Question

How do you find the maclaurin series expansion of `f(x)= x sinx`?

Answer 1

Derive or recall the Maclaurin expansion for `sin(x)` and multiply it by `x`.

Explanation:

If you know the Maclaurin series for `sin(x)` just multiply it by `x`

Let's derive the Maclaurin expansion for `sin(x)` in case you don't:

Let `s(x) = sin(x)`

Then:

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

`s^((0))(0) = sin(0) = 0`

`s^((1))(0) = cos(0) = 1`

`s^((2))(0) = -sin(0) = 0`

`s^((3))(0) = -cos(0) = -1`

Then `s^((4k+n))(x) = s^((n))(x)`

Hence:

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

`=x/(1!)-x^3/(3!)+x^5/(5!)-x^7/(7!)+...`

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

So:

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

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