Title: Half order derivative of $ {1 \over 1-x }$

Body:
I'm new to this "fractional derivative" concept and try, using wikipedia, to solve a problem with the half-derivative of the zeta at zero, in this instance with the help of the zeta's Laurent-expansion.

Part of this fiddling is now to find the half-derivative  $$ {d^{1/2}\over dx^{1/2}}{1 \over 1-x}$$

First I would like to understand, whether there is a short/closed form for this ot whether I have to express the fraction as a power series first and then to differentiate termwise.

Next I would like to know the value at $x=0$.

Body:
$$\frac{d^k}{dx^k} \left(\frac{1}{1-x} \right)=\frac{d^k}{dx^k} (1+x+x^2+...)=\Gamma(k+1) \sum_{n=k}^{\infty} {n\choose k}x^{n-k}\\
\sum_{n=k}^{\infty} {n\choose k}x^{n-k}=\frac{1}{(1-x)^{k+1}}\\
\frac{d^k}{dx^k} \frac{1}{1-x}=\frac{\Gamma(k+1)}{(1-x)^{k+1}}$$


问:
Expected Value : Poker
You are dealt $4$ cards face down on a table. You know that $2$ are red and $2$ are black. Let’s say you are given $100 to bet on this game and you are able to bet after each card is dealt and before the next one is on whether the outcome will be red or black. If you play this game optimally, what’s your expected value?

Thinking along the lines that at first there is 1/2 probability of red or black. If first is red then second round has probability of picking black 2/3. If second round is black then 1/2 probability that 3 rd round has black or red. Final round will have probability 1. I was thinking that this was optimal play.

Edit - actually I think I messed up round 3 probability.

答:
This is an extension to a classical dynamic programming problem that involves optimal stopping. I will present a solution that works in the general case.

Let $f(r, b, x)$ denote the optimal expected value for $r$ and $b$ remaining red and blue cards, respectively when we have $x$ remaining dollars. Under this notation, we want to compute $f(2, 2, 100)$. Note that we have the following recurrence:

$$f(r, b, x) = \max_{0 \leq y \leq x}\left\{\frac{r}{r + b} \cdot (y + f(r - 1, b, x + y) + \frac{b}{r + b} \cdot f(r, b - 1, x - y), \frac{b}{r + b} \cdot (y + f(r, b - 1, x + y) + \frac{r}{r + b} \cdot f(r - 1, b, x - y)\right\}.$$

since we have the option to bet anywhere between $0$ and $x$ dollars (inclusive) at any given state, and it doesn't make sense to bet on the less probable option. Moreover, we have the base cases $f(1, 0, x) = 2x$ and $f(0, 1, x) = 2x$ since we're certain to make money when there's only one card remaining (just bet that color).

You can use a dynamic programming algorithm that runs in $\mathcal{O}(RBM)$ time, where $R$ denotes the initial number of red cards, $B$ denotes the initial number of blue cards, and $M$ denotes the initial bankroll.

ContentLicense: CC BY-SA 4.0

Title: Expected Value : Poker


问:
How to convert $\lim_{n\to\infty}\frac{\pi}{n}\sum_{k=1}^{n}\sin\left(\frac{\pi}{2}+\frac{5\pi}{2}\cdot\frac{k}{n}\right)$ into an integral?
The question asked is:

$\displaystyle \lim_{n\to \infty} \frac{\pi}{n} \sum_{k=1}^{n} \sin\left(\frac{\pi}{2}+\frac{5\pi}{2}\cdot\frac{k}{n}\right) = ? $

a) $\dfrac{2\pi}{5} \hspace{3em}$
b) $\dfrac{5}{2} \hspace{3em}$
c) $\dfrac{2}{5} \hspace{3em}$
d) $\dfrac{5\pi}{2}$

The solution is given as:src)

\begin{align*}
&\lim_{n\to \infty} \frac{\pi}{n} \sum_{k=1}^{n} \sin\left(\frac{\pi}{2}+\frac{5\pi}{2}\cdot\frac{k}{n}\right)  \\
&= \frac{2}{5} \int_{0}^{5\pi/2} \sin\left(\frac{\pi}{2}+x\right)\,\mathrm{d}x
= \frac{2}{5} \int_{0}^{5\pi/2} \cos x \,\mathrm{d}x \\
&= \frac{2}{5} \left[ \sin x \right]_{0}^{5\pi/2}
= \frac{2}{5} \left[ \sin\frac{5\pi}{2} - \sin 0 \right] \\
&= \frac{2}{5}\sin\left(2\pi + \frac{\pi}{2}\right)
= \frac{2}{5}\cdot\sin\frac{\pi}{2}
= \frac{2}{5}
\end{align*}

My question is how they got

$$ \lim_{n\to \infty} \frac{\pi}{n} \sum_{k=1}^{n} \sin\left(\frac{\pi}{2}+\frac{5\pi}{2}\cdot\frac{k}{n}\right) = \frac{2}{5} \int_{0}^{5\pi/2} \sin\left(\frac{\pi}{2}+x\right)\,\mathrm{d}x. $$

Thanks in advance!

答:
$$ \lim_{n \to \infty} \frac{b-a}{n} \sum_{k=1}^n f( a+ k( \frac{b-a}{n})) = \int_a^b  f(x) dx$$

$$\displaystyle \lim_{n\to \infty} \frac{\pi}{n} \sum_{k=1}^{n}
\sin\left(\frac{\pi}{2}+\frac{5\pi}{2}\cdot\frac{k}{n}\right) = ? $$

Taking $f(x) = \sin(x)$,

By direct comparison,

$$ a= \frac{\pi}{2}$$

And,

$$ \frac{ 5 \pi}{2} \frac{k}{n} = \frac{k}{n} (b-a)  $$

Or,

$$  \frac{5 \pi}{2} = b -a$$

Or,

$$  b = 3\pi$$

Hence, this is equal to:

$$\frac{2}{5} \int_0^{3\pi} \sin x = \frac{2}{5}$$

Note: Note I multiplied by $ \frac{2}{5}$ to make it $ \frac{b-a}{n}$ form

The solution

If instead of taking,

$$ f(x) = \sin(x)$$

You took,

$$ f(x) = \sin( \frac{\pi}{2} + x)$$

Then,

$$ f(a + k \frac{b-a}{n}) = \sin( \frac{\pi}{2} + a +k \frac{b-a}{n} )$$

Compare this with the question,

$$\displaystyle \lim_{n\to \infty} \frac{\pi}{n} \sum_{k=1}^{n}
\sin\left(\frac{\pi}{2}+\frac{5\pi}{2}\cdot\frac{k}{n}\right) = ? $$

hence,

$$ a= 0 $$

$$ \frac{5\pi}{2}  = b$$


问:
Prove $\sum_{k = 0}^{n}(-1)^{n - k} \binom{n}{k} \cdot k^n = n!$ and $\sum_{k = 0}^{n}(-1)^{n - k} \binom{n}{k} \cdot k^m = 0$

Prove
$$\sum_{k = 0}^{n}(-1)^{n - k} \binom{n}{k} \cdot k^n = n! \qquad\text{and}\qquad \sum_{k = 0}^{n}(-1)^{n - k} \binom{n}{k} \cdot k^m = 0,$$
where $m \in \{0, 1, 2, \cdots, n - 1\}$.

To be honest, I tried to use some combinatorial ideas but was not able to see interconnections. Also, I had an idea to prove it using induction but failed to do that either.

Help will be much appreciated! I just got stuck and don't see any continuation of solution for this problem

答:
Here are some hints:

The $(-1)^{n - k}$ suggests to interpret this as inclusion exclusion.

$k^n$ is the number of functions from an $n$ element set to a $k$ element set

$n!$ is the number of functions from an $n$ element set to an $n$ element set which are injections/surjections/bijections

So you can think of the term $\binom{n}{k}k^n$ as the number of functions from $\{1,\dots,n\} \to \{1,\dots,n\}$ whose image is contained in a chosen $k$-element subset of the codomain.


问:
Decompose into simple fractions over $\mathbb{C}$

Decompose into simple fractions over $\mathbb{C}$:
$$f(x) = \frac{1}{(x^2-1)^n}$$

I know how to decompose some fractions by $\mathbb{R}$, but I have no idea how analyze this over $\mathbb{C}$ and solve dependency from $n$.

Body: If, for instance, $n=3$, you can do it by writing $\frac1{(x^2-1)^3}$ as$$\frac{A_1}{x+1}+\frac{A_2}{x-1}+\frac{A_3}{(x+1)^2}+\frac{A_4}{(x-1)^2}+\frac{A_5}{(x+1)^3}+\frac{A_6}{(x-1)^3},$$This will lead you to the system$$\left\{\begin{array}{l}-A_1+A_2-A_3-A_4-A_5+A_6=1\\A_1+A_2+2A_3-2A_4+3A_5+3A_6=0\\2A_1-2A_2-3A_5+3A_6=0\\-2A_1-2A_2-2A_3+2A_4+A_5+A_6=0\\-A_1+A_2+A_3+A_4=0\\A_1+A_2=0.\end{array}\right.$$Its solution is$$A_1=-\frac3{16},\ A_2=\frac3{16},\ A_3=-\frac3{16},\,A_4=-\frac3{16},\ A_5=\frac18,\text{ and }A_6=-\frac18.$$


问:
An abelian group proof with $g*g=e$ for all $g$.
I have to show that the following group
$$ (G, * , e) $$
with its operation $*$, which is defined through $ g*g = e$  for every $g \in G $ is an abelian group.

In order to do that one have only to show that the group is commutative.

How can one prove it whereas the operation is defined always between an Element and itself?

I reckon it is not so simple as it seems

Thanks in advance for your help :)

答: Consider $$(a\ast b)\ast (a\ast b)=e=(a\ast a)\ast(b\ast b)$$ then cancel $a$ on the left and $b$ on the right.


问:
The number of words that can be made by permuting the letters of _MATHEMATICS_ is
The number of words that can be made by permuting the letters of MATHEMATICS is

$1) 5040$

$2) 11!$

$3) 8!$

$4) 4989600$

First of all I do not understand the statement of the problem, I would like if some one tell me with an example.

答: I am describing solution.MATHEMATICS contains 11 letters so they can be arranged by $11!$ ways but in this word there is some repeating letters M is 2 times,A is 2 times, T is 2 times.So in $11!$ ways there are some permutations presents which are repeat so we have to remove those permutations that's why divide the $11!$ by $2!\cdot 2!\cdot 2!$ (for 2 M,2 A and 2 T).suppose 1 M is on 5th place and 2nd M is on 6th place than if we swap their place the word is remain same.</p>

ContentLicense: CC BY-SA 3.0


答:
to answer your actual question.

When the book says $*$ is defined to be that $g*g=e$ for all $g\in G$, the book is abusing language.  That isn't the definition of $*$; it is a property (a very significant property) of $*$ but not the definition of $*$.

If $G$, the set, is $\{e,a,b,c,d,e,f,g\}$ for example we know that $a*a=e$ and $c*c = e$ but we don't know what $ab$ is equal to;  It could by $ab = c$ or it could be $ab = f$.  But it doesn't matter.

......

We can prove that for all $a,b \in G$ that whatever $ab$ is equal to, that $ba$ must also be equal to the same value.

Note that if $ab = k$ and $k*k = e$ we have $(ab)(ab) = e$.  But we also have $(ab)(ba) = a(bb)a = aea =aa = e$.  So we have both $(ab)(ab) = e$ and $(ab)(ba) = e$.

If we rely on the proof we should have proven earlier that inverses are unique that means that $ab = (ab)^{-1}$ and $ba = (ab)^{-1}$ so $ab = ba$.

.....

If we want to really convince ourselves and we don't mind reinventing the wheel:

$(ab)(ab) = e$

$(ab)(ab)(ba) = (ba)$

$(ab)a(bb)a = ba$

$(ab)aa = ba$

$(ab) = ba$

=====

The thing is that other than examples to teach the concept of groups the actual operation $*$ doesn't matter and we almost never define what $a*b$ is equal to. And if $ab = c$ or $ab=f$ it won't matter.  That's just labelling.  What matters is only certain properties.

Title: An abelian group proof with $g*g=e$ for all $g$.

Body:
I have to show that the following group
$$ (G, * , e) $$
with its operation $*$, which is defined through $ g*g = e$  for every $g \in G $ is an abelian group.

In order to do that one have only to show that the group is commutative.

How can one prove it whereas the operation is defined always between an Element and itself?

I reckon it is not so simple as it seems

Thanks in advance for your help :)


问:
equivalence of Rep$(Q)$ and Mod($KQ$). Does it preserve length/dimension?
I am studying representation theory at the moment and I was wondering the following (Since I found no source discussing it):

Let $Q$ be a quiver and let $KQ$ be its path algebra. Let Rep($Q$) be the category of representations of $Q$ and Mod($KQ$) be the category of modules over $KQ$.

Now in representation theory one finds that there is an equivalence between Rep($Q$) and Mod($KQ$).

Now my question is: Given a representation, can you determine by looking at its dimension vector the length of the module that is associated to it?

Here is an example:

Let $Q$ be the quiver $1 \rightarrow 2 \rightarrow 3$ and $K \stackrel{id}{\longrightarrow} K \longrightarrow 0$ a representation of $Q$. Can one say that the module associated to it has length $2$?

答:
When the quiver has no oriented cycles, then the simple modules of $KQ$ are all one dimensional, and are indexed by the vertices. (In fact, the Jacobson radical is the ideal generated by all the arrows, and the semisimple quotient $KQ/J(KQ)$ is isomorphic to $K^n$, where $n$ is the number of vertices.) Thus the length of a module equals its dimension as a $K$-vector space, which in turn equals the dimension of the corresponding representation.

This breaks down when the quiver has oriented cycles, since then there are simple modules which are not one dimensional. For example, when we have one vertex and one loop, so the path algebra is $K[t]$, then simple modules correspond to monic irreducible polynomials, and the dimension equals the degree of the polynomial. This is always one if and only if the field $K$ is algebraically closed.

If the quiver has two vertices $1,2$ and arrows $1\to2$ and $2\to1$, then there is always a 2-dimensional simple module where both vector spaces are $K$ and both maps are non-zero.


问:
Orthogonal matrix with usual dot product
I know the following matrix
$$A=\begin{bmatrix} \sin(x) & x \\ 1 & \cos(x) \end{bmatrix}\quad B=\begin{bmatrix} \sin(x) & x+1 \\ -1 & \cos(x) \end{bmatrix}$$
with $x\in\mathbb{R}$ are orthogonal with the usual dot product and I have to find the value of $x$.

I tried to make $A^{T}\cdot B=B\cdot A^{T}=Id$ but I don't think is the right way to do it cause my options for $x$ are integer numbers.

答:
You have misunderstood the question. You are not asked to find the values of $x$ that make $A$ and $B$ orthogonal matrices. Instead, you are asked to find the values of $x$ such that, when $A$ and $B$ are viewed as vectors in the vector space $\mathbb R^{2\times2}$, the two "vectors" will be orthogonal to each other with respect to the "usual dot product", which is the dot product on $\mathbb R^4$ when you treat $A$ and $B$ as two $4$-vectors. That is, you are asked what values of $x$ would make
$$
(\sin(x),1,x,\cos(x))\cdot(\sin(x),-1,x+1,\cos(x))
$$
vanishes, so that the two vectors above are orthogonal to each other. The answer should be obvious.

This inner product above is known as Frobenius inner product in the literature. It is usually expressed in equivalent form as $\langle A,B\rangle=\operatorname{tr}(AB^T)$.


答:
Consider the exponential generating function

$$ f(x) = \sum_{m=0}^{\infty} \frac{A_m}{m!} x^m, \qquad A_m := \sum_{k=0}^{n} (-1)^{n-k}\binom{n}{k} k^m.  $$

Then it follows that

\begin{align*}
f(x)
&= \sum_{k=0}^{n} (-1)^{n-k}\binom{n}{k} \sum_{m=0}^{\infty} \frac{(kx)^m}{m!} \\
&= \sum_{k=0}^{n} (-1)^{n-k}\binom{n}{k} e^{kx} \\
&= (e^x - 1)^n.
\end{align*}

Writing $e^x - 1 = x g(x)$ for $g(x) = \sum_{j=0}^{\infty} \frac{x^j}{(j+1)!}$,

$$ f(x) = x^n g(x)^n . $$

In particular, the Maclaurin series for $f$ about $x = 0$ begins with $ f(x) = x^n + \cdots $, which translates to the desired fact:

$$ A_m
= \begin{cases}
0, &\text{if $m < n$}, \\
n!, &\text{if $m = n$}.
\end{cases} $$

In general, this computation shows that

$$ A_m = \sum_{\substack{j_1\geq1,\dots,j_n\geq1 \\ j_1+\cdots+j_n=m}} \binom{m}{j_1,\dots,j_n} \tag{1} $$

for all $m = 0, 1, 2, \dots $.

Alternatively, here is an extended version of @Trevor Gunn's answer: Let $[n] = \{1, \dots, n\}$. Then define

$$ X_i = \{ f \in [n]^{[m]} : i \notin \operatorname{range}(f) \} = ([n]\setminus\{i\})^{[m]} $$

for each $i = 1, 2, \dots, n$ and

$$ X_{\varnothing} = [n]^{[m]}, \qquad X_I = \bigcap_{i\in I} X_i, $$

for each non-empty $I \subseteq [n]$. Then $A_m$ can be written as

$$ A_m
= \sum_{k=0}^{n} (-1)^k \binom{n}{k} (n-k)^m
= \sum_{I\subseteq[n]} (-1)^{\left| I \right|} \left| X_I \right|. $$

Now by the inclusion-exclusion principle, it follows that

\begin{align*}
A_m
&= \left| \{ f \in [n]^{[m]} : f \notin X_1 \cup \cdots \cup X_n \} \right| \\
&= \text{[# of surjections from $[m]$ to $[n]$]}. \tag{2}
\end{align*}

Note that this formula also matches the formula $\text{(1)}$ in the first solution. Now the desired answer easily follows from $\text{(2)}$.

Title: Prove $\sum_{k = 0}^{n}(-1)^{n - k} \binom{n}{k} \cdot k^n = n!$ and $\sum_{k = 0}^{n}(-1)^{n - k} \binom{n}{k} \cdot k^m = 0$

Body:
Prove
$$\sum_{k = 0}^{n}(-1)^{n - k} \binom{n}{k} \cdot k^n = n! \qquad\text{and}\qquad \sum_{k = 0}^{n}(-1)^{n - k} \binom{n}{k} \cdot k^m = 0,$$
where $m \in \{0, 1, 2, \cdots, n - 1\}$.

To be honest, I tried to use some combinatorial ideas but was not able to see interconnections. Also, I had an idea to prove it using induction but failed to do that either.

Help will be much appreciated! I just got stuck and don't see any continuation of solution for this problem


问:
proof verification on a sequence convergence
let $(X,d)$ be a metric space and $(x_{n})$ a Cauchy sequence on $X$ ,

$A=\left \{ x_{n}:n=1,2... \right \}$ if ${A}'\neq \varnothing $ (where ${A}'$ is the set of the limit
points of $A$)
prove that there exist a $x\epsilon X : x_{n}\rightarrow x$

my solution :

let $ \epsilon >0$
and a $N_{0}\epsilon \mathbb{N}$

since $(x_{n})$ is  Cauchy for  every $n,m\geq N_{0}$  $d(x_{n},x_{m})<\varepsilon $ (1) and ${A}'\neq \varnothing \Rightarrow $ it exist a $x\epsilon X$ such that

$B(x,\varepsilon )\cap A\setminus \left \{ x \right \}\neq \varnothing \Leftrightarrow B(x,\varepsilon )\cap \left \{ x_{n} \right \}\neq \varnothing$ (2)

from (1) ,(2) can i say that for every $n\geq N_{0}$

$d(x,x_{n})<\varepsilon  \Rightarrow x_{n}\rightarrow x$

i feel something is missing to complete 100% the proof.

答:
There are things missing indeed. For instance, you did not tell us what $N_0$ is. Or what that $\varepsilon$ is.

You can do it as follows. Take $\varepsilon>0$. Since $x\in A'$, there is an increasing sequence $(n_k)_{k\in\Bbb N}$ of natural numbers such that $\lim_{k\to\infty}x_{n_k}=x$ and therefore there is some $K_0\in\Bbb N$ such that$$k\geqslant K_0\implies\left|x-x_{n_k}\right|<\frac\varepsilon2.$$And there is a $N_0\in\Bbb N$ such that$$m,n\geqslant N_0\implies|x_m-x_n|<\frac\varepsilon2.$$But then, if $n\geqslant\max\{n_{K_0},N_0\}$, you take some $k\in\Bbb N$ such that $k\geqslant K_0$ and that $n_k\geqslant N_0$ and then\begin{align}|x-x_n|&\leqslant|x-x_{n_k}|+|x_{n_k}-x_n|\\&<\frac\varepsilon2+\frac\varepsilon2\\&=\varepsilon.\end{align}


问:
What is the complexity of $T(n) =3T((n-2)/2) + n^2$?
I have to find the complexity of a recursive function that has this form:
$$T(n) =\left\{\begin{aligned}
       &3T((n-2)/2) + k_2n^2&&\text{ when }n > 1\\
       &k_1                     &&\text{ when } n = 1
\end{aligned}\right.$$
I have met $ T(n) = a*T(n/b) + n $, which resolves with Master Method, and $ T(n) = T(n-1) + n $ which resolves with iterative, but this is mixed, and I can't find a solution to it.

I have discovered that if I try it the iterative way starting from $T(n-2)$ and continuing with $ 3^k T( n/(2^k) )$, I end up with $ T(n-2) = 3^k*3T( n / (2^k2) ) + \sum_{i=0}^k( 3^i  ( n /2^k - 2))$ but this leads nowhere. How can I find this solution?

答:
$$\bullet \quad \text{Note that: } T(n+2)=3T\bigg(\cfrac{n}{2} \bigg) +(n+2)^2=3T\bigg(\cfrac{n}{2} \bigg)  +\Theta(n^2)$$
$$\text{Master Theorem } \implies T(n+2)=\Theta (n^2 )\iff T(n)=\Theta \big( (n-2)^2\big)=\Theta (n^2)$$
$$\implies \bbox[5px,border:1.5px solid red] {\color{black}{ T(n)=\Theta(n^2)}}$$

(note: $\log_2(3)<\log_2(4)=2$)


问:
Computing a complex derivative

Let $\gamma(t) = te^{it}$. What is $\gamma'(2\pi)$?

My incorrect answer was: $2+4\pi$.

Please, tell me what I did wrong and how to solve this correctly

答:
$$
\gamma'(t) = \frac{d}{dt} \left[te^{it}\right]
 = t \frac{d}{dt} \left[e^{it}\right] + e^{it} \frac{d}{dt}[t]
 = ite^{it} + e^{it} = e^{it}[1+it]
$$
What happens when you evaluate at $t=2\pi$?

Title: Computing a complex derivative


问:
USAMO problem hint.
Prove that for every positive integer n there exists an n-digit number divisible by 5$^n$ all of whose digits are odd.

USAMO 2003.

This is the first time i have seen a problem like this, so i am not sure what to do, induction, construction, checking small cases, contradiction are some of the things i have tried.

I know I can easily find a solution anywhere but i don't want to look at a solution so please give HINTS.

I HAVE POSTED A SOLUTION USAMO problem solution HERE, PLEASE DO CHECK IT OUT.

Please don't give the full solution, any hints would be appreciated.

答:
Hint: following lulu's comment, let's suppose you have formed a number $N$ with $n-1$ odd digits divisible by $5^{n-1}$. Let's write this number as $N = p\cdot5^{n-1}$. Then you want to find an odd digit $a$ such that $a\cdot10^{n-1}+ p\cdot5^{n-1} = k\cdot5^n$ for some integer $k > 0$. This is true iff $5 | (a\cdot2^{n-1}+p)$. Writing $a = 2m+1$, can you prove that we can always find $m$? Also $m$ is mod $5$, and hence $a$ is a digit.

The base case is obvious.

Title: USAMO problem hint.


答: It'll need at least an $n^2$ behaviour, to match the last term on the right. Try $T(n)\sim cn^2$ so, for large $n$, $3T(n/2-1)\sim 3cn^2/4$. Equating $n^2$ coefficients, $c=3c^2/4+1$ i.e. $c=4$. On the other hand, a nonzero $n^p$ coeffiient can't be matched for $p>3$. So $T(n)\sim 4n^2\in\Theta(n^2)$.

Title: What is the complexity of $T(n) =3T((n-2)/2) + n^2$?

Body:
I have to find the complexity of a recursive function that has this form:
$$T(n) =\left\{\begin{aligned}
       &3T((n-2)/2) + k_2n^2&&\text{ when }n > 1\\
       &k_1                     &&\text{ when } n = 1
\end{aligned}\right.$$
I have met $ T(n) = a*T(n/b) + n $, which resolves with Master Method, and $ T(n) = T(n-1) + n $ which resolves with iterative, but this is mixed, and I can't find a solution to it.

I have discovered that if I try it the iterative way starting from $T(n-2)$ and continuing with $ 3^k T( n/(2^k) )$, I end up with $ T(n-2) = 3^k*3T( n / (2^k2) ) + \sum_{i=0}^k( 3^i  ( n /2^k - 2))$ but this leads nowhere. How can I find this solution?


问:
Homology of quotient space where points are identified along homeomorphism
Let $X$ be topological space and $f:X \rightarrow X$ be a homeomorphism. Define $M=X \times [0,1] / \sim$, where $\sim$ is the equivalence relation given by identifying each $(x,1)$ to $(f(x),0)$.

 So for example, if $X=S^1$ and $f$ is the identity map, then $M$ is the torus  and if $f$ is a reflection, then $M$ is the Klein bottle.

 I would like to compute the homology groups of $M$.

 Writing $M$ as the union $M_1 \cup M_2$, where $M_1=X \times [0,\frac{1}{2}]$ and $M_2=X \times [\frac{1}{2},1]$ (so $M_1 \cap M_2$ is two disjoint copies of $X$), applying Mayer-Vietoris, we obtain a long exact sequence
$$\dots \rightarrow H_{n+1}(M) \rightarrow H_n(M_1 \cap M_2) \rightarrow H_n(M_1) \oplus H_n(M_2) \rightarrow H_n(M) \rightarrow H_{n-1}(M_1 \cap M_2) \rightarrow  \dots$$
I know I need to express the homomorphism $H_n(M_1 \cap M_2) \rightarrow H_n(M_1) \oplus H_n(M_2)$ in terms of the induced homomorphism $H_n(f):H_n(X) \rightarrow H_n(X)$, but I am not sure how to proceed next.

答:
The space $M$ is homeomorphic to the pushout of the triad
$$X\xleftarrow{(1,f)}X\sqcup X\xrightarrow{(in_0,in_1)}X\times I$$
where $(in_0,in_1):X\sqcup X\hookrightarrow X\times I$ includes each of the factors into a different end of the cylinder. This map is a cofibration, so the resulting pushout square is excisive and you get a Mayer-Vietoris sequence
$$\dots\rightarrow H_{*+1}(M)\rightarrow H^*(X\sqcup X)\xrightarrow{(1,f)_*+(in_0,in_1)_*}H_*(X)\oplus H_*(X\times I)\rightarrow H_*(M)\rightarrow\dots$$
This is of course exactly the sequence you have already obtained.

Now the two inclusions $X\xrightarrow{i_1}X\sqcup X\xleftarrow{i_2}X$ induce an isomorphism
$$i_{1*}+i_{2*}: H_*(X)\oplus H_*(X)\xrightarrow\cong H^*(X\sqcup X).$$
Similarly the projection $pr_X:X\times I\rightarrow X$ is a homotopy equivalence which induces an isomorphism $H_*(X\times I)\xrightarrow\cong H_*(X)$. Note that $pr_X\circ(in_0,in_1)=(id_X,id_X)$, and that $(in_0,in_1)\circ i_1=in_0$ and $(in_0,in_1)\circ i_2=in_1$.

When we use these isomorphisms to identify the groups in the Mayer-Vietoris sequence above we see that it becomes
$$\dots\rightarrow H_{*+1}(M)\rightarrow H_*(X)\oplus H_*(X)\xrightarrow{\theta}H_*(X)\oplus H_*(X)\rightarrow H_*(M)\rightarrow\dots$$
where $\theta$ is the map
$$\theta(x,y)=(x+y,f_*x+y).$$
The kernel of $\theta$ is isomorphic to the subgroup $Fix(f_*)_*=\{x\in H_*(X)\mid f_*(x)=x\}\subseteq H_*(X)$ which consists of the fixed points of $f_*$. The cokernel of $\theta$ is isomorphic to the quotient of $H_*(X)$ by the subgroup $\{x+f_*x\in H_*(X)\mid x\in H_*(X)\}$.

Thus we obtain a long exact sequence
$$\dots\rightarrow H_{*+1}(M)\rightarrow H_*(X)\xrightarrow{f_*-1_*}H_*(X)\xrightarrow{j} H_*(M)\rightarrow\dots$$
where the map $j$ is induced by the inclusion of $X$ into $M$ at level $0$. The other map is the composite $H_{n+1}(M)\rightarrow H_n(X)\oplus H_n(X)\xrightarrow{pr_1}H_nX$ and is more difficult to describe explicitly.

You may find it more convenient to split this up into short exact sequences
$$0\rightarrow \text{coker}\left(H_n(X)\xrightarrow{(f_*-1_*)}H_n(X)\right)\rightarrow H_n(M)\rightarrow\ker\left(H_{n-1}(X)\xrightarrow{(f_*-1_*)}H_{n-1}(X)\right)\rightarrow 0$$


问:
On some doubts on tangent space of immersed submanifold
In Lee's book "Introduction to Smooth manifolds", he following lemma can be found.
Lemma 8.26 Let $M$ be a smooth manifold, let $S\subseteq M$ be an embedded submanifold, and let $Y$ be a smooth vector field on $M$. Then $Y$ is tangent to $S$ iff $Yf$ vanishes on $S$ for all $f\in C^\infty(M)$ such that $f|_S=0.$

My question is what if $S$ just an immersed submanifold? Is the same conclusion true? It should be because earlier Lee proves that any immersed submanifold is locally embedded submanifold. But why does Lee state the theorem only for embedded submanifod?

答:
The lemma is not true for immersed submanifolds. A counterexample is the irrational winding of the torus (see Example 4.20 in Lee). This is the image of the map
$$
\gamma:\mathbb{R}\rightarrow\mathbb{T}^{2}:t\mapsto (e^{2\pi it},e^{2\pi i\alpha t}),
$$
where $\alpha$ is an irrational number. It is an immersed, but not an embedded submanifold.

Since $\gamma(\mathbb{R})$ is dense in $\mathbb{T}^{2}$, the zero function is the only function that vanishes on $\gamma(\mathbb{R})$. So for any vector field $Y\in\mathfrak{X}(\mathbb{T}^{2})$, we have $f|_{\gamma(\mathbb{R})}=0 \Rightarrow Y(f)|_{\gamma(\mathbb{R})}=0$. But not all vector fields on $\mathbb{T}^{2}$ are tangent to $\gamma(\mathbb{R})$.


问:
Elliptic integral into usual form
How can I show that
$$\int {(2 -\cos t)^{-3/2}} dt= \frac23 \left( (E(t/2|-2) + \frac{\sin t}{\sqrt{2 -\cos t}}\right)+C$$
?

I just want to know how to kill the $\frac32$-th power and put it into usual elliptic integral.
Having known the formula, I tried to use integral by part. Let
$${\sqrt{2 -\cos t}}dt=-vdu,\quad \frac{\sin t}{\sqrt{2 -\cos t}}=uv.$$

Thus $du/u=-(2-\cos t)/\sin t$. But it does not work well.
Any help is appreciated.

答:
To get the given form (which is returned by Wolfram Alpha), rewrite $2-\cos t=1+2\sin^2\frac t2$ and then substitute $u=\sin\frac t2,\frac{du}{dt}=\frac12\sqrt{1-u^2}$:
$$\int_0^\varphi(2-\cos t)^{-3/2}\,dt=2\int_0^{\sin\varphi/2}\frac1{(1+2u^2)\sqrt{(1+2u^2)(1-u^2)}}\,du$$
This is an elliptic integral of the third kind (argument convention as in Mathematica/mpmath – arguments $n,\varphi,m$ from left to right):
$$=2\Pi(-2,\varphi/2,-2)+K$$
Since $m=n=-2$, though, this can be simplified into an elliptic integral of the second kind plus another term (Byrd and Friedman 111.06, DLMF 19.6.13):
$$=\frac2{1-(-2)}\left(E(\varphi/2,-2)-\frac{-2\sin\varphi/2\cos\varphi/2}{\sqrt{1-(-2)\sin^2\varphi/2}}\right)+K$$
$$=\frac23\left(E(\varphi/2,-2)+\frac{\sin\varphi}{\sqrt{2-\cos\varphi}}\right)+K$$

However, I would not be inclined to use this form since it has a negative parameter. B&F 291.01 directly gives the result as
$$\int_0^\varphi(2-\cos t)^{-3/2}\,dt=\frac2{\sqrt3}E\left(\sin^{-1}\sqrt{\frac{3(1-\cos\varphi)}{2(2-\cos\varphi)}},\frac23\right)+K$$

Title: Elliptic integral into usual form


问:
What is the minimal area under the curve of parabola $y=ax-bx^2$ between $x = 0$ and $x=\frac{a}{b}$ such that it passes through (1,1)?

Let $a > b >0$. The area under a parabola $y=ax−bx^2$ between its two roots, $x=0$ and $x=\frac{a}{b}$, is given by $\frac{a^3}{6b^2}$. Determine the $a$ and $b$ such that the parabola passes through the point $(1,1)$ and this area is minimised.

So far I have calculated the maximum value of this parabola to be at $x = \frac{a}{2b}$ giving $y = \frac{a^2}{4b^2}$. The restriction of $(1,1)$ gives that $1 = a-b$.

I don't know where to go from here so that the area under $y$ is minimised. Any help?

答:
Hint. Since $b=a-1>0$, you have to minimize the function
$$f(a)=\frac{a^3}{6b^2}=\frac{a^3}{6(a-1)^2}$$
over $(1,+\infty)$.
You may also let $a=b+1$ and minimize  the function
$$g(b)=\frac{a^3}{6b^2}=\frac{(b+1)^3}{6b^2}$$
over $(0,+\infty)$.


问:
$\xi$ be a primitive cube root of unity
Could anyone tell me first of all the below problem is wrong or okay?I am not able to figure out how to solve

$\xi$ be a primitive cube root of unity define $A=\begin{pmatrix}\xi^{-1}&0\\0&\xi\end{pmatrix}$, for a vector $v=(v_1,v_2)\in\mathbb{R}^2$ define $$|v|_A=\sqrt{|vAv^T|}$$ if $w=(1,1)$ what is the value of $|w|_A$?

答:
This problem involves plugging numbers into an expression. Where are you having issues?

Did you get as far as to write it down? That is,

$$wAw^T=\begin{pmatrix}1 & 1\end{pmatrix}\begin{pmatrix}\xi^{-1}&0\\0&\xi\end{pmatrix}\begin{pmatrix}1\\1\end{pmatrix}~?$$

If so, did you evaluate it by using matrix multiplication? Do you know how to simplify the resulting expression into a real number? After that, you just take absolute value and square root.

To simplify the resulting complex number expression into the real number it denotes, consider either de Moivre's formula for complex exponentials (and how to write roots of unity as complex exponentials), or much more simply, use the relation $\xi^2+\xi+1=0$ (this follows from rewriting the expression $\xi^3=1$ as $(\xi-1)(\xi^2+\xi+1)=0$ and then dividing by $\xi-1$).

Title: $\xi$ be a primitive cube root of unity


问:
Showing function from interval to $\mathbb{N}$ is measurable
Suppose $\{f_n\}_{n=0}^{\infty}$ is a sequence of measurable functions defined on the interval $[0,1]$.

 Let $G(x)=card (\{n\in \mathbb{N}\mid f_n(x)=0\})$ and assume $G(x)$ is finite for all $x$ in $[0,1]$.

 I need to show that $G$ is measurable.

 So $G(x)$ gives the number of functions from $\{f_n\}_{n=0}^{\infty}$ which take the value $0$ at $x$. How can I prove that the inverse image of $G$ of every element in the $\sigma$-algebra on $\mathbb{N}$ is in the $\sigma$-algebra on $[0,1]$? In particualar, I am not sure what the $\sigma$-algebra on $\mathbb{N}$ is.

答:
The $\sigma$-algebra is not on $\Bbb N$, but on the domain of $G$, which is $[0,1]$. And it's, of course, Lebesgue's $\sigma$-algebra (though it could really be any $\sigma$-algebra making all the $f_n$-s measurable).

The goal is to show that $G^{-1}[\alpha,\infty)$ is measurable for all $\alpha\in\Bbb R$.

If $\alpha\le0$, that set is obviously the whole domain $[0,1]$. For $\alpha>0$, notice that $x\in G^{-1}\left[\alpha,\infty\right)$ if and only if there is $S\subseteq \Bbb N$ such that $\operatorname{card}S=\left\lceil \alpha\right\rceil$ and $x\in f_s^{-1}(0)$ for all $s\in S$. Therefore, call $\mathcal P_n(\Bbb N)=\{S\subseteq \Bbb N\,:\, \operatorname{card}S=n\}$, and we see that $$G^{-1}[\alpha,\infty)=\bigcup_{S\in\mathcal P_{\left\lceil\alpha\right\rceil}(\Bbb N)}\bigcap_{s\in S}f_s^{-1}(0)$$

which is measurable because $\mathcal P_{\left\lceil\alpha\right\rceil}(\Bbb N)$ is countable, each $S$ is finite (hence countable) and all the $f_s^{-1}(0)$ are measurable.


问:
Prove that $I = \bigcap_{n \in \mathbb{N}}(0, \frac{n + 2}{n + 1}) \subseteq \mathbb{R}$ is not open

In the Euclidean metric space R, the subset
$$
I = \bigcap_{n \in \mathbb{N}}\left(0, \frac{n + 2}{n + 1}\right)
  \subseteq \mathbb{R}
$$ is not open.
Compute I and prove this fact.

Can someone start listing the elements of $I$? I'm having trouble figuring them out.

Body: Hint : For all $n \in \mathbb{N}$, $$\frac{n+2}{n+1} = 1 + \frac{1}{n+1}$$


问:
Tail $\sigma$-algebra and Kolmogorov's 0-1 law
Oliver Knill's book proves the famous Kolmogorov's 0-1 law in section 2.2. The statemente is as follows. Let $(\Omega, \mathcal{A},p)$ be a probability space and $\{\mathcal{A}_{\alpha}\}_{\alpha \in I}$ a family of sub-$\sigma$-algebras of $\mathcal{A}$. To each non-empty $J\subset I$, set $\mathcal{A}_{J}$ to be the $\sigma$-algebra generated by the union $\bigcup_{\alpha\in J}\mathcal{A}_{\alpha}$ and, if $J=\emptyset$ then $\mathcal{A}_{\emptyset}=\{\emptyset, \Omega\}$. The tail $\sigma$-algebra $\mathcal{T}$ is defined by $\mathcal{T}:= \bigcap_{J\subset_{f}I}\mathcal{A}_{J^{c}}$, where $\subset_{f}$ indicates the set $J$ is finite. Then:

Kolmogorov's 0-1 Law: If $\{\mathcal{A}_{\alpha}\}_{\alpha\in I}$ is a family of independent sub-$\sigma$-algebras then $\mathcal{T}$ is trivial, in the sense that $p(A) = 0$ or $p(A) = 1$ for every $A \in \mathcal{T}$.

I might be wrong, but I guess a more common version of Kolmogorov's 0-1 law is the following.

Kolmogovov's 0-1 law: [Version 2]: A sequence $\{X_{n}\}_{n\in \mathbb{N}}$ of independent random variables has a trivial tail $\sigma$-algebra.

The tail $\sigma$-algebra, in version two, is usually defined in the following way. If $\sigma(X_{n},X_{n+1},...)$ is the smalles $\sigma$-algebra in which every random variable $X_{n},X_{n+1},...$ is measurable, then the tail $\sigma$-algebra $\mathcal{T}':= \bigcap_{m=n}^{\infty}\sigma(X_{m},X_{m+1},...)$.

Here's my problem. I've been following Knill's book and understoof his version of Kolmogorov's 0-1 law. However, I want to show that this version implies version two (but does it?). Here's the tricky part for me. The two tail-$\sigma$-algebras considered seems a little different to me. For instance, if $I=\mathbb{N}$ the tail-$\sigma$-algebra on Knill's book considers every finite subset of $\mathbb{N}$ while on the second version only those $J$ like $\{1,...,n\}$ are being considered. So, are these two results equivalent (as I'd expect) or is Knill's version weaker than the second version (in the sense that it does not imply the second vesion)?

答: It is trivially seen that $\bigcap_{J\subseteq_{f}I} A_{J^c}=\mathcal{T}\subseteq \mathcal{T}'.$ To see that the two sigma-algebras are equal, notice that given a finite $J,$ there exists $n$ such that $J\subseteq [n].$ In other words, $A_{[n]^c}\subseteq A_{J^c}$ and hence $A_{[n]^c}\subseteq \bigcap_{J: J\subseteq [n]}A_{J^c}.$ We therefore obtain that $\mathcal{T}'\subseteq\bigcap_{n}A_{[n]^c}\subseteq \bigcap_{J\subseteq_{f}I}A_{J^c}=\mathcal{T}.$


问:
Prove $\sum_{k = 0}^{n}(-1)^{n - k} \binom{n}{k} \cdot k^n = n!$ and $\sum_{k = 0}^{n}(-1)^{n - k} \binom{n}{k} \cdot k^m = 0$

Prove
$$\sum_{k = 0}^{n}(-1)^{n - k} \binom{n}{k} \cdot k^n = n! \qquad\text{and}\qquad \sum_{k = 0}^{n}(-1)^{n - k} \binom{n}{k} \cdot k^m = 0,$$
where $m \in \{0, 1, 2, \cdots, n - 1\}$.

To be honest, I tried to use some combinatorial ideas but was not able to see interconnections. Also, I had an idea to prove it using induction but failed to do that either.

Help will be much appreciated! I just got stuck and don't see any continuation of solution for this problem

答:
This problem has appeared a number of times. Using coefficient
extractors we have

$$\sum_{k=0}^n (-1)^{n-k} {n\choose k} k^m
\\ = m! [z^m] \sum_{k=0}^n (-1)^{n-k} {n\choose k} \exp(kz)
\\ = m! [z^m] (\exp(z)-1)^n.$$

Now $\exp(z)-1 = z+\cdots$ so $(\exp(z)-1)^n = z^n + \cdots$ and
therefore the coefficient extractor returns zero when $m\lt n$ and one
when $m=n$, which is the claim.


问:
Should $\mathbb{N}$ contain $0$?
This is a classical question, that has led to many a heated argument:

Should the symbol $\mathbb{N}$ stand for $0,1,2,3,\dots$ or $1,2,3,\dots$?

It is immediately obvious that the question is not quite well posed. This convention, as many others, are not carved in stone, and there is nothing to prevent mathematician $A$ define $\mathbb{N}$ to be the positive integers including $0$, and mathematician $B$ to define $\mathbb{N}$ to be the positive integers excluding $0$. It does not seem that one definition is accepted widely enough for it to be "the right definition", and even if this was the case, the fashion might change in the future.

I am, however, hoping that there might be a semi-mathematical reason to prefer one notion over the other. For example, I have spent much of my mathematical life believing that $0 \in \mathbb{N}$ because: 1) morally, $\mathbb{N}$ is the cardinalities of finite sets 2) the empty set is a set with $0$ elements. However, recently I realised that this reasoning applies to $\omega$ rather than $\mathbb{N}$, and - much to my horror - I saw $\omega$ and $\mathbb{N}$ used side by side with the only distinction being that $0 \in \omega$ while $0 \not \in \mathbb{N}$. For another example, $\mathbb{N}$ seems to be a much nicer semigroup if $0 \not \in \mathbb{N}$ (and in any case, adding $0$ to a semigroup is a more natural operation than removing it), which would an argument for taking $0 \not \in \mathbb{N}$.

The arguments mentioned above are, of course, rather weak, but perhaps just enough to tip the scale. In any case, this is the general type of argument I am looking for.

Question: Does there exists a convincing argument for deciding if $0 \in \mathbb{N}$ ?

(I consider it quite possible that the answer is negative because in some context one convention is preferable, and in other context the other one. The problem could be dismissed by using $\mathbb{Z}_+$ (or even $\mathbb{Z}_{>0}$ and $\mathbb{Z}_{\geq 0}$) to avoid confusion, but note that in some context one definitely does not want to do this. I would be interested in an argument that is universal in the sense that it makes overall mathematical landscape more elegant, and does not spoil any detail too much. I do not hope that the argument would be convincing to every mathematician, especially one working in a very specific and narrow area.)

答:
Let me ask a similar question in rebuttal:

Are finite sets countable?

The answer is simple. It depends on the context. Sometimes it's easier to have finite sets included in the definition of "countable" and sometimes it's easier to have just the finite set, and have "at most countable" for the term which includes finite sets as well.

I will give an argument why $\Bbb N$ should include $0$, though.

One can consider $0$ and $1$ as the basic atoms of the numbers we know. $\Bbb N$ is the set generated by $0,1$ using addition; and then $\Bbb Z$ is generated by adding additive inverses, and $\Bbb Q$ and then $\Bbb R$ and $\Bbb C$.

Of course that if you take your atomic set of numbers to be $\Bbb C$ or something else, then it might as well be redundant, but it's still a reasonable argument. With only some naive set theory, and axioms for addition and multiplication, we can create all the numbers we need! That's an incredible thing. And all just form the assumption that $0$ and $1$ exist.

On the other hand, in analysis it's often more convenient to have $0\notin\Bbb N$. For example when we say that $x^n$ is well defined for every $x\in\Bbb R$ and $n\in\Bbb N$. Or if we often talk about the sequence $\frac1n$, then we find it easier to write $\frac1n$ for $n\in\Bbb N$, rather than adding "...and $n>0$".

Title: Should $\mathbb{N}$ contain $0$?


答:
Note that the Finite difference (forward, with unit step) of a function
is defined as
$$
\Delta \,f(x) = f(x + 1) - f(x)
$$
its iteration being
$$
\Delta ^{\,n} f(x) = \Delta \left( {\Delta ^{\,n - 1} f(x)} \right) = \sum\limits_{0\, \le \,k\, \le \,n} {\left( { - 1} \right)^{\,n - k} \left( \matrix{
  n \cr
  k \cr}  \right)f(x + k)}
$$

Now, since any polynomial can be expressed as a Newton series
that implies that the $n$-th difference of a polynomial of degree $n$ is constant and equal to $n!$ times the leading coefficient, i.e.
$$
\Delta ^{\,n} \left( {a_{\,n} x^{\,n}  + a_{\,n - 1} x^{\,n - 1}  +  \cdots  + a_{\,0} } \right) = n! a_{\,n}
$$
as well as that the $n$-th difference of a polynomial of degree $m < n$ is null.

Therefore
$$
\eqalign{
  & \sum\limits_{0\, \le \,k\, \le \,n} {\left( { - 1} \right)^{\,n - k} \left( \matrix{
  n \cr
  k \cr}  \right)\left( {x + k} \right)^{\,n} }  = \Delta ^{\,n} (x^{\,n} )
 = n!\quad \left| {\;\forall x \in C} \right.  \cr
  & \sum\limits_{0\, \le \,k\, \le \,n} {\left( { - 1} \right)^{\,n - k} \left( \matrix{
  n \cr
  k \cr}  \right)\left( {x + k} \right)^{\,m} }  = \Delta ^{\,n} (x^{\,m} )
 = 0\quad \left| {\;m < n,\;\forall x \in C} \right. \cr}
$$


问:
Probability-using expectation value
A coin which has probability 0.75 of coming up heads is tossed three times.
Let X be the number of heads observed.
Find E(X^2), give your answer to 2 decimal places.</p>

答:
My guess -

You’d do
$$
=1^2 \cdot 0.75^1 + 2^2 \cdot 0.75^2 + 3^2 \cdot 0.75^3
$$
Which would give you an expected value, but this seems a little simple

My only concern is that it doesn’t take into consideration say, if I rolled a head once, this could happen three times, ie htt or tth or tht - would this take any significance in the answer of this question.

Title: Probability-using expectation value


问:
A lattice of circles and a line which does not cut any of them.

Each lattice point is a center of a circle, all with radius $d$. Let line $y={2\over 5}x+n$ doesn't cut or touch any circle, for some $n$. Find the supremum for $d$.

I was trying to maximize $$d={|2x-5y+5n|\over \sqrt{29}}$$ where $x,y$ runs over all integers and we can assume that $n$ is in $(0,1)$, but I don't know how to do it. Any help?

答:
Is this correct? (after sugesstion of @Intelligenti pauca)

So, since everything is ''repeating'' (with period $5$ in direction of $x$ axsis and $2$ in $y$ axsis) we can watch only lattice points in $[0,5)\times [0,2)$. So we are, for fixed $n$, searching for minimum in:

\begin{align}d(0,0) &= {5n\over \sqrt{29}}\\
d(1,0) &= {2+5n\over \sqrt{29}}\\
d(2,0) &= {4+5n\over \sqrt{29}}\\
d(3,0) &= {6+5n\over \sqrt{29}}\\
d(4,0) &= {8+5n\over \sqrt{29}}
\end{align}

\begin{align}
d(0,1) &= {|5n-5|\over \sqrt{29}}\\
d(1,1) &= {|5n-3|\over \sqrt{29}}\\
d(2,1) &= {|5n-1|\over \sqrt{29}}\\
d(3,1) &= {|5n+1|\over \sqrt{29}}\\
d(4,1) &= {|5n+3|\over \sqrt{29}}\\
\end{align}

The minimum among first 5 is $5n\over \sqrt{29}$ and among other 5 is ${|5n-1|\over \sqrt{29}}$ or ${|5n-3|\over \sqrt{29}}$ (since we are looking for minimum of $f(t)= |2t-a|$, where $a= 5n-1$ and $t\in \{-2,-1,0,1,2\}$ which is clearly at $t=0$ or $t=1$).

First case: $5n = |5n-1| \implies n={1\over 10}\implies d={1\over 2\sqrt{29}}$

Second case: $5n = |5n-3| \implies n={3\over 10}\implies d={3\over 2\sqrt{29}}$

Third case: $|5n-3| = |5n-1| \implies n={4\over 10}\implies d={1\over \sqrt{29}}$

So $\boxed{d_{sup} = {1\over 2\sqrt{29}}}$.


问:
What is the expected performance of a team in a contest?

Suppose $n$ people are competing in a contest. Among them is a team of $m$ people ($m < n$), performance of which is defined as the maximal rank achieved by any of its neighbours. What is the expected performance of the group if all participants are equally skilled?

The problem can be alternatively formulated as:

Suppose $\sigma$ is a random permutation uniformly distributed on $Sym(\{1, ... , n\})$. What is $E[min\{\sigma(i)| i \leq m\}]$?

The probability of any given outcome is $\frac{1}{n!}$. For any $k \leq i \leq n$, the number of outcomes in which a given member of the team takes $k$-th place and everyone else in the team performs even worse is $C_{n-k}^{m-1}$. Thus the number of outcomes in which the performance of the team is $k$ equals to $m C_{n-k}^{m-1}$. So the expectation of the outcome is:

$$\frac{\sum_{k=1}^{n-m} mC_{n-k}^{m-1}}{n!}$$

However, I wonder, if there is a closed form to this expression. I tried to simplify it anyhow, but failed.

答:
Using the Hockey-stick formula, you get that
$$\sum _{k=1}^{n-m}mC_{n-k}^{m-1}=\sum _{k=m}^{n-1}mC_{k}^{m-1}=m(C_{n}^m-1).$$
Just a comment for the confused reader: $C_n^m=\binom{n}{m}.$


答: Besides problems coming from notation (these are the only ones which distinguish $\mathbb{N}$ from $\omega$, the former used when treated as a monoid and the latter when thought as a ordinal) the reality is this: $0 \in \omega$ and $0 \subset \omega$ are both true ( remember that $0:= \emptyset$), so it's obviously contained, and it belongs to $\omega$ for lots of reasons (transitivity plus the fact that both are cardinals or by definition of $\omega $ as the set of natural numbers). The last is a good definition not given in an intuitive and inexistent way.</p>

Title: Should $\mathbb{N}$ contain $0$?

Body:
This is a classical question, that has led to many a heated argument:

Should the symbol $\mathbb{N}$ stand for $0,1,2,3,\dots$ or $1,2,3,\dots$?

It is immediately obvious that the question is not quite well posed. This convention, as many others, are not carved in stone, and there is nothing to prevent mathematician $A$ define $\mathbb{N}$ to be the positive integers including $0$, and mathematician $B$ to define $\mathbb{N}$ to be the positive integers excluding $0$. It does not seem that one definition is accepted widely enough for it to be "the right definition", and even if this was the case, the fashion might change in the future.

I am, however, hoping that there might be a semi-mathematical reason to prefer one notion over the other. For example, I have spent much of my mathematical life believing that $0 \in \mathbb{N}$ because: 1) morally, $\mathbb{N}$ is the cardinalities of finite sets 2) the empty set is a set with $0$ elements. However, recently I realised that this reasoning applies to $\omega$ rather than $\mathbb{N}$, and - much to my horror - I saw $\omega$ and $\mathbb{N}$ used side by side with the only distinction being that $0 \in \omega$ while $0 \not \in \mathbb{N}$. For another example, $\mathbb{N}$ seems to be a much nicer semigroup if $0 \not \in \mathbb{N}$ (and in any case, adding $0$ to a semigroup is a more natural operation than removing it), which would an argument for taking $0 \not \in \mathbb{N}$.

The arguments mentioned above are, of course, rather weak, but perhaps just enough to tip the scale. In any case, this is the general type of argument I am looking for.

Question: Does there exists a convincing argument for deciding if $0 \in \mathbb{N}$ ?

(I consider it quite possible that the answer is negative because in some context one convention is preferable, and in other context the other one. The problem could be dismissed by using $\mathbb{Z}_+$ (or even $\mathbb{Z}_{>0}$ and $\mathbb{Z}_{\geq 0}$) to avoid confusion, but note that in some context one definitely does not want to do this. I would be interested in an argument that is universal in the sense that it makes overall mathematical landscape more elegant, and does not spoil any detail too much. I do not hope that the argument would be convincing to every mathematician, especially one working in a very specific and narrow area.)


问:
Interpretation of the Weil algebra
Cartan introduced the Weil algebra $W(\mathfrak{g})$ associated to a Lie algebra $\mathfrak{g}$ over a field $k$ of characteristic $0$. This consists of a dga whose underlying graded abelian group is given by $Sym(\mathfrak{g}^*)\otimes \Lambda^* \mathfrak{g}^*$ with a differential which can be constructed as follows.
We first define the graded derivation $h \colon W(\mathfrak{g}) \rightarrow W(\mathfrak{g})$ giveny by
\begin{equation}
h(V \otimes x_1\wedge \dots \wedge x_n)=\sum_{i=1}^n(-1)^{i-1} (x \otimes V) \otimes x_1 \wedge \dots \wedge \hat{x_i} \wedge \dots \wedge x_n \qquad h(V \otimes 1)=0
\end{equation}
where $V \in Sym(\mathfrak{g}^*)$ and $x_1\wedge \dots \wedge x_n \in \Lambda^n \mathfrak{g}^*$. Then we add this $h$ to the usual differential defined on the Chevalley-Eilenberg complex of $\mathfrak{g}$ where we consider $Sym(\mathfrak{g}^*)$ as coefficient module. The wiki page gives a detailed formula https://en.wikipedia.org/wiki/Lie_algebra_cohomology.

It is a core result that the cohomology of this complex reduces to the base field $k$ in degree $0$. From this we can produce the Cartan map, which is a function going from the invariant elements of $Sym(\mathfrak{g}^*)$ to the space of primitive elements of $H^*(\mathfrak{g}; k)$. If the Lie algebra is reductive it can be see this is a bijection.

I found a complete reference in "Cohomology of principal bundle and homogeneous" by Greub, Halperin and Vanstone, so I am not asking for clarification about a particular point of this theory. What I am looking for is some kind of explanation of what's the idea behind this cochain complex. Everything in this formulas and constructions works but I have no idea why or how these came to be.

I read somewhere that the Weil algebra is some kind of link between the "connections" (I do not know what these should be) and the cohomology of $\mathfrak{g}$. If someone could explain this I would really be grateful.

答:
I am personally not quite sure on the precise history, but the earliest paper mentioning the Weil algebra seems to be "Cohomologie réelle d’un espace fibré principal différentiable." by Cartan. I am going to present to you the story that convinced me that it is a very natural concept that one could reasonably come up with oneself, if you accept that certain other geometric concepts are worthwhile studying. I also believe this is closely related to what Cartan originally did in his paper.

I'm not going to say anything new, all of this is contained in the book by Greub, Halperin and Vanstone that you cite, but I'll try to streamline the idea. I just hope you're comfortable with some differential geometry :)

Let $P$ and $M$ be smooth manifolds, and $P \to M$ a principal $G$-bundle. Recall that that means that you have a free and transitive action of the Lie group $G$ on the fibres of the bundle. This allows you to define, for every $\xi \in \mathfrak{g}$, the fundamental vector field $\xi_P \in \Gamma(TP)$

Then a $\mathfrak{g}$-valued differential form $\omega \in \Omega^1(P,\mathfrak{g}) = \Omega^1(P) \otimes \mathfrak{g} = \Gamma(T^* P) \otimes \mathfrak{g}$ is called a connection form for the principal bundle if we have:

It is equivariant with respect to the induced action of $G$ on vector fields $\Gamma(TP)$ and the adjoint action of $G$ on $\mathfrak{g}$, viewing $\omega$ as a map $\Gamma(TP) \to \mathfrak{g}$,

It maps fundamental vector fields to its generators, meaning that for every $\xi \in \mathfrak{g}$, it fulfils $\omega(\xi_P) = \xi$.

Geometrically, this notion allows one to split up the principal bundle $P$ into vertical and horizontal directions with respect to the bundle structure, so it is definitely useful, but it may be a bit tricky to remember the precise definition.

We can see such a $\mathfrak{g}$-valued differential form as an element of $\Gamma(T^*P) \otimes \mathfrak{g}$, as a map $\Gamma(TP) \to C^\infty(P,\mathfrak{g})$, but to understand the connection to the Weil algebra, let us take on the perspective that $\omega$ is a $G$-equivariant map $$\omega: \mathfrak{g}^* \to \Omega^1(P).$$

Now, naively, both sides of this map could be understood as naturally sitting inside two well-known cochain complexes, the left-hand side $\mathfrak{g}^*$ sitting in the first degree of the Chevalley-Eilenberg complex $C^\bullet_{\text{CE}}(\mathfrak{g})$, and $\Omega^1(P)$ in the first degree of the de Rham complex $\Omega^\bullet(P)$. An instinctive reaction should then be to extend this map to a map of complexes $C^\bullet_{\text{CE}}(\mathfrak{g}) \to \Omega^\bullet(P)$ by setting for all $\alpha_1,\dots,\alpha_n \in \mathfrak{g}^*$:
$$\omega (\alpha_1 \wedge \dots \wedge \alpha_n) := \omega(\alpha_1) \wedge \dots \wedge \omega(\alpha_n) \in \Omega^n(P).$$

Turns out: This cannot be done in general! This map does not always respect the differential on both sides. The failure to do so is characterized by what geometers call the curvature $\Omega$ of the connection $\omega$, which, without going into too much detail, can be defined as the difference "$(\omega \wedge \omega) \circ d_{\text{CE}} - d_{\text{dR}} \circ \omega$", and turns out to just be a $\mathfrak{g}$-valued 2-form $\Omega \in \Omega^2(P, \mathfrak{g})$, or, equivalently, a map
$ \Omega : \mathfrak{g}^* \to \Omega^2(P)$. If this curvature is zero, we call $\omega$ a flat connection, and your map of complexes does indeed extend to a map between the Chevalley-Eilenberg and the de Rham complex, but this is not always the case.

This is now why the Weil algebra appears to be a natural concept: What ends up being possible, even when your connection is not flat, is to extend the connection to a map from the Weil algebra to the de Rham complex, by including the curvature of your connection in your considerations: If $W(\mathfrak{g}) = S^\bullet \mathfrak{g}^* \otimes \Lambda^\bullet \mathfrak{g}^*$, then define

$$\zeta: W(\mathfrak{g}) \to \Omega^\bullet(P)$$

by setting for $\alpha \in \mathfrak{g}^*$:

$$\zeta (1 \otimes \alpha) := \omega(\alpha), \quad \zeta (\alpha \otimes 1) := \Omega(\alpha),$$
and extending in a natural, graded manner to all of $W(\mathfrak{g})$.

From this point of view, giving a connection on a principal fibre bundle $P \to M$ is equivalent to giving a morphism $\zeta : W(\mathfrak{g}) \to \Omega^\bullet(P)$ intertwining the $G$-action, cochain differential and insertion of Lie algebra elements/fundamental vector fields on both sides. The paper that inspired me to learn about this, "The cohomology of vector fields on a manifold" by Bott & Segal (Section 3), calls such a $\zeta$ a "morphism of $G$-cochain algebras". The name for this structure in the Greub/Halperin/Vanstone book is "operation of a Lie algebra".

No matter what you call it, I personally find it quite impressive that a simple morphism like this encodes so much geometric information about a bundle. And all it takes to come up with it is the acknowledgment of curvature, and the extension of $C^\bullet_{\text{CE}}(\mathfrak{g})$ to a complex that can deal with curvature. Perhaps this is a standard story to a lot of people, if they are more familiar with this whole Chern-Weil theory business, but I hadn't seen it up until very recently, and found it too cool not to post. :) Okay, I'm done now!

Title: Interpretation of the Weil algebra


问:
Show that $\int^{\infty}_{-\infty}\frac{dx}{(x^2-4x+5)^2} = \frac{\pi}{2}$ using residue theory
I'm trying to evaluate some complex integrals using residue theory. I've read a number of articles with different examples here on Stack Exchange, but I'm still really lost and could use some help.

Show that $\int^{\infty}_{-\infty}\frac{dx}{(x^2-4x+5)^2} = \frac{\pi}{2}$.

First, I found the poles of the function as $2-i$ and $2+i$. Both of these poles are of order $2$. I am also considering my region as the semicircle of radius $R$ in the upper half-plane with the line segment between $x=−R$ and $x=R$ on the real axis. So I know that I need to evaulate

$$\int_{\Gamma}f(z)dz = 2\pi iRes(f,2+i) + 2\pi iRes(f,2-i)$$

At this point, I'm really stuck on what to do from here. I tried evaluating the residues and was getting some really weird answers. My work was really messy and likely completely wrong, so hopefully it's OK if I don't reproduce it here. I'm not sure how to finish solving this.

Similarly, I'm having trouble with this integral:

$\int_{|z| = 1}z^3e^{1/z}\sin(\frac{1}{z}) dz$

I know that there is a pole at $z = 0$. So that means I need to evaluate

$$ \int_{|z| = 1}z^3e^{1/z}\sin(\frac{1}{z}) dz = 2\pi i Res(f,0)$$

Do I consider this pole of order $1$ or order $2$ and how do I find this residue?

答:
For the first problem, to compute the resides of $\frac1{(z^2-4x+5)^2}$, we evaluate the limits

$$\lim_{z\to (2\pm i)}\frac{d}{dz}\left(\frac{z-(2\pm i)}{(z^2-4x+5)^2}\right)$$

But only the pole at $z=2+i$ is in the upper half plane.  Can you finish now?

For the second problem, there is no pole at $z=0$.  Rather, the point $z=0$ is an essential singularity.  Expand both $e^{1/z}$ and $\sin(1/z)$ in Taylor series of powers of $1/z$ (the Laurent series) and use their Cauchy Product to determine the coefficient on the term $\frac1{z^4}$.  Can you wrap this up now?


问:
Three inequalities for number $e$: $\frac{e}{2n+2}<e-\left(1+\frac1n\right)^n<\frac{e}{2n+1}<\frac3{2n}$
I have to prove the following inequalities for number $e$:

$$\frac{e}{2n+2}<e-\left(1+\frac1n\right)^n<\frac{e}{2n+1}<\frac3{2n}$$

Here's what I have tried:

$\textbf{First inequality:}$ $\frac{e}{2n+2}<e-\left(1+\frac1n\right)^n$

I use that $\log{\left(\frac{n+1}n\right)}=\int_n^{n+1}\frac1xdx\leq \frac12\left(\frac1n+\frac1{n+1}\right)$.

Operating we have: $n\log{\left(\frac{n+1}n\right)}\leq 1-\frac1{2n+2}$

Exponentiating and using $\frac1{1+x}\geq e^{-x}$, we have:

$$\frac{e}{2n+3}\leq e-\left(1+\frac1n\right)^n$$

This is similar to what i need to prove, but not the same...

$\textbf{Second inequality:}$ $e-\left(1+\frac1n\right)^n<\frac{e}{2n+1}$

I don't know even how to start for this inequality.

$\textbf{Third inequality:}$ $\frac{e}{2n+1}<\frac3{2n}$

I think this is trivial as $e<3$ and $2n<2n+1$ (correct me if I am wrong).

答:
Update: The book I mentioned below provides more than what you need. Here I give only the things you need to prove the two inequalities.

Step 1: We prove $a_n=\left(1+\frac 1n \right)^{n+\frac 12}$ is decreasing. You can check out this post Is $\left(1+\frac1n\right)^{n+1/2}$ decreasing? but the book's method is different and simpler.

Note that $$\log \frac{1+x}{1-x} = 2 \left( \frac x1 + \frac{x^3}{3}+\frac{x^5}{5} + \cdots \right)$$

Let $x=\frac{1}{2n+1}$ then
$$
\log a_n = \left(n+\frac 12\right)\log \left(1+\frac 1n \right) = (2n+1)\left(\frac{1}{2n+1}+\frac{1}{3(2n+1)^3}+\frac{1}{5(2n+1)^5}+ \cdots \right)\\
=1+\frac{1}{3(2n+1)^2}+\frac{1}{5(2n+1)^4}+ \cdots
$$
Therefore $a_n$ is (strictly) decreasing.

Step 2: We prove $b_n=\left(1+\frac 1n\right)^n\left(1+\frac{1}{2n}\right)$ is monotone decreasing.

Note that $$b_n = a_n \frac{1+\frac{1}{2n}}{\left(1+\frac 1n\right)^{\frac 12}} = a_n \left( \frac{(1+\frac{1}{2n})^2}{1+\frac 1n}\right)^{\frac 12} = a_n \left(1+\frac{1}{4n(n+1)} \right)^{\frac 12}$$

Since $a_n$ and $1+\frac{1}{4n(n+1)}$ are both strictly decreasing, so is $b_n$.

Step 3: Now we go back to prove $$\frac{e}{2n+2}<e-\left(1+\frac 1n\right)^n < \frac{e}{2n+1}$$

The first inequality $ \iff \left( 1+\frac 1n\right)^{n+1} < e\left( 1+\frac{1}{2n}\right)\tag 1$

Denote $f(x)=x+x \log \left( 1+\frac x2 \right)-(1+x)\log(1+x), 0 < x \leqslant \frac 1n.$

$$f'(x)=\frac{x}{x+2} - \log \frac{1+x}{1+\frac x2}>\frac{x}{x+2}-\frac{1+x}{1+\frac x2}+1=0, f(0)=0.$$

Therefore $$f\left(\frac 1n\right) >0 \implies (1).$$

The second inequality $$\iff e < \left(1+\frac 1n\right)^n \left(1+\frac{1}{2n}\right) \tag 2$$

From step 2, $\forall m>n$
$$
\left(1+\frac 1n\right)^n\left(1+\frac{1}{2n}\right) > \left(1+\frac{1}{ n+1} \right)^{n+1}\left(1+\frac{1}{2n+2}\right) \geqslant \left(1+\frac 1m\right)^m\left(1+\frac{1}{2m}\right)\\
\implies \left(1+\frac 1n\right)^n\left(1+\frac{1}{2n}\right) > \left(1+\frac{1}{ n+1} \right)^{n+1}\left(1+\frac{1}{2{n+1}}\right) \geqslant e, m\to \infty. \blacksquare
$$

These inequalities are not trivial. They are a series of problems from Problems and Theorems in Analysis I by Polya and Szego.

167. We define $$x_n = y_n e^{-\frac{1}{12n}}, y_n = n! n^{-n-\frac 12}e^n,  n=1,2,3,\ldots$$

Then each interval $(x_k, y_k), k=1, 2, 3, \ldots, $ contains the interval $(x_{k+1}, y_{k+1})$ as a subinterval.

168. The sequence $$a_n=\left(1+\frac 1n \right)^{n+p}, n=1, 2, 3, \ldots$$ is monotone decreasing if and only if $p\geqslant \frac 12$.

169. The sequence $$a_n=\left(1+\frac 1n\right)^n\left(1+\frac xn\right), n=1, 2, 3, \ldots$$ is monotone decreasing if and only if $x\geqslant \frac 12.$

170. Let $n$ be a positive integer. Then we have $$\frac{e}{2n+2}<e-\left(1+\frac 1n\right)^n < \frac{e}{2n+1}.$$


答:
Alright, so your curve is the semi-circle of radius $R$ in the upper half-plane, so here it is:

Since the poles are $2 \pm i$ (pink points in the pic), it only contains one of them, so the integral over your curve $\Gamma$ will simply be:

$\displaystyle\int_\Gamma f(z) \mathrm{d}z = 2\pi i \text{Res}(f, 2 + i)$.

Now, your curve $\Gamma$ can be split into $\gamma_1$ and $\gamma_2$, where $\gamma_1 = [-R, R]$, and $\gamma_2$ is the half-circle.

So $\displaystyle\int_\Gamma f(z)\mathrm{d}z = \displaystyle\int_{\gamma_1} f(z)\mathrm{d}z + \displaystyle\int_{\gamma_2} f(z)\mathrm{d}z = \displaystyle\int_{-R}^R f(x)\mathrm{d}x + \displaystyle\int_{\gamma_2} f(z)\mathrm{d}z$

So you want to know the integral over the real-axis, you can just isolate it:

$\displaystyle\int_{-R}^R f(x)\mathrm{d}x = \displaystyle\int_\Gamma f(z) \mathrm{d}z - \displaystyle\int_{\gamma_2} f(z) \mathrm{d}z = 2\pi i \text{Res}(f, 2 + i) - \displaystyle\int_{\gamma_2} f(z) \mathrm{d}z$.

So... all you have to do now is compute the integral over $\gamma_2$ (hint: it's going to be zero, can you see why?)

Title: Show that $\int^{\infty}_{-\infty}\frac{dx}{(x^2-4x+5)^2} = \frac{\pi}{2}$ using residue theory

Body:
I'm trying to evaluate some complex integrals using residue theory. I've read a number of articles with different examples here on Stack Exchange, but I'm still really lost and could use some help.

Show that $\int^{\infty}_{-\infty}\frac{dx}{(x^2-4x+5)^2} = \frac{\pi}{2}$.

First, I found the poles of the function as $2-i$ and $2+i$. Both of these poles are of order $2$. I am also considering my region as the semicircle of radius $R$ in the upper half-plane with the line segment between $x=−R$ and $x=R$ on the real axis. So I know that I need to evaulate

$$\int_{\Gamma}f(z)dz = 2\pi iRes(f,2+i) + 2\pi iRes(f,2-i)$$

At this point, I'm really stuck on what to do from here. I tried evaluating the residues and was getting some really weird answers. My work was really messy and likely completely wrong, so hopefully it's OK if I don't reproduce it here. I'm not sure how to finish solving this.

Similarly, I'm having trouble with this integral:

$\int_{|z| = 1}z^3e^{1/z}\sin(\frac{1}{z}) dz$

I know that there is a pole at $z = 0$. So that means I need to evaluate

$$ \int_{|z| = 1}z^3e^{1/z}\sin(\frac{1}{z}) dz = 2\pi i Res(f,0)$$

Do I consider this pole of order $1$ or order $2$ and how do I find this residue?


问:
The number of words that can be made by permuting the letters of _MATHEMATICS_ is
The number of words that can be made by permuting the letters of MATHEMATICS is

$1) 5040$

$2) 11!$

$3) 8!$

$4) 4989600$

First of all I do not understand the statement of the problem, I would like if some one tell me with an example.

答:
The problem is asking how many distinguishable ways you can rearrange the letters of the word MATHEMATICS. Let me give you a parallel example to see what is going on, here. Consider the word $$\text{MISSOURIANS}$$ let me rewrite it as $$\text{MIS}\color{blue}{\text{S}}\text{OUR}\color{red}{\text{I}}\text{AN}\color{green}{\text{S}}.$$ This has $11$ distinct objects in it, now that I've colored the repeated letters. There are $11!$ different ways we could arrange those ($11$ options for the first letter, then $10$ options for the second, then $9$ options for the third, and so on). Now, what if all of the letters letters $\text S$ were black, instead of different colors? Well, when they were different colors, each given arrangement of letters could be made into a completely different one simply by rearranging the $\text S$s (there are $3!$ ways to do this), but this makes no difference if they are all the same color. Hence, there are $$\frac{11!}{3!}$$ distinguishable ways to rearrange the letters of $$\text{MISSOUR}\color{red}{\text{I}}\text{ANS}.$$ From there, we see in a similar fashion that there are $$\frac{11!}{3!2!}$$ distinguishable ways to rearrange the letters of $$\text{MISSOURIANS}$$

See also this wikipedia article. Can you adapt this to your example?


问:
Arc length when the curve intersect itself
I came a cross a theorem saying that the arc length of a smooth non self-intersecting parametric curve is  given by

$ L= \int_{0}^{2\pi} \sqrt(y’^2+x’^2)dx$

Why we specify that the curve should be smooth and not intersecting itself is that means we should split the interval and integrate each loop separately

Also if a particle travelled arround a circle 3 times from t=0 to $t=3\pi$  when we find the arc length is it giving us the whole distance that the particle travelled or it is only the length of the circle once?

答:
If it is not smooth, the derivatives are undefined.

If it is self-intersecting, you can run around the loop forever.


问:
Catalan Numbers and Gambling
Working on a random walk gambling problem, $a_n$ denotes a certain probability that a player wins a prize where the coin is fair. Variable $n$ denotes the number of the coin flip, and the probability of winning on an individual coin flip is $1/2$, and losing is also $1/2$.

 Now, I have computed the correct individual probabilities as the following: $a_{2n} =\frac{1}{n+1}{2n\choose n}(1/2)^{2n}$.

The problem is, I need the sum of all these probabilities, because I need the probability of winning in the general case for all $n$, $n$ to $\infty$. I have no clue how to proceed. The Catalan numbers expand exponentially, but the $(1/2)^{2n}$ decays exponentially.

答:
The generating function for Catalan numbers is
$$C(x)=\sum _{n=0}^{\infty}C_nx^n=\frac{1-\sqrt{1-4x}}{2x},$$ this comes from the relation $c(x)=1+xc(x)^2.$

Notice that $x=1/4$ gives what you want.

Title: Catalan Numbers and Gambling


答:
Note that $\phi_n(x) = 1_{ \{ 0 \} } (f_n(x))$ is measurable since $f_n$ is measurable. ($1_{ \{ 0 \} }$ is the characteristic function of the set $\{0\}$.)

Hence $G = \sum_n \phi_n$ is measurable. This holds whether or not $G$ is finite.

Title: Showing function from interval to $\mathbb{N}$ is measurable

Body:
Suppose $\{f_n\}_{n=0}^{\infty}$ is a sequence of measurable functions defined on the interval $[0,1]$.

 Let $G(x)=card (\{n\in \mathbb{N}\mid f_n(x)=0\})$ and assume $G(x)$ is finite for all $x$ in $[0,1]$.

 I need to show that $G$ is measurable.

 So $G(x)$ gives the number of functions from $\{f_n\}_{n=0}^{\infty}$ which take the value $0$ at $x$. How can I prove that the inverse image of $G$ of every element in the $\sigma$-algebra on $\mathbb{N}$ is in the $\sigma$-algebra on $[0,1]$? In particualar, I am not sure what the $\sigma$-algebra on $\mathbb{N}$ is.


问:
How can I compute $\tan(.5\arctan(x))$?
The plot for this function appears to be in the form of $\alpha*\arctan(\beta*x)$ but I've no clue how to go about simplifying the expression.</p>

答:
As shown in this answer, $\tan(x/2)=\frac{\sin(x)}{1+\cos(x)}$. Then, since $\sin\left(\tan^{-1}(x)\right)=\frac{x}{\sqrt{1+x^2}}$ and $\cos\left(\tan^{-1}(x)\right)=\frac1{\sqrt{1+x^2}}$, we get
$$
\tan\left(\tfrac12\tan^{-1}(x)\right)=\frac{x}{\sqrt{1+x^2}+1}
$$
This works for all $x\in\mathbb{R}$.

Title: How can I compute $\tan(.5\arctan(x))$?


问:
Arithmetic rank and height
I've been looking for an example of an ideal whose height and Arithmetic rank is different.

I think that it must differ just in 1, but i can't imagine an example.

$\mathbb{ara}(I)=\mathbb{ht}(I)+1$

答:
The difference ${\rm ara}(I) - {\rm ht}(I)$ can be arbitrarily large even for monomial edge ideals, i.e. monomial ideals generated in degree two. In fact, for a squarefree monomial ideal $I$ generated in degree 2 in a polynomial ring $R$, the following inequalities hold:
$$
{\rm ht}(I) \leq {\rm bight}(I) \leq {\rm pd}(R/I) \leq {\rm ara}(I),
$$
where ${\rm bight}(I)$ is the big height of $I$, the maximum height of the minimal primes of $I$.

Let $n \geq 2$ and consider the ideal $I_n = (x_0 x_i : i=1,\dots,n)$ in $R=K[x_0,\dots,x_n]$. Now, it can be easily shown that $1 = {\rm ht}(I_n) < {\rm bight}(I_n) = {\rm pd}(R/I_n) = {\rm ara}(I_n) = n$.

Title: Arithmetic rank and height


问:
How to find the coefficient of a term in this expression
How to determine the coefficient of z3q100 in

 I stumbled upon this problem while trying to solve this type of partition problem:
Find the number of integer solutions to x + y + z = 100 such that 3 ≤ x ≤ y ≤ z ≤ 60.

Thanks in advance.

答:
A possible solution to your problem:

First, count all partitions $x+y+z=91$ where $0\leq x,y,z\leq 57$. That would be the same as the number of distributions of 91 balls to 3 cells, and each cell can hold at most $57$ balls:

Number of distributions without top limitation:  $\binom{91+3-1}{3-1}=\binom{93}{2}=4278$.

Number of distributions with one of the cells holding more than $57$ balls: $3\cdot \binom{91-58+3-1}{3-1}=3\cdot\binom{35}{2}=1785$.

So, number of good distributions will be $4278-1785=2493$ (since it's impossible to have two cells with more than $57$ balls).

Each such solution where all $x,y,z$ are distinct gives you a unique partition in your problem and is counted $3!=6$ times, and a solution where two are equal is counted 3 times. (There is no solustion where all 3 are equal since 91 is not divisible by 3).

It's left to find the number of solutions to $x+y+z=91$ where $x,y,z\geq 0$ and two are equal. This is exactly 3 times (choice of which two are equal out of 3) the number of solutions to $x+2y=91$, where $x,y\geq 0$ (i.e. when $y=z$). Here $x$ could be only $1,3,5,...,57$, which is $29$ options (since $90/2=45<57$, we already take into account $x,y,z\leq 57$).

Hence there are $29\cdot 3=87$ solutions where two are equal.

So, we have a total of $(2493-87)/6+29=430$ partitions.


问:
Proof $||A \underline x|| > 0 \Leftrightarrow \underline x \neq \underline 0$
If  $n \geq m, A \in M(n,m)$ and  $rg(A)=m$

Proof $||A \underline x|| > 0  \Leftrightarrow \underline x \neq \underline 0$

a)If $m \neq 0 \leftrightarrow A \neq 0_M$

Suppose that $\underline x = \underline 0 \Leftrightarrow ||A \cdot \underline 0||=||0||$

which is a contradiction to $||A \underline x|| > 0 $

 therefore  $||A \underline x|| > 0  \Leftrightarrow \underline x \neq \underline 0$

b)If $m \neq 0 \Leftrightarrow A \neq 0_M \Leftrightarrow A > 0$

per definition:
$||A|| > \frac{||A \underline x||}{|| \underline x||}$ and $ \forall \underline x \neq \underline 0$

 then  $||A \underline x|| > 0  \Leftrightarrow \underline x \neq \underline 0$

$\Box$

Is any of these ways a correct proof?

If not where are the mistakes or what can I improve?

答:
A rigorous proof would be as follows:

a matrix $A\in M(n,m)$ has rank $m$ (i.e., maximal rank) if and only if it represents an injective homomorphism from a vector space of dimension $m$ to a vector space of dimension $n$. So, if you start with a non-zero vector $\underline x$, its image $A\underline x$ will remain different from zero, thus $||A\underline x||>0$. It is also clear that $A\underline 0=\underline 0$. \\\


问:
Confusion with notation of square bracket and round bracket of indices of a tensor
Refer to the following picture:

I am confused with the last notation. Say if I got a tensor ${T^{abc}}_{de}$ and I would like to denote a new tensor which is defined by permuting the indices $a$ and $c$. But you just can't add an open bracket before $a$ and a closed bracket after $c$, because in the notation, it means permuting $a,b,c$ instead of just $a,c$ only.

So how should I do? Thank you.

答:
Suppose, $t^{abc}_{de}=U^{ac}⊗V^{b}_{de}$ with $U∈⊗^{2}T_{x}$, $V∈⊗T_{x}⊗^{2}T^{*}_{x}$ with $x∈M$, $M$ some underlying manifold. Then, $\widetilde{t}≑ U^{(ac)}⊗V^{b}_{de}$ and,
$\overline{t}≑ U^{[ac]}⊗V^{b}_{de}$.


问:
Decompose into simple fractions over $\mathbb{C}$

Decompose into simple fractions over $\mathbb{C}$:
$$f(x) = \frac{1}{(x^2-1)^n}$$

I know how to decompose some fractions by $\mathbb{R}$, but I have no idea how analyze this over $\mathbb{C}$ and solve dependency from $n$.

答:
$x^2-1 = (x+1)(x-1)$, so you want a partial fraction decomposition of the form
$$ \dfrac{1}{(x+1)^n (x-1)^n} = \sum_{j=1}^n \frac{a_j}{(x+1)^j} + \sum_{j=1}^n \frac{b_j}{(x-1)^j}$$
If $ t = x+1$, we have by the binomial series
$$ \frac{1}{t^n(t-2)^n} = (-1)^n \sum_{k=0}^\infty {k+n-1 \choose k} t^{k-n} 2^{-n-k} $$
so that $$a_j =(-1)^n {2n-1-j \choose n-j} 2^{j-2n},\ j=1 \ldots n$$
and similarly
$$b_j = (-1)^{n+j} {2n-1-j \choose n-j} 2^{j-2n},\ j=1 \ldots n$$


问:
Automorphism of $\mathbb{F}_p[x]/(\Phi_r(x))$
I want to prove that $x\mapsto x^i$, $i\in [r-1]$ gives automorphism over the ring $\mathbb{F}_p[x]/(\Phi_r(x))$, $\Phi_r(x)$ being the $r$th cyclotomic polynomial, $r$ being a prime.

I was able to derive that this gives an endomorphism, as $\Phi_r(x)\mid \Phi_r(x^i)$, so root moves to root and $1,x$ being a generator, of the $\mathbb{F}_p$ algebra, we can derive the endomorphism. But why it will be an automorphism that is my question.

Note- $\Phi_r(x)$ is assumed reducible in $\mathbb{F}_p$ as if it is irreducible then the question is trivial.

答: It is an endomorphism with an inverse: $x^r=1$ so $(x^i)^j=x$ where $ij=1\bmod r$


问:
How do I prove $\vDash \forall x\alpha \to \alpha(t/x)$?

Prove that $\vDash \forall x\alpha \to \alpha(t/x)$, if $t$ is free for $x$ in $\alpha$. Here, $\alpha$ is a wff in FOL (first order logic). In other words, we must show that $\forall x\alpha \to \alpha(t/x)$ is valid, i.e. it is true in every interpretation $\mathcal{I}$ (satisfied by all valuations in $\mathcal{I}$). $\alpha(t/x)$ refers to the wff obtained by replacing all free occurences of $x\in\alpha$ by the term $t$.

How do I go about this?

Using the definition, $\vDash \forall x\alpha \to \alpha(t/x)$ holds iff $\mathcal{I}\vDash \forall x\alpha \to \alpha(t/x)$ for all interpretations $\mathcal{I}$. $\mathcal{I}\vDash \forall x\alpha \to \alpha(t/x)$ holds iff $\mathcal{I},v\vDash \forall x\alpha \to \alpha(t/x)$ holds for all valuations $v$ in $\mathcal{I}$. Also,
$\mathcal{I},v\vDash \forall x\alpha \to \alpha(t/x)$ holds iff whenever $\mathcal{I},v\vDash \forall x\alpha$ holds we have $\mathcal{I},v\vDash \alpha(t/x)$. Lastly, $\mathcal{I},v\vDash \forall x\alpha$ holds iff $\mathcal{I},v'\vDash \alpha$ holds for all valuations $v'$ that are $x$-equivalent to $v$ (i.e. $v,v'$ are same on all variables except $x$).

What do I do next?
I'm a beginner in FOL, and I'd appreciate any help.

答:
Assume not, i.e. there is a term $t$ and an interpretation $\mathcal I$ with domain $D$ such that:
$\mathcal I \vDash \forall x \alpha$ and $\mathcal I \nvDash \alpha (t/x)$.

The second one means that there is $d \in D$ and a valuation $v$ such that $v$ maps $t$ in $d$ and $\mathcal I \nvDash \alpha [v(x|d)]$.

But $\mathcal I \vDash \forall x \alpha$ implies that for every $d \in D$ and every valuation $v$ we have: $\mathcal I \vDash \alpha [v(x| d)]$.


问:
Simple proof of "maximum number of right angles in a convex $n$-gon is 3 for $n\geq 5$" for a 8th grade student?
I know a proof of "maximum number of right angles in a convex $n$-polygon is 3  for $n\geq 5$" as follows:

Suppose $k$ is the number of right angles. Then $180(n-2)-90k$ is the sum of other $n-k$ interior angles. Now we can perform these $n-k$ angles such that all have equal angle i.e. equal to their average $\frac{180(n-2)-90k}{n-k}=\frac{180(n-k)+90k-360}{n-k}=180+\frac{90(k-4)}{n-k}$ that $\frac{90(k-4)}{n-k}$ must be negative since each angle $<180$ so $90(k-4)<0$ and $k<4$.

But this proof is a bit cumbersome for 8$^{th}$ grade school student. (The bolded part is also dubious to me and hard to accept it! Let alone the 8$^{th}$ grade school student). Is there any more simple argument?

答: Sum of exterior angles of a convex n-gon is $360^{\circ} = 4\cdot 90^{\circ}$. We conclude for $n > 4$, at most $3$ right angles are allowed. In that case, sum of rest $n-3$ exterior angles is $90^{\circ}$.


问:
$p * z - q * n = 1 \implies \gcd{(z,n)} = 1$
Prove that:

let n $\in \mathbb{N} $ and $ z \leq n \in \mathbb{Z}$

$\exists p,q \in \mathbb{Z}: p * z - q * n = 1 \implies \gcd{(z,n)} = 1$

答:
Let $ c$ be a common divisor of $  z $ and $ n $.

$$c|z \wedge c|n \implies $$

$$c|pz \wedge c|qn \implies$$

$$c|(pz-qn) \implies $$
$$c|1 \implies $$
$$|c|=1 \implies$$
$$gcd(z,n)=1$$


答:
Note that $\gcd(n,z)|n$ and $\gcd(n,z)|z$ so $\gcd(n,z)| pz - qn$ for all $p,q\in \mathbb Z$.

SO if there is any $p,q \in \mathbb Z$ so that $pn-qz = 1$ then $\gcd(n,z)|1$.

Title: $p * z - q * n = 1 \implies \gcd{(z,n)} = 1$

Body:
Prove that:

let n $\in \mathbb{N} $ and $ z \leq n \in \mathbb{Z}$

$\exists p,q \in \mathbb{Z}: p * z - q * n = 1 \implies \gcd{(z,n)} = 1$


问:
How to show that $\mathbb{R} [i]$ is not contained in the center of $\mathbb{H}?$
Prove that $ \{a + bi | a, b \in \mathbb{R}\}$ is
a subring of $\mathbb{H}$ which is a field but is not contained in the center of $\mathbb{H}$
where  $\mathbb{H}$  denote  Hamilton Quaternions

My attempt : A non-empty subset S of R is a subring if $a, b \in S ⇒ a - b,ab \in S$

take $R=\mathbb{H}$, $S=\mathbb{R} [i]=\{a + bi | a, b \in \mathbb{R}\}$

$a= a_1 + b_1 i , b= a_2 +b_2 i $

now  $a-b = (a_1 -a_2) + (b_1 -b_2)i\in S \tag 1$

$ab= a_1a_2-b_1b_2 + (a_1b_2 + a_2b_1)i \in S \tag2$

From $(1)$ and $(2)$  we can say that $S=\mathbb{R} [i]=\{a + bi | a, b \in \mathbb{R}\}$ is  subring of $\mathbb{H}$

My doubt :How to show that $\mathbb{R} [i]$ is not contained in the center of $\mathbb{H}?$

答:
Your proof of the fact that it is a subring is fine.

And it is not contained in the center of $\Bbb H$ because $i\in\Bbb R[i]$ and $ij=-ji\ne ji$.


问:
Is it possible to calculate a probability by plugging a random variable into the cumulative distribution function?
I was calculating $p = {\rm Pr}(X < Y)$ where both $X$ and $Y$ are independent exponentially distributed with rate parameters $\lambda_X$ and $\lambda_Y$ respectively. I had the idea that, since the $p$ is just a constant, $p = \mathbb{E}[p]$, and since ${\rm Pr}(k < Y) = 1 - F_Y(k) = e^{-k\lambda_Y}$, then maybe \begin{align}p = \mathbb{E}[1-F(X)] = \mathbb{E}[e^{-X\lambda_Y}] = \int_0^\infty e^{-x\lambda_Y}\lambda_Xe^{-x\lambda_X} dx  = \frac{\lambda_X}{\lambda_X+\lambda_Y}\end{align} which is the correct answer as $$\int_0^\infty (\lambda_X+\lambda_Y)e^{-(\lambda_X+\lambda_Y)x}dx = \frac{\lambda_X}{\lambda_X+\lambda_Y}.$$ However, when I tried using the same logic to calculate $${\rm Pr}(Z < X + Y) = 1 - \frac{\lambda_X\lambda_Y}{(\lambda_X + \lambda_Z)(\lambda_Y + \lambda_Z)} $$
using the fact that $X$ and $Y$ are independent to separate the expected values however, some preliminary simulation results suggest this is not true. Is there something I am missing?

答:
If I were you, I would double check your simulation. I think the result you are claiming is true. You can formalize the argument with conditional expectations:

$$\mathsf{P}\left(Z<X+Y\right)=\mathsf{E}\left[\mathbf{1}_{Z<X+Y}\right]=\mathsf{E}\left[\mathsf{E} \left[\mathbf{1}_{Z<X+Y} \mid(X,Y)\right]\right]$$

using the law of total expectation. Now as you noticed,

$$\mathsf{E} \left[\mathbf{1}_{Z<X+Y} \mid(X,Y)\right]=\mathsf{E} \left[1-e^{-\lambda_z(X+Y)} \mid(X,Y)\right]=1-\mathsf{E} \left[e^{-\lambda_z X} \mid X\right]\mathsf{E} \left[e^{-\lambda_z Y} \mid Y\right].$$

Note that the last equality is always true, even if $X$ and $Y$ were not independent. It follows that:

$$\mathsf{P}\left(Z<X+Y\right)=1-\mathsf{E} \left[\mathsf{E} \left[e^{-\lambda_z X} \mid X\right]\mathsf{E} \left[e^{-\lambda_z Y} \mid Y\right]\right]=\mathsf{E} \left[\mathsf{E} \left[e^{-\lambda_z X} \mid X\right]\right]\mathsf{E}\left[\mathsf{E} \left[e^{-\lambda_z Y} \mid Y\right]\right]$$

where this time we used that $X$ and $Y$ are independent. Therefore

$$\mathsf{P}\left(Z<X+Y\right)=1-\frac{\lambda_X\lambda_Y}{(\lambda_X+\lambda_Z)(\lambda_Y+\lambda_Z)}.$$

Here is a "simulation" in Python:

```<code>from numpy.random import exponential

N = 50000
lx, ly, lz = 4.434, 2.343, 3.789
s = 0
for i in range(N):
    x, y, z = exponential(1 / lx), exponential(1 / ly), exponential(1 / lz)
    if x < y + z:
        s += 1
print(s / N, 1 - ly * lz / (lx + ly) / (lx + lz))
</code>```

Body:
If I were you, I would double check your simulation. I think the result you are claiming is true. You can formalize the argument with conditional expectations:

$$\mathsf{P}\left(Z<X+Y\right)=\mathsf{E}\left[\mathbf{1}_{Z<X+Y}\right]=\mathsf{E}\left[\mathsf{E} \left[\mathbf{1}_{Z<X+Y} \mid(X,Y)\right]\right]$$

using the law of total expectation. Now as you noticed,

$$\mathsf{E} \left[\mathbf{1}_{Z<X+Y} \mid(X,Y)\right]=\mathsf{E} \left[1-e^{-\lambda_z(X+Y)} \mid(X,Y)\right]=1-\mathsf{E} \left[e^{-\lambda_z X} \mid X\right]\mathsf{E} \left[e^{-\lambda_z Y} \mid Y\right].$$

Note that the last equality is always true, even if $X$ and $Y$ were not independent. It follows that:

$$\mathsf{P}\left(Z<X+Y\right)=1-\mathsf{E} \left[\mathsf{E} \left[e^{-\lambda_z X} \mid X\right]\mathsf{E} \left[e^{-\lambda_z Y} \mid Y\right]\right]=\mathsf{E} \left[\mathsf{E} \left[e^{-\lambda_z X} \mid X\right]\right]\mathsf{E}\left[\mathsf{E} \left[e^{-\lambda_z Y} \mid Y\right]\right]$$

where this time we used that $X$ and $Y$ are independent. Therefore

$$\mathsf{P}\left(Z<X+Y\right)=1-\frac{\lambda_X\lambda_Y}{(\lambda_X+\lambda_Z)(\lambda_Y+\lambda_Z)}.$$

Here is a "simulation" in Python:

from numpy.random import exponential

N = 50000
lx, ly, lz = 4.434, 2.343, 3.789
s = 0
for i in range(N):
    x, y, z = exponential(1 / lx), exponential(1 / ly), exponential(1 / lz)
    if x < y + z:
        s += 1
print(s / N, 1 - ly * lz / (lx + ly) / (lx + lz))


问:
Show that $\int_{B(x, 2r)} |y|^\alpha dy \lesssim \int_{B(x, r)} |y|^\alpha dy$ uniformly in $x$ and $r$
I want to show that
\begin{equation*}
\int_{B(x, 2r)} |y|^\alpha dy \lesssim \int_{B(x, r)} |y|^\alpha dy
\end{equation*}
uniformly in $x \in \mathbb R^d$ and $r > 0$ if and only if $\alpha > -d$. The forward direction follows from the fact $y \mapsto |y|^\alpha$ has a non-integrable singularity at the origin whenever $\alpha \leq -d$, however I am having trouble proving the converse.

答:
I'll treat the case $-d<\alpha <0.$ Note in this case $|y|^\alpha$ is a decreasing function of $|y|.$ Note also $|y|^\alpha$ is locally integrable on $\mathbb R^d.$

Lemma: There is a positive constant $C$ such that

$$\tag 1\int_{B(x,2)}|y|^\alpha\,dy \le C\int_{B(x,1)}|y|^\alpha\,dy\,\text{ for all } x\in \mathbb R^d.
$$

Proof: The functions

$$f(x) = \int_{B(x,1)}|y|^\alpha\,dy,\,\, g(x) = \int_{B(x,2)}|y|^\alpha\,dy $$

are positive and continuous. This follows from the local integrability of $|y|^\alpha$ and the DCT. It follows that on, say, $\{|x|\le 3\},$ $f$ has a positive minimum $m$ and $g$ has a positive maximum $M.$ Therefore

$$\frac{\int_{B(x,2)}|y|^\alpha\,dy}{\int_{B(x,1)}|y|^\alpha\,dy} \le \frac{M}{m}\,\,\text{for }|x|\le 3.$$

This gives $(1)$ for $|x|\le 3.$

Now suppose $|x|>3.$ Then

$$ \int_{B(x,2)}|y|^\alpha\,dy \le  \int_{B(x,2)} (|x|-2)^\alpha\,dy =(|x|-2)^\alpha 2^d V,$$

where $V$ is the volume of the open unit ball. Similarly,

$$\int_{B(x,1)}|y|^\alpha\,dy \ge (|x|+1)^\alpha V.$$

Thus

$$\frac{\int_{B(x,2)}|y|^\alpha\,dy}{\int_{B(x,1)}|y|^\alpha\,dy} \le 2^d\left (\frac{|x|-2}{|x|+1}\right )^\alpha\,\,\text{for }|x|> 3.$$

Now the function $t\to [(t-2)/(t+1)]^\alpha$ is positive and bounded on $(3,\infty).$ That implies $(1)$ holds for $|x|>3,$ and the lemma is proved.

Now suppose $x\in \mathbb R^d, r>0.$ Then the substitution $y=rz$ gives

$$\int_{B(x,r)}|y|^\alpha\,dy = r^{\alpha + d}\int_{B(x/r,1)}|z|^\alpha\,dz.$$

Similarly,

$$\int_{B(x,2r)}|y|^\alpha\,dy = r^{\alpha + d}\int_{B(x/r,2)}|z|^\alpha\,dz.$$

So

$$\tag 2 \frac{\int_{B(x,2r)}|y|^\alpha\,dy}{\int_{B(x,1)}|y|^\alpha\,dy} = \frac{\int_{B(x/r,1)}|z|^\alpha\,dz}{\int_{B(x/r,2)}|z|^\alpha\,dz}.$$

By the lemma, the right side of $(2)$ is uniformly bounded for all $x,r.$

Thus we're done for the case $-d<\alpha<0.$ The proof for $\alpha \ge 0$ is much the same.


问:
Should $\mathbb{N}$ contain $0$?
This is a classical question, that has led to many a heated argument:

Should the symbol $\mathbb{N}$ stand for $0,1,2,3,\dots$ or $1,2,3,\dots$?

It is immediately obvious that the question is not quite well posed. This convention, as many others, are not carved in stone, and there is nothing to prevent mathematician $A$ define $\mathbb{N}$ to be the positive integers including $0$, and mathematician $B$ to define $\mathbb{N}$ to be the positive integers excluding $0$. It does not seem that one definition is accepted widely enough for it to be "the right definition", and even if this was the case, the fashion might change in the future.

I am, however, hoping that there might be a semi-mathematical reason to prefer one notion over the other. For example, I have spent much of my mathematical life believing that $0 \in \mathbb{N}$ because: 1) morally, $\mathbb{N}$ is the cardinalities of finite sets 2) the empty set is a set with $0$ elements. However, recently I realised that this reasoning applies to $\omega$ rather than $\mathbb{N}$, and - much to my horror - I saw $\omega$ and $\mathbb{N}$ used side by side with the only distinction being that $0 \in \omega$ while $0 \not \in \mathbb{N}$. For another example, $\mathbb{N}$ seems to be a much nicer semigroup if $0 \not \in \mathbb{N}$ (and in any case, adding $0$ to a semigroup is a more natural operation than removing it), which would an argument for taking $0 \not \in \mathbb{N}$.

The arguments mentioned above are, of course, rather weak, but perhaps just enough to tip the scale. In any case, this is the general type of argument I am looking for.

Question: Does there exists a convincing argument for deciding if $0 \in \mathbb{N}$ ?

(I consider it quite possible that the answer is negative because in some context one convention is preferable, and in other context the other one. The problem could be dismissed by using $\mathbb{Z}_+$ (or even $\mathbb{Z}_{>0}$ and $\mathbb{Z}_{\geq 0}$) to avoid confusion, but note that in some context one definitely does not want to do this. I would be interested in an argument that is universal in the sense that it makes overall mathematical landscape more elegant, and does not spoil any detail too much. I do not hope that the argument would be convincing to every mathematician, especially one working in a very specific and narrow area.)

答:
They should arguably include zero so that they are a monoid that gives rise to the integers, and all the other numbers, etc, like in Asaf's great response. </p>

<p>In my day to day life though only the cardinality of them matters, as they are really used purely as an index set. In this context it really is purely a matter of aesthetics. I would go back and forth including zero if I want to divide by them or not. </p>

Title: Should $\mathbb{N}$ contain $0$?

Body:
They should arguably include zero so that they are a monoid that gives rise to the integers, and all the other numbers, etc, like in Asaf's great response.

In my day to day life though only the cardinality of them matters, as they are really used purely as an index set. In this context it really is purely a matter of aesthetics. I would go back and forth including zero if I want to divide by them or not.


问:
Integral of shifted function equal to the original function
Why is it that $\int_0^Lf(x)\ \mathrm dx$ is the same as $\int_0^Lf(L-x)\ \mathrm dx$.

I was just messing around and found this to be the case for various functions I tested.

答:
Let $$ g(x)=f(L-x)$$

Observe that
$$g(\frac L2)=f(\frac L2)$$

the curve $ C_g $ of the function $ g $  at $ [0,L] $, is obtained from $ C_f $  by the symetry with respect to the vertical line whose equation is $ x=\frac L2 $.

So, the two curves $ C_f $ and $ C_g $ give the same area, or the same integral.

In other words, with $ t=L-x ,$

$$\int_0^Lf(x)dx=\int_L^0f(L-t)(-1)dt$$
$$=\int_0^Lg(t)dt=\int_0^Lg(x)dx$$


答: We have $$\int_0^L f(x) ~ dx.$$ By the fact that the Lebesgue measure is invariant under translation, this is equal to $$-\int_0^{-L} f(-x) ~ dx.$$ Now define $g(x) = f(-x),$ so $$-\int_0^{-L} f(-x) ~ dx = -\int_0^{-L} g(x) ~ dx = -\int_L^{0} g(x + L) ~ dx = \int_0^{L} g(x + L) ~ dx.$$ But notice that $g(x + L) = f(L - x)$ from our definition of $g$, so $$\int_0^{L} g(x + L) ~ dx = \int_0^L f(L - x) ~ dx.$$

Title: Integral of shifted function equal to the original function

Body:
Why is it that $\int_0^Lf(x)\ \mathrm dx$ is the same as $\int_0^Lf(L-x)\ \mathrm dx$.

I was just messing around and found this to be the case for various functions I tested.


问:
Normed space inequality
I have this inequality for elements $x,y$ in a Normed space. How do I deduce from this that $\frac{4}{5}||x||\leqslant y\leqslant \frac{6}{5}||x||$

I have tried to use the triangle inequality but I’m getting the negative of the result I am after.

Title: Normed space inequality


问:
Roll a die repeatedly until you roll a 5. What is the expected value of the minimum rolled value?
I came across this interview question recently, but can't quite find a generalized way of setting this up.

Now, we know that if $X$ denotes the minimum value rolled, then $X$ cannot be 6, since the rolls do not finish until a 5 is rolled. Therefore $E[X] = \sum_{i=1}^{5} xP(X=x)$.

For $P(X=5)$, it is required that a 5 appeared before a 1, 2, 3, or a 4. Since the probability of this occurring is $\frac{1}{5}$, we can say that $P(X=5) = \frac{1}{5}$.

How would I go about calculating $P(X=x)$ for the remaining numbers? I have seen the following formula with respect to this question, but cannot understand how it's derived:

$P(X=x) = \frac{1}{x(x+1)}$.

Thanks in advance!

答: The minimum value is $1$ if and only if the first $1$ comes before the first $5$, and the probability of this is $\frac12$. The minimum value is $2$ if and only if the first $1$, $2$, and $5$ come in the order $2,5,1$ which has probability $\frac1{3!}$  The minimum value is $3$ if and only if the first $1,2,3,5$ come in the order $3,5,1,2$ or the order $3,5,2,1$.   Can you continue now?


答:
$X=x$ if and only if, out of the set $\{1,\ldots, x,5\}$, $x$ is the first number to be rolled and 5 the second. (If something else in the set came before 5, it would replace $x$ as the minimum.) Every number in the set is equally likely to be first, and the set has size $x+1$ for each $x<5$, so $x$ has probability $\frac1{x+1}$ of being first. If it is, the remaining $x$ numbers are equally likely to be second, so the probability 5 is second is $\frac1x$. Overall that gives $\frac1{x(x+1)}$.

For example, to have $X=2$ we need 2 to be rolled before a 1 or 5 ($1/2$ chance), then 5 to be rolled before 1 ($1/3$ chance given that 2 came first).

Title: Roll a die repeatedly until you roll a 5. What is the expected value of the minimum rolled value?

Body:
I came across this interview question recently, but can't quite find a generalized way of setting this up.

Now, we know that if $X$ denotes the minimum value rolled, then $X$ cannot be 6, since the rolls do not finish until a 5 is rolled. Therefore $E[X] = \sum_{i=1}^{5} xP(X=x)$.

For $P(X=5)$, it is required that a 5 appeared before a 1, 2, 3, or a 4. Since the probability of this occurring is $\frac{1}{5}$, we can say that $P(X=5) = \frac{1}{5}$.

How would I go about calculating $P(X=x)$ for the remaining numbers? I have seen the following formula with respect to this question, but cannot understand how it's derived:

$P(X=x) = \frac{1}{x(x+1)}$.

Thanks in advance!


问:
How to expand equation inside the L2-norm?
I want expand an L2-norm with some matrix operation inside.

Assume I have a regression $Y=X\beta+\epsilon$.

I want to solve (meaning expand),
$$\displaystyle\|Y-X\beta \|_{2}^2$$
Should I do:

1)
$$\displaystyle\|Y\|_{2}^2+2\beta^TX^TX\beta+\|X\|_{2}^2$$
or 2) $$\sum_{i=1}^{k}(y_i-x_i\beta_i)^2$$
$$\sum_{i=1}^{k}(y_i^2-2x_i\beta_i+x_i^2)$$

X: n by k matrix

$\beta$: k by 1 vector

$Y$: n by 1 vector

I think I have seen both: keeping the norm notation as I did in 1), and expand in summation for as I did in 2).

Could someone show me how to do it?

答: It's hard to know what you're trying to achieve. When possible, avoid coordinates until the end. What do you know here and what do you not know? What do you mean by "solve"? Your first equation would be my preference, dotting $Y-X\beta$ with itself, but you have to get the formula correct.</p>


问:
can I finish this solution by this method, if I can please tell me how
$m^2+m=n^3+n^2+n$, find solutions in natural numbers. so this was the problem and answer is that it doesn't have solutions in natural numbers.

It was pretty hard to think about this problem, I wanted to analyze the ends
of the numbers, for example, if $m$ ends on 2, then $m^2$ ends on 4, so their sum ends on 6.

First of all, it is easy to prove that n is even. since the left side is even, then right side must be the even as well, so $n$ must be even.  So I analyzed when n is even, at what number sum n^3+n^2+n ends on, for example if $n$ ends on 2 then $n^2$ ends on 4.

Then I analyzed when $n$ ends on 4,6,8. sum ends on 4 or 8. Then I analyzed ending of $m$ and $m^2$ then I got that $m^2+m$ doesn't end on 4 or on 8, it ends on 0 in 3 cases, when m ends on 4,5,9  $m+m^2$ ends on 0. In other cases $m$ doesn't end on 4 or on 8 so I supposed that $m+m^2$ doesn't equal $n+n^2+n^3$, for natural numbers, but unfortunately and I am very sad right now, I forgot to analyze when $n$ ends on 0. when $n$ ends on 0, it sum $n+n^2+n^3$ ends on 0 and unfortunately, I have to analyze 3 cases for $m$, when sum $m+m^2$ ends on 0.

So I really need your opinion about this can I finish this problem after analyzing ending 0 situations, please answer me it will be very helpful for me, I really want to defend this solution.

答:
Another approach:

$$m^2+m=\frac{m^3-1}{m-1}-1$$

$$n^3+n^2+n=\frac{n^4-1}{n-1}-1$$

Therefore:

$$\frac{m^3-1}{m-1}=\frac{n^4-1}{n-1}$$

Due to Fermat's little theorem:

$n^4-1 \equiv 0 \mod (5)$,$\rightarrow n^4-1=5t$

$m^3-1$ can be a multiple of 5 with some condition:

$m\equiv (0, 1, 2, 3, 4)\ mod (5)$

If $m \equiv 1 \ mod (5)$, then:

$m^3-1 \equiv 0 \ mod (5)$

But not all multiples of 5 are of the form $n^4-1$ unless $m=a^4$ such that $(a^4)^3-1= (a^3)^4-1$. I n this case $n=a^3$.But this does not satisfy the equation, for example let $a=2$ we have:

$m=2^4=16$, $n=2^3=8$

$m^3-1=2^{12}-1= 8^4-1$

But:

$m^2+m=16^2+16=272$

$n^3+n^2+n=8^3+8^2+8=584$

That is this equation can not have integer solution.


问:
Is it possible to prove from the definition of big $O$ that $5n^3+7n+1$ is $O(n^3)$?
Is it possible to prove from the definition of big O that $5n^3+7n+1$ is $O(n^3)$? Can this be generalised to any case where you have to (and what is the procedure for  working it out?) I guess the main question is there a specific term for this kind of procedure?

from the workings I've been given in class given $T\le O(f(n))$ if $n=1$ then $5n^3+7n+1 = 5^3+7+1 (23)$. Although I'm not sure how that relates to proving the above.

Thanks,

P.L.G.

答: Hint: You must show that there is some positive $M$ such that for all $n$ (of sufficiently large absolute value), you have $$|f(n)|\le M|n^3|.$$ Use properties of absolute value, and the fact that $|n^k|\ge |n|$ for all $k\ge1$ and all $n$.


问:
ODE $y(x)=xy'(x)-\sqrt{y'(x)-1}$
I've following differential equation...

$$y(x)=xy'(x)-\sqrt{y'(x)-1}$$

I recognised that it's an Clairaut's equation and then I wanted to find the general and the singular solution.

First I reformed the ODE like that...

$$\frac{d}{dx}y=\frac{d}{dx}dy'-\frac{d}{dx}\sqrt{y'-1}$$
$$y'=xy''+y'-\frac{1}{2\sqrt{y'-1}}*y''$$
$$0=y''(x-\frac{1}{2\sqrt{y'-1}})$$

I can then deduce, that the equation is true if

$y''=0$ or

$x-\frac{1}{2\sqrt{y'-1}}=0$

I solved 1) like that...

$$\int y'' dx = 0 \Leftrightarrow y'=C$$

In our original equation, that gives us...

$$y=Cx-\sqrt{C-1}$$

Then I solved 2)

$$x-\frac{1}{2\sqrt{y'-1}}=0 \Leftrightarrow y'=\frac{1}{4x^2}+1$$
$$\int y'\, dx=\int \frac{1}{4x^2}+1 \, dx$$
$$y = -\frac{1}{2x}+x+C$$

But I've drawn graphs of these two equations and 1), aka the general solution, doesn't touch 2), aka the singular solution.

So i probably have a mistake somewhere, don't I? Can someone point it out to me?

答:
You can insert the expressions
$$
\sqrt{y'-1}=\frac1{2x}
$$
and
$$
y'=\frac1{4x^2}+1
$$
directly into the original equation to get
$$
y=\frac1{4x}+x-\frac1{2x}=x-\frac1{4x}
$$
This is the only singular solution.

You have a constant error in integrating $\frac1{4x^2}$, and you did not compare with the original equation to get $C=0$ in the singular solution.

Title: ODE $y(x)=xy'(x)-\sqrt{y'(x)-1}$


问:
A lattice of circles and a line which does not cut any of them.

Each lattice point is a center of a circle, all with radius $d$. Let line $y={2\over 5}x+n$ doesn't cut or touch any circle, for some $n$. Find the supremum for $d$.

I was trying to maximize $$d={|2x-5y+5n|\over \sqrt{29}}$$ where $x,y$ runs over all integers and we can assume that $n$ is in $(0,1)$, but I don't know how to do it. Any help?

答:
Consider the axis (A) passing through the origin, orthogonal to the family of lines $y=\tfrac25x+p$. All projections of lattice points on (A) are at mutual distances which are integer multiples of $1/\sqrt{29}$ (see explanation below).

If we consider now the projection of the circles on axis (A), we get little segments centered in the projections of lttice points; as they musn't overlap, their maximal radius is therfore $1/(2 \sqrt{29})$

Explanation: If we consider the axis as directed by unit vector $\vec{U}=\binom{-2a}{5a}$ (represented in red) where $a=\dfrac{1}{\sqrt{29}}$, the abscissa of the projection of point $(p;q)$ on axis (A) (in green) is the dot product: $\vec{U}.\binom{p}{q}$ for any $p,q \in \mathbb{Z}$, otherwise said:

$$\dfrac{-2p+5q}{\sqrt{29}},$$

showing that all these abscissas are integer multiples of $\dfrac{1}{\sqrt{29}}$.

Remark: In terms of lines, these $\dfrac{1}{\sqrt{29}}$ "gaps" become, by oblique projection, as $\dfrac{1}{\sqrt{29}}\dfrac{1}{\cos \alpha}=1/5$, where $\alpha$ is the angle between axis (A) and $y$ axis.

Henceforth, the lines passing through the centers of the circles have intercepts multiple of $1/5$, i.e., have equations:

$$y=\dfrac{2}{5}x+\dfrac{k}{5}, \ \ \ k \in \mathbb{Z}$$


问:
Remark about stem field containing two roots
In Intro to Galois Theory on coursera the lecturer gives the definition "Let $P \in K[x]$ irreducible, monic polynomial. A stem field for P is an extension $E$ of $K$ such that $\alpha \in E$ is a root of $P$ and $E = K[\alpha]$."

After proving any two stem fields are isomorphic, the lecturer remarks "In particular, if a stem field contains two roots of $P$, there is an automorphism which takes one to the other."

$\text{Question}$: How can a stem field contain two roots? Based on the definition the only way a stem field can contain two roots is if $\alpha$ is a root of degree $2$, in which case the remark doesn't achieve anything.

答:
There are lots of examples where $K(\alpha)$ contains more than one root of the minimal polynomial of $\alpha$. For example, if $\zeta_n$ is a primitive $n$-th root of unity (meaning $\zeta_n^n = 1$, and $\zeta_n^d \neq 1$ for all $0 \leq d < n$), then the minimal polynomial of $\zeta_n$ over $\mathbb{Q}$ is the $n$-th cyclotomic polynomial $\Phi_n$, and all of the other roots of $\Phi_n$ are also primitive $n$-th roots of unity and therefore are integer powers of $\zeta_n$. (In this case, $\zeta_n \mapsto \zeta_n^a$ induces an automorphism of the cyclotomic field $\mathbb{Q}(\zeta_n)$ for each integer $a$ with $\gcd(a, n) = 1$.)

Much more generally, if $E/K$ is any Galois extension of fields, then $E = K[\alpha]$ for some $\alpha \in E$ by the primitive element theorem, and if $P \in K[x]$ is the minimal polynomial of $\alpha$, then $K[\alpha]$ contains all roots of $P$. If you want specific examples, you can find plenty in the LMFDB's database of number fields.

(By the way, I've never heard the term "stem field" before—I don't think that's standard terminology.)