Question

(previous one did not work)

You've been hired by PI Throwers to write a C++ console application that approximates PI. Use a validation loop to prompt and get from the user the number of terms to approximate PI to that is at least 1. Use the following Leibniz formula:PI approximation = 4 * (1/1 – 1/3 + 1/5 – 1/7 + 1/9 – 1/11 + ...)The terms appear within the parentheses in the formula. A PI approximation to ...

● One term is 4 * (1/1) = 4

● Four terms = 4 * (1/1 – 1/3 + 1/5 – 1/7) = 2.8952380952

● Seven terms = 4 * (1/1 – 1/3 + 1/5 – 1/7 + 1/9 – 1/11 + 1/13) = 3.2837384837

Use a for statement to calculate the number that will be multiplied by four. The loop will add and subtract terms to get the number. If i is the indexing variable, use condition i % 2 == 0 to determine whether each loop iteration is even or odd and whether an add or subtract is done. In each term, the dividend is always 1, and the divisor is an odd number starting with 1 that goes up two at a time. Format the approximation to ten decimal places. Use number of terms 1,000 for your last run.

Answer 1

#include<iostream>
#include<iomanip>

using namespace std;

int main() 
{
   int n;
   double PI = 0;
   cout << "Enter number of terms: ";
   cin >> n;//Input number of terms
   while(n < 1)//Input validation
   {
      cout << "Enter valid number of terms(atleast 1): ";
      cin >> n;
   }
   int divisor = 1;
   for(int i = 1; i <= n; i++)
   {
      if(i % 2 == 0)//Even iteration
         PI -= (1.0/(divisor));
      else//Odd iteration
         PI += (1.0/(divisor));
      divisor += 2;
   }
   PI *= 4;//Multiply summed values by 4
   cout << "Approximation of PI = " << setprecision(10) << fixed << showpoint << PI;
   return 0;
}

SAMPLE OUTPUT:

CAUsersShubham\Desktop ProgramTwo.exe Enter number of terms: 0 Enter valid number of terms (atleast 1): -1 Enter valid number of terms (atleast 1): 1000 Approximation of PI3.1405926538 Process exited after 4.792 seconds with return value Press any key to continue .. .