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?

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.