问:
Fields with no nontrivial discrete valuation
I am currently learning about DVRs and one can show that given a valuation ring $R$ of $(K,v)$ where $K$ is a field and $v$ a discrete valuation on $K$ that $K\cong F(R)$. I know that not all fields are fields of fractions of some nontrivial ring(trivial in the sense of $R=K$ or $R={0}$), e.g. finite fields. So there have to be fields with no nontrivial discrete valuation, e.g. finite fields. Now i was wondering wheter there is a different proof for showing that there is no nontrivial discrete valuation on a field besides showing that it is not the localization of a nontrivial ring and wheter this is simpler to show and if there are further examples of such fields besides the finite ones?

Thanks for answering in advance!

LG

答: A simple type of argument is that of divisibility: if every element in your field has an $n$th root of unity (for some $n>1$), then suppose $x\in K$ has minimal positive valuation. Since $x=y^n$, we have $v(x)=nv(y)$, which is contradictory unless $v(x)=v(y)=0$. So you cannot have a nontrivial discrete valuation on such a field. This covers the case of algebraically closed fields, and also of finite fields since you have the Frobenius.


问:
Landmarks of subjects of mathematics
In order to have a good view of the whole mathematical landscape one might want to know a deep theorem from the main subjects (I think my view is too narrow so I want to extend it).

For example in natural number theory it is good to know quadratic reciprocity and in linear algebra it's good to know the Cayley-Hamilton theorem (to give two examples).

So, what is one (per post) deep and representative theorem of each subject that one can spend a couple of months or so to learn about? (In Combinatorics, Graph theory, Real Analysis, Logic, Differential Geometry, etc.)

答:
In Finite Group Theory, while the Odd Order Theorem and the Classification are major results, I would put a major landmark at the Sylow Theorems and Hall's Theorem (a generalization of the Sylow Theorems). Especially the former come up all the time, and there are many interesting corollaries that often are not discussed.

Another good possibility is the O'Nan-Scott Theorem for the study of permutation groups.

(Also, I think it would take a lot longer than "a couple of months or so" to really learn and understand the Classification of Finite Simple Groups...)

ContentLicense: CC BY-SA 3.0

Title: Landmarks of subjects of mathematics


问:
Proving $ \cos ^{2} \frac{\theta}{2} \equiv \frac{1+\cos \theta}{2} $
I need help proving LHS is identical to RHS.
$$
\cos ^{2} \frac{\theta}{2} \equiv \frac{1+\cos \theta}{2}
$$

答:
It's equivalent to proving $\cos ^2 \theta = \frac{1+\cos 2\theta}{2}$, which is equivalent to $$\cos2\theta = 2\cos ^2 \theta -1$$

And we know : $$\cos 2\theta = \cos (\theta + \theta) = \cos \theta \cos \theta - \sin \theta \sin \theta = \cos ^2 \theta - \sin ^2 \theta$$
$$= \cos^2\theta - (1-\cos ^2\theta) = 2\cos ^2 \theta -1$$

This gives the answer.

ContentLicense: CC BY-SA 4.0


问:
line integral of a gradient vector field
I'm having trouble finding the line integral of this problem.
So they give me the gradient vector field $$\nabla f(x,y,z) = 2xyze^{x^2}i + ze^{x^2}j+ye^{x^2}k$$
they say that $f(0,0,0) = 5$ and i want to know what is the value of line integral $f(1,1,2)$

i know that the line integral of a gradient vector field is given by $$\int_{a}^{b} \nabla f(x,y,z)ds = f(c(b)) -f(c(a))$$
But i'm stuck by how to find $f$, can someone explain to me

答:
I think that the gradient should be
$$\nabla f(x,y,z)=2xyze^{x^2}i + \mathbf{z}e^{x^2}j+ye^{x^2}k$$
where the variable $z$ has been inadvertently replaced by the number $2$ (otherwise it cannot be a gradient of some function $f$).
Then by integration we obtain $f(x,y,z)=e^{x^2}yz+\text{constant}$. Can you take it from here?


答:
$$\frac{\partial f}{\partial x}=2xyze^{x^2}\implies f(x,y,z)=\int\frac{\partial f}{\partial x}\,dx= yze^{x^2}+g(y,z)$$
and
$$\frac{\partial f}{\partial y}=ze^{x^2} \implies f(x,y,z)=\int\frac{\partial f}{\partial y}\,dy=yze^{x^2}+h(x,z)$$
and
$$\frac{\partial f}{\partial z}=ye^{x^2} \implies f(x,y,z)=\int\frac{\partial f}{\partial z}\,dz=yze^{x^2}+m(x,y)$$
So you may take $f(x,y,z)=yze^{x^2}+5$ (choose $g,h,m=5$ to ensure $f(0,0,0)=5$) and then use the formula you know for the line integral.

Body:
I'm having trouble finding the line integral of this problem.
So they give me the gradient vector field $$\nabla f(x,y,z) = 2xyze^{x^2}i + ze^{x^2}j+ye^{x^2}k$$
they say that $f(0,0,0) = 5$ and i want to know what is the value of line integral $f(1,1,2)$

i know that the line integral of a gradient vector field is given by $$\int_{a}^{b} \nabla f(x,y,z)ds = f(c(b)) -f(c(a))$$
But i'm stuck by how to find $f$, can someone explain to me


问:
How to integrate $\frac{\cos(x)}{e^x}$?
Surprisingly I haven't found such integral in the web, so I may ask this here bluntly.

A solution that Matlab suggest sounds like $\frac{{\mathrm{e}}^{-x}\,\left(\cos\left(x\right)-\sin\left(x\right)\right)}{2}$

答:
In general, $\int e^{-x}[f’(x)-f(x)]dx= e^{-x}f(x)$.
Then
$$\int e^{-x}(\cos x-\sin x)dx= e^{-x}\sin x,\>\>\>\>\>
 \int e^{-x}(\cos x+\sin x)dx= -e^{-x}\cos x
$$
Add them to obtain
$$\int e^{-x}\cos x dx = \frac12e^{-x}(\sin x -\cos x) + C$$

Title: How to integrate $\frac{\cos(x)}{e^x}$?


问:
Motivation for the one-point compactification
I just read about the one-point, or Alexandroff, compactification of a topological space.  It seems to be nice in its simplicity and easy description, but does suffer from some unfortunate properties like not being Hausdorff unless the original space is Hausdorff and locally compact.

My questions are: why study the one-point compactification?  What is it useful for?  Are there situations where we can learn more about the original space by instead passing to its one-point compactification and studying that?

I see the interest in the one-point compactification, and think it's a nice, intuitive idea.  Normally I don't need a lot of motivation for various ideas and constructions and can appreciate them for what they are.  In this case, however, I happened to find myself questioning what its real value in the 'big picture' is.

答:
Consider a topological space . Then  may or may not be compact depending on the topology . If  is not compact, then we might want to construct a compact topological space from  while preserving as much of the topology  on  as possible. This process can be achieved by compactification.

Please refer to the below link for visual example

http://mathonline.wikidot.com/one-point-compactification-of-a-topological-space


答:
This is just the one of the linearisation formulæ deduced from the duplication formulæ :
$$\cos 2\theta=2\cos^2\theta-1=1-2\sin^2\theta,$$
from which you get instantly:
$$\cos^2\theta=\frac{1+\cos2\theta}2,\qquad \sin^2\theta=\frac{1-\cos2\theta}2.$$

Title: Proving $ \cos ^{2} \frac{\theta}{2} \equiv \frac{1+\cos \theta}{2} $

Body:
I need help proving LHS is identical to RHS.
$$
\cos ^{2} \frac{\theta}{2} \equiv \frac{1+\cos \theta}{2}
$$


问:
How to calculate conditional expectation with inequality condition
How to transition conditional expectation where the conditioned item is not an equation into a simpler form for calculation?

As a simple example, how to show that (from Introduction to Probability Models by S.M. Ross) $$E[R_1|R_1<R_2] = E[\min(R_1,R_2)]$$

where $R_1, R_2$ are continuous RVs following exponential distributions with rate $\lambda_1, \lambda_2$ respectively.

I did find:
How to calculate conditional expectation $E[X|X \geq 0]$?,
How to calculate conditional probability with inequality, and Conditional Expectation Multivariate Normal Distribution with inequality condition three questions, but it looks like they are either dealing with probabilities or not providing a concrete answer. I know there might not exist an explicit rule, but are there any general rules, or more examples or references?

Update: Please refer to the following image (Example 5.8 from the book). As is stated in @John Dawkins' answer, it holds when the two RVs follow a continuous distribution with finite mean.

Ross book Ex 5.8

Body:
Your claim is not true.

Consider independent $R_1, R_2 \sim Bern\frac{1}2$. We have $P(R_i = 0) = P(R_i =1) = \frac12$. Thus $ R_1 < R_2$ iff $R_1 = 0$ and $R_2 = 1$.

Thus $$E(R_1 | R_1 < R_2) = E(R_1 | ( \{R_1 = 0\} \cap \{ 1 = R_2\})) = 0$$
and $E\min(R_1, R_2) = P(R_1 = 1, R_2 = 1) > 0$.


问:
Exercise of interior of a closed ball
Let $E$ a normed spaces and $a\in E$. How to prove that $$[\overline{B}(a, r)]^{\circ}\subseteq B(a, r)$$ where $\overline{B}(a, r)=\{x\in E: \|x-a\|\leq r\}$ and  $B(a, r)=\{x\in E: \|x-a\|< r\}$ </p>

答: Without loss of generality, we can assume $a=0$. Let $x \in \text{int} \, \overline{B}(0,r)$ and $\delta>0$ such that $B(x,\delta) \subseteq \overline{B}(0,r)$. For $y :=t \cdot x$, $t>1$ we have $$y \in B(x,\delta) \Leftrightarrow \|\underbrace{y}_{t \cdot x}-x\|<\delta \Leftrightarrow \|x\| \cdot (t-1)< \delta$$ i.e. we can choose $t>1$ such that $t \cdot x \in B(x,\delta) \subseteq \overline{B}(0,r)$. This implies $\|t \cdot x\| \leq r$ and so (since $t>1$) we conclude $$\|x\| \leq \frac{r}{t} < r.$$</p>

Title: Exercise of interior of a closed ball


答:
I don't think the inequality is true in general.

The LHS is the same as $\mathbb E[\min(R_1,R_2)\mid R_1<R_2]$, which in general is not the same as $\mathbb E[\min(R_1,R_2)]$.

Consider, for instance, $R_1$ is uniform on $[0,1]$ and $R_2$ is almost surely $\frac{1}{2}$. The LHS is $\frac{1}{4}$ but the RHS is $\frac{1}{2}\left(\frac{1}{4}+\frac{1}{2}\right)=\frac{3}{8}$.

The equation you write will be true if, e.g. $\min(R_1,R_2)$ is independent of the event $R_1<R_2$. This is the case when, for example, $R_1$ and $R_2$ are independent exponential RVs.

Title: How to calculate conditional expectation with inequality condition

Body:
How to transition conditional expectation where the conditioned item is not an equation into a simpler form for calculation?

As a simple example, how to show that (from Introduction to Probability Models by S.M. Ross) $$E[R_1|R_1<R_2] = E[\min(R_1,R_2)]$$

where $R_1, R_2$ are continuous RVs following exponential distributions with rate $\lambda_1, \lambda_2$ respectively.

I did find:
How to calculate conditional expectation $E[X|X \geq 0]$?,
How to calculate conditional probability with inequality, and Conditional Expectation Multivariate Normal Distribution with inequality condition three questions, but it looks like they are either dealing with probabilities or not providing a concrete answer. I know there might not exist an explicit rule, but are there any general rules, or more examples or references?

Update: Please refer to the following image (Example 5.8 from the book). As is stated in @John Dawkins' answer, it holds when the two RVs follow a continuous distribution with finite mean.

Ross book Ex 5.8


Body: $$\frac{(e^{i\phi/2}+e^{-i\phi/2})^2}{2^2}=\frac{e^{2i\phi/2}+2+e^{-2i\phi/2}}4=\frac12\left(\frac{e^{i\phi}+e^{-i\phi}}2+1\right).$$


答:
Take the derivative:
$$
\begin{align}\left(\cos ^{2} \frac{\theta}{2} - \frac{1+\cos \theta}{2}\right)'&=2\cos(\theta/2)\bigl(-\sin(\theta/2)\bigr)\cdot\frac12-\frac12\sin(\theta)\\
&=\cos(\theta/2)\sin(\theta/2)-\frac12\cdot 2\cos(\theta/2)\sin(\theta/2)\\
&=0.
\end{align}
$$
Now substitute $\theta=0$.


答:
I think there's been a misunderstanding, the vector field you gave isn't conservative. I'm pretty sure the $y$ term should be $ze^{x^2}$, not $2e^{x^2}.$ Was the problem handwritten? In any case, I'm going to assume that was what you meant because otherwise the problem isn't solvable. (Use Clairaut's Theorem to figure out why)

The method I was taught to do this was to take kind of "partial integrals", like partial derivatives but in reverse. So essentially it's just like a normal integral, except you treat any other variables as constants and your constant is actually a function of the variables you hold constant.

So first, we have that $f_x = 2xyze^{x^2}$, so $f(x,y,z) = \int 2xyze^{x^2} dx = yze^{x^2} + g(y,z)$.

 $g$ being a function of $y$ and $z$ is necessary because any function of $y$ and $z$ will disappear when we differentiate with respect to $x$, not just a constant.

Now that we have an expression for $f$, we can differentiate it with respect to $y$ and compare it to what we have in the gradient: $f_y = ze^{x^2} + g_y = ze^{x^2} \to g_y = 0$. This is where the problem would become impossible with the original question: we would get $g_y = (2-z)e^{x^2}$, which is impossible because $g$ is not dependent on $x$.

Now we can do our partial integration again: $g(y,z) = \int 0 dy = h(z)$. Plugging this back into our expression for $f$ yields $f(x,y,z) = yze^{x^2} + h(z)$. This step is a bit overkill here because we already see that the $y$-derivatives match, but it's included for completeness because that's not always the case.

I've excluded the last bit for brevity's sake but repeating this one more time will give you that $h$ is just a constant, so the solution is $f(x,y,z) = yze^{x^2} + C$.

Hope this helps!

Edit: I see that you've addressed the y component issue now, so sorry if it seemed like I was laboring the point. It took a bit to write this.

Title: line integral of a gradient vector field


问:
Given $x''+x'+x+2x^3=0$. Prove that any solution can be extended to the entire line.

Given $x''+x'+x+2x^3=0$. Prove that any solution can be extended to the entire line.

I think maybe multiply $x'$ both side of the equation is useful. But I can not handle the $x'^2$

答:
Yes, that idea was correct. You find an energy functional
$$
E(x,x')=\frac12(x'^2+x^2+x^4)
$$
with
$$
\frac{d}{dt}E(x,x')=-x'^2
$$
so that in forward time the solution will cross the energy levels downwards towards the minimum at $x=x'=0$.

In the other time direction one finds increasing energy levels, however the increase is linearly bounded
$$
|e'|\le x'^2\le 2e,~~~~e(t)=E(x(t),x'(t))
$$
so that you still get à la Grönwall $e(t)\le e(0)e^{2|t|}$. This bound shows that there is no divergence to infinity in finite time, so that the solution also exists everywhere on the negative time half-axis.


问:
Effective methods for this qustion.
In the beginning, the numbers $1, 2, 3, ..., 57$ were written on the board. In each process, the numbers a and b are selected and deleted from the board so that the number $a$ divides the number $b$ completely. The number $\frac{b}{a}$ is written instead of the deleted numbers (For example, in the first operation, the numbers $a = 3$ and $b = 9$ can be deleted and the number $\frac{9}{3} = 3$ can be written on the board again). At least how many numbers can be left on the board after a few actions?

It is clear that  the remaining numbers are primes $29,31,37,41,43,47,53$.
In each of my attempts, I find $2$ different numbers in addition to these prime.  But I am waiting for your answers to see more different points of view.

答:
(Fill in the gaps as needed. If you're stuck, show your work and explain what you've tried.)

Hint: Consider the prime factorization of $ 57! $.

Focus on those whose power is odd.

As you've realized, the primes $> 28$ must always occur. These have power of 1 in the prime factorization.

Claim: The product of the numbers that are left can be written in the form $ \frac{ \prod_{a \in A } a } { \prod_{b \not \in A } b } $, where $ A \subset [57]$.

Claim: If $p ^k \mid \mid 57$ and $k$ is odd, then we're always left with a term that is a multiple of $p$.

Corollary: The product of the numbers at the end is

 $$ 2 \times 3 \times 5 \times 7 \times 11 \times 17 \times 19 \times 29 \times 31 \times 37 \times 41 \times 43 \times 47 \times 53.$$

Corollary: We must have at least $X$ numbers left behind. Recall that each of these numbers is less than 57.

 $X = 11$

Corollary: The minimum left behind is $X$, and we need to find a contraction that yields $X$.

(Disclaimer: I've not done this yet.)

 One possible set of numbers to be left is $ 7, 38, 55, 57, 29 , 31 ,37 , 41 , 43 , 47 , 53$.
 We do not use any of these in the steps, other than possibly 7.

 However, thinking through it further, because "b/a is an integer" is a very strong condition, the invariant might be too weak. So, this part is worth thinking about.

Title: Effective methods for this qustion.


问:
A Lambert W-function with no real solution?
Recently I have created this question:

$$(x+1)^xe^{x+1}=(2x)^x$$

And the solution

$$(x+1)^xe^{x+1}=(2x)^x$$$$(x+1)e^{1+\frac 1 x}=2x$$$$(1+\frac 1 x)e^{1+\frac 1 x}=2$$$$1+\frac 1 x=W(2)$$
$$x=\frac 1 {W(2)-1}$$

But Desmos shows no solution, can anyone explain what is wrong here?

答:
Wolfy says the solution to
$ze^z=2$ is about
$z≈0.85260550201372549135$
so
$x≈-6.7845137617899467654$.

The problem is defining the specified powers.


问:
Does $\widehat{M}\ (\vec{a}\times\vec{b})=\widehat{M}\ \vec{a}\times\widehat{M}\ \vec{b}$?
Is it true to write that $\widehat{M}\ (\vec{a}\times\vec{b})=\widehat{M}\ \vec{a}\times\widehat{M}\ \vec{b}$? If so, how can this be proven?

答:
This is not true: consider
$M = \begin{bmatrix}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{bmatrix}$, $\vec{a} = \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}$, $\vec{b} = \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}$.


问:
Path of X, Path space of X, and Product sigma algebra $\sigma^I$
I am reading the stochastic differential equations and came across some definitions.

Path of X :    for $\omega \in \Omega$  $X(.,\omega) : I \to S $

What I understood is that it is that, we fix one $\omega$ and calculate the value for each $X_t$ so the elements will be of the form $(1,X_1(\omega)) $ , $(2,X_2(\omega))$ ...... so if we consider I as time axis then it will be a plot between time and $X(\omega)$. And as we  change $\omega $
we get different plots.
Is this correct?

Path space : $ F \subset S^I$ such that for all $\omega \in \Omega $,  $X(.,\omega) \in F$

Here my previous understanding is a bit contradictory since the elements in $S^I$ will be of the form $(S_1, S_2,....)$ an index(I) dimensional space. Where am I wrong in these definitions? If possible can you explain with some figures?

And finally does $\sigma^{I} $ same as the product sigma algebra generated by $\sigma_1 \times \sigma_2 \times ....$

答:
Path of X

For each $\omega$ the path of $X$ is a function from $I$ to $S$. In particular it is the function $f(t)=X(t,\omega)$ (with $\omega$ fixed). You seem however to be confusing the path (which is a function) with the graph of the path (whose elements are on the form $(t,f(t)))$. Functions and graphs are closely related, but not the same thing.

Path space

Usually the notation $S^I$ refers to the set of all functions $f$ from the set $I$ to the set $S$, and in particular for each $\omega \in \Omega$ we have that the path $t\mapsto X(t,\omega)$ is an element of the path space $S^I$. Sometimes we want to restrict the path space to only allow certain types of functions, for instance we may restrict the path space to only allow continuous functions, and thus instead of working with the full path space $S^I$ we could work with a smaller path space, such as $F=C^0(I)\subseteq S^I$.

Countable vs. Uncountable case

It seems to me, that you have mostly concerned yourself with the countable case, since expressions such as $(S_1,S_2, \dots)$ or $\sigma_1 \times \sigma_2 \times \dots$ have no meaning if $I$ is uncountable. However when $I$ is countable, then there is a natural identification between the set $S^I$ (as i defined it above) and the set of all sequences indexed by $I$. But as stochastic calculus and stochastic differential equations are concerned, it is mostly the uncountable case that matters, and thus i would encourage that you get used to thinking of paths as functions rather than sequences.

$\sigma$-algebra

Recall that the product $\sigma$-algebra on $\mathbb{R}^2$ is the smallest $\sigma$-algebra, which makes the coordinate projections $(x,y)\mapsto x$ and $(x,y)\mapsto y$ measurable. The concept is similar for $\sigma^I$, which can be defined as the smallest $\sigma$-algebra, which makes the coordinate projections
\begin{align*}
\pi_t: S^I &\rightarrow S \\
     f &\mapsto f(t)
\end{align*}
measurable for all $t \in I$. When $I$ is countable, this corresponds to the sigma algebra generated by $\sigma \times \sigma \times \dots$, but as mentioned before, this does not even make sense when $I$ is uncountable.

Why do we care about path spaces?

We have just defined the path-space $S^I$ and the corresponding $\sigma$-algebra $\sigma^I$. This means, that $(S^I,\sigma^I)$ is a measurable space and it is thus possible to speak of $S^I$-valued random variables, i.e. measurable functions $\mathbf{Y}:\Omega \rightarrow S^I$. Try to verify for yourself, that $\mathbf{Y}:\Omega \rightarrow S^I$ is $\sigma^I$ measurable if and only if $\pi_t(\mathbf{Y})$ is measurable for all $t\in I$ in particular we have the following:

$$(X_t)_{t\in I} \text{ is an $S$-valued stochastic proces } \Leftrightarrow
\mathbf{X} \text{ is } \sigma^I \text{-measurable},$$
where $\mathbf{X}:\Omega \rightarrow S^I$ is defined as
$$\mathbf{X}(\omega) = X(\cdot , \omega) \quad \text{(This is the path of $X$ as defined above)}.$$
Because of this last characterisation, some authors actually prefer to think of stochastic processes as $S^I$-valued random variables, rather than collections of $S$-valued random variables.


问:
Proving that Euclidean space having the infinity metric is a complete metric space (stuck)
I am trying to prove that the space ${\mathbb{R}}^k$ with the $\infty$-metric is a complete metric space.

I know that I need to show that every Cauchy sequence in the metric space ${\mathbb{R}}^k$ with the $\infty$-metric converges to a point $x\in{\mathbb{R}}^k$. Part of my work so far involved proving that the space ${\mathbb{R}}^k$ with the old Pythagorean norm is a complete metric space, but I’m not sure if I should be using that at all in this proof. Here goes:

Proof:

Suppose that $k\in{\mathbb{N}}$ and that $\left(x_{n}\right)$ is a Cauchy sequence in the space ${\mathbb{R}}^k$ with the $\infty$-metric.
Suppose that $\epsilon>0$.
Using the fact that $\left(x_{n}\right)$ is a Cauchy sequence, choose an integer $N$ such that the inequality $||x_{m}-x_{n}||_\infty<\frac{\epsilon}{2}$ holds for all integers $m$ and $n$ such that $m\ge{N}$ and $n\ge{N}$.

This is where I’m stuck. I feel like I want to use the triangle inequality from here…

Since the space ${\mathbb{R}}^k$ is a metric space, we know that the inequality $$||{x_n}-{x_m}||_{\infty}\le||{x_n}-{x}||_{\infty}+||{x_m}-{x}||_{\infty}\lt\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$ holds for all positive integers $n$ and $m$.

However, I know that this isn’t a good thing to state. I think that I’m assuming that there is some point $x$ in the space that the sequence converges to… this is exactly what I'm trying to prove!
I’m totally stuck. Any advice would be GREATLY appreciated!

EDIT: I'm including the updated version of my proof based on advice from Brian M. Scott. Here it is:

Proof:

Suppose that $k\in{\mathbb{N}}$ and that $\left(x_{n}\right)$ is a Cauchy sequence in the space ${\mathbb{R}}^k$ with the $\infty$-metric.

Suppose that $\epsilon>0$.

For each $n\in\mathbb{Z}^{+}$ we have $x_{n}= \left(x_{n,1},x_{n,2},\ldots,x_{n,k}\right)$.

Consider the real-valued sequences $\left(x_{n,j}\right)$ for each $j=1,2,\ldots,k$.

Using the fact that $\left(x_{n}\right)$ is a Cauchy sequence, choose $N\in\mathbb{Z}$ such that the inequality $||x_{m} – x_{n}||_{\infty}\lt \frac{\epsilon}{2}$ for all $n\ge N$ and $m\ge N$.

Since the inequality $||x_{m,j} – x_{n,j}||_{\infty}\le||x_{m} – x_{n}||_{\infty}$ holds for all $m$ and $n$, we have the inequality $||x_{m,j} – x_{n,j}||_{\infty}\lt\frac{\epsilon}{2}$ for all $m\ge N$ and $n\ge N$.

Thus for $j=1,2,\ldots,k$ and all $n\in\mathbb{Z}^{+}$ the sequence $\left( x_{n,j} \right)$ is a Cauchy sequence in $\mathbb{R}$.

Using the fact that $\mathbb{R}$ is complete, define $$x\left(j\right)=\lim_{n\to\infty}x_{n,j}$$ for $j=1,2,\ldots,k$.

The point $x=\left(x_{1},x_{2},\cdots,x_{k}\right)$ is the limit of $\left(x_{n}\right)$ under the Pythagorean metric.

From the inequality $$||x-x_{n}||_\infty\le||x-x_{m}||_{\infty}+||x_{m}-x_{n}||_{\infty}\lt\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$ that holds for all $n\ge N$ and $m\ge N$ we see that the sequence $\left(x_{n}\right)$ converges to the point $x\in\mathbb{R}^{k}$ and the space $\mathbb{R}^{k}$ is a complete metric space.

答:
HINT: There are at least two reasonable approaches.

Start with a Cauchy sequence $\langle x^n:n\in\Bbb N\rangle$, where $x^n=\langle x_1^n,\dots,x_k^n\rangle$ for each $n\in\Bbb N$. Now look at the real-valued sequences $\langle x_i^n:n\in\Bbb N\rangle$ for $i=1,\dots,k$. Show that each is Cauchy in the usual metric on $\Bbb R$, let $x_i$ be the limit (since $\Bbb R$ is complete in the usual metric), and show that $\langle x^n:n\in\Bbb N\rangle$ converges to $\langle x_1,\dots,x_k\rangle$ in $\Bbb R^k$.

Show that the $\infty$ metric is equivalent to the Euclidean metric. Then a sequence is Cauchy in one of these metrics if and only if it’s Cauchy in the other, it converges to a point $x$ in one if and only if it converges to $x$ in the other, and you already know that the Euclidean metric is complete.


问:
Let $B$ be a matrix such that..
$B=\begin{bmatrix}1&2x&-13\\1&2x+2&2x-13\\1&2x+6&7x+6\end{bmatrix}$ , $x$ is a parameter.

there exists a $y,z\in\mathbb{R}$ such that for this system theres a solution:

$B^{43}\begin{bmatrix}-285\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix} $

$x=?$

I can't somehow get around how to solve this question, I know that it has something to do with inverse matrices and I know that if the matrix is invertible then $r(B)=3$, which means theres a unique answer for the equation $Bx=0$.

Any help or hints are really appreciated.

Thanks in advance.

答:
Since $y$ and $z$ can be anything, the condition is equivalent to the fact that $B^{43}$ has nontrivial kernel, which means it is noninvertible. Now $B^{43}$ is noninvertible if and only if $B$ is noninvertible so you need precisely those $x \in \Bbb{R}$ such that $\det B = 0$.

We have $$0 = \det B = 2x+38 \implies x = -19.$$

Title: Let $B$ be a matrix such that..


问:
Log inequalities and asymptotic expansions
I'm dealing with the probabilistic method and how it applies to distinct sums. In the notes I'm using I encounter the following inequality:

$2^k\le nk$, which the notes simplify into this form:

$k\le \log_2n +\log_2\log_2n + O(1)$, and I have no idea how this is arrived at.

Clearly taking a $\log_2$ on both sides gives $\log_2 2^k\le \log_2 nk \implies k\le \log_2 n +\log_2 k$

But how do we get from $\log_2k$ to $\log_2\log_2n + O(1)$?

I'm sure I'm missing something obvious but any help would be appreciated.

答:
Take the $\log_2$ of the initial inequality:
$$
k \le \log _2 n + \log _2 k.
$$
Taking the $\log_2$ again gives
$$
\log _2 k \le \log _2 (\log _2 n + \log _2 k) = \log _2 \log _2 n + \log _2 \left( {1 + \frac{{\log _2 k}}{{\log _2 n}}} \right).
$$
Substituting this back into the first inequality gives
$$
k \le \log _2 n + \log _2 \log _2 n + \log _2 \left( {1 + \frac{{\log _2 k}}{{\log _2 n}}} \right).
$$
Thus, it remains to show that the last term is $\mathcal{O}(1)$. But
$$
k \le \log _2 n + \log _2 k \le \log _2 n + \frac{k}{2} \Rightarrow k \le 2\log _2 n \Rightarrow \log _2 k \le 2\log _2 n
$$
whenever $k\geq 4$, and for $k=1,2,3$ the statement is obvious.


问:
Construct the smallest graph homeomorphic to a given graph by smoothing
The homeomorphism class $ \mathcal{H}(G) $ of a graph $G$ is the set of isomorphism classes of graphs that are topologically homeomorphic to $G$. It is natural to ask: Is there a "smallest" representative in the homeomorphism class? If yes, how to find it? Unfortunately, I found no useful result on this problem after a quick google search.

Nevertheless, guided by intuition, I have the following hypothesis:

The smallest graph homeomorphic to a graph is obtained by smoothing
every maximal ear.

In this post I attempt to sketch a proof, but there is a gap in the proof, so I don't know whether my hypothesis is actually correct. I would appreciate anyone for pointing out my mistakes and filling out the gap.

Warning: this would be a long post

First, let's clear up some terminology. The term "ear" means different things in different graph theory textbooks. In this post, we adopt the following definition:

Definition 1

An ear in a graph is either:

a cycle whose all except possibly one vertices are of degree $2$, or

a path whose all internal vertices are of degree $2$.

A maximal ear is an ear that is not a proper subgraph of a larger ear.
Equivalently, a maximal ear is one of the following:

a cycle that is a whole connected component on its own

a cycle in which exactly one vertex is of degree $ \geq 3 $, while all other vertices are of degree $2$

a path in which all internal vertices are of degree $2$, while both end vertices are of degree $ \neq 2 $

Two common operations that preserve topology on graphs are subdividing and smoothing:

Definition 2

Subdividing an edge means replacing it by an ear. More precisely, let
$e = uv$ be an edge.

If $u = v$, then subdividing the self-loop $e$ means replacing it by a
cycle $C$, and $u = v$ becomes a vertex on $C$, which may or may not
have degree $2$, depending on whether $e$ is isolated.

On the other hand, if $u \neq v$, then subdividing $e$ means replacing
it by a path $P$, and $u, v$ become the end vertices of $P$.

Subdividing a graph means preforming a sequence of subdividing on
edges.

Definition 3

Smoothing an ear means replacing it by a single edge. More precisely,
let $C$ be an ear.

If $C$ is a cycle, then smoothing $C$ means replacing it by a
self-loop $e$, and the vertex of degree $ \neq 2 $ on $C$ becomes the
only vertex incident on $e$ (if all vertices on $C$ are of degree $2$,
just choose any vertex).

On the other hand, If $C$ is actually a path $P$, then smoothing $P$
means replacing it by a loopless edge $e$, and the end vertices of $P$
become the end vertices of $e$.

Smoothing a graph means preforming a sequence of smoothing on ears.

Next, we have the following classic result on topology of graphs:

Theorem 1

Two graphs are homeomorphic if and only if one of them can be obtained
from a sequence of subdividing and smoothing operations on the another.

Proof: See this post.

Theorem 2

Let $G$ and $H$ be two homeomorphic graphs. Then $ |V(G)| = |V(H)| $
if and only if $ |E(G)| = |E(H)| $.

Sketch of proof: Subdividing (resp. smoothing) always increases (resp. decreases) the number of vertices and edges by the same number. $\square$

In light of Theorem 2, we can define an ordering on the homeomorphism class of a graph:

Definition 4

Let $ \mathcal{H}(G) $ be the homeomorphism class of a graph $G$.
Define the ordering $\preceq$ on $ \mathcal{H}(G) $ by:
$$ G_1 \preceq G_2 \iff |V(G_1)| \leq |V(G_2)| $$
for any $ G_1, G_2 \in \mathcal{H}(G) $.

If $ G_1 \preceq G_2 $ and $ G_2 \preceq G_1 $, then we denote
$ G_1 \sim G_2 $.

The ordering $\preceq$ is a total preorder, meaning it is transitive and any two homeomorphic graphs are comparable. Unfortunately it is not a total order, since $ G_1 \sim G_2 $ does not imply $ G_1, G_2 $ are isomorphic, even through Theorem 2 implies $ |E(G_1)| = |E(G_2)| $.

Theorem 3

Any graph without isolated vertex can be uniquely decomposed into an
edge-disjoint union of maximal ears.

Sketch of proof:

Let $G$ be a graph without isolated vertex. Define the relation $R$ on $E(G)$ by:
$$ eRf \iff \exists \text{ ear } C \subseteq{G} \text{ s.t. } e, f \in E(C) $$
for any $ e, f \in E(G) $.

Then $R$ is an equivalence relation on $E(G)$, in which each equivalence class contains the edges of exactly one maximal ear. Thus, $R$ induces a decomposition of $G$ into an edge-disjoint union of maximal ears. Conversely, any such decomposition must be induced by $R$, so the decomposition is unique. $\square$

Based on the above decomposition, we can define the following:

Definition 5

A graph without isolated vertex is called smooth if every maximal ear
is of length $1$. For a graph $G$ without isolated vertex, the smooth
graph obtained from smoothing every maximal ear in $G$ is denoted as
$ \text{Smooth} (G) $.

The term "smooth graph" is not standard, but I could not find any existing term for such a graph, so I just make it up on my own.

Theorem 4

Let $G$ be a smooth graph without isolated vertex and
$ H \in \mathcal{H}(G) $, then $ G \preceq H $. Moreover,
$ G \sim H $ if and only if $H$ is a smooth graph.

Sketch of proof:

By Theorem 1, $H$ can obtained from a sequence of subdividing and smoothing operations on $G$. Each step of the operations can only change one of the maximal ear to another maximal ear of possibly different length.

On the other hand, in a smooth graph all the maximal ears already have the shortest possible length (namely, $1$), so any sequence of subdividing and smoothing can never further decrease the number of vertices. Thus, $ |V(G)| \leq |V(H)| $ and the equality holds if and only if $H$ is smooth. $\square$

The following claim is based on intuition, but I don't know how to prove it. It is where the gap of my whole proof lies.

Claim 0

Let $G$ and $H$ be two smooth graphs without isolated vertex. If
they are homeomorphic, then they are isomorphic.

Finally, assuming the above claim, we can prove the main hypothesis:

Main Hypothesis

Assume Claim 0 is correct and let $G$ be a graph without isolated
vertex. Then $ \text{Smooth} (G) $ is the unique smallest graph in
$ \mathcal{H} (G) $ with respect to the ordering $ \preceq $.

Proof:

The fact that $ \text{Smooth} (G) \preceq H $ for all $ H \in \mathcal{H} (G) $ follows from Theorem 4.

To prove uniqueness, let $ H \in \mathcal{H} (G) $ be such that $ \text{Smooth} (G) \sim H $. Since $ \text{Smooth} (G) $ is smooth and $ H \in \mathcal{H} (\text{Smooth} (G)) $, by Theorem 4, $H$ is smooth as well. Claim 0 then implies $H$ is isomorphic to $ \text{Smooth} (G) $. $\square$

Questions:

Is Claim 0 correct? How to prove it?

Even if Claim 0 is wrong, is my main hypothesis still correct?

Are there any other mistakes in my proof?

Is there a better term for graphs whose every maximal ear is of length $1$, other than "smooth graphs"?

答:
Your proof appears correct to me. I don't see why you assume that the graph has no isolated vertices – does it make a difference anywhere? Also, "smooth graph" seems to be a fancy name for a graph with no vertices of degree two. (More precisely, the only vertices of degree two are isolated vertices with a loop.)

I'll give a proof of your claim. We may assume that the graphs in question are connected and that they have at least one edge. To any graph $G$, associate a vertex-colored graph $Ear(G)$ in the following way:

The vertices of $Ear(G)$ correspond to the ears in the unique decomposition of $G$ into maximal ears. They are colored blue and red according to whether the ear is a path or a cycle.

Two vertices are adjacent if the corresponding ears have a common vertex; if they have two common vertices, then we draw two parallel edges. (This can only happen if the corresponding ears are paths.)

There are two observations to be made which are more or less implicit in your proof of Theorem 4:

If $G$ and $H$ are homeomorphic, then $Ear(G)$ and $Ear(H)$ are isomorphic, with the isomorphism preserving the vertex colors. This follows from Theorem 1 after checking that both smoothing and subdividing preserve $Ear(G)$.

If $G$ is smooth, then (disregarding the coloring) $Ear(G)$ is just the line graph of $G$, defined appropriately for graphs with loops and multiple edges.

Conveniently, a theorem of Whitney states that if the line graphs of two connected simple graphs isomorphic, then the graphs themselves are isomorphic, except if the graphs are the triangle $K_3$ and the claw $K_{1,3}$. Note that the triangle is not smooth. In the case of graphs with loops and parallel edges, the situation is more complicated (though not terribly so, according to this article* to which I could only find a paywalled link; funnily enough, the name of Whitney is misspelled in the title), but in our case the vertex-coloring and Theorem 4 give us enough information to uniquely reconstruct the original graph. You could probably sort this out yourself, but I'll give the details for completeness.

So suppose that $G$ and $H$ are smooth and that $Ear(G)$ and $Ear(H)$ are isomorphic. First, we deal with loops: these correspond precisely to the red vertices of $Ear(G)$ and $Ear(H)$. It follows that if we denote by $G'$ and $H'$ the graphs obtained by deleting the loops in $G$ and $H$, then $Ear(G')$ and $Ear(H')$ are obtained by deleting the red vertices from $Ear(G)$ and $Ear(H)$; in particular, they are isomorphic. Now it is enough to show that $G'$ and $H'$ are isomorphic, since then the positions of the loops are uniquely determined by $Ear(G)$: a vertex in $G'$ has a loop if and only if there is an edge incident to it which is adjacent to a red vertex in $Ear(G)$, or if $G'$ consists of this single vertex (since we assumed that our graphs have at least one edge).

Thus, we may assume that $G$ and $H$ contain no loops. Now we just have to take care of parallel edges. If two edges are parallel in $G$, then by our construction the corresponding vertices in $Ear(G)$ are connected by two parallel edges. More generally, two or more parallel edges in $G$ correspond to a clique in $Ear(G)$ in which every edge is doubled. Every vertex in $Ear(G)$ is contained in a unique maximal such "double clique" (potentially of size one), and by contracting these cliques and replacing newly formed parallel edges by single ones, we obtain the line graph of the simple graph underlying $G$. Since this works backwards as well (i.e. from the simple graph and $Ear(G)$ we can recover $G$), we may assume that $G$ and $H$ are simple.

So we are done by Whitney's theorem, right? Well, not so fast. It could happen that after leaving loops and parallel edges from $G$ and $H$, we are left with a triangle and $K_{1,3}$: after all, a triangle with doubled edges is smooth. But this is ruled out by Theorem 4: the original $G$ and $H$ had the same number of vertices, and leaving edges doesn't change that. So $G$ and $H$ are indeed isomorphic.

*Note that, as far as I can tell, the notion of line graph used in the article is different from the one used here, in that the vertices corresponding to parallel edges are still connected with only one edge.


问:
How can we make it easy to solve quadratics without the Quadratic Formula?
For example ,

$$5x^2 - 7x-2=0$$

Then, we only need to find is when
$$5x^2-7x=2$$
$$x(5x-7)=2$$
Since that when we get $2-2=0$.

Is there any way we can find them easily without solving quadratics using formula?

I am just getting very enthusiastic in this topic.

答:
All ways of solving quadratic equations are essentially the same, but here's another tack:

You can "depress" the quadratic $ax^2+bx+c$ using the substitution $x=y - b/2a$.  This substitution makes the linear term disappear.  Then it's easy to solve for $y$ and then plug that answer into the substitution and recover $x$.

For your example: $ 5x^2-7x-2 =0$, let $x = y - \frac{-7}{2\cdot 5} = y+\frac{7}{10}.$  The equation becomes

$$5\left(y+\frac{7}{10}\right)^2 -7\left(y+\frac{7}{10}\right) -2=0$$

which simplifies to

$$5y^2 -\frac{89}{20}=0.$$

Solve for $y$ (which is easy because there's no linear term):

$$y^2=\frac{89}{100}$$

$$y = \pm \frac{\sqrt{89}}{10}.$$

Which means

$$x = \pm \frac{\sqrt{89}}{10} +\frac{7}{10}.$$

So all the steps are easy and obvious as long as you remember the substitution $x = y-b/2a.$


问:
Triple integral over volume bounded by two cylinders
I'm given the following triple integral $$\iiint_W y^2|z|\,dx\,dy\,dz$$ where $W=\left\{(x,y,z)\in\mathbb{R}^3, \ 1\le x^2+y^2\le2x, \ 0\le z\le \displaystyle\frac{2}{x}  \right\}$.

The two cylinders have equations $x^2+y^2=1$ and $(x-1)^2+ y^2=1$. In the $xy-$plane, the two circles intersect at $x=1/2$. Since $x$ varies from $1/2$ to $2$, $2/x$ varies from $1$ to $4$, therefore $z$ is nonnegative and the modululs disappears. Evaluating the triple integral, I get
$$\iiint_W y^2|z|\,dx\,dy\,dz=\iint_D \left(\int_0^{\frac{2}{x}} y^2z\,dz \right)\,dx\,dy=2\iint_D \frac{y^2}{x^2}\,dx\,dy$$ where $D=\{(x,y)\in\mathbb{R}^2, x^2+y^2\ge1, (x-1)^2+y^2\le1 \}$.

Using the polar change of coordinates
\begin{cases}
x=\rho\cos\theta \\
y=\rho\sin\theta
\end{cases}
I find that $\rho\ge1$ and $\rho\le2\cos\theta$. Knowing that in the $xy-$plane $x\ge0$, I consider that $-\displaystyle\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$. However, my textbook says that $\displaystyle-\frac{\pi}{3}\le\theta\le\frac{\pi}{3}$, and I can't figure out why this restriction.

答:
That is because you are interested only in the shaded area ($XY$ plane). For $|\theta| \gt \frac{\pi}{3}$, you are in region which is not of intersection.

As you said, $1 \leq \rho \leq 2 \cos \theta$. For $|\theta| \gt \frac{\pi}{3}$, $2 \cos \theta \lt 1$


问:
$10$ object in $2$ equal queues
I'm working on the following.

Ten objects shall be arranged in two queues of equal length. The order matters.

a) A and B must be in different queues.

b) A and B must be next to each other in the same queue.

Objects = $A,B,C,D,E,F,G,H,I,J$

Queues = $1,2$

a)
If we first place A, then there are 10 possible positions. If we then place B, then there are 5 possible positions. Then we will place the other eight objects in 8 possible positions. Then the same thing, but B is placed first, then A.

$$10\cdot 5 \cdot \binom{8}{8} \cdot 2=4032000$$

But if we place all then objects without restraints, we get a lower number. What am I doing wrong?

$$\binom{10}{10}=3628800$$

b) Calculate if A and B are next to each other in 1. Then the same thing for 2.

$$2\cdot \binom{5}{2}\cdot \binom{8}{8}=1612800$$

EDIT:

a) Removed the double counting. $$10\cdot 5 \cdot \binom{8}{8} =2016000$$

b) A and B can be placed next to eachother in each queue $2\cdot4$ possible ways (order matters). Then we've 2 queues, $2\cdot8$.

$$16\cdot \binom{8}{8} =645120$$

答:
For part a, I think you're double-counting when you consider if $A$ is placed first or if $B$ is placed first. If you like thinking in terms of cases then it may be more helpful to first think of all the cases where $A$ is in queue $1$ and double that to count all of the cases where $A$ is in queue $2$. Here it's clear that any ordering which is in one set can't be in the other, and all of the orderings must be in one set or the other. So I think the answer you want is $2016000$.

Then for part b I think the factor of $5 \choose 2$ is wrong, I think that would be correct if they only had to be in the same queue, but here they also have to be next to each other. I would count that by first looking at the number of ways two elements can be placed next to each other in a queue of $5$ (there are $4$) and then doubling that to account for the ordering, getting $8$ total. So I get $645120$ possible orderings.

Title: $10$ object in $2$ equal queues


问:
Show that $a^2+8bc=(2b+c)^2$ if $a,b$ and $c$ form an arithmetic progression
The numbers $a,b$ and $c$ form an arithmetic progression. Show that $a^2+8bc=(2b+c)^2.$ Can we say that $a,b$ and $c$ form an ap if $a^2+8bc=(2b+c)^2?$

We have that $2b=a+c$ (we know that a sequence is an arithmetic progression iff $a_n=\dfrac{a_{n-1}+a_{n+1}}{2}\text{   } \forall \text{   }n\ge2$). I am stuck here and I would be very grateful if you could give me a hint.

Body:
WLOG $a=b-d, c=b+d$

$$2b+c=2b+b+d=?$$

$$a^2+8bc=(b-d)^2+8b(b+d)=\cdots=(3b+d)^2=(2b+\underbrace{b+d})^2$$

Alternatively,

$$a^2+8bc=(2b+c)^2$$

$$\iff a^2=4b^2-4bc+c^2=(2b-c)^2$$

So, we need

either $a=2b-c\iff a+c=2b$   i.e., $a,b,c$ are in A.P.

or $a=-(2b-c)\iff 2b=c-a$  i.e., $c,b,-a$ are in A.P.


问:
how to find maximal chain and antichain of a given partial order
I have a partially ordered set , and I'm asked to find the maximal chain and antichain of :

$A=${$1,2,\dots,7$} with a relation $R$ s.t.

$R=I_A \cup ${$(a,b)|a,b \in {1,2,\dots,5} \space and \space a < b $}$\cup${$(6,7)$}

答:
Note if you look at $A=\{1,2,3,4,5\}$ and $B=\{6,7\}$, none of the elements in $A$ is comparable with any of the elements in $B$, while all elements within $A$ are comparable with each other, and all elements within $B$ are comparable with each other.

Maximal chains: $A$ and $B$. Both are chains, and you cannot extend them further: say you want to add an element $b\not\in A$ into $A$ to get a bigger chain, but then $b\in B$ and is not comparable with any elements from $A$. On the other hand, any chain is fully contained in either $A$ or $B$ (because otherwise it will contain incomparable elements) and so can be extended to either $A$ or $B$.

Maximal antichains: $\{a,b\}$ where $a\in A$ and $b\in B$ (all $10$ of those). These are antichains, and you cannot add any additional elements, because if you try to add another element from $A$, it will be comparable with $a$, and if you try to add another element from $B$, it will be comparable with $b$. On the other hand, as $A$ and $B$ are chains, any antichain can have at most one element from $A$ and at most one element from $B$, and, if not already one of those $10$ antichains above, it can be obviously extended to one of those.


问:
Generating function of total sales on a given day
Given the problem below:

The number of customers that shop at Elmo’s World of Food on a given day is geometrically distributed with parameter $p_1$. The number of items a customer buys at Elmo’s is geometrically distributed with parameter $p_2$. The distribution of the cost of each item is exponentially distributed with parameter $\lambda$. Assume that the amount customers buy and the amount of prices of items are independent. Let $X$ denote the total sales on a given day. Find the generating function of $X$.

In my attempt on solving this I let $N_c$ denote the number of customers that shop on a given fay, $N_i$ denote the number of items a customer buys, and $C$ cost of each item. Then we can write $X = \sum_{k=1}^{N_c} N_i\cdot C$. To find the generating function I used the MGF: $E[e^{tX}]$ and to compute it I conditioned on $N_c$ first and then on $C$. However, my work became very tedious and I'm stuck at simplifying the expression I got. Therefore, I was wondering if there is a different approach I could use in finding the generating function of $X$? This is just a practice problem as I study for my exam.

答:
We need to distinguish between two different kinds of probability generating functions: the ordinary power series generating function (OPSGF) and the moment-generating function (MGF).  Suppose $X$ is a discrete random variable that takes on only values in $0,1,2,3,\dots$; then we define the OPSGF of $X$ as
$$f(z) = \sum_{x=0}^{\infty} p(x) z^n$$
where $p(x)$ is the probability density function of $X$.  Equivalently, we may write $f(z) = E[z^x]$.  The MGF of $X$ is defined by $g(t) = E[e^{tx}]$.  The MGF has the advantage that it can be applied to continuous random variables as well as discrete.  There is a simple relation between the two types of generating functions: if $f(z)$ is the OPSGF and $g(z)$ is the MGF, then $g(t) = f(e^t)$.

A useful property of the OPSGF applies to sums of a random number of random variables.  If $X_1, X_2, X_3,...$ are independent random variables all having the OPSGF $f(z)$ and $N$ is an integer random variable with OPSGF $g(z)$, then the OPSGF of $Y= \sum_{i=1}^N X_i$ is $g(f(z))$.  If $f(z)$ is a MGF, then $g(f(z))$ is the MGF of $Y$.

Applying this property to our question, we see that if $f(z)$ is the OPSGF of a geometric random variable with parameter $p_1$, $g(z)$ is the OPSGF of a geometric random variable with parameter $p_2$, and $h(t)$ is the MGF of an exponential random variable with parameter $\lambda$, then the MGF of the total sales is $f(g(h(t)))$.  If you carry out the computation, I think you will find that the total sales have an exponential distribution.

Hint: The OPSGF of a geometric random variable with pdf $(1-p)^{k-1}p$ is
$$f(z) = \frac{pz}{1-(1-p)z}$$

If you are interested in more information about the uses of the OPSGF in probability, one source is An Introduction to Probability Theory and Its Applications, Volume I, Third Edition by William Feller.


答:
In the RHS, you don't have $a$.

So using $a=  2b -c$,

$\begin{align}a^2+8bc &= (2b-c)^2+8bc \\&= (2b)^2\color{blue}{- 2(2b)(c)}+c^2+\color{blue}{8bc} \\&= (2b)^2+\color{blue}{2(2b)(c)}+c^2 \\&= (2b+c)^2\end{align}$

Edit: Just now saw the second part

If $a^2+8bc = (2b+c)^2$, then

$\begin{align}a^2 + 8bc = (4b^2)+8bc +c^2 &\Rightarrow a^2 = 4b^2-4bc+c^2 = (2b-c)^2 \\&\Rightarrow a = \pm(2b-c) \\&\Rightarrow b = \frac{a+c}{2} \text{ or } b = \frac{c-a}{2}\end{align}$

So, they may or may not be in AP.

For example take, $1,3,5$ they are in AP and satisfy $1^2 = (2(3) -5)^2$

But $-1,3,5$ are not in AP even when they satisfy the condition $(-1)^2 = (2(3)-5)^2 = 1$

Title: Show that $a^2+8bc=(2b+c)^2$ if $a,b$ and $c$ form an arithmetic progression

Body:
The numbers $a,b$ and $c$ form an arithmetic progression. Show that $a^2+8bc=(2b+c)^2.$ Can we say that $a,b$ and $c$ form an ap if $a^2+8bc=(2b+c)^2?$

We have that $2b=a+c$ (we know that a sequence is an arithmetic progression iff $a_n=\dfrac{a_{n-1}+a_{n+1}}{2}\text{   } \forall \text{   }n\ge2$). I am stuck here and I would be very grateful if you could give me a hint.


问:
Given a concrete day, probability that two people have birthday on the same day.
What is the probability that on a specific day (e.g. December 24) two people have birthday in a room with 23 people?

The probability, given a specific day, one person has birthday is in my opinion $\frac{23}{365}$. So the probability for two birthdays is $(\frac{23}{365})^2$.

I believe that my calculation is not correct. If it is wrong, what is calculated by $(\frac{23}{365})^2$ in reality?

答:
If you mean "exactly two people" the probability is
$$\binom {23}2\left (\frac {1}{365}\right)^{2}\left (\frac {364}{365}\right)^{21}.$$

If you mean "at least two people" the probability is
$$1-\left (\frac {364}{365}\right)^{23}-\binom {23}1\left (\frac {1}{365}\right)^{1}\left (\frac {364}{365}\right)^{22}.$$

Above no correction for leap years and deviations from the uniform distribution are taken into account.


问:
explicit formula for sequence with multiple rules? e.g.: $a_n=\{1,2,3, 5,10,15, 25,50,75, … 5^kn,5^k(n+1),5^k(n+2), 5^{k+1}n,..\}$
What is the formula for the sequence obeying that pattern? A base-10 analog could be $\{1,2,3,4,5,6,7,8,9, 10,20,30,.., … 10^kn,10^k(n+1),10^k(n+2),..10^{k+1}n, 10^{k+1}(n+1),10^{k+1}(n+2), . . .\}$, where k is an integer starting with 0 stepping by 1.

What if the sequence doesn't increase or decrease locally over the entire domain, such as with a subsequence of {4,5,3} or a rule using sign-flipping or periodicity (e.g. $(-k)^n$ for subsequences with both even and odd or non-integer or $(-k)^{n+1}$ for sequence otherwise without) and it could feature duplicate values; is there a way to apply a rule admitting only the first (or for that matter, only the second, third, or or any rank of choice, preculuding more and more) instances of a value? E.g.: $b_n$={4,5,3, <s>3,5,</s>1,-3,<s>1</s>} with a rule of |2n|-5 of $n_i$ and simply eliminating duplicate value then continuing to $n_{i+1}$ (instead of continuing to iterate until a valid value arises from operating on $n_i$. The possibility exists also to include omitted(duplicate/sub-duplicate) in the input sequence of (could call it) $n_j$. What is the mathematical or programming way to formulate this? What if you want to take a discrete sequence with rules such as these and regress it to a smooth curve, is that possible?

答:
There are several ways how to achieve what you ask. It depends on what you want from such representation.

If you want your reader to understand how the sequence behaves, then nothing really beats the piecewise bracket:
$$
a_k = \begin{cases}
1\cdot 5^{\lfloor k/3\rfloor}&\text{if }k\equiv 0\mod 3,\\
2\cdot 5^{\lfloor k/3\rfloor}&\text{if }k\equiv 1\mod 3,\\
3\cdot 5^{\lfloor k/3\rfloor}&\text{if }k\equiv 2\mod 3,
\end{cases}
$$

If you want to compress this to a single line, you might use unit function $\delta[n]$ or $\mathbb 1_{cond}$:
$$
a_k=5^{\lfloor k/3\rfloor}\big(\delta[k\mathop{\mathrm{mod}} 3]+2\delta[(k\mathop{\mathrm{mod}} 3)-1]+3\delta[(k\mathop{\mathrm{mod}} 3)-2]\big)

If you want something that looks continuous, then you can use the following trick. If your sequence has a period of 3, take 3 roots of $x^3=1$ and solve the equation:
$$
\begin{cases}
a +b+c=1,\\
a x_1^1+bx_2^1+cx_3^1=2,\\
a x_1^2+bx_2^2+cx_3^2=3.
\end{cases}
$$
with $x_1=1$, $x_2=e^{i2\pi/3}$ and $x_3=e^{-i2\pi/3}$, we get $a=2$, $b=-(i/\sqrt3 + 1)/2$, $c=(i/\sqrt3 - 1)/2$. Thus the sequence:
$$
u_k=2-\left(\frac{i}{2\sqrt3}+\frac12\right)e^{\frac{2\pi i}3 k}+\left(\frac{i}{2\sqrt3}-\frac12\right)e^{-\frac{2\pi i}3 k}
$$
will be periodic: 1,2,3,1,2,3...

You can do a similar trick with powers of 5. They go 0,0,0,1,1,1,2,2,2... By subtracting the asymptote $k/3$, you will get $0,-\frac13,-\frac23,0,-\frac13,-\frac23,0,-\frac13,-\frac23...$ a periodic sequence again. You can find that this sequence is:
$$
v_k = -\frac13+\frac16\left(1-\frac i{\sqrt3}\right)e^{\frac{2\pi i}3 k}+\frac16\left(1+\frac i{\sqrt3}\right)e^{-\frac{2\pi i}3 k}.
$$

And finally $a_k=u_k 5^{v_k+k/3}$. If you don't like complex numbers, you can express this with trigonometric functions.


问:
Proof of Bienayme Inequality
I have a bit of trouble about the proof of Bienayme Inequality.

Bienayme Inequality is as follows:

If X has mean $\mu$ and variance $\sigma^2$, then
$$\mathbb{P}\left(\frac{|X-\mu|}{\sigma}\ge k\right)\le\frac{1}{k^2}.$$

Bienayme's Proof:

Let $B = \{|X-\mu|\ge k\sigma\}$ and $\mathbb{1}_B$ be the indicator random variable. Then $(X-\mu)^2\mathbb{1}_B\ge\sigma^2k^2\mathbb{1}_B$ and $\mathbb{E}(\sigma^2k^2\mathbb{1}_B)=\sigma^2k^2\mathbb{P}(B)$, hence

$$\sigma^2=\mathbb{E}[(X-\mu)^2]$$
$$\ge\mathbb{E}((X-\mu)^2\mathbb{1_B})$$
$$\ge\mathbb{E}(\sigma^2k^2\mathbb{1}_B)$$
$$=\sigma^2k^2\mathbb{E}(\mathbb{1}_B)$$
$$=\sigma^2k^2\mathbb{P}(\mathbb{B}).$$

There is only one thing that I don't understand:

Why is $\mathbb{E}[(X-\mu)^2]\ge \mathbb{E}((X-\mu)^2\mathbb{1_B})$. How can we explain /derive that?

Many different explanations and interpretations are appreciated.

Thanks.

答:
$1_B$ is an indicator random variable. It can take ane of two values, $0$ or $1$. So for any non-negative random variable $X$ we have $X1_B \leq X$ with probability one.

Therefore $E(X1_B)\leq E(X)$ for every non-negative random variable $X$ and every event $E$.

Title: Proof of Bienayme Inequality


问:
Continuity of function defined by a series
I have solved the following exercise and I would like to know if I have made any mistakes:

Define $g(x)=\sum_{n=0}^{\infty}\frac{x^{2n}}{1+x^{2n}}$.

Find the values of $x$ where the series converges and show that we get a continuos function on this set.

My solution:

$|x|<1\Rightarrow 0<x^{2n}<1\Rightarrow 1<1+x^{2n}<2\Rightarrow \frac{1}{2}<\frac{1}{1+x^{2n}}<2\Rightarrow 0<\frac{x^{2n}}{2}<\frac{x^{2n}}{1+x^{2n}}<x^{2n}$ so by Comparison Test $\sum_{n=0}^{\infty}\frac{x^{2n}}{1+x^{2n}}$ converges for all $x\in (-1,1)$ (since $\sum_{n=0}^{\infty} x^{2n}$ converges in that interval to $\frac{1}{1-x^2}$).
If $|x|=1$ then $\lim_{n\to\infty}\frac{x^{2n}}{1+x^{2n}}=\frac{1}{2}\neq 0$ so the series does not converge and if $|x|>1$ then $\lim_{n\to\infty}\frac{x^{2n}}{1+x^{2n}}=\lim_{n\to\infty}(1-\frac{1}{1+x^{2n}})=1\neq 0$ so the series does not converge here too; the series thus converges only if $|x|<1$.

If $x\in [-M,M], |M|<1$ then $\frac{x^{2n}}{1+x^{2n}}\leq x^{2n}\leq M^{2n}$ and since $\sum_{n=0}^{\infty} M^{2n}=\frac{1}{1-M^2}$ by Weierstrass M-Test the series converges uniformly on each interval $[-M,M], |M|<1$ and so by Term-by-Term Continuity Theorem $g(x)$ is continuous on every such interval.

答: It is correct, except that you wrote nothing about what happens when $x=-1$. In fact, in that case we have $(\forall n\in\Bbb N):\frac{x^{2n}}{1+x^{2n}}=\frac12$, and therefore the series diverges then too. Actually, that argument also works when $x=1$ too.


问:
Spherical coordinates triple integral, help
I have the following triple integral with the instructions

By using spherical coordinates, evaluate the integral:
$$L = \iiint_{D} \frac{1}{\sqrt(x^2+y^2+z^2)} dxdydz$$

Where D is the portion of the unit cylinder $x^2 + y^2 \leq 1$ which lies between $z =0 $ and $z = 1$

My thoughts were that we use the bounds and use the Jacobian:

$$0 \leq \rho \leq \sqrt2, $$
$$0 \leq \theta \leq 2 \pi  $$
$$0 \leq \varphi \leq \frac{\pi}{4}$$

To which after computing, we end up with the result
$$\pi(\sqrt2 - 2)$$

I was hoping someone could either correct me, or verify. And if you do verify it, is there a method which is easier or different, because I would want to learn it thank you!

答:
That answer that you got is negative, and therefore it cannot possibly be correct.

You can do it in cylindrical coordinates:\begin{align}\iiint_D\frac{\mathrm dx\,\mathrm dy\,\mathrm dz}{\sqrt{x^2+y^2+z^2}}&=\int_0^{2\pi}\int_0^1\int_0^1\frac\rho{\sqrt{\rho^2+z^2}}\,\mathrm d\rho\,\mathrm dz\,\mathrm d\theta\\&=2\pi\int_0^1\sqrt{1+z^2}-z\,\mathrm dz\\&=\pi\left(\sqrt2-1+\operatorname{arcsinh}(1)\right).\end{align}


问:
how to proof the inequality $n > \ln^3(n)\ln^2(\ln(n))$ for n > 1
how to prove the inequality $n > \ln^3(n)\ln^2(\ln(n))$ , n > 1 and n $\in${natural numbers}

I do not know if this holds, but I would like to prove that it holds, or if it doesn't, then with contradiction.

I tried to put both sides into exponent form with base $e$, but got nowhere. What should I do?

答:
Not a 'real' answer, but it was too big for a comment. I think that you're looking for a solution without using a calculator or PC but maybe this gives some insight. I did only a quick search with the following bound: $2\le\text{n}\le10^3$.

I wrote and ran some Mathematica-code:

```<code>In[1]:=Clear["Global`*"];
\[Alpha] = 10^3;
ParallelTable[
  If[TrueQ[n > Log[n]^3*Log[Log[n]]^2], n, Nothing], {n,
   2, \[Alpha]}] //. {} -> Nothing
</code>```

Running the code gives:

```<code>Out[1]={2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
</code>```

So, for all the other values of $\text{n}$ between the bounds $2\le\text{n}\le10^3$ is your statement wrong, this implies that when $\text{n}\in\mathbb{N}\smallsetminus\left\{1,2,3,4,5,6,7,8,9,10,11\right\}$:

$$\text{n}<\ln^3(\text{n})\ln^2(\ln(\text{n}))\tag1$$

Body:
Not a 'real' answer, but it was too big for a comment. I think that you're looking for a solution without using a calculator or PC but maybe this gives some insight. I did only a quick search with the following bound: $2\le\text{n}\le10^3$.

I wrote and ran some Mathematica-code:

In[1]:=Clear["Global`*"];
\[Alpha] = 10^3;
ParallelTable[
  If[TrueQ[n > Log[n]^3*Log[Log[n]]^2], n, Nothing], {n,
   2, \[Alpha]}] //. {} -> Nothing

Running the code gives:

Out[1]={2, 3, 4, 5, 6, 7, 8, 9, 10, 11}

So, for all the other values of $\text{n}$ between the bounds $2\le\text{n}\le10^3$ is your statement wrong, this implies that when $\text{n}\in\mathbb{N}\smallsetminus\left\{1,2,3,4,5,6,7,8,9,10,11\right\}$:

$$\text{n}<\ln^3(\text{n})\ln^2(\ln(\text{n}))\tag1$$


答:
The region you are integrating over is cylinder $x^2 + y^2 \leq 1, 0 \leq z \leq 1$

So it is easier to find the limits of integration in cylindrical coordinates.

$x = r \cos \theta, y = r \sin \theta, z$

Your integrand becomes $\frac{1}{\sqrt{x^2 + y^2 + z^2}} = \frac{1}{\sqrt{r^2 + z^2}}$

And your integral is $\displaystyle \int_0^{2\pi}\int_0^1 \int_0^1 \frac{r} {\sqrt{r^2+z^2}} \, dr \, dz \, d\theta$

EDIT: In spherical coordinates,

Your integral is $\displaystyle \int_0^{2\pi} \int_0^{\pi/4} \int_0^{1 / \cos \phi} \rho \sin \phi\, d \rho \, d\phi \, d\theta + \int_0^{2\pi} \int_{\pi/4}^{\pi/2} \int_0^{1 / \sin \phi} \rho \sin \phi\, d \rho \, d\phi \, d\theta$

Here is a diagram that shows why $\rho$ is $\frac{1}{\cos \phi}$ (as $z \leq 1$) for $0 \leq \phi \leq \frac{\pi}{4}$.

Also why $\rho$ is $\frac{1}{\cos (\frac{\pi}{2} - \phi)} = \frac{1}{\sin \phi}$ (as radius of cylinder is $1$) for $\frac{\pi}{4} \leq \phi \leq \frac{\pi}{2}$.

Title: Spherical coordinates triple integral, help

Body:
I have the following triple integral with the instructions

By using spherical coordinates, evaluate the integral:
$$L = \iiint_{D} \frac{1}{\sqrt(x^2+y^2+z^2)} dxdydz$$

Where D is the portion of the unit cylinder $x^2 + y^2 \leq 1$ which lies between $z =0 $ and $z = 1$

My thoughts were that we use the bounds and use the Jacobian:

$$0 \leq \rho \leq \sqrt2, $$
$$0 \leq \theta \leq 2 \pi  $$
$$0 \leq \varphi \leq \frac{\pi}{4}$$

To which after computing, we end up with the result
$$\pi(\sqrt2 - 2)$$

I was hoping someone could either correct me, or verify. And if you do verify it, is there a method which is easier or different, because I would want to learn it thank you!


问:
Induction of topological space by an $L^2$-space
Let $(L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R}),\langle\cdot,\cdot\rangle)$ be a Hilbert space, where $L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})$ denotes the set of all (equivalence classes of) $\boldsymbol{\Xi}$-measurable $\mathbb{R}$-valued functions that are square $\mu$-integrable on $\Xi$, and $\langle\cdot,\cdot\rangle:L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})\times L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})\to\mathbb{R}$ denotes its standard inner product given by
$$\langle f,g\rangle=\int fg\,\mathrm{d}\mu.$$
I know that this Hilbert space induces a topological space whose induced standard topology $\mathcal{O}$ is the one generated by all open balls whose radii are measured using the induced standard metric $d:L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})\times L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})\to\mathbb{R}_0^+$ given by
$$d(f,g)=\langle f-g,f-g\rangle^{\tfrac{1}{2}}.$$
My question is how to write this standard topology symbolically.

At the moment I have defined the topological basis:
$$\mathscr{B}=\{B_r(f)\subset L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})\,\mid\, r\in\mathbb{R}^+,f\in L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})\},$$
where $B_r(f):=\{g\in L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})\mid d(f,g)<r\}$ is the open ball of radius $r$ centered at $f$.

Then, we define $\mathcal{O}=\varphi(\mathscr{B})$. Here $\varphi$ represents a map which takes a topological basis as an input and returns a topology as an output. In measure theory, we have the map $\sigma$ which generates a $\sigma$-algebra from a given collection of subsets. I thought that the same could be done here. However, I haven't found the actual symbol for performing such an operation on a topological basis yet.

So, is this the correct way to write the induced standard topology symbolically? Also, is there a better way to write the same thing?

Thank you, Frederick.

答:
Given a set $X$, a collection $\mathscr{B}$ of subsets of $X$ must satisfy two conditions to be a topological basis:

$1$. For any $x \in X$, there is $B\in \mathscr{B}$ such that $x \in B$.

$2$. For any $B_1, B_2 \in \mathscr{B}$ and for any $x \in B_1 \cap B_2$, there is a $B_3 \in  \mathscr{B}$ such that $x \in B_3$.

The topology generated by $\mathscr{B}$ is usually indicated by $\tau(\mathscr{B})$ (but this notation is not as standard as the $\sigma$ for the generated $\sigma$-algebra). Using this notation, we have:

$\tau(\mathscr{B})$ is the collection of arbitrary union of sets is $\mathscr{B}$.

So, while any collection of subsets can be used to generate a $\sigma$-algebra, it is not true that any collection of subsets is a basis for a topology.

If you want to mimic the concept of "generated $\sigma$-algebra" closer, you should use the concept of topological subbase:

Given a set $X$ and any collection $\mathscr{S}$ of subsets of $X$, the generated topology, $\tau(\mathscr{S})$, is defined as:

$\tau(\mathscr{S})$ is the collection of arbitrary unions of finite intersections of sets in $\mathscr{S}$.

Note that a partiular case of  finite intersection of sets in $\mathscr{S}$ is the nullary intersection who result is $X$.  So $X \in \mathscr{S}$.

In your specific example, $X= L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})$ and we have that

$$\mathscr{B}=\{B_r(f)\subset L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})\,\mid\, r\in\mathbb{R}^+,f\in L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})\}$$
is actually a basis for the topology (the standard Hilbert space topology).

You could write $\tau(\mathscr{B})$ to indicate the generated topology.


问:
Finding the point of contact between an axisymmetric surface and a tangent line
this is my first time posting on this website. I tried to search for this exact problem here but I couldn't get any results.

Anyways. I'm doing research concerning filament winding over containers/vessels and I'm confronted to the following mathematical problem:

1 - Consider an orthonormal basis $(\vec{e_x}, \vec{e_y}, \vec{e_z})$ and, on the $(xOz)$ plane we draw a curve $z(x)$ defined for $x\in[x_t, x_b]$ (with $x_t<x_b$) such that :

$z(x_b)=z_b=0$ and $z(x_t)=z_t$ and $z(x_t)>z(x_b)$ (i.e. $z(x_t)$ is strictly positive),

$z(x)$ is $C^1$ continuous, and (for the moment) strictly monotone, i.e. strictly decreasing. Since $z(x)$ is bijective, we can easily find $x(z)$.

Now, the curve $z(x)$ is going to revolve around the $z$-axis for a full rotation and creates a surface $S(x,y,z)$. Using polar coordinates for the axisymmetric surface, the expression of $S$ will be reduced to the notation $r(z)$, so that:

$S(x,y,z)=S(r\cdot cos\theta,r\cdot sin\theta,z)$ with $\theta\in]-\pi, \pi]$. For $\theta=0$, we find the original 2D profile $x(z)$.

We put $r_b=x_b$, so that $r_b$ is the radius of the bottom surface formed by $S$.

Consider now a circle $\mathscr C_1$ centered at the origin $(0,0,0)$, with a radius $r_m$, with $r_m>r_b$ and consider the point $A(r_m, 0, 0)$ (note: point A can be anywhere on $\mathscr C_1$, but for simplicity, I'll put it at this location since the base can always rotate to find this disposition).

Now, this is when it gets complicated for me. A fiber, exemplified as a mathematical line (or segment if you will), will begin at point A and forms an angle $\alpha_{att}$, called the angle of attack, with the meridian plan $\mathscr M$, the plane crossing the three point $A$, $O$ (origin) and $(0,0,1)$. In this case, $\mathscr M$ would be $(xOz)$ plane. The fiber is supposed to come into contact with the curved surface $S$ at point $B$. My problem is then trying to find the coordinates for this point $B$.

Here is a sketch I made using GeoGebra to make things clear :
3D intersection of surface S and fiber path

Here is what I tried to do :

Obviously, there are cases where there is no intersection whatsoever. There is a condition on the value of the angle of attack $\alpha_{att}$ which ensures that a contact occurs : $$\alpha<arctan\left(\cfrac{r_b}{r_m}\right)$$We will consider that this is always the case. Now, what we don't know about the fiber path is really the $z$-elevation, since its direction projected on the $xOy$ plane is defined by $\alpha$. Therefore the fiber line belongs to the plane that passes through $A$ and and oriented along the normal vector $(sin\alpha_{att}, cos\alpha_{att},0)$ (shown in red in the figure above). The equation of the plane is: $$x\cdot sin\alpha_{att}+y\cdot cos\alpha_{att}-r_msin\alpha_{att}=0$$
However, the parametric equation is probably more suitable for this problem : \begin{equation}\begin{cases} u \\ (r_m-u)\cdot tan\alpha_{att} & \text{          } & \forall u,v\\ v \end{cases}        \end{equation}
I tried to find the intersection between this plane and the surface $S$, which is a curve that should be determined. So, to parameterize curve $S$, we have : $$\begin{equation}\begin{cases} r cos\theta \\ r sin\theta & \text{for} & \theta \in ]0, \pi] \text{     and      } r\in[r_t, r_b]\\ z(r) & \text{          } & \ \text{z is known}\end{cases}        \end{equation}$$
I restricted the domain for $\theta$ to be positive because, for all practical purposes, $\alpha_{att}>0$

Now here is where I am stuck. The parameterization for the intersection between the plane and $S$ is : \begin{equation}\begin{cases} u \\ (r_m-u)\cdot tan\alpha_{att} & \text{          } & \text{for} & u\in \text{?}\\ z\left[\sqrt{u^2+(r_m-u)^2\cdot tan^2(\alpha_{att})}\right] \end{cases}        \end{equation}
the last equation for $z$ stems from identifying $r$ as : $r = \sqrt{(rcos\theta)^2+(rsin\theta)^2} = \sqrt{u^2+(r_m-u)^2\cdot tan^2(\alpha_{att})}$. I can't work out the exact domain for $u$.

My second problem is to find a way to identify the point of contact $B$ so that the fiber path is tangent to the parameterized curve (if multiple peaks are present - I think there is only one because of the monotonic nature of $z(r)$ but I'm not sure, the fiber line must be tangent near the highest peak, so that there are no intersections between the fiber path and the surface $S$.

I hope I was clear enough in stating my problem.

答:
Giving a revolution surface around the $z$ axis

$$
G(x,y,z) = f\left(\sqrt{x^2+y^2}\right)-z=0
$$

and a plane parallel to the $z$ axis

$$
(p-p_0)\cdot\vec n = 0
$$

with

$$
\cases{
p = (x,y,z)\\
p_0 = (x_0,y_0,z_9)\\
\vec n = (a,b,0)
}
$$

the line representing the contact fiber can be described as

$$
L\to p = p_0+\lambda \vec v
$$

This line and the contact point $p^*$ should obey some conditions as follows

$$
\cases{
G(p^*) = 0\\
p^* = p_0 + \lambda^* \vec v^*\\
(p^*-p_0)\cdot \vec n=0\\
\vec v^*\cdot\vec n = 0\\
\nabla G(p^*) \cdot \vec v^* = 0
}
$$

Those are $7$ equations and $7$ unknowns $(p^*,\vec v^*, \lambda^*)$. Attached a MATHEMATICA script solving this system for the case study

$$
\cases{
f(x) = 3\cos x\\
p_0 = (2,-2,0)\\
\vec n = (2,1,0)
}
$$

```<code>[x_] := 3 Cos[x]
G[x_, y_, z_] := f[Sqrt[x^2 + y^2]] - z
gradG = Grad[G[x, y, z], {x, y, z}]
n = {2, 1, 0};
p = {x, y, z};
p0 = {2, -2, 0};
h = G[x, y, z];
g = (p - p0).n;
gr0 = ContourPlot3D[{h == 0, g == 0}, {x, -2, 2}, {y, -2, 2}, {z, 0, 4}, MeshFunctions -> {Function[{x, y, z, f}, h - g]}, MeshStyle -> {{Thick, Blue}}, Mesh -> {{0}}, ContourStyle -> Directive[Orange, Opacity[0.5], Specularity[White, 30]], SphericalRegion -> True]
v = {vx, vy, vz};
equs = Thread[p - p0 - lambda v];
equs = Join[equs, {G[x, y, z]}];
equs = Join[equs, {(p - p0).n}];
equs = Join[equs, {v.n}];
equs = Join[equs, {gradG.v}];
obj = equs.equs
sol = NMinimize[{obj, {x, y, z, vx, vy, vz, lambda} \[Element] Reals}, {x, y, z, vx, vy, vz, lambda}]
line = p0 + t v /. sol[[2]]
grline = ParametricPlot3D[line, {t, 0, 3}, PlotStyle -> {Thick, Red}];
Show[gr0, grline]
</code>```

Body:
Giving a revolution surface around the $z$ axis

$$
G(x,y,z) = f\left(\sqrt{x^2+y^2}\right)-z=0
$$

and a plane parallel to the $z$ axis

$$
(p-p_0)\cdot\vec n = 0
$$

with

$$
\cases{
p = (x,y,z)\\
p_0 = (x_0,y_0,z_9)\\
\vec n = (a,b,0)
}
$$

the line representing the contact fiber can be described as

$$
L\to p = p_0+\lambda \vec v
$$

This line and the contact point $p^*$ should obey some conditions as follows

$$
\cases{
G(p^*) = 0\\
p^* = p_0 + \lambda^* \vec v^*\\
(p^*-p_0)\cdot \vec n=0\\
\vec v^*\cdot\vec n = 0\\
\nabla G(p^*) \cdot \vec v^* = 0
}
$$

Those are $7$ equations and $7$ unknowns $(p^*,\vec v^*, \lambda^*)$. Attached a MATHEMATICA script solving this system for the case study

$$
\cases{
f(x) = 3\cos x\\
p_0 = (2,-2,0)\\
\vec n = (2,1,0)
}
$$

[x_] := 3 Cos[x]
G[x_, y_, z_] := f[Sqrt[x^2 + y^2]] - z
gradG = Grad[G[x, y, z], {x, y, z}]
n = {2, 1, 0};
p = {x, y, z};
p0 = {2, -2, 0};
h = G[x, y, z];
g = (p - p0).n;
gr0 = ContourPlot3D[{h == 0, g == 0}, {x, -2, 2}, {y, -2, 2}, {z, 0, 4}, MeshFunctions -> {Function[{x, y, z, f}, h - g]}, MeshStyle -> {{Thick, Blue}}, Mesh -> {{0}}, ContourStyle -> Directive[Orange, Opacity[0.5], Specularity[White, 30]], SphericalRegion -> True]
v = {vx, vy, vz};
equs = Thread[p - p0 - lambda v];
equs = Join[equs, {G[x, y, z]}];
equs = Join[equs, {(p - p0).n}];
equs = Join[equs, {v.n}];
equs = Join[equs, {gradG.v}];
obj = equs.equs
sol = NMinimize[{obj, {x, y, z, vx, vy, vz, lambda} \[Element] Reals}, {x, y, z, vx, vy, vz, lambda}]
line = p0 + t v /. sol[[2]]
grline = ParametricPlot3D[line, {t, 0, 3}, PlotStyle -> {Thick, Red}];
Show[gr0, grline]


答:
Your bounds aren't right.  To use spherical coordinates, you'll have to breach the region into two pieces.  One piece has bounds:

$$0\leq \rho \leq \sec\phi,\; 0\leq \phi \leq \pi/4,\; 0\leq \theta \leq 2\pi.$$

The other has:

$$0\leq \rho \leq \csc\phi,\; \pi/4\leq \phi \leq \pi/2,\; 0\leq \theta \leq 2\pi.$$

The equations $\rho = \sec\phi$ and $\rho = \csc\phi$ are the equations of horizontal and vertical lines in the $\rho\phi$-plane.


答:
A proof.

For large $n$,
$$
\log n < n,\\
\log\log n < \log n\\
(\log\log n)^2 < (\log n)^2\\
(\log n)^3(\log\log n)^2 < (\log n)^5
 < (n^{1/5})^5 = n
$$
The last step used $\log n < n^{1/5}$.  This is true for large enough $n$.

Proof that $\log n < n^{1/5}$ for large $n$.  Using l'Hopital's rule,

$$
\lim_{n\to\infty}\frac{\log n}{n^{1/5}} =
\lim_{n\to\infty}\frac{x^{-1}}{(1/5)n^{-4/5}} =
\lim_{n\to\infty}\frac{5}{n^{1/5}} = 0 .
$$

Title: how to proof the inequality $n > \ln^3(n)\ln^2(\ln(n))$ for n > 1

Body:
how to prove the inequality $n > \ln^3(n)\ln^2(\ln(n))$ , n > 1 and n $\in${natural numbers}

I do not know if this holds, but I would like to prove that it holds, or if it doesn't, then with contradiction.

I tried to put both sides into exponent form with base $e$, but got nowhere. What should I do?


问:
Calculating the Riemann sum $\lim_{n \to \infty} { \sum_{k=1}^{n} { (\frac{nk-1}{n^3})\sin(\frac{k}{n}) } }$
We need to calculate this:

$$\lim_{n \to \infty} { \sum_{k=1}^{n} { (\frac{nk-1}{n^3})\sin(\frac{k}{n}) }       }$$

So I know this is Riemann sum.
This is what I started doing:

$$\sum_{k=1}^{n} { (\frac{nk-1}{n^3})\sin(\frac{k}{n})}  = \sum_{k=1}^{n} { (\frac{k}{n^2 } - \frac{1}{n^3})\sin(\frac{k}{n})}$$

But then I noticed that we can not get $\frac{1}{n}$ outside. How do we exactly know what is the partition?

答:
Hint:

The last expression equals: $\sum_1^n \dfrac{k}{n^2} \sin \dfrac{k}{n}- \sum_1^n \dfrac{1}{n^3}\sin \dfrac{k}{n}$, now what is $\lim_{n \to \infty} \dfrac{1}{n^3} \sum_1^n \sin \dfrac{k}{n}$? And use Riemann tranformation on $\sum_1^n \dfrac{k}{n^2} \sin \dfrac{k}{n}$.

Use this : $\int_0^1 f(x) =\lim_{n \to \infty}\dfrac{1}{n} \sum_1^n f(x)$


问:
How to prove there are $\frac{3^n-1}{2}$ couples $A,B \in \mathcal{P}([n])$ such that $A \cup B = [n]$, $A \neq B$.
Consider the unordered couple of sets $A,B \in \mathcal{P}([n])$ such that $A \cup B = [n]$, $A \neq B$. I would like to prove that the number of such couples is:

$$\frac{3^n-1}{2}$$

I derived that expression by enumerating the solutions for the first values of $n$, and also considering the sum of the number of those couples for all different values of $n$ over $\mathcal{P}([m])$:

$${2^m \choose 2} = \sum_{n=0}^m {m \choose n}\frac{3^n-1}{2}$$

(this is easy to show using the binomial expansion).

I have checked OEIS A003462 where they count the number of couples $A,B \in \mathcal{P}([n])$ with $A \cap B = \emptyset$ and $A \neq \emptyset$ or $B \neq \emptyset$, with the same $\frac{3^n-1}{2}$ result, but it's not the same thing and I was not able to adapt that reasoning.

答:
You seem to treat $A,B$ as unordered... that for instance $A=\{1,2\}$ and $B=\{3,4,5,6\}$ to be the same result than $A=\{3,4,5,6\}$ and $B=\{1,2\}$.  For the time being let us go with the other interpretation that these are distinct outcomes as it makes the math far simpler.

Now... for each element $x$ in $[n]=\{1,2,3,\dots,n\}$ for it to be that $A\cup B=[n]$ as well as $A\subseteq [n]$ and $B\subseteq [n]$, exactly one of the following must be true:

$x\in A$ and $x\notin B$

$x\in B$ and $x\notin A$

$x\in A$ and $x\in B$

For each element choose which of these three it is.  Apply rule of product.  That gives $3\times 3\times 3\times \cdots \times 3 = 3^n$ different choices of $A,B$.

This included in it however the possibility that we answered $x\in A$ and $x\in B$ for every $x$ which would have meant that $A=B$.  You explicitly wanted to avoid this case, so subtracting $1$ will correct the count.

Finally, you seemed to be wanting to talk about unlabled pairs of sets rather than labeled pairs of sets.  Dividing by $2$ corrects that aspect of the count.  (Note that every outcome was counted twice.  The only outcome that might have been counted only once was the case where $A=B$ but that was already removed)

This gives, as expected, the final result of:

$$\frac{3^n-1}{2}$$


答: Note that $A\cup B=(A\setminus B)\cup(A\cap B)\cup(B\setminus A)$ is a disjoint union. Assign to each unordered couple $\{A,B\}$ the $n$-length ternary string $(a_1,a_2,\ldots,a_n)$ defined by $$a_j=\begin{cases}0\quad\text{if $a_j\in A\setminus B$}\\1\quad\text{if $a_j\in A\cap B$}\\2\quad\text{if $a_j\in B\setminus A$}\end{cases}$$

Title: How to prove there are $\frac{3^n-1}{2}$ couples $A,B \in \mathcal{P}([n])$ such that $A \cup B = [n]$, $A \neq B$.

Body:
Consider the unordered couple of sets $A,B \in \mathcal{P}([n])$ such that $A \cup B = [n]$, $A \neq B$. I would like to prove that the number of such couples is:

$$\frac{3^n-1}{2}$$

I derived that expression by enumerating the solutions for the first values of $n$, and also considering the sum of the number of those couples for all different values of $n$ over $\mathcal{P}([m])$:

$${2^m \choose 2} = \sum_{n=0}^m {m \choose n}\frac{3^n-1}{2}$$

(this is easy to show using the binomial expansion).

I have checked OEIS A003462 where they count the number of couples $A,B \in \mathcal{P}([n])$ with $A \cap B = \emptyset$ and $A \neq \emptyset$ or $B \neq \emptyset$, with the same $\frac{3^n-1}{2}$ result, but it's not the same thing and I was not able to adapt that reasoning.


问:
Unique question about packing problem
I added the related pages from part 3 of the book: combinatorial geometry by János Pach,Pankaj K.Agarwal (1995) (which is not available on net so I added them as pictures).

A. Prove that one can always find a packing $Ç$ of the plane with congruent copies of a convex disc $C$ whose density $d(Ç,R^2)$ exists and is equal to $\delta(C)$.
Similarly show that there is a Lattice packing $Ç$ with $d(Ç,R^2)=\delta_L(C)$.

B.construct a packing $Ç$ of unit discs and two convex discs $P,P'$ such that $lim_{r\to \infty} d(Ç,P(r)) \neq lim_{r\to \infty} d(Ç,P'(r))$.

Hints:
Part A:To prove the first statement, construct a packing $C_n$ of congruent copies of $C$ in the disc $D(n)$ with density $d(C_n,D(n))\geq \delta(C)-O(1)/n$

For every n choose a subsequence $n_1,n_2,...$ such that $C_{n_i}$ converges when restricted to $D(1)$ as $i\to \infty$ .
From this choose a subsequence for which $C_n$ converges when restricted to $D(2)$ and so on.
Show that the limit packing meets the requirements.

For part B: Let $Ç$ be the densest packing of unit discs in a cone.
Choose $P$ and $P'$ to be circular disc and a regular triangle respectively.

Can you kindly help me to understand this (I'm a beginner in this field).

1.(https://i.stack.imgur.com/2k1mz.jpg)

2.(https://i.stack.imgur.com/ZEdCp.jpg)

3.(https://i.stack.imgur.com/fhhvM.jpg)

4.(https://i.stack.imgur.com/WIVsz.jpg)

5.(https://i.stack.imgur.com/B2O4m.jpg)

6.(https://i.stack.imgur.com/ZuXvX.jpg)

Many thanks.

答:
EDIT: I'm leaving this post as context. See my comment to Usermat about my mistake.

For A:

I assume you have verified all of the results on the pages, or else are comfortable using them.

Note that page 5 gives us $$\delta(P) = \max_{P}d(P,\mathbb{R}^2)$$

for $P$ a packing. Hence for any packing $P$, it follows that $d(P,\mathbb{R}^2) \leqslant \delta(P)$. We wish to find a packing $X$ which satisfies $d(X,\mathbb{R}^2) \geqslant \delta(X)$, which will prove the result. This implies we need an optimal packing, which the textbook has supplied us with in the form of the packing in Figure 3.1. After all, the result of Thue says that $\overline{d}(P,\mathbb{R}^2) \leqslant \pi/\sqrt{12}$ for any packing, and so $d(P,\mathbb{R}^2) \leqslant d(X,\mathbb{R}^2) = \pi/\sqrt{12}$ for the greatest density packing $X$. Then $\max_Pd(P,\mathbb{R}^2) \leqslant d(X,\mathbb{R}^2)$ follows from this inequality holding for all $P$ packings, which is the reverse inequality we sought to show.

The second result of A follows from Corollary 3.6.

EDIT: This was a proof for discs in the normal usage of the term. The response to my comment in the post has a convex disc defined as a convex compact set with nonempty interior, so the proof does not strictly apply. In particular, the step where we show that $X$ is maximal no longer counts, because I chose a circle packing.

In general, I think that a packing of centrally symmetric hexagons $H$ will suffice. In fact, as this tiles the plane then the density will be 1, hence is easily seen to be maximal.

SECOND EDIT (In response to OP): Here is how I would write the proof.

Let $H$ be a collection of centrally symmetric hexagons which tile the plane. We wish to show:

$$d(H,R^2) = \delta(H)$$

which we will demonstrate via double inclusion (this method has many names).

The first direction is to show $d(H,R^2) \leqslant \delta(H)$. Indeed, it can be shown that $\delta(H) = \max d(H, R^2)$, where the maximum ranges over all packings with the elements of our packing. As our particular packing $H$ is one such packing, its density is at most $\delta(H)$.

The second direction is to show $d(H, R^2) \geqslant \delta(H)$. Here the choice of $H$ is relevant (notice that it was not used in the first direction), and in fact since $H$ is a plane tiling, the density of $H$ is $1$. It is easy to see that $d(Q,R^2)$ for any packing $Q$ is bounded above by $1$, hence $\delta(Q) \leqslant 1$ for any such collection of convex discs. Picking $H$ in particular, it follows that $\delta(H) \leqslant 1 = d(H, R^2)$.

Then as $d(H,R^2) \leqslant \delta(H) \leqslant d(H,R^2)$, by double inclusion we have what we wished to prove.

For a lattice packing, we note that since the elements of $H$ regularly tile the plane, the packing $H$ forms a lattice packing as well, and by Corollary 3.6, $\delta(H) = \delta_L(H)$. Hence:

$$d(H,R^2) = \delta(H) = \delta_L(H)$$

which is what we had to show.

Title: Unique question about packing problem


问:
A hard inequality (for me)
For $n \in N$; $n \geq 2$ and $x \in R$

Prove:
$-(1+x^2)^n \leq (2x)^n+(1-x^2)^n \leq (1+x^2)^n$

I have no clue to do.
Help please, thank you so much

答:
Inequality $(2x)^n + (1-x^2)^n \le (1+x^2)^n$ is equivalent to
$$(\frac{2x}{1+x^2})^n + (\frac{1-x^2}{1+x^2})^n  \le 1.$$
Put $x = \tan \frac{t}2$. We have to prove that
$$sin^n t + \cos^n t \le 1.$$

As $|\sin t|, |\cos t| \le 1$ we have
$$sin^n t + \cos^n t \le |sin^n t| + |\cos^n t| \le |sin^2 t| + |\cos^2 t| = 1.$$

It's easy to show that $(2x)^n + (1-x^2)^n \ge -(1+x^2)^n$ by the same argument.

Title: A hard inequality (for me)


问:
On Lyapunov stability of a solution
I have recently encountered the following exercise: prove or disprove that the solution of the ODE $\dot x = -\frac{2x}{t} + t^2$ that satisfies $x(1)=1$ is Lyapunov stable. I have proved that the solution is stable, however, the correct answer (as per my teacher) is the opposite.

My proof is as follows:

$$\begin{aligned}
\dot x &= -\frac{2x}{t} + t^2 \\
t^2\dot x + 2xt &= t^4 \\
t^2\dot x + x(t^2)' &= t^4 \\
(t^2 x)' &= t^4 \\
t^2 x &= \frac{t^5}{5} + C \\
x &= \frac{t^3}{5} + \frac{C}{t^2} \\
\end{aligned}$$

The solution that satisfies $x(1)=1$ is $x=\varphi(t)=\cfrac{t^3}{5} + \cfrac{4}{5t^2}$.

The definition of Lyapunov stability:
$$\forall \varepsilon > 0 \ \ \exists \delta > 0 \ \ |x(t_0) - \varphi(t_0)|< \delta \implies \forall t > t_0 \ \ |x(t) - \varphi(t)| < \varepsilon$$
$$|x(t_0) - \varphi(t_0)|< \delta \implies \left|\frac{1}{5} + \frac{C}{t^2} - \frac{1}{5} - \frac{4}{5}\right| < \delta \implies \left|C - \frac{4}{5}\right| < \delta$$
$$|x(t) - \varphi(t)| < \varepsilon \implies \left|\frac{t^3}{5} + \frac{C}{t^2} - \frac{t^3}{5} - \frac{4}{5t^2}\right| < \varepsilon \implies \left|\frac{1}{t^2}\left(C-\frac{4}{5}\right)\right| < \varepsilon$$
Let $\delta=\varepsilon$. Then
$$\left|\frac{1}{t^2}\left(C-\frac{4}{5}\right)\right| < \varepsilon \implies \varepsilon\left|\frac{1}{t^2}\right| < \varepsilon$$
, which holds.

答:
You obtained the following equation
$$\frac{d}{dt}(t^2 x)  = t^4 \quad (1) $$
then you found the following general solution:
$$t^2 x(t) = \frac{t^5}{5} + C $$
which is also true, but you can not treat $C$ as an initial condition of the original differential equation. To use the defintion of stability, you sould integrate both sides of Equaiton (1) form $t_0$ to $t\geq t_0$. Integrating (1) gives:

$$t^2x(t)-t_0^2x(t_0)=\frac{1}{5}t^5-\frac{1}{5}t_0^5$$

or equivilanetly:
$$x(t)=\frac{t_0^2}{t^2}x(t_0)+\frac{1}{5}t^3-\frac{1}{5t^2}t_0^5$$

To prove the stability, the following should hold for any   $\epsilon$ .....
$$\|x(t)-x(t_0)||=\bigg\|(\frac{t_0^2}{t^2}-1)x(t_0)+\frac{1}{5}t^3-\frac{1}{5t^2}t_0^5\bigg\|\leq\epsilon \qquad (2)$$
But if there exists at least one $\epsilon$ such that Equaiton (2) is not satisfied for all initial conditions then the system is not stable. Slecet $\epsilon =1$, then one can see that $\bigg\|(\frac{t_0^2}{t^2}-1)x(t_0)+\frac{1}{5}t^3-\frac{1}{5t^2}t_0^5\bigg\|$ is not bounded because of $t^3$.

Title: On Lyapunov stability of a solution


答: Alternatively, using what you found in OEIS, take $\{A,B\}$ with $A\cap B=\emptyset,$ $A,B\neq \emptyset$ and consider $C=[n]\setminus (A\cup B)$ counted by the expression. Now, consider mapping that set to $\{A\cup C,B\cup C\},$ check that the properties that you required are satisfied and, moreover, this mapping is a bijection. So both problems have the same number of solutions.


问:
Solving a Recursive Relation using Generating Functions
Consider the following recursive relation.
$$a_0=0$$
$$a_i = (1-p)a_{i-1} + pa_{i+1} \text{ }\forall \text{ natural numbers }i \text{ }\epsilon\text{ } [1, m - 1] $$
$$a_m=1$$
$$0<p<1$$
I'm trying to find an explicit formula for $a_i$. I have tried using generating functions to do so.

Define $A(x) = \sum_{k=0}^{m} a_k x^k$. Now,
$$\sum_{k=1}^{m-1}[a_i-(1-p)a_{i-1}-pa_{i+1}]x^k=0$$
$$\implies\sum_{k=1}^{m-1}a_kx^k - (1-p)x\sum_{k=1}^{m-1}a_{k-1}x^{k-1}-px^{-1}\sum_{k=1}^{m-1}a_{i+1}x^{i+1}=0$$
$$\implies[A(x) - a_0-a_mx^m]-(1-p)x[A(x)-a_mx^m-a_{m-1}x^{m-1}]-px^{-1}[A(x)-a_1x-a_0]=0$$
$$\implies [1-(1-p)x-px^{-1}]A(x) - x^m+(1-p)x^{m+1}+(1-p)a_{m-1}x^m+a_1p=0$$
$$\implies A(x) = \frac{(1-p)x^{m+2}-pa_{m-1}x^{m+1}+a_1px}{(1-p)x^2-x+p}$$
I am unsure of how to proceed from here. Kindly help.

答:
The recursion is of second order.

Going through a generating function,
when you put
$$
A(x) = \sum\limits_{k = 0}^m {a_{\,k} x^{\,k} }
$$
you are assuming that $a_k=0 $ for $k<0$ , which is due for unilateral transform, but you are also taking
$a_k=0 $ for $m<k$ . Together with $a_0 =0, \; a_m =1$ you are going to impose too many initial conditions.

So let's reconduct the recursion onto standard track.

We put
$$
a_{\,k}  = b_{\,k + 1} \quad \left| {\;b_{\,m < 0}  = 0,\;\,b_{\,0}  = b_{\,0} ,\;\;b_{\,1}  = b_{\,1} } \right.
$$
so that the recursion becomes
$$
pb_{\,k}  - b_{\,k - 1}  + qb_{\,k - 2}  = 0
$$
Now we add the initial conditions such as to make the recursion valid for all the values of $k$
$$
pb_{\,k}  - b_{\,k - 1}  + qb_{\,k - 2}  - pb_{\,0} \left[ {k = 0} \right] - \left( {pb_{\,1}  - b_{\,0} } \right)\left[ {k = 1} \right] = 0\quad \left| {\;\forall k} \right.
$$
where $[P]$ denotes the Iverson bracket

Then we put
$$
B(x) = \sum\limits_{0\, \le \,k} {b_{\,k} x^{\,k} }
$$
to obtain
$$
\eqalign{
  & 0 = p\sum\limits_{0\, \le \,k} {b_{\,k} x^{\,k} }  - \sum\limits_{0\, \le \,k} {b_{\,k - 1} x^{\,k} }
  + q\sum\limits_{0\, \le \,k} {b_{\,k - 2} x^{\,k} }  - pb_{\,0} \sum\limits_{0\, \le \,k} {\left[ {k = 0} \right]x^{\,k} }
  - \left( {pb_{\,1}  - b_{\,0} } \right)\sum\limits_{0\, \le \,k} {\left[ {k = 1} \right]x^{\,k} }  =   \cr
  &  = p\sum\limits_{0\, \le \,k} {b_{\,k} x^{\,k} }  - \sum\limits_{0\, \le \,k} {b_{\,k} x^{\,k + 1} }
  + q\sum\limits_{0\, \le \,k} {b_{\,k} x^{\,k + 2} }  - pb_{\,0} x^{\,0}  - \left( {pb_{\,1}  - b_{\,0} } \right)x^{\,1}  =   \cr
  &  = pB(x) - xB(x) + qx^{\,2} B(x) - pb_{\,0}  - \left( {pb_{\,1}  - b_{\,0} } \right)x\quad  \Rightarrow   \cr
  &  \Rightarrow \quad B(x) = {{pb_{\,0}  + \left( {pb_{\,1}  - b_{\,0} } \right)x} \over {qx^{\,2}  - x + p}}
 = {{pb_{\,0}  + \left( {pb_{\,1}  - b_{\,0} } \right)x} \over {q\left( {x - s - c} \right)\left( {x - s + c} \right)}} =   \cr
  &  = {R \over {\left( {x - s - c} \right)}} + {T \over {\left( {x - s - c} \right)}} \cr}
$$
and then adjust to make $b_1 = a_0 = 0$ and $b_0$ in such a way as to make $b_{m+1} = a_m =1$.

As already suggested by @TomTom the best way would be to go through a matrix relation
$$
\left( {\matrix{
   {a_0 }  \cr
   {a_1 }  \cr
   {a_2 }  \cr
    \vdots   \cr
   {a_{m - 1} }  \cr
   {a_m }  \cr
 } } \right) = \left( {\matrix{
   1 & {} & {} & {} & {} & {}  \cr
   q & 0 & p & {} & {} & {}  \cr
   {} & q & 0 & p & {} & {}  \cr
   {} & {} &  \ddots  &  \ddots  &  \ddots  & {}  \cr
   {} & {} & {} & q & 0 & p  \cr
   {} & {} & {} & {} & {} & 1  \cr
 } } \right)\left( {\matrix{
   {a_0 }  \cr
   {a_1 }  \cr
   {a_2 }  \cr
    \vdots   \cr
   {a_{m - 1} }  \cr
   {a_m }  \cr
 } } \right)\quad {\bf a} = {\bf M}\;{\bf a}
$$
which means that
$$
\left( {{\bf M} - {\bf I}} \right){\bf a} = {\bf 0}\quad  \Rightarrow \quad
 {\bf a} \in {\rm nullspace}\left( {{\bf M} - {\bf I}} \right)
$$
and since the rank of $\left( {{\bf M} - {\bf I}} \right)$ is $m-1$, $\bf a$ is the linear combination of two vectors,
with coefficients to be determined by the initial conditions.

But $\bf M$ is a Markov matrix with two absorbing states at $0$ and $m$.

Therefore if the initial conditions are $a_0 = 0, \; a_m =1$ you  end up with a steady state (the recurrence solution)
which is $$a_0 =0, \, a_1 =0, \, a_2 =0,\, \cdots , \, a_{m-1} =0, \, a_m =1$$.


问:
How to find the midpoint between 2 point in the number-circle?
There is a number circle numbered from 0..N and have 2 points in the circle, where I need to find the middle point between 2 points on the circle where the the middle point will be on the circle and the distance should be minimal. In the below image the answer is between N-1 and 0 = N-0.5 (as integer it's N-1).

Is there any formula or algorithm to find the middle point? (I need the answer to solve a programming/algorithm problem)

答:
I believe you can calculate $m = \dfrac{(a+b)}{2}$ to get one of the midpoints, then add $\dfrac{N}{2}$ and mod $N$ again to get the other solution, then you can just check which solution is closer.

Edit: correction from comments


问:
Landmarks of subjects of mathematics
In order to have a good view of the whole mathematical landscape one might want to know a deep theorem from the main subjects (I think my view is too narrow so I want to extend it).

For example in natural number theory it is good to know quadratic reciprocity and in linear algebra it's good to know the Cayley-Hamilton theorem (to give two examples).

So, what is one (per post) deep and representative theorem of each subject that one can spend a couple of months or so to learn about? (In Combinatorics, Graph theory, Real Analysis, Logic, Differential Geometry, etc.)

答: In Complex Analysis the Riemann Mapping Theorem.</p>

Title: Landmarks of subjects of mathematics


问:
How can I prove that $\sin (10^\circ), \sin(1^\circ), \sin(2^\circ), \sin(3^\circ), \tan(10^\circ)$ are irrational

Prove that $\sin (10^\circ)$, $\sin(1^\circ)$, $\sin(2^\circ)$, $\sin(3^\circ)$, and $\tan(10^\circ)$ are irrational.

My Attempt:

Let $x = 10^\circ$. Then

$$
\begin{align}
x &= 10^\circ\\
3x &= 30^\circ\\
\sin (3x) &= \sin (30^\circ)\\
3\sin (10^\circ)-4\sin^3(0^\circ) &= \frac{1}{2}
\end{align}
$$

Now let $y = \sin (10^\circ)$. Then

$$
\begin{align}
3y-4y^3 &= \frac{1}{2}\\
6y-8y^3 &= 1\\
\tag18y^3-6y+1 &= 0
\end{align}
$$

How can I calculate the roots of $(1)$?

答:
One can use a feature of complex numbers, and the span of a finite set.

Consider the set of cyclotomic numbers, ie $C(n) = \operatorname{cis}(2\pi/n)^m$, where $\operatorname{cis}(x)=\cos(x)+i\sin(x)$.  Such a set is closed to multiplication.  The 'Z-span' of the set is the set of values $\sum(a_m \operatorname{cis}(2\pi/n)$, over n, is also closed to multiplication.

We now begin with the observation that a span of a finite set, closed to multiplication, can not include the fractions.  This is proved by showing that if a rational number, not an integer, is in the set, so must all of its powers. (ie if $1/2$ is constructible by steps at multiples of $N°$, (eg a random walk of unit-size steps at exact degrees), so must all values of $1/2^a$).

Since this means that that the intersection of the cyclotomic numbers $\mathbb{C}_n$ and the rationals $\mathbb{F}$ can not include any fractions, and thus must give $\mathbb{Z}$.

The double-cosine of the half-angles, are given by $1-\operatorname{cis}(2\pi/n)$, and therefore we see that the only rational numbers that can occur in the sines and cosines, is $1/2$.  The chord, and the supplement-chords are entirely free of rationals, and further, no product of such numbers can be rational.


问:
$L^p$ norm of tensor
Let $T_{ab}$ be a $(0,2)$ tensor on a Riemannian manifold $M$ with component $T_{ab} = \nabla_a S_b - \nabla_b S_a -
g_{ab} \text{ div $S$}$. Here, $S$ is a 1-form and $g_{ab}$ is the metric tensor. By definition,  the tensor norm of $T_{ab}$ is given by $\lvert T \rvert^2 = T_{ab}T^{ab}$. However, I have some trouble computing the $L^p$ norm of $T$. In particular, I'm wondering if $\lVert T \rVert_p$ is equivalent to $\lVert \nabla S \rVert_p + \lVert \text{ div $S$} \rVert_p$. I tried to express the norm in components but it gets too tedious at some point. Any insight would be appreciated.

答:
By definition,

$$ \|A\|_{L^p} = \left( \int_M|A|^p d\mu\right)^{1/p}$$
and $|T|^2 = T^{ab}T_{ab}$. Define

$$\langle A, B\rangle = A^{ab}B_{ab}, $$
then $|A|^2 = \langle A, A\rangle$ and $\langle \cdot, \cdot \rangle$ is an inner product.

Write $B_{ab} = \nabla_a S_b - \nabla_b S_a$. The calculation in your case is easier: we have
$$\langle B_{ab} ,
g_{ab}\rangle = g^{ab} B_{ab} = 0$$

since $g^{ab}$ is symmetric and $B$ is antisymmetric. Thus

$$|T|^2 = |B_{ab} - g_{ab} \operatorname{div}(S)|^2 = |B_{ab}|^2 + |g_{ab} \operatorname{div}(S)|^2= |B_{ab}|^2 + n|\operatorname{div}(S)|^2$$

This implies
$$|B|^2 + |\operatorname{div}(S)|^2 \le |T|^2 \le n(|B|^2 + |\operatorname{div}(S)|^2). $$

To see a bit more, note that $B= dS$ and $\operatorname{div} S = d^*S$.From here we can see that $|T|$ is not equivalent to $|\nabla S|^2 + |\operatorname{div}S|^2$: Take $S$ to be a non-zero harmonic one form on a compact manifold with negative Ricci curvature (e.g. a surface of genus $g\ge 2$ with hyperbolic metric). Then $dS =d^*S = 0$, but $|\nabla S| \neq 0$. To see why $|\nabla S|^2 \neq 0$, note from the Ricci identity,

\begin{align}
\int_M (|B|^2 + | \operatorname{div}(S)|^2)d\mu &= \int_M (\langle dS, dS\rangle + (d^*S)^2)d\mu\\
&= \int_M ( \langle S , d^*dS \rangle + \langle S, dd^*S\rangle) \\
&= \int_M \langle S , \Delta_d S\rangle d\mu \\
&= \int_M \langle \langle S , -\nabla^* \nabla S\rangle + \langle S, Rc(S)\rangle d\mu \\
&= \int_M |\nabla S|^2 d\mu + \int_M Rc(S, S) d\mu.
\end{align}
If $S$ is harmonic, we have

$$\int_M |\nabla S|^2 d\mu = - \int_M Rc(S, S) d\mu.$$

So if $M$ has negative Ricci curvature then $\|\nabla S\|_{L^2} \neq 0$.

Title: $L^p$ norm of tensor


问:
Prove that a function $f$ is injective if and only if it respects intersection.
Let $f:X \to Y$ be a function. Prove that $f$ is injective if and only if $f(A\cap B)=f(A)\cap f(B),$ for any two sets $A, B \subseteq X.$

答:
Let $f(a)=x = f(b)$ and write $A=\{a\}$, $B=\{b\}$. Then

$\{x\} = f(A) \cap f(B) = f(A\cap B)$ by hypothesis.

This means that $A\cap B\ne \emptyset$ and $A\cap B = \{a\}$ with $a=b$ as required.


问:
showing that $\lim_{x\to 0}\frac{2x\sin(\frac{1}{x})-\cos(\frac{1}{x})}{x}$ does not exist
$$\lim_{x\to 0}\frac{2x\sin(\frac{1}{x})-\cos(\frac{1}{x})}{x}=\lim_{x\to 0}2\sin(\frac{1}{x})-\frac{\cos(\frac{1}{x})}{x}$$ ... I know that $$\lim_{x\to 0}2\sin(\frac{1}{x})$$ and $$\lim_{x\to 0}\frac{\cos(\frac{1}{x})}{x}$$ do not exist, how can I argue then that $\lim_{x\to 0}\frac{2x\sin(\frac{1}{x})-\cos(\frac{1}{x})}{x}$ does not exist?

答: If $x=\frac1{2n\pi}$, for some $n\in\Bbb N$, then$$\frac{2x\sin\left(\frac1x\right)-\cos\left(\frac1x\right)}x=-2n\pi\to_{n\to\infty}-\infty,$$and if $x=\frac1{(2n-1)\pi}$, for some $n\in\Bbb N$, then$$\frac{2x\sin\left(\frac1x\right)-\cos\left(\frac1x\right)}x=(2n-1)\pi\to_{n\to\infty}\infty.$$Therefore, your limit doesn't exist.


问:
How do I find the transitive closure of a relation and the partition?
Set =
{a,b,c,d,e}

R = {(a, c),(b, d),(c, e),(d, b),(e, a)}

Q:Determine the transitive closure R∗ of R. Express your answer as a set of ordered pairs.

Attempt =

```<code>{a,c}{c,e} = a,e
{a,e}{e,a} = a,a
{b,d}{d,b} = b,b
{c,e}{e,a} = {c,a}
{e,a}{a,c} = {a,c}

Transitive Closure = (a,e), (a,a), (b,b), (c,a), (a,c)
</code>```

I'm not sure this is right though

Q2 is to find the partition of the set A into equivalence classes induced by R∗

My attempt for this is  {a,c,e},{b},{d}

Not sure though

答:
You missed $(c,c),\ (d,d),\ (e,e)$ and $(e,c)$ and you also need to include the original pairs of $R$ in the transitive closure.

You can systematically write up all the paths, on each possible starting point.

The equivalence classes are $\{a,c,e\}$ and $\{b,d\}$.