11. Calculating the Euler number, e, using Loops

The Euler's number, e, is calculated for the first 17 Taylor Series.

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The recursion formulas, given here, are done. The initial equation sets the first value of innermost level. Then, there are 14 recursive equations, as we go to outer levels, for N = 17. Finally, the final equation gives the approximation to Euler's number.

 * *** 1. Start (recursion)
 * e = 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + ...
 * e = 2 + 1/2 + 1/(2*3) + 1(2*3*4) + 1/(2*3*4*5) + ...
 * e = 2 + 1/2(1 + 1/3 + 1/(3*4) + 1/(3*4*5)) + ...
 * e = 2 + 1/2(1 + 1/3(1 + 1/4 + 1/(4*5))) + ...
 * e = 2 + 1/2(1 + 1/3(1 + 1/4(1 + 1/5))) + ...
 * *** 1. End

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The factorial, is based on a different recursive equation. The numbers in the factorial become very large, very quickly, and thus it returns a long integer.

 // *** 2. Start (factorial)
 static long factorial(int n) {
  if (n==0) return 1;
  else return n*factorial(n-1);
 }
 // *** 2. End

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The Taylor Series is known for all differentiable functions. For exponent, all derivatives are the same.

  // *** 3. Start (for - Taylor series)
  double e = Math.E; // reference
  int N = 17; // Number of Taylor terms
  double a = 0.0; // approximation
  int i; // iteration term
  for (i = 0; i<=N; i++) a = a + 1.0/factorial(i);
  double error = e - a;
  System.out.println("error = " + error);
  // *** 3. End

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The algorithm is done using the while loop.

  // *** 4. Start (while - recursion)
  a = 1 + 1.0/N; // initial
  i = 0;
  while ((N-i)>3) {
   i++;
   a = 1 + a/(N-i); // recursion
  }
  a = 2 + 0.5*a; // final
  error = e - a; // reference - final
  System.out.println("error = " + error);
  // *** 4. End 

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The algorithm is done using the for loop.

  // *** 5. Start (for - recursion)
  a = 1 + 1.0/N; // initial
  for (i=1; i<=(N-3); i++) {
   a = 1 + a/(N-i); // recursion
  }
  a = 2 + 0.5*a; // final
  error = e - a; // reference - final
  System.out.println("error = " + error);
  // *** 5. End

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The algorithm is done using the do-while loop. Even though the while condition is false in the 14th time, the check is done after loop, so this only means that there is no 15th time.

  // *** 6. Start (do while - recursion)
  i = 0;
  a = 1 + 1.0/N; // initial
  do {
   i++;
   a = 1 + a/(N-i); // recursion
  } while ((N-i)>3);
  a = 2 + 0.5*a; // final
  error = e - a; // reference - final
  System.out.println("error = " + error);
  // *** 6. End

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// Ex11.java
/*
 * *** 1. Start (recursion)
 * e = 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + ...
 * e = 2 + 1/2 + 1/(2*3) + 1(2*3*4) + 1/(2*3*4*5) + ...
 * e = 2 + 1/2(1 + 1/3 + 1/(3*4) + 1/(3*4*5)) + ...
 * e = 2 + 1/2(1 + 1/3(1 + 1/4 + 1/(4*5))) + ...
 * e = 2 + 1/2(1 + 1/3(1 + 1/4(1 + 1/5))) + ...
 * *** 1. End
 */
package com.javaAndroid.ex11;

public class Ex11 {
 // *** 2. Start (factorial)
 static long factorial(int n) {
  if (n==0) return 1;
  else return n*factorial(n-1);
 }
 // *** 2. End
 public static void main(String[] args) {
  // *** 3. Start (for - Taylor series)
  double e = Math.E; // reference
  int N = 17; // Number of Taylor terms
  double a = 0.0; // approximation
  int i; // iteration term
  for (i = 0; i<=N; i++) a = a + 1.0/factorial(i);
  double error = e - a;
  System.out.println("error = " + error);
  // *** 3. End
  // *** 4. Start (while - recursion)
  a = 1 + 1.0/N; // initial
  i = 0;
  while ((N-i)>3) {
   i++;
   a = 1 + a/(N-i); // recursion
  }
  a = 2 + 0.5*a; // final
  error = e - a; // reference - final
  System.out.println("error = " + error);
  // *** 4. End 
  // *** 5. Start (for - recursion)
  a = 1 + 1.0/N; // initial
  for (i=1; i<=(N-3); i++) {
   a = 1 + a/(N-i); // recursion
  }
  a = 2 + 0.5*a; // final
  error = e - a; // reference - final
  System.out.println("error = " + error);
  // *** 5. End
  // *** 6. Start (do while - recursion)
  i = 0;
  a = 1 + 1.0/N; // initial
  do {
   i++;
   a = 1 + a/(N-i); // recursion
  } while ((N-i)>3);
  a = 2 + 0.5*a; // final
  error = e - a; // reference - final
  System.out.println("error = " + error);
  // *** 6. End
 }
}

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