# Show that there exists an infinite subsequence $(x_{a_i})_{i \geq 1}$ such that $\displaystyle\lim_{i \to \infty}x_{a_i} = 0$

Let $x_1 = \dfrac{5}{13}$ and $x_{n+1} = 2x_n\sqrt{1-x_n^2}$ for $n \geq 1$. Show that there exists an infinite subsequence $(x_{a_i})_{i \geq 1}$ such that $\displaystyle\lim_{i \to \infty}x_{a_i} = 0$.

I didn't see how to construct an infinite subsequence such that $\displaystyle\lim_{i \to \infty}x_{a_i} = 0$, so can we solve this by contradiction? That is, assume there does not exist such a subsequence with $\displaystyle\lim_{i \to \infty}x_{a_i} = 0$ and find a contradiction.

• Hint: the recursion has 2 stationary points, $0$ and $\frac{\sqrt{3}}{2}$. Prove that the second one is not stable. – user58697 Feb 3 '17 at 2:20
• In what context did you come across this problem? – H. H. Rugh Feb 3 '17 at 2:33
• @H.H.Rugh In a number theory context. – user19405892 Feb 3 '17 at 2:41
• Ah, I sort of had the feeling that elliptic curves may get into the picture. Seems that iterating the rational point x1 you keep within the rational field. Unfortunately I don't know much about EC. Could perhaps be related to the 13-adic field? – H. H. Rugh Feb 3 '17 at 2:47
• @H.H.Rugh I wasn't using elliptic curves in this context, so they may not be necessary. – user19405892 Feb 3 '17 at 2:51

Just some general remarks: The map $$x=\psi(t)=\sin(\pi t/2)$$ conjugates $$f(x)=2x\sqrt{1-x^2}$$ to the tent map on $$[0,1]$$: $$g(t)=\left\{ \begin{matrix} 2t, & 0\leq t \leq 1/2\\ 1-2t, & 1/2\leq t \leq 1\end{matrix} \right.$$ since: $$f(\psi(t))=\sin(\pi t)=\psi(g(t))$$ The map $$g$$ is uniformly expanding on $$[0,1]$$ and has nice ergodic properties, like Lebesgue almost every point has a dense forward orbit. However, it is rarely possible given an initial point to see if it has a dense orbit, or if the orbit accumulates on e.g. zero (corresponding to the stated problem). Usually one has to resort to some number theoretic arguments to reach such conclusions. Whence my comment. Anyway I don't have any particular ideas about how to proceed. (And the other given suggestions won't lead you any further either, unless I have missed something evident). So number theory is probably needed somehow.
• Not really any idea. You are right, the same problem. And I see along other linked posts that it boils down to the same question. Ultimately, it could actually have to do with iterations of the rational 5/13 along the (almost) elliptic curve: $y^2=4x^2(1-x^2)$. But admittedly, this is beyond my knowledge. – H. H. Rugh Feb 3 '17 at 4:04
• What is the initial point for the map $g$ here? – user19405892 Mar 1 '17 at 22:20
• It is the value of $0\leq t_1\leq 1$ for which $\sin(\pi t_1/2) = x_1=5/13$. You may then calculate that $\tan (\pi t_1 /2) =5/12$ so that $t_1=2/\pi \tan^{-1}(5/12)$. This shows that your problem is the same as the other problem you cite (a part from your initial point being one more iterate, and that the other considers the Bernoulli map, rather than the tent). But the difficulty is the same (and probably very high!). – H. H. Rugh Mar 3 '17 at 6:29