问:
Integral of product of Bessel functions of the first kind
I would like to solve the integral:

$$\int_0^{+\infty}\quad rJ_n(ar)J_n(br)\quad dr$$

Is there any closed form for it?
Thanks!

What happens if I have instead:

$$\int_0^{c}\quad rJ_n(ar)J_n(br)\quad dr$$

with $c>0$.

(Since the first part has already a proper answer, the bounty is offered actually for the second part!)

答:
According to Wolfram functions (at the bottom) this is simply (for any $n$ in $\mathbb{R}$) :
$$\int_0^{+\infty}r\,J_n(ar)J_n(br)\ dr=\frac {\delta(a-b)}a$$

The same formula appears in DLMF where this closure equation appears with the constraints $\ \Re(n)>-1,\;a>0,\;b>0$ and additional references (A & W $11.59$ for example).

It may be proved using the Hankel transform : see Wikipedia's 'Hankel transform'.

If we replace the upper (infinite) bound by a finite bound $c$ we get the result kindly indicated by J.M. (changing merely the sign since we want the integral over $(0,c)$ and in conformity with Alpha's result $\not =$ and result $=$ for some specific value of $n$) :
\begin{align}
\tag{1}\int_0^c r\,J_n(a r) J_n(b r) \,dr&=\frac c{a^2-b^2}\left[b \;J_n(a\,c)\,J_{n-1}(b\,c)-a\;J_{n-1}(a\,c)\,J_n(b\,c)\right]\quad\text{if}\ a\not =b\\
\tag{2}\int_0^c r\,J_n(a r)^2 \,dr&=\frac {c^2}2\left[J_n(a\,c)^2-J_{n-1}(a\,c)\,J_{n+1}(a c)\right]\\
\end{align}

The first result appears as entry $03.01.21.0064.01$ at Wolfram functions while the second is entry $03.01.21.0055.01$. They are usually called 'first and second Lommel integrals'.

For a really venerable reference you may consult Whittaker & Watson p.$381$ with this interesting exercise (using our previous notations) :

"Shew that, if $a\not =b$ and $n>-1$",
$$(a^2-b^2)\int_0^c r\,J_n(a r) J_n(b r) \,dr=c\left[J_n(a\,c)\frac d{dc}J_n(b\,c)-J_n(b\,c)\frac d{dc}J_n(a\,c)\,\right],$$
$$2a^2\int_0^c r\,J_n(a r)^2 \,dr=(a^2c^2-n^2)J_n(a\,c)^2-\left[c\frac d{dc}J_n(a\,c)\right]^2$$

For a proof see page $33$ and $34$ of Culham's lecture or Bowman's book page $101$.

The recurrence relations (Wikipedia) :
$$\frac d{dc}J_n(a\,c)=\frac a2\left[J_{n-1}(a\,c)-J_{n+1}(a\,c)\right]= a\,J_{n-1}(a\,c)-\frac ncJ_n(a\,c)$$
should allow you to obtain the $(1)$ and $(2)$ formulas.

(the exercise $19$ from W & W with $a$ and $b$ two different zeros of $J_n$ seems of interest too)


问:
How to sketch $\ln|\sin x|$
The question is to find the number of solutions of
$$\ln|\sin x|=-x^2+2x$$ but I just don't know how to sketch a logarithmic function like $\ln|\sin x|$.

A step by step method would be highly appreciated.
Thanks in advance.

答:
We are looking for solutions to $\sin{x} = \pm e^{-x^{2}+2x}$.

Now, (just taking one part of the right hand side), the graph of $e^{-x^{2}+2x}$ approaches $0$ as $x \to \infty$, so the graph of $y=\sin{x}$ will intersect the graph of $y=e^{-x^{2}+2x}$ infinitely many times. This will give infinitely many solutions to $\ln{|\sin{x}|} = -x^{2}+2x$.

ContentLicense: CC BY-SA 4.0

Title: How to sketch $\ln|\sin x|$


问:
how can I prove that if I have a regular language L, that L' is a regular language
how can I prove that if  I have a regular language L
and I create a new language L'

where L' = (L but with last letter repeated)
(i.e. if ab is in language L then abb is in language L')

that L' is a regular language

I tried solving the question but I have no way of being sure of
my solutions so I'll post it here:

L is a regular language therefor there is a regular expression for it,
we create a new language x where x =L*($\Sigma$),
$\Sigma$ is all the letters in language L
,x is a regular language because it is regular language chained with a letter
,and letters are a regular expression
the language L' is a subset of x,
there for x can be written as x=L' $\cup$ t (t=x/L'),
and therefor L' has to be a regular expression
because if it wasn't then x would not be a a regular language

答:
Let $A$ be the alphabet. For each letter $a \in A$, let
$$
  La^{-1} = \{u \in A^* \mid ua \in L\}
$$
It is a well known fact that if $L$ is regular, then so is $La^{-1}$. Now your language $L'$ can be written as
$$
  L' = \bigcup_{a \in A}\ (La^{-1})aa
$$
For each $a \in A$, $(La^{-1})aa$ is the product of the two regular languages $La^{-1}$ and $aa$ and this is regular. It follows that $L'$ is a finite union of regular languages and this is regular.


问:
A question about the property of covering numbers
I am reading the book "Multivariate approximation"(Page 321 - 322) written by V. Temlyakov and thinking about the following property of covering numbers:

Let $X$ be a Banach space and let $B_X$ denote the unit ball of $X$ whose center is at $0$.

Denote $N_{\epsilon}(A,X):=\min\{n: \exists  y^1, y^2, ... ,y^n: A\subset \cup_{j=1}^n B_X(y^j,\epsilon)\}$.

Prove that:
$$N_{\epsilon_1\epsilon_2}(A,X)\leq N_{\epsilon_1}(A, X)\cdot N_{\epsilon_2}(B_X, X).$$

The author says this is obvious but I am a bit of confusion.

My idea here is to use other balls in $X$ to cover the balls $B_X(y^j,\epsilon)$ again but I have no idea why the right-hand side is the product of two covering numbers.

Any suggestions would be welcome and thank you !!

答:
Your idea is on the right track. We start off by covering $A$ with $N_{\epsilon_1}(A,X)$ balls of radius $\epsilon_1$, and then cover each of those balls by $N_{\epsilon_2}(B_X,X)$ balls of radius $\epsilon_1\epsilon_2$. This produces a cover of $A$ by $N_{\epsilon_1}(A,X)\cdot N_{\epsilon_2}(B_X,X)$ balls of radius $\epsilon_1\epsilon_2$ and proves the covering number inequality.

Now, to prove we can cover a radius $\epsilon_1$ ball by $N(B_X,X)$ balls of radius $\epsilon_1\epsilon_2$:

Since $B(y,\epsilon_1)= y+B(0,\epsilon_1)$, we can WLOG work with a ball at the origin. Note that $\delta B_X = B_X(0,\delta)$ for $\delta>0$ by homogeneity, where by $\delta A$ I mean the set of scalings $\{\delta a:a\in A\}$. If $m:=N_{\epsilon_2}(B_X,X)$, then there are $m$ balls centered at points $y^1,\dotsc, y^m$ such that
$$B_X\subset B(y^1,\epsilon_2)\cup\dotsm \cup B(y^m,\epsilon_2),$$ so
$$B(0,\epsilon_1) = \epsilon_1 B_X \subset \epsilon_1B(y^1,\epsilon_2)\cup\dotsm\cup\epsilon_1 B(y^m,\epsilon_2)$$
where I use the fact that $\delta(A\cup B)\subset \delta A\cup \delta B$ for $A,B$ subsets of a Banach space. We also have $\epsilon_1B(y,\epsilon_2) = B(\epsilon_1y,\epsilon_1\epsilon_2)$, which is an easy exercise in definitions. Hence we have covered $B(0,\epsilon_1)$ by $m$ balls of radius $\epsilon_1\epsilon_2$, as desired.


问:
Inequality form olympiad
Let a , b , c be real numbers such that $a^2+b^2+c^2=2$

Show that $a+b+c \leq abc+2$

Please help. I'm stuck on this one. I tried a lot of things. The problem is I can't use standard inequalities because they're not necessarily positive.

答:
Proof 1. Using the Cauchy-Schwarz inequality, we have
$$[a(1-bc) +(b+c)]^2 \leqslant [a^2+(b+c)^2][(1-bc)^2+1] = 2(1+bc)[(1-bc)^2+1].$$
Thefore, we need to prove
$$(1+bc)[(1-bc)^2+1] \leqslant 2,$$
or
$$b^2c^2(bc-1) \leqslant 0.$$
Which is true because
$$bc \leqslant \frac{b^2+c^2}{2} \leqslant \frac{a^2+b^2+c^2}{2}=1.$$
Done.

Proof 2. We need to prove
$$4-(a+b+c-abc)^2 \geqslant 0.$$
equivalent to
$$4-(a+b+c)^2+2abc(a+b+c)-4a^2b^2c^2 \geqslant 0.$$
Setting $A=a^2+b^2+c^2=2,$ the inequality equivalent to
$$4[4-(a+b+c)^2+2abc(a+b+c)-a^2b^2c^2] \ge 0$$
or
$$2^3-2^2 \cdot [(a+b+c)^2-2]+2 \cdot 4abc \cdot (a+b+c)-4a^2b^2c^2\geqslant 0,$$
$$A^3-A^2\cdot\left[(a+b+c)^2-A\right]+4 \cdot A \cdot abc \cdot (a+b+c)-4a^2b^2c^2 \geqslant  0,$$
or
$$(2-2ab)(2-2bc)(2-2ca)+4a^2b^2c^2 \geqslant 0.$$
Which is true because
$$2-2ab=(a^2-2ab+b^2)+c^2=(a-b)^2+c^2 \geqslant 0.$$
The proof is completed.

Title: Inequality form olympiad


问:
Eigenvalues of a DAE and stability analysis
I have a nonlinear system of differential algebraic equations (DAE) of index -1,  and I was wondering: what is the "meaning" of eigenvalues in that case?

I know they do not indicate stability, as with regular ODEs, unless the Algebraic terms are eliminated, which is not an option in my case.

What do they mean, then?

How do I investigate the stability, in such a DAE?

答:
There are two ways that I can imagine, not necessarily the best ways.

Way I:

A DAE of index one is easily transformable to an ODE system, i.e. one simply eliminates the algebraic equations and substitute them within the differential equations. Then the eigenvalue-based stability analysis can be performed on the resulting ODE system.

Way II:

Differentiate the algebraic equations w.r.t. time, so that the algebraic variables become state variables. Introduce consistent initial conditions using the algebraic equations. Then eigenvalue stability analysis can be applied to the resulting ODE system.

A generalization to DAE systems of arbitrary index can be followed by a combination of Pantelides algorithm and method of dummy derivatives for index reduction.

OwnerDisplayName: Aladdin


问:
An example of matrices $A_1$ and $A_2$, for which the rank of sum $A_1+A_2$ is greater than $max(r_1, r_2)$
Let's assume that the matrices $A_1$, $A2$ $M_{n×n}(\mathbb{Q})$ are of rank $r_1$ and $r_2$ respectively. Give an example of matrices $A_1$ and $A_2$, for which the rank of sum $A_1+A_2$ is greater than $max(r_1, r_2)$. Give an example of $A_1$ and $A_2$ for which the rank of sum $A_1+A_2$ is smaller than $min(r_1, r_2)$.

Is that possible? I tried to do the first part and I allways get the rank equal to $max(r_1, r_2)$. So I am not sure if there is an answer to that question.

答: Hint: Consider $A_1+A_2=I$. For the second one, consider subtraction rather than addition.


问:
Conditions for function to be discontinuous at 0
I'm supposed to find out the set of values of $m$ for which f is discontinuous at 0, for $f(x)=x^m\sin\frac{1}{x}$. The solutions obviously mention the interval $(-\infty, 0]$. However;

In the left-side limit, where the $x$ is negative, i.e.
$$
\lim_{x \to 0^-}x^m\sin\frac{1}{x},
$$
if $m$ is of the form $\frac{1}{2^n}$ where $n$ is any natural number, wouldn't the resultant number become complex? Wouldn't the function only be defined in 0 and $\mathbf R^+$? The left-hand limit for $0$ wouldn't exist.

Wouldn't this be considered a discontinuity? If so, why isn't this set included with set $(-\infty, 0]$?

答:
I think the problem formulation is incorrect. Indeed, the domain and the range of $f$ (and the topologies on them) are not specified. In this setting there is no much sense to speak about continuity of $f$. Trying to fix this we assume that $f$ has natural domain $\operatorname{dom} f$ and range, endowed with natural topologies.

The value of $\tfrac 1x$ is undefined when $x=0$. The value of $x^m$ can be complex and non-unique when $x<0$ and $m$ is non-integer. So it is natural to put $\operatorname{dom} f=\Bbb R\setminus\{0\}$, when $m$ is integer and $\operatorname{dom} f=(0,\infty)$, otherwise.

But in both cases $0\not\in\operatorname{dom} f$. So I guess that a corrected question should be whether we can extend a function $f(x)= x^m\sin\tfrac 1x$  to a continuous function, whose domain contains a point $0$. That is, whether  there exists $L=\lim_{x\to 0,\, x\in\operatorname{dom} f} f(x)$.

Suppose firtst that $m$ is integer. If $m=0$ then $f(x)=\sin\tfrac 1x$ and so $L$ does not exist. If $m>0$ then $|f(x)|\le |x|^m\le |x|$, so $L=0$. If $m<0$ then it is easy to check that $L$ does not exist.

Suppose now that $m$ is not an integer. Then $\operatorname{dom} f=(0,\infty)$. If $m=0$ then $f(x)=\sin\tfrac 1x$ and so $L$ does not exist. If $m>0$ then $|f(x)|\le x^m$, and since $|\sin x|\le 1$ for each $x$ and $\lim_{x\to 0^+} x^m=0$, we have $L=0$. If $m<0$ then it is easy to check that $L$ does not exist.


问:
If origin is not included in an LP problem how do we get a basic feasible solution?
Given the following constraints for an LP problem:
\begin{align}
x_1 + 2x_2 &\ge 4 \\
-3x_1 + 4x_2 &\ge 5 \\
2x_1 + x_2 &\le 6 \\
 x_1 , x_2 &\ge 0
\end{align}
When I draw the feasible region I see the origin is not included.

Since we can't get an identity matrix by just adding slack variables, artificial variables are introduced as well.
\begin{align}
x_1 + 2x_2  -x_3 + x_6 &= 4 \\
-3x_1 + 4x_2 - x_4 + x_7 &= 5 \\
2x_1 + x_2  + x_5 &= 6 \\
x_1 , x_2, x_3,x_4,x_5,x_6,x_7 &\ge 0
\end{align}

Here $x_6$ and $x_7$ are artificial variables. The textbook states an initial basic feasible solution is  $(0,0,0,0,6,4,5)$. However I don't understand when the origin is not included in the original problem how did adding artificial variables include the origin?

答: Introducing the artificial variables relaxes the problem, making the feasible region larger. The initial feasible solution for the relaxed problem is not feasible to the original problem, but that’s OK. The purpose is just to get the algorithm started. After a few steps, the artificial variables become $0$, and you obtain a feasible solution to the original problem.


问:
What does generic immersion mean?
I have been looking for the meaning of generic immersion

In the textbook I am reading, a theorem involves a curve with y coordinate satisfying $y'(0)=0$ says at some stage the following: "...Since $y''(0)\ne0$ by the genericity condition..." so I thought those are immersions without inflection points.

I googled it in vain. The article among those I found on Google which came closest to defining a generic immersion said that the generic open subset of immersions is contained in the embedded curves whose curvature never vanishes and if the torsion at a point P on the curve vanishes then the derivative of the torsion doesn't vanish.

So the above gives only a necessary condition to be generic.

What really defines it?

答: Generic means dense in the appropriate topological space of functions being considered. Given what I can guess your context is, we might look at the $C^2$ topology. </p>

ContentLicense: CC BY-SA 3.0

Title: What does generic immersion mean?


问:
Joint PMF Probability Question!
I am struggling with this question and I really need some support.

A and B are 2 independent Bernoulli random variables. A has the probability of being 1 is 0.8 and B has the probability of being 1 is 0.4. We create M and N as M = 3A - B and N = A + 2B.
What is the joint PMF of M and N?

I have created tables with the value of M and N, corresponding with the value of A and B. However, I don't know how to find the joint PMF of these two. Can you help me with that?

答:
If your table contains all the non-zero probability events, which will be $\{(M,N)=(0,0),~~(M,N)=(-1,2),~~(M,N)=(3,1),~~(M,N)=(2,3)\}$ then this is the joint probability distribution (equivalently called the joint mass function) for $M$ and $N$.

Read https://en.wikipedia.org/wiki/Joint_probability_distribution

Title: Joint PMF Probability Question!


问:
Limit of an $L^1$ sequence
We have functions $f_n\in L^1$ such that $f_n(x)$ tends to some $f(x)$ for almost all $x$. Does this mean that $f_n\to f$ in $L^1$? A necessary condition is $\|f_n\|<M$, is it sufficient?</p>

答:
Hint:

Consider $L^1(0,1)$, look at $f_n=n\cdot\chi_{(0,1/n)}$ (here $\chi_A$ denotes the characteristic function on $A$).

$f_n\to f$ where $f=...$.

$\|f_n\|\le ...$ and $\|f\|=...$

Title: Limit of an $L^1$ sequence


问:
If $AB-BA = \begin{pmatrix} 1 & 0 &0 \\ 0 & x& 0\\ 0&0& -1 \end{pmatrix}$, find $x \in \mathbb{R}$.
Determine all real numbers $x$ for which there exists $3 \times 3$ real matrices $A$ and $B$ such that

$$AB-BA = \begin{pmatrix}
1 & 0 &0 \\
0 & x& 0\\
0&0& -1
\end{pmatrix}$$

答:
Using this fact, we have
$$\begin{align}\text{tr}(AB - BA) &= \text{tr}(AB) - \text{tr}(BA)\\ &= \text{tr}(AB) - \text{tr}(AB) \\&= 0 \end{align}$$
But
$$\text{tr}\begin{pmatrix}
1 & 0 &0 \\
0 & x& 0\\
0&0& -1
\end{pmatrix} = 1+x-1 = x $$
Hence, $x = 0$


问:
Show that if $f\in L^p$, then there is a Ball $B$ such that $\sup_{x\not\in B}|f(x)|<\varepsilon$
Let $E$ be a normed $\mathbb R$-vector space, $B_r:=\{x\in E:\left\|x\right\|_E<r\}$ for $r>0$, $\mu$ be a measure on $\mathcal B(E)$ with $\mu(E)=\infty$, $p\in[1,\infty)$ and $f\in\mathcal L^p(\mu)$.

Are we able to show that $$\forall\varepsilon>0:\exists r>0:\sup_{x\in E\setminus B_r}\left|f(x)\right|<\varepsilon\tag1?$$

Maybe we can show that, more generally, if $(E,\mathcal E,\mu)$ is any measure space and $\mathcal B\subseteq\mathcal E$ is any system containing a sequence $(B_n)_{n\in\mathbb N}\subseteq\mathcal B$ with $B_1\subseteq B_2\subseteq\cdots$ and $\bigcup_{n\in\mathbb N}B_n=E$, then $$\forall\varepsilon>0:\exists B\in\mathcal B:\sup_{x\in E\setminus B}\left|f(x)\right|<\varepsilon\tag2.$$

The idea is clearly to  use that $\mu(B_n)\xrightarrow{n\to\infty}\mu(E)=\infty$. Moreover, we would somehow need to bound $\left\|f\right\|_{L^p(\mu)}^p$ from below by something which tends to $\infty$.

答: No, that is false. Take $\mu$ to be Lebesgue measure on $\mathbb R$, $p=1$ and $f=\sum_n n \chi_{(n, n+\frac 1  {n^{3}})}$.


答:
Since $tr(AB) = tr(BA)$, you see that $tr(AB-BA) = 0$. So, the trace of the rhs must also be zero, i.e. $x = 0$.

We have proven so far that, if the equality holds, we must have $x=0$. To complete the proof we must check that when $x=0$, it is possible to find such matrices. We can take for instance
$$
A=\left(
\begin{array}{ccc}
 0 & 0 & 1 \\
 0 & 0 & 0 \\
 0 & 0 & 0 \\
\end{array}
\right), \quad B = \left(
\begin{array}{ccc}
 0 & 0 & 0 \\
 0 & 0 & 0 \\
 1 & 0 & 0 \\
\end{array}
\right).
$$

Title: If $AB-BA = \begin{pmatrix} 1 & 0 &0 \\ 0 & x& 0\\ 0&0& -1 \end{pmatrix}$, find $x \in \mathbb{R}$.

Body:
Determine all real numbers $x$ for which there exists $3 \times 3$ real matrices $A$ and $B$ such that

$$AB-BA = \begin{pmatrix}
1 & 0 &0 \\
0 & x& 0\\
0&0& -1
\end{pmatrix}$$


问:
Questions about $y'_x=-\frac{f'_x}{f'_y}$ for implicit differentiation
I was reading in my book that if we have $f(x,y)=0$ and want to find $\frac{dy}{dx}$ with respect to $x$ we use the formula $y'_x=-\cfrac{f'_x}{f'_y}$.
I am not very familiar with this formula so I have these questions:

Question $1$:How we can prove this formula works?

Question$2$:Is it really necessary to memorize this formula to find $\frac{dy}{dx}$ ? I want to know your experience in finding derivative in this form. I think it is easier to differentiate both sides of the equation instead of memorizing some formula. so is it really useful formula?(do you recommend to ignore this formula or use it?)

答:
There are two ways to pull out this formula. One is using the 'implicit function theorem' and the second is using the concept of 'gradient'. The first requires a little multivariable calculus and the second requires a little vector calculus.

For the gradient way, see here

For the multivariable way (implicit function theorem), see here


问:
To show all the coefficients $c_i, \ 1 \leq i \leq n-1$ lie in the maximal ideal $p \mathbb{Z}_p$
Consider the irreducible polynomial $h(t)=t^n+c_{n-1}t^{n-1}+\cdots +c_1t+p \in \mathbb{Z}_p[t]$. Consider the ring of $p$-adic integers and maximal ideal $p \mathbb{Z}_p$.

I want to show all the coefficients $c_i, \ 1 \leq i \leq n-1$ lie in the maximal ideal $p \mathbb{Z}_p$.

I am thinking about Hensel's lemma.

If possible let, one of the coefficients $c_i$, say $c_k$ doesn't belong to $p \mathbb{Z}_p$, then reduction mod $p$ produces $$\bar{h}(t)=t^n+c_kt^k=t^k(t^{n-k}+c_k),$$ as a factorisation. Hence by Hensel's lemma the lift $h(t)$ of $\bar h(t)$  is reducible, which is a contradiction.

Thus all the coefficients $c_i$ lie in $p \mathbb{Z}_p$.

Am I correct ?

答:
Your answer is ultimately correct, the only modification you need to make is that there could be arbitrarily many coefficients that aren't 0 mod p, in particular,

$$\bar h(t) = t^n + c_{i_1}t^{i_1} + \cdots + c_{i_k}t^{i_k} =  t^{i_k}(t^{n-i_k} + c_{i_1}t^{i_1-i_k} + \cdots + c_{i_k})$$

Your factorization will still lift by Hensel's lemma like you describe because they satisfy the condition that they are relatively prime polynomials.


问:
Prove that the function is Riemann Integrable
Question: Let $S$ be the sawtooth function defined by

$$
S(u) = \left\{
        \begin{array}{ll}
            u & \quad 0 \leq u \leq 1 \\
            u - 1 & \quad 1 < u \leq 2
        \end{array}
    \right.
$$

Prove that $S$ is Riemann integrable.

My Attempt

If $S : [0, 2]\rightarrow\mathbb{R}$, we want to show that $S$ is Riemann integrable.

We first need to show that $S_1 : [0, 1] \rightarrow\mathbb{R}$ where $S(u) = u$ is Riemann integrable. Suppose we let $P_n$ be a partition of the interval $[0, 1]$. Then we will have

$$P_n = \left\{[0, \frac{1}{n}],[\frac{1}{n}, \frac{2}{n}],...,[\frac{n-1}{n}, 1]\right\}$$

Then it follows that
\begin{align*}
u_i - u_{i - 1} &= \frac{1}{n} \\
m_i &= \frac{i - 1}{n} \\
M_i &= \frac{i}{n}
\end{align*}

Then from here, we can say that for a given $\epsilon > 0$
\begin{align*}
U(S_1, P) - L(S_1, P) &= \sum_{i = 1}^n M_i(u_i - u_{i - 1}) - \sum_{i = 1}^n m_i(u_i - u_{i - 1}) < \epsilon \\
\end{align*}

I am not exactly sure if I am on the right track, and I am not exactly sure how I should proceed from here.

答:
Here is an easy way to do this.

Just observe that for any $x,y\in [0,2]$ for any $\epsilon >0$ there is a $\delta >0$ such that
$$
|S(x) - S(y)| < \epsilon/2
$$ whenever $|x - y| < \delta $

Consider the partition $P_n$ as you considered in question such that $1/n <\delta$. Since $S$ is continous in each interval it attains maximum and minimum there, so $M_i = S(a_i)$ and $m_i = S(b_i)$. Now just observe that $|a_i - b_i| < \frac{1}{n} < \delta$ which implies that $M_i - m_i < \epsilon /2$.

So $U(S,P_n) - L(S,P_n) = \sum_{i=1}^{n} (M_i - m_i)(u_i - u_{i-1}) < \frac{\epsilon}{2} (u_n - u_0) = \epsilon$


问:
questions about using the exponential distribution
i can't seem to find any good examples i can adapt to solve this problem.

a certain place experiences an earthquake once every 10 years.

what's the probability that there will be an earthquake within the next 5 years? 20 years? i have to use the exponential distribution.

i think i am supposed to look at this problem in terms of 1 year, so the probability that there will be an earthquake in any particular year is

$\mu = \frac{1}{10}$.

i also think this implies that the probability of an earthquake within the next five years is

$\mu = \frac{1}{20}$

because this period is half as long as the above one.

for 20 years, i think it's

$\mu = \frac{1}{5}$

because this period is twice as long as the first period.

am i right about all of this?

and to calculate the probability of an earthquake within these periods, would i just calculate a left tail probability using the cumulative exponential distribution?

thanks in advance for any insight.

答:
Exponential distribution works.

$$P(X\leq 5)=1-e^{-\frac{5}{10}}=1-\frac{1}{\sqrt{e}}\approx 39.35\%$$

and so on...

poisson distribution properly set will work too

I suggest you to use Poisson to get the same result


问:
sum of ten squares
You are given an unlimited supply of $1\times 1,2\times 2,3\times 3,4\times 4,5\times 5,6\times 6$ squares.Find a set of ten squares whose areas add up to $48$.If not the whole solution,even a little prod in the right direction would help.</p>

答:
Here is another, more sophisticated answer: $23=1+4+9+9$ is the sum of four non-zero squares and so $48=23+23+1+1=4\times 1 + 2\times 4 + 4\times 9$ is the sum of ten squares.

The relevant result is Lagrange's four-square theorem: Every number is a sum of four squares; every number not of the form $4^k(8m + 7)$ is a sum of four non-zero squares.

I chose $23$ because it is of the correct form and is near $48/2$. I was lucky.

Title: sum of ten squares


问:
Existence of a unitary matrix $U$
Prove that, for any complex matrix $A$ of order $3$, there exists a unitary matrix $U$ of order $3$, such that $UAU^{-1}$ is a matrix of the form $\left( \begin{matrix}
 *&  0&  *\\
 *&  *&  0\\
 *&  0&  *\\
\end{matrix} \right)
$.

We know unitary matrix is a matrix that satisfies $UU^*=U^*U=I$. However, I think this is far from the claim above, and I don't know how to proceed further.

答:
$A$ has some eigenvector $v_1$. Its perpendicular space $v_1^\perp$ has dimension $2$.

If $\dim(Av_1^\perp)=2$, then pick any non-zero vector $w\in v_1^\perp\cap Av_1^\perp$ so that $w=Av_2$ for some $v_2\ne0$.

If $\dim(Av_1^\perp)\le1$ then pick $v_2\in v_1^\perp$ such that $Av_2=0$.

In both cases, $v_2\in v_1^\perp$ and $Av_2\in v_1^\perp$.

Finally, add a final vector $v_3$ perpendicular to $v_1,v_2$, and normalize all vectors, so that $U:=[v_3,v_1,v_2]^*$ is unitary. Thus $U^*j=v_1$, $U^*k=v_2$, etc., and $$UAU^*=[v_3,v_1,v_2]^{-1}A[v_3,v_1,v_2]$$ is of the required form since \begin{align}UAU^*j&=UAv_1=\lambda Uv_1=\begin{pmatrix}0\\*\\0\end{pmatrix}\\UAU^*k&=UAv_2=U(\alpha v_2+\beta v_3)=\begin{pmatrix}*\\0\\*\end{pmatrix}.\end{align}

Title: Existence of a unitary matrix $U$


问:
How to sketch $\ln|\sin x|$
The question is to find the number of solutions of
$$\ln|\sin x|=-x^2+2x$$ but I just don't know how to sketch a logarithmic function like $\ln|\sin x|$.

A step by step method would be highly appreciated.
Thanks in advance.

答:
First you draw $y=|\sin x|$ which is nothing else that the graph of the sine with negative part mirrored wrt $x$-axis.

Then remember that sine function is less than $1$. Logarithm of numbers in $(0,1)$ are negative ($\ln 1=0$ so at $x=\pi/2$ logarithm is zero), and that $\ln x\to -\infty$ as $x\to 0^+$ and get the graph of $y=\ln|\sin x|$.

Finally draw the parabola $y=-x^2+2x$ and see where the two curves intersect in the interval $[0,\pi]$:$x_1\approx 2$ and $x_2\approx 3$.

There are infinitely many solutions
$$2 x-x^2=\log (\left| \sin x\right| )$$
$$|\sin x |=e^{2x-x^2}$$
$$\sin x=\pm e^{2x-x^2}$$

For the limitation on $\sin$ we must have $e^{2 x-x^2}\leq 1\to 2x-x^2\leq 0\to x\le 0\lor x\geq 2$

$$...$$

Title: How to sketch $\ln|\sin x|$


问:
Finding rank of linear transformation from
Here is the problem I was trying to solve :

I thought this was easy. For calculating rank , as usual I row reduced the matrix $B$ and rank turned out be $1$ but the null space turned out to be $2$. Clearly, the rank nullity theorem was violated. I Googled the problem and found out that rank was $2$, and they proved it by taking $4\times 4$ coordinate matrix of basis elements instead. Why does this need to be solved differently? What did I do wrong? Simply row reducing matrix worked out until now for me.

答:
The matrix $B$ is not the matrix of $T$ with respect to some basis. In order to find such a matrix, you have to fix a basis of $\Bbb C^{2\times2}$ and then to work with it.

Or you can see that a basis of $\Bbb C^{2\times2}$ is $\left\{\left[\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right],\left[\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right],\left[\begin{smallmatrix}0&0\\1&0\end{smallmatrix}\right],\left[\begin{smallmatrix}0&0\\0&1\end{smallmatrix}\right]\right\}$ and to see what's the dimension of$$\operatorname{span}\left(\left\{T\left(\begin{bmatrix}1&0\\0&0\end{bmatrix}\right),T\left(\begin{bmatrix}0&1\\0&0\end{bmatrix}\right),T\left(\begin{bmatrix}0&0\\1&0\end{bmatrix}\right),T\left(\begin{bmatrix}0&0\\0&1\end{bmatrix}\right)\right\}\right).$$


问:
Feller transition kernel gives rise to ${C}_0$-semigroup
I am sorry for any inconveniences arising from my inexperience in writing questions on this site.
First I will give some context for my then succeeding question. I am working with the following definition of Feller transition kernels:

A probability transition kernel $p(t,x,A)$ is said to be a $Feller$ $transition$ $kernel$ if the following conditions hold:

(1) $p(t,x,\mathbb{R}^d)=1$ $\forall$ $t,x$ (some say this is superfluous, are they correct?)

(2) $\int_{\mathbb{R}^d}p(t,\cdot,dy)f(y) \in {C}_0(\mathbb{R}^d)\text{ for } f \in {C}_0(\mathbb{R}^d)$

(3) $\text{lim}_{t\downarrow0}\int_{\mathbb{R}^d}p(t,x,dy)f(y)=f(x),\text{ for every }f \in {C}_0(\mathbb{R}^d), x \in \mathbb{R}^d$.

Given a Feller transition kernel $p(t,x,A)$, one may define

$(P_tf)(x):=\int_{\mathbb{R}^d}f(y)p(t,x,dy)$.

This gives rise to a semigroup $\{P_t|t\geq0\}$ acting on $C_0(\mathbb{R}^d)$. Now letting $\alpha>0$ one may define

$f_\alpha(x):=\int_0^\infty\alpha e^{-\alpha t}(P_tf)(x)dt$,

then $f_\alpha$ $\in$ $C_0(\mathbb{R}^d)$ by the definition of Feller transition kernels as $f$ is assumed to be in $C_0(\mathbb{R}^d)$. Furthermore one may define $f_{\alpha,h}(x):=\int_h^\infty\alpha e^{-\alpha t}(P_tf)(x)dt$. Then clearly $f_{\alpha,h}(x)\uparrow f_{\alpha}(x)$ for $h\downarrow0$  by monotone convergence.

My question now lies in showing that

$\lim_{t\rightarrow0}(P_tf_{\alpha,h}(x))=\lim_{t\rightarrow0}P_t\int_h^\infty\alpha e^{-\alpha s}(P_sf)(x)ds=f_{\alpha,h}(x)$.

The second equality to be precise. My own thoughts go as follows: A defining property of Feller transition kernels is that $\text{lim}_{t\downarrow0}\int_{\mathbb{R}^d}f(y)p(t,x,dy)=f(x),\text{ for every }f \in {C}_0^{\infty}(\mathbb{R}^d), x \in \mathbb{R}^d$. This implies, if one is able to show that $\int_h^\infty\alpha e^{-\alpha s}(P_sf)(\cdot)ds\in C_0(\mathbb{R}^d)$ by the definition of Feller transition kernels (I am not that sure of this actually, maybe someone else can convince me), that $\lim_{t\rightarrow0}P_t\int_h^{\infty}\alpha e^{-\alpha s}(P_sf)(x)ds=\int_h^{\infty}\alpha e^{-\alpha s}(P_sf)(x)ds=f_{\alpha,h}$, which, at least in my eyes, would already answer the question. I am however led to believe that the proof is not as simple as seen above.

Yours truly!

答:
I am now more convinced that the proof is quite immediate. I will supply the necessary details in order to conclude what I have already alluded to in my finalizing remarks above.

Let $\alpha \geq 0$ and $f_\alpha(x) :=\int_0^{\infty}\alpha e^{-\alpha t}P_tf(x)dt$. Since $P_tf\in{C}_0(\mathbb{R}^d)$ by the definition above (2), as $f \in {C}_0(\mathbb{R}^d)$ and since every function continuous and tending to $0$ at $\infty$ is bounded, one has $\alpha e^{-\alpha t}||P_tf(\cdot)||_\infty<C\cdot\alpha e^{-\alpha t}$ for some constant $C>0$. Thus one may apply dominated convergence to obtain

$\lim_{x\rightarrow0}f_{\alpha}(x)=\int_0^{\infty}\alpha e^{-\alpha t}\lim_{x\rightarrow0}P_tf(x)dt=0$

showing $f_\alpha \in {C}_0(\mathbb{R}^d)$. Now let $f\geq0$ and define $f_{\alpha,h}(x)=\int_h^{\infty}\alpha e^{-\alpha t}P_tf(x)dt$. Setting $f(t,x):=\alpha e^{-\alpha t}(P_tf)(x)$, one has that $\lim_{x\rightarrow0}f(t,x)=0$ $\forall$ $t\in(h,\infty)$ by the definition above point (2) and $f(t,x)$ is measurable in $t$ $\forall$ $x\in\mathbb{R}^d$. Thus one has

$\lim_{x\rightarrow\infty}f_{\alpha,h}(x)=\lim_{x\rightarrow\infty}\int_h^{\infty}\alpha e^{-\alpha t}(P_tf)(x)dt=\int_h^{\infty}\lim_{x\rightarrow\infty}\big(\alpha e^{-\alpha t}(P_tf)(x)\big)dt$

by Lebesgues theorem as the integrand is bounded in $x$, showing that $f_{\alpha,h}\in{C}_0(\mathbb{R}^d)$. Now by the definition point (3)

$\lim_{t\downarrow0}(P_tf_{\alpha,h})(x)=\lim_{t\downarrow0}P_t\int_{h}^\infty\alpha e^{-\alpha s}(P_sf)(x)ds=f_{\alpha,h}(x)$,

finally showing the claim.


问:
Quadratic surfaces and intersection curves
You are given a second order surface $Ax^2+By^2+Cz^2=α$ and a plane $x+z=β$. Determine the type of the surface and the type of the curve at the intersection of the surface and the plane.

Basically, I need to write a program, that will do task, getting as parameters $A, B, C, \alpha,\beta$.
So, the best way I found to make this, is to just check every equation by coefficients signs to determine quadratic surface. But what about curves of intersection? Is there a simple way to determine their type only by given parameters?

答:
One of the ways would be to parametrize your plane and all points on the plane can be then represented as $(t, y, \beta - t)$.

Plugging it into the equation of your quadratic surface, you get parametrized equation of the intersection curve.

$r(t,y) = At^2+By^2+C(\beta - t)^2 - α = 0$

i.e $ \,r(t,y) = (A+C)t^2+By^2 -2 C \beta \,t - α + \beta^2C = 0$


问:
On the sharpness of the Harada-Sai lemma
Let $A$ be a finite dimensional $\mathbb{K}$-algebra, where $\mathbb{K}$ is an algebraically closed field. The following result is a well known (see for example Assem & Coelho - Basic representation theory of algebras):

Lemma (Harada-Sai). Let $m>0$ and
$$M_1\xrightarrow[]{f_1} M_2\xrightarrow[]{f_2} \dots \xrightarrow[]{f_{2^m-1}} M_{2^m}$$
be a radical path where each $M_i$ has composition length at most equal to $m$. Then, $f_{2^m-1}\dots f_2 f_1=0$.

Another well known result is that this bound on the length of non-zero radical paths is a sharp bound (see exercise VI.1.2 of Assem & Coelho).

My question is the following : Is the bound given by Harada-Sai's lemma sharp when we restrict ourselves to representation finite algebras? The exercise pointed out above gives an example of the sharpness for the algebra $\mathbb{K}[t_1,t_2]/\langle t_1^2,t_2^2\rangle$ which is representation infinite.

More precisely, I am looking for an example of a representation finite algebra $A$ such that the composition of $2^{m}-2$ radical morphisms (from modules which the lengths are bound by $m$) is non-zero.

答:
Bongartz gives an example, for any $m$, in Section A.1 on page 326 of

Bongartz, Klaus, Treue einfach zusammenhaengende Algebren. I, Comment. Math. Helv. 57, 282-330 (1982). ZBL0502.16022.

The algebra is given by the quiver
$$
1
\begin{array}{c}
\xrightarrow{a}\\
\xleftarrow[b]\\
\end{array}
2
\begin{array}{c}
\xrightarrow{a}\\
\xleftarrow[b]\\
\end{array}
3
\cdots
m-1
\begin{array}{c}
\xrightarrow{a}\\
\xleftarrow[b]\\
\end{array}
m
$$
with relations $ab=ba=0$.

As an example I'll list the modules in a chain of $14$ morphisms for $m=4$. Arrows to the right indicate the action of $a$; arrows to the left indicate the action of $b$.

$$1\to 2\to3\to4$$
$$1\to2\to3$$
$$1\to2\to3\leftarrow4$$
$$1\to2$$
$$1\to2\leftarrow3\to4$$
$$1\to2\leftarrow3$$
$$1\to2\leftarrow3\leftarrow4$$
$$1$$
$$1\leftarrow2\to3\to4$$
$$1\leftarrow2\to3$$
$$1\leftarrow2\to3\leftarrow4$$
$$1\leftarrow2$$
$$1\leftarrow2\leftarrow3\to4$$
$$1\leftarrow2\leftarrow3$$
$$1\leftarrow2\leftarrow3\leftarrow4$$


问:
Independent vectors and linear map

Let $L\colon (x_1, x_2, x_3, x_4, x_5, x_6, x_7) \mapsto 2x_3 + 3x_4 + 8x_6 + x_7$: Is $L$ a linear map?
If it is, how to find independent vectors in  $\operatorname{Ker}(L)$?

答:
(I assume $L$ goes from $\mathbb{R}^7$ to $\mathbb{R}$, for instance)

Hint: Try to write $L(\vec{x})$ as $\langle \vec{a}, \vec{x} \rangle = \sum_{k=1}^7 a_kx_k$ for some vector $\vec{a}\in\mathbb{R}^7$. $L$ is a linear map iff you can (if you don't see why, show it — it is always useful to know).

Then, if it is the case to find a basis (maximal set of independent vectors) in $\operatorname{Ker}L$, you first have to solve (for $\vec{x}=(x_k)_{1\leq k\leq 7}$) the equation $\sum_{k=1}^7 a_kx_k = 0$. Then,"skim" them to get a basis of the (now explicit) space of solutions.

Title: Independent vectors and linear map


问:
How to prove this inequality that i found while proving convergence of steepest descent for quadratic case
Let $0<y_1\le y_2  \ldots \le y_n$.

Also, $\sum_{i=1}^n x_iy_i=\alpha y_1+(1-\alpha)y_n$ where $\alpha\ge 0$, $x_i \ge 0$ and $\sum x_i=1$.

To show $\sum_{i=1}^n x_i/y_i \le \alpha/y_1 +(1-\alpha)/y_n$

答:
(Fill in the gaps as needed. If you're stuck, show your work)

Define $\alpha_i$ such that $y_i = \alpha_i y_1 + (1-\alpha_i) y_n$. Note that $ 0 \leq \alpha_i \leq 1$.

Show that $$ \alpha = \sum x_i \alpha_i.$$

 Substitute in  and show that it holds.

Show that $$ \frac{1}{ y_i} = \frac{1}{ \alpha_i y_1 + (1-\alpha_i) y_n}  \leq \frac{ \alpha_i} { y_1 } + \frac{ (1-\alpha_i) } { y_n}.$$

 Apart from expanding everything, this follows from $ f(x) = \frac{1}{x} $ is a convex function on $x> 0$.

Show that $$\sum \frac{x_i}{ y_i} \leq \frac{ \alpha } { y_1 } + \frac{ (1-\alpha) } { y_n}.$$

 Substitute in  and show that it holds.


问:
Birthday probability question
I came across a variation of the birthday problem asking "in a room of $4$ people what is the probability that at least $3$ of them share the same birthday".

I was unsure of the answer and thought that it would be P($3$ share the same birthday) + P($4$ share the same birthday), which equals: $1\cdot\frac{1}{365^2} + 1\cdot\frac{1}{365^3}$, and this comes out to be around $0.0000075$%.

However my friend said that he thinks to correctly calculate the answer, the probability of $4$th person not having the same birthday should be included in the calculation somewhere.

What would be the correct probability of at least $3$ out of $4$ people sharing the same birthday, and how could you extend the problem to work out the probability of at least "$x$" out of "$y$" people having the same birthday?

答:
i) $3$ share the same birthday

Number of ways to choose $3$ people out of $4$ having the same birthday $ = { 4 \choose 3}$

The probability of $3$ of them having the same birthday (and $4$th must have different birthday otherwise this will include cases when all $4$ of them have the same birthday).

So $\displaystyle P(3) = { 4 \choose 3} \frac{364}{365^3}$

ii) All $4$ share the same birthday

$\displaystyle P(4) = \frac{1}{365^3}$

Desired probability $ = P(3) + P(4)$.

Title: Birthday probability question


问:
WolframAlpha functionality: Computing nested integrals
Can wolframpalpha support the numerical calculation of the following type:

$$
\int_{x=0}^\infty f(x) e^{\int_{z=0}^\infty g(x,z)dz} dx
$$

Note that the inside integral is in such form where parameter x cannot be separated from it.

Thanks in advance

PS: The respective command would have been:

int(f(x)*exp[int(g(x,z),{z,0,Infinity})],{x,0,Infinity})

答:
Sure. Syntactically "Integrate[]" isn't different than any other function. So, leaving out all parameter specifications, it's just

```<code>Integrate[f[] Exp[Integrate[]]]
</code>```

or in more detail

```<code>Integrate[f[x] Exp[Integrate[g[x,z],{z,0,Infinity}]],{x,0,Infinity}]
</code>```

I don't know if you mean "numerical calculation" literally here, but for any given function $g$ and fixed parameter $x$, the command Integrate[g[x,z],{z,0,Infinity}] will try to compute the integral. NIntegrate is a numerical integration command in Mathematica btw.

Body:
Sure. Syntactically "Integrate[]" isn't different than any other function. So, leaving out all parameter specifications, it's just

Integrate[f[] Exp[Integrate[]]]

or in more detail

Integrate[f[x] Exp[Integrate[g[x,z],{z,0,Infinity}]],{x,0,Infinity}]

I don't know if you mean "numerical calculation" literally here, but for any given function $g$ and fixed parameter $x$, the command Integrate[g[x,z],{z,0,Infinity}] will try to compute the integral. NIntegrate is a numerical integration command in Mathematica btw.


问:
why k times differential of functional $u:U\to \mathbb{R}$ lies in $R^{n^k}$
Let $u$ be a smooth functional $u:U\to \mathbb{R}$ ,with $U\subset \mathbb{R}^n$.

Why k times differential $D^k u$ can be identified as a vector in $\mathbb{R}^{n^k}$.

For $Du$ I know how to do it that is $Du_x :T_xU\to \mathbb{R}$  hence $Du_x \in \text{Hom}(T_xU,\mathbb{R}) \cong \mathbb{R}^n$.

I don't know how to show the result for higher order?

答:
The situation is made apparent by looking at the second total derivative (the Hessian). To evaluate $D(Du_x)$ you need to specify a point $x_1\in T_xU\cong\mathbb{R}^n$ and a tangent vector $x_2\in \mathbb{R}^n$. Indeed, the Hessian is defined as $H_x(x_1,x_2) = D(Du_x)_{x_1}(x_2)$, which is easily seen to be a bilinear map $H_x:\mathbb{R}^n\times\mathbb{R}^n\to\mathbb{R}$. Taking another total derivative defines a trilinear ($3$-multilinear) map, and in general $D^ku: U\to L^k(\mathbb{R}^n\times\dotsm\times\mathbb{R}^n,\mathbb{R})$ associates to each $x\in U$ a $k$-multilinear map defined by the formula
$$(x_1,\dotsc, x_k)\mapsto D(D(\dotsm(Du_x)_{x_1})\dotsm)_{x_{k-2}})_{x_{k-1}}(x_k).$$

Now the problem reduces to proving that the dimension of $L^k(\mathbb{R}^n\times\dotsm\times\mathbb{R}^n,\mathbb{R})$ over the field $\mathbb{R}$ is $n^k$. To do this, fix a basis $e_1,\dotsc, e_n$ of $\mathbb{R}^n$ . Then the set of $A_{j_1\dotsc j_k}:(\mathbb{R}^n)^k\to \mathbb{R}$ defined by $A_{j_1\dotsc j_k}(e_{i_1},\dotsc, e_{i_k}) = \delta_{i_1j_1}\dotsm\delta_{i_kj_k}$ for $i_r,j_s\in \{1,\dotsc,n\}$ and $r,s\in\{1,\dotsc, k\}$ forms a basis for the set of $k$-multilinear maps; for each $j_s$ you choose from among $n$ options, and we do this $k$ times ($s=1,\dotsc, k$), so the dimension is $n^k$ as claimed.


问:
Why cos(a-b) =cos(a+(-b))?
Why sin(A-B) is same as sin[A+(-B)]?
i mean,we have derived the formula  of  sin(a+b) geometrically in  a uniqe  way..then how can we replace +b with -b in identity?
To sum up, doesn't changing the sign of variable in a identity ,change the identity itself specially when we are dealing with coordinate system?
( NOTE- I KNOW THAT sin(a-b)  yeilds us same formula as sin[a+(-b)] geometrically too)</p>

答:
It's more fundamental than that. We have
$$
a-b=a+(-b)
$$
because that's the definition of the binary operation $-$. And since those two are equal, clearly they must have the same sine and the same cosine.

As for why we can use the formula you derived for $\sin(a+b)$ to evaluate $\sin(a+(-b))$, that's just a matter of variable names. Ignore the meaning of $-$ for a second and consider $-b$ a single symbol. Then you will see that the name of a variable, whether it's $b$ or $-b$, changes nothing of the argument, and the formula that pops out in the end of just as valid.

Title: Why cos(a-b) =cos(a+(-b))?


Title: Independent vectors and linear map

Body:
Let $L\colon (x_1, x_2, x_3, x_4, x_5, x_6, x_7) \mapsto 2x_3 + 3x_4 + 8x_6 + x_7$: Is $L$ a linear map?
If it is, how to find independent vectors in  $\operatorname{Ker}(L)$?

Body:
$L$ is linear:
$$
L(\alpha(x_1,x_2,x_3,x_4,x_5,x_6,x_7) + \beta (y_1, y_2,y_3,y_4,y_5,y_6,y_7)) = L(\alpha x_1+\beta y_1,\dots,\alpha x_7 +\beta y_7 ) \\
$$
and
\begin{align*}
L(\alpha x_1+\beta y_1,\dots,\alpha x_7 +\beta y_7 )  &= 2(\alpha x_3 + \beta y_3) + 3(\alpha x_4+\beta y_4) + 8(\alpha x_6 + \beta y_6) + (\alpha x_7+\beta y_7)\\
& = \alpha(2x_1 + 3x_4 + 8x_6 +x_7) + \beta (2y_1 + 3y_4+8y_6+y_7) \\
&= \alpha L(x_1,x_2,x_3,x_4,x_5,x_6,x_7) + \beta L(y_1, y_2,y_3,y_4,y_5,y_6,y_7)
\end{align*}

$$
Ker(L) = \{(x_1,x_2,x_3,x_4,x_5,x_6,x_7): 2x_3 + 3x_4+8x_6+x_7 = 0  \}
$$
or
$$
Ker(L) = \{(x_1,x_2,x_3,x_4,x_5,x_6,x_7): x_7 = -2x_3 - 3x_4 - 8x_6
  \}
$$
meaning that
$$
Ker(L) =  \{ (x_1,x_2,x_3,x_4,x_5,x_6,-2x_3 - 3x_4 - 8x_6); x_1,x_2,x_3,x_4,x_5,x_6 \in \mathbb{R}  \}
$$


问:
Limits, counter-example
Regarding this question here:

If there exists an $L ∈ R$ such that $\lim_{x→x_0} f(x) = L$ for every $x_0 ∈ R$, then $f (x) = L$ for every $x ∈ R$.

I think this is false, and I am trying to find a counter example for that.

Can someone please give me a hint for how to find the right $f(x)$ that counter this statement?

Thanks a lot!

答:
Thanks a lot @fleablood.

Let me see If I got it:

Let $f(x) = \begin{cases}L&x\ne b\\L/2 &x=b\end{cases}$

For $lim_{x->b}f(x)=L$

Let $\varepsilon >0$. We choose $δ=\varepsilon$

Let $x \in R$ such that $0<|x-b|<δ$

So $|f(x)-L| = |L-L| = 0 < \varepsilon$

And that's because $0<|x-b|$ so $x\ne b$ and $f(x)=L$.

For $x_{0} \ne b$ , $lim_{x->x_{0}}f(x)=L$

Let $\varepsilon >0$. We choose $δ=|x_{0}-b|$

Let $x \in R$ such that $0<|x-x_{0}|<δ$

So $|x-x_{0}| < |x_{0}-b|$

So $x \ne b$ and so $h(x)=L$

$|h(x)-L| = |L-L| = 0 < \varepsilon $

Is that correct? I think that I missed something here.
Thanks!!

Title: Limits, counter-example


问:
Is the set of all Holder functions closed in uniform topology?

Let $H \subset C[0,1]$ be the set of all Holder-continuous functions under the uniform norm $||\cdot||_\infty$, i.e.
$$H = \bigcup_{\alpha > 0}C^{0, \alpha}[0,1]. $$
Is $H$ closed in the uniform norm?

I don't believe that this set is closed, as if $f_n \to f$ in $||\cdot||_\infty$ norm and $f_n$ is Holder-continuous with
$$|f_n(x) - f_n(y)| \leq C_n \cdot |x-y|^{\alpha_n}, \forall x,y \in [0,1], $$
then I don't think we must have that $C_n$ or $\alpha_n$ converge.
However, I have not been able to come up with a counter example.

答: Every polynomial is Holder continuous since it has  a bounded derivative. Every continuous function is the uniform limit of a sequence of polynomials. So take any continuous function which is not Holder continuous and you will see why the set is not closed.</p>


问:
derivative of function to see when it stops decreasing in order to check condition
$f(n) = 3n^3 - 39n^2 + 360n + 20$. We need constants $c, n_0 > 0$ such that $f(n) \geq$ c$n^3$ $\,$ $\forall n\geq n_0$. Fix $c=2.25$. The smallest integer value that works can be found using calculus to find the range where the functions have $0$'s. The factors are equal to (n - 39.9762)(n - 12.0791)(n + 0.055248). Upon further inspection that for $-0.055248 \leq n \leq 12.0791$ and $n\geq 39.9762$ we have that $f(n) \geq 2.25n^3$. In order to satisfy $\forall n\geq n_0$ $f(n) \geq 2.25n^3$ we can't take $n_0$ in the range -0.55248 to 12.0791 since, for instance $n=13$ (greater than any of those numbers) would not satisfy the inequality. Thus, we can set $n_0 = 40$, the smallest integer value that satisfies are constraints.

My question is that you could also solve this by a sort of guess and check, but you would need to know the derivative of $f(n) - cg(n)$ (note that $g(n)$ would be equal to $n^3$ in this case) to see when the function stops decreasing in order to know that your $n_0$ is correct, and so how would you do it?

答:
$f(n) = 3n^3 - 39n^2 + 360n + 20$. We need constants $c, n_0 > 0$ such that $f(n) \geq$ c$n^3$ $\,$ $\forall n\geq n_0$.

I think your approach to this problem is too complicated. It is easy to see that $f(n)-cn^3\ge 0$ when $n\ge 0$ and $(3-c)n^3\ge 39 n^2$, so for each $0<c<3$ we can put $n_0=\tfrac {39}{3-c}$. I guess that to find the smallest integer $n_0$ such that $f(n)\ge cn^3$ for each $n\ge n_0$ usually we have to find the largest root of the cubic equation
$f(n)-cn^3=0$. This is a rater complicated task, but it can be done analytically, using Cardano formula.

In a general case, given $f(n)=a_dn^d+a_{d-1}n^{d-1}+\dots+ a_1n+a_0$ with $a_d>0$, for each $n\ge 1$ we have $f(n)\ge a_dn^d-Mn^{d-1}$, where $M=\sum_{i=0}^{d-1} |a_i|$, so for each $0<c<a_d$ we can
we have $f(n)\ge cn^{d}$ for each $n\ge \tfrac M{a_d-c}$.


问:
Curve that is not an immersion
Given the curve $\alpha: \mathbb R \rightarrow \mathbb R^2, \alpha(t)=(t^3,t^2)\,$I need to prove the mapping is not an immersion. We know the map is differentiable on $\mathbb R$. $d\alpha=(3t^2,2t).$ Even though I am expecting that the differential should be linear, this map can not be linear on $t.$ What I am missing here ? Once we have a linear map, the kernel should be different zero in order to prove $d\alpha$ is not injective.

Can somebody prove, why $\alpha$ is not an immersion (according to the book)?

答: The map $d\alpha(t)$ is a linear map from $T_t\mathbb R$ to $T_{\alpha(t)}\mathbb R^2$ (and it is what you get), it is not necessarily linear in $t$. Moreover, $f: M \to N$ is an immersion if, for all $p \in M$, $df(p): T_pM \to T_{f(p)}N$ is injective (i.e. the associate matrix is injective). Do you see a $t \in \mathbb R$ where $d\alpha(t)$ is not injective ?

Title: Curve that is not an immersion


答:
Let me go for the general case.

For $i=1,2\dots,365$ let $E_i$ denote the event that at least $x$ persons have birthday on day $i$.

Then to be found is: $$P\left(\bigcup_{i=1}^{365}E_i\right)$$ and we can do that with inclusion/exclusion and symmetry, leading to:

$$\cdots=\sum_{k=1}^{365}(-1)^{k-1}\binom{365}kP\left(\bigcap_{i=1}^kE_i\right)$$
Then what remains is finding expressions for $P\left(\bigcap_{i=1}^kE_i\right)$ for $k=1,2,\dots,\lfloor\frac{y}{x}\rfloor$.

This because, if $k$ exceeds $\lfloor\frac{y}{x}\rfloor$ then $P\left(\bigcap_{i=1}^kE_i\right)=0$, so we could also write:$$\cdots=\sum_{k=1}^{\lfloor\frac{y}{x}\rfloor}(-1)^{k-1}\binom{365}kP\left(\bigcap_{i=1}^kE_i\right)$$

In your case ($x=3$ and $y=4$) we have $\lfloor\frac{y}{x}\rfloor=1$ so it reduces to:
$$365P(E_1)=365\left[\binom43\left(\frac1{365}\right)^3\frac{364}{365}+\left(\frac1{365}\right)^4\right]=\binom43\left(\frac1{365}\right)^2\frac{364}{365}+\left(\frac1{365}\right)^3$$

Quite a job to find these probabilities $P\left(\bigcap_{i=1}^kE_i\right)$ but by small $\lfloor\frac{y}{x}\rfloor$ there is hope.

Title: Birthday probability question

Body:
I came across a variation of the birthday problem asking "in a room of $4$ people what is the probability that at least $3$ of them share the same birthday".

I was unsure of the answer and thought that it would be P($3$ share the same birthday) + P($4$ share the same birthday), which equals: $1\cdot\frac{1}{365^2} + 1\cdot\frac{1}{365^3}$, and this comes out to be around $0.0000075$%.

However my friend said that he thinks to correctly calculate the answer, the probability of $4$th person not having the same birthday should be included in the calculation somewhere.

What would be the correct probability of at least $3$ out of $4$ people sharing the same birthday, and how could you extend the problem to work out the probability of at least "$x$" out of "$y$" people having the same birthday?


Title: Independent vectors and linear map

Body:
Yes, $L$ is a linear map. To prove it all you have to do is take $(x_1,...,x_7),(y_1,...,y_7)\in \Bbb R ^7, \lambda \in \Bbb R$ and prove $L(x_1+\lambda y_1,...,x_7+\lambda y_7)=L(x_1,...,x_7)+\lambda L(y_1,...,y_7)$.

$Ker(L)=\left\{(x_1,...,x_7)\in \Bbb R^7 \mid L(x_1,...,x_7)=0\right\}=\left\{(x_1,...,x_7)\in \Bbb R^7 \mid 2x_3+3x_4+8x_6+x_7=0\right\}=\left\{(x_1,...,x_7)\in \Bbb R^7 \mid x_7=-2x_3+-3x_4+-8x_6\right\}=\left\{(x_1,...,-2x_3+-3x_4+-8x_6) \mid (x_1,...,x_6)\in\Bbb R^6\right\}=\left\{x_1(1,0,0,0,0,0,0)+x_2(0,1,0,0,0,0,0)+x_3(0,0,1,0,0,0,-2)+x_4(0,0,0,1,0,0,-3)+x_5(0,0,0,0,1,0,0)+x_6(0,0,0,0,0,0,1,-8) \mid (x_1,...,x_6)\in\Bbb R^6\right\}$


问:
Proving the composition of a function with an $H^1$ function is in $H^1$.
Let $\Omega \subseteq \mathbb{R}^n$ be open, bounded with smooth boundary. Does $u \in H^1(\Omega)$ imply that $(1+u^2)^{\frac{1}{2}}\in H^1(\Omega)$?

If yes, I would expect a more general result holds about composing $H^1(\Omega)$ with $C^1$ sufficiently slowly increasing functions with sufficiently slowly increasing derivatives. I am not really sure about how one goes about proving this kind of result.

答: Define $f(x)=(1+x^2)^{1/2}$. Note that $f \in C^1(\mathbb{R})$ and $f'(x)=x/(1+x^2)^{1/2}$ is bounded in the sense $|f'(x)|\leq 1$ for all $x \in \mathbb{R}$. Then by the result in this question we have $f(u) \in H^1(\Omega)$, exactly as you wanted.


问:
What does generic immersion mean?
I have been looking for the meaning of generic immersion

In the textbook I am reading, a theorem involves a curve with y coordinate satisfying $y'(0)=0$ says at some stage the following: "...Since $y''(0)\ne0$ by the genericity condition..." so I thought those are immersions without inflection points.

I googled it in vain. The article among those I found on Google which came closest to defining a generic immersion said that the generic open subset of immersions is contained in the embedded curves whose curvature never vanishes and if the torsion at a point P on the curve vanishes then the derivative of the torsion doesn't vanish.

So the above gives only a necessary condition to be generic.

What really defines it?

答:
There is no the generic immersion or even a generic immersion, just properties generic for immersions.

In the context of differential topology, a property is generic if it is satisfied by a comeager set.

In most cases, genericity of some property is shown using Thom's transversality theorem.

The reason why there's no a generic function is simple: a conjunction of uncountably many generic conditions need not be generic; in fact, it might be contradictory. For instance, having a nonzero derivative at a particular point is generic in $C^\infty({\bf R},{\bf R})$, as is having a nonzero derivative at a countable set of points, but having nonzero derivative everywhere is not: only injective functions have nonzero derivative everwhere. But this condition is not even dense: there's no injective function anywhere near $y=x^2$.

Title: What does generic immersion mean?


问:
Let $P$ be a $30$-sided polygon inscribed in a circle. Find the value of $\frac{N}{100}$.

Let $P$ be a $30$-sided polygon inscribed in a circle. There are $N$ number of triangles whose vertices are the vertices of $P$ such that any two vertices of each triangle are separated by at least three other vertices at $P$. Find the value of $\frac{N}{100}$.

What I Tried: This is more like a Combinatorics problem rather than a geometry problem, so here is what I think.

First, fix a point of a triangle. The next point can be chosen in $23$ ways. But I am not sure how to choose the $3$rd point, as for choosing the $2$nd point there are slight variations as well, which dosen't follow the rule.

I thought before of fixing one point, and then the next $2$ points can be chosen in ${23}\choose{2}$ ways, but then I realised that is wrong since those $2$ points might not have a $3$ point gap, and I couldn't get on how to progress on this.

As usual, I also know that the number of triangles on an $n$-sided polygon with no shared sides is given by the formula :-
$$\rightarrow\frac{n(n-4)(n-5)}{6}$$
So the total number of triangles is $3250$, but I am not sure on how this fact will help in this problem.

Can anyone help me? Thank You.

答:
Choose any point and call it $A_1$. Label the points in counterclockwise manner $A_2,\ldots,A_{30}$ .

Second vertex can be any from $A_5$ to $A_{27}$.

When second is $A_5$, third vertex can be any from $A_9$ to $A_{27}$. That's $19$ ways.

When second is $A_6$, third vertex can be any from $A_{10}$ to $A_{27}$. That's $18$ ways.

And so on. Number of triangles $= 19+18+17+\ldots+1$

We could start on any point as first vertex, so desired is $$\dfrac{19\cdot20}{2} \cdot \dfrac{30}{3}$$

If we were to leave atleast $k$ points between adjacent vertices, by the same logic we'll get $$\dfrac{n(n-3k-1)(n-3k-2)}{6}$$

for appropriate $k$. Since $3k+2$ number of points are left out first when second vertex is $A_{k+2}$.


问:
Coordinates on 3d Circle
Problem:

Given is is the centerpoint $C$ of a circle and two extra reference points $R$ and $E$, which are laying on the circle.
While as the line $\overline{CE}$ is defining the zero point on the circle of a given degree $α$ (clockwise rotation). Point $F$ is also on the very same circle. Therefore all of the points are on one in space rotated plane, on the circle.
Line $\overline{CF}$ is $α$ degrees away from line $\overline{CE}$. How can I get the XYZ coordinates of Point $F$?
($R ≠ E$)
Fig. 1

Fig. 1: Illustration of described problem

Note that $\overline{CR}$ is NOT necessarily parallel to $\overline{CF}$

$\overline{CR}$ to $\overline{CE}$ not necessarily orthogonal

all Points lay in one 3d circular plane

in Fig. 1 a second circle is drawn orthogonal to the XY Plane just for better relation understanding (not a projection)

The points $E$, $C$ and $R$ are defining the circles rotation in space, while $F$ is the point to move around on this circle (with a specific degree $\alpha$)

答:
Hint:

In 3D there can be infinite points like F such that $\angle ECF=\alpha$ and also  $\angle RCF=\beta$. They are on two conic surface vertex on C with axis  CE and CR respectively. The intersection of these two surfaces makes two lines. Any point like F on these lines is such  that $\angle ECF=\alpha$ and $\angle RCF=\beta$.So you have to find equations of these conic surfaces.For this take for example xy plane. Draw a line angle with EC as $\alpha$ . Now revolve it about CE and find the equation of produced surface. Now draw a line which make angle $\beta$ with RC and revolve it about RC anf find it' equation. The system of thses two equations give the equations of lines resulting by intersection.Now to find coordinates of F you solve this system of equations:

$\begin{cases} y=mx \\x^2+y^2=r^2 \end {cases}$

Where $m=tan (\alpha)$ and r is  length of CF. So in adition to the angle you have you must also have r.Find x or y from first equation, put it in second equation and find x and y of point F.

Title: Coordinates on 3d Circle


问:
Calculate distance between two points on concentric circles
I am trying to find the shortest distance between two concentric circles. I already know the angle between the two points and radii of the circles, but I am not sure how to calculate the distance between the two.

Imagine the example below, where $R_1 = 2$ and $R_2 = 3$ and the points are separated by 45 degrees. I would like to calculate the distance between these two points. Any help is much appreciated. Thanks!

EDIT:

Let me be more clear. I do not need the linear distance, I can do that no problem. I need the distance to be of the arc that connects these two points. The arc should "rise" at a constant rate between $R_1$ and $R_2$. This connection should also not intersect with the innermost circle.

Basically I am trying to simplify my problem. I currently have the great circle distance of two points on a globe, but this assumes that the two points are at the same radius. This is not always true when you are trying find the distance between a ground station and a satellite or something like that. I know the latitude, longitude, and altitude of both objects, so I can find this their angular separation, but don't know how to calculate the distance of the arcs that connect them.

Assume: $R_1 < R_2$

答:
It’s the cosine rule. If you know the angle between the points and you know the radii, the distance d is given by:

$$d^2 = r_1^2 + r_2^2 - 2r_1r_2\cos(\theta)$$


问:
Is continuous spectrum closed?
I’m learning spectrum theory. I know the definition of point or continuous spectrum, knowing that spectrum of a bounded (or unbounded but self-adjoint) operator consisting only these two type. I have seen several operator with a closed continuous spectrum, but I don’t think it has to be closed when the point spectrum isn’t empty. Can someone show me a counter-example? And, does a continuous spectral point have to be a limit point of continuous spectral points?
Thank you so much !</p>

答:
No, the continuous spectrum need not be closed. Let $f$ be defined on $[-1,1]$ by
$$
  f(x) = \left\{\matrix{
  0, & \text{if } -1\leq x\leq0,\cr
  x, & \text{if } 0\leq x\leq1,\hfill
  }\right.
  $$
and let $T$ be the corresponding multiplication operator on $L^2[-1,1]$, namely
$$
  T(\xi)|_x = f(x)\xi(x).
  $$
Then the spectrum of $T$ is $[0,1]$, the point spectrum is $\{0\}$, and
the continuous spectrum is $(0,1]$.

The answer to the second question is also no.  If $T$ is the diagonal operator on $\ell^2$, with diagonal entries $\{1/n: n\geq1\}$, then the point spectrum is $\{1/n: n\geq1\}$ and the continuous spectrum is $\{0\}$, and clearly 0 is not the limit of continuous spectral points (different from itself).

Title: Is continuous spectrum closed?


问:
$\pi: E \rightarrow M$ ($E$ is a vector bundle) admits a global frame iff $E$ is a trivial vector bundle over $M$.
(Proof of backward direction is clear: if $F$ is a diffeomorphism between $E$ and $M \times \mathbb R^k$, then $p \mapsto (p, e_i) \mapsto F^{-1}(p, e_i)$ forms a basis of $E$ for $i = 1, \cdots, k$.)

Proof of forward direction: Let $E$ be a vector bundle over $M$ of dimension $k$, and let $s_1, \cdots, s_k$ be a global frame. For all $ p \in M$, there exists a local trivialization $\Phi: \pi^{-1}(U_p) \rightarrow U_p \times \mathbb R^k$. Since $\Phi|_{E_p}$ is a linear isomorphism of vector spaces, and $s_1(p), \cdots, s_k(p)$ forms a basis of $E_p$, $\Phi(s_1(p)), \cdots, \Phi(s_k(p))$ forms a basis of $\{p\} \times \mathbb R^k$.

Note that $(p, e_1), \cdots, (p,e_k)$ is a basis of $\{p\} \times \mathbb R^k$. Hence, I can form a linear mapping  between basis $\Phi(s_1(p)), \cdots, \Phi(s_k(p))$ and another basis $(p, e_1), \cdots, (p,e_k)$,

$$(p, e_i) = \sum_{j=1}^k a_j^i(p) \Phi(s_j(p))$$

It suffices to prove that $a_j^i$ is smooth, but how do I prove so?
Any suggestions?

答:
By a partition of unity argument, one can show that any vector bundle admits an inner product, that is a smooth symmetric $2$-tensor that is symmetric positive definite: for any smooth sections $v$ and $w$, the function
$$
p \mapsto \langle v,w\rangle_p
$$ is smooth.

Suppose $E$ is a vector bundle over $M$, of rank $k$, and chose an inner product. Suppose $E$ admits a global frame $(v_1,\ldots,v_k)$. Then
\begin{align*}
\Phi : E &\to M \times \mathbb{R}^k \\
(p,u) &\mapsto \left(p,(\langle u,v_1\rangle_p,\ldots,\langle u , v_k \rangle_p)\right)
\end{align*}
is a global trivialization.

Here is a construction of a smooth inner product on any finite rank smooth vector bundle $E$. Choose a locally finite open cover $\{U_i\}_{i\in I}$ such that $E$ is locally trivial on each $U_i$:
$$
\forall i \in I,~ \exists \Phi_i : E \overset{\sim}{\to} U_i\times \mathbb{R}^k
$$
where $\Phi_i$ is smooth. Define on $E|_{U_i}$ the smooth inner product
$$
\langle u,w\rangle_i = \langle {\Phi_i}_*u,{\Phi_i}_*w\rangle_{\mathbb{R}^k}
$$
that is $\langle \cdot,\cdot \rangle_i = (\Phi_i)^* \langle\cdot,\cdot\rangle_{\mathbb{R}^k}$. Then $\langle\cdot,\cdot \rangle_i$ is smooth over $U_i$ because so is $\Phi_i$ and the natural inner product of $\mathbb{R}^k$.

Chose a smooth partition of unity $\{\varphi_i\}_{i\in I}$ with respect to the locally finite open cover $\{U_i\}_{i\in I}$, and define, for $u$ and $w$ sections of $E$:

$$
\langle u,v \rangle = \sum_{i\in I} \varphi_i \cdot \langle u,w\rangle_i
$$

It is clearly smooth because it is a locally finite sum of smooth sections. It is clearly bilinear and symmetric. Moreover, at a point $p$:
$$
\langle u,u\rangle_p = \sum_{i \in I} \varphi_i(p) \langle u,u \rangle_i
$$
As all terms are nonnegative, it is nonnegative. Moreover, it is zero if and only if in every $U_i$, $u|_{U_i} = 0$ (this is because $\Phi_i$ are diffeomorphisms and $\langle\cdot, \cdot \rangle_{\mathbb{R}^k}$ is positive definite). Thus, it is an inner product on $E$.


问:
Adjoint functor theorem finite version
Take a pair of locally finite categories $\mathbf A,\mathbf B$ ($\mathbf C$ is locally finite if there are a finite set of morphisms $A\rightarrow B$, for every fixed pair $A,B$). Suppose $B$ is finitely complete, finitely well-powered (that is, subojects form a finite poset) and has a cogenerator, does every limit-preserving functor $R:\mathbf B\rightarrow\mathbf A$ have a left adjoint? I would say yes (and prove it basically the same way as the special adjoint functor theorem), is it correct?

答:
Sure, this works fine. Let's take a quick look. As usual $R$ has a left adjoint if and only if each comma category $a/R$ has an initial object, giving the unit of the pending adjunction at $a$. Now since the forgetful functor $a/R\to \mathbf B$ is faithful and creates limits (using continuity of $R$), including monomorphisms, we have that $a/R$ is finitely complete, finitely well-powered, and locally finite. Furthermore if $b_0$ is a cogenerator of $\mathbf B$ then the set of all objects of the form $a\to R(b_0)$ (there are only finitely many, using local finitude of $\mathbf A$!) is a finite cogenerating set of $a/R$, by the same argument as usual. So $a/R$ is finitely complete, locally finite and finitely well-powered with a finite cogenerating set. Then the minimal subobject of $a\to \prod_{f:a\to R(b_0)} R(b_0)$ is initial in $a/R$, again as usual.

From a higher level, the usual SAFT is proved for categories which have hom-sets, subobject lattices, limits, and a cogenerating set (equivalently given completeness, a cogenerator), all of which are bounded in size by the cardinal of the set-theoretic universe, call it $\Omega$. Now set-theoretically the function of $\Omega$ is that it's an inaccessible cardinal. $\aleph_0$ is also an inaccessible cardinal, although it's sometimes manually ruled out from the definitions. Thus it would be very strange if this argument wouldn't work: the proof would have to depend somehow on the uncountability of the class of all sets.

Title: Adjoint functor theorem finite version


答:
Let us consider the more general case of a smooth function $f : U \to \mathbb R^m$. Then $D_x f : T_x U \to T_{f(x)} \mathbb R^m$.

To get $D^2f$ it is not sufficient to separately consider the linear maps $D_x f: T_xU \to T_{f(x)} \mathbb R^m = \mathbb R^m$, $x \in U$, but the function $D f : TU \to \mathbb R^m, Tu(x,v) = D_xf(v))$, on the tangent bundle of $U$. We have $TU = U \times \mathbb R^n$, thus
$$D f : U \times \mathbb R^n \to \mathbb R^m .$$
This can be identfied with
$$\hat D f : U \to \text{Hom}(\mathbb R^n,\mathbb R ^m) \cong \mathbb{R}^{n \cdot m} , \hat D f (x) (v) = D_xf(v) .$$
Now you see that $D^2_x f : T_x U \to T_{f(x)} \mathbb{R}^{n \cdot m}  = \mathbb{R}^{n \cdot m}$. Iterating this yields
$$D^k_x f : T_x U \to \mathbb{R}^{n^k \cdot m} .$$

Title: why k times differential of functional $u:U\to \mathbb{R}$ lies in $R^{n^k}$

Body:
Let $u$ be a smooth functional $u:U\to \mathbb{R}$ ,with $U\subset \mathbb{R}^n$.

Why k times differential $D^k u$ can be identified as a vector in $\mathbb{R}^{n^k}$.

For $Du$ I know how to do it that is $Du_x :T_xU\to \mathbb{R}$  hence $Du_x \in \text{Hom}(T_xU,\mathbb{R}) \cong \mathbb{R}^n$.

I don't know how to show the result for higher order?


问:
Limit of an $L^1$ sequence
We have functions $f_n\in L^1$ such that $f_n(x)$ tends to some $f(x)$ for almost all $x$. Does this mean that $f_n\to f$ in $L^1$? A necessary condition is $\|f_n\|<M$, is it sufficient?</p>

答: No. For example, if $f_n=\chi_{[n,n+1]}$ then $(f_n)_{n\ge1}$ has pointwise limit $0$ but $\|f_n\|_1=1$ for every $n$ so $f_n\not\to 0$ in $L^1(\mathbb{R})$.</p>

Title: Limit of an $L^1$ sequence


问:
Combination with repetitions.
The formula for computing a k-combination with repetitions from n elements is:
$$\binom{n + k - 1}{k} = \binom{n + k - 1}{n - 1}$$

I would like if someone can give me a simple basic proof that a beginner can understand easily.

答:
To make the things more clear let there are $n$ different kinds of drinks available in a restaurant, and there are $m$ people including yourself, at your birthday party. Each one is to be served only one 'drink' as one desires/orders without restriction that two or more can ask for different or the same drink as per individual wish.
Obviously, there can be at the most $m-1$ people asking for a brand of the 'drink' which have already been desired by other people. Thus at most $m-1$ copies of one or more brands. Let $i$ people desired different brands out of the available $n$ brands which they can ask for in $n\choose i$ ways, and $i$ can vary from $1$ to $m$. Further more, the remaining $m-i$ people must have desired the same brand(s) as already opted by the $i$ people. This can be done in $m-1\choose m-i$ combinations. Thus, the drinks that can be served, have ${n\choose i}{m-1\choose m-i}$ options. Hence total number of options is the summation of ${n\choose i}{m-1\choose m-i}$ over $i$. After using the Vandermonde's Identity.
This is equal to $n+m-1\choose m$.
Therefore, the total number of options is ${n+m-1\choose m}= \frac{(n+m-1)!}{m!(n-1)!}$.
So, if $n=5$ and $m=3$ then total number of options is given by
$= \frac{(5+3-1)!}{3!(5-1)!} = \frac{7!}{3!4!} = 35$

Title: Combination with repetitions.


问:
Prove general Euclid's Lemma in a UFD using prime factorization
I have seen plenty of proofs for this theorem:
In a UFD if $(a,b)=1$ and $a|bc$ then $a|c$.
They mostly use the gcd distributive law e.g. here.
Well I wanted to prove this just by relying on the properties that the UFD had.

My attempt:
Since $a|bc$ then for some $r$ we have $ar=bc$.
Now by the existence, since we know that any non unit element like $a$ can be rewritten as $t_1×....t_n$ where $t_i$ are irreducible, we can do this:

$$p_1^{α_1}...p_n^{α_n} g_1^{ε_1}...g_m^{ε_m} =q_1^{β_1}...q_k^{β_k}h_1^{ψ_1}...h_i^{ψ_i}$$
(Where $p_i$, $g_i$, $q_i$ and $h_i$ are primes.)
By uniqueness the set that is on right should be on left too, Am I right?
But since $(a,b)=1$ then $a$ and $b$ shouldn't share prime elements. Somehow it's like $A$ is a subset of $C$. I can't really manage this but it's becoming like a problem in set theory.

Can you please help me with my own approach??

答:
You're very close. Let's look back at this equation you stated:'

$$
p_1^{α_1}...p_n^{α_n} g_1^{ε_1}...g_m^{ε_m} =q_1^{β_1}...q_k^{β_k}h_1^{ψ_1}...h_i^{ψ_i}
$$

corresponding to $ar=bc$. As you said, because we are in a UFD the set of primes, counted with multiplicity, is the same on both sides (up to units). Furthermore, as $(a,b)=1$ then no $p_i$ can divide $b$. Once again by uniqueness, that means that no $p_i$ can divide $q_j$. In fact, we can go farther and say that no $p_i^{\alpha_i}$ can divide $q_j$. Putting this together, all the $p_i^{\alpha_i}$ must appear in the factorization on the right hand side (up to units). Furthermore, the $p_i^{\alpha_i}$ cannot divide the $q_j$. Thus, up to units, the $p_i^{\alpha_i}$ must each divide some $h_j^{\psi_j}$. Hence, all the prime factors of $a$ counted with multiplicity divide $c$. Hence, $a \mid c$.


问:
Determine if the polynomial f(x) = $2x^5 + 5x^4 + 4x^3 + 7x^2 + 7x + 2$ is reducible to $\mathbb Q $
Determine if the polynomial $f (x) =  2x ^ 5 + 5x ^ 4 + 4x ^ 3 + 7x ^ 2 + 7x + 2 $ is reducible in $ \mathbb Q $. In that case, write it as a product of irreducible polynomials.

Using Eisenstein's criterion we have $ a_0 = a_5 = 2 $ so there is no prime that divides $a_0$ and does not divide $a_5$, so $f(x)$ is not irreducible.

Since all coefficients are integers, we apply the rational zeros theorem

The trailing coefficient is $2$ so $p$ can assume the following values: $1,-1, 2, -2$.

The leading coefficient  is $2$ so $q$ can assume the following values: $1,-1, 2, -2$.

So $\frac{p}{q}$ can assume: $\frac{1}{1}$, $-\frac{1}{1}$, $\frac{1}{2}$, $-\frac{1}{2}$, $\frac{2}{1}$, $-\frac{2}{1}$, $\frac{2}{2}$, $-\frac{2}{2}$

remove duplicates:$\ 1, -1, 2, -2, \frac{1}{2}, -\frac{1}{2}$

If $a$ is a root of the polynomial $P(x)$, then the remainder from the division of $P(x)$ by $x−a$ should equal $0$. What happens in the case of $-\frac{1}{2}$ and we have: divide $2x^4+x^3+2x^2+3x+1$ by $x + \frac{1}{2}$.

The quotient is $2x^3+2x+2$, and the remainder is $0$.

Therefore we have $2x^4+x^3+2x^2+3x+1=(x+\frac{1}{2})(2x^3+2x+2)$

And

$2x^5+5x^4+4x^3+7x^2+7x+2=(x+2)(2x+1)(x^3+x+1)$

That's what I did, any comment would be useful.

Thanks in advance.

答:
You seem to be fully factoring the polynomial over $\Bbb{Q}$, which is a fair bit more work than simply determining irreducibility. From the fact that it has a root at $x = -1/2$, and therefore a factor of $x + 1/2$, you can already conclude that the polynomial is indeed reducible.

What you can't do, however, is use Eisenstein's criterion to show reducibility. Eisenstein's criterion is a sufficient, but not necessary condition for irreducibility over $\Bbb{Q}$. For example, $x^2 + x + 1$ fails Eisenstein's criterion, but is still irreducible over $\Bbb{Q}$.

Finally, there's a fair number of little issues with the write up, most of which can be chalked up to hastily typesetting your working. For example:

The leading coefficient is 2 so p can assume the following values: +1,-1, 2, -2.

should be

The leading coefficient is 2 so q can assume the following values: +1,-1, 2, -2.

More concerning is the following:

remove duplicates: +1, -1, $\frac{1}{2}$, $-\frac{1}{2}$

You seem to have misplaced $2$ and $-2$. That said, you did factor out $x + 2$ (though your working makes it unclear when that was), so I'm guessing these possible roots weren't completely forgotten.


问:
Cumulative Distribution Function given mean, std dev, skew and kurtosis?
Is there a function that calculates the cumulative distribution function (CDF) of a Gaussian distribution given the mean, std dev, skew, and kurtosis?  Does anyone know of one written in a c-like language (c, c++, java, c#)?
Thanks</p>

答:
If the given CDF is Gaussian, then it has no skewness nor excess kurtosis by definition.

In general, programming functions do not take statistical moments of the distribution in as inputs in order to construct the CDF, since an infinitely many different distribution families could share the exact same moments. For example, a t-distribution or beta distribution CDF, if parameterized right, could have the exact same $\mu, \sigma, \xi, \kappa$ as a Gaussian CDF.

The CDF is usually constructed by sorting the empirical data by rank and some additional normalization step applied to the frequencies of samples/outcomes/indices.

This is more of something to consider regarding the connection between moments and your distribution family of interest, in terms of the specificity of your question. What are you more interested in: the Gaussian distribution, or targeted values for moments (i.e. non-normality), when modeling CDFs?

Besides, I thought skewness and kurtosis were not visible in the CDF, only the PDF. maybe someone can correct me on that


问:
How to prove that $50^{44}+30! \cdot44^{50}-24 $ is divisible by $31$?
I am struggling with this example. This is a new topic for me. I would really appreciate some hints to understand it. I need part a) and b) to conclude the excercise.

I was thinking about Fermat, Euler or Wilson theorems but I am not sure if it is correct, of course I can use properties of congruences in number theory.

Part a): show that $50^{44}+30! \cdot 44^{50}-24 $ is divisible by $31$

Part b) : Conclude that $ (50^{44}+30!\cdot 44^{50})^{\varphi(31)} \equiv 1 \pmod{ 31}$

Body:
Here's a handful of hints:

$50^{44} = 2^{44}5^{88}$

$44^{50} = 2^{100}11^{50}$

$2^5\equiv 1 \pmod{31}$

$5^3 \equiv 1 \pmod{31}$

$11^2 \equiv -3 \pmod{31}$


问:
Linear combination of vectors in $\mathbb{R}^3$
Show that any linear combination of $\pmatrix{1\\\frac{3}{2}\\0}$ and $\pmatrix{0\\3\\6}$ is also a linear combination of $\pmatrix{2\\3\\0}$ and $\pmatrix{0\\1\\2}$

I'm not sure how to do this. I have a proof sketch for showing that any 2-dimensional vector is a linear combination of any other two nonparallel 2d vectors, but as far as I understand, this is not the case for 3d vectors. So how should I go about showing this?

Thanks.

答: Hint: Your third and fourth vectors are just scaled versions of the first two vectors.


问:
Find the direction in which an object with linear motion should be thrown in order to touch with another object moving linearly
Suppose we have two objects in a 3D space:

The first object, A, is at the given position P1 moving at a constant 3D Vector speed of V1

The second object, B, is at a given position that we will call P2

Knowing that B can only move at a speed of magnitude M1, in which direction should B move in order to touch A? The result must be expressed as a Unit 3D Vector

Here's a visual image I put up in Paint to help visualize the problem:
Visual Example

Thank you.

答:
Let the position of the target be a vector-valued function
$$\vec{p}(t) = \vec{b} + t \vec{v} \tag{1a}\label{1a}$$
and the position of the interceptor
$$\vec{P}(t) = \vec{B} + t \vec{V}, \quad \lVert \vec{V} \rVert = M \tag{1b}\label{1b}$$
so that at some time $\tau \gt 0$, the two collide:
$$\vec{p}(\tau) = \vec{P}(\tau) \tag{2}\label{2}$$
Expanding and simplifying $\eqref{2}$ we get
$$\begin{aligned}
\vec{b} + \tau \vec{v} &= \vec{B} + \tau \vec{V} \\
\tau \vec{V} &= \vec{b} - \vec{B} + \tau \vec{v} \\
\vec{V} &= \vec{v} + \frac{\vec{b} - \vec{B}}{\tau} \\
\end{aligned} \tag{3}\label{3}$$
In Cartesian component form, we have four equations and four unknowns ($V_x$, $V_y$, $V_z$, and $\tau$):
$$\left\lbrace ~ \begin{aligned}
M^2 &= V_x^2 + V_y^2 + V_z^2 \\
V_x &= v_x + \frac{b_x - B_x}{\tau} \\
V_y &= v_y + \frac{b_y - B_y}{\tau} \\
V_z &= v_z + \frac{b_z - B_z}{\tau} \\
\end{aligned} \right. \tag{4a}\label{4a}$$
This can have zero, one, or two sets of real solutions:
$$\begin{cases}
\tau = \frac{\vec{v} \cdot (\vec{B} - \vec{b})}{\lVert\vec{v}\rVert^2 - M^2} \pm \frac{\sqrt{ M^2 \lVert \vec{B} - \vec{b} \rVert^2 - \lVert \vec{v}\times(\vec{B} - \vec{b})\rVert^2}}{\lVert \vec{v} \rVert^2 - M^2}, & \lVert\vec{v}\rVert \ne M \\
\tau = \frac{\lVert \vec{B} - \vec{b} \rVert^2}{2 \vec{v} \cdot (\vec{B} - \vec{b}) }, & \lVert\vec{v}\rVert = M \\
\end{cases} \tag{4b}\label{4b}$$
where only $\tau \gt 0$ makes sense. Note that the second only has a positive real solution when $\vec{v}$ is (in the half-space) towards $\vec{B}$ from $\vec{b}$.  Then,
$$\vec{V} = \frac{\vec{v} + \vec{b} - \vec{B}}{M \tau} \tag{4c}\label{4c}$$
and $\lVert\vec{V}\rVert = 1$ within rounding error.

You can obtain these using Maxima or wxMaxima (free, available for all operating systems) or some other Computer Algebra System like Maple, via e.g.

```<code>solve([ V_x^2 + V_y^2 + V_z^2 = M^2,
        V_x = v_x + (b_x - B_x)/tau,
        V_y = v_y + (b_y - B_y)/tau,
        V_z = v_z + (b_z - B_z)/tau ], [ tau, V_x, V_y, V_z ]);
solve([ V_x^2 + V_y^2 + V_z^2 = v_x^2 + v_y^2 + v_z^2,
        V_x = v_x + (b_x - B_x)/tau,
        V_y = v_y + (b_y - B_y)/tau,
        V_z = v_z + (b_z - B_z)/tau ], [ tau, V_x, V_y, V_z ]);
</code>```

Here is an example Python 3 implementation of the above:

```<code>from math import sqrt

def intercept(m, dx, dy, dz, vx, vy, vz):
    tdiv = m*m - vx*vx - vy*vy - vz*vz
    while True:
        if tdiv != 0:
            tsqr = m*m*(dx*dx + dy*dy + dz*dz) - (dy*vz-dz*vy)**2 - (dz*vx-dx*vz)**2 - (dx*vy-dy*vx)**2
            if tsqr < 0:
                break
            tdif = sqrt(tsqr) / tdiv
            tsum = (vx*dx + vy*dy + vz*dz) / tdiv
            t1 = tsum + tdif
            t2 = tsum - tdif
            if t1 > 0 and t2 > 0:
                t = min(t1, t2)
            elif t1 > 0:
                t = t1
            elif t2 > 0:
                t = t2
            else:
                break
        else:
            tdiv = vx*dx + vy*dy + vz*dz
            if tdiv == 0:
                break
            t = 0.5 * (dx*dx + dy*dy + dz*dz) / tdiv
            if t <= 0:
                break

        return (dx+t*vx)/(t*m), (dy+t*vy)/(t*m), (dz+t*vz)/(t*m), t

    return None, None, None, None

def V(x, y, z):
    return "(%.6f, %.6f, %.6f)" % (x, y, z)

if __name__ == '__main__':
    from random import Random
    rng = Random()

    # Interceptor launch site
    origin_x = rng.uniform(-10, +10)
    origin_y = rng.uniform(-10, +10)
    origin_z = rng.uniform(-10, +10)

    # Interceptor speed
    speed = rng.uniform(2.0, 5.0)

    # Target starting location
    target_x0 = rng.uniform(-10, +10)
    target_y0 = rng.uniform(-10, +10)
    target_z0 = rng.uniform(-10, +10)

    # Target direction and speed
    target_vx = rng.uniform(-1, +1)
    target_vy = rng.uniform(-1, +1)
    target_vz = rng.uniform(-1, +1)
    target_v = sqrt(target_vx*target_vx + target_vy*target_vy + target_vz*target_vz)

    print("Interceptor launches from %s" % V(origin_x, origin_y, origin_z))
    print("               with speed %.6f" % speed)
    print("     Target launches from %s" % V(target_x0, target_y0, target_z0))
    print("     with velocity vector %s" % V(target_vx, target_vy, target_vz))
    print("                 at speed %.6f" % target_v)
    print("             in direction %s" % V(target_vx/target_v, target_vy/target_v, target_vz/target_v))

    dir_x, dir_y, dir_z, when = intercept(speed,
                                          target_x0 - origin_x,
                                          target_y0 - origin_y,
                                          target_z0 - origin_z,
                                          target_vx,
                                          target_vy,
                                          target_vz)
    if when is None:
        print("Interception is not possible.")
    else:
        print("   Interception occurs at %s" % V(target_x0 + when*target_vx,
                                                 target_y0 + when*target_vy,
                                                 target_z0 + when*target_vz))
        print("                        = %s" % V(origin_x + when*speed*dir_x,
                                                 origin_y + when*speed*dir_y,
                                                 origin_z + when*speed*dir_z))
        print("  if launced in direction %s" % V(dir_x, dir_y, dir_z))
        print("  whose Euclidean norm is %.9f" % sqrt(dir_x*dir_x + dir_y*dir_y + dir_z*dir_z))
</code>```

When run, the above generates a random origin, random target, random target velocity, random interceptor speed, and finds the interceptor direction.

Body:
Let the position of the target be a vector-valued function
$$\vec{p}(t) = \vec{b} + t \vec{v} \tag{1a}\label{1a}$$
and the position of the interceptor
$$\vec{P}(t) = \vec{B} + t \vec{V}, \quad \lVert \vec{V} \rVert = M \tag{1b}\label{1b}$$
so that at some time $\tau \gt 0$, the two collide:
$$\vec{p}(\tau) = \vec{P}(\tau) \tag{2}\label{2}$$
Expanding and simplifying $\eqref{2}$ we get
$$\begin{aligned}
\vec{b} + \tau \vec{v} &= \vec{B} + \tau \vec{V} \\
\tau \vec{V} &= \vec{b} - \vec{B} + \tau \vec{v} \\
\vec{V} &= \vec{v} + \frac{\vec{b} - \vec{B}}{\tau} \\
\end{aligned} \tag{3}\label{3}$$
In Cartesian component form, we have four equations and four unknowns ($V_x$, $V_y$, $V_z$, and $\tau$):
$$\left\lbrace ~ \begin{aligned}
M^2 &= V_x^2 + V_y^2 + V_z^2 \\
V_x &= v_x + \frac{b_x - B_x}{\tau} \\
V_y &= v_y + \frac{b_y - B_y}{\tau} \\
V_z &= v_z + \frac{b_z - B_z}{\tau} \\
\end{aligned} \right. \tag{4a}\label{4a}$$
This can have zero, one, or two sets of real solutions:
$$\begin{cases}
\tau = \frac{\vec{v} \cdot (\vec{B} - \vec{b})}{\lVert\vec{v}\rVert^2 - M^2} \pm \frac{\sqrt{ M^2 \lVert \vec{B} - \vec{b} \rVert^2 - \lVert \vec{v}\times(\vec{B} - \vec{b})\rVert^2}}{\lVert \vec{v} \rVert^2 - M^2}, & \lVert\vec{v}\rVert \ne M \\
\tau = \frac{\lVert \vec{B} - \vec{b} \rVert^2}{2 \vec{v} \cdot (\vec{B} - \vec{b}) }, & \lVert\vec{v}\rVert = M \\
\end{cases} \tag{4b}\label{4b}$$
where only $\tau \gt 0$ makes sense. Note that the second only has a positive real solution when $\vec{v}$ is (in the half-space) towards $\vec{B}$ from $\vec{b}$.  Then,
$$\vec{V} = \frac{\vec{v} + \vec{b} - \vec{B}}{M \tau} \tag{4c}\label{4c}$$
and $\lVert\vec{V}\rVert = 1$ within rounding error.

You can obtain these using Maxima or wxMaxima (free, available for all operating systems) or some other Computer Algebra System like Maple, via e.g.

solve([ V_x^2 + V_y^2 + V_z^2 = M^2,
        V_x = v_x + (b_x - B_x)/tau,
        V_y = v_y + (b_y - B_y)/tau,
        V_z = v_z + (b_z - B_z)/tau ], [ tau, V_x, V_y, V_z ]);
solve([ V_x^2 + V_y^2 + V_z^2 = v_x^2 + v_y^2 + v_z^2,
        V_x = v_x + (b_x - B_x)/tau,
        V_y = v_y + (b_y - B_y)/tau,
        V_z = v_z + (b_z - B_z)/tau ], [ tau, V_x, V_y, V_z ]);

Here is an example Python 3 implementation of the above:

from math import sqrt

def intercept(m, dx, dy, dz, vx, vy, vz):
    tdiv = m*m - vx*vx - vy*vy - vz*vz
    while True:
        if tdiv != 0:
            tsqr = m*m*(dx*dx + dy*dy + dz*dz) - (dy*vz-dz*vy)**2 - (dz*vx-dx*vz)**2 - (dx*vy-dy*vx)**2
            if tsqr < 0:
                break
            tdif = sqrt(tsqr) / tdiv
            tsum = (vx*dx + vy*dy + vz*dz) / tdiv
            t1 = tsum + tdif
            t2 = tsum - tdif
            if t1 > 0 and t2 > 0:
                t = min(t1, t2)
            elif t1 > 0:
                t = t1
            elif t2 > 0:
                t = t2
            else:
                break
        else:
            tdiv = vx*dx + vy*dy + vz*dz
            if tdiv == 0:
                break
            t = 0.5 * (dx*dx + dy*dy + dz*dz) / tdiv
            if t <= 0:
                break

        return (dx+t*vx)/(t*m), (dy+t*vy)/(t*m), (dz+t*vz)/(t*m), t

    return None, None, None, None

def V(x, y, z):
    return "(%.6f, %.6f, %.6f)" % (x, y, z)

if __name__ == '__main__':
    from random import Random
    rng = Random()

    # Interceptor launch site
    origin_x = rng.uniform(-10, +10)
    origin_y = rng.uniform(-10, +10)
    origin_z = rng.uniform(-10, +10)

    # Interceptor speed
    speed = rng.uniform(2.0, 5.0)

    # Target starting location
    target_x0 = rng.uniform(-10, +10)
    target_y0 = rng.uniform(-10, +10)
    target_z0 = rng.uniform(-10, +10)

    # Target direction and speed
    target_vx = rng.uniform(-1, +1)
    target_vy = rng.uniform(-1, +1)
    target_vz = rng.uniform(-1, +1)
    target_v = sqrt(target_vx*target_vx + target_vy*target_vy + target_vz*target_vz)

    print("Interceptor launches from %s" % V(origin_x, origin_y, origin_z))
    print("               with speed %.6f" % speed)
    print("     Target launches from %s" % V(target_x0, target_y0, target_z0))
    print("     with velocity vector %s" % V(target_vx, target_vy, target_vz))
    print("                 at speed %.6f" % target_v)
    print("             in direction %s" % V(target_vx/target_v, target_vy/target_v, target_vz/target_v))

    dir_x, dir_y, dir_z, when = intercept(speed,
                                          target_x0 - origin_x,
                                          target_y0 - origin_y,
                                          target_z0 - origin_z,
                                          target_vx,
                                          target_vy,
                                          target_vz)
    if when is None:
        print("Interception is not possible.")
    else:
        print("   Interception occurs at %s" % V(target_x0 + when*target_vx,
                                                 target_y0 + when*target_vy,
                                                 target_z0 + when*target_vz))
        print("                        = %s" % V(origin_x + when*speed*dir_x,
                                                 origin_y + when*speed*dir_y,
                                                 origin_z + when*speed*dir_z))
        print("  if launced in direction %s" % V(dir_x, dir_y, dir_z))
        print("  whose Euclidean norm is %.9f" % sqrt(dir_x*dir_x + dir_y*dir_y + dir_z*dir_z))

When run, the above generates a random origin, random target, random target velocity, random interceptor speed, and finds the interceptor direction.


问:
How to integrate this integral using Green's theorem?
$\int_{L} (x+2xy)dy + y^{2}dx$ in the circle $L=\{(x,y)|x^2+y^2=y\}$. I already know that $P_y'=2y$ and that $Q_x'=1+2y$ where $P$ is the function multiplied by $dx$ and $Q$ is the function multiplied by $dy$. So I figured that since the given circle has $x$ in $[-0.5,0.5]$ and $y$ is in $[0,1]$ then I should integrate twice $\int_{0}^{1}(\int_{-0.5}^{0.5}1+4ydx)dy$. Is this correct?

答:
It'll look cluttered in a comment, so I'll try to address the main points here.

You're correct in deducing that $L$ traces out a circle with radius $1/2$ and center $(0,1/2)$. Green's theorem can be used for your integral. There is a sign error in your expression for the curl of your vector field $\mathbf{F}(x,y) = (y^2, x+2xy)$. It should be

$$ \text{curl}(\mathbf{F}) = \left( \dfrac{\partial Q}{\partial x} \ {\color{red}-} \ \dfrac{\partial P}{\partial y} \right) = 1+2y-2y = 1 $$

Let $R$ denote the region enclosed by $L$. By Green's theorem, we can express our integral as

$$ \oint_{L} \mathbf{F} \cdot \text{d}s = \iint_{R} \text{curl}(\mathbf{F}) \text{ d}A $$

For the purpose of this question, it should be easy to see that $ \iint_{R} \text{d}A = \text{area}(R) $.

 If you wanted to actually compute such an integral, the bounds you've selected, namely $x \in [-1/2,1/2]$ and $y \in [0,1]$, actually trace out a square in $\mathbb{R}^{2}$. You can try parameterising a circular region with center $(0,0)$ and radius $1$ using Cartesian coordinates. The you can just shift and scale this region to $R$.)


答:
An alternative approach is to use the stars and bars method.

We can generalize and consider instead of triangles, $k$-sided polygons. Also let $d$ be the minimum "distance" among vertices of those $k$-sided polygons, where "distance" is the number of inner vertices plus one. In our case we have $k = 3$ and $d = 4$. So the problem becomes finding the number of solutions of:

$$ x_1 + x_2 + \ldots + x_{k-1} + x_k = n$$

where $x_i, i=1,\ldots,k$ are the "distances" among vertices of the $k$-sided polygons, with the constraint:

$$x_i \ge d, i=1,\ldots,k$$

We can define $y_i = x_i+d, i=1,\ldots,k$, and then the first equation becomes:

$$y_1 + y_2 + \ldots + y_{k-1} + y_k = n-kd$$

with $y_i \ge 0, i=1,\ldots,k$. Therefore, by the stars and bars method, the solutions for each vertex are:

$${n-kd+k-1 \choose k-1}$$

and there are $n$ vertices, but every $k$-sided polygon is in common with $k$ of them, so the final solution is:

$${n-kd+k-1 \choose k-1}\frac{n}{k}={30-3\cdot4+3-1 \choose 3-1}\frac{30}{3}={20  \choose 2}\frac{30}{3}=1900$$

Title: Let $P$ be a $30$-sided polygon inscribed in a circle. Find the value of $\frac{N}{100}$.

Body:
Let $P$ be a $30$-sided polygon inscribed in a circle. There are $N$ number of triangles whose vertices are the vertices of $P$ such that any two vertices of each triangle are separated by at least three other vertices at $P$. Find the value of $\frac{N}{100}$.

What I Tried: This is more like a Combinatorics problem rather than a geometry problem, so here is what I think.

First, fix a point of a triangle. The next point can be chosen in $23$ ways. But I am not sure how to choose the $3$rd point, as for choosing the $2$nd point there are slight variations as well, which dosen't follow the rule.

I thought before of fixing one point, and then the next $2$ points can be chosen in ${23}\choose{2}$ ways, but then I realised that is wrong since those $2$ points might not have a $3$ point gap, and I couldn't get on how to progress on this.

As usual, I also know that the number of triangles on an $n$-sided polygon with no shared sides is given by the formula :-
$$\rightarrow\frac{n(n-4)(n-5)}{6}$$
So the total number of triangles is $3250$, but I am not sure on how this fact will help in this problem.

Can anyone help me? Thank You.


问:
Is the linear algebra a way to represent the linear equations?
If yes, then what is the representation of $f(x)=2x+1$?

What I believe is $2x+1=0$ is a linear equation.

If that is correct, then why when we represent vectors we have to draw a line from the origin to where the vector points?

While if we want to represent the earlier equation in coordinate system, then we will not reach the origin.

I'm not good at math, I just started my journey, so excuse me if I made a big mistake.

答:
Linear algebra deals with vector spaces and linear mappings between them, in particular with vectors, matrices, linear equations and systems of linear equations.

$2x+1=0$ is a linear equation regarding $x$ because it is a homogenous equation and its term containing $x$ is an algebraic term regarding $x$ of degree $1$ regarding $x$.

We have to distinguish vectors and the kinds of vectors that are position vectors.


问:
Open-closed topology on the real numbers
I am reading a paper which refers to

the topological space
$(\mathbb{R}^+, \tau)$ where $\tau$ is the closed-open topology on $\mathbb{R}^+$ generated by the set of intervals $[s, t)$ with $s \leq t$ as its basis.

What does it mean to call something a closed-open topology, as opposed to a topology? I can't find any definition of a closed-open topology.

What does it mean to say that $s \leq t$ is its basis? According to wikipedia, ''a basis for a topology on a set $X$ is a family $B$ of open subsets of $X$ such that every open set is equal to a union of some sub-family of $B$''. But I can't see how defining the basis as $s \leq t$ fits this definition.

答:
In general, suppose $X$ is a set, and $\mathcal{B}$ is a collection of subsets of $X$ such that $X=\cup_{B\in\mathcal{B}} B$, and which satisfy the following property: for all $B_1, B_2\in\mathcal{B}$ and for all $x\in B_1\cap B_2$ there is some $B\in\mathcal{B}$ such that $x\in B\subseteq B_1\cap B_2$. In this case we can define $\tau\subseteq 2^X$ like this:

$\tau=\{U\subseteq X: \forall x\in U \ \ \exists B\in\mathcal{B} $ such that $x\in B\subseteq U\}$

It's not hard to check that $\tau$ is a topology on $X$. And then note that $\mathcal{B}$ is its basis. Indeed, if $U\in\tau$ then by definition for every $x\in U$ there is some $B_x\in\mathcal{B}$ such that $x\in B_x\subseteq U$. Thus $U=\cup_{x\in U} B_x$.

So in your case we have $X=\mathbb{R^+}$ and $\mathcal{B}=\{[s,t): 0\leq s\leq t\}$. This collection indeed satisfies the required property, so it generates a topology on $\mathbb{R^+}$. The author calls it the closed-open topology.


问:
Is the following map continuous?
Let $C(X,\mathbb{R})$ and $C(Y,\mathbb{R})$ be the space of continuous functions with $X$ is compact metric space and $Y$ be closed subset of $X$. The spaces are using the supremum norm. Is the map $R:C(X)\to C(Y)$ with $Rf=f|_Y$ a continuous map?
If it is how do I show it, thank you.

答: I think you can manage to show that when $g: X \rightarrow Y$ is a continuous map between compact metric spaces, then the application $C(Y, \mathbb{R}) \rightarrow C(X, \mathbb{R})$ obtained by precomposing by $g$ is continuous with your topology (sup norm). Then, your case follows with $g$ the inclusion $X \subset Y$.

Title: Is the following map continuous?


问:
Issues with normal of a line.
I think I am doing something wrong here but think about this simple thought if we have a 2D line in a Cartesian plane, say $2x+y=5$, and $P=(0,5)$ is a point on the line. We can say that
$$
n·x = n·p
$$
Saying that $n$ is the normal to the line $2x+y=5$. If $n = [a,b]$ and $x = [x,y]$ (point X and P being on the line $2x+y=5$), we say that
$$
[a,b] ·[x,y] = n·p
$$
and both n·p and n·x are 0 as both p and x are orthogonal to the normal N. So from the equation above we can make
$$
[a,b] · [x,y] = 0
$$
or
$$
ax+by = 0
$$
so does this mean that $ax+by$ always $= 0$? That wouldn't make sense, would it? What am I doing wrong?
Thank you.

答:
This is confusion between position vectors and free vectors. Points $p$ and $x$ are position vectors; with reference to the origin $(0,0)$. They are fixed on the plane.

$n$ is a free vector. It can be moved around (but its direction stays the same).

So $n\cdot x = n\cdot p \Rightarrow n\cdot(x-p)=0$

and not $n\cdot x =0= n\cdot p$! $n\cdot x$ is non-zero in general, as is $n\cdot p$. Position vector $x$ is pointing from $(0,0)$ to point $x$, which need not be orthogonal to $n$.

$n\cdot(x-p)=0$ means $x-p$ is a vector along the given line and $n$ is of course, orthogonal to it.

Title: Issues with normal of a line.


问:
tower of separable extension
Let $K⊂L⊂M$ be a tower of fields.
Let $L/K$ and $M/L$ be separable, is it true that $M/K$ is separable?

I guess there are counterexamples, but I cannot point out them.
Thank you for your kind help.[I know if we substitute separable to normal, there are easy example.]

答:
Yes, it is! Let us first show that it is true for finite extensions and then reduce to the finite case later:

Assume that $M/L$ and $L/K$ are finite separable extensions. Recall that a finite extension is separable if and only if its degree of separability is equal to the degree of the extension. This leaves us with:
$$[M:K]_s=[M:L]_s\cdot [L:K]_s=[M:L]\cdot [L:K]=[M:K]$$
So we are done with the finite case.

Let's assume that $M/L$ and $L/K$ are (not necessarily finite) separable extensions. Let $\alpha\in M$. Assume that $a_0,...,a_n\in L$ are the coefficients of the minimal polynomial of $\alpha$ over $L$. We know that the extension $K(\alpha,a_0,\dots,a_n)/K$ is finite. Define
$$F:=K(\alpha,a_0,\dots,a_n)\cap L$$
Since $F\subset L$, we know that $F/K$ is separable. Also the minimal polynomial of $\alpha$ over $F$ is the same as that over $L$ and hence separable. Thus the extension $K(\alpha,a_0,\dots,a_n)/K$ is seperable by the first paragraph since all extensions are finite.

Title: tower of separable extension


答:
You can use matrices and vectors to solve a system of linear equations in multiple variables.

Specifically, the generic system
$$a_{11}x_1+a_{12}x_2+\dots a_{1n}x_n=b_1\\
a_{21}x_1+a_{22}x_2+\dots a_{2n}x_n=b_2\\
\vdots\\
a_{m1}x_1+a_{m2}x_2+\dots a_{mn}x_n=b_m$$
where the coefficients $a_{ij}$ and also $b_j$ are fixed numbers,
can be simply expressed as
$$Ax=b$$
where $A$ is the matrix of the coefficients $a_{ij}$, $b$ is the column vector with coordinates $b_1,\dots,b_m$ and $x$ is the $n$ dimensional column vector with the indeterminates $x_i$ as its coordinates.

Title: Is the linear algebra a way to represent the linear equations?

Body:
If yes, then what is the representation of $f(x)=2x+1$?

What I believe is $2x+1=0$ is a linear equation.

If that is correct, then why when we represent vectors we have to draw a line from the origin to where the vector points?

While if we want to represent the earlier equation in coordinate system, then we will not reach the origin.

I'm not good at math, I just started my journey, so excuse me if I made a big mistake.