How can the coefficients of $e^{f(x)}$ be related to the coefficients of $f(x)$?

[Chris L T521]

Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: If $\displaystyle f(x)=\sum_{m=0}^{\infty} c_mx^m$ has positive radius of convergence and $e^{f(x)}= \displaystyle \sum_{n=0}^{\infty} d_n x^n$, show that
\[n d_n = \sum_{i=1}^n i c_i d_{n-i},\qquad n\geq 1.\]

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[Chris L T521]

This week's problem was correctly answered by Opalg. You can find his solution below.

Spoiler

If \(\displaystyle e^{f(x)} =\sum_{n=0}^\infty d_nx^n\), then (differentiating term by term, which is legitimate within the radius of convergence) $$\sum_{n=1}^\infty nd_nx^{n-1} = f'(x)e^{f(x)} = \sum_{m=1}^\infty mc_mx^{m-1}\sum_{k=0}^\infty d_kx^k.$$ Since all these series are absolutely convergent within the radius of convergence, we can equate the coefficients of $x^{n-1}$ on both sides, using Mertens' theorem, to get $$nd_n = \sum_{\substack{1\leqslant m \leqslant n \\ m+k=n}}mc_md_k = \sum_{m=1}^n mc_md_{n-m}.$$ Those sums only makes sense when $n\geqslant 1$. To find $d_0$ you need to put $x=0$ in the original equation \(\displaystyle e^{f(x)} =\sum_{n=0}^\infty d_nx^n\), getting $d_0 = e^{c_0}.$