Talk:Faà di Bruno's formula

Formal power series version
I went through the calculations of this paragraph and there seems to be a sum over $$k$$ missing in the coefficient $$c_n$$. Plus a confusion I have about the coefficient in the composition of "exponential"-like series, cf. https://math.stackexchange.com/q/4283116/92038 — Preceding unsigned comment added by Noix07 (talk • contribs) 14:21, 21 October 2021 (UTC)

Reciprocal generating function
In the paragraph of special cases of application of Faà di Bruno to get the reciprocal, I think f(x)=1/(1-x) is ok but g(x) is given as just g(x) = -ogf(x), where ogf(x) is the ordinary generating function one wants to (multiplicatively) invert, also called the reciprocal generating function. But it should have been g(x) = 1-ogf(x), so f(g(x)) = 1/ogf(x) and it equates to Faà formula. Note the lacking 1 term for g(x). This is clearly put in https://unapologetic.wordpress.com/2008/09/24/inverses-of-power-series. Jesuslop (talk) 15:44, 9 September 2023 (UTC)

Faa di Bruno polynomials
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&mdash;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! ? —Preceding unsigned comment added by 76.217.237.122 (talk) 16:26, 6 July 2010 (UTC)

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

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)

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 <br \> 0 1 1 0 0 <br \> 1 0 0 1 0 <br \> 0 0 0 0 1 <br \> <br \> 6 0 0 0 0 0 <br \> 4 1 0 0 0 0 <br \> 2 2 0 0 0 0 <br \> 0 3 0 0 0 0 <br \> 3 0 1 0 0 0 <br \> 1 1 1 0 0 0 <br \> 0 0 2 0 0 0 <br \> 2 0 0 1 0 0 <br \> 0 1 0 1 0 0 <br \> 1 0 0 0 1 0 <br \> 0 0 0 0 0 1 <br \> <br \> 7 0 0 0 0 0 0 <br \> 5 1 0 0 0 0 0 <br \> 3 2 0 0 0 0 0 <br \> 1 3 0 0 0 0 0 <br \> 4 0 1 0 0 0 0 <br \> 2 1 1 0 0 0 0 <br \> 0 2 1 0 0 0 0 <br \> 1 0 2 0 0 0 0 <br \> 3 0 0 1 0 0 0 <br \> 1 1 0 1 0 0 0 <br \> 0 0 1 1 0 0 0 <br \> 2 0 0 0 1 0 0 <br \> 0 1 0 0 1 0 0 <br \> 1 0 0 0 0 1 0 <br \> 0 0 0 0 0 0 1 <br \> <br \> 8 0 0 0 0 0 0 0 <br \> 6 1 0 0 0 0 0 0 <br \> 4 2 0 0 0 0 0 0 <br \> 2 3 0 0 0 0 0 0 <br \> 0 4 0 0 0 0 0 0 <br \> 5 0 1 0 0 0 0 0 <br \> 3 1 1 0 0 0 0 0 <br \> 1 2 1 0 0 0 0 0 <br \> 2 0 2 0 0 0 0 0 <br \> 0 1 2 0 0 0 0 0 <br \> 4 0 0 1 0 0 0 0 <br \> 2 1 0 1 0 0 0 0 <br \> 0 2 0 1 0 0 0 0 <br \> 1 0 1 1 0 0 0 0 <br \> 0 0 0 2 0 0 0 0 <br \> 3 0 0 0 1 0 0 0 <br \> 1 1 0 0 1 0 0 0 <br \> 0 0 1 0 1 0 0 0 <br \> 2 0 0 0 0 1 0 0 <br \> 0 1 0 0 0 1 0 0 <br \> 1 0 0 0 0 0 1 0 <br \> 0 0 0 0 0 0 0 1 <br \> <br \> 9 0 0 0 0 0 0 0 0 <br \> 7 1 0 0 0 0 0 0 0 <br \> 5 2 0 0 0 0 0 0 0 <br \> 3 3 0 0 0 0 0 0 0 <br \> 1 4 0 0 0 0 0 0 0 <br \> 6 0 1 0 0 0 0 0 0 <br \> 4 1 1 0 0 0 0 0 0 <br \> 2 2 1 0 0 0 0 0 0 <br \> 0 3 1 0 0 0 0 0 0 <br \> 3 0 2 0 0 0 0 0 0 <br \> 1 1 2 0 0 0 0 0 0 <br \> 0 0 3 0 0 0 0 0 0 <br \> 5 0 0 1 0 0 0 0 0 <br \> 3 1 0 1 0 0 0 0 0 <br \> 1 2 0 1 0 0 0 0 0 <br \> 2 0 1 1 0 0 0 0 0 <br \> 0 1 1 1 0 0 0 0 0 <br \> 1 0 0 2 0 0 0 0 0 <br \> 4 0 0 0 1 0 0 0 0 <br \> 2 1 0 0 1 0 0 0 0 <br \> 0 2 0 0 1 0 0 0 0 <br \> 1 0 1 0 1 0 0 0 0 <br \> 0 0 0 1 1 0 0 0 0 <br \> 3 0 0 0 0 1 0 0 0 <br \> 1 1 0 0 0 1 0 0 0 <br \> 0 0 1 0 0 1 0 0 0 <br \> 2 0 0 0 0 0 1 0 0 <br \> 0 1 0 0 0 0 1 0 0 <br \> 1 0 0 0 0 0 0 1 0 <br \> 0 0 0 0 0 0 0 0 1 <br \> <br \>

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

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

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)

a better, memorizable representation of the "explicte example"
I've found a very nice decomposition of the Faa di Bruno-formulae as given in the "example"-section. I cannot put it in wikipedia-html-math-style; so I added the explication as screenshot from my word-formula-editor. Perhaps someone with a kind mind (and enough time and patience :-) ) might adapt this to the wikipedia-standard... Gotti

'special example'
The trivial example should be removed and replaced with an actual automorphic class.

Terrible article
The article is truly bad because, neither in the introduction nor anywhere else is it stated what f and g are.

They are obviously some kind of functions. But what is their domain? What is their codomain? What if anything is assumed about f and g? (Certainly they seem to be assumed differentiable to all orders.)

But none of these things are mentioned in the article, so it is entirely unclear what functions the formula applies to.
 * Then help to fix it... Also sign your comments with the four tildes. — Moops  ⋠ T ⋡ 17:37, 14 December 2022 (UTC)