牛顿平方根Java程序

Question

我写这个程序是为了展示牛顿求平方根的方法，然后运行数学函数来求平方根，然后把它们都打印出来。数学部分工作得很好，但是我创建的循环没有给出牛顿法的正确平方根。有什么想法吗？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。当absolutex比epsilon大时，您希望循环，而不是当它小时。当它足够小时，您希望停止

do {
} while(absolutex > .000001);

Answer 2

看起来数学部分需要一些调整，循环条件也是如此。试试看，它很管用

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);

    }

}