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Is there a function that does not depend on $a$ such that $\sum_{x=1}^\infty \frac{a^x}{x!}f(x) = \mathrm e^{-a}$?

Just to be clear, the summation starting from 1 is intentional, otherwise the solution would be trivial.

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So, what happens when you take $k$ derivatives (with respect to $a$) and evaluate at $a=0$? –  Gerry Myerson Mar 10 '13 at 23:07
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Or how about just evaluate at $a = 0$? –  Qiaochu Yuan Mar 10 '13 at 23:09
    
Sorry for the vagueness, please assume $a \in \mathbb{R^+}$ –  emaster70 Mar 10 '13 at 23:23

1 Answer 1

up vote 3 down vote accepted

This is impossible because $e^{-a}$ already has a power series centered at $a=0$ with a nonzero constant term and power series are unique.

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