Question

In my quest to learn C I’ve come across a task which is causing me a few problems. I need to make an equation for the approximate value of the formulae n!, which can be described as:

n! = n^n*e^(-n)*sqrt(2(2*n+1/3)*PI), however I simply cannot get my values to corrospond with the actual value. 5! = 120ish

I can get a value of some 148ish

Can’t figure out where my code is wrong:

#include <stdio.h>
#include <math.h>

#define PI 3.14156
#define E_CONST 2.7828

int main ()
{

double num;
double calc, first, second, third, fourth;

printf("Give an int: ");
scanf("%lf", &num);


first = pow(num , num);

second = pow(E_CONST, -num);

third = (2 * num + 1/3);

fourth = sqrt(2*third*PI);

//calc = first * second * fourth;

calc = pow(num, num) * pow(E_CONST, -num) * sqrt(2*(2*num+(1/3))*PI);
printf("Input: %f", num);

printf("1: %.2f\n2: %.10f\n3: %.8f\n4: %.2f\n", first, second, third, fourth);

printf("\nInt was: %.2f\n\nApproximate number: %.5f", num, calc);

return 0;
}

Feel like i have tried everything. The code is a bit messy, but it’s because I’ve scrambled so much with it now.

Answer 1

3.14156 is a bad value for PI: it’s better to use 3.1416, or 3.14159, or 4 * atan(1), or, for POSIX implementations, M_PI.

2.7828 is a very bad value for e: it’s better to use 2.7183, or exp(1), or, for POSIX implementations, M_E.

1/3 is integer division, the result is 0: it’s better to use 1.0/3.

Also your approximation is incorrect. The correct approximation is

n^n * e^(-n) * sqrt((2*n+1/3)*PI)