中文

若$(A+A)/(A+A)$较小则比率集较大

组合数学 2017-05-17 v2 数论

摘要

本文中,我们考虑和积问题,即对任意有限实数集AA,获得集合A+AA+A:={a+bc+d:a,b,c,dA,c+d0},\frac{A+A}{A+A}:=\left \{ \frac{a+b}{c+d} : a,b,c,d \in A, c+d \neq 0 \right\},大小的下界。主要结果是界A+AA+AA2+225A:A125logA,\left| \frac{A+A}{A+A} \right| \gg \frac{|A|^{2+\frac{2}{25}}}{|A:A|^{\frac{1}{25}}\log |A|},其中A:AA:A表示AA的比率集。在比率集大小关于A|A|为次二次的条件下,这改进了Balog和作者(arXiv:1402.5775)的结果。即,我们建立了蕴含关系A+AA+AA2A:AA2log25A.\left| \frac{A+A}{A+A} \right| \ll |A|^{2} \Rightarrow |A:A| \gg \frac{ |A|^2}{\log^{25}|A|} . 这一极值结果回答了作者与Zhelezov近期论文(arXiv:1410.1156)中类似的一些猜想的问题。

关键词

引用

@article{arxiv.1507.07672,
  title  = {If $(A+A)/(A+A)$ is small then the ratio set is large},
  author = {Oliver Roche-Newton},
  journal= {arXiv preprint arXiv:1507.07672},
  year   = {2017}
}

备注

In this version, Lemma 3.2 has been improved, and the new version of Lemma 3.2 is tight up to multiplicative constants. This results in a small improvement to the main result of the paper. To appear in JLMS. With thanks to Noga Alon, for providing the proof of the new and improved Lemma 3.2 via a private communication