# Find the Maclaurin series for the function f(x)=cos4x. Use the table of power series for elementary functions

Question
Series
Find the Maclaurin series for the function $$f(x)=\cos4x$$. Use the table of power series for elementary functions

2021-01-23
Given function is $$f(x)=\cos4x$$
To find the Maclaurin series for given function.
Solution:
Power series for $$\cos x$$ is given as:
$$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}-...+\frac{(-1)^nx^{2n}}{(2n)!}+...$$
Now, replace x with 4x then we get,
$$\cos(4x)=1-\frac{(4x)^2}{2!}+\frac{(4x)^4}{4!}-\frac{(4x)^6}{6}+...+\frac{(-1)^n(4x)^{2n}}{(2n)!}+...$$
$$=1-\frac{16x^2}{2!}+\frac{256x^4}{4!}-\frac{4096x^6}{6!}+...+\frac{(-1)^n(4x)^{2n}}{(2n)!}+...$$
$$=\sum_{n=0}^\infty(-1)^n\frac{(4x)^{2n}}{(2n)!}$$
Therefore, Maclaurin series of $$f(x)=\cos(4x)$$ is $$\sum_{n=0}^\infty(-1)^n\frac{(4x)^{2n}}{(2n)!}$$
Hence, Maclaurin series of $$\cos(4x)$$ is $$\sum_{n=0}^\infty(-1)^n\frac{(4x)^{2n}}{(2n)!}$$

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