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/*****************************************************
Project: **Find e to the Nth Digit**
Author: David Hua (Github: dhd12391)
Date: 8/26/14
Description: Just like the previous problem, but with e instead of PI.
Enter a number and have the program generate e up to that many decimal places.
Keep a limit to how far the program will go.
Program Details: If program is run on command prompt, user can input an integer >= 0
as the first argument N in order for the program to run.
Else, program will prompt user to do so during execution.
Assumption: When N = 0, result is 2.0; When N = 1, result is 2.7; When N = 2, result is 2.71; etc.
******************************************************/
import java.math.*;;
import java.util.Scanner;
import java.io.IOException;
import java.util.InputMismatchException;
public class EToNthDigit{
public static void main(String[]args) throws IOException{
final double E = Math.E;
double eToNthDigit = 0.0;
Scanner input = new Scanner(System.in);
int N;
try{
N = Integer.parseInt(args[0]); //digit limit
}catch(NumberFormatException e){
System.out.println("NumberFormatException: The argument given is not an Integer.");
return;
}catch(ArrayIndexOutOfBoundsException e){
System.out.print("Please provide an Integer as an argument for the program: ");
try{
N = input.nextInt();
}catch(InputMismatchException ex){
System.out.println("InputMismatchException: The argument given is not an Integer.");
return;
}
}
if(N < 0){
System.out.println("ERROR: The number given is a negative number");
return;
}
else{
eToNthDigit = ( Math.floor(E * Math.pow(10, N)) ) / Math.pow(10, N);
System.out.println( "e with " + N + " decimal places is: " + eToNthDigit);
}
//e.g. For N=3: 2.7182... * (10 ^ 3) = 2718.2... ;
//Math.floor() -> 2718.2 becomes 2718.0; 2718.0 / 10^3 = 2.718
}
}