8. 0/1 points details my notes scalcet9 11.10.010. use the definition of a taylor series to find the first…

Explanation:

Step1: Recall double - angle formula

We know that $\cos^{2}x=\frac{1 + \cos(2x)}{2}$, so $f(x)=5\cos^{2}x=\frac{5}{2}+\frac{5}{2}\cos(2x)$.

Step2: Recall Taylor series of $\cos t$

The Taylor series of $\cos t$ centered at $a = 0$ is $\cos t=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!}t^{2n}=1-\frac{t^{2}}{2!}+\frac{t^{4}}{4!}-\frac{t^{6}}{6!}+\cdots$. Let $t = 2x$. Then $\cos(2x)=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!}(2x)^{2n}=1-\frac{(2x)^{2}}{2!}+\frac{(2x)^{4}}{4!}-\frac{(2x)^{6}}{6!}+\cdots=1 - 2x^{2}+\frac{2}{3}x^{4}-\frac{4}{45}x^{6}+\cdots$.

Step3: Find Taylor series of $f(x)$

Since $f(x)=\frac{5}{2}+\frac{5}{2}\cos(2x)$, substituting the Taylor series of $\cos(2x)$ we get $f(x)=\frac{5}{2}+\frac{5}{2}(1 - 2x^{2}+\frac{2}{3}x^{4}-\frac{4}{45}x^{6}+\cdots)$. [ \begin{align*} f(x)&=\frac{5}{2}+\frac{5}{2}-\ 5x^{2}+\frac{5}{3}x^{4}-\frac{2}{9}x^{6}+\cdots\ &=5-5x^{2}+\frac{5}{3}x^{4}-\frac{2}{9}x^{6}+\cdots \end{align*} ] The first four non - zero terms are $5,-5x^{2},\frac{5}{3}x^{4},-\frac{2}{9}x^{6}$.

Answer:

$5,-5x^{2},\frac{5}{3}x^{4},-\frac{2}{9}x^{6}$