C++ Homework project, need help (simple problem)

[Zopharr]

We were assigned the following problem as a large project in my Freshman intro to C++ programming class:

We have to determine the number of terms needed to calculate Pi to a specific number (3.0, 3.1, 3.14, 3.141, 3.1415, and 3.14159), using the formula:

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

I thought I had the code figured out pretty easily, but I have having a problem. I was going to use the following (incomplete) code as a model:

// Date: 10/30/06
// Homework #07
// Student: Jim Hummel
// Course:  CSE 121
// This program calculates the number of terms required for an approximation of pi.

#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
double pif(double q);
int terms=0;
int main()
{double q=3.1;
cout << "Pi Value    Terms\n--------------------------\n";
cout << fixed << setprecision(2);
cout << pif(q)<< "        "<< terms << endl; 
return 0;
}
double pif (double q)
{double n=1.0;
double pi=0;
    while (pi<(q-(1.0/q)) && pi>(q+(1.0/q)))
    {
        pi = pow(-1.0,n)/(2.0*n-1.0);
        n++;
            terms++;
    }
    pi*=4.0;
    return pi;
}

In the main function there, I was going to try to have it call the function pif() once for each value q that I set to it, then cout the results, and the number of terms.

I've been toying around with possible options (as seen in the while statement there), but no matter what I do, it returns a value of o terms and 0.00 for pi each time.

What am I doing wrong?

Thanks in advance, this is killing me.

[WolfPack]

What is the meaning of q?

[Zopharr]

Sorry, q was to represent the question.

I was going to have it so that I would set q to 3.0 and send it to pif()... then cout the results, and then set q to 3.1... repeat with 3.14, 3.141, all the way to 3.14159, which is as far as I need to go. I need to know the number of terms used for each of those values (how many fractions were necessary to get pi to equal that number.)

Thanks.

[Zopharr]

Maybe this could help: This is the exact problem:

Problem 2. [The Number p] The number p may be calculated using the following infinite series:
p = 4(1-1/3+1/5-1/7+1/9-1/11+1/3
How many terms of this series you need to use before you get the approximation p of p, where:
a) p = 3.0
b) p = 3.1
c) p = 3.14
d) p = 3.141
e) p = 3.1415
f) p = 3.14159
Write a C++ program to answer this question.

[WolfPack]

I don't understand why you are testing how close the value calculated is to 1/q. But anyway, here is a more refined way of writing the source. The algorithm is the same, so any errors are due to that.
Do not use global variables unless absolutely necessary.

#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
double pif(double q);
int terms=0;
int main()
{
    double q=3.1;
    cout << "Pi Value    Terms\n--------------------------\n";
    cout.setf( ios::fixed );
    cout.precision (2);
    cout << pif(q)<< "        ";
    cout << terms << endl;  // Since you are changing the value of terms in the above line, 
                            // do not output it in the same line. The result is undefined.
                            // Use a seperate line to output terms.
                            // But a better way will be to output the value of n in pif()
    return 0;
}
double pif (double approx) // Use meaningful parameter names rather than just p, q, r
{
    long n=0;
    double pi=0.0;
    double error = fabs( pi - approx );
    double tolerance =  1 / approx; // Assuming approx is positive, this is also +ve
//     while (pi<(q-(1.0/q)) && pi>(q+(1.0/q)))// Actually this should be 
                                               // while (pi<(q-(1.0/q)) || pi>(q+(1.0/q))).
                                               // It is the same as | pi -q | > ( 1.0 / q )
                                               // |x| = absolute value of x
    while ( error > tolerance ) 
    {
        n++;
        term++;
        pi = pi + pow( -1, n + 1)/(2 * n - 1 );
        error = fabs( pi - approx); 
    }
    pi*=4.0;
    // cout << pi << "        "<< n << "         " << error << endl;
    return pi;
}

[Zopharr]

Thanks for the reply, although I didn't read it in time before I had to hand it in.

I worked on building an entirely new code without the rediculous while statements... it led to a very long, messy, and unoganized code, but it worked.

I spent the remainder of my time trying to turn it into a while loop, using another variable for the setprecision (as seen below) and the q-(number) statements, but this code here did work and got the job done:

// Date: 10/30/06
// Homework #07
// Student: Jim Hummel
// Course:  CSE 121
// This program calculates the number of terms required for an approximation of pi.

#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
int main()
{double q=3.0;
double n=1;
    bool check=false;
double pi = 0;
     int terms=0;
     while (!check) 
     {
         pi += (double) pow(-1, n+1)/(2*n-1);
cout << fixed << setprecision(1);
             if (q-.1<=pi*4 && q+.1>=pi*4)
                 check=true;
             else
                 terms++;
             n++;
     } 
pi *= 4; 
cout << terms << "        " << pi << endl; 



//Part 2//



q=3.1;
n=1;
    check=false;
pi = 0;
    terms=0;
     while (!check) 
     {
         pi += (double) pow(-1, n+1)/(2*n-1);
cout << fixed << setprecision(1);
             if (q-.01<=pi*4 && q+.01>=pi*4)
                 check=true;
             else
                 terms++;
             n++;
     } 
pi *= 4; 
cout << terms << "        " << pi << endl;



//Part 3//



q=3.14;
n=1;
    check=false;
pi = 0;
    terms=0;
     while (!check) 
     {
         pi += (double) pow(-1, n+1)/(2*n-1);
cout << fixed << setprecision(2);
             if (q-.001<=pi*4 && q+.001>=pi*4)
                 check=true;
             else
                 terms++;
             n++;
     } 
pi *= 4; 
cout << terms << "        " << pi << endl;



//Part 4//



q=3.141;
n=1;
    check=false;
pi = 0;
    terms=0;
     while (!check) 
     {
         pi += (double) pow(-1, n+1)/(2*n-1);
cout << fixed << setprecision(3);
             if (q-.0001<=pi*4 && q+.0001>=pi*4)
                 check=true;
             else
                 terms++;
             n++;
     } 
pi *= 4; 
cout << terms << "        " << pi << endl;



//Part 5//



q=3.1415;
n=1;
    check=false;
pi = 0;
    terms=0;
     while (!check) 
     {
         pi += (double) pow(-1, n+1)/(2*n-1);
cout << fixed << setprecision(4);
             if (q-.00001<=pi*4 && q+.00001>=pi*4)
                 check=true;
             else
                 terms++;
             n++;
     } 
pi *= 4; 
cout << terms << "        " << pi << endl;



//Part 6//



q=3.14159;
n=1;
    check=false;
pi = 0;
    terms=0;
     while (!check) 
     {
         pi += (double) pow(-1, n+1)/(2*n-1);
cout << fixed << setprecision(5);
             if (q-.000001<=pi*4 && q+.000001>=pi*4)
                 check=true;
             else
                 terms++;
             n++;
     } 
pi *= 4; 
cout << terms << "        " << pi << endl;
return 0;
}