Composition of power series

Question

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 composition 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.

score 6 · Accepted Answer · edited Feb 23 '23 at 11:51

6

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.

edited Feb 23 '23 at 11:51

Martin Sleziak

56,060

answered Mar 05 '11 at 19:54

Bill Dubuque

282,220

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 05 '11 at 20:20
1

@aukie: That and much more should be in the classic Schwatt: Operations with series – Bill Dubuque Mar 05 '11 at 20:29
aplologies, but i'm finding it hard to source this reference, is there any others you could provide? – aukie Mar 05 '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 05 '11 at 21:12
You have to have a deeper understanding of the function to know if it can be factored or simplified in some way beyond Faà di Bruno's formula and Bell polynomials . This is what computer algebra packages try to do. – CarP24 May 01 '23 at 04:21

score 3 · Answer 2 · answered Mar 05 '11 at 19:05

3

Please refer this link:

http://unapologetic.wordpress.com/2008/09/24/composition-of-power-series/

answered Mar 05 '11 at 19:05

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 05 '11 at 19:27
They are infinite only if the inner series has a non-zero coefficient $b_0$ at $x^0$. – Vladimir Reshetnikov Aug 05 '19 at 22:49

score 3 · Answer 3 · edited Feb 23 '23 at 11:49

3

There is a closed form, but it is kind of complicated: https://en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno%27s_formula

edited Feb 23 '23 at 11:49

Martin Sleziak

56,060

answered Mar 05 '11 at 19:38

Zarrax

46,591

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 05 '11 at 20:00
1

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 05 '11 at 20:26

score 2 · Answer 4 · edited Feb 23 '23 at 11:50

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

score 0 · Answer 5 · answered Oct 27 '15 at 01:29

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.

MR3324698
Towse, Christopher
Iteration of sine and related power series.
Math. Mag. 87 (2014), no. 5, 338–349.

score 0 · Answer 6 · answered May 01 '23 at 03:18

If for $g(f(x))$, $f(x)$ can be factored like of the form $(1+c*x)$, it is possible by applying partial fraction decomposition and then Taylor series to it, but otherwise no nice expression exists. An example where this is possible are the Chebyshev polynomials, because these can be factored nicely.

Composition of power series

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