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

1 Answer
Sep 12, 2015

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

Explanation:

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

First, we'll take note that (-x)^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^((-x)^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^((-x)^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!)

And there's our answer.