Expert Answer to "Find the Maclaurin series for the function f(x)=cos⁡4x...."

Reeves

Answered question

2021-01-22

Find the Maclaurin series for the function

f (x) = \cos 4 x

. Use the table of power series for elementary functions

Answer & Explanation

au4gsf

Skilled2021-01-23Added 95 answers

$\cos x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} - . . . + \frac{(- 1)^{n} x^{2 n}}{(2 n)!} + . . .$
$\cos (4 x) = 1 - \frac{(4 x)^{2}}{2!} + \frac{(4 x)^{4}}{4!} - \frac{(4 x)^{6}}{6} + . . . + \frac{(- 1)^{n} (4 x)^{2 n}}{(2 n)!} + . . .$
$= 1 - \frac{16 x^{2}}{2!} + \frac{256 x^{4}}{4!} - \frac{4096 x^{6}}{6!} + . . . + \frac{(- 1)^{n} (4 x)^{2 n}}{(2 n)!} + . . .$ $= \sum_{n = 0}^{\infty} (- 1)^{n} \frac{(4 x)^{2 n}}{(2 n)!}$
Maclaurin series of $f (x) = \cos (4 x)$ is $\sum_{n = 0}^{\infty} (- 1)^{n} \frac{(4 x)^{2 n}}{(2 n)!}$

Jeffrey Jordon

Expert2021-10-14Added 2605 answers

$\cos (x) = \sum_{k = 0}^{\infty} (- 1)^{k} \frac{x^{2 k}}{(2 k)!}$

for all $x \in R$ ,

$\cos (4 x) = \sum_{k = 0}^{\infty} (- 1)^{k} \frac{4^{2 k} x^{2 k}}{(2 k)!}$ $= 1 - 2 x^{2} + \frac{2}{3} x^{4} - \frac{4}{45} x^{6} + \dots$

Answer $\cos (4 x) = \sum_{k = 0}^{\infty} (- 1)^{k} \frac{4^{2 k} x^{2 k}}{(2 k)!}$

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

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