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

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)!)#