问:
Is the Cauchy–Schwarz inequality valid for the Lebesgue space $L^2(0,T,L^2(\Omega))$?
Let $n \in \mathbb{N}$ and $\Omega \subset \mathbb{R}^n$ be open and bounded set. It is known that if $f, g \in L^2(\Omega)$ then by Cauchy–Schwarz inequality we have
$$
\Bigg| \int_{\Omega} f(x)\overline{g(x)}\; dx\Bigg|^2 \leq \int_{\Omega} |f(x)|^2 \cdot  \int_{\Omega} |g(x)|^2
$$

Question. For $T>0$, is an analogous result valid for the Lebesgue space $L^2(0,T,L^2(\Omega))$ ? That is, if $h \in L^2(0,T,L^2(\Omega))$ then
$$
\Bigg\|\int_0^t h(s) \; ds\Bigg\|_{L^2(\Omega)}^2 \leq  c \cdot \int_0^t \|h(s)\|_{L^2(\Omega)}^2 \; ds, \: t \in (0,T)
$$
for some constant $c>0$?

I believe it is valid, but having little knowledge of these Lebesgue spaces $L^p(0,T, X)$, with $X$ a Banach Space, I was unable to prove. For more details of the this Lebesgue Spaces, see for instance the Eberhard Zeidler book's Nonlinear Functional Analysis and Its Applications(2A)-Linear Monotone Operators .

答:
Yes, for the first question, since $L^2(0,T;L^2(\Omega))$ is a Hilbert space. For the second question, note that $\left\| \int_{0}^{t} h(s)ds\right\|_{L^2(\Omega)} \leq \int_{0}^{t} \left\|h(s)\right\|_{L^2(\Omega)}ds$ and $$\int_{0}^{t}\|h(s)\|_{L^2(\Omega}ds \leq \left( \int_{0}^{t}1^2ds \right)^{\frac{1}{2}}\left( \int_{0}^{t}\|h(s)\|_{L^2(\Omega)}^2 ds\right)^{\frac{1}{2}}.$$ Then,
$$\left\| \int_{0}^{t} h(s)ds\right\|^{2} \leq t \int_{0}^{t}\|h(s)\|_{L^2(\Omega)}^2 ds.$$


问:
A simple quadratic reformulation of the twin prime conjecture?
Due to https://oeis.org/A024702 we have that $p^2 - 1 ≡ 0$ (mod 24).

For twin primes, we then must also have that in this case $(p+2)^2 - 1 ≡ 0$ (mod 24), which is the same as saying that $p^2 + 4p + 3 ≡ 0$ (mod 24).

Adding these and dividing both sides by 2, one would then have a simple quadratic reformulation of the twin primes conjecture like so:

There are infinitely many primes p such that $p^2 + 2p + 1 ≡ 0$ (mod 12).

Is that correct? Would it possibly be of any practical benefit, like extending it to prime gaps of arbitrary size that have been proven already, and then "proving backwards" towards a gap of size 2?

答: The issue here is that all numbers of the form $n=6k\pm 1$, not just primes, satisfy the relationship $n^2-1 \equiv 0 \bmod 24$. Such pairs of numbers may contain $0,1,\text{ or }2$ primes, e.g. $(119,121),(23,25),(47,49),(11,13)$. So even if there are infinitely many numbers such that $n^2+2n+1\equiv 0 \bmod 12$, this is not equivalent to the assertion that there are infinitely many prime numbers with that property.

ContentLicense: CC BY-SA 4.0


问:
Prove 1 + 4 Product of fibonacci numbers is a perfect square
Let $n  \in \mathbb N$ . Prove that

$1 + 4 F_{2n}F_{2n+1}F_{2n+2}F_{2n+3}$ is the square of an integer.

I first tried to used Cassini's identities but failed.I tried then by induction and again failed.

$P(1) = 1 + 4F_2 F_3 F_4 F_5 = 121 = 11^2 $

Assuming $P(k)$ to be true and equal to $I^2$

$P(k+1) = 1 + (I^2 - 1) \frac{F_{2n + 4}F_{2n+5}}{F_{2n}F_{2n+1}}$ but how to proceed from here?

Body:
Using the fact that $F_{2n+1}F_{2n+2} - F_{2n}F_{2n+3}= 1$, you can write this as:
$$1+4F_{2n}F_{2n+1}F_{2n+2}F_{2n+3} = 1+4F_{2n+1}F_{2n+2}(F_{2n+1}F_{2n+2}-1)$$ Let $x = F_{2n+1}F_{2n+2}$ and this expression becomes:
$$1+4x(x-1) = 4x^2-4x+1 = (2x-1)^2$$


问:
Is this operator T compact on $L^2([0,1])$?
Consider $L^2([0,1])$ and the operator $$Tu(t) = tu(t)$$for a.e. $t \in [0,1]$

How can I prove or disprove that T is compact?
Thanks!

答:
Hint

Let $f_n(x)=n^{3/2}\boldsymbol 1_{[0,\frac{1}{n}]}(x)$ for all $n\in\mathbb N$. Set $K=\{f_n\mid n\in\mathbb N\}\subset L^2(0,1)$. Is $K$ bounded ? Is $T(K)$ relatively compact ?


答: Hint:  $T$ is self-adjoint and has no eigenvectors. Use the spectral theorem.

Title: Is this operator T compact on $L^2([0,1])$?

Body:
Consider $L^2([0,1])$ and the operator $$Tu(t) = tu(t)$$for a.e. $t \in [0,1]$

How can I prove or disprove that T is compact?
Thanks!


问:
Use epsilon-delta definition of limits to show that $\lim_{(x,y) \to (1,2)}(x^2+3x-4y)=-4$ and $\lim_{(x,y) \to (1,-1)}(x^2+y^2)=2$
Use epsilon-delta definition of limits to show that

$$\lim_{(x,y) \to (1,2)}(x^2+3x-4y)=-4$$
$$\lim_{(x,y) \to (1,-1)}(x^2+y^2)=2$$

We need to show that $$\forall \epsilon >0( \exists \delta >0( \forall (x,y) \in \mathbb R^2 (0<\sqrt{\left(x-1\right)^{2}+\left(y-2\right)^{2}}<\delta \implies \left|\left(x^2+3x-4y\right)+4\right|<\epsilon)))$$

Taking $\left|x-1\right|,\left|y-2\right|<\delta$ follows :
$$\left|\left(x^2+3x-4y\right)+4\right|\le \left|x+4\right|\left|x-1\right|+4\left|y-2\right|< (\delta+5)\left|x-1\right|+4\left|y-2\right|$$

Let $ \delta \le 1$:

$$\left|\left(x^2+3x-4y\right)+4\right|< 6\left|x-1\right|+4\left|y-2\right|$$

So taking $\delta \le \epsilon/10$ proves the claim.

We need to show that $$\forall \epsilon >0( \exists \delta >0( \forall (x,y) \in \mathbb R^2 (0<\sqrt{\left(x-1\right)^{2}+\left(y+1\right)^{2}}<\delta \implies \left|\left(x^2+y^2\right)-2\right|<\epsilon)))$$

$$\left|\left(x^2+y^2\right)-2\right| \le \left|x-1\right|\left|x+1\right| +\left|y-1\right|\left|y+1\right|$$

Taking $\left|x-1\right|,\left|y+1\right|<\delta$ and $\delta \le 1$ follows:
$$\left|\left(x^2+y^2\right)-2\right| <3\delta +2\delta$$

So we only need to choose $\delta \le \epsilon/5$.

I want to know if my solution is true,so if you have an alternative solution that would be nice to see that,but please first check mine.

答:
Seems okay; the way you take $\delta$ is not very clear.

I would write the first one as follows. (Similarly for the second one.)

Let $\epsilon>0$. The estimate
$$
\begin{align}
|x^2+3x-4y-(-4)|
&=|x^2+3x-4y-(1^2+3\cdot 1-4\cdot 2)|\\
&\le |x+1|\cdot |x-1|+3|x-1|+4|y-2|\\
&=|x-1+2|\cdot |x-1|+3|x-1|+4|y-2|
\end{align}
$$
suggests that if you take
$$\sqrt{|x-1|^2+|y-2|^2}<\delta\tag{1}$$ then $|x-1|<\delta$ and $|y-1|<\delta$ and thus
$$
|x^2+3x-4y-(-4)|\le (\delta+2)\delta+3\delta+4\delta=(\delta+9)\delta\tag{2}
$$
So if you take $\delta=\min(1,\epsilon/10)$, then (1) implies
$$
|x^2+3x-4y-(-4)|\le (\delta+2)\delta+3\delta+4\delta=(\delta+9)\delta \le 10\delta
\le\epsilon.
$$


问:
Derivative of a composite function with respect to a matrix
I would like some hints on how to derive the following function with respect to the matrix $X$:

$$ f(X)=\left\| y - \sum_k (h_k X^k)x \right\|_2^2$$

where $X$ is a matrix, $x$ and $y$ are two given vectors and $h_k$ is a (given) scalar coefficient that weights the matrix. This is the composition of a convex function (the norm) with a non affine function. I want to linearize the function with respect to X, so I would like to compute the derivative. Some hints?

Thank you.

答:
Let's use a convention where a greek letter denotes a scalar, lowercase latin a vector, and uppercase latin a matrix. So replace your original variable names with
$$\eqalign{
\alpha_k = h_k
,\qquad
\phi(X) = f(X)
,\qquad
x = z
}$$
and define two new matrices which will be used later
$$\eqalign{
P &= \sum_{k=1}^n \alpha_kX^k \\
M &= 2z(Pz-y)^T\\
}$$
Let's also use a colon to denote the matrix inner product, i.e.
$$\eqalign{
A:B &= {\rm Tr}(A^TB) \\
A:A &= \big\|A\big\|_F^2 \\
}$$
Write the objective function in terms of the above definitions.

Then calculate its differential and gradient.
$$\eqalign{
\phi &= (Pz-y):(Pz-y) \\
d\phi &= 2(Pz-y):dP\,z \\
  &= 2(Pz-y)z^T:dP \\
  &= M^T:\sum_{k=1}^n \alpha_k\sum_{j=0}^{k-1} X^jdX\,X^{k-j-1} \\
  &= \sum_{k=1}^n\sum_{j=0}^{k-1}\left(\alpha_kX^jMX^{k-j-1}\right)^T:dX \\
\frac{\partial \phi}{\partial X}
  &= \sum_{k=1}^n\sum_{j=0}^{k-1}\left(\alpha_kX^jMX^{k-j-1}\right)^T \\
}$$


问:
Extension of real bounded analytic function to a holomorphic bounded function.
Let $f(x)$ be a real-valued analytic bounded function. Can I find a open set $U\subset\mathbb{C}$, containing $\mathbb{R}$, and a holomorphic complex-valued function $\bar{f}(z)$ such that $\bar{f}(z)$ is bounded on $U$ and $\bar{f}(x)=f(x)$ for every $x \in \mathbb{R}$? Is there any class of real-valued function with such properties? I just remark that the open set $U$ is not necessarily a strip.

答:
Yes always; let $M$ be a bound for $f$; since $f$ is analytic at $x$ there is a small plane disc $U_x$ centered at $x$ for which $f$ extends to a complex analytic function $f_x$ on $U_x$ and by continuity at $x$ we can shrink it so $|f_x| \le 2M$ there; by analytic continuation $f_x=f_y$ whenever $U_x \cap U_y$ non empty since the intersection is connected (and contains the real segment joining $x,y$) as $U_x, U_y$ are discs centered at $x,y$. In particular this means that we can define a global complex analytic function $F$ on $U=\cup U_x$ and $F=f$ on the reals, while $|F| \le 2M$ on $U$

($U$ is not a strip in general unless $f$ is the quasi analytic class $1/n!$ which has one of the equivalency definition precisely strip extension; there are bounded analytic functions on the reals that are not in there - see my answer HERE)


问:
Explicit formula of a difference equation
I have a difference equation I want to convert to a explicit formula and I'm currently stuck.

The difference equation is as follows:
$$
N_x = N_{x-1} + N_{x-3},  (f(x) = f(x-1) + f(x-3)), where: N_3 = 1, N_4 = 2, N_5 = 3
$$

I want to find the explicit formula for this difference equation.

I know how to solve it's characteristic equation and roots:
$$
λ^3 = λ + 1 \,\,\,\,\\
λ^3 - λ - 1 = 0
$$
This equation will give us one real solution and two imaginary solutions. The solutions of the equation are so convoluted that it is easier to give the approximate answers to illustrate the solutions:
$$
λ_1 ≈ 1.4656 \\
λ_2 ≈ -0.23279 - 0.79255i \\
λ_3 ≈ -0.23279 + 0.79255i
$$
This is the point at which I get stuck. I've solved difference equations in which there are either only real parts, or only imaginary parts. But I do not know how to solve for this case in which both imaginary- and real solutions are present. If someone could help me out, that would be amazing.

If you want to help me, you can just use $λ_1, λ_2\ and\ λ_3$ as variables for further calculations. Thanks in advance!

答:
What you did is very correct but the problem is far to be simple.

For your information :

This is sequence $A000930$ in $OEIS$. If you look at the documentation you will notice that, in $192$, Simon Plouffe found that its generating function is
$$\frac{1}{1-x-x^3}$$

A nice formula (given in the documentation) is
$$N_x=\sum_{i=0}^{\left\lfloor \frac{x}{3}\right\rfloor}\binom{x-2 i}{i}$$

More complex (but nice) is the hypergeometric function
$$a_n=\,
   _3F_2\left(\frac{1-n}{3},\frac{2-n}{3},-\frac{n}{3};\frac{1-n}
   {2},-\frac{n}{2};-\frac{27}{4}\right)$$


问:
Which one result in mathematics has surprised you the most?
A large part of my fascination in mathematics is because of some very surprising results that I have seen there.

I remember one I found very hard to swallow when I first encountered it, was what is known as the Banach Tarski Paradox. It states that you can separate a ball $x^2+y^2+z^2 \le 1$ into finitely many disjoint parts, rotate and translate them and rejoin (by taking disjoint union), and you end up with exactly two complete balls of the same radius!

So I ask you which are your most surprising moments in maths?

Chances are you will have more than one. May I request post multiple answers in that case, so the voting system will bring the ones most people think as surprising up. Thanks!

答: The infinitude of primes! – and the simplicity of its proof!</p>

ContentLicense: CC BY-SA 3.0

LastEditorDisplayName: anon


问:
Determine the number of digits in $(2^{120})(5^{125})$.
Determine the number of digits in $(2^{120})(5^{125})$.

This is a bonus question I found in a Grade 12 math textbook, and I'm curious on how to solve it. What I find strange is that this question is in the  "Power Functions" section of "Polynomial Functions"; yet, this seems to have little to do with functions.

Any ideas as to how to solve this? I have tried to, but can't find a solution without a computer or calculator.

答:
The base-$10$ logarithm of $2$ is $0.30103$, so the base-$10$ logarithm of $5$ is $0.69897$ (at least four exact decimal digits). So the base-$10$ logarithm of $2^{120}\cdot 5^{125}$ is
$$120\cdot0.30103+125\cdot0.69897=123.49485$$
Thus the number has 124 digits.

Of course, noticing that the number is also $5^5\cdot10^{120}$ is faster and gives the exact result.


问:
How to establish this inequality: $(1-a)(1-b)(1-c) \geq 8abc$ for $a+b+c=1$?
Let $a$, $b$, and $c$ be positive real numbers such that $a+b+c = 1$. Then how to establish the following inequality? $$ (1-a)(1-b)(1-c) \geq 8abc.$$

My effort:

Since $a+b+c =1$, we can write $$  (1-a)(1-b)(1-c) = 1 - (a+b+c) +(ab+bc+ca) - abc = 1 - 1  +(ab+bc+ca) - abc =  (ab+bc+ca) - abc = abc(a^{-1} + b^{-1} + c^{-1} ) - abc = 3abc (\frac{a^{-1} + b^{-1} + c^{-1}}{3}) - abc.$$ What next?

答:
@Saaqib Mahmood, I'm expanding on @5xum's answer here. Using $AM\geq HM$,
$$$$${\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\over{3}$$\geq$${3}\over{a+b+c}$
$$$$So, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq9$


问:
Understanding the weak finite basis theorem
I'm working my way though the proof of the weak finite basis theorem in Chapter 15 of Cassel's book "Lectures on Elliptic Curves".

I'm stuck on the proof Lemma 2, that "The kernel of $\mu$ is $2(C(\mathbb{Q}))$".

The parts I don't understand are:

(1) I don't understand the why "since $M$ has exponent 2, the kernel certainly contains $2(C(\mathbb{Q}))$". (Note $M$ is defined as a subset of $\mathbb{Q}^*(\Theta)/(\mathbb{Q}^*(\Theta)^2$). I'm guessing $M$ has "exponent 2" is referring to the $(\mathbb{Q}^*(\Theta)^2$ in the quotient of $M$? But I'm not sure why that shows the kernel contains $2(C(\mathbb{Q}))$.

(2) Further down the proof, why are we allowed to replace $\Theta$ by an 'indeterminate X'? And why does the right hand side of the equation that follows $= F(X)$?

Any help with any of these questions would be much appreciated! Thank you.

答:
Let $f: A \longrightarrow B$ be a group homomorphism. Then $B$ has exponent $2$ if and only if $2b = 0$ (if the group is written additively). In this case, $2A$ is in the kernel:
$$ f(2a) = f(a+a) = f(a) + f(a) = 2f(a) = 0 $$
since the image of $f$ has exponent $2$.

For your second question: the quadratic extension $K(\sqrt{m})$ is isomorphic to $K[x]/(x^2-m)$. Thus an equation in $K(\sqrt{m})$ is equivalent to an equation in $K[x]$ up to multiples of $x^2-m$. The same thing works for cubic extensions.

Here are the details. Let $\Theta$ be the root of an irreducible cubic monic polynomial $F$ and assume that $f(\Theta) = 0$ for some monic polynomial of degree $3$. The irreducibility of $F$ implies that $f$ is divisible by $F$. Since both polynomials have degree $3$, we must have $f(X) = c F(x)$. Since both $f$ and $F$ have leading coefficient $1$, we must have $c = 1$.

Proofs of Mordell's Theorem are a lot simpler if the roots are rational. See Silverman-Tate or any of the many introductions to elliptic curves for undergraduates.


问:
how to find the line with minimum sum of distances to multiple Points?
I am programming on a little physics simulation right now. The mathematical problem i am facing now, is pretty hard for me, a 17 year old high school student. But since i really need to solve this one, i thought i might just ask if somebody knows something about the problem.

The Problem:

All in 2D.

Given a certain number of points, find a line that approximates these points the closest.

I know this sounds much like a regression line, but a regression line minimizes the y distances from line to points. I want to approximate not in a statistical sense like regression line, but a geometrical sense. I want to minimize the actual right angled shortest distances from points to the line. distances to minimize (red)

The only solution i could think off was to make a function (f(m , h)) that takes the parameters of a line (y = mx + h) and gives the sum of distances squared. But finding a minimum of such a long multivariable function was not possible for me.

Since this seems like basic stuff to me i thought this problem is well documented, but i could not find anything about it. Please redirect me if you know the name of this problem.

Thanks.

答: This is Orthogonal Distance Regression. If you'd like to, there exists a scipy solution to it</p>


问:
Convergence of $\sum_{n=0}^{\infty}a_n/n$ if $\sum_{n=0}^{\infty}a_n^2$ exists
Let $\sum_{n=0}^{\infty}a_n^2$ a convergent series. Prove that $\sum_{n=0}^{\infty}a_n/n$ converges or give a counter-example.

I know that there is already a post concerning this statement with a lot of answers, but i did't find a similar approach to mine in that post and an answer to my question.

After a long searching for a counter-example (failed), i started to prove the statement. First, i tried to prove it by comparaison but didn't succeed. So, i tried to pass by Cauchy definition. I'm not sure at all if it holds what i did. If someone could give a feedback and point to my errors, i would really appreciate it.

Proof

As $\sum_{n=0}^{\infty}a_n^2$ exists, then $\lim_{n\to \infty}a_n^2=0$. Thus, $\forall \epsilon>0 \ \exists N \ \forall n\ge N:$ $|a_n|<\sqrt{\epsilon}$

Let's proof that $\sum_{n=0}^{\infty}\dfrac{a_n}{n}$ is Cauchy. By definition of Cauchy we have to satisfy the following: $\forall \epsilon>0 \ \exists N' \ \forall n\ge N'$:

$\Big|\sum_{k=N'}^{n}\dfrac{a_k}{k}\Big|<\epsilon$

But,

$\Big|\sum_{k=N'}^{n}\dfrac{a_k}{k}\Big|\le \sum_{k=N'}^{n}\dfrac{|a_k|}{k}\le \sqrt{\epsilon}\sum_{k=N'}^{n}\dfrac{1}{k}$

Let $(x_n)=\frac{1}{n}$. As $\lim_{n\to\infty}x_n=0$ we have by definition that: $\forall \epsilon>0 \ \exists N'' \ \forall n\ge N'':$ $|x_n|<\sqrt{\epsilon}$. Therefore,

$\sqrt{\epsilon}\sum_{k=N'}^{n}\dfrac{1}{k}<\sqrt{\epsilon}\sqrt{\epsilon}=\epsilon$.

So, if we choose $N'=\max\{N,N''\}$, we satisfy Cauchy definition.

Edit: As $x_n$ is a converging sequence: $\exists M>0$: $|x_n|\le M \ \forall n \in \mathbf{N}$. We could then take $|a_n|<\frac{\epsilon}{M(n-N)}$.

Thus,

$\Big|\sum_{k=N'}^{n}\dfrac{a_k}{k}\Big|\le \frac{\epsilon}{M(n-N)}\sum_{k=N'}^{n}\dfrac{1}{k}\le \frac{\epsilon}{M(n-N)}\cdot M(n-N')$

If we take $N'=N$, we have then:

$\Big|\sum_{k=N'}^{n}\dfrac{a_k}{k}\Big|\le\epsilon$

Does it hold this way?

答: From $x_n<\sqrt \epsilon$, you unfortunately only get $$\sqrt \epsilon\sum_{k=N'}^n\frac1n=\sqrt \epsilon\sum_{k=N'}^nx_n<\sqrt \epsilon\sum_{k=N'}^n\sqrt \epsilon =(n-N+1)\epsilon\not<\epsilon. $$


问:
Show that $K \subset \mathbb{R}^d$ is compact $\iff \forall f\in C^0(K, \mathbb{R})$, $f$ is bounded on $K$

Show that $K \subset \mathbb{R}^d$ is compact $\iff \forall f\in C^0(K, \mathbb{R})$, $f$ is bounded on $K$.

The direction "$\implies$" is given by Weierstrass theorem.

For the other direction, I want to show that $K$ is bounded and closed and thus compact.

We have in particular for $f=\pi_i$ (the $i$-th projection), that $f$ is bounded for any $1 \leq i \leq d$, which implies $K$ is bounded.

But I'm not sure how to show that $K$ is closed.

I thought about maybe looking at something like $f(x) = ||x||$ and then we have $f^{-1}[\mathbb{R}]=K$, and by the thorem that says that the preimage of a closed set by a continuous function is closed we should get that $K$ is closed?

But I'm pretty sure I'm using it wrong since it seems I can prove some open sets are closed this way, which doesn't make sense. So my questions are:

What is the mistake in the last part?

How do I correctly show $K$ is closed?

答:
Your mistake is that the preimage will be a closed set in the induced topology on $K$, not in $\mathbb{R^d}$.

Anyway, suppose $K$ is not closed in $\mathbb{R^d}$. Then there is some point $y\in\overline{K}\setminus K$. Now define $f:K\to\mathbb{R}$ by $f(x)=\frac{1}{||x-y||}$. It is well defined, because $y\notin K$, and it's easy to see it is continuous. However, $y\in\overline{K}$, and so there is some sequence $(x_n)$ of elements of $K$ such that $x_n\to y$. For such a sequence we have $f(x_n)\to\infty$, and so $f$ is not bounded, a contradiction.


答:
By Cauchy-Schwarz inequality, we have that $\sum_{n=1}^{N}|a_{n}|\cdot|\frac{1}{n}|\leq\left\{ \sum_{n=1}^{N}a_{n}^{2}\right\} ^{\frac{1}{2}}\left\{ \sum_{n=1}^{N}\frac{1}{n^{2}}\right\} ^{\frac{1}{2}}\leq\left\{ \sum_{n=1}^{\infty}a_{n}^{2}\right\} ^{\frac{1}{2}}\left\{ \sum_{n=1}^{\infty}\frac{1}{n^{2}}\right\} ^{\frac{1}{2}}<\infty$.
Therefore, $\left(\sum_{n=1}^{N}|a_{n}|\cdot|\frac{1}{n}|\right)_{N}$
is a bounded increasing sequence and hence $\lim_{N\rightarrow\infty}\sum_{n=1}^{N}|a_{n}|\cdot|\frac{1}{n}|$
exists. That is, the infinite series $\sum_{n=1}^{\infty}\frac{a_{n}}{n}$
converges absolutely.

Title: Convergence of $\sum_{n=0}^{\infty}a_n/n$ if $\sum_{n=0}^{\infty}a_n^2$ exists

Body:
Let $\sum_{n=0}^{\infty}a_n^2$ a convergent series. Prove that $\sum_{n=0}^{\infty}a_n/n$ converges or give a counter-example.

I know that there is already a post concerning this statement with a lot of answers, but i did't find a similar approach to mine in that post and an answer to my question.

After a long searching for a counter-example (failed), i started to prove the statement. First, i tried to prove it by comparaison but didn't succeed. So, i tried to pass by Cauchy definition. I'm not sure at all if it holds what i did. If someone could give a feedback and point to my errors, i would really appreciate it.

Proof

As $\sum_{n=0}^{\infty}a_n^2$ exists, then $\lim_{n\to \infty}a_n^2=0$. Thus, $\forall \epsilon>0 \ \exists N \ \forall n\ge N:$ $|a_n|<\sqrt{\epsilon}$

Let's proof that $\sum_{n=0}^{\infty}\dfrac{a_n}{n}$ is Cauchy. By definition of Cauchy we have to satisfy the following: $\forall \epsilon>0 \ \exists N' \ \forall n\ge N'$:

$\Big|\sum_{k=N'}^{n}\dfrac{a_k}{k}\Big|<\epsilon$

But,

$\Big|\sum_{k=N'}^{n}\dfrac{a_k}{k}\Big|\le \sum_{k=N'}^{n}\dfrac{|a_k|}{k}\le \sqrt{\epsilon}\sum_{k=N'}^{n}\dfrac{1}{k}$

Let $(x_n)=\frac{1}{n}$. As $\lim_{n\to\infty}x_n=0$ we have by definition that: $\forall \epsilon>0 \ \exists N'' \ \forall n\ge N'':$ $|x_n|<\sqrt{\epsilon}$. Therefore,

$\sqrt{\epsilon}\sum_{k=N'}^{n}\dfrac{1}{k}<\sqrt{\epsilon}\sqrt{\epsilon}=\epsilon$.

So, if we choose $N'=\max\{N,N''\}$, we satisfy Cauchy definition.

Edit: As $x_n$ is a converging sequence: $\exists M>0$: $|x_n|\le M \ \forall n \in \mathbf{N}$. We could then take $|a_n|<\frac{\epsilon}{M(n-N)}$.

Thus,

$\Big|\sum_{k=N'}^{n}\dfrac{a_k}{k}\Big|\le \frac{\epsilon}{M(n-N)}\sum_{k=N'}^{n}\dfrac{1}{k}\le \frac{\epsilon}{M(n-N)}\cdot M(n-N')$

If we take $N'=N$, we have then:

$\Big|\sum_{k=N'}^{n}\dfrac{a_k}{k}\Big|\le\epsilon$

Does it hold this way?


问:
How to prove $ E(|X-Y|) \le E(|X+Y|)$ when $X,Y$ are i.i.d variables?
Let $X$ and $Y$ be independent random variables having the same distribution and the finite mathematical expectation. How to prove the inequality $$ E(|X-Y|) \le E(|X+Y|)?$$</p>

答:
Below is a set of remarks that’s too long to be put in a comment.

Conjecture. The inequality becomes an equality iff $-X$ has the same
distribution as $X$.

Remark 1. The “if” part of the conjecture is easy : if $X$ and $-X$ have the same distribution, then by the independence hypothesis $(X,Y)$ and $(X,-Y)$ have the same joint distribution, therefore $|X+Y|$ and $|X-Y|$ share the same distribution, so
they will certainly share the same expectation.

Remark 2. Let $\phi_n(t)=t$ if $|t| \leq n$ and $0$ otherwise. If the inequality
holds for any $(\phi_n(X),\phi_n(Y))$ for any $n$, then it will hold for $(X,Y)$ also,
by a dominated convergence argument. So we may assume without loss of generality that
the support of $X$ is bounded.


问:
Use epsilon-delta definition of limits to show that $\lim_{(x,y) \to (0,0)}(\sqrt{-x^2-y^2})=0$
Use epsilon-delta definition of limits to show that

$$\lim_{(x,y) \to (0,0)}(\sqrt{-x^2-y^2})=0$$

We need to show that $$\forall \epsilon >0( \exists \delta >0( \forall (x,y) \in \mathbb R^2 (0<\sqrt{x^{2}+y^{2}}<\delta \implies \left|\sqrt{-(x^{2}+y^{2})}\right|<\epsilon)))$$

If $0<\sqrt{x^{2}+y^{2}}<\delta$ then $\sqrt{-(x^{2}+y^{2})}<0<\delta$

So I think if $\delta \le \epsilon$ then the claim does hold,is that true?

答:
You have to be careful with the definition of limits in this case. The function
$$
f(x,y)=\sqrt{-x^2-y^2}
$$
is only defined on the set $E=\{(0,0)\}$.

So the notion of limits must be restricted to the (trivial) metric subspace $E$ of $\mathbb{R}^2$.

In the $\epsilon$-$\delta$ definition of your limit, you should have

for every $\epsilon$, there exists $\delta>0$ such that the condition $0<\sqrt{x^2+y^2}<\delta$ and $(x,y)\in E$ implies that $|f(x,y)|<\epsilon$ .

But this is vacuously true since $0<\sqrt{x^2+y^2}<\delta$ is never satisfied.

Added notes.

In general, let $X$ and $Y$ be two metric spaces and $E\subset X$, $f:E\to Y$. To avoid the vacuously true case, one only defines the notion of limits $\lim_{x\to p}f(x)$ when $p$ is a limit point $E$, meaning that every neighborhood of $p$ contains a point $q\ne p$ such that $q\in E$. (See for instance Rudin's Principles of Mathematical Analysis.) Your example of $E$ has no limit points.


问:
If a function is biholomorphic, why is its $n$th power an $n$-to-$1$ function?
Let $f:U\longrightarrow \mathbb{C}$ be an injective holomorphic function (so a biholomorphism). Say $f(0)=0$. Define $F(z)=f(z)^n$. Then why is $F$ an $n$-to-$1$ function on $U \backslash 0$?

I am guessing this is something very simple, but I have not understood this.

答:
Take some nonzero $w_0 \in F(U)$ and notice it has $n$ different complex $n$'th roots:

There is at least one thanks to the fundamental theorem of algebra - the polynomial $w^n - w_0$ reduces to a product of linear factors over $\Bbb{C}$.

All the roots are different from each other, because there are $n$ different roots of unity, and you get the other roots by multiplying by the different roots of unity.

For any $w$ which is an $n$'th root of $w_0$ there is a unique $z \in U$ such that $f(z) = w$, and then $F(z) = w_0$. So the equation $F(z) = w_0$ has exactly $n$ solutions.


问:
If b is rational and c is real, can $x^2 + bx + c$ have one rational root and one irrational root? Use only definition of rational numbers to prove.

If $b$ is rational and $c$ is real, can $x^2 + bx + c$ have one rational root and one irrational root? Use only the definition of rational numbers to prove.

I think that it can't because since $b$ is rational, if two real roots exist then both are either rational or irrational, however, I'm not sure how to prove this using only the definition of rational numbers i.e. it is a number that can be expressed as $\frac{x}{y} $ where $x$ and $y$ are integers $y$ being nonzero. So you can't assume stuff like rational + irrational = irrational.

答:
Of course, note that all rational numbers are indeed real.

I suppose you wanted to ask "Is it possible that one root is rational and the other is not". The answer to that is then "no".

Note that the sum of the two roots is $-b \in \Bbb Q$. Using the definition of rationals, it follows that the difference of two rationals is rational. Conclude that $\text{rational} + \text{irrational} = \text{rational}$ is not possible.


问:
$2\text{D}$ Fourier Transform of Laplacian in polar coordinates
Consider a typical function written in standard $2\text{D}$ polar form:

\begin{equation}
f(\underline{r})=f(r,\theta)=\sum_{n=-\infty}^{\infty} f_n(r) e^{in\theta}
\end{equation}

executing the Laplacian of f, we have:

\begin{equation}
\begin{split}
\nabla^2 f(\underline{r})=& \Bigg(\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r} + \frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}\Bigg) \sum_{n=-\infty}^{\infty} f_n(r) e^{in\theta}\\
=& \sum_{n=-\infty}^{\infty}  \Bigg(\frac{d^2 f_n}{d r^2}+\frac{1}{r}\frac{df_n}{d r}-\frac{n^2 f_n}{r^2}\Bigg) e^{in\theta}\\
=&\sum_{n=-\infty}^{\infty} \nabla^2_n f_n\ e^{in\theta}
\end{split}
\end{equation}

where:
\begin{equation}
\nabla_n^2=\frac{d^2}{d r^2}+\frac{1}{r}\frac{d}{d r}-\frac{n^2}{r^2}
\end{equation}

the 2D Fourier Transform of $\nabla^2 f(\underline{r})$ is given by the following series:

\begin{equation}
\mathbb{F}_{2D}\Big\{ \nabla^2 f(\underline{r})\Big\}=\sum_{n=-\infty}^{\infty} 2\pi i^{-n} e^{in\psi} \int_{0}^{\infty} \Bigg(\frac{d^2 f_n}{d r^2}+\frac{1}{r}\frac{df_n}{d r}-\frac{n^2 f_n}{r^2}\Bigg)J_n(\rho r)\ rdr
\end{equation}

A simple application of integration by parts along with the definition of a
Bessel function gives:

\begin{equation}
\int_{0}^{\infty} \nabla^2_n f_n J_n(\rho r)\ rdr = -\rho^2 \int_{0}^{\infty} f_n J_n(\rho r)\ rdr\ \ \ \ \ (*)
\end{equation}

how could i proof the equation (*)?

答:
The differential equation for the Bessel functions $J_n$,
$$x^2 J_n''(x) + x J_n'(x) + (x^2-n^2) J_n(x) = 0$$
can be written as
$$\left( x J_n'(x) \right)' + (x - \frac{n^2}{x}) J_n(x) = 0,$$
which we will use later.

The given left hand expression can be split into a couple of integrals:
$$\begin{align}
\int_{0}^{\infty} \nabla^2_n f_n \, J_n(\rho r) \, r \, dr
&= \int_{0}^{\infty} \left( f_n''(r) + r^{-1} f_n'(r) - n^2 r^{-2} f_n(r) \right) \, J_n(\rho r) \, r \, dr \\
&= \int_{0}^{\infty} \left( r f_n''(r) + f_n'(r) - n^2 r^{-1} f_n(r) \right) \, J_n(\rho r) \, dr \\
&= \int_{0}^{\infty} \left( (r f_n'(r))' - n^2 r^{-1} f_n(r) \right) \, J_n(\rho r) \, dr \\
&= \int_{0}^{\infty} (r f_n'(r))'  \, J_n(\rho r) \, dr - n^2 \int_{0}^{\infty} r^{-1} f_n(r) \, J_n(\rho r) \, dr
\end{align}$$

For the first integral we use integration by parts. The boundary terms all vanish so
$$\begin{align}
\int_{0}^{\infty} (r f_n'(r))'  \, J_n(\rho r) \, dr
&= - \int_{0}^{\infty} r f_n'(r)  \, \frac{\partial}{\partial r} \left(J_n(\rho r)\right) \, dr
\\
&= - \int_{0}^{\infty} r f_n'(r)  \, \rho J_n'(\rho r) \, dr
\\
&= - \int_{0}^{\infty} f_n'(r)  \, \rho r J_n'(\rho r) \, dr
\\
&=   \int_{0}^{\infty} f_n(r) \, \frac{\partial}{\partial r} \left(\rho r J_n'(\rho r)\right) \, dr
\\
&=   \int_{0}^{\infty} f_n(r) \, \rho \frac{\partial}{\partial (\rho r)} \left(\rho r J_n'(\rho r)\right) \, dr
\\
&=   \int_{0}^{\infty} f_n(r) \, \rho ( \frac{n^2}{\rho r} - \rho r) J_n(\rho r) \, dr
\\
&=   \int_{0}^{\infty} f_n(r) \, ( \frac{n^2}{r} - \rho^2 r) J_n(\rho r) \, dr
\end{align}$$
where the rewritten differential equation for $J_n$ has been used.

Adding the second integral from the first calculation results in
$$\int_{0}^{\infty} \nabla^2_n f_n \, J_n(\rho r) \, r \, dr = -\rho^2 \int_{0}^{\infty} f_n J_n(\rho r)\, r \, dr.$$


答:
Hint

Suppose $r$ is a rational root and $j$ is a irrational root.

$$r+j=-b\to j=-(b+r).$$

Can you see the problem?

Title: If b is rational and c is real, can $x^2 + bx + c$ have one rational root and one irrational root? Use only definition of rational numbers to prove.

Body:
If $b$ is rational and $c$ is real, can $x^2 + bx + c$ have one rational root and one irrational root? Use only the definition of rational numbers to prove.

I think that it can't because since $b$ is rational, if two real roots exist then both are either rational or irrational, however, I'm not sure how to prove this using only the definition of rational numbers i.e. it is a number that can be expressed as $\frac{x}{y} $ where $x$ and $y$ are integers $y$ being nonzero. So you can't assume stuff like rational + irrational = irrational.


问:
What kind of differential equation is $y'=2(\frac{y+2}{x+y-1})^2$
I have an equation in the form $y'=2(\frac{y+2}{x+y-1})^2$. Does this type of equation have a name? I'm struggling to find any information about it or the solution method. My book says this can be solved similarly to the equation $y'=\frac{y+2}{x+y-1}$. That is, with substitutions we can turn this into a homogenous equation and then separate the variables. However, since the expression is inside a function this makes things difficult.

答:
$$y'=2\left(\frac{y+2}{x+y-1}\right)^2$$
$$2x'=\left(\frac{x+y-1}{y+2}\right )^2$$
$$2x'\color{red}{+2}=\left(\frac{x+y-1}{y+2}\right)^2\color{red}{+2}$$
Substitute $w=x+y-1$ and $z=y+2$:
$$2w'=\left(\frac{w}{z}\right)^2+2$$
The differential equation is homogeneous then substitute $w=tz$
$$2(t'z+t)=t^2+2$$
$$2(t-1)'z=(t-1)^2+1$$
This is separable.


问:
How many $(a, b) \in \mathbb{Z}_p^2$ are there such that $ax + b = y \mod p$ for some $x, y \in \mathbb{Z}_p$?
Background: I am trying to convince myself that the family of hash functions $H = \{h_{(a, b)} : \mathbb{Z}_p \rightarrow \mathbb{Z_p}, (a, b) \in \mathbb{Z}_p^2\}$, where $h_{(a, b)}(x) = ax + b \mod{p}$ with prime $p$, is a pairwise-independent family. While I realize there are simpler ways to do this, I can accomplish this if I show for any $x, y \in \mathbb{Z}$, $\Pr_{(a, b) \in \mathbb{Z}_p^2}(ax + b = y) = \frac{1}{p}$, which necessitates showing that there are exactly $p$ distinct pairs $(a, b) \in \mathbb{Z}$ such that $ax + b = y \mod p$. Owing to my weak understanding of field theory, I am unsure if this is true, let alone how to prove it. This leads to my question.

Question: Given $x, y \in \mathbb{Z}_p$, how many $(a, b) \in \mathbb{Z}_p^2$ are there such that $ax + b = y \mod p$?

答:
If $x=0$  you have only one choice for $b$: $b=y$, and $p$ choiches for $a$, in total $p$ choiches:  $(a,y)$, $a\in \mathbb Z_p$.

If $x\neq 0$, it is invertible, and for each choiche of $b$ you must take $a=(y-b)x^{-1}$, so also in this case you have $p$ choiches: $((y-b)x^{-1},b)$, $b\in \mathbb Z_p$.


问:
Bijection between $\mathbb{R}$ and $\{(X,Y), X\subset Y\subset \mathbb{N}\}$
Does anyone know how to construct a bijection between $\mathbb{R}$ and $\{(X,Y), X\subset Y\subset \mathbb{N}\}$?($\subset$ means "lies in or equals to")

I'll  appreciate any help!

Thanks in advance.

答:
No, you can't give an "explicit" bijection. What you can do is showing that your set $S=\{(X,Y):X\subset Y\subset \mathbb{N}\}$ has the same cardinality as $\mathcal{P}(\mathbb{N})$ (the set of subsets of $\mathbb{N}$ which is known to be equipotent to $\mathbb{R}$.

You have an obvious injection $S\to \mathcal{P}(\mathbb{N})\times \mathcal{P}(\mathbb{N})$, so that $|S|\le|\mathcal{P}(\mathbb{N})\times\mathcal{P}(\mathbb{N})|=|\mathcal{P}(\mathbb{N})|$.

You have also an injection $\mathcal{P}(\mathbb{N})\to S$, defined by $Y\mapsto(\emptyset,Y)$, so $|\mathcal{P}(\mathbb{N})|\le|S|$.

Showing an explicit bijection $S\to\mathcal{P}(\mathbb{N})\times\mathcal{P}(\mathbb{N})$ would already be quite complicated, but the Cantor-Schröder-Bernstein theorem gives the conclusion.

Some slight modifications are needed if with $\subset$ you mean “strict inclusion”.


问:
Infinite Products
I am confronted with infinite products. I've tried to read several textbooks on this subject, but most advanced math texts give it short shrift or else delve into its complexities. My question is basic.

Why is the infinite product typically written, using sigma notation, as $1 + a_{k}$. I get it has to do with the fact that convergence demands it approach the multiplicative inverse (which is $1$). Ok. But I don't really really get that. Is it because the $a_{k}$ is actually approaching zero, so you need "$1 + $" to get to "$1$" as the term approaches zero.

In other words, is there a basic for dummies explanation of why you have "$1 + $" in your expression?

Thanks so much.

答:
Let $(z_k)_{k \ge 1}$ be a sequence of non-zero complex numbers and define $$p_n = \prod_{k = 1}^nz_k$$
for $n \ge 1$.

Let us say that $$\prod_{k = 1}^\infty z_k$$
converges if $(p_n)_{n \ge 1}$ converges to a non-zero complex number.

The benefit of this definition now is that we have the usual similarity with series that the product converges only if $z_k \to 1$. Now, the point is that instead of looking at $(z_k)$, we can always consider $w_k:= z_k - 1$. In this case, the product becomes $$\prod_{k = 1}^\infty (1 + w_k)$$
and that converges only if $w_k \to 0$. This is even more familiar with what we see in the case of series.

Title: Infinite Products


问:
Question on Relations
Each of the following defines a relation on the positive integers $\mathbb{N}$:

1-) "$x$ is greater than $y$", 2-) "$xy$ is the square of an integer", 3-) $x$ + $y$ = $10$, 4-) $x$ + $4y$ = $10$

Determine which of the following relations are (a) Reflexive, (b) Symmetric, (c) Antisymmetric, (d) Transitive.

My thoughts:

For part (a), 1 isn't reflexive, 2 is reflexive as $x \cdot x = x^2$ gives us a square of an integer, 3 is also reflexive since $x + x = 10$ is true for $x = 5$(Or is it false because it's only true for $x=5$ and not the other positive integers?), 4 is also reflexive since $x + 4x = 10$ is true for $x=2$(Or is it false because it's only true for $x=2$ and not the other positive integers?).

For part (b) 1 is not symmetric, 2 is symmetric, 3 is symmetric, 4 is symmetric because if we have $x + 4y = 10$, then we must also have $y + 4x = 10$, solving them gives us $x=2,y=2$. Or is it false because it is not true for other positive integers?

For part (c) For 1, $x > y$ and $y > x$ cannot be true together and so it is antisymmetric. ( Since for antisymmetric, we have something of the form p implies q. But since p is always false in this case, we have that p implies q is always true and so we have that  the relation $x > y$ is always antisymmetric ). For 2, since $xy = m^{2}$, where $m$ $\in \mathbb{Z}$, and $yx = m^{2}$, $xy = yx$ does not imply $x=y$.  Therefore, 2 is not antisymmetric. For 3, if $x + y = 10$, and $y + x = 10$, then $x + y = y + x$ does not imply $x=y$. Therefore, 3 is not antisymmetric. For 4, if $x + 4y = 10$ and $y + 4x = 10$, then solving these equations gives us $y=2$ and $x=2$, which does mean $x=y$. But, it is only true for $x=y=2$, so would it be antisymmetric since it does satisfy the requirement for antisymmetric(but only for $x=y=2$), or is not antisymmetric, because it does not fulfill the requirement for antisymmetric for all positive integers x and y?

For (d), 1 is transitive, 2 is not transitive, because if we have $xy = m^2$, $yz = n^2$, then $xz = ( \frac{m \cdot n}{y})^{2}$, it is not necessary that $\frac{mn}{y}$ is an integer. For 3, If we have $x + y = 10$, $y + z = 10$, then $x + z = (10-y) + (10-y) = 20 - 2y$, and it is equal to 10 if $y = 5$. So is it transitive because the requirement is fulfilled but only for $y=5$, or is it not transitive because the requirement is not fulfilled for all positive integers y? For 4, if we have $x + 4y = 10$, $y + 4z = 10$, then $x + 4z = 20 - 5y$ and it is equal to 10 if $y=2$. So is it transitive because the requirement is fulfilled but only for $y=2$, or is it not transitive because the requirement is not fulfilled for all positive integers y?

I tried my best, thank you for reading this.

Please correct any mistakes I made or any misunderstandings I have about relations. Thank you!

答:
(a.1) correct.  (a.2) correct.  (a.3)  No, this is not reflexive.  Reflexive relations would have satisfied that every element in the domain is related to itself.  While it happens to be true that $5$ is related to $5$, it is not true that $4$ is related to $4$ since $4+4\neq 10$.  It fails to be true that every element in the domain is related to itself so it is not reflexive.  (The related property of being non-reflexive where every element is not related to itself also fails since 5 actually is related to itself).  (a.4)  Same issue as (a.3).  This also is not reflexive since again $4$ is not related to itself since $4+4\cdot 4 \neq 10$.

(b.1) correct.  (b.2) correct.  (b.3) correct.  (b.4) No.  Recall that a relation is symmetric iff whenever we have the case that one thing is related to another we must also have that the relation holds in the other direction as well.  That is, if $x\sim y$ then we must also have $y\sim x$.  Here, note that $6$ is related to $1$ since $6+4\cdot 1 = 10$ is true.  However, reversing the roles looking at if $1$ is related to $6$ they are not related since $1+4\cdot 6 \neq 10$.

(c.1) correct.  (c.2) correct.  (c.3) correct.  (c.4)  The relation does happen to be antisymmetric since the only case where both $x+4y=10$ and $y+4x=10$ are true simultaneously are when $x=y=2$.  All other cases we would only have exactly one or none of $x$ related to $y$ or $y$ related to $x$, never both.

(d.1) correct.  (d.2) Incorrect.  When you have $xy$ is a square and $yz$ is a square, the question is if $xz$ is a square.  It will happen to be true that $xz$ is a square and this can be seen by considering things by looking at the prime decomposition of the numbers involved.  $xy$ being a square can be rephrased as "For every prime $p$, the exponent of $p$ in the prime decompositions of $x$ and $y$ must be the same parity.  They are either both even, or both odd."  We see then that for each prime $p$ the exponent of $p$ in $x$ is the same parity as the exponent of $p$ in $y$ as well as the exponent of $p$ in $y$ is the same parity as the exponent of $p$ in $z$ that it must be true that the exponent of $p$ in $x$ is the same parity as the exponent of $p$ in $z$.

(d.3) This is not transitive.  Note that $\color{red}{8}$ (playing the role of $x$) is related to $2$ (playing the role of $y$) since $\color{red}{8}+2=10$.  Further, $2$ is related to $\color{blue}{8}$ (playing the role of $z$) since $2+\color{blue}{8}=10$.  It is not true however that $\color{red}{8}$ is related to $\color{blue}{8}$ since $\color{red}{8}+\color{blue}{8}\neq 10$.

(d.4) correct.  It is indeed transitive since by your analysis if it happened to be true that $x$ is related to $y$ and $y$ is related to $z$ then it follows that $y$ must have been equal to $2$ which implies that all three are equal to $2$ and we do in this case have $x$ is related to $z$ since $2$ is indeed related to $2$.  Since there are no other cases of $x$ being related to $y$ while simultaneously $y$ being related to $z$ to check, we are done.

Title: Question on Relations


问:
How might a Mathematician reformulate this Conjecture?
I have a question pertained to the below conjecture here. However, from the comments, it is clear that the question could be improved using math jargons and symbols for clarity and conciseness. How might a mathematician reformulate this conjecture?

Given an iterative piecewise linear function, $f$, with domain $\mathbb{Z}$ and range $\mathbb{Z}$ :

under iteration by $f$, if the trajectories/iterates of the first integers in all the subdomains converge to a cycle;

then, the trajectories of all integers in the domain of $f$ will converge to a cycle, that is, no trajectory diverges.

Special Conditions: (1) Zero can be included or excluded from $f$. (2) It is assumed that the subdomains and subranges are in arithmetic progression containing non-negative or non-positive integers but not a mix. (3) It is also assumed that if $f$ has only two subfunctions then the cycle must be of length 1 and, of course, $f(x)\neq x$ unless $x$ is the element to which all the trajectories of the first elements converge. (4) Importantly, if one element in a subdomain is greater than its corresponding element in the subrange, then all elements in that subdomain must be greater than  their corresponding elements in the subrange. The same is true if one element in a subdomain is less than its corresponding element in the subrange.

$\color{black}{\text{Example 1}}$

$$f(x) =
\begin{cases}
x+1,  & \text{subdoman: 1, 3, 5, 7,...Subrange: 2, 4, 6, 8,...} \\
-x+2, & \text{subdoman: 2, 4, 6, 8,...Subrange: 0,-2,-4,-6...}\\
-x, & \text{subdoman: -1, -3, -5,...Subrange: 1, 3, 5, ...}\\
x+1, & \text{subdoman: -2, -4, -6,...Subrange: -1, -3, -5,...}\\
0, & \text{subdoman: 0 $\hspace{1.6cm}$ Subrange: 0}
\end{cases}$$
Note that the trajectories of the first elements $(1, 2, -1$ and $-2)$ of the subdomains all converge to $0$ implying that no trajectory will diverge.
For example, the trajectory of $6$ under $f$ is: $6\to-4 \to-3\to3\to4\to-2\to-1\to1\to2\to0.$

$\color{black}{\text{Example 2}}$

Here is a second example in which the first elements of the subdomains converge to the cycle $-1, 1$ and $0$ is excluded:
$$g(x) =
\begin{cases}
\frac{3}{2}x,  & \text{subdoman:2, 4, 6, 8,...Subrange: 3, 6, 9, 12...} \\
\frac{-3}{2}x+\frac{1}{2}, & \text{subdoman:1, 3, 5, 7,...Subrange:-1,-4,-7, -10...}\\
\frac{-1}{2}x+\frac{1}{2}, & \text{subdoman:-1, -3, -5,...Subrange:1, 2, 3, ...}\\
\frac{1}{2}x, & \text{subdoman:-2, -4, -6,...Subrange:-1, -2, -3,...}
\end{cases}$$
In this example, the trajectory of 6 is $6\to9\to-13\to7\to-10\to-5\to3\to-4\to-2\to-1\to1.$

答:
If I understand you correctly, you make the following

Conjecture: Let $f\colon \Bbb Z\to\Bbb Z$ be a function.
Assume that for some index set $I$, there are $a_i,b_i\in\Bbb Z$, $c_i,d_i\in\Bbb Q$ such that

$\Bbb Z=\bigcup_{i\in I}(a_i+b_i\Bbb N_0)$

For all $i\in I$, $n\in\Bbb N_0$, we have $f(a_i+b_in)=c_i+d_in$

For all $i\in I$, we have $a_ib_i\ge 0$ and $c_id_i\ge 0$

For all $i\in I$, if $b_i\ne d_i$, then $\frac{a_i-c_i}{b_i-d_i}>0$

For all $i\in I$, the set $\{\,f^{\circ k}(a_i)\mid k\in\Bbb N_0\,\}$ is finite

Then for all $x\in \Bbb Z$, the set $\{\,f^{\circ k}(x)\mid k\in\Bbb N_0\,\}$ is finite.

The first bullet point describes that we partition $\Bbb Z$ into arithmetic progressions (with singletons allowed); I do note specify that the parts be disjoint for I am sure that the conjecture with and without this condition are equivalent.

The second bullet point expresses the linearity on each of the arithmetic progressions. Note that $b_i=0$ leads to a singleton subdomain (for which necessarily $d_i=0$)

The third bullet point codifies that the subdomains shall be non-negative or non-positive, but not mixed; same for the subranges

The fourth bullet point encodes your requirement (4) (do you see why?)

In the last bullet point speaks about the trajectories of the first elements of subdomains. Note that I use finiteness instead of boundedness or eventual periodicity; all these are equivalent.

In the claim, I also use finiteness instead of boundedness or eventual periodicity.


问:
Parametric Equation of circle parallel to equator on a sphere
Here they say that a circle on a unit sphere obtained by fixing $\theta=\frac{\pi}{4}$ and letting $\phi$ vary can be described parametrically as follows
$$
\left(\cos\frac{\pi}{4}\cos\pi t, \cos\frac{\pi}{4}\sin\pi t, \sin\frac{\pi}{4}\right) \qquad \text{ where } t\in[-1, 1]
$$
How does one obtain this?

Idea 1

I thought maybe they simply computed the radius of the circle
$$
r_{\text{circle}} = r_{\text{sphere}} \cdot \sin\theta = \sin \frac{\pi}{4}
$$
and then used the parametric form of a circle
\begin{align}
    x &= r_{\text{circle}} \cos \phi \\
    y &= r_{\text{circle}} \sin \phi
\end{align}

which would give

\begin{align}
x &= \sin \frac{\pi}{4}\cos\phi \\
y &= \sin\frac{\pi}{4}\sin\phi \\
z &= r_{\text{sphere}}\cos\theta = \cos\frac{\pi}{4}
\end{align}

and setting $\phi = \pi t$ for $t\in[-1, 1]$. However this is different from what they have..

Idea 2

I also tried simply using $\theta = \frac{\pi}{4}$ and $\phi=\pi t$ for $t\in[-1, 1]$ in the parametric equation for a sphere but I get the same result as in "Idea 1".

答: They do seem to have $\sin$ and $\cos$ mixed up, and your Idea 1 does work. But for what it's worth, $\cos \pi / 4 = \sin \pi / 4$ so technically they're not wrong!


问:
Morrey space and Campanato space.
I'd like to know a lot about Morrey space and Campanato spaces. For example, I'd like to know how can I see the details presents here. I'd like some reference about this. I thank you very much. </p>

答: The book Kufner A, John O, Fucik S- Function space is very useful resource for your study. And in addition the book, the article " what is a ...Morrey space? What is... a campanato space?" Melanie rupflin might be useful.</p>

Title: Morrey space and Campanato space.


问:
Why are the only associative division algebras over the real numbers the real numbers, the complex numbers, and the quaternions?
Why are the only (associative) division algebras over the real numbers the real numbers, the complex numbers, and the quaternions?

Here a division algebra is an associative algebra where every nonzero number is invertible (like a field, but without assuming commutativity of multiplication).

This is an old result proved by Frobenius, but I can't remember how the argument goes.  Anyone have a quick proof?

答:
There is another proof using the theory of central simple algebras and quaternion algebras.

Denote by $D$ such a division algebra. Note that a division algebra is always simple. Moreover, since the center of a division algebra is a field, $D$ is central simple over either $\mathbb{C}$ or $\mathbb{R}$.

As there is no nontrivial division algebra over an algebraically closed field, $D$ must be $\mathbb{C}$ if $Z(D)=\mathbb{C}$.

Otherwise, $D$ is central simple over $\mathbb{R}$, then $\mathbb{C}$ splits $D$. Hence we obtain the following:
$$
\sqrt{\mathbf{dim}_{\mathbb{R}}(D)} = \mathbf{ind}_{\mathbb{R}}(D)|[\mathbb{C}:\mathbb{R}]=2
$$
$D$ is nontrivial, hence the dimension of $D$ is 4. Then $D$ must be $\mathbb{H}$ by Wedderburn's theorem.


答:
For $a_n \gt 0$:
$$\prod\limits_{n=1}^{\infty}p_n = \prod\limits_{n=1}^{\infty}(1+a_n) \lt \infty \Leftrightarrow \sum\limits_{n=1}^{\infty}a_n\lt \infty$$
Now all appropriate knowledge for series can be used for products.

Title: Infinite Products

Body:
I am confronted with infinite products. I've tried to read several textbooks on this subject, but most advanced math texts give it short shrift or else delve into its complexities. My question is basic.

Why is the infinite product typically written, using sigma notation, as $1 + a_{k}$. I get it has to do with the fact that convergence demands it approach the multiplicative inverse (which is $1$). Ok. But I don't really really get that. Is it because the $a_{k}$ is actually approaching zero, so you need "$1 + $" to get to "$1$" as the term approaches zero.

In other words, is there a basic for dummies explanation of why you have "$1 + $" in your expression?

Thanks so much.


问:
Explain why it is not necessary to use numerical method to find the two solutions of the equation.
This question I havent been able to understand. I am self studying and have no teacher to ask so I have been stuck for the past hour and don't get this is the full question the part I don't get is 14 c

Any help is much appreciated. Thank you!

答: For part (a) of the problem: $p(x)=20x^3+8x^2-7x-3$ then $p(.5)=-2$ and $p(1)=+18.$ So since the sign is opposite at $.5$ and $1,$ there must be a root between them.


问:
Pythagoras on Mercator's Map
Very stupid question, but I cannot help but wonder whether measuring (and finding true distance after correcting scale distortion) of the horizontal and vertical displacement, and using the Pythagoras theorem will give the length of the path on the Mercator's map? Why/Why not.

答:
At a local level (e.g. looking at a map of your town), using the Pythagorean theorem with the right scaling factor will be a perfectly acceptable method in practice.

However, on a larger scale like the example you've shown, there are two major issues.

(1) There is not a single consistent scale distortion on the map. The top two corners of that square are closer together than the bottom two corners are.

(2) The straight-line path from one point to another on the Earth is not a straight line on the Mercator projection, unless the two points are both on the equator or have the same longitude. Here is an image I found that shows an example of the straight-line path on a Mercator projection versus the shortest distance on the sphere:

To find the true distance between two points given their latitude and longitude, this site has a calculator, relevant formulas, and example code to compute the distance.

Title: Pythagoras on Mercator's Map


问:
Do not understand the solution to this orthogonal projection finding question
There are a number of things I do not understand about this solution:

How do we know $v-x$ is colinear with the normal vector? In an earlier question, we were given that $x$ is a point in $n$-dimensional space. That is all the information we have for $x$. As far as I can understand, $v$ and $x$ are just two random points in space, so how do we know the vector between them is colinear?

So I get that $\lambda$ is the length of the orthogonal projection and $\theta$ gives the direction.But why is $\theta x+\theta_0=0$? It said '$x$ lies in the plane', but again the only info we were given was that $x$ is a point in space, so where does this come from? (In fact what does lying in the plane mean? I get how a point lies on a plane, but not lying in a plane)

Apology for these seemingly amateur questions as I am still trying to learn this stuff.

(These come from MITx 6.86x)

答:
(1) I'm guessing that $x$ is their notation for the projection of $v$ onto the plane $\cal P$. In that case, $v - x$ would be pointing in the normal direction and hence collinear with $\theta$.

(2) Again, now that $x$, the projection of $v$ onto the plane, lies in the plane, it must satisfying the plane's equation: $\theta \cdot x + \theta_0 = 0$.


问:
Generating functions and the Riemann Zeta Function
The generating function for the terms of the harmonic series:

$\frac{1}{n}$

is $-\ln(1 - x)$.

Does an ordinary generating function exist for the terms of the zeta function $\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}$
for any $ s > 1$?

That is, does there exist any function $f(x)$ that can be expressed in terms of elementary functions such that $f(x) = \sum_{n=1}^\infty \frac{1}{n^s}x^n$ for some $s > 1$?

I'm assuming that such a function in fact does not exist.  Can this be proven to be the case?

答: The function you are looking for is the Li$_s(z)$, the polylogarithm.</p>


问:
Show that the limit can be moved inside the integral for this function
For $x \in [-1,1]$ and $n \in \mathbb{N}$, consider

$f_n(x) = \frac{x^{2n}}{(1 + x^{2n})}$

I have found that the function converges pointwise to

$f(x) = \begin{cases}
\frac{1}{2} & \mathrm{if} x = 1,-1 \\
0 & \mathrm{if} x \in (-1,1) \end{cases}$

The convergence is not uniform, since $f$ is discontinuous at the endpoints, but I know that
$\lim_{n \to \infty} \int_{-1}^{1} f_n(x) dx = \int_{-1}^{1} f(x) dx$ is true (because my instructor told me).

I know that $\int_{-1}^{1} f(x) dx = 0$ because the function is similar to $g(x) = 0$ except on a set of measure $0$ (the set of discontinuities.) However, I am not sure how to calculate the limit of the integral of the sequence of functions. I've tried doing it with some elementary integration methods from Calculus, but I'm not getting anywhere.

My next thought to prove they are equal was to find the exact condition (as minimal as it maybe) regarding when I can move the limit inside of the integral. However, I have not had much luck finding this.

Due to that, my next thought was possibly to prove that if a function is uniformly convergent everywhere except on a set of measure $0$ (which it is), then the limit can be moved inside of the integral. Would this be the best way to go about this problem? And if so, how would one prove this? I'm not really sure where to start this proof.

Note: My class has not covered Borel sets, sigma algebras, or any Lebesgue convergence theorems. My instructor said this is doable with techniques done before this.

Body:
HINT:

Note that $0\le \frac{x^{2n}}{1+x^{2n}}\le x^{2n}$

What is $\int_{-1}^1 x^{2n}\,dx$?


问:
What is an example of an etale morphism which does not locally factor as a composition of an open immersion and a finite etale map?
Let $f:X\rightarrow Y$ be an etale morphism of schemes. By $f$ factoring locally as a composition of an open immersion and a finite etale map, I mean that for every $x \in X$ there exists an open $U$ of $x$ and an open $V \subset Y$ containing $f(U)$ such that the restriction $f:U \rightarrow V$ factors as $U \hookrightarrow W \rightarrow V$, where $U \hookrightarrow W$ is an open immersion and $W \rightarrow V$ is finite etale.

According to section 7.5 of the "Berkeley lectures" by Scholze and Weinstein, it is not true that every etale $f$ factors locally as a composition of an open immersion and a finite etale map in the category of schemes (though it is true in the category of analytic adic spaces). What is an example of this failure for schemes?

Thanks!

答:
Ariyan Javanpeykar's answer to a related question gives an example that works. Since you're asking for something a priori slightly stronger than in that question (no local factorization into open immersion + finite étale, not just no global factorization), I'll elaborate. The example is as follows: Let $X = \mathbb{A}^1 \setminus \{0, 2/3\}$, and define a morphism $f\colon X \to \mathbb{A}^1$ by $f(x) = x^2 (x - 1)$. Note that $f$ is étale.

Let $U \subseteq X$ be any open subset containing $1 \in X$, and let $V \subseteq \mathbb{A}^1$ be any open subset such that $f(U) \subseteq V$. Note that $0 = f(1) \in V$. Suppose we have a scheme $W$, an open immersion $j\colon U \to W$, and a finite morphism $h\colon W \to V$ such that $f\rvert_U = h \circ j$. Let $Z \subseteq W$ be the irreducible component containing $j(U)$ (with the reduced induced closed subscheme structure). Irreducible components are closed, and closed immersions are finite, so $g := h\rvert_Z\colon Z \to V$ is finite. Finite morphisms are affine, so $Z$ is affine, irreducible, reduced, 1-dimensional, and birational to $\mathbb{A}^1$, which means $Z$ is an open subscheme of $\mathbb{A}^1$. So $g$ is given by a rational function, but since $g$ agrees with $f$ on a dense open subset, this rational function must be $x^2 (x - 1)$.

But since $g$ is finite, all the fibers have the same degree, so $g^{-1}(0)$ consists of three points, counting multiplicity. Since $x^2 (x - 1)$ is ramified above $0$, this means $g$ (and hence $h$) cannot be étale.

Actually, now that I've written this up, I'm a little confused why this doesn't also give a counterexample for analytic adic spaces. I guess it has to do with the more flexible nature of open immersions in the analytic setting?


问:
Doubt in handling double sums of the type: $\sum_{n,k\in\mathbb{N}_1}\frac{1}{n^2k^2(n+k)}$
I'm having a little trouble calculating the double series given by the proposition below:

$$\int_0^1\frac{\left[\text{Li}_2(x)\right]^2}{x}\mathrm dx=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{1}{n^2k^2(n+k)}=2\zeta(2)\zeta(3)-3\zeta(5).\tag{1}$$

In fact, this integral has already been evaluated here and here, but i have no intention of evaluating it by integration or by using harmonic series. My interest is to attack the double sum in $(1)$ only using the manipulation of sums. Having clarified my interest in this topic, i naively proceeded as follows:
$$\begin{align*}\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{1}{n^2k^2(n+k)}&=\sum_{n,k\in\mathbb{N}_1}\frac{1}{n^2k^2(n+k)}\\
&=\sum_{\substack{{n,k\in\mathbb{N}_1}\\{n=k}}}\frac{1}{n^2k^2(n+k)}+\sum_{\substack{{n,k\in\mathbb{N}_1}\\{n>k}}}\frac{1}{n^2k^2(n+k)}+\sum_{\substack{{n,k\in\mathbb{N}_1}\\{n<k}}}\frac{1}{n^2k^2(n+k)}\\
&=\frac{\zeta(5)}{2}+\sum_{\substack{{n,k\in\mathbb{N}_1}\\{n>k}}}\frac{1}{n^2k^2(n+k)}+\sum_{\substack{{n,k\in\mathbb{N}_1}\\{n<k}}}\frac{1}{n^2k^2(n+k)}\tag{a}\end{align*}$$

Here's my difficulty with that approach! How can i solve the last two double sums in $(\text a)$ with the indexes $ n <k $ and $ n> k $ ?. Studying other posts (links at the end of the question) related to my question, i found here another possible approach using the fact that: $ \frac{1}{n^2k^2}=\frac{1}{(n+k)^2n^2}+\frac{1}{(n+k)^2k^2}+\frac{2}{(n+k)^3k}+\frac{2}{(n+k)^3n}$, then multiplying by $\frac{1}{n+k}$ on both sides and taking the double sum, we get:
$$\sum_{n,k\in\mathbb{N}_1}\frac{1}{n^2k^2(n+k)}=2\sum_{n,k\in\mathbb{N}_1}\frac{1}{(n+k)^3n^2}+4\sum_{n,k\in\mathbb{N}_1}\frac{1}{(n+k)^4n}.\tag{b} $$

According to this article, $(\text b)$ is equal to $T(2,2,1)=2T(2,0,3)+4T(1,0,4)$, where $T(a,b,c)$ is Tornheim’s double series defined by $T(a,b,c)=\sum_{n,m=1}\frac{1}{n^am^b(n+m)^c}$. In this second approach, the expression in $(\text b)$ seems to me a little less aggressive, but if i apply the concepts of the mentioned article, i would feel a little difficult to understand, since it starts with already established formulas. In short, two questions hang in my mind: How to work $(\text a)$ and $(\text b)$ without using integrals or harmonic numbers?

Related links:

A. $\displaystyle \sum_{m,n=1\, m\neq n}^\infty{\frac{m^2+n^2}{mn(m^2-n^2)^2}}$

B. $\displaystyle \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2}$

C. $\displaystyle \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$

答:
One way of doing the sum is as follows

\begin{align} \sum\limits_{n,k = 1}^\infty  \frac{1}{n^2k^2(n+k)} &= \sum\limits_{n,k = 1}^\infty  \frac{k(n+k) - k^2}{n^3k^3(n+k)} \\&= \sum\limits_{n,k = 1}^\infty  \frac{1}{n^3k^2} - \sum\limits_{n,k = 1}^\infty  \frac{1}{n^3k(n+k)} \\&= \zeta(2)\zeta(3) - \sum\limits_{n=1}^\infty \frac{1}{n^3} \sum\limits_{k=1}^\infty \frac{1}{k(n+k)} \\&= \zeta(2)\zeta(3) - \sum\limits_{n=1}^\infty \frac{H_n}{n^4}  \end{align}

where, we used the identity $\displaystyle \sum\limits_{k=1}^\infty \frac{1}{k(n+k)} = \frac{H_n}{n}$ in the second last line. The last sum is the well known Euler sum, which itself can be derived from partial fraction decomposition of $\displaystyle \frac{1}{n^3k(n+k)}$. Hope this helps! :)

As to the last Euler sum you may see Rob's answer here Generalized Euler sum $\sum_{n=1}^\infty \frac{H_n}{n^q}$.


问:
X is a direct summand of $X\otimes DX\otimes X$ in a rigid tensor-triangulated category.
I am reading Balmer's "Supports and filtrations in algebraic geometry and modular representation theory". In Prop. 2.4 he claims that $X$ (an object of a rigid tensor triangulated category) is a direct summand of $X\otimes DX\otimes X$ (see this topic All Ideals are Radical in Rigid Categories). However, my impression is that the proof only shows that $X$ is a rectract of $X\otimes DX\otimes X$. It looks like the distinction between those two notions is quite crucial for the later applications (thickness of some ideals). Later in this paper he assumes additionally that the category is also idempotent complete, in which case retract happens to be the same as a direct summand.

Coproducts are distributive under the tensor product (since they are adjoints), so one could try to show that $1$ is a direct summand of $X\otimes DX$, but I was unable to show that.

Any help would be highly appreciated.

答:
In a triangulated category a retract is a direct summand.

Suppose $X$ is a retract of $Y$. So there are maps $\alpha: X\to Y$ and $\beta: Y\to X$ with $\beta\alpha=\text{id}_X$. Complete $\alpha$ to a distinguished triangle
$$Z[-1]\xrightarrow{\gamma}X\xrightarrow{\alpha}Y\rightarrow Z.$$
Then $\gamma=\beta\alpha\gamma=0$, so the triangle is isomorphic to
$$Z[-1]\xrightarrow{0}X\rightarrow X\oplus Z\rightarrow Z.$$


问:
Let A be $\subset \mathbb{R}$ and $A_1........A_n$ be $\subset \mathbb{R}$, such that. A = $\bigcup_{i=1}^{N}A_i$
If each $A_i$ is closed in $A$, show that a given function $f:A \to B$ is continuous, if and only if the restrictions $f_{A_i} \to B$ is continuous.

I'm not sure how to go about solving this question. I know that a set is only closed if its complement is open but I don't see how that helps me here unfortunately.

答:
If $f:A\longrightarrow B$ is continuous, obviously $f|_{A_i}$ is continuous for all $i\in\{1,\ldots,n\}$.

Now assume that all the $f|_{A_i}$, $i\in\{1,\ldots,n\}$ are continuous, and let $x_0\in A:=\bigcup_{i=1}^n A_i$. Let $i_1,\ldots,i_k\in\{1,\ldots,n\}$ be the indexes of all the sets $A_{i_j}$ such that $x_0\in A_{i_j}$.

Also, because the $A_i$ are all closed, $\mathbb{R}\setminus A_i$ are all open.
Then, if $x_0\not\in A_i$ for $i\not\in\{i_1,\ldots,i_k\}$,  we can take $\delta_i\in\mathbb{R}^+$ such that $B(x_0,\delta_i)\cap A_i=\emptyset$. Let then $\delta_M=\min\{\delta_i :i\not\in\{i_1,\ldots,i_k\}\}$. We have that $B(x_0,\delta)\cap A_i=\emptyset$ for all $i\not\in\{i_1,\ldots,i_k\}$.

Let $\epsilon\in\mathbb{R}^+$, and let $\delta_{i_j}\in\mathbb{R}^+$ such that $|f|_{A_{i_j}}(x)-f(x_0)|<\epsilon$ for all $x\in B(x_0,\delta_{i_j})\cap A_{i_j}$ and $j\in\{1,\ldots,k\}$.

If we choose $\delta:=\min\{\delta_M,\delta_{i_j} : j\in\{1,\ldots,k\}\}$, then, if $x\in B(x_0,\delta)\cap A$,  $x\not\in A_i$ for all $i\not\in\{i_1,\ldots,i_j\}$ because $B(x_0,\delta)\cap A_i=\emptyset$. As $x\in A=\bigcup_{i=1}^n A_i$, there must exist $i_j\in\{i_1,\ldots,i_k\}$ such that $x\in A_{i_j}$. Then, $x\in B(x_0,\delta)\cap A_{i_j}\subset B(x_0,\delta_{i_j})\cap A_{i_j}$, and we get that: $$|f(x)-f(x_0)|=|f|_{A_{i_j}}(x)-f(x_0)|<\epsilon$$
Which concludes the proof.


问:
Let $\{a_n\}$ be a positive sequence and $\sum_{n=1}^\infty a_n^2=A$ . Calculate $\lim_{N\rightarrow\infty}\frac{1}{\sqrt{N}}\sum_{n=1}^N a_n.$
Let $\{a_n\}$ be a positive sequence and $\sum_{n=1}^\infty a_n^2=A$ . Calculate $$\lim_{N\rightarrow\infty}\frac{1}{\sqrt{N}}\sum_{n=1}^N a_n.$$

I have found that $$\Big(\frac{1}{\sqrt{N}}\sum_{n=1}^N a_n\Big)^2=\Big(\sum_{n=1}^N \frac{a_n}{\sqrt{N}}\Big)^2\leqslant\Big(\sum_{n=1}^N\frac{1}{N}  \Big)\Big( \sum_{n=1}^N a_n^2 \Big)=\sum_{n=1}^N a_n^2 \underset{N\rightarrow\infty}{\longrightarrow} A $$
by Cauchy Inequalty. But if I assume $a_n=\frac{1}{n}$ and $A=\frac{\pi^2}{6}$, then $\lim_{N\rightarrow\infty}\frac{1}{\sqrt{N}}\sum_{n=1}^N a_n=0$ is not $A$. I cannot find a way to prove the limit is $0$.

Thank you for helping!

答:
Motivation: If it is correct that $\sum_{n=1}^\infty a_n^2 = A < \infty $ implies $\lim_{N\to\infty}\frac{1}{\sqrt{N}}\sum_{n=1}^N a_n=0$, then the result does not depend on the actual value $A$, only on the fact that the series converges, and that is equivalent to the fact that the series remainder $\sum_{n=m+1}^\infty a_n^2$ converge to zero for $m \to \infty$.

This idea leads to the following proof:
For $1 \le m < N$ we split the sum in two parts
$$
\frac{1}{\sqrt{N}}\sum_{n=1}^N a_n = \frac{1}{\sqrt{N}}\sum_{n=1}^m a_n + \frac{1}{\sqrt{N}}\sum_{n=m+1}^N a_n
$$
and apply the Cauchy-Schwarz inequality only to the second part:
$$
\frac{1}{\sqrt{N}}\sum_{n=1}^N a_n \le \frac{1}{\sqrt{N}}\sum_{n=1}^m a_n
+ \sqrt{\frac{N-m}{N} \sum_{n=m+1}^N a_n^2 } \, .
$$
Now fix $m$ and take the $\limsup$ for $N \to \infty$:
$$
 \limsup_{N \to \infty} \frac{1}{\sqrt{N}}\sum_{n=1}^N a_n \le  \sqrt{ \sum_{n=m+1}^\infty a_n^2 } \, .
$$
This holds for all $m$. The right-hand side converges to zero for $m \to \infty$ because the series $\sum_{n=1}^\infty a_n^2$ converges. It follows that
$$
  \lim_{N \to \infty} \frac{1}{\sqrt{N}}\sum_{n=1}^N a_n = 0 \, .
$$


问:
What is the advantage of Runge Kutta (or any other) method over just solving the o.d.e?
An example in task had a model with "runge kutta RK4 approximation applied" so I naturally went to Google what Runge Kutta method is. I found how to use it and why it's better over euler method but I didn't find why/what it is used for.

From what I read, Runge-Kutta is used to approximate ordinary differential equation solutions.
And here's my question: Why use Runge-Kutta, if we can just solve the differential equation using integral and get infinitely more precise solution?
Question is not limited to Runge Kutta, it is just the one I'm currently working with.

Perhaps my understanding of RK or o.d.e is totally wrong.

答:
Why use Runge-Kutta, if we can just solve the differential equation using integral and get infinitely more precise solution?

That's a pretty darn big "if" you have there. In fact, most differential equations, ordinary or not, cannot be solved. And in practical use, we may not even know the true form of the equation itself, just a few discrete measurements of the coefficient functions. That's what numerical methods like Euler or RK are really for.


问:
Derivative of $h(t,x)=x\cdot x=x_1^2+\cdots+x_n^2$
Suppose that $x'=A(t)x=-A(t)^{T}x$, where $A(t)$ is a matricial function. What is the derivative of $h(t,x)=x\cdot x=x_1^2+\cdots+x_n^2$. I'm trying to prove that it is $0$, this way $h$ is a first integral for the system.

Attempt: The derivative of $h$ is $$\frac{\partial h}{\partial x}(t,x)x'+\frac{\partial h}{\partial t}(t,x)=2x \cdot Ax + 0 \neq 0 $$ what I'm doing wrong?

答: Just to make it clear: since $x^T Ax\in \mathbb R$ it holds  $x^T Ax=(x^T Ax)^T= x^T A^Tx= -x^T Ax$, hence  $x^T Ax=0$.


问:
Comparison test for $\sum_{n=1}^{\infty} \frac{\sqrt{n^5 + n^3 - n}}{2n^3 + 3n^2 + 1}$
I'm trying to use comparison test to show that this series diverges:

$$\sum_{n=1}^{\infty} \frac{\sqrt{n^5 + n^3 - n}}{2n^3 + 3n^2 + 1}$$

Here are the steps I have come up with but I am not sure if it is right.

$$\sum_{n=1}^{\infty} \frac{\sqrt{n^5}}{3n^3}\ \text{diverges (by integral test)}$$

$$0 < \frac{\sqrt{n^5}}{3n^3} \leq a_n\ \text{(for large n)}$$

$$\therefore a_n\ \text{diverges}$$

Note: I added "for large n" since according to Wolfram|Alpha, it is only true for n > 2.671... is this permissible or does it mean I have made an error?

Body:
$$\frac{\sqrt{n^5 + n^3 - n}}{2n^3 + 3n^2 + 1}\ge\frac{\sqrt{n^5}}{2n^3+2n^3+2n^3}=\frac{n^{5/2}}{6n^3}=\frac16\frac1{\sqrt n}$$

and since $\;\sum\frac1{\sqrt n}\;$ diverges ...


问:
Alternative version of linear regression models
I am learning regression this semester and we are just starting out on simple linear regression. The textbook that my lecturer is using teaches that the most basic model is of the form $$Y_i = \beta_0 + \beta_1 X_i + \epsilon_i.$$ The textbook goes on to explain the terms in the equation, which I shall not write here because it is irrelevant to my question.

I understand the above just fine.

Then, the textbook goes on to mention that an alternative form of the model can be written as $$Y_i = \beta_0X_0 + \beta_1 X_i + \epsilon_i$$ where $X_0 \equiv 1.$

The textbook just gives a one-line elaboration for the alternative model, saying that it associates an $X$ variable with each regression coefficient.

I have two questions:

I am just unclear as to what context would the alternative model be applicable/relevant.

As an aside (just a trivial question, really), why did they use $\equiv$ and not $=$? I understand that $\equiv$ means identically equal and $=$ just means equal and I have also read some posts regarding the two signs but I still have some confusion as to when which is used.

Note also that everything written above is as per the textbook; I have not rephrased anything.

I know that this may be very basic for the geniuses here, but I feel that it is important for me to get my foundations right before moving on, so any explanations will be greatly appreciated! :)

答:
In effect, the alternative model is exactly the same as the basic one. It could be helpful for some software out there though, where all $ \beta $ terms have to have an accompanying $ X $.

It is not just stating that $ X_0 $ is equal to 1, but rather, it is defining $ X_0 $ as 1.


答:
Yes, it is permissible, and instead of worrying about a direct comparison you should use the limit comparison test: since
$$\lim_{n \to \infty} \frac{\dfrac{\sqrt{n^5+n^3-n}}{2n^3 + 3n^2 + 1}}{\dfrac{\sqrt{n^5}}{2n^3}} = 1$$
the series $$\sum_{n=1}^\infty \dfrac{\sqrt{n^5+n^3-n}}{2n^3 + 3n^2 + 1} \quad \text{and} \quad \sum_{n=1}^\infty \dfrac{\sqrt{n^5}}{2n^3}$$ either both converge or both diverge.

Title: Comparison test for $\sum_{n=1}^{\infty} \frac{\sqrt{n^5 + n^3 - n}}{2n^3 + 3n^2 + 1}$

Body:
I'm trying to use comparison test to show that this series diverges:

$$\sum_{n=1}^{\infty} \frac{\sqrt{n^5 + n^3 - n}}{2n^3 + 3n^2 + 1}$$

Here are the steps I have come up with but I am not sure if it is right.

$$\sum_{n=1}^{\infty} \frac{\sqrt{n^5}}{3n^3}\ \text{diverges (by integral test)}$$

$$0 < \frac{\sqrt{n^5}}{3n^3} \leq a_n\ \text{(for large n)}$$

$$\therefore a_n\ \text{diverges}$$

Note: I added "for large n" since according to Wolfram|Alpha, it is only true for n > 2.671... is this permissible or does it mean I have made an error?


问:
Which one result in mathematics has surprised you the most?
A large part of my fascination in mathematics is because of some very surprising results that I have seen there.

I remember one I found very hard to swallow when I first encountered it, was what is known as the Banach Tarski Paradox. It states that you can separate a ball $x^2+y^2+z^2 \le 1$ into finitely many disjoint parts, rotate and translate them and rejoin (by taking disjoint union), and you end up with exactly two complete balls of the same radius!

So I ask you which are your most surprising moments in maths?

Chances are you will have more than one. May I request post multiple answers in that case, so the voting system will bring the ones most people think as surprising up. Thanks!

答: The fact that the curve of fastest descent (i.e., the brachistochrone) dips beneath its target!</p>


问:
Prove that the function f(x) = (x-3)^2 is convex on [1,5]
A function f is defined to be convex on the closed interval $[1,5]$ if and only if $f(t+(1-t)5) \le tf(1) + (1-t)f(5)$ for any $t$ between $0$ and $1$ inclusive $(0\le t\le 1).$

Please help me prove this or at least where to start and what I am looking for at the end. I am lost.

I would really appreciate it!

Thanks!

答:
In this case,
the function has a second derivative
and $f''(x) = 2 > 0$
so it is convex.


问:
Motivation for the importance of topology
Starting from tomorrow, I will be tutoring some undergraduate students following a course in general topology. I am looking for examples motivating the importance of topology in mathematics which can be explained without too much difficulty using concepts of other areas of mathematics (or physics) they have already treated (those areas would be mainly analysis, complex analysis, linear algebra, a little graph theory, some numerical methods for maths, and classical mechanics, electromagnetism, special relativity, some QM and a little statistical physics for physics). I have tried looking around, but I have found little that would motivate me to follow such a course. Do anybody have some nice example?

Note: I will of course explain to them that without topology they'll be able to do very little advanced mathematics (e.g. functional analysis, differential geometry, ...)

EDIT: Ok, I gave as examples Tychonoff's theorem, Brower's fixed point theorem and the Jordan curve theorem.

I would like to keep this question alive, for personal interest. What are interesting (not too hard) applications of topology in other areas of mathematics?

答: One useful approach to teaching a new subject is to connect it to things the students are already familiar with. For example, you mentioned that they have had some treatment of graph theory. Perhaps you could take some mathematical facts from this thesis which connects topology with graph theory.</p>


问:
Prove $x=\int_{1}^{x}\cos{\frac{1}{t}}\cdot\ln\left(3-\frac{1}{t}\right) \mathrm{d}t $ for some $x>1$
Prove that $$x=\int_{1}^{x}\cos{\frac{1}{t}}\cdot\ln\left(3-\frac{1}{t}\right) \mathrm{d}t $$ for some $x>1.$

The only idea I have is to calculate the derivative of $\int_{1}^{x}\cos{\frac{1}{t}}\cdot\ln\left(3-\frac{1}{t}\right) \ \mathrm{d}t $. First, I let $y=\int_{1}^{x}\cos{\frac{1}{t}}\cdot\ln\left(3-\frac{1}{t}\right) \mathrm{d}t -x$ , then I calculate $y'(x)$ , but $y(x)$ isn't a monotonous function, so I got stuck.

Can anyone help or give me a hint on how to approach this problem?

Thank you very much guys

答:
At x = 1 then

$$\int_{1}^{x} cos \frac{1}{t} ⋅ ln(3 - \frac{1}{t})dt = 0 \lt x $$

What does the integrand approach as t approaches infinity?

Can you see what the integral approaches in terms of x by considering the graph of this?

The integral is continuous for x $\geq$ 1. Can you see how to prove it from here?


答:
It is much simpler to use this result from asymptotic analysis:

Let $\sum_n u_n$ and $\sum_n v_n$ two  series with (ultimately) positive terms, such that $u_n\sim_\infty v_n$. Then $\sum _n u_n$ and $\sum_n v_n$ both converge or both diverge.

Now a polynomial is asymptotically equivalent to its leading term, so
$$\frac{\sqrt{n^5 + n^3 - n}}{2n^3 + 3n^2 + 1}\sim_\infty\frac{\sqrt{n^5}}{2n^3}=\frac 1{2\sqrt n},$$
and the latter is a divergent $p$-series.


答:
When I started taking Differential Equations (18.03) at MIT, I quickly got the impression that Diff Eq is not a "finished" class.  They haven't finished discovering it yet, they haven't figured out how to solve huge categories of equations, and so forth.  There was very little by way of general techniques, and a whole bunch of specialized tricks: "If you see this, try that."  It was like integration but squared.  "Derivatives are a science; Integrals are an art; Differential Equations are a black art."

So I found myself wondering why they were bothering to teach it when it wasn't done yet.  The answer is simple: our understanding of Nature is currently written in the language of differential equations.  We don't have time to wait for a finished, perfect theory.  If you go into physics or any sort of engineering, you are probably going to be fighting with differential equations at every turn.  So they teach us as best they can the tools that have been developed so far.
Differential equations are HARD.  If you find an exact solution of the Einstein equations, for example, you usually get it named after you.

Numerical approximation is the approach when all else fails.  And all else fails far more often than not.  Different techniques of approximation have different efficiencies in terms of computation time and memory usage and so forth, and it makes sense to pick the technique that works most efficiently for your problem.  Runge-Kutta is one of those techniques.

Title: What is the advantage of Runge Kutta (or any other) method over just solving the o.d.e?

Body:
An example in task had a model with "runge kutta RK4 approximation applied" so I naturally went to Google what Runge Kutta method is. I found how to use it and why it's better over euler method but I didn't find why/what it is used for.

From what I read, Runge-Kutta is used to approximate ordinary differential equation solutions.
And here's my question: Why use Runge-Kutta, if we can just solve the differential equation using integral and get infinitely more precise solution?
Question is not limited to Runge Kutta, it is just the one I'm currently working with.

Perhaps my understanding of RK or o.d.e is totally wrong.


问:
Question on Relations
Each of the following defines a relation on the positive integers $\mathbb{N}$:

1-) "$x$ is greater than $y$", 2-) "$xy$ is the square of an integer", 3-) $x$ + $y$ = $10$, 4-) $x$ + $4y$ = $10$

Determine which of the following relations are (a) Reflexive, (b) Symmetric, (c) Antisymmetric, (d) Transitive.

My thoughts:

For part (a), 1 isn't reflexive, 2 is reflexive as $x \cdot x = x^2$ gives us a square of an integer, 3 is also reflexive since $x + x = 10$ is true for $x = 5$(Or is it false because it's only true for $x=5$ and not the other positive integers?), 4 is also reflexive since $x + 4x = 10$ is true for $x=2$(Or is it false because it's only true for $x=2$ and not the other positive integers?).

For part (b) 1 is not symmetric, 2 is symmetric, 3 is symmetric, 4 is symmetric because if we have $x + 4y = 10$, then we must also have $y + 4x = 10$, solving them gives us $x=2,y=2$. Or is it false because it is not true for other positive integers?

For part (c) For 1, $x > y$ and $y > x$ cannot be true together and so it is antisymmetric. ( Since for antisymmetric, we have something of the form p implies q. But since p is always false in this case, we have that p implies q is always true and so we have that  the relation $x > y$ is always antisymmetric ). For 2, since $xy = m^{2}$, where $m$ $\in \mathbb{Z}$, and $yx = m^{2}$, $xy = yx$ does not imply $x=y$.  Therefore, 2 is not antisymmetric. For 3, if $x + y = 10$, and $y + x = 10$, then $x + y = y + x$ does not imply $x=y$. Therefore, 3 is not antisymmetric. For 4, if $x + 4y = 10$ and $y + 4x = 10$, then solving these equations gives us $y=2$ and $x=2$, which does mean $x=y$. But, it is only true for $x=y=2$, so would it be antisymmetric since it does satisfy the requirement for antisymmetric(but only for $x=y=2$), or is not antisymmetric, because it does not fulfill the requirement for antisymmetric for all positive integers x and y?

For (d), 1 is transitive, 2 is not transitive, because if we have $xy = m^2$, $yz = n^2$, then $xz = ( \frac{m \cdot n}{y})^{2}$, it is not necessary that $\frac{mn}{y}$ is an integer. For 3, If we have $x + y = 10$, $y + z = 10$, then $x + z = (10-y) + (10-y) = 20 - 2y$, and it is equal to 10 if $y = 5$. So is it transitive because the requirement is fulfilled but only for $y=5$, or is it not transitive because the requirement is not fulfilled for all positive integers y? For 4, if we have $x + 4y = 10$, $y + 4z = 10$, then $x + 4z = 20 - 5y$ and it is equal to 10 if $y=2$. So is it transitive because the requirement is fulfilled but only for $y=2$, or is it not transitive because the requirement is not fulfilled for all positive integers y?

I tried my best, thank you for reading this.

Please correct any mistakes I made or any misunderstandings I have about relations. Thank you!

答:
For part (a), 1 isn't reflexive,

Correct. As $n > n$, is never true it is never reflexive.

2 is reflexive as x⋅x=x2 gives us a square of an integer,

Correct.  For every $n\in\mathbb N, n\cdot n = n^2$ a perfect square.

3 is also reflexive since x+x=10 is true for x=5(Or is it false because it's only true for x=5 and not the other positive integers?),

The latter.  To be reflexive it must be that ALL $n \in \mathbb N$ we must have $n + n = 10$.  That's obviously not true for all $n$.  It doesn't matter if it were true for all $n$ but $1$.  So long as there is a single counter example (and we have an infinite number of counter examples) is it not reflexive.

Ditto for 4)

For part (b) 1 is not symmetric,

Correct.  If $n > m$ then we do not ever have $m> n$.

2 is symmetric

Correct.  Multiplication is commutative so if $nm = k^2$ is a perfect square then $mn = k^2$ as well.

, 3 is symmetric

Correct.  Addition is commutative so if $n+m = 10$ then $m +n = n+m=10$.

4 is symmetric because if we have x+4y=10, then we must also have y+4x=10, solving them gives us x=2,y=2. Or is it false because it is not true for other positive integers?

The later.  Symmetric means true for all pairs of natural numbers.  A single counter-example will dispute it.  So not symmetric.

This is a bit of a tricky relation as we will see later.  The only positive integer solutions to $x + 4y = 10$ are $(x=2, y=2)$ or $(x=6, y=1)$.

So we have $6 + 4*1=10$ but $1+ 4*6 \ne 10$ so that is the one and only counter example.

Not symmetric.  (A weird thing will happen when we test for transitivity)

For part (c) For 1, x>y
and y>x cannot be true together and so it is antisymmetric. ( Since for antisymmetric, we have something of the form p implies q. But since p is always false in this case, we have that p implies q is always true and so we have that the relation x>y is always antisymmetric ).

Not sure what your statement $p$ or $q$ are supposed to be.

The way I like to think of "antisymmetric" is this:

Given that $a R B$ for an arbitrary pair $a,b$ we might be able to conclude three possible results:

$bRa$ is always true.  If so this is symmetric.

$bRa$ might sometimes be true, but sometimes not.  Then this is... boring.  (We don't have a term for this.

$a$ and $b$ were arbitrary.  If by chance we picked an $a$ and $b$ so that $a = b$ than $aRb$ is the same statement as $aRa$ is the same as $bRa$ and $bRa$ will occur trivially.  But if we ignore these trivial exceptional cases and assume we picked an arbitrary $a,b$ but $a\ne b$.   Then if $bRa$ is impossible then $R$ is antisymmetric.

Anyway.  If $n > m$ then $m>n$ is impossible.  SO this relation is anti-symmetric.

For 2, since xy=m2, where m ∈Z, and yx=m2, xy=yx does not imply x=y. Therefore, 2 is not antisymmetric.

Correct.  Again multiplication is commutative so $xy = m^2$ where $x \ne y$ would mean $yx = m^2$.  SO non-trivial symmetry is not impossible.

Of course an example of $xy = m^2$ but $x\ne y$ could support your case.  (But there are many counter examples.)

For 3, if x+y=10, and y+x=10, then x+y=y+x does not imply x=y. Therefore, 3 is not antisymmetric.

Correct.  But counter examples are good idea.  Sometimes we can do relationships are many small sets or with conditions.

It's possible to have a relation on small sets.  Say for instance if our set wasn't $\mathbb N$ but the set $\{1,2,3,5,6\}$ and we had the relation $x + y = 10$.  That relation IS anti-symmetric because in that case $x + y = y+x = 10\implies x=y=5$.

For 4, if x+4y=10 and y+4x=10, then solving these equations gives us y=2 and x=2, which does mean x=y. But, it is only true for x=y=2, so would it be antisymmetric since it does satisfy the requirement for antisymmetric(but only for x=y=2

For (d), 1 is transitive,

Order is defined to be transitive.  $a> b; b> c \implies a>c$.  That's a given axiom.  (It is the only reason we are allow to write shorthand such as $a > b > c$.  That would be nonsense and wrong if we didn't know order must be transitive.

So ... correct.

2 is not transitive, because if we have xy=m2
, yz=n2, then xz=(m⋅ny)2, it is not necessary that mny is an integer.

Actually, yes, it is necessary that if $(\frac {mn}y)^2 = xy \in \mathbb N$ is an integer that $\frac {mn}y$ is an integer.

The square root of an integer is either an integer itself or irrational.  And $\frac {mn}y$ is rational.  And $(\frac {mn}y)^2=xy$ is an integer.  Ergo: $\frac {mn}y$ is an integer.

SO is transitive.

[A HUGE lash to me with a wet noodle for me for not seeing that right away.]

Note if If you have $ab = k^2$ and $a$ is not a perfect square then it has prime factors to an odd power.  But that means be has those same factors to an odd power. But the if $bc = m^2$ then $b$ has those prime factors to an odd power so $c$ must have the same odd factors to an odd power.  So $ac$ will have those prime factors to an even power.

So transitivity makes sense.

More formally: suppose $ab = k^2$ and $bc = m^2$.  Rewrite $a$ as $a = a'j^2$ we $j^2$ is factoring out all square factors of $a$ and $a'$ is a square free factor of whats left. ($a'=1$ if $a$ is a perfect square).  Rewrite $b$ as $b=b'h^2$ with the same conditions.

$ab = a'b'(jh)^2 = k^2$ which means $a'b'$ must be a perfect square.  But any prime factor of $a'$ is to a single power so it must be a factor of $b'$ and vice versa.  So $a' = b'$.

And if we rewrite $c$ ac $c=c'g^2$ with the same conditions then $bc = m^2$ by the same argument would imply $c' = b' =a'$.

So we have $ac = a'c'(jg)^2 = (a'jg)^2$.

So it transitive.

For 3, If we have x+y=10, y+z=10, then x+z=(10−y)+(10−y)=20−2y, and it is equal to 10 if y=5. So is it transitive because the requirement is fulfilled but only for y=5, or is it not transitive because the requirement is not fulfilled for all positive integers y?

To be transitive it must be true for all pairs of pairs.

If $a + b = 10$ and $b+c = 10$ it's easy to see that $a = 10-b$ and $c =10-b = a$.  There $a +c = a+a = 2a$ and $2a$ does not need to be equal to $10$.  (Unless $a$ is "forced" to be $5$.  But as we can have $a$ be any value from $1$ to $9$ we don't have to have $a=5$.)  SO not transitive.

For 4, if we have x+4y=10, y+4z=10, then x+4z=20−5y and it is equal to 10 if y=2. So is it transitive because the requirement is fulfilled but only for y=2, or is it not transitive because the requirement is not fulfilled for all positive integers y?

This is tricky.  A single counterexample would disprove transitivity.  And if our set were the reals, or the rationals, and (I think) even the integers (including negatives) counterexamples would abound.

But $x + 4y = 10$ and $y + 4z = 10$ implies $x= 10-4y$ which has positive integer solutions if $y=1, x=6$  or $y=2, x=2$.

But $y=1,x=6$ can not be a solution to $y + 4z = 10$.

So the only solution possible at all where $x$ is related to $y$ and $y$ is related to $z$ is $x=y=z=2$.  That the only example where $x+4y=10; y+4z = 10$.

And if that were the case then $x + 4z =10$.

So it is transitive.

But not because there is one case where it holds.  But because in every case it holds.  (but every case is only one case).

There are no counterexamples.  Normally we would expect counter-examples but here there are none.

It is transitive.

Title: Question on Relations


问:
How do you get hyperbola by squaring a bell?
This is a Gaussian bell (aka normal distribution).

Its square, I belive looks the same. Yet, I see that chi-square distribution, which is a sum of k such bell squares, looks like

Take a look at yellow chi-squared, with k=1. It should be just single bell squared. But, it looks like a hyperbola instead of Gaussian. Why?

PS I read that normally-distributed signal generates a uniformly distributed power. Is it the same kind of magic?

答:
Let $Z$ be standard normal, or more generally a normal with mean $0$. But in fact we will use almost no properties of the normal.

Let $Y=Z^2$. We want to show that the density function $f_Y(y)$ of $Y$ is very large for $y$ near $0$. This density function is approximately $\frac{F_Y(y)}{y}$.

We have $F_Y(y)=2\Pr(Z\le \sqrt{y})$. For $y$ close to $0$, the density function of $Z$ is close to a non-zero constant $c$.

Thus $F_Y(y)$ is approximately $2c\sqrt{y}$. It follows that near $0$, $\frac{F_Y(y)}{y}$ is very large.


问:
Why do we believe the Church-Turing Thesis?
The Church-Turing Thesis, which says that the Turing Machine model is at least as powerful as any computer that can be built in practice, seems to be pretty unquestioningly accepted in my exposure to computer science.  Why?  Do we have any more evidence for its truth than "nobody has been able to think of a more powerful model than Turing's yet?"  If so, why do we accept that argument as truth?  It seems to me that in any other mathematical discipline, a claim like the Church-Turing thesis whose support amounts to "we haven't yet found a counterexample" would not be taken seriously by the community.</p>

答:
The Church-Turing thesis is credible because every singe model of computation that anyone has come up with so far has been proven to be equivalent to turing machines (well, or strictly weaker, but those aren't interesting here). Those models include

Recursive functions over the integers

The lambda calculus

Combinatory logic

and many more. It's indeed hard to imagine anything algorithm-like which cannot be formalized in any of these models.

The difference between the Church-Turing thesis and real theorems is that it seems impossible to formalize the Church-Turing thesis. Any such formalization would need to formalize what an arbitrary computable function is, which requires a model of computation to begin with. You can think of the Church-Turing thesis as a kind of meta-theorem which states

Whenever you pick some intuitive notation of computable and formalize it, the resulting model of computation is either equivalent to turing machines, or strictly weaker.

All concrete instances of this meta-theorem that haven been encountered haven been proven. We do know that all the models of computation listed above are equivalent. Since we cannot prove the meta-theorem, for the reason stated above, we thus cannot do any better than say "Well, we don't see how it could not be true, so we'll assume it's true".


问:
Eigenvalues of a matrix $A$ where $a_{ij} = 1/a_{ji}$ and $a_{ii}=1$
Let $A$ be an $n \times n$ matrix whose diagonal entries are $a_{ii} = 1$ and whose off-diagonal entries are $$a_{ij} = \frac{1}{a_{ji}}$$ where $a_{ij} > 0$. Are the eigenvalues of $A$ always real? What more can we say about them?

I was thinking about it but I can't figure out anything. Can you help me?

EDIT: and what if in the upper triangle we have only values greather than 1?

答:
\begin{align*}
0 =  \begin{vmatrix}
1 - \lambda & a & b\\
\frac{1}{a} & 1 - \lambda & c \\
\frac{1}{b} & \frac{1}{c} & 1 - \lambda
\end{vmatrix} &= (1 - \lambda)((1 - \lambda)^2 - 1) + a \left ( \frac{c}{b} - \frac{1-\lambda}{a} \right) + b \left(\frac{1}{ac} - \frac{1-\lambda}{b} \right )\\
&= (1 - \lambda)^3 - (1 - \lambda) +  \frac{ac}{b} - (1-\lambda) + \frac{b}{ac} - (1-\lambda )\\
 &= (1 - \lambda)^3 - 3(1-\lambda ) + \frac{ac}{b} + \frac{b}{ac} \\
 &= -\lambda^3 + 3 \lambda^2 + \frac{a^2c^2 + b^2}{abc} - 2 \\
\end{align*}

No reason for that to have real roots.


问:
$xABAx^T = xBx^T$ where $B\succeq A\succ0$
Assume $xABAx^T = xBx^T$ for some vector $x\neq0$ with $B\succeq A\succ0$.

Is there a relation between the matrices? their eigenvalues?

答:
I will assume that "positive definite" here means symmetric (real) positive definite. The Loewner order can as easily be imposed on Hermitian matrices however.

The condition that a single nonzero vector $x$ satisfies:

$$ xABAx^T = xBx^T $$

is satisfied if the symmetric matrix $ABA - B$ is singular.

This can be arranged in many ways, and in particular if it were true of some pos. definite matrices $A,B$, then $B$ could as well be replaced by a scalar multiple $rB$.  So the (positive real) eigenvalues of $B$ can be made arbitrarily large.

Thus it seems to me that the condition $B\succeq A\succ 0$ will not add much to restrict the choice of $A$.  I'm not clear what the goal of the Question is, but I thought these few remarks might help the OP clarify thinking about what kind of inference about the matrices $A,B$ is desired.

Example:

A simple illustration of these ideas is possible with $2\times 2$ matrices.  Any real symmetric matrix can be diagonalized by suitable choice of an orthogonal basis, though of course we cannot generally diagonalize both $A,B$ at the same time (that would require quite a special relationship). But there is no loss of generality in assuming $A$ is diagonal with positive entries $x,y$ and $B$ is an arbitrary symmetric positive definite matrix.

For simplicity I'll take:

$$ A = \begin{pmatrix} x & 0 \\ 0 & y \end{pmatrix} \\
   B = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} $$

Then:

$$ ABA - B = \begin{pmatrix} 2x^2-2 & xy-1 \\ xy-1 & 2y^2 -2 \end{pmatrix} $$

Since all we want is for this matrix to be singular, it suffices to pick $x,y\gt 0$ such that its determinant is zero:

$$ (2x^2 - 1)(2y^2 - 1) - (xy - 1)^2 = 0 $$

To avoid having an example where the entries $x,y$ are equal, and then the eigenvectors of $A$ and $B$ and not necessarily distinct, I first picked $x = 2$ and solved for $y$:

$$ 7(2y^2 - 1) - (2y - 1)^2 = 0 $$
$$ 10y^2 + 4y - 8 = 0 $$

This has two real roots, opposite in sign, and the positive root is:

$$ y = \frac{-1 + \sqrt{21}}{5} \approx 0.716515139 $$

Now this makes $ABA - B$ singular, but it doesn't satisfy $B \succeq A$.  So instead of $B$ as chosen above, we pick a sufficiently large multiple $rB$ so that $rB-A$ is positive definite.

In fact $3B - A$ is (real symmetric) positive definite because it's strictly diagonally dominant:

$$ 3B - A = \begin{pmatrix} 4 & 3 \\ 3 & \frac{31 - \sqrt{21}}{5} \end{pmatrix} $$


问:
Is there a strategy to win this "move color" game?
My girlfriend has been playing a game on her smartphone for a while. The rules are:

there are $n+2$ tubes (with $n\ge 1$) and $n$ colored liquids.

Two tubes are left empty and the others are filled with four doses of liquid.

A tube cannot contain more than $4$ doses of liquid.

There are exactly $4$ doses of each liquid.

Liquids of different colors are non-miscible.

The goal is to move liquids so that the $4$ doses of each liquid are in a single tube.

Only the doses of liquid on the top of a tube can be moved. They have to be moved on top of a liquid of the same color.

All doses of the same color at the top have to be moved at the same time. The only exception is when there is not enough space available when moving into a new tube. Only the doses of the same color that can be moved are then moved. For example, imagine that tube 1 contains three red doses and tube 2 contains a red dose above a blue one. In that case, two red doses from tube 1 can be moved to tube 2.

Here is a picture that illustrates the problem (with $n=9$):

For example, the pink liquid at the top of the fifth tube can be moved to an empty tube. After that, the two doses of red liquid from the same tube can be moved on the remaining empty tube. It is not possible to move only one dose of red liquid.

Is there a general strategy to solve this problem?

My intuition tells me that I could use induction (like the Hanoi towers problem), or graphs, or invariants (defining the degree of disorder of a configuration, the solution of the problem corresponding to a degree of disorder of $0$), but I don't really know how to get started in any of these directions.

Note: I am not sure that it is possible to win from any starting configuration.

答:
This one is impossible to solve:

It is easy to see that the best thing you can do here is to take two pairs of the same color from the top row and put those in the two extra tubes. This either exposes four new and all different colors, so you are stuck immediately, or you expose two pairs of tiles of the same color on the second row. For example, we can take the two red top tiles and the two grey top tiles and put them in the extra tubes, thereby exposing two brown tiles as well as two blue tiles. So, we can put a brown one of top of another, and also for blue, but after that we expose one orange and one purple one, of which no counterparts have been exposed yet, and hence we are stuck. By symmetry of the configuration, this is true for whatever pairs of colors you choose to start with.

Note that the placement of the tiles on the bottom row doesn't matter, since you'll never get to the bottom row.


问:
Variance and expected value of maximum?
How to compute the expected value and variance of

\begin{equation}
Max(N-B,B)
\end{equation}

where $N$ is a constant and $B$ is a binomial variable

$$B\sim B(N,\lambda)$$

also interested in bounds !

答:
Hint: left $A= \max(N-B,B)$, define $X=1$ if $B\ge N/2$, $X=0$ otherwise.

Compute $E[A|X]$ and apply $E[A] = E[E[A|X]]$

Edited:

We have
$$
A = \begin{cases}
B & \text{if }  X=1\\
N-B & \text{if } X=0
\end{cases}
$$

Hence we can write $A= X B + (1-X) (N-B)$ and $E[A | X ] = E[B] X + (1-X)(N-E[B])= \lambda X + (1-X) (N-\lambda)$

Then, letting $\mu = E[X] = P(B \ge N/2)$ we have

$$E[A] =  E[E[A|X]] = \mu \lambda +(1- \mu)(N-\lambda)$$

To compute $\mu$ you need to evaluate the tail of a Binomial, which has no simple form.

Title: Variance and expected value of maximum?