Aequationes Mathematicae ( IF 0.9 ) Pub Date : 2021-07-01 , DOI: 10.1007/s00010-021-00826-6 Bruce Ebanks , André E. Kézdy
A derivation (of order 1) satisfies the reduction formula \(f(x^k) = kx^{k-1}f(x)\) for any integer k. In this article we find corresponding reduction formulas for derivations of higher order on commutative rings. More precisely, for every derivation f of order n and every positive integer k we find an explicit formula for \(f(x^k)\) as a linear combination of \(x^{k-1}f(x),x^{k-2}f(x^2), \ldots , x^{k-n}f(x^n)\). The proof hinges on the hypergeometric identity
$$\begin{aligned} \sum _{k \ge 0} (-1)^k \left( {\begin{array}{c}n\\ k\end{array}}\right) \left( {\begin{array}{c}n+d+1-k\\ n-j\end{array}}\right) \left( {\begin{array}{c}d+j-k\\ j\end{array}}\right) = \left( {\begin{array}{c}n\\ j\end{array}}\right) \end{aligned}$$for any positive integer n, nonnegative integer d, and integer j satisfying \(0 \le j \le n\). We prove this identity via the WZ-method.
中文翻译：
高阶导数和超几何恒等式的约简公式
推导（1 阶）满足任何整数k的约简公式\(f(x^k) = kx^{k-1}f(x)\)。在本文中，我们找到了可交换环上高阶导数的相应约简公式。更准确地说，对于n阶的每个推导f和每个正整数k，我们找到了一个明确的公式\(f(x^k)\)作为\(x^{k-1}f(x)的线性组合， x^{k-2}f(x^2), \ldots , x^{kn}f(x^n)\)。证明取决于超几何恒等式
$$\begin{aligned} \sum _{k \ge 0} (-1)^k \left( {\begin{array}{c}n\\ k\end{array}}\right) \left( {\begin{array}{c}n+d+1-k\\ nj\end{array}}\right) \left( {\begin{array}{c}d+jk\\ j\end{array }}\right) = \left( {\begin{array}{c}n\\ j\end{array}}\right) \end{aligned}$$对于任何满足\(0 \le j \le n\) 的正整数n、非负整数d和整数j。我们通过 WZ 方法证明了这个身份。
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