Talk:Faà di Bruno's formula

It would be useful to know more about the Faa di Bruno polynomials. They seem to be known as Faber polynomials although F di B's precedence is assured.

John McKay

129.199.98.69 06:27, 16 July 2007 (UTC)

Error in "explicite example"

The number of derivative-symbols for f',f, f' seems to be messy. ~~ —Preceding unsigned comment added by Druseltal2005 (talk • contribs) 08:11, 2 November 2007 (UTC)

Possible conflict of interest

Is the User:Michael Hardy who started this article and has been a prolific contributor to it any relation to the Michael Hardy whose paper in Electronic Journal of Combinatorics is reference [1]? Dewey process (talk) 22:34, 31 October 2009 (UTC)

The same. Apparently that was put there in this edit. Michael Hardy (talk) 00:07, 1 November 2009 (UTC)

Not really. My edit was just reformatting of that reference. But originally it was introduced by you in 2006. Anyway, I don't see any problem with it. Maxal (talk) 14:43, 1 November 2009 (UTC)

So this is one of those. I no longer remember which is which. Michael Hardy (talk) 00:53, 2 November 2009 (UTC)

.....OK, now I've found this item I wrote three months ago. Note that I wrote:

And I have added one myself—maybe two or three years ago.

and

See if you can guess which article.

If you'd asked me a couple of days ago, I'm not sure I'd have remember which one, although I could probably have figured it out with a bit of thought. Anyway, now I know. Michael Hardy (talk) 03:31, 2 November 2009 (UTC)

Agreed: At most a trivial appearance of coi, but adding this public-access and clear paper is a much greater good. This paper should have been added by other editors already, and I would defend its continued placement here. (Hardy's paper has been reviewed by Mathematical Reviews and is being cited by other researchers, so it deserves a place here.) Kiefer.Wolfowitz (talk) 17:40, 17 April 2010 (UTC)

Is this right?

This is a pretty bulky formula and I'm trying to wrap my head around it. I'm confused by this statement in the opening section:

"where the sum is over all n-tuples (m1, ..., mn) satisfying the constraint:" 1m1 + 2m2 + 3m3 + ... + nmn = n

If all the m's are positive integers this is impossible. Is it meant to = n! ?