加速不定超几何求和算法

E. Zima
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引用次数: 1

摘要

设K为特征为零的域,x为自变量,E为关于x的移位算子,即对于任意f(x), Ef(x) = f(x + 1)。回想一下,如果存在一个有理函数r(x)∈K(x)使得F(x + 1)/F(x) = r(x),则非零表达式F(x)称为K上的超几何项。通常将r(x)称为F(x)的有理证明。不定超几何求和(反微分)问题是:给定一个超几何项F(x),求一个满足一阶线性差分方程(E−1)G(x) = F(x)的超几何项G(x)。(1)如果找到,写Σx F(x) = G(x) + c,其中c是任意常数。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
Accelerating indefinite hypergeometric summation algorithms
Let K be a field of characteristic zero, x an independent variable, E the shift operator with respect to x, i.e., Ef(x) = f(x + 1) for an arbitrary f(x). Recall that a nonzero expression F(x) is called a hypergeometric term over K if there exists a rational function r(x) ∈ K(x) such that F(x + 1)/F(x) = r(x). Usually r(x) is called the rational certificate of F(x). The problem of indefinite hypergeometric summation (anti-differencing) is: given a hypergeometric term F(x), find a hypergeometric term G(x) which satisfies the first order linear difference equation (E − 1)G(x) = F(x). (1) If found, write Σx F(x) = G(x) + c, where c is an arbitrary constant.
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