拉格朗日反演公式(Lagrange Inversion)分析证明

The original form

\[w(t)=t\phi(w(t)) \]

\[[t^n]w^k=\frac{1}{2\pi in}\oint\frac{(w^k)'}{t^n}dt \]

\[=\frac{1}{2\pi in}\oint\frac{kw^{k-1}\phi(w)^n}{w^n}dw \]

\[=\frac{k}{n}[t^{n-k}]\phi(w)^n \]

A useful variant.

\[\mathcal{G}([t^n]F(t)\phi(t)^n)=\frac{F(w)}{1-t\phi'(w)}|w=t\phi(w) \]

which means

\[\sum_n([x^n]F(x)\phi(x)^n)t^n=\frac{F(w)}{1-t\phi'(w)}|w=t\phi(w) \]

In Lagrange_Inversion_When_and_How Sprugnoli, this is called diagonalization. because if we have

\[F(w,z)=\sum_{m,n} a_{m,n}w^mz^n \]

and we wish to compute a generating function for diagonal elements,

\[G(z)=\sum_na_{n,n}z^n \]

\[\frac{1}{2\pi i}\oint F(t,z/t)\frac{dt}{t}=\frac{1}{2\pi i}\oint\sum_{m,n}a_{m,n}t^{m-n}z^n \frac{dt}{t}=G(z) \]

here

\[\sum_n([x^n]F(x)\phi(x)^n)t^n \]

\[=\frac{1}{2\pi i}\oint\frac{F(w)}{(1-\phi(w)\frac{z}{w})w}dw \]

note that here we choose \(|\phi(w)\frac{z}{w}|<1\)

\[=\frac{1}{2\pi i}\oint\frac{F(w(t))}{(t-z)(1-\phi'(w(t))t)}dt \]

here \(|t|>z\) and the only pole is \(t=z\) with order 1

\[=\lim_{t\to z}(t-z)\frac{F(w(t))}{(t-z)(1-\phi'(w(t))t)dt}=\frac{F(w(z))}{(1-\phi'(w(z))z)} \]

substituting \(z\) with \(t\) gives the answer.