Find n-th Root of a Number 求一個數的 n 次方根

要求一個數的 n 次方根, 有很多 programing language 都有 Math.pow(x,y) 的 library. 如要求 x 的 n 次方根, 在 Java 只需用 Math.pow(x, 1.0/n) 的 function 就得了.

To find the n-th root of a number, usually many programming languages consist the Math.pow(x,y) library. In Java, simply type Math.pow(x, 1.0/n) can find the n-th root of x.

public class Root { public static void main(String[] args) { // read in the command-line argument double c = Double.parseDouble(args[0]); // a number double n = Integer.parseInt(args[1]); // n-th root double t; t = Math.pow(c, 1.0/n); System.out.println(t); } }

但如果要用 Numerical Method 來找, 筆者在 Wikipedia 的 nth root algorithm 找到一個 algorithm 是由 Newton's Method derive 出來的. 這是我 RootNewton.java 的 source code:

On the other hand, to find it using Numerical Method, I found the n-th root algorithm on Wikipedia which is derived using Newton's Method. This is my source code RootNewton.java

public class RootNewton { public static double n_algorithm (double t, double n, double c) { double t_new; t_new = 1/n*((n-1.0)*t+c/Math.pow(t, (n-1.0))); return t_new; } public static void main(String[] args) { // read in the command-line argument double c = Double.parseDouble(args[0]); double n = Double.parseDouble(args[1]); // n double epsilon = 1e-15; // relative error tolerance double t = c; // estimate of the n-th root of c double t_new; // repeatedly apply Newton update step until desired precision is achieved t_new = n_algorithm (t, n, c); while (Math.abs(t-t_new) > epsilon) { t = t_new; t_new = n_algorithm (t, n, c); } // print out the estimate of the n-th root of c System.out.println(t); } }

我覺得我的 iteration 寫得比較長氣, 如果讀者有一些比較好的辦法, 請你告訴我吧.
But I find this code is a little bit verbose, if you know how to make it better, please let me know.