问:
Showing series in Banach space is convergent
Suppoes $(X,\|\cdot \|)$ is a Banach space and $g \geq 0$ satisfies $\sum_{i=1}^{\infty} g(i) < \infty$. Suppose $f,f_n: \mathbb{N} \rightarrow X$ such that $f_n \rightarrow f$ pointwise, meaning that $\| f_n(i)-f(i)\| \rightarrow 0$ as $n \rightarrow \infty$. Also assume that $\|f_n(i)\| \leq g(i)$ for all $n,i \geq 1$.

 I have proved that $\sum_{i\geq 1} f_n(i)$ converges for fixed $n$. I did this by showing that the sequence of partial sums $s_k=\sum_{i=1}^k f_n(i)$ is Cauchy (and so converges since $X$ is Banach):

 Since $\sum_{i=1}^{\infty} g(i) < \infty$, there is $N \in \mathbb{N}$ such that for all $k_1 \geq k_2 \geq N$, $\sum_{i=k_1}^{k_2} g(i) < \epsilon$ for any given $\epsilon$. Then
$$\| s_{k_1} - s_{k_2}\| = \|\sum_{i=k_1}^{k_2} f_n(i)\| \leq \sum_{i=k_1}^{k_2} \|f_n(i)\| \leq \sum_{i=k_1}^{k_2} g(i) < \epsilon,$$
so $s_k$ is Cauchy and thus converges, so the series $\sum_{i\geq 1} f_n(i)$ converges.

 I also need to show that the series $\sum_{i\geq 1} f(i)$ converges and that
$$\lim_{n \rightarrow \infty} \sum_{i\geq 1} f_n(i) = \sum_{i\geq 1} f(i).$$
I am not sure how to prove this.

答:
It is pretty straightforward, just using the same remark you have for $g$
More precisely, for any $N>0$ we have:
$$ \| \sum_{i \ge 1} f_n(i)-f(i) \| \le  \| \sum_{i \ge 1}^N f_n(i)-f(i) \|  + \| \sum_{i \ge N+1} f_n(i)-f(i) \| \le \underbrace{\| \sum_{i \ge 1}^N f_n(i)-f(i) \|}_{ \longrightarrow 0 \text{ when } n \rightarrow +\infty}+2 \sum_{i \ge N+1} \| g(i) \|  $$

Thus,
$$ \limsup_n  \| \sum_{i \ge 1} f_n(i)-f(i) \| \le 2 \sum_{i \ge N+1} \| g(i) \|  \quad \forall N$$
Hence the conclusion.


问:
Number of $3$-letter words than can be formed from 'ABCDEA'?
I think I have a vague idea of what needs to be done, but I am stuck.

There are $5$ given letters, with $1$ letter repeated twice.

So to select $3$-letter words, we can select either $3$  letters out of $5$, or select $1$ letter out of $4$ with the repeated letter appearing twice.

$3$ distinct letters out of $5$ can be selected with $P(5,3) = 5!/2! = 5 \cdot 4 \cdot 3 = 60$.

How do I proceed from here? Or is my approach completely wrong?

答:
If A show up once the are $^5 P_3$ possibilities (you need to arranje three letters from a pool of five: A, B, C, D, E with no repeated letters;

If A show up twice u have 4*3 combinations (1 of 4 letters and 3 possible places);

Just add the result of each of the three cases.

ContentLicense: CC BY-SA 4.0


问:
What is the maximum value of $\frac{x-x^3}{x^4+2x^2+1}$?
What is the maximum value of $\frac{x-x^3}{x^4+2x^2+1}$  ?

I've got the hint that this problem can be solved by replacing $x$ with tangent $K$.

So it became $\frac{\tan K(1-\tan^2K)}{(\tan^2K+1)^2}=\frac{\tan K}{\tan^2K+1}\cdot\frac{1-\tan^2 K}{\tan^2K+1}=\frac{\sin(2K)}{2(1-\sin^2K)}$
then I have no idea what to do next.

Any help is appreciated, thank you.^^

Body: Another way$$\frac{x-x^3}{{(x^2+1)}^2}-\frac{1}{4}=-\left(\frac{x^4+2x^2+1-4x+4x^3}{{(x^2+1)}^2}\right)=-\left(\frac{{(x^2+2x-1)}^2}{{(x^2+1)}^2}\right)\le 0$$


问:
Find the area of the region enclosed by the curves $y = \tan(x)$, $y = -3\tan(x)$ and $x = \pi/4$
Find the area enclosed by the curves $y = \tan(x)$, $y = -3\tan(x)$ and $x = \pi/4$.

What can possibly be the area, $\tan(x)$ and $-3\tan(x)$ never intersect unless it is $\frac{\pi}{2}k$. There is no upper bound to find because the two curve goes to different direction, so confused about this question.

Body: $$\int\limits_{x=0}^{\pi/4} 4 \tan (x)\ dx = \log 4 \approx 1.38629$$


问:
Is projection epimorphism?
Let $\mathscr{C}$ be category with finite product, and $A,B$ be objects of $\mathscr{C}$.

And $p: A\times B \to A$ is projection.

Now, is  $p$ epimorphism?

I think it is true, but I can’t proof. Help me, thanks.

答:
Not necessarily. For example take $C = \text{Set}$ and take $B$ to be empty (and $A$ to be non-empty).

For a more interesting example, take $C = \text{Aff} = \text{CRing}^{op}$ to be the category of affine schemes and take $A = \text{Spec } \mathbb{F}_2, B = \text{Spec } \mathbb{F}_3$; then the fiber product $A \times B \cong \text{Spec } (\mathbb{F}_2 \otimes \mathbb{F}_3) \cong \text{Spec } 0$ is empty even though $A$ and $B$ themselves aren't, and $p : \emptyset \to \text{Spec } \mathbb{F}_2$ is not an epimorphism because the dual map on commutative rings $\mathbb{F}_2 \to 0$ is not a monomorphism (monomorphisms in commutative rings are precisely the injective maps).

What is true is that if $B$ has a global point (a map $1 \to B$ where $1$ is the terminal object) then $p$ has a section and so is a split epimorphism. In particular it's an epimorphism in $\text{Set}$ for any non-empty $B$, and it's always an epimorphism in any category with a zero object (an object which is both initial and terminal), for example the category of groups.

Title: Is projection epimorphism?


问:
showing $\mu(\lim\sup_{n\to\infty} E_n) = 0.$
Let $(X,\mathcal{M},\mu)$ be a measure space. Suppose $E_n\in \mathcal{M}$ such that $$\sum_{n=1}^\infty \mu(E_n) < \infty$$ show $\mu(\lim\sup_{n\to\infty} E_n) = 0.$

Also, how can I prove or give a counterexample of the conclusion if the hypothesis is replaced with $$\sum_{n=1}^\infty \mu(E_n)^2 < \infty$$

Can I prove this by Fatou's Lemma? Thank you

答:
As Bernard comments, the first part is actually the statement of Borel-Cantelli lemma.

For the second part, as Stella mentions, we may want to find a series which is not convergent but the series consists of square of its terms is convergent. The series $\sum_{n=1}^\infty \frac{1}{n}$ will satisfy this condition.

So we need to construct a sequence of sets $\{E_n\}$ s.t. $\mu(E_n)=\frac{1}{n}, \forall n\ge1.$ Also, we want that $\mu(\lim\sup_{n\to\infty} E_n) \neq 0.$

To construct a concrete example, we may need a proposition.

Proposition: For every real number x, there exists a sequence $x_1,x_2,x_3,...$ of integers such that $$x=x_1+\frac{x_2}{2!}+\frac{x_3}{3!}+\cdot\cdot\cdot \,,$$ where $x_1$ can be any integer, but for $n\ge2,x_n\in\{0,1,...,n-1\}.$ Furthermore, if we require that the partial sums be strictly smaller than x, then such a representation is unique.

A proof can be found here: https://blogs.ams.org/mathgradblog/2017/09/20/real-numbers-base-factorials-by-product/

From the proposition and its proof, it's easy to see that, if $x\in(0,1)$, being an irrational number, then we can find an unique sequence $x_1,x_2,x_3,...$ s.t. $x_1=0$ and $$x=\frac{x_2}{2!}+\frac{x_3}{3!}+\cdot\cdot\cdot \,,$$ with $x_n\in\{0,1,...,n-1\},\forall n\ge2.$

Also note that, if $x$ is an irrational number, then the corresponding series will consist of infinitely many terms. (N.B. This is crucial to our construction, which makes sure the construction process will not terminate in finite steps.)

Our example will be given in the language of probability space, but in essence, it can be easily converted into the language of general measure theory.

We define the event $E_1=\{\text{pick a random irrational number $x$ in $(0,1)$, $x_1$ is $0$}. \}$. Then it's clear that $P(E_1)=1.$

For each $n\ge 2,$ define event $E_n=\{\text{pick a random irrational number $x$ in $(0,1)$, $x_n$ is $0$}. \}$. Then it's also clear that $P(E_n)=\frac{1}{n}.$

Note that these events are mutually independent.

By the second Borel-Cantelli lemma(c.f. https://en.wikipedia.org/wiki/Borel–Cantelli_lemma#Converse_result), we conclude that $P(\lim\sup_{n\to\infty} E_n)=1\neq 0.$

So, even though $\sum_{n=1}^\infty \mu(E_n)^2 < \infty$, we may still have $\mu(\lim\sup_{n\to\infty} E_n)=1\neq 0.$

$$\tag*{$\blacksquare$}$$


问:
Matrix Norm Inequality $\lVert A\rVert_\infty \leq \sqrt{n} \lVert A\rVert_2$
So I'm trying to prove that

$\lVert A\rVert_\infty \leq \sqrt{n} \lVert A\rVert_2$.

I've written the right hand side in terms of rows, but this method doesn't seem to be getting me anywhere.

Where else should I go?

答:
Writing $A=(A_1,\dots,A_n)^\mathrm{T}$ with $A_i$ being the $i$-th row of the matrix, let $A_j$ be the row for which
$$
\lVert A\rVert_\infty = \max_{1\leq i\leq n }\lVert A_i\rVert_1 = \lVert A_j\rVert_1 = \sum_{k=1}^n \left|A_{i,j}\right|
$$
Then
$$
n\lVert A\rVert_2^2 = n\sum_{i=1}^n \lVert A_i\rVert_2^2 \geq  n\lVert A_j\rVert_2^2 \geq \lVert A_j\rVert_1^2 = \lVert A\rVert_\infty^2
$$
where the last inequality is "standard" (relation between 1 and 2-norm on $\mathbb{R}^n$, can be proven with Cauchy-Schwarz).</p>

ContentLicense: CC BY-SA 3.0


问:
Prove the sum of integrals of function equals to the integrals of sum of functions
Let $(f_n)$ be a countable sequence of functions in $M^+(S,S)$, and let $g=\sum_nf_n$. Prove $\int gd\mu=\sum_n\int f_nd\mu$

答:
The monotone convergence theorem is the standard approach, just for giggles here is another way:

Note that $\sum_i f_i$ is the pointwise limit of a measureable function and hence is measureable.

Let $|\cdot|$ denote the counting measure on $\mathbb{N}$, then it is straightforward to check that the map $(x,i) \mapsto f_i(x)$ is measurable (and non negative) on $X \times \mathbb{N}$. Hence Tonelli shows that
$\sum_i \int f_i d \mu = \int ( \int f_i(x) d\mu(x)) d |\cdot|(i) = \int ( \int f_i(x) d |\cdot|(i) )  d\mu(x)  = \int (\sum_i f_i) d \mu$.


问:
Naive way to calculate the minimum of the Gamma function .
Inspired by some question due to user Claude Leibovici I propose a simple method to calculate the first digit of the minimum of the Gamma function for positive value ($x>0$).

The first idea is to transform a little bit the function Gamma and work with the function :

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

The second natural question is : As the minimum is relatively near of the absissca $x=0.5$ how to transform it to get $x_{min}\simeq 0.5$ ?

After some hours I have found the good candidate wich is :

$$g(x)=\Gamma(x+1)^{x^x}$$

Now the third step is :how to find a relatively good approximation of the function $g(x)$ around $x=0.5$ ?

A natural way for me is to use the exponential function and we have :

$$g(x)\simeq \exp\left(\alpha x^2-\alpha x\right)=h(x)$$ around $x=0.5$ where $\alpha$ verifying $e^{-0.25\alpha}=\Gamma(1.5)^{\sqrt{0.5}}$

So we have as approximation :

$$f(x)\simeq h(x)^{\frac{1}{x^x}}$$

Around the minimum of the Gamma function.

Question :

Can we improve the situation using naive or more complex method ?

Thanks in advance!

答:
I think that we can do it without approximation using the Taylor expansion of a function
$$F(x)=f(x)^{x^x} \qquad \text{where} \qquad f(x)=\Gamma(1+x)$$
up to $O\left(\left(x-\frac{1}{2}\right)^3\right)$.

Developed as a series around $x=a$,we are left with
$$A+a^aB(x-a)+O\left(\left(x-a\right)^2\right)$$ with
$$A=\frac{a^a f'(a)}{f(a)}+a^a (\log (a)+1) \log (f(a))$$
$$B=\frac{\log (f(a)) \left(a \left(a^a (\log (a)+1)^2 \log (f(a))+\log (a) (\log
   (a)+2)\right)+a+1\right)}{a}$$
$$+\frac{\left(a^a-1\right) f'(a)^2+f(a) \left(2 (\log (a)+1) f'(a) \left(a^a \log
   (f(a))+1\right)+f''(a)\right)}{f(a)^2}$$

For the present case where $f(x)=\Gamma(x+1)$, after simplification, the problem is to find the zero of
$$g(x)=\psi (x+1)+(\log (x)+1) \log (\Gamma (x+1))$$

Using the first iteration of Newton method with $x_0=\frac 12$, we have
$$x_1=\frac 12+\frac{-4+2 \gamma +\log \left(\frac{64}{\pi }\right)+\log (2) \log \left(\frac{\pi
   }{4}\right)}{-4+\pi ^2+4\log ^2(2)+2\gamma  (\log (2)-1)+2 \log \left(\frac{\pi
   }{64}\right)}$$
Evaluated numerically, this is
$x_{\text{min}}=\color{red}{0.500812}76$ while its value computed by a full optimization is $x_{\text{min}}=\color{red}{0.50081252}$.

Using the approximation, the minimum value of $F(x)$ is numerically
$F_{\text{min}}=\color{red}{0.9181393488698}94$ while, from optimization, it is
$F_{\text{min}}=\color{red}{0.918139348869881}$.

We can still do  better using one iteration of Newton-type methods of order $n$. This would give
$$\left(
\begin{array}{ccc}
n & x_1^{(n)} & \text{method} \\
 2 &  0.5008125609311657357119337 & \text{Newton}      \\
 3 &  0.5008125200962838796408148 & \text{Halley}      \\
 4 &  0.5008125195407779789522950 & \text{Householder} \\
 5 &  0.5008125195413028999125796 & \text{no name}     \\
 6 &  0.5008125195413028258583701 & \text{no name}     \\
 7 &  0.5008125195413028255893431 & \text{no name}     \\
\cdots &\cdots &\cdots \\
\infty &0.5008125195413028255895594 & \text{no name}
\end{array}
\right)$$


问:
How to find the angle by a triangle and a median and bisector?
The problem is as follows:

In a triangle $\triangle{ABC}$, the measure of angle $\angle ABC = \angle 105^{\circ}$. On $AC$ it is located a point $M$, such that $AB = MC$. The mediatrices of $BC$ and $AM$ intersect at $Q$. Find the measure of $\angle BAC$, if the measure of $\angle BCA = \angle ACQ$.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&30^{\circ}\\
2.&40^{\circ}\\
3.&45^{\circ}\\
4.&35^{\circ}\\
4.&50^{\circ}\\
\end{array}$

Can someone help me with the right interpretation of this problem?. I'm having difficulties with the right drawing.

I found this problem in my book in the chapter of triangle congruency thus I believe the approach for this problem is intended to follow such route. Can someone help me with that?. Please include a drawing in the answer so I can understand. I'm lost if this requires some sort of construction. Please try to include a solution explained step-by-step.

答:
Before we dig into this question, note that I don't draw the mediatrices to avoid making the picture too complicated.

Connect $\overline{BQ}$. Let $\angle BCA=\angle ACQ=\theta$.

Since $Q$ is on the perpendicular bisector of $\overline{BC}$, we have $\color{blue}{\overline{BQ}=\overline{CQ}}$.

Since $Q$ is on the perpendicular bisector of $\overline{AM}$, we have $\color{red}{\overline{AQ}=\overline{MQ}}$.

Also from the question we already know $\color{purple}{\overline{AB}=\overline{MC}}$.

Hence $$\triangle BAQ\cong\triangle CMQ\quad(S.S.S.)$$
so $$\angle ABQ=\angle MCQ=\theta$$

We also have $$\angle QBC=\angle QCB=2\theta$$ since $\triangle BCQ$ is an isosceles. Therefore we can say $$\theta+2\theta=105^{\circ}\implies \theta =35^{\circ}$$

Finally $$\angle BAC=180^{\circ}-105^{\circ}-\theta=\color{red}{40^{\circ}}$$


问:
How come such different methods result in the same number, $e$?
I guess the proof of the identity
$$
\sum_{n = 0}^{\infty} \frac{1}{n!} \equiv \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x
$$
explains the connection between such different calculations. How is it done?</p>

答: Another way of viewing $e^{x}$, is if suppose $f : \mathbb{R} \to \mathbb{R}$ is a differentiable function which satisfies $f'(x) = f(x)$ for all $x \in \mathbb{R}$, then $f(x) = e^{x} + C$. To see this we have $f'(x) = f(x) \Longrightarrow e^{-x}f'(x) = e^{-x} f(x) \Longrightarrow e^{x} \cdot [f'(x) - f(x)]= 0 \Longrightarrow \frac{\rm d}{\rm dx } \Bigl[e^{-x}f(x)\Bigr]=0$ which says that $e^{-x} f(x) = C$. Hence $f(x) = Ce^{x}$.</p>


问:
Principal Divisors on a projective non-singular irreducible curve if and only if degree is 0
Let $C$ be a a projective non-singular irreducible curve under a ground field $k$. Let $D$ be a divisor on $C$ with degree 0. Does this necessarily imply that $D$ is principal? That is, $D=\text{div}f$ for some $f\in k(X)$? Also, since $C$ is birational to a curve in $\mathbb{P}^2$, can we assume WLOG that $C$ is such a curve? How does "irreducible nonsingular" help in this case?

I know that there are finitely many points $P_1, ..., P_r\in C$ that gives a nonzero contribution to $D$.

Thanks for the help.

答:
The group of divisors modulo principal divisors is known as the Picard group; the subgroup of degree zero divisors modulo principal divisors is denoted $\newcommand{\Pic}{\operatorname{Pic}}\Pic^0(C)$. When $C$ is a smooth projective curve over a field, $\Pic^0(C)$ is isomorphic to the Jacobian variety of $C$, which is an abelian variety of dimension equal to the genus $g$ of $C$, and is birational to $C^{(g)}$, the $g$-th symmetric power of $C$.

So for a curve of genus at least $1$, it's very far from true that degree zero divisors are always principal. For example, if $P, Q \in C$ such that $[P] - [Q]$ is principal, then $P = Q$.

As for your question about curves being birational to curves in $\mathbb{P}^2$, note that most curves not of genus $0$, $1$, or $3$ are only birational to a singular plane curve, not a smooth plane curve.


问:
Dot product and determinant?
Why is it true that
$$
(\mathbf{u} \times \mathbf{v}) \cdot \mathbf{u} =
        \begin{array}{|rrr|}
             u_1&u_2&u_3\\
             u_1&u_2&u_3\\
             v_1&v2&v3
        \end{array}
$$? That is, the dot product of the cross product between $\mathbf{u}$ and $\mathbf{v}$ with $\mathbf{u}$ is equal to the above determinant. Why is that true?

I know that you can see the cross product as the determinant:
$$
        \begin{array}{|rrr|}
             \mathbf{i}&\mathbf{j}&\mathbf{k}\\
             u_1&u_2&u_3\\
             v_1&v2&v3
        \end{array}
$$
but still I cannot see how the above is true.

答:
The cross product $\mathbf u\times \mathbf v$ is $(u_2v_3-u_3v_2)\mathbf i+(u_3v_1-u_1v_3)\mathbf j+(u_1v_2-u_2v_1)\mathbf  k$,

so $(\mathbf u\times \mathbf v )\cdot \mathbf u=(u_2v_3-u_3v_2)u_1+(u_3v_1-u_1v_3)u_2+(u_1v_2-u_2v_1)u_3,$

and the determinant is $u_1u_2v_3+u_2u_3v_1+u_3u_1v_2-u_3u_2v_1-u_2u_1v_3-u_1u_3v_2$,

so they're the same.

Title: Dot product and determinant?


问:
How to properly work with Leibniz notation
This question regards the manipulation of derivatives as if they were fractions. But more generally it also regards doing calculus in a "Leibnizian" way.

Before asking I checked out the current poll of question regarding this topic: link, link and another one for good measure; my objective with this question is to fill in the holes that I think are left open in the discussion of this topic on this site.

In the cited questions is clarified why is incorrect to define the derivate not as a limit but as a fraction, and the consequent importance of knowing the proper definition of the derivative as a limit. This is clear. But when working whit derivatives and integrals I think the usefulness of working with Leibniz notation, and interpreting derivatives as fractions, is beyond any doubt, as this answer points out.

Problem is: when working with imprecise assumptions things can go wrong really fast, and for this reason interpreting the derivative as a fraction has the nomea of a pretty dangerous gamble.

Here are some examples on how this can go terribly wrong:

Example 1:
$$\int \nabla \phi \cdot d\vec{x}=\int \frac{d\phi}{dx}dx+\frac{d\phi}{dy}dy+\frac{d\phi}{dz}dz=\int \frac{d\phi}{dx}dx+\int\frac{d\phi}{dy}dy+\int\frac{d\phi}{dz}dz=\phi+\phi+\phi=3\phi$$
but on the other hand if we state that: $d\phi+d\phi+d\phi=d\phi$ we get the correct result:
$$\int \nabla \phi \cdot d\vec{x}=\int \frac{d\phi}{dx}dx+\frac{d\phi}{dy}dy+\frac{d\phi}{dz}dz=\int d\phi+d\phi+d\phi=\int d\phi=\phi$$
But if we write it with the proper notation of partial derivative then we should elide $\partial x$ with $dx$, even more chaos.

Example 2:
$$\frac{\partial \phi}{\partial x}/\frac{\partial \phi}{\partial y}=\frac{\partial y}{\partial x}$$
by "eliding the $\partial \phi$".

And I am sure I could go on with more example, but I think there is no need.

Question is: is there a ruleset to follow to ensure to not fall for mistakes like this when working with Leibniz notation? I feel like this question hasn't been properly answered in the previous discussion on this topic. And if there is such ruleset, why it works?

Also it's often said that this way of handling derivatives and integrals no longer works when dealing with multivariable calculus, but from my experience it seems to give the correct answer even in this case, what is up with this exactly? Why should it no longer work? For example let's take the integral:
$$\int \frac{\partial \vec{A}}{\partial t} \cdot d\vec{x}$$
It would seem a bit difficult to solve this integral "rigorously", but by interpreting it in a somewhat Leibnizian way we get:
$$\int \frac{\partial \vec{A}}{\partial t} \cdot d\vec{x}=\int \frac{\partial A_x}{\partial t}dx+\frac{\partial A_y}{\partial t}dy+\frac{\partial A_z}{\partial t}dz=\int dA_x \frac{dx}{dt}+dA_y\frac{dy}{dt}+dA_z\frac{dz}{dt}=\int \vec{v} \cdot d\vec{A}=\vec{v}\cdot \vec{A}$$
This seems like a miracle, should we not use this way of working with things like this?

答:
The thing to keep in mind is that partial differentials, as they are commonly notated, are not distinct.  That is, $\partial \phi$ can refer to any number of values.  This is an unfortunate development in the history of mathematics.  One way (but not the only way!) to do it is to notate on the differential which other variables were free (i.e., independent) when taking the derivative.

In this way, if $z = f(x, y)$, then $dz = \partial_x z + \partial_y z$.  If I want to know the partial derivative of $z$ with respect to $x$, that is $\frac{\partial_x z}{dx}$.  In your example above, $d\phi = \partial_x\phi + \partial_y\phi + \partial_z\phi$.  That's why the integral is $\phi$.  This is also why you can't just multiply partials and get rid of one of them.  But it is why Ian's example does work.

Let's say you have $f(x, y) = c$.  Then, $df = \partial_x f + \partial_yf$.  The two partial derivatives, then, are $\frac{\partial_xf}{dx}$ and $\frac{\partial_yf}{dy}$.  If $f(x, y) = c$, then $df = 0$, which means that $\partial_x f + \partial_yf = 0 $ which means that $\partial_yf = -\partial_x f$.  Then we can rewrite the second partial derivative as $\frac{-\partial_x f}{dy}$.  Then, if we put them in ratio with each other, we get:
$$ \frac{\frac{\partial_xf}{dx}}{\frac{-\partial_x f}{dy}} = \frac{\partial_xf}{dx}\frac{dy}{-\partial_xf} = -\frac{dy}{dx}$$

If you rigorously keep track of the differentials, it all works out.  Unfortunately, most mathematical work does not keep track so rigorously.  As I mentioned in my comment, in order for all this to work out as well, you need to deal with higher-order differentials properly as well.  My article "Extending the Algebraic Manipulability of Differentials" covers this.  So, when taking the second derivative, you are actually taking the derivative of $\frac{dy}{dx}$.  If you take differentials seriously, then this is a fraction, and therefore the quotient rule has to be applied.  When you do this, you get the following formula for the second derivative:
$$\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$$
This is not what most people use for the second derivative, but you can algebraically manipulate the differentials in this equation.  Note that if you use this formula, then the "chain rule for the second derivative" becomes algebraically self-evident.

So, the question was "how do I know when I'm doing it right".  I'm sorry I don't have a definitive answer, but, in general, think about what differentials are, what they mean, and where they are coming from, and only trust formulas if you know how they were derived, since many of them have historically been derived being careless with differentials.

My own hunch is that, since, in the 1800s, the validity of differential thinking was in doubt, and they all thought that derivatives were the only "actual" things, they stopped being rigorous with lone differentials.


问:
Does an exponential series expansion exist?
I know we can expand a function $f:\mathbb{R} \to \mathbb{R}$ with a power series
$$
f(x) = \sum_{n=0}^\infty a_n x^n,
$$
where we can find find $\{a_n\}$ given $f$ using its derivatives about $0$. What I would like to know is whether an "exponential series" exists for $f$. That is, is there a sequence $\{b_n\}$ such that
$$
f(x) = \sum_{n=0}^\infty b_n e^{x n},
$$
and if so, how would we find $b_n$ given a general $f$? Are there conditions on $f$ for which no sequence $\{b_n\}$ can exist?

Edit: To be clear, $x$ here is real, not imaginary, so this is not a Fourier series, and the $e^{xn}$ functions are not an orthonormal basis. Obviously some trivial series do exist, such as $f(x) = e^x$ for $b_n = \delta_{n1}$, but what about more interesting functions such as $f(x) = e^{x^2}$?

答:
Such an expansion exists whenever $f$ is periodic. That is, say $f(x + 2\pi) = f(x)$ for every $x$. Obviously $2\pi$ is just a convention here. By rescaling, you can make it be whatever you want.

Then the Fourier Series associated to $f$ is the sum

$$f(x) = \sum_{-\infty}^\infty a_n e^{2 \pi i n x}$$

Under relatively general conditions (that are still technical enough for me to not want to mention) such a series exists, but periodicity is the important one. We can see periodicity is necessary since each of the $e^{2 \pi i n x}$ are periodic.

In much the same way that a taylor series lets you write $f$ in terms of its derivatives, a fourier series lets your write $f$ in terms of its "harmonics". Importantly, we still have a nice formula for the coefficients:

$$a_n = \int_{-\infty}^\infty e^{-2 \pi i n x} \cdot f ~ \mathrm{d}x$$

In many contexts, these $a_n$ are called $\hat{f}(n)$, and you'll see that notation scattered through the literature. This is a field that is very deep and very broad, so it should be exciting to get some exposure to it!
You can find a relatively polite treatment of the subject in these notes by Tom Leinster

I hope this helps ^_^


问:
Orthogonal transformation in $\mathbb{R}^2$
Given the following inner product in $\mathbb{R}^2$

$\langle(x_1,y_1), (x_2,y_2)\rangle= 2x_1x_2-2x_1y_2-2x_2y_1+5y_1y_2$.

Find an orthogonal linear transformation from $\mathbb{R}^2$ on $\mathbb{R}^2$.

There is a process to find such transformation?

答:
Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and $z = (x, y)$. Solve for $a, b, c, d$ such that $<Az, z>$ = 0.

With some algebra, you can see that $\begin{bmatrix} \alpha & \frac{5}{2} \alpha \\ \alpha & -\alpha \end{bmatrix}$ for any non-zero $\alpha$ works.


问:
Balls intersection problem
Let $B(a,r)$ be the ball of center $a$ and radius $r$ in $\mathbb{R}^n$. I have to show two obvious things, but I am having trouble formalising it:

i) If $B(a,r) \subset B(b,r') \Rightarrow ||a-b|| < r'-r$.

ii) Using i), prove that if $B(a_{n+1},r_{n+1}) \subset B(a_n,r_n) \; \forall n\Rightarrow \bigcap_{n} B(a_n,r_n) \neq \emptyset.$

Thank you in advance!

答:
For part (i), consider the line segment between $a$ and $b$.  This line segment can be extended to a radius $L$ of $B(b,r')$.  Let $x$ be the point where $L$ intersects the boundary of $B(b,r')$.  Then we can denote $L$ as $\overline{b x}$.  We note that $\overline{b x}$ is the union of the two segments $\overline{b a}$ and $\overline{a x}$.  What we know:

the length of $\overline{b x}$ is $r'$

the length of $\overline{b a}$ is $\|a - b\|$

the length of $\overline{a x}$ is greater than $r$, since $x$ is on the boundary of $B(b,r') \supset B(a,r)$

So we get that $r' > \|a-b\| + r$.

For part (ii), assume that $B(a_{n+1}, r_{n+1}) \subset B(a_n, r_n)$ for all $n$.  We consider the sequence $\{r_i\}_{n \in \mathbb{N}}$.  By assumption, this sequence is decreasing.  If $R>0$ is a lower bound for the sequence, then there exists $b \in \mathbb{R}^n$ such that $B(b,R) \subset \cap_n B(a_n,r_n)$ and so the intersection is non-empty.

Assume that no such $R$ exists.  Then $\{r_i\}$ is a decreasing sequence bounded below by 0.  So $\lim_{i \rightarrow \infty} r_i = 0$.  We will use this and part (i) to show that $\lim_{i \rightarrow \infty} a_i$ exists and is in our intersection.

Let $\epsilon > 0$.  Since $\lim_{i \rightarrow \infty} r_i = 0$, there exists $N$ such that $r_n < \epsilon$ for all $n > N$.  If $n,m > N$, then $\|a_n - a_m\| < |r_n - r_m| < \epsilon$, which shows that $\{a_i\}$ is Cauchy.  Since $\mathbb{R}^n$ is complete, this proves that the sequence converges.  Let $A = \lim_{i \rightarrow \infty} a_i$.  By construction, $A \in B(a_n,r_n)$ for all $n$, so $A \in \bigcap_n B(a_n,r_n)$.

Title: Balls intersection problem


问:
A geometry problem to prove three line segments intersect at the same point
$D$ is a point within $\triangle ABC$ such that $\angle DBA = \angle DCA$. $I$ is midpoint of $BC$. $DE \perp AB$, $DF \perp AC$. $E,F$ are on $AB$, $AC$ respectively. $G$ is a point such that $AG \perp GD$, $AG$ and $IF$ meet at $H$.

Prove that $EF, BH, DG$ intersect at the same point.

Through software, apparently we have $\angle EIF = 2\angle EBD$, though I can't prove it easily (would have been trivial if $CDE$ and $BDF$ are both colinear.

Also what are the arsenals to prove three line segments meet at the same point? Perhaps Pascal's theorem?

答:
Any theorem worth using once is worth using four times!

Observe that $A, D, E, F, G$ lie on a circle $Γ$ centered at the midpoint $J$ of $AD$. Let $BD, CD$ meet $Γ$ again at $K, L$.  Let the tangents to $Γ$ at $E, K$ meet at $M$, and the tangents at $F, L$ meet at $N$.

Since

$$\begin{multline*}\frac{∠KME}{2} = \frac{180° - ∠EJK}{2} = ∠KDE - 90° = ∠KBE \\ = ∠FCL = ∠FDL - 90° = \frac{180° - ∠LJF}{2} = \frac{∠FNL}{2},\end{multline*}$$

we see that

$B, E, K$ lie on a circle $Γ_1$ centered at $M$;

$C, F, L$ lie on a circle $Γ_2$ centered at $N$;

there’s a diameter $\ell$ of $Γ$ that reflects $E, K, M, Γ_1$ to $F, L, N, Γ_2$.

Let $EL, FK, \ell$ meet at $I'$.  Using Pascal’s theorem twice on $EEFKKL$ and $EFFKLL$, we find that $I'$ lies on $MN$, so it is in fact the midpoint of $MN$.  Using Pascal’s theorem on $AELDKF$, we find that $I'$ lies on $BC$.  This is enough to establish that $I' = I$, the midpoint of $BC$.

Finally, the result follows using Pascal’s theorem on $AEFKDG$.


问:
Relating factors of a polynomial over $GF(q)$ and $GF(q^m)$
I am reading Richard Blahut's book "Algebraic Codes for Data Transmission" and have come to an impasse in my understanding in section 5.3.  In this section, the author wants to relate the factors of a polynomial over $GF(q)$ and the factors of that same polynomial over a GF-extension field $GF(q^m)$.

There is a good example in the book that I will use here to ask my question.  Specifically the $GF$ fields in this example use $q=2$ and $m=4$, so there is the prime field $GF(2)$ and the extension field $GF(2^4)$.  The polynomial to be factored over both of these fields is one with particularly special properties, of the form $x^{q^m-1}-1$.  So, the polynomial for this example is $x^{15}-1$.

The author first factors $x^{15}-1$ over $GF(2)$.  The prime factors are:
$$
x^{15}-1 = (x+1)(x^2+x+1)(x^4+x+1)(x^4+x^3+1)(x^4+x^3+x^2+x+1)
$$
This factorization over $GF(2)$ makes sense to me and I can easily verify it in MATLAB.

The author then chooses two of the above prime factors and forms a new polynomial:
$$
g(x) = (x^4+x^3+1)(x^4+x^3+x^2+x+1)\\
= x^8+x^4+x^2+x+1
$$
The above multiplication over $GF(2)$ also makes sense to me and I can easily verify it in MATLAB.

However, the author then changes to analyzing the same polynomial $g(x)=x^8+x^4+x^2+x+1$ in the extension field $GF(2^4)$.  It seems to be implied that because $GF(2^4)$ in an extension field of $GF(2)$, and because the polynomial in question is comprised of prime factors (over $GF(2)$) of the special composite polynomial $x^{15}-1$, that this switch to the extension field $GF(2^4)$ is justified.

To analyze $g(x)$ in the extension field, we agree to represent the 16 elements of the extension field as the usual powers of the primitive field element alpha:
$$
0, 1, \alpha, \alpha^2, \alpha^3, \alpha^4, \alpha^5, \alpha^6, \alpha^7, \alpha^8, \alpha^9, \alpha^{10}, \alpha^{11}, \alpha^{12}, \alpha^{13}, \alpha^{14}
$$

The author claims that the aforementioned polynomial $g(x) = x^8+x^4+x^2+x+1$ has roots/zeroes over $GF(2^4)$ at $g(\alpha)=0$ and $g(\alpha^3)=0$.

I am completely unable to make sense of this claim.  It raises numerous questions in my mind, and the most basic of these questions is, how can I verify that $g(\alpha)=0$ and that $g(\alpha^3)=0$?

To show how I've tried to verify that $g(\alpha)=0$ for $g(x)=x^8+x^4+x^2+x+1$, the author uses the following detailed representation of the elements of extension field $GF(2^4)$.  Of course to describe the elements of any GF extension field, we must choose an irreducible polynomial for the reducing modulus, and throughout the book the author uses the modulus $\alpha^4+\alpha+1$, yielding the 16 field elements:
$$
0=\alpha^{-\infty}\\
1=\alpha^0\\
\alpha^1=\alpha\\
\alpha^2=\alpha^2\\
\alpha^3=\alpha^3\\
\alpha^4=\alpha+1\\
\alpha^5=\alpha^2+\alpha\\
\alpha^6=\alpha^3+\alpha^2\\
\alpha^7=\alpha^3+\alpha+1\\
\alpha^8=\alpha^2+1\\
\alpha^9=\alpha^3+\alpha\\
\alpha^{10}=\alpha^2+\alpha+1\\
\alpha^{11}=\alpha^3+\alpha^2+\alpha\\
\alpha^{12}=\alpha^3+\alpha^2+\alpha+1\\
\alpha^{13}=\alpha^3+\alpha^2+1\\
\alpha^{14}=\alpha^3+1
$$
Given the above element representation of $GF(2^4)$, we return to the enigmatic claim that for $g(x)=x^8+x^4+x^2+x+1$, there is the root $g(\alpha)=0$ over $GF(2^4)$.  We have:
$$
g(\alpha)=\alpha^8+\alpha^4+\alpha^2+\alpha+1\\
= (\alpha^2+1) + (\alpha+1) + (\alpha^2) + (\alpha) + (1)\\
= (\alpha^2+\alpha^2) + (\alpha+\alpha) + (1+1+1)\\
= 1
$$
So it seems that for $g(x)=x^8+x^4+x^2+x+1$ polynomial, $g(\alpha)\neq0$.

Can anyone show me how to demonstrate that $\alpha$ (as a $GF(16)$ element) is a root of a polynomial that was computed over $GF(2)$?

If it would be helpful, I can scan the 3 pages of the book and you can see the author's own words for the above ideas.  He throws this information at you very quickly, so it's fairly difficult for me (as a non-mathematician) to follow.

答:
It’s really difficult to see what your author was trying to get at here. In this particular example, where the polynomial is $X^{15}-1$ over $k=GF(2)$, the way it factors over $K=GF(16)$ is that it splits completely into linear factors, ’cause the elements of $K$ are exactly the roots of $X^{16}-X$.

He has factored $X^{15}-1$ over $GF(2)$ correctly: the linear factors account for the elements of $GF(2)$, the quadratic factor accounts for the two additional elements of $GF(4)$, and the three quartic factors account for the twelve additional elements of $GF(16)$. You’ve used one of these quartics, the last one, $X^4+X+1=f(X)$, for constructing your $GF(16)$, and we have $0=f(\alpha)=f(\alpha^2)=f(\alpha^4)=f(\alpha^8)$, ’cause any time $\beta$ is a root of a $GF(2)$ polynomial $h(X)$, $\beta^2$ will also be a root of $h$.

His $g$ is product of the other two quartics you’ve written down, namely $X^4+X^3+1$ and $X^4+X^3+X^2+X+1$. It follows that every element of $K$ that’s neither in $GF(4)$ nor a root of $f$ will be a root of $g$. Since the elements of $GF(4)$ are $0$, $1$, $\alpha^5=\alpha+\alpha^2$, and $\alpha^{10}=1+\alpha+\alpha^2$, the roots of $g$ should be $\{\alpha^3,\alpha^6,\alpha^7,\alpha^9,\alpha^{11},\alpha^{12},\alpha^{13},\alpha^{14}\}$, of which there are fortunately eight.

I don’t know whether this will help your understanding, nor whether it will explain what your author was trying to convey. If you have questions, I can try to answer them in an addendum.


问:
Characters on $\ell^\infty$ and characteristic functions
Let $\ell^\infty$ be the set of bounded complex-valued sequences, which we regard as a commutative $C^*$ algebra under pointwise multiplication. Write $\Phi$ for the set of characters of $\ell^\infty$, i.e. non-zero algebra homomorphisms $\ell^\infty\to \mathbb{C}$. Note that, if $B\subset\mathbb{N}$ and $\chi_B$ is its characteristic function, then $\varphi(\chi_B)\in \{0,1\}$ for any $\varphi\in\Phi$ because
$$
\varphi(\chi_B)=\varphi(\chi_B^2)=(\varphi(\chi_B))^2
$$

Fix $\varphi\in \Phi$, $\varepsilon>0$, and $x\in \ell^\infty$. Put
$$
A=\{n\in\mathbb{N}:|x_n-\varphi(x)|<\varepsilon\}
$$
I'd like to show that $\varphi(\chi_A)=1$.

Some context: By the Gelfand-Naimark Theorem, $C(\Phi)\cong \ell^\infty$, and $n\in \mathbb{N}$ can be identified with the projection $x\mapsto x_n$, which obviously belongs to $\Phi$. Under this identification, $\mathbb N$ is dense in $\Phi$ and so $\Phi$ can be regarded as the Stone-Cech compacfication of $\mathbb N$. If we let $\beta \mathbb N$ be the set of ultrafilters on $\mathbb N$, then we can find explicit bijections between $\Phi$ and $\beta \mathbb N$ by considering
$$
F: \Phi \to \beta\mathbb N\\
\varphi\mapsto \{A\subset \mathbb N:\varphi(\chi_A)=1\}
$$
and
$$
G:\beta\mathbb{N}\to \Phi \\
\mathcal{U}\mapsto \left(x\mapsto \lim_\mathcal{U} x\right)
$$
The identity I'm trying to prove arises when checking that $(G\circ F)(\varphi)=\varphi$ for every $\varphi\in\Phi$.  Of course, one can also verify that they are continuous and then check that the compositions are the identity on $\mathbb{N}$, which as a by product yields what we want. However, I suspect there's a more direct, elementary argument that I'm not seeing.

答:
I'm not sure if this is what you're looking for, but I tried.

The evaluation characters $x\mapsto x(n)$ are of course $w^*$-dense in the set $\beta \mathbb{N} \subset (\ell^\infty)^*$ of all characters.
Therefore, given any $\varphi\in\beta\mathbb{N}$ and $x,y \in \ell^\infty$, there exists a sequence $n_k$ such that $\varphi(x) = \lim_k x(n_k)$ and $\varphi(y) = \lim_k y(n_k)$.

In particular, given $\varphi$ and $x$, suppose $\varphi(x) = c$, and let $y$ be the characteristic function of  $A = \{n : |x(n) - c| < \varepsilon\}$.
Since $c = \lim_{k} x(n_k)$, eventually the sequence $n_k$ lies inside $A$.
Therefore $\varphi(y) = \lim_k y(n_k) = 1$.


问:
Why are the eigenvectors of $A$ orthogonal when $AA^T=A^T A$?
Prof. Gilbert Strang mentions, in this lecture, that a matrix $A$ has orthogonal eigenvectors when $A^T A=AA^T$; i.e., when it is normal. Could someone provide me a (preferably intuitive) proof of this?

答:
This result holds for every symmetric matrix S. The basic idea is that column space and row space of a symmetric matrix are equal and that row space and null space are orthogonal.

Given a pair of eigenvectors $x, y$ corresponding to non-zero eigenvalues $\alpha, \beta$, we have $Sx = \alpha x \Rightarrow (S-\alpha I)x = 0$ and $(S - \alpha I)y = (\beta - \alpha)y$. Therefore, $x$ is in the null space of $S-\alpha I$ and $y$ is in the column space of $S-\alpha I$. Since column space is equal to row space ($S-\alpha I$ is also symmetric), we have $x^Ty = 0$. The case of zero eigen values is trivial.


问:
Laplace transform of a gaussian function?
How do I find the Laplace transform of this Gaussian function $f(t)=\sqrt{t}e^{-\frac{1}{t}}$.

I have from a table that $L\Big[{\frac{1}{\sqrt{t}} e^{-\frac{a^2}{4t}}}\Big ]=\sqrt{\frac{\pi}{s}}e^{-a\sqrt{s}}$ for a Gaussian Normal Distribution.

If I set $a=2$ then the formula are alike apart for $\frac{1}{\sqrt{t}}$. How do I deal with that?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}\root{t}\expo{-1/t}
\expo{-st}\dd t}
\\[5mm] \stackrel{st\ \mapsto\ t}{=}\,\,\,&
{1 \over s^{3/2}}\int_{0}^{\infty}
\exp\pars{-t - {\bracks{\color{red}{2\root{s}}}^{2} \over 4t}}\,
{\dd t \over t^{\color{red}{-3/2} + 1}}
\\[5mm] = &\
{1 \over s^{3/2}}\bracks{\on{K}_{\color{red}{-3/2}}\,\pars{\color{red}{2\root{s}}} \over \pars{1/2}\pars{\color{red}{2\root{s}}/2}^{\color{red}{-3/2}}} =
{2 \over s^{3/4}}\on{K}_{3/2}\,\pars{2\root{s}}
\end{align}

See this link. $\ds{\on{K}_{\nu}}$ is a Modified Bessel Function. Note that $\ds{\on{K}_{\nu}\pars{z} = \on{K}_{-\nu}\pars{z}}$

However, $\ds{K_{3/2}\pars{z} = \root{\pi \over 2}\expo{-z}\
{z + 1 \over z^{3/2}}}$. See ${\bf\color{black}{10.2.17}}$ in A & S Table.

Then,
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}\root{t}\expo{-1/t}
\expo{-st}\dd t} =
{2 \over s^{3/4}}\bracks{\root{\pi \over 2}\expo{-2\root{s}}
{2\root{s} + 1 \over \pars{2\root{s}}^{3/2}}}
\\[5mm] = &
\bbx{{\root{\pi} \over 2}\pars{s^{-3/2} + 2s^{-1}}\expo{-2\root{s}}} \\ &
\end{align}


问:
Problem 1.2.14 (b) in Symbolic Dynamics and Coding
Given the alphabet $\mathcal{A}$ of size 3, let $X=\{x\in\mathcal{A}^{\mathbb{Z}}: x_{i+n^2}\neq x_{i} \forall i\in\mathbb{Z} \forall n\in\mathbb{N}\}$. Here $x_i$ is shorthand for $x(i)$. Show that $X=\emptyset$ I tried to use Pythagorean triples $a^2+b^2=c^2$ and concluded that $x_{a^2}=x_{b^2}$ if such an $x$ existed. So now all I need to do is prove $x_{a^2}\neq x_{b^2}$ and I'll have a proof by contradiction.

答:
(Fill in the gaps as needed. If you're stuck, write out your working and thought process to demonstrate where you're at.)

Suppose that such a sequence exists. Let $\mathcal{A} = \{R, B, G \}$.

WLOG let $ x_1 = R$.

We will focus on number of the form $ x_{1 + n^2}$, and label them $y_n$. Clearly $y_n = B$ or $G$.

Claim: If $ a^2 + b^2  = c^2$, then $ y_a, y_b$ have the same color, which is distinct from $y_c$.

WLOG let $ y_3 = y_4 = B$ so $y_5 = G$.

So $y_5 = y_{12} = G, y_{13}  = B$.

So $y_{12}=y_{16} = G, y_{20} = B$.

Using the pythagorean triples (and their multiples), keep on adding in known terms till you get a contradiction. (I get a contradiction with $y_{16}$, as it eventually needs to be $B$. Of course, you might reach another contradiction.)

 $ y_{12} = y_{9} = G, y_{15}  = B$

 $y_{15} = y_8 = B, y_{17} = G$

$y_8 = y_6 = B, y_{10} = G$

$y_{10} = y_{24} = G, y_{26} = B $

$y_{24} = y_{18} = G, y_{30} = B $

$y_{30} = y_{16} = B, y_{34} = G  $


问:
How do I show that $\int _0^\infty \! \frac{x\cos^3(x)}{t^2+x^2} \, \mathrm{d}x$ converges uniformly for all t > 0?
Until now, I've tried to solve by Dirichlet Criteria and Weierstrass M-test, but couldn't find an answer. Also tried to do inequalities, but didn't get anywhere.
Can somebody help me on this please?</p>

答:
Around $x=0$
$$\frac{x\cos^3(x)}{t^2+x^2}=\frac{x}{t^2}-\left(\frac{1}{t^4}+\frac{3}{2 t^2}\right) x^3+O\left(x^4\right)$$ so no problem at the lower bound.

You can also compute the antiderivative writing
$$\frac{x}{t^2+x^2}=\frac{x}{(x+it)(x-it)}=\frac{1}{2 (x+i t)}+\frac{1}{2 (x-i t)}$$ and use
$$4\cos^3(x)=\cos(3x)+3\cos(x)$$ to face rather simple integrals.


问:
Does uniformly bounded second moment imply Lindeberg's condition>
Suppose an independent sequence $X_i \in \mathbb{R}^k$ is such that $\sup_n E(\Vert X_i \Vert^2) \leq M$ for some constant $M$. Does this imply that $X_n$ satisfies the following Lindeberg-type condition:
\begin{align*}
\frac{1}{n}\sum_{i=1}^nE\left( \Vert X_i \Vert^2 I\{\Vert X_n \Vert \geq \epsilon \sqrt{n}\} \right) \rightarrow 0
\end{align*}

It seems clear to me that the sum does not diverge, since it is bounded by $M$, but is it true that it goes to zero?

答:
The answer is No.

Try $X_n= n.\mathcal{Bernoulli}(1/n^2)$


问:
Laurent Series of a Complex Function at $z = 4$
I am asked to find the Laurent series for several functions, but I am having trouble. I think that if I get help for one of my functions, I should be good to figure out the others.

Find the Laurent series for $\frac{6}{z(z-2)(z-3)}$ for all annuli with center at $z = 4$.

I know that this function has $3$ singular points: $0$, $2$, and $3$. My first step is to figure out all the possible annuli I needed to consider. I think I have found these regions to be $|z-4| < 1$, $2 < |z-4| < 3$, $2 < |z-4| < 4$, and $|z - 4| > 4$. However, I am not sure if I have determined these regions correctly. I keep getting confused on how to figure out these regions.

Next, I broke down the function using partial fractions to get $f(z) = \frac{1}{6}\frac{1}{z} - \frac{1}{2}\frac{1}{z-2} + \frac{1}{3}\frac{1}{z-3}$.

I know that the idea is to find the series for each piece in of this decomposition in each region specified previously. To illustrate my process for one part, I began with the region $|z-4| < 1$ and $\frac{1}{z}$ to obtain the series as follows:

$\frac{1}{z} = \frac{1}{(z-4) - 4} = \frac{1}{4 - -(z-4)} = \frac{1}{4}(\frac{1}{1 - - [(z-4)/4]}) = \frac{1}{4}\sum^{\infty}_{n=0}(-\frac{(z-4)}{4})^n$

$\implies \frac{1}{6}\frac{1}{z} = \frac{1}{24}\sum^{\infty}_{n=0}(-\frac{(z-4)}{4})^n$

I would then repeat this process within the given region for the other two pieces of the function to find their series, and then repeat this process for all the possible annuli.

I mainly looking to clarify

whether I have identified the correct annuli and

whether my process and idea on how to proceed to find the Laurent series for each annuli is correct.

答:
Trying to sort this out:  I'm getting one annulus, $1\lt|z-4|\lt2$, on which there is a convergent Laurent series.  And one more, namely $|z-4|\gt4$.  And finally a third, $2\lt|z-4|\lt4$.

The one you did would converge on $|z-4|\lt4$, which is a disk.  Also I think you have lost a minus sign.  There's a Laurent series, which is actually a Taylor series, for $|z-4|\lt4$, and one for $|z-4|\gt4$, which is the complement of a disk, both for $1/z$.

Then you do $1/(z-2)$ and $1/(z-3)$, and see where they each converge.

So, for $1/(z-2)$ we have $1/(z-4+2)=1/(z-4)\cdot1/(1+2/(z-4))=1/(z-4)\sum_{n\ge0}(-2/(z-4))^n=\sum_{n\ge0}(-2)^n(z-4)^{-n-1}$ for $|z-4|\gt2$.

There's also one  for $1/(z-2)$ that converges for $|z-4|\lt2$.

For $1/(z-3)$, we can get one for $|z-4|\lt1$, and one for $|z-4|\gt1$.

So putting these together, we get one on $1\lt|z-4|\lt2$.  Note that this annulus is contained in the disk $|z-4|\lt4$.

Another is the intersection of $|z-4|\gt1,|z-4|\gt2$ and $|z-4|\gt4$.  That's $|z-4|\gt4$.

Finally, there is the annulus $2\lt|z-4|\lt4$.

So, the first one is $1/24\sum_{n\ge0}((z-4)/4)^n+1/12\sum_{n\ge0}(-1)^n((z-4)/2)^n+1/3\sum(-1)^n/(z-4)^{n+1}$.

I leave it to you to write down each of the remaining two series.


问:
Is this always true (stochastic independence)?

Let $A,B,C,D$ be events in a probability space. Suppose $A\perp \!\!\!\!\!\! \perp B$, $B\perp \!\!\!\!\!\! \perp C$, $C\perp \!\!\!\!\!\! \perp D$.
It is always true that $$A\perp \!\!\!\!\!\! \perp D$$

This has me totally stumped!

答:
It's false. Stochastic independence is not transitive.

Look at the following example (3 events A,B,C)

Consider the experiment of tossing two fair dice. Let A denote the event of an odd total, B the event of an 1 on the first die, and C the event of a 7 total.

We obtain

$$\mathbb{P}[A|B]=\frac{1}{2}=\mathbb{P}[A]$$

$$\mathbb{P}[C|B]=\frac{1}{6}=\mathbb{P}[C]$$

But

$$\mathbb{P}[A|C]=1 \ne \mathbb{P}[A]$$

Thus A and B are independent, the same is true for B and C but it is false for A and C

...try yourself as an exercise to find a couterexample with 4 events


问:
How can we show/prove that a rank-$1$ matrix is non-convex?
In my optimization problem, I have a matrix $X = v v^H$ where $H$ denotes the complex conjugate transpose and $v \in \mathbb C^n$. This is a rank-$1$ matrix. In the published literature, it is mentioned that rank-$1$ matrix is also non-convex but I am unable to understand this. Can someone explain this to me why rank-$1$ matrix is also non-convex?</p>

答:
Let me give this a go since I was wondering the same thing. Are you asking why the set of rank 1 matrices is not convex? Let's consider the following:

$x_1 = [0,1]^T$, $x_2 = [1,0]^T$. Therefore $x_1x_1^T = \begin{bmatrix}
0 & 0 \\
0 & 1
\end{bmatrix}$ and $x_2x_2^T = \begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}$.

We note at this point that both $x_1x_1^T$ and $x_2x_2^T$ fall into the set of rank-1 matrices. However, their combination does not, and we can demonstrate this:

$\theta x_1x_1^T + (1-\theta)x_2x_2^T = \begin{bmatrix}
1-\theta & 0 \\
0 & \theta
\end{bmatrix}$, which is a rank-2 matrix. We cannot express this in the form $x_0 x_0^T$, (because what would $x_0$ be to allow me to get $x_0 x_0^T$ as seen above?) and it does not fall into the same set as the first two matrices $x_1x_1^T$ and $x_2x_2^T$.

Therefore the set of rank-1 matrices is not convex.


问:
Upper and Lower Games in Zero-Sum Games
I am working on some theory related to controls in the context of stochastic games, and I am a bit confused on some terminologies for zero-sum games.

Suppose we have a zero-sum game with two players.
Let the first players' inputs be $U \in \mathbb{R}^{m}$, whose goal is to minimize the payoff function $J(U,V)$.
Let the second players' inputs be $V \in \mathbb{R}^{\ell}$, whose goal is to maximize the payoff function $J(U,V)$.
In my context, the first player is called the "controller" and the second player is called the "stopper".

I am currently reading a paper that states the following:

The upper game is a scheme in which the stopper chooses $V$ based on the information it has on the control $U$, and the upper value is defined as

$$
\mathcal{V}^+ = \inf_{U\in\mathbb{R}^{m}}\sup_{V\in\mathbb{R}^{\ell}} J(U,V).
$$

Similarly, the lower game is a scheme in which the controller chooses $U$ based on the information it has on the control $V$, and the lower value is defined by

$$
\mathcal{V}^{-} = \sup_{V\in\mathbb{R}^{\ell}}\inf_{U\in\mathbb{R}^{m}} J(U,V).
$$

Now I know from the minimax theorem that $\mathcal{V}^{-} \leq \mathcal{V}^{+}$ and they are only equal to each other at a saddle point (equilibrium).

My question is regarding the definitions of the upper and lower games.
Looking at the equation for the upper value, how does it make sense that the upper game is a game in which the stopper plays based on the information it has on the controller if the first optimization is over $V$ and the second optimization is over $U$.
The way I'm interpreting the upper value is that the stopper optimizes the cost, and then based on the optimal stopper input, the controller then optimizes the resulting payoff, so the stopper acts first, and the controller best follows after.

Can someone please explain this to me?

答:
You have the order of the player's choices the wrong way round. In the upper game, the stopper is allowed to  make his or her choice of $\ V\ $ depend on the controller's choice of $\ U\ $. In practice, he or she would only be able to do this if the controller had already chosen $\ U\ $.  That is, it is the controller, not the stopper, who makes the first choice.

If the controller chooses $\ U\ $, then the stopper should choose $\ V^*(U)\ $ to make the payoff $\ J\big(U,V^*(U)\big)\ $ as close as possible to $\
\sup_\limits{V\in\mathbb{R}^\ell}J(U,V)\ $. When the controller chooses $\ U\ $, however, he or she can reasonably suppose that the the stopper, knowing the value of $\ U\ $, will choose $\ V^*(U)\ $, so the best value for him or her to choose would be one which makes $\
\sup_\limits{V\in\mathbb{R}^\ell}J(U,V)\ $ as close as possible to $\ \inf_\limits{U\in\mathbb{R}^m}\sup_\limits{V\in\mathbb{R}^\ell}J(U,V)\ $.

Title: Upper and Lower Games in Zero-Sum Games


问:
any divisor of a projective non-singular irreducible curve is equivalent to an effective divisor minus a multiple of a point
Here is a problem I need help getting started with. I am brand new to algebraic geometry.

Let $C$ be a projective non-singular irreducible curve of genus $g$. Let $P_0\in C$. Prove
that any divisor $D \in Div(C)$ is linearly equivalent to a divisor of the form $D_0 - nP_0$,
where $D_0\ge 0$ has degree $g$. Of course, $n=g-\deg D$. Would $D_0$ have to be canonical?

I am given the Riemann-Roch Theorem but I don't know how to connect this question to the dimensions of the Riemann-Roch spaces.

What theorems should I be thinking about? Thanks.

答:
Write $D'=D+(g-deg(D))P_0$ so the problem is to show that $D'$ is equivalente to a effective divisor, note that $deg(D')=g$ by Riemann-Roch theorem we have
$$
\dim L(D')\geq 1
$$

take $f\in L(D')$ so by definition $D_0=div(f)+D'\geq 0$ and $D_0$ is equivalent to $D'$.

rmk$_1$: $L(A)=\{f\in \mathcal{M}(X):\textrm{ }ord_p(f)\geq A(p)\}$, and $div(f)=\sum_p ord_p(f)p$ is the divisor associated to a meromorphic function.

rmk$_2$: Of course that $deg(D_0)=g$ because the $deg(div(f))=0$.

rmk$_3$: $X$ just need to be a compact Riemann surface of genus $g$.


问:
Point spectrum (eigenvalues) of two operators in $l^2$.
Let the following operators of $l^2$ to $l^2$

$S(x_0,x_1, x_2,...)=(0,x_0, x_1/2, x_2/3,...)$

$T(x_0,x_1, x_2,...)=(x_1/2, x_2/3, x_3/4,...)$

I need to find the point spectrum(eigenvalues set) $\sigma_p(S)$ and $\sigma_p(T)$.

I showed that $\sigma_p(S)=\varnothing$.

For $\sigma_p(T)$, I have this:

\begin{align}
T(x)=\lambda x\,\,\,&\Leftrightarrow\,\,\,(x_1/2, x_2/3, x_3/4,...)=\lambda(x_0,x_1, x_2,...)\\
&\Leftrightarrow\,\,\, x_n=(n+1)!\lambda x_0,\,\,\,\,\,\, n\geq 1
\end{align} Then $(x_0,x_1, x_2,x_4...)=x_0(1, 2\lambda, 6\lambda^2, 24\lambda^3,...)$.

But from here I do not know how to conclude which are the eigenvalues of T.

I appreciate the help you can give me.

答: In order that the sequence $(x_n)$ you obtained is in $l^{2}$ we must have $\sum ((n+1)!)^{2}|\lambda|^{2} <\infty$. This is possible only for $\lambda =0$. But $\lambda =0$ gives $x_n=0$ for all $n$. Hence $T$ has no eigen values.


问:
How many functions are possible?
I would appreciable any insight you give to solve this type of question:

How many $f: \{1,2,3\}\to \{1,2,3\}$ satisfy $f(f(x))=f(f(f(x)))$ for all $x$?

Normally questions of this type I used try by guessing, until I can't see any more possibilities. But I don't like this way of solving, do you know a theorem or some tool to solve this type of question with certainty?

答:
(Fill in the gaps as needed. If you're stuck, explain what you've tried and why/where you are stuck.)

Hint: Show that we cannot have $f(a) = b, f(b) = a$.

Hence, we must either have A) $ f(a) = a$ or B) $f(a) = b, f(b) = c$.

Hence, there are $X$ such functions.

Title: How many functions are possible?


问:
How many 8-digit numbers do not contain a sub-string of 3 repeated digits?
I am struggling quite a bit with the above question. I am unsure what formula to use to determine this, mainly due to the scale of the problem.

I tried 9x10x9x9x9x9x9X9, but I am unsure about this as a solution. My understanding is that the substring requirement would see 10,001,223 as a number with three repeated digits in a substring, but 10,100,100 does not contain a substring with three repeated digits.

I am very confused as to how to calculate this, any help determining a formula would be greatly appreciated.

答:
(Fill in the gaps as needed. If you're stuck, show your work and explain where/why you are stuck.)

Hint: Let $f_{k, 1}$ be the number of $k-$ digit numbers that do not contain a substring of 3 repeated digits, and end with only 1 repeated digit.

Let $f_{k, 2}$ be the number of $k-$ digit numbers that do not contain a substring of 3 repeated digits, and end with only 2 repeated digits.

Prove the recurrence relation

$$ \begin{align} f_{k+1, 1 } & = 9 f_{k,1} + 9 f_{k,2} \\
f_{k+1, 2 } &= f_{k+1} \\ \end{align}$$

Determine suitable starting values.

Hence, find $f_{8,1} + f_{8,2}$.


问:
Conjecture: Are there infinitely many triangular numbers which are of the form $qp$ , where $p$,$q$ are distinct primes?

Show that there are infinitely many positive integers $n$ such that the number of distinct odd prime factors of $n(n + 3)$ is a multiple of $3$.

I couldn't get much progress, I took $n= 3k$, and then was trying to show, that there are infinitely many  many positive integers $k$ such that the number of distinct odd prime factors of $k(k + 1)$ is $1\mod 3$.

So if I can show that

There are infinitely many triangular numbers which are of the form $qp$ , where $p,q$ is a prime

This looks true seeing the OEIS Link , the first term  is $55$, then $91$ , then $231$ and so on ..
then I will be done.

However, I think I am in a wrong path, because it's a contest problem.

Thanks in advance!

Here's the link of the question

答:
Here's a different method to solve the contest problem. Assume there's only a finite number of positive integers $n$ where the number of distinct odd prime factors of $n(n + 3)$ is a multiple of $3$. Thus, there's a maximum integer $n_0$ where this holds so, for all $n \gt n_0$, the number of distinct odd prime factors of $n(n + 3)$ is not a multiple of $3$. Note all of the integers used below are considered to be $\gt n_0$. Next, define

$$f(i) = \text{the number of distinct prime factors } \ge 5 \text{ of } i \tag{1}\label{eq1A}$$

One other thing to note is there is no prime factor $\ge 5$ in common among any two integers in a group of $4$ consecutive integers.

Similar to what you did, the product of any $2$ consecutive integers, say $m(m + 1)$, can be multiplied by $9$ to get $3m(3m + 3)$, which is of the form of $n(n + 3)$ with $n = 3m$. The finiteness assumption means that $1$ (for the factor of $3$) plus the number of distinct odd prime factors $\ge 5$ of $n(n + 3)$ (and also $m(m + 1)$) can't be a multiple of $3$. Thus, this means for any $2$ consecutive integers $m$ and $m + 1$, since the $f(i)$ function doesn't include the factor of $3$, we get

$$f(m(m + 1)) = f(m) + f(m + 1) \not\equiv 2 \pmod{3} \tag{2}\label{eq2A}$$

Squaring doesn't change the number of distinct prime factors, so $f(j^2) = f(j)$. Thus,

$$f((j^2 - 1)j^2) = f(j^2 - 1) + f(j^2) = f(j - 1) + f(j) + f(j + 1) \tag{3}\label{eq3A}$$

Using this, along $m = j^2 - 1$ in \eqref{eq2A}, gives

$$f(j - 1) + f(j) + f(j + 1) \not\equiv 2 \pmod{3} \tag{4}\label{eq4A}$$

Choose an $n_1$ where $3 \mid n_1$ and $f(n_1) \equiv 2 \pmod{3}$ (e.g., $n_1$ is $3$ times the product of $2$ large primes). Next, for somewhat simpler algebra, define

$$d_i = f(n_1 + i), \; i \ge 0 \tag{5}\label{eq5A}$$

which means

$$d_0 \equiv 2 \pmod{3} \tag{6}\label{eq6A}$$

Using \eqref{eq2A} (with $m = n_1$ in \eqref{eq7A} and $m = n_1 + 1$ in \eqref{eq9A}), \eqref{eq4A} (with $j - 1 = n_1$ in \eqref{eq8A}) and \eqref{eq5A} gives

$$d_0 + d_1 \not\equiv 2 \pmod{3} \tag{7}\label{eq7A}$$

$$d_0 + d_1 + d_2 \not\equiv 2 \pmod{3} \tag{8}\label{eq8A}$$

$$d_1 + d_2 \not\equiv 2 \pmod{3} \tag{9}\label{eq9A}$$

Using \eqref{eq6A} in \eqref{eq8A} gives $d_1 + d_2 \not\equiv 0 \pmod{3}$. Combined with \eqref{eq9A}, this gives

$$d_1 + d_2 \equiv 1 \pmod{3} \tag{10}\label{eq10A}$$

Using \eqref{eq6A} in \eqref{eq7A} gives $d_1 \not\equiv 0 \pmod{3}$. If $d_1 \equiv 2 \pmod{3}$, then \eqref{eq10A} gives $d_2 \equiv 2 \pmod{3}$. Note, though, in this case, we can then repeatedly use \eqref{eq8A}, \eqref{eq9A} and \eqref{eq10A}, with the indices being incremented by $1$ each time, to get that $d_i \equiv 2 \pmod{3}$ for all $i \ge 0$. However, this is not possible, e.g., where a $n_1 + i$ value is a prime number. Thus, this means we must instead have

$$d_1 \equiv 1 \pmod{3} \tag{11}\label{eq11A}$$

Therefore, \eqref{eq10A} gives

$$d_2 \equiv 0 \pmod{3} \tag{12}\label{eq12A}$$

Reusing \eqref{eq8A} and \eqref{eq9A} with the indices increased by $1$ results in

$$d_1 + d_2 + d_3 \not\equiv 2 \pmod{3} \tag{13}\label{eq13A}$$

$$d_2 + d_3 \not\equiv 2 \pmod{3} \tag{14}\label{eq14A}$$

Using \eqref{eq11A} in \eqref{eq13A} gives $d_2 + d_3 \not\equiv 1 \pmod{3}$. Combined with \eqref{eq14A} gives

$$d_2 + d_3 \equiv 0 \pmod{3} \tag{15}\label{eq15A}$$

Using \eqref{eq12A} in \eqref{eq15A} gives

$$d_3 \equiv 0 \pmod{3} \tag{16}\label{eq16A}$$

Using $3 \mid n_1$ with $f(n_1(n_1 + 3))$ and our finiteness assumption means that

$$d_0 + d_3 \not\equiv 2 \pmod{3} \tag{17}\label{eq17A}$$

However, using \eqref{eq6A} in \eqref{eq17A} gives

$$d_3 \not\equiv 0 \pmod{3} \tag{18}\label{eq18A}$$

This contradicts \eqref{eq16A}. Since we have shown that both of the $2$ allowed cases for the congruence of $d_1 \pmod{3}$ can't hold, this means the original finiteness assumption, i.e., there's only a finite number of $n$ which work, must be incorrect. This proves there are an infinite number of positive integers $n$ where the number of distinct odd prime factors of $n(n + 3)$ is a multiple of $3$.


问:
What is the maximum number of T-shaped polyominos (shown below) that can be put into a 6x6 grid without any overlaps? The blocks can be rotated.

I just drew the figure and manually tried the question but I am wondering is there a way to do this problem via permutations and combinations.

PS: I got answer as 7.

答:
(Fill in the gaps as needed. IF you're stuck, explain where you are stuck and show your work.)

Hint: The answer is

 8

Find a configuration that has that many T's.

To prove that we can't do 9 T's, color the grid like a chessboard. Notice that there are 18 Black and 18 White squares. Notice that every T either covers 3 Black 1 White or 3 White 1 Black squares. Find a contradiction.

 There is no non-negative integer solution to $a+b = 9, 3a+b = 18$, hence we can't place 9 tiles.


问:
Showing $E(X + a) = E(X) + a$ for Non-Negative Random Variables
I have been working through the lecture notes given here. Specifically I have been looking at page 4

Theorem 1.2. part a) and part b)

Let $X$ and $Y$ be a non-negative random variable.

a) $X$ $\leq$ $Y$, then $E(X)$ $\leq$ $E(Y)$

b) Then for $a\geq 0$ $E(X+a) = E(X) +a$.

The proof given in the text says given $E(Z + a) = E(Z) + a$ for simple random variables $Z$.
We need to simply follow the proof of part a). However, I cannot make the jump from the proof in part a) to the proof in part b). The proof in part a) is as follows:

Let $\epsilon>0$. Since $X \geq 0$, there exists a simple random variable $Z$ s.t $Z \leq X$ and $E(Z) \geq E(X) - \epsilon$. Since $X \leq Y$, we have that $Z \leq Y$. Hence, $E(X) - \epsilon \leq E(Z) \leq E(Y)$. Taking $\epsilon$ to zero gives us the desired result.

Thank you.

答:
Statement b) is true in general, not only for non-negative rv

$$\mathbb{E}[X+a]=\int(x+a)dF=\int xdF+a\int dF=\mathbb{E}[X]+a$$

Same proof for $\mathbb{E}[aX]=a\mathbb{E}[X]$


问:
In how many ways can n couples (husband and wife) be arranged on a bench so no wife would sit next to her husband?
In how many ways can n couples (husband and wife) be arranged on a bench so no wife would sit next to her husband?

I thought about this:

(Total amount of ways to sit 2n people in 2n sits)-(Using inclusion and exclusion to find that at least 1 wife sits next to her husband) And i get:

Let $A_1$ be the attribute where at least 1 wife sits with her husband, Then we "merge" up the husband and wife into a one person. We have $\binom n1$ ways to choose $1$ couple out of $n$ couples, And we are left with $2n-1$ to place $2n-1$ 'people' so we get $(2n-1)!$ and so on, And on a general note:

$(2n)!-(2\binom n1(2n-1)!-2^2\binom n2(2n-2)!+...2^k(-1)^k\binom nk(2n-k)!)$ And a bit simplified:

$(2n)!-(\sum_{k=1}^n2^k(-1)^k\binom nk(2n-k)!$)

Since i don't have answers to this question i wanna know if i did something wrong? Did i even look at the question right?

答:
As far as I am able to tell, the answer is correct.

After some unsuccessful attempts at simplifying the resulting expression, I went ahead and asked Wolfram for the answer. He came up with the formula:

$$ 4^n \frac{\Gamma(n+\frac{1}{2})}{\sqrt{\pi}} \operatorname{Hypergeometric1F1}(-n,-2n,-2)$$

The weird looking term $\frac{\Gamma(n+\frac{1}{2})}{\sqrt{\pi}} $ is just $(2n-1)!!/2^n$, so this can also be written as:

$$ 2^n (2n-1)!! \cdot \operatorname{Hypergeometri1cF1}(-n,-2n,-2)$$

Unfortunately, it seems that $\operatorname{Hypergeometri1cF1}(-n,-2n,-2)$ is not anything nicer. Of course, I only have heuristic justification of the type "if it was something nicer, it wouldn't be called with such scary name, and Wolfram would be able to simplify it".


问:
Characters on $\ell^\infty$ and characteristic functions
Let $\ell^\infty$ be the set of bounded complex-valued sequences, which we regard as a commutative $C^*$ algebra under pointwise multiplication. Write $\Phi$ for the set of characters of $\ell^\infty$, i.e. non-zero algebra homomorphisms $\ell^\infty\to \mathbb{C}$. Note that, if $B\subset\mathbb{N}$ and $\chi_B$ is its characteristic function, then $\varphi(\chi_B)\in \{0,1\}$ for any $\varphi\in\Phi$ because
$$
\varphi(\chi_B)=\varphi(\chi_B^2)=(\varphi(\chi_B))^2
$$

Fix $\varphi\in \Phi$, $\varepsilon>0$, and $x\in \ell^\infty$. Put
$$
A=\{n\in\mathbb{N}:|x_n-\varphi(x)|<\varepsilon\}
$$
I'd like to show that $\varphi(\chi_A)=1$.

Some context: By the Gelfand-Naimark Theorem, $C(\Phi)\cong \ell^\infty$, and $n\in \mathbb{N}$ can be identified with the projection $x\mapsto x_n$, which obviously belongs to $\Phi$. Under this identification, $\mathbb N$ is dense in $\Phi$ and so $\Phi$ can be regarded as the Stone-Cech compacfication of $\mathbb N$. If we let $\beta \mathbb N$ be the set of ultrafilters on $\mathbb N$, then we can find explicit bijections between $\Phi$ and $\beta \mathbb N$ by considering
$$
F: \Phi \to \beta\mathbb N\\
\varphi\mapsto \{A\subset \mathbb N:\varphi(\chi_A)=1\}
$$
and
$$
G:\beta\mathbb{N}\to \Phi \\
\mathcal{U}\mapsto \left(x\mapsto \lim_\mathcal{U} x\right)
$$
The identity I'm trying to prove arises when checking that $(G\circ F)(\varphi)=\varphi$ for every $\varphi\in\Phi$.  Of course, one can also verify that they are continuous and then check that the compositions are the identity on $\mathbb{N}$, which as a by product yields what we want. However, I suspect there's a more direct, elementary argument that I'm not seeing.

答:
Let $$y_n=\begin{cases} \frac1{x_n-\varphi(x)}\quad n\notin A,\\0 \quad\,\qquad n\in A.\end{cases}$$
Then $|y_n|\leq \frac1\epsilon$ for all $n$ so $y\in \ell^\infty$.

Then
\begin{eqnarray*}1-\varphi(\chi_A)&=&\varphi(1-\chi_A)\\&=&\varphi((x-\varphi(x))y)\\&=&\varphi(x-\varphi(x))\varphi(y)\\&=&(\varphi(x)-\varphi(x))\varphi(y)\\&=&0.\end{eqnarray*}


答: I filled the 6x6 square manually withou any calculation or considering possibility of other cases.

Title: What is the maximum number of T-shaped polyominos (shown below) that can be put into a 6x6 grid without any overlaps? The blocks can be rotated.

Body:
I just drew the figure and manually tried the question but I am wondering is there a way to do this problem via permutations and combinations.

PS: I got answer as 7.


问:
Product of derivations
Let $A$ be an algebra over a field $K$. If $D_1$ and $D_2$ are derivations of $A$, show that $D_1 \circ D_2$ is not necessarily a derivation (it is if $D_1$ or $D_2 = 0$), but $D_1 \circ D_2−D_2 \circ D_1$ is always a derivation of $A$.

It's a problem from Tu, Loring W's An introduction to manifolds. Obviously, the composition of linear maps is linear, but how can I check the Leibniz rule?

Meanwhile, why it defines to be $D_1 \circ D_2−D_2 \circ D_1$? Are there any motivations or backgrounds?

答:
1. For the first part (why $D_1 \circ D_2$ is not a derivation), just try to come up with a counter-example.

For example, let $A = K[x,y]$ be polynomials in two variables. The partial derivative operators $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ are derivations of $A$ (by the standard product rule from calculus). But their composition $\frac{\partial}{\partial y} \frac{\partial}{\partial x}$ is not. Just plug in something simple, like $xy$:

$$ \frac{\partial}{\partial y} \frac{\partial}{\partial x} xy = 1 $$
But on the other hand,
$$ \left( \frac{\partial}{\partial y} \frac{\partial}{\partial x} x \right) \cdot y + x \cdot \left( \frac{\partial}{\partial y} \frac{\partial}{\partial x} y \right) = 0 $$
So it does not satisfy the Leibniz rule.

2. For the second part (why $D_1 \circ D_2 - D_2 \circ D_1$ is a derivation), you can just compute "by-hand", and check it satisfies the Leibniz rule. Let $a,b \in A$, and evaluate:

$$
D_1(D_2(ab))-D_2(D_1(ab)) = D_1(D_2(a) \cdot b + a \cdot D_2(b)) - D_2( D_1(a) \cdot b + a \cdot D_1(b)) $$
Here we just used the Leibniz rule for $D_2$ and $D_1$. Now since $D_1$ and $D_2$ are linear, you can "distribute", so this gives
$$ D_1(D_2(a) \cdot b) + D_1(a \cdot D_2(b)) - D_2(D_1(a) \cdot b) - D_2(a \cdot D_1(b)) $$
Now just use the Liebniz identity again for each of the four terms, and a bunch of stuff should cancel. What is left over should give the Leibniz identity for $D_1 \circ D_2 - D_2 \circ D_1$.

3. As for why this is the "correct" definition: you said this comes from a book about manifolds. If the derivations are thought of as vector fields on a manifold, then this "commutator" $XY-YX$ corresponds to the Lie derivative.

Title: Product of derivations


问:
Inverses of Multidimensional Vector-Valued Functions
I'm working on a Topology assignment, and we were asked to show that the annulus $A$ of inner radius $1$ and outer radius $2$ is homeomorphic to a cylinder $C$ of radius $1$ and height $1$. Now, I already have the two homeomorphisms, but I thought to myself, I really should be able to just invert the first function. The problem is I'm a little more than confused on how to do this if we are considering the function $f:\mathbb{R^3}\rightarrow\mathbb{R^2}$.

Here is the homeomorphism for both directions:
$$f:A\rightarrow C\,\,\,\,\text{where}\,\,\,\,f(x,y,z)=(x(z+1),y(z+1))$$
$$g:C\rightarrow A\,\,\,\,\text{where}\,\,\,\,f(x,y)=\bigg(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}},\sqrt{x^2+y^2}-1 \bigg)$$

A quick computation can be done to verify these are inverse functions. How might I go about finding the other mapping once I have one of these continuous functions?

答:
Set $(x(z+1),y(z+1))=(A,B).$ Then, $A/x=z+1=B/y.\ $ Now, $x^2+y^2=1$ so

$A/x=B/\sqrt{1-x^2}\Rightarrow A^2(1-x^2)=B^2x^2\Rightarrow (A^2+B^2)x^2=A^2\Rightarrow x=\pm\frac{A}{\sqrt{A^2+B^2}}.$

Similarly, $y=\pm\frac{B}{\sqrt{A^2+B^2}}$ and then

$z+1=\pm A(\sqrt{A^2+B^2})/A=\pm\sqrt{A^2+B^2}\Rightarrow z=\pm\sqrt{A^2+B^2}-1.$

I'll leave it to you to determine the signs, and invert the other function.


问:
Prove sum of the lengths of intervals in a finite covering of $\mathbb{Q}\cap [0,1]$ is $\geq 1$
I have the following proof, but I don't think it's right. Could somebody please tell me what's wrong with it, and how to fix it?  Thanks :)

Let $B$ be the set of rational numbers in the interval $[0,1]$ Let $I_{k}$, $k=1,2,\cdots, n$ be a finite collection of open intervals that covers $B$.  Prove that $\sum_{k=1}^{n}l(I_{k})\geq 1$.

($l(I)$ denotes the length of the interval $I$).

$\underline{\mathbf{\text{Proof}}}$: Suppose $a,b \in [0,1]$ such that $a < b$. By the density of the rational numbers in $\mathbb{R}$, $\exists q_{i} \in \mathbb{Q}$ such that $\forall a_{i} < b_{i}$, $q_{i}\in(a_{i},b_{i})$.

Since the rationals are dense in $\mathbb{R}$ and hence in $[0,1]$, we can cover $[0,1]$ as follows:

$[0,1] \subseteq \cup_{k=1}^{\infty}I_{k}=\cup_{k=1}^{\infty} (a_{k},b_{k}) $.

(In the case where the $(a_{k},b_{k})$ do not completely cover $[0,1]$ completely, $\exists$ a point between $(a_{i},b_{i})$ and $(a_{i+1}, b_{i+1})$ that remains uncovered.  However, because there are countably many intervals, there will be only countably many such points.  Thus, the set of points, call it $S$, being a countable set will have $m^{*}(S)=0$.  So, these points will be negligible in terms of their effect on the sum of the lengths of the intervals.)

By Heine-Borel, since $[0,1]$ is closed and bounded, every such open cover has a finite subcover. So, $[0,1] \subset \cup_{k=1}^{n}(a_{k},b_{k}) = \cup_{k=1}^{n}I_{k}$.

Now, the outer measure of an interval is its length.  So $m^{*}([0,1])=l([0,1]) = 1$, and by the definition of the outer measure, $1 = m^{*}([0,1]) = \inf_{k=1}^{n}l(I_{k}) \leq \sum_{k=1}^{n}l(I_{k})$

Q.E.D.

I also realize this question has been asked before on here, but I am terribly dissatisfied with the answers that have thus far been given.

答: Consider the corresponding closed intervals. Their union covers $\mathbb{Q}\cap[0,1]$ and is closed  ( a finite union of closed sets). Therefore their union contains $[0,1]$. So the sum of the length of the intervals is $\ge 1$.


问:
Definition of continuous map
Let $X,Y$ be topological spaces, let $f:X\longrightarrow Y$ a function. There are several (equivalent) ways to define continuity of $f$, one of these says: $f$ is continuous if
$$(1)\qquad \forall A\subseteq X,\ \forall x\in X, \Big(x\in \overline{A}\Rightarrow f(x)\in\overline{f(A)}\Big)$$
where $\overline{A}$ denotes the closure of $A$ in $X$, and similarly for $f(A)$ in $Y$.

I was wondering if reversing this condition, i.e.
$$(2)\qquad \forall A\subseteq X,\ \forall x\in X, \Big(f(x)\in \overline{f(A)}\Rightarrow x\in\overline{A}\Big)$$
also defines a continuous map. So, i'm asking if $(1)$ and $(2)$ are equivalent, if not, i'm looking for a counterexample, i.e. a continuous map $f$ not satisfying $(2)$.
Any help appreciated!

答:
Non-injective functions are prone to this.

For example, $f\colon\mathbb R\to\mathbb R$, $x\mapsto 0$ is continuous, but does not have $(2)$, e.g. with $A=[0,1]$ and $x=2$.

On the other hand, the function $x\mapsto x$ from a set with at least two elements endowed with the indiscrete topology to the same set endowed with discrete topology has property $(2)$, but is not continuous.

Title: Definition of continuous map


问:
Help me understand the meaning of this question about sets
From this probability book page 50, question 13.

Let $X$ be any set and let $A \subset X$. Define the real valued function $f$ by

$$
f(x) :=
\begin{cases}
1, x\in A \\
0, x\notin A
\end{cases}
$$

Thus, $f:X\rightarrow\mathbb{R}$ where $\mathbb{R}:= (-\infty, \infty)$ denotes real numbers. For any arbitrary $B \subset \mathbb{R}$ find $f^{-1}(B)$. Hint: There are four cases to consider depending on if 0 or 1 belong to $B$.

What I don't understand about this question is the part where it says that $f(x)$ either returns a 1 or a 0 depending on the membership to the set $A$ but it then states in the next line that the output is the whole real line. I must be reading something wrong, how is this possible?

答:
Guide: The $4$ cases are

$1 \in B$, $0 \in B$.

$1 \in B$, $0 \notin B$.

$1 \notin B$, $0 \in B$.

$1 \notin B$, $0 \notin B$.

Evaluate the set for each case.

It is possible that $B$ contain elements other than $0$ and $1$, of which these elements will be ignored in this particular computation.

For example $f^{-1}(\{0,1,2\})=f^{-1}(\{0,1\})$ for our $f$.

$$f^{-1}(B)=\{x \in X:f(x) \in B\}$$


问:
Solution of third order ODE
I consider the following ODE
$$
f'''(x) = B f''(x) + (Ce^x+D)f'(x) + (Ee^x+F)f(x)
$$
for some constants $B$, $C$, $D$, $E$, $F$ and $x\in [0, \infty)$.

This ODE seems to be complicated so I used WolframAlpha (https://www.wolframalpha.com/input/?i=f%27%27%27%28x%29+%3D+B+f%27%27%28x%29+%2B+%28C+e%5Ex+%2B+D%29f%27%28x%29+%2B+%28E+e%5Ex+%2B+F%29+f%28x%29) to find the general form of its solution.

My problem is that I am not sure if the solution that WolframAlpha returns is correct.
Can anyone suggest me how can I find the general form of this solution or which software (similar to WolframAlpha) I can use to find this solution.

Title: Solution of third order ODE

Body:
Let $t=e^x$ ,

Then $\dfrac{df}{dx}=\dfrac{df}{dt}\dfrac{dt}{dx}=e^x\dfrac{df}{dt}=t\dfrac{df}{dt}$

$\dfrac{d^2f}{dx^2}=\dfrac{d}{dx}\left(t\dfrac{df}{dt}\right)=\dfrac{d}{dt}\left(t\dfrac{df}{dt}\right)\dfrac{dt}{dx}=\left(t\dfrac{d^2f}{dt^2}+\dfrac{df}{dt}\right)t=t^2\dfrac{d^2f}{dt^2}+t\dfrac{df}{dt}$

$\dfrac{d^3f}{dx^3}=\dfrac{d}{dx}\left(t^2\dfrac{d^2f}{dt^2}+t\dfrac{df}{dt}\right)=\dfrac{d}{dt}\left(t^2\dfrac{d^2f}{dt^2}+t\dfrac{df}{dt}\right)\dfrac{dt}{dx}=\left(t^2\dfrac{d^3f}{dt^3}+3t\dfrac{d^2f}{dt^2}+\dfrac{df}{dt}\right)t=t^3\dfrac{d^3f}{dt^3}+3t^2\dfrac{d^2f}{dt^2}+t\dfrac{df}{dt}$

$\therefore t^3\dfrac{d^3f}{dt^3}+3t^2\dfrac{d^2f}{dt^2}+t\dfrac{df}{dt}=Bt^2\dfrac{d^2f}{dt^2}+Bt\dfrac{df}{dt}+(Ct+D)t\dfrac{df}{dt}+(Et+F)f$

$t^3\dfrac{d^3f}{dt^3}+(3-B)t^2\dfrac{d^2f}{dt^2}-(Ct^2+(B+D-1)t)\dfrac{df}{dt}-(Et+F)f=0$

Seek for power series solution


问:
Basic logic for the interior of a set
I'm currently painstakingly working my way through the basics of Topology and I'm on interior sets. My textbook provides this basic question for me to prove

Question to solve

Show that interior satisfies $\operatorname{int}A=(\bar{A'})'$

So, the interior of a set $A$ is the largest open set contained inside $A$, and this question wants me to relate that definition to the idea that the closed complement of A forms an interior set when again made to a complement.

My first idea to proving this is to simplify the question by moving the second complement over the equal so the LHS is now of the form $(\operatorname{int}A)'$. After that...I'm not sure how to progress. Any hints would be appreciated.

答:
Observe that the interior of $A$ is also the union of all open sets contained in $A$. So, to check that $x\in \textrm{int}(A)$, it suffices to obtain an open $U$ such that $x\in U\subseteq A$.

If an element $x$ is in the interior, then there is an open set $U$ such that $x\in U\subseteq A$, hence $x$ is not in the closure of the complement of $A$, therefore $\textrm{int}(A)\subseteq (\bar{A^c})^c$. Conversely, if $x\in(\bar{A^c})^c$, then there is a neighbourhood $U$ of $x$ disjoint from $A^c$ for $x$ is not in the closure of $A^c$, therefore $U\subset A$, hence $x\in \textrm{int}(A)$.

Title: Basic logic for the interior of a set


问:
if $ \beta $ is one to one then $\alpha$ and $ \beta $ have inverses and are inverses to one another
Consider the mappings

$ \alpha : A \rightarrow B$ ,
$ \beta : B \rightarrow C$

Where $\beta \alpha = 1_{A}$

How would one go about proving if $ \beta $ is one to one then $\alpha$ and $ \beta $ have inverses and are inverses to one another.

My issue here is just showing that the inverse exists. Showing that they are inverses to each other i understand

答: What you want to show is that $\alpha \beta = 1_B$. So, since $\beta$ is injective, this is the same thing as asking $\beta \alpha \beta = \beta$ (if $h$ is injective, then $f = g$ if and only if $hf = hg$. Prove this formally if you need to). But we can simplify the LHS to the RHS because of our assumption $\beta \alpha = 1_A$.


问:
Property of the average digits of a number
I thought (out of curiosity) about this limit:

$$ \lim_{n \rightarrow \infty} \frac{d(k^n)}{ \lfloor{\log_{10}(k^n)} \rfloor + 1} $$

Where $d(k)$ is the sum of the digits of $k \in \mathbb{N}$ and the $\log$ part is the length of the number. Basically, it is the limit of the algebraic average of all the digits in that number.

While this limit does not seem to converge ( checked by a program, not mathematically), it does seem to get bounded after $n>N \in \mathbb{N}$.

I took for example: $k = 12$ and saw that when $n \in [300, 1000] $ the value of the expression is bounded:

$$ 4 \leq \frac{d(k^n)}{ \lfloor{\log_{10}} \rfloor + 1}  \leq 5 $$

(May not be needed, this is my python code):

```<code> import math
 def average_digits(n, k):
     return sum( map(int, list(str(n ** k))) ) / (int(math.log10(n ** k)) + 1)
</code>```

When we choose $k= 300$ for example, after some iterations it gets bounded by $0$ and $1$.

Is there a reason for this to happen? Is there a way to estimate the bounds?

Unfortunately I lack this kind of mathematical knowledge, but this is still very mysterious to me.

Thank you!

答:
We know remarkably little about the digits of powers. It is conjectured that these digits are uniformly distributed among $0$ to $9$, leading to an estimate of

$$d(k^n)\sim 4.5n\log_{10}k$$

for $10 \nmid k$. This leads to the first "bound" of $12^k$: the value should be around $4.5$. However this is unproven, see http://oeis.org/A001370, for example.

In the case of $k=300$, there are too many zeros in $300^n$. If the conjecture holds true:

$$\lim_{n\to\infty}\frac {d(300^n)}{\lfloor\log_{10}(300^n)\rfloor+1}=\lim_{n\to\infty}\frac {d(3^n)}{2n+\lfloor\log_{10}(3^n)\rfloor+1}\sim \lim_{n\to\infty}\frac {4.5n\log_{10}3}{2n+n\log_{10}3}=\frac {4.5\log_{10}3}{2+\log_{10}3}$$

which is approximately equal to $0.866750$.

Without the conjecture, the coarse upper bound

$$\lim_{n\to\infty}\frac {d(3^n)}{2n+\lfloor\log_{10}(3^n)\rfloor+1}\le \lim_{n\to\infty}\frac {9(\lfloor\log_{10}(3^n)\rfloor+1)}{2n+\lfloor\log_{10}(3^n)\rfloor+1}\le \lim_{n\to\infty}\frac {9n(\log_{10}3+1)}{2n+n \log_{10}3}=\frac {9\log_{10}3}{2+\log_{10}3}$$

holds, but this value is double that of the previous limit ($1.733500$).

We don't even know if this average is bounded or not. For example for $k=9$, we know that $9 \mid d(9^n)$. However it seems that the question "how many powers of $9$ has digit sum $9m$" is open for $m > 1$, and we only know that they are finite for each $m$ (see http://oeis.org/A056888). This means the the bounds we obtained for the average could be broken at any time, however improbable this is (do note that this is different from showing that the limit exists).


问:
MGF and Compound Poisson Distribution
I have $N_{t}$ governed by a Compound Poisson$(t\lambda$) distribution with expected value $t\lambda$ and a sequence of $X_{i}$ where each $X_{i}$ is independent of $N_{t}$. If $Y = \sum^{N_{t}}_{i=1} X_{i}$, how can I find the mgf and first moment of $Y$?

I am used to the continuous case where I can integrate and differentiate, but I don't think I can apply that here due to the discrete nature of this distribution.

答:
First off, saying "$N_t$ governed by a compound $\mathsf{Poisson}(t\lambda)$ distribution" and defining $Y=\sum_{i=1}^{N_t} X_i$ is a bit confusing - I believe what you meant is that there is a given value $\lambda>0$, and for each $t>0$ you define the random variable
$$
Y_t = \sum_{i=1}^{N_t} X_i,
$$
so that $\{Y_t : t>0\}$ is a stochastic process - that is, a collection of random variables indexed on time - where $Y_t$ has compound $\mathsf{Poisson}(t\lambda)$ distribution.

In this case, assuming the $X_i$ are i.i.d., the expected value of $Y_t$ can be computed using the law of total expectation:
\begin{align}
\mathbb E[Y_t] &= \mathbb E[\mathbb E[Y_t\mid N_t]]\\
&= \mathbb E[N_t\mathbb E[X_1]]\\
&= \mathbb E[N_t]\mathbb E[X_1]\\
&= t\lambda\mathbb E[X_1].
\end{align}
The moment-generating function of $Y_t$ is given by
\begin{align}
M_{Y_t}(\theta) :&= \mathbb E[\exp(\theta Y_t)]\\
&= \mathbb E\left[\exp\left(\theta\sum_{i=1}^{N_t} X_i\right) \right]\\
&= \mathbb E\left[\mathbb E\left[\exp\left(\theta X_i\right)^{N_t}\mid N_t \right]\right]\\
&= \mathbb E\left[M_{X_1}(\theta)^{N_t} \right].
\end{align}
Recall that for a random variable $W$ with $\mathsf{Poisson}(\mu)$ distribution, the probability-generating function of $W$ is
\begin{align}
\mathbb E[s^W] &= \sum_{k=0}^\infty \mathbb P(W=k)\cdot s^k\\
&= \sum_{k=0}^\infty e^{-\mu}\frac{\mu^k}{k!}\cdot s^k\\
&= e^{-\mu}\sum_{k=0}^\infty \frac{(s\mu)^k}{k!}\\
&= e^{-\mu} e^{s\mu}\\
&= e^{s(\mu - 1)}.
\end{align}
It follows then that the probability-generating function of $N_t$ is $e^{s(t\lambda -1)}$, and so
$$
M_{Y_t}(\theta) = e^{t\lambda (M_{X_1}(\theta)-1)}.
$$
For example, if $X_1\sim\mathsf{Ber}(p)$ then
$$
M_{X_1}(\theta) = e^{\theta\cdot 0}(1-p) + e^{\theta\cdot 1}p = 1 - p + pe^{\theta},
$$
in which case
\begin{align}
M_{Y_t}(\theta) &= e^{t\lambda p\left(1-e^\theta\right)}.
\end{align}

Title: MGF and Compound Poisson Distribution


问:
Function field of $\mathbb{P}^1$
What is the function field of $\mathbb{P}^1$ when $\mathbb{P}^1$ is viewed as a projective curve? From what I read, I think it should be $k(x)$, but I don't understand how.

答: The only regular functions on $\mathbb P^1$ are constants, so you have to pass to nontrivial open subsets to get anything interesting. But once you have deleted a single point, you are now looking at $\mathbb A^1$ or a subset thereof, and the ring of regular functions on $\mathbb A^1 \setminus \left\{ a_1,a_2,...,a_n \right\}$ is $k\left[x,\frac{1}{x-a_1}, \frac{1}{x-a_2},...,\frac{1}{x-a_n}\right]$. Now as you vary over all possible finite subsets you get a directed system of rings starting with $k \subset k[x]$ (and the other possible ways to include $k$ into a one variable polynomial ring by deleting a point from $\mathbb P^1$). Taking the colimit over this system you obtain $k(x)$, essentially as a consequence of the universal property of localization.

Title: Function field of $\mathbb{P}^1$


问:
help with a sec order differential equation
I've been having trouble with a second order differential equation and would love to get some help,
$y''*(y*(y')^2-3)=(y')^4$ and $y(2)=1,y'(2)=-1$

I've tried defining $y'=p(y)$ and then $y''=p'*p$ and subtituting that into the equation but it get's me nowhere.

$p'(y*p^2-3)=p^3$

Thank you

答:
$$p'(yp^2-3)=p^3$$
You can rewrite the DE as:
$$yp^2-3=y'p^3$$
$$y'p-y=-\dfrac 3 {p^2}$$
Is a first order linear differential equation.
$$\left ( \dfrac yp \right )'=-\dfrac 3 {p^4}$$
Integrate.
$$ \dfrac yp =-3\int \dfrac  { dp } {p^4}$$


问:
How to simplify function to $\log (1-\theta)+\frac{\theta}{2-\theta}$?
\begin{align}
f(\theta)& = \log \frac{1}{1-\theta}-\frac{\theta}{2-\theta}+\frac{\theta^{2}}{(2-\theta)^{2}} \\
& = \log (1-\theta)+\frac{\theta}{2-\theta}
\end{align}
How are the two equations above equal to each other, i.e. reduce the first into the second?

How can one term be outside of the logarithm on the second line, whereas all 3 terms fall inside the logarithm on the first line? Actually, I'm not even sure if all the terms in the first line are supposed to all fall inside the logarithm, no brackets were given, so it could be either or

Source of first equation:

Formula 22 in Mercier 2005

Source of second equation:

Formula 4.4 in Kumar 2011

If they don't equal each other like how the second author claims, which author made a mistake?

Edit

I found yet another equality presented by the same second author. This time,

Formula 3.41 of Kumar 2014:

$$
f(\theta) = \log (1-\theta)+\frac{2 \theta(1-\theta)}{(2-\theta)^{2}}
$$

答:
The first line is not equal to the second line:

$$f(\theta) = \log \left( \frac{1}{1-\theta} \right)-\frac{\theta}{2-\theta}+\frac{\theta^{2}}{(2-\theta)^{2}}$$
$$= \log( (1 - \theta)^{-1}) - \frac{\theta (2 - \theta)}{(2 - \theta)^2} + \frac{\theta^{2}}{(2-\theta)^{2}}$$
$$= -\log(1 - \theta) + \frac{-(2\theta - \theta^2) + \theta^2}{(2 - \theta)^2}$$
$$= -\log(1 - \theta) + \frac{2 \theta^2 - 2 \theta}{(2 - \theta)^2}$$
$$= -\log(1 - \theta) + \frac{(2\theta)(\theta - 1)}{(2 - \theta)^2}$$


问:
Prove that in any binary tree with $n \geq 1$ nodes, there exists a node with at least $\frac{n}{3}$ and at most $\frac{(2n + 1)}{3}$ descendants
Prove that in any binary tree with $n \geq 1$ nodes, there exists a node with at least $\frac{n}{3}$  and at most $\frac{(2n + 1)}{3}$ descendants.

I'm a beginner in graph theory and I've no idea how to approach this problem, so any help would be appreciated.

答:
Suppose that no such node exists. Since the number of descendants of a node decreases as we go down in the graph, we must eventually hit a node that has $>\frac{2n+1}{3}$ descendants, but its children have $<\frac{n}{3}$ descendants each. That is, the children have at most $\left\lceil\frac{n}{3}\right\rceil-1$ descendants and the parent has at least $\left\lfloor\frac{2n+1}{3}\right\rfloor+1$ descendants.

Using the former number, the parent has at most $2\left(\left\lceil\frac{n}{3}\right\rceil-1\right)+1$ descendants. For this to make sense, we must have

$$2\left(\left\lceil\frac{n}{3}\right\rceil-1\right)+1\geq\left\lfloor\frac{2n+1}{3}\right\rfloor+1.$$

Try showing that this is impossible for any value of $n$.


问:
What causes the convergence of Iterated continued fractions from convergents?
Here is a small discovery I stumbled across a few weeks ago. I hope at least one person will find it interesting enough to help me.
The iterated continued fractions from convergents (or convergents constants) are explained at https://oeis.org/wiki/Convergents_constant . Among other things, there you find out that a randomly selected number between 2 and 3 has a convergents constant of 2.3484074702792306..., but the same is not true for 2.1, 2.2, 2.5 and perhaps for a few other values. Also for most 0 < x < 1 the same process of iterated continued fractions returns 0.5557531042780459..., but it is not so for x =0.1, 0.11, 0.12, 0.2, 0.25, 0.34, 0.35, 0.43, 0.45, 049, 0.5, 0.65, 0.75, some values < 66/1477 and perhaps for a few other values.

I would like to know why these numbers and those like them are exceptions to the rule.

答:
Possibly a bit more light occurs, if we consider to express that iteration using the matrix-representation of the continued-fractions-evaluation.

Here we insert the coefficients of the cf in a sequence of matrices, whose form is

$ \qquad M_0(a)= \begin{bmatrix}  0 & 1\\1 & a \end{bmatrix}$

and for the product of matrices according to the coefficients in the cf allow the generalized notation

$ \qquad M_0(a,b,c,...,h)= M(a)*M(b)*...*M(h)$

The iteration refers to the partial products, used for a new continued fraction:

$ \qquad M_1(a) = M_0(a); M_1(b) = M_0(a)*M_0(b) = \begin{bmatrix}  1 & b\\a & a*b+1 \end{bmatrix} ; \ldots $

This is simple to program, for instance in Pari/GP. Unfortunately this does not hold exactly: the $M_1()$ created by this procedure are not compatible with the form $ \qquad  \begin{bmatrix}  0 & 1\\1 & a \end{bmatrix}$

, they have general (integer) values in all four entries; so do not reflect the required "simple-continued-fraction" representation. So for to model the process as you described it in your question we need some normalization.

What I tried next was to insert the evaluated partial continued fractions as rational (or real) numbers instead of integer numbers in the positions of the a in $M_1(a)$ - so $M_1(a) = M_0(a); M_1(b) = M_0( {ab+1 \over b}), \ldots  $ according to the evaluation of the partial convergents. But still this requires then another type of re-normalization (it does not lead to the same limit value), because the fractional parts of the coefficients must now be "shifted" to the remaining cont-frac-expression.

But perhaps we find here the effect, why the leading coefficients of the iterated continued fractions converge to some constant: because after evaluations of subsequent coefficients a certain threshold for the fractional part of a coefficient can no more be overcome (in the next iteration).But I do not see really clear here...


答: Take $f\colon x\in\mathbb{R}\mapsto x^2\in\mathbb{R}_+$, and $A=\mathbb{R}_+$, $x=-1$. $f(x)=f(1)\in \overline{f(A)}=f(A)$, but $x\notin\overline{A}=A$.</p>

Title: Definition of continuous map

Body:
Let $X,Y$ be topological spaces, let $f:X\longrightarrow Y$ a function. There are several (equivalent) ways to define continuity of $f$, one of these says: $f$ is continuous if
$$(1)\qquad \forall A\subseteq X,\ \forall x\in X, \Big(x\in \overline{A}\Rightarrow f(x)\in\overline{f(A)}\Big)$$
where $\overline{A}$ denotes the closure of $A$ in $X$, and similarly for $f(A)$ in $Y$.

I was wondering if reversing this condition, i.e.
$$(2)\qquad \forall A\subseteq X,\ \forall x\in X, \Big(f(x)\in \overline{f(A)}\Rightarrow x\in\overline{A}\Big)$$
also defines a continuous map. So, i'm asking if $(1)$ and $(2)$ are equivalent, if not, i'm looking for a counterexample, i.e. a continuous map $f$ not satisfying $(2)$.
Any help appreciated!


问:
Set of vertices intersecting all faces of a planar graph
Does anyone know if the following problem exists in the literature: Given a planar graph, find a minimum set of vertices intersecting all of its faces.</p>

答:
Such a set is called a face hitting set of the planar graph. (Note that in general, this depends not only on the graph itself, but also on the embedding you choose.) This terminology does not come out of nowhere: a "hitting set" for any other kind of object is a set that intersects every instance of that object.

In particular, a cycle hitting set is also called a feedback vertex set, which intersects every cycle of a graph. If the graph is planar, then any feedback vertex set is of course a face hitting set, since all faces are cycles. I am given to understand that the paper Connected feedback vertex set in planar graphs by A. Grigoriev and R. Sitters proves a half-converse: the size of the smallest feedback vertex set is at most twice the size of the smallest face hitting set. (But I can't check, since I don't have access to the paper from home.)

For some reason, lots of people are interested in finding connected face hitting sets. See for instance Connecting face hitting sets in planar graphs by Pascal Schweitzer and Patrick Schweitzer. This paper is where I got my citation for the at-most-twice-the-size fact above. It also proves a different interesting result: in a connected planar graph with minimum degree $3$, any face hitting set of size $n$ is contained in a connected face hitting set of size $5n-6$. (A factor of $5$ is the price you pay for connectivity.)

Why minimum degree $3$, by the way? Because if degree-$2$ vertices are allowed, you could subdivide each edge a billion times, inflating the size of a connected face hitting set tremendously while not affecting the general problem at all. But that's maybe more detail than you wanted to know.


问:
Laurent series for $\frac{1}{(z-1)(z+1)}$ centered at $z = 1$
practice problem got us stuck -- please help us!!! Thank you..<3

Find a Laurent series for $\frac{1}{(z-1)(z+1)}$ centered at $z=1$ and specify the region in which it converges.

What we did was looking at the region $|z-1|>2$, where the function is holomorphic. For this region we can use the formulae for Laurent series and $c_k=\frac{1}{2\pi i}\int_\gamma\frac{(w-1)^{-k-2}}{w+1}dw$. We are stuck at for which values of $k$ the integral goes to 0 and for other values of $k$ how to evaluate the integral...

Thanks!

Body:
For $\;\,z\,\;$ "close enough" to $\,1\,$ :

$$\frac1{(z-1)(z+1)}=\frac1{z-1}\cdot\frac12\cdot\frac1{1+\frac{z-1}2}=$$

$$=\frac1{2(z-1)}\left(1-\frac{z-1}2+\frac{(z-1)^2}4-\frac{(z-1)^3}8+\ldots\right)=$$

$$=\frac1{2(z-1)}-\frac14+\frac{z-1}8-\ldots$$


问:
Equation of Circle touching a straight line and passing through the centroid of a triangle and a particular point
If the equation $3{x^2}y + 2x{y^2} - 18xy = 0$  represent the sides of a triangle whose centroid is G, then the equation of the circle  which touches the line $x-y+1=0$ and passing through G and $(1,1)$ is

A:  ${x^2} + {y^2} - x - y = 0$

B:  ${x^2} + {y^2} - 9x - y + 8 = 0$

C:  ${x^2} + {y^2} - 4x + 2y = 0$

D:  ${x^2} + {y^2} + 9x + y + 8 = 0$

I got the answer but had the choice not been provided how would I find the solution.

My steps are elaborated below

$3{x^2}y + 2x{y^2} - 18xy = 0 \Rightarrow xy\left( {3x + 2y - 18} \right) = 0$

The triangle is formed by line x=0, y=0 and 3x+2y-18=0
Centroid  $G = \left( {\frac{{6 + 0 + 0}}{3},\frac{{9 + 0 + 0}}{3}} \right) = \left( {2,3} \right)$

Let the centre of the circle is (h,k)

Circle passes through (2,3) and (1,1), hence the radius is

${r^2} = {\left( {h - 1} \right)^2} + {\left( {k - 1} \right)^2} = {\left( {h - 2} \right)^2} + {\left( {k - 3} \right)^2} \Rightarrow 2h + 4k = 11$

${x^2} + {y^2} + 2gx + 2fy + c + \lambda \left( {x - y + 1} \right) = 0$, represent the equation of the circle touching the line x-y+1=0 and passing through (2,3) and (1,1)

${x^2} + {y^2} + 2x\left( {g + \frac{\lambda }{2}} \right) + 2y\left( {f - \frac{\lambda }{2}} \right) + c + \lambda  = 0$

$Centre:\left( { - \left( {g - \frac{\lambda }{2}} \right), - \left( {f - \frac{\lambda }{2}} \right)} \right)$

Putting it in 2h+4k=1

$ - 2\left( {g - \frac{\lambda }{2}} \right) - 4\left( {f - \frac{\lambda }{2}} \right) = 11 \Rightarrow  - 2g - 4f + \lambda  + 2\lambda  = 11 \Rightarrow 2g + 4f = 3\lambda  - 11$

By Hit and Trial I am getting choice (B) as it satisfies $2h+4k=11$ where $h=\frac{9}{2}$ and $k=\frac{1}{2}$.

Had the option not been given how do I proceed

答:
Let $(h,k)$ be the centre then

$(h,k)$ must lie on the perpendicular bisector of line joining $(2,3),(1,1)$

$(h,k)$ must satisfy the locus given in this result

You have the ingredients to finsih..

Method -2

let circle be $S=x^2+y^2+2gx+2fy+c=0$

distance of point from$(-g,-f)$ to line $x-y+1=0$ must be $\sqrt{g^2+f^2-c}$

$(1,1),(2,3)$ must satisfy $S=0$

You have 3 equations 3 variables ...


答:
Hint:

$k=\dfrac{11-2h}4$

For any circle, the distance of a tangent from the center $(h,\dfrac{11-2h}4)=$radius

$$r^2=\dfrac{(h-k+1)^2}{1^2+1^2}=\dfrac{\left(h+1-\dfrac{11-2h}4\right)^2}2$$

Again $r^2=(h-1)^2+(k-1)^2=(h-1)^2+\left(\dfrac{11-2h}4-1\right)^2$

Equate the two values of $r^2$ to find $h$

Title: Equation of Circle touching a straight line and passing through the centroid of a triangle and a particular point

Body:
If the equation $3{x^2}y + 2x{y^2} - 18xy = 0$  represent the sides of a triangle whose centroid is G, then the equation of the circle  which touches the line $x-y+1=0$ and passing through G and $(1,1)$ is

A:  ${x^2} + {y^2} - x - y = 0$

B:  ${x^2} + {y^2} - 9x - y + 8 = 0$

C:  ${x^2} + {y^2} - 4x + 2y = 0$

D:  ${x^2} + {y^2} + 9x + y + 8 = 0$

I got the answer but had the choice not been provided how would I find the solution.

My steps are elaborated below

$3{x^2}y + 2x{y^2} - 18xy = 0 \Rightarrow xy\left( {3x + 2y - 18} \right) = 0$

The triangle is formed by line x=0, y=0 and 3x+2y-18=0
Centroid  $G = \left( {\frac{{6 + 0 + 0}}{3},\frac{{9 + 0 + 0}}{3}} \right) = \left( {2,3} \right)$

Let the centre of the circle is (h,k)

Circle passes through (2,3) and (1,1), hence the radius is

${r^2} = {\left( {h - 1} \right)^2} + {\left( {k - 1} \right)^2} = {\left( {h - 2} \right)^2} + {\left( {k - 3} \right)^2} \Rightarrow 2h + 4k = 11$

${x^2} + {y^2} + 2gx + 2fy + c + \lambda \left( {x - y + 1} \right) = 0$, represent the equation of the circle touching the line x-y+1=0 and passing through (2,3) and (1,1)

${x^2} + {y^2} + 2x\left( {g + \frac{\lambda }{2}} \right) + 2y\left( {f - \frac{\lambda }{2}} \right) + c + \lambda  = 0$

$Centre:\left( { - \left( {g - \frac{\lambda }{2}} \right), - \left( {f - \frac{\lambda }{2}} \right)} \right)$

Putting it in 2h+4k=1

$ - 2\left( {g - \frac{\lambda }{2}} \right) - 4\left( {f - \frac{\lambda }{2}} \right) = 11 \Rightarrow  - 2g - 4f + \lambda  + 2\lambda  = 11 \Rightarrow 2g + 4f = 3\lambda  - 11$

By Hit and Trial I am getting choice (B) as it satisfies $2h+4k=11$ where $h=\frac{9}{2}$ and $k=\frac{1}{2}$.

Had the option not been given how do I proceed