Power of Series

Question

In taking the power of a series

$$\left(\sum_{k=0}^{\infty} a_k x^k \right)^n = \sum_{k=0}^{\infty} c_k x^k$$

do you know an expression for $c_k$ solely in terms of the coefficients $a_k$?

@Simple Art: how do you do a multinomial expansion on an infinite series? — Michael McGovern, Dec 28 '16 at 23:21
Monstrous. I can hardly describe it without giving you the link. — Simply Beautiful Art, Dec 28 '16 at 23:21
@MichaelMcGovern Clearly, each $c_k$ depends only on a finite amount of $a_k$, since higher powers will never decrease. — Simply Beautiful Art, Dec 28 '16 at 23:22
May I write that as $c_k = \sum_{i_1 + \ldots + i_k = k} {{k} \choose {i_1, \ldots, i_k}} \prod_{s=1}^k a_s$? — Jennifer, Dec 28 '16 at 23:36
Why would you have that coefficient?Also, $\prod_{s=1}^k a_s$ doesn't seem to depend on the $i_1,\dot,i_n$. Also, The $i_1,\dots,i_n$ should end in a suffix that is not $k$. @Jennifer — Thomas Andrews, Dec 28 '16 at 23:38
You can also compute the coefficients $c_k$ successively using Taylor series arithmetic, essentially for the cost of one power series multiplication. See https://math.stackexchange.com/q/1471438/115115 and the there linked wikipedia article or any other references on automatic differentiation. — Lutz Lehmann, Jul 23 '17 at 09:01

score 3 · Answer 1 · answered Dec 28 '16 at 23:24

3

$c_k$ is the sum of $a_{k_1} a_{k_2} \ldots a_{k_n}$ over all ordered $n$-tuples of nonnegative integers $(k_1, \ldots, k_n)$ whose sum is $k$.

answered Dec 28 '16 at 23:24

Robert Israel

448,999

score 2 · Answer 2 · answered Dec 29 '16 at 00:01

2

The form you seem to be seeking is $$c_k=\sum_{\substack i_0+i_1+i_2+\cdots+i_k=n\\i_1+2i_2+\cdots+ki_k=k}\binom{n}{i_0,\cdots,i_k}\prod_{j=0}^k a_j^{i_j}$$

answered Dec 29 '16 at 00:01

Thomas Andrews

177,126

score 0 · Answer 3 · answered Dec 28 '16 at 23:48

0

This is repeated application of Cauchy Product : https://en.wikipedia.org/wiki/Cauchy_product

answered Dec 28 '16 at 23:48

jimjim

9,675

Power of Series

3 Answers3