# Is $x_{n+1}=\frac{x_n}{2}-\frac{2}{x_n}$ bounded? [duplicate]

Consider the sequence: $x_1=3, x_{n+1}=\frac{x_n}{2}-\frac{2}{x_n}$. This sequence is bounded or is unbounded?

Attempt

Checking a few terms we get $x_1 = 3, x_2 = \dfrac{5}{6}, x_3 = -\dfrac{119}{60},x_4 = \dfrac{239}{14280},\cdots$. I will prove by contradiction that it is bounded. Suppose that the sequence is bounded. Then there exists some $x_k$ such that $x_k \geq \dfrac{x_n}{2}-\dfrac{2}{x_n}$ or $x_k \leq \dfrac{x_n}{2}-\dfrac{2}{x_n}$ for all $n$.

I seem to get stuck here.

• Appears to be bounded? After looking at four values? Every finite sequence "seems to be bounded" by this definition. – Thomas Andrews Jan 17 '16 at 15:57
• Sorry, I was guessing. It is not mathematical at all – Puzzled417 Jan 17 '16 at 15:59
• The very next value is $x_5\approx -119.5$. :) – Thomas Andrews Jan 17 '16 at 16:05
• @ThomasAndrews I deleted my comment about it being bounded. – Puzzled417 Jan 17 '16 at 16:07
• You say you are going to prove by contradiction that it is bounded, then assume it is bounded. That is backwards. Contradiction would require you to assume it is unbounded to prove it is bounded. – Thomas Andrews Jan 17 '16 at 16:08

The map $x \mapsto x/2 - 2/x$ is a degree $2$ map on $\Bbb P^1(\Bbb C)$ with two superattractive fixed points ($2i$ and $-2i$) where the derivative is $0$, and a repulsive fixed point at $\infty$ where the derivative is $2$.

Dynamically, this really looks just like the map $y \mapsto y^2$ which also has two superattractive fixed points $0$ and $\infty$, and one repulsive fixed point at $1$, with the same behaviour near those points.

Hence, you should look at $$y_n = \frac {x_n-2i}{x_n+2i}.$$ Using this change of variable, you get $$y_{n+1} = y_n^2.$$ Also, the real line is transformed into the unit circle in $\Bbb C$ ($x$ is real iff $|y|=1$).

Now the behaviour of a point on the unit circle under the squaring map is well-understood : write $y_0 = \exp(\lambda i\pi)$. If $\lambda$ is rational then the sequence is ultimately periodic, and in general the behaviour of the sequence is obtained by looking at the binary digits of $\lambda$

Since $y_0 = \frac{3-2i}{3+2i} = \frac{5-12i}{13}$, you need to look at $\frac 1 \pi \arctan \frac{12}{5}$. Since it is not rational, the sequence $(y_n)$ is not ultimately periodic. (you can also deduce this from the fact that $\Bbb Z[i]$ is a unique factorisation domain and $3\pm 2i$ are prime, so the prime factorisations of $y_n = y_0^{2^n}$ are obviously all different)

To show it is not bounded you need to prove that the binary development of that constant has strings of $0$s or $1$s of arbitrary length.

This also gives a geometrical interpretation of the recursion :

• That is quite a thing. Such an interesting dynamics and such a method of solution from a seemingly innocent formulation. – TZakrevskiy Jan 17 '16 at 16:59
• (+1) This also seems to answer my question about a rational orbit dense in the reals. (I would accept it there.) – r.e.s. Jan 17 '16 at 17:08
• I'm sorry @r.e.s. in fact I don't even have a proof that it's unbounded. – mercio Jan 17 '16 at 17:45
• Ah, yes ... this answer encounters the very same difficulty as the one attempted there. For my question, so far there is only the partial answer in the positive reals (as I noted). – r.e.s. Jan 17 '16 at 17:52