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