Euler's number question

[dpw02]

Hello, I am having a problem with my following lab assignment. I’ve included what I’ve completed so far. I am not sure if it is correct. If anyone can help me out, I would greatly appreciate it.

This is the problem in question:

Euler's number, "e" is used as the basic of natural logarithms. It can be approximated using the following formula:

e = 1 + 1/1! + 1/2! + 1/3! + 1/4! 1/5! + 1/6! + ... + 1/(n-1)! + 1/n!

Write a program that approximates "e" using a loop that terminates when difference between two successive values of "e" differ by less than 0.0000001.

This is what I’ve started/completed so far.

***start***

main()
{
/*repeat until euler(n) - euler(n-1)<=0.0000001 */
}

float euler(int n)
{
if(n==0)
return 1;
else
return (1/factorial(n) + euler(n-1));
}

int factorial(int n)
{
.
.
.
}

***is this correct?***

euler(0) => 1

euler(1) => 1/1! + euler(0)
1/1! + 1

euler(2) => 1/2! + euler(1)
1/2! + 1/1! + 1

euler(3) => 1/3! + euler(2)
1/3! + 1/2! + 1/1! + 1

euler(4) => 1/4! + 1/3! + 1/2! + 1/1! + 1
1/4! + euler(3)

***end***

[Nick]

Don't do it recursivly. Have a running sum, add 1/k! to the sum and then if 1/k! is less than 0.0000001 stop else repeat.

[dpw02]

I am still not sure how to pill this all together...

[Shiro]

It is correct. I understand you need help with the factorial? I saw you understand recursion, so the factorial shouldn't be too hard.

Factorial:

n! = n x (n - 1) x (n - 2) x ... x 3 x 2 x 1

So the end-step is 1 and the recursion step is n x (n - 1).

int fac (int n)
{
    if (n == 1)
        return 1;
    else
        return (n * fac (n - 1));
}