Java: When dividing by factorials, how do I store the factorial? Its too large to be double

Question

I am new to java, and my program is likely nowhere near as efficient as it could be, but here it is:

public class Compute {
public static void main(String[] args) {
    for(double i = 10000; i <= 100000; i += 10000)
    {
        System.out.println("The value for the series when i = " + i + " is " + e(i));
    }
}
public static double e(double input) {
    double e = 0;
    for(double i = 0; i <= input; i++)
    {
        e += 1 / factorial(input);
    }
    return e;
}
public static double factorial(double input) {
    double factorial = 1;
    for(int i = 1; i <= input; i++)
    {
        factorial *= i;
    }
    return factorial;
}
}

I believe this calculates the value e for i = 10000, 20000, ..., & 100000.
Where e = 1 + (1/1!) + (2/2!) + ... + (1/i!)
It takes about 47 seconds to do so, but I believe it works.

My issue is, for every i, the result is always 0.0
I believe this is because whenever the method factorial is called, the return value is too big to be stored which somehow causes a problem.

What can I do to store the value returned by the method Factorial?

Answer 1

Although you can calculate arbitrary precision results with BigDecimal , there is no need to calculate to 100000! for the series expansion of e . Consider that the 20th term in the series ( 20/20! ) has a magnitude of about 10 ^-19 , so its contribution to the overall total is insignificant.

In other words, the contribution of any terms after the 20th would change only digits after the 19th decimal place.

Answer 2

您可能应该使用java.math.BigInteger来存储阶乘。

Answer 3

Change this

e += 1 / factorial(input);

to

e += 1 / factorial(i);

Lots to do to speed up the code. Think about (i+1)! vs i!, don't recalc the whole factorial every time.

Also stop calculating when the answer will change less than the required precision like Jim said.