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I am trying to find the periodic points for $x^3-x$ and find out whether or not they are repelling or attracting. I have been trying to use orbit diagrams to figure this out.

I know there are three fixed points $0,\sqrt{2}, -\sqrt{2}$

I can use the orbit diagram to find if are repelling or attracting. (I believe they are 0 is attracting and the other two are repelling)

But I am unsure how to tell if there are other periodic points.

I am also having trouble finding their stable set. I believe for the fixed points there is nothing in their stable set except for themselves. Since the spiral at 0 will only reach 0 at the value of 0

zaask billard
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  • Just check the value of the derivative at those fixed points and be sure to consider the possibility that $|f'(x_0)|=1$ so that the fixed point is neutral. Of course, a cubic can have at most three fixed point, since the equation $f(x)=x$ can have at most three solutions. – Mark McClure Mar 17 '17 at 13:01
  • What is a periodic point? Do you mean a fixed point as in @Mark McClure's comment? – mlc Mar 17 '17 at 13:07
  • @mic Ahh - I misread thanks! – Mark McClure Mar 17 '17 at 13:10
  • Note that the sequence $(x_n)$ of iterates starting from a given $x_0$ is such that $|x_{n+1}|=|x_n|\cdot|1-x_n^2|$. Thus, if $|x_0|>\sqrt2$, then $|x_n|\to\infty$. If $1<|x_0|<\sqrt2$, then there exists some $n$ such that $|x_n|\leqslant1$. If $|x_0|<1$, then $|x_n|\to0$. Thus, there is no periodic point except the three fixed points. – Did Mar 17 '17 at 13:10
  • Ok, I see. That makes sense. 0 is a neutral point. While $\sqrt{2}, -\sqrt{2}$ are repelling. and the way to get the periodic points is too get the equation in a form to see the general pattern of the iteration, which in this case is none. That makes sense. Thank you guys very much – zaask billard Mar 17 '17 at 13:21

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