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Why are numbers of the form $\frac{k}{3^n}$ for $0<k<3^n$ non recurrent under times $3$ map?

I am sure there is an easy answer to this question but I just cannot see it.

If you do not know, $x$ is a recurrent point if there is a subsequence $n_k$ s.t $f^{n_k}(x)\to x$. In this case $f(x)=3x$ and we are on $S^1$

2132123
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1 Answers1

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Given a natural number $k$, $\frac{k}{3^n}$ will map to $\frac{k}{3^{n-1}}$. So, after $n$ iterations of the map, this takes us to an integer, hence $0$ modulo $1$, which is a fixed point of this map (the only other such being $\frac12$).

To sketch out an explicit example:

$$\frac{13}{27}\to \frac{13}9=\frac49\to\frac43=\frac13\to1=0\to0\to0\to\ldots$$

Note how the denominator decreases each time.

RavenclawPrefect
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