(Approximate e) Programming Exercise 5.26 approximates e using the following series:

/*********************************************************************************
* (Approximate e) Programming Exercise 5.26 approximates e using the following   *
* series:                                                                        *
*                                                                                *
*       e = 1 + (1 / 1!) + (1 / 2!) + (1 / 3!) + (1 / 4!) + ... + (1 / i!)       *
*                                                                                *
* In order to get better precision, use BigDecimal with 25 digits of precision   *
* in the computation. Write a program that displays the e value for              *
* i = 100, 200, . . . , and 1000.                                                *
*********************************************************************************/
import java.math.BigDecimal;

public class Exercise_10_20 {
	/** Main method */
	public static void main(String[] args) {
		// Displays the e value for i = 100, 200, . . . , and 1000.
		System.out.println("\n The e values for i = 100 to 1000:");
		System.out.println("-----------------------------------");
		System.out.println(" i                e");
		System.out.println("-----------------------------------");
		for (BigDecimal i = new BigDecimal("100"); 
			i.compareTo(new BigDecimal("1000")) <= 0; 
			i = i.add(new BigDecimal("100"))) {
			System.out.println(i + "    " + getE(i));
		}
	}

	/** Return e value for i */
	public static BigDecimal getE(BigDecimal v) {
		BigDecimal one = new BigDecimal("1");
		BigDecimal e = new BigDecimal("0.0");
		for (BigDecimal i = one; i.compareTo(v) <= 0; i = i.add(one)) {
			BigDecimal denominator = i;
			for (BigDecimal k = i.subtract(one); 
				k.compareTo(one) >= 1; 
				k = k.subtract(one)) {
				denominator = denominator.multiply(k);
			}
			// Use 25 digits of precision
			e = e.add(one.divide(denominator, 25, BigDecimal.ROUND_UP));  
		}
		return e;
	}
}