write a java method named sqrt() that will calculate the square root of a positi

Question

write a java method named sqrt() that will calculate the square root of a positive double to within a specified level of tolerance. The signature of the method should be:

public static double sqrt(double number, double tolerance)

The basic solution logic, is that sqrt() will call a private recursive method that has two additional parameters, lowGuess and highGuess, it will have the signature:

private static double sqrt( double number, double lowGuess, double highGuess, double tolerance)

Each time the private method is called, it will calculate a newGuess that is the mid-point between the lowGuess and the highGuess, There are three possible cases:

1) The base case is satisfied when:

(highGuess - newGuess) / newGuess < tolerance, and newGuess is returned

2) otherwise, if newGuess*newGuess < number, make a recursive call substituting newGuess for lowGuess.

3) Or, if newGuess * newGuess < number, make a recursive call substituting newGuess for lowGuess.

The public sqrt method will provide 0 and number as the original lowGuess and highGuess values to the private recursive version, and it will return the result eventually returned by its private helper.

Write a public java class named Lab4A to contain the two overloaded sqrt methods, and a main() method that will, Using console I/O, promp the user in a continuous loop to enter a positive number and a tolerance, and display the final square root found from the methods above. The input loop should be exited if the number entered is zero or negative.

Explanation / Answer

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import java.util.*;
public class Lab4A

{public static void main(String args[])
{double number,tolerance;
Scanner in =new Scanner(System.in);
System.out.print("Enter a number whose square root you want(<=0 to exit): ");
number=in.nextDouble();
while(number>=0)
{System.out.print("Enter tolerance: ");
tolerance=in.nextDouble();
System.out.println("sqrt("+number+")="+sqrt(number, tolerance));
System.out.print("Enter a number whose square root you want(<=0 to exit): ");
number=in.nextDouble();
}
}

public static double sqrt(double number, double tolerance)
{return sqrt(number, 0, number, tolerance);
}

public static double sqrt(double number,double lowGuess,double highGuess,double tolerance)
{double newGuess=(lowGuess + highGuess) / 2;
if ((highGuess-newGuess) / newGuess < tolerance)
return newGuess;
else if (newGuess * newGuess > number)
return sqrt (number, lowGuess, newGuess, tolerance);
else if (newGuess * newGuess < number)
return sqrt (number, newGuess, highGuess , tolerance);
else
return newGuess;
}
}

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