# 指数函数

$e^x=\sum _{i=0}^\infty \frac {x^i}{i!}$

## 自然数幂求和

$e^{nx}=\sum _{i=0}^\infty \frac {n^i}{i!}x^i$

$S_d(n)=\sum _{i=0}^{n-1}i^d$

\begin{aligned} \sum _{k=0}^{n-1}e^{kx}&=\sum _{k=0}^{n-1}\sum _{i=0}^\infty \frac {k^i}{i!}x^i \\ &=\sum _{i=0}^\infty \frac {x^i}{i!}\sum _{k=0}^{n-1}k^i \\ &=\sum _{i=0}^\infty S_i(n)\frac {x^i}{i!} \end{aligned}

$\sum _{i=0}^\infty S_i(n)\frac {x^i}{i!}=\frac {e^{nx}-1}{e^x-1}$

$e^{nx}-1=\sum _{i=1}^\infty \frac {n^ix^i}{i!}$

\begin{aligned} S_d(n)&=d![x^d]\frac {e^{nx}-1}{e^x-1} \\ &=d![x^d](\frac {e^{nx}-1}{x}*\frac {x}{e^x-1}) \end{aligned}

$\frac {e^{nx}-1}{x}=\sum _{i=0}^\infty \frac {n^{i+1}x^i}{(i+1)!}$

\begin{aligned} S_d(n)&=d!\sum _{i=0}^d\frac {B_i}{i!}*\frac {n^{d+1-i}}{(d+1-i)!} \\ &=\frac 1 {d+1}\sum _{i=0}^d\binom {d+1} iB_in^{d+1-i} \end{aligned}

$S_d(n)=\sum _{i=0}^{n-1}(ai+b)^d$

## 第二类斯特林数与 Bell 数

$n$ 个不同元素分成 $m$ 个有序集合，每个集合都有元素，相当于是：

\begin{aligned} \text{ans}&=\sum _{\sum a_i=n,a_i>0}\binom n {a_1}\binom {n-a_1}{a_2}\cdots\binom {n-\sum _{i=1}^{m-1}a_i}{a_m} \\ &=\frac {n!}{\prod (a_i!)} \end{aligned}

\begin{aligned} ans&=n![x^n](\sum _{i=1}^\infty \frac {x^i}{i!})^m \\ &=n![x^n](e^x-1)^m \\ &=n![x^n]\sum _{i=0}^m\binom m i e^{ix}(-1)^{m-i} \\ &=n!\sum _{i=0}^m\binom m i \frac {i^n}{n!}(-1)^{m-i} \\ &=\sum _{i=0}^m\binom m i i^n (-1)^{m-i} \end{aligned}

$\genfrac \{\}00 n m =\frac 1 {m!}\sum _{i=0}^m\binom m i i^n(-1)^{m-i}$

$\genfrac \{\}00 n m=\frac {n!} {m!}[x^n](e^x-1)^m$

$\sum _{i=0}^\infty \genfrac \{\}00 i m\frac {x^i}{i!}=\frac {(e^x-1)^m}{m!}$

\begin{aligned} B_n&=\sum _{i=0}^n\genfrac \{\}00 n i \\ &=\sum _{i=0}^n\frac {n!}{i!}[x^n](e^x-1)^i \\ & =n![x^n]\sum _{i=0}^n\frac {(e^x-1)^i} {i!} \end{aligned}

\begin{aligned} B_n=n![x^n]e^{e^x-1} \\ \frac {B_n}{n!}=[x^n]e^{e^x-1} \\ \end{aligned}

$B(x)=\sum _{i=0}^\infty B_n\frac {x^n}{n!} =e^{e^x-1}$

posted @ 2018-04-03 14:40  permui  阅读(1303)  评论(0编辑  收藏  举报