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Show Taylor Series for $f(x) = e^{-x^2}$ converges to $f$

I am stuck because when taking the (n+1) th derivative of f, I do not see a general pattern. Meaning I am having difficulty in bounding $|f^{n+1}(x)|$

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    The Taylor series for $e^{-x^2}$ is exactly what you get when you plug $z = -x^2$ into the taylor series for $e^z$. – Ben Grossmann Jul 30 '14 at 22:37
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    @Omnomnomnom The OP doesn't want the Taylor Series, S/he wamt to prove that it converges. S/he wants the $n^{\mathrm{th}}$ derivative as a function of $x$. – Fly by Night Jul 30 '14 at 22:40
  • http://math.stackexchange.com/questions/193702/find-an-expression-for-the-n-th-derivative-of-fx-ex2 – lemon Jul 30 '14 at 22:55
  • Please look at my answer – Varun Iyer Jul 30 '14 at 23:01
  • I think the easiest way to do this is to use the fact that the composition of two real analytic functions is real analytic, and that $f(x)=e^x$ and $g(x)=-x^2$ are real analytic. – user84413 Jul 31 '14 at 00:20

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The series $e^x$ can be represented as:

$$ e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}$$ To find $e^{-x^2}$ we substitute $-x^2$ for $x$:

$$ e^{-x^2} = \sum_{n=0}^{\infty}\frac{(-x^2)^n}{n!}$$ $$ = \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{n!}$$

We can conduct the root test to determine convergence:

$$\lim_{n\to\infty} \left|\frac{(-1)^nx^{2n}}{n!}\right|^{1/n}$$ $$=\lim_{n\to\infty} \left|\frac{x^{2}}{(n!)^{1/n}}\right|$$ $$= 0 < 1$$

Therefore, by the root test, the series absolutely converges.

Varun Iyer
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