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Feb 28, 2023 at 4:02 comment added Alapan Das @StevenStadnicki Oh, I see. Thank you for clearing my doubt.
Feb 27, 2023 at 20:15 comment added Steven Stadnicki @AlapanDas Even if they do converge to such a solution, that doesn't mean that the sequence 'leads to' periodicity in the sense OP means — they're specifically asking (AFAICT) if 1/3 is a preperiodic point.
Feb 27, 2023 at 18:05 comment added Alapan Das I may be navie, but how to ensure that $x_{lr}, l=1,2....$ for some $r$ (period) don't tend to an irrational solution of $f^{(r)}(x)=x$ where $f(x)=4x(1-x)$?
Feb 27, 2023 at 16:02 vote accept Vincent Granville
Feb 27, 2023 at 14:42 comment added Noam D. Elkies That's all correct; it can also be explained in more elementary (and general) fashion: if x is rational with odd denominator q, then 4x(1-x) has denominator q^2, so by induction the n-th iterate has denominator q^(2^n), and in particular x is never periodic once q>1.
S Feb 27, 2023 at 13:26 review First answers
Feb 27, 2023 at 13:28
S Feb 27, 2023 at 13:26 history edited Bill Bradley CC BY-SA 4.0
Minor typo.
Feb 27, 2023 at 12:13 history edited user500150 CC BY-SA 4.0
fixed indexing error
Feb 27, 2023 at 12:10 history edited user500150 CC BY-SA 4.0
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Feb 27, 2023 at 12:09 history edited user500150 CC BY-SA 4.0
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S Feb 27, 2023 at 12:07 review First answers
Feb 27, 2023 at 13:06
S Feb 27, 2023 at 12:07 history answered user500150 CC BY-SA 4.0