The Euler’s number e is a mathematical constant approximated at 2.718. The formu

Question

The Euler’s number e is a mathematical constant approximated at 2.718. The formula for computing it is as follows:

e = 1 + 1/1! + 1/2! + 1/3! + 1/4! + ...
where n! = n×(n-1)×(n-2)×...×1, e.g., 4!=4×3×2×1

Write a program in C named euler.c that approximates the value of e by using the following equation:

e = 1 + 1/1! + 1/2! + 1/3! + 1/4! + ... + 1/n! where n is a number entered by the user.

Your program will first compute e as 1, then as 1 + 1/1!, then as 1 + 1/1! + 1/2! and so on. In each step, print the step number and the value of e, as follows:

1: 2.000000

2: 2.500000

3: 2.666666

... ...

Hint:
(1) First ask the user for an input integer n (n should be no less than 1);
(2) Your program should consist of two for loops: The first loop should have a counting variable i that counts from 1 to n. Each value of i corresponds to one step as we just mentioned above. The second loop, which should be inside the body of the first loop, computes i! for a given i. For example,

int product = 1;
for(int j = i; j > 0; j--){

product *= j; }

Explanation / Answer

#include <stdio.h>

double eCalc(int terms, double e);

int main()
{
   int dummy;
   int terms;
   double e;
  
   do
   {
       printf("Enter number of terms: ");
       scanf_s("%d", &terms);
   } while (terms <= 0);

   while ((eCalc(terms, e) - eCalc(terms - 1, e)) < 0.0000001)
   {
       printf("An approximation of Euler's number is: %e", e);
   }

   printf(" The program has finished. Press enter to exit. ");
   scanf_s("%d", &dummy);
   return 0;
}

double eCalc(int terms)
{
   if (terms == 1)
       return = 0;
   else
       return = 1 / (terms*(terms - 1));
}

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