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

Question

7545 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-31T16:14:41

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

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)!)#