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如果需求求平方根，即设函数f(x)=x^2 - num
即求 f(x) = 0 的解
这个时候可以先定义一个初始化的解，init_solve = num/2
定义一个可以接受的误差范围： double tolerance = 1e-10
牛顿迭代法，x_n+1 = x_n - f(x_n)/f`(x_n)

f(x_n) =x_n^2 - num  
f`(x_n)=2*x_n 

x_n+1 = x_n - (x_n^2 - num)/ 2*x_n

Java实现

  public static double sqrt_newton(double num) {
        double guess = num / 2.0;
        double tolerance = 1e-10;
        if (tolerance < 1.0) {
            System.out.print("right");
        } else {
            System.out.print("wrong");
        }
        while (true) {
            double nextGuess = guess - (guess * guess - num) / (2 * guess);
            if (Math.abs(nextGuess * nextGuess - num) < tolerance) {
                break;
            }
            guess = nextGuess;
        }
        return guess;
    }