问:
Sequence such that $x_{n+m} \geq x_n+x_m$, show that $(\frac{n}{x_n})$ converges (or not)
Let $(x_n)$ be a sequence such that $x_n \geq  0$ $\forall n \in \mathbb N$ and that $$x_{n+m} \geq x_n+x_m \quad \forall n,m \in \mathbb N^* $$
I have the feeling that $(\frac{n}{x_n})$ converges but I have a little problem with the proof's end.
Can someone help me (the statement can also be wrong but I do not find counterexamples) ?

We have $x_{n+m} \geq x_n+x_m \quad \forall n,m \in \mathbb N^*$, we want to show that $(\frac{n}{x_n})$ converges :

Let's show that $ \lim_{n\to\infty} \frac{n}{x_n} = a$ where $a = inf\{\frac{n}{x_n} | n \in \mathbb N^*\}$ :

We have that $x_{pq+r} \geq x_{pq}+x_r \geq px_q+x_r$ with $p,q,r \in \mathbb N^*$ that gives $\frac{1}{x_{pq+r}} \leq \frac{1}{px_q+x_r}$.

Let $\epsilon > 0$, as $a$ is an inf, we have that $\exists q \in \mathbb N^*$ such that $ a \leq \frac{q}{x_q} \leq a + \frac{\epsilon}{2} $

Let $N>q$, so we have for $n > N$ and by Euclidean division on $n$, $$a \leq \frac{n}{x_n} \leq \frac{pq+r}{px_q+x_r} = \frac{pq}{px_q+x_r}+\frac{r}{px_q+x_r} \leq \frac{q}{x_q} + \frac{q}{px_q+x_r} \leq \frac{q}{x_q} + \frac{q}{px_q} = (1+\frac{1}{p})\frac{q}{x_q} \leq 2 \frac{q}{x_q} $$
Now I can write $a \leq \frac{n}{x_n}\leq 2a+\epsilon$ but we want $a \leq \frac{n}{x_n}\leq a+\epsilon$, is there a way to fix it ?

答:
Instead of $N > q$, pick $N >q\left(2+\dfrac{2a}\epsilon\right)$. If $n > N$, then
$$p > \dfrac nq - 1 > 1 + \dfrac {2a}\epsilon\\
p\dfrac \epsilon 2 > \dfrac \epsilon 2 + a\\\dfrac \epsilon 2 > \dfrac{a + \frac \epsilon 2}p$$

From your calculation,

$$\begin{align}a \le \dfrac n{x_n} &\le \left(1 + \dfrac1p \right)\dfrac q{x_q} \\&\le \left(1 + \dfrac1p\right)\left(a + \dfrac \epsilon 2\right)\\
&\le a + \dfrac \epsilon 2 + \dfrac{a + \frac \epsilon 2}p\\&\le a + \dfrac \epsilon 2 + \dfrac \epsilon 2\end{align}$$


问:
For which values of $\alpha \in \mathbb R$ two improper integrals converge
Question is: For which values of $\alpha \in \mathbb R$ the following improper integrals converge:

a.$$\int_0^1\!\left|\ln(x)\right|^\alpha\,dx$$ b.$$\int_{-\pi/2}^{\pi/2}\!\left|\tan(x)\right|^\alpha\,dx$$ I don't have idea how to solve b but I thought maybe to write $$\int_0^1\!\left|\ln(x)\right|^\alpha\,dx \leq \int_0^1\!\left|x\right|^\alpha\,dx=\int_0^1\!x^\alpha\,dx$$ and then if $\alpha\ge 0$ the integral converges, but I don't have idea for $\alpha<0$.

答:
Hint for part a:

If $\alpha > 0$, the integrand is unbounded near $0$. Use the substitution $u = 1/x$ to solve this case.

If $\alpha < 0$, the integrand is unbounded near $1$. Use the approximation $\ln x \approx x-1$ (valid near $x=1$) to solve this case.

Hint for part b:

If $\alpha > 0$, the integrand is unbounded near $\pm \pi/2$. Taking $x = +\pi/2$ for definiteness, use the approximation $\tan x \approx 1/(\pi/2-x)$ to solve this case.

If $\alpha < 0$, the integrand is unbounded near $0$. Use the approximation $\tan x \approx x$ to solve this case.

ContentLicense: CC BY-SA 3.0


问:
Inequality of a recursive function
Let $x_1=1$ and $x_{n+1}=\tfrac{1}{1+x_n}$.
I want to proof that for every $n,m\in \mathbb{N}$ the inequality $|x_{n+1}-x_{m+1}|\leq \tfrac{1}{2}|x_n-x_m|$ is true. My idea is to somehow exchange $x_{n+1}-x_{m+1}$ with $\tfrac{1}{1+x_n}-\tfrac{1}{1+x_m}$. Maybe one can argue with Cauchy and the Epsilon-criteria?

答:
We can trivially get
$$|x_{n + 1} - x_{m + 1}| = \frac{|x_m - x_n|}{(1 + x_n)(1 + x_m)},$$
so it is left to prove that $(1 + x_n)(1 + x_m) \geq 2$. You can notice that
$$x_1 = 1, x_2 = \frac{1}{2}, x_3 = \frac{2}{3}, x_4 = \frac{3}{5}, x_5 = \frac{5}{8},\ ... $$
and in general $x_n = \frac{F_n}{F_{n+1}}$, where $F_n$ is the $n$-th term of the Fibonacci sequence. From $F_{n+1} = F_{n-1} + F_n$ and $F_n \geq F_{n-1} > 0$ we deduce that $x_n = \frac{F_n}{F_{n+1}} \geq \frac{1}{2}$. So $(1 + x_n)(1 + x_m) \geq (1 + \frac{1}{2})^2 > 2$, which is what we wanted.

ContentLicense: CC BY-SA 4.0

Title: Inequality of a recursive function


问:
What does this mean? "If an angle is inscribed into each of the four segments outside the quadrilateral [that is inscribed in a circle] ..."
I don't know if this should be on English Language Learners or this site, but what does this mean?

A quadrilateral is inscribed in a circle. If an angle is inscribed into each of the four segments outside the quadrilateral, the sum of these four angles, expressed in degrees, is: ..."

I've highlighted in bold what I don't understand.

答:
The angles they are considering are those highlighted in green. Their sum is $540°$, by the way.

They had to specify like this:

"If an angle is inscribed into each of the four circular segments outside the quadrilateral[...]"


问:
Can you use two kinds of logic for a given proof?
Can you use two kinds of logic for a given proof? Can you use classical and paraconsistent logic at the same time in order to prove a theorem or you can only use one? Why is it allowed or not allowed?</p>

答:
The theorems you prove using logic or math are things that are true for some kind of domain or, if you want, for some (class of) world.

Thus, for example, using the axioms of Euclidian geometry, you prove theorems that hold for Euclidian worlds. Theorems proven from non-Euclidian axioms would hold for non-Euclidian worlds.

Likewise, different logics pertain to worlds that, logically, behave differently. Any theorems you prove are therefore also implicitly constrained to those kinds of worlds, and the meaning of the theorem should always be understood as such. So no, as soon as you use two different kinds of logic, with different kinds of assumptions built in their inference rules or semantics, then we are really no longer talking about the same result.


问:
Solving $9x \equiv 21 \bmod 30$

Solving $9x \equiv 21 \bmod 30$

My approach:

Finding $gcd(9,30)$:
\begin{align}
30 &= 9 \cdot 3 + 3 \\ 9 &= 3 \cdot 3 + 0
\end{align}

So $gcd(9,30)=3$ and since $3 \mid 21$ so the congruence has $3$ incongruent solutions. Express $3$ as a linear combination of $9$ and $30$, we get: $$3 = 30-9 \cdot 3 \ \ \ \ \ \ (1)$$

We have to solve this equation to find $x_0$:
$$9x-30y=21 \ \ \ \ \ \ (2)$$

Multiplying $(1)$ by $7$ we get:
$$21=30 \cdot7-9 \cdot 21$$

Rewritten in form of $(2)$:
$$-9 \cdot (21) + 30 \cdot (7) = 21 $$

So $x_0=21$, the incongruent solutions are:
$$21 + 10n, \ \ \ \ \ n = 0,1,2$$

Then $x=21,31,41$. But my answer is wrong, the answer is: $x=9,19,29$. Can you tell me which step is wrong. Thanks for your help!

答:
if $9x = 21 + 30 t,$   we see that $3x = 7 + 10 t,$  the reverse also holding. Solve
$$   3x \equiv 7 \pmod {10}  $$
for which you need a reciprocal of $3.$  Since $21 \equiv 1 \pmod {10} $ we see that
$$  \frac{1}{3} \equiv 7 \pmod {10} ,  $$

and.......

Title: Solving $9x \equiv 21 \bmod 30$


答:
$$9x-30y=21$$
to
$$21=30 \cdot7-9 \cdot 21$$

is the issue, you reverse the terms, so you should get $x \equiv -21 \equiv 9 \pmod {30}$

In other words, when you are solving the initial equation, you are solving it for a positive x less than 30, but you got a negative value.

Title: Solving $9x \equiv 21 \bmod 30$

Body:
Solving $9x \equiv 21 \bmod 30$

My approach:

Finding $gcd(9,30)$:
\begin{align}
30 &= 9 \cdot 3 + 3 \\ 9 &= 3 \cdot 3 + 0
\end{align}

So $gcd(9,30)=3$ and since $3 \mid 21$ so the congruence has $3$ incongruent solutions. Express $3$ as a linear combination of $9$ and $30$, we get: $$3 = 30-9 \cdot 3 \ \ \ \ \ \ (1)$$

We have to solve this equation to find $x_0$:
$$9x-30y=21 \ \ \ \ \ \ (2)$$

Multiplying $(1)$ by $7$ we get:
$$21=30 \cdot7-9 \cdot 21$$

Rewritten in form of $(2)$:
$$-9 \cdot (21) + 30 \cdot (7) = 21 $$

So $x_0=21$, the incongruent solutions are:
$$21 + 10n, \ \ \ \ \ n = 0,1,2$$

Then $x=21,31,41$. But my answer is wrong, the answer is: $x=9,19,29$. Can you tell me which step is wrong. Thanks for your help!


问:
How to prove or statements
How do I prove statements of the following types:

$A \text{ or } B \implies$ C

$A \implies B \text{ or } C$

I don't know how to go about proving statements like this when they have "or". Can someone tell me different ways to prove such statements? And maybe give me some basic examples to help?

Thank-you.

答:
For the first, you can prove both $A\Rightarrow C$ and $B\Rightarrow C$.

Example. If $x=1$ or $x=2$ then $x^2-3x+2=0$.

Proof:
First assume $x=1$. Then $x^2-3x+2=1^2-3\cdot 1+2=1-3+2=0$ as was to be shown.
Next assume $x=2$. Then $x^2-3x+2=4-3\cdot 2+2=4-6+2=0$ as was to be shown.
Since both cases $x=1$ and $x=2$ lead to $x^2-3x+2$, we are done. $_\square$

For the second, you can for example prove $(A\land \neg B)\Rightarrow C$.

Example. If $a\cdot b=0$ then $a=0$ or $b=0$.

Proof: Assume $ab=0$ and $a\ne 0$. Since $a\ne 0$, we are allowed to divide both sides of $ab=0$ by $a$. By this we obtain $b=0$ as was to be shown. $_\square$

Title: How to prove or statements


问:
Can convergent series have a non-exist or infinity entry?
In my advanced mathematical analysis class I came across this question: determine whether the below series converges.
$$\sum\frac{1}{n \sqrt{\log n}}$$
As you can see when $n=1$, the sequence $a_n$ does not exist. Or should we say it's infinity?
So my question is, when one of the entries in a series is infinity, can the series still be convergent? The given solution is that this series is divergent, by the integral test. But I still don't understand how, if you have $$\sum\frac{1}{n \sqrt{\log n}} = \infty + \frac{1}{2 \sqrt{\log2}} + ...$$ the result can be finite.

答:
The series is reasonably intended to start at some $n_0$ such that all the terms are actually defined.

Of course
$$ \sum_{n\geq 2}\frac{1}{n\sqrt{\log n}} $$
is divergent, either by applying Cauchy's condensation test or by noticing that

$$ 2\sqrt{\log(n+1)}-2\sqrt{\log n}=2\frac{\log\left(1+\frac{1}{n}\right)}{\sqrt{\log(n+1)}+\sqrt{\log n}}\leq \frac{1}{n\sqrt{\log n}} $$
such that
$$ \sum_{n=2}^{N}\frac{1}{n\sqrt{\log n}}\geq 2\sqrt{\log(N+1)}-2\sqrt{\log 2} . $$


问:
If $A=\begin{bmatrix} 1 & 2 \\ 5 & 4 \end{bmatrix}$, find a non singular matrix $P$ such that $P^{-1}AP=\begin{bmatrix} 6 & 0 \\ 0 & -1\end{bmatrix}$.

If $A=\begin{bmatrix} 1 & 2 \\ 5 & 4 \end{bmatrix}$, find a non singular matrix $P$ such that $P^{-1}AP=\begin{bmatrix} 6 & 0 \\ 0 & -1\end{bmatrix}$.

I let $P=\begin{bmatrix}a & b \\ c & d\end{bmatrix}$, so that $P^{-1}=\dfrac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ and I put it in the equation, but as expected, it will get tedious because of two degree terms. I couldn't think of any other method to solve it. I then tried to pre-muliply with $P$ in the equation to get $AP=P\begin{bmatrix} 6 & 0 \\ 0 & -1\end{bmatrix}$ and then letting again $P=\begin{bmatrix}a & b \\ c & d\end{bmatrix}$, but here though I am getting simple linear equations, but they are only two instead of four, namely $5a=2c$ and $b=-d$. So, this too fails.

I wanted to express $P$ solely in terms of $A$, $A^{-1}$ by some kind of manipulation in the original equation but couldn't think of any. I am sure there would exist a simple method for this. So, any hints?

答:
Why not use  brute force method for a more general answer.
Let $$P=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$
Let $$A=\begin{bmatrix} 1 & 2 \\ 5 & 4 \end{bmatrix},~~ B=\begin{bmatrix} 6 & 0 \\ 0 & -1 \end{bmatrix}$$
Then $$P^{-1}AP=B\implies AP-PB=O \implies \begin{bmatrix} -5a+2c & 2b+2d \\ 5a-2c & 5b+5d \end{bmatrix}=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.$$
$$\implies c=5a/2, d=-b.$$
So, finally there are infinitely many $P$ matrices given as
$$P=\begin{bmatrix} a & b \\ 5a/2 & -b \end{bmatrix}, a,b \ne 0.$$


问:
Contour integration with logarithm
I want to calculate $\int\limits_{0}^{\infty} \dfrac{\log(x)}{1+x^2}$. To do this, I consider $\int\limits_{\Gamma} \dfrac{\log^2(z)}{1+z^2}$ where $\Gamma$ is the keyhole contour. Basically all I need to do is to calculate $\text{Res}\Big(\dfrac{\log^2(z)}{1+z^2}, i\Big)$ and $\text{Res}\Big(\dfrac{\log^2(z)}{1+z^2}, -i\Big)$. For the first case we have $\text{Res}\Big(\dfrac{\log^2(z)}{1+z^2}, i\Big) = \dfrac{\log^2(z)}{2z}\Big|_{z=i} $ and $\log(i) = i\pi/2$. I'm struggling with the second case, namely with $\text{Res}\Big(\dfrac{\log^2(z)}{1+z^2}, -i\Big)$. Again, we have $\text{Res}\Big(\dfrac{\log^2(z)}{1+z^2}, -i\Big) = \dfrac{\log^2(z)}{2z}\Big|_{z=-i} $ but what is $\log(-i)$? It seems to me that since we're working with the principal branch of logarithm we should have $\log(-i) = \log(|-i|) + i \arg(-i) = 3 \pi i/2$ since for the principal branch of logarithm one has $0 < \arg(z) < 2 \pi$. The "residue calculator" says that it should be $-i\pi/2$ and not $3i\pi/2$. Why?

答: I think the principal branch of logarithm is defined on the plane with the branch cut along the negative real axis (including zero), so the arguments lie in $(-\pi,\pi)$, directed from the positive real axis, hence $\log(-i) = -i\pi/2$

Title: Contour integration with logarithm


问:
Bracing a polygon without triangles
The following Laman graph braces a square without triangles. Stated another way, this is a unit-distance rigid graph without 3-cycles. It seems to be the smallest example of a triangle-free braced polygon.  This happens to be a subgraph of the unit sticks cube graph.

What other regular polygons have triangle-free rigid constructions?

答:
Any rigid framework, hence all regular polygons, can be converted to a triangle-free equivalent. Simply chaining copies of the $12$-vertex triangle-free braced square shown in the question (which I discovered) along the two collinear edges gives a rigid line segment of arbitrary whole number length without triangles:

Then any triangular grid can be mimicked without triangles as follows (all straight fuchsia edges are made with the graph chaining construction above, all black edges are single sticks):

For example, to brace the hexagon without triangles:

However, the above hexagon bracing is quite big. Another approach to triangle-free bracing is the virtual edge: in any embedding of the cubical graph with one edge removed, the distance between the two degree-$2$ vertices (incident to the missing edge) must always be $1$. This leads to the following triangle-free rigid regular hexagon in $16$ vertices and $29$ edges (Shibuya commit proof):

The two versions shown above are graph-theoretically isomorphic; their coordinates have the same minimal polynomials. In particular, using the parametrisation in Shibuya, the $x$-coordinate of vertex $7$ satisfies
$$12x^2-6(\alpha+2)x+(\alpha^2+4\alpha+1)=0,\ \alpha=\sqrt[3]3$$
$$(864x^6-2592x^5+2808x^4-1296x^3+342x^2-207x+83=0)$$
(Thanks Hulpke for pointing me to the GAP function DecomPoly that allowed me to get the first polynomial.) The faint lines in the second version show that the rigid graph is related to the order-$4$ hypercube graph.

Title: Bracing a polygon without triangles


问:
How do I continue from here? (calculating expected value)
In a course in uni there is $185$ students, every students has birthday and his birthday is uniformly distributed on the $365$ days in a year. All the birthdays are independent from eachother. During the semster there is $14$ lectures. Whenever there is a lecture on the same date that  at least one of the students has birthday The teacher give a cake to the class. Let $X$ be the number of cakes brought to the class. Calculate $\mathbb{E}[X]$

Well I dont really understand how to approach this problem because I feel there is too many givens here.
I understand that $0\leq X \leq 14$ and I want to use the formula $$
\mathbb{E}(X)=\sum_{t \in R_{X}} t \cdot \mathbb{P}(X=t)
$$
My problem is to calculate $\mathbb{P}(X=t)$ for  $t$ between $0$ to $14$

答:
Let $X_j$ be a random variable that is equal to $1$ if at least one student has a birthday on   the $j_{th}\;$"lecture day", (so a cake is brought to class) and $0$ otherwise

$\Bbb P(X_j=1) = 1 - \left(\frac{364}{365}\right)^{185}$

Now the expectation of an indicator r.v. is just the probability of the event that it indicates,

so $\Bbb E(X_j) = \Bbb P(X_j)$, and by linearity of expectation,

$\Bbb E(X) = \Bbb E(X_1) + \Bbb E(X_2) + ...... \Bbb E(X_{14})$


问:
Cyclic inequality with $a+b+c=1$

Let $a;b;c$ be positive real numbers such that $a+b+c=1$
Prove: $$a^2+b^2+c^2+a^2b+b^2c+c^2a+4abc \geq \dfrac{16}{27}$$

First, I tried with $abc \geq (a+b-c)(b+c-a)(c+a-b)$
But it gave nothing.

Then I tried to add: $a^2b+\dfrac{b}{9}+b^2c+\dfrac{c}{9}+c^2a+\dfrac{a}{9}-\dfrac{1}{9}$

Then AM-GM, but from it, I got: $\dfrac{2}{3}(a^2+b^2+c^2)+4abc \geq \dfrac{10}{27}$

I think it's wrong, I'm stuck here.
Can someone help? Thanks.

答:
Claim:$$27(a^2c+b^2a+c^2b+abc)\le 4{(a+b+c)}^3$$
Proof:WLOG $a=\min(a,b,c)$ then take $b=a+u,c=a+w$ where $u,w\ge 0$ then we have to prove $$-9au^2+9auw-9aw^2-4u^2+15u^2w-12uw^2-4w^3\le 0$$ $$-9a(u^2-uw+w^2)-{(u-2w)}^2(4u+w)\le 0$$ which is true

thus $$a^2c+b^2a+c^2b+abc\le \frac{4}{27}\tag 1$$

we have to prove  $$a^2+b^2+c^2+4abc+a^2b+b^2c+c^2a\ge \frac{16}{27}$$ using (1) it suffices to prove:$$a^2+b^2+c^2+ab(a+b)+bc(b+c)+ac(a+c)+5abc\ge \frac{16}{27}+\frac{4}{27}$$ using $a+b+c=1$ and some mainipulation we have to prove $$ab+bc+ca-2abc\le \frac{7}{27}$$ WLOG $c=\min(a,b,c)$ then $c\le 1/3$ or $1-2c>0$ $$ab+bc+ca-2abc-\frac{7}{27}=\color{red}{ab}(1-2c)+c(1-c)-\frac{7}{27}$$ $$\le \color{red}{\frac{{(a+b)}^2}{4}}(1-2c)+c(1-c)-\frac{7}{27}\tag{AM-GM}$$ $$=\frac{{(1-c)}^2}{4}(1-2c)+c(1-c)-\frac{7}{27}$$ $$={(3c-1)}^2(6c+1)\frac{-1}{108}\le 0$$ which is true and proof is completed(phew!)

Acknowledgement:Earlier I only had a partial proof (I could prove only  LHS$>15/27$) ,But tbanks to River-Li who proposed a stronger Claim(see comments)  and made some corrections the proof is done!

Title: Cyclic inequality with $a+b+c=1$


问:
Computing the Hardy-Ramanujan asymptotic formula using method of steepest descent/saddle point method
I am trying to obtain and prove the Hardy-Ramanujan asymptotic approximation formula given by:
$$p(n) \sim \frac{1}{4n\sqrt{3}}e^{\pi\sqrt{\frac{2n}{3}}},$$

by using Dedekind's eta function

$$\eta(z)=e^{\frac{i\pi z}{12}}\prod_{n=1}^{\infty}(1-e^{2\pi inz})=\prod_{n=1}^{\infty}\frac{1}{1-z^{n}}$$

and the method of steepest descent (or saddle point method). I am stuck on formulating the integral to which this method can be applied.

I understand that we can use the generating function:
$$f(z)=1+\sum_{n=1}^{\infty}p(n)z^{n}$$

to represent $p(n)$ as the integral

$$p(n)=\frac{1}{2\pi i}\int_{C}\frac{f(z)}{z^{n+1}} \, dz$$

around a closed path $C$ entirely within the unit circle enclosing the origin. Using Dedekind's eta function gives me that

$$f(z)=e^{\frac{i\pi z}{12}}(\eta(z))^{-1}.$$

Since $\eta(-\frac{1}{z})=\sqrt{\frac{z}{i}}\eta(z)$ I have that

$$f(z)=\sqrt{\frac{z}{i}}e^{\frac{i\pi}{12z}}e^{\frac{i\pi z}{12}}f\left(-\frac{1}{z}\right).$$

However, if $z$ is restricted appropriately and $z \to 0$, then $\Im(-\frac{1}{z}) \to \infty$ implies that $f(-\frac{1}{z}) \to 1$ since $f(z)=1+\mathcal{O}(e^{-2\pi y})$, where $z=x+iy$, $y \geq 1$.

So now the integral I am interested in is
$$p_{1}(n)=\int_{\gamma}\sqrt{\frac{z}{i}}e^{\frac{i\pi}{12z}}e^{\frac{i\pi z}{12}}e^{-2\pi inz} \, dz.$$

Is this the correct integral to consider? How do I now apply the method of steepest descent/saddle point method to obtain the Hardy-Ramanujan asymptotic approximation formula?

答:
OP is basically correct. Recall that the Euler function is

$$ \varphi(\tau)~:=~\prod_{m\in\mathbb{N}}(1-q^m)=\frac{1}{\sum_{n\in\mathbb{N}_0}p(n)q^n},
\qquad q~\equiv~e^{2\pi i \tau},$$

and that the Dedekind eta function is

$$ \eta(\tau)~:=~q^{\frac{1}{24}}\varphi(\tau). $$

We calculate the partition function

$$\begin{align} p(n)~=~&\oint_{|q|<1}\frac{dq}{2\pi i}\! \frac{q^{-(n+1)}}{\varphi(\tau)}\cr
~=~&\int_{i\delta+\left[-\frac{1}{2},\frac{1}{2}\right]}  \! d\tau~ \frac{q^{\frac{1}{24}-n}}{\eta(\tau)},
\qquad \delta~>~0, \cr
~=~&\int_{i\delta+\left[-\frac{1}{2},\frac{1}{2}\right]} \!  d\tau~ \sqrt{-i\tau}\frac{q^{\frac{1}{24}-n}}{\eta(\tilde{\tau})}, \qquad
\tilde{\tau}~\equiv~-\tau^{-1}, \cr
~=~&P^{\prime}(n),&\end{align} $$

where

$$\begin{align} P(n)
~:=~&\int_{i\delta+\left[-\frac{1}{2},\frac{1}{2}\right]}  \! \frac{d\tau}{2\pi}~  \frac{q^{\frac{1}{24}-n}\tilde{q}^{-\frac{1}{24}} }{\sqrt{-i\tau}\varphi(\tilde{\tau})}, \qquad
\tilde{q}~\equiv~e^{2\pi i \tilde{\tau}},\cr
~=~&\frac{1}{\sqrt{\lambda}}\int_{i\lambda\delta+\left[-\frac{\lambda}{2},\frac{\lambda}{2}\right]}  \! \frac{d\bar{\tau}}{2\pi}~  \frac{e^{-\lambda S(\bar{\tau})}}{\sqrt{-i\bar{\tau}\rule[0pt]{0pt}{6pt}}\varphi(-\lambda\bar{\tau}^{-1})}, \qquad
\bar{\tau}~\equiv~\lambda\tau,
\end{align}$$

where

$$ \lambda~\equiv~\sqrt{n-\frac{1}{24}}~>~0 $$

is large, and where

$$\begin{align} S(\bar{\tau})~:=~&2\pi i\left(\bar{\tau}-\frac{1}{24\bar{\tau}}\right),\cr
S^{\prime}(\bar{\tau})~=~&2\pi i\left(\frac{1}{24\bar{\tau}^2}+1\right),\cr
S^{\prime \prime}(\bar{\tau})~=~&-\frac{\pi i}{6\bar{\tau}^3}.\end{align}$$

Note that

$$ \varphi(-\lambda\bar{\tau}^{-1}) ~\to ~1
\qquad \text{for}\qquad\lambda~\to~\infty.$$

There are 2 stationary points:

$$\begin{align}
\bar{\tau}_{\pm}~=~& \frac{\pm i}{2\sqrt{6}} \cr
\Updownarrow&\cr
\tau_{\pm}~=~& \frac{\pm i}{2\sqrt{6}\lambda} \cr
\Updownarrow&\cr
\tilde\tau_{\pm}~=~& \pm i 2\sqrt{6}\lambda. \cr
\end{align}$$

We calculate

$$\begin{align} S(\bar{\tau}_{\pm})~=~&\mp\pi \sqrt{\frac{2}{3}}, \cr
S^{\prime \prime}(\bar{\tau}_{\pm})~=~&\pm 8\sqrt{6}\pi.\end{align}$$

Clearly only the stationary point $\bar{\tau}_+$ in the upper halfplane contributes. The method of steepest descent yields

$$\begin{align} P(n)
~\sim~&\frac{1}{\lambda\sqrt{2\pi S^{\prime \prime}(\bar{\tau}_+)}}    \frac{e^{-\lambda S(\bar{\tau}_+)}}{\sqrt{-i\bar{\tau}_+}} \cr
~=~&\frac{e^{\lambda \pi \sqrt{\frac{2}{3}}}}{\lambda2\pi\sqrt{2}}\qquad \text{for}\qquad\lambda~\to~\infty,
\end{align}$$

and therefore

$$ p(n)~=~P^{\prime}(n)
~\sim~\frac{e^{\lambda \pi \sqrt{\frac{2}{3}}}}{\lambda^2 4\sqrt{3}} \qquad \text{for}\qquad\lambda~\to~\infty.$$

For more details, see Ref. 1.

References:

E.M. Stein & R. Shakarchi, Complex Analysis, 2003; Appendix A.4, Theorem 4.1, p. 334.


问:
Two groups overlapping by a fixed number
A travel agency offers the chance to its customers to win vouchers for 2 free excursions during the Covid crisis. Out of the x total customers, y will be randomly chosen to participate in the first excursion and z in the second. In how many of all possible draws we will have a total of w customers who will participate in both excursions?

I am not sure about the exact interpretation: Let's say we have 25 customers and we randomly choose 8 for the first and 7 for the second. If 3 of them are in both excursions (that is, if the 2 groups have 3 people in common), then we have a total of 3 people who will participate in both excursions.
The first draw gives us $xCy$ and the second $xCz$. Obviously it must be $w \leq y+z$.

Can anyone help? I am not good at combinatorics!
(This is a puzzle just for challenging between friends - not an assignment or anything.)

答:
You can choose the lucky $w$ first in $\binom{x}{w}$ ways, then from the remainder $x-w$ you choose the ones that were in the first excursion in $\binom{x-w}{y-w}$ and now from the $x-y$ people that are not going to the first excursion, you pick the $z-w$ in $\binom{x-y}{z-w}$ ways so
$$\binom{x}{w}\binom{x-w}{y-w}\binom{x-y}{z-w}.$$


问:
Greatest number you can write using 3 symbols of a standard keyboard.
Light problem. I think I have two possible answers: $9^9!$ or is $9^{9!}$ greater ? Unfortunately I haven't been able to find a calculator that can display either answer.

Is there something that is totally different that is larger ?

Happy holidays!

答:
You will need logarithms for this; Stirling's approximation gives
$$ \log n! = n \log n -n + \frac{1}{2} \log 2 \pi n + \frac{1}{12n} + \frac{1}{288n^2}  - O(n^{-3} )$$
so we get
$$  \log 9^{9^9}  = 9^9 \log 9,$$

$$  \log 9^9! \approx 9^9 (9 \log 9 - 1 )  $$
which is larger

Put briefly
$$ 9^9 \log 9^9 = 9^9 \cdot 9 \log 9 = 9^{10} \log 9 $$


问:
If the matrix $A$ that is $m\times n$, has a colum that has only zeros then $Ax=0$ Has infinite solutions.
Can anyone help me understand why is this true?

I took a couple of examples and it somehow sorts out to be true, but I don't know how to somehow prove it and generalize.

any help is appreciated :)

答: If the $j$th column of $A$ is "all zeros", then all vectors $x=(0,0,\ldots,\alpha,\ldots,0)^T$ ($\alpha$ is at the $j$th place) are solutions, for every $\alpha\in\mathbb R$. This is because in each row, while multiplying $Ax$, $\alpha$ will be multiplied with the zero from $j$th column in $A$.


答:
Since $A$ has a column with only zeros, $\text{rank} A < n$, so $\text{Ker} A \neq \emptyset$ which means there exists a $x$ such that $Ax = 0$.

Then by multiplying $x$ by any $\lambda$, you have a new $y$ such that $Ay = 0$.

Title: If the matrix $A$ that is $m\times n$, has a colum that has only zeros then $Ax=0$ Has infinite solutions.

Body:
Can anyone help me understand why is this true?

I took a couple of examples and it somehow sorts out to be true, but I don't know how to somehow prove it and generalize.

any help is appreciated :)


问:
Conditional series convergence guess; Prove/ Disprove
I ran into this question:

Prove or disprove:

If $\sum_{n=1}^{\infty}a_{n}$ is a converging series, but the series $\sum_{n=1}^{\infty}a_{n}^2$ diverges, then $\sum_{n=1}^{\infty}a_{n}$ is conditionally convergent.

I'm pretty sure it's true because I couldn't find any example that shows otherwise, but still, i can't find the proof.

Thanks in advance,

Yaron.

答:
We want to show that: if the series $\sum_{n=1}^{\infty}a_{n}^2$ diverges  then the $\sum_{n=1}^{\infty}|a_{n}|$ diverges.

To prove this use that $a^2_n<1$ for all $n>N$ for some $N\in\mathbb N$ and therefore $|a_n|>a_n^2$ for all $n>N$...


问:
Finding $a>0$ such that $a^x\geq x^a$ for all positive $x$
I want to find all real numbers $a>0$ such that for all $x>0$ we have
$$a^x\geq x^a.$$
My guess is that the only possible solution is $a=e$. But that's really all I have so far.

答:
Find all real numbers $a>0$ such that for all $x>0$ we have $\;a^x\geq x^a\;.$

Without using derivatives, I will prove that

$e^x>x^e\quad\forall x>0\;\land\;x\ne e\;.\quad\color{blue}{(*)}$

Let $\;x\;$ be any positive real number such that $\;x\ne e\;.$

Since $\;e^t>1+t\;$ for any $\;t\in\mathbb{R}\;\land\;t\ne0\;,\;$ by letting $\;t=\frac xe-1\ne0\;,\;$ we get that

$e^{\frac xe-1}>1+\frac xe-1\;,$

$e^{\frac xe-1}>\frac xe\;,$

$e^{\frac xe}>x\;,$

$e^x>x^e\;.$

I have proved the inequality $(*)$ and a trivial consequence of it is that $\;e^x\geqslant x^e\;$ for all $\;x>0\;.$

Now, I will prove that there does not exist another real number $\;a>0\;\;(a\ne e)\;$ such that $\;a^x\geqslant x^a\;$ for all $\;x>0\;.$

If there existed such real number $\;a\;$, it would result that $\;a^e\geqslant e^a\;,\;$ but from $(*)$ for $\;x=a\;$, we get that $\;e^a>a^e\;$ that is a contradiction.

Hence the only positive real number $\;a\;$ such that $\;a^x\geqslant x^a\;$ for all $\;x>0\;,\;$ is $\;a=e\;.$


问:
Invertible lattices in are well behaved: IJ is invertible iff I, J are
Context: I'm reading John Voight's book on quaternion algebras (https://math.dartmouth.edu/~jvoight/quat-book.pdf) with the hopes that Brandt matrices will shed some light on a white whale problem of mine; and have a question about what feels like it ought to be a very simple exercise, namely exercise 16.10. (Note that this exercise, or indeed even this chapter, makes no appearance in Voight's hints and solutions document: https://math.dartmouth.edu/~jvoight/hints-solns.pdf.)

Let $B$ be a $\mathbb Q$-algebra, such as a division algebra over $\mathbb Q$.

For a ($\mathbb Z$-)lattice $I \subset B$, we can consider the left and right orders of $I$: namely $\mathcal O_L(I):= \{ \alpha \in B: \alpha I \subset I \}$ and $\mathcal O_R(I) := \{ \alpha \in B: I \alpha \subset I \}$. Note that $I$ is then a left $\mathcal O_L(I)$ and a right $\mathcal O_R(I)$ module. Two lattices $I$ and $J$, and their product lattice $IJ$, are said to be compatible if $\mathcal O_R(I) = \mathcal O_L(J)$: that is, if multiplication in $B$ gives an isomorphism $I \otimes_{\mathcal O} J \cong IJ$, where $\mathcal O = \mathcal O_R(I) = \mathcal O_L(J)$.

A lattice $I$ is said to be invertible if there is another lattice $I'$ so that $II' = \mathcal O_L(I) = \mathcal O_R(I')$ and $I'I = \mathcal O_L(I') = \mathcal O_R(I)$ (so in particular both products $II'$ and $I' I$ are compatible). If $I$ is invertible, then its inverse $I'$ is unique so we call it $I^{-1}$; one can show that $I^{-1} = \{ \alpha \in B: I \alpha I \subset I \}$.

I am trying to show that two compatible lattices $I$ and $J$ are invertible if and only if their product $IJ$ is invertible. One direction is straightforward: if $I$ and $J$ are invertible, then $IJ$ is compatible with $K = J^{-1} I^{-1}$ (which itself is a compatible product) on both sides, and all is as expected. Note that the key appears to be the following statement: if $I$ and $J$ are compatible and each is invertible, then $\mathcal O_L(IJ) = \mathcal O_L(I)$ and $\mathcal O_R(IJ) = \mathcal O_R(J)$. In fact, only left invertibility of $I$ is required for the first statement, right invertibility of $J$ for the second.

However, despite spending several hours, with a collaborator no less!, on the converse, we have made no progress in the other direction. If $IJ$ is invertible, I can certainly propose $K = J (IJ)^{-1}$ as an (a priori, right) inverse for $I$, but although I can compute that $IK = \mathcal O_L(IJ)$, I do not know how to deduce that this last is the same as $\mathcal O_L(I)$ without a priori knowing that $J$ is right invertible. I run into the same problem no matter how I approach. For example, if I knew that the product $K = J (IJ)^{-1}$ is compatible, I would be able to conclude $\mathcal O_L(K) = \mathcal O_L(J (IJ)^{-1}) = \mathcal O_L(J) = \mathcal O_R(I)$, so that the product $IK$ would be compatible, which would be a start. But the compatibility of $J$ and $(IJ)^{-1}$ is equality of $\mathcal O_R(J)$ and $\mathcal O_L( (IJ)^{-1}) = \mathcal O_R(IJ)$ -- same problem yet again!

To be fair, in my quaternion algebra setting, everything would appear to follow from Main Theorem 16.7.7 (in this setting there's a standard involution, so invertible is the same thing as locally principal, etc.). But Voight's exercise 16.10 on p. 265 is stated quite generally, and clearly I am missing something key in the theory. Any help would be much appreciated.

答:
The statement is false!  A counterexample is indicated by Proposition 16.6.15(a), taking $J=\overline{I}$: the product $I\overline{I}$ is compatible by Lemma 16.6.7 but can have larger left (or right) order.  More explicitly, see Example 16.5.12.

I'm very sorry for this mistake!  I traced it back to a very old version of the book, where the extra hypothesis ($\mathcal{O}_{\mathsf{L}}(IJ)=\mathcal{O}_{\mathsf{L}}(I)$) was dropped.  I will make sure it is fixed before publication.  I recall thinking that this extra hypothesis made for a lame exercise, but then I failed to delete the other implication!

Two other comments.

All lattices with left (or right) order maximal are invertible (Proposition 16.6.15).   I think this is what @Kimball is using, and it would finish the job above: your containments of orders become equalities, by maximality.

About terminology: Reiner in his Maximal Orders indeed uses the term "proper", but I try to resist this overloaded term whenever possible.  Anyway, Kaplansky [Kap69] calls it "concordant", and this predates Reiner (1975).  The farthest I can trace the notion back is to Brandt's composition of quaternary quadratic forms when it is 'allowed', and this ultimately gives rise to groupoids (Chapter 19).  What is the best word to use for the situation where you have two morphisms in a small category and the source of one agrees with the target of the second?  (Is "compatible" not a good word to use?)


问:
Show that if $f$ and $g$ are both convex and $g$ decreasing, then $g \circ f$ is concave
Show that if $f$ and $g$ are both convex and $g$ decreasing, then $g \circ f$ is concave given the necessary conditions that $g \circ f$ is defined.
I tried using the basic definition of convex functions using the inequality but it doesn't quite pan out.
I'm starting to doubt that this result is true but can't come up with an easy counterexample.

Any help is appreciated!

答:
$f:x \mapsto x^2$ is convex on $\mathbb{R}^{+*}$, and $g:x \mapsto 1/x $ is convex and decreasing on the same interval.

But $x\mapsto g\circ f(x) = 1/x^2$ is still convex.


问:
Maple calls absolute convergent sum divergent, but why?

Determine whether the sum

is divergent, absolute convergent or conditionally convergent.

First, I tried to solve this by hand. I tried to find a convergent majorant series for the series in question.

$$\bigg|\cos(n^2+1) \frac{2+n^2}{1+n!} \bigg| \leq\frac{n^2}{n!}$$

It turns out that $\sum_{n=1}^{\infty}\frac{n^2}{n!}=2e$, according to Maple that is. This result shows that the sum in the question is absolute convergent.

I also tried to use Maple to double check my answer, and it gave me this.

When Maple gives this sort of output, it usually means that the sum is not convergent. But didn't I just show that it was, or have I made a mistake?

Can a "Maple man" tell me why Maple gives this result, and perhaps how I can avoid it?

答:
As it was already mentioned by Timur B., your proof is correct except a moment $\bigg|\cos(n^2+1) \frac{2+n^2}{1+n!} \bigg| \leq\frac{2+n^2}{n!}$ instead of $\frac{n^2}{n!}$.

I will add that if we see that we shouldn't trust to Maple, so it's better not to use Maple to find  $\sum_{n=1}^{\infty}\frac{n^2}{n!}=2e$ in order to finish the proof)
Although this time the statement is correct and indeed $\sum_{n=1}^{\infty}\frac{n^2}{n!}=2e$,

it's easier to notice that for $n > 100$ we have $\bigg|\cos(n^2+1) \frac{2+n^2}{1+n!} \bigg| \leq\frac{2+n^2}{n!} \le \frac{2n^2}{100 n(n-1)(n-2)(n-3)} <$$
$$ < \frac{2n^2}{100n \cdot \frac{n}2 \cdot \frac{n}2 \cdot \frac{n}2} \le \frac{1}{n^2} \le \frac{1}{n(n-1) } = \frac1{n-1} - \frac{1}n,$
and the series $\sum_{n \ge 100} ( \frac1{n-1} - \frac{1}n ) = (\frac{1}{99} - \frac{1}{100}) + (\frac{1}{100} - \frac{1}{101}) + \ldots ... $ converges (to $\frac{1}{99}$) obviously.


答:
Using tetration, you can get quite a large number. ${^{n}a}$ is defined as ${{a^a}^a}^{...}$. In other words, it is repeated exponentiation.

Therefore, using just ^, 9, 9 you can get:

${^{9}9}$

which is veeeery big.

Title: Greatest number you can write using 3 symbols of a standard keyboard.

Body:
Light problem. I think I have two possible answers: $9^9!$ or is $9^{9!}$ greater ? Unfortunately I haven't been able to find a calculator that can display either answer.

Is there something that is totally different that is larger ?

Happy holidays!


问:
Why use the Fiedler vector?
The Fiedler vector is defined as $$\vec{x_2}=\arg\min_{||\vec{x}||=1,\  \vec{x}^t\vec{x_1}=0}\ \sum_{(ij)\in E}(x_i-x_j)^2,$$ where $\vec{x_1}$ is the smallest eigenvalue's eigenvector. It can be used to find an embedding or partition of a graph. My question is, what is the advantage of having the additional constraint that $\vec{x}^t\vec{x_1}=0$ (and thus obtaining the Fiedler vector) versus just minimizing the expression without that constraint, for the same goal of finding a good partition or embedding?

答:
In general, when minimizing a positive semidefinite quadratic form subject to $\|\vec x\|=1$, you'll get the first eigenvector $\vec{x_1}$. If you add the additional constraint that $\vec x$ should be orthogonal to $\vec{x_1}$, you'll get the second eigenvector $\vec{x_2}$. If you ask for $\vec x$ to be orthogonal to the first two eigenvectors, you'll get the third, and so on.

Here, if we drop the constraint $\vec x^{\mathsf T} \vec{x_1} = 0$, we'll get back $\vec{ x_1}$ as the solution, and $\vec{x_1}$ is just $(1,1,1,\dots,1,1)$ in this case. It's not very useful for finding an embedding (it just gives every vertex the same coordinate) or a partition (it treats every vertex equally).

For connected graphs, the Fiedler eigenvector $\vec{x_2}$ gives the first nontrivial eigenvector - the first one that gives us any interesting information about the graph. (If the graph is not connected, the second eigenvector will give some information about connected components. This is sometimes helpful, but not very good for finding an embedding - you'd do better to embed each component separately.)

Title: Why use the Fiedler vector?


问:
Finding the limit of $\frac{(2n)!!}{(2n+1)!!}$ without using Stirling’s approximatation
I have proven that the sequence is convergent.

Let $x_n=\frac{(2n)!!}{(2n+1)!!}$

$\frac{x_{n+1}}{x_n}$ $=\frac{2n+2}{2n+3}<1$ Therefore the sequence decreases. On the other hand
$x_n>0$. Therefore the sequence is bounded below.
Since we have a sequence that is bounded below and decreases the sequence must converge.

But from here I don’t know how to find the limit. Could you please help me? And could you please say if what I’ve done is necessary or not?

Thanks in advance!

答:
Similar to another answer, you have that $$x_n = \prod_{k=1}^{n}\frac{2k}{2k+1}$$

Then
$$\ln(x_n) = \sum_{k=1}^{n}\ln\left( \frac{2k}{2k+1} \right)$$

Each term will be negative and with the comparison test, you have that $$\sum_{k=1}^{n}\ln\left( \frac{2k}{2k+1} \right) < -\sum_{k=1}^n \frac{1}{2k+1} \to -\infty$$

Then $\ln(x_n)$ must go to $-\infty$ as well, and $x_n = \exp(\ln(x_n)) \to 0$.


问:
Integrating functions with respect to different variables
When going through calculus and physics, I'm finding many integrals that seem to be integrating functions with respect to variables that aren't the dummy variable of the function. One example is the integral for the surface area of a solid of revolution being written as $2\pi\int_{a}^{b} f(x) ds$ instead of $2\pi\int_{a}^{b}f\left(x\right)\sqrt{1+\left(f'\left(x\right)\right)^{2}}dx$ or $2\pi\int_{a}^{b}f\left(x\right)\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}dx$. The textbook I'm working out of for calculus says you can just "substitute" $ds=\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}dx$ whenever you have $ds$ within the integral. My physics book says a similar thing about the integrals for moment of inertia and center of mass. My question is, can you treat differentials like $ds$ and $dx$ like variables when integrating and "substitute" values? I thought that the $dx$ within the integral came from the infinitely small width of an interval in a Darboux or a Reimann sum, meaning it's in some sense tied to the independent variable of the function you're integrating.

I'm aware you often change the integration variable when solving integrals using u-substitution, but as far as I know this is just a notational shortcut for using the chain rule. Is a similar thing at play here?

答: Yes, the $ds$ in this case is functioning as shorthand for $\sqrt {1 +f'(x)} \ dx$, but most calculus books will clarify what "ds" means rather than telling you to "just substitute it in,", i.e. $$2\pi \int_a^{b} f(x) \ ds \text {, where } ds = \sqrt {1 +f'(x)} \ dx.$$


Body: Well, $9^9!$ is a product of $9^9$ terms ($1, 2, 3, ...$), $9^9 - 9$ of which are greater than $9$. Since $9^9 - 9 > 9!$, $$9^9! > 9^{9!}$$


答:
If we allow tetration notation, then $\displaystyle {}^{{}^{9}9}9$ is larger than the other suggestions.
For clarity's sake. this means an iterated exponentiation of the form $$9^{9^{9^{.^{.^{.}}}}}$$ where the height of the tower is $$9^{9^{9^{9^{9^{9^{9^{9^{9^{9}}}}}}}}}$$
I'm not sure what this looks like in standard form, however.


问:
Finding Jordan Canonical form but equation for finding the generalized eigenvector becomes inconsistent
I have been given a problem to find the Jordan Canonical form of the following $ 3 \times 3 $ matrix

$$ A=
\begin{bmatrix}
3 & 0 & 1 \\
-4 & 1 & -2 \\
-4 & 0 & -1
\end{bmatrix}
$$

I have been told to do this using the transformation matrix way, wherein $A$ can be written as

$$
A = TJT^{-1}
$$
where $T$ is the modal matrix analogue which contains the eigenvectors and/or the generalized eigenvectors as columns and $J$ is Jordan Canonical form of A. Doing some matrix operations on the above equation, $J$ can be obtained as

$$
J = T^{-1}AT
$$

Coming back to the problem, I have found the characteristic equation and got all three eigenvalues as $1$. I found eigenvectors by

$$
Ax_1 = \lambda x_1
$$

Solving this I got two linearly independent eigenvectors, since I was getting only 1 pivot after row transformation operations on $A$ the above equation must give me two solutions. I got these

$$
x_1 = \begin{bmatrix}
0 \\ 1 \\ 0
\end{bmatrix}, \begin{bmatrix}
1 \\ 0 \\ -2
\end{bmatrix}
$$

For the next step, according to my textbook, I should get a generalized eigenvector by solving for $x_2$

$$
Ax_2 = \lambda x_2 + x_1
$$

The textbook example only shows one possible value for $x_1$, so clearly I've got a problem regarding which one of the two eigenvectors I got, should be chosen for the next step.

But even without considering that, I have yet another problem

$$
Ax_2 = \lambda x_2 + x_1 \\
(A - \lambda I)x_2 = x_1
$$

If I find out the matrix $A - \lambda I = A - I$, I get

$$
\begin{bmatrix}
2 & 0 & 1\\
-4 & 0 & -2\\
-4 & 0 & -2
\end{bmatrix}
$$

which forms an inconsistent set of equations to solve, because of the repeated row but different values in the rows of (either) $x_1$. So I can't proceed further to get $x_2$.

Have I done something wrong, or is there some pre-condition to be checked on $A$ that I missed?

答:
I like the backwards method: choose the generalized eigenvector(s) with integer elements and see what is forced. Since   $A-I$ gives two genuine eigenvectors, we hold off on that... The minimal polynomial gives the size of the largest Jordan block (always!).  That is, $(A- I)^2 = 0, $  so we look for any nice looking vector for which $(A-I)^2 w = 0$ but $(A-I) w \neq 0.$  I like $w=(0,0,1)^T$
Next we are forced to use $v= (A-I)w = (1,-2,-2)^T.$ A genuine eigenvector that is independent of $v$ could be $u = (0,1,0)^T$

The resulting matrix, your $T,$ is those three as columns in order $u,v,w$

This method allows us to force $T$ to have all integers, with the likelihood of some rational entries in $T^{-1}$ because $\det T$ is most likely not $\pm 1$

Alright, they set this one up with determinant $-1.$

$$
T=
\left(
\begin{array}{rrr}
0&1&0 \\
1&-2&0 \\
0&-2&1 \\
\end{array}
\right)
$$

$$
T^{-1}=
\left(
\begin{array}{rrr}
2&1&0 \\
1&0&0 \\
2&0&1 \\
\end{array}
\right)
$$


问:
Estimation of expected mean value E(x)
I am estimating the probability of the distribution of a discrete random variable $X$ which is defined $P(X=x)=cx(5−x)$ where $X∈{1,2,3,4}$.I already estimated value $c = 1/20 = 0.5$ So next step is estimation of mean expected value $E(x)$.
Can anybody help me to find $E(x)$?

Title: Estimation of expected mean value E(x)

Body:
In fact $c = \frac{1}{20} =0.05$.

As $P(X = 1) = \frac{4}{20}$, $P(X=2) = \frac{6}{20}$, $P(X=3) = \frac{6}{20}$, $P(X=4) = \frac{4}{20}$ we get that
$$EX = \sum_{k=1}^4 k P(X=k) =\frac{4}{20} + 2 \frac{6}{20} + 3 \frac{6}{20} + 4 \frac{4}{20}=2.5. $$


问:
What does $F^{\times 2}$ mean, for $F$ Field
I would love some explanation regarding this notation, seen in "Quaternion Algebras" by John Voight, https://math.dartmouth.edu/~jvoight/quat-book.pdf
in Prop. 2.2.10 and 2.2.14

Let $F$ be a field of $char \neq 2$ and $F^\times$ it's set of invertable elements, what is $F^{\times 2}$ ?

答: Sorry, yes, in general if $G$ is an abelian group written multiplicatively and $n$ is a positive integer, I write $G^n := \{g^n : g \in G\} \leq G$ for the subgroup of $n$th powers.  Here, $G=F^\times$ and we drop the parentheses $F^{\times 2} = (F^\times)^2$.


问:
Harris corner detection vs. PCA
A lot of classes shows that Harris corner detector is actually PCA (example, page 5), but I'm having trouble with that since Harris doesn't have mean subtraction in the algorithm (not in the paper and not in the OpenCV implementation).

This means that if, for example, there is a part of the image which is a slanted plane in both $x$ and $y$ directions, then Harris can recognize it as a corner since $\Sigma I_x^2$, $\Sigma I_y^2$ and $\Sigma I_xI_y$ are all large.

However, with PCA, since we do mean subtraction, the covariance matrix would be small in all elements.

So what don't I get in this nuance of Harris?

答:
The structure tensor is the locally averaged outer product of the gradient with itself:

$$\sum \nabla I \nabla^T I =
\left[\begin{array}{r}
             \sum {I_x}^2 & \sum I_xI_y\\
             \sum I_xI_y & \sum{I_y}^2
        \end{array}\right]$$

When averaging the gradient directly, the opposing vectors at each side of a line will cancel out. One approach to avoid this cancelling is doubling the angle (such that vectors pointing in opposing directions become identical). This is a mapping of $\mathbb{R}^2 \to \mathbb{R}^2$. After applying this mapping, however, vectors originally at 90 degrees will cancel out when averaged. The structure tensor is a different mapping $\mathbb{R}^2 \to \mathbb{R}^3$, namely $(I_x,I_y) \to (I_x^2,I_y^2,I_xI_y)$, which also allows averaging vectors. The first two components of the mapping accomplish approximately the same thing as doubling the angle. The third component, $I_xI_y$, preserves the information about the spread in angles within the set of averaged vectors.

In short, the structure tensor is not the covariance matrix of the gradient in a neighborhood, though it has similarities to it. And so the eigen analysis of the structure tensor is not a PCA of the gradient. However, I do understand why some people like to bring up PCA when explaining Harris. I just wish they didn't claim equality.

Title: Harris corner detection vs. PCA


问:
Product of random numbers.
Find the constant $p$ such that the product of any (positive) number $N_0$ multiplied by successive random numbers between $0$ and $p$ will, on average, neither diverge to infinity nor converge to zero.

(This is not for homework or anything. This is a problem I "invented" and solved, and now want to see how real mathematicians would think about this. Take your time.)

(Thanks md2perpe for fixing up my tag.)

(Thanks Forester for making the variables render nicely.)

答:
Here is my attempt. I leave you to fill missing details.

Call $X_n$ the random variable which is uniformly distributed in the interval $(0,p)$.
Then
$$\prod_{n=1}^N X_n \mbox{ is bounded and stays away from zero} \Longleftrightarrow \sum_{n=1}^N \log X_n \mbox{ is bounded }$$
The random variables $\log X_n$ has values in the interval $(- \infty , \log p)$, and it is not uniformly distributed.
Clearly, a necessary condition of $$\sum_{n=1}^N \log X_n \mbox{ is bounded }$$
is that the mean of $\log X_n$ is zero.

Hence we need
$$\int_{0}^{ p} \log t \  \mathrm d t =0 $$
this condition is equivalent to
$$p \log p -p =0$$

This equation in the variable $p >0$ is easily solved and the only positive solution is $p=e$.

Title: Product of random numbers.


问:
Interchanging limits and integration for improper Riemann integrals
Let $f(x)$ be a function such that the improper Riemann integral $\int_{\mathbb{R}}f(x)\text{d}x$ exists, but where $f$ is not Lebesgue integrable (for example $f(x)=cos(x^2))$. How can I show that for a (smooth) partition of unity $(\beta_n)_{n\in\mathbb{N}}$, meanining that $\sum_{n=0}^{\infty}\beta_n(x)=1$ for all $x\in\mathbb{R}$, we have
$$\int_{\mathbb{R}}f(x)\text{d}x=\sum_{n=0}^{\infty}\int_{\mathbb{R}} f(x)\beta_n(x)\text{d}x$$
You can assume that all $f, \beta_n$ are smooth and bounded. And that each $\beta_n$ has compact support. I have trouble using statements like Fubini or DCT here because $f$ is not integrable.

答:
You can't. Take $f(x) = \sum_{n \ge 1} (-1)^n \frac{1}{n}1_{[n,n+1)}(x)$ and $\beta_k = 1_{[x_k,x_k+1)}$, where $x_0,x_1,\dots = 2,4,6,8,10,12,14,16,18,20,22,1,24,26,\dots,120,122,3,\dots$, where we take odds after we have enough evens to make the sum at least $1/2$. I.e., $$\left(\sum_{k=1}^{11} \frac{1}{2k}\right)-1 > \frac{1}{2}$$ $$\left(\sum_{k=12}^{61} \frac{1}{2k}\right)-\frac{1}{3} > \frac{1}{2},$$ etc.. Then the improper integral of $f$, i.e. $\lim_{L \to \infty} \int_{-L}^L f(x)dx$, exists (since it is just $\lim_{L \to \infty} \sum_{1 \le n \le L} (-1)^n\frac{1}{n}$), but $\sum_{n=0}^\infty \int_\mathbb{R} f(x)\beta_n(x)dx = \lim_{N \to \infty} \sum_{n=0}^N \int_\mathbb{R} f(x)\beta_n(x)dx$ does not exist, as we can group consecutive terms to sum to at least $1/2$.

It's easy to make $f$ and the $\beta_k$'s smooth; just make them smooth near the endpoints of the intervals.


答:
On reflection, I'm not sure the question is correct as stated.

If one takes the arithmetic average of the answer, even with p=e, this will tend to become very large - because you will get the odd case where you are multiplying by numbers bigger than 1 consistently.  You only need a few such numbers for the arithmetic average to blow up.

Perhaps as you multiply by more and more such numbers, the average will come down(?) but I’m not sure.

Alternatively, the question could be reformulated to be about some kind of geometric average?

Title: Product of random numbers.

Body:
Find the constant $p$ such that the product of any (positive) number $N_0$ multiplied by successive random numbers between $0$ and $p$ will, on average, neither diverge to infinity nor converge to zero.

(This is not for homework or anything. This is a problem I "invented" and solved, and now want to see how real mathematicians would think about this. Take your time.)

(Thanks md2perpe for fixing up my tag.)

(Thanks Forester for making the variables render nicely.)


问:
Proof that the axiom of choice implies the dense ordering principle
The dense ordering principle sates that, for every infinite set X, there is a relation $\leq$ in X such that (X, $\leq$) is a dense linear order.
I'm not sure if this idea of proof is correct:
I sarted by stating that we can construct a choice function:

For X a countably infinite set (and therefore such that there is a bijection from X into the natural numbers).

f: P($X \times X$) $\rightarrow$ $\bigcup P(X)$
such that if Z $\subseteq$ Y $\in$ P($X \times X$), then f(Z) $\leq$ f(Y), and therefore the choice function f induces a linear order in $P(X \times X)$ (as I hope we can show).

Now we let A $\subseteq$ P($X \times X$) be the set of equivalence classes of $X \times X$ such that $[a,b]_≈= (c,d)$ and a/b= c/d.

Then, because P($X \times X$) can be linearly ordered, then its subset A can also be linearly ordered. Therefore, because A can be shown to be bijective to a subset of the rationals, then A is densely ordered.

答: One way is to use upward Lowenheim-Skolem / compactness theorem. Since the property of being a dense linear order is first-order axiomatizable, there are dense linear orders of every cardinality. Thus any set can be put in one to one correspondence with a dense linear order and thus has a dense linear order.</p>


问:
Can we multiply between two Derivative operator?
I don't know what should be the tile of my question sorry for that ..

I will explain here clearly ...

Suppose this is my integral :

$\int \frac{d^2x}{dt^2}dx$

I have to show that the integral is equal to $(\frac{dx}{dt})^2$

My steps :

$=\int \frac{d}{dt}(\frac{dx}{dt})dx$

$=\int d\frac{d^2x^2}{d^2t^2}$  Note:  Here I am multiplying  between the two operator like $dx \cdot dx=d^2x^2 $ and $dt \cdot dt=d^2t^2 $

$=(\frac{dx}{dt})^2$

So now my question is can we multiply like the way I did according to my note . If my procedure is wrong how can i show $(\frac{dx}{dt})^2$ from the integral

答:
Maybe this will help. Let $q = dx/dt$ such that your integral becomes $$I = \int \frac{d^2x}{dt^2} q \: dt \:.$$ Then, integrate by parts with: $$u = q \hspace{2.54cm} dv = \frac{d^2x}{dt^2} \: dt$$ $$du = dq \hspace{2.54cm} v = \frac{dx}{dt}$$

So then the standard recipe $I = uv \big| - \int v \: du$ tells us $$I = q \frac{dx}{dt} \bigg| - \int \frac{dx}{dt} \: dq \:,$$ or more simply, \begin{align*}I &= \left(\frac{dx}{dt}\right)^2 \bigg| - \int q \: dq \\ &= \left(\frac{dx}{dt}\right)^2 - \frac{q^2}{2} + C \\ &= \frac{1}{2} \left(\frac{dx}{dt}\right)^2 + C \:.\end{align*}

Note: The factor of $1/2$ belongs there, as your original integral is just Newton's second law applied across a distance - also known as work. The final answer is the kinetic energy plus a constant.


问:
For what $n$ and $m$ is this number a perfect square?
Find all positive integers $n$ and $m$ such that $$(2mn+1)^2-4mn(m+n)+n^2+m^2+(n-1)^2+(m-1)^2-3$$ is a perfect square.

Of course $n=m=1$ is a trivial solution and if you put $n=m+1$ (assume WLOG that $n \geq m$), the above expression is equal to $4n^4$ which is a perfect square - But I don't know if this is the only solution. I would really appreciate any help.

Body:
(Too long for a comment) I think the question can be made clearer with a rearrangement. The $2(2mn)(m+n)$ smells like a quadratic:
$$((m+n)-(2mn+1))^2=(2mn+1)^2-2(2mn+1)(m+n)+m^2+n^2+2mn$$
$$=(2mn+1)^2-4mn(m+n)-2(m+n)+m^2+n^2+2mn$$
Rearranging,
$$=(2mn+1)^2-4mn(m+n)+m^2+n^2+2mn-2(m+n)$$
$$=[(2mn+1)^2-4mn(m+n)+m^2+n^2]+2mn-2(m+n)$$
And substituting,
$$(2mn+1)^2-4mn(m+n)+m^2+n^2+(n-1)^2+(m-1)^2-3$$
$$=((m+n)-(2mn+1))^2-2mn+2(m+n)+(n-1)^2+(m-1)^2-3$$
$$=((m+n)-(2mn+1))^2-2mn+2m+2n+n^2-2n+m^2-2m-1$$
$$=((m+n)-(2mn+1))^2-2mn+n^2+m^2-1$$
$$=(m+n-2mn-1)^2+(n-m)^2-1$$

It's clear now why $n=m+1$ worked. It's equally clear that $m+n-2mn-1=1 \Rightarrow n=\frac{m-2}{2m-1}$ is a solution (although no positive integer solutions exist).


问:
Line integral using substitution
Using Green formula calculate the area that is bounded with curve:$(x^2+y^2)^2=2a^2(x^2+y^2)$. My main problem is how to find $x,y$ in polar coordinates. When i worked with double integrals i would use $x=r\cos\phi, y=r\sin\phi$ but here in solution it is used $x=a\sqrt{2}\cos t\sqrt{\cos 2t}$ and $y=a\sqrt{2}\sin t\sqrt{\cos 2t}$. So I'm not sure how to think in this situation and on what it depends choose od $x,y$.

答:
As you said, $x = r \cos t, y = r \sin t$

So we have,

$(x^2+y^2)^2 = 2a^2(x^2-y^2)$

Or, $(r^2 \cos^2 t + r^2 \sin^2t)^2 = 2a^2(r^2 \cos^2 t - r^2 \sin^2t)$

$r^4 = 2a^2 r^2 (\cos^2 t - \sin^2t)$

$r = \sqrt2 \, a \, \sqrt{\cos 2t}$

Now plug this back in $x, y$ and you have,

$x = \sqrt2 \, a \, \sqrt{\cos 2t} \cos t, y = \sqrt2 \, a \, \sqrt{\cos 2t} \sin t$

Title: Line integral using substitution


问:
Dedekind Cuts in Rudin' analysis - Step 4 - Follow-Up
This question is a continuation of this answer, but I don't have enough reputation to comment, so asking as a standalone question.

In that answer, Brian M. Scott takes pains to point out that j>0.

Why does it matter that j>0?

答:
Okay, I think I've figured it out.  We know $j>0$ and $j\in\mathbb{N}$.  Thus $j \geq 1$.  Recall that $n=j-1-k$.

If $j=1$, then $n=-k$ and $nw=-kw$, and we know $-kw\in\alpha$.  However, $(n+1)w = (j-k)w$.  By definition of $j$ and $S$ we know $(j-k)w\notin \alpha$.  Thus $n$ satisfies the conditions.

If $j > 1$, then we also have $j-1\in\mathbb{N}$.  Then $nw=((j-1)-k))w$.  If $nw\notin\alpha$,  then $j-1\in S$.  However, $j$ is the least member of $S$.  So we must have $nw\in\alpha$.  Of course, $(n+1)w=(j-k)w$, so $(n+1)w\notin\alpha$, and $n$ satisfies the conditions.

(I realize this is a bit pedantic, I just didn't want success to depend on the technicality of whether $\mathbb{N}$ includes $0$.)


答: If you're allowed the full axiom of choice, well-order $X$ with $\prec$, and consider $(X\times\Bbb Q,\leq_{\rm Lex})$ as the lexicographic product of $\prec$ and $\leq_\Bbb Q$. Now use the axiom of choice again, and prove that $X\times\Bbb Q$ is the same size as $X$.

Title: Proof that the axiom of choice implies the dense ordering principle

Body:
The dense ordering principle sates that, for every infinite set X, there is a relation $\leq$ in X such that (X, $\leq$) is a dense linear order.
I'm not sure if this idea of proof is correct:
I sarted by stating that we can construct a choice function:

For X a countably infinite set (and therefore such that there is a bijection from X into the natural numbers).

f: P($X \times X$) $\rightarrow$ $\bigcup P(X)$
such that if Z $\subseteq$ Y $\in$ P($X \times X$), then f(Z) $\leq$ f(Y), and therefore the choice function f induces a linear order in $P(X \times X)$ (as I hope we can show).

Now we let A $\subseteq$ P($X \times X$) be the set of equivalence classes of $X \times X$ such that $[a,b]_≈= (c,d)$ and a/b= c/d.

Then, because P($X \times X$) can be linearly ordered, then its subset A can also be linearly ordered. Therefore, because A can be shown to be bijective to a subset of the rationals, then A is densely ordered.


问:
Does this equation hold? $ P(A \cap B \cap C | D \cap E) = P(A \cap B \cap (C | D \cap E)) $
Does this equation hold? Can we treat conditioning and intersection operation associative?</p>

答:
Conditioning is not an operation on events. Let $\Omega$ denote the sample space and $\mathcal F\subseteq\mathcal P(\Omega)$ denote the set of events (a $\sigma$-algebra on $\Omega$). While intersection takes two events $A,B\in\mathcal F$ to a new event $A\cap B\in\mathcal F$, conditioning is an operation on probability measures:

Given any event $B\in\mathcal F$ with $P(B)\neq 0$, it takes the probability measure $P\colon \mathcal F\to[0,1]$ to a new probability measure $P({-}\mid B)\colon \mathcal F\to [0,1]$ given by
$$
P(A\mid B) = \frac{P(A\cap B)}{P(B)}.
$$

Here "$A\mid B$" is not an event or anything considered on its own. The notation $P(A\mid B)$ should be read as the probability of $A$ under the conditioned measure $P({-}\mid B)$.

Hence, an expression of the kind $P(X\cap(Y\mid Z))$ is meaningless, since it is not of the kind $P(\ldots\mid B)$ and the notation "$Y\mid Z$" on its own has no meaning.


问:
Distribution of two random variables made up of the coordinates of a point
I want to prove that random variables $$\sqrt{-\ln{X}}\cos{Y}, \sqrt{-\ln{X}}\sin{Y}$$ have the same distribution, where $(X,Y)$ is randomly selected from the square $[0;1]^2$ point. I don't know where to start. I would be very grateful for any help

答: Define the transformation $T:(0,1)^2\longrightarrow \mathbb{R}^2$ by $$T(x,y)=\Big(\sqrt{-\ln(x)}\cos(y),\sqrt{-\ln(x)}\sin(y)\Big)=(u,v)$$ The image of $(0,1)^2$ under $T$ is the region in the first quadrant of the $(u,v)-$ plane bounded by the curves $v=0$ and $v=\tan(1)u$ which we'll represent as $\tilde{\Omega}$. Notice $$T^{-1}(u,v)=\Bigg(e^{-(u^2+v^2)},\arcsin\bigg(\frac{v}{\sqrt{u^2+v^2}}\bigg)\Bigg)$$ Using $T^{-1}$ we establish the joint density of $(U,V)$ with support $\tilde{\Omega}$ to be $$f_{UV}(u,v)=f_{XY}\Big(T^{-1}(u,v)\Big)\bigg|\frac{\partial(x,y)}{\partial(u,v)}\bigg|=2e^{-(u^2+v^2)}$$ Let's examine the marginal pdfs of $U$ and $V$: $$f_{U}(u)=\int_{-\infty}^{\infty}f_{UV}(u,v)dv=2e^{-u^2}\int_0^{\tan(1)u}e^{-v^2}dv$$ $$f_{V}(v)=\int_{-\infty}^{\infty}f_{UV}(u,v)du=2e^{-v^2}\int_{v/\tan(1)}^{\infty}e^{-u^2}du$$ These are not the same, so I would deduce that the the distributions of $$U=\sqrt{-\ln(X)}\cos(Y) \\ V=\sqrt{-\ln(X)}\sin(Y)$$ are different.


答:
In general, the way to solve such problems is by considering the follow formulae:

$E(X)=\sum\limits_{x \in \mathcal{X}}x\mathbb{P}(X=x)$, if $X$ is a discrete random variable and $E(X)=\int\limits_{x \in \mathcal{X}} f_X(x) \ \mathrm{d}x$, if $X$ is a continuous random variable. Notice that $f_X(x)$ here is the probability density function (pdf) of $X$. In both formulae, $\mathcal{X}$ is the support of $X$.

Title: Estimation of expected mean value E(x)

Body:
I am estimating the probability of the distribution of a discrete random variable $X$ which is defined $P(X=x)=cx(5−x)$ where $X∈{1,2,3,4}$.I already estimated value $c = 1/20 = 0.5$ So next step is estimation of mean expected value $E(x)$.
Can anybody help me to find $E(x)$?


问:
Probability about cards
Eight hands of five cards each are dealt from a shuffled deck of cards without replacement so that there are twelve undealt cards. Find the probability that at least one of these hands has no diamonds.

The solution uses the inclusion-exclusion theorem. Is there any other way to solve this problem without using that theorem?

Thanks.

答:
You can do it by computing the probability that each hand has at least one diamond, and subtracting from $1$.  While this doesn't use inclusion-exclusion, I don't think it's easier, and I do think it's more error- prone.

You'd have to start by figuring out the ways that the $13$ Diamonds can be distributed.  We have $9$ numbers that add up to $13$, that is is we add up the number of Diamonds in each hand, and the number remaining in the undealt portion of the deck.  The numbers in the hands must be positive, but the number in the deck can be zero.  That is, we must consider the $5$ partitions of $13$ into $9$ parts, and the $7$ partitions of $13$ into $5$ parts.  Also, we must take cognizance that a partition like $5,1,1,1,1,1,1,1,1$ corresponds to $2$ cases: one where each hand is dealt $1$ Diamond, and one where $7$ hands are dealt $1$ Diamond and the fifth is dealt $5$.  The partition $4,2,1,1,1,1,1,1,1$ corresponds to $3$ different cases.

I think you'll agree that inclusion-exclusion is a superior method.

This assumes that you want the theoretical exact answer.  Simulation is a practical alternative.  Personally, if I did it by inclusion-exclusion, I would check my answer by simulation.

Title: Probability about cards


问:
Proof of Deuring's Correspondence
Let $E$ be a supersingular elliptic curve over $\overline{\mathbb{F}_q}$ where $q=p^n$ and $p$ is prime. Then $B:=\text{End}(E) \otimes \mathbb{Q}$ is a unique quaternion algebra over $\mathbb{Q}$ ramified exactly at $p$ and $\infty$, and $\text{End}(E)$ is a maximal order in $B$.

I want the proof of the following statement:

For every maximal order $O' \subseteq B$, there exists $E'$ such that $O' \simeq \text{End}(E')$.

This is given in Voight 42.2.21 (p.790): For every isogeny $\phi: E \rightarrow E'$, there exists a left $O$-ideal $I$ and an isomorphism $\rho:E_I \rightarrow E'$ such that $\phi=\rho \phi_I$. Moreover, for every maximal order $O' \subseteq B$, there exists $E'$ such that $O' \simeq \text{End}(E')$.

Here $E$ is supersingular and $O:=\text{End}(E)$. Definition of $E_I=E/E[I]$ is given in 42.2.1.

The proof just says use a connecting ideal between orders to prove the second statement, but I don't get how we can introduce the connecting ideal here. I suppose he means connecting ideal between $O:=\text{End}(E)$ and $O'$, but how do we know that they are connected in the first place? According to his definition two orders are connected if and only if they are locally isomorphic, i.e. $O_\mathfrak{p} \simeq O'_\mathfrak{p}$ for all primes $\mathfrak{p}$, and I don't think this can happen if we just choose a random maximal order $O' \subseteq B$. Can anyone give me a full proof of this?

答: Apologies if this was too brief.  We are given a maximal order $\mathcal{O}' \subset B$.  Since $\mathcal{O}$ is also a maximal ideal, we may apply Lemma 17.4.7 (in the most recent version, v.0.9.23) to get $I' := \mathcal{O}\mathcal{O}'$ a connecting fractional ideal (indeed, there is a unique genus of maximal orders, 17.4.10), and in particular $\mathcal{O}_{\mathsf{R}}(I') = \mathcal{O}'$.  Clearing denominators (section 9.3), there exists $a \in \mathbb{Z}$ such that $a\mathcal{O}' \subseteq \mathcal{O}$, so letting $I := a I' \subseteq \mathcal{O}$ we still have $\mathcal{O}_{\mathsf{R}}(I) = \mathcal{O'}$.  Now apply Lemma 42.2.9.

Title: Proof of Deuring's Correspondence


问:
Every simple group of order $60$ is isomorphic to $A_5$ - proof by contradiction
I'm trying to prove the theorem that states that

Every simple group of order $60$ is isomorphic to $A_5$.

I'm trying to do it by assuming by contradiction that it doesn't hold and reach
a contradiction.

So I was able to prove the following:

$G$ doesn't have any subgroup with index less than $5.$

If $G$ has a subgroup with index $5$ then it must be isomorphic to $A_5$.

The number of $5$-sylow groups is $6.$

The number of $2$-sylow groups is $15$ (if it's $5$ then $G$ is isomorphic to $A_5$ by previous claim - contradiction).

Now, in order to reach the contradiction, I want to show that the aforementioned conclusions result in $G$ having more than $60$ elements, but I need to show that each $P,Q$ $2$-sylow groups have trivial intersection, which I wasn't able to do (of-course if the intersection is trivial then we get the contradiction). Any help would be helpful.

答:
You are almost there. If any two distinct $2$-Sylows intersect trivially, then you are done. If $P$ and $Q$ are different $2$-Sylows which intersect non-trivially, we have $|P \cap Q| = 2$.

Observe that, since $|P| = |Q| = 4$, we obtain that $P$ and $Q$ are Abelian groups. Hence $P$ and $Q$ are both contained in the centralizer $C$ of $P \cap Q$,

$$C = \{g \in G \ | \ gh = hg \ \ \forall h \in P \cap Q\}.$$

Hence, $C$ contains the subgroup $H$ generated by $P$ and $Q$. Note that $H \neq G$, since else the group $P \cap Q$ of order $2$ would be contained in the center of $G$, a contradiction to the simplicity.

So $4$ divides the order of $H$, and $H \neq G$. Moreover, since $P$ and $Q$ are distinct, $|H| > 4$. Hence the index $(G:H)$ is either $3$ or $5$.


答:
@WilliamBarnes has the correct answer, but I wanted to take some time to correct issues with your understandings of higher order differentials.  Additionally, I'll give a way to do this without the weird substitution.

The $d()$ is an operator, not a multiplier.  $dx$ is equivalent to saying $d(x)$, like you would say $\sin(x)$ or something.  $d^2(x)$ actually means applying the operator twice: $d(d(x))$.  Also, $dx^2$ actually means $(d(x))^2$.

However, the standard notation for the second derivative is actually misleading (or, I would say, wrong).  You shouldn't take the second derivative as meaning what it says.  Most people write it the second derivative of $x$ with respect to $t$ as you did: $\frac{d^2x}{dt^2}$.  However, this is fairly misleading as far as actual differentials go.  If you want manipulable differentials, you have to write the second derivative of $x$ with respect to $t$ as:

$$  \frac{d^2x}{dt^2} - \frac{dx}{dt}\frac{d^2t}{dt^2} $$

Using this, you can now treat these as algebraically manipulable objects, keeping in mind the shorthands we are using (discussed in the first paragraph).  See my paper "Extending the Algebraic Manipulability of Differentials" for more information.

So, to do the integral, let's start (using the manipulable notation) by doing:

$$
I = \left(\frac{d^2x}{dt^2} - \frac{dx}{dt}\frac{d^2t}{dt^2}\right)\,dx
$$

We can move one of the $dt$s to be with the $dx$, giving:

$$
I = \left(\frac{d^2x}{dt} - \frac{dx}{dt}\frac{d^2t}{dt}\right)\,\frac{dx}{dt}
$$

The left side (in parenthesis) is the differential of the first derivative ($\frac{dx}{dt}$).  Therefore, we will set it to be $dv$ for integration by parts. $\frac{dx}{dt}$ will be $u$ as well as $v$.

This means we can rewrite this as:

$$
I = \frac{dx}{dt}\frac{dx}{dt} - \int \frac{dx}{dt} \left(\frac{d^2x}{dt} - \frac{dx}{dt}\frac{d^2t}{dt}\right)
$$

You'll notice that the right-hand side of this is equivalent to our previous definition of $I$.  Therefore, we can rewrite this as:

$$
I = \frac{dx}{dt}\frac{dx}{dt} - I
$$
Adding $I$ to both sides yields:
$$
2I = \frac{dx}{dt}\frac{dx}{dt} \\
I = \frac{1}{2} \left(\frac{dx}{dt}\right)^2
$$

Title: Can we multiply between two Derivative operator?

Body:
I don't know what should be the tile of my question sorry for that ..

I will explain here clearly ...

Suppose this is my integral :

$\int \frac{d^2x}{dt^2}dx$

I have to show that the integral is equal to $(\frac{dx}{dt})^2$

My steps :

$=\int \frac{d}{dt}(\frac{dx}{dt})dx$

$=\int d\frac{d^2x^2}{d^2t^2}$  Note:  Here I am multiplying  between the two operator like $dx \cdot dx=d^2x^2 $ and $dt \cdot dt=d^2t^2 $

$=(\frac{dx}{dt})^2$

So now my question is can we multiply like the way I did according to my note . If my procedure is wrong how can i show $(\frac{dx}{dt})^2$ from the integral


问:
How many different sandwiches are possible?
The canteen sells sandwiches on white, brown, or grain bread. The filling can either be egg, cheese, chicken, or ham. These can be served with tomato sauce, BBQ, or no sauce. How many different types of sandwiches are available at the canteen?</p>

答:
HINT:

It is sometimes easier to see the solution if you work on a problem like this backwards.

Suppose you already have a white bread sandwich with egg filling, then how many sauces can you choose from? Three, right? If you had chosen cheese for example, you would still have three sauce choices. So this means that if you choose white bread, then you have $4\times 3$ choices of fillings and sauce.

Now, how many types of bread can you choose from?


问:
$\operatorname{dim} \mathbb{Q}[2^{1/n},3^{1/m}] / \mathbb{Q} = nm$
I'm trying to prove that $d := \operatorname{dim} \mathbb{Q}[2^{1/n},3^{1/m}] / \mathbb{Q} = nm, n \geq 1, m \geq 1$. My attempts:

It is obvious when $\operatorname{gcd}(n,m) = 1$ (because $n | d, m | d$).

When $m \neq 0 \mod 2$  it is possible (by Hensel's lemma) to embed $\mathbb{Q}[3^{1/m}]$ into field of $2$-adic numbers $Q_2$, and then we can apply Eisenstein criteria to show that $x^n - 2$ is irreducible over $Q_2$, and, hence, irreducible over $\mathbb{Q}[3^{1/m}]$.

Computer experiments show that $\mathbb{Q}[2^{1/n},3^{1/m}] = \mathbb{Q}[2^{1/n} + 3^{1/m}]$ and for every concrete pair of numbers $n,m$ it is possible to compute minimal polynomial of $2^{1/n} + 3^{1/m}$ and show that it have degree $nm$, but I don't see how to write minimal polynomial in general (one closed formula for any $n,m \geq 1$).

Hope for your help!

答:
Since $\sqrt[n]{2}$ has degree $n$ over $\mathbf Q$ and $\sqrt[m]{3}$ has degree $m$ over $\mathbf Q$, the field $\mathbf Q(\sqrt[n]{2},\sqrt[m]{3})$ has degree at most $mn$ over $\mathbf Q$. You want to show "at most" is "equal to".

First, we reduce to the case that $m$ and $n$ are the same by the following argument.  When $m$ and $n$ are arbitrary, let $d = mn$, $K = \mathbf Q(\sqrt[n]{2},\sqrt[m]{3})$, and $L = \mathbf Q(\sqrt[d]{2},\sqrt[d]{3})$, where the $d$th roots of $2$ and $3$ are built from the chosen $n$th root of $2$ and $m$th root of $3$ by extracting further roots from them (that is, $\sqrt[d]{2}^{d/n} = \sqrt[n]{2}$ and $\sqrt[d]{3}^{d/m} = \sqrt[m]{3}$), so $\sqrt[d]{2}$ has degree at most $d/n$ over $K$ and $\sqrt[d]{3}$ has degree at most $d/m$ over $K$.  This makes
$$
\mathbf Q \subset K = \mathbf Q(\sqrt[n]{2},\sqrt[m]{3}) \subset L = \mathbf Q(\sqrt[d]{2},\sqrt[d]{3})
$$
where $[L:K] \leq (d/n)(d/m) = d^2/(mn)$. Since $[K:\mathbf Q] \leq mn$, we have $[L:\mathbf Q] \leq d^2$. Therefore if we can show $[L:\mathbf Q] = d^2$, all the field degree inequalities we have obtained up to this point have to be equalities. Proving $[L:\mathbf Q] = d^2$ is an instance of what you want when using the same amount of root extractions on $2$ and $3$.
Thus if we can prove the result when $m = n$ we will have proved the result in general.

Note that there is nothing here so far that is special about $2$ and $3$. If we have an arbitrary base field $F$ in place of $\mathbf Q$ and pick nonzero $a$ and $b$ in $F$ in place of $2$ and $3$ in $\mathbf Q$, the conditions

(i) $[F(\sqrt[m]{a},\sqrt[n]{b}):F] = mn$ for all positive integers $m$ and $n$

and

(ii) $[F(\sqrt[d]{a},\sqrt[d]{b}):F] = d^2$ for all positive integers $d$

are equivalent: (ii) is a special case of (i), and the argument above shows how (i) can be deduced from (ii) using $d = mn$. Of course determining whether or not such conditions are actually true is another matter entirely from their equivalence, but at least this lets us focus on the case where we are taking the same level of root extraction on both numbers.

In fact, it is not even important that we are adjoining roots of just two elements of the base field.  For $a_1, \ldots, a_k$ in $F^\times$, the conditions

(i) $[F(\sqrt[n_1]{a_1},\ldots,\sqrt[n_k]{a_k}):F] = n_1\cdots n_k$ for all positive integers $n_1, \ldots, n_k$

and

(ii) $[F(\sqrt[d]{a_1},\ldots,\sqrt[d]{a_k}):F] = d^k$ for all positive integers $d$

are equivalent.  Again, this equivalence merely lets us reduce the proof of (i) to the case that $n_1, \ldots, n_k$ are all equal.  There is still a lot of work ahead in actually proving (ii) in some example, such as your question with $F = \mathbf Q$, $k=2$, $a_1 = 2$, and $a_2 = 3$.

As Serge Lang once wrote, there is a feeling in most people that the degree
$[F(\sqrt[d]{a_1},\ldots,\sqrt[d]{a_k}):F]$ should be affected by multiplicative relations among $a_1, \ldots, a_k$, or more precisely relations among them in $F^\times/(F^\times)^d$. For example, $\mathbf Q(\sqrt[3]{2})$ and $\mathbf Q(\sqrt[3]{4})$ each have degree $3$ over $\mathbf Q$ but $[\mathbf Q(\sqrt[3]{2},\sqrt[3]{4}):\mathbf Q]$ is $3$ rather than $9$ because $\mathbf Q(\sqrt[3]{2}) = \mathbf Q(\sqrt[3]{4})$ due to $\sqrt[3]{2}\sqrt[3]{4} = 2$. Even that is subtle: I had in mind real cube roots in both cases, but if $\sqrt[3]{2}$ were a real cube root of $2$ and $\sqrt[3]{4}$ were a complex (i.e., non-real) cube root of $4$, then that cube root of $4$ would be quadratic over $\mathbf Q(\sqrt[3]{2})$ (since $x^3 - 4$ would split over that field into a linear factor involving the real cube root of $4$ and a quadratic factor involving the non-real cube roots of $4$). So $[\mathbf Q(\sqrt[3]{2},\sqrt[3]{4}):\mathbf Q]$ would then be $6$ rather than $3$: the field degree depends not just on multiplicative relations among the numbers whose $d$th roots are being extracted, but also on the interaction between the choice of $d$th roots and $d$th roots of unity.

If the base field $F$ contains a full set of $d$ different $d$th roots of unity, then $F(\sqrt[d]{a_1}, \ldots,\sqrt[d]{a_k})/F$ is Galois and Kummer theory implies $[F(\sqrt[d]{a_1}, \ldots,\sqrt[d]{a_k}):F]$ equals the size of the group $\langle a_1, \ldots, a_k\rangle$ inside $F^\times/(F^\times)^d$, so having the field degree be $d^k$ (the maximal possible value) amounts to saying $a_1, \ldots, a_k$ are multiplicatively independent modulo $d$th powers (the only way a product of powers of them can be a $d$th power in $F^\times$ is for all the exponents to be multiples of $d$).

Your question is about base field $\mathbf Q$, which only has roots of unity $\pm 1$. That means the above paragraph doesn't immediately help us when $d > 2$. Still, the idea that when $a_1, \ldots, a_k$ are multiplicatively independent in $\mathbf Q^\times$ we should have $[\mathbf Q(\sqrt[d]{a_1}, \ldots,\sqrt[d]{a_k}):\mathbf Q] = d^k$ should feel reasonable. Since you are asking about the case where $a_i$'s are prime, I will now simply refer to a paper of Ian Richards that confirms the result in this case: for distinct primes $p_1, \ldots, p_k$, $[\mathbf Q(\sqrt[d]{p_1}, \ldots,\sqrt[d]{p_k}):\mathbf Q] = d^k$ for all positive integers $d$, where we use positive real $d$th roots each time.

The paper is

I. Richards, "An application of Galois theory to elementary arithmetic,"
Advances in Math. 13 (1974), 268-273.

A link to the paper is here.


问:
Kelly Criterion for simultaneous independent bets
I'm trying to obtain a more generic version of the Kelly criterion for when we have simultaneous independent events to bet on,

I'm going to focus on the case where we just have 2 different events.

In the case where we have just one event the objective is to maximize the function:

$$f(x) := p\log(1+bx-x) + (1-p)\log(1-x)$$

$b$ are the odds received and $x$ the fraction of our total wealth that we bet.

In the case where we have two events to bet on, I suppose that the function to maximize would be of the form:

$$p_1p_2\log(1+x_1b_1+x_2b_2 - x_1-x_2) + p_1(1-p_2)\log(1+x_1b_1- x_1-x_2)+(1-p_1)p_2\log(1+x_2b_2 - x_1-x_2)+(1-p_1)(1-p_2)\log(1- x_1-x_2)$$

As it is differentiable, the problem is equivalent to find the solution of the system of equations:

$$p_1p_2\frac{b_1-1}{1+x_1b_1+x_2b_2 - x_1-x_2} + p_1(1-p_2)\frac{b_1-1}{1+x_1b_1- x_1-x_2}=(1-p_1)p_2\frac{1}{1+x_2b_2 - x_1-x_2}+(1-p_1)(1-p_2)\frac{1}{1- x_1-x_2}$$

$$p_1p_2\frac{b_2-1}{1+x_1b_1+x_2b_2 - x_1-x_2} + (1-p_1)p_2\frac{b_2-1}{1+x_2b_2 - x_1-x_2}=p_1(1-p_2)\frac{1}{1+x_1b_1- x_1-x_2}+(1-p_1)(1-p_2)\frac{1}{1- x_1-x_2}$$

However, this seems completely intractable to solve explicitly. I'm asking for your help to do it. Maybe there is some trick to solve it, or I'm not attacking it in the right way.

Thank you in advance.

答:
Approximately proportional to p-b, spread across all bets, but always lower than the sequential, one-event optimal kelly bet. Pick the sequential kelly bet if there aren't enough bets.
Where b is the implied odds and p is what you think the odds are.

Via https://vegapit.com/article/numerically_solve_kelly_criterion_multiple_simultaneous_bets

To add my intuition:
We bet less than single event kelly to avoid ruin. If there are say only two bets, each with optimal sequential kelly of 0.1 (let's say they are 0.1 chance of winning a big lottery prize), you don't wanna bet half your money on each, because you could very likely lose everything.
We bet proportional to the probabilistic edge because that's how much better our odds are versus the game.

Example:
We have 3 independent bets

```<code>100:1 odds 0.1 chance of winning -- kelly & net odds around 0.091
500:1 odds 0.1 chance of winning - kelly around 0.098
0.1:1 odds 0.99 chance of winning (say this is a corporate bond) -- kelly around 0.890
</code>```

In this case we bet about 9% of our money on each bet (slightly more for the higher implied odds bet), and the rest in the "risk free" (almost) bond.

I'm pretty sure the actual odds and payouts of the bets if they're say, option contracts, are gonna be rough estimates anyway, so a rough estimate of the optimal bet would probably suffice.

Otherwise you could do a numerical regression with gradient descent like described here since the function is convex
https://vegapit.com/article/numerically_solve_kelly_criterion_multiple_simultaneous_bets

Body:
Approximately proportional to p-b, spread across all bets, but always lower than the sequential, one-event optimal kelly bet. Pick the sequential kelly bet if there aren't enough bets.
Where b is the implied odds and p is what you think the odds are.

Via https://vegapit.com/article/numerically_solve_kelly_criterion_multiple_simultaneous_bets

To add my intuition:
We bet less than single event kelly to avoid ruin. If there are say only two bets, each with optimal sequential kelly of 0.1 (let's say they are 0.1 chance of winning a big lottery prize), you don't wanna bet half your money on each, because you could very likely lose everything.
We bet proportional to the probabilistic edge because that's how much better our odds are versus the game.

Example:
We have 3 independent bets

100:1 odds 0.1 chance of winning -- kelly & net odds around 0.091
500:1 odds 0.1 chance of winning - kelly around 0.098
0.1:1 odds 0.99 chance of winning (say this is a corporate bond) -- kelly around 0.890

In this case we bet about 9% of our money on each bet (slightly more for the higher implied odds bet), and the rest in the "risk free" (almost) bond.

I'm pretty sure the actual odds and payouts of the bets if they're say, option contracts, are gonna be rough estimates anyway, so a rough estimate of the optimal bet would probably suffice.

Otherwise you could do a numerical regression with gradient descent like described here since the function is convex
https://vegapit.com/article/numerically_solve_kelly_criterion_multiple_simultaneous_bets


问:
Prove that $\dot {\gamma_w} (0) = w.$

Let $\Bbb E^n$ be the Euclidean $n$-space. Let $p \in \Bbb S^{n-1},$ the $(n-1)$-sphere in $\Bbb E^n.$ Let $w \in T_p \Bbb S^{n-1},$ the subspace of $\Bbb E^n$ parallel to the affine subspace tangent to $\Bbb S^{n-1}$ at $p.$ Consider the map $\gamma_w : (-1,1) \longrightarrow \Bbb E^n$ defined by $$\gamma_w (t) = \frac {p+wt} {\sqrt {\left \langle p +wt, p+ wt \right \rangle}},\ t \in (-1,1).$$ Prove that $\dot {\gamma_w} (0) = w.$

Consider the map $g : \Bbb E^n \longrightarrow \Bbb E^n$ defined by $g(v) = \frac {v} {\sqrt {\left \langle v, v \right \rangle}},\ v \in \Bbb E^n.$ and the map $f : (-1,1) \longrightarrow \Bbb E^n$ defined by $t \mapsto p+tw,\ t \in (-1,1).$ Then $\gamma_w = g \circ f.$ So by Chain rule we have $$\dot {\gamma_w} (0) = D\gamma_w (0) (1) = Dg(f(0)) (w) = Dg(p) (w).$$

How do I compute $Dg(p)\ $? Can anybody please help me in this regard?

Thanks for your time.

答:
The differential $Dg$ is just given by the Jacobian matrix. You can compute it in coordinates easily. If $v = (x_1,\dots,x_n)$, then

$$ g(v) = (g_1,\dots,g_n) = \left( \frac{x_1}{\sqrt{x_1^2+\cdots+x_n^2}}, \; \dots, \; \frac{x_n}{\sqrt{x_1^2+\cdots+x_n^2}} \right) $$

You can compute the partial derivates using the quotient rule. You get
$$ \frac{\partial g_i}{\partial x_i} = \frac{|v|^2-x_i^2}{|v|^3} \quad \text{and} \quad \frac{\partial g_i}{\partial x_j} = \frac{-x_ix_j}{|v|^3} $$
where $i \neq j$. Putting it all together, you get
$$ Dg(v) = \frac{1}{|v|^3}\begin{pmatrix} |v|^2-x_1^2 & -x_1x_2 & \cdots & -x_1x_n \\
-x_2x_1 & |v|^2-x_2^2 & \cdots & -x_2x_n \\
\vdots & \vdots & \ddots & \vdots \\
-x_nx_1 & -x_nx_2 & \cdots & |v|^2-x_n^2 \end{pmatrix} = \frac{1}{|v|} \mathrm{Id}_n - \frac{1}{|v|^3} v \, v^\top $$
Here, $v$ is a column vector, and $v^\top$ is the transpose row vector, so $v \, v^\top$ is the matrix whose entries are $x_ix_j$.

When $p$ is on the unit sphere, $|v|=1$, and this simplifies to:
$$ Dg(p) = \mathrm{Id}_n - p \, p^\top $$

Now plug in $w$. The first term gives $\mathrm{Id}_n w = w$, and since $w$ is in the tangent plane to $p$, you have $p^\top w = 0$, so the second term vanishes.

Title: Prove that $\dot {\gamma_w} (0) = w.$


问:
Find minimum of $2x^3+x+\frac3{x^2}$ for $x>0$
Using AM-GM inequality we have:
$$2x^3+x+\frac3{x^2}=2x^3+x+\frac3{2x^2}+\frac3{2x^2}\geq 4\sqrt[4]{2x^3x\frac3{2x^2}\frac3{2x^2}}=4\sqrt[4]{\frac92}$$

As it turns out it is not the minimum, can you spot the mistake?

答: The issue is it assumes both $2x^3=x$ and $x={3\over 2x^3}$ which cannot be both true simultaneously.


答:
There is no flaw in your work - it is true that $$2x^3 + x + \frac{3}{x^2} \ge 4 \sqrt[4]{\frac{9}{2}}$$ for $x > 0$. However, AM-GM inequality won't be enough to find the minimum, as it only attains equality for $$2x^3 = x = \frac{3}{2x^2} = \frac{3}{2x^2}$$

which never happens.

Title: Find minimum of $2x^3+x+\frac3{x^2}$ for $x>0$

Body:
Using AM-GM inequality we have:
$$2x^3+x+\frac3{x^2}=2x^3+x+\frac3{2x^2}+\frac3{2x^2}\geq 4\sqrt[4]{2x^3x\frac3{2x^2}\frac3{2x^2}}=4\sqrt[4]{\frac92}$$

As it turns out it is not the minimum, can you spot the mistake?


问:
Contour integration with logarithm
I want to calculate $\int\limits_{0}^{\infty} \dfrac{\log(x)}{1+x^2}$. To do this, I consider $\int\limits_{\Gamma} \dfrac{\log^2(z)}{1+z^2}$ where $\Gamma$ is the keyhole contour. Basically all I need to do is to calculate $\text{Res}\Big(\dfrac{\log^2(z)}{1+z^2}, i\Big)$ and $\text{Res}\Big(\dfrac{\log^2(z)}{1+z^2}, -i\Big)$. For the first case we have $\text{Res}\Big(\dfrac{\log^2(z)}{1+z^2}, i\Big) = \dfrac{\log^2(z)}{2z}\Big|_{z=i} $ and $\log(i) = i\pi/2$. I'm struggling with the second case, namely with $\text{Res}\Big(\dfrac{\log^2(z)}{1+z^2}, -i\Big)$. Again, we have $\text{Res}\Big(\dfrac{\log^2(z)}{1+z^2}, -i\Big) = \dfrac{\log^2(z)}{2z}\Big|_{z=-i} $ but what is $\log(-i)$? It seems to me that since we're working with the principal branch of logarithm we should have $\log(-i) = \log(|-i|) + i \arg(-i) = 3 \pi i/2$ since for the principal branch of logarithm one has $0 < \arg(z) < 2 \pi$. The "residue calculator" says that it should be $-i\pi/2$ and not $3i\pi/2$. Why?

答:
Your analysis is correct.  If the branch cut is taken along the real axis, then the complex logarithm is defined by

$$\log(z)=\text{Log}(|z|)+i\arg(z)$$

where $\text{Log}$ is the logarithm function for real analysis and

$$0\le \arg(z)<2\pi$$

Hence, $\log(i)=i\pi/2$ and $\log(-i)=i3\pi/2$.

The residue calculator uses the principal branch of the logarithm, which is defined such which $-\pi<\arg(z)\le \pi$.

Title: Contour integration with logarithm


问:
A prime problem in Fibonacci sequence
This problem can be a little too wild but have at it anyways:

Prove that for every positive integer $n$ $\geqslant$ $7$ , $f_{n+1}$ has a prime divisor that doesn't divide $f_{n}-1$ where $f_{n}$ is the n-th number in the fibbonachi sequence.

答:
Starting from Cassini's identity:

$$F_{n+1}F_{n-1}-F_{n}^2=(-1)^n$$

and adding $1$ to both sides we get:

$$F_{n+1}F_{n-1}-(F_{n}+1)(F_{n}-1)=(-1)^n+1$$

When $n$ is even, the RHS is $2$. Since the only fibonacci numbers which are a power of $2$ are $F_3=2$ and $F_6 = 8$ (see e.g. here), then $F_{n+1}$ ($n+1 \gt 6$ for $n \ge 7$) has one or more odd prime factors, but none of them can divide $F_n-1$, otherwise it would divide $2$.

When $n$ is odd, the RHS is $0$. $F_n+1$ and $F_n-1$ cannot have any common factor other than $2$. $F_{n+1}$ and $F_{n-1}$ are coprime. Then, if all prime factors of $F_{n+1}$ divide $F_{n}-1$ we have:

$$F_{n+1}=2^{k}(F_n-1)$$

for some $k \gt 0$. If $k = 1$ we have $F_{n+1}=F_n+F_{n-1}=2F_n-2$ and then $F_{n-1}=F_n-2$, impossible, otherwise still an absurd:

$$F_n = \frac{F_{n-1}+2^k}{2^k-1} \lt F_{n-1}$$

because:

$$F_{n-1} \gt \frac{2^k}{2^k-2}$$

Title: A prime problem in Fibonacci sequence


问:
Produce the linear and quadratic Taylor polynomials for $f(x)=e^{\cos x}$
$$f(x)=e^{\cos x},x_0=0$$

I wrote first two terms of Taylor Polynomial of $f(x)$:

$$P_2(x)=e-\frac{e}{2}x^2$$

It is quadratic Taylor Polynomial. but what about linear one? Is it simply $P_1(x)=e$ or I should say it doesn't exist because we don't have $x^1$ in the expression?

答:
Your function $ f $ is an even one, so its second degree Taylor expansion arround $ x_0=0 $,  will be of the form

$$f(x)=e^{\cos(x)}=$$
$$P_2(x)+x^2\epsilon(x)=$$
$$a_0+a_2x^2+x^2\epsilon(x)$$

The tangent line at $ x_0=0 $ is horizontal and its equation is
$$y=P_1(x)=f(0)=e$$


答:
Let us suppose that we have $\sqrt{-\ln X} \cos 2\pi Y$ and $\sqrt{-\ln X} \sin 2\pi Y $, otherwise claim is not true.

It's easy to see that $X$ and $Y$ are independent.

It's sufficient to show that $\cos 2\pi Y $ and $\sin 2\pi Y$ have the same distribuion in case $Y \sim U[-1,1]$.

We have $\cos 2\pi Y = \sin (\frac{\pi}2 - 2\pi Y) $ and $\frac{\pi}2 -  2\pi Y = \frac{\pi}2 - U[0, 2 \pi] \sim U[-\frac{3\pi}2, \frac{\pi}2] = - \pi +U[-\frac{\pi}2, \frac{\pi}2] =  - \pi  - U[-\frac{\pi}2, \frac{\pi}2]$, hence
$$cos 2\pi Y = \sin (\frac{\pi}2 - 2\pi Y) = \sin (- \pi  -U[-\frac{\pi}2, \frac{\pi}2])  = sin(U[-\frac{\pi}2, \frac{\pi}2]) $$
Similarly $$\sin (2\pi Y)  \sim sin(U[-\frac{\pi}2, \frac{\pi}2]).$$
Thus $\cos 2\pi Y = \sin 2\pi Y$ in distribution, q.e.d.

Title: Distribution of two random variables made up of the coordinates of a point

Body: I want to prove that random variables $$\sqrt{-\ln{X}}\cos{Y}, \sqrt{-\ln{X}}\sin{Y}$$ have the same distribution, where $(X,Y)$ is randomly selected from the square $[0;1]^2$ point. I don't know where to start. I would be very grateful for any help


问:
Endomorphisms and K-Vector spaces
I want to proof (i) and (ii)

(i) Let $V$ be a finite-dimensional K-Vector Space and $f\in End(V)=Hom(V,V)$. If there is a $g\in End(V)$ with $f\circ g=Id_V$, then $f$ is an Isomorphism with $g=f^{-1}$

(ii) Statement (i) is not true for an infinite-dimensional K-Vectorspace $V=Abb(\mathbb{N},K)$ with $Abb(\mathbb{N},K):=\{f|f:\mathbb{N}\rightarrow K\}$.

My ideas:
(i) Let's assume that $\exists g \in End(V): f\circ g=Id_V$. If $f\circ g=Id_V$ then $g\circ f=Id_V$ because $f: V\rightarrow V$. So $g$ is the inverse function of $f$ and $f$ is an automorphism, as it is an isomorphism and an endomorphism.

I am not entirely sure about this proof.

(ii) Here I need help, I know that if $K=\mathbb{R}$ then there cannot be a surjective function $f$ but what if $K=\mathbb{N}$?

答:
I do not think that your proof for (i) is complete:

First observe that $f \circ g = id_V$ implies that $f$ is surjective: Indeed, any $v \in V$ lies in the image of $f$ as $v = f(w)$ for $w := f(v) \in V$. (Similarly, one can deduce from $f \circ g = id_V$ that $g$ is injective, but this fact is not necessary to prove (i).)

However, it is well-known that surjective $K$-linear maps between finite dimensional vector spaces are necessarily bijective, hence isomorphisms. (Actually, this is not true for infinite dimensional vector spaces, and your task in (ii) is to provide a counterexample.) As a consequence, there exists a $K$-linear map $f^{-1} \colon V \to V$ such that $f \circ f^{-1} = f^{-1} \circ f = id_V$. But from this, we may easily conclude that
$$f^{-1} = f^{-1} \circ id_V = f^{-1} \circ (f \circ g) = (f^{-1} \circ f) \circ g = id_V \circ g = g,$$
proving that $g$ is the inverse of $f$, completing the proof of (i).

Concerning (ii), you may consider the example provided by Severin Schranven in his comment, i.e., the left shift $(a_0,a_1,a_2,...) \mapsto (a_1,a_2,...)$. Note here that one may regard any function $f \colon \mathbb{N} = \{0,1,2,3,...\} \to K$ as a sequence $(f(0),f(1),f(2),f(3),...)$.
If the left shift was an isomorphism, then it would in particular be (surjective and) injective. Can you see why the left shift fails to be injective?
And concerning the map $g$, you may think about how one could define a right shift.

P.S.: Note that it is impossible to choose $K = \mathbb{N}$, because (when speaking about $K$-vector spaces) $K$ is always assumed to be a field.

Title: Endomorphisms and K-Vector spaces


问:
What is the name of the quadrilateral shape described by two radii and two arcs?
What is the name of the four sided shape described by two radii and two concentric arcs?
Like each black and white section taken individually on this image

Or the double score / triple score areas on a dart board.

I'm attempting to create each section individually, programmatically and need help on the best thing to google to get help actually making this.

答:
A quadrilateral has straight sides, so I assume you are referring to the quadrilateral that connects the 4 relevant points.

It's a trapezoid. More specifically, it is an isosceles trapezoid

Since the circles from which you got the arcs are concentric, we know we get overlapping isosceles triangles when we take, separately, each of the two pairs of arc points together with the center of the circle. For the large isosceles triangle, this implies that the far side of the quadrilateral (relative to the center of the circle) makes angles with each of the radius that are each equal to the other. Similarly, the near side creates equal angles with the radii.

The overlapping isosceles triangles have the same central angle, so the 4 base angles are all (180-A)/2 degrees. This same configuration (ultimately that all corresponding angles where a radius intersects the near and far sides are equal) is defined whenever we have parallel lines intersecting a third line. Thus, we see that the near and far sides are parallel. This means we have a trapezoid. [Note that the bases are of different sizes to match the isosceles triangles which are definitely not congruent, so we don't have a rectangle.]

So a trapezoid with opposite base angles being congruent is an isosceles trapezoid.


问:
If $p$ is a prime number, prove that there exist infinitely many whole numbers $n$, such that $p$ divides $n \cdot 2^n+1$
Let $p$ be a prime number and let $n$ be an integer.
If $p$ divides $n \cdot 2^n+1$ we get $n \cdot 2^n+1 \equiv 0 \pmod p$ or $n \cdot 2^n \equiv p-1 \pmod p $. Now if we let $n=p-1$

$(p-1)2^{\varphi(p)} \equiv p-1 \pmod p$, because $2^{\varphi(p)} \equiv 1 \pmod p$. So $n=p-1$ is one solution, how do I prove there are infinitely many?

答:
Actually $p$ needs to be odd.

Now, as you said, if $k \equiv 0 \mod{p-1}$ we have $2^k \equiv 1 \mod p$.
We can therefore choose $k$ such that $k \equiv 0 \mod{p-1}$ and $k \equiv -1 \mod{p}$.

Can you finish now?


问:
Negative Combination
If given $0<r<n$, then $C_r^n=\frac{n!}{(n-r)!r!}$. But what if $r<0$ since $r!$ will fail?

答:
For integral (even complex) $n$ and integral $r$ the following definition holds:

\begin{align*}
\binom{n}{r}=
\begin{cases}
\frac{n(n-1)\cdots (n-r+1)}{r!}=\frac{n^{\underline{r}}}{r!}&\quad r\geq 0\\
0&\quad r<0
\end{cases}
\end{align*}

See for instance formula (5.1) in chapter Binomial Coefficients of Concrete Mathematics
by D.E. Knuth, R.L. Graham and O. Patashnik.

Title: Negative Combination


问:
Proving or disproving facts about Cauchy Sequences
Exercise 2.6.2 in Understanding Analysis by Stephen Abbott
$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$

I'd like someone to verify my solution, and check if my arguments are technically correct. Do I need to add a short proof for (c)?

Give an example of each of the following, or argue that such a request is impossible.

(a) A Cauchy sequence that is not monotone.

(b) A Cauchy sequence with an unbounded subsequence.

(c) A divergent monotone sequence with a Cauchy subsequence.

(d) An unbounded sequence containing a subsequence that is Cauchy.

Proof.

(a) By CC(Cauchy Criterion),
\begin{align*}
 \text{Cauchy sequences} \iff \text{Convergent sequences}
\end{align*}
A convergent sequence could be monotone or oscillating. Consider the sequence $(a_n)$ given by,
\begin{align*}
 a_n = \frac{1}{n} \sin \left(\frac{n\pi}{4}\right)
\end{align*}

is an example of a Cauchy sequence that is not monotone.

(b) This request is impossible. Cauchy sequences are bounded. All subsequences of a bounded sequence must be bounded.

(c) This request is impossible. By MCT, a monotone and bounded sequence is convergent. The contrapositive of this statement is; if a sequence is divergent, atleast one of the two possibilities must hold (1) the sequence is oscillating (2) the sequence is unbounded. As we are told, the sequence is monotone, it must be the case, that the sequence is unbounded. An unbounded monotone sequence $(a_n)$ cannot nest inside itself, a bounded subsequence $(a_{2n})$ or $(a_{f(n)})$.

(d) This request is impossible. An unbounded sequence cannot contain a bounded subsequence. And Cauchy sequences are always bounded.

答: Besides (d) this is correct (see the comment). For (c) your proof makes sense but is a little unclear. A self contained proof would be as follows: If $x_n$ were increasing then of course $x_n \rightarrow \infty.$ Suppose there was a subsequence with $x_{n_k} \rightarrow x.$ (This is equivalent to $x_{n_k}$ being Cauchy). Then there exists $N \in \mathbb{N}$ such that $x_N > x.$ Further there exists $k \in \mathbb{N}$ such that $n_k > N.$ Using the monotoneness we have $x \geq x_{n_k} \geq x_N > x.$ You would then need to handle the case when $x_n$ were decreasing too but this is essentially the same argument with the inequality signs swapped.


问:
In a unit circle $x^2 + y^2 = 1$, why can you substitute the x and y with cos and sin?
In this video he talks about a point $(x,y)$ on the unit circle can be rewritten as $(\cos(\theta), \sin(\theta))$.

However, to my understanding, $\cos$ is the adjacent side over the hypotenuse ($\frac{x}{r}$) and sin is the opposite side over the hypotenuse ($\frac{y}{r}$).

Why is it okay to substitute the values of $(x,y)$ with the values of ($\frac{x}{r}$, $\frac{y}{r}$)?

答:
If you write $x = cos(t)$ (it is possible since $0\leqslant x \leqslant 1$), then you can remark that $y^2 = 1-x^2 = 1-\cos ^2 (t)=\sin ^2 t$, so that you can write $y=\pm \sin t$.

If $y= \sin t$ then you're finished.

If $y= -\sin t$, then you just need to note t' = -t (we have $\cos -t = \cos t$).

Hope it helped.


答: Given a point on the circumference of center $(0,0)$ and radius $1$ individuated by the angle $\phi$ measured with radians, by definition $\cos \phi$ is the first coordinate and $\sin \phi$ is the second coordinate of that point; that is $x=\cos \phi$ and $y= \sin \phi$.

Title: In a unit circle $x^2 + y^2 = 1$, why can you substitute the x and y with cos and sin?

Body:
In this video he talks about a point $(x,y)$ on the unit circle can be rewritten as $(\cos(\theta), \sin(\theta))$.

However, to my understanding, $\cos$ is the adjacent side over the hypotenuse ($\frac{x}{r}$) and sin is the opposite side over the hypotenuse ($\frac{y}{r}$).

Why is it okay to substitute the values of $(x,y)$ with the values of ($\frac{x}{r}$, $\frac{y}{r}$)?


答:
Note that choices are independent of each other, so you can have any choice of bread with any choice of filling and any sauce.

That means that we can work with choices of bread etc independently.
Suppose you only choose bread, how many choices are there? There are three. After you've chosen bread you can choose fillings, how many choices of fillings are there? There are four. So how many choices of bread AND fillings are there? How can you proceed to sauces?

Title: How many different sandwiches are possible?

Body: The canteen sells sandwiches on white, brown, or grain bread. The filling can either be egg, cheese, chicken, or ham. These can be served with tomato sauce, BBQ, or no sauce. How many different types of sandwiches are available at the canteen?</p>