Java Implementation of the method example for solving the n-degree polynomial, java polynomial example

Java Implementation of the method example for solving the n-degree polynomial, java polynomial example

The example in this article describes how Java implements the solution to a polynomial of n times. We will share this with you for your reference. The details are as follows:

Trend Prediction is required for the project. linear fitting, second-order curve fitting, and exponential fitting algorithms are used. Various linear fitting algorithms are written as matrices in this form:

X indicates the abscissa sample value, y indicates the ordinate sample value, I indicates the serial number of the sample point, a indicates the coefficient, N indicates the number of sample points, and n indicates the order, therefore, linear fitting finally becomes a problem for solving the higher order equations.

I don't know if there are any useful java matrix computing packages. I am not good at collecting such materials, so I had to pick up linear algebra that has been put down for many years, I wrote a java program using the augmented matrix algorithm to solve the high-order equations. Paste the Code directly:

Package commonAlgorithm; public class PolynomialSoluter {private double [] [] matrix; private double [] result; private int order; public PolynomialSoluter () {} // check the length of the input item and generate the Augmented Matrix private boolean init (double [] [] matrixA, double [] arrayB) {order = arrayB. length; if (matrixA. length! = Order) return false; matrix = new double [order] [order + 1]; for (int I = 0; I <order; I ++) {if (matrixA [I]. length! = Order) return false; for (int j = 0; j <order; j ++) {matrix [I] [j] = matrixA [I] [j];} matrix [I] [order] = arrayB [I];} result = new double [order]; return true;} public double [] getResult (double [] [] matrixA, double [] arrayB) {if (! Init (matrixA, arrayB) return null; // Gaussian elimination-forward for (int I = 0; I <order; I ++) {// if the diagonal line of the current row is 0, exchange if (! SwithIfZero (I) return null; // decode for (int j = I + 1; j <order; j ++) {if (matrix [j] [I] = 0) continue; double factor = matrix [j] [I]/matrix [I] [I]; for (int l = I; l <order + 1; l ++) matrix [j] [l] = matrix [j] [l]-matrix [I] [l] * factor ;}} // Gaussian deyuan-reverse-removed redundant Calculation for (int I = order-1; I> = 0; I --) {result [I] = matrix [I] [order]/matrix [I] [I]; for (int j = I-1; j>-1; j --) matrix [j] [order] = matrix [j] [order]-result [I] * matrix [j] [I];} return result ;} private boolean swithIfZero (int I) {if (matrix [I] [I] = 0) {int j = I + 1; // locate the column where the position is not 0 while (j <order & matrix [j] [I] = 0) j ++; // if the corresponding position is all 0, there is no solution if (j = order) return false; else switchRows (I, j);} return true;} private void switchRows (int I, int j) {double [] tmp = matrix [I]; matrix [I] = matrix [j]; matrix [j] = tmp ;}}

If you have better algorithms or a suitable matrix calculation package, please contact us.

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