Example of expensive computation

Sometimes I need to simulate some expensive computation, or long-running task, without using Thread.sleep or Object.wait. A real busy waiting. The following example of computing PI is adapted from this RMI tutorial:

package test;
import java.math.BigDecimal;

public class Pi implements java.io.Serializable {
   private static final long serialVersionUID = 227L;
   private static final BigDecimal FOUR = BigDecimal.valueOf(4);
   private static final int roundingMode = BigDecimal.ROUND_HALF_EVEN;
   public static void main(String[] args) {
       BigDecimal result = computePi(Integer.parseInt(args[0]));
       System.out.println("Pi is " + result);
   }

   /**
    * Compute the value of pi to the specified number of digits after the
    * decimal point. The value is computed using Machin's formula:
    * pi/4 = 4*arctan(1/5) - arctan(1/239)
    * and a power series expansion of arctan(x) to sufficient precision.
    */
   public static BigDecimal computePi(int digits) {
       int scale = digits + 5;
       BigDecimal arctan1_5 = arctan(5, scale);
       BigDecimal arctan1_239 = arctan(239, scale);
       BigDecimal pi = arctan1_5.multiply(FOUR).subtract(arctan1_239).multiply(FOUR);
       return pi.setScale(digits, BigDecimal.ROUND_HALF_UP);
   }

   /**
    * Compute the value, in radians, of the arctangent of the inverse of the
    * supplied integer to the specified number of digits after the decimal
    * point. The value is computed using the power series expansion for the arc
    * tangent:
    * arctan(x) = x - (x^3)/3 + (x^5)/5 - (x^7)/7 + (x^9)/9 ...
    */
   public static BigDecimal arctan(int inverseX, int scale) {
       BigDecimal result, numer, term;
       BigDecimal invX = BigDecimal.valueOf(inverseX);
       BigDecimal invX2 = BigDecimal.valueOf(inverseX * inverseX);
       numer = BigDecimal.ONE.divide(invX, scale, roundingMode);
       result = numer;
       int i = 1;
       do {
           numer = numer.divide(invX2, scale, roundingMode);
           int denom = 2 * i + 1;
           term = numer.divide(BigDecimal.valueOf(denom), scale, roundingMode);
           if ((i % 2) != 0) {
               result = result.subtract(term);
           } else {
               result = result.add(term);
           }
           i++;
       } while (term.compareTo(BigDecimal.ZERO) != 0);
       return result;
   }
}

To run and time the task:

$ time java test.Pi 50000
Pi is 3.14159265358979323846264338327950288419716939937510 ...
9.263u 0.187s 0:09.22 102.3% 0+0k 0+6io 8pf+0w

It takes about 10 seconds to compute PI with 5000 digits after the decimal point.