# Approximate $\ln2$ using the Maclaurin series expansion of $\ln\cos x$

Until now I have that $\ln\cos x\approx-\dfrac{x^2}{2}-\dfrac{x^4}{12}$.

Since $\cos^{-1}2$ does not exist I do not know what value of $x$ to take. I suppose I need to play around with the original function but have no idea.

• Use ln $\frac{1}{2}=-$ln$2$. Jun 3 '15 at 20:02
• @NoahOlander Thanks! Jun 3 '15 at 20:03
• Since $\cos^{-1}(1/2)=\pi/3$, this would seem to assume the value of $\pi$ is known, and that's certainly a tougher problem than finding $\log(1/2)$ from scratch. ${}\qquad{}$ Jun 3 '15 at 20:04
• @MichaelHardy Perhaps it would be tougher to derive from elementary axioms, but aren't the fastest known converging series for $\pi$ much faster than convergence of $\ln x$ for $0 < x < 1$? In that sense, it seems that at least the computational complexity would be the same whether $\pi$ is assumed to need to be approximated or not. Jun 3 '15 at 20:15
• Computing more than the first few terms of the series for $\ln(\cos x)$ is not pleasant. And there are nice tricks to speed up the convergence for the log series. Jun 3 '15 at 21:00