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

complete parts a through c for the function below. f(x)=5 sin 3x a. find the first four nonzero terms of the maclaurin series for the given function. the first nonzero term is 15x. the second nonzero term is -45x³/2. the third nonzero term is 81x⁵/8. the fourth nonzero term is -243x⁷/112. b. write the power series using summation notation. 5 sin 3x=∑(k = 0 to ∞)()

complete parts a through c for the function below. f(x)=5 sin 3x a. find the first four nonzero terms of the maclaurin series for the given function. the first nonzero term is 15x. the second nonzero term is -45x³/2. the third nonzero term is 81x⁵/8. the fourth nonzero term is -243x⁷/112. b. write the power series using summation notation. 5 sin 3x=∑(k = 0 to ∞)()

Answer

Explanation:

Step1: Recall Maclaurin series of $\sin t$

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

Step2: Substitute $t = 3x$ and multiply by 5

We have $5\sin(3x)=5\sum_{k = 0}^{\infty}\frac{(-1)^{k}}{(2k + 1)!}(3x)^{2k+1}=\sum_{k = 0}^{\infty}\frac{5(-1)^{k}3^{2k + 1}}{(2k+1)!}x^{2k+1}$.

Answer:

$5\sin(3x)=\sum_{k = 0}^{\infty}\frac{5(-1)^{k}3^{2k + 1}}{(2k+1)!}x^{2k+1}$