中文

黎曼假设成立的概率等于 1

综合数学 2018-04-27 v2

摘要

PP 为所有素数的集合,q1,q2,,qmP{q_1},{q_2}, \cdots ,{q_m} \in PPkP_kPP 中按大小升序排列的第 kk (k=1,2,m)(k = 1,2, \cdots m) 个元素,α1,α2,,αm{\alpha _1},{\alpha _2}, \cdots ,{\alpha _m} 为正整数,β1,β2,,βm{\beta _1},{\beta _2}, \cdots ,{\beta _m}α1,α2,,αm{\alpha _1},{\alpha _2}, \cdots ,{\alpha _m} 的一个满足 β1β2βm{\beta _1} \ge {\beta _2} \ge \cdots \ge {\beta _m} 的排列。本文给出以下结果: 该不等式成立:eγloglogk=1mqkαkk=1mqk1qkαkqk1eγloglogk=1mpkβkk=1mpk1pkβkpk1{e^\gamma }\log \log \prod\limits_{k = 1}^m {q_k^{{\alpha _k}}} - \prod\limits_{k = 1}^m {\frac{{{q_k} - {\textstyle{1 \over {q_k^{{\alpha _k}}}}}}}{{{q_k} - 1}}} \ge {e^\gamma }\log \log \prod\limits_{k = 1}^m {p_k^{{\beta _k}}} - \prod\limits_{k = 1}^m {\frac{{{p_k} - {\textstyle{1 \over {p_k^{{\beta _k}}}}}}}{{{p_k} - 1}}}。 若 n=k=1mpkβk=(k=1mpk)1+εm(n)n = \prod\limits_{k = 1}^m {p_k^{{\beta _k}}}= {\left( {\prod\limits_{k = 1}^m {{p_k}} } \right)^{1 + {\varepsilon _m}(n)}}limmεm(n)>0\mathop {\lim }\limits_{m \to \infty } {\varepsilon _m}(n) > 0limmεm(n)=+\mathop {\lim }\limits_{m \to \infty } {\varepsilon _m}(n) = + \infty,则 limm(eγnloglognσ(n))>0\mathop {\lim }\limits_{m \to \infty } ({e^\gamma }n\log \log n - \sigma (n)) > 0。其中 {βk}\{ {\beta _k}\} 为一序列,βkN{\beta _k} \in Nβ1β2βm{\beta _1} \ge {\beta _2} \ge \cdots \ge {\beta _m}σ(n)=dnd\sigma (n) = \sum\limits_{\left. d \right|n} dγ\gamma 为欧拉常数。 黎曼假设成立的概率等于 1。此外,还给出了当 limmεm(n)=0\mathop {\lim }\limits_{m \to \infty } {\varepsilon _m}(n) = 0 时的两个结果。

关键词

引用

@article{arxiv.1609.07555,
  title  = {The probability of Riemann's hypothesis being true is equal to 1},
  author = {Yuyang Zhu},
  journal= {arXiv preprint arXiv:1609.07555},
  year   = {2018}
}

备注

25 pages