I've been trying to prove that the following equation has a unique solution in interval 0 < x < 1 :
$$ x = \Big(\frac{1 - (1-x^2)^K}{1 - (1-x)^K}\Big)^2 $$
Where K is a number (integer, if it helps) greater than 1. I have checked it numerically and, in addition to x=0 and x=1, there is always a solution in interval (0,1). (for instance, see this for K=6: http://goo.gl/a7OSSn) Does anyone have any idea of how I can prove the existence and uniqueness of such a fixed point?
Consider the function $f(x):=(1-(1-x)^k)^2 $, a strictly increasing homeo of $ (0,1)$ onto itself, with $f'(0)=f'(1)=0$. Looking at the sign of $f''$ we see that $f$ is initially strictly convex, then strictly concave (the inflection point being at $1-\big(\frac{k-1}{2k-1}\big)^{1/k}$). This implies that $f(x)/x=f'(x)$ has a unique solution $x_0$ in $(0,1)$, and in fact this means that $g(x):=f(x)/x$ is unimodal (with $g(0)=0$, $g(1)=1$, and maximum point at $x_0$). Finally, the initial equation can be written as $g(x)=g(x^2)$, which of course has a unique solution $x=c\in(0,1)$ due to the unmorality of $g$, in fact $0< c^2 < x_0 < c <1 $.
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edit. Incidentally, the same computation works for the generalization suggested by guest in a comment : $x=\frac{ (1-(1-x^{a+1})^b )^c }{ (1-(1-x^a)^b )^c }$, with $a>0, b>1, c>1:$ we can write it in the form $g(x^{a+1})=g(x^a)$ with $g(x):=f(x)/x$ and $f(x):=(1-(1-x )^b )^c$, and the same conclusion follows.
