代数点上的多对数是线性无关的吗?

Q4 Mathematics
Sinnou David, Noriko Hirata-Kohno, M. Kawashima
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引用次数: 9

摘要

设$r,m$为正整数。设$0\lex<1$为有理数。设$\Phi_s(x,z)$为第$s$个Lerch函数$\sum_{k=0}^{infty}\tfrac{z^{k+1}}{(k+x+1)^s}$,其中$s=1,2,\ldots,r$。当$x=0$时,这是一个多对数函数。设$\alpha_1,\ldots,\alpha_m$是成对不同的代数数,$0<|\alpha_j|<1$$(1\le j\le m)$。在本文中,我们陈述了所有$rm+1$数字$的代数数域的线性独立性准则:$$\Phi_1(x,\alpha_1),\Phi_2(x,\alpha_1 1美元。这是第一个结果,它给出了$r$Lerch函数$\Phi_1(x,z),\Phi_2(x,z),\ldots,\Phi_r(x,z)$的值在$m$不同代数点处线性独立的充分条件,而没有对$r$和$m$的任何假设,即使对于$x=0$的情况,也没有多对数。我们给出了我们的证明的大纲,并解释了基本思想。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
Can polylogarithms at algebraic points be linearly independent?
Let $r,m$ be positive integers. Let $0\le x <1$ be a rational number. Let $\Phi_s(x,z)$ be the $s$-th Lerch function $\sum_{k=0}^{\infty}\tfrac{z^{k+1}}{(k+x+1)^s}$ with $s=1,2,\ldots ,r$. When $x=0$, this is the polylogarithmic function. Let $\alpha_1,\ldots ,\alpha_m$ be pairwise distinct algebraic numbers with $0<|\alpha_j|<1$ $(1 \le j \le m)$. In this article, we state a linear independence criterion over algebraic number fields of all the $rm+1$ numbers $:$ $\Phi_1(x,\alpha_1),\Phi_2(x,\alpha_1),\ldots, \Phi_r(x,\alpha_1),\Phi_1(x,\alpha_2),\Phi_2(x,\alpha_2),\ldots, \Phi_r(x,\alpha_2),\ldots,\Phi_1(x,\alpha_m),\Phi_2(x,\alpha_m),\ldots, \Phi_r(x,\alpha_m)$ and $1$. This is the first result that gives a sufficient condition for the linear independence of values of the $r$ Lerch functions $\Phi_1(x,z),\Phi_2(x,z),\ldots, \Phi_r(x,z)$ at $m$ distinct algebraic points without any assumption for $r$ and $m$, even for the case $x=0$, the polylogarithms. We give an outline of our proof and explain basic idea.
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来源期刊
Moscow Journal of Combinatorics and Number Theory
Moscow Journal of Combinatorics and Number Theory Mathematics-Algebra and Number Theory
CiteScore
0.80
自引率
0.00%
发文量
21
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