问:
Prove U is convex $U=\{y\in\mathbb{R}^m:\phi_1(x)\leq y_1,\ldots,\phi_m(x)\leq y_m,\mbox{for some x in C}\}$
Let $C\subset\mathbb{R}^n$ be a convex set and let $\phi_1(x),\ldots,\phi_m(x)$ be convex functions over C. Let U
be the following subset of $\mathbb{R}^m$:

$U=\{y\in\mathbb{R}^m:\phi_1(x)\leq y_1,\ldots,\phi_m(x)\leq y_m,\mbox{for some x in C}\}$

Prove $U$ is convex.

My thoughts were that $U$ is an intersection of $Lev(\phi_i,y_i)$ $(Lev(f,a)=\{x:\in\mathbb{R}^n:f(x)\leq a\}$ and we know that the level set of convex function is convex and intersection of convex sets is convex.

答:
Let $y_1=(y_1^1, \dots, y_m^1), y_2=(y_1^2, \dots, y_m^2)$ be elements of $U$ and $\lambda \in  [0,1]$. By definition, it exists $x^1,x^2 \in C$ such that:

$$\begin{cases}
\phi_1(x^1)\leq y_1^1,\ldots,\phi_m(x^1)\leq y_m^1\\
\phi_1(x^2)\leq y_1^2,\ldots,\phi_m(x^2)\leq y_m^2
\end{cases}$$

As all the $\phi_i$ are supposed to be convex, for $1 \le i \le m$:

$$\phi_i((1-\lambda)x^1 + \lambda x^2) \le (1-\lambda)\phi_i(x^1) + \lambda \phi_i(x^2) \le (1-\lambda)y_i^1 + \lambda y_i^2.$$

As $C$ is supposed to be convex, $(1-\lambda)x^1 + \lambda x^2$ belongs to $C$. Which implies that $(1-\lambda)y^1 + \lambda y^2$ belong to $U$.

So by the very definition of convexity, $U$ is convex.


问:
Why is $\left[\mathbb{R}(t):\mathbb{R} \right]$ infinite?
When I was reading about extensions over $\mathbb{R}$, I've saw somebody affirming this. Why is this true? How can I prove it? Thank you!

答: $1, t, t^2, \dots, t^n \dots$ is a linearly independent family of polynomials of $\mathbb R[t]$. And $\mathbb R[t] \subsetneq \mathbb R(t)$.

ContentLicense: CC BY-SA 4.0


问:
Determine bijectivity of a linear map from matrix product

Define two linear maps $L: \mathbf{R}^3 \to \mathbf{R}^3$ and $K :\mathbf{R}^3 \to \mathbf{R}^3$ such that $$L = \begin{bmatrix} 1 && 0 && 1 \\ 0 && 2 && 0 \\ 1 &&0 && 3 \end{bmatrix} \text{ and } KL = \begin{bmatrix} 2 && 2 && 3 \\ 2 && 4 && 0 \\ 4 &&6 && 9 \end{bmatrix}$$ Why is $K$ a bijection? Determine the map $K$.

How can I get any information about $K$ here? I cannot just determine it from the product, right? For $L$ I have that $\det(L) = 4 \ne 0$, thus at least $L$ is bijective.

答:
$L$ is as you said bijective. So is $KL$. As

$$K = (KL) L^{-1},$$

$K$ is also a bijection (the composition of two bijections is a bijection).


Title: Determine bijectivity of a linear map from matrix product

Body:
Define two linear maps $L: \mathbf{R}^3 \to \mathbf{R}^3$ and $K :\mathbf{R}^3 \to \mathbf{R}^3$ such that $$L = \begin{bmatrix} 1 && 0 && 1 \\ 0 && 2 && 0 \\ 1 &&0 && 3 \end{bmatrix} \text{ and } KL = \begin{bmatrix} 2 && 2 && 3 \\ 2 && 4 && 0 \\ 4 &&6 && 9 \end{bmatrix}$$ Why is $K$ a bijection? Determine the map $K$.

How can I get any information about $K$ here? I cannot just determine it from the product, right? For $L$ I have that $\det(L) = 4 \ne 0$, thus at least $L$ is bijective.

Body:
From $KL = \begin{bmatrix} 2 && 2 && 3 \\ 2 && 4 && 0 \\ 4 &&6 && 9 \end{bmatrix}$ we get, since $L$ is invertible:

$$K=KLL^{-1}=\begin{bmatrix} 2 && 2 && 3 \\ 2 && 4 && 0 \\ 4 &&6 && 9 \end{bmatrix} \cdot L^{-1}.$$

Hence compute $L^{-1}$ and then

$$\begin{bmatrix} 2 && 2 && 3 \\ 2 && 4 && 0 \\ 4 &&6 && 9 \end{bmatrix} \cdot L^{-1}.$$


问:
Group of order $p^2q^n$, $q\neq 3$, $p,q$ prime.
Let $p,q$ be prime numbers, $q \neq 3$, and $G$ be a group with order $p^2q^n$. Prove that $G$ has a normal subgroup with order $q^n$. Prove that $G$ is a semidrect product of a group with order $q^n$ and a group with order $p^2$.

My attempt:
For $n \geq 2$, $q^n>p^2$ and we have a unique $q$-sylow group, thus, we have our normal subgroup. I'm having some problems to show that the group is normal when $n=1$, but I think I can handle this.

For the semidirect product I don't know what I can do, any hint?

答:
This is not true when $p \gt q$: take $G=S_3 \times C_6$, so a group of order $36=3^2 \cdot 2^2$, hence $p=3, q=2=n$. A Sylow $2$-subgroup of $G$ is not normal.

But it is true when $p \lt q$. Let $Q \in Syl_q(G)$, then $n_q(G) \in \{1,p,p^2\}$.
If $n_q(G)=p$, then $p \equiv 1$ mod $q$, hence $q \leq p-1 \lt p \lt q$, which is absurd.

If $n_q(G)=p^2$, then $q \mid p^2-1=(p-1)(p+1)$, hence by the previous argument $q \mid p+1$, but $p \lt q$, so this can only happen when $q=p+1$, meaning $p=2$ and $q=3$, contradicting $q \neq 3$.

So $Q \lhd G$, and if $P \in Syl_p(G)$, $PQ$ is actually a subgroup, but then $G=PQ$ and $P \cap Q=1$, so $G$ is a semi-direct product $P \ltimes Q$.


问:
Faulhaber formula from geometric series and operators?
As shown in this post,
$$ \sum_{k=1}^n x^k = x \sum_{k=1}^{n}  \binom{n}{k} (x-1)^{k-1}$$

For RHS, notice $x= \left(1+( x-1) \right)$ and using this we get,

$$ \sum_{k=1}^n x^k = \sum_{k=1}^{n} \binom{n}{k} (x-1)^{k-1} + \sum_{k=1}^{n} \binom{n}{k} (x-1)^{k} \tag{1}$$

For first term,

$$ \sum_{k=1}^{n} \binom{n}{k} (x-1)^{k-1}  \to \binom{n}{1} +\sum_{k=2}^{n} \binom{n}{k} (x-1)^{k-1} \tag{2} $$

Sub, $k-1 \to j$

$$\sum_{k=2}^{n} \binom{n}{k} (x-1)^{k-1} \to  + \sum_{j=1}^{n-1} \binom{n}{j+1} (x-1)^j  \to   + \sum_{k=1}^{n-1} \binom{n}{k+1} (x-1)^k  \tag{3}$$

Using (1), (2), and  (3)

$$ \sum_{k=1}^n x^k = \binom{n}{1}  +  \sum_{k=1}^{n-1} \binom{n}{k+1} (x-1)^k + \sum_{k=1}^{n} \binom{n}{k} (x-1)^{k} $$

Or,

$$ \sum_{k=1}^n x^k= \binom{n}{1}+ \sum_{k=1}^{n-1} \binom{n+1}{k+1} (x-1)^k + (x-1)^{n}$$ =

Now apply the $P^j$ to both sides (4) where $P$ is an operator defined as $x \frac{d}{dx}$ and evaluate at x=1, see this post for more details. For LHS,

$$ \sum_{k=1}^n x^k \xrightarrow[]{P^j , x=1}  \sum_{k=1}^n k^j $$

From this answer here,

$$P^j =\sum_{i=1}^j S(j,i)  D_{1}^i$$

Where $D_1^i = \frac{d^i}{dx^i}|_{x=1}$  and S(n,k) is stirling number of second kind

Writing (4) out explicitly,

$$ \sum_{k=1}^n k^j =  \sum_{i=1}^j S(j,i)  D_{1}^i \left[ \binom{n}{1}+ \sum_{k=1}^{n-1} \binom{n+1}{k+1} (x-1)^k + (x-1)^n \right]$$

Now, consider

$$  D_{1}^i \left[\binom{n}{1}+ \sum_{k=1}^{n-1} \binom{n+1}{k+1} (x-1)^k + (x-1)^n \right] $$

We can easily evaluate this by considering taylor series of the inside term, call:

$$ f= \binom{n}{1}+ \sum_{k=1}^{n-1} \binom{n+1}{k+1} (x-1)^k + (x-1)^n  $$

Then, the taylor polynomial of $f$ around $x=1$ is given as:

$$ f = \sum_{k=0}^{n+1} \frac{d^k f}{dx^k}|_1 \frac{(x-1)^k}{k!}$$

By comparing coefficients we can easily evaluate the derivative,

$$  D_{1}^i \left[\binom{n}{1}+ \sum_{k=1}^{n-1} \binom{n+1}{k+1} (x-1)^k + (x-1)^n   \right] = \begin{cases} \binom{n}{0} , i=0 \\  i! \binom{n+1}{i+1} , i>0 \end{cases}$$

For $i \in \mathbb{N}$, hence:

$$ \sum_{k=1}^n k^j =  \sum_{i=1}^j S(j,i)   i! \binom{n+1}{i+1}  $$

With all of that in mind,

Is my proof right?

What ways can I make it better?

Are there any more simplification applicable?

Note: I am evaluating the quantity operator on by $P^j$ at x=1

答:
A quick read of the proof did not show any error, although I would need to take a good look into it. The result is indeed correct, so I hope there is no major error.

For the second question, there is indeed a more efficient proof of this identity.

\begin{align*}
 \sum_{k=0}^nk^j&=\sum_{k=0}^n\sum_{i=0}^j S(j,i)(k)_i\\&=\sum_{i=0}^j i!S(j,i)\sum_{k=0}^n\frac{(k)_i}{i!}\\&=\sum_{i=0}^j i!S(j,i)\sum_{k=0}^n\binom{k}{i}\\&=\sum_{i=0}^j i!S(j,i)\binom{n+1}{i+1}
\end{align*}

where $(k)_i$ is the falling factorial and in the first equality, I have used the fact that $\sum_{i=0}^nS(n,i)(k)_i=k^j$.

As for simplifications, nothing really jumps out beyond this. Given that there are alternate representations of this summation using Euler numbers and Bernoulli numbers (which are special numbers with no known closed forms), it is highly suggestive that one cannot get this into any good closed form.

I will keep working on this problem and add anything if I find more.


问:
Field of fractions of a finite $\mathbb{Z}$-module is finite extension of $\mathbb{Q}$
Let $A$ be a ring which is also a finitely generated $\mathbb{Z}$-module. If $A$ is an integral domain and $K$ is its field of fractions and $K$ has characteristic zero, then why is $K$ a finite dimensional vector space over $\mathbb{Q}$?

I see that $K$ contains $\mathbb{Q}$ but why is the extension finite?

答: Let $a_1,\ldots,a_n$ generate $A$ as a $\mathbf{Z}$-module, and let $F=\mathbf{Q}(a_1,\ldots,a_n)\subseteq K$. Note that $\mathbf{Q}(a_1,\ldots,a_n)$ contains $A$, so in fact we must have $\mathbf{Q}(a_1,\ldots,a_n)=K$. Now, because $A$ is finite over $\mathbf{Z}$, it is in particular integral over $\mathbf{Z}$, which means that every element of $A$ is the root of a monic polynomial with coefficients in $\mathbf{Z}$. In particular, this is true of the $a_i$, and so each $a_i$ is algebraic over $\mathbf{Q}$ (integrality over a field is the same as algebraicity). So $K$ is generated over $\mathbf{Q}$ by finitely many algebraic elements, and therefore is a finite extension of $\mathbf{Q}$.

ContentLicense: CC BY-SA 3.0


问:
Poisson Process Continuous stochastic process
Duronto Express arrives at the Bombay Central station according to a Poisson process of rate 3 trains/hour. Local Line trains arrive according to a Poisson process of rate 4 trains/hour.

Conditionally on the event that 8 trains arrive from 9 am to 9:40 am, what is the probability that no trains arrived between 9:10 and 9:20?

So far I used the Superposition Lemma and so the arrival of all the trains at the station is a Poisson process of rate 7 trains/hr

Now I want to find $P(Z(20)-Z(10)=0 | Z(40) = 8)$, where $Z(t) = X(t)+Y(t)$, but I am stuck at this point and don't know how to solve it from here

答:
Hint : $P[Z(t+h) - Z(t)=k] =\frac {e^{\lambda h}{(\lambda h)}^k}{k}$

Note : Make sure you convert $h$ to the same time units as in $\lambda$


问:
Find a vector in a dot product such that an inequation is satisfied
The problem I'm dealing with is as follows. Given the inequation:

$c_1 \langle \boldsymbol{a}, \boldsymbol{x} \rangle - c_2 \ge 0$

where $\langle *, * \rangle$ is the dot product and $c_1, c_2, \boldsymbol{a}$ are known and $c_1, c_2 > 0$, find $\min \boldsymbol{x}$ such that the above inequation is satisfied.

I've tried to solve $c_1 \langle \boldsymbol{a}, \boldsymbol{x} \rangle - c_2 = 0$ but it results in infinite solutions, which is hard to program (I'm writing Python code). So I think maybe an inequation will relax the problem. Is there any way to estimate $\boldsymbol{x}$? I wonder if we can have a minimization solution that produces an expression to compute $\boldsymbol{x}$.

As I'm not very strong in math (I have basic knowledge in linear algebra), a detailed explanation or pointers/references are very much appreciated!

Thank you very much!

答:
Due to the positivity of $c_1$, your inequality condition is equivalent to $$\langle \boldsymbol{a}, \boldsymbol{x} \rangle \ge  \dfrac{c_2}{c_1} \ \iff \ a_1x_1+a_2x_2 \ge  \dfrac{c_2}{c_1}\tag{1}$$

which is one of the two closed half-planes with boundary the line having equation:

$$a_1x_1+a_2x_2 =  \dfrac{c_2}{c_1} \tag{2}$$

Which (affine) half-plane ? Answer: the one which doesn't contain the origin, because if you take $(x_1,x_2)=(0,0)$, condition (1) isn't fulfilled.

Remark: I have assumed we were dealing with a planar issue with coordinates $(x_1,x_2)$. If we are in an $n$-dimensional context, we will write (1) under the form:

$$\sum_k a_kx_k \ge  \dfrac{c_2}{c_1}$$

with "half-plane" replaced by "half-space" delimited by an affine hyperplane.

Edit: if you are looking for an $x=(x_1;x_2)$ satisfying (1), here is a deterministic way to obtain a solution.

Let $a=(a_1,a_2)$:

if $a=0$: clearly, no $x$ can exist.

if $a \ne 0$, either : $a_1 \ne 0$ ; in this case take $x=(x_1;x_2)=(\tfrac{c_2}{c_1a_1};0)$ ; otherwise $a_2 \ne 0$, in this case take $x=(x_1;x_2)=(0;\tfrac{c_2}{c_1a_2})$;


问:
Let $X$ be a regular space and let $h:X→Y$ a homeomorphism. We will prove that $Y$ is regular.
Let $X$ be a regular space and let $h:X→Y$ a homeomorphism. We will prove that $Y$ is regular.
Suppose that $ C $ is a closed subset of $ Y $ and that $ y \in Y \setminus C $ is an arbitrary point. Then, there exists $ x \in X $ such that $ h(x) = y $ and $ x \not\in h^{-1}(C) $. Since $ h $ is continuous, $ h^{-1}(C) $ is closed in $ X $.

Since $ X $ is regular for $x \in X$ and $x \notin h^{-1}(C)$,there are $\exists{U},{V}$ such that $U\cap V=\emptyset$ and $h^{-1}(C)\subset V$ but i couldnt keep going i dont know what to do next.

答: $C \subset h(V)$ , $y=h(x) \in h(U)$ and $h(U),h(V)$ are open. Also $h(U)\cap h(V)=h(U\cap V)=\emptyset$.


问:
The continuity of a function of two variables, how can we determine it exists?
if we find that limit for x-axis and y-axis exist does is it enough to say there is continuity?

or we have always to look in circle way around the point?

for example:
$\lim_{(x,y)\to (0,0) } \frac{xy^{3}}{x^{2}+y^{6}}$ , when (x,y)=(0,0) the value is zero of the function

for partition limits of x and y the limit is 0.

then we check when y=mx (I also i need an explanation on that as to how it works) also the limit is 0.

so far we also can't conclude that the function is continuous right?

then we do $x=ky^{3}$ that we $k=1$ the limit is 0.5, so that's when we can say the function isn't continuous.

but how did they determine the $x=ky^{3}$ part? based on what? and what is the meaning of $x=ky^{3}$ graphical wise?

Body:
Let $f(x,y)=\frac{xy^{3}}{x^{2}+y^{6}}$ if $(x,y) \ne (0,0)$ and $f(0,0)=0.$

show that , for $x \ne 0$, we have $f(x,mx)=\frac{m^3x^2}{1+m^6x^4}$. Hence $f(x,mx) \to 0$ as $x \to 0.$

show that , for $y \ne 0$, we have $f(y^3,y)= 1/2.$. Hence $f(y^3,y) \to 1/2$ as $y \to 0.$


答: Show first that $K$ is an algebraic extension by considering the minimal polynomial over $\mathbb{Z}$ satisfied by any element of $A$.  Then show that any finite generating set for $A$ as a $\mathbb{Z}$-module generates $K$ as a field.  A finitely generated algebraic field extension is a finite field extension.  </p>

Title: Field of fractions of a finite $\mathbb{Z}$-module is finite extension of $\mathbb{Q}$

Body:
Let $A$ be a ring which is also a finitely generated $\mathbb{Z}$-module. If $A$ is an integral domain and $K$ is its field of fractions and $K$ has characteristic zero, then why is $K$ a finite dimensional vector space over $\mathbb{Q}$?

I see that $K$ contains $\mathbb{Q}$ but why is the extension finite?


问:
Checking my work on combinatorics
This is the question:

Given the letters of the word "CORONA", find all the arrangements of 4 letters given that it must begin with a C

I divided into 2 cases:

Case 1: no repetitions, we omit a O

$4P3=4!$

Case 2: one double O. $3P2/2!$ or in other words, $(3!/2!)/2!$ to arrange the 2 O's into the 3 remaining spaces and $*3$ because of the last remaining space on the remaining letters, R, N and A

$\frac{3!/2!}{2!}*3$

So my final answer:

$4!+\frac{3!/2!}{2!}*3$

results to be something with decimals, which isn't exactly good, what have i done wrong here?

答:
The only error is that $^3P_2$ is not $3!/2!$ but $3!/1!$; the formula for $^nP_r$ is $n!/(n-r)!$.

You could also write $^3P_2/2!$, which is the number of ways to choose two ordered positions and then forget about the order, more simply as $^3C_2$.

Title: Checking my work on combinatorics


问:
How to get to $\psi=Ae^{\xi^2/2}+Be^{-\xi^2/2}$
I saw proof that this is the solution, but not how to find it. Just in the Eisberg's book assuming thar for

\begin{equation}
\frac{d^2\psi}{d\xi^2}=\xi^2\psi
\end{equation}

Will have:

\begin{equation}
\psi=Ae^{\xi^2/2}+Be^{-\xi^2/2}
\end{equation}

答:
I checked the book, and found that the author is proposing that form

$$ A e^{\xi^2/2} + B e^{-\xi^2/2} $$

merely as an asymptotic formula as $\left|\xi\right|\to\infty$. Indeed, a precise form of the general solution is

$$ \psi(\xi) = A \sqrt{\xi}\, I_{-1/4} \left(\xi^2/2\right) + B \sqrt{\xi} \, I_{1/4} \left(\xi^2/2\right), $$

where $I_{\nu}$ is the modified Bessel function of the first kind. This can be obtained via series solution method, as illustrated in this answer.

Also note that this can also be used to justify the proposed asymptotic formula, since

\begin{gather*}
\sqrt{\xi}\, I_{-1/4} \left(\xi^2/2\right) + \sqrt{\xi} \, I_{1/4} \left(\xi^2/2\right)
\sim \sqrt{\frac{2}{\pi\smash[b]{\xi}}} \, e^{\xi^2/2}, \\
\sqrt{\xi}\, I_{-1/4} \left(\xi^2/2\right) - \sqrt{\xi} \, I_{1/4} \left(\xi^2/2\right)
\sim \sqrt{\frac{1}{\pi\smash[b]{\xi}}} \, e^{-\xi^2/2}
\end{gather*}

holds as $\xi \to \infty$.


问:
Show that the extreme points of the unit simplex $\Delta_n$ are the unit vectors $e_1,\ldots ,e_n$
Show that the extreme points of the unit simplex $\Delta_n$ are the unit vectors $e_1,\ldots ,e_n$.

I want to show that $ext(\Delta_n)=\{e_1,\ldots,e_n\}$

I've thought that proving $\rightarrow$ this part i would take some $x\in ext(\Delta_n)$ and assume that $x\neq e_i$ for all i but i got stuck.

any hints please?

答:
So take some vector $x=\sum_{j=1}^n \alpha_je_j\in \Delta_n$ such that $x\not \in \{e_1,...,e_n\}$. Then, there exists $j_0$ such that $\alpha_{j_0}\in (0,1)$. Define $y=\frac{1}{\sum_{j\neq j_o}\alpha_j}(x-\alpha_{j_0} e_{j_0})$. Then, $y$ is well-defined since $\sum_{j\neq j_0} \alpha_j=1-\alpha_{j_0}\neq 0$ and $y\neq x$, since $\alpha_{j_0}\neq 0$. I'll leave it to you to check that $y$, indeed, lies in the simplex.

Furthermore,

$$
x= \alpha_{j_0} e_{j_0}+(1-\alpha_{j_0})y,
$$
so $x$ is not an extreme point.

Now, let $1\leq i \leq n$ and assume that $e_i=tx+(1-t)y$ for some $x=\sum_{j=1}^n \alpha_j e_j,y=\sum_{j=1}^n \beta_j e_j\in \Delta_n$. Then, projecting on the $i$'th coordinate, we get that $1=t\alpha_i+(1-t) \beta_i$. But since $\alpha_i$ and $\beta_i$ are positive and smaller than $1$, this implies $\alpha_i=\beta_i=1$. Now, we simply observe that $\sum_{j=1}^n \alpha_j=\sum_{j=1}^n\beta_j=1,$ to get that $\alpha_j=\beta_j=0$ for all $j\neq i$. Thus, $x=y=e_i$. In conclusion, the canonical unit vectors are, indeed, extreme points of the $n$-simplex.


问:
prove a matrix UDU* is unitary
I'm trying to prove the following:

Prove that if $V=UDU^{\dagger}$ with D having only modulus one values, then $V$ must be unitary.

Could someone please explain to me what is exactly meant by "modulus one values"? and how do I prove it?

I was thinking to prove it by proving that $UDU^{\dagger}$  $(UDU^{\dagger})^{\dagger}$  = $(UDU^{\dagger})^{\dagger} UDU^{\dagger} = I$      but I'm not sure what D is...

Thanks in advance

答: I suppose that $D$ is a diagonal matrix such that all entries from the main diagonal have absolute value equal to $1$.

Title: prove a matrix UDU* is unitary


答:
If $D= diag(d_1,...,d_n)$ with $|d_j|=1$ for $j=1,...,n,$ then

$$ d_j \bar d_j=|d_j|^2=1$$

for $j=1,...,n,$ hence

$$D^*D=DD^*=I.$$

Now show that $V$ is unitary.

Title: prove a matrix UDU* is unitary

Body:
I'm trying to prove the following:

Prove that if $V=UDU^{\dagger}$ with D having only modulus one values, then $V$ must be unitary.

Could someone please explain to me what is exactly meant by "modulus one values"? and how do I prove it?

I was thinking to prove it by proving that $UDU^{\dagger}$  $(UDU^{\dagger})^{\dagger}$  = $(UDU^{\dagger})^{\dagger} UDU^{\dagger} = I$      but I'm not sure what D is...

Thanks in advance


问:
Find the value of $\mathop {\lim }\limits_{x \to 0} \frac{{4{{\cos }^2}\left( {f\left( x \right)} \right) - 1}}{{1 - {x^2}}}$
Let  $ f\left( x \right) = f\left( {2x} \right)\forall x \in R$ is continuous and $f(2017)=\frac{\pi}{4}$ then $\mathop {\lim }\limits_{x \to 0} \frac{{4{{\cos }^2}\left( {f\left( x \right)} \right) - 1}}{{1 - {x^2}}}$is=_________.

How do I proceed this type of question

There needs to be a functi0n that satisfies  $ f\left( x \right) = f\left( {2x} \right)\forall x \in R$ and $f(2017)=\frac{\pi}{4}$

答:
Hint:Prove inductively that $$f(2017)=f(\frac{2017}{2^n})$$ Thus taking limits $n\to \infty$, $f(0)=\frac{\pi}{4}$

Finding the limit now is a piece of cake....


问:
Integral, change of variables into rectangle
Let D be bounded by the following lines:

$x+y=c$, $x+y=d$, $y=kx$, $y=lx$
with $ 0<c<d$ and $0<k<l$
and I need to transform this region into a rectangle

What I have done so far,
the rectangle would be $R=[c,d] \times [k,l]$

I need to transform $x=x(u,v)$ and $y=y(u,v)$

I have defined $u=x+y$ and $v=\frac{x}{y}$

Now $x,y$ need to be expressed by $u$ and $v$.

I have $y=\frac{u}{v+1}$ and $x=\frac{u \cdot v}{v+1}$

Now to the Jacobian,
$det\left(\frac{\partial(x,y)}{\partial(u,v)}\right)$=$\frac{-u}{(v+1)^2}$

then for the integral I have
$\int\int_{D} f(x,y) d(x,y)= \int \int_{R} f(x(u,v), y(u,v))| \text{det}\frac{\partial(x,y)}{\partial(u,v)}| d(u,v)$

so it would be $=\int_{k}^{l} \int_{c}^{d} f(x(u,v), y(u,v)) | \text{det}\frac{\partial(x,y)}{\partial(u,v)}| dv du = \int_{k}^{l} \int_{c}^{d} \frac{-u}{(v+1)^2} du dv$
for the answer I get $\frac{-{(l-k)}{(d^2-c^2)}}{2(l+1)(k+1)}$

Could someone please tell me whether this was right or completely wrong?

答:
Yes looks alright, other than a couple of minor things. The sign of your result is minus as  $(l - k \gt 0, d^2 - c^2 \gt 0)$.

While $J(u,v) = -\frac{u}{(v+1)^2}$, $|J(u,v)| = \frac{u}{(v+1)^2}$ so you should not have negative sign inside the integral.

You set up $\frac{x}{y} = v$ and so the bounds of $v$ should be $(\frac{1}{l} \leq v \leq \frac{1}{k})$. You could have instead set up $\frac{y}{x} = v$ and then the bound of $v$ would have been $k \leq v \leq l$.

So $\int_{1/l}^{l/k} \int_{c}^{d} \frac{u}{(v+1)^2} du dv = \frac{{(l-k)}{(d^2-c^2)}}{2(l+1)(k+1)}$


问:
Help with an interesting reciprocal prime sum
How can you evaluate $\sum_{n = 1}^\infty \left( \frac{1}{\sum (n)} \right)$ where $\sum (n)$ is the sum of the first $n$ prime numbers? I was able to find numerical bounds by running code to give a lower bound of $1.02345$, and I found an upper bound by taking the reciprocal of the Prime Sum Function approximation $\frac{\ln(n)\cdot n^2}{2}$ and evaluating a sum from there, but I would like to figure out how to evaluate this exactly.

答:
Exactly would be very difficult.

You want to compute $$S=\sum_{n = 1}^\infty  \frac{1}{a_n} \qquad \text{where} \qquad a_n=\sum _{i=1}^n p_i$$ What I should do it to write it as
$$S=\sum_{n = 1}^p \frac{1}{a_n}+\sum_{n = p+1}^\infty  \frac{1}{a_n}$$

Concerning
$$s_p=\sum_{n = 1}^p \frac{1}{a_n}$$ they form the sequence
$$\left\{\frac{1}{2},\frac{7}{10},\frac{4}{5},\frac{73}{85},\frac{2129}{2380},\frac
   {89669}{97580},\frac{2649191}{2829820},\frac{29545361}{31128020},\frac{7464160
   3}{77820050}\right\}$$ For the second summation, I should use as an approximation of $a_n$
$$\frac{n^2}{2}\left[\ln n + \ln\ln n - \frac{3}{2} + \frac{\ln\ln n}{\ln n} - \frac{5}{2\ln n} - \frac{\ln^2\ln n}{2\ln^2 n} + \frac{7\ln\ln n}{2\ln^2 n} - \frac{29}{4\ln^2 n} + \cdots\right]$$ as proposed by Sinha in $2011$.


问:
Zariski closure of infinite subset of the circle is the circle
Let $S^1 \subseteq \mathbb{R}^2$ be the unit circle circle $\mathbb{V}(X^2+Y^2-1)$. Let $S$ be an infinite subset of $S^1$. I want to show that
$$\overline{S}^{\text{Zariski}}= \mathbb{V}(\mathbb{I}(S))= S^1$$

Is there an easy way to see this?

Clearly $S^1$ itself is Zariski closed, so it contains the Zariski-closure of $S$. However, the other inclusion seems more delicate. Any ideas how I can show this?

答:
The ring $\mathbb{R}[x,y]$ has dimension 2 and consequently a prime ideal containing $x^2+y^2-1$ must be a maximal one.

Moreover, the ring  $\mathbb{R}[x,y]$ is Noetherian and every ideal has a finite number of associated primes.

Geometrically this two facts  mean exactly what you want: a Zarisky closed proper subset of the Circle can only be a finite union of (closed) points.


问:
$\mathbb{Q}(\sqrt[p_1]{q_1}+\sqrt[p_2]{q_2}) = \mathbb{Q}(\sqrt[p_1]{q_1},\sqrt[p_2]{q_2})$

$\mathbb{Q}(\sqrt[p_1]{q_1}+\sqrt[p_2]{q_2}) = \mathbb{Q}(\sqrt[p_1]{q_1},\sqrt[p_2]{q_2})$ where $p_1,p_2,q_1,q_2$ are all distinct primes

My original question is actually one more: $\mathbb{Q}(\sqrt[p_1]{q_1}+\sqrt[p_2]{q_2}+\sqrt[p_3]{q_3}) = \mathbb{Q}(\sqrt[p_1]{q_1},\sqrt[p_2]{q_2},\sqrt[p_3]{q_3})$. But I think first proving the above would help the original problem. I'm pretty sure we use degree argument for this. Could you give any hints?

答:
EDIT: As pointed out in the comments, I swapped $q_i$ and $p_i$. After this edit, this change is at least consistent throughout.

First, let $K = \mathbb Q(\sqrt[q_1]{p_1}, \dots, \sqrt[q_n]{p_n})$, $\alpha = \sum \sqrt[q_i]{p_i}$. $\mathbb Q$ has characteristic 0, so this extension $K/\mathbb Q$ is separable. We can then consider a field extension $L/K$ called the Galois closure of $K/\mathbb Q$. We require that $L/\mathbb Q$ is the smallest Galois extension containing $\mathbb Q$. Essentially, we just adjoin all the conjugates of the $\sqrt[q_i]{p_i}$. We can describe it even more explicitly, as the conjugates of $\sqrt[q_i]{p_i}$ are of the form $\zeta^k \sqrt[q_i]{p_i}$ where $\zeta$ is a primitive $q_i^{th}$ root of unity. Then $L = \mathbb Q(\sqrt[q_1]{p_1}, \dots, \sqrt[q_n]{p_n}, \zeta_1, \dots, \zeta_n)$ where $\zeta_i$ is a primitive $q_i^{th}$ root of unity.

Anyways, the point of doing this is that our initial extension $K/\mathbb Q$ was not Galois (unless all $q_i = 2$), so we are trying to replace it with the Galois closure and work within that. The reason we want a Galois extension is the following:

Lemma. Let $K/F$ be a finite Galois extension and let $\alpha \in K$. Then $K=F(\alpha)$ iff $\sigma(\alpha) \neq \alpha$ for all $\sigma \in G(K/F)$ other than $\sigma = id$.

Proof: Let $H = G(K/F(\alpha))$. For $\sigma \in G(K/F)$, $\sigma(\alpha)=\alpha$ iff $\sigma \in H$. Then $[K:F(\alpha)] = |H|$ and $F(\alpha) = K$ iff $|H| = 1$.

Now, our extension $K/\mathbb Q$ is not Galois so we cannot immediately try to apply this result. However, it gives us the idea of applying every element of $G(L/\mathbb Q)$ to $\alpha$ and observing where it is fixed. To do this, we need to understand this Galois group $G = G(L/\mathbb Q)$ a bit better so let me first define $K_i = \mathbb Q(\sqrt[q_i]{p_i})$ and $L_i = K_i(\zeta_i)$ be the Galois closure. Then $K = K_1 \cdots K_n$ and $L = L_1 \cdots L_n$. We'll work first with these slices $L_i/K_i/\mathbb Q$.

Let's first consider $G_i = G(L_i/\mathbb Q)$. I won't show this, but $G_i$ is generated by the following two elements:

$$
\sigma_i: \begin{cases}
\sqrt[q_i]{p_i} \mapsto \zeta_i \sqrt[q_i]{p_i}\\
\zeta_i \mapsto \zeta_i
\end{cases}
\\
\tau_i:
\begin{cases}
\sqrt[q_i]{p_i} \mapsto \sqrt[q_i]{p_i}\\
\zeta_i \mapsto \zeta_i^{k_i}
\end{cases}
$$

where $k_i$ and $q_i$ are relatively prime. In fact, by doing some conjugation computations you can show that every element of $G_i$ is of the form $\sigma_i^{a_i} \tau_i^{b_i}$ and that this expression is unique for $a_i$ modulo $q_i$ and $b_i$ modulo $\phi(q_i) = q_i - 1$. Essentially, we are realizing $G_i$ as a semidirect product of $\langle \sigma_i \rangle$ and $\langle \tau_i \rangle$.

What then is the fixed field of $K$? Well if we let $\phi \in G(L/\mathbb Q)$ then $\phi$ is determined by its restrictions $\phi|_{L_i} \in G_i$ as $L = L_1 \cdots L_n$. Each $\phi|_{L_i} \in G_i$, so it can be written uniquely as $\sigma_i^{a_i} \tau_i^{b_i}$. With this representation, we can see that $\phi$ fixes $K_i$ iff $\phi|_{L_i}$ is purely a power of $\tau_i$. Thus, $\phi \in G(L/K)$ iff $\phi|_{L_i}$ is a power of $\tau_i$ for all $i$.

We are trying to show that $\mathbb Q(\alpha) = K$. By Galois theory, this is the same as showing that $G(L/K) = G(L/\mathbb Q(\alpha))$. All the work we did above allowed us to figure out what $G(L/K)$ is. Furthermore, as $\mathbb Q(\alpha) \subseteq K$, we have $G(L/K) \subseteq G(L/\mathbb Q(\alpha))$. Take now some $\phi \in G(L/\mathbb Q(\alpha))$. Then by definition, $\phi(\alpha) = \alpha$. We can rewrite this as $\sum \zeta_i^{m_i}\sqrt[q_i]{p_i} = \sum \sqrt[q_i]{p_i}$ for some $m_i \in \mathbb Z$. Hence, $\sum \sqrt[q_i]{p_i} = \sum Re(\zeta_i^{m_i} \sqrt[q_i]{p_i}) = \sum \sqrt[q_i]{p_i} Re(\zeta_i^{m_i})$. For this to happen, each $\zeta_i^{m_i} = 1$. In other words, $\phi|_{L_i}$ contains no powers of $\sigma_i$ and therefore fixes $K_i$. Hence, $\phi$ fixes $K$ and $G(L/K) = G(L/\mathbb Q(\alpha))$.

Bonus: Using the same ideas as above, you can show that $L = \mathbb Q(\sum \sqrt[q_i]{p_i} + \sum \zeta_i)$.


问:
Special linear group is an irreducible algebraic set
Let $k$ be an infinite field (possibly algebraically closed).

I'm trying to prove that the set of $n\times n$ matrices with entries on $k$ and determinant $1$, $\operatorname{SL}_n(k)$, is an affine variety (by affine variety I will refer to an irreducible algebraic set of $\mathbb{A}_k^{n^2}$). I've been instructed to prove it using these facts:

Morphisms between quasi-affine varieties are continuous (which respect to the Zariski topologies).

Continuity preserves irreducibility (that is, a continuous function between irreducible topological spaces has irreducible image).

Open sets of an irreducible topological space are also irreducible.

I've been given also the hint to look for a morphism between $\operatorname{GL}_n(k)$ (the general linear group of degree $n$ over $k$) and $\operatorname{SL}_n(k)$.

I already know that $\operatorname{GL}_n(k)$ is a quasi-affine variety (i.e., an open set of an affine variety) and that $\operatorname{SL}_n(k)$ is an algebraic set. It remains for me to see that $\operatorname{SL}_n(k)$ is irreducible. By the third previous fact, $\operatorname{GL}_n(k)=\operatorname{M}_n(k)\setminus\det^{-1}(0)$ is an irreducible topological space (for $\operatorname{M}_n(k)\cong\mathbb{A}_k^{n^2}$, the set of $n\times n$ matrices with entries in $k$, is irreducible since $k$ is infinite). Thus, by the second previous fact, to give a morphism $\phi:\operatorname{GL}_n(k)\to\operatorname{GL}_n(k)$ whose image coincides with $\operatorname{SL}_n(k)$ suffices for the proof. However, I can only come up with this definition for the morphism: $\phi(A)=\frac{A^n}{\det A}$ (recall that $\det$ is homogeneous of degree $n$). $\phi$ is indeed a quasi-affine varieties morphism and by construction $\operatorname{Im}\phi\subset\operatorname{SL}_n(k)$. Nevertheless, proving the reverse inclusion would require to prove that the $n$-th root for a matrix of $\operatorname{SL}_n(k)$ always exists. For what I've been looking up on the internet and MSE, this may be non-trivial or even not true in general. From here, I don't see how to continue.

Edit: Thanks to KReiser's comments I've found a proof for the irreducibility of $\operatorname{SL}_n(k)$ which passes through proving that $\det X-1$ is irreducible (where $X=(x_{ij})$ is an $n\times n$ matrix on the indeterminates $x_{ij}$). Problem is this proof is purely algebraic, and I'm still interested on finding a proof which uses the technique stated above: to look for a quasi-affine varieties morphism $\operatorname{GL}_n(k)\to \operatorname{GL}_n(k)$ with image equal to $\operatorname{SL}_n(k)$. Still, I have no clue about how to come up with the morphism.

Any help on the topic will be appreciated.

答: Kreiser comments give the answer: one could just simply prove that $\det -1$ is irreducible (proof here and here). The other way, following the original path, is to consider the morphism $\operatorname{GL}_n(k)\to \operatorname{GL}_n(k)$ which takes each matrix $A\in \operatorname{GL}_n(k)$ and maps it to a matrix equal to $A$ except for the first row, which is equal to that of $A$ divided by the determinant of $A$. It's straighforward to check that the image of this morphism is indeed $\operatorname{SL}_n(k)$.


问:
Pullback of a vector field under a surjective submersion.
Let $\pi: M \rightarrow N$ be a smooth map. If we consider a vector field $Y$ on $N$, I know that, if $\pi$ is a local diffeomorphism, there exists a unique vector field $X$ on $M$ such that $\pi_*X = Y$. I was wondering if we can weaken this hypothesis: Is there a vector field $X$ with $\pi_*X = Y$ if $\pi$ is only surjective submersion, not a local diffeomorphism? It seems quite intuitive because for each $p$, as $\pi_{*p}$ is surjective, there is at least one $X_p$ such that $\pi_{*p}X_p = Y(p)$, but I have some difficulties to show that we can construct a smooth $X$ with $X_p = X(p)$.

答:
Yes, there's at least one such vector field, provided that $\pi$ is a smooth submersion (not necessarily surjective).

By the rank's theorem, we can always find local coordinates $(U,\varphi),(V,\vartheta)$ such that  $\vartheta\circ\pi\circ\varphi^{-1}(x_1,\dots,x_m)=(x_1,\dots,x_n)$. Locally, $Y=\sum Y^i\partial_{\vartheta^i}$. If we define $X$ as $\sum_{i=1}^n Y^i(\pi(x))\partial_{\varphi^i}$, it is then clear that
$d\pi (X_p)=Y_{\pi(p)}$: the problem is that $X$, defined this way, is not a global vector field.

In order to obtain one, we must use a slighlty different construction: let $(U_\alpha,\varphi)$ be a open cover of $M$ (such that , together with coordinates $(V_\alpha,\vartheta_\alpha)$ of $N$, $\pi$ has the required form) and let $\sum\tau_\alpha$ be the partition of unity subordinated to that cover: we define
$$X=\sum_\alpha\sum_{i=1}^n \left(\tau_\alpha Y^i_\alpha\right)(\pi(x))(\partial_{\varphi_\alpha^i}$$
This is clearly a vector field. To see that it satisfies our requirements, it suffices to observe that
$d\pi(X_p)=\sum_\alpha \tau_\alpha Y_{\pi(p)}=Y_{\pi(p)}$.


问:
Will differentiation always give us the maxima and minima
If a given a curve with a maxima and a minima will differentiation always give us the maxima and minima values ?

Is it possible that the curve will have the critical points and derivatives won't give us those values ?

Sorry if this seems to simple ,I am a newbie at calculus.

Edit : my question does not seem to be clear ," if a function has a maxima and minima will differentiation ALWAYS give us the critical points .

答:
Here is an extract from an educational source on this topic:

Since the derivative is zero or undefined at both local maximum and local minimum points, we need a way to determine which, if either, actually occurs. The most elementary approach, but one that is often tedious or difficult, is to test directly whether the y coordinates "near" the potential maximum or minimum are above or below the y coordinate at the point of interest. Of course, there are too many points "near'' the point to test, but a little thought shows we need only test two provided we know that f is continuous (recall that this means that the graph of f has no jumps or gaps).

So apparently, one of the key points for being able to ascertain that the inflexion point(s) constitutes the existence of local maximum (minimum) points is based on a continuity assumption.

Note, one of this thread's questions is "Is it possible that the curve will have the critical points and derivatives won't give us those values ?" and there is nowhere use of the term "differentiable function".

Whereupon, per Wikipedia, to quote:

In particular, any differentiable function must be continuous at every point in its domain. The converse does not hold: a continuous function need not be differentiable. For example, a function with a bend, cusp, or vertical tangent may be continuous, but fails to be differentiable at the location of the anomaly.

So, I would answer yes, as in the event the derivative cannot be computed at the anomaly (may not exist), one needs to inspect to the right and left of the curve (permissible per the continuity assumption) to determine the true nature of the inflexion point.


问:
Convex continuous piecewise linear function form
Let $n\in \mathbb{N}$ and $\mathcal{E}_n$ the set of functions $f$ of the form
$$f(x)=b_0+b_1x+\sum_{i=1}^n a_i\left\lvert x-r_i\right\rvert$$
for $b_0,b_1\in\mathbb{R},a_1,\dots,a_n>0$ and $r_1,\dots,r_n\in\mathbb{R}$.

Show that if $f$ is a convex continuous piecewise linear function, then there exists $n\in\mathbb{N}$ such that $f\in\mathcal{E}_n$.

This result does not seem clear to me, and I do not know how to prove it.

答:
Here is my proof for this.

Let $k_1<\dots k_n\in\mathbb{R}$ such that $f$ is linear on $\left]k_i,k_{i+1}\right[$ for all $1\leqslant i\leqslant n-1$ and $f$ is linear on $\left]-\infty,k_1\right[$ and on $\left]k_n,+\infty\right[$.

Let $c_i$ be the slope of $f$ on $\left[ k_i,k_{i+1}\right[$ for all $0\leqslant i\leqslant n$ with the convention $k_0=-\infty$ and $k_{n+1}=+\infty$.

As $f$ is convex, we have $c_0\leqslant c_1\leqslant\dots\leqslant c_{n+1}$. Let
$$a_i=\frac{1}{2}\left(c_i-c_{i-1}\right)$$
for $1\leqslant i\leqslant n$. Indeed, in the formula $x\mapsto b_0+b_1+\sum_i\left\lvert x-k_i\right\rvert$, the jump of the derivative in $x=k_i$ is $2a_i$.

Let $b_1$ such that
$$b_i+\sum_{i=1}^na_i=c_n$$
that is
$b_1=\frac{1}{2}\left(c_n+c_o\right)$.

As such, the two applications $f$ and $x\mapsto b_1x+\sum_{i=1}^na_i\left\lvert x-k_i\right\rvert$ are two linear functions on every intervals $\left]k_{i-1},k_i\right[$ for $1\leqslant i\leqslant n+1$, continuous and with the same slope on every interval by construction.

That is the two applications differ by a constant, named $b_0$. Thus, $f\in \mathcal{E}_n$.


问:
Difficulty in solving a question regarding the sum of two arithmetic sequences.

Hey guys, I am having a problem solving a question regarding the sum of two linear sequeneces. Using the equation $S_n=\frac{n}{2}[2a(n-1)d] = 15=5n$, whenever I do try to solve it, I get $25-5n=\frac{30-10n}{n}$. Further simplifying this into a quadratic equation, I get $n=1\space and \space n=6$. However, this is misleading and instead, leaves me even more confused. Can someone point out what I am doing wrong, thanks.

答: I think you have misunderstood the question. It just wants you to add together two terms: the $n$th term of A plus the $n$th term of B. You do not need to work out the sum of either sequence.


答:
You should be adding the expression for the general terms of each sequence, not the sums of each.

$T_{A, n} = 7 + (n-1)(-2) = 9-2n$

$T_{B, n} = 3 + (n-1)(-3) = 6-3n$

$\therefore T_{A+B, n} = 9-2n+6-3n \\= 15-5n$

Title: Difficulty in solving a question regarding the sum of two arithmetic sequences.

Body: Hey guys, I am having a problem solving a question regarding the sum of two linear sequeneces. Using the equation $S_n=\frac{n}{2}[2a(n-1)d] = 15=5n$, whenever I do try to solve it, I get $25-5n=\frac{30-10n}{n}$. Further simplifying this into a quadratic equation, I get $n=1\space and \space n=6$. However, this is misleading and instead, leaves me even more confused. Can someone point out what I am doing wrong, thanks.


Body:
$$T_1=[7+(n-1)(-2)]=(9-2n)$$
$$T_2=[3+(n-1)(-3)]=[6-3n]$$
Hence $$T_1+T_2=[15-5n]$$


问:
Finding inverse modulo
I'm trying to find "the smallest positive multiple of 100" that leaves remainder 9 when divided by 19.

Here is what I have done before I got stuck:

$x ≡ 9 \mod 19$

$\gcd(9,19) = \gcd(19,9)\\
          19 = 9*2 + 1\\
          9 = 1*9 + 0$

Bezout's Theorm

$\gcd (19,9) = m*19 + n*9\\
      1 = 19 - 9*2\\
        = 1*19 -2*9$

THUS, x = -2 mod 19 = 17 mod 19 = 17

BUT! 17 is not a multiple of 100, so from here, what do I do to find "the smallest positive multiple of 100" that leaves remainder 9 when divided by 19?

Thanks!

答: You seek a multiple of $19$, which is prime and not a divisor of $100$.  Hence the smallest multiple of $100$ that includes $19$ as a divisor is $1900$.</p>

Title: Finding inverse modulo


问:
Is it true that the Parabolid is not compact?
I have given $M=\left\{(x,y,z)\in\mathbb{R^{3}}:z=\frac{ax^{2}}{2}+\frac{y^2}{2}\right\}$ with $a \in \mathbb{R}$.
As I understood this would be a pseudoparabolid in $\mathbb{R^3}$ dependent also on which $a$ we pick.

I need to know if this set is compact. I struggle so much with compactness of sets. I understand that a subset of $\mathbb{R^n}$ needs to be bounded and closed to have compactness but I have no clue about a practical way to solve this things.

I tried to see the graph in Geogebra and try to understand closeness and boundness from the graph, but am not very good at it. For $a>0$ I would assume that since the graph goes on upwards forever, it isn't bounded? so it can't be compact?
I also had to do the excercise of compactness in the case of the Cylinder and I went with the same argumentation, but don't know if it makes sense.

Also if the argumentation is true and I don't have any 3D-Plotting programm available I don't know if there is some easy and practical way to check these things out.

Thanks in advance for the help

答: Your set is unbounded, and therefore it is not compact. And it is unbounded because, for instance, it contains every point of $\Bbb R^3$ of the form $\left(x,0,\frac{ax^2}2\right)$ and the distance of such a point to $(0,0,0)$ is greater than or equal to $|x|$.


问:
USAMO problem solution
I had asked for a hint here USAMO problem hint. i had tried induction once but i thought it won't work so i left it, but after seeing @lulu 's comment, i decided to give it a go again. Please see if my solution is correct.

[USAMO 2003] Prove that for every positive integer n there exists an n-digit number divisible by $5^n$ all of whose digits are odd.

MY SOLUTION: So first thing, i checked some small cases and figured we could generate the number with (n+1) digits satisfying the property by adding a number to it's front, ie. adding b 10$^n$ to the number with n digits.

We'll proceed with induction, let P(n) mean there exists an n-digit number divisible by $5^n$ all of whose digits are odd.

P(1) is true as 5|5.

Let P(k) be true, ie. let 5$^k$ | $a_ka_{k-1}...a_1$  with $a_i$  $\neq$ 2l for i $\in$ {1,2...k} .

I'll attempt to prove that by adding $ b \cdot 10^k $  with $ b \in {1,3,5,7,9} $ . we can have a number that is divisible by $5^{k+1}$.

So we want 5$^{k+1}$ $|$ $ b \cdot 10^k $ +  $a_ka_{k-1}...a_1$ . --> eq.1

Let $a_ka_{k-1}...a_1$ = $5^km  $

So from eq. 1, inputting $a_ka_{k-1}...a_1$ = $5^km  $ , we will get

5$^{k+1}$ $|$ $ b \cdot 10^k $ +  $5^k$m, then dividing throught by $5^k$ , we need

5 $|$ $2^k \cdot b + m$

as  $ b \in {1,3,5,7,9} $ , $\equiv$ 0,1,2,3,4 (mod 5)

So now m $\equiv$ 0,1,2,3,4 (mod 5) , let m $\equiv$ r (mod 5),

We need $2^k \cdot b + r  =0 (mod 5)$

now,$2^k \equiv$ 1,2,3,4 (mod 5)

so painstakingly going through each and every case of possible values of $2^k$ and m (mod 5) (there are 16 cases), we prove that we can find a $ b \in {1,3,5,7,9} $ such that 5 $|$ $2^k \cdot b + m$ .

This is the first time i have written so much in latex, so i am sorry if there is any mistake.

If you were a grader, out of 7, how many points would you give me?

答:
There is a $\,x\in\Bbb Z\,$ with $\,5\mid 2^k x - m\!\iff\! \bmod 5\!:\ 2^k x \equiv m\,$ has a root $\,x.\,$ By general theory

$$\begin{align} \color{#c00}c\ x &\equiv \, d\!\!\pmod{\!\color{#0a0}n}\ \ \text{has a root}\ x\!\iff\! \gcd(\color{#c00}c,\,\color{#0a0}n)\mid d\\[.3em]
{\rm thus}\ \  \color{#c00}{2^k} x&\equiv m\!\!\!\pmod{\!\color{#0a0}5}\ \ \text{has a root}\ x,\ \, {\rm by}\ \ \gcd(\color{#c00}{2^k},\color{#0a0}5)\!=\!1\end{align}\qquad$$

and we can choose a root $\,x\in \{1, 3, 5, 7, 9\}\,$ since it is a complete residue system $\!\bmod 5;\,$ alternatively: $ $ if $\,0\le x < 5\,$ is even then  $\,x':= x\!+\!5\,$ is odd $< 10,\,$ and $\,x'$ remains a root by $\,x'\equiv x\pmod{\!5}$.

Title: USAMO problem solution


答:
$M$ is certainly not compact. The sequence $\{m_n\}$ is an unbounded sequence of $M$ where $n \in \mathbb N$ and

$$m_n=(2n,2n,2(a+1)n^2).$$

Title: Is it true that the Parabolid is not compact?

Body:
I have given $M=\left\{(x,y,z)\in\mathbb{R^{3}}:z=\frac{ax^{2}}{2}+\frac{y^2}{2}\right\}$ with $a \in \mathbb{R}$.
As I understood this would be a pseudoparabolid in $\mathbb{R^3}$ dependent also on which $a$ we pick.

I need to know if this set is compact. I struggle so much with compactness of sets. I understand that a subset of $\mathbb{R^n}$ needs to be bounded and closed to have compactness but I have no clue about a practical way to solve this things.

I tried to see the graph in Geogebra and try to understand closeness and boundness from the graph, but am not very good at it. For $a>0$ I would assume that since the graph goes on upwards forever, it isn't bounded? so it can't be compact?
I also had to do the excercise of compactness in the case of the Cylinder and I went with the same argumentation, but don't know if it makes sense.

Also if the argumentation is true and I don't have any 3D-Plotting programm available I don't know if there is some easy and practical way to check these things out.

Thanks in advance for the help


答: For any given $a$, this is the graph $\Gamma$ of a function $\Bbb R^2\to \Bbb R$, namely $f(x,y)=\frac{ax^2}2+\frac{y^2}2$. Therefore it is unbounded becasue for all $x,y$ there is some $z$ such that $(x,y,z)\in\Gamma$.


问:
How to work with random variables?
If $X$ and $Y$ are independent random variables described by standard normal distribution could you please explain how to formally evaluate probabilities of occurrences such as $X-Y>0$ (intuitively it's $0.5$ of course) or $X^2-Y^2>0$?

Ultimately I'd like to be able to tell how likely is it that parabola $z(t)=X^2+2Yt+t^2$ has a root in positive semi-axis of $t$.

答:
If $X$ and $Y$ are jointly normal (and not necessarily independent), then $X-Y$ is a normal random variable. More generally, $aX+bY$ is a normal random variable with mean $a\mu_X+b\mu_Y$ and variance $a^2\sigma_X^2+b^2\sigma_Y^2+2ab\rho\sigma_X\sigma_Y$ where $\rho$ is the correlation coefficient. For your special case, $X-Y \sim \mathcal N(0,2)$ and so, with $\Phi(\cdot)$ denoting the cumulative probability density function of the standard normal random variable,
$$P\{X-Y > 0\} = 1-\Phi\left(\frac{0}{\sqrt{2}}\right) = 1 - \frac{1}{2} = \frac{1}{2}$$ just as you intuited, presumably via the symmetry argument that
since $X$ and $Y$ are independent identically distributed random variables,
then $P\{X<Y\}=P\{X>Y\}$.  But the formal approach will work
in more general cases too. For example, more generally,
$$P\{aX+bY \leq c\} = \Phi\left(\frac{c-(a\mu_X+b\mu_Y)}{\sqrt{a^2\sigma_X^2+b^2\sigma_Y^2+2ab\rho\sigma_X\sigma_Y}}\right).$$
Finding $P\{X^2-Y^2 < 0\}$ requires more computation in the general case,
but for the case when $X$ and $Y$ are independent identically distributed
normal random variables, $X^2$ and $Y^2$ also are independent random
variables with identical (though non-normal) distribution, and so
$$\{X^2-Y^2 < 0\} = \frac{1}{2}$$ by symmetry. Note that it is
not necessary to
find the distributions of $X^2$ and $Y^2$ and nor is integration
of the joint density over a region necessary.  An alternative
calculation is to note that $\{X^2 - Y^2 < 0\}$ if and only if
$X+Y$ and $X-Y$ have opposite signs, and since $X+Y$ and $X-Y$
happen to be independent $\mathcal N(0,2)$ random variables,
the probability is
$\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2} = \frac{1}{2}$.

With regard to your third question, the parabola $z = X^2+2Yt+t^2$ has
value $X^2 \geq 0$ at $t=0$. If the slope at $t=0$, $\left.\frac{dz}{dt}\right|_{t=0} = 2Y+2t\bigr|_{t=0} = 2Y $ is also positive, then the parabola will not cross
the positive $t$ axis. So, $Y < 0$ is a necessary condition for getting
a crossing on the positive $t$ axis. Also, $z$ has a minimum value
of $X^2-Y^2$ at $t = -Y$. Thus, the parabola crosses the positive $t$
axis (twice) exactly when $\{Y < 0\}$ and $\{X^2-Y^2 < 0\}$. The probability
is thus $\frac{1}{4}$ via the circular symmetry of the joint density of
two independent $\mathcal N(0,1)$ random variables.

Title: How to work with random variables?


问:
On a local existence theorem for ODEs
I have a question about the proof of Theorem 3.10 in Applied Analysis by Hunter and Nachtergaele. The theorem is stated as follows:

Theorem.
Let $f: I \times \overline{B}_R(u_0) \to \mathbb{R}^n,$ where
$$
I = \{t \in \mathbb{R}: |t- t_0| \leq T \}
$$
is an interval in $\mathbb{R}$ and
$$
\overline{B}_R(u_0) = \{u \in \mathbb{R}^n : ||u - u_0|| \leq R \}
$$
is the closed ball of radius $R > 0$ centered at $u_0 \in \mathbb{R}^n.$ Suppose $f(t,u)$ is continuous on $I \times \overline{B}_R(u_0)$ and Lipschitz continuous with respect to $u$ uniformly in $t.$ Set
$$
M = \sup \{||f(t,u)|| : t \in I \text{ and } u \in \overline{B}_R(u_0) \} < \infty.
$$
Then the initial value problem
\begin{align*}
    u'(t) &= f(u)\\
    u(t_0) &= u_0.
\end{align*}
has a unique continuously differentiable local solution $u(t),$ defined in the time interval $|t-t_0| < \delta,$ where
$$
\delta = \min(T, R/M).
$$

The proof involves showing that the map defined by
$$
Tu(t) = u_0 + \int_{t_0}^t f(s, u(s)) ds
$$
is a contraction that maps $X \to X$, where
$$
X = \{ u : \lbrack t_0 - \eta, t_0 + \eta \rbrack \to \overline{B}_R(u_0) : u \text{ is continuous}\}
$$
for some $\eta \in (0, \delta).$ I understand how to show that $T$ maps $X \to X$, but I don't understand the last parts of the proof that $T$ contracts. I understand how we obtain the estimate
$$
|| Tu - Tv ||_\infty \leq C \eta ||u-v||_\infty,
$$
where $C$ is the Lipschitz constant for $f$. Next, the authors write that we should choose $\eta = C/2,$ and this proves $T$ is a contraction. How does this show $T$ contracts? So far as I know, the only information I have about about $C$, being the Lipschitz constant for $f$, is that $C \geq 0.$ How does it follow that $C^2/2 < 1?$ (which is what I believe we need for $T$ to be a contraction).

Lastly, the authors write that "since $\eta$ depends only on the Lipschitz constant of $f$ and on the distance $R$ of the initial data from the boundary of $\overline{B}_R(u_0)$, repeated application of this result gives a unique local solution defined for $|t-t_0| < \delta.$" How am I to understand what is meant by this last part, the "repeated application of this result?"

Thanks in advance for any and all replies.

答:
Obviously, for a contraction you want
$$
Cη\le\frac12
$$
(or any other value smaller $1$). Transcription of manuscripts and editing of books may be carried out by non-mathematicians, so such a mangling of small details in formulas is not uncommon.

As to the second part, you got a solution $u$ on $(t_0-η,t_0+η)$. Now consider the IVP at the point $t_1=t_0+\frac12η$ with value $u(t_1)$. By using the same sets and constants, you can conclude that a local solution on $(t_1-η,t_1+η)=(t_0-\frac12η,t_0+\frac32η)$ exists and can be used to extend the first solution. Now apply the same centered on $t_{-1}=t_0-\frac12η$, $t_2=t_0+η$ etc.

If you treat the end point $η$ dynamically, you might conclude that for any $t\in[t_0-δ,t_0+δ]$ you get
\begin{align}
\|(T^2u)(t)-(T^2v)(t)\|&\le\int_{t_0}^tC\|(Tu)(s)-(Tv)(s)\|\,ds
\\
&\le\int_{t_0}^tC^2|s-t_0|\|u(s)-v(s)\|\,ds
\\
&\le \frac{|C(t-t_0)|^2}2\|u-v\|_\infty
\\&\vdots\\
\|(T^nu)(t)-(T^nv)(t)\|&\le \frac{|C(t-t_0)|^n}{n!}\|u-v\|_\infty
\end{align}
This allows to get contractivity for some $T^n$ on the full interval $[t_0-δ,t_0+δ]$ at once, as $\frac{(Cδ)^n}{n!}$ converges to zero. Some minimal manipulation is required to show that the fixed point of $T^n$ is indeed also a fixed point of $T$, essentially if $u_*$ is the fixed point of $T^n$ then $Tu_*$ has exactly the same defining properties, uniqueness then requires $Tu_*=u_*$.

Title: On a local existence theorem for ODEs


问:
Commutation and Sharing Same Eigenvalues
If we have the following eigenvalues and eigenvectors
$$A \psi = a\psi$$ and $$B\psi = b\psi$$
we can easily show that they have same eigenvector if they commute.

$$AB\psi=Ab\psi=bA\psi=ba\psi$$
$$BA\psi=Ba\psi=aB\psi=ab\psi$$
So they commute! They satisfy the proposal.
$$[A,B]=0$$

But what if our eigenvectors are different instead of being same with corresponding eigenvalues. Can we show it again so that they have same eigenvector?

$$A\alpha_n = a_n\alpha_n$$
$$B\beta_n = b_n\beta_n$$

答:
I'm this as the following concern:

Basically, I am trying to show that if $[A,B]=0$, then I can pick the same set of eigenvectors since $α$ must be proportional to $β$ Is it possible?

There is a theorem which says :

If $\Omega$ and $\Lambda$ are two commuting Hermitian operators, there exists (at least) a basis of common eigenvectors that diagonalizes them both.

The main concern is a complete set of eigenbasis. Note the operator should be Hermitian so that there exists at least a basis of its orthonormal eigenvectors.

The proof for theorem in case of nondegenerate given by:

$$\Omega|\omega_i\rangle=\omega_i|\omega_i\rangle$$
$$\Lambda\Omega|\omega_i\rangle=\Lambda\omega_i|\omega_i\rangle$$
$$\Omega\Lambda|\omega_i\rangle=\omega_i\Lambda|\omega_i\rangle$$
i.e. $\Lambda|\omega_i\rangle$ is an eigenvector of $\Omega $ with eigenvalue $\omega_i$. Since this vector is unique up to a scale
$$\Lambda|\omega_i\rangle=\lambda_i|\omega_i\rangle$$
Thus $|\omega_i\rangle$ is also a eigenvector of $\Lambda$ with eigenvalue $\lambda_i$.

OwnerDisplayName: Malachi

OwnerDisplayName: Young Kindaichi


问:
Properties of matrix equation CA=C
I'm interested in the properties of the matric equation $CA=C$, where $C$ is an $m\times n$ matrix with real entries, and $A$ is a square $n\times n$ matrix with real entries, with $m\leq n$. What can I say about $A$, when $C$ is assigned? I know the following holds true:

Let $A=C^*C$, where $C^*$ is any generalized inverse of $C$. Then
the system above always admits at least one solution.

Whichever $C$, $A$ must be an orthogonal projection mapping the
rows of $C$ into themselves.

As for (1), can I go the other way round, i.e. $CA=C$ implies that $A$ be some linear function of (one of the) generalized inverses of $C$?

As for (2), does it imply that $A$ be symmetric (since real) and idempotent?

Thanks

答:
In fact you can only solve this question by this approach:

$CA=C$

$CA-C=0$

$C(A-I)=0$

Let $B=A-I$ ,

Then $CB=0$

Suppose $C=\left(\begin{array}{rrrr}c_{11}&c_{12}&\cdots&c_{1n}\\c_{21}&c_{22}&\cdots&c_{2n}\\\vdots&\vdots&\ddots&\vdots\\c_{m1}&c_{m2}&\cdots&c_{mn}\end{array}\right)$ and let $B=\left(\begin{array}{rrrr}b_{11}&b_{12}&\cdots&b_{1n}\\b_{21}&b_{22}&\cdots&b_{2n}\\\vdots&\vdots&\ddots&\vdots\\b_{n1}&b_{n2}&\cdots&b_{nn}\end{array}\right)$ ,

Then $\left(\begin{array}{rrrr}c_{11}&c_{12}&\cdots&c_{1n}\\c_{21}&c_{22}&\cdots&c_{2n}\\\vdots&\vdots&\ddots&\vdots\\c_{m1}&c_{m2}&\cdots&c_{mn}\end{array}\right)\left(\begin{array}{rrrr}b_{11}&b_{12}&\cdots&b_{1n}\\b_{21}&b_{22}&\cdots&b_{2n}\\\vdots&\vdots&\ddots&\vdots\\b_{n1}&b_{n2}&\cdots&b_{nn}\end{array}\right)=\left(\begin{array}{rrrr}0&0&\cdots&0\\0&0&\cdots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\cdots&0\end{array}\right)$

$\left(\begin{array}{rrrr}\sum_{k=1}^nc_{1k}b_{k1}&\sum_{k=1}^nc_{1k}b_{k2}&\cdots&\sum_{k=1}^nc_{1k}b_{kn}\\\sum_{k=1}^nc_{2k}b_{k1}&\sum_{k=1}^nc_{2k}b_{k2}&\cdots&\sum_{k=1}^nc_{2k}b_{kn}\\\vdots&\vdots&\ddots&\vdots\\\sum_{k=1}^nc_{mk}b_{k1}&\sum_{k=1}^nc_{mk}b_{k2}&\cdots&\sum_{k=1}^nc_{mk}b_{kn}\end{array}\right)=\left(\begin{array}{rrrr}0&0&\cdots&0\\0&0&\cdots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\cdots&0\end{array}\right)$

$\therefore\begin{cases}\sum_{k=1}^nc_{1k}b_{k1}=0\\\sum_{k=1}^nc_{2k}b_{k1}=0\\\vdots\\\sum_{k=1}^nc_{mk}b_{k1}=0\\\sum_{k=1}^nc_{1k}b_{k2}=0\\\sum_{k=1}^nc_{2k}b_{k2}=0\\\vdots\\\sum_{k=1}^nc_{mk}b_{k2}=0\\\vdots\\\sum_{k=1}^nc_{1k}b_{kn}=0\\\sum_{k=1}^nc_{2k}b_{kn}=0\\\vdots\\\sum_{k=1}^nc_{mk}b_{kn}=0\end{cases}$

You should handle this system of linear equations.

Title: Properties of matrix equation CA=C


问:
Show that $x^2 + 3xy - 2y^2 = 1992$ has no solutions in integers.
Here's what I did:
I managed to write $x^2 + 3xy - 2y^2 = 1992$ as
$$\Big(2x+y\big(3 + \sqrt{17}\big)\Big)\Big(2x+y\big(3 - \sqrt{17}\big)\Big) = 7968$$
My doubt is that I have got a $\sqrt{17}$ in my factorisation as $y$ has to be $\sqrt{17},\sqrt{4,913}$ and so on which are not perfect roots, does this prove that $x$,$y$ are not integers?

答:
For any integers $x,y$, the product on RHS is always an integer, since it is equal to $(2x+3y)^2 - 17y^2$. Hence the factorization trick may not always work. For this question, consider modulo $3$.

Since $1992$ is divisible by $3$, by taking modulo $3$ we get:

$$x^2-2y^2 \equiv 0 \pmod 3$$

Now $a^2 \equiv 0$ or $1 \pmod 3$. For the above difference to be divisible by $3$, both $x,y$ must be divisible by $3$.

But then $x^2 + 3xy-2y^2$ must be divisible by $9$, contradicting $9 \nmid 1992$.


问:
Dimensions of vector spaces in an exact sequence
I've read the following formula in wikipedia:

Given finite dimensional vector spaces $V_i$ and an exact sequence
$\cdots\rightarrow V_i\rightarrow V_{i+1}\rightarrow\cdots$, we have

$$
\sum_{n\in 2\mathbb{Z}}\dim V_n = \sum_{n\in 2\mathbb{Z}+1}\dim V_n
$$

Is there a name for this theorem? Could anyone please tell me where to find a proof of this in the literature?

Thank you!

答:
For a finite exact sequence of finite dimensional vector spaces, the claimed
equality is also provable by induction. For the induction anchor ($n = 1$),
observe that if $0 \rightarrow V_1 \rightarrow 0$ exact, then the second arrow
is injective, and thus $\dim(V_1) = 0$, such that the claimed equality holds.
For the induction step, the exact sequence in question is (up to adjustments
for $n = 2$ or $n = 3$):
$$
    0 \rightarrow V_1 \rightarrow V_2 \rightarrow V_3
    \rightarrow \ldots \rightarrow V_n \rightarrow 0
$$
from which the following, shorter sequence can be naturally constructed
(by considering the domain of the the third arrow, from $V_2$ to $V_3$, modulo
its kernel, which is, by exactness of the sequence in question, isomorphic to
$V_1$, and thus identified with $V_1$):
$$
    0 \rightarrow V_2 \mathbin{/} V_1 \rightarrow V_3
    \rightarrow \ldots \rightarrow V_n \rightarrow 0
$$
which is still exact, because the second arrow is now an injective map, while
it has, by construction, the same image in $V_3$ as the corresponding arrow of
the original exact sequence. Combining
$\dim(V_2 \mathbin{/} V_1) = \dim(V_2) - \dim(V_1)$ and the induction
hypothesis yields the claimed equality.


问:
Question regarding finding fixed field of extension
This question was part of an assignment which i am solving of a university in which I am not  student.

Let $\alpha = {{2}^{1/5}} \in \mathbb{R}$ and e =  $exp([2\pi i]/S)$ . Let $K=\mathbb{Q}(\alpha e)$. Then pick Correct statements.

1 There exists a field automorphisms $\sigma$ of $\mathbb{C}$ such that $\sigma(K)= K$ and $\sigma \neq id$.

There exists a field automorphisms $\sigma$ of $\mathbb{C}$ such that $\sigma(K) \neq  K$

3.For all field automorphisms $\sigma$ of K , $\sigma(\alpha e)=\alpha e$.

4 There exists a finite extension E  of $\mathbb{Q}$ such that $K\subseteq E$ and $\sigma (K) \subseteq E$for every field automorphism $\sigma$ of E.

I was confused what S is but then I took S as 5 to solve the problem as I really wanna learn how to do it. I think taking S doesn't makes sense.

I have proved 4 correct. But rest i am unable to do. They are related to finding field to K in $\mathbb{Q}$ and I am unable to find them . SO, can you please tell how to do that .

thank you !!

答:
I am sure that $S=5$, it is the most likely answer to that conundrum.

Note that $(\alpha e)^5=2$, so the minimal polynomial of this particular element is $x^5 - 2$ (shown irreducible by Eisenstein on the shifted polynomial) i.e. the extension $\mathbb Q(\alpha e)$ has degree $5$.

Furthermore, the other roots of $x^5-2$ are $\alpha,\alpha e^2,\alpha e^3$ and $\alpha e^4$.

The key ideas used in the proofs of all the above questions is : every automorphism sends a root of a polynomial, to another root of the same polynomial.

1

Suppose that $\sigma(K) = K$ and $\sigma \neq id$. Since $\sigma(\mathbb Q) = \mathbb Q$, we cannot have $\sigma(\alpha e) = \alpha e$ since the set fixed by $\sigma$ would then be the whole of $K$ , since it is a field containing $\mathbb Q$ and $\alpha e$.

Therefore, $\sigma(\alpha e) \neq \alpha e$, but it must be a root of $x^5-2=0$, hence is one of $e,\alpha e^2,\alpha e^3,\alpha e^4$. We show that none of these can belong in $K$.

Indeed, suppose that $\alpha e^n$ belongs in $K$ for $n \not \equiv 1 \mod 5$. Then, we know that $\alpha e \in K$, so taking the quotient, $e^l \in K$ for some $l$ not a multiple of $5$. By Bezout's lemma there is a $d$ with $dl \equiv 1 \mod 5$, so $(e^{l})^d = e^{ld} = e$ is also in $K$.

But then, $K$ is an extension of $\mathbb Q(e)$ so $[K:\mathbb Q]$ must be a multiple of $[\mathbb Q(e) : \mathbb Q]$. The former is $5$, and the latter is $4$, by the minimal polynomial of $e$ being $x^4+x^3+x^2+x+1$, so this is not true.

Consequently, no other root of $x^5-2$ belongs in $K$, therefore $\sigma(K) \neq K$ for any $\sigma \neq id$.

2

This one is pretty clear : any field homomorphism from $K$ is defined by what it does with $\alpha e$ because it always fixes the rational numbers. So simply define $\sigma : K \to \mathbb Q(\alpha)$ by $\alpha e \to \alpha$. This is well defined and extends to $\mathbb C$, via fixing all other elements of $\mathbb C$, and this qualifies as an automorphism of $\mathbb C$ which sends $K$ to a field not equal to $K$.

3

The image of $\sigma : K \to \mathbb C$ must contain a root of $x^5-2 = 0$. If the image equals $K$ then this root can only be $\alpha e$ as shown in part $1$. Consequently $\sigma(\alpha e) = \alpha e$.

4

This one is easy : consider the field consisting of all roots of $x^5-2 = 0$ i.e. $K(\alpha,\alpha e,\alpha e^2,\alpha e^3,\alpha e^4)$. Any automorphism must send $\alpha e$ to a root of $x^5-2 = 0$ but this extension contains all the roots, so it will contain the image of $K$ under any automorphism of $\mathbb C$.


问:
Building a tower using colorful blocks
How many possibilities are there to build a tower of n height, using colorful blocks, where:

white block is of 1 height

black, red, blue, green, yellow, pink blocks are equal to 2 heights

I need to find the generating function formula for this.

So, for n = 1 I get 1 possibility,

for n = 2 I get 2 possibilities,

for n = 3 I get 3 possibilities

for n = 4 I get > 4 possibilities etc.

The generating function at this moment would be $1 + 2x + 3x^{2} + ...$. But I have no idea how can I find the general formula calculating this.

Could you give me any suggestions, or solutions (;-)) ?

答:
Let the number of towers of height $n$ be $a_n$. To build a tower of height $n$, you start with a tower of height $n - 1$ and add a white block ($a_{n - 1}$ ways) or with a tower of height $n - 2$ and add one of 6 height-2 blocks. In all, you can write:
$$
a_{n + 2} = a_{n + 1} + 6 a_n
$$
Clearly $a_0 = a_1 = 1$.

Define $A(z) = \sum_{n \ge 0} a_n z^n$, multiply the recurrence by $z^n$ and sum over $n \ge 0$ to get:
$$
\frac{A(z) - a_0 - a_1 z}{z^2} = \frac{A(z) - a_0}{z} + 6 A(z)
$$
You get:
$$
A(z) = \frac{1}{1 - z - 6 z^2}
     = \frac{3}{5} \cdot \frac{1}{1 - 3 z} + \frac{2}{5} \cdot \frac{1}{1 + 2 z}
$$
This is just geometric series:
$$
a_n = \frac{3}{5} \cdot 3^n + \frac{2}{3} \cdot (-2)^n
    = \frac{3^{n + 1} - (-2)^{n + 1}}{5}
$$

Title: Building a tower using colorful blocks


问:
Chosing the right setup for stochastic process
Consider a stochastic process $Y_k:=x+k-(X_1+\dots +X_k)$ for $k=1,2,...$, where $x \in \mathbb{N} $ and $X_1$ are i.i.d. random variables which are integer valued and non negative.
The process starts at $Y_0=x$.
$P_x$ is the measure under the condition that $Y_0:=x$

Let $(\Omega, \mathbb{F},\mathcal{F}, P_x)$ a filtered probability space where $\mathbb{F}:= \{ \mathcal{F_k}: k \in \mathbb{N}\}$,  is the natural filtration generated by $X_k$, i.e. $\mathcal{F_k}= \sigma(X_s \mid s \leq k)$

Is $R_0 $ measurable, although my sigma-Algebra starts at $k=1$?

Since my process starts at $k=1$, I know  $R_0$ by computing $P_x(Y_k=s)$ for any $k=1,2,...$

Is this sufficient?

答:
It depends on what you are attempting to model.

For example, I've come across some stochastic processes where the paths have to be cadlag and the process predictable with regards to its underlying filtered sigma algebra.

edit

@Sarah; It sounds like your comments are along the right lines. I can't be more specific since its been about a decade since I've looked at this kind of material. I'd add though, that often the measure space is usually the Borel algebra of the space in question; that is one takes the minimal Borel algebra generated by the opens of the underlying topological space of the process in question.


问:
linearly dependent linear tranformations
Let $E$ be a vector space over $k$ of finite dimension $n$. Let $f_1, \ldots, f_p \in \mathcal{L}(E)$ with $p \leq n$ and assume that the $f_i$ are simultaneously diagonalizable (put it simply, there exists a basis of $E$, where the matrices representing the $f_i$ are all diagonal).

Assume that for all $x \in E$, the vectors $f_1(x), \ldots, f_p(x)$ are linearly dependent. Then the vectors $f_1, \ldots, f_p$ are linearly dependent in $\mathcal{L}(E)$.

I have a complicated proof of that, and I'd like to know if there is a simple argument proving this result. Furthermore, I feel like the diagonalizability assumption could be withdrawn and one could only assume that for any $i,j$ we have $f_i \circ f_j = f_j \circ f_i$, but I have no clue on how to prove that last assertion.

答:
Without loss of generality $E=k^n$ and the $f_i$ are identified with diagonal matrices $D_i$ with $f_i\colon x\mapsto D_i x$. Now note that for $D=\operatorname{diag}(\lambda_1,\dots,\lambda_n)$ and $\mathbf 1=(1,\dots,1)^t$ we have $D\mathbf 1 = (\lambda_1,\dots,\lambda_n)^t$.

So taking $x=\mathbf 1$ you get that the vectors $D_i\mathbf 1$, which consist of the diagonal entries of the $D_i$, are linearly dependent. Then of course the corresponding diagonal matrices satisfy the same linear relation.


问:
Connection between subgroups
Let M be any maximal subgroup of a finite group $G$ and $N$ is a normal subgroup of $G$ and $N$ is not a subgroup of $M$. So, $NM \leq G$. Does that necessarily mean $|NM|=|G|$? why yes or no?

My guessing is no for example if $G=Z_{12}$, $N=\langle2\rangle$,$M=\langle3\rangle$. Am I right?

Thanks

Edit: Im interested to this problem because I want to see if there is any case that $|NM|$ is not equal $|G|$ under stated conditions.

答: Hint: since $N$ is normal, $MN$ is a subgroup with $M \subseteq MN \subseteq G$. Now use that $M$ is maximal and $N \not\subseteq M$ to infer that $G=MN$.

Title: Connection between subgroups


问:
Does it exist a way to find one side of a triangle in polygon other than using similarity?
The problem is as follows:

Using the figure from below find $EB$. Assume $\triangle ABC$ is isosceles where $AB=AC$ and $\overline{BN} \parallel \overline{AC}$. Find $EB$.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&6\,cm\\
2.&8\,cm\\
3.&10\,cm\\
4.&12\,cm\\
5.&11\,cm\\
\end{array}$

What I attempted to do is sumarized in the figure from below:

As it can be seen I redrawn the figure so it can describe better what it is mentioned in the passage of the problem.

From this part. I relied on triangle similarity.

This would meant:

$\triangle ABE \sim \triangle BED$

Hence:

$\frac{6}{x}=\frac{x}{24}$

Solving this:

$x^2=6\times 24$

Which makes $x=12\,cm$

This later answer does check with the official answer from my book. But since this question is placed before introducing the concept of triangle similarity I am wondering if it could be solved relying using fundamentals of bisector angles and perpendicular bisector segment concepts. Can it be solved only using those?.

As it can be noted, some information is unused which makes me to believe was intentionally put there to use the above mentioned concepts.

All and all, is my solution correct?. I'm not good in this topic thus I need assistance on it. Please, include a drawing in your answer as it could require some construction.

答:
From your diagram, since $\Delta ABN$ is isosceles, $BN = 24$ cm.

We can find $DE$ using Pythaogras, which is $\sqrt{x^2-6^2}$. Similarly, $AE = \sqrt{24^2 - x^2}$.

Now since $\Delta ABE = \Delta NBE$, then we have:

$$\frac{1}{2} BN \cdot DE = \frac{1}{2} BE \cdot AE$$
$$\Rightarrow 24 \sqrt{x^2 - 6^2} = x \sqrt{24^2- x^2}$$
$$\Rightarrow 24^2 (x^2 - 6^2) = x^2 (24^2 - x^2)$$
$$\Rightarrow 24^2 x^2 - 24^2 \cdot 6^2 = 24^2 x^2 = -x^4$$
$$\Rightarrow x^4 = 24^2 \cdot 6^2 \Rightarrow x = ±12.$$

$x$ cannot be $-12$. Now for $x = 12$, the LHS equals the RHS when we plug it in, so $\boxed{x = 12}$.


答:
I like player3236's solution a lot.  Here's an alternative.

It's $x^2+3xy+\frac94y^2-\frac{17}4y^2=1992$

or $\left(x+\frac32y\right)^2-\frac{17}4y^2=1992$

or $(2x+3y)^2-17y^2=7968\equiv12\bmod17$.

But $12$ is not a quadratic residue mod $17$, so there are no solutions.

Title: Show that $x^2 + 3xy - 2y^2 = 1992$ has no solutions in integers.

Body:
Here's what I did:
I managed to write $x^2 + 3xy - 2y^2 = 1992$ as
$$\Big(2x+y\big(3 + \sqrt{17}\big)\Big)\Big(2x+y\big(3 - \sqrt{17}\big)\Big) = 7968$$
My doubt is that I have got a $\sqrt{17}$ in my factorisation as $y$ has to be $\sqrt{17},\sqrt{4,913}$ and so on which are not perfect roots, does this prove that $x$,$y$ are not integers?


答:
Directly or indirectly trigonometry will get in.

Due to the parallelism of sides and equality of forces ( makes for an isosceles triangle) in a parallelogram we can conclude that action of force vectors AB,AC
is now a special case ... of a rhombus.

The rhombus diagonals cut at $90^{\circ}...$ convenient trigonometry of a triangle.

$$ BE=\dfrac{6}{\sin \theta}= 24 \sin \theta  $$
$$ \sin ^2\theta = \dfrac14 $$
$$ \sin \theta = \dfrac12\to \theta = 30^{\circ}$$

We can verify that $BD$ is half of its half in length computed from a $(30,60,90) $ degrees standard right triangle:

$$ 24 \cos 60^{\circ}\cos 60^{\circ} =6 $$

So there are two equilateral triangles of side $24$ in the diagram shown below:

$BE= 12, $ the answer is Option 4

( Please ignore the circular arcs drawn during construction.)

Title: Does it exist a way to find one side of a triangle in polygon other than using similarity?

Body:
The problem is as follows:

Using the figure from below find $EB$. Assume $\triangle ABC$ is isosceles where $AB=AC$ and $\overline{BN} \parallel \overline{AC}$. Find $EB$.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&6\,cm\\
2.&8\,cm\\
3.&10\,cm\\
4.&12\,cm\\
5.&11\,cm\\
\end{array}$

What I attempted to do is sumarized in the figure from below:

As it can be seen I redrawn the figure so it can describe better what it is mentioned in the passage of the problem.

From this part. I relied on triangle similarity.

This would meant:

$\triangle ABE \sim \triangle BED$

Hence:

$\frac{6}{x}=\frac{x}{24}$

Solving this:

$x^2=6\times 24$

Which makes $x=12\,cm$

This later answer does check with the official answer from my book. But since this question is placed before introducing the concept of triangle similarity I am wondering if it could be solved relying using fundamentals of bisector angles and perpendicular bisector segment concepts. Can it be solved only using those?.

As it can be noted, some information is unused which makes me to believe was intentionally put there to use the above mentioned concepts.

All and all, is my solution correct?. I'm not good in this topic thus I need assistance on it. Please, include a drawing in your answer as it could require some construction.


问:
Representable functors preserve limits.
Theorem : Representable functors preserve limits.

I'm struggling to see why this is true. It's not obvious to me where I should be actually using the fact that functor in question is representable.

Any help would be much appreciated.

答:
Firstly we have to prove that $\hom(a,-)$ preserves limits.

Let $A$ be a category, $a\in A$ be its object and $\hom_A(a,-)\colon A\to\mathbf{Set}$ be a $\hom$-functor, corresponding to this object, let $B$ is also a category, $T\colon B\to A$ be a functor. Let's assume that $a_0$ is a limit of $T$, thus, we can consider a limiting cone $\varphi\colon\Delta_{a_0}\to T$. We want to prove that $\hom_A(a,a_0)$ is a limit of the functor $\hom_A(a,T(-))$ and that the induced cone $\varphi_*\colon\Delta_{\hom_A(a,a_0)}\to \hom_A(a,T(-))$ is also limiting. Let $X$ be an arbitrary set and $\psi\colon\Delta_X\to \hom_A(a,T(-))$ be an arbitrary natural transformation. Now we want to find a mapping $f\colon X\to \hom_A(a,a_0)$, such that $\varphi_*\circ\Delta_f=\psi$. For any element $x\in X$, let's construct a natural transformation $\psi_x\colon\Delta_a\to T$, such that $\psi_x(b)=(\psi(b))(x)$(check that this is a natural transformation). This construction will give you an arrow $f_x\colon a\to a_0$, such that $\varphi\circ\Delta_{f_x}=\psi_x$. Define the mapping $f$ by putting $f(x)=f_x$. Check that such mapping is unique which satisfies the property $\varphi_*\circ\Delta_f=\psi$.

Secondly, if $A$ and $B$ are categories, $T,S\colon A\to B$ are naturally isomorphic functors, then $T$ is continious(=preserves all limits) iff $S$ is continious(it is an easy exercise).

Now, if you have a representable functor, then it is isomorphic to some hom-functor, and therefore it is also continious - QED.

Title: Representable functors preserve limits.


问:
How to prove that $ a^{p_1} + b^{p_2}\geq \frac{1}{2^{p-1}} (a+b)^p -1$ holds?
Let $p_1, p_2>1$ and $a, b\in\mathbb{R}^+$ such that $a+b\geq2$. Could anyone please help me to prove that
$$ a^{p_1} + b^{p_2}\geq \frac{1}{2^{p-1}} (a+b)^p -1?$$
Here $p=\min(p_1, p_2)$. I am trying by using the inequality
$$(a+b)^p\leq 2^{p-1}(a^p + b^p),$$
but I am not able to obtain the desired result. Could anyone please help me or give me a hint?

Thank you in advance!

答:
Hint : The case $a\geq 1$ and $b\leq 1$ is proved now the case $a\geq 1$ and $b\geq 1$

It's sufficient to prove the inequality :

$$a^p+b^p\geq \frac{1}{2^{p-1}} (a+b)^p -1$$

For that you can use the inequality you mention .


问:
Finding subgroup of order $30$ from $S_n$
For what value of $n$, guarantee $S_n$ has a subgroup of order $30$

5

6

7

8

Its very clear that from the given option there is no $n$ for which $S_n$ contain any element of order $30$. Here I stuck to think how can I get the subgroup . Please help me

答:
Every group of order $30$ has an element of order $15$ (see here). This rules out options 1,2,3.

Option 4, which is $n=8$, works: take $H=AB$ where $A=\langle (12),(123) \rangle \cong S_3$ and $B=\langle (45678) \rangle \cong C_5$. Note that $A$ and $B$ commute elementwise.


问:
Why is $k[x,y]$ not finitely generated as $k[x]$-module?
Let $k$ a field. I want to prove that $k[x,y]$ is not finitely generated as $k[x]$-module, that is I would like to prove that there are no $b_1, ... , b_n \in k[x,y]$ such that $k[x,y] = b_1 k[x] + ... + b_n k[x]$. I tried looking for a counterexample but I wasn’t able to find any. Any hints?

答: Hint: Let $p_1,\cdots, p_n\in k[x]$. Can you find an upper bound on $\deg_y(p_1b_1+\cdots+p_nb_n)$ which is independent on $p_1,\cdots, p_n$?


问:
Elegant proof of $(n+1)^3-n^3$ is not divisible by $3$

By doing $(n+1)^3 - n^3$, I obtain $3n^2+3n+1$. Since the $+1$ prevents a multiple of 3 from forming, I understand the answer to this problem. However, what would be a more elegant manner to address why the +1 results in two consecutive terms never being a multiple of 3.

答:
$3n^2 + 3n + 1 = 3(n^2 + n) + 1$.

Explanation: If $n$ is an integer, this will never be divisible by $3$ as $3(n^2 + n)$ is always divisible by $3$, but $1$ is not divisible by $3$. Thus the sum of these two is also never divisible by $3$.

In your proof, you can just mention the first line, but you have to add that this is never divisible by $3$ for all $n \in \mathbb Z$.


问:
How many commutative rings of order $p^2$
This question was part of my algebra quiz and I was unable to solve 1 option out of 4. So, I am asking it here.

There exists exactly 2 commutative rings of order $p^2$ are there?

I have a intuition that it's true as only two groups exists and as we are adding a multiplicative operation so that must be true but it's given false.

So, kindly tell what mistake I am making. I studied algebra from Joseph Gallian.

答:
There are (at least) $3$ commutative rings of order $4$.

Aside from $\mathbb F_4$ and $\mathbb Z\big/2\mathbb Z\oplus \mathbb Z\big/2\mathbb Z$ we have the matrices of the form
$$
    \begin{bmatrix}
    x & y  \\
    0 & x  \\
  \end{bmatrix}
$$

Where $x,y \in \mathbb F_2$, the field of order $2$.

Shouldn't be difficult to demonstrate that these are all of them, though that doesn't matter for your problem.


问:
Bartle Real Analysis Theorem 8.2.3
Need help understanding how the author combined the 3 inequalities towards the end of the proof.

答:
This is just coming from

$$\begin{aligned}
\left\vert \frac{f(x)-f(c)}{x-c} - g(c) \right\vert &=\left\vert \left(\frac{f(x)-f(c)}{x-c} - \frac{f_K(x)-f_K(c)}{x-c}\right)+ \left( \frac{f_K(x)-f_K(c)}{x-c} - f^\prime_K(c)\right) + \left( f^\prime_K(c) - g(c)\right)\right\vert\\
&\le \left\vert\frac{f(x)-f(c)}{x-c} - \frac{f_K(x)-f_K(c)}{x-c}\right\vert + \left\vert \frac{f_K(x)-f_K(c)}{x-c} - f^\prime_K(c)\right\vert + \left\vert f^\prime_K(c) - g(c)\right\vert
\end{aligned}$$

Title: Bartle Real Analysis Theorem 8.2.3


问:
Find the value of $a,b$ such that given limit exists

Let $f(x)$ be a function such that $\lim_{x\to 0} \frac{f(x)}{x} =1$. If $\lim_{x\to 0} \frac{x(1+a\sin x)-b\sin x}{(f(x))^3}$ is finite, then find $a,b$

Let $f(x)=\sin x$

$$\lim_{x\to 0} \frac{ \sin x+ a\sin ^2x -b\sin x}{\sin^3 x}$$
$$\implies \lim_{x\to 0} \frac{ \sin x(1-a) +a\sin ^2 x}{\sin^3 x}$$

Which leads to nowhere. If the assumption of $f(x)$ is wrong, then I don’t know what else to do. Can I get a hint?

答:
Hint

$$\lim_{x\to 0} \frac{x(1+a\sin (x))-b\sin (x)}{(f(x))^3}=\lim_{x\to 0} \frac 1{x^3}\frac{x(1+a\sin (x))-b\sin (x)}{\left(\frac{f(x)}{x}\right)^3}$$

Now, consider Taylor series for
$$\frac{x(1+a\sin (x))-b\sin (x)}{x^3}$$


答:
Hint: The given limit exist and is finite if and only if $\frac {x(1+a \sin x)-b \sin x} {x^{3}}$ exist and is finite.  Using the expansion  of $\sin x$ as $x-\frac {x^{3}} {3!}+...$ we see that the limit exist if and only if $b=1$ and $a=0$.

More details: $\frac {1-b+ax} {x^{2}}$ is bounded. If $|1-b+ax| \leq Mx^{2}$ the letting $x \to 0$ we get $1-b=0$. And then we get $|a|\leq M|x|$ so we can letting $x \to 0$ in this to get $a=0$.

Title: Find the value of $a,b$ such that given limit exists

Body:
Let $f(x)$ be a function such that $\lim_{x\to 0} \frac{f(x)}{x} =1$. If $\lim_{x\to 0} \frac{x(1+a\sin x)-b\sin x}{(f(x))^3}$ is finite, then find $a,b$

Let $f(x)=\sin x$

$$\lim_{x\to 0} \frac{ \sin x+ a\sin ^2x -b\sin x}{\sin^3 x}$$
$$\implies \lim_{x\to 0} \frac{ \sin x(1-a) +a\sin ^2 x}{\sin^3 x}$$

Which leads to nowhere. If the assumption of $f(x)$ is wrong, then I don’t know what else to do. Can I get a hint?


答:
You can use lil' Fermat:

As $3$ is prime, we have $n^3\equiv n\mod 3$ for any $n$, hence
$$(n+1)^3-n^3\equiv (n+1)-n=1\mod 3.$$

Title: Elegant proof of $(n+1)^3-n^3$ is not divisible by $3$

Body: By doing $(n+1)^3 - n^3$, I obtain $3n^2+3n+1$. Since the $+1$ prevents a multiple of 3 from forming, I understand the answer to this problem. However, what would be a more elegant manner to address why the +1 results in two consecutive terms never being a multiple of 3.


答: By absurd, suppose that $3(n^2+n)+1=3k$ for some $k \in \mathbb{Z}$. Then, $1=3(k-n^2-n)$, that is, $1$ is multiple of $3$.


问:
Existance of function of norm one on normed spaces
On a past exam in a course on functional analysis the following problem is given:

Let $X,Y$ be normed spaces and let $x\in X$ be nonzero. Show that there exist some $f\in X^\ast$ s.t. $f(x) = \|x\|$ and $\|f\|=1$.

Here $X^\ast$ denotes the dual of $X$.

I have a proof, although it uses a corollary of the Hahn-Banach theorem which, on an exam, is hard to quote. Is there any simple proofs to this?

答:
At the moment no one knows the proof of this fact without sladgehammer like Hahn-Banach or something equivalent. On the other hand, to prove it is not only known but even impossible to find a direct proof is a difficult task. Such a proof that proof does not exist requires digging into foundations of mathematics.

Finally Hahn-Banach is so much important theorem that you must to learn it. This is one of three main theorems of functional analysis.


答: If the ring $R$ has characteristic $p^2$, then it must be $\mathbb Z/(p^2)$. Otherwise $R$ has characteristic $p$, so there’s a ring homomorphism $\mathbb F_p\to R$ (and so $R$ is an $\mathbb F_p$-algebra). Take any element $x$ not in the image of $\mathbb F_p$. Then there is an algebra homomorphism $\mathbb F_p[t]\to R$, $t\mapsto x$, necessarily surjective by comparing vector space dimensions. The kernel is generated by a monic polynomial $f$ of degree 2, and we see that there are three possibilities. Either $f$ is irreducible, or is a square, or it has two distinct factors. Up to isomorphism the quotient ring depends only on these three possibilities, not on the particular choice of $f$. This gives three more rings: a field, the ring $\mathbb F_p[t]/(t^2)$, or $\mathbb F_p\times\mathbb F_p$.

Title: How many commutative rings of order $p^2$

Body:
This question was part of my algebra quiz and I was unable to solve 1 option out of 4. So, I am asking it here.

There exists exactly 2 commutative rings of order $p^2$ are there?

I have a intuition that it's true as only two groups exists and as we are adding a multiplicative operation so that must be true but it's given false.

So, kindly tell what mistake I am making. I studied algebra from Joseph Gallian.


问:
ABCD is a cyclic quadrilateral whose diagonals intersect at point P.
$ABCD$ is a cyclic quadrilateral whose diagonals intersect at point $P$. Let the centres of the circumcircles of $ABCD$, $\triangle ABP$, $\triangle BCP$, $\triangle CDP$, $\triangle DAP$ be $O$, $O$$1$, $O$$2$, $O$$3$, $O$$4$ in the given order. Prove that $OP$, $O$$1$$O$$3$, $O$$2$$O$$4$ intersect at one point.

By connecting $O$$1$, $O$$2$, $O$$3$, $O$$4$, we can see that $O$$1$$O$$2$ is perpendicular to common cord $BP$. Similarly, $O$$3$$O$$4$ is perpendicular to common cord $DP$.

Since $D$, $P$, and $B$ are on the same line; $O$$1$$O$$2$ is parallel to $O$$3$$O$$4$. Similarly, $O$$1$$O$$4$ is parallel to $O$$2$$O$$3$.
So, $O$$1$$O$$2$$O$$3$$O$$4$ is a parallelogram

答:
Let us consider the inversion $*$ with center in $P$ and power $PA\cdot PC=PB\cdot PD$.
(The image of a point $X$ is thus denoted by $X^*$.)

Then we have by the choice of the center and the power $A^*=C$, $B^*=D$, $C^*=A$, $D^*=B$.

The circle $(O_1)=(PAB)$ is transformed in the line $A^*B^*=CD$, and $PO_1\perp CD$.

Since $O,O_3$ are both on the perpedicular bisector of $CD$, we have $OO_3\perp CD$.

This gives $PO_1\| OO_3$.

Similarly, $PO_3\| OO_1$.

So $PO_1OO_3$ is a parallelogram, its diagonals intersect in the midpoint $Q$, say, of $OP$.

So $O_1O_3$ passes through $Q$. Similarly, $O_2O_4$ passes through $Q$.

$\square$


问:
Analytic continuation of a logarithm function
What is the analytic continuation of
$$f(z) = \sum_{n=1}^{\infty} -\frac{z^n}n, \text{ where |z| < 1} $$

For real values of $z$, this function of course corresponds to the Taylor expansion of log(1-x). However, the complex logarithm function is multi-valued - so how do we choose an
analytic continuation of $f(x)$ for $|z| > 1$, since the analytic continuation is supposed unique?
Do we simply take the principial branch of the complex log function?

答:
Analytic continuation is not as universally applicable as we might sometimes want.  The identity theorem says that if $f$ and $g$ are analytic on a open connected domain $D$ and agree on an open subset of $D$ (or even on a set of points that has am accumulation point in $D$) then $f=g$ throughout $D$.  But notice that $f$ and $g$ must first both be analytic on $D$.  You cannot always guarantee that you can extend $f$ from a smaller to larger domain unless you first know such an extension will work.

An example is as follows.  We know one version of $\log z$ can be defined that is analytic on the domain
\begin{align}
D_1 = \{ z: 1 < |z| < 3, |\arg(z)| > \pi/8\}.
\end{align}

Using a power series centred at $z = 2$ we can create a second function analytic in the disc $D_2 = \{z: |z-2| < 2\}$ and which agrees with our $\log z$ on the part of $D_1 \cap D_2$ for which $\Im z > 0$.  But this will not then agree with our  $\log z$ on the intersection $D_1 \cap D_2$ where $\Im z < 0$.  That is because there is no analytic function on a domain that circles the origin and at the same time matches our $\log z$ throughout $D_1$.  In this example the identity theorem cannot be used as such an extension does not exist.
<img src="https://i.stack.imgur.com/RR0Zr.png" alt="Picture of two domains $D_1$ and $D_2$" />

Turning to your example, you can extend your power series into a larger domain, but you may end up with different branches depending on how your extension progresses around the singularity, which in your case is at $z=1$.  But as soon as you try to make your extension circle back to the first domain, the identity theorem ceased to apply.


问:
Is there a notion of $\mathfrak{g}$-$\mathfrak{h}$-bimodules?
Let $\mathfrak{g}$ and $\mathfrak{h}$ be two Lie algebras.

Is there a notion of $\mathfrak{g}$-$\mathfrak{h}$-bimodules such that $\mathfrak{g}$ becomes a $\mathfrak{g}$-$\mathfrak{g}$-bmodule via the adjoint actions of $\mathfrak{g}$ on itself from the left and from the right?

My thoughts so far:

One could try to define a $\mathfrak{g}$-$\mathfrak{h}$-bimodule as a $\operatorname{U}(\mathfrak{g})$-$\operatorname{U}(\mathfrak{h})$-bimodule, or equivalently as a vector space $M$ together with a left action of $\mathfrak{g}$ on $M$ and a right action of $\mathfrak{h}$ on $M$, such that these two actions commute. But this does not apply to the adjoint actions, so it doesn’t make $\mathfrak{g}$ into a $\mathfrak{g}$-$\mathfrak{g}$-bimodule.

There seems to exist the notion of a $\mathfrak{g}$-bimodule (coming from the more general notion of bimodules of a Leibniz algebra). For this, $\mathfrak{g}$ acts both from the left and from the right on $M$ (compatible with commutators) such that
$$
  (x \cdot m) \cdot y
  - x \cdot (m \cdot y)
  = [x, y] \cdot m \,.
$$
(Or some variation of this formula.)
But I don’t see how this definition could be generalized to a definition of $\mathfrak{g}$-$\mathfrak{h}$-bimodules.

答:
A left representation of $\mathfrak{g}$ is a vector space $V$ together with a linear map $\rho \colon \mathfrak{g} \to V$ such that $\rho([x, y]) = \rho(x) \rho(y) - \rho(y) \rho(x)$ for all $x, y \in \mathfrak{g}$. I would guess that a right representation of $\mathfrak{g}$ on $V$ is then a linear map $\pi \colon \mathfrak{g} \to V$ such that $\pi([x, y]) = \pi(y) \pi(x) - \pi(x) \pi(y)$.

If $\pi$ is a right representation, then $-\pi$ is a left representation, since $$(- \pi)([x, y]) = (-\pi)(x) (-\pi)(y) - (-\pi)(y) (-\pi)(x).$$
This shows that there is almost no difference between left and right representations of Lie algebras. Furthermore, the actions of two left representations $\rho, \eta \colon \mathfrak{g} \to V$ commute ($\rho(x) \eta(y) = \eta(y) \rho(x)$ for all $x, y \in \mathfrak{g}$) if and only if the actions of $\rho$ and $(-\eta)$ commute and conversely.

Therefore, any $(\mathfrak{g}, \mathfrak{h})$-bimodule $V$ with actions $(\rho, \pi)$ could just as well be considered a $\mathfrak{g} \times \mathfrak{h}$-module with the actions $(\rho, -\pi)$. So the answer to your question is that there is a notion of bimodules, but it is not more interesting than just considering modules over products of Lie algebras.


问:
Let $p$ be a prime number. Prove that for $a \in\mathbb Z, 0 \le a \le p−1$: $C(a, p - 1) \equiv (-1)^a\pmod p$
$C(a, p-1) = \frac{(p-1)!}{a!(p-1-a)!}$

If $a$ is an odd number, then $C(a, p-1)  \equiv p - 1 \pmod p$

If $a$ is an even number, then $C(a, p-1)  \equiv 1 \pmod p$

I'm stuck and I don't know how to proceed, I would be grateful for any help.

Body:
Hint: $a! = 1\cdot2\cdots a = (p-(p-1))(p-(p-2)
\cdots (p-(p-a)) \equiv (-1)^a (p-a)
\cdots (p-1)$


答: For the same reason as, for any commutative ring $R$, $R[X] $ is a free $R$-module, but not  a finitely generated one.

Title: Why is $k[x,y]$ not finitely generated as $k[x]$-module?

Body: Let $k$ a field. I want to prove that $k[x,y]$ is not finitely generated as $k[x]$-module, that is I would like to prove that there are no $b_1, ... , b_n \in k[x,y]$ such that $k[x,y] = b_1 k[x] + ... + b_n k[x]$. I tried looking for a counterexample but I wasn’t able to find any. Any hints?


问:
Solution set of $\frac x{x+2}>0\land\frac{x+1}{x+2}<1$
Okay people, $\Im$ stupid. I have been working on this for literal 2 days but still I can't understand.

Here is the question:
$$\frac x{x+2}>0\land\frac{x+1}{x+2}<1$$
What's the solution set? [Answer is $(0,\infty)$]

I found first inequality means $x > 0$ and second one says $x > -2$ but that doesn't make the solution set given by my lecturer which is $(0,\infty)$.

答:
The issue with your reasoning is when you multiply by a negative number, the inequality sign changes. Therefore, it is not true that $x > 0$ for all real $x$, but only when $x + 2 > 0$.

For the first part, I recommend that you split into cases. When $x + 2 > 0$, you do get $x > 0$. But when $x + 2 < 0$, then multiplying by $x+2$ on both sides gives:

$$x \color{red}{<} x+2 $$

which is true for all $x$ in the condition. Therefore, the possible values of $x$ are $x > 0, x < -2$.

For the second part, $-\frac{1}{x+2} < 0$ is correct so you can continue. From here, multiply by $-1$ to get:

$$\frac{1}{x+2} \color{red}{>} 0$$

and now use a similar method to find the possible values of $x$.


问:
Partitioning $\{1,2,...,1998\}$ such that $|a_i-b_i|$ is $1$ or $6$

Suppose that the set $\{1,2,3,\cdots, 1998\}$ has been partitioned into disjoint pairs $\{a_i,b_i\}(1\le i \le 999)$ so that for all $i, \mid a_i-b_i\mid$ equals $1$ or $6$. Then the sum $\mid a_1-b_1\mid +\mid a_2-b_2\mid+\cdots +\mid a_{999}-b_{999}\mid$ = ?

I found this problem on AoPS. I found out that the answer is 999 by partitioning the set into $\{1,2\},\{3,4\}, ... , \{1997, 1998\}$, but I am not sure about the proper way of solving it

答:
Let $C$ be the number of Consecutive pairs in the partition, and $S$ be the number of pairs that differ by Six. We have $C+S=999$ and the sum of the differences is thus

$$C+6S=999+5S$$

Now each pair that differs by $6$ consists of either two odd numbers or two even numbers. It's easy to see that the first such pair must consist of two odd numbers and the last such pair must consist of two even numbers -- e.g., you can't have $\{2,8\}$ without having $\{1,7\}$. What's a little harder to see is that the pairs that differ by $6$ must alternate between both odd and both even (when ordered, say according to the smaller number in each pair), but this can be shown by induction. It follows that $S$ must be an even number. Every even value of $S$ is possible between $0$ and $996$: Start with the "dense" set of pairs that differ by $6$ that pairs the numbers $1$ through $6$ with $7$ through $12$, $13$ through $18$ with $19$ through $24$, etc., and then two at time change $\{1,7\}$ and $\{2,8\}$ to $\{1,2\}$ and $\{7,8\}$, then $\{3,9\}$ and $\{4,10\}$ to $\{3,4\}$ and $\{9,10\}$, etc. It follows that every number ending in a $9$, from $999$ to $5979$, is a possible sum.

Remark: If all one wants to show is that the sum, whatever it is, ends in a $9$ (which is what the original AoPS problem asks for), it's enough to note that $999+5S\equiv4$ mod $5$ and that the sum and difference of any two numbers have the same parity, so the parity of the sum of the differences is

$$1+2+\cdots+1998={1998\cdot1999\over2}=999\cdot1999\equiv1\mod2$$

and therefore (combining $4$ mod $5$ and $1$ mod $2$) the sum of the differences is $\equiv9$ mod $10$. (The AoPS solution gives a different argument for the $4$ mod $5$ part of the proof; I might have thought of the parity part myself, but I saw it there first.)


问:
The difference between 'weak limit point' and 'converge weakly'
For the following theorem.

Let $S$ be a nonempty subset of $H$ and let $x:[0,+ \infty) \rightarrow H$. Assume that

$\quad$ (i) for every $z\in S$, $\lim_{t\rightarrow \infty} \left\|x(t)-z\right\|$ exists;

$\quad$ (ii) every weak sequential limit point of $x(t)$, as $t\rightarrow \infty$, belongs to $S$.

Then $x(t)$ converges weakly as $k\rightarrow \infty$ to a point in $S$.

My questions are:

What is the difference between 'weak limit point' and 'converge weakly'?

What is the difference between 'weak limit point' and 'weak sequential limit point'?

The first one is the main question, thanks!

答: A weak sequential limit point is a point $h$ such that $x(t_n) \to h$ weakly for some sequence $t_n \to \infty$. $x(t)$ converges weakly to $h$ if $x(t) \to h$ weakly as $t \to \infty$ which means $F(x(t)) \to F(h) $ as $t  \to \infty$ for every continuous linear map $F$.


问:
Probability of infinite cluster on this specific random geometry.

Context

We consider the plane, $\mathbb{R}^2$, which we split into squares of size $1$. Each square has a probability $p$ of being black and $1-p$ of being white, independently for each square. Let $F$ be the area covered by the black squares; this is a random subset of $\mathbb{R}^2$.

For description, we say that a set $K \subset F$ is connected if for any pair of squares in $K$ there exists a path of touching black squares in $K$ connecting them. Obviously, $F$ is a union of separate connected components, which are called clusters.

It is well known (see for example Percolation Theory for Mathematicians. Birkhauser, Basel, Boston, 1982.
Khruslov, E.Ya.)  that there exists a threshold probability $p_c$ such that

for $p<p_c$, $F$ is made of finite clusters (a.s.)

for $p>p_c$, $F$ contains an infinite cluster (a.s.),

and it was proved that $p_c \approx 0.41.$ This phenomenon is known as percolation.

My question

I would like to study a similar problem in $\mathbb{R}^2$ but a little bit different. In my case, I have a selection of spheres $B_i=B(x_i,1),\ i \in \mathbb{N}$, where the family of the sphere's center $(x_i)_{i \in \mathbb{N}}$ is made of random variables (see commentary, for example it could be a random Poisson Hardcore Point Process) valued in $\mathbb{R}^2$ and such as $B_i \cap B_j = \oslash,\  i \neq j$. The spheres $F=\cup_{i \in \mathbb{N}} B_i$ are colored in black and the rest $\mathbb{R}^2 \setminus \cup_{i \in \mathbb{N}} B_i$ is colored in white.

As with the random checkerboard studied above, it is possible to define the notion of a cluster in $F$ for touching spheres, and it is possible that an infinite cluster of spheres appears.

I was wondering if there were references about an equivalent of this probability threshold ? I would especially like a reference on this topic if it exists.

(Feel free to retag this question as appropriate.)

答:
I have to go with Steven Stadnicki's comment and say that your distribution for the point process has to be very weird for the spheres to be non-intersecting, yet touching each other often enough to build large clusters.
I do not know if this is helpful in your setting, but maybe you should look into percolation theory on more general geometries,
for instance on isoradial graphs.</p>


答:
Start with the following polynomial identity in $\mathbb Z[x]$:

$$x^p+1=(x+1)(x^{p-1}-x^{p-2}+\ldots+(-1)^{p-2}x+(-1)^{p-1})$$

This identity is also valid in $F_p[x]$ (coefficients reduced modulo $p$. $F_p$ is the field of remainders modulo $p$).

However, in $F_p[x]$ we also have that:

$$x^p+1=(x+1)^p$$

Namely, in $\mathbb Z[x]$, take all the coefficients modulo $p$, and note that the coefficients on the right side are all (except for the first and the last) divisible by $p$, as they are of the form $C(a,p)={p \choose a}=\frac{p!}{a!(p-a)!}$ where $p$ is the factor in the numerator but not in the denominator. So, in $F_p$ all the coefficients barring the first and the last one must be zero.

Thus, in $F_p[x]$ we have that:

$$(x+1)^{p-1}=x^{p-1}-x^{p-2}+\ldots+(-1)^{p-2}x+(-1)^{p-1}$$

and comparing the coefficients and using binomial formula, you get that:

$$C(a,p-1)={p-1\choose a}=(-1)^a$$

in $F_p$, which is the same as the claim you are trying to prove.

Title: Let $p$ be a prime number. Prove that for $a \in\mathbb Z, 0 \le a \le p−1$: $C(a, p - 1) \equiv (-1)^a\pmod p$

Body:
$C(a, p-1) = \frac{(p-1)!}{a!(p-1-a)!}$

If $a$ is an odd number, then $C(a, p-1)  \equiv p - 1 \pmod p$

If $a$ is an even number, then $C(a, p-1)  \equiv 1 \pmod p$

I'm stuck and I don't know how to proceed, I would be grateful for any help.


问:
How should I set the integral route for this complex integral?
\begin{equation}
\int_0^{\infty} \dfrac{x^{\lambda -1}}{x^2+1} dx
=
\dfrac{\pi}{2 \sin{\dfrac{\pi \lambda}{2}}} \quad (0< \lambda < 2)
\end{equation}

I will prove this using complex integral, but I don't come up with how I should set the integral route.

I tried prove the equality using the integral route below and using residue theorem.

$
C_{R} : \text{semicircle in first and second quadrant whose radius is R }\\
C_1 : [-R, \epsilon]\\
C_{\epsilon} : \text{semicircle in first and second quadrant whose radius is $\epsilon$ }\\
C_2 : [\epsilon, R]
$

However, I failed to prove. It seems that it is impossible to calculate the integral with this integral route.

I would like you to teach me how I should set the integral route.

答:
One may use $\beta$ function.

$$I=\int_{0}^{\infty} \frac{x^{k-1}}{1+x^2} dx$$
Let $x=\tan t$, then
$$I=\int_{0}^{\pi/2} \sin^{k-1}t ~\cos^{1-k}t ~dt=\frac{1}{2} \frac{\Gamma(k/2)\Gamma(1-k/2)}{\Gamma(1)}= \frac{\pi}{2}\csc(\pi k/2), 0<k<2$$

See for $\beta$ function:  https://en.wikipedia.org/wiki/Beta_function