I have to find the first two non-vanishing terms in the Maclaurin series of $$g(x) = \log(3 − \cos(x^2))$$
and that prove $x=0$ is a stationary point.
What is a quick way of working out the Maclaurin series?
I know that $\log(1+x)= x- x^2/2! + x^3/3!- x^4/4!+\ldots $.
How can you relate $\log(x+1)$ to $\log(3-\cos(x^2))$?
Since in a neighbourhood of zero we have: $$1-\cos(x) = \frac{x^2}{2}-\frac{x^4}{4!}+\ldots $$ it happens that: $$1-\cos(x^2) = \frac{x^4}{2}-\frac{x^8}{4!}+\ldots $$ and: $$\begin{eqnarray*}\log(3-\cos(x^2)) &=& \log(2+(1-\cos(x^2))) = \log 2 + \log\left(1+\frac{1-\cos x^2}{2}\right)\\&=&\color{red}{\log 2+\frac{x^4}{4}+\ldots}.\end{eqnarray*} $$ Obviously $x=0$ is a stationary point since $g(x)$ is an even smooth function in a neighbourhood of zero.