Java.lang.StrictMath.expm1() Method



The java.lang.StrictMath.expm1() method returns ex -1. 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).It include some cases:

  • 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.


Following is the declaration for java.lang.StrictMath.expm1() method

public static double expm1(double x)


  • x -- This is the exponent to raise e to in the computation of ex -1.

Return Value

This method returns the value ex - 1.


  • NA


The following example shows the usage of java.lang.StrictMath.expm1() method.

package com.tutorialspoint;

import java.lang.*;

public class StrictMathDemo {

   public static void main(String[] args) {
   double d1 = 0.0 , d2 = -(1.0/0.0), d3 = (1.0/0.0);

   /*  returns (Euler's number e raised to the power of a double value)-1 
   i.e ex - 1  */

   double eulerValue = StrictMath.expm1(d1); 
   System.out.println("Value = " + eulerValue);
   eulerValue = StrictMath.expm1(d2); 
   System.out.println("Value = " + eulerValue);
   eulerValue = StrictMath.expm1(d3);
   System.out.println("Value = " + eulerValue);

Let us compile and run the above program, this will produce the following result:

Value = 0.0
Value = -1.0
Value = Infinity