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Let $\{ a_n \}_{n=0}^\infty$ be the sequence given by $a_0 = 3$ and $$a_{n+1} = a_n - \frac{1}{a_n}, \quad n\ge 0.$$(you can easily check that the sequence is well defined).

Question: Is this sequence bounded or not?

I was only able to show that there are infinitely many $n \geq 0$ such that $a_n > 0$ and infinitely many $n \geq 0$ such that $a_n < 0$ in this way:

suppose for example that we have eventually $a_n > 0$. But then $a_n$ is eventually strictly decreasing and so, using the fact that eventually $a_n > 0$ together with the monotone convergence theorem for sequences we get that $a_n$ is convergent, which is absurd because this would imply, together with the fact that $a_{n+1} a_n = a_n^2 - 1$ for all $n$, that it's limit $L \in \mathbb{R}$ satisfies the equation $L^2 = L^2 - 1$. Using the same reasoning you can show that $a_n < 0$ eventually leads to a contradiction.

I think the boundedness question is much harder to understand because this sequence behaves in a very chaotic way.

River Li
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Rick Does Math
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    Related. – José Carlos Santos Nov 25 '21 at 16:18
  • If there are infinitely many positive iterates and infinitely many negative iterates, and the sequence converges, then it must converge to $0$. (If it converges to nonzero, then some (infinite) tail of the sequence of iterates has the same sign as the convergent so only finitely many iterates can have the opposite sign.) So what does this iteration do to $|a_n| < 1/10$? So once $\varepsilon$ is small, say less than $1/10$, can there be a tail of the sequence of iterates with magnitudes all in $(-\varepsilon, \varepsilon)$? – Eric Towers Nov 25 '21 at 16:23
  • sequence looking pretty cool – Snoop Nov 27 '21 at 16:52
  • This the "Boole map". See this answer for some references. Also, as José noted, your question was asked with $a_0 = 2$ here. Is there any reason your question is more interesting than that question?

    Also, it was shown the Boole map is ergodic with respect to the Lebesgue measure. Do you care about the behavior of the iterates for a "typical" initial value of $a_0$?

     – mathworker21 Nov 28 '21 at 02:39
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    Most likely this question is impossible to answer. As remarked, the map $f(x) = x-1/x$ is chaotic on the real line, so there are points with dense orbits (necessarily unbounded), but also a dense set of periodic points (all with bounded orbits). Questions about orbits of "typical points" can be answered, but most questions about orbits of particular points are not amenable to either theoretical or numerical tools. – Lukas Geyer Nov 29 '21 at 20:30
  • How do you easily check that the sequence is well defined? If you start at $a_0=1$ it certainly isn't. So how do you know that some iterate when starting at $3$ doesn't land you on $1$? – Barry Cipra Nov 30 '21 at 17:25
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    Well, if $a_n = \frac{p}{q}$, where $p, q$ are coprime positive integers, then $p^2 - q^2$ and $pq$ are coprime integers and therefore $a_{n+1} = \frac{p^2 - q^2}{pq}$ is not an integer. In other words, the first term is the only integer term of the sequence. Thus we cannot get to $1$. – Rick Does Math Nov 30 '21 at 17:45
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    @BarryCipra To get $a_n=0$ implies $a_{n-1}=\pm 1$. But that implies $a_{n-2}$ is irrational which is impossible. – QC_QAOA Nov 30 '21 at 18:14
  • @QC_QAOA, of course! Thanks! – Barry Cipra Nov 30 '21 at 18:53
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    @RickDoesMath, good point as well. Why I couldn't figure this out on my own is beyond me. Many thanks. – Barry Cipra Nov 30 '21 at 18:55
  • In fact, if $a_n = a$, then $a_{n-1} = \frac{a}{2} \pm \sqrt{\left(\frac{a}{2}\right)^2 + 1}$, which shows that every $a_{n}$ could be obtained by at most two distinct $a_{n-1}$. By induction, there's an enumerable set of points in the real line which, if you set $a_0$ (or equivalently, any $a_n$) equal to an element of this set, then we eventually have $a_n = 1$ for some $n$. – matheusbm98 Dec 01 '21 at 12:27
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    @matheusbm98 Yes, but $a_0=3$ isn't one of these points. By induction you can show that if $a_n$ is a rational other than ${-1,0,1}$ then $a_{n+1}$ is a rational other than ${-1,0,1}$ – QC_QAOA Dec 01 '21 at 17:14
  • @QC_QAOA Oh, yes, sorry, I misread this. I don't know where was my mind when I said this. I'll exclude my first comment. – matheusbm98 Dec 01 '21 at 21:00
  • That inequality is completely false, in fact $a_7 \approx -1.23$ – Rick Does Math Dec 02 '21 at 15:40
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    Also, note that the sequence comes from Newton's formula for approximation of roots for the function $e^\frac{x^2}{2}$. Probably meaningless comment but gives a nice geometric interpretation for $a_n$. – Salcio Dec 02 '21 at 18:28
  • Are there only two possibilities for a given starting value? (1) the iterates are eventually periodic, or (2) they are dense on the real line? – mjqxxxx Dec 02 '21 at 20:53
  • The sequence cannot converge, because as previously mentioned if it converges it must converge to zero; this is precluded by inspection of the asymptote of the map $x \mapsto x - 1/x$ at $x = 0$. If $|a_n| < 0.1$ then $|a_{n+1}| > 9$, so convergence to zero is impossible – Rob Dec 04 '21 at 03:58
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    The closer $|a_n|$ is to the golden ratio $\varphi$, the closer $|a_{n+1}|$ is to $1$ and the larger $|a_{n+2}|$ is. Thus if for every $\varepsilon>0$ there exists $n\in\mathbb N$ such that $|a_n|\in(\varphi-\varepsilon,\varphi+\varepsilon)$ then the sequence is unbounded.

    Let $a_k$ have the same sign for $n\leq k\leq m$. If $|a_k|-\varphi>0$ for $n\leq k\leq m-1$ then $|a_{k+1}|-\varphi<|a_k|-\varphi$. If $|a_m|-\varphi<0$ then $a_{m+2}$ has an opposite sign.

    So there are infinitely many $a_n$ that are in the abs. v. close to $\varphi$. We have to show whether they are close enough.

     – Pavel R. Dec 04 '21 at 23:02
  • @RickDoesMath Is there a link with the Mandelbrot set or with some fractals ? – Miss and Mister cassoulet char Feb 08 '22 at 20:09
  • @RickDoesMath You should look to Julia set see https://en.wikipedia.org/wiki/Julia_set – Miss and Mister cassoulet char Feb 09 '22 at 09:10

1 Answers1

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$\color{brown}{\textbf{Infinity points.}}$

Easily to check that the functions $$f_n(x)=\,\underbrace {f(f(f(\dots(f(x)))))}_n,\quad\text{where}\quad f(x)=x-\frac1x=2\sinh \ln x,\quad f_0(x)=x,\tag1$$ map $\;\mathbb Q\mapsto \mathbb Q.\;$

On the other hand, there are exactly two functions $$g_\pm(x)= \dfrac{x\pm\sqrt{4+x^2}}2=\dfrac2{x\mp\sqrt{4+x^2}},\quad\text{such as}\quad f(g_\pm(x))=x,\quad\text{wherein}$$ $$ g_\pm(+\infty)=\dbinom{-0}{+\infty},\quad g_\pm(-\infty)=\dbinom{+0}{-\infty},\quad g_\pm(\pm0)=\dbinom{1}{-1},\quad g_\pm(\pm1)=\frac{\pm\sqrt5\pm1}2.$$ If $\;a_n=\pm\infty,\;$ then $$a_{n-2}\in \left(\pm\infty \bigcup \frac{\pm\sqrt5\pm1}2\right),\quad a_{n-k}=\frac{\pm\sqrt5\pm1}2\not\in\mathbb Q.$$ Therefore, $\;\forall (N)\, \forall (n\le N)\; a_n\not=\pm\infty.\;$

I.e. the given sequence does not contain infinity as a value.

$\color{brown}{\textbf{Periodic sequences.}}$

Let us define periodic sequences via the equation $\;f_T(\tilde x)=\tilde x,\;$ where $\,\tilde x\,$ is a base and $\,T\,$ i a period.
For example, $\;\dbinom {\tilde x}T=\dbinom {\sqrt2^{\,-1}}2.\;$

Rewriting the equation in the form of $\;f_{k-1}(x)=g_\pm(x)\;$ and taking in account, that $\;g_\pm(3)=\dfrac{3\pm\sqrt{13}}2\not\in\mathbb Q,\;$ easily to prove that the given sequence is not periodic.

At the same time, $$f'(x)=1+\dfrac1{x^2},\quad f'_k(x)=\prod\limits_{j=1}^{k-1} f'(f_j(x)) =\prod\limits_{j=1}^k\left(1+\dfrac1{f^2_{k-1}(x)}\right)>f'_{k-1}>1,\quad (k\in\mathbb N),$$ so should be a pole between the neighbor bases with the given period.

Plot f[k], k=1, 2, 3

Therefore, $\,f_k(x)\,$ has negative infinity gaps in the poles and increasing pieces with $\,f'_k(x)>1\,$ between the poles.

If to consider the equation $\;x_{n+1}=f_k(x_n)\;$ as the simple iteration, then there are not such neighborhood of $\,x=3\,$ which can be compressed via the iteration. I.e. sufficient condition of the simple iteration convergency is not satisfied in the neighborhoods of $\;x=3.$

$\color{brown}{\textbf{Heuristic model of the process.}}$

Let us consider the model with a pair of repeated steps:

  • Decreasing of $\;x=|a_n|\;$ from $\;M_k>1\;$ to $\;m_k<1,\;$ with the homogenius distribution law.
  • The gap of $\,x\,$ from $\,m_k\,$ to $\;M_{k+1}=\dfrac1{m_k}-m_k.\;$

Modeling the first step via the ODE $$\dfrac{\text dx}{\text dt}=-\dfrac1x,\quad x(0)=M_k,\tag2$$ one can get $\;x=\sqrt{M^2_k-2t},\quad t_k =\dfrac12(M_k^2-1)\;\text{iterations},\;$ and with 2 additonal iterations per step we have $$N_s(M)=\dfrac12(M^2+3).\tag3$$

The average quantity of iterations in the step with $\;M_k<M,\;m_k>|g_-(M)|=\frac1{g_+(M)}\;$ equals to $$\bar N_s=\dfrac {g_+(M)}{2(g_+(M)+1)}\int\limits_{\frac1{g_+(M)}}^1 \left(\left(\dfrac1m-m\right)^2+3\right)\,\text dm =\dfrac {g_+(M)}{2(g_+(M)+1)}\left(-\dfrac1m+m+\dfrac13m^3\right)\bigg|_{\frac1{g_+(M)}}^1$$ $$ =\dfrac {g_+(M)\,m}{6(g_+(M)+1)}\left(-\dfrac4{m^2}+1+\left(\dfrac1m-m\right)^2\right)\bigg|_{\frac1{g_+(M)}}^1 =\dfrac {4g^2_+(M)-3g_+(M)-1-M^2}{6(g_+(M)+1)},$$ $$\bar N_s(M) =\dfrac{2M^2-3M+6+(4M-3)\sqrt{4+M^2}}{12(g_+(M)+1)}. \tag4$$ Assuming the average quantity of the required steps before $M_k>M$ to $\;g_+(M)+1\;$ gives $$\bar N_\Sigma(M)\approx\dfrac{2M^2-3M+6+(4M-3)\sqrt{4+M^2}}{12}\tag5$$ iterations totally.

The table of N(M).

$\color{brown}{\mathbf{Tables\;a_{0\dots49},\;a_{50\dots99},\;a_{100\dots149},\;a_{150\dots189}.}}$

Table a(0-49)

Table a[50-99]

Table a[100-149]

Table a[150-189]

$\color{brown}{\textbf{Conclusions.}}$

There are the reasons for the next hypotheses:

  • the given sequence is unbounded.
  • the average quantity of iterations before $|a_{N+1}|>M$ is defined by the expression $(5)$.
Yuri Negometyanov
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