MacLaurin expansion from f(x)=x^(t-m)\sin^2(4x)?

I tried to get differentials but the first is already hard enough...

First differential uses product rule. What's the second...?

1 Answer
May 26, 2018

f(x)=sum_(n=1)^oo(-1)^(n+1)2^(6n-1)x^(2n+t-m)/((2n)!)

Explanation:

We'll need to first find a series representation for sin^2(4x). However, this cannot be done right away due to the square.

Recall the identity

sin^2x=1/2(1-cos2x)

From this, we can see that

sin^2(4x)=1/2(1-cos8x)

Recalling the Maclaurin series for cosine,

cosx=sum_(n=0)^oo(-1)^nx^(2n)/((2n)!), we see that

cos8x=sum_(n=0)^oo(-1)^n(8x)^(2n)/((2n)!)

=sum_(n=0)^oo(-1)^n64^nx^(2n)/((2n)!)

So,

sin^2(4x)=1/2(1-sum_(n=0)^oo(-1)^n64^nx^(2n)/((2n)!))

However, writing out the 0th term for the series we have, we see

sum_(n=0)^oo(-1)^n64^nx^(2n)/((2n)!)=1+sum_(n=1)^oo(-1)^n64^nx^(2n)/((2n)!)

So,

sin^2(4x)=1/2(cancel1-(cancel1+sum_(n=1)^oo(-1)^n64^nx^(2n)/((2n)!)))

=-1/2sum_(n=1)^oo(-1)^n64^nx^(2n)/((2n)!)

64^n=(2^6)^n=2^(6n) -- we rewrite here because we want to be able to multiply in the 1/2=2^-1 outside.

We multiply in the - by seeing that -1*(-1)^n=(-1)^(n+1).

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

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

Thus,

f(x)=x^(t-m)sum_(n=1)^oo(-1)^(n+1)2^(6n-1)x^(2n)/((2n)!)

We can multiply in the x^(t-m), as t-m is really just some constant, and x^(t-m)*x^(2n)=x^(2n+t-m)

f(x)=sum_(n=1)^oo(-1)^(n+1)2^(6n-1)x^(2n+t-m)/((2n)!)