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Let $f$ be an interval exchange transformation of $[0,1]$. Is there an explicit formula giving $f^k(0)$ in function of $k$?

If not, are there particular cases where this formula is simple? (except when $f$ is a rotation).

I am mostly interested in the case when $f$ is periodic, and with the permutation ABC -> CBA where A,B,C are intervals of different lengths (with the interval B being translated).

For this particular interval exchange, we have: $f^k(0)=k_1 \lambda_1+k_2 \lambda_2 +k_3 \lambda_3$. where the $\lambda_i$ are the translation vector of $f$ on each of the three intervals. Is it possible to have an estimate the $k_i$?

share|cite|improve this question
What's an interval exchange transformation? – Felix Goldberg Feb 28 '14 at 13:56
What do you mean $f$ is periodic? that $f^n(x)=x$ for all $x$ for some $n$? – Anthony Quas Feb 28 '14 at 17:34
that there exists $n$ such that for any $x$, $f^n(x)=x$. it implies that the lengths of intervals are rational. – user8991 Feb 28 '14 at 23:28

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