Java Math expm1() method with Example

The java.lang.Math.expm1() 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 e^x than exp(x).

• If the argument is NaN, the result is NaN.
• If the argument is positive infinity, then the result is positive infinity.
• If the argument is negative infinity, then the result is -1.0.
• If the argument is zero, then the result is a zero with the same sign as the argument.

Syntax:

public static double expm1(double x) 
Parameter: 
x-the exponent part which raises to e. 

Returns:
The method returns the value e^x-1, where e is the base of the natural logarithms.

Example : To show working of java.lang.Math.expm1() function

// Java program to demonstrate working
// of java.lang.Math.expm1() method
import java.lang.Math;

class Gfg {

    // driver code
    public static void main(String args[])
    {
        double x = 3;

        // when both are not infinity
        double result = Math.expm1(x);
        System.out.println(result);

        double positiveInfinity = Double.POSITIVE_INFINITY;
        double negativeInfinity = Double.NEGATIVE_INFINITY;
        double nan = Double.NaN;

        // when x is NAN
        result = Math.expm1(nan);
        System.out.println(result);

        // when argument is +INF
        result = Math.expm1(positiveInfinity);
        System.out.println(result);

        // when  argument is -INF
        result = Math.expm1(negativeInfinity);
        System.out.println(result);

        x = -0;
        result = Math.expm1(x);
        // same sign as 0
        System.out.println(result);
    }
}

Output:

19.085536923187668
NaN
Infinity
-1.0
0.0