I need to write a quadratic taylor series for the expression e^(-x^2). I have no clue. It was assumed that we could figure this one out for ourselves and I can't. Any help would be greatly appreciated!!
Centered around what?
I will assume it is a zero.
Since $\displaystyle e^{x}=1+x+\frac{x^2}{2!}+...+\frac{x^n}{n!}=\sum_{ n=0}^{\infty}\frac{x^n}{n!}$
then $\displaystyle e^{-x^2}=1-x^2+\frac{x^4}{2!}-...+\frac{(-x^2)^n}{n!}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{n!}$
There is the quadratic
if you forget
remember that the taylor series for $\displaystyle f(x)$ centered around c is
$\displaystyle f(x)=f(c)+f'(c)(x-c)+\frac{f''(c)(x-c)^2}{2!}+...=\sum_{n=0}^{\infty}\frac{f^{(n)}(c)( x-c)^n}{n!}$
assuming $\displaystyle f(x)$ is analytic at $\displaystyle x$