问:
Constructing a CDF from a discontinuous PDF
I have a PDF that looks like the following: .25 for $0 \leq  x< .5$ and .75 for $.5 \leq x \leq 1$ (and 0 elsewhere). How do I construct a CDF from this discontinuous PDF? My instinct is to say that it should be $.25x$ for $0 \leq  x< .5$ and $.75x$ for $.5 \leq x \leq 1$, but when I integrate that over (0,1), I get a very small number (<.2) -- so I think that it must be incorrect. Any hints?

答:
I'm not sure the PDF you have specified is valid. To the best of my knowledge, a PDF must satistfy the constraint that $\int^{\infty}_{-\infty} f(x) dx = 1$ for a continuous r.v.

Your PDF is of the form:

$$f_X(x) =
\begin{cases}
0.25 \quad & 0 \leq x \leq 0.5 \\
0.75 \quad & 0.5 \leq x \leq 1 \\
0 \quad &\text{otherwise} \\
\end{cases}$$

Integrating the PDF where it is non-zero, we have that:

$$\int^{\infty}_{-\infty} f_X(x) dx = \int^{0.5}_0 0.25 dx + \int^{1}_{0.5} 0.75 dx = \frac{1}{8} + \frac{3}{8} = \frac{1}{2} \neq 1$$

So this isn't a valid PDF.


问:
Subobject Classifier of a Topos is Injective
Let $\mathcal{E}$ be a topos with subobject classifier $1\overset{t}{\rightarrow}\Omega$, then I want to show that $\Omega$ is injective.

So let $f:A\rightarrow\Omega$ and $g:A\rightarrow B$ be two maps in $\mathcal{E}$ with $g$ monic. Then I want to show that there exists a map $h:B\rightarrow\Omega$ such that $h\circ g = f$. Notice that by the property of a subobject classifier we have that there exists a unique map $\phi:B\rightarrow\Omega$ such that there is a pullback square
$\require{AMScd}$
\begin{CD}
A @>{\psi}>> 1\\
@VV{g}V @VV{t}V\\
B @>{\phi}>> \Omega
\end{CD}
Now I claim that $\phi$ does the job for $h$. But I don't see why $\phi\circ g= f$ (so maybe my claim is wrong). Any help would be appreciated!

答:
Thinking about maps to $\Omega$ as subobjects, this is just saying that if you have a subobject $C\subseteq A$, then there is a subobject $D\subseteq B$ such that $C=A\cap D$ as subobjects of $B$.  How do you find such a $D$?  Well, you can just take $D=C$!

To make this precise, let $i:C\to A$ be the subobject classified by $f$.  Then $gi:C\to B$ is monic since $i$ and $g$ are, so it is classified by a map $h:B\to\Omega$.  Now in the diagram
$$\require{AMScd}
\begin{CD}
C @>{1}>> C @>{}>> 1\\
@V{i}VV @VV{gi}V @VV{t}V\\
A @>{g}>> B @>{h}>> \Omega
\end{CD}
$$
the right square is a pullback by definition of $h$ and the left square is a pullback since $g$ is monic. This implies the outer rectangle is a pullback, i.e. that $hg$ classifies the subobject $i:C\to A$, so $hg=f$.

ContentLicense: CC BY-SA 4.0


问:
Question on a triple integral
I have a triple integral $\iiint_E \sqrt{16-x^2-z^2}dV$, where $E$ is the region above the $xy$-plane, outside the cylinder $x^2+z^2=4$, and inside the sphere $x^2+y^2+z^2=16$.

I rewrite the region $E$ by using cylindrical coordinates:
$$x=r\cos(\theta),\quad y=y,\quad z=r\sin(\theta).$$
Then it turns out that $E$ can be represented by
$$E=\{(r,y,\theta): 0\leq \theta\leq \pi,\quad 2\leq r\leq 4,\quad -\sqrt{16-r^2}\leq y\leq \sqrt{16-r^2}\},$$
or
$$E=\{(r,y,\theta): 0\leq \theta\leq \pi,\quad 2\leq r\leq \sqrt{16-y^2},\quad -2\sqrt{3}\leq y\leq 2\sqrt{3}\}.$$
With the first representation, I got
$$\begin{aligned}&\iiint_E \sqrt{16-x^2-z^2}dV=\int_0^{\pi}\int_2^4\int_{-\sqrt{16-r^2}}^{\sqrt{16-r^2}}r\sqrt{16-r^2}dydrd\theta\\
=&\int_0^{\pi}\int_2^42r(16-r^2)drd\theta=\pi\int_2^42r(16-r^2)dr=\left.-\frac{\pi}{2}(16-r^2)^2\right|_{r=2}^{r=4}\\
=&-\frac{\pi}{2}(0-144)=72\pi.
\end{aligned}$$
However, with the second representation, I got
$$\begin{aligned}&\iiint_E \sqrt{16-x^2-z^2}dV=\int_{-2\sqrt{3}}^{2\sqrt{3}}\int_0^{\pi}\int_2^{\sqrt{16-y^2}}r\sqrt{16-r^2}drd\theta dy\\
=&\int_{-2\sqrt{3}}^{2\sqrt{3}}\int_0^{\pi}\left.-\frac{1}{3}(16-r^2)^{3/2}\right|_{r=2}^{r=\sqrt{16-y^2}}d\theta dy=\int_{-2\sqrt{3}}^{2\sqrt{3}}\int_0^{\pi}\frac{1}{3}(12^{3/2}-y^3)d\theta dy\\
=&\frac{\pi}{3}\int_{-2\sqrt{3}}^{2\sqrt{3}}24\sqrt{3}-y^3dy=\frac{\pi}{3}\int_{-2\sqrt{3}}^{2\sqrt{3}}24\sqrt{3}dy=\frac{\pi}{3}(24\sqrt{3})(4\sqrt{3})=96\pi.
\end{aligned}$$

Two different methods, two different answers. I am confused about which one is correct, and why the other one is wrong. Any suggestions?

答:
In your computations by the second method, you used (implicitely) that $(y^2)^{3/2}=y^3$, but this holds only if $y\geqslant0$.

So, instead of computing$$\int_{-2\sqrt3}^{2\sqrt3}\int_0^\pi-\frac13\left.(16-r^2)^{3/2}\right|_{r=2}^{r=\sqrt{16-y^2}}\,\mathrm d\theta\,\mathrm dy,$$compute$$2\int_0^{2\sqrt3}\int_0^\pi-\frac13\left.(16-r^2)^{3/2}\right|_{r=2}^{r=\sqrt{16-y^2}}\,\mathrm d\theta\,\mathrm dy,$$and then you will get $72\pi$.

Title: Question on a triple integral


问:
How to prove that a collection of functions is uniformly absolutely continuous if $\sup_n \int |f_n|^{1+\gamma}\ d\mu < \infty$?
Suppose $f_1, f_2, \dotsc$ is a collection of measurable functions which satisfy $\sup_n \int |f_n|^{1 + \gamma}\ d\mu< \infty$ for some $\gamma > 0$, and $\mu$ is a finite measure. I am being asked to show that the $\{f_n\}$ are uniformly absolutely continuous. That is, for each $\epsilon > 0$ there exists a $\delta > 0$ such that for all $A$ with $\mu(A) < \delta$ $$\int_{A} |f_n|\ d\mu < \epsilon$$ for all $n \geq 1$.

I'm not sure how to use the fact that $\sup_n \int |f_n|^{1 + \gamma}\ d\mu< \infty$. If I assume only that $\sup_n \int |f_n|\ d\mu< \infty$, it seems I can prove something weaker:

Suppose $A_1, A_2, \cdots$ is any sequence of sets for which $\mu(A_n) \to 0$ as $n \to \infty$, and let $m \geq 1$. Since $f_m \cdot 1_{A_n} \to 0$ a.e. and $|f_m \cdot 1_{A_n}| \leq |f_m|$ and $\int |f_m| < \infty$, the dominated convergence theorem implies that $\int_{A_n} |f_m| \to 0$ as $n \to \infty$.

But I'm not sure where to go from here. I think this can be done by first proving that the $f_n$ are uniformly integrable, but I would like to find a direct proof if possible.

答:
Let $p=1+\gamma$. Then $\frac1q=\frac{p-1}{p}=\frac{\gamma}{1+\gamma}$ so $q=\frac{1+\gamma}{\gamma}$. By Holder's inequality,

$\int_A |f_n|\,d\mu \leq \|f_n\|_p\|1\|_q=\left[\int_A |f_n|^{1+\gamma}\,d\mu\right]^{\frac{1}{1+\gamma}}\left[\mu(A)\right]^{\frac{\gamma}{1+\gamma}}\leq \left[\sup_n\int_X |f_n|^{1+\gamma}\,d\mu\right]^{\frac{1}{1+\gamma}}\left[\mu(A)\right]^{\frac{\gamma}{1+\gamma}}  .$
It should be clear how to pick $\delta(\epsilon)$ so that the above is bounded above by $\epsilon$.


问:
Linear iterability for set models of $\mathsf{ZFC}$
As I found some problems in my writing and proof, I have heavily edited my questions.

So I am reading this corollary in Steel's notes, found here, and I am confusing myself over and over again. I hope someone can clarify the situation for me. First some context:

When I say $\omega$-complete ultrafilter, I mean that the intersection of countable sequences of members of that ultrafilter are nonempty. And we may assume that in $\mathsf{ZFC}$, we have linear iterability for normal ultrafilters. Also whenever $M \models \mathsf{ZFC} + `` \mathcal{E} \mbox{ is a family of normal ultrafilters''}$ and $I = \langle U_\alpha: \alpha \lt \lambda\rangle$ is an iteration of $(M, \mathcal{E})$, then $M^I_\infty$ denotes the direct limit of $I$ and $i_{\alpha\infty}$ denote the direct limit embeddings.

Now this is corollary $2.8$ of the above notes, together with the following exercise:

Corollary $2.8.$ Let $M \models \mathsf{ZFC} + `` \mathcal{E} \mbox{ is a family of normal ultrafilters''}$, with $M$ transitive, and $\omega_1 \in M$. Then $(M, \mathcal{E})$ is linearly iterable.

Proof. Working inside $M$, construct a $``\mbox{universal''}$ linear iteration $I = \langle U_\alpha: \alpha \lt \lambda\rangle$ with $(a)$ $\operatorname{cof}(\lambda) = \omega$, and $(b)$ whenever $W \in i_{0\alpha}(\mathcal{E})$, then $i_{\alpha\beta}(W) = U_\beta$ for cofinally many $\beta$. This implies that every $W \in i_{0\infty}(\mathcal{E})$ is $\omega$-complete. Thus $(M^I_\infty, i_{0\infty}(\mathcal{E}))$ is linearly iterable in $V$. Since $i_{0\infty}:M\rightarrow M^I_\infty$, $(M, \mathcal{E})$ is linearly iterable in $V$. $\Box$

Exercise. Prove that every $W \in i_{0\infty}(\mathcal{E})$ is $\omega$-complete, granted $(a)$ and $(b)$.

Questions:

Now I have partially worked my way through the corollary and have constructed an $I$ with some difference to what is suggested and I think I do understand what's going on, except for the questions I have below:

Firstly, I can only complete the proof with the crucial assumption that $M^I_\infty$ is actually well-founded in $V$. But I don't see how we can show this. It is well-founded from $M$'s perspective, but $V$ may not agree. So what can we say about this?

Secondly, I have proved the exercise but with one key difference. When building $I$, I made sure that $\operatorname{cof}(\lambda) = \omega_1$, instead of the suggested $\omega$! That's because when looking at our sequences from $^\omega W$ like $\langle A_n: n \in \omega \rangle$, we can bound their representatives given by $(b)$ in the iteration and then find some $\langle B_n: n\in\omega\rangle$ and $\alpha$, such that $i_{\alpha\infty}(B_n)=A_n$ and $B_n \in U_\alpha$ for each $n$. Then by normality we would have $i_{0\alpha}(\kappa) \in i_{\alpha{\alpha+1}}(B_n)$ for each $n$ and then we would have $i_{{\alpha+1}\infty}(i_{0\alpha}(\kappa))\in A_n$ for every $n$; and so $i_{{\alpha+1}\infty}(i_{0\alpha}(\kappa))\in \bigcap_n A_n$.

And as you can see from the above proof, I think $\operatorname{cof}(\lambda) = \omega$ is probably a typo right? Also the thing I have written is the only way I can think of that we can use the hypothesis $\omega_1\in M$.

I would really appreciate any thoughts and help, so I can make some progress. Also, sorry for the long post.

答:
After consulting with a friend on twitter, we came to the following conclusion for the first question:

As $M \models \mathsf{ZFC}$, then each ultrapower is contained within $M$ by the virtue of $M \models \mathsf{Powerset}$. And in general, models of $\mathsf{ZFC}$ trap all iterations in themselves(unless of course when the length of the iteration is $\ge \operatorname{Ord}^M$). And since $I$ is iterated in $M$, $M$ can see $M^I_\infty$ as a subclass! At which point it must be well-founded.

This solves most of my confusion. I would still welcome any confirmation about my second question.


问:
Philosophy of a Mathematician.
Introduction

I don't study Mathematics at university, and probably I don't have any chances to have a little understanding of what mathematics in all its aspects.

But I love to find structures and links betwen ideas, and to ask myself philosophical questions about foundations of mathematics .If I have understood well is what is happening with the "categorical" point of view: there is some kind of "great unification" of the mathematics from a structural point of view.

Obviously I understand that for humans it is impossible to reach a total understanding of something, and that moves me is a mirage. I must give up until I'm sane and study modestly mathematics, in a relaxed way: playing with my easy problems, reading books, and learning from the basics all I can, without eccesive expectations maybe.

More I understand and more new interesting mathematical topics I discover... but the facts that overwhelm me are the acceleration of this process and the fact that there are a lot of young theories; and that this implies that this growth will be even faster.

Probably I, as an amateur mathematician, don't have a chance to come at new ideas in the foundational research (or even understand the actual point of view since it is changing so fast)...

Questions

$\mathcal Q_1$ From your($*^1$) experience how can mathematicians deal (or how they usually chose to behave) with this huge mathematical universe and be satisfied from a philosophical($*^2$)  point of view?

$\mathcal Q_2$ How can a mathematician interested in the foundations of mathematics be satisfied by an always partial knowledge of the mathematics?

Personal observation and conclusion

Is maybe possible that in 100 years tha amount of mathematical knowledge will be so huge that even Mathematicians will be totally overwhelmed from it? In other words, is possible that will be impossible for anyone to research new things because the knowledge required will take like 80 or 90 years of hard study? I say this because I think that humans has a limit to the speed of learning.

If this is possible,  is in the destiny of mathematicians to abandon the hope of understanding of mathematics?

I remember a quote of John von Neumann:

"Young man, in mathematics you don't understand things. You just get used to them."

I have read many stories about his math skills, and how he was a genius... anyways from my point of view... I feel very sad when I read this quote.

Notes

($*^1$) I ask this question HERE on SEMath because I'm looking for human experiences of real mathematicians, that is very important for me in this moment, and I do not think is fair that this question was closed in 40 seconds...is a soft question, as many others questions, but it has philosophical contenents too and I think it deserves at least a chance.

I must quote Andres Caicedo that was able to express way better than me, maybe, the meaning of my question:

"I think that there may be insights that only working mathematicians
  could provide (as opposed to philosophers), and even if there are
  wildly differing points of view, seeing them described may be useful.
  [...] Of course, it may be that answering the question in detail,
  considering as many of its subtleties as possible would just be too
  long and unfeasible. That's fine; even providing a few references and
  ideas that can potentially be fleshed out would be more useful that
  simply dismissing it."

Improved version 05/14/'13:

($*^2$) Following the advice of Gustavo Bandeira I explain better the meaning of "satisfied from a philosophical point of view" in my fisrt question; What I mean is only linked to the mathematicians interested in the foundations, in other words who is interested in that process of generalization of all the mathematics to easier concepts ( like "$\in$" for set theorists or the structuralists for example)

Here the same question on Phylosphy.SE

How can a mathematician interested in the foundation of mathematics
  be satisfied by an always partial knowledge of the mathematics?

答:
I'm a student of mathematics at the third year undergraduate level.

Lets begin with what's unsatisfying. Example are when:

you perceive that a branch of math "isn't set up right", or

a theorem is clumsily worded or clumsily proved, possibly because the writer hasn't had enough exposure to adjacent fields, or

an author's definition is suboptimal for ease of expression, or

terminology/notation in the literature is inconsistent and there seems to no way to fix the problem - its like you can't win,

etc.

On the other hand, the existence of a huge mathematical universe is not unsatisfying. It's wondrous. I love the fact that there's always more to explore, always harder problems to face...  in fact, at any level of difficulty, there's always progress to be made - its like a video game set in an enormously massive world that is just full of secrets, and it just stretches forever.

It can become pretty addictive. I'll leave you with the following quote.

Yeah, I used to think it was just recreational... then I started doin'
  it during the week... you know, simple stuff: differentiation,
  kinematics. Then I got into integration by parts... I started doin' it
  every night: path integrals, holomorphic functions. Now I'm on
  diophantine equations and sinking deeper into transfinite analysis.
  Don't let them tell you it's just recreational.

Fortunately, I can quit any time I want.

ContentLicense: CC BY-SA 3.0

Title: Philosophy of a Mathematician.


问:
is quantifier domain tied to predicate argument
I have trouble understanding the domains of predicate quantifiers.

P(x) =“x knows calculus”.

Domain for x is all rabbits.

$ \forall x\neg P(x) $

To me it seems like for all rabbits no all rabbits knows calculus. Which makes no sense. So if I were to provide an actual value for the predicate P(pygmy rabbit) the statement would not make sense. So the actual x is merely the domain I suppose.

If we were to make something that could actually provide some different outcomes based on the inputs

P(x) =“x is a mammal”.

Domain for x a set of animals

$ \forall x P(x) $

let x be $\{ Rabbit, Human, Whale \}$

$ \forall x P(x) \equiv true$

The above observations leads me to believe that the predicate variable has to be presented as an argument inside of the predicate, I would like for someone to clarify this for me

答:
A One variable predicate of the form $ P(x) $ is a sentence, not a proposition, which depends on $ x $. The parameter $ x $ lies in a certain set  $ E $.
To become a proposition, we must either specify the value of $ x $, or use quantifiers.

Example :

$$E=\Bbb R\;\; P(x)\;:\; x>3$$

$x>3$ is not a proposition.

If $ x=2 $, $ P(x) $ is a false proposition.

$(\forall x\in \Bbb R) \; P(x)$ is a false proposition.

$(\exists x\in \Bbb R)\;:\; P(x)$ is a true proposition.


答:
I don’t understand what you mean by your last paragraph, but you do seem to have some misunderstandings about domains and the translation of $\forall x\big(\neg P(x)\big)$ into English.

The domain in your first example is the set of all rabbits. The expression $\forall x\big(\neg P(x)\big)$ means for any rabbit it is not the case that that rabbit knows calculus; in ordinary English this is simply no rabbit knows calculus, there is no rabbit that knows calculus, or the like. Note that pygmy rabbit is not in the domain: it is not a single specific rabbit.

Similarly, in your second example Rabbit, Human, and Whale are not in the domain: they are not single, specific animals. Jonas Grønbek and Brian M. Scott are in the domain, as is Shaman, my partner’s cat, and substituting any of them for $x$ in $P(x)$ results in a true statement.

Title: is quantifier domain tied to predicate argument

Body:
I have trouble understanding the domains of predicate quantifiers.

P(x) =“x knows calculus”.

Domain for x is all rabbits.

$ \forall x\neg P(x) $

To me it seems like for all rabbits no all rabbits knows calculus. Which makes no sense. So if I were to provide an actual value for the predicate P(pygmy rabbit) the statement would not make sense. So the actual x is merely the domain I suppose.

If we were to make something that could actually provide some different outcomes based on the inputs

P(x) =“x is a mammal”.

Domain for x a set of animals

$ \forall x P(x) $

let x be $\{ Rabbit, Human, Whale \}$

$ \forall x P(x) \equiv true$

The above observations leads me to believe that the predicate variable has to be presented as an argument inside of the predicate, I would like for someone to clarify this for me


问:
Is this Hessian matrix positive semidefinite?
I'm trying to prove the convexity of a function from its Hessian matrix.

He = $\begin{bmatrix}
\frac{1}{x_1} - \frac{1}{\sum_{i=1}^n x_i} & & \frac{-1}{\sum_{i=1}^n x_i} & \cdots & \frac{-1}{\sum_{i=1}^n x_i}\\
\frac{-1}{\sum_{i=1}^n x_i} & & \frac{1}{x_2} - \frac{1}{\sum_{i=1}^n x_i} & \cdots & \frac{-1}{\sum_{i=1}^n x_i}\\
\vdots & & \ddots & & \vdots \\
\frac{-1}{\sum_{i=1}^n x_i} & & \cdots & & \frac{1}{x_n} - \frac{1}{\sum_{i=1}^n x_i}\\
\end{bmatrix}$

My idea was to separate this into two matrices and see if we could get something out of that:

He = $\begin{bmatrix}
\frac{1}{x_1} & 0 & 0 & \cdots & 0\\
0 & \frac{1}{x_2} & 0 & \cdots & 0\\
\vdots & \ddots & \ddots & \ddots & 0\\
0 & 0 & \cdots & 0 & \frac{1}{x_n}\\
\end{bmatrix}$
+
$\begin{bmatrix}
\frac{-1}{\sum_{i=1}^n x_i} & \frac{-1}{\sum_{i=1}^n x_i} & \cdots & \frac{-1}{\sum_{i=1}^n x_i}\\
\frac{-1}{\sum_{i=1}^n x_i} & \ddots & \ddots & \frac{-1}{\sum_{i=1}^n x_i}\\
\vdots & \ddots & \ddots & \vdots\\
\frac{-1}{\sum_{i=1}^n x_i} & \frac{-1}{\sum_{i=1}^n x_i} & \cdots & \frac{-1}{\sum_{i=1}^n x_i}\\
\end{bmatrix}$

Knowing that:  $x \in \mathbb{R}_{++}^n$

we can see that the first matrix is positive definite and the second matrix consists of the same value in each position, and this value must be negative.

Also, since $\dfrac{1}{x_i} > \dfrac{1}{\sum_{i=1}^n x_i}$ for all $i$, the entries on the diagonal of the original Hessian must all be positive.

However, I don't know what else I can do with this.

答:
$D:= \text{diag}\big(x_1,x_2,...,x_n\big)$

$\alpha := \frac{1}{\sum_{k=1}^n x_k}$

$H :=D^{-1} -\alpha\mathbf 1\mathbf 1^T$

1.) argument via congruence

$\mathbf v:= D^\frac{1}{2}\mathbf 1$

$B:=D^\frac{1}{2}HD^\frac{1}{2} = I -\alpha \mathbf v \mathbf v^T= I +\big(-\alpha \mathbf v \mathbf v^T\big)$

note that $\text{trace}\Big(-\alpha \mathbf v \mathbf v^T\Big) = -\alpha\cdot\text{trace}\Big( \mathbf v \mathbf v^T\Big)= -\alpha\cdot \mathbf v^T\mathbf v = -\alpha \cdot \alpha^{-1}=-1$

so

$-\alpha \mathbf v \mathbf v^T = Q\Lambda Q^T$

where $Q^{-1}=Q^T$ and $\Lambda$ is diagonal with a $-1$ in bottom right corner and everything else is zero.  Then

$B = I + -\alpha \mathbf v \mathbf v^T = QIQ^T +  Q\Lambda Q^T = Q\big(I+\Lambda\big) Q^T$

so $H$ is congruent to $B$ which is congruent to $\big(I+\Lambda\big)$

thus they all have signature of $\big(n-1,0\big)$

2.) argument over paths

Using topological continuity of eigenvalues you can verify this as follows.  For $\tau \in [0,1]$

$H(\tau) :=D^{-1} + \tau\cdot \big(-\alpha\mathbf 1\big)\mathbf 1^T$

where $H(\tau)$ is real symmetric for all $\tau \in[0,1]$ and clearly $H(0)\succ \mathbf 0$  and $H(1) = H$.

now for $\tau \in [0,1]$, by matrix determinant lemma

$\det\Big(H(\tau)\Big) $

$= \det\Big(D^{-1} + \tau\cdot \big(-\alpha\mathbf 1\big)\mathbf 1^T\Big)$

$= \det\Big(D^{-1}\Big)\cdot \Big(1+ \tau\cdot \mathbf 1^T D\big(-\alpha\mathbf 1\big)\Big)$

$= \det\Big(D\Big)^{-1}\cdot \Big(1+ \tau\cdot \alpha^{-1} \cdot(-\alpha)\big)\Big)$

$= \det\Big(D\Big)^{-1}\cdot \Big(1- \tau\Big)$

$\geq 0$

and the inequality is strict for $\tau \in [0,1)$

This means $H(1)=H\succeq \mathbf 0$.  I.e. If $H$ was indefinite, then where must be some $\tau^* \in [0,1)$ where $\det\big(H(\tau^*)\big) =0$ via eigenvalue continuity, but there isn't so $H$'s signature is $\big(?,0\big)$.  We can actually determine that $?=n-1$ by observing

$\text{rank}\Big(H\Big) = \text{rank}\Big(D^{-1} -  \big(\alpha\mathbf 1\big)\mathbf 1^T\Big)\geq \text{rank}\Big(D^{-1}\Big) -  \text{rank}\Big(\alpha\mathbf 1\mathbf 1^T\Big) =n-1$, which must be met with equality because $\det\big(H\big) =0$.  Thus $H$ has signature $\big(n-1,0\big)$.


问:
An asymptotic connection between nth prime number and prime counting function
I am trying to prove an asymptotic between the nth prime number and the prime-counting function. Numerically it seems right but I don't know if every step is correct.
So here we go:
We know that
$$\pi(n)\sim\frac{n}{\log n} \quad \mathrm{and} \quad p_n \sim n\log n \qquad (n\rightarrow\infty).$$
The first statement can be rewritten as
$$\pi(n) \sim \frac{n\log n}{\log^2 n} \qquad (n\rightarrow\infty).$$
Now, is the next step valid?
$$\pi(n)\log^2 n\sim n\log n \qquad (n\rightarrow\infty).$$
If this step is valid we can continue and say that
$$\pi(n)\log^2 n\sim p_n \qquad (n\rightarrow\infty),$$
which is my desired outcome.

答:
Yes, all the steps are justified. In general,
$$f(n) \sim g(n) \implies h(n) \cdot f(n) \sim h(n) \cdot g(n),$$
and in your case, you can just multiply the prime number theorem by $\log^2 n$ and substitute $p_n$ for  $n\log n$ (since $\sim$ is also transitive).

Just remember the definition of $f\sim g$, i.e., that $\lim\frac fg = 1$.


问:
Construction of a connection on a vector bundle $E\to M$ with trivial determinant bundle
I read the following statement, and I am trying to see why it is true.

Let $E\to M$ be a rank 2 complex vector bundle over a smooth manifold $M$, such that the determinant bundle $\det(E)=\Lambda^2(E)$ is the trivial line bundle over $M$, with trivialization $\phi:\det(E)\to M\times \mathbb{C}$. Then there exists a connection $\nabla^E$ on $E$ which induces the trivial connection on $\det(E)$, under $\phi$.

I wanted to construct $\nabla^E$ by defining local connections, which I then glue together by a partition of unity. Let $\{U_\alpha\}$ be a trivializing open cover of $M$. Over each $U_\alpha$, I have a local frame $\{e_1^\alpha,e_2^\alpha\}$ such that $e_1^\alpha\wedge e_2^\alpha$ corresponds to 1 under the diffeomorphism $\phi$. On each $U_\alpha$, I can define a flat connection $\nabla^\alpha$ (so by declaring the sections $e^\alpha_i$ to be flat). If $(\rho_\alpha)$ is a partition of unity subordinated to the trivializing open cover, I define $\nabla^E(s)=\sum\limits_\alpha \nabla^\alpha(\rho_\alpha s)$, which is indeed a connection. Next, I want to check that it induces the trivial connection on $\det(E)$ under $\phi$. Note that $\phi^{-1}(1)=\sum\limits_\alpha \rho_\alpha e_1^\alpha\wedge e_2^\alpha.$ Then
\begin{align*}
\nabla^{\det(E)}\big(\sum_\alpha \rho_\alpha e_1^\alpha\wedge e_2^\alpha\big)&=\sum_\alpha \nabla^{\det(E)}\big(\rho_\alpha e_1^\alpha\wedge e_2^\alpha)\\&=\sum_\alpha \nabla^E(\rho_\alpha e^\alpha_1)\wedge e_2^\alpha+\rho_\alpha e_1^\alpha\wedge \nabla^E(e_2^\alpha)\\&=\sum_\alpha \rho_\alpha\big(\nabla^E(e_1^\alpha)\wedge e_2^\alpha+e_1^\alpha\wedge \nabla^E(e_2^\alpha)\big).
\end{align*}
I want this to be zero, but the problem I encounter is that $\nabla^E(e_i^\alpha)\neq\nabla^\alpha(e_i^\alpha)=0$. Nevertheless, is this the correct approach to take and how can I fix it? Does summing over all $\alpha$ make sure it is correct again?

EDIT: This question seems to have been partially asked here: Connections on Bundles with Trivial Determinant
But it seems that an answer is missing and I am very curious, it seems to take a different approach, by introducing a Hermitean metric on $E$.

答:
Maybe I'm missing something, but I think it goes like this. We have

$$ \nabla^E s = \sum_\alpha \nabla^\alpha(\rho_\alpha s) = \left( \sum_\alpha \rm{d} \rho_\alpha \right) \otimes s + \sum_\alpha \rho_\alpha \nabla^\alpha s  = \sum_\alpha \rho_\alpha \nabla^\alpha s $$

since $\sum \rho_\alpha =1$ implies $\sum \rm{d} \rho_\alpha = 0$. Now I can continue where you left off. We have

$$  \nabla^E (e_1^\alpha) \wedge e_2 ^\alpha + e_1^\alpha \wedge  \nabla^E (e_2^\alpha) = \sum_\beta \rho_\beta (\nabla^\beta (e_1^\alpha) \wedge e_2^\alpha + e_1^\alpha \wedge \nabla^\beta (e_2^\alpha) ) = \sum_\beta \rho_\beta \nabla^\beta (e_1^\alpha \wedge e_2^\alpha)  =0 ,$$

where the last equality comes from the fact that $ \nabla^\beta (e_1^\alpha \wedge e_2 ^\alpha)=0 $ where defined, by construction. (I am abusing notation, writing $\nabla^\beta$ for the connection on the restriction of $\rm{det}(E)$ induced by $\nabla^\beta$.)


问:
How to find covering space of wedge sum?
Consider the wedge sum of 2 CW complices $X\vee Y$. Let $p:C\rightarrow X\vee Y$ be a m-sheeted covering map. Then what do we know in general about the space $C$? Is that true $C$ is also a wedge of CW complices in some natural sense?

It seems that $C$ should be a wedge sum of several CW complices and each of them is a covering space of $X$ or $Y$. But I don't know if exhausted.

In general, is there any method to find all the covering space of a given index of $X\vee Y$?

答:
For me it's simpler to think about a mild variation of the wedge sum where instead of just identifying two points we insert a path between them. Then we can describe the covering spaces in quite a nice way: they are "graphs of spaces" where the "edges" correspond to lifts of the path connecting the wedge points and the "vertices" are covering spaces of the spaces we're wedging. The rule for how to connect "vertices" up by "edges" is that if a given "vertex" is an $n$-fold cover then it contains $n$ lifts of its wedge point which means it must have "degree" $n$ in this "graph." (There is also some auxiliary data needed to uniquely determine a covering map.) Furthermore this covering space is an $m$-fold cover iff there are $m$ "edges" total.

It's much easier to explain this with pictures. You can find some in this blog post where $X = B C_2$ and $Y = B C_3$, so that the wedge sum is the classifying space of the modular group $\Gamma = PSL_2(\mathbb{Z}) \cong C_2 \ast C_3$.

This observation, together with the classification of covering spaces and the Seifert-van Kampen theorem, implies the Kurosh subgroup theorem classifying subgroups of free products, which is a generalization of the classical fact that subgroups of free groups are free.

I think the story should be the same for the ordinary wedge sum of CW complexes (but probably not more general spaces); we should be able to shrink the "edge" without affecting the covering spaces. But I'm not 100% confident.


问:
Show that $24 \mid (n - 1)n(n + 1)(n + 2)$ for all positive integers $n$.

Show that $24 \mid (n - 1)n(n + 1)(n + 2)$ for all positive integers $n$.

Since $24 = 2^3 \cdot 3$ it follows that $(n - 1)n(n + 1)(n + 2)$ is divisible by $3$ since it contains the product of three consecutive numbers.

If $n$ is odd then $n-1$ and $n+1$ are even and other one of them is divisible by $2$ and the other by $4$, thus the product is divisible by $2^3$.

If $n$ is even then $n$ and $n+2$ are both even and similar reasoning follows that other one is divisible by $2$ and the other by $4$.

Is this correct? For the odd case if I have that $n=1$ then $n-1=0$ and $n+1=2$, but neither one is divisible by $4$ here? Or does $0$ count as being divisible by $4$?

答:
hint

Other approach, by induction.

For $ n=1 $, it is true.

Let $ n\ge 1 $ such that
$$24|\color{red}{(n-1)n(n+1)(n+2)}.$$

Replacing $ n $, by $ n+1$,

$$n(n+1)(n+2)(n+3)=\color{red}{(n-1)n(n+1)(n+2)}+4n(n+1)(n+2)$$

So, you just need to prove that
$$6|n(n+1)(n+2)$$


问:
evaluating expectation of $\mathbb{E}[f(x)f(x)^T]$
I am trying to evaluate $E[f(x)f(x)^T]$ where $f(x) = Nx$ and $N \in \mathbb{R}^{N \times N}$, and also $p(x) = \mathbb{N} (x|μ, \sigma)$

So, as we know,

\begin{align*}
   E[f(x)f(x)^T] &= \int_{-\infty}^{\infty} f(x)f(x)^{T} p(x)dx \\
                 &= \int_{-\infty}^{\infty} Nx(Nx)^{T} p(x) dx \\
                 &= \int_{-\infty}^{\infty} Nxx^{T}N^{T} p(x) dx
 \end{align*}

Now, I am confused about the next step as I haven't evaluated $\mathbb{E}$ with a matrix before. Can I drag the $NN^T$ part outside as it's constant with respect to $x$? Can I break the order? What to do next?

答:
Assuming the matrix doesn't depend on the random vector $x$, you can indeed take the matrix outside the expectation, since matrix multiplication is a linear operation.  You just have to be careful to remember that matrix multiplication is "order dependent" i.e. not commutative:

$$
\mathbb{E}[(Nx)(Nx)^T] = \mathbb{E}[Nxx^T N^T] = N\mathbb{E}[xx^T]N^T = NC N^T
$$ where $C = \mathbb{E}[xx^T]$.

Hope that helps!


答:
Yes, you can pull out the $N$. Doing so leaves you with
$$
N \left( \int_{-\infty}^\infty xx^T\,p(x)\,dx\right)N^T = N \,\mathbb{E}(xx^T)N^T.
$$
One way to proceed is to note that $\mathbb{E}(xx^T) - \mathbb{E}(x)\mathbb E(x)^T$ is the covariance matrix of $x$.

Title: evaluating expectation of $\mathbb{E}[f(x)f(x)^T]$

Body:
I am trying to evaluate $E[f(x)f(x)^T]$ where $f(x) = Nx$ and $N \in \mathbb{R}^{N \times N}$, and also $p(x) = \mathbb{N} (x|μ, \sigma)$

So, as we know,

\begin{align*}
   E[f(x)f(x)^T] &= \int_{-\infty}^{\infty} f(x)f(x)^{T} p(x)dx \\
                 &= \int_{-\infty}^{\infty} Nx(Nx)^{T} p(x) dx \\
                 &= \int_{-\infty}^{\infty} Nxx^{T}N^{T} p(x) dx
 \end{align*}

Now, I am confused about the next step as I haven't evaluated $\mathbb{E}$ with a matrix before. Can I drag the $NN^T$ part outside as it's constant with respect to $x$? Can I break the order? What to do next?


答:
You have four consecutive terms, so exactly one of them is divisible by $4$, another is divisible by $2$, and at least one other is divisible by $3$.

QED.

Body:
Show that $24 \mid (n - 1)n(n + 1)(n + 2)$ for all positive integers $n$.

Since $24 = 2^3 \cdot 3$ it follows that $(n - 1)n(n + 1)(n + 2)$ is divisible by $3$ since it contains the product of three consecutive numbers.

If $n$ is odd then $n-1$ and $n+1$ are even and other one of them is divisible by $2$ and the other by $4$, thus the product is divisible by $2^3$.

If $n$ is even then $n$ and $n+2$ are both even and similar reasoning follows that other one is divisible by $2$ and the other by $4$.

Is this correct? For the odd case if I have that $n=1$ then $n-1=0$ and $n+1=2$, but neither one is divisible by $4$ here? Or does $0$ count as being divisible by $4$?


问:
Partial ordering with only reflexive relation.
I have a problem understanding orders.

For know, how i understand partial order is: pair of set and relation that orders it.

From that pair, let's say, we can create some kind or 'ordered list'.
For example, let's say that we have set $A=\{5, 3, 1, 2\}$ and relation $R=\{(1, 1), (2, 2), (3, 3), (5, 5), (1, 2), (1, 3), (1, 5), (2, 3), (2, 5), (3, 5)\}$, so we can, on this basis create 'ordered list' like: $$L = [1, 2, 3, 5]$$ because we are able to compare all elements with each other.
But let's try to consider partial order $R = \{(1, 1), (2, 2), (3, 3), (5, 5)\}$ which is reflexive, transitive and antisymmetric, so it fulfills all properties needed to be named partial order. How do I proceed with such relation? How do I create such 'ordered list'? Or is there something not ok with my understanding of orders?

答:
The definition of partially ordered set does not guarantee that all of its elements are comparable, so you can only order certain subsets(one element subsets in your case).
If all elements of a partially ordered set are comparable you have a total order and the set is called totally ordered wiki article


问:
Infinite Geometric Series Formula Derivation
We know that the formula for computing a geometric series is:$$\sum_{i=1}^{\infty}{a_0r^{i-1}} = \frac{a_0}{1-r}$$
Out of curiosity, I would like ask: Is there any ways the formula can be derived other than the following two ways?

Method 1 (The way I found on my own):
$$\sum_{i=1}^{\infty}{a_0r^{i-1}} \equiv S$$
$$S = a_0r^0+a_0r^1+a_0r^2+\cdots$$
$$S = r\left(a_0r^{-1} + a_0r^{0} + a_0r^1+\cdots\right)$$
$$S = r\left(a_0r^{-1} + S\right)$$
$$S = a_0 + rS$$
$$(1-r)S = a_0$$
$$S = \frac{a_0}{(1-r)}$$
Note that for this to work, you must first confirm this: $$\lim_{n\to\infty} a_n = 0$$

Method 2 (The way I found on the web):
$$\sum_{i=1}^{n}{a_0r^{i-1}} \equiv S_n$$
$$S_n = a_0r^0+a_0r^1+a_0r^2+\cdots + a_0 r^{n-2} + a_0 r^{n-1}$$
$$rS_n = r\left(a_0r^0+a_0r^1+a_0r^2+\cdots + a_0 r^{n-2} + a_0 r^{n-1}\right)$$
$$rS_n = a_0r^1 + a_0r^2 + a_0r^3 + \cdots + a_0 r^{n-1} + a_0 r^{n}$$
$$S_n-rS_n = a_0r^0 - a_0r^n$$
$$(1-r)S_n = a_0 - a_0 r^n$$
$$S_n = \frac{a_0(1 - r^n)}{1-r}$$
Given:$$\left|r\right| < 1,$$
$$\lim_{n\to \infty} S_n = \lim_{n\to \infty}\frac{a_0(1 - r^n)}{1-r} = \frac{a_0}{1-r}$$

I personally prefer Method 1 because it is faster and more intuitive, as we don't have to multiply by $r$.

Method 1 for formula of partial sums:
$$\sum_{i=1}^{n}{a_0r^{i-1}} \equiv S_n$$
$$S_n = a_0r^0+a_0r^1+a_0r^2+\cdots+a_0r^{n-2}+a_0r^{n-1}$$
$$S_n = r\left(a_0r^{-1} + a_0r^{0} + a_0r^1+\cdots+a_0r^{n-3}+a_0r^{n-2}\right)$$
$$S_n = r\left(a_0r^{-1} + S_n - a_0r^{n-1}\right)$$
$$S_n = a_0 + rS_n - a_0r^{n}$$
$$(1-r)S_n = a_0 - a_0r^n$$
$$S_n = \frac{a_0(1 - r^n)}{(1-r)}$$

Body:
$a+ar+a(r^{2})...=s$

$a+rs=s$

$a+rs-s=0$

$a+(r-1)s=0$

$-a=(r-1)s$

$\frac{-a}{r-1}=s$

$r-1=-(1-r)$

$\frac{-a}{-(1-r)}=s$

$s=\frac{a}{1-r}$


问:
Book recommendation for linear optimisation
I am currently in a linear optimisation class that uses a lot of the simplex algorithm and things relating to it. The book we are using is called Linear Optimisation The Simplex Workbook but it is beyond confusing and seems very poorly written to introduce someone to the area. I am looking for a book on linear optimisation that is better written and explains stuff well (i.e. talks about feasibility, dual-simplex, simplex, etc). I have asked my professor but he told me he likes this book and did not give me another recommendation, so I am stuck and frustrated trying to learn from the book and his notes (based off the book).

答:
I would recommend Linear Programming 1: Introduction by Dantzig and Thapa. Dantzig created the simplex method, and the book serves as both an introduction to operations research and linear programming.

https://www.springer.com/gp/book/9780387948331


问:
Alternative ways to say "if and only if"?
There are some scenarios about which I would like to get some confirmation:

when defining a concept A,

We call A, if ... [definition of
  concept A]

Does "if" here mean equivalence
instead of just sufficiency? Is it
incorrect to replace "if" with "if
and only if"?

For precisely what condition is
   B satisfied?

Does "precisely" mean asking for
necessary and sufficient condition
for B?

Some other ways to say "if and only
if" / "necessary and sufficient"?

Some references that summarize some standard terminologies in Mathematics  such as this one?

Thanks and regards!

答:
In definitions, it is very common practice to use "if" even though "if and only if" is meant. (Personally, I always use "if and only if" explicitly). So you would not have trouble finding a book that said something like

Definition. A group $G$ is simple if $G$ is nontrivial and whenever $N\triangleleft G$, either $N=\{1\}$ or $N=G$.

But such a definition is meant to be understood to be saying that $G$ is simple if and only if the condition is met.

"Precisely" is asking for a condition that is both necessary and sufficient.

"Exactly when"; "if ... then ... and conversely", among others.


问:
Proving that $L^p \subset L^q$ when $1 \le q \le p$
Let $(E,\mathcal{F},\mu)$ be a measure space such that $\mu(E)=1$ and let $L^p=L^p(E, \mathcal{F},\mu)$. Prove that
$L^p \subset L^q\text{ if } 1 \le q \le p$.

I let $f \in L^p$. Then $(\int_E |f|^pd\mu)^{1/p} < \infty$. To prove that $f \in L^q$, I should prove that $(\int_E |f|^q d\mu)^{1/q} < \infty $ but I'm unable to do that. I read somewhere that this proof can be done using Holder's inequality but I couldn't do it.

答:
Since $1 \le q \le p < \infty$, $\varphi(x) = x^{p/q}$ is a convex function. Apply Jensen's inequality to get:

$$
\varphi\left(\int \left|f\right|^q \,d\mu\right) \le \int \varphi\left(\left|f\right|^q\right) \,d\mu
$$

Hence:
$$
\left(\int |f|^q \,d\mu\right)^{p/q} \le \int \left|f\right|^p \,d\mu
$$

It follows that $\lVert f \rVert_q \le \lVert f \rVert_p$ and $L^p \subset L^q$.


问:
Finding differential of inverse?
Let $f(x)=6x^3-11x^2-2$, $x\geqslant 1.
5$. Find $\frac{\text{d}f^{-1}}{\text{d}x}$ at the point $x=1517=f(7)$.

I thought you should first find the inverse and then differentiate but from looking at the inverse, that's clearly not what I should do. Any ideas? Also, why did they give me x>=1.5?

答:
They gave you $x\ge 1.5$ because the function is invertible in the interval $[1.5,+\infty)$
as you can see in the graph below the function is invertible in the intervals $(-\infty,0]$, $\left[0,\frac{11}{9}\right]$ and $\left[\frac{11}{9},+\infty\right)$. The values are the $x$ of the critical points of the function.

$f'(x)=18 x^2-22 x$ and $f'(7)=728$ thus the derivative of the inverse is $\left(f^{-1}\right)'(1517)=\frac{1}{f'(7)}=\frac{1}{728}\approx 0.0014$. The value is so small because $f(x)$ grows very fast around $x=7$, therefore its inverse grows very slowly.

Title: Finding differential of inverse?


问:
Clarification on locally finite graphs
I'm looking for some clarification regarding infinite graphs.

Assuming a graph could be undirected or directed, the concept
of locally finite seems not clear to me. It was told to me that
a graph is locally finite if the valency for every node in the graph is
a finite set.

Consider the following two graphs:

The graph in a) is a directed locally finite graph. While the graph in b)
is a undirected graph which it's not locally finite, the center of this star graph has infinite valency.

Is the definition of locally finiteness still valid as I wrote above? even
when the graph is directed? Certainly the graph in a) is an infinite graph.
In other words, an infinite graphs could be a locally finite graph.

答:
I find the definitions rather clear:

A locally finite graph is a graph in which the degree of each vertex is finite.

A locally finite digraph is a digraph in which the indegree and the outdegree of each vertex is finite.

In your examples, graph $b$ is not locally finite because it has a vertex with infinite degree (this is what I assumed you meant by your drawing), while digraph $a$ is not locally finite because it has a vertex with infinite indegree.

Every digraph $D$ which is an orientation of a graph $G$ not locally finite, will be a digraph not locally finite : Take the vertex $v\in V(G)$ with infinite degree, and look at the orientation of it edges in $D$. In $D$, either its indegree or its outdegree must be infinite, therefore the digraph is not locally finite.

Title: Clarification on locally finite graphs


问:
Approximation in series
How to approximate $[1 - 1/(2x)]$ to $[1/(1 + (1/2x))]$ when $x$ very large. What power series should be used?. This problem is used to find the precession velocity with the Yukawa potential.

OwnerDisplayName: Cesar Antonio Hoyos Pelaez

Title: Approximation in series

Body:
$$1-\frac{1}{2 x}\approx 1-\frac{1}{2 x+1};\; x\to\infty$$
$$1-\frac{1}{2 x+1}=\frac{2 x}{2 x+1}=\frac{1}{1+\frac{1}{2 x}}$$
Thus
$$1-\frac{1}{2 x}\approx \frac{1}{1+\frac{1}{2 x}}$$


Title: Infinite Geometric Series Formula Derivation

Body:
We know that the formula for computing a geometric series is:$$\sum_{i=1}^{\infty}{a_0r^{i-1}} = \frac{a_0}{1-r}$$
Out of curiosity, I would like ask: Is there any ways the formula can be derived other than the following two ways?

Method 1 (The way I found on my own):
$$\sum_{i=1}^{\infty}{a_0r^{i-1}} \equiv S$$
$$S = a_0r^0+a_0r^1+a_0r^2+\cdots$$
$$S = r\left(a_0r^{-1} + a_0r^{0} + a_0r^1+\cdots\right)$$
$$S = r\left(a_0r^{-1} + S\right)$$
$$S = a_0 + rS$$
$$(1-r)S = a_0$$
$$S = \frac{a_0}{(1-r)}$$
Note that for this to work, you must first confirm this: $$\lim_{n\to\infty} a_n = 0$$

Method 2 (The way I found on the web):
$$\sum_{i=1}^{n}{a_0r^{i-1}} \equiv S_n$$
$$S_n = a_0r^0+a_0r^1+a_0r^2+\cdots + a_0 r^{n-2} + a_0 r^{n-1}$$
$$rS_n = r\left(a_0r^0+a_0r^1+a_0r^2+\cdots + a_0 r^{n-2} + a_0 r^{n-1}\right)$$
$$rS_n = a_0r^1 + a_0r^2 + a_0r^3 + \cdots + a_0 r^{n-1} + a_0 r^{n}$$
$$S_n-rS_n = a_0r^0 - a_0r^n$$
$$(1-r)S_n = a_0 - a_0 r^n$$
$$S_n = \frac{a_0(1 - r^n)}{1-r}$$
Given:$$\left|r\right| < 1,$$
$$\lim_{n\to \infty} S_n = \lim_{n\to \infty}\frac{a_0(1 - r^n)}{1-r} = \frac{a_0}{1-r}$$

I personally prefer Method 1 because it is faster and more intuitive, as we don't have to multiply by $r$.

Method 1 for formula of partial sums:
$$\sum_{i=1}^{n}{a_0r^{i-1}} \equiv S_n$$
$$S_n = a_0r^0+a_0r^1+a_0r^2+\cdots+a_0r^{n-2}+a_0r^{n-1}$$
$$S_n = r\left(a_0r^{-1} + a_0r^{0} + a_0r^1+\cdots+a_0r^{n-3}+a_0r^{n-2}\right)$$
$$S_n = r\left(a_0r^{-1} + S_n - a_0r^{n-1}\right)$$
$$S_n = a_0 + rS_n - a_0r^{n}$$
$$(1-r)S_n = a_0 - a_0r^n$$
$$S_n = \frac{a_0(1 - r^n)}{(1-r)}$$

Body:
By the Taylor development:

$$f(r)=\frac a{1-r}, f(0)=a$$

$$f'(r)=\frac a{(1-r)^2}, f'(0)=a$$

$$f''(r)=\frac{2a}{(1-r)^3}, f'(0)=2a$$

$$f'''(r)=\frac{6a}{(1-r)^4}, f'(0)=6a$$
$$\cdots$$
$$f^{(n)}(r)=\frac{n!a}{(1-r)^{n+1}}, f'(0)=n!a$$
$$\cdots$$

Finally,

$$f(r)=a+ar+a\frac22 r^2+
a\frac{3!}{3!} r^3+\cdots
a\frac{n!}{n!} r^3+\cdots$$


问:
Counting leaf nodes from tree instructions
A tree with degree 7 has

7 nodes with degree 2,

6 nodes with degree 3,

5 nodes with degree 4,

4 nodes with degree 5,

3 nodes with degree 6, and

2 nodes with degree 7.

How many leaf nodes does the tree have?
a) 35
b) 28
c) 77
d) 78

By doing it the brute way (just counting), I found that there supposed to be 84 leaf nodes. What am I missing here?

答:
The correct answer is $52$. Let $n$ be the number of leaves and $e$ the number of edges. There are $\sum_{k=2}^7k=27$ non-leaf nodes, so $e=(27+n)-1=26+n$. On the other hand, if we sum the degrees of the the $27+n$ nodes we get $104+n$, so $2e=104+n$. Thus, $2(26+n)=104+n$, and $n=52$.

Alternatively, we could use the formula from this question: we have all of the information needed to calculate the righthand side, and we find that $v_1=52$.


问:
Give an example of subset which is not a subgroup
Let $G$ be an abelian group with Identity $e$. Let $H$ be a subset of $G$ such that each of its elements square is the identity element.

We now prove that $H$ is a subgroup of $G$: Clearly $e \in H$ since $e^2=e$. Also if $a \in H$ then $(a^{-1})^2=e$ which implies that inverse of each element of $H$ will be in $H$. There is one more thing to show - closure, to do that we observe that for any $a$ and $b$ in $H$, $(ab)^2= (ab)(ab)= a(ba)b = a(ab)b =(aa)(bb)=e$ . So the operation is closed in $H$.

It is a good observation to make note how the condition- $G$ is Abelian  helped us to prove that $H$ is a subgroup.

Now what if $G$ is not Abelian?

A quick search yields an example of non Abelian group which doesn't satisfy this condition. If we consider $G$ as set of all $2 \times2$ matrices with nonzero determinant is sufficient to show this. Although I didn't prove it, I believe that if $G$ is any group of $n \times n$ matrices with nonzero determinant, then $H$ will not be a subgroup of $G$.

Is there any group which is non isomorphic to these group of matrices such that $H$ is not a subgroup?

答:
Hint: Try the smallest nonabelian group: the dihedral group of order $6$.

NB: This group is isomorphic to the subgroup of ${\rm GL}_2(\Bbb Z)$ generated by $$\begin{pmatrix} 0 & 1\\ -1 & -1\end{pmatrix}\text{ and }\begin{pmatrix} 1 & 0\\ -1 & -1\end{pmatrix},$$ so it might not be in the spirit of your question.

 Let $H$ be the subset of all flips of the equilateral triangle.


问:
Describe regular language
I have an homework question that it let me out of my depth. It say: Given the next regular expression (1 + 0)*(01)0* describe with your own words the language which denote and give four examples with more than 6 characters. I could give the 4 examples but i don't know how describe the language

The examples that i given are:

string1 := {1010000}

string2 := {1010010}

string3 := {0101000}

string4 := {10010100}

答: The language is the set of all words that contain the string $01$ and have a $0$ immediately before the last $1$ in the string.

Title: Describe regular language


问:
Orbit-Stabiliser Theorem Question.
Consider the action of a group $G$ of order $900$ on a set $X$ with cardinality $15$. By using the Orbit-Stabiliser Theorem, what is the value for the order of the stabiliser of a point of $X$?

So I know that the Orbit-Stabiliser Theorem states that:
$$
|G|=|\operatorname{Stab}(x)|\times |\operatorname{Orb}(x)|
$$
But I'm not sure this can be applied to the problem at hand, nor how to approach this question. Any help would be great thanks!

答: Stabilizers are subgroups, so $80$ and $120$ are ruled out by Lagrange's theorem. Then, $45$ would lead to $|O(x)|=20$, which can't be because the orbits partiton $X$, whose size is $15$. So you are left with $150$ as possible order of a stabilizer.

Title: Orbit-Stabiliser Theorem Question.


问:
Rational map $f: \mathbb{A}_\mathbb{C}^2 \dashrightarrow \mathbb{A}_\mathbb{C}^1$ defined on a cofinite set is everywhere defined
I'm trying to prove the following: Any rational map $f: \mathbb{A}_\mathbb{C}^2 \dashrightarrow \mathbb{A}_\mathbb{C}^1$ which is defined on a cofinite set of $\mathbb{A}_\mathbb{C}^2$ (and thus open for the Zariski topology being $T_1$) must be also defined on the whole $\mathbb{A}_\mathbb{C}^2$ (and thus, is a regular function).

The result isn't true for rational maps $\mathbb{A}_\mathbb{C}^1 \to \mathbb{A}_\mathbb{C}^1$. The rational map given by $(\mathbb{A}_\mathbb{C}^1\setminus\{0\},\frac{1}{x})$ cannot be defined in any open Zariski neighborhood of $0$. Neither is true if the cofiniteness hypothesis of some set of definition of $f$ is abandoned. Both hypothesis, $f$ having two variables and the cofiniteness one, must be exploited.

Two results come to my mind which I think could come handy for this problem, but I'm not sure how exactly. They are the following:

Any non-constant polynomial of $k[x_1,...,x_n]$ with $k$ an algebraically closed field and $n\geq 2$  has an infinite number of roots
(see https://math.stackexchange.com/a/355227/394668).

The identity theorem from the complex variable theory: any two holomorphic functions defined on a common open connected subset $D\subset\mathbb{C}$ and that coincide on a subset $S\subset D$ which has an accumulation point on $D$ must be the same function.

I cannot think of any potential proof which uses the hypothesis of the two variables of $f$. I have no clue on how this could be used.

Any help will be appreciated.

答:
A rational function $\Bbb A^2\to \Bbb A^1$ can be represented by an element $f$ of the function field of $\Bbb A^2$. The condition that $f$ is defined everywhere except for a finite set means that $f \in k[x,y]_{\mathfrak{p}}$ for all height one primes $\mathfrak{p}\subset k[x,y]$. As $R=\bigcap_{\mathfrak{p} \text{ of ht 1}} R_\mathfrak{p}$ for $R$ a noetherian normal domain, this implies that $f\in k[x,y]$ or that $f$ is actually regular on $\Bbb A^2$.

The result about the intersection of localization of height one primes can be found in Matsumura's Commutative Algebra, Theorem 38, or as "Algebraic Hartog's Lemma" (11.3.11) in Vakil's notes.

If you're looking for a more elementary proof, one option in this case is to write our representative of the rational map as an element $g/h$ for $g,h\in \Bbb C[x,y]$ and $h$ nonzero. As $\Bbb C[x,y]$ is a UFD, we may assume that $g,h$ are coprime. If $h$ is nonconstant, then every point on $V(h)$, a curve in $\Bbb A^2_\Bbb C$ and thus an infinite set gives that $g/h$ is undefined. But we assumed that $g/h$ was undefined in finitely many places, so $h$ must be constant after all and our map is regular.


问:
Combinatorics - Distributing 5 distinct candies to 3 different children making sure each child gets at least one candy
I've been preparing even more for the AMC8, and I've found a new type of question I'm not sure how to do:

There are $5$ distinct candies that are to be distributed to $3$ different children. However, to make sure no child gets upset, each child should receive at least one candy. How many ways are there to do this?

Now I know how we can do a similar problem, where the candies are all the same as each other, but I'm not sure how to approach this if they are distinct, especially with the condition that each child should receive one candy. But I do know that if there were only $4$ candies instead, then that would mean that one kid would have to get $2$ candies, while the rest only get $1$. And from there, I could simply computer that the kid who gets a second candy can be chosen in $3$ ways and the two candies they get would be chosen in $5$ choose $2$ ways. And then multiplying this by $2!$ for the rest of the children would get the answer. But my question is, is there an easy way to expand this logic to $5$ candies (like the question above asks), and is there perhaps a formula I could use for even larger cases under the same restrictions?

答:
"Is there a formula I could use for even larger cases under the same restrictions?"

Yes.

Let ${n\brace k}$ denote the Stirling Number of the Second Kind which counts the number of ways $n$ distinct objects can be partitioned into $k$ non-empty unlabeled subsets.  Multiplying by $k!$ then gives the number of ways to partition a collection of $n$ distinct objects into $k$ labeled non-empty subsets.

The number of ways to distribute $n$ distinct candies to $k$ children such that every child gets at least one candy is then $${n\brace k}\cdot k!$$

As for counting this... you could do so directly as N.F.Taussig suggests, breaking into cases based on the distribution patterns... there being $\binom{5}{3}\cdot 3! + 5\cdot 3\cdot 3!=150$ different outcomes.

Alternatively (and in my opinion simpler and more easily generalizable) you could approach by inclusion-exclusion over whether or not any of the children didn't get a candy.  There are by this approach $3^5 - 3\cdot 2^5 + 3\cdot 1^5=150$ distributions.

All of these should arrive at the same answer.


答:
We can consider two cases:  one child receives three candies and the others each receive one, or two children each receive two candies and the other child receives one.

One child receives three candies and the others each receive one:  There are three ways to select the select who receives three candies, $\binom{5}{3}$ ways to select which three of the five candies that child receives, and $2!$ ways to distribute the remaining candies to the remaining children so each of them receives one candy.  Hence, there are
$$\binom{3}{1}\binom{5}{3}2! = 60$$
such distributions, as you found in the comments.

Two children each receive two candies and the other child receives one:  There are $\binom{3}{2}$ ways to select the children who will receive two candies.  There are $\binom{5}{2}$ ways to give two of the five candies to the younger of those two children and $\binom{3}{2}$ ways to give two of the remaining three candies to the older of those two children.  The remaining child must receive the remaining candy.  Hence, there are
$$\binom{3}{2}\binom{5}{2}\binom{3}{2} = 90$$
such distributions.

Total:  Since the above cases are mutually exclusive and exhaustive, there are
$$\binom{3}{1}\binom{5}{3}2! + \binom{3}{2}\binom{5}{2}\binom{3}{2} = 150$$
ways to distribute five candies to three children so that each child receives at least one candy.

This problem can also be solved using the Inclusion-Exclusion Principle.  There are three possible recipients for each candy, so there are $3^5$ possible distributions.

From these, we subtract those in which a child does not receive a candy.  There are three ways to exclude one of the children from receiving a candy and $2^5$ ways to distribute the candies to the remaining children.  There are
$3 \cdot 2^5$ such cases.

However, if we subtract $3 \cdot 2^5$ from the total, we will have subtracted too much since we will have subtracted the three cases in which all the candies are given to one child twice, once for each way of designating one of the children who has not received a candy as the child who does not receive a candy.

Therefore, we must add those three cases to our running total, which yields
$$3^5 - \binom{3}{1}2^5 + \binom{3}{2}1^5 = 150$$
ways to distribute five candies to three children so that each child receives at least one candy.

Title: Combinatorics - Distributing 5 distinct candies to 3 different children making sure each child gets at least one candy

Body:
I've been preparing even more for the AMC8, and I've found a new type of question I'm not sure how to do:

There are $5$ distinct candies that are to be distributed to $3$ different children. However, to make sure no child gets upset, each child should receive at least one candy. How many ways are there to do this?

Now I know how we can do a similar problem, where the candies are all the same as each other, but I'm not sure how to approach this if they are distinct, especially with the condition that each child should receive one candy. But I do know that if there were only $4$ candies instead, then that would mean that one kid would have to get $2$ candies, while the rest only get $1$. And from there, I could simply computer that the kid who gets a second candy can be chosen in $3$ ways and the two candies they get would be chosen in $5$ choose $2$ ways. And then multiplying this by $2!$ for the rest of the children would get the answer. But my question is, is there an easy way to expand this logic to $5$ candies (like the question above asks), and is there perhaps a formula I could use for even larger cases under the same restrictions?


问:
Generalisation of strong law of large numbers for kth moments
I have a quick question regarding the SLLN, and haven't been lucky enough to find the answer to this after searching on the internet but will ask on this forum:

I am aware that the SLLN states that for a sequence of R.V.s the sample average converges a.s. to its expected value as $n \rightarrow \infty$i.e.
$\frac{1}{n}\sum_{i=1}^{n}X_i \rightarrow \mathbb{E}[X]$

I just wanted to know if this generalises to the kth raw moment? i.e.
$\frac{1}{n}\sum_{i=1}^{n}X_i^k \rightarrow \mathbb{E}[X^k]$

assuming that the k-th moment actually exists and is finite?

Any responses to the above would be very much appreciated.

答:
Let $Y = X^k$

then apply apply the law of large numbers to $Y$


问:
Power series $\sin$
Suppose $f$ is holomorphic and $|f(\sin z)|<\infty.$  Then can I say $f(\sin z)=\sum_n^\infty a_n(\sin z)^n$ for $a_n\in\mathbb{C}$? Where does $|f(\sin z)|<\infty$ assumption play role here?

Edit: Domain of $f$ is not entire $\mathbb{C}$ plane.

答:
If $f$ is analytic in a neighbourhood of $0$, it has a Maclaurin series: there is $r > 0$ such that
$$ f(z) = \sum_{n=0}^\infty a_n z^n \ \text{for}\ |z| < r $$

That implies
$$ f(\sin(x)) = \sum_{n=0}^\infty a_n \sin(x)^n \ \text{whenever}\ |\sin(x)| < r $$

Since $\sin$ is continuous, there is some $\epsilon$ such that this is true when $|x| < \epsilon$.

The condition $|f(\sin(x))| < \infty$ plays no role.  Indeed, without specifying for what $x$ this is true, it is essentially meaningless.

Title: Power series $\sin$


答:
Consider applying Holder's Inequality to
$$\int_E |f|^q \cdot 1.$$
You'll have to be a little clever about the choice of conjugate exponents.  As a hint: you're right that you'll want a term of $\int |f|^p$.  Can you think of an exponent that will give you that?</p>

Title: Proving that $L^p \subset L^q$ when $1 \le q \le p$

Body:
Let $(E,\mathcal{F},\mu)$ be a measure space such that $\mu(E)=1$ and let $L^p=L^p(E, \mathcal{F},\mu)$. Prove that
$L^p \subset L^q\text{ if } 1 \le q \le p$.

I let $f \in L^p$. Then $(\int_E |f|^pd\mu)^{1/p} < \infty$. To prove that $f \in L^q$, I should prove that $(\int_E |f|^q d\mu)^{1/q} < \infty $ but I'm unable to do that. I read somewhere that this proof can be done using Holder's inequality but I couldn't do it.


问:
help finding n-th partial sum ... of converging sum
need help finding the nth partial sum of

1 + 2(1/2) + 3(1/4) + 4(1/8) + 5(1/16) + 6(1/32) + 7(1/64) + ... +x(1/2^(x-1)) + ...

converges by the ratio test

Thanks

Body:
Hint:

$$\sum_{k=1}^\infty k2^{1-k}-\frac12\sum_{k=1}^\infty k2^{1-k}=\sum_{k=1}^\infty k2^{1-k}-\sum_{k=2}^\infty(k-1)2^{1-k}=1+\sum_{k=2}^\infty2^{1-k}=2.$$


问:
Showing an orthogonal complement is closed using Cauchy-Schwarz
Given the space $H_0 = \{f \in L^2(P) : f(\omega) = f(T\omega)\}$ where $T$ is a bijection from $\Omega \rightarrow \Omega$, the space in question is the orthogonal complement:
$$
H_0^\perp := \{g \in L^2(P) : E[gf] = 0 \forall f \in H_0\}.
$$
My goal is to show that $H_0^\perp$ is closed. If we take a sequence $g_1,g_2,\dots \in H_0,$ which is convergent to $g$ in $L^2$, we need to show that $E[gf] = 0$. My idea is to apply Cauchy-Scwarz on the scalar product, which in this case is the expectation. This gives us $E[g_nf] \leq \sqrt{E[g_n^2]E[f^2]}$ for every $n$ and for $g$. This upper bound however does not seem very helpful given that $E[g_nf] = 0$. Can this bound somehow be used to show $E[gf] = 0$ when paired with how we defined $f$, or is there a better way to go about this problem? Perhaps I incorrectly applied Cauchy-Schwarz?

答:
It turns out to be quite straightforward once you find the trick:

$$E[f(g - g_n)] \leq \sqrt{E[f^2]E[(g - g_n)^2]} \rightarrow 0$$

since $g_m \rightarrow g$. The left hand side is equal to $E[fg] - E[fg_n] = E[fg]$.


问:
Consider $x^4 \pmod {pq}$, with $p = q = 3 \pmod4$.
Consider $x^4 \pmod {pq}$, with $p = q = 3 \pmod 4$.

Would someone explain to me why exactly one of the four square roots of $x^4 \pmod {pq}$ is also a square?

This result was given without proof and I do not understand.

What I have tried:
Edit: This boils down to proving that some $\exists$ $a: a^2 \cong x^4 \pmod {pq}$, $a$ is a quadratic residue $\pmod {pq}$

Since $p = q = 3 \pmod 4$, I know that the value of the Legendre symbol ($\frac{-1}{pq})$ is $-1$.

So, if the four candidates are $x,-x,y,-y$, one of $x,-x$ and one of $y,-y$ is not a quadratic residue, due to the multiplicity of the Legendre Symbol.

Without loss of generality, let's assume $x$ and $y$ are left as candidates. Without loss of generality, how can we prove $y$ is not a square/quadratic residue $\pmod {pq}$?

答:
We have equations $y^4 \equiv x^4 \pmod p $   if p|x then 0 is the soln. which is $0^2$

otherwise $(yx^{-1})^4 \equiv 1 \pmod p$.

As gcd(4,p-1)= 2 as($p  \equiv 3 \pmod 4$),by fermat's little theorem it reduces to solve $(yx^{-1})^2 \equiv 1 \pmod  p$ i.e. $y^2 \equiv x^2 \pmod p $ i.e. $y = (+1 or -1)x \pmod p$

Now it is easy to observe that either x or -x is a square mod p is a square. Because $p \equiv 3 \pmod 4 $
Let g be generator of $(\Bbb Z / p \Bbb Z)^*$ . Then $g^{(p-1)/2} = -1$ and (p-1)/2 is odd. Take i s.t. $g^i = x$, if i is even we are done. If i is odd $-x = (-1)*x = g^{i + (p-1)/2}$ is a square. Thus it is a square mod p.

Similarly either x or -x is a square mod q . And by chinese remainder thm we are done!


问:
Prove that $f(x)=\frac{1}{1+x}$ is uniformly continuous on $(-\infty,-\frac{3}{2})$

Using the $\epsilon,\delta$ definition of uniform continuity, prove that $f(x)=\frac{1}{1+x}$ is uniformly continuous on $(-\infty,-\frac{3}{2})$.

Attempt:

We must show that for any $\epsilon>0$ we can find a $\delta > 0$, such that if $|x-a|<\delta$ and $|\frac{1}{1+x}-\frac{1}{1+a}|<\epsilon$ for all $a \in (-\infty,-\frac{3}{2})$.

$\quad$Scratchwork for $\delta$:.

$\begin{align}
\quad \quad \Bigg|\frac{1}{1+x}-\frac{1}{1+a} \Bigg|=\Bigg|\frac{1+a-1\color{blue}{-}x}{(1+x)(1+a)}\Bigg|&=\fbox{|$\frac{1}{(1+x)(1+a)}|$}|x-a|\\&\color{blue}{<}4|x-a|<\epsilon \\ &\Rightarrow|x-a|<\frac{\epsilon}{4}
\end{align}$

$\quad \quad $ **In order to bound $\Bigg|\frac{1}{(1+x)(1+a)}\Bigg|$, I used the fact that $x,a<-\frac{3}{2}$.

$\quad \quad $Which would result in $\frac{1}{(1+x)(1+a)}<\frac{1}{(1-\frac{3}{2})(1-\frac{3}{2})}=\frac{1}{\frac{1}{4}}=4$.

$\quad$Choose $\delta=\frac{\epsilon}{4}\\$

$\quad$Proof:
Let $\epsilon>0$ be given and let $\delta=\frac{\epsilon}{4}$. If $|x-a|<\delta$ we have:

$\begin{align}
\quad \quad \Bigg|\frac{1}{1+x}-\frac{1}{1+a}\Bigg|&= \Bigg|\frac{1}{(1+x)(1+a)}\Bigg||x-a|\\&\leq4|x-a|\\ &<4\delta=4\frac{\epsilon}{4}=\epsilon.
\end{align}$

$\rule{18cm}{.03cm}$

Is it okay to use the fact that that $x,a<-\frac{3}{2}$, in order to place bounds on the function in order to reach a $\delta$? I am not sure if this technique holds for intervals involving infinity.

答:
Since $x$ and $a$ are both assumed to satisfy the inequalities $x,a < -\frac{3}{2}$ then sure, you may use those inequalities in your proof.

The proof looks fine. Your scratchwork has a mistaken inequality and a mistaken implication sign, although that seems not to have affected the proof.


问:
The Inverse of a Rational Section is a Rational Section of the Dual
This question comes after reading the last paragraph of Vakil's FOAG, p. 400.

We consider the set of $\{(\mathcal L, s) \}$, where $\mathcal L$ is an invertible sheaf on a Noetherian, reduced, regular in codimension 1 (in case any of that matters) scheme $X$, and $s$ is a nonzero rational section of $\mathcal L$.

The claim is that once you mod out by isomorphism, this set is an abelian group under $\otimes$, with inverse $\{(\mathcal L^*, 1/s) \}$. Thematically, this all works out well given that we know $\mathcal L \otimes \mathcal L^* \simeq \mathcal O_X$, but why is $1/s$ a nonzero rational section of the dual?

The best I can say is that both sheaves are locally isomorphic to $\mathcal O_X$, so perhaps we mean to consider $s$ and $1/s$ as rational sections of $\mathcal O$, and then we glue to produce $s$ and $1/s$ as rational sections of $\mathcal L$?

答:
The idea here is to find a dense affine open subset $\operatorname{Spec} A = U\subset X$ where

$\mathcal{L}|_U\cong \mathcal{O}_X|_U$,

$s$ is an honest section of $\mathcal{L}(U)$,

$s$ is invertible.

Then the image of $s$ under the isomorphism $\mathcal{L}|_U\cong \mathcal{O}_X|_U$ is a unit in $A$, so $s^{-1}$ is an element of $A$ and therefore a section of $\mathcal{L}^{-1}(U)$ and thus a rational section of $\mathcal{L}^{-1}$.

To find such a $U$, it suffices to go irreducible component by irreducible component: since $X$ is noetherian, it has finitely many irreducible components, so for any irreducible component the set of points belonging to only that irreducible component is open. Now, for irreducible $X$, take the intersection of the open sets where $s$ is an honest section and an open subset where $\mathcal{L}|_U\cong\mathcal{O}_X|_U$: this is again a nonempty open subset by irreducibility, and since the affine opens form a basis for the topology we can find an affine open subset $\operatorname{Spec} A$ inside it. We're done.


问:
Why do random variables exist?
We define a random variable to be a measurable function $X$ from a probability space $(\Omega,\mathcal{F},P)$ to a measurable space $(S,\mathcal{B})$.
Given a probability measure $\mu$ on $S$ we say that $X$ follows the probability law $\mu$ if the pushfoward of $P$ via $X$ is equal to $\mu$ $$X\sim\mu\text{ if and only if. } X_*(P)=\mu.$$
At this point it seems to me that the probability space $\Omega$ is forgotten, and all reasoning are done thinking to their law.

So my question is: Are there situations where it is relevant to distinguish random variables from their law?
If it is just a matter of notation, are there problems that are dealt with significantly better in one formalism with respect to the other?

答:
There is a lot that can be said about a random variable
$$
  X: (\Omega,\mathcal{F},P) \to  (S,\mathcal{B}),
  $$
just by looking at $\mu:=X_*(P)$.  For example, if $S=\mathbb R$, then the expectation $\mathbb E(X)$ may be computed
in terms of $\mu $ as follows,  thanks to the change of variable formula applied to $x = X(ω)$:
$$
  \mathbb E(X) =
  \int_\Omega  X(\omega ) \, dP(\omega ) =
  \int_{\mathbb R} x \, d\mu (x).
  $$
Similar facts hold for the standard deviation and most other important concepts referring to a  SINGLE random
variable.

However, statistical notions that depend of  TWO random variables,  say
$$
  X_1, X_2: \Omega \to  \mathbb R,
  $$
often cannot be expressed in terms of the probability laws $\mu_1:={X_1}_*(P)$ and $\mu_2:={X_2}_*(P)$,  an example being
the covariance
$$
  \text{cov}(X_1,X_2) =  \int_\Omega  \big (X_1(\omega )-\mathbb E(X_1)\big )(X_2(\omega )-\mathbb E(X_2)\big ) \, dP(\omega ).
  $$

A clever way to study   two random variables $X_1$ and $X_2$,  as above,  without too much emphasis on the space
$\Omega $, is to  consider the function
$$
  T:\omega \in \Omega \mapsto (X_1(\omega ),X_2(\omega ))\in \mathbb R^2,
  $$
together with the two projections
$$
  \pi _i:(x_1,x_2)\in \mathbb R^2 \mapsto  x_i \in  \mathbb R.
  $$

Besides the fact that $X_i=\pi _i\circ T$, notice that if   $Q$ is the probability measure on  $\mathbb R^2$ given by
$$
  Q=T_*(P),
  $$
we have that
$$
  {\pi _i}_*(Q) =   {\pi _i}_*(T_*(P)) = (\pi _i\circ T)_*(P) = {X_i}_*(P)=\mu _i.
  $$

This said we may safely replace $(\Omega ,P)$ by $(\mathbb R^2,Q)$,
$$
  \matrix{
    && \kern30pt\mathbb R \cr
    & & \quad \nearrow_{π_1}   \cr
    Ω & \buildrel T \over \longrightarrow & \mathbb R^2\kern 20pt \cr
    & & \quad \searrow^{π_2}   \cr
    && \kern30pt\mathbb R
    }
  $$
while replacing $X_i$ by $\pi _i$, in the sense that any
statistical information one could ask about $X_1$ and $X_2$ can be easily computed in terms of the new random variables  $\pi _1$
and $\pi _2$ by means of the change of variables $(x_1,x_2)=T(\omega )$.

Regarding expectation, for example, we have
$$
  \mathbb E(X_i) =
  \int_\Omega  X_i(\omega ) \, dP(\omega ) =
  \int_{\mathbb R^2} x_i\, dQ(x_1, x_2) =
  \mathbb E(\pi _i).
  $$

Speaking of covariance,  we have
$$
  \text{cov}(X_1,X_2) =
  \int_\Omega  \big (X_1(\omega )-\mathbb E(X_1)\big )(X_2(\omega )-\mathbb E(X_2)\big ) \, dP(\omega ) = $$ $$ =
  \int_{\mathbb R^2} \big (x_1-\mathbb E(\pi _1)\big )(x_2-\mathbb E(\pi _2)\big ) \, dQ(x_1, x_2) =
  \text{cov}(\pi _1,\pi _2).
  $$

Finally, recall that $X_1$ and $X_2$ are said to be independent if
$$
  P\big (\{\omega \in \Omega : X_1(\omega )\in A,\ X_2(\omega )\in B\}\big ) = $$ $$ =
  P\big (\{\omega \in \Omega : X_1(\omega )\in A\}\big )\
  P\big (\{\omega \in \Omega : X_2(\omega )\in B\}\big ),
  \tag 1
  $$
for any two Borel sets $A,B\subseteq \mathbb R$.  Notice that the  LHS above  is precisely equal to
$$
  P\big (T^{-1}(A\times B)\big ) = T_*(P)(A\times B) = Q(A\times B),
  $$
while the RHS is
$$
  P\big (X_1^{-1}(A)\big ) \  P\big (X_2^{-1}(B)\big ) =
  {X_1}_*(P)(A)\   {X_2}_*(P)(B) =
  \mu _1(A)\   \mu _1(B).
  $$

Therefore (1) may be expressed as
$$
  Q(A\times B) =   \mu _1(A)\   \mu _1(B),
  $$
which precisely says that $Q$ is the product measure of $\mu _1$ and $\mu _2$.

Title: Why do random variables exist?


问:
Prove $\sqrt[3]{ab}+\sqrt[3]{ac}+\sqrt[3]{bc} <0$, where $a,b,c$ are roots of $x^3 +3x^2-24x +1 = 0$
I am working on this problem:

$a, b$ and $c$ are the roots of $x^3 +3x^2-24x +1 = 0$. Prove that
$$S = \sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0$$

After I used Vietta's Formula,expanded $S^3$ and used many algebraic tricks I have

$S[S^2 -3(a'b' +a'c +b'c')] = 0$, wich leads to $S = 0 $ or $ S^2 -3(a'b' +a'c +b'c') = 0$, where $a' =\sqrt[3]{a}$,  $b' =\sqrt[3]{b}$ and  $c' =\sqrt[3]{c}$

So the final step is to prove the impossibility of the second case by proving that
$$a'b' +a'c +b'c' < 0$$
(which I verified numerically). It's here that I have trouble.
I also noticed that I didn't use the fact that $ab+ac+bc = -24$ (given by Vietta's Formula), so I guess it should be used here.

答:
Note $X^3+Y^3+Z^3 - 3XYZ = (X+Y+Z)(X^2+Y^2+Z^2-XY-YZ-ZX)$.

Using $X^3=ab, Y^3=bc, Z^3 = ca$, as $X, Y, Z$ are distinct (why?), $X^2+Y^2+Z^2> XY+YZ+ZX$ and hence the sign of $X+Y+Z = \sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ca}$ is the same as the sign of $X^3+Y^3+Z^3-3XYZ = -24-3=-27$


问:
Graph theory: Finding the number of different paths through a vertex on a complete graph
If G is a complete graph on n vertices and u,v,w are three distinct vertices in the vertex set of G, then how many different paths are there from u to v passing through w?

For 3 vertices it is simple, each vertex has degree 2, one of the edges leading directly v - so there is only 1 path.

For 4 vertices, it looks as if there are 3 distinct paths going through any fixed vertex w.

For 5 vertices it is becoming tricky and since there are 3 edges not going directly to v, but they can go to each other... well I have counted at least 8 distinct paths so far and am thinking maybe that is correct - that somehow maybe for 3 vertices, there is one edge not going directly to v, and (inventing a relation) 1^2 - 1 = 1 which is true.  For 4 vertices it would be 2 distinct edges not going to v so 2^2 - 1 = 3 which works.  Then for 5 vertices, the 3 distinct edges not going to v would give me 3^2 - 1 = 8.

So in general I could come up with a formula of
$$(n-2)^{2} -1.$$
But this is a shot in the dark and I have no idea.  Any thoughts?

Thanks,

Brian

答:
Let $p_n$ denote the total number of paths between $u$ and $v$ through some $w$. Your path can have anywhere between $3$ and $n$ vertices. If your path has $k$ vertices, then there are $k-3$ choices of intermediate vertices from the $n-3$ free vertices together with $(k-2)!$ choices of rearrangements. Therefore in total we have
$$p_n=\sum_{k=3}^n\binom{n-3}{k-3}(k-2)! = \sum_{k=3}^n\frac{(n-3)!}{(n-k)!}(k-2)$$
This is essentially the formula, but there is a "closed form" in terms of the incomplete gamma function if you want.

We rewrite the sum with $\ell = n-k$ as
$$p_n=(n-3)!\sum_{\ell=0}^{n-3}\frac{n-\ell-2}{\ell!}=(n-2)!\sum_{\ell=0}^{n-3}\frac{1}{\ell!}-(n-3)!\sum_{\ell=0}^{n-4}\frac{1}{\ell!}$$
Each of the summands can be expressed in terms of the incomplete gamma function, namely
$$k!\sum_{\ell=0}^{k-1}\frac{1}{\ell!} = ek\Gamma(k,\ 1)=ek(k-1)\Gamma(k-1,\ 1) + k$$
where $e$ is the base of the natural logarithm and $\Gamma(k,\ 1)$ is the incomplete gamma function. Using the above, we can simplify as
$$\begin{align}p_n &= e(n-2)\Gamma(n-2,\ 1)-e(n-3)\Gamma(n-3,\ 1)
\\&=e(n-2)(n-3)\Gamma(n-3,\ 1)-e(n-3)\Gamma(n-3,\ 1) + n-2
\\&=e(n-3)^2\Gamma(n-3,\ 1)+n-2\end{align}$$

The sequence begins from $n=3$ as
$$1,\ 3,\ 11,\ 49,\ 261,\ 1631,\ \cdots$$


问:
Evaluating $\frac{1}{(4\pi t)^{\frac{d}{2}}} \int_{\mathbb{R}^{n}} e^{-\frac{|x-y|^2}{4t}}e^{-y^2}dy$
I wish to find the function $u(x,t)$ which solves the following PDE:

$$\begin{cases}
\partial_{t}u - \Delta u = 0, \; t \in \mathbb{R}_{+}, \; x \in \mathbb{R}^{d} \\
u(0,x) = e^{-x^{2}}, \; x\in \mathbb{R}^{d}
\end{cases}$$

After several steps related to applying the Fourier transform I arrive at
$$u(x,t) = \frac{1}{(4\pi t)^{\frac{d}{2}}} \int_{\mathbb{R}^{n}} e^{-\frac{|x-y|^2}{4t}}e^{-y^2}dy$$
To evaluate this integral, I know I must

Complete the square

Apply Fubini's theorem

Note that $\int_{-\infty}^{\infty} e^{-x^2}dx = \sqrt{\pi}$

However, after several attempts at doing this I cannot arrive at a function $u$ which satisfies the PDE. My latest attempt yielded: $$u = \frac{e^{\frac{-x^2}{4t}}}{(4\pi t)^{\frac{d}{2}}} e^{\frac{(4t+1)x^2}{4^3 t^3}} \Bigg(\frac{\pi \cdot 4t}{4t+1}\Bigg)^{\frac{d}{2}}$$ which, upon inspecting the boundary condition, results in dividing by zero since $t = 0$ is in the denominator of many terms.

Is my computation of the integral correct? If so, that will mean my error lies elsewhere in my work.

Newest attempt:

\begin{align*}
\frac{1}{(4\pi t)^{\frac{d}{2}}}\prod_{j=1}^{d}\int_{-\infty}^{\infty}e^{-\frac{(x_j-y_j)^2}{4t}}e^{-y_j^2}dy_j &= \frac{1}{(4\pi t)^{\frac{d}{2}}}\prod_{j=1}^{d}\int_{-\infty}^{\infty}e^{\frac{-x_j^2 +2x_j y_j -y_j^2}{4t}}e^{-y_j^2}dy_j \\
&= \frac{e^{-\frac{x^2}{4t}}}{(4\pi t)^{\frac{d}{2}}}\prod_{j=1}^{d}\int_{-\infty}^{\infty}e^{\frac{2x_j y_j -y_j^2}{4t}}e^{-y_j^2}dy_j \\
&=\frac{e^{-\frac{x^2}{4t}}}{(4\pi t)^{\frac{d}{2}}}\prod_{j=1}^{d}\int_{-\infty}^{\infty}e^{-(\frac{1}{4t}+1)y_j^2 + \frac{x_j y_j}{2t}}dy_j
\end{align*}

Now complete the square of: $-(\frac{1}{4t}+1)y_j^2 + \frac{x_j y_j}{2t}$.

Factor out the coefficient of the squared term: $-(\frac{1}{4t}+1)\left ( y_j^2 - \frac{x_jy_j}{2t\cdot (\frac{1}{4t}+1)}\right )$

\begin{align*}y_j^2 - \frac{x_jy_j}{2t\cdot (\frac{1}{4t}+1)}&= (y_j -\frac{\frac{x_j}{2t\cdot (\frac{1}{4t}+1)}}{2})^2 - \frac{x_j^2}{t^2(\frac{1}{4t}+1)^{2}}\end{align*}

$-(\frac{1}{4t}+1)\left( (y_j -\frac{\frac{x_j}{2t\cdot (\frac{1}{4t}+1)}}{2})^2 - \frac{x_j^2}{t^2(\frac{1}{4t}+1)^{2}} \right) =\\\boxed{ -(\frac{1}{4t}+1)\left (y_j -\frac{x_j}{t\cdot (\frac{1}{4t}+1)} \right)^2 + \frac{x_j^2}{t^2(\frac{1}{4t}+1)}}$

Now substitute the above into the integrand and manipulate to obtain the gaussian integral.

\begin{align*}
\frac{e^{-\frac{x^2}{4t}}}{(4\pi t)^{\frac{d}{2}}}\prod_{j=1}^{d}\int_{-\infty}^{\infty}e^{-(\frac{1}{4t}+1)y_j^2 + \frac{x_j y_j}{2t}}dy_j &= \frac{e^{-\frac{x^2}{4t}}}{(4\pi t)^{\frac{d}{2}}}\prod_{j=1}^{d}\int_{-\infty}^{\infty}e^{ -(\frac{1}{4t}+1)\left (y_j -\frac{x_j}{t\cdot (\frac{1}{4t}+1)} \right)^2 + \frac{x_j^2}{t^2(\frac{1}{4t}+1)}}dy_j\\
&= \frac{e^{-\frac{x^2}{4t}}}{(4\pi t)^{\frac{d}{2}}}\cdot e^{\frac{x^2}{t^2(\frac{1}{4t}+1)}}\prod_{j=1}^{d}\int_{-\infty}^{\infty}e^{ -(\frac{1}{4t}+1)\left (y_j -\frac{x_j}{t\cdot (\frac{1}{4t}+1)} \right)^2}dy_j
\end{align*}
Let $z = \sqrt{(\frac{1}{4t}+1)}\left (y_j -\frac{x_j}{t\cdot (\frac{1}{4t}+1)} \right) \implies dz = \left (\sqrt{(\frac{1}{4t}+1)} \right ) dy_j $
\begin{align*}
&= \frac{e^{-\frac{x^2}{4t}}}{(4\pi t)^{\frac{d}{2}}}\cdot e^{\frac{x^2}{t^2(\frac{1}{4t}+1)}}\prod_{j=1}^{d}\frac{1}{\sqrt{(\frac{1}{4t}+1)}}\int_{-\infty}^{\infty}e^{ -z^2}dz \\
&= \frac{e^{-\frac{x^2}{4t}}}{(4\pi t)^{\frac{d}{2}}}\cdot e^{\frac{x^2}{t^2(\frac{1}{4t}+1)}}\prod_{j=1}^{d}\frac{1}{\sqrt{(\frac{1}{4t}+1)}}\sqrt{\pi}\\
&= \frac{e^{-\frac{x^2}{4t}}}{(4\pi t)^{\frac{d}{2}}}\cdot e^{\frac{x^2}{t^2(\frac{1}{4t}+1)}} \cdot \left (\frac{\pi}{(\frac{1}{4t}+1)} \right)^{\frac{d}{2}}\\
&=\frac{e^{-\frac{x^2}{4t}}}{(4\pi t)^{\frac{d}{2}}}\cdot e^{\frac{x^2}{t^2(\frac{1}{4t}+1)}} \cdot \left (\frac{4\pi t}{4t+1} \right)^{\frac{d}{2}} \\
&=\boxed{ e^{-\frac{x^2}{4t}}\cdot e^{\frac{x^2}{t^2(\frac{1}{4t}+1)}} \cdot \left (\frac{1}{4t+1} \right)^{\frac{d}{2}}}
\end{align*}

But again, this seems to be incorrect.

solved: $$u(x,t) = \boxed{\left (\frac{1}{4t+1} \right)^{\frac{d}{2}}e^{-\frac{x^2}{4t+1}}}$$

Everyone take my experience as a caution to check your arithmetic!

答:
Getting the integral right is just be being really careful in all steps. To be sure you have the correct answer you should first check that you get the correct limit as $t\to 0^+$ and if so the best test is to insert it into the PDE and verify that it satisfy it.

There are also other ways you can find the solution to check it. For example from completing the square its easy to see that the solution should be on the form $u(x,t) = a(t)e^{-b(t)x^2}$ where $a(0) = b(0) = 1$ to match the initial condition. Using this as an ansatz the PDE gives us
$$(\dot{a} - a\dot{b}x^2)e^{-b(t)x^2} = \sum_i (-2ab + 4ab^2 x_i^2)e^{-b(t)x^2} = (-2abd + 4ab^2 x^2)e^{-b(t)x^2}$$
which gives you the simple ODEs $\dot{b} = -4b^2$ and $\frac{\dot{a}}{a} = -2db$. Solving for $b$ gives $b = \frac{1}{4t+1}$. Then solving for $a$ gives $a = (1+4t)^{-d/2}$ so the solution is $u =  (1+4t)^{-d/2}e^{-x^2/(1-4t)}$ which agrees with what you found.

Another way would be to take the Fourier transform of the PDE to find
$$\dot{\hat{u}} + k^2\hat{u} = 0\to \hat{u}(k,t) = \hat{u}(k,0)e^{-k^2t}$$ and then take the Fourier transform of the initial condition and finally the inverse transform of $\hat{u}(k,t)$. However in this case this is not any simpler as you will have to evaluate the (complex) Fourier integrals (unless you have some transform tables you can look up the answer from).


问:
Solve $ y^2y’’ = y’, y(0) = 1, y’(0) = 1 $
Solve the ODE with the initial conditions:

$$
y^2y’’ = y’, y(0) = 1, y’(0) = 1
$$

I did the substitution:

$$
y’ = z
$$
$$
y’’ = z’ = \frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{dx} = z \frac{dz}{dy}
$$

Putting in the ODE:

$$
y^2z \frac{dz}{dy} = z \Rightarrow y^2 \frac{dz}{dy} = 1 \Rightarrow \frac{dz}{dy} = \frac{1}{y^2} \Rightarrow dz = \frac{1}{y^2}dy
$$

By integration we get:

$$
z = -\frac{1}{y} + C \Rightarrow y’ = - \frac{1}{y} + C
$$

Using the initial conditions we get:

$$
1 = -\frac{1}{1} + C \Rightarrow C = 2
$$

Therefore:

$$
y’ = -\frac{1}{y} + 2
$$

Now I am not sure how to continue.

答:
$$y’ = -\frac{1}{y} + 2$$
$$y’ = -\frac{1}{y} + \dfrac {2y}y$$
$$y’ = \frac{2y-1}{y}$$
$$\dfrac {y}{2y-1}dy = dx$$
Then integrate both sides.
$$\int \dfrac {y \, dy}{2y-1} = \int dx$$
Substitute $u=2y-1 \implies du=2dy$
and $y=\dfrac {u+1}2$
$$\dfrac 14 \int \dfrac {u+1}{u}du = \int dx$$
$$\dfrac 14 \int \left (1+\dfrac {1}{u} \right ) du = \int dx$$


问:
How to show that the limit is $l$?
Let $(a_n)_{n\in \mathbb{N^*}}$ a strictly positive sequence such that $\displaystyle : \lim_{n\to +\infty} \sum _{k=1}^n a_k=+\infty$. Show that for all sequence $(b_n)_{n\in \mathbb{N^*}}$ of real numbers such as $\displaystyle \lim_{n \to +\infty} b_n = l \in \mathbb{\bar{R}}$, we have $:$
$$\lim_{n\to +\infty} \frac{a_1b_1+a_2b_2+a_3b_3+...+a_nb_n}{a_1+a_2+a_3+...+a_n}=l$$

I tried to use some inequalities such as the Cauchy inequality which gave me $\sum a_kb_k\le \sqrt{\sum a_k^2\ \cdot \ \sum b_k^2 }$

I also tried the definition of the limit with Cesaro lemma yet it doesnt work

Any idea ? thanks

答:
Let $u_n = \sum_{k=1}^n a_k$, and $v_n = b_{n+1}-b_n$. You have that $(u_n)$ increases to $+\infty$, and $v_n$ is the general term of a convergent series, hence by Kronecker's lemma, you deduce
$$\lim_{n \rightarrow +\infty}\frac{\sum_{k=1}^n u_kv_k}{u_n}=0 \quad \quad (*)$$

But $$\sum_{k=1}^n u_kv_k = \sum_{k=1}^n u_k(b_{k+1}-b_k)=u_nb_{n+1} - u_1b_1 + \sum_{k=2}^{n} (u_{k-1}-u_k)b_{k}=u_nb_{n+1} - u_1b_1 - \sum_{k=2}^{n} a_kb_{k}$$

so $$ \frac{\sum_{k=1}^n u_kv_k}{u_n} = b_{n+1} - \frac{u_1b_1}{u_n}-\frac{\sum_{k=2}^{n} a_kb_{k}}{u_n}=b_{n+1} - \frac{u_1b_1}{u_n}-\frac{\sum_{k=2}^{n} a_kb_{k}}{\sum_{k=1}^n a_k}$$

So using $(*)$,$$\lim_{n \rightarrow +\infty} \frac{\sum_{k=2}^{n} a_kb_{k}}{\sum_{k=1}^n a_k} = \lim_{n \rightarrow +\infty} b_{n+1} = l$$

Title: How to show that the limit is $l$?


问:
How to show that if two Platonic solids have the same number of edges, vertices, and faces, then they are similar in $\mathbb{R}^{3}$?
Note: It appears that some of the terms here do not have standardized definitions, so some sources may give conflicting info.

I was looking into the proof that there are only five Platonic solids in Basic Concepts of Algebraic Topology by F.H. Croom at page 29, Theorem 2.7. To clarify,

We define a Platonic solid as a simple, regular polyhedron homeomorphic to $S^{2}$.

We define a simple polyhedron to be a polyhedron that does not self-intersect itself.

We define a regular polyhedron to be a polyhedron whose faces are regular polygons all congruent to each other and whose local regions near the vertices are all congruent to each other.

Using homology theory, one can prove that the Euler formula $V-E+F=2$ must hold for Platonic solids. Then by using Euler's formula and invoking a counting argument, we find that there are five possible tuples $(V, E, F)$.
This is a beautiful proof, but I am unsatisfied with a question: How do we know there can't be two non-similar Platonic solids that have the same $(V, E, F)$-tuple?

Almost all sources I've looked at seem to assume it is obvious that two Platonic solids with the same $(V, E, F)$-tuple are similar, and it is not obvious to me.

Does anyone have any suggestions for how to prove this? Alternatively, does anyone know of a reference where this is proved rigorously?

Edit 1: It's not completely clear, but it seems like the definition I used for "regular polyhedra" is different than the one commonly used. Note that I am not assuming any global symmetry, so if any global symmetry is to be invoked, it needs to be proven.

Edit 2: I've been made aware of Cauchy's rigidity theorem, which is proven in, e.g., Proofs From the BOOK by Aigner & Zeigler. One can show that any two Platonic solids that have the same $(V, E, F)$-tuple must be combinatorically equivalent. However, in order for the theorem to apply, we need to show that our Platonic solids are convex. I can't seem to think of any rigorous argument for why the Platonic solids have to be convex.

And actually, you don't need to show that the entire polyhedron is convex. If I'm not mistaken, the proof for Cauchy's rigidity theorem only relies on the vertices of the polyhedron being locally convex. So really it suffices to show the vertices are convex.

答:
Edit: This answer invokes rotational symmetry about the vertices, which isn't required in the OP, so this is not a complete answer without a proof that such symmetry ought to hold.

Given $V,E,F$, you can compute the number of sides $n$ of each face ($2E/F$) and the degree $k$ of each vertex ($2E/V$).

By regularity, a Platonic solid with degree $k$ should be invariant under a rotation by $360/k$ degrees about a vertex. This forces the arrangement of regular $n$-gons at a vertex to be "rigid", because this rotational symmetry along with a specific internal angle of each polygonal face forces a unique relative position of the edges from a vertex, up to rotation of the whole configuration in space. See Eric Wofsey's comment for more elaboration. (Also note that the uniqueness of such a configuration is reliant on there being no self-intersections between the polygons around the vertex; if you allow this, the Kepler-Poinsot polyhedra result.)

Furthermore, this rigid configuration about a vertex is fully determined by the position of three of the edges in the configuration; two edges are enough to determine the configuration up to a reflection, and the third tells us whether the vertex is convex or not (whether the corner points "out" or "in").

So, if we start with a vertex and an orientation of its local neighborhood, we get a rigid structure of the polygons meeting at a single vertex. This gives us a ring of "boundary" vertices, where the neighboring polygons have not yet been fully specified. We repeatedly choose a boundary vertex where two boundary polygons meet, which has at least 3 fixed edges attached to it and therefore forces a unique attachment of polygons from that vertex by the above paragraph. By adding in these polygons to our rigid structure, this ring of boundary vertices grows outward, and we can repeat the process. So long as we perform this extension on the boundary vertices closest (in the graph-theoretic sense) to our starting vertex, we will eventually specify the local neighborhood of every vertex in the graph (possibly after infinitely many operations; we aren't assuming that things will naturally close up here.)

The result of this procedure is a surface without boundary, so if the polyhedron is homeomorphic to the $2$-sphere, this process must have terminated after finitely many points and closed up into a unique finite solid.

(Of course, this does not show that any such polyhedron does exist - see this question for an algebraic topology proof of existence. Essentially, the vertex-extending process described above forms a covering space of $S^2$, and so must actually yield the sphere.)


答:
The proof is very similar to showing that if $b_n \to l$, then $\overline{b_n}=\frac{b_1+\ldots+b_n}{n} \to l.$ In fact, this proof is just a special case of your problem, but with $a_k=1$ for all $k$.

Given $\epsilon>0$, choose $n_0$ sufficiently large so that $|b_n-l|<\frac{\epsilon}{2}$ for all $n> n_0$.

Write
$\left|\frac{a_1b_1+a_2b_2+a_3b_3+...+a_nb_n}{a_1+a_2+a_3+...+a_n}-l\right|=\left|\frac{a_1(b_1-l)+a_2(b_2-l)+a_3(b_3-l)+...+a_n(b_n-l)}{a_1+a_2+a_3+...+a_n}\right|\leq A(n) + B(n)$, where
\begin{align}
A(n)&=\frac{\sum_{k=1}^{n_0}a_k|b_k-l|}{a_1+a_2+a_3+...+a_n}, \\
B(n)&=\frac{\sum_{k=n_0+1}^{n}a_k|b_k-l|}{a_1+a_2+a_3+...+a_n}
\end{align}
Since $b_k \to l$, see if you can prove that the numerator in $A(n)$ is bounded and the denominator blows up.

For $B(n)$, see if you can prove $B(n)<\frac{\epsilon}{2}$.

Title: How to show that the limit is $l$?

Body:
Let $(a_n)_{n\in \mathbb{N^*}}$ a strictly positive sequence such that $\displaystyle : \lim_{n\to +\infty} \sum _{k=1}^n a_k=+\infty$. Show that for all sequence $(b_n)_{n\in \mathbb{N^*}}$ of real numbers such as $\displaystyle \lim_{n \to +\infty} b_n = l \in \mathbb{\bar{R}}$, we have $:$
$$\lim_{n\to +\infty} \frac{a_1b_1+a_2b_2+a_3b_3+...+a_nb_n}{a_1+a_2+a_3+...+a_n}=l$$

I tried to use some inequalities such as the Cauchy inequality which gave me $\sum a_kb_k\le \sqrt{\sum a_k^2\ \cdot \ \sum b_k^2 }$

I also tried the definition of the limit with Cesaro lemma yet it doesnt work

Any idea ? thanks


答:
$y^2y''=y'$

Write as
$y''=\frac{y'}{y^2}$

Integrate both sides

$y'=-1/y +C$

$y'(0)=1,y(0)=1\to C=2$

$y'=-1/y+2$

$y=\frac{1}{2} \left(W\left(e^{4 x+1}\right)+1\right)$

Where $W(z)$ is Lambert $W$ function which solves $we^w=z$

The implicit form is

$x=\frac{1}{4} (2 y+\log (2 y-1)-1)$

Title: Solve $ y^2y’’ = y’, y(0) = 1, y’(0) = 1 $

Body:
Solve the ODE with the initial conditions:

$$
y^2y’’ = y’, y(0) = 1, y’(0) = 1
$$

I did the substitution:

$$
y’ = z
$$
$$
y’’ = z’ = \frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{dx} = z \frac{dz}{dy}
$$

Putting in the ODE:

$$
y^2z \frac{dz}{dy} = z \Rightarrow y^2 \frac{dz}{dy} = 1 \Rightarrow \frac{dz}{dy} = \frac{1}{y^2} \Rightarrow dz = \frac{1}{y^2}dy
$$

By integration we get:

$$
z = -\frac{1}{y} + C \Rightarrow y’ = - \frac{1}{y} + C
$$

Using the initial conditions we get:

$$
1 = -\frac{1}{1} + C \Rightarrow C = 2
$$

Therefore:

$$
y’ = -\frac{1}{y} + 2
$$

Now I am not sure how to continue.


问:
As rings, $\mathbb{Z}[x]$ is not isomorphic to $\mathbb{Q}[x]$

Theorem: $\mathbb{Z}[x]$ is not isomorphic to $\mathbb{Q}[x]$.

My attempt:

For a contradiction, suppose $\mathbb{Z}[x] \cong \mathbb{Q}[x]$. So there exists an isomorphism $f: \mathbb{Q}[x] \to \mathbb{Z}[x]$. Since $\mathbb{Q}$ and $\mathbb{Z}$ are integral domains, we can make the following assertions about the units of $\mathbb{Q}[x]$ and $\mathbb{Z}[x]$, denoted $\overline{\mathbb{Q}[x]}$ and $\overline{\mathbb{Z}[x]}$, respectively:
\begin{align*}
\overline{\mathbb{Q}[x]} = \overline{\mathbb{Q}} = \{q \in \mathbb{Q} \mid q \neq 0\} \\
\overline{\mathbb{Z}[x]} = \overline{\mathbb{Z}} = \{-1,1\}.
\end{align*}
Since $f$ is a homomorphism, it maps units to units, so $f(\overline{\mathbb{Q}[x]}) \subset \overline{\mathbb{Z}[x]}$. Take $q_1, q_2, q_3$ distinct in $\overline{\mathbb{Q}[x]}$. Then $f(q_1), f(q_2), f(q_3)$ are units in $\mathbb{Z}[x]$, and since $f$ is injective, $f(q_1) \neq f(q_2) \neq f(q_3)$, meaning $\mathbb{Z}[x]$ has three distinct units, a contradiction.

How does this proof look?

答:
Your proof looks good. Isomorphic rings have isomorphic groups of units.

Alternatively, the equation $2t=1$ has a solution in $\mathbb{Q}[x]$ but not in $\mathbb{Z}[x]$ and so the two rings cannot be isomorphic.


问:
For polynomial prove that $r(x)=0$ or $\deg [r_k(x)]<\deg [b(x)]$
Suppose $a(x),b(x)\in \mathbb R[x]$, $\deg(b(x)) \geq 1$. Show that there exists $m=0,1,\dots$ and $r_0(x),\dots,r_m(x)\in\mathbb R[x]$ such that

$$ a(x)=r_0(x)+r_1(x)b(x) + \cdots + r_m(x)b(x)^m$$

with the property that either $r_k(x)=0$ or $\deg[r_k(x)] < \deg[b(x)]$ for all $k = 0,1,\dots$.

Is it possible to use induction for this question? Thank you.

答:
Let $d$ be the degree of $b(x)$. We show by induction on $n$ that the result is true for all polynomials of degree $\lt nd$.

For the induction step, we need to show that if the result is true for all $a(x)$ of degree $\lt nd$, then it is true for all $a(x)$ of degree $\lt (n+1)d$. So we only need to deal with $a(x)$ of degree $\ge nd$, and $\lt (n+1)d$.

For the proof of the induction step. we can proceed as follows. By the ordinary "Division Algorithm" for polynomials, there is an $r_0(x)$ with the right degree restriction, and a polynomial $q(x)$, such that $a(x)=r_0(x)+q(x)b(x)$.

Show that $q(x)$ has degree $\lt nd$. Then by the induction hypothesis $q(x)$ can be expressed in the right form, and now from $a(x)=r_0(x)+q(x)b(x)$ we get the right form for $a(x)$.

Note that this is an analogue for polynomials of the ordinary base $b$ expansion of positive integers.


问:
Diagonalization of Linear Mapping Associated to a Matrix
Assume T to be the linear operator on $R^4$ whose matrix relative to the standard basis is

$$Q=\begin{bmatrix} 0 & 0 & 0 & 0 \\ a & 0 & 0 & 0 \\ 0 & b & 0 & 0\\0 & 0 & c & 0  \end{bmatrix}$$

When would T be diagonalisable?

I know that Q is not diagonalisable because the diagonal is all 0's and I thought the associated matrix is diagonalisable if and only if linear operator is diagonalisable.

答: The characteristic polynomial of $T$ is $x^4$, so, $0$ is the only eigenvalue of $T$. Now note that $v \in \mathbb R^4 \setminus \{0\}$ is an eigenvector of $T$ if and only if $v \in \ker T$. Thus, $T$ is diagonalizable if and only if there are four linearly independent vectors in $\ker T$, that is, $T$ is diagonalizable if and only if $\ker T = \mathbb R^4$, which means, $T$ is diagonalizable if and only if $T=0$ $(a=b=c=0)$.


问:
Sum of cube roots of two conjugate quadratic integers makes an integer.
Consider the following expression:

$$(20-\sqrt{392})^{1/3}+(20+\sqrt{392})^{1/3}.$$

This equals $4$, but how can I show this?

Note that I do not want to make use of the following line of reasoning: 4 is a solution to $x^3-6x-40=0$, that this cubic has exactly one solution, and that all solutions to the cubic
$x^3+px +q = 0$ are given by
$$(-q/2-((q/2)^2+(p/3)^3)^{0.5})^{1/3}+(-q/2+((q/2)^2+(p/3)^3)^{0.5})^{1/3}.$$

I have noted the method set out here, How can I show that this complicated expression with square and cube roots reduces to the value 7? , but implementing it does not work very well.

I attempted this and got more complicated expressions than I started off with! It doesn't result in a unique solution for $a$ and $b$.

I do think that this is the right approach, though, i.e. making use of the fact that if $\sqrt c$ is irrational and $a+b\sqrt c=d+e \sqrt f$ then $a=d$ and $b=e$. (How do you prove this fact, by the way?)

答:
First note that $392=2^3\times7^2$ and so $\sqrt{392}=14\sqrt{2}$, and hence
$$\sqrt[3]{20+\sqrt{392}}=\sqrt[3]{20+14\sqrt{2}}.$$
Next, in the hope of finding a simple expression for the cube root, we compute
$$(20+14\sqrt{2})(20-14\sqrt{2})=20^2-2\times14^2=8,$$
which shows that
$$\sqrt[3]{20+14\sqrt{2}}\sqrt[3]{20-14\sqrt{2}}=\sqrt[3]{8}=2.$$
This suggests that perhaps there exist integers $a$ and $b$ such that
$$(a+b\sqrt{2})^3=20+14\sqrt{2},$$
and then certainly $a$ and $b$ should satisfy $a^2-2b^2=2$. An obvious choice is $a=2$ and $b=1$, and indeed
$$(2+\sqrt{2})^3=2^3+3\times2^2\times\sqrt{2}+3\times2\times\sqrt{2}^2+\sqrt{2}^3=20+14\sqrt{2},$$
and entirely analogously $(2-\sqrt{2})^3=20-14\sqrt{2}$. It follows that
$$\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}=(2+\sqrt{2})+(2-\sqrt{2})=4.$$

Alternatively, we could directly try to find integers $a$ and $b$ such that
$$(a+b\sqrt{2})^3=20+14\sqrt{2}.$$
This boils down to the two simultaneous cubic equations
\begin{eqnarray*}
20&=&a^3+6ab^2&=&a(a^2+6b^2),\\
14&=&3a^2b+2b^3&=&b(3a^2+2b^2),
\end{eqnarray*}
from which it quickly follows that $a$ and $b$ are positive divisors of $20$ and $14$, respectively, and it is not hard to check that $a=2$, $b=1$ is the only solution.


问:
Finding $\lim_{n\to\infty} \mathbb{P} \{Z_{1,n} + Z_{2,n} + \cdots + Z_{n,n} = k\}.$
Can someone please help solve part (iv) of the problem? I am lost on how to do it. The following work before it may serve to help. Thanks.

A random variable $X$ has a Poisson distribution with parameter $\lambda > 0$ if $X$ takes on only nonnegative integer values and $$\mathbb{P}\{X=k\} = e^{-\lambda}\frac{\lambda^k}{k!}, \hspace{20 pt} k =0,1,2,\dots.$$ In this problem, suppose that $X,Y$ are independent Poisson random variables on a probability space with parameters $\lambda_X, \lambda_Y,$ respectively.

(i) Find $\mathbb{E}[X], \text{Var}[X],\mathbb{E}[Y], \text{Var}[Y].$

(ii) Show that $X+Y$ has a Poisson distribution with parameter $\lambda_X + \lambda_Y$ by computing directly $$\mathbb{P}\{X+Y = k\}.$$

(iii) Find the characteristic function for $X,Y$ and use this to find an alternative derivation of (ii).

(iv) Suppose that for each integer $n>\lambda$ we have independent random variables $Z_{1,n}, Z_{2,n},\dots, Z_{n,n}$ with $$\mathbb{P}\{Z_{j,n} = 1\} = 1 - \mathbb{P}\{Z_{j,n} = 0\} = \frac{\lambda}{n}.$$

Find $$\lim_{n\to\infty} \mathbb{P} \{Z_{1,n} + Z_{2,n} + \cdots + Z_{n,n} = k\}.$$

$\textit{Proof.}$ Give that $\displaystyle{\mathbb{P}\{X=k\} = e^{-\lambda}\frac{\lambda^k}{k!}, k =0,1,2,\dots},$ $X\sim \text{ Poisson}(\lambda X),$ and $Y\sim \text{ Poisson}(\lambda Y),$ then $$\mathbb{E}[X] = \sum_{k=0}^\infty k\cdot \mathbb{P}\{X=k\} = \sum_{k=0}^\infty k \cdot e^{-\lambda}\frac{\lambda X^k}{k!}$$ $$=e^{-\lambda} \sum_{k=1}^\infty \frac{\lambda X^k}{(k-1)!} = e^{-\lambda}\cdot \lambda X \cdot \sum_{k=1}^\infty  \frac{\lambda X^{k-1}}{(k-1)!}$$ $$=\lambda Xe^{-\lambda} \sum_{k-1=0}^\infty \frac{\lambda X^{k-1}}{(k-1)!}$$ $$=\lambda X e^{-\lambda}e^{\lambda} \text{ since } e^X = \sum_{k=0}^\infty \frac{X^k}{k!}$$ thus $\mathbb{E}[X] = \lambda X,$ and similarly $\mathbb{E}[Y] = \lambda Y.$ Next, $$\mathbb{E}[X^2] = \sum_{k=0}^\infty k^2\cdot e^{-\lambda}\frac{\lambda X^k}{k!}$$ $$=\lambda_X^2 + \lambda_X.$$ $$\text{Var}[X] = M_2 - M_1^2 = \mathbb{E}[X^2] - \mathbb{E}[X]^2 = \lambda X^2 + \lambda X - \lambda X^2 = \lambda X.$$ Similarly, $\text{Var}[Y] = \lambda Y.$

For (ii), $\mathbb{P}[X+Y = k] = \displaystyle{\sum_{i=0}^k \mathbb{P}(X+Y=k, Y=i)}$ $$=\sum_{i=0}^k \mathbb{P}(X=k-i) \mathbb{P}(Y=i)$$ $$=\sum_{i=0}^k \frac{e^{-\lambda x} \lambda X^{k-i}}{(k-i)!}\cdot \frac{e^{-\lambda Y}\lambda Y^i}{i!}$$ $$=\frac{1}{k!}\left[ \sum_{i=0}^k \frac{k!}{(k-i)!i!} \lambda X^{k-i} \lambda Y^i \right]e^{-(\lambda X + \lambda Y)}$$ $$=\frac{1}{k!}e^{-(\lambda X + \lambda Y)}\cdot (\lambda X + \lambda Y)^k.$$ So $X+Y$ is also a Poisson random variable with parameter $\lambda X + \lambda Y.$

For (iii) the characteristic function of Poisson random variable $X$ is $$M_X(Z) = \sum_{k=0}^\infty \mathbb{P}(X=k)\cdot e^{kZ}$$ $$=\sum_{k=0}^\infty e^{kZ} \frac{e^{-\lambda X}\lambda X^k}{k!}$$ $$=e^{-\lambda X} \sum_{k=0}^\infty \frac{(\lambda X e^Z)^k}{k!} = e^{-\lambda X} e^{e^Z\lambda X} = e^{\lambda X(e^Z - 1)}.$$ And $M_Y(Z) = e^{\lambda Y(e^Z-1)}.$ And $M_{X+Y}(Z)= M_X(Z) \cdot M_Y(Z)$ since $X$ and $Y$ are independent and equals $e^{\lambda X(e^Z - 1)}\cdot e^{\lambda Y(e^Z - 1)} = e^{(\lambda X + \lambda Y)e^Z-1}.$ Thus, $X+Y$ is also a Poisson random variable with parameter $\lambda X + \lambda Y.$

答:
For $n$ fixed we have $Z_{1,n}$, $Z_{2,n}$,$\dots$, $Z_{n,n}$ are iid Beornoulli randoms variables with paremeter $p=\frac{\lambda}{n}$, then we have that $Z_{1,n} + Z_{2,n} + \cdots + Z_{n,n}$ have Binomial distribution where
\begin{align*} \mathbb{P} \{Z_{1,n} + Z_{2,n} + \cdots + Z_{n,n} = k\}&= \binom{n}{k}p^{k}(1-p)^{n-k}\\
&=\binom{n}{k}\frac{\lambda^{k}}{n^{k}}\left(1-\frac{\lambda}{n}\right)^{n-k}\\
&=\binom{n}{k}\frac{\lambda^{k}(n-\lambda)^{n-k}}{n^{n}}.
\end{align*}
Therefore

\begin{align*}\lim_{n\to\infty} \mathbb{P} \{Z_{1,n} + Z_{2,n} + \cdots + Z_{n,n} = k\}&= \lim_{n\rightarrow\infty} \binom{n}{k}\frac{\lambda^{k}(n-\lambda)^{n-k}}{n^{n}}\\
&= \lim_{n\rightarrow\infty} \left(\frac{n(n-1)\cdots(n-k+1)}{n^{k}}\right)\left(\frac{1}{k!}\right)\lambda^{k}\left(\frac{n-\lambda}{n}\right)^{n-k}\\
&=\lim_{n\rightarrow\infty} \left(\frac{n(n-1)\cdots(n-k+1)}{n^{k}}\right)\left(\frac{1}{k!}\right)\frac{\lambda^{k}}{\left(\frac{n-\lambda}{n}\right)^{k}}\left(\frac{n-\lambda}{n}\right)^{n}\\
&=\lim_{n\rightarrow\infty} \color{red}{\left(\frac{n(n-1)\cdots(n-k+1)}{n^{k}}\right)}\left(\frac{1}{k!}\right)\color{blue}{\left(\frac{\lambda n}{n-\lambda}\right)^{k}}\color{brown}{\left(1-\frac{\lambda}{n}\right)^{n}}\\
& \qquad\qquad \mbox{Note that }\\
& \qquad\qquad\quad \color{red}{\lim_{n\rightarrow\infty} \left(\frac{n(n-1)\cdots(n-k+1)}{n^{k}}\right)=1,}\\
& \qquad\qquad\quad \color{blue}{\lim_{n\rightarrow\infty} \left(\frac{\lambda n}{n-\lambda}\right)^{k}=\lambda^{k}} \:\:\mbox{ and }\\
& \qquad\qquad\quad \color{brown}{\lim_{n\rightarrow\infty} \left(1-\frac{\lambda}{n}\right)^{n} =e^{-\lambda} }\\
&=\frac{\lambda^{k}}{k!}e^{-\lambda}.
\end{align*}