help with this calc prob pls?

Question

Find a Maclaurin series for the function xe^(-x^2)

Answer 1

f(x) = xe^(-x^2)
(Amit Y's answer is incorrect; only alternate terms are negative.)

The easiest way to do it is indeed to use the well-known Maclaurin series for e^x.
xe^(-x^2) = x(1 + (-x^2) + (-x^2)^2/2! + (-x^2)^3/3! + ....)
= x - x^3 + x^5/2 - x^7/6 + ... + (-1)^n x^(2n+1) / n! + ...

If you were to do it directly you would have
f'(x) = e^(-x^2) + xe^(-x^2).(-2x) = e^(-x^2) (1 - 2x^2)
f"(x) = e^(-x^2) ((1 - 2x^2)(-2x) + (-4x))
= e^(-x^2) (-6x + 4x^3)
f'"(x) = e^(-x^2) ((-6x + 4x^3)(-2x) + (-6 + 12x^2))
= e^(-x^2) (-6 + 24x^2 - 8x^4)
f""(x) = e^(-x^2) (60x - 64x^3 + 16x^5)
f""'(x) = e^(-x^2) (60 - 312x^2 + 208x^4 - 32x^6)
The even-number derivatives will have polynomial parts with all odd powers of x, and the odd-number derivatives will have only even powers of x. So the even-numbered derivatives will vanish at x = 0, and we need to establish the constant terms in the polynomials for the odd-numbered derivatives. It should be possible to prove that the coefficient for the (2n+1)th derivative is (-1)^n (2n+1)! / n!, but I am not going to try it here!

Answer 2

Use a shortcut:(don't derive the whole expression)

e^x = sum(n from 0 to inf) of x^n/n!

e^(-x^2) = sum(n from 0 to inf) of (-x^2)^n/n! =
= sum(n from 0 to inf) of -x^2n/n!

xe(-x^2) = x* sum(n from 0 to inf) of -x^2n/n! =
= sum(n from 0 to inf) of -(x^2n * x)/n! =
= sum(n from 0 to inf) of -x^(2n+1)/n!