Trying to approximate the value of natural log e and the number of term use to calculate e

Question

package homework1C;

public class Homework1C {

public static void main(String[] args){
    double term =2,sum;
    int n;
    final double difference = 0.0000000001;
    double x;
    for(sum=0.0,n=0;term > difference;n++){


        x = find_n_fact(n);
        term=1.0/x;
         sum+=term;
         n++;
    }

        System.out.printf("e : %f\n", sum);
        System.out.printf("term : %d\n", n);

    }

public static int find_n_fact(int n){
int i;
int fact = 2;
for(i = n; i>2;i--){

    fact *= i;
}
    return fact;
}

}

this is what i was being asked to do : Write another Java application program to find and display an approximation of e (natural logarithm). Use the following approximation formula starting with n as 2, incrementing by 1 until two successive values of e differ by less than 0.0000000001 and display not only the approximation, but how many terms of n were used in the last approximation. The formula is: approximation of e = 1/0! + 1/1! + 1/2! + 1/3! + ... , where n! is n factorial

This is my present output for this program

e : 1.043081
term : 20

what am i doing wrong ? the answer was suppose to be

e: 2.71828
term: 15

How to solve this?

Answer 1

Several mistakes you have done:

• Your factorial method was wrong. Although it could be done in iterative manner as you tried, I suggest you the recursive version.
• You are incrementing n in the for-loop of main() twice, that's nonsense.

Here is the fully working code for you:

public class Homework1C {
    public static void main(String[] args) {
        double term = 2, sum = 0;
        final double difference = 0.0000000001;
        int n;

        for (n = 0; term > difference; n++) {
            term = 1.0 / find_n_fact(n);
            sum += term;
        }

        System.out.printf("e : %f\n", sum);
        System.out.printf("term : %d\n", n);
    }

    public static double find_n_fact(int n) {

        if (n == 0 || n == 1)
            return 1.0;

        return n * find_n_fact(n - 1);
    }
}

And iterative version of factorial method is here:

public static double find_n_fact(int n) {
    double i, fact = 1;

    if(n < 0) // for negative numbers, factorial is nonsense.
        return -1;

    for (i = n; i > 1; i--)
        fact *= i;

    return fact;
}

Answer 2

It looks like your factorial function find_n_fact is not correct when n is 0 or 1.

Answer 3

In the sum, the term that follows 1/n! is 1/(n+1)!. That means there is no reason to start all over again (computing (n+1)! from scratch), but instead just divide the current term value by the next n value; ie the loop content only needs to be

    term /= n;
    sum += term;

where you initialize n, term and sum to 1 (since 1/0! is 1) before the loop starts. Of course take n++ out of the loop, because the for statement includes an n++ itself. This approach gets rid of your find_n_fact() function and the errors in it. (Minor note: 1e-10 is more convenient to write than 0.0000000001, and has the same value.) One more suggestion: Add a statement like

System.out.printf("sum : %12.10f   term: %12.10f  1/t: %12.10f,  n: %d\n", sum, term, 1/term, n);

inside your loop when debugging; this will make errors like the extra n++ and the error in the factorial function obvious.