Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Does anyone know how to derive a formula for the coefficients.

That is if, $f(x)=\sum _{n=0}^{\infty } a_nx^n$ and $g(x)=\sum _{n=0}^{\infty } b_nx^n$

suppose the compostion is an analytic function, $h(x)=f(g(x))=\sum _{n=0}^{\infty } c_nx^n$

Is there an expression we can find for the coefficients $c_n$ in terms of $a_n$ and $b_n$? Can someone show me how its derived. I know we could substitute $g$ into $f$ and collect powers of $x$. But I believe a formula for general n may be written down.

share|cite|improve this question
up vote 2 down vote accepted

There are (rather unwieldly) "closed-forms" in terms of Bell polynomials and other closely related combinatorial objects. However, if you are really interested in efficiently calculating compositions of power series then there are better algorithms, dating back at least to the work of Brent and Kung, from which you can find links to recent work in this area.

share|cite|improve this answer
But these unwieldly formulas refer to exponential generating functions, do any results exist for standard generating functions? None the less can you point me to any references that derive such formula for the coefficients... Not sure where to look, books on combinatronics, analytic functions or ... ? thanks in advance. I'll look at some of the mentioned algorithms once i've investigated the formulas. I'm particularly curious about their derivation. – aukie Mar 5 '11 at 20:20
@aukie: That and much more should be in the classic Schwatt: Operations with series – Bill Dubuque Mar 5 '11 at 20:29
aplologies, but i'm finding it hard to source this reference, is there any others you could provide? – aukie Mar 5 '11 at 21:00
@auki: In general there's not much more that can be said beyond Faà di Bruno's formula and Bell polynomials and related results. You should be able to find such results in most combinatorics textbooks, e.g. the classic introduction by Riordan. Perhaps if you say more about your motivation then we could say more. – Bill Dubuque Mar 5 '11 at 21:12

Please refer this link:

share|cite|improve this answer
But the coefficients themselves in this post are infintite series ... is there no recurrent or closed form relationship between $a_n$, $b_n$ and $c_n$? so that the unkown coefficients can be computed... – aukie Mar 5 '11 at 19:27

There is a closed form, but it is kind of complicated:

share|cite|improve this answer
Interesting, so we can in terms of bell polynomials... do you know how this formula is derived ... or can you cite me any references? This doesn't directly answer my question since the series involved are exponential power series. but still i would be interested in seeing its proof. – aukie Mar 5 '11 at 20:00
The Faa di Bruno formula does give explicit formulas for $c_n$ in terms of $a_n$ and $b_n$ (if you look at that wikipedia page it gives it at the beginning as a complicated but finite sum). I don't believe there's any simple formula. Steven Krantz's book "A primer of real-analytic functions" gives a proof of it, that's where I've seen it. – Zarrax Mar 5 '11 at 20:26

So, today I was looking into the exact same thing. What I gathered from the above internet references seemed to imply nothing for when $g(x)$ has a constant term. This article indicates that this is an open problem in the theory and lays out conditions for when the composition is defined when $g(x)$ has a constant term. However, it offers no formula for the coefficient structure of the composition.

Does anyone have experience/ideas on how to adapt the Faà di Bruno result

$$f(g(x)) = \sum_{n=1}^\infty {\sum_{k=1}^{n} a_k B_{n,k}(b_1,\dots,b_{n-k+1}) \over n!} x^n$$

where $B_{n,k}$ are Bell polynomials, to accommodate this constant term?

share|cite|improve this answer
Welcome to MSE! Was this intended to be answer to the question posed by the OP or a new question altogether? Regards – Amzoti Aug 1 '13 at 20:30

The paper Composita and its properties by V.V. Kruchinin and D.V. Kruchinin presents techniques to obtain the coefficients of compositae of formal power series.

They start with a given generating function $F(x)=\sum_{n\geq 1}f(n)x^n$, consider various other functions $G(x)=\sum_{n\geq 0}g(n)x^n$ and analyze the composition of $G$ with $F$. \begin{align*} G(F(x))&=\sum_{k\geq 0}g(k)\left[F(x)\right]^k\\ &=\sum_{k\geq 0}g(k)\sum_{n\geq k}F^\triangle(n,k)x^n \end{align*} with the so-called Composita $F^\triangle(n,k)$ \begin{align*} F^\triangle(n,k)=\sum_{{\lambda_1+\cdots\lambda_k=n}\atop{ \lambda_1,\ldots,\lambda_k\geq 1}}f(\lambda_1)\cdots f(\lambda_k) \end{align*}

Note the constant term of $F(x)$ is zero, so that the composition of generating functions is valid.

Other related papers by V.V. Kruchinin are

share|cite|improve this answer

Here is another reference on this topic for those interested in pursuing it further. It's in the Mathematics Magazine and so intended for undergraduate level.

Towse, Christopher
Iteration of sine and related power series.
Math. Mag. 87 (2014), no. 5, 338–349.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.