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Find the 100th Derivative of e^(-x^2)

  1. Dec 3, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the exact value for the 100th Derivative of [tex]f(x) = e^{-x^{2}}[/tex] evaluated at x = 0

    2. Relevant equations
    [tex]e^{x} = \sum^{\infty}_{0} \frac{x^{n}}{n!}[/tex]

    [tex]e^{-x^{2}} = \sum^{\infty}_{0} \frac{(-1)^{n}x^{2n}}{n!}[/tex]

    3. The attempt at a solution
    The only thing I can think of to do is to get the 100th term of the series which is
    factor out an [tex]x^{n}[/tex] to make it look like the general term of a MacLaurin Series

    and I thought that since
    [tex]C_{n} = \frac{f^{(n)}(0)}{n!}[/tex]

    I would just solve for the derivative
    [tex]C_{100} = \frac{x^{100}}{100!} = \frac{f^{(100)}(0)}{100!}[/tex]

    But..I can't do anything with that.

    I even attempted to use n = 50, so the coefficient would not have an X in it, but that just gives me the derivative is 1...in fact if I do it like that it says the derivative is always 1.
    I know that to be not the case because if you evaluate the 2nd derivative by hand you get -2, and the 4th is 12
  2. jcsd
  3. Dec 3, 2008 #2
    you are on the right track...
    the only term which gives you something nonzero at the 100th derivative it n=50
    so you need to evaluate the 100th derivative of x^100/50!, but this is simply 100!/50!, if you want to get the exact number you would need to plug that into some CAS. You could also just use Stirling's formula to get a very good approximation.
  4. Dec 3, 2008 #3
    Thanks very much. It's a little awkward to think that the 100th term in the series is not the 100th term in the original Taylor/MacLaurin series but I suppose that's what I didn't realize.
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