# 证明

## 盒异，球同，盒子允许空

### 做法1

$$f(x)=(\sum_{i=0}^{\infty} x^i)^m=(\frac{1}{1-x})^m=(1-x)^{-m}$$

\begin{aligned} &C_{-m}^{n}(-x)^n \&=\frac{\prod _{k=-m-n+1}^{-m}k}{n!}(-x)^n \&=\frac{\prod {k=m}^{ m+n-1}k}{n!}x^n \&=C {m+n-1}^{m-1}x^n \end{aligned}

\begin{aligned} &f’(x) = -m\cdot(1-x)^{m-1} \cdot (-1) \&f’’(x) = -m\cdot (-m-1)\cdot(1-x)^{m-2} \cdot (-1)^2 \&f’’’(x) = -m\cdot (-m-1)\cdot(-m-2)\cdot(1-x)^{m-3} \cdot (-1)^3 \&… \&f^{(n)}(x)= (\prod_{i=0}^{n-1}(-m-i))\cdot(1-x)^{m-n} \cdot (-1)^n \end{aligned}

\begin{aligned} &\frac{(\prod_{j=0}^{n-1}(-m-i))\cdot(1-0)^{m-n}}{n!}(𝑥−0)^𝑛\cdot (-1)^n \&=\frac{\prod_{j=0}^{n-1}(m+i)}{n!}𝑥^𝑛 \&=\frac{\frac{(m+n-1)!}{(m-1)!}}{n!}𝑥^𝑛 \&=C_{m+n-1}^{m-1} x^n \end{aligned}

## 盒同，球同，盒子允许空

\begin{aligned} &f(x) \&=(\space(x^m)^0+(x^m)^1+(x^m)^2+…)(\space(x^{m-1})^0+(x^{m-1})^1+(x^{m-1})^2+…)… \&=\frac{1}{1-x^m}\frac{1}{1-x^{m-1}}\frac{1}{1-x^{m-2}}…\frac{1}{1-x} \&=\prod _{j=1}^{m}\frac{1}{1-x^{j}} \end{aligned}

## 盒同，球同，盒子不允许空

\begin{aligned} &f(x) \&=(\space(x^m)^1+(x^m)^2+…)(\space(x^{m-1})^0+(x^{m-1})^1+(x^{m-1})^2+…)… \&=\frac{x^m}{1-x^m}\frac{1}{1-x^{m-1}}\frac{1}{1-x^{m-2}}…\frac{1}{1-x} \&=x^m\prod _{j=1}^{m}\frac{1}{1-x^{j}} \end{aligned}

## 盒异，球异，盒子不允许空

\begin{aligned} &f(x) \&=n!\cdot (\frac{x^{1}}{1!}+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+…)^{m} \&=n!\cdot (e^{x}-1)^{m} \&=n!\cdot\sum {k=0}^{m}(C{m}^{k}(-1)^{m-k}(e^{x})^{k}) \&=n!\cdot\sum {k=0}^{m}(C{m}^{k}(-1)^{m-k}(1+\frac{(kx)^{1}}{1!}+\frac{(kx)^{2}}{2!}+\frac{(kx)^{3}}{3!}+…)) \end{aligned}

\begin{aligned} &n!\cdot\sum {k=0}^{m}(C{m}^{k}(-1)^{m-k}\frac{(kx)^{n}}{n!}) \&=(\sum_{k=0}^m C_m^k\cdot (-1)^{m-k}\cdot k^n)\cdot x^n \end{aligned}

## 盒同，球异，盒子允许空

\begin{aligned} \sum_{i=0}^{m}\sum _{k=0}^i\frac{(C_i^k(-1)^{i-k}k^n)}{i!} \end{aligned}

# 第二类斯特林数

S(n,m)=\lbrace^n_m\rbrace=\begin{aligned}\sum _{k=0}^m\frac{(C_m^k(-1)^{m-k}k^n)}{m!}\end{aligned}

## 第二类斯特林数递推

$$S(n,m)=S(n-1,m-1)+S(n-1,m)\cdot m$$

## 计算斯特林数的某一整行

\begin{aligned} &\sum _{k=0}^m\frac{C_m^k(-1)^{m-k}k^n}{m!} \&=\sum _{k=0}^m\frac{(-1)^{m-k}k^n}{k!(m-k)!} \&=\sum _{k=0}^m\frac{(-1)^{m-k}}{(m-k)!}\frac{k^n}{k!} \end{aligned}

http://fightinggg.github.io/fluid/PGMNAB.html

fightinggg

2018年10月15日