Construct a continuous function $F:[0,1] \to [0,1]$ that has a point with period 2015.

I think I should do it with Sharkovskii's theorem, but then? Where can I start with?

Or try to find a point of period 3 and by LiYorke theorem?


How about something as simple as $$F(x)=\begin{cases}x-\frac{1}{2014}& \text{for }x>\frac{1}{2014}\\ 1-2014x & \text{for } x\leq \frac{1}{2014} \end{cases}$$

Then $f^{2015}(1)=1$, and $f^n(1)\neq 1$ for $n<2015.$


Of course this works for every natural period. Here's what the cobweb plot looks like when when the period is $15$.

enter image description here

  • $\begingroup$ hmm, but f^2015(1)=0, actually. $\endgroup$
    – Vito Chou
    Jan 12 '15 at 15:42
  • $\begingroup$ Ah, yes Fixed that. $\endgroup$ Jan 12 '15 at 15:43
  • $\begingroup$ Nice! I hope you don't mind that I added a cobweb plot - feel free to roll back, if so. $\endgroup$ Jan 19 '15 at 0:38

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