问:
Structure of vector bundles over $(\mathbb P^1)^r$
Fix an algebraically closed field $k$ of characteristic zero. I would like to know a classification of all vector bundles on $(\mathbb P_k^1)^r$, or at least of semistable vector bundles.

A theorem of Grothendieck states that a vector bundle $E$ of rank $m$ on $\mathbb P^1$ decomposes as a direct sum of line bundles: $E \cong \mathcal O(a_1) \oplus \dots \oplus \mathcal O(a_r)$.

Let $X$ be the $r$-fold fibre product of $\mathbb P^1$'s. We know (Hartshorne III Ex 12.6) that $\operatorname{Pic}(X) = \mathbb Z^r$, so every line bundle on $X$ is of the form $\pi_1^*\mathcal O_{\mathbb P^1}(a_1) \otimes \cdots \otimes \pi_r^*\mathcal O_{\mathbb P^1}(a_r)$, with $\pi_i$ the projection maps.

For $r=2$, we can find vector bundles of rank $2$ by computing $\operatorname{Ext}^1(\pi_i^*\mathcal O_{\mathbb P^1}(a_i), \pi_j^*\mathcal O_{\mathbb P^1}(a_j))$, which is non-trivial if $a_i \leq -2$ and $a_j \geq 0$ or if $a_i \geq 0$ and $a_j \leq -2$.

Does every rank $2$ vector bundle on $(\mathbb P^1)^2$ arise as such an extension? What about vector bundles of rank $2$ on $(\mathbb P^1)^r$ for $r> 2$? What about bundles of higher rank?
What do vector bundles corresponding to non-trivial extensions look like in terms of transition functions?

By the Harder-Narasimhan filtration, every vector bundle on a projective scheme arises as an iterated extension of semistable vector bundles.
Of the line bundles which arise as direct sums of tensor products of pullbacks of line bundles on $\mathbb P^1$ as above, it is not hard to show that the only semistable such are those of the form $\bigoplus_{i=1}^m (\pi_1^*\mathcal O_{\mathbb P^1}(a_1) \otimes \cdots \otimes \pi_r^*\mathcal O_{\mathbb P^1}(a_r))$. Are there any other semistable vector bundles on $X$?

答: Vector bundles on these are much more complicated. Consider $r=2$. We have a two to one map to  the projective plane and we can pull back vector bundles from the plane. If $E,F$ are bundles on the plane and they become isomorphic when you pull back, then $E\oplus E(-1)$ and $F\oplus F(-1)$ must be isomorphic.


问:
Can a connected closed manifold strictly contain a closed manifold of the same dimension?
A connected closed manifold can contain another one as a proper subset: for instance, the $1$-sphere (circle) is contained in the $2$-sphere. Is it possible with manifolds of the same dimension?

Precisely, is there a connected closed $n$-manifold $M$ containing a closed $n$-manifold $N\subsetneq M$?

答: No. $N$ is open (because of the same dimension), and closed (by assumption), therefore a connected component of $M,$ therefore equal to $M$ by assumption.

ContentLicense: CC BY-SA 4.0


问:
Rank of matrix verification using row echelon form
[![][1]][1]

so in the sum we have to find the rank of matrix using echleon form.i have solved the sum and got rank of matrix 3.but the book from which i am solving the sum is showing rank 2.There are also few sums in the book where its the same situation. I just want to know whether the rank is 3 or 2 of this sum and if i am wrong then need hints to find rank 2
[1]: https://i.stack.imgur.com/MFBpx.jpg

答: I do not understand why you keep referring to the sum. The matrix you have does have rank $3$.


问:
How do I find the minimum of a quartic function with a missing coefficient?
$$ f(x) = 7x^4 + kx^3 + 8 $$
I have a question asking me to state the range of this function. To do this it seems I have to find the relationship between k and the minimum of the function. I've been told that there is a way to do this with calculus, but this is an algebra class and I haven't learned any calculus yet. I've tried coming up with random factors and then using the remainder theorem to solve for k, but that just gives me an identity.

How do I find the range of this function without knowing k?
Alternatively, is it possible to find k using the given information?

答:
Using calculus min is at $x=\frac{-3k}{28}$

How to get this without calculus is a question.


问:
Examining recurrence relation for its limit and
I have got the following problem:

Examine why the following sequence is well-defined, determine whether it is a monotonous sequence and calculate its limit:

$$a_0 = 3, a_{n+1} = \sqrt{2a_n - 1}$$

Now just by looking at this sequence and checking its first few entries it seems obvious that it is strictly decreasing. I then proved this with induction. Now I'm trying to find the limit, it seems obvious that its $1$, but I have been struggling to prove this. Since I know the sequence is strictly decreasing I looked at the following:

$$a_n > a_{n+1}$$
$$a_n > \sqrt{2a_n - 1}$$
$$a_n^2 > 2a_n - 1$$
$$a_n^2 - 2a_n + 1 > 0$$
$$(a_n - 1)^2 > 0$$

From this we get that $a_n > 1$ for all n. Would this be a correct proof that the limit of the sequence is 1? And as for whether the sequence is well-defined, since all entries are bigger that $1$ the square root will never have negative input, so it would be well-defined. Would this also be correct?

答:
First you should prove that the sequence is bounded, and thus converge (decreasing + bounded).

Then to find the limit you can
set $lim_{n\to\infty}  \space a_n = L$ and since $lim_{n\to\infty}  \space a_n=lim_{n\to\infty}  \space a_{n+1}$

$$ a_{n+1} = \sqrt{2a_n-1} \Longrightarrow lim_{n\to\infty}  \space a_{n+1} = \sqrt{2(lim_{n\to\infty}  \space a_n)-1}$$
$$\Longrightarrow L=\sqrt{2L-1} \Longrightarrow L^2-2L+1=0 \Longrightarrow L=1$$


问:
Does this vanish? $\left[\left((12\cos^2t+K^2\sin^2t)^{(S+3)/2}-(12\cos^2t)^{(S+3)/2}\right)(-\cot t)\right]^\pi_0$
I have a problem that has kept me awake for several nights XD

If I integrate this (where $K$ and $S$ are constants;$K$ is small and $S$ can take values between 0 and 1):
$$\int_{0}^{\pi}\frac{1}{\sin^2t}\left[\left(12\cos^2t+K^2\sin^2t\right)^{\frac{S+3}{2}}-\left(12\cos^2t\right)^{\frac{S+3}{2}}\right]\mathrm{d}t$$
by parts, I get this:
$$=\left[ \left(\left(12\cos^2t+K^2\sin^2t\right)^{\frac{S+3}{2}}-\left(12\cos^2t\right)^{\frac{S+3}{2}}\right)(-\cot t)\right]^\pi_0-\int_{0}^{\pi}[...]$$
My question is: does the term
$$\left[ \left(\left(12\cos^2t+K^2\sin^2t\right)^{\frac{S+3}{2}}-\left(12\cos^2t\right)^{\frac{S+3}{2}}\right)(-\cot t)\right]^\pi_0$$ vanish when the limits of integration are applied? Someone$^*$ says yes, but why?

Thanks in advance!

Victor

$^*$ here is a statement about it from a French-speaking mathematician:

It reads (more or less)

One can integrate by parts $$\int\frac{dt}{\sin^2t}[\cdots]=$$
This expression simplifies.

Indeed, the terms in brackets in the first line cancel, due to $\sin^2t/\sin t$.

Edit: Can be used L'Hospital's rule to check that?

答:
I've worked this problem out and found, it's the result of an improper integral, where L'Hospital's rule can be applied. That is, if $\mathrm{cot}\;t=\frac{\mathrm{cos}\;t}{\mathrm{sin}\;t}$, we'll get:
$$\underset{t\to \pi }{\mathrm{lim}} \left\lbrack {\left(12\;{\mathrm{cos}}^2 \;t+K^2 {\mathrm{sin}}^2 \;t\right)}^{\frac{s+3}{2}} -{\left(12\;{\mathrm{cos}}^2 \;t\right)}^{\frac{s+3}{2}} \right\rbrack \left(-\frac{\mathrm{cos}\;t}{\mathrm{sin}\;t}\right)$$
$$=\underset{t\to \pi }{\mathrm{lim}} \frac{\left\lbrack {\left(12\;{\mathrm{cos}}^2 \;t+K^2 {\mathrm{sin}}^2 \;t\right)}^{\frac{s+3}{2}} -{\left(12\;{\mathrm{cos}}^2 \;t\right)}^{\frac{s+3}{2}} \right\rbrack \left(-\mathrm{cos}\;t\right)}{\mathrm{sin}\;t}$$
$$\overset{L\prime \mathrm{Hôpital}}{=} \underset{t\to \pi }{\mathrm{lim}} \frac{\left(\frac{s+3}{2}\right)\left\lbrack {\left(12{\mathrm{cos}}^2 \;t+K^2 {\mathrm{sin}}^2 \;t\right)}^{\frac{s+1}{2}} \left(K^2 -12\right)+12{\left(12{\mathrm{cos}}^2 \;t\right)}^{\frac{s+1}{2}} \right\rbrack \left(\mathrm{sin}\;2t\right)\left(-\mathrm{cos}\;t\right)+\cdots }{\mathrm{cos}\;t}$$
$$\cdots \frac{\left\lbrack {\left(12{\mathrm{cos}}^2 \;t+K^2 {\mathrm{sin}}^2 \;t\right)}^{\frac{s+3}{2}} -{\left(12{\mathrm{cos}}^2 \;t\right)}^{\frac{s+3}{2}} \right\rbrack \left(\mathrm{sin}\;t\right)}{\mathrm{cos}\;t}$$
Now, if $\mathrm{sin}\;2\pi =0$, we'll get:
$$=\frac{\left(\frac{s+3}{2}\right)\left\lbrack {\cdots} \right\rbrack \left(0\right)\left(1\right)+\left\lbrack {\cdots} \right\rbrack \left(0\right)}{-1}=0$$
Similarly, if $\mathrm{sin}\;0 =0$, we'll get:
$$=\frac{\left(\frac{s+3}{2}\right)\left\lbrack {\cdots} \right\rbrack \left(0\right)\left(-1\right)+\left\lbrack {\cdots} \right\rbrack \left(0\right)}{1}=0$$
:-)

P.S. In case that somebody needs it, I can explain how the derivative was made.


问:
Confused about partitions in Riemann Integrability
Let $f(x)$ be a function on the closed interval $[a,b]$.  We define the limit of Riemann Sums $S_{P}=\sum_{k=1}^{n} f(c_{k})\Delta x_{k}$ on $[a,b]$ as the mesh $\mu{(P)}\rightarrow 0$ to be the number $I$ if

Given $\epsilon>0$, $\exists\delta$ such that for every partition $P$ of the interval $[a,b]$

$$\mu{(P)}<\delta \Longrightarrow |S_{P}-I|<\epsilon$$

for any $c_{k}$ we choose in the subintervals $[x_{k-1},x_{k}]$.}

\noindent If the definition holds, then the limit of a Riemann Sum would exist and we can write

$$\lim_{\mu{(P)}\to 0} \sum_{k=1}^{n} f(c_{k})\Delta x_{k}=I$$

How am I supposed to interpret the definition of the definite above? The norm/mesh $\mu{(P)}$ of the partition approaches zero, and we get a finer partition. But wouldn't that create a new partition with its own Riemann Sum? Am I supposed to interpret this as a sequence of partitions where the norm/mesh $\mu{(P)}$ is approaching zero?

Thanks.

答:
The definition is the following: We write $\int_a^b f(x)\,dx=I$ if for every $\epsilon>0$ there exists a $\delta>0$ such that for every partition $P=\{x_0,x_1,\dots x_n\}$ of $[a,b]$ with $\mu(P)<\delta,$ we have

$$|\sum_{k=1}^{n}f(c_k)\Delta x_k - I|<\epsilon$$

for all choices of $c_k\in [x_{k-1},x_k], k= 1,\dots, n.$


问:
Necessary and sufficient condition for a degree 4 polynomial to be a sum of 4th powers
This is a problem I encountered in Hall and Knight's Higher Algebra. Suppose we have a polynomial $p(x,y)=a_0x^4+4a_1x^3y+6a_2x^2y^2+4a_3xy^3+a_4y^4.$
I want to find a necessary and sufficient condition on the coefficients such that $p(x,y)$ is equal to the sum of the fourth powers of 2 polynomials linear in $x$ and $y$.

I can find a necessary condition as follows:
Suppose $p(x,y)=(ax+by)^4+(cx+dy)^4$. Then, comparing coefficients,
$$a^4+c^4=a_0,$$
$$a^3b+c^3d=a_1,$$
$$a^2b^2+c^2d^2=a_2,$$
$$ab^3+cd^3=a_3,$$
$$b^4+d^4=a_4.$$

Then, doing some simple computations:
$$a_0a_2-a_1^2=a^2c^2(ad-bc)^2,$$
$$a_1a_3-a_2^2=abcd(ad-bc)^2,$$
$$a_2a_4-a_3^2=b^2d^2(ad-bc)^2.$$
Hence,
$$(a_0a_2-a_1^2)(a_2a_4-a_3^2)=(a_1a_3-a_2)^2.$$

Therefore, this condition is definitely necessary. I have a hunch that this should also be sufficient, but I don't know how to proceed with that. Are there any more conditions needed? If not, can we prove that if this condition is satisfied, one can solve for $a,b,c,d$? Any hints/suggestions would be appreciated.

答:
The necessary condition given in the question is indeed sufficient. The following proof essentially follows from arguments in J. J. Sylvester, “On a remarkable discovery in the theory of canonical forms and of hyperdeterminants":

Using the same notation as in the question, suppose
$$p(x,y)=(ax+by)^4+(cx+dy)^4.$$

Let $b=a\lambda_1$, $d=c\lambda_2$, $s_1=\lambda_1+\lambda_2$ and $s_2=\lambda_1\lambda_2$. Then, we get the equations:
$$a^4+c^4=a_0$$
$$a^4\lambda_1+c^4\lambda_2=a_1$$
$$a^4\lambda_1^2+c^4\lambda_2^2=a_2$$
$$a^4\lambda_1^3+c^4\lambda_2^3=a_3$$
$$a^4\lambda_1^4+c^4\lambda_2^4=a_4.$$

Eliminating $a^4$ and $c^4$ from the first three equations, we get:
$$\begin{vmatrix}
1 & 1 & a_0 \\
\lambda_1 & \lambda_2 & a_1 \\
\lambda_1^2 & \lambda_2^2 & a_2
\end{vmatrix}=0,$$
which simplifies to $a_2-a_1s_1+a_0s_2=0$.
Similarly, we can try to eliminate $a^4$ and $c^4$ from the second, third and fourth equations, as well as the last three equations. Thus, we end up with three equations linear in $s_1$ and $s_2$:
$$a_2-a_1s_1+a_0s_2=0$$
$$a_3-a_2s_1+a_1s_2=0$$
$$a_4-a_3s_1+a_2s_2=0.$$
Eliminating $s_1$ and $s_2$ from these three equations, we get:
$$\begin{vmatrix}
a_0 & a_1 & a_2 \\
a_1 & a_2 & a_3 \\
a_2 & a_3 & a_4
\end{vmatrix}=0,$$
which is exactly the necessary condition stated in the problem.

Assuming that this condition is satisfied, we can solve the above system of equations to find $s_1$ and $s_2$. Then, solving the quadratic equation $\lambda^2-s_1\lambda+s_2=0$ allows us to find $\lambda_1$ and $\lambda_2$. Finally, we can use the first five equations, that are linear in $a^4$ and $c^4$, to solve for them. This determines $a,b,c$ and $d$ and proves the claim.


问:
Finding inverse of polynomial in a field
I'm having trouble with the procedure to find an inverse of a polynomial in a field. For example, take:

In $\frac{\mathbb{Z}_3[x]}{m(x)}$, where $m(x) = x^3 + 2x +1$, find the inverse of $x^2 + 1$.

My understanding is that one needs to use the (Extended?) Euclidean Algorithm and Bezout's Identity. Here's what I currently have:

Proceeding with Euclid's algorithm:

$$ x^3 + 2x + 1  =(x^2 + 1)(x) + (x + 1) \\\\
  x^2 + 1 = (x + 1)(2 + x) + 2$$

We stop here because 2 is invertible in $\mathbb{Z}_3[x]$. We rewrite it using a congruence:

$$(x+1)(2+x) \equiv 2 \mod{(x^2+1)}$$

I don't understand the high level concepts sufficiently well and I'm lost from here. Thoughts?

Wikipedia has a page on this we a decent explanation, but it's still not clear in my mind.

Note that this question has almost the same title, but it's a level of abstraction higher. It doesn't help me, as I don't understand the basic concepts.

Thanks.

答:
Another way to do this, is by representing the elements of your quotient field (which is a three-dimensional vector space with base field GF(3)) as matrices with respect to the basis 1, x, x^2. Multiplication by x sends 1 to x, x to x^2 and x^2 to x^3 = -2x - 1 = x + 2 (modulo 3). So (multiplication by) x is represented by the following matrix m:

    \begin{pmatrix}
    0 & 0 & 2 \\
    1 & 0 & 1  \\
    0 & 1 & 0
\end{pmatrix}

m^0 (corresponding to multiplication by x^0 = 1) is the identity matrix:

    \begin{pmatrix}
    1 & 0 & 0 \\
    0 & 1 & 0  \\
    0 & 0 & 1
\end{pmatrix}

and m^2 (multiplication by x^2) equals:

    \begin{pmatrix}
    0 & 2 & 0 \\
    0 & 1 & 2  \\
    1 & 0 & 1
\end{pmatrix}

The element (x^2 + 1)^-1 is represented by (m^2 + m^0)^-1, which equals

    \begin{pmatrix}
    2 & 1 & 2 \\
    1 & 1 & 2  \\
    2 & 1 & 1
\end{pmatrix}

The coefficients of (x^2 + 1)^-1 can be read off from the first column of this matrix: (m^2 + m^0)^-1 = 2m^0 + m + 2m^2, so (x^2 + 1)^-1 = 2 + x + 2x^2. Summarizing: your problem can be solved by adding, multiplying and inverting matrices.

ContentLicense: CC BY-SA 3.0


问:
I would like to know an intuitive way to understand a Cauchy sequence and the Cauchy criterion.
My understanding from the definition in my book (Rudin) is this.

A seq. $\{p_n\}$ in a metric space $X$ (I only really know $\mathbb R^k$) is said to be a Cauchy sequence if for any given $\epsilon > 0$, $\exists N\in \mathbb N$ such that $\forall n,m\ge N$, $d(p_n,p_m)<\epsilon$.

(1) I see it as, given any tiny value $\epsilon$, we can find a natural number $N$ large enough so that the distance between $p_n$ and $p_m$ is less than $\epsilon$. Am I right ?

The reason I'm asking this is because I was trying to understand the proof of how
$$\sum a_nb_n$$
can converge, and the book said this

$$\left\lvert \sum_{n=p}^{q}a_nb_n\right\rvert \leq \epsilon$$

satisfies the Cauchy criterion and therefore it converges.

I read other questions and answers about the Cauchy sequence, but it didn't really help me…

Can someone explain me what's going on?

Edit:

Suppose

a) the partial sums of $A_n = \Sigma a_n$ form a bounded sequence

b) $b_0 \geq b_1 \geq \dotsb$

c) $\lim_{b \to \infty} b_n = 0$

Using the partial summation formula, algebraically the equation in the bottom is proved

$$\left\lvert \sum_{n=p}^{q}a_nb_n \right\rvert \leq \epsilon$$

Algebraically I had no problem, but I don't know why this proves convergence. I thought to show that a sequence is Cauchy, we need to find the distance between two terms in a sequence. That's where I'm confused.

答:
Let's start from scratch.  We say that the infinite series $\sum_{i = 0}^\infty a_i$ converges if the sequence of partial sums $\{S_n\}$ converges, where $S_n := \sum_{i =0}^n a_i$.

Suppose $\{S_n\}$ is a Cauchy sequence.  Then by definition, for every $\epsilon > 0$, there exists $N$ such that for all $n, m \ge N$, $|S_m - S_n| \le \epsilon$.

To be more clear, this is equivalent to saying that for all $\epsilon > 0$, there exists $N$ such that $m > n \ge N$ implies

$$\bigg|\sum_{i = 0}^m a_i - \sum_{i=0}^n a_i\bigg| = \bigg|\sum_{i = n + 1}^m a_i \bigg| \le \epsilon$$

which is exactly the Cauchy criterion.  Now, it only suffices to show that $\{S_n\}$ was Cauchy to show that $\sum_{i = 0}^\infty a_i$ converges.  Why?


问:
Expansion in power series of a fraction
My question is related to a problem in asymptotic analysis. The answer is given, but I cannot verify it.

Given the fraction
$$
\frac{-\left(x+y\right)}{a+ib\left(x+y\right)^2},\quad i=\sqrt{-1},\quad a,b,x,y\in\mathbb{R}.
$$
Then it holds that
$$
\frac{-\left(x+y\right)}{a+ib\left(x+y\right)^2}=\frac{1}{ibx}\sum^\infty_{k=0}\left(-\frac{y}{x}\right)^k\sum^\infty_{k=0}\left(\frac{ia}{2x^2b}\right)^k=-\frac{1}{ibx}+\frac{O\left(1\right)}{x^3}.
$$
I tried to crack the problem by first rewriting the fraction as
$$
\frac{i}{b\left(x+y\right)\left[1-\dfrac{ia}{b\left(x+y\right)^2}\right]}=\frac{i}{b\left(x+y\right)}\cdot \frac{1}{1-\dfrac{ia}{b\left(x+y\right)^2}}
$$
For the first term in the product I get (on expanding in $y$)
$$
\frac{i}{b\left(x+y\right)}=-\frac{1}{ibx}\sum_{k=0}^\infty\left(-\frac{y}{x}\right)^k.
$$
This would verify the first series in the answer above (without the minus sign, however). But I am lost with the remaining part. Any help is appreciated.

答:
The statement is false.  If we take $x=b=1$, $a=y=\frac12$, then the assertion is that $$\frac{-3/2}{1/2+i(3/2)^2} =\frac1i\sum_{k=0}^\infty\left(-\frac12\right)^k\sum_{k=0}^\infty\left(\frac{i}4\right)^k$$ and it is easy to verify that this isn't so.

$$\frac{-3/2}{1/2+i(3/2)^2} =\frac1i\sum_{k=0}^\infty\left(-\frac12\right)^k\sum_{k=0}^\infty\left(\frac{i}4\right)^k\implies\\
\frac{-3/2}{1/2+9i/4}=-i\frac1{1+1/2}\frac1{1-i/4}\implies
\\
\frac{-6}{2+9i}=\frac{-2i}3\frac4{4-i}\implies\\
6(12-3i)=8i(2+9i)\implies\\
72-18i=-72+16i
$$
I haven't been able to come close to the formula given.

Title: Expansion in power series of a fraction


问:
Is it possible to get a frequency equation from limited power expansion of differential equation solution?
I have a system of coupled differential equations (an example )in the form of,

$ x''+ax'+bx-cy=0$

$ y''-ay'+by-cx=0$

The solution to the above system looks like,

$x=Ae^{w_1t}+Be^{w_2t}+Ce^{w_3t}+De^{w_4t}$

$y=Ee^{w_1t}+Fe^{w_2t}+Ge^{w_3t}+He^{w_4t}$

the frequencies $w_1$,$w_2$,$w_3$ and $w_4$ are functions of $a$,$b$ and $c$  My goal is to determine that values of $a$,$b$ and $c$ such that all $w$ are $Real$.  However, my original system of differential equations has nonlinear terms.  I managed to derive the solution but in power expansion form or,

$x=\sum_{i=0}^{n}{\frac{a_i}{i!}t^i}$

$y=\sum_{i=0}^{n}{\frac{a_i}{i!}t^i}$

I need to determine the frequencies so that I can find the values of $a$,$b$ and $c$ such that all $w$ are $Real$.  Is it possible to derive $w$ from limited power expansion (say 8th term) of $x$ and $y$?

My first attempt:

I did a test of my method by taking Fourier Sine Transform of analytical solution which give me the answer in form of $\frac{f(w)}{g(w)}$ where w is the frequency.  Then solved the $g(w)$ for $w$ which gives me four answers for $w$ which are the frequencies $w_1$,$w_2$,$w_3$ and $w_4$.  However when I attempted with Fourier Sine Transform of series solutions, the answer is different.  This is due to the limited power expansion.  Is here a way to improve this?

My second attempt:

I linearize the nonlinear terms of differential equations and used the matrix to calculate Determinant of (A-wI) where A is the matrix of the system of differential equation.  I managed to calculate the values of $a$,$b$ and $c$  but they were incorrect because of linear terms I made.

A last method that I am considering is Monte Carlo.
Is it possible to get a frequency equation from limited power expansion of differential equation solution?  Any ideas for other methods that I missed?

答:
Regarding the linear system of ode's, applying the Laplace transform on

$$
\cases{ x''+ax'+bx-cy=0\\
y''-ay'+by-cx=0}
$$

we have

$$
\cases{
(s^2+a s+b)X(s)=cY(s)+c_1 x_0+c_2\dot x_0\\
(s^2-as+b)Y(s)=cX(s)+c_3y_0+c_4\dot y_0
}
$$

and solving for $X(s)$ we have

$$
X(s)=-\frac{(s (s-a)+b) (c_1 x_0+c_2 \dot x_0)+c (c_3 y_0+c_4 \dot y_0)}{a^2 s^2-\left(b+s^2\right)^2+c^2}
$$

now focusing on the denominator, the condition to have only real roots is

$$
a^2-2 b\pm\sqrt{a^4-4 a^2 b+4 c^2}\ge 0
$$


问:
Non linear functions where the kernel is a subspace
If $f:V\rightarrow V$ is linear function where $V$ is vectorial space over a field $F$, it is true that the $ker(f)$ is a subspace of $V$. My question is, is a there function where the kernel is a subspace but the function is not linear?

The $\text{sin}:\mathbb{R}\rightarrow\mathbb{R}$ almost satisfies it, but unfortunately it fails on $\lambda\in F$ and $v\in\ker(\text{sin})$ then $\lambda v\in\ker(\text{sin})$ unless $F=\mathbb{Z}$.

答: Try $x \mapsto x^2 : \mathbb{R} \to \mathbb{R}$. The kernel of this function is $\{0\}$, which is a subspace. But clearly the function is not linear.


问:
Most efficient way to resolve generating equations using pen and paper
I want to learn if there is any easier way to resolve generating equations.

Here is how I do it :

E.g. I have a recurrence $T_n-T_{n-1}-T_{n-3}=1.$

I calculated the generating function as $A(x)=\frac{1}{(1-x)(1-x-x^3)}$.

Now I need a formula from here, so that I can put a value of $n$ in that formula and the value of $T_n$.

For this, I assume, $A(x)=\frac{a_0}{1-x}+\frac{a_1}{b_1-x}+\frac{a_2}{b_2-x}+\frac{a_3}{b_3-x}$

Now, I equate $\frac{1}{(1-x)(1-x-x^3)} = \frac{a_0}{1-x}+\frac{a_1}{b_1-x}+\frac{a_2}{b_2-x}+\frac{a_3}{b_3-x}$

$(1-x)(b_1-x)(b_2-x)(b_3-x)=(1-x)(1-x-x^3)(a_0(b_1-x)(b_2-x)(b_3-x)+a_1(....)+a_2(....))$

Then break all those terms, and equate equal powers of x, then get 5 complicated equations from there, and do a lengthy calculation to get values of a and bs. Put them in the formula.

If there is a way to resolve generating equations in an easier way, please suggest, with details.

答:
Note that finding $a$’s and $b$’s in

$$\frac{1}{(1-x)(1-x-x^3)} = \frac{a_0}{1-x}+\frac{a_1}{b_1-x}+\frac{a_2}{b_2-x}+\frac{a_3}{b_3-x}$$

is equivalent to performing fractional decomposition. Recognize that

$$1-x-x^3=(b_1-x)(b_2-x)(b_3-x)$$

Then, $b$’s are obtained first by solving the cubic equation analytically. (Two roots are complex numbers, though) Afterwards,
$a$’s are evaluated with

$$a_0=\frac{1}{(b_1-1)(b_2-1)(b_3-1)}$$

$$a_k=\lim_{x\to b_k} \frac{b_k-x}{(1-x)(b_1-x)(b_2-x)(b_3-x)},\>\>\>k=1,2,3$$


问:
subring not graded
Definition 1. A pair $(A, \{A_{i}\}_{i\in\mathbb{Z}})$ is a graded ring if $A$ is a commutative ring with unit, and $\{A_{i}\}_{i\in\mathbb{Z}}$ is a family of $\mathbb{Z}$-submodules of $A$, such that

(1) $A = \bigoplus_{i}A_{i}$ as $\mathbb{Z}$-module.

(2) $\forall i$, $j$, $A_{i}A_{j}\subseteq A_{i + j}$.

Definition 2. Let $A = \bigoplus_{i}A_{i}$ be a graded ring. A graded ring $B = \bigoplus_{i}B_{i}$ is a graded subring of $A$, if $B$ is a subset of $A$, and the inclusion satisfies $\forall i$, $B_{i}\subseteq A_{i}$.

Remark. The definition of graded subring above is equivalent to that, a subring $B$ of $A$ is graded if and only if $B = \bigoplus_{i}(B\cap A_{i})$ as $\mathbb{Z}$-module. It seems that the only difference between subrings and graded subrings is whether $B\subseteq \bigoplus_{i}(B\cap A_{i})$ or not, i.e. closed or not in homogeneous decomposition.

Question. Is there an example of a subring of a graded ring that is not graded?

答:
Take $A = k[x,y]$ and $B = k[x]$, with $A_0 = k$, $y \in A_1$ and $x+y \in A_2$. In other words, we take the total degree grading on $k[z^2,y]$, and define the graded components of $A$ as the preimages of those of $k[z^2,y]$, under the $k$-isomorphism from $A$ to $k[z^2,y]$ defined by $x \mapsto z^2-y$ and $y \mapsto y$.

$x \in B$ can be written uniquely as a sum of homogeneous elements of different degrees, namely $x+y$ and $-y$, but these elements are not in $B$. So the grading of $A$ does not restrict to $B$. This is an example where $A = B[y]$ is the ring of polynomials over $B$. In particular, $B$ is integrally closed in $A$.

Title: subring not graded


问:
For each abelian group $G$ let $T (G)$ be the set of all elements of finite order show that $T (G / T (G)) = 0.$
For each abelian group $G$ let $T (G)$ be the set of all elements of finite order show that $T (G / T (G)) = 0.$

I already show that $T(G)$ is a subgroup of $G$

But I'm stuck here and I don't know what equals to $0$ means if empty set of identity.

Any hint on how to do it?

答:
Saying that an abelian group $A = 0$ is an abuse of notation saying that it is isomorphic to the zero group, or equivalently that $A = \{e\}$ consisting of only the identity element. Thus, to show that $T(G/T(G)) = 0$ really means to show that it contains only the identity element. Every group has an identity, so it is impossible for them to be empty. As elements of $G/T(G)$ are cosets, the identity element of this group is $T(G)$.

Now, let's take some $a \in T(G/T(G))$ and pick some $g \in G$ such that $a = gT(G)$. This means that $g T(G)$ has finite order, so $(g T(G))^n = T(G)$ for some $n \geq 1$. By definition of coset multiplication, this means that $g^n T(G) = T(G)$. As these two cosets are equal, we necessarily have $g^n \in T(G)$. Hence, there is some $m \geq 1$ such that $(g^n)^m = e$, where $e \in G$ is the identity element. But then $g^{mn} = e$, so $g$ itself had finite order, i.e. $g \in T(G)$. But if $g \in T(G)$ then $g T(G) = T(G)$, so $a$ is the identity element.

We started with an arbitrary $a \in T(G/T(G))$ and proved that $a = T(G)$ the identity element. Thus, the only element of $T(G/T(G))$ is the identity, so as defined above, $T(G/T(G)) = 0$.


问:
Do functions with vertical asymptotes necessarily have infinite derivatives for all degrees?
Suppose I have a function $f(x)$ where $f:\mathbb R^n\to\mathbb R$.

First take $n = 1$. Now suppose that there exists a point $\bar x$ where the $d$th derivative $f^d(\bar x)$ is defined and is finite for some $d \geq 1$. Does this necessarily imply that $f(\bar x)$ is defined and is finite, or is it possible that it is undefined or infinite?

(The example I am trying to construct is something like: $f(0) = 0$, $\lim_{x\to 1} f(x) = +\infty$. (So it is defined, it just isn't finite.) We can also assume that $f(x) \geq 0$ for $0 \leq x < 1$, to help at least with the $d=1$ case. Now $f^d(x)$ is defined and finite for all $x < 1$. But, is it necessary that $f^d(x)\overset{x\to 1}{\to} +\infty$ for all $d$ as well?)

Now suppose $n > 1$, and suppose $\mathcal C\subset \mathbb R^n$ is a closed connected set.
I know that for some $\bar x\in \mathcal C$, $f(\bar x) = 0$, and I also know that for all $x\in \mathcal C$,  for some $u$, $u^T\nabla f(x) u$ is finite. Does this necessarily mean that $f(x)$ is finite for all $x\in \mathcal C$?

This seems like a basic result, but I'm fearful of potential counterexamples.

Edit: the more that I think about it, the more I think that it is true in $n = 1$, but can't be true in $n > 1$.

For $n = 1$, it seems like if $f(x)$ is finite-valued for all $x \in [0,1)$, then the only way that $\lim_{x\to 1}f(x) = +\infty$ is if all its derivatives go to $+\infty$ at $x\to 1$.

For $n > 1$, the simple counterexample of
$$
\nabla^2 f(x) = \left[\begin{matrix} 1 & 0 \\ 0 & 1/(1-x_2) \end{matrix}\right]
$$
seems to satisfy the bill.

答:
No, if $d=1$ it is possible for $\lim_{x\to 1^-}f(x)= \infty,$
while $\lim_{x\to 1^-}f'(x)=\infty$ fails. Consider the example

$$f(x) = \frac{1}{1-x} + \sin(1/(1-x)).$$

Then $\lim_{x\to 1^-}f(x)= \infty.$ But

$$f'(x) = \frac{1}{(1-x)^2}(1+\cos(1/(1-x)).$$

Along the sequence $x_n= 1-1/[(2n+1)\pi]$ we have $f'(x_n)=0$ for all $n.$


问:
Limit of a quotient sequence, where the denominator isn't required to be non-zero
So I have been trying to prove the following question from my textbook:

Let $(a_n)_{n \geq 0}$ and $(b_n)_{n \geq 0}$ be two real convergent sequences, with $\lim a_n = a$ and $\lim b_n = b \neq 0$. Prove that $\lim \frac{a_n}{b_n} = \frac{a}{b}$. Why doesn't the requirement "$b_n \neq 0$ for all n" play a big role here?

Now I was able to prove that the limit of the quotient sequence is indeed the quotient of the limits. However, I just can't quite understand why the requirement "$b_n \neq 0$ for all n" doesn't have a big role here. Isn't that requirement necessary for the existence of the quotient sequence?

答: We don't need $b_n$ to be nonzero for all $n$, just all sufficiently large $n$. It's ok for $b_n$ to be zero when $n$ is small, but we only care about the limit when $n \to \infty$.


答: Strictly speaking, yes. However, as $b\ne 0$, we can at least be sure that only finitely many terms need to be removed from the sequence. And the convergence behaviour (including its possible limit) does not change when we modify the sequence in only finitely many terms. In that light, it doesn't even really matter  when some (but only finitely many!) terms are not even defined in the first place.

Title: Limit of a quotient sequence, where the denominator isn't required to be non-zero

Body:
So I have been trying to prove the following question from my textbook:

Let $(a_n)_{n \geq 0}$ and $(b_n)_{n \geq 0}$ be two real convergent sequences, with $\lim a_n = a$ and $\lim b_n = b \neq 0$. Prove that $\lim \frac{a_n}{b_n} = \frac{a}{b}$. Why doesn't the requirement "$b_n \neq 0$ for all n" play a big role here?

Now I was able to prove that the limit of the quotient sequence is indeed the quotient of the limits. However, I just can't quite understand why the requirement "$b_n \neq 0$ for all n" doesn't have a big role here. Isn't that requirement necessary for the existence of the quotient sequence?


问:
Queueing Theory: How to estimate steady-state queue length for single queue, N servers?
I have a real-life situation that can be solved using Queueing Theory.

This should be easy for someone in the field.  Any pointers would be appreciated.

Scenario:

There is a single Queue and N Servers.

When a server becomes free, the Task at the front of the queue gets serviced.

The mean service time is T seconds.

The mean inter-Task arrival time is K * T (where K is a fraction < 1)

(assume Poisson or Gaussian distributions, whichever is easier to analyze.)

Question:

At steady state, what is the length of the queue?  (in terms of N, K).

Related Question:

What is the expected delay for a Task to be completed?

Here is the real-life situation I am trying to model:

I have an Apache web server with 25 worker processes.

At steady-state there are 125 requests in the queue.

I want to have a theoretical basis to help me optimize resources
and understand quantitatively how adding more worker processes
affects the queue length and delay.

I know the single queue, single server, Poisson distribution is well analyzed.

I don't know the more general solution for N servers.

thanks in advance,

-- David Jones

dxjones@gmail.com

答:
You describe a model describe a model similar to the M/M/c queue. In this queueing model service times are exponentially distributed with parameter $\mu$ and inter-arrival times are exponentially distributed with parameter $\lambda$. $c=25$ in your situation.

The steady state distribution for such a queueing system ($\pi_k$ is the steady state probability of there being $k$ customers) is

$$\pi_k = \begin{cases}
  \pi_0\dfrac{(c\rho)^k}{k!},  & \mbox{if }0 < k < c \\[10pt]
  \pi_0\dfrac{\rho^k c^c}{c!}, & \mbox{if } c \le k.
\end{cases}$$

where

$$\pi_0 = \left[\sum_{k=0}^{c-1}\frac{(c\rho)^k}{k!} + \frac{(c\rho)^c}{c!}\frac{1}{1-\rho}\right]^{-1}.$$

The average response time in such a model is
$$\frac{\text{ C}(c,\lambda/\mu)}{c \mu - \lambda} + \frac{1}{\mu}$$

where $\text{C}(c,\lambda/\mu)$ is Erlang's C formula

$$\text{ C}(c,\lambda/\mu)=\frac{\left( \frac{(c\rho)^c}{c!}\right) \left( \frac{1}{1-\rho} \right)}{\sum_{k=0}^{c-1} \frac{(c\rho)^k}{k!} + \left( \frac{(c\rho)^c}{c!} \right) \left( \frac{1}{1-\rho} \right)}.$$

ContentLicense: CC BY-SA 2.5


问:
Order the functions below by increasing order
I am confused by a homework problem I have concerning asymptotic growth rates.

A subset of the problem is to order:

$n^a(0 < a < 1)$

and $ln(n)$

by increasing growth rates

It was proposed to me that as n approaches $infinity$, then $n^a$ also approaches $infinity$, but I argue that $a$ can grow closer and closer to zero, keeping n close to one.  As $a$ grows closer to 1 then n^a approaches infinity.

How do I order this function in regards to $ln(n)$ since the value of $n^a$ can be either above or below $ln(n)$ based on the value of $a$?

答:
The homework  question asks for  ranking  asymptotic growth rates, not ranking function themselves.

Derivatives of these two functions:

$\frac{dn^a}{dn}$=$a(n^{a-1})$, $\frac{dln(n)}{dn}$=$\frac{1}{n}$

set S = $a(x^{a-1})$ -$\frac{1}{x}$ =0, $x^a$=$\frac{1}{a}$,x= $a^{-\frac{1}{a}}$

When n > $a^{-\frac{1}{a}}$, $n^a$ grows faster, the order is ln(n),$n^a$;

When   n < $a^{-\frac{1}{a}}$, $n^a$ grows slower, the order is $n^a$,ln(n).


问:
Determining independence of two variables and their covariance
Let X$_1$,X$_2$ be two independent random variables on the same probability space (Ω,$\mathcal{F}$,$\mathbb{P}$) that take values 1 and 0 with probability equal to 1/2. Calculate the covariance of X$_1$ + X$_2$ and |X$_1$ - X$_2$|.
Are these two random variables independent? Justify your answer

I have written an explanation for the linearity properties of the covariance but wanted to check the calculation for the covariance and need some help determining whether the variables are independent.

For clarification, I believe the question is trying to ask whether the covariance of $X_1$ + $X_2$ is independent to the covariance of |$X_1$ - $X_2$|, rather than whether $X_1$ and $X_2$ are independent.

答: You are told that $X_1$ takes on values 0 and 1 with probability 1/2 each.  Nothing is said about $X_2$!  That is, for $X_2$, $P(0)= \frac{1}{2}$ and $P(1)= \frac{1}{2}$, irrespective of what $X_2$.  That is precisely what "independent" means!


问:
ODE theorem with Lipschitz condition, understanding the definition of the solution of a first-order PDE and other questions

We had a theorem given during the lecture, do you know what the name of this theorem is, and why it is true? It goes as follows:

If $\: \forall _{i=1,...,n} \:\: f_i=f_i\left(x,\:y_1,\:....,\:y_n\right)$ is continuous in the neighborhood of $\left(x_o,\:y^o_1,\:...,\:y^o_n\right)$ and (also in that neighborhood) fulfills the Lipschitz condition with regard to $y_1,\:...,\:y_n$, then in the neighborhood of $x_o$ there exists exactly one solution for $\begin{cases}
      y'_1 = f_1\left(x,\:y_1,\:....,\:y_n\right) \\ ... \\
      y'_n = f_n\left(x,\:y_1,\:....,\:y_n\right)
    \end{cases} $  with the conditions $\begin{cases}
      y_1\left(x_o\right)=y^o_1\\ ... \\
      y_n\left(x_o\right)=y^o_n
    \end{cases} $

If we have our first integrals $\Phi _1,\:...,\:\Phi _n$ (from the general solution), then why are those first integrals linearly independent if $\:\nabla \Phi _1,\:...,\:\:\nabla \Phi _n$ are linearly independent?

If we have $\Phi =\Phi \left(x,\:y_{1},\:...,\:y_n\right)$ as a first integral, then why should it be that: $$ \frac{\partial \Phi }{\partial x}+f_1\left(x,\:y_1,\:...,\:y_n\right)\frac{\partial \Phi \:}{\partial y_1}+...+f_n\left(x,\:y_1,\:...,\:y_n\right)\frac{\partial \Phi}{\partial y_n}=0 $$

That's more of an arithmetical question I believe, but why exactly: $$(1)  \:\:\:\:\:\:\:\:\:\:\:\:\:a_1\left(x\right)\frac{\partial u}{\partial x_1}+...+a_n\left(x\right)\frac{\partial u}{\partial x_n}=\:0\:\Leftrightarrow \:\frac{dx_1}{a_1\left(x\right)}=...=\frac{dx_n}{a_n\left(x\right)} $$ (And also why $a_1\left(x,u\right)\frac{\partial u}{\partial x_1}+...+a_n\left(x,u\right)\frac{\partial u}{\partial x_n}=\:b\left(x,u\right)\:\Leftrightarrow \:\frac{dx_1}{a_1}=...=\frac{dx_n}{a_n}=\frac{du}{b}$)

If $\:\: \begin{cases}
      \Psi _1\left(x_1,\:...,\:x_n\right)=C_1 \\ ... \\
      \Psi _{n-1}\left(x_1,\:...,\:x_n\right)=C_{n-1}
    \end{cases} $ is a system of first integrals for (1), our solution would be $u=\Phi \left(\Psi _1,\:...,\:\Psi _{n-1}\right)$ where $\Phi$ is an arbritary function. I understand how we are able to calculate those $\Psi_{i}$, but why can $\Phi$ be arbritary?

We also had a geometric interpretation of 1st order PDEs given. The conditions we define on a three-dimensional first-order PDE define a curve, but the curve cannot intersect with itself. Why?

I'm sorry that I'm asking you a lot of questions right now, but Office Hours only last 1 hour and are once a week, and as far as I know other students from group also plan to go this week

答:
Picard-Lindelöf in its most local (and most common) version. By local tradition the names Lipschitz and Cauchy might also be used.

The $n$ first integrals provide a coordinate system for the solution curves. That is, $\Phi_k(x,y(t))=C_k$, and in principle the system $\Phi(x,z)=C$ should have, locally, only $z=y(x)$ as solution. The gradient condition is now nothing other than establishing the assumption of the implicit function theorem. If these are satisfied one has indeed a complete system of first integrals. Any other function that is constant on the solutions can then be expressed as a function of the values $C$.

First integrals are constant along solutions. Now apply the chain rule for the differentiation of $C=Φ(x,y(x))$ by $x$, giving
$$
0=∂_xΦ(x,y(x))·1+∂_{y_1}Φ(x,y(x))y_1'(x)+...+∂_{y_n}Φ(x,y(x))y_n'(x)
$$
and insert the differential equation.

Now you are switching to the first order PDE?

Take any curve $x(t)$ and apply the composition $u(x(t))$. Now apply the chain rule while differentiating for $t$,
$$
\frac{du(x(t))}{dt}=∂_{x_1}u(x(t))\dot x_1(t)+...+∂_{x_n}u(x(t))\dot x_n(t)
$$
Then compare with the PDE. So if the ODE system $\dot x_k(t)=\lambda(t)a_k(x(t))$ is satisfied, then $\frac{du(x(t))}{dt}=0$, so that $u(x(t))$ is a constant. Writing the fraction equation chain is just another way to express this,
$$
\frac{dx_k}{a_k(x)}=\lambda(t)dt=\frac{dx_j}{a_j(x)},
$$
but removing the emphasis from the parametrization by $t$, or now removing the free scaling function $\lambda(t)$, as any other parametrization of the same curves will lead to the same conclusions.

The value of $u$ is constant along the characteristic curves, and these curves are uniquely addressed by the values of the first integrals, $C=Φ(x(t))\implies u(x(t))=const$. Note that here you only have $n-1$ first integrals as functions of $x$ without $t$, so $Φ^{-1}(C)$ is a curve or a collection of them. So $u$ is a function of these first integrals. What that function precisely is, is determined by the boundary condition.

The system of first order ODE has only one solution per initial condition. This precludes any double points, that is crossings with different tangents. But one could for example have circles as characteristic curves. I think this point only applies to linear PDE of exactly the given type, one could easily expand to
$$
a_1(x,u(x))∂_{x_1}u(x)+...+a_n(x,u(x))∂_{x_n}u(x)=b(x,u(x))
$$
without losing too much of the theory of characteristic curves.


问:
Convergence of Fourier series - strange graph in proof
I am reading a text that states the following related to convergence of Fourier series:

$$g_K(x) =
> \frac{1}{2\pi}\int_{x-2\pi}^x\frac{\sin\left[\left(K+\frac{1}{2}\right)t\right]}{\sin\left(\frac{1}{2}t\right)}f(x-t)\,dt$$

The kernel $\sin(K+\frac{1}{2})t/\sin(\frac{1}{2}t)$ of the previous
  integral is plotted for several values of $K$ in the following figure.

However, the caption of the corresponding figure says "a plot of $\sin[(K+\frac{1}{2})t]/[(K+\frac{1}{2})t]$", and indeed, the plots are of

$$\frac{\sin[(K+\frac{1}{2})t]}{(K+\frac{1}{2})t}$$

for several $K$.  Is the author then plotting a completely wrong function, or could I be missing something?  It seems too wrong to simply be a typo.  Thanks all.

答:
Indeed, the combination of this text

and this figure

is clearly a mistake. It's not uncommon for a student to be charged with making plots for a book, and Artem gave a plausible guess for how a mistake like this could happen.


问:
Does scalar multiplication of a matrix change its eigenvalues/singular values
If Av=λv then cAv=cλv. If we group the right side like (cλ)v, then a scalar multiplication does change the eigenvalue. Now for the singular value, we need to find the eigenvalues of $(cA)^{T}(cA) = c^{2}A^{T}A$ because c is just a scalar. Then the eigenvalues of cA are $c^{2}λ$ so the singular values are just $cλ$. Is that correct?

答: The singular values are the square roots of the eigenvalues of the covariance matrix. If you multiply a matrix by a constant then the eigenvalues are multiplied by a constant but you need an absolute value $|c|$ or in the complex case it'd be $\sqrt{c\cdot c^{*}}$


问:
Prove that $\mathbb{Q}$ belongs to the Borel $\sigma$-algebra of $\mathbb{R}$
Prove that the set of rational numbers $\mathbb{Q}$ belongs to the Borel $\sigma$-algebra of $\mathbb{R}$.

I'm unsure about how to prove this. I'm also unsure about the difference between a σ-algebra and a Borel σ-algebra.

答:
As mentioned in the coomments, a $\sigma$ algebra is a set of subsets containing the empty set and closed under countable unions and complements. The Borel $\sigma$ algebra is a specific one attached to $\mathbb R$. It is described abstractly as the smallest $\sigma$ algebra containing all the open subsets of $\mathbb R$. You can check that this makes sense by proving that the arbitrary intersection of $\sigma$- algebras must be a $\sigma$-algebra. Then define the Borel $\sigma$-algebra as the intersection of all $\sigma$-algebras which contain the open subsets of $\mathbb R$.

This is an elegant but somewhat intractable definition. Can we actually name any Borel sets (Borel sets are elements of the Borel $\sigma$ algebra)? Well all we know to start with is that any open set must be Borel. Thus, the complement of any open set must be Borel, i.e. closed sets are Borel. But as closed sets are Borel, any countable union of closed sets must be Borel. In particular, any singleton $\{a\}$ is closed, therefore Borel. Any countable set is a countable union of points, which are all Borel, so any countable set must therefore also be Borel. As $\mathbb Q$ is countable, it is therefore Borel.


问:
Using an expected value inequality to show convergence
Let (X$_n)_{n\in\mathbb{N}}$ be a sequence of random variables defined on a probability space $(\Omega,\mathcal{F},\mathbb{P})$.
Suppose that $\mathbb{E}(|X_n|^\frac{3}{2}) < 1/n^\frac{3}{2}$
for all $n \in\mathbb{N}$. Show that $X_n$ converges almost surely
to $0$ as $n \to \infty$.

I'm really confused about how to do this question. I can't see how we use the inequality to show convergence? Any help would be much appreciated.

答:
You can inspire yourself from the demonstration of the $L^4$ version of the strong law of large number. Here consider the series $\sum | X_n | ^{2/3}$, its sum is well defined as a random variable at values in $[0,+\infty]$, by the Beppo-Levi lemma one can swap the expectation and the series symbol :
$$ \mathbb{E} \left( \sum_{n=1}^{+\infty} | X_n | ^{3/2} \right) =\sum_{n=1}^{+\infty} \mathbb{E} \left(  | X_n | ^{3/2} \right) \leq \sum_{n=1}^{+\infty} \frac{1}{n^{3/2}}$$
where the last series is convergent. Thus, for almost every $\omega \in \Omega $, $\sum_{n=1}^{+\infty} | X_n(\omega) | ^{3/2} $ is real and $X_n (\omega)\longrightarrow 0$.


问:
Tandem queue - response time distribution
In tandem queue with two queuing system, each server has exp(mu0) and exp(mu1) service time distribution and arrival rate is poisson(lambda). Scheduling policy is FCFS.
What would be the response time distribution in the system?

Using laplace transform, I know how to calculate for single queue (M/M/1) but how to calculate for 2 queues in tandem network. What would be the conditional laplace transform of response time given n jobs n the system?

Thanks.

答:
It is known that such a model has a product-form solution. Indeed, the response time for a tandem queue was the subject of a paper by Edgar Reich in 1957 (open access).

Reich, Edgar. Waiting times when queues are in tandem. The Annals of Mathematical Statistics 28.3 (1957): 768-773.

Therefore, as the solution by bkd.online notes, you compute the response time distribution by summing the independent response times for each of the nodes.


问:
Project onto overcomplete basis
How can I project a vector onto an overcomplete basis? The reason for over-complete basis is to have multiple projections to choose from. I am not sure how I can calculate the projections though.</p>

答:
Introduce a matrix $W$ whose columns are the vectors in your overcomplete basis. To represent a vector $b$ as a linear combination of a small number of vectors in your overcomplete basis, you can set $x^\star = \arg \min_x \frac12 \| Wx - b \|_2^2 + \gamma \|x\|_1$.  Then $Wx^\star$ is a linear combination of your basis vectors that approximates $b$. The $\ell_1$-regularization term encourages $x^\star$ to be sparse (so most components of $x^\star$ are $0$). The parameter $\gamma > 0$ controls the sparsity of $x^\star$.

The optimization problem I wrote above is called a "Lasso" problem. Many algorithms are available to solve a Lasso problem. I think the proximal gradient method is the simplest one.

Title: Project onto overcomplete basis


问:
Minimal polynomial of $\alpha \in \mathbb{Q}(\sqrt[3]{d})$
Given some cubefree $d$, is there a general way to determine the minimal polynomial over $\mathbb{Q}$ of $\alpha=1+\sqrt[3]{d}+\sqrt[3]{d^2} \in \mathbb{Q}(\sqrt[3]{d})$? Or more generally, $\alpha=a+b\sqrt[3]{d}+c\sqrt[3]{d^2}, a,b,c \in \mathbb{Q}$?

答: Yes, it's the minimal polynomial of the $\mathbb Q$-linear transformation $x \mapsto \alpha x$. Use the basis $\{1,\sqrt[3]{d},\sqrt[3]{d^2}\}$.


问:
Examining recurrence relation for its limit and
I have got the following problem:

Examine why the following sequence is well-defined, determine whether it is a monotonous sequence and calculate its limit:

$$a_0 = 3, a_{n+1} = \sqrt{2a_n - 1}$$

Now just by looking at this sequence and checking its first few entries it seems obvious that it is strictly decreasing. I then proved this with induction. Now I'm trying to find the limit, it seems obvious that its $1$, but I have been struggling to prove this. Since I know the sequence is strictly decreasing I looked at the following:

$$a_n > a_{n+1}$$
$$a_n > \sqrt{2a_n - 1}$$
$$a_n^2 > 2a_n - 1$$
$$a_n^2 - 2a_n + 1 > 0$$
$$(a_n - 1)^2 > 0$$

From this we get that $a_n > 1$ for all n. Would this be a correct proof that the limit of the sequence is 1? And as for whether the sequence is well-defined, since all entries are bigger that $1$ the square root will never have negative input, so it would be well-defined. Would this also be correct?

答:
The other answer explain you the computation of the limit, but let me add little bit intuition to it. There are plenty of similar problems you can find here with different recurrence relations and almost all of them are actually examples of fixed-point iterations. This is a very useful computational/approximation tool in numerical analysis. In particular, in you problem, we have $x_{n+1}=g(x_n),\quad x_0=3$ with $g(x)=\sqrt{2x-1}.$

So, you start with (given) $x_0,$ feed it in to $g$ and compute $x_1.$ Then feed it again to $g$ and compute $x_2,$ and continue this process forever. For a "nice enough" function $g$ the sequence $x_0, x_1, x_2, \cdots$ converge a limit $L.$ The picture looks roughly like this:

Though $y=x$ is actually the tangent to $y=g(x)$ at $x=1,$ this attracting behavior still holds.

Since the sequence is decreasing and seems converges to $1,$ we can focus on the restricted function $g : [1, 3]\to [1,\sqrt{5}].$  Clearly $g$ is increasing and $g'(x)<1$ for all $x>1,$ which is "nice enough" for the existence of a fixed point.


问:
definite integral $\int_{-1}^1x^2\sqrt{4-x^2}dx$
I have solved the problem and obtained the value of the integral $2\pi/3-\sqrt{3}/2$.

In the actual solution of this problem I have seen that $x$ was substituted by $2\sin z$ and upper limit $1$ was substituted by $\pi/6$ and lower limit $-1$ by $-\pi/6$.

Now the question is why was the upper limit substituted by $\pi/6$ instead of $5\pi/6$ or anything equivalent to $\arcsin\tfrac12$?

答:
Actually,$$\sin(\pi-y)=\sin y\implies\sin\tfrac{5\pi}{6}=\sin\tfrac{\pi}{6}\ne-\sin\tfrac{\pi}{6}.$$We need a $z$ satisfying $\sin z=-\sin\tfrac{\pi}{6}$. Since the sine function is odd, $z=-\tfrac{\pi}{6}$ will do.

Alternative argument: since the original integrand is even,$$\begin{align}\int_{-1}^1x^2\sqrt{4-x^2}dx&=2\int_0^1x^2\sqrt{4-x^2}dx\\&=2\int_0^{\pi/6}16\sin^2z\cos^2zdz\\&=\int_{-\pi/6}^{\pi/6}16\sin^2z\cos^2zdz,\end{align}$$as the new integrand is even too.


问:
Definition of Lie Algebra
I have just start studying Lie Algebra. I want to know the motivation for the conditions on the Lie Bracket in the Definition of Lie Algebra. Please explain me or tell me some references. Thanks. </p>

答:
Lie algebras arise naturally in multiple contexts. Suppose you have an associative algebra $A$ which is not necessarily commutative. Then a natural thing to do is study how far it deviates from commutativity. That is, it is natural to consider the bracket $[X,Y]=XY-YX$ which measures the failure of commutativity. But now that we've defined this object, we might wonder what properties it satisfies. You can verify through direct computation that this bracket satisfies the Jacobi identity $[X,[Y,Z]]+[Z,[X,Y]]+[Y,[Z,X]]=0$ and also that $[X,X]=0$. So any associative algebra can be made into a Lie algebra.

Another reason the Jacobi identity is natural is that it is an example of an operation satisfying the product rule familiar from differentiation (the Leibniz rule.) You know that the derivative satisfies $D[fg]=(Df)g+f(Dg)$ and any operator $D$ on an algebra satisfying this rule is said to be a derivation. Now one way to write the Jacobi identity is: $[X,[Y,Z]=[[X,Y],Z]+[Y,[X,Z]]$. That is, the function $D_X$ defined by $D_X(Y)=[X,Y]$ satisfies the Leibniz rule when applied to $[Y,Z]$: $D_X[Y,Z]=[D_XY,Z]+[X,D_YZ]$.

Title: Definition of Lie Algebra


答:
It must be $2 \sin{z} = \pm1 \Rightarrow z = \arcsin{\pm\frac{1}{2}} = \pm\frac{\pi}{6}$
That's all.

Title: definite integral $\int_{-1}^1x^2\sqrt{4-x^2}dx$

Body:
I have solved the problem and obtained the value of the integral $2\pi/3-\sqrt{3}/2$.

In the actual solution of this problem I have seen that $x$ was substituted by $2\sin z$ and upper limit $1$ was substituted by $\pi/6$ and lower limit $-1$ by $-\pi/6$.

Now the question is why was the upper limit substituted by $\pi/6$ instead of $5\pi/6$ or anything equivalent to $\arcsin\tfrac12$?


问:
Continuity at point $x_0$
Let $h(x) = \begin{cases}
      f(x), & x \in \mathbb{Q} \\
      g(x), & x \not \in \mathbb{Q}
\end{cases}$

where $\mathbb{Q}$ is the set of rationals.

Show that, if $f$ and $g$ are continuous functions at $x_0$ and $f(x_0) =g(x_0)$, then h is continuous at $x_0$.

My thoughts are if we can prove $|h(x) - h(x_0) | < \epsilon$ whenever $x \in \mathbb{R}$ and $|x-x_0| < \delta$ we can do so.

I'm just stuck on proving this, if anyone could help I'd appreciate it thank you!

答:
You are given an arbitrary $\varepsilon>0$ and want to find a $\delta>0$ that works for $h$ with that $\varepsilon$.

Continuity of $f$ gives you a $\delta_f>0$ that works for $f$ with that $\varepsilon$.

Continuity of $g$ gives you a $\delta_g>0$ that works for $g$ with that $\varepsilon$.

Can you think of a way to combine $\delta_f$ and $\delta_g$ to find the $\delta$ you are looking for? It can help to reflect on what the $\delta$s  really are: more or less strict restrictions on the input values of $x$ you care about.

Title: Continuity at point $x_0$


答: When you think about the image of the interval $x \in [-1,1]$ under the mapping $$x = 2 \sin z,$$ we don't just want to find values of $z$ corresponding to the endpoints of the interval.  We need to consider how the mapping affects the entire interval.  If you plot $2 \sin z$, you can see that there are infinitely many possible ways that the interval $x \in [-1,1]$ can be mapped to some $z$-interval; for instance, you could choose $z \in [5 \pi/6, 7\pi/6]$, but this also reverses the order of integration; or you could choose $z \in [11 \pi/6, 13 \pi/6]$, etc.  But you cannot choose, for instance, $z \in [-\pi/6, 5\pi/6]$, because this interval contains points that are not mapped to $x \in [-1,1]$, e.g., $z = \pi/2$ maps to $x = 2$.


答:
Let $\epsilon>0$ be arbitrary. Then since $f$ is continuous, there is a $\delta_1>0$ such that for each $x\ne x_0$ with $|x-x_0|<\delta_1$, we have $|f(x)-h(x_0)|<\epsilon$.

Similarly, since $g$ is continuous, so there is a $\delta_2>0$ such that for each $x\ne x_0$ with $|x-x_0|<\delta_2$ we have $|g(x)-h(x_0)|<\epsilon$.

Now, let $\delta=\min(\delta_1,\delta_2)$. For any $x\ne x_0$ with $|x-x_0|<\delta$, if $x\in\mathbb Q$ we have $|x-x_0|<\delta\le\delta_1$, so $|h(x)-h(x_0)|<\epsilon$. A similar argument works when $x\notin\mathbb Q$, so that again, $|h(x)-h(x_0)|<\epsilon$.

Title: Continuity at point $x_0$

Body:
Let $h(x) = \begin{cases}
      f(x), & x \in \mathbb{Q} \\
      g(x), & x \not \in \mathbb{Q}
\end{cases}$

where $\mathbb{Q}$ is the set of rationals.

Show that, if $f$ and $g$ are continuous functions at $x_0$ and $f(x_0) =g(x_0)$, then h is continuous at $x_0$.

My thoughts are if we can prove $|h(x) - h(x_0) | < \epsilon$ whenever $x \in \mathbb{R}$ and $|x-x_0| < \delta$ we can do so.

I'm just stuck on proving this, if anyone could help I'd appreciate it thank you!


问:
Plot of the algebraic variety of the ideal $I=(x^2+y^2-1)$ of the ring $\mathbb{R}[x,y]$
So, in the class of commutative algebra that I attend, we have defined the following:

If $k$ is a field and $I$ is an ideal of the polynomial ring $k[x_1, \ldots , x_n]$, we define the set $V(I)$ as

$V(I)= \{P=(a_1, \ldots,a_n) \in k^n \mid f(P)=0 \ \ \  \forall \ f(x_1,...,x_n) \in I \} $

So, if $k=\mathbb{R}$ and $n=2$, I need to plot the set $V(I)$, for $I=(x^2+y^2-1)$ the ideal generated by the polynomial $x^2+y^2-1$ of $\mathbb{R}[x,y]$.

I know that $f(x,y) \in I \implies \exists \ \ g(x,y) \in \mathbb{R}[x,y] : f(x,y) = g(x,y)(x^2+y^2-1)$, so, if $(a_1,a_2) \in V(I)$, then $g(a_1,a_2)(a_{1}^2+a_{2}^2-1)=0$. If $g(a_1,a_2) \neq 0 $, it's a circle, but i have no clue what happens when $g(a_1,a_2)=0$.

答: $P \in V(I) \iff f(P) = 0$ for all $f \in I$. Given $P = (a_1,a_2) \in V(I)$, since $x^2 + y^2 - 1 \in I$, then in particular we must have $a_1^2 + a_2^2 - 1 =0$. Letting $C$ be the unit circle, this shows that $V(I) \subseteq C$. Now you still have to show the reverse inclusion; for that you can use your observation about what elements of $I$ look like.


问:
Equivalent definition of primitive Dirichlet character
This is likely to be a stupid question, but I can't tell which part is going wrong.

Let $\chi:(\mathbb{Z}/N)^\times\to S^1$ be a Dirichlet character mod $N$. I found the definition is that $\chi$ is primitive if it is not induced by another character $(\mathbb{Z}/M)^\times\to S^1$ with  $M|N$ and $M<N$. I also see another definition says $\chi$ is primitive if the lifted arithmetic function (sending $x$ to $0$ if $(x,N)\neq 1$) has conductor $N$.

Now let's consider $\chi: (\mathbb{Z}/5)^\times\to S^1$, $2\mapsto i,3\mapsto -i,4\mapsto -1$. It induces a Dirichlet character $\chi':(\mathbb{Z}/10)^\times\to S^1$, $\chi'(3)=-i,\chi'(7)=i,\chi'(9)=-1$. But isn't this implies the minimal natural number such that $\chi'(1+q)=\chi'(1)$ is $10$, that is $\chi'$ has conductor $10$, hence being primitive? What is wrong with my steps?

Any help is much appreciated.

答:
$\chi'$ is $N$-periodic (here $N=10$).

Its conductor is the minimal $q$, which divides $N$, such that $\chi'(n)= f(n)$ for all $\gcd(n,N)=1$, with $f$ a $q$-periodic function.

ie. $\gcd(a,N)=\gcd(b,N)=1$ and $q|a-b$ implies $\chi'(a)=\chi'(b)$.

It is really a rephrasing of your first definition!

We can set $f(n)=0$ for $\gcd(n,q)=1$ in which case $f$ is a Dirichlet character: the unique primitive character underlying $\chi'$.

Here it is $f=\chi$, $q=5$.

A character is non-primitive iff it is of the form $1_{\gcd(n,k)=1} \psi(n)$ with $\psi$ a character $\bmod m$ coprime with $k$. A character $\bmod p^2$ can be primitive with conductor $p$.


问:
Divisibility by $12$ of $\overline{abcdef}$
How can one find all six digits numbers formed only with even number that are divisible by $12$?

My attempt was to find all six digits numbers $\overline{abcdef}$ with all digits from $\{0,2,4,6,8\}$ that are divisible with $3$ or $4$. For doing this I used inclusion-exclusion principle. How can I continue?

答:
A trick is to reduce the restriction "must be divisible by $12$" to a restriction on as few digits as possible and let the other digits "run free".

Must be divisible by $12$ means must be divisible by $3$ and $4$.

Being divisible by $3$ means $a+b+c+d+e+f \equiv 0 \pmod 3$.  And that means we can take any one of the digits, say $f$, and note $f \equiv-(a+b+c+d+e)\pmod 3$.  So $a,b,c,d,e$ can "run free" and we have reduced the restriction "is divisible by $3$" to a restriction on only one digit.

Being divisible by $4$ means that as $\overline{abcdef} = \overline{abcd}\times 100 + \overline {ef}$ that $\overline {ef} =10e + f$ be divisible by $4$.  But if $e$ is already being restricted to being even then $10e +f = 20\cdot \frac e2 + f$ and this reduces to a single restriction on $f$; that $f$ is divisible by $4$.  Or that $f = 0,4,8$.

So the restriction that $a,b,c,d,e,f$ are all even and that $\overline{abcdef}$ is divisible by $12$ reduces to a single restriction on $f$ (assuming $a,b,c,d,e,$ are all even); that  $f=0,4,8$ and that $f\equiv -(a+b+c+d+e)$.

But $0\equiv 0\pmod 3$ and $4\equiv 1\pmod 3$ and $8\equiv 2\pmod 3$ so if $a,b,c,d,e$ "run free" then our single restriction on $f$ can be expressed precisely as: $f=\begin{cases}0&a+b+c+d+e \equiv 0 \pmod 3\\4&a+b+c+d+e\equiv -1\equiv 2\pmod 3\\8&a+b+c+d+e\equiv -2 \equiv 1 \pmod 3\end{cases}$

So our restrictions are:

$\overline{abcdef}$ is six digits.  So $a \ne 0$ and $a$ is even.  There are $4$ options.

$b,c,d,e$ are even.  There are $5$ options each or $5^4$ total.

$f=\begin{cases}0&a+b+c+d+e \equiv 0 \pmod 3\\4&a+b+c+d+e\equiv -1\equiv 2\pmod 3\\8&a+b+c+d+e\equiv -2 \equiv 1 \pmod 3\end{cases}$. There is $1$ option.

So there are $4\cdot 5^4 = 2500$ such numbers.


问:
Alternative proofs that $A_5$ is simple
What different ways are there to prove that the group $A_5$ is simple?

I've collected these so far:

By directly working with the cycles: page 483 of http://www.math.uiowa.edu/~goodman/algebrabook.dir/algebrabook.html

Because it has order 60 and two distinct 5-Sylow groups: https://crypto.stanford.edu/pbc/notes/group/sylow.xhtml

There's no way to sum up the conjugacy classes 1, 15, 20, 12, 12, to get a normal subgroup.

By Iwasawa's lemma on $PSL_2(5) \simeq A_5$.

It's not solvable since it's perfect, and every group of order $<60$ is solvable, so it cant have normal subgroups.

答:
Viewing $A_5$ as the rotational symmetries of an icosahedron, one can easily verify that each conjugacy class generates the group. There are really only three cases, as while there are two conjugacy classes of five cycles, they are evidently power equivalent, so we lump them together. It then suffices to see that any vertex can be carried to any other by rotations through vertex axes, any face can be taken to any other by rotations through face axes, and any edge can be taken to any other by rotations through edge axes (as each respective case clearly has the relevant stabilizers, just transitivity of the action establishes that everything is generated.

But each of those cases is very straightforward: once you're familiar with it, there's no need for even a model!

Of course, you may want to prove that this group of rotations is $A_5$, but using this established simplicity and the action on the compound of five cubes (which I think is beautiful enough to warrant thinking about), that can be established without issue.

Title: Alternative proofs that $A_5$ is simple


问:
Determining partial derivatives from the set of implicit equations.
I have a function
$$\vec{y} = F(\vec{p},\vec{x})$$
where $\vec{x}$ and $\vec{y}$ $\in \mathbb{R}^6$ and $\vec{p} \in \mathbb{R}^3$.

Derivatives of $F$ are not known but there are relationships inside $\vec{y}$:

$$p_1 y_1 = y_2 y_4$$
$$p_2 y_1 y_3 = y_5$$
$$p_3 y_1 y_4 = y_6$$

And I want to determine all the partial derivatives $\frac{\displaystyle\partial{y_i}}{\displaystyle\partial{p_j}}$.

My attempt:

For y's, which are in direct relationship with $p_j$ i rewrite the condition to explicit form, eg. $$y_2 = \frac{\displaystyle p_1 y_1}{\displaystyle y_4}$$
thus
$$\frac{\displaystyle\partial{y_2}}{\displaystyle\partial{p_1}} = \frac{\displaystyle y_1}{\displaystyle y_4}$$

and when there is no direct relationship, i combine two or more conditions:
$$y_1 = \frac{\displaystyle y_6}{\displaystyle p_3 y_4}$$

$$\frac{\displaystyle\partial{y_2}}{\displaystyle\partial{p_3}} = \frac{\displaystyle\partial}{\displaystyle\partial{p_3}}(\frac{\displaystyle p_1 y_1}{\displaystyle y_4}) = \frac{\displaystyle\partial}{\displaystyle\partial{p_3}}(\frac{\displaystyle p_1 y_6}{\displaystyle p_3 y_4^2}) = -\frac{\displaystyle p_1 y_6}{\displaystyle p_3^2 y_4^2}$$

Is this correct? I don't feel sure with this.
Please, correct me if there is any mistake.

答:
I guess it is more convinient to use total differentiation such as
$$p_1 y_1 - y_2 y_4=0 \Rightarrow dp_1y_1+p_1dy_1-dy_2y_4-y_2dy_4=0$$
Taking a partial derivative means that you have to take all other variables as fixed. In above example if you fix $y_2$ and $y_4$ it follows that $dy_2=0$ and $dy_4=0$
$$\Rightarrow dp_1y_1+p_1dy_1-0\,y_4-y_2\,0=0\Rightarrow \frac{\partial y_1}{\partial p_1}=-\frac{y_1}{p_1}$$
$$\text{by fixing $y_1$ and $y_4$}\Rightarrow \frac{\partial y_2}{\partial p_1}=\frac{y_1}{y_4}$$
$$\text{by fixing $y_1$ and $y_2$}\Rightarrow \frac{\partial y_4}{\partial p_1}=\frac{y_1}{y_2}$$
And by repeating this steps in all relations you can eliminate (or find explicit solutions for) all the partial derivatives wrt $p_1$, $p_2$ and $p_3$.</p>


问:
How do I find the minimum of a quartic function with a missing coefficient?
$$ f(x) = 7x^4 + kx^3 + 8 $$
I have a question asking me to state the range of this function. To do this it seems I have to find the relationship between k and the minimum of the function. I've been told that there is a way to do this with calculus, but this is an algebra class and I haven't learned any calculus yet. I've tried coming up with random factors and then using the remainder theorem to solve for k, but that just gives me an identity.

How do I find the range of this function without knowing k?
Alternatively, is it possible to find k using the given information?

答:
$$ f(x) = 7x^4 + kx^3 + 8 $$ has a stationary point $(\alpha,f(\alpha))$ whenever for some real $v,$ $$ g(x) = 7x^4 + kx^3 + 8 + v $$ has a repeated zero $\alpha,$ i.e., $$g(x)=7(x-\alpha)^2(x^2+\beta x+\gamma)\\=7x^4+(7\beta-14\alpha)x^3+(7\alpha-14\alpha\beta+7\alpha^2)x^2+(-14\alpha\gamma+7\alpha^2\beta)x+7\alpha^2\gamma.\\$$

Comparing coefficients gives $$\alpha=0$$ or $$\gamma=\frac12\alpha\beta\\\beta=\frac23\alpha\\k=7(\frac23\alpha)-14\alpha\\\alpha=-\frac{3}{28}k.\\$$

Now, $f(x)=x^3(7x+k)+8,\ $ so $f(x)\rightarrow\infty$ as $x\rightarrow\pm\infty.$

Therefore,

if $k=0,$ $f(x)$ has one stationary point, the minimum point $(0,8),$ so $$f(x)\in[8,\infty);$$

otherwise, $f(x)$ has two stationary points, the inflection point $(0,8)$ and the minimum point $(-\frac{3}{28}k,8-\frac{27}{87808}k^4),$ so $$f(x)\in[8-\frac{27}{87808}k^4,\infty).$$


问:
Stochastic processes, indistinguishability
This may be a fundamental question on a theory of stochastic processes.

Let $X=\{X_t\}_{t \ge 0}$ be an $\mathbb{R}^d$-valued stochastic process on a (sufficiently nice) filtered probability space (we denote by $P$ the probability measure). We assume that $X$ is right-continuous and has finite left limits. We moreover assume that for any bounded continuous function $u\colon \mathbb{R}^d \to \mathbb{R}$ and any $t>0$
\begin{align}
P\left[ u(X_t)=u(X_{t-}) \right]=1.
\end{align}
Here, $u(X_{t-})=\lim_{s \nearrow t}u(X_s)(=u(\lim_{s \nearrow t}X_s)$).

My question.

Can we show that $P\left[X_t=X_{t-}\text{ for any }t\ge0 \right]=1$?

We note that $\{u(X_t)\}_{t \ge 0}$ is right continuous and $\{u(X_{t-})\}_{t \ge 0}$ is left continuous.

答:
Take $(\Omega,\mathcal{F},\mathsf{P})=([0,1],\mathcal{B}_{[0,1]},\lambda)$, where $\lambda$ is the Lebesgue measure and let $X_t(\omega):=1\{t\ge \omega\}$, $t\in [0,1]$. Then
$$
\mathsf{P}(X_t\ne X_{t-})=\lambda(\{t\})=0
$$
for any $t\in [0,1]$. However,
$$
\mathsf{P}(X_t \text{ is continuous on } [0,1])=0.
$$


问:
Functions satisfying $f^2+g^2=1$
I am asked to prove that if $f,g$ are differentiable functions and $f^2(x)+g^2(x)=1$ $\forall x$, then there exists a differentiable function $\theta$ so that $f(x)=\cos(\theta(x))$ and $g(x)=\sin(\theta(x))$, $\forall x$.

The proof can't use complex numbers. I'm suggested to use $\theta' = fg'-gf'$, but I haven't been able to make any considerable progress.

答:
Pick $\theta_0$ such that $f(0) = \cos(\theta_0)$ and $g(0) = \sin(\theta_0)$.

If you don't assume that $f$ and $g$ are continuously differentiable, things get a little bit hairy and require a high-level result. I will provide this proof now. Define $S = \{(x, y) \in \mathbb{R}^2 : x^2 + y^2 = 1\}$ to be the unit circle. Then since $f$ and $g$ are continuous, we have a continuous map $h : \mathbb{R} \to S$ defined by $h(x) = (f(x), g(x))$.

Now consider the map $\pi : \mathbb{R} \to S$ defined by $\pi(x) = (\cos(x), \sin(x))$. It is a well-known fact that this is a covering space.

Because $\pi : \mathbb{R} \to S$ is a covering, $\pi(\theta_0) = h(0)$, and $h$ is continuous, there exists a unique continuous map $\theta : \mathbb{R} \to \mathbb{R}$ such that $h = \pi \circ \theta$ and $\theta_0 = \theta(0)$. This follows easily from the path lifting property of covering spaces. This is therefore a $\theta$ such that $\cos \circ \theta = f$ and $\sin \circ \theta = g$.

It now remains to show that $\theta$ is a differentiable function. Consider some $x \in \mathbb{R}$. Since $\pi$ is a local diffeomorphism, we can take some neighbourhood $U$ of $\theta(x)$ such that $\pi|_U : U \to \pi(U)$ is a diffeomorphism. Let $\tau : \pi(U) \to U$ be the inverse diffeomorphism of $\pi|_U$. Let $V = \theta^{-1}(U)$; then $V$ is an open set, and $x \in V$.

We see that $h|_V = \pi|_U \circ \theta|_V$. Then $\tau \circ h|_V = \tau \circ \pi|_U \circ \theta|_V = \theta|_V$. Since $\tau$ and $h|_V$ are both differentiable, so is $\theta|_V$. That is, $\theta$ is differentiable in an an open neighbourhood of $x$. Then $\theta$ is differentiable at $x$.

Then $\theta$ is differentiable.

Title: Functions satisfying $f^2+g^2=1$


问:
combinatorics equality proof
if $k , n ,q $ are  positive integers and $n \geq k $. I want to prove that:

$ {\begin{bmatrix} n+1 \\ k \end{bmatrix}}_q =   {\begin{bmatrix} n \\ k \end{bmatrix}}_q + q^{n+1-k} {\begin{bmatrix} n \\ k-1 \end{bmatrix}}_q$ .

I think it may done buy using schur theorem.

答:
Jonathan Azose’s senior thesis [PDF] at Harvey Mudd College, Applications of the q-Binomial Coefficients to Counting Problems, using a combinatorial interpretation of the $q$-binomial coefficients, equivalent to the one in Wikipedia, gives a straightforward proof (p. $8$). It also has a great many other standard results.

If you want to try it yourself using the combinatorial interpretation in Wikipedia, consider separately the words of length $n+1$ with $k$ ones that begin with $0$ and the words that begin with $1$; these two subsets are ‘counted’ by the two terms on the righthand side of the identity.

Title: combinatorics equality proof


问:
Beta and Gamma functions relation proof.
I found a proof of the Beta-Gamma functions relation, but I don't understand what happened between 3rd and 2nd last line. There was s^z and in the next line there is s^(z-1). Where did it come from? Or is the proof incorrect?

Can anyone explain it, please?

See the proof here

$$\int_0^\infty s^z \ e^{-s}$$
becomes
$$\int_0^\infty s^{z-1} \ e^{-s}$$

答:
I believe you are asking about the line
$$
\begin{align}
&\phantom{=\ }\int_0^\infty s^ze^{-s}\int_0^\infty(us)^{w-1}e^{-su}\,\mathrm{d}u\,\mathrm{d}s\\
&=\int_0^\infty s^{z-1}e^{-s}\,\Gamma(w)\,\mathrm{d}s
\end{align}
$$
That is because, using the substitution $u\mapsto\frac ts$, we get
$$
\begin{align}
\int_0^\infty(us)^{w-1}e^{-su}\,\mathrm{d}u
&=\int_0^\infty t^{w-1}e^{-t}\frac1s\,\mathrm{d}t\\
&=\frac1s\,\Gamma(w)
\end{align}
$$


问:
Show that $A\cap B = \{0\} \;\;\;\text{and}\;\;\; A+B\;\text{closed} \implies\|a\| + \|b\| \leq \|a+b\|$
This is the problem I am working on, the actual question is in bold:

Let $(X,\|\cdot\|)$ be Banach and $A,B\subset X$ closed subspaces such that $A\cap B = \{0\}$. Consider, $$A+B:=\{a+b : a\in A, b\in B\} \quad \text{with norm} \quad \|\cdot\|_d: a+b \to \|a\|+\|b\|$$
as subspace of $X$.

Prove that $A+B$ is closed if and only if  $\|\cdot\|$ is equivalent to $\|\cdot\|_d$ on $A+B$.

My attempt, we know immediately that $A,B$ are Banach. Suppose $A+B$ is closed. Let $a+b\in A+B$. Then,
$$\|a+b\| \leq \|a\| + \|b\| = \|a+b\|_d$$

now my problem is, I don't know how to do this next part,
$$A\cap B = \{0\} \;\;\;\text{and}\;\;\; A+B\;\text{closed} \implies\|a\| + \|b\| \leq \|a+b\|$$
I am not sure how to proceed with this. Obviously $A$ and $B$ are linearly independent but we are working with infinite dimensions. Ideally I would like a proof that does not involve decomposition into basis vectors.

To prove the other direction, suppose the norms are equivalent on $A+B$, then, let $x\in X$ be a limit point of $A+B$, i.e. we have $(a_n+b_n)_n \subset A+B$ approaching $x$ with respect to $\|\cdot\|$. But $a_n\to a\in A$ and $b_n\to b\in B$. It remains to prove that $x=a+b$
$$\|x - (a+b)\| = \lim_{n \to \infty}\|(a_n+b_n) - (a + b)\| \leq
\lim_{n \to \infty}C\|a_n - a\|
+
\lim_{n \to \infty}C\|b_n - b\|
\to 0$$
by equivalence of norms.

答:
Here is a useful corollary to the open mapping theorem (see Rudin, "Functional Analysis", 2nd ed, Corollary 2.12 (c)) which (in this case) states that
if $\Lambda: W \to Z$ is continuous, linear, bijective then there exist $r_1,r_2>0$ such that $r_1 \| w \| \le \|\Lambda w\| \le r_2 \|w\|$. (Symbols changed to protect the innocent.)

In the following, let $\Lambda: A \times B \to A+B$ defined by $\Lambda((a,b)) = a+b$, with the norm $(a,b) \mapsto \|a\|+\|b\|$ on $A \times B$.

Suppose $A,B, A+B$ are closed, then we see that $A\times B$ and $A+B$ are Banach spaces and $\Lambda$ satisfies the relevant conditions. Hence there are $r_1,r_2$ such that
$r_1 \|a+b\| \le \|\Lambda ((a,b))\| = \|a\|+\|b\| \le r_2 \|a+b\|$.


问:
How to prove this Stirling related equation
Here is what I need to prove, but have no idea were to start. I know there is some connection with the Stirling theorem.

$$
\sum_{i=0}^{d}\binom{m}{i} \leq \left ( \frac{em}{d} \right )^{d}
$$

for $$ m \geq d$$

I tried to open the sum and start from the right side of the equation but I achieved nothing. I think there is also some connection to Sauer Shelah Lemma.

Thanks in advance.

答:
I think it is the Sauer Shelah Lemma isn't it?

There are proofs at the link below; one of which uses a "shifting technique" and the other induction on $m+d$.

http://www.cse.buffalo.edu/~hungngo/classes/2011/Fall-694/lectures/vcd.pdf

See references therein to trace the history.


问:
Property of a covariant derivative of a vector field along a curve
I am studying the covariant derivative of a vector field $X$ along a curve $\gamma$ ( $\nabla_{\dot \gamma}X:=\frac{\nabla X}{dt}$) and I have read the following property:
$$\nabla_{\dot \gamma}X=\Big(\frac{d \xi^k}{dt}+\Gamma_{ij}^k\frac{dx^i}{dt}\xi^j\Big)\frac{\partial}{\partial_k}$$
In order to prove this above I have thought that if $X=\xi^j \frac{\partial}{\partial x^j}$ and $\dot \gamma= \frac{dx^i}{dt} \frac{\partial}{\partial x^i}$, we have that:
since $\nabla X=\Big(\frac{\partial \xi^k}{\partial x^j}+\Gamma_{ji}^k\xi^j\Big)\frac{\partial}{\partial_k}$ we obtain:
$$\nabla_{\dot \gamma}X=\nabla_{\frac{dx^i}{dt}\frac{\partial}{\partial x^i}}(\xi^j \frac{\partial}{\partial x^j})=\frac{dx^i}{dt}\Big(\nabla_{\frac{\partial}{\partial x^k}}(\xi^j \frac{\partial}{\partial x^j})\Big)=\frac{dx^i}{dt}\Big(\xi^j\nabla_{\frac{\partial}{\partial x^i}}( \frac{\partial}{\partial x^j})\Big)+\frac{dx^i}{dt}\Big(\xi^j\frac{\partial}{\partial x^i}\Big)\frac{\partial}{\partial x^j}=\Big(\frac{dx^i}{dt}\xi^j\Gamma_{ij}^k\frac{\partial}{\partial x^k}+\color{red}{\frac{dx^i}{dt}(\frac{\partial}{\partial x^i})}\frac{\partial}{\partial x^j}\Big)=\Big(\frac{dx^i}{dt}\xi^j\Gamma_{ij}^k\frac{\partial}{\partial x^k}+\color{red}{\frac{d\xi^j}{dt}}\frac{\partial}{\partial x^j}\Big)=\Big(\frac{dx^i}{dt}\xi^j\Gamma_{ij}^k+{\frac{d\xi^k}{dt}}\Big)\frac{\partial}{\partial x^k}$$
where the last equality is obtained considering the index $j$ in the second term as $k$.

$\textbf{My question:}$ do you think that the passages that I have done are correct? In particular it is valid the equality that involves the red quantities (I am not sure that is mathematically rigorous)?

答:
Here is the correct computation (which you can find in any number of texts).
\begin{align*}
\nabla_{\dot\gamma}X &= \nabla_{\dot\gamma}\big(\xi^j\frac{\partial}{\partial x^j}\big) \\
&= \big(\nabla_{\dot\gamma}\xi^j\big)\frac{\partial}{\partial x^j} + \xi^j\nabla_{\dot\gamma}\frac{\partial}{\partial x^j} \\
&= \frac{d\xi^j}{dt}\frac{\partial}{\partial x^j} + \xi^j\nabla_{\frac{dx^i}{dt}\frac{\partial}{\partial x^i}} \frac{\partial}{\partial x^j} \\
&= \frac{dx^j}{dt}\frac{\partial}{\partial x^j} + \xi^j\frac{dx^i}{dt} \Gamma^k_{ij}\frac{\partial}{\partial x^k}.
\end{align*}
Can you finish now?

Note that because a priori $X$ is a vector field only along the curve (this can be fixed, but it's not worth the trouble), $\xi^j$ are functions only of $t$, not of $x^1,\dots,x^n$. So this is why we cannot change $\nabla_{\dot\gamma}$ immediately. Indeed, $d\xi^j/dt$ is typical shorthand for $d(\xi^j\circ\gamma)/dt$.


问:
Example of a rank-one operator to satisfy conditions about the spectrum
$U$ is defined to be a bilateral shift operator such that $Ue_n= u_ne_{n+1}$ for $n\in \mathbb{Z}$, where $u_n \ne 1$.

I am looking for an example of $U$ and a rank-one operator $T \in \mathcal{B}(\ell^2(\mathbb{Z}))$ so that $\sigma (U) = \{ z \in \mathbb{C} : |z| = 1 \}$ and $\sigma(U+T)=  \{ z \in \mathbb{C} : |z| \le 1 \}$.

Some thoughts:

I showed that for any bilateral shift $U$ with $Ue_n=u_ne_{n+1}$ for $n \in \mathbb{Z}$ such that $|u_n|=1$, we have $ \sigma (U) = \{ z \in \mathbb{C} : |z| = 1 \}$.

Also, I saw it from here The spectrum of the operators $\sigma(U+T) = \sigma(U) \cup \sigma(T)  $.

So I wonder if I could find some rank one operator to satisfy the condition based on that.

But other suggestions will also be great.

Thank you.

答:
Let $V$ be the standard bilateral shift on $\ell ^2(\mathbb Z)$,  namely the unitary operator $V$ such that $Ve_n=e_{n+1}$,  and let $T$ be the rank-one operator
given by
$$
  T(x)= -\langle x,e_0\rangle e_1, \quad\forall x\in  \ell ^2(\mathbb Z).
  $$
Then   $R:= V+T$ becomes the operator given by
$$
  R(e_n) = \left\{\matrix{
  e_{n+1}, & \text { if } n\neq 1, \cr
  0 , & \text { if } n=0.
  }\right.
  $$
One then sees that $R$ leaves invariant the decomposition
$$
  \ell ^2(\mathbb Z)=\ell ^2(\mathbb Z_-)\oplus \ell ^2(\mathbb Z_+^*),
  $$
where $\mathbb Z_-$ stands for the non-positive integers and $\mathbb Z_+^*$,  for the strictly positive ones.

The restriction of $R$ to $\ell ^2(\mathbb Z_+^*)$ is then  the so called unilateral shift $S$, while the restriction of $R$
to $\ell ^2(\mathbb Z_-)$
is seen to be unitarily equivalent to $S^*$.  We then conclude that $R$ is unitarily equivalent to $S\oplus S^*$, whose spectrum
is given by
$$
  \sigma (R) = \sigma (S) \cup  \sigma (S^*) = \sigma (S) \cup  \overline{ \sigma (S)}.
  $$
Since $\sigma (S)$ is the unit disk,  we see that   $\sigma (R)$ is the unit disk as well.

Now,  the OP has asked for a version of the bilateral shift where $Ue_n=u_ne_{n+1}$,  with $u_n\neq 1$,  so we may take
$U=-V$,  i.e.  take all of the $u_n=-1$,  the requested rank-one operator being $-T$.


问:
Prove: $T \in L(Y,\ell^\infty(\mathbb K))$ is bounded $\implies$ $\forall n \in \mathbb N:g_n \in L(Y,\mathbb K)$ is bounded
Assume $Y$ is a Banach space over $\mathbb{K}$ and $T : Y → \ell^{\infty}$ is a linear operator, such that $Ty = (g_1(y), g_2(y), g_3(y), . . .)$, where  $g_n \in L(Y, \mathbb{K})$.

I am trying to prove/disprove the following statements:

($T$ is bounded) $\rightarrow$ ($\forall n \in \mathbb{N}:g_n$ is bounded )

($\forall n \in \mathbb{N}:g_n$ is bounded ) $\rightarrow$  ($T$ is bounded)

If it is bounded then $\exists$M $\geq 0$ such that $\|Ty\|_{\infty} \leq$M$\|y\|_{\infty}$ $\forall$y$\in Y$.

Then. $\|Ty\| = \|g_n(y)\|$$ = \sup_{n\in\mathbb{N}}|g_n(y)| \forall n \geq1$, so we can easily take $M=1$

Hence, $\|Ty\|_{\infty} \leq$$\|y\|_{\infty}$ and so $\|T\| \leq 1$.

I am not convinced that I am on the right track. Any help would be grateful for both "arrows" in the question. Thanks in advance.

答:
($T$ is bounded) $\rightarrow$ ($\forall n \in \mathbb{N}:g_n$ is bounded )

Assume $T$ is bounded, i.e.
$$
\exists M \geq 0:\forall y\in Y: \|Ty\|_{\infty} \leq M\|y\|_{\infty}.
$$
Since $\|Ty\|_\infty  = \sup_{n\in\mathbb{N}}|g_n(y)| \leq M$, it follows $\exists M \geq 0:\forall y\in Y: \sup_{n\in\mathbb{N}}|g_n(y)| \leq M\|y\|_{\infty}$. But $\forall n \in \mathbb N: |g_n(y)| \le \sup_{n\in\mathbb{N}}|g_n(y)|$, that yields
\begin{align}
\exists M \geq 0:\forall n \in \mathbb N:\forall y\in Y: |g_n(y)| \leq M\|y\|_{\infty}
\end{align}
which is the same as saying each $g_n$ is bounded

Added:

($\forall n \in \mathbb{N}:g_n$ is bounded ) $\rightarrow$  ($T$ is bounded)

Since for every $y$ in $Y$ we have $Ty \in \ell^\infty$, then $\forall y \in Y: \|Ty\|_\infty = \sup_{n\in \mathbb N} |g_n(y)| < \infty$. Also since $Y$ is Banach and $(g_n)$ are bounded we have by Banach-Steinhaus theorem (unifrom boundedness principle) that $\sup_{n \in \mathbb N}\|g_n\| < \infty$. So
$$
\sup_{y:\|y\| =1} \|Ty\|_\infty = \sup_{y:\|y\| =1} (\sup_{n \in \mathbb N}|g_n(y)|) = \sup_{n \in \mathbb N} (\sup_{y:\|y\| =1}|g_n(y)|) = \sup_{n \in \mathbb N} \|g_n\| < \infty
$$
(since both suprema exist,) so it follows $T$ is bounded.


问:
finite dimensional group is locally a manifold or product with a cantor set
Is it the case that every compact topological group with finite topological dimension is locally homeomoprhic to a cantor set or $ \mathbb{R}^n $ or a product of a cantor set with  $ \mathbb{R}^n $?

This is a follow up to:

Is every locally compact group/ homogeneous space locally homeomorphic to the Cantor set or an n-torus?

答:
To complete the story in the case of general (Hausdorff) locally compact finite dimensional groups $G$. In section 4.9.3 of

Montgomery, Deane; Zippin, Leo, Topological transformation groups, Mineola, NY: Dover Publications (ISBN 978-0-486-82449-9). xi, 289 p. (2018). ZBL1418.57024.

it is proven that there exists a local (continuous) isomorphism of $G$ to the product of a compact totally disconnected group and a local Lie group.


问:
Find probabilities of getting the right area code of a subscriber-number-only phone contact
Situation: I have 100 contacts on my phonebook.  We can tell what area by looking at a number.  For example +61 8 1234 5678, this is an Australian phone number where

+61  country code

8  area code (WA for Western Australia)

1234 5678 subscriber number

Problem:
If I have a contact that contains 1234 5678 only, assuming that all my contacts are in Australia and more than half of the contacts are from WA, what are the chances that this number falls on WA?

Question: What branch in mathematics do I need to read? I'm guessing that this falls on probability and statistics? If my guess is right, what topic in probability and statistics do I need to focus?

I'm looking for the right branch/topic in math to read on.

答: Have a look at conditional probabilities. </p>


问:
A representation of the symmetric group on tensor space.
While answering this question on common eigenvectors  I wondered about the natural generalisation. Although I can deal with small cases I cannot find the general answer.

Let  $V$ be an $n$-dimensional vector space over $\mathbb{C}$, and let $W$ be the $k$-fold tensor power $W=V\otimes \dots\otimes V$.

We have an obvious representation $\Psi$ of the symmetric group $S_k$ on $W$ given by
$$
\Psi(\pi):v_1\otimes v_2\otimes\dots\otimes v_k \mapsto v_{1\pi}\otimes v_{2\pi}\otimes\dots\otimes v_{k\pi}
$$
where linear maps and permutations are being written on the right.

The original question was the case $k=3$ and asked about simultaneous eigenvectors and eigenvalues of $S_3$.

In the more general case  a natural question is  (i) what is the multiplicity of the trivial representation in  $\Psi$; (ii) what is the multiplicity of the sign representation in $\Psi$.

I believe that the answers can be expressed in terms of binomial coefficients; if so, is there an "easy"  combinatorial explanation? (I say "easy", because the representation theory of the symmetric group is all combinatorics anyway - I am wondering if there's an immediate way to see what the answers to the questions are.)

答:
Well as I suggested in the comments, Schur-Weyl duality gives a complete answer for how to decompose $V^{\otimes k}$ as an $S_k$ representation.

If you just want to understand the space of invariants, there is a purely combinatorial argument.

Let $N$ be an $n$ element set, and let's think of $V$ as $\mathbb{C} \langle N \rangle$ (the vector space of formal linear combinations of elements of $N$).  The tensor power $V^{\otimes k}$ is then isomorphic to $\mathbb{C} \langle N^k \rangle$, with the action of $S_k$ coming from the action on $N^k$ by permuting the coordinates.

Here is a general lemma about combinatorial representation theory:  If $G$ is a finite group acting on a finite set $X$, then the space of $G$-invariant vectors in $\mathbb{C}\langle X \rangle$ has a basis given by elements $\sum_{x\in O} x$ where $O$ is a G-orbit in $X$.  In particular the dimension of the space of invariants is equal to the number of orbits.

Applying this to our setting where $G = S_k$ and $X = N^k$, we want to count the $S_k$ orbits on $N^k$.   Two $k$-tuples in $N^k$ are in the same orbit iff for each $n \in N$ that $n$ appears the same number of times in the two $k$-tuples. Keeping track of the multiplicity of each $n \in N$, we can recognize this as a stars-and-bars type counting problem, which gives $\binom{n+k-1}{n-1} = \binom{n+k-1}{k}$ orbits -- and hence a that dimensional space of invariants.


问:
Is it more common to use $n \times m$ or $m \times n$ for matrices?
I've seen both versions for representing the number of rows by the number of columns of a matrix. I've always used $n \times m$, but I feel like I more often see $m \times n$. Is there some well known standard that people typically follow?

答: I always use $r\times c$ personally. It stops the confusion!


答:
This is a pretty straightforward application of the properties of the covariant derivative. Remember that the Christoffel symbols are defined as a measure of how the covariant derivative acts on the basis vectors:

$$\nabla_{j}\mathrm{e}_i=\Gamma^k_{ij}\mathrm{e}_k$$

And also that the covariant derivative is linear:

$$\nabla_\mathrm{u}\mathrm{v}=\nabla_{u^j\mathrm{e}_j}\mathrm{v}=u^j\nabla_{j}\mathrm{v}=u^j\nabla_j(v^i\mathrm{e}_i)$$

Now we use the other important property of the covariant derivative, the product rule

$$u^j\nabla_j(v^i\mathrm{e}_i)=u^j(\mathrm{e}_i\nabla_j(v^i)+v^i\nabla_j(\mathrm{e}_i))$$

Since covariant differentiation on a scalar is the same as partial differentiation, we get

$$=u^j(\mathrm{e}_i\partial_jv^i+v^i\Gamma^k_{ij}\mathrm{e}_k)$$
$$=\left(u^j\partial_jv^k+v^iu^j\Gamma^k_{ij}\right)\mathrm{e}_k$$

Replacing $\mathrm{u}$ with $\dot{\gamma}$ with components $\{\dot{x}^i\}$ and replacing $\mathrm{v}$ with $\mathrm{X}$ with components $\{\xi^j\}$, using $\frac{\partial}{\partial x^k}$ instead of the more linear algebra type notation $\mathrm{e}_k$ and renaming some indices will give your desired expression.

ADDITION:

It is indeed correct that
$$\frac{\mathrm{d}\xi^i}{\mathrm{d}t}=\frac{\partial \xi^i}{\partial x^j}\frac{\mathrm{d}x^j}{\mathrm{d}t}$$
See exact differential.

Title: Property of a covariant derivative of a vector field along a curve

Body:
I am studying the covariant derivative of a vector field $X$ along a curve $\gamma$ ( $\nabla_{\dot \gamma}X:=\frac{\nabla X}{dt}$) and I have read the following property:
$$\nabla_{\dot \gamma}X=\Big(\frac{d \xi^k}{dt}+\Gamma_{ij}^k\frac{dx^i}{dt}\xi^j\Big)\frac{\partial}{\partial_k}$$
In order to prove this above I have thought that if $X=\xi^j \frac{\partial}{\partial x^j}$ and $\dot \gamma= \frac{dx^i}{dt} \frac{\partial}{\partial x^i}$, we have that:
since $\nabla X=\Big(\frac{\partial \xi^k}{\partial x^j}+\Gamma_{ji}^k\xi^j\Big)\frac{\partial}{\partial_k}$ we obtain:
$$\nabla_{\dot \gamma}X=\nabla_{\frac{dx^i}{dt}\frac{\partial}{\partial x^i}}(\xi^j \frac{\partial}{\partial x^j})=\frac{dx^i}{dt}\Big(\nabla_{\frac{\partial}{\partial x^k}}(\xi^j \frac{\partial}{\partial x^j})\Big)=\frac{dx^i}{dt}\Big(\xi^j\nabla_{\frac{\partial}{\partial x^i}}( \frac{\partial}{\partial x^j})\Big)+\frac{dx^i}{dt}\Big(\xi^j\frac{\partial}{\partial x^i}\Big)\frac{\partial}{\partial x^j}=\Big(\frac{dx^i}{dt}\xi^j\Gamma_{ij}^k\frac{\partial}{\partial x^k}+\color{red}{\frac{dx^i}{dt}(\frac{\partial}{\partial x^i})}\frac{\partial}{\partial x^j}\Big)=\Big(\frac{dx^i}{dt}\xi^j\Gamma_{ij}^k\frac{\partial}{\partial x^k}+\color{red}{\frac{d\xi^j}{dt}}\frac{\partial}{\partial x^j}\Big)=\Big(\frac{dx^i}{dt}\xi^j\Gamma_{ij}^k+{\frac{d\xi^k}{dt}}\Big)\frac{\partial}{\partial x^k}$$
where the last equality is obtained considering the index $j$ in the second term as $k$.

$\textbf{My question:}$ do you think that the passages that I have done are correct? In particular it is valid the equality that involves the red quantities (I am not sure that is mathematically rigorous)?


问:
Find the slope of the tangent line to the graph of the given function at the given value of x.
Find the slope of the tangent line to the graph of the given function at the given value of x. Find the equation of the tangent line. : $y=x^4-5x^3+2; x=2$

I understand that the slope of the line follows the equation of

$$\lim_{h\to 0} \frac{f(a+h)-f(a)}h.$$

But I don't know how to go about the rest

答:
The easiest equation is point-slope form because (1) as you observe, the derivative gives this slope and (2) it is very easy to get the $y$-coordinate of a point on the curve given its $x$-coordinate.  So, the line through the point $(x_0,y_0)$ of slope $m$ has the equation
$$  y - y_0 = m(x - x_0)  \text{.}  $$

The above should be enough for you to finish your problem.  If you need more...

 $\mathrm{d}y/\mathrm{d}x = 4x^3 - 15x^2$.  At x = 2, this evaluates to $4 \cdot 8 - 15 \cdot 4 = -28$, so the slope of the line in question is $-28$.
 Then, evaluating for $y$ when $x = x_0 = 2$, we have $y_0 = x_0^4 - 5 x_0^3 + 2 = 16 - 5 \cdot 8 + 2 = -22 \text{.}$

Finally, should you still not see how to assemble these three numbers into the equation...

 Then an equation of the line is $$  y - (-22) = -28(x - 2)  \text{,}  $$ although it is easy to clean this up a little, to $$  y + 22 = -28(x - 2)  \text{.}  $$


问:
Mistake in OEIS A103904?
The sequence OEIS A103904 is described as

Number of perfect matchings of an $n \times (n+1)$ Aztec rectangle with the third vertex in the topmost row removed.

Definition of $M \times N $ Aztec rectangle (one can see such a definition by C. Krattenthaler):

Consider a $(2M+1) \times (2N+1)$ rectangular chessboard and suppose that the corners are black. Then an $M \times N$ Aztec rectangle is the graph whose vertices are the white squares and whose edges connect precisely those pairs of white squares that are diagonally adjacent. So an $n \times (n+1)$ Aztec rectangle is when $M=n$ and $N=n+1$.

see the following figures for $n \times (n+1)$ Aztec rectangles as examples:

When $n=2$, I try to calculate the number of perfect matchings of $2 \times (2+1)$ Aztec rectangle with the third vertex in the topmost row removed:

The number of perfect matching is actually $8$, which is not in A103904.  Is there any mistake in this sequence or the definition is wrong? Or maybe I misunderstand it. I am very happy to hear from anyone and thank you very much in advance!

答:
@Marie, I agree with you that the sequence A103904 is mistaken.

The first term is plainly wrong (incidentally the formula section on the OEIS page has every entry marked with $n>1$, as none of them produces $1$ when $n=1$), but here are two ways to fix the rest:

(1) Keep the definition.  The correct sequence is then given by
$$
\frac{n(n-1)}{2}\times2^{n(n+1)/2},
$$
or
$$
0, 8, 192, 6144, 327680, \dots
$$
i.e., the OEIS entry has the exponent wrong: it should be $n+1\choose2$ rather than $n\choose 2$, so the sequence is off by a factor of $2^n$.  Note $n=1$ does not need to be singled out any more.

(2) Change the definition to the following:

Number of perfect matchings of an $n\times(n+1)$ Aztec rectangle, where all the vertices in the top-most row and the second top-most row have been removed, except the third and the last vertex of the second top-most row.

Then the sequence can remain as it is, except the first term needs to be changed to $0$:
$$
0, 2, 24, 384, 10240, \dots
$$

These results follow directly from Lemma 1 and 2 in Krattenthaler that you have quoted above.  Let me know if you have any questions.

Title: Mistake in OEIS A103904?


问:
Can we show that a local martingale is closed when it is stopped at one of the localising stopping times?
Let $M$ be a $(\mathcal F_{t}^{B})_{t\geq 0}-$ local martingale starting in $0$ where  $(\mathcal F_{t}^{B})_{t\geq 0}$ denotes the completed filtration of the Brownian motion. Now consider the reducing sequence $(\tau_{n})_{n}$ so that $M^{\tau_{n}}$ is a martingale. It is then stated that I may assume that $(M_{\tau_{n} \land t})_{t\geq 0}$ is a closed martingale (considering either $\tau_{n}$ or $\tau_{n}\land n$), $n \in \mathbb N$. I am not sure why closedness of the martingale $M^{\tau_{n}}$ or $M^{\tau_{n}\land n}$ holds. Could anyone help me to explain why this is the case?

Martingale $(M_{t})_{t\geq 0}$ is closed iff $M_{\infty}\in \mathcal{F}_{\infty}$ and integrable exists s.t. $E[M_{\infty}\lvert \mathcal{F}_{t}]=M_{t}$

答:
First, we have that $\tau_n \wedge n$ is a bounded stopping time so $N_t := M_{(\tau_n \wedge n) \wedge t}$ is a closed martingale because $N_t = \mathbb{E}[N_n | \mathcal F_t]$ for all $t$, i.e. $N_\infty = N_n = M_{\tau_n \wedge n}$.

The reason that we may assume $(M_{\tau_n \wedge t})$ is closed is because if it isn't, we can replace $\tau_n$ with $\tilde{\tau}_n := \tau_n \wedge n$.  Then $\tilde{\tau}_n$ is still a localizing sequence, and we showed above that $(M_{\tilde{\tau}_n \wedge t})$ is a closed martingale.  Since we can always replace a localizing sequence with one that makes $(M_{\tau_n \wedge t})$ closed, we might as well just assume that the original localizing sequence does too.


问:
Is the decimal component of the square root of 2 (i.e. sqrt(2) - 1) a named constant? Is it broadly significant?
I was doing some geometry with circles locked on a grid, and I noticed that the distance between a unit circle and the corner of a square that it is inscribed in is of length $\sqrt{2} - 1$, or roughly $.41421356237$. Does this number have a widely-recognized name, and does it pop up in seemingly unrelated situations in the way that pi or the golden ratio do?

Granted, the primary purpose of such names are to create shorthand, and $\sqrt{2} - 1$ isn't very long to begin with. Still, it's not much less cumbersome than the mathematical expressions that define the first few metallic ratios, which are nonetheless named. Furthermore, if we re-scale the geometry to involve a unit square instead of unit circle, then the similar line segment is of length $\frac{\sqrt{2}-1}{2}$, which is closely reminiscent of the metallic ratios. I can easily imagine that constant being named rather than the first one.

答:
There is some significance to this constant in the theory of quasilattices.  While the best-known cases involve fivefold symmetry, eightfold examples have also been realized in quasicrystalline materials in the laboratory.  So this eightfold quasilattice concept is a thing indeed.

We may then render the vertices of a regular octagon as quasilattice points, and then identify a quasiperiodic ratio by comparing the length of a side of the octagon with that of a parallel diagonal.  If we divide the diagonal by the side, we get a ratio of $\sqrt2+1$; with the reverse division it comes out to $\sqrt2-1$ (in the fivefold case we would use a pentagon instead, in which case the ratio between side and parallel diagonal would be $(\sqrt5\pm1)/2$).  Both period ratios are consistent with the same quasilattice structure because they differ by a whole number.


问:
Index vector at multiple indices
I want to define a map, which maps a vector $x \in \mathbb{R}^n$ to a new vector $y \in \mathbb{R}^m$, such that every entry of $y$ is a fixed entry of $x$ (duplicates are allowed). My current approach looks like this:

Let $I \subseteq \{1, \dots, n\}^m$. We define
$$x \mapsto (x_i)_{i \in I}$$

but I'm not sure, whether this captures the ordering of the elements correctly. I think that I could fix this issue by writing

$$x \mapsto (x_{i_j})_{j = 1, \dots, m}$$

but this seems too messy. Now, I am asking whether there es a simple way to define this map? Does the first definition work or am I missing something? Something like $x_I$ is great, but I don't think that this is standard notation and sadly it is not contained in the multi-index notation as well (even though I think it would be a nice extension).

答:
My understanding is that you want to specify some indices (say $1, 3, 4, 1$ for example) and then get the map (if $n = 5$)
$$
  (x_1, x_2, x_3, x_4, x_5) \mapsto (x_1, x_3, x_4, x_1).
$$
Your second version captures this.

The first version does not work: If $I \subseteq \{1, \dots, n\}^m$, then an element $i \in I$ is a sequence $(i_1, \dots, i_m)$ and $x_i$ does not make sense.

Depending on how much you use this operation, you might consider defining $x_I := (x_{i_1}, \dots, x_{i_m})$ for $I = (i_1, \dots, i_m) \in \{ 1, \dots, n \}^m$ (note: element, not subset). This notation is not standard (as you seem to know) but introducing custom notation is fine as long as you clearly state what it means; it certainly seems like a natural choice.

Title: Index vector at multiple indices


问:
definite integral $\int_{-1}^1x^2\sqrt{4-x^2}dx$
I have solved the problem and obtained the value of the integral $2\pi/3-\sqrt{3}/2$.

In the actual solution of this problem I have seen that $x$ was substituted by $2\sin z$ and upper limit $1$ was substituted by $\pi/6$ and lower limit $-1$ by $-\pi/6$.

Now the question is why was the upper limit substituted by $\pi/6$ instead of $5\pi/6$ or anything equivalent to $\arcsin\tfrac12$?

答:
In general, if $f(x)$ is any function defined on $[-1,1]$ and $\sin \theta_1=\sin\theta_2$ then $$\int_{\theta_1}^{\theta_2} f(\sin \theta)\cos(\theta)\,d\theta=0$$ So it doesn’t matter which $\arcsin 1/2$ you choose.

If $f$ was only defined on $[-1/2,1/2],$ you’d have to pick a careful pair of integral bounds. But any pair such that $f(\sin\theta)$ is defined between the pairs would work.

Or you could cheat and define $f(x)=0$ outside $[-1/2,1/2].$


问:
Lambda Calculus Question: If for some λ-terms M and N we have Mx =β Nx. Does it necessarily imply that M =β N?
I have a homework question that I can't figure out. Hope someone can help.

Assume that for some $\lambda$-terms $M$ and $N$ we have $Mx =_\beta  Nx$. Does it necessarily imply that $M =_\beta N$?
Here $=_\beta$ stands for $\beta$-equivalent. It is in the paragraph on the Church-Rosser property

Intuitively I would say the answer is yes, but I'm not sure how to show this.

答:
No, the fact that two $\lambda$-terms $M$ and $N$ are such that $Mx =_\beta Nx$ does not imply that $M =_\beta N$.

As a counterexample, consider the $\lambda$-terms $M = \lambda z.yz$ and $N = y$. We have that $Mx = (\lambda z. yz)x \to_\beta  yx = Nx$, and so $Mx =_\beta Nx$.
But $M \not=_\beta N$, because both $M$ and $N$ are normal terms, but they are different and hence they cannot be $\beta$-equivalent, by the Church-Rosser property.

The fact that $Mx \simeq Nx$ implies $M \simeq N$ holds if $\simeq$ stands for $\beta\eta$-equivalence, that is, the union of $\beta$-equivalence and $\eta$-equivalence. $\eta$-equivalence is the equivalence relation generated by $\lambda x. M x =_\eta M$ when $x$ is not free in $M$.


问:
How to change duty cycle of $\sin()$ function
Is there a way to change the duty cycle of the sine function?

I am trying to use $\operatorname{sgn}(\sin(x))$ functions and I would love to change the width of pulses.

Thank you in advance!

答:
Originally i thought you wanted to change the frequency of the function:

Think about how the sine function is defined. If i put something in its argument i will get the sine of that thing. Now if i want the function to lengthen the function on the $x$-axis you need to put half of it to get the same function.

See: https://www.desmos.com/calculator/zkx76calgy

For a numerical example:
$$\sin\left(a\cdot\pi\right)=0$$

If i say:

$$\sin\left(\frac{b}{2}\pi\right)=0$$

$b$ needs to be twice the value of $a$ to get the same output. Essentially stretching the function sideways.

Manipulating the duty cycle is a digital systems problem and can be done formally easily in that manner. If you want to specifically calculate a specific function like that then the easiest but also the most tedious way to calculate it is to figure out your function between two oscillations and just take the Fourier series of it. But i don't think this is a very useful method for understanding the specific ratio.