Assuming I have a function f(x) in the form of a Taylor series $f(x)=\sum_{n=0}^\infty a_n x^n$, I want to compute a power of it, i.e. $g(x)=[f(x)]^m$ where $m$ is an integer bigger than one.

I would assume that $g(x)$ also has a Taylor series of a form $g(x)=\sum_{n=0}^\infty b_n x^n$ (I'd guess there exists a proof of this?). One would want to determine $b_n$ coefficients from $\{a_n\}$ and for a given $m$:

$$ b_n = \frac{1}{n!}\frac{d^n}{dx^n}g(x)\Big|_{x=0} =\frac{1}{n!}\frac{d^n}{dx^n}\big[f(x)\big]^m\Big|_{x=0}.$$

Is there an explicit formula for $b_n$?

I've tried calculating it from scratch, but it looks quite difficult. Can someone point me to some literature where this is discussed?

My question is in part similar to this one.

Update: The solution is given by the Faà di Bruno's formula.

  • $\begingroup$ Since power series converge absolutely, you can inductively form Cauchy products for power. $\endgroup$ Dec 19 '16 at 5:36
  • $\begingroup$ This seems to lead to the similar answer as by following Pat's suggestion below, i.e. Multinomial theorem. The problem is then how to extract the $x^n$ coefficient from the multiple sums I obtain? $\endgroup$
    – z.v.
    Dec 19 '16 at 6:26

Hint: try to square a Taylor series and figure out the coefficients. Try to figure out the new coefficients in terms of the old ones without doing any derivatives. Just do what you might do if you were multiplying two big polynomials.

If you figure this out, try to cube the thing, et cetera. I think you'll be able to see the pattern.

  • $\begingroup$ Well, wouldn't this just give me a result of the Multinomial theorem? Then it is still very non-trivial to extract all coefficients such that $\prod_i x^{k_i} = x^n$... This seems to be a translation of the problem above but it is not clear to me if it makes it easier? $\endgroup$
    – z.v.
    Dec 19 '16 at 6:21
  • $\begingroup$ Square the thing. You don't get binomial theorem. $\endgroup$
    – Pat Devlin
    Dec 19 '16 at 17:13
  • $\begingroup$ If it helps, square this finite polynomial $(a_0 + x a_1 + x^2 a_2 + x^3 a_3 + x^4 a_4).$ You'll see the pattern. $\endgroup$
    – Pat Devlin
    Dec 19 '16 at 17:14
  • $\begingroup$ Thanks! I found the Faà di Bruno's formula which seems to give the solution in general case. $\endgroup$
    – z.v.
    Dec 19 '16 at 19:03
  • $\begingroup$ I don't see how that formula relates to this problem except by use of similar notation. $\endgroup$
    – Pat Devlin
    Dec 20 '16 at 1:01

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