整数字母表上的广义黄金比例
Generalised golden ratios over integer alphabets
摘要：
这是一个众所周知的结果，对于$β\in(1,\frac{1+\sqrt{5}}{2})$和$x\in(0,\frac{1}{β-1})$存在不可数多的$(ε_{i})_{i=1}^{\infty}\in {0,1}^{\mathbb{N}}$，使$x=\sum_{i=1}^{\infty}ε_{i}β^{-i}.$当$β\in(\frac{1+\sqrt{5}}{2},2]$存在$x\in (0,\frac{1}{β-1})$，其中存在一个唯一的$(ε_{i})_{i=1}^{\infty}\in {0,1}^{\mathbb{N}}$，使$x=\sum_{i=1}^{\infty}ε_{i}β^{-i}.$在本文中，我们考虑更一般的情况下，当我们的序列是元素的${0,...,m}^{\mathbb{N}}.$我们表明，类似的黄金比例存在，并给出了一个明确的公式。
英文摘要：
It is a well known result that for $β\in(1,\frac{1+\sqrt{5}}{2})$ and $x\in(0,\frac{1}{β-1})$ there exists uncountably many $(ε_{i})_{i=1}^{\infty}\in {0,1}^{\mathbb{N}}$ such that $x=\sum_{i=1}^{\infty}ε_{i}β^{-i}.$ When $β\in(\frac{1+\sqrt{5}}{2},2]$ there exists $x\in (0,\frac{1}{β-1})$ for which there exists a unique $(ε_{i})_{i=1}^{\infty}\in {0,1}^{\mathbb{N}}$ such that $x=\sum_{i=1}^{\infty}ε_{i}β^{-i}.$ In this paper we consider the more general case when our sequences are elements of ${0,...,m}^{\mathbb{N}}.$ We show that an analogue of the golden ratio exists and give an explicit formula for it.
机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。
