How do you find the Maclaurin Series for #f(x)=sin^2(4x)#?

1 Answer
Jun 23, 2017

#sin^2(4x) = sum_(n=1)^oo (-1)^(n+1)2^(6n-1)x^(2n)/((2n)!)#

Explanation:

Use the trigonometric identity:

#sin^2(4x) = (1-cos(8x))/2#

We know the MacLaurin series of #cost#:

#cost = sum_(n=0)^oo (-1)^nt^(2n)/((2n)!)#

Substituting #t=8x# we have:

#cos 8x = sum_(n=0)^oo (-1)^n(8x)^(2n)/((2n)!) = sum_(n=0)^oo (-1)^n2^(6n)x^(2n)/((2n)!)#

and then:

#sin^2(4x) = 1/2 (1-sum_(n=0)^oo (-1)^n2^(6n)x^(2n)/((2n)!))#

Extract the term for #n=0# from the sum:

#sin^2(4x) = 1/2 (1-1 - sum_(n=1)^oo (-1)^n2^(6n)x^(2n)/((2n)!))#

and then bring the common factor #-1/2# inside the sum:

#sin^2(4x) = sum_(n=1)^oo (-1)^(n+1)2^(6n-1)x^(2n)/((2n)!)#

We can also change the index with #k=n-1# to have:

#sin^2(4x) = sum_(k=0)^oo (-1)^(k+2) 2^(6(k+1)) x^(2(k+1))/((2(k+1))!)#

#sin^2(4x) = 64x^2 sum_(k=0)^oo (-1)^k 2^(6k) x^(2k)/((2k+2)!)#