# 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...?

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 ${\sin}^{2} \left(4 x\right) .$ However, this cannot be done right away due to the square.

Recall the identity

${\sin}^{2} x = \frac{1}{2} \left(1 - \cos 2 x\right)$

From this, we can see that

${\sin}^{2} \left(4 x\right) = \frac{1}{2} \left(1 - \cos 8 x\right)$

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 $0 t h$ 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)!)

${64}^{n} = {\left({2}^{6}\right)}^{n} = {2}^{6 n}$ -- we rewrite here because we want to be able to multiply in the $\frac{1}{2} = {2}^{-} 1$ outside.

We multiply in the $-$ by seeing that $- 1 \cdot {\left(- 1\right)}^{n} = {\left(- 1\right)}^{n + 1}$.

=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 ${x}^{t - m} \cdot {x}^{2 n} = {x}^{2 n + t - m}$

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