Run time errors while computing e^x - c

Question

Continuing from the question I posted before How to increase precision for function e^x
I made few changes in my code by taking advices given
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
long double exponential(long double x, long double n, long double p)
{
long double i = 1;
while (n > 0) // loop stops when becomes less than 0
i = i + (x / p) * (exponential(x, n - 0.0001, p + 1));
if (n < 0) // when n reaches 0 or becomes less than zero value of i will be returned
return i;
}
int main()
{
long double p, x, n;
scanf("%Lf", &x);
printf("math.h e^x = %lf\n", exp(x));
printf("calculated e^x = %Lf\n", exponential(x, 1, 1));
return 0;
}
But I am not getting any output its just giving run time error(http://codepad.org/jIKoYGFC) and I don't know why . Please can some one help me why I am getting these errors

Answer 1

That loop is completely bogus. You're not writing an iterative function (where it would make more sense). Further you have an edge case of zero returning undefined content.
Though I do not recommend floating point for loop control, nor do I advise thousands of invocations into recursive calls, your code should be more like
long double exponential(long double x, long double n, long double p)
{
long double i = 1;
if (n > 0)
i += (x / p) * (exponential(x, n - 0.0001, p + 1));
return i;
}
which ultimately is just this:
long double exponential(long double x, long double n, long double p)
{
return 1 + ((n > 0) ? (x / p) * exponential(x, n - 0.0001, p + 1) : 0);
}
Fixing that (either way):
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
long double exponential(long double x, long double n, long double p)
{
return 1 + ((n > 0) ? (x / p) * exponential(x, n - 0.0001, p + 1) : 0);
}
int main()
{
long double x = 5;
printf("math.h e^x = %Lf\n", expl(x));
printf("calculated e^x = %Lf\n", exponential(x, 1, 1));
return 0;
}
Output
math.h e^x = 148.413159
calculated e^x = 148.413159

Answer 2

Created a version using LibGMP out of curiosity to see how this will improve the outcome.
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <gmp.h>
#define BITS 128
The line above mean mantissa is 128 bits and exponent of 64 bits. This is my GMP based function:
void gmp_exponential(mpf_t * i, double x, double n, int p)
{
mpf_t ii;
mpf_init(ii);
mpf_set_d(ii,1);
if (n > 0){
mpf_t a,b,c;
gmp_exponential(&ii, x, n - 0.0001, p + 1);
mpf_inits (a, b, c, NULL);
mpf_set_d(a,x);
mpf_div_ui(b, a, p) ;
mpf_mul(c, b, ii);
mpf_add (*i,*i,c);
mpf_clears(a,b,c,NULL);
}
mpf_clear(ii);
}
Reusing the function from WhozCraig for comparison:
long double exponential(long double x, long double n, long double p)
{
return 1 + ((n > 0) ? (x / p) * exponential(x, n - 0.0001, p + 1) : 0);
}
And code to run it all:
int main()
{
double x = 30.0;
mpf_t i;
mpf_init2 (i, BITS);
mpf_set_d(i,1);
gmp_exponential(&i, x, 1, 1);
printf ("math.h e^x = %Lf\n", expl(x));
gmp_printf ("calculated e^x with mpf = %.*Ff\n", 6, i);
printf ("calculated e^x = %Lf\n", exponential(x, 1, 1));
mpf_clear(i);
return 0;
}
Output:
math.h e^x = 10686474581524.462147
calculated e^x with mpf = 10686474581524.462143
calculated e^x = 10686474581524.462149