怎样调用下面最小二乘法的类，并运行测试

public class LeastSquare//最小二乘法
    {
        ///

        ///用最小二乘法拟合二元多次曲线（任意次）
        ///
        ///已知点的x坐标集合
        ///已知点的y坐标集合
        ///已知点的个数
        ///方程的最高次数
        #region 总计算方法
        public static double[] MultiLine(double[] arrX, double[] arrY, int length, int dimension)//二元多次线性方程拟合曲线
        {
            int n = dimension + 1;                  //dimension次方程需要求 dimension+1个 系数
            double[,] Guass = new double[n, n + 1];      //高斯矩阵 例如：y=a0+a1*x+a2*x*x
            for (int i = 0; i < n; i++)
            {
                int j;
                for (j = 0; j < n; j++)
                {
                    Guass[i, j] = SumArr(arrX, j + i, length);
                }
                Guass[i, j] = SumArr(arrX, i, arrY, 1, length);
            }
            return ComputGauss(Guass, n);//返回值是函数的系数
        }
        #endregion
        #region 求数组的元素的n次方的和1(中间方法)
        public static double SumArr(double[] arr, int n, int length) 
        {
            double s = 0;
            for (int i = 0; i < length; i++)
            {
                if (arr[i] != 0 || n != 0)
                    s = s + Math.Pow(arr[i], n);
                else
                    s = s + 1;
            }
            return s;
        }
        #endregion
        #region 求数组的元素的n次方的和2(中间方法)
        public static double SumArr(double[] arr1, int n1, double[] arr2, int n2, int length)
        {
            double s = 0;
            for (int i = 0; i < length; i++)
            {
                if ((arr1[i] != 0 || n1 != 0) && (arr2[i] != 0 || n2 != 0))
                    s = s + Math.Pow(arr1[i], n1) * Math.Pow(arr2[i], n2);
                else
                    s = s + 1;
            }
            return s;
 
        }
        #endregion
        #region 返回值是函数的系数(中间方法)
        public static double[] ComputGauss(double[,] Guass, int n)
        {
            int i, j;
            int k, m;
            double temp;
            double max;
            double s;
            double[] x = new double[n];
            for (i = 0; i < n; i++) x[i] = 0.0;//初始化
 
            for (j = 0; j < n; j++)
            {
                max = 0;
                k = j;
                for (i = j; i < n; i++)
                {
                    if (Math.Abs(Guass[i, j]) > max)
                    {
                        max = Guass[i, j];
                        k = i;
                    }
                }
 
 
                if (k != j)
                {
                    for (m = j; m < n + 1; m++)
                    {
                        temp = Guass[j, m];
                        Guass[j, m] = Guass[k, m];
                        Guass[k, m] = temp;
                    }
                }
                if (0 == max)
                {
                    // "此线性方程为奇异线性方程" 
                    return x;
                }
 
                for (i = j + 1; i < n; i++)
                {
                    s = Guass[i, j];
                    for (m = j; m < n + 1; m++)
                    {
                        Guass[i, m] = Guass[i, m] - Guass[j, m] * s / (Guass[j, j]);
                    }
                }
 
            }//结束for (j=0;j
 
            for (i = n - 1; i >= 0; i--)
            {
                s = 0;
                for (j = i + 1; j < n; j++)
                {
                    s = s + Guass[i, j] * x[j];
                }
                x[i] = (Guass[i, n] - s) / Guass[i, i];
            }
            return x;
        }
#endregion
    }

参考于C#利用最小二乘法拟合任意次函数曲线
作者：未来无限

这里全是静态方法
直接
LeastSquare.方法(参数)
这样调用就行了

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