Question

How do you find the Maclaurin series for `f(t) =t^3(e^(-t^2))` centered at 0?

Answer 1

`t^3e^(-t^2) = sum_(n=0)^oo (-1)^n t^(2n+3)/(n!)`

Explanation:

Start with the MacLaurin series of the exponential function:

`e^x = sum_(n=0)^oo x^n/(n!)`

Substitute `x =-t^2`

`e^(-t^2) = sum_(n=0)^oo (-t^2)^n/(n!) = sum_(n=0)^oo (-1)^nt^(2n)/(n!)`

Then multiply by `t^3` term by term:

`t^3e^(-t^2) =sum_(n=0)^oo (-1)^n t^3t^(2n)/(n!) = sum_(n=0)^oo (-1)^n t^(2n+3)/(n!)`

I would also note that a MacLaurin series is always centered at `x=0`, otherwise it is called a Taylor series.