Question

我编写了这个程序，用于显示找到平方根的牛顿方法，然后运行数学函数找到平方根并打印出它们。数学部分运行良好，但我创建的循环没有为牛顿方法提供正确的平方根。有任何想法吗？提前谢谢。

package newton_sqrt;

import java.util.Scanner;

public class Newton_sqrt {

    public static void main(String[] args) {

        double guess, new_guess, last_guess, accuracy, n, x, absolutex;

        Scanner keyboard = new Scanner(System.in);
        System.out.println("Enter in N for Newton: ");
        n = keyboard.nextDouble();
        last_guess = n / 2;
        do {
            new_guess = ((n/last_guess) + last_guess)/2;
            x = new_guess - last_guess;
            if(x>=0)
                absolutex=x;
            else
            absolutex=-x;
        } while(absolutex < .000001);
        System.out.println("Newton = " +new_guess);
        double mth = Math.sqrt(n);
        System.out.println("Math.sqrt = " +mth);    
    }
}

Answer 1

反转循环条件。你希望在absolutex大于epsilon时循环，而不是在它变小的时候循环。当它足够小时，你想停止。

do {
} while(absolutex > .000001);

Answer 2

看起来像数学部分需要很少的调整，循环条件也是如此。在这里尝试，它的工作原理

package newton_sqrt;

import java.util.Scanner;
public class Newton_sqrt {
    public static void main(String[] args) {
        // TODO code application logic here
        double guess, new_guess = 0, last_guess, accuracy, n, x, absolutex;

        Scanner keyboard = new Scanner(System.in);
        System.out.println("Enter in N for Newton: ");
        n = keyboard.nextDouble();
        last_guess = n / 2;

        do {
            new_guess = last_guess - (last_guess*last_guess-n) / (2*last_guess);
            x = Math.abs(last_guess - new_guess);
            if (x < .000001) {
                break ;
            } else {
                last_guess = new_guess;
            }
        } while (n >= .00);

        System.out.println("Newton = " + new_guess);
        double mth = Math.sqrt(n);
        System.out.println("Math.sqrt = " + mth);

    }

}

牛顿平方根Java程序

2 个答案: