# The sequence $x_{n+1}=\frac{x_n}{2}-\frac{2}{x_n}, x_0>0$ is bounded?

Consider the sequence $$x_{n+1}=\frac{x_n}{2}-\frac{2}{x_n}, x_0>0.$$ How can you prove that this sequence is bounded or unbounded for those values ​​of $x_0$ for which it is defined. With a number generator I noticed that all terms of the positive out of range $[-2,2]$ form finite sets in strictly decreasing order. We found that it can be periodically. For $x_0=\frac{2}{\sqrt{3}}$ its terms repeat of 2 by 2.

I tried to prove boundedness to the method of reduction to absurdity but we did.

Thanks so much for any suggestion

• How do you ensure that no $x_n$ is ever zero? – copper.hat Oct 12 '14 at 19:41
• If you let $\phi(x) = x+ \sqrt{x^2+4}$, then if we let $x_0=\phi^n(0)$ (meaning composition), we will have $x_n = 0$. Any proof must somehow avoid all of the increasing list of points $\phi^n(0)$. – copper.hat Oct 13 '14 at 2:40
• @Did The link isn't working. Do you remember the solution? – Puzzled417 Jan 2 '17 at 18:30
• @Puzzled417 See below. – Did Jan 2 '17 at 18:39

In the comments above, one year ago, I was mentioning my answer to a nearly duplicate question, "for a full solution when $x_0=3$ and (for) clear indications about the other cases". The other page is now unavailable to users with not enough reputation hence I reproduce the question and my answer below. (Revised version following a remark by @Puzzled417.)

Is the sequence defined by $x_1=3$ and $x_{n+1}=\frac{x_n}{2}-\frac{2}{x_n}$ for every $n\geqslant1$, bounded?

A key remark is that the transformation $x_n\to x_{n+1}$ has two fixed points $\pm2i$. Even though these are not real, one can make use of them, introducing $$z_n=\frac{x_{n}-2i}{x_{n}+2i}.$$ Then, a simple computation shows that, as long as $x_{n+1}$ exists, $z_n$ is on the unit circle and $z_{n+1}=z_n^2$, hence $$z_n=(z_0)^{2^n}.$$ Note that, if the sequence $(x_n)$ stops to be defined because some $x_n$ is $0$, then the corresponding $z_n$ is $-1$. Furthermore, the sequence $(x_n)$ is bounded if and only if $(z_n)$ stays away from $1$, that is, the sequence $(x_n)$ is unbounded if and only if $1$ is a limit point of $(z_n)$.

To solve the case $x_0=3$, note that $z_0=(5-12i)/13$ hence $z_0=e^{i \theta_0}$ with $$\theta_0=\arccos\left(5/13\right).$$ One knows that the only rational multiples of $\pi$ with a rational cosine have cosine $0$ or $\pm1/2$ or $\pm1$. Thus, $\theta_0/\pi$ is irrational, hence $z_n\ne1$ for every $n$ and every $x_n$ is well defined.

As is well known, for every irrational number $\alpha$, the set of fractional parts of $n\alpha$ is dense in $[0,1]$, however this does not say that the set of fractional parts of $2^n\alpha$ is dense in $[0,1]$, and actually there exists some irrational numbers $\alpha$ such that this set is not dense in $[0,1]$, hence we cannot conclude only using the fact that $\theta/\pi$ is irrational.

However... Consider some neighborhood $N_\epsilon=(-\epsilon,\epsilon)$ of $0$ in $\mathbb R/\mathbb Z$ of Lebesgue measure $2\epsilon$. Its preimage $d^{-1}(N_\epsilon)$ by the doubling map $d:\alpha\mapsto2\alpha$ is made of two intervals of length $\epsilon$, one of them included in $N_\epsilon$, hence the measure of $N_\epsilon\cup d^{-1}(N_\epsilon)$ is at most $2\epsilon+\epsilon$. Likewise, the measure of $N_\epsilon\cup d^{-1}(N_\epsilon)\cup\cdots\cup d^{-k}(N_\epsilon)$ is at most $2\epsilon+\epsilon+\cdots+2^{1-k}\epsilon$ and the measure of the set of points whose orbit passes by $N_\epsilon$ is at most $4\epsilon$. This shows that the set of points $\alpha$ in $\mathbb R/\mathbb Z$ whose orbit's closure contains $0$ has Lebesgue measure $0$ (or course this set is not empty since it contains all the dyadic numbers, whose orbit contains $0$).

...Which is of course still not enough to rule the case of our $\theta_0/\pi$.

• You seem to be saying that since $\dfrac{\theta}{\pi}$ is irrational, then $2^n \cdot \dfrac{\theta}{\pi}$ is dense in $[0,1]$. I don't see how this follows, since there are irrational numbers that don't have this property. – Puzzled417 Jan 5 '17 at 4:25
• @Puzzled417 Quite true. Thanks for the input, I modified the post. – Did Jan 5 '17 at 12:25
• Have you found a way to finish the proof? – Puzzled417 Jan 26 '17 at 2:54
• @Puzzled417 Unfortunately, I have not. Have you? – Did Jan 26 '17 at 9:28
• @Puzzled417 Not sure these should be posted as comments to my answer. If you wish to solve this question, please post the result of your thinking as a separate answer. – Did Feb 3 '17 at 21:53

Let $$x_n=au_n$$ ,

Then $$au_{n+1}=\dfrac{au_n}{2}-\dfrac{2}{au_n}$$

$$u_{n+1}=\dfrac{u_n}{2}-\dfrac{2}{a^2u_n}$$

$$u_{n+1}=\dfrac{1}{2}\left(u_n-\dfrac{4}{a^2u_n}\right)$$

Choose $$a=2$$ , the recurrence becomes

$$u_{n+1}=\dfrac{1}{2}\left(u_n-\dfrac{1}{u_n}\right)$$

Consider $$\cot2\theta=\dfrac{1}{2}\left(\cot\theta-\dfrac{1}{\cot\theta}\right)$$

$$\cot2^{n+1}=\dfrac{1}{2}\left(\cot2^n-\dfrac{1}{\cot2^n}\right)$$

$$\therefore u_n=\cot(x_02^n)$$

$$x_n=2\cot(x_02^n)$$