Yes I know, but it illurstates the point nicely.

(though the Stone-Weiestrauss theorem applies, I don't think this is adequate

to do the analysis with polynomial approximations instead, though it would be

nice if it did).

RonL

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- May 31st 2007, 06:12 AMCaptainBlack
- Jun 4th 2007, 10:00 AMThePerfectHacker
The only answer is the trivial answer $\displaystyle f(x)=0 \mbox{ on }\mathbb{R}$.

Follow, ecMathGeek's approach and arrive at:

$\displaystyle A^2e^x = A^2(e^x-1)$

Hence, $\displaystyle A=0$ is the only possible constant.

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This problem was taken from my math book, which said it appeared as a problem in the 1990 Putnam. The problem I posed was a variation of that problem, not exactly the same.