Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
Or how about just evaluate at $a = 0$? – Qiaochu Yuan Mar 10 '13 at 23:09
Sorry for the vagueness, please assume $a \in \mathbb{R^+}$ – em70 Mar 10 '13 at 23:23
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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