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Homework Help: Prove that this function differentiable endles times

  1. Feb 9, 2009 #1
    prove that this function differentiable endles times on x=0 ?? http://img502.imageshack.us/img502/6778/83126617mm0.th.gif [Broken]

    i was told
    "once we can express the function as a power series around zero and it is differentiable at zero, we know it is infinitely differentiable"
    differentiable around zero part:
    f'(0)=\lim_{x->0}\frac{e^\frac{-1}{x^2}-0}{x-0} \\
    =\lim_{x->0}\frac x{2e^{1/x^2}}
    i know that exponentials grow faster then polinomials but the power of the exp is 1/x^2
    so the denominator goes to infinity but faster then the numenator so the expression
    goes to 0.
    power series around zero part:
    the power series of e^x
    i put -1/x^2 instead of x

    what now??
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 9, 2009 #2


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    The first derivative of [tex]e^{-1/x^2}[/itex] is, of course, [itex]e^{-1/x^2}(2x^{-3})[/itex]. The second derivative, using the product rule is [itex]e^{-1/x^2}(4x^{-6})+ e^{-1/x^2}(-6x^{-4})[/itex].

    Can you use "proof by induction" to show that every derivative is [itex]e^{-1/x^2}[/itex] times terms with negative powers of x? It doesn't really matter what those powers are because, as x goes to 0, the exponential will "dominate" and the derivative will go to 0 as x goes to 0 so we can take every derivative to be 0 at x= 0.

    That means, by the way, that we could form its Taylor series at x= 0 (its MacLaurin series) as [itex]0+ 0x+ 0x^2+ \cdot\cdot\cdot= 0[/itex]. But obviously f(x) is NOT identically 0. This is an example of a function that is infinitely differentiable so has Taylor series at 0 but is NOT equal to that series except at 0.
  4. Feb 9, 2009 #3
    my power series is fine
    i took the power series for e^t and substitutes t=-1/x^2

    so i got a power series and differentiation of the first around 0

    is it a solution

    how to do it by induction?
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