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 \(\sum_{n=0}^\infty(-1)^n\frac{(4x)^{2n}}{(2n)!}\)

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 \(\sum_{n=0}^\infty(-1)^n\frac{(4x)^{2n}}{(2n)!}\)