Talk:Lagrange inversion theorem

[Untitled]

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)

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)

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)

Proof of the theorem is not given

The proof of the theorem is not to be found anywhere. Not even in Lagrange's original paper. — Preceding unsigned comment added by 2601:9:8180:1029:C5A7:2EC4:2CEB:1843 (talk) 18:19, 30 April 2014 (UTC)

You may check the link to the wiki article on formal power series, where a few lines elementary proof is given. --pma 14:07, 3 May 2014 (UTC)

An addition that needs discussion (and a source)

A couple of weeks ago, Mathematical.Analysis made the following addition to the article, at the end of the section Lagrange_inversion_theorem#Derivation:

In fact, integration by parts doesn't work for:

$n=0\rightarrow {\frac {1}{2\pi i}}\oint _{C}{\frac {\omega f'(\omega )}{f(\omega )-f(a)}}d\omega ={\frac {1}{2\pi i}}\oint _{f(C)}{\frac {f^{-1}(u)}{u-f(a)}}du=f^{-1}(f(a))=a$

Fortunately, writing the integral as a residue in the last line fixes the problem.

I've reverted this because of the odd formating and the apparent editorialization, without making any attempt to determine the validity. The whole derivation is (sadly) unsourced; does anyone have access to a handy reference that would allow a sourced derivation, rather than the apparent WP:OR we have here? --JBL (talk) 20:51, 16 June 2024 (UTC)

IMHO

The whole article should be rewritten, starting from eliminating awkward formulas like

$f^{-1}(z)=\sum _{n=0}^{\infty }{\frac {({z-f(a)})^{n}}{n!}}\lim _{\omega \to a}{\frac {d^{n-1}}{d\omega ^{n-1}}}\left({\frac {\omega -a}{f(\omega )-f(a)}}\right)^{n},$

that no mathematician would write. One should avoid keeping throughout these proofs, that are algebraic in nature, an analytic/topologic language (limits, path integration,..) here useless and misleading. The point is that, dealing with analytic functions (like, in fact, we do in all mathematics), whenever possible computations are made at a formal level, relying on suitable general representation theorems that allows to do so (here enters the topology, once and for all). Here we can map germs of meromorphic functions at 0, to their formal Laurent series, and perform all computations in that algebraic setting. pma 15:42, 21 June 2024 (UTC)

@PMajer: The formula you've quoted here is in the (uncited) section Lagrange_inversion_theorem#Derivation, which was added about six months ago by Mathematical.Analysis and is also the subject of the discussion section just above this one. Are the points you're concerned about largely limited to that section? If so, perhaps it should be removed. --JBL (talk) 19:59, 21 June 2024 (UTC)