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.

  • 1
    $\begingroup$ You can use Cauchy products for $\Delta^{\varepsilon}$, viewed as a power series. $\endgroup$ Jun 24, 2012 at 16:11
  • 2
    $\begingroup$ You want the composition of formal power series. You can find it everywhere. en.wikipedia.org/wiki/Formal_power_series#Composition_of_series $\endgroup$ Jun 24, 2012 at 20:21
  • $\begingroup$ Oh...I just noticed the comment by Pietro Majer. Thank you. That is helpful. $\endgroup$
    – psyduck
    Jun 25, 2012 at 2:20

1 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 $c_0=0$.

  • $\begingroup$ Davide, This is incredibly helpful to me! If it is okay with you, I would like to add an acknowledgement of your help in my paper. I can also send you a draft the paper before I submit it if you like. $\endgroup$
    – psyduck
    Jun 25, 2012 at 0:03
  • $\begingroup$ No problem. What is your paper about? $\endgroup$ Jun 25, 2012 at 9:40
  • $\begingroup$ I study financial mathematics. The paper is about deriving the implied volatility smile for exponential Levy models...not sure if that makes any sense to somebody that doesn't study financial math. $\endgroup$
    – psyduck
    Jun 28, 2012 at 0:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.