surely you can do this one. Show some effort!
f = e^-x^2
f' = -2x e^(-x^2)
f" = (4x^2-2)e^(-x^2)
f"' = -x(2x^2-3)e^(-x^2)
and so on.
All the odd-order terms contain an x factor, so they go away, leaving
e^(-x^2) =∑ (-x^2)^k/k!
makes sense, since e^x = ∑x^k/k!
Find the Maclaurin series for f(x) = e^(-x^2)
Note:
I don't know how to solve or work out so show all your work. And give the answer in EXACT FORM example 3pi, sqrt(2), ln(2) not decimal approximations like 9.424,1.4242,1232
1 answer