使用 Math.Commons 库的 Newton-Raphson 方法 [英] Newton-Raphson method using the Math.Commons library

问题描述

我做了一个测试程序来尝试 NewtonRaphsonSolver 类通过 Apache Commons Math>图书馆.牛顿法用于为给定函数求根.

I made a test program to try the NewtonRaphsonSolver class through the Apache Commons Math library. Newton's method is used to find roots for a given function.

我写的测试程序引用了 cos(x) 函数(我有一个更难分析的函数，我先看 cos(x) 函数).

The test program that I wrote references the cos(x) function (I have a more difficult function to analyze and am looking at the cos(x) function first).

测试程序的代码是

import org.apache.commons.math3.analysis.differentiation.DerivativeStructure;
import org.apache.commons.math3.analysis.differentiation.UnivariateDifferentiableFunction;
import org.apache.commons.math3.analysis.solvers.*;
import org.apache.commons.math3.exception.DimensionMismatchException;

public class Test3 {

    public static void main(String args[]) {
        NewtonRaphsonSolver test = new NewtonRaphsonSolver();
        UnivariateDifferentiableFunction f = new UnivariateDifferentiableFunction() {

            public double value(double x) {
                return Math.cos(x);
            }

            @Override
            public DerivativeStructure value(DerivativeStructure t) throws DimensionMismatchException {
                return t.cos();
            }
        };

        for (int i = 1; i <= 500; i++) {
            System.out.println(test.solve(1000, f, i, i+0.1));
        }
    }
}

不确定是否需要引用 Math.cos(x) 和 t.cos() 两次

Not certain if I needed to reference Math.cos(x) and t.cos() twice

public double value(double x) {
                return Math.cos(x);
            }

            @Override
            public DerivativeStructure value(DerivativeStructure t) throws DimensionMismatchException {
                return t.cos();
            }

Newton 的方法找到所有的零并将它们显示给用户.

Newton's method finds all of the zeroes and displays them to the user.

1.5707963267948966
1.5707963267948966
-7.853981633974483
4.71238898038469
4.71238898038469
1.5707963267948966
7.853981633974483
7.853981633974483
10.995574287564276
10.995574287564276
10.995574287564276
10.995574287564276
14.137166941154069
14.137166941154069
14.137166941154069
127.23450247038663
17.278759594743864
17.278759594743864
23.56194490192345
20.420352248333657
20.420352248333657
39.269908169872416
23.56194490192345
23.56194490192345
14.137166941154069
26.703537555513243
26.703537555513243
23.56194490192345
29.845130209103036
29.845130209103036
26.703537555513243
32.98672286269283
32.98672286269283
32.98672286269283
36.12831551628262
36.12831551628262
36.12831551628262
23.56194490192345
39.269908169872416
39.269908169872416
45.553093477052
42.411500823462205
42.411500823462205

有什么方法可以防止打印出重复的零?例如，上面的输出将读取

Is there some way to prevent printing out zeroes that are duplicates? For example, the above output would read

1.5707963267948966
4.71238898038469
7.853981633974483
10.995574287564276
14.137166941154069
17.278759594743864
20.420352248333657
23.56194490192345
26.703537555513243
29.845130209103036
32.98672286269283
36.12831551628262
39.269908169872416
42.411500823462205
45.553093477052

这可以在 for 循环内完成还是通过只打印不重复值的数组来完成?

Can this be done inside a for loop or through an array which only prints out values which are not duplicates?

推荐答案

import java.util.TreeSet;
import org.apache.commons.math3.analysis.differentiation.DerivativeStructure;
import org.apache.commons.math3.analysis.differentiation.UnivariateDifferentiableFunction;
import org.apache.commons.math3.analysis.solvers.*;
import org.apache.commons.math3.exception.DimensionMismatchException;

public class Test5 {

    public static void main(String args[]) {
        NewtonRaphsonSolver test = new NewtonRaphsonSolver(1E-10);

        UnivariateDifferentiableFunction f = new UnivariateDifferentiableFunction() {

            public double value(double x) {
                return Math.sin(x);
            }

            public DerivativeStructure value(DerivativeStructure t) throws
                    DimensionMismatchException {
                return t.sin();
            }
        };

        double EPSILON = 1e-6;
        TreeSet<Double> set = new TreeSet<>();
        for (int i = 1; i <= 5000; i++) {
            set.add(test.solve(1000, f, i, i + EPSILON));
        }
        for (Double s : set) {
            if (s > 0) {
                System.out.println(s);
            }
        }
    }
}

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