Question

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...?

Answer 1

`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)!)`