Question

How do you find the Maclaurin series of `f(x)=cos(x)` ?

Answer 1

The Maclaurin series of `f(x)=cosx` is
`f(x)=sum_{n=0}^infty (-1)^nx^{2n}/{(2n)!}`.

Let us look at some details.

The Maclaurin series for `f(x)` in general can be found by
`f(x)=sum_{n=0}^infty {f^{(n)}(0)}/{n!}x^n`

Let us find the Maclaurin series for `f(x)=cosx`.

By taking the derivatives,
`f(x)=cosx Rightarrow f(0)=cos(0)=1`
`f'(x)=-sinx Rightarrow f'(0)=-sin(0)=0`
`f''(x)=-cosx Rightarrow f''(0)=-cos(0)=-1`
`f'''(x)=sinx Rightarrow f'''(0)=sin(0)=0`
`f^{(4)}(x)=cosx Rightarrow f^{(4)}(0)=cos(0)=1`

Since `f(x)=f^{(4)}(x)`, the cycle of `{1,0,-1,0}` repeats itself.

So, we have the series
`f(x)=1-{x^2}/{2!}+x^4/{4!}-cdots=sum_{n=0}^infty(-1)^n x^{2n}/{(2n)!}`