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 .

  • 2
    $\begingroup$ I changed the title so that it better represents your question. Hope this did not spoil your intention and is to your liking. $\endgroup$ Commented Feb 6, 2019 at 22:38

2 Answers 2


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 $f^{-7}([1,2])\times[1,2]$.


  • 1
    $\begingroup$ 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 $\bigcap_{n=0}^{\infty} f^{-n}([1,2])$ $\endgroup$
    – John
    Commented Feb 7, 2019 at 2:35
  • $\begingroup$ @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. $\endgroup$ Commented Feb 7, 2019 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 $f(1)=2$.


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