# Thread: Maclaurin series

1. ## Maclaurin series

use a maclaurin series to obtain the maclaurin series for the given function f(x)= (sinx)^2 hint use (sinx)^2 = 1/2(1-cos2x)

2. Originally Posted by twilightstr
use a maclaurin series to obtain the maclaurin series for the given function f(x)= (sinx)^2 hint use (sinx)^2 = 1/2(1-cos2x)
$\displaystyle \cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}$

$\displaystyle \cos(2x)=\sum_{n=0}^{\infty}\frac{(-1)^n(2x)^{2n}}{(2n)!}=\sum_{n=0}^{\infty}\frac{(-1)^n2^{2n}(x)^{2n}}{(2n)!}$

Just use this with the hint and you are done

3. $\displaystyle f(x)=sin^2(x)=\frac{1}{2}-\frac{1}{2}cos(2x)$

Basically this amounts to finding the maclaurin series for cos(u) since then you can just put in a 2x for u (then multiply by your 1/2 scalar and subtract from 1/2). Note this trick works because the radius of convergence for this series turns out to be infinite, more care should in general be used.

let
f(u)=cos(u) f(0)=1
f'(u)=-sin(u) f'(0)=0
f"(u)=-cos(u) f"(0)=-1
f"'(u)=sin(u) f"'(0)=0
and repeat.

so:
$\displaystyle cos(u)=1 + 0u -\frac{1}{2!}u^2 + 0u^3 + \frac{1}{4!}u^4... =\sum_{n=0}^{\infty}\frac{(-1)^nu^{2n}}{2n!}$

so finally you see:
$\displaystyle sin^2(x)=\frac{1}{2}(1-cos(2x))=\frac{1}{2}\left(1-\sum_{n=0}^{\infty}\frac{(-1)^n(2x)^{2n}}{2n!}\right)$

Think you can take her from here?