right) ^2}
\\
\left( \left( x-2 \right) ^2+y^2+x^2+\left( y-3 \right) ^2-5 \right) ^2=4\left( \left( x-2 \right) ^2+y^2 \right) \left( x^2+\left( y-3 \right) ^2 \right)
\\
-4x^2-48xy+80x+16y^2-80=0
\\
x^2+12xy-20x-4y^2+20=0
$$
Thus the visualized curve (of this complex equation) on $\mathbb R^2$ is $x^2+12xy-20x-4y^2+20=0$.

Then can you continue from here?

Hint. $$-4\left( y-\frac{3}{2}x \right) ^2+10\left( x-1 \right) ^2=-10
$$
and substitue $t=y-\dfrac{3}{2}x$.

Title: Can complex numbers be visualized


问:
Show uniform convergence of $\sum\limits_n(1-x^2)^2 x^n$ on $[0,1]$
I'm trying to show that if $u_n(x) = (1-x^2)^2 x^n$ then $\sum u_n$ is uniformly convergent on $[0,1]$.

The only thing I can think to use is Weierstraß M-test to show that $u_n < M_n$ where $\sum  M_n$ converges, but I can't find a way to make the $M_n$ independent of $x$.

答: HINT: We can see the limit $u(x)=(1-x^2)^2\frac1{1-x}=(1+x)^2(1-x)$ from the geometric series. Use that the limit is continuous and all $u_n$ are $\ge 0$.


问:
Showing that $\mathcal{L}(n)\cong U(n)/O(n)$
I have been reading the book Introduction to Symplectic Topology, page $50$, and now we want to see that $\mathcal{L}(n)=\mathcal{L}(\mathbb{R}^{2n},\omega_0)\cong U(n)/O(n)$. To see this we have the following lemma :

i) If $\Lambda \in \mathcal{L}(n)$ and $\Psi\in Sp(2n)$ then $\Psi \Lambda\in \mathcal{L}(n)$

This was easy to prove .

ii) For any two lagrangian subspaces $\Lambda,\Lambda'\in \mathcal{L}(n)$ there exists $\Psi\in Sp(2n)\cap O(2n)$ such that $\Lambda'=\Psi\Lambda$.

To prove this the author fixes a lagrangian subspace $\Lambda\in \mathbb{R}^{2n}$ and chooses an unitary Lagrangian frame. My first doubt is why is this always possible ?

Then we have that $\Psi \Lambda_{hor}=\Lambda$, and they end the proof here. How does this give us what we want ? I don't see how we relate this to an arbritary $\Lambda'\in \mathcal{L}(n)$.

iii) There is a natural diffeomorphism $\mathcal{L}(n)\cong U(n)/O(n)$.

Here they just mention that an unitary matrix $U=X+iY$ determined by a lagrangian frame is uniquely determined by $\Lambda$ up to right multiplication by a matrix in $O(n)$. I don't see why this would be the case , but I guess it's related to the fact that we can associate to $\Lambda$ a Lagrangian Frame and I am not sure how this construction is done.

Any help is aprpreciated. Thanks in advance.

答:
To prove ii), it suffices to show that given $\Lambda\in\mathcal{L}(n)$, there exists $\Psi_\Lambda\in\operatorname{Sp}(2n)\cap O(2n)$ such that $\Psi_\Lambda\Lambda_{\textrm{hor}}=\Lambda$, where $\Lambda_{\textrm{hor}}=\{(x,y);y=0\}$ is the horizontal Lagrangian.

Given $\Lambda,\Lambda'\in\mathcal{L}(n)$, define $\Psi=\Psi_{\Lambda'}{\Psi_{\Lambda}}^{-1}\in\operatorname{Sp}(2n)\cap O(2n)$ (recall that $\operatorname{Sp}(2n)\cap O(2n)$ is a group), it satisfies $$\Psi\Lambda=\Psi_{\Lambda'}{\Psi_\Lambda}^{-1}\Lambda=\Psi_{\Lambda'}\Lambda_{\textrm{hor}}=\Lambda',$$
which is the required result.

Let $\Lambda\in\mathcal{L}(n)$, then $\Lambda$ admits a unitary frame.

Since $\Lambda$ is a vector subspace, using Gram-Schmidt orthonormalization, it admits an orthonormal basis $\mathcal{B}=(e_1,\ldots,e_n)$, letting $e_i=(x_i,y_i)$, $X=(x_1\vert\ldots\vert x_n)$, $Y=(y_1\vert\ldots\vert y_n)$ and $Z=\begin{pmatrix}X\\Y\end{pmatrix}$, it follows that
$$\Lambda=\operatorname{range}(Z),$$
and $U=X+iY$ is unitary, indeed a straighforward computation gives:
$$\begin{align*}UU^*&=(X+iY)({}^tX-i{}^tY),\\
&=X{}^tX+Y{}^tY+i(Y{}^tX-X{}^tY),\\
&=I,\end{align*}$$
since $X{}^tX+Y{}^tY=Z{}^tZ=I$ (recall that $Z$ is orthonormal, since $\mathcal{B}$ is orthonormal) and $Y{}^tX=X{}^tY$ (recall that $\Lambda$ is Lagrangian). Finally, $Z$ is a unitary Lagrangian subspace for $\Lambda$.

There is a diffeomorphism $\Lambda(n)=U(n)/O(n)$.

As explained above, a unitary Lagrangian frame for $\Lambda$ amounts to a choice of an orthonormal basis for it, whence the result.


问:
On finding matrices of linear transformations
I'm trying to understand the meaning behind matrix of a linear transformation. To illustrate what I don't get consider the example below.
Suppose $t: \mathbb{R}^2 \rightarrow \mathbb{R}^3 $ is a linear transformation given by
$$(x,y) \mapsto (2x, 3x+y, y)$$

Suppose also that the domain basis is $E = \{(1,1), (1,0) \}$ and the codomain basis is $F= \{(1,1,1), (0,1,1),(0,0,1) \}$. Then in order to find the matrix of this transformation all we have to do is to find the transformation of the basis vectors, so in this case:

$$t(1,1) = (2,4,1) $$ $$t(1,0) = (2,3,0)$$

But then we need to convert these transformations in the codomain basis and we have:

$$(2,4,1) = (2,2,-3)_F$$
$$(2,3,0) = (2,1,-3)_F$$

From here we deduce that the matrix of the linear transformation is given by

$$\begin{bmatrix}
2 & 2 \\ 2 & 1 \\ -3 & -3
\end{bmatrix}$$

Now I understand how to do computations like the one above. However, I don't understand what does the intermediate  transformation "basis vectors" represent (i.e. what is the meaning of $t(1,1) = (2,4,1)$)? What do these vectors represent in the codomain? And how does converting these vectors to their representation in the codomain basis give the correct columns for the matrix we are looking for? Why do we need to convert in the first place?

More concisely given $\vec{v} \in V$ the matrix $A$ of the transformation satisifies $[T(\vec{v})]_F = A[\vec{v}]_E$, but what does $T(\vec{v})$ represent (without conversion to basis $F$)?

答:
As a general case, where you have different bases in both the domain and the codomain, there has to be something in between the raw vectors and the transformation matrix in order to change the vectors you want to transform from their original basis to the basis in which you have defined your linear transformation. Your linear transformation will only correctly transform vectors that are expressed in its same basis.

In order to do this in a simple manner, what we use is a change of basis matrix. The columns of the change of basis matrices are the vectors of the original basis expressed in coordinates of the basis of the transformation. These change of basis matrices are sometimes denoted as $M(B_2,B_1)$, where $B_1$ is the original basis in which the vectors are expressed (input), and $B_2$ is the basis of the linear transformation (output). This means that the "new" change of basis matrix $A(B_2,B_1)$ will be:

$$A(B_2,B_1)=A(B_2,B_2)\cdot M(B_2,B_1)$$

Where $A(B_2,B_2)$ is your original change of basis matrix that takes vectors from the basis $B_2$ and outputs transformed vectors also in $B_2$.

Note that here the output vectors will be in the basis $B_2$. If you want them back in the basis $B_1$, then you need another change of basis matrix to take those vectors back into $B_1$:

$$A(B_1,B_1)=M(B_1,B_2)\cdot A(B_2,B_2)\cdot M(B_2,B_1)$$

$M(B_1,B_2)$ can be computed easily as the inverse of $M(B_2,B_1)$, as it is doing the inverse operation.

I hope this helps to understand the use of change of basis matrices for linear transformations.


答:
Let $\;f\;$ be a function that is continuous at $\;x_0\in\mathbb{R}\;.$ Then $\;g=f\cdot D\;$ is continuous at $\;x_0\;$ if and only if $\;f(x_0)=0\;,\;$ where $\;D(x)=\begin{cases}1\quad x\notin\mathbb{Q}\\0\quad x\in\mathbb{Q}\end{cases}.$

First I prove that

$g(x)\;$ continuous at $\;x_0\;\implies\;f(x_0)=0\;.$

If $\;f(x_0)\;$ were $\;\neq0\;,\;$ since $\;f(x)\;$ and $\;g(x)\;$ are continuous at $\;x_0\;$, the function $\;D(x)=\dfrac{g(x)}{f(x)}\;$ would be continuous at $\;x_0\;$ too, but it is impossible because $\;D(x)\;$ is a nowhere continuous function.

Now, I prove that

$f(x_0)=0\;\implies\;g(x)\;$ continuous at $\;x_0\;.$

$\begin{align}\bigl|g(x)-g(x_0)\bigr|&=\bigl|f(x)D(x)-f(x_0)D(x_0)\bigr|=\\&=\bigl|f(x)D(x)\bigr|\leqslant\bigl|f(x)\bigr|= \bigl|f(x)-f(x_0)\bigr|\;.\end{align}$

Since $\;f(x)\;$ is continuous at $\;x_0\;$, then

$\forall\;\varepsilon>0\;$ there exists $\;\delta>0\;$ such that

$\;\bigl|x-x_0\bigr|<\delta\implies \bigl|g(x)-g(x_0)\bigr|\leqslant\bigl|f(x)-f(x_0)\bigr|<\varepsilon\;.$

Hence, by definition, $\;g(x)\;$ is continuous at $\;x_0\;.$

Title: Proving or disproving a statement about discontinuity

Body:
I'm trying to disprove this statement, but as spend more time on it I'm beginning to think it might actually be true.
How should I approach on proving/disproving this statement?