Question

How do you find the maclaurin series expansion of `f(x) =sin(3x)`?

Answer 1

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

Explanation:

First off the Maclaurin series is a special case of the Taylor series where `a=0`

So a Maclaurin series would be of the form `sum_(n=0)^oo ((d^n f)/(dx^n)(0)x^n)/(n!)`

Since `sin(x)` is `0` if `x=0`, but its derivative `cos(x)` is `1` when `x=0` we need to consider the odd derivatives

So that means we'll have a series like this

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

and since the derivatives alternate between positive and negative `1` we have

`sum_(n=1)^oo ((-1)^nx^(2n-1))/((2n-1)!)`

Also since we are finding the maclaurin series for `sin(3x)` instead of `sin(x)` we'll get

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