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Power series of a strange function

  1. Dec 5, 2014 #1
    1. Write ∫e^(-t^2)dt with 0<=t<=x , as power series around 0. For what values of x this series converge ?

    attempt at a solution:

    f' = e^(-x^2) => f'(0) = 1
    f''= -2x*e^(-x^2) => f''(0)= 0
    f'''= -2e^(−x2) +4*x^2*e^(−x^2) => f'''(0)=-2

    I tried to find a general rule for the derivatives but with no sucess.
     
  2. jcsd
  3. Dec 5, 2014 #2

    vela

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    You'd have to take more derivatives to hope to find a pattern. I would try an alternative approach, however. Expand the integrand as a series and then integrate term by term.
     
  4. Dec 6, 2014 #3

    Mark44

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    You're off to a bad start here. f(x) = e-x2, not f'(x).
    Still, I would follow vela's advice on this.
     
  5. Dec 6, 2014 #4

    HallsofIvy

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    No, he is taking f to be the integral itself so f' is the integrand.

    Agreed. bonildo, write the Taylor's series for e^x, replace x with -x^2, then integrate term by term.
     
  6. Dec 6, 2014 #5
  7. Dec 6, 2014 #6

    Ray Vickson

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    Last edited: Dec 6, 2014
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