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)[reply]

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)[reply]

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)[reply]

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

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)[reply]

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

.....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)[reply]

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)[reply]

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! ? —Preceding unsigned comment added by 76.217.237.122 (talk) 16:26, 6 July 2010 (UTC)[reply]

     m_i can be 0. 72.48.121.4 (talk) 15:19, 24 September 2010 (UTC)[reply]

Suppose n = 4. Then we have

1×(4) + 2×(0) + 3×(0) + 4×(0) = 4, so (m₁, m₂, m₃, m₄) = (4, 0, 0, 0)

1×(2) + 2×(1) + 3×(0) + 4×(0) = 4, so (m₁, m₂, m₃, m₄) = (2, 1, 0, 0)

1×(0) + 2×(2) + 3×(0) + 4×(0) = 4, so (m₁, m₂, m₃, m₄) = (0, 2, 0, 0)

1×(1) + 2×(0) + 3×(1) + 4×(0) = 4, so (m₁, m₂, m₃, m₄) = (1, 0, 1, 0)

1×(0) + 2×(0) + 3×(0) + 4×(1) = 4, so (m₁, m₂, m₃, m₄) = (0, 0, 0, 1)

So it is far from impossible. Michael Hardy (talk) 18:57, 6 December 2010 (UTC)[reply]

2 0
0 1

3 0 0
1 1 0
0 0 1

4 0 0 0
2 1 0 0
0 2 0 0
1 0 1 0
0 0 0 1

5 0 0 0 0
3 1 0 0 0
1 2 0 0 0
2 0 1 0 0
0 1 1 0 0
1 0 0 1 0
0 0 0 0 1

6 0 0 0 0 0
4 1 0 0 0 0
2 2 0 0 0 0
0 3 0 0 0 0
3 0 1 0 0 0
1 1 1 0 0 0
0 0 2 0 0 0
2 0 0 1 0 0
0 1 0 1 0 0
1 0 0 0 1 0
0 0 0 0 0 1

7 0 0 0 0 0 0
5 1 0 0 0 0 0
3 2 0 0 0 0 0
1 3 0 0 0 0 0
4 0 1 0 0 0 0
2 1 1 0 0 0 0
0 2 1 0 0 0 0
1 0 2 0 0 0 0
3 0 0 1 0 0 0
1 1 0 1 0 0 0
0 0 1 1 0 0 0
2 0 0 0 1 0 0
0 1 0 0 1 0 0
1 0 0 0 0 1 0
0 0 0 0 0 0 1

8 0 0 0 0 0 0 0
6 1 0 0 0 0 0 0
4 2 0 0 0 0 0 0
2 3 0 0 0 0 0 0
0 4 0 0 0 0 0 0
5 0 1 0 0 0 0 0
3 1 1 0 0 0 0 0
1 2 1 0 0 0 0 0
2 0 2 0 0 0 0 0
0 1 2 0 0 0 0 0
4 0 0 1 0 0 0 0
2 1 0 1 0 0 0 0
0 2 0 1 0 0 0 0
1 0 1 1 0 0 0 0
0 0 0 2 0 0 0 0
3 0 0 0 1 0 0 0
1 1 0 0 1 0 0 0
0 0 1 0 1 0 0 0
2 0 0 0 0 1 0 0
0 1 0 0 0 1 0 0
1 0 0 0 0 0 1 0
0 0 0 0 0 0 0 1

9 0 0 0 0 0 0 0 0
7 1 0 0 0 0 0 0 0
5 2 0 0 0 0 0 0 0
3 3 0 0 0 0 0 0 0
1 4 0 0 0 0 0 0 0
6 0 1 0 0 0 0 0 0
4 1 1 0 0 0 0 0 0
2 2 1 0 0 0 0 0 0
0 3 1 0 0 0 0 0 0
3 0 2 0 0 0 0 0 0
1 1 2 0 0 0 0 0 0
0 0 3 0 0 0 0 0 0
5 0 0 1 0 0 0 0 0
3 1 0 1 0 0 0 0 0
1 2 0 1 0 0 0 0 0
2 0 1 1 0 0 0 0 0
0 1 1 1 0 0 0 0 0
1 0 0 2 0 0 0 0 0
4 0 0 0 1 0 0 0 0
2 1 0 0 1 0 0 0 0
0 2 0 0 1 0 0 0 0
1 0 1 0 1 0 0 0 0
0 0 0 1 1 0 0 0 0
3 0 0 0 0 1 0 0 0
1 1 0 0 0 1 0 0 0
0 0 1 0 0 1 0 0 0
2 0 0 0 0 0 1 0 0
0 1 0 0 0 0 1 0 0
1 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 1

Is this correct? --ExcessPhase (talk) 04:07, 7 December 2010 (UTC)[reply]

request

I've some serious interest in how this formula looks like for some F(x(z)) with the result of F being a scalar and x being a vector and z being an even larger vector (larger number of elements). In the end I want d^nF/dz^n but I want to avoid having to carry so many derivatives (since the number of elements in z is larger than the number of elements in x).

Thanks to everybody contributing here! --ExcessPhase (talk) 22:20, 5 December 2010 (UTC)[reply]

Maybe this is the vector form of the formula that you're looking for? Michael Hardy (talk) 18:41, 6 December 2010 (UTC)[reply]

Yes -- thanks --- this is what I want. Could you explain the formula to me? I'm not a mathematician -- I don't follow what every index means in this paper. I tried to contact the author (Tsoy-Wo Ma) by email, but he cannot be reached anymore via the email address in this paper. Or even better, please publish his formula here on wikipedia. Thanks --ExcessPhase (talk) 05:14, 9 December 2010 (UTC)[reply]