MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $x_s = \sin(\theta+\frac{2\pi s}{3})$ and $y_s = 1+\cos(\frac{2\pi s}{3})$, $s=0,1,2$.

Define $f(\theta) = \sum_{s=0}^2 x_s\ln y_s$.

Is there any method to derive roots of $f(\theta)$. I have run a simulation on it, and found that $\theta=0$ is a solution. But I am unable to see how to analytically obtain it.

share|cite|improve this question
This would be better suited to – Did Jun 19 '11 at 8:52
up vote 4 down vote accepted

By trigonometry, $$f(\theta)= \log (2)\left( \sin\theta+\cos(\pi/6-\theta)-\cos(\pi/6+\theta) \right)=\log(4)\sin\theta.$$

For the revised problem, we have this: $f(\theta)$ is $2\pi/3$-periodic, and $f(\theta)$ is odd, so it suffices to find the roots between $0$ and $\pi/3$ (both of which are themselves roots). Plotting indicates that $f(\theta)$ is unimodal on this interval, $f'(\theta)$ is strictly increasing, and $f''(\theta)$ is strictly increasing, and $f'''(\theta)\geq 6$. Each of these observations follows from the one after it (sometimes needing to also evaluate at $\theta=0$), and the last one seems easiest to prove. Not elegant, certainly, but it should get the job done.

share|cite|improve this answer
Thank you. But I didn't express my question correctly. The $\theta$ is in $y_s$ too. Anyway, thanks for your reply. – Victor Jun 21 '11 at 4:08
In $y_s$ where? – Kevin O'Bryant Jun 21 '11 at 11:42
It should be $y_s=1+\cos(\theta+\frac{2\pi s}{3})$ – Victor Jun 30 '11 at 16:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.