Question

find the maclaurin series for the given function

\\( \sin 8 x \\)

\\( \sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { 2 n + 1 } 8 ^ { 2 n + 1 } x ^ { 2 n + 1 } } { ( 2 n + 1 )! } \\)

\\( \sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } 8 ^ { 2 n + 1 } x ^ { 2 n + 1 } } { ( 2 n + 1 )! } \\)

\\( \sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { 2 n + 1 } 8 ^ { 2 n + 1 } x ^ { 2 n + 1 } } { n! } \\)

\\( \sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } 8 ^ { 2 n + 1 } x ^ { 2 n + 1 } } { n! } \\)

Explanation:

Step1: Recall the Maclaurin series for \(\sin t\)

The Maclaurin series for \(\sin t=\sum_{n = 0}^{\infty}\frac{(-1)^{n}t^{2n + 1}}{(2n+1)!}\), where \(t\) is a variable.

Step2: Substitute \(t = 8x\) into the series

When \(t = 8x\), we have \(\sin(8x)=\sum_{n = 0}^{\infty}\frac{(-1)^{n}(8x)^{2n + 1}}{(2n+1)!}\).
Using the property \((ab)^{m}=a^{m}b^{m}\), \((8x)^{2n+1}=8^{2n + 1}x^{2n+1}\). So \(\sin(8x)=\sum_{n = 0}^{\infty}\frac{(-1)^{n}8^{2n+1}x^{2n + 1}}{(2n+1)!}\)

Answer:

B. \(\sum_{n = 0}^{\infty}\frac{(-1)^{n}8^{2n+1}x^{2n + 1}}{(2n+1)!}\)