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I am working on a problem to find the Feigenbaum constant. In this problem I have to get $f^{2^n}(1/2)$ with $f(x) = kx(1-x)$. I use Nest to get $f^{2^n}(1/2)$ but it takes Mathematica forever to run with $n>5$ but my goal is to get to $n=14$.

Does any body know any function or trick to get $f^{2^n}(1/2)$ with large value of $n$?

Thank you very much.

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Ha! another stupid decision from SO 10K users stackoverflow.com/q/16261682/353410 –  belisarius Apr 29 '13 at 14:38
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You do know that iterating the logistic equation eventually results in a periodic sequence, no? Find the period, and then you can extrapolate on what the $2^n$-th iterate ought to be. –  J. M. Apr 29 '13 at 14:43
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@belisarius Try k=2.9 or 3.1 or 3.4. You happen to be choosing a value of k right at a bifurcation, where the convergence to the fixed point is absurdly slow. But, of course, the iteration need not always converge to a periodic orbit; that's the point behind chaos. –  Mark McClure Apr 29 '13 at 15:01
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@belisarius No, I don't think you can analytically find the attractive orbit in terms of k in a simple way. What I think J.M. is suggesting is extrapolate from when the bifurcations occur. In fact, the existence of the Feigenbaum constant is really an assertion that such an extrapolation works. –  Mark McClure Apr 29 '13 at 15:10
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@thewanderer I assume that you mean when n>5, but note that for n=5, your polynomial has degree 2^2^5. and it's simply not feasible to solve this using simple techniques. You should look into some of the papers of Keith Briggs to learn his extrapolation techniques: keithbriggs.info –  Mark McClure Apr 29 '13 at 21:26

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