Suppose you want to compute the square root of x (x > 0) using the following met

Question

Suppose you want to compute the square root of x (x > 0) using the following method:

- guess a number g (for example guess = 7) for the square root of x,

- a better new guess is (g + x/g)/2 until |x – g^2| is less than or equal to a given epsilon

Write a recursive helper method:

public static squareRootGuess(double x, double g). If g2 is approximately equal to x (i.e. |x – g^2| < 0.0001), return g, otherwise, return squareRootGuess with the better new guess ((g + x/g)/2). Then write a method public static squareRoot(double x) that uses the helper method.

Example: Calculate the square root of x = 10; initial guess = 7

squareRootGuess(10, 7) -> |10 – 7^2| > 0.0001;

return new guess = (7 + 10/7)/2 = 4.21428

squareRootGuess(10, 4.21428) -> |10 – (4.21428)^2| > 0.0001;

return new guess = (4.21428 + 10/4.21428)/2 = 3.29358

squareRootGuess(10, 3.29358) -> . . .

. . .

. . .

Java program. Please, the program can compile and run. Thanks!!

Explanation / Answer

import java.util.Scanner;
/** method to approximate the square root of a given number.
*/
public class CalculateSquareRoot
{
    public static void main(String[] args)
    {
        //prompt the user for n
        Scanner in = new Scanner(System.in);
        System.out.print("Enter a number: ");
        double x = in.nextDouble();
        System.out.println(" The square root is " + squareRoot(x));
    }
  
    //function which calls guess square root method
    public static double squareRoot(double x)
    {
        return squareRootGuess(x, 5);
    }
  
    //function which calculates approx square root value
    private static double squareRootGuess(double x, double g)
    {
        //initialize approx diff value
        final double epsilon = .0001;
        //if difference < diff value
        if (Math.abs(x - g * g) < epsilon)
        {
            return g;
        }
      
        //otherwise recursively cal square root
        else
        {
            return squareRootGuess(x, 0.5 * (g + x / g));
        }
    }
}

Sample Output:

Enter a number: 16                                                                                                                                              

The square root is 4.0000001858445895

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