Below is the syntax highlighted version of HarmonicSum.java
from §9.1 Scientific Computation.
/****************************************************************************** * Compilation: javac HarmonicSum.java * Execution: java HarmonicSum N * * Compute the first N terms of the harmonic sum: 1/1 + 1/2 + ... + 1/N. * Compares single precision vs. double precision. Compares summing from * left-to-right vs. right-to-left. * * * % java HarmonicSum 10000 * 9.787613 * 9.787604 * 9.787606036044348 * 9.787606036044386 * * % java HarmonicSum 100000 * 12.090851 * 12.090153 * 12.090146129863335 * 12.090146129863408 // true answer = 12.090146129863428 * * % java HarmonicSum 1000000 * 14.357358 * 14.392652 * 14.392726722864989 * 14.392726722865772 // true answer = 14.392726722865724 * * % java HarmonicSum 10000000 * 15.403683 * 16.686031 * 16.695311365857272 * 16.695311365859965 // true answer = 16.695311365859852 * * % java HarmonicSum 100000000 * 15.403683 * 18.807919 * 18.997896413852555 * 18.997896413853447 // true answer = 18.997896413853898 * * % java HarmonicSum 1000000000 * 15.403683 * 18.807919 * 21.30048150234855 * 21.30048150234615 // true answer 21.30048150234794401668510 * * java HarmonicSum 2000000000 * 15.403683 * 18.807919 * 21.993628682662845 * 21.99362868265598 * ******************************************************************************/ public class HarmonicSum { public static void main(String[] args) { int N = Integer.parseInt(args[0]); // using single precision, left-to-right float sum1 = 0.0f; for (int i = 1; i <= N; i++) sum1 = sum1 + 1.0f / i; StdOut.println(sum1); // using single precision, right-to-left float sum2 = 0.0f; for (int i = N; i >= 1; i--) sum2 = sum2 + 1.0f / i; StdOut.println(sum2); // using double precision, left-to-right double sum3 = 0.0; for (int i = 1; i <= N; i++) sum3 = sum3 + 1.0 / i; StdOut.println(sum3); // using double precision, right-to-left double sum4 = 0.0; for (int i = N; i >= 1; i--) sum4 = sum4 + 1.0 / i; StdOut.println(sum4); } }