# Finding Maclaurin expansion and interval of convergence

1. Jul 23, 2017

### Mr Davis 97

1. The problem statement, all variables and given/known data
Find the Maclaurin series and inverval of convergence for $f(x) = \log (\cos x)$

2. Relevant equations

3. The attempt at a solution
I used the fact that $\log (\cos x) = \log (1+ (\cos x - 1))$, and the standard expansions for $\cos x$ and $\log (x+1)$ to get that $\displaystyle \log (\cos x) = -\frac{x^2}{2} + \frac{x^4}{12} - \frac{x^6}{720} + O (x^8)$. How do I find the interval of convergence for this? Also, how do I know that this is the valid expansion?

2. Jul 23, 2017

### haruspex

Try working from the outside in: what is the interval for log( 1+y)?

3. Jul 24, 2017

### vela

Staff Emeritus
According to Mathematica, the series is
$$-\frac{x^2}{2}-\frac{x^4}{12}-\frac{x^6}{45}-\frac{17 x^8}{2520}-\frac{31 x^{10}}{14175}+O\left(x^{11}\right)$$
It might be easier to find the series for $f'(x) = -\tan x$ and then integrate term by term.