Calculating e constant using while loops in C - c

Question

I recently started programming and I was doing some exercises when I bumped into one that said:
Write a program that can calculate an approximate value of the e constant with the formula e=1+1/1!+1/2!+1/3!+... using while and if if necessary. You cannot use do...while or for.
I wrote my code and I could almost swear the program needs two while loops, but, as you may be guessing, it doesn't work properly. Here's my code:
#include<stdio.h>
int main()
{
float number=3, factorial=1, constant=0, counter=3, variable=0;
float euler=0;
while(counter>1)
{
variable = number;
while(number>1)
{
factorial = factorial*number;
number--;
} // number is now 0, or that's what I think
constant = (1/factorial)+constant;
counter--;
variable = variable-1; // variable is still number?
number = variable; // to have a variable called number again?
}
euler = constant+1; // the 1 in the original formula...
printf("e = : %f\n", euler);
return 0;
}
It doesn't display the correct answer, hope you can help me. Thanks a lot!

Answer 1

Your iteration is too few times. Iterate more to get more accurate value.
You will have to initialize factorial in each loop to calculate factorial in this way.
You forgot to add 1/1!.
Try this:
#include<stdio.h>
int main(void)
{
float number=30, factorial=1, constant=0, counter=30, variable=0;
float euler=0;
while(counter>1)
{
variable=number;
factorial = 1;
while(number>1)
{
factorial=factorial*number;
number--;
}//number is now 1
constant=(1/factorial)+constant;
counter--;
variable=variable-1;
number=variable;
}
euler=constant+1+(1/1.f);//the 1 and 1/1! in the original formula...
printf("e = : %f\n", euler);
return 0;
}

Answer 2

As pointed by #MikeCAT, various coding errors.
As OP's iteration count was low: 3 resulting in low accuracy. As all the terms are eventually added to 1.0 (missed by OP), once a term plus 1.0 is still 1.0, it is about time to quit searching for smaller terms. Typically about 18 iterations with typical double.
When computing the sum of a series, a slightly more accurate answer is available by summing the smallest terms first, in this case, the last terms as done by OP. This can be done using a recursive summation to avoid lots of factorial recalculation.
double e_helper(unsigned n, double term) {
double next_term = term/n;
if (next_term + 1.0 == 1.0) return next_term;
return next_term + e_helper(n+1, next_term);
}
double e(void) {
return 1.0 + e_helper(1, 1.0);
}
#include <stdio.h>
#include <float.h>
#include <math.h>
int main(void) {
printf("%.*f\n", DBL_DECIMAL_DIG - 1, e());
printf("%.*f\n", DBL_DECIMAL_DIG - 1, exp(1));
puts("2.71828182845904523536028747135266249775724709369995...");
}
Output
2.7182818284590451
2.7182818284590451
2.71828182845904523536028747135266249775724709369995...

Answer 3

#include <stdio.h>
#include <math.h>
int main()
{
printf("e = : %.20lf\n", M_E );
return 0;
}
C math library have constant M_E as Euler's number. But, this may not be what you want.

Answer 4

Finding eulers constant 'e' with the formula e = 1 + 1/1! + 1 /2! ....
only using while loop
For beginners
#include <stdio.h>
int main (void) {
float n =5 , fact = 1 , f , x = 0 , e ,i ; //taking input or n as 5 ,our formula now is e = 1+ 1/1! + 1/2! + 1/3! + 1/4! + 1/5! ; as the formula depends upon the input value of n
while(n >=1 ) { /* We need to find factorial of n no of values, so keeping n in a loop starting with 5....1 , atm n is 5 */
f = n ; // let f = n , i.e, f = 5
fact = 1 ;
while(f >= 1 ){ //This while loops finds the factorial of current n value , ie. 5 ;
fact = fact * f ;
f -- ;
}
i = 1 / fact ; // i finds the 1/fact! of the formula
x = i + x ; // x = i + x ; where x = 0 ; , This eq adds all the 1/fact values , atm its adding 1/ 5 !
n-- ; // n decrements to 4 , and the loop repeats till n = 1 , and finally x has all the 1/factorial part of the eulers formula
}
//exiting all the loops since, we now have the values of all the 1/factorial,i.e x : part of the eulers formula
e = 1 + x ; // eulers e = 1 + x , where x represents all the addition of 1/factorial part of the eulers formula
printf ("e : %f",e ); //Finally printing e
return 0 ;
}
Output :
e : 2.716667