问:
How to show a null-set is a $\sigma$-algebra if and only if $\mu(\Omega)=0$?

I am having a bit of trouble with my assigment on Measures and Integrals. First I have been giving the measure space $(\Omega, \mathcal{A}, \mu)$ and the $\mu$-null set (or empty-set):

$\mathcal{N}_\mu=\{N\subseteq \Omega| \quad \exists A \in \mathcal{A}: N \subseteq A \text{ and } \mu(A)=0\}$.

I have proved the following properties:

$\emptyset \in \mathcal{N}_\mu$ and

If $N_n\in \mathcal{N}_\mu, \quad n=1,2,... ,$ then $\bigcup_{n=1}^\infty N_n \in \mathcal{N}_\mu$.

The last thing I need to show is that:

$\mathcal{N}_\mu$ is a $\sigma$-algebra if and only if $\mu(\Omega)=0$.

Maybe some of you guys can give me a hint on what to do, to get started on the proof? Thank you.

答:
If $\mathcal{N}_{\mu}$ is a $\sigma$-algebra then $\Omega$ is $\mathcal{N}_{\mu}$. By definition of $\sigma$-algebra there is some $A \in \mathcal{A}$ such that $\Omega \subseteq A$ and $\mu(A)=0$. In order for $\Omega \subseteq A$ to hold we must have $A = \Omega$. Hence $\mu(\Omega)=0$.

Suppose now $\mu(\Omega)=0$. The only things that you still needs to prove in order to show $\mathcal{N}_{\mu}$ is a $\sigma$-algebra is that $\Omega \in \mathcal{N}_{\mu}$ and that $\mathcal{N}_{\mu}$ is closed under complementation. Since you already noticed that $\emptyset \in \mathcal{N}_{\mu}$ we just need to prove the second property. Let $N \in \mathcal{N}_{\mu}$. Choose $A=\Omega \in \mathcal{A}$ (this is possible because $\mathcal{A}$ is a $\sigma$-algebra). Then $\Omega \setminus N \subseteq \Omega=A$ and $\mu(A)=\mu(\Omega)=0$. Hence $\Omega \setminus N \in \mathcal{N}_{\mu}$ and $\mathcal{N}_{\mu}$ is a $\sigma$-algebra.


问:
Can a one-sided limit exist at a discontinuity point of a derivative?
Let $f \colon \mathbb{R} \to \mathbb{R}$ be a differentiable function and fix $a \in \mathbb{R}.$ By L'Hôpital's rule, since
$$ \frac{(f(x)-f(a))'}{(x-a)'} = f'(x),$$
if either limit
$$ \lim_{x \to a^+} f'(x), \qquad \lim_{x \to a^-} f'(x)$$
exists, then it equals $f'(a).$ Hence, if both of them exist, $f'(x) \to f'(a)$ as $x \to a$ and $f'$ is continuous at $a.$ Equivalently, if $f'$ is not continuous at $a,$ then at least one of the one-sided limits does not exist. This is enough to show that every discontinuity of $f'$ is essential, but I'd like to know whether it can be assured that neither one-sided limit exists.

I appreciate any help.

答: No. Define $f$ by $f(x)=0$ for $ x\le0$ and $f(x)=x^2\sin(1/x)$ for $x>0$. One can show $f$ is differentiable and that $f'$ is not continuous at 0. Obviously, the left-hand limit of $f'$ at 0 exists.

ContentLicense: CC BY-SA 4.0


问:
does a Taylor series converge to f(x) if the radius of convergence is $R=\infty$
If I want to check whether a Taylor series converges to its function f(x). Is it enough to check if the radius of convergence of the Taylor series is infinite?

Or do I have to use the remainder theorem and show the remainder converges to zero?

Edit: I'm still confused about convergence of taylor series.
If I have a function f(x) which is defined on $x\in ]-R,R[$, and I found it's taylor series T(x) with a radius of convergence R.
Doesn't the taylor series converge to f(x)?

Why do I necessarily need to check that the remainder $R_n(x)$ converges to zero for it to be true?

In other words, what does the taylor series converge to, if it doesn't converge to f(x)? (Because I thought if a taylor series was convergent it would always be towards f(x)? But that's probably where the whole issue is?).

答:
An infinitely differentiable function of a real variable is not necessarily the sum of its Taylor series. I'll give  an example below. It it is the sum of its Taylor series, it is called an analytic function. Contrary to the real case, a function  of a complex variable which is differentiable is  infinitely differentiable and analytic.

Example of a non-analytic infinitely differentiable function of a real variable:
$$f(x)=\begin{cases}
\mathrm e^{-1/x^2}&\;\text{ if }x\ne 0,\\ 0&\;\text{ if }x=0.
\end{cases}$$
It is easy to show that $f^{(n)}(0)=0$ for all $n$, hence its Taylor series at $0$ is $0,\:\ne f(x)$ if $x\ne 0$.


问:
Log problem, $u$ substitution the only way?
Okay so basically I want to know if you can solve this log equation without the use of u substitution:

$${\log_4{\log_3{x}}} = 1$$

I believe that u substitution is the only way to solve this problem, but please prove me wrong if theres another way to do so.

答:
There is basic identity of log is:   $\log_ab=n\implies a^n=b $

so in your question : $\log_4\log_3x=1\,$, we put $\log_3x=u$

$$\log_4u=1\implies u=4^1\implies u=4\implies\log_3x=4\implies x=3^4\implies x=81$$

ContentLicense: CC BY-SA 3.0


问:
How to solve this trigonomectric equation $2^{\lvert\sin x\rvert} + 2^{\lvert\cos x\rvert} = 3$?
I thought its solutions are $(\sin x;\cos x) = \{(\pm1;0),(0;\pm 1)\}$

But how can I solve it?

Please help :(

答:
Clearly the function $f:\Bbb R\to\Bbb R:x\mapsto f(x):=2^{|\!\cos x|}+2^{|\!\sin x|}$ is periodic with period $\frac12\pi$. Also clear is that $f(0)=f(\frac12\pi)=3$, and $f(x)\equiv f(\frac12\pi-x)$. So we look to see if there are any solutions to $f(x)=3$ in the range $0<x<\frac12\pi$. It is convenient to transform the variable by writing $t:=x-\frac14\pi$ and $\alpha:=\frac12\surd2\ln2$, so that our function takes the form $$f(x)=g(t):=2\exp(\alpha\cos t)\cosh(\alpha\sin t)\quad(-\tfrac14\pi\leqslant t\leqslant\tfrac14\pi).$$Differentiation gives $$g'(t)=2\alpha[\cos t\sinh(\alpha\sin t)-\sin t\cosh(\alpha\sin t)]\exp(\alpha\cos t),$$from which it is clear that $f'(t)=0$ only when $$\tan t=\tanh(\alpha\sin t).$$Now $\sin$ is concave in the interval $[0\;\pmb,\,\frac14\pi]$ and $\tanh$ is concave in $[0\;\pmb,\,\infty)$, so the RHS of the above equation is concave in $[0\;\pmb,\,\frac14\pi]$ too ; at the same time, $\tan$ is convex in the same interval. So there is no solution to the equation beyond $0$ in the interval. By a similar argument, with convexity and concavity interchanged, there is no solution in $[-\frac14\pi\;\pmb,\,0)$ either. Thus $g$ has just the one stationary point at $0$, namely when $g(0)=2\mathrm e^\alpha\approx3.265>3$. Therefore $g$ decreases from this maximum at $0$ to a minimum value of $3$ at $\pm\frac14\pi$.

It follows that the only solutions to the equation $f(x)=3$ are of the form $x=\frac12k\pi\; (k\in\Bbb Z).$


问:
Prove $\left(\frac{a+1}{a+b}\right)^a+\left(\frac{b+1}{b+c}\right)^b+\left(\frac{c+1}{c+a}\right)^c \geqslant 3$
$a,b,c \geqslant 0,$$ a+b+c=3$, and $(a+b)(b+c)(c+a) \neq 0$ , prove

$$\left(\frac{a+1}{a+b}\right)^a+\left(\frac{b+1}{b+c}\right)^b+\left(\frac{c+1}{c+a}\right)^c \geqslant 3$$

I try Bernouli's inequality but checking the case of $a>1$ and $0<a<1$ is truly complicated. I try to come up with some estimations (i.e., $x^x \geqslant \frac12 (1+x^2$)) but has not yet been successful.

答:
Yuri Negometyanov gave
$$\left(\frac{a+1}{a+b}\right)^a = \left(1 + \frac{b-1}{a+1}\right)^{-a} \ge 1 - a\cdot \frac{b-1}{a+1}
= 2 - b + \frac{b-1}{a+1}. \tag{1}$$
This follows from the Bernoulli inequality $(1+x)^r \ge 1 + rx$ for $x > -1$ and $r \le 0$.

Using (1), it suffices to prove that
$$\frac{b-1}{1+a}+ \frac{c-1}{1+b}+ \frac{a-1}{1+c}\ge 0. \tag{2}$$

We can prove (2) in another way as follows.

After clearing the denominators, it suffices to prove that
$$a^2b + b^2c + c^2a + a^2 + b^2 + c^2 - a - b - c - 3 \ge 0$$
or (since $a+b+c=3$)
$$a^2b + b^2c + c^2a - 2(ab+bc+ca) + 3 \ge 0.$$
By Cauchy-Bunyakovsky-Schwarz inequality, we have
$$a^2b + b^2c + c^2a \ge \frac{(ab + bc + ca)^2}{b + c + a} = \frac{(ab + bc + ca)^2}{3}.$$
Thus, we have
\begin{align}
a^2b + b^2c + c^2a - 2(ab+bc+ca) + 3 &\ge \frac{(ab + bc + ca)^2}{3} - 2(ab+bc+ca) + 3\\
&= \frac{1}{3}(ab + bc + ca - 3)^2\\
&\ge 0.
\end{align}

We are done.


问:
Approximating $\ln(2)$ with error of $5*10^{-3}$?
How many terms of $ln(2)=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}$ are needed to approximate it with an error of less than $5*10^{-3}$?

答:
It's a general feature of alternating series whose terms are monotonically decreasing in absolute value that the error in taking the first $n$ terms is bounded by the next term (the $n+1$ term). So it suffices to take $\boxed{ n = 200 }$.

This is a lot of terms. A different series that converges much faster is

$$\ln 2 = \ln \left( \frac{1}{1 - \frac{1}{2}} \right) = \sum_{k \ge 1} \frac{1}{k \cdot 2^k}$$

which comes from the Taylor series

$$\ln \left( \frac{1}{1 - x} \right) = \sum_{k \ge 1} \frac{x^k}{k}.$$

The error in taking the first $n$ terms is bounded by

$$\sum_{k \ge n+1} \frac{1}{k \cdot 2^k} \lt \sum_{k \ge n+1} \frac{1}{(n+1) 2^k} = \frac{1}{(n+1) 2^n}$$

which gives that this time it suffices to take $\boxed{ n = 7 }$. Much better!

Recently I tried computing $\ln 2$ to $3$ digits of accuracy by hand for fun and I found out that there is yet another series that converges even faster. We have

$$\begin{eqnarray} \ln 2 &=& \ln \frac{3}{2} + \ln \frac{4}{3} \\
 &=& \ln \left( \frac{1}{1 - \frac{1}{3}} \right) + \ln \left( 1 + \frac{1}{3} \right) \\
 &=& \sum_{k \ge 1} \frac{1 + (-1)^{k-1}}{k \cdot 3^k} \\
 &=& \boxed{ \sum_{k \ge 0} \frac{2}{(2k+1) 3^{2k+1}}. } \end{eqnarray}$$

The terms of this series decay like $\frac{1}{9^k}$ which is more than $3$ times faster (as measured by the number of digits of accuracy you gain per term) than the $\frac{1}{2^k}$ we had before. This series comes from the Taylor series

$$\frac{1}{2} \ln \frac{1+x}{1-x} = \text{arctanh } x = \sum_{k \ge 0} \frac{x^{2k+1}}{2k+1}$$

for the hyperbolic arctangent. The error in taking the first $n$ terms is bounded by

$$\sum_{k \ge n} \frac{2}{(2k+1) 3^{2k+1}} \le \sum_{k \ge n} \frac{2}{(2n+1) 3^{2k+1}} = \frac{1}{4(2n+1) 3^{2n-1}}$$

which gives that this time it suffices to take $\boxed{ n = 2 }$, which gives us an accuracy of better than $\frac{1}{540}$. And in fact the sum of the first two terms is

$$\frac{2}{3} + \frac{2}{81} = \frac{56}{81} = 0.6913 \dots $$

and the sum of the first two terms plus the error bound above is

$$\frac{2}{3} + \frac{2}{81} + \frac{1}{540} = \frac{1123}{1640} = 0.6932 \dots $$

which is accurate to better than $0.000063$.


问:
How can I prove discontinuity of $f(x)$?
$f$ : $\mathbb{R} \rightarrow \mathbb{R}$

$f(x)=x$ $\quad$ If $x \leqq 1$

$f(x)=x+2$ $\quad$ If $x>1$

If I draw the graph, I can realize $f(x)$ is not continuous.

But I want to prove this discontinuity by showing that an open set $O$ in $\mathbb{R}$ such that $f^{-1} (O)$ is not open in $\mathbb{R}$ exists.

How should I decide $O$ ?

答: Let $O = (0, 2)$. Then, $f^{-1}[O] = (0, 1]$, which is not open. The strategy is that when you want to show the discontinuity by inverse image, let your open interval include the function value at the discontinuous point. In the example, we see that $f(1) = 1 \in O$.


问:
Substituting $u=\ln$ when lower bound is $0$
The following substitution came up in the solution set when I was practicing Diff. Geom., and it seems wrong to me:
$$\int_0^t\cos(\ln x) dx = \int_0^{\ln(t)} \cos(u)e^u du$$
When $u$ is evaluated at the lower bound, it is not defined! Additionally, the limit $\ln s\downarrow 0 = -\infty$.

Is this just very sloppy notation in the solution set, or am I missing something else?

答:
You are correct:
$$x\in[0,t],\,u=\ln(x)\Rightarrow u\in(-\infty,\ln t]$$
and so:
$$\int_0^t\cos(\ln(x))dx=\int_{-\infty}^{\ln(t)}e^u\cos(u)du$$
you may want to split this integral up into two parts: $I=I_1+I_2$ if that is clearer:
\begin{align}I=\int_{-\infty}^0e^u\cos(u)du=\Re\int_{-\infty}^{\ln(t)}e^{(1+i)u}du=\Re\left[\frac{e^{(1+i)u}}{1+i}\right]_{-\infty}^{\ln(t)}\end{align}
\begin{align}=\Re\left(\frac{(1-i)e^{(1+i)\ln(t)}}{2}\right)=\Re\left(\frac{(1-i)t\left(\cos(\ln(t))+i\sin(\ln(t))\right)}{2}\right)\end{align}
$$=\frac t2\Re\left(\left[\cos(\ln(t))+\sin(\ln(t))\right]+i\left[\sin(\ln(t))-\cos(\ln(t))\right]\right)$$
$$=\frac t2\left[\cos(\ln(t))+\sin(\ln(t))\right]=\frac t{\sqrt{2}}\cos\left(\ln(t)-\frac{\pi}{4}\right)$$


问:
Does $\lim_{n\to\infty} \frac{(-1)^{n}}{n}$ converge to $0$?
I was solving a problem in real analysis where the author used the below to support his argument.

$$\lim_{n\to\infty} \frac{(-1)^{n}}{n} = 0$$

Is the above limit valid ?

It struck me that, as $1^\infty$ is an indeterminate form, can we say that the above limit holds?
Possible known values to which $1^\infty$ can go are $1$, $e$, and $\infty$ . Can we substantiate and say that the above limit holds?

答:
Here it is a proof which is not based on the sandwich theorem for the sake of curiosity.

Proposition

Let $(a_{n})_{n=1}^{\infty}$ be a sequence of real numbers. Then $a_{n}$ converges to $0$ iff $|a_{n}|$ converges to zero.

Proof

We shall prove the implication $(\Rightarrow)$ first.

Let $\varepsilon > 0$. Then there exists $n_{\varepsilon}\in\mathbb{N}$ such that
\begin{align*}
n\geq n_{\varepsilon} \Rightarrow ||a_{n}| - 0| = |a_{n} - 0| \leq \varepsilon
\end{align*}
from which we conclude that $|a_{n}|$ converges to zero.

Similar reasoning applies to the implication $(\Leftarrow)$.

Solution

Since $|a_{n}| = 1/n \xrightarrow{n\to\infty} 0$, we conclude that $a_{n}$ converges to zero as well.

Hopefully this helps!


问:
$B_0(H_1)B_0(H_2, H_1)$ is dense in $B_0(H_2,H_1)$
Let $H_1$ and $H_2$ be Hilbert spaces. Is it true that
$$B_0(H_1)B_0(H_2, H_1)= \overline{\operatorname{span}}\{ST\mid S \in B_0(H_1), T \in B_0(H_2, H_1)\}= B_0(H_2,H_1)?$$

If so, how can I prove this?

One possible attempt I can think of is to try to show that the left hand side contains all finite-rank operators $H_2 \to H_1$, but I'm not sure these are dense in $B_0(H_2,H_1)?$

答:
Let $F: H_2\to H_1$ be finite rank. Let $P$ denote the orthogonal onto the image of $F$. Since $F$ is finite rank the image of $F$ is finite dimensional (and so closed - hence $P$ is actually well defined) and $P$ is also a finite rank map.

Further $F = P\circ F$ by definition. But this is in $B_0(H_1)\cdot B(H_2,H_1)$. Hence the finite rank maps all lie in this product. Since the finite rank maps are dense in the compacts (we are in the Hilbert space setting) your claim follows.


问:
Chromatic number of a simple graph
Let $G$ be a simple graph such that $$\chi(G-x-y)=\chi(G)-2,$$ for any distinct vertices $x$ and $y$, then $G$ is a complete graph.

I know that if $G$ is a simple graph we have that $\chi(G)\leq n$, where $n$ is the number vertices of $G$. Moreover, if we consider two distinct vertices of $G$, we have that $\chi(G-x-y)\leq n-2$. Therefore, $\chi(G-x-y)\leq 2n-2-\chi(G)$. For me it's clear that equality is true for a complete graph, because if $G$ is a complete graph we have $\chi(G) = n$ and consequently $\chi(G-x-y)=n-2=\chi(G)-2$.

I'm not sure if this thinking is correct, can anyone help me to justify that $G$ is a complete graph?

答:
Assume $G$ is not complete. Then there are vertices $x,y$ not joined by an edge.
Consider a colouring of $G-x-y$ with $m=\chi(G-x-y)$ colours. Then by assigning a new $(m+1)$st colour to $x$ and $y$, we obtain a valid colouring of $G$ and conclude that
$$ \chi(G)\le \chi(G-x-y)+1.$$

Title: Chromatic number of a simple graph


问:
Model the constraint ‘Only if decision 1 is yes and decision 2 is no, then decision 3 is allowed to be yes’
Model the constraint ‘Only if decision 1 is yes and decision 2 is no, then decision 3 is allowed to be yes’ as a set of linear constraints that should simultaneously be satisfied. Add binary variable(s) where needed.

Image Of Problem

答:
Via conjunctive normal form:
$$
y_3 \implies (y_1 \land \lnot y_2) \\
\lnot y_3 \lor (y_1 \land \lnot y_2) \\
(\lnot y_3 \lor y_1) \land (\lnot y_3 \lor \lnot y_2) \\
(1- y_3 + y_1 \ge 1) \land (1- y_3 + 1- y_2 \ge 1) \\
(y_3 \le y_1) \land (y_3 \le 1 - y_2) \\
$$


问:
Calculate probability to win in a contest?
The probability to win contest $A$ is $0.6$, and for contest $B$ it's $0.8$, then what's the probability to win at least one of them?

If I sum the probabilities I get $1.4$ which doesn't make sense at all.

答:
As I assume the events are independent of each other, you shouldn't be summing the probabilities of winning each individual contest. Note that the probability of winning both $A$ and $B$ is $0.6\times 0.8 = 0.48$. Then the probability of winning $A$ and losing $B$ is $0.6\times (1-0.8) = 0.6 \times 0.2 = 0.12$, and the probability of winning $B$ and losing $A$ is $0.8\times (1-0.6) = 0.8\times 0.4 = 0.32$. Summing these three probabilities we end up with $0.92$, or $92\%$ chance of winning at least one contest.

Alternatively, note that the only way to not win at least one of the contests is to lose both of them. The probability of that happening is $(1-0.8)\times (1-0.6) = 0.2\times 0.4 = 0.08$. So we subtract $1-0.08 = 0.92$, to get the same answer as before.


问:
Proving that the Flag Variety $Fl(n;m_1,m_2)$ is connected.
I wish to prove that the flag variety $Fl(n;m_1,m_2) = \{ W_1 \subset W_2 \subset V | dimW_i = m_i \}$, for $0 \le m_1 \le m_2 \le n$ where V is an n-dimensional vector space over $\mathbb{C}$ and $W_1, W_2$ are vector subspaces of V, is connected. I'm following a similar method we have been using in lectures for other Lie groups.

I have already shown that $U(n)$ acts transitively on $Fl(n;m_1,m_2)$ and that $U(n)$ is connected. I'm wondering if what follows is proof that $Fl(n;m_1,m_2)$ is connected.

$Fl(n;m_1,m_2)=U(n)/U(n)_{W_1 \subset W_2}$,since $U(n)$ acts transitively on $Fl(n;m_1,m_2)$. Thus, since $U(n)$ is connected, if we can show that $U(n)_{W_1 \subset W_2}$ is connected then it follows that $Fl(n;m_1,m_2)$ is connected.

Consider $U(n)_{W_1 \subset W_2}=\{F\in U(n) |F(W_1)=W_1, F(W_2)=W_2 \}$. It consists of maps with matrices of the form:

A=$ \left( \begin{array}{ccc}
A_{m_1,m_1} & 0 & 0 \\
0 & B_{m_2-m_1,m_2-m_1} & 0 \\
0 & 0 & C_{n-m_2,n-m_2} \end{array} \right)$

Where the index of the block denotes its size. This is because $F(W_2)$ will be orthogonal to $W_1$ and will be linear multiples of $e_{m_1+1},...,e_{m_2}$, and $F(W_1)$ will be orthogonal to $W_2$ and will be linear multiples of $e_{1},...,e_{m_1}$, where $e_1,...,e_n$ is the orthonormal basis of $Fl(n;m_1,m_2)$.

Now I think that I have to show that the blocks in the matrix are themselves contained in a connected component of a Lie group, which would mean that I could connect each of them to the identity matrix of the respective size and thus connect any matrix in the stabiliser subgroup I'm considering to the identity, but I'm not quite sure how to proceed. Could anyone please give me any hints as to the next step?

Many thanks.

答:
I think you were already done after the introductory remark:

If the (continuous) action of a connected group $G$ on a space $X$ is transitive, the $X$ is connected.

In fact, assume $X=U_1\cup U_2$ with $U_1,U_2$ open and $U_1\cap U_2=\emptyset$.
Select $x\in X$. Then the map $f\colon G\to X$, $g\mapsto gx$ is continuous, hence $f^{-1}(U_1), f^{-1}(U_2)$ are open and are disjoint and $f^{-1}(U_1)\cup f^{-1}(U_2)=G$.
By connectedness of $G$, one of $f^{-1}(U_1), f^{-1}(U_2)$ is empty. But by transitivity, $f$ is onto, hence $f^{-1}(U_i)=\emptyset$ implies $U_i=\emptyset$ and we conclude that $X$ is connected.


问:
minimal rank of the matrix on the picture
In an excercise I am asked about a minimal rank of that matrix. "*" are elements that are not equal to ?$0$. I think the minimal rank is 3, am I right? My argumentation is shown on the picture by red parts.

答:
I'm not sure what your argument is just from the red boxes, but here is mine. Call row $n$ by $R_n$.

If the second and fifth columns of $R_4$ and $R_5$ happen to contain opposite values from each other, then $R_6 = R_4 + R_5$ is possible. And obviously $R_1 = R_2 = R_3$ is possible. Thus $3$ is the highest possible minimum rank.

But $R_3, R_4, R_5$ cannot be linearly dependent. Each has a non-zero value in a column where the other two have $0$. Any linear combination with non-zero coefficients would have to be non-zero in these three columns. Hence the lowest possible rank is $3$.


问:
Invariant subspace $U$ of direct sum $V\oplus V$ of irreducible representation $V$ is isomorphic to $V$
I am studying from Hall's book "Lie Groups, Lie Algebras, and Representations" and I'm stumped on the following question:

Suppose that $V$ is an irreducible finite-dimensional representation of a group or Lie algebra over $\mathbb{C}$, and consider the associated representation $V\oplus V$. Show that every nontrivial invariant subspace $U$ of $V\oplus V$ is isomorphic to $V$ and is of the form $$U=\{(\lambda_1v,\lambda_2v)|v\in V\}$$ for some constants $\lambda_1$ and $\lambda_2$, not both zero.

From the looks of it, I think I should be using Schur's lemma, which tells me that if $V$ is an irreducible complex representation of a group or Lie algebra and $\phi:V\to V$ is an intertwining map of $V$ with itself, then $\phi=\lambda I$ for some $\lambda\in\mathbb{C}$. An intertwining map is a linear map which commutes with the action. My understanding is that other authors just call this a homomorphism of representations.

However, I do not see how to connect the end result with Schur's lemma. Perhaps the projection maps will come in handy here? Any help is appreciated.

答:
Step 1:  Assume $U$ is irreducible. For $i: U \hookrightarrow V \oplus V$ the inclusion and $pr_1 , pr_2: V \oplus V \twoheadrightarrow V$ the projections on the first and second component, repectively, look at the maps $$pr_j \circ i: U \rightarrow V.$$ Use Schur's Lemma. Notice what happens if both are zero.

Step 2: Now, assume $U$ contains two different irreducible subrepresentations $U_1 \neq U_2$. Using the result of step 1, show that $U_1+U_2$ contains both $V \oplus 0$ and $0 \oplus V$, hence is all of $V \oplus V$.


问:
Using Stokes' theorem around a circular loop

I know that I have to use:

$$\oint F\cdot dr = \iint \operatorname{curl}(F)\cdot n \;dS,$$

with $\operatorname{curl}(F)= -x j -yk$ and $n\;dS = dA\; k$

But the dot product of the curl and $dS$ leaves me with
$$\iint -y \; dA$$
but I am unsure how to proceed from here. Do I need to change coordinate systems as I am dealing with a cicle? Any help is appreciated!

答: In order to apply Stoke's Theorem to evaluate $\int_CF\cdot d\vec{r}$ where $C$ is the circular loop $x^2+y^2=a^2$ you need to first establish a surface $S$ whose boundary is $C$ and is equipped with a normal vector $\vec{n}$ that induces the original orientation prescribed to $C$. Let's take $S$ to be the circular disc $$\{x^2+y^2\leq a^2,z=0\}$$ which we'll parameterize by $$\vec{r}(u,v)=(u\cos(v),u\sin(v),0)$$ on the domain $0\leq u\leq a$ and $0\leq v\leq 2\pi$. Clearly the vector $\vec{n}=\big<0,0,1\big>$ is perpendiular to $S$ and induces a counterclockwise orientation on its boundary $C$. (I'm assuming you're prescribing a counterclockwise orientation to $C$.) A little bit of calculation also reveals that $$\text{curl}(F)=\big<0,-1,-1\big>$$ $$\vec{r}_u \times \vec{r}_v=\big<0,0,u\big>$$ $$dS=||\vec{r}_u \times \vec{r}_v||dudv=ududv$$ So with Stoke's Theorem, $$\int_C F \cdot d\vec{r}=\int_{S}\big[\text{curl}(F)\cdot \vec{n}\big]dS=\int_0^{2\pi} \int_0^a -ududv=-\pi a^2$$


问:
How to solve this equation for $r$?
I have a problem....
I have to express unknown "r" from from this equation:

$$Y\times r=(-Z\times 2r\times L)+(z\times K^2\times 0.5L)$$

Can someone help me ?

答: Hint:  move the term containing $r$ to the left by adding $2ZLr$ to both sides, then use the distributive law to write the left as (something)$r$, then divide both sides by (something)</p>

Title: How to solve this equation for $r$?


问:
Proving $f:[a,b) \longrightarrow (a,b)$ cannot be bijective and continuous (verification)
I have come up with a proof for the expression, which I am confident in, but if there is anything wrong with it, please do let me know where exactly I went wrong and what incorrect assumption was made; I would like to be able to come up with the solution on my own. Thank you!

Assume $f:[a,b) \longrightarrow (a,b)$ is both continuous and bijective. By the bijectivity of $f$, there exists one, and only one $x \in [a,b)$ such that $f(x) = y \in (a,b)$. As such, let $f(a) \in (a,b)$. Moreover, let $U \subset (a,b)$ be an open neighborhood of $f(a)$. Now, by continuity of $f$ at $0$, there exists some $V \in [a,b)$ open such that $a \in V$ and $f(V) \subset U$. However, $V$ cannot be open since for any $r > 0$, $B_r(a) \not\subset V \implies f$ is not continuous. We conclude that $f$ cannot be bijective and continuous.

The intervals are regarded with the subset topology induced by the standard topology of $\mathbb R$

答:
Hint) The definition of (open) ball:
$$
B_r(c) =\{x \in [a,b)~|~d(x,c) < r \}
$$
where $d$ is a metric given that $[a,c)$ is a metric space. That is, it is always true that
$$
B_r(c) \subseteq [a,h)
$$
for all $r >0 $ and all $c \in [a,c)$ where $h \in (a,c)$.


问:
Combinations of $(0,1)$-Matrices with equal row and colum sum
Consider a $(2n\times2n)$ - Matrix with elements from $\{0,1\}$. Row and colum sums have to be equal to $n$ for each row and sum respectively.
Here is an exampel for $n=2$:

$$
\begin{pmatrix}
1 & 0 & 1 & 0 \\
0 & 1 & 0 & 1 \\
1 & 1 & 0 & 0 \\
0 & 0 & 1 & 1 \\
\end{pmatrix}
$$
How to find the number of all these matrices dependent on $n$?

答:
Such matrices have the nice property to be the sum of $n$ permutation matrices  see here. This result being in connection with results on bipartite graphs.

Using this principle, I have built a Matlab program that computes the number of such matrices in the cases $n=2$ and $n=3$ ($6 \times 6$ matrices). See at the bottom.

for $n=1$: 2 matrices.

for $n=2: 90 $ matrices (computation time with my program: a millisecond), a value that can be found here: http://oeis.org/A001499.

for $n=3: 297200 $ matrices! (computing time with my program: about an hour on my slow PC). This value can be found here: http://oeis.org/A001501.

The general case is evidently out of reach with this kind of methods. One finds in the "litterature":

for $n=4$: $116963796250$ matrices (see http://oeis.org/A058528).

for $n=5$: $6736218287430460752$ matrices (see http://oeis.org/A075754).

The following references on OEIS site are connected to this issue:

http://oeis.org/A008300, and http://oeis.org/wiki/Index_to_OEIS:_Section_Mat#binmat

Later on, I found this document authored by Odama, Yumi and Musiker, Gregg: "Enumeration of (0,1) and Integer Doubly Stochastic Matrices" (December 2001), on Science Direct giving a general formula based on partitions of integer $N=2n$. One finds (page 2) understandable particular cases whereas the interpretation of the general formula is pretty hard.

For further reading, in particular a classification of (0-1) matrices up to row and column permutations: see here and here ; see as well the very dense document here.

Matlab program:

```<code>clear all;
n=3;m=2*n;%
p=m^2-1;
f=factorial(m);
I=eye(m);
Pe=perms(1:m);% all permutations of 1...2n.
for P=1:f
    M=I(:,Pe(P,:));% The P-th permutation matrix, stored
    T(:,P)=M(:); % vertically as the P-th column of T
end;
S=[];
V=2.^(0:p);% V*M yields a (binary) coding (see below)
for P=1:f;
  for Q=P+1:f;
    U=T(:,P)+T(:,Q);
    if all(U<2)
       for R=Q+1:f
          M=U+T(:,R);
          if all(M<2);
              S=[S,V*M];
          end;
       end;
    end;
end;
S=unique(S);
end;
L=length(S)
</code>```

Body:
Such matrices have the nice property to be the sum of $n$ permutation matrices  see here. This result being in connection with results on bipartite graphs.

Using this principle, I have built a Matlab program that computes the number of such matrices in the cases $n=2$ and $n=3$ ($6 \times 6$ matrices). See at the bottom.

for $n=1$: 2 matrices.

for $n=2: 90 $ matrices (computation time with my program: a millisecond), a value that can be found here: http://oeis.org/A001499.

for $n=3: 297200 $ matrices! (computing time with my program: about an hour on my slow PC). This value can be found here: http://oeis.org/A001501.

The general case is evidently out of reach with this kind of methods. One finds in the "litterature":

for $n=4$: $116963796250$ matrices (see http://oeis.org/A058528).

for $n=5$: $6736218287430460752$ matrices (see http://oeis.org/A075754).

The following references on OEIS site are connected to this issue:

http://oeis.org/A008300, and http://oeis.org/wiki/Index_to_OEIS:_Section_Mat#binmat

Later on, I found this document authored by Odama, Yumi and Musiker, Gregg: "Enumeration of (0,1) and Integer Doubly Stochastic Matrices" (December 2001), on Science Direct giving a general formula based on partitions of integer $N=2n$. One finds (page 2) understandable particular cases whereas the interpretation of the general formula is pretty hard.

For further reading, in particular a classification of (0-1) matrices up to row and column permutations: see here and here ; see as well the very dense document here.

Matlab program:

clear all;
n=3;m=2*n;%
p=m^2-1;
f=factorial(m);
I=eye(m);
Pe=perms(1:m);% all permutations of 1...2n.
for P=1:f
    M=I(:,Pe(P,:));% The P-th permutation matrix, stored
    T(:,P)=M(:); % vertically as the P-th column of T
end;
S=[];
V=2.^(0:p);% V*M yields a (binary) coding (see below)
for P=1:f;
  for Q=P+1:f;
    U=T(:,P)+T(:,Q);
    if all(U<2)
       for R=Q+1:f
          M=U+T(:,R);
          if all(M<2);
              S=[S,V*M];
          end;
       end;
    end;
end;
S=unique(S);
end;
L=length(S)


问:
Expected earliest point at which a sock matches an earlier sock
The University of Waterloo CEMC's Problem of the Month for November, the problem statement for which can be seen here, was authored by me. (As far as I know, it is original, though I have come to expect that such low hanging fruits have already been plucked. Forgive me if someone authored a similar problem earlier.)

In essence, I was inspired by the fact that I had $n$ pairs of matching socks in a laundry bag and I was pulling socks out one at a time and hoping that, at some point, I would obtain a sock that matched a previously drawn sock (this was before I had degenerated to the point that I did not care about my socks matching). The main part of the problem asks for the pulled sock number $k+1$ at which the probability is highest that the earliest match occurs. The official solution can be read here and it interestingly has to do with triangular numbers.

Given $2n$ total socks (so $n$ pairs of matching socks), the probability of the first match occurring upon removing the $(k+1)^{\text{th}}$ sock for $1\le k\le n$ is $$P(n,k) = \frac{k2^k n! (2n-k-1)!}{(2n)!(n-k)!}.$$ For those wondering, this probability is the same regardless of whether the socks within a single pair are distinguishable or indistinguishable, though in reality, they are of course indistinguishable.

This means the expected sock number at which the first match occurs is $$E(n)=\sum_{k=1}^{n}{k\cdot P(n,k)} = \sum_{k=1}^{n}{\frac{k^2 2^k n! (2n-k-1)!}{(2n)!(n-k)!}}.$$
According to Wolfram Alpha, the first $9$ values of this function are: $$1,\frac{5}{3},\frac{11}{5},\frac{93}{35},\frac{193}{63},\frac{793}{231},\frac{1619}{429},\frac{26333}{6435},\frac{53381}{12155}.$$
Does it seem to grow like a square root? I'm not sure.

My question is: Can this expected value function be written in a closed form, or be approximated by bounds or by an asymptotic formula?

答:
First of all, your "expected sock number" is $1$ off, because $P(n,k)$ is the probability for the $(k+1)^{\text{th}}$ sock. I am making this correction because the formula when you add $1$ is nicer. So I will give a formula for the sequence $$2, \frac83, \frac{16}{5}, \frac{128}{35}, \dots$$ which is shifted by $1$ from your sequence. (Just looking at this might lead you to some guesses about what the formula is!)

It is often easier to compute the expected value of a
(nonnegative integer-valued) random variable $\mathbf X$ by the formula
$$
   \mathbb E[\mathbf X] = \sum_{k \ge 0} \Pr[\mathbf X > k].
$$
(You can check that $\Pr[\mathbf X = k]$ is counted exactly $k$ times in this sum.) In this case, $\Pr[\mathbf X > k]$ is the probability that the first $k$ socks are all different. This has a nicer formula: $$\Pr[\mathbf X > k] = \frac{\binom{n}{k} 2^k}{\binom{2n}{k}} = \frac{n!(2n-k)!2^k}{(2n)!(n-k)!} = \frac{\binom{2n-k}{k}2^k}{\binom{2n}{n}}.$$ So we now want to evaluate the sum
$$
   \mathbb E[\mathbf X] = \frac{1}{\binom{2n}{n}} \sum_{k=0}^n \binom{2n-k}{n}2^{k}.
$$
This sum has a combinatorial interpretation: $\binom{2n-k}{n}2^{k}$ is A276418, the number of paths of length $2n$ that start at $0$, go up or down at each step, and return to $0$ exactly $k$ times. (I don't know if there's an easy argument that counts these.) Summing these over all $k$ gives $4^n$: the number of paths of length $2n$ that start at $0$ and go up or down at each step.

Therefore $\mathbb E[\mathbf X] = 4^n/\binom{2n}{n}$. Asymptotically as $n \to \infty$, $\mathbb E[\mathbf X] \sim \sqrt{\pi n}$.


问:
inequality $\frac{2(x + y)^2}{2x^2 + y^2} \leq 3$
I'm looking for a "nice" way to show that the following inequality holds, i.e. without differentiating and determining the maximum:

$$
\frac{2(x + y)^2}{2x^2+ y^2} \leq 3 \quad \forall x,y \in \mathbb{R}
$$

It's rather easy to show

$$
\frac{4xy}{x^2 + y^2} \le 2
$$

and obviously

$$
\frac{2x^2 + 2y^2}{2x^2+y^2} \le 2
$$

but combining these two does not give me a sufficient low bound.

Body: $$\frac{2(x + y)²}{2x² + y²} \le 3 \iff 2x^2+4xy+2y^2\le 6x^2+3y^2 \\\iff0 \le 4x^2-4xy+y^2=(2x-y)^2.\blacksquare$$


Title: inequality $\frac{2(x + y)^2}{2x^2 + y^2} \leq 3$

Body:
I'm looking for a "nice" way to show that the following inequality holds, i.e. without differentiating and determining the maximum:

$$
\frac{2(x + y)^2}{2x^2+ y^2} \leq 3 \quad \forall x,y \in \mathbb{R}
$$

It's rather easy to show

$$
\frac{4xy}{x^2 + y^2} \le 2
$$

and obviously

$$
\frac{2x^2 + 2y^2}{2x^2+y^2} \le 2
$$

but combining these two does not give me a sufficient low bound.

Body: by Cauchy-Schwarz inequality $$(2x^2+y^2)(1+2)\ge {(\sqrt{2}x+\sqrt{2}y)}^2=2{(x+y)}^2$$


问:
Show that controllability and observability are not affected by replacing A with (A+αI )
I have been asked to show that controllability and observability are not affected by replacing $A$ with $(A+αI)$. And also to show that this is not necessarily true for stabilizability.

I have thought about different approaches:

Popov-Belevitch-Hautus test

If the system is controllable must be true that
$$Rank [(A - \lambda I) B] = n$$

If we replace $A$ with $(A+αI)$ then
$$Rank [((A +αI) - \lambda I) B] $$
$$Rank [(A  - (\lambda -α)I) B] $$
$$Rank [(A  - gI) B]  $$ where $g = (\lambda -α) \in C$

$(A  - gI)$ has already $rank = n$ for all $g \in C$ except for those which are eigenvalues for $A$.

For those values, we need to prove that the concatenation of $B$ will guarantee the rank to be n.

The issue is that since B is the same we can't guarantee it.

Controllability Matrix

Given an LTI if and only if the control matrix $C$ has full column rank, then the system is controllable.
$$ C = [B ,AB, A²B ... A^{n-1} B]$$
but then altering the diagonal of $A$ can affect the rank of C, or if not, how come?

Controllability Gramian

I tried to plug $(A+αI)$ in the integral, but doing that then I don't know how to prove that the matrix $W$ is still nonsingular for any t > 0.

答:
Controllability: there does not exist a left-eigenvector for $A$ that is orthogonal for $B$.

For your problem,
for any $x$ such that $xA = \lambda x$, $xB\neq0$.

Since the eigenvectors of $A$ are also the eigenvecors of $A+\alpha I$, i.e.,
$x(A+I\alpha) = x(\lambda+\alpha)$, you have $xB\neq0$.


问:
I am having trouble with a probability review problem I found online.
The question is:

There are 3 ways a CEO can commit accounting fraud: (A) he can overstate
revenue by booking fake orders, (B) he can overstate profits by recording near-term expenses as longterm capital expenses, and (C) he can understate liabilities by making overly optimistic assumptions
about the growth prospects of the company’s pension funds. In the United States, 50% of CEOs commit fraud (A), 20% of CEOs commit fraud (B), 15% of CEOs commit both fraud (A) and fraud (B),
and all CEOs commit fraud (C).
The probability of getting caught if a CEO does (A) is 0.01, independent of whether or not he commits
other kinds of fraud. Similarly, the probability of getting caught if he does (B) is 0.05, and of getting
caught if he does (C) is 0.001.

a) What are the chances (rounded to 2 decimal places) that a CEO who perpetrates all 3 kinds of
fraud will get caught?

I feel like this is supposed to be a very simple question but I am struggling with it despite having an easy time with what I assumed were harder questions in the same place I found this one.

The problem I keep running into is that, any solution I can come up with requires I figure out the P(caught ∩ A ∩ B ∩ C). I have not been able to figure out P(caught ∩ A ∩ B ∩ C) so either I am looking in the wrong direction or there is a trick to solving for it that has not yet hit me.

答:
Let $A, B, C$ denote committing crimes and $A_0,B_0,C_0$ denote caught with crimes.

a) What are the chances (rounded to 2 decimal places) that a CEO who perpetrates all 3 kinds of fraud will get caught?

In the language of my events, this question is asking for
$$
P(A_0\cup B_0\cup C_0 |ABC).
$$
The words "who perpetrates" is telling us the given and "caught" is a vague word for caught with one of the crimes.

By complement law, we have
$$
P(A_0\cup B_0\cup C_0 |ABC)=1-P(A_0^c B_0^c C_0^c|ABC).
$$
Now the trick is that regardless of what crimes were committed, it's always the case that the process of getting caught is independent. Hence, above equals
$$
1-P(A_0^c|ABC)P(B_0^c|ABC)P(C_0^c|ABC).
$$
It's also the case that knowing other crimes were committed is irrelevant so this simplifies to
$$
1-P(A_0^c|A)P(B_0^c|B)P(C_0^c|C).
$$


问:
Prove that $\forall\, x>0$ exists $n\in\mathbb{N}$ s.t. $ \frac{1}{2^n}<x$
My teacher gave me this question and I don't really know where to start, she said we might need to use the Archimedes' principle but I don't know how to do it.
I would be great if you could help me.</p>

答:
The question seems incorrectly stated, for example if you take $x=1$ there is no $n$ such that $1/2^n > x$, you probably meant $<$.

For that problem, for every $x>0$ show that there exist an $n \in \mathbb{N}$ such that $\frac{1}{2^n} < x$, we need to find an $n$ such that $\frac{1}{2^n}$ is as small as we like. Turning it around, given $x >0$ we want to find $n$ such that $2^n > \frac{1}{x}$.

A little useful trick here may be to notice that $2^n > n$ for all $n \in \mathbb{N}$. So in other words if we can find an $n$ such that $n > \frac{1}{x}$, then $2^n > n > \frac{1}{x}$. Can you see how to use the Archimedean property to find such an $n$?


问:
How to solve modular arithmetics problems, that are not linear?
$2x^{20} + 3x + 4 \equiv 0  $(mod 176)

I know the answer is $x = 176k + 20$,   $k \in \mathbb{Z}$ , but I don't know, how to solve this problem.

答:
$176=2^411$.

Mod $2^4=16$:

If $\gcd(16,x)>1$, it's $3x+4\equiv0\bmod16$ or $3x\equiv12\bmod 16$ or $x\equiv4\bmod 16.$

If $\gcd(16,x)=1,$ $2x^{20}+3x+4\equiv0\bmod 16$ has no solutions,

because $3x\equiv0\bmod2$ has no solutions with odd $x$.

Mod $11$:

If $\gcd(11,x)>1,$ it would be $4\equiv0\bmod11$ and there are no solutions.

If $\gcd(11,x)=1$, then by Fermat's little theorem it $x^{20}\equiv x^{10}\equiv 1\bmod11$,

and it would be $2+3x+4\equiv0\bmod11$ or $3x\equiv-6\bmod11$ or $x\equiv-2\bmod11$.

Can you solve $x\equiv4 \bmod16$ and $x\equiv9\bmod11$ using the Chinese remainder theorem?


问:
find if 2 planes are parallel or not
I have 2 parallel planes with equation :

a.x + b.y + c.z +d = 0
a'.x + b'.y + c'.z +d = 0

And I need to demonstrate they are parallel.

I could read that they are parallel if they have "2 collinear normal vectors"
But I couldn't understand what it means?
Well, collinear mean parallel, but does it mean I need minimu 2 true sentences on these ones ?

a=k.a'
b=k.b'
c=k.c'

Excuse me if this is an easy issue, but just starting to learn back mathematics...

答:
With the equation of the plane in standard form $ax+by+cz = d,  (a,b,c)$ is the normal vector  to the plane.

If two planes are parallel then these normal vectors are scalar multiples of one another.

So, if $(a,b,c) = (ka',kb',kc')$ the normal vectors are parallel.  However, if also $d = kd'$ then they are the same plane.

Title: find if 2 planes are parallel or not


问:
How do convergent sequences in a metric space determine the topology?
It is often stated that a metric topology is 'fully described  by its convergent sequences'. How can this be expressed in a more formal way?</p>

答:
The formal statement would be

Theorem. Two metrics on a set determine the same topology if and only if they have the same convergent sequences.


问:
How will we solve this differentiation?
How will we solve this differentiation? I am not not understanding from where to start.
$\frac{Root 3}{4}*a^2 *2 + 3a$ $\frac{4}{root 3* a^2}$

We have solve dA/da [That whole equation is equal to A]

What I am not getting is that from which function to start.
Like there is already root present and then there is a^2.

What I did

I solved the first part using product rule.
u=root 3*4 and v = 4.

答:
If we simplify  $A = \frac {\sqrt 3}{4}(a^2)(2) + (3a)\frac 4{(\sqrt 3)(a^2)},$ we get $A = \frac {\sqrt 3}{2}a^2 + \frac {4\sqrt 3}{a}.$

Now differentiate.  Each term is pretty simple, just apply the power rule.

$\frac {dA}{da} = \sqrt 3a - \frac {4\sqrt 3}{a^2}$

The $\sqrt 3$ is just a constant, and should be treated like any other constant.  It just rides along.

Title: How will we solve this differentiation?


答:
Fermats little theorem and Eulers theorem and/or chinese remainder can reduce $x^{20}$ to something smaller and factoring/quadratic formula is still valid.

$176 = 11*16$.

To solve $2x^{20}+3x +4 \equiv 0 \pmod{11}$ we have $x\not \equiv 0 \pmod {11}$ (obviously) so $x^{10} \equiv 1$ and so $3x + 6 \equiv 0$ and it is linear $3x \equiv -6 = \pmod{11}$ and $x \equiv -2\equiv 9 \pmod {11}$.

To $x^{20}+3x + 4\equiv 0 \pmod{16}$ if $x$ is even then $x^4|x^{20}$ so $2x^{20}+3x +4\equiv 3x + 4 \equiv 0\pmod {16}$.  $3x \equiv -4 \equiv 12 \pmod{16}$ and $\gcd(3,12)=1$ so $3$ is invertible so $x \equiv 4\pmod {16}$.

If $x$ is odd then $\phi(16)= 8$ so $x^{20}\equiv x^4$ and we have $2x^4+3x +4\equiv 0\pmod{16}$ and,  well, $3x$ is odd but $2x^4 + 4$ is even so there is not solution.

So we one solution

$x\equiv 9 \pmod {11}; x\equiv 4 \pmod 16$ or the one unique solution (its unique because CRT tells us so) $x\equiv 20\pmod {176}$

Title: How to solve modular arithmetics problems, that are not linear?

Body:
$2x^{20} + 3x + 4 \equiv 0  $(mod 176)

I know the answer is $x = 176k + 20$,   $k \in \mathbb{Z}$ , but I don't know, how to solve this problem.


问:
Find a representation of the field $\mathbb{F}_{16}$ as $\frac{\mathbb{Z}_2}{(f)}$
I have to find representation of the field $\mathbb{F}_{16}$ as $\frac{\mathbb{Z}_2}{(f)}$ where $f \in \mathbb{Z}_2[X]$ a monic irreducible polynomial over $\mathbb{Z}$.

I don't really have a clue how to solve this, but I think it has to do with finding the irreducible polynomials that divide $X^{16}-X$ and have a degree that divide 4.

I also have to find a unique subfield of $\mathbb{F}_{16}$ with 4 elements starting from a representation that I found, which I only can do after finding the specific representation, so I'm also stuck on this one.

Thanks in advance!

答:
$x^2+x+1$ is the unique irreducible polynomial of degree $2$ over $\Bbb F_2$.
$x^4+x+1$ has no roots, so no irreducible factor of degree $1$, it is also distinct from $(x^2+x+1)^2=x^4+x^2+1$, so it's not the product of two irreducible factors of degree $2$ either. Thus $x^4+x+1 \in \Bbb F_2[x]$ is irreducible and we have $\Bbb F_2[x]/(x^4+x+1) \cong \Bbb F_{16}$


问:
How to calculate $ \frac{d}{dt}(\int_{0}^{t} \sin(x^2)\,dx)$
$$ \frac{d}{dt}(\int_{0}^{t} \sin(x^2)\,dx)$$

I only need to know if it is zero or not at $$t=0$$ $$t=\pm \pi$$ $$t=\pm 2\pi$$ ... Don't need to calculate the integral

Hint says use Leibniz rule but I couldn't use it.

答: You can use FTOC, and you will obtain $\sin(t^2)$, then just evaluate at your test points.


问:
On the eigenvalues of a linear transformation $\tau$ such that $\tau^3 = \mathrm{id}$
I am reading  the book on representation theory by Fulton and Harris in GTM.  I came across this paragraph.

[..] we will start our analysis of an arbitrary representation $W$ of $S_3$ by looking just at the action of the abelian subgroup $\mathfrak A_3 = \mathbb Z/3 \subset \mathfrak S_3$ on $W$. This yields a very simple decomposition: if we take $\tau$ to be any generator of $\mathfrak A_3$ (that is, any three-cycle), the space $W$ is spanned by eigenvectors $v_i$ for the action of $\tau$, whose eivenvalues are of course all powers of a cube root of unity $\omega = e^{2\pi i/3}.$

I know the eigenvalues of $\tau$ are some cube root of unity, but I don't know why all of the cube roots of unity are eigenvalues of $\tau$.

I tried to calculate the minimal polynomial $f$ of $\tau$.  It is easily seen that $f$ divides $X^3 - 1$, but I cannot go further in this regard, since I do not know anything about $W$ except the fact that it is finite-dimensional.

I would be most grateful if you could help me understand the paragraph.

答: "All eigenvalues are powers of $\omega$" is not the same as "All powers of $\omega$ are eigenvalues".</p>


问:
Skew a bell-shaped curve
Consider symmetric bell-shaped 2-dimensional functions like
$$f((x,y))=\exp(-(x^2+y^2)), \ (x,y) \in \mathbb{R}^2.$$

A transformation to skew $f$ in direction $-(1,1)$ is
$$\hat{f}((x,y))=\frac{f((x,y))}{1+\exp( a(x+y)+b)},$$
looking like that:

The problem is that $f$ gets elongated (green) perpendicular to the skew direction.

Is there a better transformation/method to skew without elongation? Ideas appreciated!

I thought about transforming the space like $\hat{f}=f \circ h$ for a mapping $h : \mathbb{R}^2 \longrightarrow \mathbb{R}^2$ but that caused elongations as well.

答:
Multiplying the $f$ with
$$\begin{cases}
\exp(-2a(x^2+y^2)) & x+y <0\\
\exp(-a(x-y)^2) & \text{otherwise}
\end{cases}$$
allows to skew without elongation. The level sets are fairly circular as well.

A similar transformation without separate cases is given by
$$\frac{\exp(-\sqrt{a}(x-y)^2)}{1+\exp(-a(x+y))}, \ (x,y) \in \mathbb{R}^2.$$

Title: Skew a bell-shaped curve


问:
Prove that $\frac{d}{dx} (f(x))^2=2*f(x)f'(x)$
A problem in my textbook (page $280$), http://classicalrealanalysis.info/com/documents/TBB-AllChapters-Portrait.pdf, said:

Prove that $\frac{d}{dx} (f(x))^2=2*f(x)f'(x)$ using Theorem 7.7 and also directly from the definition of derivative.

So I proved it.

Proof

Note that $\frac{d}{dx} (f(x))^2= \frac{d}{dx} [f(x)f(x)]$

Then $\frac{d}{dx} [f(x)f(x)]$ = $f(x)f'(x)+f'(x)f(x)=2f(x)f'(x)$

$\square$

This proof seems too obvious given the way that the question is worded. Theorem 7.7 basically just gives you your algebraic operations under differentation (sum, product, quotient rules, etc.) and the limit definition of derivative doesn't seem necessary here.

My proof is correct, but does it satisfy the question? It seems too easy.

Body:
As hint:$$\lim_{t \to 0}\frac{f^2(x+t)-f^2(x)}{t}=\\
\lim_{t \to 0}\frac{f^2(x+t)-f(x+t)f(x)+f(x+t)f(x)-f^2(x)}{t}=\\
\lim_{t \to 0}\frac{f(x+t)(f(x+t)-f(x))+f(x)(f(x+t)-f(x))}{t}=\\
\lim_{t \to 0}\frac{f(x+t)(f(x+t)-f(x))}{t}+\lim_{t \to 0}\frac{f(x)(f(x+t)-f(x))}{t}=\\$$can you take over now ?


Body:
Let $y=l.x$; we have:

$\frac{2[(l+1)x]^2}{x^2(l^2+2)}=\frac {2l^2+4l+2}{l62+2}=2+\frac{4l-2}{l^2+2}$

$\frac{4l-2}{l^2+2}<1$

because:

$(l-2)^2>0$

$l^2-4l+2+2>0$

$4l-2<l^2+2$

If $l=2$ the equality holds.


问:
Minimal polynomial of $e^{\frac{2 \pi i}{p^2}}$ (p is prime number)
$X^{p^2}-1 = (X-1)(X^{p-1}+X^{p-2}+..+X+1)(X^{p(p-1)}+X^{p(p-2)}+..X^{p}+1)$

Then, $e^{\frac{2 \pi i}{p^2}}$ is a root of $f(X)=X^{p(p-1)}+X^{p(p-2)}+..X^{p}+1$

Hence, $g(X)=\frac{X^{p^2}-1}{X-1}$ for $X\neq 1$

thus, for $X \neq 0$ we have

$$g(X+1)=\frac{(X+1)^{p^2}-1}{X}=\sum_{i=1}^{p^2} \binom{p^2}{i}X^{i-1}$$

But $g(X)$ isn't irreducible and minimal polynomial

So, How can I prove it by Eisenstein Criteria?

答: The polynomial $$\frac{(x+1)^{p^2}-1}{(x+1)^p-1}$$ is Eisenstein at $p$.


问:
Cubic bezier get $y$ as a function of $x$ (not $t$)
I'm trying to write a program to display curves.

I have 4 control points $ A, B, C $ and $ D $ and a value for $x$. From that, I want to find the value for $y$.

Where I've gotten so far

We can use De Casteljau's formula to describe a cubic bezier:

$$
P = (1-t)^3A+3(1-t)^2tB + 3(1-t)t^2C+t^2D
$$

Every point is a vector so

$$
x = (1-t)^3ax+3(1-t)^2tbx + 3(1-t)t^2cx+t^2dx
$$

If I solve this for $t$ I should be able to then plug that $t$ value into
$$
y = (1-t)^3ay+3(1-t)^2tby + 3(1-t)t^2cy+t^2dy
$$
To get my desired y value.

The problem is that I haven't found a way to solve this equation for t. I plugged it into wolfram alpha but it seems to be only an approximation.

Is there a more precise / simpler way to achieve what I want?

EDIT: https://github.com/gre/bezier-easing

答:
Wolfram gives you the exact expressions, the same that Cardano would give. They can involve complex numbers. It is possible to implement them in a simpler form, by using intermediate variables.

Beware that depending on the positions of the control points, the value of $t$ might not be unique.

An alternative solution is to sample $t$ roughly, find an interval that contains your $x$ and use a few iterations of regula falsi.

Drawing the parametric curve $(x(t),y(t))$ is million times easier.


答:
Here is a proof directly from the definition of the derivative. We have:

$$\begin{align}\frac{d}{dx}f(x)^{2} &= \lim_{h\to 0}\frac{f(x+h)^{2} - f(x)^{2}}{h} \\&= \lim_{h\to 0}\frac{\big(f(x + h) - f(x)\big)\big(f(x + h) + f(x)\big)}{h}\\&=\lim_{h\to 0}\frac{f(x + h) - f(x)}{h}\big(f(x+h) + f(x)\big)\\&=\lim_{h\to 0}\frac{f(x + h) - f(x)}{h}\lim_{h\to 0}\big(f(x + h) + f(x)\big)\\&=2f(x)\lim_{h\to 0}\frac{f(x + h) - f(x)}{h}\end{align}$$

Using the definition of a derivative:

$$\boxed{\frac{d}{dx}f(x)^{2} = 2f(x)f'(x)}$$

Body:
A problem in my textbook (page $280$), http://classicalrealanalysis.info/com/documents/TBB-AllChapters-Portrait.pdf, said:

Prove that $\frac{d}{dx} (f(x))^2=2*f(x)f'(x)$ using Theorem 7.7 and also directly from the definition of derivative.

So I proved it.

Proof

Note that $\frac{d}{dx} (f(x))^2= \frac{d}{dx} [f(x)f(x)]$

Then $\frac{d}{dx} [f(x)f(x)]$ = $f(x)f'(x)+f'(x)f(x)=2f(x)f'(x)$

$\square$

This proof seems too obvious given the way that the question is worded. Theorem 7.7 basically just gives you your algebraic operations under differentation (sum, product, quotient rules, etc.) and the limit definition of derivative doesn't seem necessary here.

My proof is correct, but does it satisfy the question? It seems too easy.


问:
Possible arrangements of the word ARRANGEMENT
In how many the alphabets of the word ARRANGEMENT can be arranged such that :

Exactly two pairs of consecutive identical alphabets appear in the arrangement?

At least two pairs of consecutive identical alphabets appear in the arrangement?

What I've tried so far:

The alphabets that can be repeated should appear more than once in the given word, these are $$\text{AA,EE,NN,RR}$$

The answer to the first part is given by :
$$\binom{4}{2}\left[\frac{9!}{\left(2!\right)^{2}}-\binom{2}{1}\frac{8!}{\left(2!\right)^{1}}+\binom{2}{2}\frac{7!}{\left(2!\right)^{0}}\right]$$

For the second question it's needed to subtract the cases where less than two pair of consecutive alphabets appear in the arrangement from the total arrangements of the given word:

$$\frac{11!}{\left(2!\right)^{4}}$$
$$-\left[\underbrace{\frac{11!}{\left(2!\right)^{4}}-\binom{4}{1}\frac{10!}{\left(2!\right)^{3}}+\binom{4}{2}\frac{9!}{\left(2!\right)^{2}}-\binom{4}{3}\frac{8!}{\left(2!\right)^{1}}+\binom{4}{4}\frac{7!}{\left(2!\right)^{0}}}_{\text{Arrangements with no pair of consecutive identical alphabets}}\right]$$
$$-\binom{4}{1}\left[\underbrace{\frac{10!}{\left(2!\right)^{3}}-\binom{3}{1}\frac{9!}{\left(2!\right)^{2}}+\binom{3}{2}\frac{8!}{\left(2!\right)^{1}}-\binom{3}{3}\frac{7!}{\left(2!\right)^{0}}}_{\text{Arrangements with only one pair of consecutive identical alphabets}}\right]$$
$$=\color{green}{398160}$$

But the other way I came with was to compute the number of arrangements with two pairs of consecutive identical alphabets which is given by $$\binom{4}{2}\frac{9!}{\left(2!\right)^{2}}$$

I thought that these are the arrangements that at least two such pairs will appear but I don't understand why this way gives the wrong answer?

答:
An easier way to get the answer to the second part, having already computed $n(exactly\; 2\;pairs) = 332,640 \;$would be to add $n(exactly\; 3\;pairs) + n(exactly\; 4\;pairs)\\
= \binom 43\left[\frac{8!}{(2!)^1}-\binom 4 4 \cdot\frac{7!}{(2!)^0}\right] + \binom 4 4\cdot \frac{7!}{(2!)^0}$

This would also have maintained the inclusion-exclusion formula used in the same pattern as in the first part.


问:
Small prime not dividing any among a "sparse" set of numbers
Let $n\in\mathbb{N}$ and $S\subseteq\mathbb{N}_{\le n^2}$ be such that $\forall_{a,b\in S, a\neq b} |a-b|\ge n$. Is there always a $p=O(n)$ dividing only $O(1)$ of numbers from $S$?

答: No, as it turns out. There are $O(n/\log n)$ such primes. Divide $\{1,\dots,n^2\}$ into $O(n/\log n)$ intervals of size $O(n\log n)$. For each such interval, choose one of the primes that is $O(n)$, and pick $\gg \log n$ multiples of that prime from the interval with minimal separation exceeding $n$. (Hopefully this outline is enough for you to work out the details.)


问:
inequality $\frac{2(x + y)^2}{2x^2 + y^2} \leq 3$
I'm looking for a "nice" way to show that the following inequality holds, i.e. without differentiating and determining the maximum:

$$
\frac{2(x + y)^2}{2x^2+ y^2} \leq 3 \quad \forall x,y \in \mathbb{R}
$$

It's rather easy to show

$$
\frac{4xy}{x^2 + y^2} \le 2
$$

and obviously

$$
\frac{2x^2 + 2y^2}{2x^2+y^2} \le 2
$$

but combining these two does not give me a sufficient low bound.

答:
Well $2x^2+y^2> 0$ and $2(x+y)^2\ge 0$ so

So $\frac{2(x + y)^2}{2x^2+ y^2} \leq 3\iff$

$2(x+y)^2 \leq3(2x^2 + y^2) \iff$

$2x^2 + 4xy +2y^2 \leq 6x^2 + 3y^2 \iff$

$0 \leq 4x^2 -4xy +y^2 \iff$

$0 \leq(2x -y)^2 \iff$ God's in his heaven and all's right in the world.


问:
What is this "union"?

The book is Cisinski HCHA 2019.

The notation suggests that that image does not depend on $X$ and $T$, does it?

答: That is true, as you can recover $X$ as the pullback of the cospan $Y\to T \leftarrow Z$.

Title: What is this "union"?


问:
Does this series converge
I need to check by Cauchy's convergence test if this series converges:
$$a_{n}=\frac{\sin(5)}{1*2} - \frac{\sin(5^2)}{2*3}+\ldots+(-1)^{n+1}\frac{\sin(5^n)}{n(n+1)}$$
I started to value
$$|a_{n+p} - a_{n}| < \epsilon$$
But don't know how to continue from:
$$\Biggl|(-1)^{n+2}*\frac{\sin(5^{n+1})}{(n+1)(n+2)}+\ldots+(-1)^{n+p+1}\frac{\sin(5^{n+p})}{(n+p)(n+p+1)}\Biggr|=\ldots$$

答:
Let's the call the expression you have obtained as  $S$. Applying triangle inequality to $S$ and using the fact that $|\sin(x)|\leq 1$ we get,
\begin{align*}
S \leq \frac{1}{(n+1)(n+2)} + \frac{1}{(n+2)(n+3)}+ \cdots + \frac{1}{(n+p)(n+p+1)}
\end{align*}
By using the well known identity$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$, the above expression simplifies to
\begin{align*}
\frac{1}{n+1} - \frac{1}{n+2}+\cdots+\frac{1}{n+p}-\frac{1}{n+p+1} = \frac{p}{(n+1)(n+p+1)}.
\end{align*}
Now choose an $n$ such that $\frac{1}{n+1}<\frac{\epsilon}{2} $. This implies
\begin{align*}
&&0 < \frac{1}{n+p+1}\leq\frac{1}{n+1}<\frac{\epsilon}{2}\\
\implies && S\leq \frac{p}{(n+p+1)(n+1)}<\epsilon
\end{align*}
Hence the above series is convergent

Title: Does this series converge


问:
Trying to find the sum of cosines
I have been trying to calculate this sum, could someone confirm if all my working is correct please :) :
$$S_k(x)=\sum_{n=0}^k\cos(nx)=\Re\sum_{n=0}^ke^{inx}=\Re\sum_{n=0}^k\left(e^{ix}\right)^n=\Re\left(\frac{1-e^{i(k+1)x}}{1-e^{ix}}\right)=\Re\left(\frac{1-\cos((k+1)x)-i\sin((k+1)x)}{1-\cos(x)-i\sin(x)}\right)\tag{1}$$
$$=\Re\left(\frac{\left[1-\cos((k+1)x)-i\sin((k+1)x)\right]\left[(1-\cos x)+i\sin x\right]}{(1-\cos x)^2+\sin^2x}\right)=\frac{(1-\cos x)(1-\cos((k+1)x))+\sin((k+1)x)\sin(x)}{1-2\cos(x)}$$
$$=\frac{1-\cos((k+1)x)-\cos (x)+\cos((k+1)x)\cos (x)+\sin((k+1)x)\sin(x)}{1-2\cos(x)}$$

$$\sin((k+1)x)\sin(x)=\frac{\cos(kx)-\cos((k+2)x)}{2}$$
$$\cos((k+1)x)\cos(x)=\frac{\cos(kx)+\cos((k+2)x)}{2}$$
$$\therefore\sin((k+1)x)\sin(x)+\cos((k+1)x)\cos(x)=\cos(kx)$$

$$S_k(x)=\frac{1-\cos((k+1)x)-\cos(x)+\cos(kx)}{1-2\cos(x)}\tag{2}$$

$$\cos(kx)-\cos((k+1)x)=-2\sin\left(\frac{(2k+1)x}{2}\right)\sin\left(\frac{-x}{2}\right)=2\sin\left(\left(k+\frac12\right)x\right)\sin\left(\frac x2\right)$$

$$S_k(x)=\frac{1-\cos(x)+2\sin\left(\left(k+\frac12\right)x\right)\sin\left(\frac x2\right)}{1-2\cos(x)}$$

答:
You can use this :

$$S_n(x)= \mathcal{Re}\left( \frac{1-e^{i(k+1)x}}{1-e^{ix}}\right)\\
=  \mathcal{Re}\left(e^{ikx/2}\frac{e^{-i(k+1)x/2}-e^{i(k+1)x/2}}{e^{-ix/2}-e^{ix/2}} \right)\\
=\mathcal{Re}\left( e^{ikx/2}\frac{\sin((k+1)x/2)}{\sin{(x/2)}}\right)\\
= \cos\left(\frac{kx}{2}\right)\frac{\sin\left(\frac{(k+1)x}{2}\right)}{\sin\left(\frac{x}{2}\right)}$$

Title: Trying to find the sum of cosines


问:
Given $G$, when can we find a division ring $R$ with $R^*=G$?
This is motivated by a characterization of finite cyclic groups, in which one proves

Let $G$ be a finite group. If $\#\{g\in G\colon g^d=e\}$ is at most $d$, then $G$ is cyclic.

The proof is not actually difficult, but a unnecessarily complicated idea occurred to me (maybe because the first time I saw such a result, it was used to prove that finite subgroups of the multiplicative group of a field are cyclic).

If we can construct this $G$ as the multiplicative group of a certain division ring, that is, $R^*\colon=\{r\in R\colon r\neq 1\}$. Then we know $G$ is abelian since finite division rings are commutative, and the abelian case follows from a direct use of the structure theorem of finite abelian groups.

Of course such a proof is unnecessarily complicated and very likely results in some cyclic arguments since it uses two very big structure theorems. But I guess it would still be nice to know what kind of groups are multiplicative groups of division rings.

Thanks very much!

答:
"The answer" for classifying which finite groups are subgroups of units of division rings was given by Amitsur in Finite subgroups of division rings. Trans. Amer. Math. Soc. 80 (1955), 361–386. This is generally considered to be the analogue of the "finite subgroups cyclic" theorem for the unit groups of fields.

(In Lam's paper on the quaternions, he describes an interesting lead-up to that discovery about Coexeter's work determining the finite subgroups of $\Bbb H^\ast$.)

This clearly acts as "local" restrictions on elements of finite order.


问:
Solution to $y′=x+y,\, y(0)=1$
I am trying to learn numerical method for differentials. It is key for me to understand this stuff right now, but I've come across an obstacle.

The books I read about these all seem to often say 'exact solution is easily found to be' in more or less the same way. I clearly have an issue with this, because I do not know any practical method, or even the name of the method I am suppose to learn, to find this sort of solution 'easily' or sometimes at all.

We shall illustrate our methods by applying them to the simple problem
$y′=x+y, y(0)=1, (3)$ which we call our benchmark problem. This
differential equation in $(3)$ is clearly linear, and the exact solution
is easily found to be $y=2e^x − x − 1$.

I look at this, and I don't understand. I guess I can say I covered one of these ideas which is that $y'=y, y(0) = 1$ because it is glearingly obvious that the only thing that differentiates to itself is $e^x$.

Now, $y' = x+y$, how would I approach that? What is the method behind the solution to get there? I tried to look at the solution instead, but $y'$ of the solution for me is $2e^x - 1$.

This must be something really trivial, but I just don't get it.

答:
This is trivial... when you know the theory of differential equations. I will lead you to a solution by an ad-hoc trial and error process.

You correctly observe that $(e^x)'=e^x$. If we plug this tentative solution in the equation, we have

$$e^x\color{red}=x+e^x$$ which is not what we want because of the term $x$. So let us try $y=e^x-x$, and we get

$$e^x-1\color{red}=e^x.$$ Then to get rid of the $1$, $y=e^x-x-1$, and

$$e^x-1=e^x-1.$$

But we are not done yet, because of the condition $y(0)=1$, while our solution says $y(0)=0$. But recall that $(e^x)'=e^x$, so that we can freely add terms $e^x$ without violating the equation. As we are off by one unit and $e^0=1$, the fix is

$$y=e^x+e^x-x-1.$$

The real way would be:

characterize the equation as first order linear with constant coefficients.

solve the "homogeneous" part of the equation, $y'-y=0$, giving $y=ce^x$ for some $c$;

find any solution of the initial equation, $y'-y=x$. As the RHS is a linear polynomial, we can try a linear polynomial, let $ax+b$. Then $a-ax-b=x$ yields $y=-x-1$.

combine these two solutions $y=ce^x-x-1$ and solve for $c$ after plugging the condition $y(0)=1$, which gives $c=2$.

Title: Solution to $y′=x+y,\, y(0)=1$


问:
Why is it that the congruence relations usually correspond to some type of subobject?
From the perspective of universal algebra, quotient structures of algebraic structures are built using congruence relations. If $A$ is an algebraic structure (a set with a bunch of operations on the set) und $R$ congruence relation on a set, then the quotient $A/R$ is well-defined and it will be an algebraic structure of the same type.

Now, as it turns out, in particular algebraic categories, these congruence relations on $A$ correspond exactly to some type of subobject of $A$. For instance, the congruence relations on a ring correspond precisely to the ideals of that ring; the congruence relations on a group correspond precisely to the normal subgroups of that group; the congruence relations on a module correspond precisely to the submodules of that module.

Why is it that the congruence relations usually correspond to some type of subobject? Is this a general phenomenon that can be generalized to all algebraic structures (as studied in this generality by universal algebra)?

答:
This is not true "usually"; for example, it's not true for monoids or semirings. It's a very special fact that it's true for groups and rings, and in both cases it's true for the same reason: the presence of inverses allows you to replace thinking about an equivalence relation $a \equiv b$ with thinking about $b^{-1} a \equiv 1$ for groups (producing normal subgroups) and $a - b \equiv 0$ for rings (producing ideals).

Note also that, strictly speaking, ideals are not subobjects in the category of rings (with identity).


问:
Name and reference for "ultra holomorphic" functions
Any holomorphic function $f : U\to \mathbb{C}$ from a domain $U\subset\mathbb{C}$ induces a real-analytic mapping $f(x+iy)=u(x,y)+iv(x,y)$ which as such induces by complexification of $x,y$ and $u,v$ a holomorphic mapping $f^* : U^* \to \mathbb{C}^2$ for some domain $U^*\subset  \mathbb{C}^2$. Of course the construction can be iterated to produce mappings $f^{*\cdots *}$ in $2^n$ complex variables for any $n\in \mathbb N$.

I guess that  such "ultraholomorphic" functions have a standard name in the litterature, perhaps related to terms like Cauchy-Riemann but I wasn't able to find anything in my search engine. Is there some classic reference I should be aware of? Is there some nice/surprising property of $f^*$ (except the obvious real+Cauchy-Riemann)?

答:
Given $R$ a subring of $M_n(\Bbb{C})$ let $$\eta(R)=\left\{\pmatrix{A&B\\-B&A},A,B\in R\right\}\subset M_{2n}(\Bbb{C}).$$
$\eta(R)$ is isomorphic to $R[t]/(t^2+1)$ and $\eta^2(R)$ is isomorphic to $R[t,u]/(t^2+1,u^2+1)$ and so on.

For $f(x)=\sum_{k\ge 0} c_k x^k$ an entire power series with real coefficients then you are just looking at $$f^{\overbrace{*\cdots *}^n} : \eta^n(\Bbb{C})\to \eta^n(\Bbb{C}), \qquad f^{\overbrace{*\cdots *}^n} (\alpha)= \sum_{k\ge 0} c_k \alpha^k$$
For $f(x)=\sum_{k\ge 0} (c_k+id_k) x^k$ then you are looking at
$$f^{\overbrace{*\cdots *}^n} (\alpha)= \sum_{k\ge 0} (c_k+t d_k) \alpha^k$$ where $t$ is the $2^{n-1}\times 2^{n-1}$ diagonal matrix of $M_{2^{n-1}}(\eta(\Bbb{C}))$ where each diagonal entry is $\pmatrix{0&1\\-1&0} \in \eta(\Bbb{C})$.

Then your analytic function $\Bbb{C}^{2^n}\to \Bbb{C}^{2^n}$ is recovered from the natural vector space isomorphism $\Bbb{C}^{2^n}\to \eta^n(\Bbb{C})$.


问:
A company hires $11$ new employees, and they will be assigned to four different departments, A, B, C, D

A company hires $11$ new employees, and they will be assigned to four different departments, A, B, C, D. Each department has at least one new employee. In how many ways can these assignments be done?

I know that for each section (A,B,C,D) I should add a () and as long as every section must get a new employee
we should start like this:
$$(x+x^2/2!+x^3/3!+...)^4$$
then if we look  it's $(e^x-1)^4$.
After this step I don't know what to do.

答:
Using balls and boxes, we know that there are $11$ distinct objects and $4$ distinct bins.

First, ignoring the restriction that each bin must get at least $1$ object, you get $4^{11}$ ways to arrange since each object have $4$ ways, and there are $11$ objects.

To take care of the restriction, we subtract all cases where only 1 bin is empty, then add the cases where only 2 bins are empty, etc (PIE). Therefore, we have $11$ objects into $3$ bins, and one of the bins is guaranteed empty, and there are $\binom41$ ways to choose that bin. Therefore, the restricted cases for $1$ box being empty are $\binom41 \cdot (4-1)^{11}$.

For the case where $2$ bin are empty, do the same reasoning as above, and get $\binom42\cdot(4-2)^{11}$.

For the case where $3$ bin are empty, you can get $\binom43\cdot(4-3)^{11}$

And for the $4$ bins empty case you can get $\binom44\cdot(4-4)^{11}$

Alternate adding and subtracting: $4^{11}-\binom41 \cdot (4-1)^{11}+\binom42\cdot(4-2)^{11}-\binom43\cdot(4-3)^{11}+\binom44\cdot(4-4)^{11}=3,498,000$

Sorry, since I am only a high schooler taking precalc right now, I need to research how to do generating functions for combinatorics. The answer to that might take a while.


问:
Intuition behind binomial variance
Suppose that I perform a stochastic task $n$ times (like tossing a coin) and that $p$ is the probability that one of the possible outcomes occurs. If $K$ is the stochastic variable that measures how many times this outcome occurred during the whole experiment, and if all the events are mutually independent, then the probability that $K$ is equal to a specific $k$, with $0\leq k\leq n$, is
$$\mathrm{Pr}\{K=k\} = \binom n k p^k (1-p)^{n-k} =: b_n(k), $$
which makes intuitive sense to me.

Now suppose that I want to know what the average, or expected, value of $K$ is going to be: the formulae tell me that
$$\begin{split}
\langle K\rangle &= \sum_{k=0}^n k b_n(k) = \sum_{k=0}^n k \binom n k p^k (1-p)^{n-k} \\
&= \sum_{k=0}^n k \frac{n(n-1)!}{k(k-1)!(n-k)!} p p^{k-1} (1-p)^{n-k} \\
&= np \sum_{\kappa=1}^\nu \frac{\nu!}{\kappa!(\nu-\kappa)!} p^\kappa (1-p)^{\nu-\kappa} = np(p + 1- p)^{\nu} \\ &= np,
\end{split}$$
where I’ve made the substitutions $\kappa = k-1$ and $\nu = n-1$. Notwithstanding the mathematical certainty of this derivation, it also makes perfect intuitive sense to me that $np$ should be the expected value of $K$, since it is the product of the probability of the outcome times the number of trials performed: if there’s a $1/6$ chance that I roll a 5 on a fair dice, and I throw it $600$ times, then I expect to see a 5 about $100$ of those times.

If instead I want to know how I should expect the outcomes to vary around the expected value, I may compute the variance of $K$: with the same substitutions as before,
$$\begin{split}
 \mathrm{Var}[K] &= \langle K^2\rangle - \langle K \rangle^2 = \left(\sum_{k=0}^n k^2 b_n(k) \right) -n^2p^2 \\
&= \left( \sum_{k=1}^n k^2 \binom n k p^k (1-p)^{n-k}\right) -n^2p^2 \\
&= \left( np \sum_{\kappa = 0}^\nu (\kappa +1) \binom \nu \kappa p^\kappa (1-p)^{\nu-\kappa} \right)-n^2p^2 \\
&= \left(np \Big(\sum_{\kappa=0}^\nu \kappa b_\nu(\kappa) +(p + 1-p)^\nu \Big) \right) -n^2p^2 \\
&= np(\nu p + 1) - n^2 p^2 = np( np - p + 1 - np) \\ &= np(1-p).
\end{split}$$
Again, the derivation is mathematically crystalline; but why should I expect that this be the formula for the variance of $K$? Why does multiplying the expected value of $K$ times the probability that my outcome doesn’t occur give me a measure of the dispersion of $K$? In other words, how can I justify this formula for variance intuitively in a similar way as I can with the formula for the mean?

EDIT. Up until now, I’ve received answers that are just perfectly good explanations of how to derive the formula for the variance of $K$ in ways that differ from the one presented above. That’s not what I’m asking for. The ideal answer should contain as few formulae as possible, and use simple enough words to explain not why the formula is mathematically true, but why it’s reasonable and couldn’t possibly be otherwise – something like the intuitive explanation for $\langle K\rangle = np$ that I gave above.

答:
I am not a statistician and never got into calculus stuff. I have always used intuition to make sense of statistics in social science and I will try to explain how I conceptualize the formula for variance. Also a short thank you @Mathem-antics whose answer helped me write this out better.

If P is the probability of an event occurring (i.e. my response variable being, Y=1) and we already established that the mean of the distribution is E(Y)= P, what is the variance?

To intuitively answer that let's think of two examples:

When is variance at its maximum?

Randomness from the toss of a coin, when the probability of getting head/tail is equal= 0.5 and 0.5

When is it at its lowest?

When we always get same result. Such as probability of a Cambridge grad passing an elementary school exam, here probability of passing is 100 and failing is 0. Y=1 (which is passing) will always happen. Therefore the variance ought to be 0 here.

Now let's understand the spectrum in which this occurs

When we move away from toss of a coin scenario (where both P and 1- P is 0) and go in either direction, there is a pattern and variance starts to decrease.
For all combinations of P and 1- P, (from a spectrum of 1-0) the maximum possible variance occurs during the 50/50 scenario and lowest in 100/0 scenario. So the formula of variance must be designed such that it captures this information. And the best way to do that is through multiplication, if you look at the following table:

Therefore, maximum variance is captured at p=0.5, which is expected, and starts decreasing and drops to 0 when p=1 or 0. Therefore, intuitively it makes sense to multiply them, because the moment you see 0.25 you know it is completely random or when you 0.18 you know that there is definitely a pattern in the distribution, one of the events is more likely to happen and the distribution is dense around that. This is how I see it anyways.

Since I don't have deep understanding of stats there may be some technical error in the phrasing of my sentences, I apologize for that in advance.

Title: Intuition behind binomial variance


答:
Recall that congruences on $A$ can be viewed as certain subalgebras of its square  $A^2,\,$ e.g. see here.

In algebras like groups and rings, where we can normalize $\,a = b\,$ to $\,a\!-\!b = \color{#c00}0\,$ congruences are determined by a single congruence class (e.g. an ideal in a ring). This has the effect of collapsing said relationship between congruences with subalgebras from  $A^2$ down to  $A.\,$  Such algebras are called ideal determined varieties and they have been much studied.

One answer to your question is that ideal-determined varieties are characterized by two properties of their congruences, namely being $\,\rm\color{#c00}{0\text{-regular}}\,$ and $\rm\color{#c00}{0\text{-permutable}}$. Below is an excerpt of one paper on related topics that yields a nice entry point into literature on this and related topics.

On subtractive varieties iv: Definability of principal ideals.

Paolo Agliano and Aldo Ursini

Foreword

We have been asked the following questions:

(a) What are ideals in universal algebra good for?

(b) What are subtractive varieties good for?

(c) Is there a reason to study definability of principal ideals?

Being in the middle of a project in subtractive varieties,
this seems the right place to address them.

To (a). The notion of ideal in general algebra [13], [17], [22] aims
at recapturing some essential properties of the congruence classes of $0$,
for some given constant $0$. It encompasses: normal subgroups, ideals
in rings or operator groups, filters in Boolean or Heyting algebras,
ideals in Banach algebra, in l-groups and in many more classical
settings. In a sense it is a luxury, if one is satisfied with the
notion of "congruence class of $0$". Thus in part this question might
become: Why ideals in rings? Why normal subgroups in groups? Why filters
in Boolean algebras?, and many more. We do not feel like attempting any
answer to those questions. In another sense, question (a) suggests similar
questions: What are subalgebras in universal algebra good for? and many
more. Possibly, the whole enterprise called "universal algebra" is
there to answer such questions?

Having said that, it is clear that the most proper setting for a theory
of ideals is that of ideal determined classes (namely, when mapping a
congruence E to its $0$-class $\,0/E$ establishes a lattice isomorphism between
the congruence lattice and the ideal lattice). The first paper in this
direction [22] bore that in its title.

It comes out that -- for a variety V -- being ideal determined is the
conjunction of two independent features:

V has $\,\rm\color{#c00}{0\text{-regular}}\,$ congruences, namely for any congruences $\rm\,E,E'$
of any member of $V,$ from  $\,\rm 0/E = 0/E'$ it follows  $\rm\,E = E'$.

V has $\,\rm\color{#c00}{0\text{-permutable}}\,$ congruences, namely for any congruences $\,\rm E,E'$
of any member of $V,$ if  $\,\rm 0 \ E\ y \ E'\, x,\,$ then for some $\rm z,\  0\ E'\, z\ E\ x.$

Body:
From the perspective of universal algebra, quotient structures of algebraic structures are built using congruence relations. If $A$ is an algebraic structure (a set with a bunch of operations on the set) und $R$ congruence relation on a set, then the quotient $A/R$ is well-defined and it will be an algebraic structure of the same type.

Now, as it turns out, in particular algebraic categories, these congruence relations on $A$ correspond exactly to some type of subobject of $A$. For instance, the congruence relations on a ring correspond precisely to the ideals of that ring; the congruence relations on a group correspond precisely to the normal subgroups of that group; the congruence relations on a module correspond precisely to the submodules of that module.

Why is it that the congruence relations usually correspond to some type of subobject? Is this a general phenomenon that can be generalized to all algebraic structures (as studied in this generality by universal algebra)?


问:
Find a representation of the field $\mathbb{F}_{16}$ as $\frac{\mathbb{Z}_2}{(f)}$
I have to find representation of the field $\mathbb{F}_{16}$ as $\frac{\mathbb{Z}_2}{(f)}$ where $f \in \mathbb{Z}_2[X]$ a monic irreducible polynomial over $\mathbb{Z}$.

I don't really have a clue how to solve this, but I think it has to do with finding the irreducible polynomials that divide $X^{16}-X$ and have a degree that divide 4.

I also have to find a unique subfield of $\mathbb{F}_{16}$ with 4 elements starting from a representation that I found, which I only can do after finding the specific representation, so I'm also stuck on this one.

Thanks in advance!

答:
Well, referring to the comment, the decomposition of $X^{16}-X$ into irreducibles over $GF(2)$ is

$X\cdot (X+1) \cdot (X^2+X+1)\cdot (X^4+X+1) \cdot (X^4+X^3+1)\cdot (X^4+X^3+X^2+X+1)$$

where $X^4+X+1$ and $X^4+X^3+1$ are primitive and conjugate, while $X^4+X^3+X^2+X+1$ is not primitive as it divides $X^5+1$.


答:
This is generally false. The fact that it holds in groups and rings is due to the fact that in each case we have an operation with identity and inverses, namely the group operation or the ring addition respectively.

As long as we have such an operation, every congruence is indeed determined by a single class. To see this, suppose $S$ is a structure with an invertible operation $*$, $a\in S$, and $\sim,\approx$ are congruences on $S$ with $[a]_\sim=[a]_\approx$. Fix $b\in S$; we want to show $[b]_\sim\subseteq[b]_\approx$ (from which by symmetry and universal generalization we'll get $\sim=\approx$).

Suppose $b\sim c$. Then $b b'a\sim cb'a$, where $x'$ denotes the $*$-inverse of $x$. This means $a\sim cb'a$, and so since $[a]_\sim=[a]_\approx$ we get $a\approx cb'a$. Now we undo the previous step: we multiply on the right by $a'b$ to get $b\approx c$ as desired.

Title: Why is it that the congruence relations usually correspond to some type of subobject?


问:
Is the Euclidean norm unique norm of two-dimensional real vector space?
Let $V$ is a two-dimensional vector space over a field $\mathbb{R}$ with a norm $\| \cdot  \|: V \rightarrow \mathbb{R}$ which, in addition, satisfies these axioms:
$$
\| (x, 0) \| = \| (0, x) \| = | x | \;\;\; \forall x \in \mathbb{R} \tag{1}\label{ax1}
$$
$$
\| (x, x) \| = \sqrt{2}\: | x | \;\;\; \forall x \in \mathbb{R} \tag{2}\label{ax2}
$$

Is it possible to prove that the Euclidean norm $\| (x, y) \| := \sqrt{x^2+y^2}$ is the only norm satisfying these additional axioms?

Or find a example of a norm, that is not Euclidean?

A naive attempt can be a function:

$$
\| (x, y) \| = \left\{ \begin{array}{rc1}
(|x|^p+|y|^p)^{\frac{1}{p}} & \mbox{for} & x \ne y\\
\sqrt{2}\;|x| & \mbox{for} & x = y \\
\end{array}\right.
$$
Which satisfies these axioms, however is discontinuous in case $p \ne 2$.

Solution

Based on the answer from @Qiaochu Yuan, I found an example of the non-euclidean norm satisfying these additional axioms:

$$
\| (x, y) \| = \left\{ \begin{array}{rc1}
(\sqrt{2}-1)|x|+|y| & \mbox{for} & |x| \leq |y| \\
|x|+(\sqrt{2}-1)|y| & \mbox{for} & |x| \geq |y|
\end{array}\right.
$$

Another question is whether the enhanced of the requirement from the continuity to the differentiability of the norm will be sufficient for the unique solution in the form of the Euclidean norm.

答:
(Norms on $\mathbb{R}^n$ are necessarily continuous wrt the Euclidean topology.)

A norm on $\mathbb{R}^n$ is completely specified by specifying its unit ball $\{ x \in \mathbb{R}^n : \| x \| \le 1 \}$, which must be symmetric (meaning closed under $x \mapsto -x$), convex, and compact (equivalently, closed and bounded (wrt the Euclidean norm)). Conversely, any symmetric convex compact subset of $\mathbb{R}^n$ is the unit ball of a unique norm (see, for example, Theorem 1.17 in these notes).

Your conditions specify six points on the unit sphere, namely the points $(\pm 1, 0), (0, \pm 1), \left( \pm \frac{1}{\sqrt{2}}, \pm \frac{1}{\sqrt{2}} \right)$. These are six of the eight points of the regular octagon inscribed in the Euclidean unit circle. The solid regular octagon is a symmetric convex compact subset of $\mathbb{R}^2$, so we can take the norm with this unit ball.


问:
Counting exercise
Three players a,b,c take turns in a game according to the following rules:

At the start A and B play (so C does not play). The winner of the first trial plays against C and so on until one of the players wins two trials in a row.
  Possible outcomes are
  aa,acc,acbb,acbaa,
  bb,bcc,bcaa,bcabb etc.

We have to prove that probability of A winning equals p(A) = 5/14, p(B) = 5/14, p(C) = 2/7

I have been stuck on this problem for a long time. The only thing I have been able to find out so far is that C can never win on even turns.

Re-edit

Players continue to play until one of them wins two times consecutively and all players are equally good at playing the game.

Apologies for leaving such crucial information.

I have finally solved this problem with a different approach

Sample space

aa ,acc ,acbb ,acbaa ,acbacc ,acbacbb ,acbabcaa ,...

bb ,bcc ,bcaa ,bcabb ,bcabcc ,bcabcaa ,bcabcabb ,...

Each point in the sample space has a probability (1/2^k) associated with it where k is the number of turns.

For example probability of point (aa) = 1/4

Now let us construct a table enumerating probabilities -

Let us consider the event that C wins overall. But C can only win if k = 3,6,9,12,...

P(C) = P(C3) + P(C6) + P(C9) ......

where P(Ck) = Probability of C winning overall after k turns.

P(C) = 1/4 + 1/32  .....

P(C) = a / (1 - r)       ( Sum of an infinite GP )

a = 1/4                  ( First Term )

r = 1/8                  ( Ratio )

P(C) = 2/7

P(A) = 5/14 = P(B)

答:
Here's my crack at the question:

Since both players $A,B$ satisfy:

$$M_k(p) = \{\{1, 1\},\{1,0,1,1\},\{1,0,1,0,1,1\},...\}$$ where $M_k(p)$ is the $k$th way for player $p$ to win the game. $\{1\}$ indicates winning the trial, and $\{0\}$ indicates loss.

As player C, and only $C$, satisfies;
$$N_k(p) = \{\{0,1, 1\},\{0,1,0,1,1\},\{0,1,0,1,0,1\},...\}$$

Trials needed to have elapsed for $C$ to win the game is $\mid N_1(C)\mid = (\mid M_1(A) \mid +\mid M_1(B)\mid)-1 = 3$

Thus, chances of player $p$ winning the game in this case is:

Define players $A,B,C$ as $0,1,2$

$t=$Trials needed to win the game.

$n=$Number of players.

$k=$Number of players with the same chances of winning($A,B$), .

$$
P(p) = \left\{\begin{matrix}
{kt-1 \over k((M_1(p)\;t)+1)} & p < 2\\
{t \over (N_1(p)\;t)+1} & p = 2
\end{matrix}\right.
,\quad -1 < p < n
$$

Where $P(p)$ is the probability of player $p$ winning.

Edit:

To clarify: The reason that the chance of player $C$ winning is highest, is because there is a greater chance that player ($A,B$) won't win because they share the same set of combinations needed to win.

Disclaimer:  I have not tested this thoroughly in any way, and drafting was started prior to question being answered.

Title: Counting exercise


问:
Finite fields of even cardinality have characteristic $2$
Computer science studies-

as for now we have dealt with the set-theory and residue class rings with modular arithmetic, although we did not really define the connection between the two so rigorously

So I have to prove the following property for every $x \in \mathbb{F}$, whereas $\mathbb{F}$ is a finite field with even cardinality:
$$ x+x = 0$$

As far as I considered it could be satisfied only with modular arithmetic, but apart of that I do not have a concrete approach..

Thank you for your help

答:
First, let's consider $F$ as an abelian group under addition.  Since $\vert F \vert$ is even, let $P$ be a $2$-Sylow subgroup of $F$.  Then $P$ is non-trivial and any non-zero element $z \in P$ has order $2^k$ for some $k \in \Bbb N$.  If $k \gt 1,$ let $y = 2^{k-1}z$.  If $k=1$, let $y = z$.  In either case, $F$ must have some element $y$ with order $2$; in other words, for that specific $y, y \neq 0 \text{ and } y+y=0$.

There's another way to see this.  Assume that $F$ has no element of additive order $2$.  Then you'd be able to partition $F$ by pairing up its non-zero elements,$\{x_n, -x_n\}$.  But $F$ has an odd number of non-zero elements, so that's not possible.  Thus, $F$ must have an element of order $2$.

Now choose an arbitrary $x \in F$ with $x \neq 0$.  Then $\frac xy$ exists and is nonzero and $x+x=y \frac xy+y \frac xy=(y+y) \frac xy=0$.

Alternatively, $0=y+y=1y+1y=(1+1)y \Rightarrow 1+1=0.$  Thus, for any $x \in F$ we have $x+x=1x+1x=(1+1)x=0x=0$.

Incidentally, this also proves that $\vert F \vert$ must be a power of $2$ because if an odd prime divided $\vert F \vert$, then $F$ would have to have an element of odd (additive) order, which we just proved can't happen.


问:
In a unital division algebra, does the unit have norm 1?
Suppose that $\mathbb{R}^n$ is a unital division algebra.
That is, $\mathbb{R}^n$ is furnished with a bilinear map
$$ *: \mathbb{R}^n\times \mathbb{R}^n\to \mathbb{R}^n $$
such that
$$ x*y = 0 \implies x=0 \text{ or }y=0. $$
In addition, there is a distinguished element $\mathbf{1}\in \mathbb{R}^n$ where
$$ \mathbf{1}*x = x*\mathbf{1} = x $$
for all $x\in\mathbb{R}^n$.

Is it now true that $$ \| \mathbf{1} \| =1 $$
where $\|\cdot\|$ is the usual euclidean norm on $\mathbb{R}^n$? In other words, is the unit of $\mathbb{R}^n$ in the unit sphere $S^{n-1}$?

I have tried proving that
$$ \|x*y\| =\|x\|\|y\| \quad \forall x,y\in \mathbb{R}^n $$
without success. Of course, if this above equation holds, then $\|\mathbf{1}\|=1$ is an immediate consequence.

A positive answer to this question would help me prove the following assertion: If $\mathbb{R}^n$ is a unitial division algebra, then the sphere $S^{n-1}$ is an $H$-space.

Of course, I cannot use the classification of division algebras over $\mathbb{R}$, to just say "well in all four cases, $\mathbb{R}, \mathbb{C}, \mathbb{H}, \mathbb{O}$ the unit clearly has norm $1$."

EDIT: Apparently the result $\|\mathbf{1}\|=1$ is false, which puts me back at square one.

答:
A positive answer to this question would help me prove the following assertion: If $\mathbb{R}^n$ is a unitial division algebra, then the sphere $S^{n-1}$ is an $H$-space.

You don't need the fact you're trying to prove to prove this. $S^{n-1}$ can be defined without any reference to the Euclidean norm, as the quotient of $\mathbb{R}^n \setminus \{ 0 \}$ by scalar multiplication by positive reals (very similar to the construction of real projective space). The multiplication on an $n$-dimensional unital division algebra induces a multiplication on this quotient because, by hypothesis, the product of two nonzero elements is nonzero, and the multiplication is bilinear so respects scalars. And the image of $1$ under this quotient map continues to act as an identity.

Said more concretely, instead of using the multiplication $\ast$ you started with, you can consider "normalized multiplication" $\frac{x \ast y}{\| x \ast y \|}$, which by construction has the property that if $\| x \| = 1$ and $\| y \| = 1$ then $ \| x \ast y \| = 1$, and also that $1$ continues to act as an identity (on the unit sphere).

The recognition that we can scale multiplication arbitrarily (in a way that depends on the vector, even) shows that the fact you're trying to prove can't possibly be true, since e.g. we can consider $\mathbb{R}$ with the scaled multiplication $x \ast y = 2xy$, which has an identity given by $\frac{1}{2}$ which does not have Euclidean norm $1$.


问:
$\mathbb{Q}(\pi) \subsetneq \mathbb{Q}(\sqrt {\pi})$?
I'm studying transcendental numbers and I have a question, $\mathbb{Q}(\pi) \subsetneq \mathbb{Q}(\sqrt{\pi})$??

I know that $\sqrt{\pi} \notin \mathbb{Q}(\pi),$ because if $\sqrt{\pi}$ is in $\mathbb{Q}(\pi)$ then there exists $p,q \in \mathbb{Q}[x]$ s.t $\sqrt \pi = \frac{p(\pi)}{q(\pi)} \Rightarrow \pi$ is root of $x(q(x))^2 - (p(x))^2$ (Actually is not clear why this polynomial is nonzero).

My question is: How can I prove that $\pi \in \mathbb{Q}(\sqrt{\pi})$?

答:
$\pi\in\Bbb Q(\sqrt \pi)$ because $\pi=(\sqrt \pi)^2$.

Also, from $\sqrt\pi=\frac{p(\pi)}{q(\pi)}$ you obtain that $\sqrt \pi$ is a root of $xq(x^2)-p(x^2)$, which is non-zero because in $xq(x^2)$ has all monomials of odd degree (and it isn't the zero polynomial) and $p(x^2)$ has only of even degree. $\sqrt\pi$ can't be algebraic because then $\pi$ would be too.


问:
ideal polynomial
Let $R$ be a commutative ring and denote $A := R[x,y]$, $A_x$ le localization of $A$ by $\{ x^n \mid n \geq0\}$ and $A_y$ the same for $y$.

Let $I_x \subset A_x$, $I_y \subset A_y$  ideals s.t $I_x = I_y$ in $A_{xy}$, so $I_x$ and $I_y$ become equal in the localization by $xy$.

Does there exist a unique ideal $I$ of $A$ s.t. $I = I_x$ in $A_x$ and $I = I_y$ in $A_y$ ?

There is no uniqueness; see comment of Mohan. So the question is existence.

答:
Yes. See here for a solution to a generalization of this problem. In this specific case, let $\iota_x:A\rightarrow A_x$ and $\iota_y:A\rightarrow A_y$, and likewise $\tau_x:A_x\rightarrow A_{xy}$ and $\tau_y:A_y\rightarrow A_{xy}$, be the canonical localization maps. The equality condition that you express is then that $\tau_x(I_x)A_{xy}=\tau_y(I_y)A_{xy}$. Note also that $$\ker{\tau_x}=\{f\in A_x:\exists k\in\mathbb{N}\text{ such that }(y^k\big/1)f=0\big/1\in A_x\},$$ and that likewise $$\ker{\tau_y}=\{f\in A_y:\exists k\in\mathbb{N}\text{ such that }(x^k\big/1)f=0\big/1\in A_y\}$$

Now, let $I=\iota_x^{-1}(I_x)\cap\iota_y^{-1}(I_y)$. We claim that the image of $I$ in $A_x$ generates $I_x$. Thus let $J=\iota_x(I)A_x$ be the ideal of $A_x$ generated by the image of $I$. Because $I\subseteq\iota_x^{-1}(I_x)$, we have $\iota_x(I)\subseteq I_x$ and so certainly $J\subseteq I_x$. To show the other direction, let $a\big/x^m\in I_x$ be arbitrary, for some $a\in A$. Now, to show $a\big/x^m\in J$, it suffices to show $a\big/1\in J$, so we just need to find $b\in I$ and $n\in\mathbb{N}$ such that $a\big/1=b\big/x^n\in A_x$.

To do this, note that – because $a\big/1\in I_x$ – we have $a\big/1\in\tau_x(I_x)A_{xy}=\tau_y(I_y)A_{xy}$, so there is $f\in I_y$ and $l\in\mathbb{N}$ such that $a\big/1=f\big/x^l\in A_{xy}$. This means that $f-x^la\big/1\in\ker{\tau_y}$, and so there is some $k\in\mathbb{N}$ such that $(x^k\big/1)(f-x^la\big/1)=0\big/1\in A_y$. But this means $x^{k+l}a\big/1=(x^k\big/1)f\in I_y$, since $f\in I_y$, and so $x^{k+l}a\in\iota_y^{-1}(I_y)$. Also, because $a\big/1\in I_x$, we have $a\in\iota_x^{-1}(I_x)$, and so certainly $x^{k+l}a\in\iota_x^{-1}(I_x)$. Putting these two facts together gives $x^{k+l}a\in I$, and so letting $b=x^{k+l}a$ and $n=k+l$ gives $a\big/1=b\big/x^n\in J=\iota_x(I)A_x$ as desired.

Thus $\iota_x(I)A_x=I_x$, and by exactly the same argument we can show $\iota_y(I)A_y=I_y$, so we have proven existence. However, as Mohan points out, $I$ is certainly not unique, for the images of $I\cdot(x,y)<I$ in $A_x$ and $A_y$ will clearly generate the same ideals as the images of $I$. The problem is essentially that the ideal $(x, y)$ generated by $x$ and $y$ is not all of $A$; the answer I've linked to above shows how a related condition in the general case would allow for a proof of uniqueness.

Title: ideal polynomial


问:
Evaluate $\lim_{x \rightarrow \pi/2}\left(\arctan(x/(\tan(x)-x))/(x-\pi/2)\right)?$
I need to evaluate this limit without L'Hospital's Rule but I haven't found a method. $$\lim_{x \rightarrow \pi/2}\left(\arctan(x/(\tan(x)-x))/(x-\pi/2)\right)$$

答:
Note that $\lim_{x\to \pi/2}{\arctan\left(\frac{x}{\tan(x)-x}\right)}=0$. Then this is just the limit definition of the derivative:

$$
\lim_{x\to \pi/2}\frac{\arctan\left(\frac{x}{\tan(x)-x}\right)-0}{x-\pi/2}
$$
$$
= \lim_{x\to \pi/2}\frac{d}{dx}{\arctan\left(\frac{x}{\tan(x)-x}\right)}
$$A tedious chain-rule calculation produces:
$$
= \lim_{x\to \pi/2}\frac{- x \sec^2(x)+\tan(x) }{\tan^2(x)+2 x^2  - 2 x \tan(x)}
$$
Multiply through by $\cos^2(x)$ to evaluate the limit.
$$
= \lim_{x\to \pi/2}\frac{-x+\sin(x)\cos(x)}{\sin^2(x) +2 x^2\cos^2(x) - 2 x \sin(x)\cos(x)} = \frac{-\pi/2+0}{1+0+0}=\frac{-\pi}{2}
$$


答: You could easily use $$\cos(nx)=\frac{e^{inx}+e^{-inx}}{2}$$ to make calculations bit easier. Let me propose you an alternative less messy solution, which uses telescoping method.  Note that $$\sin((n+1)x)-\sin((n-1)x)=2\cos(nx)\sin(x).$$ Now telescoping gives $$\sin((n+1)x)+\sin(nx)-\sin(x)=2\sin(x)\sum_{k=1}^n\cos(kx)$$

Title: Trying to find the sum of cosines

Body:
I have been trying to calculate this sum, could someone confirm if all my working is correct please :) :
$$S_k(x)=\sum_{n=0}^k\cos(nx)=\Re\sum_{n=0}^ke^{inx}=\Re\sum_{n=0}^k\left(e^{ix}\right)^n=\Re\left(\frac{1-e^{i(k+1)x}}{1-e^{ix}}\right)=\Re\left(\frac{1-\cos((k+1)x)-i\sin((k+1)x)}{1-\cos(x)-i\sin(x)}\right)\tag{1}$$
$$=\Re\left(\frac{\left[1-\cos((k+1)x)-i\sin((k+1)x)\right]\left[(1-\cos x)+i\sin x\right]}{(1-\cos x)^2+\sin^2x}\right)=\frac{(1-\cos x)(1-\cos((k+1)x))+\sin((k+1)x)\sin(x)}{1-2\cos(x)}$$
$$=\frac{1-\cos((k+1)x)-\cos (x)+\cos((k+1)x)\cos (x)+\sin((k+1)x)\sin(x)}{1-2\cos(x)}$$

$$\sin((k+1)x)\sin(x)=\frac{\cos(kx)-\cos((k+2)x)}{2}$$
$$\cos((k+1)x)\cos(x)=\frac{\cos(kx)+\cos((k+2)x)}{2}$$
$$\therefore\sin((k+1)x)\sin(x)+\cos((k+1)x)\cos(x)=\cos(kx)$$

$$S_k(x)=\frac{1-\cos((k+1)x)-\cos(x)+\cos(kx)}{1-2\cos(x)}\tag{2}$$

$$\cos(kx)-\cos((k+1)x)=-2\sin\left(\frac{(2k+1)x}{2}\right)\sin\left(\frac{-x}{2}\right)=2\sin\left(\left(k+\frac12\right)x\right)\sin\left(\frac x2\right)$$

$$S_k(x)=\frac{1-\cos(x)+2\sin\left(\left(k+\frac12\right)x\right)\sin\left(\frac x2\right)}{1-2\cos(x)}$$


问:
Is the tensor algebra a Lie algebra?
I found that the reverse is true. To show that it is a Lie algebra I just have to show that the Jacobi identity holds, so I was wontering if it could be true either for the tensor algebra of a vector space or for a subalgebra of that space.</p>

答:
As Torsten says in the comments, every associative algebra is a Lie algebra with bracket the commutator bracket $[a, b] = ab - ba$.

The tensor algebra $T(V)$, as a Lie algebra, has a subalgebra generated by $V$ (which is not all of $T(V)$), and $T(V)$ has a special relationship to this subalgebra: at least in characteristic $0$, it is precisely the universal enveloping algebra of this subalgebra, and moreover this subalgebra is the free Lie algebra $L(V)$ on $V$. This mostly follows from the fact that the two have the same universal property, except that there's a little work to be done explaining why the natural map $L(V) \to U(L(V))$ is injective. But this follows from the Poincare-Birkhoff-Witt theorem.

Title: Is the tensor algebra a Lie algebra?


问:
Inequality with convex functions

A function $f : I\to \mathbb{R}$ is convex on an interval $I \subseteq \mathbb{R}$ when $\lambda f(x) + (1 − \lambda)f (y) \ge f (\lambda x+(1 − \lambda)y )$ holds for all $x,y \in I$. Prove
$$
\sum_{i=0}^n \lambda_i f(x_i) \ge  f\left(\sum_{i=0}^n \lambda_i x_i\right)
$$
holds for all $n\ge 2, x_1, \ldots, x_n \in I$ and $\lambda_1, \ldots, \lambda_n \ge 0$ satisfying $1 =  \sum_{i=0}^n \lambda_i$.

My Attempt:

If $\lambda_1>0$ then I take $\lambda_0 = \lambda_1 + \lambda_2$
and $\Lambda_1 = \lambda_1/\lambda_0, \Lambda_2 = \lambda_2/\lambda_0, x_0 = \Lambda_1 x_1 + \Lambda_2 x_2$
and
$$
\lambda_0(\Lambda_1 f(x_1) + \Lambda_2 f(x_2)) + \sum_{j=1}^n \lambda_i f(x_i).
$$
Pls help me on what to do next and how to complete the proof

答:
HINT
I would proceed by Induction. The base case $n=2$ is already given, so let's inductively assume this holds for all $n$ up to $k-1$, and then prove it for $n=k$.

Assume WLOG that $\lambda_0 \ne 0$ (otherwise, you must have some non-zero coefficient since they have to sum up to 1, so pick any non-zero coefficient and renumber them).

Let $x = \sum_{i=0}^{k-1} x_i \lambda_i/1-\lambda_k$. Then by the inductive assumption, we have
$$
\sum_{i=0}^{k-1} f(x_i) \frac{\lambda_i}{1-\lambda_k}
 \ge f\left(\sum_{i=0}^{k-1} x_i \frac{\lambda_i}{1-\lambda_k} \right)
 = f(x)
$$
and since $I$ is convex, $x \in I$. Then,
$$
f\left(\sum_{i=0}^k x_i \lambda_i \right)
 = f\left((1-\lambda_k) x + \lambda_k x_k \right)
 \le (1-\lambda_k) f(x) + \lambda_k f(x_k)
$$

Can you finish?

Title: Inequality with convex functions


问:
$f:X \to Y$ a homeomorphism, prove $\forall x\in X$ $f$ from $X \setminus \lbrace x\rbrace$ to $Y \setminus \lbrace f(x) \rbrace$ is a homeomorphism

let $f:X \to Y$ a homeomorphism, prove that for all $x\in X$ $f$ induces an homeomorphism between $X \setminus \lbrace x\rbrace$ and $Y \setminus \lbrace f(x) \rbrace$

I always had a idea to prove some thing,but in this case the unique idea that comes to my mind is use the definition of homeomorphism, the problem is ,that when i writte the  proof i use the fact that

since $f:X \to Y$ a homeomorphism then their restriction to a set  $X \setminus \lbrace x\rbrace$ will be continuous,and here i cant argue some thing. Any hint or comment was very helful.

答:
Fix $x\in X$, and for convenience let $\hat f=f\upharpoonright X\setminus\{x\}$. Since $f$ is a bijection from $X$ to $Y$, it’s clear that $\hat f$ is a bijection from $X\setminus\{x\}$ to $Y\setminus\{f(x)\}$. Let $U$ be an open set in $X\setminus\{x\}$; then there is an open $V\subseteq X$ such that $U=V\setminus\{x\}$. Then

$$\hat f[U]=f[V\setminus\{x\}]=f[V]\setminus\{f(x)\}\,,$$

and $f[V]$ is open in $Y$, so $\hat f[U]$ is open in $Y\setminus\{f(x)\}$. Now use a very similar argument to show that if $U$ is open in $Y\setminus\{f(x)\}$, then $\hat f^{-1}[U]$ is open in $X\setminus\{x\}$.


问:
length of sum of two submodule
Let $M$ be a $R$-module with finite length and $K$ and $N$ be a submodule of $M$. Prove that
$l(K+N)+l(K\cap N)=l(K)+l(N)$.

My proof: First, by assuming that $K\cap N=\{0\}$, we can conclude that $l(K+N)=l(K)+l(N)$.

The detail: Let $n=l(K)$ and $m=l(N)$.
Let $\{0\}\subseteq K_0\subsetneq K_1 \subsetneq \cdots \subsetneq K_n=K$ be a composition series of $K$. Let $\{0\}\subseteq N_0\subsetneq N_1 \subsetneq \cdots \subsetneq N_m=N$ be a composition series of $N$.
Define $\phi: K \rightarrow K+N $ by $\phi(k)=k$ and $\psi:K+N \rightarrow N$ by $\psi(k+n)=n$. For $0 \leq i \leq n$, let $M_i=\phi(K_i)$ and for $n+1\leq i\leq n+m$, let $M_i=\psi^{-1}(N_i)$, then $\{M_i\}_{i=0}^{n+m}$ be a composition series for $K+N$. Hence, $l(K+N)=n+m=l(K)+l(N)$.

Could you help me proving the general cases, that is $K\cap N \neq \{0\}.$

答:
Hint 1: Consider $(K+N)/K$. If you have a composition series for this module, then it can "stack on top of" a composition series of $K$ to make a composition series for $K+N$.

Hint 2: By an isomorphism theorem, $(N+K)/K\cong N/(N\cap K)$.

Hint 3: Convince yourself that $\ell(M/N)=\ell(M)-\ell(N)$, (at least when all the lengths are finite.)

Title: length of sum of two submodule