Periodic Points and One Dimensional Maps Homework Help

Question

Let f be the tripling map $f(x)=3x\mod(1)$. I need to make a table that includes the following for $n\le6$: number of points in Fix($f^n$), number of points in Fix($f^n$) of lower period, number of points in Per(n,f), and number of period-n orbits. How do I find each of these things? Can someone help me make the map for each of these n's? My professor was unable to help me with this question earlier today and am in pure desperation.

Answer 1

A little geometric thinking goes a long way. Here are the plots of $y=f^k(x)$ together with $y=x$ for $k=1,2,3$. The points of intersection represent the fixed points. Do you see the relationship between the number of iterates and the number of fixed points at level $k$?

I should emphasize that, while I generated this picture with a computer, it's fairly easy to come up with, if you understand what's going on. Furthermore, once you understand it, it's easy to prove that the pattern you see persists.

Answer 2

Rather than address the tripling map directly, let me discuss an analogue which may clarify my comment to the OP; I'll leave it to the reader to see how to apply this to the case of the OP.

Consider the $10$-fold map $f(x)=10x\mod 1$. If we apply it to $x=\frac{5}{36}=0.13\overline{8}$, then $f(1.3\overline{8})=0.3\overline{8}=\frac{7}{18}$. More generally, we have

$$x=0.b_1b_2b_3\ldots\mapsto f(x)=0.b_2b_3b_4\ldots$$

so that $f(x)$ shifts the decimal once to the right and drops the integer part.The set of fixed points of this map are readily seen to be $\{0.\overline{0},0.\overline{1},0.\overline{2},\ldots\overline 0.\overline{8}\}=\{0,\dfrac{1}{9},\dfrac{2}{9},\ldots, \dfrac{8}{9}\}$. By similar logic, the 2-cycles are $0.\overline{b_1b_2}$ for $b_1,b_2\in\{0,1,2,\ldots 9\}$ with $b_1\neq b_2$. The generalization to $n$-cycles follows in the obvious way.