# Are Periodic Orbit all Iterations?

We have,

If $x$ is a periodic point of a continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$ and the period is $k$ i.e. $f^k(x)=x$ but $f^n(x)\not=x, \forall n: 0<n<k$.

(The statement up to this is slightly different in the text. This is my generalization but the main idea is same.)

Next the text says,

The periodic orbit of $x$ is $O(x)=\{x,f(x),f^1(x),f^2(x),\dots,f^{k-1}(x)\}$.

My question is, does it not have to be true that $f(x)=x$ to be periodic orbit? if $f^1(x)\not=x$ is it still in the periodic orbit?

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If $f(x)=x$ then the orbit of $x$ is $O(x)=\{x\}$. Why would you need $f(x)=x$ in general? Can you explain your reasoning? –  jathd Jan 18 '13 at 15:15

I am guessing $f^k(x)$ means that you are iterating the function $f$ $k$ times. If that is the case, and $k=1$, then the function is of period one, actually the fundamental or the minimum period is one, then it is also 2,3,... periodic. Nothing would be wrong with the definition (saying that if $x=f(x)$ and you say is 1-periodic).
Next $O(x)$ refers precisely to the orbit, this is, the points in one fundamental period, that's why it says: $O(x)=\left\{x,f(x),f1(x),f^2(x),…,f^{k−1}(x)\right\}$, these are all the points within one period, and since, by definition, $x=f^k(x)$, then $f^k(x)$ is not included in the orbit.
Well it depends on wether your dynamical system is discrete or continuous, although the idea is exactly the same. In your question, I guess you refer to a discrete one since you are iterating the function $f$. Fixed points are "equilibria" and are also periodic points, of period one. The orbit is "the result of applying the iteration" ... orbits are not necesarily periodic, just in case you find a function such that your definition applies, then you call it "periodic orbit", if not, just orbit ... there are some other types of orbits by the way. –  user58533 Jan 18 '13 at 15:40
In general the orbit of $x$ is the set of all points traced when starting at $x$ and repeatedly applying $f$, $O(x)=\{f^n(x)\mid n\in\mathbb N_0\}$. However, for a periodic point, this infinite set becomes finite (of size precisely the period length) –  Hagen von Eitzen Jan 18 '13 at 15:44