事实上,
\[e^{(e^t-1)x}=\sum_{k=0}^{\infty}\frac{B_k(x)}{k!}.\]
\[B_n(x)=x\sum_{k=1}^{n}\binom{n-1}{k-1}B_{k-1}(x),\]
其中$B_0(x)=1$.
%http://mathworld.wolfram.com/BellPolynomial.html
\[B_n=\sum_{k=0}^{n-1}\binom{n-1}{k}B_k
=\frac{1}{e}\sum_{k=0}^{\infty}\frac{k^n}{k!},\]
\[e^{e^x-1}=\sum_{n=0}^{\infty}\frac{B_n}{n!}x^n.\]
\[\frac{{\ln {B_n}}}{n} = \ln n - \ln \ln n - 1 + \frac{{\ln \ln n}}{{\ln n}} + \frac{1}{{\ln n}} + \frac{1}{2}{\left( {\frac{{\ln \ln n}}{{\ln n}}} \right)^2} + O\left( {\frac{{\ln \ln n}}{{{{\ln }^2}n}}} \right)\]
%de Bruijn, N. G. Asymptotic Methods in Analysis. New York: Dover, pp. 102-109, 1981.
\[{B_n} \sim \frac{1}{{\sqrt n }}{\left[ {\lambda \left( n \right)} \right]^{n + \frac{1}{2}}}{e^{\lambda \left( n \right) - n - 1}},\]
其中$\lambda \left( n \right) = \frac{n}{{W\left( n \right)}}$,其中$W(n)$为 the Lambert W-function.
%Lovász, L. Combinatorial Problems and Exercises, 2nd ed. Amsterdam, Netherlands: North-Holland, 1993.
Odlyzko (1995) gave
\[{B_n} \sim \frac{{n!}}{{\sqrt {2\pi {W^2}\left( n \right){e^{W\left( n \right)}}} }}\frac{{{e^{{e^{W\left( n \right)}} - 1}}}}{{{W^n}\left( n \right)}}.\]
%http://mathworld.wolfram.com/BellNumber.html
$$
a_n=e\frac{B_n}{n!}=\frac{1}{n!}\sum_{k=0}^{\infty}{\frac{k^n}{k!}}\ge e\left( \gamma \ln n \right) ^{-n}
$$
\item[B-3] 已知
\[E(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!},\quad
T(x)=\frac{E(x)-E(-x)}{E(x)+E(-x)}.\]
\begin{enumerate}
\item 求证$T'(x)+T^2(x)=1$.
\item 求$T$的反函数.
\end{enumerate}
\item[B-4] 对任意自然数$m$, $f^{(m+1)}(x)$的级数展式中$x^m$项系数为$1$,求$f(x)$.
\end{enumerate}
Tangss同学面试问题:面试65人,有5个面试室,每个好像风格不太一样.我那个教室老师先问我学了些什么大学内容,然后问了一些相关方面的知识.最后考了点拓扑的东西(曲面的分类,欧拉示性数等)