most recent 30 from http://mathoverflow.net 2013-05-21T17:40:05Z http://mathoverflow.net/feeds/question/68202 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/68202/roots-of-an-entropy-like-function Victor 2011-06-19T08:25:51Z 2011-06-30T19:22:34Z

<p>Let $x_s = \sin(\theta+\frac{2\pi s}{3})$ and $y_s = 1+\cos(\frac{2\pi s}{3})$, $s=0,1,2$.</p> <p>Define $f(\theta) = \sum_{s=0}^2 x_s\ln y_s$.</p> <p>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.</p>

http://mathoverflow.net/questions/68202/roots-of-an-entropy-like-function/68215#68215 Kevin O'Bryant 2011-06-19T15:15:15Z 2011-06-30T19:22:34Z

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