# Math Help - Problem 24

1. Originally Posted by ThePerfectHacker
Not every continously differenciable function is analytic.*

*)By analytic I mean it has a power series expansion on $\mathbb{R}$.
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

2. The only answer is the trivial answer $f(x)=0 \mbox{ on }\mathbb{R}$.
Follow, ecMathGeek's approach and arrive at:
$A^2e^x = A^2(e^x-1)$
Hence, $A=0$ 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.

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