拉格朗日反演

拉格朗日反演

拉格朗日反演用于从隐式生成函数中提取系数

拉格朗日反演定理

如果生成函数\(A(z)\)满足函数方程\(z=f(A(z))\),其中\(f(z)\)满足\(f(0)=0\),且\(f’(0) \neq 0\)，则\[a_n = [z^n]A(z) = \frac{1}{n}[u^{n-1}](\frac{u}{f(u)})^n\] \[[z^n](A(z))^m = \frac{m}{n} [u^{n-m}](\frac{u}{f(u)})^n\] \[[z^n]g(A(z)) =\frac{1}{n}[u^{n-1}]g’(u)(\frac{u}{f(u)})^n \]

比如标号树计数生成函数\(C(z)\)满足：\[C(z) = ze^{C(z)}\] 令\(f(u) = \frac{u}{e^u}\),则\[[z^n]C(z) = \frac{1}{n} [u^{n-1}]e^{un} = \frac{1}{n} \frac{n^{n-1}}{(n-1)!}=\frac{n^{n-1}}{n!}\]