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(Deceptively?) Simple question about Taylor series expansions

  1. Jan 30, 2012 #1
    Under what circumstances is it correct to say of the function [itex]u(x) \in L^2(-\infty,\infty)[/itex] that

    [tex]
    u(x-t) = u(x) - \frac{du}{dx}t + \frac 12 \frac{d^2u}{dx^2}t^2 - \cdots = \sum_{n=0}^\infty \frac{u^{(n)}(x)}{n!}(-t)^n.
    [/tex]
     
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  3. Jan 30, 2012 #2

    micromass

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    We can do that if the function is analytic. I don't really know if there are specific conditions which I could give.

    Some notes:
    1) Saying that u is differentiable an infinite number of times is not enough. A bump function is a counterexample. http://en.wikipedia.org/wiki/Bump_function These functions have the property that theTaylor expansion doesn't converge to the function.

    2) If we were talking about complex functions instead of real functions then the answer is simpler. A complex function is analytic if and only if it is differentiable. However, these functions will rarely be in [itex]L^2[/itex].
     
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