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 02222013, 05:03 AM #1Member
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Newton's Square Root Method, need somebody to help see the problem.
Java Code: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)); } }
Enter a value for (n)
4
Newton(4.0) = 2.5
Math.sqrt(4.0) = 2.0
Last edited by LetsG0Blue; 02222013 at 05:14 AM.
 02222013, 05:25 AM #2Senior Member
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Re: Newton's Square Root Method, need somebody to help see the problem.
Well, initially, xy == 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.
JimThe Java^{TM} Tutorials  SSCCE  Java Naming Conventions
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 02222013, 05:31 AM #3Member
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Re: Newton's Square Root Method, need somebody to help see the problem.
But if I put a > there it equals 1 for the output of Newton
 02222013, 05:38 AM #4Senior Member
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Re: Newton's Square Root Method, need somebody to help see the problem.
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.
JimLast edited by jim829; 02222013 at 05:47 AM.
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