F. Amoroso, M. Omarjee
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引用次数: 0

摘要

设α为无理数;SN (α):=(−1)[α] +(−1)[2α] +···+(−1)[Nα]的性质取决于α/2的连分式展开。由于2/2 \sqrt 2/2的连分式展开式具有有界的部分商,因此SN(2)=O(log({N)) S_N}\left ({\sqrt 2 }\right)=O \left ({\log\left (N \right) }\right),此界是最佳可能。e的连分式展开的偏商增长缓慢,因此SN(2e)=O(log(N)2log log(N)2) {S_N}\left ({2e}\right)=O \left ({{{\log{{\left (N \right)}^2}}\over{\log \, \log{{\left (N \right)}^2}}}}\right),再次为最佳可能。e/2的连分式展开式的部分商的行为与e的相似。令人惊讶的是SN(e)=O(log(N)log log(N)) 1188。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
A curiosity About (−1)[e] +(−1)[2e] + ··· +(−1)[Ne]
Abstract Let α be an irrational real number; the behaviour of the sum SN (α):= (−1)[α] +(−1)[2α] + ··· +(−1)[Nα] depends on the continued fraction expansion of α/2. Since the continued fraction expansion of 2/2 \sqrt 2 /2 has bounded partial quotients, SN(2)=O(log(N)) {S_N}\left( {\sqrt 2 } \right) = O\left( {\log \left( N \right)} \right) and this bound is best possible. The partial quotients of the continued fraction expansion of e grow slowly and thus SN(2e)=O(log(N)2log log(N)2) {S_N}\left( {2e} \right) = O\left( {{{\log {{\left( N \right)}^2}} \over {\log \,\log {{\left( N \right)}^2}}}} \right) , again best possible. The partial quotients of the continued fraction expansion of e/2 behave similarly as those of e. Surprisingly enough SN(e)=O(log(N)log log(N)) 1188 .
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