零极限迭代数列的收敛阶

The Order of Convergence of Iterative Sequence with Zero Limit

给出了迭代数列x_(n+1)=f(x_n)极限的一般性结论.在早期文献[1],[2]结论limn→∞nx_n^q=1/cq的基础上,通过函数f(x)在x=0处的Taylor展式,给出了无穷小量nx_n^q-1/(cq)等价量的一般计算方法.此等价量的阶的推导和估计在本文的最后一节给出.

A general result of iterative sequence x（n＋1）=f（xn）is obtained.Based on the earlier results in[1]and[2]that limn→∞nxn^q = 1/cq,the equivalent infinitesimal of nxn^q-1/（cq）is found by Taylor expansion of f（x）at x = 0.The method to estimate the order of nxn^q-1/（cq）is also given.