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I'm new to dynamical systems, but I've stumbled onto an interesting system and I'm wondering if the orbit is bounded or not for a particular initial condition. The function in question is:
$f(x) = \begin{cases} \sqrt{x} &\quad\text{if }x \geq 2\\ \frac{x}{2-x} &\quad \text{if } x < 2 \end{cases} $
and initial condition $x_0 = \frac{1 + \sqrt{5}}{2}$. Does anyone know if this has a bounded or dense orbit?

Thank you!

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  • $\begingroup$ Do you mean if $x \ge 2$? $\endgroup$
    – Robert Israel
    Feb 19, 2018 at 19:42
  • $\begingroup$ Math experiment done with Maple produces first 21 terms of the sequence $$\left[{\frac {1/2\,\sqrt {5}+1/2}{-1/2\,\sqrt {5}+3/2}},\sqrt {2},{\frac { \sqrt {2}}{2-\sqrt {2}}},\sqrt {2},{\frac {\sqrt {2}}{2-\sqrt {2}}}, \sqrt {2},{\frac {\sqrt {2}}{2-\sqrt {2}}},\sqrt {2},{\frac {\sqrt {2} }{2-\sqrt {2}}},\sqrt {2},{\frac {\sqrt {2}}{2-\sqrt {2}}},\sqrt {2},{ \frac {\sqrt {2}}{2-\sqrt {2}}},\sqrt {2},{\frac {\sqrt {2}}{2-\sqrt { 2}}},\sqrt {2},{\frac {\sqrt {2}}{2-\sqrt {2}}},\sqrt {2},{\frac { \sqrt {2}}{2-\sqrt {2}}},\sqrt {2}\right] $$ $\endgroup$
    – user64494
    Feb 19, 2018 at 20:31
  • $\begingroup$ @user64494 The first term simplified is 2 + sqrt(5), so the second term would need to be the square root of that and not sqrt(2). $\endgroup$
    – Corey DeGagne
    Feb 19, 2018 at 21:46
  • $\begingroup$ @user64494 This sequence cannot have dense orbit as the function $\frac{x}{2-x}$ has zero as its attracting fixed point.Maybe you had in mind dense in the axis $[\sqrt{2},+\infty)$? $\endgroup$
    – Taras Banakh
    Feb 20, 2018 at 7:22
  • $\begingroup$ You are right. The adjusted result is $$[4.23606797749986, 2.05817102727151, 1.43463271511266, 2.53752340020618, 1.59296057710358, 3.91352897900738, 1.97826413277079, 91.0138119592614, 9.54011593007451, 3.08870780911282, -2.83704018953422, -.586524006079720, -.226761477837079, -.101834650946692, -0.484503625919501e-1, -0.236522024046728e-1, -0.116878791605432e-1, -0.580998637095753e-2, -0.289657864425599e-2, -0.144619481362171e-2] $$ up to 15 digits. In my previous comment I started from a different point $x_0$ than in the question: explanation, but not justification. $\endgroup$
    – user64494
    Feb 20, 2018 at 12:24

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