4.25 (computing ) You can approximate by using thefollowing series: = 4(1-1/3 +

Question

4.25 (computing ) You can approximate by using thefollowing series:

= 4(1-1/3 + 1/5-1/7 + 1/9-1/11 + 1/13-.........- 1/(2i-1) +1/(2i+1) )

Display the value for i = 10000,20000,...... and 100000.

please help with this code , cuz i cannot proceed

this is my code


#include <iostream>
using namespace std;

int main()
{
  
    int number;
    double pi , series_neg = 0,series_pos = 0;
    double j = 3;
  
    cout <<"Enter the number ::";
    cin >> number;
    while(j <= number)
    {

    for(int i = 3; i <= number ;i++)
    {
    if((i % 2) != 0)
    {
    series_neg += (1/(2*j-1));
    series_pos += (1/(2*j+1));

    }
  
    }
    j++;
    }
    pi = (1 - (1/ ( series_pos + series_neg)));
    cout <<"The value of pi is " << pi<<" ";
   
    system("pause");
    return 0;
  
  
}

Explanation / Answer

please rate - thanks be patient-it runs very slowly #include using namespace std; int main() {        double pi, series_neg,series_pos;    int i, j;        for(j=10000;j

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