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Thread: Java program Square root

  1. #1
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    Cool Java program Square root

    Hi can someone post a java program that finds the square root of a number and please explain what each line of code does. Also add a method (for example Public Static void).


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    Default Re: Java program Square root

    Math.sqrt(n);

    Easy I'm sure you can work it out

    Regards,
    Chris
    chris[at]javaprogrammingforums[dot]com

    Prifysgol Bangor University, North Wales

  3. #3
    mmm.. coffee JavaPF's Avatar
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    Default Re: Java program Square root

    Hey, Hey, lol Welcome to the Java Programming Forums.

    Like Chris says, this is an easy one! Check the Math API

    Math (Java 2 Platform SE 5.0)

    import java.util.Scanner;
     
    public class Hey {
     
        /**
         * JavaProgrammingForums.com
         */
        public static double number, answer;
     
        public static void calculateSquare(double number){
     
            answer = Math.sqrt(number);
            System.out.println("The square root of " + number + " is " + answer);        
        }
     
        public static void main(String[] args) {
     
            Hey hey = new Hey();
            Scanner sc = new Scanner(System.in);
            System.out.println("Enter a double: ");
            number = sc.nextDouble();
            hey.calculateSquare(number);
     
        }
     
    }
    Please use [highlight=Java] code [/highlight] tags when posting your code.
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  4. #4
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    Default Re: Java program Square root

    public class SquareRoot{
                  public static void main(String[] args){
                 Scanner sc = new Scanner(System.in()); //this creates an object of the scanner class to input from keyboard
                 int someNumber;
     
                someNumber = sc.nextInt(); //reads an integer from the keyboard
     
                //calls the method to calculate the root and prints on screen//
                System.out.println("The square root is: " + sRoot(someNumber)); 
     
    }
    private static double sRoot(int a){
             int z = a;
             int root = (double) a * a;
             return root;
           }
    }
    As they've already pointed out on the other threads, there are predefined classes to calculate square roots and most mathematical functions. But if you want to do it manually and know how it works, then this is one way.

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    JavaPF (August 15th, 2009)

  6. #5
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    Default Re: Java program Square root

    Quote Originally Posted by hajidel View Post
    private static double sRoot(int a){
             int z = a;
             int root = (double) a * a;
             return root;
           }
    }
    That code is going to only put out the square but not the square root... the following will do the root, had to do it for class earlier this year. Spent about four hours trying to figure out the convergent infinite series math and what not and then realized that trail and error by digit was a lot easier.
    import java.util.*;
     
    public class SRoot
    {
     
    public static void main (String[] args)
    {
           Scanner k = new Scanner(System.in);
    	System.out.println("Insert Number");
    	int number = k.nextInt();
    	int lowBound=34404;
    	for (int i=0; i<100; i++)
    	{
    		if ( i*i <= number )
    		{
    			lowBound = i;
    		}
    	}
    	System.out.println(lowBound);
    	double estimate = lowBound;
    	double ten=0;
    	for (double i=0; i<1; i+=.1)
    	{
    		if ((estimate+i)*(estimate+i) <=number)
    		{
    			ten = i;	
    		}
    	}
    	estimate = estimate + ten;
    	double hund = 0;
    	for (double i=0; i<.1; i+=.01)
    	{
    		if ((estimate+i)*(estimate+i) <=number)
    		{
    			hund = i;
    		}
    	}
       estimate = estimate + hund;
    	double thou = 0;
    	for (double i=0; i<.01; i+=.001)
    	{
    		if ((estimate+i)*(estimate+i) <=number)
    		{
    			thou = i;
    		}
    	}
    	estimate = estimate + thou;
    	System.out.println(estimate);
    }}

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    Default Re: Java program Square root

    You can use
    3.1622776601683793319988935444327*log(x)
    which is actually root 10 * log(x) to find it also. or read up on babylonian Method, the true iterative method, for those numbers whose square root is not an integer such as 8.

    Chris
    chris[at]javaprogrammingforums[dot]com

    Prifysgol Bangor University, North Wales

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