Java Math expm1() method with Example
The java.lang.Math.expm1() returns ex -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).
- 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 ex-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() methodimport 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
