Question

Replace <missing code> with your answer. Your code should approximate the value of π (~3.141592653589793). You can approximate pi using the following formula:

π = 4 * (1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + 1/13 - 1/15 + …)

For this problem, let n be the number of terms to add in the series. For example, if n = 5, your code should compute the following: 4 * (1 - 1/3 + 1/5 - 1/7 + 1/9). If n = 3, your code should compute 4 * (1 - 1/3 + 1/5)

public static void main(String[] args)
{
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();

int piApprox = <missing value>;

<missing code>

System.out.println("piApprox = " + piApprox);

}

Answer 1

import java.util.Scanner;

public class CalcPiApprox {

   public static void main(String[] args) {
       Scanner scanner = new Scanner(System.in);
       int n = scanner.nextInt();

       // we need to make piApprox double
       // otherwise, values will be rounded off
       double piApprox = 0;

       // Initially, set denominator for term to 1
       double denom = 1;
       int terms = 0;

       while (terms < n) {
           // Increment denom by 2 and term by 1 after calculating one term
           piApprox = piApprox + 1 / denom;
           denom += 2;
           terms += 1;

           // Check if term = n, if yes break
           // This could be the case when n is odd
           if (terms == n) {
               break;
           }
           // Increment denom by 2 and term by 1 after calculating another term
           piApprox = piApprox - 1 / denom;
           denom += 2;
           terms += 1;
       }

       // Multiply piApprox by 4 and print the result
       piApprox = piApprox * 4;
       System.out.println("piApprox = " + piApprox);
   }
}

OUTPUT