抛物线法、牛顿法、弦截法求根实例

计算方法第四次计算实习题

7.用下列方法求 f ( x ) = x 3 3 x 1 = 0 x 0 = 2 x = 1.87938524.. , f(x)=x^3-3x-1=0在x_0=2附近的根,根的准确值x^{*}=1.87938524..,要求计算结果准确到四位有效数字

(1)用牛顿法

(2)用弦截法,取 x 0 = 2 , x 1 = 1.9 x_0=2,x_1=1.9

(3)用抛物线法,取 x 0 = 1 , x 1 = 3 , x 2 = 2 x_0=1,x_1=3,x_2=2

解题思路:按部就班,套公式编写程序即可注意控制精度,要求准确到四位有效数字,即要求准确解和所得近似解误差不超过 0.5 1 0 4 0.5*10^{-4} ,同时要注意迭代时的变量关系,以下是源代码:

(1)牛顿法:

/**
 * @Title: newton.java
 * @Desc: TODO
 * @Package: root
 * @author: glm233
 * @date: 2020年5月25日 下午10:53:16
 * @version 1.0
 */

package root;

/**
 * @ClassName: newton
 * @Desc: TODO
 * @author: glm233
 * @date: 2020年5月25日 下午10:53:16
 * @version 1.0
 */
import java.util.Scanner;
public class newton {
	 public static void main(String[] args) {
	        Scanner scanner = new Scanner(System.in);
	        System.out.print("请输入起始点:");
	        double x = scanner.nextDouble();
	        System.out.print("请输入误差允许范围:");
	        double e = scanner.nextDouble();
	        System.out.print("请输入最大迭代次数:");
	        int N = scanner.nextInt();
	        scanner.close();
	        double res = getEistimate(x,e,N);
	        System.out.println("牛顿法得到的解为:" + res);
	    }

	    private static double getEistimate(double x,double e,int N) {
	        double result,x1;
	        for (int k = 1; true ; k++ ) {
	            if(f(x) == 0){
	                throw new RuntimeException("出现奇异情况....");
	            }
	            x1 = x - F(x)/f(x);
	            if(Math.abs(x1 - x) < e ){
	                System.out.println("迭代" + k + "次得到结果");
	                return x1;
	            }
	            if(k == N){
	                throw new RuntimeException("迭代失败,已达最大迭代次数N....");
	            }
	            x = x1;
	        }
	    }

	    private static double F(double x) {
	        return x*x*x -3*x-1;
	    }

	    private static double f(double x) {
	        return 3*x*x -3;
	    }

}


运行结果:

tPQqXV.png

(2)用弦截法,取 x 0 = 2 , x 1 = 1.9 x_0=2,x_1=1.9

/**
 * @Title: secant.java

 * @Desc: TODO
 * @Package: root
 * @author: glm233
 * @date: 2020年5月25日 下午11:05:56
 * @version 1.0
 */

package root;
import java.util.Scanner;
/**
 * @ClassName: secant
 * @Desc: TODO
 * @author: glm233
 * @date: 2020年5月25日 下午11:05:56
 * @version 1.0
 */
public class secant {
	   public static void main(String[] args) {
	        Scanner scanner = new Scanner(System.in);
	        System.out.print("请输入初值1:");
	        double x0 = scanner.nextDouble();
	        System.out.print("请输入初值2:");
	        double x1 = scanner.nextDouble();
	        System.out.print("请输入误差允许范围:");
	        double e = scanner.nextDouble();
	        System.out.print("请输入最大迭代次数:");
	        int N = scanner.nextInt();
	        scanner.close();
	        double res = getEistimate(x0,x1,e,N);
	        System.out.println("快速弦截法得到的解为:" + res);
	    }

	    private static double getEistimate(double x0,double x1,double e,int N) {
	        double x2;
	        for (int k = 0; true ; k++ ) {
	            if(f(x0,x1) == 0){
	                throw new RuntimeException("出现奇异情况....");
	            }
	            x2 = x1 - F(x1)/f(x0,x1);
	            if(Math.abs(x2 - x1) < e){
	                System.out.println("迭代" + k + "次得到结果");
	                return x2;
	            }
	            if(k == N){
	                throw new RuntimeException("迭代失败,已达最大迭代次数N....");
	            }
	            x0 = x1;
	            x1 = x2;
	        }
	    }

	    private static double f(double x0, double x1) {
	        return (F(x1) - F(x0))/(x1 - x0);
	    }

	    private static double F(double x) {
	    	return x*x*x -3*x-1;
	    }
}



运行结果:

tPQHlq.png]

(3)用抛物线法,取 x 0 = 1 , x 1 = 3 , x 2 = 2 x_0=1,x_1=3,x_2=2

/**
 * @Title: parabolic.java

 * @Desc: TODO
 * @Package: root
 * @author: glm233
 * @date: 2020年5月25日 下午11:10:01
 * @version 1.0
 */

package root;

import java.util.Scanner;

/**
 * @ClassName: parabolic
 * @Desc: TODO
 * @author: glm233
 * @date: 2020年5月25日 下午11:10:01
 * @version 1.0
 */
public class parabolic {
	
	public final static double check=1.87938524;
	public static void main(String[] args) {
		
        Scanner scanner = new Scanner(System.in);
        System.out.print("请输入初值1:");
        double x0 = scanner.nextDouble();
        System.out.print("请输入初值2:");
        double x1 = scanner.nextDouble();
        System.out.print("请输入初值3:");
        double x2 = scanner.nextDouble();
        System.out.print("请输入误差允许范围:");
        double e = scanner.nextDouble();
        System.out.print("请输入最大迭代次数:");
        int N = scanner.nextInt();
        scanner.close();
        double res = getEistimate(x0,x1,x2,e,N);
        System.out.println("精确解为:" + check);
        System.out.println("抛物线法得到的解为:" + res);
    }

    private static double getEistimate(double x0,double x1,double x2,double e,int N) {
        double x3,w=omiga(x0,x1,x2);
        for (int k = 0; true ; k++ ) {
            if(w*w<4*F(x2)*f(x0,x1,x2)){
                throw new RuntimeException("出现奇异情况....");
            }
            if(w>0) {
            	x3 = x2 - 2*F(x2)/(w+Math.sqrt(w*w-4*F(x2)*f(x0,x1,x2)));
            }
            else	x3 = x2 - 2*F(x2)/(w-Math.sqrt(w*w-4*F(x2)*f(x0,x1,x2)));
            if(Math.abs(x3 - check) < e){
                System.out.println("迭代" + k + "次得到结果");
                //System.out.println(x3-check);
                return x2;
            }
            if(k == N){
                throw new RuntimeException("迭代失败,已达最大迭代次数N....");
            }
            //System.out.println(x3);
            x0 = x1;
            x1 = x2;
            x2 = x3;
        }
    }

    private static double omiga(double x0,double x1,double x2){
    	return f(x0,x1,x2)*(x2-x1)+f(x2,x1);
    	
    	
    }
    private static double f(double x0, double x1) {
        return (F(x1) - F(x0))/(x1 - x0);
    }
    
    private static double f(double x0, double x1,double x2) {
        return (f(x1,x2)-f(x1,x0))/(x2-x0);
    }

    private static double F(double x) {
    	return x*x*x -3*x-1;
    }
}

运行结果:

tPQb60.png

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转载自blog.csdn.net/weixin_43798170/article/details/106348650