Not every continously differenciable function is analytic.*
*)By analytic I mean it has a power series expansion on .
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
__________________ Truth does not change because it is, or is not, believed by a majority of the people.
The only answer is the trivial answer .
Follow, ecMathGeek's approach and arrive at:
Hence, 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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