Calculating nth root of a positive integer in Java

Question

My task is to write a program which prompts the user to enter a positive double a and an integer n greater than 2, and print the value of the nth root of positive integer a to the screen with accuracy to 100 places. I've used Math.pow to be able to get the root, and I feel as though I've done everything right. The only problem is that every time I run my program, the output is 1.0, no matter what numbers I input for a and n. What is the problem with my code?

import java.util.Scanner;
import java.lang.Math;

public class Q8 {

    public static void main(String[] args) {

    System.out.println("Enter a positive number: ");
    Scanner in = new Scanner(System.in);
    double a = in.nextDouble();

    System.out.println("Enter an integer greater than 2: ");
    Scanner in2 = new Scanner(System.in);
    int n = in.nextInt();

    System.out.println(pow (a,n));
    }

private static double pow(double a, int n) {
      if (n == 0){
            return 1;
      }

      else{
         double sum = Math.pow(a,(1/n));
         return sum;
}

Why is the answer always 1.0?

I don't see `root` defined anywhere used in this expression: `Math.pow(a,(1/root))`. What is it? — icza, Nov 10 '14 at 08:54
@icza It's a typo. Either `root` should be `n` or `n` should be `root`. — Dawood ibn Kareem, Nov 10 '14 at 08:56
You're never going to get 100 decimal places using `double`. May I interest you in a `BigDecimal`? — Dawood ibn Kareem, Nov 10 '14 at 08:59
I'm just looking into it now. It looks like the JDK doesn't give you a way of doing it (as far as I can find). You'd have to either use some kind of mathematics library, or write your own implementation of one of the classic algorithms. The Newton Raphson method should do it, but it's going to require quite a lot of coding. — Dawood ibn Kareem, Nov 10 '14 at 09:02
I don't think the logic for `n == 0` is correct. Besides `n` is meant to be positive according to the question. — weston, Nov 10 '14 at 09:03
I can't find a way of doing exponents with `BigDecimal` or anything equivalent to it in either Guava or Apache. I can't believe this is so hard to find. Perhaps that's why your teacher has asked you to look into this. — Dawood ibn Kareem, Nov 10 '14 at 09:08
Is it homework or a programming challenge like project euler? — weston, Nov 10 '14 at 09:10

Dawood ibn Kareem · Answer 1 · 2014-11-10T08:57:24.943

8

Replace 1/n with 1.0/n.

You're getting integer division, so no matter what n is, if it's 2 or higher, then 1/n is coming out zero. Then you're raising your number to the zeroeth power, which gives 1.

Replacing 1 with 1.0 makes the division into a floating point division - that is, the result won't be truncated to an integer. This is what you want.

edited Nov 10 '14 at 08:57

answered Nov 10 '14 at 08:52

Dawood ibn Kareem

77,785
15
98
110

1

Provided that `root` is an integer as well, otherwise it's a floating point operation regardless of the literal's type. – Thomas Nov 10 '14 at 08:54
Right. It looks to me like the OP intended `root` and `n` to be the same, and just typed one of them wrong. – Dawood ibn Kareem Nov 10 '14 at 08:55
the root should be n, not root. but nonetheless, that worked! – Ibrewster Nov 10 '14 at 08:56
OK, since you just edited your question, I edited my answer to match it. – Dawood ibn Kareem Nov 10 '14 at 08:57
yeah i did! thanks for the help, such a silly mistake! – Ibrewster Nov 10 '14 at 08:59

score 5 · Answer 2 · answered Nov 10 '14 at 08:54

First of all, I'm assuming that

double sum = Math.pow(a,(1/root));

should be

double sum = Math.pow(a,(1/n));

since there is no root variable in your code.

Second of all, 1/n would give you 0 for every integer n > 1. Therefore sum would be 1.0. You should replace it with :

double sum = Math.pow(a,(1.0/n));

or

double sum = Math.pow(a,(1/(double)n));

In order to get a division of double variables.

Calculating nth root of a positive integer in Java

2 Answers2