Newton's Square Root Method, need somebody to help see the problem.

[LetsG0Blue]

import java.util.Scanner;

public class SquareRoot {

	public static void main(String[] args) {

		Scanner in = new Scanner(System.in);
		System.out.println("Enter a value for (n)");
		double n = in.nextDouble();
		double x = 1, y = 1;
		x = y;

			while(Math.abs(x - y) <= 0.00001)
			{
				x=y;
				x = ((n / x) + x) / 2.0;
			}
			
			System.out.println("Newton("+n+") = "+x);
			System.out.println("Math.sqrt("+n+") = "+Math.sqrt(n));
	}
}

Here's my output.

Enter a value for (n)
4
Newton(4.0) = 2.5
Math.sqrt(4.0) = 2.0

Where's the extra .5 in 2.5 coming from?

[jim829]

Well, initially, x-y == 0 which is less than .00001 so the loop goes thru once.

Then x = 2.5 and since 2.5 -1 > .00001 the loop exits.

Jim

[LetsG0Blue]

But if I put a > there it equals 1 for the output of Newton

[jim829]

That is because you chose x and y to be the same values to start so the loop would never enter if x = y = 1. So you could either chose different values for x and y. Or just change to a do while loop to force at least one iteration of the loop.

And I believe you want to set y = x.

Jim