问:
Find $\mathbb E(X | X + 2Y)$, where $X$ and $Y$ ~ $U[0,1]$
Let $X$ and $Y$ two independent random variables with unifrom distribution $U[0,1]$. Need to find $\mathbb E(X | X + 2Y)$. Is my calculation below correct?

$\mathbb E(X | X + 2Y) = \mathbb E(X|X) + \mathbb E(X | 2Y) = X + 2\mathbb E(X|Y) = X + 2*\frac{1}{2} = X +1$

答:
Maybe that example can hint you :
\begin{align*}
\mathbb E[X|X+Y]+\mathbb E[Y|X+Y]=\mathbb E[X+2Y|X+Y]=X+Y
\end{align*}
But by symmetry $\mathbb E[X|X+Y]=\mathbb E[Y|X+Y]$ which means that $$\mathbb E[X|X+Y]=\frac{X+Y}{2}$$


问:
Ray Disk intersection
So if I have a ray parameterized as $O + tD$ where $O$ is the origin, $D$ is the direction and $t$ is the parameter variable and a flat circular disk with a center point $P$ in 3D space and a radius $r$, how do I determine if they intersect?</p>

答: First, find the equation for the plane that contains the circular disk.  Second, find the intersection of the line and the plane.  Third, determine if the intersection point is in the right direction from the origin (the ray points only one way).  Lastly, determine if the intersection point is distance at most $r$ from the center of the disk.</p>

ContentLicense: CC BY-SA 3.0

Title: Ray Disk intersection


问:
Prove $\frac{( 4n+5 ) ( ( 2n+2 )! ) ^2}{2}( \int_{-1}^1{f( x ) \text{d}x} ) ^2\le \int_{-1}^1{( f^{( 2n+2 )}( x ) )^2 \text{d}x}$
Let $f \in C^{2n+2}[-1,1]$, and
$$
f\left( 0 \right) =f''\left( 0 \right) =\cdots =f^{\left( 2n+2 \right)}\left( 0 \right) =0
$$
Prove that
$$
\frac{\left( 4n+5 \right) \left( \left( 2n+2 \right) ! \right) ^2}{2}\left( \int_{-1}^1{f\left( x \right) \text{d}x} \right) ^2\le \int_{-1}^1{\left( f^{\left( 2n+2 \right)}\left( x \right) \right)^2 \text{d}x}
$$

I even don't know how to deal with the basic circumstances, such as when $n=0$, we have
$$
10\left( \int_{-1}^1{f\left( x \right) \text{d}x} \right) ^2\le \int_{-1}^1{\left( f''\left( x \right) \right) ^2\text{d}x}
$$
I tried to apply Cauchy–Schwarz inequality and Integration by parts, but they didn't work.

Also I think the general formula may relate to Taylor series since it has factorials on the left side.

Can anyone help?

答:
I will do the case $n = 0$ by considering @r9m's suggestion.
This means we have $f(0) = f''(0) = 0$ and also $g'(x) = f(x). g''(x) = f'(x)$ and $g'''(x) = f''(x).$ Therefore, expanding $g$ around the point $a = 0$ by Taylor's theorem:

$$g(x) = g(0) + g'(0)x + g''(0)\frac{x^2}{2} + \frac 12\int_0^x g'''(t)(x-t)^2dt = f'(0)\frac{x^2}{2} + \frac 12\int_0^xf''(t)(x-t)^2dt$$
Therefore,
$$\int_{-1}^1f(x)dx = g(1) - g(-1) = \frac 12\int_{0}^1f''(t)(1-t)^2dt+\frac 12\int_{-1}^0f''(t)(1+t)^2dt$$
and so the C-S on each term of the RHS gives:

$$\int_{0}^1f''(t)(1-t)^2dt\leq\left(\int_0^1 (f''(t))^2dt\right)^{1/2}\left(\int_0^1(1-t)^4dt\right)^{1/2} =\left(\int_0^1 (f''(t))^2dt\right)^{1/2}\dfrac{1}{\sqrt{5}} $$

and

$$\int_{-1}^0f''(t)(1+t)^2dt\leq\left(\int_{-1}^0 (f''(t))^2dt\right)^{1/2}\left(\int_{-1}^0(1+t)^4dt\right)^{1/2} =\left(\int_{-1}^0 (f''(t))^2dt\right)^{1/2}\dfrac{1}{\sqrt{5}}.$$

Finally then we have by another application of C-S:

$$\int_{-1}^1f(x)dx\leq\dfrac{1}{2\sqrt{5}}\left( \left(\int_0^1 (f''(t))^2dt\right)^{1/2} +\left(\int_{-1}^0 (f''(t))^2dt\right)^{1/2} \right)\leq\dfrac{1}{2\sqrt{5}}\sqrt{2\int_{-1}^1(f''(t))^2dt}$$
where the last inequality is just:
$$\sqrt{a}+\sqrt{b}\leq\sqrt{2(a+b)},$$
which is just Cauchy-Schwartz.

You can clearly see from here that this should be immediatly generalizable to any $n>0.$

Interesting to note here is that you will have equality when:
$$f''(x) =
\begin{cases}
1-x,\quad 0\leq t\leq 1 \\
1+x,\quad -1\leq t\leq 0
\end{cases}$$

ContentLicense: CC BY-SA 4.0


答:
Let $z=x+2y$.  Draw a graph with $x$ on the horizontal and $z$ on the vertical.  Find the region in the plane that can be reached by $(x,z)$.  It should be a parallelogram.

Draw horizontal lines through the region, finding the intervals of allowed $x$ values for each $z$ value.

The midpoint of an interval is the expected $x$-value for that $z$-value.

Title: Find $\mathbb E(X | X + 2Y)$, where $X$ and $Y$ ~ $U[0,1]$

Body:
Let $X$ and $Y$ two independent random variables with unifrom distribution $U[0,1]$. Need to find $\mathbb E(X | X + 2Y)$. Is my calculation below correct?

$\mathbb E(X | X + 2Y) = \mathbb E(X|X) + \mathbb E(X | 2Y) = X + 2\mathbb E(X|Y) = X + 2*\frac{1}{2} = X +1$


问:
A union of proper filters in a chain is a proper filter
I was looking into the proof of the Ultrafilter lemma by Zorn's lemma, and I wasn't able to find a sufficient explanation for the following claim:

If $\mathcal{C}\subseteq P$ is a chain of proper filters, then $F:=\cup \mathcal{C}$ is also a proper filter.

I'm okay with showing that $F$ is a filter, but it's not clear to me why it is proper.

答:
Recall that a filter $\cal F$ is proper if and only if $\varnothing\notin\cal F$.

Now suppose that $\cal F$ is the union of any family of filters, $\bigcup\cal C$. If $\varnothing\in\cal F$, then by the definition of $\bigcup\cal C$, there is some $X\in\cal C$ such that $\varnothing\in X$. Therefore $X$ could not have been a proper filter.


答:
$$\mathbb{E}[X|X+2Y]+\mathbb{E}[Y|X+2Y]+\mathbb{E}[Y|X+2Y]=\mathbb{E}[X+2Y|X+2Y]=X+2Y$$

Being

$$X \perp\!\!\!\!\!\! \perp Y$$

they are exchangeable too, thus

$$\mathbb{E}[X|X+2Y]=\frac{X+2Y}{3}$$

Alternative solution:

Looking at the following drawing

and setting $U=X$ and $Z=X+2Y$

If $0<z<1$ $X$ is uniform in $(0;z)$ with expectation $\frac{z}{2}$

If $1<z<2$ $X$ is uniform in $(0;1)$ with expectation $\frac{1}{2}$

If $2<z<3$ $X$ is uniform in $(z-2;1)$ with expectation $\frac{z-1}{2}$

Thus the expected value is

$$\frac{z}{2}\cdot\frac{1}{4}+\frac{1}{2}\cdot\frac{1}{2}+\frac{z-1}{2}\cdot\frac{1}{4}=\frac{2z+1}{8}$$


问:
Suppose we flipped 3 coins simultaneously, till we get the first head, what would be the mean number of tosses(including the last one)?
I tried to evaluate it by multiplying the probability of each scenario with the number of flips it took to get there but this didn't work, and now I'm really confused, so if someone could provide an answer, I would really appreciate it if they could go into detail.</p>

答: The chance of one of the coins being heads is .875. The mean of a geometric distribution is 1/p which means it would take on average 1.142857 flips.


问:
If $A\subset B$ are definable sets, and $A$ is open in $B$, then there exists a definable open $U$ with $U\cap B=A$
This is Lemma 3.4, chapter 1 in the book "Tame Topology and O-minimal Structures." Consider an o-minimal structure $\mathcal{S}=(\mathcal{S}_{n})$ on a dense linearly ordered nonempty set without endpoints, $R$, which is equipped with the interval topology. Let $A\subset B\subset R^{m}$ are definable sets, and $A$ is open in $B$, prove that there is a definable open $U\subset R^{m}$ with $U\cap B=A$.

The book tells that we can take for $U$ the union of all boxes (a Cartesian product of intervals) in $R^{m}$ whose intersection with $B$ is contained in $A$. However, since $\mathcal{S}_{n}$ is a Boolean algebra instead of topological space, I don't know why the union of these boxes which might be arbitrary many is definable. Can someone explain to me a little bit?

答:
I have an answer now:

note the following equivalence:

\begin{align}
(x_1,\dots , x_m)\in U&\Leftrightarrow \exists y_1\dots\exists y_m\exists z_1\dots\exists z_m \\
[&(y_1<x_1<z_1)\wedge\dots\wedge (y_m<x_m<z_m)] \wedge\
\\ \big[\forall w_1\dots\forall w_m[&(y_1<w_1<z_1)\wedge\dots\wedge (y_m<w_m<z_m) \wedge (w_1,\dots, w_m)\in B]
\\ \rightarrow[&(w_1,\dots, w_m)\in A]\big]
\end{align}

Therefore, we have that $U$ is a definable open set.


问:
difference in the definiton of modular arithmetic
When i look at my textbook , i realized that there is a little difference in the definiton of modular artihmetic among books.

For example, wikipedia says that $a\equiv b\pmod n$ where $a,b \in Z,n \in Z^+$ and $n \gt1$ .

On the other hand, some books and websites says that $a\equiv b\pmod n$ where $a,b \in Z,n \in Z^+$ but there is not any restriction for $n$ ,i.e, $n$ can be $1$.

I know that it is not a big difference and does not affect results generally.However, i want to know the exact definiton of it.

Briefly, which definiton is correct ? If they are same , then why do books ignore $n=1$

答:
The modulus $n=1$ is not interesting because $a \equiv b \bmod 1$ is always true.

I prefer the general definition that allows all $n \in \mathbb Z$ as modulus.

The only other singular modulus is $n=0$ because $a \equiv b \bmod 0$ iff $a=b$.


问:
Maximum and minimum cardinality of range and preimage
I found the following question related to understanding of images and preimages:

Let A and B be nonempty sets and $f$ be a function $f: A \rightarrow  B$. In terms of $|A|$ and $|B|$, what is the maximum possible cardinality of
$f[A]$? What is the minimum possible cardinality of $f[A]$? How about $f^{-1}[B]$?

My thoughts are the following:

The maximum possible cardinality of $f[A]$ is $|B|$, since $f[A]$ is the range of the function and the function might be surjective, so it will map to all the elements of $B$

Minimum cardinality of $f[A]$ is $1$, because $f[A]$ might be a function which maps all the elements to a single element of $B$.

Maximum and minimum possible cardinality of $f^{-1}[B]$ is $|A|$, because we have a preimage of entire codomain, which must be an entire domain for $f$ to be a well-defined function.

I'm almost sure of my answers, but I'd be very grateful if someone could check and criticise/correct me if I'm wrong somewhere. And I also doubt about the second point - my answer states that cardinality is $1$, but the question asks me to define the cardinalities in terms of $|A|$ and $|B|$. Is this possible to do in the second point?

答:
I agree with your statements 2 and 3.

For statement 1, there might be a ``trap".

Well, here are two things we know about the maximum cardinality of $f[A]$, say $|f[A]|_{MAX}$.

$|f[A]|_{MAX} \le |A|$ by the definition of a function.

$|f[A]|_{MAX} \le |B|$ by the fact $f[A]$ is a subset of $B$.

Since we don't know which is larger between $|A|$ and $|B|$, it is better to say $|f[A]|_{MAX} = \min\left\{ |A|, |B|\right\}$


问:
Is the set of primes at even positions a Large Set?
If we start numbering the prime numbers from 2, and we consider the subset of prime numbers at even positions, is it known is that set is large or not?

A subset of natural numbers is called "large" if the sum of reciprocals of its elements diverges. For example it is well known that the set of natural numbers and set of prime numbers are large sets. Surprisingly I could not find any reference to whether the set of primes at alternating positions is large or not.

I tried using Bertrands Postulate but failed. I intuitively believe the set primes at odd positions must be large as the but I cant seem to be able to prove it rigorously.

Any guidance/help would be appreciated. I am also looking for amateurish resources for studies on large subsets.

答:
You know that $\sum_{k=1}^{\infty}\frac1{p_k}=\infty$. By Cauchy condensation, we have that $\sum_{k=1}^\infty \frac{2^k}{p_{2^k}}=\infty$ as well. But we know that $$\frac{2^k}{p_{2^k}}=4\cdot 2^{k-2}\frac1{p_{2^k}}\le 4\sum\limits_{2^{k-1}< j \le 2^k\\ j\text{ even}}\frac1{p_{2j}}$$

And therefore $$\infty=\frac14\sum_{k=1}^\infty \frac{2^k}{p_{2^k}}\le\sum_{k=1}^\infty \frac1{p_{2k}}$$


问:
Gravitational fields
Could anyone help me with the following question?

Consider a right circular cone of constant mass density sigma, height h, and semi-vertical angle alpha. By dissecting the cone into discs, show that find the gravitational field at the apex of the cone.

I have shown that for a disc at distance b, the gravitational potential is $(2\pi G\sigma)\cdot\dfrac{(a^2+b^2)^1}{2-b}$. But I dont know how to take the grad of this to find what the gravitational field is...

答:
Let $\sigma$ the density of the cone, $\alpha$ its half-angle, $H$ its height, $x$ the coordinate of a point $P$ located in the symmetry axis of the cone (which origin is at the apex of the cone), $G$ the gravitational constant, and $R$ the radius of a infinitesimal disc of width $dh$ which center is at a distance of $h$ units from the origin. See the figure below:

Let's calculate the gravitational potential $U_{\text{disc}} (x)$ at point P due to the presence of the infinitesimal disc.
Let $\mathrm {d}M$ the mass of a infinitesimal piece of the disc and $d$ its distance from point $P$, then $U_{\text{disc}} (x)$ can be calculated by the following equation:
$$U_{\text{disc}} (x) = \iint \limits_{\text{disc}}{\frac{-G}{d} \mathrm {d}M} \tag{1}$$
But $\mathrm {d}M = \sigma \, \mathrm {d}h \, r \, \mathrm {d}r \, \mathrm {d}\theta$, so we get the following equation:
$$U_{\text{disc}} (x) = \int_0^{2\pi} \int_0^R{\frac{-G \sigma}{\sqrt{(x+h)^2+r^2}} \, \mathrm {d}h \, r \, \mathrm {d}r \, \mathrm {d}\theta} \tag{2}$$
We know that
$$R=h \tan{\alpha} \tag{3}$$
Therefore evaluating the double integral and substituting $(3)$ in $(2)$, we get:
$$U_{\text{disc}} (x) = -2\pi G \sigma\, \mathrm {d}h \left(\sqrt{(x+h)^2+\left(h\tan{\alpha}\right)^2}-|x+h|\right) \tag{4}$$
The gravitational field $\vec{g}_{\text{disc}}(x)$ at point $P$  can be calculated by:
$$\vec{g}_{\text{disc}}(x)= - \mathrm {grad}\, U_{\text{disc}} (x) \tag{5}$$
But we know that $\mathrm {grad} \, U_{\text{disc}} (x)=\dfrac{\mathrm {d}U_{\text{disc}}}{\mathrm {d}x}\vec{i}$, therefore for $x>-h$ we get:
$$\vec{g}_{\text{disc}}(x)= 2\pi G \sigma\, \mathrm {d}h \left(\frac{x+h}{\sqrt{(x+h)^2+(h\tan{\alpha})^2}}-1\right)\vec{i} \tag{6}$$

After that calculate the integral:
$$\vec{g}_{\text{cone}}(x)=2\pi G \sigma\, \int_0^{H} {\left(\frac{x+h}{\sqrt{(x+h)^2+(h\tan{\alpha})^2}}-1\right)\vec{i}} \, \mathrm {d}h \tag{7}$$
where $\vec{g}_{\text{cone}}(x)$ is the gravitational field at point P due to the presence of the cone.
Finally substitute $x=0$ and we are done.

Title: Gravitational fields


问:
Hint for Spivak's Calculus Problem 2-4-a
(note: this is very similar to a related question but as I'm trying to solve it without looking at the answer yet, I hope the gods may humor me anyways)

I'm self-learning math, and an answer to an /r/math post about self-learning was finally enough to motivate me to try getting feedback on this site :) . After reading throughout the internet, I've decided to start with Spivak's Calculus. I'm loving the book thankfully, but I'm stuck on this problem and I don't want to look at the answer quite yet.

4 . (a) Prove that
$$\sum_{k=0}^l \binom{n}{k} \binom{m}{l-k} = \binom{n+m}{l}$$.
Hint: Apply the binomal theorem to $(1+x)^n(1+x)^m$ .

I've done all the prior problems, including Problem 3 (proving the Binomial Theorem) which is obviously closely tied to this, but even with the hint it feels like there's too much I don't know. I applied the hint and found:

$$\left(\sum_{i=0}^n \binom{n}{i} x^i\right)\left(\sum_{j=0}^m \binom{m}{j} x^j\right) = \sum_{k=0}^{n+m} \binom{n+m}{k} x^k$$

And, setting $x=1$ got something even more interesting:

$$\left(\sum_{i=0}^n \binom{n}{i}\right)\left(\sum_{j=0}^m \binom{m}{j}\right) = \sum_{k=0}^{n+m} \binom{n+m}{k}$$

However, I don't know where to go after that...is there some property of multiplying sums that I need to prove first, to relate the multiplication of two sums of binomials with the sum of the multiplication of two binomials?

Thank you!

答:
The solution uses some facts about polynomials that, unfortunately, Spivak hasn't yet proven.

He hasn't formally introduced polynomials yet. This happens in Chapter 3 (3rd Ed.). You may wish to return to this problem after reading chapter 3.

A polynomial function $f$ (functions are also introduced in chapter 3) has the form:
$$f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0$$
where the $a_i$ coefficients are constants.

This polynomial is said to have degree $n$, where $n$ is the largest power of $x$ with a nonzero coefficient.

A number $x_i$ is said to be a root of the polynomial if $f(x_i) = 0$.

An $n$-th degree polynomial can have at most $n$ roots. That is, there can be no more than $n$ different numbers $x_1, x_2, \dots, x_n$ such that $f(x_i) = 0$

This fact comes from Chapter 3 exercise 7 (3rd Ed.)

Because an $n$ degree polynomial has at most $n$ distinct roots, if a polynomial is zero for all $x$ then all of its coefficients (the $a_i$'s) must be zero. (If any $a_i \neq 0$, this forces $f$ to have a finite number of roots, which contradicts $f$ being zero for all values of $x$.)

Restating,
$$0 = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0 \text{ for all }x \text{ if and only if } a_n = 0, a_{n-1} = 0, \dots a_0 = 0$$

Closely related to this fact is the following: If two polynomials are equal for all values of $x$, they must have the same coefficients.

To write it out explicitly, suppose

$$f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0$$
$$g(x) = b_nx^n + b_{n-1}x^{n-1} + \dots + b_1x + b_0$$
and
$$f(x) = g(x) \text{ for all }x$$

Then $a_i = b_i$ for all $i = n, n-1, \dots, 1, 0.$

(You can show this by looking at $f-g$ and using what we previously learned about polynomials that are $0$ for all $x$.)

Now let's return to the exercise.

We have
$$(1+x)^n(1+x)^m = (1+x)^{n+m}$$

We can expand each side using the binomial coefficients developed in preceding problems.

When we do so, we will have polynomials on the left hand side and right hand side of the equation. These polynomials will have different looking coefficients for each power of $x$.

However, using what we know about polynomials, we know that these different looking coefficients must actually be equal. The coefficient for $x^l$ on the LHS must equal the coefficient for $x^l$ on the RHS. You can use this fact to prove the desired result.

Edit: When $a < b$, $\binom{a}{b}$ is defined to be $0$. IIRC this is also used in the problem (despite not being defined in the text).


答:
This follows immediately from the fact that the set of primes is large. More generally, if $A=\{a_1<a_2<a_3<...\}$ is large, then so is $\{a_1<a_3<a_5<...\}$. The same holds for even positions (just think about $A\setminus\{a_1\}$) or indeed any arithmetic sequence of positions.

This is just because we have ${1\over a_{2n-1}}>{1\over a_{2n}}$. So we get $${1\over a_1}+\color{red}{1\over a_1}+{1\over a_3}+\color{red}{1\over a_3}+...>{1\over a_1}+\color{red}{1\over a_2}+{1\over a_3}+\color{red}{1\over a_4}+...$$ or more snappily $$\sum_{n\in\mathbb{N}} {2\over a_{2n-1}}>\sum_{n\in\mathbb{N}}{1\over a_n}.$$ If $A$ is large then the right hand side diverges, so the left hand side also diverges. Now just divide by $2$.

Title: Is the set of primes at even positions a Large Set?

Body:
If we start numbering the prime numbers from 2, and we consider the subset of prime numbers at even positions, is it known is that set is large or not?

A subset of natural numbers is called "large" if the sum of reciprocals of its elements diverges. For example it is well known that the set of natural numbers and set of prime numbers are large sets. Surprisingly I could not find any reference to whether the set of primes at alternating positions is large or not.

I tried using Bertrands Postulate but failed. I intuitively believe the set primes at odd positions must be large as the but I cant seem to be able to prove it rigorously.

Any guidance/help would be appreciated. I am also looking for amateurish resources for studies on large subsets.


问:
Optional stopping with a nonnegative supermartingale
I encountered the following problem:

Let $ X_n $ be a nonnegative supermartingale. For two stopping times $ M \le N \le \infty $, prove that $$ \mathbb{E}[ X_{N} | \mathcal{F}_{M}] \le X_{M} . $$

At first I thought this is not true, because to have the optional stopping theorem we need something as powerful as the uniform integrability. Since a nonnegative supermartingale is not necessarily uniformly integrable, I started there and tried to construct a counterexample where $ \mathbb{E} X_M < \mathbb{E} X_N $. However it seems it's not that simple.

So my question is

Is this true?

If it is true, how can I prove it? If not, how can I get to a counterexample? Even a small hint would be very much appreciated.

答:
This is Theorem 28 in Chapter V of Meyer's classical "Probability and potentials", but, candidly, I don't like the proof very much, because the pertinent ideas are obscured by technicalities.

First of all: the condition $M \le N \color{red}{\le} \infty$ does make sense, because non-negative supermartingales converge so some integrable $X_\infty$, a.s. Still better: we can extend our supermartingale $(X_n)_{n\in\mathbb{N}}$ to a supermartingale $(X_n)_{n\in\mathbb{N}\cup\{\infty\}}$. That's because the supermartingale property
$$X_m\ge\mathbb{E}(X_n|\mathcal{F}_m)\quad\mbox{for}\quad m<n\quad\mbox{a.s.}$$ is equivalent to $$\int_{A}X_m\,dP\ge\int_{A}X_n\,dP\quad\mbox{for}\quad m<n\quad\mbox{and all}\quad A\in\mathcal{F}_m.\tag{1}$$ This implies
$$\int_{A}X_m\,dP\ge\liminf_{n\to\infty}\int_{A}X_n\,dP\ge\int_{A}X_\infty\,dP\quad\mbox{for all}\quad A\in\mathcal{F}_m$$ by Fatou's lemma, and that's equivalent to
$$X_m\ge\mathbb{E}(X_\infty|\mathcal{F}_m)\quad\mbox{a.s.}\tag{2}$$

So we proved the proposition for $N=\infty$ and a deterministic $M=m$. Let's extend that to a non-deterministic stopping time $M$: we have to prove
$$X_M\ge\mathbb{E}(X_\infty|\mathcal{F}_M)\quad\mbox{a.s.}\tag{3}$$ and that's equivalent to
$$\int_{A}X_M\,dP\ge\int_{A}X_\infty\,dP\quad\mbox{for all}\quad A\in\mathcal{F}_M.$$
Here, we have to remember the definition of $\mathcal{F}_M$:
$$\mathcal{F}_M=\{A\subset\Omega: A\cap\{M=m\}\in\mathcal{F}_m\quad\mbox{for all}\quad m\in\mathbb{N}\}$$ (if I were a textbook author, I'd add "the proof that this is indeed a $\sigma$-algebra is left as an exercise to the reader"). In fact, you may more often find $\{M\le m\}$ instead of $\{M=m\}$ in definitions, but for discrete time, both definitions are equvalent.

Now we can split $A$ like $\displaystyle A=\bigcup_{m\in\mathbb{N}\cup\{\infty\}} A\cap\{M=m\}$, and $A\in\mathcal{F}_M$ means $A\cap\{M=m\}\in\mathcal{F}_m$, and $X_M=X_m$ on $A\cap\{M=m\}$, so
$$\int_{A}X_M\,dP=\sum_{m\in\mathbb{N}\cup\{\infty\}}\int_{A\cap\{M=m\}}X_m\,dP\ge\sum_{m\in\mathbb{N}\cup\{\infty\}}\int_{A\cap\{M=m\}}X_\infty\,dP=\int_{A}X_\infty\,dP,$$ indeed, due to (2).

The rest is easy: if $X_n$ is a non-negative supermartingale and $N\le\infty$ a stopping time, $Y_n=X_{N\wedge n}$ is a non-negative supermartingale as well, $Y_\infty=X_N$, $Y_M=X_M$ for a stopping time $M\le N$, so (3) becomes
$$X_M\ge\mathbb{E}(X_N|\mathcal{F}_M)\quad\mbox{a.s.}$$


问:
Assuming the earth is a sphere, what is the shortest distance between 2 points through the earth?
Assuming the earth is a perfect sphere, is there a way to find out the straight line distance (which goes through the earth) between any 2 points on the surface of the earth?
I've tried searching for this on Google, but it keeps showing me the Great Circle Distance which is the shortest distance on the surface of the earth, but not through the earth.</p>

答: The sphere is actually a red herring here. Because your movement is not impeded by the sphere, this question is equivalent to finding the shortest distance between two points $P(a,b,c)$ and $Q(x,y,z)$. Do you know what to do from here?


问:
Evaluate the double integral $\int_{0}^{16} \int_{\sqrt[4]{x}}^{2} \sqrt{2 y^{5}+1} d y d x$
$\int_{0}^{16} \int_{\sqrt[4]{x}}^{2} \sqrt{2 y^{5}+1} d y d x$

I considered changing the order of integration but that I do not think it does anything as it just leads me to evaluating dy still.

Another possibility I considered is converting the integral to polar coordinates but again the square root bogs me in solving this integral. Any help or hint is appreciated. Thank you.

Body:
$$I=\int_{0}^{16} \int_{x^{1/4}}^{2} \sqrt{2x^5+1} dy~ dx$$
$$\implies I=\int_{0}^{2} dy \sqrt{2y^5+1} \int_{0}^{y^4} dx$$
$$\implies I=\int_{0}^{2} y^4 \sqrt{2y^5+1} dy=\int_{1}^{65} \frac{\sqrt{t}}{10} dt.$$
$$I=\frac{(65\sqrt{65}-1)}{15}.$$


答:
We can split it into two parts:

Probability of having at least one head when fliping 3 coins

When will the first head shows up?

Part 1 is binomial distributed, assume all 3 coins are fair, $p=0.5$

So, $P(at\ least \ one \ head) = 1 - P(all \ tail) = 1-0.5^{3}=0.875$

Part 2 is geometrically distributed, you can find the mean by $\frac{1}{p}$

Title: Suppose we flipped 3 coins simultaneously, till we get the first head, what would be the mean number of tosses(including the last one)?

Body: I tried to evaluate it by multiplying the probability of each scenario with the number of flips it took to get there but this didn't work, and now I'm really confused, so if someone could provide an answer, I would really appreciate it if they could go into detail.</p>


答:
If your points are given by longitude $\varphi$ and latitude $\theta$, and if $r$ is the radius of the Earth, then this gives you the spherical co-ordinates of the point on the surface.

The Cartesian co-ordinates of that point (assuming a suitably placed co-ordinate system so that the origin is in the centre of the Earth and the $z$ axis goes through the poles) can be then obtained from the spherical co-ordinates using the formula(e):

$$(x,y,z)=(r\cos\varphi\cos\theta, r\sin\varphi\cos\theta, r\sin\theta)$$

and now you can use the ordinary (Euclidean) distance between two points:

$$d((x_1,y_1,z_1),(x_2,y_2,z_2))=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$

Title: Assuming the earth is a sphere, what is the shortest distance between 2 points through the earth?

Body:
Assuming the earth is a perfect sphere, is there a way to find out the straight line distance (which goes through the earth) between any 2 points on the surface of the earth?
I've tried searching for this on Google, but it keeps showing me the Great Circle Distance which is the shortest distance on the surface of the earth, but not through the earth.</p>


问:
How to draw Hasse diagram of following relation
Consider this relation;
$$R = \{(1,1),(1,3),(1,4),(1,5),(2,2),(2,4),(2,3),(3,3),(3,4),(3,5),(4,4),(4,5),(5,5)\}$$

R is reflexive, transitive and antisymmetric. Therefore R is partially ordering relation. I want to draw a Hasse diagram for this relation.
2 and 5 are not related. So I placed them at same level. But 3 and 5 are related and 2 and 3 are related which makes me place 2 below 3 and 3 below 5 and in turn it shows 2 is related to 5 but it is not! . Where did I make mistake? How to place 2, 3, 5 in correct order? Any help would be greatly appreciated.

Edit: This relation is not transitive. If (2,3) and (3,5) are elements in R, then (2,5) should be! Otherwise transitive condition would fail. I couldn't see that at first place.

答:
The relation given is neither transitive nor reflexive.

For a start it does not include $(5, 5)$.

For another thing, while $(2, 3)$ and $(3,5)$ are there, $(2, 5)$ is not.

If you were to create the reflexive closure and transitive closure, you should be able to work with it, but at the moment it is not a partial ordering.

Maybe you transcribed it wrong, and it should be $(2,5)$ instead of $(2, 3)$.


问:
Loaded die two roll probability
There is a loaded die for which the probability of rolling a 6 is 1/2. The probability for each other face appearing is 1/10. What is the probability of rolling a 4 and then a 6 on two tosses of the die?</p>

答:
The probability of two independent events $A$ and $B$ both occurring is $P(A \cap B)=P(A)*P(B)$

So in this case we get $\frac{1}{2}\cdot \frac{1}{10}=\frac{1}{20}$

Title: Loaded die two roll probability


答:
If $n>1$ then the ring $\mathbb{Z}_n$ is a "ring with unity" which means that it has a multiplicative identity distinct from the additive identity.   That is not the case if $n=1$.  I can't think of an example off the top of my head, but there must be some theorems about rings with unity that need the fact that $0 \neq 1$.  So you allow $n=1$, then every time to need to invoke some properties of a ring with unity, then you have to say "$n>1$".

So it might make sense to just specify $n>1$ at the outset and thereby exclude the "bo-ring."

Title: difference in the definiton of modular arithmetic

Body:
When i look at my textbook , i realized that there is a little difference in the definiton of modular artihmetic among books.

For example, wikipedia says that $a\equiv b\pmod n$ where $a,b \in Z,n \in Z^+$ and $n \gt1$ .

On the other hand, some books and websites says that $a\equiv b\pmod n$ where $a,b \in Z,n \in Z^+$ but there is not any restriction for $n$ ,i.e, $n$ can be $1$.

I know that it is not a big difference and does not affect results generally.However, i want to know the exact definiton of it.

Briefly, which definiton is correct ? If they are same , then why do books ignore $n=1$


问:
Does this series that resembles the Maclaurin series of $\cosh x$ converge?
Does this series converge?

$$\left(\frac{f_{i}^2}{2!} +\frac{f_{i}^4}{4!}\left(2\kappa c\right)+\frac{f_{i}^6}{6!}\left(2\kappa c\right)^2 + \frac{f_{i}^8}{8!}\left(2\kappa c\right)^3+ \cdots\right), \tag{1}$$

where $k, c \geq 0$. I am trying to establish a relation with the Maclaurin series of $\cosh x$:

$$ 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \cdots=\sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}. \tag{2}$$

答:
Multiply the series given by Eq. (1) by $\frac{2 \kappa c}{2\kappa c}$

$$\frac{1}{2\kappa c}\left(\frac{f_{i}^2}{2!}\left(2\kappa c \right) +\frac{f_{i}^4}{4!}\left(2\kappa c\right)2+\frac{f_{i}^6}{6!}\left(2\kappa c\right)^3 + \frac{f_{i}^8}{8!}\left(2\kappa c\right)^4+ \cdots\right), $$

Add a zero to the above expression

$$\frac{1}{2\kappa c}\left(1+\frac{f_{i}^2}{2!}\left(2\kappa c \right) +\frac{f_{i}^4}{4!}\left(2\kappa c\right)2+\frac{f_{i}^6}{6!}\left(2\kappa c\right)^3 + \frac{f_{i}^8}{8!}\left(2\kappa c\right)^4+ \cdots -1\right), $$

comparing with Eq. (2) of the question, we can establish the relation with $\cosh x$

$$\frac{1}{2\kappa c}\left(\sum_{n=0}^{\infty} \frac{\left[f_{i}\left(2\kappa c\right)^{\frac{1}{2}}\right]^{2n}}{2n!}- 1\right) =\frac{\cosh \left[f_{i}\left(2\kappa c\right)^{\frac{1}{2}} \right]-1}{2\kappa c}.$$

Which is the result given by @Sangchul Lee.


问:
$n \ge \left(\frac{z_{\alpha/2}}{2}\right)^2p(1-p) = \left(\frac{1.96}{0.03}\right)^2 \frac{19}{20}\frac{1}{20} =202.75$?

A polling firm proposes to estimate proportions in a population by the
corresponding proportions in the samples so that the margin of error
is at most $3$%, and this $19$ times out of $20$ on average.

What is the minimum sample size here? The answer is $1068$ here.

For me that answer is wrong because

$$n \ge \left(\frac{z_{\alpha/2}}{\text{margin_error}}\right)^2p(1-p) = \left(\frac{1.96}{0.03}\right)^2 \frac{19}{20}\frac{1}{20} =202.75.$$

So the minimum size must be $203$. Where am I wrong?

答:
The wording of the question is not especially clear, but I understand the 19 out of 20 as giving you only the value of $\alpha$. Consequently the problem does not explicitly tell you what value of $p$ to use to estimate the standard deviation. This would imply that the one asking you the question doesn't have any idea about $p$ and that thus you as the analyst must make the worst possible assumption about it (the one that maximizes the standard deviation).

Thankfully in the case of the binomial distribution this is actually possible: you just estimate $p$ as $1/2$. Doing that seems to get you the desired answer here.


问:
Better method to show $-x^3 + 1 = (-x + 1)(x^2 + x + 1)$
Is there a rule, algorithm, or theorem used for this equality below:

$$-x^3 + 1 = (-x + 1)(x^2 + x + 1)$$

I know the RHS can be established using trial and error distributing the first term through the second, or alternatively same with FOIL.  I'm curious if there is a better method?

Image below from a CAS.

答:
In the base-$10$ numeration,

$$\color{green}{9\cdot111=999}.$$

This translates

$$(10-1)(10^2+10+1)=10^3-1.$$

More generally, in the base $x=w+1$,

$$\color{green}{w\cdot111=www}$$ or $$(x-1)(x^2+x+1)=x^3-1.$$

This immediately generalizes to higher degrees.


答:
The modulus can be any integer - no restrictions are needed. The definition is

$$ a\equiv b\!\!\!\pmod{\!n}\!\!\overset{\rm def\!\!}\iff n\mid a-b\!\!\overset{\rm def\!\!}\iff a- b = kn,\ \text{for some } k\in\Bbb Z$$

Thus congruence and divisibility is defined for all $\,n\,$  (note $\,0\mid c\iff c = 0)\,$

Such generality can prove useful, e.g. a fundamental number theory reduction technique relies on the property that congruences persist mod divisors of the modulus, so we can attempt to deduce solutions of congruences by piecing together solutions mod smaller moduli (e.g. by CRT). With the general defintion this reduction method also applies the same way to the modulus $\,n = 0,\,$ where $\,a\equiv b\pmod{\!0}\iff 0\mid a-b\iff a = b,\,$ so the above reduction says that integer equalities persist (as congruences $\!\bmod n$ for every integer $n\,$ (since every $n$ divides $0).$

We can use such reductions to help study solutions of integer (Diophantine) equations, one of the most fundamental problem solving methods in number theory (and algebra). For example by reducing $\bmod 2\,$ we can often deduce some integer equations are unsolvable due to a parity contradiction. This parity analysis works for any even modulus, including $\,n=0.\,$ It is unnatural to separate out the modulus $\,n = 0\,$ from other even moduli. The reduction applies uniformly to all even moduli.


问:
Another Information Theory Riddle
The following nice riddle is a quote from the excellent, free-to-download book: Information Theory, Inference, and Learning Algorithms, written by David J.C. MacKay.

In a magic trick, there are three participants: the magician, an assistant, and a volunteer.

The assistant, who claims to have paranormal abilities, is in a soundproof room. The magician gives the volunteer six blank cards, five white and one blue. The volunteer writes a different integer from 1 to 100 on each card, as the magician is watching. The volunteer keeps the blue card. The magician arranges the five white cards in some order and passes them to the assistant.

The assistant then announces the number on the blue card.

How does the trick work?

答: The five cards are uniquely identified by their numbers (low to high).  There are $5!=120$ possible orderings for the five cards, which is more than enough to encode the number on the sixth card.  In fact, by orienting the cards carefully, there are 4 ways to orient each card and still make a pile (180 degree rotation, turn upside down), there could be $4^55!=122880$ possibilities.  I'd only try this variation with a lot of practice, though.</p>

Title: Another Information Theory Riddle


问:
Gaussian space generated by Gaussian process
I have a somewhat silly question after reading the following definition in Brownian motion, martingales and stochastic calculus by Jean-François Le Gall

Proposition 1.7 If $(X_t)_{t\in T}$ is a Gaussian process, the closed linear subspace of
$L^2$ spanned by the variables $X_t, t\in T$, is a Gaussian space, which is called the
Gaussian space generated by the process $X$.

My question is: how do we prove that this space is closed? If I denote this space by $G$, and I take a sequence $(Y_n)_{n\in\mathbb{N}}$ in $G$ that converges to some $Y\in L^2$, I can show that $Y$ is a centered Gaussian. But how do I show that $Y$ is a linear combination of the $(X_t)_{t\in T}$?

答: $G$ is the closed linear span of $(X_t)$, i.e., the elements of $G$ need not belong to the linear span of $(X_t)$.


问:
Estimating the integral
I'm trying to find an estimate for the modulus of the integral given below.

|$\int_0^r\frac{w^{z-1}}{w^2+1}dw$|, where $r\in[0,1]$ and $z\in\mathbb{C}$ such that $Re(z)>0$.

Any hint will be greatly appreciated. Thanks!

Title: Estimating the integral

Body:
Write
$1/(w^2+1)
=\sum_{n=0}^{\infty} (-1)^n w^{2n}
$,
multiply, and integrate term by term.
The result will be a series with alternating lower and upper bounds.

(added later)

Since
$\dfrac{1-x^n}{1-x}
=\sum_{k=0}^{n-1} x^k
$,
putting $-x$ for $x$ gives
$\dfrac{1-(-1)^nx^n}{1+x}
=\sum_{k=0}^{n-1} (-1)^kx^k
$,
so
$\dfrac{1-(-1)^nx^{2n}}{1+x^2}
=\sum_{k=0}^{n-1} (-1)^kx^{2k}
$
so
$\dfrac{1}{1+x^2}
=\dfrac{(-1)^nx^{2n}}{1+x^2}+\sum_{k=0}^{n-1} (-1)^kx^{2k}
$.

Now we can integrate.

$\begin{array}\\
\int_0^r\dfrac{w^{z-1}}{w^2+1}dw
&=\int_0^rw^{z-1}\left(\sum_{k=0}^{n-1} (-1)^kw^{2k}+\dfrac{(-1)^nw^{2n}}{1+w^2}\right)dw\\
&=\sum_{k=0}^{n-1} (-1)^k\int_0^rw^{z-1}w^{2k}dw+\int_0^rw^{z-1}\dfrac{(-1)^nw^{2n}}{1+w^2}dw\\
&=\sum_{k=0}^{n-1} (-1)^k\int_0^rw^{2k+z-1}dw+\int_0^r\dfrac{(-1)^nw^{2n+z-1}}{1+w^2}dw\\
&=\sum_{k=0}^{n-1} (-1)^k\dfrac{w^{2k+z}}{2k+z}|_0^r+(-1)^nE_{2n+z-1}(r)
\qquad E_{v}(r)=\int_0^r\dfrac{w^{v}}{1+w^2}dw\\
&=\sum_{k=0}^{n-1} (-1)^k\dfrac{r^{2k+z}}{2k+z}+(-1)^nE_{2n+z-1}(r)\\
\end{array}
$

Here are some easy bounds for
$E_v(r)$.

$\begin{array}\\
E_{v}(r)
&=\int_0^r\dfrac{w^{v}}{1+w^2}dw\\
&<\int_0^rw^{v}\\
&=\dfrac{r^{v+1}}{v+1}\\
E_{v}(r)
&=\int_0^r\dfrac{w^{v}}{1+w^2}dw\\
&>\int_0^r\dfrac{w^{v}}{1+r^2}dw\\
&=\dfrac{r^{v+1}}{(v+1)(1+r^2)}\\
\end{array}
$


问:
Cubing a simple thing
I am trying to expand $\quad (x + 2)^3 $

I am actually not to sure what to do from here, the rules are confusing. To square something is simple, you just foil it. It is easy to memorize and execute. Here though I am not sure if I need to do it like multiplication where I take one $(x + 2)$ term and multiply by another or if I need to multiply all $(x + 2)$ terms by it.

I want to treat it like how I would square it so I just square it and then I am left with the result and the $(x + 2)$ term. This is wrong and I do not know why. I get this

$$(x^2 + 4x + 4)(x + 2)$$

This is wrong and I am not sure why. So not I try the other way, multiplying everything by everything. This leaves me with

$$(x^2 + 4x + 4)(x^2 + 4x + 4)$$

Which is again wrong. I have exhausted all my options and nothing results in a correct answer and I am not sure why.

Title: Cubing a simple thing

Body:
You had a correct step, when you had $(x^2 + 4x+4)(x+2)$. Now just distribute.

$$(x^2 + 4x + 4)x + (x^2 + 4x + 4)2 = x^3 + 4x^2 + 4x + 2x^2 + 8x + 8 = x^3 + 6x^2 + 12x + 8$$


问:
Unable to prove or find an counterexample for a question related to analytic continuation
This question is from Ponnusamy and silvermann Complex variables with applications , chapter- Analytic Continuation.

Note: If a function f is analytic in domain D then author represent them as (f,D).

Is it possible that function elements $(f, D_1) $ and $(g ,D_2) $ can be connected by an infinite chain of functional elements , but by no finite subchain.

I am not sure but intution says that it will not be possible.

Can you please help me prove it or find an counterexample.I am sorry , I can't provide anything as attempt in this question.

答:
This is based on a lot of ignorance but, suppose the nth domain has size 1/2^n.
Can the limit of these domains reach a domain which is 1 away?</p>


问:
Please give me a hint for Exercise 2 in Section 13 in "Analysis on Manifolds" by James R. Munkres. I have no idea now.
I am reading "Analysis on Manifolds" by James R. Munkres.

On p.111, there is the following exercise(exercise 2) for section 13("The Integral over a Bounded Set").

Let $A$ be a rectangle in $\mathbb{R}^k$; let $B$ be a rectangle in $\mathbb{R}^n$; let $Q=A\times B$. Let $f:Q\to\mathbb{R}$ be a bounded function. Show that if $\int_Q f$ exists, then $$\int_{\mathbf{y}\in B}f(\mathbf{x},\mathbf{y})$$ exists for $\mathbf{x}\in A-D$, where $D$ is a set of measure zero in $\mathbb{R}^k$.

I have no idea.

The following propositions are all the propositions in section 13. (I added exercise 1 in this picture.)

I guess I must use one of these propositions or exercise 1 to solve this exercise.

But I don't know which proposition I must use.

RRL, Thank you very much for your elegant answer. RRL used Theorem 12.2(Fubini's theorem) on p.100 in Section 12 and Theorem 11.3 on p.96 in Section 11 and Definition(integrable over Q) on pp.84-85 in Section 10.

RRL didn't use any proposition in Section 13.

I wonder why Munkres set Exercise 2 for Section 13 and did not set Exercise 2 for Section 12.

I noticed that RRL used Theorem 13.3(a) on p.106 in Section 13.

So Munkres set Exercise 2 for Section 13.

I like Munkres.

答:
Since $f$ is bounded, the upper and lower integrals $\overline{\int_B} f(x,y) \, dy,\,\,\,\underline{\int}_B f(x,y) \, dy,$ exist for each $x \in A$. If $f$ is Riemann integrable on $Q$, then by Fubini's theorem

$$\int_A \left(\overline{\int_B} f(x,y) \, dy \right) \, dx = \int_A \left(\underline{\int}_B f(x,y) \, dy \right) \, dx,$$

and, hence,

$$\int_A \left(\overline{\int_B} f(x,y) \, dy - \underline{\int}_B f(x,y) \, dy \right) \, dx = 0.$$

Since the difference of upper and lower Darboux integrals is nonnegative, it follows that

$$\overline{\int}_B f(x,y) \, dy = \underline{\int}_B f(x,y) \, dy ,$$

except possibly on a set $D \subset A$ of measure zero, and the function $f(x, \cdot) :y \mapsto f(x,y)$ is Riemann integrable for every $x \in A-D$.


问:
Extremal distribution of random variable that averages to a given value
Denote by $\mathcal M$ the set of probability measures over $[0,1]$ that average to $p$. This is a convex set and I am trying to characterize its extreme points. For any $x$, $y$ such that $0\leq x< p< y\leq 1$ define
\begin{align*}
\lambda_x^y = \frac{p-y}{x-y}\cdot\delta_x+\frac{x-p}{x-y}\cdot\delta_y
\end{align*}
where for any measurable $A\subseteq [0,1]$ and $a\in[0,1]$, $\delta_a(A)=\mathbf 1(a\in A)$. Since $\delta_x$ (resp. $\delta_y$) averages to $x$ (resp. $y$) we get that $\lambda_x^y$ averages to $\frac{p-y}{x-y}\cdot x+\frac{x-p}{x-y}\cdot y=p$ and so is in $\mathcal M$.

I am very tempted to say that the extreme points of $\mathcal M$ is the set $\{ \lambda_x^y : 0\leq x< p < y \leq 1 \}\cup \{ \delta_p \}$ without being able to prove it. If I am not mistaking I am supposed to prove that (ignoring the $\delta_x$ component)

If for some $x$ and $y$ and $a\in[0,1]$, $\lambda_x^y=a \mu+(1-a) \nu$ then $\mu=\nu=\lambda_x^y$.

For any $\mu\in\mathcal M$ there is a probability measure $\kappa$ over $[0,p]\times[p,1]$ such that $\mu(A)=\int \lambda_x^y(A) d\kappa(x,y)+\mu(\{ p \})\delta_p(A)$ for all measurable $A\subseteq [0,1]$.

For the first point, it is quite clear by positivity of the measures $\mu$ and $\nu$ that their support $\{ x,y \}$, then there is only one probability measure with such a support that averages to $p$ and it is $\mu_x^y$.

For the second statement however it is less clear to me, Maybe one way to proceed is computing the integral
\begin{align*}
\int \lambda_x^y(A) d\kappa(x,y)&=\int \left( \frac{p-y}{x-y}\cdot\delta_x(A)+\frac{x-p}{x-y}\cdot\delta_y(A) \right) d\kappa(x,y)\\
&=\int_{\left(A\cap [0,p]\right)\times [p,1]}\frac{p-y}{x-y} d\kappa(x,y)+\int_{[0,p]\times\left(A\cap [p,1]\right)}\frac{x-p}{x-y}d\kappa(x,y)
\end{align*}
It now feel easier to separate cases $A\subseteq [0,p]$ and $A\subseteq [p,1]$ to set one of the two integral to $0$ but I still don't know how to proceed.

Actually thinking about how I could use it in my research it would be much more powerful to not enforce $x\leq p \leq y$ but just that $p$ is in the convex hull of $\{ x,y \}$ and then we get repetition $\lambda_x^y=\lambda_y^x$ but the of extreme points is the same. This makes $\kappa$ a probability measure over $[0,1]^2$ where we have to be careful assigning $0$ probability to sets $A\times B\subseteq [0,1]^2$ where the convex hull of $A\cup B$ does not contain $p$.

Here is the last thing I tried :
Denote $a_x^y=\frac{p-y}{x-y}$, for $\mu \in\mathcal M$ and $A$ measurable
\begin{align*}
\mu(A)&=\int_{[0,p)} \delta_x(A) d\mu(x)+\int_{(p,1]} \delta_y(A) d\mu(y)+\mu(\{ p \})\delta_p(A)\\
&=\int_{[0,p)\times (p,1]} \left(\frac{a_x^y}{a_x^y}\delta_x(A) + \frac{1-a_x^y}{1-a_x^y}\delta_y(A) \right)d\mu\otimes\mu(x,y)+\mu(\{ p \})\delta_p(A)\\
\end{align*}
It feels like we can define $\kappa$ as a function of $\mu\otimes \mu$ and $a_x^y$, I cannot finish the argument though.

答:
Indeed the second claim is true. In order to prove this, we introduce some notation:

1. We will use binary splitting process to establish the desired fact. This will involve splitting the interval $[0, 1]$ into pieces, and we will track this process using binary trees. More precisely, we will consider the binary trees $\mathsf{T}$ such that

Each node of $\mathsf{T}$ is a tagged interval of the form $(I, a)$, where $a \in I$.

Each internal node $(I, a)$ of $\mathsf{T}$ has exactly two childs $(I_0, a_0)$ and $(I_1, a_1)$, where

$I_0 = I \cap (-\infty, a)$ and $I_1 = I \cap [a, \infty)$, and

$a_0, a_1 \in I$ such that $a_0 < a < a_1$.

So, the tagged point $a$ is used to split the interval $I$.

Here is an example of a binary tree:

$$
\small
\begin{gathered}
([0, 1], 0.7) \\
\swarrow \hspace{4em} \searrow \\
([0, 0.7), 0.3) \quad ([0.7, 1], 0.8) \\
\hspace{5.5em} \swarrow \hspace{4em} \searrow \\
\hspace{5.5em} ([0.7, 0.8), 0.72) \quad ([0.8, 1], 0.95)
\end{gathered}
$$

Since knowing the initial interval (the interval in the root) and all the tagged points in $\mathsf{T}$ is enough to reconstruct the entire $\mathsf{T}$, we will usually abbreviate by omitting the interval part whenever no confusion arises. For instance, the above example can be abbreviated as

$$
\small
\begin{gathered}
0.7 \\
\swarrow \quad \searrow \\
0.3 \qquad 0.8 \\
\hspace{3.25em} \swarrow \quad \searrow \\
\hspace{3.5em} 0.72 \qquad 0.95
\end{gathered}
$$

2. Now we recursively define $\nu[\mathsf{T}]$ as follows:

$\nu[a] := \delta_a$.

If $\mathsf{T} = \Bigl( {\scriptsize\begin{gathered} a \\[-5pt] \swarrow \ \searrow \\[-5pt] \mathsf{T}_0 \qquad \mathsf{T}_1 \end{gathered}} \Bigr) $ such that $\mathsf{T}_i$ has root $a_i$ for each $i = 0, 1,$ (in particular, $a_0 < a < a_1$), then

$$ \nu[\mathsf{T}] = \frac{a_1 - a}{a_1 - a_0} \nu[\mathsf{T}_0] + \frac{a - a_0}{a_1 - a_0} \nu[\mathsf{T}_1]. $$

This notation is related to OP's notation in that, if $x < p < y$, then

$$ \nu \Bigl[ {\scriptsize\begin{gathered} p \\[-5pt] \swarrow \ \searrow \\[-5pt] x \qquad y \end{gathered}} \Bigr] = \lambda^{y}_{x}. $$

Then the following lemma holds:

Lemma. For each binary tree $\mathsf{T}$, there exists a probability measure $\kappa$ on $([0,p)\times(p,1])\cup\{p\}$ such that

$$ \nu[\mathsf{T}](\cdot) = \int_{[0,p)\times(p,1]} \lambda_{x}^{y}(\cdot) \, \kappa(\mathrm{d}x\mathrm{d}y) + \delta_p(\cdot) \, \kappa(\{p\}). $$

This follows by the structural induction together with the following reduction formula

$$ \nu \Biggl[ {\scriptsize\begin{gathered} a \\[-5pt] \swarrow \ \searrow \\[-5pt] a_0 \qquad a_1 \\[-5pt] \swarrow \ \searrow \hspace{3em} \\[-5pt] a_{00} \qquad a_{01} \hspace{3em} \end{gathered}} \hspace{-1.8em} \Biggr]
= \alpha \nu \Bigl[ {\scriptsize\begin{gathered} a \\[-5pt] \swarrow \ \searrow \\[-5pt] a_{00} \qquad a_1 \end{gathered}} \Bigr]
+ (1-\alpha) \nu \Bigl[ {\scriptsize\begin{gathered} a \\[-5pt] \swarrow \ \searrow \\[-5pt] a_{01} \qquad a_1 \end{gathered}} \Bigr], $$

where $\alpha$ is given by

$$ \alpha = \frac{(a_1 - a_{00})(a_{01} - a_0)}{(a_1 - a_0)(a_{01} - a_{00})} \in (0, 1). $$

3. The final ingredient is the observation that any probability measure $\mu \in \mathcal{M}$ on $[0, 1]$ that averages to $p$ can be approximated by $\nu[\mathsf{T}]$. This follows from the binary splitting process (see this, for instance), which is often adopted in the proof of the Skorokhod's embedding theorem.

Here, we only outline the key idea. We will construct a sequence of binary trees $(\mathsf{T}_n)_{n=0}^{N}$ by running the following algorithm:

We begin with the binary tree $\mathsf{T}_0$ which consists only of the root $([0, 1], p)$.

Suppose that $\mathsf{T}_n$ is defined and satisfies $a = \mu[x \mid x \in I] = \frac{1}{\mu(I)} \int_{I} x \, \mu(\mathrm{d}x)$ for each node $(I, a)$ of $\mathsf{T}_n$.

If $\mu(I) = \mu(\{a\})$ holds for all leaves $(I, a)$ of $\mathsf{T}_n$, we set $N = n$ and halt the algorithm.

Otherwise, pick a leaf $(I, a)$ of $\mathsf{T}_n$ and set

$$ a_0 = \mu[x \mid x \in I \cap [0, a)] \qquad\text{and}\qquad a_1 = \mu[x \mid x \in I \cap [a, 1]]. $$

The two assumptions, $a = \mu[x \mid x \in I]$ and $\mu(I) \neq \mu(\{a\})$, shows that both $\mu(I \cap [0, a)) > 0$ and $\mu(I \cap [a, 1]) > 0$ hold true, and so, both $a_0$ and $a_1$ are well-defined. Moreover, with this choice, we have

$$ \nu \Bigl[ {\scriptsize\begin{gathered} a \\[-5pt] \swarrow \ \searrow \\[-5pt] a_{0} \qquad a_{1} \end{gathered}} \Bigr] = \frac{\mu(I \cap [0, a))}{\mu(I)} \delta_{a_0} + \frac{\mu(I \cap [a, 1])}{\mu(I)} \delta_{a_1}. $$

Then we set $\mathsf{T}_{n+1}$ as the binary tree obtained by inserting two children $(I \cap [0, a), a_0)$ and $(I \cap [a, 1], a_1)$ under the node $(I, a)$.

If the algorithm does not halt in finite time, set $N = \infty$.

Then it can be proved that $\nu[\mathsf{T}_n]$ converges to $\mu$ weakly as $n \to N$. If we denote $\kappa_n$ for the probability measure obtained by applying the lemma to $\nu[\mathsf{T}_n]$, then any limit point of $(\kappa_n)_{n=0}^{N}$ as $n \to N$ will provide the desired probability measure associated with $\mu$.


问:
Is there any reference for the optimal bound $3|M|-1$ in Simon factorization forest theorem?
Let $S$ be a finite semigroup. A factorization tree of a product $w_1\cdots w_n$ in $S$ is a finite ordered tree labeled by elements of $S$, such that each leaf is labeled by the $w_i$ in their order in the product, and such that each internal node is labeled by the product of their leaves (in their order). A factorization tree is Ramseyan if each internal node of arity $3$ or more has all its children labeled by the same idempotent.

Theorem. (Simon) Each product in a finite semigroup $S$ has a Ramseyan factorization tree of height at most $9|S|$.

The bound was later improved to at least $3|S|$. I found in the two papers [1] and [2] the mention that this bound can be improved to $3|S|-1$. Here is a quote from [2]:

The improved bound $3 |S| − 1$ can be obtained from the current proof by some additional rotation technique as in the proof of Lemma 2. We postpone this proof to a forthcoming journal version of this article.

However, I found no such journal version and was unable to reconstruct the proof for the bound $3|S|-1$. We can make a recurrence from the lowest Green $\mathcal{J}$-classes to the highest instead of the other way around but I still obtained the bound $3|S|$ instead of $3|S|-1$.

Can someone give a proof of this result or a reference?

[1] Factorization forests for infinite words and applications to countable scattered linear orderings (Colcombet, 2010)

[2] The Height of Factorization Forests (Kufleitner, 2008)

答:
The published version of [2] is

M. Kufleitner, The height of factorization forests. In E. Ochmanski and J. Tyszkiewicz, editors, Proc. 33rd Symp., Mathematical Foundations of Computer Science 2008, volume 5162 of Lecture Notes in Comput. Sci., pages 443–454. Springer, 2008.

It contains a detailed proof of the result you are looking for.


问:
Does the group growth rate limit the number of edges going out of a vertex in its Cayley graph?
The growth rate of a group $B_n(G, T)$ is based on the number of vertices that can be reached from a given one by $n$ steps along an edge in the Cayley graph of the group, where $G$ is the group (or its graph) and $T$ is a set of generators of the group or the respective edges in the graph.

I learned here that $\mathbb{Z}^3$ has a growth rate of the order of $n^3$. Looking at graphs (not necessarily Cayley), I wonder if the following exists for an arbitrary but fixed $n_0\in\mathbb{N}$:

The graph is infinite.

The graph is symmetric.

The growth rate is of order $n^3$.

Each vertex has $m>=n_0$ edges.

This exists for $m=n_0=6$ per the tiling of three dimensional space with cubes.

Question: Is the following proof that I can find an $m$ for any $n_0$ correct? (Risking a yes/no question as per this meta post.)

Define a graph $G_1 = (V, E_1)$ such that $V=\mathbb{Z}^3$. The vertices can be considered centers of cubes that tile $\mathbb{R}^3$. Define an edge of the graph for each two cubes that "touch" directly, either on sides, edges or corners. Consider a Rubik's Cube, where center cube has an edge to all surrounding cubes. More formally, let $v, w\in V$ be connected, i.e. $\{v, w\}\in E_1$, if they are "direct neighbors" along any coordinate combination, i.e. $v-w \in \{-1,0,1\}^3$ and $v\neq w$.

The ball $B_n(G_1, v)\subset V$ shall be the set of nodes reachable from $v$ with a minimal path length of $\leq n$. For $n=1$ this is again like looking at Rubik's Cube and $|B_1(G_1, v)| = 3^3 = 27$. In general the number of elements in the ball $B_n$ is an ever bigger "Rubik's Cubes" albeit always with an odd number of cubes along one dimension:
$$|B_n(G_1, v)| = (1+2n)^3$$
So the growth rate is of the order of $n^3$, but we don't yet have an arbitrary large number of neighbors for a given vertex.

Now we define the graph $G_k=(V,E_k)$ based on $G_1$ such that we add edges to $E_1$ from $v$ to every vertex $w\in B_k(G_1, v)\setminus E_1$, so that now all vertices of that ball are direct neighbors of $v$.

In the new graph, we have
$$ |B_n(G_k, v)| = (1+kn)^3$$
which still is a growth rate of order $n^3$, but since we are free to choose $k$, we can create a symmetric graph of order $n^3$ were each vertex has arbitrary many edges going out.

Specific head-scratchers

Is $|B_n(G_k, v)|$ correct?

Is the constructed graph $G_k$ really symmetric?

答:
Yes, this is a fine construction. (Or: no, the group growth rate doesn't limit the degree of vertices.) A generalization of this: if you find an infinite graph $G$ which is symmetric, connected, and has a growth rate $|B_n(G,v)| = O(f(n))$, then we can let $G^k$ be the graph with an edge $vw$ whenever $d(v,w) \le k$ in $G$. We can make $G^k$ have arbitrarily large minimum degree, and still have$ |B_n(G^k,v)| = O(f(n))$.

We can even find a Cayley graph that will have the property you want. Take the group $\mathbb Z^3 \times \mathbb Z_2^k$, and take $T$ to be a set of $3+k$ generators corresponding to each of the factors. Then each vertex of the Cayley graph will have degree $6+2k$, and the growth rate will be $O(n^3)$. (The idea is that after $n$ steps, there are $O(n^3)$ possibilities for the element of $\mathbb Z^3$ we have, and at most $2^k = O(1)$ possibilities for the element of $\mathbb Z_2^k$.)

Or, we could even take $\mathbb Z^3$, but with a different, larger generating set. The growth rate will still be $O(n^3)$, because if no generator lets you change any coordinate by more than $M$, then after $n$ steps we are limited to a cube with $(2Mn+1)^3$ vertices in it. The degree of each vertex is twice the number of generators.


问:
difference in the definiton of modular arithmetic
When i look at my textbook , i realized that there is a little difference in the definiton of modular artihmetic among books.

For example, wikipedia says that $a\equiv b\pmod n$ where $a,b \in Z,n \in Z^+$ and $n \gt1$ .

On the other hand, some books and websites says that $a\equiv b\pmod n$ where $a,b \in Z,n \in Z^+$ but there is not any restriction for $n$ ,i.e, $n$ can be $1$.

I know that it is not a big difference and does not affect results generally.However, i want to know the exact definiton of it.

Briefly, which definiton is correct ? If they are same , then why do books ignore $n=1$

答:
The biggest difference is that if we use the standard definition and apply it to $\pmod 1$ the result is utterly trivial.

$\pmod 1$ we have $a \equiv 0 \pmod 1$ and $a \equiv b\pmod 1$ for any possible integers.

(Think about it.  $a \equiv b \pmod 1 \iff 1|a-b \iff \exists k\in \mathbb Z: a= b+k\cdot 1$ (or in other words $k=a-b$) $\iff a$ and $b$ will both have the same remainder when divided by $1$.)

[  And as the only residue class of $k \in \mathbb Z$ and $0 \le k < n = 1$ is ...  $k=0$, $\pmod 1$ arithmetic is an arithmetic with exactly one element so that $[0]+[0] \equiv [0]\pmod 1$ and  $[0]\cdot [0]\equiv 0\pmod 1$ and $[0]^k \equiv [0]\equiv 1 \pmod 1$.  You can't get much more boring than that!]

So ... it's pointless to consider $\pmod 1$.

.........

However, i want to know the exact definiton of it. Briefly, which definiton is correct ?

You think mathematicians can ever agree on definitions (or anything) and there are always correct definitions of everything (or even anything)?

Well, bless your heart!

....

(To explain my last somewhat facetious comment:  One thing I've found all mathematician have is a love to quibble over precision and semantics and picayune details and, with only one exception, disagree on everything.  That sounds frustrating and futile but 1) we enjoy it and 2) the one thing we all agree on is if we are logically precise and consistent arbitrary choices and definitions won't actually matter.)

(In other words: the is no "correct" definition.  IMO, I'd allow $\pmod 1$ because... well, although it's pointless and trivial,t it is consistent and follows the definitions precisely.  I suspect its only necessary to exclude $n=1$ when considering certain result of abstract ring theory or algebra but examples and details elude me.  I suspect the reason some texts exclude it is they want to avoid having to make a detour and confusing beginners about something that won't really matter.  Image if the very first time I introduce modular arithmetic I immediately took a detour to explain my third paragraph above...)


答:
The actual numbers written on the white cards don't matter, let's call them, in increasing order, $C_1,C_2,\dots,C_5$. There are $5!$ permutations of $\{1,2,\dots,5\}$, enough to encode all the numbers from $1$ to $120$.

The code (for $120$) could go as follows. If $C_1$ is the top card, then the hidden card is in the range $1$ to $24$; if it is $2$, the hidden card is $25$ to $48$, and so on. Then the range identified by the top card is narrowed down by considering the next card. And so on. With some practice the "reading" would be quick.

Title: Another Information Theory Riddle

Body:
The following nice riddle is a quote from the excellent, free-to-download book: Information Theory, Inference, and Learning Algorithms, written by David J.C. MacKay.

In a magic trick, there are three participants: the magician, an assistant, and a volunteer.

The assistant, who claims to have paranormal abilities, is in a soundproof room. The magician gives the volunteer six blank cards, five white and one blue. The volunteer writes a different integer from 1 to 100 on each card, as the magician is watching. The volunteer keeps the blue card. The magician arranges the five white cards in some order and passes them to the assistant.

The assistant then announces the number on the blue card.

How does the trick work?


问:
Prove that recursively defined function is injective and equal to subset
This is my question from one of the past paper exams, I am solving it it but I am not sure.
The question is following:

A certain subset A of the set {0,1}* is defined recursively as follows:

Base case: λ ∈ A.

Constructor case 1: If s ∈ A then 0s ∈ A.

Constructor case 2: If s ∈ A then 10s ∈ A.

Constructor case 3: If s ∈ A then 11s ∈ A.

A certain function f : {x,y,z}* --> {0,1}* is defined recursively as follows:

Base case: f(λ) = λ.

Constructor case 1: f(xs) = 0f(s) = <0, f(s)>

Constructor case 2: f(ys) = 10f(s) = <1, <0, f(s)>>

Constructor case 3: f(zs) = 11f(s) = <1, <1, f(s)>>

Prove that f is injective.

Prove that f({x,y,z}*) = A.

This is my solution:

In injective function if f(x)=f(y), then x=y.

Let's define a predicate P(s) ---> {0,1}*
Where for every t  ∈ {x,y,z}*

If we can prove that s=t when f(s)=f(t)
Then we can conclude that function f is injective.

Base Case:

In Base case λ ∈ A
As well as f(λ) = λ, therefore
f(s) = f(t) as both equal to λ

Constructor Case:

Case 1;

f(xs) = 0f(s) = <0, f(s)>

f(xt) = 0f(t) = <0, f(t)>

f(xs)=f(xt) => s = t

Case 2 and 3 the same way.
Therefore, we conclude that the function f is injective.

I do not understand how to do this

答:
Actually, surjectivity is easier:
We need to show that

$f(s)=\lambda$ for some $s\in \{x,y,z\}^*$

If $s\in f(\{x,y,z\}^*)$ then $0s\in f(\{x,y,z\}^*)$.

If $s\in f(\{x,y,z\}^*)$ then $10s\in f(\{x,y,z\}^*)$.

If $s\in f(\{x,y,z\}^*)$ then $11s\in f(\{x,y,z\}^*)$.

The first is clear because $f(\lambda)=\lambda$.
For the other three parts, note that $s\in f(\{x,y,z\}^*)$ means $s=f(t)$ for some $t\in \{x,y,z\}^*$. But then $0s=f(xt)$, $10s=f(yt)$, $11s=f(zt)$. $\square$

For injectivity, you may need to be a bit more careful.
As you suggest, we let
$$ P(s)\equiv \forall t\in\{x,y,z\}^*\colon f(s)=f(t)\to s=t$$

We have $P(\lambda)$ because if $t\ne \lambda$ then $t$ begins with $x$, $y$, or $z$, hence $f(t)$ begins with $0$, $10$, or $11$. At any rate $f(t)\ne \lambda= f(\lambda)$.

If $s=xs'$ and $f(s)=f(t)$ then $t=\lambda$ or $t=xt'$ or $t=yt'$ or $t=zt'$. As $f(s)$ begins with $0$, we can exclude $t=\lambda$, $t=yt'$, and $t=zt'$. Hence $t=xt'$ and $0f(s')=f(s)=f(t)=0f(t')$ implies $f(s')=f(t')$. By induction hypothesis, $s'=t'$ and ultimately $s=xs'=xt'=t$.

If $s=ys'$ and $f(s)=f(t)$ then $t=\lambda$ or $t=xt'$ or $t=yt'$ or $t=zt'$. As $f(s)$ begins with $10$, we can exclude $t=\lambda$, $t=xt'$, and $t=zt'$. Hence $t=yt'$ and $10f(s')=f(s)=f(t)=10f(t')$ implies $f(s')=f(t')$. By induction hypothesis, $s'=t'$ and ultimately $s=ys'=yt'=t$.

If $s=zs'$ and $f(s)=f(t)$ then $t=\lambda$ or $t=xt'$ or $t=yt'$ or $t=zt'$. As $f(s)$ begins with $11$, we can exclude $t=\lambda$, $t=xt'$, and $t=yt'$. Hence $t=zt'$ and $11f(s')=f(s)=f(t)=11f(t')$ implies $f(s')=f(t')$. By induction hypothesis, $s'=t'$ and ultimately $s=zs'=zt'=t$.


问:
Sign change of a polynomial between roots
Let $p(x)$ be a polynomial with $n$ distinct real roots. Label the roots of $p(x)$ as $x_1, x_2, ...,x_n,$ such that $x_1 < x_2 < ... < x_n$. Suppose that $p(x)$ has a root that is also a stationary point, i.e. $\exists \ i \in \{1,2, ... ,n\}$ such that $p(x_i) = p'(x_i)=0$. Is it then true that the sign of $p(x)$ over the interval $\left(x_{i-1},x_i\right)$ is the same as the sign of $p(x)$ over $\left(x_{i},x_{i+1}\right)?$ Intuitively this seems true, but does a rigorous proof for this exist?

Edit: As commented in one of the answers, this claim is not always true. In particular, if one has the polynomial $p(x)=x^3(x-1)(x+1)$, where the root $x=0$ is a point of inflection, then there is a sign change. Thus a root which is a stationary point does not imply that there is a sign change. However, would the converse be true then? If there is no sign change at a root, does that mean that the root is a stationary point?

答: $x^3(x-1)(x+1)$ is a counterexample.

Body: $x^3(x-1)(x+1)$ is a counterexample.


问:
Proving $E[X^4]=3σ^4$
Given a random variable $X\sim\mathcal N(0,\sigma^2)$, how can we prove that $E[X^4]=3\sigma^4$? I am having trouble even starting with the proof.</p>

答:
@Yves Daoust already has a great answer, but we can omit a round of integration by parts, by just working it out in the most straight-forward way.

Applying integration by parts once,
$$
\mathbb{E} (x^4) = \int x^4 \varphi(x) dx = 0 + 3\sigma^2 \int x^2 \varphi(x) dx
$$

where $\varphi(x)$ is normal PDF and $\varphi(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{x^2}{2\sigma^2}}$

We recognise $\int x^2 \varphi(x) dx = \sigma^2$. So,

$$
\mathbb{E} (x^4) = 3\sigma^2 \sigma^2 = 3 \sigma^4
$$

Title: Proving $E[X^4]=3σ^4$


问:
Further Stirling number series resummation
\begin{equation}
\sum_{m=1}^\infty\sum_{n=1}^\infty (-1)^{n } \frac{S_m^{(3)}}{m! n}(-1 + u)^{(m + n - 1)} (\frac{x}{-1 + x})^m
\end{equation}

Note: $S^{(3)}_m$ belongs to the Stirling number of the first kind https://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind If you see the wiki page "Table of values" $S^{(3)}_m$ corresponds to the third column.

Please let me know if addition information is needed.

答:
The double series can be factored
\begin{eqnarray*}
& \sum_{m=1}^\infty\sum_{n=1}^\infty (-1)^{n } \frac{S_m^{(3)}}{m! n}(-1 + u)^{(m + n - 1)} (\frac{x}{-1 + x})^m \\ &=  \left( \sum_{m=1}^\infty \frac{S_m^{(3)}}{m! }(-1 + u)^{(m  - 1)} (\frac{x}{-1 + x})^m \right) \left( \sum_{n=1}^\infty (-1)^{n } \frac{1}{ n} (-1 + u)^{ n } \right)
\end{eqnarray*}
The second sum is the power series for $ \ln $ and the first sum can be calculated using
\begin{eqnarray*}
\sum_{n=k}^{\infty} (-1)^{n-k} { n \brack k } \frac{z^n}{n!} = \frac{( \ln(1+z))^k}{k!}.
\end{eqnarray*}


问:
Given $A$ and $B$ are $n \times n$ matrices, show that every vector $x$ in null space of $A$ is also in null space of $AB$
For the null space equation $Ax = 0$ and for $A$ and $B$ having the same dimensions, I am currently trying to prove whether or not every vector x in $Nul(A)$ is also in $Nul(AB)$. Also trying to prove whether for every vector $x$ in $Nul(A)$ and is also in $Nul(BA)$. I have to use definite proofs, not specific examples. I have thought about the order of multiplying $A$ and $B$ and how they transform $x$ but I can't see how that helps in proving. I am pretty stuck here.

答:
If $x$ is in the nullspace of $A$, then it is certainly in the nullspace of $BA$ since
$$
BAx = B(Ax) = B0 = 0
$$
But not necessarily in the nullspace of $AB$. I'm not sure what you mean by wanting definite proofs, rather than 'specific examples' as giving a counterexample to the claim that $\ker(A)\subseteq \ker(AB)$ would absolutely constitute a definite proof.


问:
Generators of level $2$ modular forms
The ring of modular forms for $\Gamma_1=\text{SL}(2,\mathbf{Z})$
$$M(\Gamma_1)\ =\ \bigoplus_{k\ge 0} M(\Gamma_1)_k\ =\ k[E_4,E_6]$$
is a free ring, with a generator in weight $4$ and weight $6$. (The reason is that $\mathfrak{H}/\Gamma_1=\mathbf{P}(4,6)$ is a weighted projective space, and the above is $\bigoplus_{k\ge 0}H^0(\mathbf{P}(4,6),\mathcal{O}(k))$). The generators are Eisenstein series.

Question: what is the ring of modular forms of level $2$
$$M(\Gamma_2)\ =\ \bigoplus_{k\ge 0} M(\Gamma_2)_k\ ?$$
Since $\mathfrak{H}/\Gamma_2=\mathbf{P}(2,2)$ I believe it should be free on two generators of weight $2$, but I don't know what they are explicitly as functions on $\mathfrak{H}$.

答:
Yes, the ring of holomorphic modular forms on $\Gamma(2) = \Gamma_2 \triangleleft \mathrm{SL}_2(\mathbb{Z})$ is equal to $\mathbb{C}[X,Y]$ where $X = \Theta_2^4$ and $Y= \Theta_3^4$ and where

$$
\Theta_2(z) = \sum_{n \in \mathbb{Z}}{e^{\pi i (n+1/2)^2 z}}, \quad \Theta_3(z)  = \sum_{n \in \mathbb{Z}}{e^{\pi i n^2 z}}, \qquad  \Theta_4(z) = \sum_{n \in \mathbb{Z}}{(-1)^ne^{\pi i n^2 z}}
$$

are the classical theta constants (sometimes also denoted $\Theta_2 =\theta_{10}, \Theta_3 = \theta_{00}, \Theta_4 =\theta_{01}$)

Thus, for example $$
M_2(\Gamma_2)= \mathbb{C} \Theta_2^4 \oplus \mathbb{C} \Theta_3^4, \quad M_4(\Gamma_2) = \mathbb{C} \Theta_2^8 \oplus \mathbb{C} \Theta_2^4\Theta_3^4  \oplus \mathbb{C}\Theta_3^8.
$$

You can prove this description using the following valency formula (which is in some sense simpler than the one for level 1 forms and also be proved via contour integration, as is done (in level 1) in Serre's course of arithmetic, for example)

$$
\nu_{[1]}(f) + \nu_{[0]}(f) + \nu_{[\infty]}(f) + \sum_{[p] \in \Gamma_2 \backslash \mathbb{H}}{\nu_{[p]}(f)}= \frac{k}{2}
$$

valid for any nonzero meromorphic modular form $f$ of even integral weight $k$. It implies first that none of the theta constants $\Theta_2, \Theta_3, \Theta_4$ has a zero on the upper half plane. Another input is that multiplication by $\Theta_3^4$ defines an isomorphism from $M_k(\Gamma_2)$ onto $S_{k+2}^{[1]}(\Gamma_2) \subset M_{k+2}(\Gamma_2)$ where $S_{k+2}^{[1]}(\Gamma_2)$ denotes the subspace of forms vanishing at the cusp 1.  (This can be seen as an analogue of multiplication by $\Delta \in S_{12}(\Gamma_1)$, shifting the weight by 12)

Title: Generators of level $2$ modular forms


问:
Couple problems and classic wolf/goat/cabbage and abstraction
I was reviewing Dijkstra's approach on the problem/puzzle of how 2 married couple can cross a river with 1 boat that can carry 2 people. The original problem's restriction is that the wife can't be in the presence of another husband without hers being present as well.

His approach was to change the original restriction to be that the husbands have to stay together and the result of this abstraction is that there is no longer a need to distinguish between husband/wife so that the problem is simplified.

Although what he says is clear, I don't understand why the restriction is required in the first place.

The problem seems identical to the goat/cabbage/wolf where the cabbage and wolf had a symmetry with the goat (neither could be left alone with the goat) and hence replaced with an abstract concept "a".

Isn't this also applicable in the case of the couples? Is the approach of having an extra restriction redundant or not?

答:
Of course if you can solve the problem with the husbands being apart at some point, that's just fine. That is, there is clearly no a priori need for imposing this additional restriction: the problem is stated the way it is ... why would you be required to add another restriction? That makes little sense.

Indeed, there is obviously a danger in trying to impose restrictions: maybe in doing so the problem actually becomes more difficult, if not impossible to solve. So, imposing further restrictions is not only unnecessary, but it could even be dangerous.

Now, one possible explanation for Dijkstra making this additional restriction  is the following. Maybe Dijkstra analyzed the puzzle, and came to the conclusion that there is no solution possible if the husbands get separated. And note, depending on the nature of the puzzle, one can come to such a conclusion without actually finding all solutions to the puzzle, or even just a single one. And, if that is so, then yes, you might as well treat the two husbands as one item, thus eliminating the need to distinguish between the two, and treating them ,as you say, as one 'object' '$a$'.

Indeed, if you do treat the two husbands as one object '$a$', with the understanding that $a$ alone fills up the whole boat, it is easy to find a solution to this problem: Have the two wives row across, and have one of them return. Now send over '$a$', and have the second wive return, and finally have the two wives go across together again.

However, as others have pointed out, it turns out that there is also a solution with you keep the two husbands together: have one couple row across, after which the husband rows back to fetch the other husband, and then the wife rows back to fetch the other wife.

So, we know that the restriction that Dijkstra imposes is not necessitated at all. In fact, I would say Dijkstra was really rather 'lucky' in creating the additional restriction. How did he know that there was a solution where you could keep the two husbands together?


问:
When can we ignore small-oh inside of a limit and when we cannot?
Here's an example.
$$\lim_{x \to 0} \frac {\ln\left(1 + x + \sqrt{1 + x^2} - 1\right) - x}{x^3}.$$
$\ln(1 + x + \sqrt{1 + x^2} - 1) \sim  x + \sqrt{1 + x^2} - 1$ (the limit of their division is equal to 1). And so the left side we can replace with $x + \sqrt{1 + x^2} - 1 + o(x + \sqrt{1 + x^2} - 1)$. Now if we write it inside of the limit we'll have
$$
\lim_{x\to0} \frac {x + \sqrt{1 + x^2} - 1 + o\left(x + \sqrt{1 + x^2} - 1\right) - x}{x^3}
$$
Here we can cancel the two x-es. Now what I'm wondering about is why can't I ignore the small-oh in this limit? In fact the limit of the expression is not the same with and without the small-oh (with small-oh I mean the $o(x + \sqrt{1 + x^2} - 1)$).
And more generally where can we ignore and when we cannot?
For example
If we have
$\lim_{x \to 0} \dfrac {\ln(1+x)} {x}$ here
$\ln(1+x) = x + o(x)$ and so if we ignore $o(x)$ then we'll have this limit $\lim_{x \to 0} x/x = 1$ which's correct. So what's the difference and how can I detect it in problems to avoid errors?

答:
I will use the example. The denominator is $x^3$. So it is enough to know both numerator and denominator modulo $o(x^3)$. Well, since i prefer the big-oh-notation, let us please work modulo $O(x^4)$, it is possible here.

In the numerator, the term $-x$ is clear, so we need to know $\ln(x+\sqrt{1+x^2})+O(x^4)$. I.e. the Taylor polynomial of $\ln(x+\sqrt{1+x^2})$ in $0$ with the corresponding "precision". Note that "building Taylor polynomials" is an operation compatible with composition in a certain sense. In our case, we have the compositions $U_0\to V_1\overset\ln\to W_0$, where $U_0$ is some ("good small")
neighbourhood of $0$, $V_1$ of $1$, and $W_0$ of $0$. The function from $U_0\to V_1$ is $x\to x+\sqrt{1+x^2}$. Now we can get even more involved in the bureaucracy of Taylor polynomials, and separate the $\sqrt{1+x^2}$ part, since it involves $x^2$. I will write $h$ for $x^2$. And i will go with $h$ till $O(h^4)$ to see that later we only need $O(h^2)$, which may by an answer by example for the OP.)

Now one can formally write:
$$
\begin{aligned}
\sqrt{1+h}
&= (1+h)^{1/2} = \binom{1/2}0 + \binom{1/2}1 h+ \binom{1/2}2 h^2 + \binom{1/2}3 h^3 + O(h^4)\\
&=1 +\frac 12h-\frac 18h^2+\frac 1{16}h^3+O(h^4)
\ ,\\
&\qquad\text{(Do we really/minimally need $O(h^4)$ for the later $O(x^4)$-"precision"?!)}\\
\sqrt{1+x^2}
&=1 +\frac 12x^2-\frac 18x^4+\frac 1{16}x^6+O(x^8)\\
&=1 +\frac 12x^2+O(x^4)\ .\\
&\qquad\text{(No, $O(h^2)$ would have been enough above for the $O(x^4)$-"precision"...)}\\
x +\sqrt{1+x^2}
&=1 +x+\frac 12x^2+O(x^4)\ .\\
\ln(x +\sqrt{1+x^2})
&=\ln\Bigg(1 +\underbrace{\left(x+\frac 12x^2\right)}_{=:w\text{ and }O(x)=O(w)}\Bigg)+O(x^4)
\\
&=w-\frac 12w^2+\frac 13w^3+O(w^4)
\\
&=\left(x+\frac 12x^2\right)
-\frac 12\left(x+\frac 12x^2\right)^2
+\frac 13\left(x+\frac 12x^2\right)^3+O(x^4)
\\
&=\left(x+\frac 12x^2\right)
-\frac 12\left(x^2+2\cdot\frac 12x^3+O(x^4)\right)
+\frac 13\left(x^3+O(x^4)\right)+O(x^4)
\\
&=x-\frac 16x^3+O(x^4)\ .
\\
\ln(x +\sqrt{1+x^2})- x
&=
-\frac 16x^3+O(x^4)
\\
\frac{\ln(x +\sqrt{1+x^2})- x}{x^3}
&=
-\frac 16+O(x)\ .
\end{aligned}
$$
So the answer may be for short the following one. When the involved "pieces" (w.r.t. the operations addition, subtraction, multiplication, composition, division) are analytic functions with known Taylor expansions, one has to follow the degree needed for each piece. Sometimes compositions and multiplicaiont may make the "needed precision" become smaller. In the above example, the "degree-two-map" $x\to h=x^2$ is such an expample, when the precision "in $h$" is not $O(h^4)$, but the smaller $O(h^2)$.

(So even shorter, compute the valuation in the corresponding ring of formal power series modulo the needed precision.)


答:
The following is correct, though not fully expanded:

$$(x^2 + 4x + 4)(x + 2) = (x+2)^2(x+2) = (x+2)^3$$

Consider writing this as $(x + 2)(x^2 + 4x + 4)$, and then distribute (multiply) each term in the first factor, with each term of the second factor.

$$
\begin{align} (\color{blue}{\bf x + 2})(x^2 + 4x + 4)
& = \color{blue}{\bf x}(x^2 + 4x + 4) + \color{blue}{\bf 2}(x^2 + 4x + 4) \\ \\
& = (x^3 + {\bf 4x^2} + \color{blue}{\bf 4x}) + ({\bf 2x^2 }+ \color{blue}{\bf 8x} + 8) \\ \\
& = x^3 + {\bf 6x^2} + \color{blue}{\bf 12x} + 8\\ \\
\end{align}
$$

Title: Cubing a simple thing

Body:
I am trying to expand $\quad (x + 2)^3 $

I am actually not to sure what to do from here, the rules are confusing. To square something is simple, you just foil it. It is easy to memorize and execute. Here though I am not sure if I need to do it like multiplication where I take one $(x + 2)$ term and multiply by another or if I need to multiply all $(x + 2)$ terms by it.

I want to treat it like how I would square it so I just square it and then I am left with the result and the $(x + 2)$ term. This is wrong and I do not know why. I get this

$$(x^2 + 4x + 4)(x + 2)$$

This is wrong and I am not sure why. So not I try the other way, multiplying everything by everything. This leaves me with

$$(x^2 + 4x + 4)(x^2 + 4x + 4)$$

Which is again wrong. I have exhausted all my options and nothing results in a correct answer and I am not sure why.


问:
Proving onto function with strings
I am trying to prove that this function is onto:

Let $S$ be the set of all strings of $0$’s and $1$’s, and define $D: S \to \Bbb{Z}$ as
follows: For all $s \in S$, $D(s) =$ the number of $1$’s in $s$ minus the
number of $0$’s in $s$.

I know that it’s onto, I understand more or less why. I know we should have three cases, when $x = 0$, when $x > 1$ and when $x < 1$. But for the last case $(x < 0)$, I don't understand why it's the image of string $0^x$ instead of the image of $0^{-x}$ (because in this case $x$ is negative and $-x$ is positive, so how can we have  a negative number of zeroes)?

Thank you in advance!

答:
There is indeed at least a typo in the given proof. The result is actually easy to prove. Indeed, for all $a, b \geqslant 0$, one has by definition
$D(1^a0^b) = a-b$. It now suffices to prove that for any integer $c \in \Bbb{Z}$, there exist two nonnegative integers $a$ and $b$ such that $c = a - b$, which should not be too difficult to establish...

Title: Proving onto function with strings


问:
Eigenvalues and eigenvectors of $T^*T$
I am working on the following problem:

Define a bounded linear operator $T$ on $L^2[0,1]$ as $$Tf(x)=\int_x^1 f(t)dt, \; f \in L^2[0,1], \; x \in [0,1].$$ Find the range of $T^*+T$ and the eigenvalues and eigenvectors of $T^*T$.

I started by noticing that
$$
T^*g(x)= \int_{0}^{x}g(t)dt.
$$

So I got $$(T^*+T)f=\int_0^1f(t)dt.$$ At this point, by the Cauchy-Schwarz inequelity I have $$(T^*+T)f \leq \|f\|_2,$$ and so I want to say that $\text{range}(T^*+T)=\{c \in \mathbb{R} \text{ such that } c=\|f\|_2 \text{ for some } f \in L^2[0,1] \}$.

Finally, I think $T^*T$ has no eigenvalues, since neither $T$ nor $T^*$ have some. Am I correct? Any help is appreciated.

Edit: as Aaron pointed out, we have $(T^*T)^*=T^*T^{**}=T^*T$. So the operator is self-adjoint and therefore it will have non-negative eigenvalues.

答:
As you found, $T+T^*$ is a rank-one operator, and its range are the constant functions.

As $T$ is compact, so is $T^*T$. So its spectrum will consist of zero and positive eigenvalues.

If $\lambda>0$ and $T^*Tf=\lambda f$, we have
$$\tag1
\lambda f=\int_0^x\int_t^1f(s)\,ds\,dt.
$$
Since $f$ is in $L^2$, it is also in $L^1$ (use Cauchy-Schwarz) and so $Tf$ is continuous. It follows that $\lambda f$ is differentiable, and going back and forth we get that $f$ is C$^\infty$. Differentiating $(1)$ we obtain
$$\tag2
\lambda f'=\int_x^1f(s)\,ds,
$$
and differentiating once more
$$\tag3
\lambda f''=-f.
$$
The solutions to $(3)$ are of the form
$$
f(t)=\alpha\,\cos\frac t{\sqrt\lambda}+\beta\,\sin\frac t{\sqrt\lambda}.
$$
From $(1)$ we have the initial condition $f(0)=0$, which gives $\alpha=0$. And from $(2)$ we have $f'(1)=0$, which gives
$$
0=\beta\,\cos\frac1{\sqrt\lambda}.
$$
As the sine is zero on $\frac\pi2+k\pi$, $k\in\mathbb Z$, we get
$$
\lambda_k=\frac4{(2k+1)^2\pi^2},\qquad k\in\mathbb N,
$$
with eigenvectors
$$
f_k(t)=\sin 2k\pi t.
$$

Title: Eigenvalues and eigenvectors of $T^*T$


问:
prove a non-strict inequality using the triangle inequality
In an $\varepsilon - \  \delta $ delta proof I had to show that  $$|x^3+y^3|\leq 2(x^2+y^2)^{3/2}$$ i tried some things already such as splitting it but i cannot seem to figure out how to show this. Note: i had to write from the left side untill i arrive at the right side.

Body:
What about squaring both sides?
\begin{align*}
|x^{3} + y^{3}| \leq 2(x^{2} + y^{2})^{3/2} & \Longleftrightarrow x^{6} + 2x^{3}y^{3} + y^{6} \leq 4(x^{2} + y^{2})^{3}\\\\
& \Longleftrightarrow x^{6} + 2x^{3}y^{3} + y^{6} \leq 4(x^{6} + 3x^{4}y^{2} + 3x^{2}y^{4} + y^{6})\\\\
& \Longleftrightarrow 3(x^{6} + y^{6}) + 10x^{4}y^{2} + 10x^{2}y^{4} + 2x^{2}y^{2}(x^{2} - xy + y^{2})\geq 0
\end{align*}

where the last inequality is true indeed.


问:
All possible angles of triangle such that the feet of the angle bisectors determine a right angle

Triangle $ABC$ has angle bisectors $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$. Given that $\angle EDF = 90^\circ$, enter all possible values of $\angle BAC$ (in degrees), separated by commas.

I plan to use vectors here; that is what the hint says. We can see that $\angle{EDF} = 90^\circ$, so $\overrightarrow{DE}\cdot \overrightarrow{DF} = 0$.

However, here I'm stuck. We are dealing with incenters here (intersection of angle bisectors), but I'm not sure how to incorporate angle bisectors or incenters here. Perhaps I could use the fact that since $\angle{BAD} = \angle{CAD}$, then $$\frac{\overrightarrow{AB}\cdot \overrightarrow{AD}}{|AB||AD|} = \frac{\overrightarrow{AC}\cdot \overrightarrow{AD}}{|AC||AD|}$$

However, I don't really see that being useful. Help is greatly appreciated, thanks!

答:
I entered $120^\circ$ and it was correct, but without much reason to it. Here is their solution.

As usual, let $a = BC,$ $b = AC,$ and $c = AB.$

By the Angle Bisector Theorem, $BD:DC = c:b,$ so
$\overrightarrow{D} = \frac{b}{b + c} \overrightarrow{B} + \frac{c}{b + c} \overrightarrow{C} = \frac{b \overrightarrow{B} + c \overrightarrow{C}}{b + c}.$ Similarly,
\begin{align*}
\overrightarrow{E} &= \frac{a \overrightarrow{A} + c \overrightarrow{C}}{a + c}, \\
\overrightarrow{F} &= \frac{a \overrightarrow{A} + b \overrightarrow{B}}{a + b}.
\end{align*} If we let $A$ be the origin, then we get
$\overrightarrow{E} = \frac{c \overrightarrow{C}}{a + c}, \quad \overrightarrow{F} = \frac{b \overrightarrow{B}}{a + b}.$ Therefore,
\begin{align*}
\overrightarrow{DE} &= \overrightarrow{E} - \overrightarrow{D} \\
&= \frac{c \overrightarrow{C}}{a + c} - \frac{b \overrightarrow{B} + c \overrightarrow{C}}{b + c} \\
&= \frac{- b(a + c) \overrightarrow{B} + c(b - a) \overrightarrow{C}}{(a + c)(b + c)},
\end{align*}and
\begin{align*}
\overrightarrow{DF} &= \overrightarrow{F} - \overrightarrow{D} \\
&= \frac{b \overrightarrow{B}}{a + b} - \frac{b \overrightarrow{B} + c \overrightarrow{C}}{b + c} \\
&= \frac{b(c - a) \overrightarrow{B} - c(a + b) \overrightarrow{C}}{(a + b)(b + c)}.
\end{align*} Since $A$ is the origin, $|\overrightarrow{B}| = c$, $|\overrightarrow{C}| = b$, and by the Law of Cosines,
$\overrightarrow{B} \cdot \overrightarrow{C} = |\overrightarrow{B}| |\overrightarrow{C}| \cos A = bc \cdot \frac{b^2 + c^2 - a^2}{2bc} = \frac{b^2 + c^2 - a^2}{2}.$ We have that $\angle EDF = 90^\circ$ if and only if $\overrightarrow{DE} \cdot \overrightarrow{DF} = 0$, or equivalently,
\begin{align*}
&[-b(a + c) \overrightarrow{B} + c(b - a) \overrightarrow{C}] \cdot [b(c - a) \overrightarrow{B} - c(a + b) \overrightarrow{C}] \\
&= -b^2 (a + c)(c - a) |\overrightarrow{B}|^2 + bc(a + c)(a + b) \overrightarrow{B} \cdot \overrightarrow{C} \\
&\quad + bc(b - a)(c - a) \overrightarrow{B} \cdot \overrightarrow{C} - c^2 (b - a)(a + b) |\overrightarrow{C}|^2 \\
&= -b^2 c^2 (c^2 - a^2) + 2bc(a^2 + bc) \cdot \frac{b^2 + c^2 - a^2}{2} - b^2 c^2 (b^2 - a^2) \\
&= a^2 bc(b^2 + bc + c^2 - a^2) \\
&= 0,
\end{align*}so $a^2 = b^2 + bc + c^2$. Then by the Law of Cosines,
$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{-bc}{2bc} = -\frac{1}{2}.$ Therefore, $$A = \boxed{120^\circ}.$$


问:
Prove that : 1. $\sum_{n=1}^{\infty}\int_{0}^{\infty}|f_n|dx=\infty$ 2. $\sum_{n=1}^{\infty}\int_{0}^{\infty}f_ndx=0$
Given the function sequence $f_n(x)=ae^{-anx}-be^{-bnx}$ where $0<a<b$ Prove that :

$\sum_{n=1}^{\infty}\int_{0}^{\infty}|f_n|dx=\infty$

$\sum_{n=1}^{\infty}\int_{0}^{\infty}f_ndx=0$

The second part I've managed to prove by just integrating and then it's a sum of zeroes($\sum_{n=1}^{\infty}\int_{0}^{\infty}f_ndx=\sum_{1}^{\infty}(-\frac{e^{anx}}{n}+\frac{e^{-bnx}}{n})|_{0}^{\infty}=\sum_{n=1}^{\infty}0=0$ but for the first part I've got no idea. any hint please

答:
Let $n\in \mathbb{N}$ fixed, now note that  $$a\exp(-ax)=b\exp(-bx) \iff x={{\ln\left(\frac{b}{a}\right)}\over {b-a}}$$Call this positive number $t$, then for $x\in [0,t/n]$ we have that $$b\exp(-bnx)\ge a\exp(-anx)$$
Therefore, e have \begin{eqnarray*}
\int_0^\infty |a\exp(-anx)-b\exp(-bnx)|dx &\ge& \int_0^{t/n} (b\exp(-bnx)-a\exp(-anx))dx\\
&=&{1\over n} \int_0^t(be^{-bt}-ae^{-at})dt
\end{eqnarray*} This integral is constant and positive, and the result follows


问:
Conjecture : A lower bound for the prime counting function .
Well it take my a little bit of time to find it but now I think that my conjecture is ready .
Conjecture :

Let $n\geq 100$ and then define the sum :

$$S(n)=\sum_{k=1}^{n}\frac{1}{\operatorname{argtanh}\left(\frac{k}{k+1}\right)}$$

Then we have :

$$\lceil{S(n)}\rceil\leq 2\pi(n)$$

Where we see the ceiling function and the prime counting function .Nicely the equality case is $n=100$ .
I have checked my conjecture with wolfram alpha up to $n=1000$ and extra values of $n$ up to $n=1000000$

Obviously I'm inspired by the prime number theorem and it's not a chance if I choose $\operatorname{argtanh}$ in my conjecture .

I have several questions:

Well first I would like to know if it's true for larger $n$ . Is my conjecture equivalent to other conjecture (stronger or weaker conjecture) . Can we improve the conjecture by adding (by example) a fixed exponent ? What my conjecture involves ? If it's not true for small $n$ can we take a larger $n$ to start with ?

Thanks!

答:
$$\tanh(x)= \frac{e^{2x}-1}{e^{2x}+1}, \qquad e^{2x}=\frac{1+\tanh(x)}{1-\tanh(x)}$$
$$x = \frac12 \log(\frac{1+\tanh(x)}{1-\tanh(x)}), \quad tanh^{-1}(y)=\frac12 \log(\frac{1+y}{1-y}), \quad tanh^{-1}(1-t)=\frac12 \log(\frac{2-t}{t})$$ Thus for $t$ small enough $$tanh^{-1}(1-t)\ge \frac12\log(1/t)+ \frac14$$

$$\sum_{k=1}^n \frac1{tanh^{-1}(1-\frac1{k+1})}\le O(1)+\sum_{k=1}^n \frac1{\frac12\log(k+1)+\frac14}$$ $$\le O(1)+ 2\sum_{k=1}^n \frac1{\log(k)} (1-\frac14\frac1{\log(k)})\le O(1)+2 Li(n)-\frac18 \frac{n}{\log^2 n}$$

On the other hand the PNT says that $$\pi(n) = Li(n)+O(\frac{n}{\log^3 n})$$
thus for $n$ large enough $$2\pi(n)\ge 1+\sum_{k=1}^n \frac1{tanh^{-1}(1-\frac1{k+1})}$$

It remains to make all the $O$ constants effective and to do a numerical check for $n\le N$.


问:
Is the series $\sum \frac{(n!)^2}{(2n)!} (2x + 1)^n$ convergent? absolutely?
Is the series $\sum \frac{(n!)^2}{(2n)!} (2x + 1)^n$ convergent? absolutely?

Can someone hint me which theorem should I use to analyse that? Thanks in advance

答:
As said in the comments, by the ratio test you have $$ \lim_{n \to \infty} \frac{(2n)! ((n+1)!)^2 } {(2(n+1))! (n!) ^2 } = \lim_{n \to \infty} \frac{(n+1)^2 } {(2n +1)(2n+2) } = \frac {1} {4}$$
Thus the series converges uniformly and so absolutely (and pointwise) when $ 2x+1 \in (-4,4)$, which is for $x \in (-\frac{5}{2}, \frac{3}{2}) $
The series does not converge when $2x+1 = +4$ or $-4$, because using Stirling formula $$ \lim_{n \to \infty} \frac{(n!)^2 (4^n)} {(2n)! } =  \lim_{n \to \infty} \frac {(4^n)(2\pi n) (\frac{n} {e})^{2n} } {( \sqrt{4 \pi n} ) )(\frac{2n} {e} )^{2n}} = +\infty $$


问:
Vieta's formulae - looking for a plain English explanation of the general-case formulae
I am working with a set of equations within which arise functions of the coefficients of a polynomial expressed in terms of the roots. The expressions get rather complicated as the order of the polynomial goes above 3, and so I want to write these coefficients using Vieta's formulae:
\begin{equation}
\sum_{1\leq i_1 < i_2 \cdots i_k \leq n}\left(\prod_{j=1}^k \alpha_{i_j}\right)
\end{equation}
However, I am struggling to disentangle this notation, in particular the expression beneath the summation and the notation subscripts for $\alpha$. I know what the result should be for a polynomial of order 1, 2, 3 4 and so on, but I would appreciate a plain-language description of how the notation describes this - and any suggestions as to how the expression could be made more straightforward to interpret?

答:
This summation merely expresses the sum of all possible products of $k$ roots.

The roots, $\alpha_1$, $\alpha_2$, $\ldots,$ $\alpha_n$ have indices $1$, $2$, $\ldots$, $n$, so call the set $\{1,2,\ldots,n\}$ the index set. The summation,
$$
\sum_{1\le i_1<i_2<\ldots<i_k\le n},
$$
which could also be written
$$
\sum_{i_1=1}^{n-k+1}\sum_{i_2=i_1+1}^{n-k+2}\ldots\sum_{i_k=i_{k-1}+1}^n
$$
effectively runs over all subsets of the index set that are composed of $k$ elements from that set. So the first subset will be $\{1,2,\ldots,k\}$, the second will be $\{1,2,\ldots,k-1,k+1\}$, and the last will be $\{n-k+1,n-k+2,\ldots,n\}$. In the particular case $n=5$, $k=3$, the complete list of subsets is
\begin{align}
&\{1,2,3\},\ \{1,2,4\},\ \{1,2,5\},\ \{1,3,4\},\ \{1,3,5\},\\
&\{1,4,5\},\ \{2,3,4\},\ \{2,3,5\},\ \{2,4,5\},\ \{3,4,5\}
\end{align}
In the summation, the fourth of these subsets, $\{1,3,4\}$, to take an example, corresponds to $i_1=1$, $i_2=3$, $i_3=4$ and the summand will be the product $\alpha_1\alpha_3\alpha_4$.

You could express the summation as
$$
\sum_{\text{all size-$k$ subsets $S$ of the index set}}(\text{product of roots indexed by elements of $S$}).
$$

If you multiply out
$$
(x-\alpha_1)(x-\alpha_2)(x-\alpha_3)(x-\alpha_4)(x-\alpha_5),
$$
you will get $2^5=32$ terms:
\begin{align}
&x x x x x+x x x x (-\alpha_5)+xxx(-\alpha_4)x+xxx(-\alpha_4)(-\alpha_5)\\
&+xx(-\alpha_3)xx+xx(-\alpha_3)x(-\alpha_5)+xx(-\alpha_3)(-\alpha_4)x+xx(-\alpha_3)(-\alpha_4)(-\alpha_5)\\
&+x(-\alpha_2)xxx+\ldots\\
&\vdots\\
&+\ldots+(-\alpha_1)(-\alpha_2)(-\alpha_3)(-\alpha_4)(-\alpha_5).
\end{align}
If you think about collecting the $x^2=x^{5-3}$ terms in this expansion, you'll realize that, apart from a minus-sign issue, the resulting coefficient of $x^2$ is precisely the sum of products indexed by the ten three-element subsets listed above. I hope this suffices to illustrate the general principle.


问:
Show that two $ (a\times b) \times (c \times d) = b - a $ expressions are equal
$$ (a\times b) \times (c \times d) = \times c, d>b - a $$

$a,b,c,d$ .... unknown vectors

Show that the expression is equal.

I used the formula to get on the left side: $c((a \times b) d)-d((a\times b)c)$

I don't know how to continue.

EDIT:
$$ ... means scalar product of a and b

答:
$$V=(a \times b)\times (c \times d)=e \times (c \times d)= (e.d)c-(e.c)d=[a,b,d]c-[a,b,c]d$$
$p\times(q\times r)=(p.r)q-(p.q)r$ and $p.(q\times r)=[p,q,r]$

Title: Show that two $ (a\times b) \times (c \times d) = <a \times c, d>b - <b\times c,d>a $ expressions are equal


答:
The Triple Product Formula reduces the RHS as follows
$$(d\cdot (a\times c))b -(d\cdot(b\times c))a = (a\cdot (c\times d))b - (b\cdot (c\times d))a$$
Next, we can let $v = c\times d$, and we now must show that
$$(a\times b) \times v = (a\cdot v)b - (b\cdot v)a$$
This should look easier to prove. For one, $(a\times b)$ is perpendicular to the plane formed by $a$ and $b$, and so $(a\times b)\times v$ will again lie on that plane. This means we can write $(a\times b) \times v$ as
$\alpha a + \beta b$.

To find $\alpha$ and $\beta$, first dot each side by $v$. The LHS is $0$ as $x\times v$ is perpendicular to $v$. Thus we know
$$0=\alpha a\cdot v + \beta b \cdot v$$
And so our formula holds up to scaling. Now just try it out for some combination of vectors and you can find the constant!

Body:
$$ (a\times b) \times (c \times d) = \times c, d>b - a $$

$a,b,c,d$ .... unknown vectors

Show that the expression is equal.

I used the formula to get on the left side: $c((a \times b) d)-d((a\times b)c)$

I don't know how to continue.

EDIT:
$$ ... means scalar product of a and b


问:
A good book for learning mathematical trickery
I've seen several question here on what book to read to learn writing and reading proofs. This question is not about that. I've been doing that for a while, and I'm quite comfortable with proofs. I am looking for resources (books, ideally) that can teach not the concept of proofs, but rather some of the specific mathematical tricks that are commonly employed in proofs: those that mostly include clever number manipulation, ad-hoc integration techniques, numerical methods and other thing you are likely never to learn in theory-oriented books. I come mainly from applied math and engineering, and when I look at proofs from Stochastic Processes, Digital Signal Processing, Non-Linear Systems and other applied subjects, I feel like I need to learn a new method to understand every proof I read. Is there any good literature on such mathematical tricks?

答: I enjoyed Mahajan's Street-Fighting Mathematics.  It has a strongly "applied" bent, and is freely available.


问:
Order of Convergence Proof with p>1 and M>0
So I am struggling with a problem on my homework, the problem statement is Assuming $x_{n}\rightarrow x^*$, show that any sequence that satisfies $$\lim\limits_{n \to \infty} \frac{|x_{n+1}-x^*|}{|x_{n}-x^*|^p}= M$$ with $p>1$ for some $M>0$ also satisfies $$\lim\limits_{n \to \infty} \frac{|x_{n+1}-x^*|}{|x_{n}-x^*|}= 0$$.

I looked at this problem for a bit and think it is as simple as seeing that $x_{n+1}$ goes to $x^*$. This would make the numerator of the limit $= 0$, but if I use this methodology, then I would have the bottom of the limit $=0$ as well which would cause an error in my methodology. I would appreciate any hints or nudges in the correct direction. This is my first experience with this so a lighter explanation would help.

答:
First of all note that $\lim_{n\to+\infty}|x_{n}-x^{*}|=0$ since the sequence is convergent. Then $\forall p>1$ the following holds:
$$\lim_{n\to+\infty}|x_{n}-x^{*}|^{p-1}=0$$
Now multiply and divide by $|x_n-x^{*}|^p$ the argument of the limit you want to show to hold to get
$$\lim_{n\to+\infty}\frac{|x_{n+1}-x^{*}|}{|x_{n}-x^{*}|^p}|x_{n}-x^{*}|^{p-1}=M\cdot 0=0$$