complete parts a through c for the function below. f(x)=5 sin 3x a. find the first four nonzero terms of the…

Explanation:

Step1: Recall Maclaurin series of $\sin t$

The Maclaurin series of $\sin t=\sum_{n = 0}^{\infty}\frac{(- 1)^{n}}{(2n+1)!}t^{2n + 1}=t-\frac{t^{3}}{3!}+\frac{t^{5}}{5!}-\frac{t^{7}}{7!}+\cdots$

Step2: Substitute $t = 3x$ into $\sin t$ series

We get $\sin(3x)=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}(3x)^{2n+1}=(3x)-\frac{(3x)^{3}}{3!}+\frac{(3x)^{5}}{5!}-\frac{(3x)^{7}}{7!}+\cdots=3x-\frac{27x^{3}}{6}+\frac{243x^{5}}{120}-\frac{2187x^{7}}{5040}+\cdots=3x-\frac{9x^{3}}{2}+\frac{81x^{5}}{40}-\frac{243x^{7}}{560}+\cdots$

Step3: Multiply by 5

Since $f(x)=5\sin(3x)$, then $f(x)=5\left(3x-\frac{9x^{3}}{2}+\frac{81x^{5}}{40}-\frac{243x^{7}}{560}\right)=15x-\frac{45x^{3}}{2}+\frac{405x^{5}}{40}-\frac{1215x^{7}}{560}=15x-\frac{45x^{3}}{2}+\frac{81x^{5}}{8}-\frac{243x^{7}}{112}$

Answer:

$15x-\frac{45x^{3}}{2}+\frac{81x^{5}}{8}-\frac{243x^{7}}{112}$