问:
Helmholtz equation - FEA error estimate for optimal convergence order
I am currently trying the following problem. Unfortunately, I haven't got any idea for days. Would be really grateful for help.

Let $\Omega \subseteq \mathbb{R}^2$ be a bounded polygon and consider the eigenvalues of the problem $ -\Delta u = \lambda u, u = 0 \quad \text{on} \quad \partial \Omega.$ Weyl’s asymptotic formula shows that $\lim_{\lambda \to \infty}  \# \{n \in \mathbb{N} | \lambda_n \leq \lambda \}/ \lambda = |\Omega|/(4 \pi)$. With $E_n := P^0(\mathcal{T}_h)$, where $\mathcal{T}_h$ is a uniform mesh with mesh size $h > 0$ and $n = \# \mathcal{T}_h$, what is the convergence rate (error vs. degrees of freedom) of $\inf_{v \in E_n} \Vert u - v \Vert_{L^2(\Omega)}$. Show that this rate can, in general, not be improved. (The only information you have about $u$ is that $u \in H_0^1(\Omega).$)

I got the hint to use the following result (I could prove it):

Consider an ONB $(u_n)_{n \in \mathbb{N}}$ of a Hilbert space $X$ and a sequence $0 < \lambda_1 \leq \lambda_2 \leq \dots \to \infty $.
Define the set
$$ B:= \left \{ \sum_{j=1}^{\infty}\alpha_j u_j : \alpha_j \in \mathbb{R}, \sum_{j=1}^{\infty} |\alpha_j|^2 \lambda_j \leq 1 \right \}. $$
It holds:
$$ \inf_{\substack{E_n \subseteq X \\ dim(E_n)=n}} \sup_{u \in B} \inf_{v \in E_n} \Vert u-v \Vert_X = \lambda_{n+1}^{-1/2}. $$
So I want to show that the following applies to the Helmholtz problem:

$\inf_{v \in E_n} \Vert u - v \Vert_{L^2(\Omega)} = C*\frac{1}{\sqrt{n}}$

If I understood that correctly.
Bramble-Hilbert should simply give the "upper estimate".

The "downward estimation" becomes problematic, which is probably the crux of the matter.
It is not clear to me how I can use the hint (especially set B) here.

Would be very grateful for ideas.

答:
If someone still needs the solution:

First we will prove
$$ \inf_{\substack{E_n \subseteq X \\ dim(E_n)=n}} \sup_{u \in B} \inf_{v \in E_n} \Vert u-v \Vert_X \leq \lambda_{n+1}^{-1/2}. $$ For that we choose $E_n = V_n = span\{u_1, \dots, u_n\}$, this implies that for every $v \in V_n$ there exists a unique vector $(\mu_1, \dots, \mu_n) \in \mathbb{R}^n$, so that
$$ v = \sum_{i=1}^{n} \mu_i u_i.$$
Now we compute $$ \Vert u-v \Vert_X^2 = \left \Vert \sum_{i=1}^{\infty} \alpha_i u_i - \sum_{i=1}^{n} \mu_i u_i \right \Vert_X^2 = \left \Vert \sum_{i=n+1}^{\infty} \alpha_i u_i + \sum_{i=1}^{n} (\alpha_i - \mu_i) u_i \right \Vert_X^2 = \sum_{i=1}^{n} (\alpha_i - \mu_i)^2 + \sum_{i=n+1}^{\infty} \alpha_i^2,$$ the last equality holds by Parsevals identity.
This yields $$ \inf_{v \in V_n} \Vert u-v \Vert_X = \inf_{(\mu_i) \in \mathbb{R}^n} \left (\sum_{i=1}^{n} (\alpha_i - \mu_i)^2 + \sum_{i=n+1}^{\infty} \alpha_i^2\right )^{1/2} = \left (\sum_{i=n+1}^{\infty} \alpha_i^2\right )^{1/2}.$$
Eventually $$\sup_{u \in B} \inf_{v \in E_n} \Vert u-v \Vert_X = \sup_{u \in B} \left (\sum_{i=n+1}^{\infty} \alpha_i^2\right )^{1/2} = \left (\sup_{u \in B} \sum_{i=n+1}^{\infty} \alpha_i^2\right )^{1/2} = \left (\frac{1}{\lambda_{n+1}} \right)^{1/2}.$$
The last equality is true because
$$ \sum_{j=1}^{\infty} \alpha_j^2 \lambda_j \leq 1 $$ implies
$$ \sum_{j=1}^{\infty} \alpha_j^2 \frac{\lambda_j}{\lambda_{n+1}} \leq \frac{1}{\lambda_{n+1}} $$ and the monotonicity of $(\lambda_j)_{j \in \mathbb{N}}$ implies $$ \sum_{j=n+1}^{\infty} \alpha_j^2 \leq \frac{1}{\lambda_{n+1}}. $$
Choosing $(\alpha_j)_{j \in \mathbb{N}} = (0,0, \dots,0,0, \lambda_{n+1}^{-1/2},0,0,0,0\dots)$ yields
$$ \sup_{u \in B} \sum_{i=n+1}^{\infty} \alpha_i^2 = \frac{1}{\lambda_{n+1}}.$$

It remains to prove
$$ \inf_{\substack{E_n \subseteq X \\ dim(E_n)=n}} \sup_{u \in B} \inf_{v \in E_n} \Vert u-v \Vert_X \geq \lambda_{n+1}^{-1/2}.$$ Let $E_n \subset X$ an arbitrary finite dimensional subspace with $dim(E_n)=n \in \mathbb{N}.$
We have to show that
$$ \exists u \in B \forall v \in E_n : \frac{1}{\lambda_{n+1}} \leq \Vert u-v \Vert_X^2.$$ $u \in B$ has the representation $$ u=\sum_{i=1}^{\infty}\alpha_iu_i, \quad \sum_{i=1}^{\infty} \alpha_i^2 \lambda_i \leq 1.$$ We choose $(\alpha_1,\dots,\alpha_{n+1})$ so that
$$ \hat{u} := \sum_{i=1}^{n+1} \alpha_i u_i \in E_n^{\perp}, \quad \frac{1}{\lambda_{n+1}} \leq \sum_{i=1}^{n+1} \alpha_{i}^2, \quad \text{and} \quad \alpha_i = 0 \quad \text{for} \quad i > n+1.$$
This yields
$$ \Vert u-v \Vert_X^2 = \Vert \hat{u} - v \Vert_X^2 = \Vert \hat{u} \Vert_X^2 + \Vert v \Vert_X^2 \geq \Vert \hat{u} \Vert_X^2 = \sum_{i=1}^{n+1} \alpha_{i}^2 \geq \frac{1}{\lambda_{n+1}}.$$


问:
Spectrum of a self-adjoint integral operator
Trying to figure out the ingredients for the spectral representation of the following self-adjoint integral operator:

$(Tf)(s)=\int_0^1 (s-t)^2 f(t) dt$

A hint in the exercise was to consider $im T$ and make an ansatz for the possible eigenfunctions.

I tried using a solution like the one in: How to Find the Spectrum of an Integral Operator but it didn't quite work so now I'm stuck.

答:
Hints: Since $Tf(s)=s^{2}\int_0^{1} f(t)dt-2s\int_0^{1} f(t)dt+\int_0^{1} t^{2}f(t)dt$ the range of $T$ is contained in $span \{1,s,s^{2}\}$. Hence $T$ is a compact operator. So all the non-zero points in its spectrum are eigen values.

Consider $Tf(s)=\lambda f(s)$ with $\lambda \neq 0$. Since LHS is smooth it follows that $f$ is also smooth. Differentiate twice to find eigen values and eigen functions. I hope these hints help.

ContentLicense: CC BY-SA 4.0


答:
Any linear map between finite dimensional vector spaces can be expressed by matrix multiplication (from the left, if vectors are coordinated into columns).

In this case, $a$ and $b$ are linear maps $\Bbb C^4\to\Bbb C$, and by letting
$$a_{11}:=a(1,0,0,0),\ \ a_{12}:=a(0,1,0,0),\ \ a_{21}:=a(0,0,1,0),\ \ a_{22}:=a(0,0,0,1)$$
linearity ensures that
$$a(v_{11},\,v_{12},\,v_{21},\,v_{22})=v_{11}a_{11}+v_{12}a_{12}+v_{21}a_{21}+v_{22}a_{22}\,,$$
in other words, $a$ must be a fixed linear combination of its input coordinates.

Title: Most general $2$ linear map $(\mathbb{C}^2)^2\to \mathbb{C}^2$

Body:
Which is most general $2$ linear map
$$
\phi(v_1,v_2)=a(v_{1,1},v_{1,2},v_{2,1},v_{2,2})e_1
+b(v_{1,1},v_{1,2},v_{2,1},v_{2,2})e_2,
$$
where $v_1,v_2\in\mathbb{C}^2$, and $e_1,e_2$ is the standard basis of $\mathbb{C}^2$?
By applying linearity and using differentiation, I see that $a$ and $b$ can at most be linear in $v_{1,1},v_{1,2},v_{2,1},v_{2,2}$ separately
$$
\phi(v_1,v_2)=\sum_{i,j} v_{1,i}v_{2,i}\phi(e_i,e_j)
$$

What are the most general expression for $a$ and $b$ for $\phi$ to be $2$-linear?

EDIT 1.  Example
$$
\phi(v_1,v_2)=v_1-2v_2,\\
\psi(v_1,v_2):=\phi(v_1,v_2)-\phi(v_2,v_1) =3v_1-3v_2\\
\psi\left(v_1,v_2\right)
=3(v_{1,1}v_{2,2}-v_{1,2}v_{2,1})\left(e_1 - e_2 \right)
=3\det\big(v_1;v_2\big)\left(e_1 - e_2 \right).
$$
in this example the coefficients are a product of the components

EDIT 2.

\begin{align*}
&\psi\left(v_1,v_2\right)
=\psi\left(v_{1,1}e_1+v_{1,2}e_2 , v_{2,1}e_1+v_{2,2}e_2 \right)\\
&=v_{1,1}v_{2,1}\psi\left(e_1 , e_1 \right)
+v_{1,1}v_{2,2}\psi\left(e_1 , e_2 \right)
+v_{1,2}v_{2,1}\psi\left(e_2 , e_1 \right)
+v_{1,2}v_{2,2}\psi\left(e_2 , e_2 \right)\\
&=(v_{1,1}v_{2,2}-v_{1,2}v_{2,1})\psi\left(e_1 , e_2 \right)
=3(v_{1,1}v_{2,2}-v_{1,2}v_{2,1})\left(e_1 - e_2 \right)\\
&=3\det\big(v_1;v_2\big)\left(e_1 - e_2 \right).
\end{align*}

$e_1=(1,0)^T$, $e_2=(0,1)^T$, $v_i=(v_{i,1},v_{i,2})^T$


问:
Proving $\sum_{cyc}\frac{\sqrt{s(s-a)}}{a}\geq\frac{3\sqrt{3}}2$ for $a$, $b$, $c$, $s$ the sides and semi-perimeter of a triangle

If $a, b, c$ are the lengths of the sides of a triangle and s is the semiperimeter, prove that:
$$\sum_{cyc}\frac{\sqrt{s(s-a)}}{a}\geq \frac{3\sqrt{3}}{2}$$

My attempt: $$\sum_{cyc} cos \,\frac{A}{2}=\sum _{cyc} \sqrt{\frac{s(s-a)}{bc}}\leq \frac{3 \sqrt{3}}{2}\leq \sum_{cyc}\frac{\sqrt{s(s-a)}}{a}$$. As the first inequality is obvious, it is enough to show that $$\sum _{cyc} \sqrt{\frac{s(s-a)}{bc}}\leq \sum_{cyc}\frac{\sqrt{s(s-a)}}{a}$$. We prove that $$\sqrt{\frac{s(s-a)}{bc}} \leq\frac{\sqrt{s(s-a)}}{a}$$ or equivalently $a^{2}\leq bc$. Similary $b^{2}\leq ac$ and $c^{2}\leq ab$ and adding the inequalities we get $(a-b)^{2}+(b-c)^{2}+ (c-a)^{2}\leq 0$.

答: After using Ravi's substituition $a=x+y,b=y+z,c=x+z$ as nguyenhuyen_ag did we have to prove $$\sum \frac{\sqrt{z(x+y+z)}}{y+x} \geq \frac{3\sqrt 3}{2}.$$  Now we will use the method of Isolated fudging. We guess that we might be able to prove the follwing inequality $$\frac{\sqrt{z(x+y+z)}}{y+x}\ge \frac{3\sqrt{3}}{2}\frac{z}{x+y+z}$$ Indeed after squaring this inequality is equivalent to   $$\frac{{(x+y+z)}^3}{27}\ge z(\frac{x+y}{2})(\frac{x+y}{2})$$ which is evident by AM_GM. Hence $$\sum \frac{\sqrt{x(x+y+z)}}{y+z} \ge\frac{3\sqrt{3}}{2} \sum \frac{z}{x+y+z}=\frac{3\sqrt{3}}{2}$$ done


问:
Can I find the asymptotes of a function knowing the derivative?
Given for example
$$F(x)=\int_{0}^{x}\frac{t^3+2t+e^t}{2t^2+2e^t}e^{-t^2}dt$$ for $x \to +\infty$ can I say if it has asymptotes and if so, which ones, without integrating?

答: To answer the question of the title : the answer is no. For example, $f(x)=0$ (and thus $f'(x)\to 0$ whenever $x\to \infty $) has an asymptote at $+\infty $ which is given by the line $y=0$, whereas, $g(x)=\ln(x)$ is s.t. $g'(x)\to 0$ whenever $x\to \infty $, but $g$ has no asymptote at $+\infty $.


问:
Limit over a path
I don't know how to compute a limit over a path when it is not a specific point I have to look for.

What is the procedure in this case? Do I have to find a point in which the function is not continous, or do I have to parametrize the curve in some way. No idea. Anything will help.

答: This function is continuous at $(x,y)$ if $x \neq -y^{2}$ and discontinuous $(x,y)$ if $x = -y^{2}$. For discontinuity at $(-y^{2},y)$ with $y \neq 0$ consider  limit of $f(\frac 1 n -y^{2}, y)=1-ny^{2}$ as $n \to \infty$ This limit is $-\infty$ so $f$ is not continuous at $(x,y)$. Now consider the case $y=0$ and $x=-y^{2}=0$. In this case the limit of $f(\frac 1 n -y^{2}, y)=1-ny^{2}$ is $1$ whereas the limit of $f(x,y)$ along $x=0$ is $0$. Hence, $s$ is not continuous at $(0,0)$ either.

Title: Limit over a path


问:
Transformations of functions at GCSE - enlargement vs stretch vs translation (ordering)

What happens when you have a function f(x) and apply a transformation af(x/a)?
Is this just equivalent to an enlargement of f(x) with scale factor a with centre of enlargement (0,0)? (Because we have stretch in each of x and y directions by the same scale factor - this is equivalent to an enlargement with centre the origin I'm assuming...)

I can't seem to find this described anywhere in books at GCSE or even online.

Also, to confirm, if we have another function f(x) and apply the transformation af(x)+b, is it first a stretch of scale factor a in the y direction and then a translation by vector (0 b)? And not vice versa? Similar to how we apply BIDMAS with multiplication first?

Finally, if we had the transformation af(bx), then it doesn't matter which order we say stretch in y -direction by scale factor a and stretch in x- direction by scale factor 1/b?

If there is a whole list of such transformations which summarises these, it would be great if you can point me to these:) As only the "standard" ones seem to exist from wherever I've been looking.

Thanks very much in advance!

答:
Welcome to Math Stackexchange. It’s very good that you’re thinking about these generalisations.

You are correct.

Your first assumption is correct. To get the “other order”, you can try $a(f(x)+b)$. Think about it.

Your observation is again correct.

If operations can be done in either order (as in number 3), we say that the operations commute.

The observation at (3) is a special case of the fact that two “purely stretching” linear maps that have “the same axes of stretching” always commute. In technical language, one can say that two matrices that are diagonalised in the same basis will commute.

In general, the rules you have stated are all you need. You might want to throw in $f(x-a)$. You don’t need a “whole list” of such transformations, because you can always compose these transformations as you need, or decipher them when given one.


问:
Setting $P\left(\{k\}\right)=\frac{1}{2^k}$ when picking number from the naturals
We randomly peak a number from the natural numbers $\mathbb{N}$. What is the probability to get $13$? My prof explained that if for every $k$, the probability is $0$ then we get:
$$
1=P\left(\mathbb{N}\right)=P\left(\bigcup_{k=1}^{\infty} \{k\}\right)=\sum_{k=1}^{\infty} P\left(\{k\}\right)=0
$$
Then it's impossible to have $p=0$. Also it's impossible to have $p>0$ because then you get $1=P\left(\mathbb{N}\right)=\infty$.

I understand this but then he said that for $P\left(\{k\}\right)=\frac{1}{2^k}$ (sequence so the sum is 1) it works. I don't understand this part. What should work? Why can we choose it to be $P\left(\{k\}\right)=\frac{1}{2^k}$? I also looked at Probability of picking a random natural number but didn't see something that explain this part.

答:
In the latter case , we have countable infinite many events from which exactly one occurs.

In this case, the probability that any event occurs is the sum of the probabilities, so we need a sequence of countable infinite many entries summing up to $1$.

The given sequence does the job, hence is a suitable probability distribution.


问:
Number of non isomorphic groups of order p(p prime)
What is the number of non isomorphic groups of order p?

I tried searching through various textbooks but couldn't find the solution.
Can we use Euler-phi function to calculate the number somehow?

答:
Take group $G$ of order $p$, where $p$ is prime. Only divisors of $p$ are 1 and $p$ itself.  By Lagrange Theorem, any subgroup of $G$ has to have either one or $p$ elements. So the only subgroups of $G$ are the one-element subgroup - containing only neutral element e and the whole $G.$

So take $a\in G$, that is not equal to the neutral element. Then subgroup $\langle a\rangle_G$ generated by $a$ has to be the whole $G$. So I can define group isomorphism $\phi: G  \to {\Bbb Z}_p$ by sending $\phi(a)=1$.

Hence all groups of order $p$ are isomorphic to (additive) group $\Bbb Z_p$. So there is only one group of order $p$ up to isomorphism.


问:
Decomposition of 3D rotations into products
I really need help with this question, thank you in advance.

Let
$e_1=
  \begin{pmatrix}
  1 \\
  0 \\
  0 \\
  \end{pmatrix}
$,
$e_2=
  \begin{pmatrix}
  0 \\
  1 \\
  0 \\
  \end{pmatrix}
  \in
  \mathbb{R}^3.
$
For a non-zero vector v $ \in \mathbb{R}^3$ and a real number $\theta \in  \mathbb{R}$, let us denote by $R(v,\theta)$ the space rotation around the axis v with the angle $\theta$.

(1) Let u $\in  \mathbb{R}$ be a vector with |u| $= 1$. Show that there exist $ \alpha, \beta \in \mathbb{R}$ such that $$R(e_2,\beta)R(e_1,\alpha)u=e_1.$$

答:
Answer to Question (1):

Relationship

$$R(e_2,\beta)R(e_1,\alpha)u=e_1$$

can be written into an equivalent form:

$$u=\underbrace{R(e_1,\alpha)^{-1}}_{R_1}\underbrace{R(e_2,\beta)^{-1}}_{R_2}e_1$$

which can be transformed into:

$$\begin{pmatrix}u_1\\u_2\\u_3\end{pmatrix}=\underbrace{\begin{pmatrix}1&0&0\\0&\cos(\alpha)&\sin(\alpha)\\0&-\sin(\alpha)&\cos(\alpha)\end{pmatrix}}_{R_1}\underbrace{\begin{pmatrix}\cos(\beta)&0&\sin(\beta)\\0&1&0\\-\sin(\beta)&0&\cos(\beta)\end{pmatrix}}_{R_2}
\begin{pmatrix}1\\0\\0\end{pmatrix}$$

$$\begin{pmatrix}u_1\\u_2\\u_3\end{pmatrix}=\begin{pmatrix}\cos(\beta)\\ -\sin(\alpha)\sin(\beta)\\-\cos(\alpha)\sin(\beta)\end{pmatrix}$$

(do you recognize spherical coordinates, but not in the usual order ? In fact $\beta$ can be at once recognized as the classical longitude).

As coordinate $|u|=1 \implies u_1 \in (-1,1)$, we can deduce that :

$$\beta = \operatorname{acos}u_1 \ \ \text{and} \ \ \alpha=\operatorname{atan}\tfrac{u_2}{u_3}$$

with a particular case for the forbidden value $u_3=0$ for which $\alpha=\pi/2.$

Up to you for the other questions.


问:
Exhibit the complete table of multiplication for $(\mathbb{Z} / 15 \mathbb{Z})^*$, and use this to solve $7X \equiv 3 \pmod{15}$.
I have no idea how to start this question? I know how to build a multiplication table for $(\mathbb{Z} / 15 \mathbb{Z})$, but not for $(\mathbb{Z} / 15 \mathbb{Z})^*$? Can anyone help?

答:
Use the extended Euclidean algorithm, which has only one step:

$15 = 2\cdot 7 + 1$.

Then $1 = 1\cdot 15 + (-2)\cdot 7$ and so $(-2)\cdot 7 \equiv 1\mod 15$ or $(-2)\cdot 7 =1$ in $\Bbb Z_{15}$.

Thus multiplying $7x=3$ with $-2$ gives $x=-6$ or $x= 9$ in $\Bbb Z_{15}$


问:
Find the limit without use of L'Hôpital or Taylor series: $\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)$
Find the limit without the use of L'Hôpital's rule or Taylor series

$$\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2x}\right)$$

答:
Well, you have
$$\left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right) = \left(\frac{1}{x} - \frac{1}{\sin x}\right)\left(\frac{1}{x} + \frac{1}{\sin x}\right)$$
$$=\left(\frac{1}{x^2} - \frac{1}{x\sin x}\right)\left(1+ \frac{x}{\sin x}\right)$$
Since the limit of $\left(1+ \frac{x}{\sin x}\right)$ is $2$, your answer will be
$$2\lim_{x \rightarrow 0}\left(\frac{1}{x^2} - \frac{1}{x\sin x}\right)$$
$$=2\lim_{x \rightarrow 0}\left(\frac{x}{\sin x}\right)\left(\frac{\sin x - x}{x^3}\right)$$
$$=2\lim_{x \rightarrow 0}\frac{\sin x - x}{x^3}$$
An elementary calculus way of showing this limit is $-{1 \over 6}$ can be found here.

So the overall answer is $-\frac{1}{3}$.

ContentLicense: CC BY-SA 3.0


答:
If the improper integral is convergent there is an horizontal asymptote. If it isn't, the slope of oblique asymptotes $y=mx$, can be compute $m$ using L'Hopital's rule:
$$
m = \lim_{x\to +\infty} \frac{I(x)}{x} = \lim_{x\to +\infty}\dfrac{\frac{x^3+2x+e^x}{2x^2+2 e^x}\cdot e^{-x^2}}{1}=0
$$

So, either there is an horizontal asymptote or no asymptote at all.

Bottom line: Check if the integral is convergent. If it is, there is a horizontal asymptote; If it isn't, there is no asymptote.

Title: Can I find the asymptotes of a function knowing the derivative?

Body:
Given for example
$$F(x)=\int_{0}^{x}\frac{t^3+2t+e^t}{2t^2+2e^t}e^{-t^2}dt$$ for $x \to +\infty$ can I say if it has asymptotes and if so, which ones, without integrating?


问:
If $X$ is a set, what does $S(X) = \{f\colon X \to X | \text{where f is a bijection}\}$ mean?
If $X$ is a set, what does $S(X) = \{f\colon X \to X | \text{where $f$ is a bijection}\}$ mean.
From my understanding it means the Set containing the function $f$ that maps elements of $X$ onto elements of $X$, where $f$ is a bijection.
However, in the proof to see whether $S(X)$ is a group under composition, they said that composition is a closed binary operation for $S(X)$ because
if $f$ is a bijection, then $f(g)$ is also a bijection so $f(g)$ is in $S(X)$

But where did they get $g$ from? Why can you randomly talk about other functions if $S(X)$ only contains the function $f$ that is bijective.

Thanks in advance

答:
This notation means $S(X)$ is the set of all bijective mappings from $X$ to $X$.

You can easily check that this is a group:

closed under $\circ$, since compositions of bijective mappings are bijective

neutral element: $id\colon X\to X, x\mapsto x$.

inverse element: $f\circ f^{-1} = id$


问:
Prime and semiprime ideals of $A=T_3(D)$, the ring of $3\times 3$ upper triangular matrices over $D$
Let $D$ be a division ring. Could anyone tell me which are the prime and semiprime ideals of $A=T_{3}(D)$, where $A=T_{3}(D)$ the ring of $3\times 3$ upper triangular matrices with coefficients in $D$?</p>

答:
Since $A$ is left (and right) Artinian, the prime ideals are the maximal ideals.

The set $J$ of strictly upper triangular matrices is nilpotent, and hence it is contained in all prime ideals.

$A/J\cong D\times D\times D$ is semisimple, showing that $J$ is the Jacobson radical of $A$. Hence the maximal ideals of $A$ correspond to those of $A/J$.

These maximal ideals are easy to catalogue: $I_1\lhd A$ corresponding to the matrices zero on the $1,1$ position, $I_2$ the set of matrices zero on the $2,2$ position, and $I_3$ the matrices zero on the $3,3$ position. This is a complete list of the prime ideals.

A complete list of semiprime ideals is given by all possible intersections of the prime ideals, namely: $I_1,\ \ I_2,\ \ I_3,\ \  I_1\cap I_2,\ \ I_1\cap I_3,\ \ I_2\cap I_3,\ \ I_1\cap I_2\cap I_3=J$.


答: The set $S(X)$ is defined by set-builder notation: it is the set of all bijections from $X$ to $X$. The letter $f$ is a dummy variable used in the definition. Typically (when $X$ has more than one element) there will be more than one bijection $X \to X$, and so $S(X)$ will have more than one element.

Title: If $X$ is a set, what does $S(X) = \{f\colon X \to X | \text{where f is a bijection}\}$ mean?

Body:
If $X$ is a set, what does $S(X) = \{f\colon X \to X | \text{where $f$ is a bijection}\}$ mean.
From my understanding it means the Set containing the function $f$ that maps elements of $X$ onto elements of $X$, where $f$ is a bijection.
However, in the proof to see whether $S(X)$ is a group under composition, they said that composition is a closed binary operation for $S(X)$ because
if $f$ is a bijection, then $f(g)$ is also a bijection so $f(g)$ is in $S(X)$

But where did they get $g$ from? Why can you randomly talk about other functions if $S(X)$ only contains the function $f$ that is bijective.

Thanks in advance


答:
$S(X)$ is a group with the operation of function composition

$(f\circ g)(x) = f(g(x)),\quad x\in X$,

and identity element $id_X :X\rightarrow X:x\mapsto x$.


问:
A problem about Sylow Theorem
I'm learning Sylow Theorem. My textbook uses the following result directly and does not tell me why. I don't know how to prove it.

$C_{p^rn}^{p^r}\equiv n \mod pn$. Here $p$ is a prime number and $r,n$ are positive integers.

答:
We assume $n$ and $p$ are coprime. Then $\binom{p^rn}{p^r}=n\binom{p^rn-1}{p^r-1}$, so it’s enough to show that $N=\binom{p^rn-1}{p^r-1}$ is congruent to $1$ mod $p$.

Now, in $\mathbb{F}_p$, $N$ is the coefficient of $X^{p^r-1}$ in $(X+1)^{p^rn-1}$. But (mod $p$ by Fermat’s little theorem, $(X+1)^{p^rn-1}=(X+1)^{p^r(n-1)}(X+1)^{p^r-1}=(X^{p^r}+1)^{n-1}(X+1)^{p^r-1}$, and it’s easy to show that the coefficient of $X^{p^r-1}$ in the polynomial in the right hand side is $1$.

Title: A problem about Sylow Theorem


答:
Given a set $X$ and some property elements of $X$ might satisfy, say is green. We can filter out the subset of all elements of $X$, which are green by writing
$$\{x \in X \mid x \text{ is green}\}.$$
The appearance of $x$ does not mean that there can only be one element of $X$, which is green. Rather, if we want to speak of an element which is green, we have to give it a name: In this case the variable $x$.

So in your specific instance we have the set $\operatorname{Fun}(X,X)$ of functions from $X$ to $X$ and want to filter out the subset of all elements (which are functions!), which are bijective. This is what $S(X)$ does:
$$S(X) = \{f \in \operatorname{Fun}(X,X) \mid f \text{ bijective}\}$$


问:
Try to choose a special number $t$ which satisify $\eta(f(x))=f(t)$
$\eta : \mathbb{R}[x]\rightarrow \mathbb{R}$ is a ring homomorphism with  $\eta(1)=1$. Prove there exist $ t \in\mathbb{R}$ s.t.$\eta(f(x))=f(t)$ for all $f(x)\in \mathbb{R}[x]$.

My idea:

$\eta$ is surjective, then there exist a $g(x)$ s.t. $\mathbb{R}[x]/\left\langle g(x)\right\rangle $ is isomorphism to $ \mathbb{R}$. But what can I do next, find the root of $g(x)$? It may not a real number. Could someone give me some details, thank you!

答:
Hint: Show that the only ring homomorphism $\Bbb R\to \Bbb R$ is the identity map (this is necessary, in general if we would consider arbitrary rings $S$ and homomorphisms $S[x]\to S$ we would need the extra hypothesis that $\eta$ is a morphism of $S$-algebras) . Then let $t:=\eta(x)$ and show that this $t$ satisfies the required property.

(Here I assumed that ring homomorphisms preserve the unit, but the claim is wrong anyways if we wouldn't require that)


问:
What is the product of category "2" by itself?
Saunders Mac Lane, at page 10 from "Categories for the Working Mathematician" names $\mathbb{2}$ the category object of a previous question of mine:

$\mathbb{2}$ is the category $↻ \to ↻$ with two objects $a$, $b$, and just one arrow $a \to b$ not the identity;

What is $\mathbb{2} \times \mathbb{2}$ like?

My understanding is that it should contain

4 objects, $(a,a)$, $(a,b)$, $(b,a)$, $(b,b)$,

some morphisms. But which morphisms?

At least one identity morphism for each of the 4 objects, and they all come from combinations of $id_a$ and $id_b$; one of them, for instance, is $id_{(a,b)} = (id_a,id_b)$.

What else? Having called $f$ the only non-identity morphism in $\mathbb{2}$, $f: a \to b$, I can only think of other 5 morphisms:

$f_1 = (f,f) : (a,a) \to (b,b)$,

$f_2 = (id,f) : (a,a) \to (a,b)$

$f_3 = (f,id) : (a,a) \to (b,a)$

$f_4 = (id,f) : (b,a) \to (b,b)$

$f_5 = (f,id) : (a,b) \to (b,b)$

Is this all? Please also correct me if I've abused the notation.

答:
Yes, I think that's all!

You could also say that the arrows of $2\times 2$ are $f_2,\cdots, f_5$ plus their compositions, which will yield $f_1$ as the composition $(f_5\circ f_2)$ or $(f_4\circ f_3)$.


问:
Why the probability of single element event is "zero" on continuous interval?
Why the probability of a single element event is "zero" on the continuous model?, the explanation given is based on probability additivity axioms. but how? some more explanation with the source would be helpful.

cool thanks!

答:
Suppose that there is a biased random number generator that can produce numbers between $0$ and $1$. The probability density function corresponding to this generator is $f(x)=2x$:

To find the probability of falling into a particular interval $[a,b]$, you have to compute the integral between $a$ and $b$ of the probability density function. In our example, that means finding $\int_{a}^{b} 2x \, dx$:
$$
\int_{a}^{b} 2x \, dx = \left[x^2\right]_{a}^{b} = b^2 - a^2
$$
In fact, you need not use integration. The area of the trapezium shown below in below is given by
$$
\text{area} = \text{base} \times \text{average height} = (b-a) \times \frac{2a+2b}{2} = b^2 - a^2
$$

Since the probability of an event corresponds to area, we know that
$$
P(a \leq x \leq b) = b^2 - a^2 \, .
$$
The probability of a single event happening corresponds to an interval of width zero. In other words,
$$
P(x=a) = P(a \leq x \leq a) = a^2 - a^2 = 0
$$
This means that in a continuous probability distribution, the chance of a single event happening is equal to $0$.

You probably still find this answer a little unconvincing. In a discrete setting, the sum of the probabilities of all the events is equal to one. In a continuous setting, not only is this untrue, it doesn't make any sense: there is no sensible way to define the sum of uncountably many terms. One way to reconcile the differences between discrete and continuous settings is to realise that a continuous probability distribution is in a sense the 'limit' of a discrete one, as user2661923 has already mentioned. Imagine we approximate the probability density function $f(x) = 2x$ with $10$ rectangles of width $0.1$:

These rectangles give us an upper bound for the chance of falling into an interval, with
\begin{align}
P(x \in [a,a+0.1]) &\leq \text{interval width} \times \text{height of rectangle at end of interval}\\
&= 0.1 \times 2(a+0.1) = \frac{a+0.1}{5}
\end{align}
What about the general case? If we split the probability distribution function into $n$ intervals of width $1/n$, then
$$
P(x \in [a,a+1/n]) \leq 1/n \times 2(a+1/n) = \frac{2(a+1/n)}{n}
$$

Furthermore, since $a$ is a subset of $[a,a+1/n]$
$$
P(x = a) \leq P(x \in [a,a+1/n]) \, .
$$
Hence,
$$
P(x=a) \leq P(x \in [a,a+1/n]) \leq \frac{2(a+1/n)}{n} \, .
$$
For any interval of width $1/n$, the above inequality must be satisfied. If we make the interval arbitrarily small, then $n$ tends to infinity, and
$$
\frac{2(a+1/n)}{n}
$$
tends to $0$. This means for every $\varepsilon > 0$, $P(x=a) < \varepsilon$. But since probabilities must be nonnegative, we are left with one option:
$$
P(x = a) = 0 \, .
$$


问:
Projective bundle formula for sheaf cohomology
Given a projective bundle $p:P(E)\rightarrow X$ associated to a vector bundle $E$ and a line bundle on $P(E)$ of the form $\mathcal{O}_{P(E)}(1)\otimes p^*L$ where $L$ is a line bundle on $X$, is there a description of the cohomology groups $H^i(P(E),\mathcal{O}_{P(E)}(1)\otimes p^*L )$ in terms of the cohomology groups of $L$, similar to the projective bundle formulas?

答:
One has
$$
Rp_*(\mathcal{O}_{\mathbb{P}(E)}(1) \otimes p^*L) \cong
Rp_*(\mathcal{O}_{\mathbb{P}(E)}(1)) \otimes L \cong
E^\vee \otimes L,
$$
therefore
$$
H^\bullet(\mathbb{P}(E), \mathcal{O}_{\mathbb{P}(E)}(1) \otimes p^*L) \cong
H^\bullet(X, E^\vee \otimes L).
$$


问:
Dense orbit in real linear fractional transformation
Consider $f:\mathbb{R} \backslash \{-2\}\to \mathbb{R}$ defined as $f(x)=\frac{x-1}{x+2}$. Is there a real number that the orbit $\{f(r), f\circ f(r), \cdots\}$is dense in some interval? If there is, can you classify them?

答: Identically we have $f^7=f$, hence all orbits are finite.


问:
Examples of almost compact spaces.
This question is a follow up question to this one

Definition 1. A non-compact Hausdorff topological space X is called almost compact if its Stone-Cech compactification coincides with its one point compactification.

The only two examples of almost compact spaces I know are from the book Pseudocompact topological spaces. M. Hrusak, A. Tamariz-Mascarua, M. Tkachenko. On the page 17 authors say that $[0,\omega_1)$ is almost compact and Mrowka-Isbell space $\Psi(\mathcal{A})$ is almost compact for some specific maximal almost disjoint family $\mathcal{A}\subset 2^\omega$.

I would like to know more on almost compact spaces, but I found almost nothing on this subject.

Questions:

What are other examples of almost compact spaces?

Is true that $\beta\mathbb{N}\setminus\{p\}$ is almost compact for $p\in\beta\mathbb{N}\setminus\mathbb{N}$ ?

Is it true that $X\setminus \{p\}$ is almost compact whenever $X$ is extremally disconnected and $p\in X$.

答:
Concerning the third bullet:
It should definitely read as "whenever X is extremally disconnected and compact". Otherwise, $\mathbb{N}$ is extremally disconnected, but $\mathbb{N} \setminus \{1\}$ is not almost compact.
Furthermore, if $p$ is isolated in $X$, then $X \setminus \{p\}$ is compact, hence not almost compact as defined above.
(Although, almost compactness should include compactness, which actually is, to my knowledge, the most common understanding.)

Anyway, if $X$ is extremally disconnected, compact and $p\in X$, $p$ not isolated in $X$, then
$X \setminus \{p\}$ is almost compact:

We have to show, that $\beta (X \setminus \{p\}) = X$, i.e. that if $A, B \subset X \setminus \{p\}$ are completely separated in $X \setminus \{p\}$,
then $A, B$ are completely separated in $X$:

There exist open, disjoint subsets $U, V$ of $X \setminus \{p\}$ such that $A \subset U$ and $B \subset V$.
$U, V$ are also open in $X$. Since $X$ is extremally disconnected, the closures of $U, V$ are disjoint and compact, hence completely separated.
Hence also $A, B$ are completely separated in $X$.

Title: Examples of almost compact spaces.


答: It's one.  There's only one group of prime order $p$, the cyclic one.

Title: Number of non isomorphic groups of order p(p prime)

Body:
What is the number of non isomorphic groups of order p?

I tried searching through various textbooks but couldn't find the solution.
Can we use Euler-phi function to calculate the number somehow?


问:
How to complete this proof process of Arzelà-Ascoli theorem?

Theorem. Let $X$ be a compact space and $Y$ be a metric space. A set $\mathscr F \subseteq C(X,Y)$ is precompact if and only if it is
pointwise precompact and equi-continuous.

Definitions. A subset $\mathscr F \subseteq C(X,Y)$  is called:

equi-continuous if for every $x ∈ X$ and every $ε > 0$, there exists an
open neighborhood $U \subseteq X$ of $x$ such that $d_Y(f(x), f(x')) < ε$ for all $ x'∈ U$ and all $f ∈ \mathscr F$.

precompact  if the closure of $\mathscr F$ is compact.

pointwise precompact if for each $x∈X$, $\mathscr F(x) = \{f(x): f∈\mathscr F \}$ is precompact.

Lemma. Let $(X,d)$ be a metric space and let $K \subseteq X$. Then every sequence in $K$ has a Cauchy subsequence if and only if $K $is totally bounded.

I know how to prove it from left, the way is similar with $X$ is compact metric space. But from right: auth give me some hint.
The key is is to prove every sequence has Cauchy subsequence. By the lemma, we only need to prove $\mathscr F$ is totally bounded. He gives me two hint:

The set $F = \{f(x) :  x ∈ X, f ∈ \mathscr F\} \subseteq Y$ is precompact so  is totally bounded.

Let $ε > 0$. Cover $F$ by finitely many open balls $V_1,\dots,V_n$ of radius $ε/3$ and cover $X$ by finitely many open sets $U_1,\dots,U_m$ such that
$$\sup_{x,x' ∈ U_i} \sup_{f ∈ \mathscr F} d_Y(f(x),f(x'))＜\frac ε3.$$
Let $\alpha $ be a any map from $\{1,\dots,m\}$ to $\{1,\dots,n\}$.
Define $$\mathscr{F}_α = \{f ∈ \mathscr F : f(U_i) ∩ V_{\alpha(i)} ≠ \varnothing \} .$$  Let $A$ be
the set of all $α$ such that $\mathscr F_α ≠ \varnothing$. Prove that $\mathscr F = \bigcup_{α∈A} \mathscr F_α$.

I can following the hint 2 to achieve it, but I don't how to prove hint 1.

My try: Originally, I choose  $f_n(x_n)$ be sequence of $F$, I use diagonal method to find a subsequence $g_n$ of $f_n$ such that $f_{n_k}$ converges at each $x_i$, and I want to prove that $f_{n_k}(x_{n_k})$ converges to subsequence of $f_n(x_n)$. However I think this way is not ture now.

答:
Let's prove that $F$ is totally bounded

Take $\epsilon \gt 0$. As $\mathcal F$ is supposed to be equicontinuous, for all $x \in X$, it exists an open neighborhood $\mathcal O_x$ such that $d(f(x),f(y)) \lt \epsilon/4$ for all $ f \in \mathcal F$ and all $y \in \mathcal O_x$.

As $X$ is compact, we can extract from $\{\mathcal O_x \mid x \in X\}$ a finite cover $\mathcal O_{x_1}, \dots \mathcal O_{x_m}$.

For all $i \in \{1, \dots, m\}$ $\mathcal F(x_i)$ is precompact  and therefore so is their union $\mathcal U$ (a finite union of precompact subsets is precompact). Let $\{u_1, \dots, u_n\}$ be a finite subset of $\mathcal U$ such that $\mathcal U \subseteq \bigcup_{j=1}^n B(u_j, \epsilon/4)$ where $B(x, r)$ stands for the open ball centered on $x$ of radius equal to $r$.

Now take $f(x) \in  F$. It exists $x_i$ such that $x \in \mathcal O_{x_i}$ and $u_j$ such that $f(x_i) \in B(u_j, \epsilon/4)$. Therefore

$$d(f(x),u_j) \le \underbrace{d(f(x),f(x_i))}_{x \in \mathcal O_{x_i}} + \underbrace{d(f(x_i),u_j)}_{f(x_i) \in B(u_j, \epsilon/4)} \le \epsilon/4 + \epsilon/4 \le \epsilon/2$$ which proves that $$F \subseteq \bigcup_{j=1}^n B(u_j, \epsilon/2)$$

As the diameter of each of the $B(u_j, \epsilon/2)$ is less or equal to $\epsilon$, we can conclude that $\mathcal F$ is totally bounded and to the desired result.


问:
Div-Curl lemma and precompactness in $H^{-1}$
I am trying to understand $\operatorname{div-curl}$ lemma. An important requirement to apply $\operatorname{div-curl}$ lemma is the precomapctness of the sequences $\operatorname{div}(A_n)$ and $\operatorname{curl}(B_n)$ in $H^{-1}(\Omega).$

What are the important compactness results which help in checking whether a sequence is precompact in $H^{-1}(\Omega)$ or not? Where can I find the details?

Are there any books which illustrate the application of $\operatorname{div-curl}$ through some examples?

Any help is appreciated..

答:
Usually, div curl lemma is applied for functions $A_n$ and $B_n$ that verifies $div(A_n)$ and $curl(B_n)$ are bounded sequences in $L^2(\Omega)$, which implies the precompactness in $H^{-1}(\Omega)$.

There are several applications to the Div Curl lemma, the most famous is the application to Homogeneization of PDE using Tartar's oscillating test function method. You can find informations easily on the web on this methid. Even if not using Tartar's method, div curl lemma is very often used in Homogeneization.

Hope this help a little.


问:
Positive semidefinite inequality under tensor product
Let $\rho, \sigma$ be finite dimensional positive semidefinite matrices with trace less than or equal to 1. Let $\lambda(1)$ be the smallest nonnegative number (possibly infinity) such that

$$\rho \preceq \lambda(1)\sigma,$$

where $A\preceq B$ means that $B-A$ is positive semidefinite. We are now required to find $\lambda(n)$ which is the smallest nonnegative number such that

$$\rho^{\otimes n} \preceq \lambda(n)\sigma^{\otimes n},$$

where $A^{\otimes n} = A\otimes A... \otimes A$ and there are $n$ terms in the tensor product. For $n\rightarrow\infty$, can we say anything about $\lambda(n)$ in terms of $\lambda(1)$? It seems like $\lambda(n)$ cannot be smaller than $\lambda(1)^n$ (one can construct trivial examples to see this) but are there any matrices $\rho$ and $\sigma$ for which $\lambda(n) < \lambda(1)^n$?

答:
No such matrices exist and indeed $\lambda(n) = \lambda(1)^n$. To see this note that we should have $\mathrm{supp}(\rho) \subseteq \mathrm{supp}(\sigma)$ when $\lambda(1) < \infty$, where $\mathrm{supp}$ denotes the orthogonal complement of the kernel. For such cases we see that $\rho \leq \lambda(1) \sigma$ is equivalent to the condition (the inverse is taken on the support of $\sigma$)
$$
\sigma^{-1/2} \rho \sigma^{-1/2} \leq \lambda(1) I
$$
which is in turn equivalent to
$$
\|\sigma^{-1/2} \rho \sigma^{-1/2}\| \leq \lambda(1)
$$
where $\| \cdot \|$ is the operator norm. Then for PSD matrices it is straightforward to show via the spectral theorem that $\|A^{\otimes n}\| =  \|A\|^n$. If we take the smallest such $\lambda(1)$, i.e., $\lambda(1) = \|\sigma^{-1/2} \rho \sigma^{-1/2}\|$ then,
$$
\lambda(n) = \|(\sigma^{\otimes n})^{-1/2} \rho^{\otimes n} (\sigma^{\otimes n})^{-1/2}\| = \|(\sigma^{-1/2} \rho \sigma^{-1/2})^{\otimes n}\| = \lambda(1)^n.
$$


问:
Interpolation at Tschebyschff nodes
We are given a function $f:[-1,1]\rightarrow \mathbb{R}$ and the Tschebyscheff nodes, $x_j=  $ cos$(\frac{2j+1}{2n}\pi), \,j=0, 1, ....n-1.$ We require that lim$_{n\to \infty} {\left\lVert f- p_n\right\rVert}_{C[-1,1]}=0,\,\,p_n$ being the interpolation polynomial related to the Tschebyscheff nodes. I need to state the conditions on $f$ for the requirement to be fulfilled. We know that the Weierstrass approximation theorem requires only continuity of $f$ to find a polynomial of degree $\leq n$ in order to achieve an approximation with a given error $\epsilon. $ What does the additional condition on Tschebyscheff nodes change in respect to the requirements on $f$ ? Is the continuity condition on $f$ sufficient ?

答:
Remember that the interpolation error at $x$ when interpolating at points $x_0, \dotsc, x_n$ in $[a, b]$ is given for some $\zeta \in [a, b]$ (depending on $x_0, \dotsc, x_n$ and $x$) by:

$\begin{align*}
  f(x) - p_n(x)
     &= \frac{1}{(n + 1)!} f^{(n+1)}(\zeta)
              \prod_{0 \le j \le n}(x - x_j)
\end{align*}$

(this assumes $f$ has enough derivatives). The Chebyshev nodes minimize the maximal value of the product.

Title: Interpolation at Tschebyschff nodes


问:
If $d(x,A)=0\forall x\in X$ for some subset $A$ of $X$, does it follow that $A$ is dense?
If $d(x,A)=0 \:\:\forall x\in X$ for some subset $A$ of $X$ then $A$ is dense in $X$, right?

Once I did one problem which says $d(x,A)=0\Leftrightarrow x\in \bar{A}$ so by the condition here we get $\bar{A}=X$.

Other options were

$2$. $A$ is closed

$3$. $A=X$

$4$. $A$ is compact

A counterexample for $2, 3$ and $4$ is $A=\mathbb{Q}\subseteq\mathbb{R}=X$

答: Yes, if all points of $X$ have distance $0$ to $A$ then $A$ is dense, because then all points are in the closure of $A$.</p>


问:
Convergence of 1 over sine serie, a hard one
I want to find the radius of this series, I tried my best I proved that r<1.
$$ \sum _ { n = 1 } ^ \infty \frac { x ^ n } { \sin ( n \theta ) } $$
is convergent for $ \theta = t \pi $ such that $ t $ is irrational, e.g.  $ \theta = \sqrt 3 \pi $.

It is a really interesting question because it will tell something about how the $ \sin \left( \sqrt 3 \pi n \right)$ goes to zero for some sequence goes to infinity (since $ \sqrt 3 $ is irrational sin will have a dense image in $ [ 0 , 1 ] $).P

答:
tl;dr

$r=1$ for $t=\sqrt3$ and pretty much any other number you've ever heard of.
Then there are just as many "bad" numbers for which $r<1$ and can be anything, all the way down to $0$.

Now come the details.

Roth's theorem basically says that an algebraic number $\alpha$ can't be approximated by the rationals "too well" - that is, $\left|\alpha-{\frac {p}{q}}\right|>\frac1{q^{2+\varepsilon}}$ for all but finitely many $\frac pq$. With that in mind, we have:

$$\left|{1\over\sin n\theta}\right|=\left|{1\over\sin(n\theta-p\pi)}\right| \leqslant{\pi/2\over|n\theta-p\pi|}={n\cdot const\over\left|{\theta\over\pi}-{p\over n}\right|}\leqslant n^{3+\varepsilon}\cdot const$$

and of course the series $\sum x^nn^4$ (here I chose $\varepsilon=1$) has the radius of convergence equal to $1$.

The same goes for all other numbers with finite irrationality measure.

Now the "bad" part. One can construct numbers with infinite irrationality measure, and they (though not even all of them) can make our radius a bit shorter. Say we have $x={1\over2}$; to break the convergence, we need $\left|{1\over\sin n\theta}\right|>2^n$ for infinitely many $n$. Can that be done? Why, that's easy. We know that for the continued fraction convergents of $\alpha$, $\left|\alpha-{p_n\over q_n}\right|<{1\over q_nq_{n+1}}$. Well, just make that $q_{n+1}$ bigger. How big can it be made? As big as you wish and even bigger. Can it be greater than $2^{q_n}$? Sure, and the extra topping is on the house.

So it goes.


问:
The rank of the power of a symmetric matrix

The rank of the power of a symmetric matrix.

The task is to show that $g(S) = S^3-S^2+S$ has the same rank as $S$. Here $S$ is a symmetric matrix.

I think, that the key to the solution of the problem is to show that the powers of a symmetric matrix $S$ have the same rank as $S$. I don't know how to show that. (Eigenvalue decomposition is not allowed).

答:
For real $S$.

$S^3-S^2+S=S(S^2-S+I)$ and so it suffices to prove that $S^2-S+I$ is invertible. Suppose, on the contrary, that $W=$ Kernel$(S^2-S+I)$ is non-empty.

$S$ is symmetric on $W$ and so has an eigenvector, $v\in W$. Let $S(v)=\lambda v$, where $\lambda $ is real. But then $(S^2-S+I)v=(\lambda^2-\lambda+1)v\ne 0$, a contradiction.


问:
Shortest time to travel around quicksand
You want to travel from one side of a quicksand $(1,0)$ pit to another side of the quicksand pit $(-1,0)$. The speed you can run is determined by how far you are away from the quicksand $|v(x,y)| = \sqrt{x^2+y^2}$. What is the optimal path you should travel such that the time taken to travel is minimized?

In my attempted solution:

Use cylindrical coordinates $(r,\theta)$ and find an expression for the total time taken $T$

Attempt to minimize $T$ using Beltrami's identity

Solve the resulting differential equation to find $r(\theta)$

However, when I do this, I find $r = c_2 e^{c_1 \theta}$. This is strange because I would have expected a symmetrical path around the y axis. Also, when I plug in the start and end points I get nonsensical values for the constants $c_1$ and $c_2$. Can someone help me figure out what I'm doing wrong?

答:
From

$$
T = \int_{0}^{\pi}\sqrt{1+\left(\frac{r'(\theta)}{r(\theta)}\right)^2}d\theta
$$

we obtain the Euler-Lagrange  equations

$$
r(\theta ) r''(\theta )-r'(\theta )^2=0
$$

with solution

$$
r(\theta) = c_2e^{c_1\theta}
$$

now applying the boundary conditions

$$
\cases{
r(0) = c_2 = 1\\
r(\pi) = e^{\pi c_1}= 1
}
$$

we determine $c_1 = 0, c_2 = 1$ so the minimum time orbit is the semi-circumference with radius $r = 1$


问:
Conic equation: ${4x^{2}−4xy+7y^{2}+12x+6y−9=0}$ Converting this conic to "standard form"
This is a conic equation:

${4x^{2}−4xy+7y^{2}+12x+6y−9=0}$

I tried a way like this:

1.)${B^{2}-4AC <0} --> ((-4)^2-4x4x7 = 16-112 = -96 < 0)$ "So this is an ellipse"

2.)${Tan2θ = \frac{B}{A-C} = \frac{4}{3} ->θ = 27}$ "degree"

3.) (I saw them using these formulas on a site) There are formulas for ${A^{'}, B^{'}, C^{'}, D^{'}, E^{'}, F^{'}}$ Taking B = 0 to get rid of the xy expression, and the others are:

${A^{'} = Acos^{2}θ+Bcosθsinθ+Csin^{2}θ}$ Plug in the angle "θ" 27 degrees and A 'is about 2,96.

${B^{'} = 0 }$

${C^{'} = Asin^{2}θ+Bsinθcosθ+Ccos^{2}θ}$ Plug in the angle "θ" 27 degrees and A 'is about 4,73

${D^{'} = Dcosθ+Esinθ}$ Plug in the angle "θ" 27 degrees and A 'is about 13,38

${E^{'} = -Dsinθ+Ecosθ}$ Plug in the angle "θ" 27 degrees and A 'is about -0,06

${F^{'} = F = -9}$

4.)Since this is an ellipse according to 1, the standard form of the general equation given is:

${\frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1}$

it must be.

Finally,
I've got this:

${2,96x^{2}+4,73y^{2}+13,38x-0,06y=9}$

Well, what should I do now? As a result of my various researches, there are completely different solutions and I am confused.

I request your help

答:
Let $2x-y=z$, then the given equation becomes:
\begin{align}
&z^2+6y^2+6z+12y-9=0\\
\implies &(z+3)^2+6(y+1)^2-9-6-9=0\\
\implies &(z+3)^2+6(y+1)^2=24\\
\implies&\boxed{\dfrac{(2x-y+3)^2}{24}+\dfrac{(y+1)^2}{4}=1}
\end{align}
This is the required standard form (note that the major axis is not parallel to the $x$-axis in this case).


问:
How compute overall reliability based on two related (dependent) inter-rater reliability coefficients

The Situation

In many disability programs, a physician, psychologist, or other health professional conducts an exam with the claimant. Although it's not easy, a research study could determine an inter-rater reliability estimate1 for these medical2 examinations.

After the medical exam, lay3 adjudicators review all relevant records, including questionnaires and statements from the claimant, along with the medical examiner's report, and apply regulatory standards to reach an administrative/legal decision regarding the disability claim. Researchers could conduct a study to estimate the inter-rater reliability for these adjudicative decisions.

As you can see, the medical examiner's report influences the adjudicative decision, at least to some extent. (In most systems, the examiner's report significantly influences adjudicative decisions.) Thus, the two inter-rater reliability estimates are not independent.

My Question

What is/are the best way(s) to calculate the overall reliability of such a disability determination process?

Brief Background

I'm asking this question because some organizations conduct research on the inter-rater reliability of the adjudicative decisions and describe the results as "the accuracy of our disability determination decisions". The unspoken assumption is that the inter-rater reliability of the medical examinations is 1.00.

Notes

I wasn't sure if "coefficient" would be a better term than "estimate".

Most disability programs refer to all exams as "medical" even though some are actually psychological, audiological, etc.

lay, adj. - Not of or belonging to a particular profession; nonprofessional.

Prior research

I searched SE-Math but found just one related thread: combined reliability

I reviewed the following texts/tutorials (not that I couldn't have missed something relevant).

Gwet, Kilem L. Handbook of Inter-Rater Reliability. 4th ed. Gaithersburg, MD: Advanced Analytics, 2014.

Khan Academy, Statistics and probability.

Stat Trek, Statistics and Probability.

Trochim, William M., James P. Donnelly, and Kanika Arora. Research Methods: The Essential Knowledge Base. 2nd ed. Boston: Cengage Learning, 2016.

答:
I'm not a statistician, but here are some starting thoughts. I've made this into an answer rather than a comment due to its length.

Before you calculate anything, you need to model the situation. There is rarely a single, god-given model. Modeling requires making choices, and these choices are informed by the experience of subject experts.

Modeling I: Patients

You want to understand the inter-reliability of medical and adjucative examinations. It seems likely that this reliability is a function of the underlying medical condition,  as well as a host of other socio-economic factors. If you don't include this information in your model, you will get average estimates of inter-reliability that don't take into account the variance seen in different patient groups. It's up to you to decide what level of granularity to adopt.

Modeling II: Evaluations

Just as there are different ways of modeling the patient pool, there are different ways of modeling the evaluations. One possibility is binary: approve or deny claim. Binary outcomes make the most sense for the legal decisions. However, I'm imagining that the medical examiner's report contains more information, perhaps a value between $0$ and $5$ indicating the severity of the condition or their conviction for requesting approval.

Modeling III: Connections

What is the relationship between healthcare professionals and legal adjucators? Does an adjucator treat every healthcare professional's opinion equally? How many healthcare professionals does a given claimant see? Are they random?

A Simple Model

A model that is too complicated is impossible to use in practice, due to the lack of infinite data. Let me suggest the following simple model. Let $P$ be the set of patients. $P$ could contain all patients, or could be narrowed down to a group of patients with similar characteristics. Assuming every patient sees $d$ doctors, and each doctor gives a "yes-no" answer, we have a function $f: P \to \{0, 1, \cdots, d\}$ that records the total number of "approve" responses. Assume every patient has $n$ adjucators, we then have a function $g: \{0, 1, \cdots, d\} \to \{0, 1, \cdots, n\}$ that takes the total number of medical "approve" responses and outputs the total number of adjucator "approve" responses.

If you look at the variance of the function $g \circ f$, you get the total variance of the process over the set $P$. If you look at the variance of $f$, you get the total variance of the medical evaluations. Lastly, to get the variance of the legal adjucators for a fixed medical assessment, you need to take the variances of $g$ conditioned on particular outputs of $f$ (you can then average these quantities).

To my mind, variance is a good statistic to measure reliability. If the collections of patients in $P$ are all similar, an assumption of reliability would predict a fairly consistent medical and legal evaluation. The larger the variance of the various functions described in the prior paragraph, the more disagreement between medical and legal experts. Assuming patient homogeneity, this implies a lack of reliability.


问:
Determining the minimum degree if $δ(G) \leq k, k \in N^* $
Let be $G = (V,E)$ a graph with $n$ nodes and $δ(G)  \leq k, k \in N^* $.
Describe an algorithm with complexity $O (n) $ for determining the minimum degree of this graph.

Please help with some hints or solutions.
Thanks.

答:
Def. 1: Let $G=(V,E)$ with incidence mapping $p$ where $V=\{x_1,\cdots,x_n\}$ is a graph. The $n \times n$ matrix $A(G)=(a_{ij})_{i,j=1,\cdots,n}$ defined by
$$a_{ij}:=|\{e \in E\ |\ p(e)=(x_i,x_j) \}|$$
is called the adjacency matrix of $G$.

If $A(G)$ is symmetric, then $G$ is undirected. If $G$ is simple, i.e. no multiple edges, then we can define $A(G)$ by
$$a_{ij} :=
     \left\{\begin{array}{ll} 1, & \text{if } (x_i,x_j) \in E, \\
         0, & \text{otherwise}\end{array}\right.
  $$

Proposition 1: $\forall x_i \in V$ with $A(G)=(a_{ij})_{i,j\in|V|=n}$, we have
$$indeg(x_i)=\sum_{j=1}^n a_{ji}$$

$$outdeg(x_i)= \sum_{j=1}^n a_{ij}$$

whereby $indeg$ is the in-degree and $outdeg$ is the outdegree. In the symmetric case, we of course have

$$deg(x_i)=\sum_{j=1}^n a_{ij}=\sum_{j=1}^n a_{ji}$$

Hint: Now, implement an algorithm building the degree sequence of a given graph using its adjacency matrix. What is its running time complexity?


问:
Could someone clarify this picture in Spivak, illustrating $|x - a| < \frac{|a|}{2}$?

It's not really clear to me what the picture is trying to illustrate. If we mark $a$ to be the largest vertical bar, then how many of these unit bars is $a$ supposed to be? $4$?

答:
The largest (middle) vertical bar is $0$ and each unit bar has length $\dfrac{|a|}2$.

If $a>0$, $a$ is the second vertical bar on the right starting from the middle.

If $a<0$, $a$ is the second vertical bar on the left starting from the middle.

The values that $x$ can take is contained in one of the greyed areas, depending on the sign of $a$.


问:
Why might one be inclined to think that polynomials of the form $\cos(n\arccos{x})$ would minimize error in Lagrange interpolation?
I was first introduced to Chebyshev polynomials (of the first kind) in the form $T_n(x)=\cos\left(n \operatorname{arccos}(x)\right)$. The usual recurrence relation was then derived from using trig identities.

The purpose of Chebyshev polynomials, in this context, is that their roots constitute the best set of points to use in Lagrange interpolation, i.e. they minimize the error function $E(x;f)\triangleq f(x)-P_n(x)$.  I am left with an unsatisfied feeling, however, as I realize that although I can verify that this works (and the proof is not hard), I have no idea why Chebyshev might have thought to look at these particular polynomials in the first place to solve this particular problem.

Why might one be inclined to think that polynomials of the form $\cos\left(n \operatorname{arccos}(x)\right)$ would provide the best set of points for Lagrange interpolation?

答:
The error interpolating over $[a, b]$ by a polynomial $p_n$ of degree $n$ at points $x_0, \dotsc, x_n$ is, for some $\zeta \in [a, b]$ (depending on $x, x_0, \dotsc, x_n$) is:

$\begin{align*}
   f(x) - p_n(x)
     &= \frac{1}{(n + 1)!} f^{(n+1)}(\zeta)
              \prod_{0 \le j \le n}(x - x_j)
\end{align*}$

The Chebyshev polynomial $T_n(x)$ is the monic polynomial that minimizes the maximal value of $\lvert T_n(x) \rvert$  over $[-1, 1]$, and thus minimizes the product in the error.


问:
Complex Exponential Expansion
I'm currently working through a problem using exponentials with complex numbers!
I am simply wondering what the steps are to go from

$$\pi \cdot (e^{i\frac{\pi}{3}} - e^{i\pi})  = \frac{\pi}{2}\cdot(3 + i\sqrt{3})$$

If someone could point me in the direction of the correct expansion that would be very helpful!

Thank you

Title: Complex Exponential Expansion

Body: Hint: Use $e^{i\theta}=cos(\theta)+i\sin(\theta)$.


问:
Prove that the unitary PCP is decidable.

Prove that the unitary PCP (Post Correspondence Problem where $\vert \Sigma \vert = 1$) is decidable.

Case $1$:
If a tuple $(x_i, y_i)$ with $\vert x_i\vert = \vert y_i\vert$ exists, then the answer is "yes" and the proof is trivial.

Case $2$:
If a tuple $(x_i, y_i)$ with $\vert x_i\vert > \vert y_i\vert$ and a tuple $(x_j, y_j)$ with $\vert x_j\vert < \vert y_j\vert$ exists, then the answer is "yes". Let $a:= \vert x_i\vert - \vert y_i\vert$ and $b:= \vert y_j\vert - \vert x_j\vert$. Then you get:

\begin{align}
\vert x_i^b \cdot x_j^a\vert &= b \cdot \vert x_i\vert + a \cdot \vert x_j\vert \\
&= (\vert y_j\vert - \vert x_j\vert) \cdot \vert x_i\vert + (\vert x_i\vert - \vert y_i\vert) \cdot \vert x_j\vert \\
&= \vert y_i\vert \cdot \vert x_i\vert - \vert x_j\vert \cdot \vert y_i\vert \\
&= (\vert y_j\vert - \vert x_j\vert) \cdot \vert y_i \vert + (\vert x_i\vert - \vert y_i\vert) \cdot \vert y_j \vert \\
&= b \cdot \vert y_i \vert + a \cdot \vert y_j \vert \\
&= \vert y_i^b \cdot y_j^a\vert
\end{align}

Case $3$: If every tuple we have that $\vert x_i \vert > \vert y_i\vert$ or $\vert x_i \vert < \vert y_i\vert$, then the answer is "no" because then one of the strings is always longer than the other and there will be no solution.

I am having trouble with case $2$ because I cannot seem to understand how we went from the left to the right side.

答:
We need to show that $|x_i^bx_j^a| = |y_i^by_j^a|$.

Here, $s_1s_2$ means concatenation of the two strings $s_1$ and $s_2$, and $s^k$ means concatenating $k$ times the string $s$ with itself: $\underbrace{ss\ldots s}_{k~\text{times}}$

Also, keep in mind that $(*)~a=|x_i|−|y_i|~~\text{and}~~b=|y_j|−|x_j|$.

So, we have:
\begin{align}
|x_i^bx_j^a| &= |x_i^b| + |x_j^a| & \text{(string concatenation)}\\
&=b|x_i|+ a|x_j| &\text{(string concatenation)}\\
&=(|y_j|−|x_j|)|x_i|+ (|x_i|−|y_i|)|x_j| &\text{(using the equations in}~(*))\\
&= |y_j||x_i| - |y_i||x_j| &\text{(distributivity and deletion of opposites)}\\
&= |y_j||x_i| - |y_i||y_j| + |y_i||y_j| - |y_i||x_j| &\text{(introduction of opposites)}\\
&= (|x_i| - |y_i|)|y_j| + (|y_j| - |x_j|)|y_i|&\text{(factorization)}\\
&= a|y_j| + b|y_i| &\text{(using the equations in}~(*))\\
&= |y_j^a| + |y_i^b| = |y_j^ay_i^b| & \text{(string concatenation)}
\end{align}


问:
Sum of square roots inequality

For all $a, b, c, d > 0$, prove that
$$2\sqrt{a+b+c+d} ≥ \sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$$

The idea would be to use AM-GM, but $\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$ is hard to expand. I also tried squaring both sides, but that hasn't worked either. Using two terms at a time doesn't really work as well. How can I solve this question? Any help is appreciated.

答:
Alternative solution using Cauchy-Schwarz, which finishes the problem off immediately. By C-S, we have:
\begin{align}
& (a+b+c+d)(1+1+1+1) \geq (\sqrt{a}+ \sqrt{b}+ \sqrt{c}+ \sqrt{d} )^2 \\
& \Rightarrow 4(a+b+c+d) \geq (\sqrt{a}+ \sqrt{b}+ \sqrt{c}+ \sqrt{d} )^2  \\
& \Rightarrow 2\sqrt{a+b+c+d} \geq \sqrt{a}+ \sqrt{b}+ \sqrt{c}+ \sqrt{d}.
\end{align}

Title: Sum of square roots inequality


问:
What is the product of category "2" by itself?
Saunders Mac Lane, at page 10 from "Categories for the Working Mathematician" names $\mathbb{2}$ the category object of a previous question of mine:

$\mathbb{2}$ is the category $↻ \to ↻$ with two objects $a$, $b$, and just one arrow $a \to b$ not the identity;

What is $\mathbb{2} \times \mathbb{2}$ like?

My understanding is that it should contain

4 objects, $(a,a)$, $(a,b)$, $(b,a)$, $(b,b)$,

some morphisms. But which morphisms?

At least one identity morphism for each of the 4 objects, and they all come from combinations of $id_a$ and $id_b$; one of them, for instance, is $id_{(a,b)} = (id_a,id_b)$.

What else? Having called $f$ the only non-identity morphism in $\mathbb{2}$, $f: a \to b$, I can only think of other 5 morphisms:

$f_1 = (f,f) : (a,a) \to (b,b)$,

$f_2 = (id,f) : (a,a) \to (a,b)$

$f_3 = (f,id) : (a,a) \to (b,a)$

$f_4 = (id,f) : (b,a) \to (b,b)$

$f_5 = (f,id) : (a,b) \to (b,b)$

Is this all? Please also correct me if I've abused the notation.

答: An arrow in the product category is a pair of arrows, one from each factor category (that is, $\text{Arr}(C\times D) = \text{Arr}(C)\times \text{Arr}(D)$). Since $\mathbb{2}$ has $3$ arrows, $\mathbb{2}\times\mathbb{2}$ has $3^2=9$ arrows. So yes, you found them all.


问:
closed subspace in $W[0,1]$
Let $W[0,1]$ be the space of all continuously differentiable functions on $[0,1]$ with values in $\mathbb{C}$, with the following inner product

$(f,g)_W =\int_0^1 {f(t)\overline{g(t)}+f'(t)\overline{g'(t)}}dt$.

We need to prove that $(f,\cosh)_W=f(1)\sinh1$ and then use this fact to prove that the subspace $\{f \in W[0,1]: f(1)=0 \}$ is $\textbf{closed}$.

I was able to prove the identity, but I couldn't find  how can I apply it to prove the subspace is closed.

I know it is closed if it contains all its limits points. But couldn't start the proof.

答:
Hint: rewriting the identity you proved, we have

$$
f(1) = \left(f,\frac{\cosh}{\sinh(1)}\right)_W,
$$
and in general for any $u$ we know that $(-,u)_W$ is a linear operator, continuous with respect to the topology induced by $(-,-)_W$ (in fact, Riesz's representation theorem tells us all continuous functionals are of this form).

 The subspace of interest is $\ker(ev_1)$; hence its closedness is equivalent to the continuity of $ev_1$, as the latter is a functional on $W[0,1]$. But this stems from the fact that $ev_1$ is of the form $(-,u)_W$ with $u  = \frac{\cosh}{\sinh(1)}$.

Title: closed subspace in $W[0,1]$


问:
How to constructing a binary regular expression for 8 bits, where "1 1 1" occurs twice in any configuration and the other two bits can be whatever?
I'm trying to make a regex for my language with alphabet $\Sigma = \{"0","1"\}$ that captures any 8 bits where the sequence "1 1 1" occurs twice without overlap. These two sequences can be adjacent, but don't have to be. The remaining two bits can be either "0" or "1". It doesn't matter what they are, but it needs to be possible for them to be anything.

Mainly the length limitation is what I'm struggling with. If the length was irrelevant it could have easily been
$$ \Sigma^* \circ (111) \circ \Sigma^* \circ (111) \circ \Sigma^* $$
since $\Sigma^*$ includes empty. However, the size of each realization $\Sigma^*$ depends on the realizations of the others. I have no clue how to capture this in the expression.

Any help would be greatly appreciated!

答:
As $111$ is together, we can consider this as $4$ characters of two {111} and other two as $1$ or $0$.

If other two are both $0$, we have $4 \choose 2$ ways to place two {111} them in $4$ spaces and other two places are then taken by $0$.

If other two characters are both $1$ then there is only one way to arrange as all are $1$.

If other two characters are one $1$ and one $0$, we again have $4 \choose 2$ ways to arrange. If you think of total arrangements, it is $12$ but half of them are duplicates.

So in total, we have $13$ distinct arrangements.


问:
Is the set of essentially bounded and not-necessarily-measurable functions a Banach space?
Suppose that $(\Omega, \mathcal{F}, \mu)$ is a probability space and for $f : \Omega \to \mathbb{R}$ define
$$\DeclareMathOperator{esssup}{ess\,sup}
\esssup f = \inf_{A : \mu(A) = 0} \sup_{x \notin A} f(x).
$$
To me it seems that a short elementary argument shows that for measurable $f$ this agrees with the usual definition of the essential supremum, yet there is nothing in the definition that requires $f$ to be measurable.

Is it true that the set of equivalences classes of a.e. equal, bounded $\mathbb{R}$-valued functions on $\Omega$ is a Banach space when equipped with the norm $\|f\| = \esssup |f|$?

答:
The set $B$ of ALL bounded functions is a Banach space for the sup norm.  The subset $N$ formed by the functions vanishing off a null set is a closed subspace.  Your space is exactly the quotient space $B/N$ with the quotient norm, hence, yes, it is a Banach space.

The reason it isn't studied in the literature is that there is not much one can say about it.  In particular there is no well defined linear functional given by integration.


问:
Let $X$ be a set of $5$ elements. Find the number of ordered pairs $(A,B)$ of subsets of $X$ such that $A, B\neq \emptyset$ and $A\cap B=\emptyset$
I know this question involves use of combinatorics, but all the solutions I have seen just give the equation without explaining it. Can I get an explanation, because I know the answer and I know the method, I just don’t know how to apply.</p>

答:
Let us first find all possibilities such that $A \cap B = \varnothing$.

To create such a pair, each element in $X$ has exactly $3$ possibilities: It is in only $A$, or only $B$, or in neither.

This gives you that there are $3^5 = 243$ ordered pairs $(A, B)$ such that $A \cap B = \varnothing$.

Now, we count the possibilities where one of them is empty. If $A = \varnothing$, then there are exactly $2^5$ possibilities for $B$. ($2^5$ is the total number of subsets of $X$.)

Similarly, if $B = \varnothing$, then there are $2^5$ choices for $A$.

Thus, in total there are $2^5 + 2^5 - 1 = 63$ cases where $A = \varnothing$ or $B = \varnothing$. (The $-1$ is for the double counting of the case $A = B = \varnothing$.)

Thus, the desired answer is $243 - 63 = \boxed{180}.$


问:
Spivak Calculus Chapter 5 Problem 10 (b)
I am unsure about how to solve this problem and unfortunately I do not understand the given solution. The problem goes as follows (Spivak, Calculus, 3rd edition, pg. 107, Problem 10-(b)):

Prove that  $\lim\limits_{x \to 0} f(x)=\lim\limits_{x \to a} f(x-a)$.

My attempt:

"$\Rightarrow$": Let $\lim\limits_{x \to 0} f(x)=m$. Then for every $\varepsilon>0$ there is a $\delta>0$ such that, for all $x$, if $0<\lvert x-0\rvert<\delta$, then $\lvert f(x)-m\rvert<\varepsilon$. Now let $0<\lvert (x-a)-0\rvert<\delta$ (which can als be written $0<\lvert x-a\rvert<\delta$), then $\lvert f(x-a)-m\rvert<\varepsilon$. Thus $\lim\limits_{x \to a} f(x-a)=m$.

"$\Leftarrow$": Let $\lim\limits_{x \to a} f(x-a)=m$. Then for every $\varepsilon>0$ there is a $\delta>0$ such that, for all $x$, if $0<\lvert x-a\rvert<\delta$, then $\lvert f(x-a)-m\rvert<\varepsilon$. Now let $0<\lvert y\rvert<\delta$. ($0<\lvert y-0\rvert<\delta$.) This inequality can be written $0<\lvert (y+a)-a\rvert<\delta$. It follows that $\lvert f((y+a)-a)-m\rvert<\varepsilon$, which is $\lvert f(y)-m\rvert<\varepsilon$. Thus $\lim\limits_{y \to 0} f(y)=m$ (which is equivalent to $\lim\limits_{x \to 0} f(x)=m$).

Spivak's solution:

Suppose that $\lim\limits_{x \to a} f(x)=l$, and let $g(x)=f(x-a)$. Then for all $\varepsilon>0$ there is a $\delta>0$ such that, for all $x$, if $0<\lvert x-a\rvert<\delta$, then $\lvert f(x)-l\rvert<\varepsilon$. Now, if $0<\lvert y\rvert<\delta$, then $0<\lvert (y+a)-a\rvert<\delta$, so $\lvert f(y+a)-l\rvert<\varepsilon$. But this last inequality can be written $\lvert g(y)-l\rvert<\varepsilon$. So $\lim\limits_{y \to 0} g(x)=l$. The argument in the reverse direction is similar.

My troubles:
Why does he start with $\lim\limits_{x \to a} f(x)=l$? Why can $\lvert f(y+a)-l\rvert<\varepsilon$ be written $\lvert g(y)-l\rvert<\varepsilon$? (If $g(x)=f(x-a)$, then shouldn't $f(y+a)=g(y+2a)$?) It looks like he is trying to prove $\lim\limits_{x \to a} f(x)=\lim\limits_{x \to 0} f(x-a)$, which is very weird...

答:
Indeed, Spivak makes a mistake in his argument: the inequality
$$
|f(y+a)-l|<\epsilon
$$
can not be written as
$$
|g(y)-l|<\epsilon
$$
The definition of $g$ is $g(y)=f(y-a)$, not $g(y)=f(y+a)$.

This error is also in the fourth edition of his book.

Once you understand the underlying ideas and can write down a formal proof correctly, the "solution" is not so important. In textbooks, authors occasionally make mistakes; sometimes, when they make an argument $X$ to show statement $Y$, they may really mean to write argument $X'$.

The spirit of the proof for this problem is in Spivak's comment in his solution:

Intuitively, making $x$ close to $a$ is the same as making $x-a$ close to $0$.

Your attempt is correct. I may change your wording slightly though.

Suppose $\lim_{x\to 0}f(x)=m$. We want to show that $\lim_{x\to a}f(x-a)=m$. Let $\epsilon>0$. There exists $\delta>0$ such that
$$
\textrm{for every $x$ with  } 0<|x-0|<\delta,\quad|f(x)-m|<\epsilon. \tag{1}
$$
If $0<|y-a|=|(y-a)-0|<\delta$, then (1) implies that $|f(y-a)-m|<\epsilon$. Thus
$
\lim_{y\to a}f(y-a)=m,
$
i.e.
$
\lim_{x\to a}f(x-a)=m.
$

Conversely, suppose $
\lim_{x\to a}f(x-a)=m.
$
We want to show that $\lim_{x\to 0}f(x)=m$. Let $\epsilon>0$. There exists $\delta>0$ such that
$$
\textrm{for every $x$ with  } 0<|x-a|<\delta,\quad|f(x-a)-m|<\epsilon. \tag{2}
$$
If $0<|y|=|(y+a)-a|<\delta$, then (2) implies that $|f(y)-m|=|f((y+a)-a)-m|<\epsilon$. Hence $\lim_{y\to 0}f(y)=m$.


答:
Let $f(x,y) =4x^{2}−4xy+7y^{2}+12x+6y−9$. The center of the ellipse is $(-2,-1)$, obtained from $f’_x=f’_y=0$. Then, its major and minor axes can  be obtained as follows

$$\frac{f’_y}{f’_x}= \frac{y-(-1)}{x-(-2)}\implies
y= \frac12x,\>\>\>y =-2x-5$$
They are at an angle $\theta=\tan^{-1} \frac12 $ with the $x$-axis. Thus, with $\cos\theta =\frac2{\sqrt5}$ and $\sin\theta =\frac1{\sqrt5}$, apply the transformation below
$$ x=-2+\frac2{\sqrt5} u + \frac1{\sqrt5} v,\>\>\>\>\>
 y=-1+\frac1{\sqrt5} u - \frac2{\sqrt5} v$$
to convert $f(x,y)=0$ to the standard form
$$ \frac{u^2}8 +\frac{v^2}3=1$$

Title: Conic equation: ${4x^{2}−4xy+7y^{2}+12x+6y−9=0}$ Converting this conic to "standard form"

Body:
This is a conic equation:

${4x^{2}−4xy+7y^{2}+12x+6y−9=0}$

I tried a way like this:

1.)${B^{2}-4AC <0} --> ((-4)^2-4x4x7 = 16-112 = -96 < 0)$ "So this is an ellipse"

2.)${Tan2θ = \frac{B}{A-C} = \frac{4}{3} ->θ = 27}$ "degree"

3.) (I saw them using these formulas on a site) There are formulas for ${A^{'}, B^{'}, C^{'}, D^{'}, E^{'}, F^{'}}$ Taking B = 0 to get rid of the xy expression, and the others are:

${A^{'} = Acos^{2}θ+Bcosθsinθ+Csin^{2}θ}$ Plug in the angle "θ" 27 degrees and A 'is about 2,96.

${B^{'} = 0 }$

${C^{'} = Asin^{2}θ+Bsinθcosθ+Ccos^{2}θ}$ Plug in the angle "θ" 27 degrees and A 'is about 4,73

${D^{'} = Dcosθ+Esinθ}$ Plug in the angle "θ" 27 degrees and A 'is about 13,38

${E^{'} = -Dsinθ+Ecosθ}$ Plug in the angle "θ" 27 degrees and A 'is about -0,06

${F^{'} = F = -9}$

4.)Since this is an ellipse according to 1, the standard form of the general equation given is:

${\frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1}$

it must be.

Finally,
I've got this:

${2,96x^{2}+4,73y^{2}+13,38x-0,06y=9}$

Well, what should I do now? As a result of my various researches, there are completely different solutions and I am confused.

I request your help


答:
If $|A| = k>0$ then you have ${5\choose k}$ possibilities for $A$ and $2^{5-k}-1$ possibilities for $B$ since $B\subseteq X\setminus A$ and $B\ne \emptyset$. So you have to sum (use binomial theorem): \begin{align}\sum _{k=1}^5{5\choose k}(2^{5-k}-1) &= \color{red}{\sum _{k=1}^5{5\choose k}2^{5-k}}
-\color{blue}{\sum _{k=1}^5{5\choose k}}\\
&=\color{red}{(2+1)^5-2^5} -\color{blue}{(1+1)^5+1} \\&= 3^5-2^6+1\\ &= 243-64+1 =180 \end{align}

Title: Let $X$ be a set of $5$ elements. Find the number of ordered pairs $(A,B)$ of subsets of $X$ such that $A, B\neq \emptyset$ and $A\cap B=\emptyset$

Body: I know this question involves use of combinatorics, but all the solutions I have seen just give the equation without explaining it. Can I get an explanation, because I know the answer and I know the method, I just don’t know how to apply.</p>


问:
Definition of Frobenius automorphism
I quote the following problem from the chapter Hilbert's Ramification Theory of Jurgen Neukirch

Let $L/K$ is a Galois extension with prime ideal $\mathfrak{P}$, unramified over $K$, then there is one and only one automorphism $\Phi_{\mathfrak{P}}\in\operatorname{Gal}L/K$, s.t. $\Phi_{\mathfrak{P}}(a)\equiv a^q \pmod{\mathfrak{P}}$, where $q$ is the size of the residue field $O_K/\mathfrak{P}\cap O_K$. It is called the Frobenius Automorphism.

Now, if we take $K=\mathbb{Q}$ and $p=\mathfrak{P}\cap \mathbb{Z}$, I know that the residue field $O_L/\mathfrak{P}$ is a finite extension over $\mathbb{F}_p$,  so, $a\mapsto a^p$ is a frobenius automorphism in $O_L/\mathfrak{P}$ and it generates all the other frobenius automorphisms. But the definition of the automorphism from the exercise seems different. Am I missing anything? How can it be a frobenius automorphism? The frobenius automorphism in the sense of finite field is not matching with this.

答:
For a number field $L$ and a prime ideal $\mathfrak{P}\subset O_L$ and $L/K$ Galois. There is some automorphism $\in Gal(L/K)$ whose reduction is the Frobenius. If the extension is unramified at $\mathfrak{P}$ then it is unique and we can call it the Frobenius.

It is not true that $q$ is the degree of the extension. It is the cardinality of $O_K/(\mathfrak{P}\cap O_K)$, which is also $p^f$ where $f=[O_K/(\mathfrak{P}\cap O_K):\Bbb{F}_p]$ and $p$ is the characteristic. The order of the Frobenius (here I assume the extension is unramified at $\mathfrak{P}$) is $[O_L/\mathfrak{P}:O_K/(\mathfrak{P}\cap O_K)]$.

Title: Definition of Frobenius automorphism


问:
How can i find residues for this function?
I have a contour integral $\int_{0}^{\infty} \frac{x^3}{x^5-a^5}dx $ where $ a>0$
Here I tried to find residue, however I failed since the roots are $z=a,-(-1)^{1/5},(-1)^{2/5},-(-1)^{3/5},(-1)^{4/5}$
It's not possible to calculate $\lim_{x->x_0} (x-x_0)\frac{x^3}{(x-a)(x^4+x^3a+x^2a^2+xa^3+a^4)}$

So is there any suggested method to calculate this integral with contour integral method?

答: The roots are $ae^{2\pi i/5}$, $ae^{4\pi i/5}$, $ae^{6\pi i/5}$, and $ae^{8\pi i/5}$. If $r$ is any of them, then it is a simple root of the denominator, and therefore\begin{align}\operatorname{res}_{z=r}\left(\frac{z^3}{z^5-a^5}\right)&=\frac{r^3}{5r^4}\\&=\frac1{5r}.\end{align}


问:
Is there a way to exactly solve this integral?
Is there a way to determine $\theta_c$ in the following integral equation:

$$\int_{-\theta_c}^{\theta_c} d\theta \, \exp (a \cos\theta)=1,$$

where $a$ belongs to positive reals?

答:
For $a>0$, the integrand has a maximum at $\theta=0$. Writing the series for $\cos$ around zero, we find$^\dagger$

$$ e^a \int_{-\theta_0}^{\theta_0} d\theta \ \exp( -a \theta^2 /2)=e^a \sqrt{\frac{2\pi}{a}} \operatorname{erf}\left(\theta_0 \sqrt{\frac{a}{2}} \right)=1$$

Rearranging

$$\operatorname{erf}\left(\theta_0 \sqrt{\frac{a}{2}} \right)=e^{-a} \sqrt{\frac{a}{2\pi}} $$

Apply inverse $\operatorname{erf}$

$$ \theta_0= \sqrt{\frac{2}{a}} \operatorname{erf}^{-1} \left(e^{-a} \sqrt{\frac{a}{2\pi}}\right) $$

The argument of $\operatorname{erf}^{-1}$ above goes to zero exponentially fast as $a\rightarrow\infty$. The series for $\operatorname{erf}^{-1}(z)$ around $z=0$ is

$$ \operatorname{erf}^{-1}(z)=\frac{\sqrt{\pi}}{2}z+O(z^2) ,\  z\rightarrow0$$

We are left with

$$ \theta_0 \sim \frac{e^{-a}}{2} \ \ , \ \ a\rightarrow\infty$$

Here is a plot of the original integral: $\int_{-\theta_0}^{\theta_0}d\theta \exp(a \cos\theta)$, numerically evaluated with $\theta_0=e^{-a}/2$. We see that 'large $a$' means modest values of $a>1$.

$\dagger$ The justification of what we are doing is the same as the justification for Laplace's method.


问:
Is polynomial algebra the only minimal dense subalgebra of $C^{\infty}(U)$
Let $U$ be connected open set of $\mathbb{R}^{n}$. Consider $C^{\infty}(U)$ with sup norm, we said $A\subset C^{\infty}(U)$ is a minimal dense subalgebra of $C^{\infty}(U)$ if and only if any subalgebra contain in $A$ is not dense in $C^{\infty}(U)$. I want to ask are the polynomial algebra of $U$, either with coefficients in $\mathbb{Q}$ or $\mathbb{R}$\ $\mathbb{Q}$, the only two minimal dense subalgebra of $C^{\infty}(U)$?

答:
In orded for the sup norm  to be well defined for polinomials one needs to assume that $U$ is bounded.  On the other hand the connectedness of $U$ seems irrelevant here.

In that case, the algebra of polynomials is dense in the sup norm by Stone-Weierstrass' Theorem, but it is not minimal.  The reason is that, for every odd number $k$, the set of polynomials of the form
$$
  p(x) = \sum_{n=0}^d a_n x^{kn},
  $$
is dense (again by Stone-Weierstrass) and it is a proper subalgebra of the polinomial algebra (this works also if $k$ is even and $U\subseteq (0,+\infty)$).


问:
Is $f:\mathbb{R}\rightarrow\mathbb{R}, f(x)=\left \lfloor{x}\right \rfloor -x$ a Darboux function?
Let $f:\mathbb{R}\rightarrow\mathbb{R}, f(x)=\left \lfloor{x}\right \rfloor -x$

My book claims this function is not a Darboux function.

Is this true ? I cannot find a counterexample to prove that this is not a Darboux function.

Thanks in advance !

答: On the interval $(\frac{1}{2}, 1]$ the image of your function is equivalently $-x$ away from $1$ and $f(1) = 0$.  Thus the image of this interval under $f$ is $(-1, -\frac{1}{2}) \cup \lbrace 0 \rbrace$ which is not connected, but the interval $(\frac{1}{2}, 1]$ is connected, contradicting the Darboux property.


Title: How can i find residues for this function?

Body:
I have a contour integral $\int_{0}^{\infty} \frac{x^3}{x^5-a^5}dx $ where $ a>0$
Here I tried to find residue, however I failed since the roots are $z=a,-(-1)^{1/5},(-1)^{2/5},-(-1)^{3/5},(-1)^{4/5}$
It's not possible to calculate $\lim_{x->x_0} (x-x_0)\frac{x^3}{(x-a)(x^4+x^3a+x^2a^2+xa^3+a^4)}$

So is there any suggested method to calculate this integral with contour integral method?

Body:
Solutions of $x^5=a^5$ are

$a,-\sqrt[5]{-1} a,(-1)^{2/5} a,-(-1)^{3/5} a,(-1)^{4/5} a$

Residue at $x=a$ is, for instance
$$\underset{x\to a}{\text{lim}}\frac{x^3 (x-a)}{x^5-a^5}=\underset{x\to a}{\text{lim}}\frac{x^3 (x-a)}{(x-a)(x^4+ax^3+a^2x^2+a^3x+a^4)}=\frac{a^3}{5a^4}=\frac{1}{5 a}$$
At $x=(-1)^{2/5} a$ it is
$$\underset{x\to (-1)^{2/5} a}{\text{lim}}\frac{x^3 \left(x-(-1)^{2/5} a\right)}{x^5-a^5}=$$
$$=\underset{x\to (-1)^{2/5} a}{\text{lim}}\frac{x^3 \left(x-(-1)^{2/5} a\right)}{\left(x-(-1)^{2/5} a\right)(-(-1)^{3/5} a^4-\sqrt[5]{-1} a^3 x+(-1)^{4/5} a^2 x^2+(-1)^{2/5} a x^3+x^4)}=$$
$$=-\frac{(-1)^{3/5}}{5 a}$$

Hope this helps


问:
An upper bound of product of two inner products
The question is,

Let $A \in M_{n \times n}(\Bbb C)$ be a self-adjoint matrix. Arrange the eigenvalues of $A$ as $0 < \lambda_1 \le \lambda_2 \le \cdots \le \lambda_n. $

Show that $ \langle Ax, x \rangle  \langle A^{-1}x,x \rangle \le \frac{(\lambda_1+\lambda_n)^2}{4 \lambda_1 \lambda_n} $ for all $||x||=1, x\in V$

My approach:

Since $A$ is self-adjoint, $V$ has an orthonormal basis consisting of eigenvectors of $A.$ Let $\{v_1,v_2, \cdots, v_n\}$ be an orthonormal basis of $V$ where $Av_i=\lambda_iv_i, \forall i \in \Bbb N.$

Any arbitrary $x \in V$ with $||x||=1$ can be written as $x=\sum_{i=1}^n c_iv_i$ such that $\sum_{i=1}^n|c_i|^2=1$ where $c_i \in \Bbb C, \forall i \in \Bbb N.$ Therefore, we can write,

$$\langle Ax, x \rangle \langle A^{-1}x, x \rangle=\langle A(\sum_{i=1}^n{c_iv_i}),\sum_{i=1}^n{c_iv_i}\rangle  \langle A^{-1}(\sum_{i=1}^n{c_iv_i}),\sum_{i=1}^n{c_iv_i}\rangle$$
$$=\left(\sum_{i=1}^n|c_i|^2\lambda_i\right)\left(\sum_{i=1}^n\frac{1}{\lambda_i}|c_i|^2 \right)$$

And I could not figure out how to proceed after this. I managed to show the inequality holds when $A$ is a $2\times 2$ matrix. Also clearly if $c_1=c_n= \sqrt{\frac{1}{2}}, c_2=\cdots=c_{n-1}=0,$ then the equality holds. But apart from these, I could not do much. I tried using Lagrange multiplier but haven't had much luck with that either. So any help would be really appreciated.

答:
Remark: As pointed out by @leslie townes, this is the so-called the Kantorovich inequality. Years ago, I came up with a proof myself (However, perhaps it is not new).

Problem: Let $a_k \ge 0$, $k = 1, 2, \cdots, n$ with $\sum_{k=1}^n a_k = 1$.
Let $0 < \lambda_1 \le \lambda_2 \le \cdots \le \lambda_n$. Prove that
$$\Big(\sum_{k=1}^n \lambda_k a_k\Big)
\Big(\sum_{k=1}^n \frac{a_k}{\lambda_k}\Big)\le \frac{(\lambda_1+\lambda_n)^2}{4\lambda_1\lambda_n}.$$

Proof:

If $\lambda_1 = \lambda_n$, the inequality is clearly true.
In the following, assume $\lambda_1 < \lambda_n$.

Consider the optimization problem
$$\max_{a_k\ge 0, \forall k; \ \sum_{k=1}^n a_k = 1} \Big(\sum_{k=1}^n \lambda_k a_k\Big)
\Big(\sum_{k=1}^n \frac{a_k}{\lambda_k}\Big).$$
Let $(a_1^\ast, a_2^\ast, \cdots, a_n^\ast)$ be a global maximizer.

We claim that if $\lambda_1 < \lambda_k < \lambda_n$, then $a_k^\ast = 0$.
Indeed, if $a_k^\ast > 0$, let
$$a_1' = a_1^\ast + (1 - y) a_k^\ast, \quad a_k' = 0, \quad a_n' = a_n^\ast + y a_k^\ast$$
where $\frac{\lambda_k - \lambda_1}{\lambda_n - \lambda_1} < y <
\frac{\lambda_n}{\lambda_k}\cdot \frac{\lambda_k - \lambda_1}{\lambda_n - \lambda_1}$. We have
$$a_1' + a_k' + a_n' = a_1^\ast + a_k^\ast + a_n^\ast,$$
and
\begin{align}
\lambda_1 a_1' + \lambda_k a_k' + \lambda_n a_n' &= \lambda_1 a_1^\ast + \lambda_n a_n^\ast
+ [\lambda_1 + (\lambda_n - \lambda_1) y]a_k^\ast \\
&> \lambda_1 a_1^\ast + \lambda_n a_n^\ast
+ \left(\lambda_1 + (\lambda_n - \lambda_1)\cdot \frac{\lambda_k - \lambda_1}{\lambda_n - \lambda_1}\right)a_k^\ast\\
&= \lambda_1 a_1^\ast + \lambda_n a_n^\ast + \lambda_k a_k^\ast,
\end{align}
and
\begin{align}
\frac{a_1'}{\lambda_1} + \frac{a_k'}{\lambda_k} + \frac{a_n'}{\lambda_n}
&= \frac{a_1^\ast}{\lambda_1} + \frac{a_n^\ast}{\lambda_n}
+ \left(\frac{1}{\lambda_1} - \frac{\lambda_n - \lambda_1}{\lambda_1 \lambda_n}y\right)a_k^\ast \\
&> \frac{a_1^\ast}{\lambda_1} + \frac{a_n^\ast}{\lambda_n}
+ \left(\frac{1}{\lambda_1} - \frac{\lambda_n - \lambda_1}{\lambda_1 \lambda_n}
\cdot \frac{\lambda_n}{\lambda_k}\cdot \frac{\lambda_k - \lambda_1}{\lambda_n - \lambda_1}\right)a_k^\ast \\
&= \frac{a_1^\ast}{\lambda_1} + \frac{a_n^\ast}{\lambda_n}
+ \frac{a_k^\ast}{\lambda_k}.
\end{align}
However, this contradicts the optimality of $(a_1^\ast, a_2^\ast, \cdots, a_n^\ast)$.

Now, assume that, among $\lambda_1, \lambda_2, \cdots, \lambda_n$, there are $p$ elements
equal to $\lambda_1$, and there are $q$ elements equal to $\lambda_n$, where $p + q \le n$.
Denote $A = a_1^\ast + a_2^\ast + \cdots + a_p^\ast$ and $B = a_{n-q+1}^\ast + a_{n-q+2}^\ast + \cdots + a_n^\ast$.
We have $A + B \le 1$. We have
\begin{align}
&\Big(\sum_{k=1}^n \lambda_k a_k^\ast\Big)
\Big(\sum_{k=1}^n \frac{a_k^\ast}{\lambda_k}\Big)\\
=\ & \Big(\lambda_1 A + \lambda_n B\Big)\left(\frac{A}{\lambda_1} + \frac{B}{\lambda_n}\right)\\
=\ & (A + B)^2 + \left(\frac{\lambda_1}{\lambda_n} + \frac{\lambda_n}{\lambda_1} - 2\right)AB\\
\le\ & 1 + \left(\frac{\lambda_1}{\lambda_n} + \frac{\lambda_n}{\lambda_1} - 2\right)\cdot \frac{1}{4}\\
=\ & \frac{(\lambda_1+\lambda_n)^2}{4\lambda_1\lambda_n}.
\end{align}
We are done.


答: $f(\frac34)=-\frac34$ while $f(1)=0$. But the intermediate value $-\frac12$ is not assumed on $(-\frac34,1)$.

Title: Is $f:\mathbb{R}\rightarrow\mathbb{R}, f(x)=\left \lfloor{x}\right \rfloor -x$ a Darboux function?

Body:
Let $f:\mathbb{R}\rightarrow\mathbb{R}, f(x)=\left \lfloor{x}\right \rfloor -x$

My book claims this function is not a Darboux function.

Is this true ? I cannot find a counterexample to prove that this is not a Darboux function.

Thanks in advance !


问:
Finer Polish topology which turns countably many given Borel sets to clopen sets
Let $(X,\tau)$ be a Polish space and $(B_n)_{n \in \omega}$ a sequence of Borel sets in $(X,\tau)$. I would like to know if this implies that there is a Polish topology ${\tau}_{\omega}$ on $X$ such that $\tau \subseteq {\tau}_{\omega}$, the Borel $\sigma$-algebra generated by ${\tau}_{\omega}$ is equal to the Borel $\sigma$-algebra generated by $\tau$ and $B_n$ is clopen in $(X,{\tau}_{\omega})$ for every $n \in \omega$.

I know the following theorem:

If $(X,\tau)$ is a Polish space and $B \subseteq X$ is a Borel set in $(X,\tau)$, then there exists a Polish topology $\tau'$ on $X$ such that $\tau \subseteq \tau'$, the Borel $\sigma$-algebra generated by $\tau'$ is equal to the Borel $\sigma$-algebra generated by $\tau$ and $B$ is clopen in $(X,\tau')$.

Using this theorem I would be able to solve the above problem if the sequence of Borel sets was finite (I would just use the theorem finitely many times). But what to do in the case of an infinite sequence?

答:
Let $\tau \cup \{B_n, B_n^\complement: n \in \omega\}$ be the subbase for a new topology on $X$.

It's clearly finer, and still second countable as $\tau$ is. Because we only add $\tau$-Borel sets, the new Borel sets are the same. By definition the $B_n$ are clopen.


问:
Vectors in Physics vs. Vectors in Abstract Linear Algebra
In physics class, students are often shown vectors in $\mathbb{R}^n$ (for $n=2$ or $n=3$ of course) via arrow diagrams and they are told that a vector is an entity with magnitude and direction.

But then in abstract linear algebra, we learn that yes, $\mathbb{R}^n$ is one example of a vector space but there are others like $C^\infty$ functions for example. However, if $C^\infty$ functions are to have magnitude and direction, we must have a norm and an inner product defined on the space as well. It doesn't come automatically with one.

Therefore, we have two definitions of vector:

(1) An entity with magnitude and direction
(2) An entity which satisfies the vector space axioms

These are very different because a vector space can't be assumed as endowed with something that gives its members magnitude and direction. Doesn't this mean the physics lessons are technically giving a wrong definition?

答:
The first definition is closer to elementary geometry and is more intuitive, but the intuition seems to be limited to "simple" vector spaces like $\mathbb R^2$ and $\mathbb R^3$. However, it is not that far from (2) than one might believe.

Let us first think about the concept of direction. What is a direction in the plane $\mathbb R^2$? This seems to be very clear, but if you want to a make it precise you will see that it is by no means trivial.

Let us agree that, in a naive sense, a vector $v$ in $\mathbb R^2$ is an arrow starting at $(0,0)$ and ending at $(x,y)$. [In physics sometimes a vector is also regarded as an arrow connecting any two points in the plane - this leads to the concept of an affine space which we shall not discuss here.] Since the starting point $(0,0)$ is fixed, a vector is uniquely determined by its end point and we can simply write $v = (x,y)$.

How do we know that that the vectors $v_1= (1,0)$ and $v_2 = (2,0)$ have the same direction and $w_1= (0,-1)$ and $w_2 = (1,3)$ do not? The answer is that $v_2$ is a positive multiple of $v_1$, but $w_2$ is not a multiple of $w_1$. Thus a formal definition of direction which works is any vector space is this:

A direction is an equivalence class of non-zero vectors, where $v_1 \sim v_2$ if $v_2 = t v_1$ for some $t > 0$.

Note that $u_1= (1,0)$ and $u_2 = (-1,0)$ do not have the same direction because $u_2$ is a negative multiple of $u_1$. In fact, they have the opposite direction. We can also assign to $0 \in V$ a direction (the "zero direction"), but it is a philosophical question whether it makes sense to regard this as a "real" direction.

Let us next think about the concept of magnitude. In $\mathbb R^2$ (and also in $\mathbb R^3$) the meaning is clear, but it requires an additional structural component beyond addition and scalar multiplication. This component is the standard Euclidean norm given by $\lVert (x,y) \rVert = \sqrt{x^2+y^2}$. It associates to each vector a magnitude (or length) and thereby allows to compare the magnitude of non-collinear vectors $v_1, v_2$. Note that if $v_1,v_2$ are collinear, i.e. lie on a single line through the origin, the $v_1,v_2$ have the same magnitude if and only if $v_2 = \pm v_1$.

We conclude that speaking both about direction and magnitude is only possible in normed vector spaces. This is an adequate setting for physics, and there is nothing wrong about it. In a normed vector space $(V,\lVert - \rVert)$ there is a $1$-$1$-correspondence between vectors $v \ne 0$ and pairs $(d,l)$ where $d$ is a direction and $l > 0$. Precisely in that sense (1) and (2) are equivalent. The issue with (1) is that is not completely transparent how the addition of vectors in the form $(d_1,l_1)$ and $(d_2,l_2)$ is working.


Title: Evaluating $48^{322} \pmod{25}$

Body: How do I find $48^{322} \pmod{25}$?</p>

Body: mod $\,25\!:\ 48^{322}\! \equiv (-48)^{322}\!\equiv 2^{322}\! \equiv 2^2\!\cdot (2^{10})^{32} = 4\cdot 1024^{32}\equiv 4\,(-1)^{32}\! \equiv 4$


问:
Number theory olympiad problem
This is a problem I have been tackling recently, but I am unsure how to address it.

A positive integer $n$ is good if there exists a set of divisors of $n$ whose members sum to $n$ and include $1$. Prove that every positive integer has a multiple which is good.

Any help with a solution appreciated.

答:
Hint: try for multiples of the form $2^pn$ where $n$ is odd and $p$ is large.

Full solution:

Let $n$ be an odd integer; there are $a_1 > a_2 > \ldots > a_k=0$ such that $n=\sum_{l=1}^k{2^{a_l}}$. Let $p \geq a_1$, then $2^pn=(2^p-1)n+n=\sum_{l=0}^{p-1}{2^ln}+\sum_{l=1}^k{2^{a_l}}$ so $n2^p$ is good.

Let $A$ be any integer, then $A|B$ for some integer $B$ of the form $2^ab$ where $a \geq 1$, $b \geq 3$ is odd. We know that for large enough $p$, $2^pb$ is good. But for such large enough $p$, $2^ab|2^pb$ so $A|2^pb$ and $2^pb$ is good.

Title: Number theory olympiad problem


问:
Fixpoints of gradient of monotone increasing, concave function
Suppose we have a system of equations:

$$
x_1 = f_1(x_1, ..., x_n) \\
x_2 = f_2(x_1, ..., x_n) \\
... \\
x_n = f_n(x_1, ..., x_n)
$$

where the $f_i$ are components of the gradient ${\nabla}f$ for a monotone increasing, concave function $f$, so $f_i > 0$ and $\frac{{\partial}f_i}{{\partial}x_i}\leq 0$.

Does the system of equations necessarily have a solution? And is it unique if it exists?

I feel like there must be some application of a multivariate version of the Intermediate Value Theorem or Brouwer's Fixed Point Theorem that would be useful here...

答:
I think I have the answer to the uniqueness question.

Let $g = f + h$ where $h(x_1, ..., x_n) = -\frac{1}{2}\Sigma_i x_i^2$

Then $g$ is a strictly concave function, because it is the sum of a concave and a strictly concave function. And stationary points of concave functions are in general global maxima, so uniqueness follows by taking the derivative of $g$.


问:
Taylor expansion in complex analysis. Why $f(z)=\operatorname{Log} (z+1)$?
$f(z)=\text{log} 2+ \sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{n} \left( \dfrac{z}{2}-\dfrac{1}{2} \right)^n.$

Why $f(z)=\operatorname{Log} (z+1)$ at $|z-1| < 2 $ ?

At $|z-1|<2,$ $\operatorname{Log} (z+1)=\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{n} z^n.$

But why this is equal to $f(z)$ ?

I don't know how should I deformate $f(z)$ or $\text{Log} (z+1).$

答: Assume that $|z-1|<2$. Then we can write $z=1+2u$ where $|u| < 1$, so that $\frac{z}{2}-\frac{1}{2}=u$. Thus $f(z)=\log{2}+\log(1+u)=\log(2+2u)=\log(1+z)$.


问:
Bijection between a set of injections and a union of bijection
Consider $A$ a set with $n$ elements and $k\leq n$.

Show that
$$
\phi:\left\{  f:\left\{  1,2,\ldots,k\right\}  \rightarrow A\mid f\text{
injective}\right\}  \rightarrow
{\bigsqcup\limits_{B\subset A,card(B)=k}}
\left\{  g:\left\{  1,2,\ldots,k\right\}  \rightarrow B\mid g\text{
bijective}\right\}
$$
by constructing an explicit bijection.

By generalizing the construction above, show that
$$
\Phi:\left\{  f:A\rightarrow C\mid f\text{ injective}\right\}  \rightarrow
{\bigsqcup\limits_{B\subset C.card(C)=n}}
\left\{  g:B\rightarrow C\mid g\text{ bijective}\right\}  .
$$

I attempted to start from the observation that every injective function $f$
has an image $\operatorname{Im}f$ as subset of $A$ with exactly $k$ elements.
Reciprocally every subset of $A$ with $k$ elements is the image of some
injection. But there are $k!$ injections that maps a fixed set $C$ into a
subset of $A$ with $k$ elements. Now, every
$$
f:\left\{  1,\ldots,k\right\}  \rightarrow A
$$
gives us a bijection $g:\left\{ 1,\ldots,k\right\}  \rightarrow B,B\subset A$
and $card(B)=k$. From this point, I don't know how to move forward. How can I
find an explicit bijection?

答:
Hint

Denote $E=\left\{  f:\left\{  1,2,\ldots,k\right\}  \rightarrow A\mid f\text{
injective}\right\}$ and $F={\bigsqcup\limits_{B\subset A,card(B)=k}}
\left\{  g:\left\{  1,2,\ldots,k\right\}  \rightarrow B\mid g\text{
bijective}\right\}$.

Let

$$\begin{array}{l|rcl}
\Phi : & E & \longrightarrow & F \\
    & f & \longmapsto & \begin{array}{l|rcl}
\Phi(f) : & \left\{  1,2,\ldots,k\right\} & \longrightarrow & f[\left\{  1,2,\ldots,k\right\}] \\
    & x & \longmapsto & f(x)  \end{array}  \end{array}
$$

$\Phi$ is a formal definition of the bijection you're looking for.


问:
Lie algebra that cannot be realised from the commutator bracket of an algebra?
What is an example of a Lie algebra that cannot be realised from the commutator bracket of an algebra?

Edit: More precisely, does every Lie algebra admit a faithful representation on a vector space. From the comment below, Ado's theorem says this is true for finite dimensional algebras. So are there infinite dimensional Lie algebras not admitting faithful representations?

答:
A semi-simple Lie algebra cannot be realized as the commutator bracket of an associative algebra. An associative algebra is a particular case of a left-symmetric algebra: these are algebra defined by:
$x(yz)-(xy)z=y(xz)-(yx)z$. For a left-semmetric algebra $A$, the bracket defined by $[x,y]=xy-yx$ is a Lie algebra. The Lie algebra associated to $A$ is not semi-simple. This is a corollary of the Whitehead Lemma.

The proof works like so: if $A$ is a left-symmetric, let $L(A)$ be the associated Lie algebra. Define $L_a(x)=ax$, the map $a\rightarrow L_a$ is a representation of $L_a$ and $Id_A$ is a $1$-cocycle for this representation since $ax-xa=[a,x]$. Withehead lemma implies that there exists $e\in A$ such that $ae=a$ for every $a\in A$.

Let $ad_e(x)=ex-xe=ex-x$, we have $Id_A=L_e+ad_e$, we deduce that $Trace(Id_A)=Trace(L_e)+Trace(ad_e)=0$ since $L(A)$ is semi-simple, the image of $L$ is contained in $sl(n,\mathbb{R})$.


答:
You have$\DeclareMathOperator{\Log}{Log}$
$$\Log(1+u)=\sum_{n\ge 1}(-1)^{n-1}\frac{u^n}n,\quad u\in\mathbf C, |u|<1.$$
Now set $\dfrac{z-1}2=u:\quad |u|<1\iff |z-1|<2$, and therefore
$$\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{n} \Bigl( \dfrac{z-1}{2} \Bigr)^n=\Log\Bigl(1+\frac{z-1}2\Bigr)=\Log\frac{z+1}2=\Log(z+1)-\Log 2.$$

Title: Taylor expansion in complex analysis. Why $f(z)=\operatorname{Log} (z+1)$?

Body:
$f(z)=\text{log} 2+ \sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{n} \left( \dfrac{z}{2}-\dfrac{1}{2} \right)^n.$

Why $f(z)=\operatorname{Log} (z+1)$ at $|z-1| < 2 $ ?

At $|z-1|<2,$ $\operatorname{Log} (z+1)=\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{n} z^n.$

But why this is equal to $f(z)$ ?

I don't know how should I deformate $f(z)$ or $\text{Log} (z+1).$


问:
How does $ 0 \le a \lt p^\alpha $ in Theorem 5.30 of Introduction to Analytic Number Theory (Apostol)?
Theorem 5.30 from Introduction to Analytic Number Theory (Apostol) has a long proof. I am unable to understand one specific step in it.

First it assumes that $ \alpha \ge 2 $ and $ r $ is a solution of the congruence $$ f(x) \equiv 0 \pmod{p^{\alpha - 1}} $$ where $ r $ lies in the interval $ 0 \le r < p^{\alpha - 1}$.

In one of the steps of the proof it defines a variable $ a $ as $$ a = r + qp^{\alpha - 1} $$ and shows that $ a $ satisfies $ f(x) \equiv 0 \pmod{p^\alpha} $ if and only if $ q $ satisfies the congruence $ qf'(r) + k \equiv 0 \pmod{p} $. The variable $ q $ was described to be "an integer to be specified presently".

Here is the step I am unable to understand:

If $ f'(r) \equiv 0 \pmod{p} $ this congruence ($qf'(r) + k \equiv 0 \pmod{p}$) has a unique solution $ q $ mod $ p $, and if we choose $ q $ in the interval $ 0 \le q \lt p $ then the number $ a $ given by (22) will satisfy (20) and will lie in the interval $ 0 \le a < p^\alpha $.

How will $ a $ lie in the interval $ 0 \le a < p^\alpha $?

The equation (22) just refers to $ a = r + qp^{\alpha - 1} $.

Only two constraints have been specified $ 0 \le r < p^{\alpha - 1} $ and $ 0 \le a \lt p^\alpha $. If I combine these two inequalities, I get

$$
a = r + qp^{\alpha - 1} \lt p^{\alpha - 1} + p \cdot p^{\alpha - 1} = p^{\alpha - 1} + p^\alpha.
$$

I don't get $ 0 \le a < p^\alpha $. What am I missing?

答:
You were almost there. Use the $\le$ instead of the $<$ to get sharper bounds (thanks to the fact that we are using integers, there is an actual gap between the two). You get

$$ a= r +qp^{\alpha-1} \le (p^{\alpha-1} -1 )+ (p-1) p^{\alpha-1} = (p^{\alpha-1} -1) + (p^{\alpha} - p^{\alpha-1}) = p^{\alpha} -1 < p^{\alpha} $$


问:
Moore–Penrose inverse definition as a vector form
As an excercise I am trying to prove that $AA^{+}\vec{a}^{\,}_{i}=\vec{a}^{\,}_{i}$,
$\forall{i=1..n}$ is equal
to the more popular version $AA^{+}A=A$. I started by setting up a system of equations as follows:

\begin{equation}
    \begin{cases}
      AA^{+}\vec{a}^{\,}_{1}=\vec{a}^{\,}_{1}\\
      \vdots\\
      AA^{+}\vec{a}^{\,}_{n}=\vec{a}^{\,}_{n}\\
    \end{cases}
\end{equation}

Then I turned it into a matrix form:
$$
\begin{bmatrix}
      AA^{+}\vec{a}^{\,}_{1}\\
      \vdots\\
      AA^{+}\vec{a}^{\,}_{n}\\
\end{bmatrix}
=
\begin{bmatrix}
      \vec{a}^{\,}_{1}\\
      \vdots\\
      \vec{a}^{\,}_{n}\\
\end{bmatrix}
$$

$$
AA^{+}
\begin{bmatrix}
      \vec{a}^{\,}_{1}\\
      \vdots\\
      \vec{a}^{\,}_{n}\\
\end{bmatrix}
=
\begin{bmatrix}
      \vec{a}^{\,}_{1}\\
      \vdots\\
      \vec{a}^{\,}_{n}\\
\end{bmatrix}
$$

Since $\vec{a}^{\,}_{i}$ is a column vector it has shape $(m\space x\space1)$ which seems to produce weird vector in the last equation. Is this part of a "proof" any good or is it a bad idea from the begining?

答:
I suppose $\vec{a}_j$ means the $j$-th column of $A$. Then $A$ is equal to $\begin{bmatrix}\vec{a}_1&\cdots&\vec{a}_n\end{bmatrix}$. So, the correct identity should be
$$
\begin{bmatrix}AA^{+}\vec{a}_1&\cdots&AA^{+}\vec{a}_n\end{bmatrix}
=AA^{+}\begin{bmatrix}\vec{a}_1&\cdots&\vec{a}_n\end{bmatrix}
=AA^+A=A
=\begin{bmatrix}\vec{a}_1&\cdots&\vec{a}_n\end{bmatrix}.
$$
The matrix-vector product $AA^{+}
\begin{bmatrix}\vec{a}_1\\ \vdots\\\vec{a}_n\end{bmatrix}$ does not make sense because the size of $AA^+$ (which is, for example, $n\times n$ when $A$ is $n\times n$) does not match the size of the the column vector $\begin{bmatrix}\vec{a}_1\\ \vdots\\\vec{a}_n\end{bmatrix}$ (which is $n^2\times1$ in this example).


问:
Has the set of all ideals of a ring a ring-like structure?
After learning about ideals of a ring and the basic operations that can be made with them, I started wondering about the following situation.

Let $R$ be a commutative ring with unity (for simplicity), and denote by $I(R)$ the set of all ideals of $R$. The sum of two ideals, $I$ and $J$, is given by: $I+J=\{i+j:i \in I,j \in J\}$, and their product is: $IJ=\{\sum_{i=1}^n x_iy_i:x_i \in I, y_i \in J, n \in \mathbb{N}^\ast \}$. Then:

$i)$ $(I(R),+)$ is a commutative monoid, with identity element $(0)$;

$ii)$ $(I(R),\cdot)$ is also a commutative monoid, with identity element $R$;

$iii)$ We have distributivity: $I(J+K)=IJ+IK$ and $(I+J)K=IK+JK$, for $I,J,K \in I(R)$.

Now my questions:

$1)$ Are all of the above, $i),ii),iii)$ correct? I am quite unsure about the product being associative (I failed to prove for myself, so any proof/indication would be appreciated). The other axioms seem to hold.

$2)$ If $1)$ is true, then what type of structure is $(I(R),+,\cdot)$?

Thank you in advance!

答:
''ii)  (I(R),⋅) is also a commutative monoid, with identity element, again, (0);''

Identity element $R$.

''2)  If 1) is true, then what type of structure is (I(R),+,⋅)?''

Its a semiring.