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On Two Exponents of Approximation Related to a Real Number and Its Square

Published online by Cambridge University Press:  20 November 2018

Damien Roy*
Affiliation:
Département de Mathématiques, Université d'Ottawa, 585 King Edward, Ottawa, ON, K1N 6N5 e-mail: [email protected]
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Abstract

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For each real number $\xi$, let $\widehat{{{\lambda }_{2}}}\left( \xi \right)$ denote the supremum of all real numbers $\text{ }\!\!\lambda\!\!\text{ }$ such that, for each sufficiently large $X$ , the inequalities $\left| {{x}_{0}} \right|\,\le \,X,\,\left| {{x}_{0}}\xi \,-\,{{x}_{1}} \right|\,\le \,{{X}^{-\lambda \text{ }}}$ and $\left| {{x}_{0}}{{\xi }^{2}}\,-\,{{x}_{2}} \right|\,\le \,{{X}^{-\lambda \text{ }}}$ admit a solution in integers ${{x}_{0}},\,{{x}_{1}}$ and ${{x}_{2}}$ not all zero, and let $\widehat{{{\omega }_{2}}}\left( \xi \right)$ denote the supremum of all real numbers $\omega $ such that, for each sufficiently large $X$, the dual inequalities $\left| {{x}_{0}}\,+\,{{x}_{1}}\xi \,+\,{{x}_{2}}{{\xi }^{2}} \right|\,\le \,{{X}^{-\omega }}$, $\left| {{x}_{1}} \right|\,\le \,X$ and $\left| {{x}_{2}} \right|\,\le \,X$ admit a solution in integers ${{x}_{0}},\,{{x}_{1}}$ and ${{x}_{2}}$ not all zero. Answering a question of Y. Bugeaud and M. Laurent, we show that the exponents $\widehat{{{\lambda }_{2}}}\left( \xi \right)$ where $\xi$ ranges through all real numbers with $[\mathbb{Q}(\xi )\,:\mathbb{Q}]\,>\,2$ form a dense subset of the interval $\left[ 1/2,\,\left( \sqrt{5}\,-\,1 \right)/2 \right]$ while, for the same values of $\xi$, the dual exponents $\widehat{{{\omega }_{2}}}\left( \xi \right)$ form a dense subset of $\left[ 2,\,\left( \sqrt{5}\,+\,3 \right)/2 \right]$. Part of the proof rests on a result of V. Jarník showing that $\widehat{{{\lambda }_{2}}}\left( \xi \right)=1-{{\hat{\omega }}_{2}}{{\left( \xi \right)}^{-1}}$ for any real number $\xi$ with $[\mathbb{Q}(\xi )\,:\mathbb{Q}]\,>\,2$.

Keywords

Type
Research Article
Copyright
Copyright © Canadian Mathematical Society 2007

References

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