# Does the sequence $(x_n)$ given by $x_{n+1} = -16+6x_n+\frac{12}{x_n}$ converge?

Question. If $$x_0$$ is sufficiently close to $$2$$, then will the sequence obtained as

$$x_{n+1} = -16+6x_n+\frac{12}{x_n}$$

converge to 2 ?

My attempt : I have shown that if $$x_0$$ is close to $$2$$ , then $$x_n > 0$$ for all $$n \in \mathbb{N}$$ . Also , if $$x_n > 2$$ for some $$n \in \mathbb{N}$$ then , $$x_n \rightarrow \infty$$ as $$n \rightarrow \infty$$ . I strongly suspect that this sequence will diverge , but I am not able to proceed further .

• I changed the title so that it better represents your question. Hope this did not spoil your intention and is to your liking. – Sangchul Lee Feb 6 '19 at 22:38

Summary. Let $$f(x) = -16 + 6x + \frac{12}{x}$$ and consider $$(x_n)_{n=0}^{\infty}$$ defined by $$x_{n+1} = f(x_n)$$.

1. $$(x_n)$$ remains bounded if and only if $$x_0 \in \bigcap_{n=0}^{\infty} f^{-n}([1,2])$$.

2. $$(x_n)$$ converges if and only if it is eventually constant.

A more detailed explanation.

The problem is much more complicated than it may seem. First, note that

• If $$(x_n)$$ converges, then the limit must solve $$x = f(x)$$, and so, $$x = 2$$ or $$x = 6/5$$.

• If $$x_0 > 2$$, then it is easy to check that $$(x_n)$$ is strictly increasing, and so, $$(x_n)$$ cannot converge. In fact, $$x_n \to \infty$$ in this case.

• If $$0 < x_0 < 1$$, then $$x_1 > 2$$ and the previous observation applies to show that $$x_n \to \infty$$.

• If $$x_0 < 0$$, then $$x_n < 0$$ for all $$n$$, and so, $$(x_n)$$ cannot converge. Moreover, we can in fact prove that $$x_n \to -\infty$$.

So, a necessary condition for $$(x_n)$$ to remain bounded is that $$x_n \in [1, 2]$$ for all $$n$$. In other words, $$x_0$$ must lie in $$K := \bigcap_{n=0}^{\infty} f^{-n}([1,2])$$. And as it turns out, this construction is just a 'distorted' version of that of Cantor set.

Indeed, both $$f_0 : [1, \frac{4}{3}] \to [1, 2]$$ and $$f_1 : [\frac{3}{2}, 2] \to [1, 2]$$ given by $$f_i(x) = f(x)$$ are diffeomorphisms. So, for each $$L \subseteq [1, 2]$$, we have $$f^{-1}(L) = f_0^{-1}(L) \cup f_1^{-1}(L)$$ and $$f_i^{-1}(L)$$ is homeomorphic to $$L$$. Using this, we may track the history of the construction of $$K$$ by introducing

$$\forall \overline{a} = (a_1, \cdots, a_n) \in \{0, 1\}^n, \quad I_{\overline{a}} := (f_{a_1}^{-1} \circ \cdots \circ f_{a_n}^{-1}) ([1, 2]).$$

If we write $$0\overline{a} = (0, a_1,\cdots,a_n)$$ and $$1\overline{a} = (1,a_1,\cdots,a_n)$$, then it follows that $$f^{-1}(I_{\overline{a}}) = I_{0\overline{a}} \cup I_{1\overline{a}}$$. So, $$f^{-n}([1,2])$$ is simply the disjoint union of $$2^n$$ closed intervals $$I_{\overline{a}}$$ indexed by $$\overline{a} \in \{0,1\}^n$$. This can be pushed further by defining the map $$\varphi : K \to \{0, 1\}^{\mathbb{N}}$$ as

$$\varphi(x) = (a_1, \cdots, a_n, \cdots) \quad \text{whenever} \quad x \in I_{(a_1, \cdots, a_n)}$$

This $$\varphi$$ is in fact a homeomorphism between $$K$$ and the Cantor space $$\{0, 1\}^{\mathbb{N}}$$.

Next, since $$f(K) \subseteq K$$, we can consider the function $$\theta = \varphi \circ f \circ \varphi^{-1}$$. Then this is simply the Bernoulli shift on $$\{0, 1\}^{\mathbb{N}}$$, and so, $$(x_n)$$ with $$x_0 \in K$$ converges if and only if $$\varphi(x_n) = \theta^{\circ n}(\varphi(x_0))$$ converges in $$\{0, 1\}^{\mathbb{N}}$$. But the latter takes place only when $$\varphi(x_0)$$ is eventually constant, hence the same is true for $$(x_n)$$.

The following figure is a visualization of this phenomenon. It is the plot of both $$y = x$$ and $$y = f(x)$$ within the region $$[1,2]\times[1,2]$$, together with thin blue rectangles representing the set $$f^{-7}([1,2])\times[1,2]$$.

$$\hspace{7em}$$ • So , this problem looks deceptively simple , but it turns out that it is quite a hard nut to crack . After running $C$ codes to check behaviour of this sequence , I find it highly likely that $x_0$ is not in $\bigcap_{n=0}^{\infty} f^{-n}([1,2])$ – John Feb 7 '19 at 2:35
• @John Indeed I also suspect that $K$ has zero Lebesgue-measure, much like the original Cantor set, so that for almost every starting point $x_0$ in $[1, 2]$ the sequence will diverge to $+\infty$. But I have no good idea for proving (or disproving) this. – Sangchul Lee Feb 7 '19 at 3:32

Please regard this as a comment that is too long to fit in a comment. Obviously @SangchulLee's answer is quite comprehensive and the right way to go here. But I just wanted to comment that if this problem came from a textbook, I suspect that the answer that the original author of the problem was looking for is the following.

Let $$f(x)=-16+6x+12/x$$

Since $$f(2) = 2$$, $$2$$ is a fixed point of $$f$$, but the question is whether it is a stable fixed point. Suppose that $$x_0 = 2 + \delta$$ for small $$\delta$$. Then $$f(x_0) \approx f(2) + f'(2) \delta$$. $$f'(x) = 6 - 12/x^2$$, and $$f'(2) = 3$$, we have $$f(x_0) \approx 2 + 3 \delta$$ or, in other words, $$\vert f(x_0) - 2 \vert \approx 3 \vert x_0 - 2 \vert$$ Since the distance from $$2$$ increases when you apply $$f$$, then $$2$$ is not a stable fixed point, and there is no obvious reason to believe that this sequence will converge to $$2$$.

Note that, as @SangchulLee has pointed out, this is very different than saying that we have proved that it will not converge. I think the main point here is that it can converge by hitting $$2$$ by first reaching $$1$$, since $$f(1)=2$$.