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
