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Thread: Sqrt Program, skips over method

  1. #1
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    Default Sqrt Program, skips over method

    Hey guys , I have to do a program where you must find the square root of a number by repeatedly divide a number by the number and past guess untill the diffrence is withing .000001. My code doesnt output , and when it did it would give me 0. If anyoene can help me out itd be much appricated. thanks!


    import java.util.*;
    import java.lang.Math;
    public class sqrt
    {
    public static void main(String[] args)
    {
    double value;
    double guess;
    Scanner in= new Scanner(System.in);
    System.out.println("Please enter the number of which you want to squareroot: ");
    value=in.nextDouble();
    System.out.println("Please enter youre first estimate: ");
    guess=in.nextDouble();
    while(guess<(value*.5))
    {
    System.out.print("Please input a higher value: ");
    guess=in.nextDouble();
    }
    System.out.println("The squareroot is: "+ sqrt(value,guess));

    }

    private static double sqrt(double value, double guess)
    {
    double nextGuess=guess*2;
    while( ((nextGuess-guess)!=.00001)||((nextGuess-guess)!=-.00001) )
    {
    nextGuess=(guess+(value/guess))/2;
    guess=nextGuess;
    }
    return guess;

    }

    }


  2. #2
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    Default Re: Sqrt Program, skips over method

    Please post your code in code tags.

    Your problem statement and description of the problem are not helpful. Here's my sample run:
    Please enter the number of which you want to squareroot: 
    51
    Please enter youre first estimate: 
    7
    Please input a higher value: 8
    Please input a higher value: 8
    Please input a higher value: 8
    Please input a higher value: 8
    Please input a higher value:
    Seems to be running, even outputting, though it's annoying and I have no idea what it's trying to do. Why don't you post a sample run and describe what it should be doing.

    Oh, and there's no "skips over method" that I can see.

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