Let $\alpha, \beta \in \mathbb{R}$. Let $\{x\}$ denote the fractional part of $x$ and let $\|x\| = \min(\{x\}, 1-\{x\})$.

If we assume that $\alpha$ is irrational, then there exists an increasing sequence of integers $(n_k)_{k \in \mathbb{N}}$ such that $\|n_k \alpha - \beta\| \to 0$. Is it possible that the rate of convergence is exponential, that is, that there exists some $\eta > 1$ such that $\eta^{n_k} \|n_k \alpha - \beta\| \to 0$?

I believe that the fact that the sequence $\{n \alpha\}_{n \in \mathbb{N}}$ is equidistributed implies that if this is true for some $\eta > 1$, then it must be true for every $\eta > 1$, so this seems to be a rather strong condition.

If the rate may be exponential for general $\alpha$ and $\beta$, what about the case when $e^{2\pi i \alpha}$ and $e^{2\pi i \beta}$ are algebraic?


2 Answers 2


For the algebraic case a lot is known. There is a deep result of Gelfond (see A. O. Gelfond, Transcendental and Algebraic Numbers, Dover, New York, 1960) that if $\lambda$ is an algebraic number of absolute value 1, then for every $\epsilon>0$ there is a constant $C>0$ such that $$|\lambda^n-1|>Ce^{-n\epsilon}$$ for all $n\ge1$. More recent work of Feldman (see Chapter 9 in Stolarsky's book Algebraic Numbers and Diophantine Approximation, Dekker, 1974) actually shows that there are effectively computable numbers $C=C(\lambda)>0$ and $N=N(\lambda)\ge1$ such that $$|\lambda^n-1|>Cn^{-N}$$ for all $n\ge1$.

These diophantine results are very useful while investigating the dynamical properties of maps defined algebraically, e.g. automorphisms of tori. See for example my paper "Dynamical Properties of Quasihyperbolic Toral Automorphisms", Ergod. Th. Dynam. Sys. 2 (1982), 49--68.

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    $\begingroup$ This answer has been accepted and all, but isn't it completely different from the original question? Here, powers of $\lambda$; there, multiples of $\alpha$. $\endgroup$ Nov 15, 2014 at 22:49
  • $\begingroup$ @KevinO'Bryant: multiplicative group of unit complex numbers is isomorphic to the additive group of reals modulo 1, so it's exactly the same thing, isn't it? $\endgroup$
    – tomasz
    Nov 16, 2014 at 1:45
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    $\begingroup$ @KevinO'Bryant: Yes, powers of $\lambda = e^{2 \pi i\alpha}$ are the same as integral multiples of $\alpha$ (mod 1), so this was precisely the question asked. $\endgroup$ Nov 16, 2014 at 6:17
  • $\begingroup$ Ah, I see. But still a little weird as distance isn't respected by the isomorphism (but boundedly so), and certainly algebraic $\alpha$ is not the same as algebraic $\lambda$ (but that wasn't part of the original question). $\endgroup$ Nov 16, 2014 at 21:43
  • $\begingroup$ Note that the algebraic case is part of the original question, and it asks what happens when $e^{2\pi i \alpha}$ is algebraic, not $\alpha$. The distance is almost respected in the sense that the rate of convergence is exponential in one case if and only it's exponential for the other case. $\endgroup$ Nov 17, 2014 at 4:07

Certainly it is possible for the rate to be exponential: for example, if we define a sequence $(d_n)$ by $d_1:=2$, $d_{n+1}:=2^{d_n}$ then the number $$\alpha:=\sum_{n=1}^\infty \frac{1}{d_n}$$ satisfies $$\{d_n\alpha\}=\sum_{k=n+1}^\infty \frac{1}{d_k} \in \left(\frac{1}{d_{n+1}},\frac{2}{d_{n+1}}\right)$$ and we have $d_{n+1}=2^{d_n}$ so the size of this fractional part is exponentially small relative to $d_n$. In the case $\beta=0$ you can construct more general examples using the theory of continued fractions.

  • $\begingroup$ Thank you. And what about the algebraic case? It seems Baker's theorem might be useful, but I'm not sure. $\endgroup$ Nov 15, 2014 at 19:22

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