Question

How do you find the maclaurin series expansion of `x^2sin(x)`?

`x^2 sin(x)`

Answer 1

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

Explanation:

In most general form, the Maclaurin series for a function `f(x)` is given by:

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

In particular for `sin(x)` we have `d/(dx) sin x = cos x` and `d/(dx) cos x = -sin x`. Hence, the even terms are `0` and the odd terms alternate in sign to give:

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

To get the Maclaurin series for `x^2 sin(x)` just multiply by `x^2` to get:

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