# Rewriting a series $\sum_{n=0}^\infty \frac{1}{n!}(\Delta^\varepsilon)^n a_n$ in the form $\sum_{n=0}^\infty c_n \varepsilon^n$

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

Obviously, I can do this term-by-term. But the general case seems quite difficult.

• You can use Cauchy products for $\Delta^{\varepsilon}$, viewed as a power series. 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 Jun 24 '12 at 20:21
• Oh...I just noticed the comment by Pietro Majer. Thank you. That is helpful. 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 $c_0=0$.