Editing Lagrange inversion theorem (section)

===Lagrange–Bürmann formula===

There is a special case of Lagrange inversion theorem that is used in [[combinatorics]] and applies when <math>f(w)=w/\phi(w)</math> for some analytic <math>\phi(w)</math> with <math>\phi(0)\ne 0.</math> Take <math>a=0</math> to obtain <math>f(a)=f(0)=0.</math> Then for the inverse <math>g(z)</math> (satisfying <math>f(g(z))\equiv z</math>), we have

:<math>\begin{align}
  g(z) &= \sum_{n=1}^{\infty} \left[ \lim_{w \to 0} \frac {d^{n-1}}{dw^{n-1}} \left(\left( \frac{w}{w/\phi(w)} \right)^n \right)\right] \frac{z^n}{n!} \\
  {} &= \sum_{n=1}^{\infty} \frac{1}{n} \left[\frac{1}{(n-1)!} \lim_{w \to 0} \frac{d^{n-1}}{dw^{n-1}} (\phi(w)^n) \right] z^n,
 \end{align}</math>

which can be written alternatively as

:<math>[z^n] g(z) = \frac{1}{n} [w^{n-1}] \phi(w)^n,</math>

where <math>[w^r]</math> is an operator which extracts the coefficient of <math>w^r</math> in the Taylor series of a function of {{mvar|w}}.

A generalization of the formula is known as the '''Lagrange–Bürmann formula''':
:<math>[z^n] H (g(z)) = \frac{1}{n} [w^{n-1}] (H' (w) \phi(w)^n)</math>

where {{math|''H''}} is an arbitrary analytic function.

Sometimes, the derivative {{math|''{{prime|H}}''(''w'')}} can be quite complicated. A simpler version of the formula replaces {{math|''{{prime|H}}''(''w'')}} with {{math|''H''(''w'')(1 &minus; ''{{prime|φ}}''(''w'')/''φ''(''w''))}} to get 

:<math> [z^n] H (g(z)) = [w^n] H(w) \phi(w)^{n-1} (\phi(w) - w \phi'(w)),  </math>

which involves {{math|''{{prime|φ}}''(''w'')}} instead of {{math|''{{prime|H}}''(''w'')}}.