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Let us consider the tent map: $f [0,1] \rightarrow [0,1]$ where $f(x) = 2x$ if $0\leq x \leq \frac{1}{2}$ and $f(x) = 2(1-x)$ if $\frac{1}{2}\leq x \leq 1$.

Show that $x$ is a periodic point IFF it is a rational number of the form $\frac {m}{p}$ where $m$ is even and $p$ is odd.

I have a proof of this here, but it seems overly complicated with Euler's functions and such. So I was wondering whether any of you can provide a simpler proof that is more natural and avoids using specialized functions.

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    $\begingroup$ Well, this proof you have there is not that awful. Not sure it can be made significantly easier. $\endgroup$ – Julien Mar 31 '13 at 20:23
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    $\begingroup$ As mentioned before, there is not a whiff of real-analysis in the proof of this result, which boils down to showing that every $(2m)/(2n-1)$ can be written as some $k/(2^i-1)$. $\endgroup$ – Did Mar 31 '13 at 21:04
  • $\begingroup$ As I mentioned before, this IS from a real analysis text book which deals slightly with Dynamical Systems. If you have a chauvinist viewpoint of what REAL ANALYSIS is, please keep it to yourself, and I say it very respectfully. I just do not agree with you, and will keep on doing so. So if you would kindly not drop in to every post I make, it will be much appreciated. $\endgroup$ – user43901 Mar 31 '13 at 23:32
  • $\begingroup$ @julien: Can we do without Euler's function? Also, how did the author arrive at the periodic points being of the form $x- \frac {i}{2^{n-1}}$, for I could not see it. I got $ x- \frac{i}{2^n-1}$. Any ideas on that? I just wish I had a more fleshed out version of it. I can follow the logic, but am failing to produce the algebraic expressions. $\endgroup$ – user43901 Mar 31 '13 at 23:36
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    $\begingroup$ @user43901 You seem to be mistaken about the ways the site works. To keep it simple, let me mention this: the page is not yours, in particular comments on a post are not only, are not even mainly, addressed to the author of the question. In the present case, I was expecting you to yell and to avoid reading what I actually wrote (you do not even pretend to address it)--but all this does not even concern me. Hence the comment. (Unrelated: if you want a comment of yours to be signaled to the user @nameofuser, insert @nameofuser at the beginning of said comment. If you don't, don't.) $\endgroup$ – Did Apr 1 '13 at 17:00
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There is a shorter proof if you are willing to concede certain facts about binary expansions.

Let $D\colon[0,1]\to[0,1]$ be the doubling map: $$ D(x) \;=\; \begin{cases}2x & \text{if }x < 1/2, \\ 2x-1 & \text{if }x \geq 1/2.\end{cases} $$ Then $D$ acts as a bitwise left shift on the binary digits of a number. In particular, a point is periodic under $D$ if and only if its binary expansion is repeating, i.e. if and only if it is a rational number with odd denominator. (We will assume this bit of knowledge about binary expansions.)

Now, observe that $T\circ D = T \circ T$, where $T$ is the tent map. (This is a semiconjugacy between $T$ and $D$.) Then $T\circ D^n = T^{n+1}$ for all $n$, which makes it easy to classify periodic points for $T$.

Theorem. Let $p\in[0,1]$. Then $p$ is periodic under $T$ if and only if $p$ is a rational number with even numerator and odd denominator.

Proof: Suppose first that $p$ is a fraction with odd denominator and even numerator. Then $p/2$ is a fraction with odd denominator, so $D^n(p/2) = p/2$ for some $n$. Then $$ T^n(p) \;=\; T^{n+1}(p/2) \;=\; T(D^n(p/2)) \;=\; T(p/2) \;=\; p $$ so $p$ is periodic under $T$.

For the converse, suppose that $p$ is periodic under $T$. Then $T^n(p) = p$ for some $n$. Since $D(p/2) = p$, it follows that $$ T(D^n(p/2)) = T(D^{n-1}(p)) = T^n(p) = p = T(p/2). $$ Therefore, either $D^n(p/2) = p/2$, or $D^n(p/2) = 1-(p/2)$.

  • If $D^n(p/2) = p/2$, then $p/2$ is periodic under $D$. Thus $p/2$ is a fraction with odd denominator, so $p$ is a fraction with odd denominator and even numerator.

  • If $D^n(p/2) = 1-(p/2)$, then $D^{2n}(p/2) = p/2$ (since $D(1-x) = 1-D(x)$ for all $x$), so again $p/2$ is periodic under $D$, and hence $p$ is a fraction with odd denominator and even numerator.

Q.E.D.

Of course, this sweeps most of the number theory details under the rug of binary expansions, but even if you aren't content to assume the required fact, it's probably slightly easier to sort through the number theory of binary expansions and then apply the results to tent map than to work through everything at once.

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