Question

How do you find the maclaurin series expansion of `f(x)= (2x^2)sin(4x)`?

Answer 1

Substitute `4x` into the well-known Maclaurin series for `sin(x)` and then multiply everything by `2x^2` to get: `f(x)=2x^2sin(4x)=8x^3-64/3 x^5+256/15 x^7-2048/315 x^9+cdots` for all `x`.

Explanation:

We know `sin(x)=x-x^3/(3!)+x^5/(5!)-x^7/(7!)+cdots` for all `x`. Therefore,

`sin(4x)=4x-(4x)^3/(3!)+(4x)^5/(5!)-(4x)^7/(7!)+cdots`

`=4x-64/6 x^3+1024/120 x^5-16384/5040 x^7+cdots`

`=4x-32/3 x^3+128/15 x^5-1024/315x^7+cdots` for all `x`.

Hence,

`f(x)=2x^2sin(4x)=8x^3-64/3 x^5+256/15 x^7-2048/315 x^9+cdots` for all `x`.