Is the sequence $a_{n+1}=a_n-\frac{1}{a_n}$, $a_0=2$ bounded?

The sequence again for convenience $$a_{n+1}=a_n-\frac{1}{a_n},\;a_0=2$$ My friend asked me this question and I do not know how to tackle it. It's clear it does not have a limit, but I am not sure whether it is unbounded; it seems to oscillate with a large amplitude when you simulate it numerically.

I also can't seem to get anywhere with generating functions, but I also don't know how to use them for nonlinear recurrence relations.

• Wow, anyone able to explain why this question has so many upvotes? Commented Apr 9, 2017 at 23:56
• Commented Apr 10, 2017 at 0:20
• Note that the recursion can also be written $$\frac pq \mapsto \frac{p^2-q^2}{pq}$$ and if $p/q$ is in lowest terms, then the right-hand side will also be. Commented Apr 10, 2017 at 0:42
• Thinking that the mapping $x \mapsto x - 1/x$ is an infinite ergodic transform, I would be astounded if $a_n$ is bounded. But to be honest, I have absolutely no idea how this iteration behaves for this particular initial value. Commented Apr 10, 2017 at 3:10

Note that there is a 2-cycle if we begin with $$x = \pm \frac{1}{\sqrt 2} \approx \pm 0.7071,$$ as we then get $$x - \frac{1}{x} = -x.$$ $$f(x) = x - \frac{1}{x}.$$ Important: $f$ is odd, $$f(-x) = -f(x).$$ $$\mbox{If} \; \; \; f(x) = -x, \; \; \; \mbox{then} \; \; \; f(f(x)) = x.$$

We also get 4-cycles at the roots of $$2 x^4 - 4 x^2 + 1,$$ as then $$f(f(x)) = -x,$$ $$\mbox{If} \; \; \; f(f(x)) = -x, \; \; \; \mbox{then} \; \; \; f(f(f(f(x)))) = x.$$

The presence of cycles of larger and larger degree tends to support the hypothesis of chaotic behavior elsewhere.

6-cycles at the roots of $$2x^8 - 11x^6 + 17x^4 - 8x^2 + 1$$ $$\mbox{If} \; \; \; f(f(f(x))) = -x, \; \; \; \mbox{then} \; \; \; f(f(f(f(f(f(x)))))) = x.$$ Now that I think of it, this diagram also shows some 3-cycles.

Seems to me that these points involved in $2k$-cycles might be dense in the real line. If so, that would be strong evidence.

Wildly unstable. I did the first 25 steps with the given $x$ seed value in rational number arithmetic,

    1: 1.5 =                       3/2

2: 0.8333333333333333 =                       5/6

3: -0.3666666666666666 =                    -11/30

4: 2.36060606060606 =                   779/330

5: 1.93698603493212 =             497941/257070

6: 1.420720051612811 = 181860254581/128005692870

7: 0.7168516161213897 = 16687694789137362648661/23279147893155496537470

8: -0.6781372177053623 = -263439569256003706800705587722279993788907979/388475314992168993748220639081347493631827670

9: 0.7964905919638025 = 81512663708476146329709015825571064954724426915346799560162522434680208602364731247764459/102339769648127358726761918460732576814168548432921287355299929744910591862606847215978930

10: -0.4590170186589809 = -3829114106780645548860005128366999929762127231121785938212801344907004576697195186639451276755270737038953266911319233153169994507036998538685318746477046140028242936033060382219/8341987227330719589550045299118644973579016226280953918450224539222226868008655704094557970730123521406125932252315167316321272393191426438953060083145494237274901280505746848870

11: 1.719551442531198 =

12: 1.138004432499333 =

13: 0.2592732330050055 =

14: -3.597661740227243 =

15: -3.319703423907924 =

16: -3.018471695555874 =

17: -2.687178213005645 =

18: -2.315040631969854 =

19: -1.883082770759831 =

20: -1.352038668223384 =

21: -0.6124148516101889 =

22: 1.020465208974159 =

23: 0.0405199926102738 =

24: -24.63865528804984 =

25: -24.5980686574765 =

• @qbert $f(x) = -x,$ then $f(f(x)) = -x,$ then $f(f(f(x))) = -x$ Commented Apr 10, 2017 at 1:44
• So ... where is it proved that the sequence is bounded? Commented Nov 30, 2021 at 23:58
• @Salcio appears I decided the sequence was unbounded. Commented Dec 1, 2021 at 0:05