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

Question

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 $|f^{n+1}(x)|$

Answer 1

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.