Use power series operations to find the Taylor series at x = 0 for the given function.
-$\cos ^ { 2 } ( 2 x )$
A) $\frac { 1 } { 2 } + \sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } ( 4 ) ^ { 2 n _ { x } 2 n } } { 2 ( 2 n ) ! }$
B) $\frac { 1 } { 2 } + \sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n } ( 4 ) ^ { 2 n _ { x } 2 n } } { ( 2 n ) ! }$
C) $\frac { 1 } { 2 } + \sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } ( 4 ) ^ { 2 n _ { x } 2 n } } { ( 2 n ) ! }$
D) $\frac { 1 } { 2 } + \sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n } ( 4 ) ^ { 2 n _ { x } 2 n } } { 2 ( 2 n ) ! }$

Question

Quizplus · Accepted Answer

The Taylor series for $\cos x$ is $\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}$. Substituting $2x$ for $x$ gives the Taylor series for $\cos 2x$:$$\cos 2x = \sum_{n=0}^\infty \frac{(-1)^n (2x)^{2n}}{(2n)!} = \sum_{n=0}^\infty \frac{(-1)^n 4^n x^{2n}}{(2n)!}$$Squaring this series gives the Taylor series for $\cos^2 2x$:$$\cos^2 2x = \left(\sum_{n=0}^\infty \frac{(-1)^n 4^n x^{2n}}{(2n)!}\right)^2 = \frac{1}{2} + \sum_{n=0}^\infty \frac{(-1)^n 4^{2n} x^{2n}}{(2n)!}$$

Use Power Series Operations to Find the Taylor Series at X