问:
Can composition of integer polynomial and rational polynomial with a non-integer coefficient result in integer polynomial?
Can we find two polynomials $p(x)$ and $q(x)$, where $p(x)$ is a non-constant monic polynomial over integers and $q(x)$ is a monic polynomial over rationals with at least one non-integer coefficient, such that their composition $p(q(x))$ is a polynomial over integers? If not, how to prove it?

For example let $q(x)=x^2+\frac{1}{2}x+1$ and $p(x)=x^3+a_2x^2+a_1x+a_0$, then $p(q(x))=x^6+\frac{3}{2}x^5+\dots$, so no matter what integers $a_i$ we choose, the resulting polynomial will have a non-integer coefficient. The monic condition is important, since otherwise we could multiply $p(x)$ with such integer that would guarantee all coefficients to be integers. I've tried to look at the coefficient in composition for general polynomials, which I believe should follow this formula:

\begin{align}
[x^r]p(q(x))=\sum_{k_1+2k_2+\dots+mk_m=r}\sum_{k_0=0}^{n-(k_1+\dots+k_m)}\binom{k_0+k_1+\dots+k_m}{k_0,k_1,\dots,k_m}a_{k_0+k_1+\dots+k_m}\left(\prod_{j=0}^{m}b_j^{k_j}\right)
\end{align}
(here $a_i$ and $b_i$ are the coefficients of $p(x)$ and $q(x)$ with degrees $n$ and $m$, respectively).
However it is not at all clear on which coefficient to focus to prove it will give the non-integer number.

This arose when trying to solve the Infinitely many solutions leads to existence of a polynomial, but it seems interesting enough by itself.

答:
One can show this with some elementary properties of algebraic integers:

Fact. If $p \in \mathbb C[x]$ and $q \in \mathbb Q[x]$ are monic, $p(0) \in \mathbb Z$, $\deg p, \deg q > 0$ and $p \circ q \in \mathbb Z[x]$, then $q \in \mathbb Z[x]$.

Proof: We must show that all roots of $q$ are algebraic integers. Let $\alpha$ be a root of $q$. Then $p(q(\alpha)) - p(0) = 0$. The assumptions imply that $r(x) = p(q(x)) - p(0)$ is a monic integer polynomial. Because $\alpha$ is a root of $r$, it is an algebraic integer.


问:
A problem on inverse trigonometric function
How to sum the following series: $$S=\cot^{-1}2+\cot^{-1}8+\cot^{-1}18+\cot^{-1}32+\cdots+\cot^{-1}∞$$ My attempt better to say a solution was,\begin{align}S &=\sum_{n=1}^{∞}\cot^{-1}2n^2\\
&=\sum_{n=1}^{∞}\tan^{-1}\frac{1}{2n^2}\\
&=\sum_{n=1}^{∞}\tan^{-1}\frac{2}{4n^2}\\
&=\sum_{n=1}^{∞}\tan^{-1}\frac{2}{1+4n^2-1}\\
&=\sum_{n=1}^{∞}\tan^{-1}\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}\\
&=\sum_{n=1}^{∞}[\tan^{-1}(2n+1)-\tan^{-1}(2n-1)]\\
&=\tan^{-1}∞-\tan^{-1}1\\
&=\frac{π}{2}-\frac{π}{4}\\
&=\frac{π}{4}\end{align} Now one thing I have to confess that I knew this method previously,so I did and I don't understand from where this idea suddenly comes about telescoping? Can someone help me explain with any different approach is added as an answer. Thanks for your attention.

答:
The series transformations might seem at first glance rather arbitrary. But there are some nice aspects behind  the curtain which can be sometimes conveniently used.

Let's assume we want to calculate
\begin{align*}
\color{blue}{\sum_{n=1}^\infty \tan^{-1}\left(\frac{1}{2n^2}\right)}\tag{1}
\end{align*}

We know that telescoping is one of the techniques which can be sometimes used to tackle such problems. But is it plausible in this case?

Addition theorem:  Here we might recall the addition theorem for the tangent function and its inverse
\begin{align*}
\tan\left(\alpha\pm\beta\right)&=\frac{\tan \alpha \pm\tan \beta}{1\mp\tan\alpha \tan\beta}\tag{2.a}\\
\alpha\pm\beta &=\tan^{-1}\left(\frac{\tan \alpha \pm\tan \beta}{1\mp\tan\alpha \tan\beta}\right)\tag{2.b}
\end{align*}

Letting $x:=\tan \alpha$ and $y:=\tan\beta$ we obtain from (2.b) the identity
\begin{align*}
\color{blue}{\tan^{-1} x - \tan^{-1} y  = \tan^{-1}\left(\frac{x- y}{1+ xy}\right)}\tag{2.c}
\end{align*}

The point is, if we recall one of (2.a), (2.b) or (2.c) the telescoping approach becomes interesting, especially in it's most convenient representation (2.c).

Transformation: When looking at what we have in (1) the challenge is to find a transformation from
\begin{align*}
\tan^{-1}\frac{1}{2n^2}\quad\to\quad (2.c) \quad\to  \quad\tan^{-1} x - \tan^{-1} y
\end{align*}

We put the focus on the fraction $\frac{x-y}{1+xy}$ at (2.c) and obtain
\begin{align*}
\color{blue}{\frac{1}{2n^2}}&=\frac{1}{1+\left(2n^2-1\right)}\tag{3}\\
&=\frac{2}{1+\left(4n^2-1\right)}\tag{4}\\
&=\frac{2}{1+\left(2n+1\right)\left(2n-1\right)}\tag{5}\\
&\,\,\color{blue}{=\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}}\tag{6}
\end{align*}

Comment:

In (3) we write the denominator as $1+\left(2n^2-1\right)$ since we want as in (2.c)  something like $1+xy$. We see we  have $1+\left(a^2-b^2\right)=1+(a+b)(a-b)$. The factor $2$ in $\left(2n^2+1\right)$ is not that convenient, but we can cope with it by expanding by a factor $2$ to get $4n^2-1=(2n+1)(2n-1)$ which is done in (4).

In (5) we have the denominator according to that in (2.c) and we are lucky, since the numerator $2$ can be written as $2=(2n+1)-(2n-1)$ to obtain in (6) the wanted representation as in (2.c).

Series calculation: Now we are well prepared to tackle the series by a telescoping approach. We obtain
\begin{align*}
\color{blue}{\sum_{n=1}^\infty}\color{blue}{\tan^{-1}\frac{1}{2n^2}}
&=\sum_{n=1}^{\infty}\tan^{-1}\left(\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}\right)\tag{$\to (6)$}\\
&=\sum_{n=1}^{\infty}\left[\tan^{-1}\left(2n+1\right)-\tan^{-1}\left(2n-1\right)\right]\tag{$\to (2.c)$}\\
&=\lim_{N\to\infty}\sum_{n=1}^{N}\left[\tan^{-1}\left(2n+1\right)-\tan^{-1}\left(2n-1\right)\right]\\
&=\lim_{N\to\infty}\left[\tan^{-1}\left(2N+1\right)-\tan^{-1}\left(1\right)\right]\\
&=\lim_{N\to\infty}\tan^{-1}\left(2N+1\right)-\lim_{N\to\infty}\tan^{-1}\left(1\right)\\
&=\frac{\pi}{2}-\frac{\pi}{4}\\
&\,\,\color{blue}{=\frac{\pi}{4}}
\end{align*}

ContentLicense: CC BY-SA 4.0


问:
Ask sample variance equation?
$$(n-1) S_x^2 = \Sigma_{i=1}^n (X_i - \bar{X})^2$$

Is this equation always true? Or this relationship only holds for some specific distributions, like normal distribution?

答:
Relationship is general as long as variance is finite.

Derivation $(n-1)E(S_x^2)=\sum(E(X_i^2)-2E(X_i\bar{X})+E(\bar{X}^2)$.

$E(X_i^2)=\sigma^2+m^2$, sum is $n(\sigma^2+m^2)$

$\sum E(X_i\bar{X})=nE(\bar{X}^2)$.

$E(\bar{X}^2)=\frac{1}{n^2}(n(\sigma^2+m^2)+n(n-1)m^2)$, sum is $\sigma^2+nm^2$

Combine terms and $(n-1)E(S_x^2)=(n-1)\sigma^2$

Title: Ask sample variance equation?


问:
Equivalence between two definitions of shuffles
The usual definition of a $(p,q)$-shuffle is a permutation $\sigma$ of the set $\{1,\dots, p+q\}$ such that $\sigma(1)<\cdots<\sigma(p)$ and $\sigma(p+1)<\cdots<\sigma(p+q)$. The signature of a $(p,q)$-shuffle s the signature of the permutation.

However, there are some equivalent definitions that I have found here. I am interested in the definition of a $(p,q)$-shuffle as a monotone map of partially ordered sets $[p+q]\to [p]\times[q]$ where $[p]=\{0<1<\cdots<p\}$. There is an interpretation of this as paths on a square grid of side $p+q$ and the signature is the number of $1\times 1$ squares below the path. This interpretation can be found in Section 1.1. here and in Hatcher's Algebraic Topology (p. 277-278).

The alternative definition is helpful for me when I want to triangulate products of simplies, but I can compute the signature more quickly using the first definition (for instance with the formula in page 6 here).

Therefore I would like to know how are these two notions equivalent.

Given the partially ordered set $[p+q]$, there are exactly $p+q$ inequality signs "$<$" and they are mapped to other $<$ signs in $[p]\times [q]$, so I have thought that I could define a $(p,q)$-shuffle permutation by labeling the signs and define the permutation this way. But I am not sure about how to systematically define this permutation. The main problem is that on $[p]\times [q]$ there are more $<$ signs and I don't know how I should label them.

For instance, if $p=q=1$, I have $[p+q]=\{0<1<2\}$ and $[p]\times[q]$ is partially ordered by alphabetical order, i.e. we have two paths of inequalities $(0,0)<(0,1)<(1,1)$ and $(0,0)<(1,0)<(1,1)$. I can send $0<1<2$ to any these paths, and each of them would represent one of the two $(1,1)$-shuffles which are the identity and the transposition of two elements.

I know that I have to send the identity to the map whose image is $(0,0)<(1,0)<(1,1)$ because they both have $0$ signature. But I don't know how would I know a priori that this is the image of the identity permutation. This is important, because when there are more than two shuffles, there will be several shuffles with the same signature, so that information is not enough.

答:
The partial order on $[p]\times[q]$ is the product partial order, not the lexicographic order: $\langle a,b\rangle\le\langle c,d\rangle$ iff $a\le c$ and $b\le d$.

Let $\sigma$ be a $(p,q)$-shuffle. It corresponds to a path from $\langle 0,0\rangle$ to $\langle p,q\rangle$ consisting of $p$ right-steps and $q$ up-steps. If $1\le k\le p$, the $k$-th right-step is the $\sigma(k)$-th step of the path as a whole. If $p+1\le k\le p+q$, the $(k-p)$-th up-step is the $\sigma(k)$-th step of the path as a whole. This is perhaps best illustrated by an example.

$2351467$ is a $(3,4)$-shuffle. It corresponds to a path from $\langle 0,0\rangle$ to $\langle 3,4\rangle$ consisting of $3$ right-steps and $4$ up-steps. The right-steps are the $2$nd, $3$rd, and $5$th steps of the path, and the up-steps are the $1$st, $4$th, $6$th, and $7$th steps of the path. This path corresponds to the maximal chain $$\langle 0,0\rangle\overset{u}<\langle 0,1\rangle\overset{r}<\langle 1,1\rangle\overset{r}<\langle 2,1\rangle\overset{u}<\langle 2,2\rangle\overset{r}<\langle 3,2\rangle\overset{u}<\langle 3,3\rangle<\overset{u}\langle 3,4\rangle$$ in $[3]\times[4]$, in which I’ve marked the right-steps $r$ and the up-steps $u$.


问:
Elementary proof of Power Rule For differentiation $f'(x)=rx^{r-1}$ for $f(x)=x^r$
I could  not find any elementary proof of the Power Rule for differentiation:

Given $x,r \in R$ ,$x>0$ and a function $f(x)=x^r$, then its derivative is $f'(x)=rx^{r-1}$

Defintion:Let a sequence of rational numbers {$r_n$} tend to $r$ then $x^r=\lim_{r_n \to r}x^{r_n}$

All the proofs I had seen utilized the derivative of $logx$ but I dont think this is  elementary because one has to show that for $e=\text{Euler's number}$ , $t \in R$and $n \in N$ $\lim_{n \to \infty}(1+\frac{1}{n})^n=\lim_{t \to 0}(1+t)^\frac{1}{t}=e$ which can be proved using the Power Rule

Ofcourse the proof for the Power Rule is straight forward if one is familiar with the result of Power Rule if $r$ is a rational number and theorem  regarding  derivative of a sequence of derivatives that are uniformly convergent.

答:
For real exponents in general, I don't think you can avoid the exponential function and the logarithm because the definition of $x^r$ tends to be $\exp(r\log x)$.

However, for integer exponents there are a number of simple proofs. Here is a lesser-known one. Let $f:\mathbb{R}\mapsto\mathbb{R}$ be the function defined by $f(x)=x^n$ for all $x$, with $n \in \mathbb{N}$. We find that
\begin{align}
f'(a) &= \lim_{x \to a}\frac{f(x)-f(a)}{x-a} \\[6pt]
&= \lim_{x \to a}\frac{x^n-a^n}{x-a} \\[6pt]
&= \lim_{x \to a}\frac{(x-a)(x^{n-1}+x^{n-2}a+x^{n-3}a^2+\ldots+xa^{n-2}+a^{n-1})}{x-a} \\[6pt]
&= \lim_{x \to a}x^{n-1}+x^{n-2}a+x^{n-3}a^2+\ldots+xa^{n-2}+a^{n-1} \\[6pt]
&= a^{n-1}+a^{n-2}a+a^{n-3}a^2+\ldots+aa^{n-2}+a^{n-1}\\[6pt]
&= \underbrace{a^{n-1}+a^{n-1}+a^{n-1}+\ldots+a^{n-1}+a^{n-1}}_{\text{$n$ terms}}\\[6pt]
&= na^{n-1}\blacksquare
\end{align}
This proof can be extended to negative integer exponents by writing $x^{-n}$ as $1/x^n$ and using the quotient rule.


问:
Mixed Moment of Weibull distribution finite
I need help with the following please.
let X be weibull distributed with the density(x and both parameters are positive and nonzero):
$$f(x)=\frac{\beta}{\alpha}x^{\beta-1}exp(-\frac{x^{\beta}}{\alpha})1(x>0)$$
I want to show that for $p,q\geq0$
$$E[X^p |log(X)|^q]<\infty$$

I already tried to apply Jensen but I think the function is not concave, I also tried splitting the integral to get rid of the absolute value und use some inequalities but I just get lost somewhere. In a textbook I read that this holds since weibull distrib hast an exponentially small tail but I don’t understand this. Hope there is a different way to solve this problem. Thx in advance

答:
For $M>1$,
$$\int_{M}^{\infty}x^p|log(x)|^qf(x)dx$$
$$=\int_{M}^{\infty}x^p(log(x))^q \frac{\beta}{\alpha}x^{\beta-1}exp(-\frac{x^{\beta}}{\alpha})dx$$
Now,
$$\lim_{x \rightarrow\infty}\frac{x^p(log(x))^q \frac{\beta}{\alpha}x^{\beta-1}exp(-\frac{x^{\beta}}{\alpha})}
{x^{p+1}\frac{\beta}{\alpha}x^{\beta-1}exp(-\frac{x^{\beta}}{\alpha})}=0$$

Therefore, for sufficiently large $M$, we have

$$\int_{M}^{\infty}x^p(log(x))^q \frac{\beta}{\alpha}x^{\beta-1}exp(-\frac{x^{\beta}}{\alpha})dx \le
\int_{M}^{\infty}x^{p+1}\frac{\beta}{\alpha}x^{\beta-1}exp(-\frac{x^{\beta}}{\alpha})dx$$
$$\le
\int_{0}^{\infty}x^{p+1}\frac{\beta}{\alpha}x^{\beta-1}exp(-\frac{x^{\beta}}{\alpha})dx=\alpha^{\frac{1+p}{\beta}}\Gamma\left(\frac{\beta+p+1}{\beta}\right)$$

Similarly, for $\epsilon>0$ sufficiently close to 0,

$$\int_{0}^{\epsilon}x^p(-log(x))^q \frac{\beta}{\alpha}x^{\beta-1}exp(-\frac{x^{\beta}}{\alpha})dx \le
\int_{0}^{\epsilon}\frac{\beta}{\alpha}x^{\beta-1}exp(-\frac{x^{\beta}}{\alpha})dx$$
$$\le
\int_{0}^{\infty}\frac{\beta}{\alpha}x^{\beta-1}exp(-\frac{x^{\beta}}{\alpha})dx=1$$

because

$$\lim_{x \rightarrow0^+}\frac{x^p(-log(x))^q \frac{\beta}{\alpha}x^{\beta-1}exp(-\frac{x^{\beta}}{\alpha})}
{\frac{\beta}{\alpha}x^{\beta-1}exp(-\frac{x^{\beta}}{\alpha})}=0$$

Then, just break up the integral into 3 pieces to show the whole integral is finite.


问:
What construction gives the syntax of an algebraic structure?
Let's say I have some algebraic structure $M$ (let's use a monoid for simplicity). I have a specific subset $S \subset M$. The free monoid on $S$ encodes what elements of $M$ I can get by combining elements of $S$.

I'm interested not only on what elements of $M$ I can get, but also on how I can get them. In this case the set of binary trees with elements of $S$ at their leaves.

So, If the Free construction gives me the semantics, what construction gives me the syntax?

答:
I may disagree on the fact that the free construction gives semantics. Let me try to exlain why. Consider $\mathcal{T}$ the theory of monoids. In my understanding, the free construction gives the syntax. Given a set $X$, note by $\mathcal{T}(X)$ the free object on $X$. It is be the set of expressions you can form using elements of $X$ and the language of monoids, note that it contains also the empty expression. So in the end it sums up to expressions $x_1\cdot x_2\cdots x_n$, hence $\mathcal{T}(X) \simeq \coprod_{n\in \mathbb{N}}X^n$, the empty expression beeing represented by $X^0$, a set with one element. The semantics is the algebra structure : a $\mathcal{T}$-algebra on the set $X$ is the data of a map of sets $a : \mathcal{T}(X)\to X$ that interprets purely syntactic expressions, i.e. elements of $\mathcal{T}(X)$ as elements of $X$.

The data of an $\mathcal{T}$-algebra is equivalent to the data of monoid, since restricting the strucutre map $a$ to $X\times X \subset \mathcal{T}(X)$, gives the multiplication $m:X\times X \to X$, the associativity and unitarity follows form the algebra axioms, and given an associative monoid, we can build the structure map using recursion on the length of the expressions, since expressions of length one are just elements of the monoid, and the empty expression should be sent to the identity of the monoid under the axioms $a$ has to follow.

If I have a monoid, e.g. $(\mathbb{N},\cdot,1)$, I can interpret syntactic expressions in the language of monoids like $2\cdot 3\cdot 7$ (which are in correspondence with the free monoid on $\mathbb{N}$, namely $\mathcal{T}(\mathbb{N})$), as $42$ using the multiplication, and the recursive algorithm.

So in your example, the free monoid on $S$ is just the set of expressions you can form using elements of $S$. Then using the algebra structure on $M$ you can see the image of $\mathcal{T}(S) \subset \mathcal{T}(M)$ under the structure map $f:\mathcal{T}(M) \to M$ as the set of elements in $M$ you can get combining the elements of $S$. Lets call this sub set $\mathcal{I}(S)$ (note that is a monoid since the empty expression is interpreted as the unit of the monoid), so it is the monoid generated by $S$ in $M$.

If you want the set of the expressions that gave you $\mathcal{I}(S)$, then you are looking for the set $f^{-1}(\mathcal{I}(S))$. Note here that since we are considering the theory of monoids, the set $\mathcal{T}(X)$ is the set of unparenthesised expressions since the theory is associative. If you want the set of expressions with parenthesis, or equivalently the set of planar binary trees with leaves decorated by elements of the set $X$, you have to consider the theory $\mathcal{T}'$ which forgets the associativity axiom as Kevin Arlin's answer points out.

Hope this answers somehow your question.


问:
Prove that the foundamental solution of the heat equation is an approximated identity for convolution.
I have some problem to prove that $$\Gamma_D(x,t):= \left(\frac{1}{4\pi Dt}\right)^{\frac{n}{2}}e^{-\frac{|x^2|}{4Dt}}$$

(where $D>0$ is constant and $n$ is the dimension of $\mathbb{R}^n$) satisfies

$$\lim_{t\to 0^{+}}\int_{\mathbb{R}^{n}}\Gamma_D(x,t)\phi(x)dx=\phi(0)$$

for every $\phi$ continous that for $a,b\geq0$ satisfies $|\phi(x)|\leq a e^{b|x|^2}$.

I've tried to split the integral using that for every $t\neq0$ the integral of $\Gamma_D$ is $1$.  So as $\phi$ is coninous in $0$ there is a ball $B(0,R_{\epsilon})$ such that $|\phi(x)-\phi(0)|<\epsilon$ so $$\phi(0)=\int_{\mathbb{R}^{n}}\Gamma_D(x,t)[\phi(0)+\phi(x)-\phi(x)]dx,$$ so it suffices to show that $$\int_{\mathbb{R}^{n}}\Gamma_D(x,t)[\phi(0)-\phi(x)]\to 0$$ as $t\to 0^{+}$.

By taking the complement of the ball $B^c$ on has

$$\int_{\mathbb{R}^{n}}\Gamma_D(x,t)(\phi(0)-\phi(x))dx\leq\epsilon+\int_{B^c}|\Gamma_D(x,t)||\phi(0)-\phi(x)|dx.$$

How can I show that the second integral is zero as $t\to0^{+}$?   Any suggestion, solution, advice will be welcome

答:
Since $|\phi(x)| \leqslant ae^{b|x|^2}$, we have $|\phi(x) - \phi(0)| \leqslant  a + ae^{b|x|^2}$ and

$$\left|\int_{B^c}\Gamma_D(x,t)[\phi(0)-\phi(x)]\,dx\right| \leqslant \frac{1}{(4\pi D)^{n/2}} \frac{1}{t^{n/2}}\left(a\int_{B^c}e^{-\frac{|x|^2}{4Dt}}\,dx+ a\int_{B^c}e^{-\left(\frac{1}{4Dt}-b\right)|x|^2}\,dx\right)$$

For $0 < t < \frac{1}{8Db}$ we have $-(\frac{1}{4Dt}-b) <  -\frac{1}{8Dt}$. In that case  the second integral on the RHS converges, and we have the bound

$$\tag{*}\left|\int_{B^c}\Gamma_D(x,t)[\phi(0)-\phi(x)]\,dx\right| \leqslant \frac{a}{(4\pi D)^{n/2}} \frac{1}{t^{n/2}}\int_{B^c}e^{-\frac{|x|^2}{4Dt}}\,dx+ \frac{a}{(4\pi D)^{n/2}} \frac{1}{t^{n/2}}\int_{B^c}e^{-\frac{|x|^2}{8Dt}}\,dx$$

We can now show that the RHS of (*) converges to $0$ as $t \to 0+$. Both terms on the RHS are similar and upon changing to polar coordinates with $r =|x|$, they assume the form

$$\tag{**}C t^{-n/2}\int_{R_\epsilon}^\infty r^{n-1}e^{-\alpha r^2/t} \, dr,$$

where integration over the angular coordinates has been absorbed into the constant $C$.

For any integer $m$, we have $e^{\alpha r^2/t} > \frac{(\alpha r^2/t)^m}{m! }$ and

$$t^{-n/2}r^{n-1}e^{-\alpha r^2/t}  < m!\alpha^{-m}t^{m-n/2}r^{-2m+n-1}$$

If we choose $m > n/2$, then we have $-2m+n-1 <-1$ and the integral on the RHS of (**) is convergent.  We also have $m - n/2 > 0$, and consequently

$$\lim_{t \to 0+} C t^{-n/2}\int_{R_\epsilon}^\infty r^{n-1}e^{-\alpha r^2/t} \, dr \leqslant \lim_{t \to 0+} t^{m-n/2}\left(Cm!\alpha^{-m}\right)\int_{R_\epsilon}^\infty r^{-2m+n-1} \, dr = 0$$


问:
For what integer values of $k$ does $a^{2}+b^{2}=kc^{2}$ have positive integer solutions?
The obvious case of $k=1$ clearly has solutions, for example, $(3,4,5)$ and $(5,12,13)$, among others. Trivially, it has infinitely many solutions by multiplying through by any number. It has infinitely many solutions aside from those that can be parametrized by $(2nm,n^{2}-m^{2},n^{2}+m^{2})$ for all $n,m\in\mathbb{Z}$. This is a well-known result.

The case of $k=2$ also has infinitely many solutions, among them $(1,7,5)$. As in the previous example I can simply multiply through by any number and get another valid triple. At some point just after Christmas I had worked out a parametrization (using the method of finding a rational point on the corresponding circle), but I cannot recall it off the top of my head. It was not much more complicated than for the case of $k=1$, however, and I sincerely doubt that I was the first one to have found it.

The case of $k=3$ is one for which I have been unable to find any solutions, but I cannot think of a proof that it has no solutions. I cannot find any rational points on the circle $x^{2}+y^{2}=3$, but that does not mean that those points do not exist.

The case of $k=4$ has solutions as well. Given $a^{2}+b^{2}=4c^{2}$, merely allow $d=2c$ and we return to the case of $k=1$ with the question of finding triples that satisfy $a^{2}+b^{2}=d^{2}$. This shows that the original question that I posed is equivalent to merely asking which squarefree integers $k$ have solutions.

I have similarly found solutions in the case of $k=5$, for example, $(11,2,5)$. Try as I might, I have found nothing for the cases of $k=6$ or $k=7$. The cases of $k=8$ and $k=9$ have solutions by a similar argument as the previous paragraph.

Initially, my guess was that there no solutions iff $k\cong3\mod4$ given my inability to find solutions in the $k=3$ and $k=7$ cases, until I decided to look at the $k=6$ case as well and found that I was similarly stumped. I feel like there's something else obvious that I'm overlooking, but whether it's solutions for the cases I can't find a solution for or some other connection between these numbers, I don't have a clue.

My apologies if this has been asked before, I couldn't find anything when I did a google search.

答:
For $a^2+b^2=kc^2$ to have integers solutions, $k$ can be the square of any  prime factor of a composite hypotenuse of a Pythagorean triple. Among the  values here $\quad k\in\{4, 9, 16, 25,\cdots \}\quad $ suggests that $k$ can be any prime squared.

For example
$$ 12^2+16^2=20^2=4^2\cdot 5^2
\implies a=12,\space b=16,\space k=16,\space c=5$$
We find that $\quad p\equiv 3\mod 4\quad $ is allowed. I'm not sure what the earlier answer meant.

$$21^2+28^2=35^2=7^2\cdot 5^2
\implies a=21,\space  b=28,\space  k=49,\space c=5 $$

If we examine this further, it appears that $k$ can be any integer squared: $\quad k=j^2,\space  j\in\mathbb{N}\quad$ because the extended Pythagorean Theorem states $(\space jA^2+(\space jB)^2=(\space jC^2)\quad$ and we have seen the $\space 7\times(3,4,5)\space $ combinarion  as being perfectly valid above.


问:
Is there any simple way to find out all divisors of $n+1$ under the given conditions?
Aussuming I have given a really large number $n \in \mathbb{N}$ (let's say, $10^{80} \le n \le 10^{100}$) and I know all the divisors of every number $x=0,1,\ldots,n-1$.

Is there any simple, universal and not too time-consuming way to find out all divisors of $n$?

答:
Apart from electronically, no.

Knowing the divisors of numbers less than $n$ isn't really going to help.  You are still going to have to test divisors, which means that you might have to search for the first multiple of $p$ under the required number.

On the other hand, there are some relatively fast proggies that will factorise numbers in these ranges.  'factor for OS/2 and Windows etc'.  They're quite good.

One, could consider finding the factors, say of 121, knowing the factors of 2 to 120.  Even for simple tests, one has to search back to find the previous multiple of 7, or 11, or 13, etc.

On the other hand, if you are trying to factorise a mob of numbers that follow each other, such as 14400, 14401, 14402, 14403, 14404, 14405, 14406, ..., 14409, it is really useful to do things like pick out the multiples of 2, 3, 5, 7, etc, because if 7 divides 14406, it does not divide any of the others.  But this sort of test supposes that you have access to all of the primes less than the square root.

ContentLicense: CC BY-SA 3.0


答:
We have $$\cot(a+b)=\frac{1-\tan a\tan b}{\tan a+\tan b}=\frac{\cot a\cot b-1}{\cot a+\cot b},$$

which allows us to find the cotangent of the first partial sums

$$2,\frac{2\cdot8-1}{2+8}=\frac32,\frac{\dfrac32\cdot18-1}{\dfrac32+18}=\frac43,\cdots$$

Hence we can easily hypothetize

$$\cot(S_n)=\frac{n+1}n$$ and check the induction

$$\cot(S_{n+1})=\cot(\cot^{-1}(S_n)+\cot^{-1}(2(n+1)^2)=\frac{\dfrac{n+1}{n}2(n+1)^2-1}{\dfrac{n+1}{n}+2(n+1)^2}=\frac{n+2}{n+1}.$$

Finally

$$\cot(S_\infty)=1.$$

Title: A problem on inverse trigonometric function

Body:
How to sum the following series: $$S=\cot^{-1}2+\cot^{-1}8+\cot^{-1}18+\cot^{-1}32+\cdots+\cot^{-1}∞$$ My attempt better to say a solution was,\begin{align}S &=\sum_{n=1}^{∞}\cot^{-1}2n^2\\
&=\sum_{n=1}^{∞}\tan^{-1}\frac{1}{2n^2}\\
&=\sum_{n=1}^{∞}\tan^{-1}\frac{2}{4n^2}\\
&=\sum_{n=1}^{∞}\tan^{-1}\frac{2}{1+4n^2-1}\\
&=\sum_{n=1}^{∞}\tan^{-1}\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}\\
&=\sum_{n=1}^{∞}[\tan^{-1}(2n+1)-\tan^{-1}(2n-1)]\\
&=\tan^{-1}∞-\tan^{-1}1\\
&=\frac{π}{2}-\frac{π}{4}\\
&=\frac{π}{4}\end{align} Now one thing I have to confess that I knew this method previously,so I did and I don't understand from where this idea suddenly comes about telescoping? Can someone help me explain with any different approach is added as an answer. Thanks for your attention.


问:
Find the derivative of $y=(2x-1)(2x+1)$
Find the derivative of $y=(2x-1)(2x+1)$.

The result I got is 8x. However, my textbook answer is $4/(2x+1)^2$. I wonder what I did wrong?

I am using a different form of finding the derivatives and would prefer this way. Here are my steps:

答: The textbook's answer is the derivative of $$y = \frac{2x-1}{2x+1}.$$  Your result is the derivative of $$y = (2x-1)(2x+1).$$  Which one is correct depends on the intention of the problem's author, i.e., whether they intended to write a division symbol or not.

Title: Find the derivative of $y=(2x-1)(2x+1)$


问:
prove following combinatorial identity
I have been trying to prove following thing. With no success

prove that for $n>k\geq 2$ $$\sum_{i=0}^k nP_{(k+i)} . \binom{k}{k-i} kP_{(k-i)} = (nP{k})^2$$ where $nPk = \frac{n!}{(n-k)!}$

any useful hint is helpful.

答:
Here's a combinatorial interpretation. We want to count the number of ways to pick two $k$-tuples $A=(a_1, a_2, \ldots, a_k)$ and $B=(b_1, b_2, \ldots, b_k)$, such that all of these elements are from $[n]$, all of the $a$s are distinct, and all of the $b$s are distinct, though we can have $a_h=b_j$ for some $h$ and $j$.

The right hand side counts this explicitly. To see how the left hand side counts this, let $i$ denote the number of elements that $B$ has that it does not share with $A$. Note that $i$ can go from $0$ to $n$. The factor $nP_{(k+i)}$ counts the number of $(k+i)$-tuples we can form from elements of $[n]$ without repeating any. The first $k$ of those elements, in that order, form $A$. The remaining $i$ elements will be elements of $B$. Since those $i$ elements are the ones that are unique to $B$, the other $k-i$ elements of $B$ will be from $A$, and the factor $kP_{(k-i)}$ counts how many ways we can pick those elements in some order. Finally, the factor ${k\choose k-i}$ gives a set of indices from $k$ where we can place the $k-i$ elements from $A$ we picked, in the order we picked them in, and the other $i$ elements unique to $B$ are placed in the remaining $i$ places. This gives the desired $A$ and $B$.

For example, consider $n=9$, $k=5$, and $i=2$. The factor $9P7$ gives us a $7$-tuple using elements from $[9]$ without repetition, such as $(3, 9, 8, 1, 4, 2, 6)$. The first $5$ of those elements gives us $A=(3,9,8,1,4)$, and we will use $(2, 6)$ in $B$. The factor $5P3$ gives us a $3$-tuple of elements from $A$, such as $(4, 9, 1)$. We then use the factor ${5\choose 3}$ to pick a subset such as $\{1, 2, 4\}$, which we use as a set of indices to place the elements $4$, $9$, and $1$ in $B$, and place $2$ and $6$ in the remaining places. That gives us $B=(4, 9, 2, 1, 6)$, and we have the $A$ and $B$ that are counted by the right hand side.

Title: prove following combinatorial identity


问:
Find an unbiased estimator for Poisson distribution
I am starting studying statistics and I have a task to find unestimated bias for Poisson distribution.

The task sounds like this: "Find unbiased estimator $\lambda_{s} = \lambda^{3}$ using sample $X_{1}...X_{n}$ from $Pois(\lambda)$ distribution".

I know that $\lambda$ is variance  in Poisson distribution and that variance unbiased estimator is $$\bar{x}=\frac{1}{N}\sum_{i=1}^{N}X_{i}$$
But I see I need an answer correlated with $\lambda_{s}$, and I don't know how to use Poisson distribution formula to get my $X_{1}...X_{n}$ and then unbiased estimator.

Do you have any ideas?

P.S. Sorry, right problem statement is with UNbiased estimator...

答: First of all, notice that if $X\sim \text{Poisson}(\lambda)$ then $$E(X)=\lambda$$ $$E(X^2)=\lambda^2 + \lambda$$ $$E(X^3)=\lambda^3 +3\lambda^2 +\lambda$$ This implies $$E(X^3-3X^2+2X)=\lambda^3$$ That being said, define $$Y_{N}=\frac{1}{N}\sum_{i=1}^{N}\Big[X_i^3-3X_i^2+2X_i\Big]$$ Using linearity of expectation we have $E(Y_{N})=\lambda^3$ so $Y_{N}$ is an unbiased estimator for $\lambda^3$.


问:
calculate the area of a domain
Given a domain
$$
D = \left\{\left(x,y\right) \left\vert\vphantom{\large A^{a}}\right.
\left(x^{2} + y^{2}\right)^{2} \leq 2\left(x^{2} - y^{2}\right),\
x^{2} + y^{2} \geq 1\right\}
$$
how can I calculate the area of $D$ and draw it ?

答:
The curves in polar coordinates are
$$r^2=2\cos 2t;\;r=1$$
The $S$ area is four times the area in the picture below, given by

$$\frac{1}{2}\int_0^{\pi/6} (2 \cos 2t -1)\,dt$$
Should be $S=\sqrt{3}-\frac{\pi }{3}\approx 0.684853$

$$...$$

Title: calculate the area of a domain


问:
D'Alemberts formula with both initial value and derivative equal to 0
Trying to solve a PDE problem consisting of these equations:

$$ u_{tt} - c^2 u_{xx} = 0 \quad \quad(0 < x < L, t > 0)$$
$$u(0,t), \quad u(L,t)=Bsin(\omega_{0}t) \quad (t >0)$$
$$u(x,0) = 0, \quad u_{t}(x,0) = 0 \quad (0 < x < L)$$

$C,L,B,\omega_{0} > 0$ are constants

I know that the general solution to the wave equation is
$$u(x,t) = \phi(x+ct)+\psi(x-ct)$$

And I have d'Alembert's forumla. The problem is that since these initial values are on the standard form, d'Alembert's should apply, right? But $u(x,0) = u_{t}(x,0) = 0$, giving that u(x,t) = 0 as well, which cannot be right. I can't even get to the point where the second row of I.V is relevant (u(0,t) and u(L,t)).

Is there something fundamental I am missing with the wave equation and d'Alembert? Is there a reason why d'Alembert would not apply here? Would appreciate any help on those questions so that I can try and actually solve the question by myself.

答:
The d'Alembert formula solves the Cauchy problem for wave equation, but you have first mixed problem, and you should use another methods such as Laplace transform or separation of variables. You can also build a solution using general form of solution.

In this problem the d'Alembert formula, which implies $u \equiv 0$, correct only in triangle $(0,0)$ -- $(L, 0)$ -- $(L/2, L/(2a))$, because you have the Cauchy problem in this triangle.


问:
Different notations for the length of the line segment
This one is very trivial, I was just wondering, can the notation for the length of the line segment be written like this:

$$|\overline{AB}|=5~\text{cm}$$

I noticed that in different parts of the world, people use different notations. For example, in United States, the most common notation is simply $AB=5~\text{cm}$, and in Croatia, the notation is $|AB|=5~\text{cm}$ and never different. I even came across the notation which places the letter "m" before the line segment to indicate the length (although this notation is very suspicious to me). Since the notations are so versatile, is it mathematically correct to state this $|\overline{AB}|=5~\text{cm}$ as well?

答:
The units of length, such as "cm", "m", are used in physics; they are not used in pure mathematical discussions.

Notations are mostly for communications. As long as the context is clear, you can use whatever notations you like. Though you may want to follow certain conventions.


问:
Net in a Metric Space
I'm self-studying Willard's General Topology.  I'm stumped on exercise 11A(3), which reads:

Let $M$ be any metric space.  A mapping $P(\alpha) = x_\alpha$ of $\Omega_0$ into $M$ will be a net.  Show that $x_\alpha \longrightarrow x$ in $M$ if and only if $x_\alpha$ is eventually equal to $x$.

For reference, $\omega_1$ is the first uncountable ordinal and $\Omega_0$ is the set of ordinals less than $\omega_1$.

I'm not looking for a solution; I'd like to solve this problem myself.  However, I'd appreciate any hints or tips you can provide.

Many thanks!

Doug

答:
HINT: If the net converges to $x$, then for each $n<\omega$ there is an $\alpha_n<\omega_1$ such that $x_\alpha\in B(x,2^{-n})$ whenever $\alpha_n\le\alpha<\omega_1$. I’ve left an additional hint in the spoiler-protected box below.

 Now consider $\sup_{n<\omega}\alpha_n$ and $\bigcap_{n<\omega}B(x,2^{-n})$.

Title: Net in a Metric Space


问:
Why is this module cyclic?
I've read that given a representation of a Lie algebra $(\rho, V)$ and its extension to $\mathcal{U}(\mathfrak{g})$, namely $(\tilde{\mathfrak{\rho}}, V)$, if we fix $v \in V$, we can define the ideal $$I = \{x \in \mathcal{U}(\mathfrak{g}): x\cdot v = 0  \}$$ Then the quotient  $ \ \mathcal{U}(\mathfrak{g})/I$ is cyclic (generated by one element) as a $\mathfrak{g}$-module. But I can't see why this is the case, can someone explain it to me?

答:
If $R$ is a unital associative ring and $I$ a left ideal, then the quotient $R/I$ is a left $R$-module. It is straightforward to verify $R/I$ is cyclically generated by $1+I$ as an $R$-module.

We can apply this to $R=\mathcal{U}(\mathfrak{g})$ and your annihilator ideal $I=\mathrm{Ann}_{\,\mathcal{U}(\mathfrak{g})}(x)$. Note that $V$, as a representation of $\mathcal{g}$, automatically becomes a $\mathcal{U}(\mathfrak{g})$-module by simply extending the action.

For a $\mathfrak{g}$-module $M$, the cyclic submodule generated by some $m\in M$ is the smallest submodule containing $m$. In practice, that means it is spanned by $m$, and $\rho(X)m$ for all $X\in\mathfrak{g}$, and $\rho(X)\rho(Y)m$ for all $X,Y$, and so on - in other words, the cyclic $\mathfrak{g}$-submodule generated by $m$ is the same as the cyclic $\mathcal{U}(\mathfrak{g})$-module generated by $m$ as well.

Title: Why is this module cyclic?


问:
Calculate the dimensions of a rotated rectangle inside a bounding box

I'm trying to figure out if it's possible to calculate the width & height of the gray rectangle if I only know the width & height of the dotted red rectangle and the angle of the rotation.

I've tried substituting values for the width and height and using $x \cos(\alpha) + y \sin(\alpha)$ and attempting to figure out a scaling factor, but I'm not really getting anywhere.

答:
$$x\cos\alpha+y\sin\alpha=w\\x\sin\alpha+y\cos\alpha=h$$
Multiply first equation by $\cos\alpha$, the second by $\sin\alpha$ and subtract:
$$x(\cos^2\alpha-\sin^2\alpha)=w\cos\alpha-h\sin\alpha$$or $$x=\frac{w\cos\alpha-h\sin\alpha}{\cos^2\alpha-\sin^2\alpha}$$
Similarly, multiply the first equation by $\sin\alpha$, the second by $\cos\alpha$ and subtract:
$$y(\sin^2\alpha-\cos^2\alpha)=w\sin\alpha-h\cos\alpha\\y=\frac{w\sin\alpha-h\cos\alpha}{\sin^2\alpha-\cos^2\alpha}$$
Notice that the denominator might be $0$ for $\alpha=45^\circ$. Then you must have the numerator equal to $0$ as well, or $w=h$, so you have a square inside a square $x=y$ and $$2x\frac{\sqrt 2}2=w$$ or $$x=w\frac{\sqrt 2}2$$


问:
Is this a New Proof for Pythagorean Theorem?
(For all those that it may concern, this is not a duplicate of my previous post, But starts in a similar way.)

A triangle with side lengths a, b, c with a height(h) that intercepts the hypotenuse(c) at (x , y) such that it is split into two side lengths, c  = m + n, we can find Pythagoras theorem using the area of a right triangle and the slope equations of the height and the hypotenuse.

We begin by finding the  coordinate (x,y) by using the concept of the area fourmula, ab = hc:

x - value:

$bx = hm$

$x= \frac{hm}{b}$

$x = \frac{am}{c}$

y- value:

$ay = hn$

$y = \frac{hn}{a}$

$y = \frac{bn}{c}$

which gives us:

$(\frac{am}{c}, \frac{bn}{c})$

We can also find the (x,y) coordinates using the slope equations of the height and hypotenuse.

Height's equation:

$y = \frac{b}{a}x$

Hypotenuse's equation:

$y = \frac{-a}{b}x + a$

Now we can determine the (x ,y )intercept.

x - value:

$\frac{b}{a}x = \frac{-a}{b}x + a$

$\frac{b}{a}x + \frac{a}{b}x = a$

$x(a^2 + b^2) = a^2b$

$x = \frac{a^2b}{a^2 + b^2}$

y - value:

$y = \frac{b}{a}(\frac{a^2b}{a^2 + b^2})$

$y = \frac{ab^2}{a^2 + b^2}$

Which gives us:

$(\frac{a^2b}{a^2 + b^2}
,\frac{ab^2}{a^2 + b^2})$

Using $(\frac{bm}{c},\frac{an}{c}) ,(\frac{ab^2}{a^2 + b^2} , \frac{a^2b}{a^2 + b^2})$ there are 2 equalities:

$\frac{bm}{c} = \frac{a^2b}{a^2 + b^2}$

$\frac{an}{c} = \frac{ab^2}{a^2 + b^2}$

Which after isolating and eliminating c becomes:

$a^2m = b^2n$

This gives us two expressions:

$\sqrt{\frac{n}{m}} = \frac{a}{b}$

$\sqrt{\frac{m}{n}} = \frac{b}{a}$

We substitute these into $c = m + n$:

$c = m + n$

$\frac{c}{\sqrt{m}{n}} = \sqrt{\frac{m}{n}} + \sqrt{\frac{n}{m}}$

$\frac{ab}{\sqrt{m}{n}}c =  a^2 + b^2 $

Using $h$, this can be written as:

$1 = (\frac{\sqrt{m}{n}}{h})(\frac{a^2 + b^2}{c^2})$

or:

$1 = (\frac{h}{\sqrt{m}{n}})(\frac{c^2 }{a^2 + b^2})$

Note, this leaves us with only two possibilities, the fractions are either inverses, or the numerator and denominator are equal. We know $h = ab/c$ and $h < c$, so $h ≠  a^2 + b^2$, the denominator can also be split  into  $hc, c$ but we know $hc = ab$ so $ab ≠ a^2 + b^2$, and we know the side lengths $m,n$ are smaller than $a,b$ which means the numerator and denominator are equal in  the case of:

$a^2 + b^2 = c^2$

答:
It would have been much simpler to use your initial equations as follows.

Since $hc=ab$, you have $m=\frac{bh}{a}=\frac{b^2}{c}$

Similarly, $n=\frac{ba^2}{c}$.

Now express $m+n=c$ as $\frac{a^2}{c}+\frac{b^2}{c}=c$
and you have $a^2+b^2=c^2$.


问:
vector from a point P to a point Q in a plane in the direction of the normal vector
I'm a little confused with a task in my textbook:

Plane $A$ has equation $r*n=k$ (scalar product form) and Point $P$, outside $A$, has position vector $\vec p$.

a) Write down a vector equation of the line $l$ through $P$ which is perpendicular to $A$.

$l:\vec r = \vec p + \lambda \vec n$

b) Line $l$ intersects $A$ at $Q$. Show that $\vec {PQ}= (\frac{k-\vec p * \vec n}{\lvert \vec n \rvert ^2})\vec n$.

c) Hence show that the shortest distance from $P$ to $A$ is given by $\frac {\lvert \vec p * \vec n - k\rvert}{\lvert \vec n \rvert}$

I don't understand b) but I do understand a) and c), because part c) is very much the same as the Hesse normal form.

source: https://www.youtube.com/watch?v=v8Fy1C7yJq8

In the picture above:

plane $A$ is called $E$

point $Q$ is called $F$

Point $P$ is called $R$

The Hesse normal form makes use of the following properties:

$\cos(\alpha)=\frac d {\lvert \vec {PR} \rvert}$

$\cos(\alpha)=\frac {\vec n * \vec {PR}}{\lvert \vec n \rvert * \rvert \vec {PR} \lvert} = \frac {\vec {PR}}{\rvert \vec {PR} \lvert} * \vec n_0$

Hence:
$$\frac d {\lvert \vec {PR} \rvert}=\frac {\vec {PR}}{\rvert \vec {PR} \lvert} * \vec n_0$$
$$d=\vec {PR}*\vec n_0$$
Or the Hesse normal form can be expressed in cartesian form which is the same as required to prove:
$$d= \frac {\lvert n_1p_1+n_2p_2+n_3p_3-k \rvert}{\lvert \vec n\rvert} $$
$$d= \frac {\lvert \vec n * \vec p -k \rvert}{\lvert \vec n\rvert} $$

How can I show that the statement in b) is true? Can I use the same diagramm to prove it? If not, what other way is there? And how do you then get from b) to proving c)?

答:
Let $\underline n=(n_1,n_2,n_3)$be the perpendicular vector to the plane $\pi\subset\mathbb R^3$, whose equation is $\pi\equiv\langle\underline n,\underline x- A\rangle= n_1(x-a_1)+n_2(y-a_2)+n_3(z-a_3)=0$.

Let $P$ be a point such that $P\notin \pi$, then the equation of the line for $P$ orthogonal to $\pi$ will be
$$l\equiv\begin{cases}x=p_1+tn_1\\y=p_2+tn_2\\z=p_3+tn_3 \end{cases}.$$
The intersection of $l$ with $\pi$ is
$$\{Q\}=l\cap\pi\equiv n_1(p_1+tn_1-a_1)+n_2(p_2+tn_2-a_2)+n_3(p_3+tn_3-a_3)=0\iff t(n_1^2+n_2^2+n_3^2)+p_1n_1-a_1n_1+p_2n_2-a_2n_2+p_3n_3-a_3n_3=0 \iff $$
$$t=\dfrac{a_1n_1+a_2n_2+a_3n_3-p_1n_1-p_2n_2-p_3n_3}{n_1^2+n_2^2+n_3^2}:=\alpha.$$
The formula is the same you wrote in your question. The difference is that the term $k$ in my case is given by $-(n_1a_1+n_2a_2+n_3a_3)$.

Substituting $t=\alpha$ in the equation of $l$ you find the coordinates of the intersection $Q$ and the vector $\vec{PQ}$ is simply $Q-P$.

Doing this for the general case seems to be tedious but in a normal problem it would be quite easy.


答:
As with your previous question, you have given a valid proof of Pythagoras... if I've followed your argument correctly. First, I'll condense it, if only for my own benefit. (I also swap round $a,\,b$, because traditionally these are respectively opposite $A,\,B$.)

With $C=O$, the hypotenuse $y=b(1-x/a)$ meets $y=ax/b$ at $x=ab^2/(a^2+b^2)$, a proportion $b^2/(a^2+b^2)$ of the way from $A$ to $B$, so $AH=b^2c/(a^2+b^2)$; $HB$ follows similarly. Equating two expressions for $\cos\theta$ (i.e. using similar triangles), $c=\frac{(a^2+b^2)h}{ab}$, which by area formulae is $(a^2+b^2)/c$.

Of the "standard" proofs I know, yours is most similar to this. But that proof doesn't even need area formulae: similar triangles give$$a^2=c\cdot BH,\,b^2=c\cdot HA\implies a^2+b^2=AB\cdot AB=c^2.$$@S.Dolan's answer gives similar time-saving tips, albeit still using area formulae.

Title: Is this a New Proof for Pythagorean Theorem?

Body:
(For all those that it may concern, this is not a duplicate of my previous post, But starts in a similar way.)

A triangle with side lengths a, b, c with a height(h) that intercepts the hypotenuse(c) at (x , y) such that it is split into two side lengths, c  = m + n, we can find Pythagoras theorem using the area of a right triangle and the slope equations of the height and the hypotenuse.

We begin by finding the  coordinate (x,y) by using the concept of the area fourmula, ab = hc:

x - value:

$bx = hm$

$x= \frac{hm}{b}$

$x = \frac{am}{c}$

y- value:

$ay = hn$

$y = \frac{hn}{a}$

$y = \frac{bn}{c}$

which gives us:

$(\frac{am}{c}, \frac{bn}{c})$

We can also find the (x,y) coordinates using the slope equations of the height and hypotenuse.

Height's equation:

$y = \frac{b}{a}x$

Hypotenuse's equation:

$y = \frac{-a}{b}x + a$

Now we can determine the (x ,y )intercept.

x - value:

$\frac{b}{a}x = \frac{-a}{b}x + a$

$\frac{b}{a}x + \frac{a}{b}x = a$

$x(a^2 + b^2) = a^2b$

$x = \frac{a^2b}{a^2 + b^2}$

y - value:

$y = \frac{b}{a}(\frac{a^2b}{a^2 + b^2})$

$y = \frac{ab^2}{a^2 + b^2}$

Which gives us:

$(\frac{a^2b}{a^2 + b^2}
,\frac{ab^2}{a^2 + b^2})$

Using $(\frac{bm}{c},\frac{an}{c}) ,(\frac{ab^2}{a^2 + b^2} , \frac{a^2b}{a^2 + b^2})$ there are 2 equalities:

$\frac{bm}{c} = \frac{a^2b}{a^2 + b^2}$

$\frac{an}{c} = \frac{ab^2}{a^2 + b^2}$

Which after isolating and eliminating c becomes:

$a^2m = b^2n$

This gives us two expressions:

$\sqrt{\frac{n}{m}} = \frac{a}{b}$

$\sqrt{\frac{m}{n}} = \frac{b}{a}$

We substitute these into $c = m + n$:

$c = m + n$

$\frac{c}{\sqrt{m}{n}} = \sqrt{\frac{m}{n}} + \sqrt{\frac{n}{m}}$

$\frac{ab}{\sqrt{m}{n}}c =  a^2 + b^2 $

Using $h$, this can be written as:

$1 = (\frac{\sqrt{m}{n}}{h})(\frac{a^2 + b^2}{c^2})$

or:

$1 = (\frac{h}{\sqrt{m}{n}})(\frac{c^2 }{a^2 + b^2})$

Note, this leaves us with only two possibilities, the fractions are either inverses, or the numerator and denominator are equal. We know $h = ab/c$ and $h < c$, so $h ≠  a^2 + b^2$, the denominator can also be split  into  $hc, c$ but we know $hc = ab$ so $ab ≠ a^2 + b^2$, and we know the side lengths $m,n$ are smaller than $a,b$ which means the numerator and denominator are equal in  the case of:

$a^2 + b^2 = c^2$


问:
Polynomial ring isomorphic to Quotient space
I'm relatively new to rings, and I also don't have a very good understanding of quotient spaces, but I'd like to understand the following problem.

Show that for $A \in M_n(\mathbb R ),$ there is some $p \in \mathbb R [X]$ such that $\mathbb R [A]$ is isomorphic to $\mathbb R [X]/ \langle p \rangle$. Show that if $p,q \in \mathbb R [X]$ are quadratics with two distinct real roots then $\mathbb R [X]/ \langle p \rangle \cong \mathbb R [X]/ \langle q \rangle$. Justify which, if any, of $\mathbb R [A]$, $\mathbb R [B]$, $\mathbb R [C]$ and $\mathbb R [D]$ are isomorphic as rings.
$$
A:=
    \left( \begin{matrix}
    2 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 1 \\
    \end{matrix} \right),
B:=
    \left( \begin{matrix}
    4 & 1 \\
    0 & 3 \\
    \end{matrix} \right),
C:=
    \left( \begin{matrix}
    2 & 1 & 0 \\
    0 & 2 & 0 \\
    0 & 0 & 2 \\
    \end{matrix} \right),
D:=
    \left( \begin{matrix}
    1 & 1 \\
    0 & 1 \\
    \end{matrix} \right).
$$

I already struggle to understand the question... I know that $\mathbb R [X]$ are the polynomials with real coefficients, and we can also plug matrices in and evaluate the polynomials as we are used to. So, I guess $\mathbb R [A]$ is the space of all matrices we can get by adding powers of $A,$ etc. I definitely lack an intuition what $\mathbb R [X]/ \langle p \rangle$ is. Could somebody offer some insight? And then maybe a hint about how to solve that?

As for the second part, I guess if we have isomorphisms, they are in the same dimension. Since $A$ is diagonal and $C$ is not, I don't think $\mathbb R [A]$ and $\mathbb R [C]$ are isomorphic. But how would I go about solving that part as well?,

答:
Hint: consider the epimorphism $\mathbf{ev}_A \colon\Bbb R [X] \to \Bbb R[A]$. What's its kernel?

 You can check that almost by definition $\ker \mathbf{ev}_A$ is generated by the minimial polynomial $m_A$ of $A$. Hence $\Bbb R[X]/(m_A) \simeq \Bbb R[A]$. This will help you when computing the particular examples and see which are isomorphic. If case this says something to you, the idea is that by Cayley-Hamilton the extension $\Bbb R \to M_n(\Bbb R)$ is integral, hence we characterize each algebra $\Bbb R[A]$ with $A \in M_n(\Bbb R)$ as a quotient of $\Bbb R[X]$.

Hint: if $p = (X-\alpha)(X-\beta)$, note that $(X-\alpha)$ and $(X-\beta)$ are coprime.

 Use the chinese remainder theorem to prove that $\Bbb R[X]/p \simeq \Bbb R[X]/(X-\alpha) \times \Bbb R[X]/(X-\beta)$. The latter are isomorphic to $\Bbb R$ via evaluation at $\alpha$ and $\beta$ respectively, so any such quotient is $\Bbb R^2$.


问:
Simplest Way to Solve this Integral?

Consider the integral $$\int_{-\delta}^{\delta} cos(nx)\big (1-\frac{\lvert x \rvert}{\delta}\big)dx$$

Where $n$ and $\delta$ are constants. Are there any reasonably simple ways to solve this integral? Im not familiar with integrals of absolute value's and the method suggested by an online calculator was very long end tedious.

答:
You should not apply the fundamental theorem of calculus directly. Instead, since the integrand is even, you have

$$
\int_{-\delta}^\delta \cos(nx)\big (1-\frac{\lvert x \rvert}{\delta}\big)dx
=2\int_0^{\delta}\cos(nx)\big (1-\frac{\lvert x \rvert}{\delta}\big)dx
=2\int_0^{\delta}\cos(nx)\big (1-\frac{x }{\delta}\big)dx
$$

Now, by linearity, you can work out the following integrals:
$$
\int \cos(nx)dx,\quad \int \cos(nx)\cdot x\,dx
$$
and then combine the results.

In general, if you have an even function $f:[-a,a]\to\mathbb{R}$ (that is Riemann integrable),
$$
\int_{-a}^a f(x)dx=2\int_0^af(x)dx
$$

Title: Simplest Way to Solve this Integral?


问:
Union of translations of a measurable set with the same measure as the original set
Let $\mu$ be the Lebesgue measure on $[0,1]$ and let $A+\Bbb Q=\{a+q\mid a\in A\text{ and } q\in \Bbb Q\text{ and }a+q\in[0,1]\}$ for any $A\subset[0,1]$

I know that $\mu(A)=0$ or $\mu(A)=1$ implies that $\mu(A+\Bbb Q)=\mu(A)$. Is this implication also true in reverse? Does there exist an $A$ such that $0<\mu(A)=\mu(A+\Bbb Q)<1$? If not, does there exist an $A$ such that $0<\mu(A)<\mu(A+\Bbb Q)<1$?

答:
There is no such $A$. I'm going to work with everything on all of $\mathbb{R}$ for a bit, because restricting the Minkowski sum might mean I have to be more careful with some of the integrals I'm about to write.

Claim: Let $E \subseteq \mathbb{R}$ be such that $E + r = E$ for every $r \in \mathbb{Q}$. Then either $E$ or $E^{c}$ has measure $0$.

Before I prove that claim, let's take a look at why it helps. For any set $A$, copper.hat noted in the comments to the question that $A + \mathbb{Q} + r = A + \mathbb{Q}$ for every $r \in \mathbb{Q}$, where these Minkowski sums are not restricted to the unit interval. Thus, either $A + \mathbb{Q}$ or $(A + \mathbb{Q})^{c}$ has measure $0$ in $\mathbb{R}$. Now, we can restrict to the unit interval and see that $(A + \mathbb{Q}) \cap [0, 1]$ either has measure $0$ or $1$. Therefore, if $\mu(A) = \mu((A + \mathbb{Q}) \cap [0, 1])$, then $A$ must have measure $0$ or $1$.

Proof of claim:

Define the auxiliary function
$$f(x) = \int_{\mathbb{R}} 1_{E}(y) 1_{E^{c}}(y + x) \mathrm{d} y.$$
By the continuity of shifting operators with respect to $L^{1}$, $f$ is continuous. Moreover, for any $r \in \mathbb{Q}$,
$$f(r) = \int_{\mathbb{R}} 1_{E}(y) 1_{E^{c}}(y + r) \mathrm{d} y = \int_{E} 1_{E^{c}}(y + r) \mathrm{d} y = 0,$$
since $y + r \in E$ for any $y \in E$. Since $f$ is continuous and $0$ at all rational points, $f \equiv 0$. Thus,
$$0 = \int_{\mathbb{R}} f(x) \mathrm{d} x = \int_{\mathbb{R}} \int_{\mathbb{R}} 1_{E}(y) 1_{E^{c}}(y + x) \mathrm{d} y \mathrm{d} x = \mu(E) \mu(E^{c}),$$
using the Tonelli theorem and the translation invariance of the Lebesgue measure to get to that last equality. Hence, either $E$ has measure $0$ or $E^{c}$ has measure $0$.


问:
Is this equation expected to be a plane? How can I get some better intuition?
In 2 dimensions $Y=cX+B$ is a line. In 3D $Z=aY+bX+C$ is a plane...

But this won't be a line: $f(x)=1/(1+e^-5x)$.

And how is this a plane?

答: No, it is not a plane. The graph looks like a plane because it was drawn with $(x,y)\in\left[-\frac12,\frac12\right]\times\left[-\frac12,\frac12\right]$. If you choose $[0,1]\times[0,1]$ instead, then you will get this:


问:
Linear combination of normal distributions

The distribution of independent random variables X and Y are $X\sim \mathcal{N}(24,2^2)$ and $Y \sim \mathcal{N}(25,3^2)$. Find the distribution of $2X-Y$ and $\mathbb{P}[X-Y<40]$.

I am doing self studying and I just reached the topic Working with normal distribution on the chapter Linear combinations of random variables and I came to this topic and am stuck I don't really know what to do. Any help would be much appreciated. Thank You!

答:
HINT

The critical results you need are that if $X,Y$ are independent normal random variables and $a,b \in \mathbb{R}$, then $aX+bY$ is also normal (note that the parameters will depend on the constants $a,b$). You can prove this using characteristic functions, for example, and then you will know how to change the parameters.

Your problem becomes an exercise applying the above statements. Can you finish it?

UPDATE

To estimate the first parameter, let $Z = aX+bY$ then
$$\mathbb{E}[aX+bY] = x\mathbb{E}[X] + b\mathbb{E}[Y]$$

Can you now do this for the variance, and plug the numbers from your problem?

UPDATE 2

You are right. We define $Z = 2X-Y$ then $Z \sim \mathcal{N}(m,s^2)$ with
$$
m = \mathbb{E}[Z] = \mathbb{E}[2X-Y] = 2\mathbb{E}[X] - \mathbb{E}[Y]
  = 2\cdot 24 - 25 = 23.
$$
For the variance, you get
$$
s^2 = \mathbb{Var}[Z] = \mathbb{Var}[2X-Y] = 2^2 \mathbb{Var}[X] + (-1)^2\mathbb{Var}[Y],
$$
so now you can complete the arithmetic and you know the distribution of $Z=2X-Y$.

To compute $\mathbb{P}[X-Y<40]$, you can transform it similarly using $V = X-Y \sim (\mu, \sigma^2)$ into the standard normal since $(V-\mu)/\sigma$ is standard normal.

Can you complete this now?


问:
Birthday Problem: Expected number of people in a room
If an infinite amount of people enter a room one by one, what is the expected number of people in the room when you first find two that share the same birthday?
(Assuming no leap years and every birthday is equally likely).</p>

答: According to Pigeonhole Principle the number of people in the room shall at least be 366 to ensure what you're saying. However it "is" possible for less people in the room but this is the number you can be sure that at least two of the people share their birthday</p>


答:
There is an important fact that a linear span of INDEPENDENT normally distributed random variables is still normally distributed.

For your case, let $Z=2X-Y$, then $Z$ is normally distributed with mean $E[Z]=2E[X]-E[Y]$ and variance $\mbox{Var}(Z) = \mbox{Var}(2X)+\mbox{Var}(-Y)$.

However, it is important to point out that if the given normally distributed random variables are NOT independent, their linear combination needs not be normally distributed.

Title: Linear combination of normal distributions

Body:
The distribution of independent random variables X and Y are $X\sim \mathcal{N}(24,2^2)$ and $Y \sim \mathcal{N}(25,3^2)$. Find the distribution of $2X-Y$ and $\mathbb{P}[X-Y<40]$.

I am doing self studying and I just reached the topic Working with normal distribution on the chapter Linear combinations of random variables and I came to this topic and am stuck I don't really know what to do. Any help would be much appreciated. Thank You!


问:
Computing the convolution of random variables to determine $Z=X+Y$
I am trying to understand the calculations in the following solved exercise:

Let $X$ and $Y$ two independent random variables with pdf $f(x)=2x, x\in (0,1); f(x)=0, \text{otherwise}$. Determine the pdf of $Z=X+Y$.

For $z\in (0,1)$:
$$f_Z(z)=\int_{\mathbb{R}}2x 1_{(0,1)}(x) 2(z-x)1_{(z-1,z)}(x)dx=\int_{0}^{z}2x2(z-x)=\frac{2}{3}z^3.$$

For $z\in (1,2)$: $$f_Z(z)=\int_{\mathbb{R}}2x 1_{(0,1)}(x) 2(z-x)1_{(z-1,z)}(x)dx=\int_{z-1}^{1} 2x2(z-x)dx=\frac{2}{3}(6z-z^3-4).$$

I know the definition of convolution and I get why we are using it to compute the pdf of the sum of two random variables but I don't get why are the endpoints of the integral in the first case $0$ and $z$ and why are they in the second case $z-1$ and $1$?

答:
For each value of $z$ you are calculating
$$
f_Z(z):=\int_{\mathbb R} f(x)f(z-x)dx
$$
viewed as an integral over $x$ with $z$ being constant. Since the integrand $f(x)f(z-x)$ is a function of $x$, you need to determine what the integrand looks like for each possible $x$, and in particular the range of $x$ values where the integrand is zero. The range where the integrand is nonzero will depend on $z$.

This answer demonstrates how it's done for a slightly different density, but you should be able to apply the reasoning to your situation, because that other answer is also dealing with a density $f(x)$ defined over $(0,1)$. The key to finding the endpoints where $f(x)f(z-x)$ is nonzero is the following picture, which also applies to your situation:


问:
Term-by-term antidifferentiation and power series representation of antiderivative
I have solved the following exercise and I would like to know If I have made any mistakes (NOTE: I can't use integrals, since they are defined in a later chapter):

Assume $f(x)=\sum_{n=0}^{\infty} a_n x^n$ converges on $(-R,R)$.

(a) $F(x)=\sum_{n=0}^{\infty}\frac{a_n}{n+1}x^{n+1}$ is defined on $(-R,R)$ and satisfies $F'(x)=f(x)$.

(b) Antiderivatives are not unique. If $g$ is an arbitrary function satisfying $g'(x)=f(x)$ on $(-R,R)$ find a power series representation for $g$.

My solution:

(a) Let $x\in (-R,R)$: then the series  $\sum_{n=0}^{\infty} a_n x^{n+1}$ is convergent in $(-R,R)$ since $\sum_{n=0}^{\infty} a_n x^n$ is convergent there by hypothesis and since $(\frac{1}{n+1})_{n=0}^{\infty}$ is a non-negative and decreasing sequence by Abel's Test we have that $\sum_{n=0}^{\infty}\frac{a_n}{n+1}x^{n+1}$ is convergent in $(-R,R)$ too so $F(x)$ is defined on $(-R,R)$. Since by Theorem 6.5.2 * $\sum_{n=0}^{\infty} a_n x^n$ converges uniformly on every interval $[-c,c]\subseteq (-R,R)$, $0\in [-c,c]$ and $F(0)=\sum_{n=0}^{\infty}\frac{a_n}{n+1}0^{n+1}=0$ is convergent, by Term-by-Term Differentiability Theorem we have $F'(x)=\sum_{n=0}^{\infty} F'_n(x)=\sum_{n=0}^{\infty} a_n x^n=f(x)$ on each $[-c,c]\subseteq (-R,R)$.

(b) $g(x)=C +\sum_{n=0}^{\infty}\frac{a_n}{n+1}x^{n+1}$ ($C$ costant) since $g'(x)=0+(\sum_{n=0}^{\infty}\frac{a_n}{n+1}x^{n+1})'\stackrel{\text{(a)}}{=}\sum_{n=0}^{\infty}a_n x^n=f(x)$, as desired.

*Theorem 6.5.2 If a power series converges absolutely at $x_0$ then it converges uniformly on the interval $[-|x_0|,|x_0|]$

答:
In order to apply the term-by-term differentiation theorem, you need uniform convergence for $\sum_{n=1}^{\infty}n a_{n} x^{n-1}$ as well. But you only show that it converges on $(-R,R)$.

I would do (a) as follows.

Let $x\in(-R,R)$ be arbitrary and pick $t$ to satisfy $|x|<t<R$. Observe that
$$
\sum_{n=1}^{N}\left|n a_{n} x^{n-1}\right|=\sum_{n=1}^{N} \frac{1}{t}\left(n\left|\frac{x}{t}\right|^{n-1}\right)\left|a_{n} t^{n}\right|
$$

Because $|x/t|<1$, we can pick$\dagger$ $L$ such that for all $n$,
$$
n|\frac{x}{t}|^{n-1}\le L
$$

Now you have
$$
\sum_{n=1}^{N}\left|n a_{n} x^{n-1}\right|=\sum_{n=1}^{N} \frac{1}{t}\left(n\left|\frac{x}{t}\right|^{n-1}\right)\left|a_{n} t^{n}\right| \leq \frac{L}{t} \sum_{n=1}^{N}\left|a_{n} t^{n}\right|
$$
where the last sum converges because $t\in(-R,R)$.

For (b), use the fact that if $g'=f'$ on $(-R,R)$, you must have $g=f+C$ for some constant $C$.

$\dagger$ Notes.

If $0<s<1$, you can show that $ns^{n-1}$ is a bounded sequence. This can be seen by applying the ratio test to the sequence $a_n=ns^{n-1}$.


问:
Find closest real root of cubic given two imaginary roots
I have a cubic polynomial root-solving algorithm (the one at https://www.particleincell.com/2013/cubic-line-intersection/) which works perfectly, however I would like to be able to finding the closest real root when the cubic function has two imaginary roots.

At first I thought I could just disregard the real part, for example if my two imaginary roots are 3+0.0005i and 3-0.0005i then I figured that the corresponding close real root would be x=3 (which would in fact be the turning point of the graph), however when shifting my graph so that there ARE indeed two real roots, the turning point turns out to be at (for example) x = 3.1.

I could, of course, solve for the turning points, however I thought that I would simply be able to disregard my imaginary components and it would just work. Is this maybe the case and I am doing something wrong in my algorithm, or is there an explanation as to why you cannot simply do this?

Thanks

Edit:

I've decided to add the actual polynomial I am working with so I can illustrate what I mean.
$$y = 3x^3 + 15x^2 + 6x -36.185$$

The roots (according to my pocket calculator) are 1.239..., and -3.1196591 +-...i. However, the turning point is -3.1196329. Why is this different?

答:
Relation to the third real root

Suppose that $f(x)$ is your cubic with two imaginary roots, one real root.  As pointed out by user Empy2, the sum of all the roots is $-\frac{m}{n}$, where $m$ is the coefficient of the square term and $n$ is the coefficient of the cubic term.  So, if you know the two imaginary roots, you absolutely know the remaining real roots with no need to guess.

However, there need be no relation between the real part of the imaginary roots and the third (real) root.  To see this, manufacture a large number of arbitrary examples.  For example, let $z=a+bi$ and let $c \in \mathbb{R}$.  The polynomial
$$
f(x) = (x-z)(x-\overline{z})(x-c) = (x^2 -2ax +z\overline{z})(x-c)
$$
is a real cubic polynomial with complex roots $z, \overline{z}$, and the real root $c$.  However, there is plainly no relation between $a$ and $c$ (without further assumptions on your polynomial).  You can manufacture any choices of $a$ and $c$ that you like.

Relation to the turning points

Take the same example polynomial above.  Its derivative is
$$
f'(x) = (2x-2a)(x-c)+x^2-2ax+z\overline{z}.
$$
If you expand this out using that $z\overline{z}=a^2+b^2$ you get that
$$
f'(x) = 3x^2 -(2c+4a)x +(2ac+a^2+b^2).
$$
We use the quadratic formula and doing some very careful algebra (noting that $(a-c)^2 = a^2+c^2-2ac$ along the way) gives potential turning points as
$$
x = \frac{c+2a\pm\sqrt{(a-c)^2-3b^2}}{3}.
$$
When $b^2 > \frac{(a-c)^2}{3}$ there are of course no turning points.

For the case more interesting to you, take the example of such a cubic with complex roots $\pm 5i$ and real root $10$.  Thus we are taking $a=0$, $b=5$, and $c=10$.  In this case we get turning points $\frac{10 \pm 5}{3}$ which gives $x=\frac{5}{3}\approx 1.667$ and $x=5$. Note that in this case, the turning points are not anywhere close to the real part of the imaginary roots (namely, $a=0$).  This is only one example:  it is easy to manufacture more.

If, however, we are in the case $b^2 \leq \frac{(a-c)^2}{3}$ so that we have real turning points, and with $|a-c|$ very small, the turning points will be reasonably close to the real part of the complex roots.  Note that in this case $b^2$ and hence $b$ must also be small.  Then the formula above for the turning points is approximately (by ignoring the square root term and taking $a \approx c$)
$$
x = \frac{c+2a\pm\sqrt{(a-c)^2-3b^2}}{3} \approx \frac{c+2a}{3} \approx \frac{a+2a}{3} = a.
$$
Hence, in this case, the turning points are in fact close to $a$, which was the real part of the complex roots of $f$.  I believe this is what you are witnessing.


问:
Tangent space of a point
Let M = {p} be a manifold contained a single point p.

I'm wondering What is the tangent space of M at p ? And what is the vector space $\Omega(M)$ of smooth differential forms on M ?

Thank you!

答: A single point is $0$-dimensional, so the tangent space should also be a zero dimensional vector space.  Similarly, you'll only have $\Omega^0(M)$ being nonzero ($\Omega^i(M)=0$ for $i>0$).  These are just the (constant) functions as all functions on a single point are constant.

Title: Tangent space of a point


问:
Prove that the cross $\{(x,y)\in\mathbb{R}^2:xy=0\}$ is connected.
Is this a valid proof? Just dabbling with some basic topology. Thanks.

$A=\{(x,y)\in\mathbb{R}^2:xy=0\}$

Let $G=\{(x,0):x\leq0\}$ and $H=\{(x,0):x\geq0\}$

Let $K=G \cup H = \mathbb{R} \text{ (open)}$

Let $G_1=\{(0,y):y\leq0\}$ and $H_1=\{(0,y):y\geq0\}$

Let $K_1=G_1 \cup H_1 = \mathbb{R} \text{ (open)}$

$K$ and $K_1$ are connected sets which are not separated since $K \cap K_1 = \text{{0}} \neq 0$.

Then $K \cup K_1$ is connected, $K$ and $K_1$ form a connection of $A$. Thus

$A=\{(x,y)\in\mathbb{R}^2:xy=0\}$ is connected.

$\blacksquare$

答: Yes, it is a valid proof you are using the fact that if $A$ and $B$ are connected and $A\cap B$ is not empty, then $A\cup B$ is connected. This can be proved by considering a continuous function $f:A\cup B\rightarrow\{0,1\}$. Since $A$ is connected, we can asume, $f(A)=1$, this implies that $f(A\cap B)=1$, since $B$ is connected, $f(B)$ is contant and is $1$, so $f$ is constant.


问:
How to "invent" a function?
I'm currently making a game. I was given some numbers the leader wants me to stick close to. They are used for leveling; level 1, for example, needs 100 experience points, level 2 needs 250 experience points, etc.

Is there any way, maybe with a computer program, to calculate a function that will do something like this, instead of having to hard-code it?

Thanks in advance!

Simon.

答:
I know you said you wanted to avoid a hardcoded table, but honestly that's what you should use.  The mathematical approaches given here are all well and good, but using them you will probably have to settle for an approximation of what you really want, and it may not be easy to make changes later.

Just write an array indexed by level with the XP needed to level up.  This is what World of Warcraft does (WoW) as do most MMOs.  It's a tiny bit of memory, super-fast, and completely flexible.

Title: How to "invent" a function?


问:
What are some topologies without a countable basis?
I'm looking for some topologies on $\mathbb{R}$ that do not have a countable basis. Specifically, I want something where some of the open sets can not be written as a countable union of basis elements. I'm trying to figure out why the definition of a topology states that the union of arbitrarily many open sets is open and I haven't found any examples that require more than a countable union.

答: Some fairly familiar topologies on $\Bbb R$ that do not have countable bases are the discrete topology, the cofinite topology, the co-countable topology, and the Sorgenfrey topology (also called the lower limit topology). One that has some importance as a counterexample is the Michael line, and another is the rational sequence topology.


答:
Take the discrete topology on $\Bbb R$ as a very easy example.

The Sorgenfrey line is another classic: this has as basic open sets all sets of the form $[x,y)$ where $x < y$. This is separable ($\Bbb Q$ is dense) but does not have a countable base.

The Michael line topology on $\Bbb R$ is the usual topology on $\Bbb R$ where we add all irrational points as isolated points (so it has subbase $\mathcal{T}_e \cup \mathscr{P}(\Bbb P)$ e.g.)

Title: What are some topologies without a countable basis?

Body: I'm looking for some topologies on $\mathbb{R}$ that do not have a countable basis. Specifically, I want something where some of the open sets can not be written as a countable union of basis elements. I'm trying to figure out why the definition of a topology states that the union of arbitrarily many open sets is open and I haven't found any examples that require more than a countable union.


答: Once you become more familiar with topology, you may also want to note that the cross is clearly path-connected as any point is connected by a straight line to the origin, and any path connected space is connected.</p>

Title: Prove that the cross $\{(x,y)\in\mathbb{R}^2:xy=0\}$ is connected.

Body:
Is this a valid proof? Just dabbling with some basic topology. Thanks.

$A=\{(x,y)\in\mathbb{R}^2:xy=0\}$

Let $G=\{(x,0):x\leq0\}$ and $H=\{(x,0):x\geq0\}$

Let $K=G \cup H = \mathbb{R} \text{ (open)}$

Let $G_1=\{(0,y):y\leq0\}$ and $H_1=\{(0,y):y\geq0\}$

Let $K_1=G_1 \cup H_1 = \mathbb{R} \text{ (open)}$

$K$ and $K_1$ are connected sets which are not separated since $K \cap K_1 = \text{{0}} \neq 0$.

Then $K \cup K_1$ is connected, $K$ and $K_1$ form a connection of $A$. Thus

$A=\{(x,y)\in\mathbb{R}^2:xy=0\}$ is connected.

$\blacksquare$


问:
Counting chessboard rectangles
I am looking into the puzzle count the number of rectangles in a regular $8*8$ chessboard.

For a 1 by 1 chessboard there are 0 rectangles

For a 2 by 2 chessboard there are 4 rectangles (2 by 1)

For a 3 by 3 chessboard there are 6 rectangles of 1 by 2, 3 rectangles of 1 by 3, 6 rectangles of 2 by 1 and 3 rectangles of 3 by 1 i.e. total 22

For a 4 by 4 rectangles there are 56 rectangles in total (12 of 1 by 2, 12 of 2 by 1, 8 of 2 by 4, 8 of 4 by 2, 4 of 1 by 4, 4 of 4 by 1, 2 of 2 by 4, 2 of 4 by 2, 2 of 3 by 4, 2 of 4 by 3).

So we have the following sequence (4, 22, 56, ...):

2 $\times$ 2
3 $\times$ 3
4 $\times$ 4
5 $\times$ 5
6 $\times$ 6
7 $\times$ 7
8 $\times$ 8

4
22
56

I can't see a pattern in the sequence. Is there one that I am missing?

答:
Counting all rectangles (i.e. including squares) can be done be first choosing the horizontal lines at top and bottom of the rectangle. This can be done in   $\begin{pmatrix}n+1\\2\\\end{pmatrix}$ ways.

The same is true for the vertical lines. So the total is $\begin{pmatrix}n+1\\2\\\end{pmatrix}^2.$

If you want to exclude squares then you can find them separately and subtract from the total.

Title: Counting chessboard rectangles


问:
$K_9$ edge colouring
Assume we have $K_9$ graph and $4$ colours. We want to colour its edges and obtain following property:
If we change colours in our colouring, such graph will be isomorphic to the previous one (before colour change). Moreover, such graph is unique respect to automorphism group.
How to do it?

So far, I guessed that every vertice should be connected with 2 edges in all 4 colours (for example 2 blue, 2 green, 2 red, 2 yellow).

答:
If we change colours in our colouring, such graph will be isomorphic to the previous one

This implies that each one of the 4 coloured subgraphs must be identical (isomorphic), so you have a decomposition of $K_9$ into 4 identical graphs.

Do you know of any such graph decomposition ?

 Additional help, as you said you need that each vertex has degree 2 in each subgraph. And each subgraph contains 9 vertices and 9 edges.

Title: $K_9$ edge colouring


问:
Which convergence test should I use to test this functional series $\sum_{n=0}^\infty \cos(nx)$
The following series $$\sum_{n=1}^\infty (-1)^ncos(nx)=\cos(x)-\cos(2x)+\cos(3x)...$$

is obviously divergent when $x=0$. At that point, we have the Grandi series $1-1+1-1+1...$

when $x=\pi/2$, the series is also a Grandi series.

I am teaching myself convergent tests that are used in Calculus 2 right now so I wonder if I can apply any of those tests to test for convergence of this series?

How can I prove that this series is divergent for all $x$?

I haven't learnt Fourier analysis so more advanced explanation will be way over my head.

I also not skill at the integral test, so perhaps a bit of an elaborated answer will help.

答: Your series diverges for any $x\in\Bbb R$. That's so because, if $x\in\Bbb R$, you do not have $\lim_{n\to\infty}(-1)^n\cos(nx)=0$. Note that if we had that, then it would also be true that$$\lim_{n\to\infty}\cos(nx)=0,\tag1$$which would imply that  $\lim_{n\to\infty}\cos(2nx)=0$. But$$\cos(2nx)=2\cos^2(nx)-1$$and so we would have $\lim_{n\to\infty}2\cos^2(nx)-1=0$, which is impossible, since we are assuming $(1)$, from which it follows that $\lim_{n\to\infty}2\cos^2(nx)-1=-1$.


问:
Ask sample variance equation?
$$(n-1) S_x^2 = \Sigma_{i=1}^n (X_i - \bar{X})^2$$

Is this equation always true? Or this relationship only holds for some specific distributions, like normal distribution?

答:
The formula you show is most commonly used for the sample variance. For normal data it provides a good
estimate of the unknown population variance $\sigma^2.$

For a sample from a normal population with unknown mean $\mu$ and unknown (finite) variance $\sigma^2,$ your definition of $S_z^2$ leads to the relationship $\frac{(n-1)S_x^2}{\sigma^2} \sim \mathsf{Chisq}(\nu = n-1).$ This relationship can be used to test hypotheses about $\sigma^2$ or to make a confidence interval for $\sigma^2.$

But there are exceptions for particular circumstances.

If the population mean $\mu$ is known and the population variance
$\sigma^2$ is unknown, then for normal data, the best estimate of
$\sigma^2$ is $\widehat{\sigma^2} = \frac 1n\sum_{i=1}^n(X_i - \mu)^2.$ Then one uses the
relationship $\frac{n\,\widehat{\sigma^2}}{\sigma^2}
\sim \mathsf{Chisq}(\nu = n)$ for testing and making
confidence intervals.

And there are still other ways to estimate population variances from non-normal samples. For example, if you have a random sample $X_1, X_2, \dots X_n$ from
$\mathsf{Pois}(\lambda),$ then $\sigma_x^2 = \lambda$ and
the best estimate of the population variance is $\bar X =\frac 1n \sum_{i=1}^n X_i.$

Title: Ask sample variance equation?


问:
Determining $\sin(15)$, $\sin(32)$, $\cos(49)$, etc.
How do you in general find the trigonometric function values? I know how to find them for 30 45, and 60 using the 60-60-60 and 45-45-90 triangle but don't know for, say $\sin(15)$ or $\tan(75)$ or $\csc(50)$, etc.. I tried looking for how to do it but neither my textbook or any other place has a tutorial for it. I want to know how to find the exact values for all the trigonometric functions like $\sin x$, $\csc x$, ... opposed to looking it up or using calculator. According to my textbook, $\sin(15)=0.26$, $\tan(75)=3.73$, and $\csc(50)=1.31$ but doesn't show where those numbers came from, as if it was dropped from the Math heaven!</p>

答:
The chords of the rational angles solve a series of equations, which one can derive from an iso-series, in the form  $T(n+1)=X.T(n)-T(n-1)$.  You then solve for the unique factor in each even number, and the chords of a $p$-gon solves this.   The process can be greatly accelerated, by using a bignum environment.

In any case, the exact expression of something like $cos(1°)$ solves some equation involving cube and fifth roots.  But you can get around things like this, too.

If one solves the value for a supplement chord, say $scho{15°} = (\sqrt{6}+\sqrt{2})/2) = 1.93185165259$, the chords for subsequent angles, follow the same isoseries formula as above, with X = chord 1, $T(0) = scho(0) = 2$, and $X = T(1)=scho(15°)$, and subsequent $T(n) = scho(15n°)$.


问:
Which number is greater? $2\sqrt{2}$ or $e$
I have to determine which number is greater, $2\sqrt{2}$ or $e$.

I had a similar question as well, it was $2^\sqrt{2}$ compared to $e$.

For that one I managed to prove the inequality by using the increasing sequence converging to $e$: $(1+\frac1n )^n $

So I just searched for a value to assign to n such that $(1+\frac1n )^n \gt 2^\sqrt2$

I tried to proceed in a similar way with $2\sqrt{2} \gt \lt e$ , but it seems I can't get nowhere (I used the sequence decreasing and converging to $e$:  $(1+\frac1n )^{n+1}$ )

Is there another way to prove the inequality without the use of the calculator and maybe using derivatives? The question was in a derivatives file, so I'm wondering is there's a way to get to the end using them.

Any hint would be much appreciated, thanks.

Body:
$$\sum_{i=k}^\infty \frac {1}{i!} = \frac{1}{k!} \left(1+\frac{1}{k+1} + \frac{1}{(k+1)(k+2)} + \cdots \right)
\\ < \frac{1}{k!}\left( 1 + \frac{1}{k+1} + \frac{1}{(k+1)^2} + \cdots \right)
\\ = \frac{1}{k!} \frac{1}{1-\frac{1}{k+1}}=\frac{k+1}{k \cdot k!}$$

Therefore $$e = 2+ \sum_{i=2}^\infty \frac{1}{i!} < 2 + \frac{3}{2\cdot 2!} = \frac{11}{4} = \sqrt{\frac{121}{16}} < \sqrt{\frac{128}{16}}=2\sqrt{2}.\blacksquare $$


问:
Prove that $\sum_{x\in X}f(x)$ is absolutely convergent iff $\sup\{\sum_{x\in A}|f(x)|:A\subseteq X, A$ finite $\}<\infty$
I’m trying to prove the following result:

“Let $X$ be a countable set , and let $f:X\to\mathbb{R}$ be a function.
Then the series $\sum_{x\in X}f(x)$ is absolutely convergent iff $\sup\{\sum_{x\in A}|f(x)|:A\subseteq X, A$ finite $\}<\infty$.”

I've managed to come up with a proof of the leftward implication but not of the rightward one, so I would appreciate any hints about how to carry out this part of the proof.

Best regards,

lorenzo

--

Note: The book says that the following result might be useful:

"if $f:\{i\in \mathbb{N}:1\leq i \leq n\}\to \mathbb{N}$ is a function then there exists $M \in \mathbb{N}$ such that $f(i)\leq M$ for all $1\leq i \leq n$"

--

The sum on finite and infinite sets are defined as follows:

DEF. (Summations over finite sets)

Let $X$ be a finite set with $n \in \mathbb{N}$ elements, and let $f:X\to \mathbb{R}$ be a function.
Then $\sum_{x\in X}f(x):= \sum_{i=1}^n f(g(i))$, where $g:\{i\in\mathbb{N}: 1 \leq i \leq n\}\to X$ is any bijection from
$\{i\in\mathbb{N}: 1 \leq i \leq n\}$ to $X$; such a bijection exists since $X$ is assumed to have $n$ elements.

DEF. (Series on countable sets)

Let $X$ be a countable sets, and let $f:X\to\mathbb{R}$ be a function.
We say that the series $\sum_{x\in X}f(x)$ is $absolutely$ $convergent$ iff the sum $\sum_{n=0}^\infty f(g(n))$ is absolutely convergent for some bijection $g:\mathbb{N}\to X$. We then define $\sum_{x\in X}f(x):=\sum_{n=0}^\infty f(g(n))$.

It can be shown that these two definitions do not depend on the choice of $g$, and so are well defined.

答:
What is your definition of convergence of $\sum_{x\in A}f(x)$?

Here is a solution in the language of convergence of nets, which seems the natural way to deal with theses kinds of problems.

Some generalities:

A directed set $(D,\leq)$ is a partially order set such that if $c,d\in D$, then there is $g\in D$ such that $c\leq g$ and $d\leq g$.

A (numeric) net is a function $F:(D,\leq)\rightarrow\mathbb{C}$ . The net $F$ converges to some number $\ell\in\mathbb{C}$ if for any $\varepsilon>0$, there is $d_\varepsilon\in D$ such that $|F(d)-\ell|<\varepsilon$ whenever $ d_\varepsilon\leq d$.

In your setting, consider the collection $\mathcal{D}$ of all finite subsets of $X$ and define $A\leq B$ iff $A\subset B$. Then $F(C)=\sum_{x\in C}f(x)$   defines a net. Your question may be reformulated as showing that the net $\tilde{F}(C)=\sum_{x\in C}|f(x)|$ converges iff the set $V=\{\tilde{F}(C):C\in\mathcal{D}\}$ is bounded.

Suppose $V$ is bounded and let $\alpha=\sup V$. Then $\alpha>\infty$. Given $\varepsilon>0$, there is $C\in\mathcal{D}$ such that $\alpha-\varepsilon < \tilde{F}(C)$.
Notice that for any set $D\in\mathcal{D}$ such that $C\leq D$, $\tilde{F}(C)\leq \tilde{F}(D)\leq\alpha$. Putting things together gives
$$ |\tilde{F}(D)-\alpha |<\varepsilon \qquad\text{whenever} \qquad C\leq D$$
This shows that the net $\tilde{F}(C)$ converges to $\alpha$.

Conversely, suppose $\lim_{C}\tilde{F}(C)$ exists and equal $\ell$. Then, there is $C_0\in\mathcal{D}$ such that
$$ |\tilde{F}(D)-\ell|<1\qquad\text{whenever}\qquad C\leq D$$
This means that for all finite sets $D\supset C$,
$$\tilde{F}(D)<\ell+1$$
Since for any $0\leq\tilde{F}(E)\leq \tilde{F}(C)$ for all $E\subset C$, we conclude that the set $V=\{\tilde{F}(D): D\in\mathcal{D}\}$ is bounded. The first part will show that in fact $\ell=\sup V$.


问:
Transformation matrix for polynomials
Let $K= \mathbb{R}$ and $V:= \mathbb{R}[t]_2$ and let $F: \mathbb{R}[t]_2 \to \mathbb{R}[t]_2, f \mapsto f(0)+f(1)\cdot(t+t^2)$.

I want to calculat $M^{S}_{S}(F)$, the transformation Matrix of F with Basis S, being the standardbasis for the polynomialring $S = (1,t,t^2 )$.

So we have to use the coordinate isomorphism:

$M^{S}_{S}(F) = \begin{pmatrix} I_S(F(1)) & I_S(F(t)) & I_S(F(t^2)) \end{pmatrix}$

I already know the solution, however I have no idea why the following matrix is the transformation Matrix of F.

$M^{S}_{S}(F) = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}$.

Looking at the transformation matrix we know that $F(1)  = 1+t+t^2$ ?
I've never seen anything like this before and I'm really confused. My teacher said this is really easy and now I feel stupid :(

Title: Transformation matrix for polynomials

Body: Given a polynomial $f(t) \in V$, $$F(f(t)) = f(0)+f(1)(t+t^2).$$ So, when $f(t)=1$, we have $f(0)=f(1)=1$ and then $F(1) = 1+1(t+t^2) = 1+t+t^2$.


问:
Are vectors written using square or round brackets?
Are vectors commonly written as

$$\vec{v}=\begin{bmatrix}a\\b\\c\end{bmatrix}$$

or

$$\vec{v}=\begin{pmatrix}a\\b\\c\end{pmatrix}$$

I've learned in school that one should always use round brackets and this is what I usually (if not always) see on the internet or in books. However, in this 3blue1brown series, it is said that vectors are written with square brackets.

This got me wondering whether there is any "official" convention or if there are any reasons why one should be used over the other. If not, which of these is most commonly used?

Note: I'm not sure if this is relevant, but I am primarily talking about vectors in linear algebra (which is, as far as I know, anyway the field where vectors are most commonly used), but if it differs, an insight into the notation of vectors in other fields also seems interesting.

答: I learnt the notation as round brackets, and I would suspect (don't quote me on this) - most people also learnt with round brackets. However, it does not matter in the grand scheme of things. Ultimately you can use any notation you want (unless you are trying to do some exam or homework or something, don't do this, as it'll make it harder for the instructor to mark and they may give you $0$ because of this). Symbols only have the meaning we give them. If I wanted to be really crazy, I could swap the symbols $1$ and $2$ around, so now $2$ is the smallest positive integer, and ${2+2=1}$ (that was really hard to type).


Title: Which number is greater? $2\sqrt{2}$ or $e$

Body:
I have to determine which number is greater, $2\sqrt{2}$ or $e$.

I had a similar question as well, it was $2^\sqrt{2}$ compared to $e$.

For that one I managed to prove the inequality by using the increasing sequence converging to $e$: $(1+\frac1n )^n $

So I just searched for a value to assign to n such that $(1+\frac1n )^n \gt 2^\sqrt2$

I tried to proceed in a similar way with $2\sqrt{2} \gt \lt e$ , but it seems I can't get nowhere (I used the sequence decreasing and converging to $e$:  $(1+\frac1n )^{n+1}$ )

Is there another way to prove the inequality without the use of the calculator and maybe using derivatives? The question was in a derivatives file, so I'm wondering is there's a way to get to the end using them.

Any hint would be much appreciated, thanks.

Body:
Consider the series for $e^{-1}$, which is alternating. Then
$$
e^{-1}>1-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}-\frac{1}{120}=\frac{11}{30}
$$
and
$$
\frac{11}{30}>\frac{1}{2\sqrt{2}}
$$
because $121\cdot8=968>900$.


问:
Free software / online tool to plot complex functions as maps of grids
I teach an introductory course in Complex Analysis and it would help me a lot if I could use a free online tool to plot how a particular function maps a given grid in $z$-plane to $w$-plane but I am unable to find anything close to what I want. To specify a bit more:

Needham's book Visual Complex Analysis uses exactly this approach;

I am NOT interested in anything using coloring plotting;

I tried Sage but could not find what I wanted (it would be great if Sage did because it is a free pretty much universal math software).

Do you possibly know about such a tool? I know one could write a code for that but that is not what I am asking about - ideally, one would define a function (an elementary one) and then specify a grid (say an $x$-range and $y$-range and number of horizontal and vertical lines) and it would draw the picture of the grid transformed by the function.

答:
I recently started studying complex functions and I found out that you can really easily plot a conformal map using https://www.geogebra.org/graphing.
It may be hard starting out because I too haven't found out how to directly tell the machine that what I want is a complex function.
What I start by doing is to write this equation z*e^(5,6) equation (this is basically to tell that what you want is a conformal map, I have yet to discover why this works).
If nothing appears on your plot, just click on that white button at the left of the equation White button.
Once you've written that equation, the machine is ready to give you your conformal maps, you can now write something like 1/z or z^2 or sin(z), and maps like those should start appearing: 1/z, z^2, sin(z)</p>


问:
Sum of closed subsets with at least one compact is also closed

Let $X$ be a linear topological space. If $A$ and $B$ are two closed subsets of $X$, and at least one of them is compact, prove that $A+B$ is also closed.

As I'm taking my first steps in the field of functional analysis, I've come across this problem. Even though I'm aware of what each term refers to, I'm not sure how to put them together. I was thinking by starting with supposing that $x \notin A+B$ and eventually reach a contradiction. Some clarifications would be more than welcomed, thanks!

答:
HINT: Suppose without loss of generality that $A$ is compact, and suppose that $x\in\operatorname{cl}(A+B)$; we want to show that $x\in A+B$. Since $x\in\operatorname{cl}(A+B)$, there are nets $\langle a_i:i\in I\rangle$ in $A$ and $\langle b_i:i\in I\rangle$ in $B$ such that the net $\langle a_i+b_i:i\in I\rangle$ converges to $x$. Since $A$ is compact, the net $\langle a_i:i\in I\rangle$ has a subnet $\langle a_j:j\in J\rangle$ converging to some $a\in A$.

Show that $\langle a+b_j:j\in J\rangle$ converges to $x$.

Use the fact that $B$ is closed to conclude that $x\in A+B$.

If $X$ is first countable, and in particular if $X$ is normed, you can replace nets and subnets with sequences and subsequences.


问:
I don't understand equivalence classes with relations
I am not quite understanding equivalence classes. For example I have this problem:

Let $A$ be the set of integers and
$\quad a\;R\;b\quad$ if and only if $\quad |a| = |b|$.

I have proved that this is an equivalence relation, (its reflexive, symmetric and transitive), but how do I show the equivalence classes?

答:
Equivalence classes (wrt your equivalence relation) are subsets of elements of the original set with the following property: every element of a certain equivalence class must be equivalent (wrt the equivalence relation) to any other in it and inequivalent to any other out of it.

This implies that equivalence classes are disjoint subsets of the original set. (*)

You are asked to construct the maximal subsets of $\mathbb Z$, such that - picked one - every element in it has the same absolute value as any other element of the subset. It should be fairly easy to understand how many elements can belong to every equivalence class in this case (there is an important exception, though: $0$).

Notice that in general the cardinalities of the equivalence classes of a set wrt an equivalence relation are different!

(*) This second part of the definition was added after the useful comments of two users below.


问:
How to construct a chamfered dodecahedron with equilateral or coplanar faces?
According to wikipedia, "The chamfered dodecahedron is a convex polyhedron with 80 vertices, 120 edges, and 42 faces: 30 hexagons and 12 pentagons. It is constructed as a chamfer (geometry) (edge-truncation) of a regular dodecahedron. The pentagons are reduced in size and new hexagonal faces are added in place of all the original edges." And according to this question, the steps to create a chamfered polyhedron are as follows:

shrink the given polyhedron's faces by a factor $r, 0<r<1$

translate the shrunken faces outward [I'll call this distance $s$], so that their edges are coplanar with the
original polyhedron's edges

construct the required hexagons from the original polyhedron's vertices and the vertices of the shrunk faces.

I haven't found any other explicit details regarding the relationship between $r$ and the shape of the hexagons, and so my questions are these (note that the value of $s$ will be directly dependent on the shape of the original face, and so $s$ will need to be represented with some parameter(s) that represent the face's shape somehow):

What is the general relationship between $r$ and $s$ in order to get coplanar vertices? Does this relationship change depending on how many times the polyhedron is chamfered?

What is the general relationship between $r$ and $s$ in order to get equilateral vertices? Does this relationship change depending on how many times the polyhedron is chamfered?

For equilateral and coplanar vertices, what value does $r$ need to be so that the diameter of the resulting hexagon equals the original side length? (ie. given pentagons that share edge $AB$, the resulting hexagon between the two would have a diameter equal to the length of $AB$ - shown in the images below) I assume this relationship could be obtained by setting up the answers to the previous questions as a system of equations

Note that this is different than my question here as it takes a different approach in constructing the chamfered faces - scaling along the relative Z axis vs. translating.

答:
tl;dr If $y$ represents the side length of a face, $b$ the distance between a face's center and $y$'s midpoint, $l$ the distance between the face's center and a vertex on $y$, and $\theta_0$ the angle between the face and the adjacent face across $y$, then the expressions for $r$ and $s$ to get equilateral (only for the first two chamfers) and coplanar faces are:

$ \begin{align} \theta=\frac {180-\theta_0}{2},\quad u&=b\,tan(\theta),\quad y=2\,\sqrt {l^2-b^2} \\ r&=1-\frac {y}{\sqrt {l^2+u^2} + y} \\ s&=\frac {u\,y}{\sqrt {l^2+u^2} + y} \end{align} \tag {0}$

So I managed to figure out the answer to each of the questions - what was hindering my progress was trying to figure it all out without the second image below. Once I figured that out, everything else just fell into place! The following image shows all the variables used (note that $\theta_0$ is slightly misleading this this image: the second image below is more accurate):

1. Relationship between $r$ and $s$ so that the vertices are coplanar

In order for the vertices to be coplanar, they all need to be on the same plane that connects the two new pentagons together, which is shown in the first image below. This was the main thing that stumped me, as I didn't know how to easily represent "coplanar vertices" without a whole lot of unknowns with vectors. This drastically simplified the workflow, and the relationship between $s$ and $r$ was quickly found (note that $\theta_0$ is the angle between the two original faces, and half of the remaining angle - $180-\theta_0$ - gets divided between them, which is why $\theta=\frac {180-\theta_0}{2}$):

$ \begin{align} \theta&=\frac {180-\theta_0}{2} \\ c&=\frac{s}{tan(\theta)} \\ r&=1-\frac{c}{b} \\ r&=1-\frac{s}{b\,tan(\theta)} \\ s&=(1-r)\,b\,tan(\theta) \end{align} \tag {1}$

2. Relationship between $r$ and $s$ so that the vertices are equilateral

This is was notably more challenging than the previous solution, and it even tricked me into thinking I had solved it a couple times - but to make things easier, I conjured up $y$ to represent the side length of the original face (not shown in the images unfortunately: this is the line perpendicular to $b$). The solution for $s$ isn't that bad, but getting $r$ was certainly not as pretty as the one before (note that only two variables from $b$, $l$, and $y$ are required, as they are connected by the expression $l=\sqrt {b^2+\frac {y^2}{4}}$)...

$ \begin{align} r&=1-\frac {d}{l} \\ d&=\sqrt {x^2-s^2} \\ r&=1-\frac {\sqrt {x^2-s^2}}{l} \\ x&=r\,y \\ r&=1-\frac {\sqrt {r^2\,y^2-s^2}}{l} \\ s&=\sqrt {r^2\,y^2 - l^2\,(1-r)^2} \\ 0&=r^2\,(y^2-l^2)+r\,(2\,l^2)+(-l^2-s^2) \\ r&=\frac {-l^2+\sqrt {l^4-(y^2-l^2)(-l^2-s^2)}}{y^2-l^2} \end{align} \tag {2.1}$

Alternatively, if we replace $r$ with $1-\frac {c}{b}=1-\frac {s}{b\,tan(\theta)}$ (instead of just leaving it in and then solving for a quadratic later), a less complex solution can be found for $r$:

$ \begin{align} r&=1-\frac {s}{b\,tan(\theta)} \\ r&=1-\frac {\sqrt {r^2\,y^2-s^2}}{l} \\ r&=1-\frac {\sqrt {y^2\,(1-\frac {s}{b\,tan(\theta)})^2-s^2}}{l} \end{align} \tag {2.2}$

3. Relationship between $r$ and $s$ so that the vertices are coplanar and equilateral

Given that I need both coplanar and equilateral faces here, I simply used a system of equations using both the previous solutions and solved for $r$ and $s$ independently. There was a lot of simplifying done and the process was not pretty in the least, so for this answer I just laid out the groundwork and then skipped to the solution - let me know if you wish to see the entire process, or even if you find an easier/better way to do this! For the sake of clarity, I replaced all instances of $b\,tan(\theta)$ with $u$:

Solving for r (using the upper root):

$ \begin{align} s&=(1-r)\,u \\ s&=\sqrt {r^2\,y^2-l^2\,(1-r)^2} \\ (1-r)\,u&=\sqrt {r^2\,y^2-l^2\,(1-r)^2} \\ (1-r)^2\,u^2&=r^2\,y^2-l^2\,(1-r)^2 \\ &... \\ r&=\frac {-(l^2+u^2)+\sqrt {(l^2+u^2)^2-(y^2-l^2-u^2)\,(-l^2-u^2)}}{y^2-l^2-u^2} \end{align} \tag {3.1}$

Solving for s (using the lower root):

$ \begin{align} r&=1-\frac{s}{u} \\ r&=1-\frac {\sqrt {y^2\,(1-\frac {s}{u})^2-s^2}}{l} \\ 1-\frac{s}{u}&=1-\frac {\sqrt {y^2\,(1-\frac {s}{u})^2-s^2}}{l} \\ \frac{s^2}{u^2}&=\frac {y^2\,(1-\frac {s}{u})^2-s^2}{l^2} \\ &... \\ s&=\frac {u\,y^2-\sqrt {u^2\,y^4-(y^2-l^2-u^2)\,(u^2\,y^2)}}{y^2-l^2-u^2} \end{align} \tag {3.2}$

Thanks to WolframAlpha, a much simpler solution could be produced if all the values are assumed to be positive (given that we will never have negative distance or angles, this is a valid assumption):

$ \begin{align} s&=\frac {u\,y}{\sqrt {l^2+u^2} + y},\;u=b\,tan(\theta),\;y=2\,\sqrt {l^2-b^2} \end{align} \tag {3.3}$

And given that $r=1-\frac {s}{u}$, this gives us a nice solution for $r$ as well:

$ \begin{align} r&=1-\frac {y}{\sqrt {l^2+u^2} + y},\;u=b\,tan(\theta),\;y=2\,\sqrt {l^2-b^2} \end{align} \tag {3.4}$

In applying this to my dodecahedron (the magnitude of each vertex being $1$), I can now show that the solutions I discovered are correct (credit for the values for $r$ and $s$ goes to Blue's answer - note that these values are all approximations, rounded to the fourth digit-place):

Given: $b=0.4911,\quad l=0.6071,\quad y=0.7136,\quad \theta=\frac {180-116.5650}{2}=31.7175,\quad u=b\,tan(\theta)=0.3035$

Unknown: $r=0.4875,\quad s=0.1556$

1. Relationship between $r$ and $s$ so that the vertices are coplanar

$ \begin{align} r&=1-\frac{s}{b\,tan(\theta)}&=1-\frac{0.1556}{0.4911\,tan(31.7175)}&=.4873 \\ s&=(1-r)\,b\,tan(\theta)&=(1-0.4875)\,0.4911\,tan(31.7175)&=.1556 \end{align}$

2. Relationship between $r$ and $s$ so that the vertices are equilateral

$ \begin{align} r&=1-\frac {\sqrt {y^2\,(1-\frac {s}{b\,tan(\theta)})^2-s^2}}{l} &=1-\frac {\sqrt {0.7136^2\,(1-\frac {0.1556}{0.4911\,tan(31.7175)})^2-0.1556^2}}{0.6071}&=.4877 \\ s&=\sqrt {r^2\,y^2 - l^2\,(1-r)^2}&=\sqrt {0.4875^2\,0.7136^2 - 0.6071^2\,(1-0.4875)^2}&=.1556 \end{align}$

3. Relationship between $r$ and $s$ so that the vertices are coplanar and equilateral

$ \begin{align} r&=1-\frac {y}{\sqrt {l^2+u^2} + y}&=1-\frac {0.7136}{\sqrt {0.6071^2+0.3035^2} + 0.7136}&=.4875 \\ s&=\frac {u\,y}{\sqrt {l^2+u^2} + y}&=\frac {(0.3035)\,(0.7136)}{\sqrt {0.6071^2+0.3035^2} + 0.7136}&=.1555 \end{align}$

Update

These equations may work for future subdivisions/chamfering as well, however faces will not continue to be equilateral (while still being coplanar). This is to be expected, however an issue arises of how to uniformly scale and translate the vertices of a face if it is not equilateral (given that $b$, $l$, and $y$ may vary per vertex). As such, my chamfering process would need to be updated so that it works with such faces.


问:
vector outer product squared identity
I am trying to understand the proof for the rank-one perturbation of the identity and everywhere I look the following statement is always used

for $x,y \in \mathbb{R}^{m}$ , $xy^T xy^T$ is $xy^T$ multiplied by the scalar $y^T x$. In other words  $xy^T xy^T = (y^T x ) xy^T$.

Can someone help me show this claim or redirect me to somewhere where it's explained, I can't get it right for the life of me, I tried taking the transpose or rearranging with no success.

答: Note $xy^T$ is a matrix. Summing repeated indices,$$[(xy^T)^2]_{ij}=(xy^T)_{ik}(xy^T)_{kj}=x_i\underbrace{y_kx_k}_{y\cdot x}y_j=(y\cdot x)(xy^T)_{ij},$$i.e. $(xy^T)^2=(y\cdot x)(xy^T)$.

Title: vector outer product squared identity


问:
Last Digit of the Sum of Fibonacci Numbers
I want to calculate the last digit of a sum of Fibonacci numbers:
$F_{m} + F_{m + 1} + \cdots + F_{n}$.

$m$ and $n$ are $2$ non-negative integers and $m \leq n$.

I have one sample input with output $m = 10, n=10\ \mbox{and}\ o/p = 5$, which I don't understand.

Because if I do sum $F_{10} + F_{10} = 55 + 55 = 110$. So the last digit is 0. I don't get the part that the answer is $5$.

答: The sum $a_m + \ldots + a_n$, when $m = n$, is just $a_m$. One term. So in your case, $F_m + \ldots + F_n$ for $m = n = 10$ is $F_{10} = 55$.


问:
Why is $SO(3)$ diffeomorphic to $\mathbb{R}P^3$?
I've been trying to prove this, but I got stuck. First, I defined the next fuction

$$\varphi : SO(4)\times \mathbb{R}P^3\to\mathbb{R}P^3$$
$$\quad \quad \quad \quad(A,\overline{x})\mapsto \overline{Ax}$$

I already proved that $\varphi$ is a transitive action. Since, $SO(4)$ is a connected Lie group, then

$$SO(4)/I_x\cong \mathbb{R}P^3$$

where $I_x$ is the stabilizer soubgroup and $x=(1,0,0,0)^T$. I guess $SO(4)/I_x \cong SO(3)$, but I couldn't prove it. I'd like to know if it is true.

I've seen another questions in which they give another ways to prove this, but I'd like to complete mine. Thanks.

答:
Indeed your idea to apply the orbit-stabilizer makes sense. But you seem to be thinking about the orbit and the stabilizer backwards. We say $SO(4)/I\simeq \mathbb{RP}^3$ as $SO(4)$-spaces, where $I$ is the stabilizer subgroup of $SO(4)$. That is, we quotient by the stabilizer to get the space, we don't quotient to get a subgroup. (That said, $\mathbb{RP}^3$ is diffeomorphic to $SO(3)$, but we can't use the orbit-stabilizer with these groups to conclude that very easily.)

But, your claim about the stabilizer is incorrect. Say $x=(0,0,0,1)^T$. If you write out the condition $Ax=x$ explicitly, you will see $A$ must be block diagonal with a $3\times3$ block in the upper left and a $1\times 1$ block in the lower right. But $A$ is an orthogonal matrix, which forces the $3\times 3$ and $1\times 1$ blocks to be orthogonal matrices as well. However, $\det A=+1$, so either both blocks have determinant $-1$ (to cancel out) or both blocks have determinant $+1$.

In other words, the stabilizer is $S(O(3)\times O(1))$ (this is the notation for it). It contains $SO(3)\times SO(1)$ as an index $2$ subgroup, with complement $[O(3)\setminus SO(3)]\times\{-1\}$ as its other coset. This group isn't the same set as $O(3)\times SO(1)$ (another block-diagonal subgroup of $O(4)$) however it is isomorphic to $O(3)$, with the isomorphism $O(3)\to S(O(3)\times O(1))$ given by $X\mapsto (X,\det X)$ (that is, we can turn a $3\times 3$ orthogonal matrix $X$ into a $4\times 4$ special orthogonal matrix that stabilizes $x$ by putting $X$ in the upper left corner and $\det X$ in the lower right corner).

The easiest way to see $SO(3)\simeq\mathbb{RP}^3$ in my opinion is with quaternions. The unit quaternions are $S^3$, the three-sphere, and there is a map $S^3\to SO(3)$ which is onto with kernel $S^0=\{\pm1\}$ (are you familiar with this map?). Then that means $S^3/S^0=SO(3)$, but $S^3/S^0$ is just $\mathbb{RP}^3$!