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Math Help - Real-Analytic Function

  1. #1
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    Lightbulb Real-Analytic Function

    If a function f(x) \in C^{\infty}(\mathbb{R}),

    can we get the conclusion: f(x) is a real-analytic function?


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  2. #2
    Junior Member nimon's Avatar
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    No this isn't true. The following example is from 'Loring W. Tu, An Introduction to Manifolds, Springer (2008).'

    Consider the function
    f(x)=\left\{ \begin{array}{cl} e^{-\frac{1}{x}}, & x>0 \\ 0 & x\leq 0. \end{array} \right.
    Then f\in C^{\infty}(\mathbb{R}) but f^{n}(0) = 0 \,\,\forall n\in\mathbb{N}. So the taylor series is 0 in the neighbourhood of the origin, but f(\epsilon) \neq 0 \,\, \forall \epsilon>0.

    The converse is true, however: all analytic functions are C^{\infty} on the appropriate domain.
    Last edited by nimon; October 18th 2009 at 10:00 AM. Reason: Acknowledging Sources
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