The logistic map $x_{n+1} = r(x_{n}(1-x_{n+1}))$ has clopsed form solutions for $r=2$, $r=4$, and $r=-2$. It is conjectured that these are the only values of $r$ for which closed form solutions exist. How can one solve this difference equation for these three values of $r$?
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1Isn't it $x_{n+1} = r x_n (1-x_n)$? – EditPiAf Mar 14 '17 at 01:01
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You may have a look at this post and this website – EditPiAf Mar 14 '17 at 01:09
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Thanks Harry! Unless I am mistaken, we are seeking an expression of the form $x_n = f(x_0, n)$. So the $n^{th}$ iterate of this mapping can be expressed as a function of the initial point $x_0$, and $n$. This is what I ws thinking of when I asked for a closed form solution. Sincere thanks for the post, and the website! I will look into them. – student Mar 15 '17 at 04:27
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Just looked at the Wolfram website that you directed me to. I did see the Wolfram website after my post yesterday, and the closed form solutions on it. I also derived closed form solutions, but could not reconcile my answers with theirs. My answers seem to be correct though-because they do generate the first few iterates accurately. I cannot really understand my own reasoning clearly I am afraid! Hence my queries... – student Mar 15 '17 at 04:32
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You may update your question by detailing your trials below the original text (+ please correct the original text). – EditPiAf Mar 15 '17 at 14:17