Exponents for three-dimensional simultaneous Diophantine approximations
Number Theory
2010-12-09 v2
Abstract
Let Θ = ( θ 1 , θ 2 , θ 3 ) ∈ R 3 \Theta = (\theta_1,\theta_2,\theta_3)\in \mathbb{R}^3 Θ = ( θ 1 , θ 2 , θ 3 ) ∈ R 3 . Suppose that 1 , θ 1 , θ 2 , θ 3 1,\theta_1,\theta_2,\theta_3 1 , θ 1 , θ 2 , θ 3 are linearly independent over Z \mathbb{Z} Z . For Diophantine exponents α ( Θ ) = sup { γ > 0 : lim sup t → + ∞ t γ ψ Θ ( t ) < + ∞ } , \alpha(\Theta) = \sup \{\gamma >0:\,\,\, \limsup_{t\to +\infty} t^\gamma \psi_\Theta (t) <+\infty \} , α ( Θ ) = sup { γ > 0 : t → + ∞ lim sup t γ ψ Θ ( t ) < + ∞ } , β ( Θ ) = sup { γ > 0 : lim inf t → + ∞ t γ ψ Θ ( t ) < + ∞ } \beta(\Theta) = \sup \{\gamma >0:\,\,\, \liminf_{t\to +\infty} t^\gamma \psi_\Theta (t) <+\infty\} β ( Θ ) = sup { γ > 0 : t → + ∞ lim inf t γ ψ Θ ( t ) < + ∞ } we prove β ( Θ ) ≥ 1 / 2 ( α ( Θ ) / 1 − α ( Θ ) + α ( Θ ) / 1 − α ( Θ ) ) 2 + 4 α ( Θ ) / 1 − α ( Θ ) ) α ( Θ ) \beta (\Theta) \ge {1/2} ({\alpha (\Theta)}/{1-\alpha(\Theta)} +\sqrt{{\alpha(\Theta)}/{1-\alpha(\Theta)})^2 +{4\alpha(\Theta)}/{1-\alpha(\Theta)}}) \alpha (\Theta) β ( Θ ) ≥ 1/2 ( α ( Θ ) / 1 − α ( Θ ) + α ( Θ ) / 1 − α ( Θ ) ) 2 + 4 α ( Θ ) / 1 − α ( Θ ) ) α ( Θ )
Cite
@article{arxiv.1009.0987,
title = {Exponents for three-dimensional simultaneous Diophantine approximations},
author = {Nikolay Moshchevitin},
journal= {arXiv preprint arXiv:1009.0987},
year = {2010}
}
Comments
8 pages, correction of misprints, submitted to Czechoslovak Mathematical Journal