Taylor series expansion of $ f(x)=e^{-x^2}$

Question

How to find Taylor series expansion of $f(x)=e^{-x^2}$ What I'm stuck at is proving that the error $$R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}$$ of the expansion tends to zero.

Answer 1

$$ f^{(n+1)}(x)=H_{n+1}(x)e^{-x^2}\ , $$ where $H_n(x)$ is the $n$-th Hermite polynomial https://en.wikipedia.org/wiki/Gaussian_function . So it is a polynomial of degree $n+1$. The remainder then $$ R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}\sim (cx)^{n+1}e^{-c^2}/(n+1)!\to 0\ , $$ as factorials grow faster than exponentials.

Answer 2

The simplest thing to do in such cases is to write first the series of $\textrm e ^x$ and then replace $x$ by $-x^2$. This means that in

$$\textrm e ^x = \sum _{n=0} ^\infty \frac {x^n} {n!}$$

you plug $-x^2$, getting

$$\textrm e ^{-x^2} = \sum _{n=0} ^\infty (-1)^n \frac {x^{2n}} {n!} .$$