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Old May 31st, 2007, 08:12 AM
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Originally Posted by ThePerfectHacker View Post
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
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Old June 4th, 2007, 12:00 PM
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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.

---
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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