# How do you write the first 4 nonzero terms AND the general term of the taylor series for e^((-x)^2) centered at x=0?

Sep 12, 2015

${e}^{{\left(- x\right)}^{2}} = 1 + {x}^{2} + {x}^{4} / 2 + {x}^{6} / 6 + \ldots$
= sum_(n=0)^infty x^(2n)/(n!)

#### Explanation:

e^((-x)^2) = ?

First, we'll take note that ${\left(- x\right)}^{2} = {x}^{2}$. Replacing $x$ with $- x$ reflects the function about the $y$-axis, but since ${e}^{{x}^{2}}$ is already symmetric about the $y$-axis, it has no effect. They're equivalent.

${e}^{{\left(- x\right)}^{2}} = {e}^{{x}^{2}}$

Alright, now there are two ways to find the Taylor series of this function. The first is the standard of calculating the derivatives of ${e}^{{x}^{2}}$, finding a general pattern, et cetera .

While this would work just fine, it would be time-consuming. There's a much faster way. We can actually make a substitution into a Taylor series that resembles this function which we already know.

What do I mean? Well, imagine just for a second that we have a variable $u$, where $u = {x}^{2}$. Now we have

${e}^{{\left(- x\right)}^{2}} = {e}^{{x}^{2}} = {e}^{u}$

Clearly, the Taylor series of ${e}^{u}$ equals the Taylor series for ${e}^{{x}^{2}}$.

Now, you should be able to see that the Taylor series of ${e}^{u}$ will be

e^u = 1 + u + u^2/2 + u^3/6 = sum_(n=0)^infty u^(n)/(n!)

(if not, see here. Most calculus students are expected to have memorized some of these basic series)

Okay, now, $u = {x}^{2}$ so let's just substitute back for $x$:

e^((-x)^2) = 1 + x^2 + x^4/2 + x^6/6 = sum_(n=0)^infty (x^2)^(n)/(n!)
=sum_(n=0)^infty (x^(2n))/(n!)