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Let $a_n=\left(\frac{1+\sqrt{5}}{2}\right)^n.$ For a real number $r$, denote by $\langle r\rangle$ the fractional part of $r$.

Why is the sequence $\langle a_n\rangle$ not equidistributed in $[0,1]$?

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    I am wondering why $\{ \mbox{frac} (x^n)\}$ is not equidistributed if $\phi^n\to 0\ (\mbox{mod}\ 1)$ holds?2012-11-15

2 Answers 2

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Let \phi'=(1-\sqrt 5)/2 denote the Galois conjugate of the golden mean $\phi$. Then \phi^n+\phi'^{n} is an integer for every $n\in\mathbb N$, i.e. \phi^n+\phi'^{n}\equiv 0\ (\mbox{mod}\ 1).

But |\phi'|<1, so \phi'^{n}\to 0. This implies that $\phi^n\to 0\ (\mbox{mod}\ 1)$.

The property that the sequence $\{ \mbox{frac} (x^n)\}$ is not equidistributed is shared by other Pisot numbers. There is quite a lot of research publications devoted to them.

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If you just try it, the fractional part becomes very close to 0 or 1 quickly. This is because $\phi ^2=\phi +1$.