问:
Determining the solid angle of 3 overlapping spherical caps of same angular radius
Please consider figure 1 which displays 3 spherical caps slightly overlapping on the unit sphere $S2$ with a spherical triangle intersection area hightlighted in green. Let $\vec{U} = [u_x,u_y,u_z]$; $\vec{V}=[v_x,v_y,v_z]$ and $\vec{W}=[w_x,w_y,w_z]$ be the 3 unit vectors giving the positions of the spherical caps, which have a height $h$ and a spherical angular radius of $r_a$ where $h$ is inferred to be the distance from the origin $O$ to the center of the circle creating the cap. (Note: $\arccos(h) = r_a$)

We have three angles to consider, $\angle\alpha$ is the angle between $\vec{U},\vec{V}$; $\angle\beta$ between $\vec{U},\vec{W}$ and $\angle\gamma$ between $\vec{V},\vec{W}$, as shown in Fig 1.

Can we find a formula giving the solid angle of the highlighted green area in terms of $h$, $\vec{U}$, $\vec{V}$, $\vec{W}$, $\alpha$, $\beta$ and $\gamma$ and/or their dot products and square root radicals involving those? I am trying to find a formula which can be differentiated.

For the area of a digon, created by 2 overlapping spherical caps, I found a nice differentiable formula:

$ area(\vec{A},\vec{B},h) = 2\pi - 4\pi h - 2\: \text{acos} \left(\frac{\vec{A} \bullet \vec{B} - h^2}{1-h^2}\right) + 4 h\: \text{acos} \left(\frac{h \vec{A} \bullet \vec{B} - h}{\sqrt{1-h^2} \sqrt{1 - (\vec{A} \bullet \vec{B})^2 }}\right)$ where $\vec{A}\bullet\vec{B}$ denotes the usual dot product.

I am aware of a reference 3 spherical caps overlap which derives this solid angle on pages 2046-2048, but unfortunately even the simple two caps overlap formuli on pg 2046 seems to be in error.

It should be possible to find this formula, using the 4 terms $h$, $U$, $V$, and $W$ alone since the angles $\alpha$, $\beta$ and $\gamma$ can be found from the dot products of the vectors. It is necessary to find a diffentiable formula, since this is needed to create a Jacobian later on for optimization.

答: The reference (after correcting for 2 typographical errors in the equation for Aij) has the correct formula for determining the solid angle of 3 overlapping spherical caps. See https://royalsocietypublishing.org/doi/10.1098/rspa.1996.0108 however the equation on page 2046 and 2047 must read $\alpha_{ij} = \text{arctan}\left(\frac{1}{\tan{(\frac{\beta_{ij}}{2})\cdot\cos{(r)}}}\right)$ for the $\alpha_{ij}$ to be correct. I have alerted the publisher about the 3 typos. If a person uses a symbolic calculator to fully list out the final equation, it is 17,781 chars in size, for a GP-Pari expression, a very complex expression.


问:
Cayley-Hamilton theorem on square matrices
Can anyone help me by giving the proof of the Cayley-Hamilton theorem? It states that every square matrix $A$ satisfies its own characteristic equation:

$$p_{A}(A) = 0$$

I could prove it when $A$ has distinct eigenvalues, because then it will have $n$ linearly independent eigenvectors, but I couldn't prove the general case.

答:
For any diagonal matrix $D$, it is clear that $p_D(D)=0$. Therefore, for any $P \in  GL_n(\mathbb{C})$, $p_{PDP^{-1}}(PDP^{-1})=p_D(PDP^{-1})=Pp_D(D)P^{-1}=0$: Cayley theorem holds for diagonalizable matrices. Moreover:

Lemma: the set of diagonalizable matrices is dense in $M_n(\mathbb{C})$.

Proof: Let $M \in M_n(\mathbb{C})$. For some $P \in GL_n(\mathbb{C})$, $M=PT(\lambda_1,\dots,\lambda_n)P^{-1}$ where $T(\lambda_1,\dots,\lambda_n)$ is an upper triangular matrix with diagonal $(\lambda_1,\dots,\lambda_n)$. Let $\epsilon_{ij}$ such that $\lambda_i+\epsilon_{ij} \neq \lambda_k + \epsilon_{kj}$ for $i \neq k$ and $\epsilon_{ij} \underset{j \to + \infty}{\longrightarrow} 0$. Then $PT(\lambda_1+ \epsilon_{1j},\dots,\lambda_n+ \epsilon_{nj})P^{-1}$ is diagonalizable (all their eigenvalues are distincts) and converges to $M$. $\square$

Now $\det$ is continuous, so if $M \in M_n(\mathbb{C})$ and $(M_n)$ is a sequence of diagonalizable matrices converging to $M$, $p_M(M)=\lim\limits_{n \to + \infty} p_{M_n}(M_n)=0$.

ContentLicense: CC BY-SA 4.0

ContentLicense: CC BY-SA 3.0


问:
Show a trigonometric function is a contraction
I honestly feel like this function is not a contraction, but the exercise I am doing explicitly relies on it being a contraction and using the Banach fixed point theorem.

Show that $f(x)=\sin x+1$ is a contraction in a neighbourhood of $\pi$.

It being around $\pi$ actually makes it worse because I cannot put an upper bound on $\cos x$ less than $1$. Using the mean value theorem all I can show is that $f$ is a Lipschitz function of constant $1$. Any help would be greatly appreciated.

答:
The result is wrong.

If $f$ was a contraction in a neighborhood of $\pi$, it would exist $0\le k \lt 1$ such that

$$\vert f(x) -f(\pi) \vert = \vert \sin x \vert \le k \vert x -\pi\vert $$ for $x$ in that neighborhood. A contradiction as

$$\lim\limits_{x \to \pi} \frac{\sin x}{x-\pi}=-1.$$


问:
Presentation of Rubik's Cube group
The Rubik's Cube group is the group of permutations of the 20 cubes at the edges and vertices of a Rubik's group (taking into account their specific rotation) which are attainable by succesive rotations of its sides (the cubes in the middle of the sides are considered as fixed). Wikipedia says that this group is given as $(\mathbb Z_3^7 \times \mathbb Z_2^{11}) \rtimes \,((A_8 \times A_{12}) \rtimes \mathbb Z_2)$.

Is there also a nice presentation of this group in terms of generators and relations (i. e. at most as many relations as one can write down on a single sheet of paper)? In particular, are 6 generators (e. g.  the six standard rotations) the minimum number needed for a presentation or is it even possible to obtain a rotation around one side as a composition of rotations around the other five faces?

答:
This belongs as a comment, but due to being new here I can't comment yet. The link Dominik mentioned (129.81.170.14/~erowland/courses/2009-2/projects/Cordell.pdf) is indeed an interesting paper, but I noticed several things wrong, and would advise skepticism. For example, I don't know if the two elements mentioned are indeed generators, but I do know that the orders he claims they have (253 and 60) are wrong; they are in fact 77 and 12. In another section, he claims that "Any permutation of two adjacent sides (for example, F R) has order 105." This is a false generalization. F R does in fact have order 105, but F R' has order 63.

Nevertheless I was quite impressed with the paper; I just felt like mentioning that not everything there is accurate. I have very little understanding of the group theory of the Rubik's Cube, and it's possible there are many more errors I didn't recognize.

Title: Presentation of Rubik's Cube group


问:
Show that a group of that order exists having this property.

Let $G$ be a finite group having an element of order $k$, for each $1 \leq k \leq 10$.
What is the smallest possible order of $G$? Show that a group of that order exists
having this property.

My attempt:

Since for each $a\in G$ order of $a$ divides order of $G$,  $|G|=[1,2,3,4,5,6,7,8,9,10]=2520$. This is the smallets possible order of $G$ (it is right?). Now

Show that a group of that order exists
having this property.

I think one way to prove this would be by considering cyclic subgroups, but I'm not sure. my idea is this: If $|G|=2520$ and $a\in G$ has order $k$ $(1\leq k\leq 10)$, then $|\left \langle a \right \rangle| \in \{1,2,3,4,5,6,7,8,9,10\}$, so the group of that order exists having this property.

答:
As $2520=2^3\,3^25\,7$, you simply have to consider the product of cyclic groups
$$\mathbf Z/8\mathbf Z\times\mathbf Z/9\mathbf Z\times\mathbf Z/5\mathbf Z\times\mathbf Z/7\mathbf Z.$$


问:
Superfluous submodule of a quotient ring
Let $I$ be a proper nontrivial ideal of a ring $R$.

Show that $I/I^{2}$ is a superfluous submodule of $R/I^{2}$.

What I've tried:

If $J/I^{2}$ is such that $J/I^{2} + I/I^{2} = R/I^{2}$, then $\forall\ r \in R$, $\exists\ i \in I$, $\exists\ j \in J$ such that $r + I^{2} = i + j + I^{2}$, that is, $r - (i + j) \in I^{2}$, but I cannot prove that $r - j \in I^{2}$ because $i$ doesn't have to be in $I^2$.

Do you have any clue?

答:
Every nilpotent ideal is going to be superfluous because it’s contained in the Jacobson radical, which is always a superfluous submodule of the ring (on the right and on the left.)

Another way to see it is that if $A^2=\{0\}$  and $A+B=R$, then $R=R^2=(A+B)^2\subseteq B$, so $B=R$.

Either of these is enough to see $I/I^2$ is superfluous in $R/I^2$.


问:
Another way to prove that $81^{3^{n}}+4^{2n+1}$ is not prime for any positive integer $n$
The question is to prove that $81^{3^{n}}+4^{2n+1}$ is not prime for any positive integer $n$.

The easy way to solve this is by taking the last digit of each number:

$81^{3^{n}}$ ends in $1$ because $81$ also ends in $1$.

$4^{2n+1}$ ends in $4$ as $2n + 1$ is odd.

So the sum will end in $4+1=5$ and because $81^{3^{n}}+4^{2n+1}>5$, it is $5|81^{3^{n}}+4^{2n+1}$, thus the number is not prime.

Are/is there any other proof(s) which do not reuse the same idea?

Body:
\begin{align}
81^{3n} + 4^{2n+1} &= (3^4)^{3n} + (2^2)^{2n+1} \\
&= (3^{6n})^2 + (2^{2n+1})^2 \\
&= (3^{6n})^2 + (2^{2n+1})^2 + 2 \times 3^{6n} 2^{2n+1}  - 2 \times 3^{6n} 2^{2n+1} \\
&= (3^{6n} + 2^{2n+1})^2 - 3^{6n} 2^{2n+2} \\
&= (3^{6n} + 2^{2n+1})^2 - (3^{3n} 2^{n+1})^2 \\
&= (3^{6n} + 2^{2n+1} - 3^{3n} 2^{n+1}) (3^{6n} + 2^{2n+1} + 3^{3n} 2^{n+1})
\end{align}

As you see $81^{3n} + 4^{2n+1}$ can be written as a product of two integers and none of them are $1$ for positive integers1, so it is not prime.

 1. Because both of them are increasing functions and above $1$ for $n=1$.


问:
$n_1n_2...n_{11}-1$ is a product of two consecutive natural numbers.

Prove that there exist distinct, pairwise coprime natural numbers $n_1,n_2,...,n_{11}$ such that $n_1n_2...n_{11}-1$ is a product of two consecutive natural numbers.

Let $N = n_1n_2...n_{11}$ My approach was to firstly notice that none of the $n_1n_2...n_{11}$ is an even integer, otherwise $n_1n_2...n_{11}-1$ would be odd. Then I thought about using $CRT$ \begin{align} N&=1\mod x, \\ N&=1 \mod x+1,\end{align} $x,x+1$ to conssecutive natural numbers. By $CRT$ we know that this system of congruences has a solution. Now I want to prove that the sol. is $0$ $\mod x(x+1)$ and deduce that $n_1n_2...n_{11}-1 = x(x+1)$ is true. How to proceed? Any help appreciated.

答:
One of (not too large) examples:
$$
n= 303750157,
$$
$$
n(n+1) = 3 \times 7 \times 13 \times 19 \times 31 \times 37 \times 43 \times 61 \times 67 \times 79 \times 1117 -1,
$$
where all the factors in RHS are prime numbers.

Another examples:

$n = 540178291$,

$n = 876805855$,

$n = 956495446$,

$\ldots$


问:
Prove that: $ \sum_{i=0}^{\infty} \frac{\tan \frac{\theta}{2^i}}{2^i}= \frac{1}{\theta} - 2 \cot 2 \theta$

My attempt:

Consider the following series:

$$ S = \sum_{i=0}^n \ln( \sec \frac{x}{2^i} )$$

Notice that $ \lim_{n \to \infty} \frac{dS}{dx}$ is the required sum.

Simplfying S,

$$ S = - \ln \left(  \cos x \cdot \cos \frac{x}{2} ... \cos \frac{x}{2^n} \right)$$

or,

$$ S = - \ln \left(  \frac{ \sin(2x)}{2^{n+1} \sin (\frac{x}{2^n})} \right)= \ln(2^{n+1}) + \ln( \sin \frac{x}{2^n}) - \ln( \sin(2x) )$$

Now,

$$ \frac{dS}{dx}  = \frac{1}{2^n} \cot \frac{x}{2^n} - 2 \cot(2x)$$

The problem I'm having is proving that

$$ \lim_{ n \to \infty} \frac{1}{2^n} \cot \frac{x}{2^n} = \frac{1}{x}$$

答:
Thanks to @gary,

$ \frac{1}{2^n} = u$,

$$ \lim_{u \to 0 } u \cot ( ux)=  \lim_{u \to 0} \cos(ux) \frac{ux}{x\sin (ux)} = \frac{1}{x}$$


问:
Can a function be uniformly continuous on an open interval?
I am learning analysis and all the uniformly continuous functions I have seen are over a closed interval. So, can a uniformly continuous function be defined on an open interval?</p>

答:
This result may help you:

Let ${F}:(a, b)\rightarrow \mathbb{R}$ that is continuous on the bounded open interval $(a, b)$ then the two limits given by

$F(a +) = \lim_{x\to a^{+}} F(x)$,  $F(b -) = \lim_{x\to b^{-}} F(x)$ exists iff $F$ is uniformly continuous on $(a, b)$.

This result has been given in the book "The calculus integral by Brian S. Thomson".


答:
As has already been noted, the problem is much easier modulo $5$:

$81^{3^{n}}+4^{2n+1}\equiv1^{3n}+(-1)^{2n+1} \equiv 0\pmod 5$.

Since $81^{3^{n}}+4^{2n+1} > 5$ for all $n$ we are done.

Title: Another way to prove that $81^{3^{n}}+4^{2n+1}$ is not prime for any positive integer $n$

Body:
The question is to prove that $81^{3^{n}}+4^{2n+1}$ is not prime for any positive integer $n$.

The easy way to solve this is by taking the last digit of each number:

$81^{3^{n}}$ ends in $1$ because $81$ also ends in $1$.

$4^{2n+1}$ ends in $4$ as $2n + 1$ is odd.

So the sum will end in $4+1=5$ and because $81^{3^{n}}+4^{2n+1}>5$, it is $5|81^{3^{n}}+4^{2n+1}$, thus the number is not prime.

Are/is there any other proof(s) which do not reuse the same idea?


问:
Any finite subgroup of $GL_n(\Bbb Z)$ is isomorphic to a subgroup of $GL_n(\Bbb F_p)$ for $p \ge 3$.

Any finite subgroup of $GL_n(\Bbb Z)$ is isomorphic to a subgroup of $GL_n(\Bbb F_p)$ for $p \ge 3$.

I have two questions related to this problem:

I found this proof but it doesn't give any reference about an article or a book where this problem is from nor where it is useful.
Does someone know any reference (textbook, ...) that could help me find more informations about this result?

Does someone know another proof that uses basic group theory?

答:
The statement is much more precise: if $G$ is a finite subgroup of $GL_n(\mathbb{Z})$ then the quotient map to $GL_n(\mathbb{F}_p)$ for an odd prime $p$ is an embedding. This follows from the claim that the kernel of this quotient map, namely the congruence subgroup consisting of invertible matrices congruent to $I \bmod p$, is torsion-free (and so does not contain any elements of finite order, and so intersects trivially with $G$), which is essentially what the post you linked proves.

I think the proof can be cleaned up by working $p$-adically: more specifically, passing to $GL_n(\mathbb{Z}_p)$ I think we can show that for $p \ge 3$, $1$ is the unique root of unity (in the ring of algebraic integers over $\mathbb{Z}_p$) congruent to $1 \bmod p$, and so if an element of the congruence subgroup has finite order then its eigenvalues are roots of unity which must then all be equal to $1$.

As for a reference, you can consult Serre's Bounds for the order of the finite subgroups of $G(k)$ which contains information on this result (although he calls the torsion-freeness above an "easy exercise") and many generalizations of it, with references. It's not written for students but you may find it at least a bit helpful.

Edit: Here are some details on a somewhat different argument from the blog post, inspired by $p$-adic ideas but not explicitly working $p$-adically. Let $\Gamma(d) = \text{ker}(GL_n(\mathbb{Z}) \to GL_n(\mathbb{Z}/d))$ denote the congruence subgroup

$$\{ M \in GL_n(\mathbb{Z}) : M \equiv I \bmod d \}.$$

We want to show that for $p$ an odd prime, $\Gamma(p)$ is torsion-free. It suffices to show that there is no $q$-torsion where $q$ is prime. So suppose $M \in \Gamma(p)$ satisfies $M^q = I$. Then the eigenvalues of $M$ must be $q^{th}$ roots of unity, and in fact the characteristic polynomial of $M$ must be a product of copies of the cyclotomic polynomials $\Phi_1(x) = x - 1$ and

$$\Phi_q(x) = \frac{x^q - 1}{x - 1} = x^{q-1} + x^{q-2} + \dots + 1.$$

On the other hand, since $M \equiv I \bmod p$, the characteristic polynomial $\varphi(x)$ of $M$ must be congruent to $(x - 1)^n$ mod $p$. So a necessary condition is that $\Phi_q(x) \equiv (x - 1)^{q-1} \bmod p$, or

$$x^q - 1 \equiv (x - 1)^q \bmod p.$$

Taking the coefficient of $x^{q-1}$ on both sides gives $q \equiv 0 \bmod p$ so $q = p$; in other words, the only possible torsion is $p$-torsion so now it suffices to rule out this possibility. (Note that so far we have not used the hypothesis that $p$ is odd, and that $\Gamma(2)$ does have $2$-torsion, since it contains $-I$.)

Write $M = I + p^k N$ where $k \ge 1$ and $N \not\equiv 0 \bmod p$. Then $M^p - I$ is equal to

$$\sum_{i=1}^p {p \choose i} p^{ki} N^i.$$

Taking both sides $\bmod p^{k+2}$ gives $p^{k+1} N + p^{kp} N^p \bmod p^{k+2}$ (all the other terms are divisible at least by $p^{2k+1}$). If either $k \ge 2$ or $p \ge 3$ then $kp \ge k+2$ so this simplifies to $p^{k+1} N \bmod p^{k+2}$ which is nonzero by hypothesis. So $M^p \neq I$ and $\Gamma(p)$ has no $p$-torsion for odd $p$ as desired.

If $k = 1$ and $p = 2$ we get $p^2 N + p^2 N^2 \bmod p^3$ which could be zero, and in fact this happens if we take $N = -I$. What we learn from this argument is that $\Gamma(2)$ can have $2$-torsion but $\Gamma(4)$ cannot (this corresponds to $k \ge 2$), and so cannot have any torsion. This kind of exceptional behavior at $p = 2$ is common $p$-adically which is a useful hint here.


问:
How to prove being conservative vector field of $R^3$
How to prove that $\mathbb{R}^3$ a conservative vector field $\mathbb{F}$ = (P, Q, R) satisfies
these 3 conditions:

$P_y = Q_x$

$P_z = R_x$

$Q_z = R_y$

答:
If a vector field is conservative, then it is the gradient of a function $f$. That is
$$
F=\nabla f
$$
for some function $f$. Since $\mathrm{curl} \nabla g=0$ regardless of the function $g$, you get that if $F$ is conservative then
$$\mathrm{curl} F=\mathrm{curl} \nabla f=0\, .
$$
This gives you the necessary conditions that $F$ has to satisfy and that you were looking for.


问:
If $a_n=100a_{n−1}+134$ , find least value of $n$ for which $a_n$ is divisible by $99$

Let $a_{1}=24$ and form the sequence $a_{n}, n \geq 2$ by $a_{n}=100 a_{n-1}+134 .$ The first few terms are
$$
24,2534,253534,25353534, \ldots
$$
What is the least value of $\mathrm{n}$ for which $\mathrm{a}_{\mathrm{n}}$ is divisible by $99 ?$

So, In this post,the present answer there get to the result that $a_n \equiv a_1 + (n - 1)35 \equiv 35n - 11 \equiv 0 \pmod{99} \tag{1}\label{eq1A}$

But after that what i did is to multiply both sides by $9$ in the congruence

$35n - 11 \equiv 0 \pmod{99}$

$315n \equiv 0 \pmod{99}$

$18n \equiv 0 \pmod{99}$

dividing by $18$ ,we get
$n \equiv 0 \pmod{11}$ but I did not get the other part of the congruence which was arrived in that answer by this process ?

thankyou

答:
It's an easy one-liner using a fractional form of mDL = mod Distributive Law, viz.

Notice  $\,\ \dfrac{\color{#c00}{11}}{35}\bmod \color{#c00}{11}(9)\,=\, \color{#c00}{11}(\color{#0a0}8)\,$ by $\color{#0a0}{\bmod 9\!:\ \dfrac{1}{35}\equiv \dfrac{1}{-1}\equiv 8},\ $ via

Theorem $\ \ \dfrac{\color{#c00}ab}d\bmod \color{#c00}ac\, =\, \color{#c00}a\left(\color{#0a0}{\dfrac{b}d\bmod c}\right)\ \ $ if $\ \ (d,ac) = 1$

Proof $\, $ Bezout $\Rightarrow$ exists $\, d' \equiv d^{-1}\pmod{\!ac}.\,$ Factoring out $\,\color{#c00}a\,$ by mDL

$$\color{#c00}abd'\bmod \color{#c00}ac\, =\ \color{#c00}a(bd'\bmod c)\qquad\qquad\qquad$$

and $\,dd' \equiv 1\pmod{\!ac}\Rightarrow dd' \equiv 1\pmod{\!c},\,$ so $\,d'\bmod c = d^{-1}\bmod c$


问:
Proof: F is isometric if and only if its matrix is orthogonal/unitary
I'd like to show that $F \in End(V)$ isometric $<=> M_{\beta \beta} (F)$ orthogonal/unitary

But it seems as if I still have some trouble doing that ;/

"=>"
$<v_i, v_j> = <f(v_i),f(v_j)> = <K_{\beta}^{-1} (f(v_i)), K_{\beta}^{-1} (f(v_j))> $.
So, the rows of $M_{\beta \beta}$ form an orthogonal basis of $K^n$.

"<="
The rows of the matrix form an orthogonal basis, so we have $\delta_{ij} = <f(v_i), f(v_j)>$ = ??

答:
I only prove the converse part. first consider the lemma below:

Lemma.

if $T$ is a linear mapping over the field $\mathbb{C}$, and we know that:
\begin{align*}
\langle v,Tv \rangle = 0
\end{align*}
then $T = 0$

proof

simply write:

\begin{align*}
\langle Tu,w \rangle = \frac{\langle T(u+w), u+w \rangle - \langle T(u - w), u - w \rangle}{4} + \frac{\langle T(u + iw), u + iw \rangle - \langle T(u - iw) , u-iw \rangle}{4}i
\end{align*}
note that each term on the right hand side is in the form $\langle Tv,v \rangle$ and is therefore $0$. so we have that $\langle Tu,v \rangle = 0$ for every $u,v$.this yields:
\begin{align*}
\langle Tu,Tu \rangle = ||Tu||^2 = 0 \Rightarrow Tu = 0 \hspace{0.5cm}(\forall u)
\end{align*}

now! for the main problem. Assume that $U$ is a square isometry matrix:
\begin{align*}
 &\langle v,v \rangle = ||v||^2 = ||Uv||^2 = \langle Uv,Uv\rangle = \langle U^*Uv,v \rangle \\
 &\Rightarrow \langle v,v \rangle = \langle U^*Uv,v \rangle \Rightarrow \langle (U^*U-I)v,v \rangle = 0 \hspace{0.5cm}(\forall v)
\end{align*}

then using the proven lemma, you can deduce the result.


问:
How can i show this inequality?
Let $n>1$ and $a_1,...,a_n \in \mathbb{R}^+$ be such that $\sum a_i=1$. For evey $i$, define $b_i=\sum_{j=1,j\neq i}a_j$. Show that

$\sum_{k=1}^n \dfrac{a_k}{1+b_k}\ge \dfrac{n}{2n-1}$

Thanks a lot for any suggestion.

答:
By Cauchy-Schwarz:

$\sum_{k=1}^{n}{\dfrac{a_k}{1+b_k}}\sum_{k=1}^{n}((1+b_k)a_k)=\sum_{k=1}^{n}{\dfrac{a_k}{1+b_k}}\sum_{k=1}^{n}((2-a_k)a_k)\geq (\sum_{k=1}^{n}a_i)^2=1 \ \ \ \ \ \ \ \ \ \ \ \ (1)$

By Cauchy-Schwarz:

$\sum (a_i)^2\sum 1\geq \sum (a_i)=1 \Rightarrow -\sum (a_i)^2\leq -\dfrac{1}{n} \ \ \ \ \ \ \ \ \ (2)$

Dividing both sides in the last inequality of $(1)$ by $\sum_{k=1}^{n}((2-a_k)a_k)$ and applying $(2)$ we are done.

Title: How can i show this inequality?


问:
bijective $C^{1}$-image
Let $F: \mathbb{R^{3}}\to \mathbb{ R}$ a $C^{1}$-image and assume that $(dF) (x, y, z)\neq 0$ as soon as $F (x, y, z) = 0$.

Define $O= \{(x, y, z) ∈ \mathbb{R^{3}}| F (x, y, z) = 0\}$.

Prove that every point of $O$ has an environment that looks like a piece of the xy plane in $\mathbb{R^{3}}$.

I think this means that I have to prove that for every point $(a, b, c) \in O$ there open parts $U, V \subset \mathbb{R^{3}}$ exist and a bijective $C^{1}$- image $\phi: U → V$ with $C^{1}$-inverse such that $(0, 0, 0) \in U, (a, b, c) \in V$ and $O \cap V = \phi (U \cap (\mathbb{R^{2}} \times {0})$

My attempt:

Let's take $(a,b,c) \in O$ random. Let's assume that (1x3)- matric $(dF)(x,y,z)$ is of rang $1$.
Because of the implicit function theorem there exist

an open part $V\subset \mathbb{R^{3}}$ so $(a,b,c)\in V$

an open part $U \subset \mathbb{R^{2}}$ with $(0,0) \in U$

a $C^{1}$-image $\phi:U\to \mathbb{R}$ with $\phi((0,0))=(a,b,c)$

and $ V \cap O=\phi(U)$ and the image of $(d\phi)(y)$ equals de kern of $(dF)(\phi(y))$ for all $y\in U$.

Because the condition that $(dF)(x,y,z) \neq 0$, $U$ needs to be a part of $\mathbb{R^{3}}$. This means that $(0,0,0) \in U$ and $O \cap V=\phi(U\cap (\mathbb{R^{2} \times {0} ))}$

I'm not that sure that the conclusions I made about the condition $(dF)(x,y,z) \neq 0$ are correct. Or maybe I started the question wrong. Can someone check my answer and see what I do wrong or right.

答: I have written an answer here, but am unsure if it is correct. Hence I am waiting there for verification.</p>

Title: bijective $C^{1}$-image


问:
Finding a closed form expression for the recurrence relation $a(n,k+1)=(k+1)^n+\sum_{m=1}^k {k+1\choose m}a(n,m)$
I originally stumbled upon this beast whilst trying to find the number of surjective functions from $[n]=\{ 1,...,n\}$ to $[k]$.

Here is my idea, denote this number $a(n,k)$ hence $a(n,k) =k^n- \#\{{f:[n]\to [k]\; | \; f \; isn't \; surjective}\}$. For every $1\leq m\leq k-1$ we have ${m\choose k}$ ways of choosing a set of $m$ elements from $[k]$ and $a(n,m)$ functions in $\{{f:[n]\to [k]\; | \; f \; isn't \; surjective}\}$ with $|im(f)|=m$ and so:

$a(n,k+1)=(k+1)^n-\sum_{m=1}^k {k+1\choose m}a(n,m)$

From here I tried to use induction to get $a(n,k+1)$ to the form of a polynomial in $a(n,1)$ which is clearly $1$, but had no luck as to finding the polynomial itself.

Solutions to the original problem or to the recurrence relation would by much appreciated.

EDIT: from the inclusion-exclusion prinicple I got the follwing expression for $a(n,k)$ which is:

$a(n,k) = \sum_{m=0}^k (-1)^m{k\choose m}(k-m)^n$

and plugging it into $a(n,k+1)=(k+1)^n-\sum_{m=1}^k {k+1\choose m}a(n,m)$ yeilds:

$a(n,k+1)=(k+1)^n-\sum_{m=1}^k{\sum_{i=0}^m {k+1 \choose k+1-m,i,m-i}(m-i)^n}$

which, to be honest, looks terrible.

My original question still stands, what is the closed-form expression (if there even is one) for this monster double sum?

答:
The Stirling number $n\brace k$ of the second kind is the number of ways to partition $[n]$ into $k$ non-empty sets. The fibres of a surjective function from $[n]$ to $[k]$ are such a partition of $[n]$, and each such partition is generated by $k!$ different surjections, one for each way of assigning the parts to the elements of $[k]$. Thus, there are $k!{n\brace k}$ surjections from $[n]$ to $[k]$.

These numbers satisfy a nice recurrence:

$${{n+1}\brace k}=k{n\brace k}+{n\brace{k-1}}$$

for $k>0$, with initial values ${0\brace 0}=1$ and ${n\brace 0}={0\brace n}=0$ for $n>0$.

They also have an explicit expression as a summation:

$${n\brace k}=\frac1{k!}\sum_{i=0}^k(-1)^{k-i}\binom{k}ii^n\,,$$

which can be derived by an inclusion-exclusion solution to your problem.

For $i\in[k]$ let $F_i$ be the set of functions $f:[n]\to [k]$ such that $i\notin\operatorname{ran}f$; clearly $|F_i|=(k-1)^n$. Thus,

$$\begin{align*}
\left|\bigcup_{i\in[k]}F_i\right|&=\sum_{\varnothing\ne I\subseteq[k]}(-1)^{|I|+1}\left|\bigcap_{i\in I}F_i\right|\\
&=\sum_{i=1}^k(-1)^{i+1}\binom{k}i(k-i)^n\,,
\end{align*}$$

and the number of surjections is therefore

$$\begin{align*}
k!{n\brace k}&=k^n-\sum_{i=1}^k(-1)^{i+1}\binom{k}i(k-i)^n\\
&=\sum_{i=0}^k(-1)^i\binom{k}i(k-i)^n\\
&=\sum_{i=0}^k(-1)^i\binom{k}{k-i}(k-i)^n\\
&=\sum_{i=0}^k(-1)^{k-i}\binom{k}ii^n\,.
\end{align*}$$


问:
Probability: Shoes in a row

20 shoes, from 10 pairs of shoes, are lined randomly. What is the
probability that there is a set of 10 consecutive shoes with 5 shoes
left and 5 right shoes?

I thought that would be a good idea imagine 10 shoes as one unique object, as follow:
Instead enumerate 20 objects, let's separate 10 shoes and think them as another object (call it X), so that there is 11 in total.

There is 11! ways to permute each objects. The problem is to know how to count X. I mean, the first way i thought was: $|X| = \frac{10!}{5!}\frac{10!}{5!}$, so that the answer would be: $\frac{\frac{10!}{5!}\frac{10!}{5!}*11!}{20!}$. But so i realized that |X| that i counted was wrong, what actually i counted for |X| was counting first the 5 possible right shoes, and after this, the 5 possible let shoes.

So i am stuck in how to count |X| right. I thought that we could first choose $C_{10,5} = 10!/(5!5!)$ right shoes, $C_{10,5}$ left shoes. Now the probability for one aleatory choice of X is: $\frac{\frac{10!}{5!}*11!}{20!}$, but X have more than one option, so the probability will be the sum of all possibility for X,
namely $C_{10,5}*C_{10,5}*\frac{{10!}*11!}{20!}$. But this is greater than 1 (...), i have no idea what to do now.

答:
The probability is $1$.

Label the shoes from $1$ to $20$ according to their position. For $n \in \lbrace 1, \dotsc, 20 \rbrace$, set $$\varepsilon_{n} = \begin{cases} 0 & \text{if the } n \text{th shoe is left}\\ 1 & \text{if the } n \text{th shoe is right} \end{cases} \, \text{.}$$ For $N \in \lbrace 1, \dotsc, 11 \rbrace$, define $$\ell_{N} = \sum_{n = N}^{N +9} \varepsilon_{n} \in \lbrace 0, \dotsc, 10 \rbrace$$ to be the number of left shoes whose label is in $\lbrace N, \dotsc, N +9 \rbrace$.

I pretend that there exists $N_{0} \in \lbrace 1, \dotsc, 11 \rbrace$ such that $\ell_{N_{0}} = 5$. We have $$\ell_{1} +\ell_{11} = \sum_{n = 1}^{20} \varepsilon_{n} = 10 \, \text{,}$$ and hence we have $$\ell_{1} \leq 5 \leq \ell_{11} \quad \text{or} \quad \ell_{11} \leq 5 \leq \ell_{1} \, \text{.}$$ Now, note that, for all $N \in \lbrace 1, \dotsc, 10 \rbrace$, we have $$\ell_{N +1} = \ell_{N} +\varepsilon_{N +10} -\varepsilon_{N} \in \left\lbrace \ell_{N} -1, \ell_{N}, \ell_{N} +1 \right\rbrace \, \text{.}$$ Therefore, by some kind of discrete intermediate value theorem, there exists $N_{0} \in \lbrace 1, \dotsc, 11 \rbrace$ such that $\ell_{N_{0}} = 5$. More precisely, we can take $$N_{0} = \begin{cases} \max\left\lbrace N \in \lbrace 1, \dotsc, 11 \rbrace : \ell_{N} \leq 5 \right\rbrace & \text{if } \ell_{1} \leq 5 \leq \ell_{11}\\ \min\left\lbrace N \in \lbrace 1, \dotsc, 11 \rbrace : \ell_{N} \leq 5 \right\rbrace & \text{if } \ell_{11} \leq 5 \leq \ell_{1} \end{cases} \, \text{.}$$

Title: Probability: Shoes in a row


问:
Expectation of function of stochast
I've got a general question regarding a certain sticking point I often encounter. When tackling questions where for example an UMVUE (uniformly minimum-variance unbiased estimator) has to found I get stuck at the part where an expectation has to be determined. Maybe I'm overlooking certain theorems.

Example 1.  Consider a random sample of size $n$ with $X_i\tilde{}Poi(\mu)$.

I am able to find the UMVUE of $\mu$ and the MLE (Maximum Likelihood Estimator) of $\theta=e^{-\mu}$. But then for example follows the question: is this MLE unbiased for $\theta$?

If I want to determine the expectation, should I just write out the sum as follows from the definition of the expected value of a discrete distributed variable? This seems to lead to quite an ugly expression...

(For this particular problem I found that Jensen's Inequality could come in handy to show biasedness (or actually non-unbiasedness))

But if I could summarize the problem I've got: how to rewrite expected values of functions of stochast in general?
Which theorems or properties should I use... for example I do use $E(aX)=aE(X)$ for $a$ a constant and $var(X)=E(X^2)-E(X)^2$ to rewrite certain expectations to a combination of simpler/known ones.

Example 2. Take a random sample $n$ of a distribution with pdf $f(x;\theta)=\theta x^{\theta -1}$ if $0<x<1$ and else $0$ and with $\theta > 0$.

Find the UMVUE of $\theta$

Here I get stuck again at the expectation. I've got a sufficient and complete statistic in
$\sum ln(X_i)$.

The expected value, i'd figure, is equal to $nE(ln(X_i))=-n/ \theta$  $\,\,\,\,$(where $[-ln(X)]=\frac{1}{\theta}$ was given as a hint).

But what if I want to determine the expected value of $\frac{1}{\sum_{i}ln(X_i)}$ ?

In general: how to determine the expectations of functions of stochasts?
For example $E[ln(X)]$or $E[e^{\bar{X}}]$...?

答:
If $X$ has density $f_X$, then for every function $u$, $E[u(X)]=\displaystyle\int u(x)f_X(x)\mathrm dx$. For example, in your last case,
$$
E_\theta\left[\frac1{\sum\limits_i\ln X_i}\right]=\iint\cdots\int\frac1{\sum\limits_i\ln x_i}\prod_if(x_i;\theta)\prod_i\mathrm dx_i.
$$
If $X$ is discrete and $p_x=P[X=x]$ for every $x$, this should read
$E[u(X)]=\displaystyle\sum\limits_xu(x)p_x$.</p>

Title: Expectation of function of stochast


问:
Linear transformation of rotation
I was doing exercises for my algebra class and i stucked. I have to check wheter the L transformation is linear. With transformation like for eg. $\Bbb R^3 → \Bbb R^3, L(x,y,z) =(3x+2y-z,z+x,5y-7z)$ i don't have any troubles. Problems start for example this one: : $L: \Bbb R^2 → \Bbb R^2$, where $L$ is a rotation by angle $\frac{\pi}{4}$ relative to $(0,0)$. How to check if given tranformation is linear?

答:
HINT

The rotation by angle $\theta$ can be written as (draw it if you need):
\begin{align*}
L_{\theta}(x,y) = (\cos(\theta)x - \sin(\theta)y,\sin(\theta)x + \cos(\theta)y)
\end{align*}

Can you take it from here?

Title: Linear transformation of rotation


问:
Counting number of way to choose ordered distinct quadruples $(A,B,C,D)\subseteq\{1,2,...,n\}(n\geq 2)$

Counting number of way to choose ordered distinct quadruples
$(A,B,C,D)\subseteq\{1,2,...,n\}(n\geq 2)$ with additional condition $A\cap B\subseteq C\cup D$.

In order to solve this problem,i set length of $|A\cap B|=k$ then distribute those $k$ elements in $3$ region of Venn diagram of $B\cup C$ and i can do this in $3^k$ different way and we have $\binom{n}{k}$ choice for select those $k$ elements.
At the last step for remaining $n-k$ we have $7^{n-k}$ choice,because for each $n-k$ elements we can push it in $A $ or $B$ (not in intersect region) and $4$ why for push it in $C$ or $D$. Summing over all we have:

$\sum_{k=0}^{n}\binom{n}{k}3^k 7^{n-k}=10^n$

But i think my solution is not true!.Any one can help me?

答:
For any $x \in \{1,2,\ldots,n\}$ there are $16 = 2 \cdot 2 \cdot 2 \cdot 2$ cases:

$2$ cases for $A$: $x \in A$ or $x \notin A$;

$2$ cases for $B$: $x \in B$ or $x \notin B$;

$2$ cases for $C$: $x \in C$ or $x \notin C$;

$2$ cases for $D$: $x \in D$ or $x \notin D$;

However, due to the additional condition $A \cap B \subseteq C \cup D$, we cannot have $x \in A \land x \in B \land x \notin C \land x \notin D$. Therefore the $16$ cases are reduced to $15$ and considering all elements we have $15^n$ ordered distinct quadruples (where any two sets can be equal).

The result agrees with your reasoning once that you correct it replacing $7 = 3 + 4$ ($3$ for $A, B$ and $4$ for $C, D$) with $12 = 3 \cdot 4$.


问:
Does a truly random sequence in the range of x..y average to the average of x and y?
I recently started wondering if the average of a truly random sequence between numbers x and y has to be the average of x and y itself.

This little JavaScript function seems to prove it (uses Math.random() which should be a pretty good source of random numbers). The bigger precision is, the closer is the result to 0.5:

```<code>function averageOfRandom(precision) { // returns the average of `precision` random numbers between 0 and 1
  const sequence = [];
  for(let i = 0; i < precision; i++) { // make the sequence
    sequence[i] = Math.random();
  }

  let totalSum = sequence.reduce((a, b) => a + b);

  return totalSum / sequence.length; // average
}
</code>```

Then, is a random sequence the only sequence that can have such trait without knowing its own length? Maybe a sequence simply alternating x and y is also good?

答:
The notion "truly random sequence" is not a well-known rigorous concept.

However, an independent sequence of random variables, each uniformly distributed on $[x,y]$, does have mean converging to the midpoint $\frac{a+b}{2}$.  This is (a special case of what is) known as the "law of large numbers".

Your guess is correct that there are many different sequences with this same limiting behavior.


问:
Surface area using line integrals
Using line integrals and formula to calculate the surface area($\int zds$ on some curve $c$) I need to find area of cylinder between the plane $z=0$ and surface $z=R+\frac{x^2}{R^2}$.
After parametric equation $x=R\cos t, y=R\sin t$ i get that $ds=R$ and $z=R(1+\cos t)$. Then, since we have symmetry on XY plane the area is $P=4R^2\int\limits_{0}^{\frac{\pi}{2}}(1+\cos t)$ and I get $2R^2(\pi+2)$ but correct solution should be $3\pi R^2$. Where am I mistaking?

Title: Surface area using line integrals

Body:
As you wrote, you will get

$$P=4R^2\int_0^{\frac{\pi}{2}}(1+\cos^2(t))dt$$

$$=4R^2\int_0^{\frac{\pi}{2}}(1+\frac{1+\cos(2t)}{2})dt$$

$$=4R^2(\frac 32\frac{\pi}{2}+0)=3\pi.R^2$$


问:
Prove that any circuit contains a cycle
This is a practice question (not HW)

Prove that any circuit in a graph must contain a cycle AND that any circuit that is not a cycle contains at least two cycles.

Note : This is for a  first course in Graph theory

I have answer to this question but the answer raises more questions.
ANSWER: Suppose the vertices of the circuit are $v_0,v_1,...v_n,v_0$. Consider all subcircuits of the form $v_i,...v_n,v_i$. The subcircuit that uses the fewest number of vertices is a cycle (my Q: BUT WHY).

答:
Suppose this smallest circuit $C$ is not a cycle.  Then it must repeat a vertex $v_j$.  So write the circuit as $v_i C v_j C v_j C v_i$ (that is, follow the cycle from $v_i$ to $v_j$ around back to $v_j$ and back to $v_i$).

$v_j C v_j$ is a smaller circuit, contradicting the choice of $C$.

Your approach to the proof was correct: you just needed to look for a contradiction.  The combination of looking for a graph satisfying some minimal or maximal property together with a proof by contradiction is extremely powerful in graph theory.

Title: Prove that any circuit contains a cycle


问:
Computing the resolvent of a rank one projector
I'm reading through this paper in dynamical systems, that's using a bit of perturbation theory and operator theory. The authors make the following claim:

Now, since $P_0$ is a projection of rank one, it is easy to compute explicitly its resolvent, namely
$$ R_0(ζ) = (P_0 − ζ)^{−1} = \frac{P_0}{1−ζ} − \frac{I−P_0}{ζ} $$

I have no background in operator theory, but I've skimmed through Kato's Perturbation Theory for Linear Operators and done a quick Google search and it's not at all obvious to me where this formula comes from. Does anyone have a source for the resolvent of a rank one projection?

答:
For nonzero $a,b$,
$$(aP + b(I-P))^{-1} = \frac{P}{a} + \frac{I-P}{b} .$$
Setting $a = 1-\zeta$ and $b = -\zeta$ gives the result.


问:
If I can prove $Y$ from "$X$ is true" and from "$X$ is false", can I prove $Y$ without using $X$ at all?
Suppose I have a statement $X$, for which I do not know whether it is true or false. And suppose further that I want to prove a statement $Y$:

I first assume that $X$ is true, and I construct an involved argument that shows that $Y$ follows.

I then assume that $X$ is false, and another involved argument shows that $Y$ still follows.

I have then proven $Y$ (assuming excluded middle). Now, is there any kind of meta-theorem in mathematics that tells me that there should be a proof of $Y$ that is not mentioning $X$ at all (whatever that means formally)?

You can think about this as follows:

Was it necessary to have formulated $X$ in order to give a proof for $Y$, or could we have proven $Y$ right away without the "halt in between"?

For example, the Riemann hypothesis seems to be such an important "halt in the middle" for some statements.

I know, this is a rather informal question.
I do not want to formulate it in terms of mathematical logic and proof theory, simply beacuse I do not know enough about these subjects.
Maybe the question has a trivial counterexample if we go formal enough.
But then again, maybe someone with enough knowledge still sees an interesting variant of the question.
I am also happy to receive quite technical answers.

答:
Proposition: The set $c\mathbb R = \{cx : x\in\mathbb R\}$ is closed.

You can prove this if you assume $c=0,$ and also if you assume $c\ne0.$ But I will be surprised if you can do it without considering those two cases.


问:
Number of elements of order $p$ in a (possibly infinite) group
If a group is finite, then it's easy to see by group action that $n_p\equiv-1\pmod p$.

However, is the result true for an infinite group? My conjecture is that if $n_p$ is non-zero and finite, then $n_p\equiv-1\pmod p$ of infinite groups. I proved the result for $n_2$, but don't know whether it is true for all primes.

答:
Suppose $G$ is a group with finitely many elements of order $p$.  We may assume that $G$ is generated by its elements of order $p$ (otherwise, just look at the subgroup they generate).  Since every conjugate of an element of order $p$ has order $p$, each element of order $p$ must have centralizer of finite index.  So, the intersection $Z$ of the centralizers of elements of order $p$ is finite index.  Since $G$ is generated by its elements of order $p$, $Z$ is actually the center of $G$.

So, the center of $G$ has finite index.  This implies that the commutator subgroup $[G,G]$ is finite (see this question for instance).  But $G/[G,G]$ is an abelian group generated by finitely many elements of finite order, so it is finite as well.  Thus $G$ is finite, and so the result for finite groups applies.


问:
Find surface area of cylinder $x^2+z^2=1$ between two planes $x+y-4=0, y-z+4=0$
Find surface area of cylinder $x^2+z^2=1$ between two planes $x+y-4=0, y-z+4=0$.

I drew the picture but I can't seem to figure out the way to start.. If someone could give me a hint on how to start it would mean a lot..

答: Projects everything on the yz plane and works as a solid of revolution. I think it is the first idea that comes to mind, I hope it helps you</p>


问:
Finitely generated module as an (in)finite sum of submodules
Let $R$ be a ring, $M$ be a nontrivial proper left $R$-module and $(M_\lambda)_{\lambda \in \Lambda}$ be a family of submodules of $M$ such that $M = \sum_{\lambda \in \Lambda} M_\lambda$.

If $M$ is finitely generated, show that $M$ is a finite sum of those $M_\lambda$.

I really don't know what to do, this seems intuitive, but I have no clue on how to proceed.

Any help would be welcome.

答:
This one is easy. Suppose that a set of generators for $M$ is given by $m_1, \ldots, m_n \in M$. For each $j \in \{1, \ldots, n\}$, write
$$
m_j = r_{j, \lambda_1} x_{j, \lambda_1} + \cdots + r_{j, \lambda_{k_j}} x_{j, \lambda_{k_j}}
$$
with $x_{j, \lambda_l} \in M_{\lambda_l}$. Then the elements $x_{j, \lambda_l}$ obviously generate all of $M$, and there are finitely many of them, so that there are finitely many $M_{\lambda_l}$'s which are necessary to generate $M$.


问:
probability theory: question about conditional means
have a little trouble here with understanding one thing.

Let us assume $X,Y,Z$ three random variables $X$ is independent of $Z$. My question is if the following holds:

$$\mathbb E[XY|Z]=\mathbb E[X]\cdot \mathbb E[Y|Z]$$

答: No, this is not true in general.  Suppose $X = Y$.  Then since $X$ and $Y$ are independent of $Z$ we have $\mathbb{E}[XY|Z] = \mathbb{E}[X^2]$ but $\mathbb{E}[X]\mathbb{E}[Y|Z] = \mathbb{E}[X]^2$ so $\mathbb{E}[XY|Z] \ne \mathbb{E}[X]\mathbb{E}[Y|Z]$ unless $X$ is constant.


问:
Prove $\sin \alpha+\sin \beta+\sin \gamma \geq\sin 2\alpha+\sin 2\beta+\sin 2\gamma $
Prove that $\sin \alpha+\sin \beta+\sin \gamma \geq\sin 2\alpha+\sin 2\beta+\sin 2\gamma $ where $\alpha$ $,\beta$ $,\gamma$ are the angles of a triangle</p>

Body:
use
$$\sin{2A}+\sin{2B}=2\sin{C}\cos{(A-C)}\le 2\sin{C}$$
$$\sin{2B}+\sin{2C}\le 2\sin{A},$$
$$\sin{2C}+\sin{2A}\le 2\sin{B}$$


问:
Is $\int_{0}^{\infty}\frac1{\sqrt{x^{2n}+1}}dx$ finite?
I want to evaluate the integral $\int_{0}^{\infty}\frac1{\sqrt{x^{2n}+1}}dx$, where $n\in \mathbb{N},n\ge2$ in order to show the integrability of another function. However, this seems rather difficult, but I need only to know whether it has a finite value, not necessarily how to solve it. WolframAlpha didn't help, so I'm gonna ask here. :)

答:
If you break your integral into two pieces, you can see that

$$\int_0^1 \frac{1}{\sqrt{x^{2n}+1}} \; dx $$

Is a bounded function over a finite interval and hence finite.

The other piece is

$$\int_1^\infty \frac{1}{\sqrt{x^{2n}+1}} \; dx \leq \int_1^\infty \frac{1}{x^n} \; dx $$

which is convergent for $n\geq 2$ on that interval.

The sum of two finites is finite.


答:
When $0\leq x \leq 1$, the function is bounded, and thus has a finite integral.  When $1\leq x$, we have that $0\leq 1\leq x^{2n}$, and so

$$\int_{1}^{\infty}\frac{1}{\sqrt{x^{2n}+x^{2n}}}dx<\int_{1}^{\infty}\frac{1}{\sqrt{x^{2n}+1}}dx<\int_{1}^{\infty}\frac{1}{\sqrt{x^{2n}}}dx.$$

If $n>1$, the outside integrals both converge by the $p$-test (or you can simply compute their anti-derivative, evaluate, and take the limit).  Similarly, if $n\leq 1$ both outside integrals diverge.  So by the comparison test, the integral converges if and only if $n>1$.

This can be done slightly more simply by using the limit comparison test, since $\lim_{x\to \infty} \frac{1}{\sqrt{1+x^{2n}}} / \frac{1}{x^n} =1$, although care must be taken to break the interval up into two parts, since the function we are comparing with has a singularity at $x=0$, and the limit comparison test only helps us as we go to infinity.

Title: Is $\int_{0}^{\infty}\frac1{\sqrt{x^{2n}+1}}dx$ finite?

Body: I want to evaluate the integral $\int_{0}^{\infty}\frac1{\sqrt{x^{2n}+1}}dx$, where $n\in \mathbb{N},n\ge2$ in order to show the integrability of another function. However, this seems rather difficult, but I need only to know whether it has a finite value, not necessarily how to solve it. WolframAlpha didn't help, so I'm gonna ask here. :)


问:
Simple Monte Carlo integration about b-a

I don't understand why we need to * (b-a). Here b-a = 4-2=2.I think we already make random x from 2 to 4. We've already considered the interval (not 0,1). Why do we times (b-a) again?

答:
They are approaching this integration by approximating the average value of $e^{-x}$ over $[2,4]$ (I.e., the average y value) and multiplying it by the length of the x interval to get the estimate of the integral.

More precisely:

$$ \bar{f}_{[a,b]}= \frac{\int_a^b f dx}{b-a}$$

The code mean(exp(-x)) is estimating $\bar{f}_{[a,b]}$ so you need to multiply by the “base” of the rectangle you are making.


问:
Continuity of Taxicab Metric on Metric Space Product
W.A. Sutherland's "Introduction to Metric and Topological Spaces" (1st edition, 1975, Oxford Science Publications) has this puzzling question in Chapter $2$: Continuity Generalized: Metric Spaces, Exercises $2.6: 17$:

"Given a metric space $M = \{A, d\}$, let $M \times M = \{A \times A, d_1\}$ be as in Example $2.2.7$. Show that $d: A \times A \to \mathbb R$ is a continuous mapping from $M \times M$ to $\mathbb R$."

In this context, example $2.2.7$ defines the product space:

"Given two metric spaces $M = \{A, d\}$, $M' = \{A', D'\}$, we define several metrics on $A \times A'$ ... If $x = (x_1, x_2)$ and $y = (y_1, y_2)$ are in $A \times A'$, let:
$$d_1(x, y) = d (x_1, y_1) + d'(x_2, y_2)$$"

... which Sutherland defines as a the "taxicab metric".

The expression for what is presumed to be the object of the exercise seems to be wrong.

What is given $d: A \times A \to \mathbb R$ is merely the metric $d$ on $M$.

Whether $d$ is or is not continuous is independent of what $d_1$ may happen to be.

If the exercise were to prove that $d$ is continuous, then there would be no point in mentioning that it is a factor space of a Cartesian product.

Thus it is supposed that what we are expecting to prove is that the metric $d_1: (A \times A) \times (A \times A): \mathbb R$ is continuous.

As it happens, a metric is always a "continuous mapping" (that is, a "continuous real-valued function") no matter what $A$ or $d$ actually are.

Hence it is trivial to apply that result to $d_1$.

Is there anyone out there who may have an idea of what Sutherland was trying to get at here?

答:
Given metric spaces $(X,\rho_X)$ and $(Y,\rho_Y)$, and a function $f : X \to Y$, we say that $f$ is continuous with respect to $\rho_X$ and $\rho_Y$ (in shorthand, $f : (X,\rho_X) \to (Y,\rho_Y)$ is continuous) if for any $x_0 \in X$ and any $\varepsilon>0$ there exists $\delta>0$ (that depends on $x_0$ and $\varepsilon$) such that
$$(\forall x \in X) \quad \rho_X(x,x_0)<\delta \!\implies\! \rho_Y(f(x),f(x_0)) < \varepsilon.$$
So, you need to prove that if $(A,d)$ is a metric space, then $d : (A \times A,d_1) \to (\mathbb R,|\cdot|)$ is continuous, that is, given $(x_0,y_0) \in A \times A$ and $\varepsilon>0$, you need to find $\delta>0$ such that $$(\forall x,y \in A) \quad d_1((x,y),(x_0,y_0)) < \delta \!\implies\! |d(x,y)-d(x_0,y_0)| < \varepsilon.$$


问:
Proposition 3.11 in Wainwright's High-Dimensional Statistics
I was unable to follow the proof of Proposition 3.11 in Wainwright's book High-dimensional statistics: A non-asymptotic viewpoint. If $\rho$ is a metric, $\rho(x,A)=\inf_{y\in A}\rho(x,y)$, $\epsilon\ge 0$, $P[A]\ge 1/2$, random variable $X\sim P$, and $f(x)=\min\{\rho(x,A),\epsilon\}$, the proof (which is on page 71, in Section 3.2) says that $$E[f(X)]\le (1-P(A))\epsilon\,.$$

I did not see why this is true. Thank you for your help!

答:
Hint: Show that
$$f(x):= \min\{\rho(x, A), \epsilon\} \le (1 - \mathbf{1}_A(x)) \cdot \epsilon.$$
(Think about how $f(x)$ behaves when $x \in A$ and $x \notin A$.)


问:
cyclical points on a conic section
$P_1, P_2, P_3, P_4$ are four arbitrary points on $xy = 1$. $P_1P_4$ intersects $P_2P_3$ at $D_1$, and similarly define $D_2, D_3$. Prove that $O, D_1, D_2, D_3$ are cyclical, where $O$ is the origin.

I have never seen a geometry problem asking to prove cyclical points on a conic section. So frankly not sure where to start. I suspect bashing coordinates would work, but that's quite a bit of efforts (and I am not sure how to go from coordinates to proving points are cyclical).

Is there some shortcut I could use by some geometric properties of $xy=1$?

答:
Here's a proof without coordinates. I will relabel $D_1=P_1P_4\cap P_2P_3$, $D_2=P_2P_4\cap P_1P_3$ and $D_3=P_3P_4\cap P_1P_2$.

Lemma 1. The centre of any rectangular hyperbola $\mathcal H$ through points $A$, $B$, $C$ lies on the nine-point circle of $\triangle ABC$.

Proof. Let $(ABC)$ meet $\mathcal H$ again at $D$, and let $H_A$, $H_B$, $H_C$, $H_D$ be the orthocentres of $\triangle BCD$, $\triangle CDA$, $\triangle DAB$, $\triangle ABC$, which all lie on $\mathcal H$. There is a $180^{\circ}$ rotation mapping $ABCD$ to $H_AH_BH_CH_D$, and the centre of this rotation is both the centre of $\mathcal H$ and the midpoint of $\overline{DH_D}$, so it lies on the nine-point circle of $\triangle ABC$. $\square$

Lemma 2. The incentre $I$ and excentres $I_1, I_2, I_3$ (opposite $D_1$, $D_2$, $D_3$) of $\triangle D_1D_2D_3$ lie on the hyperbola.

Proof. Take a projective transformation sending $P_1P_2P_3P_4$ to a square. Then $D_1$ and $D_3$ go to infinity, so $I_1I_2I_3I$ becomes a rectangle with sides parallel to $P_1P_2P_3P_4$. Then $D_2$ maps to the common centre of rectangle $I_1I_2I_3I$ and square $P_1P_2P_3P_4$, so these eight points lie on a common conic. $\square$

Now $(D_1D_2D_3)$ is the nine-point circle of $\triangle I_1I_2I_3$, so our two lemmas imply the result.

Title: cyclical points on a conic section


问:
About the universal bundle $EG\rightarrow BG$
For a topological group $G$, we define $EG$ to be the infinite join of $G$, and $B$ to be the quotient of $EG$ by the left action of $G$. Explicitly $EG$ can be expressed, as a set, as $EG=\lbrace(g_{0},t_{0},g_{1},t_{1},\cdots)\in(G\times I)^{\infty}\rbrace/\sim$ such that at most finitely many $t_{i}$ are nonzero, $\sum t_{i}=1$, and the equivalence relation is induced from the join. Milnor proved that this $EG\rightarrow BG$ is a numerable principal $G$-bundle, and this is a universal bundle.

I am now studying the fact using some modern references (especially Paul Selick's "Introduction to Homotopy Theory" and Tammo tom Dieck's "Algebraic Topology"), and I can understand most of the proof, but one thing matters in my mind. Specifically, the proofs in both books say that this bundle $EG\rightarrow BG$ is numerable since $\lbrace U_{i}\rbrace_{i=0}^{\infty}$ forms a numerable cover for the bundle where $V_{i}=\lbrace(g_{0},t_{0},\cdots)\in EG\mid t_{i}>0\rbrace$ and $U_{i}:=V_{i}/G\subset BG$. But I think this cover is not locally finite, but just point-finite. Indeed I can modify the definition of $V_{i}$ to be, for instance, something like $\lbrace(g_{0},t_{0},\cdots)\in EG\mid t_{i}>2^{-(i+2)}\rbrace$, and then I can believe that $U_{i}$'s form a numerable cover. But I am afraid of the possibility that I have some misunderstandings about the topology of joins. Am I wrong? Please clarify this issue for me.

答:
This is a great question and I have been struggling with exactly the same thing recently. Given what is stated in Dieck's proof of 14.4.3 (or rather, what is not), you are absolutely right to be worried (though it will all work out in the end). The functions $t_i:EG\rightarrow [0,1]$ do not form a partition of unity in the usual sense, since the sets $t_i^{-1}(0,1]$ only form a point-finite cover. Worse yet, the conditions in Proposition 14.4.5 seem to imply that a non-numerable bundle might possess a $G$-map into $EG$, which would only be possible if the universal bundle $EG\rightarrow BG$ were itself non-numerable. The difficulty lies in the fact that Dieck is implicitly using an equivalent definition of "numerable" that initially seems weaker. But if we look to the previous chapter of the book, we can find a magnificent fix! Namely, Lemma 13.1.7 essentially says the following (actually, it says something even stronger):

Lemma. Consider a set of continuous functions $\{u_j:X\rightarrow [0,1]\mid j\in J\}$ on a topological space $X$, such that $\{j\in J\mid u_j(x)>0\}$ is finite for every $x\in X$ and $\sum_{j\in J}u_j=1$. Then the open cover $\{u_j^{-1}(0,1]\mid j\in J\}$ is numerable. (Notice, this cover is point-finite, not locally finite! The lemma says that we can find new functions, which shrink the supports sufficiently to get a partition of unity in the usual, stronger sense.)

If you have a bundle $p:E\rightarrow X$ such that $p$ is trivializable over each $u_j^{-1}(0,1]$, then this proves that $p$ is a numerable bundle, since it is trivializable over a numerable cover. In particular, this fixes your issue with the numerability of the open cover of $EG$. However, in the course of proving Lemma 13.1.7 (which also uses the proofs of 13.1.4 and 13.1.5), you will get a new collection of functions, different from the original coordinate projections $t_i$ on $EG$. Indeed, these coordinate functions are not a numeration of the bundle and you cannot just refine the cover. You really do need to create new functions with smaller supports.


问:
determinants of 2 matrices with given property
I have two $3\times3$ integer matrices $A$ and $B$ such that $AB=A+B$. I need to find all possibe values of $\det(A-E)$, where $E$ denotes the identity matrix. Any help is appreciated.</p>

答:
Hints:

$$\bullet\;\;\;\;\;AB=A+B\implies (A-I)(B-I)=I\;,\;\;I:=\text{the unit (identity) matrix)}$$

$$\bullet\;\;\;\;\text{For square matrices $\,X,Y\,$ of the same order, it's true that}\;\;\det(XY)=\det X\cdot\det Y$$


问:
Turning a sequence of "zigzags" into a prespectrum
Suppose that we are given a sequence of spaces $\{C_j\}$ and a family of diagrams

$$C_j \rightarrow D_j \xleftarrow{\sim} E_j \rightarrow \Omega C_{j+1}$$

Then, if we let

$$ F_j:= \operatorname{Holim}( C_j \rightarrow D_j \xleftarrow{\sim} E_j \rightarrow \Omega C_{j+1} \rightarrow \Omega D_{j+1} \xleftarrow{\sim} \Omega E_{j+1} \to \Omega^2 C_{j+2} \rightarrow \cdots)$$

Why is $F_j$ (weakly) equivalent to $C_j$?

The point is to replace this sequence of diagrams into a prespectrum, by using the fact that loops commute with homotopy limits and a morphism of diagrams in the right direction. I just can't see why we get the equivalence.

答:
We may compose the composable arrows in your diagram to obtain a zigzag of the form $$A_0 \rightarrow B_0 \leftarrow  A_1 \rightarrow B_1 \leftarrow \cdots$$ for which the directions of the arrows strictly alternate, which is initial in the original diagram.  So it suffices to give a formula to compute the (homotopy) limit of this reduced diagram.

We claim that $$\lim (A_0 \rightarrow B_0 \leftarrow  A_1 \rightarrow B_1 \leftarrow \cdots) \simeq \varprojlim_n (A_0 \times_{B_0} A_1 \times_{B_1} \cdots \times_{B_n} A_{n+1}).$$

Given this, we see that the limit in the question is equivalent to $$\varprojlim_n (C_j \times_{D_j} E_j \times_{\Omega D_{j+1}} \Omega E_{j+1} \times_{\Omega^2 D_{j+2}} \cdots \times_{\Omega^n D_{j+n}} \Omega^n E_{j+n}).$$  However, since we have $\Omega^k E_{j+k} \xrightarrow{\sim} \Omega^k D_{j+k},$ this iterated pullback collapses to just $C_j$ at each level.  So this inverse limit diagram is equivalent to a constant diagram at $C_j$, so the limit is just $C_j$.

We sketch a proof of the claim.  The essential ingredient is the technique to compute limits by decomposing the indexing diagram; see Corollary 4.2.3.10 in Higher Topos Theory for example.

First, write the infinite zigzag diagram as an increasing union of finite zigzags starting at the left.  This shows that the limit can be computed as an inverse limit of limits of composable spans.

Second, decompose the $n$-fold span into individual spans to show that the limit of iterated spans is computed by iterated pullbacks.

The formula in the claim follows by combining these two observations.


问:
What is the smallest nonempty subset U of $R^2$ that is not a subspace of $R^2$ but is closed under addition and takes additive inverses?
This is a take on 1.C.7 of Axler's "Linear Algebra Done Right", but changed to "the smallest nonempty subset" instead of "an example of a nonempty subset".

When I was first doing this problem I thought of the subset ${(0,0),(1,1)}$ where $(1,1)+(1,1)=(0,0)$. This trick makes sense to me after reading the Digression on Fields section in Chapter 1. However, I am not sure how to prove that this subset is the smallest that satisfies the requirements or if my answer violates some rule of math.

The solutions manual gives $Z^2$ as the answer because it fails the scalar multiplication requirement but had consistent addition and additive inverses.

答:
There is no smallest: consider $ ... 2k \mathbb{Z} \times \{0\} \subsetneq ... \subsetneq 4 \mathbb{Z} \times \{0\} \subsetneq 2  \mathbb{Z} \times \{0\} \subsetneq  \mathbb{Z} \times \{0\}$. None of these sets is even minimal.

A smallest subset of $\mathbb{R}$  closed under addition and "minus" must be contained by every one of these sets, so it's contained by their intersection $S_0 = \{(0, 0)\}$. Since you're looking for nonempty suitably closed sets, the smallest has to equal $S_0$. However, that's a subspace.


答:
It is hard to say about what does a proof "contains" in here, indeed if $T∪\{a\}⊢x$ and $T∪\{¬a\}⊢x$ then $T⊢x$ because in any model of $T$ either $a$ or $¬a$ are true, hence $T⊨x⇒T⊢x$, but how do you define "not mentioning"?

One interpretation you can think of is if $a$ is not in the language of $T$, that is:

Given a theory $T$ in the language $\cal{L}$, and given a language $\cal{L}'\supset\cal{L}$, is it possible to have a formula $a$ in the language of $\cal{L}'$ such that $T∪\{a\}⊢_{\cal{L}'}x$ and $T∪\{¬a\}⊢_{\cal{L}'}x$ and $T⊢_{\cal{L}}x$ is false?

***In this interpretation the answer is false:

Let $\cal{L}$ be empty and $\{f,g\}=\cal{L}'$ be 1-ary functional symbols, and let $a$ be the following theory:

$$(f\text{ is injective}∧f\text{ is not surjective})∧(g\text{ is injective}∧g\text{ is surjective})$$
And let $x$ be:
$$∃z,y\ (z≠y)$$

Notice that $a$ does prove $x$, as well as $¬a$ proves $x$, but $x$ is not tautology.***

***As Noah said, this example is wrong, $¬a$ does not prove $x$.

Although the above answer is wrong, I leave this post up and not deleting it because it is yet another possible interpretation of the question.

Title: If I can prove $Y$ from "$X$ is true" and from "$X$ is false", can I prove $Y$ without using $X$ at all?

Body:
Suppose I have a statement $X$, for which I do not know whether it is true or false. And suppose further that I want to prove a statement $Y$:

I first assume that $X$ is true, and I construct an involved argument that shows that $Y$ follows.

I then assume that $X$ is false, and another involved argument shows that $Y$ still follows.

I have then proven $Y$ (assuming excluded middle). Now, is there any kind of meta-theorem in mathematics that tells me that there should be a proof of $Y$ that is not mentioning $X$ at all (whatever that means formally)?

You can think about this as follows:

Was it necessary to have formulated $X$ in order to give a proof for $Y$, or could we have proven $Y$ right away without the "halt in between"?

For example, the Riemann hypothesis seems to be such an important "halt in the middle" for some statements.

I know, this is a rather informal question.
I do not want to formulate it in terms of mathematical logic and proof theory, simply beacuse I do not know enough about these subjects.
Maybe the question has a trivial counterexample if we go formal enough.
But then again, maybe someone with enough knowledge still sees an interesting variant of the question.
I am also happy to receive quite technical answers.


问:
Which one is bigger, $\sum_{i\in I}\kappa_i$ or $\prod_{i\in I}\kappa_i$?
Let $(\kappa_i)_{i\in I}$ be a family of cardinals. If $I=\{1, 2\}$ and at least one $\kappa_i$ is infinite and they are all non-empty, then
$$\sum_{i\in I}\kappa_i = \prod_{i\in I}\kappa_i.$$
But what if $I$ is a set, which one is bigger:
$$\sum_{i\in I}\kappa_i \qquad\text{or}\qquad  \prod_{i\in I}\kappa_i\qquad?$$
I know that if $(\kappa'_i)$ is another family indexed by the same set $I$ and $\kappa_i<\kappa'_i$ for all $i$, then
$$\sum_{i\in I}\kappa_i < \prod_{i\in I}\kappa'_i.$$
Does it also hold that
$$\prod_{i\in I}\kappa_i < \prod_{i\in I}\kappa'_i$$
and
$$\sum_{i\in I}\kappa_i < \sum_{i\in I}\kappa'_i\qquad ?$$

答:
$\sum_{\kappa\in\varnothing}\kappa=0$, and $\prod_{\kappa\in\varnothing}\kappa=1$. The first of these is easily derived from the definition of cardinal addition; the second is a bit less obvious, but it also follows from the definition.

Your last two inequalities do not hold in general. Let $\kappa_n=\omega_n$ and $\kappa_n'=\omega_{n+1}$ for $n\in\omega$; then $\kappa_n<\kappa_n'$ for all $n\in\omega$, but $\sum_{n\in\omega}\kappa_n=\sum_{n\in\omega}\kappa_n'$ and $\prod_{n\in\omega}\kappa_n=\prod_{n\in\omega}\kappa_n'$.


问:
If $\alpha,\beta,\gamma$ are the roots of equation $x^3-7x^2+12x-11=0$, find the last four digits of $\alpha^{21007}+\beta^{21007}+\gamma^{21007}$

If $\alpha,\beta,\gamma$ are roots of the equation $x^3-7x^2+12x-11=0$, then find the last four digits of $\alpha^{21007}+\beta^{21007}+\gamma^{21007}$.

I know that $\alpha+\beta+\gamma=7, \alpha\beta+\beta\gamma+\gamma\alpha=12,\alpha\beta\gamma=11$. From this how I can proceed. please give some idea

答:
This turned out to be interesting. If we let $x_n = \alpha^n + \beta^n + \gamma^n\bmod 10000,$ then one easily has the recurrence:

$$x_{n+3} = 7x_{n+2} - 12x_{n+1} + 11x_{n}$$
with $x_0 = 3, x_1 = 7, x_2 = 25.$ The idea is obviously to find the period of the sequence $\{x_n\},$ which turns out to be $T = 21000$, after a short Python simulation. This gives the answer:

$$x_{21007} = 4637.$$
For reference, the indices $n\leq 21007$ for which we get the same four digits are:

```<code>[    7,    11,  2261,  3007,  3761,  5257,  5261,  7511,  8257,
   10507, 10511, 12761, 13507, 15757, 15761, 17257, 18011, 18757,
   21007]
</code>```

I will be eagerly waiting for an analytic solution but it looks to be beyond the elementary methods that usually tackle this kind of problems. Something to be consider would be maybe an extenstion of Euler/Fermat theorems for complex primes but I am not well-versed in that area.

Body:
This turned out to be interesting. If we let $x_n = \alpha^n + \beta^n + \gamma^n\bmod 10000,$ then one easily has the recurrence:

$$x_{n+3} = 7x_{n+2} - 12x_{n+1} + 11x_{n}$$
with $x_0 = 3, x_1 = 7, x_2 = 25.$ The idea is obviously to find the period of the sequence $\{x_n\},$ which turns out to be $T = 21000$, after a short Python simulation. This gives the answer:

$$x_{21007} = 4637.$$
For reference, the indices $n\leq 21007$ for which we get the same four digits are:

[    7,    11,  2261,  3007,  3761,  5257,  5261,  7511,  8257,
   10507, 10511, 12761, 13507, 15757, 15761, 17257, 18011, 18757,
   21007]

I will be eagerly waiting for an analytic solution but it looks to be beyond the elementary methods that usually tackle this kind of problems. Something to be consider would be maybe an extenstion of Euler/Fermat theorems for complex primes but I am not well-versed in that area.


答:
Maybe comparing Monte Carlo integration with Riemann approximation
will help you understand the former.

Suppose we want to approximate the integral $J = \int_a^b e^{-x}\,dx$
by $n=1000$ rectangles of widths $w$ and heights $h = e^{-m},$ where $m$ is the midpoint of the base of a rectangle. In R,

```<code>a = 2;  b = 4
n = 1000
w = (b-a)/n
m = seq(a+w/2, b-w/2, length=n)
h = exp(-m)
sum(w*h)
[1] 0.1170196
</code>```

This is a deterministic computation, so each run will give the same answer.

You are integrating the density function of the distribution $\mathsf{Exp}(\mathrm{rate\,}=\lambda = 1).$ In R, the CDF of
this distribution is pexp (with default $\lambda=1).$ So a Riemann approximation with 1000 evenly spaced grid points as centers of rectangles has given us four-place accuracy.

```<code>diff(pexp(c(2,4)))
[1] 0.1170196
</code>```

A basic Monte Carlo integration substitutes for the evenly spaced points above, randomly chosen $m$ distributed as $\mathsf{Unif}(a,b).$
For this one-dimensional integration, randomly chosen grid
points are not quite as efficient as a precisely evenly spaced
grid.  So I will use $n = 10\,000$ random points (uniformly distributed).

```<code>set.seed(2021)
a = 2;  b = 4
n = 10000
w = (b-a)/n
m = runif(m, a, b)
h = exp(-m)
sum(w*h)
[1] 0.01189166
</code>```

This is a random procedure, so (for different seeds, or no set seed)
the result will be slightly different on each run.

The average widths are $w = (b-a)/n.$ If I had used mean instead of
sum as you did, that would take care of the $1/n,$ but the
length of the interval, over which the integration takes place, is still needed.

Note: Sampling method. For Monte Carlo integration in one dimension, a 'sampling
method' is sometimes used (often more for its ease of programming than for its efficiently). If we use the R function rexp to sample a million observations from $\mathsf{Exp}(\lambda=1),$ then we can ask what proportion of them
lie between $2$ and $4.$

```<code>set.seed(120)
x = rexp(10^6)
mean((x>2)&(x<4))
[1] 0.117494
</code>```

The vector (x>2)&(x<4) is a logical vector with a million
elements TRUE or FALSE; the mean of a logical vector is its proportion of TRUEs.

This method is especially convenient when it would be
some trouble to find the density function of the distribution of interest. Suppose time to completion of a 2-phase process is the sum of $Z \sim \mathsf{Norm}(\mu=30,\sigma = 5)$ and $X \sim \mathsf{Exp}(\lambda = .05).$ The the waiting time to completion is $W = Z+X.$ The average waiting time $E(W) = E(Z)+E(X) = 30+20 = 50,$ and $P(W > 60) \approx 0.22935 \pm 0.00084.$

```<code>set.seed(109)
z = rnorm(10^6, 30,5)
x = rexp(10^6, .05)
w = z + x
mean(w > 60)
[1] 0.229349
2*sd(w > 60)/1000
[1] 0.0008408287

hist(w, prob=T, br=60, col="skyblue2",
      main="Time to Completion")
 abline(v = 60, col="red", lwd=2)
</code>```

Note: Higher dimensions. The accuracy of a basic Monte Carlo integration
depends in part on the "wiggliness" of the function $f(x)$
being  integrated. However, generally speaking, Monte Carlo integrations in two (or higher) dimensions with $n$ randomly chosen points are about as good as (or better than) a Riemann approximation with a grid of $n$ evenly spaced points.

Example: The integral of a bivariate standard normal
distribution (correlation $0)$ over first quadrant of the unit circle can be shown to be $J = 0.0984.$

```<code>pchisq(1,2)/4
[1] 0.09836734
</code>```

We show Monte Carlo integration
with about $100^2\pi/4$ points randomly distributed in the
triangle. Notice we average the random heights and multiply
by the area of the quarter circle.

```<code>set.seed(1234)  # 1234
n = 100;  u1 = runif(n);  u2 = runif(n)
h.sq = dnorm(u1)*dnorm(u2)
h.qd = h.sq[u1^2 + u2^2 < 1]
(pi/4)*mean(h.qd)
[1] 0.09813584
</code>```

Title: Simple Monte Carlo integration about b-a

Body: I don't understand why we need to * (b-a). Here b-a = 4-2=2.I think we already make random x from 2 to 4. We've already considered the interval (not 0,1). Why do we times (b-a) again?

Body:
Maybe comparing Monte Carlo integration with Riemann approximation
will help you understand the former.

Suppose we want to approximate the integral $J = \int_a^b e^{-x}\,dx$
by $n=1000$ rectangles of widths $w$ and heights $h = e^{-m},$ where $m$ is the midpoint of the base of a rectangle. In R,

a = 2;  b = 4
n = 1000
w = (b-a)/n
m = seq(a+w/2, b-w/2, length=n)
h = exp(-m)
sum(w*h)
[1] 0.1170196

This is a deterministic computation, so each run will give the same answer.

You are integrating the density function of the distribution $\mathsf{Exp}(\mathrm{rate\,}=\lambda = 1).$ In R, the CDF of
this distribution is pexp (with default $\lambda=1).$ So a Riemann approximation with 1000 evenly spaced grid points as centers of rectangles has given us four-place accuracy.

diff(pexp(c(2,4)))
[1] 0.1170196

A basic Monte Carlo integration substitutes for the evenly spaced points above, randomly chosen $m$ distributed as $\mathsf{Unif}(a,b).$
For this one-dimensional integration, randomly chosen grid
points are not quite as efficient as a precisely evenly spaced
grid.  So I will use $n = 10\,000$ random points (uniformly distributed).

set.seed(2021)
a = 2;  b = 4
n = 10000
w = (b-a)/n
m = runif(m, a, b)
h = exp(-m)
sum(w*h)
[1] 0.01189166

This is a random procedure, so (for different seeds, or no set seed)
the result will be slightly different on each run.

The average widths are $w = (b-a)/n.$ If I had used mean instead of
sum as you did, that would take care of the $1/n,$ but the
length of the interval, over which the integration takes place, is still needed.

Note: Sampling method. For Monte Carlo integration in one dimension, a 'sampling
method' is sometimes used (often more for its ease of programming than for its efficiently). If we use the R function rexp to sample a million observations from $\mathsf{Exp}(\lambda=1),$ then we can ask what proportion of them
lie between $2$ and $4.$

set.seed(120)
x = rexp(10^6)
mean((x>2)&(x<4))
[1] 0.117494

The vector (x>2)&(x<4) is a logical vector with a million
elements TRUE or FALSE; the mean of a logical vector is its proportion of TRUEs.

This method is especially convenient when it would be
some trouble to find the density function of the distribution of interest. Suppose time to completion of a 2-phase process is the sum of $Z \sim \mathsf{Norm}(\mu=30,\sigma = 5)$ and $X \sim \mathsf{Exp}(\lambda = .05).$ The the waiting time to completion is $W = Z+X.$ The average waiting time $E(W) = E(Z)+E(X) = 30+20 = 50,$ and $P(W > 60) \approx 0.22935 \pm 0.00084.$

set.seed(109)
z = rnorm(10^6, 30,5)
x = rexp(10^6, .05)
w = z + x
mean(w > 60)
[1] 0.229349
2*sd(w > 60)/1000
[1] 0.0008408287

hist(w, prob=T, br=60, col="skyblue2",
      main="Time to Completion")
 abline(v = 60, col="red", lwd=2)

Note: Higher dimensions. The accuracy of a basic Monte Carlo integration
depends in part on the "wiggliness" of the function $f(x)$
being  integrated. However, generally speaking, Monte Carlo integrations in two (or higher) dimensions with $n$ randomly chosen points are about as good as (or better than) a Riemann approximation with a grid of $n$ evenly spaced points.

Example: The integral of a bivariate standard normal
distribution (correlation $0)$ over first quadrant of the unit circle can be shown to be $J = 0.0984.$

pchisq(1,2)/4
[1] 0.09836734

We show Monte Carlo integration
with about $100^2\pi/4$ points randomly distributed in the
triangle. Notice we average the random heights and multiply
by the area of the quarter circle.

set.seed(1234)  # 1234
n = 100;  u1 = runif(n);  u2 = runif(n)
h.sq = dnorm(u1)*dnorm(u2)
h.qd = h.sq[u1^2 + u2^2 < 1]
(pi/4)*mean(h.qd)
[1] 0.09813584


答: Hint: Notice that $A=(A+B)-B=AB-B=(A-E)B$.

Title: determinants of 2 matrices with given property

Body: I have two $3\times3$ integer matrices $A$ and $B$ such that $AB=A+B$. I need to find all possibe values of $\det(A-E)$, where $E$ denotes the identity matrix. Any help is appreciated.</p>


答:
Let $(x_0, y_0) \in A \times A$. If $d_1((x, y), (x_0, y_0)) = d(x, x_0) + d(y, y_0) < \epsilon$, then
$$
|d(x, y) - d(x_0, y_0)| \le |d(x, y) - d(x, y_0)| + |d(x, y_0) - d(x_0, y_0)| \le 2\epsilon,
$$
since in a general metric space it holds that if $d(x, y) < \epsilon$, then $d(z, x) - d(z, y) < \epsilon$; this is because $d(z, x) \le d(z, y) + d(x, y)$, and we may assume without loss of generality that $d(z, x) \ge d(z, y)$.

Title: Continuity of Taxicab Metric on Metric Space Product

Body:
W.A. Sutherland's "Introduction to Metric and Topological Spaces" (1st edition, 1975, Oxford Science Publications) has this puzzling question in Chapter $2$: Continuity Generalized: Metric Spaces, Exercises $2.6: 17$:

"Given a metric space $M = \{A, d\}$, let $M \times M = \{A \times A, d_1\}$ be as in Example $2.2.7$. Show that $d: A \times A \to \mathbb R$ is a continuous mapping from $M \times M$ to $\mathbb R$."

In this context, example $2.2.7$ defines the product space:

"Given two metric spaces $M = \{A, d\}$, $M' = \{A', D'\}$, we define several metrics on $A \times A'$ ... If $x = (x_1, x_2)$ and $y = (y_1, y_2)$ are in $A \times A'$, let:
$$d_1(x, y) = d (x_1, y_1) + d'(x_2, y_2)$$"

... which Sutherland defines as a the "taxicab metric".

The expression for what is presumed to be the object of the exercise seems to be wrong.

What is given $d: A \times A \to \mathbb R$ is merely the metric $d$ on $M$.

Whether $d$ is or is not continuous is independent of what $d_1$ may happen to be.

If the exercise were to prove that $d$ is continuous, then there would be no point in mentioning that it is a factor space of a Cartesian product.

Thus it is supposed that what we are expecting to prove is that the metric $d_1: (A \times A) \times (A \times A): \mathbb R$ is continuous.

As it happens, a metric is always a "continuous mapping" (that is, a "continuous real-valued function") no matter what $A$ or $d$ actually are.

Hence it is trivial to apply that result to $d_1$.

Is there anyone out there who may have an idea of what Sutherland was trying to get at here?


问:
Is the space of real analytic functions a freely generated algebra?
Let us consider the space $C^{\omega}(\mathbb{R})$ of all the functions $f \colon \mathbb{R} \to \mathbb{R}$ which are analytic on the whole real line. It is clear that $\mathcal{C}^\omega(\mathbb{R})$ is an algebra, because it is closed under addition, multiplication by scalars and inner multiplication. However, is it true that $\mathcal{C}^\omega(\mathbb{R})$ is freely $\mathfrak{c}$-generated? That is:

Is there a set $S \subseteq \mathcal{C}^\omega(\mathbb{R})$ of cardinality $\mathfrak{c}$ such that the algebra generated by $S$ is $\mathcal{C}^\omega(\mathbb{R})$ and whenever a polynomial $P \in \mathbb{R}[X_1, \dotsc, X_p]$ with $P(0, \dotsc, 0) = 0$ satisfies $P(s_1, \dotsc, s_p) = 0$ for some $s_1, \dotsc, s_p \in S,$ then $P = 0?$

If so, how can it be proven?

答:
This is certainly false but it seems surprisingly difficult to disprove. Golasinski and Henriksen's Residue Class Rings of Real-Analytic and Entire Functions proves that every residue field of $C^{\omega}(\mathbb{R})$ (quotient by a maximal ideal) is a real closed field. But the ring of polynomials in $\mathfrak{c}$ variables over $\mathbb{R}$ does not have this property: its residue fields consist of every field extension of $\mathbb{R}$ which is generated as an algebra by at most $\mathfrak{c}$ variables, and in particular contains $\mathbb{R}(x)$ (which is $\mathfrak{c}$-dimensional as a vector space), which is not real closed.

Edit: Never mind, here's a much easier argument. The ring of polynomials in any number of variables over $\mathbb{R}$ has the property that its group of units is exactly the invertible scalar multiples of the identity $\mathbb{R}^{\times}$ (because a nonconstant polynomial has positive degree and so its product with another nonzero polynomial also has positive degree). But there are many invertible real analytic functions which are not scalar multiples of the identity, e.g. $\exp(x)$.


答:
Claim:  if $n$ is composite, it shares a common factor with some $x$ in the interval $[n/2, n/2 + \sqrt{n}/2]$.

Proof:  Let $p$ be the smallest prime divisor of $n$.  Then $p \leq \sqrt{n}$, since otherwise $n$ would be prime.  Any multiple of $p$ shares a common factor (namely $p$) with $n$.  The closest multiple of $p$ to $n/2$ is either $n/2$ itself (if $n$ is even) or $\frac{n \pm p}{2}$ (if $n$ is odd).  Since $0 < p \leq \sqrt{n}$, both cases produce an integer in $[n/2, n/2 + \sqrt{n}/2]$ that shares the factor $p$ with $n$.

So it's enough to go from $x = n/2$ up to $n/2 + \sqrt{n}/2$.  With this improvement, your algorithm will still be somewhat slower than your isPrimeNumber routine--just because calculating a gcd is slower than calculating divisibility.  This will be feasible for testing numbers with maybe 15-20 digits, but you would need completely different methods to test something much larger, like the 183-digit number you mention.

Title: Speedup Primality Test

Body:
I'm trying to speedup my primality test where every $n \in \mathbb{Z}$ is prime if all lattice points on $x+y=n$ are visible from the origin.

It's really inefficient at the moment because it requires $\approx\frac {n}2$ $gcd$ computations.

For example, to determine primality of $n = 249999991$, it takes $\approx 12$ seconds on my computer!

To speed things up, I tried reducing the $\frac {n}2$ steps by running the C program below to see if there's a way to calculate a maximum bound on the number of steps needed to determine primality.

The max steps needed for each $10,000,000$ range follows:

```<code>Range [0,10000000] Max step 1568
Range [10000000,20000000] Max step 2231
Range [20000000,30000000] Max step 2738
Range [30000000,40000000] Max step 3161
Range [40000000,50000000] Max step 3534
Range [50000000,60000000] Max step 3870
Range [60000000,70000000] Max step 4181
Range [70000000,80000000] Max step 4470
Range [80000000,90000000] Max step 4739
Range [90000000,100000000] Max step 4986
Range [100000000,110000000] Max step 5243
Range [110000000,120000000] Max step 5474
Range [120000000,130000000] Max step 5699
Range [130000000,140000000] Max step 5915
Range [140000000,150000000] Max step 6120
Range [150000000,160000000] Max step 6323
Range [160000000,170000000] Max step 6518
Range [170000000,180000000] Max step 6705
Range [180000000,190000000] Max step 6890
Range [190000000,200000000] Max step 7053
Range [200000000,210000000] Max step 7244
Range [210000000,220000000] Max step 7415
Range [220000000,230000000] Max step 7580
Range [230000000,240000000] Max step 7736
Range [240000000,250000000] Max step 7904
Range [250000000,260000000] Max step 8055
Range [260000000,270000000] Max step 8213
Range [270000000,280000000] Max step 8364
Range [280000000,290000000] Max step 8514
Range [290000000,300000000] Max step 8658
Range [300000000,310000000] Max step 8799
Range [310000000,320000000] Max step 8940
Range [320000000,330000000] Max step 9074
Range [330000000,340000000] Max step 9219
Range [340000000,350000000] Max step 9350
Range [350000000,360000000] Max step 9486
Range [360000000,370000000] Max step 9615
Range [370000000,380000000] Max step 9744
Range [380000000,390000000] Max step 9869
Range [390000000,400000000] Max step 9998
Range [400000000,410000000] Max step 10116
Range [410000000,420000000] Max step 10241
Range [420000000,430000000] Max step 10365
Range [430000000,440000000] Max step 10481
Range [440000000,450000000] Max step 10605
Range [450000000,460000000] Max step 10716
Range [460000000,470000000] Max step 10836
Range [470000000,480000000] Max step 10946
Range [480000000,490000000] Max step 11066
Range [490000000,500000000] Max step 11174
Range [500000000,510000000] Max step 11286
Range [510000000,520000000] Max step 11393
Range [520000000,530000000] Max step 11510
Range [530000000,540000000] Max step 11613
Range [540000000,550000000] Max step 11723
Range [550000000,560000000] Max step 11831
Range [560000000,570000000] Max step 11936
Range [570000000,580000000] Max step 12041
Range [580000000,590000000] Max step 12140
Range [590000000,600000000] Max step 12240
Range [600000000,610000000] Max step 12348
Range [610000000,620000000] Max step 12444
Range [620000000,630000000] Max step 12548
Range [630000000,640000000] Max step 12630
Range [640000000,650000000] Max step 12735
Range [650000000,660000000] Max step 12839
Range [660000000,670000000] Max step 12936
Range [670000000,680000000] Max step 13026
Range [680000000,690000000] Max step 13133
Range [690000000,700000000] Max step 13224
Range [700000000,710000000] Max step 13320
Range [710000000,720000000] Max step 13410
...
</code>```

Notice the range $[240000000,250000000]$ has max steps $7904$.

I modified the program to impose that $7904$ bound.

Running the primality test again for $n = 249999991$, it now takes $\approx \frac {1}{1000}$ second. (compared to $12$ seconds)

Question: Is there a way to calculate a maximum bound on the number of steps required to determine primality?

For example, I could apply the "bound method" (if such a thing exists) to say it will require $123456$ steps to determine primality of

$5311379928167670986895882065524686273295931177$

$27031923199444138200403559860852242739162502265$

$22928566888932948624650101534657933765270723940$

$9519978766587351943831270835393219031728127$

Thanks.

```<code>#include <stdio.h>
#include <stdint.h>
#include <math.h>

uint64_t gcd(uint64_t a, uint64_t b)
{
    return (b != 0) ? gcd(b, a % b) : a;
}

int isPrimeNumber(uint64_t num)
{
    uint64_t i;

    for (i = 2; i <= sqrt(num); i++)
    {
        if (num % i == 0)
        {
            // not prime
            return 0;
        }
    }

    // prime
    return 1;
}

int main(int argc, char* argv)
{
    uint64_t x, y, n;
    uint64_t i, mPrime;
    uint64_t step, maxStep = 0;
    uint64_t begin, end;
    uint64_t interval = 10000000;

    for (i = 0; i < 200; i++)
    {
        begin = i * interval;
        end = begin + interval;

        printf("Range [%llu,%llu] ", begin, end);

        for (n= begin; n < end; n++)
        {
            mPrime = n % 10;

            if (mPrime == 1 || mPrime == 3 || mPrime == 7 || mPrime == 9)
            {
                if (isPrimeNumber(n) != 1)
                {
                    // Start near line x=y
                    x = (n / 2) + 2;
                    y = n - x;

                    step = 0;

                    do {
                        step++;

                        if (gcd(x, y) != 1)
                        {
                            // Lattice point not visible
                            break;
                        }

                        x++;
                        y--;
                    } while (x < n - 1);

                    if (step > maxStep) { maxStep = step; }
                }
            }
        }

        printf("Max step %llu\n", maxStep);
    }

    return 0;
}
</code>```

EDIT:

Many thanks to Ravi's answer!

He was able to determine the test just needs to step thru this interval $[n/2, n/2 + \sqrt{n}/2]$ to determine primality.

I've updated the C program with his suggested interval.

```<code>// Primality Test
// Every n is prime if all lattice points on x+y=n are visible from the origin.

#include <stdio.h>
#include <stdint.h>
#include <math.h>

uint64_t gcd(uint64_t a, uint64_t b)
{
    return (b != 0) ? gcd(b, a % b) : a;
}

int isPrimeNumber(uint64_t n)
{
    if (n == 1) return 0;
    if (n == 2 || n == 3) return 1;
    if (n % 2 == 0) return 0;

    // Start near line x=y.
    uint64_t x = (n / 2) + 2;
    uint64_t y = n - x;
    uint64_t count = sqrt(n) / 2;

    for (uint64_t i = 0; i < count; ++i) {
        // Check lattice point visibility...
        if (gcd(x, y) != 1) return 0;
        x++; y--;
    }

    return 1;
}

int main(int argc, char* argv)
{
    uint64_t n = 1000000007;

    if (isPrimeNumber(n) == 1)
    {
        printf("%llu prime.", n);
    }
    else
    {
        printf("%llu not prime.", n);
    }

    return 0;
}
</code>```


问:
Create a number by multi by $2$ and divide by $3$ (integer part)
How can I create a given postitive integer $N$ by multi by $2$ and divide by $3$ (integer part) ?

(Write a computer program is allowed)
For example:
$$100 = 2*2*2*2*2*2*2*2*2*2*2/3/3/3/3*2*2$$

(The numbers of $2, 3$ need to be minimum)

($/3$ mean $\text{div}\ 3$, eg: $2/3 = 0, 4/3 = 1...$)

Is solving equation

$$\left \lfloor{\frac {2^a}{3^b} }\right \rfloor *2^c = N$$
help ?

If it help then, how to solve
$$\left \lfloor{\frac {2^a}{3^b} }\right \rfloor *2^c = N$$

such that $a > c, a+b+c$ is minimum.

答:
Lessons were learnt from numerical experiments. The basic rules with my last attempt are:

It's easier to work the other way around: define sequences of operations in order of terseness
and see what numbers come out, rather than define a number and find out what its optimal sequence is.

Optimal even numbers always end with (*2) . All odd numbers end with (/3) .

Even numbers can always be constructed from previous ones by multiplying the latter with 2 .

All patterns are generated in order, namely with binary counting and iterpreting bits as:
0 → (*2) , 1 → (/3) .

Interpretation of the binary coding shall end at the leftmost bit up, hence with (/3) as the last operation.

The above rules ( 4 , 5 ) result in even numbers as well; discard them and use rule ( 3 ) instead.

Our method is brute force in every respect; it's greedy with respect to memory as well as computing time.

An enormous amount of duplicates will be found, most of them being 0 . Get rid of them as soon as possible.

Quite fortunately, the first pattern found with odd numbers is always the best. Keep it and discard the rest.

Finding a contiguous sequence of numbers with their optimal patterns has become a matter of sorting now.

Given a maximal number of operations, (large) gaps in the contiguous sequence of numbers are expected to occur soon.

The last attempt (22 bits) was interrupted by the first gap (at 51) in the contiguous sequence of numbers.

Notes. If rule (6) is not taken into account, then sub-optimal results like the one below will be encountered:```
2*2*2*2/3*2/3*2*2*2/3 = 8    <i>instead of</i>   2*2*2 = 8```
Accompanying program sources and outputs are found at:
MSE publications / references 2021.

Most operations are not commutative nor associative. Concerning rule (6) the following is noticed:
$$
  2^{\mbox{ even}}/3 = \mbox{odd} \qquad ; \qquad 2^{\mbox{ odd}}/3 = \mbox{even} = 2^{\mbox{ even}}/3\times 2
$$
Output of my last attempt:
```
  1 = 2*2/3 (#2)
  2 = 2 (#0)
  3 = 2*2*2*2/3*2/3 (#6)
  4 = 2*2 (#1)
  5 = 2*2*2*2/3 (#4)
  6 = 2*2*2*2/3*2/3*2 (#7)
  7 = 2*2*2*2*2*2/3/3 (#7)
  8 = 2*2*2 (#2)
  9 = 2*2*2*2*2*2/3/3*2*2/3 (#10)
 10 = 2*2*2*2/3*2 (#5)
 11 = 2*2*2*2/3*2*2*2/3*2*2/3*2/3 (#13)
 12 = 2*2*2*2/3*2/3*2*2 (#8)
 13 = 2*2*2*2/3*2*2*2/3 (#8)
 14 = 2*2*2*2*2*2/3/3*2 (#8)
 15 = 2*2*2*2/3*2*2*2*2*2/3*2/3*2/3*2/3 (#16)
 16 = 2*2*2*2 (#3)
 17 = 2*2*2*2/3*2*2*2/3*2*2/3 (#11)
 18 = 2*2*2*2*2*2/3/3*2*2/3*2 (#11)
 19 = 2*2*2*2/3*2*2*2/3*2*2/3*2/3*2*2*2/3*2/3 (#19)
 20 = 2*2*2*2/3*2*2 (#6)
 21 = 2*2*2*2*2*2/3 (#6)
 22 = 2*2*2*2/3*2*2*2/3*2*2/3*2/3*2 (#14)
 23 = 2*2*2*2/3*2*2*2*2*2/3*2/3*2/3 (#14)
 24 = 2*2*2*2/3*2/3*2*2*2 (#9)
 25 = 2*2*2*2*2*2*2*2/3*2*2/3*2/3/3 (#14)
 26 = 2*2*2*2/3*2*2*2/3*2 (#9)
 27 = 2*2*2*2/3*2*2*2*2*2*2*2/3/3*2/3*2/3*2*2/3*2/3 (#22)
 28 = 2*2*2*2*2*2/3/3*2*2 (#9)
 29 = 2*2*2*2/3*2*2*2/3*2*2/3*2/3*2*2*2/3 (#17)
 30 = 2*2*2*2/3*2*2*2*2*2/3*2/3*2/3*2/3*2 (#17)
 31 = 2*2*2*2/3*2*2*2*2*2*2*2/3/3*2/3*2/3 (#17)
 32 = 2*2*2*2*2 (#4)
 33 = 2*2*2*2*2*2*2*2/3*2*2/3*2/3/3*2*2/3 (#17)
 34 = 2*2*2*2/3*2*2*2/3*2*2/3*2 (#12)
 35 = 2*2*2*2/3*2*2*2*2*2/3*2/3 (#12)
 36 = 2*2*2*2*2*2/3/3*2*2/3*2*2 (#12)
 37 = 2*2*2*2*2*2/3/3*2*2*2*2/3 (#12)
 38 = 2*2*2*2/3*2*2*2/3*2*2/3*2/3*2*2*2/3*2/3*2 (#20)
 39 = 2*2*2*2/3*2*2*2/3*2*2/3*2/3*2*2*2*2*2/3/3 (#20)
 40 = 2*2*2*2/3*2*2*2 (#7)
 41 = 2*2*2*2/3*2*2*2*2*2*2*2/3/3*2/3*2/3*2*2/3 (#20)
 42 = 2*2*2*2*2*2/3*2 (#7)
 43 = 2*2*2*2*2*2/3/3*2*2*2*2/3*2*2/3*2*2/3*2/3 (#20)
 44 = 2*2*2*2/3*2*2*2/3*2*2/3*2/3*2*2 (#15)
 45 = 2*2*2*2/3*2*2*2/3*2*2/3*2*2*2/3 (#15)
 46 = 2*2*2*2/3*2*2*2*2*2/3*2/3*2/3*2 (#15)
 47 = 2*2*2*2/3*2*2*2*2*2*2*2/3/3*2/3 (#15)
 48 = 2*2*2*2/3*2/3*2*2*2*2 (#10)
 49 = 2*2*2*2*2*2/3/3*2*2*2*2/3*2*2/3 (#15)
 50 = 2*2*2*2*2*2*2*2/3*2*2/3*2/3/3*2 (#15)
```

Especially take notice of the number $27$, which has been mentioned by Christian Blatter in a comment.

Body:
Lessons were learnt from numerical experiments. The basic rules with my last attempt are:

It's easier to work the other way around: define sequences of operations in order of terseness
and see what numbers come out, rather than define a number and find out what its optimal sequence is.

Optimal even numbers always end with (*2) . All odd numbers end with (/3) .

Even numbers can always be constructed from previous ones by multiplying the latter with 2 .

All patterns are generated in order, namely with binary counting and iterpreting bits as:
0 → (*2) , 1 → (/3) .

Interpretation of the binary coding shall end at the leftmost bit up, hence with (/3) as the last operation.

The above rules ( 4 , 5 ) result in even numbers as well; discard them and use rule ( 3 ) instead.

Our method is brute force in every respect; it's greedy with respect to memory as well as computing time.

An enormous amount of duplicates will be found, most of them being 0 . Get rid of them as soon as possible.

Quite fortunately, the first pattern found with odd numbers is always the best. Keep it and discard the rest.

Finding a contiguous sequence of numbers with their optimal patterns has become a matter of sorting now.

Given a maximal number of operations, (large) gaps in the contiguous sequence of numbers are expected to occur soon.

The last attempt (22 bits) was interrupted by the first gap (at 51) in the contiguous sequence of numbers.

Notes. If rule (6) is not taken into account, then sub-optimal results like the one below will be encountered:

2*2*2*2/3*2/3*2*2*2/3 = 8    instead of   2*2*2 = 8

Accompanying program sources and outputs are found at:
MSE publications / references 2021.

Most operations are not commutative nor associative. Concerning rule (6) the following is noticed:
$$
  2^{\mbox{ even}}/3 = \mbox{odd} \qquad ; \qquad 2^{\mbox{ odd}}/3 = \mbox{even} = 2^{\mbox{ even}}/3\times 2
$$
Output of my last attempt:

  1 = 2*2/3 (#2)
  2 = 2 (#0)
  3 = 2*2*2*2/3*2/3 (#6)
  4 = 2*2 (#1)
  5 = 2*2*2*2/3 (#4)
  6 = 2*2*2*2/3*2/3*2 (#7)
  7 = 2*2*2*2*2*2/3/3 (#7)
  8 = 2*2*2 (#2)
  9 = 2*2*2*2*2*2/3/3*2*2/3 (#10)
 10 = 2*2*2*2/3*2 (#5)
 11 = 2*2*2*2/3*2*2*2/3*2*2/3*2/3 (#13)
 12 = 2*2*2*2/3*2/3*2*2 (#8)
 13 = 2*2*2*2/3*2*2*2/3 (#8)
 14 = 2*2*2*2*2*2/3/3*2 (#8)
 15 = 2*2*2*2/3*2*2*2*2*2/3*2/3*2/3*2/3 (#16)
 16 = 2*2*2*2 (#3)
 17 = 2*2*2*2/3*2*2*2/3*2*2/3 (#11)
 18 = 2*2*2*2*2*2/3/3*2*2/3*2 (#11)
 19 = 2*2*2*2/3*2*2*2/3*2*2/3*2/3*2*2*2/3*2/3 (#19)
 20 = 2*2*2*2/3*2*2 (#6)
 21 = 2*2*2*2*2*2/3 (#6)
 22 = 2*2*2*2/3*2*2*2/3*2*2/3*2/3*2 (#14)
 23 = 2*2*2*2/3*2*2*2*2*2/3*2/3*2/3 (#14)
 24 = 2*2*2*2/3*2/3*2*2*2 (#9)
 25 = 2*2*2*2*2*2*2*2/3*2*2/3*2/3/3 (#14)
 26 = 2*2*2*2/3*2*2*2/3*2 (#9)
 27 = 2*2*2*2/3*2*2*2*2*2*2*2/3/3*2/3*2/3*2*2/3*2/3 (#22)
 28 = 2*2*2*2*2*2/3/3*2*2 (#9)
 29 = 2*2*2*2/3*2*2*2/3*2*2/3*2/3*2*2*2/3 (#17)
 30 = 2*2*2*2/3*2*2*2*2*2/3*2/3*2/3*2/3*2 (#17)
 31 = 2*2*2*2/3*2*2*2*2*2*2*2/3/3*2/3*2/3 (#17)
 32 = 2*2*2*2*2 (#4)
 33 = 2*2*2*2*2*2*2*2/3*2*2/3*2/3/3*2*2/3 (#17)
 34 = 2*2*2*2/3*2*2*2/3*2*2/3*2 (#12)
 35 = 2*2*2*2/3*2*2*2*2*2/3*2/3 (#12)
 36 = 2*2*2*2*2*2/3/3*2*2/3*2*2 (#12)
 37 = 2*2*2*2*2*2/3/3*2*2*2*2/3 (#12)
 38 = 2*2*2*2/3*2*2*2/3*2*2/3*2/3*2*2*2/3*2/3*2 (#20)
 39 = 2*2*2*2/3*2*2*2/3*2*2/3*2/3*2*2*2*2*2/3/3 (#20)
 40 = 2*2*2*2/3*2*2*2 (#7)
 41 = 2*2*2*2/3*2*2*2*2*2*2*2/3/3*2/3*2/3*2*2/3 (#20)
 42 = 2*2*2*2*2*2/3*2 (#7)
 43 = 2*2*2*2*2*2/3/3*2*2*2*2/3*2*2/3*2*2/3*2/3 (#20)
 44 = 2*2*2*2/3*2*2*2/3*2*2/3*2/3*2*2 (#15)
 45 = 2*2*2*2/3*2*2*2/3*2*2/3*2*2*2/3 (#15)
 46 = 2*2*2*2/3*2*2*2*2*2/3*2/3*2/3*2 (#15)
 47 = 2*2*2*2/3*2*2*2*2*2*2*2/3/3*2/3 (#15)
 48 = 2*2*2*2/3*2/3*2*2*2*2 (#10)
 49 = 2*2*2*2*2*2/3/3*2*2*2*2/3*2*2/3 (#15)
 50 = 2*2*2*2*2*2*2*2/3*2*2/3*2/3/3*2 (#15)

Especially take notice of the number $27$, which has been mentioned by Christian Blatter in a comment.


问:
Dimension of the Cantor set in the context of cellular automata
I'm reading the book "Cellular Automata and Complexity" by Wolfram. On page 426 he gives a very intuitive picture showing why any cellular automaton state $\{0,1\}^\infty$, can be uniquely identified with a point in the Cantor set:

The above construction applies to any generalized Cantor set, not just the familiar ternary one. The screenshot from wikipedia below shows an intuitive formula giving the fractal dimension of a generalized Cantor set.

So my question: all of this seems to imply that the fractal dimension of $\{0,1\}^\infty$ completely depends on your choice of $\gamma$? Why is the fractal dimension not fixed? The reason I am asking this, is because on the next page Wolfram goes on to say: "The dimension of the Cantor set of all possible configurations for an infinite one-dimensional cellular automaton is 1. A disordered ensemble, in which each possible configuration occurs with equal probability, thus has dimension 1."
How can $\{0,1\}^\infty$ have dimension 1? Does this mean that Wolfram arbitrarily sets $\gamma=0$?

Furthermore he says that irreversible time evolution causes the cellular automaton's state space to contract to an attractor that is a Cantor set with fractal dimension $<1$ and gives an example, where the top line represents $\{0,1\}^\infty$ and the lines below represent the ensemble of cellular automaton configurations on each subsequent time step:

Does this imply that time evolution of the cellular automaton simply alters the value of $\gamma$?

To summarize: the notion of fractal dimension is essential in the theory of dynamical systems, but in the case of cellular automata, it seems that this concept is meaningless since it depends of your choice of $\gamma$. The point of an attractor is that after a certain time, certain configurations in the state space are no longer accessible. However, as the attractor is a Cantor space and each Cantor space (regardless of its dimension) can be identified with $\{0,1\}^\infty$, it seems that even in the infinite time limit, all configurations remain accessible.

答:
Hausdorff dimension is an invariant of a metric space, not a topological space. So it depends on a choice of metric. The Cantor set, as a topological space, embeds into $\mathbb{R}$ in a number of different ways and each of these gives it a different induced metric and so a priori a potentially different Hausdorff dimension.

In other words, there is no such thing as "the fractal dimension of the Cantor set," only the fractal dimension of the Cantor set equipped with a metric, and in particular the fractal dimension of the Cantor set equipped with a particular embedding into $\mathbb{R}$ (or any other metric space).

I have no idea what notion of dimension Wolfram is referring to in the excerpt you quote.

Edit:

The point of an attractor is that after a certain time, certain configurations in the state space are no longer accessible. However, as the attractor is a Cantor space and each Cantor space (regardless of its dimension) can be identified with $\{0,1\}^\infty$, it seems that even in the infinite time limit, all configurations remain accessible.

I don't understand the details of this example but this doesn't follow. The attractor may be a smaller and different Cantor set, which is not the entire original state space.


问:
Prove or disprove: If $H$ is a normal subgroup and cyclic and $G/H$ are cyclic, then $G$ is cyclic.

Prove or disprove: If $H$ is a normal subgroup and cyclic and $G/H$ are cyclic, then $G$ is cyclic.

I don't understand the quotient group $G/H$ being cyclic. What does it mean? From what I understand, $G/H$ is a group that includes groups and not elements, right?

I know that the reverse is right (If $G$ is cyclic then $G/N$ is cyclic.)

How do I disprove it?

答:
This is a fun one!

To address the question in the title, note that:

$D_4$ is not cyclic,

the group $C_4 = \{\rho_0, \rho_{90}, \rho_{180}, \rho_{270} \}$ (the rotation subgroup) is cyclic, and a normal subgroup of $D_4$

$D_4/C_4$ is of order $2$ and is thus cyclic.


问:
How to find the angle formed below a triangle joined by another?
The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&30^{\circ}\\
2.&40^{\circ}\\
3.&20^{\circ}\\
4.&45^{\circ}\\
\end{array}$

I'm stuck with this problem. What exactly should it be done here?. The thing is that I was only able to spot these relationships:

$\angle ABD = \angle ADB$

I'm assuming the intended approach is to draw a line between $DC$. But I can't really find a case of congruence to find the requested angle. Can someone help me here?.

How to solve this using only euclidean geometry?. The thing here is that. I'm stuck. I guess that the approach maybe is to spot an isosceles?. But other than the ones stated there I can't find others.

I've attempted to look if this figure could be inscribed in a circle but it doesn't seem the case. So I don't know what else to do?.

Can someone help me here please?. Please include a drawing in the answer so I can understand what to do.

答: The triangle ABC is equilateral from the givens and, in turn, A, C and D are cyclic with their circumcenter B. Thus, $x$ = $\frac12\angle$ABC = $\frac12\cdot 60^\circ = 30^\circ$.