问:
logistic differential equation, carrying capacity.
Assume that a population grows according to the below logistic differential equation $$\frac{\mathrm{dP} }{\mathrm{d} t}=0.01P-0.0002P^2$$
Then what is the maximum population that this model holds?

I think the answer is 50000(I can be wrong!!). Can anyone show me the steps of how to do this? using direction fields? or solve the differential equation directly? and then how do you get the maximum population? Thank you.

答:
This equation is separable. It is also a Ricatti equation (thus linearisable) if you are interested. We have
$$\frac{dP}{dt} = aP(1-bP) \implies \int\frac{dP}{P(1-bP)} = \int a\,dt$$
where $a=\frac{1}{100}$ and $b=\frac{1}{50}$. By partial fractions
$$\frac{1}{P(1-bP)} = \frac{1}{P} + \frac{b}{1-bP}$$
and substituting into the integral
$$\ln|P|-\ln|1-bP| = at+c \implies \frac{P}{1-bP} = Ce^{at}$$
where $c$ is a constant of integration and $C=\pm e^c$. Rearranging we get
$$P=\frac{Ce^{at}}{1+bCe^{at}} = \frac{Ce^{\frac{1}{100}t}}{1+\frac{1}{50}Ce^{\frac{1}{100}t}} = 50\frac{Ce^{\frac{1}{100}t}}{50+Ce^{\frac{1}{100}t}} = \frac{50}{Ae^{-\frac{1}{100}t}+1}$$
for some arbitrary $A=\frac{50}{C}$. Thus the long run population is $P\to 50$.

Note also that $P\equiv 0$ is a solution to the equation. Assuming that $0<P_0<50$, then the solution monotone increases from $P_0$ to $50$.


问:
Prove that $\frac{(1-x_1)(1-x_2) \cdots (1-x_n)}{x_1 x_2 \cdots x_n} \geq (n-1)^n$ for $x_i > 0$ and $x_1 + \ldots + x_n = 1$
It is the problem 2.3 in Algebraic Inequalities by Sedrakyan & Sendrakyan.

Solution

I can't figure out how to prove the key inequality in the solution:

$$\frac{(1-x_1)(1-x_2)}{x_1x_2} \geq \frac{(1-\frac{1}{n})(1-x_1 - x_2 + \frac{1}{n})}{\frac{1}{n}(x_1 + x_2 - \frac{1}{n})}$$

The problem 1.17 referenced in the solution had the following inequality:

$$\lambda(x_1 + x_2 - \lambda) \geq x_1x_2  \text{        for    } x_1 \leq \lambda \leq x_2$$

Substituting $\frac{1}{n}$ for $\lambda$ we have $$\frac{1}{n}(x_1 + x_2 - \frac{1}{n}) \geq x_1x_2 \iff \frac{1}{x_1x_2} \geq \frac{1}{\frac{1}{n}(x_1 + x_2 - \frac{1}{n})}$$

The denominators are fine but we still don't have the full inequality and I'm not sure how to get to it. I imagine I must be missing something simple because otherwise the solution would spend more time on this inequality.

ContentLicense: CC BY-SA 4.0

Body:
$$
\frac{1}{n}(x_1 + x_2 - \frac{1}{n}) \geq x_1x_2 \\
\Rightarrow 1+\frac{1-x_1-x_2}{x_1 x_2} \geq 1+ \frac{1-x_1-x_2}{\frac{1}{n} (x_1+x_2-\frac{1}{n})} \\
\Rightarrow \frac{(1-x_1)(1-x_2)}{x_1 x_2} \geq \frac{(1-\frac 1n) (1-x_1-x_2+\frac 1n)}{\frac 1n (x_1+x_2-\frac 1n)}.
$$


问:
Let $G$ be a group of order $pq$, with $p$ and $q$ prime. Prove that the order of the center of $G$ is 1 or $pq$.

Let $G$ be a group of order $pq$, with $p$ and $q$ prime. Prove that
  the order of the center of $G$ is 1 or $pq$.

Let me start off with what I did:

Assume $G$ is abelian. Then we know $Z(G)=G$, thus the center must have order $pq$.

Now assume $G$ is non-abelian. By Cauchy's theorem we know there must exist elements $x$ and $y$ of respectively order $p$ and $q$. Without loss of generality we can assume $p \leq q$. Because $[G: \langle y \rangle ] = p$, and $p$ is the smallest prime divisor of $G$, $\langle y \rangle$ must be a normal subgroup of $G$.

I claim that $\langle x \rangle$ is a complete set of representatives for $\langle y \rangle$ in $G$. We know, because $p$ is prime, that $\langle x \rangle$ is cyclic and therefore, $\langle x \rangle = \{ x^k : k \in \{ 0, \dots , p-1 \} \}$. We can assume $i < j$. If $x^i \langle y \rangle = x^j \langle y \rangle$, then $x^{-i}x^j = x^{j-i} \in \langle y \rangle$. But $x^{j-i} \in \langle x \rangle$, thus of order p, hence $x^{j-i}$ can't be in $\langle y \rangle$. As so, $\langle x \rangle$ gives a full set of representatives.

Now we can write every element in $G$ as $x^m y^n$, with $m \in \{ 0 , \dots , p-1\}$ and $n \in \{ 0, \dots , q-1\}$. Furthermore, every element other than the identity element has either order $p$ or $q$, or else $G$ would be cyclic, and thus abelian.

Now I'm kind of stuck, but I want to prove that there exists an element in $G$ which doesn't commute with anything in $G$, except for the identity element.

I know there is an easier solution to this, namely by saying that the order of $Z(G)$ must divide the order of the group $G$. So either it must be $1, p, q$ or $pq$. However it can't be $pq$, or else the group would be cyclic and thus abelian. Nor can it be $p$, since then $[G : Z(G) ] = q$, and thus it's cyclic, so by a specific theorem $G$ must be abelian. The same argument for $q$.

I think the proof of that specific theorem has to do with the proof I started. However I can't connect the dots. But I do want to know how I should finish it this way (if that is possible).

Thanks in advance for any help.

答:
Hint:

For any group $\,G\,$ , the quotient group $\,G/Z(G)\,$ cannot be cyclic non-trivial.

ContentLicense: CC BY-SA 3.0


问:
How can I show that $F(x,y)$ is not continuous?
I want to show
$F(x,y)=\frac{xy}{x^2+y^2}$ is not continuous at the origin. Here is my attempt:

Choose a path $y=mx$. Then, $F(x,mx)=\dfrac{xmx}{x^2+m^2x^2}=\dfrac{mx^2}{x^2(1+m^2)}=\dfrac{m^2}{1+m^2}$ if $(x,mx)\neq0$. Thus, $\lim_\limits{(x,y)\to(0,0)}F(x,y)$ does not exist, so $F$ is not continuous.

答:
If your function was $F(x,y)=\dfrac{xy}{x^2+y^2}$, you're correct indeed, $F$ is not continuous at $(0,0)$, since we have a different limit for each path $y=mx$.

To be more specific, for each $m$ we can consider two paths, $y=mx$ and $y=-mx$, which will give the same limit, but limits are distinct for another path $y=nx$ if $|m|\neq|n|$.

Strictly speaking, to talk about the continuity of $F$ at $(0,0)$ you should have defined $F$ there, as it was pointed out in the comments. But you could reformulate the question as: Could $F$ de defined at $\mathit{(0,0)}$ so it is continuous there? And the answer would be no because of what you showed: different paths have different limits, so we can't make it continuous.


问:
Proof about compactness
Suposse $(X, \mathcal{T}_1)$ and  $(X, \mathcal{T}_2)$ are compact Hausdorff Spaces. Prove that either $\mathcal{T}_1= \mathcal{T}_2$  or $\mathcal{T}_1 \not\subset \mathcal{T}_2$ or $\mathcal{T}_2 \not\subset \mathcal{T}_1$.

My idea: Let $U \in \mathcal{T}_1$ then $X-U$ is closed in $\mathcal{T}_1$ and since $X$ is compact then $X-U$ is compact. But I don't know how to continue. I need an idea

答:
As stated this is an exercise in basic logic. P or Q or R is equivalent to (not Q and not R) implies P.

Here not Q and not R means $\mathcal T_1 \subset \mathcal T_2$ And $\mathcal T_2 \subset \mathcal T_1$. Then this implies P.

Title: Proof about compactness


问:
$T(n)=4T(n/3)+n\log_2(n)$
I was wondering if i could find a closed form for this function:

$$T(n)=4T(n/3)+n\log_2(n) \text{ for $ n\in \mathbb N$}$$
$$\text{ & }T(0)=1$$
or at least some function $g$ s.t $T(n)=Θ\big(g(n) \big)$,

(I've tried Master theorem and it does not work here).

any ideas?

EDIT:$\text{ it turns out that: $\bbox[1px,border:1px solid red] {T(n)=Θ\big( n^{\log_3(4)} \big)}$}$

By using the strong version of Master theorem:

$$ \bbox[7px,border:2px solid red] {\text{MASTER THEOREM}}: $$

$$T(n)=aT(n/b)+f(n).$$
$$\bbox[1px,border:1px solid green] {a\ge 1,b>1,f(n)=Θ\big( n^k\cdot \log^p(n)\big)}$$
$$\text{CASE 1}:$$

$$ \text{if } \log_b(a)>k\text{ then: }\quad \bbox[1px,border:1px solid black] {T(n)=Θ\big( n^{\log_b(a)}\big)}   $$

$$\text{CASE 2}:$$

$$\text{if } \log_b(a)=k:\begin{cases}
\text{if $p>-1:\quad \bbox[1px,border:1px solid black] {T(n)=Θ\big( n^k\cdot \log^{p+1}(n)\big)}$} \\
\text{if $p=-1:\quad \bbox[1px,border:1px solid black] {T(n)=Θ\big( n^k\cdot \log\big( \log(n)\big)\big)}
$ }\\
\text{if $p<-1:\quad \bbox[1px,border:1.5px solid black] {T(n)=Θ\big( n^k\big)}$ }\\
\end{cases} $$

$$\text{CASE 3}:$$

$$\text{if $\log_b(a)<k:\begin{cases}
\text{if $p\ge 0:\quad \bbox[1px,border:1px solid black] {T(n)=Θ\big( n^k\cdot\log^p(n)\big)}
$} \\
\text{if $p<0:\quad \bbox[0.5px,border:1px solid black] {T(n)=\mathcal O \big( n^k\big)}
$}
\end{cases}$}$$

答:
Note that the sequence $T(3x), T(3^{2}x), T(3^{3}x), ...$ with $3\nmid x$ can be derived if and only if $T(x)$ is known. (In other words, all 'chains' like this are independent.) Thus, for all values of $T(n)$ to be known, all initial values $T(n)$ with $3\nmid n$ must be specified. In fact, $no$ values of $T(n)$ other than $T(0)$ can be derived right now because $\frac{0}{3} = 0$. We cannot even use continuity conditions because of the domain restriction to natural numbers. So, $\boxed{\text{No, we cannot derive a closed form for }T(n).}$

I will work on this under the condition that $n\in\mathbb{R}$ and that $T$ is continuous, but I do not think I will get very far.

Title: $T(n)=4T(n/3)+n\log_2(n)$


答:
Suppose $\mathcal{T}_1 \subseteq \mathcal{T}_2$. We wish to show that $\mathcal{T}_2 \subseteq \mathcal{T}_1$. Taking your idea on board, let $U \in \mathcal{T}_2$, and hence $X \setminus U$ is compact with respect to $\mathcal{T}_2$.

Since $\mathcal{T}_1 \subseteq \mathcal{T}_2$, any open cover from $\mathcal{T}_1$ of $X \setminus U$ is also an open cover from $\mathcal{T}_2$. By compactness with respect to $\mathcal{T}_2$, such a cover has a finite subcover. Hence, $X \setminus U$ is compact with respect to $\mathcal{T}_1$. Thus $X \setminus U$ is closed with respect to $\mathcal{T}_1$, and so $U \in \mathcal{T}_1$.

Title: Proof about compactness

Body:
Suposse $(X, \mathcal{T}_1)$ and  $(X, \mathcal{T}_2)$ are compact Hausdorff Spaces. Prove that either $\mathcal{T}_1= \mathcal{T}_2$  or $\mathcal{T}_1 \not\subset \mathcal{T}_2$ or $\mathcal{T}_2 \not\subset \mathcal{T}_1$.

My idea: Let $U \in \mathcal{T}_1$ then $X-U$ is closed in $\mathcal{T}_1$ and since $X$ is compact then $X-U$ is compact. But I don't know how to continue. I need an idea


问:
How to write an algorithm as an equation?
Alright, it's with trepidation that I ask this, because dear god do I feel stupid. A question asked:
Whenever we run this formula we double x, then subtract 3 from the result. x = 5. What's the result if we do this 4 times? What's the result if we do it 12 times?
5*2 = 10, 10 - 3 = 7. 7*2 = 14, 14-3 = 11 and so on.

At first I thought it would be simple: ((x*2)-3)*12; but that really doesn't come close to the correct value. Then I thought "aha, exponentiation!" but that's clearly not correct either.

Truth be told, I've never felt so bad at math as I do right now. Right now I just wonder: how would you write that? Where can I learn more on this? It's obvious that I'm going to need to refresh my math skills (and I want to learn more, because math is cool but it's been a while since I did any).

Additionally I feel a bit ashamed asking such a simple question here, because judging by the "similar questions" list I'm the village idiot. The village itself seem to consist of people who are amazingly good at math, so yeah -- it sucks to have to admit to being lost at something like this.

In fact, I'm so lost I don't even know what tags I should use.

Edit: I know the answer (I wrote a short for loop), I just want to know how I'd approach it mathematically instead.

答:
You’re starting with the function $f(x)=2x-3$. I’ll write $f^n(x)$ for the result of applying $f$ repeatedly $n$ times. Thus, you’re looking for $f^{12}(5)$ and more generally $f^n(x)$.

One elementary approach is to calculate $f^n(x)$ for some small values of $n$ and look for a pattern:

$$\begin{align*}
f(x)&=2x-3\\
f^2(x)&=2f(x)-3\\
&=2(2x-3)-3\\
&=2^2x-2\cdot3-3\\
f^3(x)&=2f^2(x)-3\\
&=2\left(2^2x-2\cdot3-3\right)-3\\
&=2^3x-2^2\cdot3-2\cdot3-3\\
f^4(x)&=2f^3(x)-3\\
&=2\left(2^3x-2^2\cdot3-2\cdot3-3\right)-3\\
&=2^4x-2^3\cdot3-2^2\cdot3-2\cdot3-3\;.
\end{align*}$$

At this point you can pick up a clear pattern:

$$f^n(x)=2^nx-2^{n-1}\cdot3-2^{n-2}\cdot3-\ldots-2\cdot3-3=2^nx-3\sum_{k=0}^{n-1}2^k\;.$$

If you know about geometric progressions (or just have some experience working with binary numbers) you can work out that

$$\sum_{k=0}^{n-1}2^k=1+2+2^2+\ldots+2^{n-2}+2^{n-1}=2^n-1\;,$$

so it appears that

$$f^n(x)=2^nx-3\left(2^n-1\right)=2^n(x-3)+3\;.\tag{1}$$

You can now check this: it certainly works when $n=1$, since it says that $$f(x)=2(x-3)+3=2x-3\;.$$ And if it’s right for $f^n(x)$, then it’s right for $f^{n+1}(x)$:

$$\begin{align*}
f^{n+1}(x)&=f\big(f^n(x)\big)\\
&=2f^n(x)-3\\
&=2\Big(2^n(x-3)+3\Big)-3\\
&=2^{n+1}(x-3)+2\cdot3-3\\
&=2^{n+1}(x-3)+3\;,
\end{align*}$$

exactly as $(1)$ would have it. (If you’re familiar with proof by induction, you’ll recognize this argument as an example.)

In particular, we can now compute $f^{12}(5)=2^{12}(5-3)+3=4096\cdot2+3=8195$.


答:
Let $C$ be the solutions. Since $C = -C$ so we can look for solutions with non negative real part.

Note that $2 = |2i| = |z+i - (z-i)| \le |z+i|+|z-i|$.

Let $f(x) = |x+iy+i| + |x+iy-i|$ and note that $f$ is continuous everywhere, differentiable for $ x > 0$ and $f'(x) >0$ for such $x$. It follows from the mean value theorem that $f(x) > f(0) \ge 2$ for $x>0$. Hence it follows that $x=0$.

It is not hard to see that $|iy+i| + |iy-i| = |y+1|+|y-1| = \max(2,2|y|)$, hence
we must have $|y| \le 1$, that is $C \subset [-i,i]$.

It is straightforward to verify that $[-i,i] \subset C$.

Title: Need help with complex equation: |z−i|+|z+i|=2

Body:
I am trying to solve this equation: |z−i|+|z+i|=2 and don't know how to do it. This what I have:
$$\sqrt{(x+1)^2+y^2}+ \sqrt{(x-1)^2+y^2} = 2 /^2$$
$$(x+1)^2+y^2+(x-1)^2+y^2 + 2\sqrt{[(x+1)^2+y^2][(x-1)^2+y^2]} = 4$$
$$x^2 +2x + 1+y^2+x^2-2x + 1+y^2 + 2\sqrt{[(x^2 +2x + 1)+y^2][(x^2-2x + 1)+y^2]}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{[(x^2 +2x + 1)+y^2][(x^2-2x + 1)+y^2]}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{(x^2 +2x + 1)(x^2-2x + 1)+x^2y^2-2xy^2+y^2+x^2y^2-2xy^2+y^2+y^4}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{(x^2 +2x + 1)(x^2-2x + 1)+2x^2y^2-4xy^2+2y^2+y^4}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{x^4+2x^3+x^2-2x^3+4x^2-2x+x^2+2x+1+2x^2y^2-4xy^2+2y^2+y^4}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{x^4+6x^2+1+2x^2y^2-4xy^2+2y^2+y^4}=4$$

And I have no idea what to do next.
Any help would be much appreciated


问:
there are 4 places available to sit, how many ways can 10 people sit if...
there are 4 places available to sit, how many ways can 10 people sit if:

There are 2 people who don't want to sit next to each other

There are 2 people who want to sit next to each other

I know that this problem uses permutation(order matters) or combination(order doesn't matters) or maybe both, also i'm lost because i don't know if this uses repetition or not and  how to approach the problem.

I know that there are 5040 ways for the people if the 4 places are available, but i don't know where to go from here.

答:
Based on the comments from the question, I'm going to assume the following:

There is one pair, Person A and Person B, such that Person A is selected if and only if Person B is selected, and whenever A and B are selected they sit next to each other.

There is also a third person, Person C, such that whenever Person A is selected, either Person C is not selected or Person C is selected but does not sit adjacent to Person A.

If Person D and Person E are seated in spots 1 and 4, respectively, they are not sitting adjacent to each other, i.e. the table is not a circle.

Both adjacency and handedness matter in the count; i.e. order matters, and flipping the order backwards gives a different solution.

Then let's do some case work!

Case 1: Person A is not selected. Then Person B is also not selected. Any of the other 8 people can be seated with no restrictions. There are $\frac {8!}{4!}=1680$ ways to do this.

Case 2: Person A is selected, but Person C is not selected. We know that Person B must be selected, and then we must choose two other people (out of the 7 remaining) to sit. There are 21 ways to choose the other two people, then 2 ways to choose whether A is on the left or the right of B, then 3 places to choose where A sits (not four, because we already picked what side B was on), and then 2 ways to fill in the other two seats. This
gives a total of $252$ solutions.

Case 3: Person A and Person C are selected. Then Person B is also selected. Pick the last person (7 ways to do this) and call that person Person D. Now, if A is on the edge (fully to the left or to the right), we have 2 ways to pick what side, and then B must be next to A, and then C and D can be in either seat (2 ways to pick). But if A is in one of the two center seats, then it's 2 ways to pick whether it's right-center or left-center, then 2 ways to pick what side B is on, and then C and D only have one option. So that's $7(2\cdot2+2\cdot2)=56$ solutions.

In total, with the assumptions given, there are $1680+252+56=1988$ ways for the people to sit.


问:
"Weaker continuum hypothesis:" $2^{\aleph_0} < 2^{\aleph_1}$?
For finite cardinals $A < B$, it's clear that the powerset $2^A$ is smaller than the powerset $2^B$.

However, it seems that when $A < B$ are infinite cardinals, the strict inequality cannot be proven from ZFC: Easton's Theorem tells us that in general, unless something is known about the cofinalities $\text{cf}(A), \text{cf}(B)$, the most we can say is that $2^A \leq 2^B$. (That is, there are models for which both $2^A = 2^B$ and $2^A < 2^B$.)

My question is about the smallest possible case: $A = \aleph_0$, $B = \aleph_1$. If we had that $\aleph_1 = 2^{\aleph_0}$ (Continuum Hypothesis) then obviously we would have $2^{\aleph_0} < 2^{(2^{\aleph_0})} = 2^{\aleph_1}$, but it seems like that's probably a lot stronger than we would need to show $2^{\aleph_0} < 2^{\aleph_1}$. Contrapositively, if $2^{\aleph_0} = 2^{\aleph_1}$, that would imply the negation of the Continuum Hypothesis, but again, this seems like a very strong sufficient condition for $\neg CH$ that wouldn't be necessary.

Can we show for sure that $2^{\aleph_0} < 2^{\aleph_1}$ or $2^{\aleph_0} = 2^{\aleph_1}$? Is there simply no canonical way to answer this (any more than for the Continuum Hypothesis)? What is the status of this question, and are there any well-known open statements to which it is equivalent? I played around with it in a desultory way to discover why approaches like diagonalization or Konig's theorem wouldn't work to establish an injection $2^{\aleph_1} \to 2^{\aleph_0}$ or lack of same, but maybe there are some cleverer techniques out there I'm not familiar with.

答:
Nope, it's consistent with $\mathsf{ZFC}$ that $2^{\aleph_0}=2^{\aleph_1}$. And Easton's theorem already tells us this: for example, the function sending $\kappa$ to $\kappa^+$ for $\kappa>\aleph_0$ and sending $\aleph_0$ to $\aleph_2$ satisfies the requirements in Easton's theorem.

(Of course Easton is overkill here: just start with a model of $\mathsf{CH}$ and add $\aleph_2$-many Cohen reals. This is what Cohen did in his original model of $\mathsf{ZFC+\neg CH}$ if memory serves.)

Amusingly, $2^{\aleph_0}=2^{\aleph_1}$ can happen for non-silly reasons: the proper forcing axiom $\mathsf{PFA}$ proves that $2^{\aleph_0}=2^{\aleph_1}=\aleph_2$, for example.)


问:
How to represent the floor function using mathematical notation?
I'm curious as to how the floor function can be defined using mathematical notation.

What I mean by this, is, instead of a word-based explanation (i.e. "The closest integer that is not greater than x"), I'm curious to see the mathematical equivalent of the definition, if that is even possible.

For example, a word-based explanation of the factorial function would be "multiply the number by all integers below it", whereas a mathematical equivalent of the definition would be $n! = \prod^n_{i=1}i$.

So, in summary: how can I show what $\lfloor x \rfloor$ means without words?

答:
$\lfloor x\rfloor=n\leftrightarrow \{n\}=\{y\in\mathbb{Z}:\forall z\in\mathbb{Z}~ z\le x \rightarrow z\le y\}$

This game is fun!  Now let's prove 1+1=2.


问:
Can I use argmin in a set definition?
If I have a set $S \subseteq \mathbb{R}^n$ and some $t\in \mathbb{R}^n$, how should I write "all vectors $s \in S$ such that the magnitude of their sum $s+t$ is minimal"?

My attempt is: $$\{s \in S \mid \underset{s}{\mathrm{argmin}}(|s+t|)\}$$ but it doesn't really seem right to me.

EDIT:

I had originally posed this question in terms of natural numbers to make it simpler, but failed to realise this meant there was only ever one unique solution..

Another attempt:$$\{s\in S : |s+t| \le |u+t|, \forall u \in S\}$$does this work?

答:
Set builder notation is supposed to go something like this:
$$\{s \in S \mid P(s)\},$$
where $P(s)$ is a predicate, i.e. a true or false statement depending on $s$ (e.g. $s^\top \mathbf{1} = 0$). What you have written doesn't fit this mould. The expression $\underset{s}{\mathrm{argmin}}(|s+t|)$ refers to a set, not a true/false statement. If you are looking for the set of vectors $s \in S$ which minimise the expression $|s + t|$, then you actually want,
$$\underset{s \in S}{\mathrm{argmin}}(|s+t|).$$
The point of argmin is to return the set of minimising vectors $s$, whereas min returns the minimal numeric value of $|s + t|$ (i.e. it will be a non-negative real number, not a scalar).

Title: Can I use argmin in a set definition?


问:
Integral of a function times a Fourier transform is zero
This comes from Hall's Quantum Theory for Mathematicians, Lemma 9.33. I'm having trouble with one of the arguments in the proof. I believe it boils down to the following:

Let $f\colon \mathbb{R}^n \to \mathbb{R}$ such that $\int f(x) \hat g(x) \,\mathrm{d}x = 0$ for all smooth $g$ with compact support, where $\hat g$ denotes the Fourier transform of $g$. I then want to show that $f = 0$. In the proof $f$ is a difference of an $L^2$-function and an $L^2$-function times a coordinate function.

Hall cites the Stone-Weierstrass theorem and the theorem about density of continuous functions with compact support in $L^p$, but he's not super clear. (He also starts talking about the functions as if they are defined on $\mathbb{R}$, so I don't know what is going on.)

Also, do feel free to change the title to something more descriptive, I wasn't sure how to title my question.

答:
Since $g\mapsto \hat g:L^2(\mathbb R^n)\to L^2(\mathbb R^n)$ is an isometric isomorphism, we can choose an approximate identity $(\hat \phi_n).$ Now $\phi_n$ is smooth for each integer $n$ and  there are compactly supported smooth functions $(\psi_{jn})$ such that $\|\psi_{jn}-\phi_n\|\to 0$ and so $\|\widehat{\psi_{jn}-\phi_n}\|=\|\hat\psi_{jn}-\hat\phi_n\|\to 0$ as well. Then,

$\tag1 \|\hat\psi_{jn}*f-f\|\le \|\hat \psi_{jn}*f-\hat \phi_n*f\|+\|\hat \phi_n*f-f\|.$

We have, by Cauchy-Schwarz,

$\|\hat \psi_{jn}*f-\hat \phi_n*f\|\le \|\hat \psi_{jn}-\hat \phi_n\|\cdot \|f\|$ and so $\|\hat \psi_{jn}*f-\hat \phi_n*f\|<\epsilon$ if $j$ is large enough. On the other hand, $\|\hat \phi_n*f-f\|<\epsilon$ if $n$ is large enough, because $(\hat\phi_n)$ is an approximate identity.

But $\hat \psi_{jn}*f(x)=\int \hat \psi_{jn}(x-y)f(y)dy=0$ by hypothesis and so $(1)$ becomes

$\tag2 \|f\|\le \|\hat \psi_{jn}*f-\hat \phi_n*f\|+\|\hat \phi_n*f-f\|<2\epsilon.$

It follows that $f=0$ almost everywhere.


问:
Benford's Law and ice-cream parlor
I have a question which might be remotely related to Benford's Law or perhaps the three busses problem (https://www.newscientist.com/article/dn18074-why-three-buses-come-at-once-and-how-to-avoid-it/).

In an ice cream parlor they have the different ice cream flavors in containers which can be half empty, almost empty, completely full, etc. Whenever a container is empty, it is replaced by a new one of the same flavor. The question is: Is it possible to deduce probabilistically which is the customer's favorite flavor based on the amount left in each container?

Can we deduce any valid statistical information with just one snapshot of the state of the containers?

答: Yes. There is only one reason a container could be almost empty (it's been repeatedly ordered by customers), but two reasons a container could be full, so it's more likely that an almost empty container is more popular than a full one.</p>

Title: Benford's Law and ice-cream parlor


答:
I am an engineer, perhaps this helps if you do not expect too much.

$x-(x$ mod $1$)

Title: How to represent the floor function using mathematical notation?

Body:
I'm curious as to how the floor function can be defined using mathematical notation.

What I mean by this, is, instead of a word-based explanation (i.e. "The closest integer that is not greater than x"), I'm curious to see the mathematical equivalent of the definition, if that is even possible.

For example, a word-based explanation of the factorial function would be "multiply the number by all integers below it", whereas a mathematical equivalent of the definition would be $n! = \prod^n_{i=1}i$.

So, in summary: how can I show what $\lfloor x \rfloor$ means without words?


答:
Triangle inequality tells us $|z+i| + |z-i| \ge |(z+i)-(z-i)| = |2i| = 2$ with equality holding if and only if geometrically of $0$ is colinear and between $z+i$ and $z-i$.

The only values colinear between $z + i$ and $z-i$ must all have the same real value so the real part as $z$ has and if $0$ is one such number, the real part of $z$ must be $0$.

If $Im(z) = b$ then $z = bi$ and we have $|bi+i| + |bi - i| = |b+1|+|b-1|=2$.  By cases we can verify then this will hold if and only if $b \in [-1,1]$.

(If $b > 1$ then $|b+1| > 0$ and $|b-1| > 2$ and if $b < -1$ then $|b+1| > 2$ and $|b-1| > 0$; but if $-1 < b < 1$ then $|b+1| = b+1$ and $|b-1|= 1-b$ and $|b+1|+|b-1| = (b+1)+(1-b)=2$.)

So $z$ can be any $bi; -1 \le b \le 1$.

======

Alternatively (in hindsight) note:

If $z = a + bi$ then $|z\pm i| = \sqrt{a^2 + (b\pm 1)^2}$ so

$|z+i| +|z-i|$

$=\sqrt {a^2 + (b+ 1)^2} + \sqrt {a^2 + (b-1)^2}$

$[*]\ge\sqrt {(b+1)^2} +\sqrt {(b-1)^2}$

$ =|b+1|+|b-1|$

$=\begin{cases}(b+1)+(b-1)=2b>2& b> 1\\ (b+1)+(1-b)=2 &-1\le b \le 1\\(-b-1)+(1-b)=-2b>2 & b < -1\end{cases}$

$[**] \ge 2$

with equality holding if and only if $a = 0$ and $-1 \le b \le 1$.

[*] Equality holding if and only if $a = 0$.

[**] Equality holding if and only if $-1 \le b \le 1$.


问:
Learning by the Moore method - Books for self-study
I recently read about the Moore method for learning mathematics (Moore method Wikipedia) and wanted to apply it to my own learning (undergraduate level). However, I am unable to find any books that follow such a method or something similar.

Specifically, I'm looking for books that introduce the reader to important mathematical concepts through problems/questions rather than blocks of text (for instance, instead of the book proving Lagrange's theorem in group theory, the reader is presented with a question which guides the reader by introducing the definition of cosets etc., so that the reader can prove it themself). I am not very picky about the topic, for I have not been able to find such books on any topic so far. Any recommendations? Thanks.

答:
I found the recently published book "Topology Through Inquiry" here: https://math.hmc.edu/su/topology-through-inquiry/

I actually don't know anything about this book (yet) but am posting this answer in case someone else like me finds this question. I continued my internet search and just stopped here to post what I've found.


问:
If an homomorphism maps elements of the same orders to each other, is then an isomorphism?
I know that, given $f:G_1 \rightarrow G_2$ homomorphism, if $g \in G_1$ has order $n$, then $f(g)$ has order that divides $n$, and that if $f$ is an isomorphism then $f(g)$ has order equals to $n$.

Now my question is, if $f$ is an homomorphism such that $\text{ord}(g)=\text{ord}(f(g))$ for every $g \in G$, can we say that $f$ is an isomorphism?

答:
Note that an element has order $1$ if and only if it is the identity. If you have a homomorphism that respects orders, then if $g\in G_1$, $g\neq e$, it follows that $f(g)\neq e$. Thus, $f$ has trivial kernel, and hence is one-to-one.

In particular, if $G_1$ and $G_2$ are finite of the same order (size), then $f$ must be a bijection. In that case, it is an isomorphism. In even more particular, if $G=G_1=G_2$ and it is a finite group, then $f$ will be an automorphism/isomorphism from $G$ to itself.

If $G_1$ and $G_2$ have different orders, then you must have $|G_1|\lt |G_2|$; this is certainly possible and $f$ need not be an isomorphism. For example, the map from $C_2$ (cyclic of order two) to $C_2\times C_2$ sending $x$ to $(x,e)$ respects order but is not an isomorphism, because it is not surjective.

For infinite groups, it is also not the case that $f$ needs to be an isomorphism, even if they have the same cardinality: $f\colon\mathbb{Z}\to\mathbb{Z}$ given by $f(x)=2x$ respect order but is not surjective. If you want one in which elements have finite order, just take a direct product of infinitely many copies of $C_2$, and then let $f$ be the “right shift” morphism.


答:
I am sharing my calculations with hope to be useful. Taking $n=3^k$ we have
$$T(n)=4\left( \frac{n}{3} \right)+n\log n = 4\left[4T\left( \frac{n}{3^2}\right) + \frac{n}{3}\log \frac{n}{3}\right]+n\log n = \\
 = 4^2T\left( \frac{n}{3^2}\right)+4 \frac{n}{3}\log \frac{n}{3} +n\log n=\\
=4^2\left[ 4T\left( \frac{n}{3^3}\right) + \frac{n}{3^2}\log \frac{n}{3^2} \right] +4 \frac{n}{3}\log \frac{n}{3} +n\log n=\\
= 4^3T\left( \frac{n}{3^3}\right)+4^2\frac{n}{3^2}\log \frac{n}{3^2}+4 \frac{n}{3}\log \frac{n}{3} +n\log n = \\ = \cdots =\\
=4^kT(1)+4^{k-1}\frac{n}{3^{k-1}}\log \frac{n}{3^{k-1}}+\cdots+4 \frac{n}{3}\log \frac{n}{3} +n\log n$$
Now we can take some rough estimation, for example $T(n) \in O(n\log_3 n4^{\log_3 n})$, but I don't know how convenient it is for you.

Title: $T(n)=4T(n/3)+n\log_2(n)$

Body:
I was wondering if i could find a closed form for this function:

$$T(n)=4T(n/3)+n\log_2(n) \text{ for $ n\in \mathbb N$}$$
$$\text{ & }T(0)=1$$
or at least some function $g$ s.t $T(n)=Θ\big(g(n) \big)$,

(I've tried Master theorem and it does not work here).

any ideas?

EDIT:$\text{ it turns out that: $\bbox[1px,border:1px solid red] {T(n)=Θ\big( n^{\log_3(4)} \big)}$}$

By using the strong version of Master theorem:

$$ \bbox[7px,border:2px solid red] {\text{MASTER THEOREM}}: $$

$$T(n)=aT(n/b)+f(n).$$
$$\bbox[1px,border:1px solid green] {a\ge 1,b>1,f(n)=Θ\big( n^k\cdot \log^p(n)\big)}$$
$$\text{CASE 1}:$$

$$ \text{if } \log_b(a)>k\text{ then: }\quad \bbox[1px,border:1px solid black] {T(n)=Θ\big( n^{\log_b(a)}\big)}   $$

$$\text{CASE 2}:$$

$$\text{if } \log_b(a)=k:\begin{cases}
\text{if $p>-1:\quad \bbox[1px,border:1px solid black] {T(n)=Θ\big( n^k\cdot \log^{p+1}(n)\big)}$} \\
\text{if $p=-1:\quad \bbox[1px,border:1px solid black] {T(n)=Θ\big( n^k\cdot \log\big( \log(n)\big)\big)}
$ }\\
\text{if $p<-1:\quad \bbox[1px,border:1.5px solid black] {T(n)=Θ\big( n^k\big)}$ }\\
\end{cases} $$

$$\text{CASE 3}:$$

$$\text{if $\log_b(a)<k:\begin{cases}
\text{if $p\ge 0:\quad \bbox[1px,border:1px solid black] {T(n)=Θ\big( n^k\cdot\log^p(n)\big)}
$} \\
\text{if $p<0:\quad \bbox[0.5px,border:1px solid black] {T(n)=\mathcal O \big( n^k\big)}
$}
\end{cases}$}$$


问:
Evaluate $\int_{0}^{\pi/4}x\ln^{2}(\sin(x))dx$
Here is a some what challenging log sine integral.

$$I=\int_{0}^{\pi/4}x\ln^{2}(\sin(x))dx$$

The upper limit of integration is $\frac{\pi}{4}$ instead of the usual $\frac{\pi}{2}$.

I tried the famous identity $\displaystyle\ln(\sin(x))=-\sum_{k=1}^{\infty}\frac{\cos(2kx)}{k}-\ln(2)$

But, the log is squared and I do not think Parseval is applicable here. But, I may be wrong.  By Parseval, I mean $\displaystyle\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{\cos(kx)\cos(nx)}{kn}$.

I have tried other means.  I made the sub $t=\sin(x)$. This led to $\displaystyle\int_{0}^{\frac{1}{\sqrt{2}}}\frac{\sin^{-1}(t)\ln^{2}(t)}{\sqrt{1-t^{2}}}dt$

Now, I thought perhaps the famous sum $\displaystyle\frac{\sin^{-1}(t)}{\sqrt{1-t^{2}}}=\sum_{k=1}^{\infty}\frac{(2t)^{2k-1}}{k\binom{2k}{k}}$ could be used.

So, I integrated and got the form

$$I=\frac{\ln^{2}(2)}{16}\sum_{k=1}^{\infty}\frac{(k!)^{2}2^{k}}{k^{2}(2k)!}+\frac{\ln(2)}{8}\sum_{k=1}^{\infty}\frac{(k!)^{2}2^{k}}{k^{3}(2k)!}+\frac{1}{8}\sum_{k=1}^{\infty}\frac{(k!)^{2}2^{k}}{k^{4}(2k)!}$$

These sums actually evaluate to the correct integral result, so it would appear they are equivalent to said integral. But evaluating them may be even more daunting. Though the first one from the left evaluates to $\frac{{\pi}^{2}}{8}$ and the center one evaluates to

$$\frac{\ln(2)}{8}\left(\pi G-\frac{35}{16}\zeta(3)+\frac{{\pi}^{2}}{8}\ln(2)\right)\;.$$

This center one looks to be related to the integral:

$\displaystyle8\ln(2)\int_{0}^{\frac{\pi}{4}}\ln(\sin(x))dx=\frac{35}{16}\ln(2)\zeta(3)-\pi\ln(2)G-\frac{{\pi}^{2}\ln^{2}(2)}{4}$, which I got by expanding the square of the log sine and integrating.

This just leaves the sum to the right.  It is tougher and I can find no closed form.

Alas, If I could evaluate it then I think the integral would be solved.

I also thought...how about a closed form of some sort for

$$\int_{0}^{\frac{\pi}{4}}x\sin^{a}(x)dx$$

that could be differentiated w.r.t 'a' and applied?. I do not know of a closed form for
this, though.

I got an encouraging looking solution by expanding, making the sub $t=x/2$, then solving for the integral on the far right (which is the integral in question):

$\displaystyle  \int_{0}^{\frac{\pi}{2}}x\ln^{2}(2\sin(x/2))dx$
$=\displaystyle \ln^{2}(2)\int_{0}^{\frac{\pi}{2}}xdx+8\ln(2)\int_{0}^{\frac{\pi}{4}}x\ln(\sin(x))dx+\color{red}4\int_{0}^{\frac{\pi}{4}}x\ln^{2}(\sin(x))dx$

But, it did not check numerically. (Error has now been fixed. - Editor) Numerically, the integral in question should evaluate to $I = 0.263605....$

Then again, I may just have a computation error in all of that.

So, does anyone know of a method for evaluating this log-sine integral?.

EDIT:  Here is what Mathematica gave for the solution.  It may be too involved to be worth the effort.

$\displaystyle I=\frac{37{\pi}^{4}}{92160}-\frac{\pi}{4}\sum_{k=1}^{\infty}\frac{\sin(\pi k/4)}{2^{k/2}k^{3}}$ $\displaystyle +\frac{\pi}{8}\ln(2)G+\frac{25}{768}{\pi}^{2}\ln^{2}(2)+\frac{\ln^{4}(2)}{384}+\sum_{k=1}^{\infty}\frac{\cos(\pi k/4)}{2^{k/2}k^{4}}-\frac{35}{128}\ln(2)\zeta(3)$.

Sometimes with these math engines the solution can be simplified down. But, maybe not in this case. Those complex polylogs may pose a problem.

Body:
Here is what I got from Mathematica:

$$\frac{1}{8} \, _5F_4\left(1,1,1,1,1;\frac{3}{2},2,2,2;\frac{1}{2}\right)+\frac{1}{8} \log (2) \, _4F_3\left(1,1,1,1;\frac{3}{2},2,2;\frac{1}{2}\right)+\frac{1}{128} \pi ^2 \log ^2(2) = 0.263605$$


问:
why Rudin has been taken $\mu(S_n^c)=0$ and why $\mu(E^c)=0 ?$
I have some  confusion in  Rudin RCA Book

My confusion have been marked  in red box , given  below

My attempt  : Here its given that  $S_n$ is the set   on $f_n$ and $\mu(S_n^c)=0$

Then it will  implies  $ S_n$ is infinite and $S_n^c$ is finite

Then  it   should be  $\int_S \varphi d\mu  > \infty$

Another  its given that $E= \{ x \in S : \varphi(x)  < \infty\}$ this implies $E $ is  finite set and $\mu(E^c)=\infty$ since $E^c= \{ x \in S : \varphi(x)  > \infty\}$

My doubt : $1.$ why Rudin has been taken   $\mu(S_n^c)=0$ ? what is the  logics behind this ?

$2.$why  $\mu(E^c)=0 ?$

答:
a.e. means almost everywhere, so when something happens a.e. we're saying that the set where it doesn't happen has zero measure.

Since each $f_n$ is defined a.e. on $X$, if we denote $S_n$ as the set where $f_n$ is defined, then the set where it doesn't happen (where $f_n$ isn't defined) is $S_n^\mathsf c$, so $\mu(S_n^\mathsf c)=0$.

Now consider $S=\bigcap S_n$ and $E=\{x\in S:\varphi(x)=\sum|f_n(x)|<\infty\}$. Note that $x\in E^\mathsf c\Leftrightarrow x\notin E\Leftrightarrow x\notin S\lor (x\in S\land \varphi(x)=\infty)\Leftrightarrow x\in S^\mathsf c\lor(x\in S\land\varphi(x)=\infty)\Leftrightarrow x\in\bigcup S_n^\mathsf c\,\lor(x\in S\land\varphi(x)=\infty)$

so $E^\mathsf c=\big(\bigcup S_n^\mathsf c\big)\sqcup\{x\in S:\varphi(x)=\infty\}$ (the union is clearly disjoint, since the first part is $S^\mathsf c$ and the second part is in $S$).

Since $\mu(\bigcup S_n^\mathsf c)=\mu(S^\mathsf c)=0$, then $\mu(E^\mathsf c)=\mu\big(\bigcup S_n^\mathsf c\big)+\mu\big(\{x\in S:\varphi(x)=\infty\}\big)=\mu\big(\{x\in S:\varphi(x)=\infty\}\big)$.

Now imagine $\mu(E^\mathsf c)\neq0\Rightarrow\mu(E^\mathsf c)>0$. Then, since $\mu(S^\mathsf c)=0$ we have that $\mu\big(\{x\in S:\varphi(x)=\infty\}\big)>0$, and knowing that $\mu\big(\{x\in S:\varphi(x)=\infty\}\big)\subseteq S$ and $\phi(x)\ge0$ on $S$,

$$\int_S\varphi\,\text{d}\mu\ge\int_{\{x\in S:\varphi(x)=\infty\}}\varphi\,\text{d}\mu=\infty$$ since $\varphi(x)=\infty$ on $\{x\in S:\varphi(x)=\infty\}$ and $\mu\big(\{x\in S:\varphi(x)=\infty\}\big)>0$.

Therfore $\displaystyle\int_S\varphi\,\text{d}\mu=\infty$, which is a contradiction, so we conclude that $\mu(E^\mathsf c)=0$.


问:
Complexity of matrix multiplication with different size
I have three matrices $A, ~B, \text{and} ~C$ which are of size $M \times N, ~ N \times P, ~\text{and}~ P \times M $, respectively. What is the computational complexity of the multiplication $A\times B \times C$?

答:
The "naive" matrix multiplication for $A\times B$ involves multiplying and adding $N$ terms for each of $MP$ entries in $AB$. So the complexity is $O(NMP)$. And then multiplying this $M\times P$ matrix by $C$ requires multiplying and adding $P$ terms for each of $MN$ entries. So the total complexity is $O(M^2N^2P^2)$. (EDIT: This conclusion was incorrect and based on a silly arithmetic mistake that I made. The correct answer, as explained in the comments, is $O(MNP+M^2P)$. Many thanks to @HoldenLee and @Rami Zouari for catching this.)

However, this may not be optimal algorithm. But unfortunately, there's very little information online about the efficiency of non-square matrix multiplication. If all three matrices were square, then the fastest known algorithm for multiplying two of them has complexity $\approx O(N^{2.3729})$; this means that multiplying three $N\times N$ matrices will have complexity just under $O(N^{4.75})$. If the matrices have dimensions that are multiples of each other (or close to multiples) then we can use the square algorithms and block multiplication to speed up the implementation.

I managed to find a paper from 2012 which gets better than $O(N^3)$ results for multiplying $N\times M$ by $M\times N$ matrices, for those values $M<N^{0.30298}$. Assuming that either $N$ or $P$ is less than or equal to $M^{0.30298}$, you could perform one multiplication naively and then use the results of the paper to perform the other multiplication quickly. Depending on the exact values in question, that might be able to get you a result that is better by maybe a factor of $\sqrt{M}$. But in general, I don't think that there are any known algorithms for your specific case that get any better than $O(M^2N^2P^2)$.


问:
Finding $c_n$ so that $\int f_n(x) dm =1$ and $f$ is a valid pdf.
We have $f_n(x) = c_n(1-\frac{x}{n})^n \mathbb{1}_{[0,n]}$ and $x \in \mathbb{R}, n\geq 1$. I want to find $c_n$ so that $\int f_n\, dm = 1$ where $m$ is Lebesgue measure and show that $\lim_{n\to\infty} f_n(x) = f(x)$ exists for all $x \in \mathbb{R}$ and that $f$ is a probability density function (pdf).

So $$1 = \int f_n \, dm = \int c_n(1-\frac{x}{n})^n \mathbb{1}_{[0,n]}, \, dm = c_n(1-\frac{x}{n})^n *n $$

So $c_n =\frac{1}{n(1-\frac{x}{n})^n}$. But then $$ \lim_{n\to\infty} f_n(x) = \lim_{n\to\infty} \frac{1}{n(1-\frac{x}{n})^n} *(1-\frac{x}{n})^n \mathbb{1}_{[0,n]} = 0 = f?$$

So how could $f$ be a probability density function if it is 0? I believe that my $c_n$ is wrong and I don't really know where to make use of $\lim_{n\to\infty} (1-\frac{x}{n})^n =e^{-x}$ or if I am calculating the integral correctly. I'll appreciate any pointers for how to go about the problem.

答:
You need to calculate $\int_{\mathbb{R}} (1-\frac{x}{n})^n 1_{[0,n]}dx=\int_0^n (1-\frac{x}{n})^n dx=n\int_0^1 u^n du=\frac{n}{n+1}$. In your calculation, you never computed the integral correctly.

As far as what does $f_n$ converge to: Look up the definition of $e^x$ with limits.


问:
Is there a shortcut to calculate eigendecomposition of a singular matrix?
I know that at least one of the eigenvalues are zero, but is there any other way to ease the eigendecomposition?</p>

答:
Here is a useful trick for singular matrices which works particularly well if $A$ has "very low" rank. If $A$ is of size $n$ and rank $r<n$, then we can compute a rank-factorization $A = CF$ where $C,F$ are of full rank with sizes $n \times r, r \times n$ respectively. Note that such a factorization can be computed using only row-reduction.

We find that the non-zero eigenvalues of $A$ are precisely equal to the eigenvalues of the $r \times r$ matrix $FC$. Moreover, if $x$ is an eigenvector of $FC$ with $FC = \lambda x$, then $Cx$ is an eigenvector of $A$ with
$$
A(Cx) = C(FC)x = C (\lambda x) = \lambda Cx.
$$
If $A$ is diagonalizable, then it can be diagonalized as follows: let $P$ be any matrix such that $[C \ \ P]$ is invertible. Taking $S = [C \ \ P]$, we find that $S^{-1}AS$ is diagonal.


问:
Verifying Convolution Identities
Note:  I don't yet have a solution to my main issue yet which I have elaborated on in the edit.  Further attention is deeply appreciated.  :>

$\bf{\text{Original Question}}$:

Let $G$ be a locally compact group, $\mu$ a Haar measure for $G$ and $f,g\in L^{1}(G)$.

Then by definition $(f*g)(x) = \int_{G}f(y)g(y^{-1}x)\mu(dy)$.

The author remarks in the book I am reading that we also have the following identities:

\begin{eqnarray*}
(f*g)(x) &=& \int_{G}f(xy)g(y^{-1})\mu(dy)\\
         &=& \int_{G}\delta(y^{-1})f(y^{-1})g(yx)\mu(dy)\\
         &=& \int_{G}\delta(y^{-1})f(xy)g(y)\mu(dy)
\end{eqnarray*}

where $\delta:G\to(0,\infty)$ is the unique modular function such that $\delta(x)\mu(B) = \mu(Bx)$ for all Borel sets $B$ and $x\in G$.

I cannot seem to get verify any of them.  Can anyone help me get started on this?  The smallest suggestion possible is best.

I was trying to use the property that $\int_{G}f(xy)\mu(dy) = \int_{G}f(y)\mu(x\cdot dy)$ but I couldn't get anywhere.  Any suggestions on what the correct trick is?

$\bf{\text{Edit }}\text{(Far more basic question)}:$

I get very confused with some of the measure theory notation, so I've been trying to neurotically stick to one convention so as not to be confused.

For a $\mu$-integrable $f$, I've been writing $$\int_{G}fd\mu = \int_{G}f(x)\mu(dx)$$ to mean that $x$ is my variable of integration.

Then based on this convention, if we define $\lambda(E) = \mu(yE)$, then I've been writing

$$\int_{G}fd\lambda = \int_{G}f(x)\lambda(dx) = \int_{G}f(x)\mu(y\cdot dx)$$

It this the quantity you are referring to when you write $\int_{G}f(x)d(yx)$?

If so, am I correct to perform the following calculation (returning to the case where $\mu$ is a Haar measure)?

$\begin{eqnarray*}
\int_{G}fd\mu &=& \int_{G}f(x)\mu(dx)\\
              &=& \int_{G}f(x)\mu(y\cdot dx)\\
              &=& \int_{G}f(y^{-1}yx)\mu(y\cdot dx)\\
              &=& \int_{G}f(y^{-1}x)\mu(dx)\\
              &=& \int_{G}f(y^{-1}\cdot)d\mu
\end{eqnarray*}$

Please be mercilessly honest if anything is even slightly incorrect.  I've had a "vague" understanding of these technicalities for far too long and I want to finally really nail them down.

答:
To get all of these properties, use the defining property of the Haar measure: It is left-invariant under the action of $G$. In other words, $\int f(x)dx=\int f(x) d(yx)$ for any $y\in G$. Once you have this, you can change the measure in that fashion without affecting the value of the integral, and then change variables to get your old measure back: for instance, in my previous example, replace $x$ with $y^{-1}x,$ so you get $\int f(x) d(yx)=\int f(y^{-1}x) d(yy^{-1}x)=\int f(y^{-1}x)dx.$ [By the way, I'm suppressing the $\mu$ in my notation to make things look cleaner.]

Thus, we can get the first equality you ask about by replacing $y$ with $xy,$ so we get \begin{align*}(f * g)(x)&=\int_G f(y)g(y^{-1}x)dy\\&=\int_Gf(xy)g((xy)^{-1}x)d(xy)\\&=\int_Gf(xy)g(y^{-1})d(xy)\\&=\int_G f(xy) g(y^{-1})dy,\end{align*} where the final equality comes from the left-invariance of Haar measure.

The next identity you ask about can be shown in a similar fashion. For the final two, remember that the defining property of the modular function is that it reverses invariance of Haar measure: multiplying by the modular function will change your Haar measure from left-invariant to right-invariant, so those two identities will require you to use the property of right-invariance, but are otherwise the same as the above two.

Title: Verifying Convolution Identities


问:
Prove that $\sqrt[3]{2}+\sqrt[3]{4}$ is irrational
As the question says, how can I prove that $\sqrt[3]{2}+\sqrt[3]{4}$ is irrational?

I have tried setting it to be equal to $a$, and $\sqrt[3]{4}$ equal to $a^2$, but I haven't gotten anywhere. The solution does not have to use the above. Any help is appreciated. (I know this is a duplicate, but I haven't seen any answers that I have gotten a full solution from)

答:
Let $\alpha=\sqrt[3]2+\sqrt[3]4$.

Then $\alpha^3=2+4+3\sqrt[3]{32}+3\sqrt[3]{16}=6+6\sqrt[3]4+6\sqrt[3]2=6+6\alpha.$

Apply the rational root theorem to $x^3-6x-6$.


问:
How to solve $\ddot{f}=f^{-2}$
I am facing a question where I need to find the solution for this differential equation:

$$\frac{d^2}{dt^2}f(t)=\frac{1}{f(t)^2}$$

What type of differential equation is this? What is the standard method for solving it?

答:
We want to solve $f''=f^{-2}$. To that end, we multiply both sides by $f'$:

$$f''f'=f^{-2}f'\implies \frac{df'}{dx}f'=f^{-2}\frac{df}{dx}$$

So if we integrate with respect to $x$ and simplify the differentials:

\begin{align*}
\int f'df'&=\int f^{-2}df\\
(f')^2&=-f^{-1}+C_1\\
f'&=\pm\sqrt{C_1-f^{-1}}\\
\int\frac{df}{\sqrt{C_1-f^{-1}}}&=\pm\int dx
\end{align*}
Using our good friend WolframAlpha we can simplify the left-hand integral:
$$C_1^{-\frac32}\tanh^{-1}\left(\frac{\sqrt{C_1-f^{-1}}}{\sqrt{C_1}}\right)+C_1^{-1}f\sqrt{C_1-f^{-1}}=C_2\pm x$$
Where $f^{-1}$ denotes $\frac1{f(x)}$ and not the inverse of the function $f$. This is an implicit description of $f$, but I think it's the best we can do.

Title: How to solve $\ddot{f}=f^{-2}$


问:
Simple solution to combinations with limited repetition
Suppose that you are going to prepare a fruit salad with oranges, apples, and bananas. The salad should consist of 10 pieces. No fruit should be used more than 5 times.

Without the constraint, the solution would be

with $n=10$ and $k=3$, I believe.

However, the problem has a constraint. A straigthforward approach would be: If there are zero bananas, then there is one possible distribution of apples and oranges (only (5,5) would yield 10). If there are 5 bananas, there are 6 possible distributions (5,5), (4,1), (3,2) ...).

This yields $6 + 5 + 4 + 3 + 2 + 1 = 21$ in total.

Is there a more efficient and more generalizable approach that would easily work with more fruits?

I am not looking for solving this with the inclusion-exclusion approach, which is possible for any constraint (right?). I am also not looking for dynamic programming or anything like that.

I found the exercise in a textbook about the basics of combinatorics, so I assume that there must be a simple solution using the typical combinatorics formulas that works in the special case where the constraint is the same for all the fruits involved (i.e. in contrast to different limits per fruit).

Any advice is appreciated!

答:
I visualise three containers , containing $5$ oranges, $5$ apples, and $5$ bananas.

If I remove a total of $5$ fruits from them any which way, I will be left with the $10$ fruits I need, and thus $5+3-1\choose 2$ ways


问:
$T(n)=4T(n/3)+n\log_2(n)$
I was wondering if i could find a closed form for this function:

$$T(n)=4T(n/3)+n\log_2(n) \text{ for $ n\in \mathbb N$}$$
$$\text{ & }T(0)=1$$
or at least some function $g$ s.t $T(n)=Θ\big(g(n) \big)$,

(I've tried Master theorem and it does not work here).

any ideas?

EDIT:$\text{ it turns out that: $\bbox[1px,border:1px solid red] {T(n)=Θ\big( n^{\log_3(4)} \big)}$}$

By using the strong version of Master theorem:

$$ \bbox[7px,border:2px solid red] {\text{MASTER THEOREM}}: $$

$$T(n)=aT(n/b)+f(n).$$
$$\bbox[1px,border:1px solid green] {a\ge 1,b>1,f(n)=Θ\big( n^k\cdot \log^p(n)\big)}$$
$$\text{CASE 1}:$$

$$ \text{if } \log_b(a)>k\text{ then: }\quad \bbox[1px,border:1px solid black] {T(n)=Θ\big( n^{\log_b(a)}\big)}   $$

$$\text{CASE 2}:$$

$$\text{if } \log_b(a)=k:\begin{cases}
\text{if $p>-1:\quad \bbox[1px,border:1px solid black] {T(n)=Θ\big( n^k\cdot \log^{p+1}(n)\big)}$} \\
\text{if $p=-1:\quad \bbox[1px,border:1px solid black] {T(n)=Θ\big( n^k\cdot \log\big( \log(n)\big)\big)}
$ }\\
\text{if $p<-1:\quad \bbox[1px,border:1.5px solid black] {T(n)=Θ\big( n^k\big)}$ }\\
\end{cases} $$

$$\text{CASE 3}:$$

$$\text{if $\log_b(a)<k:\begin{cases}
\text{if $p\ge 0:\quad \bbox[1px,border:1px solid black] {T(n)=Θ\big( n^k\cdot\log^p(n)\big)}
$} \\
\text{if $p<0:\quad \bbox[0.5px,border:1px solid black] {T(n)=\mathcal O \big( n^k\big)}
$}
\end{cases}$}$$

答:
This might not be what you're looking for, but there's always the method of power series.

Let $T(x)=\sum_{k\ge0}a_kx^k$, so that
$$\sum_{k\ge0}\left(1-4/3^k\right)a_kx^k=-\frac{1}{\log2}+\frac{x}{\log2}+\frac{1}{\log2}\sum_{k\ge2}\frac{(-1)^k(x-1)^k}{k(k-1)}.$$
Since $v>u\Rightarrow{u\choose v}=0$, we have from the binomial theorem that
$$(x-1)^k=\sum_{r\ge0}(-1)^{k-r}{k\choose r}x^r,$$
and thus
$$\sum_{k\ge0}\left(1-4/3^k\right)a_kx^k=-\frac{1}{\log2}+\frac{x}{\log2}+\frac{1}{\log2}\sum_{r\ge0}x^r\underbrace{\sum_{k\ge2}\frac{(-1)^k}{k(k-1)}{k\choose r}}_{\alpha_{r}}.$$
This is
$$\sum_{k\ge0}\left(1-4/3^k\right)a_kx^k=2-\frac2{\log2}+\left(1+\frac1{\log2}\right)x+\sum_{k\ge2}\frac{\alpha_k}{\log2}x^k,\qquad |x-1|<1.$$
Therefore
$$\begin{align}
a_0&=\frac13\left(\frac1{\log2}-1\right),\\
a_1&=-3\left(1+\frac1{\log2}\right),\\
a_k&=\frac{3^k}{(3^k-4)\log2}\sum_{n\ge2}\frac{(-1)^n}{n(n-1)}{n\choose k},\qquad k\ge2.
\end{align}
$$
This being the case, I highly doubt there is a closed form.

Title: $T(n)=4T(n/3)+n\log_2(n)$


问:
Proof about sets with no maximum: show the distance between two functions is a metric
My topology book asks me to prove that the distance property defined on pairs of functions below is a metric. The definition of the property involves the concept of least upper bound, presumably in order to encompass functions which do not have a maximum or minimum. THe first three properties of a metric are easy to demonstrate. I have difficulty with the last one.

For functions bounded on [a,b], the distance is defined as $d(f,g)$ =l.u.b. $\cup_{x\in [a,b]} \{|f(x)-g(x)|\}$

It is required to prove that $d(f,h)\le d(f,g)+d(g,h)$.

If $|f(x)-h(x)|$ has a maximum then the proof is easy:

$d(f,h)=|f(c)-h(c)|\le |f(c)-g(c)|+|g(c)-h(c)|\le d(f,g)+d(g,h)$

Actually, I don't have an example of a function pair whose absolute difference has no maximum on a closed interval.

Here is my attempt at proof using the least upper bound concept:

$\forall \epsilon\gt 0,\exists x\in [a,b]$ such that $|f(x)-h(x)| + \epsilon \gt d(f,h)$

$|f(x)-h(x)|\le |f(x)-g(x)|+|g(x)-h(x)| \le d(f,g)+d(g,h)$

$\forall \epsilon \gt 0$,  $d(f,h)-\epsilon \lt d(f,g)+d(g,h)$

$d(f,h)\le d(f,g)+d(g,h)$

Is this a correct way of dealing with least upper bound and solving the problem?

答:
Aside from a $<$ that should be a $\le$ in your third last line of mathematics, this is all perfectly valid. If you want to expand the implication from the second last line to your last line, suppose for the sake of contradiction that $d(f, h) > d(f, g) + d(g, h)$, and let $\varepsilon = d(f, h) - d(f, g) - d(g, h) > 0$. This will lead to a contradiction.

If you want an example where this maximum is not achieved, take $f(x) = \arctan(x)$, and $g(x) = 0$. Then the supremal value of $|f(x) - g(x)|$ is $\pi/2$, but it is never achieved (and only approached via the two horizontal asymptotes).


问:
Convergence of quadratic variations
This may be a fundamental question on a martingale theory.

Let $n \in \mathbb{N}$ and $M^n=(M^{n,1},\ldots,M^{n,d})$ be a $d$-dimensional square integrable martingale on a probability space with probability measure $P_n$. Each $M^n$ may not be continuous.

We denote by $E_n$ the expectation under $P_n$. We assume that for any $t \in [0,1]$
\begin{align*}
\lim_{n \to \infty}E_n\left[\left|[M^{n,i},M^{n,j}]_t - \delta_{i,j}t\right|\right]=0.
\end{align*}
Then, can we show that the law of $M^n$ converges weakly to that of a $d$-dimensional Brownian motion in $D([0,1])$? Here, $D([0,1])$ is the space of right continuous functions with finite left limits with Skorohod topology.

This may be true. However, I do not the proof.

Please tell me a reference.

答:
For a sequence $ M^n=\{M_t^n, t\in [0,1] \} $ of square integrable martingales,
the conclusion of $ M^n\overset{D([0,1])}{\longrightarrow}\mathrm{BM} $ from $ [M^n]_t\to t, t\in [0,1] $ may not be true.
In p.476 of the book:  Jacod, J. and  A. N. Shiryayev, Limit Theory for Stochastic Processes, 2ed. Springer, 2003, there is an example to show the condition ($ [M^n]_t\to t, t\in [0,1] $) is not sufficient. Also in p.473 of same book, the Theorem 3.11 explains
that if $ |\Delta M^n|\le K $, then the conlusion is OK. Generally, to guarantee
$ M^n\overset{D([0,1])}{\longrightarrow}\mathrm{BM} $, further restriction on  $|\Delta M^n|$ (similar to Lindeberg's condition) is nessesary.

Title: Convergence of quadratic variations


问:
Show $ \sum_{i=0}^n X_{i} $ is a complete sufficient statistic.
We're given a random sample of $X_{1}, X_{2}, .... , X_{n}$ from a distribution with a pdf of

$f(x;\theta) = \theta^x(1-\theta), x = 0,1,2,.... $ and $ 0 < \theta < 1 $

We're asked to show that that $Tn =  \sum_{i=0}^n X_{i} $ is a complete sufficient statistic for $\theta$.

I am able to show that $Tn$ is a sufficient statistic for $\theta$, I am just stuck on how to show that it is complete.

To show that $Tn$ is complete, we must show $E(g(Tn)) = 0 $. Where I am stuck is I do not know the distribution of Tn so I am not sure what to use for $f_{Tn}(t)$ for this expectation.

答:
$X_i$ is a geometric random variable. Thus $T_n$ is the number of tails that appear before the $n$th heads, when repeatedly flipping a coin that has a $\theta$ chance of being tails. You can show that the PMF is
$$P(T_n = k) = \binom{n+k-1}{k}\theta^k (1-\theta)^n$$


问:
How to multiply a vector in vector form by a scalar?
Is this possible to do in linear algebra?

答:
$$
\begin{align*}
-2 \cdot \left( \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}
+ t \begin{bmatrix} x_1 \\ x_2 \\ x_3-1 \end{bmatrix} \right)
&=
(-2) \cdot \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}
+ (-2)t \cdot \begin{bmatrix} x_1 \\ x_2 \\ x_3-1 \end{bmatrix} \\
&= \begin{bmatrix} 0 \\ 0 \\ -2
\end{bmatrix}
+ \begin{bmatrix}
-2tx_1 \\ -2tx_2 \\ -2tx_3 + 2t
\end{bmatrix} \\
&=\begin{bmatrix}
-2tx_1 \\
-2tx_2 \\
-2 -2tx_3 + 2t
\end{bmatrix}
\end{align*}
$$


问:
An interesting problem with "decomposing" natural numbers.
Consider the following one-person game:

A player starts with score $0$ and writes the
number $20$ on an empty whiteboard. At each step, she may erase any one integer (call it $a$)
and writes two positive integers (call them $b$ and $c$) such that $b + c = a$. The player then
adds $b × c$ to her score. She repeats the step several times until she ends up with all $1$s on
the whiteboard. Then the game is over, and the final score is calculated.

Example: At the first step, a player erases $20$ and writes $14$ and $6$, and gets a score of
$14 × 6 = 84$. In the next step, she erases $14$, writes $9$ and $5$, and adds $9 × 5 = 45$ to her
score. Her score is now $84 + 45 = 129$. In the next step, she may erase any of the remaining
numbers on the whiteboard: $5$, $6$ or $9$. She continues until the game is over.

Alya and Bob play the game separately. Alya manages to get the highest possible final score.
Bob, however, manages to get the lowest possible final score. What is the difference between
Alya’s and Bob’s final scores?

I tried to "decompose" into a few numbers and I get the same scores. I am not sure how to prove the conjecture that any numbers will yield the same score no matter which path is taken.

答:
Compared to Misha Larov's answer, my solution has essentially the same idea but a different interpretation.

Let's say the number we start off with is $n$. At any stage of the game, we assign a complete graph $K_i$ to any number $i$ written on the board.

The action of splitting a number $a$ into $b$ and $c$ can be reinterpreted as

Choosing disjoint vertex subsets $B$ and $C$ of $K_a$ with respectively $b$ and $c$ elements

Deleting every edge connecting $i \in B$ and $j \in C$

Obtaining new complete graphs $K_b$ and $K_c$.

The score the player gets after this splitting is the number of edges deleted in step 2. Throughout the game, we actually count the total number of edges deleted.

At the final condition, where all graphs are $K_1$, which are individual vertices, we have eliminated all edges of the initial $K_n$. Hence the final score is always the number of edges in $K_n$, $\frac{n(n-1)}{2}$.


问:
Prove that if $x^p - a$ is reducible over a field $F,$ then it has a root in $F.$
This question was asked on my abstract algebra quiz, and I was unable to solve it.

Consider a field $F$ with an element $a \in F$ and a prime number $p.$ Suppose that the polynomial $x^{p} - a$ is reducible in $F[x].$ Prove that this polynomial has a root in $F.$

Attempt: Reducibility of the polynomial implies that $x^{p}-a =q(x)r(x)$ with $q(x)$ and $r(x)$ both nonzero non-units. Now, I assumed that neither $q(x)$ nor $r(x)$ has a root in $F,$ so $\frac{1}{q(x)}$ and $\frac{1}{r(x)}$ are well-defined for all $x$ in $F.$ But I need to prove that $\frac{1}{q(x)}$ and $\frac{1}{r(x)}$ belong to $F[x].$ They can be defined using the Newton Binomial Theorem, but there is no surety that they would necessarily belong to $F[x].$

So, I think there is a problem in my approach. Can anyone please tell me how to approach the problem? Thanks.

答:
Let $K$ denote $F(\zeta)$, where $\zeta$ is the $p$th root of unity $e^{2\pi i/p}$. Since $f=x^{p}-a$ is reducible over $F$, it's also reducible over $K$. So $f$ splits completely over $K$.

{$s, \zeta s..., \zeta^{p}s$} are the $p$ roots of $f$. Suppose none of them is in $F$. We can choose an element $\sigma$ of $G=G(K/F)$ different from the identity. Note that $K = F(\zeta)$ is also generated by $s$ and $\zeta s$, so $\sigma$ cannot fix both $s$ and $\zeta s$. WLOG, we assume $\sigma(s) = \zeta^{k}s$, $\sigma^{n}(s)=\zeta^{nk}(s)$. Because $k<p$ and p is a prime, $nk$ can run through all residues modulo $p$. This shows that $G$ operates transitively on the $p$ roots of $f$. However $|G|\leq p-1$, which leads to a contradiction. Therefore some of {$s, \zeta s..., \zeta^{p}s$} must be in $F$.


问:
Closed-form analytical solutions to Optimal Transport/Wasserstein distance
Kuang and Tabak (2017) mentions that:

"closed-form solutions of the multidimensional optimal transport problems are relatively rare, a number of numerical algorithms have been proposed."

I'm wondering if there are some resources (lecture notes, papers, etc.) that collect/contain known solutions to optimal transport and/or Wasserstein distance between two distributions in dimensions greater than 1. For example, let $ \mathcal{N_1}(\mu_1, \Sigma_1) $ and $ \mathcal{N_2}(\mu_2, \Sigma_2) $ denote two Gaussian distributions with different means and covariances matrices. Then the optimal transport map between them is:

$$ x \longrightarrow \mu_2 + A( x - \mu_1 ) $$ where $ A = \Sigma_1^{- 1/2} (\Sigma_1^{1/2} \Sigma_2 \Sigma_1^{1/2})^{1/2} \Sigma_1^{- 1/2}$. And so the Wasserstein 2 distance is

$$ W_2 ( \mathcal{N_1}(\mu_1, \Sigma_1), \mathcal{N_2}(\mu_2, \Sigma_2) ) = || \mu_1 - \mu_2 ||^2_2 + \mathrm{Tr}( \Sigma_1 + \Sigma_2 - 2( \Sigma_1^{1/2} \Sigma_2 \Sigma_1^{1/2} )^{1/2} ) $$ where $\mathrm{Tr}$ is the trace operator.

It will be nice to know more worked out examples of optimal transport, such as uniform distributions between different geometric objects, e.g. concentric and overlapping balls, between rectangles, etc.

答:
Optimal transport (OT) problems admit closed-form analytical solutions in a very few notable cases, e.g. in 1D or between Gaussians. Below I cite articles providing analytical solutions for the 1-dimensional case only (does 1D mean univariate?)

Formula 3 in the following gives a closed-form analytical solution for Wasserstein distance in the case of 1-D probability distributions, but a source for the formula isn't given and I wonder how to convert it to a discretized linear programming model:

Kolouri et al (2019) "Generalized Sliced Wasserstein Distances" https://arxiv.org/pdf/1902.00434.pdf

Formula 9 in the following also gives a closed-form solution:

Kolouri et al (2019) "Sliced-Wasserstein Auto-encoders" https://openreview.net/pdf?id=H1xaJn05FQ

Formula 7 in the article below does as well:

Kolouri et al (2017) "Optimal Mass Transport: Signal-processing and machine learning applications" https://www.math.ucdavis.edu/~saito/data/acha.read.s19/kolouri-etal_optimal-mass-transport.pdf


答:
The definiton of Completeness is not the one you posted. It is a very difficult property to prove.

A statistic $T$ is complete if, $\forall \theta$ and $\forall g$ misurable function

$$\mathbb{E}_{\theta}[g(T)]=0$$

implies

$$\mathbb{P}_{\theta}[g(T)=0]=1$$

Good for you that in your case it is not necessary to show completeness with the definition.

It is enough to get a sufficient estimator by showing that the density belongs to the exponential family

$$f(x;\theta)=(1-\theta)Exp(x\cdot log\theta)$$

$T(x)=x$ and thus $\Sigma_i X_i$ is Sufficient and Complete (Exponential Family Canonical Statistic)

Title: Show $ \sum_{i=0}^n X_{i} $ is a complete sufficient statistic.

Body:
We're given a random sample of $X_{1}, X_{2}, .... , X_{n}$ from a distribution with a pdf of

$f(x;\theta) = \theta^x(1-\theta), x = 0,1,2,.... $ and $ 0 < \theta < 1 $

We're asked to show that that $Tn =  \sum_{i=0}^n X_{i} $ is a complete sufficient statistic for $\theta$.

I am able to show that $Tn$ is a sufficient statistic for $\theta$, I am just stuck on how to show that it is complete.

To show that $Tn$ is complete, we must show $E(g(Tn)) = 0 $. Where I am stuck is I do not know the distribution of Tn so I am not sure what to use for $f_{Tn}(t)$ for this expectation.


问:
Problem with showing that every Contractive Sequence is Cauchy
Here's the proof which shows that every contractive sequence satisfies the Cauchy criterion, and I don't seem to be convinced with it.

My doubts:

Why is the fact that the RHS tends to zero on approaching larger $n$, sufficient to say that we can always find a natural number (say $K$) such that the Cauchy criterion is met for $m,n \ge K$

What exactly is the epsilon we choose, here? Is it the expression on the RHS? If so, we should be able to show that the RHS ranges over all real numbers, because we're concerned with all $\varepsilon >0$

答:
Take the sequence
$$h_n=C^{n-1}\left(\frac{1}{1-C}\right)|x_2-x_1|$$
Now, $h_n\rightarrow 0$ as $n\rightarrow\infty$, which in other sense, means that $h_n$ converges to $0$.

Thus, $\forall\epsilon>0,$ $\exists$ $k\in\mathbb{N}$ such that $|h_n-0|<\epsilon$ $\forall n\geq k$.... $(i)$

So combining the result that $|x_m-x_n|\leq h_n \forall n\in\mathbb{N}$ and $(i)$, we get,

$\forall\epsilon>0, \exists k\in\mathbb{N}$ such that $|x_m-x_n|\leq|h_n|<\epsilon$ $\forall n\geq k$$ $$\implies \forall\epsilon>0, \exists k\in\mathbb{N}$ such that $|x_m-x_n|<\epsilon$ $\forall m,n\geq k$ which proves that this sequence is a Cauchy sequence.


问:
Infinite product involving powers of 2
I have the following infinite product:

$2/1 * 3/2 * 5/4 * 9/8 * 17/16 * 33/32 * 65/64...$

What does it converge to?

I can take its $\ln()$ to get

$\ln(2) + \ln(3/2) + \ln(5/4)....$

Which using the quotient difference rule resolves to

$S(n) = \ln(2^n + 1) - n\ln(2)$ for $n = 0$ to $\infty$

At this point I don't know how to evaluate this: so I took its derivative to see if I could recognize anything:

$S'(n) = 2^n * ln(2) / (2^n + 1) - ln(2)$

Which really looks like it converges to 0...

Integrating $S(n)$ hasn't been much to avail unless its because I lack knowledge.

What's this come out to?

答:
There is no nice answer to this. Here is a way to approximate it.
$$\sum_{n=0}^{\infty} \ln\left(1+\dfrac1{2^n}\right) = \sum_{n=0}^{\infty} \sum_{k=1}^{\infty}(-1)^{k-1} \dfrac{1}{k2^{kn}} = \sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}k \sum_{n=0}^{\infty} \dfrac1{2^{kn}} = \sum_{k=1}^{\infty} \left(\dfrac{(-1)^{k-1}}k \dfrac{2^k}{2^k-1} \right)$$</p>

Title: Infinite product involving powers of 2


问:
How can I show that $F(x,y)$ is not continuous?
I want to show
$F(x,y)=\frac{xy}{x^2+y^2}$ is not continuous at the origin. Here is my attempt:

Choose a path $y=mx$. Then, $F(x,mx)=\dfrac{xmx}{x^2+m^2x^2}=\dfrac{mx^2}{x^2(1+m^2)}=\dfrac{m^2}{1+m^2}$ if $(x,mx)\neq0$. Thus, $\lim_\limits{(x,y)\to(0,0)}F(x,y)$ does not exist, so $F$ is not continuous.

答:
Hint:

Consider first the two sequences $(x_n)_{n \in \mathbb N^\times}$ and $(y_n)_{n \in \mathbb N^\times}$ defined as
$$
x_n = \frac 1 n \quad \text{and} \quad y_n = \frac 1 n \; .
$$
Note that these are both zero sequences. Now consider $F(x_n, y_n)$ and calculate
the limit
$$
\lim_{n \to \infty} F(x_n, y_n) \; .
$$
Next, consider the two sequences $(x'_n)_{n \in \mathbb N^\times}$ and $(y'_n)_{n \in \mathbb N^\times}$ defined as
$$
x'_n = \frac 1 n \quad \text{and} \quad y'_n = -\frac 1 n \; .
$$
Again, calculate the limit
$$
\lim_{n \to \infty} F(x_n', y_n') \; .
$$
Deduce from that, that the function $F$ is not continuous at $(0,0)$.


问:
Let $a \in \Bbb{Z} _{\ge2}$ . Suppose that for all m,n $\in \Bbb{Z}^+$ if $a | mn$, then $a | m$ or $a | n$. Prove that a is a prime number
to prove that a is a prime number, I need to show that a only has 2 divisors: 1 and itself. My approach to this question is to use the definition of divisibility and the hint in the question that since $a = k(mn)$, a will either be $a = i(m)$ or $a = j(n)$ (where i, j, k are constants) to show that either m and n either has to be a or 1. I'm not sure if the approach I'm taking is a correct/logical one.

答: You're on the right track, thinking about the divisors of $a$. Let me provide a proof by contradiction. Let $a\in\mathbb{Z}_{\geq 2}$ have the property given in the question, and assume $a$ is not prime. Well, then $a$ is composite, and so there exist integers $b,c$ greater than or equal to $2$ with $a=bc$. But then clearly, $a|a$ and so $a|bc$, but also clearly, $a\not|b$ and $a\not|c$. So something is wrong; $a$ can't be composite and it must be prime.


问:
Divergence of series - Counter example
I am attempting to disprove following statement with counter example.

If $\sum_{n=1}^{\infty} a_{n}$ converges and lim $b_{n}$ = 0, then $\sum_{n=1}^{\infty} a_{n}b_{n}$
converges

My work: If a series converges then lim $ a_{n} $ = 0. All of the examples I am trying to construct are turning out to be convergent. Can I get some help here?

Title: Divergence of series - Counter example

Body: [Counter-Example] : $a_{n}=b_{n}=\frac{(-1)^n}{\sqrt{n}}$


问:
Find the infimum of the set $\{x+\frac 1x\}$
I am pretty new to the website, so I apologize in advance if I ask my question incorrectly.

I am trying to find the infimum of the Set
$$A=\left\{x+\frac{1}{x} \text{ for all }x \in \mathbb N\right\}.$$

Here's how I started: Using the definition of infimum, there is a $u$ in the set $A$ such that $u<x+\frac{1}{x}$.

From the Archimedean property, $\epsilon>0$ implies $\frac{1}{\epsilon}>0$.

I am not sure how to continue. Any advice?

答: We can use AM-GM: $$n + \frac{1}{n} \ge 2\sqrt{1} = 2.$$


问:
Show that an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1 is a smooth curve
Show that the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1 is a smooth curve by producing an admissable parametrization.

I know the parametrization is $a*cos(t) + i*b*sin(t)$. It looks right but I don't know how to work towards this answer as I suck at parametrization so I don't know how to show

答:
You are on the right track. The parameterization of $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is:

$$(a\cos(\theta), b\sin(\theta))\text{ for }0\leq 0 < 2\pi$$

Both components of the parameterization are smooth, so it remains to check that the endpoints coincide and are smooth. We have:

$$\lim_{\theta\rightarrow 0^{+}}(a\cos(\theta), b\sin(\theta)) = \lim_{\theta\rightarrow 2\pi^{-}}(a\cos(\theta), b\sin(\theta)) = (a, 0)$$

The directional derivative of this curve is:

$$\frac{d}{d\theta}(a\cos(\theta), b\sin(\theta)) = (-a\sin(\theta), b\cos(\theta))$$

We have:

$$\lim_{\theta\rightarrow 0^{+}}(-a\sin(\theta), b\cos(\theta)) = \lim_{\theta\rightarrow 2\pi^{-}}(-a\sin(\theta), b\cos(\theta)) = (0, b)$$

Thus, $\boxed{\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \text{ is smooth.}}$


问:
Proving $\int_0^1 \frac{(\ln(x))^5}{1+x} \mathrm{d}x = -\frac{31\pi^6}{252}$
I would like to show the following identity:

$$\boxed{
I :=
\int_0^1 \dfrac{(\ln(x))^5}{1+x} \mathrm{d}x =
-\dfrac{31\pi^6}{252}
}$$

Here is what I tried.
The change of variables $u=1/x$ yields
$$I=
\int_1^{\infty} \dfrac{(\ln(x))^5}{1+1/u} \dfrac{1}{u^2} \mathrm{d}u =
\int_1^{\infty} \dfrac{(\ln(x))^5}{u^2+u} \mathrm{d}u$$
Then $z=u-1$ gives
$$I = \int_{0}^{\infty} \dfrac{(\ln(z+1))^5}{z^2+3z+2} \mathrm{d}z $$
with $z^2+3z+2=(z+1)(z+2)$.
I wanted to use contour integration like here, but I was not sure how to proceed in this case. Anyway, the computations of the residues (of which "well-chosen" function? Maybe something like this?) seem to be difficult.

I believe that we can generalize to $\frac{(\ln(x))^n}{1+x}$, or maybe even more (e.g. $\frac{(\ln(x))^n}{ax^2+bx+c}$).
Related computations are: (1), (2), (3), (4).

Thank you for your detailed help.

Body:
Let $\Re(s)>0$ and
$$J(s)=\int_0^1\frac{\ln^sx}{1+x}dx.$$
Letting $x=e^{-u}$,
$$\begin{align}
J(s)&=(-1)^s\int_0^\infty\frac{u^se^{-u}}{1+e^{-u}}du\\
&=(-1)^s\int_0^\infty u^s\sum_{n\ge1}(-1)^ne^{-nu}du\\
&=(-1)^s\sum_{n\ge1}(-1)^n\int_0^\infty u^se^{-nu}du\\
&=(-1)^s\sum_{n\ge1}\frac{(-1)^n}{n^{s+1}}\int_0^\infty u^se^{-u}du\\
&=(-1)^s\Gamma(s+1)\sum_{n\ge1}\frac{(-1)^n}{n^{s+1}}\\
&=(-1)^s(2^{-s}-1)\Gamma(s+1)\zeta(s+1).
\end{align}$$


问:
$c$-dense set with perfect set
A set  $D\subset\mathbb R$ is called continuum dense if $|D\cap(a,b)|=c$ ($c$ is cardinality of $\mathbb R$) for every $a,b\in\mathbb R$ and $a<b.$ Let $P\subset\mathbb R$ be perfect set (uncountable closed set without isolated set). How about $$D\cap P \ ?$$  can be empty set ? what is the cardinality?  Any will be appreciated greatly

答:
There's nothing nontrivial we can say.

Removing a nowhere-dense set won't affect the "density character" of a given set. In particular, if we let $C$ be a perfect nowhere-dense set (e.g. the usual Cantor set) then $\mathbb{R}\setminus C$ is still continuum-dense but of course $C\cap (\mathbb{R}\setminus C)=\emptyset$.

Title: $c$-dense set with perfect set


问:
Bound from above on covariance matrix of Gaussian-like distribution
Assume that $\rho$ is a probability density such that
$$
\rho(x) \propto \exp \left(-\frac{|x|^2}{2} - f(x) \right).
$$
This means
$$
\rho(x) = \frac{\exp \left(-\frac{|x|^2}{2} - f(x) \right)}{\int_{\mathbb R^n} \exp \left(-\frac{|y|^2}{2} - f(y) \right) \, \mathrm d y}.
$$
Here $f$ is a smooth function such that $f(0) = \nabla f(0) = 0$ and ${\rm Hess} f(x) \geq 0$ for all $x \in \mathbb R^n$. Is there a standard result asserting that the covariance matrix of $\rho$ is bounded from above, using the usual order on positive semidefinite matrices, by the identity matrix, i.e. that the presence of the function $f$ can only reduce the covariance compared to the case where $f = 0$? I have a proof for this in dimension one (see below), but I am interested in a proof for the multidimensional case.

A natural way of proceeding would be to find a bound on
$$
    \int_{\mathbb R^n} xx^T \, \rho(x) \, \mathrm d x,
$$
which is an upper bound for the covariance matrix.
This is the approach I take below, in dimension $n = 1$.

Proof in the case $n = 1$.
First note that
$$
\mathrm{Cov} (\rho) = \mathbb E_{X \sim \rho}(X^2) - |\mathbb E_{X \sim \rho}(X)|^2 \leq \mathbb E_{X \sim \rho}(X^2).
$$
Since
$$
    \mathbb E_{X \sim \rho} (X^2) = \int_{0}^{\infty} \mathbb P_{X \sim  \rho} (X^2 \geq y) \, \mathrm d y,
$$
it is sufficient to show $\mathbb P_{X \sim \rho} (X^2 \geq y) \leq \mathbb P_{X \sim g} (X^2 \geq y)$ for all $y \geq 0$,
where $g$ denotes the density of the standard normal distribution.
Showing this is equivalent, in view of the fact that the function $r \mapsto \frac{r}{1-r}$ is increasing for $y \in [0, 1]$, to proving the following:
$$
    \forall y > 0, \qquad
    \frac{\mathbb P_{X \sim \rho} (X^2 \geq y)}{\mathbb P_{X \sim \rho} (X^2 \leq y)} \leq \frac{\mathbb P_{X \sim g} (X^2 \geq y)}{\mathbb P_{X \sim g} (X^2 \leq y)}.
$$
But this is obvious because,
introducing the function $u(y) =  \exp \left( - f\left(y\right) \right) + \exp \left( - f\left(-y\right) \right)$,
which is positive and nonincreasing over $[0, \infty)$,
we have
\begin{align*}
    \frac{\mathbb P_{X \sim \rho} (X^2 \geq y)}{\mathbb P_{X \sim \rho} (X^2 \leq y)}
    &= \frac{\int_{y}^{\infty} \exp \left( - \frac{x^2}{2} \right)  u(x) \mathrm d x}
    {\int_{0}^{y} \exp \left( - \frac{x^2}{2} \right)  u(x) \mathrm d x}
    \leq
    \left( \frac{\max_{\{x^2 \geq y\}} u(x)}
    {\min_{\{x^2 \leq y\}} u(x)} \right)
    \frac{\int_{\{x^2 \geq y\}} \exp \left( - \frac{x^2}{2} \right) \mathrm d x}
    {\int_{\{x^2 \leq y\} } \exp \left( - \frac{x^2}{2} \right) \mathrm d x},
\end{align*}
and, since $f$ is smooth, the minimum and the maximum coincide.

Ideas for the multi-dimensional case. In the multi-dimensional case,
the approach above can be applied,
by passing to radial coordinates,
to obtain a bound of the form
$$
\mathrm{Trace}(\mathrm{Cov} (\rho)) = \int_{\mathbb R^n} |x|^2 \, \rho(x) \, \mathrm d x \leq n,
$$
where $n$ is the dimension.
However, this gives only that
$$
    \mathrm{Cov} (\rho) \leq n I,
$$
i.e. there is an undesired factor $n$ on the right-hand side.
Here $I$ is the identity matrix

Note. In Bayesian terms, the function $e^{-x^2/2}/\sqrt{2\pi}$ can be viewed as a prior distribution,
the function $e^{-f}$ as the likelihood and $\rho$ as the posterior distribution.
The question can then be formulated as follows, approximately: is the posterior covariance bounded from above by the prior covariance given that the log-likelihood is convex.

答:
I believe this follows from the Poincare inequality. See for example Theorem 3.1 in https://arxiv.org/pdf/1202.1510.pdf

For a $\rho$-convex function $V$, let $\mu$ be the distribution $\frac{1}{Z}e^{-V}$. For all smooth function $f$, we have
\begin{equation}
\int \bigg(f - \int f d\mu\bigg)^2 d\mu \leq \frac{1}{\rho} \int |\nabla f|^2 d\mu .
\end{equation}

Take $f$ to be a linear function and $\rho = 1$ would give you the desired result.


问:
Unique 2-Factorization of a K9 graph
Does there exist a 2-factorization of K9 in which no two 2-factors are isomorphic? How would I do this?My thought was to split the graph into disjoint cycles but that was pretty much it. I couldn't think of where to go from there</p>

答:
Well, you should begin by thinking what a $2$-factor of $K_9$ can look like. A $2$-regular graph is a disjoint union of cycles, and there are only four ways to break up $9$ vertices into cycles:

A single $C_9$.

$C_4$ and $C_5$.

$C_3$ and $C_6$.

Three $C_3$'s.

A $2$-factorization of $K_9$ must have $4$ factors, because $K_9$ is regular of degree $8$, so each of these must be used once.

The remainder of this problem wouldn't be out of place on Puzzling StackExchange. Here's one way to do it:

The four factors are in different colors: a black $C_9$, a red $C_4$ and $C_5$, a purple $C_3$ and $C_6$, and three blue $C_3$'s.

Title: Unique 2-Factorization of a K9 graph


问:
gluing ideals together
While pondering basic facts about closed subschemes the following claim occurred to me . I think it must be true but I have trouble proving it algebraically.

Let $ R$ be a commutative ring with 1.  Let $f_1,...,f_k\in R$ with $ (f_1,...,f_k)=R$  and let $I_1\subset R_{f_1}, ..., I_k\subset R_{f_k}$ be ideals. Suppose that for each $i,j$, the ideal generated by the image of $I_i$ in $ R_{f_if_j}$ is the same as the ideal generated by the image of $I_j$ in $ R_{f_if_j}$.  Then there exists a unique ideal $I\subset R$ whose image in each $R_{f_i}$ generates $I_i$.

答:
I suspect there is possibly a much more elegant solution than this, but here is a (slightly ugly) proof.

For each $i, j$, let $\iota_i:R\rightarrow R_{f_i}$ and $\tau_{ij}:R_{f_i}\rightarrow R_{f_if_j}$ be the canonical localization maps. (Recall in particular that $\text{ker}(\tau_{ij})=\{x\in R_{f_i}:\exists b\in\mathbb{N}\text{ such that }xf_j^b\big/1=0\big/1\in R_{f_i}\}$.) Now, define $I=\bigcap_{i=1}^k\iota_i^{-1}(I_i)$. Certainly $I$ is an ideal of $R$, and each $\iota_i(I)R_{f_i}\leqslant I_i$. I claim that this inclusion is an equality.

To see this, let $r\big/{f_i^m}\in I_i$, where $r\in R$. Now, to show $r\big/f_i^m\in\iota_i(I)R_{f_i}$ it certainly suffices to show $r\big/1\in\iota_i(I)R_{f_i}$, so really we just need to find $s\in I$ and $n\in\mathbb{N}$ such that $r\big/1=s\big/f_i^n\in R_{f_i}$.

To do this, fix some $j\neq i\in\{1,\dots,k\}$. Note that $r\big/1\in I_i$, so $r\big/1\in \tau_{ij}(I_i)R_{f_if_j}=\tau_{ji}(I_j)R_{f_if_j}$, so there are $f\in I_j$ and $a\in\mathbb{N}$ such that $r\big/1=f\big/f_i^a\in R_{f_if_j}$. Hence $(rf_i^a-f)\big/1\in\text{ker}(\tau_{ji})$, so there is $b\in\mathbb{N}$ such that $(rf_i^a-f)f_i^b\big/1=0\big/1\in R_{f_j}$, ie such that $rf_i^{a+b}\big/1=ff_i^b\big/1\in R_{f_j}$. But then, because $f\in I_j$, we have $rf_i^{a+b}\big/1\in I_j$. Denote therefore $n_j=a+b$, so that $rf_i^{n_j}\big/1\in I_j$.

Now, do this for all $j$, let $n=\text{max}_{j\neq i}n_j$, and define $s=rf_i^n\in R$. Clearly $s\in I$, since (for all $j$) $\iota_j(s)=s\big/1=(rf_i^{n_j})f_i^{n-n_j}\big/1\in I_j$ by construction of the $n_j$. Furthermore, we have $s\big/f_i^n=rf_i^n\big/f_i^n=r\big/1\in R_{f_i}$, so this $s\in I$ and $n\in\mathbb{N}$ give us exactly the witnesses of $r\big/1\in\iota_i(I)R_{f_i}$ that we desire.

Oops, forgot to show uniqueness. First note that $I$ is certainly maximal with the property you desire; if $r\in R\setminus I$, then by construction of $I$ there is some $i\in\{1,\dots,k\}$ with $\iota_i(r)\notin I_i$. But then clearly the image of any ideal containing $r$ in $R_{f_i}$ is strictly larger than $I_i$, so there can be no such ideal with the property we desire.

Conversely, suppose that $J\leqslant I$ has the property we desire. So $\iota_i(J)R_{f_i}=I_i$ for each $i$. Let $r\in I$, and fix some $i\in\{1,\dots,k\}$. By construction we have $r\big/1\in I_i=\iota_i(J)R_{f_i}$, so there must be $s_i\in J$ and $a\in\mathbb{N}$ such that $r\big/1=s_i\big/f_i^a\in R_{f_i}$. Thus $rf_i^a-s_i\in\text{ker}(\iota_i)$, so there is $b\in\mathbb{N}$ such that $(rf_i^a-s_i)f_i^b=0\in R$. But then, since $s_i\in J$, we have $rf_i^{a+b}\in J$, so let $m_i=a+b$.

Repeat this process for all $i$, and let $m=\sum_{i=1}^k m_i$. Now, this is where we need the hypothesis that $\langle f_1,\dots,f_k\rangle=R$. Indeed, in particular, there are $\lambda_i\in R$ such that $\lambda_1 f_1+\dots+\lambda_k f_k=1$. This gives $r=r1=r1^m=r(\lambda_1 f_1+\dots+\lambda_k f_k)^m$. Every monomial term after expanding the right hand side will have a factor of $rf_i^{m_i}$ for some $i$, and hence lie in $J$ by construction. Thus the entire right hand side is an element of $J$, so $r$ is too, and we have $I=J$ as desired.

Title: gluing ideals together


问:
Cleverly showing that $\lim_{x\to 0}\frac{x^{(\sin x)^x}-(\sin x)^{x^{\sin x}}}{x^3}=\frac{1}{6}$
$$\lim_{x\to 0}\frac{\textstyle x^{\textstyle(\sin  x)^{\textstyle x}}-(\textstyle \sin  x)^{\textstyle x^{\textstyle \sin  x}}}{\textstyle x^3}=\frac{1}{6}$$

The limit is easy to get results, but how to rigorously prove it without using Taylor formula?

At first, I guess the numerator is equivalent to $x-\sin x$

And I also find the following several limits have the same results:

$$\lim_{x\to 0}\frac{\textstyle x^{\textstyle x^{\textstyle x}}-(\textstyle \sin  x)^{\textstyle (\sin x)^{\textstyle \sin  x}}}{\textstyle x^3}=\frac{1}{6}$$

$$\lim_{x\to 0}\frac{\textstyle x^{\textstyle x^{\textstyle \tan x}}-(\textstyle \sin  x)^{\textstyle x^{\textstyle \tan  x}}}{\textstyle x^3}=\frac{1}{6}$$

I guess if $f(x)\sim g(x)\sim h(x)\sim O(x)^k$ when $x\to 0$ , then $$\lim_{x\to 0}\textstyle f(x)^{\textstyle g(x)^{\textstyle h(x)}}=f(x)$$

答:
Note the limit is only defined as $x\to 0^+$, so I'll answer as such. Not sure how 'clever' this is either and the solution is a little fast and loose, but oh well.

Using LHR or other methods we have
$$
\lim_{x\to0^+} f_1(x)^{f_2(x)} = 1,
$$for $f_1,f_2\in \{\sin,\operatorname{id}\}$. This means we can interchange some of the exponents with mild impunity. In particular, I claim
$$
\lim_{x\to0^+} \frac{x^{(\sin x)^x}-(\sin x)^{x^{\sin x}}}{x^3}
=\lim_{x\to0^+} \frac{x^{x^x}-(\sin x)^{(\sin x)^{\sin x}}}{x^3}
$$

Introduce $x-\sin x$:
$$
=\lim_{x\to0^+} \frac{x^{x^x}-(\sin x)^{(\sin x)^{\sin x}}}{x-\sin x}\cdot \frac{x-\sin x}{x^3}
$$We will break into two limits and justify this by showing the first equals $1$; the second clearly equals $1/6$ by LHR. Then we are interested in
$$
\lim_{x\to0^+} \frac{x^{x^x}-(\sin x)^{(\sin x)^{\sin x}}}{x-\sin x}
$$

Let $q:[0,1]\to\mathbb{R}$, $q(z)=z^{z^z}$ and $q(0)=0$. By the MVT, for some $y$ between $\sin(x)$ and $x$ we have
$$
\frac{x^{x^x}-(\sin x)^{(\sin x)^{\sin x}}}{x-\sin x} = q'(y)
$$Since $q'$ is continuous, it suffices to show $\lim_{y\to 0^+}q'(y) = 1$.

We have
$$
q'(y) = y^{y^y+y-1} \left(y \log ^2 y+y \log y+1\right)
$$The second term is easily seen to approach $1$ in the limit; for example, put $y=e^w$ or play with LHR. The first term isn't as apparent, but using the bound $1>y^y>1+y\log y$ on $(0,1)$ reduces the problem to evaluating $\lim_{y\to 0^+}y^{y+y\log y},$ which is just $1$ (exercise). One could also show it directly.


答:
4.768462058062755

According to Wolfram Alpha.  Look it up it is explained there.

Title: Infinite product involving powers of 2

Body:
I have the following infinite product:

$2/1 * 3/2 * 5/4 * 9/8 * 17/16 * 33/32 * 65/64...$

What does it converge to?

I can take its $\ln()$ to get

$\ln(2) + \ln(3/2) + \ln(5/4)....$

Which using the quotient difference rule resolves to

$S(n) = \ln(2^n + 1) - n\ln(2)$ for $n = 0$ to $\infty$

At this point I don't know how to evaluate this: so I took its derivative to see if I could recognize anything:

$S'(n) = 2^n * ln(2) / (2^n + 1) - ln(2)$

Which really looks like it converges to 0...

Integrating $S(n)$ hasn't been much to avail unless its because I lack knowledge.

What's this come out to?


问:
Proof of the special case of Moore-Penrose Inverse $A^+=A^*(AA^*)^{-1}$
What's the simplest proof to the equations $A^+=A^*(AA^*)^{-1}$ and $A^+=(A^*A)^{-1}A^*$ for the Moore-Penrose inverse? I am not taking any course about generalized inverses and I can not seem to find any reference which explains the proof clearly, or maybe it's just because I lack the searching capability as English is not my main language. Thanks in advance!

答:
The Moore-Penrose pseudoinverse of $A$ is defined by the following four equations:

$ABA = A$

$BAB = B$

$(AB)^* = AB$

$(BA)^* = BA$

There is one and only one $B$ that satisfies all of these properties, and that matrix we denote by $A^+$.

Now, let us suppose $AA^*$ is invertible. Note that $(AA^*)^* = AA^*$, and consequently, $$((AA^*)^{-1})^* = ((AA^*)^*)^{-1} = (AA^*)^{-1}.$$
Let $B = A^*(AA^*)^{-1}$. Note that,
$$AB = AA^*(AA^*)^{-1} = I,$$
hence $ABA = IA = A$ and $BAB = BI = B$, meaning both 1 and 2 are satisfied. Also $AB = I$ implies $(AB)^* = I^* = I = AB$, hence 3 is also satisfied.

The only condition left is 4. We have
$$(BA)^* = (A^*(AA^*)^{-1}A)^* = A^*((AA^{-1})^{-1})^*(A^*)^* = A^*(AA^*)^{-1}A = BA.$$

Thus, $B = A^*(AA^*)^{-1}$ is the one and only matrix satisfying the conditions 1-4, making $B = A^+$.

If $A^*A$ is invertible, then the other formula can be proven similarly.


问:
Determine $k$ in PDF.

Let $X$ and $Y$ random variables with PDF given by
\begin{eqnarray*}
f(x,y)=\left\{\begin{aligned}kxy, \quad x\geq 0, y\geq 0, x+y\leq 1\\ 0, \quad \text{elsewhere} \end{aligned} \right.
\end{eqnarray*}
$1.$ Calculate $k$.

$2.$ Calculate the marginals density of $X$ and $Y$.

$3.$ Are $X$ and $Y$ independent random variables?

$4.$ Calculate $$\mathbb{P}(X\geq Y), \quad \mathbb{P}(X\geq 1/2 | X+Y\leq 3/4), \quad \text{and} \quad \mathbb{P}(X^{2}+Y^{2}\leq 1)$$

$5.$ Calcule the joint density function of $U=X+Y$ and $V=X-Y$ with their respective marginals.

My approach:

Since that $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x,y)=1 \implies \int_{0}^{1}\int_{0}^{1-x}kxydydx=1 \implies k=24.$$

By definition, we have that $$f_{X}(x)=\int_{0}^{1-x}24xydy=12x(x-1)^{2}, \quad x\in [0,1]$$and $$f_{Y}(y)=\int_{0}^{1-y}24xydx=12y(y-1)^{2}, \quad y \in [0,1]$$

Since that $$f(x,y)\not=f_{X}(x)f_{Y}(y)$$so $X$ and $Y$ are not independents.

Now, here I'm problem in how can I calculate $\mathbb{P}(X\geq 1/2| X+Y\leq 3/4)$ and $\mathbb{P}(X^{2}+Y^{2}\leq 1)$ Can you help me here?. I could find that $\mathbb{P}(X\geq Y)=1/2$.

Here, I don't know how can I solve this part. I was trying to apply the Jacobian transformation, but I don't know how to use this method well. So, I'm not getting a good solution focus.

答:
Indeed, $k=24$, and the marginals are as you calculated: $f_{\small X}(x)=12x(1-x)^2\mathbf 1_{0\leqslant x\leqslant 1}$ and $f_{\small Y}(x)=f_{\small X}(x)$ by symmetry.

Also by symmetry, $\Bbb P(X\leq Y)=\Bbb P(X\geq Y)$, so (since $\Bbb P(X=Y)=0)$ then $\Bbb P(X\geq Y)=1/2$.

Then you just need to use the definition of conditional probability:$$\begin{align}\Bbb P(X\geq 1/2\mid X+Y\leq 3/4)&=\dfrac{\Bbb P(1/2\leq X, X+Y\leq 3/4)}{\Bbb P(X+Y\leq 3/4)}\\[1ex]&=\dfrac{\Bbb P(1/2\leq X\leq 3/4-Y)}{\Bbb P(X\leq 3/4-Y)}\\[1ex]&=\dfrac{\displaystyle\int_0^{1/4}\int_{1/2}^{3/4-y}24xy\,\mathrm dx\,\mathrm d y}{\displaystyle\int_0^{3/4}\int_0^{3/4-y}24 xy\,\mathrm d x\,\mathrm d y}\end{align}$$

For the next problem, notice that for $0\leq x\leq 1$ that $1-x\leq \surd(1-x^2)$

We have $U=X+Y$ and $V=X-Y$, so $U+V=2X$ and $U-V=2Y$.

Thus the Jacobian matrix is: $$\begin{align}\mathsf J_{\small U,V}^{\small X,Y}(u,v)&=\begin{bmatrix}\dfrac{\partial (u+v)/2}{\partial u}&\dfrac{\partial (u+v)/2}{\partial v}\\\dfrac{\partial (u-v)/2}{\partial u}&\dfrac{\partial (u-v)/2}{\partial v}\end{bmatrix}\\[1ex]&=\begin{bmatrix}1/2&1/2\\1/2&-1/2\end{bmatrix}\end{align}$$
Now just apply the Jacobian Transformation: $$ f_{\small U,V}(u,v)=\lVert\mathsf J_{\small U,V}^{\small X,Y}(u,v)\rVert\, f_{\small X,Y}\left(\tfrac {u+v}{2},\tfrac {u-v}2\right)$$

$\blacksquare$

Title: Determine $k$ in PDF.


问:
Help me decide if I'm ready or not for this research topic
so a bit of background info on me.

I'm a math grad student. My undergrad classes were mainly math-theory based stuff. I took Number Theory, Intro to Analysis, Intro to Abstract algebra, and such.

I did take some computationally sweaty courses such as Financial Math for Actuaries(sweaty, but fun!). I also took Intro to Probability. And of course, I have the basics such as differential equations, linear algebra, complex variables...

My professor told me he could assign me stuff related to optimization(for example a minimal-time control problem). He said if I knew my stuff, he could assign me something on single-cell RNA sequencing data analysis, which is basically a study on large scale sparse matrix. He also proposed a topic on stochastic control which would be challenging to both him and me.

Can someone let me know what I'd be getting myself into if I chose to do research with him? I honestly have no idea what I'd be getting myself into.

Appreciate any feedback. !~

答:
To be honest, it is hard to say whether or not you are ready for this research topic. It doesn’t really have anything to do with your background, your background should be fine for a topic like optimization. I am by no means an expert in optimization, but I can say that for applications, all you really need is a working knowledge of numerical linear algebra, vector calculus, and experience with programming (in Matlab for example). Some depth in topics like convex and set-valued analysis would be a plus, but are not essential.

That being said, being ready for a research topic is not necessarily dependent on your mathematical background alone. It also has a lot to do with your attitudes towards learning as a whole, and having a growth (as opposed to a fixed) mindset. In other words, things like:

How well do you learn from your mistakes?

How do you view failure/rejection?

How well do you work with others?

Are you open to learning new subjects/topics?

Can you synthesize prior information and come up with new approaches to solving problems?

It is really subjective, so a definitive answer is hard to give. I should also note that in practice, you will learn what you need to do research on the go. I haven’t met (yet) an advisor who expects their mentees to know absolutely everything about their research the first semester.

These are my 2 cents, take with a grain of salt.


问:
How to prove that: let $n$ a natural odd, then prove that $x^n<y^n$ iff $x<y$
I tried it in many ways but I couldn't prove it.
  TASK:
 Let $n$ a natural odd, $\ \ x,y\in\mathbb{R},$ then prove that $x^n<y^n$ iff $x<y$

My Attempt(s): $$\begin{align} & x^n<y^n \\ \Longleftrightarrow{} &  x^n-y^n<0 \\ \Longleftrightarrow{} & (x-y)\sum_{k=0}^{n-1}x^{n-k-1}y^k<0 \end{align}$$ And I'm stuck here.

Thus, I can't solve it with induction either. Any help would be appreciated :)

答: It is elementary to show that $\sum_{k=0}^{n−1}x^{n−k−1}y^k\geq \frac{x^{n-1}+y^{n-1}}{2},$ provided $n$ is an odd natural number. With this, the desired result follows.


问:
On the quantity $I(q^k) + I(n^2)$ where $q^k n^2$ is an odd perfect number with special prime $q$
The topic of odd perfect numbers likely needs no introduction.

In what follows, we let $\sigma(x)$ denote the sum of divisors of the positive integer $x$.  Let
$$D(x) = 2x - \sigma(x)$$
denote the deficiency of $x$, and let
$$s(x) = \sigma(x) - x$$
denote the sum of aliquot/proper divisors of $x$.  Finally, denote the abundancy index of $x$ by
$$I(x) = \frac{\sigma(x)}{x}.$$

Euler proved that a hypothetical odd perfect number must have the form $N = q^k n^2$ where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

Since $q$ is prime and $N$ is perfect, we obtain
$$\frac{q+1}{q} \leq I(q^k) < \frac{q}{q-1}$$
so that we get
$$\frac{2(q-1)}{q} < I(n^2) = \frac{2}{I(q^k)} \leq \frac{2q}{q+1}.$$
Using the identity
$$\frac{D(x)}{x} = 2 - I(x)$$
we obtain the bounds
$${q^k}\bigg(\frac{q-2}{q-1}\bigg) < D(q^k) \leq {q^{k-1}}\bigg(q-1\bigg)$$
and
$$\frac{2n^2}{q+1} \leq D(n^2) < \frac{2n^2}{q}.$$
This implies that
$${2q^k n^2}\cdot\bigg(\frac{q-2}{(q-1)(q+1)}\bigg) < D(q^k)D(n^2) < {2q^k n^2}\cdot\bigg(\frac{q-1}{q^2}\bigg).$$
Dividing both sides of the last inequality by $2q^k n^2$, we get
$$\frac{q-2}{(q-1)(q+1)} < \frac{D(q^k)D(n^2)}{2q^k n^2} < \frac{q-1}{q^2}.$$
But we know that
$$D(q^k)D(n^2)=2s(q^k)s(n^2),$$
which can be verified by a direct, brute-force computation.
Thus, the fraction in the middle of the last inequality simplifies to
$$\frac{D(q^k)D(n^2)}{2q^k n^2}=\bigg(\frac{s(q^k)}{q^k}\bigg)\bigg(\frac{s(n^2)}{n^2}\bigg)=\bigg(I(q^k) - 1\bigg)\bigg(I(n^2) - 1\bigg) = 3 - \bigg(I(q^k) + I(n^2)\bigg).$$

We therefore finally have the bounds
$$3 - \bigg(\frac{q-1}{q^2}\bigg) < I(q^k) + I(n^2) < 3 - \bigg(\frac{q-2}{(q-1)(q+1)}\bigg)$$
which does not improve on the known bounds
$$3 - \bigg(\frac{q-2}{q(q-1)}\bigg) < I(q^k) + I(n^2) \leq 3 - \bigg(\frac{q-1}{q(q+1)}\bigg).$$
(See this paper for a proof.)

Here are my:

QUESTIONS Is it possible to improve on the bounds for $D(q^k)$ and $D(n^2)$ (where $q^k n^2$ is an odd perfect number with special prime $q$) to hopefully produce stronger bounds for $I(q^k) + I(n^2)$?  If so, how could this be done?

答:
On OP's request, I am converting my comment into an answer.

Let $f(k):=I(q^k) + I(n^2)$.

Then, we have
$$f'(k)=\frac{-( q^{2 k + 2}- 4 q^{2 k + 1}+2 q^{k + 1}   + 2 q^{2 k} - 1) \log(q)}{(q - 1)q^k (q^{k + 1} - 1)^2}$$
which is negative, so we see that $f(k)$ is decreasing.

It follows that
$$\lim_{k\to\infty}f(k)\lt f(k)\le f(1)\tag1$$
which is the known bounds
$$3 - \bigg(\frac{q-2}{q(q-1)}\bigg) < I(q^k) + I(n^2) \leq 3 - \bigg(\frac{q-1}{q(q+1)}\bigg)\tag2$$

So, I think that this means the following :

If you prove that $k\not=1$, then you can get an improved upper bound for $f(k)$.

If you get an improved upper bound for $f(k)$, then you can say that $k\not=1$.

If you prove that there is an integer $a$ such that $k\le a$, then you can get an improved lower bound for $f(k)$.

If you get an improved lower bound for $f(k)$, then you can say that there is an integer $a$ such that $k\le a$.

(I'm not saying that if you can't improve the bounds for $k$, you can't improve the bounds for $f(k)$.)


问:
Given the expected inflation and the random shock to inflation, how to do we obtain the inflation at a given time?
I am confused as to how in this example the following steps occurred:

$$\pi_1=\pi^*+0.5(-0.5(\pi_1-\pi^* ))+v_1$$

where $\pi_1$ is inflation (at time 1), $\pi^*$ is expected inflation, and $v_1$ is a random shock to inflation. How from that equation do we arrive at

$$\pi_1=\pi^*+0.8v_1$$

I don't understand how to get $0.8$ nor how the $v_1$ becomes a part of $0.8$.

答:
I’ll assume you meant

$$\pi_1=\pi^*+0.5(-0.5(\pi_1-\pi^* ))+v_1$$
in which case this is just straightforward algebra. Multiply out the parentheses to get

$$\pi_1=\pi^*+0.25\pi^*-0.25\pi_1+v_1$$
then combine and transpose to get

$$1.25\pi_1=1.25+pi^*+v_1$$
and finally divide by 1.25 to get the desired result.


问:
For Abelian group $G$, show that $W_n = \{a \in G : \text{ord}(a) \mid n\}$ is a subgroup of $G$ for any positive integer $n$

It seems to me that we are trying to show that $W_n$ is a subgroup of $(G,+)$
So I started with let $a1,a2$ be arbitrary elements in $W_n$.
And trying to show $a_1+a_2$ are still in $W_n$.
Then, I have $\text{ord}(a_1)= x$, where $n=xj_1$ for some integer $j_1$
And $\text{ord}(a_2)=y$, where $n=yj_2$ for some integer $j_2$.
Then I compute $a_1+a_2$ but I don't know how to make use of the previous information to finish the proof.

答:
Subgroup criterion.  $e\in W_n$.  Let $x,y\in W_n$.  Then $(xy^{-1})^n\stackrel{\text{G abelian}}=x^ny^{-n}=e\cdot e=e$.  Thus $xy^{-1}\in W_n$.

Here I have used repeatedly that $x^n=e\iff$ the order of $x$ divides $n$.


答: Well, assume that $a,b \in W_n$. We want to show that $a+b \in W_n$ as well, but you know that $$\underbrace{a+\cdots+a}_{\mbox{$n$ times}} = 0 \quad \mbox{and}\quad \underbrace{b+\cdots+b}_{\mbox{$n$ times}} = 0.$$So $$\underbrace{(a+b)+\cdots+(a+b)}_{\mbox{$n$ times}} = \underbrace{a+\cdots+a}_{\mbox{$n$ times}} + \underbrace{b+\cdots+b}_{\mbox{$n$ times}} = 0+0 = 0,$$by associativity and commutativity of $+$. Thus ${\rm ord}(a+b) \mid n$, showing that $a+b \in W_n$. Also, note that $$\underbrace{0+\cdots + 0}_{\mbox{$n$ times}} = 0,$$so ${\rm ord}(0) \mid n$ and $0 \in W_n$. Lastly, if $a \in W_n$, we have $$\underbrace{(-a)+\cdots+(-a)}_{\mbox{$n$ times}} = -\underbrace{\big(a+\cdots+a\big)}_{\mbox{$n$ times}} = -0 = 0,$$so ${\rm ord}(-a) \mid n$ and $-a \in W_n$.

Title: For Abelian group $G$, show that $W_n = \{a \in G : \text{ord}(a) \mid n\}$ is a subgroup of $G$ for any positive integer $n$

Body:
It seems to me that we are trying to show that $W_n$ is a subgroup of $(G,+)$
So I started with let $a1,a2$ be arbitrary elements in $W_n$.
And trying to show $a_1+a_2$ are still in $W_n$.
Then, I have $\text{ord}(a_1)= x$, where $n=xj_1$ for some integer $j_1$
And $\text{ord}(a_2)=y$, where $n=yj_2$ for some integer $j_2$.
Then I compute $a_1+a_2$ but I don't know how to make use of the previous information to finish the proof.


答:
There's no need for any epsilon arguments or fancy inequalities here. One can simply observe that for all $n \in \mathbb N$,
$$n + \frac{1}{n} > n.$$
Therefore for $n \geq 2$, we have
$$n + \frac{1}{n} > 2.$$
On the other hand, if $n=1$, we have
$$n + \frac{1}{n} = 2.$$
Therefore the sequence has a minimum of $2$ which occurs at $n=1$.

Title: Find the infimum of the set $\{x+\frac 1x\}$

Body:
I am pretty new to the website, so I apologize in advance if I ask my question incorrectly.

I am trying to find the infimum of the Set
$$A=\left\{x+\frac{1}{x} \text{ for all }x \in \mathbb N\right\}.$$

Here's how I started: Using the definition of infimum, there is a $u$ in the set $A$ such that $u<x+\frac{1}{x}$.

From the Archimedean property, $\epsilon>0$ implies $\frac{1}{\epsilon}>0$.

I am not sure how to continue. Any advice?


问:
Field of fractions of a local ring.
Consider the following local ring:

$\mathbb{Z}_{(2)}=\{\frac{a}{b} \in \mathbb{Q}: gcd(a,b)=1 \land   2\nmid b \}$.

Describe is its field of fractions.

I guess this field of fractions is $\mathbb{Q}$ itself because for example $\frac{\frac{6}{11}}{\frac{8}{13}}=\frac{39}{44}$.

I don have any clue where to start. Any hint?

Thanks in advance.

答:
There is an obvious injective ring homomorphism $\mathbb{Z}_{(2)} \rightarrow \mathbb{Q}$, so by the universal property of fields of fractions, there is a ring homomorphism $Quot(\mathbb{Z}_{(2)}) \rightarrow \mathbb{Q}$. However, any ring homomorphism between fields must be injective, so $Quot(\mathbb{Z}_{(2)})$ is a subfield of $\mathbb{Q}$. As $\mathbb{Q}$ is a prime field, the only subfield of $\mathbb{Q}$ is $\mathbb{Q}$ itself.

Therefore, $Quot(\mathbb{Z}_{(2)}) = \mathbb{Q}$

Title: Field of fractions of a local ring.