**5.26 (Compute e) You can approximate e using the following series:
e=1+11!+12!+13!+14!+...+1i!
Write a program that displays the e value for i = 10000, 20000, …, and 100000. (Hint: Because i! = i * (i - 1) * ... * 2 * 1, then i/i! is 1/i*(i-1)
Initialize e and item to be 1 and keep adding a new item to e. The new item is the previous item divided by i for i = 2, 3, 4, . . . .)
e=1+11!+12!+13!+14!+...+1i!
Write a program that displays the e value for i = 10000, 20000, …, and 100000. (Hint: Because i! = i * (i - 1) * ... * 2 * 1, then i/i! is 1/i*(i-1)
Initialize e and item to be 1 and keep adding a new item to e. The new item is the previous item divided by i for i = 2, 3, 4, . . . .)
public class ProgrammingEx5_26 { public static void main(String[] args) { double e = 1,p=1; for (int i = 1; i <= 100000; i++) { p=p/i; e += p; if (i == 10000) { System.out.println("e at i = 10000 is " + e); } else if (i == 20000) { System.out.println("e at i = 20000 is " + e); } else if (i == 100000) { System.out.println("e at i = 100000 is " + e); } } System.out.println("Java's e is " + Math.E); } }
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