一族恒等式的一致性证明方法
A Method for Uniformly Proving a Family of Identities
摘要：
本文给出了证明方法和结果。所提出的证明方法特别适合于一致证明一族递归序列所满足的恒等式。为了说明该方法，我们研究了递归序列族$F^{(k)}_n = \sum_{i=1}^k F^{(k)}_{n-i}, n \ge 0, k \ge 2,$，其中$n$是随整数变化的参数，$k$是族中成员的参数索引。主要定理陈述了$ F^{(k)}_n = \sum_{j=1}^k P_{k,j} F^{(k)}_{n-jk},$与$P$是一个递归三角形，满足三角递归$P_{i,j}=2P_{i-1,j}- P_{i-1,j-1},$，并具有适当的初始条件。定理的证明利用了这样一个事实，即恒等式的特征多项式可以被生成底层序列的递归的特征多项式整除。
英文摘要：
This paper presents both a proof method and a result. The proof method presented is particularly suitable for uniformly proving families of identities satisfied by a family of recursive sequences. To illustrate the method, we study the family of recursive sequences $F^{(k)}_n = \sum_{i=1}^k F^{(k)}_{n-i}, n \ge 0, k \ge 2,$ with $n$ a parameter varying over integers, and $k$ a parameter indexing members of the family. The main theorem states $ F^{(k)}_n = \sum_{j=1}^k P_{k,j} F^{(k)}_{n-jk},$ with $P$ a recursive triangle satisfying the triangle recursion $P_{i,j}=2P_{i-1,j}- P_{i-1,j-1},$ with appropriate initial conditions. The proof of the theorem exploits the fact that characteristic polynomials of identities are divisible by the characteristic polynomial of the recursion generating the underlying sequence.
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