expansion of $\log(2 \cos(\theta/2))$

Question

If $-\pi<\theta<\pi$ how to show that $$\log(2 \cos (\theta/2)) = \sum_{n=1}^\infty \frac{(-1)^{n+1} \cos (n\theta)}{n}$$

Fourier expansion seems to give this result but I am looking if I could get result with expansion of $\log$. Any hints?

Answer 1

Recall that for $\vert z \vert \leq1$ and $z \neq 1$, we have $$\log(1+z) = \sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}z^k}k$$ Hence, setting $z=e^{it}$, where $t \neq 0$, we get $$\sum_{k=1}^{\infty} \dfrac{(-1)^{k+1} e^{ikt}}k = \log(1+e^{it}) = \log \left(1+\cos(t) + i \sin(t) \right)$$ What you want is the real part of the above identity. Hence, $$\sum_{k=1}^{\infty} \dfrac{(-1)^{k+1} \cos(kt)}k = \log \left(\left \vert 1+\cos(t) + i \sin(t) \right \vert\right) = \dfrac{\log \left(2+2\cos(t) \right)}2 = \log\left(\vert2\cos(t/2)\vert\right)$$ where in the last step we made use of the identity $1+\cos(t) = 2 \vert \cos(t/2) \vert^2$.

Answer 2

Let, $$S_1 = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}\cos (n\theta)}{n}\\ S_2 = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}\sin (n\theta)}{n}$$

Then $$S_1 + iS_2 = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}[\cos(n\theta)+i\sin(n\theta)]}{n}=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}e^{in\theta}}{n}$$

Now, from the Taylor expansion, $$\ln (1+x) = x -\frac{x^2}{2}+\frac{x^3}{3} ...=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}$$

$$\begin{align} \therefore S_1+iS_2 &= \ln(1+e^{i\theta}) \\&=\ln(1+\cos\theta+i\sin \theta) \\ &=\ln(2\cos^2(\theta/2) + 2i\sin(\theta/2)\cos(\theta/2)) \\ &=\ln(2\cos(\theta/2))+\ln(\cos\theta/2+i\sin\theta/2) \\ &=\ln(2\cos(\theta/2))+\ln(e^{i\theta /2}) \\ &=\ln(2\cos(\theta/2))+i\theta/2 \end{align} $$

Taking the real part of both sides, $$S_1 = \ln(2\cos(\theta /2))$$