# Periodic orbits

Let $x\in\mathbb R$ be a periodic point with lenght 2 of the recursion $x_{n+1}=f(x_n)$

My book about Dynamic systems says that this recursion has a fixed point now, because a periodic point with length 2 exists.

May you could help me to prove this statement?

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What does length 2 mean? – Amr Nov 9 '12 at 0:49
Does it mean that there exists n such that f(f(x n))=x n – Amr Nov 9 '12 at 0:50
Yes, exactly. $f(f(x_2))=x_2$ – Montaigne Nov 9 '12 at 0:51

In general if $f$ is continuous, and there exists an integer k>1 and a real number $a$ such that $f^k(a)=a$ then there exists a real number $r$ such that $f(r)=r$.

Proof: Let $g(x)=f(x)-x$. Since: $g(a)+g(f(a))+g(f^2(a))+...+g(f^{k-1}(a))=f^k(a)-a=0$, therefore either one of the numbers $g(a),g(f(a)),g(f^2(a)),...,g(f^{k-1}(a))$ is zero (in this case we are done) or one of these numbers is positive and another number is negative. Assume WLOG that the second case holds. Let $g(f^i(a))g(f^j(a))<0$ (it means that they have different signs). Since, $g$ has a sign change in the interval $[\min(f^i(a),f^j(a)),\max(f^i(a),f^j(a))]$, thus $g$ has a root.

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This works because g(a)=-g(f(a)). Thus if a is different from f(a) from a, we know that g(x) has a sign change on the interval [min{a,f(a)},max{a,f(a)}]. Thus g has a root r in that interval. Therefore g(r)=0, hence f(r)=r – Amr Nov 9 '12 at 0:57
Thank you for your answer. Why do we know that g(a)=-g(f(a)) ? – Montaigne Nov 9 '12 at 1:01
Because a-f(a)=-(f(a)-a) – Amr Nov 9 '12 at 1:02
Thanks, now everything is clear. – Montaigne Nov 9 '12 at 1:02

Also, if you like hitting flies with sledgehammers, this directly follows from Sharkovskii's theorem.

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This is a nice theorem. – Amr Dec 29 '12 at 15:47