Question

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.

Answer 1

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 $c_0=0$.