For the function, locate the Maclaurin series&nbsp;f(x)=sin⁡5x. Use the table of power series for elementary functions

Given:&nbsp;f(x)=sin⁡5x&nbsp;The Maclaurin series for the aforementioned function using the table of power series is derived from the table of power series.sin⁡x=x−x33!+x55!+x77!+...=∑n=0∞(−1)nx2n+1(2n+1)!&nbsp;Replacing x by 5x,&nbsp;sin⁡5x=5x−(5x)33!+(5x)55!+(5x)77!+...&nbsp;sin⁡5x=5x−125x36+325x524+15625x71008+...&nbsp;sin⁡5x=∑n=0∞(−1)n(5x)2n+1(2n+1)!&nbsp;Hence,&nbsp;sin⁡5x=∑n=0∞(−1)n(5x)2n+1(2n+1)!

Find out the answer to "For the function, locate the Maclaurin seriesf(x)=sin⁡5x. Use..."

vestirme4

Answered question

2021-02-14

For the function, locate the Maclaurin series $f (x) = \sin 5 x$ . Use the table of power series for elementary functions

Answer & Explanation

dessinemoie

Skilled2021-02-15Added 90 answers

Given:
$f (x) = \sin 5 x$
The Maclaurin series for the aforementioned function using the table of power series is derived from the table of power series.
$\sin x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + \frac{x^{7}}{7!} + . . . = \sum_{n = 0}^{\infty} \frac{(- 1)^{n} x^{2 n + 1}}{(2 n + 1)!}$
Replacing x by 5x,
$\sin 5 x = 5 x - \frac{(5 x)^{3}}{3!} + \frac{(5 x)^{5}}{5!} + \frac{(5 x)^{7}}{7!} + . . .$
$\sin 5 x = 5 x - \frac{125 x^{3}}{6} + \frac{325 x^{5}}{24} + \frac{15625 x^{7}}{1008} + . . .$
$\sin 5 x = \sum_{n = 0}^{\infty} \frac{(- 1)^{n} (5 x)^{2 n + 1}}{(2 n + 1)!}$
Hence, $\sin 5 x = \sum_{n = 0}^{\infty} \frac{(- 1)^{n} (5 x)^{2 n + 1}}{(2 n + 1)!}$

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Calculus 2

Didn't find what you were looking for?