事实上,
[e^{(e^t-1)x}=sum_{k=0}^{infty}frac{B_k(x)}{k!}.]
[B_n(x)=xsum_{k=1}^{n}inom{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}inom{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}}}
ight)^2} + Oleft( {frac{{ln ln n}}{{{{ln }^2}n}}}
ight)]
%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
ight)}
ight]^{n + frac{1}{2}}}{e^{lambda left( n
ight) - n - 1}},]
其中$lambda left( n
ight) = frac{n}{{Wleft( n
ight)}}$,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 {2pi {W^2}left( n
ight){e^{Wleft( n
ight)}}} }}frac{{{e^{{e^{Wleft( n
ight)}} - 1}}}}{{{W^n}left( n
ight)}}.]
%http://mathworld.wolfram.com/BellNumber.html
$$
a_n=efrac{B_n}{n!}=frac{1}{n!}sum_{k=0}^{infty}{frac{k^n}{k!}}ge eleft( gamma ln n
ight) ^{-n}
$$
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)}.]
egin{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)