13. Calculating pi

We will calculate pi using one of its series approximation. We can only calculate upto 16-decimal positions with 64-bit double. This is done after about 25 terms. The user has to type the number of terms.

---

This is the arctan series. It converges for x less than 1. It converges at 1, but extremely slowly. The graph, or convergence, is shown in a later slide. As the angle x decreases, the rate of convergence increases rapidly. It is based on the geometric series, as the derivative can be written as an infinite series.

---

For angle of x = 1 radians (57.3 degrees), we have a slowly converging series. It is too slow to be useful.

---

Here, two small angles are chosen according to this right triangle. Both alpha and beta are smaller than 45 degrees (pi/4=0.785). In fact, the sum of alpha and beta is exactly 45 degrees.

---

These are equations to find the two angles, and the identity is shown here, as the last equation.

---

In the blue line, we can see the slow convergence for x = 1. The faster convergence is shown by red line, which is based on the angles atan(1/2) = 26.565 degrees, and atan(1/3) = 18.435 degrees. As angles decrease, we have even faster convergence, that is, it will go to machine limit, at smaller N.

---

This shows the recursive equations in for loop. The program prompts for N, corresponding to the number of terms. Should the user type a letter or anything which is not-a-number, an Exception is raised and a default value of N used.

/*
 * Using identity: arctan(1/2)+arctan(1/3) = pi/4
 * As angles become smaller, the seies converge at 
 * faster rate. Many better approximations
 * with smaller angles exist.
 * arctan(x) = x - x^3/3 + x^5/5 - x^7/7 + x^9/9 - x^11/11 + ...
 * = x - x^2(x/3 - x^3/5 + x^5/7 - x^7/9 + x^9/11)
 * = x - x^2(x/3 - x^2(x/5 - x^3/7 + x^5/9 - x^7/11))
 * = x - x^2(x/3 - x^2(x/5 - x^2(x/7 - x^3/9 + x^5/11)))
 * = x - x^2(x/3 - x^2(x/5 - x^2(x/7 - x^2(x/9 - x^3/11))))
 * = x - x^2(x/3 - x^2(x/5 - x^2(x/7 - x^2(x/9 - x^2(x/11))))) + ...
 * arctan(1/2) = 1/2 - 1/4(1/6 - 1/4(1/10 - 1/4 (1/14 - 1/4 (1/18 - 1/4(1/22))))) + ...
 * arctan(1/3) = 1/3 - 1/9(1/9 - 1/9(1/15 - 1/9 (1/21 - 1/9 (1/27 - 1/9(1/33))))) + ...
 */

package com.javaAndroid.ex13;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Ex13 {
 public static void main(String[] args) {
  double pi = Math.PI; // reference
  int N = 0; // Number of terms
  BufferedReader br = new BufferedReader( new InputStreamReader(System.in));
  System.out.println("How many terms?");
  try {
   N = Integer.parseInt(br.readLine());
  } catch (IOException e) {
   e.printStackTrace();
  } catch (NumberFormatException e) {
   System.out.println("You probably want N = 20");
   N = 20;
  }
  System.out.print("For " + N + " terms, the error is: ");
  double a = 1.0/(4*N+2); // initial
  double b = 1.0/(6*N+3); // initial
  for (int i = N-1; i>=0; i--) {
   a = 1.0/(4*i+2) - a/4.0;
   b = 1.0/(6*i+3) - b/9.0;
  }
  double error = 4*a+4*b-pi;
  System.out.println(error);
 }
}

---

---