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Math Help - Taylor series...help me.

  1. #1
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    Taylor series...help me.

    If f(x)= e^x^2 show that f(2n)(0)= (2n)!/ n! . (f(2n)(x) is Differentiation of f(x)--2nth).
    Thanks.
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  2. #2
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    Quote Originally Posted by Bugati View Post
    If f(x)= e^{x^2} show that f^{(2n)}(0)= (2n)!/ n! .
    The power series for e^{x^2} is \sum_{k=0}^\infty\frac{x^{2k}}{k!} (just substitute x^2 for x in the usual power series for e^x). Compare that with the Taylor series formula f(x) = \sum_{r=0}^\infty\frac{f^{(r)}(0)}{r!}x^r, and you see that, for the function f(x) = e^{x^2}, f^{(r)}(0) = 0 if r is odd. If r=2k is even, then compare the coefficients of x^{2k} in the two series to see that \frac1{k!} = \frac{f^{(2k)}(0)}{(2k)!}.
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  3. #3
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    Oh...Thank a lot.That's simple!
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