Verify that e^-x² = ∑ ((-1)ⁿ x²ⁿ) / n!

by RoRi

April 3, 2016

Assuming that $e^{-x^2}$ has a power-series representation in terms of powers of $x$ , verify that it has the form

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

valid for all real $x$ .

We know the expansion

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

valid for all real $x$ . So in this case have,

$\begin{align*} e^{-x^2} &= \sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!} \\[9pt] &= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}, \end{align*}$

valid for all $x \in \mathbb{R}$ .

Stumbling Robot