AI 中文总结
研究针对哈代函数\(Z(t)\)，通过利用广义三次高斯和子序列给出渐近表达式，采用类似二次和的递归方案快速计算，得到运算量降低且精度高的计算算法，适用于\(t\in[10^{23},10^{35}]\) 。
AI 中文摘要
2011年G. A. Hiary为哈代函数\(Z(t)\)设计了一种计算算法，运算量仅为\(O(t^{1/3}(\log(t))^\kappa)\)。与使用经典黎曼 - 西格尔公式计算\(Z(t)\)所需的\(O(\sqrt{t})\)运算量相比有优势。其方法是将黎曼 - 西格尔公式细分为不同长度\(N\)的二次高斯/指数和序列，可通过标准递归方案快速计算。最近主要作者开发了类似算法，运算量为\(O((t/\varepsilon_t)^{1/3}(\log(t))^2)\)且相对误差精确到\(\epsilon_t\)。本文对这些想法进行了重大扩展。主要理论结果是用长度逐渐增加的广义\(m\)阶高斯和的子序列给出\(Z(t)\)的渐近表达式。计算上主要关注三次高斯和公式，通过类似二次和的递归方案可快速计算。最终得到\(Z(t)\)的计算算法，对于\(t\in[10^{23},10^{35}]\)，运算量降低为\(O\bigl((t/\varepsilon_t)^{[.25,.3]}(\log t)^{2 + o(1)}\bigr)\)且精度高。样本计算为这些结果提供了实际支持。
英文摘要
In 2011 G. A. Hiary devised a computational algorithm for the Hardy function $Z(t)$, requiring just $O(t^{1/3} (\log(t))^κ)$ operations. This compares to $O(\sqrt t)$ operations necessary for computing $Z(t)$ using the classical Riemann-Siegel formula. The methodology involved the sub-division of the Riemann-Siegel formula into sequences of quadratic Gauss/exponential sums of various lengths $N$. Such sums can be computed rapidly, in order $\log(N)$ operations, using standard recursive schemes. More recently, the principal author developed a similar algorithm with an $O((t/\varepsilon_t)^{1/3} (\log(t))^2)$ operational count, accurate to $ε_t$ in the relative error. Although constructively analogous, the sub-division into quadratic sums was applied to a different asymptotic formula for $Z(t)$, giving the new algorithm an original formulation. This paper presents a significant extension of these ideas. The main theoretical result is an asymptotic expression for $Z(t)$ in terms of sub-sequences of generalised, $m^{\rm th}$-order, Gauss sums of progressively increasing length. Computationally, the main focus falls upon the cubic Gauss sum formulation. The particular parameterisation of these cubic sums makes them amenable to rapid computation, utilising a recursive scheme similar to those implemented for quadratic sums. The net result is a computational algorithm for $Z(t)$ with a reduced $O\bigl((t/\varepsilon_t)^{[.25,.3]}(\log t)^{2+o(1)}\bigr)$ operational count for $t \in [10^{23},10^{35}]$, to high accuracy. Sample computations lend practical support to these findings.
Comments66 pages, 3 figures, 5 tables
