Talk:Lagrange inversion theorem

What about this:

$\left.g(z)=a+\sum _{n=1}^{\infty }{\frac {d^{n-1}}{(dw)^{n-1}}}\left({\frac {(w-a)^{n}}{(f(w)-b)^{n}}}\right)\right|_{w=a}{\frac {(z-b)^{n}}{n!}}$

                ∞    dn-1  /  (w - a)n    \ |      (z - b)n 
   g(z) = a  +  ∑  ------ | -----------  | |      --------                      
               n=1 (dw)n-1 \ (f(w) - b)n  / |         n!
                                           | w=a

--Edmund 02:23 Feb 22, 2003 (UTC)

The d is typically italicised. The TeX is not correct. The formula says: differentiate the stuff in the parentheses n-1 times with respect to w, then plug in w=a, then multiply the whole enchilada with (z-b)^n/n!. AxelBoldt 01:12 Feb 22, 2003 (UTC)