# combinatorics problem: $$\sum_n a_n (\sum_k b_k \epsilon^k )^n$$

I would like to rewrite the following series $$\sum_{n=0}^\infty \frac{1}{n!}(\Delta^\epsilon)^n a_n , \qquad \Delta^\epsilon=\sum_{k=1}^\infty \epsilon^k b_k$$ As a series in $\epsilon$ $$\sum_{n=0}^\infty c_n \epsilon^n$$ (i.e. I need to find the c_n).

Obviously, I can do this term-by-term. But the general case seems quite difficult. Any help would be greatly appreciated.

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You can use Cauchy products for $\Delta^{\varepsilon}$, viewed as a power series. – Davide Giraudo Jun 24 '12 at 16:11
You want the composition of formal power series. You can find it everywhere. en.wikipedia.org/wiki/Formal_power_series#Composition_of_series – Pietro Majer Jun 24 '12 at 20:21
Oh...I just noticed the comment by Pietro Majer. Thank you. That is helpful. – psyduck Jun 25 '12 at 2:20

We have, using Cauchy products, at least for $\varepsilon$ small enough, that $$(\Delta^{\varepsilon})^n=\sum_{k=1}^{+\infty}\left(\sum_{j_1+\dots+j_n=k}\prod_{i=1}^nb_{j_i}\right)\varepsilon^k,$$ hence, if the coefficients have a good behaviour
\begin{align} \sum_{n=0}^{+\infty}\frac 1{n!}(\Delta^{\varepsilon})^na_n&=\sum_{n=0}^{+\infty}\frac 1{n!}a_n\sum_{k=1}^{+\infty}\left(\sum_{j_1+\dots+j_n=k}\prod_{i=1}^nb_{j_i}\right)\varepsilon^k\\\ &=\sum_{k=1}^{+\infty}\sum_{n=0}^{+\infty}\frac 1{n!}a_n\left(\sum_{j_1+\dots+j_n=k}\prod_{i=1}^nb_{j_i}\right)\varepsilon^k.\\\ \end{align} So we can take $$c_k:=\sum_{n=0}^{+\infty}\frac 1{n!}a_n\left(\sum_{j_1+\dots+j_n=k}\prod_{i=1}^nb_{j_i}\right)$$ for $k\geq 1$ and $c_0=0$.