Thomas ha scritto:I didn't say there is a minimum in 0, but that there is a minimum (positive) in $[-\pi,\pi]$... it can be in zero (anyway, in 0 the function is not defined but can be extended by continuity), but we don't care...
But I had to check that the limit in zero existed!! ... in fact Weiestrass theorem tells us that there is a minimum only if the function is continuos...
done that, I observede that the minimum is positive and this proved the existence of such a $c$...
Also I have made the same demonstration and the conclusion is
$c<=1/2*(ln(1-2*p*q*(1-cosx))/(-x^2))$
Now because $0<p<=1$, $0<q<=1$, then $0<=2p*q*(1-cosx)<=1$ and this implicates that $0<=1-2p*q*(1-cosx)<=1$ and this implicates that $(ln(1-2*p*q*(1-cosx))<=0 AAx in [-pi,pi]$. But also $-1/x^2<0 AAx in [-pi,0)$ U$(0,pi]$ and for this reason $1/2*(ln(1-2*p*q*(1-cosx))/(-x^2))>0 AAx in [-pi,pi]$ and this proves our thesis.
Then also in $x=0$ there aren't problems because $lim_(x->0)1/2*(ln(1-2*p*q*(1-cosx))/(-x^2))=(pq)/2$
