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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?

freedome321
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  • If $F$ denotes the rhs of your map, then first notice that $F(I) \subset I$, where $I=(0,1)$ --- so existence is easy. You may wish to consider the map $F'$ also... – Suvrit Sep 13 '14 at 19:38
  • Maybe rewrite this as polynomial equation and then use Sturm's theorem. It should be easy to get exact symbolic solutions (instead of numerical solutions) this way. – Moritz Firsching Sep 13 '14 at 19:53
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    Suvrit, $x\mapsto x^2$ sends (0, 1) to (0, 1) yet has no fixed point in that interval, so I don't understand your comment. – Noah Schweber Sep 13 '14 at 20:19
  • Survit, I think Noah is right. Moritz, do you have any idea of how to rewrite it as a polynomial for general K? Thanks. – freedome321 Sep 13 '14 at 20:40
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    Existence is easy since $\textrm{RHS}<x$ for small $x>0$ and $\textrm{RHS}>x$ near $x=1$. – Christian Remling Sep 13 '14 at 21:08
  • Yes. I think existence should be doable that way, uniqueness is the trickier part. – freedome321 Sep 13 '14 at 21:13
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    Why the votes to close? Is there a simple argument for uniqueness that some of us are missing? – Michael Renardy Sep 13 '14 at 22:19
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    @MichaelRenardy I agree; despite the fact that the language needed to ask the question is elementary, the problem doesn't look trivial. – Todd Trimble Sep 13 '14 at 22:21
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    I'm curious, why the restriction to integral $K$? It'd be interesting to see what happens to the fixed point(s) as $K$ varies continuously. – Noah Schweber Sep 13 '14 at 22:56
  • Can you show that it is sigmoidal with f'(0)=0, f'(1)=0, f'(x) increasing between x=0 and x=c for some inflection point c, and f'(x) decreasing between x=c and x=1? Even if c is not at the fixed point this would still convince me. – guest Sep 14 '14 at 01:05
  • The restriction to integer K's is not important. We can simply relax it if it helps. – freedome321 Sep 14 '14 at 02:07
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    Is this the cdf of a known unimodal distribution on the unit interval? It looks kind of like http://en.wikipedia.org/wiki/Kumaraswamy_distribution#Cumulative_distribution_function. – guest Sep 14 '14 at 02:33
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    What about $$ x = \frac{(1 - (1-x^{a+1})^b)^c}{(1 - (1-x^a)^b)^c} $$ with $0<a, 1<b, 1<c$? – guest Sep 14 '14 at 03:12
  • @Noah as I read the question, there is no restriction of $K$ to integers; it is "a number greater than $1$", which only implies that the numbers must be real (because of the ordering relation); th restriction to natural numbers is an option, if that would make things easier. Using $K$ to denote real numbers is however a bit misleading; maybe replacing it by $t$ would help. – Manfred Weis Sep 14 '14 at 06:03
  • @NoahS: Ok, I take back my comment; I implicitly assumed a closed interval, while writing an open one. However, it seems that the map $F$ is strictly monotonic... – Suvrit Sep 14 '14 at 06:12
  • @ManfredWeis, the original version of the question specifically took $K$ to be an integer. – Noah Schweber Sep 14 '14 at 06:24
  • @Noah please accept my apologies; I didn't see the unedited version of the question. – Manfred Weis Sep 14 '14 at 06:28
  • @MohammadAkbarpour In the case where $K$ is not an integer I don't see how this can be a polynomial (except if $2K$ is an integer of course). – Moritz Firsching Sep 14 '14 at 12:44
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    If $K$ is reasonably large then the fixed point seems very close to $K^{-2/3}$ -- even for $K=6$ this is a pretty close approximation. Some asymptotic analysis around here (which might be unpleasant but straightforward) should settle the issue -- first show that the fixed points have to be reasonably close to $K^{-2/3}$ and then use the derivative to check uniqueness. – Lucia Sep 14 '14 at 16:19
  • @MoritzFirsching In the problem that I face, K is an integer. Whichever makes the proof of uniqueness easier (integer or real) is fine... – freedome321 Sep 14 '14 at 16:28
  • Is there an intuition as to why the fixed point behaves like $K^{-2/3}$ asymptotically? – Eckhard Sep 14 '14 at 18:10
  • @guest: the conclusion holds true in the generality you said, by the same computation. Write the equation in the form $g(x^a)=g(x^{a+1})$. – Pietro Majer Sep 17 '14 at 22:36

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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 $.

$${*}$$

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.

Pietro Majer
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