# 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?

#### Solutions Collecting From Web of "Periodic orbits"

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.

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