Question

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?

Answer 1

`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.