Question

How do you find the Maclaurin series for `f(x) = x^2e^(-x)` centered at 0?

Answer 1

`x^2e^(-x) = sum_(n=0)^oo (-1)^n/(n!) x^(n+2)`

Explanation:

The Maclaurin series for `e^t` is:

`sum_(n=0)^oo 1/(n!) t^n`

So, substituting `t = -x` and multiplying by `x^2` we find:

`x^2e^(-x) = x^2 sum_(n=0)^oo 1/(n!) (-x)^n=sum_(n=0)^oo (-1)^n/(n!) x^(n+2)`