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!}}

                ∞    d^n-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)

Yup, the TeX is correct now. AxelBoldt 20:45 Mar 2, 2003 (UTC)

Examples

I think the use of the "group" in binary tree example unnecessary complicates things. If I didn't know what this is all about I wouldn't have guessed what is meant here. Can someone rewrite this in plain language, like "removing the root splits a binary tree into two subtrees, so accounting for the missing root vertex we have..."

Also, while I am here - the example with enumeration of labeled trees giving Cayley's formula is much more interesting, I say. Mhym 23:14, 22 July 2006 (UTC)[reply]

Generalizations? How?

The article states that the theorem is readily generalized to functions of several variables, but I don't see how, not the least of which is because the dimensions of the domain need not be the same as the dimensions of the range (for example, a vector-valued function of a scalar would have what kind of series, or vice versa?).--Jasper Deng (talk) 06:03, 20 January 2014 (UTC)[reply]