java math: expm1: return e power x minus 1

Java math 'expm1' method returns  e^x -1. Note that for values of x near 0, the exact sum of expm1(x) + 1 is much closer to the true result of ex than exp(x).

 

Signature

public static double expm1(double x)

 

Special cases to consider

a.   If the argument is NaN, the result is NaN.

b.   If the argument is positive infinity, then the result is positive infinity.

c.    If the argument is negative infinity, then the result is -1.0.

d.   If the argument is zero, then the result is a zero with the same sign as the argument.

 

Find the below working application.

 

ExpmOneDemo.java

package com.sample.app;

public class ExpmOneDemo {

	public static void main(String[] args) {
		double d1 = 10.0;

		System.out.printf("expm1(%f) : %f\n", d1, Math.expm1(d1));

		// If the argument is NaN, the result is NaN.
		System.out.printf("expm1(%f) : %f\n", Double.NaN, Math.expm1(Double.NaN));

		// If the argument is positive infinity, then the result is positive infinity.
		System.out.printf("expm1(%f) : %f\n", Double.POSITIVE_INFINITY, Math.expm1(Double.POSITIVE_INFINITY));

		// If the argument is negative infinity, then the result is -1.0
		System.out.printf("expm1(%f) : %f\n", Double.NEGATIVE_INFINITY, Math.expm1(Double.NEGATIVE_INFINITY));

		// If the argument is zero, then the result is a zero with the same sign as the
		// argument.
		System.out.printf("expm1(%f) : %f\n", 0.0, Math.expm1(0.0));
		System.out.printf("expm1(%f) : %f\n", -0.0, Math.expm1(-0.0));

	}

}

 

Output

 

expm1(10.000000) : 22025.465795
expm1(NaN) : NaN
expm1(Infinity) : Infinity
expm1(-Infinity) : -1.000000
expm1(0.000000) : 0.000000
expm1(-0.000000) : -0.000000

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