How does one obtain the expansion of $e^{-x^2}$ in a power series?

Question

So I know that the Power Series $y = \displaystyle\sum_{m=0}^\infty\displaystyle\frac{(-1)^m}{m!} x^{2m}$ is equivalent to $e^{-x^2}$. Could someone show me why this is?

Answer 1

You know that:

$$e^x=\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{x^n}{n!}$$

is a valid equation for every $-\infty<x<\infty$. For every $x\in \mathbb{R}$ we know that $\infty<-x^2\leq 0$ then if we replace $x$ by $-x^2$ in the equation above we get

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

which is valid for every $x\in \mathbb{R}$.