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Let $\alpha \in \mathbb{S}^1$ be an algebraic number with $\mathop{\mathrm{arg}}(\alpha)/\pi$ irrational. Is it possible for the rotation by $\alpha$ to converge exponentially fast to a $\xi \in \mathbb{S}^1$? That is, may there exist $\xi \in \mathbb{S}^1$ and $A > 1$ such that $\liminf_{n \to \infty} A^n |\alpha^n - \xi| = 0$?

This answer cites a result of Feldman that implies that if $\xi$ is algebraic and $\xi \neq \alpha^q$ for every $q \in \mathbb{Q}$, then $|n \log \alpha - \log \xi| > C n^{-N}$, where $C > 0$ and $N \geq 1$ are effectively computable numbers depending only on $\alpha$, showing that the rate cannot be exponential in this case.

What about the case where $\xi$ is not algebraic and the case where $\xi = \alpha^q$ for some $q \in \mathbb{Q}$?

Thanks in advance.

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  • $\begingroup$ You can reduce the case of different $\xi$ to the case $\xi=1$. If $\alpha^{n_i}$ converges exponentially to $\xi$, then $\alpha^{n_i-n_{i-1}}$ converges exponentially to 1. $\endgroup$ May 31, 2015 at 6:17
  • $\begingroup$ No, you only have $|\alpha^{n_i-n_{i-1}} - 1|<A^{-n_{i-1}}$. Since it might be that the sequence of the $n_i$ is extremely sparse, this bound is a lot weaker than the required bound $A^{-(n_i-n_{i-1})}$. $\endgroup$ May 31, 2015 at 13:01

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