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Math Help - Problem 24

  1. #16
    Grand Panjandrum
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    Quote 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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  2. #17
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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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