How to calculate pi with 3 decimals precision in C++? - c++

Question

hey guys I am trying to calculate pi using this formula:
pi = 4 · [ 1 – 1/3 + 1/5 – 1/7 + 1/9 ... + (–1)^n/(2n + 1) ]
yet i always get a zero for my output pi value and I am really confused as to where I had gone wrong. Here is my code:
#include <cmath>
#include <iostream>
using namespace std;
int main()
{
int n;
double b = 0;
char c = 'Y';
int s = 1;
while (c == 'Y') {
cout << "Enter the value of the parameter 'n' in the Leibniz formula (or -1 to quit):" << endl;
cin >> n;
if (n != -1) {
c = 'Y';
for (int a = 1; a <= n; a++) {
s = -s;
b += 4 * (s/ (2 * a + 1));
}
cout << "The approximate value of pi using 1 term is:" << b << endl;
}
else {
c = 'N';
}
}
return 0;
}

Answer 1

In both C and C++, mathematical operations on integers result in an integer even if the result would be fractional in conventional mathematics. Change your int to a float or double and I suspect that it will work better.
The result is truncated to the integer value and has an integer type.
So for example: 2 / 4 results in 0 and 5 / 2 would result in 2.
NOTE if you perform an operation between a floating point value and an integer value, the result is a floating point value. So:
2.0 / 4 == 0.5

Answer 2

Your code seems to be complicated and int type is used in places where floating operations are expected.
Consider the following simplified example:
#include <cmath>
#include <iostream>
using namespace std;
int main()
{
int n = 0;
double b = 0;
double s = 1; // Tytpe is changed
while (n != -1) { // there is no need for char c
cout << "Enter the value of the parameter 'n' in the Leibniz formula (or -1 to quit):" << endl;
cin >> n;
b = 0; // init b before starting the loop
s = 1; // the same for s (it can be -1 from the next user input)
// there is no need for if (n != -1) because for has condition
for (int a = 1; a <= n; a++) {
s = -s;
b += 4 * (s / (2.0 * a + 1));
}
cout << "The approximate value of pi using 1 term is:" << b << endl;
}
return 0;
}
IMPORTANT UPDATE:
To make your calculation correct (in terms of Leibniz's formula) I suggest the following changes in the for loop:
for (int a = 0; a <= n; a+=2) { // start from 0 with step 2
b += 4.0 * (s / (a + 1.0));
s = -s; // change the sign for next calculation
}
and further, consider some kind of optimization
b = 0; // do not forget about reseting b to 0 before making sum
s = 1; // set 1 in the sign
for (int a = 0; a <= n; a+=2) { // start from 0 with step 2
b += s / (a + 1.0); // no multiplication on each iteration
s = -s; // because s was initialized with 1
}
b *= 4.0; // multiply once for the whole sum
UPDATE 2
For case if precision is really important for output, final snippet can be like:
#define _USE_MATH_DEFINES
#include <cmath>
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
int n = 0;
double b = 0;
double s = 1;
int prec = 0;
cout << "What precision should be used for output? (Value from 1 to 10): ";
while (prec< 1 || prec > 10)
{
cin >> prec;
}
while (true) {
cout << "Enter the value of the parameter 'n' in the Leibniz formula (or -1 to quit):" << endl;
cin >> n;
if (n == -1)
{
break; // go out the loop if user enter -1 (want to exit)
}
else if (n <= 0)
{
cout << "'n' have to be 1 or greater" << endl;
continue; // go to the next iteration to ask new 'n'
}
s = 1;
b = 1.0; // we can start from 1 (no need to claculate the first term) and make loop from 2
for (int a = 2; a < n*2; a+=2) { // start from 2 with step 2 (so n should be doubled)
s = -s; // change the sign for this iteration, because now loop started from a = 2
b += s / (a + 1.0);
}
b *= 4.0;
cout << "The approximate value of pi using 1 term is: " << setprecision(prec+1) << b << " (PI = " << M_PI << ")" << endl;
}
return 0;
}
Note:
In this version b initialized with 1.0 because the first item in the Leibniz series is always 1 (we can skip calculation, but we should change the logic for sign changes - make s = -1; or move s = -s; before summation - I choose the 2nd option).
Also I'am not sure what is "parameter 'n' in the Leibniz formula", so pay attention to condition of for loop - now (with a < n*2) it is correct for case if n is number of items in the Leibniz series to be calculated.

Answer 3

Along with doing integer math, you have a few other minor problems.
First, the formula is [1 - ...], not [0 - ...], so you need to initialize b to 1.0, not 0.
Second, it's supposed to be 4 * [...], but you're multiplying by 4 on every iteration of the loop, so you're getting `[0 - b1 * 4 + b2 * 4 -b3 * 4 ....].
You can distribute the multiplication if you want to, but if you do you'll need to distribute it correctly (e.g., the starting value of 1.0 would also need to be multiplied by 4).
Also note that you're not re-initializing correctly, so the second (and subsequent) times you attempt to re-compute the value, you'll get completely incorrect answers (until you fix more stuff).

Answer 4

You've been burned by integer division.
b += 4 * (s/ (2 * a + 1));
a is an int so the division result is an int.
A cast to double will fix it:
b += 4 * (s/ (2 * double(a) + 1));