CODEWITHSUNDEEP
cwhile-loop
Lea
Lea

Reputation: 31

Calculating e constant using while loops in C

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!

Upvotes: 0

Views: 2800

Answers (4)

kami
kami

Reputation: 1

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

Upvotes: 0

Kwanghyun Valken Lee
Kwanghyun Valken Lee

Reputation: 1

#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.

Upvotes: 0

chux
chux

Reputation: 154174

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...

Upvotes: 1

MikeCAT
MikeCAT

Reputation: 75062

  • 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;
}

Upvotes: 2

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