Maclurin Series. (Approximation)

Question

Given that $y=\ln \cos x$, show that the first non-zero terms of Maclurin's series for $y=-\frac{x^2}{2}-\frac{x^4}{12}$. Use this series to find the approximation in terms of $\pi$ for $\ln 2$.

My question is how to determine value of $x$ which is suitable?

Answer 1

A suitable value is a value for which $\log\cos x=\log\frac{1}{2}$, for instance $x=\frac{\pi}{3}$.

That gives: $$ \log 2 \approx \frac{\pi^2}{18}+\frac{\pi^4}{972}.$$

A better choice is $x=\frac{\pi}{4}$, that leads to: $$ \log 2 \approx \frac{\pi^2}{16}+\frac{\pi^4}{1536}.$$

Answer 2

$$\cos\left(\dfrac{\pi}{3}\right)=\dfrac{1}{2}$$ $$\ln\cos(\pi/3)=\ln(1/2)=\ln(1)-\ln(2)=-\ln(2)$$

Answer 3

We have \begin{align} \ln(\cos(x)) & = \dfrac{\ln(\cos^2(x))}2 = \dfrac12 \cdot \ln(1-\sin^2(x)) = - \dfrac12 \sum_{k=1}^{\infty} \dfrac{\sin^{2k}(x)}k\\ & = -\dfrac12 \sin^2(x) - \dfrac14 \sin^4(x) - \cdots\\ & = -\dfrac12 \left(x-\dfrac{x^3}{3!} + \mathcal{O}(x^5)\right)^2-\dfrac14 \left(x + \mathcal{O}(x^3)\right)^4 + \mathcal{O}(x^6)\\ & = -\dfrac12\left(x^2 - \dfrac{x^4}6\right) - \dfrac{x^4}4 + \mathcal{O}(x^6)\\ & = -\dfrac{x^2}2 - \dfrac{x^4}{12} + \mathcal{O}(x^6) \end{align} We need $\ln(2)$ to be expressed in terms of $\pi$, this means we need $\ln(\cos(x))$ to be some value related to $\ln(2)$. One such value is when $\cos(x) = 1/2$, which implies $$\ln(1/2) = \ln(\cos(2n\pi \pm \pi/3)) = -\dfrac{(2n\pi \pm \pi/3)^2}2-\dfrac{(2n\pi \pm \pi/3)^4}{12} + \mathcal{O}\left((2n\pi\pm \pi/3)^6\right)$$ Since typically we are after good approximation, we would like to have the least possible error term, which in turn forces $n=0$, which gives us that $$\ln(2) \approx \dfrac{\pi^2}{18} + \dfrac{\pi^4}{972}$$

---

Another way is to take $x=\pi/4$, which gives us that $$-\dfrac12\ln(2) = \ln(1/\sqrt{2}) = \ln(\cos(\pi/4))) \approx - \dfrac{\pi^2}{32} - \dfrac{\pi^4}{3072}$$ This gives us that $$\ln(2) \approx \dfrac{\pi^2}{16} + \dfrac{\pi^4}{1536} + \cdots$$