问:
Homotopy problem
Let $X$ and $Y$ metric space with $X$ compact and $f,g: X \rightarrow Y$ homotopic. Then for each $\epsilon>0$, exists continuous functions $f_0,\ldots,f_k: X \rightarrow Y$ such that:

$$ d(f_i(x),f_{i-1}(x))<\epsilon, \quad \forall x\in X $$

and $f_0=f$, $f_k=g$.

I was thinking of using the fact that $H: X \times [0,1] \rightarrow Y$ is continuous in a compact, so is uniformly continuous. But I can't prove what it asks for

答:
The homotopy $H$ tells you how to move continuously from $f$ to $g$, and the problem asks you to move from $f$ to $g$ in small, discrete steps. Getting discrete steps is just a matter of discretizing the continuous motion that you already have. Subdivide the parameter interval $[0,1]$ into little pieces, say $[\frac{i-1}N,\frac iN]$ for some big $N$ and $i=1,2,\dots,N$. Then take snapshots of the motion $H$ at those subdivision points; that is, set $f_i(x)=H(x,\frac iN)$. (Another rough description of this: Watch the motion $H$ with a stroboscope.) That gets you from $f_0=f$ to $f_N=g$ in discrete steps.

But the steps are also required to be small. To achieve that, you want to take $N$ big enough so that the intervals $[\frac{i-1}N,\frac iN]$ are very short. Here, your idea of using the uniform continuity of $H$ solves the problem. For any desired $\varepsilon$ in the problem, find a $\delta$ as in the definition of uniform continuity, and take $N>\frac1\delta$.

Title: Homotopy problem


答:
You are not setting up your induction very well. You should not start with the equality you want to establish, namely that $(-1)^{n+1}-10^{n+1}$ is a multiple of $11$. Instead, you should start with the Induction Hypothesis, which is that $(-1)^n - 10^n$ is a multiple of $11$.

So: the Inductive Step is to show that if $(-1)^n - 10^n$ is a multiple of $11$, then $(-1)^{n+1} - 10^{n+1}$ is also a multiple of $11$.

Let's write out our Induction Hypothesis: it says that
$$\text{There exists an integer }k\text{ such that }(-1)^n - 10^n = 11k.$$

What we want to prove is that:
$$\text{there exists an integer }\ell\text{ such that }(-1)^{n+1}-10^{n+1}=11\ell.$$
(Note that the multiple may be different, that's why I used a different letter).

So now we can try manipulating the expression we want. One possibility is to use the following identity:
$$a^{n+1}-b^{n+1} = (a-b)(a^n + a^{n-1}b + a^{n-2}b^2 + \cdots + ab^{n-1}+b^n),$$
if you already know this identity.

So we have, with $a=-1$ and $b=10$,
$$
(-1)^{n+1} - 10^{n+1} = \Bigl( (-1) - 10\Bigr)\Bigl( (-1)^n + (-1)^{n-1}(10) + \cdots + (-1)10^{n-1} + 10^n\Bigr).$$
Now notice that you don't even need to use the induction hypothesis to conclude that $(-1)^{n+1}-10^{n+1}$ is a multiple of $11$ (as could be seen in mac's answer).

If you don't know the identity, then you can perform some purely algebraic manipulations. E.g.,
$$\begin{align*}
(-1)^{n+1} - 10^{n+1} &= -1\left( (-1)^n + 10^{n+1}\right)\\
 &= -\left( (-1)^n -10^n + 10^n + 10^{n+1}\right)\\
&= -\left( \Bigl((-1)^n - 10^n\Bigr) + 10^n\Bigl(1 + 10\Bigr)\right)\\
&= -\left( 11k + 10^n(11)\right) &\quad&\text{(by the induction hypothesis)}\\
&= -\left( 11(k+10^n)\right)\\
&= 11\left( -(k+10^n)\right),
\end{align*}$$
which gives that $(-1)^{n+1} - 10^{n+1}$ is a multiple of $11$, as desired, from the assumption that $(-1)^n - 10^n$ is a multiple of $11$.

But easier still is to use the following property of congruences:

Proposition. Let $a,b,c,d,k$ be integers. If
$$a\equiv b\pmod{k}\qquad\text{and}\qquad c\equiv d\pmod{k}$$
then $ac\equiv bd\pmod{k}$.

Proof. Since $a\equiv b\pmod{k}$, then $k|a-b$, so $k$ divides any multiple of $a-b$; for example, $k|(a-b)c = ac-bc$. Since $k$ divides $ac-bc$, then $ac\equiv bc\pmod{k}$.

Since $c\equiv d\pmod{k}$, then $k|c-d$, so $k|(c-d)b = cb-db$, hence $bc\equiv bd\pmod{k}$.

Since $ac\equiv bc\pmod{k}$ and $bc\equiv bd\pmod{k}$, then $ac\equiv bd\pmod{k}$. QED

Corollary. If $a_1\equiv b_1\pmod{k}$, $a_2\equiv b_2\pmod{k},\ldots, a_n\equiv b_n\pmod{k}$, then
$$a_1\cdots a_n\equiv b_1\cdots b_n\pmod{k}.$$

Proof. Induction on $n$. QED

(This is where you would want to use induction, rather than the specific case you are looking at).

Corollary. If $a\equiv b\pmod{k}$, then for all positive integers $n$, $a^n\equiv b^n\pmod{k}$.

Proof. Apply previous corollary with $a_i=a$ and $b_i=b$ for all $i$. QED

ContentLicense: CC BY-SA 3.0

Title: Complete induction of $10^n \equiv (-1)^n \pmod{11}$

Body:
To prove $10^n \equiv (-1)^n\pmod{11}$, $n\geq 0$, I started an induction.

It's $$11|((-1)^n - 10^n) \Longrightarrow (-1)^n -10^n = k*11,\quad k \in \mathbb{Z}. $$

For $n = 0$:
$$ (-1)^0 - (10)^0 = 0*11 $$

$n\Rightarrow n+1$
$$\begin{align*}
 (-1) ^{n+1} - (10) ^{n+1} &= k*11\\
(-1)*(-1)^n - 10*(10)^n &= k*11
\end{align*}$$
But I don't get the next step.


问:
Dominating set in an $r$-uniform hypergraph
I'm trying to solve an exercise from "Probability and computing: Randomized algorithms and probabilistic analysis" by Michael Mitzenmacher and Eli Upfal:

A hyper graph $H$ is a pair $(V,E)$ where $V$ is the set of vertices and $E$ is the set of hyper edges (i.e. a certain set of subsets of $V$). We call $H$ "$r$-uniform" if each hyper edge has size $r$. A "dominating set" $S$ in $H$ is a set of vertices such that every hyper edge $e$ intersects $S$.

Let $H=(V,E)$ be an $r$-uniform hyper graph with $n$ vertices and $m$ edges. Why must there be a dominating set of size $\leq (m+n \ln r)/r$? Why must there be a dominating set of size $\leq np+(1-p)^r m$ for every real number $0 \leq p \leq 1$?

Does anybody have an idea for a solution? I'm pretty sure it should be solved using the probabilistic method but I'm just really having a hard time of pinpointing a specific solution.

Also if anybody knows of a solution manual for the entire book I'd like a reference please.

Thanks

EDIT (October 2016): I am a professor who teaches a course using this book. I have edited the question to make it harder for my students to google, since this is the only place online currently where an answer to this exercise can be found. In general, it's very hard to find good, accessible problems about randomized algorithms, so I'd be really happy if people would stop putting solutions online for the few good problems we have.

答:
Let $T$ be a set of vertices constructed as follows:

For each vertex $v$ in $V$, $v$ is included in $T$ independently with probability $p$, $p$ to be determined.

Let $X$ = $| T |$ . It is obvious that $E[X]$=$np$.

For each edge $S$ in $E$, let $Y_{s}$ be the random variable which indicates whether $S\cap{T}=\emptyset$. $Y_{s}=\begin{cases}1 & S\cap{T}=\emptyset\\
0 &otherwise\end{cases}$.  It is obvious that $E[Y_{s}]=\left(1-p\right)^r$

Let random variable $Y$ be the number of such edges, by linearity of expectation, it holds that $E[Y]=\sum_{e\in{E}}{E[Y_{s}]} =m\left(1-p\right)^r$.

Note that although $T$ is not necessarily an blocking set, it can be modified to one if we add one endpoint of those edges that have no endpoint in $T$ into $T$. Let $T'$ be the resulting set. It is obvious that $T'$ is an blocking set since every edge has a endpoint in it.

Since there are $Y$ such edges, there are at most $Y$ vertices that are added to make it become $T'$ . Therefore, $|T'|<= X+Y$. By linearity of expectation, $E|T'|<= E[X+Y]=np+m(1-p)^{r}<=np+me^{-pr}$;

The right hand side is minimized at $p=\frac{\ln{\frac{mr}{n}}}{r}$, and the minimum is $\frac{n\ln{\left(\frac{mr}{n}+1\right)}}{r}$. Notice that $\frac{n\ln{\left(\frac{mr}{n}+1\right)}}{r}<=\frac{\left(m+n\ln{r}\right)/r}{r}$, paying attention that $\frac{m}{n}<=1$.


问:
Solution to abstract linear equation system
I want to calculate all possible solution sets of this linear equation System $\alpha,\beta \in \mathbb R$:
\begin{align}
x+y+\alpha z&=\beta \\
x+\alpha y+z&=\beta \\
\alpha x+y+z&=\beta
\end{align}

答:
Let's consider the system you wrote $$
\left[\begin{array}{ccc|c}
1 & 1 & \alpha & \beta\\
0 & \alpha-1 & 1-\alpha & 0 \\
0 & 0 & 2-\alpha^2-\alpha & \beta(1-\alpha)\\
\end{array}\right]
.$$
If $\alpha=1$ the second and the third row become the zero row, so the rank of the matrix is $1$ and in this case we'll have $\infty^2$ solution.

If $\beta\ne0$ and $\alpha^2+\alpha-2=0\implies \alpha=\dfrac{-1\pm3}{2}\implies\alpha=1\text{ or }\alpha=-2$. The system has no solutions for $\alpha=2$ because we would have $0=3\beta$, with $\beta\ne 0$.For $\alpha=1$ we are in the first case that we analised.

Supposing $\alpha\ne 1,-2$ and $\beta\ne0$ you can find $z=\dfrac{(\alpha-1)\beta}{\alpha^2+\alpha-2}=\dfrac{\beta}{\alpha+2}$ and then, substituting in the second and in the first row, you find
$$(\alpha-1)y-(\alpha-1)z=0\underset{\alpha\ne1}{\implies} y=z=\dfrac{\beta}{\alpha+2}\text{ and }$$
$$x=\beta-\alpha z-y=\beta-\dfrac{\beta(\alpha+1)}{\alpha+2}\in\mathbb R.$$
If $\alpha=1$ we have
$$x=\beta-\alpha t-s,\text{ where s and t are two real free parameters}.$$
Try to think about the case in which $\beta$ is zero and $\alpha$ is $-2$ and let me know if it's now more clear for you.

ContentLicense: CC BY-SA 4.0


问:
Real numbers $a$ and $b$ satisfy $a^3+b^3-6ab=-11$. Prove that $-\frac{7}{3}<a+b<-2$

Real numbers $a$ and $b$ satisfy $a^3+b^3-6ab=-11$. Prove that $$-\frac{7}{3}<a+b<-2$$

I have shown that $a+b<-2$. My approach: $-3=8-11=a^3+b^3-6ab+2^3=\frac{1}{2}(a+b+2)((a-b)^2+(a-2)^2+(b-2)^2)$. From this we must have that $a+b<-2$.

Please give some idea/hint for the other part.

Body:
Showing that $a+b > - \frac73$:
$$(a-2)^2+(b-2)^2+(a-b)^2\geq(a-2)^2+(b-2)^2 \geq\frac{(a+b)^2}{2}-4(a+b)+8$$
Now using the previous result $a + b < -2$ we get:
$$\frac{(a+b)^2}{2}-4(a+b)+8> 18$$
From OP's equation in the question, this gives

$$a+b=-\frac{6}{(a-2)^2+(b-2)^2+(a-b)^2}-2>-\frac{6}{18} -2 = -\frac{7}{3}. \qquad \Box$$


问:
What does this sine wave say exactly?
I am currently starting to learn about sine waves and I was just wondering why this sine waves amplitude is the way it is? Like why exactly does it start at 2, peak at negative 1 and peak at 3 instead of starting at 0, peaking at positive then negative 2? id assume it starts just above 2 because its offset from 0 by a certain degrees?Demodulator signal</p>

答:
The most general "sine wave" is $y=y_0+A\sin(2\pi f(x-x_0))$, where $y$ obtains its mean value $y_0$ at $x=x_0$, but in general fluctuates around this mean with amplitude $A>0$, frequency $f$. Two pernickety points:

We can impose $A>0$ because $A\mapsto-A,\,x_0\mapsto x_0+1/(2f)$ preserves $y$;

We call $f$ the linear frequency, while $\omega:=2\pi f$ is the circular frequency. You'll often see the sine's argument written as $\omega x-\phi$ with $\phi:=\omega x_0$.

The sine function is $x_0=y_0=0,\,A=\omega=1$.

Title: What does this sine wave say exactly?


问:
Convergence of $\sum_{n=1}^{\infty} \frac{1}{n^3}$
I know the alternating series $$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^3}$$ converges by doing the alternating series test, but I can't seem to find a way to prove $$\sum_{n=1}^{\infty} \frac{1}{n^3}$$ converges. I tried it with  $\lim_{n \to +\infty}\sqrt[n] \frac{1}{n^3} $ and $\lim_{n \to +\infty}\frac{a_{n+1}}{a_n}$, but both of these result in 1, which means you can't conclude anything from them.

Any help would be appreciated!

答:
I thought it might be instructive to present an approach the relies on straightforward elementary analysis only.  To that end, we now proceed.

Note that we can write

$$\begin{align}
\frac1{(n-1)^2}-\frac1{n^2}&=\frac1{n^2}\left(\frac1{\left(1-\frac1n\right)^2}-1\right)\\\\
&\ge \frac2{n^3}\tag1
\end{align}$$

Summing the telescoping sum and applying the inequality in $(1)$ reveals

$$\begin{align}
1-\frac1{N^2}&=\sum_{n=2}^N \left(\frac1{(n-1)^2}-\frac1{n^2}\right)\\\\
&\ge 2\sum_{n=2}^N \frac1{n^3}\tag2
\end{align}$$

Therefore, we see that

$$\sum_{n=1}^N\frac1{n^3}\le \frac32-\frac1{2N^2}$$

whereupon application of the squeeze theorem we see that the series of interest converges to a value of less than $3/2$.

TOOLS USED:  Telescoping series, the squeeze theorem, elementary inequalities


问:
Calculate $\lim_{x \to 0} \frac {2e^{-1/x^2}}{x^3}$
Got this question and was wondering why the limit is $0$ ?

I saw a few people that mentioned that it can be written when $\frac {2e^{-1/x^2}}{x^3}$ and such limits is always $0$.

Is there any way to explain it or to show it specifically somehow ?

Thanks in advance.

答:
Just note that for $t>0$ you have $e^t = 1+ t +\frac{t^2}{2} +\cdots > \frac{t^2}{2}$.

Hence,

\begin{eqnarray*} \left| \frac {2e^ \frac {-1}{x^2}}{x^3} \right|
& = & \frac 2{e^\frac 1{x^2}\cdot |x|^3} \\
& < & \frac 2{\frac 1{2x^4}\cdot |x|^3} \\
& = & 4|x| \stackrel{x\to 0}{\longrightarrow} 0
\end{eqnarray*}


问:
How long will it take Marie to saw another board into 3 pieces?
So this is supposed to be really simple, and it's taken from the following picture:

Text-only:

It took Marie $10$ minutes to saw a board into $2$ pieces. If she works just as fast, how long will it take for her to saw another board into
  $3$ pieces?

I don't understand what's wrong with this question.  I think the student answered the question wrong, yet my friend insists the student got the question right.

I feel like I'm missing something critical here. What am I getting wrong here?

答:
Well, i have something else in my mind.I know it is not that practical still i want to share my views about the question.

Assuming the board is of 1 meter long and we have to cut it into 3 pieces of equal length i.e. finally each of the pieces should be length of 1/3 meter long (see the picture given in the question). So we have to make two cuts at length 1/3 and at length 2/3. Now note that after cutting it first time(which will take 10 minutes), we will have two piece. One is of 1/3 meter long and the other portion is of 2/3 meter long.Now we have to make a cut to the last portion(which is of 2/3 meter long) and make it in two pieces.

Now the interesting part comes. If we assume the board has a uniform resistance against the saw, then after losing its one third end, board will lose its resistance uniformly (assuming resistance depends on length proportionally). In that case, it will take  another $10*(2/3)=6.67$ minutes to get another two pieces.

hence  we need total $16.67$ minutes.

I know it is not practical, still...


问:
Volume of fundamental domain in lattice of $\mathbb{Z}^2$
I'm trying to prove exercise 6.1 on Algebraic number theory by Stewart and Tall and essentially it states:

If $L$ is a lattice in $\mathbb{R^2}$ with basis $u, v\in \mathbb{Z^2}$, $T$ its fundamental domain, i.e.
$$T=\{xu+yv: 0\leq x,y <1\}$$
and $n$ is the number of points of $\mathbb{Z}^2$ that are in $T$ then
$$\operatorname{vol}(T)=n$$

I know Pick's theorem but the exercise is stated without any knowledge on that topic and I don't know how to prove it just knowing the definitions I gave. I've tried to prove it inductively on $n$ but I can't even prove the base case and I don't know how to relate in any way the volume and $n$. Any ideas?

答:
(I'm posting this as a separate answer to avoid cofusion with my previous reply).

The corrected form of your question is a simple consequence of Hermit normal form of integer matrices, aka row reduction.

Fix a basis $(u1, u2), (v1,v2)\in \mathbf Z^2$ of your sublattice, and consider the matrix
$
M = \bigl(
\begin{smallmatrix}
  u1 & u2 \\
  v1 & v2
\end{smallmatrix}
\bigr)
$.
To say that your sublattice has index n is to say that
$
\det(M)=n
$.
By row reduction over $\mathbf Z$, which is equivalent to mutiplying $M$ on the left by a matrix in $SL_2(\mathbf Z)$, we can turn $M$ into its unique Hermite normal form, i.e. a matrix of the form
$
H = \bigl(
\begin{smallmatrix}
  a & b \\
  0 & d
\end{smallmatrix}
\bigr)
$
where $a, b, d\in \mathbf Z$ with $a|d$, and of course $ad=n$.  That means
$(a,b), (0,d)$ is a basis of your sublattice, and your exercise follows immediately (draw a picture).


问:
Partial derivates and integral $\int_{\mathbb{R}^n}\partial_ifdx=0$
If $f\in C_c^1(\mathbb{R}^n)$, we have $\int_{\mathbb{R}^n}\partial_ifdx=0$ for all $1\le i\le n$.
I started like this. But I am definitely missing something. Can someone help me finish the proof?$$\int_{\mathbb{R}^n}\,\partial _i f(x) \mathrm{d}x=\int \limits_{-r}^{r}\cdots \int \limits_{-r}^{r}\partial_if(x_1,...,x_i,...,x_n)\, \mathrm{d}x_1\ldots \mathrm{d}x_i\ldots \mathrm{d}x_n=\cdots = 0$$

答:
To follow leoli1's comment, suppose your function is supported on $B=[-r/2,r/2]^n$.

By Fubini's theorem,
$$
\int_B f=\int_{U}\left(\int_{-r}^r\partial_if(x)dx_i\right)dx_1\cdots\hat{dx_i}\cdots dx_nn
$$
where $\hat{dx_i}$ means obmitting $dx_i$ and $U\times[0,1]^{n-1}=B$. But
$$
\int_{-r}^r\partial_if(x)dx_i=f(x_1,\cdots,x_i=r,\cdots,x_n)-f(x_1,\cdots,x_i=-r,\cdots,x_n)=0-0=0
$$


问:
Christoffel symbols manipulation help
We have $\sum_l\Gamma^{l}_{ij}g_{lk}=\frac{1}{2}(\frac{\partial(g_{jk})}{\partial{x}_i}$+$\frac{\partial(g_{ki})}{\partial{x_j}}$-$\frac{\partial(g_{ij})}{\partial{x}_k})$

Since $(g^{lk})$ is the inverse of $(g_{lk})$,

$\Gamma^{m}_{ij}=\frac{1}{2}\sum_k(\frac{\partial(g_{jk})}{\partial{x_i}}$+$\frac{\partial(g_{ki})}{\partial{x_j}}$-$\frac{\partial(g_{ij})}{\partial_{x_j}})g^{km}$

May someone please supply the linear algebra argument above ? the indices are slightly confusing me every time I attempt to understand the linear algebra argument.

答: $\Gamma^{l}_{ij}g_{lk}g^{km}=\Gamma^{l}_{ij}\delta^m_{l}=\Gamma^{m}_{ij}$ (i'm using the summation convention)

Title: Christoffel symbols manipulation help


问:
How to use maximum likelihood method to find $\mu$ and $\sigma$ from erroneous observations.
Let $X$ be a random variable having a normal distribution with mean $\mu$ and standard deviation $\sigma$. The observations $(x_i; i=1,2,\cdots,n)$ have measurement errors that are normally distributed with zero mean and known standard deviation $\delta$. Use the method of maximum likelihood to estimate $\mu$ and $\sigma^2$ from erroneous observations.

答:
You can't estimate $\mu$ or $\sigma^2$ by maximum likelihood unless you assume that $\mu$ and $\sigma^2$ are parameters that determine some distribution. For example, if it were assumed that the things whose expected value is $\mu$ and whose variance is $\sigma^2$ are normally distributed, and if in addition it were assumed that they are independent, then it can be done.

In effect you have $x_i\sim\operatorname N(\mu, \sigma^2 + \delta^2)$ and $\delta^2$ is known.

It is tempting to say the m.l.e. for $\sigma^2+\delta^2$ is the familiar $s^2$ and then by equivariance of m.l.e.s, say that the m.l.e. of $\sigma^2$ is $s^2-\delta^2.$ But what about cases where $s^2-\delta^2<0\text{ ?}$

Here we need to say that the parameter space for $\sigma^2+\delta^2$ is the interval $[\delta^2,+\infty),$ not $[0,+\infty).$

As a function of $\sigma^2+\delta^2$, the likelihood function decreases on the interval $\Big(\max\{s^2,\delta^2\},+\infty\Big).$ It increases on the interval $\Big[\delta^2, \max\{s^2,\delta^2\}\Big)$ if that is not empty.

Therefore the m.l.e. of $\sigma^2+\delta^2$ is $\max\{s^2,\delta^2\}.$

By equivariance of m.l.e.s, the m.l.e. of $\sigma^2$ is therefore
$$
\widehat\sigma^2 = \max\{s^2,\delta^2\} - \delta ^2 = \max\{s^2-\delta^2, 0\}.
$$


问:
Compute $\iint_Y F.N \ dS$
The question is: Find $\iint_Y F.N \ dS $
$$ F=(x^4+yz-x^5,5x^4y,z),\quad  \text{The surface }Y=x^2+y^2-z^2=1, \ \ 0\leq z\leq 1 \quad N=\text{
the normal points away from z axis}$$

Here is how I thought:

I parameterized the surface with $r(x,y)=(x,y,\sqrt{x^2+y^2-1}) $ and i think the normal is: $(r'_x\times r'_y)= (\frac{x}{\sqrt{x^2+y^2-1}},\frac{y}{\sqrt{x^2+y^2-1}},-1)$ and then I calculated the integral in this way but the integral looks very complicated and awful, have I calculated correctly? and what is the limit of this integration?
$$\iint \frac{x^5-x^6+5x^4y^2+x^2+y^2-1+\sqrt{x^2+y^2-1}xy}{\sqrt{x^2+y^2-1}}\ dx\,dy $$
Any suggestion would be great, thanks.

答:
One of the ways would be to use divergence theorem and find the flux through entire closed surface which consists of the hyperboloid between $0 \leq z \leq 1, $ and circular top and bottom surfaces at $z = 0, z = 1$. We can then subtract the flux through circular surfaces at $z = 0$ and $z = 1$ which is an easier calculation (in fact you can see why the flux through the circular bottom at $z = 0 \ $ for the given vector field would be zero. So it is only the flux through the circular top that we may have to find and subtract).

$div\vec{F} = \nabla \cdot \vec{F} = 4x^3-5x^4 + 5x^4 + 1 = 1 + 4x^3$

Also as $x^3$ is an odd function and there is symmetry along $x-$plane ($YZ$), the integral simplifies further.

So, Flux $ = \displaystyle \int_V div\vec{F} \ dV = \int_0^{2\pi} \int_0^1 \int_0^\sqrt{1+z^2} \ r \ dr \ dz \ d\theta$

Now calculate the flux through the circular top and subtract from the above.

Also to comment on your working, the parametrization and your steps for double integral are correct. It is a tedious one! For limits,

From here on, you can see that for $0 \leq z \leq 1, 1 \leq x^2+y^2 \leq 2$. So you can convert it into polar coordinates as $x = r \cos\theta, y = r\sin\theta, z$.

$\int_0^{2\pi} \int_1^\sqrt2 \vec{F}\cdot (r_x \times r_y) \ r \ dr \ d\theta$

Another way to parametrize the hyperboloid would be,

$r(\theta, z) = (\sqrt{1+z^2} \cos \theta, \sqrt{1+z^2} \sin \theta, z)$ and

$(r'_\theta \times r'_z) = (\sqrt{1+z^2} \cos \theta, \sqrt{1+z^2} \sin \theta, z)$

But it does not make the integral any easier.

Title: Compute $\iint_Y F.N \ dS$


问:
The existence of a graph homomorphism
Is there a graph homomorphism between the 6-hypercube (https://en.wikipedia.org/wiki/File:6-cube_graph.svg) and the cube (https://en.wikipedia.org/wiki/File:3-cube_graph.svg)?</p>

答: Yes. For example, with $A=\{-1,1\}$, associate the vertices of the graphs with vectors in $A^6$ and $A^3$, respectively; then the function that maps $a\in A^6$ to $(\prod_ia_i,1,1)$ is a homomorphism.</p>

Title: The existence of a graph homomorphism


问:
Examples of surjective sheaf morphisms which are not surjective on sections
Let $X$ be a topological space, and let $\mathscr{F}, \mathscr{G}$ be sheaves of sets on $X$. It is well-known that a morphism $\varphi : \mathscr{F} \to \mathscr{G}$ is epic (in the category of sheaves on $X$) if and only if the induced map of stalks $\varphi_P : \mathscr{F}_P \to \mathscr{G}_P$ is surjective for every point $P$ in $X$, but the section maps $\varphi_U : \mathscr{F}(U) \to \mathscr{G}(U)$ need not be surjective. I know of a couple of examples from complex analysis:

Let $X$ be the punctured complex plane, $\mathscr{F}$ the sheaf of meromorphic functions, $\mathscr{G}$ the sheaf of differential $1$-forms, and $\varphi$ the differential map; then $\varphi$ is epic and indeed the sequence $0 \to \mathscr{F} \to \mathscr{G} \to 0$ is even exact, but there are global sections of $\mathscr{G}$ which are not the image of a global section of $\mathscr{F}$, e.g. $z \mapsto \frac{1}{z} \, \mathrm{d}z$.

Let $X$ be the punctured complex plane again, $\mathscr{F}$ the sheaf of meromorphic functions, $\mathscr{G}$ the sheaf of nowhere-zero meromorphic functions, and let $\varphi : \mathscr{F} \to \mathscr{G}$ be composition with $\exp : \mathbb{C} \to \mathbb{C}$; then $\varphi$ is epic but again fails to be surjective on (global) sections: after all, there is no holomorphic function $f : X \to \mathbb{C}$ such that $\exp f(z) = z$ for all non-zero $z$.

Question. Are there simpler examples which do not require much background knowledge beyond knowing the definition of sheaves and stalks?

答:
We seek two surjective local homeomorphisms
$\pi_{\mathcal E}\colon\mathcal E\rightarrow\mathcal X$ and
$\pi_{\mathcal H}\colon\mathcal H\rightarrow\mathcal X$ and a surjective
sheaf morphism $f\colon\mathcal E\rightarrow\mathcal H$ such that some
(continuous) section of $\pi_{\mathcal H}$ is not $f\circ s$ for any
(continuous) section $s$ of $\pi_{\mathcal E}$. If the viewpoint of a presheaf
is preferred (assignment of sets to open sets where restriction is respected)
then translate to the presheaves of continuous sections of $\mathcal E$
and $\mathcal H$. Those automatically satisfy the usual sheaf gluing
condition i.e. they are sheaves in the second sense.
$
\newcommand\bbR{{\mathbb R}}
\newcommand\clE{{\mathcal E}} \newcommand\clF{{\mathcal F}}
\newcommand\clH{{\mathcal H}} \newcommand\clX{{\mathcal X}}
\newcommand{\set}[2]{\{\mskip1mu #1\mid#2\mskip1mu\}}
\newcommand{\sset}[1]{\{\mskip1.25mu#1\mskip1.25mu\}}
\newcommand{\rto}[1]{\mathchoice{\bigl|_{#1}}{\vert_{#1}}{|_{#1}}{|_{#1}}}
$

A simple example is the presheaf of real valued continuously differentiable
functions $s(x)$ of a real variable $x$, such that $s(0)=s(1)=0$, this presheaf
being mapped by differentiation to the presheaf of continuous functions
of $x$. The constant section with value $1$ is not hit because if $s'=1$ then
$s(0)$ and $s(1)$ cannot both be zero, by the fundamental theorem of
calculus. But it would be better to find an example that does not use even
undergraduate calculus.

Let $\clX=\sset{A,B,C}$ with topology
$\sset{\emptyset,\sset{B},\sset{A,B},\sset{B,C},\clX}$, and let
$E=\sset{+1,-1}$ with the discrete topology, so that $\clE=\clX\times E$ is a
sheaf (see below). The cartesian products of the open sets of $\clX$ with
$\sset{-1}$ and $\sset{+1}$ is a basis for the $\clE$-topology by the
definition of the product topology. The equivalence relation that collapses
$(B,1)$ and $(B,-1)$ to $(B,0)$ provides the quotient
$f\colon\clE\rightarrow\clH$, where
$$
\clH=\sset{(A,+1),(A,-1),(B,0),(C,1),(C,-1)}.
$$
The images under $f$ of the $\clE$ basis are closed under intersection and they
form a basis of $\clH$ such that $f\colon\clE\rightarrow\clH$ is an open map,
so the topology defined by that basis, and the quotient topology, are the same,
$\clH$ is a sheaf, and $f$ is a sheaf epimorphism. There are two global
constant sections in $\clE$, but $\clH$ has four global sections: the sections
of $\clE$ composed with $f$, and also the two global sections
$$
A\mapsto(A,+1),\; B\mapsto(B,0),\; C\mapsto(C,-1),\quad\mbox{and}\quad
A\mapsto(A,-1),\; B\mapsto(B,0),\; C\mapsto(C,+1).
$$
In the graphic, the kinked green sections are $f\circ s$ for the two constant
sections $s$ of $\clE$, but the two slanted magenta sections of $\pi_\clH$
are not hit.

The example is easily understood: the sections of $\pi_\clE$ cannot go
vertically along the fiber at $B$ even though the singleton $\sset{B}$ is open
because sections are single valued maps. But when $(B,+1)$ and $(B,-1)$ are
collapsed then the germ at $B$ provides no control because it has no extent in
$\clX$, so the sections of $\pi_\clH$ can ``change tracks'' and two more of
them appear.

Below are two background items that place this example in a more general
setting.

Constant sheaves: Let $\clX$ and $E$ be topological spaces and
suppose $E$ has the discrete topology. Then $\clE=\clX\times E$ with
$\pi\colon\clE\rightarrow\clX$ the projection is a sheaf: $\pi$ is continuous
and surjective, and for any $(x,e)\in\clE$ and any open $U\ni x$,
$U'=U\times\sset{e}$ is open because it is the product of open sets, and
$\pi\rto{U'}\rightarrow U$ is continuous with continuous
inverse $x\mapsto(x,e)$. The continuous sections of $\clE$ are the
maps $x\mapsto (x,e(x))$ where $e\colon U\rightarrow E$ is continuous, so the
presheaf of sections of $\clE$ is isomorphic with the presheaf of continuous
functions on $\clX$ with values in $E$. Singletons of $E$ are open-closed so
they are pulled back by any continuous $e\colon U\rightarrow E$ to open-closed
subsets of $U$. Thus the sections of $\clE$ correspond to the functions on
$\clX$ which are constant on connected subsets of $\clX$.

A fiberwise quotient of a sheaf is a sheaf if the quotient map is open:
Given (possibly trivial) equivalence relations in each fiber of a sheaf $\clE$
(equivalently, a fiberwise equivalence relation on $\clE$), the quotient space
$\clH$ is a topological space, the (surjective) quotient map
$f\colon\clE\rightarrow\clH$ is continuous, and there is a unique continuous
surjective $\pi_\clH\colon\clH\rightarrow\clX$ such that
$\pi_\clH\,f=\pi_\clE$.  If $f$ is an open map then $\clH$ is a sheaf and
$f\colon\clE\rightarrow\clH$ is a sheaf epimorphism: Let $\delta\in\clH$ and
$\pi_\clH(\delta)=x$, choose $\gamma\in\clE$ such that $f(\gamma)=\delta$, and
choose open $U'\ni\gamma$, $U\ni x$, and a section $s\colon U\rightarrow\clF$,
such that $\pi_\clE\rto U'\colon U'\rightarrow U$ and $s\colon U\rightarrow U'$
are inverses. Then $V'\equiv f(U')$ is open and $t\equiv f\circ s\colon
U\rightarrow V'$ is surjective. If $y\in U$ then
$\pi_\clH\,t(y)=\pi_\clH(f(s(y)))=\pi_\clE\,s(y)=y$ so $\pi_\clH$ is a
local diffeomorphism.


问:
Why does the Hamiltonian vanish for $\int f(x,y)\sqrt{\dot{x}^2+\dot{y}^2} \text{ d}{t}$?
So I have this integral
$$
\int_{t_1}^{t_2} f(x,y)\sqrt{\dot{x}^2+\dot{y}^2} \text{ d}t,
$$

where $x,y$ are functions of $t$ and $f$ is a function of just $x,y$. Also $\dot{x}$ denotes the first
derivative of $x$ with respect to $t$. I am trying to study this type of integral and look for
its extremas, depending on what $f$ is. The first thing I am wondering though is what is the reason for
its vanishing Hamiltonian?

First we have the Lagrangian $L = f(x,y)\sqrt{\dot{x}^2+\dot{y}^2}$ which is
just the integrand. Then the generalized momenta we get from that are
$$
p_x = \frac{f(x,y)\dot{x}}{\sqrt{\dot{x}^2+\dot{y}^2}} \quad\text{and}\quad
p_y = \frac{f(x,y)\dot{y}}{\sqrt{\dot{x}^2+\dot{y}^2}}.
$$
And now we calculate the Hamiltonian to see that it is zero
$$
H = p_x\dot{x} + p_y\dot{y} - L = 0.
$$
I want to know what is the reason for this? I think it has something to do with the integrand's independence from $t$. But I'm not sure where to go from there.

答:
This is due to the reparametrization invariance of the Lagrangian.  Specifically, your Lagrangian has the property that if we rewrite it in terms of a new "time function" $\tau(t)$, then we have
\begin{align*}
\sqrt{ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dx}{dt}\right)^2 } \, dt &= \left[ \frac{d\tau}{dt} \sqrt{ \left(\frac{dx}{d\tau}\right)^2 + \left(\frac{dx}{d\tau}\right)^2 } \right] \left[ \frac{dt}{d\tau} \, d\tau \right] \\&= \sqrt{ \left(\frac{dx}{d\tau}\right)^2 + \left(\frac{dx}{d\tau}\right)^2 } \, d\tau,
\end{align*}
since $(dt/d\tau)(d\tau/dt) = 1$.  Thus, the Lagrangian does not change under this reparametrization  $t \to \tau$.

The Hamiltonian for any Lagrangian which is parametrization-invariant is identically equal to zero.  A nice proof of this can be found in this answer over on Physics.SE.


答:
$$\frac{2 e^{-\frac{1}{x^2}}}{x^3};\;x=\frac{1}{t}$$

we have
$$\underset{x\to 0}{\text{lim}}\frac{2 e^{-\frac{1}{x^2}}}{x^3}=\underset{t\to \infty }{\text{lim}}\frac{2 t^3}{e^{t^2}}=\underset{t\to \infty }{\text{lim}}\frac{6 t^2}{2 te^{t^2} }=\underset{t\to \infty }{\text{lim}}\frac{3t}{e^{t^2}}=\underset{t\to \infty }{\text{lim}}\frac{3}{2te^{t^2}}=0$$

Title: Calculate $\lim_{x \to 0} \frac {2e^{-1/x^2}}{x^3}$

Body:
Got this question and was wondering why the limit is $0$ ?

I saw a few people that mentioned that it can be written when $\frac {2e^{-1/x^2}}{x^3}$ and such limits is always $0$.

Is there any way to explain it or to show it specifically somehow ?

Thanks in advance.


问:
Why is $\mathbb{Z}[i \sqrt{2}]$ a Unique Factorization Domain
Why is $\mathbb{Z}[i \sqrt{2}]$ a Unique Factorization Domain?

We know that $\mathbb{Z}[i \sqrt{5}]$ is not a UFD as
$$(1 + i \sqrt{5})(1 - i \sqrt{5}) = 6$$
and $6$ is also equal to $2 \times 3$.

Now $\mathbb{Z}[i \sqrt{2}]$ is a UFD since $2$ is a Heegner number, however the simple factorization

$$(2 + i \sqrt{2})(2 - i \sqrt{2}) = 4 + 2 = 6 $$

which is also equal to $2 \times 3$, and therefore not unique.

What is wrong here?

答:
The fallacy is that none of the four numbers
\begin{align*}
2+i\sqrt2 &= i\sqrt2 (1-i\sqrt2) \\
2-i\sqrt2 &= -i\sqrt2 (1+i\sqrt2) \\
2 &= -i\sqrt 2 (i\sqrt 2) \\
3 &= (1+i\sqrt2)(1-i\sqrt2)
\end{align*}
is irreducible in $\Bbb Z[i\sqrt2]$. Once we observe this, the error is the same as if we asserted that $14\times15=210=10\times21$ shows that $\Bbb Z$ is not a unique factorization domain.


问:
Basic probability questions
How is P(A , B) different from P(A $\cap$ B)? I'm genuinely curious as to why one might prefer one over the other.

Also as far as some probability P (A | B , C) goes, what is the order of operations i.e., is it P (A | (B , C)) or P ((A | B) , C)? Or are they both similar?

答:
The expression $\Pr(A|B,C)$ is an abbreviation for $\Pr(A|(B\cap C))$. Since $B\cap C=C\cap B$, there is no precedence issue.

Sometimes $\Pr(A,B)$ is used as an abbreviation for $\Pr(A\cap B)$, but that is an uncommon practice. By way of contrast, one bumps often into the conditional probability version mentioned in the first paragraph.

Title: Basic probability questions


问:
showing covariant derivative of skew-symmetric rank $2$ tensor vanish

If $A_{ij}$ is a skew-symmetric tensor such that $A_{ij \ , \ k}=A_{ik
 \ , \ j}$ , prove that $A_{ij \ , \ k}=0$ .

We have $$A_{ij,k}=\frac{\partial A_{ij}}{\partial x^k}-\Gamma^l_{ik}A_{lj}-\Gamma^l_{jk}A_{il}=-\frac{\partial A_{ji}}{\partial x^k}+\Gamma^l_{ki}A_{jl}+\Gamma^l_{kj}A_{li}=-A_{ji,k}$$ which shows that $A_{ij,k}$ is skew-symmetric in $i,j$ . Again $$A_{ij,k}=A_{ik,j}=-A_{ki,j}=-A_{kj,i}$$ which shows $A_{ij,k}$ is skew-symmetric in $i,k$ . But how to use these to show $A_{ij,k}=0$ I don't understand . Any help is appreciated .

答:
The tensor $A_{ij,k}$ satisfies two identities, namely $A_{ij,k} = A_{ik,j}$ and $A_{ij,k} = -A_{ji,k}$. If we alternate between the two, we find that

$$A_{ij, k} = A_{ik,j} = -A_{ki,j} = -A_{kj,i} = A_{jk,i} = A_{ji,k} = -A_{ij,k}$$

so $A_{ij,k} = 0$.


问:
Differentiability and partial derivatives at the origin
I'm interested in checking if the following function
$$
f(x,y) = \begin{cases}
\dfrac{\sin(x+y)}{\sqrt{x^2+y^2}},& \text{if } (x,y) \neq (0,0),\\
0, &\text{if } (x,y) = (0,0).
\end{cases}
$$
is differentiable at the origin, and also I would like to compute its partial derivatives at the origin, if possible. For the differentiability, my claim is that $f$ is not differentiable at $(0,0)$ since it's not continuous at $(0,0)$. Indeed, if we compute the limit in polar coordinates, we obtain that
$$
\lim_{r \to 0} \dfrac{\sin(r(\cos(\theta)+ \sin(\theta)))}{r} = \begin{cases}
\cos(\theta) + \sin(\theta),& \text{if } \cos(\theta) + \sin(\theta) \neq 0,\\
0, &\text{otherwise}.
\end{cases}
$$
therefore the limit
$$
\lim_{(x,y) \to (0,0)} \dfrac{\sin(x+y)}{\sqrt{x^2 + y^2}}
$$
doesn't exist. For the partial derivatives at the origin, I obtain the following
$$
\dfrac{\partial f}{\partial x}(0,0) = \lim_{h \to 0} \dfrac{\sin(h)}{|h|h} = +\infty
$$
and the same for the partial derivative with respect to $y$ at $(0,0)$. However, Wolfram Alpha says that the partial derivatives at the origin are both equal to $0$.

QUESTION:
Is my reasoning right or have I done a mistake?

答: The limit doesn't exist. Hence there can be no derivative. For let $x=0.$ Then you have $$\frac{\sin y}{|y|},$$ with limit $\pm 1$ according as $y>0$ or $<0$ as $y\to 0.$


问:
Is every normal subgroup of $G$ of this form?

Let $G = H \times K$. If $N$ is a normal subgroup of $H$ and $L$ is a normal subgroup
of $K$, show that $N \times L$ is a normal subgroup of $G$. Is every normal subgroup of $G$ of
this form?

My attempt:

Let $G = H \times K$. If $N$ is a normal subgroup of $H$ and $L$ is a normal subgroup
of $K$, show that $N \times L$ is a normal subgroup of $G$

Since $e\in N$ and $e\in L$, then $(e,e)\in N\times L$. If $(n_1,l_1),(n_2,l_2) \in N\times L$, then $(n_1,l_1)(n_2,l_2)=(n_1n_2,l_1l_2)\in N\times L$. Also, if $n\in N$, then $n^{-1}\in N$ and if $l\in L$, then $l^{-1}\in L$, so $(n,l)(n^{-1},l^{-1})=(nn^{-1},ll^{-1})=(e,e)$. Thus $N\times L$ has inverses and therefore, $N\times L \leq G$.

For normality, if $(h,k)$ is an element of $G$ and $(n,l)\in N\times L$, then $(h^{-1},k^{-1})(n,l)(h,k)=(h^{-1}nh,k^{-1}lk)$. Since $N$ is normal in $H$ and $L$ is normal in $K$, we have $h^{-1}nh\in N$ and $k^{-1}lk\in L$, so $(h^{-1}nh,k^{-1}lk) \in N\times L $ and therefore $N\times L$ is a normal subgroup of $G$.

Is every normal subgroup of $G$ of this form?

In this part, my intuition says yes, but I don't know how to formalize this, could you help me to formalize the answer, either positive or negative?

答: No. Consider for example the group $G=\Bbb Z/2\Bbb Z\times \Bbb Z/2\Bbb Z$. It has the (normal) subgroup $\{(0,0),(1,1)\}$ which is not of the desired form. However under certain conditions it is true, for example when $H,K$ are finite of coprime order.


问:
Does a function exist that maps the natural numbers to the natural numbers excluding multiples of 2 and 3 and starting at an arbitrary point?
This question is a generalization of this other question. To recap, I was looking for a function that mapped the set of natural numbers to the set of natural numbers excluding multiples of 2 or 3 or both. So a function $f: x \mapsto y$ where $x = \{0, 1, 2, 3,...,m\}$ and $y=\{1,5,7,11,13,17,...,n\}$ for some arbitrary limit $m$. @A.J. gave a great solution

$y=\qquad 6\left \lceil \dfrac{x}{2} \right \rceil + (-1)^x$

(Erratum: the independent set $y$ should have been $y=\{5,7,11,13,...,n\}$ however since an answer was given for $y=\{1,5,...,n\}$ I didn't want to change the original and the difference is trivial.)

Now I'm wondering about a generalization that allows for the set $y$ to start at an arbitrary point (that is not a multiple of 2 nor 3). For example instead of $x=0$ mapping to $y=1$ it would instead map to $y=5$ and the formula would then be $y=6\left \lceil \dfrac{x+3}{2} \right \rceil + (-1)^{x+3}$ in other words translated 3 units. Or if we wanted $x=0$ to may to $y=41$ then it would be $y=6\left \lceil \dfrac{x+13}{2} \right \rceil + (-1)^{x+13}$. But I only know this because I precomputed the set starting at 1 and translated the number per the number of skipped members. If the number is bigger this would be non-trivial. For example, $101$ is not a multiple of 2 nor 3 so what function would start with $x=0$ mapping to $y=101$ and then continue by skipping multiples of 2 and 3 e.g. $103, 107, 109$.

Is there a generalization for a function that given the "starting point" $a$ that is not a multiple of 2 nor 3, and the parameter $x$, it maps $x=\{0,1,2,3,...,m\}$ to $y$ where each member of $y$ is $\ge$ $a$ and each subsequent term is the next non-multiple of 2 nor 3?

答:
For a map from the natural numbers, consider the function $$f(n)=3n+a-3+D$$ where

$D=1$ if $n$ is even and $a \equiv 1 \bmod 6 $

$D=-1$ if $n$ is even and $a \equiv 5 \bmod 6 $

$D=0$ otherwise.

If $n$ is odd then $f(n)\equiv a \bmod 6$. If $n$ is even then $f(n)\equiv a-3+D \bmod 6$. Thus no $f(n)$ is divisible by $2$ or $3$.

Also, for any positive integer $n$,  $f(n+1)-f(n)=3\pm 1$ and so the sequence $f(n)$ is monotonic increasing with differences of $2$ or $4$. Thus no non-multiple of $2$ and $3$ greater than $a=f(1)$ is omitted and $f$ is the required function.

(For a map from the non-negative integers replace $n$ by $n+1$ in the formulae.)

A continuous function of $n$

Although you are only interested in integer values of $n$ it is amusing to extend the function to all $n$ and also to give it as a definition without different cases.

$$f(n)=3n+a-3-\frac{1+\cos n\pi}{\sqrt3}\sin\frac{a+3}{3}\pi.$$


问:
We have 100 letters: 81-$M$, 10-$B$, 7-$S$ , 2-$H$. How many ways we can choose 4 letters so we can write : $SBBH$
I'm struggling with this question and here's what I did:

for example the $S$ letter, we need just one, so what I thought about is doing $100 \choose 7$$7 \choose 1$. (choosing the $7$ letters $S$ then choosing one from them). and it goes like that for the rest of the letters then I add the results up.
 I feel like I'm making mistakes and I'm not sure of my solution, would appreciate any feedback and tips.

答: You need an S $7 \choose 1$, two B $10 \choose 2$ and one H $2 \choose 1$.


问:
Solution to abstract linear equation system
I want to calculate all possible solution sets of this linear equation System $\alpha,\beta \in \mathbb R$:
\begin{align}
x+y+\alpha z&=\beta \\
x+\alpha y+z&=\beta \\
\alpha x+y+z&=\beta
\end{align}

答:
$$A=\left(
\begin{array}{ccc}
 1 & 1 & \alpha  \\
 1 & \alpha  & 1 \\
 \alpha  & 1 & 1 \\
\end{array}
\right)$$
$\det A=-\alpha ^3+3 \alpha -2=-(\alpha -1)^2 (\alpha +2)$

The system has one and only one solution if $\det A\ne 0 \to \alpha\ne 1;\;\alpha\ne -2;\;\forall\beta\in\mathbb{R}$
If $\alpha=1$ then $\text{rank}(A)=1$
$$A|B=\left(
\begin{array}{ccc|c}
 1 & 1 & 1 &\beta \\
 1 & 1  & 1&\beta \\
 1  & 1 & 1&\beta \\
\end{array}
\right)$$
In order to have $\text{rank}(A|B)=1$ must be $\beta=1$

If $\alpha=1$ and $\beta=1$ the system has infinite solution which depend on two parameters.

If $\alpha=-2$ then  $\text{rank}(A)=2$
$$A|B=\left(
\begin{array}{ccc|c}
 1 & 1 & -2 &\beta \\
 1 & -2  & 1&\beta \\
 -2  & 1 & 1&\beta \\
\end{array}
\right)$$
In this case we have $\text{rank}(A|B)=2$ for any $\beta$, thus the system has infinite solutions dependent from one parameter.

TL;DR

$\alpha\ne 1;\;\alpha\ne -2;\;\forall\beta\in\mathbb{R}$ one and only
one solution
$$x= \frac{\beta }{\alpha +2},y= \frac{\beta }{\alpha +2},z= \frac{\beta }{\alpha +2}$$

$\alpha=1$ and $\beta=1$ infinite$^2$ solutions
$$(h,k,1 - h - k);\;\forall h,k\in\mathbb{R}$$

$\alpha=-2$ infinite solutions for any $\beta$
$$(h,h-\beta ,h-\beta);\;\forall h\in\mathbb{R}$$


问:
Is the Pumping Lemma properly applied to prove $L = \{ 0^i1^{i+j}0^i \mid i, j > 0 \}$
The problem is to prove whether $L = \{0^i1^{i+j}0^i \mid i, j > 0\}$ is regular or not.

I've used the Pumping Lemma with a string $s = 0^p1^{2p}0^p$, however I do not know I can choose such string (yet the language description does not state that $i \neq j$ nor the opposite). With this string, we can say that $\vert s \vert \geq p$, so $s$ can be split in three pieces as $s = xyz$.

To satisfy the second ($\vert y \vert > 0$) and third ($\vert xy \vert \leq p$) condition of the lemma, $y$ must only contain $0$'s, and therefore: $x = \varepsilon$, $y = 0^p$ and $z = 1^{2p}0^p$.

Finally, if we pump $y$ and check whether it is still in $L$, we clearly check that $s \notin L$, for instance:
$$s = xyyz = 0^p0^p1^{2p}0^p = 0^{2p}1^{2p}0^p,\;s \notin L$$

To sum up, $L$ is not regular.

Back to the question, is this a proper way of using the Lemma?

答:
Your basic idea is fine, but some of the details are wrong.

Yes, you can use the string $s=0^p1^{2p}0^p$. You could also use the string $0^p1^{p+1}0^p$: $L$ can also be described as the set of all words $0^k1^\ell0^k$ such that $0<k<\ell$. And when you split $s$ as $s=xyz$ with $|xy|\le p$ and $|y|>0$, it is true that $xy$ must be contained in the first block of zeroes. However, you cannot conclude that $x=\epsilon$ and $y=0^p$. All that you can conclude is that there are non-negative integers $r$ and $s$ such that $x=0^r$, $y=0^s$, $r+s\le p$, and $s\ge 1$. Then $z=0^{p-r-s}1^{2p}0^p$, and

$$xy^kz=0^r0^{ks}0^{p-r-s}1^{2p}0^p=0^{p+(k-1)s}1^{2p}0^p$$

for all $k\ge 0$. This word is in $L$ if and only if the two blocks of zeroes are the same size, i.e, if and only if $p+(k-1)s=p$, which is the case precisely when $k=1$, so pumping the word up (by making $k>1$) or down (by making $k=0$) will give you a word not in $L$. In particular, you can — as you did — take $k=2$ to get

$$xy^2z=0^{p+s}1^{2p}0^p\notin L\,,$$

since $p+s\ne p$, and conclude that $L$ is not regular.


问:
Graphical explanation for the equation used to calculate points of a vector in a base in $R^3$
I am currently taking a course in linear algebra and I am having difficulty understanding the point and visual/graphical derivation of the following in my course book:

If $B=(u_1,u_2,u_3)$ is an orthogonal base in the room ($R^3$) and $v$ is a vector in the room. Then it follows that
$$v=\begin{pmatrix}\frac{u_1\cdot v}{|u_1|^2} \end{pmatrix}u_1 + \begin{pmatrix}\frac{u_2\cdot v}{|u_2|^2} \end{pmatrix}u_2 +\begin{pmatrix}\frac{u_3\cdot v}{|u_3|^2} \end{pmatrix}u_3$$

The reason for this equation is given by the book as follows:

Because $B$ is a base, we get $v=a_1u_1+a_2+u_2+a_3u_3$ for adequate choices of real values for $a_1,a_2,a_3$. If we multiply this equation with $u_1$ we get $v\cdot u_1=a_1u_1\cdot u_1+a_2u_2\cdot u_1+a_3u_3\cdot u_1=a_1|u_1|^2$

I am having a difficult time understanding here when it is multiplication and when it is the dot product. Here it looks as if it is the dot product since everything except $a_1|u_1|^2$ cancels out (My guess is that $a_1\cdot a_x, x \in \mathbb{R}$ is orthogonal and therefor results in 0). However I have recently learned that "regular" algebraic simplification applies rarely in linear algebra. Is what is happening here the same as doing the dot product the following way: $u_1\cdot (v_1+v_2)=v_1\cdot u_1+v_2 \cdot u_1$?

Then the book simplifies the last equation to be $a_1=\frac{v\cdot u_1}{|u_1|^2}$. I follow that simplification and the fact that we do the same for $a_2$ and $a_3$ and put them back in our equation to get the first equation $v=...$.

The questions I am trying to get across and get answered is both the question in bold above and also the question I got confused with in the first place:

What does this equation actually mean? Graphically speaking.

What is the dot product of two vectors divided by the length of one of them squared times the first vector, besides mathematical nonsense? Obviously it is not nonsense and that is what I am trying to see.

答:
Let's try to unpack this problem step by step.

Since $B$ is a basis we can write any vector $v \in \mathbb{R}^3$ uniquely as $a_1u_1 + a_2u_2 + a_3u_3$ where $v = a_1, a_2, a_3 \in \mathbb{R}$. In that equation the multiplication is just scalar multiplication since $a_i$ is a scalar and $u_i$ is a vector. Next we can dot product $v$ with $u_1$, $v \cdot u_i$. And we can substitute in $v$ to get:

$$ v \cdot u_1 = (a_1u_1 + a_2u_2 + a_3u_3) \cdot u_1$$

and simplify using the linearity of the dot product to get:
$$ = (a_1 u_1) \cdot u_1 + (a_2u_2) \cdot u_1 + (a_3 u_3) \cdot u_1 $$
$$ = a_1(u_1 \cdot u_1) + a_2 (u_2 \cdot u_1) + a_3(u_3 \cdot u_1) $$

Since the basis is orthogonal we know that $u_2 \cdot u_1 = 0$ and $u_3 \cdot u_1 = 0$ (since the dot product of orthogonal vectors is $0$). This allows us to further simplify to:

$$ v \cdot u_1 = a_1|u_1|^2 $$
$$a_1=\frac{v\cdot u_1}{|u_1|^2}$$

We could have done the same for $u_2$ and $u_3$ so we have similar equations for $a_2$ and $a_3$.

Graphically what we've done is project $v$ onto each of the basis vectors. The term $\frac{v\cdot u_1}{|u_1|^2}u_1$ is the projection of $v$ onto $u_1$.


问:
Is there a way to find the sum of an infinite series (not geometric)
I want to calculate

$$\sum_{k=0}^\infty\binom{k+3}k(0.2)^k$$

to get the exact value of it. I have excel and other tools to help me so it is fine if it is computationally expensive. Is there a clear and repeatable way to solve this infinite series? Thank you. This is my first post and be sure to give me some suggestions as well.

答:
It’s a general fact that

$$\sum_{k\ge 0}\binom{k+n}nx^k=\frac1{(1-x)^{n+1}}\;.$$

You can prove this by induction on $n$, starting with the geometric series

$$\frac1{1-x}=\sum_{k\ge 0}x^k$$

and differentiating repeatedly with respect to $x$. You want the case $n=3$:

$$\sum_{k\ge 0}\binom{k+3}kx^k=\sum_{k\ge 0}\binom{k+3}3x^k=\frac1{(1-x)^4}\,.$$

Now just substitute $x=0.2$.


问:
Check whether the following polynomial is irreducible over $\mathbf Q$
I was trying this problem from my Abstract Algebra book exercise that says:

Show that the polynomial $x^2+\frac 13x-\frac 25$ is irreducible in $\mathbf Q[x]$.

What I tried:  $x^2+\frac 13x-\frac 25 \equiv 15x^2+5x-6=f(x)$,say.

Now I compute ,$f(x+1)=15x^2+35x+14$. Now, $ 7 \mid 35, 7 \mid 14; 7\nmid 15,7^2 \nmid 14$. Hence using Eisenstein's criterion ,$f(x+1)$ is irreducible  over $\mathbf Q$. Hence $f(x)$ is irreducible over $\mathbf Q$.

Am I right?

I want to add another problem similar to the above :

Show that the polynomial $2x^3-x^2+4x-2$ is not irreducible in $\mathbf Z[x]$.

Here,$f(x+1)=2x^3+5x^2+8x+3$ where $f(x)=2x^3-x^2+4x-2$. Here ,I can not apply Eisenstein's criterion for $f(x+1)$. I can not also factorize $f(x)$.Then how Can I prove it?

Can someone point me in the right direction? Thanks in advance for your time.

答:
Your answer is:

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Post question's edit:

You want to prove that the polynomial is reducible in $\Bbb Z[x]$. By (the second) Gauss's lemma, if it is reducible in $\Bbb Q[x]$, then it also is in $\Bbb Z[x]$. Since its degree is $3$, it is reducible (over $\Bbb Q$) if, and only if, there is a rational root. Now note that in case Eistein's criterion doesn't work, you still won't know wether it is reducible or not, so that shouldn't be the way to go. You want to find a rational root. Try finding roots using the rational root theorem.

Something else you can try is to factor your polynomial. If you can't do this, you can always put the polynomial as input in a very well known on-line software.


问:
Ultralimits vs limits of subsequences
Let $(x_n)_{n \geq 1} \subset \mathbb{R}$ be a sequence of real numbers, and let $\omega \subset \mathcal{P}(\mathbb{N})$ be a non-principal ultrafilter. We say that $x$ is an ultralimit of $(x_n)_{n \geq 1}$ along $\omega$ if for all $\varepsilon > 0$ we have $\{ n \geq 1 : |x - x_n| < \varepsilon \} \in \omega$.

Suppose that there exists a set $\{ n_k : k \geq 1 \} \in \omega$ such that the subsequence $(x_{n_k})_{k \geq 1}$ has converges to $x$ (in the standard sense). Then it follows that $x$ is the ultralimit $(x_n)_{n \geq 1}$ along $\omega$, since $\omega$ is non-principal and so taking away finitely many elements from $(n_k)_{k \geq 1}$ we are still in $\omega$.

Conversely, if $x$ is the ultralimit of $(x_n)_{n \geq 1}$ along $\omega$, we can find a subsequence $(x_{n_k})_{k \geq 1}$ converging to $x$, for instance we can choose inductively $n_k > n_{k-1}$ such that $n_k \in \{ n \geq 1 : |x_n - x| < 1/k \}$. Although this subsequence is constructed using sets in $\omega$, it is not clear to me that the resulting set $\{ n_k : k \geq 1 \}$ should be in $\omega$.

So my question is: if $x$ is the ultralimit of $(x_n)_{n \geq 1}$ along $\omega$, can we always choose a subsequence $(x_{n_k})_{k \geq 1}$ converging to $x$ so that $\{ n_k : k \geq 1 \} \in \omega$?

答:
Consider the following counterexample: Let $(N_n)_n$ be a sequence in $\mathcal{P}(\mathbb{N})$ such that
$$ \mathbb{N} = N_1 \supseteq N_2 \supseteq N_3 \supseteq \dots $$
and for every $n \in \mathbb{N}$ we have
$$ | N_n \setminus N_{n+1} | = \aleph_0, \quad | N_n | = \aleph_0 \quad \text{ and } \quad \forall m \in \mathbb{N} \; \exists n \in \mathbb{N}: m \not \in N_n. $$
(Such a sequence can easily be obtained, for example by inductively throwing away half of the elements of $N_n$ starting from $N_1$ in a systematic way.)

Set $\Delta_n := N_n \setminus N_{n+1}$ for every $n \in \mathbb{N}$ and consider the family
$$ U := \{V \subseteq \mathbb{N}: \exists N \in \mathbb{N} \; \forall n \geq N: | \Delta_n \setminus V | < \aleph_0 \}. $$
Intuitively, $U$ contains all $V \subseteq \mathbb{N}$ such that for almost all $n \in \mathbb{N}$ the set $V$ contains almost all points in $\Delta_n$. It is clear that $U$ has the finite intersection property since the intersection of sets $V_1, \dots, V_m$, which for almost all $n \in \mathbb{N}$ contain almost all $k \in \Delta_n$, still satisfies this property. Thus, by a general theorem there is an ultrafilter $\mathcal{U} \subseteq \mathcal{P}(\mathbb{N})$ extending $U$.

Now, define the sequence $(x_n)_n \subseteq \mathbb{R}$ such that
$$ x_n = \frac{1}{k} \quad \text{if } n \in \Delta_k. $$
First of all it is to be shown that $0$ is an ultralimit of $(x_n)_n$ with respect to $\mathcal{U}$. For that let $\epsilon > 0$ and choose $n \in \mathbb{N}$ with $\frac{1}{n} < \epsilon$. Then
$$  \{n \in\mathbb{N}: |x_n| = x_n \leq \epsilon\} \supseteq
\left\{n \in \mathbb{N}: x_n \leq \frac{1}{n}\right\} = N_n \in \mathcal{U}, $$
where $N_n \in \mathcal{U}$ holds since for all $m \geq n$ the set $N_n$ contains every element from $\Delta_m = N_m \setminus N_{m+1}$ as $N_n \supseteq N_m$.

We show that for every subsequence $(x_{n_k})_k$ converging to $0$ in the standard sense we have
$$ S := \{n_k: k \in \mathbb{N}\} \not \in \mathcal{U}. $$
In fact for every $n \in \mathbb{N}$ there are only finitely many $k \in \mathbb{N}$ such that
$$ \frac{1}{n+1} < x_{n_k} \leq \frac{1}{n} \quad \Leftrightarrow \quad n_k \in \Delta_n. $$
Hence for $S^C := \mathbb{N} \setminus S$ we find that for all $n \in \mathbb{N}$ the set $\Delta_n \setminus S^C = \Delta_n \cap S$ is finite, thus $S^C \in U \subseteq \mathcal{U}$. It follows that $S \not \in \mathcal{U}$.

Title: Ultralimits vs limits of subsequences


问:
Asymptotics of the Lattice Green Function
Let
$$
\mu_k=\sum_{i=1}^d (2\sin (k_i/2))^2, \quad k\in \mathbb{R}^d
$$
Then it's not hard to show that if $d>2$, then
$$
\int_{(-\pi,\pi]^d} \frac{e^{ikx}}{\mu_k} dk =\frac{1}{|x|^{d-2}}C_d(x), \quad x\in \mathbb{Z}^d
$$
where $C_d(x)\to\text{const}$ as $x\to \infty$, or formally,
$$
C_d(x)\to \int \frac{e^{ip_1}}{p^2} dp,  \quad |x|\to\infty
$$
It then seems reasonable to say that $C_d(x) = \text{const} +O(|x|^{-\epsilon})$ for some $\epsilon >0$. However, how would one go about to prove this?

答:
I think I found a nice solution using the local central limit theorem. Let me provide a sketch of the proof, since it's a lot to write down.

Theorem. Let $d>2$. Then the lattice Green function $C(x)$ satisfies
$$
C(x) \equiv\int_{(-\pi,\pi]^d} \frac{dk}{(2\pi)^d} \frac{e^{ikx}}{\mu_k} =\frac{c_d}{|x|^{d-2}}+O\left( \frac{1}{|x|^d} \right)
$$
where $c_d$ is a constant such that $c_d/|x|^{d-2}$ is the Fourier transform (as tempered distribution) of $1/k^2$, i.e., the continuum Green function $G(x)$, formally written as,
$$
G(x) \equiv\int\frac{dk}{(2\pi)^d}\frac{e^{ikx}}{k^2} =\frac{c_d}{|x|^{d-2}}
$$

To prove this, we will need the local central limit theorem, i.e.,

Lemma (LCLT). Consider a simple random walk on $\mathbb{Z}^d$ with transition matrix $p$ and let $p_n(x) =p^n(0,x)$ be the probability of starting at $0$ and ending at $x$ after $n$ steps. Define $\bar{p}_0(x)=\delta(x)$ and
$$
\bar{p}_n(x) = 2 \left( \frac{d}{2\pi n} \right)^{d/2} e^{-dx^2/2n}
$$
Then the error $E_n(x) = p_n(x) -\bar{p}_n(x)$ satisfies
$$
E_n(x)=\frac{1}{|x|^{2m}} O\left( \frac{1}{n^{(d+2-2m)/2}}\right), \quad m\in \mathbb{N}
$$

Also notice that the lattice Green function is equal to
$$
C(x) =\sum_{n=0}^\infty p_n(x)
$$
which you can show using the fact that the characteristic function $\varphi_n(k)$ of $p_n(x)$ is equal to $=\varphi(x)^n$ where $\varphi(x)$ is the characteristic function of $p(x)$.  From my answer here, you see that the continuum lattice Green function can be written as
$$
G(x)= \frac{1}{(4\pi)^{d/2}}\int_\epsilon^\infty t^{-d/2} e^{-x^2/4t} dt
$$
Notice the similarity between $G(x)$ and $\sum \bar{p}_n(x)$. Using the lemma, we can do some approximations that will ultimately lead to an error $O(|x|^{-d})$. I might try to fill in the details later on.


问:
Continuum mechanics- Deformation of circle
Diameter of a circle is a.b before deformation.

What will be new form of circle under deformation:

$$x_1 = 2 X_1+{1\over 2}X_2$$

$$x_2 = 2 X_1-{1\over 2}X_2$$

I tried to substitute $$ X_1^2+X_2^2=({a.b\over2})^2$$ using $$x_1+x_2=4X_1$$ and $$x_1-x_2=X_2$$
I ended up with $$17x_1^2+17x_2^2-30x_1x_2=4a^2b^2$$
and it doesn't feel right, can't tell whats the new form after deformation by looking that. Am i on the right track? Searched for a similar example on books and on the internet but couldn't find any.

答:
Hint :

Write
$$17x_1^2+17x_2^2-30x_1x_2=4a^2b^2$$ as

$$ \left(\dfrac{x_1-x_0}{h_1}\right)^2+\left(\dfrac{x_2-y_0}{h_2}\right)^2=1 $$

which is possible because $x_1$ and $x_2$ have the same coefficient, and
which is an ellipse equation.


问:
geometry question-semicircle inscribed in quadrilateral.
there is a line AD whose midpoint is O (AO=OD). a semicircle is drawn with centre O and any radius < AO with its straight edge being part of line AD. lines AB and CD are drawn tangent to the semicircle's arc such that BC is also a tangent to the arc. the question is to prove that $AO^2=AB·CD$

答:
Triangles with the same color are congruent, and at the point $O$ we have:

$2α+2ß+2\gamma=180°$

$\alpha+\beta+\gamma=90°$

In the right-Angled triangle $OMB$, we have:

$\angle OBM=\alpha+\gamma$

Therefore the two triangles $CDO$ and $OAB$ are similar:

$\dfrac{CD}{OD}=\dfrac{OA}{AB}$

$OA\times OD=AB\times CD$

Since $OA=OD$

Thus: $OA^2=AB \times CD$


问:
Integration by partial fractions; how and why does it work?
Could someone take me through the steps of decomposing
$$\frac{2x^2+11x}{x^2+11x+30}$$
into partial fractions?

More generally, how does one use partial fractions to compute integrals
$$\int\frac{P(x)}{Q(x)}\,dx$$
of rational functions ($P(x)$ and $Q(x)$ are polynomials) ?

This question is asked in an effort to cut down on duplicates. See Coping with *abstract* duplicate questions.

Also see List of Generalizations of Common Questions.

答:
Here are some formal aspects of the delightful answer ( by Arturo Magidin ) above.

Let $ f(x), g(x) $ be polynomials ( $ g \neq 0 $ ). Our goal now is to integrate $ \frac{f}{g} $.

Dividing $ f $ by $ g $ gives that $ f = qg + r $ with $ \deg(r) < \deg(g) $, so $ \frac{f}{g} $ is a polynomial $ q $ plus a proper fraction $ \frac{r}{g} $ [ We'll call a  fraction $ \frac{p}{q} $ of polynomials  "proper" if $ \deg{p} < \deg{q} $ ]. Since the $ q, r $ here are unique, we'll call $ q $ the "polynomial part" and $ \frac{r}{g} $ the "proper fraction part" of $ \frac{f}{g} $.

Now we need only figure out how to integrate proper fractions, so let's take $ \deg(f) < \deg(g) $ to begin with.

$ g $ can be broken into a product consisting of linear factors, and quadratic factors with negative discriminant [ Proved here, for instance ]. Also, any proper fraction $ \displaystyle \frac{f}{g_1 \ldots g_k} $ where $ g_i $s are mutually coprime can be written as a sum of proper fractions with denominators $ g_1, \ldots, g_k $ [ Let's show it for $ k = 2 $, the general case follows by induction. Since $ g_1, g_2 $ are coprime, by Euclidean algorithm $ g_1 u_1 + g_2 u_2 = 1 $ for some polynomials $ u_1, u_2 $. Now $ \displaystyle \frac{f u_1}{g_2} + \frac{f u_2}{g_1} = \frac{f}{g_1 g_2} $, and since right hand side is proper the polynomial parts of $ \displaystyle \frac{f u_1}{g_2}, \frac{f u_2}{g_1} $ sum up to $ 0 $. So ( proper fraction part of $ f u_1 / g_2 $ ) + ( proper fraction part of $ f u_2 / g_1 $ ) = $ f / g_1 g_2 $ , done. ]

So $ \displaystyle \frac{f}{g} $ breaks into a sum of proper fractions of the form $ \displaystyle \frac{s(x)}{t(x) ^l} $ and $ \displaystyle \frac{p(x)}{h(x) ^k} $, where the $ t(x) $s are linear polynomials and $ h(x) $s are negative discriminant quadratics .

The terms here can be broken down further. Any proper fraction of the form $ \displaystyle \frac{f}{g^k} $ is a sum of proper fractions, with denominators $ g, \ldots, g^k $ and numerators having lower degree than $ g $ ( The $ f, g $ here are separate from the original problem. Notation repeated to try keep things more readable ). This is because : Division gives $ f = qg + r $ with $ \deg(r) < \deg(g) $, so $ \displaystyle \frac{f}{g^k} = \frac{q}{g^{k-1}} + \frac{r}{g^k} $. All the terms here are proper and $ \deg(r) < \deg(g) $. Now continuing onto $ \displaystyle \frac{q}{g^{k-1}} $, we're done.

So our original proper fraction $ \displaystyle \frac{f}{g} $ is a sum of proper fractions, of the form $ \displaystyle \frac{m}{(x+p)^l} $, and of the form $ \displaystyle \frac{ax+b}{(x^2 + cx + d)^k} $ where $ x^2 + cx + d $ has negative discriminant.

Proper fractions of the form $ \displaystyle \frac{m}{(x+p)^l} $ are easily integrated, for instance by substituting $ u = x + p $.

So finally, the problem is reduced to integrating fractions of the form $ \displaystyle \frac{ax+b}{(x^2 + cx + d)^k} $ where $ x^2 + cx + d $ has negative discriminant. Writing $ x^2 + cx + d $ as $ \displaystyle (x + \frac{c}{2})^2 + (d - \frac{c^2}{4}) $ and substituting $ u = x + \frac{c}{2} $ makes the integral take the form $ \displaystyle \int \frac{Au + B}{(u^2 + C^2)^k} du $. We can integrate the $ \displaystyle \frac{Au}{(u^2 + C^2)^k} $ part by substituting $ v = u^2 + C^2 $. And the $ \displaystyle \frac{B}{(u^2 + C^2)^k} $ part, as mentioned in the above answer, can be handled using integration by parts.


问:
Linear system $Ax=b$ is inconsistent implies $\det(A)=0$
Consider a matrix equation $Ax=b$.

If $\det(A)=0$, it follows that the equation has a non-unique solution or it does not have solutions at all.

I have proven that if the rows/columns of $A$ are linearly dependent, then $\det(A)=0$ and $Ax=0$ does not have a unique solution. Therefore the original equation $Ax=b$ does not have a unique solution.

However what I haven't been able to prove is that if the system of linear equations is inconsistent, then the determinant is zero. If the system is inconsistent, does it always imply that row operations on the matrix A lead to a zero row?

答: Let's prove the contrapositive. If $\det A\ne0$, $A^{-1}$ exists, so $Ax=b$ has exactly one solution, $x=A^{-1}b$, i.e. is not inconsistent.


问:
Calculate heat flow through the sphere
$T(x,y,z)=x$, $S=\{(x,y,z): x^2+y^2+z^2=1\}$, I also got a formula for heat flow $\int \int_S (-k \nabla T)\circ dS$ for some parameter $k$.

My attempt:
$x=sin\phi cos\theta$, $y=sin\phi sin\theta$, $z=cos\phi$ for $0\leq \theta \leq 2\pi$ and $0\leq \phi \leq \pi$,
$\nabla T=\hat{i}$, $S(\phi, \theta)=\hat{i}sin\phi cos\theta +\hat{j}sin\phi sin\theta+\hat{k}cos\theta$,
$dS=\frac{\partial S}{\partial \phi}d\phi \times \frac{\partial S}{\partial \theta}d\theta=(\hat{i}cos\phi cos\theta+\hat{j} cos\phi sin\theta - \hat{k}sin\phi)d\phi \times (-\hat{i}sin\phi sin\theta+\hat{j}sin\phi cos\theta)d\theta=$
$=(\hat{i}sin^2\phi cos\theta + \hat{j} sin^2 \phi sin\theta+\hat{k}sin\phi cos\phi)d\phi d\theta$.

$\nabla T \circ dS=sin^2\phi cos\theta d\phi d\theta$, $\int \int_S (-k \nabla T)\circ dS=k\int_0^{2\pi}\int_0^{\pi}sin^2\phi cos\theta d\phi d\theta=k\int_0^{2\pi}\frac{\pi}{2} cos\theta d\theta=0$
Is this correct? Thanks in advance.

答:
Yes there are two ways to look at it without direct computation

$\textbf{Option 1}$: Surface integral

Using $n = (x,y,z)$ for the unit sphere we have that

$$\iint -k\nabla T \cdot n dS = -k \iint x dS$$

$x$ is an odd function and the sphere has $x$ symmetry i.e. $(-x)^2 + y^2 +z^2 = 1 \iff x^2+y^2+z^2=1$ so the integral is $0$ automatically.

$\textbf{Option 2}$: Divergence theorem

By Divergence theorem we have that

$$-k\iint \nabla T \cdot n dS = -k \iiint \Delta T dV = -k \iiint 0 \:dV = 0$$

Title: Calculate heat flow through the sphere


问:
Upper bound for chromatic number ($\chi(G) \leq \Delta(G)+1$)
i am new in graph theory. I saw in some text book this "obvious" upper bound $$\chi(G) \leq \Delta(G)+1$$ i can't see how to demonstrate it

答: Given $k$ colors where $k$  is one larger than the maximal degree, Order the vertices arbitrarily, and color them one by one in this order; for each vertex use a color that has not been used for any of its previously colored neighbors.


问:
Representation of elements of fundamental group by shortest loops
I have some questions from the following text:

$\bullet$ It is clear to me that $f_\alpha$ is a loop and represents a generator of $\pi_1(M_\alpha)$ as covering induces injective map on fundamental groups, but why $f_\alpha$ is shortest?

Recall that  a loop on a surface equipped with a Riemannian metric is
called shortest if it has the shortest length in its free homotopy
class.

$\bullet$ I think here $l:\Bbb R\to \widetilde M$ is obtained as the lifting of the following composition map $\Bbb R\xrightarrow{\exp}\Bbb S^1\xrightarrow{f_\alpha} M$ w.r.t. the universal covering $p:\widetilde M\to M$ using the fact $\Bbb R$ is simply-connected. Now, $l$ is simple as $f_\alpha$ is simple. Am I right?

$\bullet$ Why the map $l:\Bbb R\to \widetilde M$ is properly embedded in $\widetilde M$? I don't know the meaning of "proper" here. I think the term embedding comes to form the fact that $l$ is simple.

$\bullet$ What will be the full pre-image of $f_\alpha(\Bbb S^1)$ in universal cover i.e. what is $p^{-1}\big(f_\alpha(\Bbb S^1)\big)$? I guess  it will be $\pi_1(M)$-tanslates of $l$, but have no proof.

Any help will be appreciated. It will be better if someone tells me some text containing all these basic facts.

答:
In the first bullet, suppose that $M_\alpha$ contained a loop $f'_\alpha : \mathcal S^1 \to M_\alpha$ in the same free homotopy class as the lift $f_\alpha$ but with length shorter than $f_\alpha$. Then the loop $f' : \mathcal S^1 \to M$ obtained by composing $f'_\alpha$ with the covering map $M_\alpha \mapsto M$ would be a shorter loop in the same homotopy class as $f$, which is a contradiction (the homotopy from $f_\alpha$ to $f'_\alpha$, when composed with the covering map $M_\alpha \mapsto M$, would be a homotopy from $f$ to $f'$).

In the second bullet, some care is needed. First one needs to verify that $\alpha$ has infinite order, but this is true in the fundamental group of any oriented surface: every nontrivial element has infintie order. Also, however, your interpretation does not quite make sense because the statement says that $l$ is the preimage of $f_\alpha(S^1)$, so $l$ itself is a set, not a function. However, if instead you state your interpretation to say that $l$ is the image in $\widetilde M$ of the map $\mathbb R \mapsto \widetilde M$ obtained by lifting the map $\mathbb R \mapsto \mathbb S^1 \mapsto M$, then your interpretation does make sense and is correct.

In your third bullet, the meaning of "properly embedded" is that the parameterization $\mathbb R \mapsto \widetilde M$ of $l$ is a proper function, meaning one for which the inverse image of every compact subset of $\widetilde M$ is a compact subset of $\mathbb R$. Intuitively this means that as points in $\mathbb R$ go to $-\infty$ or to $+\infty$, their images in $\widetilde M$ also "go to infinity".

In your fourth bullet, your guess is correct, if you restate it slightly: the full preimage will be the union of the $\pi_1(M)$ translates of $l$.


答: If they wouldn't, then Elementary Row Operations would bring it to its Reduced Row Echelon Form which would have no zero rows. This can only happen if the RRE form is in fact $I_n$. This implies that $A$ is invertible so $\mathrm{det}(A) \neq 0$.

Title: Linear system $Ax=b$ is inconsistent implies $\det(A)=0$

Body:
Consider a matrix equation $Ax=b$.

If $\det(A)=0$, it follows that the equation has a non-unique solution or it does not have solutions at all.

I have proven that if the rows/columns of $A$ are linearly dependent, then $\det(A)=0$ and $Ax=0$ does not have a unique solution. Therefore the original equation $Ax=b$ does not have a unique solution.

However what I haven't been able to prove is that if the system of linear equations is inconsistent, then the determinant is zero. If the system is inconsistent, does it always imply that row operations on the matrix A lead to a zero row?


问:
Calculate $\lim_{x \to 0} \frac {2e^{-1/x^2}}{x^3}$
Got this question and was wondering why the limit is $0$ ?

I saw a few people that mentioned that it can be written when $\frac {2e^{-1/x^2}}{x^3}$ and such limits is always $0$.

Is there any way to explain it or to show it specifically somehow ?

Thanks in advance.

答: You have that $$\frac{2}{x^3e^{1/x^2}}=\frac{2}{x^3\left(1+\frac{1}{x^2}+\frac12\frac{1}{x^4}+\cdots\right)}=\frac{2}{x^3+x+\frac12 \frac 1x+\cdots},$$ so that as $x\to 0,$ we see that the term in $1/x$ and the higher terms go to $\infty,$ so that the quotient goes to $0.$


问:
Representation Theory. Why does $a^{r}b^{s}a^{t}b^{u} = a^{i}b^{j}$?

I want to know that why we have $a^{r}b^{s}a^{t}b^{u} = a^{i}b^{j}$. Please let me know.

答:
Suppose that $x,y\in D_{2n}$. Then for some $r,t \in \{0,1,...,n-1\}$ and for some $s,u \in \{0,1\}$, we know that $x=a^rb^s$ and $y=a^tb^u$. Recall that $D_{2n}$ is a group, and is thus closed under multiplication. Thus, $xy\in D_{2n}$ so that for some $i \in \{0,1,...,n-1\}$ and for some $j \in \{0,1\}$, we have $xy = a^ib^j$. Hence, as desired:
$$
a^rb^sa^tb^u=xy=a^ib^j
$$</p>


问:
Using Cauchy product for an integral
To evaluate
$$
\int_0^1 e^x \ln(x+1)dx
$$
I was thinking about using the cauchy product of the taylor series of $e^x$ and $\ln(x+1)$. We know that
$$
e^x = \sum_{n=0}^\infty \frac{x^n}{n!}
$$
and
$$
\ln(x+1) = \sum_{m=0}^\infty \frac{(-1)^{m+1}x^m}{m}
$$
So using:
$$
\left(\sum_{n=0}^\infty a_n x^n\right) \cdot \left(\sum_{m=0}^\infty b_m x^m \right) = \sum_{n=0}^\infty \sum_{m=0}^n (a_m b_{n-m}) x^n
$$
We have:
$$
\int_0^1 e^x \ln(x+1)dx = \int_0^1  \sum_{n=0}^\infty \sum_{m=0}^n \left(\frac{(-1)^{1 + n - m}}{m\cdot m!} b_{n-m}\right) x^n dx
$$
I switch the integration and summation bounds but I do not know how to justify this:
$$
\int_0^1 e^x \ln(x+1)dx =  \sum_{n=0}^\infty \sum_{m=0}^n \left(\frac{(-1)^{1 + n - m}}{(n-m)\cdot m!}\right) \int_0^1 x^n dx
$$
Obtaining:
$$
\int_0^1 e^x \ln(x+1)dx =  \sum_{n=0}^\infty \sum_{m=0}^n \left(\frac{(-1)^{1 + n - m}}{(n-m)(n+1)\cdot m!} \right)
$$
I tried to plug this into desmos but it does not work.

答:
Your integral is

$$ \int_0^1 e^x \ln(x+1) dx=e\ln(2)-\int_0^1 \dfrac{e^x}{x+1}dx=e\ln(2)-\int_1^2e^{-1}\dfrac{e^u}{u}du$$

Which is  $$ e\ln(2)-e^1 (Ei(1)-Ei(2))$$

Where $Ei$ is the exponential integral :

https://mathworld.wolfram.com/En-Function.html

Title: Using Cauchy product for an integral


问:
Is every normal subgroup of $G$ of this form?

Let $G = H \times K$. If $N$ is a normal subgroup of $H$ and $L$ is a normal subgroup
of $K$, show that $N \times L$ is a normal subgroup of $G$. Is every normal subgroup of $G$ of
this form?

My attempt:

Let $G = H \times K$. If $N$ is a normal subgroup of $H$ and $L$ is a normal subgroup
of $K$, show that $N \times L$ is a normal subgroup of $G$

Since $e\in N$ and $e\in L$, then $(e,e)\in N\times L$. If $(n_1,l_1),(n_2,l_2) \in N\times L$, then $(n_1,l_1)(n_2,l_2)=(n_1n_2,l_1l_2)\in N\times L$. Also, if $n\in N$, then $n^{-1}\in N$ and if $l\in L$, then $l^{-1}\in L$, so $(n,l)(n^{-1},l^{-1})=(nn^{-1},ll^{-1})=(e,e)$. Thus $N\times L$ has inverses and therefore, $N\times L \leq G$.

For normality, if $(h,k)$ is an element of $G$ and $(n,l)\in N\times L$, then $(h^{-1},k^{-1})(n,l)(h,k)=(h^{-1}nh,k^{-1}lk)$. Since $N$ is normal in $H$ and $L$ is normal in $K$, we have $h^{-1}nh\in N$ and $k^{-1}lk\in L$, so $(h^{-1}nh,k^{-1}lk) \in N\times L $ and therefore $N\times L$ is a normal subgroup of $G$.

Is every normal subgroup of $G$ of this form?

In this part, my intuition says yes, but I don't know how to formalize this, could you help me to formalize the answer, either positive or negative?

答: No.  For any abelian group $G$, you can consider the diagonal subgroup of $G\times G$, given by $\Delta=\{(g,g)|g\in G\}$.  $\Delta\not\cong H\times L$ for any nontrivial $H,L\lt G$.


答:
Using the negative binomial coefficient $\binom{-4}{k}$, we have
$$
\begin{align}
\sum_{k=0}^\infty\binom{k+3}{k}(0.2)^k
&=\sum_{k=0}^\infty\binom{-4}k(-1)^k(0.2)^k\\
&=(1-0.2)^{-4}\\[6pt]
&=\frac{625}{256}
\end{align}
$$

Title: Is there a way to find the sum of an infinite series (not geometric)

Body:
I want to calculate

$$\sum_{k=0}^\infty\binom{k+3}k(0.2)^k$$

to get the exact value of it. I have excel and other tools to help me so it is fine if it is computationally expensive. Is there a clear and repeatable way to solve this infinite series? Thank you. This is my first post and be sure to give me some suggestions as well.


问:
A probability question
Suppose $X_1, X_2, ...,$ are IID random variables with $P(X_n=1)=p$ and $P(X_n=2)=1-p$. Let $S_n=\sum_{i=1}^n X_i$.

I was wondering how to find $P(S_n \neq z, \forall n \in \mathbb{N})$ for some particular natural number $z$?

How about the case when $P(X_n=1) = p_1, P(X_n=2)=p_2, P(X_n=3)=1-p_1-p_2$?

Does this problem belong to some kind of problems of random process?

Thanks!

答:
Here's another approach, which may be more sophisticated than the asker wanted, but is a bit slicker.

Let $T = \inf\{n : S_n \ge z\}$.  Note $S_T$ is either $z$ or $z+1$, and if $S_T = z+1$ then $z$ is never reached.  So we are seeking $P(S_T = z+1)$.  Call this value $q_z$ for short.

$T$ is a stopping time, so we'll use optional stopping on an exponential martingale.  We seek $\theta$ such that $Y_n := \theta^{S_n}$ is a martingale; this will happen iff $E[\theta^{X_n}] = p \theta + (1-p) \theta^2 = 1$.  The solutions of this equation are $\theta = 1$ (not useful) and $\theta = -1/(1-p)$.  One easily sees that ${Y_{n \wedge T}}$ is bounded, so by the optional stopping theorem we have
$$ 1 = E[Y_0] = E[Y_T] = (1-q_z) \theta^z + q_z\theta^{z+1}.$$
Taking $\theta = -1/(1-p)$ and solving for $q_z$ gives the result.

Incidentally, I "borrowed" a variant of this problem to use on an exam.

Title: A probability question


答: It looks fine to me. You might also consider applying the rational root theorem to show that the polynomial has no roots in $\mathbb{Q}$, and therefore it is irreducible since it has degree less than three.</p>

Title: Check whether the following polynomial is irreducible over $\mathbf Q$

Body:
I was trying this problem from my Abstract Algebra book exercise that says:

Show that the polynomial $x^2+\frac 13x-\frac 25$ is irreducible in $\mathbf Q[x]$.

What I tried:  $x^2+\frac 13x-\frac 25 \equiv 15x^2+5x-6=f(x)$,say.

Now I compute ,$f(x+1)=15x^2+35x+14$. Now, $ 7 \mid 35, 7 \mid 14; 7\nmid 15,7^2 \nmid 14$. Hence using Eisenstein's criterion ,$f(x+1)$ is irreducible  over $\mathbf Q$. Hence $f(x)$ is irreducible over $\mathbf Q$.

Am I right?

I want to add another problem similar to the above :

Show that the polynomial $2x^3-x^2+4x-2$ is not irreducible in $\mathbf Z[x]$.

Here,$f(x+1)=2x^3+5x^2+8x+3$ where $f(x)=2x^3-x^2+4x-2$. Here ,I can not apply Eisenstein's criterion for $f(x+1)$. I can not also factorize $f(x)$.Then how Can I prove it?

Can someone point me in the right direction? Thanks in advance for your time.


问:
Proving a partial differential equation of a U(x1,x2) function depends on x1-x2
this is my first time doing differential equations instead of just learning about it and i have no idea where to start with this one.

with this equation :

$$
\frac{\partial U}{\partial x1} + \frac{\partial U}{\partial x2} = 0
$$

I need to prove that U is only dependent on x1-x2

I have these following conditions:

-x1 and x2 are Cartesian coordinates.

-The following variables exist.

$$
\epsilon = x1 + x2
$$

$$
\eta = x1 - x2
$$

both of them are independent and correspond to oblique, 45-degree coordinates on the (x1,x2) plane.

答: $f(t)=U(x_1+t,x_2+t)$ has $f'(t)=0$ so $f$ is a constant. But if $z_1-z_2=x_1-x_2$ then there is a $t$ such that $(z_1,z_2)=(x_1+t,x_2+t)$.


问:
Lower Sum Inequality
For every partiton P of [a,b], and functions f and g bounded on [a,b],
why does the sum of lower sums of f and g is less than or equal to the lower sum of f+g?

i.e), L(P, f) + L(P, g) <= L(P, f+g)

答:
For any interval $[a,b]$, if we let $m_1=\mathrm{inf}_{[a,b]}f$, $m_2=\mathrm{inf}_{[a,b]}g$ then $\forall x \in [a,b]$:
$$(f+g)(x)=f(x)+g(x) \geq m_1 + m_2. $$

Hence, since the infimum is at least as big as any lower bound, $\mathrm{inf}_{[a,b]}(f+g) \geq m_1 + m_2$.
Summing these up over all the subintervals of our partition gives the required result.

Title: Lower Sum Inequality


问:
Upper bound of non-negative concave function on interval

Let $f(x)$ be a non-negative and upper convex (concave) function
defined on the interval $[a,b]$. Suppose
$f\left(\frac{a+b}{2}\right)\leq2$. Show that $f(x)\leq4$ for all
$x\in[a,b]$.

How can I show the statement above? Note that we do not assume $f$ being continuous or differentiable.

I can't come up with any trick rather than either to use a non-negativity like
$$f(x)\leq f(x) + f\left(\frac{a+b}{2}\right)\leq f(x) + 2f\left(\frac{a+b}{2}\right)\leq f(x) + 4$$
or to use a subadditivity along the lines of
$$f\left(x+\frac{a+b}{2}\right)\leq f(x) + f\left(\frac{a+b}{2}\right)\leq f(x) + 2$$
which doesn't lead to the desired upper bound. A complete solution or at least a hint will be appreciated.

答:
In the case $\frac{a+b}{2} \le x \le b$ we can use the concavity condition for $a < \frac{a+b}{2} < x$, which gives
$$
 2 \ge f(\frac{a+b}{2}) \ge \frac{x-(a+b)/2}{x-a}f(a) + \frac{(a+b)/2-a}{x-a} f(x) \, .
$$
Now use that $f(a) \ge 0$ and $x-a \le b-a$.

Graphically: Let $l$ be the line joining $(a, f(a))$ and $(\frac{a+b}{2}, f(\frac{a+b}{2}))$. Then $f(x) \le l(x) \le 4$ for $\frac{a+b}{2} \le x \le b$.

The case $a \le x \le \frac{a+b}{2}$ works similarly, or can be reduced to the first case because of the symmetry of the problem.


问:
Understanding the connection between completion of rings and completions of values fields
I would like to understand if there is a connection between these definitions of completion:

1. Completion of a ring with respect to an ideal (which I found in the German Wikipedia article)

Let $R$ be a ring, and $I \subseteq R$ be an ideal. Then $R^{\Bbb{N}}$ is ring of sequences in $R$. We call a sequence $(a_i) \in R^{\Bbb{N}}$ a Cauchy sequence if
$$
\forall n \in \Bbb{N} \: \exists k \in \Bbb{N} \: \forall i,j > k \: : \: a_i-a_j \in I^n,
$$
and call it a zero sequence if
$$
\forall n \in \Bbb{N} \: \exists k \in \Bbb{N} \: \forall i > k \: : \: a_i \in I^n.
$$
The set $\textbf{CF}$ of all Cauchy sequences in $R$ constitute a subring in $R^\Bbb{N}$, and the set $\textbf{NF}$ of all zero sequences in $R$ is an ideal of $\textbf{CF}$. Therefore, we can define
$$
\hat{R}_I := \textbf{CF}/\textbf{NF}
$$
which is called the completion of $R$ with respect to $I$.

2. Completions of rings as an inverse limit (also taken from the cited Wikipedia article above)

Now it can be shown that
$$
\hat{R}_I \simeq \varprojlim_n (R_n,f_n)_{n \in \Bbb{N}} := \left\{ (x_n) \in \prod_n R_n \: \middle| \: x_n \in R_n, \: f_n(x_n) = x_{n-1} \right\}
$$
where $R_n = R/I^n$ and $f_n : R_n \to R_{n-1}$, $\bar{x} \mapsto \bar{x}$ (in their respective equivalence classes, of course).

An example I know for that are the $p$-adic integers $\Bbb{Z}_p$ which is the completion of $R = \Bbb{Z}$ wrt. the ideal $I = (p)$, i.e. $\hat{\Bbb{Z}}_{(p)} = \Bbb{Z}_p$.

3. Completions of valued fields

Let $(K,|\cdot|)$ be a valued field, i.e. a field $K$ with an absolute value $|\cdot|: K \to \Bbb{R}$. Here, we can define Cauchy sequences and zero sequences in the usual sense for metric spaces. Also here, we can take the quotient $\tilde{K}$ of all Cauchy sequences in $K$ and all zero sequences in $K$, and call it the completion of $K$ wrt. $| \cdot |$.

I am also familiar with one particular example: For instance, I know that if we take $K=\mathbb{Q}$, and denote $|\cdot | = | \cdot |_p$ to be the $p$-adic absolute value on $K$, the completion of $K$ wrt. $|\cdot |_p$ would be $\tilde{K} = \mathbb{Q}_p$.

What do I want to know?

Now according to the discussion in this post of mine (in the comments of the answer and which I have a hard time understanding), there seems to be a connection between these concepts. I am more familiar with 3., while I found 1. and 2. just recently.

In particular, I am interested if there is a connection from 1./2. to 3.. In 1./2., there is only a ring while in 3., we have a field (where all ideals are the zero ideal or the field itself). Also, the definition is 1./2. in quite algebraic while the one in 3. is rather analytical.

To make this more concrete: Let $R$ be an integral domain and $I \subseteq R$ be an ideal. Let $K = Q(R)$ be the field of fractions of $R$ where I can define a valuation $|\cdot|: K \to \Bbb{R}$ to be $|x| := e^{-v(x)}$ where $v(a) = n \Leftrightarrow a \in I^n \setminus I^{n+1}$ (defined over $R$) and $v\left( \frac{a}{b}\right) = v(a) - v(b)$ (where $a,b \in R$).

Question In the setting I just described, do we have $Q\left( \hat{R}_I \right) = \tilde{K}$?

答:
Yes it is true.in fact you can do better than that: take any $w\in I$, then $\tilde K=\tilde R[\frac{1}{w}]$.(I assume that you really have a valuation field which means that $R$ is local and $I$ is principal and maximal, otherwise you don't get a valuation but see the third paragraph for more general setting)

this is true because you can easily see that components of a Cauchy sequence($a_k$) in $K$ have bounded valuations(the whole sequence is bounded) hence there is a power $n$ of $W$ such that $(w^n a_k)=(b_k)$ is a cauchy sequence in $R$ and hence an element of $\tilde R$. in this way you get a map $\tilde K\to \tilde R[\frac{1}{w}]$ and it is not hard to see that this is an isomorphism.

in fact if you are interested there is a more general theory of this things called tate rings: a Tate ring is a topological ring $R$ and a subring $R_0$ and an ideal $w\in I$ of $R_0$ such that $R=R_0[\frac{1}{w}]$ and the base for the topology is given by $I^n R_0$. in your language $R=K,R_0=R,I=I$ is an example of a tate ring. for a tate you can always define $\tilde{R}=\tilde{R_0}[\frac{1}{w}]$


答: First use $b^2=1$ to reduce $b^s$ to at most one $b$. Then write $b^{-1}ab=a^{-1}$ as $ba=a^{-1}b$ (again using $b^2=1$), and use that to successively move the single $b$ to the right.</p>

Title: Representation Theory. Why does $a^{r}b^{s}a^{t}b^{u} = a^{i}b^{j}$?

Body: I want to know that why we have $a^{r}b^{s}a^{t}b^{u} = a^{i}b^{j}$. Please let me know.


问:
Show that $n_1=n_2=...=n_k=1$
I've tried to reduce to the case $k=2$, and have also tried to use the Chinese Remainder Theorem. However, it is not possible, since the terms in each equivalent relation are different. Could anyone give me some help?

The entire exercise:

Let $k\geq 2$ and $n_1,n_2, ..., n_k \geq 1$ natural numbers such that

$$
n_2 \mid (2^{n_1}-1)\\
n_3 \mid (2^{n_2}-1)\\
\dots\\
n_k \mid (2^{n_{k-1}}-1)\\
n_1 \mid (2^{n_k}-1).
$$

Show that $n_1=n_2=...=n_k=1$.

答: For all $j \in \lbrace 1, \dotsc, k \rbrace$, we have $2^{n_{j}} -1 \mid 2^{\operatorname{lcm}\left( n_{1}, \dotsc, n_{k} \right)} -1$ since $n_{j} \mid \operatorname{lcm}\left( n_{1}, \dotsc, n_{k} \right)$, and hence $$n_{(j +1) \bmod k} \mid 2^{\operatorname{lcm}\left( n_{1}, \dotsc, n_{k} \right)} -1 \, \text{.}$$ It follows that $$\operatorname{lcm}\left( n_{1}, \dotsc, n_{k} \right) \mid 2^{\operatorname{lcm}\left( n_{1}, \dotsc, n_{k} \right)} -1 \, \text{,}$$ and hence $\operatorname{lcm}\left( n_{1}, \dotsc, n_{k} \right) = 1$ since $n$ does not divide $2^{n} -1$ whenever $n \geq 2$ (see For $n \geq 2$, show that $n \nmid 2^{n}-1$). Therefore, we have $$n_{1} = \dotso = n_{k} = 1 \, \text{.}$$

Title: Show that $n_1=n_2=...=n_k=1$


答:
The following proof is related to $n$ divides $2^n-1$ $\implies n=1$. First, if any $n_i$ is $1$, the right side of the $i$'th expression becomes $1$, so the corresponding value on the left side must be $1$, with this then extending to all of the rest of the values. Thus, assume none of the $n_i$ values is $1$.

Let $p_i$ be the least prime factor of $n_i$ for $1 \le i \le k$. Now, consider your first expression, i.e.,

$$n_2 \mid (2^{n_1} - 1) \tag{1}\label{eq1A}$$

The multiplicative order of $2$ modulo $p_2$ must divide $n_1$. However, $\operatorname{ord}_2{p_2}$ also divides the Euler's totient function of $p_2$, i.e., $p_2 - 1$, so

$$\operatorname{ord}_2{p_2} \le p_2 - 1 \tag{2}\label{eq2A}$$

Since $2^1 - 1 = 1$, all of the multiplicative orders must be $\gt 1$. Thus, $\operatorname{ord}_2{p_2}$ must have at least one prime factor less than $p_2$. Since $\operatorname{ord}_2{p_2} \mid n_1$, this means

$$p_1 \lt p_2 \tag{3}\label{eq3A}$$

Repeating this with your next expression gives

$$p_2 \lt p_3 \tag{4}\label{eq4A}$$

Combining with \eqref{eq3A}, this then becomes

$$p_1 \lt p_3 \tag{5}\label{eq5A}$$

Continuing this until the second last expression results with

$$p_1 \lt p_k \tag{6}\label{eq6A}$$

However, using the original argument (i.e., that which resulted in \eqref{eq3A}) with your last expression gives

$$p_k \lt p_1 \tag{7}\label{eq7A}$$

This contradicts \eqref{eq6A}, which means the original assumption all of the $n_i$ are not $1$ must be incorrect. Thus, all of the $n_i$ must be $1$.

Title: Show that $n_1=n_2=...=n_k=1$

Body:
I've tried to reduce to the case $k=2$, and have also tried to use the Chinese Remainder Theorem. However, it is not possible, since the terms in each equivalent relation are different. Could anyone give me some help?

The entire exercise:

Let $k\geq 2$ and $n_1,n_2, ..., n_k \geq 1$ natural numbers such that

$$
n_2 \mid (2^{n_1}-1)\\
n_3 \mid (2^{n_2}-1)\\
\dots\\
n_k \mid (2^{n_{k-1}}-1)\\
n_1 \mid (2^{n_k}-1).
$$

Show that $n_1=n_2=...=n_k=1$.


问:
How many sequences we can make from 4 digits, such that they don't have more than one zero.
My attempt: I tried to count the number of ways that the four digits don't have one zero, which is $9^4$, then count the number of ways that they have exactly one zero, I got confused a little on how to count them, but I split this into cases where the zero appears on the first digit, I have $9^3$ ways, and shifted it to the left one by one and added all of them up and got $4*9^3$. so my answer is $9^4+4*9^3$

Would appreciate if someone can approve my work, or if there's other easier ways to solve this.
 Thanks in advance to everyone.

答:
Yes, you're correct assuming the problem asks for number of $4$-strings.

To make it concrete, the case of one zero appearing can be handled as : choosing one of the four spots for zero, and filling rest three spots with non-zero digits.
$$\binom{4}{1}\cdot9^3$$


答:
Hint:

Let $V(x,y):=U(x+y,x-y)$.

Then

$$\dfrac{\partial V}{\partial x}=\dfrac{\partial U}{\partial x}+\dfrac{\partial U}{\partial y}=0$$ and $V$ is a function of $y$ alone.

Title: Proving a partial differential equation of a U(x1,x2) function depends on x1-x2

Body:
this is my first time doing differential equations instead of just learning about it and i have no idea where to start with this one.

with this equation :

$$
\frac{\partial U}{\partial x1} + \frac{\partial U}{\partial x2} = 0
$$

I need to prove that U is only dependent on x1-x2

I have these following conditions:

-x1 and x2 are Cartesian coordinates.

-The following variables exist.

$$
\epsilon = x1 + x2
$$

$$
\eta = x1 - x2
$$

both of them are independent and correspond to oblique, 45-degree coordinates on the (x1,x2) plane.


问:
Mathematical Systems Question Help
Alright, this is another question for my math for teachers course. This question is not actually in the homework, but there are problems similar to it. I'd really like to learn how to do problems like this one, so it would be much appreciated if I was given a solid explanation/demonstration. I don't really have a clue where to start, so I've defined the terms as best as I can under each question... The question should be answered using the data from the table.

A mathematical system is defined by the following table:

Is there an identity element in this system? If so, what is it?
Identity is: aXe=a and eXa=a
Is closure satisfied by this system? Explain.
Closure is: aXb
Is this system commutative? Explain.
Commutative is: aXb=bXa
Does every element have an inverse? Explain.
Inverse is: aXx=e and xXa=e
Is this system a group? Explain.
Not sure what they mean by this...
Are they referring to another property?.

答:
To demonstrate that you have an identity here, you need to show that for all a, there exists an e such that (aXe)=a=(eXa).  So, you need to list a set of equations which show this true.  Some of the equations you need are (bXa)=a, (bXb)=b, and (bXc)=c.  Do you see the other ones you need to list also?

To spot an identity element, if it exists, on a table look at the leftmost column.  Then look at the other columns and see if any of them matches the leftmost column (outside of the cross).  Call this column C.  The element at the top of the column in the cross can be the identity element, so long as it exists.  An identity element will exist if there exists a similar row.

To show closure you just need to show that for all x, for all y, (xXy) is a member of the same set that x and y belong to.  Since x and y in your example come from the set {a, b, c}, this means that you just need to show that for all x, for all y, (xXy) belongs to {a, b, c}.

To show commutativity you need to show that for all x, for all y, (xXy)=(yXx).  So since x and y belong to {a, b, c}, that means you need to show that (aXb)=(bXa), (aXc)=(cXa), and (bXc)=(cXb).  You can usually fairly easily tell if commutativity holds or not if there exists a symmetry over the diagonal.

To show that an inverse exists, you need to show that for all x, there exist a y such that (xXy)=e=(yXx), where e is the identity element of the system.  There exists a geometrical way to spot if an inverse element exists, given an identity element.  Here that means you need to list equations for a, for b, and for c, such as (aXc)=b=(cXa).  If the system commutes, this cuts down the number of equations you need to show.

A group consists of a mathematical system (G, X) where G is the carrier set, and X a binary function such that

G satisfies closure under function X.

The system has an identity element.

For all elements of the system, there exists an inverse.

Associativity holds for the system, or that for all x, for all y, for all z
(xX(yXz))=((xXy)Xz).

So you need to show that 1.-4. hold for your system if it's a group.

This requires (3x(3x3))=27 cases to check for your system for associativity.  You can use something similar to truth tables to simplify the work here.  Suppose we have a set {a, b}, and operation Y such that (aYb)=(bYa)=(aYa)=a and (bYb)=b.  But, let's put operations at the end of expressions instead of in the middle.  So, we have abY=baY=aaY=a and bbY=b.  Associativity now means that for all x, y, z xyzYY=xyYzY.  Then the following table shows that associativity holds:

```<code> x  y  z  x yzY Y  xyY z Y
 a  a  a      a a    a   a
 a  a  b      a a    a   a
 a  b  a      a a    a   a
 a  b  b      b a    a   a
 b  a  a      a a    a   a
 b  a  b      a a    a   a
 b  b  a      a a    b   a
 b  b  b      b b    b   b
</code>```

since the final column for xyzYY and xyYzY match.

Title: Mathematical Systems Question Help

Body:
To demonstrate that you have an identity here, you need to show that for all a, there exists an e such that (aXe)=a=(eXa).  So, you need to list a set of equations which show this true.  Some of the equations you need are (bXa)=a, (bXb)=b, and (bXc)=c.  Do you see the other ones you need to list also?

To spot an identity element, if it exists, on a table look at the leftmost column.  Then look at the other columns and see if any of them matches the leftmost column (outside of the cross).  Call this column C.  The element at the top of the column in the cross can be the identity element, so long as it exists.  An identity element will exist if there exists a similar row.

To show closure you just need to show that for all x, for all y, (xXy) is a member of the same set that x and y belong to.  Since x and y in your example come from the set {a, b, c}, this means that you just need to show that for all x, for all y, (xXy) belongs to {a, b, c}.

To show commutativity you need to show that for all x, for all y, (xXy)=(yXx).  So since x and y belong to {a, b, c}, that means you need to show that (aXb)=(bXa), (aXc)=(cXa), and (bXc)=(cXb).  You can usually fairly easily tell if commutativity holds or not if there exists a symmetry over the diagonal.

To show that an inverse exists, you need to show that for all x, there exist a y such that (xXy)=e=(yXx), where e is the identity element of the system.  There exists a geometrical way to spot if an inverse element exists, given an identity element.  Here that means you need to list equations for a, for b, and for c, such as (aXc)=b=(cXa).  If the system commutes, this cuts down the number of equations you need to show.

A group consists of a mathematical system (G, X) where G is the carrier set, and X a binary function such that

G satisfies closure under function X.

The system has an identity element.

For all elements of the system, there exists an inverse.

Associativity holds for the system, or that for all x, for all y, for all z
(xX(yXz))=((xXy)Xz).

So you need to show that 1.-4. hold for your system if it's a group.

This requires (3x(3x3))=27 cases to check for your system for associativity.  You can use something similar to truth tables to simplify the work here.  Suppose we have a set {a, b}, and operation Y such that (aYb)=(bYa)=(aYa)=a and (bYb)=b.  But, let's put operations at the end of expressions instead of in the middle.  So, we have abY=baY=aaY=a and bbY=b.  Associativity now means that for all x, y, z xyzYY=xyYzY.  Then the following table shows that associativity holds:

 x  y  z  x yzY Y  xyY z Y
 a  a  a      a a    a   a
 a  a  b      a a    a   a
 a  b  a      a a    a   a
 a  b  b      b a    a   a
 b  a  a      a a    a   a
 b  a  b      a a    a   a
 b  b  a      a a    b   a
 b  b  b      b b    b   b

since the final column for xyzYY and xyYzY match.


问:
The degree of vector bundle on integral projective curve is the degree of its determinant bundle.
This is exercise 18.4.J in Vakil's "The Rising Sea  Foundations of Algebraic Geometry".
Here the degree of a coherent sheaf $\mathscr{F} $ on integral projective curve $ C $ is defined to be
$$ \mathrm{deg} \mathscr{F}=\chi(C,\mathscr{F})-(\mathrm{rank} \mathscr{F})\cdot \chi(C,\mathscr{O}_C).$$

Here the curve may be singular, and thus the local ring at a closed point is not a discrete valuation ring. I have no idea whether we can define the order of zeros or poles of a rational section in this case.

A hint is given: Exercise 13.5.H, i.e., given an exact sequence
$$0\to \mathscr{F}_1\to \cdots \to \mathscr{F}_n \to 0 $$
of finite rank locally free sheaves on $ X $, the alternating product of determinant bundle is trivial.

My idea is that since $\mathscr{F}(m) $ is gloally generated for $m>>0 $, we get an exact sequence of locally free sheaf
$$ 0\to \mathscr{F_0} \to \oplus\mathscr{O}(-m)\to \mathscr{F}\to 0,$$
but I don't know whether $\mathscr{F_0}$ is direct sum of line bundles, and I'm not sure if we still have
$$\mathrm{deg}(\mathscr{F}\otimes\mathscr{G})=\mathrm{deg} \mathscr{F}+\mathrm{deg} \mathscr{G}$$
in this case. When C is regular, this is done by computing the divisor of zeros and poles of a rational section, but here $ C $ may be singular now.

答:
Let $\mathcal{F}$ be a rank $n$ vector bundle (i.e. locally free sheaf) on such a curve $C$. Note that $\det(\mathcal{F})$ is a locally free sheaf of rank 1 (an invertible sheaf). We want to show that $\deg \mathcal{F} = \deg (\det \mathcal{F} ) $, i.e.
$$ \chi(C, \mathcal{F}) - n \cdot \chi(C, \mathcal{O}_C) = \chi(C, \det(\mathcal{F})) - \chi(C, \mathcal{O}_C), $$
or equivalently,
$$ \chi(C, \mathcal{F}) = \chi(C, \det(\mathcal{F})) + (n-1) \chi(C, \mathcal{O}_C). $$
Recall that Euler characteristic is additive in exact sequences, so we would be done if we can find some exact sequence
$$ 0 \to ? \to \mathcal{F} \to \det ( \mathcal{F} ) \to 0 $$
or
$$ 0 \to \det ( \mathcal{F} ) \to \mathcal{F} \to ? \to 0 $$
where $\chi(C, ?) = (n-1) \chi(C, \mathcal{O}_C)$. Thinking about the definition of $\det{\mathcal{F}}$, can you find such a sequence? The hint given in the text might give you some guidance but Vakil's hints can also be distracting at times.

Another route: use a result like https://stacks.math.columbia.edu/tag/0AYT, bullet point (4).


问:
If solutions to an expression are just another expression, why are they considered solutions?
Questions in mathematics are posed as:

```<code>Solve for y = yadda yadda yadda
</code>```

The yadda yadda yadda is some expression involving variables and integrals and so forth. “Solving” for y often leads to another expression, rather than some exact number.

If the result is just another expression, why is this considered a solution? It makes sense to consider it a “simplification” (although I’m not sure what the measure of reduced difficulty is...number of symbols?) but a solution, one might think, should be more than just another way to express the same formula seen in the question.

Further, there are often many different expressions one could come up with as a “solution.” In other words, rarely do 2 people find the same final expression as a solution to the posed mathematical question.

This again begs the question, what exactly is the definition of a mathematical solution given the immense number of possible ways to express it?

答:
You're right that figuring out a property of a solution or just re-writing the solution in another language may seem useless. But a lot of the great problems in mathematics arise from attempting to build connections between different theories as to enable further inspection of one theory by the results provided by the other.

For instance, consider the number $e$. There are many ways to characterize it which are all equivalent, but looking at a certain definition of this number leads to reasoning that are not possible under other definitions.

You can define $e^x$ as the solution to the differential equation $y' = y$ with initial conditions $y(0) = 1$, and you can define $e \overset{def}= y(1)$. With this definition, the number $e$ is not as interesting as the function $e^x$ it defines, and the number itself doesn't initially look very special.

You can also define $e$ as the limit $\lim_{n \to \infty} \left( 1 + \frac 1n \right)^n$. This gives you the opportunity to approximate the value of $e$ with a concrete expression, something that was not a priori very easy to do with the differential equation.

You can also define $e$ as the value to the series $\sum_{k \ge 0} \frac 1{k!}$. Again, you get a good approximation, but now the $k!$ tells you that maybe the number $e$ has something to do with permutations, since $k!$ is also the order of the symmetric group $S_k$ on $k$ symbols.

It is not directly obvious from the last two definitions that if you look at $\lim_{n \to \infty} \left( 1 + \frac xn \right)^n$  and $\sum_{k \ge 0} \frac{x^k}{k!}$ that these two functions are in fact all equal to $e^x$. If you find a relationship between the symmetric group and $e^x$, because $e^x$ is also the solution to the differential equation $y' = y$, now you can use a property of differential equations to study the symmetric groups.

Maybe my example wasn't the best, but I was trying to indicate that finding connections between fields gives you more tools to study all these fields, and thus proves very useful in practice.

Hope that helps,