问:
Property on average orders of multiplicative functions
I read the statement that, for a multiplicative function $h(n)$ we have
$$\sum_{n<x} h(n)n^a \sim R \frac{x^{a+1}}{a+1}$$
where
$$R := \prod_p \left(1 - \frac 1p \right) \left( 1 + \frac{h(p)}{p} + \frac{h(p^2)}{p^2} + \cdots \right).$$

Where does it come from and under which conditions on $h$? Is there a way to see the result in "integration" terms? (there is a residue/arithmetic part, but the remainder of the result looks like direct integration)

答:
It is not true in general, try with $h(n)=n^2$ (which gives $R=\infty$), $h(n) = n^{-1/2}$ (which gives $R=0$)

and $\sum_n h(n)n^{-s} =\zeta(s)\zeta(s+i)$ which gives $R=\zeta(1+i)$ (yes $\prod_p \frac1{1-p^{-s}}$ converges for $\Re(s)=1,s\ne 1$ by the PNT because $\sum_p p^{-s}$ looks like $\sum_n \frac{n^{-s}}{\log n}$ and $\sum_{n> x} \frac{n^{-s}}{\log n}$ looks like $\frac{x^{1-s}}{(1-s)\log x}$). If we had $\sum_{n\le x}h(n) = Rx+ o(x)$ then $\sum_n (h(n)-1)n^{-s}$ would be $o(1/\Re(s-1))$ for $\Re(s)>1$ which is absurd. In this example $\sum_{n\le x} h(n)= \zeta(s+i) x+\frac{x^{1+i}}{1+i}+o(x)$.

The conditions are those of the Tauberian theorems proving the PNT (often satisfied for the functions appearing naturally in analytic number theory).

Under much stronger assumptions there is a one line proof, see How to determine the residue of an (arithmetic) Dirichlet series. For $a> -1$ the asymptotic for $\sum_{n\le x} h(n)$ gives the asymptotic for $\sum_{n\le x} h(n) n^a$ by partial summation.


问:
$\mathbb{E}[X \mid X \leq n], X \sim$ Geom($p$)
Let $X$ denote a geometric distribution with success probability $p$, I want to compute the conditional expectation :
$$
\mathbb{E}[X \mid X \leq n].
$$
So intuitively we have a coin toss where we count the number of tosses until we throw heads, but we know that we are going to throw head at least once in at most $n$ tosses.

答:
There are two parametrization of Geometric: one that counts the trials and another one that counts the failures before the first success

Let's assume we are interested in the one counting the trials (the solution is not very different in the other case)

The conditional distribution is the following:

$$\mathbb{P}[X=x|X\leq n]=\frac{1}{1-q^n}p\cdot q^{x-1}$$

Where $p$ is the probability of success and $q=1-p$ is the probability of failure.

Thus

$$\mathbb{E}[X=x|X\leq n]=\frac{p}{1-q^n}\sum_{x=1}^{n}x q^{x-1}$$

can you conclude? It is not difficult

hint:

$$x q^{x-1}=\frac{d}{dq}q^x$$

after some calculations (using derivatives and geometric series) it results to me

$$\mathbb{P}[X=x|X\leq n]=\frac{1}{p}-\frac{n q^n}{(1-q^n)}$$

ContentLicense: CC BY-SA 4.0


问:
Determining the number of solutions of a system of linear inequalities.
I want to determine the number of solutions of a system of linear inequalities, and I was wondering if there was a simple way to to that. I know that linear programming is often used to check whether there are a zero or non-zero number of solutions, i.e. if the system/bounds is/are feasible, but is it possible to distinguish between there being a finite amount of unique solutions or infinitely many solutions? For instance, the system
$$
\begin{array}{lcl}
x & \leq & y \\
x & \geq & y \\
x + y & \leq & 1 \\
x + y & \geq & 1
\end{array}
$$

has 1 unique solution, namely $x=y=\frac{1}{2}$, while the system

$$
\begin{array}{lcl}
x & \geq & y \\
x + y & \leq & 1 \\
x + y & \geq & 1
\end{array}
$$

has infinitely many solutions. Is there a away to find out how many solutions a system of linear inequalities has, if any?

答:
The techniques for solving systems of linear inequalities differ from those for linear equations because the inequality signs do not allow us to perform substitution as we do with equations. Many of the concepts we learned when studying systems of linear equations translate to solving a system of linear inequalities, but the process can be somewhat difficult. Perhaps the most lucid way to simultaneously solve a set of linear inequalities is through the use of graphs. The solution to the system is all the points that satisfy both inequalities or the region in which the shading overlaps.

A system of linear inequalities in two variables consists of at least two linear inequalities in the same variables. It involves several expressions that, when solved, may yield a range of solutions. The solution of a linear inequality is the ordered pair that is a solution to all inequalities in the system. For a system of linear inequalities, there is only one solution set which can contain any number of solutions, or no solution.

To find the number of solution sets, we use the graphical representation of the inequalities, and shades in the values which satisfy each separate inequality. By visually representing the potential values of each one, we shall quickly notice if there be an overlap. Wherever the shading overlaps, it is said to be the solution set for the system.  If they do not overlap, there is no solution to the system. For example, consider two parallel lines. If the solution to one are the values above the line, and the solution to the other one are the values below the other line, there is no intersection and therefore there is also no solution to the system.

The following links (and the links therein) may give the idea of some algorithm to solve the linear inequality.
 Algorithm for finding integer solutions for linear inequalities

Solving a system of linear inequalities — what is the dimension of the solution set?


问:
Using the hypothesis $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ to prove something else
Assuming that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$$

Is it possible to use this fact to prove something like:

$$\frac{1}{a^{2013}}+\frac{1}{b^{2013}}+\frac{1}{c^{2013}}=\frac{1}{a^{2013}+b^{2013}+c^{2013}}$$

Just curious.  Thanks for the help!

ContentLicense: CC BY-SA 3.0

Body:
Well, assuming that $a,b,c\neq 0$, we can see that:

\begin{align}
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}&=\frac{1}{a+b+c}
\\ \\
\frac{ab+bc+ac}{abc}&=\frac{1}{a+b+c}
\\ \\
a^2b+abc+a^2c+ab^2+b^2c+abc+abc+bc^2+ac^2&=abc
\\ \\
(ab+b^2+ac+bc)c+(ab+b^2+ac+bc)a&=0
\\ \\
(ab+b^2+ac+bc)(c+a)&=0
\\ \\
(a+b)(b+c)(c+a)&=0
\end{align}

Now we know that $(a+b)(b+c)(c+a)=0$

$\therefore a=-b\text{ or }b=-c\text{ or }a=-c$.

We can use the fact $a=-b$ to our LHS:

$$\frac{1}{(-b)^{2013}}+\frac{1}{b^{2013}}+\frac{1}{c^{2013}}=-\frac{1}{b^{2013}}+\frac{1}{b^{2013}}+\frac{1}{c^{2013}}=\frac{1}{c^{2013}}$$

$$\text{RHS: }\frac{1}{-b^{2013}+b^{2013}+c^{2013}}=\frac{1}{c^{2013}}=LHS$$


问:
Show $R/(fg)\simeq R/(f) \oplus R/(g)$
Let R be a principal ideal domain and let $f,g\in R$ have no irreducible elements in common. Id like to show prove the isomorphism of rings above.

I already know the following isomorphism holds: $$R/(I_1\cap I_2) \simeq R/I_1 \oplus R/I_2$$

I'm clueless on where to even begin from here.

答:
Recall that PIDs are factorial, ie. have a unique factorization into primes/irreducibles. By assumption the greatest common divisor of $f$ and $g$ is 1, which means that there are $r,s\in R$ with $rf+sg=1$. Now consider the homomorphism
$$\begin{array}{rcl}
R \times R &\longrightarrow &R/(fg)\\
(x,y) & \longmapsto & (xg + yf + (fg))
\end{array}$$
It is welldefined (needs to be checked) and surjective: given $z + (fg)$ we can write $z = z1 = z(rf+sg) = zrf + zsg$, so $(zs,zr)$ is a preimage. If we know that the kernel of this map is $(f)\times (g)$ then we are done by the isomorphism theorem. But $(x,y) \in \ker$  is equivalent to $xg + yf \in (fg)$ and with $f,g$ sharing no common primes this is equivalent to $x = fa \in (f)$ and $y = gb \in (g)$ for some $a,b\in R$. So indeed $\ker = (f)\times (g)$.

Title: Show $R/(fg)\simeq R/(f) \oplus R/(g)$


问:
Doubt on proof of Lebesgue's Theorem
I am following the proof of Theorem 3.2 on this page. Towards the end the author states "By using the Vitali covering theorem and the definitions for $D^+f$ and $D^-f$ it is easy to check the following inequalities $$q|E_q| \leq |f(E_q)|$$ if $D^f(x) > q$ at every $x \in E_q$. But I cannot see why this hold, would anyone be able to explain why?

答:
I think your goal is to prove the following lemma:

Let $F:[a,b]\mapsto\mathbb{R}$ be an increasing function on bounded interval $[a,b]$. For real numbers $r<R$, define $E=\{x\in(a,b)\,|\,D_-F(x)<r<R<D^+F(x)\}$. Prove that $m^*(E)=0$.

I can give a very detailed (hence long) proof that may be easier for you to understand:

Part 1:

Let $m^*(E)=s$. For any $\epsilon>0$, there exists open set $O\supset E$ s.t. $m(O)<s+\epsilon$. Let $x\in E$. Then $D_-F(x)<r$ implies for all $\delta>0$, there exists $0<h<\delta$ s.t. $\frac{F(x)-F(x-h)}{h}<r$. Collect all of such intervals $[x-h,x]\subset O$, we obtain a Vitali covering of $E$. By Vitali covering theorem, there exists disjoint intervals $I_1,\ldots,I_N$, where $I_k=[x_k-h_k,x_k]$ s.t. $m^*(E\setminus\bigcup_{k=1}^NI_k)<\epsilon$. Denote $I_k^{o}=(x_k-h_k,x_k)$, then $m^*(E\setminus\bigcup_{k=1}^NI_k^o)<\epsilon$. Define $A=E\cap(\bigcup_{k=1}^NI_k^{o})$, then $m^*(A)>s-\epsilon$. Moreover, we have
$$\begin{equation}
\sum_{k=1}^N(F(x_k)-F(x_k-h_k))<r\sum_{k=1}^Nh_k<rm(O)<r(s+\epsilon)\tag{1}
\end{equation}$$

Part 2:

Let $y\in A$. Then $D^+F(y)>R$ implies there exists arbitrarily small $k>0$ s.t. $[y,y+k]\subset I_k$ for some $k$ and $\frac{F(y+k)-F(y)}{k}>R$. The collection of such intervals is a Vitali covering of $A$, so by Vitali covering theorem again, there exists disjoint $J_1,\ldots,J_M$ with $J_j=[y_j,y_j+k_j]$ s.t. $m^*(A\setminus \bigcup_{j=1}^MJ_j)<\epsilon$. It further implies $m^*(A\cap(\bigcup_{j=1}^MJ_j))>s-2\epsilon$. Moreover,
$$\sum_{j=1}^M(F(y_j+k_j)-F(y_j))>R\sum_{j=1}^Mk_j>R(s-2\epsilon)$$
Also, notice that each $J_j$ is contained in some $I_n$, so for each fixed $n$, by increasing property of $F$,
$$\sum_{j:J_j\subset I_n}(F(y_j+k_j)-F(y_j))\leq F(x_n)-F(x_n-h_n)$$
Sum both sides over $n=1,\ldots,N$,
\begin{align*}
  \sum_{n=1}^N(F(x_n)-F(x_n-h_n))&\geq\sum_{n=1}^N\sum_{j:J_j\subset I_n}(F(y_j+k_j)-F(y_j)) \\
  &=\sum_{j=1}^M(F(y_j+k_j)-F(y_j))>R(s-2\epsilon)\tag{2}
\end{align*}

Part 3:

Combining (1) and (2), we have $r(s+\epsilon)>R(s-2\epsilon)$ for all $\epsilon>0$. Take $\epsilon\to 0$, we obtain $rs\geq Rs$. Since $r<R$, $s=0$ and we are done.

Title: Doubt on proof of Lebesgue's Theorem


问:
Find the number of orthogonal matrices of order $2\times 2$ whose entries are $0,1$
Is there some method to solve these type of questions, or do I have to manually try out every possible matrix and check if it’s orthogonal?</p>

答: Hint. Each row or column of an orthogonal matrix is a unit vector. If the matrix's entries are zeroes and ones, how many ones are there on each row or each column?


答:
let the matrix have its elements $a_{11}=a,a_{12}=b,a_{21}=c,a_{22}=d $

Now easy to see $ad,bc\in${$0,1$},

As it is orthogonal $ad-bc=\pm 1$ if $ad-bc=1$ then $ad=1,bc=0$ (how many cases?) else $ad-bc=-1$  then $ad=0,bc=1$ (how many cases?)

Title: Find the number of orthogonal matrices of order $2\times 2$ whose entries are $0,1$

Body: Is there some method to solve these type of questions, or do I have to manually try out every possible matrix and check if it’s orthogonal?</p>


问:
Standard error of the mean for root mean square of data
If I have a distribution of data, X, representing N samples taken during one measurement, then the mean square of X is $\bar{X^2} = \langle X^2 \rangle$, the variance of $X^2$, $\mathrm{var}(X^2)$ is $\langle\langle X^4 \rangle - \langle X^2 \rangle \rangle$, and the standard error of the mean square is $\sqrt{\frac{\mathrm{var}(X^2)}{N}}$.

Thus, the standard error of the mean square represents one standard deviation of the distribution that would be produced by repeating the measurement (taking N samples each time), assuming that $X^2$ is normally distributed with the variance of the original measurement.

What is the standard error of the quantity $\sqrt{\langle X^2 \rangle}$ (the standard error of the root mean square?)?

A rephrasing:  assume Y is normally distributed with mean $\mu$ and variance $\sigma^2$.  Define Z = $\sqrt{\frac{1}{N}\sum_1^N Y_i}$ What is the variance of Z?

答:
If $Y$ is normally distributed with mean $\mu$ and variance $\sigma^2$, $U = \frac{1}{N} \sum_1^N Y_i$ is normally distributed with mean $\mu$ and variance $\sigma^2/N$.  Note that $P\{U < 0\} > 0$, so $Z$ will have complex values.
Anyway, $E[|Z|] = \sqrt{\frac{N}{2\pi \sigma^2}} \int_{-\infty}^\infty \exp(-N(t-\mu)^2/(2 \sigma^2)) \sqrt{|t|}\, dt$.  I doubt there is a closed form for this integral. $E[|Z|^2] = E|U| = \sqrt{\frac{2}{\pi N}} \sigma \exp(-\mu^2 N/(2 \sigma^2)) + \mu\, {\rm erf}(\mu \sqrt{N/2}/\sigma)$, and ${\rm Var}(Z) = E[|Z|^2] - E[|Z|]^2$.</p>


问:
how to find the domain of this function:$f(x)=\frac{\ln(4-x^2)}{\ln(x+1)}$
I'm having trouble solving this exercise: (find the domain of this function) $$f(x)=\frac{\ln(4-x^2)}{\ln(x+1)}$$
Here is my attempt:

I've found the domain of the natural logarithm in the numerator, which is equal to: $$4-x^2>0$$ (it must be strictly positive),

1.1 $$-x^2>-4$$

1.2 $$x^2<4$$

1.3 (result) $$-2<x<2$$

I've found the domain of the natural logarithm in the denominator, which is equal to: $$x+1>0$$

1.1 (result) $$x>-1$$

Now, I have to find the interval (on the number line) in which each possible solution is acceptable. So, I've plotted each result on a number line, here is the result:

$$-1<x<2$$

but this question has these possible solutions (it is a multiple choice question):

(1)  $-1<x<2, except\space0$

(2)  $-\infty<x<-1$

(3) $-2<x<-1$

(4) $-\infty<x<0$

The solution is $(1)$, but I don't understand why is not 0 included in the solution's interval.

EDIT (correct answer):

In the denominator I would have had considered the equation involving the denominator: $\ln(x+1)=0$. it mustn't be equal to 0.
this error occurs because of distraction.

答: At $x=0$ we have $\ln (1-0) = \ln (1) = 0$ in the denominator, which is undefined.


问:
Question about Bishop's "constructive analysis" for people familiar with bishop style constructive analysis only
It has been a while since i read Bishop's book "constructive analysis", recently I dug it out of my book shelve and started to read. I came around this observation on the top of page 85.

"A subset Y of $\mathbb{R}$ can be bounded as a
metric space but not bounded as a subset of $\mathbb{R}$."

I have tried to figure out how this can be and came up with nothing so far. Can anybody give me a hint?

PS: As far as I understand Bishop is referring to the metric induced by the absolute value.

PPS: The definition provided:
A metric space $(X, p)$ is called bounded if there exists a real number
$C > 0$, called a bound for $(X, p)$, such that $p(x, y) \leq C \forall x,y \in X$.
A subset $Y$ of a non void metric space $X$ is bounded if, for all
(equivalently, some) $x$ in $X$, the set $Y\cup\{x\}$ with the induced metric p* is a bounded metric space.

答:
It is not an actual answer, but a bit long to be a comment.

Bishop's claim is refutable in the classical context; that is, every subset of $\mathbb{R}$ which is bounded as the metric space with the inherited metric is also a bounded subset of $\mathbb{R}$: let $X\subseteq\mathbb{R}$ is an inhabited subset which forms a bounded metric space with a bound $r$. Take any $a\in X$, then $X\subseteq \overline{B}(a,r)$.

Let $x\in \mathbb{R}$ be any real, and we will see that $X\cup\{x\}$ is a bounded metric space under the usual metric. Since $X$ is a bounded (in the classical sense), the set
$$\{|x-y|: y\in X\}$$
is also bounded, so it has the supremum $M$. Then $\max(M,r)$ bounds $X\cup\{x\}$ as a metric space.

Thus we have to understand Bishop's claim as claiming the existence of a weak counterexample.


问:
Solving Prop 86 Euclid's Data
I was given the following problem:

Solve the equations of Prop-86 of the Data algebraically. Show that the two hyperbolas defined by the equations each have their axes as the asymptotes of each other.

All I can see about equation of Prop-86 is the system: $$xy=a\\\frac{y^2-b}{x^2}=\alpha$$

I don't understand what to do here. Am I supposed to solve this system of equations?

*Edit: This is the work I did.
I understand all of it, until the end where I stated: "Therefore, two hyperbolas as defined have their axes as the asymptotes of the other." I wrote that, because based on the question I know this will be the conclusion, but I have no idea how that result actually came to be.

答:
This is not an answer to your question. What I am trying to do here is to give you an insight into it. That is why I am providing you with a diagram, which shows an example of the scenario you have described in your problem statement. If you are already aware of these facts, then you can ignore them and let me know that, so that I can delete my post.

You can visualize in the diagram that the axes and asymptotes of one hyperbola are the asymptotes and axes of the other hyperbola. The equations of the two hyperbolae are
$$xy=8 \quad \mathrm{and}\quad \frac{y^2-5}{x^2}=1.$$

In other words, I have assumed $a=8$, $\alpha=1$ and $b=5$. In general, $a$ and $b$ can have any value, but $\alpha$ can have only one value, i.e. 1.

The two asymptotes of the hyperbola $xy=a$ are given by $x=0$ and $y=0$. Its axes are $y=\pm x$. They do not depend on the value of $a$. The two axes of the other hyperbola are $x=0$ and $y=0$, Therefore they too are independent of $\alpha$ and $b$. However, its asymptotes have the equations $y=\pm\sqrt{\alpha}x$, which depend on $\alpha$. That is why $\alpha$ cannot have a value other than 1.

I found an article at link, which, I think, describes a case of two hyperbolae and Prop 86. If you have access to this article, read it to see whether you can get more insight.

Title: Solving Prop 86 Euclid's Data


问:
Topological spaces can't be characterized by sequence convergence generally. Is there a Hausdorff counterexample?
A topology can be characterized by net convergence generally.

However, none of the counterexamples I have learnt where sequence convergence does not characterize a topology is Hausdorff.

Such as cocountable topology on an uncountable space, where every set is closed under sequence convergence.

I believe there are Hausdorff counterexamples but can't find one.

答:
A good source of counterexamples are the ordinal numbers with the order topology, which is Hausdorff.

The first uncountable ordinal $\omega_1$ is what you're looking for.

Consider a sequence $(\alpha_n)$ in $\omega_1$. Then each term of the sequence is a countable ordinal and the supremum of a countable set of countable ordinals is countable. Therefore the sequence is inside $\beta+1$, where $\beta=\sup\{\alpha_n\}$. The ordinal $\beta+1$ is, like all successor ordinals, compact.

Hence the sequence has a convergent subsequence.

On the other hand, $\omega_1$ is not compact, because a limit ordinal is not compact.

More simply, as suggested by Henno Brandsma, the subspace $\omega_1$ of $\omega_1+1$ is dense, but $\omega_1$ is not a limit of a sequence in $\omega_1$ (same argument as before).


问:
Is PID cyclic? I cannot understand quotient ring is cyclic
In the textbook, it says
“If $R$ is a principal ideal domain and $r$ is an elt of $R$, then the quotient ring $R/(r)$ is a cyclic R-module.

I dont know why $R/(r)$ is cyclic.

Thank you in advance.

答:
I am just elaborating on Berci's comment.

Maybe you are confusing two different notions of being cyclic:

A cyclic group is a group $G$, such there exists $g \in G$ with $G = \langle g \rangle$, i.e., all elements of $G$ are of the form $g^m$ for $m \in \mathbb Z$.

A cyclic $R$-module (for a commutative unital ring $R$) is an $R$-module $M$ such that there exists an $m \in M$ with $M = \langle m \rangle_R$, i.e., $M$ is generated by $m$ as an $R$-module. By definition, $\langle m \rangle_R = \{rm \ | \ r \in R\}$.

Here, obviously, we are speaking about the notion of a cyclic $R$-module.
Now, as noted in the comments, $R/(r)$ is generated by $1 + (r)$ as an $R$-module. Here, one does not need to use that $R$ is a principal ideal domain (and it's hard to guess why you require this, because you don't provide much context).


问:
How to factorize equation of pair of lines?

Find two equations of lines
: $$6x^2 -xy -12y^2 -8x + 29y -14= 0$$

I tried doing it using quadratic equation but got some very much unusual value. How to solve this?

答:
you can also partially diff. the equation
F(x,y)=` 6x2−xy−12y2−8x+29y−14
strong text
dF/dx=0                            ...(i)
dF/dy=0                            ...(ii)
(i) and(ii) will give unique linear equations.


问:
Prove that nearly all positive integers are equal to $a + b + c$ where $a | b$ and $b | c$, $a \lt b \lt c$

If a positive integer $n$ is equal to $a + b + c$ where $a | b$, $b | c$ and $a \lt b \lt c$, let it be called "faithful". Prove that nearly all numbers are faithful and list the non-faithful numbers.

Here's where I am:

Let the number be $n$, $b = ap$ and $c = bq = apq$ where $p ,q \in Z_+$.
Let the set of faithful numbers be $F$.

$$n = a + b + c$$
$$\therefore n = a + ap + apq$$

Since $1 \lt p \lt q$ and $p, q \in Z_+$, $n \ge 7$. Hence $1, 2 \ldots 7 \notin F$.

Now what?

答:
Lemma $n \geq 7$ can be written in the form $1+p+pq$ where $1< p < q$ if and only if $n-1 \neq p, p^2+p$ for $p$ prime.

The proof follows immediately from:

$$n=1+p+pq \Leftrightarrow n-1=p(q+1)$$

This proves that all numbers $n \geq 7$ which are not of the forms $p+1$ or $p^2+p+1$ are faitful.

Lemma 2: If $n=p^2+p+1$ then $n$ is faithful.

Proof: $a=1, b=2, c=p^2+p-2$.

This shows that $F \subset \{ 1,2,3,4,5,6,7 \} \cup \{ p+1| p \, \mbox{prime} \}$.

To complete the proof, all you need to do is to observe that

Lemma 3: If $n=p+1 > 7$ and $n$ is faitfull then $\frac{n}{2}$ is also faitful.

Proof assume by contradiction that $\frac{n}{2}=a+b+c$ then $n=2a+2b+2c$ contradiction.

Now, all you need is brute force it. The following lemma tells you how:

Lemma 4 If for some $n$ there is no faithful number between $n$ and $2n$, then there is no faithful number greater than $n$.

Proof By contradiction. Assume there is some $m>n$ faithful. Then $m=p+1>2n$ and hence by Lemma 3  we have $\frac{m}{2}$ is also faithful. Repeat until the first time you get a number under $2n$, that number must be between $n$ and $2n$ contradiction.

The brute force Look for all the faithfull numbers between $8$ and $16$ of the type $p+1$. If none, done by lemma 4. If you find any, pick the largest of them $k$ and look for all  the faithfull numbers between $(k+1)$ and $2(k+1)$ of the type $p+1$. Repeat until you find none.

This works only if you are sure there are only finitely many faithful numbers.


问:
Question about squeeze and toast rule in limits

I was wondering if I could use the toast rule to show that the limit of $g(x)$ exists and by (c) it means limit of $f(x)$ must also exist, but I'm not entirely sure I understand the toast rule correctly. I can't seem to find any information on google, so maybe it has a bunch of different names?

It's basically a corollary of the squeeze rule, just with only one side of the inequation.

Regardless, how would you approach this question?

答:
Let's prove that the limit of $f$ is $L$.

Fix $\varepsilon>0$. Then you can find $\delta>0$ such that, for $0<|x-x_0|<\delta$, it holds

$|h(x)-L|<\varepsilon/2$

$|f(x)-g(x)|<\varepsilon/2$

From 2 and (a), we get also that $f(x)-h(x)<\varepsilon/2$. Hence, for $0<|x-x_0|<\delta$,
$$
|f(x)-L|=|f(x)-h(x)+h(x)-L|\le|f(x)-h(x)|+|h(x)-L|<\varepsilon/2+\varepsilon/2=\varepsilon
$$


问:
P + Q = P + R where P, Q and R are subspaces. Should Q = R?
Suppose that P, Q and R are subspaces of a vector space V such that
P + Q = P + R. Should Q = R necessarily?

I have an intuition that the answer is no, but I couldn't find proper examples to illustrate the same. Any help would be very appreciated.

答:
The answer is no. Take in $\mathbb{R}^2$:

$P = \{(x,0),x\in \mathbb{R}\}$ ($x$-axis) and

$Q= \{(0,y);y\in \mathbb{R}\}$ ($y$-axis) and

$R = \{(x,x);x\in \mathbb{R}\}$.


答: No. Let $V = \mathbb{R}^2$, and let $P$ be the y-axis, $Q$ be the x-axis, and $R$ be the $y=x$ line. Then $P+Q = P+R = \mathbb{R}^2$, but $Q \neq R$.

Title: P + Q = P + R where P, Q and R are subspaces. Should Q = R?

Body:
Suppose that P, Q and R are subspaces of a vector space V such that
P + Q = P + R. Should Q = R necessarily?

I have an intuition that the answer is no, but I couldn't find proper examples to illustrate the same. Any help would be very appreciated.


Body:
Counterexample:

Let $e_1$, $e_2$ - basis in $\mathbb{R}^2$.

$P = \{ a e_1, a  \in \mathbb{R}^2 \}$, $Q = \{ a e_2,  a  \in \mathbb{R}^2 \}$, $R = \{ a (e_1+e_2),  a  \in \mathbb{R}^2 \}$.


答: No. Take $P=V$, and then you will always have $P+R=Q+R$, no matter which subspaces $Q$ and $R$ are.


答:
Since it is not stated that $P$ and $Q$ resp. $P$ and $R$ intersect trivially, here is another class of counterexamples:

Let $P = V$. Then for any choice of $Q$ and $R$, one has $P+Q = P+R = V$. Hence it suffices to choose different $Q$ and $R$.


问:
What are the continuous functions $\mathbb{N} \to \mathbb{R}$ when $\mathbb{N}$ is given the coprime topology?
Consider $\mathbb{N}$ with the topology generated by sets of the form $U_{a,b}=\mathbb{N} \ \cap \ \{an+b: n \in \mathbb{Z}\} $ where $\text{gcd}(a,b)=1$.

What are the continuous functions $\mathbb{N} \to \mathbb{R}$ when $\mathbb{N}$ is given this topology and $\mathbb{R}$ has the Euclidean topology?

Some thoughts: if the range of such a continuous function were infinite, it would imply we could partition $\mathbb{N}$ into countably infinite many disjoint sets of the form $U_{a,b}$ as above, which somehow seems unlikely.

答: The space (see here) is connected and countable, so its continuous image in $\Bbb R$ is also connected and countable hence a singleton. So all such maps are constant.


问:
For field extensions $K/E/F$, $K/F$ is a radical extension if and only if $K/E$ and $E/F$ is a radical extension.
Definition. E/F is radical extension if there is $F\subset F(\alpha_1)\subset F(\alpha_1,\alpha_2)\subset\cdots\subset F(\alpha_1,...,\alpha_n) = E$ such that $\alpha_1^{n_1}\in F$ and $\alpha_i^{n_i}\in F(\alpha_1,...,\alpha_{i-1})$ for $i\geq 2$

For field extensions $K/E/F$, $K/F$ is a radical extension if and only if $K/E$ and $E/F$ is a radical extension.

If direction is clear. But the only if direction is not clear to me. Why is this true? (The fact that $K/E$ is radical extension is not trivial to me)

答:
$F \subset F(\alpha_1) \subset ... \subset F(\alpha_1, ..., \alpha_n)=K$ with $\alpha_i^{n_i}\in F(\alpha_1,...,\alpha_{i-1})$.

Then you have $E \subset E(\alpha_1) \subset ... \subset E(\alpha_1, ..., \alpha_n)$. But $F \subset E \Rightarrow F(\alpha_1,...,\alpha_{i-1}) \subset E(\alpha_1,...,\alpha_{i-1})$ and so $\alpha_i^{n_i}\in E(\alpha_1,...,\alpha_{i-1})$. For the same reason $K=F(\alpha_1, ..., \alpha_n) \subset E(\alpha_1, ..., \alpha_n)$. But $\alpha_1,...,\alpha_n \in K \Rightarrow E(\alpha_1, ..., \alpha_n)=K$. So $K/E$ is radical


问:
Show that any cyclic group of even order has exactly one element of order $2$

Show that any cyclic group of even order has exactly one element of order $2$

Attempt:

Let $G$ be a finite group of even order.

Assuming a set $A=\lbrace g \in G \vert g \neq g^{-1} \rbrace$, I can show that  $A\cup\{e\}$ has odd number of elements as $A$ has even number of elements. So $G$ has at least one element $a$ such that $a$ does not in $A$, ie $a=a^{-1}$ ie $a^2=e$. That is $G$ has atleast one element of order $2$. "More specifically, $G$ has odd number of elements of order $2$ --- ???"

But how to show that exactly one such element if $G$ be cyclic.

NB: Please don't use isomorphic properties.

答:
Let $(G.e)$ by a finite cyclic group containing an involution $v$. By Lagrange's theorem the order of $G$ is even, and for some $m \ge 1$ we  can write

$\quad |G| = 2m$

If $g$ is any generator of $G$ there is a minimal integer $k$ satisfying

$\tag 1 1 \le k \lt 2m \; \land \; \{g^1,g^2,\dots,g^k\} \cap \{e,v\} = \{v\}$

The theory found here allows us to assert that $k$ must be equal to $m$.

So by fixing the generator $g$ and 'running' the above argument against involutions $v, u \in G$, we see that $v = g^m$ and $u = g^m$. Using basic properties of the equality relation, the involutions $u$ and $v$ are equal.

We have shown that if a finite cyclic group contains an involution then its order is even and, moreover, it contains exactly one element that is an involution.


问:
Showing that the irreducible factors of the cyclotomic polynomial in $\mathbb{Q}[x]$ have the same degree.
For each $n\in \mathbb{N},\, \,$ $\phi_{n}(x):=\displaystyle \prod_{\substack{1\leq k\leq n\\ (k,n)=1}}(x-e^{\frac{2\pi i}{n}k})$
Show that the irreducible factors of $\phi_n$ in $\mathbb{Q}[x]$ have the same degree.
I have a confusion with this exercise. I understand that the cyclotomic polynomial $\phi_n$ is irreducible in $\mathbb{Q}[x]$ so $\phi_n$ has a unique irreducible factor, the same. Is this correct or am I confusing something?

答:
The argument you gave is correct, but proving irreducibility of the cyclotomic polynomials over $\mathbb Q$ takes some work. You can solve the question at hand without citing this result, which I'm guessing is the intent of the exercise. I will, however, be using some field theory.

Let $\phi_n = \prod f_i$ with each $f_i$ monic and irreducible. The $f_i$ are distinct as $\phi_n$ has no repeated roots. Let $\alpha_i$ be a root of $f_i$. These are therefore roots of $\phi_n$ so they are primitive $n^{th}$ roots of unity. Let's consider the field $\mathbb Q(\alpha_i)$. The degree of this field is the degree of the minimal polynomial of $\alpha_i$, which is $f_i$. Furthermore, as $\alpha_i$ is a primitive $n^{th}$ root of unity, the set $\{\alpha_i^k : k \in \mathbb Z\}$ is precisely the set of $n^{th}$ roots of unity. But since $\mathbb Q(\alpha_i)$ is a field, each of these powers lies in $\mathbb Q(\alpha_i)$. Hence, $\mathbb Q(\alpha_i)$ contains all the $n^{th}$ roots of unity. Furthermore, because the above result held for all $i$, we have $\alpha_j \in \mathbb Q(\alpha_i)$ for all pairs $i,j$. The definition of $\mathbb Q(\alpha_j)$ is the smallest field containing $\mathbb Q$ and $\alpha_j$, so $\mathbb Q(\alpha_j) \subseteq \mathbb Q(\alpha_i)$ for all pairs $i,j$. Thus, all of these fields $\mathbb Q(\alpha_i)$ are the same. In particular, they have the same degree over $\mathbb Q$. We have that $[\mathbb Q(\alpha_i) : \mathbb Q] = \deg f_i$, so for all $i,j$ $\deg f_i = \deg f_j$ as desired.


问:
Find polynomials $r(x)$ and $s(x)$ such that $r(x)(x^2+x+1)+s(x)(x^3-x-1)=1$.
I realize this has something to do with the Euclidean algorithm over $\mathbb{Q}[x]$, and I know how to show that the $\gcd(x^2+x+1, x^3-x-1)=1$, but I have no idea where to even begin looking for polynomials satisfying the Euclidean algorithm.

Is it largely just trial and error? Or is there some methodic process that I can go about solving for these polynomials...

答:
It's trivial using the method of simpler multiples. Let $\,g = x^3\!-\!x\!-\!1.\,$ Note $\,f = x^2\!+\!x\!+\!1\,$ has simpler multiple $\,\color{#0a0}{h = x^3\!-\!1}\,$ so $\, r\:\!f + s\,g = 1\iff \bmod f\!:\ s\,g\equiv 1\iff s\equiv 1/g,\,$ which we compute first modulo the simpler multiple $\,\color{#0a0}{h},\,$ i.e.  $\,1/g \bmod f = (1/g \bmod \color{#0a0}{x^3\!-\!1})\bmod f,\,$ i.e.

$$\bmod \color{#c00}{x^2\!+\!x}\!+\!1\!:\,\ \dfrac{1}{g}\,\equiv\, \dfrac{1}{\color{#0a0}{x^3\!-\!1}\!-\!x\bmod \color{#0a0}{x^3\!-\!1}}\,\equiv\, \dfrac{-(\color{#c00}{x^2+x})}{-x}\,\equiv\, x+1\qquad $$

Remark $ $ The method in Deepak's answer is essentially the same but instead replaces arithmetic $\bmod f\,$ by equivalent calculations using the roots of $\,f.$


问:
How to prove that $(k_{1}k_{2}...k_{n}) \equiv (z_{1}z_{2}...z_{n})\mod n$ if they are both complete system of residues modulo n.?
Prove that if $(k_{1},k_{2},...,k_{n})$ and $(z_{1},z_{2},...,z_{n})$ are both complete system of residues $\bmod n$ then
$$(k_{1}+k_{2}+...+k_{n}) \equiv (z_{1}+z_{2}+...+z_{n})\mod n$$  and
$$(k_{1}k_{2}...k_{n}) \equiv (z_{1}z_{2}...z_{n})\mod n$$
can you give me some hints to complete this proof?

I know that proofs are not obvious but in this case

$(k_{1},k_{2},...,k_{n})\mod n$  and   $(z_{1},z_{2},...,z_{n}) \mod n$

I probably need to apply some theorems in number theory for complete system of residues.

答:
A complete residue system consists of a set of integers of the form:

$$\{0+a_0 n, 1+a_1 n, \dots, (n-1)+a_{n-1} n\}$$

where it is easy to see that addition and multiplication modulo $n$ give the same result.


问:
Statistics: Why are school grades qualitative variable?
I am struggling to understand, why in descriptive statistics we say that school grades are a qualitative and not quantitative variable? I can understand why color of the hair is qualitative, but grades are numerical... </p>

答:
A finer categorization of the "level of data" is surely "nominal scaled", "ordinal scaled", "interval scaled" and "ratio scaled" (see more about this in wikipedia).

"nominal" are data for instance from your example of color of hair, or telephone numbers or ICD-codes or many labelings ...

"interval"  are data like temperature, for instance degree of Celsius or of Fahrenheit, which we give numerical values on the scale of real numbers, and "ratio" if they have a zero which are inherent/relevant with the meaning of the measure, such that relations as "double" or "half" make sense.

With grades in school it is between those concepts; while they are usually no measure related to the natural or real numbers, they still have an order: level "A" is better than "B" and that is better than "C" or so -and what is precisely what we want usually with such grades-, but what "nominal" data don't have. So the term "ordinal scale" has been introduced.

There is a lot more to say about, but I think wikipedia is a good-enough source for the beginning.


问:
expectation of functions of random variable
in genreral

can we claim that, for random variable' lets say discrete,  X

E[f(X)]=E[g(X)] means f(x)=g(x) ?

(for all x such that p(X=x) is not zero)

答:
No. Counterexample:

Let $\Omega = \{ 0, 1 \}$, $P(0) = P(1) = \frac{1}{2}$ - a symmetric coin.

f(x) = x, g(x) = 1-x.


问:
Question about aurifeullian factors

For which perfect powers $\ b>1\ $ can the expression $$\frac{b^{p^2}-1}{b^p-1}$$ be a prime number , if $\ p\ $ is prime ?

If $\ b>1\ $ is a perfect power , then $\ b=a^k\ $ for some integers $\ a,k>1\ $.

So we get $$\frac{a^{kp^2}-1}{a^{kp}-1}$$

and if the polynomial $$f(x)=\frac{x^{kp^2}-1}{x^{kp}-1}$$ is reducible we (probably) will get aurifeullian factors. It seems that in most cases, we get an irreducible polynomial. Who can shed some light on this ?

答:
The irreducible factorization of $x^m-1$ (over $\Bbb Z$) is the product of the cyclotomic polynomials $\Phi_d(x)$ over all positive integers $d$ dividing $m$. So the irreducible factorization of $f(x)$ is the product of the cyclotomic polynomials $\Phi_d(x)$ over all positive integers $d$ that divide $kp^2$ but not $kp$.

Assuming for simplicity that $p$ doesn't divide $k$: these integers $d$ are precisely those of the form $bp^2$ where $b$ divides $k$; in particular, there is more than one such integer when $k>1$, and so $f(x)$ will not be irreducible. (The assumption that $p$ doesn't divide $k$ can be weakened to $k$ not being a power of $p$.)

Title: Question about aurifeullian factors


问:
Show that the norm of Gaussian prime of form $a + bi$ is always prime.
I want to prove the partial converse of the following theorem:

Let $z \in \mathbb{Z}[i]$. If $N(z)$ - prime, then $z$ is also prime in $\mathbb{Z}[i]$.

I state the converse like this:

Let $z \in \mathbb{Z}[i]$, $z = a + bi$, $a,b \in \mathbb{Z}$ and both non-zero. If $z$ - prime in $\mathbb{Z}[i]$, then $N(z)$ is also prime.

I have tried to assume that $N(z) = a^2 + b^2 = n$ - composite, and then reach a contradiction, but unfourtunately had no success with it. Since $z$ - prime, writing $z = xy$ involves that either $x$ or $y$ is a unit. By multiplicativity of the norm in $\mathbb{Z}[i]$ I have $N(z) = N(x)N(y)$, and though $N(z) = N(x)$ or $N(z) = N(y)$. But from this point I don't know how to proceed and need some hint.

Would appreciate any help, thank you in advance.

答:
Based on the Lukas Heger answer and slightly modifying it I have the following proof.

We have $z=a+bi$ and it's conjugate $\overline{z}=a - bi$. Then $N(z) = a^2 + b^2 = (a+bi)(a-bi)=z \overline{z}$, thus $z \mid N(z)$. Since $z$ is prime in $\mathbb{Z}[i]$ and not a unit, then $N(z) \neq 1$ and we can factor it in primes as $N(z)=\prod_{i=1}^n p_i$, and so $z \mid p_j$ for some $j$. Let $p = p_j$, then $z \mid p$ implies that $\exists{y} \in \mathbb{Z}[i]$ such that $zy=p$. Since $p \in \mathbb{Z}$, we have $N(p) = p^2 = N(z)N(y)$. Since $z$ is not a unit, then $N(z) \neq 1$. And if $y$ is a unit, then it would contradict with the fact that for $z = a + bi$ we have both $a,b \neq 0$. So neither $z$ or $y$ are units, and so $N(z)=p$, as required.


问:
Proof of Problem 26 in Chapter 7 of Royden's Real Analysis 3ed.
There is a hint for Problem $26$ in Chapter $7$ of Royden's Real Analysis $3$ed., which states as below.

Let $O$ be an open subset of a complete metric space $(X, \rho)$. Hint: Let $\varphi(x) = [\rho(x, \tilde{O})]^{-1}$ for each $x \in O$, where $\tilde{O}$ is the complement of $O$ and $\rho(x, \tilde{O}) = \inf_{y \in \tilde{O}} \rho(x, y)$ is the distance of $x$ to $\tilde{O}$. Then $\{\langle x, y \rangle : x \in O, y = \varphi(x)\}$ is a closed, and therefore complete, subset of $X \times \mathbb{R}$.

I cannot figure out why the set $\{\langle x, y \rangle : x \in O, y = \varphi(x)\}$ is closed. Please provide your kind help, many thanks!

答:
If $\lim_{n\to\infty}\langle x_n,\varphi(x_n)\rangle= \langle x,y\rangle$ with $x\in\widetilde O$ then, by continuity of $\rho(\bullet,\widetilde O)$, we'd have $\rho(x_n,\widetilde O)\to\rho(x,\widetilde O)=0$ and, therefore, $\varphi(x_n)\to\infty$. Therefore no such sequence exists, which means $$\operatorname{cl}_{X\times\Bbb R}\{\langle x,y\rangle\,:\, x\in O\land y=\varphi(x)\}\subseteq O\times \Bbb R.$$ Therefore, $$\operatorname{cl}_{X\times\Bbb R}\{\langle x,y\rangle\,:\, x\in O\land y=\varphi(x)\}=\operatorname{cl}_{O\times\Bbb R}\{\langle x,y\rangle\,:\, x\in O\land y=\varphi(x)\}$$

But $\varphi:O\to\Bbb R$ is continuous and $\Bbb R$ is Hausdorff, therefore its graph is already closed in the product $O\times \Bbb R$.


问:
Complex polynomials converging in the compact-uniform topology
Suppose that $p_n$ are polynomials with degree $n$, with $p_n(0)=1$ and which converges uniformly on compact sets of $\mathbb{D}$ towards an analytic function $f$.
Now, suppose that the $p_n$ have no roots in a disk of radius strictly greater than 1, say - for simplicity - $D(0,2)$.

Is it true that one can extend the uc convergence to all compacts of the disk $D(0,2)$ ?  Or in any intermediary disk $D(0, r)$ for some $1<r<2$ ? What would be simple assumptions on the $p_n$ to ensure such a behaviour (on the coefficients, the roots, on $f$...) ?

Answer. As answered above by Sangchul, the answer to the first question is no. He gave a counterexample.

Edit: a possible other formulation. By taking the inverses $z \to p_n(z)^{-1}$, whose radii of convergence are greater than $2$,  and using the Hurwitz theorem, the question becomes: we have a sequence of rational functions with no poles in $D(0, 2)$ which takes the value $1$ at zero, and which converges uniformly on compact sets of $D(0, 1)$ towards an analytic function $g=f^{-1}$. The radius of convergence $R$ of $g$ is greater than or equal to $1$. Is it possible that $R=1$ ? If not, $R>1$; then, does the uniform convergence of $1/p_n$ towards $g$ holds (uniformly on compact sets) for some disk $D(0, s)$ with $1<s<R$ ?

答:
In this answer, we will construct a counter-example. First, note that it suffices to find a sequence $(q_k)_{k\geq 1}$ of polynomials satisfying:

$0 < \deg q_1 < \deg q_2 < \deg q_3 < \dots $;

$q_k(0) = 1$;

Each $q_k$ has no roots in $D(0, 2)$;

$q_k$ converges uniformly on compact subsets of $\mathbb{D}$;

$q_k(1)$ does not converge.

Then we can construct $(p_n)_{n\geq 1}$ by letting $p_n(z) = q_k(z) + \epsilon_n z^n$ for $\deg q_k \leq n < \deg q_{k+1}$ and for some $\epsilon_n \geq 0$ chosen sufficiently small.

Construction. For each $k \geq 1$, choose $q_k(z)$ as the Taylor polynomial of $e^{z+z^2+\dots+z^k}$ with degree large enough so that

$$ \sup_{z \in D(0,2)} \left| q_k(z) - e^{z+z^2+\dots+z^k} \right| < e^{-2^{k+1}} $$

as well as $\deg q_k > \deg q_{k-1}$ if $k \geq 2$. Then for $z \in D(0, 2)$,

$$ \left| q_k(z) \right| > \left| e^{z+z^2+\dots+z^k} \right| - e^{-2^{k+1}} \geq e^{-2-2^2-\dots-2^k} - e^{-2^{k+1}} > 0, $$

and so, $q_k$ has on zero on $D(0, 2)$. Moreover, since $e^{z+z^2+\dots+z^k}$ converges to $e^{z/(1-z)}$ uniformly on compact subsets of $\mathbb{D}$, the same is true for $q_k$. Finally, since $\left| q_k(1) - e^{k} \right| < e^{-2^{k+1}}$, it follows that $q_k(1)$ diverges as $k\to\infty$.


问:
finding parameter's values in PDF using median point
Let X be a continuous random variable, with PDF  $$ \begin{equation}
  f_X(x) =\begin{cases}
    0, & \text{ $x<20$}\\
    a, & \text{ $20 \le x<25$}\\
    \frac{1}{15}, & \text{ $25 \le x<30$}\\
    b, & \text{ $30 \le x<35$}\\
    c, & \text{ $35 \le x$}
  \end{cases}
\end{equation}$$
It is known that $$P\{X \le 30 \} = P\{X \ge 30 \}$$
I need to find the values of a, b and c.

From the given info that 30 is the median I know that $$\frac{1}{2} = P\{X \le 30 \} = \int_{20}^{25}adx + \int_{25}^{30}\frac{dx}{15}=5a + \frac{1}{3}\Longrightarrow a = \frac{1}{30}$$
But I don't know how to find b & c. I instinctively want to say that $c = 0$ because $$1 = \int_{20}^{25}adx + \int_{25}^{30}\frac{1}{15}dx + \int_{30}^{35}bdx +\int_{35}^{\infty}cdx$$ and $\int_{35}^{\infty}cdx$ doesn't converge for any $c\in \mathbb{R}$ and then $$\int_{30}^{35}bdx = \frac{1}{2}\Longrightarrow b = \frac{1}{10}$$ but I'm not sure that is correct.

答: Because the density is piecewise constant, all we need to do is compute the relevant areas under each interval.  You are correct that $c = 0$.  We then have $$(25 - 20)a + (30 - 25)\tfrac{1}{15} + (35 - 30)b = 1,$$ and $$(35 - 30)b = \tfrac{1}{2}.$$  This gives $b = \frac{1}{10}$ and $a = \frac{1}{30}$.


问:
Do the two r.v. generate the same σ-algebra?
Let $(X_{n})$ be a sequence of independent integrable random variables and $Y_{n}=\sum_{k=1}^{n}\prod_{j=1}^{k}X_{j}$. How to prove that $\sigma(Y_{1},\cdots Y_{n})=\sigma(X_{1},\cdots X_{n}),\forall n\in\mathbb{N}?$ Thank you for your help.

答:
The claim is not correct, but we can prove that $\sigma(Y_1, \ldots, Y_n) \subset \sigma(X_1, \ldots, X_n) $.

For $n \le N$ we have $Y_n = f(X_1, \ldots, X_n)$, hence $\sigma(Y_n) \subset \sigma(X_1, \ldots, X_n) \subset \sigma(X_1, \ldots, X_N) $. Thus $\sigma(Y_1, \ldots, Y_N) \subset \sigma(X_1, \ldots, X_N) $.

Example of situation when $\sigma(Y_1, \ldots, Y_n) \ne \sigma(X_1, \ldots, X_n) $:
$X_1 = 0$, $X_i$, $i \ge 2$ - nondegenerate i.i.d. Then $Y_n = 0 $, $\sigma(Y_1, \ldots, Y_n)$ is trivial and  $\sigma(X_1, \ldots, X_n) $ is not trivial.


问:
Further explanation needed :Finding all $z$ such that the modulus of $f(z)=e^{(z+1)/(z-1)}$ is equal to/at most $1$
I was solving a previous exam paper and there I got stuck on the following problem:

Let $f(z)=e^{\frac{z+1}{z-1}}$. Then find all $z \in \Bbb C$ for which

$|f(z)|=1$,

$|f(z)|\le 1$.

Any idea about how to tackle this problem will be highly appreciated. Thanks for your time.

答: If $w$ is any complex number, $|e^w| =  e^{\operatorname{Re}(w)}$, so you are trying to respectively find the $z$ for which $\operatorname{Re}({z +1 \over z- 1}) = 0$ and $\operatorname{Re}({z +1 \over z- 1}) \leq 0$. To figure out the details of which $z$ satisfy these, knowledge of Möbius transformations are helpful, but you can work them out by hand too.</p>


问:
Does an abelian group must have a degree one faithful representation?
I know that if a finite group $G$ has degree one faithful representation over a field $F$, then it must be abelian. But is the converse true?

I know that there is trivial representation but it is not faithful. Any help on this is appreciated. Thanks.

答:
If $G$ is cyclic then the question of whether or not there exists a 1D faithful rep amounts to if the field contains enough roots of unity - specifically, if it has a $|G|$th root of unity.

If $G$ is not cyclic then it's impossible. The fundamental theorem of finite abelian groups says $G$ is a direct sum of cyclic groups, and by Chinese Remainder Theorem we can say $G$ is not cyclic iff there are a pair of cyclic summands (with respect to some choice of direct sum decomposition) that share a factor. Thus, $G$ has more than $\phi(d)$ elements of order $d$ for some number $d$, but there can be at most $\phi(d)$ such elements in the group of units of a field (good exercise), so there can be no faithful 1D rep of $G$ over any field.


问:
Different ways of selecting $9$ cubes of various colors

How many different ways can I select $9$ cubes, with at least $3$ blue cubes? There are $10$ white cubes, $17$ blue cubes, $11$ black cubes and $14$ red cubes?

My answer:
$$\binom{12}9 = \frac{12!}{9!(12-9)!} = 220$$

Is correct?

答:
Your solution is wrong. The reason is the following: your solution basically answers the question without taking into account color and the total number of cubes, so it answers "In how many ways can we select $9$ cubes from $12$ cubes? While these cubes can be any color.

Solution:

First, you have $52$ cubes in total, this includes $17$ blue cubes $35$ non-blue cubes, to get the total number of ways you can get AT LEAST $3$ blue cubes out of $9$ you need to find the total number of ways you can get EXACTLY $3$ cubes, the total number of ways you can get EXACTLY $4$ cubes, etc... till the total numbers of ways you can get all $9$ cubes to be blue and add them together.

To get the total number of ways of getting $x$ blue cubes out of $9$ drawn you apply the following general formula:

$\binom{17}{x}*\binom{35}{9-x}$

Since you are choosing $x$ blue cubes out of $17$ and when you choose $x$ cubes you will have $9-x$ non-blue-cubes remaining as you need $9$ cubes in total, you multiply both combinations as for every way to choose $x$ blue cubes you have $\binom{35}{9-x}$ of choosing the rest.

So numbers of ways of choosing EXACTLY $3$ cubes you get:

$\binom{17}{3}*\binom{35}{6}$

You do this for $x=3,4,5,6,7,8,9$ and add them all up to get your final answer.


问:
Prove that $\sum \frac{i+1}{i!+1}<3$

Prove that $$\sum_{i=1}^{n}\frac{i+1}{i!+1}<3$$

My approach: If $k=1$ we can see that $$S_{1}=\sum_{k=1}^{1}\frac{k+1}{k!+1}=\frac{1+1}{1!+1}=\frac{2}{2}=1<3$$ Now, suppose that the statement is true for some $n\in \mathbb{N}$, it's to say suppose that $S_{n}=\sum_{k=1}^{n}\frac{k+1}{k!+1}<3$ is true, so we need to show that is true for $n+1$, it's to say we need to show that $S_{n+1}=\sum_{k=1}^{n+1}\frac{k+1}{k!+1}<3$ is also true.

Now, we can see that
\begin{eqnarray*}
S_{n+1}&=&\sum_{k=1}^{n+1}\frac{k+1}{k!+1}\\
&=&\sum_{k=1}^{n}\frac{k+1}{k!+1}+\sum_{k=n+1}^{n+1}\frac{k+1}{k!+1}
\end{eqnarray*}
but, how can I continue from here?

答:
As @Arthur points out we cannot directly use induction over the inequality. This is because the summation is an increasing function of $n$ and the upper bound is constant, so if $P(n)<3$ for some constant $n$ then $P(n+1)=P(n)+m<3+m\not<3$ as $m>0$.

We can prove something stronger.

Claim. For all $n\in\Bbb N$ we have $$\sum\limits_{i=1}^n\frac{i+1}{i!+1}<3-\frac1{n^2}.$$

Proof. Let $P(n)$ be the statement in the claim. Evidently $P(1)$ holds so we suppose that $P(k)$ holds for some $k\in\Bbb N$. Then \begin{align}\sum\limits_{i=1}^{k+1}\frac{i+1}{i!+1}=\frac{k+2}{(k+1)!+1}+\sum\limits_{i=1}^k\frac{i+1}{i!+1}<3+\frac{k+2}{(k+1)!+1}-\frac1{k^2}\end{align} so it suffices to show that $$\frac{k+2}{(k+1)!+1}-\frac1{k^2}<-\frac1{(k+1)^2}\impliedby\frac{k+2}{(k+1)!+1}<\frac{2k+1}{k^2(k+1)^2}.$$ Rearranging yields $$(k+1)!>\frac{k^2(k+1)^2(k+2)}{2k+1}-1\impliedby(k-1)!>\frac{(k+1)(k+2)}2$$ which is true for all $k\ge7$ as $(k-1)(k-2)>k+2$ and $k-3>(k+1)/2$. Checking $P(2)$ through $P(6)$ is straightforward and thus $P(n)$ holds for all $n\in\Bbb N$.

Title: Prove that $\sum \frac{i+1}{i!+1}<3$


问:
if $d'(x_1, x_2)=\min(1, d(x_1, x_2))$, is $d'$ a metric on $X$?
I'm given the problem:

If $d'(x_1, x_2)=\min(1, d(x_1, x_2))$, is $d'$ a metric on $X$?

I'm stuck in the triangular inequality. Since we have to show that

$$ d'(x_1, x_3) \leq d'(x_1, x_2) + d'(x_2, x_3) $$

there is a case in which I'm confused:

If $\min(1, d(x_1, x_3))$ is $1$, and $\min(1, d(x_1, x_2))$ and $\min(1, d(x_2, x_3))$ are $d(x_1, x_2)$ and $d(x_2, x_3)$ respectively. What if the values of $d(x_1, x_2)$ and $d(x_2, x_3)$ are less than $0.5$ and there sum isn't $1$ or greater than $1$?

I don't know if I am thinking the right way or there is some mistake in it. Please help me with this problem!

答:
You’re almost done.

Just remember that

$1=d’(x_1,x_3) \leq d(x_1,x_3) \leq d(x_1,x_2) + d(x_2,x_3)$.

It’s very important that you make sure that you have used every hypothesis in a exercice. Otherwise, something is likely missing in your proof or you’ve proved a more general statement which is unlikely.


问:
Show that the group $R^\times$ acts transitively on the set $\{s\in R| (s) = (r)\}$ for a fixed r.
Maybe I'm having some issue with notation or something, but this is for an Integral Domain R. Fix some $r\in R$. Then we want to show the group $R^\times$ acts transitively on the set $\{s\in R | (s)=(r)\}$. Where (s) is the ideal generated by s.

Really not sure what $R^\times$ means...

答:
For any (commutative) ring $R$, $R^\times$ denotes the group of units of $R$ ie. the group of all invertible elements
$$R^\times = \{r \in R \mid \exists s\in R : rs=sr=1\}$$
with group operation given by the multiplication of the ring. This group is fundamental to the theory of rings. Just for convenience let me remark that elements $x,y\in R$ are called associate if there is a unit $u\in R^\times$ with $x=uy$ or equivalently $y=ux$.

The exercise reduces to the following two observations:

associate elements generate the same principle ideal, so for fixed $r\in R$ there is a welldefined group action $R^\times \curvearrowright \{s \in R\mid (s)=(r)\}$

in an integral domain $R$, elements, which generate the same principal ideal, are associate. Thus $R^\times \curvearrowright \{s \in R\mid (s)=(r)\}$ is transitive.

The first observation is immediate from plugging in the definition of being associate. The second observation requires the cancellation rule of integral domains.

If you have questions, feel free to ask andI will elaborate. For now I think it is good to leave the details as an exercise.


答:
Battle plan (aka hint):

1.) First show by induction that $k!>2^{k-1}$ for $k\geq 3$.

2.) Then we have for $k\geq 1$

$$ \frac{k+1}{k!+1} \leq \frac{2k}{k!+1} \leq \frac{2k}{k!} = 2 \frac{1}{(k-1)!}.$$

3.) Now write for $n\geq 3$

$$ \sum_{k=1}^n \frac{k+1}{k!+1} = 1 + 1 +\sum_{k=3}^n \frac{k+1}{k!+1}
\leq 2+ \sum_{k=3} \frac{1}{(k-1)!}
<2 + \sum_{k=3}^n (1/2)^{k-2}.$$
Where we used 1.) in the last inequality to estimate by a geometric series. Try to conclude.

Title: Prove that $\sum \frac{i+1}{i!+1}<3$

Body:
Prove that $$\sum_{i=1}^{n}\frac{i+1}{i!+1}<3$$

My approach: If $k=1$ we can see that $$S_{1}=\sum_{k=1}^{1}\frac{k+1}{k!+1}=\frac{1+1}{1!+1}=\frac{2}{2}=1<3$$ Now, suppose that the statement is true for some $n\in \mathbb{N}$, it's to say suppose that $S_{n}=\sum_{k=1}^{n}\frac{k+1}{k!+1}<3$ is true, so we need to show that is true for $n+1$, it's to say we need to show that $S_{n+1}=\sum_{k=1}^{n+1}\frac{k+1}{k!+1}<3$ is also true.

Now, we can see that
\begin{eqnarray*}
S_{n+1}&=&\sum_{k=1}^{n+1}\frac{k+1}{k!+1}\\
&=&\sum_{k=1}^{n}\frac{k+1}{k!+1}+\sum_{k=n+1}^{n+1}\frac{k+1}{k!+1}
\end{eqnarray*}
but, how can I continue from here?


问:
Prove that $\vert \int_{B_a}f(x)dx \vert \leq \frac{a^{m+1}}{m^2+m}\omega _m \sup_{B_a} |\nabla f|$
Assume that $a\in \mathbb{R}^{+}, m\in\mathbb{N}^{+}, B_a \subset \mathbb{R}^m$. $f:\overline{B_a} \to \mathbb{R}$ is $C^1$, $f|_{\partial{B_a}}=0$.
Prove that
$$
\Big| \int_{B_a}f(x)dx \Big| \leq \frac{a^{m+1}}{m^2+m}\omega _m \sup_{B_a} |\nabla f|,
$$
where $B_a=\{x\in \mathbb{R}^m:|x| \leq a \}$, $\omega_m$ denotes the area of m-dimensional unit ball, i.e., $\omega_m=\int_{\partial{B_1}} 1d\sigma$. In fact the area of m-dimensional ball with radius $r$ is $\int _{|x|=r} d\sigma$, which equals  $\omega_m \cdot r^{m-1}$.

Hint from my professor: Apply the coarea formula(1), and the fact that if $\Omega \subset \mathbb{R}^m$ is an open region, $f \in C^1{(\Omega)}, [a,b] \subset \Omega$, then
$$
f(b)-f(a)=\int_0 ^1 \nabla f(a+t(b-a))\cdot (b-a)dt.(2)
$$
(1):Let $G \subset \mathbb{R}^m$ be bounded open region, $f \in C^1(G), \forall x \in G$ we have $\nabla f(x) \neq 0, \Omega = f^{-1}[a,b]$. Let $g\in C(\Omega)$, then
$$
\int _\Omega g=\int _a ^ b dt \int _{f^{-1}(t)} \frac{g}{|\nabla f|} d\sigma.
$$
My attempt: if we set $h:\mathbb{R}^m \to \mathbb{R}, x\mapsto |x|$, then $B_a=h^{-1}[0,a]$. Applying (1):
$$
\int_{B_a}f(x)dx=\int_{h^{-1}[0,a]}f(x)dx=\int _0 ^a dt \int _{|x|=t} \frac{f}{|\nabla h|} d\sigma,
$$
but note that $\nabla h(x) = x/|x|$, hence $|\nabla h|=1$.
$$
\int _0 ^a dt \int _{|x|=t} \frac{f}{|\nabla h|} d\sigma=\int _0 ^a dt \int _{|x|=t} f(x) d\sigma.
$$
Use absolute value inequality,
$$
\Big| \int_{B_a}f(x)dx \Big| = \Big| \int _0 ^a dt \int _{|x|=t} f(x) d\sigma \Big| \leq \int _0 ^a dt \Big| \int _{|x|=t} f(x) d\sigma \Big|.
$$
Then I use (2) to estimate RHS: let $p\in \partial B_{a}$, then by assumption $f(p)=0.$
$$
f(x)=f(x)-f(p)=\int_0 ^1 \nabla f(p+t(x-p))\cdot (x-p)dt, \\
|f(x)| \leq  \int_0 ^1 \Big| \nabla f(p+t(x-p))\cdot (x-p) \Big| dt \leq \sup |\nabla f| \cdot |x-p|.
$$
If $|x|=t$, then $|f(x)|\leq \sup |\nabla f| \cdot (a+t)$. Hence
$$
\int _0 ^a dt \Big| \int _{|x|=t} f(x) d\sigma \Big| \leq \sup |\nabla f|  \int _0 ^a dt (a+t) \int _{|x|=t} d\sigma = \\
 \sup |\nabla f|  \int _0 ^a dt (a+t) \omega_m \cdot t^{m-1}= \\ \frac{2m+1}{m^2+m} a^{m+1} \omega _m \sup_{B_a} |\nabla f|.
$$
And it's the best result I could obtain.

Thank you in advance.

答:
Given $x\in B_a$, $x\neq 0$, let $y_x := a x / |x|$, and let $y_0$ be any point of $\partial B_a$, so that $f(y_x) = 0$ for every $x\in B_a$.
Denoting $M := \sup_{B_a} |\nabla f|$, we have that
$$
f(x) = f(x) - f(y_x) = \int_0^1 \nabla f(y_x + t(x-y_x))\cdot (x-y_x)\, dt
$$
so that
$$
|f(x)| \leq (a- |x|) M, \qquad \forall x\in B_a.
$$
Hence,
$$
\int_{B_a} |f(x)|\, dx \leq M \int_{B_a} (a - |x|) \, dx=
M \omega_m \frac{a^{m+1}}{m(m+1)}\,.
$$


问:
General Solution of Diophantine equation
Having the equation:

$$35x+91y = 21$$

I need to find its general solution.

I know gcf $(35,91) = 7$, so I can solve $35x+917 = 7$ to find $x = -5, y = 2$. Hence a solution to $35x+91y = 21$ is $x = -15, y = 2$.

From here, however, how do I move on to finding the set of general solutions? Any help would be very much appreciated!

Cheers

答:
You really ought to have checked your particular solution, because it isn’t one:

$$35(-15)+92\cdot2=-343\;,$$

not $21$. Divide the original equation by $7$ to get $5x+13y=3$. By inspection $x=-2$, $y=1$ is a solution. Suppose that $x=-2+a$, $y=1+b$ is also a solution. Then

$$5(-2+a)+13(1+b)=3\;,$$

so

$$-10+5a+13+13b=3\;,$$

and therefore $5a+13b=0$. Thus, $b=-\frac5{13}a$. Since $a$ and $b$ must be integers, this says that $a$ must be a multiple of $13$. Say $a=13k$. What does that make $b$? Can you now write down the general solution in terms of $k$?


问:
Inner product on $\Bbb{R}^2$ defined using a matrix
Let $A = \begin{bmatrix}a & b\\c & d\end{bmatrix}$, where $a,b,c,d$ are real numbers.

For $x,y \in \Bbb{R}^2$ , let
$f_A(x,y) = y^T A x$ be an inner product on $\Bbb{R}^2$.

Show that $b = c$ , $a > 0$ , $d > 0$ , and $ad − bc > 0$.

答:
If I'm right, you want to know the conditions for it to be a inner product, right? If it is, you need to show the following: Use your muscles to "math this out"!

Linearity:

In this case, prove that
$$
f_{A}(ax,y) = a f_{A}(x,y), a \in \Bbb{R}
$$
and
$$
f_{A}(x+y,z) = f_{A}(x,z) + f_{A}(y,z).
$$

Symmetry

Show that
$$
f_{A}(x,y)=f_{A}(y,x).
$$

"Positive-definitedness"

Just show that, if $x \neq 0$, then
$$
f_{A}(x,x) > 0.
$$

While you prove this, you will be able to achieve the conditions that you want. To make it easiear, take $x = (x_{1},x_{2})$ and $y = (y_{1},y_{2})$.


问:
Finding equidimensional radical of an ideal
In the article by $\textit{Eisenbud D., Huneke C., Vasconcelos W.}$ titled as $\textit{Direct Methods for Primary Decomposition (1992, Inventiones Mathematicae,110, pg.207 - 235)}$ Proposition 2.3 is as follows:

If $J$ $\subset$ $I$ $\subset$ $S =$ $k[x_1,...x_n]$ are ideals of the same dimension, and $J$ is equidimensional with radical $J$´, then the equidimensional hull of the radical of $I$ is given by the formula:

equidimensional radical of $I = (J´:(J´:I))$.

This proposition is suggested to be proven by using last statement of Lemma 2.4 which is as follows:

If $J$ is an ideal in a Noetherian ring $R$, and $M$ is a finitely generated $R-module$, then $...$c)If $J$ is a radical ideal, then $(I:J)$ is radical and $(I:J)$ = $\bigcap P_j$ where $P_j$ ranges over all primes containing $I$ but not containing $J$.

To prove Prop.2.3, I have $(J´:(J´:I)) = (J´: \bigcap R_i) $ where $R_i$ are prime ideals such that $J´ \subset R_i$ but $I \not\subset R_i$. Now, if  $J´ \subset R_i$ then $R_i$ is one of the prime ideals containing $J$, by definition of radical ideal (which is intersection of all prime ideals containing the ideal in concern). So, since dimensions of $I$ and $J $ are equal and $J \subset J´ \subset R_i $, therefore $R_i$ should contain $I$, but lemma 2.4 above says the opposite. What do I miss?

Thanks in advance!

答: My supervisor's answer: $R_i$ need not contain $I$ since it does not have to be the maximal prime ideal chain that determines the dimension of the ideal $J$.


问:
Limit of $(1+z)(1+z/2)\cdots(1+z/n)$

Let $z$ be a complex number such that $\operatorname{Re}(z)<0$ and
$$z_n = (1+z)\Bigl(1+\frac{z}{2}\Bigr) \cdots \Bigl( 1+\frac{z}{n} \Bigr).$$
Prove that $ \lim_{n \to \infty} z_n = 0$.

I noticed that: If $| z_n |^2 \to 0$ then $z_n \to 0$.

\begin{align*}
\lvert z_n \rvert^2
&= z_n \overline{z_n} \\
&= (1+z)(1+\overline{z}) \cdot \Bigl(1+ \frac{z}{n}\Bigr) \Bigl(1+ \frac{\overline{z}}{n}\Bigr) \\
&= (1 + 2 \operatorname{Re}(z) + \lvert z \rvert^2) \cdots \biggl(1 + \frac{2\operatorname{Re}(z)}{n} + \frac{\lvert z \rvert^2}{n^2}\biggr).
\end{align*}

For a fixed $z$ there will be such $ n \in \mathbb{N}$ such that for every $m >n$  we have $\frac{2\left|\operatorname{Re}(z)\right|}{m} > \frac{\left|z\right|^2}{m^2}$ .
Which means $$ 1 + \frac{2\operatorname{Re}(z)}{m} + \frac{\lvert z \rvert^2}{m^2} < 1.$$
So from one point we will be multiplying by a factor always smaller than $1$ (but closer and closer to $1$).

Is that a proper approach to the problem? I understand it is not enough to conclude that the limit is $0$.

答:
Write

$$ z_n = \exp\biggl( z H_n + \sum_{k=1}^{n} \left[ \log\left(1+\frac{z}{k}\right) - \frac{z}{k} \right] \biggr), $$

where $H_n = \sum_{k=1}^{n} \frac{1}{k}$ is the $n$th harmonic number. Since $ \left| \log(1+w) - w \right| \leq 4\left|w\right|^2 $ for all $\left|w\right|\leq\frac{1}{2}$, we find that

$$ \sum_{k=1}^{\infty} \left[ \log\left(1+\frac{z}{k}\right) - \frac{z}{k} \right] $$

converges. This shows that

$$ \left| z_n \right| \leq e^{\operatorname{Re}(z_n)H_n + C} $$

for some finite constant $C$ whose value is not so important to us. Since $H_n \to \infty$ as $n\to\infty$, it follows that $\left|z_n\right| \to 0$.

Title: Limit of $(1+z)(1+z/2)\cdots(1+z/n)$


问:
Basic maths question
If I have 25 apples and multiply them zero times, why is it that I end up with zero? I understand zero times means no times, right? I mean, multiply by zero is no multiplication. Can anyone explain, please?</p>

答:
Ok, if we use some "powerfull" definitions in math, we may remember that the set of Integers number (a.k.a. "$\Bbb{Z}$") is an abelian grup (if you want to know more about this, I recommend to start with basic ring theory; you may find a good resume here). Rings are more "complete" in some sense that groups, but, in my opinion, is easier to have some touch with then before.

So, doesn't matter if you multiply $a \times b$ or $b \times a$. Although it looks a rule to everything, there are others groups, rings, etc. which this isn't true (e.g. the Matrix multiplications). So in this case, what you may think is:

$0$ times $25$ is the same as $25$ times $0$. So,

$$
0 + 0+ \dots + 0 = 0.
$$
(Trust me, there are $25$ $0$'s in the dots).

The $0$ itself is something complicated in math and is a object of trouble since ever, just look at division, for example. So, I hope you've understanded and, otherwise, just get in touch!

Title: Basic maths question


答:
On division by $7$ we get $5x+13y=3$

Now, by observation $13-5\cdot2=3$

So,  $5x+13y=3=13-5\cdot2$

$\implies 5(x+2)=13(1-y)\implies \frac{5(x+2)}{13}=1-y$ which is an integer

$\implies 13$ divides $5(x+2)$

$\implies 13$ divides $(x+2)$ as $(5,13)=1$

So, $x+2=13a$ where $a$ is any integer, $x=13a-2$

So, $13y=3-5x=3-5(13a-2)=13(1-5a)\implies y=1-5a$

We can use the convergent property to a continued fraction as follows:

$$\frac{13}5=2+\frac35=2+\frac1{\frac53}=2+\frac1{1+\frac23}=2+\frac1{1+\frac1{\frac32}}=2+\frac1{1+\frac1{1+\frac12}}$$

So, the previous convergent of $\frac{13}5$ is $2+\frac1{1+\frac11}=\frac52$

$\implies 13\cdot2-5\cdot5=1$

So, we can write $5x+13y=3=3(13\cdot2-5\cdot5)$

Can you take it from here?

Title: General Solution of Diophantine equation

Body:
Having the equation:

$$35x+91y = 21$$

I need to find its general solution.

I know gcf $(35,91) = 7$, so I can solve $35x+917 = 7$ to find $x = -5, y = 2$. Hence a solution to $35x+91y = 21$ is $x = -15, y = 2$.

From here, however, how do I move on to finding the set of general solutions? Any help would be very much appreciated!

Cheers


答:
$$\log \left(\prod _{k=1}^n \left(\frac{z}{k}+1\right)\right)=\sum _{k=1}^n \log \left(\frac{z}{k}+1\right)$$
as $k\to \infty$ we have $\log \left(\frac{z}{k}+1\right)\to \frac{z}{k}$

and $$\sum _{k=1}^n \frac{z}{k}=z H_n$$
$H_n$ being the $n$th harmonic number.

$\text{Re}(z)<0$ therefore $\text{Re}(z H_n)\to-\infty$ and so $\text{Re}\left(\prod _{k=1}^n \left(\frac{z}{k}+1\right)\right)\to 0$ as $n\to\infty$

Hope this can be useful

Title: Limit of $(1+z)(1+z/2)\cdots(1+z/n)$

Body:
Let $z$ be a complex number such that $\operatorname{Re}(z)<0$ and
$$z_n = (1+z)\Bigl(1+\frac{z}{2}\Bigr) \cdots \Bigl( 1+\frac{z}{n} \Bigr).$$
Prove that $ \lim_{n \to \infty} z_n = 0$.

I noticed that: If $| z_n |^2 \to 0$ then $z_n \to 0$.

\begin{align*}
\lvert z_n \rvert^2
&= z_n \overline{z_n} \\
&= (1+z)(1+\overline{z}) \cdot \Bigl(1+ \frac{z}{n}\Bigr) \Bigl(1+ \frac{\overline{z}}{n}\Bigr) \\
&= (1 + 2 \operatorname{Re}(z) + \lvert z \rvert^2) \cdots \biggl(1 + \frac{2\operatorname{Re}(z)}{n} + \frac{\lvert z \rvert^2}{n^2}\biggr).
\end{align*}

For a fixed $z$ there will be such $ n \in \mathbb{N}$ such that for every $m >n$  we have $\frac{2\left|\operatorname{Re}(z)\right|}{m} > \frac{\left|z\right|^2}{m^2}$ .
Which means $$ 1 + \frac{2\operatorname{Re}(z)}{m} + \frac{\lvert z \rvert^2}{m^2} < 1.$$
So from one point we will be multiplying by a factor always smaller than $1$ (but closer and closer to $1$).

Is that a proper approach to the problem? I understand it is not enough to conclude that the limit is $0$.


问:
Homology of $S^1 \times (S^1 \vee S^1)$
I'm trying to solve question 2.2.9(b) in Hatcher's Algebraic Topology.

Question: Calculate the homology groups of $X=S^1 \times (S^1 \vee S^1)$.

My attempt: I try to use the Mayer-Vietoris sequence.

Let $A=S^1 \times (S^1 \vee \text{small bit}) =S^1\times S^1$ and $B=S^1 \times ( \text{small bit} \vee S^1)=S^1 \times S^1$. (Unsure how to express this, but hopefully this is clear enough.)

Then $A\cap B=S^1\times \{\text{point}\}=S^1$ and $A\cup B=X$.

The Mayer-Vietoris sequence then gives us an exact sequence in reduced homology as follows:

$$0 \to \mathbb{Z}^2 \to \tilde{H}_2(X) \to \mathbb{Z} \to \mathbb{Z}^2 \to \tilde{H}_1(X) \to 0.$$

Is it possible from this exact sequence to determine the homology groups, in some algebraic fashion? Or do I need to determine what the maps are? How do I determine the maps?

答:
This solution relies on the long exact reduced homology sequence of a NDR pair (Hatcher's Theorem $2.13$), and provides different approach to the problem from @tsho's solution.

Let us call $$\underbrace{\Huge{\mathsf x} \normalsize\times S^1}_{A}~~\subset~~
\underbrace{\Huge{\propto}\normalsize\times S^1}_{B}~~\subset~~
\underbrace{\Huge{\infty}\normalsize \times S^1}_{X}$$ All three pairs $(B,A),~(X,A),~(X,B)$ are Neighborhood Deformation Retracts, as Hatcher puts it, "good pairs". Also, it is obvious that $A$ is homotopy equivalent to the circle, and $B$ to the torus. Let us write the long exact reduced homology sequence for the the good pairs $(X,A)$ and $(X,B)$ : the morphism of pair given by the inclusion $(X,A)\hookrightarrow (X,B)$ gives following commutative diagram
$$\begin{array}{c}0\to &0&\to&\tilde{H}_2(X)&\to&\tilde{H}_2(X/A)&\to&\tilde{H}_1(A)&\to&\tilde{H}_1(X)&\to &\tilde{H}_1(X/A)&\to 0\\
&\downarrow&&\Vert&&\downarrow&&\downarrow&&\Vert&&\downarrow\\
0\to& \tilde{H}_2(B)&\to&\tilde{H}_2(X)&\to&\tilde{H}_2(X/B)&\to&\tilde{H}_1(B)&\to&\tilde{H}_1(X)&\to &\tilde{H}_1(X/B)&\to 0
\end{array}$$
Now $X/A$ is the wedge sum of two pinched spheres $P$ (the space studied in the previous question), and $X/B$ is simply a pinched sphere, and it follows that $\tilde{H}_*(X/B)\simeq\tilde H_*(P)$ and $\tilde{H}_*(X/A)\simeq\tilde H_*(P)\bigoplus \tilde H_*(P)$ where the isomorphism is given by the map $(i_*^+,i_*^-)$ where $i^+$ (resp. $i^-$) are the inclusions of $P$ as the upper (resp. lower) pinched sphere in $X/A$.

Since a pinched sphere is homotopy equivalent to a sphere with a diameter attached to it, which in turn is homotopy equivalent to the wedge sum of a sphere and a circle, we have $\tilde H_*(P)\simeq \Bbb Z\oplus\Bbb Z$ concentrated in degree $1$ and $2$. We can now replace the above diagram by the following simpler one

$$\begin{array}{c}0\to &0&\to&\tilde{H}_2(X)&\to&\Bbb Z\oplus \Bbb Z &\stackrel{\gamma}{\to}&\Bbb Z &\to&\tilde{H}_1(X)&\to & \Bbb Z\oplus \Bbb Z &\to 0\\
&\downarrow&&\Vert&&~~\downarrow\beta&&\downarrow&&\Vert&&\downarrow\\
0\to&  \Bbb Z &\to&\tilde{H}_2(X)&\stackrel{\alpha}{\to}& \Bbb Z &\to& \Bbb Z\oplus \Bbb Z &\to&\tilde{H}_1(X)&\to & \Bbb Z &\to 0
\end{array}$$

From the left side of this diagram, it follows that $\tilde H_2(X)$ is a subgroup of $\Bbb Z\oplus \Bbb Z$ containing a copy of $\Bbb Z$, so $\tilde H_2(X)\simeq\Bbb Z$ or $\Bbb Z\oplus\Bbb Z$. Let us assume $\tilde H_2(X)\simeq \Bbb Z$ and try to derive a contradiction.

Since $\Bbb Z$ is torsion free, we must have $\alpha=0$. The vertical map $\beta$ is onto as it corresponds to collapsing the lower copy of $P$ inside $P\vee P\simeq X/A$ to a point, and thus $\beta$ is the projection onto the first factor. The commutativity of the diagram then forces the image of $\tilde H_2(X)$ to lie inside $\Bbb Z\oplus 0\subset \Bbb Z\oplus \Bbb Z$. <s>However, there is an obvious self-homeomorphism of $X$ interchanging the upper and lower toruses of $X$ which passes to the quotient, and permutes the two factors $\Bbb Z \oplus \Bbb Z=\tilde H_2(X/A)$ (and possibly adds a sign). Thus, the image of $\tilde H_2(X)$ inside $\Bbb Z \oplus \Bbb Z$ is contained in $\Bbb Z\oplus 0\cap 0\oplus \Bbb Z=0$, but this contradicts the injectivity of the top left arrow</s>. The same argument works when we replace $B$ with $B'=T(B)$ where $T$ is the self-homeomorphism of $X$ that interchanges the two circles in the wedge sum $S^1\vee S^1$. The new map $\beta'$ is the projection onto the second factor, so the map
$\tilde H_2(X)\to \Bbb Z\oplus\Bbb Z$ sends $\tilde H_2(X)$ into $\ker(\beta)\cap\ker(\beta')=\Bbb Z\oplus 0\cap 0\oplus \Bbb Z=0$ contradicting injectivity.

As a consequence, $$\tilde H_2(X)\simeq \Bbb Z\oplus\Bbb Z$$

To finish the proof, we note that by the standard theory of finitely generated abelian groups, the quotient of $\Bbb Z\oplus\Bbb Z$ by a subgroup $S$ isomorphic to $\Bbb Z\oplus\Bbb Z$ is the product of two cyclic groups, and cannot be a subgroup of $\Bbb Z$ unless the subgroup $S$ is all of $\Bbb Z\oplus\Bbb Z$. This forces the top left arrow $\tilde{H}_2(X)\to\Bbb Z\oplus \Bbb Z $ to be an isomorphism, and $\gamma=0$. The top sequence then degenerates to a short exact sequence
$$0\to\Bbb Z \to\tilde{H}_1(X)\to \Bbb Z\oplus \Bbb Z \to 0$$

Consequently, $$\tilde{H}_1(X)\simeq \Bbb Z\oplus\Bbb Z\oplus\Bbb Z$$

Title: Homology of $S^1 \times (S^1 \vee S^1)$


问:
How to find coefficient of $s^{12}t^{13}$ in the expansion of $(2s-3t)^{25}$
I am trying to find the coefficient of $s^{12}t^{13}$ in the expansion of $(2s-3t)^{25}$.

I don't know if I am doing right, but I know we have to start with $\binom{25}{12}$. When $t^{13}$ is added, becomes more confusing. Any help in handling this problem?

Body:
The general term is given by $${{25}\choose r}(2s)^r (-3t)^{25-r}$$

You need $r=12$: $${{25}\choose {12}} (2s)^{12} (-3t)^{13} = -{{25}\choose{12}}\cdot 2^{12}\cdot 3^{13} s^{12}t^{13}$$


问:
Convergence of a series of averages
Suppose, we have a sequence defined like $a_1=10,a_2=20,a_n=\sum\limits_{i=1}^{n-1}\frac{a_i}{n-1}+10\quad n\ge3$. If we restrict the maximum value of $a_i$ to be $100$, what is the maximum value sequence reach to?

Seeing the first few terms of the sequence ($10, 20, 25, 28.33,$), I guess that the sequence is slowly diverging and the sequence may well reach $99$.. Am I right? How do we solve such kind of recurrences. Will the recursion tree method help here? Any hints? Thanks beforehand.

答:
Notice that:

\begin{align}a_{n+1} &= \sum_{i=1}^n\frac {a_i}{n}+10
\\&=\frac1n a_n + \frac{n-1}n\sum_{i=1}^{n-1}\frac {a_i}{n-1} + 10
\\&=\frac {a_n}{n} + \frac{n-1}n(a_n-10)+10
\\&=\left(\frac 1n + \frac {n-1}n\right) a_n + 10\left(1-\frac {n-1}n\right)
\\&=a_n+\frac {10}n
\end{align}

so the rate of increase decreases sharply, but not sharp enough, since the harmonic series diverges.

Title: Convergence of a series of averages


问:
Integrating a second-order partial differential equation w/ boundary conditions
I haven't dealt with solving many partial differential equations before. How would I go about solving:

$$
A = \frac{\partial^{2} v}{\partial x^{2}} + \frac{\partial^{2} v}{\partial y^{2}}
$$

where A is a constant and v is a function of velocity.

The boundary conditions follow an equilateral triangular shape. Here is an image of the triangle. All of the vertices of the triangle have a velocity of zero:

$$
v(-2a,0) = v(a,a\sqrt{3}) = v(a,-a\sqrt{3}) = 0
$$

And all of the edges of the triangle have a velocity of zero:

$$
v(x,\frac{x+2a}{\sqrt{3}})=v(x,-\frac{x+2a}{\sqrt{3}})=0
$$

Ideally, I could get an equation of  v(x,y)  but I am not sure if that's possible. If there isn't a "pretty" solution to this. How could I go about approximating the solution?

Thanks

答:
By inspection, the linear form of boundary conditions draw us to try a polynomial equation as solution of the PDE. For example one can try the simplest form :
$$v=c_1x^2+c_2y^2+c_3xy+c_4x+c_5y+c_6$$
Putting it into the PDE and into the conditions leads to the system of equations :
$$\begin{cases}
2c_1+2c_2=A \\
\frac13 c_3+c1+\frac13 c_2\sqrt{3}=0\\
\frac23 c_2a\sqrt{3}+\frac43 c_3a+\frac13 c_5\sqrt{3}+c4=0\\
\frac43 c_3a^2+\frac23 c_5a\sqrt{3}+c_6=0\\
\frac13 c_3+c1-\frac13 c_2\sqrt{3}=0\\
-\frac23 c_2a\sqrt{3}+\frac43 c_3a-\frac13 c_5\sqrt{3}+c4=0\\
\frac43 c_3a^2-\frac23 c_5a\sqrt{3}+c_6=0\\
\end{cases}$$
Solving the system leads to :
$$\begin{cases}
c_1=-\frac14 A \\
c_2=0 \\
c_3=\frac34 A \\
c_4=-Aa \\
c5=0 \\
c_6=-Aa^2 \\
\end{cases}$$
Putting them into the second order polynomial and after simplification :
$$\boxed{v(x,y)=-\frac{1}{4}Ax^2+\frac{3}{4}Ay^2-aAx-aA}$$

The above method is much simpler than using the general solution (See : Integrating a second-order partial differential equation )
$$v(x,y)=\frac{A}{2}x^2+f(x+iy)+g(x+iy)$$
The determination of the functions $f$ and $g$ in order to satisfy the boundary conditions is a boring calculus leading to :
$$f(Z)=g(Z)=-\frac{3A}{8}Z^2-\frac{aA}{2}Z-\frac{aA}{2}$$
With $Z=x+iy$ and $Z=x-iy$ respectively the solution $v(x,y)$ is the same as above.


问:
Determine the area of the ellipse $A=\{(x,y) \in \mathbf{R}^² \mid 3x^2+4y^2\leqslant12 \}$

Determine the area of the ellipse $A=\{(x,y) \in \mathbf{R}^² \mid 3x^2+4y^2\leqslant12 \}$

I tried to use polar coordinates here, but couldn't get it to work.

If I have $x=r\cos(\theta)$ and $y=r\sin(\theta)$ I get that $$3r^2\cos^2(\theta)+4r^2\sin^2(\theta)\leqslant 12 \Longrightarrow \frac{r^2\cos^2(\theta)}{4}+\frac{r^2\sin^2(\theta)}{3} \leqslant 1$$

but I cannot get anywhere from here. Similarly, if I don't do the polar conversion I have that $$\frac{x^2}{4}+\frac{y^2}{3} \leqslant 1$$

but this also doens't seem to get me any furhter. What should I do here?

答:
Hint

Doing the substitution $(x,y)=(2r\cos(\theta ),\sqrt{3}r\sin(\theta ))$ yields
$$\text{Area}(A)=\iint_A\,\mathrm d x\,\mathrm d y=\int_0^{2\pi}\int_0^1 \sqrt{12}\,r\,\mathrm d r\,\mathrm d \theta =...$$


问:
A basic question on the definition of $E[X]$
My question is regarding the definition of $E[X]$ in a probability book. It starts with the definition in case of a simple random variable (a random variable which takes only finite number of values) where $E[X]$ is defined by :

$$E[X]=\sum_{i=1}^{n}x_i P(X=x_i)$$

Now a 3-dimensional figure is drawn (I am confused about how to describe the figure in words) and it is said that from that figure it can be understood that the previous sum can be viewed as an integral over $\Omega$. With this motivation the following equation is written :

$$E[X] = \int _{\Omega}X(\omega)P(d\omega)$$

I am confused about the whole method of defining $E[X]$.

Actually, my question was why a "3 dimensional figure" is needed to describe expectation ? The finite sum I have mentioned can be thought of an uncountable sum $X(\omega)P(d\omega)$ over all $\omega$. Now, if our probability measure is Lebesgue measure, then the  uncountable sum is equivalent to $X(\omega)d\omega$ over all $\omega$ which is nothing but $\int_{\Omega}X(\omega)d\omega$ which is equal to $\int_{\Omega}X(\omega)P(d\omega)$

答:
In the first case the probabilities are assigned as discrete values.

In the second case the probability of an event (outcome) is determined by a probability measure $P(d\omega)$, which can be a quite general construction.  Often it amounts to ordinary integration with respect to a (nonnegative) weight called a probability density function $p(\omega)$, so that the integral can be written:

$$ E[X] = \int_\Omega X(\omega) p(\omega) d\omega $$

where $\omega$ represents a point in the "event space" $\Omega$.

In a first year graduate course in real analysis one is typically taught the theory of integration at a level that allows discrete and continuous measures (and probablity measures in particular) to be combined.


问:
Do the two r.v. generate the same σ-algebra?
Let $(X_{n})$ be a sequence of independent integrable random variables and $Y_{n}=\sum_{k=1}^{n}\prod_{j=1}^{k}X_{j}$. How to prove that $\sigma(Y_{1},\cdots Y_{n})=\sigma(X_{1},\cdots X_{n}),\forall n\in\mathbb{N}?$ Thank you for your help.

答: Counter-example: Let $X_1=0$. Then $Y_n=0%$ for all $n$. So $\sigma (Y_1,Y_1,...,Y_n)=\{\emptyset, \Omega\}$.  But $\sigma (X_1,X_2,...,X_n)$ is non-trivial for any i.i.d sequence $(X_n)_{n\geq 2}$ of non-degenerate random variables.


问:
Solve differential equation $y=(y')^2\cdot e^{y'}$
Solve differential equation $y=(y')^2\cdot e^{y'}$.

First I substitute $\frac{\mathrm{d} y}{\mathrm{d} x}=p$  and    get $y=(p)^2\cdot e^{p}$.

Differentiate both sides by $x$.

$p=(2p+p^2)\cdot e^p\cdot \frac{\mathrm{d} p}{\mathrm{d} x}$

$dx=(p+2)\cdot e^p\cdot dp$

$x=e^p \cdot (p+1)$

What to do with that?

答:
Hint:

You have got to $$x=e^{y’}(y’+1) $$ Now notice from the original DE, we have $e^{y’}=\frac{y}{(y’)^2} $. Substitute it in to get $$x(y’)^2=yy’+y $$

$$x(y’)^2-yy’-y=0  \\ y’=\frac{y\pm \sqrt{y^2+4xy}}{2x} \\ 2y’=\frac yx \pm \sqrt{\left(\frac yx\right)^2 +4\frac yx}$$ Now substitute in $y=tx$.


问:
Proving that $\lim_{h \to 0}\frac{e^h-1}{h}=1$
I understand that this question has been asked many times, but I'm wondering if the approach I am using can be made rigorous. Say we define $e$ as $\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n$. In general, $a^b$ is defined through continuity. How can we prove that $\frac{d}{dx}(e^x)=e^x$? Here is what I have tried so far:
\begin{align}
\frac{d}{dx}(e^x) &= \lim_{h \to 0}\frac{e^xe^h-e^x}{h} \\
&= e^x \lim_{h \to 0}\frac{e^h-1}{h} \, .
\end{align}
Then, to evaluate $\lim_{h \to 0}\frac{e^h-1}{h}$ we make the substitution $h=1/n$:
\begin{align}
\lim_{h \to 0}\frac{e^h-1}{h} &= \lim_{h \to 0}\frac{\left(\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n\right)^h-1}{h} \\
&= \lim_{h \to 0}\frac{\left(\lim_{h \to 0}\left(1+h\right)^{1/h}\right)^h-1}{h} \\
&= \lim_{h \to 0}\frac{(1+h)-1}{h} \\
&= 1 \, .
\end{align}
Unfortunately, I feel like some of the steps used here were a little hand-wavey. I'm not sure the way I 'absorbed' one of the limits into  the other is actually justified, and nor do I see a way of proving that
$$
\left(\lim_{h \to 0}(1+h)^{1/h}\right)^h=\lim_{h \to 0}(1+h)^1=1 \, .
$$
So was the method I used rigorous, and if not, can it be amended?

答: This is just $e'(0)$, by the definition of the derivative.


答:
This is a common problem in elementary probability. What exactly is the definition of $E[X]$? For discrete random variables you see one thing, for continuous ones you see another. What if I am working in a problem where I have both discrete and continuous random variables at the same time? Or what if my random variables is neither discrete nor continuous? Then how do I calculate $E[X]$?

Well, the reason for this lack of definition is because the general definition requires some technical machinery from measure theory. In particular for a random variable $X:\Omega \to \mathbb {R}$  the definition is
$$
E[X] := \int_\Omega X(\omega)d\mathbb{P}(\omega)
$$
when the above Lebesgue integral with respect to the probability measure $\mathbb{P}$ exists.

Title: A basic question on the definition of $E[X]$

Body:
My question is regarding the definition of $E[X]$ in a probability book. It starts with the definition in case of a simple random variable (a random variable which takes only finite number of values) where $E[X]$ is defined by :

$$E[X]=\sum_{i=1}^{n}x_i P(X=x_i)$$

Now a 3-dimensional figure is drawn (I am confused about how to describe the figure in words) and it is said that from that figure it can be understood that the previous sum can be viewed as an integral over $\Omega$. With this motivation the following equation is written :

$$E[X] = \int _{\Omega}X(\omega)P(d\omega)$$

I am confused about the whole method of defining $E[X]$.

Actually, my question was why a "3 dimensional figure" is needed to describe expectation ? The finite sum I have mentioned can be thought of an uncountable sum $X(\omega)P(d\omega)$ over all $\omega$. Now, if our probability measure is Lebesgue measure, then the  uncountable sum is equivalent to $X(\omega)d\omega$ over all $\omega$ which is nothing but $\int_{\Omega}X(\omega)d\omega$ which is equal to $\int_{\Omega}X(\omega)P(d\omega)$


答:
Short answer: define $e$ another way first, then prove the definitions' equivalence.

Long answer, very carefully worded to avoid
circularity:

For $a>0$, define $a^x$ in the obvious way for $x\in\Bbb Q$, then impose continuity. You can similarly show $\frac{d}{dx}a^x=a^x\ln a$ with $\ln a:=\lim_{h\to0}\frac{a^h-1}{h}$. Since $(a^b)^x=a^{bx}$, by the chain rule $\ln(a^b)=b\ln a$.

Provided $a\ne1$, we can take $e:=a^{1/\ln a}$ so $\ln e=1$, and $\frac{d}{dx}e^{kx}=ke^{kx}$ for all $k\in\Bbb R$. Since $e$ is therefore unique, $\ln a:=\log_ea$.

If $y^\prime=y$ then $(y/e^x)^\prime=\frac{(y^\prime-y)e^x}{e^{2x}}=0$ so $y\propto e^x.$ But$$\frac{d}{dx}\lim_{n\to\infty}(1+x/n)^n=\lim_{n\to\infty}\frac{d}{dx}(1+x/n)^n=\lim_{n\to\infty}(1+x/n)^{n-1}=\lim_{n\to\infty}(1+x/n)^n$$(the first $=$ exchanges two limits, which I think you can justify with this), so $\lim_{n\to\infty}(1+x/n)^n=e^x$ (the proportionality constant follows from the case $x-0$). Now take $x=1$.

Title: Proving that $\lim_{h \to 0}\frac{e^h-1}{h}=1$

Body:
I understand that this question has been asked many times, but I'm wondering if the approach I am using can be made rigorous. Say we define $e$ as $\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n$. In general, $a^b$ is defined through continuity. How can we prove that $\frac{d}{dx}(e^x)=e^x$? Here is what I have tried so far:
\begin{align}
\frac{d}{dx}(e^x) &= \lim_{h \to 0}\frac{e^xe^h-e^x}{h} \\
&= e^x \lim_{h \to 0}\frac{e^h-1}{h} \, .
\end{align}
Then, to evaluate $\lim_{h \to 0}\frac{e^h-1}{h}$ we make the substitution $h=1/n$:
\begin{align}
\lim_{h \to 0}\frac{e^h-1}{h} &= \lim_{h \to 0}\frac{\left(\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n\right)^h-1}{h} \\
&= \lim_{h \to 0}\frac{\left(\lim_{h \to 0}\left(1+h\right)^{1/h}\right)^h-1}{h} \\
&= \lim_{h \to 0}\frac{(1+h)-1}{h} \\
&= 1 \, .
\end{align}
Unfortunately, I feel like some of the steps used here were a little hand-wavey. I'm not sure the way I 'absorbed' one of the limits into  the other is actually justified, and nor do I see a way of proving that
$$
\left(\lim_{h \to 0}(1+h)^{1/h}\right)^h=\lim_{h \to 0}(1+h)^1=1 \, .
$$
So was the method I used rigorous, and if not, can it be amended?


问:
Fermat Last Theorem for non Integer Exponents
We now that Fermat's last theorem is true so there are not positive integer solutions to
$$x^n+y^n=z^n$$
for $n\in\mathbb{N}$ and $n>2$.

But what about if $n\in\mathbb{R}$ or $n\in\mathbb{R}^+$?

答: Suppose $z> \max(x,y)$  then $x^0+y^0 = 2 > z^0$ but there exists some $N$ such that $x^N+y^N<z^N$.  Therefore there exists some $n\in[0,N]$ satisfying $x^n+y^n=z^n$.</p>


答:
You should note that $X(1)$, $X(2)$, and $X(5)$ are not independent random variables, since they count Poisson point-events occurring in overlapping intervals.

However, $X(1)$, $X(2){-}X(1)$, and $X(5){-}X(2)$, are independent.   Further, they are integer-valued, being Poisson distributed with rate parameters $\lambda, \lambda,$ and $3\lambda$, respectively.

To save type-space, let us define $X(s,t)$ as $X(t){-}X(s)$.

$$~~~X(1)\sim\mathcal P(\lambda)\\X(1,2)\sim\mathcal P(\lambda)\\~~X(2,5)\sim\mathcal P(3\lambda)$$

The event of $\{X(1){+}X(2){=}3\}$ is thus the event that $\{2X(1){+}X(1,2){=}3\}$ and as such, this event is :

$$\{X(1){=}0, X(1,2){=}3\}{\,\cup\,}\{X(1){=}1, X(1,2){=}1\}$$

Likewise the intersection of this event and $\{X(5){=}3\}$ will be :

$$\{X(1){=}0, X(1,2){=}3,X(2,5){=}0\}{\,\cup\,}\{X(1){=}1,X(1,2){=}1,X(2,5){=}1\}$$

Then you can now evaluate $\mathsf P(X(5){=}3\mid X(1){+}X(2){=}3)$ using Bayes' rule.

Similar reasoning will let you evaluate $\mathsf P(X(5){−}X(3){=}1\mid X(4){−}X(2){=}1)$, using the independent random variables $X(2,3)$, $X(3,4)$, and $X(4,5)$.

Title: Poisson process - calculating probabilities

Body:
Let $\{X(t), t \geq 0 \}$ be a Poisson process with  $\lambda$ as parameter. Calculate:

$\mathbb{P}(X(5)=3 | X(1)+X(2)=3)$

$\mathbb{P}(X(5)-X(3)=1 | X(4)-X(2)=1)$

I did the first one as follows:
$$
\mathbb{P}(X(5)=3 | X(1)+X(2)=3) = \\ = \frac{\mathbb{P}(X(5)=3 , X(1)+X(2)=3)}{\mathbb{P}(X(1)+X(2)=3)} = \\ = \frac{\mathbb{P}(X(5) - (X(1)+X(2))=3 - 3 , X(1)+X(2)=3)}{\mathbb{P}(X(1)+X(2)=3)} = \\ =  \frac{\mathbb{P}(X(5 - (1+2))=0)\mathbb{P}(X(1)+X(2)=3)}{\mathbb{P} =(X(1)+X(2)=3)} = \\ = \mathbb{P}(X(2)=0)  = e^{-2\lambda}
$$
Is this correct?

What about the second one? I got kinda stuck there:
$$
\mathbb{P}(X(5)-X(3)=1 | X(4)-X(2)=1) = \frac{\mathbb{P}(X(5)-X(3)=1 , X(4)-X(2)=1)}{\mathbb{P}(X(4)-X(2)=1)} = \\
= \frac{\mathbb{P}(X(5)-X(3) - (X(4)-X(2))=1 - 1 , X(4)-X(2)=1)}{\mathbb{P}(X(4)-X(2)=1)} = \\ =
\frac{\mathbb{P}(X((5-3) - (4-2))=1 - 1 , X(4)-X(2)=1)}{\mathbb{P}(X(4)-X(2)=1)} = \frac{\mathbb{P}(X(0)=0)\mathbb{P}( X(4)-X(2)=1)}{\mathbb{P}(X(4)-X(2)=1)} = \mathbb{P}(X(0)=0) = 1
$$
This seems fishy. I appreciate any help or affirmation that it is actually correct.


问:
Unicity of the equidistant set from the boundary of an open set
Suppose $F$ is a closed set in $\mathbb{R}^n$ and let $V$ be a bounded open set such that if $x\in \partial V$ is a boundary point of $V$ then dist$(x,F)=a$ ($a>0$ constant). We also define a set $A$ to be the set of points $x$ such that dist$(x, V)<a$. We know that in general $\partial A$ and $F$ can be disjoint (see below), but what if we impose that $F$ be in the complement of $V$ ($A$ is an open set that contains $V$ and $\partial A$ is in the complement of $V$)? Can we say that $\partial A\subset F$?

As an example, suppose $n=1$, $F=[-1,1]$, $V=(-2,2)$ and $A=(-3,3)$. Here $\partial A$ and $F$ are disjoint, but if we move $F$ to the complement of $V$ (for example by setting $F=[-4,-3]\cup [3,4]$), then $\partial A=\partial F$.

答:
The answer I believe is no. Let $V=(0,1)\times(0,1)$, let $F$ be the
square with corners $(-1,-1),\ (2,-1),\ (2,2)$, and $(-1,2)$. The
minimum distance between $\partial V$ and $F$ is $a=1$. However,
$\partial A$ contains the circular arcs centered at
$(0,0),\ (1,0),\ (1,1)$, and $(0,1)$ which connect $(-1,0)$ to
$(0,-1)$, $(-1,1)$ to $(2,0)$, $(2,1)$ to $(1,2)$, and $(0,2)$ to
$(-1,-1)$. Thus, $\partial A\not\subset F$.


问:
Finding maximum value of perimeter of trapezium using tan θ
I study maths as a hobby. I am struggling with this problem.

In a trapezium ABCD the sides AB, DC are parallel and $\angle ABC$ is a right angle. If AD = DB = 10 cm and $\angle DAB = \theta$ find the value of $\tan \theta$ for which the perimeter p cm is a maximum. Hence find the range of values taken by p for $0\lt\theta\lt\frac{1}{2}\pi$

My working is as follows:

$\tan \theta = \frac{DE}{EA} = \frac{CB}{EA}$

$CB = EA.\tan \theta$

$EA = 10 \cos \theta \to CB = 10 \cos \theta \tan \theta$

$ABD$ is isosceles so $BE = EA$

$CD = EA = 10 \cos \theta$

$BA = 20 \cos \theta$

So perimeter P = $20 \cos \theta + 10 \cos \theta \tan \theta + 10 \cos \theta + 10$

$= 30 \cos \theta + 10 cos \theta \tan \theta + 10$

$\frac{dP}{d\theta} = - 30 \sin \theta + 10 \cos \theta \sec^2 \theta - 10 \sin \theta \tan \theta$

But now I am stuck.

答: $$P=30\cos\theta+10\sin\theta+10 \\ P’ = -30\sin\theta+10\cos\theta = 0 \implies \tan\theta =\frac 13$$ You may use $$\sin\theta= \frac{\tan\theta}{\sqrt{1+\tan^2\theta}} \\ \cos\theta = \frac{1}{\sqrt{1+\tan^2\theta}} $$


问:
Bounded Symmetric Matrix
Assume $H \in \mathbb{R}_{\mathrm{sym}}^{n \times n}$ with $\left\|H\right\| \leq C_{H}$ with some constant $C_{H}>0$.

What can I say about
$\sup_{d \in B_{\Delta}(0)} \frac{1}{2} d^T H d$?

Does $\sup_{d \in B_{\Delta}(0)} \frac{1}{2} d^THd = \frac{1}{2} C_H \Delta^2$ hold?

Many thanks in advance.

答:
Since $H$ is symmetric, it has a full orthogonal complement of eigenvectors and so $$d^\top Hd=\sum_{i=1}^n\lambda_id_i^2$$
$$\therefore\ |d^\top Hd|\le|\lambda_{max}|\sum_id_i^2=C_H\|d\|^2$$
$$\therefore\ \max_{d\in B_\Delta(0)}|d^\top Hd|=C_H\Delta^2$$
where $C_H=|\lambda_{max}|$ and the maximum is achieved by the eigenvector of $\lambda_{max}$.

However, whether this is equal to $\sup_{d\in B_\Delta(0)}d^\top Hd$ depends on whether the right-most eigenvalue $\lambda_n$ or the left-most eigenvalue $\lambda_1$ has the larger absolute value.

For example, $H=\begin{pmatrix}1&0\\0&-2\end{pmatrix}$ satisfies the relation $$\sup_{d\in B_\Delta(0)}|d^\top Hd|=2=\|C_H\|\|d\|>1=\sup_{d\in B_\Delta(0)}d^\top Hd$$

Title: Bounded Symmetric Matrix


问:
What is the definite integral of...
$$\int^L_{-L} x \sin(\frac{\pi nx}{L})$$

I've seen something like this in Fourier theory, but I'm still not sure how to approach this integral. Wolfram Alpha gives me the answer, but no method. Integrate by parts? Substitution?

答:
For what it's worth, WolframAlpha will show you the steps to evaluate the indefinite integral associated with this definite integral.  It's then just a matter of plugging in the limits and subtracting.

Here are the steps returned by WolframAlpha via the Mathematica interface:

Title: What is the definite integral of...


问:
Non isomorphic graph and spectrum of a adjacency matrix
Are two non isomorphic graphs with the same spectrum of adjacency matrix possible?</p>

答:
It is very likely possible. They called that kind of graph "isospectral" or "cospectral".

Check the Table1 in the following paper.

https://arxiv.org/pdf/1008.3646.pdf

It says there are total 274668 non-isomorphic graphs that have 9 vertices. But 51039 of them have the same eigenvalues according to Adjacency. Btw, if you use normalized laplacian instead of Adjacency, there are fewer (just 1092 pairs) cospectral graphs.


问:
Is this set of integrals verified by infinitely many functions?
We have two bounded continuous functions $f_{1}(x,y)$ and $f_{2}(x,y)$, defined on $\mathcal{A}=[a,b]\times [c,d]$ and $\mathcal{B}=[e,f]\times [g,h]$ respectively, with no intersection between both sets. In addition, $f_{1}(x,y)$ and $f_{2}(x,y)$. Ley $r$ and $s$ be positive constants, $x'$ and $y'$ are variables that verify $x'>a$ and $y'<f$. Let $f_{1}(x)=\int_{c}^{d}f_{1}(x,y)dy$ and $f_{2}(y)=\int_{e}^{f}f_{2}(x,y)dx$. The functions verify the following integrals:
$$\int_{\mathcal{A}}f_{1}(x,y)dx dy=r$$
$$\int_{\mathcal{B}}f_{2}(x,y)dx dy=\frac{r}{s}$$
$$\int_{a}^{x'}f_{1}(x)\leq s\cdot \int_{y'}^{h}f_{2}(y)$$

Then, could we say that there are infinitely many functions $f_{1}(x,y)$ and $f_{2}(x,y)$ that verify these equations. I do not need a proof for this, I just want to confirm my intuition

答: We seek solutions of$$\int_{[a,\,b]\times[c,\,d]}f_1dxdy=\int_{[e,\,f]\times[g,\,h]}sf_2dxdy=r,\,\int_{[a,\,x^\prime]\times[c,\,d]}f_1dxdy\le\int_{[e,\,f]\times[y^\prime,\,h]}sf_2dxdy.$$One solution in terms of Iverson brackets is $f_1=\frac{r[x\le x^\prime]}{(x^\prime-a)(d-c)},\,sf_2=\frac{r[y\ge y^\prime]}{(f-e)(h-y^\prime)}$, for which the last condition's $\le$ becomes $=$, with both sides $r$. Call this solution $f_1=f_{10},\,f_2=f_{20}$, so an arbitrary solution with $g_1:=f_1-f_{10},\,g_2:=f_2-f_{20}$ is equivalent to$$\int_{[a,\,b]\times[c,\,d]}g_1dxdy=\int_{[e,\,f]\times[g,\,h]}sg_2dxdy=0,\,\int_{[a,\,x^\prime]\times[c,\,d]}g_1dxdy\le\int_{[e,\,f]\times[y^\prime,\,h]}sg_2dxdy.$$If we can find nonvanishing $g_1,\,g_2$ satisfying this, we can rescale them by any common factor, so we're done. One choice is$$g_1=(2y-c-d)f_1,\,g_2=(2x-e-f)f_2.$$(Again, the $\le$ becomes $=$ in this case, with both sides $0$.)


问:
Explicit solution for transport equation with source given by ODE
Consider the transport equation on $(0,\infty) \times (0,1)$ where the source solves an ODE:
$$
\begin{cases}
u_t(t,x)+au_x(t,x)+f(t,x)= 0\\
f_t(t,x) = u(t,x) \\
u(t,0) = g(t) & t>0 \\
u(0,x) = u_0(x) & x \in (0,1)
\end{cases}$$
Can we give an explicit solution $u$ for this equation with the method of characteristics?

答:
In the present form, the linear system of first order PDEs $$
\begin{bmatrix}
u\\
f
\end{bmatrix}_t + \begin{bmatrix}
a&0\\
0&0
\end{bmatrix}\begin{bmatrix}
u\\
f
\end{bmatrix}_x = \begin{bmatrix}
0&-1\\
1&0
\end{bmatrix}\begin{bmatrix}
u\\
f
\end{bmatrix}
$$
of the form ${\bf u}_t + {\bf A}{\bf u}_x = {\bf B}{\bf u}$
hardly decouples, since $\bf A$, $\bf B$ are not co-diagonalisable. Therefore the method of characteristics is of little use here if applied directly (except for the case $a=0$). To solve the problem, one option would be to differentiate the first line w.r.t. $t$ which gives us the second-order hyperbolic equation $$
u_{tt} + au_{xt} +u =0
$$
to be solved the standard way (change of variables, reduction to canonical form). Note that the proposed boundary conditions may not be sufficient for this problem to be well-posed.


答: Take $x=4, y=9$ and $n = 0.5$. You can solve to get $z = 25$. So this works!</p>

Title: Fermat Last Theorem for non Integer Exponents

Body:
We now that Fermat's last theorem is true so there are not positive integer solutions to
$$x^n+y^n=z^n$$
for $n\in\mathbb{N}$ and $n>2$.

But what about if $n\in\mathbb{R}$ or $n\in\mathbb{R}^+$?


问:
Why is everything geometrical modeled on $\Bbb R$?
The reals naturally arise when discussing limiting processes of rational numbers like trying to compute roots or $\pi$. Similarly I have read that axioms of Euclidean geometry (I mean those not explicitly relying on $\Bbb R$ a priori) naturally lead to the real numbers. So there is no doubt in their motivation or in their usefulness.

Nevertheless I am wondering why they are so prominent in every geometric kind of situation. To be more explicit let me draw the following analogy: In algebra it is common to consider some property of the ring of integers and turn this into an abstract definition (think integral domain, factorial domain, principal ideal domain, euclidean domain etc.). I find it particularly interesting that in these settings properties we know and love in $\Bbb Z$ continue to hold, because the corresponding abstract definition perfectly encapsulates the essence of the properties (eg. prime=irreducible in factorial domains). On the analytic / geometric side however the most abstract things derived from the reals I know of are topological spaces and sigma-algebras. Yet as soon as we try to work with them we immediately come back to the reals by forcing connections to $\Bbb R$ (metric spaces have a distance function evaluating in $\Bbb R$, similarly normed vector spaces have norms evaluating in $\Bbb R$, measures evaluate in $\Bbb R$, path-connectivity and homotopy are based on the unit interval $[0,1]\subseteq \Bbb R$ etc. etc.). Compared to the algebraic side mentioned before, this feels rather unaxiomatic to me. Why don’t we assume that evaluations take place in totally ordered groups or something along these lines? Why do we pick the unit interval in $\Bbb R$ and not some totally ordered lattice? Is it just convenience, are the more general / abstract definitions unnecessary by some sort of Lefschetz-principle or is this purely coincidental?

TLDR Everything geometrical (metric spaces, path-connectedness etc.) is inherently based on $\Bbb R$. Why is that and why don’t we see more axiomatic versions.

As always thank you very much for your time, patience and considerations.

答:
There is an axiomatic characterization of the reals: they are the unique complete, ordered, archimedic field. All of these properties are important for what we classically think of as geometry. The field operations correspond to classical constructions done with compass and straightedge. The ordering allows us to identify which of three points on a line is between the other two. Completeness guarantees that lines are connected. And the Archimedic property guarantees a certain sense of finiteness: if we go along a line in fixed increments, we will eventually pass any given point. So the reals are a very natural object to capture the essence of what a line is, and in geometry, lines are fundamental.

Of course, there are other ways to do geometry. For instance, we can define affine planes over arbitrary fields, even if they aren't ordered. For instance, the affine plane over $\mathbb F_2$ consists of four points, and its lines consist of two points. But it misses a lot of the classical properties of geometry. If we want the most important properties of classical geometry to hold (basically the ones mentioned above), we have to model it on the real line - or some other way which is then equivalent to a model based on the real line.


问:
Can a Pythagorean Triple ever have TWO identical exponents > 1?
For example, the triple

$(12, 16, 20)$

has these exponents:

$(12^1, 4^2, 20^1)$ or $(12^1, 2^4, 20^1)$

denoted by

$(1,2,1)$ or $(1,4,1)$

Here are more triples (ignore simple (1,1,1) cases):

I ran millions of triple tests looking for two identical exponents > 1, but found no matches.

Is there an existing proof or can someone prove it?

Thanks!

答:
We do know that all three exponents can assume different values greater than $1$.  Here a solution is constructed from the $(3,4,5)$ triple.

Multiply that triple by $9$ to get $(27,36,45)$.  This has one leg with exponent $2$ and the other leg with exponent $3$.  The hypotenuse has no exponent greater than $1$, but now we take advantage of the fact that multiplying the $(27,36,45)$ triple by any sixth power preserves the exponents of $2$ and $3$ in the legs.  We choose $45^6$ as the multiplier and arrive at

$(6075^\color{blue}{3}, 546750^\color{blue}{2},45^\color{blue}{7}).$


问:
Solving a functional equation over $\mathbb{R}^+$
How do we solve the functional equation $f : \mathbb{R}^{+}\rightarrow \mathbb{R}^+$ such that $f(x)f(y)+f(xy) = xy$?

答:
If $y=1$ then $f(x)f(1)+f(x)=x$

$f(x)(f(1)+1)=x$

if $f(1)+1=0$ then $0=x$ for all x which is not true so $f(1)+1\neq0$

$f(x)=\frac{x}{f(1)+1}=\frac{x}{c}$ where c is a constant for all x.

$(\frac{x}{c})(\frac{y}{c})+\frac{xy}{c}=\frac{xy}{c^2}+\frac{xy}{c}=xy$

$\frac{1}{c^2}+\frac{1}{c}=1$

$1+c=c^2$

c has to be positive as it is positive to positive so this is the well known golden ratio $\varphi$

$f(x)=\frac{x}{\varphi}$

Substitute to confirm solution.