Fitting two-yuan multiple curves with least squares method

The original source is unknown, the mathematical principle probably does not understand, the code is valid. The function is known as a pile of sample data, fitted with an approximate 2-yuan n-th function expression.

 public class Line {///<summary>///fitting two-yuan multiple curves with least squares///</summary>///<p Aram Name= "Arrx" > set of x coordinates of known points </param>///<param name= "Arry" > y-coordinate set of known points </param>///<par Am name= "Length" > number of known points </param>///<param name= "Dimension" > equation highest number of </param> public St  Atic double[] MultiLine (double[] Arrx, double[] arry, int length, int dimension)//two yuan multiple linear equation fitting curve {int n                  = dimension + 1;      The dimension equation requires dimension+1 coefficients double[,] guass = new Double[n, n + 1];
                Gaussian matrices such as: 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);

    }    public static double Sumarr (double[] arr, int n, int length)//n-th square of the elements of an array and {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; public static double Sumarr (double[] arr1, int n1, double[] arr2, int n2, int length) {Doub
            Le 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;
            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;//initialization 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) {//"This linear equation is singular linear equation"
                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]);
                }}//end for (j=0;j<n;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; }
    }