C#，数值计算——用算法加速序列的收敛的计算方法与源程序

算法对序列的收敛加速。初始化方式使用参数nmax调用构造函数，nmax是要求和的项数，以及eps，即所需的精度。然后连续调用next函数，参数为next部分和序列的。序列极限的当前估计值为next返回。检测到收敛设置标志cnvgd。

using System;

namespace Legalsoft.Truffer
{
    /// <summary>
    /// Convergence acceleration of a sequence by the algorithm.Initialize by
    /// calling the constructor with arguments nmax, an upper bound on the
    /// number of terms to be summed, and epss, the desired accuracy.Then make
    /// successive calls to the function next, with argument the next partial sum
    /// of the sequence. The current estimate of the limit of the sequence is 
    /// returned by next.The flag cnvgd is set when convergence is detected.
    /// </summary>
    public class Epsalg
    {
        private double[] e { get; set; }
        private int n { get; set; }
        private int ncv { get; set; }
        public bool cnvgd { get; set; }
        /// <summary>
        /// Numbers near machine underflow and overflow limits.
        /// </summary>
        private double eps { get; set; }
        private double small { get; set; }
        private double big { get; set; }
        private double lastval { get; set; }
        private double lasteps { get; set; }

        public Epsalg(int nmax, double epss)
        {
            this.e = new double[nmax];
            this.n = 0;
            this.ncv = 0;
            this.cnvgd = false;
            this.eps = epss;
            this.lastval = 0.0;
            small = float.MinValue * 10.0;
            big = double.MaxValue;
        }

        public double next(double sum)
        {
            e[n] = sum;
            double temp2 = 0.0;
            for (int j = n; j > 0; j--)
            {
                double temp1 = temp2;
                temp2 = e[j - 1];
                double diff = e[j] - temp2;
                if (Math.Abs(diff) <= small)
                {
                    e[j - 1] = big;
                }
                else
                {
                    e[j - 1] = temp1 + 1.0 / diff;
                }
            }
            n++;
            double val = (n & 1) != 0 ? e[0] : e[1];
            if (Math.Abs(val) > 0.01 * big)
            {
                val = lastval;
            }
            lasteps = Math.Abs(val - lastval);
            if (lasteps > eps)
            {
                ncv = 0;
            }
            else
            {
                ncv++;
            }
            if (ncv >= 3)
            {
                cnvgd = true;
            }
            return (lastval = val);
        }

    }
}